Solution Manual For Chemistry The Molecular Nature of Matter and Change, 9th Edition by Martin Silbe by StudyGuide

CHAPTER 1 KEYS TO THE STUDY OF CHEMISTRY FOLLOW–UP PROBLEMS 1.1A

Plan: The real question is “Does the substance change composition or just change form?” A change in composition is a chemical change while a change in form is a physical change. Solution: The figure on the left shows red atoms and molecules composed of one red atom and one blue atom. The figure on the right shows a change to blue atoms and molecules containing two red atoms. The change is chemical since the substances themselves have changed in composition.

1.1B

Plan: The real question is “Does the substance change composition or just change form?” A change in composition is a chemical change while a change in form is a physical change. Solution: The figure on the left shows red atoms that are close together, in the solid state. The figure on the right shows red atoms that are far apart from each other, in the gaseous state. The change is physical since the substances themselves have not changed in composition.

1.2A

Plan: The real question is “Does the substance change composition or just change form?” A change in composition is a chemical change while a change in form is a physical change. Solution: a) Both the solid and the vapor are iodine, so this must be a physical change. b) The burning of the gasoline fumes produces energy and products that are different gases. This is a chemical change. c) The scab forms due to a chemical change.

1.2B

Plan: The real question is “Does the substance change composition or just change form?” A change in composition is a chemical change while a change in form is a physical change. Solution: a) Clouds form when gaseous water (water vapor) changes to droplets of liquid water. This is a physical change. b) When old milk sours, the compounds in milk undergo a reaction to become different compounds (as indicated by a change in the smell, the taste, the texture, and the consistency of the milk). This is a chemical change. c) Both the solid and the liquid are butter, so this must be a physical change.

1.3A

Plan: We need to find the amount of time it takes for the professor to walk 10,500 m. We know how many miles she can walk in 15 min (her speed), so we can convert the distance the professor walks to miles and use her speed to calculate the amount of time it will take to walk 10,500 m. Solution:       Time (min) = 10,500 m  = 97.8869 = 98 min       Road map: Distance (m) 1000 m = 1 km Distance (km)

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1.609 km = 1 mi Distance (mi) 1 mi = 15 min Time (min) 1.3B

Plan: We need to find the number of virus particles that can line up side by side in a 1 inch distance. We know the diameter of a virus in nm units. If we convert the 1 inch distance to nm, we can use the diameter of the virus to calculate the number of virus particles we can line up over a 1 inch distance. Solution:     7  5 5 No. of virus particles = 1.0 in     = 8.4667 × 10 = 8.5 × 10 virus particles      Road map: Length (in) 1 in = 2.54 cm Length (cm) 7

1 cm = 1 × 10 nm Length (nm) 30 nm = 1 particle No. of particles 1.4A

Plan: The diameter in nm is used to obtain the radius in nm, which is converted to the radius in dm. The volume of 3 the ribosome in dm is then determined using the equation for the volume of a sphere given in the problem. This volume may then be converted to vo Solution: diameter  21.4 nm   1 m  1 dm  = 1.07 × 10–7 dm Radius (dm) = =   110 9 nm   0.1 m    2 2 3 4 3 4 3  3.141591.07107 dm = 5.13145 × 10–21 = 5.13 × 10–21 dm3 Volume (dm ) = 3 3  1 L  1  –15 –15 Volume ( L) = 5.131451021 dm 3  = 5.13145 × 10 = 5.13 × 10 L   (1 dm)3 106 L 

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Road map: Diameter (dm) diameter = 2r Radius (dm) 3

Volume (dm ) 3

1 dm = 1 L 6 1 L = 10

1.4B

Plan: We need to convert gallon units to liter units. If we first convert gallons to dm , we can then convert to L. Solution: 3    Volume (L) = 8400 gal  = 31,794 = 32,000 L 3    Road map: Volume (gal) 1 gal = 3.785 dm

Volume (dm ) 3

1 dm = 1 L Volume (L) 1.5A

Plan: The time is given in hours and the rate of delivery is in drops per second. Conversions relating hours to seconds are needed. This will give the total number of drops, which may be combined with their mass to get the total mass. The mg of drops will then be changed to kilograms. Solution: 3  60 min   60 s 1.5 drops  65 mg 10 g   1 kg   Mass (kg) = 8.0 h       3  = 2.808 = 2.8 kg  1 h  1 min  1 s  1 drop  1 mg  10 g  Road map: Time (hr) 1 hr = 60 min Time (min) 1 min = 60 s

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Time (s) 1 s = 1.5 drops No. of drops 1 drop = 65 mg Mass (mg) of solution 3

1 mg = 10 g Mass (g) of solution 3

10 g = 1 kg Mass (kg) of solution

1.5B

Plan: We have the mass of apples in kg and need to find the mass of potassium in those apples in g. The number of apples per pound and the mass of potassium per apple are given. Convert the mass of apples in kg to pounds. Then use the number of apples per pound to calculate the number of apples. Use the mass of potassium in one apple to calculate the mass (mg) of potassium in the group of apples. Finally, convert the mass in mg to g. Solution:         = 3.4177 = 3.42 g Mass (g) = 3.25 kg    3       Road map: Mass (kg) of apples 0.4536 kg = 1 lb Mass (lb) of apples 1 lb = 3 apples No. of apples 1 apple = 159 mg potassium Mass (mg) potassium 3

10 mg = 1 g Mass (g) potassium

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1.6A

Plan: We know the area of a field in m . We need to know how many bottles of herbicide will be needed to treat 2 that field. The volume of each bottle (in fl oz) and the volume of herbicide needed to treat 300 ft of field are 2 2 given. Convert the area of the field from m to ft (don’t forget to square the conversion factor when converting from squared units to squared units!). Then use the given conversion factors to calculate the number of bottles of 2 herbicide needed. Convert first from ft of field to fl oz of herbicide (because this conversion is from a squared unit to a non-squared unit, we do not need to square the conversion factor). Then use the number of fl oz per bottle to calculate the number of bottles needed. Solution: 2     2  No. of bottles = 2050 m    = 6.8956 = 7 bottles 2 2 2      Road map: 2

Area (m ) 2

(0.3048) m = 1 ft

Area (ft ) 2

300 ft = 1.5 fl oz Volume (fl oz) 16 oz = 1 bottle No. of bottles

1.6B

Plan: Calculate the mass of mercury in g. Convert the surface area of the lake from mi to ft . Find the volume of 3 2 the lake in ft by multiplying the surface area (in ft ) by the depth (in ft). Then convert the volume of the lake to 3 3 3 3 3 mL by converting first from ft to m , then from m to cm , and from cm to mL. Finally, divide the mass in g by the volume in mL to find the mass of mercury in each mL of the lake. Solution:   7  = 7.5 × 10 g Mass (g) = 75,000 kg    6 3  (5280)2 ft 2   0.02832 m 3  110 cm  1 mL      35 ft     = 1.24349 × 1014 mL Volume (mL) = 4.5 mi  2   1 m 3 1 cm 3  1 ft 3  1 mi   2

Mass (g) of mercury per mL =





 7  14

-7

–7

= 6.0314 × 10 = 6.0 × 10 g/mL

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Road map: 2

Area (mi ) 2

1 mi = (5280) ft

Area (ft ) 2

V = area (ft ) × depth (ft) 3

Volume (ft ) 3

1 ft = 0.02832 m

Volume (m ) 3

1 m = 10 cm

Volume (cm ) Mass (kg) 3

1 cm = 1 mL 3

1 kg = 10 g Volume (mL)

Mass (g)

divide mass by volume Mass (g) of mercury in 1 mL of water 1.7A

Plan: Find the mass of Venus in g. Calculate the radius of Venus by dividing its diameter by 2. Convert the radius from km to cm. Use the radius to calculate the volume of Venus. Finally, find the density of Venus by dividing 3 the mass of Venus (in g) by the volume of Venus (in cm ). Solution:  3  24 27  = 4.9 × 10 g Mass (g) = 4.9 × 10 kg      

 Radius (cm) =   3

Volume (cm ) =

4 3

Density (g/cm ) =



4 3

4.91027 g 

   

 8  = 6.05 10 cm



= 9.27587 × 10 cm

= 5.28252 = 5.3 g/cm

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Road map: Diameter (km) d = 2r Radius (km) 3

1 km = 10 m Radius (m) 2

1 m = 10 cm Mass (kg)

Radius (cm) 3

1 kg = 10 g Mass (g)

4 3

Volume (cm ) divide mass by volume 3 by volume Densitydivide (g/cmmass )

1.7B

Plan: The mass (pounds) of aluminum must be converted to grams, then dividing by the density of aluminum will give the volume of aluminum available to make cans. Finally, dividing by the volume per can will give the number of cans possible. Solution:  453.6 g  1 cm 3   1 can   Number of cans = 16.2 lb  = 533 cans  3  1 lb  2.70 g   5.1 cm  Road map:





Mass (lb) 1 lb = 453.6 g Mass (g) 2.70 g = 1 cm

Volume (cm ) 3

5.1 cm = 1 can Number of cans

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1.8A

Plan: Using the relationship between the Kelvin and Celsius scales, change the Kelvin temperature to the Celsius temperature. Then convert the Celsius temperature to the Fahrenheit value using the relationship between these two scales. Solution: T (in °C) = T (in K) – 273.15 = 234 K – 273.15 = –39.15 = –39°C 9 9 T (in °F) = T (in °C) + 32 = (–39.15°C) + 32 = –38.47 = –38°F 5 5 Check: Since the Kelvin temperature is below 273, the Celsius temperature must be negative. The low Celsius value gives a negative Fahrenheit value.

1.8B

Plan: Convert the Fahrenheit temperature to the Celsius value using the relationship between these two scales. Then use the relationship between the Kelvin and Celsius scales to change the Celsius temperature to the Kelvin temperature. Solution: 5 5 T (in °C) = (T (in °F)  32) = (2325 °F – 32) = 1273.8889 = 1274 °C 9 9 T (in K) = T (in °C) + 273.15 = 1274 °C + 273.15 = 1547.15 = 1547 K Check: Since the Fahrenheit temperature is large and positive, both the Celsius and Kelvin temperatures should also be positive. Because the Celsius temperature is greater than 273, the Kelvin temperature should be greater than 273, which it is.

1.9A

Plan: Determine the significant figures by counting the digits present and accounting for the zeros. Zeros between non-zero digits are significant, as are trailing zeros to the right of a decimal point. Trailing zeros to the left of a decimal point are only significant if the decimal point is present. Solution: a) 31.070 mg; five significant figures b) 0.06060 g; four significant figures c) 850°C; three significant figures — note the decimal point that makes the zero significant. Check: All significant zeros must come after a significant digit.

1.9B

Plan: Determine the significant figures by counting the digits present and accounting for the zeros. Zeros between non-zero digits are significant, as are trailing zeros to the right of a decimal point. Trailing zeros to the left of a decimal point are only significant if the decimal point is present. Solution: 2 a) 2.000 × 10 mL; four significant figures –6 b) 3.90 × 10 m; three significant figures –4 c) 4.01 × 10 L; three significant figures Check: All significant zeros must come after a significant digit.

1.10A

Plan: Use the rules presented in the text. Add the two values in the numerator before dividing. The time conversion is an exact conversion and, therefore, does not affect the significant figures in the answer. Solution: The addition of 25.65 mL and 37.4 mL gives an answer where the last significant figure is the one after the decimal point (giving three significant figures total): 25.65 mL + 37.4 mL = 63.05 (would round to 63.0 if not an intermediate step) When a four significant figure number divides a three significant figure number, the answer must round to three significant figures. An exact number (1 min / 60 s) will have no bearing on the number of significant figures. 63.05 mL = 51.4344 = 51.4 mL/min 1 min    73.55 s    60 s 

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1.10B

Plan: Use the rules presented in the text. Subtract the two values in the numerator and multiply the numbers in the denominator before dividing. Solution: The subtraction of 35.26 from 154.64 gives an answer in which the last significant figure is two places after the decimal point (giving five significant figures total): 154.64 g – 35.26 g = 119.38 g The multiplication of 4.20 cm (three significant figures) by 5.12 cm (three significant figures) by 6.752 cm (four significant figures) gives a number with three significant figures. 3 4.20 cm × 5.12 cm × 6.752 cm = 145.1950 (would round to 145 cm if not an intermediate step) When a three significant figure number divides a five significant figure number, the answer must round to three significant figures. 3

= 0.82220 = 0.822 g/cm

END–OF–CHAPTER PROBLEMS 1.1

Plan: If only the form of the particles has changed and not the composition of the particles, a physical change has taken place; if particles of a different composition result, a chemical change has taken place. Solution: a) The result in C represents a chemical change as the substances in A (red spheres) and B (blue spheres) have reacted to become a different substance (particles consisting of one red and one blue sphere) represented in C. There are molecules in C composed of the atoms from A and B. b) The result in D represents a chemical change as again the atoms in A and B have reacted to form molecules of a new substance. c) The change from C to D is a physical change. The substance is the same in both C and D (molecules consisting of one red sphere and one blue sphere) but is in the gas phase in C and in the liquid phase in D. d) The sample has the same chemical properties in both C and D since it is the same substance but has different physical properties.

1.2

Plan: Apply the definitions of the states of matter to a container. Next, apply these definitions to the examples. Gas molecules fill the entire container; the volume of a gas is the volume of the container. Solids and liquids have a definite volume. The volume of the container does not affect the volume of a solid or liquid. Solution: a) The helium fills the volume of the entire balloon. The addition or removal of helium will change the volume of a balloon. Helium is a gas. b) At room temperature, the mercury does not completely fill the thermometer. The surface of the liquid mercury indicates the temperature. c) The soup completely fills the bottom of the bowl, and it has a definite surface. The soup is a liquid, though it is possible that solid particles of food will be present.

1.3

Plan: Apply the definitions of the states of matter to a container. Next, apply these definitions to the examples. Gas molecules fill the entire container; the volume of a gas is the volume of the container. Solids and liquids have a definite volume. The volume of the container does not affect the volume of a solid or liquid. Solution: a) The air fills the volume of the room. Air is a gas. b) The vitamin tablets do not necessarily fill the entire bottle. The volume of the tablets is determined by the number of tablets in the bottle, not by the volume of the bottle. The tablets are solid. c) The sugar has a definite volume determined by the amount of sugar, not by the volume of the container. The sugar is a solid.

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1.4

Plan: Define the terms and apply these definitions to the examples. Solution: Physical property – A characteristic shown by a substance itself, without interacting with or changing into other substances. Chemical property – A characteristic of a substance that appears as it interacts with, or transforms into, other substances. a) The change in color (yellow–green and silvery to white), and the change in physical state (gas and metal to crystals) are examples of physical properties. The change in the physical properties indicates that a chemical change occurred. Thus, the interaction between chlorine gas and sodium metal producing sodium chloride is an example of a chemical property. b) The sand and the iron are still present. Neither sand nor iron became something else. Colors along with magnetism are physical properties. No chemical changes took place, so there are no chemical properties to observe.

1.5

Plan: Define the terms and apply these definitions to the examples. Solution: Physical change – A change in which the physical form (or state) of a substance, but not its composition, is altered. Chemical change – A change in which a substance is converted into a different substance with different composition and properties. a) The changes in the physical form are physical changes. The physical changes indicate that there is also a chemical change. Magnesium chloride has been converted to magnesium and chlorine. b) The changes in color and form are physical changes. The physical changes indicate that there is also a chemical change. Iron has been converted to a different substance, rust.

1.6

Plan: Apply the definitions of chemical and physical changes to the examples. Solution: a) Not a chemical change, but a physical change — simply cooling returns the soup to its original form. b) There is a chemical change — cooling the toast will not “un–toast” the bread. c) Even though the wood is now in smaller pieces, it is still wood. There has been no change in composition, thus this is a physical change, and not a chemical change. d) This is a chemical change converting the wood (and air) into different substances with different compositions. The wood cannot be “unburned.”

1.7

Plan: If there is a physical change, in which the composition of the substance has not been altered, the process can be reversed by a change in temperature. If there is a chemical change, in which the composition of the substance has been altered, the process cannot be reversed by changing the temperature. Solution: a) and c) can be reversed with temperature; the dew can evaporate and the ice cream can be refrozen. b) and d) involve chemical changes and cannot be reversed by changing the temperature since a chemical change has taken place.

1.8

Plan: A system has a higher potential energy before the energy is released (used). Solution: a) The exhaust is lower in energy than the fuel by an amount of energy equal to that released as the fuel burns. The fuel has a higher potential energy. b) Wood, like the fuel, is higher in energy by the amount released as the wood burns.

1.9

Plan: Kinetic energy is energy due to the motion of an object. Solution: a) The sled sliding down the hill has higher kinetic energy than the unmoving sled. b) The water falling over the dam (moving) has more kinetic energy than the water held by the dam.

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1.10

Observations are the first step in the scientific approach. The first observation is that the toast has not popped out of the toaster. The next step is a hypothesis (tentative explanation) to explain the observation. The hypothesis is that the spring mechanism is stuck. Next, there will be a test of the hypothesis. In this case, the test is an additional observation — the bread is unchanged. This observation leads to a new hypothesis — the toaster is unplugged. This hypothesis leads to additional tests — seeing if the toaster is plugged in, and if it works when plugged into a different outlet. The final test on the toaster leads to a new hypothesis — there is a problem with the power in the kitchen. This hypothesis leads to the final test concerning the light in the kitchen.

1.11

A quantitative observation is easier to characterize and reproduce. A qualitative observation may be subjective and open to interpretation. a) This is qualitative. When has the sun completely risen? b) The astronaut’s mass may be measured; thus, this is quantitative. c) This is qualitative. Measuring the fraction of the ice above or below the surface would make this a quantitative measurement. d) The depth is known (measured) so this is quantitative.

1.12

A well-designed experiment must have the following essential features: 1) There must be two variables that are expected to be related. 2) There must be a way to control all the variables, so that only one at a time may be changed. 3) The results must be reproducible.

1.13

A model begins as a simplified version of the observed phenomena, designed to account for the observed effects, explain how they take place, and to make predictions of experiments yet to be done. The model is improved by further experiments. It should be flexible enough to allow for modifications as additional experimental results are gathered.

1.14

Plan: Review the definitions of mass and weight. Solution: Mass is the quantity of material present, while weight is the interaction of gravity on mass. An object has a definite mass regardless of its location; its weight will vary with location. The lower gravitational attraction on the Moon will make an object appear to have approximately one-sixth its Earth weight. The object has the same mass on the Moon and on Earth.

1.15

The unit you begin with (feet) must be in the denominator to cancel. The unit desired (inches) must be in the numerator. The feet will cancel leaving inches. If the conversion is inverted the answer would be in units of feet squared per inch.

1.16

Plan: Density =

mass . An increase in mass or a decrease in volume will increase the density. A decrease volume in density will result if the mass is decreased or the volume increased.

Solution: a) Density increases. The mass of the chlorine gas is not changed, but its volume is smaller. b) Density remains the same. Neither the mass nor the volume of the solid has changed. c) Density decreases. Water is one of the few substances that expands on freezing. The mass is constant, but the volume increases. d) Density increases. Iron, like most materials, contracts on cooling; thus the volume decreases while the mass does not change. e) Density remains the same. The water does not alter either the mass or the volume of the diamond.

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1-11

1.17

Plan: Review the definitions of heat and temperature. The two temperature values must be compared using one temperature scale, either Celsius or Fahrenheit. Solution: Heat is the energy that flows between objects at different temperatures while temperature is the measure of how hot or cold a substance is relative to another substance. Heat is an extensive property while temperature is an intensive property. It takes more heat to boil a gallon of water than to boil a teaspoon of water. However, both water samples boil at the same temperature. 9 9 Convert 65°C to °F: T (in °F) = T (in °C) + 32 = (65°C) + 32 = 149°F 5 5 A temperature of 65°C is 149°F. Heat will flow from the hot water (65°C or 149°F) to the cooler water (65°F). The 65°C water contains more heat than the cooler water.

1.18

There are two differences in the Celsius and Fahrenheit scales (size of a degree and the zero point), so a simple one-step conversion will not work. The size of a degree is the same for the Celsius and Kelvin scales; only the zero point is different so a one-step conversion is sufficient.

1.19

Plan: Review the definitions of extensive and intensive properties. Solution: An extensive property depends on the amount of material present. An intensive property is the same regardless of how much material is present. a) Mass is an extensive property. Changing the amount of material will change the mass. b) Density is an intensive property. Changing the amount of material changes both the mass and the volume, but the ratio (density) remains fixed. c) Volume is an extensive property. Changing the amount of material will change the size (volume). d) The melting point is an intensive property. The melting point depends on the substance, not on the amount of substance.

1.20

1000 m  100 cm  2 2 ; to convert from m to cm , use 2 2 1 km  1 m c) This problem requires two conversion factors: one for distance and one for time. It does not matter which conversion is done first. Alternate methods may be used. To convert distance, mi to m, use: 2

b) To convert from km to m , use

1.609 km  1000 m  = 1.609 × 103 m/mi   1 mi  1 km 

To convert time, h to s, use:  1 h  1 min  = 1 h/3600 s   60 s   60 min 

1.609103 m  1 h    0.4469 m  h .  Therefore, the complete conversion factor is    1 mi mi  s  3600 s  Do the units cancel when you start with a measurement of mi/h? Copyright

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1-12

1000 g . 2.205 lb 3  1 ft3  3 3  1 in  = 3.531 × 10–5 ft3/cm3. To convert volume from ft to cm use,   3 12 in3  2.54 cm  d) To convert from pounds (lb) to grams (g), use

1.21

Plan: Review the table of conversions in the chapter or inside the back cover of the book. Write the conversion factor so that the unit initially given will cancel, leaving the desired unit. Solution: a) This problem requires two conversion factors: one for distance and one for time. It does not matter which conversion is done first. Alternate methods may be used.  1 in  To convert distance, cm to in, use:    2.54 cm   1 min  To convert time, min to s, use:    60 s 

100 cm  1 in 3 3 ; to convert from cm to in , use 3 3 1 m  2.54 cm  c) This problem requires two conversion factors: one for distance and one for time. It does not matter which conversion is done first. Alternate methods may be used.  1 km  To convert distance, m to km, use:    1000 m  2 2 To convert time, s to h , use: 2 2  60 s2  60 min   = 3600 s  1 min 2   1 h 2  h2 3

b) To convert from m to cm , use

d) This problem requires two conversion factors: one for volume and one for time. It does not matter which conversion is done first. Alternate methods may be used.  4 qt   1 L   ; to convert qt to L, use:   To convert volume, gal to qt, use:   1 gal   1.057 qt   1 h  To convert time, h to min, use:    60 min  –12

1.22

Plan: Use conversion factors from the inside back cover: 1 pm = 10 Solution: 1012 m  1 nm     = 1.43 nm Radius (nm) = 1430 pm   1 pm  109 m 

1.23

Plan: Use conversion factors from the inside back cover: 10 Solution:  1 pm  0.01   = 2.22 Radius ( ) = 2.22 1010 m 12   10 m  1 pm 

1.24

Plan: Use conversion factors: 0.01 m = 1 cm; 2.54 cm = 1 in. Solution:  1 cm   1 in   3 3 Length (in) = 100. m    = 3.9370 × 10 = 3.94 × 10 in  2.54 cm   0.01 m 

–12

–9

m; 10 m = 1 nm.

m = 1 pm; 1 pm = 0.01 .

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1-13

1.25

Plan: Use the conversion factor 12 in = 1 ft to convert 6 ft 10 in to height in inches. Then use the conversion factors 1 in = 2.54 cm; 1 cm = 10 mm. Solution: 12 in     10 in = 82 in Height (in) = 6 ft   1 ft   2.54 cm 10 mm     = 2.0828 × 103 = 2.1 × 103 mm Height (mm) = 82 in   1 cm   1 in 

  

1.26



Plan: Use conversion factors (1 cm) = (0.01 m) ; (1000 m) = (1 km) to express the area in km . To calculate 2 2 the cost of the patch, use the conversion factor: (2.54 cm) = (1 in) . Solution: 0.01 m2  1 km 2  2 –9 2  a) Area (km ) = 20.7 cm 2   2  = 2.07 × 10 km  1 cm2   1000 m 

 1 in 2  $3.25   = 10.4276 = $10.43  b) Cost = 20.7 cm 2   1 in 2  2.54 cm 2  1.27

–3

Plan: Use conversion factors (1 mm) = (10 m) ; (0.01 m) = (1 cm) ; (2.54 cm) = (1 in) ; (12 in) = (1 ft) to 2 express the area in ft . Solution: 2  103 m 2  2 2    2 2   1 cm   1 in   1 ft   a) Area (ft ) = 7903 mm   2   1 mm 2  0.01 m 2   12 in 2  2.54 cm     –2

–2

= 8.5067 × 10 = 8.507 × 10 ft

 45 s    = 2.634333 × 103 = 2.6 × 103 s b) Time (s) = 7903 mm 2  135 mm 2 





1.28

Plan: Use conversion factors 1 lb = 16 oz, 0.4536 kg = 1 lb, and 1000 g = 1 kg. Solution:  1 lb  0.4536 kg  1000 g      Mass (g) = 6.14 oz   1 lb  1 kg  = 174 kg 16 oz   

1.29

Plan: Use conversion factor 1 short ton = 2000 lb; 2.205 lb = 1 kg; 1000 kg = 1 metric ton. Solution:  2000 lb   1 kg   1 T   15 15 Mass (T) = 2.60 1015 ton   10 3 kg  = 2.35828 × 10 = 2.36 × 10 T  1 ton   2.205 lb  

1.30

Plan: Mass in g is converted to kg in part (a) with the conversion factor 1000 g = 1 kg; mass in g is converted to lb 3 3 in part (b) with the conversion factors 1000 g = 1 kg; 1 kg = 2.205 lb. Volume in cm is converted to m with the 3 3 3 3 3 3 3 conversion factor (1 cm) = (0.01 m) and to ft with the conversion factors (2.54 cm) = (1 in) ; (12 in) = (1 ft) . The conversions may be performed in any order. Solution:  5.52 g  1 cm3  3  1 kg  = 5.52 × 103 kg/m3 a) Density (kg/m ) =    3  3   cm 0.01 m 1000 g 

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1-14

3 3  5.52 g 2.54 cm   3 12 in   1 kg   2.205 lb  = 344.661 = 345 lb/ft3 b) Density (lb/ft ) =     3 3   1 kg  3     cm  1 in   1 ft  1000 g  

1.31

Plan: Length in m is converted to km in part (a) with the conversion factor 1000 m = 1 km; length in m is converted to mi in part (b) with the conversion factors 1000 m = 1 km; 1 km = 0.62 mi. Time is converted using the conversion factors 60 s = 1 min; 60 min = 1 h. The conversions may be performed in any order. Solution:  2.998108 m   60 s   60 min   1 km   9 9 a) Velocity (km/h) =     3  = 1.07928 × 10 = 1.079 × 10 km/h 1s 1 min   1 h  10 m    2.998108 m   60 s   1 km   0.62 mi   7 7 b) Velocity (mi/min) =    3   = 1.11526 × 10 = 1.1 × 10 mi/min 1s 1 min  10 m   1 km  

1.32

Plan:

–6

–3

= (1 × 10 m) ; (1 × 10 m) = (1 mm) to convert to mm . 3 –6 3 –2 3 3 3 = (1 × 10 m) ; (1 × 10 m) = (1 cm) ; 1 cm = 1 mL;

–3

1 mL = 1 × 10 L. Solution: 3  1106 m    1 mm 3     = 2.56 × 10–9 mm3/cell  3 3 3   1     110 m    3 3  1106 m   2.56   1 cm3    3     b) Volume (L) = 10 5 cells  3 3 3   1102 m  1 cm  1 mL  cell  1   

 2.56 a) Volume (mm ) =   cell

= 2.56 × 10

–10

= 10

L –3

1.33

  

–3

Plan: For part (a), convert from qt to mL (1 qt = 946.4 mL) to L (1 mL = 1 × 10 L) to m (1 L = 10 m ). –3 For part (b), convert from gal to qt (1 gal = 4 qt) to mL (1 qt = 946.4 mL) to L (1 mL = 10 L). Solution: 3 3 3  946.4 mL  3 10 L  10 m  = 9.464 × 10–4 m3 a) Volume (m ) = 1 qt     1 L   1 qt  1 mL   4 qt   946.4 mL 103 L   3 3 b) Volume (L) = 835 gal   = 3.160976 × 10 = 3.16 × 10 L     1 gal   1 qt   1 mL 



1.34



Plan: The mass of the mercury in the vial is the mass of the vial filled with mercury minus the mass of the empty vial. Use the density of mercury and the mass of the mercury in the vial to find the volume of mercury and thus the volume of the vial. Once the volume of the vial is known, that volume is used in part (b). The density of water is used to find the mass of the given volume of water. Add the mass of water to the mass of the empty vial. Solution: a) Mass (g) of mercury = mass of vial and mercury – mass of vial = 185.56 g – 55.32 g = 130.24 g  1 cm 3  3  = 9.626016 = 9.626 cm3 Volume (cm ) of mercury = volume of vial = 130.24 g 13.53 g  3

b) Volume (cm ) of water = volume of vial = 9.626016 cm  0.997 g  Mass (g) of water = 9.626016 cm3  = 9.59714 g water  1 cm 3  Mass (g) of vial filled with water = mass of vial + mass of water = 55.32 g + 9.59714 g = 64.91714 = 64.92 g Copyright

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1-15

1.35

Plan: The mass of the water in the flask is the mass of the flask and water minus the mass of the empty flask. Use the density of water and the mass of the water in the flask to find the volume of water and thus the volume of the flask. Once the volume of the flask is known, that volume is used in part (b). The density of chloroform is used to find the mass of the given volume of chloroform. Add the mass of the chloroform to the mass of the empty flask. Solution: a) Mass (g) of water = mass of flask and water – mass of flask = 489.1 g – 241.3 g = 247.8 g  1 cm 3   3  = 247.8 = 248 cm3 Volume (cm ) of water = volume of flask = 247.8 g  1.00 g 





b) Volume (cm ) of chloroform = volume of flask = 247.8 cm 1.48 g   = 366.744 g chloroform Mass (g) of chloroform = 247.8 cm3   cm3 





Mass (g) of flask and chloroform = mass of flask + mass of chloroform = 241.3 g + 366.744 g = 608.044 g = 608 g 1.36

Plan: Calculate the volume of the cube using the relationship Volume = (length of side) . The length of side in 3 mm must be converted to cm so that volume will have units of cm . Divide the mass of the cube by the volume to find density. Solution: 103 m  1 cm    = 1.56 cm (convert to cm to match density unit)  Side length (cm) = 15.6 mm   2  1 mm  10 m  3 3 3 3 Al cube volume (cm ) = (length of side) = (1.56 cm) = 3.7964 cm mass 10.25 g 3 Density (g/cm3 )   = 2.69993 = 2.70 g/cm volume 3.7964 cm3



1.37



Plan: Use the relationship c = 2 to find the radius of the sphere and the relationship V = 4/3 to find the 3 3 volume of the sphere. The volume in mm must be converted to cm . Divide the mass of the sphere by the volume to find density. Solution: c=2 c 32.5 mm = Radius (mm) = = 5.17254 mm 2 2 4 3  4  3 3 Volume (mm ) = =   (5.17254 mm)3 = 579.6958 mm  3 3 103 m   1 cm 3 3  = 0.5796958 cm3 Volume (cm ) = 579.6958 mm 3     1 mm  102 m  3

Density (g/cm3 )  1.38

mass 4.20 g 3  = 7.24518 = 7.25 g/cm 3 volume 0.5796958 cm

Plan: Use the equations given in the text for converting between the three temperature scales. Solution: 5 5 a) T (in °C) = [T (in °F) – 32] = [68°F – 32] = 20.°C 9 9 T (in K) = T (in °C) + 273.15 = 20.°C + 273.15 = 293.15 = 293 K

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1-16

b) T (in K) = T (in °C) + 273.15 = –164°C + 273.15 = 109.15 = 109 K 9 9 T (in °F) = T (in °C) + 32 = (–164°C) + 32 = –263.2 = –263°F 5 5 c) T (in °C) = T (in K) – 273.15 = 0 K – 273.15 = –273.15 = –273°C T (in °F) =

9 9 T (in °C) + 32 = (–273.15°C) + 32 = –459.67 = –460.°F 5 5

1.39

Plan: Use the equations given in the text for converting between the three temperature scales. Solution: 5 5 a) T (in °C) = [T (in °F) – 32] = [106°F – 32] = 41.111 = 41°C 9 9 (106 – 32) = 74 This limits the significant figures. T (in K) = T (in °C) + 273.15 = 41.111°C + 273.15 = 314.261 = 314 K 9 9 b) T (in °F) = T (in °C) + 32 = (3410°C) + 32 = 6170°F 5 5 T (in K) = T (in °C) + 273.15 = 3410°C + 273 = 3683 K 3 3 3 c) T (in °C) = T (in K) –273.15 = 6.1 × 10 K – 273 = 5.827 × 10 = 5.8 × 10 °C 9 9 4 4 T (in °F) = T (in °C) + 32 = (5827°C) + 32 = 1.0521 × 10 = 1.1 × 10 °F 5 5

1.40

Plan: Find the volume occupied by each metal by taking the difference between the volume of water and metal and the initial volume of the water (25.0 mL). Divide the mass of the metal by the volume of the metal to calculate density. Use the density value of each metal to identify the metal. Solution: Cylinder A: volume of metal = [volume of water + metal] – [volume of water] volume of metal = 28.2 mL – 25.0 mL = 3.2 mL mass 25.0 g = Density = = 7.81254 = 7.8 g/mL volume 3.2 mL Cylinder A contains iron. Cylinder B: volume of metal = [volume of water + metal] – [volume of water] volume of metal = 27.8 mL – 25.0 mL = 2.8 mL mass 25.0 g = Density = = 8.92857 = 8.9 g/mL volume 2.8 mL Cylinder B contains nickel. Cylinder C: volume of metal = [volume of water + metal] – [volume of water] volume of metal = 28.5 mL – 25.0 mL = 3.5 mL mass 25.0 g = Density = = 7.14286 = 7.1 g/mL volume 3.5 mL Cylinder C contains zinc.

1.41

Plan: Use 1 nm = 10 m to convert wavelength in nm to m. To convert wavelength in pm to , use 1 pm = 0.01 . Solution: 109 m    = 2.47 × 10–7 m a) Wavelength (m) = 247 nm   1 nm 

–9





 0.01    = 67.6 b) Wavelength ( ) = 6760 pm   1 pm   





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1-17

1.42

Plan: The liquid with the larger density will occupy the bottom of the beaker, while the liquid with the smaller density volume will be on top of the more dense liquid. Solution: a) Liquid A is more dense than water; liquids B and C are less dense than water. b) Density of liquid B could be 0.94 g/mL. Liquid B is more dense than C so its density must be greater than 0.88 g/mL. Liquid B is less dense than water so its density must be less than 1.0 g/mL.

1.43

Plan: Calculate the volume of the cylinder in cm by using the equation for the volume of a cylinder. The 3 3 diameter of the cylinder must be halved to find the radius. Convert the volume in cm to dm by using the 3 –2 3 –1 3 3 conversion factors (1 cm) = (10 m) and (10 m) = (1 dm) . Solution: Radius = diameter/2 = 0.85 cm/2 = 0.425 cm 3 2 2 3 Volume (cm ) = h = (0.425 cm) (9.5 cm) = 5.3907766 cm 3 3 102 m   1 dm   3    = 5.39078 × 10–3 = 5.4 × 10–3 dm3 Volume (dm ) = 5.3907766 cm 3   1 cm  101 m 

1.44

Plan: Use the percent of copper in the ore to find the mass of copper in 5.01 lb of ore. Convert the mass in lb to mass in g. The density of copper is used to find the volume of that mass of copper. Use the volume equation for a cylinder to calculate the height of the cylinder (the length of wire); the diameter of the wire is used to find the radius which must be expressed in units of cm. Length of wire in cm must be converted to m. Solution:  66%  Mass (lb) of copper = 5.01 lb Covellite   = 3.3066 lb copper 100% 





 1 kg  1000 g   3   Mass (g) of copper = 3.3066 lb  1 kg  = 1.49959 × 10 g  2.205 lb    cm 3 Cu   3  = 167.552 cm3 Cu Volume (cm ) of copper = 1.49959 10 3 g Cu  8.95 g Cu  2 V= rh  6.304 103 in  2.54 cm   = 8.00608 × 10–3 cm  Radius (cm) =   2  1 in 

Height (length) in cm =

= 2

167.552 cm3



  8.0060810

3

–5

cm

= 8.3207 × 10 cm

10 m   = 8.3207 × 103 = 8.32 × 103 m Length (m) = 8.320710 5 cm  1 cm  An exact number is defined to have a certain value (exactly). There is no uncertainty in an exact number. An exact number is considered to have an infinite number of significant figures and, therefore, does not limit the digits in the calculation. 2

1.45

1.46

Random error of a measurement is decreased by (1) taking the average of more measurements. More measurements allow a more precise estimate of the true value of the measurement. Calibrating the instrument will allow greater accuracy but not necessarily greater precision.

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1-18

1.47

a) If the number is an exact count then there are an infinite number of significant figures. If it is not an exact count, there are only 5 significant figures. b) Other things, such as number of tickets sold, could have been counted instead. 4 c) A value of 15,000 to two significant figures is 1.5 × 10 . 4 Values would range from 14,501 to 15,499. Both of these values round to 1.5 × 10 .

1.48

Plan: Review the rules for significant zeros. Solution: a) No significant zeros (leading zeros are not significant) b) No significant zeros (leading zeros are not significant) c) 0.0410 (terminal zeros to the right of the decimal point are significant) 4 d) 4.0100 × 10 (zeros between nonzero digits and terminal zeros to the right of the decimal point are significant)

1.49

Plan: Review the rules for significant zeros. Solution: a) 5.08 (zeros between nonzero digits are significant) b) 508 (zeros between nonzero digits are significant) 3 c) 5.080 × 10 (zeros between nonzero digits are significant; terminal zeros to the right of the decimal point are significant) d) 0.05080 (leading zeros are not significant; zeros between nonzero digits are significant; terminal zeros to the right of the decimal point are significant)

1.50

Plan: Review the rules for rounding. Solution: (significant figures are underlined) a) 0.0003554: the extra digits are 54 at the end of the number. When the digit to be removed is 5 and that 5 is followed by nonzero numbers, the last digit kept is increased by 1: 0.00036 b) 35.8348: the extra digits are 48. Since the digit to be removed (4) is less than 5, the last digit kept is unchanged: 35.83 c) 22.4555: the extra digits are 555. When the digit to be removed is 5 and that 5 is followed by nonzero numbers, the last digit kept is increased by 1: 22.5

1.51

Plan: Review the rules for rounding. Solution: (significant figures are underlined) a) 231.554: the extra digits are 54 at the end of the number. When the digit to be removed is 5 and that 5 is followed by nonzero numbers, the last digit kept is increased by 1: 231.6 b) 0.00845: the extra digit is 5 at the end of the number. When the digit to be removed is 5 and that 5 is not followed by nonzero numbers, the last digit kept remains unchanged if it is even and increased by 1 if it is odd: 0.0084 c) 144,000: the extra digits are 4000 at the end of the number. When the digit to be removed (4) is less than 5, the 5 last digit kept remains unchanged: 140,000 (or 1.4 × 10 )

1.52

Plan: Review the rules for rounding. Solution: 19 rounds to 20: the digit to be removed (9) is greater than 5 so the digit kept is increased by 1. 155 rounds to 160: the digit to be removed is 5 and the digit to be kept is an odd number, so that digit kept is increased by 1. 8.3 rounds to 8: the digit to be removed (3) is less than 5 so the digit kept remains unchanged. 3.2 rounds to 3: the digit to be removed (2) is less than 5 so the digit kept remains unchanged. 2.9 rounds to 3: the digit to be removed (9) is greater than 5 so the digit kept is increased by 1. 4.7 rounds to 5: the digit to be removed (7) is greater than 5 so the digit kept is increased by 1.

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1-19

 20 160 8  2  = 568.89 = 6 × 10  33 5  Since there are numbers in the calculation with only one significant figure, the answer can be reported only to one significant figure. (Note that the answer is 560 with the original number of significant digits.) 1.53

Plan: Review the rules for rounding. Solution: 10.8 rounds to 11: the digit to be removed (8) is greater than 5 so the digit kept is increased by 1. 6.18 rounds to 6.2: the digit to be removed (8) is greater than 5 so the digit kept is increased by 1. 2.381 rounds to 2.38: the digit to be removed (1) is less than 5 so the digit kept remains unchanged. 24.3 rounds to 24: the digit to be removed (3) is less than 5 so the digit kept remains unchanged. 1.8 rounds to 2: the digit to be removed (8) is greater than 5 so the digit kept is increased by 1. 19.5 rounds to 20: the digit to be removed is 5 and the digit to be kept is odd, so that digit kept is increased by 1. 11 6.2  2.38    24  2  20  = 0.1691 = 0.2 Since there is a number in the calculation with only one significant figure, the answer can be reported only to one significant figure. (Note that the answer is 0.19 with original number of significant figures.)

1.54

Plan: Use a calculator to obtain an initial value. Use the rules for significant figures and rounding to get the final answer. Solution: a)

2.795 m3.10 m = 1.3371 = 1.34 m (maximum of 3 significant figures allowed since two of the original 6.48 m

numbers in the calculation have only 3 significant figures) 3 4 3 b) V =   = 21,620.74 = 21,621 mm (maximum of 5 significant figures allowed)  3 c) 1.110 cm + 17.3 cm + 108.2 cm + 316 cm = 442.61 = 443 cm (no digits allowed to the right of the decimal since 316 has no digits to the right of the decimal point)



1.55



Plan: Use a calculator to obtain an initial value. Use the rules for significant figures and rounding to get the final answer. Solution: 2.420 g  15.6 g a) = 3.7542 = 3.8 (maximum of 2 significant figures allowed since one of the original 4.8 g numbers in the calculation has only 2 significant figures) b)

7.87 mL 16.1 mL  8.44 mL

= 1.0274 = 1.0 (After the subtraction, the denominator has 2 significant figures; only one

digit is allowed to the right of the decimal in the value in the denominator since 16.1 has only one digit to the right of the decimal.) 2 3 c) V = (6.23 cm) (4.630 cm) = 564.556 = 565 cm (maximum of 3 significant figures allowed since one of the original numbers in the calculation has only 3 significant figures) 1.56

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1-20

Solution: 5 a) 1.310000 × 10 –4 b) 4.7 × 10 5 c) 2.10006 × 10 3 d) 2.1605 × 10

(Note that all zeros are significant.) (No zeros are significant.)

1.57

Plan: Review the procedure for changing a number to scientific notation. There can be only 1 nonzero digit to the left of the decimal point in correct scientific notation. Moving the decimal point to the left results in a positive exponent while moving the decimal point to the right results in a negative exponent. Solution: 2 a) 2.820 × 10 (Note that the zero is significant.) –2 b) 3.80 × 10 (Note the one significant zero.) 3 c) 4.2708 × 10 4 d) 5.82009 × 10

1.58

Plan: Review the examples for changing a number from scientific notation to standard notation. If the exponent is positive, move the decimal back to the right; if the exponent is negative, move the decimal point back to the left. Solution: a) 5550 (Do not use terminal decimal point since the zero is not significant.) b) 10070. (Use terminal decimal point since final zero is significant.) c) 0.000000885 d) 0.003004

1.59

Plan: Review the examples for changing a number from scientific notation to standard notation. If the exponent is positive, move the decimal back to the right; if the exponent is negative, move the decimal point back to the left. Solution: a) 6500. (Use terminal decimal point since the final zero is significant.) b) 0.0000346 c) 750 (Do not use terminal decimal point since the zero is not significant.) d) 188.56

1.60

Plan: In most cases, this involves a simple addition or subtraction of values from the exponents. There can be only 1 nonzero digit to the left of the decimal point in correct scientific notation. Solution: 4 2 2 4 a) 8.025 × 10 (The decimal point must be moved an additional 2 places to the left: 10 + 10 = 10 ) –3 3 –6 –3 b) 1.0098 × 10 (The decimal point must be moved an additional 3 places to the left: 10 + 10 = 10 ) –11 –2 –9 –11 c) 7.7 × 10 (The decimal point must be moved an additional 2 places to the right: 10 + 10 = 10 )

1.61

(The decimal point must be moved an additional 1 place to the left: 10 + 10 = 10 ) 2 –2 0 (The decimal point must be moved an additional 2 places to the left: 10 + 10 = 10 ) 3 –3 0 (The decimal point must be moved an additional 3 places to the left: 10 + 10 = 10 )

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1-21

1.62

6.626 10

34



J/s 2.9979 10 8 m/s 9

48910 m

= 4.06 × 10

–19

 = 4.062185 × 10

–19

–9

J (489 × 10 m limits the answer to 3 significant figures; units of m and s cancel)

6.0221023 molecules/mol1.2310 2 g 46.07 g/mol 24

= 1.6078 × 10

= 1.61 × 10 molecules (1.23 × 10 g limits answer to 3 significant figures; units of mol and g cancel) 1 1 5 c) 6.0221023 atoms/mol2.181018 J/atom 2  2  = 1.82333 × 10 2 3  5 –18 = 1.82 × 10 J/mol (2.18 × 10 J/atom limits answer to 3 significant figures; unit of atoms cancels) 1.63

Plan: Calculate a temporary answer by simply entering the numbers into a calculator. Then you will need to round the value to the appropriate number of significant figures. Cancel units as you would cancel numbers, and place the remaining units after your numerical answer. Solution: 4.3210 7 g 3 a) = 1.3909 = 1.39 g/cm 3 4 2 3.1416 1.9510 cm 3 7 (4.32 × 10 g limits the answer to 3 significant figures) 1.84 102 g44.7 m/s2 5 5 2 2 b) = 1.8382 × 10 = 1.84 × 10 g·m /s 2 2 (1.84 × 10 g limits the answer to 3 significant figures)

1.07104 mol / L 3.8103 mol / L c) = 0.16072 = 0.16 L/mol 3 8.35105 mol / L 1.48102 mol / L  2

–3

(3.8 × 10 mol/L limits the answer to 2 significant figures; mol /L in the numerator cancels mol /L in the denominator to leave mol/L in the denominator or units of L/mol) 1.64

Plan: Exact numbers are those which have no uncertainty. Unit definitions and number counts of items in a group are examples of exact numbers. Solution: a) The height of Angel Falls is a measured quantity. This is not an exact number. b) The number of planets in the solar system is a number count. This is an exact number. c) The number of grams in a pound is not a unit definition. This is not an exact number. d) The number of millimeters in a meter is a definition of the prefix “milli–.” This is an exact number.

1.65

Plan: Exact numbers are those which have no uncertainty. Unit definitions and number counts of items in a group are examples of exact numbers. Solution: a) The speed of light is a measured quantity. It is not an exact number. b) The density of mercury is a measured quantity. It is not an exact number. c) The number of seconds in an hour is based on the definitions of minutes and hours. This is an exact number. d) The number of states is a counted value. This is an exact number.

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1-22

1.66

Plan: Observe the figure, and estimate a reading the best you can. Solution: The scale markings are 0.2 cm apart. The end of the metal strip falls between the mark for 7.4 cm and 7.6 cm. If we assume that one can divide the space between markings into fourths, the uncertainty is one-fourth the separation between the marks. Thus, since the end of the metal strip falls between 7.45 and 7.55 we can report its length as 7.50 ± 0.05 cm. (Note: If the assumption is that one can divide the space between markings into halves only, then the result is 7.5 ± 0.1 cm.)

1.67

Plan: You are given the density values for five solvents. Use the mass and volume given to calculate the density of the solvent in the cleaner and compare that value to the density values given to identify the solvent. Use the uncertainties in the mass and volume to recalculate the density. Solution: mass 11.775 g  a) Density (g/mL)  = 0.7850 g/mL. The closest value is isopropanol. volume 15.00 mL b) Ethanol is denser than isopropanol. Recalculating the density using the maximum mass = (11.775 + 0.003) g with the minimum volume = (15.00 – 0.02) mL, gives mass 11.778 g Density (g/mL)   = 0.7862 g/mL. This result is still clearly not ethanol. volume 14.98 mL Yes, the equipment is precise enough.

1.68

Plan: Calculate the average of each data set. Remember that accuracy refers to how close a measurement is to the actual or true value while precision refers to how close multiple measurements are to each other. Solution: 8.72 g  8.74 g  8.70 g a) Iavg = = 8.7200 = 8.72 g 3 8.56 g  8.77 g  8.83 g IIavg = = 8.7200 = 8.72 g 3 8.50 g  8.48 g  8.51 g IIIavg = = 8.4967 = 8.50 g 3 8.41 g  8.72 g  8.55 g IVavg = = 8.5600 = 8.56 g 3 Sets I and II are most accurate since their average value, 8.72 g, is closest to the true value, 8.72 g. b) To get an idea of precision, calculate the range of each set of values: largest value – smallest value. A small range is an indication of good precision since the values are close to each other. Irange = 8.74 g – 8.70 g = 0.04 g IIrange = 8.83 g – 8.56 g = 0.27 g IIIrange = 8.51 g – 8.48 g = 0.03 g IVrange = 8.72 g – 8.41 g = 0.31 g Set III is the most precise (smallest range), but is the least accurate (the average is the farthest from the actual value). c) Set I has the best combination of high accuracy (average value = actual value) and high precision (relatively small range). d) Set IV has both low accuracy (average value differs from actual value) and low precision (has the largest range). Plan: Remember that accuracy refers to how close a measurement is to the actual or true value; since the bull’seye represents the actual value, the darts that are closest to the bull’s-eye are the most accurate. Precision refers to how close multiple measurements are to each other; darts that are positioned close to each other on the target have high precision.

1.69

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1-23

Solution: a) Experiments II and IV — the averages appear to be near each other. b) Experiments III and IV — the darts are closely grouped. c) Experiment IV and perhaps Experiment II — the average is in or near the bull’s-eye. d) Experiment III — the darts are close together, but not near the bull’s-eye. 1.70

Plan: If it is necessary to force something to happen, the potential energy will be higher. Solution: a) b)

a) The balls on the relaxed spring have a lower potential energy and are more stable. The balls on the compressed spring have a higher potential energy, because the balls will move once the spring is released. This configuration is less stable. b) The two + charges apart from each other have a lower potential energy and are more stable. The two + charges near each other have a higher potential energy, because they repel one another. This arrangement is less stable. 1.71

Plan: A physical change is one in which the physical form (or state) of a substance, but not its composition, is altered. A chemical change is one in which a substance is converted into a different substance with different composition and properties. Solution: a) Bonds have been broken in three yellow diatomic molecules. Bonds have been broken in three red diatomic molecules. The six resulting yellow atoms have reacted with three of the red atoms to form three molecules of a new substance. The remaining three red atoms have reacted with three blue atoms to form a new diatomic substance. b) There has been one physical change as the blue atoms at 273 K in the liquid phase are now in the gas phase at 473 K.

1.72

Plan: Use the concentrations of bromine given. Solution: Mass bromine in Dead Sea 0.50 g/L = = 7.7/1 Mass bromine in seawater 0.065 g/L

1.73

Plan: The swimming pool is a rectangle so the volume of the water can be calculated by multiplying the three dimensions of length, width, and the depth of the water in the pool. The depth in ft must be converted to units of 3 m before calculating the volume. The volume in m is then converted to volume in gal. The density of water is used to find the mass of this volume of water. Solution: 2 12 in   2.54 cm 10 m   a) Depth of water (m) = 4.8 ft   1 in  1 cm  = 1.46304 m  1 ft    3 3 Volume (m ) = length × width × depth = 50.0 m 25.0 m 1.46304 m = 1828.8 m 3 10 L  1.057 qt  1 gal   5 5 Volume (gal) = 1828.8 m 3   3   1 L  4 qt  = 4.8326 × 10 = 4.8 × 10 gal  1 m   











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1-24

b) Using the density of water = 1.0 g/mL.  4 qt  1000 mL 1.0 g  1 kg   6 6 Mass (kg) = 4.8326 10 5 gal     = 1.8288 × 10 = 1.8 × 10 kg  1 gal     1.057 qt  mL 1000 g 





1.74

Plan: In each case, calculate the overall density of the ball and contents and compare to the density of air. The 3 volume of the ball in cm is converted to units of L to find the density of the ball itself in g/L. The densities of the ball and the gas in the ball are additive because the volume of the ball and the volume of the gas are the same. Solution: a) Density of evacuated ball: the mass is only that of the sphere itself: 3  1 mL  10 L   Volume of ball (L) = 560 cm 3    = 0.560 = 0.56 L 3   1 mL  1 cm  mass 0.12 g   0.21 g/L Density of evacuated ball = volume 0.560 The evacuated ball will float because its density is less than that of air. b) Because the density of CO2 is greater than that of air, a ball filled with CO2 will sink. c) Density of ball + density of hydrogen = 0.0899 + 0.21 g/L = 0.30 g/L The ball will float because the density of the ball filled with hydrogen is less than the density of air. d) Because the density of O2 is greater than that of air, a ball filled with O2 will sink. e) Density of ball + density of nitrogen = 0.21 g/L + 1.165 g/L = 1.38 g/L The ball will sink because the density of the ball filled with nitrogen is greater than the density of air. 1.189 g   0.66584 g f) To sink, the total mass of the ball and gas must weigh 0.560 L  1 L  For the evacuated ball: 0.66584 – 0.12 g = 0.54585 = 0.55 g. More than 0.55 g would have to be added to make the ball sink. For ball filled with hydrogen:  0.0899 g   0.0503 g Mass of hydrogen in the ball = 0.56 L  1 L  Mass of hydrogen and ball = 0.0503 g + 0.12 g = 0.17 g 0.66584 – 0.17 g = 0.4958 = 0.50 g. More than 0.50 g would have to be added to make the ball sink.

1.75

Plan: Convert the crossto mm and then use the tensile strength of grunerite to find the mass that can be held up by a strand of grunerite with that cross-sectional area. Calculate the area of aluminum and steel that can match that mass. Solution:  1106 m 2   1 mm 2  2 2    = 1.0 × 10–6 mm2 Cross-sectional area (mm ) = 1.0  2 2    1  1103 m  

–6

Calculate the mass that can be held up by grunerite with a cross-sectional area of 1.0 × 10 mm :  3.5102 kg    3.5104 kg 1106 mm 2   1 mm 2  –4

Calculate the area of aluminum required to match a mass of 3.5 × 10 kg: 2 2  2.205 lb  2.54 cm2  10 mm   1.9916105 = 2.0 × 10–5 mm2  1 in    3.5104 kg   2 2    1 kg  2.510 4 lb  1 in   1 cm  –4

Calculate the area of steel required to match a mass of 3.5 × 10 kg: 2 2  2.205 lb  2.54 cm 2  10 mm   9.9580106 = 1.0 × 10–5 mm2  1 in    3.5104 kg  2 2    1 kg  5.0 10 4 lb  1 in  1 cm  Copyright

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1-25

1.76

Plan: Convert the surface area to m and then use the surface area and the depth to determine the volume of the 3 oceans (area × depth = volume) in m . The volume is then converted to liters, and finally to the mass of gold using the density of gold in g/L. Once the mass of the gold is known, its density is used to find the volume of that amount of gold. The mass of gold is converted to troy oz and the price of gold per troy oz gives the total price. Solution: 1000 m2  2  = 3.63 × 1014 m2 a) Area of ocean (m ) = 3.63108 km 2   1 km2  3 14 2 18 3 Volume of ocean (m ) = (area)(depth) = (3.63 × 10 m )(3800 m) = 1.3794 × 10 m  1 L  5.8109 g   = 8.00052 × 1012 = 8.0 × 1012 g Mass of gold (g) = 1.3794 1018 m 3  3 3  10 m  L  b) Use the density of gold to convert mass of gold to volume of gold:  1 cm3 0.01 m3  3  = 4.14535 × 105 = 4.1 × 105 m3  Volume of gold (m ) = 8.00052 1012 g  1 cm3  19.3 g  c) Value of gold = 

1.77





  

  

  



$4.11014

Plan: The mass of zinc in the sample of yellow zinc in part (a) is found from the percent of zinc in the sample. The mass of copper is found by subtracting the mass of zinc from the total mass of yellow zinc. In part (b), subtract the mass percent of zinc from 100 to find the mass percent of copper. Solution:   34% zinc  = 62.9 g Zn a) Mass of zinc in the 34% zinc sample = 185 g yellow zinc 100% yellow zinc 

  37% zinc  = 68.45 g Zn Mass of zinc in the 37% zinc sample = 185 g yellow zinc 100% yellow zinc  Mass copper = total mass – mass zinc Mass copper (34% zinc sample) = 185 g – 62.9 g = 122.1 = 122 g Mass copper (37% zinc sample) = 185 g – 68.45 g = 116.55 = 117 g 117 to 122 g copper b) The 34% zinc sample contains 100 – 34 = 66% copper. The 37% zinc sample contains 100 – 37 = 63% copper.  34% zinc   = 23.95 = 24 g Mass of zinc = 46.5 g copper   66% copper   37% zinc   = 27.31 = 27 g Mass of zinc = 46.5 g copper   63% copper  24 to 27 g zinc 1.78

Plan: Use the equations for temperature conversion given in the chapter. The mass of nitrogen is conserved when the gas is liquefied; the mass of the nitrogen gas equals the mass of the liquid nitrogen. Use the density of nitrogen gas to find the mass of the nitrogen; then use the density of liquid nitrogen to find the volume of that mass of liquid nitrogen. Solution: a) T (in °C) = T (in K) – 273.15 = 77.36 K – 273.15 = –195.79°C 9 9 b) T (in °F) = T(in °C) + 32 = (–195.79°C) + 32 = –320.422 = –320.42°F 5 5

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1-26

 4.566 g  c) Mass of liquid nitrogen = mass of gaseous nitrogen = 895.0 L   = 4086.57 g N2  1 L 

1.79

 1 L   = 5.0514 = 5.05 L Volume of liquid N2 = 4086.57 g  809 g  Plan: For part (a), convert mi to m and h to s. For part (b), time is converted from h to min and length from mi to km. For part (c), convert the distance in ft to mi and use the average speed in mi/h to find the time necessary to cover the given distance. Solution:  5.9 mi  1000 m   1 h   a) Speed (m/s) =    = 2.643 = 2.6 m/s  h  0.62 mi   3600 s   1 h  5.9 mi  1 km   b) Distance (km) = 98 min   = 15.543 = 16 km   60 min   h  0.62 mi 





 1 mi  1 h  c) Time (h) = 4.7510 4 ft    = 1.5248 = 1.5 h  5280 ft  5.9 mi 

If she starts running at 11:15 am, 1.5 hours later the time is 12:45 pm. 1.80

Plan: A physical change is one in which the physical form (or state) of a substance, but not its composition, is altered. A chemical change is one in which a substance is converted into a different substance with different composition and properties. A physical property is a characteristic shown by a substance itself, without interacting with or changing into other substances. A chemical property is a characteristic of a substance that appears as it interacts with, or transforms into, other substances. Solution: a) Scene A shows a physical change. The substance changes from a solid to a gas but a new substance is not formed. b) Scene B shows a chemical change. Two diatomic elements form from a diatomic compound. c) Both Scenes A and B result in different physical properties. Physical and chemical changes result in different physical properties. d) Scene B is a chemical change; therefore, it results in different chemical properties. e) Scene A results in a change in state. The substance changes from a solid to a gas.

1.81

Plan: In visualizing the problem, the two scales can be set next to each other. Solution: There are 50 divisions between the freezing point and boiling point of benzene on the °X scale and 74.6 divisions  50X  o o (80.1 C – 5.5 C) on the °C scale. So °X =  °C  74.6C  This does not account for the offset of 5.5 divisions in the °C scale from the zero point on the °X scale.  50X  So °X =  (°C – 5.5°C)  74.6C  Check: Plug in 80.1°C and see if result agrees with expected value of 50°X.  50X  So °X =  (80.1°C – 5.5°C) = 50°X  74.6C  Use this formula to find the freezing and boiling points of water on the °X scale.  50X  fpwater °X =  (0.00°C – 5.5°C) = 3.68°X = –3.7°X  74.6C 

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1-27

 50X  bpwater °X =  (100.0°C – 5.5°C) = 63.3°X  74.6C 

1.82

Plan: Determine the total mass of Earth’s crust in metric tons (t) by finding the volume of crust 3 3 (surface area × depth) in km and then in cm and then using the density to find the mass of this volume, using conversions from the inside back cover. The mass of each individual element comes from the concentration of that element multiplied by the mass of the crust. Solution: 3 10 3 Volume of crust (km ) = area × depth = 35 km 5.10 108 km 2  = 1.785 × 10 km 3 1000 m3  3  1 cm  = 1.785 × 1025 cm3 Volume of crust (cm ) = 1.7851010 km 3  3   1 km  0.01 m3   2.8 g  1 kg   1 t  = 4.998 × 1019 t  Mass of crust (t) = 1.7851025 cm 3   1000 kg  3 1 cm 1000 g  

 4.55105 g oxygen   = 2.2741 × 1025 = 2.3 × 1025 g oxygen Mass of oxygen (g) = 4.9981019 t   1t   2.72 105 g silicon   = 1.3595 × 1025 = 1.4 × 1025 g silicon Mass of silicon (g) = 4.9981019 t    1t  1104 g element   Mass of ruthenium = mass of rhodium = 4.9981019 t  1t   15

= 4.998 × 10 = 5 × 10 g each of ruthenium and rhodium

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1-28

CHAPTER 2 THE COMPONENTS OF MATTER FOLLOW---UP PROBLEMS 2.1A

Plan: An element has only one kind of atom; a compound is composed of at least two kinds of atoms. A mixture consists of two or more substances mixed together in the same container. Solution: (a) There is only one type of atom (blue) present, so this is an element. (b) Two different atoms (brown and green) appear in a fixed ratio of 1/1, so this is a compound. (c) These molecules consist of one type of atom (orange), so this is an element.

2.1B

Plan: An element has only one kind of atom; a compound is composed of at least two kinds of atoms. Solution: The circle on the left contains molecules with either only orange atoms or only blue atoms. This is a mixture of two different elements. In the circle on the right, the molecules are composed of one orange atom and one blue atom so this is a compound.

2.2A

Plan: Use the mass fraction of iron in fool’s gold to find the mass of fool’s gold that contains 86.2 g of iron. Subtract the amount of iron in the mass fraction from the amount of fool’s gold in the mass fraction to obtain the mass of sulfur in that amount of fool’s gold. Find the mass fraction of sulfur in fool’s gold and multiply the amount of fool’s gold by the mass fraction of sulfur to determine the mass of sulfur in the sample. Solution:   Mass (g) of fool’s gold = 86.2 g iron ×   = 185.195 g fool’s gold   Mass (g) of sulfur in 110.0 g of fool’s gold = 110.0 g fool’s gold --- 51.2 g iron = 58.8 g sulfur    = 98.995 = 99.0 g sulfur Mass (g) of sulfur = 185.195 g fool’s gold ×   

2.2B

Plan: Subtract the amount of silver from the amount of silver bromide to find the mass of bromine in 26.8 g of silver bromide. Use the mass fraction of silver in silver bromide to find the mass of silver in 3.57 g of silver bromide. Use the mass fraction of bromine in silver bromide to find the mass of bromine in 3.57 g of silver bromide. Solution: Mass (g) of bromine in 26.8 g silver bromide = 26.8 g silver bromide --- 15.4 g silver = 11.4 g bromine    = 2.05 g silver Mass (g) of silver in 3.57 g silver bromide = 3.57 g silver bromide   

 Mass (g) of bromine in 3.57 g silver bromide = 3.57 g silver bromide   2.3A

  = 1.52 g bromine 

Plan: The law of multiple proportions states that when two elements react to form two compounds, the different masses of element B that react with a fixed mass of element A is a ratio of small whole numbers. The law of definite composition states that the elements in a compound are present in fixed parts by mass. The law of mass conversation states that the total mass before and after a reaction is the same. Solution:

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2-1

The law of mass conservation is illustrated because the number of atoms does not change as the reaction proceeds (there are 14 red spheres and 12 black spheres before and after the reaction occurs). The law of multiple proportions is illustrated because two compounds are formed as a result of the reaction. One of the compounds has a ratio of 2 red spheres to 1 black sphere. The other has a ratio of 1 red sphere to 1 black sphere. The law of definite proportions is illustrated because each compound has a fixed ratio of red-to-black atoms. 2.3B

Plan: The law of multiple proportions states that when two elements react to form two compounds, the different masses of element B that react with a fixed mass of element A is a ratio of small whole numbers. Solution: Only Sample B shows two different bromine-fluorine compounds. In one compound there are three fluorine atoms for every one bromine atom; in the other compound, there is one fluorine atom for every bromine atom.

2.4A

Plan: The subscript (atomic number = Z ) gives the number of protons, and for an atom, the number of electrons. The atomic number identifies the element. The superscript gives the mass number (A) which is the total of the protons plus neutrons. The number of neutrons is simply the mass number minus the atomic number (A --- Z ). Solution: 46 + --0 Ti Z = 22 and A = 46, there are 22 p and 22 e and 46 --- 22 = 24 n 47 + --0 Ti Z = 22 and A = 47, there are 22 p and 22 e and 46 --- 22 = 25 n 48 + --0 Ti Z = 22 and A = 48, there are 22 p and 22 e and 46 --- 22 = 26 n 49 + --0 Ti Z = 22 and A = 49, there are 22 p and 22 e and 46 --- 22 = 27 n 50 + --0 Ti Z = 22 and A = 50, there are 22 p and 22 e and 46 --- 22 = 28 n

2.4B

Plan: The subscript (atomic number = Z ) gives the number of protons, and for an atom, the number of electrons. The atomic number identifies the element. The superscript gives the mass number (A) which is the total of the protons plus neutrons. The number of neutrons is simply the mass number minus the atomic number (A --- Z ). Solution: (a) Mass number --- number of neutrons = 88 --- 50 = 38 = atomic number. The element is strontium. (b) Mass number --- atomic number = 86 --- 38 = 48 = number of neutrons.

2.5A

Plan: First, divide the percent abundance value by 100 to obtain the fractional value for each isotope. Multiply each isotopic mass by the fractional value, and add the resulting masses to obtain silicon’s atomic mass. Solution: 28 28 29 29 Atomic Mass = ( Si mass) (fractional abundance of Si) + ( Si mass) (fractional abundance of Si) + 30 30 ( Si mass) (fractional abundance of Si) Atomic mass = (27.97693 amu)(0.9223) + 28.976495 amu)(0.0467) + (29.973770 amu)(0.0310) = 28.09 amu

2.5B

Plan: To find the percent abundance of each B isotope, let x equal the fractional abundance of B and (1 --- x) 11 10 equal the fractional abundance of B. Remember that atomic mass = isotopic mass of B x fractional abundance) 11 + (isotopic mass of B x fractional abundance). Solution: 10 10 11 11 Atomic Mass = ( B mass) (fractional abundance of B) + ( B mass) (fractional abundance of B) 10 11 10 11 Amount of B + Amount B = 1 (setting B = x gives B = 1 --- x) 10.81 amu = (10.0129 amu)(x) + (11.0093 amu) (1 --- x) 10.81 amu = 11.0093 --- 11.0093x + 10.0129 x 10.81 amu = 11.0093 --- 0.9964 x ---0.1993 = ---0.9964x x = 0.20; 1 --- x = 0.80 (10.81 --- 11.0093 limits the answer to 2 significant figures) Fraction x 100% = percent abundance. 10 11 % abundance of B = 20.%; % abundance of B = 80.%

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2-2

2.6A

Plan: Use the provided atomic numbers (the Z numbers) to locate these elements on the periodic table. The name of the element is on the periodic table or on the list of elements inside the front cover of the textbook. Use the periodic table to find the group/column number (listed at the top of each column) and the period/row number (listed at the left of each row) in which the element is located. Classify the element from the color coding in the periodic table. Solution: (a) Z = 14: Silicon, Si; Group 4A(14) and Period 3; metalloid (b) Z = 55: Cesium, Cs; Group 1A(1) and Period 6; main-group metal (c) Z = 54: Xenon, Xe; Group 8A(18) and Period 5; nonmetal

2.6B

Plan: Use the provided atomic numbers (the Z numbers) to locate these elements on the periodic table. The name of the element is on the periodic table or on the list of elements inside the front cover of the textbook. Use the periodic table to find the group/column number (listed at the top of each column) and the period/row number (listed at the left of each row) in which the element is located. Classify the element from the color coding in the periodic table. Solution: (a) Z = 12: Magnesium, Mg; Group 2A(2) and Period 3; main-group metal (b) Z = 7: Nitrogen, N; Group 5A(15) and Period 2; nonmetal (c) Z = 30: Zinc, Zn; Group 2B(12) and Period 4; transition metal

2.7A

Plan: Locate these elements on the periodic table and predict what ions they will form. For A-group cations (metals), ion charge = group number; for anions (nonmetals), ion charge = group number ---8. Or, relate the element’s position to the nearest noble gas. Elements after a noble gas lose electrons to become positive ions, while those before a noble gas gain electrons to become negative ions. Solution: 2--a) 16S [Group 6A(16); 6 --- 8 = ---2]; sulfur needs to gain 2 electrons to match the number of electrons in 18Ar. + b) 37Rb [Group 1A(1)]; rubidium needs to lose 1 electron to match the number of electrons in 36Kr. 2+ c) 56Ba [Group 2A(2)]; barium needs to lose 2 electrons to match the number of electrons in 54Xe.

2.7B

Plan: Locate these elements on the periodic table and predict what ions they will form. For A-group cations (metals), ion charge = group number; for anions (nonmetals), ion charge = group number ---8. Or, relate the element’s position to the nearest noble gas. Elements after a noble gas lose electrons to become positive ions, while those before a noble gas gain electrons to become negative ions. Solution: 2+ a) 38Sr [Group 2A(2)]; strontium needs to lose 2 electrons to match the number of electrons in 36Kr. 2--b) 8O [Group 6A(16); 6 --- 8 = ---2]; oxygen needs to gain 2 electrons to match the number of electrons in 10Ne. + c) 55Cs [Group 1A(1)]; cesium needs to lose 1 electron to match the number of electrons in 54Xe.

2.8A

Plan: When dealing with ionic binary compounds, the first name is that of the metal and the second name is that of the nonmetal. If there is any doubt, refer to the periodic table. The metal name is unchanged, while the nonmetal has an -ide suffix added to the nonmetal root. Solution: a) Zinc is the metal and oxygen is the nonmetal: zinc oxide. b) Silver is the metal and bromine is the nonmetal: silver bromide. c) Lithium is the metal and chlorine is the nonmetal: lithium chloride. d) Aluminum is the metal and sulfur is the nonmetal: aluminum sulfide.

2.8B

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2-3

Solution: a) Potassium is the metal and sulfur is the nonmetal: potassium sulfide. b) Barium is the metal and iodine is the nonmetal: barium iodide. c) Cesium is the metal and nitrogen is the nonmetal: cesium nitride. d) Sodium is the metal and hydrogen is the nonmetal: sodium hydride. 2.9A

Plan: Use the charges of the ions to predict the lowest ratio leading to a neutral compound. The sum of the total charges must be 0. Solution: 2+ 2--a) Zinc should form Zn and oxygen should form O ; these will combine to give ZnO. The charges cancel [+2 + ---2 = 0], so this is an acceptable formula. + --b) Silver should form Ag and bromine should form Br ; these will combine to give AgBr. The charges cancel [+1 + ---1 = 0], so this is an acceptable formula. + --c) Lithium should form Li and chlorine should form Cl ; these will combine to give LiCl. The charges cancel [+1 + ---1 = 0], so this is an acceptable formula. 3+ 2--d) Aluminum should form Al and sulfur should form S ; to produce a neutral combination the formula is Al2S3. This way the charges will cancel [2(+3) + 3(---2) = 0], so this is an acceptable formula.

2.9B

Plan: Use the charges of the ions to predict the lowest ratio leading to a neutral compound. The sum of the total charges must be 0. Solution: + 2--a) Potassium should form K and sulfur should form S ; these will combine to give K2S. The charges cancel [2(+1) + ---2 = 0], so this is an acceptable formula. 2+ – b) Barium should form Ba and iodine should form I , these will combine to give BaI2. The charges cancel [+2 + 2(–1) = 0], so this is an acceptable formula. + 3--c) Cesium should form Cs and nitrogen should form N ; these will combine to give Cs 3N. The charges cancel [3(+1) + ---3 = 0], so this is an acceptable formula. + --d) Sodium should form Na and hydrogen should form H ; to produce a neutral combination the formula is NaH. This way the charges will cancel (+1 + ---1 = 0), so this is an acceptable formula.

2.10A

Plan: Determine the names or symbols of each of the species present. Then combine the species to produce a name or formula. The metal or positive ions are written first. Review the rules for nomenclature covered in the chapter. For metals, like many transition metals, that can form more than one ion each with a different charge, the ionic charge of the metal ion is indicated by a Roman numeral within parentheses immediately following the metal’s name. Solution: 4+ 2--a) The Roman numeral means that the lead is Pb ; oxygen produces the usual O . The neutral combination is [+4 + 2(---2) = 0], so the formula is PbO2. b) Sulfide (Group 6A(16)), like oxide, is ---2 (6 --- 8 = ---2). This is split between two copper ions, each of which must be +1. This is one of the two common charges for copper ions. The +1 charge on the copper is indicated with a Roman numeral. This gives the name copper(I) sulfide (common name = cuprous sulfide). c) Bromine (Group 7A(17)), like other elements in the same column of the periodic table, forms a ---1 ion. Two of these ions require a total of +2 to cancel them out. Thus, the iron must be +2 (indicated with a Roman numeral). This is one of the two common charges on iron ions. This gives the name iron(II) bromide (or ferrous bromide). 2+ --d) The mercuric ion is Hg , and two ---1 ions (Cl ) are needed to cancel the charge. This gives the formula HgCl2.

2.10B

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2-4

Solution: 2+ 3--a) The Roman numeral means that the copper ion is Cu ; nitride produces the usual N . The neutral combination 2+ 3--is [3(+2) + 2(---3) = 0], so the formula is Cu 3N2. Three Cu ions balance two N ions. 2+ b) The anion is iodide, I , and the formula shows two I . Therefore the cation must be Pb , lead(II) ion: PbI2 is lead(II) iodide. (The common name is plumbous iodide.) 3+ 2--c) Chromic is the common name for chromium(III) ion, Cr ; sulfide ion is S . To balance the charges, the formula is Cr2S3. [The systematic name is chromium(III) sulfide.] 2--2+ d) The anion is oxide, O , which requires that the cation be Fe . The name is iron(II) oxide. (The common name is ferrous oxide.) 2.11A

Plan: Determine the names or symbols of each of the species present. Then combine the species to produce a name or formula. The metal or positive ions always go first. Solution: 2+ --a) The cupric ion, Cu , requires two nitrate ions, NO3 , to cancel the charges. Trihydrate means three water molecules. These combine to give Cu(NO3)2 3H2O. 2+ --b) The zinc ion, Zn , requires two hydroxide ions, OH , to cancel the charges. These combine to give Zn(OH) 2. + --c) Lithium only forms the Li ion, so Roman numerals are unnecessary. The cyanide ion, CN , has the appropriate charge. These combine to give lithium cyanide.

2.11B

Plan: Determine the names or symbols of each of the species present. Then combine the species to produce a name or formula. The metal or positive ions always go first. Solution: + 2--a) Two ammonium ions, NH4 , are needed to balance the charge on one sulfate ion, SO4 . These combine to give (NH4)2SO4. --2+ b) The nickel ion is combined with two nitrate ions, NO3 , so the charge on the nickel ion is 2+, Ni . There are 6 water molecules (hexahydrate). Therefore, the name is nickel(II) nitrate hexahydrate. + --c) Potassium forms the K ion. The bicarbonate ion, HCO3 , has the appropriate charge to balance out one potassium ion. Therefore, the formula of this compound is KHCO3.

2.12A

Plan: Determine the names or symbols of each of the species present. Then combine the species to produce a name or formula. The metal or positive ions always go first. Make corrections accordingly. Solution: + 3--a) The ammonium ion is NH4 and the phosphate ion is PO4 . To give a neutral compound they should combine [3(+1) + (---3) = 0] to give the correct formula (NH4)3PO4. 3+ --b) Aluminum gives Al and the hydroxide ion is OH . To give a neutral compound they should combine [+3 + 3(---1) = 0] to give the correct formula Al(OH) 3. Parentheses are required around the polyatomic ion. 2+ c) Manganese is Mn, and Mg, in the formula, is magnesium. Magnesium only forms the Mg ion, so Roman --numerals are unnecessary. The other ion is HCO3 , which is called the hydrogen carbonate (or bicarbonate) ion. The correct name is magnesium hydrogen carbonate or magnesium bicarbonate.

2.12B

Plan: Determine the names or symbols of each of the species present. Then combine the species to produce a name or formula. The metal or positive ions always go first. Make corrections accordingly. Solution: 3----a) Either use the ‘‘-ic’’ suffix or the ‘‘(III)’’ but not both. Nitride is N , and nitrate is NO3 . This gives the correct name: chromium(III) nitrate (the common name is chromic nitrate). ----b) Cadmium is Cd, and Ca, in the formula, is calcium. Nitrate is NO3 , and nitrite is NO2 . The correct name is calcium nitrite. ----c) Potassium is K, and P, in the formula, is phosphorus. Perchlorate is ClO4 , and chlorate is ClO3 . Additionally, parentheses are not needed when there is only one of a given polyatomic ion. The correct formula is KClO3.

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2-5

2.13A

Plan: Use the name of the acid to determine the name of the anion of the acid. The name hydro______ic acid indicates that the anion is a monatomic nonmetal. The name ______ic acid indicates that the anion is an oxoanion with an --- ate ending. The name ______ous acid indicates that the anion is an oxoanion with an ---ite ending. Solution: --a) Chloric acid is derived from the chlorate ion, ClO3 . The ---1 charge on the ion requires one hydrogen. These combine to give the formula HClO3. --b) Hydrofluoric acid is derived from the fluoride ion, F . The ---1 charge on the ion requires one hydrogen. These combine to give the formula HF. ----c) Acetic acid is derived from the acetate ion, which may be written as CH3COO or as C 2H3O2 . The ---1 charge means that one H is needed. These combine to give the formula CH3COOH or HC2H3O2. --d) Nitrous acid is derived from the nitrite ion, NO2 . The ---1 charge on the ion requires one hydrogen. These combine to give the formula HNO2.

2.13B

Plan: Remove a hydrogen ion to determine the formula of the anion. Identify the corresponding name of the anion and use the name of the anion to name the acid. For the oxoanions, the -ate suffix changes to -ic acid and the -ite suffix changes to -ous acid. For the monatomic nonmetal anions, the name of the acid includes a hydro- prefix and the ---ide suffix changes to ---ic acid. Solution: --a) Removing a hydrogen ion from the formula H2SO3 gives the oxoanion HSO3 , hydrogen sulfite; removing two 2--hydrogen ions gives the oxoanion SO3 , sulfite. To name the acid, the ‘‘-ite of ‘‘sulfite’’ must be replaced with ‘‘-ous.’’ The corresponding name is sulfurous acid. --b) HBrO is an oxoacid containing the BrO ion (hypobromite ion). To name the acid, the ‘‘-ite’’ must be replaced with ‘‘-ous.’’ This gives the name: hypobromous acid. --c) HClO2 is an oxoacid containing the ClO2 ion (chlorite ion). To name the acid, the ‘‘-ite’’ must be replaced with ‘‘-ous.’’ This gives the name: chlorous acid. --d) HI is a binary acid containing the I ion (iodide ion). To name the acid, a ‘‘hydro-’’ prefix is used, and the ‘‘-ide’’ must be replaced with ‘‘-ic.’’ This gives the name: hydroiodic acid.

2.14A

Plan: Determine the names or symbols of each of the species present. Since these are binary compounds consisting of two nonmetals, the number of each type of atom is indicated with a Greek prefix. Solution: a) Sulfur trioxide ----- one sulfur and three (tri) oxygens, as oxide, are present. b) Silicon dioxide ----- one silicon and two (di) oxygens, as oxide, are present. c) N2O Nitrogen has the prefix ‘‘di’’ = 2, and oxygen has the prefix ‘‘mono’’ = 1 (understood in the formula). d) SeF6 Selenium has no prefix (understood as = 1), and the fluoride has the prefix ‘‘hexa’’ = 6.

2.14B

Plan: Determine the names or symbols of each of the species present. Since these are binary compounds consisting of two nonmetals, the number of each type of atom is indicated with a Greek prefix. Solution: a) Sulfur dichloride ----- one sulfur and two (di) chlorines, as chloride, are present. b) Dinitrogen pentoxide ----- two (di) nitrogen and five (penta) oxygens, as oxide, are present. Note that the ‘‘a’’ in ‘‘penta’’ is dropped when this prefix is combined with ‘‘oxide’’. c) BF3 Boron doesn’t have a prefix, so there is one boron atom present. Fluoride has the prefix ‘‘tri’’ = 3. d) IBr3 Iodine has no prefix (understood as = 1), and the bromide has the prefix ‘‘tri’’ = 3.

2.15A

Plan: Determine the names or symbols of each of the species present. For compounds between nonmetals, the number of atoms of each type is indicated by a Greek prefix. If both elements in the compound are in the same group, the one with the higher period number is named first. Solution: a) Suffixes are not used in the common names of the nonmetal listed first in the formula. Sulfur does not qualify for the use of a suffix. Chlorine correctly has an ‘‘ide’’ suffix. There are two of each nonmetal atom, so both names require a ‘‘di’’ prefix. This gives the name disulfur dichloride.

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2-6

b) Both elements are nonmetals, and there is just one nitrogen and one oxygen. These combine to give the formula NO. c) Br has a higher period number than Cl and should be named first. The three chlorides are correctly named. The correct name is bromine trichloride. 2.15B

Plan: Determine the names or symbols of each of the species present. For compounds between nonmetals, the number of atoms of each type is indicated by a Greek prefix. If both elements in the compound are in the same group, the one with the higher period number is named first. Solution: a) The name of the element phosphorus ends in ---us, not ---ous. Additionally, the prefix hexa- is shortened to hexbefore oxide. The correct name is tetraphosphorus hexoxide. b) Because sulfur is listed first in the formula (and has a lower group number), it should be named first. The fluorine should come second in the name, modified with an ---ide ending. The correct name is sulfur hexafluoride. c) Nitrogen’s symbol is N, not Ni. Additionally, the second letter of an element symbol should be lowercase (Br, not BR). The correct formula is NBr3.

2.16A

Plan: First, write a formula to match the name. Next, multiply the number of each type of atom by the atomic mass of that atom. Sum all the masses to get an overall mass. Solution: 2--a) The peroxide ion is O2 , which requires two hydrogen atoms to cancel the charge: H2O2. Molecular mass = (2 × 1.008 amu) + (2 × 16.00 amu) = 34.016 = 34.02 amu. +1 2--b) Two Cs ions are required to balance the charge on one CO3 ion: Cs2CO3; formula mass = (2 × 132.9 amu) + (1 × 12.01 amu) + (3 × 16.00 amu) = 325.81 = 325.8 amu.

2.16B

Plan: First, write a formula to match the name. Next, multiply the number of each type of atom by the atomic mass of that atom. Sum all the masses to get an overall mass. Solution: 2--a) Sulfuric acid contains the sulfate ion, SO4 , which requires two hydrogen atoms to cancel the charge: H2SO4; molecular mass = (2 × 1.008 amu) + 32.06 amu + (4 × 16.00 amu) = 98.076 = 98.08 amu. 2--+ b) The sulfate ion, SO4 , requires two +1 potassium ions, K , to give K2SO4; formula mass = (2 × 39.10 amu) + 32.06 amu + (4 × 16.00 amu) = 174.26 amu.

2.17A

Plan: Since the compounds only contain two elements, finding the formulas by counting each type of atom and developing a ratio. Name the compounds. Multiply the number of each type of atom by the atomic mass of that atom. Sum all the masses to get an overall mass. Solution: a) There are two brown atoms (sodium) for every red (oxygen). The compound contains a metal with a nonmetal. Thus, the compound is sodium oxide, with the formula Na2O. The formula mass is twice the mass of sodium plus the mass of oxygen: 2 (22.99 amu) + (16.00 amu) = 61.98 amu b) There is one blue (nitrogen) and two reds (oxygen) in each molecule. The compound only contains nonmetals. Thus, the compound is nitrogen dioxide, with the formula NO2. The molecular mass is the mass of nitrogen plus twice the mass of oxygen: (14.01 amu) + 2 (16.00 amu) = 46.01 amu.

2.17B

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2-7

b) There is one green (chlorine) and three golds (fluorine) in each molecule. The compound only contains nonmetals. Thus, the compound is chlorine trifluoride, with the formula ClF3. The molecular mass is the mass of chlorine plus three times the mass of fluorine: (35.45 amu) +3 (19.00 amu) = 92.45 amu. END---OF---CHAPTER PROBLEMS 2.1

Plan: Refer to the definitions of an element and a compound. Solution: Unlike compounds, elements cannot be broken down by chemical changes into simpler materials. Compounds contain different types of atoms; there is only one type of atom in an element.

2.2

Plan: Refer to the definitions of a compound and a mixture. Solution: 1) A compound has constant composition but a mixture has variable composition. 2) A compound has distinctly different properties than its component elements; the components in a mixture retain their individual properties. Plan: Recall that a substance has a fixed composition. Solution: a) The fixed mass ratio means it has constant composition, thus, it is a pure substance (compound). b) All the atoms are identical, thus, it is a pure substance (element). c) The composition can vary, thus, this is an impure substance (a mixture). d) The specific arrangement of different atoms means it has constant composition, thus, it is a pure substance (compound).

2.3

2.4

Plan: Remember that an element contains only one kind of atom while a compound contains at least two different elements (two kinds of atoms) in a fixed ratio. A mixture contains at least two different substances in a composition that can vary. Solution: a) The presence of more than one element (calcium and chlorine) makes this pure substance a compound. b) There are only atoms from one element, sulfur, so this pure substance is an element. c) This is a combination of two compounds and has a varying composition, so this is a mixture. d) The presence of more than one type of atom means it cannot be an element. The specific, not variable, arrangement means it is a compound.

2.5

Some elements, such as the noble gases (He, Ne, Ar, etc.) occur as individual atoms. Many other elements, such as most other nonmetals (O2, N2, S8, P4, etc.) occur as molecules.

2.6

Compounds contain atoms from two or more elements, thus the smallest unit must contain at least a pair of atoms in a molecule.

2.7

Mixtures have variable composition; therefore, the amounts may vary. Compounds, as pure substances, have constant composition so their composition cannot vary.

2.8

The tap water must be a mixture, since it consists of some unknown (and almost certainly variable) amount of dissolved substance in solution in the water.

2.9

Plan: Recall that an element contains only one kind of atom; the atoms in an element may occur as molecules. A compound contains two kinds of atoms (different elements). Solution: a) This scene has 3 atoms of an element, 2 molecules of one compound (with one atom each of two different elements), and 2 molecules of a second compound (with 2 atoms of one element and one atom of a second element).

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2-8

b) This scene has 2 atoms of one element, 2 molecules of a diatomic element, and 2 molecules of a compound (with one atom each of two different elements). c) This scene has 2 molecules composed of 3 atoms of one element and 3 diatomic molecules of the same element. 2.10

Plan: Recall that a mixture is composed of two or more substances physically mixed, with a composition that can vary. Solution: The street sample is a mixture. The mass of vitamin C per gram of drug sample can vary. Therefore, if several samples of the drug have the same mass of vitamin C per gram of sample, this is an indication that the samples all have a common source. Samples of the street drugs with varying amounts of vitamin C per gram of sample have different sources. The constant mass ratio of the components indicates mixtures that have the same composition by accident, not of necessity.

2.11

Separation techniques allow mixtures (with varying composition) to be separated into the pure substance components which can then be analyzed by some method. Only when there is a reliable way of determining the composition of a sample, can you determine if the composition is constant.

2.12

Plan: Restate the three laws in your own words. Solution: a) The law of mass conservation applies to all substances ----- elements, compounds, and mixtures. Matter can neither be created nor destroyed, whether it is an element, compound, or mixture. b) The law of definite composition applies to compounds only, because it refers to a constant, or definite, composition of elements within a compound. c) The law of multiple proportions applies to compounds only, because it refers to the combination of elements to form compounds.

2.13

In ordinary chemical reactions (i.e., those that do not involve nuclear transformations), mass is conserved and the law of mass conservation is still valid.

2.14

Plan: Review the three laws: law of mass conservation, law of definite composition, and law of multiple proportions. Solution: a) Law of Definite Composition ----- The compound potassium chloride, KCl, is composed of the same elements and same fraction by mass, regardless of its source (Chile or Poland). b) Law of Mass Conservation ----- The mass of the substances inside the flashbulb did not change during the chemical reaction (formation of magnesium oxide from magnesium and oxygen). c) Law of Multiple Proportions ----- Two elements, O and As, can combine to form two different compounds that have different proportions of As present.

2.15

Plan: The law of multiple proportions states that two elements can form two different compounds in which the proportions of the elements are different. Solution: Scene B illustrates the law of multiple proportions for compounds of chlorine and oxygen. The law of multiple proportions refers to the different compounds that two elements can form that have different proportions of the elements. Scene B shows that chlorine and oxygen can form both Cl2O, dichlorine monoxide, and ClO2, chlorine dioxide.

2.16

Plan: Review the definition of percent by mass. Solution: a) No, the mass percent of each element in a compound is fixed. The percentage of Na in the compound NaCl is 39.34% (22.99 amu/58.44 amu), whether the sample is 0.5000 g or 50.00 g.

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2-9

b) Yes, the mass of each element in a compound depends on the mass of the compound. A 0.5000 g sample of NaCl contains 0.1967 g of Na (39.34% of 0.5000 g), whereas a 50.00 g sample of NaCl contains 19.67 g of Na (39.34% of 50.00 g). 2.17

Generally no, the composition of a compound is determined by the elements used, not their amounts. If too much of one element is used, the excess will remain as unreacted element when the reaction is over.

2.18

Plan: Review the mass laws: law of mass conservation, law of definite composition, and law of multiple proportions. For each experiment, compare the mass values before and after each reaction and examine the ratios of the mass of white compound to the mass of colorless gas. Solution: Experiment 1: mass before reaction = 1.00 g; mass after reaction = 0.64 g + 0.36 g = 1.00 g Experiment 2: mass before reaction = 3.25 g; mass after reaction = 2.08 g + 1.17 g = 3.25 g Both experiments demonstrate the law of mass conservation since the total mass before reaction equals the total mass after reaction. Experiment 1: mass white compound/mass colorless gas = 0.64 g/0.36 g = 1.78 Experiment 2: mass white compound/mass colorless gas = 2.08 g/1.17 g = 1.78 Both Experiments 1 and 2 demonstrate the law of definite composition since the compound has the same composition by mass in each experiment.

2.19

Plan: Review the mass laws: law of mass conservation, law of definite composition, and law of multiple proportions. For each experiment, compare the mass values before and after each reaction and examine the ratios of the mass of reacted copper to the mass of reacted iodine. Solution: Experiment 1: mass before reaction = 1.27 g + 3.50 g = 4.77 g; mass after reaction = 3.81 g + 0.96 g = 4.77 g Experiment 2: mass before reaction = 2.55 g + 3.50 g = 6.05 g; mass after reaction = 5.25 g + 0.80 g = 6.05 g Both experiments demonstrate the law of mass conversation since the total mass before reaction equals the total mass after reaction. Experiment 1: mass of reacted copper = 1.27 g; mass of reacted iodine = 3.50 g --- 0.96 g = 2.54 g Mass reacted copper/mass reacted iodine = 1.27 g/2.54 g = 0.50 Experiment 2: mass of reacted copper = 2.55 g --- 0.80 g = 1.75 g; mass of reacted iodine = 3.50 g Mass reacted copper/mass reacted iodine = 1.75 g/3.50 g = 0.50 Both Experiments 1 and 2 demonstrate the law of definite composition since the compound has the same composition by mass in each experiment.

2.20

Plan: Fluorite is a mineral containing only calcium and fluorine. The difference between the mass of fluorite and the mass of calcium gives the mass of fluorine. Mass fraction is calculated by dividing the mass of element by the mass of compound (fluorite) and mass percent is obtained by multiplying the mass fraction by 100. Solution: a) Mass (g) of fluorine = mass of fluorite --- mass of calcium = 2.76 g --- 1.42 g = 1.34 g fluorine mass Ca 1.42 g Ca = b) Mass fraction of Ca = = 0.51449 = 0.514 mass fluorite 2.76 g fluorite mass F 1.34 g F Mass fraction of F = = = 0.48551 = 0.486 mass fluorite 2.76 g fluorite c) Mass percent of Ca = 0.51449 × 100 = 51.449 = 51.4% Mass percent of F = 0.48551 × 100 = 48.551 = 48.6%

2.21

Plan: Galena is a mineral containing only lead and sulfur. The difference between the mass of galena and the mass of lead gives the mass of sulfur. Mass fraction is calculated by dividing the mass of element by the mass of compound (galena) and mass percent is obtained by multiplying the mass fraction by 100.

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2-10

Solution: a) Mass (g) of sulfur = mass of galena --- mass of sulfur = 2.34 g --- 2.03 g = 0.31 g sulfur mass Pb 2.03 g Pb = b) Mass fraction of Pb = = 0.8675214 = 0.868 mass galena 2.34 g galena mass S 0.31 g S Mass fraction of S = = = 0.1324786 = 0.13 mass galena 2.34 g galena c) Mass percent of Pb = (0.8675214)(100) = 86.752 = 86.8% Mass percent of S = (0.1324786)(100) = 13.248 = 13% 2.22

Plan: Dividing the mass of magnesium by the mass of the oxide gives the ratio. Multiply the mass of the second sample of magnesium oxide by this ratio to determine the mass of magnesium. Solution: a) If 1.25 g of MgO contains 0.754 g of Mg, then the mass ratio (or fraction) of magnesium in the oxide mass Mg 0.754 g Mg = compound is = 0.6032 = 0.603. mass MgO 1.25 g MgO

 0.6032 g Mg    = 322.109 = 322 g magnesium b) Mass (g) of magnesium = 534 g MgO   1 g MgO 



2.23



Plan: Dividing the mass of zinc by the mass of the sulfide gives the ratio. Multiply the mass of the second sample of zinc sulfide by this ratio to determine the mass of zinc. Solution: a) If 2.54 g of ZnS contains 1.70 g of Zn, then the mass ratio (or fraction) of zinc in the sulfide compound is mass Zn 1.70 g Zn = = 0.66929 = 0.669. mass ZnS 2.54 g ZnS

 0.66929 kg Zn    = 2.5567 = 2.56 kg zinc b) Mass (g) of zinc = 3.82 kg ZnS   1 kg ZnS 



2.24



Plan: Since copper is a metal and sulfur is a nonmetal, the sample contains 88.39 g Cu and 44.61 g S. Calculate the mass fraction of each element in the sample by dividing the mass of element by the total mass of compound. Multiply the mass of the second sample of compound in grams by the mass fraction of each element to find the mass of each element in that sample. Solution: Mass (g) of compound = 88.39 g copper + 44.61 g sulfur = 133.00 g compound  88.39 g copper   = 0.664586 Mass fraction of copper =  133.00 g compound  103 g compound   0.664586 g copper  Mass (g) of copper = 5264 kg compound  1 g compound   1 kg compound   6

= 3.49838 × 10 = 3.498 × 10 g copper  44.61 g sulfur   = 0.335414 Mass fraction of sulfur =  133.00 g compound 

103 g compound   0.335414 g sulfur  Mass (g) of sulfur = 5264 kg compound   1 g compound   1 kg compound  6 6 = 1.76562 × 10 = 1.766 × 10 g sulfur

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2-11

2.25

Plan: Since cesium is a metal and iodine is a nonmetal, the sample contains 63.94 g Cs and 61.06 g I. Calculate the mass fraction of each element in the sample by dividing the mass of element by the total mass of compound. Multiply the mass of the second sample of compound by the mass fraction of each element to find the mass of each element in that sample. Solution: Mass of compound = 63.94 g cesium + 61.06 g iodine = 125.00 g compound  63.94 g cesium   = 0.51152 Mass fraction of cesium =  125.00 g compound 

 0.51152 g cesium    = 19.83163 = 19.83 g cesium Mass (g) of cesium = 38.77 g compound   1 g compound   61.06 g iodine   = 0.48848 Mass fraction of iodine =  125.00 g compound 





 0.48848 g iodine    = 18.9384 = 18.94 g iodine Mass (g) of iodine = 38.77 g compound   1 g compound 



2.26



Plan: The law of multiple proportions states that if two elements form two different compounds, the relative amounts of the elements in the two compounds form a whole-number ratio. To illustrate the law we must calculate the mass of one element to one gram of the other element for each compound and then compare this mass for the two compounds. The law states that the ratio of the two masses should be a small whole-number ratio such as 1:2, 3:2, 4:3, etc. Solution: 47.5 mass % S Compound 1: = 0.90476 = 0.905 52.5 mass % Cl Compound 2:

31.1 mass % S 68.9 mass % Cl

= 0.451379 = 0.451

0.90476 = 2.0044 = 2.00:1.00 0.451379 Thus, the ratio of the mass of sulfur per gram of chlorine in the two compounds is a small whole-number ratio of 2:1, which agrees with the law of multiple proportions. Ratio:

2.27

Plan: The law of multiple proportions states that if two elements form two different compounds, the relative amounts of the elements in the two compounds form a whole-number ratio. To illustrate the law we must calculate the mass of one element to one gram of the other element for each compound and then compare this mass for the two compounds. The law states that the ratio of the two masses should be a small whole-number ratio such as 1:2, 3:2, 4:3, etc. Solution: 77.6 mass % Xe Compound 1: = 3.4643 = 3.46 22.4 mass % F Compound 2:

63.3 mass % Xe 36.7 mass % F

= 1.7248 = 1.72

3.4643 = 2.0085 = 2.01:1.00 1.7248 Thus, the ratio of the mass of xenon per gram of fluorine in the two compounds is a small whole-number ratio of 2:1, which agrees with the law of multiple proportions. Ratio:

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2-12

2.28

Plan: Calculate the mass percent of calcium in dolomite by dividing the mass of calcium by the mass of the sample and multiply by 100. Compare this mass percent to that in fluorite. The compound with the larger mass percent of calcium is the richer source of calcium. Solution: 1.70 g calcium 100% = 21.767 = 21.8% Ca Mass percent calcium = 7.81 g dolomite Fluorite (51.4%) is the richer source of calcium.

2.29

Plan: Determine the mass percent of sulfur in each sample by dividing the grams of sulfur in the sample by the total mass of the sample and multiplying by 100. The coal type with the smallest mass percent of sulfur has the smallest environmental impact. Solution:  11.3 g sulfur  100% = 2.9894 = 2.99% S (by mass) Mass % in Coal A =   378 g sample 

 19.0 g sulfur  100% = 3.8384 = 3.84% S (by mass) Mass % in Coal B =   495 g sample   20.6 g sulfur  100% = 3.0519 = 3.05% S (by mass) Mass % in Coal C =   675 g sample  Coal A has the smallest environmental impact. 2.30

We now know that atoms of one element may change into atoms of another element. We also know that atoms of an element can have different masses (isotopes). Finally, we know that atoms are divisible into smaller particles. Based on the best available information in 1805, Dalton was correct. This model is still useful, since its essence (even if not its exact details) remains true today.

2.31

Plan: This question is based on the law of definite composition. If the compound contains the same types of atoms, they should combine in the same way to give the same mass percentages of each of the elements. Solution: Potassium nitrate is a compound composed of three elements ----- potassium, nitrogen, and oxygen ----- in a specific ratio. If the ratio of these elements changed, then the compound would be changed to a different compound, for example, to potassium nitrite, with different physical and chemical properties. Dalton postulated that atoms of an element are identical, regardless of whether that element is found in India or Italy. Dalton also postulated that compounds result from the chemical combination of specific ratios of different elements. Thus, Dalton’s theory explains why potassium nitrate, a compound comprised of three different elements in a specific ratio, has the same chemical composition regardless of where it is mined or how it is synthesized.

2.32

Plan: Review the discussion of the experiments in this chapter. Solution: Millikan determined the minimum charge on an oil drop and that the minimum charge was equal to the charge on one electron. Using Thomson’s value for the mass/charge ratio of the electron and the determined value for the ---28 charge on one electron, Millikan calculated the mass of an electron (charge/(charge/mass)) to be 9.109 × 10 g.

2.33

Plan: The charges on the oil droplets should be whole-number multiples of a minimum charge. Determine that minimum charge by dividing the charges by small integers to find the common factor. Solution: ---19 ---19 ---3.204 × 10 C/2 = ---1.602 × 10 C ---19 ---19 ---4.806 × 10 C/3 = ---1.602 × 10 C ---19 ---19 ---8.010 × 10 C/5 = ---1.602 × 10 C ---18 ---19 ---1.442 × 10 C/4 = ---1.602 × 10 C

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2-13

The value ---1.602 × 10

---19

C is the common factor and is the charge for the electron.

2.34

Thomson’s ‘‘plum pudding’’ model described the atom as a ‘‘blob’’ of positive charge with tiny electrons embedded in it. The electrons could be easily removed from the atoms when a current was applied and ejected as a stream of ‘‘cathode rays.’’

2.35

Rutherford and co-workers expected that the alpha particles would pass through the foil essentially unaffected, or perhaps slightly deflected or slowed down. The observed results (most passing through straight, a few deflected, a very few at large angles) were partially consistent with expectations, but the large-angle scattering could not be explained by Thomson’s model. The change was that Rutherford envisioned a small (but massive) positively charged nucleus in the atom, capable of deflecting the alpha particles as observed.

2.36

Plan: Re-examine the definitions of atomic number and the mass number. Solution: The atomic number is the number of protons in the nucleus of an atom. When the atomic number changes, the identity of the element also changes. The mass number is the total number of protons and neutrons in the nucleus of an atom. Since the identity of an element is based on the number of protons and not the number of neutrons, the mass number can vary (by a change in number of neutrons) without changing the identity of the element.

2.37

Plan: Recall that the mass number is the sum of protons and neutrons while the atomic number is the number of protons. Solution: Mass number (protons plus neutrons) --- atomic number (protons) = number of neutrons (c).

2.38

The actual masses of the protons, neutrons, and electrons are not whole numbers so their sum is not a whole number.

2.39

Plan: There is one peak for each type of Cl atom and peaks for the Cl2 molecule. The m/e ratio equals the mass divided by 1+. Solution: 35 37 a) There is one peak for the Cl atom and another peak for the Cl atom. There are three peaks for the three 35 35 37 37 35 37 possible Cl2 molecules: Cl Cl (both atoms are mass 35), Cl Cl (both atoms are mass 37), and Cl Cl (one atom is mass 35 and one is mass 37). So the mass of chlorine will have 5 peaks. b) Peak m/e ratio 35 Cl 35 lightest particle 37 Cl 37 35 35 Cl Cl 70 (35 + 35) 35 37 Cl Cl 72 (35 + 37) 37 37 Cl Cl 74 (35 + 37) heaviest particle

2.40

Plan: The superscript is the mass number, the sum of the number of protons and neutrons. Consult the periodic table to get the atomic number (the number of protons). The mass number --- the number of protons = the number of neutrons. For atoms, the number of protons and electrons are equal. Solution: Isotope Mass Number # of Protons # of Neutrons # of Electrons 36 Ar 36 18 18 18 38 Ar 38 18 20 18 40 Ar 40 18 22 18

2.41

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2-14

Solution: Isotope 35 Cl 37 Cl 2.42

Mass Number 35 37

# of Protons 17 17

# of Neutrons 18 20

# of Electrons 17 17

Plan: The superscript is the mass number (A), the sum of the number of protons and neutrons; the subscript is the atomic number (Z, number of protons). The mass number --- the number of protons = the number of neutrons. For atoms, the number of protons = the number of electrons. Solution: 16 17 a) 8 O and 8 O have the same number of protons and electrons (8), but different numbers of neutrons. 16 8

O and 178 O are isotopes of oxygen, and 168 O has 16 --- 8 = 8 neutrons whereas 178 O has 17 --- 8 = 9 neutrons. Same

Z value 40

b) 18 Ar and 19 K have the same number of neutrons (Ar: 40 --- 18 = 22; K: 41 --- 19 = 22) but different numbers of protons and electrons (Ar = 18 protons and 18 electrons; K = 19 protons and 19 electrons). Same N value 60

c) 27 Co and 28 Ni have different numbers of protons, neutrons, and electrons. Co: 27 protons, 27 electrons, and 60 --- 27 = 33 neutrons; Ni: 28 protons, 28 electrons and 60 --- 28 = 32 neutrons. However, both have a mass number of 60. Same A value 2.43

a) 1 H and 2 He have different numbers of protons, neutrons, and electrons. H: 1 proton, 1 electron, and 3 --- 1 = 2 neutrons; He: 2 protons, 2 electrons, and 3 --- 2 = 1 neutron. However, both have a mass number of 3. Same A value 14 15 b) 6 C and 7 N have the same number of neutrons (C: 14 --- 6 = 8; N: 15 --- 7 = 8) but different numbers of protons and electrons (C = 6 protons and 6 electrons; N = 7 protons and 7 electrons). Same N value c)

19 9

F and 189 F have the same number of protons and electrons (9), but different numbers of neutrons.

19 9

F and 189 F are isotopes of oxygen, and 199 F has 19 --- 9 = 10 neutrons whereas 189 F has 18 --- 9 = 9 neutrons. Same Z value 2.44

Plan: Combine the particles in the nucleus (protons + neutrons) to give the mass number (superscript, A). The number of protons gives the atomic number (subscript, Z) and identifies the element. Solution: 38 a) A = 18 + 20 = 38; Z = 18; 18 Ar 55

b) A = 25 + 30 = 55; Z = 25; 25 Mn 109

c) A = 47 + 62 = 109; Z = 47; 47 Ag 2.45

b) A = 40 + 50 = 90; Z = 40; 40 Zr 61

c) A = 28 + 33 = 61; Z = 28; 28 Ni Copyright

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2-15

2.46

Plan: Determine the number of each type of particle. The superscript is the mass number (A) and the subscript is the atomic number (Z, number of protons). The mass number --- the number of protons = the number of neutrons. For atoms, the number of protons = the number of electrons. The protons and neutrons are in the nucleus of the atom. Solution: 48 79 11 a) 22Ti b) 34 Se c) 5 B 22 protons

34 protons

5 protons

22 electrons

34 electrons

5 electrons

48 --- 22 = 26 neutrons

79 --- 34 = 45 neutrons

11 --- 5 = 6 neutrons

2.47

a) 82 Pb 82 protons 82 electrons 207 --- 82 = 125 neutrons

b) 4 Be 4 protons 4 electrons 9 --- 4 = 5 neutrons

c) 33 As 33 protons 33 electrons 75 --- 33 = 42 neutrons

2.48

Plan: To calculate the atomic mass of an element, take a weighted average based on the natural abundance of the isotopes: (isotopic mass of isotope 1 × fractional abundance) + (isotopic mass of isotope 2 × fractional abundance). Solution:  60.11%   39.89%  Atomic mass of gallium = 68.9256 amu   70.9247 amu  = 69.7230 = 69.72 amu  100%   100% 

2.49

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2-16

Solution:

 78.99%  10.00%  11.01%  Atomic mass of Mg = 23.9850 amu    24.9858 amu    25.9826 amu    100%   100%   100% 













= 24.3050 = 24.31 amu 35

2.50

Plan: To find the percent abundance of each Cl isotope, let x equal the fractional abundance of Cl and (1 --- x) 37 equal the fractional abundance of Cl since the sum of the fractional abundances must equal 1. Remember that 35 37 atomic mass = (isotopic mass of Cl × fractional abundance) + (isotopic mass of Cl × fractional abundance). Solution: 35 37 Atomic mass = (isotopic mass of Cl × fractional abundance) + (isotopic mass of Cl × fractional abundance) 35.4527 amu = 34.9689 amu(x) + 36.9659 amu(1 --- x) 35.4527 amu = 34.9689 amu(x) + 36.9659 amu --- 36.9659 amu(x) 35.4527 amu = 36.9659 amu --- 1.9970 amu(x) 1.9970 amu(x) = 1.5132 amu x = 0.75774 and 1 --- x = 1 --- 0.75774 = 0.24226 35 37 % abundance Cl = 75.774% % abundance Cl = 24.226%

2.51

Plan: To find the percent abundance of each Cu isotope, let x equal the fractional abundance of Cu and (1 --- x) 65 equal the fractional abundance of Cu since the sum of the fractional abundances must equal 1. Remember that 63 65 atomic mass = (isotopic mass of Cu × fractional abundance) + (isotopic mass of Cu × fractional abundance). Solution: 63 65 Atomic mass = (isotopic mass of Cu × fractional abundance) + (isotopic mass of Cu × fractional abundance) 63.546 amu = 62.9396 amu(x) + 64.9278 amu(1 --- x) 63.546 amu = 62.9396 amu(x) + 64.9278 amu --- 64.9278 amu(x) 63.546 amu = 64.9278 amu --- 1.9882 amu(x) 1.9882 amu(x) = 1.3818 amu x = 0.69500 and 1 --- x = 1 --- 0.69500 = 0.30500 63 65 % abundance Cu = 69.50% % abundance Cu = 30.50%

2.52

Iodine has more protons in its nucleus (higher Z), but iodine atoms must have, on average, fewer neutrons than Te atoms and thus a lower atomic mass.

2.53

Plan: Review the section in the chapter on the periodic table. Solution: a) In the modern periodic table, the elements are arranged in order of increasing atomic number. b) Elements in a column or group (or family) have similar chemical properties, not those in the same period or row. c) Elements can be classified as metals, metalloids, or nonmetals.

2.54

The metalloids lie along the ‘‘staircase’’ line, with properties intermediate between metals and nonmetals.

2.55

Plan: Review the section on the classification of elements as metals, nonmetals, or metalloids. Solution: To the left of the “staircase” are the metals, which are generally hard, shiny, malleable, ductile, good conductors of heat and electricity, and form positive ions by losing electrons. To the right of the “staircase” are the nonmetals, which are generally soft or gaseous, brittle, dull, poor conductors of heat and electricity, and form negative ions by gaining electrons.

2.56

Plan: Review the properties of these two columns in the periodic table. Solution: The alkali metals (Group 1A(1)) are metals and readily lose one electron to form cations whereas the halogens (Group 7A(17)) are nonmetals and readily gain one electron to form anions.

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2-17

2.57

Plan: Locate each element on the periodic table. The Z value is the atomic number of the element. Metals are to the left of the ‘‘staircase,’’ nonmetals are to the right of the ‘‘staircase,’’ and the metalloids are the elements that lie along the ‘‘staircase’’ line. Solution: a) Germanium Ge 4A(14) metalloid b) Phosphorus P 5A(15) nonmetal c) Helium He 8A(18) nonmetal d) Lithium Li 1A(1) metal e) Molybdenum Mo 6B(6) metal

2.58

Plan: Locate each element on the periodic table. The Z value is the atomic number of the element. Metals are to the left of the ‘‘staircase,’’ nonmetals are to the right of the ‘‘staircase,’’ and the metalloids are the elements that lie along the ‘‘staircase’’ line. Solution: a) Arsenic As 5A(15) metalloid b) Calcium Ca 2A(2) metal c) Bromine Br 7A(17) nonmetal d) Potassium K 1A(1) metal e) Aluminum Al 3A(13) metal

2.59

Plan: Review the section in the chapter on the periodic table. Remember that alkaline earth metals are in Group 2A(2), the halogens are in Group 7A(17), and the metalloids are the elements that lie along the ‘‘staircase’’ line; periods are horizontal rows. Solution: a) The symbol and atomic number of the heaviest alkaline earth metal are Ra and 88. b) The symbol and atomic number of the lightest metalloid in Group 4A(14) are Si and 14. c) The symbol and atomic mass of the coinage metal whose atoms have the fewest electrons are Cu and 63.55 amu. d) The symbol and atomic mass of the halogen in Period 4 are Br and 79.90 amu.

2.60

Plan: Review the section in the chapter on the periodic table. Remember that the noble gases are in Group 8A(18), the alkali metals are in Group 1A(1), and the transition elements are the groups of elements located between Groups 2A(s) and 3A(13); periods are horizontal rows and metals are located to the left of the ‘‘staircase’’ line. Solution: a) The symbol and atomic number of the heaviest nonradioactive noble gas are Xe and 54, respectively. b) The symbol and group number of the Period 5 transition element whose atoms have the fewest protons are Y and 3B(3). c) The symbol and atomic number of the only metallic chalcogen are Po and 84. d) The symbol and number of protons of the Period 4 alkali metal atom are K and 19.

2.61

Plan: Review the section of the chapter on the formation of ionic compounds. Solution: Reactive metals and nometals will form ionic bonds, in which one or more electrons are transferred from the metal atom to the nonmetal atom to form a cation and an anion, respectively. The oppositely charged ions attract, forming the ionic bond.

2.62

Plan: Review the section of the chapter on the formation of covalent compounds. Solution: Two nonmetals will form covalent bonds, in which the atoms share two or more electrons.

2.63

The total positive charge of the cations is balanced by the total negative charge of the anions.

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2-18

2.64

Plan: Assign charges to each of the ions. Since the sizes are similar, there are no differences due to the sizes. Solution: Coulomb’s law states the energy of attraction in an ionic bond is directly proportional to the product of charges and inversely proportional to the distance between charges. The product of charges in MgO (+2 × ---2 = ---4) is greater than the product of charges in LiF (+1 × ---1 = ---1). Thus, MgO has stronger ionic bonding.

2.65

There are no molecules; BaF2 is an ionic compound consisting of Ba and F ions.

2.66

There are no ions present; P and O are both nonmetals, and they will bond covalently to form P 4O6 molecules.

2.67

Plan: Locate these groups on the periodic table and assign charges to the ions that would form. Solution: + + + The monatomic ions of Group 1A(1) have a +1 charge (e.g., Li , Na , and K ) whereas the monatomic ions of ------Group 7A(17) have a ---1 charge (e.g., F , Cl , and Br ). Elements gain or lose electrons to form ions with the same number of electrons as the nearest noble gas. For example, Na loses one electron to form a cation with the same number of electrons as Ne. The halogen F gains one electron to form an anion with the same number of electrons as Ne.

2.68

Plan: A metal and a nonmetal will form an ionic compound. Locate these elements on the periodic table and predict their charges. Solution: Magnesium chloride (MgCl2) is an ionic compound formed from a metal (magnesium) and a nonmetal (chlorine). 2+ Magnesium atoms transfer electrons to chlorine atoms. Each magnesium atom loses two electrons to form a Mg ion and the same number of electrons (10) as the noble gas neon. Each chlorine atom gains one electron to form a --2+ --Cl ion and the same number of electrons (18) as the noble gas argon. The Mg and Cl ions attract each other to 2+ --form an ionic compound with the ratio of one Mg ion to two Cl ions. The total number of electrons lost by the magnesium atoms equals the total number of electrons gained by the chlorine atoms.

2.69

Plan: A metal and a nonmetal will form an ionic compound. Locate these elements on the periodic table and predict their charges. Solution: Potassium sulfide (K2S) is an ionic compound formed from a metal (potassium) and a nonmetal (sulfur). Potassium atoms transfer electrons to sulfur atoms. Each potassium atom loses one electron to form an ion with +1 charge and the same number of electrons (18) as the noble gas argon. Each sulfur atom gains two electrons to form an ion with a ---2 charge and the same number of electrons (18) as the noble gas argon. The oppositely + 2--+ 2--charged ions, K and S , attract each other to form an ionic compound with the ratio of two K ions to one S ion. The total number of electrons lost by the potassium atoms equals the total number of electrons gained by the sulfur atoms.

2.70

Plan: Recall that ionic bonds occur between metals and nonmetals, whereas covalent bonds occur between nonmetals. Solution: --KNO3 shows both ionic and covalent bonding, covalent bonding between the N and O in NO 3 and ionic bonding --+ between the NO3 and the K .

2.71

Plan: Locate these elements on the periodic table and predict what ions they will form. For A group cations (metals), ion charge = group number; for anions (nonmetals), ion charge = group number minus 8. Solution: + --Potassium (K) is in Group 1A(1) and forms the K ion. Bromine (Br) is in Group 7A(17) and forms the Br ion (7 --- 8 = ---1).

2.72

---

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2-19

Solution: 2+ 2--Radium in Group 2A(2) forms a +2 ion: Ra . Selenium in Group 6A(16) forms a ---2 ion: Se (6 --- 8 = ---2). 2.73

Plan: Use the number of protons (atomic number) to identify the element. Add the number of protons and neutrons together to get the mass number. Locate the element on the periodic table and assign its group and period number. Solution: a) Oxygen (atomic number = 8) mass number = 8p + 9n = 17 Group 6A(16) Period 2 b) Fluorine (atomic number = 9) mass number = 9p + 10n = 19 Group 7A(17) Period 2 c) Calcium (atomic number = 20) mass number = 20p + 20n = 40 Group 2A(2) Period 4

2.74

Plan: Use the number of protons (atomic number) to identify the element. Add the number of protons and neutrons together to get the mass number. Locate the element on the periodic table and assign its group and period number. Solution: a) Bromine (atomic number = 35) mass number = 35p + 44n = 79 Group 7A(17) Period 4 b) Nitrogen (atomic number = 7) mass number = 7p + 8n = 15 Group 5A(15) Period 2 c) Rubidium (atomic number = 37) mass number = 37p + 48n = 85 Group 1A(1) Period 5

2.75

Plan: Determine the charges of the ions based on their position on the periodic table. For A group cations (metals), ion charge = group number; for anions (nonmetals), ion charge = group number minus 8. Next, determine the ratio of the charges to get the ratio of the ions. Solution: + 2--Lithium [Group 1A(1)] forms the Li ion; oxygen [Group 6A(16)] forms the O ion (6 --- 8 = ---2). The ionic + compound that forms from the combination of these two ions must be electrically neutral, so two Li ions 2--+ 2--combine with one O ion to form the compound Li2O. There are twice as many Li ions as O ions in a sample of Li2O.  1 O2 ion  2-- = 4.2 × 10 21 O2--- ions Number of O ions = (8.4 1021 Li ions)   2 Li ions 

2.76

Plan: Determine the charges of the ions based on their position on the periodic table. For A group cations (metals), ion charge = group number; for anions (nonmetals), ion charge = group number minus 8. Next, determine the ratio of the charges to get the ratio of the ions. Solution: 2+ --Ca [Group 2A(2)] forms Ca and I [Group 7A(17)] forms I ions (7 --- 8 = ---1). The ionic compound that forms 2+ --from the combination of these two ions must be electrically neutral, so one Ca ion combines with two I ions to --2+ form the compound CaI2. There are twice as many I ions as Ca ions in a sample of CaI2.  2 I ions  -- = 1.48 × 1022 = 1.5 × 10 22 I--- ions Number of I ions = (7.4 1021 Ca 2 ions)  2 1 Ca ion 

2.77

Plan: The key is the size of the two alkali metal ions. The charges on the sodium and potassium ions are the same as both are in Group 1A(1), so there will be no difference due to the charge. The chloride ions are the same in size and charge, so there will be no difference due to the chloride ion. Solution: Coulomb’s law states that the energy of attraction in an ionic bond is directly proportional to the product of charges and inversely proportional to the distance between charges. The product of the charges is the same in both compounds because both sodium and potassium ions have a +1 charge. Attraction increases as distance decreases, + so the ion with the smaller radius, Na , will form a stronger ionic interaction (NaCl).

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2-20

2.78

Plan: The key is the charge of the two metal ions. The sizes of the lithium and magnesium ions are about the same (magnesium is slightly smaller), so there will be little difference due to ion size. The oxide ions are the same in size and charge, so there will be no difference due to the oxide ion. Solution: Coulomb’s law states the energy of attraction in an ionic bond is directly proportional to the product of charges and inversely proportional to the distance between charges. The product of charges in MgO (+2 × ---2 = ---4) is greater than the product of charges in Li2O (+1 × ---2 = ---2). Thus, MgO has stronger ionic bonding.

2.79

Plan: Review the definition of molecular formula. Solution: The subscripts in the formula, MgF2, give the number of ions in a formula unit of the ionic compound. The --2+ subscripts indicate that there are two F ions for every one Mg ion. Using this information and the mass of each element, we could calculate the percent mass of each element.

2.80

Plan: Review the definitions of molecular and structural formulas. Solution: Both the structural and molecular formulas show the actual numbers of the atoms of the molecule; in addition, the structural formula shows the arrangement of the atoms (i.e., how the atoms are connected to each other).

2.81

Plan: Review the concepts of atoms and molecules. Solution: The mixture is similar to the sample of hydrogen peroxide in that both contain 20 billion oxygen atoms and 20 billion hydrogen atoms since both O2 and H2O2 contain 2 oxygen atoms per molecule and both H2 and H2O2 contain 2 hydrogen atoms per molecule. They differ in that they contain different types of molecules: H2O2 molecules in the hydrogen peroxide sample and H2 and O2 molecules in the mixture. In addition, the mixture contains 20 billion molecules (10 billion H2 molecules + 10 billion O2 molecules) while the hydrogen peroxide sample contains 10 billion molecules.

2.82

Plan: Review the rules for naming compounds. Solution: Roman numerals are used when naming ionic compounds that contain a metal that can form more than one ion. This is generally true for the transition metals, but it can be true for some non-transition metals as well (e.g., Sn).

2.83

Plan: Review the rules for naming compounds. Solution: Greek prefixes are used only in naming covalent compounds.

2.84

Molecular formulas cannot be written for ionic compounds since they only have ions and there are no molecules.

2.85

Plan: Locate each of the individual elements on the periodic table, and assign charges to each of the ions. For A group cations (metals), ion charge = group number; for anions (nonmetals), ion charge = group number minus 8. Find the smallest number of each ion that gives a neutral compound. To name ionic compounds with metals that form only one ion, name the metal, followed by the nonmetal name with an -ide suffix. Solution: a) Sodium is a metal that forms a +1 (Group 1A) ion and nitrogen is a nonmetal that forms a ---3 ion (Group 5A, 5 --- 8 = ---3). +3 ---3 +1 ---3 +1 Na N Na3N The compound is Na3N, sodium nitride. b) Oxygen is a nonmetal that forms a ---2 ion (Group 6A, 6 --- 8 = ---2) and strontium is a metal that forms a +2 ion (Group 2A). +2 ---2 Sr O The compound is SrO, strontium oxide.

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2-21

c) Aluminum is a metal that forms a +3 ion (Group 3A) and chlorine is a nonmetal that forms a ---1 ion (Group 7A, 7--- 8 = ---1). +3 ---3 +3 ---1 +3 ---1 Al Cl AlCl3 The compound is AlCl3, aluminum chloride. 2.86

Plan: Locate each of the individual elements on the periodic table, and assign charges to each of the ions. For A group cations (metals), ion charge = group number; for anions (nonmetals), ion charge = group number minus 8. Find the smallest number of each ion that gives a neutral compound. To name ionic compounds with metals that form only one ion, name the metal, followed by the nonmetal name with an -ide suffix. Solution: a) Cesium is a metal that forms a +1 (Group 1A) ion and bromine is a nonmetal that forms a ---1 ion (Group 7A, 7 --- 8 = ---1). +1 ---1 Cs Br The compound is CsBr, cesium bromide. b) Sulfur is a nonmetal that forms a ---2 ion (Group 6A, 6 --- 8 = ---2) and barium is a metal that forms a +2 ion (Group 2A). +2 ---2 Ba S The compound is BaS, barium sulfide. c) Fluorine is a nonmetal that forms a ---1 ion (Group 7A, 7 --- 8 = ---1) and calcium is a metal that forms a +2 ion (Group 2A). +2 ---2 +2 ---1 +2 ---1 Ca F CaF2 The compound is CaF2, calcium fluoride.

2.87

Plan: Based on the atomic numbers (the subscripts) locate the elements on the periodic table. Once the atomic numbers are located, identify the element and based on its position, assign a charge. For A group cations (metals), ion charge = group number; for anions (nonmetals), ion charge = group number minus 8. Find the smallest number of each ion that gives a neutral compound. To name ionic compounds with metals that form only one ion, name the metal, followed by the nonmetal name with an -ide suffix. Solution: 2+ a) 12L is the element Mg (Z = 12). Magnesium [Group 2A(2)] forms the Mg ion. 9M is the element F (Z = 9). --Fluorine [Group 7A(17)] forms the F ion (7 --- 8 = ---1). The compound formed by the combination of these two elements is MgF2, magnesium fluoride. 2+ b) 30L is the element Zn (Z = 30). Zinc forms the Zn ion (see Table 2.3). 16M is the element S (Z = 16). 2--Sulfur [Group 6A(16)] will form the S ion (6 --- 8 = ---2). The compound formed by the combination of these two elements is ZnS, zinc sulfide. --c) 17L is the element Cl (Z = 17). Chlorine [Group 7A(17)] forms the Cl ion (7 --- 8 = ---1). 38M is the element Sr 2+ (Z = 38). Strontium [Group 2A(2)] forms the Sr ion. The compound formed by the combination of these two elements is SrCl2, strontium chloride.

2.88

Plan: Based on the atomic numbers (the subscripts) locate the elements on the periodic table. Once the atomic numbers are located, identify the element and based on its position, assign a charge. For A group cations (metals), ion charge = group number; for anions (nonmetals), ion charge = group number minus 8. Find the smallest number of each ion that gives a neutral compound. To name ionic compounds with metals that form only one ion, name the metal, followed by the nonmetal name with an -ide suffix. Solution: + a) 37Q is the element Rb (Z = 37). Rubidium [Group 1A(1)] forms the Rb ion. 35R is the element Br (Z = 35). --Bromine [Group 7A(17)] forms the Br ion (7 --- 8 = ---1). The compound formed by the combination of these two elements is RbBr, rubidium bromide. 2--b) 8Q is the O (Z = 8). Oxygen [Group 6A(16)] will form the O ion (6 --- 8 = ---2). 13R is the element Al (Z = 13). 3+ Aluminum [Group 3A(13)] forms the Al ion. The compound formed by the combination of these two elements is Al2O3, aluminum oxide. McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution

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2-22

c) 20Q is the element Ca (Z = 20). Calcium [Group 2A(2)] forms the Ca ion. 53R is the element I (Z = 53). Iodine --[Group 7A(17)] forms the I ion (7 --- 8 = ---1). The compound formed by the combination of these two elements is CaI2, calcium iodide. 2.89

Plan: Review the rules for nomenclature covered in the chapter. For ionic compounds, name the metal, followed by the nonmetal name with an -ide suffix. For metals, like many transition metals, that can form more than one ion each with a different charge, the ionic charge of the metal ion is indicated by a Roman numeral within parentheses immediately following the metal’s name. Solution: 4+ --a) tin(IV) chloride = SnCl4 The (IV) indicates that the metal ion is Sn which requires 4 Cl ions for a neutral compound. b) FeBr3 = iron(III) bromide (common name is ferric bromide); the charge on the iron ion is +3 to match the ---3 --charge of 3 Br ions. The +3 charge of the Fe is indicated by (III). +6 ---6 c) cuprous bromide = CuBr (cuprous is +1 copper ion, cupric is +2 copper ion). +3 ---2 d) Mn2O3 = manganese(III) oxide Use (III) to indicate the +3 ionic charge of Mn: Mn2O3

2.90

Plan: Review the rules for nomenclature covered in the chapter. For ionic compounds, name the metal, followed by the nonmetal name with an -ide suffix. For metals, like many transition metals, that can form more than one ion each with a different charge, the ionic charge of the metal ion is indicated by a Roman numeral within parentheses immediately following the metal’s name. Solution: a) CoO = cobalt(II) oxide Cobalt forms more than one monatomic ion so the ionic charge must be indicated with 2--a Roman numeral. Since the Co is paired with one O ion, the charge of Co is +2. b) mercury(I) chloride = Hg 2Cl2 The Roman numeral I indicates that mercury has an ionic charge of +1; mercury 2+ + is an unusual case in which the +1 ion formed is Hg2 , not Hg . 3+ c) chromic oxide = Cr2O3 ‘‘chromic’’ denotes a +3 charge (see Table 2.4), oxide has a ---2 charge. Two Cr ions 2--are required for every three O ions. (d) CuBr2 = copper(II) bromide Copper forms more than one monatomic ion so the ionic charge must be --indicated with a Roman numeral. Since the Cu is paired with two Br ions, the charge of Cu is +2.

2.91

Plan: Review the rules for nomenclature covered in the chapter. For ionic compounds containing polyatomic ions, name the metal, followed by the name of the polyatomic ion. Hydrates, compounds with a specific number of water molecules associated with them, are named with a prefix before the word hydrate to indicate the number of water molecules. Solution: + 2--a) Na2HPO4 = sodium hydrogen phosphate Sodium [Group 1A(1)] forms the Na ion; HPO4 is the hydrogen phosphate ion. + b) ammonium perchlorate = NH4ClO4 Ammonium is the polyatomic ion NH4 and perchlorate is the polyatomic --+ --ion ClO4 . One NH4 is required for every one ClO4 ion. --c) Pb(C2H3O2)2 O = lead(II) acetate trihydrate The C2H3O2 ion has a ---1 charge (see Table 2.5); since there 2 are two of these ions, the lead ion has a +2 charge which must be indicated with the Roman numeral II. The O indicates a hydrate in which the number of H2O molecules is indicated by the prefix tri-. 2 --d) NaNO2 = sodium nitrite NO2 is the nitrite polyatomic ion.

2.92

Plan: Review the rules for nomenclature covered in the chapter. For ionic compounds containing polyatomic ions, name the metal, followed by the name of the polyatomic ion. For metals, like many transition metals, that can form more than one ion each with a different charge, the ionic charge of the metal ion is indicated by a Roman numeral within parentheses immediately following the metal’s name. Solution: a) Sn(SO3)2 = tin(IV) sulfite Tin forms more than one monatomic ion so the ionic charge must be indicated with 2--a Roman numeral. Each SO3 polyatomic ion has a charge of ---2, so the ionic charge of tin is +4.

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2-23

2---

b) potassium dichromate = K2Cr2O7 Dichromate is the polyatomic ion Cr2O7 ; two K ions are required for a neutral compound. c) FeCO3 = iron(II) carbonate Iron forms more than one monatomic ion so the ionic charge must be indicated 2--with a Roman numeral. The CO3 polyatomic ion has a charge of ---2, so the ionic charge of iron is +2. + 2--d) potassium carbonate dihydrate = K2CO3 2H2O Potassium [Group 1A(1)] forms the K ion; carbonate is CO3 . + 2--Two K ions are required for every one CO3 ion. Dihydrate indicates the presence of two H2O molecules in the formula unit, indicated with O. 2 2.93

Plan: Review the rules for nomenclature covered in the chapter. For metals, like many transition metals, that can form more than one ion each with a different charge, the ionic charge of the metal ion is indicated by a Roman numeral within parentheses immediately following the metal’s name. Compounds must be neutral. Solution: 2+ 2--a) Barium [Group 2A(2)] forms Ba and oxygen [Group 6A(16)] forms O (6 --- 8 = ---2) so the neutral compound 2+ 2--forms from one Ba ion and one O ion. Correct formula is BaO. 2+ --b) Iron(II) indicates Fe and nitrate is NO3 so the neutral compound forms from one iron(II) ion and two nitrate ions. Correct formula is Fe(NO3)2. c) Mn is the symbol for manganese. Mg is the correct symbol for magnesium. Correct formula is MgS. 2--2--Sulfide is the S ion and sulfite is the SO3 ion. + d) P is the symbol for phosphorus. K is the correct symbol for potassium. Potassium [Group 1A(1)] forms K and --+ --iodine [Group 7A(17)] forms I , so the neutral compound forms from one K ion and one I ion. Correct formula is KI.

2.94

Plan: Review the rules for nomenclature covered in the chapter. For metals, like many transition metals, that can form more than one ion each with a different charge, the ionic charge of the metal ion is indicated by a Roman numeral within parentheses immediately following the metal’s name. Compounds must be neutral. Solution: --a) copper(I) iodide Cu is copper, not cobalt; since iodide is I , this must be copper(I). --b) iron(III) hydrogen sulfate HSO4 is hydrogen sulfate, and this must be iron(III) to be neutral. 2+ 2--c) magnesium dichromate Mg forms Mg and Cr2O7 is named dichromate ion. 2+ d) calcium chloride Ca [Group 2A(2)] forms a Ca ion only, so no Roman numeral is needed in the name; it also --means that there must be two chloride ions (Cl ) for the compound to be neutral, so a prefix like ‘‘di-’’ is unnecessary.

2.95

Plan: Acids donate H ion to the solution, so the acid is a combination of H and a negatively charged ion. Binary acids (H plus one other nonmetal) are named hydro- + nonmetal root + -ic acid. Oxoacids (H + an oxoanion) are named by changing the suffix of the oxoanion: -ate becomes -ic acid and -ite becomes -ous acid. Solution: --a) Hydrogen carbonate is HCO3 , so its source acid is H2CO3. The name of the acid is carbonic acid (-ate becomes ---ic acid). --b) HIO4, periodic acid. IO4 is the periodate ion: -ate becomes ---ic acid. --c) Cyanide is CN ; its source acid is HCN hydrocyanic acid (binary acid). d) H2S, hydrosulfuric acid (binary acid).

2.96

Plan: Acids donate H ion to the solution, so the acid is a combination of H and a negatively charged ion. Binary acids (H plus one other nonmetal) are named hydro- + nonmetal root + -ic acid. Oxoacids (H + an oxoanion) are named by changing the suffix of the oxoanion: -ate becomes -ic acid and -ite becomes -ous acid. Solution: --a) Perchlorate is ClO4 , so the source acid is HClO4. Name of acid is perchloric acid (-ate becomes -ic acid). --b) nitric acid, HNO 3 NO3 is the nitrate ion: -ate becomes -ic acid. --c) Bromite is BrO2 , so the source acid is HBrO2. Name of acid is bromous acid (-ite becomes -ous acid). --d) H2PO4 is dihydrogen phosphate, so its source acid is H3PO4. The name of the acid is phosphoric acid (-ate becomes ---ic acid).

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2-24

2.97

Plan: Use the formulas of the polyatomic ions. Recall that oxoacids are named by changing the suffix of the oxoanion: -ate becomes -ic acid and -ite becomes -ous acid. Compounds must be neutral. Solution: + a) ammonium ion = NH4 ammonia = NH3 b) magnesium sulfide = MgS magnesium sulfite = MgSO3 magnesium sulfate = MgSO4 2--2--2--Sulfide = S ; sulfite = SO3 ; sulfate = SO4 . c) hydrochloric acid = HCl chloric acid = HClO3 chlorous acid = HClO2 Binary acids (H plus one other nonmetal) are named hydro- + nonmetal root + -ic acid. Chloric indicates the ----polyatomic ion ClO3 while chlorous indicates the polyatomic ion ClO2 . d) cuprous bromide = CuBr cupric bromide = CuBr2 The suffix -ous indicates the lower charge, +1, while the suffix -ic indicates the higher charge, +2.

2.98

Plan: Use the formulas of the polyatomic ions. For metals, like many transition metals, that can form more than one ion each with a different charge, the ionic charge of the metal ion is indicated by a Roman numeral within parentheses immediately following the metal’s name. Compounds must be neutral. Solution: a) lead(II) oxide = PbO lead(IV) oxide = PbO2 2+ 4+ Lead(II) indicates Pb while lead(IV) indicates Pb . b) lithium nitride = Li3N lithium nitrite = LiNO2 lithium nitrate = LiNO3 3------Nitride = N ; nitrite = NO2 ; nitrate = NO3 . c) strontium hydride = SrH2 strontium hydroxide = Sr(OH) 2 ----Hydride = H ; hydroxide = OH . d) magnesium oxide = MgO manganese(II) oxide = MnO

2.99

Plan: This compound is composed of two nonmetals. The element with the lower group number is named first. Greek numerical prefixes are used to indicate the number of atoms of each element in the compound. Solution: disulfur tetrafluoride S 2F 4 Di- indicates two S atoms and tetra- indicates four F atoms.

2.100

Plan: This compound is composed of two nonmetals. When a compound contains oxygen and a halogen, the halogen is named first. Greek numerical prefixes are used to indicate the number of atoms of each element in the compound. Solution: dichlorine monoxide Cl2O Di- indicates two Cl atoms and mono- indicates one O atom.

2.101

Plan: These compounds are composed of two nonmetals. The element with the lower group number is named first. Greek numerical prefixes are used to indicate the number of atoms of each element in the compound. Solution: a) Tetraphosphorus decoxide is P4O10. Tetra- indicates four P atoms and deca- indicates ten O atoms. b) Diboron trioxide is B2O3. Di-indicates two B atoms and triindicates three O atoms. c) Phosphorus trifluoride is PF3. P has a lower group number than F and is written first.

2.102

Plan: These compounds are composed of two nonmetals. The element with the lower group number is named first. Greek numerical prefixes are used to indicate the number of atoms of each element in the compound. Solution: a) CBr4 is carbon tetrabromide. Tetra- is used to indicate that there are 4 Br atoms in the compound. b) IF7 is iodine heptafluoride. Hepta- is used to indicate that there are 7 I atoms in the compound. c) NO is nitrogen monoxide. Roman numerals are not used when naming molecular compounds. Mono- is used to indicate that there is only one O atom in the compound. Mono- is generally not used with the first element.

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2-25

2.103

Plan: Review the rules for nomenclature covered in the chapter. For ionic compounds containing polyatomic ions, name the metal, followed by the name of the polyatomic ion. The molecular (formula) mass is the sum of the atomic masses of all of the atoms. Solution: + 2--a) (NH4)2SO4 ammonium is NH4 and sulfate is SO4 N = 2(14.01 amu) = 28.02 amu H = 8(1.008 amu) = 8.064 amu S = 1(32.06 amu) = 32.06 amu O = 4(16.00 amu) = 64.00 amu 132.14 amu + --b) NaH2PO4 sodium is Na and dihydrogen phosphate is H2PO4 Na = 1(22.99 amu) = 22.99 amu H = 2(1.008 amu) = 2.016 amu P = 1(30.97 amu) = 30.97 amu O = 4(16.00 amu) = 64.00 amu 119.98 amu + --c) KHCO3 potassium is K and bicarbonate is HCO3 K = 1(39.10 amu) = 39.10 amu H = 1(1.008 amu) = 1.008 amu C = 1(12.01 amu) = 12.01 amu O = 3(16.00 amu) = 48.00 amu 100.12 amu

2.104

Plan: Review the rules for nomenclature covered in the chapter. For ionic compounds containing polyatomic ions, name the metal, followed by the name of the polyatomic ion. The molecular (formula) mass is the sum of the atomic masses of all of the atoms. Solution: + 2--a) Na2Cr2O7 sodium is Na and dichromate is Cr2O7 Na = 2(22.99 amu) = 45.98 amu Cr = 2(52.00 amu) = 104.00 amu O = 7(16.00 amu) = 112.00 amu 261.98 amu + --b) NH4ClO4 ammonium is NH4 and perchlorate is ClO4 N = 1(14.01 amu) = 14.01 amu H = 4(1.008 amu) = 4.032 amu Cl = 1(35.45 amu) = 35.45 amu O = 4(16.00 amu) = 64.00 amu 117.49 amu 2+ --c) Mg(NO2)2 3H2O magnesium is Mg , nitrite is NO2 , and trihydrate is 3H2O Mg = 1(24.31 amu) = 24.31 amu N = 2(14.01 amu) = 28.02 amu H = 6(1.008 amu) = 6.048 amu O = 7(16.00 amu) = 112.00 amu 170.38 amu

2.105

Plan: Convert the names to the appropriate chemical formulas. The molecular (formula) mass is the sum of the masses of each atom times its atomic mass. Solution: a) dinitrogen pentoxide N2O5 (di- = 2 and penta- = 5) N = 2(14.01 amu) = 28.02 amu O = 5(16.00 amu) = 80.00 amu 108.02 amu

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2-26

---

b) lead(II) nitrate Pb = N = O =

Pb(NO3)2 (lead(II) is Pb and nitrate is NO3 ) 1(207.2 amu) = 207.2 amu 2(14.01 amu) = 28.02 amu 6(16.00 amu) = 96.00 amu 331.2 amu

c) calcium peroxide Ca = O =

CaO2 (calcium is Ca and peroxide is O2 ) 1(40.08 amu) = 40.08 amu 2(16.00 amu) = 32.00 amu 72.08 amu

2---

2.106

Plan: Convert the names to the appropriate chemical formulas. The molecular (formula) mass is the sum of the masses of each atom times its atomic mass. Solution: 2+ --a) iron(II) acetate tetrahydrate Fe(C2H3O2)2 4H2O (iron(II) is Fe , acetate is C2H3O2 , and tetrahydrate is 4H2O) Fe = 1(55.85 amu) = 55.85 amu C = 4(12.01 amu) = 48.04 amu H = 14(1.008 amu) = 14.112 amu O = 8(16.00 amu) = 128.00 amu 246.00 amu b) sulfur tetrachloride SCl4 (tetra- = 4) S = 1 (32.06 amu) = 32.06 amu Cl = 4(35.45 amu) = 141.80 amu 173.86 amu + --c) potassium permanganate KMnO4 (potassium is K and permanganate is MnO4 ) K = 1(39.10 amu) = 39.10 amu Mn = 1(54.94 amu) = 54.94 amu O = 4(16.00 amu) = 64.00 amu 158.04 amu

2.107

Plan: Break down each formula to the individual elements and count the number of atoms of each element by observing the subscripts. The molecular (formula) mass is the sum of the atomic masses of all of the atoms. Solution: a) There are 12 atoms of oxygen in Al2(SO4)3. The molecular mass is: Al = 2(26.98 amu) = 53.96 amu S = 3(32.06 amu) = 96.18 amu O = 12(16.00 amu) = 192.00 amu 342.14 amu b) There are 9 atoms of hydrogen in (NH4)2HPO4. The molecular mass is: N = 2(14.01 amu) = 28.02 amu H = 9(1.008 amu) = 9.072 amu P = 1(30.97 amu) = 30.97 amu O = 4(16.00 amu) = 64.00 amu 132.06 amu c) There are 8 atoms of oxygen in Cu3(OH)2(CO3)2. The molecular mass is: Cu = 3(63.55 amu) = 190.65 amu O = 8(16.00 amu) = 128.00 amu H = 2(1.008 amu) = 2.016 amu C = 2(12.01 amu) = 24.02 amu 344.69 amu

2.108

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2-27

Solution: a) There are 9 atoms of hydrogen in C6H5COONH4. The molecular mass is: C = 7(12.01 amu) = 84.07 amu H = 9(1.008 amu) = 9.072 amu O = 2(16.00 amu) = 32.00 amu N = 1(14.01 amu) = 14.01 amu 139.15 amu b) There are 2 atoms of nitrogen in N2H6SO4. The molecular mass is: N = 2(14.01 amu) = 28.02 amu H = 6(1.008 amu) = 6.048 amu S = 1(32.06 amu) = 32.06 amu O = 4(16.00 amu) = 64.00 amu 130.13 amu c) There are 12 atoms of oxygen in Pb4SO4(CO3)2(OH)2. The molecular mass is: Pb = 4(207.2 amu) = 828.8 amu S = 1(32.06 amu) = 32.06 amu O = 12(16.00 amu) = 192.00 amu C = 2(12.01 amu) = 24.02 amu H = 2(1.008 amu) = 2.016 amu 1078.9 amu 2.109

Plan: Use the chemical symbols and count the atoms of each type to give a molecular formula. Use the nomenclature rules in the chapter to derive the name. The molecular (formula) mass is the sum of the masses of each atom times its atomic mass. Solution: a) Formula is SO3. Name is sulfur trioxide (the prefix triindicates 3 oxygen atoms). S = 1(32.06 amu) = 32.06 amu O = 3(16.00 amu) = 48.00 amu 80.06 amu b) Formula is C 3H8. Since it contains only carbon and hydrogen it is a hydrocarbon and with three carbons its name is propane. C = 3(12.01 amu) = 36.03 amu H = 8(1.008 amu) = 8.064 amu 44.09 amu

2.110

Plan: Use the chemical symbols and count the atoms of each type to give a molecular formula. Use the nomenclature rules in the chapter to derive the name. The molecular (formula) mass is the sum of the masses of each atom times its atomic mass. Solution: a) Formula is N2O. Name is dinitrogen monoxide (the prefix diindicates 2 nitrogen atoms and mono- indicates 1 oxygen atom). N = 2(14.01 amu) = 28.02 amu O = 1(16.00 amu) = 16.00 amu 44.02 amu b) Formula is C 2H6. Since it contains only carbon and hydrogen it is a hydrocarbon and with three carbons its name is ethane. C = 2(12.01 amu) = 24.02 amu H = 6(1.008 amu) = 6.048 amu 30.07 amu

2.111

Plan: Review the nomenclature rules in the chapter. For ionic compounds, name the metal, followed by the nonmetal name with an -ide suffix. For ionic compounds containing polyatomic ions, name the metal, followed by the name of the polyatomic ion. For metals, like many transition metals, that can form more than one ion each

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2-28

with a different charge, the ionic charge of the metal ion is indicated by a Roman numeral within parentheses immediately following the metal’s name. Oxoacids (H + an oxoanion) are named by changing the suffix of the oxoanion: -ate becomes -ic acid and -ite becomes -ous acid. Greek numerical prefixes are used to indicate the number of atoms of each element in a compound composed of two nonmetals. Solution: a) blue vitriol CuSO4 5H2O copper(II) sulfate pentahydrate 2--SO4 = sulfate; II is used to indicate the 2+ charge of Cu; penta- is used to indicate the 5 waters of hydration. b) slaked lime Ca(OH)2 calcium hydroxide --The anion OH is hydroxide. c) oil of vitriol H2SO4 sulfuric acid 2--SO4 is the sulfate ion; since this is an acid, -ate becomes -ic acid. d) washing soda Na2CO3 sodium carbonate 2--CO3 is the carbonate ion. e) muriatic acid HCl hydrochloric acid Binary acids (H plus one other nonmetal) are named hydro- + nonmetal root + -ic acid. f) Epsom salts MgSO4 7H2O magnesium sulfate heptahydrate 2--SO4 = sulfate; hepta- is used to indicate the 7 waters of hydration. g) chalk CaCO3 calcium carbonate 2--CO3 is the carbonate ion. h) dry ice CO2 carbon dioxide The prefix diindicates 2 oxygen atoms; since there is only one carbon atom, no prefix is used. i) baking soda NaHCO3 sodium hydrogen carbonate --HCO3 is the hydrogen carbonate ion. j) lye NaOH sodium hydroxide --The anion OH is hydroxide. 2.112

Plan: Use the chemical symbols and count the atoms of each type to give a molecular formula. Use the nomenclature rules in the chapter to derive the name. The molecular (formula) mass is the sum of the masses of each atom times its atomic mass. Solution: a) Each molecule has 2 blue spheres and 1 red sphere so the molecular formula is N2O. This compound is composed of two nonmetals. The element with the lower group number is named first. Greek numerical prefixes are used to indicate the number of atoms of each element in the compound. The prefix diindicates 2 nitrogen atoms and mono- indicates 1 oxygen atom. The name is dinitrogen monoxide. N = 2(14.01 amu) = 28.02 amu O = 1(16.00 amu) = 16.00 amu 44.02 amu b) Each molecule has 2 green spheres and 1 red sphere so the molecular formula is Cl2O. This compound is composed of two nonmetals. When a compound contains oxygen and a halogen, the halogen is named first. Greek numerical prefixes are used to indicate the number of atoms of each element in the compound. The prefix diindicates 2 chlorine atoms and mono- indicates 1 oxygen atom. The name is dichlorine monoxide. Cl = 2(35.45 amu) = 70.90 amu O = 1(16.00 amu) = 16.00 amu 86.90 amu

2.113

Plan: Review the discussion on separations. Solution: Separating the components of a mixture requires physical methods only; that is, no chemical changes (no changes in composition) take place and the components maintain their chemical identities and properties throughout. Separating the components of a compound requires a chemical change (change in composition).

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2-29

2.114

Plan: Review the definitions of homogeneous and heterogeneous. Solution: A homogeneous mixture is uniform in its macroscopic, observable properties; a heterogeneous mixture shows obvious differences in properties (density, color, state, etc.) from one part of the mixture to another.

2.115

A solution (such as salt or sugar dissolved in water) is a homogeneous mixture.

2.116

Plan: Review the definitions of homogeneous and heterogeneous. The key is that a homogeneous mixture has a uniform composition while a heterogeneous mixture does not. A mixture consists of two or more substances physically mixed together while a compound is a pure substance. Solution: a) Distilled water is a compound that consists of H2O molecules only. b) Gasoline is a homogeneous mixture of hydrocarbon compounds of uniform composition that can be separated by physical means (distillation). c) Beach sand is a heterogeneous mixture of different size particles of minerals and broken bits of shells. d) Wine is a homogeneous mixture of water, alcohol, and other compounds that can be separated by physical means (distillation). e) Air is a homogeneous mixture of different gases, mainly N2, O2, and Ar.

2.117

Plan: Review the definitions of homogeneous and heterogeneous. The key is that a homogeneous mixture has a uniform composition while a heterogeneous mixture does not. A mixture consists of two or more substances physically mixed together while a compound is a pure substance. Solution: a) Orange juice is a heterogeneous mixture of water, juice, and bits of orange pulp. b) Vegetable soup is a heterogeneous mixture of water, broth, and vegetables. c) Cement is a heterogeneous mixture of various substances. d) Calcium sulfate is a compound of calcium, sulfur, and oxygen in a fixed proportion. e) Tea is a homogeneous mixture.

2.118

Plan: Review the discussion on separations. Solution: a) Salt dissolves in water and pepper does not. Procedure: add water to mixture and filter to remove solid pepper. Evaporate water to recover solid salt. b) The water/soot mixture can be filtered; the water will flow through the filter paper, leaving the soot collected on the filter paper. c) Allow the mixture to warm up, and then pour off the melted ice (water); or, add water, and the glass will sink and the ice will float. d) Heat the mixture; the alcohol will boil off (distill), while the sugar will remain behind. e) The spinach leaves can be extracted with a solvent that dissolves the pigments. Chromatography can be used to separate one pigment from the other.

2.119

Plan: Review the discussion on separations. Solution: a) Filtration ----- separating the mixture on the basis of differences in particle size. The water moves through the holes in the colander but the larger pasta cannot. b) Extraction ----- The colored impurities are extracted into a solvent that is rinsed away from the raw sugar (or chromatography). A sugar solution is passed through a column in which the impurities stick to the stationary phase and the sugar moves through the column in the mobile phase.

2.120

Analysis time can be shortened by operating the column at a higher temperature or by increasing the rate of flow of the gaseous mobile phase.

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2-30

2.121

Plan: Use the equation for the volume of a sphere in part (a) to find the volume of the nucleus and the volume of the atom. Calculate the fraction of the atom volume that is occupied by the nucleus. For part (b), calculate the total mass of the two electrons; subtract the electron mass from the mass of the atom to find the mass of the nucleus. Then calculate the fraction of the atom’s mass contributed by the mass of the nucleus. Solution: 3 4 4 3 15 3 ---44 3 a) Volume (m ) of nucleus = r = 2.510 m = 6.54498 × 10 m 3 3 3 4 4 3 11 3 ---31 3 Volume (m ) of atom = r = 3.110 m = 1.24788 × 10 m 3 3







Fraction of volume =



volume of Nucleus volume of Atom

6.544981044 m 3 ---13 ---13 = 5.2449 × 10 = 5.2 × 10 1.247881031 m 3

b) Mass of nucleus = mass of atom --- mass of electrons = 6.64648 × 10 Fraction of mass =

---24

g --- 2(9.10939 × 10

massof Nucleus mass of Atom

---28

g) = 6.64466 × 10

6.64466 10

6.6464810

24 24

g g

---24

 = 0.99972617 = 0.999726



As expected, the volume of the nucleus relative to the volume of the atom is small while its relative mass is large. 2.122

Plan: Use Coulomb’s law which states that the energy of attraction in an ionic bond is directly proportional to the product of charges and inversely proportional to the distance between charges. Choose the largest ionic charges and smallest radii for the strongest ionic bonding and the smallest ionic charges and largest radii for the weakest ionic bonding. Solution: 2+ 2+ 2– Strongest ionic bonding: MgO. Mg , Ba , and O have the largest charges. Attraction increases as distance 2+ 2+ decreases, so the positive ion with the smaller radius, Mg , will form a stronger ionic bond than the larger ion Ba . + + ----Weakest ionic bonding: RbI. K , Rb , Cl , and I have the smallest charges. Attraction decreases as distance + --increases, so the ions with the larger radii, Rb and I , will form the weakest ionic bond.

2.123

Plan: Use the chemical symbols and count the atoms of each type to give a molecular formula. Use the nomenclature rules in the chapter to derive the name. These compounds are composed of two nonmetals. Greek numerical prefixes are used to indicate the number of atoms of each element in each compound. The molecular (formula) mass is the sum of the masses of each atom times its atomic mass. Solution: a) Formula is BrF3. When a compound is composed of two elements from the same group, the element with the higher period number is named first. The prefix triindicates 3 fluorine atoms. A prefix is used with the first word in the name only when more than one atom of that element is present. The name is bromine trifluoride. Br = 1(79.90 amu) = 79.90 amu F = 3(19.00 amu) = 57.00 amu 136.90 amu b) The formula is SCl2. The element with the lower group number is the first word in the name. The prefix diindicates 2 chlorine atoms. A prefix is used with the first word in the name only when more than one atom of that element is present. The name is sulfur dichloride. S = 1(32.06 amu) = 32.06 amu Cl = 2(35.45 amu) = 70.90 amu 102.96 amu

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2-31

c) The formula is PCl3. The element with the lower group number is the first word in the name. The prefix triindicates 3 chlorine atoms. A prefix is used with the first word in the name only when more than one atom of that element is present. The name is phosphorus trichloride. P = 1(30.97 amu) = 30.97 amu Cl = 3(35.45 amu) = 106.35 amu 137.32 amu d) The formula is N2O5. The element with the lower group number is the first word in the name. The prefix diindicates 2 nitrogen atoms and the prefix penta- indicates 5 oxygen atoms. Only the second element is named with the suffix -ide. The name is dinitrogen pentoxide. N = 2(14.01 amu) = 28.02 amu O = 5(16.00 amu) = 80.00 amu 108.02 amu 2.124

Plan: These polyatomic ions are oxoanions composed of oxygen and another nonmetal. Oxoanions with the same number of oxygen atoms and nonmetals in the same group will have the same suffix ending. Only the nonmetal root name will change. Solution: 2--2--a) SeO4 selenate ion from SO4 = sulfate ion 3--3--b) AsO4 arsenate ion from PO4 = phosphate ion ----c) BrO2 bromite ion from ClO2 = chlorite ion ----d) HSeO4 hydrogen selenate ion from HSO4 = hydrogen sulfate ion 2--2--e) TeO3 tellurite ion from SO3 = sulfite ion

2.125

Plan: Write the formula of the compound and find the molecular mass. Determine the mass percent of nitrogen or phosphorus by dividing the mass of nitrogen or phosphorus in the compound by the molecular mass and multiplying by 100. For part (b), multiply the 100. g sample of compound by the mass ratio of ammonia to compound. Solution: + --a) Ammonium is NH4 and dihydrogen phosphate is H2PO4 . The formula is NH4H2PO4. N = 1(14.01 amu) = 14.01 amu H = 6(1.008 amu) = 6.048 amu P = 1(30.97 amu) = 30.97 amu O = 4(16.00 amu) = 64.00 amu 115.03 amu 14.01 amu N 100 = 12.18% N Mass percent of N = 115.03 amu compound Mass percent of P =

30.97 amu P 100 = 26.92% P 115.03 amu compound

  17.03 amu NH3  = 14.80 g NH3 b) Mass (g) of ammonia (NH3) = 100. g NH4 H2 PO4  115.03 amu NH4 H2 PO4  2.126

Plan: Determine the percent oxygen in each oxide by subtracting the percent nitrogen from 100%. Express the percentage in amu and divide by the atomic mass of the appropriate elements. Then divide each amount by the smaller number and convert to the simplest whole-number ratio. To find the mass of oxygen per 1.00 g of nitrogen, divide the mass percentage of oxygen by the mass percentage of nitrogen.

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2-32

Solution: a) I

(100.00 --- 46.69 N)% = 53.31% O  46.69 amu N    = 3.3326 N   14.01 amu N 

3.3326 N

2.6303

= 1.0000 mol N

III

= 1.0000 O

 63.15 amu O    = 3.9469 O  16.00 amu O 

3.9469 O = 1.5001 O 2.6303

The simplest whole-number ratio is 1:1.5 N:O = 2:3 N:O. (100.00 --- 25.94 N)% = 74.06% O  25.94 amu N   74.06 amu O     = 1.8515 N  = 4.6288 O    14.01amu N   16.00 amu O 

1.8515 N = 1.0000 N 1.8515

3.3319 O

3.3319 3.3319 The simplest whole-number ratio is 1:1 N:O. (100.00 --- 36.85 N)% = 63.15% O  36.85 amu N    = 2.6303 N   14.01 amu N  2.6303N

III

= 1.0002 N

 53.31 amu O    = 3.3319 O  16.00 amu O 

4.6288 O = 2.5000 O 1.8515

The simplest whole-number ratio is 1:2.5 N:O = 2:5 N:O.  53.31 amu O    = 1.1418 = 1.14 g O   46.69 amu N   63.15 amu O    = 1.7137 = 1.71 g O   36.85 amu N   74.06 amu O    = 2.8550 = 2.86 g O   25.94 amu N 

2.127

Plan: Recall that density = mass/volume. Solution: The mass of an atom of Pb is several times that of one of Al. Thus, the density of Pb would be expected to be several times that of Al if approximately equal numbers of each atom were occupying the same volume.

2.128

Plan: Review the law of mass conservation and law of definite composition. For each experiment, compare the mass values before and after each reaction and examine the ratios of the mass of reacted sodium to the mass of reacted chlorine. Solution: In each case, the mass of the starting materials (reactants) equals the mass of the ending materials (products), so the law of mass conservation is observed. Case 1: 39.34 g + 60.66 g = 100.00 g Case 2: 39.34 g + 70.00 g = 100.00 g + 9.34 g Case 3: 50.00 g + 50.00 g = 82.43 g + 17.57 g

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2-33

Each reaction yields the product NaCl, not Na2Cl or NaCl2 or some other variation, so the law of definite composition is observed. In each case, the ratio of the mass of sodium to the mass of chlorine in the compound is the same. Case 1: Mass Na/mass Cl2 = 39.34 g/60.66 g = 0.6485 Case 2: Mass of reacted Cl2 = initial mass --- excess mass = 70.00 g --- 9.34 g = 60.66 g Cl2 Mass Na/mass Cl2 = 39.34 g/60.66 g = 0.6485 Case 3: Mass of reacted Na = initial mass --- excess mass = 50.00 g --- 17.57 g = 32.43 g Na Mass Na/mass Cl2 = 32.43 g/50.00 g = 0.6486 2.129

Plan: Recall the definitions of solid, liquid, gas (from Chapter 1), element, compound, and homogeneous and heterogeneous mixtures. Solution: a) Gas is the phase of matter that fills its container. A mixture must contain at least two different substances. B, F, G, and I each contain only one gas. D and E each contain a mixture; E is a mixture of two different gases while D is a mixture of a gas and a liquid of a second substance. b) An element is a substance that cannot be broken down into simpler substances. A, C, G, and I are elements. c) The solid phase has a very high resistance to flow since it has a fixed shape. A shows a solid element. d) A homogeneous mixture contains two or more substances and has only one phase. E and H are examples of this. E is a homogeneous mixture of two gases and H is a homogeneous mixture of two liquid substances. e) A liquid conforms to the container shape and forms a surface. C shows one element in the liquid phase. f) A diatomic particle is a molecule composed of two atoms. B and G contain diatomic molecules of gas. g) A compound can be broken down into simpler substances. B and F show molecules of a compound in the gas phase. h) The compound shown in F has molecules composed of two white atoms and one blue atom for a 2:1 atom ratio. i) Mixtures can be separated into the individual components by physical means. D, E, and H are each a mixture of two different substances. j) A heterogeneous mixture like D contains at least two different substances with a visible boundary between those substances. k) Compounds obey the law of definite composition. B and F depict compounds.

2.130

Plan: To find the mass percent divide the mass of each substance in mg by the amount of seawater in mg and multiply by 100. The percent of an ion is the mass of that ion divided by the total mass of ions. Solution: 1000 g  1000 mg  = 1106 mg a) Mass (mg) of seawater = 1 kg  1 g   1 kg   mass of substance   100% Mass % =   mass of seawater   18,980 mg Cl   --100% = 1.898% Cl--Mass % Cl =   110 6 mg seawater     10.560 mg Na    + 100% = 1.056% Na+ Mass % Na =   1106 mg seawater     2650 mg SO 2   2--4 100% = 0.265% SO 2--Mass % SO4 =  4  1106 mg seawater     1270 mg Mg 2   2+ 2+ Mass % Mg =  100% = 0.127% Mg  1106 mg seawater   

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2-34

 400 mg Ca 2    2+ 100% = 0.04% Ca2+ Mass % Ca =   1106 mg seawater        380 mg K  + + Mass % K =  100% = 0.038% K  1106 mg seawater     140 mg HCO    --3 100% = 0.014% HCO --Mass % HCO3 =  3  1106 mg seawater    The mass percents do not add to 100% since the majority of seawater is H2O. b) Total mass of ions in 1 kg of seawater = 18,980 mg + 10,560 mg + 2650 mg + 1270 mg + 400 mg + 380 mg + 140 mg = 34,380 mg  10,560 mg Na +  100 = 30.71553 = 30.72% % Na + =   34,380 mg total ions  2+

c) Alkaline earth metal ions are Mg and Ca (Group 2 ions). 2+ 2+ Total mass % = 0.127% Mg + 0.04% Ca = 0.167% + + + + Alkali metal ions are Na and K (Group 1 ions). Total mass % = 1.056% Na + 0.038% K = 1.094% Mass % of alkali metal ions 1.094% = = 6.6 Mass % of alkaline earth metal ions 0.167% Total mass percent for alkali metal ions is 6.6 times greater than the total mass percent for alkaline earth metal ions. Sodium ions (alkali metal ions) are dominant in seawater. --2----d) Anions are Cl , SO4 , and HCO3 . --2----Total mass % = 1.898% Cl + 0.265% SO4 + 0.014% HCO3 = 2.177% anions + 2+ 2+ + Cations are Na , Mg , Ca , and K . + 2+ 2+ + Total mass % = 1.056% Na + 0.127% Mg + 0.04% Ca + 0.038% K = 1.2610 = 1.26% cations The mass fraction of anions is larger than the mass fraction of cations. Is the solution neutral since the mass of anions exceeds the mass of cations? Yes, although the mass is larger, the number of positive charges equals the number of negative charges. 2.131

Plan: Review the mass laws in the chapter. Solution: The law of mass conservation is illustrated in this change. The first flask has six oxygen atoms and six nitrogen atoms. The same number of each type of atom is found in both of the subsequent flasks. The mass of the substances did not change. The law of definite composition is also illustrated. During both temperature changes, the same compound, N2O, was formed with the same composition.

2.132

Plan: Use the density values to convert volume of each element to mass. Find the mass ratio of Ba to S in the compound and compare that to the mass ratio present. Solution: For barium sulfide the barium to sulfur mass ratio is (137.3 g Ba/32.06 g S) = 4.283 g Ba/g S  3.51 g Ba  Mass (g) of barium = 2.50 cm 3 Ba   = 8.775 g Ba  1 cm3 Ba 





 2.07 g S  Mass (g) of sulfur = 1.75 cm 3 S  = 3.6225 g S  1 cm 3 S  8.775 g Ba Barium to sulfur mass ratio = = 2.4224 = 2.42 g Ba/g S 3.6225 g S No, the ratio is too low; there is insufficient barium.





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2-35

2.133

Plan: First, count each type of atom present to produce a molecular formula. The molecular (formula) mass is the sum of the atomic masses of all of the atoms. Divide the mass of each element in the compound by the molecular mass and multiply by 100 to obtain the mass percent of each element. Solution: The molecular formula of succinic acid is C 4H6O4. C = 4(12.01 amu) = 48.04 amu H = 6(1.008 amu) = 6.048 amu O = 4 (16.00 amu) = 64.00 amu 118.09 amu  48.04 amu C   100% = 40.6815 = 40.68% C % C =   118.088 amu   6.048 amu H   100% = 5.1216 = 5.122% H % H =   118.088 amu   64.00 amu O   100% = 54.1969 = 54.20% O % O =   118.088 amu  Check: Total = (40.68 + 5.122 + 54.20)% = 100.00% The answer checks.

2.134

Plan: The toxic level of fluoride ion for a 70-kg person is 0.2 g. Convert this mass to mg and use the concentration of fluoride ion in drinking water to find the volume of water that contains the toxic amount. Convert the volume of the reservoir to liters and use the concentration of 1mg of fluoride ion per liter of water to find the mass of sodium fluoride required. Solution: --A 70-kg person would have to consume 0.2 mg of F to reach the toxic level.  1 mg F     = 200 mg F--Mass (mg) of fluoride for a toxic level = 0.2 g F   0.001 g F  1 L water    = 200 = 2 × 10 2 L water Volume (L) of water = 200 mg   1 mg F 





 4 qt   1 L   8 Volume (L) of reservoir = 8.50 10 7 gal    = 3.216651 × 10 L 1 gal  1.057 qt  ---

The molecular mass of NaF = 22.99 amu Na + 19.00 amu F = 41.99 amu. There are 19.00 mg of F in every 41.99 mg of NaF. 3  1 mg F    41.99 mg NaF  10 g  1 kg NaF   8 Mass (kg) of NaF = 3.21665110 L      1 L H 2 O  19.00 mg F   1 mg 10 3 g NaF       = 710.88 = 711 kg NaF 2.135

Plan: Z = the atomic number of the element. A is the mass number. To find the percent abundance of each Sb isotope, let x equal the fractional abundance of one isotope and (1 --- x) equal the fractional abundance of the second isotope since the sum of the fractional abundances must equal 1. Remember that atomic mass = (isotopic mass of the first isotope × fractional abundance) + (isotopic mass of the second isotope × fractional abundance). Solution: 121 a) Antimony is element 51so Z = 51. Isotope of mass 120.904 amu has a mass number of 121: 51 Sb 123 Isotope of mass 122.904 amu has a mass number of 123: 51 Sb b) Let x = fractional abundance of antimony-121. This makes the fractional abundance of antimony-123 = 1 --- x × (120.904 amu) + (1 --- x) (122.904 amu) = 121.8 amu McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution

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2-36

120.904 amu(x) + 122.904 amu --- 122.904 amu(x) = 121.8 amu 2x = 1.104 x = 0.552 = 0.55 fraction of antimony-121 1 --- x = 1 --- 0.552 = 0.45 fraction of antimony-123 2.136

Plan: List all possible combinations of the isotopes. Determine the masses of each isotopic composition. The molecule consisting of the lower abundance isotopes (N-15 and O-18) is the least common, and the one containing only the more abundant isotopes (N-14 and O-16) will be the most common. Solution: a) b) Formula Mass (amu) 15 18 N2 O 2(15 amu N) + 18 amu O = 48 least common 15 16 N2 O 2(15 amu N) + 16 amu O = 46 14 18 N2 O 2(14 amu N) + 18 amu O = 46 14 16 N2 O 2(14 amu N) + 16 amu O = 44 most common 15 14 18 N N O 1(15 amu N) + 1(14 amu N) + 18 amu O = 47 15 14 16 N N O 1(15 amu N) + 1(14 amu N) + 16 amu O = 45

2.137

Plan: Review the information about the periodic table in the chapter. Solution: a) Nonmetals are located in the upper-right portion of the periodic table: Black, red, green, and purple b) Metals are located in the large left portion of the periodic table: Brown and blue c) Some nonmetals, such as oxygen, chlorine, and argon, are gases: Red, green, and purple d) Most metals, such as sodium and barium are solids; carbon is a solid: Brown, blue, and black e) Nonmetals form covalent compounds; most noble gases do not form compounds: Black and red or black and green or red and green f) Nonmetals form covalent compounds; most noble gases do not form compounds: Black and red or black and green or red and green g) Metals react with nonmetals to form ionic compounds. For a compound with a formula of MX, the + --2+ 2--ionic charges of the metals and nonmetal must be equal in magnitude like Na and Cl or Ba and O : Brown and green or blue and red h) Metals react with nonmetals to form ionic compounds. For a compound with a formula of MX, the + --2+ 2--ionic charges of the metals and nonmetal must be equal in magnitude like Na and Cl or Ba and O : Brown and green or blue and red i) Metals react with nonmetals to form ionic compounds. For a compound with a formula of M2X, the + 2--2+ 4--ionic charge of the nonmetal must be twice as large as that of the metal like Na and O or Ba and C : Brown and red or blue and black j) Metals react with nonmetals to form ionic compounds. For a compound with a formula of MX2, the 2+ --ionic charge of the metal must be twice as large as that of the nonmetal like Ba and Cl : Blue and green k) Most Group 8A(18) elements are unreactive: Purple 2--2--l) Different compounds often exist between the same two nonmetal elements. Since oxygen exists as O or O2 , metals can sometimes form more than one compound with oxygen: Black and red or red and green or black and green or brown and red or blue and red

2.138

Plan: To find the formula mass of potassium fluoride, add the atomic masses of potassium and fluorine. Fluorine has only one naturally occurring isotope, so the mass of this isotope equals the atomic mass of fluorine. The atomic mass of potassium is the weighted average of the two isotopic masses: (isotopic mass of isotope 1 × fractional abundance) + (isotopic mass of isotope 2 × fractional abundance). Solution:

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2-37

Average atomic mass of K = 39 41 (isotopic mass of K × fractional abundance) + (isotopic mass of K × fractional abundance)  93.258%   6.730% Average atomic mass of K = (38.9637 amu)   (40.9618 amu)  = 39.093 amu  100%   100%  The formula for potassium fluoride is KF, so its molecular mass is (39.093 + 18.9984) = 58.091 amu 10

2.139

Plan: List all possible combinations of the isotopes. BF3 contains either B or B. Determine the masses of each isotopic composition and also the masses of each molecule missing one, two, or all three F atoms. Solution: 10 19 B F3 = 10 amu B + 3(19 amu F) = 67 amu 10 19 B F2 = 10 amu B + 2(19 amu F) = 48 amu 10 19 B F = 10 amu B + 19 amu F = 29 amu 10 B = 10 amu B = 10. amu 11 19 B F3 = 11 amu B + 3(19 amu F) = 68 amu 11 19 B F2 = 11 amu B + 2(19 amu F) = 49 amu 11 19 B F = 11 amu B + 19 amu F = 30. amu 11 B = 11 amu B = 11 amu

2.140

Plan: One molecule of NO is released per atom of N in the medicine. Divide the total mass of NO released by the molecular mass of the medicine and multiply by 100 for mass percent. Solution: NO = (14.01 + 16.00) amu = 30.01 amu Nitroglycerin: C3H5N3O9 = 3(12.01 amu C) + 5(1.008 amu H) + 3(14.01 amu N) + 9(16.00 amu O) = 227.10 amu In C3H5N3O9 (molecular mass = 227.10 amu), there are 3 atoms of N; since 1 molecule of NO is released per atom of N, this medicine would release 3 molecules of NO. The molecular mass of NO = 30.01 amu. total mass of NO 3(30.01 amu) 100  100 = 39.6433 = 39.64% Mass percent of NO = mass of compound 227.10 amu Isoamyl nitrate: C5H11NO3 = 5(12.01 amu C) + 11(1.008 amu H) + 1(14.01 amu N) + 3(16.00 amu O) = 133.15 amu In (CH3)2CHCH2CH2ONO2 (molecular mass = 133.15 amu), there is one atom of N; since 1 molecule of NO is released per atom of N, this medicine would release 1 molecule of NO. total mass of NO 1(30.01 amu) 100  100 = 22.5385 = 22.54% Mass percent of NO = mass of compound 133.15 amu

2.141

Plan: Each peak in the mass spectrum of carbon represents a different isotope of carbon. The heights of the peaks correspond to the natural abundances of the isotopes. Solution: 12 13 14 12 Carbon has three naturally occurring isotopes: C, C, and C. C has an abundance of 98.89% and would have the tallest peak in the mass spectrum as the most abundant isotope. 13 C has an abundance of 1.11% and thus would have a significantly shorter peak; the shortest 14 peak in the mass spectrum would correspond to the least abundant isotope, C, the abundance of 12 which is less than 0.01%. Peak Y, as the tallest peak, has a m/e ratio of 12 ( C); X, the shortest 14 13 peak, has a m/e ratio of 14( C). Peak Z corresponds to C with a m/e ratio of 13.

2.142

Plan: First, count each type of atom present to produce a molecular formula. Determine the mass fraction of each total mass of the element element. Mass fraction = . The mass of TNT multiplied by the mass fraction of each molecular mass of TNT element gives the mass of that element.

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2-38

Solution: The molecular formula for TNT is C7H5O6N3. The molecular mass of TNT is: C = 7(12.01 amu) = 84.07 amu H = 5(1.008 amu) = 5.040 amu O = 6(16.00 amu) = 96.00 amu N = 3(14.01 amu) = 42.03 amu 227.14 amu The mass fraction of each element is: 84.07 amu 5.040 amu C= = 0.3701 C H= = 0.02219 H 227.14 amu 227.14 amu

96.00 amu 42.03 amu = 0.4226 O N= = 0.1850 N 227.14 amu 227.14 amu Masses of each element in 1.00 lb of TNT = mass fraction of element × 1.00 lb. Mass (lb) C = 0.3701 × 1.00 lb = 0.370 lb C Mass (lb) H = 0.02219 × 1.00 lb = 0.0222 lb H Mass (lb) O = 0.4226 × 1.00 lb = 0.423 lb O Mass (lb) N = 0.1850 × 1.00 lb = 0.185 lb N O=

2.143

Plan: The superscript is the mass number, the sum of the number of protons and neutrons. Consult the periodic table to get the atomic number (the number of protons). The mass number --- the number of protons = the number of neutrons. Divide the number of neutrons by the number of protons to obtain the N/Z ratio. For atoms, the number of protons and electrons are equal. Solution: neutrons (N) protons (Z) N/Z 144 a) 62 Sm 144 --- 62 = 82 62 82/62 = 1.3 56

b) 26 Fe

56 --- 26 = 30

30/26 = 1.2

20 10

20 --- 10 = 10

10/10 = 1.0

107 --- 47 = 60

60/47 = 1.3

neutrons

protons

electrons

107 d) 47

2.144

238 92

238 --- 92 = 146

234 92

234 --- 92 = 142

214 82

214 --- 82 = 132

210 82

210 --- 82 = 128

206 82

206 --- 82 = 124

Plan: Determine the mass percent of platinum by dividing the mass of Pt in the compound by the molecular mass of the compound and multiplying by 100. For part (b), divide the total amount of money available by the cost of Pt per gram to find the mass of Pt that can be purchased. Use the mass percent of Pt to convert from mass of Pt to mass of compound. Solution: a) The molecular formula for platinol is Pt(NH3)2Cl2. Its molecular mass is: Pt = 1(195.1 amu) = 195.1 amu N = 2 (14.01 amu) = 28.02 amu H = 6(1.008 amu) = 6.048 amu Cl = 2(35.45 amu) = 70.90 amu 300.1 amu McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution

in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

2-39

Mass % Pt =

195.1 amu mass of Pt 100 = 65.012 = 65.01% Pt 100 = 300.1 amu molecular mass of compound

   = 19,608 g Pt b) Mass (g) of Pt = $1.00106   $51  Mass (g) of platinol = 

  65.01 g Pt



  = 3.0162 × 104 = 3.0 × 104 g platinol 

2.145

Plan: Obtain the information from the periodic table. The period number of an element is its row number while the group number is its column number. Solution: a) Building-block elements: Name Symbol Atomic number Atomic mass Period number Group number Hydrogen H 1 1.008 amu 1 1A(1) Carbon C 6 12.01 amu 2 4A(14) Nitrogen N 7 14.01 amu 2 5A(15) Oxygen O 8 16.00 amu 2 6A(16) b) Macronutrients: Sodium Na 11 22.99 amu 3 1A(1) Magnesium Mg 12 24.31 amu 3 2A(2) Potassium K 19 39.10 amu 4 1A(1) Calcium Ca 20 40.08 amu 4 2A(2) Phosphorus P 15 30.97 amu 3 5A(15) Sulfur S 16 32.06 amu 3 6A(16) Chlorine Cl 17 35.45 amu 3 7A(17)

2.146

Plan: Review the definitions of pure substance, element, compound, homogeneous mixture, and heterogeneous mixture. Solution: Matter is divided into two categories: pure substances and mixtures. Pure substances are divided into elements and compounds. Mixtures are divided into solutions (homogeneous mixtures) and heterogeneous mixtures.

2.147

Plan: A change is physical when there has been a change in physical form but not a change in composition. In a chemical change, a substance is converted into a different substance. Solution: 1) Initially, all the molecules are present in blue-blue or red-red pairs. After the change, there are no red-red pairs, and there are now red-blue pairs. Changing some of the pairs means there has been a chemical change. 2) There are two blue-blue pairs and four red-blue pairs both before and after the change, thus no chemical change occurred. The different types of molecules are separated into different boxes. This is a physical change. 3) The identity of the box contents has changed from pairs to individuals. This requires a chemical change. 4) The contents have changed from all pairs to all triplets. This is a change in the identity of the particles, thus, this is a chemical change. 5) There are four red-blue pairs both before and after, thus there has been no change in the identity of the individual units. There has been a physical change.

McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution

in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

2-40

CHAPTER 3 STOICHIOMETRY OF FORMULAS AND EQUATIONS FOLLOW–UP PROBLEMS 3.1A

Plan: The mass of carbon must be changed from mg to g. The molar mass of carbon can then be used to determine the number of moles. Solution: 103 g   1 mol C   –2 –2 Moles of carbon = 315 mg C    = 2.6228 × 10 = 2.62 × 10 mol C  1 mg   12.01 g C  Road map: Mass (mg) of C 3

10 mg = 1 g Mass (g) of C Divide by

(g/mol)

Amount (moles) of C 3.1B

Plan: The number of moles of aluminum must be changed to g. Then the mass of aluminum per can can be used to calculate the number of soda cans that can be made from 52 mol of Al. Solution:     = 100.21 = 100 soda cans Number of soda cans = 52 mol Al        Road map: Amount (mol) of Al Multiply by (g/mol) (1 mol Al = 26.98 g Al) Mass (g) of Al 14 g Al = 1 soda can Number of cans

3.2A

Plan: Avogadro’s number is needed to convert the number of nitrogen molecules to moles. Since nitrogen molecules are diatomic (composed of two N atoms), the moles of molecules must be multiplied by 2 to obtain moles of atoms.

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3-1

Solution:   1 mol N 2  2 N atoms   Moles of N atoms = 9.72 10 21 N 2 molecules     6.022 10 23 N 2 molecules   1 mol N 2 





–2

= 3.2281634 × 10 = 3.23 × 10 mol N Road map: No. of N2 molecules Divide by Avogadro’s number (molecules/mol) Amount (moles) of N2 Use chemical formula (1 mol N2 = 2 mol N) Amount (moles) of N 3.2B

Plan: Avogadro’s number is needed to convert the number of moles of He to atoms. Solution:    23 26 26 Number of He atoms = 325 mol He   = 1.9572 × 10 = 1.96 × 10 He atoms   Road map: Amount (mol) of He Multiply by Avogadro’s number 23 (1 mol He = 6.022 × 10 He atoms) Number of He atoms

3.3A

Plan: Avogadro’s number is needed to convert the number of atoms to moles. The molar mass of manganese can then be used to determine the number of grams. Solution:   1 mol Mn 54.94 g Mn     Mass (g) of Mn = 3.22  10 20 Mn atoms  23  1 mol Mn   6.022  10 Mn atoms  





–2

= 2.9377 × 10 = 2.94 × 10 g Mn Road map: No. of Mn atoms Divide by Avogadro’s number (molecules/mol) Amount (moles) of Mn

3-2

Multiply by

(g/mol)

Mass (g) of Mn 3.3B

Plan: Use the molar mass of copper to calculate the number of moles of copper present in a penny. Avogadro’s number is then needed to convert the number of moles of Cu to Cu atoms. Solution:     23  Number of Cu atoms = 0.0625 g Cu        20

= 5.9225 × 10 = 5.92 × 10 Cu atoms Road map: Mass (g) of Cu Divide by (g/mol) (1 mol Cu = 63.55 g Cu) Amount (moles) of Cu Multiply by Avogadro’s number 23 (1 mol Cu = 6.022 × 10 Cu atoms) No. of Cu atoms

3.4A

Plan: Avogadro’s number is used to change the number of molecules to moles. Moles may be changed to mass by multiplying by the molar mass. The molar mass of tetraphosphorus decoxide is obtained from its chemical formula. Each molecule has four phosphorus atoms, so the total number of atoms is four times the number of molecules. Solution: a) Tetra = 4, and deca = 10 to give P4O10. The molar mass, , is the sum of the atomic weights, expressed in g/mol: P = 4(30.97) = 123.88 g/mol O = 10(16.00) = 160.00 g/mol = 283.88 g/mol of P4O10   1 mol 283.88 g     Mass (g) of P4O10 = 4.6510 22 molecules P4 O10   1 mol   6.022 10 23 molecules   = 21.9203 = 21.9 g P4O10  4 atoms P   23 b) Number of P atoms = 4.6510 22 molecules P4 O10   = 1.86 × 10 P atoms 1 P4 O10 molecule 



3.4B



Plan: The mass of calcium phosphate is converted to moles of calcium phosphate by dividing by the molar mass. Avogadro’s number is used to change the number of moles to formula units. Each formula unit has two phosphate ions, so the total number of phosphate ions is two times the number of formula units.

3-3

Solution: a) The formula of calcium phosphate is Ca3(PO4)2. The molar mass, , is the sum of the atomic weights, expressed in g/mol: = (3 × of Ca) + (2 × of P) + (8 × of O) = (3 × 40.08 g/mol Ca) + (2 × 30.97 g/mol P) + (8 × 16.00 g/mol O) = 310.18 g/mol Ca3(PO4)2  No. of formula units Ca3(PO4)2 = 75.5 g Ca3(PO4)2   

   23  3 4 2      3 4 2 3 4 2 23 23 = 1.4658 × 10 = 1.47 × 10 formula units Ca3(PO4)2 3

4 2

 3 23 b) No. of phosphate (PO4 ) ions = 1.47 × 10 formula units Ca3(PO4)2  

3– 4 3

   4 2

= 2.94 × 10 phosphate ions 3.5A

Plan: Calculate the molar mass of glucose. The total mass of carbon in the compound divided by the molar mass of the compound, multiplied by 100% gives the mass percent of C. Solution: The formula for glucose is C6H12O6. There are 6 atoms of C per each formula. Molar mass of C6H12O6 = (6 × of C) + (12 × of H) + (6 × of O) = (6 × 12.01 g/mol) + (12 × 1.008 g/mol) + (6 × 16.00 g/mol) = 180.16 g/mol



Mass % of C = 6

3.5B

(100) = 39.9978 = 40.00% C

Plan: Calculate the molar mass of CCl3F. The total mass of chlorine in the compound divided by the molar mass of the compound, multiplied by 100% gives the mass percent of Cl. Solution: The formula is CCl3F. There are 3 atoms of Cl per each formula. Molar mass of CCl3F = (1 × of C) + (3 × of Cl) + (1 × of F) = (1 × 12.01 g/mol) + (3 × 35.45 g/mol) + (1 × 19.00 g/mol) = 137.36 g/mol Mass % of Cl =

(100) =



(100) = 77.4243 = 77.42% Cl

3.6A

Plan: Multiply the mass of the sample by the mass fraction of C found in the preceding problem. Solution:    = 6.6196 = 6.620 g C Mass (g) of C = 16.55 g C6H12O6    6 12 6 

3.6B

Plan: Multiply the mass of the sample by the mass fraction of Cl found in the preceding problem. Solution:     = 1819.47 = 1820 g Cl Mass (g) of Cl = 2.35 kg CCl3F       3 

3.7A

Plan: Calculate the number of moles of each element in the sample by dividing by the molar mass of the corresponding element. The calculated numbers of moles are the fractional amounts of the elements and can be used as subscripts in a chemical formula. Convert the fractional amounts to whole numbers by dividing each number by the smallest subscripted number.

3-4

Solution:  Moles of H = 1.23 g H  

  = 1.2202 mol H 

 Moles of P = 12.64 g P  

  = 0.40814 mol P 

 Moles of O = 26.12 g O  

  = 1.6325 mol O 

Divide each subscript by the smaller value, 0.408: H 1.2202 P0.40814 O 1.6325 = H3PO4, this is phosphoric acid. 3.7B

Plan: The moles of sulfur may be calculated by dividing the mass of sulfur by the molar mass of sulfur. The moles of sulfur and the chemical formula will give the moles of M. The mass of M divided by the moles of M will give the molar mass of M. The molar mass of M can identify the element. Solution:  1 mol S    = 0.089832 mol S Moles of S = 2.88 g S   32.06 g S   2 mol M    = 0.059888 mol M Moles of M = 0.089832 mol S   3 mol S  3.12 g M Molar mass of M = = 52.0972 = 52.1 g/mol 0.059888 mol M The element is Cr (52.00 g/mol); M is Chromium and M2S3 is chromium(III) sulfide.







3.8A



Plan: If we assume there are 100 grams of this compound, then the masses of carbon and hydrogen, in grams, are numerically equivalent to the percentages. Divide the atomic mass of each element by its molar mass to obtain the moles of each element. Dividing each of the moles by the smaller value gives the simplest ratio of C and H. The smallest multiplier to convert the ratios to whole numbers gives the empirical formula. To obtain the molecular formula, divide the given molar mass of the compound by the molar mass of the empirical formula to find the whole-number by which the empirical formula is multiplied. Solution: Assuming 100 g of compound gives 95.21 g C and 4.79 g H:  1 mol C    = 7.92756 mol C Moles of C = 95.21 g C  12.01 g C   1 mol H    = 4.75198 mol H Mole of H = 4.79 g H  1.008 g H  Divide each of the moles by 4.75198, the smaller value: C 7.92756 H 4.75198 = C1.6683H1 4.75198

4.75198

The value 1.668 is 5/3, so the moles of C and H must each be multiplied by 3. If it is not obvious that the value is near 5/3, use a trial and error procedure whereby the value is multiplied by the successively larger integer until a value near an integer results. This gives C5H3 as the empirical formula. The molar mass of this formula is: (5 × 12.01 g/mol) + (3 × 1.008 g/mol) = 63.074 g/mol

3-5

Whole-number multiple =

molar mass of compound 252.30 g/mol = =4 molar mass of empirical formula 63.074 g/mol

Thus, the empirical formula must be multiplied by 4 to give 4(C5H3) = C20H12 as the molecular formula of benzo[a]pyrene. 3.8B

Plan: If we assume there are 100 grams of this compound, then the masses of carbon, hydrogen, nitrogen, and oxygen, in grams, are numerically equivalent to the percentages. Divide the atomic mass of each element by its molar mass to obtain the moles of each element. Dividing each of the moles by the smaller value gives the simplest ratio of C, H, N, and O. To obtain the molecular formula, divide the given molar mass of the compound by the molar mass of the empirical formula to find the whole-number by which the empirical formula is multiplied. Solution: Assuming 100 g of compound gives 49.47 g C, 5.19 g H, 28.86 g N, and 16.48 g O:    = 4.1191 mol C Moles of C = 49.47 g C     Moles of H = 5.19 g H  

  = 5.1488 mol H 

 Moles of N = 28.86 g N  

  = 2.0600 mol N 

 Moles of O = 16.48 g O  

  = 1.0300 mol O 

Divide each subscript by the smaller value, 1.030: C 4.1191 H 5.1488 N 2.0600 O 1.0300 = C4H5N2O This gives C4H5N2O as the empirical formula. The molar mass of this formula is: (4 × 12.01 g/mol) + (5 × 1.008 g/mol) + (2 × 14.01 g/mol) + (1 × 16.00 g/mol) = 97.10 g/mol The molar mass of caffeine is 194.2 g/mol, which is larger than the empirical formula mass of 97.10 g/mol, so the molecular formula must be a whole-number multiple of the empirical formula. Whole-number multiple =

=2 ol Thus, the empirical formula must be multiplied by 2 to give 2(C4H5N2O) = C8H10N4O2 as the molecular formula of caffeine. 3.9A

Plan: The carbon in the sample is converted to carbon dioxide, the hydrogen is converted to water, and the remaining material is chlorine. The grams of carbon dioxide and the grams of water are both converted to moles. One mole of carbon dioxide gives one mole of carbon, while one mole of water gives two moles of hydrogen. Using the molar masses of carbon and hydrogen, the grams of each of these elements in the original sample may be determined. The original mass of sample minus the masses of carbon and hydrogen gives the mass of chlorine. The mass of chlorine and the molar mass of chlorine will give the moles of chlorine. Once the moles of each of the elements have been calculated, divide by the smallest value, and, if necessary, multiply by the smallest number required to give a set of whole numbers for the empirical formula. Compare the molar mass of the empirical formula to the molar mass given in the problem to find the molecular formula. Solution: Determine the moles and the masses of carbon and hydrogen produced by combustion of the sample.  1 mol CO 2  12.01 g C   1 mol C     = 0.12307 g C 0.451 g CO 2     0.010248 mol C   44.01 g CO 2   1 mol C  1 mol CO 2 

3-6

 1 mol H 2 O  1.008 g H   2 mol H     = 0.006904 g H 0.0617 g H 2 O     0.0068495 mol H   18.016 g H 2 O   1 mol H 2 O   1 mol H 

The mass of chlorine is given by: 0.250 g sample – (0.12307 g C + 0.006904 g H) = 0.12003 g Cl The moles of chlorine are:  1 mol Cl   0.12003 g Cl   = 0.0033859 mol Cl. This is the smallest number of moles.  35.45 g Cl  Divide each mole value by the lowest value, 0.0033850: C 0.010248 H 0.0068495 Cl 0.0033859 = C3H2Cl 0.0033859

0.0033859

The empirical formula has the following molar mass: (3 × 12.01 g/mol) + (2 × 1.008 g/mol) + (35.45 g/mol) = 73.496 g/mol C3H2Cl molar mass of compound 146.99 g/mol = Whole-number multiple = =2 molar mass of empirical formula 73.496 g/mol Thus, the molecular formula is two times the empirical formula, 2(C3H2Cl) = C6H4Cl2. 3.9B

Plan: The carbon in the sample is converted to carbon dioxide, the hydrogen is converted to water, and the remaining material is oxygen. The grams of carbon dioxide and the grams of water are both converted to moles. One mole of carbon dioxide gives one mole of carbon, while one mole of water gives two moles of hydrogen. Using the molar masses of carbon and hydrogen, the grams of each of these elements in the original sample may be determined. The original mass of sample minus the masses of carbon and hydrogen gives the mass of oxygen. The mass of oxygen and the molar mass of oxygen will give the moles of oxygen. Once the moles of each of the elements have been calculated, divide by the smallest value, and, if necessary, multiply by the smallest number required to give a set of whole numbers for the empirical formula. Compare the molar mass of the empirical formula to the molar mass given in the problem to find the molecular formula. Solution: Determine the moles and the masses of carbon and hydrogen produced by combustion of the sample.  3.516 g CO2    1.007 g H2O  

2 2

   

   2

   

  

    

 = 0.95949 g C   = 0.11266 g H 

The mass of oxygen is given by: 1.200 g sample – (0.95949 g C + 0.11266 g H) = 0.12785 g O The moles of C and H are calculated above. The moles of oxygen are:    = 0.0079906mol O. This is the smallest number of moles. 0.12785 g O    Divide each subscript by the smallest value, 0.00800: C 0.079891 H 0.11176 O 0.0079906 = C10H14O 0.0079906

0.0079906

This gives C10H14O as the empirical formula. The molar mass of this formula is: (10 × 12.01 g/mol) + (14 × 1.008 g/mol) + (1 × 16.00 g/mol) = 150.21 g/mol The molar mass of the steroid is 300.42 g/mol, which is larger than the empirical formula mass of 150.21 g/mol, so the molecular formula must be a whole-number multiple of the empirical formula. Whole-number multiple =

mol

3-7

Thus, the empirical formula must be multiplied by 2 to give 2(C10H14O) = C20H28O2 as the molecular formula of the steroid. 3.10A

Plan: In each part it is necessary to determine the chemical formulas, including the physical states, for both the reactants and products. The formulas are then placed on the appropriate sides of the reaction arrow. The equation is then balanced. Solution: a) Sodium is a metal (solid) that reacts with water (liquid) to produce hydrogen (gas) and a solution of sodium hydroxide (aqueous). Sodium is Na; water is H2O; hydrogen is H2; and sodium hydroxide is NaOH. Na(s) + H2O(l)  H2(g) + NaOH(aq) is the equation. Balancing will precede one element at a time. One way to balance hydrogen gives: Na(s) + 2H2O(l)  H2(g) + 2NaOH(aq) Next, the sodium will be balanced: 2Na(s) + 2H2O(l)  H2(g) + 2NaOH(aq) On inspection, we see that the oxygen is already balanced. b) Aqueous nitric acid reacts with calcium carbonate (solid) to produce carbon dioxide (gas), water (liquid), and aqueous calcium nitrate. Nitric acid is HNO3; calcium carbonate is CaCO3; carbon dioxide is CO2; water is H2O; and calcium nitrate is Ca(NO3)2. The starting equation is HNO3(aq) + CaCO3(s)  CO2(g) + H2O(l) + Ca(NO3)2(aq) Initially, Ca and C are balanced. Proceeding to another element, such as N, or better yet the group of elements in – NO3 gives the following partially balanced equation: 2HNO3(aq) + CaCO3(s)  CO2(g) + H2O(l) + Ca(NO3)2(aq) Now, all the elements are balanced. c) We are told all the substances involved are gases. The reactants are phosphorus trichloride and hydrogen fluoride, while the products are phosphorus trifluoride and hydrogen chloride. Phosphorus trifluoride is PF 3; phosphorus trichloride is PCl3; hydrogen fluoride is HF; and hydrogen chloride is HCl. The initial equation is: PCl3(g) + HF(g)  PF3(g) + HCl(g) Initially, P and H are balanced. Proceed to another element (either F or Cl); if we will choose Cl, it balances as: PCl3(g) + HF(g)  PF3(g) + 3HCl(g) The balancing of the Cl unbalances the H, this should be corrected by balancing the H as: PCl3(g) + 3HF(g)  PF3(g) + 3HCl(g) Now, all the elements are balanced.

3.10B

3-8

One element should be picked and balanced. Any element except oxygen will work. Oxygen will not work in this case because it appears more than once on one side of the reaction arrow. We will start with carbon. Balancing C gives: C3H5N3O9(l)  3CO2(g) + H2O(g) + N2(g) + O2(g) Now balancing the hydrogen gives: C3H5N3O9(l)  3CO2(g) + 5/2H2O(g) + N2(g) + O2(g) Similarly, if we balance N we get: C3H5N3O9(l)  3CO2(g) + 5/2H2O(g) + 3/2N2(g) + O2(g) Clear the fractions by multiplying everything except the unbalanced oxygen by 2: 2C3H5N3O9(l)  6CO2(g) + 5H2O(g) + 3N2(g) + O2(g) This leaves oxygen to balance. Balancing oxygen gives: 2C3H5N3O9(l)  6CO2(g) + 5H2O(g) + 3N2(g) + 1/2O2(g) Again clearing fractions by multiplying everything by 2 gives: 4C3H5N3O9(l)  12CO2(g) + 10H2O(g) + 6N2(g) + O2(g) Now all the elements are balanced. b) Potassium superoxide (KO2) is a solid. Carbon dioxide (CO2) and oxygen (O2) are gases. Potassium carbonate (K2CO3) is a solid. The initial equation is: KO2(s) + CO2(g)  O2(g) + K2CO3(s) Counting the atoms indicates that the carbons are balanced, but none of the other atoms are balanced. One element should be picked and balanced. Any element except oxygen will work (oxygen will be more challenging to balance because it appears more than once on each side of the reaction arrow). Because the carbons are balanced, we will start with potassium. Balancing potassium gives: 2KO2(s) + CO2(g)  O2(g) + K2CO3(s) Now all elements except for oxygen are balanced. Balancing oxygen by adding a coefficient in front of the O2 gives: 2KO2(s) + CO2(g)  3/2O2(g) + K2CO3(s) Clearing the fractions by multiplying everything by 2 gives: 4KO2(s) + 2CO2(g)  3O2(g) + 2K2CO3(s) Now all the elements are balanced. c) Iron(III) oxide (Fe2O3) is a solid, as is iron metal (Fe). Carbon monoxide (CO) and carbon dioxide (CO 2) are gases. The initial equation is: Fe2O3(s) + CO(g)  Fe(s) + CO2(g) Counting the atoms indicates that the carbons are balanced, but none of the other atoms are balanced. One element should be picked and balanced. Because oxygen appears in more than one compound on one side of the reaction arrow, it is best not to start with that element. Because the carbons are balanced, we will start with iron. Balancing iron gives: Fe2O3(s) + CO(g)  2Fe(s) + CO2(g) Now all the atoms but oxygen are balanced. There are 4 oxygen atoms on the left-hand side of the reaction arrow and 2 oxygen atoms on the right-hand side of the reaction arrow. In order to balance the oxygen, we want to change the coefficients in front of the carbon-containing compounds (if we changed the coefficient in front of the iron(III) oxide, the iron atoms would no longer be balanced). To maintain the balance of carbons, the coefficients in front of the carbon monoxide and the carbon dioxide must be the same. On the left-hand side of the equation, there are 3 oxygens in Fe2O3 plus 1X oxygen atoms from the CO (where X is the coefficient in the balanced Copyright McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

3-9

equation). On the right-hand side of the equation, there are 2X oxygen atoms. The number of oxygen atoms on both sides of the equation should be the same: 3 + 1X = 2X 3=X Balancing oxygen by adding a coefficient of 3 in front of the CO and CO2 gives: Fe2O3(s) + 3CO(g)  2Fe(s) + 3CO2(g) Now all the elements are balanced. 3.11A

Plan: Count the number of each type of atom in each molecule to write the formulas of the reactants and products. Solution: 6CO(g) + 3O2(g)  6CO2(g) or,

3.11B

Plan: Count the number of each type of atom in each molecule to write the formulas of the reactants and products. Solution: 6H2(g) + 2N2(g)  4NH3(g) or,

3.12A

2CO(g) + O2(g)  2CO2(g)

3H2(g) + N2(g)  2NH3(g)

Plan: The reaction, like all reactions, needs a balanced chemical equation. The balanced equation gives the molar ratio between the moles of iron and moles of iron(III) oxide. Solution: The names and formulas of the substances involved are: iron(III) oxide, Fe2O3, and aluminum, Al, as reactants, and aluminum oxide, Al2O3, and iron, Fe, as products. The iron is formed as a liquid; all other substances are solids. The equation begins as: Fe2O3(s) + Al(s)  Al2O3(s) + Fe(l) There are 2 Fe, 3 O, and 1 Al on the reactant side and 1 Fe, 3 O, and 2 Al on the product side. Balancing aluminum:

Fe2O3(s) + 2Al(s)  Al2O3(s) + Fe(l)

Fe2O3(s) + 2Al(s)  Al2O3(s) + 2Fe(l) 1 mol Fe2 O3  3 Moles of Fe2O3 = 3.60 103 mol Fe = 1.80 × 10 mol Fe2O3  2 mol Fe  Balancing iron:

Road map: Amount (moles) of Fe Molar ratio (2 mol Fe = 1 mol Fe2O3) Amount (moles) of Fe2O3 3.12B

Plan: Divide the mass of silver sulfide by its molar mass to obtain moles of the compound. The balanced equation gives the molar ratio between moles of silver sulfide and moles of silver. Solution:  Moles of Ag = 32.6 g Ag2S  

    2

  = 0.2630 = 0.263 mol Ag  2 

3-10

Road map: Mass (g) of Ag2S Divide by (g/mol) (247.9 g Ag2S = 1 mol Ag2S) Amount (moles) of Ag2S Molar ratio (3 mol Ag2S = 6 mol Ag) Amount (moles) of Ag 3.13A

Plan: The mass of aluminum oxide must be converted to moles by dividing by its molar mass. The balanced chemical equation (follow-up problem 3.14A) shows there are two moles of aluminum for every mole of aluminum oxide. Multiply the moles of aluminum by Avogadro’s number to obtain atoms of Al. Solution: 23  1 mol Al O      2 mol Al  6.022 10 atoms Al  2 3   Atoms of Al = 1.00 g Al2 O3     101.96 g Al2 O3  1 mol Al 1 mol Al2 O3   





= 1.18125 × 10 = 1.18 × 10 atoms Al Road map: Mass (g) of Al2O3 Divide by

(g/mol)

Amount (moles) of Al2O3 Molar ratio Amount (moles) of Al Multiply by Avogadro’s number Number of Al atoms 3.13B

Plan: The mass of aluminum sulfide must be converted to moles by dividing by its molar mass. The balanced chemical equation (follow-up problem 3.14B) shows there are two moles of aluminum for every mole of aluminum sulfide. Multiply the moles of aluminum by its molar mass to obtain the mass (g) of aluminum. Solution:  Mass (g) of Al = 12.1 g Al2S3  

3 2

   

  

 = 4.3487 = 4.35 g Al 

3-11

Road map: Mass (g) of Al2S3 Divide by (g/mol) (150.14 g Al2S3 = 1 mol Al2S3) Amount (moles) of Al2S3 Molar ratio (1 mol Al2S3 = 2 mol Al) Amount (moles) of Al Multiply by (g/mol) (1 mol Al = 26.98 g Al) Mass (g) of Al 3.14A

Plan: Write the balanced chemical equation for each step. Add the equations, canceling common substances. Solution: Step 1 2SO2(g) + O2(g)  2SO3(g) Step 2 SO3(g) + H2O(l)  H2SO4(aq) Adjust the coefficients since 2 moles of SO3 are produced in Step 1 but only 1 mole of SO3 is consumed in Step 2. We have to double all of the coefficients in Step 2 so that the amount of SO 3 formed in Step 1 is used in Step 2. Step 1 2SO2(g) + O2(g)  2SO3(g) Step 2 2SO3(g) + 2H2O(l)  2H2SO4(aq) Add the two equations and cancel common substances. Step 1 2SO2(g) + O2(g)  2SO3(g) Step 2 2SO3(g) + 2H2O(l)  2H2SO4(aq) 2SO2(g) + O2(g) + 2SO3(g) + 2H2O(l)  2SO3(g) + 2H2SO4(aq) Or

3.14B

2SO2(g) + O2(g) + 2H2O(l)  2H2SO4(aq)

Plan: Write the balanced chemical equation for each step. Add the equations, canceling common substances. Solution: Step 1 N2(g) + O2(g)  2NO(g) Step 2 NO(g) + O3(g)  NO2(g) + O2(g) Adjust the coefficients since 2 moles of NO are produced in Step 1 but only 1 mole of NO is consumed in Step 2. We have to double all of the coefficients in Step 2 so that the amount of NO formed in Step 1 is used in Step 2. Step 1 N2(g) + O2(g)  2NO(g) Step 2 2NO(g) + 2O3(g)  2NO2(g) + 2O2(g)

3-12

Add the two equations and cancel common substances. Step 1 N2(g) + O2(g)  2NO(g) Step 2 2NO(g) + 2O3(g)  2NO2(g) + 2O2(g) N2(g) + O2(g) + 2NO(g) + 2O3(g)  2NO(g) + 2NO2(g) + 2O2(g) Or 3.15A

N2(g) + 2O3(g)  2NO2(g) + O2(g)

Plan: Count the molecules of each type and find the simplest ratio. The simplest ratio leads to a balanced chemical equation. The substance with no remaining particles is the limiting reagent. Solution: 4 AB molecules react with 3 B2 molecules to produce 4 molecules of AB2, with 1 B2 molecule remaining unreacted. The balanced chemical equation is 4AB(g) + 2B2(g)  4AB2(g) or 2AB(g) + B2(g)  2AB2(g) The limiting reagent is AB since there is a B2 molecule left over (excess).

3.15B

Plan: Write a balanced equation for the reaction. Use the molar ratios in the balanced equation to find the amount (molecules) of SO3 produced when each reactant is consumed. The reactant that gives the smaller amount of product is the limiting reagent. Solution: 5 SO2 molecules react with 2 O2 molecules to produce molecules of SO3. The balanced chemical equation is 2SO2(g) + O2(g)  2SO3(g)  Amount (molecules) of SO3 produced from the SO2 = 5 molecules SO2    Amount (molecules) of SO3 produced from the O2 = 2 molecules SO2  

 = 5 molecules SO3  2 3

3 2

  = 4 molecules SO3 

O2 is the limiting reagent since it produces less SO3 than the SO2 does. 3.16A

Plan: Use the molar ratios in the balanced equation to find the amount of AB2 produced when 1.5 moles of each reactant is consumed. The smaller amount of product formed is the actual amount. Solution:  2 mol AB2  Moles of AB2 from AB = 1.5 mol AB  = 1.5 mol AB2  2 mol AB   2 mol AB2   = 3.0 mol AB2 Moles of AB2 from B2 = 1.5 mol B2   1 mol B2  Thus AB is the limiting reagent and only 1.5 mol of AB2 will form.

3.16B

Plan: Use the molar ratios in the balanced equation to find the amount of SO3 produced when 4.2 moles of SO2 are consumed and, separately, the amount of SO3 produced when 3.6 moles of O2 are consumed. The smaller amount of product formed is the actual amount. Solution: The balanced chemical equation is 2SO2(g) + O2(g)  2SO3(g)

3-13

 Amount (mol) of SO3 produced from the SO2 = 4.2 mol SO2  

 = 4.2 mol SO3  2  3

 Amount (mol) of SO3 produced from the O2 = 3.6 mol SO2  

  = 7.2 mol SO3  2  3

4.2 mol of SO3 (the smaller amount) will be produced. 3.17A

Plan: First, determine the formulas of the materials in the reaction and write a balanced chemical equation. Using the molar mass of each reactant, determine the moles of each reactant. Use molar ratios from the balanced equation to determine the moles of aluminum sulfide that may be produced from each reactant. The reactant that generates the smaller number of moles is limiting. Change the moles of aluminum sulfide from the limiting reactant to the grams of product using the molar mass of aluminum sulfide. To find the excess reactant amount, find the amount of excess reactant required to react with the limiting reagent and subtract that amount from the amount given in the problem. Solution: The balanced equation is 2Al(s) + 3S(s)  Al2S3(s) Determining the moles of product from each reactant:    Moles of Al2S3 from Al = (10.0 g Al)     Moles of Al2S3 from S = (15.0 g S)  

  

  = 0.18532 mol Al2S3 

  = 0.155958 mol Al2S3 

Sulfur produces less product so it is the limiting reactant.   2 3  = 23.4155 = 23.4 g Al2S3 Mass (g) of Al2S3 = (0.155958 mol Al2S3)    2 3  The mass of aluminum used in the reaction is now determined:       Mass (g) of Al = (15.0 g S)  = 8.4155 g Al used      Subtracting the mass of aluminum used from the initial aluminum gives the mass remaining. Excess Al = Initial mass of Al – mass of Al reacted = 10.0 g – 8.4155 g = 1.5845 = 1.6 g Al 3.17B

Plan: First, determine the formulas of the materials in the reaction and write a balanced chemical equation. Using the molar mass of each reactant, determine the moles of each reactant. Use molar ratios and the molar mass of carbon dioxide from the balanced equation to determine the mass of carbon dioxide that may be produced from each reactant. The reactant that generates the smaller mass of carbon dioxide is limiting. To find the excess reactant amount, find the amount of excess reactant required to react with the limiting reagent and subtract that amount from the amount given in the problem. Solution: The balanced equation is: 2C4H10(g) + 13O2(g)  8CO2(g) + 10H2O(g) Determining the mass of product formed from each reactant:  Mass (g) of CO2 from C4H10 = 4.65 g C4H10  

10 4

   

2 4

   

  = 14.0844 = 14.1 g CO2  2  2

3-14

 Mass (g) of CO2 from O2 = 10.0 g O2  

    2

  = 8.4635 = 8.46 g CO2  2 

Oxygen produces the smallest amount of product, so it is the limiting reagent, and 8.46 gof CO2 are produced. The mass of butane used in the reaction is now determined:    2  4 10    Mass (g) of C4H10=10.0 g O2       2

  = 2.7942 = 2.79 g C4H10 used  10  10

Subtracting the mass of butane used from the initial butane gives the mass remaining. Excess butane = Initial mass of butane – mass of butane reacted = 4.65 g – 2.79 g = 1.86 g butane 3.18A

Plan: Determine the formulas, and then balance the chemical equation. The mass of marble is converted to moles, the molar ratio (from the balanced equation) gives the moles of CO2, and finally the theoretical yield of CO2 is determined from the moles of CO2 and its molar mass. To calculate percent yield, divide the given actual yield of CO2 by the theoretical yield, and multiply by 100. Solution: The balanced equation: CaCO3(s) + 2HCl(aq)  CaCl2(aq) + H2O(l) + CO2(g) Find the theoretical yield of carbon dioxide.  1 mol CaCO3   1 mol CO2   44.01 g CO2   Mass (g) of CO2 = 10.0 g CaCO3     100.09 g CaCO3  1 mol CaCO3   1 mol CO2 





= 4.39704 g CO2 The percent yield:

 3.65 g CO2   actual yield   100% = 83.0104 = 83.0%   100% =   theoretical yield   4.39704 g CO2  3.18B

Plan: Determine the formulas, and then balance the chemical equation. The mass of potassium iodide is converted to moles, the molar ratio (from the balanced equation) gives the moles of lead(II) iodide, and finally the theoretical yield of lead(II) iodide is determined from the moles of lead(II) iodide and its molar mass. To calculate actual yield, multiply the theoretical yield of lead(II) iodide by the percent yield, and divide by 100. Solution: The balanced equation: 2KI(aq) + Pb(NO3)2(aq)  PbI2(s) + 2KNO3(aq) Find the theoretical yield of lead(II) iodide.  Mass (g) of PbI2 = 112 g KI  

   

  

  = 155.518 = 156 g PbI2 2  2

The actual yield:   Actual yield (g) of PbI2 =    100% 



100%

(156 g PbI2) = 140.40 = 140. g PbI2

3-15

END–OF–CHAPTER PROBLEMS 3.1

Plan: The atomic mass of an element expressed in amu is numerically the same as the mass of 1 mole of the element expressed in grams. We know the moles of each element and have to find the mass (in g). To convert moles of element to grams of element, multiply the number of moles by the molar mass of the element. Solution: Al 26.98 amu  26.98 g/mol Al  26.98 g Al  Mass Al (g) = 3 mol Al = 80.94 g Al  1 mol Al  Cl

35.45 amu  35.45 g/mol Cl

 35.45 g Cl  Mass Cl (g) = 2 mol Cl  = 70.90 g Cl  1 mol Cl  3.2

Plan: The molecular formula of sucrose tells us that 1 mole of sucrose contains 12 moles of carbon atoms. Multiply the moles of sucrose by 12 to obtain moles of carbon atoms; multiply the moles of carbon atoms by Avogadro’s number to convert from moles to atoms. Solution:  12 mol C    = 12 mol C a) Moles of C atoms = 1 mol C12 H 22 O11  1 mol C12 H 22 O11 





23  12 mol C   6.022 10 C atoms  25 b) C atoms = 2 mol C12 H 22 O11    = 1.445 × 10 C atoms 1 mol C H O   1 mol C  12 22 11 



3.3

3.4



Plan: Review the list of elements that exist as diatomic or polyatomic molecules. Solution: “1 mol of chlorine” could be interpreted as a mole of chlorine atoms or a mole of chlorine molecules, Cl2. Specify which to avoid confusion. The same problem is possible with other diatomic or polyatomic molecules, e.g., F2, Br2, I2, H2, O2, N2, S8, and P4. For these elements, as for chlorine, it is not clear if atoms or molecules are being discussed. The molecular mass is the sum of the atomic masses of the atoms or ions in a molecule. The molar mass is the mass of 1 mole of a chemical entity. Both will have the same numeric value for a given chemical substance but molecular mass will have the units of amu and molar mass will have the units of g/mol.

3.5

A mole of a particular substance represents a fixed number of chemical entities and has a fixed mass. Therefore the mole gives us an easy way to determine the number of particles (atoms, molecules, etc.) in a sample by weighing it. The mole maintains the same mass relationship between macroscopic samples as exist between individual chemical entities. It relates the number of chemical entities (atoms, molecules, ions, electrons) to the mass.

3.6

Plan: The mass of the compound is given. Divide the given mass by the molar mass of the compound to convert from mass of compound to number of moles of compound. The molecular formula of the compound tells us that 1 mole of compound contains 2 moles of phosphorus atoms. Use the ratio between P atoms and P4 molecules (4:1) to convert moles of phosphorus atoms to moles of phosphorus molecules. Finally, multiply moles of P4 molecules by Avogadro’s number to find the number of molecules.

3-16

Solution: Roadmap Mass (g) of Ca3(PO4)2 Divide by

(g/mol)

Amount (mol) of Ca3(PO4)2 Molar ratio between Ca3(PO4)2 and P atoms Amount (moles) of P atoms Molar ratio between P atoms and P4 molecules Amount (moles) of P4 molecules 23

Multiply by 6.022 × 10 formula units/mol Number of P4 molecules

3.7

Plan: The relative atomic masses of each element can be found by counting the number of atoms of each element and comparing the overall masses of the two samples. Solution: a) The element on the left (green) has the higher molar mass because only 5 green balls are necessary to counterbalance the mass of 6 yellow balls. Since the green ball is heavier, its atomic mass is larger, and therefore its molar mass is larger. b) The element on the left (red) has more atoms per gram. This figure requires more thought because the number of red and blue balls is unequal and their masses are unequal. If each pan contained 3 balls, then the red balls would be lighter. The presence of 6 red balls means that they are that much lighter. Because the red ball is lighter, more red atoms are required to make 1 g. c) The element on the left (orange) has fewer atoms per gram. The orange balls are heavier, and it takes fewer orange balls to make 1 g. d) Neither element has more atoms per mole. Both the left and right elements have the same number of 23 atoms per mole. The number of atoms per mole (6.022 × 10 ) is constant and so is the same for every element.

3.8

Plan: Locate each of the elements on the periodic table and record its atomic mass. The atomic mass of the element multiplied by the number of atoms present in the formula gives the mass of that element in one mole of the substance. The molar mass is the sum of the masses of the elements in the substance expressed in g/mol. Solution: a) = (1 × of Sr) + (2 × of O) + (2 × of H) = (1 × 87.62 g/mol Sr) + (2 × 16.00 g/mol O) + (2 × 1.008 g/mol H) = 121.64 g/mol of Sr(OH)2 b) = (2 × of N) + (3 × of O) = (2 × 14.01 g/mol N) + (3 × 16.00 g/mol O) = 76.02 g/mol of N2O3 c) = (1 × of Na) + (1 × of Cl) + (3 × of O) = (1 × 22.99 g/mol Na) + (1 × 35.45 g/mol Cl) + (3 × 16.00 g/mol O) = 106.44 g/mol of NaClO3

3-17

= (2 × of Cr) + (3 × of O) = (2 × 52.00 g/mol Cr) + (3 × 16.00 g/mol O) = 152.00 g/mol of Cr2O3

3.9

Plan: Locate each of the elements on the periodic table and record its atomic mass. The atomic mass of the element multiplied by the number of atoms present in the formula gives the mass of that element in one mole of the substance. The molar mass is the sum of the masses of the elements in the substance expressed in g/mol. Solution: a) = (3 × of N) + (12 × of H) + (1 × of P) + (4 × of O) = (3 × 14.01 g/mol N) + (12 × 1.008 g/mol H) + (1 × 30.97 g/mol P) + (4 × 16.00 g/mol O) = 149.10 g/mol of (NH4)3PO4 b) = (1 × of C) + (2 × of H) + (2 × of Cl) = (1 × 12.01 g/mol C) + (2 × 1.008 g/mol H) + (2 × 35.45 g/mol Cl) = 84.93 g/mol of CH2Cl2 c) = (1 × of Cu) + (1 × of S) + (9 × of O) + (10 × of H) = (1 × 63.55 g/mol Cu) + (1 × 32.06 g/mol S) + (9 × 16.00 g/mol O) + (10 × 1.008 g/mol H) = 249.69 g/mol of CuSO4 5H2O d) = (1 × of Br) + (3 × of F) = (1 × 79.90 g/mol Br) + (3 × 19.00 g/mol F) = 136.90 g/mol of BrF3

3.10

Plan: Locate each of the elements on the periodic table and record its atomic mass. The atomic mass of the element multiplied by the number of atoms present in the formula gives the mass of that element in one mole of the substance. The molar mass is the sum of the masses of the elements in the substance expressed in g/mol. Solution: a) = (1 × of Sn) + (1 × of O) = (1 × 118.7 g/mol Sn) + (1 × 16.00 g/mol O) = 134.7 g/mol of SnO b) = (1 × of Ba) + (2 × of F) = (1 × 137.3 g/mol Ba) + (2 × 19.00 g/mol F) = 175.3 g/mol of BaF2 c) = (2 × of Al) + (3 × of S) + (12 × of O) = (2 × 26.98 g/mol Al) + (3 × 32.06 g/mol S) + (12 × 16.00 g/mol O) = 342.14 g/mol of Al2(SO4)3 d) = (1 × of Mn) + (2 × of Cl) = (1 × 54.94 g/mol Mn) + (2 × 35.45 g/mol Cl) = 125.84 g/mol of MnCl2

3.11

Plan: Locate each of the elements on the periodic table and record its atomic mass. The atomic mass of the element multiplied by the number of atoms present in the formula gives the mass of that element in one mole of the substance. The molar mass is the sum of the masses of the elements in the substance expressed in g/mol. Solution: a) = (2 × of N) + (4 × of O) = (2 × 14.01 g/mol N) + (4 × 16.00 g/mol O) = 92.02 g/mol of N2O4 b) = (4 × of C) + (10 × of H) + (1 × of O) = (4 × 12.01 g/mol C) + (10 × 1.008 g/mol H) + (1 × 16.00 g/mol O) = 74.12 g/mol of C4H9OH c) = (1 × of Mg) + (1 × of S) + (11 × of O) + (14 × of H) = (1 × 24.31 g/mol Mg) + (1 × 32.06 g/mol S) + (11 × 16.00 g/mol O) + (14 × 1.008 g/mol H) = 246.48 g/mol of MgSO4 7H2O

3-18

3.12

= (1 × of Ca) + (4 × of C) + (6 × of H) + (4 × of O) = (1 × 40.08 g/mol Ca) + (4 × 12.01 g/mol C) + (6 × 1.008 g/mol H) + (4 × 16.00 g/mol O) = 158.17 g/mol of Ca(C2H3O2)2

Plan: Determine the molar mass of each substance; then perform the appropriate molar conversions. To find the mass in part (a), multiply the number of moles by the molar mass of Zn. In part (b), first multiply by Avogadro’s number to obtain the number of F2 molecules. The molecular formula tells us that there are 2 F atoms in each molecule of F2; use the 2:1 ratio to convert F2 molecules to F atoms. In part (c), convert mass of Ca to moles of Ca by dividing by the molar mass of Ca. Then multiply by Avogadro’s number to obtain the number of Ca atoms. Solution:

 a) (0.346 mol Zn)    b) (2.62 mol F2)  

 c) 28.5 g Ca  

3.13

3.14

  = 22.6 g Zn  

    

2 2

  



 24  = 3.16 × 10 F atoms 

 23  = 4.28 × 10 Ca atoms

Plan: Determine the molar mass of each substance; then perform the appropriate molar conversions. In part (a), 3 convert mg units to g units by dividing by 10 ; then convert mass of Mn to moles of Mn by dividing by the molar mass of Mn. In part (b) convert number of Cu atoms to moles of Cu by dividing by Avogadro’s number. In part (c) divide by Avogadro’s number to convert number of Li atoms to moles of Li; then multiply by the molar mass of Li to find the mass. Solution:      = 1.13 × 103 mol Mn a) (62.0 mg Mn)     

 22 b) (1.36 × 10 Cu atoms)  



 24 c) (8.05 × 10 Li atoms)  



 = 0.0226 mol Cu 

   

 = 92.8 g Li 

Plan: Determine the molar mass of each substance; then perform the appropriate molar conversions. To find the mass in part (a), multiply the number of moles by the molar mass of the substance. In part (b), first convert mass of compound to moles of compound by dividing by the molar mass of the compound. The molecular formula of the compound tells us that 1 mole of compound contains 6 moles of oxygen atoms; use the 1:6 ratio to convert moles of compound to moles of oxygen atoms. In part (c), convert mass of compound to moles of compound by dividing by the molar mass of the compound. Since 1 mole of compound contains 6 moles of oxygen atoms, multiply the moles of compound by 6 to obtain moles of oxygen atoms; then multiply by Avogadro’s number to obtain the number of oxygen atoms. Solution: a) of KMnO4 = (1 × of K) + (1 × of Mn) + (4 × of O) = (1 × 39.10 g/mol K) + (1 × 54.94 g/mol Mn) + (4 × 16.00 g/mol O) = 158.04 g/mol of KMnO4 158.04 g KMnO 4    = 107.467 = 1.1 × 102 g KMnO Mass of KMnO4 = 0.68 mol KMnO 4  4  1 mol KMnO 4 





of Ba(NO3)2 = (1 × of Ba) + (2 × of N) + (6 × of O) = (1 × 137.3 g/mol Ba) + (2 × 14.01 g/mol N) + (6 × 16.00 g/mol O) = 261.3 g/mol Ba(NO3)2

3-19

 1 mol Ba(NO3 )2    = 0.031305 mol Ba(NO ) Moles of Ba(NO3)2 = 8.18 g Ba(NO3 )2  3 2  261.3 g Ba(NO3 )2 





 6 mol O atoms    = 0.18783 = 0.188 mol O atoms Moles of O atoms = 0.031305 mol Ba(NO3 )2  1 mol Ba(NO3 )2  of CaSO4 2H2O = (1 × of Ca) + (1 × of S) + (6 × of O) + (4 × of H) = (1 × 40.08 g/mol Ca) + (1 × 32.06 g/mol S) + (6 × 16.00 g/mol O) + (4 × 1.008 g/mol H) = 172.17 g/mol (Note that the waters of hydration are included in the molar mass.)





 1 mol CaSO  2H O   4 2 –5 Moles of CaSO4 2H2O = 7.3103 g CaSO 4  2H 2 O   = 4.239995 × 10 mol 172.17 g CaSO4  2H 2 O 





 6 mol O atoms    Moles of O atoms = 4.239995105 mol CaSO 4  2H 2 O  1 mol CaSO4  2H 2 O 





–5

= 2.543997 × 10 mol O atoms

 6.022 10 23 O atoms    Number of O atoms = 2.54399710 mol O atoms   1 mol O atoms  20 20 = 1.5320 × 10 = 1.5 × 10 O atoms



3.15



4

Plan: Determine the molar mass of each substance, then perform the appropriate molar conversions. To find the mass in part (a), divide the number of molecules by Avogadro’s number to find moles of compound and then multiply the mole amount by the molar mass in grams; convert from mass in g to mass in kg. In part (b), first convert mass of compound to moles of compound by dividing by the molar mass of the compound. The molecular formula of the compound tells us that 1 mole of compound contains 2 moles of chlorine atoms; use the 1:2 ratio to convert moles of compound to moles of chlorine atoms. In part (c), convert mass of compound to moles of compound by dividing by the – molar mass of the compound. Since 1 mole of compound contains 2 moles of H ions, multiply the moles of compound by – – 2 to obtain moles of H ions; then multiply by Avogadro’s number to obtain the number of H ions. Solution: a) of NO2 = (1 × of N) + (2 × of O) = (1 × 14.01 g/mol N) + (2 × 16.00 g/mol O) = 46.01g/mol of NO2   1 mol NO2   = 7.63866 × 10–3 mol NO Moles of NO2 = 4.610 21 molecules NO2  2 23  6.022 10 molecules NO2 





 46.01 g NO     1 kg  2 3 –4 –4 Mass (kg) of NO2 = 7.6386610 mol NO2   3  = 3.51455 × 10 = 3.5 × 10 kg NO2  1 mol NO2  10 g 





of C2H4Cl2 = (2 × of C) + (4 × of H) + (2 × of Cl) = (2 × 12.01g/mol C) + (4 × 1.008 g/mol H) + (2 × 35.45 g/mol Cl) = 98.95 g/mol of C2H4Cl2  1 mol C 2 H 4 Cl 2    = 6.21526 × 10–4 mol C H Cl Moles of C2H4Cl2 = 0.0615 g C 2 H 4 Cl2  2 4 2  98.95 g C 2 H 4 Cl2 





 2 mol Cl atoms   4  = 1.2431 × 10–3 Moles of Cl atoms = 6.2152610 mol C2 H 4 Cl2   1 mol C2 H 4 Cl 2 





–3

= 1.24 × 10 mol Cl atoms Copyright McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

3-20

c) of SrH2 = (1 × SrH2

of Sr) + (2 ×

of H) = (1 × 87.62 g/mol Sr) + (2 × 1.008 g/mol H) = 89.64 g/mol of

 1 mol SrH 2    = 0.0649264 mol SrH Moles of SrH2 = 5.82 g SrH 2  2  89.64 g SrH 2 





 2 mol H   –  = 0.1298528 mol H– ions Moles of H ions = 0.0649264 mol SrH 2  1 mol SrH 2 





 6.022 10 23 H ions  –  = 7.81974 × 1022 = 7.82 × 1022 H– ions Number of H ions = 0.1298528 mol H ions    1 mol H 



3.16



Plan: Determine the molar mass of each substance; then perform the appropriate molar conversions. To find the mass in part (a), multiply the number of moles by the molar mass of the substance. In part (b), first convert the mass of compound in kg to mass in g and divide by the molar mass of the compound to find moles of compound. In part (c), convert mass of compound in mg to mass in g and divide by the molar mass of the compound to find moles of compound. Since 1 mole of compound contains 2 moles of nitrogen atoms, multiply the moles of compound by 2 to obtain moles of nitrogen atoms; then multiply by Avogadro’s number to obtain the number of nitrogen atoms. Solution: a) of MnSO4 = (1 × of Mn) + (1 × of S) + (4 × of O) = (1 × 54.94 g/mol Mn) + (1 × 32.06 g/mol S) + (4 × 16.00 g/mol O) = 151.00 g/mol of MnSO4 151.00 g MnSO  2  4 Mass (g) of MnSO4 = 6.44 10 mol MnSO 4   = 9.7244 = 9.72 g MnSO4  1 mol MnSO 4  b)

of Fe(ClO4)3 = (1 × of Fe) + (3 × of Cl) + (12 × of O) = (1 × 55.85 g/mol Fe) + (3 × 35.45 g/mol S) + (12 × 16.00 g/mol O) = 354.20 g/mol of Fe(ClO4)3 103 g   = 1.58 × 104 kg Fe(ClO4)3 Mass (g) of Fe(ClO4)3 = 15.8 kg Fe(ClO 4 )3   1 kg   1 mol Fe(ClO )  4  4 3  Moles of Fe(ClO4)3 = 1.5810 g Fe(ClO 4 )3   = 44.6076 = 44.6 mol Fe(ClO4)3  354.20 g Fe(ClO 4 )3 

of NH4NO2 = (2 × of N) + (4 × of H) + (2 × of O) = (2 × 14.01 g/mol N) + (4 × 1.008 g/mol H) + (2 × 16.00 g/mol O) = 64.05 g/mol NH4NO2 103 g   = 0.0926 g NH4NO2 Mass (g) of NH4NO2 = 92.6 mg NH 4 NO2   1 mg 





 1 mol NH 4 NO2   –3 Moles of NH4NO2 = 0.0926 g NH 4 NO2   = 1.44575 × 10 mol NH4NO2  64.05 g NH 4 NO2 





 2 mol N atoms    = 2.8915 × 10–3 mol N atoms Moles of N atoms = 1.44575103 mol NH 4 NO2  1 mol NH 4 NO2  Number of N atoms = 2.891510

3

 6.022 1023 N atoms    mol N atoms  1 mol N atoms  21

= 1.74126 × 10 = 1.74 × 10 N atoms

3-21

3.17

Plan: Determine the molar mass of each substance; then perform the appropriate molar conversions. In part (a), divide the mass by the molar mass of the compound to find moles of compound. Since 1 mole of compound 2+ – contains 3 moles of ions (1 mole of Sr and 2 moles of F ), multiply the moles of compound by 3 to obtain moles of ions and then multiply by Avogadro’s number to obtain the number of ions. In part (b), multiply the number of moles by the molar mass of the substance to find the mass in g and then convert to kg. In part (c), divide the number of formula units by Avogadro’s number to find moles; multiply the number of moles by the molar mass to obtain the mass in g and then convert to mg. Solution: a) of SrF2 = (1 × of Sr) + (2 × of F) = (1 × 87.62 g/mol Sr) + (2 × 19.00 g/mol F) = 125.62 g/mol of SrF2  1 mol SrF2    = 0.303296 mol SrF Moles of SrF2 = 38.1 g SrF2  2 125.62 g SrF2 





 3 mol ions    = 0.909888 mol ions Moles of ions = 0.303296 mol SrF2  1 mol SrF2 





 6.022 1023 ions   23 23 Number of ions = 0.909888 mol ions   = 5.47935 × 10 = 5.48 × 10 ions  1 mol ions  b) of CuCl2 2H2O = (1 × of Cu) + (2 × of Cl) + (4 × of H) + (2 × of O) = (1 × 63.55 g/mol Cu) + (2 × 35.45 g/mol Cl) + (4 × 1.008 g/mol H) + (2 × 16.00 g/mol O) = 170.48 g/mol of CuCl2 2H2O (Note that the waters of hydration are included in the molar mass.) 170.48 g CuCl2  2H 2 O    = 610.32 g CuCl 2H O Mass (g) of CuCl2 2H2O = 3.58 mol CuCl2  2H 2 O  2 2  1 mol CuCl2  2H 2 O 









  Mass (kg) of CuCl2 2H2O = (610.32 g CuCl2 2H2O)  3  = 0.61032 = 0.610 kg CuCl2 2H2O   c) of Bi(NO3)3 5H2O = (1 × of Bi) + (3 × of N) + (10 × of H) + (14 × of O) = (1 × 209.0 g/mol Bi) + (3 × 14.01 g/mol N) + (10 × 1.008 g/mol H) + (14 × 16.00 g/mol H) = 485.11 g/mol of Bi(NO3)3 5H2O (Note that the waters of hydration are included in the molar mass.)  1 mol  Moles of Bi(NO3)3 5H2O = 2.8810 22 FU   = 0.047825 mol Bi(NO3)3 5H2O  6.022 10 23 FU 





 485.1 g Bi(NO3 )3  5H 2 O   = 23.1999 g Mass (g) of Bi(NO3)3 5H2O = (0.047825 mol Bi(NO3)3 5H2O)   1 mol Bi(NO3 )3  5H 2 O   1mg  Mass (mg) of Bi(NO3)3 5H2O = (23.1999 g Bi(NO3)3 5H2O)  3  10 g  4

= 23199.9 = 2.32 × 10 mg Bi(NO3)3 5H2O

3-22

3.18

a) Carbonate is a polyatomic anion with the formula, CO3 . Copper(I) indicates Cu . The correct formula for this ionic compound is Cu2CO3. of Cu2CO3 = (2 × of Cu) + (1 × of C) + (3 × of O) = (2 × 63.55 g/mol Cu) + (1 × 12.01 g/mol C) + (3 × 16.00 g/mol O) = 187.11 g/mol of Cu2CO3

187.11 g Cu2 CO3    = 1562.4 = 1.56 × 103 g Cu CO Mass (g) of Cu2CO3 = 8.35 mol Cu2 CO3  2 3  1 mol Cu2 CO3 





b) Dinitrogen pentaoxide has the formula N2O5. Di- indicates 2 N atoms and penta- indicates 5 O atoms. of N2O5 = (2 × of N) + (5 × of O) = (2 × 14.01 g/mol N) + (5 × 16.00 g/mol O) = 108.02 g/mol of N2O5

  1 mol N 2 O5   = 6.7087 × 10–4 mol N O Moles of N2O5 = 4.04 1020 N 2 O5 molecules  2 5  6.022 10 23 N 2 O5 molecules 





108.02 g N O   2 5 Mass (g) of N2O5 = 6.7087104 mol N 2 O5   = 0.072467 = 0.0725 g N2O5  1 mol N2 O5 





c) The correct formula for this ionic compound is NaClO4; Na has a charge of +1 (Group 1 ion) and the –

perchlorate ion is ClO4 . of NaClO4 = (1 × of Na) + (1 × of Cl) + (4 × of O) = (1 × 22.99 g/mol Na) + (1 × 35.45 g/mol Cl) + (4 × 16.00 g/mol O) = 122.44 g/mol of NaClO4

 1 mol NaClO 4    = 0.644397 = 0.644 mol NaClO Moles of NaClO4 = 78.9 g NaClO 4  4 122.44 g NaClO4 





FU = formula units

 6.022 10 23 FU NaClO   4 FU of NaClO4 = 0.644397 mol NaClO4   1 mol NaClO4  





= 3.88056 × 10 = 3.88 × 10 FU NaClO4

 1 Na  ion   +  = 3.88 × 1023 Na+ ions d) Number of Na ions = 3.88056 1023 FU NaClO4  1 FU NaClO4   1 ClO  ion   4 –  = 3.88 × 1023 ClO – ions Number of ClO4 ions = 3.88056 1023 FU NaClO4  4 1 FU NaClO4   1 Cl atom    = 3.88 × 1023 Cl atoms Number of Cl atoms = 3.88056 1023 FU NaClO4  1 FU NaClO4   4 O atoms    = 1.55 × 1024 O atoms Number of O atoms = 3.88056 1023 FU NaClO4  1 FU NaClO4  Copyright McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

3-23

3.19

Plan: The formula of each compound must be determined from its name. The molar mass for each formula comes from the formula and atomic masses from the periodic table. Determine the molar mass of each substance, then perform the appropriate molar conversions. In part (a), multiply the moles by the molar mass of the compound to find the mass of the sample. In part (b), divide the number of molecules by Avogadro’s number to find moles; multiply the number of moles by the molar mass to obtain the mass. In part (c), divide the mass by the molar mass to find moles of compound and multiply moles by Avogadro’s number to find the number of formula units. In part (d), use the fact that each formula unit contains 2 Li ions, 1 sulfate ion, 1 S atom, and 4 O atoms. Solution: 2– 3+ a) Sulfate is a polyatomic anion with the formula, SO4 . Chromium(III) indicates Cr . Decahydrate indicates 10 water molecules (“waters of hydration”). The correct formula for this ionic compound is Cr2(SO4)3 10H2O. of Cr2(SO4)3 10H2O = (2 × of Cr) + (3 × of S) + (22 × of O) + (20 × of H) = (2 × 52.00 g/mol Cr) + (3 × 32.06 g/mol S) + (22 × 16.00 g/mol O) + (20 × 1.008 g/mol H) =572.34 g/mol of Cr2(SO4)3 10H2O  572.34 g   Mass (g) of Cr2(SO4)3 10H2O = 8.42 mol Cr2 (SO 4 )3  10H 2 O   mol  3

= 4819.103 = 4.82 × 10 g Cr2(SO4)3 10H2O b) Dichlorine heptaoxide has the formula Cl2O7. Di- indicates 2 Cl atoms and hepta- indicates 7 O atoms. of Cl2O7 = (2 × of Cl) + (7 × of O) = (2 × 35.45 g/mol Cl) + (7 × 16.00 g/mol O) = 182.9 g/mol of Cl2O7

  1 mol   = 3.038858 mol Cl O Moles of Cl2O7 = 1.8310 24 molecules Cl2 O 7  2 7 23  6.022 10 molecules 





182.9 g Cl2 O7    = 555.807 = 5.56 × 102 g Cl O Mass (g) of Cl2O7 = 3.038858mol Cl2 O7  2 7   1 mol 





c) The correct formula for this ionic compound is Li2SO4; Li has a charge of +1 (Group 1 ion) and the 2–

sulfate ion is SO4 . of Li2SO4 = (2 × of Li) + (1 × of S) + (4 × of O) = (2 × 6.941 g/mol Li) + (1 × 32.06 g/mol S) + (4 × 16.00 g/mol O) = 109.94 g/mol of Li2SO4  1 mol Li 2 SO 4    = 0.056394 = 0.056 mol Li SO Moles of Li2SO4 = 6.2 g Li 2 SO4  2 4 109.94 g Li2 SO4 





 6.022 1023 FU   22 22 FU of Li2SO4 = 0.056394 mol Li2 SO4   = 3.3960 × 10 = 3.4 × 10 FU Li2SO4  1 mol Li 2 SO4 





 2 Li ions   +  = 6.7920 × 1022 = 6.8 × 1022 Li+ ions d) Number of Li ions = 3.39601022 FU Li2 SO4  1 FU Li 2 SO4   1 SO 2 ion   4 2–  = 3.3960 × 1022 = 3.4 × 1022 SO 2– ions Number of SO4 ions = 3.39601022 FU Li2 SO4  4 1 FU Li 2 SO4 

3-24

 1 S atom    = 3.3960 × 1022 = 3.4 × 1022 S atoms Number of S atoms = 3.39601022 FU Li2 SO4  1 FU Li 2 SO4   4 O atoms    = 1.3584 × 1023 = 1.4 × 1023 O atoms Number of O atoms = 3.39601022 FU Li2 SO4  1 FU Li 2 SO4  3.20

Plan: Determine the formula and the molar mass of each compound. The formula gives the relative number of moles of each element present. Multiply the number of moles of each element by its molar mass to find the total total mass of element 100 . mass of element in 1 mole of compound. Mass percent = molar mass of compound Solution: + – a) Ammonium bicarbonate is an ionic compound consisting of ammonium ions, NH 4 and bicarbonate ions, HCO3 . The formula of the compound is NH4HCO3. of NH4HCO3 = (1 × of N) + (5 × of H) + (1 × of C) + (3 × of O) = (1 × 14.01 g/mol N) + (5 × 1.008 g/mol H) + (1 × 12.01 g/mol C) + (3 × 16.00 g/mol O) = 79.06 g/mol of NH4HCO3 There are 5 moles of H in 1 mole of NH4HCO3. 1.008 g H  Mass (g) of H = 5 mol H = 5.040 g H  1 mol H  total mass H 5.040 g H 100 = 100 = 6.374905 = 6.375% H Mass percent = molar mass of compound 79.06 g NH 4 HCO3 + b) Sodium dihydrogen phosphate heptahydrate is a salt that consists of sodium ions, Na , dihydrogen phosphate – ions, H2PO4 , and seven waters of hydration. The formula is NaH2PO4 7H2O. Note that the waters of hydration are included in the molar mass. of NaH2PO4 7H2O = (1 × of Na) + (16 × of H) + (1 × of P) + (11 × of O) = (1 × 22.99 g/mol Na) + (16 × 1.008 g/mol H) + (1 × 30.97 g/mol P) + (11 × 16.00 g/mol O) = 246.09 g/mol NaH2PO4 7H2O There are 11 moles of O in 1 mole of NaH2PO4 7H2O. 16.00 g O  Mass (g) of O = 11 mol O = 176.00 g O  1 mol O  total mass O 176.00 g O 100 = 100 Mass percent = molar mass of compound 246.09 g NaH2 PO 4  7H2 O = 71.51855 = 71.52% O

3.21

–

a) Strontium periodate is an ionic compound consisting of strontium ions, Sr and periodate ions, IO4 . The formula of the compound is Sr(IO4)2. of Sr(IO4)2 = (1 × of Sr) + (2 × of I) + (8 × of O) = (1 × 87.62 g/mol Sr) + (2 × 126.9 g/mol I) + (8 × 16.00 g/mol O) = 469.4 g/mol of Sr(IO4)2 There are 2 moles of I in 1 mole of Sr(IO4)2. Copyright McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

3-25

126.9 g I  Mass (g) of I = 2 mol I = 253.8 g I  1 mol I  total mass I 253.8 g I 100 = 100 = 54.0690 = 54.07% I Mass percent = molar mass of compound 469.4 g Sr(IO4 )2 +

–

b) Potassium permanganate is an ionic compound consisting of potassium ions, K and permanganate ions, MnO4 . The formula of the compound is KMnO4. of KMnO4 = (1 × of K) + (1 × of Mn) + (4 × of O) = (1 × 39.10 g/mol K) + (1 × 54.94 g/mol Mn) + (4 × 16.00 g/mol O) = 158.04 g/mol of KMnO4 There is 1 mole of Mn in 1 mole of KMnO4.

 54.94 g Mn  Mass (g) of Mn = 1 mol Mn  = 54.94 g Mn  1 mol Mn  Mass percent =

3.22

total mass Mn 54.94 g Mn 100 = 100 = 34.76335 = 34.76% Mn molar mass of compound 158.04 g KMnO4

–

a) Cesium acetate is an ionic compound consisting of Cs cations and C2H3O2 anions. (Note that the formula for –

–

acetate ions can be written as either C2H3O2 or CH3COO .) The formula of the compound is CsC2H3O2. of CsC2H3O2 = (1 × of Cs) + (2 × of C) + (3 × of H) + (2 × of O) = (1 × 132.9 g/mol Cs) + (2 × 12.01 g/mol C) + (3 × 1.008 g/mol H) + (2 × 16.00 g/mol O) = 191.9 g/mol of CsC2H3O2 There are 2 moles of C in 1 mole of CsC2H3O2. 12.01 g C  Mass (g) of C = 2 mol C = 24.02 g C  1 mol C  total mass C 24.02 g C = Mass fraction = = 0.125169 = 0.1252 mass fraction C molar mass of compound 191.9 g CsC 2 H 3 O 2 2+

2–

b) Uranyl sulfate trihydrate is a salt that consists of uranyl ions, UO2 , sulfate ions, SO4 , and three waters of hydration. The formula is UO2SO4 3H2O. Note that the waters of hydration are included in the molar mass. of UO2SO4 3H2O = (1 × of U) + (9 × of O) + (1 × of S) + (6 × of H) = (1 × 238.0 g/mol U) + (9 × 16.00 g/mol O) + (1 × 32.06 g/mol S) + (6 × 1.008 g/mol H) = 420.1 g/mol of UO2SO4 3H2O There are 9 moles of O in 1 mole of UO2SO4 3H2O. 16.00 g O  Mass (g) of O = 9 mol O = 144.0 g O  1 mol O  total mass O 144.0 g O = Mass fraction = = 0.3427755 = 0.3428 mass fraction O molar mass of compound 420.1 g UO2 SO4  3H 2 O

3-26

3.23

–

a) Calcium chlorate is an ionic compound consisting of Ca cations and ClO3 anions. The formula of the compound is Ca(ClO3)2. of Ca(ClO3)2 = (1 × of Ca) + (2 × of Cl) + (6 × of O) = (1 × 40.08 g/mol Ca) + (2 × 35.45 g/mol Cl) + (6 × 16.00 g/mol O) = 206.98 g/mol of Ca(ClO3)2 There are 2 moles of Cl in 1 mole of Ca(ClO3)2.

 35.45 g Cl  Mass (g) of Cl = 2 mol Cl  = 70.92 g Cl  1 mol Cl  Mass fraction =

total mass Cl 70.90 g Cl = = 0.342545 = 0.3425 mass fraction Cl molar mass of compound 206.98 g Ca(ClO3 )2

b) Dinitrogen trioxide has the formula N2O3. Di- indicates 2 N atoms and triindicates 3 O atoms. of N2O3 = (2 × of N) + (3 × of O) = (2 × 14.01 g/mol N) + (3 × 16.00 g/mol O) = 76.02 g/mol of N2O3 There are 2 moles of N in 1 mole of N2O3. 14.01 g N  Mass (g) of N = 2 mol N = 28.02 g N  1 mol N  total mass N 28.02 g N = Mass fraction = = 0.368587 = 0.3686 mass fraction N molar mass of compound 76.02 g N 2 O3 3.24

Plan: Divide the mass given by the molar mass of O2 to find moles. Since 1 mole of oxygen molecules contains 2 moles of oxygen atoms, multiply the moles by 2 to obtain moles of atoms and then multiply by Avogadro’s number to obtain the number of atoms. Solution:  1 mol O2   Moles of O2 = 38.0 g O2   = 1.1875 mol O2  32.00 g O2 





 2 mol O atoms    = 2.375 mol O atoms Moles of O atoms = 1.1875 mol O2   1 mol O2 





 6.022 1023 O atoms    = 1.430225 × 1024 = 1.43 × 1024 O atoms Number of O atoms = 2.375 mol O atoms   1 mol O atoms 



3.25



Plan: Determine the formula of cisplatin from the figure, and then calculate the molar mass from the formula. Divide the mass given by the molar mass to find moles of cisplatin. Since 1 mole of cisplatin contains 6 moles of hydrogen atoms, multiply the moles given by 6 to obtain moles of hydrogen and then multiply by Avogadro’s number to obtain the number of atoms. Solution: The formula for cisplatin is Pt(Cl)2(NH3) 2. of Pt(Cl)2(NH3) 2 = (1 × of Pt) + (2 × of Cl) + (2 × of N) + (6 × of H) = (1 × 195.1 g/mol Pt) + (2 × 35.45 g/mol Cl) + (2 × 14.01 g/mol N) + (6 × 1.008 g/mol H) = 300.1 g/mol of Pt(Cl)2(NH3)2

3-27

 1 mol cisplatin    = 0.9506831 = 0.9507 mol cisplatin a) Moles of cisplatin = 285.3 g cisplatin   300.1 g cisplatin 





 6 mol H    = 5.88 mol H atoms b) Moles of H atoms = 0.98 mol cisplatin  1 mol cisplatin 





 6.0221023 H atoms    = 3.540936 × 1024 = 3.5 × 1024 H atoms Number of H atoms = 5.88 mol H atoms   1 mol H atoms 



3.26



Plan: Determine the formula of allyl sulfide from the figure, and then calculate the molar mass from the formula. In part (a), multiply the given amount in moles by the molar mass to find the mass of the sample. In part (b), divide the given mass by the molar mass to find moles of compound. Since 1 mole of compound contains 6 moles of carbon atoms, multiply the moles of compound by 6 to obtain moles of carbon and then multiply by Avogadro’s number to obtain the number of atoms. Solution: The formula, from the figure, is (C3H5)2S. of (C3H5)2S = (6 × of C) + (10 × of H) + (1 × of S) = (6 × 12.01 g/mol C) + (10 × 1.008 g/mol H) + (1 × 32.06 g/mol S) = 114.20 g/mol of (C3H5)2S 114.20 g allyl sulfide    = 300.3460 = 300 g allyl sulfide a) Mass (g) of allyl sulfide = 2.63 mol allyl sulfide   1 mol allyl sulfide 





 1 mol (C3 H 5 )2 S    = 0.312609 mol allyl sulfide b) Moles of allyl sulfide = 35.7 g (C3 H 5 )2 S  114.20 g (C3 H 5 )2 S 





  6 mol C   = 1.8757 mol C atoms Moles of C atoms = 0.312609 mol (C3 H 5 )2 S  1 mol (C3 H 5 )2 S 





 6.022 1023 C atoms    = 1.129546 × 1024 = 1.13 × 1024 C atoms Number of C atoms = 1.8757 mol C atoms   1 mol C atoms 



3.27



Plan: Determine the molar mass of rust. Convert mass in kg to mass in g and divide by the molar mass to find the moles of rust. Since each mole of rust contains 1 mole of Fe2O3, multiply the moles of rust by 1 to obtain moles of Fe2O3. Multiply the moles of Fe2O3 by 2 to obtain moles of Fe (1:2 Fe2O3:Fe mole ratio) and multiply by the molar mass of Fe to convert to mass. Solution: a) of Fe2O3 4H2O = (2 × of Fe) + (7 × of O) + (8 × of H) = (2 × 55.85 g/mol Fe) + (7 × 16.00 g/mol O) + (8 × 1.008 g/mol H) = 231.76 g/mol 103 g   = 4.52 × 104 g Mass (g) of rust = 45.2 kg rust   1 kg 





 1 mol rust    = 195.029 = 195 mol rust Moles of rust = 4.52 10 4 g rust   231.76 g rust 





b) The formula shows that there is 1 mole of Fe2O3 for every mole of rust, so there are also 195 mol of Fe2O3.

3-28

 2 mol Fe    = 390.058 mol Fe c) Moles of iron = 195.029 mol Fe2 O3  1 mol Fe 2 O3 





 55.85 g Fe    = 21784.74 = 2.18 × 104 g Fe Mass (g) of iron = 390.058 mol Fe   1 mol Fe 



3.28



Plan: Determine the molar mass of propane. Divide the given mass by the molar mass to find the moles. Since each mole of propane contains 3 moles of carbon, multiply the moles of propane by 3 to obtain moles of C atoms. Multiply the moles of C by its molar mass to obtain mass of carbon. Solution: a) The formula of propane is C3H8. of C3H8 = (3 × of C) + (8 × of H) = (3 × 12.01 g/mol C) + (8 × 1.008 g/mol H) = 44.09 g/mol  1 mol C3 H8   Moles of C3H8 = 85.5 g C3 H8   = 1.939215 = 1.94 mol C3H8  44.09 g C3 H8 





 3 mol C    = 5.817645 mol C b) Moles of C = 1.939215 mol C3 H8  1 mol C3 H8 





12.01 g C    = 69.86992 = 69.9 g C Mass (g) of C = 5.817645 mol C   1 mol C 



3.29



Plan: Determine the formula and the molar mass of each compound. The formula gives the relative number of moles of nitrogen present. Multiply the number of moles of nitrogen by its molar mass to find the total mass of nitrogen in 1 mole of compound. Divide the total mass of nitrogen by the molar mass of compound and multiply mol Nmolar mass N 100 . Then rank the values in by 100 to determine mass percent. Mass percent = molar mass of compound order of decreasing mass percent N. Solution: Name

Formula

Molar Mass (g/mol)

Potassium nitrate

KNO3

101.11

Ammonium nitrate

NH4NO3

80.05

Ammonium sulfate

(NH4)2SO4

132.14

Urea

CO(NH2)2

60.06

Mass % N in potassium nitrate =

1 mol N 14.01 g/mol N100

Mass % N in ammonium nitrate = Mass % N in ammonium sulfate =

Mass % N in urea =

101.11 g/mol

= 13.856196 = 13.86% N

2 mol N14.01 g/mol N 100 80.05 g/mol

= 35.003123 = 35.00% N

2 mol N  14.01 g/mol N  100 = 21.20478 = 21.20% N 132.14 g/mol

2 mol N14.01 g/mol N 100 60.06 g/mol

= 46.6533 = 46.65% N

Rank is CO(NH2)2> NH4NO3> (NH4)2SO4> KNO3 Copyright McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

3-29

3.30

Plan: The volume must be converted from cubic feet to cubic centimeters. The volume and the density will give the mass of galena which is then divided by molar mass to obtain moles. Part (b) requires a conversion from cubic decimeters to cubic centimeters. The density allows a change from volume in cubic centimeters to mass which is then divided by the molar mass to obtain moles; the amount in moles is multiplied by Avogadro’s number to obtain formula units of PbS which is also the number of Pb atoms due to the 1:1 PbS:Pb mole ratio. Solution: 2+ 2– Lead(II) sulfide is composed of Pb and S ions and has a formula of PbS. of PbS = (1 × of Pb) + (1 × of S) = (1 × 207.2 g/mol Pb) + (1 × 32.06 g/mol S) = 239.3 g/mol 3 3  12 in    3  2.54 cm   = 28316.85 cm3 a) Volume (cm ) = 1.00 ft 3 PbS  3  1 in 3   1 ft  

 7.46 g PbS  Mass (g) of PbS = 28316.85 cm 3 PbS  = 211243.7 g PbS  1 cm 3   1 mol PbS   = 882.7568 = 883 mol PbS Moles of PbS = 211243.7 g PbS  239.3 g PbS  0.1 m3  1 cm3  3 3 3   b) Volume (cm ) = 1.00 dm 3 PbS  2 3  = 1.00 × 10 cm   1 dm3  10 m     

 7.46 g PbS  Mass (g) of PbS = 1.00 103 cm 3 PbS  = 7460 g PbS  1 cm 3   1 mol PbS   = 31.17426 mol PbS Moles of PbS = 7460 g PbS  239.3 g PbS 

 1 mol Pb  Moles of Pb = 31.17426 mol PbS  = 31.17426 mol Pb 1 mol PbS   6.022 10 23 Pb atoms  25 25 Number of lead atoms = 31.17426 mol Pb   = 1.87731 × 10 = 1.88 × 10 Pb atoms  1 mol Pb  3.31

Plan: If the molecular formula for hemoglobin (Hb) were known, the number of Fe ions in a molecule of hemoglobin could be calculated. It is possible to calculate the mass of iron from the percentage of iron and the molar mass of the compound. Assuming you have 1 mole of hemoglobin, take 0.33% of its molar mass as the mass of Fe in that 1 mole. Divide the mass of Fe by its molar mass to find moles of Fe in 1 mole of hemoglobin which is also the number of ions in 1 molecule. Solution:

 0.33% Fe  6.810 4 g    Mass of Fe =   = 224.4 g Fe   100% Hb  mol   1 mol Fe    = 4.0179 = 4.0 mol Fe2+/mol Hb Moles of Fe = 224.4 g Fe  55.85 g Fe  2+ Thus, there are 4 Fe /molecule Hb. 3.32

Plan: Review the definitions of empirical and molecular formulas. Solution: An empirical formula describes the type and simplest ratio of the atoms of each element present in a compound, whereas a molecular formula describes the type and actual number of atoms of each element in a molecule of the compound. The empirical formula and the molecular formula can be the same. For example, the compound

3-30

formaldehyde has the molecular formula, CH2O. The carbon, hydrogen, and oxygen atoms are present in the ratio of 1:2:1. The ratio of elements cannot be further reduced, so formaldehyde’s empirical formula and molecular formula are the same. Acetic acid has the molecular formula, C2H4O2. The carbon, hydrogen, and oxygen atoms are present in the ratio of 2:4:2, which can be reduced to 1:2:1. Therefore, acetic acid’s empirical formula is CH2O, which is different from its molecular formula. Note that the empirical formula does not uniquely identify a compound, because acetic acid and formaldehyde share the same empirical formula but are different compounds. 3.33

1. Compositional data may be given as the mass of each element present in a sample of compound. 2. Compositional data may be provided as mass percents of each element in the compound. 3. Compositional data obtained through combustion analysis provides the mass of C and H in a compound.

3.34

Plan: Remember that the molecular formula tells the actual number of moles of each element in one mole of compound. Solution: a) No, this information does not allow you to obtain the molecular formula. You can obtain the empirical formula from the number of moles of each type of atom in a compound, but not the molecular formula. b) Yes, you can obtain the molecular formula from the mass percentages and the total number of atoms. Plan: 1) Assume a 100.0 g sample and convert masses (from the mass % of each element) to moles using molar mass. 2) Identify the element with the lowest number of moles and use this number to divide into the number of moles for each element. You now have at least one elemental mole ratio (the one with the smallest number of moles) equal to 1.00 and the remaining mole ratios that are larger than one. 3) Examine the numbers to determine if they are whole numbers. If not, multiply each number by a whole-number factor to get whole numbers for each element. You will have to use some judgment to decide when to round. Write the empirical formula using these whole numbers. 4) Check the total number of atoms in the empirical formula. If it equals the total number of atoms given then the empirical formula is also the molecular formula. If not, then divide the total number of atoms given by the total number of atoms in the empirical formula. This should give a whole number. Multiply the number of atoms of each element in the empirical formula by this whole number to get the molecular formula. If you do not get a whole number when you divide, return to step 3 and revise how you multiplied and rounded to get whole numbers for each element. Roadmap: Mass (g) of each element (express mass percent directly as grams) Divide by

(g/mol)

Amount (mol) of each element Use numbers of moles as subscripts Preliminary empirical formula Change to integer subscripts Empirical formula Divide total number of atoms in molecule by the number of atoms in the empirical formula and multiply the empirical formula by that factor Molecular formula

3-31

c) Yes, you can determine the molecular formula from the mass percent and the number of atoms of one element in a compound. Plan: 1) Follow steps 1–3 in part (b). 2) Compare the number of atoms given for the one element to the number in the empirical formula. Determine the factor the number in the empirical formula must be multiplied by to obtain the given number of atoms for that element. Multiply the empirical formula by this number to get the molecular formula. Roadmap: (Same first three steps as in b). Empirical formula Divide the number of atoms of the one element in the molecule by the number of atoms of that element in the empirical formula and multiply the empirical formula by that factor Molecular formula d) No, the mass % will only lead to the empirical formula. e) Yes, a structural formula shows all the atoms in the compound. Plan: Count the number of atoms of each type of element and record as the number for the molecular formula. Roadmap: Structural formula Count the number of atoms of each element and use these numbers as subscripts Molecular formula 3.35

MgCl2 is an empirical formula, since ionic compounds such as MgCl2 do not contain molecules.

3.36

Plan: Examine the number of atoms of each type in the compound. Divide all atom numbers by the common factor that results in the lowest whole-number values. Add the molar masses of the atoms to obtain the empirical formula mass. Solution: a) C2H4 has a ratio of 2 carbon atoms to 4 hydrogen atoms, or 2:4. This ratio can be reduced to 1:2, so that the empirical formula is CH2. The empirical formula mass is 12.01 g/mol C + 2(1.008 g/mol H) = 14.03 g/mol. b) The ratio of atoms is 2:6:2, or 1:3:1. The empirical formula is CH3O and its empirical formula mass is 12.01 g/mol C + 3(1.008 g/mol H) + 16.00 g/mol O = 31.03 g/mol. c) Since, the ratio of elements cannot be further reduced, the molecular formula and empirical formula are the same, N2O5. The formula mass is 2(14.01 g/mol N) + 5(16.00 g/mol O) = 108.02 g/mol. d) The ratio of elements is 3 atoms of barium to 2 atoms of phosphorus to 8 atoms of oxygen, or 3:2:8. This ratio cannot be further reduced, so the empirical formula is also Ba3(PO4)2, with a formula mass of 3(137.3 g/mol Ba) + 2(30.97 g/mol P) + 8(16.00 g/mol O) = 601.8 g/mol. e) The ratio of atoms is 4:16, or 1:4. The empirical formula is TeI4, and the formula mass is 127.6 g/mol Te + 4(126.9 g/mol I) = 635.2 g/mol.

3-32

3.37

Plan: Examine the number of atoms of each type in the compound. Divide all atom numbers by the common factor that results in the lowest whole-number values. Add the molar masses of the atoms to obtain the empirical formula mass. Solution: a) C4H8 has a ratio of 4 carbon atoms to 8 hydrogen atoms, or 4:8. This ratio can be reduced to 1:2, so that the empirical formula is CH2. The empirical formula mass is 12.01 g/mol C + 2(1.008 g/mol H) = 14.03 g/mol. b) C3H6O3 has a ratio of atoms of 3:6:3, or 1:2:1. The empirical formula is CH2O and its empirical formula mass is 12.01 g/mol C + 2(1.008 g/mol H) + 16.00 g/mol O = 30.03 g/mol. c) P4O10 has a ratio of 4 P atoms to 10 O atoms, or 4:10. This ratio can be reduced to 2:5, so that the empirical formula is P2O5. The empirical formula mass is 2(30.97 g/mol P) + 5(16.00 g/mol O) = 141.94 g/mol. d) Ga2(SO4)3 has a ratio of 2 atoms of gallium to 3 atoms of sulfur to 12 atoms of oxygen, or 2:3:12. This ratio cannot be further reduced, so the empirical formula is also Ga2(SO4)3, with a formula mass of 2(69.72 g/mol Ga) + 3(32.06 g/mol S) + 12(16.00 g/mol O) = 427.6 g/mol. e) Al2Br6 has a ratio of atoms of 2:6, or 1:3. The empirical formula is AlBr3, and the formula mass is 26.98 g/mol Al + 3(79.90 g/mol Br) = 266.7 g/mol.

3.38

Plan: Use the chemical symbols and count the atoms of each type to obtain the molecular formula. Divide the molecular formula by the largest common factor to give the empirical formula. Use nomenclature rules to derive the name. This compound is composed of two nonmetals. The naming rules for binary covalent compounds indicate that the element with the lower group number is named first. Greek numerical prefixes are used to indicate the number of atoms of each element in the compound. The molecular (formula) mass is the sum of the atomic masses of all of the atoms. Solution: The compound has 2 sulfur atoms and 2 chlorine atoms and a molecular formula of S 2Cl2. The compound’s name is disulfur dichloride. Sulfur is named first since it has the lower group number. The prefix di- is used for both elements since there are 2 atoms of each element. The empirical formula is S 2 Cl 2 or SCl. 2

of S2Cl2 = (2 × 3.39

of S) + (2 ×

of Cl) = (2 × 32.06 g/mol S) + (2 × 35.45 g/mol Cl) = 135.02 g/mol

Plan: Use the chemical symbols and count the atoms of each type to obtain the molecular formula. Divide the molecular formula by the largest common factor to give the empirical formula. Use nomenclature rules to derive the name. This compound is composed of two nonmetals. The naming rules for binary covalent compounds indicate that the element with the lower group number is named first. Greek numerical prefixes are used to indicate the number of atoms of each element in the compound. The molecular (formula) mass is the sum of the atomic masses of all of the atoms. Solution: The compound has 4 phosphorus atoms and 6 oxygen atoms and a molecular formula of P 4O6. The compound’s name is tetraphosphorus hexaoxide. Phosphorus is named first since it has the lower group number. The prefix tetra- is for phosphorus since there are 4 P atoms and hexa- is used for oxygen since there are 6 O atoms. The empirical formula is P4 O 6 or P2O3. 2

of P4O6 = (4 × 3.40

of P) + (6 ×

of O) = (4 × 30.97 g/mol P) + (6 × 16.00 g/mol O) = 219.88 g/mol

3-33

Solution: Only approximate whole-number values are needed. a) CH2 has empirical mass equal to 12.01 g/mol C + 2(1.008 g/mol C) = 14.03 g/mol  42.08 g/mol  molar mass of compound   = 3 =  Whole-number multiple = empirical formula mass  14.03 g/mol  Multiplying the subscripts in CH2 by 3 gives C3H6. b) NH2 has empirical mass equal to 14.01 g/mol N + 2(1.008 g/mol H) = 16.03 g/mol Whole-number multiple =

 32.05 g/mol  molar mass of compound  =   = 2  16.03 g/mol  empirical formula mass

Multiplying the subscripts in NH2 by 2 gives N2H4. c) NO2 has empirical mass equal to 14.01 g/mol N + 2(16.00 g/mol O) = 46.01 g/mol Whole-number multiple =

 92.02 g/mol  molar mass of compound   = 2 =   46.01 g/mol  empirical formula mass

Multiplying the subscripts in NO2 by 2 gives N2O4. d) CHN has empirical mass equal to 12.01 g/mol C + 1.008 g/mol H + 14.01 g/mol N = 27.03 g/mol 135.14 g/mol  molar mass of compound   = 5 =  Whole-number multiple =  27.03 g/mol  empirical formula mass Multiplying the subscripts in CHN by 5 gives C5H5N5. 3.41

Plan: Determine the molar mass of each empirical formula. The subscripts in the molecular formula are wholenumber multiples of the subscripts in the empirical formula. To find this whole number, divide the molar mass of the compound by its empirical formula mass. Multiply each subscript in the empirical formula by the whole number. Solution: Only approximate whole-number values are needed. a) CH has empirical mass equal to 12.01 g/mol C + 1.008 g/mol H = 13.02 g/mol  78.11 g/mol  molar mass of compound   = 6 =  Whole-number multiple = empirical formula mass 13.02 g/mol  Multiplying the subscripts in CH by 6 gives C6H6. b) C3H6O2 has empirical mass equal to 3(12.01 g/mol C) + 6(1.008 g/mol H) + 2(16.00 g/mol O) = 74.08 g/mol Whole-number multiple =

 74.08 g/mol  molar mass of compound   = 1 =  empirical formula mass  74.08 g/mol 

Multiplying the subscripts in C3H6O2 by 1 gives C3H6O2. c) HgCl has empirical mass equal to 200.6 g/mol Hg + 35.45 g/mol Cl = 236.0 g/mol Whole-number multiple =

 472.1 g/mol  molar mass of compound   = 2 =   236.0 g/mol  empirical formula mass

Multiplying the subscripts in HgCl by 2 gives Hg2Cl2.

3-34

d) C7H4O2 has empirical mass equal to 7(12.01 g/mol C) + 4(1.008 g/mol H) + 2(16.00 g/mol O) = 120.10 g/mol Whole-number multiple =

molar mass of compound  240.20 g/mol  = = 2  120.10 g/mol  empirical formula mass

Multiplying the subscripts in C7H4O2 by 2 gives C14H8O4. 3.42

Plan: The empirical formula is the smallest whole-number ratio of the atoms or moles in a formula. All data must be converted to moles of an element by dividing mass by the molar mass. Divide each mole number by the smallest mole number to convert the mole ratios to whole numbers. Solution: a) 0.063 mol Cl and 0.22 mol O: preliminary formula is Cl0.063O0.22 Converting to integer subscripts (dividing all by the smallest subscript): Cl 0.063 O 0.22 0.063

O3.5

0.063

The formula is Cl1O3.5, which in whole numbers (x 2) is Cl2O7. b) Find moles of elements by dividing by molar mass:

 1 mol Si    = 0.08722 mol Si Moles of Si = 2.45 g Si   28.09 g Si 





 1 mol Cl    = 0.349788 mol Cl Moles of Cl = 12.4 g Cl   35.45 g Cl 





Preliminary formula is Si0.08722Cl0.349788 Converting to integer subscripts (dividing all by the smallest subscript): Si 0.08722 Cl 0.349788 0.08722

Cl4

0.08722

The empirical formula is SiCl4. c) Assume a 100 g sample and convert the masses to moles by dividing by the molar mass:

 27.3 parts C by mass  1 mol C   Moles of C = 100 g   = 2.2731 mol C  100 parts by mass 12.01 g C 





 72.7 parts O by mass  1 mol O   = 4.5438 mol O  Moles of O = 100 g   100 parts by mass 16.00 g O 





Preliminary formula is C2.2731O4.5438 Converting to integer subscripts (dividing all by the smallest subscript): C 2.2731 O 4.5438 2.2731

C1O2

2.2731

The empirical formula is CO2. 3.43

Plan: The empirical formula is the smallest whole-number ratio of the atoms or moles in a formula. All data must be converted to moles of an element by dividing mass by the molar mass. Divide each mole number by the smallest mole number to convert the mole ratios to whole numbers. Solution: a) 0.039 mol Fe and 0.052 mol O: preliminary formula is Fe0.039O0.052 Converting to integer subscripts (dividing all by the smallest subscript):

3-35

Fe 0.039 O 0.052 0.039

O1.33

0.039

The formula is Fe1O1.33, which in whole numbers (x 3) is Fe3O4. b) Find moles of elements by dividing by molar mass:  1 mol P    = 0.029157 mol P Moles of P = 0.903 g P   30.97 g P 





 1 mol Br    = 0.087484 mol Br Moles of Br = 6.99 g Br   79.90 g Br 





Preliminary formula is P0.029157Br0.087484 Converting to integer subscripts (dividing all by the smallest subscript): P0.029157 Br0.087484 0.029157

Br3

0.029157

The empirical formula is PBr3. c) Assume a 100 g sample and convert the masses to moles by dividing by the molar mass: 79.9% C and 100 – 79.9 = 20.1% H

 79.9 parts C by mass  1 mol C   = 6.6528 mol C  Moles of C = 100 g   100 parts by mass 12.01 g C 





 20.1 parts H by mass  1 mol H   = 19.940 mol H  Moles of H = 100 g   100 parts by mass 1.008 g H 





Preliminary formula is C6.6528H19.940 Converting to integer subscripts (dividing all by the smallest subscript): C 6.6528 H 19.940 6.6528

6.6528

The empirical formula is CH3. 3.44

Plan: The percent oxygen is 100% minus the percent nitrogen. Assume 100 grams of sample, and then the moles of each element may be found by dividing the mass of each element by its molar mass. Divide each of the moles by the smaller value, and convert to whole numbers to get the empirical formula. The subscripts in the molecular formula are whole-number multiples of the subscripts in the empirical formula. To find this whole number, divide the molar mass of the compound by its empirical formula mass. Multiply each subscript in the empirical formula by the whole number. Solution: a) % O = 100% Assume a 100 g sample and convert the masses to moles by dividing by the molar mass:

 30.45 parts N by mass  1 mol N   = 2.1734 mol N  Moles of N = 100 g   100 parts by mass 14.01 g N 





 69.55 parts O by mass  1 mol O   = 4.3469 mol O Moles of O = 100 g    100 parts by mass 16.00 g O 





Preliminary formula is N2.1734O4.3469 Converting to integer subscripts (dividing all by the smallest subscript):

3-36

N 2.1734 O 4.3469 2.1734

2.1734

The empirical formula is NO2. b) Formula mass of empirical formula = 14.01 g/mol N + 2(16.00 g/mol O) = 46.01 g/mol  90 g/mol  molar mass of compound   = 2 =  Whole-number multiple =  46.01 g/mol  empirical formula mass Multiplying the subscripts in NO2 by 2 gives N2O4 as the molecular formula. Note: Only an approximate value of the molar mass is needed. 3.45

Plan: The percent silicon is 100% minus the percent chorine. Assume 100 grams of sample, and then the moles of each element may be found by dividing the mass of each element by its molar mass. Divide each of the moles by the smaller value, and convert to whole numbers to get the empirical formula. The subscripts in the molecular formula are whole-number multiples of the subscripts in the empirical formula. To find this whole number, divide the molar mass of the compound by its empirical formula mass. Multiply each subscript in the empirical formula by the whole number. Solution: a) % Si = 100% Assume a 100 g sample and convert the masses to moles by dividing by the molar mass:

 20.9 parts Si by mass  1 mol Si   = 0.7440 mol Si  Moles of Si = 100 g   100 parts by mass  28.09 g Si 





 79.1 parts Cl by mass  1 mol Cl   = 2.2313 mol Cl  Moles of Cl = 100 g   100 parts by mass  35.45 g Cl 





Preliminary formula is Si0.7440Cl2.2313 Converting to integer subscripts (dividing all by the smallest subscript): Si 0.7440 Cl 2.2313 0.7440

Cl3

0.7440

The empirical formula is SiCl3. b) Formula mass of empirical formula = 28.09 g/mol Si + 3(35.45 g/mol Cl) = 134.44 g/mol Whole-number multiple =

 269 g/mol  molar mass of compound   = 2 =  empirical formula mass 134.44 g/mol 

Multiplying the subscripts in SiCl3 by 2 gives Si2Cl6 as the molecular formula. 3.46

Plan: The moles of the metal are known, and the moles of fluorine atoms may be found in part (a) from the M:F mole ratio in the compound formula. In part (b), convert moles of F atoms to mass and subtract the mass of F from the mass of MF2 to find the mass of M. In part (c), divide the mass of M by moles of M to determine the molar mass of M which can be used to identify the element. Solution: a) Determine the moles of fluorine.

 2 mol F    = 1.20 mol F Moles of F = 0.600 mol M  1 mol M 





b) Determine the mass of M.

3-37

19.00 g F    = 22.8 g F Mass of F = 1.20 mol F   1 mol F 





Mass (g) of M = MF2(g) – F(g) = 46.8 g – 22.8 g = 24.0 g M c) The molar mass is needed to identify the element. 24.0 g M Molar mass of M = = 40.0 g/mol 0.600 mol M The metal with the closest molar mass to 40.0 g/mol is calcium. 3.47

Plan: The moles of the metal oxide are known, and the moles of oxygen atoms may be found in part (a) from the compound:oxygen mole ratio in the compound formula. In part (b), convert moles of O atoms to mass and subtract the mass of O from the mass of M2O3 to find the mass of M. In part (c), find moles of M from the compound:M mole ratio and divide the mass of M by moles of M to determine the molar mass of M which can be used to identify the element. Solution: a) Determine the moles of oxygen.

 3 mol O   Moles of O = 0.370 mol M2 O3   = 1.11 mol O 1 mol M2 O3 





b) Determine the mass of M.

16.00 g O    = 17.76 g O Mass of O = 1.11 mol O   1 mol O 





Mass(g) of M = M2O3(g) – O(g) = 55.4 g (M + O) – 17.76 = 37.64 = 37.6 g M c) First, the number of moles of M must be calculated.

 2 mol M    = 0.740 mol M Moles M = 0.370 mol M2 O3  1 mol M2 O3 





The molar mass is needed to identify the element. 37.64 g M Molar mass of M = = 50.86 g/mol 0.740 mol M The metal with the closest molar mass to 50.9 g/mol is vanadium. 3.48

Plan: In combustion analysis, finding the moles of carbon and hydrogen is relatively simple because all of the carbon present in the sample is found in the carbon of CO2, and all of the hydrogen present in the sample is found in the hydrogen of H2O. Convert the mass of CO2 to moles and use the ratio between CO2 and C to find the moles and mass of C present. Do the same to find the moles and mass of H from H2O. The moles of oxygen are more difficult to find, because additional O2 was added to cause the combustion reaction. Subtracting the masses of C and H from the mass of the sample gives the mass of O. Convert the mass of O to moles of O. Take the moles of C, H, and O and divide by the smallest value to convert to whole numbers to get the empirical formula. Determine the empirical formula mass and compare it to the molar mass given in the problem to see how the empirical and molecular formulas are related. Finally, determine the molecular formula. Solution:

 1 mol CO2   1 mol C   Moles of C = 0.449 g CO2    = 0.010202 mol C  44.01 g CO2  1 mol CO2 





3-38

12.01 g C    = 0.122526 g C Mass (g) of C = 0.010202 mol C   1 mol C 





 1 mol H 2 O  2 mol H     Moles of H = 0.184 g H 2 O  1 mol H O  = 0.020422 mol H 18.02 g H2 O  2 





1.008 g H    = 0.020585 g H Mass (g) of H = 0.020422 mol H   1 mol H 





Mass (g) of O = Sample mass – (mass of C + mass of H) = 0.1595 g – (0.122526 g C + 0.020585 g H) = 0.016389 g O

 1 mol O    = 0.0010243 mol O Moles of O = 0.016389 g O  16.00 g O 





Preliminary formula = C0.010202H0.020422O0.0010243 Converting to integer subscripts (dividing all by the smallest subscript): C 0.010202 H 0.020422 O 0.0010243 0.0010243

0.0010243

H20O1

0.0010243

Empirical formula = C10H20O Empirical formula mass = 10(12.01 g/mol C) + 20(1.008 g/mol H) + 1(16.00 g/mol O) = 156.26 g/mol The empirical formula mass is the same as the given molar mass so the empirical and molecular formulas are the same. The molecular formula is C10H20O. 3.49





12.01 g C    = 0.15910 g C Mass (g) of C = 0.013247 mol C   1 mol C 





 1 mol H2 O  2 mol H     Moles of H = 0.119 g H 2 O  1 mol H O  = 0.013208 mol H 18.02 g H 2 O  2 





1.008 g H    = 0.013314 g H Mass (g) of H = 0.013208 mol H   1 mol H 





Mass (g) of O = Sample mass – (mass of C + mass of H) = 0.2572 g – (0.15910 g C + 0.013314 g H) = 0.084786 g O Copyright McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

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 1 mol O    = 0.0052991 mol O Moles of O = 0.084786 g O  16.00 g O 

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

Preliminary formula = C0.013247H0.013208O0.0052991 Converting to integer subscripts (dividing all by the smallest subscript): C 0.013247 H 0.013208 O 0.0052991 0.0052991

0.0052991

2.50

H2.49O1

0.0052991

Multiply the subscripts by 2 to obtain integers: C(2.5x2)H(2.49×2)O(1.0×2) = C5H5O2 Empirical formula = C5H5O2 Empirical formula mass = 5(12.01 g/mol C) + 5(1.008 g/mol H) + 2(16.00 g/mol O) = 97.09 g/mol Whole-number multiple =

194.2 g/mol  Molar mass of compound   = 2 =   97.09 g/mol  Empirical formula mass

Multiplying the subscripts in the empirical formula by 2 the molecular formula: C(5×2)H(5×2)O(2×2) = C10H10O4 3.50

Plan: The empirical formula is the smallest whole-number ratio of the atoms or moles in a formula. Divide each mmole number by the smallest mmole value to convert the mmole ratios to whole numbers. Since all the values are given in millimoles, there is no need to convert to moles. Solution: Preliminary formula is C6.16H8.56N1.23 Converting to integer subscripts (dividing all by the smallest subscript): C 6.16 H 8.56 N 1.23 1.23

1.23

H 7N 1

1.23

The empirical formula is C5H7N. 3.51

Plan: The empirical formula is the smallest whole-number ratio of the atoms or moles in a formula. Assume 100 grams of cortisol so the percentages are numerically equivalent to the masses of each element. Convert each of the masses to moles by dividing by the molar mass of each element involved. Divide each mole number by the smallest mole number to convert the mole ratios to whole numbers. The subscripts in the molecular formula are whole-number multiples of the subscripts in the empirical formula. To find this whole number, divide the molar mass of the compound by its empirical formula mass. Multiply each subscript in the empirical formula by the whole number. Solution:  1 mol C    = 5.7952 mol C Moles of C = 69.6 g C  12.01 g C 

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

 1 mol H    = 8.2738 mol H Moles of H = 8.34 g H  1.008 g H 

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

 1 mol O    = 1.38125 mol O Moles of O = 22.1 g O  16.00 g O 

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Preliminary formula is C5.7952H8.2738O1.38125 Converting to integer subscripts (dividing all by the smallest subscript):

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C 5.7952 H 8.2738 O 1.38125 1.38125

1.38125

4.2

H 6O 1

1.38125

The carbon value is not close enough to a whole number to round the value. The smallest number that 4.20 may be multiplied by to get close to a whole number is 5. (You may wish to prove this to yourself.) All three ratios need to be multiplied by 5: 5(C4.2H6O1) = C21H30O5. The empirical formula mass is = 21(12.01 g/mol C) + 30(1.008 g/mol H) + 5(16.00 g/mol O) = 362.45 g/mol molar mass of compound  362.47 g/mol  = Whole-number multiple =  =1  362.45 g/mol  empirical formula mass The empirical formula mass and the molar mass given are the same, so the empirical and the molecular formulas are the same. The molecular formula is C21H30O5. 3.52

Plan: Determine the molecular formula from the figure, and the molar mass from the molecular formula. The formula gives the relative numbers of moles of each element present. Multiply the number of moles of each element by its molar mass to find the total mass of element in 1 mole of compound. total mass of element 100 . Mass percent = molar mass of compound Solution: Molecular formula = C8H9NO2 Molar mass = 8(12.01 g/mol C) + 9(1.008 g/mol H) + 1(14.01 g/mol N) + 2(16.00 g/mol O) = 151.16 g/mol There are 8 moles of C in 1 mole of C8H9NO2. 12.01 g C  Mass (g) of C = 8 mol C = 96.08 g C  1 mol C  total mass C 96.08 g C 100 = 100 = 63.5618 = 63.56% C Mass percent = molar mass of compound 151.16 g C8 H 9 NO2 There are 9 moles of H in 1 mole of C8H9NO2. 1.008 g H  Mass (g) of H = 9 mol H = 9.072 g H  1 mol H  total mass H 9.072 g H 100 = 100 = 6.00159 = 6.002% H Mass percent = molar mass of compound 151.16 g C8 H 9 NO2 There is 1 mole of N in 1 mole of C8H9NO2. 14.01 g N  Mass (g) of N = 1 mol N = 14.01 g N  1 mol N  total mass N 14.01 g N 100 = 100 = 9.2683 = 9.268% N Mass percent = molar mass of compound 151.16 g C8 H 9 NO2

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There are 2 moles of O in 1 mole of C8H9NO2. 16.00 g O  Mass (g) of O = 2 mol H = 32.00 g O  1 mol O  total mass O 32.00 g O 100 = 100 = 21.1696 = 21.17% O Mass percent = molar mass of compound 151.16 g C8 H 9 NO2 A balanced chemical equation describes: 1) The identities of the reactants and products. 2) The molar (and molecular) ratios by which reactants form products. 3) The physical states of all substances in the reaction. In a balanced equation, the total mass of the reactants is equal to the total mass of the products formed in the reaction. Thus, the law of mass conversation is obeyed.

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3.55

Students I and II are incorrect. Both students changed a given formula. Only coefficients should be changed when balancing; subscripts cannot be changed. Student I failed to identify the product correctly, writing AlCl 2 instead of AlCl3. Student II used atomic chlorine instead of molecular chlorine as a reactant. Student III followed the correct process, changing only coefficients.

3.56

Plan: Examine the diagram and label each formula. We will use A for red atoms and B for green atoms. Solution: The reaction shows A2 and B2 diatomic molecules forming AB molecules. Equal numbers of A2 and B2 combine to give twice as many molecules of AB. Thus, the reaction is A2 + B2  2 AB. This is the balanced equation in b.

3.57

Plan: Examine the diagram and label each formula. Count the number of each type of molecule to obtain the coefficients. Write the balanced reaction with smallest whole-number set of coefficients. Solution: The reaction shows two CH4 and eight F2 molecules forming two CF4 and eight HF molecules to give 2CH4(g) + 8F2(g)  2CF4(g) + 8HF(g) With the smallest whole-number coefficients, the balanced reaction is CH4(g) + 4F2(g)  CF4(g) + 4HF(g)

3.58

Plan: Balancing is a trial-and-error procedure. Balance one element at a time, placing coefficients where needed to have the same number of atoms of a particular element on each side of the equation. The smallest whole-number coefficients should be used. Solution: a) __Cu(s) + __ S8(s)  __Cu2S(s) Balance the S first, because there is an obvious deficiency of S on the right side of the equation. The 8 S atoms in S8 require the coefficient 8 in front of Cu2S: __Cu(s) + __S8(s)  8Cu2S(s) Then balance the Cu. The 16 Cu atoms in Cu2S require the coefficient 16 in front of Cu: 16Cu(s) + S8(s)  8Cu2S(s) b) __P4O10(s) + __H2O(l)  __H3PO4(l) Balance the P first, because there is an obvious deficiency of P on the right side of the equation. The 4 P atoms in P4O10 require a coefficient of 4 in front of H3PO4: ___P4O10(s) + __H2O(l)  4H3PO4(l) Balance the H next, because H is present in only one reactant and only one product. The 12 H atoms in 4H3PO4 on the right require a coefficient of 6 in front of H2O: ___P4O10(s) + 6H2O(l)  4H3PO4(l) Balance the O last, because it appears in both reactants and is harder to balance. There are 16 O atoms on each side: P4O10(s) + 6H2O(l)  4H3PO4(l) c) __B2O3(s) + __ NaOH(aq)  __Na3BO3(aq) + __H2O(l) Balance oxygen last because it is present in more than one place on each side of the reaction. The 2 B atoms in B2O3 on the left require a coefficient of 2 in front of Na3BO3 on the right: __B2O3(s) + __NaOH(aq)  2Na3BO3(aq) + __H2O(l) The 6 Na atoms in 2Na3BO3 on the right require a coefficient of 6 in front of NaOH on the left: __B2O3(s) + 6NaOH(aq)  2Na3BO3(aq) + __H2O(l)

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The 6 H atoms in 6NaOH on the left require a coefficent of 3 in front of H 2O on the right: __B2O3(s) + 6NaOH(aq)  2Na3BO3(aq) + 3H2O(l) The oxygen is now balanced with 9 O atoms on each side: B2O3(s) + 6NaOH(aq)  2Na3BO3(aq) + 3H2O(l) d) __CH3NH2(g) + __O2(g)  __CO2(g) + __H2O(g) + __N2(g) There are 2 N atoms on the right in N2 so a coefficient of 2 is required in front of CH3NH2 on the left: 2CH3NH2(g) + __O2(g)  __CO2(g) + __H2O(g) + __N2(g) There are now 10 H atoms in 2CH3NH2 on the left so a coefficient of 5 is required in front of H2O on the right: 2CH3NH2(g) + __O2(g)  __CO2(g) + 5H2O(g) + __N2(g) The 2 C atoms on the left require a coefficient of 2 in front of CO2 on the right: 2CH3NH2(g) + __O2(g)  2CO2(g) + 5H2O(g) + __N2(g) The 9 O atoms on the right (4 O atoms in 2CO2 plus 5 in 5H2O) require a coefficient of 9/2 in front of O2 on the left: 2CH3NH2(g) + 9/2O2(g)  2CO2(g) + 5H2O(g) + __N2(g) Multiply all coefficients by 2 to obtain whole numbers: 4CH3NH2(g) + 9O2(g)  4CO2(g) + 10H2O(g) + 2N2(g) 3.59

Plan: Balancing is a trial-and-error procedure. Balance one element at a time, placing coefficients where needed to have the same number of atoms of a particular element on each side of the equation. The smallest whole-number coefficients should be used. Solution: a) __Cu(NO3)2(aq) + __KOH(aq)  __Cu(OH)2(s) + __ KNO3(aq) The 2 N atoms in Cu(NO3)2 on the left require a coefficient of 2 in front of KNO3 on the right: __Cu(NO3)2(aq) + __ KOH(aq)  __Cu(OH)2(s) + 2KNO3(aq) The 2 K atoms in 2KNO3 on the right require a coefficient of 2 in front of KOH on the left: __Cu(NO3)2(aq) + 2KOH(aq)  __Cu(OH)2(s) + 2KNO3(aq) There are 8 O atoms and 2 H atoms on each side: Cu(NO3)2(aq) + 2KOH(aq)  Cu(OH)2(s) + 2KNO3(aq) b) __BCl3(g) + __H2O(l)  __H3BO3(s) + __HCl(g) The 3 Cl atoms in BCl3 on the left require a coefficient of 3 in front of HCl on the right: __BCl3(g) + __H2O(l)  __H3BO3(s) + 3HCl(g) The 6 H atoms on the right (3 in H3BO3 and 3 in HCl) require a coefficient of 3 in front of H2O on the left: __BCl3(g) + 3H2O(l)  __H3BO3(s) + 3HCl(g) There are 3 O atoms and 1 B atom on each side: BCl3(g) + 3H2O(l)  H3BO3(s) + 3HCl(g) c) __CaSiO3(s) + __HF(g)  __SiF4(g) + __CaF2(s) + __H2O(l) The 6 F atoms on the right (4 in SiF4 and 2 in CaF2) require a coefficient of 6 in front of HF on the left: __CaSiO3(s) + 6HF(g)  __SiF4(g) + __CaF2(s) + __H2O(l) The 6 H atoms in 6HF on the left require a coefficient of 3 in front of H 2O on the right: __CaSiO3(s) + 6HF(g)  __SiF4(g) + __CaF2(s) + 3H2O(l) There are 1 Ca atom, 1 Si atom, and 3 O atoms on each side: CaSiO3(s) + 6HF(g)  SiF4(g) + CaF2(s) + 3H2O(l)

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d) __ (CN)2(g) + __H2O(l)  __H2C2O4(aq) + __NH3(g) The 2 N atoms in (CN)2 on the left requires a coefficient of 2 in front of NH3 on the left: __ (CN)2(g) + __H2O(l)  __H2C2O4(aq) + 2NH3(g) The 4 O atoms in H2C2O4 on the right requires a coefficient of 4 in front of H2O on the right: __(CN)2(g) + 4H2O(l)  __H2C2O4(aq) + 2NH3(g) There are 2 C atoms and 8 H atoms on each side: (CN)2(g) + 4H2O(l)  H2C2O4(aq) + 2NH3(g) 3.60

Plan: Balance one element at a time, placing coefficients where needed to have the same number of atoms of a particular element on each side of the equation. The smallest whole-number coefficients should be used. Solution: a) __SO2(g) + __O2(g)  __SO3(g) There are 4 O atoms on the left and 3 O atoms on the right. Since there is an odd number of O atoms on the right, place a coefficient of 2 in front of SO3 for an even number of 6 O atoms on the right: __SO2(g) + __O2(g)  2SO3(g) Since there are now 2 S atoms on the right, place a coefficient of 2 in front of SO 2 on the left. There are now 6 O atoms on each side: 2SO2(g) + O2(g)  2SO3(g) b) __Sc2O3(s) + __H2O(l) __ Sc(OH)3(s) The 2 Sc atoms on the left require a coefficient of 2 in front of Sc(OH)3 on the right: __Sc2O3(s) + __H2O(l)  2Sc(OH)3(s) The 6 H atoms in 2Sc(OH)3 on the right require a coefficient of 3 in front of H2O on the left. There are now 6 O atoms on each side: Sc2O3(s) + 3H2O(l)  2Sc(OH)3(s) c) __H3PO4(aq) + __NaOH(aq)  __Na2HPO4(aq) + __H2O(l) The 2 Na atoms in Na2HPO4 on the right require a coefficient of 2 in front of NaOH on the left: __H3PO4(aq) + 2NaOH(aq)  __Na2HPO4(aq) + __H2O(l) There are 6 O atoms on the right (4 in H3PO4 and 2 in 2NaOH); there are 4 O atoms in Na2HPO4 on the right so a coefficient of 2 in front of H2O will result in 6 O atoms on the right: __H3PO4(aq) + 2NaOH(aq)  __Na2HPO4(aq) + 2H2O(l) Now there are 4 H atoms on each side: H3PO4(aq) + 2NaOH(aq)  Na2HPO4(aq) + 2H2O(l) d) __C6H10O5(s) + __O2(g)  __CO2(g) + __H2O(g) The 6 C atoms in C6H10O5 on the left require a coefficient of 6 in front of CO2 on the right: __C6H10O5(s) + __O2(g)  6CO2(g) + __H2O(g) The 10 H atoms in C6H10O5 on the left require a coefficient of 5 in front of H2O on the right: __C6H10O5(s) + __O2(g)  6CO2(g) + 5H2O(g) There are 17 O atoms on the right (12 in 6CO2 and 5 in 5H2O); there are 5 O atoms in C6H10O5 so a coefficient of 6 in front of O2 will bring the total of O atoms on the left to 17: C6H10O5(s) + 6O2(g)  6CO2(g) + 5H2O(g)

3.61

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Solution: a) __As4S6(s) + __O2(g)  __As4O6(s) + __SO2(g) The 6 S atoms in As4S6 on the left require a coefficient of 6 in front of SO2 on the right: __As4S6(s) + __O2(g)  __As4O6(s) + 6SO2(g) The 18 O atoms on the right (6 in As4O6 and 12 in 6SO2) require a coefficient of 9 in front of O2 on the left: __As4S6(s) + 9O2(g)  __As4O6(s) + 6SO2(g) There are 4 As atoms on each side: As4S6(s) + 9O2(g)  As4O6(s) + 6SO2(g) b) __Ca3(PO4)2(s) + __SiO2(s) + __C(s)  __P4(g) + __CaSiO3(l) + __CO(g) The 4 P atoms in P4 require a coefficient of 2 in front of Ca3(PO4)2 on the left: 2Ca3(PO4)2(s) + __SiO2(s) + __C(s)  __P4(g) + __CaSiO3(l) + __CO(g) The 6 Ca atoms in 2Ca3(PO4)2 on the left require a coefficient of 6 in front of CaSiO3 on the right: 2Ca3(PO4)2(s) + __SiO2(s) + __C(s)  __P4(g) + 6CaSiO3(l) + __CO(g) The 6 Si atoms in 6CaSiO3 on the right require a coefficient of 6 in front of SiO2 on the left: 2Ca3(PO4)2(s) + 6SiO2(s) + __C(s)  __P4(g) + 6CaSiO3(l) + __CO(g) There are 28 O atoms on the left (16 in 2Ca3(PO4)2 and 12 in 6SiO2); there are 18 O atoms on the right in 6CaSiO3 so a coefficient of 10 in front of CO on the right will bring the total O atoms to 18 on the right: 2Ca3(PO4)2(s) + 6SiO2(s) + __C(s)  __P4(g) + 6CaSiO3(l) + 10CO(g) The 10 C atoms in 10CO on the right require a coefficient of 10 in front of C on the left: 2Ca3(PO4)2(s) + 6SiO2(s) + 10C(s)  P4(g) + 6CaSiO3(l) + 10CO(g) c) __Fe(s) + __H2O(g)  __Fe3O4(s) + __H2(g) The 3 Fe atoms in Fe3O4 on the right require a coefficient of 3 in front of Fe on the left: 3Fe(s) + __H2O(g)  __Fe3O4(s) + __H2(g) The 4 O atoms in Fe3O4 on the right require a coefficient of 4 in front of H2O on the left: 3Fe(s) + 4H2O(g)  __Fe3O4(s) + __H2(g) The 8 H atoms on the left in 4H2O require a coefficient of 4 in front of H2 on the right: 3Fe(s) + 4H2O(g)  Fe3O4(s) + 4H2(g) d) __S2Cl2(l) + __NH3(g)  __S4N4(s) + __S8(s) + __NH4Cl(s) The 12 S atoms on the right (4 in S4N4 and 8 in S8) require a coefficient of 6 in front of S2Cl2 on the left: 6S2Cl2(l) + __NH3(g)  __S4N4(s) + __S8(s) + __NH4Cl(s) The 12 Cl atoms in 6S2Cl2 on the left require a coefficient of 12 in front of NH4Cl on the right: 6S2Cl2(l) + __NH3(g)  __S4N4(s) + __S8(s) + 12NH4Cl(s) The 16 N atoms on the right (4 in S4N4 and 12 in 12NH4Cl) require a coefficient of 16 in front of NH3 on the left: 6S2Cl2(l) + 16NH3(g)  S4N4(s) + S8(s) + 12NH4Cl(s) 3.62

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A coefficient of 2 in front of Ga on the left is needed to balance the 2 Ga atoms in Ga2O3: 2Ga(s) + __O2(g)  __Ga2O3(s) The 3 O atoms in Ga2O3 on the right require a coefficient of 3/2 in front of O2 on the left: 2Ga(s) + 3/2O2(g)  __Ga2O3(s) Multiply all coefficients by 2 to obtain whole numbers: 4Ga(s) + 3O2(g)  2Ga2O3(s) b) Liquid hexane and oxygen gas are the reactants while carbon dioxide gas and gaseous water are the products: __C6H14(l) +__O2(g)  __CO2(g) + __H2O(g) The 6 C atoms in C6H14 on the left require a coefficient of 6 in front of CO2 on the right: __C6H14(l) +__O2(g)  6CO2(g) + __H2O(g) The 14 H atoms in C6H14 on the left require a coefficient of 7 in front of H2O on the right: __C6H14(l) +__O2(g)  6CO2(g) + 7H2O(g) The 19 O atoms on the right (12 in 6CO2 and 7 in 7H2O) require a coefficient of 19/2 in front of O2 on the left: Multiply all coefficients by 2 to obtain whole numbers: 2C6H14(l) + 19O2(g)  12CO2(g) + 14H2O(g) c) Aqueous solutions of calcium chloride and sodium phosphate are the reactants; solid calcium phosphate and an aqueous solution of sodium chloride are the products: __CaCl2(aq) + __Na3PO4(aq)  __Ca3(PO4)2(s) + __NaCl(aq) The 3 Ca atoms in Ca3(PO4)2 on the right require a coefficient of 3 in front of CaCl2 on the left: 3CaCl2(aq) + __Na3PO4(aq)  __Ca3(PO4)2(s) + __NaCl(aq) The 6 Cl atoms in 3CaCl2 on the left require a coefficient of 6 in front of NaCl on the right: 3CaCl2(aq) + __Na3PO4(aq)  __Ca3(PO4)2(s) + 6NaCl(aq) The 6 Na atoms in 6NaCl on the right require a coefficient of 2 in front of Na3PO4 on the left: 3CaCl2(aq) + 2Na3PO4(aq)  __Ca3(PO4)2(s) + 6NaCl(aq) There are now 2 P atoms on each side: 3CaCl2(aq) + 2Na3PO4(aq)  Ca3(PO4)2(s) + 6NaCl(aq) 3.63

Plan: The names must first be converted to chemical formulas. Balancing is a trial-and-error procedure. Balance one element at a time, placing coefficients where needed to have the same number of atoms of a particular element on each side of the equation. The smallest whole-number coefficients should be used. Remember that oxygen is diatomic. Solution: a) Aqueous solutions of lead(II) nitrate and potassium iodide are the reactants; solid lead(II) iodide and an aqueous solution of potassium nitrate are the products: __Pb(NO3)2(aq) + __KI(aq)  __PbI2(s) + __KNO3(aq) There are 2 N atoms in Pb(NO3)2 on the left so a coefficient of 2 is required in front of KNO3 on the right: __Pb(NO3)2(aq) + __KI(aq)  __PbI2(s) + 2KNO3(aq) The 2 K atoms in 2KNO3 and the 2 I atoms in PbI2 on the right require a coefficient of 2 in front of KI on the left: Pb(NO3)2(aq) + 2KI(aq)  PbI2(s) + 2KNO3(aq) There are now 6 O atoms on each side: Pb(NO3)2(aq) + 2KI(aq)  PbI2(s) + 2KNO3(aq) b) Liquid disilicon hexachloride and water are the reactants and solid silicon dioxide, hydrogen chloride gas and hydrogen gas are the products:

3-46

__Si2Cl6(l) + __H2O(l)  __SiO2(s) + __HCl(g) + __H2(g) The 2 Si atoms in Si2Cl6 on the left require a coefficient of 2 in front of SiO2 on the right: __Si2Cl6(l) + __H2O(l)  2SiO2(s) + __HCl(g) + __H2(g) The 6 Cl atoms in Si2Cl6 on the left require a coefficient of 6 in front of HCl on the right: __Si2Cl6(l) + __H2O(l)  2SiO2(s) + 6HCl(g) + __H2(g) The 4 O atoms in 2SiO2 on the right require a coefficient of 4 in front of H2O on the left. __Si2Cl6(l) + 4H2O(l)  2SiO2(s) + 6HCl(g) + __H2(g) There are 8 H atoms in 4H2O on the left; there are 8 H atoms on the right (6 in 6HCl and 2 in H2): Si2Cl6(l) + 4H2O(l)  2SiO2(s) + 6HCl(g) + H2(g) c) Nitrogen dioxide and water are the reactants and an aqueous solution of nitric acid and nitrogen monoxide gas are the products: __NO2(g) + __H2O(l)  __HNO3(aq) + __NO(g) Start with hydrogen it occurs in only one reactant and one product: The 2 H atoms in H2O on the left require a coefficient of 2 in front of HNO3 on the right: __NO2(g) + __H2O(l)  2HNO3(aq) + __NO(g) The 3 N atoms on the right (2 in 2HNO3 and 1 in NO) require a coefficient of 3 in front of NO2 on the left; 3NO2(g) + __H2O(l)  2HNO3(aq) + __NO(g) There are now 7 O atoms on each side: 3NO2(g) + H2O(l)  2HNO3(aq) + NO(g) 3.64

Plan: The names must first be converted to chemical formulas. Balancing is a trial-and-error procedure. Balance one element at a time, placing coefficients where needed to have the same number of atoms of a particular element on each side of the equation. The smallest whole-number coefficients should be used. Remember that oxygen is diatomic. Add the three reactions to obtain the overall equation. Substances on each side of the arrow cancel out. Solution: a) Step 1 Cl(g) + O3(g)  ClO(g) + O2(g) Step 2 2ClO(g)  ClOOCl(g) Step 3 ClOOCl(g)  2Cl(g) + O2(g) b) Add the 3 steps to obtain the overall balanced equation after multiplying Step 1 by 2 so that ClO(g) and Cl(g), intermediate products, can be eliminated from the overall equation. Step 1 2Cl(g) + 2O3(g)  2ClO(g) + 2O2(g) Step 2 2ClO(g)  ClOOCl(g) Step 3 ClOOCl(g)  2Cl(g) + O2(g) 2Cl(g) + 2O3(g) + 2ClO(g) + ClOOCl(g)  2ClO(g) + 2O2(g) + ClOOCl(g) + 2Cl(g) + O2(g) 2O3(g)  3O2(g)

3.65

The stoichiometrically equivalent molar ratio is the ratio of the coefficients in the balanced equation. This can be used as a conversion factor to calculate amounts of reactants or products in a chemical reaction.

3.66

The percent yield is the ratio of the actual to the theoretical value. Both yields can be expressed as a mass or mole comparison. The percent yield will be the same since mass and moles are directly proportional.

3-47

3.67

Plan: First, write a balanced chemical equation. Since A is the limiting reagent (B is in excess), A is used to determine the amount of C formed, using the mole ratio between reactant A and product C. Solution: Plan: The balanced equation is aA + bB  cC. Divide the mass of A by its molar mass to obtain moles of A. Use the molar ratio from the balanced equation to find the moles of C. Multiply moles of C by its molar mass to obtain mass of C. Roadmap: Mass (g) of A Divide by

(g/mol)

Amount (mol) of A Molar ratio between A and C Amount (moles) of C Multiply by

(g/mol)

Mass (g) of C 3.68

Plan: First, write a balanced chemical equation. Since the amounts of both reactants are given, the limiting reactant must be determined. Solution: The balanced equation is dD + eE  fF. Divide the mass of each reactant by its molar mass to obtain moles of each reactant. Use the appropriate molar ratios from the balanced equation to find the moles of F obtained from Reactant D and Reactant E. The smaller amount of F is the amount produced. Multiply moles of F by its molar mass to obtain mass of F. Roadmap: Mass (g) of D

Mass (g) of E

Divide by

(g/mol)

Amount (mol) of D

Amount (mol) of E

Molar ratio between D and F

Molar ratio between E and F

Amount (moles) of F

Choose lower number of moles of F and multiply by (g/mol) Mass (g) of F 3.69

Plan: Always check to see if the initial equation is balanced. If the equation is not balanced, it should be balanced before proceeding. Use the mole ratio from the balanced chemical equation to determine the moles of Cl 2 produced. The equation shows that 1 mole of Cl2 is produced for every 4 moles of HCl that react. Multiply the moles of Cl2 produced by the molar mass to convert to mass in grams.

3-48

Solution: 4HCl(aq) + MnO2(s)  MnCl2(aq) + 2H2O(g) + Cl2(g)  1 mol Cl 2    = 0.455 mol Cl a) Moles of Cl2 = 1.82 mol HCl  2  4 mol HCl 





 70.90 g Cl 2    = 32.2595 = 32.3 g Cl b) Mass (g) of Cl2 = 0.455 mol Cl 2  2  1 mol Cl 2 



3.70



Plan: Always check to see if the initial equation is balanced. If the equation is not balanced, it should be balanced before proceeding. Divide the amount of reactant in grams by its molar mass to determine moles of reactant. Use the mole ratio from the balanced chemical equation to determine the moles of Bi produced. The equation shows that 2 moles of Bi are produced for every 1 mole of Bi2O3 that reacts. Solution: Bi2O3(s) + 3C(s)  2Bi(s) + 3CO(g)  2 mol Bi    = 1.214 = 1.21 mol Bi a) Moles of Bi = 0.607 mol Bi 2 O3  1 mol Bi 2 O3 





 3 mol CO  28.01 g CO     = 51.00621 = 51.0 g CO b) Grams CO = 0.607 mol Bi2 O3  1 mol Bi 2 O3   1 mol CO 



3.71



Plan: Convert the kilograms of oxygen to grams of oxygen and then moles of oxygen by dividing by its molar mass. Use the moles of oxygen and the mole ratio from the balanced chemical equation to determine the moles of KNO3 required. Multiply the moles of KNO3 by its molar mass to obtain the mass in grams. Solution:

103 g   4 a) Mass (g) of O2 = 56.6 kg O2   = 5.66 × 10 g O2  1 kg 





 1 mol O   2  3 Moles of O2 = 5.66 10 4 g O2   = 1.76875 × 10 mol O2  32.00 g O2 





 4 mol KNO3    = 1415 = 1.42 × 103 mol KNO Moles of KNO3 = 1.76875 mol O2  3  5 mol O2 





101.11 g KNO3    = 143070.65 = 1.43 × 105 g KNO b) Mass (g) of KNO3 = 1415 mol KNO3  3  1 mol KNO3 





Combining all steps gives:

103 g   1 mol O2   4 mol KNO3 101.11 g KNO3   Mass (g) of KNO3 = 56.6 kg O2       1 kg   32.00 g O2   5 mol O2  1 mol KNO3 





= 143070.65 = 1.43 × 10 g KNO3 3.72

Plan: Convert mass of Cr2S3 to moles by dividing by its molar mass. Use the mole ratio between Cr2S3 and Cr2O3 from the balanced chemical equation to determine the moles of Cr 2O3 required. Multiply the moles of Cr2O3 by its molar mass to obtain the mass in grams. Solution:

3-49

 1 mol Cr2 S3    = 2.103107 mol Cr S a) Moles of Cr2S3 = 421 g Cr2 S3  2 3  200.18 g Cr2 S3 





1 mol Cr2 O3    = 2.103107 = 2.10 mol Cr O Moles of Cr2O3 = 2.103107 mol Cr2 S3  2 3  1 mol Cr2 S3 





152.00 g Cr2 O3    = 319.672 = 3.20 × 102 g Cr O b) Mass (g) of Cr2O3 = 2.103107 mol Cr2 O3  2 3  1 mol Cr2 O3 





Combining all steps gives:

 1 mol Cr2 S3  1 mol Cr2 O3 152.00 g Cr2 O3   Mass (g) of Cr2O3 = 421 g Cr2 S3      200.18 g Cr2 S3   1 mol Cr2 S3  1 mol Cr2 O3 





= 319.672 = 3.20 × 10 g Cr2O3 3.73

Plan: First, balance the equation. Convert the grams of diborane to moles of diborane by dividing by its molar mass. Use mole ratios from the balanced chemical equation to determine the moles of the products. Multiply the mole amount of each product by its molar mass to obtain mass in grams. Solution: The balanced equation is: B2H6(g) + 6H2O(l)  2H3BO3(s) + 6H2(g).

 1 mol B2 H6    = 1.583665 mol B H Moles of B2H6 = 43.82 g B2 H6  2 6  27.67 g B2 H 6 





 2 mol H3 BO3    = 3.16733 mol H BO Moles of H3BO3 = 1.583665 mol B2 H6  3 3  1 mol B2 H6 





 61.83 g H 3 BO3    = 195.83597 = 195.8 g H BO Mass (g) of H3BO3 = 3.16733 mol H 3 BO3  3 3  1 mol H3 BO3  Combining all steps gives:  1 mol B2 H6  2 mol H3 BO3  61.83 g H3 BO3      Mass (g) of H3BO3 = 43.82 g B2 H6   1 mol B H  1 mol H BO   27.67 g B2 H6  2 6  3 3  = 195.83597 = 195.8 g H3BO3  6 mol H 2    = 9.50199 mol H Moles of H2 = 1.583665 mol B2 H6  2 1 mol B2 H 6 













 2.016 g H 2    = 19.15901 g H = 19.16 g H Mass (g) of H2 = 9.50199 mol H 2  2 2  1 mol H 2  Combining all steps gives:  1 mol B2 H6   6 mol H 2   2.016 g H 2   Mass (g) of H2 = 43.82 g B2 H6     = 19.15601 = 19.16 g H2  27.67 g B2 H6  1 mol B2 H6   1 mol H 2 

 

3.74



Plan: First, balance the equation. Convert the grams of silver sulfide to moles of silver sulfide by dividing by its molar mass. Use mole ratios from the balanced chemical equation to determine the moles of the products. Multiply the mole amount of each product by its molar mass to obtain mass in grams. Solution:

3-50

First, balance the equation: Ag2S(s) + 2 HCl(aq)  2 AgCl(s) + H2S(g)

 1 mol Ag2 S    = 0.7018959 mol Ag S Moles of Ag2S = 174 g Ag2 S  2  247.9 g Ag2 S 





 2 mol AgCl    = 1.403792 mol AgCl Moles of AgCl = 0.7018959 mol Ag2 S   1 mol Ag2 S 





143.4 g AgCl    = 201.304 = 201 g AgCl Mass (g) of AgCl = 1.403792 mol Ag2 S  1 mol AgCl  Combining all steps gives:

 1 mol Ag 2 S   2 mol AgCl  143.4 g AgCl   Mass (g) AgCl = 174 g Ag2 S     = 201.304 = 201 g AgCl  1 mol Ag 2 S   1 mol AgCl   247.9 g Ag2 S 





 1 mol H 2 S   Moles of H2S = 0.7018959 mol Ag2 S   = 0.7018959 mol H2S 1 mol Ag2 S 





 34.08 g H 2 S    = 23.9206 = 23.9 g H S Mass (g) of H2S = 0.7018959 mol H 2 S 2  1 mol H2 S  Combining all steps gives:  1 mol Ag2 S   1 mol H 2 S   34.08 g H 2 S   Mass (g) of H2S = 174 g Ag2 S    = 23.9206 = 23.9 g H2S 1 mol Ag2 S   1 mol H 2 S   247.9 g Ag2 S  3.75

Plan: Write the balanced equation by first writing the formulas for the reactants and products. Convert the mass of phosphorus to moles by dividing by the molar mass, use the mole ratio between phosphorus and chlorine from the balanced chemical equation to obtain moles of chlorine, and finally divide the moles of chlorine by its molar mass to obtain amount in grams. Solution: Reactants: formula for phosphorus is given as P4 and formula for chlorine gas is Cl2 (chlorine occurs as a diatomic molecule). Product: formula for phosphorus pentachloride (the name indicates one phosphorus atom and five chlorine atoms) is PCl5. Equation: P4 + Cl2  PCl5 P4 + 10Cl2  4PCl5

Balancing the equation:

 1 mol P4    = 3.67291 mol P Moles of P4 = 455 g P4  4 123.88 g P4 





10 mol Cl2   Moles of Cl2 = 3.67291 mol P4   = 36.7291 mol Cl2  1 mol P4 





 70.90 g Cl2    = 2604.09 = 2.60 × 103 g Cl Mass (g) of Cl2 = 36.7291 mol Cl2  2  1 mol Cl2 





Combining all steps gives:

 1 mol P4  10 mol Cl2   70.90 g Cl2   3 Mass (g) of Cl2 = 455 g P4     = 2604.09267 = 2.60 × 10 g Cl2 123.88 g P4   1 mol P4   1 mol Cl2 





3-51

3.76

Plan: Write the balanced equation by first writing the formulas for the reactants and products. Convert the mass of sulfur to moles by dividing by the molar mass, use the mole ratio between sulfur and fluorine from the balanced chemical equation to obtain moles of fluorine, and finally divide the moles of fluorine by its molar mass to obtain amount in grams. Solution: Reactants: formula for sulfur is given as S8 and formula for fluorine gas is F2 (fluorine occurs as a diatomic molecule). Product: formula for sulfur hexafluoride (the name indicates one sulfur atom and six fluoride atoms) is SCl6. Equation: S8 + F2  SF6 Balancing the equation: S8(s) + 24F2(g)  8SF6(s)

 1 mol S8    = 0.0694011 mol S Moles of S8 = 17.8 g S8  8  256.48 g S8 





 24 mol F2    = 1.665626 mol F Moles of F2 = 0.0694011 mol S8  2  1 mol S8 





 38.00 g F2    = 63.294 = 63.3 g F Mass (g) of F2 = 1.665626 mol F2  2  1 mol F2 





Combining all steps gives:

 1 mol S8   24 mol F2  38.00 g F2   Mass (g) of F2 = 17.8 g S8     = 63.29382 = 63.3 g F2  256.48 g S8   1 mol S8  1 mol F2 



3.77



Plan: Begin by writing the chemical formulas of the reactants and products in each step. Next, balance each of the equations. Combine the equations for the separate steps by adjusting the equations so the intermediate (iodine monochloride) cancels. Finally, change the mass of product from kg to grams to moles by dividing by the molar mass and use the mole ratio between iodine and product to find the moles of iodine. Multiply moles by the molar mass of iodine to obtain mass of iodine. Solution: I2(s) + Cl2(g)  2ICl(s)

a) Step 1

Step 2 ICl(s) + Cl2(g)  ICl3(s) b) Multiply the coefficients of the second equation by 2, so that ICl(s), an intermediate product, can be eliminated from the overall equation.

I 2 (s) + Cl2 (g)   2ICl(s) 2ICl( s ) + 2Cl 2 ( g )   2ICl 3 ( s )

I2(s) + Cl2(g) + 2ICl(s) + 2Cl2(g)  2ICl(s) + 2ICl3(s) Overall equation: I2(s) + 3Cl2(g)  2ICl3(s)

103 g    = 2450 g ICl c) Mass (g) of ICl3 = 2.45 kg ICl3  3  1 kg 





 1 mol ICl 3    = 10.506 mol ICl Moles of ICl3 = 2450 g ICl3  3  233.2 g ICl3 





 1 mol I 2    = 5.253 mol I Moles of I2 = 10.506 mol ICl3  2  2 mol ICl 3 





3-52

 253.8 g I 2    = 1333.211 = 1.33 × 103 g I Mass (g) of I2 = 5.253 mol I 2  2  1 mol I 2  Combining all steps gives: 103 g  1 mol ICl3  1 mol I 2  253.8 g I 2   3     Mass (g) of I2 = 2.45 kg ICl3   233.2 g ICl  2 mol ICl  1 mol I  = 1333.211 = 1.33 × 10 g I2  1 kg  3  3  2  3.78









Plan: Begin by writing the chemical formulas of the reactants and products in each step. Next, balance each of the equations. Combine the equations for the separate steps and cancel the intermediate (lead(II) oxide). Finally, change the amount of lead from metric tons to grams to moles by dividing by the molar mass and use the mole ratio between lead and sulfur dioxide to find the moles of sulfur dioxide. Multiply moles by the molar mass of sulfur dioxide to obtain mass and convert to metric tons. Solution: a) Step 1

2PbS(s) + 3O2(g)   2PbO(s) + 2SO2(g)

Step 2

2PbO(s) + PbS(s)   3Pb(l) + SO2(g)

b) Combine the two reactions:

2PbS(s) + 3O2 (g)   2PbO(s) + 2SO2 (g) 2PbO( s ) + PbS( s )   3Pb(l ) + SO 2 ( g )

2PbS(s) + 3O2(g) + 2PbO(s) + PbS(s)  2PbO(s) + 2SO2(g) + 3Pb(l) + SO2(g) Overall equation: 3PbS(s) + 3O2(g)  3Pb(l) + 3SO2(g)

 PbS(s) + O2(g)  Pb(l) + SO2(g) c) 1 metric ton = 1000 kg

103 kg  103 g   6   Mass (g) of Pb = 1 ton Pb   1 kg  = 1.000 × 10 g Pb  1 ton  





 1 mol Pb    = 4826.255 mol Pb Moles of Pb = 1106 g Pb   207.2 g Pb 





1 mol SO2    = 4826.255 mol SO Moles of SO2 = 4826.255 mol Pb  2  1 mol Pb 





 64.06 g SO2   1 kg   1 ton   Metric ton SO2 = 4826.255 mol SO2   3  3  = 0.309217 = 0.3092 ton SO2  1 mol SO2  10 g  10 kg  Combining all steps gives: 3 103 kg  10 g  1 mol Pb 1 mol SO2  64.06 g SO2  1 kg  1 ton   Metric ton SO2 = 1 ton Pb       3  3   1 ton   1 kg  207.2 g Pb  1 mol Pb  1 mol SO2 10 g 10 kg 





= 0.309217 = 0.3092 ton SO2 3.79

Plan: Write the balanced reaction between A2 and B2 to form AB3. To find the limiting reactant, find the number of molecules of product that would form from the numbers of molecules of each reactant. Solution: a) The reaction is A2 + 3B2  2AB3

3-53

 2 AB3 molecules   = 6 AB3 molecules Molecules of AB3 from A2 = 3 A 2 molecules  1 A 2 molecule   2 AB3 molecules   = 4 AB3 molecules Molecules of AB3 from B2 = 6 B2 molecules  3 B2 molecules 

B2 is the limiting reactant since fewer molecules of product form from B2,  2 AB3 molecules   = 4 AB3 molecules b) Using the molar ratio between B2 and AB3: 6 B2 molecules  3 B2 molecules 

3.80

N2(g) + 3H2(g)  2NH3(g)  2 NH 3 molecules   = 4 NH3 molecules Molecules of NH3 from N2 = 2 N 2 molecules  1 N 2 molecule   2 NH 3 molecules   = 6 NH3 molecules Molecules of NH3 from H2 = 9 H 2 molecules  3 H 2 molecules 

N2 is the limiting reactant.  3 H 2 molecules   = 6 H2 molecules Molecules of H2 reacting = 2 N 2 molecules  1 N 2 molecule  9 – 6 = 3 H2 molecules remain after reaction. The scene after reaction shows 4 NH3 molecules, 3 H2 molecules, and no N2 molecules.

3.81

Plan: Write the balanced reaction. Calculate the amount (mol) of CH3OH produced from the amount of each reactant. The smaller amount of product is the correct answer. Solution: The reaction is CO(g) + 2H2(g)  CH3OH(l). 1 mol CH3 OH  Amount (mol) of CH3OH from CO = 4.5 mol CO  = 4.5 mol CH3OH  1 mol CO 

(mol)

1 mol CH 3 OH   = 3.6 mol CH3OH Amount (mol) of CH3OH from H2 = 7.2 mol H 2   2 mol H 2 

H2 is the limiting reactant and 3.6 mol of CH3OH are produced. 3.82

The reaction is 4NH3(g) + 5O2(g)  4NO(g) + 6H2O(l).  6 mol H 2 O   = 3.525 = 3.52 mol H2O Amount (mol) of H2O from NH3 = 2.35 mol NH 3   4 mol NH 3   6 mol H 2 O   = 3.30 mol H2O Amount (mol) of H2O from O2 = 2.75 mol O 2   5 mol O 2 

O2 is the limiting reactant and 3.30 mol of H2O are produced. 3.83

Plan: Convert the given mass of each reactant to moles by dividing by the molar mass of that reactant. Use the mole ratio from the balanced chemical equation to find the moles of CaO formed from each reactant, assuming an excess of the other reactant. The reactant that produces fewer moles of CaO is the limiting reactant. Convert the moles of CaO obtained from the limiting reactant to grams using the molar mass. Solution: 2Ca(s) + O2(g)  2CaO(s)

3-54

 1 mol Ca    = 0.104790 mol Ca a) Moles of Ca = 4.20 g Ca   40.08 g Ca 





 2 mol CaO    = 0.104790 = 0.105 mol CaO Moles of CaO from Ca = 0.104790 mol Ca   2 mol Ca 

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

 1 mol O2    = 0.0875 mol O b) Moles of O2 = 2.80 g O2  2  32.00 g O2 





 2 mol CaO    = 0.17500 = 0.175 mol CaO Moles of CaO from O2 = 0.0875 mol O2   1 mol O2 





c) Calcium is the limiting reactant since it will form less calcium oxide. d) The mass of CaO formed is determined by the limiting reactant, Ca.

 56.08 g CaO    = 5.8766 = 5.88 g CaO Mass (g) of CaO = 0.104790 mol CaO   1 mol CaO  Combining all steps gives:  1 mol Ca   2 mol CaO   56.08 g CaO   Mass (g) of CaO = 4.20 g Ca     1 mol CaO  = 5.8766 = 5.88 g CaO  40.08 g Ca   2 mol Ca  







3.84



Plan: Convert the given mass of each reactant to moles by dividing by the molar mass of that reactant. Use the mole ratio from the balanced chemical equation to find the moles of H 2 formed from each reactant, assuming an excess of the other reactant. The reactant that produces fewer moles of H 2 is the limiting reactant. Convert the moles of H2 obtained from the limiting reactant to grams using the molar mass.

Solution: SrH2(s) + 2H2O(l)  Sr(OH)2(s) + 2H2(g)

 1 mol SrH 2    = 0.0635877 mol SrH a) Moles of SrH2 = 5.70 g SrH 2  2  89.64 g SrH 2 





 2 mol H 2    = 0.127175 = 0.127 mol H Moles of H2 from SrH2 = 0.0635877 mol SrH 2  2 1 mol SrH 2 





 1 mol H 2 O    = 0.263596 mol H O b) Mass (g) of H2O = 4.75 g H 2 O  2 18.02 g H 2 O 





 2 mol H 2    = 0.263596 = 0.264 mol H Moles of H2 from H2O = 0.263596 mol H 2 O  2  2 mol H2 O 





c) SrH2 is the limiting reagent since it will yield fewer moles of hydrogen gas. d) The mass of H2 formed is determined by the limiting reactant, SrH2.

 2.016 g H 2    = 0.256385 = 0.256 g H Mass (g) of H2 = 0.127175 mol H 2  2  1 mol H2 





Combining all steps gives: Copyright McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

3-55

 1 mol SrH 2   2 mol H 2   2.016 g H 2   Mass (g) of H2 = 5.70 g SrH 2     = 0.256385 = 0.256 g H2  89.64 g SrH 2  1 mol SrH 2   1 mol H 2 



3.85



Plan: First, balance the chemical equation. To determine which reactant is limiting, calculate the amount of HIO3 formed from each reactant, assuming an excess of the other reactant. The reactant that produces less product is the limiting reagent. Use the limiting reagent and the mole ratio from the balanced chemical equation to determine the amount of HIO3 formed and the amount of the excess reactant that reacts. The difference between the amount of excess reactant that reacts and the initial amount of reactant supplied gives the amount of excess reactant remaining. Solution: The balanced chemical equation for this reaction is: 2ICl3 + 3H2O  ICl + HIO3 + 5HCl Hint: Balance the equation by starting with oxygen. The other elements are in multiple reactants and/or products and are harder to balance initially. Finding the moles of HIO3 from the moles of ICl3 (if H2O is limiting):  1 mol ICl3    = 2.722985 mol ICl Moles of ICl3 = 635 g ICl3  3  233.2 g ICl3 





1 mol HIO3    = 1.361492 = 1.36 mol HIO Moles of HIO3 from ICl3 = 2.722985 mol ICl3  3  2 mol ICl3 





Finding the moles of HIO3 from the moles of H2O (if ICl3 is limiting):

 1 mol H 2 O    = 6.57603 mol H O Moles of H2O = 118.5 g H 2 O  2 18.02 g H 2 O 

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

1 mol HIO3    = 2.19201 = 2.19 mol HIO Moles HIO3 from H2O = 6.57603 mol H2 O  3  3 mol H2 O 





ICl3 is the limiting reagent and will produce 1.36 mol HIO3.

175.9 g HIO3    = 239.486 = 239 gHIO Mass (g) of HIO3 = 1.361492 mol HIO3  3  1 mol HIO3 





Combining all steps gives:

 1 mol ICl3  1 mol HIO3  175.9 g HIO3   Mass (g) of HIO3 = 635 g ICl3     1 mol HIO  = 239.486 = 239 g HIO3  233.2 g ICl3   2 mol ICl3  3 





The remaining mass of the excess reagent can be calculated from the amount of H 2O combining with the limiting reagent.  3 mol H 2 O    = 4.0844775 mol H O Moles of H2O required to react with 635 g ICl3 = 2.722985 mol ICl3  2  2 mol ICl3 





18.02 g H 2 O    Mass (g) of H2O required to react with 635 g ICl3 = 4.0844775 mol H 2 O   1 mol H 2 O  = 73.6023 = 73.6g H2O reacted Remaining H2O = 118.5 g – 73.6 g = 44.9 g H2O





3-56

3.86

Plan: First, balance the chemical equation. To determine which reactant is limiting, calculate the amount of H2S formed from each reactant, assuming an excess of the other reactant. The reactant that produces less product is the limiting reagent. Use the limiting reagent and the mole ratio from the balanced chemical equation to determine the amount of H2S formed and the amount of the excess reactant that reacts. The difference between the amount of excess reactant that reacts and the initial amount of reactant supplied gives the amount of excess reactant remaining. Solution: The balanced chemical equation for this reaction is: Al2S3 + 6H2O  2Al(OH)3 + 3H2S Finding the moles of H2S from the moles of Al2S3 (if H2O is limiting):

 1 mol Al 2 S3    = 1.05235 mol Al S Moles of Al2S3 = 158 g Al 2 S3  2 3 150.14 g Al 2 S3 





 3 mol H 2 S    = 3.15705 = 3.16 mol H S Moles of H2S from Al2S3 = 1.05235 mol Al2 S3  2 1 mol Al2 S3  Finding the moles of H2S from the moles of H2O (if Al2S3 is limiting):  1 mol H 2 O    = 7.26970 mol H O Moles of H2O = 131 g H2 O  2 18.02 g H 2 O 

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

 3 mol H 2 S    = 3.63485 = 3.63 mol H S Moles of H2S from H2O = 7.26970 mol H 2 O  2  6 mol H 2 O 





Al2S3 is the limiting reagent and 3.16 mol of H2S will form.

 34.08 g H 2 S    = 107.592 = 108 g H S Mass (g) of H2S = 3.15705 mol H 2 S  2  1 mol H 2 S 

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

Combining all steps gives:

 1 mol Al2 S3   3 mol H 2 S  34.08 g H2 S   Grams H2S = 158 g Al 2 S3     = 107.592 = 108 g H2S 1 mol Al2 S3  1 mol H2 S  150.14 g Al2 S3  The remaining mass of the excess reagent can be calculated from the amount of H 2O combining with the limiting reagent.  6 mol H 2 O    = 6.31410 mol H O Moles of H2O required to react with 158 g of Al2S3 = 1.05235 mol Al2 S3  2 1 mol Al2 S3 









18.02 g H 2 O    = 113.780 g H O Mass (g) of H2O required to react with 158 g of Al2S3 = 6.31410 mol H 2 O  2  1 mol H2 O  Remaining H2O = 131 g H2O – 113.780 g H2O = 17.220 = 17 g H2O



3.87



Plan: Write the balanced equation; the formula for carbon is C, the formula for oxygen is O 2, and the formula for carbon dioxide is CO2. To determine which reactant is limiting, calculate the amount of CO2 formed from each reactant, assuming an excess of the other reactant. The reactant that produces less product is the limiting reagent. Use the limiting reagent and the mole ratio from the balanced chemical equation to determine the amount of CO2 formed and the amount of the excess reactant that reacts. The difference between the amount of excess reactant that reacts and the initial amount of reactant supplied gives the amount of excess reactant remaining.

3-57

Solution: The balanced equation is: C(s) + O2(g)  CO2(g) Finding the moles of CO2 from the moles of carbon (if O2 is limiting): 1 mol CO2    = 0.100 mol CO Moles of CO2 from C = 0.100 mol C  2  1 mol C 





Finding the moles of CO2 from the moles of oxygen (if C is limiting):

 1 mol O2    = 0.250 mol O Moles of O2 = 8.00 g O2  2  32.00 g O2 

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

1 mol CO2    = 0.25000 = 0.250 mol CO Moles of CO2 from O2 = 0.250 mol O2  2  1 mol O2 





Carbon is the limiting reactant and will be used to determine the amount of CO 2 that will form.

 44.01 g CO2   Mass (g) of CO2 = 0.100 mol CO2   = 4.401 = 4.40 g CO2  1 mol CO2  Since carbon is limiting, the O2 is in excess. The amount remaining depends on how much combines with the limiting reagent. 1 mol O2    = 0.100 mol O Moles of O2 required to react with 0.100 mol of C = 0.100 mol C  2  1 mol C 









 32.00 mol O2    = 3.20 g O Mass (g) of O2 required to react with 0.100 mol of C = 0.100 mol O2  2  1 mol O2 

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

Remaining O2 = 8.00 g – 3.20 g = 4.80 g O2 3.88

Plan: Write the balanced equation; the formula for hydrogen is H2, the formula for oxygen is O2, and the formula for water is H2O. To determine which reactant is limiting, calculate the amount of H2O formed from each reactant, assuming an excess of the other reactant. The reactant that produces less product is the limiting reagent. Use the limiting reagent and the mole ratio from the balanced chemical equation to determine the amount of H2O formed and the amount of the excess reactant that reacts. The difference between the amount of excess reactant that reacts and the initial amount of reactant supplied gives the amount of excess reactant remaining. Solution: The balanced equation is: 2H2(g) + O2(g)  2H2O(l) Finding the moles of H2O from the moles of hydrogen (if O2 is limiting):

 1 mol H 2    = 0.01860 mol H Moles of H2 = 0.0375 g H 2  2  2.016 g H 2 

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

 2 mol H 2 O   Moles of H2O from H2 = 0.01860 mol H 2   = 0.01860 = 0.0186 mol H2O  2 mol H 2 

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

Finding the moles of H2O from the moles of oxygen (if H2 is limiting):

 2 mol H 2 O    = 0.0370 mol H O Mole of H2O from O2 = 0.0185 mol O2  2  1 mol O2 





The hydrogen is the limiting reactant and will be used to determine the amount of water that will form. Copyright McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

3-58

18.02 g H2 O    = 0.335172 = 0.335 g H O Mass (g) of H2O = 0.01860 mol H 2 O  2  1 mol H 2 O 





Since the hydrogen is limiting; the oxygen must be the excess reactant. The amount of excess reactant is determined from the limiting reactant.

 1 mol O2    = 0.00930 mol O Moles of O2 required to react with 0.0375 g of H2 = 0.01860 mol H 2  2  2 mol H 2 





 32.00 g O2   Mass (g) of O2 required to react with 0.0375 g of H2 = 0.00930 mol O2   = 0.2976 g O2  1 mol O2 





 32.00 mol O2    = 0.5920 g O Mass of O2 supplied = 0.0185 mol O2  2  1 mol O2 





Remaining O2 = 0.5920 g – 0.2976 g = 0.2944 = 0.294 g O2 3.89

Plan: The question asks for the mass of each substance present at the end of the reaction. “Substance” refers to both reactants and products. Solve this problem using multiple steps. Recognizing that this is a limiting reactant problem, first write a balanced chemical equation. To determine which reactant is limiting, calculate the amount of any product formed from each reactant, assuming an excess of the other reactant. The reactant that produces less product is the limiting reagent. Any product can be used to predict the limiting reactant; in this case, AlCl3 is used. Use the limiting reagent and the mole ratio from the balanced chemical equation to determine the amount of both products formed and the amount of the excess reactant that reacts. The difference between the amount of excess reactant that reacts and the initial amount of reactant supplied gives the amount of excess reactant remaining. Solution: The balanced chemical equation is: Al(NO2)3(aq) + 3NH4Cl(aq)  AlCl3(aq) + 3N2(g) + 6H2O(l) Now determine the limiting reagent. We will use the moles of AlCl3 produced to determine which is limiting. Finding the moles of AlCl3 from the moles of Al(NO2)3 (if NH4Cl is limiting):

 1 mol Al(NO2 )3    = 0.439367 mol Al(NO ) Moles of Al(NO2)3 = 72.5 g Al(NO2 )3  2 3 165.01 g Al(NO2 )3 





 1 mol AlCl3    = 0.439367 = 0.439 mol AlCl Moles of AlCl3 from Al(NO2)3 = 0.439367 mol Al(NO2 )3  3 1 mol Al(NO2 )3 





Finding the moles of AlCl3 from the moles of NH4Cl (if Al(NO2)3 is limiting):  1 mol NH 4 Cl    = 1.09553 mol NH Cl Moles of NH4Cl = 58.6 g NH 4 Cl  4  53.49 g NH 4 Cl 

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

 1 mol AlCl 3   Moles of AlCl3 from NH4Cl = 1.09553 mol NH 4 Cl   = 0.365177 = 0.365 mol AlCl3  3 mol NH 4 Cl  Ammonium chloride is the limiting reactant, and it is used for all subsequent calculations. Mass of substances after the reaction: Al(NO2)3:





Mass (g) of Al(NO2)3 (the excess reactant) required to react with 58.6 g of NH4Cl = Copyright McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

3-59

1 mol Al(NO2 )3 165.01 g Al(NO2 )3 

1.09553 mol NH Cl 3 mol NH Cl  1 mol Al(NO )  = 60.2579 = 60.3 g Al(NO ) 4

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

2 3



2 3

Al(NO2)3 remaining: 72.5 g – 60.3 g = 12.2 g Al(NO2)3 NH4Cl: None left since it is the limiting reagent. AlCl3:

133.33 g AlCl3    = 48.689 = 48.7 g AlCl Mass (g) of AlCl3 = 0.365177 mol AlCl3  3  1 mol AlCl3 

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N2:

 3 mol N2   28.02 g N 2   Mass (g) of N2 = 1.09553 mol NH 4 Cl    = 30.697 = 30.7 g N2  3 mol NH 4 Cl   1 mol N 2 

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

H2O:

 6 mol H 2 O  18.02 g H 2 O   Mass (g) of H2O = 1.09553 mol NH 4 Cl    = 39.483 = 39.5 g H2O  3 mol NH 4 Cl   1 mol H 2 O 

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3.90

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Plan: The question asks for the mass of each substance present at the end of the reaction. “Substance” refers to both reactants and products. Solve this problem using multiple steps. Recognizing that this is a limiting reactant problem, first write a balanced chemical equation. To determine which reactant is limiting, calculate the amount of any product formed from each reactant, assuming an excess of the other reactant. The reactant that produces less product is the limiting reagent. Any product can be used to predict the limiting reactant; in this case, CaF2 is used. Use the limiting reagent and the mole ratio from the balanced chemical equation to determine the amount of both products formed and the amount of the excess reactant that reacts. The difference between the amount of excess reactant that reacts and the initial amount of reactant supplied gives the amount of excess reactant remaining. Solution: The balanced chemical equation is: Ca(NO3)2(s) + 2NH4F(s)  CaF2(s) + 2N2O(g) + 4H2O(g) Now determine the limiting reagent. We will use the moles of CaF2 produced to determine which is limiting. Finding the moles of CaF2 from the moles of Ca(NO3)2 (if NH4F is limiting):  1 mol Ca(NO3 )2    = 0.1023766 mol Ca(NO ) Moles of Ca(NO3)2 = 16.8 g Ca(NO3 )2  3 2 164.10 g Ca(NO3 )2 

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

 1 mol CaF2    Moles of CaF2 from Ca(NO3)2 = 0.1023766 mol Ca(NO3 )2  1 mol Ca(NO3 )2 

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

= 0.1023766 = 0.102 mol CaF2 Finding the moles of CaF2 from the moles of NH4F (if Ca(NO3)2 is limiting):

 1 mol NH 4 F    = 0.47246 mol NH F Moles of NH4F = 17.50 g NH 4 F  4  37.04 g NH 4 F 

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

 1 mol CaF2   Moles of CaF2 from NH4F = 0.47246 mol NH 4 F   = 0.23623 = 0.236 mol CaF2  2 mol NH 4 F 

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

Calcium nitrate is the limiting reactant, and it is used for all subsequent calculations Copyright McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

3-60

Mass of substances after the reaction: Ca(NO3)2: None (It is the limiting reactant.) NH4F: Mass (g) of NH4F (the excess reactant) required to react with 16.8 g of Ca(NO3)2 =



 2 mol NH 4 F   37.04 g NH 4 F   0.1023766 mol Ca(NO3 )2    = 7.58406 g Ca(NO3)2  1 mol NH 4 F  1 mol Ca(NO3 )2 



NH4F remaining: 17.50 g – 7.58 g = 9.9159 = 9.92 g NH4F CaF2:

 1 mol CaF2   78.08 g CaF2   Mass (g) of CaF2 = 0.1023766 mol Ca(NO3 )2    = 7.99356 = 7.99 g CaF2 1 mol Ca(NO3 )2   1 mol CaF2 

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N2O:

 2 mol N 2 O   44.02 g N 2 O   Mass (g) of N2O = 0.1023766 mol Ca(NO3 )2    = 9.0132 = 9.01 g N2O  1 mol N 2 O  1 mol Ca(NO3 )2  H2O:  4 mol H 2 O  18.02 g H 2 O     Mass (g) of H2O = 0.1023766 mol Ca(NO3 )2   1 mol H O  = 7.3793 = 7.38 g H2O 1 mol Ca(NO3 )2  2 

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3.91

Plan: Express the yield of each step as a fraction of 1.00; multiply the fraction of the first step by that of the second step and then multiply by 100 to get the overall percent yield. Solution: 73% = 0.73; 68% = 0.68 (0.73 × 0.68) × 100 = 49.64 = 50%

3.92

Plan: Express the yield of each step as a fraction of 1.00; multiply the fraction of the first step by that of the second step and then multiply by 100 to get the overall percent yield. Solution: 48% = 0.48; 73% = 0.73 (0.48 × 0.73) × 100 = 35.04 = 35%

3.93

Plan: Write and balance the chemical equation using the formulas of the substances. Determine the theoretical yield of the reaction from the mass of tungsten(VI) oxide. To do that, convert the mass of tungsten(VI) oxide to moles by dividing by its molar mass and then use the mole ratio between tungsten(VI) oxide and water to determine the moles and then mass of water that should be produced. Use the density of water to determine the actual yield of water in grams. The actual yield divided by the theoretical yield just calculated (with the result multiplied by 100%) gives the percent yield. Solution: The balanced chemical equation is: WO3(s) + 3H2(g)  W(s) + 3H2O(l) Determining the theoretical yield of H2O:

 1 mol WO3    = 0.1962053 mol WO Moles of WO3 = 45.5 g WO3  3  231.9 g WO3 

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

 3 mol H 2 O  18.02 g H2 O   Mass (g) of H2O (theoretical yield) = 0.1962053 mol WO3    = 10.60686 g H2O 1 mol WO3   1 mol H 2 O 

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3-61

Determining the actual yield of H2O:

1.00 g H 2 O    = 9.60 g H O Mass (g) of H2O (actual yield) = 9.60 mL H 2 O  2  1 mL H 2 O 

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

 actual Yield   9.60 g H O   2 100% =  100% = 90.5075 = 90.5% % yield =    theoretical Yield  10.60686 g H 2 O  3.94

Plan: Write and balance the chemical equation using the formulas of the substances. Determine the theoretical yield of the reaction from the mass of phosphorus trichloride. To do that, convert the mass of phosphorus trichloride to moles by dividing by its molar mass and then use the mole ratio between phosphorus trichloride and HCl to determine the moles and then mass of HCl that should be produced. The actual yield of the HCl is given. The actual yield divided by the theoretical yield just calculated (with the result multiplied by 100%) gives the percent yield. Solution: The balanced chemical equation is: PCl3(l) + 3H2O(l)  H3PO3(aq) + 3HCl(g) Determining the theoretical yield of HCl:

 1 mol PCl3    = 1.456452 mol PCl Moles of PCl3 = 200. g PCl3  3 137.32 g PCl3 

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

 3 mol HCl   36.46 g HCl   Mass (g) of HCl (theoretical yield) = 1.456452 mol PCl3    = 159.3067 g HCl 1 mol PCl3   1 mol HCl 

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

Actual yield (g) of HCl is given as 128 g HCl. Calculate the percent yield:

 actual Yield   128 g HCl   100% =  100% = 80.3481586 = 80.3% % yield =     theoretical Yield  159.3067 g HCl  3.95

Plan: Write the balanced chemical equation. Since quantities of two reactants are given, we must determine which is the limiting reactant. To determine which reactant is limiting, calculate the amount of any product formed from each reactant, assuming an excess of the other reactant. The reactant that produces less product is the limiting reagent. Any product can be used to predict the limiting reactant; in this case, CH3Cl is used. Only 75.0% of the calculated amounts of products actually form, so the actual yield is 75% of the theoretical yield. Solution: The balanced equation is: CH4(g) + Cl2(g)  CH3Cl(g) + HCl(g) Determining the limiting reactant: Finding the moles of CH3Cl from the moles of CH4 (if Cl2 is limiting):

 1 mol CH 4    = 1.278055 mol CH Moles of CH4 = 20.5 g CH 4  4 16.04 g CH 4 

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

1 mol CH3 Cl    = 1.278055 mol CH Cl Moles of CH3Cl from CH4 = 1.278055 mol CH 4  3  1 mol CH 4 

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

Finding the moles of CH3Cl from the moles of Cl2 (if CH4 is limiting):

3-62

 1 mol Cl 2    = 0.634697 mol Cl Moles of Cl2 = 45.0 g Cl 2  2  70.90 g Cl 2 

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1 mol CH3 Cl    = 0.634697 mol CH Cl Moles of CH3Cl from Cl2 = 0.634697 mol Cl2  3  1 mol Cl2 

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

Chlorine is the limiting reactant and is used to determine the theoretical yield of CH 3Cl:

 50.48 g CH 3 Cl   Mass (g) of CH3Cl (theoretical yield) = 0.634697 mol CH3 Cl   = 32.0395 g CH3Cl  1 mol CH3 Cl 

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

 actual Yield   100% % yield =   theoretical Yield 

% yield 75% theoretical yield = 32.0395 g CH3Cl = 24.02962 = 24.0 g CH3Cl 100% 100%

Actual yield (g) of CH3Cl = 3.96

Plan: Write the balanced chemical equation. Since quantities of two reactants are given, we must determine which is the limiting reactant. To determine which reactant is limiting, calculate the amount of product formed from each reactant, assuming an excess of the other reactant. Only 93.0% of the calculated amount of product actually forms, so the actual yield is 93.0% of the theoretical yield. Solution: The balanced equation is: 3Ca(s) + N2(g)  Ca3N2(s) Determining the limiting reactant: Finding the moles of Ca3N2 from the moles of Ca (if N2 is limiting):  1 mol Ca    = 1.412176 mol Ca Moles of Ca = 56.6 g Ca   40.08 g Ca 

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

1 mol Ca 3 N 2    = 0.470725 mol Ca N Moles of Ca3N2 from Ca = 1.412176 mol Ca  3 2  3 mol Ca 

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

Finding the moles of Ca3N2 from the moles of N2 (if Ca is limiting):

 1 mol N2    = 1.08851 mol N Moles of N2 = 30.5 g N 2  2  28.02 g N 2 

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

1 mol Ca 3 N 2    = 1.08851 mol Ca N Moles of Ca3N2 from N2 = 1.08851 mol N 2  3 2  1 mol N 2  Ca is the limiting reactant and is used to determine the theoretical yield of Ca3N2. 148.26 g Ca 3 N 2    = 69.7897 g Ca N Mass (g) of Ca3N2 (theoretical yield) = 0.470725 mol Ca 3 N 2  3 2  1 mol Ca 3 N 2 

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

 actual Yield   100% % yield =   theoretical Yield  Actual yield (g) of Ca3N2 =

3.97

% yield 93% theoretical yield = 69.7897 g Ca3 N2  = 64.9044 = 64.9 g Ca3N2 100% 100%

Plan: Write the balanced equation; the formula for fluorine is F2, the formula for carbon tetrafluoride is CF4, and the formula for nitrogen trifluoride is NF3. To determine which reactant is limiting, calculate the amount of CF4

3-63

formed from each reactant, assuming an excess of the other reactant. The reactant that produces less product is the limiting reagent. Use the limiting reagent and the mole ratio from the balanced chemical equation to determine the mass of CF4 formed. Solution: The balanced chemical equation is: (CN)2(g) + 7F2(g)  2CF4(g) + 2NF3(g) Determining the limiting reactant: Finding the moles of CF4 from the moles of (CN)2 (if F2 is limiting):

 1 mol (CN)2  2 mol CF4     Moles of CF4 from (CN)2 = 60.0 g (CN)2  1 mol (CN)  = 2.30592 mol CF4  52.04 g (CN)2  2

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Finding the moles of CF4 from the moles of F2 (if (CN)2 is limiting):

 1 mol F2   2 mol CF4   Moles of CF4 from F2 = 60.0 g F2    = 0.4511278 mol CF4  38.00 g F2   7 mol F2 

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

F2 is the limiting reactant, and will be used to calculate the amount of CF4 produced.  1 mol F2   2 mol CF4   88.01 g CF4   Mass (g) of CF4 = 60.0 g F2     1 mol CF  = 39.70376 = 39.7 g CF4  38.00 g F2   7 mol F2  4 

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3.98

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Plan: Write and balance the chemical reaction. Remember that both chlorine and oxygen exist as diatomic molecules. Use the mole ratio between oxygen and dichlorine monoxide to find the moles of dichlorine monoxide that reacted. Multiply the amount in moles by Avogadro’s number to convert to number of molecules. Solution: a) Both oxygen and chlorine are diatomic. Scene A best represents the product mixture as there are O2 and Cl2 molecules in Scene A. Scene B shows oxygen and chlorine atoms and Scene C shows atoms and molecules. Oxygen and chlorine atoms are NOT products of this reaction. b) The balanced reaction is 2Cl2O(g)  2Cl2(g) + O2(g). c) There is a 2:1 mole ratio between Cl2 and O2. In Scene A, there are 6 green molecules and 3 red molecules. Since twice as many Cl2 molecules are produced as there are O2 molecules produced, the red molecules are the O2 molecules.  2 O atoms  0.050 mol O atoms 1 mol O 2 molecules  2 mol Cl 2 O    Moles of Cl2O = 3 O 2 molecules    2 mol O atoms  1 mol O 2  1 O atom 1 O 2 molecule  = 0.30 mol Cl2O

 6.022 10 23 Cl O molecules   2  0.30 mol Cl O Molecules of Cl2O =  2   1 mol Cl 2 O  23

= 1.8066 × 10 = 1.8 × 10 Cl2O molecules 3.99

Plan: Write a balanced equation for the reaction. Convert the given mass of each reactant to moles by dividing by the molar mass of that reactant. Use the mole ratio from the balanced chemical equation to find the moles of nitrogen monoxide formed from each reactant, assuming an excess of the other reactant. The reactant that produces fewer moles of product is the limiting reactant. Convert the moles of nitrogen monoxide obtained from the limiting reactant to grams using the molar mass. Solution: The balanced chemical equation is: 4NH3(g) + 5O2(g)  4NO(g) + 6H2O(g) Determining the limiting reactant:

3-64

Finding the moles of NO from the amount of NH3 (if O2 is limiting):

 1 mol NH3   4 mol NO   Moles of NO from NH3 = 485 g NH3    = 28.47915 mol NO 17.03 g NH3   4 mol NH3 

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Finding the moles of NO from the amount of O2 (if NH3 is limiting):

 1 mol O2   4 mol NO   Moles of NO from O2 = 792 g O2    = 19.8 mol NO  32.00 g O2   5 mol O2 

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

O2 is the limiting reactant, and will be used to calculate the amount of NO formed:  30.01 g NO    = 594.198 = 594 g NO Mass (g) of NO = 19.8 mol NO   1 mol NO 

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

Combining all of the steps gives:

 1 mol O2   4 mol NO  30.01 g NO   Mass (g) of NO = 792 g O2     = 594.198 = 594 g NO  5 mol O2  1 mol NO   32.00 g O2 

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3.100



Plan: Write a balanced equation. Use the density of butane to convert the given volume of butane to mass and divide by the molar mass of butane to convert mass to moles. Use the mole ratio between butane and oxygen to find the moles and then mass of oxygen required for the reaction. The mole ratio between butane and water is used to find the moles of water produced and the mole ratio between butane and carbon dioxide is used to find the moles of carbon dioxide produced. The total moles of product are multiplied by Avogadro’s number to find the number of product molecules. Solution: The balanced chemical equation is: 2C4H10(g) + 13O2(g)  8CO2(g) + 10H2O(g)

 0.579 g C 4 H10  1 mol C 4 H10   = 0.054792 mol C H  a) Moles of C4H10 = 5.50 mL C 4 H10  4 10  1 mL C 4 H10  58.12 g C 4 H10 

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 13 mol O2   32.00 g O2   Mass (g) of O2 = 0.054792 mol C4 H10    = 11.3967 = 11.4 g O2  1 mol O2   2 mol C 4 H10 

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

 10 mol H 2 O    = 0.27396 = 0.274 mol H O b) Moles of H2O = 0.054792 mol C4 H10  2  2 mol C 4 H10 

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

 8 mol CO2    = 0.219168 mol CO c) Moles of CO2 = 0.054792 mol C4 H10  2  2 mol C 4 H10 

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

Total moles = 0.27396 mol H2O + 0.219168 mol CO2 = 0.493128 mol  6.022 1023 molecules    = 2.96962 × 1023 = 2.97 × 1023 molecules Total molecules = 0.493128 mol   1 mol  



3.101

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Plan: Write a balanced equation for the reaction. Convert the given mass of each reactant to moles by dividing by the molar mass of that reactant. Use the mole ratio from the balanced chemical equation to find the moles of NaBH4 formed from each reactant, assuming an excess of the other reactant. The reactant that produces fewer moles of product is the limiting reactant. Convert the moles of NaBH4 obtained from the limiting reactant to grams using the molar mass. This is the theoretical yield of NaBH4. Since there is a yield of 88.5%, the amount of NaBH4 actually obtained will be 88.5% of the theoretical yield.

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Solution: The balanced chemical equation is: 2NaH(s) + B2H6(g)  2NaBH4(s) Determining the limiting reactant: Finding the moles of NaBH4 from the amount of NaH (if B2H6 is limiting):

 1 mol NaH  2 mol NaBH 4     Moles of NaBH4 from NaH = 7.98 g NaH   2 mol NaH  = 0.3325 mol NaBH4  24.00 g NaH  





Finding the moles of NaBH4 from the amount of B2H6 (if NaH is limiting):  1 mol B2 H 6  2 mol NaBH 4    = 0.58981 mol NaBH4 Moles of NaBH4 from B2H6 = 8.16 g B2 H 6   1 mol B2 H 6   27.67 g B2 H 6  NaH is the limiting reactant and will be used to calculate the theoretical yield of NaBH4.  37.83 g NaBH 4    = 12.5785 g NaBH Mass (g) of NaBH4 = 0.3325 mol NaBH 4  4  1 mol NaBH 4 





 actual Yield   100% % yield =   theoretical Yield   % yield   88.5%  Mass (g) of NaBH4 =   theoretical yield =  12.5785 g NaHB4  = 11.13197 = 11.1 g NaBH4  100%   100%  Combining all steps gives:

 1 mol NaH   2 mol NaBH 4   37.83 g NaBH 4  88.5%   Mass (g) of NaBH4 = 7.98 g NaH       2 mol NaH   1 mol NaBH 4  100%   24.00 g NaH  = 11.13197 = 11.1 g NaBH4



3.102



Plan: An atomic mass unit (amu) is exactly 1/12 the mass of a C atom and Avogadro’s number of C atoms has a mass of 12 g. Solution: 12 1 atom 12 C   1 mol 12 C   12 g C   1 amu a) Mass in grams =      6.0221023 atoms 12 C  1mol 12 C   12 amu 





–24

= 1.6605779 × 10 = 1.661 × 10

–24

Remember that all the values in this calculation except Avogadro’s number are exact numbers. b) The g/amu ratio is 1.661 × 10 3.103

–24

g/amu.

Plan: First determine the empirical formula. Convert the mass of each element to moles by dividing the mass of each element by its molar mass. Divide each of the moles by the smaller value, and convert to whole numbers to get the empirical formula. The subscripts in the molecular formula are whole-number multiples of the subscripts in the empirical formula. To find this whole number, divide the molar mass of the compound by its empirical formula mass. Multiply each subscript in the empirical formula by the whole number. Solution:

 1 mol S    = 0.0713662 mol S Moles of S = 2.288 g S   32.06 g S 





3-66

 1 mol N    = 0.0713776 mol N Moles of N = 1.000 g N  14.01 g N 





Preliminary formula is S0.0713439N0.0713662 Converting to integer subscripts (dividing all by the smallest subscript): S 0.0713662 O 0.0713776  S1N1 0.0713662

0.0713662

The empirical formula is SN. Formula mass of empirical formula = 32.06 g/mol S + 14.01 g/mol N = 46.07 g/mol Whole-number multiple =

184.27 g/mol  molar mass of compound   = 4 =   46.07 g/mol  empirical formula mass

Multiplying the subscripts in SN by 4 gives S4N4 as the molecular formula. 3.104

Plan: The first step is to write and balance the chemical equation for the reaction. Convert the mass of each reactant to moles by dividing by the molar mass, remembering that the mass of the phosphoric acid reactant is 85% of the given mass of phosphoric acid solution. To determine which reactant is limiting, calculate the amount of hydroxyapatite formed from each reactant, assuming an excess of the other reactant. The reactant that produces less product is the limiting reagent. Use the limiting reagent and the mole ratio from the balanced chemical equation to determine the amount of hydroxyapatite that forms. Solution: a) The balanced equation is: 5Ca(OH)2(aq) + 3H3PO4(aq)  Ca5(PO4)3(OH)(s) + 9H2O(l) b) Finding the moles of Ca5(PO4)3(OH) from the moles of Ca(OH)2 (if H3PO4 is limiting):

 1 mol Ca(OH)2    = 1.349528 mol Ca(OH) Moles of Ca(OH)2 = 100. g Ca(OH)2  2  74.10 g Ca(OH)2 





1 mol Ca 5 (PO 4 )3 (OH)    Moles of Ca5(PO4)3(OH) from Ca(OH)2 = 1.349528 mol Ca(OH)2   5 mol Ca(OH)2 





= 0.2699056 mol Ca5(PO4)3(OH) Finding the moles of Ca5(PO4)3(OH) from the moles of H3PO4 (if Ca(OH)2 is limiting):   85 g H 3 PO 4  1 mol H 3 PO 4  = 0.867435 mol H PO Moles of H3PO4 = 100. g H 3 PO 4 solution  3 4  97.99 g H PO  100. g H 3 PO 4 solution  3 4 1 mol Ca 5 (PO 4 )3 (OH)   Moles of Ca5(PO4)3(OH) from H3PO4 = 0.867435 mol H 3 PO 4 solution   3 mol H 3 PO 4  = 0.289145 mol Ca5(PO4)3(OH) Ca(OH)2 is the limiting reactant, and will be used to calculate the amount of Ca5(PO4)3(OH) produced.  502.32 g Ca 5 (PO 4 )3 (OH)    Mass (g) of Ca5(PO4)3(OH) = 0.2699056 mol Ca 5 (PO4 )3 (OH)   1 mol Ca 5 (PO 4 )3 (OH) 





= 135.57898 = 140 g Ca5(PO4)3(OH) 3.105

Plan: The moles of narceine and the moles of water are required. We can assume any mass of narceine hydrate (we will use 100 g), and use this mass to determine the moles of hydrate. The moles of water in the hydrate is obtained by taking 10.8% of the 100 g mass of hydrate and converting the mass to moles of water. Divide the moles of water by the moles of hydrate to find the value of x.

3-67

Solution: Assuming a 100 g sample of narceine hydrate:

 1 mol narceine hydrate   Moles of narceine hydrate = 100 g narceine hydrate  499.52 g narceine hydrate 

= 0.20019 mol narceine hydrate   10.8% H 2 O  = 10.8 g H2O Mass (g) of H2O = 100 g narceine hydrate 100% narceine hydrate   1 mol H 2 O   = 0.59933 mol H2O Moles of H2O = 10.8 g H 2 O  18.02 g H 2 O  moles of H 2 O 0.59933 mol = x= =3 moles of hydrate 0.20019 mol

Thus, there are three water molecules per mole of hydrate. The formula for narceine hydrate is narceine 3H2O. 3.106

Plan: Determine the formula and the molar mass of each compound. The formula gives the relative numbers of moles of each element present. Multiply the number of moles of each element by its molar mass to find the total total mass of element 100. List the compounds mass of element in 1 mole of compound. Mass percent = molar mass of compound from the highest %H to the lowest. Solution: Name

Chemical formula

Molar mass (g/mol)

Ethane

C2H6

30.07

Propane

C3H8

44.09

Benzene

C6H6

78.11

Ethanol

C2H5OH

46.07

Cetyl palmitate

C32H64O2

480.83

moles of H  molar mass 100 molar mass of compound 6 mol(1.008 g/mol) 100 = 20.11% H 30.07 g 8 mol(1.008 g/mol) 100 = 18.29% H 44.09 g 6 mol(1.008 g/mol) 100 = 7.743% H 78.11 g 6 mol(1.008 g/mol) 100 = 13.13% H 46.07 g 64 mol(1.008 g/mol) 100 = 13.42% H 480.83 g

Mass percent H =

The hydrogen percentage decreases in the following order: Ethane > Propane > Cetyl palmitate > Ethanol > Benzene 3.107

Plan: First determine the empirical formula. Assume 100 grams of sample, and then the moles of each element may be found by dividing the mass of each element by its molar mass. Divide each of the moles by the smallest value, and convert to whole numbers to get the empirical formula. The subscripts in the molecular formula are whole-number multiples of the subscripts in the empirical formula. To find this whole number, divide the molar mass of the compound by its empirical formula mass. Multiply each subscript in the empirical formula by the whole number. Solution: Assume a 100 g sample and convert the masses to moles by dividing by the molar mass:

3-68

 68.2 parts C by mass  1 mol C   = 5.6786 mol C Moles of C = 100 g    100 parts by mass 12.01 g C 





 6.86 parts H by mass  1 mol H   = 6.8056 mol H  Moles of H = 100 g   100 parts by mass 1.008 g H 





15.9 parts N by mass  1 mol N   = 1.1349 mol N  Moles of N = 100 g   100 parts by mass 14.01 g N 





 9.08 parts O by mass  1 mol O   = 0.5675 mol O  Moles of O = 100 g   100 parts by mass 16.00 g O 





Preliminary formula is C5.6786H6.8056N1.1349O0.5675 Converting to integer subscripts (dividing all by the smallest subscript): C 5.6786 H 6.8056 N 1.1349 O 0.5675  C10H12N2O 0.5675

0.5675

The empirical formula is C10H12N2O Formula mass of empirical formula = 10(12.01g/mol C) + 12(1.008 g/mol H) + 2(14.01 g/mol N) + 1(16.00 g/mol O) = 176.22 g/mol  176 g/mol  molar mass of compound  =  Whole-number multiple =  = 1 176.22 g/mol  empirical formula mass The empirical formula mass and the molar mass are the same, thus, the molecular and empirical formulas are the same. The molecular formula is C10H12N2O. 3.108

a) Before 1961 the amu was defined as 1/16 the average mass of an atom of naturally occurring oxygen and Avogadro’s number was fixed by the definition of the mole as about 16 g of oxygen. After 1961, Avogadro’s number was fixed by the definition of the mole as exactly 12 g of carbon-12. So Avogadro’s number changed to a smaller value in 1961. However, the difference is quite small. b) Yes, the definition of the mole changed. c) Yes, the mass of a mole of substance changed. d) Avogadro’s number to 3 significant figures will not change since the difference in values is so small.

3.109

Plan: The names must first be converted to chemical formulas. Balancing is a trial-and-error procedure. Balance one element at a time, placing coefficients where needed to have the same number of atoms of a particular element on each side of the equation. The smallest whole-number coefficients should be used. Remember that oxygen, chlorine, and hydrogen are diatomic. Solution: a) All of the substances are gases. H2S(g) + O2(g)   SO2(g) + H2O(g) There are 2 O atoms in O2 on the left and 3 O atoms in SO2 and H2O on the right; place a coefficient of 2 in front of H2O on the right and a coefficient of 2 in front of O2 on the left for a total of 4 oxygen atoms on each side: H2S(g) + 2O2(g)   SO2(g) + 2H2O(g) Now the 4 H atoms in 2H2O on the right require a coefficient of 2 in front of H2S on the left: 2H2S(g) + 2O2(g)   SO2(g) + 2H2O(g)

3-69

The 2 S atoms in 2H2S on the left require a coefficient of 2 in front of SO2 on the right: 2H2S(g) + 2O2(g)   2SO2(g) + 2H2O(g) Now the O atoms are no longer balanced; the 6 O atoms on the right (4 in 2SO 2 and 2 in 2H2O) require a coefficient of 3 in front of O2 on the left:

 2H2S(g) + 3O2(g)  2SO2(g) + 2H2O(g) b) All of the substances are solid (crystalline). KClO3(s)   KCl(s) + KClO4(s) There are 3 O atoms in KClO3 on the left and 4 O atoms in KClO4 on the right. Place a coefficient of 4 in front of KClO3 and a coefficient of 3 in front of KClO4 for a total of 12 O atoms on each side. The K and Cl atoms are balanced with 4 K atoms and 4 Cl atoms on each side:  4KClO3(s)  KCl(s) + 3KClO4(s) c) Hydrogen and water vapor are gases; iron and iron(III) oxide are solids. H2(g) + Fe2O3(s)  Fe(s) + H2O(g) The 2 Fe atoms in Fe2O3 on the left require a coefficient of 2 in front of Fe on the right: H2(g) + Fe2O3(s)  2Fe(s) + H2O(g) The 3 O atoms in Fe2O3 on the left require a coefficient of 3 in front of H2O on the right: H2(g) + Fe2O3(s)  2Fe(s) + 3H2O(g) The 6 H atoms in 3H2O on the right require a coefficient of 3 in front of H 2 on the left: 3H2(g) + Fe2O3(s)  2Fe(s) + 3H2O(g) d) All of the substances are gases; combustion required oxygen as a reactant. C2H6(g) + O2(g)   CO2(g) + H2O(g) The 2 C atoms in C2H6 on the left require a coefficient of 2 in front of CO2 on the right: C2H6(g) + O2(g)   2CO2(g) + H2O(g) The 6 H atoms in C2H6 on the left require a coefficient of 3 in front of H2O on the right: C2H6(g) + O2(g)   2CO2(g) + 3H2O(g) The 7 O atoms on the right (4 in 2CO2 and 3 in 3H2O) require a coefficient of 7/2 in front of O2 on the left: C2H6(g) + 7/2O2(g)   2CO2(g) + 3H2O(g) Double all coefficients to get whole number coefficients: 2C2H6(g) + 7O2(g)   4CO2(g) + 6H2O(g) e) Iron(II) chloride and iron(III) fluoride are solids and the other substances are gases. FeCl2(s) + ClF3(g)  FeF3(s) + Cl2(g) There are 3 Cl atoms on the left (2 in FeCl2 and 1 in ClF3) and 2 Cl atoms in Cl2 on the right. Place a coefficient of 2 in front of Cl2 and a coefficient of 2 in front of ClF3 on the left for a total of 4 Cl atoms on each side: FeCl2(s) + 2ClF3(g)  FeF3(s) + 2Cl2(g) The 6 F atoms in 2ClF3 require a coefficient of 2 in front of FeF3 on the right: FeCl2(s) + 2ClF3(g)  2FeF3(s) + 2Cl2(g) The 2 Fe atoms in FeF3 on the right require a coefficient of 2 in front of FeCl2 on the left: 2FeCl2(s) + 2ClF3(g)  2FeF3(s) + 2Cl2(g) Copyright McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

3-70

Now the Cl atoms are not balanced with 6 on the left (4 in 2FeCl2 and 2 in 2ClF3) and 4 in 2Cl2 on the right; place a coefficient of 3 in front of Cl2 on the right: 2FeCl2(s) + 2ClF3(g)  2FeF3(s) + 3Cl2(g) 3.110

 1 mol CO2   1 mol C   Moles of C = 2.657 g CO2  1 mol CO  = 0.06037 mol C  44.01 g CO2  2





 1 mol H 2 O   2 mol H   Moles of H = 1.089 g H 2 O    = 0.1209 mol H 18.02 g H 2 O  1 mol H 2 O 





Preliminary formula = C0.06037H0.1209 Converting to integer subscripts (dividing all by the smallest subscript): C 0.06037 H 0.1209  C1H2 0.06037

0.06037

This gives an empirical formula of CH2. 3.111

Plan: The key to solving this problem is determining the overall balanced equation. Each individual step must be set up and balanced first. The separate equations can then be combined to get the overall equation. The mass of iron is converted to moles of iron by dividing by the molar mass, and the mole ratio from the balanced equation is used to find the moles and then the mass of CO required to produce that number of moles of iron. Solution: 3+ a) In the first step, ferric oxide (ferric denotes Fe ) reacts with carbon monoxide to form Fe3O4 and carbon dioxide: 3Fe2O3(s) + CO(g)  2Fe3O4(s) + CO2(g) (1) In the second step, Fe3O4 reacts with more carbon monoxide to form ferrous oxide: Fe3O4(s) + CO(g)  3FeO(s) + CO2(g)

(2)

In the third step, ferrous oxide reacts with more carbon monoxide to form molten iron: FeO(s) + CO(g)  Fe(l) + CO2(g)

(3)

Common factors are needed to allow these equations to be combined. The intermediate products are Fe 3O4 and FeO, so multiply equation (2) by 2 to cancel Fe3O4 and equation (3) by 6 to cancel FeO: 3Fe2O3(s) + CO(g)  2Fe3O4(s) + CO2(g) 2Fe3O4(s) + 2CO(g)  6FeO(s) + 2CO2(g) 6FeO(s) + 6CO(g)  6Fe(l) + 6CO2(g) 3Fe2O3(s) + 9CO(g)  6Fe(l) + 9CO2(g) Then divide by 3 to obtain the smallest integer coefficients: Fe2O3(s) + 3CO(g)  2Fe(s) + 3CO2(g) b) A metric ton is equal to 1000 kg. Copyright McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

3-71

Converting 45.0 metric tons of Fe to mass in grams: 3 103 kg  10 g   7 Mass (g) of Fe = 45.0 ton Fe    = 4.50 × 10 g Fe  1 ton    1 kg   1 mol Fe    = 8.05730 × 105 mol Fe Moles of Fe = 4.50 10 7 g Fe   55.85 g Fe   3 mol CO   28.01 g CO   7 7 Mass (g) of CO = 8.05730105 mol Fe    = 3.38527 × 10 = 3.39 × 10 g CO  2 mol Fe   1 mol CO 











3.112



Plan: Write a balanced equation. Use the density of toluene to convert the given volume of toluene to mass and divide by the molar mass of toluene to convert mass to moles. Use the mole ratio between toluene and oxygen to find the moles and then mass of oxygen required for the reaction. The mole ratio between toluene and the gaseous products are used to find the moles of product produced. The moles of water are multiplied by Avogadro’s number to find the number of water molecules. Solution: The balanced chemical equation is: C7H8(l) + 9O2(g)  7CO2(g) + 4H2O(g)

 0.867 g C 7 H8   1 mol C 7 H8   a) Moles of C7H8 = 20.0 mL C 7 H8    = 0.1882123 mol C7H8  92.13 g C 7 H8   1 mL C 7 H8 





 9 mol O2   32.00 g O2   Mass (g) oxygen = 0.1882123 mol C7 H8   1 mol O  = 54.20514 = 54.2 g O2 1 mol C 7 H8  2 





11 mol product gas    = 2.07034 = 2.07 mol of gas b) Total moles of gas = 0.1882123 mol C7 H8    1 mol C 7 H8  The 11 mol of gas is an exact, not measured, number, so it does not affect the significant figures.  4 mol H2 O    = 0.7528492 mol H O c) Moles of H2O = 0.1882123 mol C7 H8  2 1 mol C7 H8 





 6.022 10 23 H O molecules   2  Molecules of H2O = 0.7528492 mol H 2 O    1 mol H 2 O  23 23 = 4.53366 × 10 = 4.53 × 10 molecules H2O



3.113



Plan: If 100.0 g of dinitrogen tetroxide reacts with 100.0 g of hydrazine (N 2H4), what is the theoretical yield of nitrogen if no side reaction takes place? First, we need to identify the limiting reactant. To determine which reactant is limiting, calculate the amount of nitrogen formed from each reactant, assuming an excess of the other reactant. The reactant that produces less product is the limiting reagent. Use the limiting reagent and the mole ratio from the balanced chemical equation to determine the theoretical yield of nitrogen. Then determine the amount of limiting reactant required to produce 10.0 grams of NO. Reduce the amount of limiting reactant by the amount used to produce NO. The reduced amount of limiting reactant is then used to calculate an “actual yield.” The “actual” and theoretical yields will give the maximum percent yield. Solution: The balanced reaction is 2N2H4(l) + N2O4(l)  3N2(g) + 4H2O(g) Determining the limiting reactant: Finding the moles of N2 from the amount of N2O4 (if N2H4 is limiting):

3-72

 1 mol N 2 O4   3 mol N 2   Moles of N2 from N2O4 = 100.0 g N 2 O4    = 3.26016 mol N2  92.02 g N2 O4  1 mol N 2 O4 





Finding the moles of N2 from the amount of N2H4 (if N2O4 is limiting):  1 mol N 2 H 4   3 mol N 2  = 4.68019 mol N N2 from N2H4 = 100.0 g N 2 H 4  2  2 mol N H   32.05 g N 2 H 4  2 4 

N2O4 is the limiting reactant.

 1 mol N2 O4   3 mol N 2  28.02 g N 2   Theoretical yield of N2 = 100.0 g N2 O4     = 91.3497 g N2  92.02 g N 2 O 4  1 mol N 2 O4  1 mol N 2 





How much of the limiting reactant is used to produce 10.0 g NO? N2H4(l) + 2N2O4(l)  6NO(g) + 2H2O(g)  1 mol NO   2 mol N 2 O 4   92.02 g N 2 O4   Mass (g) of N2O4 used = 10.0 g NO      30.01 g NO   6 mol NO   1 mol N 2 O4 





= 10.221 g N2O4 Amount of N2O4 available to produce N2 = 100.0 g N2O4 – mass of N2O4 required to produce 10.0 g NO = 100.0 g – 10.221 g = 89.779 g N2O4 Determine the “actual yield” of N2 from 89.779 g N2O4:  1 mol N 2 O 4  3 mol N 2  28.02 g N 2     “Actual yield” of N2 = 89.779 g N 2 O 4  1 mol N 2 O 4   1 mol N 2   92.02 g N 2 O 4  = 82.01285 g N2  actual yield   82.01285  100 =  Theoretical yield =  100 = 89.7790 = 89.8%  theoretical yield   91.3497  3.114

Plan: Write a balanced chemical equation, using X to represent the halogen element. Convert the mass of strontium sulfate to moles and use the mole ratio in the balanced equation to find the moles of strontium halide required to produce that number of moles of strontium sulfate. Divide the mass of strontium halide by the moles of strontium halide to determine its molar mass. Subtract out the molar mass of strontium to obtain the molar mass of X2. Divide the molar mass of X2 by 2 to determine the molar mass of X. The molar mass of X can be used to identify X, using the periodic table and examining Group 7A(17). Once the identity of X is known, the formula of the strontium halide can be written. Solution: SrX2(aq) + H2SO4(aq)  SrSO4(s) + 2HX(aq) 0.652 g

0.755 g

 1 mol SrSO4    = 0.0041104 mol SrSO Moles of SrSO4 = 0.755 g SrSO 4  4 183.68 g SrSO 4 





 1 mol SrX2    = 0.0041104 mol SrX Moles of SrX2 = 0.0041104 mol SrSO4  2 1 mol SrSO4 





The 0.652 g sample of SrX2 = 0.0041104 mol

3-73

0.652 g = 158.62 g/mol = molar mass 0.0041104 mol Molar mass of X2 = molar mass of SrX2 – molar mass of Sr = 158.62 g – 87.62 g = 71.00 g = X2 Molar mass of X = 71.00 g/2 = 35.50 = 35.5 g/mol = Cl The original halide formula is SrCl2. SrX2 =

3.115

Plan: Identify the product molecules and write the balanced equation. To determine the limiting reactant for part (b), examine the product circle to see which reactant remains in excess and which reactant was totally consumed. For part (c), use the mole ratios in the balanced equation to determine the number of moles of product formed by each reactant, assuming the other reactant is in excess. The reactant that produces fewer moles of product is the limiting reactant. Use the mole ratio between the two reactants to determine the moles of excess reactant required to react with the limiting reactant. The difference between the initial moles of excess reactant and the moles required for reaction is the moles of excess reactant that remain. Solution: a) The contents of the boxes give: AB2 + B2  AB3 Balancing the reaction gives: 2AB2 + B2  2AB3 b) Two B2 molecules remain after reaction so B2 is in excess. All of the AB2 molecules have reacted so AB2 is the limiting reactant. c) Finding the moles of AB3 from the moles of AB2 (if B2 is limiting):  2 mol AB3   = 5.0 mol AB3 Moles of AB3 from AB2 = 5.0 mol AB2   2 mol AB2  Finding the moles of AB3 from the moles of B2 (if AB2 is limiting):  2 mol AB3   = 6.0 mol AB3 Moles of AB3 from B2 = 3.0 mol B2   1 mol B2  AB2 is the limiting reagent and 5.0 mol of AB3 is formed.

 1 mol B2   = 2.5 mol B2 d) Moles of B2 that react with 5.0 mol AB2 = 5.0 mol AB2   2 mol AB2  The unreacted B2 is 3.0 mol – 2.5 mol = 0.5 mol B2.

3.116

Plan: Write the formulas in the form CxHyOz. Reduce the formulas to obtain the empirical formulas. Add the atomic masses in that empirical formula to obtain the molecular mass. Solution: Compound A: C4H10O2 = C2H5O

Compound B: C2H4O

Compound C: C4H8O2 = C2H4O

Compound D: C6H12O3 = C2H4O

Compound E: C5H8O2 Compounds B, C and D all have the same empirical formula, C2H4O. The molecular mass of this formula is (2 × 12.01 g/mol C) + (4 × 1.008 g/mol H) + (1 × 16.00 g/mol O) = 44.05 g/mol. 3.117

Plan: Write the balanced chemical equation. Since quantities of two reactants are given, we must determine which is the limiting reactant. To determine which reactant is limiting, calculate the amount of product formed from each reactant, assuming an excess of the other reactant. The reactant that produces less product is the limiting reagent. Use the limiting reactant to determine the theoretical yield of product. The actual yield is given. The actual yield divided by the theoretical yield just calculated (with the result multiplied by 100%) gives the percent yield.

3-74

Solution: Determine the balanced chemical equation: ZrOCl2 8H2O(s) + 4H2C2O4 2H2O(s) + 4KOH(aq)  K2Zr(C2O4)3(H2C2O4) H2O(s) + 2KCl(aq) + 20H2O(l) Determining the limiting reactant: Finding the moles of product from the amount of ZrOCl2 8H2O (if H2C2O4 2H2O is limiting): Moles of product from ZrOCl2 8H2O =  1 mol ZrOCl  8H O   1 mol product  2 2   = 0.00521334 mol product 1.68 g ZrOCl 2  8H 2 O    322.25 g ZrOCl 2  8H 2 O  1 mol ZrOCl 2  8H 2 O 





Finding the moles of product from the amount of H2C2O4 2H2O (if ZrOCl2

8H2O is limiting):

Moles of product from H2C2O4 2H2O =  1 mol H C O  H O   1 mol product  2 2 4 2   = 0.0103117 mol product 5.20 g H 2 C 2 O 4  2H 2 O   126.07 g H 2 C 2 O 4  2H 2 O   4 mol H 2 C 2 O 4  2H 2 O 





It is not necessary to find the moles of product from KOH because KOH is stated to be in excess. The ZrOCl2 8H2O is the limiting reactant, and will be used to calculate the theoretical yield:  541.53 g product    = 2.82318 g product Mass (g) of product = 0.00521334 mol product   1 mol product 





Calculating the percent yield:

 actual Yield   1.25 g   100% =  100% = 44.276 = 44.3% yield Percent yield =    theoretical Yield   2.82318 g  3.118

Plan: Since 85% of ions in seawater are from NaCl, take 85% of the mass percent of dissolved ions (4.0%) to find + – the mass % of NaCl in part (a). To find the mass % of Na and Cl individually in part (b), use the ratio of the mass of the two ions to the mass of NaCl. To find the molarity in part (c), use the mass of NaCl in 100 g of seawater; convert mass of NaCl to moles and mass of seawater to volume in liters, using the density. Molarity = moles of NaCl/L of seawater. Solution:

 85% NaCl    = 3.4% NaCl a) 4.0% ions  100% ions 





 22.99 g Na    +  = 1.3375 = 1.3% Na+ ions b) % Na ions = 3.4% NaCl   58.44 g NaCl 





 35.45 g Cl    = 2.062 = 2.1% Cl– ions % Cl ions = 3.4% NaCl   58.44 g NaCl  –

3.119





a) False, a mole of one substance has the same number of units as a mole of any other substance. b) True c) False, a limiting-reactant problem is present when the quantity of available material is given for more than one reactant. d) False, The empirical and molecular formulas of a compound can be the same, but they are often different (they are different when the formula masses of the molecular and empirical formulas are different).

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3.120

Plan: To convert mass to moles, divide the mass by the molar mass of the substance. To convert moles to mass, divide by the molar mass. To obtain number of particles, multiply moles by Avogadro’s number. Divide a number of particles by Avogadro’s number to obtain moles. Solution: a) Since 1 mole of any substance contains Avogadro’s number of entities, equal amounts of moles of various substances contain equal numbers of entities. The number of entities (O3 molecules) in 0.4 mol of O3 is equal to the number of entities (O atoms) in 0.4 mol of O atoms. b) O3 has a molar mass of 3(16.00 g/mol O) = 48.00 g/mol; O has a molar mass of 1(16.00 g/mol O) = 16.00 g/mol. Since O3 has a larger molar mass than O, 0.4 mol of O3 has a greater mass than 0.4 mol of O.  1 mol N 2 O4    = 0.043 mol N O c) Moles of N2O4 = 4.0 g N2 O4  2 4  92.02 g N2 O4 





 1 mol SO2    = 0.052 mol SO Moles of SO2 = 3.3 g SO2  2  64.06 g SO2 





SO2 is the larger quantity in terms of moles.

 28.05 g C2 H 4    = 17 g C H d) Mass (g) of C2H4 = 0.6 mol C2 H 4  2 4  1 mol C2 H 4 





 38.00 g F2    = 23 g F Mass (g) of F2 = 0.6 mol F2  2  1 mol F2  F2 is the greater quantity in terms of mass. Note that if each of these values is properly rounded to one significant figure, the answers are identical.  2 mol ions   = 4.6 mol ions e) Total moles of ions in 2.3 mol NaClO3 = 2.3 mol NaClO 3  1 mol NaClO 3 





 3 mol ions   = 6.6 mol ions Total moles of ions in 2.2 mol MgCl2 = 2.2 mol MgCl 2  1 mol MgCl 2  MgCl2is the greater quantity in terms of total moles of ions.

f)The compound with the lower molar mass will have more molecules in a given mass. H2O (18.02 g/mol) has a lower molar mass than H2O2 (34.02 g/mol). 1.0 g H2O has more molecules than 1.0 g H2O2.  0.500 mol  g) Moles of NaBr = 0.500 L NaBr   = 0.250 mol NaBr  1L   1 mol Na   + + Moles of Na = 0.250 mol NaBr   = 0.250 mol Na 1 mol NaBr 

103 g  1 mol NaCl   = 0.250 mol NaCl  Moles of NaCl = 0.0146 kg NaCl   1 kg  58.44 g NaCl 





 2 mol ions   +  = 0.250 mol Na+ Moles of Na = 0.250 mol NaCl  1 mol NaCl  The two quantities are equal. 238 h) The heavier atoms, U, will give a greater total mass since there is an equal number of particles of both.



3.121



Plan: Write a balanced equation. The coefficients in the balanced equation give the number of molecules or moles of each reactant and product. Moles are converted to amount in grams by multiplying by the molar masses. Solution: P4S3(s) + 8O2(g)  P4O10(s) + 3SO2(g)

3-76

a) 1 molecule of P4S3 reacts with 8 molecules of O2 to produce 1 molecule of P4O10 and 3 molecules of SO2. b) 1 mol of P4S3 reacts with 8 mol of O2 to produce 1 mol of P4O10 and 3 mol of SO2. c) 220.09 g of P4S3 react with 8(32.00 g/mol O) = 256.00 g of O2 to produce 283.88 g of P4O10 and 3(64.06 g/mol SO2) = 192.18 g of SO2. 3.122

Plan: Write a balanced equation. Use the actual yield (105 kg) and the percent yield (98.8%) to find the theoretical yield of hydrogen. Use the mole ratio between hydrogen and water in the balanced equation to obtain the amount of hydrogen required to produce that theoretical yield of water. Solution: The balanced equation is 2H2(g) + O2(g)  2H2O(g)

 actual Yield   100% % yield =   theoretical Yield  actual yield 105 kg 100 = 100 = 106.2753 kg H2O Theoretical yield (g) of H2O = % yield 98.8% 103 g   1 mol H2 O   2 mol H2   2.016 g H 2   Mass (g) of H2 = 106.2753 kg H2 O       1 kg  18.02 g H 2 O   2 mol H 2 O   1 mol H2 





= 1.18896 × 10 = 1.19 × 10 g H2 3.123

Plan: For part (a), convert 3.0 kg/s to kg/y and multiply by 50 billion years. For part (b), the mass of nitrogen is 75.5% of the mass of the atmosphere. Divide the mass of N2 in grams by its molar mass to obtain moles. Solution:  3.0 kg  60 s   60 min   24 h   365.25 d  9  a) Mass of atmosphere =      50 10 y  1 h   1 d   1 y   s 1min 





= 4.73364 × 10 = 4.7 × 10 kg atmosphere

 103 g  75.5%  = 3.573898 × 1021 g N  b) Mass (g) of N2 = 4.73364 1018 kg atmosphere  2 100% atmosphere  1 kg 





 1 mol N   2  20 20 Moles of N2 = 3.5738981021 g N 2   = 1.27548 × 10 = 1.3 × 10 mol N2  28.02 g N 2 



3.124



Plan: Divide the given mass of a substance by its molar mass to obtain moles; multiply the given moles of a substance by its molar mass to obtain mass in grams. Number of particles is obtained by multiplying an amount in moles by Avogadro’s number. Density is used to convert mass to volume. Solution:

 1 mol NH 4 Br    = 0.0060037 = 0.00600 mol NH Br a) Moles of NH4Br = 0.588 g NH 4 Br  4  97.94 g NH 4 Br 





 1 mol KNO3    = 0.875284 mol KNO b) Moles of KNO3 = 88.5 g KNO3  3 101.11 g KNO3 





23   1 mol K    6.02210 K ions   + Number of K ions = 0.875284 mol KNO3    1 mol KNO3   1 mol K  





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= 5.27096 × 10 = 5.27 × 10 K ions

 92.09 g C3 H8 O3    = 538.7265 = 539 g C H O c) Mass (g) of C3H8O3 = 5.85 mol C3 H8 O3  3 8 3  1 mol C3 H8 O3 





119.37 g CHCl3    = 340.2045 g CHCl d) Mass (g) of CHCl3 = 2.85 mol CHCl3  3  1 mol CHCl3 





  mL   = 229.868 = 230 mL CHCl Volume (mL) of CHCl3 = 340.2045 g CHCl3  3 1.48 g CHCl3 





 2 mol Na    +  = 4.22 mol Na+ e) Moles of Na = 2.11 mol Na 2 CO3  1 mol Na 2 CO3 





 6.0221023 Na  ions   + 24 24 + Number of Na = 4.22 mol Na    = 2.54128 × 10 = 2.54 × 10 Na ions   1 mol Na 





106 g   1 mol Cd   –7 f) Moles of Cd atoms = 25.0   = 2.224199 × 10 mol Cd atoms      1   6.022 1023 Cd atoms    Number of Cd atoms = 2.224199107 mol Cd    1 mol Cd 









= 1.3394126 × 10 = 1.34 × 10 Cd atoms 23  2 mol F   6.022 10 F atoms   g) Number of F atoms = 0.0015 mol F2    1 mol F2  1 mol F 





= 1.8066 × 10 = 1.8 × 10 F atoms 3.125

Neither A nor B has any XY3 molecules. Both C and D have XY3 molecules. D shows both XY3 and XY molecules. Only C has a single XY3 product, thus the answer is C.

3.126

Plan: Deal with the methane and propane separately, and combine the results. Balanced equations are needed for each hydrocarbon. The total mass and the percentages will give the mass of each hydrocarbon. The mass of each hydrocarbon is changed to moles, and through the balanced chemical equation the amount of CO2 produced by each gas may be found. Summing the amounts of CO2 gives the total from the mixture. For part (b), let x and 252 – x represent the masses of CH4 and C3H8, respectively. Solution: a) The balanced chemical equations are: Methane:

CH4(g) + 2O2(g)  CO2(g) + 2H2O(l)

Propane:

C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(l)

Mass (g) of CO2 from each:

 25.0%  1 mol CH 4  1 mol CO2   44.01 g CO2  Methane: 200.g Mixture     = 137.188 g CO2   100% 16.04 g CH 4  1 mol CH 4   1 mol CO2 





 75.0%  1 mol C3 H8   3 mol CO2   44.01 g CO2  Propane: 200.g Mixture     = 449.183 g CO2   100%  44.09 g C3 H8  1 mol C3 H8   1 mol CO2 





Total CO2 = 137.188 g + 449.183 g = 586.371 = 586 g CO2 Copyright McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

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b) Since the mass of CH4 + the mass of C3H8 = 252 g, let x = mass of CH4 in the mixture and 252 – x = mass of C3H8 in the mixture. Use mole ratios to calculate the amount of CO2 formed from x amount of CH4 and the amount of CO2 formed from 252 – x amount of C3H8. The total mass of CO2 produced = 748 g.  1 mol CO 2   = 16.996 mol CO2 The total moles of CO2 produced = 748 g CO 2   44.01 g CO 2   1 mol CH 4   3 mol CO 2  1 mol CO 2  + 252  x g C H  1 mol C 3 H 8  16.996 mol CO2 = x g CH 4  3 8       1 mol C 3 H 8  16.04 g CH 4 1 mol CH 4   44.09 g C 3 H 8 

x 3(252  x) mol CO2 + mol CO2 16.04 44.09 x 756  3x mol CO2 + mol CO2 16.996 mol CO2 = 16.04 44.09 16.996 mol CO2 = 0.06234x mol CO2 + (17.147 – 0.06804x mol CO2) = 17.147 – 0.0057x x = 26.49 g CH4 252 – x = 252 g – 26.49 g = 225.51 g C3H8 16.996 mol CO2 =

3.127

Mass % CH4 =

mass of CH 4 26.49 g CH 4 100 = 100 = 10.5% CH4 mass of mixture 252 g mixture

Mass % C3H8 =

mass of C3 H 8 225.51 g C3 H8 100 = 100 = 89.5% C3H8 mass of mixture 252 g mixture

Plan: If we assume a 100-gram sample of fertilizer, then the 30:10:10 percentages become the masses, in grams, of N, P2O5, and K2O. These masses may be changed to moles of substance, and then to moles of each element. To get the desired x:y:1.0 ratio, divide the moles of each element by the moles of potassium. Solution: A 100-gram sample of 30:10:10 fertilizer contains 30 g N, 10 g P2O5, and 10 g K2O.

 1 mol N    = 2.1413 mol N Moles of N = 30 g N  14.01 g N 

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

 1 mol P2 O5   2 mol P   Moles of P = 10 g P2 O5    = 0.14090 mol P 1 mol P2 O5  141.94 g P2 O5 

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

 1 mol K 2 O   2 mol K   Moles of K = 10 g K 2 O    = 0.21231 mol K  94.20 g K 2 O  1 mol K 2 O 

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This gives a N:P:K ratio of 2.1413:0.14090:0.21231 The ratio must be divided by the moles of K and rounded.

2.1413 mol N = 10.086 0.21231 10.086:0.66365:1.000

0.14090 mol P = 0.66365 0.21231 or

0.21231 mol K =1 0.21231

10:0.66:1.0

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3.128

Plan: If we assume a 100-gram sample of fertilizer, then the 10:10:10 percentages become the masses, in grams, of N, P2O5, and K2O. These masses may be changed to moles of substance, and then to moles of each element. Use the mole ratio between N and ammonium sulfate, P and ammonium hydrogen phsophate, and K and potassium chloride to find the mass of each compound required to provide the needed amount of the respective element. Divide the mass of each compound by the total mass of sample, 100 g, and multiply by 100 for mass %. Solution: Assume a 100 g sample. 10:10:10 indicates 10 g N, 10 g P2O5 and 10 g K2O.  1 mol N    = 0.713776 mol N Moles of N = 10 g N  14.01 g N 

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 1 mol P2 O5  2 mol P     Moles of P = 10 g P2 O5  1 mol P O  = 0.14090 mol P 141.94 g P2 O5  2 5

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 1 mol K 2 O   2 mol K   Moles of K = 10 g K 2 O    = 0.21231 mol K  94.20 g K 2 O  1 mol K 2 O 

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To obtain 0.713776 mol N from (NH4)2SO4: 1 mol (NH 4 )2 SO 4 132.14 g (NH 4 )2 SO 4   = 47.1592 g (NH4)2SO4 0.713776 mol N     2 mol N  1 mol (NH 4 ) 2 SO 4 

Mass % (NH4)2SO4 =

mass of (NH 4 )2 SO 4 47.1592 g (NH 4 )2 SO 4 100 = 100 mass of mixture 100 g mixture

= 47.1592% = 47.2% (NH4)2SO4 To obtain 0.14090 mol P from (NH4)2HPO4: 1 mol (NH 4 ) 2 HP O 4 132.06 g (NH 4 )2 HP O 4   = 18.6073 g (NH4)2HPO4 0.14090 mol P     1 mol P  1 mol (NH 4 ) 2 HP O 4  Mass % (NH4)2HPO4 =

mass of (NH 4 )2 HPO 4 18.6073 g (NH 4 )2 HPO 4 100 = 100 mass of mixture 100 g mixture

= 18.6073% = 18.6% (NH4)2HPO4 To obtain 0.21231 mol K from KCl:

1 mol KCl  74.55 g KCl  0.21231 mol K  = 15.8277 g KCl   1 mol K  1 mol KCl  mass of KCl 15.8277 g KCl 100 = 100 = 15.8277% = 15.8% KCl Mass % KCl= mass of mixture 100 g mixture 3.129

2.233 g

0.457 g

3-80

 1 mol CO2   1 mol C   Moles of C = 2.233 g CO2    = 0.050738 mol C  44.01 g CO2  1 mol CO2 

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12.01 g C    = 0.60936 g C Mass (g) of C = 0.050738 mol C   1 mol C 

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 1 mol H2 O   2 mol H   Moles of H = 0.457 g H 2 O    = 0.050721 mol H 18.02 g H 2 O  1 mol H 2 O 

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1.008 g H    = 0.051127 g H Mass (g) of H = 0.050721 mol H   1 mol H 

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Mass (g) of Fe = Sample mass – (mass of C + mass of H) = 0.9437 g – (0.60936 g C + 0.052217 g H) = 0.283213 g Fe

 1 mol Fe    = 0.005071 mol Fe Moles of Fe = 0.283213 g Fe   55.85 g Fe 

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Preliminary formula = C0.050738H0.050721Fe0.005071 Converting to integer subscripts (dividing all by the smallest subscript): C 0.050738 H 0.050721 Fe 0.005071 0.005071

0.005071

H10Fe1

0.005071

Empirical formula = C10H10Fe 3.130

Plan: Assume 100 grams of mixture. This means the mass of each compound, in grams, is the same as its percentage. Find the mass of C from CO and from CO2 and add these masses together. For mass %, divide the total mass of C by the mass of the mixture (100 g) and multiply by 100. Solution: 100 g of mixture = 35 g CO and 65 g CO2.  1 mol CO   1 mol C  12.01 g C   Mass (g) of C from CO = 35.0 g CO     = 15.007 g C 1 mol CO   1 mol C   28.01 g CO 

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 1 mol CO2   1 mol C  12.01 g C   Mass (g) of C from CO2 = 65.0 g CO2     1 mol C  = 17.738 g C  44.01 g CO2  1 mol CO2  

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

Total mass (g) of C = 15.007 g + 17.738 g = 32.745 g C mass of C 32.745 g C 100 = 100 = 32.745 = 32.7% C Mass % C = mass of mixture 100 g mixture 3.131

Plan: Write a balanced equation for the reaction. Count the molecules of each reactant to obtain the moles of each reactant present. Use the mole ratios in the equation to calculate the amount of product formed. Only 87.0% of the calculated amount of product actually forms, so the actual yield is 87.0% of the theoretical yield. Solution: The balanced equation is SiH4 + N2F4

+ N2 + 2H2.

1.25102 mol   = 0.0375 mol SiH4 Moles of SiH4 = 3 SiH4 molecules  1 molecule  1.25102 mol   = 0.0375 mol N2F4 Moles of N2F4 = 3 N2 F4 molecules  1 molecule 

3-81

Since there is an equal amount of each reactant and the ratio between each reactant and SiF 4 is 1:1, neither reactant is in excess and either may by used to calculate the amount of SiF4 produced.  1 mol SiF4  104.09 g SiF4  = 3.903375 g SiF Mass (g) of SiF4 = 0.0375 mol SiH 4  4  1 mol SiF  1 mol SiH   4

 actual Yield   100% % yield =   theoretical Yield  % yield 87% theoretical yield = Actual yield (g) of SiF4 = 3.903375 g SiF4  = 3.3959 = 3.4 g SiF4 100% 100% 3.132

Plan: Determine the molecular formula from the figure. Once the molecular formula is known, use the periodic table to determine the molar mass. Convert the volume of lemon juice in part (b) from qt to mL and use the density to convert from mL to mass in g. Take 6.82% of that mass to find the mass of citric acid and use the molar mass to convert to moles. Solution: a) The formula of citric acid obtained by counting the number of carbon atoms, oxygen atoms, and hydrogen atoms is C6H8O7. Molar mass = (6 × 12.01 g/mol C) + (8 × 1.008 g/mol H) + (7 × 16.00 g/mol O) = 192.12 g/mol b) Converting volume of lemon juice in qt to mL:

 1 L   1 mL   Volume (mL) of lemon juice = 1.50 qt   3  = 1419.111 mL 1.057 qt  10 L  Converting volume to mass in grams: 1.09 g   = 1546.831 g lemon juice Mass (g) of lemon juice = 1419.111 mL   mL 

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

 6.82% C 6 H 8 O 7   = 105.494 g C6H8O7 Mass (g) of C6H8O7= 1546.831 g lemon juice  100% lemon juice 

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

 1 mol C6 H8 O7    = 0.549104 = 0.549 mol C H O Moles of C6H8O7 = 105.494 g C6 H8 O7  6 8 7 192.12 g C6 H8 O7 

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3.133



Plan: Determine the formulas of each reactant and product, then balance the individual equations. Remember that nitrogen and oxygen are diatomic. Combine the three smaller equations to give the overall equation, where some substances serve as intermediates and will cancel. Use the mole ratio between nitrogen and nitric acid in the overall equation to find the moles and then mass of nitric acid produced. The amount of nitrogen in metric tons must be converted to mass in grams to convert the mass of nitrogen to moles. Solution: a) Nitrogen and oxygen combine to form nitrogen monoxide: N2(g) + O2(g)  2NO(g) Nitrogen monoxide reacts with oxygen to form nitrogen dioxide: 2NO(g) + O2(g)  2NO2(g) Nitrogen dioxide combines with water to form nitric acid and nitrogen monoxide: 3NO2(g) + H2O(g)  2HNO3(aq) + NO(g)

3-82

b) Combining the reactions may involve adjusting the equations in various ways to cancel out as many materials as possible other than the reactants added and the desired products. 2 × (N2(g) + O2(g)  2NO(g)) = 2N2(g) + 2O2(g)  4NO(g) 3 × (2NO(g) + O2(g)  2NO2(g)) = 6NO(g) + 3O2(g)  6NO2(g) 2 × (3NO2(g) + H2O(g)  2HNO3(aq) + NO(g)) = 6NO2(g) + 2H2O(g)  4HNO3(aq) + 2NO(g) Multiplying the above equations as shown results in the 6 moles of NO on each side and the 6 moles of NO2 on each side canceling. Adding the equations gives: 2N2(g) + 5O2(g) + 2H2O(g)  4HNO3(aq) 3 103 kg  10 g   9 c) Mass (g) of N2 = 1350 t N 2    = 1.35 × 10 g N2  1t   1 kg   1 mol N2    = 4.817987 × 107 mol N Moles of N2 = 1.35x109 g N 2  2  28.02 g N 2   4 mol HNO  63.02 g HNO3   3   = 6.072591 × 109 g HNO Mass (g) of HNO3 = 4.817987107 mol N 2   3  2 mol N 2   1 mol HNO3  Metric tons HNO3 =  1 kg   1t   3 3 6.072591109 g HNO3  3  3  = 6.072591 × 10 = 6.07 × 10 metric tons HNO3 10 g  10 kg 

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3.134

Plan: Write and balance the chemical reaction. Use the mole ratio to find the amount of product that should be produced and take 66% of that amount to obtain the actual yield. Solution: 2NO(g) + O2(g) (g) 2 With 6 molecules of NO and 3 molecules of O2 reacting, 6 molecules of NO2 can be produced. If the reaction only has a 66% yield, then (0.66)(6) = 4 molecules of NO2 will be produced. Circle A shows the formation of 4 molecules of NO2. Circle B also shows the formation of 4 molecules of NO2 but also has 2 unreacted molecules of NO and 1 unreacted molecule of O2. Since neither reactant is limiting, there will be no unreacted reactant remaining after the reaction is over.

3.135

Plan: For parts a) and b), convert the masses to moles. Take the moles and divide by the smallest value to convert to whole numbers to get the empirical formula. For part (c), write the two balanced equations and use two equations as shown. Solution:  1 mol Pt   = 0.001676 mol Pt a) Moles of Pt = 0.327 g Pt  195.1 g Pt  Mass (g) of F = mass of product – mass of Pt = 0.519 g – 0.327 g = 0.192 g F  1 mol F   = 0.010105 mol F Moles of F = 0.192 g F  19.00 g F  Preliminary formula = Pt0.001676F0.010105 Converting to integer subscripts (dividing all by the smallest subscript): Pt 0.001676 F0.0010105 0.001676

0.001676

Empirical formula = PtF6  1 mol PtF6   = 0.0008576 mol PtF6 b) Moles of PtF6 = 0.265 g PtF6   309.1 g PtF6  Copyright McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

3-83

Mass of Xe = mass of product – mass of Xe = 0.378 g – 0.265 g = 0.113 g Xe  1 mol Xe   = 0.0008606 mol Xe Moles of Xe = 0.113 g Xe 131.3 g Xe  Preliminary formula = Xe0.0008606(PtF6)0.0008576 Converting to integer subscripts (dividing all by the smallest subscript):

Xe 0.0008606 PtF6 0.0008576

0.0008576

(PtF6)1

Empirical formula = XePtF6 c) This problem can be solved as a system of two equations and two unknowns. The two equations are:

The two unknowns are:

Xe(g) + 2F2(g)  XeF4(s)

x = mol XeF4 produced

Xe(g) + 3F2(g)  XeF6(s)

y = mol XeF6 produced –4

–6

–4

Moles of Xe consumed = 1.85 × 10 mol present – 9.00 × 10 mol excess = 1.76 × 10 mol Xe Then

–4

x + y = 1.76 × 10 mol Xe consumed –4

2x + 3y = 5.00 × 10 mol F2 consumed Solve for x using the first equation and substitute the value of x into the second equation: –4

x = 1.76 × 10 – y –4

2(1.76 × 10 – y) + 3y = 5.00 × 10 –4

(3.52 × 10 ) – 2y + 3y = 5.00 × 10 –4

–4

–4 –4 –4

y = (5.00 × 10 ) – (3.52 × 10 ) = 1.48 × 10 mol XeF6 –5

x = (1.76 × 10 ) – (1.48 × 10 ) = 2.8 × 10 mol XeF4 Converting moles of each product to grams using the molar masses: Mass (g) of XeF4 = 2.810

5

Mass (g) of XeF6 = 1.4810

 207.3 g XeF  –3 4  mol XeF4   = 5.8044 × 10 g XeF4  1 mol XeF4 

4

 245.3 g XeF  –2 6 mol XeF6   = 3.63044 × 10 g XeF6  1 mol XeF6 

Calculate the percent of each compound using the total weight of the products: –3

–2

(5.8044 × 10 + 3.63044 × 10 ) g = 0.0421088 g Mass % XeF4 = Mass % XeF6 = 3.136

mass of XeF4 total mass mass of XeF6 total mass

100 = 100 =

5.8044 103 g XeF4 0.0421088 g

100 = 13.784 = 14% XeF4

3.63044 102 g XeF6 0.0421088 g

100 = 86.2157 = 86.2% XeF6

Plan: Use the mass percent to find the mass of heme in the sample; use the molar mass to convert the mass of heme to moles. Then find the mass of Fe in the sample by using the mole ratio between heme and iron. The mass of hemin is found by using the mole ratio between heme and hemoglobin. Solution:

 6.0% heme    = 0.039 g heme a) Mass (g) of heme = 0.65 g hemoglobin  100% hemoglobin 

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 1 mol heme    = 6.32614 × 10–5 = 6.3 × 10–5 mol heme b) Moles of heme = 0.039 g heme   616.49 g heme 

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3-84

 1 mol Fe  55.85 g Fe  c) Mass (g) of Fe = 6.32614 105 mol heme   1 mol heme  1 mol Fe  –3 –3 = 3.5331 × 10 = 3.5 × 10 g Fe 1 mol hemin   651.94 g hemin  d) Mass (g) of hemin = 6.32614 105 mol heme  1 mol hemin   1 mol heme  –2 –2 = 4.1243 × 10 = 4.1 × 10 g hemin 3.137

Plan: Find the Mn:O ratio in the two oxides. Write two equations to solve simultaneously; one equation shows that the sum of the ratio of Mn in the two oxides will equal the ratio of Mn in the sample and the other equation shows that the total amount of oxide in the sample is the sum of the amounts of the two oxides. The two equations will give the mole ratio of the two oxides. Convert moles of each oxide to mass to obtain the mass ratio of the two oxides from which the mass % of each can be calculated. Use that mass % of each to find the mass of each in the sample. For 3+ 2+ part (b), the moles of Mn come from the Mn2O3 and the moles of Mn come from the MnO. Solution: Mn:O ratio: In sample: 1.00:1.42 or 0.704 In braunite: 2.00:3.00 or 0.667 In manganosite: 1.00:1.00 or 1.00 a) The total amount of ore is equal to the amount of braunite (B) + the amount of manganosite (M). B + M = 1.00 M = 1.00 – B The amount of Mn is dependent on the sample’s composition. M(1.00) + B(0.667) = 0.704 (1.00 – B)(1.00) + B(0.667) = 0.704 1.00 – 1.00B + 0.667B = 0.704 0.296 = 0.333B B = 0.888889 mol braunite M = 1.00 – B = 1.00 = 0.888889 = 0.111111 mol manganosite 157.88 g  Mass (g) of braunite = 0.888889 mol  = 140.338 g braunite  1 mol   70.94 g  Mass (g) of manganosite = 0.111111 mol = 7.88221 g manganosite  1 mol  There are 140.338 g of braunite for every 7.88221 g of manganosite. Finding mass % of each: mass of braunite 140.338 g 100 = 100 = 94.6821% Mass % braunite = mass of braunite + manganosite 140.338 + 7.88221 g

mass of manganosite 7.88221 g 100 = 100 mass of braunite + manganosite 140.338 + 7.88221 g = 5.3179% In the 542.3 g sample:  94.6821 braunite   = 513.461 = 513 g braunite Mass (g) of braunite = 542.3 g sample  100% sample  Mass % manganosite =

 5.3179% manganosite   = 28.839 = 28.8 g manganosite Mass (g) of manganosite = 542.3 g sample   100% sample  3+ b) Each mole of braunite, Mn2O3, contains 2 moles of Mn while each mole of manganosite, MnO, contains 1 2+ mole of Mn .

3-85

Moles of Mn = 2(0.888889 mol braunite) = 1.777778 mol Mn 2+ 2+ Moles of Mn = 1(0.111111 mol manganosite) = 0.111111 mol Mn 3 1.777778 mol Mn 3+ 2+ Mn :Mn = = 16.000 = 16.0 0.111111 mol Mn 2 3.138

Plan: Determine the formula and the molar mass of each compound. The formula gives the relative number of moles of nitrogen present. Multiply the number of moles of nitrogen by its molar mass to find the total mass of total mass of element 100. For part (b), convert mass of nitrogen in 1 mole of compound. Mass percent = molar mass of compound ornithine to moles, use the mole ratio between ornithine and urea to find the moles of urea, and then use the ratio between moles of urea and nitrogen to find the moles and mass of nitrogen produced. Solution: a) Urea: CH4N2O, = 60.06 g/mol There are 2 moles of N in 1 mole of CH4N2O. 14.01 g N  Mass (g) of N = 2 mol N = 28.02 g N  1 mol N  total mass N 28.02 g N 100 = 100 = 46.6533 = 46.65% N in urea Mass percent = molar mass of compound 60.06 g CH4 N2 O Arginine: C6H15N4O2, = 175.22 g/mol There are 4 moles of N in 1 mole of C6H15N4O2. 14.01 g N  Mass (g) of N = 4 mol N = 56.04 g N  1 mol N  total mass N 56.04 g N 100 = 100 Mass percent = molar mass of compound 175.22 g C6 H15 N4 O2 = 31.98265 = 31.98% N in arginine Ornithine: C5H13N2O2, = 133.17 g/mol There are 2 moles of N in 1 mole of C5H13N2O2. 14.01 g N  Mass (g) of N = 2 mol N = 28.02 g N  1 mol N  total mass N 28.02 g N 100 = 100 Mass percent = molar mass of compound 133.17 g C5 H13 N2 O2 = 21.04077 = 21.04% N in ornithine

 1 mol C5 H13 N 2 O2   1 mol CH 4 N 2 O   b) Moles of urea = 135.2 g C5 H13 N 2 O2    = 1.015244 mol urea 1 mol C 5 H13 N 2 O2  133.17 g C 5 H13 N 2 O2 





 2 mol N  14.01 g N   Mass (g) of nitrogen = 1.015244 mol CH 4 N 2 O   1 mol N  = 28.447 = 28.45 g N 1 mol CH 4 N 2 O  



3.139



Plan: To determine which reactant is limiting, calculate the amount of aspirin formed from each reactant, assuming an excess of the other reactant. Use the density of acetic anhydride to determine the amount of this reactant in grams. The reactant that produces less product is the limiting reagent. Use the limiting reagent and the mole ratio from the balanced chemical equation to determine the theoretical yield of aspirin. The actual yield divided by the theoretical yield just calculated (with the result multiplied by 100%) gives the percent yield. Use the formula for percent atom economy to determine that quantity.

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Solution: a) Finding the moles of aspirin from the moles of C7H6O3 (if (CH3CO)2O is limiting):  1 mol C 7 H6 O3  1 mol C 9 H8 O4   Moles of aspirin from C7H6O3 = 3.077 g C 7 H6 O3    138.12 g C 7 H6 O3  1 mol C 7 H6 O3 





= 0.0222777 mol C9H8O4 Finding the moles of aspirin from the moles of C4H6O3 (if C7H6O3 is limiting): 1.080 g   = 5.94 g (CH3CO)2O Mass (g) of (CH3CO)2O = 5.50 mL (CH 3 CO)2 O   1 mL 

 1 mol (CH3 CO)2 O   1 mol C9 H8 O 4   Moles of aspirin from (CH3CO)2O = 5.94 g (CH3 CO)2 O    102.09 g (CH3 CO)2 O  1 mol (CH3 CO)2 O 





= 0.058183955 mol C9H8O4 The limiting reactant is C7H6O3. b) First, calculate the theoretical yield from the limiting reagent:

 1 mol C 7 H6 O3  1 mol C9 H8 O4  180.15 g C9 H8 O4   Mass (g) of C9H8O4 = 3.077 g C 7 H6 O3     138.12 g C 7 H6 O3  1 mol C 7 H6 O3   1 mol C9 H8 O 4 





= 4.01333 g C9H8O4

 actual Yield   3.281 g   100% =  100% = 81.7526 = 81.75% yield Percent yield =    theoretical Yield   4.01333 g  no. of moles x molar mass of desired products c) % atom economy = 100% sum of no. of moles x molar mass for all products



% atom economy =

3.140



1mol180.15 g / mol 1 mol180.15 g/mol + 1 mol60.05 g/mol

100% = 75.00% atom economy

Plan: Determine the molar mass of each product and use the equation for percent atom economy. Solution: Molar masses of product: N2H4: 32.05 g/mol NaCl: 58.44 g/mol H2O: 18.02 g/mol no. of moles molar mass of desired products % atom economy = 100% sum of no. of moles  molar mass for all products





% atom economy =

  100% 1 molmolar mass of N H  + 1 molmolar mass of NaCl + 1 molmolar mass of H O 1 mol molar mass of N 2 H 4

1mol32.05 g / mol 1 mol32.05 g/mol + 1 mol58.44 g/mol + 1 mol18.02 g/mol

100%

= 29.5364 = 29.54% atom economy 3.141

Plan: Use the mass percent to obtain the mass of lead(II) chromate needed. Use the mole ratio between PbCrO4 and K2CrO4 and then the mole ratio between K2CrO4 and FeCr2O4 to find the moles and then mass of FeCr2O4

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needed to produce that mass of PbCrO4. Solution:

103 g  0.511%     = 5.11 g PbCrO4 Mass (g) of PbCrO4 in 1 kg of paint = 1 kg paint   kg  100%   1 mol PbCrO4  1 mol K 2 CrO4   Moles of K2CrO4 = 5.11 g PbCrO4    = 0.015811 mol K2CrO4  323.2 g PbCrO4  1 mol PbCrO4 





 4 mol FeCr2 O 4   223.85 g FeCr2 O 4   Mass of FeCr2O4 = 0.015811 mol K 2 CrO4     1 mol FeCr2 O4   8 mol K 2 CrO4  = 1.769606 = 1.77 g FeCr2O4



3.142



Plan: Convert the mass of ethanol to moles, and use the mole ratio between ethanol and diethyl ether to determine the theoretical yield of diethyl ether. The actual yield divided by the theoretical yield just calculated (with the result multiplied by 100%) gives the percent yield. The difference between the actual and theoretical yields is related to the quantity of ethanol that did not produce diethyl ether, forty-five percent of which produces ethylene instead. Use the mole ratio between ethanol and ethylene to find the mass of ethylene produced by the forty-five percent of ethanol that did not produce diethyl ether. Solution: a) The determination of the theoretical yield: Mass (g) of diethyl ether =

 1 mol CH3 CH 2 OH 1 mol CH3 CH2 OCH 2 CH3   74.12 g CH3 CH2 OCH 2 CH3     1 mol CH3 CH 2 OCH2 CH3  3 2   2 mol CH3 CH 2 OH  = 40.2214 g diethyl ether Determining the percent yield:  actual Yield   35.9 g   100% =  100% = 89.2560 = 89.3% yield Percent yield =    theoretical Yield   40.2214 g  b) To determine the amount of ethanol not producing diethyl ether, we will use the difference between the theoretical yield and actual yield to determine the amount of diethyl ether that did not form and hence, the amount of ethanol that did not produce the desired product. Forty-five percent of this amount will be used to determine the amount of ethylene formed.

50.0 g CH CH OH 46.07 g CH CH OH  3

Mass difference = theoretical yield – actual yield = 40.2214 g – 35.9 g = 4.3214 g diethyl ether that did not form Mass (g) of ethanol not producing diethyl ether =

 1 mol CH3 CH 2 OCH 2 CH3 

 46.07 g CH3 CH 2 OH 

2 mol CH3 CH 2 OH

4.3214 g CH CH OCH CH  74.12 g CH CH OCH CH 1 mol CH CH OCH CH  1 mol CH CH OH  3







= 5.37202 g ethanol

 45.0%  Mass of ethanol producing ethylene = 5.37202 g CH3 CH2 OH  = 2.417409 g ethanol  100%  Mass (g) of ethylene =





1 mol C2 H 4

 28.05 g C2 H 4 

1 mol CH CH OH       2.417409 g CH CH OH 46.07 1 mol CH CH OH  1 mol C H  g CH CH OH  3







= 1.47185 = 1.47 g ethylene

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3.143

Plan: First balance the given chemical equation. To determine which reactant is limiting, calculate the amount of ZnS formed from each reactant, assuming an excess of the other reactant. The reactant that produces less product is the limiting reagent. Use the limiting reagent and the mole ratio from the balanced chemical equation to determine the theoretical yield of ZnS. The actual yield divided by the theoretical yield just calculated (with the result multiplied by 100%) gives the percent yield. For part (b), determine the mass of Zn that does not produce ZnS; use that amount of zinc and the mole ratio between Zn and ZnO in that reaction to determine the mass of ZnO produced. Find the moles of S8 in the reactant and the moles of S8 in the product ZnS. The difference between these two amounts is the moles of S8 in SO2. Solution: a) The balanced equation is 8Zn(s) + S8(s)  8ZnS(s). Finding the limiting reagent: Finding the moles of ZnS from the moles of Zn (if S8 is limiting):  1 mol Zn  8 mol ZnS   = 1.27256 mol ZnS  Moles of ZnS from Zn = 83.2 g Zn   65.38 g Zn  8 mol Zn  Finding the moles of ZnS from the moles of S8 (if Zn is limiting):

 1 mol S8   8 mol ZnS   Moles of ZnS from S8 = 52.4 g S8    = 1.6344 mol ZnS  256.48 g S8   1 mol S8 





The zinc will produce less zinc sulfide, thus, zinc is the limiting reactant and will first be used to determine the theoretical yield and then the percent yield.  1 mol Zn  8 mol ZnS  97.44 g ZnS     Theoretical yield (g) of ZnS = 83.2 g Zn   1 mol ZnS   65.38 g Zn  8 mol Zn  = 123.9983 g ZnS (unrounded)  actual Yield   100% =  104.4 g 100% = 84.1947 = 84.2% yield Percent yield =  123.9983 g   theoretical Yield  b) The reactions with oxygen are: 2Zn(s) + O2(g)  2ZnO(s) S8(s) + 8O2(g)  8SO2(g) The theoretical yield indicates that 84.2% of the zinc produced zinc sulfide so (100 – 84.2)%= 15.8% of the zinc became zinc oxide. This allows the calculation of the amount of zinc oxide formed. 15.8%  Mass (g) of Zn that does not produce ZnS = 83.2 g Zn  = 13.1456 g ZnS  100%   1 mol Zn  2 mol ZnO  81.38 g ZnO     Mass (g) of ZnO = 13.1456 g Zn   1 mol ZnO  = 16.3626 = 16.4 g ZnO  65.38 g Zn  2 mol Zn  The calculation is slightly different for the sulfur. We need to determine the amount of sulfur not in zinc sulfide. The sulfur not in the zinc sulfide must be in sulfur dioxide. The amount of sulfur not in zinc sulfide will be converted to the mass of sulfur dioxide.  1 mol S8   = 0.204304 mol S8 Moles of S8 in original S8 reactant = 52.4 g S8   256.48 g S8   1 mol ZnS  1 mol S8   Moles of S8 in ZnS product = 104.4 g ZnS  = 0.133928 mol S8  97.44 g ZnS  8 mol ZnS 

Moles of S8 in SO2 = 0.204304 – 0.133928 mL = 0.070376 mol S8  8 mol SO 2   64.06 g SO 2  = 36.0663 = 36.1 g SO Mass (g) of SO2 = 0.070376 mol S8  2  1 mol S8  1 mol SO 2  Copyright McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

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3.144

Plan: For part (a), use the given solubility of the salt to find the mass that is soluble in the given volume of water. For part (b), convert the mass of dissolved salt in part (a) to moles of salt and then to moles of cocaine and then to mass of cocaine. Use the solubility of cocaine to find the volume of water needed to dissolve this mass of cocaine. Solution:

103 L H2 O  2.50 kg salt 103 g      = 125 g salt a) Mass (g) of dissolved salt = 50.0 mL H 2 O   1 mL H 2 O  1 L H 2 O  1 kg 





 1 mol salt    = 0.367853 mol salt b) Moles of dissolved salt = 125 g salt   339.81 g salt 





1 mol cocaine   303.35 g cocaine   Mass (g) cocaine = 0.367853 mol salt    = 111.588 g cocaine  1 mol salt   1 mol cocaine 





  1L   = 65.64 L Volume (L) of water needed to dissolve the cocaine = 111.588 g cocaine  1.70 g cocaine 





Additional water needed = total volume needed – original volume of water = 65.64 L – 0.0500 L = 65.59 = 65.6 L H2O 3.145

Plan: Use the given values of x to find the molar mass of each compound. To determine which reactant is limiting, calculate the amount of either product formed from each reactant, assuming an excess of the other reactants. The reactant that produces the smallest amount of product is the limiting reagent. To find the mass of excess reactants, find the mass of each excess reactant that is required to react with the limiting reagent and subtract that mass from the starting mass. a) x = 0 La2Sr0CuO4 = 2(138.9 g/mol La) + 0(87.62 g/mol Sr) + 1(63.55 g/mol Cu) + 4(16.00 g/mol O) = 405.4 g/mol x=1 La1Sr1CuO4 = 1(138.9 g/mol La) + 1(87.62 g/mol Sr) + 1(63.55 g/mol Cu) + 4(16.00 g/mol O) = 354.1 g/mol x = 0.163 La(2–0.163)Sr0.163CuO4 = La1.837Sr0.163CuO4 = 1.837(138.9g/mol La) + 0.163 (87.62 g/mol Sr) + 1(63.55g/mol Cu) + 4(16.00g/mol O) = 397.0 g/mol b) Assuming x grams to be the “equal” mass leads to:  1 mol BaCO3   2 mol YBa 2 Cu3 O7   Moles of product from BaCO3 = x g BaCO3     4 mol BaCO3  197.3 g BaCO3  = 0.002534x mol product  1 mol CuO   2 mol YBa 2 Cu3 O7   Moles of product from CuO = x g CuO    = 0.004190x mol product  79.55 g CuO  6 mol CuO  





 1 mol Y2 O3   2 mol YBa 2 Cu3 O7   Moles of product from Y2O3 = x g Y2 O3    = 0.008857x mol product  225.82 g Y2 O3   1 mol Y2 O3 





BaCO3 is the limiting reactant. c) These calculations are based on the limiting reactant. BaCO3 remaining = 0% (limiting reagent)

3-90

 1 mol BaCO3   6 mol CuO   79.55 g CuO   CuO remaining = x g CuO – x g BaCO3     197.3 g BaCO3   4 mol BaCO3   1 mol CuO  = 0.39521x g CuO  0.39521 x g   100% = 39.521 = 39.52% CuO remaining Percent CuO =    xg 





 1 mol BaCO3   1 mol Y2 O3   225.82 g Y2 O3   Y2O3 remaining = x g Y2O3 – x g BaCO3     197.3 g BaCO3   4 mol BaCO3   1 mol Y2 O3 





= 0.713862x g Y2O3  0.713862 x g   100% = 71.3862 = 71.39% Y O remaining Percent Y2O3 =  2 3  xg  

3-91

CHAPTER 4 THREE MAJOR CLASSES OF CHEMICAL REACTIONS FOLLOW–UP PROBLEMS 4.1A

Plan: Examine each compound to see what ions, and how many of each, result when the compound is dissolved in water and match one compound’s ions to those in the beaker. Use the total moles of particles and the molar ratio in the compound’s formula to find moles and then mass of compound. Solution: H2 O a) LiBr(s)   Li (aq) + Br (aq) +

–

H2 O Cs2CO3(s)   2Cs (aq) + CO3 (aq) +

2–

H2 O BaCl2(s)   Ba (aq) + 2Cl (aq) Since the beaker contains +2 ions and twice as many –1 ions, the electrolyte is BaCl2.  0.05 mol Ba 2 ions  1 mol BaCl 2  208.2 g BaCl 2    b) Mass (g) of BaCl2 = 3 Ba 2 particles  1 mol BaCl   1 Ba 2  particle 1 mol Ba 2 ions  2 2+

–

= 31.23 = 31.2 g BaCl2 4.1B

Plan: Write the formula for sodium phosphate and then write a balanced equation showing the ions that result when sodium phosphate is placed in water. Use the balanced equation to determine the number of ions that result when 2 formula units of sodium phosphate are placed in water. The molar ratio from the balanced equation gives the relationship between the moles of sodium phosphate and the moles of ions produced. Use this molar ratio to calculate the moles of ions produced when 0.40 mol of sodium phosphate is placed in water. Solution: a) The formula for sodium phosphate is Na3PO4. When the compound is placed in water, 4 ions are produced for + 3– each formula unit of sodium phosphate: three sodium ions, Na , and 1 phosphate ion, PO4 . + 3– H2 O Na3PO4(s)   3Na (aq) + PO4 (aq) If two formula units of sodium phosphate are placed in water, twice as many ions should be produced: H2 O 2Na3PO4(s)   6Na (aq) + 2PO4 (aq) +

3–

Any drawing should include 2 phosphate ions (each with a 3– charge) and 6 sodium ions (each with a 1+ charge).

 b) Moles of ions = 0.40 mol Na3PO4   4.2A

  = 1.6 moles of ions   4

Plan: Write an equation showing the dissociation of one mole of compound into its ions. Use the given information to find the moles of compound; use the molar ratio between moles of compound and moles of ions in the dissociation equation to find moles of ions. Solution: a) One mole of KClO4 dissociates to form one mole of potassium ions and one mole of perchlorate ions. + – H2O KClO4(s)   K (aq) + ClO4 (aq) + – Therefore, 2 moles of solid KClO4 produce 2 mol of K ions and 2 mol of ClO4 ions. 2+ – H2O b) Mg(C2H3O2)2(s)   Mg (aq) + 2C2H3O2 (aq) First convert grams of Mg(C2H3O2)2 to moles of Mg(C2H3O2)2 and then use molar ratios to determine the moles of each ion produced.  1 mol Mg(C 2 H 3 O 2 )2    = 2.48596 mol Moles of Mg(C2H3O2)2 = 354 g Mg(C 2 H 3 O 2 )2  142.40 g Mg(C 2 H 3 O 2 )2  2+

–

The dissolution of 2.48596 mol Mg(C2H3O2)2(s) produces 2.49 mol Mg and (2 × 2.48596) = 4.97 mol C2H3O2 . Copyright McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

4-1

H2O c) (NH4)2CrO4(s)   2NH4 (aq) + CrO4 (aq) First convert formula units to moles. 1 mol (NH ) CrO   4 2 4  Moles of (NH4)2CrO4 = 1.8810 24 FU   = 3.121886 mol  6.022  10 23 FU  + 2– The dissolution of 3.121886 mol (NH4)2CrO4(s) produces (2 × 3.121886) = 6.24 mol NH4 and 3.12 mol CrO4 . +

4.2B

2–

Therefore, 4 moles of solid Li2CO3 produce 8 mol of Li ions and 4 mol of CO3 ions. H2O b) Fe2(SO4)3(s)   2Fe (aq) + 3SO4 (aq) First convert grams of Fe2(SO4)3 to moles of Fe2(SO4)3 and then use molar ratios to determine the moles of each ion produced.   2 4 3   = 0.2801 = 0.280 mol Fe2(SO4)3 Moles of Fe2(SO4)3 = (112 g Fe2(SO4)3)    2 4 3 3+

2–

The dissolution of 0.280 mol Fe2(SO4)3(s) produces 0.560 mol Fe (2 × 0.2801) and 0.840 mol SO4 (3 × 0.2801). H2O c) Al(NO3)3(s)   Al (aq) + 3NO3 (aq) First convert formula units of Al(NO3)3 to moles of Al(NO3)3 and then use molar ratios to determine the moles of each ion produced.   22 3 3  Moles of Al(NO3)3 = (8.09 × 10 formula units Al(NO3)3)  23    3 3 3+

–

= 0.1343 mol Al(NO3)3 3+

–

The dissolution of 0.134 mol Al(NO3)3 (s) produces 0.134 mol Al and 0.403 mol NO3 (3 × 0.1343). 4.3A

Plan: Convert the volume from mL to liters. Convert the mass to moles by dividing by the molar mass of KI. Divide the moles by the volume in liters to calculate molarity. Solution:    = 0.041988 mol KI Amount (moles) of KI = 6.97 g KI     M =  

4.3B

   

 = 0.41988 = 0.420 M 

Plan: Convert the volume from mL to liters. Convert the mass to moles by first converting mg to g and then dividing by the molar mass of NaNO3. Divide the moles by the volume in liters to calculate molarity. Solution:    3   Amount (moles) of NaNO3 = 175 mg NaNO3   = 0.0020588 mol NaNO3    3

 M =  

  

 = 0.13725 = 0.137 M 

4-2

4.4A

Plan: Divide the mass of sucrose by its molar mass to change the grams to moles. Divide the moles of sucrose by the molarity to obtain the volume of solution. Solution: Volume (L) of solution =  1 mol C H O   1L 135 g C12 H 22 O11  342.30 g 12C H22 O11  3.30 mol C H O  = 0.11951 = 0.120 L 12 22 11  12 22 11   Road map: Mass (g) of sucrose Divide by

(g/mol)

Amount (moles) of sucrose Divide by M (mol/L) Volume (L) of solution 4.4B

Plan: Convert the volume from mL to L; then multiply by the molarity of the solution to obtain the moles of H2SO4. Solution:   2 4  = 0.005184 = 0.00518 mol H2SO4 Amount (mol) of H2SO4 = 40.5 mL H2SO4      Road map: Volume (mL) of soln 1000 mL = 1 L Volume (L) of soln Multiply by M (1 L soln = 0.128 mol H2SO4) Amount (mol) H2SO4

4.5A

Plan: Multiply the volume and molarity to calculate the number of moles of sodium phosphate in the solution. Write the formula of sodium phosphate. Determine the number of each type of ion that is included in each formula unit. Use this information to determine the amount of each type of ion in the described solution. Solution:  3 4   = 0.7260 = 0.73 mol Na3PO4 Amount (mol) of Na3PO4 = 1.32 L    +

3–

In each formula unit of Na3PO4, there are 3 Na ions and 1 PO4 ion.

4-3

 + Amount (mol) of Na = 0.7260 mol Na3PO4  

 3– Amount (mol) of PO4 = 0.73 mol Na3PO4   4.5B

  = 0.73 mol PO43–  3 4 Plan: Write the formula of aluminum sulfate. Determine the number of aluminum ions in each formula unit. Calculate the number of aluminum ions in the sample of aluminum sulfate. Convert the volume from mL to L. Divide the number of moles of aluminum ion by the volume in L to calculate the molarity of the solution. Solution: 3+ In each formula unit of Al2(SO4)3, there are 2 Al ions. 3+   3+  = 2.50 mol Al3+ Amount (mol) of Al = 1.25 mol Al2(SO4)3    34

 M =   4.6A

  

4 3

 = 2.86 M 

Plan: Determine the new volume from the dilution equation (Mconc)(Vconc) = (Mdil)(Vdil). Solution: M V Vdil = conc conc  M dil

4.6B

  = 2.178 = 2.2 mol Na+   4

= 216 mL

Plan: Determine the new concentration from the dilution equation (Mconc)(Vconc) = (Mdil)(Vdil). Convert the molarity (mol/L) to g/mL in two steps (one step is moles to grams, and the other step is L to mL). Solution:

M dil 







7.50 M 25.0 m 3 M concVconc  = 0.375 M Vdil 500.m 3

3  0.375 mol H 2 SO 4   98.08 g H 2 SO 4  10 L   Concentration (g/mL) =     1L  1 mol H 2 SO 4   1 mL    –2

= 0.036780 = 3.68 × 10 g/mL solution 4.7A

Plan: Count the number of particles in each solution per unit volume. Solution: Solution A has 6 particles per unit volume while Solution B has 12 particles per unit volume. Solution B is more concentrated than Solution A. To obtain Solution B, the total volume of Solution A was reduced by half: 6 particles1.0 mL N V Vconc  dil dil  = 0.50 mL N conc 12 particles Solution C has 4 particles and is thus more dilute than Solution A. To obtain Solution C, the volume of solvent must be added for every volume of Solution A: 6 particles1.0 mL N V Vdil  conc conc  = 1.5 mL N dil 4 particles

4.7B

Plan: Count the number of particles in each solution per unit volume. Determine the final volume of the solution. Use the dilution equation, (Nconc)(Vconc) = (Ndil)(Vdil), to determine the number of particles that will be present when 300. mL of solvent is added to the 100. mL of solution represented in circle A. (Because M is directly proportional to the number of particles in a given solution, we can replace the molarity terms in the dilution equation with terms representing the number of particles.) Solution: There are 12 particles in circle A, 3 particles in circle B, 4 particles in circle C, and 6 particles in circle D.

4-4

The concentrated solution (circle A) has a volume of 100. mL. 300. mL of solvent are added, so the volume of the diluted solution is 400. mL. N V Ndil = conc conc  = 3 particles Vdil There are 3 particles in circle B, so circle B represents the diluted solution. 4.8A

Plan: Determine the ions present in each substance on the reactant side and write new cation-anion combinations. Use Table 4.1 to determine if either of the combinations of ions is not soluble. If a precipitate forms there will be a reaction and chemical equations may be written. The molecular equation simply includes the formulas of the substances and balancing. In the total ionic equation, all soluble substances are written as separate ions. The net ionic equation comes from the total ionic equation by eliminating all substances appearing in identical form (spectator ions) on each side of the reaction arrow. Solution: a) The resulting ion combinations that are possible are iron(III) phosphate and cesium chloride. According to Table 4.1, iron(III) phosphate is insoluble, so a reaction occurs. We see that cesium chloride is soluble. Total ionic equation: 3+ – + 3– – + Fe (aq) + 3Cl (aq) + 3Cs (aq) + PO4 (aq)  FePO4(s) + 3Cl (aq) + 3Cs (aq) Net ionic equation: 3+ 3– Fe (aq) + PO4 (aq)  FePO4(s) b) The resulting ion combinations that are possible are sodium nitrate (soluble) and cadmium hydroxide (insoluble). A reaction occurs. Total ionic equation: + – 2+ – + – 2Na (aq) + 2OH (aq) + Cd (aq) + 2NO3 (aq)  Cd(OH)2(s) + 2Na (aq) + 2NO3 (aq) + – Note: The coefficients for Na and OH are necessary to balance the reaction and must be included. Net ionic equation: 2+ – Cd (aq) + 2OH (aq)  Cd(OH)2(s) c) The resulting ion combinations that are possible are magnesium acetate (soluble) and potassium bromide (soluble). No reaction occurs.

4.8B

Plan: Determine the ions present in each substance on the reactant side and write new cation-anion combinations. Use Table 4.1 to determine if either of the combinations of ions is not soluble. If a precipitate forms there will be a reaction and chemical equations may be written. The molecular equation simply includes the formulas of the substances and balancing. In the total ionic equation, all soluble substances are written as separate ions. The net ionic equation comes from the total ionic equation by eliminating all substances appearing in identical form (spectator ions) on each side of the reaction arrow. Solution: a) The resulting ion combinations that are possible are silver chloride (insoluble, an exception) and barium nitrate (soluble). A reaction occurs. Total ionic equation: + – 2+ – 2+ – 2Ag (aq) + 2NO3 (aq) + Ba (aq) + 2Cl (aq)  2AgCl(s) + Ba (aq) + 2NO3 (aq) Net ionic equation: + – Ag (aq) + Cl (aq)  AgCl(s) b) The resulting ion combinations that are possible are ammonium sulfide (soluble) and potassium carbonate (soluble). No reaction occurs. c) The resulting ion combinations that are possible are lead(II) sulfate (insoluble, an exception) and nickel(II) nitrate (soluble). A reaction occurs. Total ionic equation: 2+ 2– 2+ – 2+ – Ni (aq) + SO4 (aq) + Pb (aq) + 2NO3 (aq)  PbSO4(s) + Ni (aq) + 2NO3 (aq) Net ionic equation: 2+ 2– Pb (aq) + SO4 (aq)  PbSO4 (s)

4-5

4.9A

Plan: Look at the ions (number and charge) produced when each of the given compounds dissolves in water and find the match to the ions shown in the beaker. Once the ions in each beaker are known, write new cation-anion combinations and use Table 4.1 to determine if any of the combination of ions is not soluble. If a precipitate forms there will be a reaction and chemical equations may be written. The molecular equation simply includes the formulas of the substances and must be balanced. In the total ionic equation, all soluble substances are written as separate ions. The net ionic equation comes from the total ionic equation by eliminating all substances appearing in identical form (spectator ions) on each side of the reaction arrow. Solution: a) Beaker A has four ions with a +2 charge and eight ions with a –1 charge. The beaker contains dissolved Zn(NO3)2 2+ – which dissolves to produce Zn and NO3 ions in a 1:2 ratio. The compound PbCl2 also has a +2 ion and –1 ion in a 1:2 ratio but PbCl2 is insoluble so ions would not result from this compound. b) Beaker B has three ions with a +2 charge and six ions with a –1 charge. The beaker contains dissolved Ba(OH)2 2+ – which dissolves to produce Ba and OH ions in a 1:2 ratio. Cd(OH)2 also has a +2 ion and a –1 ion in a 1:2 ratio but Cd(OH)2 is insoluble so ions would not result from this compound. c) The resulting ion combinations that are possible are zinc hydroxide (insoluble) and barium nitrate (soluble). The 2+ – precipitate formed is Zn(OH)2. The spectator ions are Ba and NO3 . Balanced molecular equation: Zn(NO3)2(aq) + Ba(OH)2(aq)  Zn(OH)2(s) + Ba(NO3)2(aq) Total ionic equation: Zn (aq) + 2NO3 (aq) + Ba (aq) + 2OH (aq)  Zn(OH)2(s) + Ba (aq) + 2NO3 (aq) 2+

–

Net ionic equation: Zn (aq) + 2OH (aq)  Zn(OH)2(s) 2+

–

d) Since there are only six OH ions and four Zn ions, the OH is the limiting reactant.

 – Mass of Zn(OH)2= (6 OH ions)  

 

  

2 

  

   2  2

= 14.9100 = 15 g Zn(OH)2 4.9B

Plan: Look at the ions (number and charge) produced when each of the given compounds dissolves in water and find the match to the ions shown in the beaker. Once the ions in each beaker are known, write new cation-anion combinations and use Table 4.1 to determine if any of the combination of ions is not soluble. If a precipitate forms there will be a reaction and chemical equations may be written. The molecular equation simply includes the formulas of the substances and must be balanced. In the total ionic equation, all soluble substances are written as separate ions. The net ionic equation comes from the total ionic equation by eliminating all substances appearing in identical form (spectator ions) on each side of the reaction arrow. Solution: a) Beaker A has eight ions with a +1 charge and four ions with a –2 charge. The beaker contains dissolved Li2CO3 + 2– which dissolves to produce Li and CO3 ions in a 2:1 ratio. The compound Ag2SO4 also has a +1 ion and –2 ion in a 2:1 ratio but Ag2SO4 is insoluble so ions would not result from this compound. b) Beaker B has three ions with a +2 charge and six ions with a –1 charge. The beaker contains dissolved CaCl2 2+ – which dissolves to produce Ca and Cl ions in a 1:2 ratio. Ni(OH)2 also has a +2 ion and a –1 ion in a 1:2 ratio but Cd(OH)2 is insoluble so ions would not result from this compound. c) The resulting ion combinations that are possible are calcium carbonate (insoluble) and lithium chloride (soluble). + – The precipitate formed is CaCO3. The spectator ions are Li and Cl . Balanced molecular equation: Li2CO3(aq) + CaCl2(aq)  CaCO3(s) + 2LiCl(aq) Total ionic equation: + 2– 2+ – + – 2Li (aq) + CO3 (aq) + Ca (aq) + 2Cl (aq)  CaCO3(s) + 2Li (aq) + 2Cl (aq) Net ionic equation: 2+ 2– Ca (aq) + CO3 (aq)  CaCO3 (s)

4-6

2–

d) Since there are only four CO3 ions and three Ca ions, the Ca is the limiting reactant. 2+     2+ 3 3     = 60.0540 = 60. g CaCO3 Mass of CaCO3= (3 Ca ions)   2+ 2+       3  4.10A

Plan: We are given the molarity and volume of calcium chloride solution, and we must find the volume of sodium phosphate solution that will react with this amount of calcium chloride. After writing the balanced equation, we find the amount (mol) of calcium chloride from its molarity and volume and use the molar ratio to find the amount (mol) of sodium phosphate required to react with the calcium chloride. Finally, we use the molarity of the sodium phosphate solution to convert the amount (mol) of sodium phosphate to volume (L). Solution: The balanced equation is: 3CaCl2(aq) + 2Na3PO4(aq)  Ca3(PO4)2(s) + 6NaCl(aq) Finding the volume (L) of Na3PO4 needed to react with the CaCl2:

 Volume (L) of Na3PO4 = 0.300 L CaCl2  

  

4 2

   

   4

= 0.1346 = 0.135 L Na3PO4 4.10B

Plan: We are given the mass of silver chloride produced in the reaction of silver nitrate and sodium chloride, and we must find the molarity of the silver nitrate solution. After writing the balanced equation, we find the amount (mol) of silver nitrate that produces the precipitate by dividing the mass of silver chloride produced by its molar mass and then using the molar ratios from the balanced equation. Then we calculate the molarity by dividing the moles of silver nitrate by the volume of the solution (in L). The balanced equation is: AgNO3(aq) + NaCl(aq)  AgCl(s) + NaNO3(aq)

 Amount (mol) of AgNO3 = 0.148 g AgCl  

  Molarity (M) of AgNO3 =   4.11A



   

–3 3

  

 = 0.0010321 = 1.03 × 10–3 mol AgNO 3  

 –2 –2 = 2.2935 × 10 = 2.29 × 10 M 

Plan: Multiply the volume in liters of each solution by its molarity to obtain the moles of each reactant. Write a balanced equation. Use molar ratios from the balanced equation to determine the moles of lead(II) chloride that may be produced from each reactant. The reactant that generates the smaller number of moles is limiting. Change the moles of lead(II) chloride from the limiting reactant to the grams of product using the molar mass of lead(II) chloride. Solution: (a) The balanced equation is: Pb(C2H3O2)2(aq) + 2NaCl(aq)  PbCl2(s) + 2NaC2H3O2(aq) 103 L  1.50 mol Pb(C 2 H 3 O2 )2   Moles of Pb(C2H3O2)2 = 268 mL    = 0.402 mol Pb(C2H3O2)2  1 mL  1L  





103 L   3.40 mol NaCl   Moles of NaCl = 130 mL    = 0.442 mol NaCl  1 mL  1L     1 mol PbCl 2   Moles of PbCl2 from Pb(C2H3O2)2 = 0.402 mol Pb(C 2 H 3 O 2 )2  1 mol Pb(C 2 H 3 O 2 )2  = 0.402 mol PbCl2 1 mol PbCl 2    = 0.221 mol PbCl Moles of PbCl2 from NaCl = 0.442 mol NaCl 2  2 mol NaCl 





4-7

The NaCl is limiting. The mass of PbCl2 may now be determined using the molar mass.  278.1 g PbCl 2    = 61.4601 = 61.5 g PbCl Mass (g) of PbCl2 = 0.221 mol PbCl 2  2  1 mol PbCl 2  (b) Ac is used to represent C2H3O2: Amount (mol) Pb(Ac)2 + 2NaCl + 2NaAc 2



4.11B



Initial

0.402

0.442

Change

– 0.221

– 0.442

+0.221

+0.442

Final

0.181

0.221

0.442

Plan: Write a balanced equation. Multiply the volume in liters of each solution by its molarity to obtain the moles of each reactant. Use molar ratios from the balanced equation to determine the moles of mercury(II) sulfide that may be produced from each reactant. The reactant that generates the smaller number of moles is limiting. Change the moles of mercury(II) sulfide from the limiting reactant to the grams of product using the molar mass of mercury(II) sulfide. Solution: (a) The balanced equation is: Hg(NO3)2(aq) + 2Na2S(aq)  HgS(s) + 2NaNO3(aq)  0.010 mol Hg(NO 3 )2    = 5.0 × 10–4 mol Hg(NO ) Moles of Hg(NO3)2= 0.050 L  3 2   1L   0.10 mol Na 2 S    = 2.0 × 10–3 mol Na S Moles of Na2S = 0.020 L  2   1L   1 mol HgS    = 5.0 × 10–4 mol HgS Moles of HgS from Hg(NO3)2 = 5.0 104 mol Hg(NO 3 )2  1 mol Hg(NO3 )2   1 mol HgS   –3 Moles of HgS from Na2S = 2.0 103 mol Na 2 S  = 2.0 × 10 mol HgS 1 mol Na 2 S 





The Hg(NO3)2 is limiting. The mass of HgS may now be determined using the molar mass.  232.7 g HgS    = 0.11635 = 0.12 g HgS Mass (g) of HgS = 2.0 103 mol HgS  1 mol HgS  (b) Amount (mol) Hg(NO3)2 + Na2S 3

4.12A

–4

2.0 × 10

–4

– 5.0 × 10

Initial

5.0 × 10

Change

– 5.0 × 10

Final

–3

–4

–3

1.5 × 10

+ 5.0 × 10 5.0 × 10

–4

+ 1.0 × 10 1.0 × 10

–3

Plan: Convert the given volume from mL to L and multiply by the molarity (mol/L) to find moles of Ca(OH)2. Each mole of the strong base Ca(OH)2 will produce two moles of hydroxide ions. Finally, multiply the amount (mol) of hydroxide ions by Avogadro’s number to calculate the number of hydroxide ions produced. Solution: 2O Ca(OH)2(s) H  Ca (aq) + 2OH (aq)

–

No. of OH ions produced = 451 mL Ca(OH)2  

  

   



    2



  

= 6.5182 × 10 = 6.52 × 10 OH ions 21

–

4-8

4.12B

Plan: Convert the given volume from mL to L and multiply by the molarity (mol/L) to find moles of HCl. Each mole of the strong acid HCl will produce one mole of hydrogen ions. Finally, multiply the amount (mol) of hydrogen ions by Avogadro’s number to calculate the number of hydrogen ions produced. Solution: 2O HCl(g) H  H (aq) + Cl (aq)

–

 + No. of H ions produced = 65.5 mL HCl  

  

   

  



  

+ +

= 2.8479 × 10 = 2.85 × 10 H ions 22

4.13A

Plan: The reactants are a weak acid and a strong base. The acidic species is H and a proton is transferred to OH from the acid. The only spectator ion is the cation of the base. Solution: Molecular equation: 2HNO2(aq) + Sr(OH)2(aq)  Sr(NO2)2(aq) + 2H2O(l) Ionic equation: 2+ – 2+ – 2HNO2(aq) + Sr (aq) + 2OH (aq)  Sr (aq) + 2NO2 (aq) + 2H2O(l) Net ionic equation: – – 2HNO2(aq) + 2OH (aq)  2NO2 (aq) + 2H2O(l) or – – HNO2(aq) + OH (aq)  NO2 (aq) + H2O(l)

–

HNO2(aq) + OH (aq)  NO2 (aq) + H2O(l) –

–

The salt is Sr(NO2)2, strontium nitrite, and the spectator ion is Sr . 4.13B

Plan: The reactants are a strong acid and the salt of a weak base. The acidic species is H3O and a proton is – transferred to the weak base HCO3 to form H2CO3, which then decomposes to form CO2 and water. Solution: Molecular equation: 2HBr(aq) + Ca(HCO3)2(aq)  CaBr2(aq) + 2H2CO3(aq) Ionic equation: + – 2+ – – 2+ 2H3O (aq) + 2Br (aq) + Ca (aq) + 2HCO3 (aq)  2CO2(g) + 4H2O(l) + 2Br (aq) + Ca (aq) Net ionic equation: + – + – 2H3O (aq) + 2HCO3 (aq)  2CO2(g) + 4H2O(l) or H3O (aq) + HCO3 (aq)  CO2(g) + 2H2O(l) H3O (aq) + HCO3 (aq)  CO2(g) + 2H2O(l) +

–

The salt is CaBr2, calcium bromide. 4.14A

Plan: Write a balanced equation. Determine the moles of HCl by multiplying its molarity by its volume, and, through the balanced chemical equation and the molar mass of aluminum hydroxide, determine the mass of aluminum hydroxide required for the reaction. Solution: The balanced equation is: Al(OH)3(s) + 3HCl(aq)  AlCl3(aq) + 3H2O(l)

 –2 Mass(g) of Al(OH)3 = 3.4 × 10 L HCl  

  

   

   3  3

= 0.08840 = 0.088 g Al(OH)3 4.14B

Plan: Write a balanced equation. Determine the moles of NaOH by multiplying its molarity by its volume, and, through the balanced chemical equation and the molar mass of acetylsalicylic acid, determine the mass of acetylsalicylic acid in the tablet.

4-9

Solution: The balanced equation is: HC9H7O4(aq) + NaOH(aq)  NaC9H7O4(aq) + H2O(l)

 Mass (g) of HC9H7O4= 14.10 mL NaOH  

  

   

   

9 9

7 7

4 4

  

= 0.3251 = 0.325 g HC9H7O4 4.15A

Plan: Write a balanced chemical equation. Determine the moles of HCl by multiplying its molarity by its volume, and, through the balanced chemical equation, determine the moles of Ba(OH) 2 required for the reaction. The amount of base in moles is divided by its molarity to find the volume. Solution: The molarity of the HCl solution is 0.1000 M. However, the molar ratio is not 1:1 as in the example problem. According to the balanced equation, the ratio is 2 moles of acid per 1 mole of base: 2HCl(aq) + Ba(OH)2(aq)  BaCl2(aq) + 2H2O(l) Volume (L) of Ba(OH)2 solution =  1 L   1L  0.1000 mol HCl  1 mol Ba(OH)2    50.00 mL      1L   2 mol HCl   0.1292 mol Ba(OH)2  1000 mL  





= 0.0193498 = 0.01935 L Ba(OH)2 solution 4.15B

Plan: Write a balanced chemical equation. Determine the moles of H2SO4 by multiplying its molarity by its volume, and, through the balanced chemical equation, determine the moles of KOH required for the reaction. The amount of base in moles is divided by its volume (in L) to find the molarity. Solution: The balanced equation is: KOH(aq) + HNO3(aq)  KNO3(aq) + H2O(l) Amount (mol) of KOH = 20.00 mL HNO3    

Molarity =  4.16A

   

 

  

  = 0.004904 mol KOH NO3 

 = 0.270193 = 0.2702 M 

Plan: Apply Table 4.4 to the compounds. Do not forget that the sum of the O.N.’s (oxidation numbers) for a compound must sum to zero, and for a polyatomic ion, the sum must equal the charge on the ion. Solution: a) Sc = +3 O = –2. In most compounds, oxygen has a –2 O.N., so oxygen is often a good starting point. If each oxygen atom has a –2 O.N., then each scandium must have a +3 oxidation state so that the sum of O.N.’s equals zero: 2(+3) + 3(–2) = 0. b) Ga = +3 Cl = –1. In most compounds, chlorine has a –1 O.N., so chlorine is a good starting point. If each chlorine atom has a –1 O.N., then the gallium must have a +3 oxidation state so that the sum of O.N.’s equals zero: 1(+3) + 3(–1) = 0. 2– c) H = +1 P = +5 O = –2. The hydrogen phosphate ion is HPO4 . Again, oxygen has a –2 O.N. Hydrogen has a +1 O.N. because it is combined with nonmetals. The sum of the O.N.’s must equal the ionic charge, so the following algebraic equation can be solved for P: 1(+1) + 1(P) + 4(–2) = –2; O.N. for P = +5. d) I = +3 F = –1. The formula of iodine trifluoride is IF3. In all compounds, fluorine has a –1 O.N., so fluorine is often a good starting point. If each fluorine atom has a –1 O.N., then the iodine must have a +3 oxidation state so that the sum of O.N.’s equal zero: 1(+3) + 3(–1) = 0.

4-10

4.16B

Plan: Apply Table 4.4 to the compounds. Do not forget that the sum of the O.N.’s (oxidation numbers) for a compound must sum to zero, and for a polyatomic ion, the sum must equal the charge on the ion. Solution: a) K = +1 C = +4 O = –2. In all compounds, potassium has a +1 O.N., and in most compounds, oxygen has a –2 O.N. If each potassium has a +1 O.N. and each oxygen has a –2 O.N., carbon must have a +4 oxidation state so that the sum of the O.N.’s equals zero: 2(+1) + 1(+4) + 3(–2) = 0. b) N = –3 H = +1. When it is combined with a nonmetal, like N, hydrogen has a +1 O.N. If hydrogen has a +1 O.N., nitrogen must have a –3 O.N. so the sum of the O.N.’s equals +1, the charge on the polyatomic ion: 1(–3) + 4(+1) = +1. c) Ca = +2 P = –3. Calcium, a group 2A metal, has a +2 O.N. in all compounds. If calcium has a +2 O.N., the phosphorus must have a –3 O.N. so the sum of the O.N.’s equals zero: 3(+2) + 2(–3) = 0. d) S = +4 Cl = –1. Chlorine, a group 7A nonmetal, has a –1 O.N. when it is in combination with any nonmetal except O or other halogens lower in the group. If chlorine has an O.N. of –1, the sulfur must have an O.N. of +4 so that the sum of the O.N.’s equals zero: 1(+4) + 4(–1) = 0.

4.17A

Plan: Apply Table 4.4 to determine the oxidation numbers for all the compounds in the reaction. Do not forget that the sum of the O.N.’s (oxidation numbers) for a compound must sum to zero, and for a polyatomic ion, the sum must equal the charge on the ion. After determining oxidation numbers for all atoms in the reaction, identify the atoms for which the oxidation numbers have changed from the left-hand side of the equation to the right-hand side of theequation. If the oxidation number of a particular atom increases, that atom has been oxidized, and the compound, element, or ion containing that atom on the reactant side of the equation is the reducing agent. If the oxidation number of a particular atom decreases, that atom has been reduced, and the compound, element, or ion containing that atom on the reactant side of the equation is the oxidizing agent. Solution: a) Oxidation numbers in NCl3: N = +3, Cl = –1 Oxidation numbers in H2O: H = +1, O = –2 Oxidation numbers in NH3: N = –3, H = +1 Oxidation numbers in HOCl: H = +1, O = –2, Cl = +1 The oxidation number of nitrogen decreases from +3 to –3, so N is reduced and NCl3 is the oxidizing agent. The oxidation number of chlorine increases from –1 to +1, so Cl is oxidized and NCl3 is also the reducing agent. b) Oxidation numbers in AgNO3: N = +1, N = +5, O = –2 Oxidation numbers in NH4I: N = –3, H = +1, I = –1 Oxidation numbers in AgI: Ag = +1, I = –1 + – Oxidation numbers in NH4NO3: N (in NH4 ) = –3, H = +1, N (in NO3 ) = +5, O = –2 None of the oxidation numbers change, so this is not an oxidation-reduction reaction. c) Oxidation numbers in H2S: H = +1, S = –2 Oxidation numbers in O2: O = 0 Oxidation numbers in SO2: S = +4, O = –2 Oxidation numbers in H2O: H = +1, O = –2 The oxidation number of oxygen decreases from 0 to –2, so O is reduced and O2 is the oxidizing agent. The oxidation number of sulfur increases from –2 to +4, so S is oxidized and H2S is the reducing agent.

4.17B

4-11

Solution: a) Oxidation numbers in SiO2: Si = +4, O = –2 Oxidation numbers in HF: H = +1, F = –1 Oxidation numbers in SiF4: Si = +4, F = –1 Oxidation numbers in H2O: H = +1, O = –2 None of the oxidation numbers change, so this is not an oxidation-reduction reaction. b) Oxidation numbers in C2H6: C = –3, H = +1 Oxidation numbers in O2: O = 0 Oxidation numbers in CO2: C = +4, O = –2 Oxidation numbers in H2O: H = +1, O = –2 The oxidation number of oxygen decreases from 0 to –2, so O is reduced and O2 is the oxidizing agent. The oxidation number of carbon increases from –3 to +4, so C is oxidized and C2H6 is the reducing agent. c) Oxidation numbers in CO: C = +2, O = –2 Oxidation numbers in I2O5: I = +5, O = –2 Oxidation numbers in I2: I = 0 Oxidation numbers in CO2: C = +4, O = –2 The oxidation number of iodine decreases from +5 to 0, so I is reduced and I2O5 is the oxidizing agent. The oxidation number of carbon increases from +2 to +4, so C is oxidized and CO is the reducing agent. 4.18A

Plan: Write a balanced reaction. Find moles of KMnO4 by multiplying its volume and molarity and determine the 2+ 2+ moles of Ca in the sample using the molar ratio from the reaction. Divide moles of Ca by the volume of the milk 2+ sample to obtain molarity. In part (b), moles of Ca will be converted to grams. Solution: 2KMnO4(aq) + 5CaC2O4(s) + H2SO4(aq)  2MnSO4(aq) + K2SO4(aq) + 5CaSO4(s) + 10CO2(g) + 8H2O(l)

103 L  4.56103 mol KMnO4    Moles of KMnO4 = 6.53 mL  1 mL  1L  –5

= 2.97768 × 10 mol KMnO4 2+  5 mol CaC O   1  2+ 2 4  1 mL   1 mol Ca  Molarity of Ca = 2.97768105 mol KMnO 4     103 L  1 mol CaC 2 O 4  2.50 mL   2 mol KMnO 4  –2

–2

= 2.97768 × 10 = 2.98 × 10 mol/L = 2.98 × 10 M Ca

2+  2.97768102 mol Ca 2+   40.08 g Ca  = 1.193454 = 1.19 g/L b) Concentration (g/L) =   1 mol Ca 2+   L  This concentration is consistent with the typical value in milk.

4.18B

Plan: Find moles of K2Cr2O7 by multiplying its volume and molarity, and determine the moles of Fe in the sample 2+ using the molar ratio from the reaction provided in the problem statement (remember that the Fe is part of the 2+ 2+ FeSO4 compound and that there is one Fe ion for every formula unit of FeSO4). Divide moles of Fe by the 2+ volume of the FeSO4 solution to obtain molarity. In part (b), convert the moles of Fe to grams. Then divide the mass of iron ion by the total mass of the ore and multiply by 100% to find the mass percent of iron in the ore. Solution: a) The balanced equation (provided in the problem statement) is: 6FeSO4(aq) + K2Cr2O7(aq) + 7H2SO4(aq)  3Fe2(SO4)3(aq) + Cr2(SO4)3(aq) + 7H2O(l) + K2SO4(aq)

 2+ Amount (mol) of Fe = 21.85 mL K2Cr2O7  

 

  

4 2

   

   4

= 0.0322775 mol Fe

 2+ Molarity of Fe =  

  

 = 1.0925 = 1.09 M 

4-12

2+  b) Mass (g) of iron = 0.032775 mol Fe  

4.19A

4.19B

  

  = 1.8305 = 1.83 g Fe 

   (100%) = 70.9496 = 70.9% Mass % of iron in sample =    Plan: To classify a reaction, compare the number of reactants used versus the number of products formed. Also examine the changes, if any, in the oxidation numbers. Recall the definitions of each type of reaction: Combination: X + Y  Z; Decomposition: Z  X + Y Single displacement: X + YZ  XZ + Y Solution: a) Combination: S8(s) + 16F2(g)  8SF4(g) O.N.: S=0 F=0 S = +4 F = –1 Sulfur changes from 0 to +4 oxidation state; it is oxidized and S8 is the reducing agent. Fluorine changes from 0 to –1 oxidation state; it is reduced and F2 is the oxidizing agent. b) Displacement: 2CsI(aq) + Cl2(aq)  2CsCl(aq) + I2(aq) O.N.: Cs = +1 Cl = 0 Cs = +1 I = 0 I = –1 Cl = –1 + – + – Total ionic eqn: 2Cs (aq) + 2I (aq) + Cl2(aq)  2Cs (aq) + 2Cl (aq) + I2(aq) – – Net ionic eqn: 2I (aq) + Cl2(aq)  2Cl (aq) + I2(aq) Iodine changes from –1 to 0 oxidation state; it is oxidized and CsI is the reducing agent. Chlorine changes from 0 to –1 oxidation state; it is reduced and Cl2 is the oxidizing agent. c) Displacement: 3Ni(NO3)2(aq) + 2Cr(s)  2Cr(NO3)3(aq) + 3Ni(s) O.N.: Ni = +2 Cr = 0 Cr = +3 Ni = 0 N = +5 N = +5 O = –2 O = –2 +2 – +3 – Total ionic eqn: 3Ni (aq) + 6NO3 (aq) + 2Cr(s)  2Cr (aq) + 6NO3 (aq) + 3Ni(s) +2 +3 Net ionic eqn: 3Ni (aq) + 2Cr(s)  2Cr (aq) + 3Ni(s) Nickel changes from +2 to 0 oxidation state; it is reduced and Ni(NO3)2 is the oxidizing agent. Chromium changes from 0 to +3 oxidation state; it is oxidized and Cr is the reducing agent. Plan: To classify a reaction, compare the number of reactants used versus the number of products formed. Also examine the changes, if any, in the oxidation numbers. Recall the definitions of each type of reaction: Combination: X + Y  Z; Decomposition: Z  X + Y Single displacement: X + YZ  XZ + Y Solution: a) Displacement: Co(s) + 2HCl(aq)  CoCl2(aq) + H2(g) (molecular equation) O.N.: Co = 0 H = +1, Cl = –1 Co = +2, Cl = –1 H = 0 + – 2+ – Co(s) + 2H (aq) + 2Cl (aq)  Co (aq) + 2Cl (aq) + H2(g) (total ionic equation) + 2+ Co(s) + 2H (aq)  Co (aq) + H2(g) (net ionic equation) Cobalt changes from 0 to +2 oxidation state; it is oxidized and Co is the reducing agent. Hydrogen changes from +1 to 0 oxidation state; it is reduced and HCl is the oxidizing agent. b) Combination: 2CO (g) + O2(g)  2CO2(g) O.N.: C = +2, O = –2 O=0 C = +4, O = –2 Carbon changes from +2 to +4 oxidation state; it is oxidized and CO is the reducing agent. Oxygen changes from 0 to –2 oxidation state; it is reduced and O2 is the oxidizing agent. c) Decomposition: 2N2O5(s)  4NO2(g) + O2(g) O.N.: N = +5 N = +4 O = 0 O = –2 O = –2 Nitrogen changes from +5 to +4 oxidation state; it is reduced and N2O5 is the oxidizing agent. Oxygen changes from –2 to 0 oxidation state; it is oxidized and N2O5 is also the reducing agent.

4-13

END–OF–CHAPTER PROBLEMS 4.1

Plan: Review the discussion on the polar nature of water. Solution: Water is polar because the distribution of its bonding electrons is unequal, resulting in polar bonds, and the shape of the molecule (bent) is unsymmetrical.

4.2

Plan: Review the discussion on water soluble compounds. Solution: Ionic and polar covalent compounds are most likely to be soluble in water. Because water is polar, the partial charges in its molecules are able to interact with the charges, either ionic or dipole-induced, in other substances.

4.3

Plan: Solutions that conduct an electric current contain electrolytes. Solution: Ions must be present in an aqueous solution for it to conduct an electric current. Ions come from ionic compounds or from other electrolytes such as acids and bases.

4.4

Plan: Review the discussion on ionic compounds in water. Solution: The ions on the surface of the solid attract the water molecules (cations attract the “negative” ends and anions attract the “positive” ends of the water molecules). The interaction of the solvent with the ions overcomes the attraction of the oppositely charged ions for one another, and they are released into the solution.

4.5

Plan: Recall that ionic compounds dissociate into their ions when dissolved in water. Examine the charges of the ions in each scene and the ratio of cations to anions. Solution: 2+ – a) CaCl2 dissociates to produce one Ca ion for every two Cl ions. Scene B contains four 2+ ions and twice that number of 1– ions. + 2– b) Li2SO4 dissociates to produce two Li ions for every one SO4 ion. Scene C contains eight 1+ ions and half as many 2– ions. + – c) NH4Br dissociates to produce one NH4 ion for every one Br ion. Scene A contains equal numbers of 1+ and 1– ions.

4.6

Plan: Write the formula for magnesium nitrate and note the ratio of magnesium ions to nitrate ions. Solution: 2+ Upon dissolving the salt in water, magnesium nitrate, Mg(NO3)2, would dissociate to form one Mg ion for every – two NO3 ions, thus forming twice as many nitrate ions. Scene B best represents a volume of magnesium nitrate solution. Only Scene B has twice as many nitrate ions (red circles) as magnesium ions (blue circles).

4.7

Plan: Review the discussion of ionic compounds in water. Solution: In some ionic compounds, the force of the attraction between the ions is so strong that it cannot be overcome by the interaction of the ions with the water molecules. These compounds will be insoluble in water.

4.8

Plan: Review the discussion of covalent compounds in water. Solution: The interaction with water depends on the structure of the molecule. If the interaction is strong, the substance will be soluble; otherwise, the substance will not be very soluble. Covalent compounds that contain polar groups interact well with the polar solvent water and therefore dissolve in water. Covalent compounds that do not contain polar bonds are not soluble in water.

4-14

4.9

Plan: Review the discussion of covalent compounds in water. Solution: Some covalent compounds that contain the hydrogen atom dissociate into ions when dissolved in water. These compounds form acidic solutions in water; three examples are HCl, HNO3, and HBr.

4.10

Plan: Count the total number of spheres in each box. The number in box A divided by the volume change in each part will give the number we are looking for and allow us to match boxes. Solution: The number in each box is: A = 12, B = 6, C = 4, and D = 3. a) When the volume is tripled, there should be 12/3 = 4 spheres in a box. This is box C. b) When the volume is doubled, there should be 12/2 = 6 spheres in a box. This is box B. c) When the volume is quadrupled, there should be 12/4 = 3 spheres in a box. This is box D.

4.11

Plan: Recall that molarity = moles of solute/volume (L) of solution. Solution: a) Mdil = molarity of the diluted solution Mconc = molarity of the concentrated solution Vdil = volume of the diluted solution Vconc = volume of the concentrated solution The equation works because the quantity (moles) of solute remains the same when a solution is diluted; only the amount of solvent changes. M × V = moles; Mdil × Vdil = Mconc × Vconc molesdil = molesconc moles solute b) Molarity = liters solution Moles CaCl2 = molarity × liters of solution; Mass CaCl2 = molarity × liters of solution × molar mass of CaCl2.

4.12

Plan: Recall that molarity = moles of solute/volume (L) of solution. Here you can use the number of particles in place of moles of solute. Solution: a) Solution B has the highest molarity as it has the largest number of particles, 12, in a volume of 50 mL. b) Solutions A and F both have 8 particles in a volume of 50 mL and thus the same molarity. Solutions C, D, and E all have 4 particles in a volume of 50 mL and thus have the same molarity. c) Mixing Solutions A and C results in 8 + 4 = 12 particles in a volume of 100 mL. That is a lower molarity than that of Solution B, which has 12 particles in a volume of 50 mL or 24 particles in a volume of 100 mL. d) Adding 50 mL to Solution D would result in 4 particles in a total volume of 100 mL; adding 75 mL to Solution F would result in 4 particles in a volume of 100 mL. The molarity of each solution would be the same. e) Solution A has 8 particles in a volume of 50 mL while Solution E has the equivalent of 4 particles in a volume of 50 mL. The molarity of Solution E is half that of Solution A. Therefore half of the volume, 12.5 mL, of Solution E must be evaporated. When 12.5 mL of solvent is evaporated from Solution E, the result will be 2 particles in 12.5 mL or 8 particles in 50 mL as in Solution A.

4.13

Plan: Remember that molarity is moles of solute/volume of solution. Solution: Volumes may not be additive when two different solutions are mixed, so the final volume may be slightly different from 1000.0 mL. The correct method would state, “Take 100.0 mL of the 10.0 M solution and add water until the total volume is 1000 mL.”

4.14

4-15

Solution: a) Benzene, a covalent compound, is likely to be insoluble in water because it is nonpolar and water is polar. b) Sodium hydroxide (NaOH) is an ionic compound and is therefore likely to be soluble in water. c) Ethanol (CH3CH2OH) will likely be soluble in water because it contains a polar –OH bond like water. d) Potassium acetate (KC2H3O2) is an ionic compound and will likely be soluble in water. 4.15

Plan: Compounds that are soluble in water tend to be ionic compounds or covalent compounds that have polar bonds. Many ionic compounds are soluble in water because the attractive force between the oppositely charged ions in an ionic compound are replaced with an attractive force between the polar water molecule and the ions when the compound is dissolved in water. Covalent compounds with polar bonds are often soluble in water since the polar bonds of the covalent compound interact with those in water. Solution: a) Lithium nitrate is an ionic compound and is expected to be soluble in water. b) Glycine (H2NCH2COOH) is a covalent compound, but it contains polar N–H and O–H bonds. This would make the molecule interact well with polar water molecules, and make it likely that it would be soluble. c) Pentane (C5H12) has no bonds of significant polarity, so it would be expected to be insoluble in the polar solvent water. d) Ethylene glycol (HOCH2CH2OH) molecules contain polar O–H bonds, similar to water, so it would be expected to be soluble.

4.16

Plan: Substances whose aqueous solutions conduct an electric current are electrolytes such as ionic compounds, acids, and bases. Solution: + – a) Cesium bromide, CsBr, is a soluble ionic compound, and a solution of this salt in water contains Cs and Br ions. Its solution conducts an electric current. + – b) HI is a strong acid that dissociates completely in water. Its aqueous solution contains H and I ions, so it conducts an electric current.

4.17

Plan: Substances whose aqueous solutions conduct an electric current are electrolytes such as ionic compounds, acids, and bases. Solution: + 2– a) Potassium sulfate, K2SO4, is an ionic compound that is soluble in water, producing K and SO4 ions. Its solution conducts an electric current. b) Sucrose is neither an ionic compound, an acid, nor a base, so it would be a nonelectrolyte (even though it’s soluble in water). Its solution does not conduct an electric current.

4.18

Plan: To determine the total moles of ions released, write an equation that shows the compound dissociating into ions with the correct molar ratios. Convert mass and formula units to moles of compound and use the molar ratio to convert moles of compound to moles of ions. Solution: + – a) Each mole of NH4Cl dissolves in water to form 1 mole of NH4 ions and 1 mole of Cl ions, or a total of 2 + – moles of ions: NH4Cl(s) (aq) + Cl (aq). 4  2 mol ions    = 0.64 mol of ions Moles of ions = 0.32 mol NH 4 Cl  1 mol NH 4 Cl  2+ – b) Each mole of Ba(OH)2 8H2O forms 1 mole of Ba ions and 2 moles of OH ions, or a total of 3 moles of ions: 2+ – Ba(OH)2 8H2O(s) (aq) + 2OH (aq). The waters of hydration become part of the larger bulk of water. Convert mass to moles using the molar mass.  1 mol Ba(OH)2  8H 2 O   3 mol ions    Moles of ions = 25.4 g Ba(OH)2  8H 2 O  1 mol Ba(OH)  8H O   315.4 g Ba(OH)2  8H 2 O  2 2 





= 0.2415980 = 0.242 mol of ions + – c) Each mole of LiCl produces 2 moles of ions (1 mole of Li ions and 1 mole of Cl ions): + – 23 LiCl(s) (aq) + Cl (aq). Recall that a mole contains 6.022 × 10 entities, so a mole of LiCl contains 23 6.022 × 10 units of LiCl, more easily expressed as formula units. Copyright McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

4-16

  1 mol LiCl  2 mol ions   Moles of ions = 3.551019 FU LiCl     6.022 10 23 FU LiCl  1 mol LiCl 





–4

= 1.17901 × 10 = 1.18 × 10 mol of ions 4.19

Plan: To determine the total moles of ions released, write an equation that shows the compound dissociating into ions with the correct molar ratios. Convert mass and formula units to moles of compound and use the molar ratio to convert moles of compound to moles of ions. Solution: + 2– a) Each mole of Rb2SO4 dissolves in water to form 2 moles of Rb ions and 1 mole of SO4 ions, or a total of + 2– 3 moles of ions: Rb2SO4(s) (aq) + SO4 (aq).  3 mol ions   Moles of ions = 0.805 mol Rb 2 SO 4   = 2.415 = 2.42 mol of ions 1 mol Rb 2 SO 4  2+ – b) Each mole of Ca(NO3)2 forms 1 mole of Ca ions and 2 moles of NO3 ions, or a total of 2+ – 3 moles of ions: Ca(NO3)2(s) (aq) + 2NO3 (aq). Convert mass to moles using molar mass.  1 mol Ca(NO )  3 mol ions   3 2    Moles of ions = 3.85103 g Ca(NO3 )2    164.10 g Ca(NO3 )2  1 mol Ca(NO 3 )2 









–5

= 7.03839 × 10 = 7.0 × 10 mol of ions 2+ – c) Each mole of Sr(HCO3)2 produces 3 moles of ions (1 mole of Sr ions and 2 moles of HCO3 ions): 2+ – 23 Sr(HCO3)2(s) (aq) + 2HCO3 (aq). Recall that a mole contains 6.022 × 10 entities, so a mole of Sr(HCO3)2 23 contains 6.022 × 10 units of Sr(HCO3)2, more easily expressed as formula units.   1 mol Sr(HCO3 )2  3 mol ions   Moles of ions = 4.031019 FU Sr(HCO 3 )2    23 1 mol Sr(HCO 3 )2   6.022 10 FU Sr(HCO3 )2  –4 –4 = 2.0076 × 10 = 2.01 × 10 mol of ions



4.20



Plan: To determine the total moles of ions released, write an equation that shows the compound dissociating into ions with the correct molar ratios. Convert mass and formula units to moles of compound and use the molar ratio to convert moles of compound to moles of ions. Solution: + 3– a) Each mole of K3PO4 forms 3 moles of K ions and 1 mole of PO4 ions, or a total of 4 moles of ions: + 3– K3PO4(s)  3K (aq) + PO4 (aq)  4 mol ions    = 3.0 mol of ions. Moles of ions = 0.75 mol K 3 PO 4  1 mol K 3 PO 4  2+ – b) Each mole of NiBr2 3H2O forms 1 mole of Ni ions and 2 moles of Br ions, or a total of 3 moles of ions: 2+ – NiBr2 3H2O(s)  Ni (aq) + 2Br (aq). The waters of hydration become part of the larger bulk of water. Convert mass to moles using the molar mass.  1 mol NiBr  3H O   3 mol ions   2 2  Moles of ions = 6.88103 g NiBr2  3H 2 O    272.54 g NiBr2  3H 2 O  1 mol NiBr2  3H 2 O 









–5

= 7.5732 × 10 = 7.57 × 10 mol of ions 3+ – c) Each mole of FeCl3 forms 1 mole of Fe ions and 3 moles of Cl ions, or a total of 4 moles of ions: 3+ – 23 FeCl3(s)  Fe (aq) + 3Cl (aq). Recall that a mole contains 6.022 × 10 entities, so a mole of FeCl3 contains 6.022 23 × 10 units of FeCl3, more easily expressed as formula units.   1 mol FeCl 3  4 mol ions   Moles of ions = 2.2310 22 FU FeCl3  23 1 mol FeCl   6.022 10 FU FeCl 3  3





= 0.148124 = 0.148 mol of ions

4-17

4.21





2–

b) Each mole of CuSO4 5H2O forms 1 mole of Cu ions and 1 mole of SO4 ions, or a total of 2 moles of ions: +2 2– CuSO4 5H2O(s)  Cu (aq) + SO4 (aq). The waters of hydration become part of the larger bulk of water. Convert mass to moles using the molar mass.  1 mol CuSO 4  5H 2 O    2 mol ions    Moles of ions = 3.86 g CuSO 4  5H 2 O   249.69 g CuSO  5H O  1 mol CuSO  5H O  4

–2

= 3.0918 × 10 = 3.09 × 10 mol of ions 2+

–

c) Each mole of NiCl2 forms 1 mole of Ni ions and 2 moles of Cl ions, or a total of 3 moles of ions: 2+ – 23 NiCl2(s)  Ni (aq) + 2Cl (aq). Recall that a mole contains 6.022 × 10 entities, so a mole of NiCl2 contains 23 6.022 × 10 units of NiCl2, more easily expressed as formula units.   1 mol NiCl 2  3 mol ions   Moles of ions = 8.66 10 20 FU NiCl 2    23  6.022 10 FU NiCl 2  1 mol NiCl 2 





–3

= 4.31418 × 10 = 4.31 × 10 mol of ions 4.22

 moles solute  Plan: In all cases, use the known quantities and the definition of molarity  M   to find the  L of solution  unknown quantity. Volume must be expressed in liters. The molar mass is used to convert moles to grams. The chemical formulas must be written to determine the molar mass. (a) You will need to convert milliliters to liters, multiply by the molarity to find moles, and convert moles to mass in grams. (b) Convert mass of solute to moles and volume from mL to liters. Divide the moles by the volume. (c) Multiply the molarity by the volume. Solution:

a) Calculating moles of solute in solution: 103 L   0.267 mol Ca(C 2 H 3 O2 )2   Moles of Ca(C2H3O2)2 = 185.8 mL    = 0.0496086 mol Ca(C2H3O2)2  1 mL  1L  





Converting from moles of solute to grams: 158.17 g Ca(C 2 H 3 O2 )2    Mass (g) of Ca(C2H3O2)2 = 0.0496086 mol Ca(C 2 H 3 O 2 )2   1 mol Ca(C 2 H 3 O 2 )2  = 7.84659 = 7.85 g Ca(C2H3O2)2 b) Converting grams of solute to moles:  1 mol KI   = 0.127108 moles KI Moles of KI = 21.1 g KI 166.0 g KI 





103 L   Volume (L) = 500. mL  1 mL  = 0.500 L Molarity of KI =

0.127108 mol KI = 0.254216 = 0.254 M KI 0.500 L

 0.850 mol NaCN    = 123.76 = 124 mol NaCN c) Moles of NaCN = 145.6 L   1L  





4-18

4.23

 

Plan: In all cases, use the known quantities and the definition of molarity  M 

moles solute   to find the L of solution 

unknown quantity. Volume must be expressed in liters. The molar mass is used to convert moles to grams. The chemical formulas must be written to determine the molar mass. (a) You will need to convert mass of solute to moles and divide by the molarity to obtain volume in liters, which is then converted to milliliters. (b) Multiply the volume by the molarity to obtain moles of solute. Use Avogadro’s number to determine the number of ions present. (c) Divide mmoles by milliliters; molarity may be expressed not only as moles/L, but also as mmoles/mL. Solution: a) Converting mass of solute to moles:  1 mol KOH   = 0.15006 mol KOH Moles of KOH = 8.42 g KOH  56.11 g KOH   1 L  = 0.066398 L KOH solution  2.26 mol   1 L  Volume (mL) of KOH solution = 0.066398 L KOH  3 = 66.39823 = 66.4 mL KOH solution 10 mL 

Volume (L) of KOH solution = 0.15006 mol KOH 

 2.3 mol CuCl 2  b) Moles of CuCl2 = 52 L   = 119.6 mol CuCl2  L 





 1 mol Cu 2    2+  = 119.6 mol Cu2+ ions Moles of Cu ions = 119.6 mol CuCl 2  1 mol CuCl 2  Converting moles of ions to number of ions:  6.022 10 23 Cu 2 ions   2+  = 7.2023 × 1025 = 7.2 × 1025 Cu2+ ions Number of Cu ions = 119.6 mol Cu 2 ions  1 mol Cu 2  ions  135 mmol glucose    = 0.490909 = 0.491 M glucose c) M glucose =    275 mL  –3 –3 Note: Since 1 mmol is 10 mol and 1 mL is 10 L, we can use these units instead of converting to mol and L since molarity is a ratio of mol/L. Molarity may be expressed not only as moles/L, but also as mmoles/mL.

4.24

 

Plan: In all cases, use the known quantities and the definition of molarity  M 

moles solute   to find the L of solution 

unknown quantity. Volume must be expressed in liters. The molar mass is used to convert moles to grams. The chemical formulas must be written to determine the molar mass. (a) Convert volume in milliliters to liters, multiply the volume by the molarity to obtain moles of solute, and convert moles to mass in grams. (b) The simplest way will be to convert the milligrams to millimoles. Molarity may be expressed not only as moles/L, but also as mmoles/mL. (c) Convert the milliliters to liters and find the moles of solute and moles of ions by multiplying the volume and molarity. Use Avogadro’s number to determine the number of ions present. Solution: a) Calculating moles of solute in solution: 103 L  5.62 102 mol K 2 SO 4    Moles of K2SO4 = 475 mL   = 0.026695 mol K2SO4  L  1 mL  





Converting moles of solute to mass:

4-19

174.26 g K 2 SO 4    = 4.6519 = 4.65 g K SO Mass (g) of K2SO4 = 0.026695 mol K 2 SO 4  2 4  1 mol K 2 SO 4 





b) Calculating mmoles of solute:  7.25 mg CaCl 2   1 mmol CaCl 2   mmoles of CaCl2 =    = 0.065327 mmoles CaCl2 1 mL 110.98 mg CaCl 2    Calculating molarity:  0.065327 mmol CaCl 2    = 0.065327 = 0.0653 M CaCl Molarity of CaCl2 =  2  1 mL   If you feel that molarity must be moles/liters then the calculation becomes: 3  7.25 mg CaCl 2  10 g  1 mL  1 mol CaCl 2   Molarity of CaCl2 =   1 mg 103 L 110.98 g CaCl  = 0.065327 = 0.0653 M CaCl2  1 mL 2    Notice that the two central terms cancel each other. c) Converting volume in L to mL: 103 L    = 0.001 L Volume (L) = 1 mL  1 mL  Calculating moles of solute and moles of ions:  0.184 mol MgBr2    = 1.84 × 10–4 mol MgBr Moles of MgBr2 = 0.001 L 2   1L   1 mol Mg 2    2+  = 1.84 × 10–4 mol Mg2+ ions Moles of Mg ions = 1.84 104 mol MgBr2  1 mol MgBr2   6.022 10 23 Mg 2  ions   2+ Number of Mg ions = 1.84 104 Mg 2  ions   1 mol Mg 2  ions  20 20 2+ = 1.1080 × 10 = 1.11 × 10 Mg ions



4.25



 

Plan: In all cases, use the known quantities and the definition of molarity  M 

moles solute   to find the L of solution 

unknown quantity. Volume must be expressed in liters. The molar mass is used to convert moles to grams. The chemical formulas must be written to determine the molar mass. (a) Convert mass of solute to moles and volume from mL to liters. Divide the moles by the volume. (b) You will need to convert mass of solute to moles and divide by the molarity to obtain volume in liters. (c) Divide the mmoles of solute by the molarity to obtain volume in mL. Solution: a) Calculating moles of solute:  1 mol AgNO 3    = 0.2707475 mol AgNO Moles of AgNO3 = 46.0 g AgNO3  3 169.9 g AgNO3  103 L    = 0.335 L Volume (L) = 335 mL  1 mL 





Calculating the molarity:

4-20

 0.2707475 mol AgNO3   = 0.80820 = 0.808 M AgNO3 Molarity of AgNO3 =   0.335 L  b) Calculating moles of solute:  1 mol MnSO 4    = 0.417218 mol MnSO Moles of MnSO4 = 63.0 g MnSO4  4 151.00 g MnSO 4  Calculating volume of solution:   1L   = 1.08368 = 1.08 L MnSO solution Volume (L) of solution = 0.417218 mol MnSO 4  4  0.385 mol MnSO 4    1 mL   c) Volume (mL) of ATP solution = 1.68 mmol ATP  2  6.44 10 mmol ATP 













= 26.087 = 26.1 mL ATP solution 4.26

Plan: To determine the total moles of ions released, write an equation that shows the compound dissociating into ions with the correct molar ratios. Convert the information given to moles of compound and use the molar ratio to convert moles of compound to moles of ions. Avogadro’s number is used to convert moles of ions to numbers of ions. Solution: 3+ – 3+ – a) Each mole of AlCl3 forms 1 mole of Al ions and 3 moles of Cl ions: AlCl3(s)  Al (aq) + 3Cl (aq). Molarity and volume must be converted to moles of AlCl3. 103 L  0.45 mol AlCl3    = 0.0585 mol AlCl  Moles of AlCl3 = 130. mL  3   1 mL    L   1 mol Al 3   3+  = 0.0585 = 0.058 mol Al3+ Moles of Al = 0.0585 mol AlCl 3  1 mol AlCl 3   6.022 10 23 Al3   3+  = 3.52287 × 1022 = 3.5 × 1022 Al3+ ions Number of Al ions = 0.0585 mol Al3   1 mol Al3   3 mol Cl   –  = 0.1755 = 0.18 mol Cl– Moles of Cl = 0.0585 mol AlCl 3  1 mol AlCl 3   6.022 10 23 Cl   –  = 1.05686 × 1023 = 1.1 × 1023 Cl– ions Number of Cl ions = 0.1755 mol Cl   1 mol Cl 













b) Each mole of Li2SO4 forms 2 moles of Li ions and 1 mole of SO4 ions: Li2SO4(s)  2Li (aq) + SO4 (aq). +

2–

103 L   2.59 g Li 2 SO 4  1 mol Li 2 SO 4   –4 Moles of Li2SO4 = 9.80 mL     = 2.3087 × 10 mol Li2SO4  1 mL    1L  109.94 g Li 2 SO 4 





 2 mol Li   +  = 4.6174 × 10–4 = 4.62 × 10–4 mol Li+ Moles of Li = 2.3087 104 mol Li 2 SO 4  1 mol Li 2 SO 4   6.022 10 23 Li   +  = 2.7806 × 1020 = 2.78 × 1020 Li+ ions Number of Li ions = 4.6174 104 mol Li   1 mol Li   1 mol SO 2   4 2– –4 –4 2– Moles of SO4 = 2.3087 104 mol Li 2 SO 4   = 2.3087 × 10 = 2.31 × 10 mol SO4 1 mol Li 2 SO 4 









4-21



Number of SO4 ions = 2.308710 2–

4

mol SO 4

2

 6.022 10 23 SO 2   4    2   1 mol SO 4 



2–

= 1.39030 × 10 = 1.39 × 10 SO4 ions c) Each mole of KBr forms 1 mole of K ions and 1 mole of Br ions: KBr(s)  K (aq) + Br (aq). 103 L  3.6810 22 FU KBr   1 mol KBr    = 0.01497 mol KBr  Moles of KBr = 245 mL   23    1 mL  L  6.022 10 FU KBr  +



–



 1 mol K   +  = 0.01497 = 1.50 × 10–2 mol K+ Moles of K = 0.01497 mol KBr 1 mol KBr   6.0221023 K    +  = 9.016 × 1021 = 9.02 × 1021 K+ ions Number of K ions = 0.01497 mol K    1 mol K  





1 mol Br  –  = 0.01497 = 1.50 × 10–2 mol Br– Moles of Br = 0.01497 mol KBr 1 mol KBr   6.02210 23 Br   –  = 9.016 × 1021 = 9.02 × 1021 Br– ions Number of Br ions = 0.01497 mol Br    1 mol Br  



4.27







 6.022 10 23 Mg 2   2+  = 9.27388 × 1022 = 9.3 × 1022 Mg2+ ions Number of Mg ions = 0.154 mol Mg 2  2   1 mol Mg    2 mol Cl   – Moles of Cl = 0.154 mol MgCl 2   = 0.308 = 0.31 mol Cl 1 mol MgCl 2   6.022 10 23 Cl  –    = 1.854776 × 1023 = 1.9 × 1023 Cl– ions Number of Cl ions = 0.308 mol Cl   1 mol Cl 









2–

b) Each mole of Al2(SO4)3 forms 2 moles of Al ions and 3 moles of SO4 ions: 3+ 2– Al2(SO4)3(s)  2Al (aq) + 3SO4 (aq). 103 L   0.22 g Al 2 (SO 4 )3  1 mol Al 2 (SO 4 )3   Moles of Al2(SO4)3 = 321 mL      1 mL  1L   342.14 g Al 2 (SO 4 )3  –4 = 2.06406 × 10 mol Al2(SO4)3  2 mol Al3   3+  = 4.12812 × 10–4 = 4.1 × 10–4 mol Al3+ Moles of Al = 2.06406 104 mol Al 2 (SO 4 )3  1 mol Al 2 (SO 4 )3 





4-22

 6.022 10 23 Al3   3+  = 2.4860 × 1020 = 2.5 × 1020 Al3+ ions Number of Al ions = 4.12812 104 mol Al3   1 mol Al3   3 mol SO 2   4 2–  = 6.19218 × 10–4 = 6.2 × 10–4 mol SO 2 Moles of SO4 = 2.06406 104 mol Al 2 (SO 4 )3  4 1 mol Al 2 (SO 4 )3   6.022 10 23 SO 2   4 2–  Number of SO4 ions = 6.19218104 mol SO 4 2   1 mol SO 4 2  20 20 2– = 3.7289 × 10 = 3.7 × 10 SO4 ions + – + – c) Each mole of CsNO3 forms 1 mole of Cs ions and 1 mole of NO3 ions: CsNO3(s)  Cs (aq) + NO3 (aq)













  8.8310 21 FU CsNO  1 mol CsNO 3 3   = 0.024194 mol CsNO  Moles of CsNO3 = 1.65 L  3 23    L  6.022 10 FU CsNO 3 





 1 mol Cs   = 0.024194 = 0.0242 mol Cs+ Moles of Cs = 0.024194 mol CsNO3  1 mol CsNO3 





 6.022 10 23 Cs   +  = 1.45695 × 1022 = 1.46 × 1022 Cs+ ions Number of Cs ions = 0.024194 mol Cs   1 mol Cs 





 1 mol NO3   = 0.024194 = 0.0242 mol NO3– Moles of NO3 = 0.024194 mol CsNO3  1 mol CsNO3 



–



 6.022 10 23 NO    – 22 22 – 3  Number of NO3 ions = 0.024194 mol NO3   = 1.45695 × 10 = 1.46 × 10 NO3 ions  1 mol NO   3  



4.28



Plan: These are dilution problems. Dilution problems can be solved by converting to moles and using the new volume; however, it is much easier to use M1V1 = M2V2. The dilution equation does not require a volume in liters; it only requires that the volume units match. In part (c), it is necessary to find the moles of sodium ions in each separate solution, add these two mole amounts, and divide by the total volume of the two solutions. Solution: a) M1 = 0.250 M KCl V1 = 37.00 mL M2 = ? V2 = 150.00 mL M1V1 = M2V2

M2 =

M1 V1 V2

0.250 M 37.00 mL = 0.061667 = 0.0617 M KCl 150.00 mL

b) M1 = 0.0706 M (NH4)2SO4

V1 = 25.71 mL

M2 = ?

V2 = 500.00 mL

M1V1 = M2V2

M2 =

M1 V1 V2

0.0706 M25.71 mL = 0.003630 = 0.00363 M (NH4)2SO4 500.00 mL

 103 L   0.348 mol NaCl  1 mol Na   + c) Moles of Na from NaCl solution = 3.58 mL      1 mL  1L  1 mol NaCl 





= 0.00124584 mol Na 2  103 L   6.8110 mol Na 2 SO 4  2 mol Na   + Moles of Na from Na2SO4 solution = 500. mL      1 mL  1L  1 mol Na 2 SO 4 





= 0.0681 mol Na + + + + Total moles of Na ions = 0.00124584 mol Na ions + 0.0681 mol Na ions = 0.06934584 mol Na ions Copyright McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

4-23

Total volume = 3.58 mL + 500. mL = 503.58 mL = 0.50358 L +

Molarity of Na = 4.29

total moles Na ions 0.06934584 mol Na ions + = = 0.1377057 = 0.138 M Na ions total volume 0.50358 L

V1 = ?

M2 = 0.8543 M Cu(NO3)2

V2 = 750.0 mL

M1V1= M2V2

V1 =

M 2 V2 M1

0.8543 M 750.0 mL = 312.5488 = 312.5 mL 2.050 M

b) M1 = 1.63 M CaCl2 –

M1 = 3.26 M Cl

 1.63 mol CaCl 2   2 mol Cl   – – M1 Cl =    = 3.26 M Cl ions  1 mol CaCl 2  1L

V1 = ?

–2

–

M2 = 2.86 × 10 M Cl ions

V2 = 350 mL

M1V1= M2V2

V1 =

M 2 V2 M1

2.86x102 M 350. mL 3.26 M

c) M1 = 0.155 M Li2CO3

= 3.07055= 3.07 mL

V1 = 18.0 mL

M2 = 0.0700 M Li2CO3

V2 = ?

M1V1= M2V2 M V 0.155 M 18.0 mL  V2 = 1 1 = = 39.8571 = 39.9 mL 0.0700 M  M2 4.30

Plan: Use the density of the solution to find the mass of 1 L of solution. Volume in liters must be converted to volume in mL. The 70.0% by mass translates to 70.0 g solute/100 g solution and is used to find the mass of HNO3 in 1 L of solution. Convert mass of HNO3 to moles to obtain moles/L, molarity. Solution:  1 mL  1.41 g solution   a) Mass (g) of 1 L of solution = 1 L solution  3   = 1410 g solution 1 mL  10 L    70.0 g HNO3    = 987 g HNO /L Mass (g) of HNO3 in 1 L of solution = 1410 g solution 3 100 g solution   1 mol HNO 3    = 15.6617 mol HNO b) Moles of HNO3 = 987 g HNO3  3  63.02 g HNO3  15.6617 mol HNO 3   = 15.6617 = 15.7 M HNO3 Molarity of HNO3 =   1 L solution 



4.31



Plan: Use the molarity of the solution to find the moles of H2SO4 in 1 mL. Convert moles of H2SO4 to mass of H2SO4, divide that mass by the mass of 1 mL of solution, and multiply by 100 for mass percent. Use the density of the solution to find the mass of 1 mL of solution. Solution: 3 18.3 mol H 2 SO 4  10 L   –2 a) Moles of H2SO4 in 1 mL =    = 1.83 × 10 mol H2SO4/mL 1L  1 mL  

4-24

 98.08 g H SO   2 4 b) Mass of H2SO4 in 1 mL = 1.83102 mol H 2 SO 4   = 1.79486 g H2SO4  1 mol H 2 SO 4 





1.84 g  = 1.84 g solution  1 mL 

Mass of 1 mL of solution = 1 mL  Mass percent =

4.32

1.79486 g H 2 SO 4 mass of H 2 SO 4 100 = 97.5467 = 97.5% H2SO4 by mass 100  = 1.84 g solution mass of solution  

Plan: Recall the definition of molarity  M 

moles solute   . Convert mass of solute to moles and volume from L of solution 

mL to liters. Divide the moles by the volume. Solution:  1 mol NaClO    = 0.2785896 mol NaClO Moles of NaClO = 20.5 g NaClO   74.44 g NaClO  103 L    = 0.375 L Volume (L) = 375 mL  1 mL 





 0.2753896 mol NaClO    = 0.73437 = 0.734 M NaClO Molarity of NaClO =    0.375 L 

4.33

Plan: The first part of the problem is a simple dilution problem (M1V1 = M2V2). The volume in units of gallons can be used. In part (b), convert mass of HCl to moles and use the molarity to find the volume that contains that number of moles. Solution: a) M1 = 11.7 M

V1 =

M 2 V2 M1

V1 = ?

M2 = 3.5 M

V2 = 3.0 gal

3.5 M 3.0 gal = 0.897436 gal = 0.90 gal 11.7 M

Instructions: Be sure to wear goggles to protect your eyes! Pour approximately 2.0 gal of water into the container. Add slowly and with mixing 0.90 gal of 11.7 M HCl into the water. Dilute to 3.0 gal with water. b) Converting from mass of HCl to moles of HCl:  1 mol HCl    = 0.264948 mol HCl Moles of HCl = 9.66 g HCl   36.46 g HCl  Converting from moles of HCl to volume:   1 mL  1L  Volume (mL) of solution = 0.264948 mol HCl   3  11.7 mol HCl  10 L 









= 22.64513 = 22.6 mL muriatic acid solution 4.34

Plan: Convert the mass of the seawater in kg to g and use the density to convert the mass of the seawater to volume in L. Convert mass of each compound to moles of compound and then use the molar ratio in the dissociation of the compound to find the moles of each ion. The molarity of each ion is the moles of ion divided by the volume of the seawater. To find the total molarity of the alkali metal ions [Group 1A(1)], add the moles of the alkali metal ions and divide by the volume of the seawater. Perform the same calculation to find the total molarity of the alkaline earth metal ions [Group 2A(2)] and the anions (the negatively charged ions). Solution: a) The volume of the seawater is needed.

4-25

3 10 3 g   cm 3   1 mL  10 L   Volume (L) of seawater = 1.00 kg    3  1 mL  = 0.97560976 L  1 kg  1.025 g  1 cm  





The moles of each ion are needed. If an ion comes from more than one source, the total moles are needed. NaCl: + – + – Each mole of NaCl forms 1 mole of Na ions and 1 mole of Cl ions: NaCl(s)  Na (aq) + Cl (aq)  1 mol NaCl    = 0.4534565 mol NaCl Moles of NaCl = 26.5 g NaCl   58.44 g NaCl   1 mol Na    +  = 0.4534565 mol Na+ Moles of Na = 0.4534565 mol NaCl  1 mol NaCl   1 mol Cl   –  = 0.4534565 mol Cl– Moles of Cl = 0.4534565 mol NaCl  1 mol NaCl 













MgCl2: 2+ – 2+ – Each mole of MgCl2 forms 1 mole of Mg ions and 2 moles of Cl ions: MgCl2(s)  Mg (aq) + 2Cl (aq)  1 mol MgCl 2    = 0.025207 mol MgCl Moles of MgCl2 = 2.40 g MgCl 2  2  95.21 g MgCl 2 





 1 mol Mg 2   2+  = 0.025207 mol Mg2+ Moles of Mg = 0.025207 mol MgCl 2  1 mol MgCl 2   2 mol Cl   –  = 0.050415 mol Cl– Moles of Cl = 0.025207 mol MgCl 2  1 mol MgCl 2  MgSO4: 2+ 2– 2+ 2– Each mole of MgSO4 forms 1 mole of Mg ions and 1 mole of SO4 ions: MgSO4(s)  Mg (aq) + SO4 (aq)  1 mol MgSO 4   Moles of MgSO4 = 3.35 g MgSO 4   = 0.0278308 mol MgSO4 120.37 g MgSO 4   1 mol Mg 2    2+  = 0.0278308 mol Mg2+ Moles of Mg = 0.0278308 mol MgSO 4  1 mol MgSO 4 















Moles of SO4

2–



 1 mol SO 4 2    = 0.0278308 mol SO 2– = 0.0278308 mol MgSO 4  4 1 mol MgSO 4 





CaCl2: 2+ – 2+ – Each mole of CaCl2 forms 1 mole of Ca ions and 2 moles of Cl ions: CaCl2(s)  Ca (aq) + 2Cl (aq) 2  1 mol CaCl 2   1 mol Ca   Moles of CaCl2 = 1.20 g CaCl 2    = 0.0108128 mol CaCl2 110.98 g CaCl 2  1 mol CaCl 2   1 mol Ca 2   2+  = 0.0108128 mol Ca2+ Moles of Ca = 0.0108128 mol CaCl 2  1 mol CaCl 2   2 mol Cl   – – Moles of Cl = 0.0108128 mol CaCl 2   = 0.0216255 mol Cl 1 mol CaCl 2 













KCl: Each mole of KCl forms 1 mole of K ions and 1 mole of Cl ions: KCl(s)  K (aq) + Cl (aq) +

–

4-26

 1 mol KCl    = 0.0140845 mol KCl Moles of KCl = 1.05 g KCl   74.55 g KCl   1 mol K    +  = 0.0140845 mol K+ Moles of K = 0.0140845 mol KCl  1 mol KCl   1 mol Cl   –  = 0.0140845 mol Cl– Moles of Cl = 0.0140845 mol KCl  1 mol KCl  NaHCO3: + – + – Each mole of NaHCO3 forms 1 mole of Na ions and 1 mole of HCO3 ions: NaHCO3(s)  Na (aq) + HCO3 (aq)  1 mol NaHCO 3    = 0.00374955 mol NaHCO Moles of NaHCO3 = 0.315 g NaHCO 3  3  84.01 g NaHCO3 

















 1 mol Na     = 0.00374955 mol Na+ Moles of Na = 0.00374955 mol NaHCO3  1 mol NaHCO3 





 1 mol HCO3   –  = 0.00374955 mol HCO – Moles of HCO3 = 0.00374955 mol NaHCO3  3 1 mol NaHCO3  NaBr + – + – Each mole of NaBr forms 1 mole of Na ions and 1 mole of Br ions: NaBr(s)  Na (aq) + Br (aq)  1 mol NaBr    = 0.0009524735 mol NaBr Moles of NaBr = 0.098 g NaBr  102.89 g NaBr   1 mol Na    + + Moles of Na = 0.0009524735 mol NaBr   = 0.0009524735 mol Na 1 mol NaBr   1 mol Br    –  = 0.0009524735 mol Br– Moles of Br = 0.0009524735 mol NaBr  1 mol NaBr 













Total moles of each ion: – – Cl : 0.4534565 + 0.050415 + 0.0216255 + 0.0140845 = 0.5395815 mol Cl +

0.4534565 + 0.00374955 + 0.0009524735 = 0.458158523 mol Na

0.025207 + 0.0278285 = 0.0530355 mol Mg

SO4 :

2–

0.0278308 mol SO4

0.0108128 mol Ca

Na : Mg : Ca : +

2–

0.0140845 mol K –

HCO3 : 0.00374955 mol HCO3 –

–

Br : 0.0009524735 mol Br Dividing each of the numbers of moles by the volume (0.97560976 L) and rounding to the proper number of significant figures gives the molarities. mol M= L –

M Cl = +

M Na =

0.5395815 mol Cl – = 0.55307 = 0.553 M Cl 0.97560976 L

0.45815823 mol Na + = 0.469612 = 0.470 M Na 0.97560976 L

4-27

0.0530355 mol Mg2 2+ = 0.054361 = 0.0544 M Mg 0.97560976 L 0.0278308 mol SO4 2 2– 2– M SO4 = = 0.028526 = 0.0285 M SO4 0.97560976 L 0.0108128 mol Ca2 2+ 2+ M Ca = = 0.011083 = 0.0111 M Ca 0.97560976 L 0.0140845 mol K + + MK = = 0.014437 = 0.0144 M K 0.97560976 L 0.00374955 mol HCO3 – – M HCO3 = = 0.003843 = 0.00384 M HCO3 0.97560976 L 2+

M Mg =

–

M Br =

0.0009524735 mol Br  – = 0.0009763 = 0.00098 M Br 0.97560976 L +

b) The alkali metal cations are Na and K . Add the molarities of the individual ions. + + 0.469612 M Na + 0.014437 M K = 0.484049 = 0.484 M total for alkali metal cations 2+ 2+ c) The alkaline earth metal cations are Mg and Ca . Add the molarities of the individual ions. 2+ 2+ 0.054361 M Mg + 0.011083 M Ca = 0.065444 = 0.0654 M total for alkaline earth cations – 2– – – d) The anions are Cl , SO4 , HCO3 , and Br . Add the molarities of the individual ions. – 2– – – 0.55307 M Cl + 0.028526 M SO4 + 0.003843 M HCO3 + 0.0009763 M Br = 0.5864153 = 0.586 M total for anions 4.35

Plan: Use the molarity and volume of the ions to find the moles of each ion. Multiply the moles of each ion by that ion’s charge to find the total moles of charge. Since sodium ions have a +1 charge, the total moles of charge equals the moles of sodium ions. Solution:  0.015 mol Ca 2    2+  = 15 mol Ca2+ Moles of Ca = 1.0 103 L  L    2 mol charge   2+  = 30. mol charge from Ca2+ Moles of charge from Ca = 15 mol Ca 2    1 mol Ca 2    0.0010 mol Fe 3   3+  = 1.0 mol Fe3+ Moles of Fe = 1.0 103 L    L   3 mol charge   3+  = 3.0 mol charge from Fe3+ Moles of charge from Fe = 1.0 mol Fe 3   1 mol Fe 3  Total moles of charge = 30. mol + 3.0 mol = 33 mol charge  1 mol Na    + + Moles Na = 33 mol charge   = 33 mol Na 1 mol charge  Plan: Review the definition of spectator ions. Solution: Ions in solution that do not participate in the reaction do not appear in a net ionic equation. These spectator ions remain as dissolved ions throughout the reaction. These ions are only present to balance charge.







4.36

4.37



Plan: Write the total ionic and net ionic equations for the reaction given. The total ionic equation shows all soluble ionic substances dissociated into ions. The net ionic equation eliminates the spectator ions. New equations may be written by replacing the spectator ions in the given equation by other spectator ions. Solution: The reaction given has the following total ionic and net ionic equations: 2+ – + 2– + – Total ionic equation: Ba (aq) + 2NO3 (aq) + 2Na (aq) + CO3 (aq)  BaCO3(s) + 2Na (aq) + 2NO3 (aq)

4-28

The spectator ions are underlined and are omitted: 2+ 2– Net ionic equation: Ba (aq) + CO3 (aq)  BaCO3(s) New equations will contain a soluble barium compound and a soluble carbonate compound. The “new” equations are: Molecular: BaCl2(aq) + K2CO3(aq)  BaCO3(s) + 2KCl(aq) 2+ – + 2– + – Total ionic: Ba (aq) + 2Cl (aq) + 2K (aq) + CO3 (aq)  BaCO3(s) + 2K (aq) + 2Cl (aq) Molecular: BaBr2(aq) + (NH4)2CO3(aq)  BaCO3(s) + 2NH4Br(aq) 2+ – + 2– + – Total ionic: Ba (aq) + 2Br (aq) + 2NH4 (aq) + CO3 (aq)  BaCO3(s) + 2NH4 (aq) + 2Br (aq) 4.38

If the electrostatic attraction between the ions is greater than the attraction of the ions for water molecules, the ions will form a precipitate. This is the basis for the solubility rules.

4.39

Plan: Write the new cation-anion combinations as the products of the reaction and use the solubility rules to determine if any of the new combinations are insoluble. The spectator ions are the ions that are present in the soluble ionic compound. Solution: a) Ca(NO3)2(aq) + 2NaCl(aq)  CaCl2(aq) + 2NaNO3(aq) Since the possible products (CaCl2 and NaNO3) are both soluble, no reaction would take place. b) 2KCl(aq) + Pb(NO3)2(aq)  2KNO3(aq) + PbCl2(s) According to the solubility rules, KNO3 is soluble but PbCl2 is insoluble so a precipitation reaction takes place. The + – K and NO3 would be spectator ions, because their salt is soluble.

4.40

Plan: Use the solubility rules to predict the products of this reaction. Ions not involved in the precipitate are spectator ions and are not included in the net ionic equation. Solution: + – Assuming that the left beaker is AgNO3 (because it has gray Ag ions) and the right must be NaCl, then the NO3 is + – – + blue, the Na is brown, and the Cl is green. (Cl must be green since it is present with Ag in the precipitate in the beaker on the right.) Molecular equation: AgNO3(aq) + NaCl(aq)  AgCl(s) + NaNO3(aq) + – + – + – Total ionic equation: Ag (aq) + NO3 (aq) + Na (aq) + Cl (aq)  AgCl(s) + Na (aq) + NO3 (aq) + – Net ionic equation: Ag (aq) + Cl (aq)  AgCl(s)

4.41

Plan: Write the new cation-anion combinations as the products of the reaction and use the solubility rules to determine if any of the new combinations are insoluble. The total ionic equation shows all soluble ionic substances dissociated into ions. The spectator ions are the ions that are present in the soluble ionic compound. The spectator ions are omitted from the net ionic equation. Solution: a) Molecular: Hg2(NO3)2(aq) + 2KI(aq)  Hg2I2(s) + 2KNO3(aq) 2+ – + – + – Total ionic: Hg2 (aq) + 2NO3 (aq) + 2K (aq) + 2I (aq)  Hg2I2(s) + 2K (aq) + 2NO3 (aq) 2+ – Net ionic: Hg2 (aq) + 2I (aq)  Hg2I2(s) + – Spectator ions are K and NO3 . b) Molecular: FeSO4(aq) + Sr(OH)2(aq)  Fe(OH)2(s) + SrSO4(s) 2+ 2– 2+ – Total ionic: Fe (aq) + SO4 (aq) + Sr (aq) + 2OH (aq)  Fe(OH)2(s) + SrSO4(s) Net ionic: This is the same as the total ionic equation because there are no spectator ions.

4.42

Plan: Write the new cation-anion combinations as the products of the reaction and use the solubility rules to determine if any of the new combinations are insoluble. The total ionic equation shows all soluble ionic substances dissociated into ions. The spectator ions are the ions that are present in the soluble ionic compound. The spectator ions are omitted from the net ionic equation. Solution: a) Molecular: 3CaCl2(aq) + 2Cs3PO4(aq)  Ca3(PO4)2(s) + 6CsCl(aq) 2+ – + 3– + – Total ionic: 3Ca (aq) + 6Cl (aq) + 6Cs (aq) + 2PO4 (aq)  Ca3(PO4)2(s) + 6Cs (aq) + 6Cl (aq) 2+ 3– Net ionic: 3Ca (aq) + 2PO4 (aq)  Ca3(PO4)2(s) + – Spectator ions are Cs and Cl .

4-29

b) Molecular: Na2S(aq) + ZnSO4(aq)  ZnS(s) + Na2SO4(aq) + 2– 2+ 2– + 2– Total ionic: 2Na (aq) + S (aq) + Zn (aq) + SO4 (aq)  ZnS(s) + 2Na (aq) + SO4 (aq) 2+ 2– Net ionic: Zn (aq) + S (aq)  ZnS(s) + 2– Spectator ions are Na and SO4 . 4.43

Plan: A precipitate forms if reactant ions can form combinations that are insoluble, as determined by the solubility rules in Table 4.1. Create cation-anion combinations other than the original reactants and determine if they are insoluble. Any ions not involved in a precipitate are spectator ions and are omitted from the net ionic equation. Solution: a) NaNO3(aq) + CuSO4(aq)  Na2SO4(aq) + Cu(NO3)2(aq) + 2– No precipitate will form. The ions Na and SO4 will not form an insoluble salt according to the first solubility rule 2+ – which states that all common compounds of Group 1A ions are soluble. The ions Cu and NO3 will not form an insoluble salt according to the solubility rule #2: All common nitrates are soluble. There is no reaction. + – b) A precipitate will form because silver ions, Ag , and bromide ions, Br , will combine to form a solid salt, silver bromide, AgBr. The ammonium and nitrate ions do not form a precipitate. Molecular: NH4Br(aq) + AgNO3(aq)  AgBr(s) + NH4NO3(aq) + – + – + – Total ionic: NH4 (aq) + Br (aq) + Ag (aq) + NO3 (aq)  AgBr(s) + NH4 (aq) + NO3 (aq) + – Net ionic: Ag (aq) + Br (aq)  AgBr(s)

4.44

Plan: A precipitate forms if reactant ions can form combinations that are insoluble, as determined by the solubility rules in Table 4.1. Create cation-anion combinations other than the original reactants and determine if they are insoluble. Any ions not involved in a precipitate are spectator ions and are omitted from the net ionic equation. Solution: a) Barium carbonate (BaCO3) precipitates since the solubility rules state that all common carbonates are insoluble. Molecular: K2CO3(aq) + Ba(OH)2(aq)  BaCO3(s) + 2KOH(aq) + 2– 2+ – + – Total ionic: 2K (aq) + CO3 (aq) + Ba (aq) + 2OH (aq)  BaCO3(s) + 2K (aq) + 2OH (aq) 2+ 2– Net ionic: Ba (aq) + CO3 (aq)  BaCO3(s) b) Aluminum phosphate (AlPO4) precipitates since most common phosphates are insoluble; the sodium nitrate is soluble. Molecular: Al(NO3)3(aq) + Na3PO4(aq)  AlPO4(s) + 3NaNO3(aq) 3+ – + 3– + – Total ionic: Al (aq) + 3NO3 (aq) + 3Na (aq) + PO4 (aq)  AlPO4(s) + 3Na (aq) + 3NO3 (aq) 3+ 3– Net ionic: Al (aq) + PO4 (aq)  AlPO4(s)

4.45

Plan: A precipitate forms if reactant ions can form combinations that are insoluble, as determined by the solubility rules in Table 4.1. Create cation-anion combinations other than the original reactants and determine if they are insoluble. Any ions not involved in a precipitate are spectator ions and are omitted from the net ionic equation. Solution: a) New cation-anion combinations are potassium nitrate (KNO3) and iron(III) chloride (FeCl3). The solubility rules state that all common nitrates and chlorides (with some exceptions) are soluble, so no precipitate forms. 3KCl(aq) + Fe(NO3)3(aq)  3KNO3(aq) + FeCl3(aq) b) New cation-anion combinations are ammonium chloride and barium sulfate. The solubility rules state that most chlorides are soluble; however, another rule states that sulfate compounds containing barium are insoluble. Barium sulfate is a precipitate and its formula is BaSO4. Molecular: (NH4)2SO4(aq) + BaCl2(aq)  BaSO4(s) + 2NH4Cl(aq) + 2– 2+ – + – Total ionic: 2NH4 (aq) + SO4 (aq) + Ba (aq) + 2Cl (aq)  BaSO4(s) + 2NH4 (aq) + 2Cl (aq) 2+ 2– Net ionic: Ba (aq) + SO4 (aq)  BaSO4(s)

4.46

Plan: A precipitate forms if reactant ions can form combinations that are insoluble, as determined by the solubility rules in Table 4.1. Create cation-anion combinations other than the original reactants and determine if they are insoluble. Any ions not involved in a precipitate are spectator ions and are omitted from the net ionic equation. Solution: a) New cation-anion combinations are nickel(II) sulfide (NiS) and sodium sulfate (Na2SO4). The solubility rules state that all compounds of Group 1A(1) like sodium are soluble; another rule states that common sulfide compounds are insoluble. Nickel(II) sulfide is a precipitate. Molecular: Na2S(aq) + NiSO4(aq)  NiS(s) + Na2SO4(aq) + –2 2+ 2– + 2– Total ionic: 2Na (aq) + S (aq) + Ni (aq) + SO4 (aq)  NiS(s) + 2Na (aq) + SO4 (aq) 2+ 2– Net ionic: Ni (aq) + S (aq)  NiS(s)

4-30

b) New cation-anion combinations are lead(II) bromide (PbBr2) and potassium nitrate (KNO3). The solubility rules state that all common nitrate compounds are soluble; another rule states that while most common bromide compounds are soluble, lead(II) bromide is insoluble. Lead(II) bromide is a precipitate. Molecular: Pb(NO3)2(aq) + 2KBr(aq)  PbBr2(s) + 2KNO3(aq) 2+ – + – + – Total ionic: Pb (aq) + 2NO3 (aq) + 2K (aq) + 2Br (aq)  PbBr2(s) + 2K (aq) + 2NO3 (aq) 2+ – Net ionic: Pb (aq) + 2Br (aq)  PbBr2(s) 4.47

Plan: Write a balanced equation for the chemical reaction described in the problem. By applying the solubility rules to the two possible products (NaNO3 and PbI2), determine that PbI2 is the precipitate. By using molar relationships, determine how many moles of Pb(NO3)2 are required to produce 0.628 g of PbI2. The molarity is calculated by dividing moles of Pb(NO3)2 by its volume in liters. Solution: The reaction is: Pb(NO3)2(aq) + 2NaI(aq)  PbI2(s) + 2NaNO3(aq).  1 mol PbI 2  1 mol Pb(NO3 )2   Moles of Pb(NO3)2 = 0.628 g PbI 2    = 0.001362256 mol Pb(NO3)2  1 mol PbI 2   461.0 g PbI 2 





Moles of Pb = moles of Pb(NO3)2 = 0.001362256 mol Pb

moles Pb2 0.001362256 mol  1 mL   3  = 0.035383 = 0.0354 M Pb2+ = 10 L  volume of Pb2 38.5 mL

Molarity of Pb = 4.48

Plan: Write a balanced equation for the chemical reaction described in the problem. By applying the solubility rules to the two possible products (KNO3 and AgCl), determine that AgCl is the precipitate. By using molar relationships, determine how many moles of AgNO3 are required to produce 0.842 g of AgCl. The molarity is calculated by dividing moles of AgNO3 by its volume in liters. Solution: The reaction is AgNO3(aq) + KCl(aq)  AgCl(s) + KNO3(aq).  1 mol AgCl  1 mol AgNO 3   Moles of AgNO3 = 0.842 g AgCl   1 mol AgCl  = 0.0058717 mol AgNO3 143.4 g AgCl  





Moles of Ag = moles of AgNO3 = 0.0058717 mol Ag

moles Ag 0.0058717 mol  1 mL   3  = 0.2348675 = 0.235 M Ag+ = 10 L  volume of Ag  25.0 mL

Molarity of Ag = 4.49

Plan: The first step is to write and balance the chemical equation for the reaction. Multiply the molarity and volume of each of the reactants to determine the moles of each. To determine which reactant is limiting, calculate the amount of barium sulfate formed from each reactant, assuming an excess of the other reactant. The reactant that produces less product is the limiting reagent. Use the limiting reagent and the mole ratio from the balanced chemical equation to determine the mass of barium sulfate formed. Solution: The balanced chemical equation is: BaCl2(aq) + Na2SO4(aq)  BaSO4(s) + 2NaCl(aq) 103 L   0.160 mol BaCl 2   Moles of BaCl2 = 35.0 mL    = 0.00560 mol BaCl2  1 mL  1L  





Finding the moles of BaSO4 from the moles of BaCl2 (if Na2SO4 is in excess): 1 mol BaSO 4    = 0.00560 mol BaSO Moles of BaSO4 from BaCl2 = 0.00560 moL BaCl 2  4  1 mol BaCl 2 





103 L   0.065 mol Na 2 SO 4   Moles of Na2SO4 = 58.0 mL    = 0.00377 mol Na2SO4  1 mL  1L  





4-31

Finding the moles of BaSO4 from the moles of Na2SO4 (if BaCl2 is in excess):  1 mol BaSO 4    = 0.00377 mol BaSO Moles BaSO4 from Na2SO4 = 0.00377 moL Na 2 SO 4  4 1 mol Na 2 SO 4 





Sodium sulfate is the limiting reactant. Converting from moles of BaSO4 to mass:

 233.4 g BaSO 4    = 0.879918 = 0.88 g BaSO Mass (g) of BaSO4 = 0.0377 moL BaSO 4  4  1 mol BaSO 4 



4.50



Plan: The first step is to write and balance the chemical equation for the reaction. To determine which reactant is limiting, calculate the amount of iron(III) sulfide formed from each reactant, assuming an excess of the other reactant. The reactant that produces less product is the limiting reagent. Multiply the molarity and volume (in L) of each of the reactants to determine the moles of each. Use the mole ratio from the balanced chemical equation and the molar mass of iron(III) sulfide to determine the mass of iron(III) sulfide formed. Solution: The balanced chemical equation is: 2FeCl3(aq) + 3CaS(aq)  Fe2S3(s) + 3CaCl2(aq) Finding the mass of Fe2S3 from the molarity and volume of FeCl3 (if CaS is in excess): Mass of Fe2S3 from FeCl3 =      3 2 3  2 3   = 0.870 g Fe2S3 62.0 mL FeCl3  3            3  2 3  Finding the mass of Fe2S3 from the molarity and volume of CaS (if FeCl3 is in excess): Mass of Fe2S3 from CaS =      2 3 2 3     = 0.889 g Fe2S3 45.0 mL CaS  3           2 3  FeCl3 is the limiting reactant. 0.870 g of Fe2S3 are formed.

4.51

Plan: A precipitate forms if reactant ions can form combinations that are insoluble, as determined by the solubility rules in Table 4.1. Create cation-anion combinations other than the original reactants and determine if they are insoluble. Any ions not involved in a precipitate are spectator ions and are omitted from the net ionic equation. Use the molar ratio in the balanced net ionic equation to calculate the mass of product. Solution: – – a) The yellow spheres cannot be ClO4 or NO3 as these ions form only soluble compounds. So the yellow sphere 2– must be SO4 . The only sulfate compounds possible that would be insoluble are Ag 2SO4 and PbSO4. The precipitate 2+ has a 1:1 ratio between its ions. Ag2SO4 has a 2:1 ratio between its ions. Therefore the blue spheres are Pb and the 2– yellow spheres are SO4 . The precipitate is thus PbSO4. 2+ 2– b) The net ionic equation is Pb (aq) + SO4 (aq)  PbSO4(s).  5.0104 mol Pb2+ 1 mol PbSO 4  303.3 g PbSO 4     c) Mass (g) of PbSO4 = 10 Pb2+ spheres  1 mol PbSO   1 Pb2+ sphere  1 mol Pb2+  4 = 1.5165 = 1.5 g PbSO4

4.52

4-32

Li2SO4(aq) + 2AgNO3(aq)  2LiNO3(aq) + Ag2SO4(s) Reaction 1 does not produce a precipitate since all common nitrate and chloride compounds are soluble. Reaction 2 does not produce a precipitate since all common perchlorate and chloride compounds are soluble. Reaction 4 produces a precipitate, PbBr2, but it has a cation:anion ratio of 1:2, instead of 2:1. Total ionic equation for Reaction 3 = + 2– + – + – 2Li (aq) + SO4 (aq) + 2Ag (aq) + 2NO3 (aq)  2Li (aq) + 2NO3 (aq) + Ag2SO4(s) + 2– Net ionic equation = 2Ag (aq) + SO4 (aq)  Ag2SO4(s) b) There are 4 unreacted spheres of ions. 23  2.5103 mol ions   6.02210 ions  = 6.022 × 1021 = 6.0 × 1021 ions Number of ions = 4 spheres    1 sphere  1 mol ions 

 2.5103 mol SO 2 ions 1 mol Ag SO  311.9 g Ag 2SO 4  4 2 4    c) Mass (g) of solid = 4 spheres of SO 4 2 ions  2   1 mol Ag2 SO 4  1 sphere   1 mol SO 4  = 3.119 = 3.1 g solid 4.53

Plan: Multiply the molarity of the CaCl2 solution by its volume in liters to determine the number of moles of CaCl2 reacted. Write the precipitation reaction using M to represent the alkali metal. Use the moles of CaCl2 reacted and the molar ratio in the balanced equation to find the moles of alkali metal carbonate. Divide the mass of that carbonate by the moles to obtain the molar mass of the carbonate. Subtract the mass of CO3 from the molar mass to obtain the molar mass of the alkali metal, which can be used to identify the alkali metal and the formula and name of the compound. Solution: The reaction is: M2CO3(aq) + CaCl2(aq)  CaCO3(s) + 2MCl(aq). Since the alkali metal M forms a +1 ion and carbonate is a –2 ion, the formula is M2CO3. 103 L  0.350 mol CaCl2   = 0.010885 mol CaCl2  Moles of CaCl2 = 31.10 mL  1 mL   1L

1 mol M2 CO3   = 0.010885 mol M2CO3 Moles of M2CO3 = 0.010885 mol CaCl 2   1 mol CaCl2 

mass of M2 CO3 1.50 g = = 137.804 g/mol moles of M2 CO3 0.010885 mol Molar mass (g/mol) of M2 = molar mass of M2CO3 – molar mass of CO3 = 137.804 g/mol – 60.01 g/mol = 77.79 g/mol Molar mass (g/mol) of M = 77.79 g/mol/2 = 38.895 g/mol = 38.9 g/mol This molar mass is closest to that of potassium; the formula of the compound is K2CO3, potassium carbonate. Molar mass (g/mol) of M2CO3 =

4.54

Plan: Multiply the molarity of the AgNO3 solution by its volume in liters to determine the number of moles of AgNO3 reacted. Write the precipitation reaction using M to represent the metal in the chloride compound. Use the moles of AgNO3 reacted and the molar ratio in the balanced equation to find the moles of metal chloride. Divide the mass of the chloride by the moles to obtain the molar mass of the chloride. Subtract the mass of Cl2 from the molar mass to obtain the molar mass of the metal, which can be used to identify the metal and the formula and name of the compound. Solution: The reaction is: MCl2(aq) + 2AgNO3(aq)  2AgCl(s) + M(NO3)2(aq). Since the metal M is a +2 ion and chloride is a –1 ion, the formula of the metal chloride is MCl2. 103 L  0.515 mol AgNO3   = 0.011536 mol AgNO  Moles of AgNO3 = 22.40 mL 3  1 mL   1L

 1 mol MCl 2   = 0.005768 mol MCl2 Moles of MCl2 = 0.011536 mol AgNO3   2 mol AgNO3 

4-33

mass of MCl2 0.750 g = = 130.0277 g/mol moles of MCl2 0.005768 mol Molar mass (g/mol) of M = molar mass of MCl2 – molar mass of Cl2 = 130.0277 g/mol – 70.90 g/mol = 59.1277 = 59.1 g/mol This molar mass is closest to that of nickel; the compound is NiCl2, nickel(II) chloride. Molar mass (g/mol) of MCl2 =

4.55

Plan: Write a balanced equation for the reaction. Find the moles of AgNO3 by multiplying the molarity and volume of the – AgNO3 solution; use the molar ratio in the balanced equation to find the moles of Cl present in the 25.00 mL sample. – Then, convert moles of Cl into grams, and convert the sample volume into grams using the given density. The mass – – percent of Cl is found by dividing the mass of Cl by the mass of the sample volume and multiplying by 100. Solution: – – The balanced equation is AgNO3(aq) + Cl (aq)  AgCl(s) + NO3 (aq). 103 L  0.2970 mol AgNO3    = 0.01592811 mol AgNO  Moles of AgNO3 = 53.63 mL  3  L   1 mL    1 mol Cl   35.45 g Cl   – – Mass (g) of Cl = 0.01592811 mol AgNO3  = 0.56465 g Cl    1 mol AgNO3   1 mol Cl  1.024 g   = 25.60 g sample Mass (g) of seawater sample = 25.00 mL   mL 















–

Mass % Cl = 4.56

0.56465 g Cl mass Cl – 100% = 100% = 2.20566 = 2.206% Cl mass sample 25.60 g sample

Plan: Write the reaction between aluminum sulfate and sodium hydroxide and check the solubility rules to determine the precipitate. Spectator ions are omitted from the net ionic equation. Find the moles of sodium hydroxide by multiplying its molarity by its volume in liters; find the moles of aluminum sulfate by converting grams per liter to moles per liter and multiplying by the volume of that solution. To determine which reactant is limiting, calculate the amount of precipitate formed from each reactant, assuming an excess of the other reactant, using the molar ratio from the balanced equation. The smaller amount of precipitate is the answer. Solution: a) According to the solubility rules, most common sulfate compounds are soluble, but most common hydroxides are insoluble. Aluminum hydroxide is the precipitate. Total ionic equation: Al2(SO4)3(aq) + 6NaOH(aq)  3Na2SO4(aq) + 2Al(OH)3(s) 3+ – Net ionic equation: Al (aq) + 3OH (aq)  Al(OH)3(s) 103 L 15.8 g Al 2 (SO 4 )3  1 mol Al 2 (SO 4 )3      b) Moles of Al2(SO4)3 = 627 mL    1 mL  L  342.14 g Al 2 (SO 4 )3 





= 0.028955 mol Al2(SO4)3  2 mol Al(OH)3   78.00 g Al(OH)3   Mass (g) of Al(OH)3 from Al2(SO4)3 = 0.028955 mol Al 2 (SO 4 )3   1 mol Al(OH)  1 mol Al 2 (SO 4 )3  3 





= 4.5170 g Al(OH)3 103 L  0.533 mol NaOH     = 0.0988715 mol NaOH Moles of NaOH = 185.5 mL   1 mL   L 





 2 mol Al(OH)3   78.00 g Al(OH)3   Mass (g) of Al(OH)3 from NaOH = 0.0988715 mol NaOH     6 mol NaOH   1 mol Al(OH)3  = 2.570659 = 2.57 g Al(OH)3 NaOH is the limiting reagent.





4-34

4.57

Plan: Write the chemical reaction between the two reactants. Then write the total ionic equation in which all soluble ionic substances are dissociated into ions. Omit spectator ions in the net ionic equation. Solution: The molecular equation is: H2SO4(aq) + Sr(OH)2(aq)  SrSO4(s) + 2H2O(l) The total ionic equation is: 2H (aq) + SO4 (aq) + Sr (aq) + 2OH (aq)  SrSO4(s) + 2H2O(l) According to the solubility rules, SrSO4 is insoluble and therefore does not dissociate into ions. Since there are no spectator ions, the total and net ionic equations are the same. +

2–

–

4.58

Plan: H is used to represent an acid while OH is used to represent the base. In a neutralization reaction, stoichiometric amounts of an acid and a base react to form water. Solution: + – The general equation for a neutralization reaction is H (aq) + OH (aq)  H2O(l).

4.59

Plan: Review the section on acid-base reactions. Solution: a) Any three of HCl, HBr, HI, HNO3, H2SO4, or HClO4 b) Any three of NaOH, KOH, Ca(OH)2, Sr(OH)2, Ba(OH)2 c) Strong acids and bases dissociate 100% into ions in aqueous solution.

4.60

Plan: Review the section on acid-base reactions. Solution: a) There are many possibilities including acetic acid (HC2H3O2), chlorous acid (HClO2), and nitrous acid (HNO2). All acids are weak except for the six strong acids listed in the text. b) NH3. c) Strong acids and bases dissociate 100% into ions and are therefore strong electrolytes; weak acids and bases dissociate much less than this (typically less than 10%) in aqueous solution and are therefore weak electrolytes. The electrical conductivity of a solution of a strong acid or base would be much higher than that of a weak acid or base of equal concentration.

4.61

Plan: Formation of a gaseous product or a precipitate drives a reaction to completion. Solution: a) The formation of a gas, SO2(g), and formation of water drive this reaction to completion, because both products remove reactants from solution. b) The formation of a precipitate, Ba3(PO4)2(s), will cause this reaction to go to completion. This reaction is one + – between an acid and a base, so the formation of water molecules through the combination of H and OH ions also drives the reaction.

4.62

Plan: Since strong acids and bases dissociate completely in water, these substances can be written as ions in a total ionic equation; since weak acids and bases dissociate into ions only to a small extent, these substances appear undissociated in total ionic equations. Solution: a) Acetic acid is a weak acid and sodium hydroxide is a strong base: Molecular equation: CH3COOH(aq) + NaOH(aq)  CH3COONa(aq) + H2O(l) + – + – Total ionic equation: CH3COOH (aq) + Na (aq) + OH (aq)  Na (aq) + CH3COO (aq) + H2O(l) + – – Net ionic equation (remove the spectator ion Na ): CH3COOH(aq) + OH (aq)  CH3COO (aq) + H2O(l) Hydrochloric acid is a strong acid: Molecular equation: HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l) + – + – + – Total ionic equation: H (aq) + Cl (aq) + Na (aq) + OH (aq)  Na (aq) + Cl (aq) + H2O(l)

4-35

Net ionic equation (remove the spectator ions Na and Cl ): H (aq) + OH (aq)  H2O(l) The difference in the net ionic equation is due to the fact that CH3COOH is a weak acid and dissociates very little while HCl is a strong acid and dissociates completely. b) When acetic acid dissociates in water, most of the species in the solution is un-ionized acid, CH3COOH(aq); the + – + – amounts of its ions, H and CH3COO , are equal but very small: [CH3COOH] >> [H ] = [CH3COO ]. +

4.63

–

Plan: The acids in this problem are all strong acids, so you can assume that all acid molecules dissociate completely + + to yield H ions and associated anions. One mole of HClO4, HNO3, and HCl each produce one mole of H upon + dissociation, so moles H = moles acid. Calculate the moles of acid by multiplying the molarity (moles/L) by the volume in liters. Solution: + – a) HClO4(aq) (aq) + ClO4 (aq)  0.25 mol  + + Moles H = mol HClO4 = 1.40 L   = 0.35 mol H  1 L  + – b) HNO3(aq) (aq) + NO3 (aq)

103 L  0.92 mol  +  = 6.256 × 10–3 = 6.3 × 10–3 mol H+  Moles H = mol HNO3 = 6.8 mL  1 mL  1 L  c) HCl(aq)

–

(aq) + Cl (aq)

 0.085 mol  + + Moles H = mol HCl = 2.6 L  = 0.221 = 0.22 mol H  1 L  4.64

Plan: The acids in this problem are all strong acids, so you can assume that all acid molecules dissociate completely + + to yield H ions and associated anions. One mole of HBr, HI, and HNO3 each produce one mole of H upon + dissociation, so moles H = moles acid. Calculate the moles of acid by multiplying the molarity (moles/L) by the volume in liters. Solution: + – a) HBr(aq) (aq) + Br (aq) 103 L  0.75 mol  + –3 –3 +   Moles H = mol HBr = 1.4 mL  1 mL  1 L  = 1.05 × 10 = 1.0 × 10 mol H b) HI(aq)

–

(aq) + I (aq)

103 L 1.98 mol  + –3 –3 +   Moles H = mol HI = 2.47 mL  1 mL  1 L  = 4.8906 × 10 = 4.89 × 10 mol H c) HNO3(aq)

4.65

–

(aq) + NO3 (aq)

103 L  0.270 mol  +  = 0.10665 = 0.107 mol H+  Moles H = mol HNO3 = 395 mL  1 mL  1 L  Plan: Remember that strong acids and bases can be written as ions in the total ionic equation but weak acids and bases cannot be written as ions. Omit spectator ions from the net ionic equation. Solution: a) KOH is a strong base and HBr is a strong acid; both may be written in dissociated form. KBr is a soluble compound since all Group 1A(1) compounds are soluble. Molecular equation: KOH(aq) + HBr(aq)  KBr(aq) + H2O(l) + – + – + – Total ionic equation: K (aq) + OH (aq) + H (aq) + Br (aq)  K (aq) + Br (aq) + H2O(l) – + Net ionic equation: OH (aq) + H (aq)  H2O(l) + – The spectator ions are K (aq) and Br (aq). b) NH3 is a weak base and is written in the molecular form. HCl is a strong acid and is written in the dissociated form (as ions). NH4Cl is a soluble compound, because all ammonium compounds are soluble. Molecular equation: NH3(aq) + HCl(aq)  NH4Cl(aq) + – + – Total ionic equation: NH3(aq) + H (aq) + Cl (aq)  NH4 (aq) + Cl (aq)

4-36

Net ionic equation: NH3(aq) + H (aq)  NH4 (aq) – Cl is the only spectator ion. +

4.66

Plan: Remember that strong acids and bases can be written as ions in the total ionic equation but weak acids and bases cannot be written as ions. Omit spectator ions from the net ionic equation. Solution: a) CsOH is a strong base and HNO3 is a strong acid; both may be written in dissociated form. CsNO3 is a soluble compound since all nitrate compounds are soluble. Molecular equation: CsOH(aq) + HNO3(aq)  CsNO3(aq) + H2O(l) + – + – + – Total ionic equation: Cs (aq) + OH (aq) + H (aq) + NO3 (aq)  Cs (aq) + NO3 (aq) + H2O(l) – + Net ionic equation: OH (aq) + H (aq)  H2O(l) + – Spectator ions are Cs and NO3 . b) HC2H3O2 is a weak acid and is written in the molecular form. Ca(OH)2 is a strong base and is written in the dissociated form (as ions). Ca(C2H3O2)2 is a soluble compound, because all acetate compounds are soluble. Molecular equation: Ca(OH)2(aq) + 2HC2H3O2(aq)  Ca(C2H3O2)2(aq) + 2H2O(l) 2+ – 2+ – Total ionic equation: Ca (aq) + 2OH (aq) + 2HC2H3O2 (aq)  Ca (aq) + 2C2H3O2 (aq) + 2H2O(l) – – Net ionic equation: OH (aq) + HC2H3O2(aq)  C2H3O2 (aq) + H2O(l) 2+ Spectator ion is Ca .

4.67

Plan: Write an acid-base reaction between CaCO3 and HCl. Remember that HCl is a strong acid. Solution: Calcium carbonate dissolves in HCl(aq) because the carbonate ion, a base, reacts with the acid to form H2CO3 which decomposes into CO2(g) and H2O(l). CaCO3(s) + 2HCl(aq)  CaCl2(aq) + H2CO3(aq) Total ionic equation: + – 2+ – CaCO3(s) + 2H (aq) + 2Cl (aq)  Ca (aq) + 2Cl (aq) + H2O(l) + CO2(g) Net ionic equation: + 2+ CaCO3(s) + 2H (aq)  Ca (aq) + H2O(l) + CO2(g)

4.68

Plan: Write an acid-base reaction between Zn(OH)2 and HNO3. Remember that HNO3 is a strong acid. Solution: Zinc hydroxide dissolves in HCl(aq) because the hydroxide ion, a base, reacts with the acid to form soluble zinc nitrate and water. Zn(OH)2(s) + 2HNO3(aq)  Zn(NO3)2(aq) + 2H2O(aq) Total ionic equation: + – 2+ – Zn(OH)2(s) + 2H (aq) + 2NO3 (aq)  Zn (aq) + 2NO3 (aq) + 2H2O(l) Net ionic equation: + 2+ Zn(OH)2(s) + 2H (aq)  Zn (aq) + 2H2O(l)

4.69

Plan: Convert the mass of calcium carbonate to moles, and use the mole ratio in the balanced chemical equation to find the moles of hydrochloric acid required to react with these moles of calcium carbonate. Use the molarity of HCl to find the volume that contains this number of moles. Solution: 2HCl(aq) + CaCO3(s)  CaCl2(aq) + CO2(g) + H2O(l) Converting from grams of CaCO3 to moles:  1 mol CaCO3   Moles of CaCO3 = 16.2 g CaCO3   = 0.161854 mol CaCO3 100.09 g CaCO3  Converting from moles of CaCO3 to moles of HCl:  2 mol HCl    = 0.323708 mol HCl Moles of HCl = 0.161854 mol CaCO3  1 mol CaCO3 

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

Converting from moles of HCl to volume: Copyright McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

4-37

  1L  1 mL   Volume (mL) of HCl = 0.323708 mol HCl   3  = 845.1906 = 845 mL HCl solution  0.383 mol HCl  10 L 



4.70



Plan: Convert the volume of NaOH solution to liters and multiply by the molarity of the solution to obtain moles of NaOH. Use the mole ratio in the balanced chemical equation to find the moles of NaH 2PO4 required to react with these moles of NaOH. Finally, convert moles of NaH2PO4 to moles. Solution: NaH2PO4(s) + 2NaOH(aq)  Na3PO4(aq) + 2H2O(l) 103 L    = 0.04374 mL Volume (L) = 43.74 mL   1 mL 





Finding moles of NaOH:  0.285 mol NaOH    = 0.0124659 mol NaOH Moles of NaOH = 0.04374 L    1L  Converting from moles of NaOH to moles of NaH2PO4: 1 mol NaH 2 PO 4    = 0.00623295 mol NaH PO Moles of NaH2PO4 = 0.0124659 mol NaOH  2 4  2 mol NaOH  Converting from moles of NaH2PO4 to mass: 119.98 g NaH 2 PO 4    = 0.747829 = 0.748 g NaH PO Mass (g) of NaH2PO4 = 0.00623295 mol NaH 2 PO 4  2 4  1 mol NaH 2 PO 4  Plan: Write a balanced equation. Find the moles of KOH from the molarity and volume information and use the molar ratio in the balanced equation to find the moles of acid present. Divide the moles of acid by its volume to determine the molarity. Solution: The reaction is: KOH(aq) + CH3COOH(aq)  CH3COOK(aq) + H2O(l) 103 L  0.1180 mol KOH    = 0.00306564 mol KOH Moles of KOH = 25.98 mL     1 mL  L  











4.71





1mol CH 3 COOH    = 0.00306564 mol CH COOH Moles of CH3COOH = 0.00306564 mol KOH  3  1 mol KOH   0.00306564 mol CH 3COOH  1 mL  Molarity of CH3COOH =   3  = 0.05839314 = 0.05839 M CH3COOH    52.50 mL 10 L  Plan: Write a balanced equation. Find the moles of NaOH from the molarity and volume information and use the molar ratio in the balanced equation to find the moles of acid present. Divide the moles of acid by its volume to determine the molarity. Solution: The reaction is: 2NaOH(aq) + H2SO4(aq)  Na2SO4(aq) + 2H2O(l) 103 L  0.1850 mol NaOH    = 0.00485625 mol NaOH  Moles of NaOH = 26.25 mL    L  1 mL   1mol H 2 SO 4    = 0.002428125 mol H SO Moles of H2SO4 = 0.00485625 mol NaOH  2 4  2 mol NaOH 



4.72





4-38

 0.002428125 mol H 2 SO 4  1 mL  Molarity of H2SO4 =   3  = 0.097125 = 0.09712 M H2SO4    25.00 mL 10 L  4.73

Plan: Write a balanced equation. Find the moles of H2SO4 from the molarity and volume information and use the molar ratio in the balanced equation to find the moles of NaHCO3 required to react with that amount of H2SO4. Divide the moles of NaHCO3 by its molarity to find the volume. Solution: The reaction is: 2NaHCO3(aq) + H2SO4(aq)  Na2SO4(aq) + 2 H2O(l) + 2CO2(g) 103 L  2.6 mol H 2 SO 4    = 0.2288 mol H SO  Moles of H2SO4 = 88 mL  2 4  L   1 mL    2 mol NaHCO 3    = 0.4576 mol NaHCO Moles of NaHCO3 = 0.2288 mol H 2 SO 4  3  1 mol H 2 SO 4    1L  1 mL   Volume (mL) of NaHCO3 = 0.4576 mol NaHCO3   3  1.6 mol NaHCO3  10 L  2 = 286 = 2.9 × 10 mL NaHCO3

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



4.74



Plan: Write a balanced equation. Find the moles of HCl added from the molarity and volume information and use the molar ratio in the balanced equation to find the moles of NaOH present. Find the volume of NaOH from the difference in buret readings. Divide the moles of NaOH by its volume to determine the molarity. Solution: The reaction is: NaOH(aq) + HCl(aq)  NaCl(aq) + H2O(l) 103 L  0.1528 mol HCl    = 0.00382 mol HCl  Moles of HCl = 25.00 mL    1 mL  L  1mol NaOH    = 0.00382 mol NaOH Moles of NaOH = 0.00382 mol HCl   1 mol HCl 









Volume (mL) of NaOH = final buret reading – initial buret reading = 39.21 mL – 2.24 mL = 36.97 mL NaOH  0.00382 mol NaOH  1 mL  Molarity of NaOH =  103 L  = 0.103327 = 0.1033 M NaOH   36.97 mL 4.75

Plan: Write balanced equations for the reaction of NaOH with oxalic acid, benzoic acid, and HCl. Find the moles of added NaOH from the molarity and volume information; then use the molarity and volume information for HCl to find the moles of HCl required to react with the excess NaOH. Use the molar ratio in the NaOH/HCl reaction to find the moles of excess NaOH. The moles of NaOH required to titrate the acid samples is the difference of the added NaOH and the excess NaOH. Let x = mass of benzoic acid and 0.3471 – x = mass of oxalic acid. Convert the mass of each acid to moles using the molar mass and use the molar ratios in the balanced reactions to find the amounts of each acid. Mass percent is calculated by dividing the mass of benzoic acid by the mass of the sample and multiplying by 100. Solution: Oxalic acid is H2C2O4 and benzoic acid is HC7H5O2. The reactions are: NaOH(aq) + HCl(aq)  NaCl(aq) + H2O(l) 2NaOH(aq) + H2C2O4(aq)  Na2C2O4(aq) + 2H2O(l) NaOH(aq) + HC7H5O2(aq)  NaC7H5O2(aq) + H2O(l) 103 L  0.1000 mol NaOH   = 0.01000 mol NaOH  Moles of NaOH added = 100.0 mL  1L  1 mL 

4-39

103 L  0.2000 mol HCl   = 0.004000 mol HCl  Moles of added HCl = 20.00 mL  1 mL   1L 1 mol NaOH  = 0.004000 mol NaOH  1 mol HCl 

Moles of excess NaOH = 0.004000 mol HCl 

Moles of NaOH required to titrate sample = moles NaOH added – moles excess NaOH = 0.01000 mol – 0.004000 mol = 0.006000 mol NaOH Let x = mass of HC7H5O2 and 0.3471 – x = mass of H2C2O4 Moles of NaOH required to titrate HC7H5O2 =  1 mol HC 7 H 5 O2   1 mol NaOH  = 0.0081887x x g HC 7 H 5 O2  1 mol HC H O  122.12 g HC 7 H 5 O 2  7 5 2

 1 mol H 2 C 2 O 4   2 mol NaOH  Moles of NaOH required to titrate H2C2O4 = (0.3471  x) g H 2 C 2 O 4  1 mol H C O   90.04 g H 2 C 2 O 4  2 2 4 = 0.0077099 – 0.022212x Moles of NaOH required to titrate sample = 0.006000 mol = 0.0081887x + (0.0077099 – 0.022212x ) 0.006000 = 0.0077099 – 0.014023x –0.0017099 = –0.014021x 0.12195 = x = mass of HC7H5O2 mass of HC 7 H5 O2 0.12195 g 100 = 100 = 35.1348 = 35.13% Mass % of HC7H5O2 = mass of sample 0.3471 g 4.76

Plan: Balance the reaction. Convert the amount of UO2 from kg to g to moles; use the molar ratio in the balanced reaction to find the moles of HF required to react with the moles of UO2. Divide moles of HF by its molarity to calculate the volume. Solution: The reaction is: UO2(s) + 4HF(aq)  UF4(s) + 2H2O(l) 103 g   1 mol UO 2   Moles of UO2 = 2.15 kg UO2    = 7.96296 mol UO2  1 kg   270.0 g UO2   4 mol HF    = 31.85184 mol HF Moles of HF = 7.96296 mol UO 2  1 mol UO2 

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



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  1L   = 13.2716 = 13.3 L HF Volume (L) of HF = 31.85184 mol HF   2.40 mol HF 

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4.77



Plan: Write balanced reactions between HNO3 and each of the bases. Find the moles of HNO3 from its molarity and volume. Let x = mass of Al(OH)3 and 0.4826 – x = mass of Mg(OH)2. Convert the mass of each base to moles using the molar mass and use the molar ratios in the balanced reactions to find the amounts of each base. Mass percent is calculated by dividing the mass of Al(OH)3 by the mass of the sample and multiplying by 100. Solution: The reactions are: 3HNO3(aq) + Al(OH)3(aq)  Al(NO3)3(aq) + 3H2O(l) 2HNO3(aq) + Mg(OH)2(aq)  Mg(NO3)2(aq) + 2H2O(l) 103 L 1.000 mol HNO3   = 0.0173 mol HNO3  Moles of HNO3 = 17.30 mL  1 mL   1L Let x = mass of Al(OH)3 and 0.4826 – x = mass of Mg(OH)2 Moles of HNO3 required to titrate Al(OH)3 =  1 mol Al(OH)3  3 mol HNO 3    = 0.038462x x g Al(OH)3  1 mol Al(OH)3   78.00 g Al(OH)3 

4-40

Moles of HNO3 required to titrate Mg(OH)2 =

 1 mol Mg(OH)2   2 mol HNO3  1 mol Mg(OH)   58.33 g Mg(OH)2  2

(0.4826  x) g Mg(OH)2 

= 0.016547 – 0.034288x Moles of HNO3 required to titrate sample = 0.0173 mol = 0.038462x + (0.016547 – 0.034288x) 0.0173 = 0.004174x + 0.016547 0.000753 = 0.004174x 0.18040 g = x = mass of Al(OH)3 mass of Al(OH)3 0.18040 g 100 = 100 = 37.3808 = 37.38% Mass % of Al(OH)3 = mass of sample 0.4826 g 4.78

Plan: Consult the Rules for Assigning an Oxidation Number. The sum of the O.N. values for the atoms in a molecule equals zero, while the sum of the O.N. values for the atoms in an ion equals the ion’s charge. Solution: a) Since H is bonded to a nonmetal, the O.N. of each hydrogen is +1, for a total of +2. The O.N.s must add to zero, so the O.N. of S is –2. b) The O.N. of each oxygen is –2 for a total of –6. Since the O.N.s must add to –2, the charge of the ion, the O.N. of S is +4.

4.79

Plan: Use the Rules for Assigning an Oxidation Number to assign each atom an O.N. If the O.N.s of atoms change as the reactants become products, the reaction is a redox reaction. Solution: The O.N. of hydrogen is +1 in NH3, for a total of +3. The O.N. of N is then –3. In HCl, the O.N. of H is +1 and that of Cl is –1. The oxidation numbers of N, H, and Cl are the same in the products. No, since no element changes oxidation number, this cannot be a redox reaction.

4.80

Plan: Recall that oxidation is the loss of electrons and reduction is the gain of electrons. Solution: An oxidizing agent causes something else to be oxidized; i.e., to lose electrons. The oxidizing agent accepts these electrons and is reduced.

4.81

Plan: Recall that oxidation is the loss of electrons and reduction is the gain of electrons. Solution: The electrons that a substance gains during reduction must come from somewhere. So there must be an oxidation in which electrons are lost, to provide the electrons gained during reduction.

4.82

Plan: An oxidizing agent gains electrons and therefore has an atom whose oxidation number decreases during the reaction. Use the Rules for Assigning an Oxidation Number to assign S in H2SO4 an O.N. and see if this oxidation number changes during the reaction. An acid transfers a proton during reaction. Solution: a) In H2SO4, hydrogen has an O.N. of +1, for a total of +2; oxygen has an O.N. of –2 for a total of –8. The S has an O.N. of +6. In SO2, the O.N. of oxygen is –2 for a total of –4 and S has an O.N. of +4. So the S has been reduced from +6 to +4 and is an oxidizing agent. Iodine is oxidized during the reaction. b) The oxidation number of S is +6 in H2SO4; in BaSO4, Ba has an O.N. of +2, the four oxygen atoms have a total O.N. of –8, and S is again +6. Since the oxidation number of S (or any of the other atoms) did not change, this is not a – redox reaction. H2SO4 transfers a proton to F to produce HF, so it acts as an acid.

4.83

Plan: Use the Rules for Assigning an Oxidation Number to assign each atom an oxidation number. An oxidizing agent gains electrons and therefore has an atom whose oxidation number decreases during the reaction. A reducing agent loses electrons and therefore has an atom whose oxidation number increases during the reaction.

4-41

Solution: +3

–4

–3 +1

+4 –2

8NH3(g) + 6NO2(g)

+2 0

+1 –2

(g) + 12H2O(l) 2

Oxidizing agent = NO2; Reducing agent = NH3; the NO2 is the oxidizing agent (O.N. of N decreases from +4 in NO2 to 0 in N2), and the NH3 is the reducing agent (O.N. of N increases from –3 in NH3 to 0 in N2). 4.84

Plan: Consult the Rules for Assigning an Oxidation Number. The sum of the O.N. values for the atoms in a molecule equals zero, while the sum of the O.N. values for the atoms in an ion equals the ion’s charge. Solution: a) CF2Cl2. The rules dictate that F and Cl each have an O.N. of –1; two F and two Cl yield a sum of –4, so the O.N. of C must be +4. C = +4 b) Na2C2O4. The rule dictates that Na [Group 1A(1)] has a +1 O.N.; rule 5 dictates that O has a –2 O.N.; two Na and four O yield a sum of –6 [2(+1) + 4(–2)]. Therefore, the total of the O.N.s on the two C atoms is +6 and each C is +3. C = +3 – c) HCO3 . H is combined with nonmetals and has an O.N. of +1; O has an O.N. of –2 and three O have a sum of –6. To have an overall oxidation state equal to –1, C must be +4 because (+1) + (+4) + (–6) = –1. C = +4 d) C2H6. Each H has an O.N. of +1; six H gives +6. The sum of O.N.s for the two C atoms must be –6, so each C is –3. C = –3

4.85

Plan: Consult the Rules for Assigning an Oxidation Number. The sum of the O.N. values for the atoms in a molecule equals zero. Solution: a) KBr. Potassium [Group 1A(1)] has an O.N. of +1. Therefore, the O.N. of Br must be –1. Br = –1 b) BrF3. According to Rule 6, a halogen has an O.N. of –1 when it is in combination with a halogen lower in the group. So the O.N. of F is –1 and the total for the three F atoms is –3; the O.N. of Br must be +3. Br = +3 c) HBrO3. Hydrogen has an O.N. of +1; the O.N. of oxygen is –2 and the total of three oxygen atoms is –6. To have an overall sum of O.N.s = 0, Br must be +5 since [(+1) + (+5) + (–6)] = 0. Br = +5 d) CBr4. The O.N. of halogens in combination with a nonmetal like carbon is –1; the total of four Br atoms is –4. The O.N. of C must be +4. Br = –1

4.86

Plan: Consult the Rules for Assigning an Oxidation Number. The sum of the O.N. values for the atoms in a molecule equals zero, while the sum of the O.N. values for the atoms in an ion equals the ion’s charge. Solution: a) NH2OH. Hydrogen has an O.N. of +1, for a total of +3 for the three hydrogen atoms. Oxygen has an O.N. of –2. The O.N. of N must be –1 since [(–1) + (+3) + (–2)] = 0. N = –1 b) N2F4. The O.N. of each fluorine is –1 for a total of –4; the sum of the O.N.s for the two N atoms must be +4, so each N has an O.N. of +2. N = +2 + c) NH4 . The O.N. of each hydrogen is +1 for a total of +4; the O.N. of nitrogen must be –3 since the overall sum of the O.N.s must be +1: [(–3) + (+4)] = +1. N = –3 d) HNO2. The O.N. of hydrogen is +1 and that of each oxygen is –2 for a total of –4 from the oxygens. The O.N. of nitrogen must be +3 since [(+1) + (+3) + (–4)] = 0. N = +3 Plan: Consult the Rules for Assigning an Oxidation Number. The sum of the O.N. values for the atoms in a molecule equals zero. Solution: a) SOCl2. The O.N. of oxygen is –2 and that of each chlorine is –1 for a total of –2 for the two chlorine atoms. The O.N. of sulfur must be +4 since [(+4) + (–2) + (–2)] = 0. S = +4 b) H2S2. The O.N. of each hydrogen is +1, for a total of +2. The sum of the O.N.s of the two sulfur atoms must equal –2, so the O.N. of each S atom is –1. S = –1

4.87

4-42

c) H2SO3. The O.N. of each hydrogen atom is +1 for a total of +2; the O.N. of each oxygen atom is –2 for a total of –6. The O.N. of the sulfur must be +4 since [(+2) + (+4) + (–6)] = 0. S = +4 d) Na2S. The O.N. of each sodium [Group 1A(1)] is +1, for a total of +2. The O.N. of sulfur is –2. S = –2 4.88

Plan: Consult the Rules for Assigning an Oxidation Number. The sum of the O.N. values for the atoms in a molecule equals zero, while the sum of the O.N. values for the atoms in an ion equals the ion’s charge. Solution: a) AsH3. H is combined with a nonmetal, so its O.N. is +1 (Rule 3). Three H atoms have a sum of +3. To have a sum of 0 for the molecule, As has an O.N. of –3. As = –3 – b) H2AsO4 . The O.N. of H in this compound is +1, for a total of +2. The O.N. of each oxygen is –2, for a total of –8. As has an O.N. of +5 since [(+2) + (+5) + (–8)] = –1, the charge of the ion. As = +5 c) AsCl3. Each chlorine has an O.N. of –1, for a total of –3. The O.N. of As is +3. As = +3

4.89

Plan: Consult the Rules for Assigning an Oxidation Number. The sum of the O.N. values for the atoms in a molecule equals zero, while the sum of the O.N. values for the atoms in an ion equals the ion’s charge. Solution: 2– a) H2P2O7 . The O.N. of each hydrogen is +1, for a total of +2; the O.N. of each oxygen is –2, for a total of –14. The sum of the O.N.s of the two phosphorus atoms must be +10 since [(+2) + (+10) + (–14)] = –2, the charge of the ion. Each of the two phosphorus atoms has an O.N. of +5. P = +5 + b) PH4 . The O.N. of each hydrogen is +1, for a total of +4. The O.N. of P is –3 since [(–3) + (+4)] = +1, the charge of the ion. P = –3 c) PCl5. The O.N. of each Cl is –1, for a total of –5. The O.N. of P is therefore +5. P = +5

4.90

Plan: Consult the Rules for Assigning an Oxidation Number. The sum of the O.N. values for the atoms in a molecule equals zero, while the sum of the O.N. values for the atoms in an ion equals the ion’s charge. Solution: 2– a) MnO4 . The O.N. of each oxygen is –2, for a total of –8; the O.N. of Mn must be +6 since [(+6) + (–8)] = –2, the charge of the ion. Mn = +6 b) Mn2O3. The O.N. of each oxygen is –2, for a total of –6; the sum of the O.N.s of the two Mn atoms must be +6. The O.N. of each manganese is +3. Mn = +3 c) KMnO4. The O.N. of potassium is +1 and the O.N. of each oxygen is –2, for a total of –8. The O.N. of Mn is +7 since [(+1) + (+7) + (–8)] = 0. Mn = +7

4.91

Plan: Consult the Rules for Assigning an Oxidation Number. The sum of the O.N. values for the atoms in a molecule equals zero, while the sum of the O.N. values for the atoms in an ion equals the ion’s charge. Solution: a) CrO3. The O.N. of each oxygen atom is –2, for a total of –6. The O.N. of chromium must be +6. Cr = +6 2– b) Cr2O7 . The O.N. of each oxygen is –2, for a total of –14. The sum of the O.N.s of the two chromium atoms must be +12 since [(+12) + (–14)] = –2, the charge of the ion. Each of the two chromium atoms has an O.N. of +6. Cr = +6 2– c) Cr2(SO4)3. It is convenient to treat the polyatomic ion SO4 as a unit with a –2 charge, for a total of –6 for the three sulfate ions. The sum of the two chromium atoms must be +6 and the O.N. of each chromium atom is +3. Cr = +3

4.92

Plan: First, assign oxidation numbers to all atoms following the rules. The reactant that is the reducing agent contains an atom that is oxidized (O.N. increases from the left side to the right side of the equation). The reactant that is the oxidizing agent contains an atom that is reduced (O.N. decreases from the left side to the right side of the equation). Recognize that the agent is the compound that contains the atom that is oxidized or reduced, not just the atom itself. Solution: a) +2 +6 –8

–8

+1 +3 –2

+7 –2

–4 +1

+4 –2

+2 +1 –2

5H2C2O4(aq) + 2MnO4 (aq) + 6H (aq)  2Mn (aq) + 10CO2(g) + 8H2O(l) –

4-43

–

Mn in MnO4 changes from +7 to +2 (reduction). Therefore, MnO4 is the oxidizing agent. C in H2C2O4 changes from +3 to +4 (oxidation), so H2C2O4 is the reducing agent. b) –6 +2 0

+5 –2

+2 –2

+1 –2

3Cu(s) + 8H (aq) + 2NO3 (aq)  3Cu (aq) + 2NO(g) + 4H2O(l) – Cu changes from 0 to +2 (is oxidized) and Cu is the reducing agent. N changes from +5 (in NO3 ) to +2 (in NO) – and is reduced, so NO3 is the oxidizing agent. +

4.93

–

+1 –1

+1 –2

2H (aq) + H2O2(aq) + 2Fe (aq)  2Fe (aq) + 2H2O(l) Oxygen changes from –1 in H2O2 to –2 in H2O (is reduced) so H2O2 is the oxidizing agent. Fe changes from +2 2+ to +3 (is oxidized) so Fe is the reducing agent. +

4.94

–1

–

+5 –2 –

8H (aq) + 6Cl (aq) + Sn(s) + 4NO3 (aq)

+4 –1 

+4 –2

+1 –2

2–

SnCl6 (aq) + 4NO2(g) + 4H2O(l)

–

Nitrogen changes from an O.N. of +5 in NO3 to +4 in NO2 (is reduced) so NO3 is the oxidizing agent. Sn 2– changes from an O.N. of 0 to an O.N. of +4 in SnCl6 (is oxidized) so Sn is the reducing agent. b) –8 +2 +7 –2

–1

+1 –2

2MnO4 (aq) + 10Cl (aq) + 16H (aq)  5Cl2(g) + 2Mn (aq) + 8H2O(l) –

–

Manganese changes from an O.N. of +7 in MnO4 to an O.N. of +2 in Mn (is reduced) so MnO4 is the oxidizing – – agent. Chlorine changes its O.N. from –1 in Cl to 0 as the element Cl2 (is oxidized) so Cl is the reducing agent. 4.95

+12 –14

–6

–8

4-44

+1 +6 –2 +4 –2 +3 +6 –2 +1 –2 + 2– 2– 3+ 2– 8H (aq) + Cr2O7 (aq) + 3SO3 (aq)  2Cr (aq) + 3SO4 (aq) + 2H2O(l) 2– 3+ 2– Chromium changes from an O.N. of +6 in Cr2O7 to +3 in Cr (is reduced) so Cr2O7 is the oxidizing agent. 2– 2– 2– Sulfur changes from an O.N. of +4 in SO3 to +6 in SO4 (is oxidized) so SO3 is the reducing agent. b) –6 +2 –8 +4 +3 +5 –2 0 –2 +1 +1 –2 +2 –2 +1 –3 +1 – – 2– NO3 (aq) + 4Zn(s) + 7OH (aq) + 6H2O(l)  4Zn(OH)4 (aq) + NH3(aq) – – Nitrogen changes from an O.N. of +5 in NO3 to an O.N. of –3 in NH3 (is reduced) so NO3 is the oxidizing agent. 2– Zinc changes from an O.N. of 0 to an O.N. of +2 in Zn(OH) 4 (is oxidized) so Zn is the reducing agent. 4.96

–

Plan: Find the moles of MnO4 from the molarity and volume information. Use the molar ratio in the balanced equation to find the moles of H2O2. Multiply the moles of H2O2 by its molar mass to determine the mass of H2O2 present. Mass percent is calculated by dividing the mass of H2O2 by the mass of the sample and multiplying by 100. Assign oxidation numbers to all atoms following the rules. The reactant that is the reducing agent contains an atom that is oxidized (O.N. increases from the left side to the right side of the equation). Solution: 103 L  0.105 mol MnO4   –  –3 –3 – a) Moles of MnO4 = 43.2 mL    = 4.536 × 10 = 4.54 × 10 mol MnO4  L  1 mL 





 5 mol H O   2 2  b) Moles of H2O2 = 4.536 103 MnO4   = 0.01134 = 0.0113 mol H2O2  2 mol MnO4   34.02 g H 2 O 2    = 0.3857868 = 0.386 g H O c) Mass (g) of H2O2 = 0.01134 mol H 2 O2  2 2  1 mol H 2 O 2 









d) Mass percent of H2O2 = e)

–8

mass of H 2 O2 mass of sample

100 =

0.3857868 g H 2 O2 14.8 g sample

+2 –2

100 = 2.606668 = 2.61% H2O2

+7 –2 +1 –1 +1 0 +2 +1 –2 – + 2+ 2MnO4 (aq) + 5H2O2(aq) + 6H (aq)  5O2(g) + 2Mn (aq) + 8H2O(l) The O.N. of oxygen increases from –1 in H2O2 to 0 in O2 and is therefore oxidized while the O.N. of – 2+ Mn decreases from +7 in MnO4 to +2 in Mn and is reduced. H2O2 is the reducing agent. 4.97

2–

Plan: Find the moles of Cr2O7 from the molarity and volume information. Use the molar ratio in the balanced equation to find the moles of C2H5OH and multiply the moles of C2H5OH by its molar mass to determine the mass of C2H5OH present. Mass percent is calculated by dividing the mass of C2H5OH by the mass of the sample and multiplying by 100. Solution: 2 103 L   0.05961 mol Cr2 O 7   2– 2– Moles of Cr2O7 = 35.46 mL    = 0.0021138 mol Cr2O7  1 mL   1L   1 mol C 2 H 5 OH    = 0.0010569 mol C H OH Moles of C2H5OH = 0.0021138 mol Cr2 O 72  2 5  2 mol Cr2 O 7 2   46.07 g C 2 H 5 OH    = 0.0486914 g C H OH Mass (g) of C2H5OH = 0.0010569 mol C 2 H 5 OH  2 5  1 mol C 2 H 5 OH  mass of C 2 H 5 OH 0.0486914 g C 2 H 5OH Mass percent of C2H5OH = 100 = 100 mass of sample 28.00 g sample













= 0.173898 = 0.1739% C2H5OH Copyright McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

4-45

4.98

Plan: The three types of redox reactions are combination, decomposition, and displacement. In a combination reaction, two or more reactants form one product, so the number of substances decreases. In a decomposition reaction, one reactant forms two or more products, so the number of substances increases. In a displacement reaction, the number of substances is the same, but atoms exchange places. Solution: a) decomposition b) combination c) displacement

4.99

Plan: Recall that a reactant breaks down into two or more products in a decomposition reaction, while reactants combine to form a product in a combination reaction. Solution: By definition, elements cannot decompose into anything simpler, so they could not be reactants in a decomposition reaction.

4.100

Plan: Review the types of redox reaction discussed in this section. Solution: Combination, decomposition, and displacement reactions generally produce only one compound; combustion reactions, however, often produce both carbon dioxide and water.

4.101

Plan: Recall that combustion is a reaction in which a reactant is combined with elemental oxygen. Solution: Yes, all combustion reactions are redox reactions since the oxidation number of O2 will change from zero (in the element) to some negative value (typically –2) in the product(s) during the reaction.

4.102

Plan: Recall that two reactants are combined to form one product in a combination reaction. A reaction is a redox reaction only if the oxidation numbers of some atoms change during the reaction. Solution: A common example of a combination/redox reaction is the combination of a metal and nonmetal to form an ionic salt, such as 2Mg(s) + O2(g)  2MgO(s) in which magnesium is oxidized and oxygen is reduced. A common example of a combination/non-redox reaction is the combination of a metal oxide and water to form an acid, such as CaO(s) + H2O(l)  Ca(OH)2(aq).

4.103

Plan: In a combination reaction, two or more reactants form one product. In a decomposition reaction, one reactant forms two or more products. In a displacement reaction, atoms or ions exchange places. Balance the reactions by inspection. Solution: a) Ca(s) + 2H2O(l)  Ca(OH)2(aq) + H2(g) Displacement: one Ca atom displaces 2 H atoms. b) 2NaNO3(s)  2NaNO2(s) + O2(g) Decomposition: one reactant breaks into two products. c) C2H2(g) + 2H2(g)  C2H6(g) Combination: two reactants combine to form one product.

4.104

Plan: In a combination reaction, two or more reactants form one product. In a decomposition reaction, one reactant forms two or more products. In a displacement reaction, atoms or ions exchange places. Balance the reactions by inspection. Solution: a) 2HI(g)  H2(g) + I2(g) Decomposition: one reactant breaks into two products. b) Zn(s) + 2AgNO3(aq)  Zn(NO3)2(aq) + 2Ag(s) Displacement: one Zn displaces 2 Ag atoms. c) 2NO(g) + O2(g)  N2O4(l) Combination: two reactants combine to form one product.

4-46

4.105

Plan: In a combination reaction, two or more reactants form one product. In a decomposition reaction, one reactant forms two or more products. In a displacement reaction, atoms or ions exchange places. Balance the reactions by inspection. Solution: a) 2Sb(s) + 3Cl2(g)  2SbCl3(s) Combination: two reactants combine to form one product. b) 2AsH3(g)  2As(s) + 3H2(g) Decomposition: one reactant breaks into two products. c) Zn(s) + Fe(NO3)2(aq)  Zn(NO3)2(aq) + Fe(s) Displacement: one Zn displaces one Fe atom.

4.106

Plan: In a combination reaction, two or more reactants form one product. In a decomposition reaction, one reactant forms two or more products. In a displacement reaction, atoms or ions exchange places. Balance the reactions by inspection. Solution: a) Mg(s) + 2H2O(g)  Mg(OH)2(s) + H2(g) Displacement: one Mg displaces two H atoms. b) Cr(NO3)3(aq) + Al(s)  Al(NO3)3(aq) + Cr(s) Displacement: one Al displaces one Cr atom. c) PF3(g) + F2(g)  PF5(g) Combination: two reactants combine to form one product.

4.107

c) This is a displacement reaction. Mn is a more reactive metal and displaces Cu from solution. Mn(s) + Cu(NO3)2(aq)  Mn(NO3)2(aq) + Cu(s) 4.108

4.109

4-47

Solution: a) The combination of two nonmetals gives a covalent compound. N2(g) + 3H2(g)  2NH3(g) b) Some compounds undergo thermal decomposition to simpler substances. 2NaClO3(s)  2NaCl(s) + 3O2(g) c) This is a displacement reaction. Active metals like Ba can displace hydrogen from water. Ba(s) + 2H2O(l)  Ba(OH)2(aq) + H2(g) 4.110

4.111

Plan: In a combination reaction, two or more reactants form one product. Two elements as reactants often result in a combination reaction. In a decomposition reaction, one reactant forms two or more products; one reactant only often indicates a decomposition reaction. In a displacement reaction, atoms or ions exchange places. An element and a compound as reactants often indicate a displacement reaction. Balance the reactions by inspection. Solution: a) Cs, a metal, and I2, a nonmetal, combine to form the binary ionic compound, CsI. 2Cs(s) + I2(s)  2CsI(s) b) Al is a stronger reducing agent than Mn and is able to displace Mn from solution, i.e., cause the reduction from 2+ 0 Mn (aq) to Mn (s). 2Al(s) + 3MnSO4(aq)  Al2(SO4)3(aq) + 3Mn(s) c) This is a combination reaction in which sulfur dioxide, SO2, a nonmetal oxide, combines with oxygen, O2, to form the higher oxide, SO3. 2SO2(g) + O2(g)  2SO3(g) It is not clear from the problem, but energy must be added to force this reaction to proceed. d) Butane is a four carbon hydrocarbon with the formula C4H10. It burns in the presence of oxygen, O2, to form carbon dioxide gas and water vapor. Although this is a redox reaction that could be balanced using the oxidation number method, it is easier to balance by considering only atoms on either side of the equation. First, balance carbon and hydrogen (because they only appear in one species on each side of the equation), and then balance oxygen. 2C4H10(g) + 13O2(g)  8CO2(g) + 10H2O(g) e) Total ionic equation in which soluble species are shown dissociated into ions: 2+ 2– 3+ 2– 2Al(s) + 3Mn (aq) + 3SO4 (aq)  2Al (aq) + 3SO4 (aq) + 3Mn(s) Net ionic equation in which the spectator ions are omitted: 2+ 3+ 2Al(s) + 3Mn (aq)  2Al (aq) + 3Mn(s) – Note that the molar coefficients are not simplified because the number of electrons lost (6 e ) must equal the – electrons gained (6 e ).

4-48

4.112

Plan: In a combination reaction, two or more reactants form one product. Two elements as reactants often result in a combination reaction. In a decomposition reaction, one reactant forms two or more products; one reactant only often indicates a decomposition reaction. In a displacement reaction, atoms or ions exchange places. An element and a compound as reactants often indicate a displacement reaction. Balance the reactions by inspection. Solution: a) Pentane is a five carbon hydrocarbon with the formula C5H12. It burns in the presence of oxygen, O2, to form carbon dioxide gas and water vapor. Although this is a redox reaction that could be balanced using the oxidation number method, it is easier to balance by considering only atoms on either side of the equation. First, balance carbon and hydrogen (because they only appear in one species on each side of the equation), and then balance oxygen. C5H12(l) + 8O2(g)  5CO2(g) + 6H2O(g) b) Phosphorus trichloride, PCl3, is a nonmetal halide that combines with additional halogen to form the higher halide, PCl5. PCl3(l) + Cl2(g)  PCl5(s) c) This is a displacement reaction. Active metals like Zn can displace hydrogen from acid. Zn(s) + 2HBr(aq)  ZnBr2(aq) + H2(g) d) This is a displacement reaction in which bromine displaces iodine. A halogen higher in the periodic table can displace a halogen that is lower. 2KI(aq) + Br2(l)  2KBr(aq) + I2(s) e) Total ionic equation in which soluble species are shown dissociated into ions: + – + – 2K (aq) + 2I (aq) + Br2(l)  2K (aq) + 2Br (aq) + I2(s) Net ionic equation in which the spectator ions are omitted: – – 2I (aq) + Br2(l)  I2(s) + 2Br (aq)

4.113

Plan: Write a balanced equation that shows the decomposition of HgO to its elements. Convert the mass of HgO to moles and use the molar ratio from the balanced equation to find the moles and then the mass of O 2. Perform the same calculation to find the mass of the other product. Solution:

4.114

a) The balanced chemical equation is 2HgO(s)  2Hg(l) + O2(g). 103 g   1 mol HgO   Moles of HgO = 4.27 kg HgO   = 19.71376 mol HgO  1 kg   216.6 g HgO   1 mol O 2    = 9.85688 mol O Moles of O2 = 19.71376 mol HgO 2  2 mol HgO   32.00 g O 2    = 315.420 = 315 g O Mass (g) of O2 = 9.85688 mol O 2  2  1 mol O 2  b) The other product is mercury.  2 mol Hg    = 19.71376 mol Hg Moles of Hg = 19.71376 mol HgO  2 mol HgO   200.6 g Hg   1 kg   Mass (kg) Hg = 19.71376 mol Hg  3  = 3.95458 = 3.95 kg Hg 10 g   1 mol Hg  Plan: Write a balanced equation that shows the decomposition of calcium chloride to its elements. Convert the mass of CaCl2 to moles and use the molar ratio from the balanced equation to find the moles and then the mass of Cl2. Perform the same calculation to find the mass of the other product. Solution: The balanced chemical equation is CaCl2(l) elect   Ca(l) + Cl2(g).

4-49

Note: The reaction cannot be done in the presence of water as elemental calcium would displace the hydrogen from the water.  1 mol CaCl 2    = 7.87529 mol CaCl Moles of CaCl2 = 874 g CaCl 2  2 110.98 g CaCl 2   1 mol Cl 2    = 7.87529 mol Cl Moles of Cl2 = 7.87529 mol CaCl 2  2 1 mol CaCl 2   70.90 g Cl 2    = 558.358 = 558 g Cl Mass (g) of Cl2 = 7.87529 mol Cl 2  2  1 mol Cl 2  The other product is calcium.  1 mol Ca    = 7.87529 mol Ca Moles of Ca = 7.87529 mol CaCl 2  1 mol CaCl 2   40.08 g Ca    = 315.64 = 316 g Ca Mass (g) of Ca = 7.87529 mol Ca   1 mol Ca 

4.115

Plan: To determine the reactant in excess, write the balanced equation (metal + O 2  metal oxide), convert reactant masses to moles, and use molar ratios to see which reactant makes the smaller (“limiting”) amount of product. Use the limiting reactant to calculate the amount of product formed. Use the molar ratio to find the amount of excess reactant required to react with the limiting reactant; the amount of excess reactant that remains is the initial amount of excess reactant minus the amount required for the reaction. Solution: The balanced equation is 4Li(s) + O2(g)  2Li2O(s).  1 mol Li   2 mol Li 2 O   a) Moles of Li2O if Li limiting = 1.62 g Li   = 0.1166979 mol Li2O  6.941 g Li   4 mol Li   1 mol O 2   2 mol Li 2 O   Moles of Li2O if O2 limiting = 6.50 g O 2    = 0.40625 mol Li2O  32.00 g O 2   1 mol O 2  Li is the limiting reactant since it produces the smaller amount of product; O2 is in excess. b) Using Li as the limiting reagent, 0.1166979 = 0.117 mol Li2O is formed. c) Li is limiting, thus there will be none remaining (0 g Li).  29.88 g Li 2 O    = 3.4869 = 3.49 g Li O Mass (g) of Li2O = 0.1166979 mol Li 2 O  2  1 mol Li 2 O   1 mol Li  1 mol O 2   32.00 g O 2   Mass (g) of O2 reacted = 1.62 g Li    = 1.867166 g O2  6.941 g Li   4 mol Li   1 mol O 2  Remaining O2 = initial amount – amount reacted = 6.50 g O2 – 1.867166 g O2 = 4.632834 = 4.63 g O2

4.116

Plan: To determine the reactant in excess, write the balanced equation (metal + N 2  metal nitride), convert reactant masses to moles, and use molar ratios to see which reactant makes the smaller (“limiting”) amount of product. Use the limiting reactant to calculate the amount of product formed. Use the molar ratio to find the amount of excess reactant required to react with the limiting reactant; the amount of excess reactant that remains is the initial amount of excess reactant minus the amount required for the reaction. Solution: The balanced equation is 3Mg(s) + N2(g)  Mg3N2(s).

4-50

 1 mol Mg  1 mol Mg 3 N 2   a) Moles of Mg3N2 if Mg is limiting = 2.22 g Mg   = 0.030440 mol Mg3N2  24.31 g Mg   3 mol Mg   1 mol N 2  1 mol Mg 3 N 2   Moles of Mg3N2 if N2 is limiting = 3.75 g N 2    = 0.13383 mol Mg3N2  28.02 g N 2   1 mol N 2 

Mg is the limiting reactant since it produces the smaller amount of product; N2 is present in excess. b) Using Mg as the limiting reactant, 0.030440 = 0.0304 mol Mg3N2 is formed. c) There will be 0 Mg remaining since it is the limiting reagent and will be completely consumed. 100.95 g Mg 3 N 2    = 3.07292 = 3.07 g Mg N Mass (g) of Mg3N2 = 0.030440 mol Mg 3 N 2  3 2  1 mol Mg 3 N 2   1 mol Mg   1 mol N 2   28.02 g N 2   Mass (g) of N2 reacted = 2.22 g Mg    = 0.852933 g N2  24.31 g Mg   3 mol Mg   1 mol N 2  Remaining N2 = initial amount – amount reacted = 3.75 g N2 – 0.852933 g N2 = 2.897067 = 2.90 g N2 4.117

Plan: Since mass must be conserved, the original amount of mixture – amount of residue = mass of oxygen produced. Write a balanced equation and use molar ratios to convert from the mass of oxygen produced to the amount of KClO3 reacted. Mass percent is calculated by dividing the mass of KClO3 by the mass of the sample and multiplying by 100. Solution: 2KClO3(s)  2KCl(s) + 3O2(g) Mass (g) of O2 produced = mass of mixture – mass of residue = 0.950 g – 0.700 g = 0.250 g O2  1 mol O 2   2 mol KClO 3  122.55 g KClO 3   Mass (g) of KClO3 = 0.250 g O 2     = 0.63828125 g KClO3  32.00 g O 2   3 mol O 2   1 mol KClO 3  mass of KClO3 0.63828125 g KClO3 Mass % KClO3 = 100% = 100% = 67.1875 = 67.2% KClO3 mass of sample 0.950 g sample

4.118

Plan: Since mass must be conserved, the original amount of mixture – amount of remaining solid = mass of carbon dioxide produced. Write a balanced equation and use molar ratios to convert from the mass of CO 2 produced to the amount of CaCO3 reacted. Mass percent is calculated by dividing the mass of CaCO3 by the mass of the sample and multiplying by 100. Solution: CaCO3(s)  CaO(s) + CO2(g) Mass (g) of CO2 produced = mass of mixture – mass of remaining solid = 0.693 g – 0.508 g = 0.185 g CO2  1 mol CO 2  1 mol CaCO 3  100.09 g CaCO 3   Mass (g) of CaCO3 = 0.185 g CO 2     = 0.420737 g CaCO3  44.01 g CO 2   1 mol CO 2   1 mol CaCO 3  mass of CaCO3 0.420737 g CaCO3 Mass % CaCO3 = 100% = 100% = 60.7124 = 60.7% CaCO3 mass of sample 0.693 g sample

4.119

Plan: Write the balanced equation for the displacement reaction, convert reactant masses to moles, and use molar ratios to see which reactant makes the smaller (“limiting”) amount of product. Use the limiting reactant to calculate the amount of product formed. Solution: The balanced reaction is 2Al(s) + Fe2O3(s)  2Fe(l) + Al2O3(s).

4-51

10 3 g   1 mol Al   2 mol Fe   Moles of Fe if Al is limiting = 1.50 kg Al     = 55.59674 mol Fe  1 kg   26.98 g Al   2 mol Al 





 2 mol Fe    = 50.0 mol Fe Moles of Fe if Fe2O3 is limiting = 25.0 mol Fe 2 O 3  1 mol Fe 2 O 3 

Fe2O3 is the limiting reactant since it produces the smaller amount of Fe; 50.0 moles of Fe forms.  55.85 g Fe   Mass (g) of Fe = 50.0 mol Fe   = 2792.5 = 2790 g Fe = 2.79 kg Fe  1 mol Fe  4.120

Plan: In ionic compounds, iron has two common oxidation states, +2 and +3. First, write the balanced equations for the formation and decomposition of compound A. Then, determine which reactant is limiting by converting reactant masses to moles, and using molar ratios to see which reactant makes the smaller (“limiting”) amount of product. From the amount of the limiting reactant calculate how much compound B will form. Solution: Compound A is iron chloride with iron in the higher, +3, oxidation state. Thus, the formula for compound A is FeCl3 and the correct name is iron(III) chloride. The balanced equation for formation of FeCl3 is: 2Fe(s) + 3Cl2(g)  2FeCl3(s) To find out whether 50.6 g Fe or 83.8 g Cl2 limits the amount of product, calculate the number of moles of iron(III) chloride that could form based on each reactant.  1 mol Fe   2 mol FeCl 3   Moles of FeCl3 if Fe is limiting = 50.6 g Fe   = 0.905998 mol FeCl3  55.85 g Fe   2 mol Fe   1 mol Cl 2   2 mol FeCl 3   Moles of FeCl3 if Cl2 is limiting = 83.8 g Cl 2    = 0.787964 mol FeCl3  3 mol Cl 2   70.90 g Cl 2  Since fewer moles of FeCl3 are produced from the available amount of chlorine, the chlorine is the limiting reactant, producing 0.787964 mol of FeCl3. The FeCl3 decomposes to FeCl2 with iron in the +2 oxidation state. FeCl2 is compound B. The balanced equation for the decomposition of FeCl3 is: 2FeCl3(s)  2FeCl2(s) + Cl2(g)  2 mol FeCl 2  126.75 g FeCl 2   Mass (g) of FeCl2 = 0.787964 mol FeCl 3    = 99.874 = 99.9 g FeCl2  2 mol FeCl 3   1 mol FeCl 2 

4.121

Plan: Use the volume and molarity of HCl to find the moles of HCl added to the metal. That number of moles minus the moles of HCl that remain after reaction gives the moles of HCl that actually reacted. Use the moles of reacted HCl and the mole ratio in the balanced chemical equation to find the moles of Mg that reacted. Convert moles of Mg to mass of Mg, divide that mass by the mass of impure metal sample, and multiply by 100 to find mass %. Solution: Mg(s) + 2HCl(aq)  MgCl2(aq) + H2(g)  0.750 mol HCl   0.100 L  = 0.0750 mol HCl Moles of HCl added =   1L   Moles of HCl reacting with Mg = moles of added HCl – moles of HCl remaining = 0.0750 mol – 0.0125 mol HCl = 0.0625 mol HCl  1 mol Mg    = 0.03125 mol Mg Moles of Mg reacting = 0.0625 mol HCl  2 mol HCl 

4-52

 24.31 g Mg    = 0.7596875 g Mg Mass (g) of Mg = 0.03125 mol Mg  1 mol Mg  mass of Mg 0.7596875 g Mg Mass percent Mg = 100 = 100 = 57.552 = 57.6% Mg mass of sample 1.32 g sample

4.122

The equilibrium state is dynamic, because the forward and reverse processes continue even after apparent change has ceased.

4.123

The system must be closed so the gaseous product cannot escape.

4.124

Acetic acid molecules are constantly and randomly colliding with each other and with water molecules. Occasionally, + an H is transferred from an acetic acid molecule to a water molecule because of one of these collisions. At first, this + + – process occurs at a larger rate than the reverse transfer of an H from a H3O to a CH3COO . As the concentration of + – + + H and CH3COO builds up, the rate of transfer of H between them equals the rate of transfer of H between CH3COOH and H2O and equilibrium is reached when about 2% of the CH3COOH molecules have ionized.

4.125

Plan: Review the concept of dynamic equilibrium. Solution: 2NO(g) + Br2(g) g) On a molecular scale, chemical reactions are dynamic. If NO and Br 2 are placed in a container, molecules of NO and molecules of Br2 will react to form NOBr. Some of the NOBr molecules will decompose and the resulting NO and Br2 molecules will recombine with different NO and Br2 molecules to form more NOBr. In this sense, the reaction is dynamic because the original NO and Br2 pairings do not remain permanently attached to each other. Eventually, the rate of the forward reaction (combination of NO and Br 2) will equal the rate of the reverse reaction (decomposition of NOBr) at which point the reaction is said to have reached equilibrium. If you could take “snapshot” pictures of the molecules at equilibrium, you would see a constant number of reactant (NO, Br 2) and product molecules (NOBr), but the pairings would not stay the same.

4.126

Plan: Write the balanced equation for the combination reaction to produce ammonia. Convert reactant masses to moles, and use molar ratios to see which reactant makes the smaller (“limiting”) amount of product. Use the limiting reactant to calculate the amount of product formed. This is the theoretical yield of ammonia. Divide the actual yield of ammonia by the theoretical yield and multiply by 100 to obtain the percent yield. Use molar ratios to determine the amount of N2 and H2 that reacted to form the product. The difference between the initial moles of reactants and the amounts that actually reacted will give the moles of reactant remaining at equilibrium. Solution: a) The reaction is N2(g) + 3H2(g)  2NH3(g). First, the limiting reactant must be found.  1 mol H 2   2 mol NH 3   Moles of NH3 if H2 is limiting = 10.0 g H 2    = 3.306878 mol NH3  2.016 g H 2   3 mol H 2   1 mol N 2   2 mol NH 3   Moles of NH3 if N2 is limiting = 20.0 g N 2    = 1.42755 mol NH3  28.02 g N 2   1 mol N 2  N2 is the limiting reactant since it produces the smaller amount of product. 17.03 g NH 3   Mass (g) of NH3 = 1.42755 mol NH 3   = 24.311 g NH3  1 mol NH 3  actual yield 15.0 g % yield = 100 = 100 = 61.700 = 61.7% theoretical yield 24.311 g  1 mol H 2    = 4.9603 mol H b) Moles of H2 initially present = 10.0 g H 2  2  2.016 g H 2 

4-53

 1 mol N 2    = 0.71378 mol N Moles of N2 initially present = 20.0 g N 2  2  28.02 g N 2   1 mol NH 3   3 mol H 2   Moles of H2 required to produce 15.0 g NH3 = 15.0 g NH 3    = 1.3212 mol H2 17.03 g NH 3   2 mol NH 3   1 mol NH 3   1 mol N 2   Moles of N2 required to produce 15.0 g NH3 = 15.0 g NH 3    = 0.4404 mol N2 17.03 g NH 3   2 mol NH 3  Moles of H2 at equilibrium = initial moles – reacted moles = 4.9603 mol – 1.3212 mol = 3.6391 = 3.64 mol H2 Moles of N2 at equilibrium = initial moles – reacted moles = 0.71378 mol – 0.4404 mol = 0.27338 = 0.273 mol N2 2+

4.127

Plan: Ferrous ion is Fe . Write a reaction to show the conversion of Fe to Fe . Convert the mass of Fe in a 125-g serving to the mass of Fe in a 737-g sample. Use molar mass to convert mass of Fe to moles of Fe and use Avogadro’s number to convert moles of Fe to moles of ions. Solution: 2+ + a) Fe oxidizes to Fe with a loss of 2 electrons. The H in the acidic food is reduced to H2 with a gain of 2 electrons. The balanced reaction is: + 2+ Fe(s) + 2H (aq)  Fe (aq) + H2(g) O.N.: 0 +1 +2 0 3  49 mg Fe  10 g   b) Mass (g) of Fe in the jar of tomato sauce = 737 g sauce   = 0.288904 g Fe 125 g sauce   1 mg  2  1 mol Fe  1 mol Fe   6.022 10 23 Fe 2  ions   2+ Number of Fe ions = 0.288904 g Fe      55.85 g Fe  1 mol Fe 2   1 mol Fe    21 21 2+ = 3.11509 × 10 = 3.1 × 10 Fe ions per jar of sauce

4.128

Plan: Write balanced equations for the conversion of CaCO3 to CaO, the reaction of CaO with SO2, and the combustion of sulfur to produce SO2. Add these three reactions to obtain the overall reaction. Use the mass % of sulfur in the coal sample to find the mass and then moles of sulfur present. Use the molar ratio in the overall balanced reaction to find the mass of CaCO3 required to react with this amount of sulfur, assuming a 70% yield. Solution: The reactions are: CaCO3(s)  CaO(s) + CO2(g) CaO(s) + SO2(g)  CaSO3(s) S(s) + O2(g)  SO2(g) Overall reaction: CaCO3(s) + S(s) + O2(g)  CaSO3(s) + CO2(g) 10 3 g  0.33%  1 mol S     = 8749.220 mol S  Moles of sulfur = 8.510 4 kg coal  1 kg  100%  32.06 g S  1 mol CaCO 3  100.09 g CaCO 3 100%   Mass (g) CaCO3 = 8749.220 mol S     1 mol CaCO 3  70%   1 mol S  6

= 1.25101 × 10 = 1.3 × 10 g CaCO3 needed

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4.129

Plan: Convert the mass of glucose to moles and use the molar ratios from the balanced equation to find the moles of ethanol and CO2. The amount of ethanol is converted from moles to grams using its molar mass. The amount of CO2 is converted from moles to volume in liters using the conversion factor given. Solution:  1 mol C 6 H12 O 6   2 mol C 2 H 5 OH   Moles of C2H5OH = 100. g C 6 H12 O 6    = 1.11012 mol C2H5OH 180.16 g C 6 H12 O6   1 mol C 6 H12 O 6   46.07 g C 2 H 5 OH    = 51.143 = 51.1 g C H OH Mass (g) of C2H5OH = 1.11012 mol C 2 H 5 OH  2 5  1 mol C 2 H 5 OH   1 mol C 6 H12 O 6   2 mol CO 2   Moles of CO2 = 100. g C 6 H12 O6    = 1.11012 mol CO2 180.16 g C 6 H12 O 6  1 mol C 6 H12 O 6   22.4 L CO 2    = 24.8667 = 24.9 L CO Volume (L) of CO2 = 1.11012 mol CO 2  2  1 mol CO 2 

4.130

Plan: Find the moles of KMnO4 from the molarity and volume information. Use the molar ratio in the balanced equation to find the moles and then mass of iron. Mass percent is calculated by dividing the mass of iron by the mass of the sample and multiplying by 100. Solution: 103 L  0.03190 mol MnO 4    = 0.00125431 mol KMnO  Moles of KMnO4 = 39.32 mL  4  1 mL  L   5 mol Fe 2    55.85 g Fe   Mass (g) of Fe = 0.00125431 mol MnO 4  = 0.350266 g Fe  2  1 mol MnO 4  1 mol Fe 

Mass % of Fe = 4.131

mass of Fe 0.350266 g 100 = 100 = 31.6096 = 31.61% Fe mass of sample 1.1081 g

Plan: Remember that spectator ions are omitted from net ionic equations. Assign oxidation numbers to all atoms in the titration reaction, following the rules. The reactant that is the reducing agent contains an atom that is oxidized (O.N. increases from the left side to the right side of the equation). The reactant that is the oxidizing agent contains an atom that is reduced (O.N. decreases from the left side to the right side of the equation). Find the number of moles of KMnO4 from the molarity and volume and use molar ratios in the balanced equations to find the moles and then mass of CaCl2. Mass percent is calculated by dividing the mass of CaCl2 by the mass of the sample and multiplying by 100. Solution: a) The reaction is: Na2C2O4(aq) + CaCl2(aq) O (s) + 2NaCl(aq) 2 4 The total ionic equation in which soluble substances are dissociated into ions is: + 2– 2+ – + – 2Na (aq) + C2O4 (aq) + Ca (aq) + 2Cl (aq) O (s) + 2Na (aq) + 2Cl (aq) 2 4 + – Omitting Na and Cl as spectator ions gives the net ionic equation: 2+ 2– Ca (aq) + C2O4 (aq)  CaC2O4(s) – b) You may recognize this reaction as a redox titration, because the permanganate ion, MnO 4 , is a common – 2– oxidizing agent. The MnO4 oxidizes the oxalate ion, C2O4 to CO2. Mn changes from +7 to +2 (reduction) and C changes from +3 to +4 (oxidation). The equation that describes this process is: + – 2+ + H2C2O4(aq) + K (aq) + MnO4 (aq)  Mn (aq) + CO2(g) + K (aq) H2C2O4 is a weak acid, so it cannot be written in a fully dissociated form. KMnO4 is a soluble salt, so it can be + written in its dissociated form. K (aq) is omitted in the net ionic equation because it is a spectator ion. We will balance the equation using the oxidation number method. First assign oxidation numbers to all elements in the reaction: +2 +6 –8 –8 –4 +1 +3 –2

+7 –2

+4 –2

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H2C2O4(aq) + MnO4 (aq)  Mn (aq) + 2CO2(g) Identify the oxidized and reduced species and multiply one or both species by the appropriate factors to make the electrons lost equal the electrons gained. Each C in H2C2O4 increases in O.N. from +3 to +4 for a total gain of 2 – electrons. Mn in MnO4 decreases in O.N. from +7 to +2 for a loss of 5 electrons. Multiply C by 5 and Mn by 2 to get an electron loss = electron gain of 10 electrons. – 2+ 5H2C2O4(aq) + 2MnO4 (aq)  2Mn (aq) + 10CO2(g) + Adding water and H (aq) to finish balancing the equation is appropriate since the reaction takes place in acidic + medium. Add 8H2O(l) to right side of the equation to balance the oxygen and then add 6H (aq) to the left to balance hydrogen. –

5H2C2O4(aq) + 2MnO4 (aq) + 6H (aq)  10CO2(g) + 2Mn (aq) + 8H2O(l) –

–

c) Mn in MnO4 decreases in O.N. from +7 to +2 and is reduced. KMnO4 is the oxidizing agent. d) C in H2C2O4 increases in O.N. from +3 to +4 and is oxidized. H2C2O4 is the reducing agent. e) The balanced equations provide the accurate molar ratios between species. 103 L   0.1019 mol KMnO 4   – – Moles of MnO4 = 37.68 mL   = 0.0038396 mol MnO4  1 mL   1L    5 mol H 2 C 2 O 4    = 0.009599 mol H C O Moles of H2C2O4 = 0.0038396 mol MnO 4  2 2 4  2 mol MnO 4    1 mol CaCl 2  110.98 g CaCl 2   Mass (g) CaCl2 = 0.009599 mol H 2 C 2 O 4    = 1.06530 g CaCl2 1 mol H 2 C 2 O 4   1 mol CaCl 2  mass of CaCl 2 1.06530 g CaCl 2 Mass percent CaCl2 = 100 = 100 = 55.05995 = 55.06% CaCl2 mass of sample 1.9348 g mixture 4.132

Plan: Write balanced equations for the two acid-base reactions. Find the moles of H2SO4 from the molarity and volume information and use the molar ratio in the balanced equation for the reaction of H 2SO4 and NaOH to find the moles of NaOH used in the titration. Divide the moles of NaOH by its volume to determine its molarity. Then find the moles of NaOH used in the titration of HCl by multiplying the NaOH molarity by its volume; use the molar ratio in this reaction to find moles of HCl. Dividing moles of HCl by its volume gives its molarity. Solution: Write the balanced chemical equations: NaOH(aq) + HCl(aq)  NaCl(aq) + H2O(l) 2NaOH(aq) + H2SO4(aq)  Na2SO4(aq) + 2H2O(l) Determine the NaOH concentration from the reaction of NaOH with H 2SO4: 103 L  0.0782 mol H 2 SO 4    = 0.00391 mol H SO  Moles of H2SO4 = 50.0 mL 2 4   L  1 mL    2 mol NaOH    = 0.00782 mol NaOH Moles of NaOH = 0.00391 mol H 2 SO 4  1 mol H 2 SO 4   0.00782   1 mL   Molarity of NaOH =   3  = 0.425 M NaOH 10 L  18.4 mL 

Use the NaOH concentration and the reaction of HCl with NaOH to determine HCl concentration: 103 L  0.425 mol NaOH    = 0.0116875 mol NaOH  Moles of NaOH = 27.5 mL   L   1 mL   Copyright McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

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 1 mol HCl    = 0.0116875 mol HCl Moles of HCl = 0.0116875 mol NaOH  1 mol NaOH   1   1 mL   Molarity of HCl = 0.0116875 mol HCl  3  = 0.116875 = 0.117 M HCl 100.mL  10 L 

4.133

Plan: Recall that the total ionic equation shows all soluble ionic substances dissociated into ions and the net ionic equation omits the spectator ions. Use the molar ratio in the balanced reaction to find the moles of acid and base. Divide the moles of acid and base by the volume to obtain the molarity. Solution: a) Molecular: H2SO4(aq) + 2NaOH(aq)  Na2SO4(aq) + 2H2O(l) + 2– + – + 2– Total ionic: 2H (aq) + SO4 (aq) + 2Na (aq) + 2OH (aq)  2Na (aq) + SO4 (aq) + 2H2O(l) + – Net ionic: H (aq) + OH (aq)  H2O(l) + 2– (Na and SO4 are spectator ions.)  0.010 mol SO 4 2 1 mol H 2 SO 4   = 0.020 mol H2SO4  b) Moles of H2SO4 = 2 orange spheres  1 mol SO 4 2   1 orange sphere 

 2 mol NaOH   = 0.040 mol NaOH Moles of NaOH = 0.020 mol H 2 SO4  1 mol H2 SO4   1   1 mL   c) Molarity of H2SO4 = 0.020 mol H 2 SO 4   3  = 0.80 M H2SO4 10 L   25 mL   1   1 mL   Molarity of NaOH = 0.040 mol NaOH   3  = 1.6 M NaOH 10 L   25 mL 

4.134

Plan: Write a balanced equation for the reaction. This is an acid-base reaction between HCl and the base 2– CO3 to form H2CO3 which then decomposes into CO2 and H2O. Find the moles of HCl from the molarity and volume information and use the molar ratio in the balanced equation to find the moles and mass of dolomite. To find mass %, divide the mass of dolomite by the mass of soil and multiply by 100. Solution: The balanced equation for this reaction is: 2+ 2+ – CaMg(CO3)2(s) + 4HCl(aq)  Ca (aq) + Mg (aq) + 2H2O(l) + 2CO2(g) + 4Cl (aq) 103 L  0.2516 mol HCl     = 0.0084437 mol HCl Moles of HCl = 33.56 mL    1L  1 mL   1 mol CaMg(CO 3 )2  184.41 g CaMg(CO 3 ) 2     Mass CaMg(CO3)2 = 0.0084437 mol HCl   4 mol HCl  1 mol CaMg(CO 3 ) 2  

= 0.389276 g CaMg(CO3)2 0.389276 g CaMg(CO 3 )2 mass CaMg(CO 3 )2 Mass percent CaMg(CO3)2 = 100 100 = 13.86 g soil mass soil = 2.80863 = 2.809% CaMg(CO3)2 4.135

Plan: Write balanced chemical equations for the acid-base titration reactions. To find the concentration of HA, find the moles of NaOH used for its titration by multiplying the molarity of NaOH by the volume used in the titration and using the molar ratio to find the moles of HA; dividing moles of HA by its volume gives the molarity. Multiply the molarity of HA by the volume of HA in the acid mixture to find the moles of HA in the mixture. Use the molar ratio to find the volume of NaOH required to titrate this amount of HA. The total volume of NaOH used in the titration of the mixture minus the volume required to titrate HA is the volume of NaOH required to titrate HB. Use this volume and molarity of NaOH and the molar ratio to find the moles and then molarity of HB. The volume of HB in the acid mixture is the total volume minus the volume of HA.

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Solution: The balanced chemical equations for HA or HB with sodium hydroxide are the same. For HA it is: HA(aq) + NaOH(aq)  NaA(aq) + H2O(l) To find the concentration of HA: 103 L  0.0906 mol NaOH     = 0.007909 mol NaOH Moles of NaOH = 87.3 mL  1 mL   L   1 mol HA    = 0.007909 mol HA Moles of HA = 0.007909 mol NaOH  1 mol NaOH     1 mL  1  Molarity of HA = 0.007909 mol HA   3  = 0.1818248 = 0.182 M HA  43.5 mL  10 L  The titration of the acid mixture involves the reaction of NaOH with both of the acids. 103 L  0.1818248 mol HA    = 0.0067639 mol HA  Moles of HA in the acid mixture = 37.2 mL    1 mL  L  

Volume (mL) of NaOH required to titrate HA = 1 mol NaOH  L   1 mL  = 74.6565 mL NaOH 0.0067639 mol HA 1 mol HA  0.0906 1mol  NaOH  103 L    Volume of NaOH required to titrate HB = total NaOH volume – volume of NaOH required to titrate HA = 96.4 mL – 74.6565 mL = 21.7435 mL NaOH 10 L  0.0906 mol NaOH  1 mol HB    Moles of HB = 21.7435 mL    = 0.00196996 mol HB   1 mL  L 1 mol NaOH  3

Volume (mL) of HB = Volume of mixture – volume of HA = 50.0 mL – 37.2 mL = 12.8 mL  1  1 mL  Molarity of HB = 0.00196996 mol HB  = 0.153903 = 0.154 M HB 12.8 mL 103 L  4.136

Plan: For part (a), assign oxidation numbers to each element; the oxidizing agent has an atom whose oxidation number decreases while the reducing agent has an atom whose oxidation number increases. For part (b), use the molar ratios, beginning with step 3, to find the moles of NO2, then moles of NO, then moles of NH3 required to produce the given mass of HNO3. Solution: a) Step 1 +3

–3 +1

+2 –2

+1 –2

4NH3(g) + 5O2(g)  4NO(g) + 6H2O(l) N is oxidized from –3 in NH3 to +2 in NO; O is reduced from 0 in O2 to to –2 in NO. Oxidizing agent = O2 Step 2 +2 –2

Reducing agent = NH3

–4 0

+4 –2

2NO(g) + O2(g)  2NO2(g) N is oxidized from +2 in NO to +4 in NO2; O is reduced from 0 in O2 to –2 in NO2. Copyright McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

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Oxidizing agent = O2 Step 3 –4 +2 +4 –2

+1 –2

Reducing agent = NO –6

+1+5 –2

+2 –2

3NO2(g) + H2O(l)  2HNO3(l) + NO(g) N is oxidized from +4 in NO2 to +5 in HNO3; N is reduced from +4 in NO2 to +2 in NO. Oxidizing agent = NO2

Reducing agent = NO2

103 g  1 mol HNO3   3 mol NO2  5  b) Moles of NO2 = 3.0104 kg HNO3    = 7.14059 × 10 mol NO2  1 kg  63.02 g HNO3   2 mol HNO3   2 mol NO   = 7.14059 × 105 mol NO Moles of NO = 7.14059105 mol NO2   2 mol NO2   4 mol NH 3  5 Moles of NH3 = 7.14059 10 5 mol NO = 7.14059 × 10 mol NH3  4 mol NO 

17.03 g NH  1 kg  4 4 3   Mass (kg) of NH3 = 7.14059105 mol NH3   3  = 1.21604 × 10 = 1.2 × 10 kg NH3   1 mol NH3 10 g  4.137

Plan: In part (a), use the density of the alloy to find the volume of a 0.263-g sample of alloy. That volume is the sum of the volume of Mg and Al in the alloy. Letting x = mass of Mg and 0.263 – x = mass of Al, find the volume of each metal and set that equal to the total volume of the alloy. In part (b), write balanced displacement reactions in which Mg and Al displace hydrogen from the HCl to produce H 2. Use the molar ratios to find the masses of Mg and Al that must be present to produce the given amount of H2. In part (c), write balanced reactions for the formation of MgO and Al2O3 and use molar ratios to find the masses of Mg and Al that must be present in the sample to produce the given amount of oxide. Solution: a) Let x = mass of Mg and 0.263 – x = mass of Al  1 cm 3  3  = 0.10958 cm3 Volume (cm ) of alloy = 0.263 g alloy  2.40 g alloy  Volume of alloy = volume of Mg + volume of Al  1 cm 3 Mg   1 cm 3 Al  3  + (0.263  x) g Al  0.10958 cm = x g Mg 1.74 g Mg   2.70 g Al  3 0.10958 cm = 0.574713x + 0.097407 – 0.37037x 0.012173 = 0.204343x x = 0.05957 g Mg mass of Mg 0.05957 g Mg 100 = 100 = 22.6502 = 22.7% Mg Mass percent Mg = mass of alloy sample 0.263 g sample alloy b) Mg(s) + 2HCl(aq) (aq) + H2(g) 2 2Al(s) + 6HCl(aq) (aq) + 3H2(g) 3 Let x = mass of Mg and 0.263 – x = mass of Al Moles of H2 produced = moles of H2 from Mg + moles of H2 from Al  1 mol Mg  –2  1 mol H 2  + (0.263  x) g Al 1 mol Al  3 mol H 2  1.38 × 10 mol H2 = x g Mg 1 mol Mg   26.98 g Al  2 mol Al   24.31 g Mg   –2

1.38 × 10 mol H2 = 0.041135x + 0.014622 – 0.055597x –4

8.22 × 10 = 0.014462x x = 0.05684 g Mg Copyright McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

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mass of Mg 0.05684 g Mg 100 = 100 = 21.6122 = 21.6% Mg mass of alloy sample 0.263 g sample alloy c) 2Mg(s) + O2(g) s) 4Al(s) + 3O2(g) O (s) 2 3 Let x = mass of Mg and 0.263 – x = mass of Al Mass of oxide produced = mass of MgO from Mg + mass of Al2O3 from Al 0.483 g oxide =

Mass percent Mg =

 1 mol Mg   2 mol MgO  40.31 g MgO  + (0.263  x) g Al 1 mol Al  2 mol Al2 O3 101.96 g Al2 O3     2 mol Mg  1 mol MgO    24.31 g Mg     26.98 g Al  4 mol Al  1 mol Al2 O3 

x g Mg

0.483 g = 1.6582x + 0.49695 – 1.88955x 0.01395 = 0.23135x x = 0.060298 g Mg mass of Mg 0.060298 g Mg Mass percent Mg = 100 = 100 = 22.927 = 22.9% Mg mass of alloy sample 0.263 g sample alloy 4.138

Plan: An oxidation-reduction reaction is one in which oxidation numbers of atoms change from reactant to product. If there is no change in oxidation numbers, the reaction is not an oxidation-reduction reaction. Solution: a) Reaction (1): 0 0 +1 –1 0 Cl(g) + O3(g)  ClO(g) + O2(g) Cl is oxidized (0 to +1), and O is reduced (0 to –1). This is an oxidation-reduction reaction. Reaction (2): +1 –1 +1 –1 +1 –1 ClO(g) + ClO(g)  Cl2O2(g) No atom has a change in oxidation number; this is not an oxidation-reduction reaction. Reaction (3): +1 –1 0 0 light Cl2O2(g)    2Cl(g) + O2(g) Cl is reduced (+1 to 0) and O is oxidized (–1 to 0). This is an oxidation-reduction reaction. Reaction (4): +1 –1 0 0 0 ClO(g) + O(g)  Cl(g) + O2(g) Cl is reduced (+1 to 0) and O is oxidized (–1 to 0). This is an oxidation-reduction reaction. The oxidation-reduction reactions are 1, 3, and 4. b) Combining the equations requires multiplying equation 1 by 2, and then adding the three equations. 1) 2Cl(g) + 2O3(g)  2ClO(g) + 2O2(g) 2) ClO(g) + ClO(g)  Cl2O2(g) light 3) Cl2O2(g)    2Cl(g) + O2(g)

2O3(g)  3O2(g) 4.139

Plan: Write a balanced equation and use the molar ratio between Na2O2 and CO2 to convert the amount of Na2O2 given to the amount of CO2 that reacts with that amount. Convert that amount of CO2 to liters of air. Solution: The reaction is: 2Na2O2(s) + 2CO2(g)  2Na2CO3(s) + O2(g).

4-60

 1 mol Na 2 O 2   2 mol CO 2   44.01 g CO 2   Mass (g) of CO2 = 80.0 g Na 2 O 2     = 45.150 g CO2  77.98 g Na 2 O 2   2 mol Na 2 O 2   1 mol CO 2    L air   = 627.08 = 627 L air Volume (L) of air = 45.150 g CO 2   0.0720 g CO2 

4.140

The salts are NaBr 2H2O and MgBr2 6H2O NaBr(s) + AgNO3(aq) s) + NaNO3(aq) MgBr2(s) + 2AgNO3(aq) s) + Mg(NO3)2(aq) 103 g  1 mol H 2 O   = 0.013990 mol H2O  Total moles of H2O in the sample = 252.1 mg H2 O 18.02 g H2 O   1 mg  0.013990 mol H2O = mol H2O from NaBr 2H2O + mol H2O from MgBr2 6H2O. Let x = mol H2O from NaBr 2H2O and 0.013990 – x = mol H2O from MgBr2 6H2O.

1 mol NaBr   = 0.5000x mol NaBr Moles of NaBr = x mol H 2 O  2 mol H 2 O  1 mol MgBr2   = 0.002332 – 0.166667 × mol MgBr2 Moles of MgBr2 = 0.013990  x mol H 2 O  6 mol H2 O  Moles of AgBr produced = moles of AgBr from NaBr + moles of AgBr from MgBr2

 2 mol AgBr  1 mol AgBr  –3  6.00 × 10 mol AgBr = 0.5000x mol NaBr  + (0.002332  0.166667x) mol MgBr2    1 mol NaBr  1 mol MgBr2  –3

6.00 × 10 mol AgBr = 0.50000x + 0.004664 – 0.333334x 0.001336 = 0.166666x x = 0.008016 mol Moles of NaBr 2H2O = 0.5000x = 0.5000(0.008016 mol) = 0.004008 mol NaBr 2H2O

138.93 g NaBr  2H 2 O   = 0.5568 g NaBr 2H2O Mass of NaBr 2H2O = 0.004008 mol NaBr  2H 2 O  1 mol NaBr  2H 2 O  Moles of MgBr2 6H2O = 0.002332 – 0.166667x = 0.002332 – 0.166667(0.008016 mol) –4

= 9.960 × 10 mol MgBr2 6H2O

 292.23 g MgBr  6H O  2 2  Mass of MgBr2 6H2O = 9.960104 mol MgBr2  6H2 O  = 0.2911 g MgBr2 6H2O  1 mol MgBr2  6H2 O  Total mass of hydrates = 0.5568 g NaBr 2H2O + 0.2911 g MgBr2 6H2O = 0.8479 g mass of NaBr  2H 2 O 0.5568 g NaBr  2H 2 O Mass percent of NaBr 2H2O = (100) = (100) total mass 0.8479 g = 65.668 = 65.7% NaBr 2H2O Mass percent of MgBr2 6H2O =

mass of MgBr2  6H 2 O 0.2911 g MgBr2  6H 2 O 100 = 100 total mass 0.8472 g

= 34.662 = 34.7% MgBr2 6H2O 4.141

Plan: Use the mass percents to find the mass of each component needed for 1.00 kg of glass. Write balanced reactions for the decomposition of sodium carbonate and calcium carbonate to produce an oxide and CO2. Use the molar ratios in these reactions to find the moles and mass of each carbonate required to produce the amount of oxide in the glass sample.

4-61

Solution: A 1.00-kg piece of glass of composition 75% SiO2, 15% Na2O, and 10% CaO would contain:  75% SiO 2   = 0.75 kg SiO2 Mass (kg) of SiO2 = 1.00 kg glass 100% glass 

 15% Na 2 O   = 0.15 kg Na2O Mass (kg) of Na2O = 1.00 kg glass 100% glass   10.% CaO   = 0.10 kg CaO Mass (kg) of CaO = 1.00 kg glass 100% glass  In this example, the SiO2 is added directly while the sodium oxide comes from decomposition of sodium carbonate and the calcium oxide from decomposition of calcium carbonate: Na2CO3(s)  Na2O(s) + CO2(g) CaCO3(s)  CaO(s) + CO2(g) Mass (kg) of Na2CO3 = 103 g   1 mol Na 2 O  1 mol Na 2 CO 3  105.99 g Na 2 CO 3   1 kg   0.15 kg Na O     3   2   61.98 g Na 2 O   1 mol Na 2 O   1 mol Na 2 CO 3  10 g   1 kg  = 0.25651 = 0.26 kg Na2CO3 103 g   1 mol CaO  1 mol CaCO 3  100.09 g CaCO 3   1 kg   Mass (kg) of CaCO3 = 0.10 kg CaO       10 3 g   1 kg   56.08 g CaO   1 mol CaO   1 mol CaCO 3  

= 0.178477 = 0.18 kg CaCO3 4.142

Plan: For part (a), assign oxidation numbers to each element; the substance that has reduced has a decrease in oxidation number as it gains electrons while the substance that has oxidized has an increase in oxidation number as it loses electrons. For part (b), calculate the moles of Na2S2O3 by multiplying its molarity by the volume required in the titration. Then use the molar ratios in the balanced equations to find the moles and mass of oxygen in the sample. Solution: a) Step 1 0

+2 –8 +1 –1

+1 +6 –2

O2(aq) + 4KI(aq) + 2H2SO4(aq)

+1 –2

+1 +6 –2

–8

(aq) + 2H2O(l) + 2K2SO4(aq)

The O.N. of iodine increases from –1 in KI to 0 in I2; I in KI is oxidized. Step 2 +2 +4 –6

+2 +10 –12

+1 +2.5 –2

+1 +2 –2

I2(aq) + 2Na2S2O3(aq)

+1 –1

S4O6(aq) + 2NaI(aq)

The O.N. of iodine decreases from 0 in I2 to –1 in NaI; I2 is reduced. 103 L   0.0105 mol Na 2 S2 O 3   –4 b) Moles of Na2S2O3 = 15.75 mL   = 1.65375 × 10 mol Na2S2O3  1 mL  1L     1 mol I 2   = 8.26875 × 10–5 mol I Moles of I2 (Step 2) = 1.65375 104 mol Na 2 S2 O 3  2  2 mol Na 2 S2 O 3  1 mol O   2 –5 Moles of O2 (Step 1) = 8.26875105 mol I 2   = 4.134375 × 10 mol O2  2 mol I 2 

4-62

 32.00 g O   2 Mass (g) of O2 = 4.134375105 mol O 2   = 0.001323 g = 0.00132 g O2  1 mol O  2

4.143

Plan: To determine the reactant in excess, convert reactant masses to moles, and use molar ratios to see which reactant makes the smaller (“limiting”) amount of product. Use the limiting reactant to calculate the amount of product formed. Use the molar ratio to find the amount of excess reactant required to react with the limiting reactant; the amount of excess reactant that remains is the initial amount of excess reactant minus the amount required for the reaction. Multiply moles of products and excess reactant by Avogadro’s number to obtain number of molecules. Solution: 103 g  1 mol C 2 H 4  1 mol C 2 H 5 Cl   a) Moles of C2H5Cl if C2H4 is limiting = 0.100 kg C2 H 4   28.05 g C 2 H 4   1 kg   1 mol C2 H 4  = 3.56506 mol C2H5Cl 103 g   1 mol HCl 1 mol C 2 H 5 Cl  Moles of C2H5Cl if HCl is limiting = 0.100 kg HCl  36.46 g HCl  1 mol HCl   1 kg  = 2.74273 mol C2H5Cl The HCl is limiting. Moles HCl remaining = 0 mol

103 g  1 mol C2 H 4   = 3.56506 mol C2H4  Moles of C2H4 initially present = 0.100 kg C2 H 4   28.05 g C2 H 4   1 kg  103 g   1 mol HCl 1 mol C 2 H 4  = 2.74273 mol C H Moles of C2H4 that react = 0.100 kg HCl 2 4  36.46 g HCl  1 mol HCl   1 kg   Moles of C2H4 remaining = initial moles – reacted moles = 3.56506 mol – 2.74273 mol = 0.82233 mol C2H4 Moles of C2H5Cl formed = 2.74273 mol C2H5Cl Total moles of gas = moles HCl + moles C2H4 + moles C2H5Cl = 0 mol + 0.82233 mol + 2.74273 mol = 3.56506 mol

 6.022 10 23 molecules   = 2.146879 × 1024 = 2.15 × 1024 molecules Molecules of gas = 3.56506 mol gas  1 mol gas  b) This will still be based on the HCl as the limiting reactant. 103 g   1 mol HCl  = 2.74273 mol HCl Initial moles of HCl = 0.100 kg HCl  36.46 g HCl   1 kg  Moles of HCl remaining = intial moles/2 = (2.74273 mol HCl)/2 = 1.371365 mol HCl 1 mol C 2 H 4  Moles of C2H4 reacting with half of HCl = 1.371365 mol HCl   = 1.371365 mol C2H4  1 mol HCl  Moles of C2H4 remaining = initial moles – reacted moles = 3.56506 mol – 1.371365 mol = 2.193695 mol C2H4 1 mol C 2 H 5 Cl  Moles of C2H5Cl formed = 1.371365 mol HCl   = 1.371365 mol C2H5Cl  1 mol HCl 

Total moles of gas = moles HCl + moles C2H4 + moles C2H5Cl = 1.371365 mol + 2.193695 mol + 1.371365 mol = 4.936425 = 4.94 mol total 4.144

Plan: Write balanced reactions and use molar ratios to find the moles of product. To find mass %, divide the mass of thyroxine by the mass of extract and multiply by 100. Solution:

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a) There is not enough information to write complete chemical equations, but the following equations can be written: – C15H11I4NO4(s) + Na2CO3(s)  4I (aq) + other products – – I (aq) + Br2(l) + HCl(aq)  IO3 (aq) + other products  1 mol IO3  4 mol I – –  Moles of IO3 = 1 mol C15 H11 I 4 NO 4   = 4 moles IO3 are produced 1 mol C15 H11 I 4 NO 4  1 mol I  b)

+5 –2

–1

+1 –2

IO3 (aq) + H (aq) + I (aq)  I2(aq) + H2O(l) This is a difficult equation to balance because the iodine species are both reducing and oxidizing. The O.N. of iodine – – decreases from +5 in IO3 to 0 in I2 and so gains 5 electrons; the O.N. of iodine increases from –1 in I to 0 in I2 and so – loses 1 electron. Start balancing the equation by placing a coefficient of 5 in front of I (aq), so the electrons lost equal – the electrons gained. Do not place a 5 in front of I2(aq), because not all of the I2(aq) comes from oxidation of I (aq). – Some of the I2(aq) comes from the reduction of IO3 (aq). Place a coefficient of 3 in front of I2(aq) to correctly balance iodine: – + – IO3 (aq) + H (aq) + 5I (aq)  3I2(aq) + H2O(l) The reaction is now balanced from a redox standpoint, so finish balancing the reaction by balancing the oxygen and hydrogen. – + – IO3 (aq) + 6H (aq) + 5I (aq)  3I2(aq) + 3H2O(l) –

–

IO3 is the oxidizing agent, and I is the reducing agent.

  4 mol IO3  3 mol I 2  Moles of I2 produced per mole of thyroxine = 1 mol C15 H11 I 4 NO 4  1 mol IO   1 mol C15 H11 I 4 NO 4  3  = 12 moles of I2 are produced per mole of thyroxine c) Using Thy to represent thyroxine: The balanced equation for this reaction is I2(aq) + 2S2O3 (aq)  2I (aq) + S4O6 (aq). 2–

–

2–

2 103 L   0.1000 mol S2 O 3   2– 2– Moles of S2O3 = 17.23 mL   = 0.001723 mol S2O3 L   1 mL   1 mol I 2  1 mol Thy     = 7.179167 × 10–5 mol Thy Moles of Thy = 0.001723 mol S2 O 32  2   12 mol I 2   2 mol S2 O 3   776.8 g Thy    = 0.055767769 g Thy Mass (g) of Thy = 7.179167 105 mol Thy   1 mol Thy  mass of Thy 0.055767769 g Thy Mass % Thy = 100 = 100 = 12.8734 = 12.87% thyroxine mass of extract 0.4332 g

4.145

Plan: Write the balanced equation for the reaction between the base KOH and the oleic acid. For each sample, obtain moles of added KOH by multiplying the molarity and volume of the KOH. Use the molar ratio in the equation to obtain the moles and then mass of acid. To find the mass percent, divide the mass of acid by the mass of the sample and multiply by 100. Solution: Let HA represent oleic acid. The titration reaction is HA + KOH O + KA. 2 a) Sample 1 10 3 L  0.050 mol KOH   = 9.8 × 10–4 mol KOH  Moles of KOH = 19.60 mL  1L  1 mL 

4-64

 1 mol HA  282.45 g HA  Mass (g) of oleic acid = 9.8104 mol KOH   = 0.276801 g HA  1 mol KOH  1 mol HA  mass of oleic acid 0.276801 g Mass % oleic acid = 100 = 100 = 2.76801 = 2.8% oleic acid mass of sample 10.00 g

Sample 2

10 3 L  0.050 mol KOH   = 9.9 × 10–4 mol KOH  Moles of KOH = 19.80 mL  1L  1 mL   1 mol HA  282.45 g HA  Mass (g) of oleic acid = 9.9 104 mol KOH  = 0.2796255 g HA  1 mol KOH  1 mol HA  mass of oleic acid 0.2796255 g Mass % oleic acid = 100 = 100 = 2.796255 = 2.8% oleic acid mass of sample 10.00 g

Samples 3 and 4

10 3 L  0.050 mol KOH   = 1.0 × 10–3 mol KOH  Moles of KOH = 20.00 mL  1 mL   1L  1 mol HA  282.45 g HA  Mass (g) of oleic acid = 1.0 103 mol KOH   = 0.28245 g HA  1 mol KOH  1 mol HA  mass of oleic acid 0.28245 g Mass % oleic acid = 100 = 100 = 2.8245 = 2.8% oleic acid mass of sample 10.00 g b) The variation in acidity is a systematic error. The acidity values steadily increased over time as the KOH solution warmed and its volume increased. c) The actual acidity is 2.8245% as in Samples 3 and 4. This can be demonstrated by performing the titration several more times with the ethanol solution at 25°C.

4.146

Plan: Check the solubility rules to see if the compounds are soluble or insoluble. Any insoluble ionic compound(s) will appear as solid in the correct scene, while any soluble ionic compound(s) will appear as dissociated ions. Find molarity by dividing moles by the volume of the solution in liters. Solution: a) Of the three ionic compounds, only AgCl is insoluble. CuCl2 and MgCl2 should appear as ions and AgCl should 2+ 2+ – appear as a solid. Scene A best represents the mixture. There is AgCl solid and Mg , Cu , and Cl ions in the proper mole ratio in A. Scene B shows solid MgCl2 and Scenes C and D show solid CuCl2. Since MgCl2 and CuCl2 are soluble in water, they would be present as ions only. b) There are 12 spheres representing ions in Scene A.  5.0103 mol   = 0.060 moles of ions Moles of dissolved ions = 12 spheres  1 sphere   0.060 moles of ions  moles of ions  1 mL  = 1.2 M =  103 L    50.0 mL volume of solution  5.0 103 mol Ag  1 mol AgCl  143.4 g AgCl  c) Mass (g) of AgCl(s) = 4 Ag spheres   1 mol AgCl   1 mol Ag  1 sphere   

Molarity of dissolved ions =

= 2.868 = 2.9 g AgCl 4.147

Plan: Balance the equation to obtain the correct molar ratios. Use the mass percents to find the mass of each reactant in a 1.00 g sample, convert the mass of each reactant to moles, and use the molar ratios to find the limiting reactant and the amount of CO2 produced. Convert moles of CO2 produced to volume using the given conversion factor. Solution: a) Here is a suggested method for balancing the equation. 2– — Since PO4 remains as a unit on both sides of the equation, treat it as a unit when balancing.

4-65

— On first inspection, one can see that Na needs to be balanced by adding a “2” in front of NaHCO 3. This then affects the balance of C, so add a “2” in front of CO2. — Hydrogen is not balanced, so change the coefficient of water to “2,” as this will have the least impact on the other species. — Verify that the other species are balanced. Ca(H2PO4)2(s) + 2NaHCO3(s)  2CO2(g) + 2H2O(g) + CaHPO4(s) + Na2HPO4(s) Determine whether Ca(H2PO4)2 or NaHCO3 limits the production of CO2. In each case, calculate the moles of CO2 that might form.  31.0%  Mass (g) of NaHCO3 = 1.00 g   = 0.31 g NaHCO3  100% 





 35.0%  Mass (g) of Ca(H2PO4)2 = 1.00 g   = 0.35 g Ca(H2PO4)2  100% 





 1 mol NaHCO 3   2 mol CO 2   Moles of CO2 if NaHCO3 is limiting = 0.31 g NaHCO 3     84.01 g NaHCO 3   2 mol NaHCO 3  –3

= 3.690 × 10 mol CO2  1 mol Ca(H 2 PO 4 )2   2 mol CO 2    Moles of CO2 if Ca(H2PO4)2 is limiting = 0.35 g Ca(H 2 PO 4 )2    234.05 g Ca(H 2 PO 4 )2  1 mol Ca(H 2 PO 4 ) 2  –3

4.148

= 2.9908 × 10 mol CO2 –3 Since Ca(H2PO4)2 produces the smaller amount of product, it is the limiting reactant and 3.0 × 10 mol CO2 will be produced.  37.0 L   = 0.1106596 = 0.11 L CO2 b) Volume (L) of CO2 = 2.9908103 mol CO 2  1 mol CO2  Plan: Write a balanced acid-base reaction. Find the total moles of NaOH used by multiplying its molarity and volume in liters and use the molar ratio in the reaction to find the moles of HNO 3. Divide moles of HNO3 by its volume to obtain the molarity. Use the molarity and volume information to find the moles of NaOH initially added and the moles of HNO3 initially present. The difference of these two values is the moles of excess NaOH. Solution: The chemical equation is: HNO3(g) + NaOH(aq)  NaNO3(aq) + H2O(l) a) It takes a total of (20.00 + 3.22) mL = 23.22 mL NaOH to titrate a total of (50.00 + 30.00) mL = 80.00 mL of acid. 103 L  0.0502 mol NaOH    = 0.0011656 mol NaOH  Moles of NaOH = 23.22 mL    L  1 mL   1 mol HNO 3   Moles of HNO3 = 0.0011656 mol NaOH   = 0.0011656 mol HNO3 1 mol NaOH     1 mL  1  Molarity of HNO3 = 0.0011656 mol HNO 3   3  = 0.01457055 = 0.0146 M HNO3  80.00 mL  10 L 

b) First calculate the moles of the acid and base initially present. The difference will give the excess NaOH. 103 L  0.0502 mol NaOH    = 1.004 × 10–3 mol NaOH  Moles of NaOH = 20.00 mL   L   1 mL   103 L  0.01457055 mol HNO 3    = 7.285275 × 10–4 mol HNO Moles of HNO3 = 50.00 mL   3  L   1 mL   Copyright McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

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–4

Moles of NaOH required to titrate 7.285275 × 10 mol HNO3 = 7.285275 × 10 mol NaOH Moles excess NaOH = moles of added NaOH – moles of NaOH required for reaction –3

–4

= 1.004 × 10 mol NaOH – 7.285275 × 10 mol NaOH –4

–4

= 2.754725 × 10 = 2.8 × 10 mol NaOH 4.149

Plan: To determine the empirical formula, find the moles of each element present and divide by the smallest number of moles to get the smallest ratio of atoms. To find the molecular formula, divide the molar mass by the mass of the empirical formula to find the factor by which to multiple the empirical formula. Write the balanced acid-base reaction for part (c) and use the molar ratio in that reaction to find the mass of bismuth(III) hydroxide. Solution: a) Determine the moles of each element present. The sample was burned in an unknown amount of O 2; therefore, the moles of oxygen must be found by a different method.  1 mol CO 2   1 mol C   –3 Moles of C = 0.1880 g CO 2    = 4.271756 × 10 mol C 1 mol CO 2   44.01 g CO 2   1 mol H 2 O   2 mol H   –3 Moles of H = 0.02750 g H 2 O   = 3.052164 × 10 mol H 1 mol H 2 O  18.02 g H 2 O   1 mol Bi 2 O3  2 mol Bi   –4   Moles of Bi = 0.1422 g Bi 2 O 3  1 mol Bi O  = 6.103004 × 10 mol Bi  466.0 g Bi 2 O 3  2 3 Subtracting the mass of each element present from the mass of the sample will give the mass of oxygen originally present in the sample. This mass is used to find the moles of oxygen. 12.01 g C  Mass (g) of C = 4.271756 10 3 mol C   = 0.0513038 g C  1 mol C  1.008 g H  Mass (g) of H = 3.052164 10 3 mol H   = 0.0030766 g H  1 mol H   209.0 g Bi    Mass (g) of Bi = 6.103004 104 mol Bi  1 mol Bi  = 0.127553 g Bi    

Mass (g) of O = mass of sample – (mass C + mass H + mass Bi) = 0.22105 g sample – (0.0513038 g C + 0.0030766 g H + 0.127553 g Bi) = 0.0391166 g O  1 mol O   = 2.44482 × 10–4 mol O Moles of O = 0.0391166 g O  16.00 g O  Divide each of the moles by the smallest value (moles Bi).

4.271756103 3.052164103 = 7 H = =5 6.103004 104 6.103004 104 2.4448103 6.103004 104 O= =4 Bi = =1 4 6.103004 10 6.103004 104 Empirical formula = C7H5O4Bi b) The empirical formula mass is 362 g/mol. Therefore, there are 1086/362 = 3 empirical formula units per molecular formula making the molecular formula = 3 × C7H5O4Bi = C21H15O12Bi3. c) Bi(OH)3(s) + 3HC7H5O3(aq)  Bi(C7H5O3)3(s) + 3H2O(l) C=

4-67

103 g   1 mol C 21H15 O12 Bi 3   d) Moles of C21H15O12Bi3 = 0.600 mg C 21H15 O12 Bi 3     1 mg   1086 g C 21 H15 O12 Bi 3 





–4

= 5.52486 × 10 mol C21H15O12Bi3 Mass (mg) of Bi(OH)3 = 

3 mol Bi

1 mol Bi(OH)     260.0 g Bi(OH)3  1 mg  100%  3   3   1 mol Bi   1 mol Bi(OH)3  10 g  88.0%  3 

5.5248610 mol C H O Bi 1 mol C H O Bi  7



= 0.48970 = 0.490 mg Bi(OH)3 4.150

Plan: Use the solubility rules to predict the products of this reaction. For the total ionic equation, write all soluble ionic substances as dissociated ions. Ions not involved in the precipitate are spectator ions and are not included in the net ionic equation. Find the moles of dissolved ions and divide each by the volume in liters to find the concentration. The volume of the final solution is the sum of the volumes of the two reactant solutions. Solution: a) According to the solubility rules, all chloride compounds are soluble and most common carbonate compounds are insoluble. CaCO3 is the precipitate. Molecular equation: Na2CO3(aq) + CaCl2(aq)  CaCO3(s) + 2NaCl(aq) + 2– 2+ – + – Total ionic equation: 2Na (aq) + CO3 (aq) + Ca (aq) + 2Cl (aq)  CaCO3(s) + 2Na (aq) + 2Cl (aq) 2– 2+ Net ionic equation: CO3 (aq) + Ca (aq)  CaCO3(s) 2+ 2– 2+ 2– b) Ca and CO3 combine in a 1:1 ratio in CaCO3. There are two spheres of Ca and three spheres of CO3 ion. 2+ 2+ Since there are fewer spheres of Ca , Ca is the limiting reactant.  0.050 mol Ca 2  1 mol CaCO3 100.09 g CaCO 3    Mass of CaCO3 = 2 Ca 2  spheres  1 mol CaCO   1 sphere  1 mol Ca 2   3 = 10.009 = 10. g CaCO3 c) Original moles:  0.050 mol Na   +  = 0.30 mol Na+ Moles of Na = 6 Na spheres 1 sphere  

 0.050 mol CO32  2–  = 0.15 mol CO32– Moles of CO3 = 3 CO32spheres 1 sphere    0.050 mol Ca 2  2+  = 0.10 mol Ca2+ Moles of Ca = 2 Ca 2spheres  1 sphere   0.050 mol Cl  –  = 0.20 mol Cl– Moles of Cl = 4 Clspheres  1 sphere  + – 2+ 2– The moles of Na and Cl do not change. The moles of Ca goes to zero, and removes 0.10 mol of CO3 . 2– 2– 2– Moles of remaining CO3 = 0.15 mol CO3 – 0.10 mol = 0.050 mol CO3 Volume of final solution = 250. mL + 250. mL = 500. mL 0.30 mol  1 mL  + + Molarity of Na =  = 0.60 M Na  500. mL 103 L  –

Molarity of Cl =

2–

0.20 mol  1 mL  –  = 0.40 M Cl  500. mL 103 L 

Molarity of CO3 = 4.151

0.050 mol  1 mL  2–  3  = 0.10 M CO3  500. mL 10 L 

Plan: Write balanced equations. Use the density to convert volume of fuel to mass of fuel and then use the molar ratios to convert mass of each fuel to the mass of oxygen required for the reaction. Use the conversion factor given to convert mass of oxygen to volume of oxygen.

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Solution: a) Complete combustion of hydrocarbons involves heating the hydrocarbon in the presence of oxygen to produce carbon dioxide and water. Ethanol: C2H5OH(l) + 3O2(g)  2CO2(g) + 3H2O(l) Gasoline: 2C8H18(l) + 25O2(g)  16CO2(g) + 18H2O(g) b) The mass of each fuel must be found:  90%   1 mL   0.742 g  = 667.8 g gasoline Mass (g) of gasoline = 1.00 L  3    100%  10 L  1 mL   10%   1 mL   0.789 g  = 78.9 g ethanol Mass (g) of ethanol = 1.00 L  1 mL  100% 103 L   1 mol C8 H18   25 mol O 2  32.00 g O 2   Mass (g) of O2 to react with gasoline = 667.8 g C8 H18   2 mol C H  1 mol O  114.22 g C8 H18  8 18  2 





= 2338.64 g O2

 1 mol C 2 H 5 OH   3 mol O 2   32.00 g O 2   Mass (g) of O2 to react with ethanol = 78.9 g C 2 H 5 OH      46.07 g C 2 H 5 OH  1 mol C 2 H 5OH   1 mol O 2 





= 164.41 g O2 3

Total mass (g) of O2 = 2338.64 g O2 + 164.41 g O2 = 2503.05 = 2.50 × 10 g O2

 1 mol O2   22.4 L  = 1752.135 = 1.75 × 103 L O c) Volume (L) of O2 = 2503.05 g O2  2 1 mol O   32.00 g O2  2

 100%  3 d) Volume (L) of air = 1752.135 L O2   = 8383.42 = 8.38 × 10 L air  20.9%  4.152

Plan: Write balanced reactions for the complete combustion of gasoline and for the incomplete combustion. Use molar ratios to find the moles of CO2 and moles of CO produced. Obtain the number of molecules of each gas by multiplying moles by Avogadro’s number. Solution: a) Complete combustion: 1. 2C8H18(l) + 25O2(g)  16CO2(g) + 18H2O(g) Incomplete combustion: 2. 2C8H18(l) + 17O2(g)  16CO(g) + 18H2O(g) Assuming a 100-g sample of gasoline, 95%, or 95.0 g, will react by equation 1, and 5.0%, or 5.0 g, will react by equation 2. 23  1 mol C H    16 mol CO 2   8 18   6.022 10 CO2  Molecules of CO2 = 95.0 g C8 H18    1 mol CO   2 mol C8 H18  114.22 g C8 H18   2





= 4.00693 × 10 molecules CO2  1 mol C H   23  16 mol CO   8 18   6.022 10 CO  Molecules of CO = 5.0 g C8 H18    1 mol CO   2 mol C8 H18  114.22 g C8 H18  





= 2.10891 × 10 molecules CO Ratio of CO2 to CO molecules =

4.006931024 CO2 molecules

= 18.99998 = 19 2.108911023 CO molecules b) Again, we may assume 100 g of gasoline.  1 mol C8 H18   16 mol CO 2   44.01 g CO 2   Mass (g) of CO2 = 95.0 g C8 H18     1 mol CO  114.22 g C8 H18   2 mol C8 H18  2  = 292.83 g CO2 Copyright McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

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 1 mol C8 H18   16 mol CO   28.01 g CO   Mass (g) of CO = 5.0 g C8 H18     = 9.8091 g CO 114.22 g C8 H18   2 mol C8 H18   1 mol CO 





Mass ratio of CO2 to CO =

292.83 g CO2 = 29.85289 = 30 9.8091 g CO

c) Let x = fraction of CO2 and y = fraction of CO. For a 1/1 mass ratio of CO2 to CO,

(x)(44.01) = 1, y(28.01)

where 44.01 g/mol is the molar mass of CO2 and 28.01 g/mol is the molar mass of CO. x + y = 1 or y = 1 – x Substituting:

(x)(44.01) =1 (1 x)(28.01)

44.01x = 28.01 – 28.01x 72.02x = 28.01 x = 0.39 and y = 1 – 0.39 = 0.61 Thus, 61% of the gasoline must form CO. 4.153

Plan: From the molarity and volume of the base NaOH, find the moles of NaOH and use the molar ratios from the two balanced equations to convert the moles of NaOH to moles of HBr to moles of vitamin C. Use the molar mass of vitamin C to convert moles to grams. Solution: 103 L  0.1350 mol NaOH   = 0.005832 mol NaOH  Moles of NaOH = 43.20 mL NaOH  1 mL   1L

 1 mol HBr 1 mol C6 H8 O6 176.12 g C 6 H 8O 6   1 mg   Mass (g) of vitamin C = 0.005832 mol NaOH      2 mol HBr   1 mol C H O 103 g  1 mol NaOH  6 8 6 = 513.5659 = 513.6 mg C6H8O6 Yes, the tablets have the quantity advertised. 4.154

Plan: Convert mass of NaCl to moles of salt and then to moles of ions using the molar ratio in the formula. Use the molar ratio of ions in MgCl2 to find the mass of MgCl2 that contains the same number of moles of ions as the NaCl. Do the same for the CaS. Solution:  1 mol NaCl   2 mol ions   a) Moles of ions in NaCl = 12.4 g NaCl    = 0.4243669 mol ions 1 mol NaCl   58.44 g NaCl  1 mol MgCl 2   95.21 g MgCl 2   Mass (g) of MgCl2 = 0.4243669 mol ions   = 13.46799 = 13.5 g MgCl2  1 mol MgCl 2   3 mol ions   1 mol CaS   72.14 g CaS   b) Mass (g) of CaS = 0.4243669 mol ion    = 15.3069 = 15.3 g CaS  2 mol ions   1 mol CaS 





c) The CaS solution dissolves the most protein. All three solutions have equal ion concentrations, but CaS will form two divalent ions. 4.155

Plan: For part (a), assign oxidation numbers to each element. The reactant that is the reducing agent contains an atom that is oxidized (O.N. increases from the left side to the right side of the equation). The reactant that is the oxidizing agent contains an atom that is reduced (O.N. decreases from the left side to the right side of the equation). Use the molar ratios in the balanced equation to convert mass of ammonium perchlorate to moles of product and to moles of Al required in the reaction. Use the density values to convert masses to volumes.

4-70

Solution: a) +4

–8

–3 +1+7 –2

+6 –6

–3

+3 –2

+3 –1

+2 +1 –2

+2 –2

catalyst 3NH4ClO4(s) + 3Al(s)   Al2O3(s) +AlCl3(s) + 6H2O(g) + 3NO(g) The O.N. of chlorine decreases from +7 in NH4ClO4 to –1 in AlCl3 and is reduced; the O.N. of Al increases from 0 in Al to +3 in the products and is oxidized. The oxidizing agent is ammonium perchlorate and the reducing agent is aluminum. 103 g   1 mol NH 4 ClO 4   9 mol gas   b) Moles of gas = 50.0 kg NH 4 ClO 4      1 kg  117.49 g NH 4 ClO 4   3 mol NH 4 ClO 4 





= 1276.70 = 1.28 × 10 mol gas c) Initial volume: 103 g  1 cc  1 mL 103 L       Volume (L) of NH4ClO4 = 50.0 kg  1.95 g  1 cc  1 mL  = 25.6410 L  1 kg     3 10 g   3 mol Al  1 mol NH 4 ClO 4    26.98g Al   Mass of Al = 50.0 kg NH 4 ClO 4       1 mol Al  = 11481.828 g Al  1 kg  117.49 g NH 4 ClO 4   3 mol NH 4 ClO 4  





3  1 cc  1 mL  10 L   Volume (L) of Al = 11481.828 g Al     = 4.2525 L  2.70 g Al   1 cc   1 mL 





Initial volume = 25.6410 L + 4.2525 L = 29.8935 L Final volume: 10 3 g   1 mol NH 4 ClO 4  1 mol Al 2 O 3 101.96 g Al 2 O 3   Mass (g) of Al2O3 = 50.0 kg NH 4 ClO 4       1 kg  117.49 g NH 4 ClO 4  3 mol NH 4 ClO 4  1 mol Al 2 O 3  = 14463.64 g Al2O3 3   1 cc 1 mL  10 L   Volume (L) of Al2O3 = 14463.674 g Al 2 O3     = 3.6432 L  1 cc   1 mL   3.97 g Al 2 O3  10 3 g  1 mol NH 4 ClO 4  1 mol AlCl3  133.33 g AlCl 3       Mass (g) of AlCl3 = 50.0 kg NH 4 ClO 4  117.49 g NH ClO  3 mol NH ClO  1 mol AlCl   1 kg  4 4  4 4  3 









= 18913.67 g AlCl3   1 cc 1 mL  103 L      Volume (L) of AlCl3 = 18913.67 g AlCl3   1 cc  1 mL  = 7.7515 L  2.44 g AlCl3     22.4 L   = 28598.08 L Volume (L) of gas = 1276.70 mol gas  1 mol gas  Final volume = 3.6432 L + 7.7515 L + 28598.08 L = 28609.4747 L 4 Volume change = Final volume – initial volume = (28609.4747 L) – (29.8935 L) = 28579.5812 = 2.86 × 10 L The volumes of all solids (before and after) are insignificant.



4.156



Plan: Write soluble ionic substances as dissociated ions and omit spectator ions in the net ionic equations. Solution: 1) This is a combination redox reaction. Cu(s) + Br2(aq)  CuBr2(aq) 2+ – Net ionic: Cu(s) + Br2(aq)  Cu (aq) + 2Br (aq) 2) This is a precipitation reaction; most common hydroxide compounds are insoluble. CuBr2(aq) + 2NaOH(aq)  Cu(OH)2(s) + 2NaBr(aq)

4-71

Cu (aq) + 2Br (aq) + 2Na (aq) + 2OH (aq)  Cu(OH)2(s) + 2Na (aq) + 2Br (aq) 2+ – Net ionic: Cu (aq) + 2OH (aq)  Cu(OH)2(s) 3) Cu(OH)2(s)  CuO(s) + H2O(l) Net ionic: Cu(OH)2(s)  CuO(s) + H2O(l) 4) CuO(s) + 2HNO3(aq)  Cu(NO3)2(aq) + H2O(l) + – 2+ – CuO(s) + 2H (aq) + 2NO3 (aq)  Cu (aq) + 2NO3 (aq) + H2O(l) + 2+ Net ionic: CuO(s) + 2H (aq)  Cu (aq) + H2O(l) 5) This is a precipitation reaction; most common phosphate compounds are insoluble. 3Cu(NO3)2(aq) + 2Na3PO4(aq)  Cu3(PO4)2(s) + 6NaNO3(aq) 2+ – + 3– + – 3Cu (aq) + 6NO3 (aq) + 6Na (aq) + 2PO4 (aq)  Cu3(PO4)2(s) + 6Na (aq) + 6NO3 (aq) 2+ 3– Net ionic: 3Cu (aq) + 2PO4 (aq)  Cu3(PO4)2(s) 6) Cu3(PO4)2(s) + 3H2SO4(aq)  3CuSO4(aq) + 2H3PO4(aq) + 2– 2+ 2– Cu3(PO4)2(s) + 6H (aq) + 3SO4 (aq)  3Cu (aq) + 3SO4 (aq) + 2H3PO4(aq) + 2+ Net ionic: Cu3(PO4)2(s) + 6H (aq)  3Cu (aq) + 2H3PO4(aq) 7) This is a displacement reaction. CuSO4(aq) + Zn(s)  Cu(s) + ZnSO4(aq) 2+ 2– 2+ 2– Cu (aq) + SO4 (aq) + Zn(s)  Cu(s) + Zn (aq) + SO4 (aq) 2+ 2+ Cu (aq) + Zn(s)  Cu(s) + Zn (aq) 2+

4.157

–

Plan: Write a balanced equation for this acid-base reaction. Find the moles of acid and base and use the molar ratio in the balanced reaction to determine if stoichiometric amounts of each have been combined. To determine the molarity of B, divide the moles present by the volume in liters. For part (c), find the volume of Solution B required for reaction and subtract from that the current volume to determine the additional volume needed. Solution: a) The balanced equation is Ca(OH)2(aq) + 2HCl(aq) CaCl2(aq) + 2H2O(l). 2    0.0010 mol Ca 1 mol Ca(OH)2  = 0.0040 mol Ca(OH) Moles of Ca(OH)2 = 4 spheres Ca 2   2  1 sphere   1 mol Ca 2 

 0.0010 mol H 1 mol HCl   Moles of HCl = 6 spheres H   = 0.0060 mol HCl  1 sphere  1 mol H   2 mol HCl   = 0.0080 mol HCl Moles of HCl required for the titration = 0.0040 mol Ca(OH)2  1 mol Ca(OH)2  Since 0.0060 mol of HCl has been added and 0.0080 mol HCl are required for complete titration, the equivalence point of the titration has not been reached. b) Molarity =

mol HCl 0.0060 mol HCl  1 mL   3  = 0.24 M = 10 L  volume of solution 25.0 mL

 2 mol HCl   = 0.0080 mol HCl c) Moles of HCl required = 0.0040 mol Ca(OH)2  1 mol Ca(OH)2 

 1 L solution   1 mL  = 33.333 mL Volume (L) of HCl = 0.0080 mol HCl  0.24 mol HCl 103 L  Additional volume = volume needed – volume added = 33.333 mL

= 8.333 = 8.3 mL

4-72

CHAPTER 5 GASES AND THE KINETICMOLECULAR THEORY FOLLOW–UP PROBLEMS 5.1A

Plan: Use the equation for gas pressure in an open-end manometer to calculate the pressure of the gas. Use 2 conversion factors to convert pressure in mmHg to units of torr, pascals and lb/in . Solution: Because Pgas < Patm, Pgas = Patm – h Pgas = 753.6 mmHg – 174.0 mmHg = 579.6 mmHg  1 torr    = 579.6 torr Pressure (torr) = 579.6 mmHg  1 mmHg 





 1 atm 1.01325105 Pa    = 7.727364 × 104 = 7.727 × 104 Pa  Pressure (Pa) = 579.6 mmHg   1atm   760 mmHg   1 atm 14.7 lb/in 2  2  = 11.21068 = 11.2 lb/in2  Pressure (lb/in ) = 579.6 mmHg   760 mmHg   1 atm 







5.1B



Plan: Convert the atmospheric pressure to torr. Use the equation for gas pressure in an open-end manometer to calculate 2 the pressure of the gas. Use conversion factors to convert pressure in torr to units of mmHg, pascals and lb/in . Solution: Because Pgas > Patm, Pgas = Patm + h   Pgas = (0.9475 atm)  + 25.8 torr = 745.9 torr    Pressure (mmHg) = (745.9 torr)    Pressure (Pa) = (745.9 mmHg)  

 2 Pressure (lb/in ) = (745.9 mmHg)   5.2B

 = 745.9 mmHg    5   

  

  = 9.94452 × 104 = 9.945 × 104 Pa   2  = 14.427 = 14.4 lb/in

Plan: Given in the problem is an initial volume, initial pressure, and final volume for the argon gas. The final pressure is to be calculated. The temperature and amount of gas are fixed. Rearrange the ideal gas law to the appropriate form and solve for P2. Once solved for, P2 must be converted from atm units to kPa units. Solution: P1 = 0.871 atm; V1 = 105 mL P2 = unknown V2 = 352 mL PV PV 1 1 = 2 2 At fixed n and T: n1T1 n2T2 PV 1 1 = P2V2 PV P2 (atm) = 1 1 = = 0.259815 = 0.260 atm V2

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5-1

 P2 (kPa) = (0.260 atm)   5.2A

  = 26.3 kPa 

Plan: Given in the problem is an initial volume, initial pressure, and final pressure for the oxygen gas. The final volume is to be calculated. The temperature and amount of gas are fixed. Convert the final pressure to atm units. Rearrange the ideal gas law to the appropriate form and solve for V2. Solution: P1 = 122 atm; V1 = 651 L P2 = 745 mmHg V2 = unknown PV P2V2 1 1 = At fixed n and T: n1T1 n2T2 PV 1 1 = P2V2

 P2 (atm) = (745 mmHg)   PV V2 (atm) = 1 1 = P2 5.3A

  = 0.980263 atm  4

= 8.1021 × 10 = 8.10 × 10 L

Plan: Convert the temperatures to kelvin units and the initial pressure to units of torr. Examine the ideal gas law, PV PV noting the fixed variables and those variables that change. R is always constant so 1 1 = 2 2 . In this problem, n1T1 n2T2 P and T are changing, while n and V remain fixed. Solution: T1 = 23°C T2 = 100°C P1 = 0.991 atm P2 = unknown n and V remain constant Converting T1 from °C to K: 23°C + 273.15 = 296.15 K Converting T2 from °C to K: 100°C + 273.15 = 373.15 K   P1 (torr) = (0.991 atm)  = 753.16 torr   Arranging the ideal gas law and solving for P2: P1 V1 P2 V2 P1 P2 n1 T1 n 2 T2 T1 T2 P2 (torr) = P1

T2 T1

  

  = 948.98 = 949 torr 

Because the pressure in the tank (949 torr) is less than the pressure at which the safety valve will open 3 (1.00 × 10 torr), the safety valve will not open. 5.3B

Plan: This is Charles’s law: at constant pressure and with a fixed amount of gas, the volume of a gas is directly proportional to the absolute temperature of the gas. The temperature must be lowered to reduce the volume of a gas. Arrange the ideal gas law, solving for T2 at fixed n and P. Temperature must be converted to kelvin units. Solution: V1 = 32.5 L V2 = 28.6 L T1 = 40°C (convert to K) T2 = unknown n and P remain constant Converting T from °C to K: T1 = 40 °C + 273.15 = 313.15K McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution

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5-2

Arranging the ideal gas law and solving for T2: PV PV V1 V 1 1 = 2 2 or  2 n1T1 n2 T2 T1 T2 V2   T2  T1 = (313.15 K)  = 275.572 K – 273.15 = 2.422 = 2°C   V1 5.4A

Plan: In this problem, the amount of gas is decreasing. Since the container is rigid, the volume of the gas will not change with the decrease in moles of gas. The temperature is also constant. So, the only change will be that the pressure of the gas will decrease since fewer moles of gas will be present after removal of the 5.0 g of ethylene. Rearrange the ideal gas law to the appropriate form and solve for P2. Since the ratio of moles of ethylene is equal to the ratio of grams of ethylene, there is no need to convert the grams to moles. (This is illustrated in the solution by listing the molar mass conversion twice.) Solution: P1 = 793 torr; P2 = ? mass1 = 35.0 g; mass2 = 35.0 – 5.0 = 30.0 g PV P2V2 1 1 = At fixed V and T: n1T1 n2T2 P1 P = 2 n1 n2

 1 mol C2 H 4 

Pn P2 = 1 2 = 793 torr  n1

5.4B

30.0 g C H  28.05 g C H  2

2 4  = 679.714 = 680. torr  1 mol C2 H 4    35.0 g C2 H 4    28.05 g C 2 H 4 





Plan: Examine the ideal gas law, noting the fixed variables and those variables that change. R is always constant PV PV so 1 1 = 2 2 . In this problem, n and V are changing, while P and T remain fixed. n1T1 n2T2 Solution: m1 = 1.26 g N2 m2 = 1.26 g N2 + 1.26 g He V1 = 1.12 L V2 = unknown P and T remain constant   2   = 0.044968 mol N2 = n1 Converting m1 (mass) to n1 (moles): (1.26 g N2)    2

 Converting m2 (mass) to n2 (moles): 0.044968 mol N2 + (1.26 g He)  

  

= 0.044968 mol N2 + 0.31476 mol He = 0.35973 mol gas = n2 Arranging the ideal gas law and solving for V2: P1 V1 P2 V2 V1 V2 n1 T1 n2 T2 n1 n2 V2 = V1

P2 P1

  

  = 8.9597 = 8.96 L 

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5-3

5.5A

Plan: Convert the temperatures to kelvin. Examine the ideal gas law, noting the fixed variables and those variables PV PV that change. R is always constant so 1 1 = 2 2 . In this problem, P, V, and T are changing, while n remains n1T1 n2T2 fixed. Solution: T1 = 23°C T2 = 18°C P1 = 755 mmHg P2 = unknown V1 = 2.55 L V2 = 4.10 L n remains constant Converting T1 from °C to K: 23°C + 273.15 = 296.15 K Converting T2 from °C to K: 18°C + 273.15 = 291.15 K Arranging the ideal gas law and solving for P2: PV P2V2 P1V1 P2V2 1 1 n 1 T1 n 2 T2 T1 T2 P2 (mmHg) = P1

5.5B

  

V1T2 V2T1

  = 461.645 = 462 mmHg 

Plan: Convert the temperatures to kelvin. Examine the ideal gas law, noting the fixed variables and those variables PV PV that change. R is always constant so 1 1 = 2 2 . In this problem, P, V, and T are changing, while n remains n1T1 n2T2 fixed. Solution: T1 = 28°C T2 = 21°C P1 = 0.980 atm P2 = 1.40 atm V1 = 2.2 L V2 = unknown n remains constant Converting T1 from °C to K: 28°C + 273.15 = 301.15 K Converting T2 from °C to K: 21°C + 273.15 = 294.15 K Arranging the ideal gas law and solving for V2: P1V1 P2V2 P1V1 P2V2 n 1 T1 n 2 T2 T1 T2 V2 (L) = V1

5.6A

  

P1T2 P2T1

  = 1.5042 = 1.5 L 

Plan: From Sample Problem 5.6 the temperature of 21°C and volume of 438 L are given. The pressure is 1.37 atm and the unknown is the moles of oxygen gas. Use the ideal gas equation PV = nRT to calculate the number of moles of gas. Multiply moles by molar mass to obtain mass. Solution: PV = nRT n=





 Mass (g) of O2 = (24.8475 mol O2)   Copyright



1.37 atm 438 L PV = = 24.8475 mol O2  0.0821 atm  L  RT  273.15  21 K   mol  K    = 795.12 = 795 g O2 2  2

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5-4

5.6B

Plan: Convert the mass of helium to moles, the temperature to kelvin units, and the pressure to atm units. Use the ideal gas equation PV = nRT to calculate the volume of the gas. Solution: P = 731 mmHg V = unknown m = 3950 kg He T = 20°C     = 9.8676 × 105 mol = n Converting m (mass) to n (moles): (3950 kg He)       Converting T from °C to K: 20°C + 273.15 = 293.15 K  Converting P from mmHg to atm: (731 mmHg)  

  = 0.962 atm 

PV = nRT nRT V= P

5.7A

  

  mol  K 

= 2.4687 × 10 = 2.47 × 10 L

Plan: Balance the chemical equation. The pressure is constant and, according to the picture, the volume approximately doubles. The volume change may be due to the temperature and/or a change in moles. Examine the balanced reaction for a possible change in number of moles. Rearrange the ideal gas law to the appropriate form and solve for the variable that changes. Solution: The balanced chemical equation must be 2CD  C2 + D2 Thus, the number of mole of gas does not change (2 moles both before and after the reaction). Only the temperature remains as a variable to cause the volume change. Let V1 = the initial volume and 2V1 = the final volume V2. T1 = (–73 + 273.15) K = 200.15 K PV PV 1 1 = 2 2 At fixed n and P: n1T1 n2T2 V1 V = 2 T1 T2

T2 = 5.7B



2V1 200.15 K  V2 T1 = = 400.30 K – 273.15 = 127.15 = 127°C V1 V1 

Plan: The pressure is constant and, according to the picture, the volume approximately decreases by a factor of 2 (the final volume is approximately one half the original volume). The volume change may be due to the temperature change and/or a change in moles. Consider the change in temperature. Examine the balanced reactions for a possible change in number of moles. Think about the relationships between the variables in the ideal gas law in order to determine the effect of temperature and moles on gas volume. Solution: Converting T1 from °C to K: 199°C + 273.15 = 472.15 K Converting T2 from °C to K: –155°C + 273.15 = 118.15 K According to the ideal gas law, temperature and volume are directly proportional. The temperature decreases by a factor of 4, which should cause the volume to also decrease by a factor of 4. Because the volume only decreases by a factor of 2, the number of moles of gas must have increased by a factor of 2 (moles of gas and volume are also directly proportional). 1/4 (decrease in V from the decrease in T) × 2 (increase in V from the increase in n) = 1/2 (a decrease in V by a factor of 2) Thus, we need to find a reaction in which the number of moles of gas increases by a factor of 2. McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution

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5-5

In equation (1), 3 moles of gas yield 2 moles of gas. In equation (2), 2 moles of gas yield 4 moles of gas. In equation (3), 1 mole of gas yields 3 moles of gas. In equation (4), 2 moles of gas yield 2 moles of gas. Because the number of moles of gas doubles in equation (2), that equation best describes the reaction in the figure in this problem. 5.8A

Plan: Density of a gas can be calculated using a version of the ideal gas equation, d = (PM )/(RT). Two calculations are required, one with T = 0°C = 273.15 K and P = 380 torr and the other at STP which is defined as T = 273 K and P = 1 atm. Solution: Density at T = 273 K and P = 380 torr: 380 torr 44.01 g/mol  1 atm   d=  = 0.98124 = 0.981 g/L  760 torr   0.0821 atm  L    273.15 K  mol  K 











Density at T = 273 K and P = 1 atm. (Note: The 1 atm is an exact number and does not affect the significant figures in the answer.)

44.01 g/mol1 atm

 0.0821 atm  L    273.15 K  mol  K 





= 1.96249 = 1.96 g/L

The density of a gas increases proportionally to the increase in pressure. 5.8B

Plan: Density of a gas can be calculated using a version of the ideal gas equation, d = PM /RT Solution: Density of NO2 at T = 297.15 K (24°C + 273.15) and P = 0.950 atm: d=

= 1.7917 = 1.79 g/L atm  L   mol  K Nitrogen dioxide is more dense than dry air at the same conditions (density of dry air = 1.13 g/L). 5.9A

  

Plan: Calculate the mass of the gas by subtracting the mass of the empty flask from the mass of the flask containing the condensed gas. The volume, pressure, and temperature of the gas are known. dRT mRT The relationship d = PM /RT is rearranged to give M = or M = P PV Solution: Mass (g) of gas = mass of flask + vapor – mass of flask = 68.697 – 68.322 = 0.375 g T = 95.0°C + 273.15 = 368.15 K  1 atm  P = 740 torr   = 0.973684 atm  760 torr 

V = 149 mL = 0.149 L

mRT = PV

 0.0821 atm  L  368.15 K  mol  K  = 78.125 = 78.1 g 0.973684 atm0.149 L 

0.375 g 

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5-6

5.9B

Plan: Calculate the mass of the gas by subtracting the mass of the empty glass bulb from the mass of the bulb containing the gas. The volume, pressure, and temperature of the gas are known. The relationship d = PM /RT is dRT mRT rearranged to give M = or M = . Use the molar mass of the gas to determine its identity. P PV Solution: Mass (g) of gas = mass of bulb + gas – mass of bulb = 82.786 – 82.561 = 0.225 g T = 22°C + 273.15 = 295.15 K    = 0.965 atm P = (733 mmHg)    V = 350 mL = 0.350 L  atm  L     mRT mol K M= = = 16.1426 = 16.1 g/mol PV Methane has a molar mass of 16.04 g/mol. Nitrogen monoxide has a molar mass of 30.01 g/mol. The gas that has a molar mass that matches the calculated value is methane.

5.10A

Plan: Calculate the number of moles of each gas present and then the mole fraction of each gas. The partial pressure of each gas equals the mole fraction times the total pressure. Total pressure equals 1 atm since the problem specifies STP. This pressure is an exact number, and will not affect the significant figures in the answer Solution:  1 mol He   = 1.373970 mol He Moles of He = 5.50 g He  4.003 g He   1 mol Ne   = 0.743310 mol Ne Moles of Ne = 15.0 g Ne  20.18 g Ne 

 1 mol Kr   = 0.417661 mol Kr Moles of Kr = 35.0 g Kr  83.80 g Kr  Total number of moles of gas = 1.373970 + 0.743310 + 0.417661 = 2.534941 mol PA = XA × Ptotal 1.37397 mol He  1 atm  = 0.54201 = 0.542 atm He  2.534941 mol   0.74331 mol Ne  PNe =  1 atm  = 0.29323 = 0.293 atm Ne  2.534941 mol   0.41766 mol Kr  PKr =  1 atm  = 0.16476 = 0.165 atm Kr  2.534941 mol 

PHe = 

5.10B

Plan: Use the formula PA = XA × Ptotal to calculate the mole fraction of He. Multiply the mole fraction by 100% to calculate the mole percent of He. Solution: PHe = XHe × Ptotal P     Mole percent He = XHe (100%) =  He  (100%) = 70.1%    total 

5.11A

Plan: The gas collected over the water will consist of H2 and H2O gas molecules. The partial pressure of the water can be found from the vapor pressure of water at the given temperature given in the text. Subtracting this partial pressure of water from total pressure gives the partial pressure of hydrogen gas collected over the water. Calculate

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5-7

the moles of hydrogen gas using the ideal gas equation. The mass of hydrogen can then be calculated by converting the moles of hydrogen from the ideal gas equation to grams. Solution: From the table in the text, the partial pressure of water is 13.6 torr at 16°C. P = 752 torr – 13.6 torr = 738.4 = 738 torr H2 The unrounded partial pressure (738.4 torr) will be used to avoid rounding error.  1 atm  738.4 torr 1495 mL 103 L  PV  Moles of hydrogen = n = =     760 torr   0.0821 atm  L  RT  1 mL    273.15  16 K  mol  K 









 

= 0.061186 mol H2

 2.016 g H 2   = 0.123351 = 0.123 g H2 Mass (g) of hydrogen = 0.061186 mol H 2   1 mol H 2  5.11B

Plan: The gas collected over the water will consist of O2 and H2O gas molecules. The partial pressure of the water can be found from the vapor pressure of water at the given temperature given in the text. Subtracting this partial pressure of water from total pressure gives the partial pressure of oxygen gas collected over the water. Calculate the moles of oxygen gas using the ideal gas equation. The mass of oxygen can then be calculated by converting the moles of oxygen from the ideal gas equation to grams. Solution: From the table in the text, the partial pressure of water is 17.5 torr at 20°C. P = 748 torr – 17.5 torr = 730.5 = 730 torr O2 PV      Moles of oxygen = n = =      atm  L  RT    mol  K  = 0.012252 mol O2   2  = 0.39207 = 0.392 g O2 Mass (g) of oxygen = (0.012252 mol O2)    2 

5.12A

Plan: Write a balanced equation for the reaction. Calculate the moles of CO(g) from the ideal gas equation. Then, use the stoichiometric ratio from the balanced equation and the molar mass of SiO2 to determine the mass of SiO2. Solution: The balanced equation is SiO2(s) + 2C(s)  Si(s) + 2CO(g). (0.975 atm)(44.8 L) PV Moles of CO    1.785  1.79 mol CO RT (0.0821 atm  L/mol  K)(298 K) 1 mol SiO2   60.09 g SiO2   Mass(g) of SiO2  (1.79 mol CO)     53.78  53.8 g SiO 2  2 mol CO   1 mol SiO 2 

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5-8

5.12B

Plan: Write a balanced equation for the reaction. Calculate the moles of H2(g) from the starting amount of CuO(s) using the stoichiometric ratio from the balanced equation. Find the volume of the H2(g) from the ideal gas equation. Solution: The balanced equation is CuO(s) + H2(g)  Cu(s) + H2O(g).  1 mol CuO   1 mol H 2   Amount (mol) of H 2 ( g)  (35.5 g CuO)     0.46626  0.446 mol H 2 ( g)  79.55 g CuO  1 mol CuO 

Volume of H 2 ( g) 

5.13A

(498 K)  1 atm  nRT   18.1 L H ( g)   (0.446 mol)(0.0821 atm  L/mol  K) 2 P (765 torr)  760 torr 

Plan: Balance the equation for the reaction. Determine the limiting reactant by finding the moles of each reactant from the ideal gas equation, and comparing the values. Calculate the moles of remaining excess reactant. This is the only gas left in the flask, so it is used to calculate the pressure inside the flask. Solution: The balanced equation is NH3(g) + HCl(g)  NH4Cl(s). The stoichiometric ratio of NH3 to HCl is 1:1, so the reactant present in the lower quantity of moles is the limiting reactant. Moles of ammonia =







0.452 atm 10.0 L PV = = 0.18653 mol NH3  0.0821 atm  L  RT    mol  K  273.15  22 K



Moles of hydrogen chloride =



 





103 L  7.50 atm 155 mL PV   = 0.052220 mol HCl =    0.0821 atm  L  1 mL RT     271.15 K   mol  K 





The HCl is limiting so the moles of ammonia gas left after the reaction would be 0.18653 – 0.052220 = 0.134310 mol NH3.  0.0821 atm  L   273.15  22 K 0.134310 mol   mol  K  nRT Pressure (atm) of ammonia = = V 10.0 L









 



= 0.325458 = 0.325 atm NH3 5.13B

Plan: Balance the equation for the reaction. Use the ideal gas law to calculate the moles of fluorine that react. Determine the limiting reactant by determining the moles of product that can be produced from each of the reactants and comparing the values. Use the moles of IF5 produced and the ideal gas law to calculate the volume of gas produced. Solution: The balanced equation is I2(s) + 5F2 (g)  2IF5(g). PV Amount (mol) of F2 that reacts = n = = = 0.10105 mol F2  atm  L  RT    mol  K 

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5-9

 Amount (mol) of IF5 produced from F2 = 0.10105 mol F2    Amount (mol) of IF5 produced from I2 = 4.16 g I2  

  = 0.040421 mol IF5  2  5

2 2

   

  = 0.032782 mol IF5  2  5

Because a smaller number of moles is produced from the I2, I2 is limiting and 0.032782 mol of IF5 are produced.   

nRT Volume (L) of IF5 = P

5.14A

atm  L   mol  K 

= 1.08850 = 1.09 L

Plan: Graham’s law can be used to solve for the effusion rate of the ethane since the rate and molar mass of helium are known, along with the molar mass of ethane. In the same way that running slower increases the time to go from one point to another, so the rate of effusion decreases as the time increases. The rate can be expressed as 1/time. Solution: Rate He Rate C 2 H 6



Molar mass C2 H 6 Molar m ass He

 0.010 mol He      1.25 min    0.010 mol C 2 H6     tC H   2

30.07 g/mol 4.003 g/mol

0.800 t = 2.74078 t = 3.42597 = 3.43 min 5.14B

Plan: Graham’s law can be used to solve for the molar mass of the unknown gas since the rates of both gases and the molar mass of argon are known. Rate can be expressed as the volume of gas that effuses per unit time. Solution: Rate of Ar = 13.8 mL/time Rate of unknown gas = 7.23 mL/time Mass of Ar = 39.95 g/mol  Molar mass     Molar mass   Ar

  

 Molar mass   Molar mass

  as 

 Munknown gas = (M Ar)  

  

  

 

Munknown gas = (39.95 g/mol) 

  = 146 g/mol  2

CHEMICAL CONNECTIONS BOXED READING PROBLEMS B5.1

Plan: Examine the change in density of the atmosphere as altitude changes.

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5-10

Solution: The density of the atmosphere decreases with increasing altitude. High density causes more drag on the aircraft. At high altitudes, low density means that there are relatively few gas particles present to collide with the aircraft. B5.2

Plan: The conditions that result in deviations from ideal behavior are high pressure and low temperature. At high pressure, the volume of the gas decreases, the distance between particles decreases, and attractive forces between gas particles have a greater effect. A low temperature slows the gas particles, also increasing the affect of attractive forces between particles. Solution: Since the pressure on Saturn is significantly higher and the temperature significantly lower than that on Venus, atmospheric gases would deviate more from ideal gas behavior on Saturn.

B5.3

Plan: To find the volume percent of argon, multiply its mole fraction by 100. The partial pressure of argon gas can be found by using the relationship PAr = XAr × Ptotal. The mole fraction of argon is given in Table B5.1. Solution: Volume percent = mole fraction × 100 = 0.00934 × 100 = 0.934% The total pressure at sea level is 1.00 atm = 760 torr. PAr = XAr × Ptotal = 0.00934 × 760 torr = 7.0984 = 7.10 torr

B5.4

Plan: To find the moles of gas, convert the mass of the atmosphere from t to g and divide by the molar mass of air. Knowing the moles of air, the volume can be calculated at the specified pressure and temperature by using the ideal gas law. Solution: 1000 kg 1000 g   1 mol  a) Moles of gas = 5.14 1015 t    28.8 g   1 t  1 kg   20

= 1.78472 × 10 = 1.78 × 10 mol b) PV = nRT

L  atm  1.7847210 mol0.0821 mol 273.15  25K  K  = 4.36866 × 10 = 4 × 10 L 1 atm 20

nRT V= = P

END-OF-CHAPTER PROBLEMS 5.1

Plan: Review the behavior of the gas phase vs. the liquid phase. Solution: a) The volume of the liquid remains constant, but the volume of the gas increases to the volume of the larger container. b) The volume of the container holding the gas sample increases when heated, but the volume of the container holding the liquid sample remains essentially constant when heated. c) The volume of the liquid remains essentially constant, but the volume of the gas is reduced.

5.2

The particles in a gas are further apart than those are in a liquid. a) The greater empty space between gas molecules allows gases to be more compressible than liquids. b) The greater empty space between gas molecules allows gases to flow with less resistance (hindrance) than liquids. c) The large empty space between gas molecules limits their interaction, allowing all mixtures of gases to be solutions. d) The large empty space between gas molecules increases the volume of the gas, therefore decreasing the density.

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5-11

5.3

The mercury column in the mercury barometer stays up due to the force exerted by the atmosphere on the mercury in the outer reservoir just balancing the gravitational force on the mercury in the tube. Its height adjusts according to the air pressure on the reservoir. The column of mercury is shorter on a mountaintop as there is less atmosphere to exert a force on the mercury reservoir. On a mountaintop, the air pressure is less, so the height of mercury it balances in the barometer is shorter than at sea level where there is more air pressure.

5.4

The pressure of mercury is its weight (force) per unit area. The weight, and thus the pressure, of the mercury column is directly proportional to its height.

5.5

When the mercury level in the arm attached to the flask is higher than the level in the other arm, the pressure in the flask is less than the pressure exerted in the other arm. This is an impossible situation for a closed-end manometer as the flask pressure cannot be less than the vacuum in the other arm.

5.6

Plan: The ratio of the heights of columns of mercury and water are inversely proportional to the ratio of the densities of the two liquids. Convert the height in mm to height in cm. Solution: hH2 O dHg  hHg dH2 O

13.5 g/mL  103 m   1 cm     hH O   hHg =  730 mmHg   2  = 985.5 = 990 cm H2O  2 1.00 g/mL  dH O 10 m   1 mm  2

dHg

5.7



Plan: The ratio of the heights of columns of mercury and water are inversely proportional to the ratio of the densities of the two liquids. Solution: hH2 O dHg  hHg dH2 O

hH O  2

5.8



13.5 g/mL    755 mmHg = 10,192.5 = 1.02 × 104 mm H O  hHg =  2 1.00 g/mL  dH O  2

dHg





Plan: Use the conversion factors between pressure units: 1 atm = 760 mmHg = 760 torr = 101.325 kPa = 1.01325 bar Solution:  760 mmHg    = 566.2 = 566 mmHg a) Converting from atm to mmHg: P(mmHg) = 0.745 atm   1 atm 





1.01325 bar    = 1.32256 = 1.32 bar b) Converting from torr to bar: P(bar) = 992 torr   760 torr 





 1 atm    = 3.60227 = 3.60 atm c) Converting from kPa to atm: P(atm) = 365 kPa  101.325 kPa 





101.325 kPa    = 107.191 = 107 kPa d) Converting from mmHg to kPa: P(kPa) = 804 mmHg   760 mmHg 



5.9



Plan: Use the conversion factors between pressure units: 1 atm = 760 mmHg = 760 torr = 101.325 kPa = 1.01325 bar

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5-12

Solution: a) Converting from cmHg to atm:

102 m   1 mm   1 atm   P(atm) = 76.8 cmHg   3   = 1.01053 = 1.01 atm  1 cm  10 m   760 mmHg 





101.325 kPa    = 2.786 × 103 = 2.79 × 103 kPa b) Converting from atm to kPa: P(kPa) = 27.5 atm    1 atm 





1.01325 bar    = 6.5861 = 6.59 bar c) Converting from atm to bar: P(bar) = 6.50 atm   1 atm 





 760 torr    = 7.02808 = 7.03 torr d) Converting from kPa to torr: P(torr) = 0.937 kPa  101.325 kPa 



5.10



Plan: This is an open-end manometer. Since the height of the mercury column in contact with the gas is higher than the column in contact with the air, the gas is exerting less pressure on the mercury than the air. Therefore the h) between the two arms is subtracted from the atmospheric pressure. Since the height difference is in units of cm and the barometric pressure is given in units of torr, cm must be converted to mm and then torr before the subtraction is performed. The overall pressure is then given in units of atm. Solution: 102 m   1 mm   1 torr   2.35 cm   3   = 23.5 torr  1 cm  10 m  1 mmHg 





738.5 torr – 23.5 torr = 715.0 torr

 1 atm    = 0.940789 = 0.9408 atm P(atm) = 715.0 torr   760 torr 



5.11



Plan: This is an open-end manometer. Since the height of the mercury column in contact with the gas is higher than the column in contact with the air, the gas is exerting less pressure on the mercury than the air. Therefore the pressure h) between the two arms is subtracted from the atmospheric pressure. Since the height difference is in units of cm and the barometric pressure is given in units of mmHg, cm must be converted to mm before the subtraction is performed. The overall pressure is then given in units of kPa. Solution: 102 m   1 mm   1.30 cm   3  = 13.0 mmHg  1 cm  10 m 





765.2 mmHg – 13.0 mmHg = 752.2 mmHg

101.325 kPa    = 100.285 = 100.3 kPa P(kPa) = 752.2 torr   760 torr 



5.12

Plan: This is a closed-end manometer. h) equals the gas pressure. The height difference is given in units of m and must be converted to mmHg and then to atm. Solution:  1 mmHg   1 atm   P(atm) = 0.734 mHg  3   = 0.965789 = 0.966 atm 10 mHg   760 mmHg 





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5-13

5.13

Plan: This is a closed-end manometer. h) equals the gas pressure. The height difference is given in units of cm and must be converted to mmHg and then to Pa. Solution: 102 mHg  1 mmHg  1.01325105 Pa   3 P(Pa) = 3.56 cm   3   = 4746.276 = 4.75 × 10 Pa  1 cmHg   10 m   760 mmHg 



5.14



Plan: Use the conversion factors between pressure units: 5 1 atm = 760 mmHg = 760 torr = 1.01325 × 10 Pa = 14.7 psi Solution:

 1 atm    = 0.361842 = 0.362 atm a) Converting from mmHg to atm: P(atm) = 2.75102 mmHg   760 mmHg 





 1 atm    = 5.85034 = 5.9 atm b) Converting from psi to atm: P(atm) = 86 psi  14.7 psi 





 1 atm   = 90.303 = 90.3 atm c) Converting from Pa to atm: P(atm) = 9.1510 6 Pa  1.0132510 5 Pa 





 1 atm    = 33.42105 = 33.4 atm d) Converting from torr to atm: P(atm) = 2.54 10 4 torr   760 torr 





5.15

Plan: 1 atm = 1.01325 × 10 Pa = 1.01325 × 10 N/m . So the force on 1 m of ocean is 1.01325 × 10 N, where 1N=1

kg  m 2 . Use F = mg to find the mass of the atmosphere in kg/m for part (a). For part (b), convert this s2 2

mass to g/cm and use the density of osmium to find the height of this mass of osmium. Solution: a) F = mg 5 1.01325 × 10 N = mg 1.0132510 5

kg  m 2 = (mass) (9.81 m/s ) s2 4

mass = 1.03287 × 10 = 1.03 × 10 kg 2 103 g 102 m     4 kg  3 2     b) 1.0328710 2    = 1.03287 × 10 g/cm (unrounded)  m  1 kg   1 cm  3 1 cm  g  1 mL    Height = 1.03287103   22.6 g  1 mL  = 45.702 = 45.7 cm Os  cm 2    

5.16

The statement is incomplete with respect to temperature and mass of sample. The correct statement is: At constant temperature and moles of gas, the volume of gas is inversely proportional to the pressure.

5.17

a) Charles’s law: At constant pressure, the volume of a fixed amount of gas is directly proportional to its kelvin temperature. Variable: volume and temperature; Fixed: pressure and moles b) Avogadro’s law: At fixed temperature and pressure, the volume occupied by a gas is directly proportional to the moles of gas. Variable: volume and moles; Fixed: temperature and pressure c) Amontons’s law: At constant volume, the pressure exerted by a fixed amount of gas is directly proportional to the kelvin temperature. Variable: pressure and temperature; Fixed: volume and moles

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5-14

5.18

Plan: Examine the ideal gas law; volume and temperature are constant and pressure and moles are variable. Solution: RT PV = nRT R, T, and V are constant P=n V P = n × constant At constant temperature and volume, the pressure of the gas is directly proportional to the amount of gas in moles.

5.19

Plan: Examine the ideal gas law, noting the fixed variables and those variables that change. R is always constant PV PV so 1 1 = 2 2 . n1T1 n2T2 Solution: PV PV V1 V = 2 a) P is fixed; both V and T double: 1 1 = 2 2 or n1T1 n2T2 n1T1 n2T2 T can double as V doubles only if n is fixed. PV PV b) T and n are both fixed and V doubles: 1 1 = 2 2 or P1V1 = P2V2 n1 T1 n2 T2 P and V are inversely proportional; as V doubles, P is halved. c) T is fixed and V doubles. n doubles since one mole of reactant gas produces a total of 2 moles of product gas. PV PV PV PV 1 1 1 1 = 2 2 = 2 2 or n1 T1 n2 T2 n1 n2 V and n can both double only if P is fixed. d) P is fixed and V doubles. n is fixed since 2 moles of reactant gas produce 2 moles of product gas. PV PV V1 V 1 1 = 2 2 or = 2 n1T1 n2 T2 T1 T2 V and T are directly proportional so as V is doubled, T is doubled.

5.20

Plan: Use the relationship

PV PV PV n T 1 1 = 2 2 or V2 = 1 1 2 2 . n1T1 n2T2 P2 n1T1

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5-15

V2 =

5.21

5.22

PV nT (1)(V1 )(4n1 )(1) 1 1 2 2 = P2 n1T1 (1)(n1 )(1)

Plan: Use the relationship

PV PV PV T 1 1 = 2 2 or V2 = 1 1 2 . R and n are fixed. T1 T2 P2 T1

Solution: a) As the pressure on a fixed amount of gas (n is fixed) doubles from 101 kPa to 202 kPa at constant temperature, the volume decreases by a factor of . As the temperature of a fixed amount of gas (n is fixed) decreases by a factor of (from 310 K to 155 K) at constant pressure, the volume decreases by a factor of . The changes in pressure and temperature combine to decrease the volume by a factor of 4. P1 = 760 torr = 101 kPa T1 = 37°C + 273 = 310 K PV T (101 kPa)( V )(155 K) 1 V2 = 1 1 2 = V2 = 1 4 V1 P2 T1 (202 kPa)(310 K) b) As the pressure on a fixed amount of gas (n is fixed) decreases at constant temperature (T is fixed), the molecules move farther apart, increasing the volume. When the pressure is reduced by a factor of 2, the volume increases by a factor of 2 at constant temperature (Boyle’s law). T2 = 32°C + 273 = 305 K P2 = 101 kPa = 1 atm PV (2 atm)(V1 )(305 K) 1 1T2 V2 = = V2 = 2V1 P2T1 (1 atm)(305 K) c) As the pressure on a fixed amount of gas (n is fixed) decreases at constant temperature (T is fixed), the molecules move farther apart, increasing the volume. When the pressure is reduced by a factor of 4, the volume increases by a factor of 4 at constant temperature (Boyle’s law). PV T (P )(V )(1) V2 = 1 1 2 = 1 1 V2 =4V1 P2T1 (1/ 4 P1 )(1) PV PV PV T Plan: Use the relationship 1 1 = 2 2 or V2 = 1 1 2 . R and n are fixed. T1 T2 P2 T1 Solution: a) The temperature is decreased by a factor of 2, so the volume is decreased by a factor of 2 (Charles’s law). PV T (1)(V1 )(400 K) V2 = 1 1 2 = V2 = V1 P2 T1 (1)(800 K) b) T1 = 250°C + 273.15 = 523.15 K T2 = 500°C + 273.15 = 773.15 K The temperature increases by a factor of 773.15/523.15 = 1.48, so the volume is increased by a factor of 1.48

PV (1)(V1 )(773.15 K) 1 1T2 = V2 = 1.48V1 P2T1 (1)(523.15 K) c) The pressure is increased by a factor of 3, so the volume decreases by a factor of 3 (Boyle’s law). PV T (2 atm)(V1 )(1) V2 = 1 1 2 = V2 = V1 P2T1 (6 atm)(1) PV PV PV n T Plan: Use the relationship 1 1 = 2 2 or V2 = 1 1 2 2 . n1T1 n2T2 P2 n1T1 Solution:  1 atm    = 0.950 atm a) P1 = 722 torr   760 torr  V2 =

(Charles’s law).

5.23

V2 = 4V1



T1 =



5 5 [T (in °F) – 32] = 9 9

[32°F – 32] = 0°C

T1 = 0°C + 273.15 = 273.15 K

Both P and T are fixed: P1 = P2 = 0.950 atm; T1 = T2 = 273.15 K, so the volume remains constant. Copyright

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5-16

PV (1)(V1 )(1)(1) 1 1 n2 T2 = V2 = V1 P2 n1T1 (1)(1)(1) b) Since the number of moles of gas is decreased by a factor of 2, the volume would be decreased by a factor of 2 (Avogadro’s law). 1 (1)(V1 )( n1 )(1) PV n T 2 V2 = V1 V2 = 1 1 2 2 = P2 n1T1 (1)( n1 )(1) c) If the pressure is decreased by a factor of 4, the volume will increase by a factor of 4 (Boyle’s law). If the temperature is decreased by a factor of 4, the volume will decrease by a factor of 4 (Charles’s law). These two effects offset one another and the volume remains constant. PV n T (P )(V )(1)(14 T1 ) V2 = 1 1 2 2 = 1 1 V2 = V1 P2 n1T1 ( 14 P1 )(1)(T1 ) V2 =

5.24

Plan: Examine the ideal gas law, noting the fixed variables and those variables that change. R is always constant PV PV so 1 1 = 2 2 . In this problem, P and V are changing, while n and T remain fixed. n1T1 n2T2 Solution: V1 = 1.61 L V2 = unknown P1 = 734 torr P2 = 0.844 atm n and T remain constant  

Converting P1 from torr to atm: (734 torr) 

 = 0.966 atm 

Arranging the ideal gas law and solving for V2: PV PV 1 1 = 2 2 or P1V1 = P2V2 n1 T1 n2 T2 V2 = V1 5.25

  

 = 1.84 L 

Plan: Examine the ideal gas law, noting the fixed variables and those variables that change. R is always constant PV PV so 1 1 = 2 2 . In this problem, P and V are changing, while n and T remain fixed. n1T1 n2T2 Solution: V1 = 10.0 L V2 = 7.50 L P1 = 725 mmHg P2 = unknown n and T remain constant Arranging the ideal gas law and solving for P2: PV PV 1 1 = 2 2 or P1V1 = P2V2 n1 T1 n2 T2 P2 = P1

5.26

P1 P2

V1 V2

  

 = 967 mmHg 

Plan: This is Charles’s law: at constant pressure and with a fixed amount of gas, the volume of a gas is directly proportional to the absolute temperature of the gas. The temperature must be lowered to reduce the volume of a gas. Arrange the ideal gas law, solving for T2 at fixed n and P. Temperature must be converted to kelvin. Solution: V1 = 9.10 L V2 = 2.50 L T1 = 198°C (convert to K) T2 = unknown n and P remain constant McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution

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5-17

Converting T from °C to K: T1 = 198°C + 273.15 = 471.15K Arranging the ideal gas law and solving for T2: PV PV V1 V 1 1 = 2 2 or  2 n1T1 n2 T2 T1 T2

T2  T1 5.27

Plan: This is Charles’s law: at constant pressure and with a fixed amount of gas, the volume of a gas is directly proportional to the absolute temperature of the gas. If temperature is reduced, the volume of gas will also be reduced. Arrange the ideal gas law, solving for V2 at fixed n and P. Temperature must be converted to kelvins. Solution: V1 = 93 L V2 = unknown T1 = 145°C (convert to K) T2 = –22°C n and P remain constant Converting T from °C to K: T1 = 145°C + 273.15 = 418.15 K T2 = –22°C + 273.15 = 251.15 K Arranging the ideal gas law and solving for V2: PV PV V1 V 1 1 = 2 2 or  2 n1T1 n2 T2 T1 T2

V2  V1 5.28

T2  251.15 K  = 93 L   = 55.858 = 56 L  418.15 K  T1

Plan: Examine the ideal gas law, noting the fixed variables and those variables that change. R is always constant PV PV so 1 1 = 2 2 . In this problem, P and T are changing, while n and V remain fixed. n1T1 n2T2 Solution: T1 = 25 °C T2 = 195 °C P1 = 177 atm P2 = unknown n and V remain constant Converting T1 from °C to K: 25 °C + 273.15 = 298.15 K Converting T2 from °C to K: 195 °C + 273.15 = 468.15 K Arranging the ideal gas law and solving for P2: P1 V1 P2 V2 P1 P2 n 1 T1 n 2 T2 T1 T2 P2 = P1

5.29

V2  2.50 L  = 471.15 K   = 129.437 K – 273.15 = –143.713 = –144°C  9.10 L  V1

T2  T1

  

  = 277.92 = 278 atm 

Plan: Examine the ideal gas law, noting the fixed variables and those variables that change. R is always constant PV PV so 1 1 = 2 2 . In this problem, P and T are changing, while n and V remain fixed. n1T1 n2T2 Solution: T1 = 30.0 °C T2 = unknown P1 = 110. psi P2 = 105 psi n and V remain constant Converting T1 from °C to K: 30.0 °C + 273.15 = 303.15 K Arranging the ideal gas law and solving for T2: P1 V1 P2 V2 P1 P2 n 1 T1 n 2 T2 T1 T2

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5-18

T2 = T1

  

P2  P1

  = 289.37 = 289 K 

Converting T2 from K to °C: 289.37 K  273.15 = 16.22 = 16 °C 5.30

Plan: Examine the ideal gas law, noting the fixed variables and those variables that change. R is always constant PV PV so 1 1 = 2 2 . In this problem, n and V are changing, while P and T remain fixed. n1T1 n2T2 Solution: m1 = 1.92 g He m2 = 1.92 g – 0.850 g = 1.07 g He V1 = 12.5 L V2 = unknown P and T remain constant    = 0.47964 mol He = n1 Converting m1 (mass) to n1 (moles): (1.92 g He)   

 Converting m2 (mass) to n2 (moles): (1.07 g He)  

  = 0.26730 mol He = n2 

Arranging the ideal gas law and solving for V2: P1 V1 P2 V2 V1 V2 n1 T1 n2 T2 n1 n2 V2 = V1 n 2 n1

5.31

  

 = 6.9661 = 6.97 L 

Plan: Examine the ideal gas law, noting the fixed variables and those variables that change. R is always constant PV PV so 1 1 = 2 2 . In this problem, n and V are changing, while P and T remain fixed. n1T1 n2T2 Solution: 22 n1 = 1 × 10 molecules of air* n2 = unknown V1 = 500 mL V2 = 350 mL P and T remain constant *The number of molecules of any substance is directly proportional to the moles of that substance, so we can use number of molecules in place of n in this problem. Arranging the ideal gas law and solving for n2: P1 V1 P2 V2 V1 V2 n1 T1 n2 T2 n1 n2 n2 = n1

5.32

 V2 22 = (1 × 10 molecules of air)   V1

 21  = 7 × 10 molecules of air 

Plan: Since the volume, temperature, and pressure of the gas are changing, use the combined gas law. Arrange the ideal gas law, solving for V2 at fixed n. STP is 0°C (273 K) and 1 atm (101.325 kPa) Solution: P1 = 153.3 kPa P2 = 101.325 kPa V1 = 25.5 L V2 = unknown T1 = 298 K T2 = 273 K n remains constant Arranging the ideal gas law and solving for V2: McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution

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5-19

PV PV 1 1 = 2 2 n1T1 n 2 T2

PV PV 1 1 = 2 2 T1 T2

 273 K  153.3 kPa   T  P     V2 = V1  2  1  = 25.5 L  101.325 kPa  = 35.3437 = 35.3 L  P2   T1   298 K  



5.33



Plan: Since the volume, temperature, and pressure of the gas are changing, use the combined gas law. Arrange the ideal gas law, solving for V2 at fixed n. Temperature must be converted to kelvins. Solution: P1 = 745 torr P2 = 367 torr V1 = 3.65 L V2 = unknown T1 = 298 K T2 = –14°C + 273.15 = 259.15 K n remains constant Arranging the ideal gas law and solving for V2: PV PV PV PV 1 1 1 1 = 2 2 or = 2 2 n1T1 n 2 T2 T1 T2

 T  P   259.15 K   745 torr  V2 = V1  2  1  = 3.65 L     = 6.4434 = 6.44 L  P2   298 K   367 torr   T1  5.34

Plan: Given the volume, pressure, and temperature of a gas, the number of moles of the gas can be calculated using the ideal gas law, solving for n. The gas constant, R = 0.0821 L atm/mol K, gives pressure in atmospheres and temperature in kelvin. The given pressure in torr must be converted to atmospheres and the temperature converted to kelvins. Solution: P = 328 torr (convert to atm) V = 5.0 L T = 37°C n = unknown  1 atm  Converting P from torr to atm: P = 328 torr  = 0.43158 atm  760 torr  Converting T from °C to K: T = 37°C + 273.15 = 310.15 K PV = nRT Solving for n: PV (0.43158 atm)(5.0 L) n=  = 0.084745 = 0.085 mol chlorine RT 0.0821 L  atm  (310.15 K)   mol  K 

5.35

Plan: Given the volume, moles, and temperature of a gas, the pressure of the gas can be calculated using the ideal gas law, solving for P. The gas constant, R = 0.0821 L atm/mol K, gives volume in liters and temperature in Kelvin. The given volume in mL must be converted to L and the temperature converted to kelvins. Solution: V = 75.0 mL T = 26°C –3 n = 1.47 × 10 mol P = unknown 103 L   = 0.0750 L Converting V from mL to L: V = 75.0 mL  1 mL  Converting T from °C to K: PV = nRT Solving for P:

T = 26°C + 273.15 = 299.15 K

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5-20

L  atm  1.4710 mol0.0821 mol 299.15 K   K 3

nRT P= = V

0.0750 L

= 0.48138 atm

 760 torr  Convert P to units of torr: 0.48138 atm  = 365.848 = 366 torr  1 atm 

5.36

Plan: Solve the ideal gas law for moles and convert to mass using the molar mass of ClF3. The gas constant, R = 0.0821 L atm/mol K, gives volume in liters, pressure in atmospheres, and temperature in kelvin so volume must be converted to L, pressure to atm, and temperature to K. Solution: V = 357 mL T = 45°C P = 699 mmHg n = unknown 103 L   = 0.357 L Converting V from mL to L: V = 357 mL   1 mL  Converting T from °C to K:

T = 45°C + 273.15 = 318.15 K  1 atm   = 0.91974 atm Converting P from mmHg to atm: P = 699 mmHg  760 mmHg  PV = nRT Solving for n: 0.91974 atm 0.357 L  PV n=  = 0.012571 mol ClF3  RT 0.0821 L  atm 318.15 K    mol  K   92.45 g ClF3   = 1.16216 = 1.16 g ClF3 Mass ClF3 = 0.012571 mol ClF3    1 mol ClF  3

5.37

Plan: Solve the ideal gas law for pressure; convert mass to moles using the molar mass of N2O. The gas constant, R = 0.0821 L atm/mol K, gives temperature in kelvin so the temperature must be converted to units of kelvins. Solution: V = 3.1 L T = 115°C n = 75.0 g (convert to moles) P = unknown Converting T from °C to K: T = 115°C + 273.15 = 388.15 K  1 mol N 2 O   = 1.70377 mol N2O Converting from mass of N2O to moles: n = 75.0 g N 2 O  44.02 g N 2 O  PV = nRT Solving for P: nRT P=  V

5.38

 L  atm  1.70377 mol0.0821 388.15 K   mol  K  = 17.5143 = 18 atm N2O 3.1 L 

Plan: Solve the ideal gas law for moles. The gas constant, R = 0.0821 L atm/mol K, gives pressure in atmospheres, and temperature in kelvin so pressure must be converted to atm and temperature to K. Solution: V = 1.5 L T = 23°C P = 85 + 14.7 = 99.7 psi n = unknown Converting T from °C to K: T = 23°C + 273.15 = 296.15 K

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5-21

Converting P from psi to atm:

 1 atm   = 6.7823 atm P = 99.7 psi 14.7 psi 

PV = nRT Solving for n: 6.7823 atm 1.5 L  PV n=  = 0.418421 = 0.42 mol SO2 RT 0.0821 L  atm 296.15 K    mol  K  5.39

Plan: Assuming that while rising in the atmosphere the balloon will neither gain nor lose gas molecules, the number of moles of gas calculated at sea level will be the same as the number of moles of gas at the higher altitude (n is fixed). Volume, temperature, and pressure of the gas are changing. Arrange the ideal gas law, solving for V2 at fixed n. Given the sea-level conditions of volume, pressure, and temperature, and the temperature and pressure at the higher altitude for the gas in the balloon, we can set up an equation to solve for the volume at the higher altitude. Comparing the calculated volume to the given maximum volume of 835 L will tell us if the balloon has reached its maximum volume at this altitude. Temperature must be converted to kelvins and pressure in torr must be converted to atm for unit agreement. Solution: P1 = 745 torr P2 = 0.066 atm V1 = 65 L V2 = unknown T1 = 25°C + 273.15 = 298.15 K T2 = –5°C + 273.15 = 268.15 K n remains constant  1 atm  Converting P from torr to atm: P = 745 torr  = 0.98026 atm  760 torr  Arranging the ideal gas law and solving for V2: PV PV PV PV 1 1 1 1 = 2 2 or = 2 2 n1T1 n 2 T2 T1 T2

 T  P   268.15 K   0.98026 atm  = 868.268 = 870 L V2 = V1  2  1  = 65 L  P2   T1   298.15 K  0.066 atm  The calculated volume of the gas at the higher altitude is more than the maximum volume of the balloon. Yes, the balloon will reach its maximum volume. Check: Should we expect that the volume of the gas in the balloon should increase? At the higher altitude, the pressure decreases; this increases the volume of the gas. At the higher altitude, the temperature decreases, this decreases the volume of the gas. Which of these will dominate? The pressure decreases by a factor of 0.98/0.066 = 15. If we label the initial volume V1, then the resulting volume is 15V1. The temperature decreases by a factor of 298/268 = 1.1, so the resulting volume is V1/1.1 or 0.91V1. The increase in volume due to the change in pressure is greater than the decrease in volume due to change in temperature, so the volume of gas at the higher altitude should be greater than the volume at sea level. 5.40

Air is mostly N2 (28.02 g/mol), O2 (32.00 g/mol), and argon (39.95 g/mol). These “heavy” gases dominate the density of dry air. Moist air contains H2O (18.02 g/mol). The relatively light water molecules lower the density of the moist air.

5.41

The molar mass of H2 is less than the average molar mass of air (mostly N2, O2, and Ar), so air is denser. To collect a beaker of H2(g), invert the beaker so that the air will be replaced by the lighter H2. The molar mass of CO2 is greater than the average molar mass of air, so CO2(g) is more dense. Collect the CO2 holding the beaker upright, so the lighter air will be displaced out the top of the beaker.

5.42

Gases mix to form a solution and each gas in the solution behaves as if it were the only gas present.

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5-22

5.43

PA = XA PT The partial pressure of a gas (PA) in a mixture is directly proportional to its mole fraction (XA).

5.44

Plan: Calculate the mole fraction of each gas; the partial pressure of each gas is directly proportional to its mole fraction so the gas with the highest mole fraction has the highest partial pressure. Use the relationship between partial pressure and mole fraction to calculate the partial pressure of gas D 2. Solution: 4 A particles 3 B particles n nB a) X A = A = = 0.25 XB = = = 0.1875 16 total particles 16 total particles ntotal ntotal nD2 4 D2 particles 5 C particles nC XC = = = 0.3125 X D2 = = = = 0.25 16 total particles ntotal 16 total particles ntotal Gas C has the highest mole fraction and thus the highest partial pressure. b) Gas B has the lowest mole fraction and thus the lowest partial pressure. c) PD2 = X D2 x Ptotal PD2 = 0.25 × 0.75 atm = 0.1875 = 0.19 atm

5.45

Plan: Rearrange the ideal gas law to calculate the density of xenon from its molar mass at STP. Standard temperature is 0°C (273 K) and standard pressure is 1 atm. Do not forget that the pressure at STP is exact and will not affect the significant figures. Solution: P = 1 atm T = 273 K M of Xe = 131.3 g/mol d = unknown PV = nRT Rearranging to solve for density: d = PM /RT 

5.46

Plan: Rearrange the ideal gas law to calculate the density of CFCl3 from its molar mass. Temperature must be converted to kelvins. Solution: P = 1.5 atm T = 120°C + 273.15 = 393.15 K M of CFCl3 = 137.4 g/mol d = unknown PV = nRT Rearranging to solve for density: d = PM /RT 

5.47

1 atm131.3 g/mol = 5.8581 = 5.86 g/L  L  atm  0.0821  273 K   mol  K 

1.5 atm 137.4 g/mol = 6.3852 = 6.4 g/L  L  atm  0.0821  393.15 K   mol  K 

Plan: Solve the ideal gas law for moles. Convert moles to mass using the molar mass of AsH3 and divide this mass by the volume to obtain density in g/L. Standard temperature is 0°C (273 K) and standard pressure is 1 atm. Do not forget that the pressure at STP is exact and will not affect the significant figures. Solution: V = 0.0400 L T = 273 K P = 1 atm n = unknown M of AsH3 = 77.94 g/mol PV = nRT Solving for n:

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5-23







1 atm 0.0400 L PV –3 –3 = 1.78465 × 10 = 1.78 × 10 mol AsH3   L  atm  RT 0.0821 273 K  mol  K  Converting moles of AsH3 to mass of AsH3:  77.94 g AsH  3  3 Mass (g) of AsH3 = 1.7846510 mol AsH3   = 0.1391 g AsH3  1 mol AsH3  n=

5.48



 



 

0.1391 g mass = = 3.4775 = 3.48 g/L volume 0.0400 L

Plan: Solve the density form of the ideal gas law for molar mass. Temperature must be converted to kelvins. Compare the calculated molar mass to the molar mass values of the noble gases to determine the identity of the gas. Solution: P = 3.00 atm T = 273 K d = 2.71 g/L M = unknown d = PM /RT Rearranging to solve for molar mass: L  atm   2.71 g/L 0.0821 273 K   dRT mol  K  = M = = 20.24668 = 20.2 g/mol P 3.00 atm









Therefore, the gas is Ne. 5.49

Plan: Rearrange the formula PV = (m/M )RT to solve for molar mass. Convert the mass in ng to g and volume in L to L. Temperature must be in kelvin and pressure in atm. Solution: V T = 45°C + 273.15 = 318.15 K P = 380 torr m = 206 ng M = unknown  1 atm  Converting P from torr to atm: P = 380 torr  = 0.510526 atm  760 torr  106 L   = 2.06 × 10–7 L Converting V V = 0.206 L  1 L 

109 g   = 2.06 × 10–7 g m = 206 ng  1 ng 

Converting m from ng to g: PV = (m/M )RT Solving for molar mass, M : 

mRT M =  PV

L  atm  318.15 K  mol  K  = 51.163 = 51.2 g/mol 0.510526 atm 2.06 107 L 

2.06 107 g0.0821

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5-24

5.50

Plan: Rearrange the formula PV = (m/M )RT to solve for molar mass. Compare the calculated molar mass to that of N2, Ne, and Ar to determine the identity of the gas. Convert volume to liters, pressure to atm, and temperature to kelvin. Solution: V = 63.8 mL T = 22°C + 273.15 = 295.15 K P = 747 mm Hg m = 0.103 g M = unknown

 1 atm   = 0.982895 atm Converting P from mmHg to atm: P = 747 mmHg  760 mmHg  Converting V from mL to L:

103 L   = 0.0638 L V = 63.8 mL  1 mL 

PV = (m/M )RT Solving for molar mass, M :  L  atm  295.15 K  0.103 g0.0821  mRT mol  K  M = = 39.8011 = 39.8 g/mol  PV 0.982895 atm 0.0638 L The molar masses are N2 = 28 g/mol, Ne = 20 g/mol, and Ar = 40 g/mol. Therefore, the gas is Ar. 5.51

Plan: Use the ideal gas law to determine the number of moles of Ar and of O2. The gases are combined (ntotal = nAr + n O 2 ) into a 400 mL flask (V) at 27°C (T). Use the ideal gas law again to determine the total pressure from ntotal, V, and T. Pressure must be in units of atm, volume in units of L and temperature in K. Solution: For Ar: V = 0.600 L T = 227°C + 273.15 = 500.15 K P = 1.20 atm n = unknown PV = nRT Solving for n: 1.20 atm0.600 L PV n=  = 0.01753433 mol Ar RT 0.0821 L  atm 500.15 K   mol  K  For O2: V = 0.200 L T = 127°C + 273.15 = 400.15 K P = 501 torr n = unknown  1 atm  Converting P from torr to atm: P = 501 torr  = 0.6592105 atm  760 torr  PV = nRT Solving for n: 0.6592105 atm 0.200 L PV n=  = 0.00401318 mol O2  RT 0.0821 L  atm  400.15 K    mol  K  ntotal = nAr + n O 2 = 0.01753433 mol + 0.00401318 mol = 0.02154751 mol For the mixture of Ar and O2: V = 400 mL T = 27°C + 273.15 = 300.15 K P = unknownn n = 0.02154751 mol

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5-25

103 L   = 0.400 L V = 400 mL  1 mL 

Converting V from mL to L: PV = nRT Solving for P: nRT Pmixture =  V

5.52

 L  atm  0.02154751 mol0.0821 300.15 K   mol  K  = 1.32745 = 1.33 atm 0.400 L

Plan: Use the ideal gas law, solving for n to find the total moles of gas. Convert the mass of Ne to moles and subtract moles of Ne from the total number of moles to find moles of Ar. Volume must be in units of liters, pressure in units of atm, and temperature in kelvins. Solution: V = 355 mL T = 35°C + 273.15 = 308.15 K P = 626 mmHg ntotal = unknown  1 atm   = 0.823684 atm Converting P from mmHg to atm: P = 626 mmHg  760 mmHg 

103 L   = 0.355 L V = 355 mL   1 mL 

Converting V from mL to L:

PV = nRT Solving for ntotal: 0.823684 atm0.355 L  PV ntotal =  = 0.011558029 mol Ne + mol Ar RT 0.0821 L  atm 308.15 K    mol  K 

 1 mol Ne   = 0.007234886 mol Ne Moles Ne = 0.146 g Ne  20.18 g Ne  Moles Ar = ntotal – nNe = (0.011558029 – 0.007234886) mol = 0.004323143 = 0.0043 mol Ar 5.53

Plan: Use the ideal gas law, solving for n to find the moles of O2. Use the molar ratio from the balanced equation to determine the moles (and then mass) of phosphorus that will react with the oxygen. Standard temperature is 0°C (273 K) and standard pressure is 1 atm. Solution: V = 35.5 L T = 273 K P = 1 atm n = unknown PV = nRT Solving for n: 1 atm 35.5 L  PV n=  = 1.583881 mol O2 RT 0.0821 L  atm 273 K    mol  K  P4(s) + 5O2(g)  P4O10(s)  1 mol P4  123.88 g P4   Mass P4 = 1.583881 mol O2    = 39.24224 = 39.2 g P4  5 mol O2   1 mol P4 



5.54



Plan: Use the ideal gas law, solving for n to find the moles of O2 produced. Volume must be in units of liters, pressure in atm, and temperature in kelvins. Use the molar ratio from the balanced equation to determine the moles (and then mass) of potassium chlorate that reacts.

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5-26

Solution: V = 638 mL P = 752 torr Converting P from torr to atm: Converting V from mL to L:

T = 128°C + 273.15 = 401.15 K n = unknown  1 atm  P = 752 torr  = 0.9894737 atm  760 torr  103 L   = 0.638 L V = 638 mL  1 mL 

PV = nRT Solving for n: 0.9894737 atm0.638 L PV n=  = 0.0191679 mol O2 RT 0.0821 L  atm  401.15 K   mol  K  2KClO3(s)  2KCl(s) + 3O2(g)

 2 mol KClO3  122.55 g KClO3   Mass (g) of KClO3 = 0.0191679 mol O2    = 1.5660 = 1.57 g KClO3  3 mol O2   1 mol KClO3 



5.55



Plan: Since the amounts of two reactants are given, this is a limiting reactant problem. To find the mass of PH3, write the balanced equation and use molar ratios to find the number of moles of PH3 produced by each reactant. The smaller number of moles of product indicates the limiting reagent. Solve for moles of H2 using the ideal gas law. Solution: Moles of hydrogen: V = 83.0 L T = 273 K P = 1 atm n = unknown PV = nRT Solving for n: 1 atm83.0 L PV n=  = 3.7031584 mol H2 RT 0.0821 L  atm 273 K    mol  K  P4(s) + 6H2(g)  4PH3(g)

 4 mol PH3   = 2.4687723 mol PH3 PH3 from H2 = 3.7031584 mol H2   6 mol H2   1 mol P4   4 mol PH3  = 1.21085 mol PH PH3 from P4 = 37.5 g P4  3  1 mol P4  123.88 g P4  P4 is the limiting reactant because it forms less PH3.

 1 mol P4   4 mol PH3  33.99 g PH3   Mass PH3 = 37.5 g P4     = 41.15676 = 41.2 g PH3 123.88 g P4   1 mol P4  1 mol PH3 



5.56



Plan: Since the amounts of two reactants are given, this is a limiting reactant problem. To find the mass of NO, write the balanced equation and use molar ratios to find the number of moles of NO produced by each reactant. Since the moles of gas are directly proportional to the volumes of the gases at the same temperature and pressure,

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5-27

the limiting reactant may be found by comparing the volumes of the gases. The smaller volume of product indicates the limiting reagent. Then use the ideal gas law to convert the volume of NO produced to moles and then to mass. Solution: 4NH3(g) + 5O2(g)  4NO(g) + 6H2O(l)  4 L NO   = 35.6 L NO Mol NO from NH3 = 35.6 L NH3   4 L NH  3

 4 L NO   = 32.4 L NO Mol NO from O2 = 40.5 L O2   5 L O2  O2 is the limiting reactant since it forms less NO. Converting volume of NO to moles and then mass: V = 32.4 L T = 273 K P = 1 atm n = unknown PV = nRT Solving for n: 1 atm32.4 L PV n=  = 1.44557 mol NO  RT 0.0821 L  atm 273 K    mol  K  30.01 g NO    = 43.38156 = 43.4 g NO Mass (g) of NO = 1.44557 mol NO  1 mol NO  5.57

Plan: First, write the balanced equation. The moles of hydrogen produced can be calculated from the ideal gas law. The problem specifies that the hydrogen gas is collected over water, so the partial pressure of water vapor must be subtracted from the overall pressure given. Table 5.2 reports pressure at 26°C (25.2 torr) and 28°C (28.3 torr), so take the average of the two values to obtain the partial pressure of water at 27°C. Volume must be in units of liters, pressure in atm, and temperature in kelvins. Once the moles of hydrogen produced are known, the molar ratio from the balanced equation is used to determine the moles of aluminum that reacted. Solution: V = 35.8 mL T = 27°C + 273.15 = 300.15 K Ptotal = 751 mmHg n = unknown Pwater vapor = (28.3 + 25.2) torr/2 = 26.75 torr = 26.75 mmHg Phydrogen = Ptotal – Pwater vapor = 751 mmHg – 26.75 mmHg = 724.25 mmHg  1 atm   = 0.952960526 atm Converting P from mmHg to atm: P = 724.25 mmHg  760 mmHg  Converting V from mL to L:

103 L   = 0.0358 L V = 35.8 mL   1 mL 

PV = nRT Solving for n: PV 0.952960526 atm 0.0358 L  n=  = 0.001384447 mol H2  L  atm  RT 0.0821 300.15 K    mol  K  2Al(s) + 6HCl(aq)  2AlCl3(aq) + 3H2(g)  2 mol Al   26.98 g Al   Mass (g) of Al = 0.001384447 mol H2    = 0.024902 = 0.0249 g Al  3 mol H 2   1 mol Al 

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5-28

5.58

Plan: First, write the balanced equation. Convert mass of lithium to moles and use the molar ratio from the balanced equation to find the moles of hydrogen gas produced. Use the ideal gas law to find the volume of that amount of hydrogen. The problem specifies that the hydrogen gas is collected over water, so the partial pressure of water vapor must be subtracted from the overall pressure given. Table 5.2 reports the vapor pressure of water at 18°C (15.5 torr). Pressure must be in units of atm and temperature in kelvins. Solution: 2Li(s) + 2H2O(l)  2LiOH(aq) + H2(g)  1 mol Li 1 mol H 2   = 0.0605100 mol H2  Moles H2 = 0.84 g Li  6.941 g Li  2 mol Li  Finding the volume of H2: V = unknown T = 18°C + 273.15 = 291.15 K Ptotal = 725 mmHg n = 0.0605100 mol Pwater vapor = 15.5 torr = 15.5 mmHg Phydrogen = Ptotal – Pwater vapor = 725 mmHg – 15.5 mmHg = 709.5 mmHg  1 atm   = 0.933552631 atm Converting P from mmHg to atm: P = 709.5 mmHg  760 mmHg  PV = nRT Solving for V: nRT V= = P

5.59

 L  atm  0.0605100 mol0.0821 291.15 K   mol  K  = 1.5493 = 1.5 L H2 0.933552631 atm 

Plan: Rearrange the ideal gas law to calculate the density of the air from its molar mass. Temperature must be converted to kelvins and pressure to atmospheres. Solution: P = 744 torr T = 17°C + 273.15 = 290.15 K or T = 60°C + 273.15 = 333.15 K M of air = 28.8 g/mol d = unknown  1 atm  Converting P from torr to atm: P = 744 torr  = 0.978947368 atm  760 torr  PV = nRT Rearranging to solve for density: At 17°C 0.978947368 atm 28.8 g/mol d = PM /RT  = 1.18355 = 1.18 g/L  L  atm  0.0821 290.15 K   mol  K  At 60.0°C 0.978947368 atm 28.8 g/mol d = PM /RT  = 1.03079 = 1.03 g/L  L  atm  0.0821 (333.15 K)   mol  K 

5.60

Plan: Solve the ideal gas law for molar volume, n/V. Pressure must be converted to atm and temperature to K. Solution: P = 650. torr T = –25°C + 273.15 = 248.15 K n/V = unknown

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5-29

 1 atm  P = 650 torr   = 0.855263157 atm  760 torr 



Converting P from torr to atm:



PV = nRT Solving for n/V: 0.855263157 atm  n P =  = 0.041978 = 0.0420 mol/L V RT 0.0821 L  atm 248.15 K   mol  K  5.61

Plan: The problem gives the mass, volume, temperature, and pressure of a gas; rearrange the formula PV = (m/M )RT to solve for the molar mass of the gas. Temperature must be in kelvin and pressure in atm. The problem also states that the gas is a hydrocarbon, which by, definition, contains only carbon and hydrogen atoms. We are also told that each molecule of the gas contains five carbon atoms so we can use this information and the calculated molar mass to find out how many hydrogen atoms are present and the formula of the compound. Solution: V = 0.204 L T = 101°C + 273.15 = 374.15 K P = 767 torr m = 0.482 g M = unknown  1 atm  Converting P from torr to atm: P = 767 torr  = 1.009210526 atm  760 torr  PV = (m/M )RT Solving for molar mass, M :  L  atm  374.15 K  0.482 g0.0821  mRT mol  K  M = = 71.9157 g/mol (unrounded)  PV 1.009210526 atm 0.204 L The mass of the five carbon atoms accounts for [5(12 g/mol)] = 60 g/mol; thus, the hydrogen atoms must make up the difference (72 – 60) = 12 g/mol. A value of 12 g/mol corresponds to 12 H atoms. (Since fractional atoms are not possible, rounding is acceptable.) Therefore, the molecular formula is C5H12.

5.62

Plan: Solve the ideal gas law for moles of air. Temperature must be in units of kelvins. Use Avogadro’s number to convert moles of air to molecules of air. The percent composition can be used to find the number of molecules (or atoms) of each gas in that total number of molecules. Solution: V = 1.00 L T = 25°C + 273.15 = 298.15 K P = 1.00 atm n = unknown PV = nRT Solving for n: Moles of air = n =







1.00 atm 1.00 L PV = 0.04085282 mol   L  atm  RT 0.0821 298.15 K   mol  K 





Converting moles of air to molecules of air:

 6.022 1023 molecules   = 2.4601568 × 1022 molecules Molecules of air = 0.04085282 mol  1 mol   78.08% N 2 molecules  Molecules of N2 = 2.46015681022 air molecules    100% air 22 22 = 1.920890 × 10 = 1.92 × 10 molecules N2 Copyright

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5-30

 20.94% O 2 molecules  Molecules of O2 = 2.46015681022 air molecules    100% air 21 21 = 5.151568 × 10 = 5.15 × 10 molecules O2  0.05% CO 2 molecules  Molecules of CO2 = 2.46015681022 air molecules    100% air 19

= 1.230078 × 10 = 1 × 10 molecules CO2  0.93% Ar molecules  Molecules of Ar = 2.46015681022 air molecules    100% air 20

= 2.287946 × 10 = 2.3 × 10 molecules Ar 5.63

Plan: Since you have the pressure, volume, and temperature, use the ideal gas law to solve for n, the total moles of gas. Pressure must be in units of atmospheres and temperature in units of kelvins. The partial pressure of SO2 can be found by multiplying the total pressure by the volume fraction of SO 2. Solution: a) V = 21 L T = 45°C + 273.15 = 318.15 K P = 850 torr n = unknown  1 atm  Converting P from torr to atm: P = 850 torr  = 1.118421053 atm  760 torr  PV = nRT 1.118421053 atm 21 L  PV Moles of gas = n =  = 0.89919 = 0.90 mol gas RT 0.0821 L  atm 318.15 K   mol  K  b) The equation PSO2  X SO2  Ptotal can be used to find partial pressure. The information given in ppm is a way of expressing the proportion, or fraction, of SO2 present in the mixture. Since n is directly proportional to V, the 3 volume fraction can be used in place of the mole fraction, XSO2. There are 7.95 × 10 parts SO2 in a million parts of 3 6 –3 mixture, so volume fraction = (7.95 × 10 /1 × 10 ) = 7.95 × 10 . –3 PD2 = volume fraction x Ptotal = (7.95 × 10 ) (850 torr) = 6.7575 = 6.76 torr

5.64

Plan: First, write the balanced equation. Convert mass of P4S3 to moles and use the molar ratio from the balanced equation to find the moles of SO2 gas produced. Use the ideal gas law to find the volume of that amount of SO2. Pressure must be in units of atm and temperature in kelvins. Solution: P4S3(s) + 8O2(g)  P4O10(s) + 3SO2(g)  1 mol P4 S3  3 mol SO2    = 0.010906 mol SO2 Moles SO2 = 0.800 g P4 S3  1 mol P S   220.06 g P S  4 3

Finding the volume of SO2: V = unknown P = 725 torr Converting P from torr to atm:

4 3

T = 32°C + 273.15 = 305.15 K n = 0.010906 mol  1 atm  P = 725 torr  = 0.953947368 atm  760 torr 

PV = nRT Solving for V: nRT V= = P

 L  atm  0.010906 mol0.0821 305.15 K   mol  K  = 0.2864193 L 0.953947368 atm 

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5-31

Converting V from L to mL:  1 mL   V = 0.2864193 L 3  = 286.4193 = 286 mL SO2 10 L  5.65

Plan: The moles of Freon-12 produced can be calculated from the ideal gas law. Volume must be in units of L, pressure in atm, and temperature in kelvins. Then, write the balanced equation. Once the moles of Freon-12 produced is known, the molar ratio from the balanced equation is used to determine the moles and then grams of CCl4 that reacted. Solution: 3 V = 16.0 dm T = 27°C + 273.15 = 300.15 K Ptotal = 1.20 atm n = unknown  1 L  3 Converting V from dm to L: V = 16.0 dm 3  = 16.0 L 1 dm 3  PV = nRT Solving for n:

1.20 atm16.0 L PV  = 0.7791476 mol Freon-12  RT 0.0821 L  atm 300.15 K    mol  K  CCl4(g) + 2HF(g)  CF2Cl2(g) + 2HCl(g)  1 mol CCl 4  153.81 g CCl4   Mass of Freon-12 (CF2Cl2) = 0.7791476 mol CF2 Cl2     1 mol CCl 4  1 mol CF2 Cl2  2 = 119.8407 = 1.20 × 10 g CCl4 Moles of Freon-12 = n =

5.66

Plan: First, write the balanced equation. Given the amount of xenon hexafluoride that reacts, we can find the number of moles of silicon tetrafluoride gas formed by using the molar ratio in the balanced equation. Then, using the ideal gas law with the moles of gas, the temperature, and the volume, we can calculate the pressure of the silicon tetrafluoride gas. Temperature must be in units of kelvins. Solution: 2XeF6(s) + SiO2(s)  2XeOF4(l) + SiF4(g)  1 mol XeF6   1 mol SiF4  = 0.0040766 mol SiF Moles SiF4 = n = 2.00 g XeF6  4  2 mol XeF6   245.3 g XeF6  Finding the pressure of SiF4: V = 1.00 L P = unknown PV = nRT Solving for P:

T = 25°C + 273.15 = 298.15 K n = 0.0040766 mol



nRT Pressure SiF4 = P = = V 5.67

0.0040766 mol SiF4 0.0821 1.00 L

L  atm  298.15 K  mol  K 

= 0.099788 = 0.0998 atm SiF4

Plan: Use the ideal gas law with T and P constant; then volume is directly proportional to moles. Solution: PV = nRT. At constant T and P, V n. Since the volume of the products has been decreased to the original volume, the moles (and molecules) must have been decreased by a factor of as well. Cylinder A best represents the products as there are 2 product molecules (there were 4 reactant molecules).

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5-32

5.68

Plan: Write the balanced equation. Since the amounts of 2 reactants are given, this is a limiting reactant problem. To find the volume of SO2, use the molar ratios from the balanced equation to find the number of moles of SO 2 produced by each reactant. The smaller number of moles of product indicates the limiting reagent. Solve for moles of SO2 using the ideal gas law. Solution: Moles of oxygen: V = 228 L T = 220°C + 273.15 = 493.15 K P = 2.0 atm n = unknown PV = nRT Solving for n: 2.0 atm228 L PV  Moles of O2 = n = = 11.2627 mol O2  RT 0.0821 L  atm  493.15 K    mol  K  2PbS(s) + 3O2(g)  2PbO(g) + 2SO2(g)  2 mol SO2   = 7.5085 mol SO2 Moles SO2 from O2 = 11.2627 mol O2   3 mol O  2

10 g    1 mol PbS    2 mol SO2  = 15.6707 mol SO (unrounded) Moles SO2 from PbS = 3.75 kg PbS   2    1 kg   239.3 g PbS  2 mol PbS 





O2 is the limiting reagent because it forms less SO2. Finding the volume of SO2: V = unknown T = 0°C + 273.15 = 273.15 K Ptotal = 1 atm n = 7.5107 mol PV = nRT Solving for V:  L  atm   7.5085 mol0.0821 273.15 K    nRT 2 mol  K V= = 168.38 = 1.7 × 10 L SO2 = P 1 atm  5.69

Plan: First, write the balanced equation. Given the amount of xenon HgO that reacts (20.0% of the given amount), we can find the number of moles of oxygen gas formed by using the molar ratio in the balanced equation. Then, using the ideal gas law with the moles of gas, the temperature, and the volume, we can calculate the pressure of the silicon tetrafluoride gas. Temperature must be in units of kelvins. Solution: 2HgO(s)  2Hg(l) + O2(g)  20.0%  1 mol HgO  1 mol O2   = 0.01846722 mol O2 Mole O2 = n = 40.0 g HgO     100%  216.6 g HgO  2 mol HgO  Finding the pressure of O2: V = 502 mL T = 25°C + 273.15 = 298.15 K P = unknown n = 0.01846722 mol 103 L   = 0.502 L Converting V from mL to L: V = 502 mL   1 mL 





PV = nRT Solving for P:



nRT Pressure O2 = P = = V Copyright

0.01846722 mol O2 0.0821 0.502 L

L  atm  298.15 K  mol  K 

= 0.900484 = 0.900 atm O2

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5-33

5.70

As the temperature of the gas sample increases, the most probable speed increases. This will increase both the number of collisions per unit time and the force of each collision with the sample walls. Thus, the gas pressure increases.

5.71

At STP (or any identical temperature and pressure), the volume occupied by a mole of any gas will be identical. One mole of krypton has the same number of particles as one mole of helium and, at the same temperature, all of the gas particles have the same average kinetic energy, resulting in the same pressure and volume.

5.72

The rate of effusion is much higher for a gas than its rate of diffusion. Effusion occurs into an evacuated space, whereas diffusion occurs into another gas. It is reasonable to expect that a gas will escape faster into a vacuum than it will into a space already occupied by another gas. The ratio of the rates of effusion and diffusion for two gases will be the same since both are inversely proportional to the square root of their molar masses.

5.73

a) PV = nRT Since the pressure, volume, and temperature of the two gases are identical, n must be the same for the two gases. Since the molar mass of O2 (32.0 g/mol) is greater than the molar mass of H2 (2.02 g/mol), a given number of moles of O2 has a greater mass than the same number of moles of H2. mass O2 > mass H2 b) d = PM /RT The pressure and temperature are identical and density is directly proportional to molar mass M. Since the molar mass of O2 (32.0 g/mol) is greater than the molar mass of H2 (2.02 g/mol), the density of O2 is greater than that of H2. dO > dH 2

c) The mean free path is dependent on pressure. Since the two gases have the same pressure, their mean free paths are identical. d) Kinetic energy is directly proportional to temperature. Since the two gases have the same temperature, their average moelcular kinetic energies are identical. 2 e) Kinetic energy = mass × speed . O2 and H2 have the same average kinetic energy at the same temperature and mass and speed are inversely proportional. The lighter H2 molecules have a higher speed than the heavier O2 molecules. average speed H2 > average speed O2 1 f) Rate of effusion  H2 molecules with the lower molar mass have a faster effusion time than O2 molar mass

molecules with a larger molar mass. effusion time H2 < effusion time O2 5.74

Plan: The molar masses of the three gases are 2.016 for H2 (Flask A), 4.003 for He (Flask B), and 16.04 for CH4 (Flask C). Since hydrogen has the smallest molar mass of the three gases, 4 g of H2 will contain more gas molecules (about 2 mole’s worth) than 4 g of He or 4 g of CH4. Since helium has a smaller molar mass than methane, 4 g of He will contain more gas molecules (about 1 mole’s worth) than 4 g of CH 4 (about 0.25 mole’s worth). Solution: a) PA > PB > PC. The pressure of a gas is proportional to the number of gas molecules (PV = nRT). So, the gas sample with more gas molecules will have a greater pressure. b) EA = EB = EC. Average kinetic energy depends only on temperature. The temperature of each gas sample is 273 K, so they all have the same average kinetic energy. c) rateA > rateB > rateC. When comparing the speed of two gas molecules, the one with the lower mass travels faster. d) total EA > total EB > total EC. Since the average kinetic energy for each gas is the same (part b) of this problem), the total kinetic energy would equal the average times the number of molecules. Since the hydrogen flask contains the most molecules, its total kinetic energy will be the greatest. e) dA = dB = dC. Under the conditions stated in this problem, each sample has the same volume, 5 L, and the same mass, 4 g. Thus, the density of each is 4 g/5 L = 0.8 g/L.

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5-34

f) Collision frequency (A) > collision frequency (B) > collision frequency (C). The number of collisions depends on both the speed and the distance between gas molecules. Since hydrogen is the lightest molecule it has the greatest speed and the 5 L flask of hydrogen also contains the most molecules, so collisions will occur more frequently between hydrogen molecules than between helium molecules. By the same reasoning, collisions will occur more frequently between helium molecules than between methane molecules. 5.75

Plan: To find the ratio of effusion rates, calculate the inverse of the ratio of the square roots of the molar masses (Graham’s law). Solution: Rate H2 molar mass UF6 352.0 g/mol = = = 13.2137 = 13.21 Rate UF6 molar mass H2 2.016 g/mol

5.76

Plan: To find the ratio of effusion rates, calculate the inverse of the ratio of the square roots of the molar masses (Graham’s law). Solution: Rate O2 molar mass Kr 83.80 g/mol = = = 1.618255 = 1.618 Rate Kr molar mass O2 32.00 g/mol

5.77

Plan: Recall that the heavier the gas, the slower the molecular speed. The molar mass of Ar is 39.95 g/mol while the molar mass of He is 4.003 g/mol. Solution: a) The gases have the same average kinetic energy because they are at the same temperature. The heavier Ar atoms are moving more slowly than the lighter He atoms to maintain the same average kinetic energy. Therefore, Curve 1 with the lower average molecular speed, better represents the behavior of Ar. b) A gas that has a slower molecular speed would effuse more slowly, so Curve 1 is the better choice. c) Fluorine gas exists as a diatomic molecule, F2, with M = 38.00 g/mol. Therefore, F2 is much closer in mass to Ar (39.95 g/mol) than He (4.003 g/mol), so Curve 1 more closely represents the behavior of F2.

5.78

Plan: Recall that the lower the temperature, the lower the average kinetic energy and the slower the molecular speed. Solution: a) At the lower temperature, the average molecular speed is lower so Curve 1 represents the gas at the lower temperature. b) When a gas has a higher kinetic energy, the molecules have a higher molecular speed. Curve 2 with the larger average molecular speed represents the gas when it has a higher kinetic energy. c) If a gas has a higher diffusion rate, then the gas molecules are moving with a higher molecular speed as in Curve 2.

5.79

Plan: To find the ratio of effusion rates, calculate the inverse of the ratio of the square roots of the molar masses (Graham’s law). Then use the ratio of effusion rates to find the time for the F2 effusion. Effusion rate and time required for the effusion are inversely proportional. Solution: M of He = 4.003 g/mol M of F2 = 38.00 g/mol Rate He molar mass F2 38.00 g/mol = = = 3.08105 (unrounded) Rate F2 molar mass He 4.003 g/mol

Rate He Rate F2

time F2 time He

time F2 3.08105 = 1.00 4.85 min He

Time F2 = 14.9431 = 14.9 min

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5-35

5.80

Plan: Effusion rate and time required for the effusion are inversely proportional. Therefore, time of effusion for a gas is directly proportional to the square root of its molar mass. The ratio of effusion times and the molar mass of H2 are used to find the molar mass of the unknown gas. Solution: M of H2 = 2.016 g/mol Time of effusion of H2 = 2.42 min Time of effusion of unknown = 11.1 min rate H2 time unknown molar mass unknown = = rate unknown time H2 molar mass H 2

11.1 min

molar mass unknown

2.42 min

4.586777 =

2.016 g/mol molar mass unknown 2.016 g/mol

molar mass unknown 2.016 g/mol Molar mass unknown = 42.41366 = 42.4 g/mol 21.03852196 =

5.81

Plan: White phosphorus is a molecular form of the element phosphorus consisting of some number, x, of phosphorus atoms; the number of atoms in a molecule determines the molar mass of the phosphorus molecule. Use the relative rates of effusion of white phosphorus and neon (Graham’s law) to determine the molar mass of white phosphorus. From the molar mass of white phosphorus, determine the number of phosphorus atoms, x, in one molecule of white phosphorus. Solution: M of Ne = 20.18 g/mol Rate Px molar mass Ne = 0.404 = Rate Ne molar mass Px

20.18 g/mol molar mass Px

0.404 = 2

(0.404) =

20.18 g/mol molar mass Px

0.163216 =

20.18 g/mol molar mass Px

Molar mass Px = 123.6398 g/mol 123.6398 g  1 mol P     = 3.992244 = 4 mol P/mol Px or 4 atoms P/molecule Px  mol P  30.97 g P  x

Thus, 4 atoms per molecule, so Px = P4. 5.82

Plan: Use the equation for root mean speed (urms) to find this value for He at 0°C and 30°C and for Xe at 30°C. The calculated root mean speed is then used in the kinetic energy equation to find the average kinetic energy for the two gases at 30°C. Molar mass values must be in units of kg/mol and temperature in kelvins. Solution: a) 0°C = 273 K

30°C + 273 = 303 K

R = 8.314 J/mol K

1 J = kg m /s

 4.003 g He  1 kg   = 0.004003 kg/mol M of He =    103 g  mol

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5-36

3RT Molar mass

urms =

J   38.314 273 K  2 2  mol  K   kg  m /s  = 1.3042 × 103 = 1.30 × 103 m/s  0.004003 kg/mol J 



urmsHe (at 0°C) =



J   38.314 303 K  2 2  mol  K   kg  m /s  = 1.3740 × 103 = 1.37 × 103 m /s 0.004003 kg/mol J  



urmsHe (at 30°C) =



131.3 g Xe  1 kg   = 0.1313 kg/mol M of Xe =    mol 103 g  2 2 1 J = kg m /s

b) 30°C + 273 = 303 K R = 8.314 J/mol K 3RT urms = Molar mass

J   38.314 303 K  2 2  mol  K   kg  m /s  = 239.913 m/s (unrounded) 0.1313 kg/mol J  



urmsXe (at 30°C) =



Rate He/Rate Xe = (1.3740 × 10 m/s)/(239.913 m/s) = 5.727076 = 5.73 He molecules travel at almost 6 times the speed of Xe molecules. c) Ek = EHe = EXe =

1 mu2 2





2 1   3  0.004003 kg/mol  1.3740 10 m/s 1 J/kg  m2 /s2 = 3778.58 = 3.78 × 103 J/mol   2  1    2    1 J/kg  m2 /s2 = 3778.70 = 3.78 × 103 J/mol  0.1313 kg/mol   239.913 m/s 2





 3778.58 J  1 mol  –21 –21  d)  = 6.2746 × 10 = 6.27 × 10 J/He atom  mol  6.0221023 atoms  5.83

Plan: Use Graham’s law: the rate of effusion of a gas is inversely proportional to the square root of the molar mass. When comparing the speed of gas molecules, the one with the lowest mass travels the fastest. Solution: a) M of S2F2 = 102.12 g/mol; M of N2F4 = 104.02 g/mol; M of SF4 = 108.06 g/mol SF4 has the largest molar mass and S2F2 has the smallest molar mass: rate SF4 < rate N2 F4 < rate S2 F2 molar mass N 2 F4 RateS2 F2 104.02 g/mol = b) = = 1.009260 = 1.0093:1 molar mass S2 F2 102.12 g/mol Rate N2 F4 c)

Rate X Rate SF4

0.935 =

= 0.935 =

molar mass SF4 molar mass X

108.06 g/mol molar mass X

(0.935) =

108.06 g/mol molar mass X

0.874225 =

108.06 g/mol molar mass X

Molar mass X = 123.60662 = 124 g/mol Copyright

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5-37

5.84

Plan: To find the ratio of effusion rates, calculate the inverse of the ratio of the square roots of the molar masses (Graham’s law). Solution:

Rate 235 UF6 Rate 5.85

5.86

5.87

5.88 5.89

5.90

238

UF6



Molar Mass 238 UF6 Molar Mass

235

UF6



352.04 g/mol 349.03 g/mol

 1.0043

Interparticle attractions cause the real pressure to be less than ideal pressure, so it causes a negative deviation. The size of the interparticle attraction is related to the constant a. According to Table 5.4, aN2 = 1.39, aKr = 2.32, and aCO2 = 3.59. Therefore, CO2 experiences a greater negative deviation in pressure than the other two gases: N2 < Kr < CO2. Particle volume causes a positive deviation from ideal behavior. Thus, VReal Gases > VIdeal Gases. The particle volume is related to the constant b. According to Table 5.4, bH2 = 0.0266, bO 2 = 0.0318, and bCl 2 = 0.0562. Therefore, the order is H2 < O2 < Cl2. Nitrogen gas behaves more ideally at 1 atm than at 500 atm because at lower pressures the gas molecules are farther apart. An ideal gas is defined as consisting of gas molecules that act independently of the other gas molecules. When gas molecules are far apart they act more ideally, because intermolecular attractions are less important and the volume of the molecules is a smaller fraction of the container volume. SF6 behaves more ideally at 150°C. At higher temperatures, intermolecular attractions become less important and the volume occupied by the molecules becomes less important. Plan: To find the total force, the total surface area of the can is needed. Use the dimensions of the can to find the surface area of each side of the can. Do not forget to multiply the area of each side by two. The surface area of 2 2 the can in cm must be converted to units of in . Solution: Surface area of can = 2(40.0 cm)(15.0 cm) + 2(40.0 cm)(12.5 cm) + 2(15.0 cm)(12.5 cm) 3 2 = 2.575 × 10 cm 2  1 in  14.7 lb  3 3 Total force = 2.575103 cm 2     = 5.8671 × 10 = 5.87 × 10 lbs  2.54 cm   1 in 2  Plan: Use the ideal gas law to find the number of moles of O2. Moles of O2 is divided by 4 to find moles of Hb since O2 combines with Hb in a 4:1 ratio. Divide the given mass of Hb by the number of moles of Hb to obtain molar mass, g/mol. Temperature must be in units of kelvins, pressure in atm, and volume in L. Solution: V = 1.53 mL T = 37°C + 273.15 = 310.15 K P = 743 torr n = unknown 103 L   = 1.53 × 10–3 L Converting V from mL to L: V = 1.53 mL   1 mL  Converting P from torr to atm: PV = nRT Solving for n:

 1 atm  P = 743 torr  = 0.977631578 atm  760 torr 

PV 0.977631578 atm 1.5310 L  –5 = 5.874240 × 10 mol O2  L  atm   RT 0.0821 310.15 K   mol  K  3

Moles of O2 = n =

 1 mol Hb  –5 Moles Hb = 5.874240105 mol O2   = 1.46856 × 10 mol Hb (unrounded)  4 mol O2  1.00 g Hb 4 4 Molar mass hemoglobin = = 6.80939 × 10 = 6.81 × 10 g/mol 5 1.46856 10 Hb Copyright

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5-38

5.91

Plan: First, write the balanced equations. Convert mass of NaHCO3 to moles and use the molar ratio from each balanced equation to find the moles of CO2 gas produced. Use the ideal gas law to find the volume of that amount of CO2. Temperature must be in kelvins. Solution: Reaction 1: 2NaHCO3(s)  Na2CO3(s) + H2O(l) + CO2(g)  1 mol NaHCO3   1 mol CO2  = 5.95167 × 10–3 mol CO Moles CO2 = 1.00 g NaHCO3  2    84.01 g NaHCO  2 mol NaHCO  3

Finding the volume of CO2: V = unknown P = 0.975 atm PV = nRT Solving for V:

T = 200°C + 273.15 = 473.15 K –3 n = 5.95167 × 10 mol

 L  atm  5.95167 103 mol 0.0821 473.15 K   nRT mol  K  Volume of CO2 = V = = 0.2371244 L = P 0.975 atm  Converting V from L to mL:  1 mL   V = 0.2371244 L 3  = 237.1244 = 237 mL CO2 in Reaction 1 10 L 





Reaction 2: NaHCO3(s) + H (aq)  H2O(l) + CO2(g) + Na (aq)  1 mol NaHCO3  1 mol CO2    = 1.1903 × 10–2 mol CO2 Moles CO2 = 1.00 g NaHCO3  1 mol NaHCO   84.01 g NaHCO  +

Finding the volume of CO2: V = unknown P = 0.975 atm PV = nRT Solving for V:

T = 200°C + 273.15 = 473.15 K –2 n = 1.1903 × 10 mol

 L  atm  1.1903102 mol 0.0821 473.15 K   nRT mol  K  Volume of CO2 = V = = 0.4742352 L = P 0.975 atm  Converting V from L to mL:  1 mL   V = 0.4742352 L 3  = 474.2352 = 474 mL CO2 in Reaction 2 10 L 



5.92

Plan: Use the relationship



PV PV PV n T 1 1 = 2 2 or V2 = 1 1 2 2 ; P is fixed while V and T change. n2 is 0.75n1 since n1T1 n2 T2 P2 n1T1

one-fourth of the gas leaks out. Only the initial and final conditions need to be considered. Solution: P1 = 1.01 atm P2 = 1.01 atm (Thus, P has no effect, and does not need to be included.) T1 = 305 K T2 = 250 K n1 = n1 n2 = 0.75n1 V1 = 600. L V2 = ? 600 L0.75n1 250 K Vn T V2 = 1 2 2 = = 368.852 = 369 L n1T1 n1  305 K



5.93



Plan: Convert the mass of Cl2 to moles and use the ideal gas law and van der Waals equation to find the pressure of the gas. McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution

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5-39

Solution:

103 g  1 mol Cl2   = 8.3921016 mol  a) Moles Cl2: 0.5950 kg Cl2   70.90 g Cl2   1 kg  V = 15.50 L n = 8.3921016 mol Ideal gas law: PV = nRT

T = 225°C + 273.15 = 498.15 K P = unknown

Solving for P:

nRT PIGL = = V

L  atm  498.15 K  8.3921016 mol0.0821 mol  K 

15.50 L  n2 a  b) van der Waals equation:  P  2 V  nb  nRT  V 

= 22.1433 = 22.1 atm

Solving for P: nRT n2 a atm  L2 PVDW = From Table 5.4: a = 6.49 ;  2 V  nb V mol 2 n = 8.3921016 mol from part (a)

b = 0.0562

L mol 2

PVDW



atm  L2   2  

L  atm  498.15 K  8.3921016 mol Cl  6.49 8.3921016 mol Cl 0.0821 mol mol    K =  L   15.50 L 8.3921016 mol Cl 0.0562 15.50 L   mol  2

= 20.93573 = 20.9 atm 5.94

Plan: Rearrange the formula PV = (m/M )RT to solve for molar mass. Convert the volume in mL to L. Temperature must be in kelvin. To find the molecular formulas of I, II, III, and IV, assume 100 g of each sample so the percentages are numerically equivalent to the masses of each element. Convert each of the masses to moles by using the molar mass of each element involved. Divide all moles by the lowest number of moles and convert to whole numbers to determine the empirical formula. The empirical formula mass and the calculated molar mass will then relate the empirical formula to the molecular formula. For gas IV, use Graham’s law to find the molar mass Solution: a) V = 750.0 mL T = 70.00°C + 273.15 = 343.15 K m = 0.1000 g P = 0.05951 atm (I); 0.07045 atm (II); 0.05767 atm (III) M = unknown 103 L   = 0.7500 L Converting V from mL to L: V = 750.0 mL   1 mL  PV = (m/M )RT

Solving for molar mass, M : mRT Molar mass I = M =  PV

 L  atm  343.15 K   mol  K  = 63.0905 = 63.09 g I/mol 0.05951 atm 0.7500 L 

0.1000 g0.08206

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5-40

 L  atm  343.15 K   mol  K  = 53.293 = 53.29 g II/mol 0.07045 atm 0.7500 L 

0.1000 g0.08206

mRT Molar mass II = M =  PV

 L  atm  0.1000 g0.08206 343.15 K   mRT mol  K  Molar mass III = M = = 65.10349 = 65.10 g III/mol  PV 0.05767 atm 0.7500 L  b) % H in I = 100% – 85.63% = 14.37% H % H in II = 100% – 81.10% = 18.90% H % H in III = 100% – 82.98% = 17.02% H Assume 100 g of each so the mass percentages are also the grams of the element. I  1 mol B    = 7.921369 mol B (unrounded) Moles B = 85.63 g B  10.81 g B 





 1 mol H    = 14.25595 mol H (unrounded) Moles H = 14.37 g H  1.008 g H 





 7.921369 mol B 14.25595 mol H     = 1.00   = 1.7997     7.921369 mol B   7.921369 mol B  The hydrogen value is not close enough to a whole number to round. Thus, both amounts need to be multiplied by the smallest value to get near whole numbers. This value is 5. Multiplying each value by 5 gives (1.00 × 5) = 5 for B and (1.7997 × 5) = 9 for H. The empirical formula is B5H9, which has a formula mass of 63.12 g/mol. The empirical formula mass is near the molecular mass from part (a) (63.09 g/mol). Therefore, the empirical and molecular formulas are both B5H9. II

 1 mol B    = 7.50231 mol B (unrounded) Moles B = 81.10 g B  10.81 g B 





 1 mol H    = 18.750 mol H (unrounded) Moles H = 18.90 g H  1.008 g H 





 7.50231 mol B   = 1.00   7.50231 mol B

III

 18.750 mol H    = 2.4992   7.50231 mol B

The hydrogen value is not close enough to a whole number to round. Thus, both amounts need to be multiplied by the smallest value to get near whole numbers. This value is 2. Multiplying each value by 2 gives (1.00 × 2) = 2 for B and (2.4992 × 2) = 5 for H. The empirical formula is B2H5, which has a formula mass of 26.66 g/mol. Dividing the molecular formula mass from part (a) by the empirical formula mass gives the relationship between the formulas: (53.29 g/mol)/(26.66 g/mol) = 2. The molecular formula is two times the empirical formula, or B4H10.

 1 mol B    = 7.6762 mol B (unrounded) Moles B = 82.98 g B  10.81 g B 





 1 mol H    = 16.8849 mol H (unrounded) Moles H = 17.02 g H  1.008 g H 





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5-41

 7.6762 mol B 16.8849 mol H     = 1.00  = 2.2     7.6762 mol B  7.6762 mol B  The hydrogen value is not close enough to a whole number to round. Thus, both amounts need to be multiplied by the smallest value to get near whole numbers. This value is 5. Multiplying each value by 5 gives (1.00 × 5) = 5 for B and (2.2 × 5) = 11 for H. The empirical formula is B5H11, which has a formula mass of 65.14 g/mol. The empirical formula mass is near the molecular mass from part (a). Therefore, the empirical and molecular formulas are both B5H11. Rate SO2 molar mass IV = Rate IV molar mass SO2  250.0 mL     molar mass IV 13.04 min  = 0.657318 =  350.0 mL  64.06 g/mol     12.00 min  molar mass IV 2 0.657318 = 64.06 g/mol Molar mass IV = 27.6782 = 27.68 g/mol % H in IV = 100% – 78.14% = 21.86% H  1 mol B   Moles B = 78.14 g B   = 7.22849 mol B (unrounded) 10.81 g B 





 1 mol H    = 21.6865 mol H (unrounded) Moles H = 21.86 g H  1.008 g H 





 7.22849 mol B  21.6865 mol H     = 1.00  = 3.00    7.22849 mol B  7.22849 mol B  The empirical formula is BH3, which has a formula mass of 13.83 g/mol. Dividing the molecular formula mass by the empirical formula mass gives the relationship between the formulas: (27.68 g/mol)/(13.83 g/mol) = 2. The molecular formula is two times the empirical formula, or B2H6. 5.95

Plan: Calculate the mole fraction of each gas; the partial pressure of each gas is directly proportional to its mole fraction so the gas with the highest mole fraction has the highest partial pressure. Remember that kinetic energy is directly proportional to kelvin temperature. Solution: n a) X A = A ntotal 3 A particles 4 A particles 5 A particles I. XA = = 0.33; II. XA = = 0.33; III. XA = = 0.33 9 total particles 12 total particles 15 total particles The partial pressure of A is the same in all 3 samples since the mole fraction of A is the same in all samples. 3 B particles 3 B particles 3 B particles b) I. XB = = 0.33; II. XB = = 0.25; III. XB = = 0.20 9 total particles 12 total particles 15 total particles The partial pressure of B is lowest in Sample III since the mole fraction of B is the smallest in that sample. c) All samples are at the same temperature, T, so all have the same average kinetic energy.

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5-42

5.96

Plan: Use the relationship

PV PV PV T 1 1 = 2 2 or V2 = 1 1 2 . R and n are fixed. Temperatures must be converted to T1 T2 P2 T1

kelvin. Solution: a) T1 = 200°C + 273.15 = 473.15 K; T2 = 100°C + 273.15 = 373.15 K PV T (2 atm)( V )(373.15 K) 1 V2 = 1 1 2 = = 1.577 V1 Increase P2T1 (1 atm)(473.15 K) b) T1 = 100°C + 273.15 = 373.15 K; T2 = 300°C + 273.15 = 573.15 K PV T (1 atm)( V )(573.15 K) 1 V2 = 1 1 2 = = 0.5120 V1 Decrease P2T1 (3 atm)(373.15 K) c) T1 = –73°C + 273.15 = 200.15 K; T2 = 127°C + 273.15 = 400.15 K PV (3 atm)(V1 )(400.15 K) 1 1T2 V2 = = = 0.9996V1 Unchanged P2 T1 (6 atm)(200.15 K) d) T1 = 300°C + 273.15 = 573.15 K; T2 = 150°C + 273.15 = 423.15 K PV T (0.2 atm)( V )(423.15 K) 1 V2 = 1 1 2 = = 0.3691V1 Decrease P2T1 (0.4 atm)(573.15 K) 5.97

Plan: Partial pressures and mole fractions are calculated from Dalton’s law of partial pressures: PA = XA × Ptotal. Remember that 1 atm = 760 torr. Solve the ideal gas law for moles and then convert to molecules using Avogadro’s number to calculate the number of O2 molecules in the volume of an average breath. Solution: a) Convert each mole percent to a mole fraction by dividing by 100%. Ptotal = 1 atm = 760 torr PNitrogen = XNitrogen × Ptotal = 0.786 × 760 torr = 597.36 = 597 torr N2 POxygen = XOxygen × Ptotal = 0.209 × 760 torr = 158.84 = 159 torr O2 PCarbon Dioxide = XCarbon Dioxide × Ptotal = 0.0004 × 760 torr = 0.304 = 0.3 torr CO2 PWater = XWater × Ptotal = 0.0046 × 760 torr = 3.496 = 3.5 torr H2O b) Mole fractions can be calculated by rearranging Dalton’s law of partial pressures: P XA = A and multiply by 100 to express mole fraction as percent Ptotal Ptotal = 1 atm = 760 torr 569 torr XNitrogen = × 100% = 74.8684 = 74.9 mol% N2 760 torr 104 torr XOxygen = × 100% = 13.6842 = 13.7 mol% O2 760 torr 40 torr X Carbon Dioxide = × 100% = 5.263 = 5.3 mol% CO2 760 torr 47 torr XWater = × 100% = 6.1842 = 6.2 mol% H2O 760 torr c) V = 0.50 L T = 37°C + 273.15 = 310.15 K P = 104 torr n = unknown  1 atm  Converting P from torr to atm: P = 104 torr  = 0.136842105 atm  760 torr  PV = nRT Solving for n: PV 0.136842105 atm 0.50 L  n=  = 0.0026870 mol O2 RT 0.0821 L  atm 310.15 K    mol  K  McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution

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5-43

 6.022 1023 molecules O  2 Molecules of O2 = 0.0026870 mol O 2    1 mol O 2  21 21 = 1.6181 × 10 = 1.6 × 10 molecules O2 5.98

Plan: Convert the mass of Ra to moles and then atoms using Avogadro’s number. Convert from number of Ra atoms to Rn atoms produced per second and then to Rn atoms produced per day. The number of Rn atoms is converted to moles and then the ideal gas law is used to find the volume of this amount of Rn. Solution: 23  1 mol Ra   6.022 10 Ra atoms   21 Atoms Ra = 1.0 g Ra    = 2.664602 × 10 Ra atoms  226 g Ra  1 mol Ra  





1.37310 4 Rn atoms   = 3.65849855 × 1010 Rn atoms/s Atoms Rn produced/s = 2.664602 10 21 Ra atoms   1.0 1015 Ra atoms 





 3.658498551010 Rn atoms  3600 s  24 h  1 mol Rn     Moles Rn produced/day =   23         s  h  day  6.02210 Rn atoms  –9

= 5.248992 × 10 mole Rn/day PV = nRT Solving for V (at STP):  L  atm  5.248992 109 mol 0.0821 273 K   nRT mol  K  Volume of Rn = V = = P 1 atm  –7 –7 = 1.17647 × 10 = 1.2 × 10 L Rn



5.99



Plan: For part (a), since the volume, temperature, and pressure of the gas are changing, use the combined gas law. For part (b), use the ideal gas law to solve for moles of air and then moles of N2. Solution: a) P1 = 1450. mmHg P2 = 1 atm V1 = 208 mL V2 = unknown T1 = 286 K T2 = 298 K  1 atm   = 1.9079 atm Converting P1 from mmHg to atm: P1 = 1450. mmHg  760 mmHg  Arranging the ideal gas law and solving for V2: PV PV 1 1 = 2 2 T1 T2

 T  P   298 K  1.9079 atm  2  V2 = V1  2  1  = 208 L   = 413.494 mL = 4 × 10 mL  P2   T1   286 K   1 atm  b) V = 208 mL P = 1450 mmHg = 1.9079 atm Converting V from mL to L:

T = 286 K n = unknown 103 L   = 0.208 L V = 208 mL  1 mL 

PV = nRT Solving for n: Moles of air = n =

1.9079 atm 0.208 L PV  = 0.016901 mol air RT 0.0821 L  atm 286 K   mol  K 

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5-44

 77% N 2  Mole of N2 = 0.016901 mol  = 0.01301 = 0.013 mol N2  100%  5.100

Plan: The amounts of both reactants are given, so the first step is to identify the limiting reactant. Write the balanced equation and use molar ratios to find the number of moles of NO2 produced by each reactant. The smaller number of moles of product indicates the limiting reagent. Solve for volume of NO2 using the ideal gas law. Solution: Cu(s) + 4HNO3(aq)  Cu(NO3)2(aq) + 2NO2(g) + 2H2O(l)  8.95 g Cu  1 mol Cu  2 mol NO2      Moles NO2 from Cu = 4.95 cm3   3  1 mol Cu  = 1.394256 mol NO2  cm  63.55 g Cu  





 68.0% HNO3 1 cm3 1.42 g 1 mol HNO3   2 mol NO2      Moles NO2 from HNO3 = 230.0 mL   3  4 mol HNO   100%  1 mL  cm  63.02 g  3 = 1.7620 mol NO2 Since less product can be made from the copper, it is the limiting reactant and excess nitric acid will be left after the reaction goes to completion. Use the calculated number of moles of NO 2 and the given temperature and pressure in the ideal gas law to find the volume of nitrogen dioxide produced. Note that nitrogen dioxide is the only gas involved in the reaction. V = unknown T = 28.2°C + 273.15 = 301.35 K P = 735 torr n = 1.394256 mol NO2  1 atm  Converting P from torr to atm: P = 735 torr  = 0.967105 atm  760 torr  PV = nRT Solving for V:  L  atm  1.394256 mol0.0821 301.35 K   nRT mol  K  = 35.668.6 = 35.7 L NO2 V=  P 0.967105 atm 



5.101



Plan: Solve the ideal gas law for moles of air and then convert to molecules using Avogadro’s number. Volume must be converted to liters and temperature to kelvins. Solution: a) V = 1200 mL T = 37°C + 273.15 = 310.15 K P = 1.0 atm n = unknown 103 L   = 1.2 L Converting V from mL to L: V = 1200 mL  1 mL  PV = nRT Solving for n:

1.0 atm1.2 L PV  = 0.047127 = 0.047 mol air  RT 0.0821 L  atm 310.15 K   mol  K   6.022 1023 molecules    = 2.837971 × 1022 = 2.8 × 1022 molecules air b) Molecules of air = 0.047127 mol air    1 mol air  Moles of air = n =

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5-45

5.102

Plan: The amounts of two reactants are given, so the first step is to identify the limiting reactant. Write the balanced equation and use molar ratios to find the number of moles of Br2 produced by each reactant. The smaller number of moles of product indicates the limiting reagent. Solve for volume of Br2 using the ideal gas law. Solution: 5NaBr(aq) + NaBrO3(aq) + 3H2SO4(aq)  3Br2(g) + 3Na2SO4(aq) + 3H2O(g)  1 mol NaBr   3 mol Br2   Moles Br2 from NaBr = 275 g NaBr    = 1.60365 mol Br2 (unrounded) 102.89 g NaBr   5 mol NaBr 





 1 mol NaBrO3  3 mol Br2     Moles Br2 from NaBrO3 = 175.6 g NaBrO3  1 mol NaBrO  = 3.491285 mol Br2 150.89 g NaBrO3  3 The NaBr is limiting since it produces a smaller amount of Br2. Use the ideal gas law to find the volume of Br2: V = unknown T = 300°C + 273.15 = 573.15 K P = 0.855 atm n = 1.60365 mol Br2 PV = nRT Solving for V:  L  atm  1.60365 mol0.0821 573.15 K   nRT mol  K  Volume (L) of Br2 = V = = 88.258 = 88. L Br2  P 0.855 atm  Plan: First, write the balanced equation. Convert mass of NaN3 to moles and use the molar ratio from the balanced equation to find the moles of nitrogen gas produced. Use the ideal gas law to find the volume of that amount of nitrogen. The problem specifies that the nitrogen gas is collected over water, so the partial pressure of water vapor must be subtracted from the overall pressure given. Table 5.2 reports the vapor pressure of water at 26°C (25.2 torr). Pressure must be in units of atm and temperature in kelvins. Solution: 2NaN3(s)  2Na(s) + 3N2(g)  1 mol NaN3   3 mol N2  = 1.15349 mol N Moles N2 = 50.0 g NaN3  2    65.02 g NaN  2 mol NaN 



5.103



Finding the volume of N2: V = unknown T = 26°C + 273.15 = 299.15 K Ptotal = 745.5 mmHg n = 1.15319 mol Pwater vapor = 25.2 torr = 25.2 mmHg Pnitrogen = Ptotal – Pwater vapor = 745.5 mmHg – 25.2 mmHg = 720.3 mmHg  1 atm   = 0.9477632 atm Converting P from mmHg to atm: P = 720.3 mmHg  760 mmHg  PV = nRT Solving for V: nRT V= = P

5.104

 L  atm  1.15349 mol0.0821 299.15 K   mol  K  = 29.8914 = 29.9 L N2 0.9477632 atm 

Plan: Use the percent composition information to find the empirical formula of the compound. Assume 100 g of sample so the percentages are numerically equivalent to the masses of each element. Convert each of the masses to moles by using the molar mass of each element involved. Divide all moles by the lowest number of moles and convert to whole numbers to determine the empirical formula. Rearrange the formula PV = (m/M )RT to solve for molar mass.The empirical formula mass and the calculated molar mass will then relate the empirical formula to the molecular formula. McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution

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5-46

Solution: Empirical formula: Assume 100 g of each so the mass percentages are also the grams of the element.

 1 mol C   Moles C = 64.81 g C   = 5.39634 mol C 12.01 g C 





 1 mol H    = 13.49206 mol H Moles H = 13.60 g H  1.008 g H 





 1 mol O    = 1.349375 mol O Moles O = 21.59 g O  16.00 g O 





 5.39634 mol C    = 4  1.349375 mol O 

13.749206 mol H    = 10   1.349375 mol O 

1.349375 mol O    = 1.00  1.349375 mol O 

Empirical formula = C4H10O (empirical formula mass = 74.12 g/mol) Molecular formula: V = 2.00 mL T = 25°C + 273.15 = 298.15 K m = 2.57 g P = 0.420 atm M = unknown PV = (m/M )RT Solving for molar mass, M : Molar mass = M =

mRT  PV

 L  atm  298.15 K   mol  K  = 74.89 g/mol 0.420 atm 2.00 L 

2.57 g0.0821

Since the molar mass (74.89 g/mol ) and the empirical formula mass (74.12 g/mol) are similar, the empirical and molecular formulas must both be: C4H10O 5.105

–

Plan: The empirical formula for aluminum chloride is AlCl3 (Al and Cl ). The empirical formula mass is (133.33 g/mol). Calculate the molar mass of the gaseous species from the ratio of effusion rates (Graham’s law). This molar mass, divided by the empirical weight, should give a whole-number multiple that will yield the molecular formula. Solution:

Rate unknown Rate He 0.122 =

= 0.122 =

molar mass He molar mass unknown

4.003 g/mol molar mass unknown

0.014884 =

4.003 g/mol molar mass unknown

Molar mass unknown = 268.9465 g/mol

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5-47

The whole-number multiple is 268.9465/133.33, which is about 2. Therefore, the molecular formula of the gaseous species is 2 × (AlCl3) = Al2Cl6. 5.106

Plan: First, write the balanced equation. Convert mass of C8H18 to moles and use the molar ratio from the balanced equation to find the total moles of gas produced. Use the ideal gas law to find the volume of that amount of gas. For part (b), use the molar ratio from the balanced equation to find the moles of oxygen that react with the C8H18. Use the composition of air to calculate the amount of N2 and Ar that remains after the O2 is consumed and use the ideal gas law to find the volume of that amount of gas. The volume of the gaseous exhaust is the sum of the volume of gaseous products and the residual air (N2 and Ar) that does not react. Solution: 2C8H18(l) + 25O2(g)  16CO2(g) + 18H2O (g)  1 mol C8 H18   34 mol gas   a) Moles gaseous products = 100. g C8 H18    = 14.883558 mol gas 114.22 g C8 H18   2 mol C8 H18  Finding the volume of gaseous product: V = unknown T = 350°C + 273.15 = 623.15 K Ptotal = 735 torr n = 14.883558 mol  1 atm  Converting P from torr to atm: P = 735 torr  = 0.9671053 atm  760 torr  PV = nRT





Solving for V:  L  atm  14.883558 mol0.0821 623.15 K   mol  K  = 787.352 = 787 L gas 0.9671053 atm   1 mol C H  25 mol O2  8 18     b) Moles O2 = 100 g C8 H18    2 mol C H  = 10.94379 mol O2 114.22 g C8 H18  8 18  nRT Volume gas = V = = P





 78% + 1% N 2 and Ar    = 41.1695 mol Ar + N Moles other gases = 10.94379 mol O2  2  21% O2   Finding the volume of Ar + N2: V = unknown T = 350°C + 273.15 = 623.15 K Ptotal = 735 torr = 0.9671053 atm n = 41.1695 mol PV = nRT Solving for V:  L  atm  41.1695 mol 0.0821 623.15 K   nRT mol  K  Volume gas = V = = 2177.90 L residual air = P 0.9671053 atm  3 Total volume of gaseous exhaust = 787.352 L + 2177.90 L = 2965.25 = 2.97 × 10 L



5.107



Plan: First, write the balanced equation for the reaction: 2SO2 + O2  2SO3. The total number of moles of gas will change as the reaction occurs since 3 moles of reactant gas forms 2 moles of product gas. From the volume, temperature, and pressures given, we can calculate the number of moles of gas before and after the reaction using the ideal gas law. For each mole of SO3 formed, the total number of moles of gas decreases by 1/2 mole. Thus, twice the decrease in moles of gas equals the moles of SO3 formed. Solution: Moles of gas before and after reaction: V = 2.00 L T = 800 K Ptotal = 1.90 atm n = unknown McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution

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5-48

PV = nRT Initial moles = n =

Final moles = n =







1.90 atm 2.00 L PV = = 0.05785627 mol  L  atm  RT 0.0821 800. K  mol  K 







1.65 atm 2.00 L PV = = 0.050243605 mol  L  atm  RT 0.0821 800. K   mol  K 





Moles of SO3 produced = 2 × decrease in the total number of moles = 2 × (0.05785627 mol – 0.050243605 mol) –2 = 0.01522533 = 1.52 × 10 mol Check: If the starting amount is 0.0578 total moles of SO2 and O2, then x + y = 0.0578 mol, where x = mol of SO2 and y = mol of O2. After the reaction: (x – z) + (y – 0.5z) + z = 0.0502 mol Where z = mol of SO3 formed = mol of SO2 reacted = 2(mol of O2 reacted). Subtracting the two equations gives: x – (x – z) + y – (y – 0.5z) – z = 0.0578 – 0.0502 z = 0.0152 mol SO3 The approach of setting up two equations and solving them gives the same result as above. 5.108

Plan: Use the density of C2H4 to find the volume of one mole of gas. Then use the van der Waals equation with 1.00 mol of gas to find the pressure of the gas (the mole ratio is 1:1, so the number of moles of gas remains the same). Solution: 3  28.05 g C2 H 4   1 mL  10 L  = 0.130465 L = 0.130 L a) 1 mole C2 H 4       1 mole C2 H 4   0.215 g  1 mL  V = 0.130465 L

T = 10°C + 273.15 = 283.15 K + 950 K = 1233.15 K

Ptotal = unknown

n = 1.00 mol

atm  L2 From Table 5.4 for CH4: a = 2.25 ; mol 2 2    P  n a V  nb  nRT  V 2 

Pressure of CH4 = PVDW =

b = 0.0428

L mol

nRT n2 a  2 V  nb V

atm  L2  2 L  atm   1.00 mol 2.25  1.00 mol0.0821 1233.15 K   mol 2    mol  K   PVDW = = 1022.681 atm = 1023 atm 2 0.130465 L 1.00 mol 0.0428 L/mol 0.130465 L b) n =

1022.681 atm 0.130465 L  PV = = 1.31788 = 1.32 mol  L  atm  RT 0.0821 1233.15 K    mol  K 

This value is smaller than that shown in Figure 5.23 for CH4. The temperature in this situation is very high (1233 K). At high temperatures, the gas particles have high kinetic energy. Thus the gas particles have the energy to overcome the effects of intermolecular attraction and the gas behaves more ideally.

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5-49

5.109

Plan: First, write the balanced equation. Convert mass of ammonium nitrate to moles and use the molar ratio from the balanced equation to find the moles of gas produced. Use the ideal gas law to find the volume of that amount of gas. Solution: 103 g   1 mol NH 4 NO3   7 mol gas   Moles gas = 15.0 kg NH 4 NO3      1 kg   80.05 g NH 4 NO3   2 mol NH 4 NO3  = 655.840 mol gas Finding the volume of gas: V = unknown T = 307°C + 273.15 = 580.15 K P = 1.00 atm n = 655.840 mol PV = nRT Solving for V:  L  atm  655.840 mol0.0821 580.15 K   nRT 4 4 mol  K  V= = 3.12379 × 10 = 3.12 × 10 L = P 1.00 atm



5.110



Plan: Multiply the molarity and volume of the I2 solution to find initial moles of I2 added. Multiply molarity and 2– volume of the S2O3 solution to obtain moles of that solution and use the molar ratio in the balanced equation to find moles of excess I2. Subtract the excess I2 from the initial moles of added I2 to find the moles of I2 reacted with the SO2; the molar ratio betweeen SO2 and I2 gives the moles of SO2 present. Use the ideal gas law to find the moles of air which is compared to the moles of SO2 present. Solution: 2 2 The balanced equation is: I2(aq) + 2S2O3 (aq)  2I(aq) + S4O6 (aq). 103 L  0.01017 mol I2    = 2.034 × 10–4 mol I initial  Initial moles of I2 = 20.00 mL   2  1 mL  L 





103 L  0.0105 mol S2 O32  1 mol I 2   2–    Moles I2 reacting with S2O3 = 11.37 mL     2 mol S O 2  L  1 mL  2 3   –5 2– = 5.96925 × 10 mol I2 reacting with S2O3 (not reacting with SO2) –4 –5 Moles I2 reacting with SO2 = 2.034 × 10 mol – 5.96925 × 10 mol –4 = 1.437075 × 10 mol I2 reacted with SO2 + The balanced equation is: SO2(aq) + I2(aq) + 2H2O(l)  HSO4(aq) + 2I(aq) + 3H (aq). 1 mol SO  4 –4 2 Moles SO2 = 1.43707510 mol I 2   = 1.437075 × 10 mol SO2  1 mol I2 





Moles of air: V = 500 mL = 0.500 L

T = 38°C + 273.15 = 311.15 K

Ptotal = 700 torr Converting P from torr to atm: PV = nRT Moles air = n =

n = unknown  1 atm  P = 700 torr   = 0.921052631 atm  760 torr 





 

0.921052632 atm 0.500 L PV = = 0.018028 mol air (unrounded)  L  atm  RT 0.0821 311.15 K   mol  K 



1.43707510 mol SO 2 4

Volume % SO2 = mol % SO2 =

0.018028 mol air

100 = 0.797146 = 0.797%

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5-50

5.111

Plan: First, write the balanced equation. The moles of CO that react in part (a) can be calculated from the ideal gas law. Volume must be in units of L, pressure in atm, and temperature in kelvins. Once the moles of CO that react are known, the molar ratio from the balanced equation is used to determine the mass of nickel that will react with the 3 CO. For part (b), assume the volume is 1 m . Use the ideal gas law to solve for moles of Ni(CO)4, which equals the 3 moles of Ni, and convert moles to grams using the molar mass. For part (c), the mass of Ni obtained from 1 m (part b) can be used to calculate the amount of CO released. Use the ideal gas law to calculate the volume of CO. The vapor pressure of water at 35°C (42.2 torr) must be subtracted from the overall pressure (see Table 5.2). Solution: a) Ni(s) + 4CO(g)  Ni(CO)4(g) 3 V = 3.55 m T = 50°C + 273.15 = 323.15 K P = 100.7 kPa n = unknown  1L  3 Converting V from m to L: V = 3.55 m 3  3 3  = 3550 L 10 m   1 atm  Converting P from kPa to atm: P = 100.7 kPa  = 0.993831729 atm 101.325 kPa  PV = nRT Solving for n:

PV 0.993831729 atm 3550 L   = 132.98232 mol CO RT 0.0821 L  atm 323.15 K   mol  K   1 mol Ni   58.69g Ni   3 Mass Ni = 132.98232 mol CO   = 1951.183 = 1.95 × 10 g Ni  4 mol CO   1mol Ni  Moles of CO = n =

b) Ni(s) + 4 CO(g)  Ni(CO)4(g) 3 V=1m T = 155°C + 273.15 = 428.15 K P = 21 atm n = unknown  1L  3 Converting V from m to L: V = 1 m 3  3 3  = 1000 L 10 m  PV = nRT Solving for n: 21 atm1000 L PV  Moles of Ni(CO)4 = n = = 597.42059 mol Ni(CO)4 RT 0.0821 L  atm  428.15 K   mol  K 

 1 mol Ni   58.69 g Ni   Mass Ni = 597.42059 mol Ni(CO)4    1 mol Ni(CO)4   1 mol Ni  4

= 3.50626 × 10 = 3.5 × 10 g Ni The pressure limits the significant figures.  1 mol Ni  4 mol CO   = 2389.68238 mol CO  c) Moles CO = 3.50626 10 4 g Ni  58.69 g Ni  1 mol Ni  Finding the volume of CO: V = unknown T = 35°C + 273.15 = 308.15 K Ptotal = 769 torr n = 2389.68238 mol Pwater vapor = 42.2 torr PCO = Ptotal – Pwater vapor = 769 torr – 42.2 torr = 726.8 torr  1 atm  Converting P from torr to atm: P = 726.8 torr  = 0.956315789 atm  760 torr  PV = nRT Copyright

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5-51

Solving for V: nRT V= = P

 L  atm  2389.68238 mol0.0821 308.15 K   mol  K  = 63218.4995 L CO 0.956315789 atm 

103 m 3   = 63.2184995 = 63 m3 CO V = 63218.4995 L   1 L  The answer is limited to two significant figures because the mass of Ni comes from part (b). 3

Converting V from L to m :

5.112

Plan: Use the percent composition information to find the empirical formula of the compound. Assume 100 g of sample so the percentages are numerically equivalent to the masses of each element. Convert each of the masses to moles by using the molar mass of each element involved. Divide all moles by the lowest number of moles and convert to whole numbers to determine the empirical formula. Rearrange the formula PV = (m/M )RT to solve for molar mass.The empirical formula mass and the calculated molar mass will then relate the empirical formula to the molecular formula. Solution: Empirical formula: Assume 100 g of each so the mass percentages are also the grams of the element.  1 mol Si    = 1.17515 mol Si Moles Si = 33.01 g Si   28.09 g Si 





 1 mol F   Moles F = 66.99 g F   = 3.525789 mol F 19.00 g F 





1.17515 mol Si    = 1  1.17515 mol Si 

 3.525789 mol F    = 3   1.17515 mol Si 

Empirical formula = SiF3 (empirical formula mass = 85.1 g/mol) Molecular formula: V = 0.250 L T = 27°C + 273.15 = 300.15 K m = 2.60 g P = 1.50 atm M = unknown PV = (m/M )RT Solving for molar mass, M :  L  atm  300.15 K  2.60 g0.0821  mRT mol  K  Molar mass = M = = 170.853 g/mol  PV 1.50 atm 0.250 L  The molar mass (170.853 g/mol ) is twice the empirical formula mass (85.1 g/mol), so the molecular formula must be twice the empirical formula, or 2 × SiF3 = Si2F6.

5.113

a) A preliminary equation for this reaction is 4CxHyNz + nO2  4CO2 + 2N2 + 10H2O. Since the organic compound does not contain oxygen, the only source of oxygen as a reactant is oxygen gas. To form 4 volumes of CO2 would require 4 volumes of O2 and to form 10 volumes of H2O would require 5 volumes of O2. Thus, 9 volumes of O2 was required. b) Since the volume of a gas is proportional to the number of moles of the gas we can equate volume and moles. From a volume ratio of 4CO2:2N2:10H2O we deduce a mole ratio of 4C:4N:20H or 1C:1N:5H for an empirical formula of CH5N. McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution

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5-52

5.114

a) There is a total of 6 × 10 blue particles and 6 × 10 black particles. When equilibrium is reached after opening the stopcocks, the particles will be evenly distributed among the three containers. Therefore, container B 6 6 will have 2 × 10 blue particles and 2 × 10 black particles. 6 6 b) The particles are evenly distributed so container A has 2 × 10 blue particles and 2 × 10 black particles. 6 6 6 c) There are 2 × 10 blue particles and 2 × 10 black particles in C for a total of 4 × 10 particles.

  750 torr  = 500 torr Final pressure in C = 4 106 particles 6  6 10 particles  6

d) There are 2 × 10 blue particles and 2 × 10 black particles in B for a total of 4 × 10 particles.

  750 torr  = 500 torr Final pressure in B = 4 106 particles 6  6 10 particles  5.115

Plan: Write the balanced equation for the combustion of n-hexane. For part (a), assuming a 1.00 L sample of air at STP, use the molar ratio in the balanced equation to find the volume of n-hexane required to react with the oxygen in 1.00 L of air. Convert the volume n-hexane to volume % and divide by 2 to obtain the LFL. For part (b), use the LFL calculated in part (a) to find the volume of n-hexane required to produce a flammable mixture and then use the ideal gas law to find moles of n-hexane. Convert moles of n-hexane to mass and then to volume using the density. Solution: a) 2C6H14(l) + 19O2(g) (g) + 14H2O(g) 2 For a 1.00 L sample of air at STP:  20.9 L O2  2 L C6 H14   = 0.0220 L C6H14 Volume of C6H14 vapor needed = 1.00 L air    100 L air   19 L O2  C H volume 0.0220 L C 6 H14 Volume % of C6H14 = 6 14 100 = 100 = 2.2% C6H14 air volume 1.00 L air LFL = 0.5(2.2%) = 1.1% C6H14

 1 L 1.1% C 6 H14  b) Volume of C6H14 vapor = 1.000 m 3 air  3 3   = 11.0 L C6H14 10 m  100% air 

V = 11.0 L P = 1 atm PV = nRT Solving for n:

T = 0°C + 273 = 273 K n = unknown

1 atm11.0 L PV  = 0.490780 mol C6H14 RT 0.0821 L  atm 273 K    mol  K   86.17 g C6 H14  1 mL  Volume of C6H14 liquid = 0.490780 mol C6 H14    1 mol C H  0.660 g C H

Moles of C6H14 = n =

5.116

  = 64.0765 = 64 mL C6H14  14 

Plan: To find the factor by which a diver’s lungs would expand, find the factor by which P changes from 125 ft to the surface, and apply Boyle’s law. To find that factor, calculate Pseawater at 125 ft by converting the given depth from ft-seawater to mmHg to atm and adding the surface pressure (1.00 atm). Solution:  2  12 in   2.54 cm 10 m  1 mm  = 3.81 × 104 mmH2O P(H2O) = 125 ft   1 in  1 cm 103 m   1 ft 

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5-53

P(Hg):

hH2 O hHg

dHg

3.8110 4 mmH 2 O

dH 2 O

hHg

13.5 g/mL 1.04 g/mL

hHg = 2935.1111 mmHg

 1 atm   = 3.861988 atm (unrounded) P(Hg) = 2935.11111 mmHg  760 mm Hg  Ptotal = (1.00 atm) + (3.861988 atm) = 4.861988 atm (unrounded) Use Boyle’s law to find the volume change of the diver’s lungs: P 1V 1 = P 2V 2 V2 P V2 4.861988 atm = 1 = = 4.86 V1 P2 V1 1 atm To find the depth to which the diver could ascend safely, use the given safe expansion factor (1.5) and the pressure at 125 ft, P125, to find the safest ascended pressure, Psafe. P125/Psafe = 1.5 Psafe = P125/1.5 = (4.861988 atm)/1.5 = 3.241325 atm (unrounded) Convert the pressure in atm to pressure in ft of seawater using the conversion factors above. Subtract this distance from the initial depth to find how far the diver could ascend.  760 mmHg  h(Hg): 4.861988  3.241325 atm  = 1231.7039 mmHg  1 atm  hH2 O hH2 O dHg 13.5 g/mL = = hH2 O = 15988.464 mm 1231.7039 mmHg 1.04 g/mL hHg dH 2 O

103 m 1.094 yd  3 ft    = 52.4741 ft   1 mm  1 m 1 yd 

15988.464 mmH 2 O

Therefore, the diver can safely ascend 52.5 ft to a depth of (125 – 52.4741) = 72.5259 = 73 ft. 5.117

Plan: Write a balanced equation. Convert mass of CaF2 to moles and use the molar ratio from the balanced equation to find the moles of gas produced. Use the ideal gas law to find the temperature required to store that amount of HF gas at the given conditions of temperature and pressure. Solution: CaF2(s) + H2SO4(aq)  2HF(g) + CaSO4(s)  1 mol CaF2   2 mol HF   Moles HF gas = 15.0 g CaF2  1 mol CaF  = 0.3842213 mol HF  78.08 g CaF2  2





Finding the temperature: V = 8.63 L P = 875 torr Converting P from torr to atm: PV = nRT Solving for T:



T = unknown n = 0.3842213 mol  1 atm  P = 875 torr  = 1.151312789 atm  760 torr 





1.151315789 atm 8.63 L PV = = 314.9783 K  L  atm  nR 0.3842213 mol HF 0.0821  mol  K  The gas must be heated to 315 K. T=

5.118

Plan: First, write the balanced equation. According to the description in the problem, a given volume of peroxide solution (0.100 L) will release a certain number of “volumes of oxygen gas” (20). Assume that 20 is exact.

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5-54

A 0.100 L solution will produce (20 × 0.100 L) = 2.00 L O2 gas. Use the ideal gas law to convert this volume of O2 gas to moles of O2 gas and convert to moles and then mass of H2O2 using the molar ratio in the balanced equation. Solution: 2H2O2(aq)  2H2O(l) + O2(g) V = 2.00 L T = 0°C + 273 = 273 K P = 1 atm n = unknown PV = nRT Solving for n: 1 atm2.00 L PV –2  Moles of O2 = n = = 8.92327 × 10 mol O2  L  atm  RT 0.0821 273 K   mol  K   2 mol H O  34.02 g H2 O2  2 2   = 6.071395 = 6.07 g H2O2 Mass H2O2 = 8.92327102 mol O2   1 mol H O   1 mol O2  2 2 5.119

Plan: The moles of gas may be found using the ideal gas law. Multiply moles of gas by Avogadro’s number to obtain the number of molecules. Solution: V = 1 mL = 0.001 L T = 500 K –8 P = 10 mmHg n = unknown  1 atm   = 1.315789 × 10–11 atm Converting P from mmHg to atm: P = 108 mmHg  760 mmHg  PV = nRT Solving for n:

11 PV 1.315789 10 atm 0.001 L –16 = 3.2053337 × 10 mol gas  L  atm   RT 0.0821 500 K   mol  K   6.022 10 23 molecules   8 8 Molecules = 3.20533371016 mol  = 1.93025 × 10 = 10 molecules 1 mol   –8 (The 10 mmHg limits the significant figures.)

Moles of gas = n =

5.120

Plan: Use the equation for root mean speed (urms) to find this value for O2 at 0°C. Molar mass values must be in units of kg/mol and temperature in kelvins. Divide the root mean speed by the mean free path to obtain the collision frequency. Solution:  32.00 g O2  1 kg   = 0.03200 kg/mol a) 0°C = 273 K M of O2 =    mol 103 g  2

R = 8.314 J/mol K 3RT urms = Molar mass

J   38.314 273 K  2 2  mol  K   kg  m /s  = 461.2878 = 461 m/s  0.03200 kg/mol J 



urms =

b) Collision frequency =

1 J = kg m /s



urms 461.2878 m/s 9 9 –1 = = 7.2873 × 10 = 7.29 × 10 s 8 mean free path 6.3310 m

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5-55

5.121

Plan: Convert the volume of the tubes from ft to L. Use the ideal gas law to find the moles of gas that will occupy that volume at the given conditions of pressure and temperature. From the mole fraction of propylene and the total moles of gas, the moles of propylene can be obtained. Convert moles to mass in grams and then pounds using the molar mass and scale the amount added in 1.8 s to the amount in 1 h. Solution: 3 3  3   102 m   1 L   3  12 in   2.54 cm     Volume of tubes = 100 ft   3 3  = 2831.685 L  3  3     1 in 3   1 cm   1 ft     10 m V = 2831.685 L T = 330°C + 273.15 = 603.15 K P = 2.5 atm n = unknown PV = nRT 2.5 atm 2831.685 L  PV Moles of gas = n = = = 142.9606 mol gas  L  atm  RT 0.0821 603.15 K    mol  K  moles of propylene Xpropylene = moles of mixture x moles of propylene 0.07 = 142.996 moles of mixture Moles of propylene = 10.00724 moles  42.08 g C3 H6   1 lb C3 H6  1   3600 s  Pounds of propylene = 10.00724 moles C3 H6     1 h   1 mole C H  453.6 g C H 1.8 s  3

= 1856.723 = 1900 lb/h C3H6 5.122

Plan: Use the ideal gas law to calculate the molar volume, the volume of exactly one mole of gas, at the temperature and pressure given in the problem. Solution: V = unknown P = 90 atm PV = nRT

T = 730. K n = 1.00 mol

Solving for V: nRT V= = P

5.123

 L  atm  1.00 mol0.0821 730. K   mol  K  = 0.66592 = 0.67 L/mol 90 atm 

Plan: The diagram below describes the two Hg height levels within the barometer. First, find the pressure of the N2. The PN2 is directly related to the change in column height of Hg. Then find the volume occupied by the N 2. The volume of the space occupied by the N2(g) is calculated from the length and cross-sectional area of the barometer. To find the mass of N2, use these values of P and V (T is given) in the ideal gas law to find moles which is converted to mass using the molar mass of nitrogen. vacuum (74.0 cm)

with N2 (64.0 cm)

Solution:

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5-56

102 mHg   1 mmHg   1 atm  = 0.1315789 atm Pressure of the nitrogen = 74.0 cmHg  64.0 cmHg   760 mmHg   3   10 mHg   1 cmHg    1 mL 103 L   = 0.0432 L Volume of the nitrogen = 1.00 10 2 cm  64.0 cm 1.20 cm 2   1 cm 3  1 mL  V = 0.0432 L T = 24°C + 273.15 = 297.15 K P = 0.1315789 atm n = unknown PV = nRT Solving for n: 0.1315789 atm 0.0432 L  PV –4 Moles of N2 = n = = = 2.32998 × 10 mol N2  L  atm  RT 0.0821 297.15 K   mol  K 

 28.02 g N  –3 –3 2 Mass of N2 = 2.32998104 mol N2   = 6.52859 × 10 = 6.53 × 10 g N2  1 mol N2  5.124

5.125

Plan: Use the ideal gas law to find the moles of ammonia gas in 10.0 L at this pressure and temperature. Molarity is moles per liter. Use the moles of ammonia and the final volume of solution (0.750 L) to get the molarity. Solution: V = 10.0 L T = 33°C + 273.15 = 306.15 K P = 735 torr n = unknown  1 atm  Converting P from torr to atm: P = 735 torr  = 0.9671053 atm  760 torr  PV = nRT Solving for n: 0.9671053 atm 10.0 L PV Moles of ammonia = n =  = 0.384766 mol RT 0.0821 L  atm 306.15 K   mol  K  mol ammonia 0.384766 mol Molarity = M = = 0.513021 = 0.513 M = liters of solution 0.750 L Plan: Use the ideal gas law to determine the total moles of gas produced. The total moles multiplied by the fraction of each gas gives the moles of that gas which may be converted to metric tons. Solution: 3 3 V = 1.5 × 10 m T = 298 K P = 1 atm n = unknown  1L  3 6 Converting V from m to L: V = 1.5103 m 3  3 3  = 1.5 × 10 L 10 m  PV = nRT Solving for n: Moles of gas/day = n =

1 atm 1.5106 L  PV 5 = 6.13101 × 10 mol/day  RT 0.0821 L  atm 298 K   mol  K 

 6.1310110 5 mol   365.25 day  = 2.23935 × 107 mol/yr Moles of gas/yr =   1 yr  day   

 2.23935107 mol  44.01 g CO2  1 kg    3  13 t  = 482.519 = 4.83 × 102 t CO2/yr  Mass CO2 = 0.4896    10 kg  yr   1 mol CO 10 g  2

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5-57

 2.2393510 7 mol  28.01 g CO  1 kg   1 t  = 9.15773 = 9.16 t CO/yr   Mass CO = 0.0146  10 3 kg  3   yr   1 mol CO 10 g  

 2.23935107 mol 18.02 g H2 O  1 kg    3  13 t  = 149.70995 = 1.50 × 102 t H2O/yr  Mass H2O = 0.3710 10 kg  yr   1 mol H O 10 g  2

 2.23935107 mol  64.06 g SO2  1 kg    3  13 t  = 169.992 = 1.70 × 102 t SO2/yr  Mass SO2 = 0.1185 10 kg  yr   1 mol SO 10 g  2

 2.2393510 mol  64.12 g S2  1 kg    3  13 t  = 0.4307614 = 4 × 10–1 t S2/yr  Mass S2 = 0.0003    10 kg  yr   1 mol S2 10 g  7

 2.23935107 mol  2.016 g H2  1 kg    3  13 t  = 0.21218 = 2.1 × 10–1 t H2/yr  Mass H2 = 0.0047    10 kg  yr   1 mol H 10 g  2

 2.2393510 7 mol  36.46 g HCl  1 kg  1t     3  = 0.6531736 = 6 × 10–1 t HCl/yr Mass HCl = 0.0008 3  10 kg   yr  1 mol HCl 10 g 

 2.23935107 mol  34.08 g H2 S 1 kg    3  13 t  = 0.228951 = 2 × 10–1 t H2S/yr  Mass H2S = 0.0003       yr  1 mol H2 S 10 g 10 kg  5.126

Plan: Use the molar ratio from the balanced equation to find the moles of H 2 and O2 required to form 28.0 moles of water. Then use the ideal gas law in part (a) and van der Waals equation in part (b) to find the pressure needed to provide that number of moles of each gas. Solution: a) The balanced chemical equation is: 2H2(g) + O2(g)  2H2O(l)  2 mol H2   = 28.0 mol H2 Moles H2 = 28.0 mol H2 O  2 mol H O  2

 1 mol O2   = 14.0 mol O2 Moles O2 = 28.0 mol H2 O  2 mol H2 O  V = 20.0 L P = unknown PV = nRT Solving for P:

nRT PIGL of H2 = = V

T = 23.8°C + 273.15 =296.95 K n = 28.0 mol H2; 14.0 mol O2

L  atm  28.0 mol0.0821 mol 296.95 K  K

= 34.131 = 34.1 atm H2 20.0 L  L  atm  14.0 mol 0.0821 296.95 K   nRT mol  K  PIGL of O2 = = = 17.0657 = 17.1 atm O2 V 20.0 L b) V = 20.0 L T =23.8°C + 273.15 =296.95 K P = unknown n = 28.0 mol H2; 14.0 mol O2 Van der Waals constants from Table 5.4: atm  L2 L H2: a = 0.244 ; b = 0.0266 mol 2 mol atm  L2 L O2: a = 1.36 ; b = 0.0318 mol 2 mol





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5-58

2    P  n a V  nb  nRT  V 2 

PVDW =

nRT n2 a  2 V  nb V

atm  L2  2 L  atm   28.0 mol 0.244  28.0 mol0.0821 296.95 K   mol 2    mol  K   PVDW of H2 = = 34.9734 = 35.0 atm H2 20.0 L  28.0 mol 0.0266 L/mol 20.0 L2 atm  L2  2 L  atm     14.0 mol  1.36    14.0 mol0.0821 296.95 K  mol 2    mol  K   PVDW of O2 = = 16.7878 = 16.8 atm O2 20.0 L 14.0 mol 0.0318 L/mol 20.0 L2 c) The van der Waals value for hydrogen is slightly higher than the value from the ideal gas law. The van der Waals value for oxygen is slightly lower than the value from the ideal gas law. 5.127

Plan: Deviations from ideal gas behavior are due to attractive forces between particles which reduce the pressure of the real gas and due to the size of the particle which affects the volume. Compare the size and/or attractive forces between the substances. The greater the size and/or the stronger the attractive forces, the greater the deviation from ideal gas behavior. Solution: a) Xenon would show greater deviation from ideal behavior than argon since xenon is a larger atom than argon. The electron cloud of Xe is more easily distorted so intermolecular attractions are greater. Xe’s larger size also means that the volume the gas occupies becomes a greater proportion of the container’s volume at high pressures. b) Water vapor would show greater deviation from ideal behavior than neon gas since the attractive forces between water molecules are greater than the attractive forces between neon atoms. We know the attractive forces are greater for water molecules because it remains a liquid at a higher temperature than neon (water is a liquid at room temperature while neon is a gas at room temperature). c) Mercury vapor would show greater deviation from ideal behavior than radon gas since the attractive forces between mercury atoms is greater than that between radon atoms. We know that the attractive forces for mercury are greater because it is a liquid at room temperature while radon is a gas. d) Water is a liquid at room temperature; methane is a gas at room temperature. Therefore, water molecules have stronger attractive forces than methane molecules and should deviate from ideal behavior to a greater extent than methane molecules.

5.128

Plan: Use the molarity and volume of the solution to find the moles of HBr needed to make the solution. Then use the ideal gas law to find the volume of that number of moles of HBr gas at the given conditions. Solution: 1.20 mol HBr  Moles of HBr in the hydrobromic acid:  3.50 L  = 4.20 mol HBr   L V = unknown P = 0.965 atm PV = nRT Solving for V:

5.129

T = 29°C + 273.15 =302.15 K n = 4.20 mol

 L  atm  4.20 mol0.0821 302.15 K   nRT mol  K  V= = 107.9662 = 108 L HBr = P 0.965 atm  Plan: Convert the mass of each gas to moles using the molar mass. Calculate the mole fraction of CO. Use the relationship between partial pressure and mole fraction (PA = XA × Ptotal) to calculate the partial pressure of CO. Solution:

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5-59

 1 mol CO   = 0.24991 mol CO Moles CO = 7.0 g CO  28.01 g CO 

 1 mol SO2   = 0.156104 mol SO2 Moles SO2 = 10.0 g SO2   64.06 g SO2  mol CO 0.24991 mol CO = = 0.615521 mol CO + mol SO 2 0.24991 mol CO + 0.156104 mol SO 2 PCO = XCO × Ptotal = 0.615521 × 0.33 atm = 0.20312 = 0.20 atm CO

XCO =

5.130

Plan: First, balance the equation. The pressure of N2 is found by subtracting the pressure of O2 from the total pressure. The pressure of the remaining 15% of O2 is found by taking 15% of the original O2 pressure. The molar ratio between O2 and SO2 in the balanced equation is used to find the pressure of the SO2 that is produced. Since pressure is directly proportional to moles of gas, the molar ratio may be expressed as a pressure ratio. Solution: 4FeS2(s) + 11O2(g)  8SO2(g) + 2Fe2O3(s) Pressure of N2 = 1.05 atm – 0.64 atm O2 = 0.41 atm N2 Pressure of unreacted O2 = (0.64 atm O2) [(100 – 85)%/100%] = 0.096 atm O2  8 atm SO2   = 0.46545 = 0.47 atm SO2 Pressure of SO2 produced = 0.64 atm O2   11 atm O2  Total Pressure = (0.41 atm) + (0.096 atm) + (0.46545 atm) = 0.97145 = 0.97 atm total

5.131

Plan: V and T are not given, so the ideal gas law cannot be used. The total pressure of the mixture is given. Use PA = XA × Ptotal to find the mole fraction of each gas and then the mass fraction. The total mass of the two gases is 35.0 g. Solution: Ptotal = Pkrypton + Pcarbon dioxide = 0.708 atm The NaOH absorbed the CO2 leaving the Kr, thus Pkrypton = 0.250 atm. Pcarbon dioxide = Ptotal – Pkrypton = 0.708 atm – 0.250 atm = 0.458 atm Determining mole fractions: PA = XA × Ptotal PCO2 0.458 atm Carbon dioxide: X = = = 0.64689 Ptotal 0.708 atm P 0.250 atm Krypton: X = Kr = = 0.353107 Ptotal 0.708 atm

  83.80 g Kr    0.353107   mol    Relative mass fraction =   = 1.039366  0.64689 44.01 g CO2     mol  35.0 g = x g CO2 + (1.039366 x) g Kr 35.0 g = 2.039366 x Grams CO2 = x = (35.0 g)/(2.039366) = 17.16219581 = 17.2 g CO2 Grams Kr = 35.0 g – 17.162 g CO2 = 17.83780419 = 17.8 g Kr 5.132

As the car accelerates, the air in the car is pressed toward the back. The helium balloon floats on “top” of this denser air, which pushes it toward the front of the car.

5.133

Plan: Convert molecules of OH to moles and use the ideal gas law to find the pressure of this 3 number of moles of OH in 1 m of air. The mole percent is the same as the pressure percent as long as the other conditions are the same.

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5-60

Solution: 3 3  1 mol  2.51012 molecules  10 m  –15   Moles of OH =     = 4.151445 × 10 mol/L 3 23     m  6.022 10 molecules   1 L 

V = 1.00 L P = unknown PV = nRT Solving for P:

T = 22°C + 273.15 =295.15 K –15 n = 4.151445 × 10 mol

 L  atm  4.1514451015 mol 0.0821 295.15 K   nRT mol  K  Pressure of OH = P = = V 1.00 L –13 –13 =1.005970 × 10 = 1.0 × 10 atm OH 1.005970 1013 atm –11 –11 Mole percent = 100 = 1.005459 × 10 = 1.0 × 10 % 1.00 atm



5.134



Plan: Write the balanced equations. Use the ideal gas law to find the moles of SO2 gas and then use the molar ratio between SO2 and NaOH to find moles and then molarity of the NaOH solution. Solution: The balanced chemical equations are: SO2(g) + H2O(l)  H2SO3(aq) H2SO3(aq) + 2NaOH(aq)  Na2SO3(aq) + 2H2O(l) Combining these equations gives: SO2(g) + 2NaOH(aq)  Na2SO3(aq) + H2O(l) V = 0.200 L P = 745 mmHg

T = 19°C + 273.15 =292.15 K n = unknown

Converting P from mmHg to atm:

 1 atm   = 0.980263 atm P = 745 mmHg  760 mmHg 

PV = nRT Solving for n:

0.980263 atm0.200 L PV –3  = 8.17379 × 10 mol SO2 RT 0.0821 L  atm 292.15 K   mol  K   2 mol NaOH   = 0.0163476 mol NaOH Moles of NaOH = 8.17379103 mol SO2   1 mol SO2  mol NaOH 0.0163476 mol NaOH  1 mL   3  = 1.63476 = 1.63 M NaOH M NaOH = = 10 L  volume of NaOH 10.0 mL Moles of SO2 = n =

5.135

Plan: Rearrange the formula PV = (m/M )RT to solve for molar mass. The mass of the flask and water and the density of water are used to find the volume of the flask and thus the gas. The mass of the condensed liquid equals the mass of the gas in the flask. Pressure must be in units of atmospheres, volume in liters, and temperature in kelvins. Solution: Mass of water = mass of flask filled with water – mass of flask = 327.4 g – 65.347 g = 262.053 g 3   1 mL 10 L  = 0.2628415 L Volume of flask = volume of water = 262.053 g water    0.997 g water  1 mL  Mass of condensed liquid = mass of flask and condensed liquid – mass of flask = 65.739 g – 65.347 g = 0.392 g

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5-61

V = 0.2628415 L P = 101.2 kPa M = unknown Converting P from kPa to atm:

T = 99.8°C + 273.15 = 372.95 K m = 0.392 g

 1 atm    = 0.998766 atm P = 101.2 kPa  101.325 kPa 





PV = (m/M )RT Solving for molar mass, M :  L  atm  0.392 g0.0821 372.95 K   mRT mol  K  M = = 45.72166 = 45.7 g/mol  PV 0.998766 atm 0.2628415 L  5.136

Plan: Write the balanced chemical equations. Convert mass of hydride to moles and use the molar ratio from the balanced equation to find the moles of hydrogen gas produced. Use the ideal gas law to find the volume of that amount of hydrogen. Pressure must be in units of atm and temperature in kelvins. Solution: LiH(s) + H2O(l)  LiOH(aq) + H2(g) MgH2(s) + 2H2O(l)  Mg(OH)2(s) + 2H2(g) LiOH is water soluble; however, Mg(OH)2 is not water soluble. Lithium hydride LiH:  1 kg 103 g   1 mol LiH  1 mol H 2  = 57.0746 mol H Moles H2 = 1.00 lb LiH   2  7.946 g LiH 1 mol LiH    2.205 lb  1 kg   Finding the volume of H2: V = unknown P = 750 torr Converting P from torr to atm:

T = 27°C + 273.15 = 300.15 K n = 57.0746 mol  1 atm  P = 750. torr  = 0.98684 atm  760 torr 

PV = nRT Solving for V:  L  atm  57.0746 mol0.0821 300.15 K   nRT mol  K  V= = 1425.2029 = 1430 L H2 from LiH = P 0.98684 atm  Magnesium hydride, MgH2  1 kg 103 g  1 mol MgH2   2 mol H2  = 34.44852 mol H  Moles H2 = 1.00 lb MgH2   2       2.205 lb  1 kg  26.33 g MgH2  1 mol MgH2 

Finding the volume of H2: V = unknown P = 750 Torr = 0.98684 atm PV = nRT Solving for V: nRT V= = P

T = 27°C + 273.15 = 300.15 K n = 34.44852 mol

 L  atm  34.44852 mol0.0821 300.15 K   2 mol  K  = 860.20983= 8.60 × 10 L H2 from MgH2 0.98684 atm 

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5-62

5.137

5.138

Plan: Use the equation for root mean speed (urms). Molar mass values must be in units of kg/mol and temperature in kelvins. Solution:



 



 



urms Ne =

3 8.314 J /mol  K 370 K 10 3 g  kg  m 2 /s2   = 676.24788 = 676 m/s Ne    kg  J  20.18 g/mol

urms Ar =

3 8.314 J /mol  K 370 K 10 3 g  kg  m 2 /s2   = 480.6269 = 481 m/s Ar    kg  J  39.95 g/mol

urms He =

3 8.314 J /mol  K 370 K 103 g  kg  m 2 /s2  3    = 1518.356 = 1.52 × 10 m/s He     kg J    4.003 g/mol  



 



Plan: Use the ideal gas law to find the number of moles of CO2 and H2O in part (a). The molar mass is then used to convert moles to mass. Temperature must be in units of kelvins, pressure in atm, and volume in L. For part (b), use the molar ratio in the balanced equation to find the moles and then mass of C6H12O6 that produces the number of moles of CO2 exhaled during 8 h. Solution: a) V = 300 L T = 37.0°C + 273.15 = 310.15 K P = 30.0 torr n = unknown  1 atm  Converting P from torr to atm: P = 30.0 torr  = 0.0394737 atm  760 torr  PV = nRT Solving for n:

0.0394737 atm300 L PV  = 0.4650653 mol RT 0.0821 L  atm 310.15 K   mol  K   44.01 g CO2   = 20.4675 = 20.5 g CO2 Mass (g) of CO2 = 0.4650653 mol CO2   1 mol CO2  Moles of CO2 = moles of H2O = n =

18.02 g H2 O   = 8.3805 = 8.38 g H2O Mass (g) of H2O = 0.4650653 mol H2 O  1 mol H2 O  b) C6H12O6(s) + 6 O2(g)  6 CO2(g) + 6 H2O(g)  0.4650653 mol CO 2  Moles of CO2 exhaled in 8 h =  8 h  = 3.720523 mol CO2   h

1 mol C6 H12 O6 180.16 g C6 H12 O6    Mass (g) of C6H12O6 = 3.720523 mol CO2   1 mol C6 H12 O6   6 mol CO2  2

= 111.7149 = 1 × 10 g C6H12O6 (= body mass lost) (This assumes the significant figures are limited by the 8 h.) 5.139

Plan: For part (a), the number of moles of water vapor can be found using the ideal gas law. Convert moles of water to mass using the molar mass and adjust to the 1.6% water content of the kernel. For part (b), use the ideal gas law to find the volume of water vapor at the stated set of condition.

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5-63

Solution:

 75% H 2 O   = 0.1875 mL = 1.875 × 10–4 L a) Volume of water in kernel = 0.25 mL kernel 100% kernel  –4

V = 1.875 × 10 L P = 9.0 atm PV = nRT Solving for n:

T = 170°C + 273.15 = 443.15 K n = unknown

9.0 atm 1.875 104 L PV –5 = 4.638204 × 10 mol  L  atm   RT 0.0821  443.15 K   mol  K  18.02 g H O 100%   2   = 0.052238 = 0.052 g Mass (g) of = 4.638204 105 mol H 2 O    1 mol H 2 O  1.6%  b) V = unknown T = 25°C + 273.15 = 298.15 K –5 P = 1.00 atm n = 4.639775 × 10 mol PV = nRT Solving for V:  L  atm  298.15 K  4.638204 104 mol0.0821 nRT mol  K  V= = 0.00113534 L = 1.13534 mL = 1.1 mL = P 1.00 atm

Moles of H2O = n =



5.140



Plan: For part (a), find the SO2 volume in 4 GL of flue gas and take 95% of that as the volume of SO 2 removed. The ideal gas law is used to find the number of moles of SO2 in that volume. The molar ratio in the balanced equation is used to find the moles and then mass of CaSO4 produced. For part (b), use the molar ratio in the balanced equation to calculate the moles of O2 needed to produce the amount of CaSO4 found in part (a). Use the ideal gas law to obtain the volume of that amount of O2 and adjust for the mole fraction of O2 in air. Solution: –10 –7 a) Mole fraction SO2 = 1000(2 × 10 ) = 2 × 10 9 10 L   2 107  95%  = 760 L Volume of SO2 removed = 4 GL   1 GL  100%  V = 760 L P = 1.00 atm PV = nRT Solving for n:

T = 25°C + 273.15 = 298.15 K n = unknown

1.00 atm760 L PV  = 31.04814mol  RT 0.0821 L  atm 298.15 K    mol  K  The balanced chemical equations are: CaCO3(s) + SO2(g)  CaSO3(s) + CO2(g) 2CaSO3(s) + O2(g)  2CaSO4(s) 1 mol CaSO4  136.14 g CaSO 4   1 kg   Mass (kg) of CaSO4 = 31.04814 mol SO2    3  = 4.2269 = 4 kg CaSO4  1 mol SO2   1 mol CaSO4  10 g  b) Moles of O2 = 103 g  1 mol CaSO4   1 mol O2   4.2269 kg CaSO4    = 15.52407 mol O2  1 kg 136.14 g CaSO4   2 mol CaSO4  V = unknown T = 25°C + 273.15 = 298.15 K Moles of SO2 = n =





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5-64

P = 1.00 atm PV = nRT

n = 15.52407 mol

Solving for V:  L  atm  15.52407 mol0.0821 298.15 K   mol  K  = 380 L O2 1.00 atm   1  Volume of air = 380 L O 2   = 1818.1818 = 2000 L air  0.209  nRT Volume of O2 = V = = P

5.141

Plan: Use the ideal gas law to find the moles of gas occupying the tank at 85% of the 85.0 atm ranking. Then use van der Waals equation to find the pressure of this number of moles of gas. Solution: a) V = 850. L T = 298 K  80%  P = 85.0 atm  n = unknown  = 68.0 atm 100%  PV = nRT Solving for n: Moles of Cl2 = n =

68.0 atm 850. L  PV 3 3  = 2.36248 × 10 = 2.36 × 10 mol Cl2 RT 0.0821 L  atm 298 K    mol  K 

b) V = 850 L P = unknown Van der Waals constants from Table 5.4: a = 6.49

atm  L2 ; mol 2

b = 0.0562

T =298 K 3 n = 2.36248 × 10 mol Cl2

L mol

2    P  n a V  nb  nRT V 2  

PVDW =

nRT n2 a  2 V  nb V 



L  atm  298 K  2.3624810 mol Cl 0.08206 mol  K  3

PVDW =

 L  850 L  2.3624810 mol Cl 2 0.0562   mol 











L   2.3624810 mol Cl  6.49 atm mol  3

850 L

= 30.4134 = 30.4 atm c) The engineer did not completely fill the tank. She should have filled it to (80.0%/100%)(85.0 atm) = 68 atm, but only filled it to 30.4 atm. 5.142

Plan: Solve the density form of the ideal gas law for molar mass. Temperature must be converted to kelvins and pressure to atmospheres. Solution: P = 102.5 kPa T = 10.0°C + 273.15 = 283.15 K d = 1.26 g/L M = unknown  1 atm    = 1.011596 atm Converting P from kPa to atm: P = 102.5 kPa  101.325 kPa 





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5-65

d = (PM )/(RT) Rearranging to solve for molar mass: L  atm   1.26 g/L 0.0821 283.15 K   dRT mol  K  = M = = 28.9550 = 29.0 g/mol P 1.011596 atm







5.143

Plan: Use the relationship



PV PV PV n T 1 1 = 2 2 or V2 = 1 1 2 2 . R is fixed. n1T1 n2 T2 P2 n1T1

Solution: a) As the pressure on a fixed amount of gas (n is fixed) increases at constant temperature (T is fixed), the molecules move closer together, decreasing the volume. When the pressure is increased by a factor of 2, the volume decreases by a factor of 2 at constant temperature (Boyle’s law). PV T (P )(V )(1) V2 = 1 1 2 = 1 1 V2 = V1 P2T1 (2 P1 )(1) Cylinder B has half the volume of the original cylinder. b) The temperature is decreased by a factor of 2, so the volume is decreased by a factor of 2 (Charles’s law). PV T (1)(V1 )(200 K) V2 = 1 1 2 = V2 = V1 P2T1 (1)(400 K) Cylinder B has half the volume of the original cylinder. c) T1 = 100°C + 273.15 = 373.15 K T2 = 200°C + 273.15 = 473.15 K The temperature increases by a factor of 473/373 = 1.27, so the volume is increased by a factor of 1.27 (Charles’s law). PV T (1)(V1 )(473.15 K) V2 = 1 1 2 = V2 = 1.27V1 P2T1 (1)(373.15 K) None of the cylinders show a volume increase of 1.27. d) As the number of molecules of gas increases at constant pressure and temperature (P and T are fixed), the force they exert on the container increases. This results in an increase in the volume of the container. Adding 0.1 mole of gas to 0.1 mole increases the number of moles by a factor of 2, thus the volume increases by a factor of 2 (Avogadro’s law). PV n T (1)(V1 )(0.2)(1) V2 = 1 1 2 2 = V2 = 2V1 P2 n1T1 (1)(0.1)(1) Cylinder C has a volume that is twice as great as the original cylinder. e) Adding 0.1 mole of gas to 0.1 mole increases the number of moles by a factor of 2, thus increasing the volume by a factor of 2. Increasing the pressure by a factor of 2 results in the volume decreasing by a factor of . The two volume changes cancel out so that the volume does not change. PV n T (P )(V )(0.2)(1) V2 = 1 1 2 2 = 1 1 V2 = V1 P2 n1T1 (2 P1 )(0.1)(1) Cylinder D has the same volume as the original cylinder. 5.144

Plan: Use both the ideal gas law and van der Waals equation to solve for pressure. Convert mass of ammonia to moles and temperature to kelvin. Solution: Ideal gas law: V = 3.000 L T = 0°C + 273.15 =273.15 K or 400°C + 273.15 =673.15 K

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5-66

P = unknown

 1 mol NH3   = 3.0005872 mol n = 51.1 g NH3  17.03 g NH3 

PV = nRT Solving for P:

nRT PIGL of NH3 at 0°C = = V

L  atm  3.0005872 mol0.0821 mol 273.15 K  K 3.000 L

= 22.43000 = 22.4 atm

 L  atm  3.0005872 mol 0.0821 673.15 K   nRT mol  K  PIGL of NH3 at 400°C = = = 55.27643 = 55.3 atm V 3.000 L van der Waals equation: V = 3.000 L T = 0°C + 273.15 =273.15 K or 400°C + 273.15 =673.15 K  1 mol NH3   = 3.0005872 mol P = unknown n = 51.1 g NH3  17.03 g NH 





van der Waals constants from Table 5.4:

atm  L2 L ; b = 0.0371 mol 2 mol 2    P  n a V  nb  nRT V 2   a = 4.17

PVDW =

nRT n2 a  2 V  nb V

atm  L2  2 L  atm   3.0005872 mol 4.17  273.15 K 3.0005872 mol0.0821   mol 2    mol  K   PVDW of NH3 at 0°C = 3.000 L  3.0005872 mol 0.0371 L/mol 3.000 L2 = 19.10964 = 19.1 atm NH3

atm  L2  2 L  atm    3.0005872 mol0.0821 673.15 K  3.0005872 mol 4.17  mol 2    mol  K   PVDW of NH3 at 400°C = 3.000 L  3.0005872 mol 0.0371 L/mol 3.000 L2 = 53.2350 = 53.2 atm NH3 5.145

Plan: Since the mole fractions of the three gases must add to 1, the mole fraction of methane is found by subtracting the sum of the mole fractions of helium and argon from 1. Pmethane = Xmethane Ptotal is used to calculate the pressure of methane and then the ideal gas law is used to find moles of gas. Avogadro’s number is needed to convert moles of methane to molecules of methane. Solution: Xmethane = 1.00 – (Xargon + Xhelium) = 1.00 – (0.35 + 0.25) = 0.40 Pmethane = Xmethane Ptotal = (0.40)(1.75 atm) = 0.70 atm CH4 V = 6.0 L T = 45°C + 273.15 =318.15 K P = 0.70 atm n = unknown PV = nRT Solving for n:

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5-67

0.70 atm6.0 L PV  = 0.1607956 mol  RT 0.0821 L  atm 318.15 K    mol  K   6.0221023 CH molecules  4  Molecules of CH4 = 0.1607956 mol CH 4  1 mol CH4   Moles of CH4 = n =

= 9.68311 × 10 = 9.7 × 10 molecules CH4 5.146

Plan: For part (a), convert mass of glucose to moles and use the molar ratio from the balanced equation to find the moles of CO2 gas produced. Use the ideal gas law to find the volume of that amount of CO2. Pressure must be in units of atm and temperature in kelvins. For part (b), use the molar ratios in the balanced equation to calculate the moles of each gas and then use Dalton’s law of partial pressures to determine the pressure of each gas. Solution: a) C6H12O6(s) + 6O2(g)  6CO2(g) + 6H2O(g)  1 mol C6 H12 O6  6 mol CO2    = 0.666075 mol CO2 Moles CO2: 20.0 g C6 H12 O6  1 mol C H O  180.16 g C H O  6

Finding the volume of CO2: V = unknown P = 780. torr Converting P from torr to atm:

T = 37°C + 273.15 = 310.15 K n = 0.666075 mol  1 atm  P = 780 torr   = 1.0263158 atm  760 torr 





PV = nRT Solving for V: nRT V= = P

 L  atm  0.666075 mol 0.0821 310.15 K   mol  K  = 16.5256 = 16.5 L CO2 1.0263158 atm 

This solution assumes that partial pressure of O2 does not interfere with the reaction conditions.  1 mol C6 H12 O6   6 mol   b) Moles CO2 = moles O2 = 10.0 g C6 H12 O6   1 mol C H O  180.16 g C H O  6

= 0.333037 mol CO2 = mol O2 At 37°C, the vapor pressure of water is 48.8 torr. No matter how much water is produced, the partial pressure of H2O will still be 48.8 torr. The remaining pressure, 780 torr – 48.8 torr = 731.2 torr is the sum of partial pressures for O2 and CO2. Since the mole fractions of O2 and CO2 are equal, their pressures must be equal, and must be onehalf of 731.2 torr. Pwater = 48.8 torr 2 (731.2 torr)/2 = 365.6 = 3.7 × 10 torr Poxygen = Pcarbon dioxide 5.147

Plan: Use the equations for average kinetic energy and root mean speed (urms) to find these values for N2 and H2. Molar mass values must be in units of kg/mol and temperature in kelvins. Solution: The average kinetic energies are the same for any gases at the same temperature: 3 3 Average kinetic energy = 3/2RT = (3/2)(8.314 J/mol K) (273 K) = 3.40458 × 10 = 3.40 × 10 J for both gases rms speed: T = 0°C + 273 = 273 K

 28.02 g N 2  1 kg   = 0.02802 kg/mol M of N2 =    103 g  mol

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5-68

 2.016 g H 2  1 kg   = 0.002016 kg/mol M of H2 =    103 g  mol R = 8.314 J/mol K

5.148



1 J = kg m /s





urmsN2 =

3 8.314 J/mol  K 273 K  kg  m 2 /s2  2 2   = 4.92961 × 10 = 4.93 × 10 m/s N2   J   0.02802 kg/mol

urms H2 =

3 8.314 J /mol  K 273K  kg  m 2 /s2   = 1.83781 × 103 = 1.84 × 103 m/s H  2   J   0.002016 kg/mol











Plan: Use the relationship between mole fraction and partial pressure, PA = XA Ptotal, to find the mole fraction of each gas in parts (a) and (b). For parts (c) and (d), use the ideal gas law to find the moles of air in 1000 L of air at these conditions and compare the moles of each gas to the moles of air. Mass and molecules must be converted to moles. Solution: a) Assuming the total pressure is 1 atm = 760 torr. PA = XA Ptotal PBr2 0.2 torr –4 6 XBr2 = = = 2.6315789 × 10 × (10 ) = 263.15789 = 300 ppmv Unsafe Ptotal 760 torr PCO2 0.2 torr –4 6 = b) XCO2 = = 2.6315789 × 10 × (10 ) = 263.15789 = 300 ppmv Safe Ptotal 760 torr 6 (0.2 torr CO2/760 torr)(10 ) = 263.15789 = 300 ppmv CO2 Safe  1 mol Br2   = 2.5031 × 10–6 mol Br2 (unrounded) c) Moles Br2 = 0.0004 g Br2  159.80 g Br2  Finding the moles of air: V = 1000 L T = 0°C + 273 =273 K P = 1.00 atm n = unknown PV = nRT 1.00 atm1000 L PV  Moles of air = n = = 44.616 mol air (unrounded) RT 0.0821 L  atm 273 K   mol  K  6 –6 6 Concentration of Br2 = mol Br2/mol air(10 ) = [(2.5031 × 10 mol)/(44.616 mol)] (10 ) = 0.056103 = 0.06 ppmv Br2 Safe   1 mol CO2  = 0.046496 mol CO2 d) Moles CO2 = 2.81022 molecules CO2  23  6.02210 molecules CO2  6

Concentration of CO2 = mol CO2/mol air(10 ) = [(0.046496 mol)/(44.616 mol)] (10 ) = 1042.1 3 = 1.0 × 10 ppmv CO2 Safe 5.149

Plan: For part (a), use the ideal gas law to find the moles of NO in the flue gas. The moles of NO are converted to moles of NH3 using the molar ratio in the balanced equation and the moles of NH3 are converted to volume using the ideal gas law. For part (b), the moles of NO in 1 kL of flue gas is found using the ideal gas law; the molar ratio in the balanced equation is used to convert moles of NO to moles and then mass of NH 3. Solution: McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution

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5-69

a) 4NH3(g) + 4NO(g) + O2(g)  4N2(g) + 6H2O(g) Finding the moles of NO in 1.00 L of flue gas: V = 1.00 L T = 365°C + 273.15 =638.15 K –5 P = 4.5 × 10 atm n = unknown PV = nRT Solving for n: 4.5105 atm1.00 L PV –7 Moles of NO = n = = 8.5891 × 10 mol NO  RT 0.0821 L  atm 638.15 K    mol  K 

 4 mol NH   3 –7 Moles of NH3 = 8.5891107 mol NO  = 8.5891 × 10 mol NH3  4 mol NO  Volume of NH3: V = unknown T =365°C + 273.15 =638.15 K –7 P = 1.00 atm n = 8.5891 × 10 mol PV = nRT Solving for V:  L  atm  8.5891107 mol0.0821 638.15 K  nRT –5 –5 mol  K  V= = 4.50001 × 10 = 4.5 × 10 L NH3 = P 1.00 atm  b) Finding the moles of NO in 1.00 kL of flue gas: V = 1.00 kL = 1000 L T = 365°C + 273.15 =638.15 K –5 P = 4.5 × 10 atm n = unknown PV = nRT Solving for n: 5 PV 4.5000110 atm 1000 L  –4 Moles of NO = n = = 8.5891 × 10 mol NO  L  atm   RT 0.0821 638.15 K   mol  K   4 mol NH   3 –4 Moles of NH3 = 8.5891104 mol NO  = 8.5891 × 10 mol NH3  4 mol NO 

17.03 g NH  3 Mass of NH3 = 8.5891104 mol NH3   = 0.014627 = 0.015 g NH3  1 mol NH3  5.150

Plan: Use Graham’s law to compare effusion rates. Solution: molar mass Xe 131.3 g/mol 2.55077 Rate Ne = = = enrichment factor (unrounded) molar mass Ne 20.18 g/mol 1 Rate Xe Thus XNe =

5.151

moles of Ne 2.55077 mol = 0.71837 = 0.7184 = moles of Ne + moles of Xe 2.55077 mol + 1 mol

Plan: To find the number of steps through the membrane, calculate the molar masses to find the ratio of effusion rates. This ratio is the enrichment factor for each step. Solution: Rate 235 UF 352.04 g/mol molar mass 238 UF6 6 = = 235 Rate 238 UF molar mass UF6 349.03 g/mol 6

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5-70

= 1.004302694 enrichment factor 235 Therefore, the abundance of UF6 after one membrane is 0.72% × 1.004302694 235 N Abundance of UF6 after “N” membranes = 0.72% × (1.004302694) 235 N Desired abundance of UF6 = 3.0% = 0.72% × (1.004302694) Solving for N: N 3.0% = 0.72% × (1.004302694) N 4.16667 = (1.004302694) N ln 4.16667 = ln (1.004302694) ln 4.16667 = N × ln (1.004302694) N = (ln 4.16667)/(ln 1.004302694) N = 1.4271164/0.004293464 = 332.39277 = 332 steps 5.152

Plan: Use van der Waals equation to calculate the pressure of the gas at the given conditions. Solution: V = 22.414 L T = 273.15 K P = unknown n = 1.000 mol Van der Waals constants from Table 5.4: atm  L2 L a = 1.45 ; b = 0.0395 mol 2 mol 2    P  n a V  nb  nRT  V 2  PVDW =

PVDW =

nRT n2 a  2 V  nb V L  atm  1.000 mol CO0.08206 mol 273.15 K  K L   22.414 L  1.000 mol CO 0.0395  mol 









L   1.000 mol CO 1.45 atm mol  2



22.414 L

= 0.998909977 = 0.9989 atm 5.153

Plan: The amount of each gas that leaks from the balloon is proportional to its effusion rate. Using 35% as the rate for H2, the rate for O2 can be determined from Graham’s law. Solution: molar mass H 2 2.016 g/mol rate O2 Rate O 2 = = = molar mass O2 32.00 g/mol Rate H 2 35 0.250998008 =

rate O2

35 Rate O2 = 8.78493 Amount of H2 that leaks = 35%; 100–35 = 65% H2 remains Amount of O2 that leaks = 8.78493%; 100–8.78493 = 91.21507% O2 remains O2 91.21507 = = 1.40331 = 1.4 H2 65

5.154

Plan: For part (a), put together the various combinations of the two isotopes of Cl with P and add the masses. Multiply the abundances of the isotopes in each combination to find the most abundant for part (b). For part (c), use Graham’s law to find the effusion rates. Solution: a) Options for PCl3: All values are g/mol P First Cl Second Cl Third Cl Total McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution

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5-71

31 35 35 35 136 31 37 35 35 138 31 37 37 35 140 31 37 37 37 142 35 37 b) The fraction abundances are Cl = 75%/100% = 0.75, and Cl = 25%/100% = 0.25. The relative amount of each mass comes from the product of the relative abundances of each Cl isotope. Mass 136 = (0.75) (0.75) (0.75) = 0.421875 = 0.42 (most abundant) Mass 138 = (0.25) (0.75) (0.75) = 0.140625 = 0.14 Mass 140 = (0.25) (0.25) (0.75) = 0.046875 = 0.047 Mass 142 = (0.25) (0.25) (0.25) = 0.015625 = 0.016 c)

Rate P 37 Cl3 = Rate P 35 Cl3

molar mass P 35 Cl 3 37

molar mass P Cl3

136 g/mol 142 g/mol

= 0.978645 = 0.979 5.155

Plan: Use the combined gas law for parts (a) and (b). For part (c), use the ideal gas law to find the moless of air represented by the pressure decrease and convert moles to mass using molar mass. Solution: PV PV a) 1 1 = 2 2 T1 T2

35.0 psi318 K  P1 V1T2 PT = 1 2 = = 37.7288 = 37.7 psi V2 T1 T1 295 K b) New tire volume = V2 = V1(102%/100%) = 1.02V1 PV PV 1 1 = 2 2 T1 T2 P2 =

P2 =

35.0 psiV1 318 K  PV PT 1 1T2 = 36.9890 = 37.0 psi = 1 2 = V2T1 T1 1.02V1 295 K 

 1 atm   = 0.135306 atm c) Pressure decrease = 36.9890 – 35.0 psi = 1.989 psi; 1.989 psi 14.7 psi  V = 218 L T = 295 K P = 0.135306 atm n = unknown PV = nRT Solving for n: 0.135306 atm 218 L PV Moles of air lost = n =  = 1.217891 mol  RT 0.0821 L  atm 295 K    mol  K   28.8 g 1 min   Time = 1.217891 mol = 23.384 = 23 min  1 mol  1.5  5.156

Plan: Rearrange the ideal gas law to calculate the density of O2 and O3 from their molar masses. Temperature must be converted to kelvins and pressure to atm. Solution: P = 760 torr = 1.00 atm T = 0°C + 273 = 273 K M of O2 = 32.00 g/mol d = unknown M of O3 = 48.00 g/mol PV = nRT Rearranging to solve for density:

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5-72

1.00 atm 32.00 g/mol PM  = 1.42772 = 1.43 g O2/Ls  L  atm  RT 0.0821  273 K   mol  K  1.00 atm 48.00 g/mol PM d of O3 =  = 2.141585576 = 2.14 g O3/L  L  atm  RT 0.0821 273 K    mol  K  b) dozone/doxygen = (2.141585576)/(1.42772) = 1.5 The ratio of the density is the same as the ratio of the molar masses. d of O2 =

5.157

Plan: Since the amounts of two reactants are given, this is a limiting reactant problem. Write the balanced equation and use molar ratios to find the number of moles of IF7 produced by each reactant. The mass of I2 is converted to moles using its molar mass and the moles of F2 is found using the ideal gas law. The smaller number of moles of product indicates the limiting reagent. Determine the moles of excess reactant gas and the moles of product gas and use the ideal gas law to solve for the total pressure. Solution: Moles of F2: V = 2.50 L T = 250 K P = 350 torr n = unknown  1 atm  Converting P from torr to atm: P = 350 torr   = 0.460526315 atm  760 torr  PV = nRT Solving for n: PV 0.460526315 atm 2.50 L  n=  = 0.056093339 mol F2  L  atm  RT 0.0821 250. K    mol  K  7F2(g) + I2(s)  2IF7(g)

 2 mol IF7    = 0.016026668 mol IF (unrounded) Moles IF7 from F2 = 0.056093339 mol F2  7  7 mol F2 

 1 mol I2   2 mol IF7  = 0.019700551 mol IF (unrounded) Moles IF7 from I2 = 2.50 g I2  7   1 mol I2   253.8 g I 2  F2 is limiting. All of the F2 is consumed. Mole I2 remaining = original amount of moles of I2 – number of I2 moles reacting with F2  1 mol I 2   1 mol I2   = 1.83694 × 10–3 mol I  – 0.056093339 mol F2  Mole I2 remaining = 2.50 g I2  2  253.8 g I2   7 mol F2  –3 Total moles of gas = (0 mol F2) + (0.016026668 mol IF7) + (1.83694 × 10 mol I2) = 0.0178636 mol gas V = 2.50 L T = 550. K P = unknown n = 0.0178636 mol PV = nRT Solving for P:  L  atm  0.0178636 mol0.0821 550. K   nRT mol  K  P (atm) = = = 0.322652 atm V 2.50 L  760 torr  P (torr) = 0.322652 atm  = 245.21552 = 245 torr  1 atm  –3 Piodine (torr) = Xiodine Ptotal = [(1.83694 × 10 mol I2)/(0.0178636 mol)] (245.21552 torr) = 25.215869 = 25.2 torr Copyright

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5-73

CHAPTER 6 THERMOCHEMISTRY: ENERGY FLOW AND CHEMICAL CHANGE FOLLOW–UP PROBLEMS 6.1A

Plan: The system is the liquid. Since the system absorbs heat from the surroundings, the system gains heat and q is positive. Because the system does work, w is negative. Use the equation E = q + w to calculate E. Convert E to kJ. Solution: E(kJ) = q + w = +13.5 kJ + –1.8 kJ = 11.7 kJ   4 E (J) = 11.7 kJ  = 1.17 × 10 J  

6.1B

Plan: The system is the reactant and products of the reaction. Since heat is absorbed by the surroundings, the system releases heat and q is negative. Because work is done on the system, w is positive. Use the equation E = q + w to calculate E. Both kcal and Btu must be converted to kJ. Solution:  4.184 kJ  q = 26.0 kcal  = –108.784 kJ  1 kcal 

1.055 kJ  w = +15.0 Btu  = 15.825 kJ  1 Btu  E = q + w = –108.784 kJ + 15.825 kJ = –92.959 = –93 kJ 6.2A

Plan: Convert the pressure from torr to atm units. Subtract the initial V from the final V Use w = –P V to calculate w in atm L. Convert the answer from atm L to J. Solution: Vinitial = 5.68 L

Vfinal = 2.35 L P = 732 torr

 Converting P from torr to atm: (732 torr)  

 = 0.96315 atm 

w (atm L) = –P V = –(0.96315 atm)(2.35 L – 5.68 L) = 3.2073 = 3.21 atm L

 101.3 J  w (J) = (3.2073 atm L)  = 324.901 = 325 J 1 atm  L  6.2B

Plan: Subtract the initial V from the final V answer from atm L to J. Solution: Vinitial = 10.5 L

V. Use w = –P V to calculate w in atm L. Convert the

Vfinal = 16.3 L P = 5.5 atm w (atm L) = –P V = –(5.5 atm)(16.3 L – 10.5 L) = –31.9 atm L Copyright

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6-1

 101.3 J  3 3 w (J) = (–31.9 atm L)  = –3.2315 × 10 = –3.2 × 10 J 1 atm  L  6.3A

Plan: Since heat is released in this reaction, the reaction is exothermic (H< 0) and the reactants are above the products in an enthalpy diagram. Solution:

H = –5.72 × 10 kJ 3

6.3B

Plan: Since heat is absorbed in this reaction, the reaction is endothermic (H>0) and the reactants are below the products in an enthalpy diagram. Solution:

6.4A

Plan: Heat is added to the aluminum foil, so q will be positive. The heat is calculated using the equation q = c × mass × T. Table 6.2 lists the specific heat of aluminum as 0.900 J/g K. Solution:    = 357 K T = 375°C – 18°C = (357°C)    q = c × mass × T = (0.900 J/gK) (7.65 g) (357 K) = 2.45795 × 10 = 2.46 × 10 J 3

6.4B

Plan: Heat is transferred away from the ethylene glycol as it cools so q will be negative (–177 kJ in this case). The mass is calculated using the equation q = c × mass × T, rearranged to solve for mass. Table 6.2 lists the specific heat of ethylene glycol as 2.42 J/gK. The mass of ethylene glycol is converted to volume in mL by using the density. We will convert the volume from milliliters to liters as a final step. Solution:    = –12.0 K T = 25.0°C – 37.0°C = (–12.0°C)    Mass =

 1000 J  q 177 kJ  = 6.0950 × 103  =    1 kJ  c T 2.42 J/g  K 12.0 K  

 1 mL  1 L  3  = 5.49102 = 5.49 L  Volume (L) = 6.0950 × 10 g ×  1.11 g 1000 mL 

6.5A

Plan: The heat absorbed by the water can be calculated with the equation c × mass × T; the heat absorbed by the water equals the heat lost by the hot metal. Since the mass and temperature change of the metal is known, the specific heat capacity can be calculated and used to identify the metal.

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6-2

Solution:  Tfinal – Tinitial = 27.25°C – 25.55°C = (1.70°C)  

TH2 O

 Tfinal – Tinitial = 27.25°C – 65.00°C = (–37.75°C)   qH2 O =  qmetal

Tmetal

  = 1.70 K    = –37.75 K 

cH O  mass H O  TH O = cmetal  massmetal  Tmetal 2

–cmetal = 

cH O  massH O TH O 2

massmetal Tmetal

=

 4.184 J/g  K 25.00 g 1.70 K  = 0.386738 = 0.387 J/g K 12.18 g37.75 K 

From Table 6.2, the metal with this value of specific heat is copper. 6.5B

Plan: The heat absorbed by the titanium metal can be calculated with the equation c × mass × T; the heat absorbed by the metal equals the heat lost by the water. Since the mass and temperature change of the water is known, along with the specific heat and final temperature of the titanium, initial temperature of the metal can be calculated. Solution:    = –0.70 K TH2 O Tfinal – Tinitial = 49.30°C – 50.00°C = (–0.70°C)    o

Converting Tfinal from C to K = 49.30 C + 273.15 = 322.45 K

Tmetal

Tfinal – Tinitial = 322.45 K – Tinitial

qH2 O = qmetal cH O  mass H O  TH O = cmetal  massmetal  Tmetal 2

Tmetal = 

cH O  mass H O TH O 2

massmetal  cmetal

=

 4.184 J/g  K 75.0 g-0.70 K  = 29.0187 = 29.0 K 33.2 g0.228 J/g  K 

Tmetal =29.0187 K = 322.45 K –Tinitial (using unrounded numbers to avoid rounding errors) o

Tinitial = 293.4313 K – 273.15 = 20.2813 = 20.3 C 6.6A

Plan: First write the balanced molecular, total ionic, and net ionic equations for the acid-base reaction. To find qsoln, we use the equation q = c × mass × T, so we need the mass of solution, the change in temperature, and the specific heat capacity. We know the solutions’ volumes (25.0 mL and 50.0 mL), so we find their masses with the given density (1.00 g/mL). Then, to find qsoln, we multiply the total mass by the given c (4.184 J/g K) and the change in T, which we find from Tfinal – Tinitial. The heat of reaction (qrxn) is the negative of the heat of solution (qsoln). Solution: a) The balanced molecular equation is: HNO3(aq) + KOH(aq)  KNO3(aq) + H2O(l) + – + – + – The total ionic equation is: H (aq) + NO3 (aq) + K (aq) + OH (aq)  K (aq) + NO3 (aq) + H2O(l) + – The net ionic equation is: H (aq) + OH (aq)  H2O(l) b) Total mass (g) of solution = (25.0 mL + 50.0 mL) × 1.00 g/mL = 75.0 g   o o o  = 5.15 K T = 27.05 C – 21.50 C = (5.15 C)    qsoln(J) = csoln × masssoln × Tsoln = (4.184 J/gK)(75.0 g)(5.15 K) = 1616.07 = 1620 J

 qsoln(kJ) = 1616.07 J  

  = 1.61607 = 1.62 kJ 

qrxn = –qsoln

qrxn= –1.62 kJ

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6-3

6.6B

Plan: Write a balanced equation. Multiply the volume by the molarity of each reactant solution to find moles of each reactant. Use the molar ratios in the balanced reaction to find the moles of water produced from each reactant; the smaller amount gives the limiting reactant and the actual moles of water produced. Divide the heat evolved by the moles of water produced to obtain the enthalpy in kJ/mol. Since qsoln is positive, the solution absorbed heat that was released by the reaction; qrxn H are negative. Solution: Ba(OH)2(aq) + 2HCl(aq) O(l) + BaCl2(aq) 2  10 3 L  0.500 mol Ba(OH)2   = 0.0250 mol Ba(OH)2  Moles of Ba(OH)2 = 50.0 mL  1 mL   1L

103 L  0.500 mol HCl   = 0.0250 mol HCl  Moles of HCl = 50.0 mL   1 mL   1L  2 mol H 2 O   = 0.0500 mol H2O Moles of H2O from Ba(OH)2 = 0.0250 mol Ba(OH)2  1 mol Ba(OH)2 

 2 mol H2 O  Moles of H2O from HCl = 0.0250 mol HCl  = 0.0250 mol H2O  2 mol HCl  HCl is the limiting reactant; 0.0250 mol of H2O is produced. q 1.386 kJ = H (kJ/mol) = = –55.44 = –55.4 kJ/mol moles of H2 O produced 0.0250 mol 6.7A

Plan: The bomb calorimeter gains heat from the combustion of graphite, so –qgraphite = qcalorimeter. Convert the mass of graphite from grams to moles and use the given kJ/mol to find qgraphite. The heat lost by graphite equals the heat gained by the calorimeter, or T multiplied by Ccalorimeter. Solution:  1 mol C   = 0.0720233 mol C Moles of graphite = 0.8650 g C 12.01 g C 

 393.5 kJ  qgraphite= 0.0720233 mol C  = –28.3412 kJ/mol  1 mol C  – qgraphite = Ccalorimeter Tcalorimeter 28.3412 kJ/mol = Ccalorimeter(3.116 K) Ccalorimeter = 9.09537 = 9.10 kJ/K 6.7B

Plan: The bomb calorimeter gains heat from the combustion of the dessert, so –qrxn= qcalorimeter. The amount of heat lost by the combustion of the dessert equals the amount of heat gained by the calorimeter, or T multiplied by Ccalorimeter. Convert the heat gained by the calorimeter to Calories to validate or invalidate the manufacturer’s claim. Solution:   o o o  = 4.937 K T = 26.799 C – 21.862 C = 4.937 C   

  –qrxn = qcalorimeter = (ccalorimeter × T) =   (4.937 K) = 40.24148 kJ  1 K   1 kcal  1 Calorie   q(Calories) = 40.24148 kJ    = 9.61795 = 9.618 Calories < 10 Calories  1 kcal   4.184 kJ  Yes, the claim is correct. Copyright

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6-4

6.8A

Plan: To find the heat required, write a balanced thermochemical equation and use appropriate molar ratios to solve for the required heat. Solution: C2H4(g) + H2(g)  C2H6(g) H = –137 kJ 103 g   1 mol C 2 H 6  137 kJ   4 4 Heat (kJ) = 15.0 kg C 2 H 6     = –6.83405 × 10 = –6.83 × 10 kJ  30.07 g C 2 H 6 1 mol C 2 H 6   1 kg 





6.8B

Plan: Write a balanced thermochemical equation and use appropriate molar ratios to solve for the required heat. Solution: N2(g) + O2(g) 2NO(g) H = +180.58 kJ    3   7 7   Heat (kJ) = (3.50 t NO)    = 1.05303 × 10 = 1.05 × 10 kJ         

6.9A

Plan: Manipulate the two equations so that their sum will result in the overall equation. The first equation remains the same, but the second equation must be reversed and its coefficients must be multiplied by 1/2. For the second equation, then, the sign of H must be changed and its value must be multiplied by 1/2. Solution: Equation 1 remains the same: CO(g) + 12 O2(g)  CO2(g)

H = –283.0 kJ

Reverse equation 2 and multiply by 12 : NO(g)  12 N2(g) + 12 O2(g)

H = – 12 (180.6 kJ)

= –90.3kJ CO(g) + O 2 ( g) + NO(g)  CO2(g) + 12 N2(g) + 12 O 2 ( g) 1 2

CO(g) + NO(g)  CO2(g) + 12 N 2 ( g) 6.9B

Plan: Manipulate the three equations so that their sum will result in the overall equation. Reverse the first equation (and change the sign of H) and multiply the coefficients (and H) by 1/2. Multiply the coefficients of the second equation (and H) by 1/2. Reverse the third equation (and change the sign of H). Solution: NH3(g) 1/2N2(g) + 3/2H2(g) H = –1/2(–91.8 kJ)= 45.9 kJ 1/2N2(g) + 21/2H2(g) + 1/2Cl2(g)  NH4Cl(s)

H = 1/2(–628.8 kJ)= –314.4 kJ

NH4Cl(s) NH3(g) + HCl(g)

H = –(–176.2 kJ) = 176.2 kJ

Total: 1/2H2(g) + 1/2Cl2(g)  HCl(g) 6.10A

H = –373.3 kJ

H = –92.3 kJ

Plan: Write the elements as reactants (each in its standard state), and place one mole of the substance formed on the product side. Balance the equation with the following differences from “normal” balancing — only one mole of the desired product can be on the right-hand side of the arrow (and nothing else), and fractional coefficients are allowed on the reactant side. The values for the standard heats of formation ( H f ) may be found in the appendix. Solution: a) C(graphite) + 2H2(g) + 1/2O2(g)  CH3OH(l) H f = –238.6 kJ/mol b) Ca(s) + 1/2O2(g)  CaO(s)

H f = –635.1 kJ/mol

c) C(graphite) + 1/4S8(rhombic)  CS2(l)

H f = 87.9 kJ/mol

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6-5

6.10B

6.11A

a) C(graphite) + 1/2H2(g) + 3/2Cl2(g)  CHCl3(l)

H f = –132 kJ/mol

b) 1/2N2(g) + 2H2(g)+ 1/2Cl2(g)  NH4Cl(s)

H f = –314.4 kJ/mol

c) Pb(s) + 1/8S8(rhombic) + 2O2(g)  PbSO4(s)

H f = –918.39 kJ/mol

   Plan: Look up H f values from the appendix and use the equation H rxn = m H f(products) – n H f(reactants) to  solve for H rxn .

Solution:    H rxn = m H f(products) – n H f(reactants)

= {4 H f [H3PO4(l)]} – {1 H f [P4O10(s)] + 6 H f [H2O(l)]} = (4 mol)( –1271.7 kJ/mol) – [(1 mol)( –2984 kJ/mol) + (6 mol)(–285.840 kJ/mol)] = –5086.8 kJ – [–2984 kJ + –1714.8 kJ] = –388 kJ 6.11B

 Plan: Apply the H rxn to this reaction, substitute given values, and solve for the H f (CH3OH). Solution:    H rxn = m H f(products) – n H f(reactants)  H rxn = {1 H f [CO2(g)] + 2 H f [H2O(g)]} – {1 H f [CH3OH(l)] + 3/2 H f [O2(g)]}

–638.6 kJ = [(1 mol)(–393.5 kJ/mol) + (2 mol)(–241.826 kJ/mol)] – [ H f [CH3OH(l)] + (3/2 mol)(0 kJ/mol)]  f

–638.6 kJ = (–877.152 kJ) – H [CH3OH(l)]

H f [CH3OH(l)] = –238.552 = –238.6 kJ CHEMICAL CONNECTIONS BOXED READING PROBLEMS B6.1

Plan: Convert the given mass in kg to g, divide by the molar mass to obtain moles, and convert moles to kJ of energy. Sodium sulfate decahydrate will transfer 354 kJ/mol. Solution: 10 3 g   354 kJ  1 mol Na 2 SO 4  10H 2 O     Heat (kJ) = 500.0 kg Na 2 SO 4  10H 2 O     1 kg   322.20 g Na 2 SO 4  10H 2 O  1 mol Na 2 SO 4  10H 2 O 





= –5.4935 × 10 = –5.49 × 10 kJ B6.2

Plan: Three reactions are given. Equation 1 must be multiplied by 2, and then the reactions can be added, canceling   substances that appear on both sides of the arrow. Add the H rxn values for the three reactions to get the H rxn for    the overall gasification reaction of 2 moles of coal. Use the relationship H rxn = m H f(products) – n H f(reactants) to  find the heat of combustion of 1 mole of methane. Then find the H rxn for the gasification of 1.00 kg of coal and  H rxn for the combustion of the methane produced from 1.00 kg of coal and sum these values. Solution:  a) 1) 2C(coal) + 2H2O(g)  2CO(g) + 2H2(g) H rxn = 2(129.7 kJ) 2) CO(g) + H2O(g) CO2(g) + H2(g)

 H rxn = –41 kJ

3) CO(g) + 3H2(g) CH4(g) + H2O(g)

 H rxn = –206 kJ

2C(coal) + 2H2O(g)  CH4(g) + CO2(g) Copyright

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6-6

b) The total may be determined by doubling the value for equation 1 and adding to the other two values.  H rxn = 2(129.7 kJ) + (–41 kJ) + (–206 kJ) = 12.4 = 12 kJ c) Calculating the heat of combustion of CH4: CH4(g) + 2O2(g) (g) + 2H2O(g) 2    H rxn = m H f(products) – n H f(reactants)  H rxn = [(1 mol CO2)( H f of CO2) + (2 mol H2O)( H f of H2O)]

– [(1 mol CH4)( H f of CH4) + (2 mol O2)( H f of O2)]  H rxn = [(1 mol)(–395.5 kJ/mol) + (2 mol)(–241.826 kJ/mol)]

– [(1 mol)(–74.87 kJ/mol) + (2 mol)(0.0 kJ/mol)]  rxn

H = –804.282 kJ/mol CH4 Total heat for gasification of 1.00 kg coal: 10 3 g   1 mol coal  12.4 kJ   H° = 1.00 kg coal     = 516.667 kJ 12.00 g coal  2 mol coal   1 kg 





Total heat from burning the methane formed from 1.00 kg of coal: 10 3 g   1 mol coal  1 mol CH 4   804.282 kJ   H° = 1.00 kg coal      = –33511.75 kJ 12.00 g coal   2 mol coal   1 mol CH 4   1 kg 





Total heat = 516.667 kJ + (–33511.75 kJ) = –32995.083 = –3.30 × 10 kJ END–OF–CHAPTER PROBLEMS 6.1

The sign of the energy transfer is defined from the perspective of the system. Entering the system is positive, and leaving the system is negative.

6.2

No, an increase in temperature means that heat has been transferred to the surroundings, which makes q negative.

6.3

E = q + w = w, since q = 0. Thus, the change in work equals the change in internal energy.

6.4

Plan: Remember that an increase in internal energy is a result of the system (body) gaining heat or having work done on it and a decrease in internal energy is a result of the system (body) losing heat or doing work. Solution: The internal energy of the body is the sum of the cellular and molecular activities occurring from skin level inward. The body’s internal energy can be increased by adding food, which adds energy to the body through the breaking of bonds in the food. The body’s internal energy can also be increased through addition of work and heat, like the rubbing of another person’s warm hands on the body’s cold hands. The body can lose energy if it performs work, like pushing a lawnmower, and can lose energy by losing heat to a cold room.

6.5

a) electric heater e) battery (voltaic cell)

6.6

Plan: Use the law of conservation of energy. Solution: The amount of the change in internal energy in the two cases is the same. By the law of energy conservation, the change in energy of the universe is zero. This requires that the change in energy of the system (heater or air conditioner) equals an opposite change in energy of the surroundings (room air). Since both systems consume the same amount of electrical energy, the change in energy of the heater equals that of the air conditioner.

b) sound amplifier

c) light bulb

d) automobile alternator

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6-7

6.7

Heat energy; sound energy  Kinetic energy  Potential energy  Mechanical energy  Chemical energy

(impact) (falling text) (raised text) (raising of text) (biological process to move muscles)

6.8

The system does work and thus its internal energy is decreased. This means the sign will be negative.

6.9

Plan: The change in a system’s energy is E = q + w. If the system receives heat, then its qfinal is greater than

 

qinitial so q is positive. Since the system performs work, its wfinal<winitial so w is negative. Solution: E = q + w E = (+425 J) + (–425 J) = 0 J

6.10

q + w = –255 cal + (–428 cal) = –683 cal

6.11

Plan: The change in a system’s energy is E = q + w. A system that releases thermal energy has a negative value for q and a system that has work done on it has a positive value for work. Convert work in calories to work in joules. Solution:  4.184 J  Work (J) = 530 cal = 2217.52 J  1 cal  E = q + w = –675 J + 2217.52 J = 1542.52 = 1.54 × 10 J 3

6.12

6.13

103 cal  4.184 J  103 J    3    +  0.247 kcal  E = q + w = 0.615 kJ    1 cal  = 1648.4 = 1.65 × 10 J  1 kJ    1 kcal   









Plan: Convert 6.6 × 10 J to the other units using conversion factors. Solution: 10 C(s) + O2(g)  CO2(g) + 6.6 × 10 J (2.0 tons)  1 kJ   10 7 a) E (kJ) = (6.6 × 10 J)  3  = 6.6 × 10 kJ 10 J   1 cal   1 kcal   10 7 7 b) E (kcal) = (6.6 × 10 J)   3  = 1.577 × 10 = 1.6 × 10 kcal  4.184 J   10 cal   1 Btu   10  = 6.256 × 107 = 6.3 × 107 Btu c) E (Btu) = (6.6 × 10 J)  1055 J 

6.14

CaCO3(s) + 9.0 × 10 kJ  CaO(s) + CO2(g) 6

(5.0 tons) Copyright

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6-8

103 J    = 9.0 × 109 J a) E (J) = (9.0 × 10 kJ)   1 kJ  6

103 J   1 cal   6 9 9 b) E (cal) = (9.0 × 10 kJ)    = 2.15105 × 10 = 2.2 × 10 cal  4.184 J   1 kJ  103 J  1 Btu   6 6 6   c) E (Btu) = (9.0 × 10 kJ)  1055 J  = 8.5308 × 10 = 8.5 × 10 Btu  1 kJ  

6.15

Plan: Convert the initial volume from mL to L. Subtract the initial V from the final V atm L. Convert the answer from atm L to J. Solution: Vinitial = 922 mL

V. Calculate w in

Vfinal = 1.14 L P = 2.33 atm

 Converting Vinitialfrom mL to L: (922 mL)  

 = 0.922 L 

w (atm L) = –P V = –(2.33 atm)(1.14 L – 0.922 L) = –0.50794 atm L

  w (J) = (–0.50794 atm L)  = 51.4543 = –51 J 1 atm  L  6.16

Plan: Convert the pressure from mmHg to atm. Subtract the initial V from the final V atm L. Convert the answer from atm L to kJ. Solution: Vinitial = 0.88 L Vfinal = 0.63 L P = 2660 mmHg    = 3.50 atm Converting P from mmHg to atm: (2660 mmHg)   

V. Calculate w in

w (atm L) = –P V = –(3.50 atm)(0.63 L – 0.88 L) = 0.875 atm L

    w (J) = (0.875 atm L)  103 L  = 0.0886375 = 0.089 kJ 1 atm  L 

6.17

 10 3 cal   4.184 J   3 7 7 E (J) = (4.1 × 10 calorie)    = 1.7154 × 10 = 1.7 × 10 J  1 cal  1 Calorie   10 3 cal   4.184 J   1 kJ   3 4 4 E (kJ) = (4.1 × 10 calorie)    3  = 1.7154 × 10 = 1.7 × 10 kJ 1 Calorie   1 cal  10 J 

6.18

Plan: 1.0 lb of body fat is equivalent to about 4.1 × 10 calories. Convert calories to kJ with the appropriate conversion factors. Solution:

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6-9

3  4.110 3 Cal  10 cal   4.184 J   1 kJ  h   Time = 1.0 lb     3   = 8.79713 = 8.8 h  1.0 lb   1 Cal   1 cal  10 J 1950 kJ 





6.19

Since many reactions are performed in an open flask, the reaction proceeds at constant pressure. The determination of H (constant pressure conditions) requires a measurement of heat only, whereas E requires measurement of heat and PV work.

6.20

The hot pack is releasing (producing) heat, thus H is negative, and the process is exothermic.

6.21

Plan: An exothermic process releases heat and an endothermic process absorbs heat. Solution: a) Exothermic, the system (water) is releasing heat in changing from liquid to solid. b) Endothermic, the system (water) is absorbing heat in changing from liquid to gas. c) Exothermic, the process of digestion breaks down food and releases energy. d) Exothermic, heat is released as a person runs and muscles perform work. e) Endothermic, heat is absorbed as food calories are converted to body tissue. f) Endothermic, the wood being chopped absorbs heat (and work). g) Exothermic, the furnace releases heat from fuel combustion. Alternatively, if the system is defined as the air in the house, the change is endothermic since the air’s temperature is increasing by the input of heat energy from the furnace.

6.22

The internal energy of a substance is the sum of kinetic (EK) and potential (EP) terms. EK (total) = EK (translational) + EK (rotational) + EK (vibrational) EP = EP (atom) + EP (bonds) EP (atom) has nuclear, electronic, positional, magnetic, electrical, etc., components.

6.23

H = E + PV (constant P) a) H<E, PV is negative. b) H = E, a fixed volume means PV = 0. c) H>E, PV is positive for the transformation of solid to gas.

6.24

Plan: An exothermic reaction releases heat, so the reactants have greater H (Hinitial) than the products (Hfinal). H = Hfinal – Hinitial< 0. Solution:

6.25

Plan: An endothermic reaction releases heat, so the reactants have lower H (Hinitial) than the products (Hfinal).

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6-10

6.26

Plan: Combustion of hydrocarbons and related compounds requires oxygen (and a heat catalyst) to yield carbon dioxide gas, water vapor, and heat. Combustion reactions are exothermic. The freezing of liquid water is an exothermic process as heat is removed from the water in the conversion from liquid to solid. An exothermic reaction or process releases heat, so the reactants have greater H (Hinitial) than the products (Hfinal). Solution: a) Combustion of ethane: 2C2H6(g) + 7O2(g)  4CO2(g) + 6H2O(g) + heat

b) Freezing of water: H2O(l)  H2O(s) + heat

6.27

a) Na(s) + 1/2Cl2(g)  NaCl(s) + heat

b) C6H6(l) + heat  C6H6(g)

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6-11

6.28

Plan: Combustion of hydrocarbons and related compounds requires oxygen (and a heat catalyst) to yield carbon dioxide gas, water vapor, and heat. Combustion reactions are exothermic. An exothermic reaction releases heat, so the reactants have greater H (Hinitial) than the products (Hfinal). If heat is absorbed, the reaction is endothermic and the products have greater H (Hfinal) than the reactants (Hinitial). Solution: a) 2CH3OH(l) + 3O2(g)  2CO2(g) + 4H2O(g) + heat

b) Nitrogen dioxide, NO2, forms from N2 and O2. 1/2N2(g) + O2(g) + heat  NO2(g)

6.29

a) CO2(s) + heat  CO2(g)

b) SO2(g) + 1/2O2(g)  SO3(g) + heat

6.30

Plan: Recall that qsys is positive if heat is absorbed by the system (endothermic) and negative if heat is released by the system (exothermic). Since E = q + w, the work must be considered in addition to qsys Esys. Solution: a) This is a phase change from the solid phase to the gas phase. Heat is absorbed by the system so qsys is positive (+).

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6-12

b) The system is expanding in volume as more moles of gas exist after the phase change than were present before the phase change. So the system has done work of expansion and w Esys = q + w. Since q is positive and w Esys cannot be predicted. It will be positive if q>w and negative if q<w. Euniv = 0. If the system loses energy, the surroundings gain an equal amount of energy. The sum of the energy of the system and the energy of the surroundings remains constant. 6.31

a) There is a volume decrease; Vfinal<Vinitial V is negative. Since wsys = –P V, w is positive, +. b) Hsys is – as heat has been removed from the system to liquefy the gas. c) Esys= q + w. Since q is negative and w Esys Esurr Esys will be Esurr will be negative if w>q Esys Esurr will be positive if w<q.

6.32

The molar heat capacity of a substance is larger than its specific heat capacity. The specific heat capacity of a substance is the quantity of heat required to change the temperature of 1 g of a substance by 1 K while the molar heat capacity is the quantity of heat required to change the temperature of 1 mole of a substance by 1 K. The specific heat capacity of a substance is multiplied by its molar mass to obtain the molar heat capacity.

6.33

To determine the specific heat capacity of a substance, you need its mass, the heat added (or lost), and the change in temperature.

6.34

Specific heat capacity is an intensive property; it is defined on a per gram basis. The specific heat capacity of a particular substance has the same value, regardless of the amount of substance present.

6.35

Specific heat capacity is the quantity of heat required to raise 1 g of a substance by 1 K. Molar heat capacity is the quantity of heat required to raise 1 mole of substance by 1 K. Heat capacity is also the quantity of heat required for a 1 K temperature change, but it applies to an object instead of a specified amount of a substance. Thus, specific heat capacity and molar heat capacity are used when talking about an element or compound while heat capacity is used for a calorimeter or other object.

6.36

In a coffee-cup calorimeter, reactions occur at constant pressure. qp = H. In a bomb calorimeter, reactions occur at constant volume. qv = E.

6.37

6.38 6.39

Plan: The heat required to raise the temperature of water is found by using the equation q = c × mass × T. The specific heat capacity, cwater, is found in Table 6.2. Because the Celsius degree is the same size as the Kelvin degree, T = 100°C – 25°C = 75°C = 75 K. Solution:  J  3 22.0 g75 K  = 6903.6 = 6.9 × 10 J q (J) = c × mass × T = 4.184  g  K   J  0.10 g 75  10 K = –17.7395 = –18 J q (J) = c × mass × T = 2.087  g  K 



 

Plan: Use the relationship q = c × mass × T. We know the heat (change kJ to J), the specific heat capacity, and the mass, so T can be calculated. Once T is known, that value is added to the initial temperature to find the final temperature.

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6-13

Solution: q (J) = c × mass × T

Tinitial= 13.00°C

Tfinal = ?

mass = 295 g

c = 0.900 J/g K

10 J  4 q = 75.0 kJ   = 7.50 × 10 J  1 kJ  3

7.50 × 10 J = (0.900 J/g K)(295 g)(T) T =

7.50 10 4 J



 0.900 J   295 g   g  K 



= 282.4859 K = 282.4859°C

(Because the Celsius degree is the same size as the Kelvin degree, T is the same in either temperature unit.) T = Tfinal – Tinitial TfinalT + Tinitial Tfinal = 282.4859°C + 13.00°C = 295.49 = 295°C 6.40

q (J) = c × mass × T –688 J = (2.42 J/g K)(27.7 g)(T) (T) =

688 J 

 2.42 J   27.7 g   g  K 



= –10.26345 K = –10.26345°C

T = Tfinal – Tinitial Tinitial = Tfinal – T Tinitial = 32.5°C – (–10.26345°C) = 42.76345 = 42.8°C 6.41

Plan: Since the bolts have the same mass and same specific heat capacity, and one must cool as the other heats (the heat lost by the “hot” bolt equals the heat gained by the “cold” bolt), the final temperature is an average of the two initial temperatures. Solution:  T + T   100C + 55C 2   1   = 77.5°C =     2 2    

6.42

–qlost = qgained – 2(mass)(cCu)(Tfinal – 105)°C = (mass)(cCu)(Tfinal – 45)°C – 2(Tfinal – 105)°C = (Tfinal – 45)°C 2(105°C) – 2Tfinal = Tfinal – 45°C 210°C + 45°C = Tfinal + 2Tfinal = 3Tfinal (255°C)/3 = Tfinal = 85.0°C

6.43

Plan: The heat lost by the water originally at 85°C is gained by the water that is originally at 26°C. Therefore –qlost = qgained. Both volumes are converted to mass using the density. Solution: 1.00 g  1.00 g  Mass (g) of 75 mL =  75 mL  = 75 g Mass (g) of 155 mL = 155 mL  = 155 g  1 mL   1 mL  –qlost = qgained McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution

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6-14

c × mass × T (85°C water) = c × mass × T (26°C water) – (4.184 J/g°C)(75 g)(Tfinal – 85)°C = (4.184 J/g°C)(155 g)(Tfinal – 26)°C – (75 g)(Tfinal – 85)°C = (155 g) (Tfinal – 26)°C 6375 – 75Tfinal = 155Tfinal – 4030 6375 + 4030 = 155Tfinal + 75Tfinal 10405 = 230.Tfinal Tfinal = (10405/230.) = 45.24 = 45°C 6.44

–qlost = qgained – [24.4 mL(1.00 g/mL)](4.184 J/g°C)(23.5 – 35.0)°C = (mass)(4.184 J/g°C)(23.5 – 18.2)°C – (24.4)(23.5 – 35.0) = (mass)(23.5 – 18.2) – (24.4)(–11.5) = (mass)(5.3) 280.6 = (mass)(5.3) 52.943 g = mass  1 mL   = 52.943 = 53 mL Volume (mL) = 52.943 g 1.00 g 

6.45

Plan: Heat gained by water and the container equals the heat lost by the copper tubing so qwater + qcalorimeter = –qcopper. Solution: T = Tfinal – Tinitial Specific heat capacity in units of J/g K has the same value in units of J/g°C since the Celsius and Kelvin units are the same size. –qlost = qgained = qwater + qcalorimeter – (455 g Cu)(0.387 J/g°C)(Tfinal – 89.5)°C = (159 g H2O)(4.184 J/g°C)(Tfinal – 22.8)°C + (10.0 J/°C)(Tfinal – 22.8)°C – (176.085)(Tfinal – 89.5) = (665.256)(Tfinal – 22.8) + (10.0)(Tfinal – 22.8) 15759.6075 – 176.085Tfinal = 665.256Tfinal– 15167.8368 + 10.0Tfinal – 228 15759.6075 + 15167.8368 + 228 = 176.085Tfinal + 665.256Tfinal + 10.0Tfinal 31155.4443 = 851.341Tfinal Tfinal = 31155.4443/(851.341) = 36.59573 = 36.6°C

6.46

–qlost = qgained = qwater + qcalorimeter – (30.5 g alloy)(calloy)(31.1 – 93.0)°C = (50.0 g H2O)(4.184 J/g°C)(31.1 – 22.0)°C + (9.2 J/°C)(31.1 – 22.0)°C – (30.5 g)(calloy)(–61.9°C) = (50.0 g)(4.184 J/g°C)(9.1°C) + (9.2 J/°C)(9.1°C) 1887.95(calloy) = 1903.72 + 83.72 = 1987.44 calloy = 1987.44/1887.95 = 1.052697 = 1.1 J/g°C

6.47

Plan: Find the total mass of the solution, using the density. Use Equation 6.7 to find q for the solution; q for the reaction has the opposite sign. Use molar ratios to find the amount (mol) of water formed from the given amount of reactants. Divide qrxn by the moles of water formed to obtain H per mole of water. Solution: The reaction is: 2KOH(aq) + H2SO4(aq)  K2SO4(aq) + 2H2O(l) qsoln = (csoln)(masssoln)( Tsoln) McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution

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6-15

1.00 g  4.184 J   1 kJ   30.17C  23.50C  q (kJ) = (25.0 mL  25.0 mL)    = 1.395364 kJ  1 mL  gC  1000 J 

qrxn(kJ) = ---1.395364 kJ (The temperature increased so the heat of reaction is exothermic.)

 1 L  0.500 mol H 2 SO4  Amount (moles) of H2SO4 = 25.0 mL  = 0.0125 mol H2SO4  1000 mL   1L  1 L  1.00 mol KOH  = 0.0250 mol KOH Amount (moles) of KOH = 25.0 mL  1000 mL   1L The moles show that both H2SO4 and KOH are limiting since the balanced reaction indicates that twice the moles of KOH than H2SO4 are required.  2 mol H 2 O   = 0.0250 mol H 2 O Amount (mol) of H2O formed = 0.0125 mol H 2 SO 4  1 mol H 2 SO 4   1.395364 kJ   = ---55.81456 = ---55.8 kJ/mol H 2O H =   0.0250 mol H 2 O  6.48

The reaction is: AgNO3(aq) + NaCl(aq)  AgCl(s) + NaNO3(aq) 1.00 g  4.184 J   1 kJ   25.65C  24.72C  qsoln(kJ) = (20.0 mL  30.0 mL)    = 0.194556 kJ  1 mL  gC  1000 J  qrxn(kJ) = ---0.194556 kJ (The temperature increased so the heat of reaction is exothermic.)  1 L  0.200 mol AgNO3  ---3  Amount (moles) of AgNO3 = 20.0 mL  = 4.00 × 10 mol AgNO3  1000 mL   1L

 1 L   0.100 mol NaCl  = 3.00 × 10---3 mol NaCl Amount (moles) of NaCl = 30.0 mL  1000 mL   1L The moles show that NaCl is limiting. 1 mol AgCl  Amount (mol) of AgCl formed = 3.00 103 mol NaCl  = 3.00 103 mol AgCl  1 mol NaCl   0.194556 kJ    H =  3.00  10 –3 mol AgCl  = ---64.852 = ---65 kJ/mol AgCl  



6.49



Plan: Use Equation 6.8 to find q for the calorimeter; q for the reaction has the opposite sign. Use the molar mass of glucose to convert q for the 2.150 g sample to q (E) for a mole of glucose. Solution:     qcalorimeter = C × T =  28.745C – 23.446C   = 33.473783 kJ  K    ---qrxn = qcalorimeter qrxn = ---33.473783 kJ  33.473783 kJ  180.16 g C 6 H12 O6  = ---2804.947323 = ---2805 kJ/mol C H O E (kJ/mol) =  6 12 6  1 mol C H O   2.150 g C6 H12 O 6  6 12 6 

6.50

Plan: Use Equation 6.8 to find q for the calorimeter; q for the reaction has the opposite sign. Use the molar mass of ethanol to convert q for the 1.750 g sample to q (E) for a mole of ethanol.

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6-16

 qcalorimeter = C × T =  

  4.287C   

  = 51.65835 kJ 

---qrxn = qcalorimeter qrxn = ---51.65835 kJ

 51.65835 kJ   46.07 g C 2 H 6 O  = ---1359.942963 = ---1360. kJ/mol C H O E (kJ/mol) =  2 6  1 mol C H O  1.750 g C 2 H 6 O   2 6 6.51

Benzoic acid is C6H5COOH, and will be symbolized as HBz. –qreaction = qwater + qcalorimeter  1 mol HBz  3227 kJ 10 3 J  4   –qreaction = – 1.221 g HBz = 3.226472 × 10 J 122.12 g HBz 1 mol HBz  1 kJ  qwater = c × mass × T = 4.184 J/g°C × 1200 g × T qcalorimeter = C × T = 1365 J/°C × T –qreaction = qwater + qcalorimeter 3.226472 × 10 J = 4.184 J/g°C × 1200 g × T + 1365 J/°C × T 4 4

3.226472 × 10 J = 5020.8(T) + 1365(T) 4

3.226472 × 10 J = 6385.8(T) T = 3.226472 × 10 /6385.8 = 5.052573 = 5.053°C 4

6.52

a) Energy will flow from Cu (at 100.0°C) to Fe (at 0.0°C). b) To determine the final temperature, the heat capacity of the calorimeter must be known. c) – qCu = qFe + qcalorimeter assume qcalorimeter = 0. – qCu = qFe + 0 – (20.0 g Cu)(0.387 J/g°C)(Tfinal – 100.0)°C = (30.0 g Fe)(0.450 J/g°C)(Tfinal – 0.0)°C + 0.0 J – (20.0 g)(0.387 J/g°C)(Tfinal – 100.0°C) = (30.0 g)(0.450 J/g°C)(Tfinal – 0.0°C) – (7.74)(Tfinal – 100.0) = (13.5)(Tfinal – 0.0) 774 – 7.74 Tfinal = 13.5Tfinal 774 = (13.5 + 7.74) Tfinal = 21.24Tfinal Tfinal = 774/21.24 = 36.44068 = 36.4°C

6.53

qcalorimeter = Ccalorimeter × T = 11.09 kJ/K × (23.55°C – 20.00°C) qcalorimeter = 39.3695 kJ qrxn = –qcalorimeter = –39.3695 kJ 39.3695 kJ =  25.9009868 =  25.9 kJ/g Heat released/g = 1.520 g

6.54

Reactants  Products + Energy Hrxn = (–) Thus, energy is a product.

6.55

Plan: Recall that H is positive for an endothermic reaction in which heat is absorbed, while H is negative for an exothermic reaction in which heat is released.

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6-17

Solution: The reaction has a positive Hrxn, because this reaction requires the input of energy to break the oxygen-oxygen bond in O2: O2(g) + energy  2O(g) 6.56

Plan: Recall that H is positive for an endothermic reaction in which heat is absorbed, while H is negative for an exothermic reaction in which heat is released. Solution: As a substance changes from the gaseous state to the liquid state, energy is released so H would be negative for the condensation of 1 mol of water. The value of H for the vaporization of 2 mol of water would be twice the value of H for the condensation of 1 mol of water vapor but would have an opposite sign (+H). H2O(g)  H2O(l) + Energy 2H2O(l) + Energy  2H2O(g) Hcondensation = (–) Hvaporization = (+)2[Hcondensation] The enthalpy for 1 mole of water condensing would be opposite in sign to and one-half the value for the conversion of 2 moles of liquid H2O to H2O vapor.

6.57

Plan: Recall that H is positive for an endothermic reaction in which heat is absorbed, while H is negative for an exothermic reaction in which heat is released. The Hrxn is specific for the reaction as written, meaning that 20.2 kJ is released when one-eighth of a mole of sulfur reacts. Use the ratio between moles of sulfur and H to convert between amount of sulfur and heat released. Solution: a) This reaction is exothermic because H is negative. b) Because H is a state function, the total energy required for the reverse reaction, regardless of how the change occurs, is the same magnitude but different sign of the forward reaction. Therefore, H = +20.2 kJ.  20.2 kJ   2 c) Hrxn = 2.6 mol S8   = –420.16 = –4.2 × 10 kJ  1/ 8 mol S8 





d) The mass of S8 requires conversion to moles and then a calculation identical to part (c) can be performed.  1 mol S8   20.2 kJ   Hrxn = 25.0 g S8    = –15.7517 = –15.8 kJ  256.48 g S8   1/ 8 mol S8 



6.58



MgCO3(s)  MgO(s) + CO2(g)

Hrxn = 117.3 kJ

a) Absorbed b) Hrxn (reverse) = –117.3 kJ  117.3 kJ    = –627.555 = –628 kJ c) Hrxn = 5.35 mol CO 2  1 mol CO 2   1 mol CO2   117.3 kJ   d) Hrxn = 35.5 g CO2    = –94.618 = –94.6 kJ  44.01 g CO 2  1 mol CO2 



6.59



Plan: A thermochemical equation is a balanced equation that includes the heat of reaction. Since heat is absorbed in this reaction, H will be positive. Convert the mass of NO to moles and use the ratio between NO and H to find the heat involved for this amount of NO. Solution: a) 1/2N2(g) + 1/2O2(g)  NO (g) H = 90.29 kJ

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6-18

 1 mol NO   90.29 kJ   b) Hrxn = 3.50 g NO    = –10.5303 = –10.5 kJ  30.01 g NO   1 mol NO 



6.60



a) KBr(s)  K(s) + 1/2Br2(l)

Hrxn = 394 kJ

103 g   1 mol KBr   394 kJ   4 4 b) Hrxn = 10.0 kg KBr     = –3.3109 × 10 = –3.31 × 10 kJ 119.00 g KBr  1 mol KBr   1 kg 



6.61



Plan: For the reaction written, 2 moles of H2O2 release 196.1 kJ of energy upon decomposition. Use this ratio to convert between the given amount of reactant and the amount of heat released. The amount of H 2O2 must be converted from kg to g to moles. Solution: 2H2O2(l)  2H2O(l) + O2(g) Hrxn = –196.1 kJ 10 3 g   1 mol H 2 O 2   196.1 kJ   6 6 Heat (kJ) = q = 652 kg H 2 O2     = –1.87915 × 10 = –1.88 × 10 kJ  34.02 g H 2 O2   2 mol H 2 O 2   1 kg 



6.62



For the reaction written, 1 mole of B2H6 releases 755.4 kJ of energy upon reaction. B2H6(g) + 6Cl2(g)  2BCl3(g) + 6HCl(g)

Hrxn = –755.4 kJ

10 g   1 mol B2 H 6   755.4 kJ   4 4 Heat (kJ) = q = 1 kg    = –2.73003 × 10 = –2.730 × 10 kJ/kg  1 kg   27.67 g B2 H 6  1 mol B2 H 6  3

6.63

4Fe(s) + 3O2(g)  2Fe2O3(s)

Hrxn = –1.65 × 10 kJ 3

3 103 g   1 mol Fe   1.65 10 kJ   a) Heat (kJ) = q = 0.250 kg Fe     = –1846.46 = –1850 kJ  55.85 g Fe   4 mol Fe   1 kg   2 mol Fe O    159.70 g Fe 2 O3  2 3  b) Mass (g) of Fe2O3 = 4.85 103 kJ    = 938.84 = 939 g Fe2O3 3  1 mol Fe 2 O3   1.65 10 kJ 







6.64



2HgO(s)  2Hg(l) + O2(g)

Hrxn = 181.6 kJ

 1 mol HgO   181.6 kJ   a) Heat (kJ) = q = 555 g HgO    = 232.659 = 233 kJ  216.6 g HgO   2 mol Hg   2 mol Hg   200.6 g Hg   b) Mass (g) of Hg = 275 kJ    = 607.544 = 608 g Hg  1 mol Hg   181.6 kJ 







6.65



Plan: A thermochemical equation is a balanced equation that includes the heat of reaction. Heat is released in this reaction so H is negative. Use the ratio between H and moles of C2H4 to find the amount of C2H4 that must react to produce the given quantity of heat. Solution: a) C2H4(g) + 3O2(g)  2CO2(g) + 2H2O(l) Hrxn = –1411 kJ

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6-19

1 mol C 2 H 4   28.05 g C 2 H 4   b) Mass (g) of C2H4 = 70.0 kJ    = 1.39157 = 1.39 g C2H4  1411 kJ   1 mol C 2 H 4 



6.66



a) C12H22O11(s) + 12O2(g)  12CO2(g) + 11H2O(l)

Hrxn = –5.64 × 10 kJ 3

3  1 mol C H O     5.64 10 kJ  12 22 11  b) Heat (kJ) = q = 1 g C12 H 22 O11    = –16.47677 = –16.5 kJ/g 1 mol C12 H 22 O11   342.30 g C12 H 22 O11 

6.67

Hess’s law: Hrxn is independent of the number of steps or the path of the reaction.

6.68

Hess’s law provides a useful way of calculating energy changes for reactions which are difficult or impossible to measure directly.

6.69

Plan: Two chemical equations can be written based on the description given: C(s) + O2(g)  CO2(g) H1 (1) CO(g) + 1/2O2(g)  CO2(g) H2 (2) The second reaction can be reversed and its H sign changed. In this case, no change in the coefficients is necessary since the CO2 cancels. Add the two H values together to obtain the H of the desired reaction. Solution: C(s) + O2(g)  CO2(g) H1

Total

CO2(g)  CO(g) + 1/2O2(g)

–H2

C(s) + 1/2O2(g)  CO(g)

Hrxn = H1 + – (H2)

(reaction is reversed)

How are the H values for each reaction determined? The H1 can be found by using the heats of formation in Appendix B: H1 = [Hf(CO2)] – [Hf(C) + Hf(O2)] = [–393.5 kJ/mol] – [0 + 0] = –393.5 kJ/mol. The H2 can be found by using the heats of formation in Appendix B: H2 = [Hf(CO2)] – [Hf(CO) + 1/2Hf(O2)] = [–393.5] – [–110.5 kJ/mol + 0)] = –283 kJ/mol. Hrxn = H1 + – (H2) = –393.5 kJ + – (–283.0 kJ) = –110.5 kJ 6.70

6.71

6.72

Plan: To obtain the overall reaction, add the first reaction to the reverse of the second. When the second reaction is reversed, the sign of its enthalpy change is reversed from positive to negative. Solution: Ca(s) + 1/2O2(g) CaO(s) H = –635.1 kJ CaO(s) + CO2(g)  CaCO3(s)

H = –178.3 kJ (reaction is reversed)

Ca(s) + 1/2O2(g) + CO2(g) CaCO3(s)

H = –813.4 kJ

2NOCl(g) 2NO(g) + Cl2(g)

H = –2(–38.6 kJ)

2NO(g)  N2(g) + O2(g)

H = –2(90.3 kJ)

2NOCl(g)  N2(g) + O2(g) + Cl2(g)

H = 77.2 kJ + (– 180.6 kJ) = –103.4 kJ

Plan: Add the two equations, canceling substances that appear on both sides of the arrow. When matching the equations with the arrows in the Figure, remember that a positive H corresponds to an arrow pointing up while a negative H corresponds to an arrow pointing down. McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution

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6-20

Solution: 1) N2(g) + O2(g)  2NO(g)

H = 180.6 kJ

2NO(g) + O2(g)  2NO2(g)

H = –114.2 kJ

N2(g) + 2O2(g)  2NO2(g)

Hrxn = +66.4 kJ

In Figure P6.67, A represents reaction 1 with a larger amount of energy absorbed, B represents reaction 2 with a smaller amount of energy released, and C represents reaction 3 as the sum of A and B. 6.73

P4(s) + 6Cl2(g)  4PCl3(g)

H1 = –1148 kJ

4PCl3(g) + 4Cl2(g)  4PCl5(g)

H2 = – 460 kJ

P4(s) + 10Cl2(g)  4PCl5(g)

Hoverall = –1608 kJ

Equation 1 = B, equation 2 = C, equation 3 = A 6.74

Plan: Vaporization is the change in state from a liquid to a gas: H2O(l)  H2O(g). The two equations describing the chemical reactions for the formation of gaseous and liquid water can be combined to yield the equation for vaporization. Solution: 1) Formation of H2O(g): H2(g) + 1/2O2(g)  H2O(g) H = –241.8 kJ 2) Formation of H2O(l): H2(g) + 1/2O2(g)  H2O(l) H Reverse reaction 2 (change the sign of H) and add the two reactions: H2(g) + 1/2O2(g)  H2O(g) H = –241.8 kJ

6.75

6.76

6.77

6.78

6.79

H2O(l) H2 (g) + 1/2O2(g)

H = +285.8 kJ

H2O(l)  H2O(g)

Hvap = 44.0 kJ

C(s) + 1/4S8(s) CS2(l)

H = +89.7 kJ

CS2(l)  CS2(g)

H = +27.7 kJ

C(s) + 1/4S8(s)  CS2(g)

H = +117.4 kJ

= –285.8 kJ

C (diamond) + O2(g)CO2(g)

H = –395.4 kJ

CO2(g)  C(graphite) + O2(g)

H = –(–393.5 kJ)

C(diamond)  C(graphite)

H = – 1.9 kJ

 The standard heat of reaction, H rxn , is the enthalpy change for any reaction where all substances are in their standard states. The standard heat of formation, H f , is the enthalpy change that accompanies the formation of one mole of a compound in its standard state from elements in their standard states. Standard state is 1 atm for gases, 1 M for solutes, and the most stable form for liquids and solids. Standard state does not include a specific temperature, but a temperature must be specified in a table of standard values. The standard heat of reaction is the sum of the standard heats of formation of the products minus the sum of the standard heats of formation of the reactants multiplied by their respective stoichiometric coefficients.  H rxn = m H f(products) – n H f(reactants)

Plan: H f is for the reaction that shows the formation of one mole of compound from its elements in their standard states. Solution: McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution

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6-21

a) 1/2Cl2(g) + Na(s)  NaCl(s). The element chlorine occurs as Cl2, not Cl. b) H2(g) + 1/2O2(g)  H2O(g). The element hydrogen exists as H2, not H, and the formation of water is written with water as the product. c) No changes 6.80

Plan: Formation equations show the formation of one mole of compound from its elements. The elements must be in their most stable states ( H f = 0). Solution: a) Ca(s) + Cl2(g)  CaCl2(s) b) Na(s) + 1/2H2(g) + C(graphite) + 3/2O2(g)  NaHCO3(s) c) C(graphite) + 2Cl2(g)  CCl4(l) d) 1/2H2(g) + 1/2N2(g) + 3/2O2(g)  HNO3(l)

6.81

a) 1/2H2(g) + 1/2I2(s)  HI(g) b) Si(s) + 2F2(g)  SiF4(g) c) 3/2O2(g)  O3(g) d) 3Ca(s) + 1/2P4(s) + 4O2(g)  Ca3(PO4)2(s)

6.82

= [(2 mol)(–296.8 kJ/mol) + (2 mol)(–241.826 kJ/mol)] – [(2 mol)(–20.2 kJ/mol) + (3 mol)(0.0 kJ/mol)] = –1036.852 = –1036.9 kJ b) The balanced equation is CH4(g) + 4Cl2(g)  CCl4(l) + 4HCl(g)  H rxn = {1 H f [CCl4(l)] + 4 H f [HCl(g)]} – {1 H f [CH4(g)] + 4 H f [Cl2(g)]}  H rxn = [(1 mol)(–139 kJ/mol) + (4 mol)(–92.31 kJ/mol)] – [(1 mol)(–74.87 kJ/mol) + (4 mol)(0 kJ/mol)]

= –433.37 = –433 kJ 6.83

 H rxn = m H f(products) – n H f(reactants)  a) H rxn = {1 H f [SiF4(g)] + 2 H f [H2O(l)]} – {1 H f [SiO2(s)] + 4 H f [HF(g)]}

= [(1 mol)(–1614.9 kJ/mol) + (2 mol)(–285.840 kJ/mol)] – [(1 mol)(–910.9 kJ/mol) + (4 mol)(–273 kJ/mol)] = –183.68 = –184 kJ b) 2C2H6(g) + 7O2(g)  4CO2(g) + 6H2O(g)  H rxn = {4 H f [CO2(g)] + 6 H f [H2O(g)]} – {2 H f [C2H6(g)] + 7 H f [ O2(g)]}

= [(4 mol)(–393.5 kJ/mol) + (6 mol)(–241.826 kJ/mol)] – [(2 mol)(–84.667 kJ/mol) + (7 mol)(0 kJ/mol)] = –2855.622 = –2855.6 kJ (or –1427.8 kJ for reaction of 1 mol of C2H6) 6.84

Plan: The enthalpy change of a reaction is the sum of the heats of formation of the products minus the sum of the heats of formation of the reactants. Since the H f values (Appendix B) are reported as energy per one mole, use McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution

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6-22

 the appropriate stoichiometric coefficient to reflect the higher number of moles. In this case, H rxn is known and  H f of CuO must be calculated. Solution:  H rxn = m H f(products) – n H f(reactants)  H rxn = –146.0 kJ

Cu2O(s) + 1/2O2(g)  2CuO(s)

 H rxn = {2 H f [CuO(s)]} – {1 H f [Cu2O(s)] + 1/2 H f [O2(g)]}

–146.0 kJ = {(2 mol) H f [CuO(s)]} – {(1 mol)(–168.6 kJ/mol) + (1/2 mol)(0 kJ/mol)} –146.0 kJ = 2 mol H f [CuO(s)] + 168.6 kJ 314.6 kJ H f [CuO(s)] =  = –157.3 kJ/mol 2 mol 6.85

 H rxn = m H f(products) – n H f(reactants)  H rxn = –1255.8 kJ

C2H2(g) + 5/2O2(g)  2CO2(g) + H2O(g)

 H rxn = {2 H f [CO2(g)] + 1 H f [H2O(g)]} – {1 H f [C2H2(g)] + 5/2 H f [O2(g)]}

–1255.8 kJ = {(2 mol)(–393.5 kJ/mol) + (1 mol)(–241.826 kJ/mol)} – {(1 mol) H f [C2H2(g)] + (5/2 mol)(0.0 kJ/mol)} –1255.8 kJ = –787.0 kJ – 241.8 kJ – (1 mol) H f [C2H2(g)] 227.0 kJ H f [C2H2(g)] = = 227.0 kJ/mol 1 mol 6.86

Plan: The enthalpy change of a reaction is the sum of the heats of formation of the products minus the sum of the heats of formation of the reactants. Since the H f values (Appendix B) are reported as energy per one mole, use the appropriate stoichiometric coefficient to reflect the higher number of moles. Hess’s law can also be used to calculate the enthalpy of reaction. In part (b), rearrange equations 1 and 2 to give the equation wanted.   Reverse the first equation (changing the sign of H rxn ) and multiply the coefficients (and H rxn ) of the second reaction by 2. Solution: 2PbSO4(s) + 2H2O(l)  Pb(s) + PbO2(s) + 2H2SO4(l)  H rxn = m H f(products) – n H f(reactants)  a) H rxn = {1 H f [Pb(s)] + 1 H f [PbO2(s)] + 2 H f [H2SO4(l)]}

– {2 H f [PbSO4(s)] + 2 H f [H2O(l)]} = [(1 mol)(0 kJ/mol) + (1 mol)(–276.6 kJmol) + (2 mol)(–813.989 kJ/mol)] – [(2 mol)(–918.39 kJ/mol) + (2 mol)(–285.840 kJ/mol)] = –503.882 = 503.9 kJ

6.87

b) Use Hess’s law: PbSO4(s)  Pb(s) + PbO2(s) + 2SO3(g)

 H rxn = –(–768 kJ) Equation has been reversed.

2SO3(g) + 2H2O (l)  2H2SO4(l)

 H rxn = 2(–132 kJ)

2PbSO4(s) + 2H2O(l)  Pb(s) + PbO2(s) + 2H2SO4(l)

 H rxn = 504 kJ

Plan: The enthalpy change of a reaction is the sum of the heats of formation of the products minus the sum of the heats of formation of the reactants. Since the H f values (Appendix B) are reported as energy per one mole, use the appropriate stoichiometric coefficient to reflect the higher number of moles. Convert the mass of stearic acid  to moles and use the ratio between stearic acid and H rxn to find the heat involved for this amount of acid. For McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution

in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

6-23

part (d), use the kcal/g of fat relationship calculated in part (c) to convert 11.0 g of fat to total kcal and compare to the 100. Cal amount. Solution: a) C18H36O2(s) + 26O2(g)  18CO2(g) + 18H2O(g)  b) H rxn = m H f(products) – n H f(reactants)  H rxn = {18 H f [CO2(g)] + 18 H f [H2O(g)]} – {1 H f [C18H36O2(s)] + 26 H f [O2(g)]} = [(18 mol)(–393.5 kJ/mol) + (18 mol)(–241.826 kJ/mol)] – [(1 mol)(–948 kJ/mol) + (26 mol)(0 kJ/mol)] = –10,487.868 = –10,488 kJ  1 mol C18 H36 O 2   10, 487.868 kJ   c) q (kJ) = 1.00 g C18 H 36 O2    = –36.8681 = –36.9 kJ  1 mol C18 H 36 O2   284.47 C18 H36 O 2   1 kcal    = –8.811688 = –8.81 kcal q (kcal) = 36.8681 kJ   4.184 kJ 

 8.811688 kcal   = 96.9286 = 96.9 kcal d) q (kcal) = 11.0 g fat     1.0 g fat   Since 1 kcal = 1 Cal, 96.9 kcal = 96.9 Cal. The calculated calorie content is consistent with the package information.

6.88

a) H2SO4(l)  H2SO4(aq)  H rxn = {1 H f [H2SO4(aq)]} – {1 H f [H2SO4(l)]}

= [(1 mol)(–907.51 kJ/mol)] – [(1 mol)(–813.989 kJ/mol)] = –93.52 kJ b) q (J) = c × mass × T

 3.50 J  1.060 g  3  1000 mL   25.0C 93.52 kJ × 10 J/kJ =   T  1 mL   final  g  C   3.50 J  4  1060. g Tfinal  25.0C 9.352 × 10 J =   g  C  4

9.352 × 10 J = (Tfinal)3710 J/°C – 9.2750 × 10 J Tfinal = 50.1995 = 50.2°C c) Adding the acid to a large amount of water releases the heat to a large mass of solution and thus, the potential temperature rise is minimized due to the large heat capacity of the larger volume. 6.89

Plan: Use the ideal gas law, PV = nRT, to calculate the volume of one mole of helium at each temperature. Then E to find the change in internal energy. The equation for work, w = –P V, is needed for part (c), and qP E + P V is used for part (d). For part (e), H = q P. Solution: nRT a) PV = nRT or V = P T = 273 + 15 = 288 K and T = 273 + 30 = 303 K L  atm   0.0821  288 K nRT  mol  K  Initial volume (L) = V = = = 23.6448 = 23.6 L/mol P 1.00 atm



 



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6-24

L  atm   0.0821 303 K  nRT  K mol Final volume (L) = V = = = 24.8763 = 24.9 L/mol P 1.00 atm

b) Internal energy is the sum of the potential and kinetic energies of each He atom in the system (the balloon). The energy of one mole of helium atoms can be described as a function of temperature, E = 3/2nRT, where n = 1 mole. Therefore, the internal energy at 15°C and 30°C can be calculated. The inside back cover lists values of R with different units. E = 3/2nRT = (3/2)(1.00 mol) (8.314 J/mol K)(303 – 288)K = 187.065 = 187 J c) When the balloon expands as temperature rises, the balloon performs PV work. However, the problem specifies that pressure remains constant, so work done on the surroundings by the balloon is defined by the equation: w = – PV. When pressure and volume are multiplied together, the unit is L atm, so a conversion factor is needed to convert work in units of L atm to joules.  101.3 J  2 w = –PV = 1.00 atm (24.8763  23.6448) L  = –124.75 = –1.2 × 10 J 1 L  atm  d) qP = E + PV = (187.065 J) + (124.75 J) = 311.815 = 3.1 × 10 J 2

e) H = qP = 310 J. f) When a process occurs at constant pressure, the change in heat energy of the system can be described by a state function called enthalpy. The change in enthalpy equals the heat (q) lost at constant pressure: H = E + PV = E – w = (q + w) – w = qP 6.90

a) Respiration: C6H12O6(s) + 6O2(g)  6CO2(g) + 6H2O(g)  H rxn = m H f(products) – n H f(reactants)

= {6 H f [CO2(g)] + 6 H f [H2O(g)]} – {1 H f [C6H12O6(s)] + 6 H f [O2(g)]} = [(6 mol)(–393.5 kJ/mol) + (6 mol)(–241.826 kJ/mol)] – [(1 mol)(–1273.3 kJ/mol) + (6 mol)(0.0 kJ/mol)] = –2538.656 = –2538.7 kJ Fermentation: C6H12O6(s)  2CO2(g) + 2CH3CH2OH(l)  H rxn = {2 H f [CO2(g)] + 2 H f [CH3CH2OH(l)]} – [1 H f [C6H12O6(s)]}

= [(2 mol)(–393.5 kJ/mol) + (2 mol)(–277.63 kJ/mol)] – [(1 mol)(–1273.3 kJ/mol)] = –68.96 = –69.0 kJ b) Combustion of ethanol: CH3CH2OH(l) + 3O2(g)  2CO2(g) + 3H2O(g)  H rxn = {2 H f [CO2(g)] + 3 H f [H2O(g)]} – {1 H f [CH3CH2OH(l)] + 3 H f [O2(g)]}  H rxn = [(2 mol)(–393.5 kJ/mol) + (3 mol)(–241.826 kJ/mol)] – [(1 mol)(–277.63 kJ/mol) + (3 mol)(0.0 kJ/mol)]

= –1234.848 = –1234.8 kJ Heats of combustion/mol C:  2538.656 kJ 1 mol C 6 H12 O6   = –423.1093 = –423.11 kJ/mol C  Sugar:   1 mol C6 H12 O6  6 mol C

 1234.848 kJ 1 mol CH 3 CH 2 OH   Ethanol:   = –617.424 = –617.42 kJ/mol C  1 mol CH 3 CH 2 OH  2 mol C Ethanol has a higher value.

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6-25

6.91

a) Reactions: 1) C21H44(s) + 32O2(g)  21CO2(g) + 22H2O(g) 2) C21H44(s) + 43/2O2(g)  21CO(g) + 22H2O(g) 3) C21H44(s) + 11O2(g)  21C(s) + 22H2O(g) Heats of combustion:  1) H rxn = {21 H f [CO2(g)] + 22 H f [H2O(g)]} – {[1 H f [C21H44(s)] + 32 H f [O2(g)]} = [(21 mol)(–393.5 kJ/mol) + (22 mol)(–241.826 kJ/mol)] – [(1 mol)(–476 kJ/mol) + (32 mol)(0.0 kJ/mol)] = –13,107.672= –13,108 kJ  rxn

2) H = {21 H f [CO(g)] + 22 H f [H2O(g)]} – {1 H f [C21H44(s)] + 43/2 H f [O2(g)]} = [(21 mol)(–110.5 kJ/mol) + (22 mol)(–241.826 kJ/mol)] – [(1 mol)(–476 kJ/mol) + (43/2 mol)(0.0 kJ/mol)] = –7164.672 = –7165 kJ  rxn

 f

3) H = {21 H [C(s)] + 22 H [H2O(g)]} – {1 H f [C21H44(s)] + 11 H f [O2(g)]} = [(21 mol)(0.0 kJ/mol) +(22 mol)(–241.826 kJ/mol)] – [(1 mol)(–476 kJ/mol) + (11 mol)(0.0 kJ/mol)] = –4844.172 = –4844 kJ  1 mol C 21 H 44  13107.672 kJ   4 4   b) q (kJ) = 254 g C 21H 44   1 mol C H  = –1.12266 × 10 = –1.12 × 10 kJ  296.56 g C 21 H 44  21 44 





c) The moles of C21H44 need to be calculated one time for multiple usage. It must be assumed that the remaining 87.00% of the candle undergoes complete combustion. Moles C21H44 = (254 g C21H44)(1 mol C21H44/296.56 g C21H44) = 0.856488 mol q = (0.87)(0.856488 mol)(–13107.672 kJ/mol) + (0.0800)(0.856488 mol)(–7164.672 kJ/mol) 4

+ (0.0500)(0.856488 mol)(–4844.172 kJ/mol)] = –1.04655 × 10 = –1.05 × 10 kJ 6.92

a) EO(l)

 = 569.4 J/g(44.05 g/mol)(1 kJ/1000 J) = 25.08 kJ/mol H vap

 = {1 H f [EO(g)]} – {1 H f [EO(l)]} H vap

25.08 kJ/mol = { H f [EO(g)]} – [(1 mol)(–77.4 kJ/mol)]

H f [EO(g)] = –52.32 kJ/mol EO(g)

(g) + CO(g)

 H rxn = {1 H f [CH4(g)] + 1 H f [CO(g)]} – {1 H f [EO(g)]}  H rxn = [(1 mol)(–74.87 kJ/mol) + (1 mol)(–110.5 kJ/mol)] – [(1 mol)(–52.32 kJ/mol)]  H rxn = –133.05 = –133.0 kJ/mol

b) Assume that you have 1.00 mole of EO(g). 1.00 mole of EO(g) produces 1.00 mole or 16.04 g of CH4(g) and 1.00 mole or 28.01 g of CO(g). There is a total product mass of 16.04 g + 28.01 g = 44.05 g. q = c × mass × T 1000 J  133.05 kJ  q  1 kJ  T= = c mass 2.5 J/gC 44.05 g

T = 1208.1725°C T = Tfinal – Tinitial 1208.1725°C = Tfinal – 93°C Copyright

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6-26

Tfinal = 1301.1725 = 1301°C 6.93

a) 3N2O5(g) + 3NO(g) (g) 2  H rxn = {9 H f [NO2(g)]} – {3 H f [N2O5(g)] + 3 H f [NO(g)]} = [(9 mol)(33.2 kJ/mol)] – [(3 mol)(11 kJ/mol) + (3 mol)(90.29 kJ/mol)] = –5.07 = –5 kJ 3  1.50 102 mol  5.07 kJ  10 J    = –76.05 = –76.0 J  1 molecule product  9 moles product  1 kJ 

 b) 9 molecules product 

6.94

3  4 qt   1 L   1 mL  0.692 g 1 mol C8 H18   5.44 10 kJ    a) Heat (kJ) = 20.4 gal    3    1 gal  1.057 qt  10 L  mL  114.22 g   1 mol C8 H18 





= –2.54435678 × 10 = –2.54 × 10 kJ

  1h  65 mi   1 km  = 4.84995 × 103 = 4.8 × 103 km b) Distance (km) = 2.54435678106 kJ  1 h   0.62 mi   5.510 4 kJ  c) Only a small percentage of the chemical energy in the fuel is converted to work to move the car; most of the chemical energy is lost as waste heat flowing into the surroundings. 6.95

6.96

6.97

q = c × mass × T In this situation, all of the samples have the same mass, 50 g, so mass is not a variable. All also have the same q value, 450 c × T). c T, the lower the value of specific heat capacity: T: B > D > C > A Specific heat capacity: B < D < C < A ClF(g) + 1/2O2(g)  1/2 Cl2O(g) + 1/2OF2(g)

 H rxn = 1/2(167.5 kJ) =

T are inversely

83.75 kJ

 rxn

F2(g) + 1/2O2(g)  OF2(g)

H = 1/2(–43.5 kJ) = –21.75 kJ

1/2Cl2O(g) + 3/2OF2(g)  ClF3(l) + O2(g)

 H rxn = –1/2(394.1 kJ) = –197.05 kJ

ClF(g) + F2(g)  ClF3(l)

 H rxn =

–135.1 kJ

a) AgNO3(aq) + NaI(aq)  AgI(s) + NaNO3(aq) 103 L  5.0 g AgNO3  1 mol AgNO3      Moles of AgNO3 = 50.0 mL   169.9 g AgNO  1L  1 mL  3 





–3

= 1.47145 × 10 mol AgNO3 103 L  5.0 g NaI  1 mol NaI      Moles of NaI = 50.0 mL   1 L 149.9 g NaI   1 mL   





–3

= 1.6677785 × 10 mol NaI The AgNO3 is limiting, and will be used to finish the problem:  1 mol AgI  234.8 g AgI     Mass (g) of AgI = 1.47145103 mol AgNO 3  1 mol AgNO3  1 mol AgI  = 0.345496 = 0.35 g AgI Copyright

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6-27

b) Ag (aq) + I (aq)  AgI(s) +

–

 H rxn = {1 H f [AgI(s)]} – {1 H f [Ag (aq)] + 1 H f [I (aq)]} +

–

= [(1 mol)(–62.38 kJ/mol)] – [(1 mol)(105.9 kJ/mol) + (1 mol)(–55.94 kJ/mol)] = –112.34 = –112.3 kJ/mol AgI  c) H rxn = q = c × mass × T

112.34 kJ  1 mol AgI      1.47145103 mol AgNO   3  1 mol AgNO   mol AgI   10 3 J  3 H rxn   T = =  1 kJ   4.184 J   1.00 g  c mass    50.0  50.0 mL    g  K    mL    = 0.39508 = 0.40 K







6.98



Plan: Use conversion factors to solve parts (a) and (b). For part (c), first find the heat of reaction for the combustion of methane by using the heats of formation of the reactants and products. The enthalpy change of a reaction is the sum of the heats of formation of the products minus the sum of the heats of formation of the reactants. Since the Hfo values (Appendix B) are reported as energy per one mole, use the appropriate stoichiometric coefficient to reflect the higher number of moles. For part (e), convert the amount of water in gal to mass in g and use the relationship q = c × mass × T to find the heat needed; then use the conversion factors between joules and therms and the cost per therm to determine the total cost of heating the water. Solution: 1 cal   4.184 J  453.6 g  1.0 C  1.00 lb F   3 a)       = 1054.368 = 1.1 × 10 J/Btu  g C   1 cal  1 lb   1.8 F   1 Btu  100,000 Btu 1054.368 J    = 1.054368 × 108 = 1.1 × 108 J  b) E = 1.00 therm    1 therm  Btu 





c) CH4(g) + 2O2(g)  rxn

(g) + 2H2O(g)

H = {1 H [CO2(g)] + 2 Hfo [H2O(g)]} – {1 Hfo [CH4(g)] + 2 Hfo [O2(g)]} o f

= [(1 mol)(–393.5 kJ/mol) + (2 mol)(–241.826 kJ/mol)] – [(1 mol)(–74.87 kJ/mol) + (2 mol)(0.0 kJ/mol)] = –802.282 = –802.3 kJ/mol CH4 1.054368108 J   1 kJ 1 mol CH 4   Moles of CH4 = 1.00 therm   3    1 therm 10 J  802.282 kJ  





= 131.4211 = 1.3 × 10 mol CH4  $0.66  1 therm  d) Cost =   = 0.005022 = $0.0050/mol   therm 131.4211 mol   4 qt  1 L  1 mL 1.0 g     = 1.203406 × 106 g    e) Mass (g) of = 318 gal     1 gal 1.057 qt 103 L  mL   4.184 J    1.203406 106 g 42.0  15.0 C = 1.359464 × 108 J q = c × mass × T =   g C 









 

  $0.66  1 therm  = 0.850999 = $0.85  Cost = 1.359494 108 J  8  1 therm  1.05436810 J 





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6-28

6.99

1018 J   1 kJ  1 mol CH 4  16.04 g CH 4  1 kg   14 a) Mass (kg) of = 5600 EJ   3    3  = 1.12 × 10 kg CH4  1 EJ       10 J   802 kJ   1 mol CH 4 10 g 





  1 yr   = 14 yr b) Years = 5600 EJ  2  4.0 10 EJ   1 L   1 kJ  1 mL 1.00 g  4.184 J  1 mol CH 4      100.0  25.0 C      c) Moles of CH4 = 1.00 qt   3  103 L  mL  g C   10 J 802 kJ  1.057 qt         











 

= 0.370172 mol CH4 16.04 g CH 4  103 m 3  35.3 ft 3  1L  3     Volume (ft ) of CH4 = 0.370172 mol CH 4   0.72 g CH  1 L  1 m 3   1 mol CH 4  4   





= 0.291105 = 0.29 ft

3 3 3  1 kJ   1L 1 mol CH 4  16.04 g CH 4   10 m   35.3 ft   d) Volume = 2 1013 J  3   1 mol CH  0.72 g CH  1 L  1 m 3  10 J   802 kJ    4  4 





= 1.9611 × 10 = 2 × 10 ft 6.100

The reaction is exothermic. The argon atoms in the chamber after the reaction are moving with greater kinetic energy, indicating an increase in temperature.

6.101

H2SO4(aq) + 2NaOH(aq) +

SO4(aq) + 2H2O(l)

–

2H (aq) + 2OH (aq) O(l) 2 + –  o H rxn = {2 Hf [H2O(l)]} – {2 Hfo [H (aq)] + 2 Hfo [OH (aq)]} = [(2 mol)(–285.84 kJ/mol)] – [(2 mol)(0 kJ/mol) + (2 mol)(–229.94 kJ/mol)] = –111.8 kJ 1 mole of H2SO4 reacts with 2 moles of NaOH.

 1.00 mL  1.030 g  1.00 L    Mass (g) of H2SO4 solution = 1 mol H 2 SO 4   1.00 mL   0.50 mol H 2 SO 4  103 L  = 2060 g H2SO4 solution  40.00 g NaOH 100 g solution   = 200 g NaOH solution Mass (g) of NaOH solution = 2 mol NaOH    1 mol NaOH  40 g NaOH 

q = c × mass × T 103 J  111.8 kJ    1 kJ  q T= = 11.82°C = c  mass 4.184 J/g C 2060 + 200 g

31C + 11.82C = 42.82 = 43C This temperature is above the temperature at which a flammable vapor could be formed so the temperature increase could cause the vapor to explode. 6.102

a) 2C12H26(l) + 37O2(g)  rxn

(g) + 26H2O(g)

b) H = {24 H [CO2(g)] + 26 Hfo [H2O(g)]} – {2 Hfo [C12H26(g)] + 37 Hfo [O2(g)]} o f

–1.50 × 10 kJ = [(24 mol)(–393.5 kJ/mol) + (26 mol)(–241.826 kJ/mol)] – [(2 mol) Hfo [C12H26(g)] + (37 mol)(0.0 kJ/mol)] Copyright

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6-29

–1.50 × 10 kJ = –9444.0 kJ + –6287.476 kJ – [(2 mol) Hfo [C12H26(g)] + 0.0 kJ] 4

–1.50 × 10 kJ = –15,731.476 kJ – (2 mol) Hfo [C12H26(g)] 4

–1.50 × 10 kJ + 15,731.476 kJ = –(2 mol) Hfo [C12H26(g)] 731.476 kJ = –(2 mol) Hfo [C12H26(g)] 2

Hfo [C12H26(g)] = –365.738 = – 3.66 × 10 kJ  4 qt   1 L   1 mL  0.749 g C12 H 26  1 mol C12 H 26   1.50 10 4 kJ    c) Heat (kJ) = 0.50 gal    3    mL 1.057 qt  10 L   2 mol C12 H 26  1 gal  170.33 g C12 H 26 





= –6.2403 × 10 = –6.2 × 10 kJ 1.055 kJ  0.50 gal   –2   d) Volume (gal) = 1250. Btu   6.240310 4 kJ  = 0.010566 = 1.1 × 10 gal  1 Btu  



6.103



 Plan: Heat of reaction is calculated using the relationship H rxn = m H f(products) – n H f(reactants) . The heats of  formation for all of the species, except SiCl4, are found in Appendix B. Use reaction 3, with its given H rxn , to find the heat of formation of SiCl4(g). Once the heat of formation of SiCl4 is known, the heat of reaction of the other two reactions can be calculated. When reactions 2 and 3 are added to obtain a fourth reaction, the heats of reaction of reactions 2 and 3 are also added to obtain the heat of reaction for the fourth reaction. Solution: a) (3) SiCl4(g) + 2H2O(g)  SiO2(s) + 4HCl(g)  H rxn = {1 H f [SiO2(s)] + 4 Hfo [HCl(g)]} – {1 Hfo [SiCl4(g)] + 2 Hfo [H2O(g)]}

–139.5 kJ = [(1 mol)(–910.9 kJ/mol) + (4 mol)(–92.31 kJ/mol)] – [ Hfo [SiCl4(g)] + (2 mol)(–241.826 kJ/mol)] –139.5 kJ = –1280.14 – [ Hfo [SiCl4(g)] + (–483.652 kJ)] 1140.64 kJ = – Hfo [SiCl4(g)] + 483.652 kJ

Hfo [SiCl4(g)] = –656.988 kJ/mol The heats of reaction for the first two steps can now be calculated. 1) Si(s) + 2Cl2(g)  SiCl4(g) o H rxn = {1 Hfo [SiCl4(g)]} – {1 Hfo [Si(s)] + 2 Hfo [Cl2(g)]}

= [(1 mol)(–656.988 kJ/mol)] – [(1 mol)(0 kJ/mol) + (2 mol)(0 kJ/mol)] = –656.988 = –657.0 kJ 2) SiO2(s) + 2C(graphite) + 2Cl2(g)  SiCl4(g) + 2CO(g) o H rxn = {1 Hfo [SiCl4(g)] + 2 Hfo [CO(g)]}

– {1 Hfo [SiO2(g)] + 2 Hfo [C(graphite)] +2 Hfo [Cl2(g)]} = [(1 mol)(–656.988 kJ/mol) + (2 mol)(–110.5 kJ/mol)] – [(1 mol)(–910.9 kJ/mol) + (2 mol)(0 kJ/mol) + (2 mol)(0 kJ/mol)] = 32.912 = 32.9 kJ b) Adding reactions 2 and 3 yields: o (2) SiO2(s) + 2C(graphite) + 2Cl2(g)  Si Cl4(g) + 2CO(g) H rxn = 32.912 kJ (3)

SiCl4(g) + 2H2O(g)  Si O2(s) + 4HCl(g)

2C(graphite) + 2Cl2(g) + 2H2O(g) g) + 4HCl(g) o Confirm this result by calculating H rxn using Appendix B values. 2C(graphite) + 2Cl2(g) + 2H2O(g)

o H rxn = –139.5 kJ o H rxn = –106.588 kJ = –106.6 kJ

g) + 4HCl(g)

o H rxn = {2 Hfo [CO(g)] + 4 Hfo [HCl(g)]} – {2 Hfo [C(graphite)] + 2 Hfo [Cl2(g)] + 2 Hfo [H2O(g)]}

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6-30

= [(2 mol)(–110.5 kJ/mol) + (4 mol)(–92.31 kJ) – [(2 mol)(0 kJ/mol) + (2 mol)(0 kJ/mol) + (2 mol)(–241.826 kJ/mol)] = –106.588 = –106.6 kJ 6.104

Plan: Use PV = nRT to find the initial volume of nitrogen gas at 0°C and then the final volume at 819°C. Then the relationship w = –P V can be used to calculate the work of expansion. Solution: a) PV = nRT L  atm   1 mol0.0821  273 K nRT  mol  K  Initial volume at 0°C + 273 = 273 K = V = = = 22.4133 L P 1.00 atm



nRT Final volume at 819°C + 273 = 1092 K = V = = P



L  atm   1 mol0.0821 1092 K  mol  K 



 = 89.6532 L

1.00 atm

V = Vfinal – Vinitial = 89.6532 L – 22.4133 L = 67.2399 L w = –PV = –(1 atm) × 67.2399 L = –67.2399 atm L  101.3 J  3 w (J) = 67.2399 atm  L = –6811.40187 = –6.81 × 10 J 1 L  atm  b) q = c × mass × T

 28.02 g   = 28.02 g Mass (g) of N2 = 1 mol N 2   1 mol N 2  T = 6.105

q 6.81255310 3 J = = 243.132 = 243 K = 243°C (c)(mass) 28.02 g (1.00 J/g  K)

Plan: Note the numbers of moles of the reactants and products in the target equation and manipulate equations   1-5 and their H rxn values so that these equations sum to give the target equation. Then the manipulated H rxn  values will add to give the H rxn value of the target equation. Solution: Only reaction 3 contains N2O4(g), and only reaction 1 contains N2O3(g), so we can use those reactions as a starting point. N2O5 appears in both reactions 2 and 5, but note the physical states present: solid and gas. As a rough start, adding reactions 1, 3, and 5 yields the desired reactants and products, with some undesired intermediates:

Reverse (1)

 H rxn = –(–39.8 kJ) =

N2O3(g)  NO(g) + NO2(g)

39.8 kJ

 rxn

Multiply (3) by 2

4NO2(g)  2N2O4(g)

H = 2(–57.2 kJ) = –114.4 kJ

(5)

N2O5(s)  N2O5(g)

 H rxn = (54.1 kJ)

54.1 kJ

N2O3(g) + 4NO2(g) + N2O5(s)  NO(g) + NO2(g) + 2N2O4(g) + N2O5(g) To cancel out the N2O5(g) intermediate, reverse equation 2. This also cancels out some of the undesired NO 2(g) but adds NO(g) and O2(g). Finally, add equation 4 to remove those intermediates:  Reverse (1) N2O3(g)  NO(g) + NO2(g) H rxn = –(–39.8 kJ) = 39.8 kJ Multiply (3) by 2

4NO2(g)  2N2O4(g)

 H rxn = 2(–57.2 kJ) = –114.4 kJ

(5)

N2O5(s) N2O5(g)

 H rxn =

Reverse (2) Copyright

N2O5(g) NO(g) + NO2(g) + O2(g)

54.1 kJ

 rxn

H = –(–112.5 kJ) = 112.5

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6-31

(4) Total: 6.106

2NO(g) + O2(g) 2NO2(g)

 H rxn =

–114.2 kJ

N2O3(g) + N2O5(s)  2N2O4(g)

 rxn

–22.2 kJ

H =

Plan: The enthalpy change of a reaction is the sum of the heats of formation of the products minus the sum of the heats of formation of the reactants. Since the H f values (Appendix B) are reported as energy per one mole, use  the appropriate stoichiometric coefficient to reflect the higher number of moles. In this case, H rxn of the second   reaction is known and H f of N2H4(aq) must be calculated. For part (b), calculate H rxn for the reaction between N2H4(aq) and O2, using the value of H f for N2H4(aq) found in part (a); then determine the moles of O2 present  by multiplying volume and molarity and multiply by the H rxn for the reaction. Solution: a) 2NH3(aq) + NaOCl(aq) H (aq) + NaCl(aq) + H2O(l) 2 4  H rxn = {1 H f [N2H4(aq)] + 1 H f [NaCl(aq)] + 1 H f [H2O(l)]}

– {2 H f [NH3(aq)] + 1 H f [NaOCl(aq)]} Note that the Appendix B value for N2H4 is for N2H4(l), not for N2H4(aq), so this term must be calculated. In addition, Appendix B does not list a value for NaCl(aq), so this term must be broken down into + – H f [Na (aq)] and H f [Cl (aq)]. –151 kJ = [ H f [N2H4(aq)] + (1 mol)(–239.66 kJ/mol) + (1 mol)(–167.46 kJ/mol) + (1 mol)(–285.840 kJ/mol)] – [(2 mol)(–80.83 kJ/mol) + (1 mol)(–346 kJ/mol)]  f  f

–151 kJ = [ H [N2H4(aq)] + (– 692.96 kJ)] – [–507.66 kJ] –151 kJ = [ H [N2H4(aq)] + (–185.3 kJ)

H f [N2H4(aq)] = 34.3 = 34 kJ  2.50 104 mol  b) Moles of O2 = 5.00 10 3 L  = 1.25 mol O2  1L 

N2H4(aq) + O2(g)  rxn

(g) + 2H2O(l)

 f

H = {1 H [N2(g)] + 2 H f [H2O(l)]} – {1 H f [N2H4(aq)] + 1 H f [O2(g)]} = [(1 mol)(0 kJ/mol) + (2 mol)(–285.840 kJ/mol)] – [(1 mol)(34.3 kJ/mol) + (1 mol)(0 kJ/mol] = –605.98 kJ

 605.98 kJ   = –757.475 = –757 kJ Heat (kJ) = 1.25 mol O2   1 mol O2  6.107

CO(g) + 2H2(g)

OH(l)

 H rxn = {1 H f [CH3OH(l)]} – {1 H f [CO(g)] + 2 H f [H2(g)]}

= [(1 mol)(–238.6 kJ/mol)] – [(1 mol)(–110.5 kJ mol) + (2 mol)(0.0 kJ/mol)] = –128.1 kJ Find the limiting reactant:  1 atm  112 kPa15.0 L PV   = 0.5641135 mol CO =  101.325 kPa   L  atm  RT 0.0821  273  85 K  mol  K  1 mol CH3 OH  Moles of CH3OH from CO = 0.5641135 mol CO  = 0.5641135 mol CH3OH  1 mol CO 

Moles of CO =





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6-32

 1 atm  744 torr18.5 L PV   = 0.6338824 mol H =  2   L  atm  RT 760 torr   0.0821 273  75 K     mol  K  1 mol CH 3 OH   = 0.3169412 mol CH3OH Moles of CH3OH from H2 = 0.6338824 mol H 2   2 mol H 2 

Moles of H2 =





H2 is limiting.  128.1 kJ    = –40.6002 = –40.6 kJ Heat (kJ) = 0.6338824 mol H 2   2 mol H 2  Plan: First find the heat of reaction for the combustion of methane. The enthalpy change of a reaction is the sum of the heats of formation of the products minus the sum of the heats of formation of the reactants. Since the H f values (Appendix B) are reported as energy per one mole, use the appropriate stoichiometric coefficient to reflect the higher number of moles. Convert the mass of methane to moles and multiply that mole number by the heat of combustion. Solution: a) The balanced chemical equation for this reaction is: CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)



6.108



 H rxn = {1 H f [CO2(g)] + 2 H f [H2O(g)]} – {1 H f [CH4(g)] + 2 H f [O2(g)]}

= [(1 mol)(–393.5 kJ/mol) + (2 mol)(–241.826 kJ/mol)] – [(1mol)(–74.87 kJ/mol) + (2 mol)(0.0 kJ/mol)] = –802.282 kJ  1 mol   Moles of CH4 = 25.0 g CH 4   = 1.5586 mol CH4 16.04 g CH 4 





 802.282 kJ   = –1250.4 = –1.25 × 103 kJ Heat (kJ) = 1.5586 mol CH 4   1 mol CH 4  b) The heat released by the reaction is “stored” in the gaseous molecules by virtue of their specific heat capacities, c, using the equation q = c × mass × T. The problem specifies heat capacities on a molar basis, so we modify the equation to use moles, instead of mass. The gases that remain at the end of the reaction are CO2 and H2O. All of the methane and oxygen molecules were consumed. However, the oxygen was added as a component of air, which is 78% N2 and 21% O2, and there is leftover N2. 1 mol CO2   = 1.5586 mol CO2(g) Moles of CO2(g) = 1.5586 mol CH 4  1 mol CH 4   2 mol H 2 O   = 3.1172 mol H2O(g) Moles of H2O(g) = 1.5586 mol CH 4   1 mol CH 4   2 mol O2   = 3.1172 mol O2(g) Moles of O2(g) reacted = 1.5586 mol CH 4  1 mol CH 4  Mole fraction N2 = (79%/100%) = 0.79 Mole fraction O2 = (21%/100%) = 0.21

 0.79 mol N 2   = 11.72661 mol N2 Moles of N2(g) = 3.1172 mol O 2 reacted   0.21 mol O 2  q = c × mass × T 10 3 J  6 q = 1250.4 kJ   = 1.2504 × 10 J  1 kJ  Copyright

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6-33

1.2504 × 10 J = (1.5586 mol CO2)(57.2 J/mol°C)(Tfinal – 0.0)°C + (3.1172 mol H2O)(36.0 J/mol°C)(Tfinal – 0.0)°C + (11.72661 mol N2)(30.5 J/mol°C)(Tfinal – 0.0)°C 6

1.2504 × 10 J = 89.15192 J/°C(Tfinal) + 112.2192 J/°C(Tfinal) + 357.6616 J/°C(Tfinal) 6

1.2504 × 10 J = (559.03272 J/°C)Tfinal 6

Tfinal = (1.2504 × 10 J)/(559.0324 J/°C) = 2236.72 = 2.24 × 10 °C

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6-34

CHAPTER 7 QUANTUM THEORY AND ATOMIC STRUCTURE 8

The value for the speed of light will be 3.00 × 10 m/s except when more significant figures are necessary, in which cases, 8 2.9979 × 10 m/s will be used. FOLLOW–UP PROBLEMS 7.1A

Plan: Given the frequency of the light, use the equation c =  to solve for wavelength. Solution: 3.00 108 m/s  1 nm  c  = 414.938 = 415 nm  = = 7.231014 s1 109 m  8 3.00 10 m /s  1 Å  c  = 4149.38 = 4150  = =  14 10 7.23 10 s1 10 m 

7.1B

Plan: Given the wavelength of the light, use the equation c =  to solve for frequency. Remember that wavelength must be in units of m in this equation. Solution: 8 9 10 nm  c  3.00 10 m/s  14 14 –1  = =    = 3.1915 × 10 = 3.2 × 10 s   1m  940 nm     This is infrared radiation.

7.2A

Plan: Use the formula E = h to solve for the frequency. Then use the equation c =  to solve for the wavelength. Solution: 19 E  8.2 10 J  15 15 –1 = =  = 1.2375 × 10 = 1.2 × 10 s 34  h  6.626 10 J  s  (using the unrounded number in the next calculation to avoid rounding errors) =

7.2B

9  3.00 10 8 m/s  10 nm   =   = 240 nm 1.2375 1015 s1  1 m    

Plan: To calculate the energy for each wavelength, we use the formula E = hc/ . Solution: E=

6.62610

34



J  s 3.00 108 m/s

110

6.62610

34

8



J  s 3.00 108 m/s

510

7

 = 1.9878 × 10 = 2 × 10 J –17

–17

 = 3.9756 × 10 = 4 × 10 J –19

–19

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7-1

7.3A

6.626 10 =

34

 = 1.9878 × 10 = 2 × 10 J

J  s)(3.00 10 m /s 4

–21

110 m As the wavelength of light increases from ultraviolet to visible to infrared, the energy of the light decreases.  1 1   18  2  2  to find the energy change; a photon in Plan: Use the equation relating E    nfinal ninitial  the IR (infrared) region is emitted when n has a final value of 3. Then use E = hc/ to find the wavelength of the photon. Solution:



 1 1    n2  n 2   final initial 

18



1 1 J  2  2   3 6 

– 18

E = – 2.18 × 10

E = –1.8166667 × 10

7.3B

= –1.82 × 10

– 19

b) E = 

– 19

34 8  hc 6.626 10 J  s3.00 10 m/s  1 4 4   10  1.094202 × 10 = 1.09 × 10 19 E 1.8166667 10 J 10 m 

Plan: Use the equation E = hc/ to find the energy change for this reaction. Then use the equation

E =

 1 1    n2  n2  to find the final energy level to which the electron moved.  final initial 

18



Solution: a)







 6.626 1034 J  s 3.00 108 m/s 10 9 nm         1 m   410. nm    

E = 4.8483 × 10

Because the photon is emitted, energy is being given off, so the sign of –19

–19

= 4.85 × 10

E should be negative.

–19

Therefore, E = –4.8483 × 10 or –4.85 × 10 J rounded to three significant figures. 23  4.84831019 J   6.022 10 H atoms   1 kJ   E (kJ/mol) =     = –291.9646 = –292 kJ/mol (the number of    1 H atom 1 mol H    1000 J  atoms is a positive number)

1 2 final



2.181018 J



 1 1    n2  n 2   final initial 

18





n 2initial

4.8483 1019 J 18

2.18 10



1 62

= 0.25018

2 = 3.9972 = 4 nfinal

(The final energy level is an integer, so its square is also an integer.)

nfinal = 2 Copyright

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7-2

7.4A

Plan: With the equation for the de Broglie wavelength, = h/mu and the given de Broglie wavelength, calculate the electron speed. The wavelength must be expressed in meters. Use the same formulas to calculate the speed of the golf ball. The mass of both the electron and the golf ball must be expressed in kg in their respective calculations. Solution: a) = h/mu

h u= = m

7.4B

6.62610

34

 kg  m 2 /s2    = 7273.3 = 7.27 × 103 m/s  109 m   J   31  9.1110 kg  100. nm   1 nm       1 kg    = 0.0459 kg b) Mass (kg) of the golf ball = (45.9 g)  1000 g 

h u= = m



Js





 kg  m 2 /s2    = 1.4436 × 10–25 = 1.44 × 10–25 m/s      109   J    (0.0459 kg) (100. nm)  1 nm    6.626 1034 J  s

Plan: Use the equation for the de Broglie wavelength, = h/mu with the given mass and speed to calculate the de Broglie wavelength of the racquetball. The mass of the racquetball must be expressed in kg, and the speed must be expressed in m/s in the equation for the de Broglie wavelength. Solution:  1 kg   Mass (kg) of the racquetball = (39.7 g)   = 0.0397 kg 1000 g   55 mi  1 hr 1.609 km 1000 m     = 24.5819 = 25 m/s   Speed (m/s) of the racquetball =   hr  3600 s  mi  1 km  = h/mu =

7.5A

6.626 1034 J  s h –34 –34 = = 6.67607 × 10 = 6.7 × 10 m mu (0.0397 kg)(25 m/s)

Plan: Use the equation

xm u 

h to solve for the uncertainty ( x) in position of the baseball. 4

Solution: h 4 u = 1% of u = 0.0100(44.7 m/s) = 0.447 m/s xm u 

x

h 4

x 7.5B

= – 34

× 10

6.626 1034 J  s 4 (0.142 kg)(0.447 m/s)

= 8.3070285 × 10

– 34

Plan: Use the equation

xm u 

h 4

to solve for the uncertainty ( x) in position of the neutron.

Solution: h 4 7 5 u = 1% of u = 0.0100(8 × 10 m/s) = 8 × 10 m/s xm u 

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7-3

h x  4  x

6.626 1034 J  s



4 1.6710

27





kg 8 10 m/s 5

= 3.9467 × 10

–14

4 × 10 m

7.6A

Plan: Following the rules for l (integer from 0 to n – 1) and ml (integer from –l to +l), write quantum numbers for n = 4. Solution: For n = 4 l = 0, 1, 2, 3 For l = 0, ml = 0 For l = 1, ml = –1, 0, 1 For l = 2, ml = –2, –1, 0, 1, 2 For l = 3, ml = –3, –2, –1, 0, 1, 2, 3

7.6B

Plan: Following the rules for l (integer from 0 to n – 1) and ml (integer from –l to +l), determine which value of the principal quantum number has five allowed levels of l. Solution: The number of possible l values is equal to n, so the n = 5 principal quantum number has five allowed values of l. For n = 5 l = 0, 1, 2, 3, 4 For l = 0, ml = 0 For l = 1, ml = –1, 0, 1 For l = 2, ml = –2, –1, 0, 1, 2 For l = 3, ml = –3, –2, –1, 0, 1, 2, 3 For l = 4, ml = –4, –3, –2, –1, 0, 1, 2, 3, 4

7.7A

Plan: Identify n and l from the sublevel designation. n is the integer in front of the sublevel letter. The sublevels are given a letter designation, in which s represents l = 0, p represents l = 1, d represents l = 2, and f represents l = 3. Knowing the value for l, find the ml values (integer from –l to +l). Solution: Sublevel name 4p 3d

7.7B

n value 4 3

l value 1 2

ml values –1, 0, 1 –2, –1, 0, 1, 2

n value 4 6

l value 2 0

ml values –2, –1, 0, 1, 2 0

The number of orbitals for each sublevel equals 2l + 1. Sublevel 4d should have 5 orbitals and sublevel 6s should have 1 orbital. Both of these agree with the number of ml values for the sublevel. 7.8A

Plan: Use the rules for designating quantum numbers to fill in the blanks. For a given n, l can be any integer from 0 to n–1. For a given l, ml can be any integer from – l to + l. The sublevels are given a letter designation, in which s represents l = 0, p represents l = 1, d represents l = 2, and f represents l = 3.

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7-4

Solution: The completed table is: n 4 2 3 2

a) b) c) d) 7.8B

l 1 1 2 0

ml 0 0 –2 0

Name 4p 2p 3d 2s

Plan: Use the rules for designating quantum numbers to determine what is wrong with the quantum number designations provided in the problem. For a given n, l can be any integer from 0 to n–1. For a given l, ml can be any integer from – l to + l. The sublevels are given a letter designation, in which s represents l = 0, p represents l = 1, d represents l = 2, and f represents l = 3. Solution: The provided table is: n 5 2 6

a) b) c)

l 3 2 1

ml 4 1 –1

Name 5f 2d 6s

a) For l = 3, the allowed values for ml are –3, –2, –1, 0, 1, 2, 3, not 4 b) For n = 2, l = 0 or 1 only, not 2; the sublevel is 2p, since ml = 1. c) The value l = 1 indicates the p sublevel, not the s; the sublevel name is 6p. TOOLS OF THE LABORATORY BOXED READING PROBLEMS B7.1

Plan: Plot absorbance on the y-axis and concentration on the x-axis. Since this is a linear plot, the graph is of the type y = mx + b, with m = slope and b = intercept. Any two points may be used to find the slope, and the slope is used to find the intercept. Once the equation for the line is known, the absorbance of the solution in part (b) is used to find the concentration of the diluted solution, after which the dilution equation is used to find the molarity of the original solution. Solution: a) Absorbance vs. concentration:

Absorbance

0.5 0.4 0.3 0.2 0.1 0 0

0.00001

0.00002

0.00003

0.00004

Concentration (M)

This is a linear plot, thus, using the first and last points given: Copyright

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7-5





0.396  0.131 y2  y1 4 –1 = = 13,250 = 1.3 × 10 M 5 5 x 2  x1 3.0 10  1.0 10 M





Using the slope just calculated and any of the data points, the value of the intercept may be found. –1 –5 b = y – mx = 0.396 – (13,250 M )(3.0 × 10 M) = –0.0015 = 0.00 (absorbance has no units) –1 b) Use the equation just determined: y = (13,250 M ) x + 0.00. –1 –1 –5 –5 x = (y – 0.00)/(13,250 M ) = (0.236/13,250 M ) = 1.78113 × 10 M = 1.8 × 10 M This value is Mf in a dilution problem (MiVi)= (MfVf) with Vi = 20.0 mL and Vf = 150 mL.

 Mf Vf  1.7811310 M 150 mL –4 –4 = = 1.335849 × 10 = 1.3 × 10 M 20.0 mL Vi    5

Mi = B7.2

Plan: The color of light associated with each wavelength can be found from Figure 7.3. The frequency of each c wavelength can be determined from the relationship c =  or  = . The wavelength in nm must be converted to meters. Solution: a) red

3.00 108 m /s  1 nm   = 4.4709 × 1014 = 4.47 × 1014 s–1  = 9   671 nm 10 m 

3.00 108 m /s  1 nm   = 6.6225 × 1014 = 6.62 × 1014 s–1  b) blue = 9   453 nm 10 m  3.00 108 m /s  1 nm   = 5.0933786 × 1014 = 5.09 × 1014 s–1  c) yellow-orange  = 109 m  589 nm  

END–OF–CHAPTER PROBLEMS 7.1

All types of electromagnetic radiation travel as waves at the same speed. They differ in both their frequency, wavelength, and energy.

7.2

Plan: Recall that the shorter the wavelength, the higher the frequency and the greater the energy. Figure 7.3 describes the electromagnetic spectrum by wavelength and frequency. Solution: –2 12 a) Wavelength increases from left (10 nm) to right (10 nm) in Figure 7.3. The trend in increasing wavelength is: x-ray < ultraviolet < visible < infrared < microwave < radio wave. b) Frequency is inversely proportional to wavelength according to the equation c = , so frequency has the opposite trend: radio wave < microwave < infrared < visible < ultraviolet < x-ray. c) Energy is directly proportional to frequency according to the equation E = . Therefore, the trend in increasing energy matches the trend in increasing frequency: radio wave < microwave < infrared < visible < ultraviolet < x-ray.

7.3

a) Refraction is the bending of light waves at the boundary of two media, as when light travels from air into water. b) Diffraction is the bending of light waves around an object, as when a wave passes through a slit about as wide as its wavelength. c) Dispersion is the separation of light into its component colors (wavelengths), as when light passes through a prism. d) Interference is the bending of light through a series of parallel slits to produce a diffraction pattern of brighter and darker spots. Note: Refraction leads to a dispersion effect and diffraction leads to an interference effect.

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7-6

7.4

Evidence for the wave model is seen in the phenomena of diffraction and refraction. Evidence for the particle model includes the photoelectric effect and blackbody radiation.

7.5

a) Frequency: C < B < A b) Energy: C < B < A c) Amplitude: B < C < A d) Since wave A has a higher energy and frequency than B, wave A is more likely to cause a current. e) Wave C is more likely to be infrared radiation since wave C has a longer wavelength than B.

7.6

Radiation (light energy) occurs as quanta of electromagnetic radiation, where each packet of energy is called a photon. The energy associated with this photon is fixed by its frequency, E = h . Since energy depends on frequency, a threshold (minimum) frequency is to be expected. A current will flow as soon as a photon of sufficient energy reaches the metal plate, so there is no time lag.

7.7

Plan: Wavelength is related to frequency through the equation c = . Recall that a Hz is a reciprocal second, or –1 1/s = s . Assume that the number “950” has three significant figures. Solution: c= 

c (m) = = 

3.00 108 m /s

= 315.789 = 316 m 103 Hz  1  s      950. kHz     1 kHz  Hz  c  1 nm  11 11 (nm) = = 315.789 m  9 = 3.15789 × 10 = 3.16 × 10 nm 10 m    1 Å c  12 12 ( ) = = 315.789 m  10  = 3.158 × 10 = 3.16 × 10 10 m     7.8





Wavelength and frequency relate through the equation c = . Recall that a Hz is a reciprocal second, or 1/s = s . –1

c (m) = = 

3.00 108 m/s

= 3.208556 = 3.21 m 106 Hz  1  s     93.5 MHz    1 MHz  Hz    c  1 nm  9 9 (nm) = = 3.208556 m  9 = 3.208556 × 10 = 3.21 × 10 nm 10 m    1 Å  c  10 10 ( ) = = 3.208556 m  10  = 3.208556 × 10 = 3.21 × 10  10 m  7.9

7.10





–1

Plan: Frequency is related to energy through the equation E = h. Note that 1 Hz = 1 s . Solution: E = h –34 10 –1 –23 –23 E = (6.626 × 10 J s)(3.8 × 10 s ) = 2.51788 × 10 = 2.5 × 10 J

6.626 10 =

34

J  s3.00 108 m /s  1 Å   = 1.5291 × 10–15 = 1.5 × 10–15 J   10  1.3 Å 10 m 

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7-7

7.11

Plan: Energy is inversely proportional to wavelength ( E =

). As wavelength decreases, energy increases.

Solution: In terms of increasing energy, the order is red < yellow < blue. 15

–1

7.12

Since energy is directly proportional to frequency (E = h): UV ( = 8.0 × 10 s ) > IR ( = 6.5 × 10 s ) > 11 –1 microwave ( = 9.8 × 10 s ) or UV > IR > microwave.

7.13

Plan: Wavelength is related to frequency through the equation c = . Recall that a Hz is a reciprocal second, –1 or 1/s = s . Solution: 10 9 Hz  s1   –1 10 –1   = (s ) = 22.235 GHz    = 2.2235 × 10 s  1 GHz  Hz  





2.9979 108 m /s  1 nm  c  = 1.3482797 × 107 = 1.3483 × 107 nm  (nm) = = 10 1  9   2.2235 10 s 10 m 

(

7.14

 2.9979 108 m /s  1 c  = 1.3482797 × 108 = 1.3483 × 108  =  2.2235 1010 s1 1010 m 

Frequency and wavelength can be calculated using the speed of light: c = . 3.00 108 m /s  1 m  c 13 13 –1  a)  = =  = 3.125 × 10 = 3.1 × 10 s 106 m  9.6 m  

c b) (m) = = 

7.15

 1 m    = 3.464979 = 3.465 m 1   106 m   s 13   8.652 10 Hz    Hz  2.9979 108 m/s





Frequency and energy are related by E = h, and wavelength and energy are related by E = hc/ . 106 eV  19   1.602 10 J  1.33 MeV     1 MeV  1 eV   Hz    E 20 20  (Hz) = =  1  = 3.2156 × 10 = 3.22 × 10 Hz s  h 6.626 1034 J  s



(m) =

hc = E



6.626 10

34



J  s 3.00 108 m /s

 = 9.32950 × 10 = 9.33 × 10 m –13

–13

 106 eV  1.602 1019 J   1.33 MeV    1 MeV  1 eV   





The wavelength can also be found using the frequency calculated in the equation c = . 7.16

Plan: The least energetic photon in part (a) has the longest wavelength (242 nm). The most energetic photon in part (b) has the shortest wavelength (2200 Å). Use the relationship c =  to find the frequency of the photons and relationship E =

to find the energy.

Solution: a) c = 

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7-8

=

3.00 108 m /s  1 nm   = 1.239669 × 1015 = 1.24 × 1015 s–1  = 9  242 nm 10 m 

b)  =

6.626 10

34





J  s 3.00 108 m /s  1 nm   = 8.2140 × 10–19 = 8.21 × 10–19 J  109 m  242 nm  

 3.00 108 m /s  1  = 1.3636 × 1015 = 1.4 × 1015 s–1  =  10  2200 10 m 

6.626 10

34





J  s 3.00 108 m /s  1 Å   = 9.03545 × 10–19 = 9.0 × 10–19 J   10  2200 Å 10 m 

7.17

“n” in the Rydberg equation is equal to a Bohr orbit of quantum number “n” where n = 1, 2, 3, ....

7.18

Bohr’s key assumption was that the electron in an atom does not radiate energy while in a stationary state, and the electron can move to a different orbit by absorbing or emitting a photon whose energy is equal to the difference in energy between two states. These differences in energy correspond to the wavelengths in the known spectra for the hydrogen atoms. A Solar System model does not allow for the movement of electrons between levels.

7.19

An absorption spectrum is produced when atoms absorb certain wavelengths of incoming light as electrons move from lower to higher energy levels and results in dark lines against a bright background. An emission spectrum is produced when atoms that have been excited to higher energy emit photons as their electrons return to lower energy levels and results in colored lines against a dark background. Bohr worked with emission spectra.

7.20

Plan: The quantum number n is related to the energy level of the electron. An electron absorbs energy to change from lower energy (lower n) to higher energy (higher n), giving an absorption spectrum. An electron emits energy as it drops from a higher energy level (higher n) to a lower one (lower n), giving an emission spectrum. Solution: a) The electron is moving from a lower value of n (2) to a higher value of n (4): absorption b) The electron is moving from a higher value of n (3) to a lower value of n (1): emission c) The electron is moving from a higher value of n (5) to a lower value of n (2): emission d) The electron is moving from a lower value of n (3) to a higher value of n (4): absorption

7.21

The Bohr model works only for a one-electron system. The additional attractions and repulsions in many-electron systems make it impossible to predict accurately the spectral lines.

7.22

The Bohr model has successfully predicted the line spectra for the H atom and Be ion since both are one-

electron species. The energies could be predicted from En =



 Z 2  2.18 1018 J 2

 where Z is the atomic number

n 3+ for the atom or ion. The line spectra for H would not match the line spectra for Be since the H nucleus contains 3+

one proton while the Be nucleus contains 4 protons (the Z values in the equation do not match); the force of attraction of the nucleus for the electron would be greater in the beryllium ion than in the hydrogen atom. This means that the pattern of lines would be similar, but at different wavelengths. 7.23

Plan: Calculate wavelength by substituting the given values into Equation 7.3, where n1 = 2 and n2 = 5 because n2 > n1. Although more significant figures could be used, five significant figures are adequate for this calculation.

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7-9

Solution: 1 1 1   R  2  2  n2   n1 n1 = 2

R = 1.096776 × 10 m

–1

n2 = 5

1 1 1  1  R  2  2  = 1.096776 10 7 m 1  2  2  = 2,303,229.6 m–1    n1 2 n2  5 

   1 nm  1  (nm) =   = 434.1729544 = 434.17 nm  1   2,303, 229.6 m  109 m 

7.24

Calculate wavelength by substituting the given values into the Rydberg equation, where n1 = 1 and n2 = 3 because n2 > n1. Although more significant figures could be used, five significant figures are adequate for this calculation. 1 1 1 1  1  R  2  2  = 1.096776 10 7 m1  2  2  = 9,749,120 m–1    n1  n2  1 3  (

7.25

  1 Å 1   = 1025.7336 = 1025.7  ) =  1   9, 749,120 m 1010 m   

Plan: The Rydberg equation is needed. For the infrared series of the H atom, n1 equals 3. The least energetic spectral line in this series would represent an electron moving from the next highest energy level, n2 = 4. Although more significant figures could be used, five significant figures are adequate for this calculation. Solution:

1 1 1  1  R  2  2  = 1.096776 10 7 m1  2  2  = 533,155 m–1    n1 3 n2  4 

  1 nm  1   = 1875.627 = 1875.6 nm  (nm) =  1   533,155 m 109 m 

7.26

Plan: The Rydberg equation is needed. For the visible series of the H atom, n1 equals 2. The least energetic spectral line in this series would represent an electron moving from the next highest energy level, n = 3. Although more significant figures could be used, five significant figures are adequate for this calculation. Solution: 1 1 1 1  1  R  2  2  = 1.096776 10 7 m1  2  2  = 1,523,300 m–1   2 n2  3   n1    1 nm  1  (nm) =   9  = 656.4695 = 656.47 nm 1, 523,300 m 1  10 m 

7.27

Plan: To find the transition energy, use the equation for the energy of an electron transition and multiply by Avogadro’s number to convert to energy per mole. Solution:  1 1  18 E = 2.18 10 J  2  2  ninitial   nfinal





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7-10

1 1 –19 E = 2.181018 J  2  2  = –4.578 × 10 J/photon  2 5  23  4.578 1019 J   6.022 10 photons  = –2.75687 × 105 = –2.76 × 105 J/mol E =    photon 1 mol   





The energy has a negative value since this electron transition to a lower n value is an emission of energy. 7.28

To find the transition energy, use the equation for the energy of an electron transition and multiply by Avogadro’s number.  1 1  18 E = 2.18 10 J  2  2  ninitial   nfinal





1 1 –18 E = 2.18 1018 J  2  2  = 1.93778 × 10 J/photon  3 1  23 1.93778 1018 J   6.022 10 photons  = 1.1669 × 106 = 1.17 × 106 J/mol E =    photon 1 mol   



7.29



Plan: Determine the relative energy of the electron transitions. Remember that energy is directly proportional to frequency (E = h). Solution: Looking at an energy chart will help answer this question. n=5 n=4 n=3

(d) (a)

Frequency is proportional to energy so the smallest frequency will be (d) n = 4 to n = 3; levels 3 and 4 have a smaller E than the levels in the other transitions. The largest frequency is (b) n = 2 to n = 1 since levels 1 and 2 have a larger E than the levels in the other transitions. Transition (a) n = 2 to n = 4 will be smaller than transition (c) n = 2 to n = 5 since level 5 is a higher energy than level 4. In order of increasing frequency the transitions are d < a < c < b. 7.30

b>c>a>d

7.31

Plan: Use the Rydberg equation. Since the electron is in the ground state (lowest energy level), n1 = 1. Convert the wavelength from nm to units of meters. Solution: 109 m  –8 = 97.20 nm  ground state: n1 = 1; n2 = ?  = 9.720 × 10 m  1 nm 



1

= 1.096776 10 m 7





 n1  n1  2 1

2 2

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7-11

 1  1 7 1  1  1.096776  10 m  =  12 n22  9.720 108 m 





1 1 0.93803 =  2  2  1 n2  1 = 1 – 0.93803 = 0.06197 n22 n22 = 16.14

n2 = 4 7.32

109 m  –6 = 1281 nm   = 1.281 × 10 m  1 nm 

= 1.096776 10 m 7

1

1

1

 n  n  2

2 2

 1



 1  1 7 1  1  1.096776  10 m  =   n12 52  1.281106 m





1 1 0.07118 =  2  2  5   n1 1 = 0.07118 + 0.04000 = 0.11118 n12 n12 = 8.9944 n1 = 3 hc

6.626 10

34





J  s 3.00 108 m/s  1 nm  –19 –19   = 4.55917 × 10 = 4.56 × 10 J 9   10 m  436 nm 

7.33

7.34

a) Absorptions: A, C, D; Emissions: B, E, F b) Energy of emissions: E < F < B c) Wavelength of absorption: D < A < C

7.35

If an electron occupies a circular orbit, only integral numbers of wavelengths (= 2n r) are allowed for acceptable standing waves. A wave with a fractional number of wavelengths is forbidden due to destructive interference with itself. In a musical analogy to electron waves, the only acceptable guitar string wavelengths are those that are an integral multiple of twice the guitar string length (2 L).

7.36

De Broglie’s concept is supported by the diffraction properties of electrons demonstrated in an electron microscope.

7.37

Macroscopic objects have significant mass. A large m in the denominator of = h/mu will result in a very small wavelength. Macroscopic objects do exhibit a wavelike motion, but the wavelength is too small for humans to see it.

7.38

The Heisenberg uncertainty principle states that there is fundamental limit to the accuracy of measurements. This limit is not dependent on the precision of the measuring instruments, but is inherent in nature.

7.39

Plan: Use the de Broglie equation. Mass in lb must be converted to kg and velocity in mi/h must be converted to 2 2 m/s because a joule is equivalent to kg m /s .





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7-12

Solution:  1 kg    = 105.2154 kg a) Mass (kg) = 232 lb   2.205 lb  3 19.8 mi  1 km 10 m  1 h  Velocity (m/s) =    = 8.87097 m/s    h  0.62 mi  1 km  3600 s 

h = mu

6.626 10

34

J  s

 kg  m 2 /s2    = 7.099063 × 10–37 = 7.10 × 10–37 m    J m  105.2154 kg 8.87097    s





3  0.1 mi  1 km 10 m  1 h   b) Uncertainty in velocity (m/s) =    = 0.0448029 m/s    h  0.62 mi  1 km  3600 s 

x mv  x 

7.40

h 4



6.626 10



6.62610

x 



34

–15

= 6.6 × 10

3  kg  m 2 /s2  10 g  0.62 mi  1 km  3600 s         1 kg  1 km 103 m  1 h   0.110 7 mi   J        24  6.610 g    h  

(6.6261034 J  s)





–14

 2 × 10

–14

Plan: Use the de Broglie equation. Mass in g must be converted to kg and wavelength in 2 2 m because a joule is equivalent to kg m /s . Solution:  1 kg   Mass (kg) = 56.5 g 3  = 0.0565 kg 10 g  1010 m    = 5.4 × 10–7 m Wavelength (m) = 5400 Å   1   h = mu



u= Copyright

–15

h 4

 1.783166 × 10 7.41

Js







3  kg  m 2 /s2  10 g  0.62 mi  1 km  3600 s      3       J 24  1 kg  1 km 10 m  1 h  7 mi     6.6 10 g 3.4 10   h 

= 6.59057 × 10 b) x mv 

J  s

 kg  m 2 /s2     1.11855 × 10–35  1 × 10–35 m   0.0448029 m  J   4 105.2154 kg   s  



h a) = = mu

34

must be converted to







6.626 1034 J  s  kg  m 2 /s2  h   = 2.1717 × 10–26 = 2.2 × 10–26 m/s =  7 J m  0.0565 kg 5.4 10 m





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7-13

7.42

h mu

 

 

6.626 1034 J  s  kg  m 2 /s2 103 g  1 pm  h –23 –23     u= =   12  = 4.666197 × 10 = 4.67 × 10 m/s  J m  1 kg 10 m  142 g 100. pm  7.43



Plan: The de Broglie wavelength equation will give the mass equivalent of a photon with known wavelength and velocity. The term “mass equivalent” is used instead of “mass of photon” because photons are quanta of 8 electromagnetic energy that have no mass. A light photon’s velocity is the speed of light, 3.00 × 10 m/s. Wavelength in nm must be converted to m. Solution: 109 m    = 5.89 × 10–7 m Wavelength (m) = 589 nm   1 nm  h = mu



7.44

= m=



6.626 10

34

h mu

5.8910

7



J  s



m 3.00108 m/s

 kg  m 2 /s2    = 3.7499 × 10–36 = 3.75 × 10–36 kg/photon  J 

6.626 10 J s  kg m /s  1 nm  =  = 3.2916 × 10 kg/photon   671 nm3.00 10 m /s  J 10 m  34



–36

9

23  3.2916 1036 kg   6.022 10 photons  = 1.9822 × 10–12 = 1.98 × 10–12 kg/mol     photon mol  

7.45

The quantity

7.46

Since is the probability of finding an electron within a small region or volume, electron density would represent a probability per unit volume and would more accurately be called electron probability density.

7.47

A peak in the radial probability distribution at a certain distance means that the total probability of finding the electron is greatest within a thin spherical volume having a radius very close to that distance. Since principal quantum number (n) correlates with distance from the nucleus, the peak for n = 2 would occur at a greater distance from the nucleus than 0.529 Å. Thus, the probability of finding an electron at 0.529 Å is much greater for the 1s orbital than for the 2s.

7.48

a) Principal quantum number, n, relates to the size of the orbital. More specifically, it relates to the distance from the nucleus at which the probability of finding an electron is greatest. This distance is determined by the energy of the electron. b) Angular momentum quantum number, l, relates to the shape of the orbital. It is also called the azimuthal quantum number. c) Magnetic quantum number, ml, relates to the orientation of the orbital in space in three-dimensional space.

expresses the probability of finding an electron within a specified tiny region of space.

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7-14

7.49

Plan: The following letter designations correlate with the following l quantum numbers: l = 0 = s orbital; l = 1 = p orbital; l = 2 = d orbital; l = 3 = f orbital. Remember that allowed ml values are – l to + l. The number of orbitals of a particular type is given by the number of possible ml values. Solution: a) There is only a single s orbital in any shell. l = 1 and ml = 0: one value of ml = one s orbital. b) There are five d orbitals in any shell. l = 2 and ml = –2, –1, 0, +1, +2. Five values of ml = five d orbitals. c) There are three p orbitals in any shell. l = 1 and ml = –1, 0, +1. Three values of ml = three p orbitals. d) If n = 3, l = 0(s), 1(p), and 2(d). There is a 3s (1 orbital), a 3p set (3 orbitals), and a 3d set (5 orbitals) for a total of nine orbitals (1 + 3 + 5 = 9).

7.50

a) All f orbitals consist of sets of seven (l = 3 and ml = –3, –2, –1, 0, +1, +2, +3). b) All p orbitals consist of sets of three (l = 1 and ml = –1, 0, +1). c) All d orbitals consist of sets of five (l = 2 and ml = –2, –1, 0, +1, +2). d) If n = 2, then there is a 2s (1 orbital) and a 2p set (3 orbitals) for a total of four orbitals (1 + 3 = 4).

7.51

Plan: Magnetic quantum numbers (ml) can have integer values from –l to + l. The l quantum number can have integer values from 0 to n – 1. Solution: a) l = 2 so ml = –2, –1, 0, +1, +2 b) n = 1 so l = 1 – 1 = 0 and ml = 0 c) l = 3 so ml = –3, –2, –1, 0, +1, +2, +3

7.52

Magnetic quantum numbers can have integer values from –l to +l. The l quantum number can have integer values from 0 to n – 1. a) l = 3 so ml = –3, –2, –1, 0, +1, +2, +3 b) n = 2 so l = 0 or 1; for l = 0, ml = 0; for l = l, ml = –1, 0, +1 c) l = 1 so ml = –1, 0, +1

7.53

Plan: The s orbital is spherical; p orbitals have two lobes; the subscript x indicates that this orbital lies along the x-axis. Solution: a) s: spherical b) px: 2 lobes along the x-axis

x y

The variations in coloring of the p orbital are a consequence of the quantum mechanical derivation of atomic orbitals that are beyond the scope of this course.

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7-15

7.54

a) pz: 2 lobes along the z-axis

b) dxy: 4 lobes

The variations in coloring of the p and d orbitals are a consequence of the quantum mechanical derivation of atomic orbitals that are beyond the scope of this course. 7.55

Plan: The following letter designations for the various sublevels (orbitals) correlate with the following l quantum numbers: l = 0 = s orbital; l = 1 = p orbital; l = 2 = d orbital; l = 3 = f orbital. Remember that allowed ml values are – l to + l. The number of orbitals of a particular type is given by the number of possible ml values. Solution: sublevel allowable ml # of possible orbitals a) d (l = 2) –2, –1, 0, +1, +2 5 b) p (l = 1) –1, 0, +1 3 c) f (l = 3) –3, –2, –1, 0, +1, +2, +3 7

7.56

sublevel a) s (l = 0) b) d (l = 2) c) p (l = 1)

7.57

Plan: The integer in front of the letter represents the n value. The letter designates the l value: l = 0 = s orbital; l = 1 = p orbital; l = 2 = d orbital; l = 3 = f orbital. Remember that allowed ml values are – l to + l. Solution: a) For the 5s subshell, n = 5 and l = 0. Since ml = 0, there is one orbital. b) For the 3p subshell, n = 3 and l = 1. Since ml = –1, 0, +1, there are three orbitals. c) For the 4f subshell, n = 4 and l = 3. Since ml = –3, –2, –1, 0, +1, +2, +3, there are seven orbitals.

7.58

a) n = 6; l = 4; 9 orbitals (ml = –4, –3, –2, –1, 0, +1, +2, +3, +4) b) n = 4; l = 0; 1 orbital (ml = 0) c) n = 3; l = 2; 5 orbitals (ml = –2, –1, 0, +1, +2)

7.59

Plan: Allowed values of quantum numbers: n = positive integers; l = integers from 0 to n – 1; ml = integers from – l through 0 to + l. Solution: a) n = 2; l = 0; ml = –1: With n = 2, l can be 0 or 1; with l = 0, the only allowable ml value is 0. This combination is not allowed. To correct, either change the l or ml value. Correct: n = 2; l = 1; ml = –1 or n = 2; l = 0; ml = 0. b) n = 4; l = 3; ml = –1: With n = 4, l can be 0, 1, 2, or 3; with l = 3, the allowable ml values are –3, –2, –1, 0, +1, +2, +3. Combination is allowed.

allowable ml 0 –2, –1, 0, +1, +2 –1, 0, +1

# of possible orbitals 1 5 3

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7-16

c) n = 3; l = 1; ml = 0: With n = 3, l can be 0, 1, or 2; with l = 1, the allowable ml values are –1, 0, +1. Combination is allowed. d) n = 5; l = 2; ml = +3: With n = 5, l can be 0, 1, 2, 3, or 4; with l = 2, the allowable ml values are –2, –1, 0, +1, +2. +3 is not an allowable ml value. To correct, either change l or ml value. Correct: n = 5; l = 3; ml = +3 or n = 5; l = 2; ml = 0. 7.60

a) Combination is allowed. b) No; n = 2; l = 1; ml = +1 or n = 2; l = 1; ml = 0 c) No; n = 7; l = 1; ml = +1 or n = 7; l = 3; ml = 0 d) No; n = 3; l = 1; ml = –1 or n = 3; l = 2; ml = –2

7.61

Determine the max for -carotene by measuring its absorbance in the 610-640 nm region of the visible spectrum. Prepare a series of solutions of -carotene of accurately known concentration (using benzene or chloroform as a solvent), and measure the absorbance for each solution. Prepare a graph of absorbance versus concentration for these solutions and determine its slope (assuming that this material obeys Beer’s law). Measure the absorbance of the oil expressed from orange peel (diluting with solvent if necessary). The -carotene concentration can then be either read directly from the calibration curve or calculated from the slope (A = kC, where k = slope of the line and C = concentration).

7.62

Plan: For part (a), use the values of the constants h, , me, and a0 to find the overall constant in the equation. hc Use the resulting equation to calculate E in part (b). Use the relationship E = to calculate the wavelength in 2

part (c). Remember that a joule is equivalent to kg m /s . Solution: –34 –31 –12 a) h = 6.626 × 10 J s; me = 9.1094 × 10 kg; a0 = 52.92 × 10 m  1  h2 h2   E=  2 =  2 2 2 2   2  8 8 e

E= 

6.626 10 

34

Js





8 2 9.1094 1031 kg 52.92 1012 m

 kg  m 2 /s2  1    2  2   J  n 



1 1 –18 J)  2  = –(2.180 × 10 J)  2  n  n  –18 This is identical with the result from Bohr’s theory. For the H atom, Z = 1 and Bohr’s constant = –2.18 × 10 J. For the hydrogen atom, derivation using classical principles or quantum-mechanical principles yields the same constant. b) The n = 3 energy level is higher in energy than the n = 2 level. Because the zero point of the atom’s energy is defined as an electron’s infinite distance from the nucleus, a larger negative number describes a lower energy level. Although this may be confusing, it makes sense that an energy change would be a positive number. 1 1 –18 –19 –19 E = –(2.17963 × 10 J)  2  2  = –3.02726388 × 10 = 3.027 × 10 J 2 3  hc c) E = –18

= –(2.17963 × 10

6.626 1034 J  s2.9979 108 m/s  hc –7 –7 (m) = = = 6.56173 × 10 = 6.562 × 10 m  19 E 3.0272638810 J

 1 nm  (nm) = 6.56173 107 m  9  = 656.173 = 656.2 nm 10 m  This is the wavelength for the observed red line in the hydrogen spectrum. Copyright

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7-17

7.63

Plan: When light of sufficient frequency (energy) shines on metal, electrons in the metal break free and a current flows. Solution: a) The lines do not begin at the origin because an electron must absorb a minimum amount of energy before it has enough energy to overcome the attraction of the nucleus and leave the atom. This minimum energy is the energy of photons of light at the threshold frequency. b) The lines for K and Ag do not begin at the same point. The amount of energy that an electron must absorb to leave the K atom is less than the amount of energy that an electron must absorb to leave the Ag atom, where the attraction between the nucleus and outer electron is stronger than in a K atom. c) Wavelength is inversely proportional to energy. Thus, the metal that requires a larger amount of energy to be absorbed before electrons are emitted will require a shorter wavelength of light. Electrons in Ag atoms require more energy to leave, so Ag requires a shorter wavelength of light than K to eject an electron. d) The slopes of the line show an increase in kinetic energy as the frequency (or energy) of light is increased. Since the slopes are the same, this means that for an increase of one unit of frequency (or energy) of light, the increase in kinetic energy of an electron ejected from K is the same as the increase in the kinetic energy of an electron ejected from Ag. After an electron is ejected, the energy that it absorbs above the threshold energy becomes the kinetic energy of the electron. For the same increase in energy above the threshold energy, for either K or Ag, the kinetic energy of the ejected electron will be the same.

7.64

a) E =

6.626 10



J  s)(3.00 10 8 m /s  1 nm   = 2.8397 × 10–19 J  9  700. nm 10 m 

34

This is the value for each photon, that is, J/photon.  1 photon  Number of photons = 2.0 1017 J  = 70.430 = 70 photons  19  2.8397 10 J 

6.626 10



J  s)(3.00 108 m/s  1nm   9  = 4.18484 × 10–19 J 475. nm 10 m  This is the value for each photon, that is, J/photon.   1 photon Number of photons = 2.0 1017 J  = 47.7916 = 48 photons  19  4.18484 10 J  b) E =

7.65

34

Determine the wavelength: –1

–4

= 1/(1953 cm ) = 5.1203277 × 10 cm

102 m  1 nm   = 5120.3277 = 5.120 × 103 nm  (nm) = 5.1203277 104 cm   1 cm 109 m  (

 102 m  1  10  = 51203.277 = 5.120 × 104 ) = 5.1203277 104 cm   1 cm 10 m 

 = c/ =

7.66

2.9979 108 m /s  1 cm  1 Hz  13 13   1  = 5.8548987 × 10 = 5.855 × 10 Hz 5.1203277 104 cm 102 m  1 s 

Plan: The Bohr model has been successfully applied to predict the spectral lines for one-electron species other than H. Common one-electron species are small cations with all but one electron removed. Since the problem 2+

specifies a metal ion, assume that the possible choices are Li or Be . Use the relationship E = h to convert the Copyright

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7-18

 Z2  frequency to energy and then solve Bohr’s equation E = 2.18 1018 J 2  to verify if a whole number for Z  n  can be calculated. Recall that the negative sign is a convention based on the zero point of the atom’s energy; it is deleted in this calculation to avoid taking the square root of a negative number. Solution: The highest energy line corresponds to the transition from n = 1 to n = . E = h = (6.626 × 10

–34

–1

J s) (2.961 × 10 Hz) (s /Hz) = 1.9619586 × 10

 Z2  E = 2.18 1018 J 2   n 

–17

Z = charge of the nucleus

En2 1.9619586 1017 (12 ) = = 8.99998 2.18 10 18 J 2.18 1018 J 2 Then Z = 9 and Z = 3. 2

Z =

Therefore, the ion is Li with an atomic number of 3. 7.67

h a) Electron: = = mu h Proton: = = mu b) E = 1/2mu u=

J  s

 kg  m 2 /s2    = 2.139214 × 10–10 = 2.1 × 10–10 m    J m   9.111031 kg 3.4 106    s



6.626 10

34

J  s

 kg  m 2 /s2  –13 –13   = 1.16696 × 10 = 1.2 × 10 m    J m  27 6 1.67 10 kg3.4 10 s  2

2E m

Proton: u =

2 2.7 1015 J  kg  m 2 /s2    = 7.6991 × 107 m/s  31  J  9.1110 kg





2 2.7 1015 J  kg  m 2 /s2    = 1.7982 × 106 m/s  27  J  1.67 10 kg





h Electron: = = mu

Proton: =



34

therefore u = 2E/m

Electron: u =

7.68

6.626 10

h = mu

6.626 10

34

J  s

 kg  m 2 /s2  –12 –12   = 9.44698 × 10 = 9.4 × 10 m    J m  31 7 9.1110 kg7.699110 s 

6.626 10

34



 kg  m 2 /s2    = 2.20646 × 10–13 = 2.2 × 10–13 m    J m  1.67 1027 kg1.7982 106 s  Js

Plan: The electromagnetic spectrum shows that the visible region goes from 400 to 750 nm (4000 to 7500 ). Thus, wavelengths b, c, and d are for the three transitions in the visible series with nfinal = 2. Wavelength a is in the ultraviolet region of the spectrum and the ultraviolet series has nfinal = 1. Wavelength e is in the infrared region of the spectrum and the infrared series has nfinal = 3. Use the Rydberg equation to find the ninitial for each line. Convert the wavelengths from to units of m.

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7-19

Solution: n = ?  n = 1; = 1212.7 (shortest corresponds to the largest E) 1010 m  –7 (m) = 1212.7   = 1.2127 × 10 m  1 



1

= 1.096776 10 m 7





 n1  n1  2 1

2 2

    1   = 1.096776 107 m1  1  1   7  2 2  1 n2  1.2127 10 m  1 1 0.7518456 =  2  2  n2  1  1    = 1 – 0.7518456  n2  2

 1    = 0.2481544  n2  2

n22 = 4.029749

n2 = 2 for line (a) (n = 2  n = 1) n = ?  n = 3; = 10,938 (m) = 10,938



10

101



1

= 1.096776 10 m 7

(longest corresponds to the smallest E) m  –6  = 1.0938 × 10 m 





 n1  n1  2 1

2 2

    1   = 1.096776 107 m1  1  1   6  2 2   3 n2  1.093810 m  1 1 0.083357397 =  2  2   3 n2  0.083357397 = 0.111111111 

1 n22

 1    = 0.11111111 – 0.083357396  n2  2

 1    = 0.0277537151  n2  2

n22 =36.03121

n2 = 6 for line (e) (n = 6  n = 3) For the other three lines, n1 = 2. For line (d), n2 = 3 (largest  smallest E). For line (b), n2 = 5 (smallest  largest E). For line (c), n2 = 4. Copyright

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7-20

7.69

thus =



hc E



6.626 10 J  s 3.00 108 m/s  1 nm  hc   a) (nm) = =  9  = 432.130 = 432 nm E 4.60 1019 J 10 m  34

b) (nm) =

6.626 1034 J  s3.00 108 m /s  1 nm  hc =  9  = 286.4265 = 286 nm E 6.94 1019 J 10 m 

c) (nm) =

6.626 1034 J  s3.00 108 m /s  1 nm  = 450.748 = 451 nm hc =  9  E 4.411019 J 10 m 

7.70

Index of refraction = c/v; v = c/(index of refraction) 8 8 8 a) Water v = c/(index of refraction) = (3.00 × 10 m/s)/(1.33) = 2.2556 × 10 = 2.26 × 10 m/s 8 8 8 b) Diamond v = c/(index of refraction) = (3.00 × 10 m/s)/(2.42) = 1.239669 × 10 = 1.24 × 10 m/s

7.71

Extra significant figures are necessary because of the data presented in the problem. He–Ne = 632.8 nm 14 –1 Ar  = 6.148 × 10 s –19 Ar–Kr E = 3.499 × 10 J Dye = 663.7 nm Calculating missing values: 8 14 –1 –7 –7 Ar = c/ = (2.9979 × 10 m/s)/(6.148 × 10 s ) = 4.8762199 × 10 = 4.876 × 10 m –34 8 –19 –7 –7 Ar–Kr = hc/E = (6.626 × 10 J s) (2.9979 × 10 m/s)/(3.499 × 10 J) = 5.67707 × 10 = 5.677 × 10 m Calculating missing  values: 8 –9 14 14 –1 He–Ne  = c/ = (2.9979 × 10 m/s)/[632.8 nm (10 m/nm)] = 4.7375 × 10 = 4.738 × 10 s –19 –34 14 14 –1 Ar–Kr  = E/h = (3.499 × 10 J)/(6.626 × 10 J s) = 5.28071 × 10 = 5.281 × 10 s 8 –9 14 14 –1 Dye  = c/ = (2.9979 × 10 m/s)/[663.7 nm (10 m/nm)] = 4.51695 × 10 = 4.517 × 10 s Calculating missing E values: –34 8 –9 He–Ne E = hc/ = [(6.626 × 10 J s)(2.9979 × 10 m/s)]/[632.8 nm (10 m/nm)] –19 –19 = 3.13907797 × 10 = 3.139 × 10 J –34 14 –1 –19 –19 Ar E = h = (6.626 × 10 J s)(6.148 × 10 s ) = 4.0736648 × 10 = 4.074 × 10 J –34 8 –9 Dye E = hc/ = [(6.626 × 10 J s)(2.9979 × 10 m/s)]/[663.7 nm (10 m/nm)] –19 –19 = 2.99293 × 10 = 2.993 × 10 J The colors may be predicted from Figure 7.3 and the frequencies. 14 –1 He–Ne  = 4.738 × 10 s Orange 14 –1 Ar  = 6.148 × 10 s Green 14 –1 Ar–Kr  = 5.281 × 10 s Yellow 14 –1 Dye  = 4.517 × 10 s Red

7.72

Plan: Allowed values of quantum numbers: n = positive integers; l = integers from 0 to n – 1; ml = integers from – l through 0 to + l. Solution: a) The l value must be at least 1 for ml to be – 1, but cannot be greater than n – 1 = 3 –1 = 2. Increase the l value to 1 or 2 to create an allowable combination. b) The l value must be at least 1 for ml to be +1, but cannot be greater than n – 1 = 3 – 1 = 2. Decrease the l value to 1 or 2 to create an allowable combination. c) The l value must be at least 3 for ml to be +3, but cannot be greater than n – 1 = 7 – 1 = 6. Increase the l value to 3, 4, 5, or 6 to create an allowable combination.

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7-21

d) The l value must be at least 2 for ml to be –2, but cannot be greater than n – 1 = 4 – 1 = 3. Increase the l value to 2 or 3 to create an allowable combination.

7.73



1

= 1.096776 10 m 7





 n1  n1  2 1

2 2

  1 nm  1  1 7 1  1    1.096776  10 m  =     12 94.91 nm 109 m  n22  

1 1 0.9606608 =  2  2  1 n2  1 = 0.0393392 n22 n22 = 24.41994

n2 = 5 b)

 1 nm   1  1  7 1  1   9  = 1.096776 10 m  2  2  1281 nm 10 m  5   n1

1 1 0.071175894 =  2  2   n1 5  1 = 0.111175894 n12 n12 = 8.9947556

n1 = 3 c)

1 1 = 1.096776 10 7 m 1  2  2  1 3  6

= 9.74912 × 10 m

–1

 1 nm  = 1.02573 107 m  9  = 102.573 = 102.6 nm 10 m 

7.74

Plan: Ionization occurs when the electron is completely removed from the atom, or when nfinal = . We can use the equation for the energy of an electron transition to find the quantity of energy needed to remove completely the electron, called the ionization energy (IE). To obtain the ionization energy per mole of species, multiply by Avogadro’s number. The charge on the nucleus must affect the IE because a larger nucleus would exert a greater pull on the escaping electron. The Bohr equation applies to H and other one-electron species. Use the expression 4+ to determine the ionization energy of B . Then use the expression to find the energies of the transitions listed and hc use E = to convert energy to wavelength. Solution:

Z2  a) E = 2.18 1018 J  2   n 





Z = atomic number

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7-22

 1 1  2 18 E = 2.1810 J  2  2  Z  nfinal ninitial 





 1 1   6.022 10 23  E = 2.18 1018 J  2  2  Z 2   ninitial   1 mol   





= (1.312796 × 10 ) Z for n = 1 b) In the ground state n = 1, the initial energy level for the single electron in B . Once ionized, n =  is the final energy level. 4+ Z = 5 for B . 4+

E = IE = (1.312796 × 10 ) Z = (1.312796 × 10 J/mol)(5 ) = 3.28199 × 10 = 3.28 × 10 J/mol 6

c) nfinal = , ninitial = 3, and Z = 2 for He . +

E = 2.1810

18

 1  1 1  1 J 2  2  Z 2 = 2.18x1018 J 2  2  22 = 9.68889 × 10–19 J  nfinal ninitial    3 





6.626 10 J  s 3.00 108 m/s hc –7 (m) = = = 2.051628 × 10 m 19 E 9.68889 10 J  1 nm   (nm) = 2.051628 107 m  9  = 205.1628 = 205 nm 10 m    34

d) nfinal = , ninitial = 2, and Z = 4 for Be . 3+

 1  1 1  2 1  2 18 18 –18 E = 2.1810 J  2  2  Z = 2.1810 J 2  2  4 = 8.72 × 10 J   2    ninitial 











6.626 1034 J  s 3.00 108 m /s hc –8 (m) = = = 2.279587 × 10 m E 8.72 1018 J  1 nm   (nm) = 2.279587 108 m  9  = 22.79587 = 22.8 nm 10 m   

7.75

a) Orbital D has the largest value of n, given that it is the largest orbital. b) l = 1 indicates a p orbital. Orbitals A and C are p orbitals. l = 2 indicates a d orbital. Orbitals B and D are d orbitals. c) In an atom, there would be four other orbitals with the same value of n and the same shape as orbital B. There would be two other orbitals with the same value of n and the same shape as orbital C. d) Orbital D has the highest energy and orbital C has the lowest energy.

7.76

Plan: Use the values and the equation given in the problem to calculate the appropriate values. Solution: 2 n 2 h 0 a) rn = 2 e

2 C 2   12 6.626 1034 J  s 8.854 1012  J  m   kg  m 2 /s2   = 5.2929377 × 10–11 = 5.293 × 10–11 m r1 =  2  31 19 J  9.109 10 kg 1.602 10 C











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7-23

2 C 2   10 2 6.626 1034 J  s 8.854 1012  J  m   kg  m 2 /s2   = 5.2929377 × 10–9 = 5.293 × 10–9 m b) r10 =  2   31 19 J   9.109 10 kg 1.602 10 C











7.77

n 2 h 0

a) rn =

2 e

2 C 2   32 6.626 1034 J  s 8.854 1012  J  m   kg  m 2 /s2   = 4.76364 × 10–10 = 4.764 × 10–10 m r3 =  2 31   19 J   9.109 10 kg 1.602 10 C











b) Z = 1 for an H atom

Z2   12  –19 –19 En = 2.18 1018 J  2  = 2.18 1018 J  2  = –2.42222 × 10 = –2.42 × 10 J  n   3 









c) Z = 3 for a Li atom

Z2   32  –1 En = 2.181018 J  2  = 2.181018 J  2  = –2.18 × 10  n   3  d) The greater number of protons in the Li nucleus results in a greater interaction between the Li nucleus and its electrons. Thus, the energy of an electron in a particular orbital becomes more negative with increasing atomic number.



7.78







Plan: Refer to Chapter 6 for the calculation of the amount of heat energy absorbed by a substance from its specific heat capacity and temperature change (q = c × mass × T). Using this equation, calculate the energy absorbed by the water. This energy equals the energy from the microwave photons. The energy of each photon can be calculated from its wavelength: E = hc/ . Dividing the total energy by the energy of each photon gives the number of photons absorbed by the water. Solution: q = c × mass × T 4 q = (4.184 J/g°C)(252 g)(98 – 20)°C = 8.22407 × 10 J E=

6.626 10 =

34



J  s 3.00 108 m /s

1.55 10

2

 = 1.28245 × 10 J/photon –23

  1 photon 27 27 Number of photons = 8.22407 10 4 J   = 6.41278 × 10 = 6.4 × 10 photons 1.28245 1023 J 

7.79

One sample calculation will be done using the equation in the book: 3

3  2 r  1  1  2 r a0  1  1  e a0 = 1.465532 × 10–3 er a0    =  =   e        0    52.92 pm 

For r = 50 pm: –3

r

= 1.465532 × 10 e 2

–3

–4 2

4 r

50

= 1.465532 × 10 e

= (5.69724 × 10 ) = 3.24585 × 10 2

–3/2

(pm ) –3 1.47 × 10 –3 0.570 × 10

52.92

= 5.69724 × 10

–4

–7

= 4 (50) (3.24585 × 10 ) = 1.0197 × 10

r (pm) 0 50 Copyright

–3

–2

(pm ) –6 2.15 × 10 –6 0.325 × 10

–1

4 r (pm ) 0 –2 1.02 × 10

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7-24

100 200

–3

0.221 × 10 –3 0.0335 × 10

–6

0.0491 × 10 –6 0.00112 × 10

–2

0.616 × 10 –2 0.0563 × 10

The plots are similar to Figure 7.17A in the text. 7.80

Plan: In general, to test for overlap of the two series, compare the longest wavelength in the “n” series with the shortest wavelength in the “n + 1” series. The longest wavelength in any series corresponds to the transition between the n1 level and the next level above it; the shortest wavelength corresponds to the transition between the n1 level and the n =  level. Use the relationship

1 1 = R  2  2  to calculate the wavelengths. n2   n1

Solution: 1 1 1  1 1 7 1 = R  2  2  = 1.096776 10 m  2  2   n1 n2  n2   n1





a) The overlap between the n1 = 1 series and the n1 = 2 series would occur between the longest wavelengths for n1 = 1 and the shortest wavelengths for n1 = 2. Longest wavelength in n1 = 1 series has n2 equal to 2. 1 1 1 –1 = 1.096776 10 7 m 1  2  2  = 8,225,820 m 1  2





1 –7 –7 = 1.215684272 × 10 = 1.215684 × 10 m 1 8,225,820 m Shortest wavelength in the n1 = 2 series:

1 1  –1 = 1.096776 10 7 m1  2   = 2,741,940 m  2 2  1 –7 –7 = = 3.647052817 × 10 = 3.647053 × 10 m 2,741,940 m1 Since the longest wavelength for n1 = 1 series is shorter than shortest wavelength for n1 = 2 series, there

is no overlap between the two series.

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7-25

b) The overlap between the n1 = 3 series and the n1 = 4 series would occur between the longest wavelengths for n1 = 3 and the shortest wavelengths for n1 = 4. Longest wavelength in n1 = 3 series has n2 equal to 4. 1 1 1 –1 = 1.096776 10 7 m1  2  2  = 533,155 m  3 4 





1 –6 –6 = 1.875627163 × 10 = 1.875627 × 10 m 533,155 m1

Shortest wavelength in n1 = 4 series has n2 = . 1 1  –1 = 1.096776 10 7 m 1  2   = 685,485 m  4  2  1 –6 –6 = = 1.458821127 × 10 = 1.458821 × 10 m 685,485 m1 Since the n1 = 4 series shortest wavelength is shorter than the n1 = 3 series longest wavelength, the series do overlap. c) Shortest wavelength in n1 = 5 series has n2 = .





1 1  –1 = 1.096776 10 7 m1  2   = 438,710.4 m  5 2  1 –6 –6 = = 2.27940801 × 10 = 2.279408 × 10 m 438,710.4 m1 Calculate the first few longest lines in the n1 = 4 series to determine if any overlap with the shortest wavelength in the n1 = 5 series: For n1 = 4, n2 = 5: 1 1 1 –1 = 1.096776 10 7 m1  2  2  = 246,774.6 m  4  5









1 –6 = 4.052281 × 10 m 1 246,774.6 m For n1 = 4, n2 = 6: 1 1 1 –1 = 1.096776 10 7 m1  2  2  = 380,825 m  4 6 





1 –6 = 2.625878 × 10 m 380,825 m1 For n1 = 4, n2 = 7:

1 1 –1 = 1.096776 10 7 m1  2  2  = 461,653.2 m  4 7  1 –6 = = 2.166128 × 10 m 461,653.2 m1 The wavelengths of the first two lines of the n1 = 4 series are longer than the shortest wavelength in the n1 = 5 series. Therefore, only the first two lines of the n1 = 4 series overlap the n1 = 5 series. d) At longer wavelengths (i.e., lower energies), there is increasing overlap between the lines from different series (i.e., with different n1 values). The hydrogen spectrum becomes more complex, since the lines begin to merge into a more-or-less continuous band, and much more care is needed to interpret the information.





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7-26

7.81

a) The highest frequency would correspond to the greatest energy difference. In this case, the greatest energy difference would be between E3 and E1. E = E3 – E1 = h = (–15 × 10  = E/h = (5 × 10

–19

J) – (–20 × 10

J)/(6.626 × 10

–34

–19

J) = 5 × 10

–19

–1

J s) = 7.54603 × 10 = 8 × 10 s 14

–1

–7

= c/ = (3.00 × 10 m/s)/(7.54603 × 10 s ) = 3.97560 × 10 = 4 × 10 m –19

b) The ionization energy (IE) is the same as the reverse of E1. Thus, the value of the IE is 20 × 10 J/atom. –19 3 23 3 IE = (20 × 10 J/atom)(1 kJ/10 J)(6.022 × 10 atoms/mol) = 1204.4 = 1.2 × 10 kJ/mol c) The shortest wavelength would correspond to an electron moving from the n = 4 level to the highest level available in the problem (n = 6). –19 –19 –19 E = E6 – E4 = hc/ = (–2 × 10 J) – (–11 × 10 J) = 9 × 10 J = hc/E = 7.82

6.626 10







J  s 3.00 108 m /s  1 nm   = 220.867 = 2 × 102 nm   9 19  10 m 9 10 J  

34



Plan: The energy differences sought may be determined by looking at the energy changes in steps. The hc wavelength is calculated from the relationship = . E Solution: a) The difference between levels 3 and 2 (E32) may be found by taking the difference in the energies for the 3  1 transition (E31) and the 2  1 transition (E21). –17 –17 –18 E32 = E31 – E21 = (4.854 × 10 J) – (4.098 × 10 J) = 7.56 × 10 J







6.6261034 J  s 3.00 108 m/s hc –8 –8 = = = 2.629365 × 10 = 2.63 × 10 m E 7.56 1018 J





b) The difference between levels 4 and 1 (E41) may be found by adding the energies for the 4  2 transition (E42) and the 2  1 transition (E21). –17 –17 –17 E41 = E42 + E21 = (1.024 × 10 J) + (4.098 × 10 J) = 5.122 × 10 J =







6.626 1034 J  s 3.00 108 m/s hc –9 –9 = = 3.88091 × 10 = 3.881 × 10 m 17 E 5.122 10 J





c) The difference between levels 5 and 4 (E54) may be found by taking the difference in the energies for the 5  1 transition (E51) and the 4  1 transition (see part (b)). E54 = E51 – E41 = (5.242 × 10



34

–17

J) – (5.122 × 10



J) = 1.2 × 10

–18



6.626 10 J  s 3.00 108 m/s hc –7 –7 = = = 1.6565 × 10 = 1.66 × 10 m 18 E 1.2 10 J





7.83

a) A dark green color implies that relatively few photons are being reflected from the leaf. A large fraction of the photons is being absorbed, particularly in the red region of the spectrum. A plant might adapt in this way when photons are in short supply — i.e., in conditions of low light intensity. b) An increase in the concentration of chlorophyll, the light-absorbing pigment, would lead to a darker green color (and vice versa).

7.84

Plan: For part (a), use the equation for kinetic energy, Ek = mu . For part (b), use the relationship E = hc/ to find the energy of the photon absorbed. From that energy subtract the kinetic energy of the dislodged electron to obtain the work function.

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7-27

Solution: a) The energy of the electron is a function of its speed leaving the surface of the metal. The mass of the electron is –31 9.109 × 10 kg.  2 J 1 1 2  = 1.86552 × 10–19 = 1.87 × 10–19 J Ek = mu = 9.109 1031 kg 6.40 10 5 m/s  2 2 2 2  kg  m /s  b) The minimum energy required to dislodge the electron () is a function of the incident light. In this example, the incident light is higher than the threshold frequency, so the kinetic energy of the electron, Ek, must be subtracted from the total energy of the incident light, h, to yield the work function, . (The number of significant figures given in the wavelength requires more significant figures in the speed of light.)







109 m    = 3.581 × 10–7 m (m) = 358.1 nm   1nm 

6.626 10

34

E = hc/ =



J  s 2.9979 108 m/s

3.58110 m

 = h – Ek = (5.447078 × 10

–19

7.85



7

a) = h/mu =

6.626 10



= 5.447078 × 10

J) – (1.86552 × 10 34

–19

J) = 3.581558 × 10

–19

= 3.58 × 10 J

J  s

 kg  m 2 /s2    = 1.322568 × 10–8 m     J m  9.109 1031 kg 5.5 10 4    s



–8

–9

Smallest object = /2 = (1.322568 × 10 m)/2 = 6.61284 × 10 = 6.6 × 10 m b) = h/mu =

6.626 10



34

J  s

 kg  m 2 /s2  –11   = 2.424708 × 10 m    J m 31  7 9.109 10 kg 3.0 10   s 



Smallest object = /2 = (2.424708 × 10 7.86

–11

m)/2 = 1.212354 × 10

–11

= 1.2 × 10

–11

Plan: Examine Figure 7.3 and match the given wavelengths to their colors. For each salt, convert the mass of salt to moles and multiply by Avogadro’s number to find the number of photons emitted by that amount of salt hc (assuming that each atom undergoes one-electron transition). Use the relationship E = to find the energy of one photon and multiply by the total number of photons for the total energy of emission. Solution: a) Figure 7.3 indicates that the 641 nm wavelength of Sr falls in the red region and the 493 nm wavelength of Ba falls in the green region. b) SrCl2 23  1 mol SrCl  photons  2  6.022 10   = 1.8994449 × 1022 photons Number of photons = 5.00 g SrCl2     158.52 g SrCl2  1 mol SrCl2 109 m    = 6.41 × 10–7 m (m) = 641 nm   1nm 

6.626 10 =





J  s 3.00 108 m /s  1 kJ   = 3.10109 × 10–22 kJ/photon  Ephoton = 7 3   6.4110 m 10 J   3.10109 1022 kJ   = 5.89035 = 5.89 kJ Etotal = 1.8994449 10 22 photons  1 photon 

34

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7-28

BaCl2 23  1 mol BaCl  photons  2  6.022 10   = 1.44620557 × 1022 photons Number of photons = 5.00 g BaCl2     1 mol BaCl2  208.2 g BaCl2 

109 m    = 4.93 × 10–7 m (m) = 493 nm   1nm 

Ephoton =

6.626 10





J  s 3.00 10 8 m /s  1 kJ   = 4.0320487 × 10–22 kJ/photon  7 3   4.93 10 m 10 J  34

 4.0320487 1022 kJ   = 5.83117 = 5.83 kJ Etotal = 1.44620557 10 22 photons  1 photon  7.87

a) The highest energy line corresponds to the shortest wavelength. The shortest wavelength line is given by

1 1 1  1   R  2  2  = 1.096776 107 m1  2  2   n1  n1 n2  n2 

 1 nm   1  7 1  1  1    9  = 1.096776 10 m  2  3282 nm 10 m  2   n1





1 –1 7 –1 304,692 m = (1.096776 × 10 m )  2   n1   1    = 0.0277807  n2  1

n12 = 35.9962

n1 = 6 b) The lowest energy line corresponds to the longest wavelength. The longest wavelength line is given by



= 1.096776 10 m 7

1



  1    n 2  1

   2 n1  1  1





 1  1 nm  1 7 1    = 1.096776 10 m  2    9  n 7460 nm 10 m   1





  1 134,048 m = 1.096776 10 m  2   n  1 7

–1

   2  n1  1  1





   2  n1  1  1

1





   1  1 0.0122220 =  2  2   n1 n1  1  





Rearranging and solving this equation for n1 yields n1 = 5. (You and your students may well need to resort to trialand-error solution of this equation!)

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7-29

7.88

Plan: Examine Figure 7.3 to find the region of the electromagnetic spectrum in which the wavelength lies. Compare the absorbance of the given concentration of vitamin A to the absorbance of the given amount of fishliver oil to find the concentration of vitamin A in the oil. Solution: a) At this wavelength the sensitivity to absorbance of light by vitamin A is maximized while minimizing interference due to the absorbance of light by other substances in the fish-liver oil. b) The wavelength 329 nm lies in the ultraviolet region of the electromagnetic spectrum. –3 c) A known quantity of vitamin A (1.67 × 10 g) is dissolved in a known volume of solvent (250. mL) to give a standard concentration with a known response (1.018 units). This can be used to find the unknown quantity of vitamin A that gives a response of 0.724 units. An equality can be made between the two concentration-toabsorbance ratios. 1.67 103 g    = 6.68 × 10–6 g/mL vitamin A Concentration (C1, g/mL) of vitamin A =   250. mL    Absorbance (A1) of vitamin A = 1.018 units. Absorbance (A2) of fish-liver oil = 0.724 units Concentration (g/mL) of vitamin A in fish-liver oil sample = C2 A1 A = 2 C1 C2

0.7246.6810 AC C2 = 2 1 = A1 1.018

6

g/mL

–6

= 4.7508 × 10 g/mL vitamin A

 4.7508 106 g vitamin A    = 2.3754 × 10–3 g vitamin A Mass (g) of vitamin A in oil sample = 500 mL oil    1 mL oil  





2.3754 10 g Concentration of vitamin A in oil sample = = 1.92808 × 10 = 1.93 × 10 g vitamin A/g oil 0.1232 g Oil 3

–2

6.62610

= hc/E =







J  s 3.00108 m/s  1 nm   = 261.897 = 262 nm  9  19  10 m  7.5910 J   34

7.89

–2



Silver is not a good choice for a photocell that uses visible light because 262 nm is in the ultraviolet region. 7.90

Mr. Green must be in the dining room where green light (520 nm) is reflected. Lower frequency, longer wavelength light is reflected in the lounge and study. Both yellow and red light have longer wavelengths than green light. Therefore, Col. Mustard and Ms. Scarlet must be in either the lounge or study. The shortest wavelengths are violet. Prof. Plum must be in the library. Ms. Peacock must be the murderer.

7.91

Ek = v=

1 2 mv 2

4.7110

15

 kg  m 2 /s2    = 1.01692775 × 108 m/s  J 

9.109 10 kg  kg m /s  6.626 10 J s   = 7.15304 × 10 = 7.15 × 10 m = h/mv =     J m   9.109 10 kg1.0169277510 s  1 m 2

1 2

31

34

31



–12

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7-30

7.92

Plan: First find the energy in joules from the light that shines on the text. Each watt is one joule/s for a total hc of 75 J; take 5% of that amount of joules and then 10% of that amount. Use E = to find the energy of one photon of light with a wavelength of 550 nm. Divide the energy that shines on the text by the energy of one photon to obtain the number of photons. Solution: The amount of energy is calculated from the wavelength of light: 109 m  –7 (m) = 550 nm   = 5.50 × 10 m  1 nm  E=

6.626 10 =

34



J  s 3.00 108 m /s 7

5.50 10 m

 = 3.614182 × 10 J/photon –19

1 J/s  Amount of power from the bulb = 75 W  = 75 J/s 1 W   5%  Amount of power converted to light = 75 J/s  = 3.75 J/s 100% 





 10%  Amount of light shining on book = 3.75 J/s  = 0.375 J/s 100%   0.375 J   1 photon 18 18  Number of photons:   = 1.0376 × 10 = 1.0 × 10 photons/s  s   3.614182 1019 J  7.93

a) Sodium ions emit yellow-orange light, and potassium ions emit violet light. b) The cobalt glass filter absorbs the yellow-orange light, while the violet light passes through. c) Sodium salts used as the oxidizing agents would emit intense yellow-orange light which would obscure the light emitted by other salts in the fireworks.

7.94

a) 6CO2(g) + 6H2O(l)  C6H12O6(s) + 6O2(g) Hrxn = {(1 mol)  H f [C6H12O6] + (6 mol)  H f [O2]} – {(6 mol)  H f [CO2] + (6 mol)  H f [H2O]} Hrxn = [–1273.3 kJ + 6(0.0 kJ)] – [6(–393.5 kJ) + 6(–285.840 kJ)] = 2802.74 = 2802.7 kJ 6CO2(g) + 6H2O(l)  C6H12O6(s) + 6O2(g) Hrxn = 2802.7 kJ (for 1.00 mol C6H12O6)

6.626 10 b) E = hc/ =





J  s 3.00 10 8 m /s  1 nm   = 2.9232353 × 10–19 J/photon  109 m  680. nm   10 3 J  1 photon    24 24 Number of photons = 2802.7 kJ    = 9.5877 × 10 = 9.59 × 10 photons  1 kJ  2.9232353 1019 J    



7.95

34



Plan: In the visible series with nfinal = 2, the transitions will end in either the 2s or 2p orbitals since those are the only two types of orbitals in the second main energy level. With the restriction that the angular momentum quantum number can change by only ±1, the allowable transitions are from a p orbital to 2s (l = 1 to l = 0), from an s orbital to 2p (l = 0 to l = 1), and from a d orbital to 2p (l = 2 to l = 1). The problem specifies a change in energy level, so ninit must be 3, 4, 5, etc. (Although a change from 2p to 2s would result in a +1 change in l, this is not a change in energy level.) Solution: The first four transitions are as follows: 3s 2p

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7-31

3d 4s 3p 7.96

2p 2p 2s

A = kC 3

–1

k = 4.5 × 10 M = slope 3

–1

–4

b) A/k = C = (0.55)/(4.5 × 10 M ) = 1.2222 × 10 = 1.2 × 10 M

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7-32

CHAPTER 8 ELECTRON CONFIGURATION AND CHEMICAL PERIODICITY FOLLOW–UP PROBLEMS 8.1A

Plan: The atomic number gives the number of electrons. The order of filling may be inferred by the location of the element on the periodic table. The partial orbital diagrams shows only those electrons after the preceding noble gas except those used to fill inner d and f subshells. The number of inner electrons is simply the total number of electrons minus those electrons in the partial orbital diagram. Solution: 2 2 6 2 6 2 8 a) For Ni (Z = 28), the full electron configuration is 1s 2s 2p 3s 3p 4s 3d . 2 8 The condensed configuration is [Ar]4s 3d . The partial orbital diagram for the valence electrons is

There are 28 – 10(valence) = 18 inner, 2 outer, and 10 valence electrons. 2 2 6 2 6 2 10 6 2 b) For Sr (Z = 38), the full electron configuration is 1s 2s 2p 3s 3p 4s 3d 4p 5s . 2 The condensed configuration is [Kr]5s . The partial orbital diagram is

There are 38 – 2(valence) = 36 inner, 2 outer, and 2 valence electrons. 2 2 6 2 6 2 10 6 2 10 6 2 14 10 4 c) For Po (84 electrons), the full configuration 1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f 5d 6p . 2 14 10 4 The condensed configuration is [Xe]6 s 4 f 5 d 6 p . The partial orbital diagram represents valence electrons only; the inner electrons are those in the previous noble 2 2 6 2 6 2 10 6 2 10 6 10 14 gas (Xe or 1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p ) and filled transition (5d ) and inner transition (4f ) levels.

There are 84 – 6(valence) = 78 inner, 6 outer, and 6 valence electrons. 8.1B

Plan: The partial orbital diagrams show the electrons that have been added after the preceding noble gas except for those used to fill inner d and f subshells (in most cases). Use the orbital diagrams and the periodic table to identify the elements. Write their full and condensed electron configurations. The number of inner electrons is simply the total number of electrons minus those electrons in the partial orbital diagram. a) As (Z = 33) has two 4s electrons and three 4p electrons. Its full electron configuration is 2 2 6 2 6 2 10 3 2 10 3 1s 2s 2p 3s 3p 4s 3d 4p and its condensed electron configuration is [Ar] 4s 3d 4p . There are 33 – 5(valence) = 28 inner, 5 outer, and 5 valence electrons.

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8-1

b) Zr (Z = 40) has two 5s electrons and two 4d electrons. Its full electron configuration is 2 2 6 2 6 2 10 6 2 2 2 2 1s 2s 2p 3s 3p 4s 3d 4p 5s 4d and its condensed electron configuration is [Kr] 5s 4d . There are 40 – 4(valence) = 36 inner, 2 outer, and 4 valence electrons. c) I (Z = 53) has two 5s electrons and five 5p electrons. Its full electron configuration is 2 2 6 2 6 2 10 6 2 10 5 2 10 5 1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p and its condensed electron configuration is [Kr] 5s 4d 5p . There are 53 – 7(valence) = 46 inner, 7 outer, and 7 valence electrons. 8.2A

Plan: Locate each of the elements on the periodic table. All of these are main-group elements, so their sizes increase down and to the left in the periodic table. Solution: a) Cl < Br < Se. Cl has a smaller n than Br and Se. Br experiences a higher Zeff than Se and is smaller. b) Xe < I < Ba. Xe and I have the same n, but Xe experiences a higher Zeff and is smaller. Ba has the highest n and is the largest.

8.2B

Plan: Locate each of the elements on the periodic table. All of these are main-group elements, so their sizes increase down and to the left in the periodic table. Solution: a) Cs > As > S. Cs has the largest n of the group. Additionally, it is further to the left of the table than the other two elements. Therefore, Cs is the largest. As has a larger n than S and is further to the left on the periodic table, so it is the next largest element of this group. b) K > P > F. K has the largest n of the group. Additionally, it is further to the left of the table than the other two elements. Therefore, K is the largest. P has a larger n than F and is further to the left on the periodic table, so it is the next largest element of this group.

8.3A

Plan: These main-group elements must be located on the periodic table. The value of IE1 increases toward the top (same column) and the right of the periodic table (same n). Solution: a) Sn < Sb < I. These elements have the same n, so the values increase to the right on the periodic table. Iodine has the highest IE, because its outer electron is most tightly held and hardest to remove. b) Cs < Na < Mg. Mg is farther to the right on the periodic table than either of the other two elements, so it has the largest IE. Cs and Na are in the same column, so the values decrease towards the bottom of the column. Because Cs is lower than Na in the column, it has a lower IE.

8.3B

Plan: These main-group elements must be located on the periodic table. The value of IE 1 increases toward the top (same column) and the right of the periodic table (same n). Solution: a) O > As > Rb. Oxygen is the farthest to the right on the periodic table (and closer to the top of the table), so it has the largest IE. Similarly, As is farther to the right and closer to the top of the table than Rb and, thus, has a higher IE. b) Cl > Si > Sn. Cl and Si have the same n value and are closer to the top of the periodic table than Sn. Therefore, Sn has the lowest IE. Because Cl is closer to the right-hand side of the periodic table than Si, Cl has the highest IE.

8.4A

Plan: We must look for a large “jump” in the IE values. This jump occurs after the valence electrons have been removed. The next step is to determine the element in the designated period with the proper number of valence electrons. Solution: The exceptionally large jump from IE3 to IE4 means that the fourth electron is an inner electron. Thus, Q has three 2 2 6 2 1 valence electrons. Since Q is in period 3, it must be aluminum, Al: 1s 2s 2p 3s 3p .

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8-2

8.4B

8.5A

Plan: A large jump in IE values occurs after the valence electrons have been removed. Thus, the highest IE3, for example, will be for an element that has 2 valence electrons because it will require much more energy to remove the third electron, an inner electron. Write the condensed electron configurations for the atoms, determine the number of valence electrons for each, and use this information to determine where the jump in IE values will occur. Solution: 1 Rb: [Kr] 5s 2 Sr: [Kr] 5s 2 1 Y: [Kr] 5s 4d Because Rb has one valence electron, a large jump in IE values will occur between IE 1 and IE2. Therefore, IE2 will be large for Rb. Because Sr has two valence electrons, a large jump in IE values will occur between IE 2 and IE3. Therefore, IE3 will be large for Sr. Plan: We need to locate each element on the periodic table. Elements in the first two columns on the left or the two columns to the left of the noble gases tend to adopt ions with a noble gas configuration. Elements in the remaining columns may use either their ns and np electrons or just their np electrons. Solution: 2 2+ – a) Barium loses two electrons to be isoelectronic with Xe: Ba ([Xe] 6s )  Ba ([Xe]) + 2e – 2 4 – 2– b) Oxygen gains two e to be isoelectronic with Ne: O ([He] 2s 2p + 2e  O ([Ne]) c) Lead can lose two electrons to form an “inert pair” configuration: 2 14 10 2 2+ 2 14 10 – Pb ([Xe]6s 4f 5d 6p )  Pb ([Xe]6s 4f 5d ) + 2e or lead can lose four electrons form a “pseudo–noble gas” configuration: 2 14 10 2 4+ 14 10 – Pb ([Xe]6s 4f 5d 6p )  Pb ([Xe]4f 5d ) + 4e

8.5B

Plan: We need to locate each element on the periodic table. Elements in the first two columns on the left or the two columns to the left of the noble gases tend to adopt ions with a noble gas configuration. Elements in the remaining columns may use either their ns and np electrons or just their np electrons. Solution: 2 5 – – 2 6 a) Fluorine gains one electron to be isoelectronic with Ne: F ([He] 2s 2p ) + e  F ([He] 2s 2p ) – 2 14 10 1 + 2 14 10 – b) Thallium loses 1 e to form an “inert pair” configuration: Tl ([Xe] 6s 4f 5d 6p  Tl ([Xe] 6s 4f 5d ) + e – or thallium can lose 3 e to form a “pseudo-noble gas” configuration: 2 14 10 1 3+ 14 10 – Tl ([Xe] 6s 4f 5d 6p  Tl ([Xe] 4f 5d ) + 3e 2 2+ – c) Magnesium loses two electrons to be isoelectronic with Ne: Mg ([Ne] 3s )  Mg ([Ne]) + 2e

8.6A

Plan: Write the condensed electron configuration for each atom, being careful to note those elements which are irregular. The charge on the cation tells how many electrons are to be removed. The electrons are removed beginning with the ns electrons. If any electrons in the final ion are unpaired, the ion is paramagnetic. If it is not obvious that there are unpaired electrons, a partial orbital diagram might help. Solution: 2 3 3+ 2 a) The V atom ([Ar]4s 3d ) loses the two 4s electrons and one 3d electron to form V ([Ar]3d ). There are two 3+ unpaired d electrons, so V is paramagnetic. 2 8 2+ 8 b) The Ni atom ([Ar]4s 3d ) loses the two 4s electrons to form Ni ([Ar]3d ). There are two unpaired d electrons, 2+ so Ni is paramagnetic.

3d 2

c) The La atom ([Xe]6s 5d ) loses all three valence electrons to form La ([Xe]). There are no unpaired electrons, 3+ so La is diamagnetic.

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8-3

8.6B

8.7A

8.7B

Plan: Write the condensed electron configuration for each atom, being careful to note those elements which are irregular. The charge on the cation tells how many electrons are to be removed. The electrons are removed beginning with the ns electrons. If it is not obvious that there are unpaired electrons, a partial orbital diagram might help. Solution: 2 2 2+ 2 a) The Zr atom ([Kr]5s 4d ) loses the two 5s electrons to form Zr ([Kr]4d ). There are two unpaired d electrons. 2 14 6 3+ 14 5 b) The Os atom ([Xe] 6s 4f 5d ) loses the two 6s electrons and one 5d electron to form Os ([Xe] 4f 5d ). There are five unpaired d electrons. 2 7 2+ 7 c) The Co atom ([Ar] 4s 3d ) loses the two 4s electrons to form Co ([Ar] 3d ). There are three unpaired d electrons.

3d Plan: Locate each of the elements on the periodic table. Cations are smaller than the parent atoms, and anions are larger than the parent atoms. If the electrons are equal, anions are larger than cations. The more electrons added or removed; the greater the change in size. Solution: – – – a) Ionic size increases down a group, so F < Cl < Br . b) These species are isoelectronic (all have 10 electrons), so size increases with increasing negative charge: 2+ + – Mg < Na < F 3+ 2+ c) Ionic size increases as charge decreases for different cations of the same element, so Cr < Cr . Plan: Locate each of the elements on the periodic table. Cations are smaller than the parent atoms, and anions are larger than the parent atoms. If the electrons are equal, anions are larger than cations. The more electrons added or removed; the greater the change in size. Solution: 3– 2– – a) P > S > Cl Anion size decreases with decreasing charge. +

b) Ionic size decreases up (toward the top of) a group, so Cs > Rb > K . c) These species are isoelectronic (all have 54 electrons), so size decreases with increasing positive charge: – + 2+ I > Cs > Ba . END–OF–CHAPTER PROBLEMS 8.1

Elements are listed in the periodic table in an ordered, systematic way that correlates with a periodicity of their chemical and physical properties. The theoretical basis for the table in terms of atomic number and electron configuration does not allow for an “unknown element” between Sn and Sb.

8.2

Today, the elements are listed in order of increasing atomic number. This makes a difference in the sequence of elements in only a few cases, as the larger atomic number usually has the larger atomic mass. One of these exceptions is iodine, Z = 53, which is after tellurium, Z = 52, even though tellurium has a higher atomic mass.

8.3

Plan: The value should be the average of the elements above and below the one of interest. Solution: a) Predicted atomic mass (K) = Na  Rb 22.99  85.47 = = 54.23 amu 2 2

(actual value = 39.10 amu)

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8-4

b) Predicted melting point (Br2) = Cl 2  I 2 101.0  113.6 = = 6.3°C 2 2 8.4

a) Predicted boiling point (HBr) = HCl + HI 84.9  (35.4) = = –60.15 = –60.2°C 2 2 b) Predicted boiling point (AsH3) = PH3  SbH3 87.4  (17.1) = = –52.25 = –52.2°C 2 2

(actual value = –7.2C)

(actual value = –67.0C)

(actual value = –55C)

8.5

The allowed values of n: positive integers: 1, 2, 3, 4,... The allowed values of l: integers from 0 to n – 1: 0, 1, 2, ... n – 1 The allowed values of ml: integers from –l to 0 to +l: –l, (–l + 1), ... 0, ... (l – 1), +l The allowed values of ms: –1/2 or +1/2

8.6

The quantum number ms relates to just the electron; all the others describe the orbital.

8.7

The exclusion principle states that no two electrons in the same atom may have the same four quantum numbers. Within a particular orbital, there can be only two electrons and they must have opposing spins.

8.8

In a one-electron system, all sublevels of a particular level (such as 2s and 2p) have the same energy. In many electron systems, the principal energy levels are split into sublevels of differing energies. This splitting is due to 3+ electron-electron repulsions. Be would be more like H since both have only one 1s electron.

8.9

Shielding occurs when inner electrons protect or shield outer electrons from the full nuclear attractive force. The effective nuclear charge is the nuclear charge an electron actually experiences. As the number of inner electrons increases, shielding increases, and the effective nuclear charge decreases.

8.10

Penetration occurs when the probability distribution of an orbital is large near the nucleus, which results in an increase of the overall attraction of the nucleus for the electron, lowering its energy. Shielding results in lessening this effective nuclear charge on outer shell electrons, since they spend most of their time at distances farther from the nucleus and are shielded from the nuclear charge by the inner electrons. The lower the l quantum number of an orbital, the more time the electron spends penetrating near the nucleus. This results in a lower energy for a 3p electron than for a 3d electron in the same atom.

8.11

Plan: The integer in front of the letter represents the n value. The l value designates the orbital type: l = 0 = s orbital; l = 1 = p orbital; l = 2 = d orbital; l = 3 = f orbital. Remember that a p orbital set contains 3 orbitals, a d orbital set has 5 orbitals, and an f orbital set has 7 orbitals. Any one orbital can hold a maximum of 2 electrons. Solution: a) The l = 1 quantum number can only refer to a p orbital. These quantum numbers designate the 2p orbital set (n = 2), which holds a maximum of 6 electrons, 2 electrons in each of the three 2p orbitals. b) There are five 3d orbitals, therefore a maximum of 10 electrons can have the 3d designation, 2 electrons in each of the five 3d orbitals. c) There is one 4s orbital which holds a maximum of 2 electrons.

8.12

a) The l = 1 quantum number can only refer to a p orbital, and the ml value of 0 specifies one particular p orbital, which holds a maximum of 2 electrons.

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8-5

b) The 5p orbitals, like any p orbital set, can hold a maximum of 6 electrons. c) The l = 3 quantum number can only refer to an f orbital. These quantum numbers designate the 4f orbitals, which hold a maximum of 14 electrons, 2 electrons in each of the seven 4f orbitals. 8.13

Plan: The integer in front of the letter represents the n value. The l value designates the orbital type: l = 0 = s orbital; l = 1 = p orbital; l = 2 = d orbital; l = 3 = f orbital. Remember that a p orbital set contains 3 orbitals, a d orbital set has 5 orbitals, and an f orbital set has 7 orbitals. Any one orbital can hold a maximum of 2 electrons. Solution: a) 6 electrons can be found in the three 4p orbitals, 2 in each orbital. b) The l = 1 quantum number can only refer to a p orbital, and the ml value of +1 specifies one particular p orbital, which holds a maximum of 2 electrons with the difference between the two electrons being in the ms quantum number. c) 14 electrons can be found in the 5f orbitals (l = 3 designates f orbitals; there are 7f orbitals in a set).

8.14

a) Two electrons, at most, can be found in any s orbital. b) The l = 2 quantum number can only refer to a d orbital. These quantum numbers designate the 3d orbitals, which hold a maximum of 10 electrons, 2 electrons in each of the five 3d orbitals. c) A maximum of 10 electrons can be found in the five 6d orbitals.

8.15

Properties recur periodically due to similarities in electron configurations recurring periodically. 2 2 6 1 Na: 1s 2s 2p 3s K:

1s 2s 2p 3s 3p 4s

The properties of Na and K are similar due to a similarity in their outer shell electron configuration; both have one electron in an outer shell s orbital. 8.16

Hund’s rule states that electrons will fill empty orbitals in the same sublevel before filling half-filled orbitals. This lowest-energy arrangement has the maximum number of unpaired electrons with parallel spins. In the correct electron configuration for nitrogen shown in (a), the 2p orbitals each have one unpaired electron; in the incorrect configuration shown in (b), electrons were paired in one of the 2p orbitals while leaving one 2p orbital empty. The arrows in the 2p orbitals of configuration (a) could alternatively all point down. (a) – correct (b) – incorrect

8.17

Similarities in chemical behavior are reflected in similarities in the distribution of electrons in the highest energy orbitals. The periodic table may be re-created based on these similar outer electron configurations when orbital filling in is order of increasing energy.

8.18

For elements in the same group (vertical column in periodic table), the electron configuration of the outer electrons are identical except for the n value. For elements in the same period (horizontal row in periodic table), their configurations vary because each succeeding element has one additional electron. The electron configurations are similar only in the fact that the same level (principal quantum number) is the outer level.

8.19

Outer electrons are the same as valence electrons for the main-group elements. The d electrons are often included among the valence electrons for transition elements.

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8-6

8.20

The total electron capacity for an energy level is 2n , so the n = 4 energy level holds a maximum of 2(4 ) = 32 2 6 10 14 electrons. A filled n = 4 energy level would have the following configuration: 4s 4p 4d 4f .

8.21

Plan: Write the electron configuration for the atom or ion and find the electron for which you are writing the quantum numbers. Assume that the electron is in the ground state configuration. Keep in mind the following letter orbital designation for each l value: l = 0 = s orbital, l = 1 = p orbital, l = 2 = d orbital, and l = 3 = f orbital. Solution: 1 a) Rb: [Kr]5s . The outermost electron in a rubidium atom would be in a 5s orbital (rubidium is in Row 5, Group 1). The quantum numbers for this electron are n = 5, l = 0, ml = 0, and ms = +1/2 or –1/2. – 2 5 b) The S ion would have the configuration [Ne]3s 3p . The electron added would go into a 3p orbital and is the second electron in that orbital. Possible quantum numbers are: n = 3, l = 1, ml = –1, 0 or +1, and ms = +1/2 or –1/2. 1 10 c) Ag atoms have the configuration [Kr]5s 4d . The electron lost would be from the 5s orbital with quantum numbers n = 5, l = 0, ml = 0, and ms = +1/2 or –1/2. 2 5 d) The F atom has the configuration [He]2s 2p . The electron gained would go into a 2p orbital and is the second electron in that orbital. Possible quantum numbers are n = 2, l = 1, –1, 0 or +1, and ms = +1/2 or –1/2.

8.22

a) n = 2; l = 0; ml = 0; ms = +1/2 or –1/2 b) n = 4; l = 1; ml = –1, 0 or +1; ms = +1/2 or –1/2 c) n = 6; l = 0; ml = 0; ms = +1/2 or –1/2 d) n = 2; l = 1; ml = –1, 0 or +1; ms = +1/2 or –1/2

8.23

1s 2s 2p 3s 3p 4s 3d 4p 5s 2 2 6 2 6 2 10 2 1s 2s 2p 3s 3p 4s 3d 4p 2 2 6 2 6 1s 2s 2p 3s 3p

8.24

a) Br: 1s 2s 2p 3s 3p 4s 3d 4p 2 2 6 2 b) Mg: 1s 2s 2p 3s 2 2 6 2 6 2 10 4 c) Se: 1s 2s 2p 3s 3p 4s 3d 4p

8.25

8.26

a) S: b) Kr: c) Cs:

1s 2s 2p 3s 3p 2 2 6 2 6 2 10 6 1s 2s 2p 3s 3p 4s 3d 4p 2 2 6 2 6 2 10 6 2 10 6 1 1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s

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8-7

8.27

Plan: The atomic number gives the number of electrons and the periodic table shows the order for filling sublevels. Recall that s orbitals hold a maximum of 2 electrons, a p orbital set holds 6 electrons, a d orbital set holds 10 electrons, and an f orbital set holds 14 electrons. Valence electrons are those in the highest energy level; in transition metals, the (n – 1)d electrons are also counted as valence electrons. For a condensed ground-state electron configuration, the electron configuration of the previous noble gas is shown by its element symbol in brackets, followed by the electron configuration of the energy level being filled. Solution: 2 2 a) Ti (Z = 22); [Ar]4s 3d

b) Cl (Z = 17); [Ne]3s 3p

Ne 3s 2

c) V (Z = 23); [Ar]4s 3d

3p 3

Ar 4s

8.28

a) Ba:

[Xe]6s Xe

b) Co:

[Ar]4s 3d Ar

c) Ag:

[Kr]5s 4d

8.29

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8-8

Solution: 2 5 a) Mn (Z = 25); [Ar]4s 3d Ar 4s 2

b) P (Z = 15); [Ne]3s 3p

3d 3

Ne 3s 2 6 c) Fe (Z = 26); [Ar]4s 3d

Ar 4s

8.30

a) Ga:

[Ar]4s 3d 4p

Ar 4s b) Zn:

[Ar]4s 3d

Ar 4s

c) Sc:

[Ar]4s 3d

8.31

4s 3d Plan: Add up all of the electrons in the electron configuration to obtain the atomic number of the element which is then used to identify the element and its position in the periodic table. When drawing the partial orbital diagram, only include electrons after those of the previous noble gas; remember to put one electron in each orbital in a set before pairing electrons. Solution: a) There are 8 electrons in the configuration; the element is O, Group 6A(16), Period 2.

He 2s

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8-9

b) There are 15 electrons in the configuration; the element is P, Group 5A(15), Period 3.

8.32

3s a) Cd; Group 2B(12); Period = 5

Kr 5s

b) Ni; Group 8B(10); Period = 4 Ar 4s

8.33

3d Plan: Add up all of the electrons in the electron configuration to obtain the atomic number of the element which is then used to identify the element and its position in the periodic table. When drawing the partial orbital diagram, only include electrons after those of the previous noble gas; remember to put one electron in each orbital in a set before pairing electrons. Solution:

a) There are 17 electrons in the configuration; the element is Cl; Group 7A(17); Period 3.

Ne 3s 3p b) There are 33 electrons in the configuration; the element is As; Group 5A(15); Period 4. Ar 4s

8.34

a) Mn; Group 7B(7); Period = 4 Ar 4s

b) Zr; Group 4B(4); Period = 5

8.35

5s 4d Plan: Use the periodic table and the partial orbital diagram to identify the element. Solution: a) The orbital diagram shows the element is in Period 4 (n = 4 as outer level). The configuration is 2 2 6 2 6 2 10 1 2 10 1 1s 2s 2p 3s 3p 4s 3d 4p or [Ar]4s 3d 4p . One electron in the p level indicates the element is in Group 3A(13). The element is Ga.

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8-10

b) The orbital diagram shows the 2s and 2p orbitals filled which would represent the last element in Period 2, Ne. 2 2 6 2 6 The configuration is 1s 2s 2p or [He]2s 2p . Filled s and p orbitals indicate Group 8A(18). 8.36

a) [Kr]5s 4d Nb; 5B(5) b) [He]2s 2p N; 5A(15)

8.37

Plan: Inner electrons are those seen in the previous noble gas and completed transition series (d orbitals). Outer electrons are those in the highest energy level (highest n value). Valence electrons are the outer electrons for main-group elements; for transition metals, valence electrons also include electrons in the outermost d set of orbitals. It is easiest to determine the types of electrons by writing a condensed electron configuration. Solution: 2 4 a) O (Z = 8); [He]2s 2p . There are 2 inner electrons (represented by [He]) and 6 outer electrons. The number of valence electrons (6) equals the outer electrons in this case. 2 10 2 b) Sn (Z = 50); [Kr]5s 4d 5p . There are 36 (from [Kr]) + 10 (from the filled 4d set) = 46 inner electrons. There are 4 outer electrons (highest energy level is n = 5) and 4 valence electrons. 2 c) Ca (Z = 20); [Ar]4s . There are 2 outer electrons (the 4s electrons), 2 valence electrons, and 18 inner electrons (from [Ar]). 2 6 d) Fe (Z = 26); [Ar]4s 3d . There are 2 outer electrons (from n = 4 level), 8 valence electrons (the d orbital electrons count in this case because the sublevel is not full), and 18 inner electrons (from [Ar]). 2 10 4 e) Se (Z = 34); [Ar]4s 3d 4p . There are 6 outer electrons (2 + 4 in the n = 4 level), 6 valence electrons (filled d sublevels count as inner electrons), and 28 inner electrons (18 from [Ar] and 10 from the filled 3d set).

8.38 a) Br b) Cs c) Cr d) Sr e) F

Inner electrons

Outer electrons

Valence electrons

28 54 18 36 2

7 1 1 2 7

7 1 6 2 7

8.39

Plan: Add up all of the electrons in the electron configuration to obtain the atomic number of the element which is then used to identify the element and its position in the periodic table. Solution: 2 1 a) The electron configuration [He]2s 2p has a total of 5 electrons (3 + 2 from He configuration) which is element boron with symbol B. Boron is in Group 3A(13). Other elements in this group are Al, Ga, In, and Tl. b) The electrons in this element total 16, 10 from the neon configuration plus 6 from the rest of the configuration. Element 16 is sulfur, S, in Group 6A(16). Other elements in Group 6A(16) are O, Se, Te, and Po. c) Electrons total 3 + 54 (from xenon) = 57. Element 57 is lanthanum, La, in Group 3B(3). Other elements in this group are Sc, Y, and Ac.

8.40

a) Se; other members O, S, Te, Po b) Hf; other members Ti, Zr, Rf c) Mn; other members Tc, Re, Bh

8.41

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8-11

c) The electrons in this element total 15, 10 from the neon configuration plus 5 from the rest of the configuration. Element 15 is phosphorus; other Group 5A(15) elements include N, As, Sb, and Bi. 8.42

a) Ge; other members C, Si, Sn, Pb b) Co; other members Rh, Ir, Mt c) Tc; other members Mn, Re, Bh

8.43

Plan: Write the ground-state electron configuration of sodium; for the excited state, move the outermost electron to the next orbital. Solution: 2 2 6 1 1 The ground-state configuration of Na is 1s 2s 2p 3s . Upon excitation, the 3s electron is promoted to the 3p level, 2 2 6 1 with configuration 1s 2s 2p 3p .

8.44

1s 2 a) Mg: [Ne]3s 2 5 b) Cl: [Ne]3s 3p 2 5 c) Mn: [Ar]4s 3d 2 6 d) Ne: [He]2s 2p

8.45

The size of an atom may be defined in terms of how closely it lies to a neighboring atom. The metallic radius is onehalf the distance between nuclei of adjacent atoms in a crystal of the element (typically metals). The covalent radius is one-half the distance between nuclei of identical covalently bonded atoms in molecules.

8.46

a) A = silicon; B = fluorine; C = strontium; D = sulfur b) F < S < Si < Sr c) Sr < Si < S < F

8.47

High IEs correspond to elements in the upper right of the periodic table, while relatively low IEs correspond to elements at the lower left of the periodic table.

8.48

a) For a given element, successive ionization energies always increase. As each successive electron is removed, the positive charge on the ion increases, which results in a stronger attraction between the leaving electron and the ion. b) When a large jump between successive ionization energies is observed, the subsequent electron must come from a full lower energy level. Thus, by looking at a series of successive ionization energies, we can determine 1 the number of valence electrons. For instance, the electron configuration for potassium is [Ar]4s . The first electron lost is the one from the 4s level. The second electron lost must come from the 3p level, and hence breaks into the core electrons. Thus, we see a significant jump in the amount of energy for the second ionization when compared to the first ionization. c) There is a large increase in ionization energy from IE3 to IE4, suggesting that the element has 3 valence electrons. The Period 2 element would be B; the Period 3 element would be Al; and the Period 4 element would be Ga.

8.49

The first drop occurs because the 3p sublevel is higher in energy than the 3s, so the 3p electron of Al is pulled off 4 more easily than a 3s electron of Mg. The second drop occurs because the 3p electron occupies the same orbital as another 3p electron. The resulting electron-electron repulsion raises the orbital energy and thus it is easier to 4 3 remove an electron from S (3p ) than P (3p ).

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8-12

8.50

A high, endothermic IE1 means it is very difficult to remove the first outer electron. This value would exclude any metal, because metals lose an outer electron easily. A very negative, exothermic EA 1 suggests that this element easily gains one electron. These values indicate that the element belongs to the halogens, Group 7A(17), which form –1 ions.

8.51

EA1 for oxygen is negative because energy is released when an electron is added to the neutral atom due to its attraction to the atom’s nuclear charge. The EA2 for oxygen is positive. The second electron affinity is always positive (greater energy) because it requires energy to add a (negative) electron to a (negative) anion.

8.52

After an initial shrinking for the first 2 or 3 elements, the size remains relatively constant as the shielding of the 3d electrons just counteracts the increase in the number of protons in the nucleus so the Zeff remains relatively constant.

8.53

Plan: Atomic size decreases up a main group and left to right across a period. Solution: a) Increasing atomic size: K < Rb < Cs; these three elements are all part of the same group, the alkali metals. Atomic size decreases up a main group (larger outer electron orbital), so potassium is the smallest and cesium is the largest. b) Increasing atomic size: O < C < Be; these three elements are in the same period and atomic size decreases across a period (increasing effective nuclear charge), so beryllium is the largest and oxygen the smallest. c) Increasing atomic size: Cl < S < K; chlorine and sulfur are in the same period so chlorine is smaller since it is further to the right in the period. Potassium is the first element in the next period so it is larger than either Cl or S. d) Increasing atomic size: Mg < Ca < K; calcium is larger than magnesium because Ca is further down the alkaline earth metal group on the periodic table than Mg. Potassium is larger than calcium because K is further to the left than Ca in Period 4 of the periodic table.

8.54

a) Pb > Sn > Ge

8.55

Plan: Ionization energy increases up a group and left to right across a period. Solution: a) Ba < Sr < Ca The “group” rule applies in this case. Ionization energy increases up a main group. Barium’s outer electron receives the most shielding; therefore, it is easiest to remove and has the lowest IE. b) B < N < Ne These elements have the same n, so the “period” rule applies. Ionization energy increases from left to right across a period. B experiences the lowest Zeff and has the lowest IE. Ne has the highest IE, because it’s very difficult to remove an electron from the stable noble gas configuration. c) Rb < Se < Br IE decreases with increasing atomic size, so Rb (largest atom) has the smallest IE. Se has a lower IE than Br because IE increases across a period. d) Sn < Sb < As IE increases up a group, so Sn and Sb will have smaller IEs than As. The “period” rule applies for ranking Sn and Sb.

8.56

a) Li > Na > K

8.57

Plan: When a large jump between successive ionization energies is observed, the subsequent electron must come from a full lower energy level. Thus, by looking at a series of successive ionization energies, we can determine the number of valence electrons. The number of valence electrons identifies which group the element is in. Solution: The successive ionization energies show a very significant jump between the third and fourth IEs. This indicates that the element has three valence electrons. The fourth electron must come from the core electrons and thus has a very large ionization energy. The electron configuration of the Period 2 element with three valence electrons is 2 2 1 1s 2s 2p , which represents boron, B.

b) Sr > Sn > Te

b) F > C > Be

c) Na > F > Ne

c) Ar > Cl > Na

d) Na > Mg > Be

d) Cl > Br > Se

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8-13

8.58

The successive ionization energies show a significant jump between the second and third IEs, indicating that the 2 2 6 2 element has only two valence electrons. The configuration is 1s 2s 2p 3s , Mg.

8.59

Plan: For a given element, successive ionization energies always increase. As each successive electron is removed, the positive charge on the ion increases, which results in a stronger attraction between the leaving electron and the ion. A very large jump between successive ionization energies will occur when the electron to be removed comes from a full lower energy level. Examine the electron configurations of the atoms. If the IE2 represents removing an electron from a full orbital, then the IE2 will be very large. In addition, for atoms with the same outer electron configuration, IE2 is larger for the smaller atom. Solution: a) Na would have the highest IE2 because ionization of a second electron would require breaking the stable [Ne] configuration: 1 + – First ionization: Na ([Ne]3s )  Na ([Ne]) + e (low IE) + +2 2 5 – Second ionization: Na ([Ne])  Na ([He]2s 2p ) + e (high IE) b) Na would have the highest IE2 because it has one valence electron and is smaller than K. c) You might think that Sc would have the highest IE2, because removing a second electron would require breaking the stable, filled 4s shell. However, Be has the highest IE2 because Be’s small size makes it difficult to remove a second electron.

8.60

a) Al

8.61

Three of the ways that metals and nonmetals differ are: (1) metals conduct electricity, nonmetals do not; (2) when they form stable ions, metal ions tend to have a positive charge, nonmetal ions tend to have a negative charge; and (3) metal oxides are ionic and act as bases, nonmetal oxides are covalent and act as acids.

8.62

Metallic character decreases up a group and decreases toward the right across a period. These trends are the same as those for atomic size and opposite those for ionization energy.

8.63

In general, the larger the element in Group 1A(1), the more readily that element loses an electron (oxidation) and the stronger the reducing strength of the element. Atomic size increases down Group 1A(1) and thus reducing strength increases down the group as well. When IE is high, atoms tend to gain electrons rather than to lose electrons. When an atom gains electrons, it is reduced and serves as an oxidizing agent. IE increases up Group 7A(17); thus, oxidizing strength also increases up Group 7A(17). Fluorine, which has the highest IE in Group 7A(17), is the strongest oxidizing agent in the group.

8.64

Generally, oxides of metals are basic while oxides of nonmetals are acidic. As the metallic character decreases, the oxide becomes more acidic. Thus, oxide acidity increases from left to right across a period and from bottom to top in a group.

8.65

Plan: Write the electron configurations for the two elements. Remember that these elements lose electrons to achieve pseudo-noble gas configurations. Solution: The two largest elements in Group 4A, Sn and Pb, have atomic electron configurations that look like 2 10 2 ns (n – 1)d np . Both of these elements are metals so they will form positive ions. To reach the noble gas configuration of xenon the atoms would have to lose 14 electrons, which is not likely. Instead the atoms lose either 2 or 4 electrons to attain a stable configuration with either the ns and (n – 1)d filled for the 2+ ion or the 2+ 2+ (n – 1)d orbital filled for the 4+ ion. The Sn and Pb ions form by losing the two p electrons: 2 10 2 2+ 2 10 – Sn ([Kr]5s 4d 5p )  Sn ([Kr]5s 4d ) + 2 e 2 10 2 2+ 2 10 – Pb ([Xe]6s 5d 6p )  Pb ([Xe]6s 5d ) + 2 e

b) Sc

c) Al

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8-14

The Sn and Pb ions form by losing the two p and two s electrons: 2 10 2 4+ 10 – Sn ([Kr]5s 4d 5p )  Sn ([Kr]4d ) + 4 e 2 10 2 4+ 10 – Pb ([Xe]6s 5d 6p )  Pb ([Xe]5d ) + 4 e Possible ions for tin and lead have +2 and +4 charges. The +2 ions form by loss of the outermost two p electrons, while the +4 ions form by loss of these and the outermost two s electrons. 10

8.66

An (n – 1)d ns np configuration is called a pseudo-noble gas configuration. In : [Kr]4d

8.67

Paramagnetism is the tendency of a species with unpaired electrons to be attracted by an external magnetic field. The degree of paramagnetism increases with the number of unpaired electrons, so that the number of unpaired electrons may be determined by magnetic measurements. A substance with no unpaired electrons is weakly repelled from a magnetic field and is said to be diamagnetic. 1+ 10 Cu : [Ar]3d thus, diamagnetic 2+ 9 Cu : [Ar]3d thus, paramagnetic, with one unpaired electron

8.68

3+ < 2+ < 1+ < 0 < 1– < 2– < 3–

8.69

Plan: Metallic behavior decreases up a group and decreases left to right across a period. Solution: a) Rb is more metallic because it is to the left and below Ca. b) Ra is more metallic because it lies below Mg in Group 2A(2). c) I is more metallic because it lies below Br in Group 7A(17).

8.70

a) S

8.71

Plan: Metallic behavior decreases up a group and decreases left to right across a period. Solution: a) As should be less metallic than antimony because it lies above Sb in the same group of the periodic table. b) P should be less metallic because it lies to the right of silicon in the same period of the periodic table. c) Be should be less metallic since it lies above and to the right of sodium on the periodic table.

8.72

a) Rn

8.73

Plan: Generally, oxides of metals are basic while oxides of nonmetals are acidic. Solution: The reaction of a nonmetal oxide in water produces an acidic solution. An example of a Group 6A(16) nonmetal oxide is SO2(g): SO2(g) + H2O(g)  H2SO3(aq).

8.74

A main-group metal oxide produces a basic solution SrO(s) + H2O(l)  Sr(OH)2(aq)

8.75

Plan: For main-group elements, the most stable ions have electron configurations identical to noble gas atoms. Write the electron configuration of the atom and then remove or add electrons until a noble gas configuration is achieved. Metals lose electrons and nonmetals gain electrons. Solution: 2 2 6 2 5 a) Cl: 1s 2s 2p 3s 3p ; chlorine atoms are one electron short of a noble gas configuration, so a –1 ion will form by – 2 2 6 2 6 adding an electron to have the same electron configuration as an argon atom: Cl , 1s 2s 2p 3s 3p . 2 2 6 1 b) Na: 1s 2s 2p 3s ; sodium atoms contain one more electron than the noble gas configuration of neon. Thus, a + 2 2 6 sodium atom loses one electron to form a +1 ion: Na , 1s 2s 2p . 2 2 6 2 6 2 c) Ca: 1s 2s 2p 3s 3p 4s ; calcium atoms contain two more electrons than the noble gas configuration of argon. 2+ 2 2 6 2 6 Thus, a calcium atom loses two electrons to form a +2 ion: Ca , 1s 2s 2p 3s 3p .

b) In

b) Te

c) As

c) Se

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8-15

8.76

a) Rb : 1s 2s 2p 3s 3p 4s 3d 4p 3– 2 2 6 b) N : 1s 2s 2p – 2 2 6 2 6 2 10 6 c) Br : 1s 2s 2p 3s 3p 4s 3d 4p

8.77

Plan: For main-group elements, the most stable ions have electron configurations identical to noble gas atoms. Write the electron configuration of the atom and then remove or add electrons until a noble gas configuration is achieved. Metals lose electrons and nonmetals gain electrons. Solution: 2 2 6 2 1 a) Al: 1s 2s 2p 3s 3p ; aluminum atoms contain three more electrons than the noble gas configuration of Ne. 3+ 2 2 6 Thus, an aluminum atom loses its 3 outer shell electrons to form a +3 ion: Al , 1s 2s 2p . 2 2 6 2 4 b) S: 1s 2s 2p 3s 3p ; sulfur atoms are two electrons short of the noble gas configuration of argon. Thus, a 2– 2 2 6 2 6 sulfur atom gains two electrons to form a –2 ion: S , 1s 2s 2p 3s 3p . 2 2 6 2 6 2 10 6 2 c) Sr: 1s 2s 2p 3s 3p 4s 3d 4p 5s ; strontium atoms contain two more electrons than the noble gas configuration of 2+ 2 2 6 2 6 2 10 6 krypton. Thus, a strontium atom loses two electrons to form a +2 ion: Sr , 1s 2s 2p 3s 3p 4s 3d 4p .

8.78

a) P 1s 2s 2p 3s 3p 2+ 2 2 6 b) Mg 1s 2s 2p 2– 2 2 6 2 6 2 10 6 c) Se 1s 2s 2p 3s 3p 4s 3d 4p

8.79

Plan: To find the number of unpaired electrons look at the electron configuration expanded to include the different orientations of the orbitals, such as px and py and pz. Remember that one electron will occupy every orbital in a set (p, d, or f ) before electrons will pair in an orbital in that set. In the noble gas configurations, all electrons are paired because all orbitals are filled. Solution: 2 a) Configuration of 2A(2) group elements: [noble gas]ns , no unpaired electrons. The electrons in the ns orbital are paired. 2 1 1 1 b) Configuration of 5A(15) group elements: [noble gas]ns npx npy npz . Three unpaired electrons, one each in px, py, and pz. 2 6 c) Configuration of 8A(18) group elements: noble gas configuration ns np with no half-filled orbitals, no unpaired electrons. 2 1 d) Configuration of 3A(13) group elements: [noble gas]ns np . There is one unpaired electron in one of the p orbitals.

3–

8.80

+1 –3 –1

ns ns np np c) d) To find the number of unpaired electrons look at the electron configuration expanded to include the different orientations of the orbitals, such as px and py and pz. In the noble gas configurations, all electrons are paired because all orbitals are filled. 2 1 1 0 a) Configuration of 4A(14) group elements: [noble gas]ns npx npy npz . Two unpaired electrons. 2 2 2 1 b) Configuration of 7A(17) group elements: [noble gas]ns npx npy npz . One unpaired electron. 1 c) Configuration of 1A(1) group elements: [noble gas]ns . One unpaired electron. 2 2 1 1 d) Configuration of 6A(16) group elements: [noble gas]ns npx npy npz . Two unpaired electrons.

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8-16

8.81

Plan: Substances are paramagnetic if they have unpaired electrons. To find the number of unpaired electrons look at the electron configuration expanded to include the different orientations of the orbitals, such as px and py and pz. Remember that all orbitals in a p, d, or f set will each have one electron before electrons pair in an orbital. In the noble gas configurations, all electrons are paired because all orbitals are filled. Solution: 2 10 1 a) Ga (Z = 31) = [Ar]4s 3d 4p . The s and d sublevels are filled, so all electrons are paired. The lone p electron is unpaired, so this element is paramagnetic. 2 1 1 0 b) Si (Z = 14) = [Ne]3s 3px 3py 3pz . This element is paramagnetic with two unpaired electrons.

Correct Incorrect c) Be (Z = 4) = [He]2s . The two s electrons are paired so Be is not paramagnetic. 2 10 2 1 1 d) Te (Z = 52) = [Kr]5s 4d 5px 5py 5pz is paramagnetic with two unpaired electrons in the 5p set. 2

8.82

a) Ti : [Ar]3d 2+ 10 b) Zn : [Ar]3d 2+ c) Ca : [Ar] 2+ 2 10 d) Sn : [Kr]5s 4d

paramagnetic diamagnetic diamagnetic diamagnetic

8.83

Plan: Substances are paramagnetic if they have unpaired electrons. Write the electron configuration of the atom and then remove the specified number of electrons. Remember that all orbitals in a p, d, or f set will each have one electron before electrons pair in an orbital. In the noble gas configurations, all electrons are paired because all orbitals are filled. Solution: 2 3 3+ 2 a) V: [Ar]4s 3d ; V : [Ar]3d Transition metals first lose the s electrons in forming ions, so to form the +3 ion a vanadium atom loses two 4s electrons and one 3d electron. Paramagnetic Ar 4s 3d 2 10 2+ 10 b) Cd: [Kr]5s 4d ; Cd : [Kr]4d Cadmium atoms lose two electrons from the 5s orbital to form the +2 ion. Diamagnetic Kr 5s

4d c) Co: [Ar]4s 3d ; Co : [Ar]3d Cobalt atoms lose two 4s electrons and one 3d electron to form the +3 ion. Paramagnetic 2

Ar 4s 3d 1 10 + 10 d) Ag: [Kr]5s 4d ; Ag : [Kr]4d Silver atoms lose the one electron in the 5s orbital to form the +1 ion. Diamagnetic Kr 5s Copyright

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8-17

8.84

a) Mo : [Kr]4d + 14 10 b) Au : [Xe]4f 5d 2+ 5 c) Mn : [Ar]3d 2+ 14 2 d) Hf : [Xe]4f 5d

paramagnetic diamagnetic paramagnetic paramagnetic

8.85

Plan: Substances are diamagnetic if they have no unpaired electrons. Draw the partial orbital diagrams, remembering that all orbitals in d set will each have one electron before electrons pair in an orbital. Solution: 2 8 You might first write the condensed electron configuration for Pd as [Kr]5s 4d . However, the partial orbital diagram is not consistent with diamagnetism. Kr 5p 5s 4d 1 9 Promoting an s electron into the d sublevel (as in (c) [Kr]5s 4d ) still leaves two electrons unpaired. Kr 5s 4d 10 The only configuration that supports diamagnetism is (b) [Kr]4d .

8.86

5s 4d 2 3 The expected electron configuration for Group 5B(5) elements is ns (n–1)d . 2 3 – Nb (expected): [Kr]5s 4d 3 unpaired e

Kr 5s 1

Nb (actual): [Kr]5s 4d

4d 4

5 unpaired e

–

8.87

8.88

5s 4d Plan: The size of ions increases down a group. For ions that are isoelectronic (have the same electron configuration), size decreases with the increasing atomic number. Solution: + + + a) Increasing size: Li < Na < K , size increases down Group 1A(1). + – 2– b) Increasing size: Rb < Br < Se , these three ions are isoelectronic with the same electron configuration as krypton. Size decreases with increasing atomic number in an isoelectronic series. – 2– 3– c) Increasing size: F < O < N , the three ions are isoelectronic with an electron configuration identical to neon. Size decreases with the increasing atomic number in an isoelectronic series. 2–

2–

a) Se > S > O , size increases down a group. 2– – + b) Te > I > Cs , size decreases with the increasing atomic number in an isoelectronic series. + 2+ 2+ c) Cs > Ba > Sr , both reasons as in parts (a) and (b).

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8-18

8.89

a) oxygen g) thallium m) scandium

b) cesium h) krypton n) manganese

8.90

Plan: Write the electron configuration for each atom. Remove the specified number of electrons, given by the positive ionic charge, to write the configuration for the ions. Remember that electrons with the highest n value are removed first. Solution: 2 1 1 2 7 Ce: [Xe]6s 4f 5d Eu: [Xe]6s 4f 4+ 2+ 7 Ce : [Xe] Eu : [Xe]4f 4+ 2+ Ce has a noble gas configuration and Eu has a half-filled f subshell.

8.91

Plan: Write the formula of the oxoacid. Remember that in naming oxoacids (H + polyatomic ion), the suffix of the polyatomic changes: -ate becomes -ic acid and -ite becomes -ous acid. Determine the oxidation state of the nonmetal in the oxoacid; hydrogen has an O.N. of +1 and oxygen has an O.N. of –2. Based on the oxidation state of the nonmetal, and the oxidation state of the oxide ion (–2), the formula of the nonmetal oxide may be determined. The name of the nonmetal oxide comes from the formula; remember that nonmetal compounds use prefixes to indicate the number of each type of atom in the formula. Solution: + a) hypochlorous acid = HClO has Cl so the oxide is Cl2O = dichlorine oxide or dichlorine monoxide 3+ b) chlorous acid = HClO2 has Cl so the oxide is Cl2O3 = dichlorine trioxide 5+ c) chloric acid = HClO3 has Cl so the oxide is Cl2O5 = dichlorine pentaoxide 7+ d) perchloric acid = HClO4 has Cl so the oxide is Cl2O7 = dichlorine heptaoxide 6+ e) sulfuric acid = H2SO4 has S so the oxide is SO3 = sulfur trioxide 4+ f) sulfurous acid = H2SO3 has S so the oxide is SO2 = sulfur dioxide 5+ g) nitric acid = HNO3 has N so the oxide is N2O5 = dinitrogen pentaoxide 3+ h) nitrous acid = HNO2 has N so the oxide is N2O3 = dinitrogen trioxide 4+ i) carbonic acid = H2CO3 has C so the oxide is CO2 = carbon dioxide 5+ j) phosphoric acid = H3PO4 has P so the oxide is P2O5 = diphosphorus pentaoxide

8.92

a) The distance, d, between the electron and the nucleus would be smaller for He than for H, so IE would be larger. b) You would expect that IE1 (He) to be approximately 2 IE1 (H), since ZHe = 2 ZH. However, since the effective nuclear charge of He is < 2 due to shielding, IE1 (He) < 2 IE1(H).

6.62610

34

8.93

= hc/E =



d) carbon j) ruthenium p) sulfur

e) rubidium k) vanadium q) strontium

f) bismuth l) indium r) arsenic

 = 4.59564  10 = 4.610 m

J  s 3.00108 m/s

7

1.60210 J   2.7 eV    1 eV  The absorption of light of this wavelength (blue) leads to the complimentary color (yellow) being seen. An electron in gold’s 5d subshell can absorb blue light in its transition to a 6s subshell, giving gold its characteristic “gold” color.



8.94

c) aluminum i) silicon o) lutetium



19

Plan: Remember that isoelectronic species have the same electron configuration. Atomic radius decreases up a group and left to right across a period. Solution: 2+ – a) A chemically unreactive Period 4 element would be Kr in Group 8A(18). Both the Sr ion and Br ion are isoelectronic with Kr. Their combination results in SrBr2, strontium bromide. 2+ 2– b) Ar is the Period 3 noble gas. Ca and S are isoelectronic with Ar. The resulting compound is CaS, calcium sulfide. 2+ c) The smallest filled d subshell is the 3d shell, so the element must be in Period 4. Zn forms the Zn ion by losing 10 its two s subshell electrons to achieve a pseudo–noble gas configuration ([Ar]3d ). The smallest halogen is fluorine, – whose anion is F . The resulting compound is ZnF2, zinc fluoride. d) Ne is the smallest element in Period 2, but it is not ionizable. Li is the largest atom whereas F is the smallest atom in Period 2. The resulting compound is LiF, lithium fluoride. McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution

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8-19

8.95

Plan: Recall Hess’s law: the enthalpy change of an overall process is the sum of the enthalpy changes of its individual steps. Both the ionization energies and the electron affinities of the elements are needed. Solution: a) F: ionization energy = 1681 kJ/mol electron affinity = –328 kJ/mol + – F(g)  F (g) + e H = 1681 kJ/mol – – F(g) + e  F (g) H = –328 kJ/mol – – Reverse the electron affinity reaction to give: F (g)  F(g) + e H = +328 kJ/mol Summing the ionization energy reaction with the reversed electron affinity reaction (Hess’s law): + – F(g)  F (g) + e H = 1681 kJ/mol F (g)  F(g) + e –

H = +328 kJ/mol

–

F (g)  F (g) + 2 e –

–

H = 2009 kJ/mol

b) Na: ionization energy = 496 kJ/mol Na(g)  Na (g) + e +

–

Na(g) + e  Na (g) –

–

electron affinity = –52.9 kJ/mol

H = 496 kJ/mol H = –52.9 kJ/mol

Reverse the ionization reaction to give: Na (g) + e  Na(g) H = –496 kJ/mol Summing the electron affinity reaction with the reversed ionization reaction (Hess’s law): – – Na(g) + e  Na (g) H = –52.9 kJ/mol +

–

Na (g) + e  Na(g)

H = –496 kJ/mol

Na (g) + 2 e  Na (g)

H = –548.9 = –549 kJ/mol

+ +

–

8.96

(a) For nonmetals, the lowest oxidation number is given by the A-group number minus eight. Sulfur is in Group 6A(16). 2– Therefore, the lowest oxidation sulfur can exhibit is 6 – 8 = –2. Sulfide ion, S , has the lowest oxidation number and 2– cannot gain additional electrons to form a more negative ion. Thus the S ion cannot be reduced and cannot function as an oxidizing agent. The sulfide ion can lose electrons and be oxidized and therefore can function as a reducing agent. (b) For nonmetals, the highest oxidation number is given by the A-group number. Sulfur, in Group 6A(16), has a 2– maximum oxidation number of +6, as seen in the sulfate ion, SO4 . Sulfur in the sulfate ion can gain electrons (reduction), resulting in a lower oxidation number, and can therefore function as an oxidizing agent. However, sulfur in the sulfate ion cannot lose additional electrons (oxidation) since a higher oxidation number is not possible. Since S in the sulfate ion cannot be oxidized, it cannot function as a reducing agent. (c) The highest oxidation number for S is +6 and the lowest oxidation number is –2. Sulfur in sulfur dioxide, SO2, has an intermediate oxidation number of +4. Therefore sulfur in SO 2 can function as a reducing agent by losing electrons (oxidation), resulting in a higher oxidation number, or sulfur can function as an oxidizing agent by gaining electrons (reduction), resulting in a lower oxidation number.

8.97

Plan: Determine the electron configuration for iron, and then begin removing one electron at a time. Remember that all orbitals in a d set will each have one electron before electrons pair in an orbital, and electrons with the highest n value are removed first. Ions with all electrons paired are diamagnetic. Ions with at least one unpaired electron are paramagnetic. The more unpaired electrons, the greater the attraction to a magnetic field. Solution: 2 6 Fe [Ar]4s 3d partially filled 3d = paramagnetic number of unpaired electrons = 4 + 1 6 Fe [Ar]4s 3d partially filled 3d = paramagnetic number of unpaired electrons = 5 2+ 6 Fe [Ar]3 d partially filled 3d = paramagnetic number of unpaired electrons = 4 3+ 5 Fe [Ar]3d partially filled 3d = paramagnetic number of unpaired electrons = 5 4+ 4 Fe [Ar]3d partially filled 3d = paramagnetic number of unpaired electrons = 4 5+ 3 Fe [Ar]3d partially filled 3d = paramagnetic number of unpaired electrons = 3 6+ 2 Fe [Ar]3d partially filled 3d = paramagnetic number of unpaired electrons = 2 7+ 1 Fe [Ar]3d partially filled 3d = paramagnetic number of unpaired electrons = 1 8+ Fe [Ar] filled orbitals = diamagnetic number of unpaired electrons = 0 9+ 2 5 Fe [Ne]3s 3p partially filled 3p = paramagnetic number of unpaired electrons = 1

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8-20

10+

Fe [Ne]3s 3p partially filled 3p = paramagnetic number of unpaired electrons = 2 11+ 2 3 Fe [Ne]3s 3p partially filled 3p = paramagnetic number of unpaired electrons = 3 12+ 2 2 Fe [Ne]3s 3p partially filled 3p = paramagnetic number of unpaired electrons = 2 13+ 2 1 Fe [Ne]3s 3p partially filled 3p = paramagnetic number of unpaired electrons = 1 14+ 2 Fe [Ne]3s filled orbitals = diamagnetic number of unpaired electrons = 0 + 3+ Fe and Fe would both be most attracted to a magnetic field. They each have 5 unpaired electrons. 2

8.98

a) Ga [Ar]4s 3d 4p Ge [Ar]4s 3d 4p In each case, an electron is removed from a 4p orbital. Because the Zeff of Ge is greater, IE of Ge will be higher. + 2 10 + 2 10 1 b) Ga [Ar]4s 3d Ge [Ar]4s 3d 4p + + IE2 would be higher for Ga , because an electron must be removed from a full 4s orbital. For Ge the electron is removed from the 4p orbital. 2+ 1 10 2+ 2 10 c) Ga [Ar]4s 3d Ge [Ar]4s 3d 2+ 2+ In each case, an electron is removed from a 4s orbital. Because the Zeff of Ge is greater, IE3 of Ge will be higher. 3+ 10 2+ 1 10 d) Ga [Ar]3d Ge [Ar]4s 3d 3+ 3+ IE4 would be higher for Ga , because an electron must be removed from a full 3d orbital. For Ge the electron is removed from the 4s orbital.

8.99

a) The energy required would be 376 kJ/mol (Cs). 6.626 1034 J  s 3.00 108 m /s  1 kJ   = 3.18365 × 10–7 = 3.18 × 10–7 m  = hc/E = 103 J     kJ mol      376  6.022 10 23   mol 







b) The energy required would be 503 kJ/mol (Ba). = hc/E =

6.62610





J  s 3.00 108 m /s  1 kJ   = 2.37983 × 10–7 = 2.38 × 10–7 m  10 3 J     kJ mol     503  6.022 10 23   mol  34

c) No other element, other than the alkali metals, in Figure 8.16 has a lower ionization energy than Ba so the –7 radiation, 2.39 × 10 m, cannot ionize other elements. d) Both of these photons are in the ultraviolet region of the EM spectrum. +

–

8.100

a) Rubidium atoms form +1 ions, Rb ; bromine atoms form –1 ions, Br . 1 + + b) Rb: [Kr]5s ; Rb : [Kr]; Rb is a diamagnetic ion that is isoelectronic with Kr. 2 10 5 – 2 10 6 – Br: [Ar]4s 3d 4p ; Br : [Ar]4s 3d 4p or [Kr]; Br is a diamagnetic ion that is isoelectronic with Kr. + – c) Rb is a smaller ion than Br ; B best represents the relative ionic sizes.

8.101

A: X = [Kr]4d ; X = [Kr]5s 4d . The element is palladium and the oxide is PdO. 3+ 6 2 7 B: X = [Ar]3d ; X = [Ar]4s 3d . The element is cobalt and the oxide is Co2O3. + 10 1 10 C: X = [Kr]4d ; X = [Kr]5s 4d . The element is silver and the oxide is Ag2O. +4 3 2 5 D: X = [Ar]3d ; X = [Ar]4s 3d . The element is manganese and the oxide is MnO2.

8.102

balloonium inertium allotropium brinium canium fertilium liquidium utilium crimsonium

= = = = = = = = =

helium neon sulfur sodium tin nitrogen bromine aluminum strontium

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8-21

CHAPTER 9 MODELS OF CHEMICAL BONDING FOLLOW–UP PROBLEMS 9.1A

Plan: First, write out the condensed electron configuration, partial orbital diagram, and electron-dot structure for magnesium atoms and chlorine atoms. In the formation of the ions, each magnesium atom will lose two electrons to form the +2 ion and each chlorine atom will gain one electron to form the –1 ion. Write the condensed electron configuration, partial orbital diagram, and electron-dot structure for each of the ions. The formula of the compound is found by combining the ions in a ratio that gives a neutral compound. Solution: Condensed electron configurations: 2 2 5 2+ – 2 6 Mg ([Ne]3s ) + Cl ([Ne]3s 3p )  Mg ([Ne]) + Cl ([Ne]3s 3p ) In order to balance the charge (or the number of electrons lost and gained) two chlorine atoms are needed: 2 2 5 2+ – 2 6 Mg ([Ne]3s ) + 2Cl ([Ne]3s 3p )  Mg ([Ne]) + 2Cl ([Ne]3s 3p ) Partial orbital diagrams:

Lewis electron–dot symbols:

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9-1

The formula of the compound would contain two chloride ions for each magnesium ion, MgCl2. 9.1B

Plan: First, write out the condensed electron configuration, partial orbital diagram, and electron-dot structure for calcium atoms and oxygen atoms. In the formation of the ions, each calcium atom will lose two electrons to form the +2 ion and each oxygen atom will gain two electrons to form the –2 ion. Write the condensed electron configuration, partial orbital diagram, and electron-dot structure for each of the ions. The formula of the compound is found by combining the ions in a ratio that gives a neutral compound. Solution: Condensed electron configurations: 2 2 4 2+ 2– 2 6 Ca ([Ar]4s ) + O ([He]2s 2p )  Ca ([Ar]) + O ([He]2s 2p ) The charge is balanced (the number of electrons lost equals the number of electrons gained). Partial orbital diagrams:

Lewis electron–dot symbols:

The formula of the compound would contain one oxide ion for each calcium ion, CaO. 9.2A

Plan: Examine the charge of the ions involved in the compounds. Use periodic trends in ionic radii to determine the relative size of the ions in the compounds. Then apply Coulomb’s law. According to Coulomb’s law, for ions of similar size, lower charge leads to a smaller lattice energy. For ions with the same charge, larger size leads to smaller lattice energy. Solution: The compound with the smaller lattice energy is BaF2. The only difference between these compounds is the 2+ 2+ size of the cation: the Ba ion is larger than the Sr ion (ionic size increases down the group). According to Coulomb’s law, for ions with the same charge, larger size leads to smaller lattice energy. McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or

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9-2

9.2B

Plan: Examine the charge of the ions involved in the compounds. Use periodic trends in ionic radii to determine the relative size of the ions in the compounds. Then apply Coulomb’s law. According to Coulomb’s law, for ions of similar size, higher charge leads to a larger lattice energy. For ions with the same charge, smaller size leads to larger lattice energy. Solution: + 22+ Na2O < MgF2 < CaO In Na2O, the ions are Na and O ; the ions in MgF2 are Mg and F . The products of the cation charge and anion charge are the same for these two compounds (1 × 2 and 2 × 1); + 2+ 2however, since Na is larger than Mg , and O is larger than F , Na2O has the smaller lattice energy. In 2+ 22+ + 2+ CaO, the ions are Ca and O . The Ca ion is larger than Na or Mg but the larger product of cation and anion charge (2 × 2 for CaO) results in a larger lattice energy for CaO. In nearly every case, charge is more important than size.

9.3A

Plan: a) All bonds are triple bonds from carbon to a second row element. The trend in bond length can be predicted from the fact that atomic radii decrease across a row, so oxygen will be smaller than nitrogen, which is smaller than carbon. Bond length will therefore decrease across the row while bond energy increases. b) All the bonds are single bonds from phosphorus to a group 7A element. The trend in bond length can be predicted from the fact that atomic radii increase down a column, so fluorine will be smaller than bromine, which is smaller than iodine. Bond length will therefore increase down the column while bond energy increases. Solution: a) Bond length: CC > CN > CO Bond strength: CO > CN > CC b) Bond length: P–I > P–Br > P–F Bond energy: P–F > P–Br > P–I Check: (Use the tables in the chapter.) Bond lengths from Table: CC, 121 pm > CN, 115 pm > CO, 113 pm Bond energies from Table: CO, 1070 kJ/mol > CN, 891 kJ/mol > CC, 839 kJ/mol Bond lengths from Table: P–I, 246 pm > P–Br, 222 pm > P–F, 156 pm Bond energies from Table: P–F, 490 kJ/mol > P–Br, 272 kJ/mol > P–I, 184 kJ/mol The values from the tables agree with the order predicted.

9.3B

Plan: a) All bonds are single bonds from silicon to a second row element. The trend in bond length can be predicted from the fact that atomic radii decrease across a row, so fluorine will be smaller than oxygen, which is smaller than carbon. Bond length will therefore decrease across the row while bond energy increases. b) All the bonds are between two nitrogen atoms, but differ in the number of electrons shared between the atoms. The more electrons shared the shorter the bond length and the greater the bond energy. Solution: a) Bond length: Si–F < Si–O < Si–C Bond strength: Si–C < Si–O < Si–F b) Bond length: NN < N=N < N–N Bond energy: N–N < N=N < NN Check: (Use the tables in the chapter.) Bond lengths from Table: Si–F, 156 pm < Si–O, 161 pm < Si–C, 186 pm Bond energies from Table: Si–C, 301 kJ < Si–O, 368 kJ < Si–F, 565 kJ Bond lengths from Table: NN, 110 pm < N=N, 122 pm < N–N, 146 pm Bond energies from Table: N–N, 160 kJ < N=N, 418 kJ < NN, 945 kJ The values from the tables agree with the order predicted.

9.4A

Plan: Assume that all reactant bonds break and all product bonds form. Use the bond energy table to find the values for the different bonds. Sum the values for the reactants, and sum the values for the products (these are all negative values). Add the sum of the product values to the sum of the reactant values. Solution: Calculating H  for the bonds broken:

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9-3

1 × NN = (1 mol)(945 kJ/mol) = 945 kJ 3 × H–H = (3 mol)(432 kJ/mol) = 1296 kJ H

 bonds broken

= 2241 kJ 

Calculating H for the bonds formed: 1 × NH3 = (3 mol)(–391 kJ/mol) = –1173 kJ 1 × NH3 = (3 mol)(–391 kJ/mol) = –1173 kJ H

 bonds formed

= –2346 kJ H

 rxn

H

 bonds broken

 H bonds formed

 = 2241 kJ + (–2346 kJ) = –105 kJ  H rxn

9.4B

 bonds broken

= 1314 kJ 

Calculating H for the bonds formed: 2 × H–F = (2 mol)(–565 kJ/mol) = –1130 kJ 2 × O=O = (1 mol)(–498 kJ/mol) = –498 kJ H

 bonds formed

= –1628 kJ H

 rxn

H

 bonds broken

 H bonds formed

 = 1314 kJ + (–1628 kJ) = –314 kJ  H rxn

9.5A

Plan: Bond polarity can be determined from the difference in electronegativity of the two atoms. The more electronegative atom holds the – charge and the other atom holds the + charge. Solution: a) From the electronegativity figure, EN of Cl = 3.0, EN of F = 4.0, EN of Br = 2.8. Cl–Cl will be the least polar since there is no difference between the electronegativity of the two chlorine atoms. Cl–F will be more polar than Cl–Br since the electronegativity difference between Cl and F (4.0 – 3.0 = 1.0) is greater than the electronegativity difference between Cl and Br (3.0 – 2.8 = 0.2).

b) EN of P = 2.1, EN of F = 4.0, EN of N = 3.0, EN of O = 3.5. The least polar is N–O with an EN difference of 0.5 (3.5 – 3.0 = 0.5), next is N–F with an EN difference of 1.0 (4.0 – 3.0 = 1.0), the highest is P–F with an EN difference of 1.9 (4.0 – 2.1 = 1.9) .

9.5B

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9-4

a) From the electronegativity figure, EN of Cl = 3.0, EN of F = 4.0, EN of Br = 2.8, and EN of Se = 2.4. Se–F will be more polar than Se–Cl, which will be more polar than Se–Br since the electronegativity difference between Se and F (4.0 – 2.4 = 1.6) is greater than the electronegativity difference between Se and Cl (3.0 – 2.4 = 0.6), which is greater than the electronegativity difference between Se and Br (2.8 – 2.4 = 0.4).

b) EN of Cl = 3.0, EN of B = 2.0, EN of F = 4.0, EN of S = 2.5. The most polar is F–B with an EN difference of 2.0, next is Cl–B with an EN difference of 1.0, and next is S–B with an EN of 0.5.

TOOLS OF THE LABORATORY BOXED READING PROBLEMS B9.1

Plan: Bond strength increases as the number of electrons in the bond increases. Solution: The C=C bond would show absorption of IR at shorter wavelength (higher energy) due to it being a stronger bond than C–C. The CC bond would show absorption at a shorter wavelength (higher energy) than the C=C bond since the triple bond has a larger bond energy than the double bond.

B9.2

Plan: The frequencies of the vibrations are given and are converted to wavelength with the relationship c . The energy of each vibration is calculated by using E = . Solution: c a) c = or =  c 3.00 × 108 m/s 10 9 nm  (nm) = =  = 7462.687 = 7460 nm (symmetric stretch)   4.02 × 1013 /s  1 m  (nm) =

c 3.00 × 108 m/s 10 9 nm  4 =  = 1.50 × 10 nm (bending)   2.00 × 1013 /s  1 m 

(nm) =

c 3.00 × 108 m/s 10 9 nm  =  = 4255.32 = 4260 nm (asymmetrical stretch)   7.05 × 1013 /s  1 m 

b) E = E = (6.626 × 10

–34

E = (6.626 × 10

–34

E = (6.626 × 10

–34

–1

–20

–1

–20

J s)(4.02 × 10 s ) = 2.66365 × 10 J s)(2.00 × 10 s ) = 1.3252 × 10 J s)(7.05 × 10 s ) = 4.6713 × 10

–20

= 2.66 × 10

–20

J (symmetric stretch)

= 1.33 × 10

–20

J (bending)

= 4.67 × 10

–20

J (asymmetrical stretch)

Bending requires the least amount of energy.

END–OF–CHAPTER PROBLEMS 9.1

a) Larger ionization energy decreases metallic character. b) Larger atomic radius increases metallic character. c) Larger number of outer electrons decreases metallic character. d) Larger effective nuclear charge decreases metallic character. McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or

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9-5

9.2 9.3

A has covalent bonding, B has ionic bonding, and C has metallic bonding. The tendency of main-group elements to form cations decreases from Group 1A(1) to 4A(14), and the tendency to form anions increases from Group 4A(14) to 7A(17). Group 1A(1) and 2A(2) elements form mono- and divalent cations, respectively, while Group 6A(16) and 7A(17) elements form di- and monovalent anions, respectively.

9.4

Plan: Metallic behavior increases to the left and down a group in the periodic table. Solution: a) Cs is more metallic since it is further down the alkali metal group than Na. b) Rb is more metallic since it is both to the left and down from Mg. c) As is more metallic since it is further down Group 5A(15) than N.

9.5

a) O

9.6

Plan: Ionic bonding occurs between metals and nonmetals, covalent bonding between nonmetals, and metallic bonds between metals. Solution: a) Bond in CsF is ionic because Cs is a metal and F is a nonmetal. b) Bonding in N2 is covalent because N is a nonmetal. c) Bonding in Na(s) is metallic because this is a monatomic, metal solid.

9.7

a) covalent

9.8

Plan: Ionic bonding occurs between metals and nonmetals, covalent bonding between nonmetals, and metallic bonds between metals. Solution: a) Bonding in O3 would be covalent since O is a nonmetal. b) Bonding in MgCl2 would be ionic since Mg is a metal and Cl is a nonmetal. c) Bonding in BrO2 would be covalent since both Br and O are nonmetals.

9.9

a) metallic

9.10

Plan: Lewis electron-dot symbols show valence electrons as dots. Place one dot at a time on the four sides (this method explains the structure in b) and then pair up dots until all valence electrons are used. The group number of the main-group elements (Groups 1A(1)-8A(18)) gives the number of valence electrons. Rb is Group 1A(1), Si is Group 4A(14), and I is Group 7A(17). Solution:

b) Be

c) Se

b) covalent

c) ionic

b) covalent

c) ionic

9.11 a) Ba

9.12

Plan: Lewis electron-dot symbols show valence electrons as dots. Place one dot at a time on the four sides (this method explains the structure in b) and then pair up dots until all valence electrons are used. The group number of the main main-group elements (Groups 1A(1)-8A(18)) gives the number of valence electrons. Sr is Group 2A(2), P is Group 5A(15), and S is Group 6A(16).

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9-6

Solution:

b) Se

9.13

c) Ga

9.14

Plan: Assuming X is an A-group element, the number of dots (valence electrons) equals the group number. Once the group number is known, the general electron configuration of the element can be written. Solution: a) Since there are 6 dots in the Lewis electron-dot symbol, element X has 6 valence electrons and is a 2 4 Group 6A(16) element. Its general electron configuration is [noble gas]ns np , where n is the energy level. b) Since there are 3 dots in the Lewis electron-dot symbol, element X has 3 valence electrons and is a 2 1 Group 3A(13) element with general electron configuration [noble gas]ns np .

9.15

a) 5A(15); ns np 2 2 b) 4A(14); ns np

9.16

Energy is required to form the cations and anions in ionic compounds but energy is released when the oppositely charged ions come together to form the compound. This energy is the lattice energy and more than compensates for the required energy to form ions from metals and nonmetals.

9.17

a) Because the lattice energy is the result of electrostatic attractions among the oppositely charged ions, its magnitude depends on several factors, including ionic size, ionic charge, and the arrangement of ions in the solid. For a particular arrangement of ions, the lattice energy increases as the charges on the ions increase and as their radii decrease. b) Increasing lattice energy: A < B < C

9.18

The lattice energy releases even more energy when the gas is converted to the solid.

9.19

The lattice energy drives the energetically unfavorable electron transfer resulting in solid formation.

9.20

Plan: Write condensed electron configurations and draw the Lewis electron-dot symbols for the atoms. The group number of the main-group elements (Groups 1A(1)-8A(18)) gives the number of valence electrons. Remove electrons from the metal and add electrons to the nonmetal to attain filled outer levels. The number of electrons lost by the metal must equal the number of electrons gained by the nonmetal. Solution: a) Barium is a metal and loses 2 electrons to achieve a noble gas configuration: 2 2+ – Ba ([Xe]6s )  Ba ([Xe]) + 2e

Chlorine is a nonmetal and gains 1 electron to achieve a noble gas configuration: 2 5 – – 2 6 Cl ([Ne]3s 3p ) + 1e  Cl ([Ne]3s 3p )

Two Cl atoms gain the 2 electrons lost by Ba. The ionic compound formed is BaCl2. Copyright

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9-7

b) Strontium is a metal and loses 2 electrons to achieve a noble gas configuration: 2 2+ – Sr ([Kr]5s )  Sr ([Kr]) + 2e Oxygen is a nonmetal and gains 2 electrons to achieve a noble gas configuration: 2 4 – 2– 2 6 O ([He]2s 2p ) + 2e  O ([He]2s 2p ) One O atom gains the two electrons lost by one Sr atom. The ionic compound formed is SrO.

c) Aluminum is a metal and loses 3 electrons to achieve a noble gas configuration: 2 1 3+ – Al ([Ne]3s 3p )  Al ([Ne]) + 3e Fluorine is a nonmetal and gains 1 electron to achieve a noble gas configuration: 2 5 – – 2 6 F ([He]2s 2p ) + 1e  F ([He]2s 2p ) Three F atoms gains the three electrons lost by one Al atom. The ionic compound formed is AlF3.

d) Rubidium is a metal and loses 1 electron to achieve a noble gas configuration: 1 + – Rb ([Kr]5s )  Rb ([Kr]) + 1e Oxygen is a nonmetal and gains 2 electrons to achieve a noble gas configuration: 2 4 – 2– 2 6 O ([He]2s 2p ) + 2e  O ([He]2s 2p ) One O atom gains the two electrons lost by two Rb atoms. The ionic compound formed is Rb2O.

9.21

a) Cesium loses 1 electron to achieve a noble gas configuration: 1 + – Cs ([Xe]6s )  Cs ([Xe]) + 1e Sulfur gains 2 electrons to achieve a noble gas configuration: 2 4 – 2– 2 6 S ([Ne]3s 3p ) + 2e  S ([Ne]3s 3p ) One S atom gains the two electrons lost by two Cs atoms. The ionic compound formed is Cs2S. 2 + S S 2 Cs 2 Cs + + b) Gallium loses 3 electrons to achieve a noble gas configuration: 10 2 1 3+ 10 – Ga ([Ar]3d 4s 4p )  Ga ([Ar]3d ) + 3e Oxygen gains 2 electrons to achieve a noble gas configuration: 2 4 – 2– 2 6 O ([He]2s 2p ) + 2e  O ([He]2s 2p ) Three O atoms gain the six electrons lost by two Ga atoms. The ionic compound formed is Ga2O3.

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9-8

+ 2

2 Ga

+ 3

c) Magnesium loses 2 electrons to achieve a noble gas configuration: 2 2+ – Mg ([Ne]3s )  Mg ([Ne]) + 2e Nitrogen gains 3 electrons to achieve a noble gas configuration: 2 3 – 3– 2 6 N ([He]2s 2p ) + 3e  N ([He]2s 2p ) Two N atoms gain the six electrons lost by three Mg atoms. The ionic compound formed is 3 Mg

+ 2

3 Mg

+ 2

Mg3N2.

d) Lithium loses 1 electron to achieve a noble gas configuration: 1 + – Li ([He]2s )  Li ([He]) + 1e Bromine gains 1 electron to achieve a noble gas configuration: 10 2 5 – – 10 2 6 Br ([Ar]3d 4s 4p ) + 1e  Br ([Ar]3d 4s 4p ) One Br atoms gains the one electron lost by one Li atom. The ionic compound formed is LiBr.

9.22

Plan: Find the charge of the known atom and use that charge to find the ionic charge of element X. For A-group cations, ion charge = the group number; for anions, ion charge = the group number – 8. Once the ion charge of X is known, the group number can be determined. Solution: – a) X in XF2 is a cation with +2 charge since the anion is F and there are two fluoride ions in the compound. Group 2A(2) metals form +2 ions. 2+ b) X in MgX is an anion with – 2 charge since Mg is the cation. Elements in Group 6A(16) form –2 ions (6 – 8 = –2). c) X in X2SO4 must be a cation with +1 charge since the polyatomic sulfate ion has a charge of –2. X comes from Group 1A(1).

9.23

a) 1A(1)

9.24

Plan: Find the charge of the known atom and use that charge to find the ionic charge of element X. For A-group cations, ion charge = the group number; for anions, ion charge = the group number – 8. Once the ion charge of X is known, the group number can be determined. Solution: a) X in X2O3 is a cation with +3 charge. The oxygen in this compound has a –2 charge. To produce an electrically neutral compound, 2 cations with +3 charge bond with 3 anions with –2 charge: 2(+3) + 3(–2) = 0. Elements in Group 3A(13) form +3 ions. 2– b) The carbonate ion, CO3 , has a –2 charge, so X has a +2 charge. Group 2A(2) elements form +2 ions. + c) X in Na2X has a –2 charge, balanced with the +2 overall charge from the two Na ions. Group 6A(16) elements gain 2 electrons to form –2 ions with a noble gas configuration.

9.25

a) 7A(17)

9.26

Plan: The magnitude of the lattice energy depends on ionic size and ionic charge. For a particular arrangement of ions, the lattice energy increases as the charges on the ions increase and as their radii decrease. Solution: a) BaS would have the higher lattice energy since the charge on each ion (+2 for Ba and –2 for S) is twice the charge on the ions in CsCl (+1 for Cs and –1 for Cl) and lattice energy is greater when ionic charges are larger.

b) 3A(13)

b) 6A(16)

c) 2A(2)

c) 3A(13)

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9-9

b) LiCl would have the higher lattice energy since the ionic radius of Li is smaller than that of Cs and lattice energy is greater when the distance between ions is smaller. 9.27

a) CaO: O has a smaller radius than S. b) SrO: Sr has a smaller radius than Ba.

9.28

Plan: The magnitude of the lattice energy depends on ionic size and ionic charge. For a particular arrangement of ions, the lattice energy increases as the charges on the ions increase and as their radii decrease. Solution: 2+ 2+ a) BaS has the lower lattice energy because the ionic radius of Ba is larger than Ca . A larger ionic radius results in a greater distance between ions. The lattice energy decreases with increasing distance between ions. 2+ b) NaF has the lower lattice energy since the charge on each ion (+1, –1) is half the charge on the Mg and 2– O ions. Lattice energy increases with increasing ion charge.

9.29

a) NaCl: Cl has a larger radius than F. b) K2S: S has a larger radius than O.

9.30

Plan: The lattice energy of NaCl is represented by the equation NaCl(s) (g) + Cl (g). Use Hess’s law and arrange the given equations so that they sum up to give the equation for the lattice energy. You will need to reverse the last equation (and change the sign of H°); you will also need to multiply the second equation (H°) by Solution: H° Na(s) Na(g) 109 kJ 1/2Cl2(g) Cl(g) 243/2 = 121.5 kJ + – Na(g) (g) + e 496 kJ – – Cl(g) + e (g) –349 kJ NaCl(s) Na(s) + 1/2Cl2 (g) 411 kJ (Reaction is reversed; sign of H° changed.) + – NaCl(s) (g) + Cl (g) 788.5 = 788 kJ The lattice energy for NaCl is less than that of LiF, which is expected since lithium and fluoride ions are smaller than sodium and chloride ions, resulting in a larger lattice energy for LiF.

9.31

Lattice energy: MgF2(s) (g) + 2 F (g) Use Hess’s law: H° Mg(s) Mg(g) 148 kJ F2(g) 2F(g) 159 kJ + – Mg(g) Mg (g) + e 738 kJ 2+ – Mg+(g) (g) + e 1450 kJ – – 2F(g) + 2e (g) 2(–328) = – 656 kJ MgF2(s) Mg(s) + F2(g) 1123 kJ (Reaction is reversed and the sign of H° changed.)

–

MgF2(s) (g) + 2F (g) 2962 kJ The lattice energy for MgF2 is greater than that of LiF and NaCl, which is expected since magnesium ions have twice the charge of lithium and sodium ions. Lattice energy increases with increasing ion charge. 9.32

The lattice energy in an ionic solid is directly proportional to the product of the ion charges and inversely proportional to the sum of the ion radii. The strong interactions between ions cause most ionic materials to be hard. A very large lattice energy implies a very hard material. The lattice energy is McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or

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9-10

2–

predicted to be high for Al2O3 since the ions involved, Al and O , have fairly large charges and are relatively small ions. 9.33

Plan: The electron affinity of fluorine is represented by the equation F(g) + e  F (g). Use Hess’s law and + arrange the given equations so that they sum up to give the equation for the lattice energy, KF(s)  K (g) – + F (g). You will need to reverse the last two given equations (and change the sign of H ); you will also need to multiply the third equation (H°) by . Solve for EA. Solution: An analogous Born-Haber cycle has been described in Figure 9.7 for LiF. Use Hess’s law and solve for the EA of fluorine: H° K(s)  K(g) 90 kJ + – K(g)  K (g) + e 419 kJ 1/2F2(g)  F(g) 1/2(159) = 79.5 kJ – – F(g) + e  F (g) ? = EA KF(s)  K(s) + 1/2F2(g) 569 kJ (Reverse the reaction and change the sign of H°.) –

KF(s)  K (g) + F (g) +

–

821 kJ (Reverse the reaction and change the sign of H°.)

821 kJ = 90 kJ + 419 kJ + 79.5 kJ + EA + 569 kJ EA = 821 kJ – (90 + 419 + 79.5 + 569) kJ EA = –336.5 = –336 kJ 9.34

When two chlorine atoms are far apart, there is no interaction between them. Once the two atoms move closer together, the nucleus of each atom attracts the electrons on the other atom. As the atoms move closer this attraction increases, but the repulsion of the two nuclei also increases. When the atoms are very close together the repulsion between nuclei is much stronger than the attraction between nuclei and electrons. The final internuclear distance for the chlorine molecule is the distance at which maximum attraction is achieved in spite of the repulsion. At this distance, the energy of the molecule is at its lowest value.

9.35

The bond energy is the energy required to overcome the attraction between H atoms and Cl atoms in one mole of HCl molecules in the gaseous state. Energy input is needed to break bonds, so bond energy is  always absorbed (endothermic) and H bond breaking is positive. The same amount of energy needed to break   the bond is released upon its formation, so Hbond forming has the same magnitude as H bond breaking , but

opposite in sign (always exothermic and negative). 9.36

The strength of the covalent bond is generally inversely related to the size of the bonded atoms. The bonding orbitals in larger atoms are more diffuse, so they form weaker bonds.

9.37

Bond strength increases with bond order, so CC > C=C > C–C. Two nuclei are more strongly attracted to two shared electron pairs than to one shared electron pair and to three shared electron pairs than to two. The atoms are drawn closer together with more electron pairs in the bond and the bond is stronger.

9.38

When benzene boils, the gas consists of C6H6 molecules. The energy supplied disrupts the intermolecular attractions between molecules but not the intramolecular forces of bonding within the molecule.

9.39

Plan: Bond strength increases as the atomic radii of atoms in the bond decrease; bond strength also increases as bond order increases. Solution:

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9-11

a) I–I < Br–Br < Cl–Cl. Atomic radii decrease up a group in the periodic table, so I is the largest and Cl is the smallest of the three. b) S–Br < S–Cl < S–H. H has the smallest radius and Br has the largest, so the bond strength for S–H is the greatest and that for S–Br is the weakest. c) C–N < C=N < CN. Bond strength increases as the number of electrons in the bond increases. The triple bond is the strongest and the single bond is the weakest. 9.40

a) H–F < H–Cl < H–I b) C=O < C–O < C–S c) N–H < N–O < N–S

9.41

Plan: Bond strength increases as the atomic radii of atoms in the bond decrease; bond strength also increases as bond order increases. Solution: a) The C=O bond (bond order = 2) is stronger than the C–O bond (bond order = 1). b) O is smaller than C so the O–H bond is shorter and stronger than the C–H bond. H2(g) + O2(g)  H–O–O–H(g)

9.42

 H rxn =

 H bonds broken +

 H bonds formed

 = BE H2 + BE OO  [2 (BEOH ) + BE OO ] Use negative values for the bond H rxn

energies of the products. 9.43

Reaction between molecules requires the breaking of existing bonds and the formation of new bonds. Substances with weak bonds are more reactive than are those with strong bonds because less energy is required to break weak bonds.

9.44

Bond energies are average values for a particular bond in a variety of compounds. Heats of formation are specific for a compound.

9.45

Plan: Write the combustion reactions of methane and of formaldehyde. The reactants requiring the smaller amount of energy to break bonds will have the greater heat of reaction. Examine the bonds in the reactant molecules that will be broken. In general, more energy is required to break double bonds than to break single bonds. Solution: For methane: CH4(g) + 2O2(g)  CO2(g) + 2H2O(l), which requires that 4 C–H bonds and 2 O=O bonds be broken and 2 C=O bonds and 4 O–H bonds be formed. For formaldehyde: CH2O(g) + O2(g)  CO2(g) + H2O(l), which requires that 2 C–H bonds, 1 C=O bond, and 1 O=O bond be broken and 2 C=O bonds and 2 O–H bonds be formed. Methane contains more C–H bonds and fewer C=O bonds than formaldehyde. Since C–H bonds take less energy to break than C=O bonds, more energy is released in the combustion of methane than of formaldehyde.

9.46

Ethanol has the greater heat of reaction per mole because 4 C=O bonds and 6 O–H bonds are formed from its combustion. The formation of these bonds releases a lot of energy. Less energy is released from the combustion of methanol, in which only 2 C=O bonds and 4 O–H bonds are formed.

9.47

Plan: To find the heat of reaction, add the energy required to break all the bonds in the reactants to the energy released to form all bonds in the product. Remember to use a negative sign for the energy of the bonds formed since bond formation is exothermic. The bond energy values are found in Table 9.2.

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9-12

Solution: Reactant bonds broken: 1 × C=C = (1 mol)(614 kJ/mol) = 614 kJ 4 × C–H = (4 mol)(413 kJ/mol) = 1652 kJ 1 × Cl–Cl = (1 mol)(243 kJ/mol) = 243 kJ  H bonds broken = 2509 kJ Product bonds formed: 1 × C–C = (1 mol)(–347 kJ/mol) = –347 kJ 4 × C–H = (4 mol)(–413 kJ/mol) = –1652 kJ 2 × C–Cl = (2 mol)(–339 kJ/mol = –678 kJ  H bonds formed = –2677 kJ  H rxn =

 H bonds broken +

 H bonds formed = 2509 kJ + (–2677 kJ) = –168 kJ

9.48

CO2 + 2NH3  (H2N)2CO + H2O



 H rxn =

 H bonds broken +

 H bonds formed

 H rxn = [(2 mol BEC=O + 6 BEN–H] + [4 (BEN–H) + (BEC=O) + 2 (BEC–N) + 2 (BEO–H)]

= [2 mol (799 kJ/mol) + 6 mol (391 kJ/mol)] + [4 mol (–391 kJ/mol) + 1 mol (–745 kJ/mol) + 2 mol (–305 kJ/mol) + 2 mol (–467 kJ/mol)] = 3944 kJ + (–3853 kJ) = 91 kJ 9.49

Reactant bonds broken: 1 × C–O = (1 mol)(358 kJ/mol) = 358 kJ 3 × C–H = (3 mol)(413 kJ/mol) = 1239 kJ 1 × O–H = (1 mol)(467 kJ/mol) = 467 kJ 1 × C  O = (1 mol)(1070 kJ/mol) = 1070 kJ  H bonds broken = 3134 kJ Product bonds formed: 3 × C–H = (3 mol)(–413 kJ/mol) = –1239 kJ

1 × C–C = (1 mol)(–347 kJ/mol) = –347 kJ Copyright

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9-13

1 × C=O = (1 mol)(–745 kJ/mol) = –745 kJ 1 × C–O = (1 mol )(–358 kJ/mol) = –358 kJ 1 × O–H = (1 mol)(–467 kJ/mol) = –467 kJ  H bonds formed = –3156 kJ 

 H rxn =

9.50

C H

 H bonds broken +

 H bonds formed = 3134 kJ + (–3156 kJ) = –22 kJ

Reactant bonds broken: 1 × C=C = (1 mol)(614 kJ/mol) = 614 kJ 4 × C–H = (4 mol)(413 kJ/mol) = 1652 kJ 1 × H–Br = (1 mol)(363 kJ/mol) = 363 kJ  H bonds broken = 2629 kJ

Product bonds formed: 5 × C–H = (5 mol)(–413 kJ/mol) = –2065 kJ 1 × C–C = (1 mol)(–347 kJ/mol) = –347 kJ 1 × C–Br = (1 mol)(–276 kJ/mol) = –276 kJ  H bonds formed = –2688 kJ  H rxn =

 H bonds broken +

 H bonds formed = 2629 kJ + (–2688 kJ) = –59 kJ

9.51

Electronegativity increases from left to right across a period (except for the noble gases) and increases from bottom to top within a group. Fluorine (F) and oxygen (O) are the two most electronegative elements. Cesium (Cs) and francium (Fr) are the two least electronegative elements.

9.52

In general, electronegativity and ionization energies are directly related. Electronegativity relates the strength with which an atom attracts bonding electrons and the ionization energy measures the energy needed to remove an electron. Atoms that do not require much energy to have an electron removed do not have much attraction for bonding electrons.

9.53

Ionic bonds occur between two elements of very different electronegativity, generally a metal with low electronegativity and a nonmetal with high electronegativity. Although electron sharing occurs to a very small extent in some ionic bonds, the primary force in ionic bonds is attraction of opposite charges resulting from electron transfer between the atoms. A nonpolar covalent bond occurs between two atoms with identical electronegativity values where the sharing of bonding electrons is equal. A polar covalent bond is between two atoms (generally nonmetals) of different electronegativities so that the bonding electrons are unequally shared.

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9-14

The H–O bond in water is polar covalent. The bond is between two nonmetals so it is covalent and not ionic, but atoms with different electronegativity values are involved. 9.54

Electronegativity is the tendency of a bonded atom to hold the bonding electrons more strongly. Electron affinity is the energy involved when an atom acquires an electron.

9.55

The difference in EN is a reflection of how strongly one atom in a bond attracts bonding electrons. The greater this difference is, the more likely the bond will be ionic; the smaller the EN difference, the more covalent the bond.

9.56

Plan: Electronegativity increases from left to right across a period (except for the noble gases) and increases from bottom to top within a group. Solution: a) Si < S < O, sulfur is more electronegative than silicon since it is located further to the right in the table. Oxygen is more electronegative than sulfur since it is located nearer the top of the table. b) Mg < As < P, magnesium is the least electronegative because it lies on the left side of the periodic table and phosphorus and arsenic on the right side. Phosphorus is more electronegative than arsenic because it is higher in the table.

9.57

a) I < Br < N b) Ca < H < F

9.58

Plan: Electronegativity increases from left to right across a period (except for the noble gases) and increases from bottom to top within a group. Solution: a) N > P > Si, nitrogen is above P in Group 5(A)15 and P is to the right of Si in Period 3. b) As > Ga > Ca, all three elements are in Period 4, with As the rightmost element.

9.59

a) Cl > Br > P b) F > O > I

9.60

Plan: The polar arrow points toward the more electronegative atom. Electronegativity increases from left to right across a period (except for the noble gases) and increases from bottom to top within a group. Solution: none a) b) c) N B N O C S

O –

9.61 +

a) d) Copyright

–

) and the less electronegative element is

). –

b) F

–

e) As

–

c) H

–

f) S

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9-15

9.62

Plan: The more polar bond will have a greater difference in electronegativity, EN. Solution: a) N: EN = 3.0; B: EN = 2.0; ENa = 3.0 – 2.0 = 1.0 b) N: EN = 3.0; O: EN = 3.5; ENb = 3.5 – 3.0 = 0.5 c) C: EN = 2.5; S: EN = 2.5; ENc = 2.5 – 2.5 = 0 d) S: EN = 2.5; O: EN = 3.5; ENd = 3.5 – 2.5 = 1.0 e) N: EN = 3.0; H: EN = 2.1; ENe = 3.0 – 2.1 =0.9 f) Cl: EN = 3.0; O: EN = 3.5; ENf = 3.5 – 3.0 = 0.5 (a), (d), and (e) have greater bond polarity.

9.63

b) is more polar; EN is 1.0 for F–Cl and 0.2 for Br–Cl c) is more polar; EN is 1.4 for H–O and 0.3 for Se–H f) is more polar; EN is 0.5 for S–N and 0.1 for As–H

9.64

Plan: Ionic bonds occur between two elements of very different electronegativity, generally a metal with low electronegativity and a nonmetal with high electronegativity. Although electron sharing occurs to a very small extent in some ionic bonds, the primary force in ionic bonds is attraction of opposite charges resulting from electron transfer between the atoms. A nonpolar covalent bond occurs between two atoms with identical electronegativity values where the sharing of bonding electrons is equal. A polar covalent bond is between two atoms (generally nonmetals) of different electronegativities so that the bonding electrons are unequally shared. For polar covalent bonds, the larger the EN, the more polar the bond. Solution: a) Bonds in S8 are nonpolar covalent. All the atoms are nonmetals so the substance is covalent and bonds are nonpolar because all the atoms are of the same element and thus have the same electronegativity value. EN = 0. b) Bonds in RbCl are ionic because Rb is a metal and Cl is a nonmetal. EN is large. c) Bonds in PF3 are polar covalent. All the atoms are nonmetals so the substance is covalent. The bonds between P and F are polar because their electronegativity differs (by 1.9 units for P–F). d) Bonds in SCl2 are polar covalent. S and Cl are nonmetals and differ in electronegativity (by 0.5 unit for S–Cl). e) Bonds in F2 are nonpolar covalent. F is a nonmetal. Bonds between two atoms of the same element are nonpolar since EN = 0. f) Bonds in SF2 are polar covalent. S and F are nonmetals that differ in electronegativity (by 1.5 units for S–F). Increasing bond polarity: SCl2 < SF2 < PF3

9.65

a) KCl ionic

b) P4

nonpolar covalent

c) BF3

polar covalent

d) SO2 polar covalent

e) Br2

nonpolar covalent

f) NO2 polar covalent

NO2 < SO2 < BF3

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9-16

9.66

Plan: Increasing ionic character occurs with increasing EN. Electronegativity increases from left to right across a period (except for the noble gases) and increases from bottom to top within a group. The polar arrow points toward the more electronegative atom. Solution: a) H: EN = 2.1; Cl: EN = 3.0; Br: EN = 2.8; I: EN = 2.5 ENHBr = 2.8 – 2.1 = 0.7; ENHCl = 3.0 – 2.1 = 0.9; ENHI = 2.5 – 2.1 = 0.4 b) H: EN = 2.1; O: EN = 3.5; C: EN = 2.5; F: EN = 4.0 ENHO = 3.5 – 2.1 = 1.4; ENCH = 2.5 – 2.1 = 0.4; ENHF = 4.0 – 2.1 = 1.9 c) Cl: EN = 3.0; S: EN = 2.5; P: EN = 2.1; Si: EN = 1.8 ENSCl = 3.0 – 2.5 = 0.5; ENPCl = 3.0 – 2.1 = 0.9;ENSiCl = 3.0 – 1.8 = 1.2

9.67

Increasing ionic character occurs with increasing EN. a) ENPCl = 0.9, ENPBr = 0.7, ENPF = 1.9 P–F > P–Cl > P–Br + – + – b) ENBF = 2.0, ENNF = 1.0, ENCF = 1.5

B–F > C–F > N–F + –

c) ENSeF = 1.6, ENTeF = 1.9, ENBrF = 1.2

+ –

Te–F > Se–F > Br–F + –

+ –

9.68

C–C + Cl–Cl  2 C–Cl 347 kJ/mol 243 kJ/mol d) The value should be greater than the average of the two bond energies given. This is due to the electronegativity difference.

9.69

a) A solid metal is a shiny solid that conducts heat, is malleable, and melts at high temperatures. (Other answers include relatively high boiling point and good conductor of electricity.) b) Metals lose electrons to form positive ions and metals form basic oxides.

9.70

a) Potassium is a larger atom than sodium, so its electrons are held more loosely and thus its metallic bond strength is weaker. b) Be has two valence electrons per atom compared with Li, which has one. The metallic bond strength is stronger for the Be. c) The boiling point is high due to the large amount of energy needed to separate the metal ions from each other in the electron sea.

9.71

When metallic magnesium is deformed, the atoms are displaced and pass over one another while still being tightly held by the attraction of the “sea of electrons.” When ionic MgF2 is deformed, the ions are displaced so that repulsive forces between neighboring ions of like charge cause shattering of the crystals.

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9-17

9.72

Molten rock cools from top to bottom. The most stable compound (the one with the largest lattice energy) will solidify first near the top. The less stable compounds will remain in the molten state at the bottom and eventually crystallize there later.

9.73

Plan: Write a balanced chemical reaction. The given heat of reaction is the sum of the energy required to break all the bonds in the reactants and the energy released to form all bonds in the product. Remember to use a negative sign for the energy of the bonds formed since bond formation is exothermic. The bond energy values are found in Table 9.2. Use the ratios from the balanced reaction between the heat of reaction and acetylene and between acetylene and CO2 and O2 to find the amounts needed. The ideal gas law is used to convert from moles of oxygen to volume of oxygen. Solution:  a) C2H2 + 5/2O2  2CO2 + H2O H rxn = –1259 kJ/mol H–CC–H + 5/2O=O  2O=C=O + H–O–H  H rxn =

 H bonds broken +

 H bonds formed

 H rxn = [2 BEC–H + BECC +5/2 BEO=O] + [4 (–BEC=O) + 2 (–BEO–H)]

–1259 kJ = [2(413) + BECC + 5/2(498)] + [4(–799) + 2(–467)] –1259 kJ = [826 + BECC + 1245] + [–4130)] kJ –1259 kJ = –2059 + BECC kJ BECC = 800 kJ/mol

Table 9.2 lists the value as 839 kJ/mol.

 1 mol C 2 H 2  1259 kJ     b) Heat (kJ) = 500.0 g C 2 H 2  1 mol C H   26.04 g C 2 H 2  2 2





= –2.4174347 × 10 = –2.417 × 10 kJ  1 mol C 2 H 2   2 mol CO 2   44.01 g CO 2   c) Mass (g) of CO2 = 500.0 g C 2 H 2     1 mol CO   26.04 g C 2 H 2  1 mol C 2 H 2  2 





= 1690.092 = 1690 g CO2  1 mol C 2 H 2   (5/2) mol O 2   d) Amount (mol) of O2 = 500.0 g C 2 H 2     26.04 g C 2 H 2   1 mol C 2 H 2 





= 48.0030722 mol O2 PV = nRT

L atm  48.0030722 mol O 0.08206 mol  (298 K) K 

nRT Volume (L) of O2 = = P



18.0 atm

= 65.2145 = 65.2 L O2

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9-18

9.74

a) Br

9.75

Al3+

Plan: The heat of formation of MgCl is represented by the equation Mg(s) + 1/2Cl2(g) s). Use Hess’s law and arrange the given equations so that they sum up to give the equation for the heat of formation of MgCl. You will need to multiply the second equation by ; you will need to reverse the + – equation for the lattice energy [MgCl(s) (g) + Cl (g)] and change the sign of the given lattice energy value. Negative heats of formation are energetically favored. Solution: a) 1) Mg(s)  Mg(g) H1 = 148 kJ

H 2 = 1/2(243 kJ) = 121.5 kJ

2) 1/2Cl2(g)  Cl(g) 3) Mg(g)  Mg (g) + e +

–

H3 = 738 kJ

4) Cl(g) + e  Cl (g)

H 4 = –349 kJ

5) Mg (g) + Cl (g)  MgCl(s)

 H 5 = –783.5 kJ (= – H lattice (MgCl))

–

Mg(s) + 1/2Cl2(g)  MgCl(s) H f (MgCl) = ?    H f (MgCl) = H1 + H 2 + H3 + H 4 + H 5 = 148 kJ + 121.5 kJ + 738 kJ + (–349 kJ) + (–783.5 kJ) = –125 kJ  f

b) Yes, since H for MgCl is negative, MgCl(s) is stable relative to its elements. c) 2MgCl(s)  MgCl2(s) + Mg(s)   H rxn = H rxn = m Hf(products) – n Hf(reactants)  H rxn = {1 Hfo [MgCl2(s)] + 1 Hfo [Mg(s)]} – {2 Hfo [MgCl(s)]}  H rxn = [1 mol (–641.6 kJ/mol) + 1 mol (0)] – [2 mol (–125 kJ/mol)]  H rxn = –391.6 = –392 kJ  Hfo for MgCl2 is much more negative than that for MgCl. This makes the H rxn value for the

d) No,

above reaction very negative, and the formation of MgCl2 would be favored. 9.76

 a) H rxn =

 H bonds broken +

 H bonds formed

= [1 mol (BEH-H) + 1 mol (BECl-Cl)] + [2 mol (BEH-Cl)] = [1 mol (432 kJ/mol) + 1 mol (243 kJ/mol)] + [2 mol (–427 kJ/mol)] = –179 kJ b) H

 rxn

 H bonds broken +

 H bonds formed

= [1 mol (BEH-H) + 1 mol (BEI-I)] + [2 mol (BEH-I)] = [1 mol (432 kJ/mol) + 1 mol (151 kJ/mol)] + [2 mol (–295 kJ/mol)] = –7 kJ Copyright

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9-19

 c) H rxn =

 H bonds broken +

 H bonds formed

= [2 mol (BEH-H) + 1 mol (BEO=O)] + [4 mol (BEH-O)] = [2 mol (432 kJ/mol) + 1 mol (498 kJ/mol)] + [4 mol (–467 kJ/mol)] = –506 kJ Reactions (a) and (c) are strongly exothermic and are a potential explosive hazard. Reaction (c) should occur most explosively. 9.77

Plan: Find the bond energy for an H–I bond from Table 9.2. For part (a), calculate the wavelength with this energy using the relationship from Chapter 7: E = hc/ . For part (b), calculate the energy for a wavelength of 254 nm and then subtract the energy from part (a) to get the excess energy. 2 For part (c), speed can be calculated from the excess energy since Ek = 1/2mu . Solution: a) Bond energy for H–I is 295 kJ/mol (Table 9.2).   295 kJ 103 J  1 mol  = 4.898705 × 10–19 J/photon  Bond energy (J/photon) =   23     mol  1 kJ  6.022 10 photons  E = hc/

6.626 10 (m) = hc/E =

34



 = 4.057807 × 10 m

J  s 3.00 108 m/s

4.898705 10

19

J

–7

 1 nm  (nm) = 4.057807 107 m 9  = 405.7807 = 406 nm 10 m  –19

b) E (HI) = 4.898705 × 10 J

6.626 10

E (254 nm) = hc/ =

Excess energy = 7.82598 × 10

–19





J  s 3.00 108 m /s  1 nm  –19   = 7.82598 × 10 J 109 m  254 nm  

34

J – 4.898705 × 10

–19

J = 2.92728 × 10

–19

= 2.93 × 10

–19

1.008 g H   1 kg  mol  –27  c) Mass (kg) of H =   3  = 1.67386 × 10 kg  mol  6.022 10 23   10 g  2

Ek = 1/2mu thus, u = u=

9.78

2E m

J)  kg  m 2 /s 2  4 4   = 1.8701965 × 10 = 1.87 × 10 m/s 27   J 1.67386 10 kg  

2(2.92728 10

19

“Excess bond energy” refers to the difference between the actual bond energy for an X–Y bond and the average of the energies for the X–X and the Y–Y bonds. Excess bond energy = BEX–Y – 1/2 (BEX–X + BEY–Y). The excess bond energy is zero when the atoms X and Y are identical or have the same electronegativity, as in (a), (b), and (e). ENPH = 0, ENCS = 0,ENBrCl = 0.2, ENBH = 0.1,ENSeSe = 0

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9-20

9.79

Rb ([Kr]5s ) has one valence electron, so the metallic bonding would be fairly weak, resulting in 2 10 a soft, low-melting material. Cd ([Kr]5s 4d ) has two valence electrons so the metallic bonding is stronger. 2 3 V ([Ar]4s 3d ) has five valence electrons, so its metallic bonding is the strongest, that is, its hardness, melting point, and other metallic properties would be greatest.

9.80

Plan: Find the appropriate bond energies in Table 9.2. Calculate the wavelengths using E = hc/ . Solution: C–Cl bond energy = 339 kJ/mol   339 kJ 103 J  1 mol   = 5.62936 × 10–19 J/photon Bond energy (J/photon) =      23    mol  1 kJ  6.022 10 photons  E = hc/ (m) = hc/E =



6.626 10

34



 = 3.5311296 × 10 = 3.53 × 10 m

J  s 3.00 108 m/s

5.62936 10

19

J

–7

O2 bond energy = 498 kJ/mol

  498 kJ 103 J  1 mol   = 8.269678 × 10–19 J/photon Bond energy (J/photon) =      23    mol  1 kJ  6.022 10 photons  E = hc/ (m) = hc/E = 9.81

6.626 10

34



 = 2.40372 × 10 = 2.40 × 10 m

J  s 3.00 108 m/s

8.269678 10

19

J

–7

Plan: Write balanced chemical equations for the formation of each of the compounds. Obtain the bond energy of fluorine from Table 9.2 (159 kJ/mol). Determine the average bond energy from H = bonds broken + bonds formed. Remember that the bonds formed (Xe–F) have negative values since bond formation is exothermic. Solution:    H rxn = H bonds H bonds broken + formed XeF2

Xe(g) + F2(g)  XeF2(g)  H rxn = –105 kJ/mol = [1 mol F2 (159 kJ/mol)] + [2 (–Xe–F)]

–264 kJ/mol = 2 (–Xe–F) Xe–F = 132 kJ/mol XeF4

Xe(g) + 2F2(g)  XeF4(g)  H rxn = –284 kJ/mol = [2 mol F2 (159 kJ/mol)] + [4 (–Xe–F)]

–602 kJ/mol = 4 (–Xe–F) Xe–F = 150.5 = 150 kJ/mol XeF6

Xe(g) + 3F2(g)  XeF6(g)  H rxn = –402 kJ/mol = [3 mol F2 (159 kJ/mol)] + [6 (–Xe–F)]

–879 kJ/mol = 6 (–Xe–F) Xe–F = 146.5 = 146 kJ/mol 9.82

The difference in electronegativity produces a greater than expected overlap of orbitals, which shortens the bond. As EN becomes smaller (i.e., as you proceed from HF to HI), this effect lessens and the bond lengths become more predictable. McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or

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9-21

9.83

a)The presence of the very electronegative fluorine atoms bonded to one of the carbon atoms in H3C—CF3 makes the C–C bond polar. This polar bond will tend to undergo heterolytic rather than homolytic cleavage. More energy is required to force heterolytic cleavage. b) Since one atom gets both of the bonding electrons in heterolytic bond breakage, this results in the formation of ions. In heterolytic cleavage a cation is formed, involving ionization energy; an anion is also formed, involving electron affinity. The bond energy of the O2 bond is 498 kJ/mol. H = (homolytic cleavage + electron affinity + first ionization energy) H = (498/2 kJ/mol + (–141 kJ/mol) + 1314 kJ/mol) = 1422 = 1420 kJ/mol It would require 1420 kJ to heterolytically cleave 1 mol of O2.

9.84

The bond energies are needed from Table 9.2. N2 = 945 kJ/mol; O2 = 498 kJ/mol; F2 = 159 kJ/mol N 2:



= hc/E =

O 2: 



6.626 10

34



 10 J   mol 945 kJ    23      mol  1 kJ  6.022 10  3

6.626 10 = hc/E =

34

 = 1.26672 × 10 = 1.27 × 10 m

J  s 3.00 108 m/s



J  s 3.00 108 m /s

3  10 J   mol 498 kJ    23      mol  1 kJ  6.022 10 

F2:

6.626 10

34



J  s 3.00 10 m /s



–7

= 2.40372 × 10 = 2.40 × 10 m

 = 7.528636 × 10 = 7.53 × 10 m



9.85

a) To compare the two energies, the ionization energy must be converted to the energy to remove an

= hc/E =

3   10 J   mol  159 kJ   23  mol  1 kJ   6.022 10 

–7

electron from an atom. The energy needed to remove an electron from a single gaseous Ag atom (J) =  731 kJ 10 3 J   mol  –18 –18 –19    = 1.21388 × 10 = 1.21 × 10 J > 7.59 × 10 J    23    6.022  10  mol  1 kJ   It requires less energy to remove an electron from the surface of solid silver. b) The electrons in solid silver are held less tightly than the electrons in gaseous silver because the electrons in metals are delocalized, meaning they are shared among all the metal nuclei. The delocalized attraction of many nuclei to an electron (solid silver) is weaker than the localized attraction of one nucleus to an electron (gaseous silver). 9.86

Plan: The heat of formation of SiO2 is represented by the equation Si(s) + O2(g)

(s). Use Hess’s law

and arrange the given equations so that they sum up to give the equation for the heat of formation. The lattice energy of SiO2 is represented by the equation SiO2(s)

2–

(g) + 2O (g). You will need to reverse the lattice energy equation (and change the sign of

H°); you will also need to multiply the fourth given equation by 2. Solution: Use Hess’ law. Hf of SiO2 is found in Appendix B. Si(s)  Si(g)

1) Copyright

H1 = 454 kJ

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9-22

H 2 = 9949 kJ

Si(g)  Si (g) + 4 e

O2(g)  2O(g)

2O(g) + 4e  2O (g)

H 4 = 2(737) kJ

Si (g) + 2O (g)  SiO2(s)

 H 5 = – H lattice (SiO2) = ?

–

H3 = 498 kJ 2–

2–

Si(s) + O2(g)  SiO2(s) Hf (SiO2) = –910.9 kJ      H = [ H1 + H 2 + H3 + H 4 + (– H lattice )]  f

 –910.9 kJ = [454 kJ + 9949 kJ + 498 kJ + 2(737) kJ + (– H lattice )]  – H lattice = –13,285.9 kJ  H lattice = 13,286 kJ

9.87

 H rxn =

 H bonds broken +

 H bonds formed

 For ethane: H rxn = [1 mol (BEC – C) + 6 mol (BEC – H) + 1 mol (BEH – H)] + [8 mol (BEC – H)]

–65.07 kJ = [1 mol (347 kJ/mol) + 6 mol (BEC – H) + 1 mol (432 kJ/mol)] + [8 mol (–415 kJ/mol)] BEC – H =

65.07  347  432  3320  kJ = 412.655 = 413 kJ/mol 6 mol

 For ethene: H rxn = [1 mol (BEC =C) + 4 mol (BEC – H) + 2 mol (BEH – H)] + [8 mol (BEC – H)]

–202.21 kJ = [1 mol (614 kJ/mol) + 4 mol (BEC – H) + 2 mol (432 kJ/mol]) + [8 mol (–415 kJ/mol)] 202.21 614  864  3320 kJ BEC – H = = 409.9475 = 410. kJ/mol 4 mol  For ethyne: H rxn = [1 mol (BECC) + 2 mol (BEC – H) + 3 mol (BEH – H)] + [8 mol (BEC – H)]

–376.74 kJ = [1 mol (839 kJ/mol) + 2 mol (BEC – H) + 3 mol (432 kJ/mol)] + [8 mol (– 415kJ/mol)] 376.74  839 1296  3320 kJ BEC – H = = 404.13 = 404 kJ/mol 2 mol 9.88

Plan: Convert the bond energy in kJ/mol to units of J/photon. Use the equations E = h, and E = hc/ to find the frequency and wavelength of light associated with this energy. Solution:   347 kJ 103 J  1 mol   = 5.762205 × 10–19 J/photon Bond energy (J/photon) =      23  mol  1 kJ  6.022 10 photons 

E h E 5.762205 1019 J 14 14 –1 = = = 8.6963553 × 10 = 8.70 × 10 s h 6.6261034 J s E = h or  =

E = hc/ or = hc/E

6.626 10 (m) = hc/E =

34



 = 3.44972 × 10 = 3.45 × 10 m

J  s 3.00108 m /s 19

–7

5.762205 10 J This is in the ultraviolet region of the electromagnetic spectrum. Copyright

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9-23

9.89

3   10 J   mol   467 kJ     1 kJ  6.02210 23   mol     E 15 15 –1 = = = 1.170374 × 10 = 1.17 × 10 s h 6.6261034 J  s

6.626 10 (m) = hc/E =

34



 = 2.56328 × 10 = 2.56 × 10 m

J  s 3.00 108 m/s

–7

 10 J   mol 467 kJ    23  mol  1 kJ  6.022 10    1 mol  kJ   = 7.7548987 × 10–22 = 7.75 × 10–22 kJ/photon Ephoton = 467  23   mol  6.022 10 photons  9.90

Plan: Write the balanced equations for the reactions. Determine the heat of reaction from H = bonds broken + bonds formed. Remember that the bonds formed have negative values since bond formation is exothermic. Solution: a) 2CH4(g) + O2(g)  CH3OCH3(g) + H2O(g)  H rxn =

 H bonds broken +

 H bonds formed

 H rxn = [8 × (BEC–H) + 1 × (BEO=O)] + [6 × (BEC–H) + 2 × (BEC–O) + 2 × (BEO–H)]  H rxn = [8 mol (413 kJ/mol) + 1 mol (498 kJ/mol)]

+ [6 mol (–413 kJ/mol) + 2 mol (–358 kJ/mol) + 2 mol (–467 kJ/mol)]

H

 rxn

= –326 kJ

2CH4(g) + O2(g)  CH3CH2OH(g) + H2O(g)  H rxn =

 H bonds broken +

 H bonds formed

 H rxn = [8 × (BEC–H) + 1 × (BEO=O)] + [5 × (BEC–H) + 1 × (BEC–C) + 1 × (BEC–O) + 3 × (BEO–H)]  H rxn = [8 mol (413 kJ/mol) + 1 mol (498 kJ/mol)]

+ [5 mol (–413 kJ/mol) + 1 mol (–347 kJ/mol) + 1 mol (–358 kJ/mol) + 3 mol (–467 kJ/mol)]

H

 rxn

= –369 kJ

b) The formation of gaseous ethanol is more exothermic. c) The conversion reaction is CH3CH2OH(g)  CH3OCH3(g). Use Hess’s law: CH3CH2OH(g) + H2O(g)  2CH4(g) + O2(g)

 H rxn = –(–369 kJ) = 369 kJ

2CH4(g) + O2(g)  CH3OCH3(g) + H2O(g)

 H rxn = –326 kJ  H rxn = –326 kJ + 369 kJ = 43 kJ

CH3CH2OH(g)  CH3OCH3(g) 9.91

a) CH2=CH2(g) + H2O(g)  CH3CH2OH(g)  Using bond energies: H rxn =

 H bonds broken +

 H bonds formed

 H rxn = [4 × (BEC–H) + 1 × (BEC=C) + 2 × (BEO–H)]

H

 rxn

+ [5 × (BEC–H) + 1 × (BEC–C) + 1 × (BEC–O) + 1 × (BEO–H)] = [4 mol (413 kJ/mol) + 1 mol (614 kJ/mol) + 2 mol (467 kJ/mol)]

+ [5 mol (–413 kJ/mol) + 1 mol (–347 kJ/mol) + 1 mol (–358 kJ/mol) + 1 mol (–467 kJ/mol)]  H rxn = –37 kJ

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9-24

 Using heats of formation: H rxn = m Hf(products) – n Hf(reactants)  H rxn = [1 mol ( H f of CH3CH2OH(g))]

– [1 mol ( H f of CH2=CH2(g)) + 1 mol ( H f of H2O (g))]  H rxn = [–235.1 kJ] – [52.47 kJ + –241.826 kJ]  H rxn = –45.744 = –45.7 kJ

b) C2H4O(l) + H2O(l)  C2H6O2(l)  H rxn =

 H bonds broken +

 H bonds formed

 H rxn = [4 × (BEC–H) + 1 × (BEC-C) + 2 × (BEC–O) + 2 × (BEO–H)]

+ [4 × (BEC–H) + 1 × (BEC–C) + 2 × (BEC–O) + 2 × (BEO–H)]

H

 rxn

= [4 mol (413 kJ/mol) + 1 mol (347 kJ/mol) + 2 mol (358 kJ/mol) + 2 mol (467 kJ/mol)]

+ [4 mol (–413 kJ/mol) + 1 mol (–347 kJ/mol) + 2 mol (–358 kJ/mol) + 2 mol (–467 kJ/mol)]  H rxn = 0 kJ  c) In the hydrolysis in part (b), the H rxn appears to be 0 kJ using bond energies since the number

and types of bonds broken and the number and types of bonds formed are the same. Since the bond energy values used are average values, this method does not differentiate between an O–H bond in water, for example, and an O–H bond in ethylene glycol.

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9-25

CHAPTER 10 THE SHAPES OF MOLECULES FOLLOW–UP PROBLEMS 10.1A

Plan: Count the valence electrons and follow the steps outlined in the sample problem to draw the Lewis structures. Solution: a) The sulfur is the central atom, as the hydrogen is never central. Each of the hydrogen atoms is placed around the sulfur. The actual positions of the hydrogen atoms are not important. The total number of valence electrons available is [2 × H(1e–)] + [1 × S(6e–)] = 8e–. Connect each hydrogen atom to the sulfur with a single bond. These bonds use 4 of the electrons leaving 4 electrons. The last 4e– go to the sulfur because the hydrogen atoms can take no more electrons.

H Solution: b) The aluminum has the lower group number so it is the central atom. Each of the chlorine atoms will be attached to the central aluminum. The total number of valence electrons available is [4 × F(7e–)] + [1 × Al(3e–)] + 1e– (for the negative charge) = 32e–. Connecting the four chlorine atoms to the aluminum with single bonds uses 4 × 2 = 8e–, leaving 32 – 8 = 24e–. The more electronegative chlorine atoms each need 6 electrons to complete their octets. This requires 4 × 6 = 24e–. There are no more remaining electrons at this step; however, the aluminum has 8 electrons around it.

Check: Count the electrons. Each of the five atoms has an octet. Solution: c) Both S and O have a lower group number than Cl, thus, one of these two elements must be central. Between S and O, S has the higher period number so it is the central atom. The total number of valence electrons available is – – – – [2 × Cl(7e )] + [1 × S(6e )] + [1 × O(6e )] = 26e . Begin by distributing the two chlorine atoms and the oxygen atom around the central sulfur atom. Connect each of the three outlying atoms to the central sulfur with single bonds. This uses 3 × 2 = 6e–, leaving 26 – 6 = 20e–. Each of the outlying atoms still needs 6 electrons to complete their octets. Completing these octets uses 3 × 6 = 18 electrons. The remaining 2 electrons are all the sulfur needs to complete its octet. Cl S Cl

Check: Count the electrons. Each of the four atoms has an octet.

10-1

10.1B

Plan: Count the valence electrons and follow the steps outlined in the sample problem to draw the Lewis structures. Solution: a) The oxygen has the lower group number so it is the central atom. Each of the fluorine atoms will be attached to the central oxygen. The total number of valence electrons available is [2 × F(7e–)] + [1 × O(6e–)] = 20e–. Connecting the two fluorine atoms to the oxygen with single bonds uses 2 × 2 = 4e–, leaving 20 – 4 = 16e–. The more electronegative fluorine atoms each need 6 electrons to complete their octets. This requires 2 × 6 = 12e–. The 4 remaining electrons go to the oxygen.

O F

Check: Count the electrons. Each of the three atoms has an octet. Solution: b) The carbon has the lower group number so it is the central atom (technically, hydrogen is in group 1A, but it can only form one bond and, thus, cannot be the central atom). Each of the hydrogen and bromine atoms will be attached to the central carbon. The total number of valence electrons available is [1 × C(4e–)] + [2 × H(1e–)] + [2 × Br(7e–)] = 20e–. Connecting the two hydrogen atoms and the two bromine atoms to the carbon with single bonds uses 4 × 2 = 8e–, leaving 20 – 8 = 12e–. The “octets” of the two hydrogen atoms (2 electrons) are filled through their bonds to the carbon. The more electronegative bromine atoms each need 6 electrons to complete their octets. This requires 2 × 6 = 12e–. H

Br Check: Count the electrons. Each of the five atoms has an octet. Solution: c) Both I and Br have the same group number. Between the two atoms, I has the higher period number so it is the central atom. The total number of valence electrons available is [1 × I(7e–)] + [2 × Br(7e–)] – 1e– (for the positive charge) = 20e–. Connecting the two bromine atoms to the iodine with single bonds uses 2 × 2 = 4e–, leaving 20 – 4 = 16e–. Each of the outlying atoms still needs 6 electrons to complete their octets. Completing these octets uses 2 × 6 = 12 electrons. The remaining 4 electrons are all the iodine needs to complete its octet.

I Br

Check: Count the electrons. Each of the three atoms has an octet. 10.2A

Plan: Count the valence electrons and follow the steps outlined in the sample problem to draw the Lewis structures. Solution: a) As in CH4O, the N and O both serve as “central” atoms. The N is placed next to the O and the H atoms are distributed around them. The N needs more electrons so it gets two of the three hydrogen atoms. You can try placing the N in the center with all the other atoms around it, but you will quickly see that you will have trouble with the oxygen. The number of valence electrons is [3 × H(1e–)] + [1 × N(5e–)] + [1 × O(6e–)] = 14e–. Four

10-2

single bonds are needed (4 × 2 = 8e–). This leaves 6 electrons. The oxygen needs 4 electrons, and the nitrogen needs 2. These last 6 electrons serve as three lone pairs. H H

H incorrect

correct Solution: b) The hydrogen atoms cannot be the central atoms. The problem states that there are no O–H bonds, so the oxygen must be connected to the carbon atoms. Place the O atom between the two C atoms, and then distribute the H atoms equally around each of the C atoms. The total number of valence electrons is [6 × H(1e–)] + [2 × C(4e–)] + [1 × O(6e–)] = 20e– Draw single bonds between each of the atoms. This creates six C–H bonds, and two C–O bonds, and uses 8 × 2 = 16 electrons. The four remaining electrons will become two lone pairs on the O atom to complete its octet. H H

H H Check: Count the electrons. Both the C’s and the O have octets. Each H has its pair. 10.2B

Plan: Count the valence electrons and follow the steps outlined in the sample problem to draw the Lewis structures. Solution: a) The hydrogen atoms cannot be the central atoms, so the two N atoms both serve as “central” atoms, bonded to each other, and the H atoms are distributed around them (2 hydrogens per N atom). The number of valence electrons is [4 × H(1e–)] + [2 × N(5e–)] = 14e–. Connect the two nitrogens to each other via single bonds. Connect two hydrogens to each of the two nitrogens via single bonds. These five single bonds require 5 × 2 = 10e–. This leaves 4 electrons. Each of the nitrogen atoms needs 2 electrons to complete its octet. These last 4 electrons serve as two lone pairs. H

H H Solution: b) The hydrogen atoms cannot be the central atoms. The C and the N atoms both serve as “central” atoms, bonded to each other. According to the formula, three hydrogen atoms are bonded to the C and two hydrogen atoms are bonded to the N. The number of valence electrons is [5 × H(1e–)] + [1 × N(5e–)] + [1 × C(4e–)] = 14e–. Connect the C and N atoms via a single bond. Connect the appropriate number of hydrogen atoms to the C and N. These six bonds require 6 × 2 = 12e–. This leaves 2 electrons. The nitrogen atom needs 2 electrons to complete its octet. These last 2 electrons serve as a lone pair on the nitrogen. H

H H Check: Count the electrons. Both the C and the N have octets. Each H has its pair.

10-3

10.3A

Plan: Count the valence electrons and follow the steps outlined in the sample problem to draw the Lewis structures. Solution: a) In CO there are a total of [1 × C(4e–)] + [1 × O(6e–)] = 10e–. The hint states that carbon has three bonds. Since oxygen is the only other atom present, these bonds must be between the C and the O. This uses 6 of the 10 electrons. The remaining 4 electrons become two lone pairs, one pair for each of the atoms. C

Check: Count the electrons. Both the C and the O have octets. Solution: b) In HCN there are a total of [1 × H(1e–)] + [1 × C(4e–)] + [1 × N(5e–)] = 10 electrons. Carbon has a lower group number, so it is the central atom. Place the C between the other two atoms and connect each of the atoms to the central C with a single bond. This uses 4 of the 10 electrons, leaving 6 electrons. Distribute these 6 to nitrogen to complete its octet. However, the carbon atom is 4 electrons short of an octet. Change two lone pairs on the nitrogen atom to bonding pairs to form two more bonds between carbon and nitrogen for a total of three bonds. H

Check: Count the electrons. Both the C and the N have octets. The H has its pair. Solution: c) In CO2 there are a total of [1 × C(4e–)] + [2 × O(6e–)] = 16 electrons. Carbon has a lower group number, so it is the central atom. Placing the C between the two O atoms and adding single bonds uses 4 electrons, leaving 16 – 4 = 12e–. Distributing these 12 electrons to the oxygen atoms completes those octets, but the carbon atom does not have an octet. Change one lone pair on each oxygen atom to a bonding pair to form two double bonds to the carbon atom, completing its octet.

Check: Count the electrons. Both the C and O atoms have octets. 10.3B

Plan: Count the valence electrons and follow the steps outlined in the sample problem to draw the Lewis structures. Solution: a) In NO+ there are a total of [1 × N(5e–)] + [1 × O(6e–)] – 1 e– (for the positive charge on the ion) = 10e–. Connecting the N and O atoms via a single bond uses 1 × 2 = 2e–, leaving 10 – 2 = 8e–. The more electronegative O atom needs 6 more electrons to complete its octet. The remaining 2 electrons become a lone pair on the N atom. However, this only gives the N atom 4 electrons. Change two lone pairs on the oxygen atom to bonding pairs to form a triple bond between the N and O atoms. In this way, the atoms’ octets are complete. N

Check: Count the electrons. Both the N and the O have octets. Solution: b) In H2CO there are a total of [1 × C(4e–)] + [1 × O(6e–)] + [2 × H(1e–)] = 12 electrons. Carbon has a lower group number, so it is the central atom (H has a lower group number, but it cannot be a central atom). Place the C between the other three atoms and connect each of the atoms to the central C with a single bond. This uses 6 of the 12 electrons, leaving 6 electrons. Distribute these 6 to oxygen to complete its octet. However, the carbon atom is 2 electrons short of an octet. Change one lone pair on the oxygen atom to a bonding pair between the oxygen and carbon. H

Check: Count the electrons. Both the C and the O have octets. The H atoms have 2 electrons each. Solution: c) In N2H2there are a total of [2 × N(5e–)] + [2 × H(1e–)] = 12 electrons. Hydrogen cannot be a central atom, so the nitrogen atoms are the central atoms (there is more than one central atom). Connecting the nitrogen atoms via Copyright McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

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a single bond and attaching one hydrogen to each nitrogen atom uses 6 electrons, leaving six electrons. Each nitrogen needs four more electrons to complete its octet; however, there are not enough electrons to complete both octets. Four electrons can be added to one of the nitrogen atoms, but this only leaves two electrons to add to the other nitrogen atom. The octet of the first nitrogen atom is complete, but the octet of the second nitrogen atom is not complete. Change one lone pair on the first nitrogen to a bonding pair to complete both atoms’ octets. H

Check: Count the electrons. Both N atoms have octets, and each H atom has 2 electrons. 10.4A

Plan: Count the valence electrons and follow the steps to draw the Lewis structures. Each new resonance structure is obtained by shifting the position of a multiple bond and the electron pairs. Solution: H3CNO2 has a total of [1 × C(4e–)] + [2 × O(6e–)] + [3 × H(1e–)] + [1 × N(5e–)] = 24 electrons. According to the problem, the H atoms are bonded to C, and the C atom is bonded to N, which is bonded to both O atoms. Connect all of these atoms via single bonds. This uses 6 × 2 = 12e–. There are 12 electrons remaining. Each of the oxygen atoms needs 6 more electrons to complete its octet. Placing 6 electrons on each of the oxygen atoms leaves the nitrogen two electrons short of an octet. Change one of the lone pairs on one of the oxygen atoms to a bonding pair to complete the octets. The other resonance structure is obtained by taking a lone pair from the other oxygen to create the double bond. H H O O

H O

C H

N O

1 no. of shared valence e-) 2 FCO (single-bonded) = 6 – [6 + 1/2(2)] = –1 FCO (double-bonded) = 6 – [4 + 1/2(4)] = 0

Formal charge of atom = no. of valence e- - (no. of unshared valence e- + FCN = 5 – [0 + 1/2(8)] = +1 FCC = 4 – [0 + 1/2(8)] = 0 FCH = 1 – [0 + 1/2(2)] = 0 10.4B

Plan: Count the valence electrons and follow the steps to draw the Lewis structures. Each new resonance structure is obtained by shifting the position of a multiple bond and the electron pairs. Solution: SCN– has a total of [1 × C(4e–)] + [1 × S(6e–)] + [1 × N(5e–)] + 1 e– (for the negative charge) = 16 electrons. According to the problem, the C is the central atom. Connect it via single bonds to both the S and the N atoms. This uses 2 × 2 = 4e–. There are 12 electrons remaining. The outlying sulfur and nitrogen each require 6 electrons to complete their octets. Placing 6 electrons each on the N and on the S leaves the C atom 4 electrons short of an octet. Convert two lone pairs on the outer atoms to bonding pairs to complete the octet. The lone pairs can come from the N, from the S, or from both the S and the N. Each of the generated structures is a resonance structure.

Structure I

Structure II

Structure III

1 no. of shared valence e-) 2 Structure I Structure II Structure III FCS = 6 – [4 + 1/2(4)] = 0 FCS = 6 – [6 + 1/2(2)] = -1 FCS = 6 – [2 + 1/2(6)] = +1 FCC = 4 – [0 + 1/2(8)] = 0 FCC = 4 – [0 + 1/2(8)] = 0 FCC = 4 – [0 + 1/2(8)] = 0 FCN = 5 – [4 + 1/2(4)] = –1 FCN = 5 – [2 + 1/2(6)] = 0 FCN = 5 – [6 + 1/2(2)] = –2 Structure I is the most important structure; the formal charges are lower than those in Structure III, and the negative formal charge is on the more electronegative N atom, while the negative formal charge in Structure II is on the less electronegative S atom.

Formal charge of atom = no. of valence e- - (no. of unshared valence e- +

10-5

10.5A

Plan: The presence of available d orbitals makes checking formal charges more important. Use the equation for formal charge: FC = # of valence e– – [# unshared valence e– + (# shared valence e–)] Solution: a) In POCl3, the P is the most likely central atom because all the other elements have higher group numbers. The molecule contains: [1 × P(5e–)] + [1 × O(6e–)] + [3 × Cl(7e–)] = 32 electrons. Placing the P in the center with single bonds to all the surrounding atoms uses 8 electrons and gives P an octet. The remaining 24 can be split into 12 pairs with each of the surrounding atoms receiving three pairs. At this point, in structure I below, all the atoms have an octet. The central atom is P (smallest group number, highest period number) and can have more than an octet. To see how reasonable this structure is, calculate the formal (FC) for each atom. The +1 and –1 formal charges are not too unreasonable, but 0 charges are better. If one of the lone pairs is moved from the O (the atom with the negative FC) to form a double bond to the P (the atom with the positive FC), structure II results. The calculated formal changes in structure II are all 0 so this is a better structure even though P has 10 electrons.

I II FCP = 5 – [0 + 1/2(8)] = +1 FCP = 5 – [0 + 1/2(10)] = 0 FCO = 6 – [6 + 1/2(2)] = –1 FCO = 6 – [4 + 1/2(4)] = 0 FCCl = 7 – [6 + 1/2(2)] = 0 FCCl = 7 – [6 + 1/2(2)] = 0 Solution: b) In ClO2, the Cl is probably the central atom because the O atoms have a lower period. The molecule contains [1 × Cl(7e–)] + [2 × O(6e–)] = 19 electrons. The presence of an odd number of electrons means that there will be an exception to the octet rule. Placing the O atoms around the Cl and using 4 electrons to form single bonds leaves 15 electrons, 14 of which may be separated into 7 pairs. If 3 of these pairs are given to each O, and the remaining pair plus the lone electron are given to the Cl, we have the following structure: O

Calculating formal charges: FCCl = 7 – [3 + 1/2(4)] = +2 FCO = 6 – [6 + 1/2(2)] = –1 The +2 charge on the Cl is a little high, so other structures should be tried. Moving a lone pair from one of the O atoms (negative FC) to form a double bond between the Cl and one of the oxygen atoms gives either structure I or II below. If both O atoms donate a pair of electrons to form a double bond, then structure III results. The next step is to calculate the formal charges.

III

FCCl = 7 – [3 + 1/2(6)] = +1 FCCl = 7 – [3 + 1/2(6)] = +1 FCCl = 7 – [3 + 1/2(8)] = 0 The oxygen atom on the left: FCO = 6 – [4 + 1/2(4)] = 0 FCO = 6 – [6 + 1/2(4)] = –1 FCO = 6 – [4 + 1/2(8)] = 0 The oxygen atom on the right: FCO = 6 – [6 + 1/2(2)] = –1 FCO = 6 – [4 + 1/2(2)] = 0 FCO = 6 – [4 + 1/2(8)] = 0 Pick the structure with the best distribution of formal charges (structure III). Solution: c) In IBr4-, I is most likely to be the central atom because it has a higher period than Br. The molecule contains [1 × I(7e–)] + [4 × Br(7e–)] + [1 × Cl(7e–) + 1 e– (for the negative charge) = 36 electrons. Drawing a single bond between I and each of the Br atoms uses 4 × 2 = 8e–, leaving 28 electrons. Placing 6 electrons around each of the four outer Br atoms to complete their octets uses 24 electrons. The remaining 4 electrons are placed on the I atom, expanding its octet. This gives the structure: Copyright McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

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Calculating formal charges: FCI= 7 – [4 + 1/2(8)] = –1 FCBr = 7 – [6 + 1/2(2)] = 0 The formal charges are minimized for this structure. 10.5B

Plan: The presence of available d orbitals makes checking formal charges more important. Use the equation for formal charge: FC = # of valence e– – [# unshared valence e– + (# shared valence e–)] Solution: a) In BeH2, the Be is the central atom (H cannot be a central atom). The molecule contains: [1 × Be(2 e–)] + [2 × H(1 e–)]= 4 electrons. Placing the Be in the center with single bonds to all the surrounding atoms uses all 4 electrons, leaving Be 4 atoms short of an octet. There are not any more electrons to add to Be. Nor are there any lone pairs to change to bonding pairs. Thus, BeH2 is an exception to the octet rule in that Be does not have a complete octet. H

Determine the formal charges for each atom: FCBe = 2 – [0 + 1/2(4)] = 0 FCH = 1 – [0 + 1/2(2)] = 0 Solution: b) In I3–, one of the I atoms is central, and the other two I atoms are bonded to it. The molecule contains [3 × I(7e–)] + 1e– (for the negative charge) = 22 electrons. Placing one I atom in the center and connecting each of the other two I atoms to it by single bonds uses 2 × 2 = 4e–. This leaves 18 electrons. Placing 6 electrons on each of the outer I atoms leaves 18 – 12 = 6e–. These remaining 6 electrons are placed, as three lone pairs on the central I atom.

Calculating formal charges: FCI (central)= 7 – [6 + 1/2(4)] = –1 FCI (outer)= 7 – [6 + 1/2(2)] = 0 These are low, reasonable formal charges, and we do not need to adjust the structure. Solution: c) XeO3 is a noble gas compound, thus, there will be an exception to the octet rule. The molecule contains [1 × Xe(8e–)] + [3 × O(6e–)] = 26 electrons. The Xe is in a higher period than O so Xe is the central atom. If it is placed in the center with a single bond to each of the three oxygen atoms, 6 electrons are used, and 20 electrons remain. The remaining electrons can be divided into 10 pairs with 3 pairs given to each O and thelast pair being given to the Xe. This gives the structure: O O

Determine the formal charges for each atom: FCXe = 8 – [2 + 1/2(6)] = +3 FCO= 6 – [6 + 1/2(2)] = –1 The +3 charge on the Xe is a little high, so other structures should be tried. Moving a lone pair from one of the O atoms (negative FC) to form a double bond between the Xe and one of the oxygen atoms gives structure I (or one of its resonance structures). Moving two lone pair from two of the O atoms (negative FC) gives structure II below (or one of its resonance structures). If all three O atoms donate a pair of electrons to form a double bond, then structure III results. The next step is to calculate the formal charges. Copyright McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

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O O

III

FCXe = 8 – [2 + 1/2(8)] = +2 FCXe = 8 – [2 + 1/2(10)] = +1 FCXe = 8 – [2 + 1/2(12)] = 0 The oxygen atom on the left: FCO = 6 – [6 + 1/2(2)] = –1 FCO = 6 – [4 + 1/2(4)] = 0 FCO = 6 – [4 + 1/2(4)] = 0 The oxygen atom on the right: FCO = 6 – [6 + 1/2(2)] = –1 FCO = 6 – [6 + 1/2(2)] = –1 FCO = 6 – [4 + 1/2(4)] = 0 The oxygen atom on the top: FCO = 6 – [4 + 1/2(4)] = 0 FCO = 6 – [4 + 1/2(4)] = 0 FCO = 6 – [4 + 1/2(4)] = 0 Pick the structure with the best distribution of formal charges (structure III). 10.6A

Plan: Draw a Lewis structure. Determine the electron arrangement by counting the electron pairs around the central atom. Solution: a) The Lewis structure for CS2 is shown below. The central atom, C, has two pairs (double bonds only count once). The two pair arrangement is linear with the designation, AX2. The absence of lone pairs on the C means there is no deviation in the bond angle (180°). S

Solution: b) Even though this is a combination of a metal with a nonmetal, it may be treated as a molecular species. The Lewis structure for PbCl2 is shown below. The molecule is of the AX2E type; the central atom has three pairs of electrons (1 lone pair and two bonding pairs). This means the electron-group arrangement is trigonal planar (120°), with a lone pair giving a bent or V-shaped molecule. The lone pair causes the ideal bond angle to decrease to < 120°. Cl

Pb Cl

Solution: c) The Lewis structure for the CBr4 molecule is shown below. It has the AX4 type formula which is a perfect tetrahedron (with 109.5° bond angles) because all bonds are identical, and there are no lone pairs. Br Br

Br Br

Solution: d) The SF2 molecule has the Lewis structure shown below. This is a AX2E2 molecule; the central atom is surrounded by four electron pairs, two of which are lone pairs and two of which are bonding pairs. The electron group arrangement is tetrahedral. The two lone pairs give a V-shaped or bent arrangement. The ideal tetrahedral bond angle is decreased from the ideal 109.5° value.

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S F

10.6B

Plan: Draw a Lewis structure. Determine the electron arrangement by counting the electron pairs around the central atom. Solution: a) The Lewis structure for BrO2– is shown below. The molecule is of the AX2E2 type; the central atom has four pairs of electrons (2 lone pairs and two bonding pairs; each double bond counts as one electron domain).This means the electron-group arrangement is tetrahedral (109.5°), with two lone pairs giving a bent or V-shaped molecule. The lone pairs cause the bond angle to decrease to <109.5°.

The molecular shape of BrO2–:

Solution: b) The Lewis structure for AsH3 is shown below. The molecule is of the AX3E type; the central atom has four pairs of electrons (1 lone pair and three bonding pairs). This means the electron-group arrangement is tetrahedral (109.5°), with a lone pair giving a trigonal pyramidal molecule. The lone pair causes the bond angle to decrease to <109.5°.

H H

Solution: c) The Lewis structure for N3– is shown below. The two pair electron arrangement is linear with the designation, AX2. The molecular shape is also linear. The absence of lone pairs on the central N means there is no deviation in the bond angle (180°).

Solution: d) The SeO3 molecule has the Lewis structure shown below. This is an AX3 molecule; the central atom is surrounded by three electron pairs (each double bond counts as one electron domain). The electron group arrangement and molecular shape is trigonal planar. The absence of lone pairs on the central atom means there is no deviation in the bond angle (120°). O Se O

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10.7A

Plan: Draw a Lewis structure. Determine the electron arrangement by counting the electron pairs around the central atom. Solution: a) The Lewis structure for the ICl2– is shown below. This is an AX2E3 type structure. The five pairs give a trigonal bipyramidal arrangement of electron groups. The presence of 3 lone pairs leads to a linear shape (180°). The usual distortions caused by lone pairs cancel in this case. In the trigonal bipyramidal geometry, lone pairs always occupy equatorial positions.

Cl Cl

I Cl

Solution: b) The Lewis structure for the ClF3 molecule is given below. Like ICl2– there are 5 pairs around the central atom; however, there are only 2 lone pairs. This gives a molecule that is T-shaped. The presence of the lone pairs decreases the ideal bond angles to less than 90°. F F

F Cl

F F

Solution: c) The SOF4 molecule has several possible Lewis structures, two of which are shown below. In both cases, the central atom has 5 atoms attached with no lone pairs. The formal charges work out the same in both structures. The structure on the right has an equatorial double bond. Double bonds require more room than single bonds, and equatorial positions have this extra room. O F F F F F F S O S + S O F F F F F F The molecule is trigonal bipyramidal, and the double bond causes deviation from ideal bond angles. All of the F atoms move away from the O. Thus, all angles involving the O are increased, and all other angles are decreased. 10.7B

Plan: Draw a Lewis structure. Determine the electron arrangement by counting the electron pairs around the central atom. Solution: a) The Lewis structure for the BrF4– ion is shown below. This is an AX4E2 type structure. The six pairs give an octahedral arrangement of electron groups. The presence of 2 lone pairs on the central atom leads to a square planar shape (90°). The usual distortions caused by lone pairs cancel in this case. In the square planar, lone pairs always occupy axial positions.

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Solution: b) The Lewis structure for the ClF4+ion is given below. This is an AX4E type structure. There are 5 pairs around the central atom; however, there is 1 lone pair. This gives a molecule that has a seesaw shape. The presence of the lone pair decreases the ideal bond angles to less than 90°(Fax–Cl–Feq) and less than 120° (Feq–Cl–Feq).

F F

F Cl

F F

Solution: c) The Lewis structure for the PCl6– is shown below. This is an AX6 type structure. The six pairs give an octahedral arrangement of electron groups and an octahedral shape, with 90° bond angles. The absence of lone pair electrons on the central atom means there is no deviation from the ideal bond angles.

10.8A

Plan: Draw the Lewis structure for each of the substances, and determine the molecular geometry of each. Solution: a) The Lewis structure for H2SO4 is shown below. The double bonds ease the problem of a high formal charge on the sulfur. Sulfur is allowed to exceed an octet. The S has 4 groups around it, making it tetrahedral. The ideal angles around the S are 109.5°. The double bonds move away from each other, and force the single bonds away. This opens the angle between the double-bonded oxygen atoms, and results in an angle between the single-bonded oxygen atoms that is less than ideal. Each single-bonded oxygen atom has 4 groups around it; since two of the 4 groups are lone pairs, the shape around each of these oxygen atoms is bent. The lone pairs compress the H–O–S bond angle to <109.5°. O

O H

H H

Solution: b) The hydrogen atoms cannot be central so the carbons must be attached to each other. The problem states that there is a carbon-carbon triple bond. This leaves only a single bond to connect the third carbon to a triple-bonded Copyright McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

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carbon, and give that carbon an octet. The other triple-bonded carbon needs one hydrogen to complete its octet. The remaining three hydrogen atoms are attached to the single-bonded carbon, which allows it to complete its octet. This structure is shown below. The single-bonded carbon has four groups tetrahedrally around it leading to bond angles ~109.5° (little or no deviation). The triple-bonded carbons each have two groups (the triple bond counts as one electron group) so they should be linear (180°). H H H H

Solution: c) Fluorine, like hydrogen, is never a central atom. Thus, the sulfuratoms must be bonded to each other. Each F has 3 lone pairs, and the sulfur atoms have 2 lone pairs. All 4 atoms now have an octet. This structure is shown below. Each sulfur has four groups around it, so the electron arrangement is tetrahedral. The presence of the lone pairs on the sulfur atoms results in a geometry that is V-shaped or bent and in a bond angle that is < 109.5°.

F S

F F 10.8B

Plan: Draw the Lewis structure for each of the substances, and determine the molecular geometry of each. Solution: a) The Lewis structure for CH3NH2 is shown below. The C has 4 groups around it (AX4), making it tetrahedral. The ideal angles around the C are 109.5°. The nitrogen also has 4 groups around it, but one of those groups is a lone pair. The lone pair makes the N an AX3E central atom with trigonal pyramidal geometry. The presence of the lone pair compresses the H–N–H angle to <109.5°.

H C

H H Solution: b) The Lewis structure for C2Cl4 is shown below. The carbons are central and connected to each other. There are a total of 36 valence electrons. In order for all of the atoms to have a full octet, there must be a double bond between the two carbon atoms. Each of the carbon atoms has three electron groups around it, yielding a trigonal planar geometry. The presence of the double bond compresses each Cl–C–Cl angle to <120°. Cl

Cl C

C Cl

Solution: c) The Lewis structure for Cl2O7 is shown below. The problem states that three oxygen atoms are bonded to the first Cl atom, which is then bonded to a fourth oxygen atom. That fourth oxygen atom is bonded, in turn, to a second Cl atom, which is bonded to 3 additional oxygen atoms. Each Cl has 4 electron groups (3 of which are double bonds to oxygen and one of which is a single bond to oxygen). This gives the Cl a tetrahedral geometry Copyright McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

10-12

with bond angles very close to 109.5°. The central oxygen atom, on the other hand, also has four electron pairs around it. In this case, two of the electron pairs are bonding pairs and two of the pairs are lone pairs. The presence of the lone pairs gives the central oxygen a bent or V-shaped geometry and compresses the Cl–O–Cl bond angle to <109.5°. O

O O

10.9A

Plan: Draw the Lewis structures, predict the shapes, and then examine the positions of the bond dipoles. Solution: a) Dichloromethane, CH2Cl2, has the Lewis structure shown below. It is tetrahedral, and if the outlying atoms were identical, it would be nonpolar. However, the chlorine atoms are more electronegative than hydrogen so there is a general shift in their direction resulting in the arrows shown.

H H

C H

Solution: b) Iodine oxide pentafluoride, IOF5, has the Lewis structure shown below. The overall geometry is octahedral. All six bonds are polar, with the more electronegative O and F atoms shifting electron density away from the I. The 4 equatorial fluorines counterbalance each other. The axial F is not equivalent to the axial O. The more electronegative F results in an overall polarity in the direction of the axial F. O O F F F F I I F F F F F F The lone electron pairs are left out for simplicity. Solution: c) Iodine pentafluoride, IF5, has the square pyramidal Lewis structure shown below. The fluorine is more electronegative than the I, so the shift in electron density is toward the F, resulting in the dipole indicated below.

F F F

10.9B

F F

Plan: Draw the Lewis structures, predict the shapes, and then examine the positions of the bond dipoles. Solution: a) Xenon tetrafluoride, XeF4, has the Lewis structure shown below. It is square planar, and, because the outlying atoms are identical, it is nonpolar. However, the fluorine atoms are more electronegative than xenon, so the individual bonds are polar. Because of the square planar geometry, those polar bonds cancel each other out.

10-13

F F

Solution: b) Chlorine trifluoride, ClF3, has the Lewis structure shown below. The overall geometry is T-shaped. All three bonds are polar, with the more electronegative F atoms shifting electron density away from the Cl. The more electronegative F results in an overall dipole indicated below.

F Solution: c) Sulfur monoxide tetrafluoride, SOF4, has the trigonal bipyramidal Lewis structure shown below. Both the F atoms and the O atom are more electronegative than the S, and F is more electronegative than O, so the shift in electron density is toward the F atoms, resulting in the dipole indicated below. F F O

S F F

CHEMICAL CONNECTIONS BOXED READING PROBLEMS B10.1

Plan: Examine the Lewis structure, noting the number of regions of electron density around the carbon and nitrogen atoms in the two resonance structures. The molecular shape is determined by the number of electron regions. An electron region is any type of bond (single, double, or triple) and an unshared pair of electrons. Solution: Resonance structure on the left: Carbon has three electron regions (two single bonds and one double bond); three electron regions are arranged in a trigonal planar arrangement. The molecular shape around the C atom is trigonal planar. Nitrogen has four electron regions (three single bonds and an unshared pair of electrons); the four electron regions are arranged tetrahedrally; since one corner of the tetrahedron is occupied by an unshared electron pair, the shape around N is trigonal pyramidal. Resonance structure on the right: This C atom also has three electrons regions (two single bonds and one double bond) so the molecular shape is again trigonal planar. The N atom also has three electron regions (two single bonds and one double bond); the molecular shape is trigonal planar.

10-14

B10.2

The top portion of both molecules is similar so the top portions will interact with biomolecules in a similar manner. The mescaline molecule may fit into the same nerve receptors as dopamine due to the similar molecular shape.

END–OF–CHAPTER PROBLEMS 10.1

Plan: To be the central atom in a compound, an atom must be able to simultaneously bond to at least two other atoms. Solution: He, F, and H cannot serve as central atoms in a Lewis structure. Helium (1s2) is a noble gas, and as such, it does not need to bond to any other atoms. Hydrogen (1s1) and fluorine (1s22s22p5) only need one electron to complete their valence shells. Thus, they can only bond to one other atom, and they do not have d orbitals available to expand their valence shells.

10.2

Resonance must be present any time that a single Lewis structure is inadequate in explaining one or more aspects of a molecule or ion. The two N–O bonds in NO2 are equivalent in bond length and bond energy; no single Lewis structure can account for this. The following Lewis structures may be drawn for NO2: O

The average of all of these structures gives equivalent N–O bonds with a bond length that is between N–O and N=O. 10.3

Plan: For an element to obey the octet rule it must be surrounded by eight electrons. To determine the number of electrons present, (1) count the individual electrons actually shown adjacent to a particular atom (lone pairs), and (2) add two times the number of bonds to that atom: number of electrons = individual electrons + 2(number of bonds). Solution: (a) 0 + 2(4) = 8; (b) 2 + 2(3) = 8; (c) 0 + 2(5) = 10; (d) 2 + 2(3) = 8; (e) 0 + 2(4) = 8; (f) 2 + 2(3) = 8; (g) 0 + 2(3) = 6; (h) 8 + 2(0) = 8. All the structures obey the octet rule except: c and g.

10.4

For an atom to expand its valence shell, it must have readily available d orbitals. The d orbitals do not become readily available until the third period or below on the periodic table. For the elements in the problem F, S, H, Al, Se, and Cl, the period numbers are 2, 3, 1, 3, 4, and 3, respectively. All of these elements, except those in the first two periods (H and F), can expand their valence shells.

10.5

Plan: Count the valence electrons and draw Lewis structures. Solution: Total valence electrons: SiF4: [1 × Si(4e–)] + [4 × F(7e–)] = 32; SeCl2: [1 × Se(6e–)] + [2 × Cl(7e–)] = 20; COF2: [1 × C(4e–)] + [1 × O(6e–)] + [2 × F(7e–)] = 24. The Si, Se, and the C are the central atoms, because these are the elements in their respective compounds with the lower group number (in addition, we are told C is central). Place the other atoms around the central atoms and connect each to the central atom with a single bond. SiF4: At this point, eight electrons (2e– in four Si–F bonds) have been used with 32 – 8 = 24 remaining; the remaining electrons are placed around the fluorine atoms (three pairs each). All atoms have an octet. SeCl2: The two bonds use 4e– (2e– in two Se–Cl bonds) leaving 20 – 4 = 16e–. These 16e– are used to complete the octets on Se and the Cl atoms. COF2: The three bonds to the C use 6e– (2e– in three bonds) leaving 24 – 6 = 18 e–. These 18e– are distributed to the surrounding atoms first to complete their octets. After the 18e– are used, the central C is two electrons short of an octet. Forming a double bond to the O (change a lone pair on O to a bonding pair on C) completes the C octet. (a) SiF4 (b) SeCl2

10-15

F F

O 10.6

O +

Total valence electrons: PH4 has 8; C2F4 has 36; and SbH3 has 8. Ignoring H, the atom in the lower group number is central: P, C, and Sb. Added proof: H and F are never central. The two central C atoms must be adjacent. Place all the other atoms around the central atom. Split the F atoms so that each C gets two. Connect all the atoms with single bonds. This uses all the electrons in PH4+, and gives P an octet. The H atoms need no additional electrons. The C atoms have six electrons each, but can achieve an octet by forming a double bond. Splitting the twenty-four remaining electrons in C2F4 into twelve pairs and giving three pairs to each F leaves each F with an octet. The last two electrons in SbH3 end as a lone pair on the Sb, and complete its octet. (a) (b) (c)

H H

F C

10.7

Plan: Count the valence electrons and draw Lewis structures. Solution: a) PF3: [1 × P(5 e–)] + [3 × F(7e–)] = 26 valence electrons. P is the central atom. Draw single bonds from P to the three F atoms, using 2e– × 3 bonds = 6 e–. Remaining e–: 26 – 6 = 20 e–. Distribute the 20 e– around the P and F atoms to complete their octets. b) H2CO3: [2 × H(1e–)] + [1 × C(4e–)] + 3 × O(6e–)] = 24 valence electrons. C is the central atom with the H – – atoms attached to the O atoms. Place appropriate single bonds between all atoms using 2e × 5 bonds = 10e so – – that 24 – 10 = 14e remain. Use these 14e to complete the octets of the O atoms (the H atoms already have their two electrons). After the 14e– are used, the central C is two electrons short of an octet. Forming a double bond to the O that does not have an H bonded to it (change a lone pair on O to a bonding pair on C) completes the C octet. c) CS2: [1 × C(4e–)] + [2 × S(6e–)] = 16 valence electrons. C is the central atom. Draw single bonds from C to the two S atoms, using 2e– × 2 bonds = 4e–. Remaining e–: 16 – 4 = 12e–. Use these 12e– to complete the octets of the surrounding S atoms; this leaves C four electrons short of an octet. Form a double bond from each S to the C by changing a lone pair on each S to a bonding pair on C. a) PF3 (26 valence e–) b) H2CO3 (24 valence e–)

P F

O H

C O

O H

C O

O H

10-16

c) CS2 (16 valence e–) S

10.8

The C and S atoms are central. The S in part (a) is attached to an H and the C. All atoms are attached with single bonds and the remaining electrons are divided into lone pairs. All the atoms, except H, have octets. a) CH4S H

b) S2Cl2

c) CHCl3 H

Cl Cl

H Cl

10.9

Plan: The problem asks for resonance structures, so there must be more than one answer for each part. Solution: a) NO2+ has [1 × N(5e–)] + [2 × O(6e–)] – 1e– (+ charge) = 16 valence electrons. Draw a single bond from N to each O, using 2e– × 2 bonds = 4e–; 16 – 4 = 12e– remain. Distribute these 12e– to the O atoms to complete their octets. This leaves N 4e– short of an octet. Form a double bond from each O to the N by changing a lone pair on each O to a bonding pair on N. No resonance is required as all atoms can achieve an octet with double bonds. O

b) NO2F has [1 × N(5e–)] + [2 × O(6e–)] + [1 × F(7e–)] = 24 valence electrons. Draw a single bond from N to each surrounding atom, using 2e– × 3 bonds = 6e–; 24 – 6 = 18e– remain. Distribute these 18e– to the O and F atoms to complete their octets. This leaves N 2e– short of an octet. Form a double bond from either O to the N by changing a lone pair on O to a bonding pair on N. There are two resonance structures since a lone pair from either of the two O atoms can be moved to a bonding pair with N:

O 10.10

F N O

a) O

N O

10-17

10.11

Plan: Count the valence electrons and draw Lewis structures. Additional structures are needed to show resonance. Solution: a) N3– has [3 × N(5e–)] + [1 e–(from charge)] = 16 valence electrons. Place a single bond between the nitrogen atoms. This uses 2e– × 2 bonds = 4 electrons, leaving 16 – 4 = 12 electrons (6 pairs). Giving three pairs on each end nitrogen gives them an octet, but leaves the central N with only four electrons as shown below: N

The central N needs four electrons. There are three options to do this: (1) each of the end N atoms could form a double bond to the central N by sharing one of its pairs; (2) one of the end N atoms could form a triple bond by sharing two of its lone pairs; and (3) the other end N atom could form the triple bond instead. N

b) NO2– has [1 × N(5e–)] + [2 × O(6e–)] + [1 e– (from charge)] = 18 valence electrons. The nitrogen should be the central atom with each of the oxygen atoms attached to it by a single bond (2e– × 2 bonds = 4 electrons). This leaves 18 – 4 = 14 electrons (seven pairs). If three pairs are given to each O and one pair is given to the N, then both O atoms have an octet, but the N atom only has six. To complete an octet the N atom needs to gain a pair of electrons from one O atom or the other (form a double bond). The resonance structures are: O

10.12

a) HCO2– has 18 valence electrons.

10-18

b) HBrO4 has 32 valence electrons. O

O O

O O H

10.13

O H

Plan: Initially, the method used in the preceding problems may be used to establish a Lewis structure. The total of the formal charges must equal the charge on an ion or be equal to 0 for a compound. The formal charge only needs to be calculated once for a set of identical atoms. Formal charge (FC) = no. of valence electrons – [no. of unshared valence electrons + no. of shared valence electrons]. Solution: a) IF5 has [1 × I(7e–)] + [5 × F(7e–)] = 42 valence electrons. The presence of five F atoms around the central I means that the I atom will have a minimum of 10 electrons; thus, this is an exception to the octet rule. The five I–F bonds use 2e– × 5 bonds = 10 electrons leaving 42 – 10 = 32 electrons (16 pairs). Each F needs three pairs to complete an octet. The five F atoms use 15 of the 16 pairs, so there is one pair left for the central I. This gives:

Calculating formal charges: FC = no. of valence electrons – [no. of unshared valence electrons + no. of shared valence electrons]. For iodine: FCI = 7 – [2 + (10)] = 0 For each fluorine: FCF = 7 – [6 + (2)] = 0 Total formal charge = 0 = charge on the compound. b) AlH4– has [1 × Al(3e–)] + [4 × H(1e–)] + [1e– (from charge)] = 8 valence electrons. The four Al–H bonds use all the electrons and Al has an octet.

FC = no. of valence electrons – [no. of unshared valence electrons + For aluminum: FCAl = 3 – [0 + (8)] = –1 For each hydrogen: FCH = 1 – [0 + (2)] = 0 10.14

no. of shared valence electrons].

a) OCS has sixteen valence electrons.

FCS = 6 – [4 + (4)] = 0 FCC = 4 – [0 + (8)] = 0 FCO = 6 – [4 + (4)] = 0 Copyright McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

10-19

b) NO (has 11 valence electrons); the odd number means there will be an exception to the octet rule. O

FCO = 6 – [4 + (4)] = 0 FCO = 6 – [3 + (4)] = +1 FCN = 5 – [3 + (4)] = 0 FCN = 5 – [4 + (4)] = –1 The first resonance structure has a better distribution of formal charges. 10.15

Plan: Initially, the method used in the preceding problems may be used to establish a Lewis structure. The total of the formal charges must equal the charge on an ion or be equal to 0 for a compound. The formal charge only needs to be calculated once for a set of identical atoms. Formal charge (FC) = no. of valence electrons – [no. of unshared valence electrons + no. of shared valence electrons]. Solution: a) CN–: [1 × C(4e–)] + [1 × N(5e–)] + [1 e– from charge] = 10 valence electrons. Place a single bond between the carbon and nitrogen atoms. This uses 2e– × 1 bond = 2 electrons, leaving 10 – 2 = 8 electrons (four pairs). Giving three pairs of electrons to the nitrogen atom completes its octet but that leaves only one pair of electrons for the carbon atom which will not have an octet. The nitrogen could form a triple bond by sharing two of its lone pairs with the carbon atom. A triple bond between the two atoms plus a lone pair on each atom satisfies the octet rule and uses all ten electrons.

FC = no. of valence electrons – [no. of unshared valence electrons + no. of shared valence electrons]. FCC = 4 – [2 + (6)] = –1; FCN = 5 – [2 + (6)] = 0 Check: The total formal charge equals the charge on the ion (–1). b) ClO–: [1 × Cl(7e–)] + [1 × O(6e–)] + [1e– from charge] = 14 valence electrons. Place a single bond between the chlorine and oxygen atoms. This uses 2e– × 1 bond = 2 electrons, leaving 14 – 2 = 12 electrons (six pairs). Giving three pairs of electrons each to the carbon and oxygen atoms completes their octets. Cl

FC = no. of valence electrons – [no. of unshared valence electrons + no. of shared valence electrons]. FCCl = 7 – [6 + 1/2(2)] = 0 FCO = 6 – [6 + 1/2(2)] = –1 Check: The total formal charge equals the charge on the ion (–1). 10.16

a) ClF2+ has 20 valence electrons.

FCF = 7 – [6 + 1/2(2)] = 0 FCCl = 7 – [4 + 1/2(4)] = –1 b) ClNO has 18 valence electrons. Cl

FCCl = 7 – [6 + 1/2(2)] = 0; 10.17

FCN = 5 – [2 + 1/2(6)] = 0;

FCO = 6 – [4 + 1/2(4)] = 0

Plan: The general procedure is similar to the preceding problems, plus the oxidation number determination. Solution: a) BrO3– has [1 × Br(7e–)] + 3 × O(6e–)] + [1e– (from charge)] = 26 valence electrons. Placing the O atoms around the central Br and forming three Br–O bonds uses 2e– × 3 bonds = 6 electrons and leaves 26 – 6 = 20 electrons (10 pairs). Placing three pairs on each O (3 × 3 = 9 total pairs) leaves one pair for the Br and yields structure I below. In structure I, all the atoms have a complete octet. Calculating formal charges: FCBr = 7 – [2 + 1/2(6)] = +2 FCO = 6 – [6 + 1/2(2)] = –1

10-20

The FCO is acceptable, but FCBr is larger than is usually acceptable. Forming a double bond between any one of the O atoms gives structure II. Calculating formal charges: FCBr = 7 – [2 + (8)] = +1 FCO = 6 – [6 + (2)] = –1 FCO = 6 – [4 + (4)] = 0 (Double-bonded O) The FCBr can be improved further by forming a second double bond to one of the other O atoms (structure III). FCBr = 7 – [2 + (10)] = 0 FCO = 6 – [6 + (2)] = –1 FCO = 6 – [4 + (4)] = 0 (Double-bonded O atoms) Structure III has the most reasonable distribution of formal charges. O

III

–6 The oxidation numbers (O.N.) are: O.N.Br = +5 and O.N.O = –2. +5 –2 Check: The total formal charge equals the charge on the ion (–1). BrO3– 2– – – – b) SO3 has [1 × S(6e )] + [3 × O(6e )] + [2e (from charge)] = 26 valence electrons. Placing the O atoms around the central S and forming three S–O bonds uses 2e– × 3 bonds = 6 electrons and leaves 26 – 6 = 20 electrons (10 pairs). Placing three pairs on each O (3 × 3 = 9 total pairs) leaves one pair for the S and yields structure I below. In structure I all the atoms have a complete octet. Calculating formal charges: FCS = 6 – [2 + (6)] = +1; FCO = 6 – [6 + (2)] = –1 The FCO is acceptable, but FCS is larger than is usually acceptable. Forming a double bond between any one of the O atoms (structure II) gives: FCS = 6 – [2 + (8)] = 0 FCO = 6 – [6 + (2)] = –1 FCO = 6 – [4 + (4)] = 0 (Double-bonded O)

Structure II has the more reasonable distribution of formal charges. The oxidation numbers (O.N.) are: O.N.S = +4 and O.N.O = –2. Check: The total formal charge equals the charge on the ion (–2). 10.18

+4 –2 SO32–

a) AsO43– has 32 valence electrons. See structure I. FCAs = 5 – [0 + (8)] = +1 FCO = 6 – [6 + (2)] = –1 Net formal charge (+1 – 4) = –3 The octet rule is followed by all atoms.

10-21

For more reasonable formal charges, move a lone pair from an O to a bonded pair on As (structure II):

FCAs = 5 – [0 + (10)] = 0 FCO(single bond) = 6 – [6 + (2)] = –1 FCO(double bond) = 6 – [4 + (4)] = 0 Net formal charge: (0 + 3(–1)) + 0 = –3 Improved formal charge distribution O.N.: O –2 each × 4 = –8 total; As +5 b) ClO2– has 20 valence electrons. For structure I in which all atoms have an octet: FCCl = 7 – [4 + (4)] = +1 FCO = 6 – [6 + (2)] = –1 For more reasonable formal charges, see structure II: O

I Formal charges in structure II: FCCl = 7 – [4 + (6)] = 0 FCO (single bond) = 6 – [6 –

(2)] = –1

Plan: The octet rule states that when atoms bond, they share electrons to attain a filled outer shell of eight electrons. If an atom has fewer than eight electrons, it is electron deficient; if an atom has more than eight electrons around it, the atom has an expanded octet. Solution: a) BH3 has [1 × B(3e–)] + [3 × H(1e–)] = 6 valence electrons. These are used in three B–H bonds. The B has six electrons instead of an octet; this molecule is electron deficient. b) AsF4– has [1 × As(5e–)] +[4 × F(7e–)] + [1e– (from charge)] = 34 valence electrons. Four As–F bonds use eight electrons leaving 34 – 8 = 26 electrons (13 pairs). Each F needs three pairs to complete its octet and the remaining pair goes to the As. The As has an expanded octet with ten electrons. The F cannot expand its octet. c) SeCl4 has [1 × Se(6e–)] + 4 × Cl(7e–)] = 34 valence electrons. The SeCl4 is isoelectronic (has the same electron structure) as AsF4–, and so its Lewis structure looks the same. Se has an expanded octet of 10 electrons.

B H

H (a)

10.20

FCO(double bond) = 6 – [4 + (4)] = 0 O.N.: O –2 each × 2 = –4 total; Cl +3 10.19

a) PF6– has 48 valence electrons.

F (b)

Cl (c)

P has an expanded octet of 12 e–.

10-22

F F

P F

F F

b) ClO3 has 25 valence electrons. The odd number means that there will be an exception. This is a radical: the chlorine or one of the oxygen atoms will lack an e– to complete its octet. Cl

O There are two additional resonance structures where the other O atoms are the ones lacking the octet. The FC predicts that Cl will end with the odd electron. c) H3PO3 has 26 valence electrons. To balance the formal charges; the O lacking an H will form a double bond to the P. This compound is an exception in that one of the H atoms is attached to the central P. P has an expanded octet of 10 e–. H

H 10.21

O H

Plan: The octet rule states that when atoms bond, they share electrons to attain a filled outer shell of eight electrons. If an atom has fewer than eight electrons, it is electron deficient; if an atom has more than eight electrons around it, the atom has an expanded octet. Solution: a) BrF3 has [1 × Br(7e–)] + [3 × F(7e–)] = 28 valence electrons. Placing a single bond between Br and each F uses 2e– × 3 bonds = 6e–, leaving 28 – 6 = 22 electrons (11 pairs). After the F atoms complete their octets with three pairs each, the Br gets the last two lone pairs. The Br has an expanded octet of ten electrons. b) ICl2– has [1 × I(7e–)] + [2 × Cl(7e–)] + [1e– (from charge)] = 22 valence electrons. Placing a single bond between I and each Cl uses 2e– × 2 bond = 4e–, leaving 22 – 4 = 18 electrons (nine pairs). After the Cl atoms complete their octets with three pairs each, the iodine finishes with the last three lone pairs. The iodine has an expanded octet of ten electrons. c) BeF2 has [1 × Be(2e–)] + [2 × F(7e–)] = 16 valence electrons. Placing a single bond between Be and each of the F atoms uses 2e– × 2 bonds = 4e–, leaving 16 – 4 = 12 electrons (six pairs).The F atoms complete their octets with three pairs each, and there are no electrons left for the Be. Formal charges work against the formation of double bonds. Be, with only four electrons, is electron deficient.

F a) 10.22

a) O3– has 19 valence electrons (note the odd number). There are several resonance structures possible; only one is necessary for the answer. One of the O atoms has the odd electron (seven total).

10-23

c) SbF4– has 34 valence electrons. F

Xe has an expanded octet of 10e–.

b) XeF2 has 22 valence electrons. F

Sb has an expanded octet of 10e–.

10.23

Plan: Draw Lewis structures for the reactants and products. Solution: Beryllium chloride has the formula BeCl2. BeCl2 has [1 × Be(2e–)] + [2 × Cl(7e–)] = 16 valence electrons. Four of these electrons are used to place a single bond between Be and each of the Cl atoms, leaving 16 – 4 = 12 electrons (six pairs). These six pairs are used to complete the octets of the Cl atoms, but Be does not have an octet – it is electron deficient. Chloride ion has the formula Cl– with an octet of electrons. BeCl42– has [1 × Be(2e–)] + [4 × Cl(7e–)] + [2e– (from charge)] = 32 valence electrons. Eight of these electrons are used to place a single bond between Be and each Cl atom, leaving 32 – 8 = 24 electrons (12 pairs). These twelve pairs complete the octet of the Cl atoms (Be already has an octet).

10.24

Draw a Lewis structure. If the formal charges are not ideal, a second structure may be needed. BrO4– has 32 valence electrons. O O

O O

In the structure on the left, all atoms have octets. The formal charges are: FCBr = 7 – [0 + (8)] = +3 FCO = 6 – [6 + (2)] = –1 The structure on the right expands the valence shell of the Br to give more favorable formal charges. FCBr = 7 – [0 + (14)] = 0 FCO(single bonded) = 6 – [6 + (2)] = –1 FCO (double bonded) = 6 – [4 + (4)] = 0

10-24

10.25

Count the total valence electrons and draw a Lewis structure. AlF63– has 48 valence electrons.

10.26

Plan: Use the structures in the text to determine the formal charges. Formal charge (FC) = no. of valence electrons – [no. of unshared valence electrons + no. of shared valence electrons]. Solution: Structure A: FCC = 4 – [0 + (8)] = 0; FCO = 6 – [4 + (4)] = 0; FCCl = 7 – [6 + (2)] = 0 Total FC = 0 Structure B: FCC = 4 – [0 + (8)] = 0; FCO = 6 – [6 + (2)] = –1; FCCl(double bonded) = 7 – [4 + (4)] = +1; FCCl(single bonded) = 7 – [6 + (2)] = 0 Total FC = 0 Structure C: FCC = 4 – [0 + (8)] = 0; FCO = 6 – [6 + (2)] = –1; FCCl(double bonded) = 7 – [4 + (4)] = +1; FCCl(single bonded) = 7 – [6 + (2)] = 0 Total FC = 0 Structure A has the most reasonable set of formal charges.

10.27

Determine the total number of valence electrons present. Next, draw a Lewis structure. Finally, use VSEPR or valence bond theory to predict the shape.

10.28

The molecular shape and the electron-group arrangement are the same when there are no lone pairs on the central atom.

10.29

A bent (V-shaped) molecule will have the stoichiometry AX2, so only AX2En geometries result in a bent molecule. The presence of one or two lone pairs in the three and four electron-group arrangements can produce a bent (V-shaped) molecule as either AX2E or AX2E2. Examples are: NO2– and H2O.

N 120° AX2E

109.5° AX2E2

10.30

Plan: Examine a list of all possible structures, and choose the ones with four electron groups since the tetrahedral electron-group arrangement has four electron groups. Solution: Tetrahedral AX4 Trigonal pyramidal AX3E Bent or V shaped AX2E2

10.31

a) A, which has a square planar molecular geometry, has the most electron pairs. There are four shared pairs and two unshared pairs for a total of six pairs of electrons. The six electron pairs are arranged in an octahedral arrangement with the four bonds in a square planar geometry. B and C have five electron pairs and D has four electron pairs. b) A has the most unshared pairs around the central atom with two unshared pairs. B has only one unshared pair on the central atom and C and D have no unshared pairs on the central atom. c) C and D have only shared pairs around the central atom.

10-25

10.32

Plan: Begin with the basic structures and redraw them. Solution: a) A molecule that is V shaped has two bonds and generally has either one (AX2E) or two (AX2E2) lone electron pairs. b) A trigonal planar molecule follows the formula AX3 with three bonds and no lone electron pairs. c) A trigonal bipyramidal molecule contains five bonding pairs (single bonds) and no lone pairs (AX 5). d) A T-shaped molecule has three bonding groups and two lone pairs (AX3E2). e) A trigonal pyramidal molecule follows the formula AX3E with three bonding pairs and one lone pair. f) A square pyramidal molecule shape follows the formula AX5E with five bonding pairs and one lone pair. X X X X A A A X A X X X X X X X

(a)

(c)

(b)

X X

X A

X (e)

(d)

(f)

10.33

Determine the geometry from the lone pairs and the number of groups attached to the central atom. a) AX3E tetrahedral 109.5° smaller b) AX2 linear 180° none c) AX3 trigonal planar 120° none d) AX2E2 tetrahedral 109.5° smaller e) AX2 linear 180° none f) AX4E trigonal bipyramidal 180°, 120°, 90° smaller

10.34

Plan: First, draw a Lewis structure, and then apply VSEPR. Solution: a) O3: The molecule has [3 × O(6e–)] = 18 valence electrons. Four electrons are used to place single bonds between the oxygen atoms, leaving 18 – 4 = 14e– (seven pairs). Six pairs are required to give the end oxygen atoms an octet; the last pair is distributed to the central oxygen, leaving this atom two electrons short of an octet. Form a double bond from one of the end O atoms to the central O by changing a lone pair on the end O to a bonding pair on the central O. This gives the following Lewis structure: O

O O

There are three electron groups around the central O, one of which is a lone pair. This gives a trigonal planar electron-group arrangement (AX2E), a bent molecular shape, and an ideal bond angle of 120°. b) H3O+: This ion has [3 × H(1e–)] + [1 × O(6e–)] – [1e– (due to + charge)] = eight valence electrons. Six electrons are used to place a single bond between O and each H, leaving 8 – 6 = 2e– (one pair). Distribute this pair to the O atom, giving it an octet (the H atoms only get two electrons). This gives the following Lewis structure:

10-26

O H

H H

There are four electron groups around the O, one of which is a lone pair. This gives a tetrahedral electron-group arrangement (AX3E), a trigonal pyramidal molecular shape, and an ideal bond angle of 109.5°. c) NF3: The molecule has [1 × N(5e–)] + [3 × F(7e–)] = 26 valence electrons. Six electrons are used to place a single bond between N and each F, leaving 26 – 6 = 20 e– (10 pairs). These 10 pairs are distributed to all of the F atoms and the N atoms to give each atom an octet. This gives the following Lewis structure: F

N F

F F

There are four electron groups around the N, one of which is a lone pair. This gives a tetrahedral electron-group arrangement (AX3E), a trigonal pyramidal molecular shape, and an ideal bond angle of 109.5°. 10.35

Lewis structure (a)

Electron-group arrangement Tetrahedral

(b)

Trigonal planar O

10.36

Trigonal pyramidal

109.5°

H P

In addition, there are other resonance forms. Bent 120°

In addition, there are other resonance forms. (c) Tetrahedral P

Tetrahedral

Ideal bond angle 109.5°

O O

Molecular shape

H H

Plan: First, draw a Lewis structure, and then apply VSEPR. Solution: (a) CO32–: This ion has [1 × C(4e–)] + [3 × O(6e–)] + [2e– (from charge)] = 24 valence electrons. Six electrons are used to place single bonds between C and each O atom, leaving 24 – 6 = 18 e– (nine pairs). These nine pairs are used to complete the octets of the three O atoms, leaving C two electrons short of an octet. Form a double bond from one of the O atoms to C by changing a lone pair on an O to a bonding pair on C. This gives the following Lewis structure:

10-27

There are two additional resonance forms. There are three groups of electrons around the C, none of which are lone pairs. This gives a trigonal planar electron-group arrangement (AX3), a trigonal planar molecular shape, and an ideal bond angle of 120°. (b) SO2: This molecule has [1 × S(6e–)] + [2 × S(6e–)] = 18 valence electrons. Four electrons are used to place a single bond between S and each O atom, leaving 18 – 4 = 14e– (seven pairs). Six pairs are needed to complete the octets of the O atoms, leaving a pair of electrons for S. S needs one more pair to complete its octet. Form a double bond from one of the end O atoms to the S by changing a lone pair on the O to a bonding pair on the S. This gives the following Lewis structure: O

There are three groups of electrons around the C, one of which is a lone pair. This gives a trigonal planar electron-group arrangement (AX2E), a bent (V-shaped) molecular shape, and an ideal bond angle of 120°. (c) CF4: This molecule has [1 × C(4e–)] + [4 × F(7e–)] = 32 valence electrons. Eight electrons are used to place a single bond between C and each F, leaving 32 – 8 = 24 e– (12 pairs). Use these 12 pairs to complete the octets of the F atoms (C already has an octet). This gives the following Lewis structure: F F

F F

There are four groups of electrons around the C, none of which is a lone pair. This gives a tetrahedral electron-group arrangement (AX4), a tetrahedral molecular shape, and an ideal bond angle of 109.5°. 10.37

Lewis structure

Electron-group arrangement Trigonal planar O

Molecular shape Trigonal planar

Ideal bond angle 120°

Linear

180°

Tetrahedral

109.5°

S O

Linear N

Tetrahedral

10-28

H Cl

Cl Cl

C Cl

H H

10.38

Plan: Examine the structure shown, and then apply VSEPR. Solution: a) This structure shows three electron groups with three bonds around the central atom. There appears to be no distortion of the bond angles so the shape is trigonal planar, the classification is AX3, with an ideal bond angle of 120°. b) This structure shows three electron groups with three bonds around the central atom. The bonds are distorted down indicating the presence of a lone pair. The shape of the molecule is trigonal pyramidal and the classification is AX3E, with an ideal bond angle of 109.5°. c) This structure shows five electron groups with five bonds around the central atom. There appears to be no distortion of the bond angles so the shape is trigonal bipyramidal and the classification is AX5, with ideal bond angles of 90° and 120°.

10.39

a) This structure shows five electron groups with five bonds around the central atom. There appears to be no distortion of the bond angles so the shape is square pyramidal (in reality square pyramidal structures have a slight distortion of the bond angles because there is a lone pair across from the atom at the apex of the pyramid). The classification is AX5E, with an ideal bond angle of 90°. b) This structure shows three electron groups with three bonds around the central atom. There appears to be no distortion of the bond angles so the shape is T shaped (in reality T-shaped structures have a slight distortion of the bond angles to the apical bonds because there are two equatorial lone pairs). The classification is AX3E2, with an ideal bond angle of 90°. c) This structure shows four electron groups with four bonds around the central atom. There appears to be no distortion of the bond angles so the shape is tetrahedral and the classification is AX4, with an ideal bond angle of 109.5°.

10.40

Plan: The Lewis structures must be drawn, and VSEPR applied to the structures. Lone pairs on the central atom generally result in a deviation of the ideal bond angle. Solution: a) The ClO2– ion has [1 × Cl(7e–)] + [2 × O(6e–)] + [1e– (from charge)] = 20 valence electrons. Four electrons are used to place a single bond between the Cl and each O, leaving 20 – 4 = 16 electrons (eight pairs). All eight pairs are used to complete the octets of the Cl and O atoms. There are two bonds (to the O atoms) and two lone pairs on the Cl for a total of four electron groups (AX2E2). The structure is based on a tetrahedral electron-group arrangement with an ideal bond angle of 109.5°. The shape is bent (or V shaped). The presence of the lone pairs will cause the remaining angles to be less than 109.5°. b) The PF5 molecule has [1 × P(5 e–)] + [5 × F(7 e–)] = 40 valence electrons. Ten electrons are used to place single bonds between P and each F atom, leaving 40 – 10 = 30 e– (15 pairs). The 15 pairs are used to complete the octets of the F atoms. There are five bonds to the P and no lone pairs (AX 5). The electron-group arrangement and the shape is trigonal bipyramidal. The ideal bond angles are 90° and 120°. The absence of lone pairs means the angles are ideal. c) The SeF4 molecule has [1 × Se(6e–)] + [4 × F(7e–)] = 34 valence electrons. Eight electrons are used to place single bonds between Se and each F atom, leaving 34 – 8 = 26e– (13 pairs). Twelve pairs are used to complete the octets of the F atoms which leaves one pair of electrons. This pair is placed on the central Se atom. There are four bonds to the Se which also has a lone pair (AX4E). The structure is based on a trigonal bipyramidal structure with ideal angles of 90° and 120°. The shape is seesaw. The presence of the lone pairs means the angles are less than ideal.

10-29

d) The KrF2 molecule has [1 × Kr(8e–)] + [2 × F(7e–)] = 22 valence electrons. Four electrons are used to place a single bond between the Kr atom and each F atom, leaving 22 – 4 = 18 e– (nine pairs). Six pairs are used to complete the octets of the F atoms. The remaining three pairs of electrons are placed on the Kr atom. The Kr is the central atom. There are two bonds to the Kr and three lone pairs (AX2E3). The structure is based on a trigonal bipyramidal structure with ideal angles of 90° and 120°. The shape is linear. The placement of the F atoms makes their ideal bond angle to be 2 × 90° = 180°. The placement of the lone pairs is such that they cancel each other’s repulsion, thus the actual bond angle is ideal.

F F

Se F

10.41

F F

a) The ClO3– ion has 26 valence electrons. The Cl is the central atom. There are three bonds (to the O atoms) and one lone pair on the Cl (AX3E). The shape is trigonal pyramidal. The structure is based on a tetrahedral electrongroup arrangement with an ideal bond angle of 109.5°. The presence of the lone pair will cause the remaining angles to be less than 109.5°. b) The IF4– ion has 36 valence electrons. The I is the central atom. There are four bonds to the I and two lone pairs (AX4E2). The shape is square planar. The structure is based on an octahedral electron-group arrangement with ideal bond angles of 90°. The repulsion from the two lone pairs cancels so the angles are ideal. c) The SeOF2 molecule has 26 valence electrons. The Se is the central atom. There are three bonds to the Se which also has a lone pair (AX3E). The shape is trigonal pyramidal. The structure is based on a tetrahedral structure with ideal angles of 109.5°. The presence of the lone pair means the angles are less than ideal. d) The TeF5– ion has 42 valence electrons. The Te is the central atom. There are five bonds to the Te which also has one lone pair (AX5E). The shape is square pyramidal. The structure is based on an octahedral with ideal angles of 90°. The presence of the lone pair means the angles are less than ideal. F O

I F

F F

10-30

F Cl

F Se

10.42

O F

Plan: The Lewis structures must be drawn, and VSEPR applied to the structures. Solution: a) CH3OH: This molecule has [1 × C(4e–)] + [4 × H(1e–)] + [1 × O(6e–)] = 14 valence electrons. In the CH3OH molecule, both carbon and oxygen serve as central atoms. (H can never be central.) Use eight electrons to place a single bond between the C and the O atom and three of the H atoms and another two electrons to place a single bond between the O and the last H atom. This leaves 14 – 10 = 4 e– (two pairs). Use these two pairs to complete the octet of the O atom. C already has an octet and each H only gets two electrons. The carbon has four bonds and no lone pairs (AX4), so it is tetrahedral with no deviation (no lone pairs) from the ideal angle of 109.5°. The oxygen has two bonds and two lone pairs (AX2E2), so it is V shaped or bent with the angles less than the ideal angle of 109.5°. H H H

H H

O H

H b) N2O4: This molecule has [2 × N(5e–)] + [4 × O(6e–)] = 34 valence electrons. Use ten electrons to place a single bond between the two N atoms and between each N and two of the O atoms. This leaves 34 – 10 = 24e– (twelve pairs). Use the twelve pairs to complete the octets of the oxygen atoms. Neither N atom has an octet, however. Form a double bond from one O atom to one N atom by changing a lone pair on the O to a bonding pair on the N. Do this for the other N atom as well. In the N2O4 molecule, both nitrogen atoms serve as central atoms. This is the arrangement given in the problem. Both nitrogen atoms are equivalent with three groups and no lone pairs (AX3), so the arrangement is trigonal planar with no deviation (no lone pairs) from the ideal angle of 120°. The same results arise from the other resonance structures.

O O

10.43

O N

a) In the H3PO4 molecule the P and each of the O atoms with an H attached serve as central atoms. The P has four groups and no lone pairs (AX4), so it is tetrahedral with no deviation from the ideal angle of 109.5°. The H bearing O atoms have two bonds and two lone pairs (AX2E2), so the arrangement is V shaped or bent with angles less than the ideal value of 109.5°. O O

O H

P H

O O

H Copyright McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

10-31

b) In the CH3OCH2CH3 molecule, all atoms except the hydrogen atoms serve as central atoms. All the carbons have four bonds and no lone pairs (AX4), so they are tetrahedral with no deviation from the ideal bond angle of 109.5°. The oxygen has two bonds and two lone pairs (AX2E2), so the arrangement is V shaped or bent with angles less than the ideal value of 109.5°. H H H H H O H C C H C O C C H H H C H H H H H H

10.44

Plan: The Lewis structures must be drawn, and VSEPR applied to the structures. Solution: a) CH3COOH has [2 × C(4e–)] + [4 × H(1e–)] + [2 × O(6e–)] = 24 valence electrons. Use fourteen electrons to place a single bond between all of the atoms. This leaves 24 – 14 = 10 e– (five pairs). Use these five pairs to complete the octets of the O atoms; the C atom bonded to the H atoms has an octet but the other C atom does not have a complete octet. Form a double bond from the O atom (not bonded to H) to the C by changing a lone pair on the O to a bonding pair on the C. In the CH3COOH molecule, the carbons and the O with H attached serve as central atoms. The carbon bonded to the H atoms has four groups and no lone pairs (AX4), so it is tetrahedral with no deviation from the ideal angle of 109.5°. The carbon bonded to the O atoms has three groups and no lone pairs (AX3), so it is trigonal planar with no deviation from the ideal angle of 120°. The H bearing O has two bonds and two lone pairs (AX2E2), so the arrangement is V shaped or bent with an angle less than the ideal value of 109.5°.

O H

C C

H O

H b) H2O2 has [2 × H(1e–)] + [2 × O(6e–)] = 14 valence electrons. Use six electrons to place single bonds between the O atoms and between each O atom and an H atom. This leaves 14 – 6 = 8 e– (four pairs). Use these four pairs to complete the octets of the O atoms. In the H2O2 molecule, both oxygen atoms serve as central atoms. Both O atoms have two bonds and two lone pairs (AX2E2), so they are V shaped or bent with angles less than the ideal value of 109.5°.

10.45

O H

a) In the H2SO3 molecule, the S and the O atoms with an H attached serve as central atoms. The S has three groups and one lone pair (AX3E), so it is trigonal pyramidal with angles less than the ideal angle of 109.5°. The H bearing O atoms each have two bonds and two lone pairs (AX2E2), so the arrangement is V shaped or bent with an angle less than the ideal value of 109.5°. O H S O O S O H O O H H

10-32

b) The N2O3 molecule has the structure indicated in the problem with the N atoms serving as central atoms. The nitrogen labeled N1 has two groups and a lone pair (AX2E), so it is V shaped or bent with angles less than the ideal value of 120°. The nitrogen labeled N2 has three bonds and no lone pairs (AX3), so it is trigonal planar with no deviation from the ideal angle of 120°. O

O 10.46

N O

Plan: First, draw a Lewis structure, and then apply VSEPR. The presence of lone pairs on the central atom generally results in a smaller than ideal bond angle. Solution: F F

B F

C F

N F

O F

120o 180o 109.5o< 109.5o<< 109.5o Bond angles: OF2< NF3< CF4< BF3< BeF2 BeF2 is an AX2 type molecule, so the angle is the ideal 180°. BF3 is an AX3 molecule, so the angle is the ideal 120°. CF4, NF3, and OF2 all have tetrahedral electron-group arrangements of the following types: AX4, AX3E, and AX2E2, respectively. The ideal tetrahedral bond angle is 109.5°, which is present in CF4. The one lone pair in NF3 decreases the angle a little. The two lone pairs in OF2 decrease the angle even more. 10.47

Bond angles: SiCl4> PCl3> SCl2> OCl2> SiCl62– All the species except SiCl62– are based on a tetrahedral electron-group arrangement. SiCl62– has an octahedral electron arrangement with an ideal angle of 90°. The tetrahedral arrangement has an ideal bond angle of 109.5°, which is present in AX4 species like SiCl4. The ideal tetrahedral bond angle is reduced slightly by the lone pair in AX3E species such as PCl3. A greater reduction in the ideal tetrahedral bond angle is present in AX2E2 species such as SCl2 and OCl2 with two lone pairs. The angle is reduced less around the larger S atom. 10.48

Plan: The ideal bond angles depend on the electron-group arrangement. Deviations depend on lone pairs. Solution: a) The C and N have three groups, so they are ideally 120°, and the O has four groups, so ideally the angle is 109.5°. The N and O have lone pairs, so the angles are less than ideal. b) All central atoms have four pairs, so ideally all the angles are 109.5°. The lone pairs on the O reduce this value.

10-33

c) The B has three groups (no lone pairs) leading to an ideal bond angle of 120°. All the O atoms have four pairs (ideally 109.5°), two of which are lone, and reduce the angle. 10.49

a) The N has three groups, no lone pairs, so the angle is ideal, and equal to 120°. The O, attached to the H, has four groups (ideally 109.5°); the lone pairs reduce the bond angle from ideal. b) The C, attached to the O, has three groups and no lone pairs so the angle will be the ideal 120°. The remaining C has four groups, and with no lone pairs the angle will be ideal and equal to 109.5°. c) The C with three groups will have angles that are ideal (120°). The O, with the H attached, has four groups. The presence of four groups gives an ideal angle of 109.5°, which is reduced by the lone pairs.

10.50

a) Type: AX2E

Shape: bent

Ideal angle: 120° H

Actual angle: <120 (because of the lone pair)

H H

Shape: trigonal pyramidal

Ideal angle: 109.5° Cl

H H b) Type: AX3E

Actual angle: <109.5 (because of the lone pair) Cl

Sn Cl

Cl Cl

c) Type: AX4

Shape: tetrahedral

Ideal angle: 109.5°

Actual angle: 109.5 (there are no lone pairs) CH3

CH3 H3C

CH3

Sn H3C

CH3

CH3 d) Type: AX5 Ideal angles: 120° and 90°

F F

Shape: trigonal pyramidal Actual angle: 120° and 90° (there are no lone pairs)

F Sn

CH3

F F

e) Type: AX6 Ideal angles: 90°

Shape: octahedral Actual angle: 90° (there are no lone pairs)

10-34

10.51

Plan: The Lewis structures are needed to predict the ideal bond angles. Solution: The P atoms have no lone pairs in any case so the angles are ideal. PCl4+:

PCl5:

Cl P

Cl Cl

PCl6–:

Cl Cl

P Cl

Cl Cl

The original PCl5 is AX5, so the shape is trigonal bipyramidal, and the angles are 120° and 90°. The PCl4+ is AX4, so the shape is tetrahedral, and the angles are 109.5°. The PCl6– is AX6, so the shape is octahedral, and the angles are 90°. Half the PCl5 (trigonal bipyramidal, 120° and 90°) become tetrahedral PCl4+ (tetrahedral, 109.5°), and the other half become octahedral PCl6– (octahedral, 90°). 10.52

Molecules are polar if they have polar bonds that are not arranged to cancel each other. A polar bond is present any time there is a bond between elements with differing electronegativities.

10.53

Molecules are polar if they have polar bonds that are not arranged to cancel each other. If the polar covalent bonds are arranged in such a way as to cancel each other, the molecule will be nonpolar. An example of a molecule with polar covalent bonds that is not polar is SO3. The trigonal planar shape causes the three polar S–O bonds to cancel. O

S O 10.54

Molecules must come together to react. This becomes difficult for large molecules. Biomolecules are generally large molecules and have difficulty reacting if the shapes of the molecules are not compatible.

10-35

10.55

Plan: To determine if a bond is polar, determine the electronegativity difference of the atoms participating in the bond. The greater the electronegativity difference, the more polar the bond. To determine if a molecule is polar (has a dipole moment), it must have polar bonds, and a certain shape determined by VSEPR. Solution: a)Molecule Bond Electronegativities Electronegativity difference SCl2 S–Cl S = 2.5 Cl = 3.0 3.0 – 2.5 = 0.5 F2 F–F F = 4.0 F = 4.0 4.0 – 4.0 = 0.0 CS2 C–S C = 2.5 S = 2.5 2.5 – 2.5 = 0.0 CF4 C–F C = 2.5 F = 4.0 4.0 – 2.5 = 1.5 BrCl Br–Cl Br = 2.8 Cl = 3.0 3.0 – 2.8 = 0.2 The polarities of the bonds increase in the order: F–F = C–S < Br–Cl < S–Cl < C–F. Thus, CF4 has the most polar bonds. b) The F2 and CS2 cannot be polar since they do not have polar bonds. CF4 is an AX4 molecule, so it is tetrahedral with the four polar C–F bonds arranged to cancel each other giving an overall nonpolar molecule. BrCl has a dipole moment since there are no other bonds to cancel the polar Br–Cl bond. SCl2 has a dipole moment (is polar) because it is a bent molecule, AX2E2, and the electron density in both S–Cl bonds is pulled toward the more electronegative chlorine atoms.

F S C

F F

nonpolar 10.56

Cl polar

a) The greater the difference in electronegativity the more polar the bond: Molecule Bond Electronegativities Electronegativity difference BF3 B–F B = 2.0 F = 4.0 4.0 – 2.0 = 2.0 PF3 P–F P = 2.1 F = 4.0 4.0 – 2.1 = 1.9 BrF3 Br–F Br = 2.8 F = 4.0 4.0 – 2.8 = 1.2 SF4 S–F S = 2.5 F = 4.0 4.0 – 2.5 = 1.5 SF6 S–F S = 2.5 F = 4.0 4.0 – 2.5 = 1.5 The polarities of the bonds are increasing in the order: Br–F < S–F < P–F < B–F. Thus, BF3 has the most polar bonds. b) All the molecules meet the requirement of having polar bonds. The arrangement of the bonds must be considered in each case. BF3 is trigonal planar, AX3, so it is nonpolar because the polarities of the bonds cancel. PF3 has a dipole moment (is polar) because it has a trigonal pyramidal geometry, AX3E. BrF3 has a dipole moment because it has a T-shaped geometry, AX3E2. SF4 has a dipole moment because it has a see-saw geometry, AX4E. SF6 is nonpolar because it is octahedral, AX6, and the bonds are arranged so they cancel.

10-36

B F

P F

F F

S F 10.57

S F

Plan: If only two atoms are involved, only an electronegativity difference is needed. The greater the difference in electronegativity, the more polar the bond. If there are more than two atoms, the molecular geometry must be determined. Solution: a) All the bonds are polar covalent. The SO3 molecule is trigonal planar, AX3, so the bond dipoles cancel leading to a nonpolar molecule (no dipole moment). The SO2 molecule is bent, AX2E, so the polar bonds result in electron density being pulled toward one side of the molecule. SO2 has a greater dipole moment because it is the only one of the pair that is polar. O S S O

b) ICl and IF are polar, as are all diatomic molecules composed of atoms with differing electronegativities. The electronegativity difference for ICl (3.0 – 2.5 = 0.5) is less than that for IF (4.0 – 2.5 = 1.5). The greater difference means that IF has a greater dipole moment. c) All the bonds are polar covalent. The SiF4 molecule is nonpolar (has no dipole moment) because the bonds are arranged tetrahedrally, AX4. SF4 is AX4E, so it has a see-saw shape, where the bond dipoles do not cancel. SF4 has the greater dipole moment.

F F

Si F

F F

S F

F F

Si F

F F

d) H2O and H2S have the same basic structure. They are both bent molecules, AX2E2, and as such, they are polar. The electronegativity difference in H2O (3.5 – 2.1 = 1.4) is greater than the electronegativity difference in H2S (2.5 – 2.1 = 0.4) so H2O has a greater dipole moment. 10.58

a) All the bonds are polar covalent. Both the molecules are bent (SO2 and ClO2 are AX2E2). The difference in electronegativity is greater in SO2 than in ClO2 so SO2 has a greater dipole moment.

10-37

b) HBr and HCl are polar, as are all diatomic molecules composed of atoms with differing electronegativities. The electronegativity difference for HBr is less than that for HCl. The greater difference means that HCl has a greater dipole moment. c) All the bonds are polar covalent. The BeCl2 molecule is nonpolar (has no dipole moment) because the bonds are arranged linearly, AX2. SCl2 is AX2E2, so it has a bent shape, where the bond dipoles do not cancel. SCl2 has the greater dipole moment. d) All the bonds are polar covalent. AsF5 is AX5, so it is trigonal bipyramidal and nonpolar. AsF3 is AX3E, so it is trigonal pyramidal and polar. AsF3 has a greater dipole moment. 10.59

Plan: Draw Lewis structures, and then apply VSEPR. A molecule has a dipole moment if polar bonds do not cancel. Solution: C2H2Cl2 has [2 × C(4e–)] + [2 × H(1e–)] + [2 × Cl(7e–)] = 24 valence electrons. The two carbon atoms are bonded to each other. The H atoms and Cl atoms are bonded to the C atoms. Use 10 electrons to place a single bond between all of the atoms. This leaves 24 – 10 = 14e– (seven pairs). Use these seven pairs to complete the octets of the Cl atoms and one of the C atoms; the other C atom does not have a complete octet. Form a double bond between the carbon atoms by changing the lone pair on one C atom to a bonding pair. There are three possible structures for the compound C2H2Cl2: H

Cl C

C Cl

I II III The presence of the double bond prevents rotation about the C=C bond, so the structures are “fixed.” The C–Cl bonds are more polar than the C–H bonds, so the key to predicting the polarity is the positioning of the C–Cl bonds. Structure I has the C–Cl bonds arranged so that they cancel leaving I as a nonpolar molecule. Both II and III have C–Cl bonds on the same side so the bonds work together making both molecules polar. Both I and II will react with H2 to give a compound with a Cl attached to each C (same product). Structure III will react with H2 to give a compound with two Cl atoms on one C and none on the other (different product). Structure I must be X as it is the only one that is nonpolar (has no dipole moment). Structure II must be Z because it is polar and gives the same product as compound X. This means that Structure III must be the remaining compound, Y. Compound Y (III) has a dipole moment. 10.60

The possible structures for the compounds differ only in the positions of the N–F bonds. These structures are “fixed” because the N=N bond does not allow rotation. The N–F bonds are polar. a) F N

N F

(trans) (cis) b) In the trans-form the N–F bonds pull equally in opposite directions, thus, they cancel and the molecule is nonpolar. The N–F bonds in the cis-form pull in the same general direction resulting in a polar molecule. 10.61

Plan: The Lewis structures are needed to do this problem. A single bond (bond order = 1) is weaker and longer than a double bond (bond order = 2) which is weaker and longer than a triple bond (bond order = 3). To find the heat of reaction, add the energy required to break all the bonds in the reactants to the energy released to form all bonds in the product. Remember to use a negative sign for the energy of the bonds formed since bond formation is exothermic. The bond energy values are found in Table 9.2.

10-38

Solution: a) The H atoms cannot be central, and they are evenly distributed on the N atoms. N2H4 has [2 × N(5e–)] + [4 × H(1e–)] = 14 valence electrons, 10 of which are used in the bonds between the atoms. The remaining two pairs are used to complete the octets of the N atoms. N2H2 has [2 × N(5e–)] + (2 × H(1e–)] = 12 valence electrons, six of which are used in the bonds between the atoms. The remaining three pairs of electrons are not enough to complete the octets of both N atoms, so one lone pair is moved to a bonding pair between the N atoms. N2 has [2 × N(5 e–)] = 10 valence electrons, two of which are used to place a single bond between the two N atoms. Since only four pairs of electrons remain and six pairs are required to complete the octets, two lone pairs become bonding pairs to form a triple bond.

Nitrogen Diazene Hydrazine The single (bond order = 1)N–N bond is weaker and longer than any of the others are. The triple bond (bond order = 3) is stronger and shorter than any of the others. The double bond (bond order = 2) has an intermediate strength and length. b) N4H4 has [4 × N(5e–)] + [4 × H(1e–)] = 24 valence electrons, 14 of which are used for single bonds between the atoms. When the remaining five pairs are distributed to complete the octets, one N atom lacks two electrons. A lone pair is moved to a bonding pair for a double bond. H

 H bonds broken +

H +

H H Reactant bonds broken: 4 N–H = 4 mol (391 kJ/mol) = 1564 kJ 2 N–N = 2 mol (160 kJ/mol) = 320 kJ 1 N=N = 1 mol (418 kJ/mol) = 418 kJ  H bonds broken = 2302 kJ  H rxn =

Product bonds formed: 4 N–H = 4 mol (–391 kJ/mol) = –1564 kJ 1 N–N = 1 mol (–160 kJ/mol) = –160 kJ 1 NN = 1 mol (–945 kJ/mol) = –945 kJ  H bonds formed = –2669 kJ

 H bonds formed = 2302 kJ + (–2669 kJ) = –367 kJ

10.62

F F

F P

(a)

(b)

(c)

Cl (d)

H (e)

(f)

10-39

Br Br

Br (g)

(h)

(i)

F F

10.63

(j)

(k)

(l)

a) AX5 = trigonal bipyramidal

b) AX4 = tetrahedral

F F

F F c) AX3E = trigonal pyramidal

Cl Cl

d) AX3E2 = T shaped

e) AX2 = linear H

C Cl

f) AX2E2 = bent H

g) AX4 = tetrahedral

h) AX3E = trigonal pyramidal

Br Ge Br

Br Br

C H

H H

10-40

i) AX3 = trigonal planar

j) AX4E = seesaw

F Br

B Cl

k) AX3E = trigonal pyramidal

l) AX4E = seesaw

F F

Xe O

10.64

a) SiF4 with its 32 valence electrons is an AX4 molecule and has a tetrahedral molecular shape. SiF5– with its 40 valence electrons is an AX5 ion and has a trigonal bipyramidal molecular shape. B best represents the change in molecular shape from tetrahedral to trigonal bipyramidal. b) SiF4: tetrahedral, AX4; SiF5–: trigonal bipyramidal, AX5.

10.65

Plan: Use the Lewis structures shown in the text. The equation for formal charge (FC) is FC = no. of valence electrons – [no. of unshared valence electrons + no. of shared valence electrons]. Solution: a) Formal charges for Al2Cl6: FCAl = 3 – [0 + (8)] = –1 FCCl, ends = 7 – [6 + (2)] = 0 FCCl, bridging = 7 – [4 + (4)] = +1 (Check: Formal charges add to zero, the charge on the compound.) Formal charges for I2Cl6: FCI = 7 – [4 + (8)] = –1 FCCl, ends = 7 – [6 + (2)] = 0 FCCl, bridging = 7 – [4 + (4)] = +1 (Check: Formal charges add to zero, the charge on the compound.) b) The aluminum atoms have no lone pairs and are AX4, so they are tetrahedral. The two tetrahedral Al atoms cannot give a planar structure. The iodine atoms in I2Cl6 have two lone pairs each and are AX4E2 so they are square planar. Placing the square planar I atoms adjacent can give a planar molecule.

10.66

The Lewis structure for each is required. Compound Lewis structure

Molecular geometry

XeF2

Linear (AX2E3) F

10-41

XeF4

Square planar (AX4E2)

F F

F XeF6

Distorted octahedral (AX6E) F F

F Xe

F F 2–

10.67

a) SO3 is an AX3 molecule and has a trigonal planar shape. SO3 is an AX3E species and has a trigonal pyramidal molecular shape. C best illustrates the change in molecular shape from trigonal planar to trigonal pyramidal. b) Yes, there is a change in polarity during the reaction as the nonpolar SO3 molecule becomes the polar SO32– ion.

10.68

From the Lewis structures, both are AX2E which has an ideal bond angle of 120o. But the “lone pair” on N in NO2 is only half a pair, so it only exerts “half” the repulsion. This allows the bond angle to open to a larger than normal bond angle. The “complete” lone pair in NO2–, like other lone pairs, forces the bonding pairs together to give a smaller than normal bond angle.

10.69

Xe(g) + 3F2(g)  XeF6(g)    H rxn = H bonds H bonds broken + formed The three F–F bonds must be broken, and six Xe–F bonds are formed.  H rxn = 3 BEF–F + 6 BEXe–F –402 kJ/mol = (3 mol)(159 kJ/mol) + (6 mol)(–BEXe–F) –879 kJ/mol = 6 (–BEXe–F) 146.5 = 146 kJ/mol = BEXe–F

10.70

Plan: Draw the Lewis structures, and then use VSEPR to describe propylene oxide. Solution: a) H H C

CH2

H2O2

H3C

CH2

H2O

O H

H C

C O

10-42

In propylene oxide, the C atoms are all AX4. The C atoms do not have any unshared (lone) pairs. All of the ideal bond angles for the C atoms in propylene oxide are 109.5° and the molecular shape around each carbon atom is tetrahedral. b) In propylene oxide, the C that is not part of the three-membered ring should have an ideal angle. The atoms in the ring form an equilateral triangle. The angles in an equilateral triangle are 60°. The angles around the two carbons in the rings are reduced from the ideal 109.5° to 60°. 10.71

Cl Cl

O C

Cl Cl

H O

H Cl

O O C

C H

The C with the chlorine atoms attached does not change shape. That C is tetrahedral in both compounds. The other C changes from trigonal planar (AX3) to tetrahedral (AX4).

10.72

a) C

Bond order (avg.) 3.0

Each C–O bond is a single bond two-thirds of the time and a double bond the rest of the time. The average is [(1 + 1 + 2)/3] = 4/3 = 1.33 c) H C H

2.0

H H

C H

H 1.0

10-43

The resonating double bond means the average bond length is [(1 + 2)/2] = 1.5 The C–O bond for the O attached to the H does not resonate and remains 1.0 Bond length a<c<e<b<d ignoring O attached to H in part (e) Bond strength d<b<e<c<a 10.73

The reaction is balanced as usual: ClCH2CH2SCH2CH2Cl + 2H2O  HOCH2CH2SCH2CH2OH + 2HCl Most of the molecule remains the same. Reactant bonds broken: 2 × C–C = (2 mol)(347 kJ/mol) = 694 kJ 8 × C–H = (8 mol)(413 kJ/mol) = 3304 kJ 2 × C–Cl = (2 mol)(339 kJ/mol) = 678 kJ 2 × C–S = (2 mol)(259 kJ/mol) = 518 kJ 4 × O–H = (4 mol)(467 kJ/mol) = 1868 kJ  H bonds broken = 7062 kJ

Product bonds formed: 2 × C–C = (2 mol)(–347 kJ/mol) = –694 kJ 8 × C–H = (8 mol)(–413 kJ/mol) = –3304 kJ 2 × H–Cl = (2 mol)(–427 kJ/mol) = –854 kJ 2 × C–S = (2 mol)(–259 kJ/mol) = –518 kJ 2 × C–O = (2 mol)(–358 kJ/mol) = –716 kJ 2 × O–H = (2 mol)(–467 kJ/mol) = –934 kJ  H bonds formed = –7020 kJ  H rxn =

 H bonds broken +

 H bonds formed = 7062 kJ + (–7020 kJ) = 42 kJ

10.74

CBr4< CH2Br2< CH2Cl2< CF2Cl2< CF2Br2< CH2F2

10.75

Plan: Ethanol burns (combusts) with O2 to produce CO2 and H2O. To find the heat of reaction in part (a), add the energy required to break all the bonds in the reactants to the energy released to form all bonds in the product. Remember to use a negative sign for the energy of the bonds formed since bond formation is exothermic. The bond energy values are found in Table 9.2. The heat of vaporization of ethanol must be included for part (b). The enthalpy change in part (c) is the sum of the heats of formation of the products minus the sum of the heats of formation of the reactants. The calculation for part (d) is the same as in part (a).

10-44

Solution: a) CH3CH2OH(g) + 3O2(g) H H

+ 3

(g) + 3H2O(g)

2 O

+ 3

Reactant bonds broken: 1 × C–C = (1 mol)(347 kJ/mol) = 347 kJ 5 × C–H = (5 mol)(413 kJ/mol) = 2065 kJ 1 × C–O = (1 mol)(358 kJ/mol) = 358 kJ 1 × O–H = (1 mol)(467 kJ/mol) = 467 kJ 3 × O=O = (3 mol)(498 kJ/mol) = 1494 kJ  H bonds broken = 4731 kJ Product bonds formed: 4 × C=O = (4 mol)(–799 kJ/mol) = –3196 kJ 6 × O–H = (6 mol)(–467 kJ/mol) = –2802 kJ  H bonds formed = –5998 kJ  H rxn =

 H bonds broken +

 H bonds formed = 4731 kJ + (–5998 kJ) = –1267 kJ for each mole of ethanol burned

b) If it takes 40.5 kJ/mol to vaporize the ethanol, part of the heat of combustion must be used to convert liquid ethanol to gaseous ethanol. The new value becomes:  40.5 kJ   Hcombustion  = –1226.5 = –1226 kJ per mole of liquid ethanol burned (liquid) = –1267 kJ + 1 mol    1 mol   c) H rxn = m H f(products) – n H f (reactants)  H rxn = {2 H f [CO2(g)] + 3 H f [H2O(g)]} – {1 H f [C2H5OH(l)] + 3 H f [O2(g)]}

= [(2 mol)(–393.5 kJ/mol) + (3 mol)(–241.826 kJ/mol)] – [(1 mol)(–277.63 kJ/mol) + 3 mol(0 kJ/mol)] = –1234.848 = –1234.8 kJ The two answers differ by less than 10 kJ. This is a very good agreement since average bond energies were used to calculate the answers in (a) and (b). d) C2H4(g) + H2O(g) CH2OH(g) 3 The Lewis structures for the reaction are: H H H H O C C H C C O + H H H H H H H Reactant bonds broken: 1 × C=C = (1 mol)(614 kJ/mol) = 614 kJ 4 × C–H = (4 mol)(413 kJ/mol) = 1652 kJ 2 × O–H = (2 mol)(467 kJ/mol) = 934 kJ  H bonds broken = 3200 kJ

10-45

Product bonds formed: 1 × C–C = (1 mol)(–347 kJ/mol) = –347 kJ 5 × C–H = (5 mol)(–413 kJ/mol) = –2065 kJ 1 × C–O = (1 mol)(–358 kJ/mol) = –358 kJ 1 × O–H = (1 mol)(–467 kJ/mol) = –467 kJ  H bonds formed = –3237 kJ  H rxn =

10.76

 H bonds broken +

 H bonds formed = 3200 kJ + (–3237 kJ) = –37 kJ

a) C3H4: 16 valence electrons. H H

C Bond Order 1 C

Bond Order 2 b) C3H6: 18 valence electrons. H H C Bond Order 1 (all) H C C H

H H c) C4H6: 22 valence electrons. H H H

H Bond Order 1

Bond Order 2 d) C4H4: 20 valence electrons. Resonance exists. H H H C

H C

C Bond Order 1.5 (all)

C H

10-46

e) C6H6: 30 valence electrons. Resonance exists. H

C C

H C Bond Order 1.5 (all)

C H

H 10.77

C C

Plan: Determine the empirical formula from the percent composition (assuming 100 g of compound). Use the titration data to determine the mole ratio of acid to the NaOH. This ratio gives the number of acidic H atoms in the formula of the acid. Finally, combine this information to construct the Lewis structure. Solution:  1 mol   = 2.222 mol H Moles of H = 2.24 g H 1.008 g H   1 mol   = 2.223 mol C Moles of C = 26.7 g C 12.01 g C   1 mol   = 4.444 mol O Moles of O =  71.1 g O 16.00 g O  The preliminary formula is H2.222C2.223O4.444. Dividing all subscripts by the smallest subscript to obtain integer subscripts: H 2.222 C 2.223 O 4.444 = HCO2 2.222

2.222

The empirical formula is HCO2. Calculate the amount of NaOH required for the titration:  1 L   0.040 mol NaOH   1 mmol  = 2.0 mmol NaOH mmoles of NaOH = 50.0 mL    0.001 mol   1000 mL   L Thus, the ratio is 2.0 mmole base/1.0 mmole acid, or each acid molecule has two hydrogen atoms to react (diprotic). The empirical formula indicates a monoprotic acid, so the formula must be doubled to: H 2C2O4. H2C2O4 has [2 × H(1e–)] + [2 × C(4e–)] + [4 × O(6e–)] = 34 valence electrons to be used in the Lewis structure. Fourteen of these electrons are used to bond the atoms with single bonds, leaving 34 – 14 = 20 electrons or 10 pairs of electrons. When these 10 pairs of electrons are distributed to the atoms to complete octets, neither C atom has an octet; a lone pair from the oxygen without hydrogen is changed to a bonding pair on C.

O C

O 10.78

C O

Determine the empirical formula from the percent composition (assuming 100 g of compound). The molar mass may be determined from the density of the gas. The empirical formula and the molar mass may then be used to determine the molecular formula. Count the valence electrons in the empirical formula and then construct the Lewis structure.

10-47

 1 mol   = 2.065 mol C Moles of C = 24.8 g C 12.01 g C   1 mol   = 2.063 mol H Moles of H = 2.08 g H 1.008 g H   1 mol   = 2.062 mol Cl Moles of Cl =  73.1 g Cl  35.45 g Cl  The preliminary formula is C2.065H2.063Cl2.062. Dividing all subscripts by the smallest subscript to obtain integer subscripts: C 2.065 H 2.063 Cl 2.062 = CHCl 2.062

2.062

Thus, the empirical formula is CHCl, and its molar mass is 48.47 g/mol. The density is 4.3 g/L at STP:

d=P

/RT

Rearranging to solve for molar mass:  L  atm  4.3 g/L 0.0821 273 K  dRT  mol  K  = = = 96.37719 = 96 g/mol P 1.00 atm









The molar mass (96 g/mol) is double the empirical formula mass (48.47 g/mol), so the empirical formula must be doubled to get the molecular formula: C2H2Cl2. The formula contains 24 valence electrons. A variety of structures may be drawn: H

Cl C

10.79

H C

Cl or

H C

There are 32 valence electrons present. Begin four Lewis structures by placing a Cl in the center and four O atoms around it. Connect all the O atoms to the central Cl with single bonds. In the second structure, convert one of the single bonds to a double bond. In the third structure, two of the bonds are double, and in the last, three of the bonds are double. (It does not matter which bonds are chosen to become double bonds.) O O

O O

FCCl = 7 – [0 + (8)] FCCl = +3

FCCl = 7 – [0 + (10)] FCCl = +2

FCCl = 7 – [0 + (12)] FCCl = +1

FCCl = 7 – [0 + (14)] FCCl = 0

(most important)

10-48

Average bond order of the last structure: 10.80

(3 2)  (11) = 1.75 4

Plan: Write the balanced chemical equations for the reactions and draw the Lewis structures. To find the heat of reaction, add the energy required to break all the bonds in the reactants to the energy released to form all bonds in the product. Remember to use a negative sign for the energy of the bonds formed since bond formation is exothermic. The bond energy values are found in Table 9.2. Divide the heat of reaction by the number of moles of oxygen gas appearing in each reaction to get the heat of reaction per mole of oxygen. Solution: CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) H H

+ 2

H Reactant bonds broken:

4 × C–H = (4 mol)(413 kJ/mol) = 1652 kJ 2 × O=O = (2 mol)(498 kJ/mol) = 996 kJ  H bonds broken = 2648 kJ

Product bonds formed: 2 × C=O = (2 mol)(–799 kJ/mol) = –1598 kJ 4 × O–H = (4 mol)(–467 kJ/mol) = –1868 kJ  H bonds formed = –3466 kJ  H rxn =

 H bonds broken +

 H bonds formed = 2648 kJ + (–3466 kJ) = –818 kJ for 2 mol O2

Per mole of O2 = –818/2 = –409 kJ/mol O2 2H2S(g) + 3O2(g)  2SO2(g) + 2H2O(g)

S O

Reactant bonds broken: 4 × S–H = (4 mol)(347 kJ/mol) = 1388 kJ 3 × O=O = (3 mol)(498 kJ/mol) = 1494 kJ  H bonds broken = 2882 kJ Product bonds formed: 4 × S=O = (4 mol)(–552 kJ/mol) = –2208 kJ 4 × O–H = (4 mol)(–467 kJ/mol) = –1868 kJ  H bonds formed = –4076 kJ  H rxn =

 H bonds broken +

 H bonds formed = 2882 kJ + (–4076 kJ) = –1194 kJ for 3 mol O2

Per mole of O2 = –1194/3 = –398 kJ/mol O2 Copyright McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

10-49

10.81 F F

F (a)

(b)

(c)

F S

F F

(d)

(e)

Stable: a, c, and e Unstable radicals: b and d 10.82

Plan: Draw the Lewis structure of the OH species. The standard enthalpy of formation is the sum of the energy required to break all the bonds in the reactants and the energy released to form all bonds in the product. Remember to use a negative sign for the energy of the bonds formed since bond formation is exothermic. The bond energy values are found in Table 9.2. Solution: a) The OH molecule has [1 × O(6e–)] + [1 × H(1e–)] = 7 valence electrons to be used in the Lewis structure. Two of these electrons are used to bond the atoms with a single bond, leaving 7 – 2 = 5 electrons. Those five electrons are given to oxygen. But no atom can have an octet, and one electron is left unpaired. The Lewis structure is: O

b) The formation reaction is: 1/2O2(g) + 1/2H2(g)  OH(g). The heat of reaction is:  H rxn =

[

 H bonds broken +

(BEO=O) +

 H bonds formed = 39.0 kJ

(BEH–H)] + [BEO–H] = 39.0 kJ

[( mol)(498 kJ/mol) + ( mol)(432 kJ/mol)] + [BEO–H] = 39.0 kJ 465 kJ + [BEO–H] = 39.0 kJ BEO–H = –426 kJ or 426 kJ c) The average bond energy (from the bond energy table) is 467 kJ/mol. There are two O–H bonds in water for a total of 2 × 467 kJ/mol = 934 kJ. The answer to part (b) accounts for 426 kJ of this, leaving: 934 kJ – 426 kJ = 508 kJ 10.83

Both N3– and HN3 have 16 valence electrons. Azide ion: N

10-50

There are three resonance structures for the N3– ion. The formal charges in the first structure are, from left to right, –1, +1, and –1. In the other two Lewis structures the single-bonded N has a formal charge of –2, making both of these less stable than the first structure. The central N is +1 and the triple bonded N is 0. The first resonance structure is more important; the structure should have two equal bonds with a bond order of 2. Hydrazoic acid:

HN3 also has three resonance structures. The formal charge for the H is 0 in all the structures. In the structure with two double bonds, the formal charges for the N atoms are, left to right: 0, +1, and –1. The structure where the H is attached to the single-bonded N, has N atoms with the following formal charges: –1, +1, and 0. In the final Lewis structure, the formal charges on the N atoms are: +1, +1, and –2. The third structure is clearly not as good as the other two. The first two structures should be averaged to give, starting at the H–end, a bond order of 1.5 then a bond order of 2.5. Thus, the two bonds are unequal. 10.84

Plan: The basic Lewis structure will be the same for all species. The Cl atoms are larger than the F atoms. All of the molecules are of the type AX5 and have trigonal bipyramidal molecular shape. The equatorial positions are in the plane of the triangle and the axial positions above and below the plane of the triangle. In this molecular shape, there is more room in the equatorial positions. Solution: a) The F atoms will occupy the smaller axial positions first so that the larger Cl atoms can occupy the equatorial positions which are less crowded. b) The molecule containing only F atoms is nonpolar (has no dipole moment), as all the polar bonds would cancel. The molecules with one F or one Cl would be polar since the P–F and P–Cl bonds are not equal in polarity and thus do not cancel each other. The presence of two axial F atoms means that their polarities will cancel (as would the three Cl atoms) giving a nonpolar molecule. The molecule with three F atoms is also polar. F

F Cl

Cl Cl

10.85

F F

Polar

Nonpolar No dipole moment

Cl F

F F

Cl F

Polar

F F

Nonpolar No dipole moment

N2O has 16 valence electrons; there are three resonance structures. N

FC –1 +1 0 –2 +1 +1 0 +1 –1 The third structure has a more reasonable distribution of formal charges. The third form has a strong triple bond between the N atoms and a weak N–O bond. It is easy to break the N–O bond which is why this compound easily decomposes to support combustion. 10.86

Plan: Count the valence electrons and draw Lewis structures for the resonance forms. Solution: The H2C2O4 molecule has [2 × H(1e–)] + [2 × C(4e–)] + [4 × O(6e–)] = 34 valence electrons to be used in the Lewis structure. Fourteen of these electrons are used to bond the atoms with a single bond, leaving 34 –14 = 20

10-51

electrons. If these 20 electrons are given to the oxygen atoms to complete their octet, the carbon atoms do not have octets. A lone pair from each of the oxygen atoms without hydrogen is changed to a bonding pair on C. The HC2O4– ion has [1 × H(1e–)] + [2 × C(4e–)] + [4 × O(6e–)] + [1e– (from the charge)] = 34 valence electrons to be used in the Lewis structure. Twelve of these electrons are used to bond the atoms with a single bond, leaving 34 –12 = 22 electrons. If these 22 electrons are given to the oxygen atoms to complete their octet, the carbon atoms do not have octets. A lone pair from two of the oxygen atoms without hydrogen is changed to a bonding pair on C. There are two resonance structures. The C2O42– ion has [2 × C(4e–)] + [4 × O(6e–)] + [2e– (from the charge)] = 34 valence electrons to be used in the Lewis structure. Ten of these electrons are used to bond the atoms with a single bond, leaving 34 –10 = 24 electrons. If these 24 electrons are given to the oxygen atoms to complete their octets, the carbon atoms do not have octets. A lone pair from two oxygen atoms is changed to a bonding pair on C. There are four resonance structures. H2C2O4: HC2O4–:

In H2C2O4, there are two shorter C=O bonds and two longer, weaker C—O bonds. In HC2O4–, the C—O bonds on the side retaining the H remain as one long C—O bond and one shorter, stronger C=O bond. The C—O bonds on the other side of the molecule have resonance forms with an average bond order of 1.5, so they are intermediate in length and strength. In C2O42–, all the carbon to oxygen bonds are resonating and have an average bond order of 1.5. 10.87

The molecule has 42 valence electrons. Thirty electrons are already accounted for in the skeleton structure in the bonds. 42 – 30 = 12 valence electrons remain. If these 12 electrons are given to the two oxygen atoms to complete their octets, the carbon atom that is bonded to the two oxygen atoms does not have an octet. A lone pair from one of the oxygen atoms is changed to a bonding pair on the C. All the atoms have 0 formal charge except the N (FC = +1), and the single bonded O (FC = –1)

10-52

O H

H H

C H

H H

10.88

N2H4(g) + O2(g)  N2(g) + 2H2O(g) H H

Reactant bonds broken: 4 × N–H = (4 mol)(391 kJ/mol) = 1564 kJ 1 × N–N = (1 mol)(160 kJ/mol) = 160 kJ 1 × O=O = (1 mol)(498 kJ/mol) = 498 kJ  H bonds broken = 2222 kJ

Product bonds formed: 1×N –945 kJ/mol) =

–945 kJ

4 × O–H = (4 mol)(–467 kJ/mol) = –1868 kJ  H bonds formed = –2813 kJ  H rxn =

 H bonds broken +

 H bonds formed = 2222 kJ + (–2813 kJ) = –591 kJ per mole of N2H4 reacted

10.89

a) Yes, the black sphere can represent selenium. SeF4 has 34 valence electrons. Eight of these electrons are used in the four Se-F single bonds and 24 electrons are used to complete the octets of the F atoms. The remaining electron pair goes to selenium and the molecule is AX4E. The molecular geometry is the seesaw molecular shape shown. b) Yes, the black sphere can represent nitrogen if the species is an anion with a –1 charge. The NF4– ion has 34 valence electrons and would have the seesaw molecular shape as an AX4E species. c) For BrF4 to have the 34 valence electrons needed for this seesaw molecular geometry, the charge of the species must be +1. BrF4+ would have [1 × Br(7e–)] + [4 × F(7e–)] – [1e– from + charge] = 34 valence electrons.

10.90

Plan: Draw the Lewis structures. Calculate the heat of reaction using the bond energies in Table 9.2. Solution: SO3(g) + H2SO4(l)  H2S2O7(l)

10-53

O S

O O

O H

S O

O H

Reactant bonds broken: 5 × S=O = (5 mol)(552 kJ/mol) = 2760 kJ 2 × S–O = (2 mol)(265 kJ/mol) = 530 kJ 2 × O–H = (2 mol)(467 kJ/mol) = 934 kJ  H bonds broken = 4224 kJ

Product bonds formed: 4 × S=O = (4 mol)(–552 kJ/mol) = –2208 kJ 4 × S–O = (4 mol)(–265 kJ/mol) = –1060 kJ 2 × O–H = (2 mol)(–467 kJ/mol) = –934 kJ  H bonds formed = –4202 kJ  H rxn =

10.91

 H bonds broken +

 H bonds formed = 4224 kJ + (–4202 kJ) = 22 kJ

Plan: Pick the VSEPR structures for AY3 substances. Then determine which are polar. Solution: The molecular shapes that have a central atom bonded to three other atoms are trigonal planar, trigonal pyramidal, and T shaped: Y Y

A Y

Y a) b) c) three groups four groups five groups (AX3) (AX3E) (AX3E2) trigonal planar trigonal pyramidal T shaped Trigonal planar molecules, such as (a), are nonpolar, so it cannot be AY3. Trigonal pyramidal molecules (b) and T-shaped molecules (c) are polar, so either could represent AY3. 10.92

a) Shape A is T-shaped (AX3E2); Shape B is trigonal planar (AX3); Shape C is trigonal pyramid (AX3E). XeF3+, with 28 valence electrons, has two unshared pairs on Xe and is AX3E2 and is the T-shaped molecular shape in A. SbBr3, with 26 valence electrons, has one unshared pair on Sb; thus it is AX3E and is the trigonal pyramidal molecular shape in C. GaCl3, with 24 valence electrons, has no unshared pairs on Ga; thus it is AX3 and is the trigonal planar shape in B. b) Shapes A and C are polar. c) Shape A, which is T-shaped, has the most valence electrons (10) around the central atom.

10-54

10.93

The simplified Lewis structures for the reaction are: H

C N + 2H H

Reactant bonds broken: 1 × C–H = (1 mol)(413 kJ/mol) = 413 kJ 1 × CN = (1 mol)(891 kJ/mol) = 891 kJ 2 × H–H = (2 mol)(432 kJ/mol) = 864 kJ  H bonds broken = 2168 kJ

Product bonds formed: 3 × C–H = (3 mol)(–413 kJ/mol) = –1239 kJ 1 × C–N = (1 mol)(–305 kJ/mol) = –305 kJ 2 × N–H = (2 mol)(–391 kJ/mol) = –782 kJ  H bonds formed = –2326 kJ

 H rxn =

10.94

 H bonds broken +

 H bonds formed = 2168 kJ + (–2326 kJ) = –158 kJ

a) H

H C

F C

All the carbons are trigonal planar so the ideal angles should all be 120°. b) The observed angles are slightly less than ideal because the C=C bond repels better than the single bonds. The larger F atoms cannot get as close together as the smaller H atoms, so the angles in tetrafluoroethylene are not reduced as much. 10.95

Plan: Draw the Lewis structure of each compound. Atoms 180° apart are separated by the sum of the bond’s length. Atoms not at 180° apart must have their distances determined by geometrical relationships. Solution: (a)

(b) F

(c)

F S

F F

a) C2H2 has [2 × C(4e–)] + [2 × H(1e–)] = 10 valence electrons to be used in the Lewis structure. Six of these electrons are used to bond the atoms with a single bond, leaving 10 – 6 = 4 electrons. Giving one carbon atom the four electrons to complete its octet results in the other carbon atom not having an octet. The two lone pairs from the carbon with an octet are changed to two bonding pairs for a triple bond between the two carbon atoms. The Copyright McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

10-55

molecular shape is linear. The H atoms are separated by two carbon-hydrogen bonds (109 pm) and a carboncarbon triple bond (121 pm). Total separation = 2(109 pm) + 121 pm = 339 pm b) SF6 has [1 × S(6e–)] + [6 × F(7e–)] = 48 valence electrons to be used in the Lewis structure. Twelve of these electrons are used to bond the atoms with a single bond, leaving 48 – 12= 36 electrons. These 36 electrons are given to the fluorine atoms to complete their octets. The molecular shape is octahedral. The fluorine atoms on opposite sides of the S are separated by twice the sulfur-fluorine bond length (158 pm). Total separation = 2(158 pm) = 316 pm Adjacent fluorines are at two corners of a right triangle, with the sulfur at the 90° angle. Two sides of the triangle are equal to the sulfur-fluorine bond length (158 pm). The separation of the fluorine atoms is at a distance equal to the hypotenuse of this triangle. This length of the hypotenuse may be found using the Pythagorean Theorem (a2 + b2 = c2). In this case a = b = 158 pm. Thus, c2 = (158 pm)2 + (158 pm)2, and so c = 223.4457 = 223 pm. c) PF5 has [1 × P(5e–)] + [5 × F(7e–)] = 40 valence electrons to be used in the Lewis structure. Ten of these electrons are used to bond the atoms with a single bond, leaving 40 – 10= 30 electrons. These 30 electrons are given to the fluorine atoms to complete their octets. The molecular shape is trigonal bipyramidal. Adjacent equatorial fluorine atoms are at two corners of a triangle with an F-P-F bond angle of 120o. The length of the P-F bond is 156 pm. If the 120o bond angle is A, then the F-F bond distance is a and the P-F bond distances are b and c. The F-F bond distance can be found using the Law of Cosines: a2 = b2 + c2 – 2bc (cos A). a2 = (156)2 + (156)2 – 2(156)(156)cos 120o. a = 270.1999 = 270 pm. 10.96

PCl5(l) + SO2(g)  POCl3(l) + SOCl2(l)

O Cl

+ Cl

S O

Cl Cl

PCl5:

AX5

trigonal bipyramidal

SO2:

AX2E

bent

POCl3: AX4

tetrahedral

SOCl2: AX3E

trigonal pyramidal

+ S Cl

10-56

CHAPTER 11 THEORIES OF COVALENT BONDING FOLLOW–UP PROBLEMS 11.1A

Plan: Draw a Lewis structure. Determine the number and arrangement of the electron pairs about the central atom. From this, determine the type of hybrid orbitals involved. Write the partial orbital diagram of the central atoms before and after the orbitals are hybridized. Solution: a) Be is surrounded by two electron groups (two single bonds) so hybridization around Be is sp.

Isolated Be atom Hybridized Be atom b) In SiCl4, Si is surrounded by four electron groups (four single bonds) so its hybridization is sp3.

Cl Cl

sp3

Isolated Si atom Hybridized Si atom c) In XeF4, xenon is surrounded by 6 electron groups (4 bonds and 2 lone pairs) so the hybridization is sp3d2.

F F

sp3d2

Isolated Xe atom

Hybridized Xe atom

Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

11-1

11.1B

Plan: Draw a Lewis structure. Determine the number and arrangement of the electron pairs about the central atom. From this, determine the type of hybrid orbitals involved. Write the partial orbital diagram of the central atoms before and after the orbitals are hybridized. Solution: a) NO2 has 17 valence electrons and the resonance structures are shown below:

The electron group arrangement is trigonal planar, so the central N atom is sp2 hybridized, which means one 2s and two 2p orbital are mixed. One hybrid orbital is filled with a lone pair, and two are half-filled. One electron remains in the unhybridized p orbital to form the π bond between the central N and one of the neighboring O atoms.

2p mix Isolated N atom sp2

Hybridized N atom

b) PCl3 has 26 valence electrons and the Lewis structure is shown below: Cl

The electron-group arrangement is tetrahedral, so the central P atom is sp3 hybridized, which means one 3s and three 3p orbital are mixed. One hybrid orbital is filled with a lone pair, and three are half-filled.

mix Isolated P atom

sp3

Hybridized P atom

11-2

c) KrF2 has 22 valence electrons and the Lewis structure is shown below:

The central Kr atom has five electron groups (two bonds and three lone pairs); the electron-group arrangement is trigonal bipyramidal and the molecular shape is linear, so Kr is sp3d hybridized:

4d mix

sp3d2

4p Isolated Br atom

Hybridized Br atom

11.2A

Plan: First, determine the Lewis structure for the molecule. Then count the number of electron groups around each atom. Hybridization is sp if there are two groups, sp2 if there are three groups, and sp3 if there are four groups. No hybridization occurs with only one group of electrons. The bonds are then designated as sigma or pi. A single bond is a sigma bond. A double bond consists of one sigma and one pi bond. A triple bond includes one sigma bond and two pi bonds. Hybridized orbitals overlap head on to form sigma bonds whereas pi bonds form through the sideways overlap of p or d orbitals. Solution: a) Hydrogen cyanide has H–C≡N: as its Lewis structure. The single bond between carbon and hydrogen is a sigma bond formed by the overlap of a hybridized sp orbital on carbon with the 1s orbital from hydrogen. Between carbon and nitrogen are three bonds. One is a sigma bond formed by the overlap of a hybridized sp orbital on carbon with a hybridized sp orbital on nitrogen. The other two bonds between carbon and nitrogen are pi bonds formed by the overlap of p orbitals from carbon and nitrogen. One sp orbital on nitrogen is filled with a lone pair of electrons. b) The Lewis structure of urea, H2NCONH2, is O C N H

N H

11-3

Both nitrogen atoms are sp3 hybridized (the N atoms have four electron groups – three single bonds and one lone pair). The carbon and the oxygen are sp2 hybridized (the C atom has three electron groups – two single bonds to nitrogen and one double bond to oxygen; the O atom also has three electron groups – two lone pairs and one double bond to carbon). One sp3 hybrid orbital of each nitrogen forms a sigma bond with a sp2 hybrid orbital of carbon. Two sp3 hybrid orbitals of each nitrogen form two sigma bonds with two s orbitals from the hydrogen atoms (one s orbital from each hydrogen atom). The fourth sp3 hybrid orbital on each of the nitrogens is filled with a lone pair of electrons. The remaining sp2 hybrid orbital on carbon forms a sigma bond with an sp2 hybrid orbital on the oxygen. The remaining sp2 hybrid orbitals on oxygen are filled with lone pair electrons. Unhybridized p orbitals on the C and O overlap to form a pi bond. 11.2B

Plan: Use the Lewis structure of the molecule to count the number of electron groups around each atom. Hybridization is sp if there are two groups, sp2 if there are three groups, and sp3 if there are four groups. No hybridization occurs with only one group of electrons. The bonds are then designated as sigma or pi. A single bond is a sigma bond. A double bond consists of one sigma and one pi bond. A triple bond includes one sigma bond and two pi bonds. Hybridized orbitals overlap head-on to form sigma bonds whereas pi bonds form through the sideways overlap of p or d orbitals. Solution: a) C1 has 3 electron groups, so it is sp2 hybridized. C2 has 4 electrons groups, so it is sp3 hybridized. N3 has 4 electrons groups, so it is sp3 hybridized. N4 has 3 electrons groups, so it is sp2 hybridized. O5 has 3 electrons groups, so it is sp2 hybridized. b) There are 25 σ bonds and four π bonds.

11.3A

Plan: Draw the molecular orbital diagram. Determine the bond order from calculation: BO = 1/2(# e– in bonding orbitals – #e– in antibonding orbitals). A bond order of zero indicates the molecule will not exist. A bond order greater than zero indicates that the molecule is at least somewhat stable and is likely to exist. H22– has 4 electrons (1 from each hydrogen and two from the –2 charge). Solution:

Configuration for H22– is (σ1s)2(σ*1s)2. Bond order of H22– is 1/2(2 – 2) = 0. Thus, it is not likely that two H– ions would combine to form the ion H22–.

11-4

11.3B

Plan: Draw the molecular orbital diagram. Determine the bond order from calculation: BO = 1/2(# e– in bonding orbitals – #e– in antibonding orbitals). A bond order of zero indicates the molecule will not exist. A bond order greater than zero indicates that the molecule is at least somewhat stable and is likely to exist. He22+ has 2 electrons (2 from each helium less two from the +2 charge). Solution:

Configuration for He22+ is (σ1s)2. Bond order of He22+ is 1/2(2 – 0) = 1. Thus, we predict that He22+ does exist. 11.4A

Plan: To find bond order, it is necessary to determine the molecular orbital electron configuration from the total number of electrons. Bond order is calculated from the configuration as 1/2(# e– in bonding orbitals – #e– in antibonding orbitals). Solution: N22+: total electrons = 7 + 7 – 2 = 12 Configuration: (σ1s)2(σ*1s)2(σ2s)2(σ*2s)2(π2p)4 Bond order = 1/2(8 – 4) = 2 N2–: total electrons = 7 + 7 + 1 = 15 Configuration: (σ1s)2(σ*1s)2(σ2s)2(σ*2s)2(π2p)4(σ2p)2(π*2p)1 Bond order = 1/2(10 – 5) = 2.5 2–

N2 : total electrons = 7 + 7 + 2 = 16 Configuration: (σ1s)2(σ*1s)2(σ2s)2(σ*2s)2(π2p)4(σ2p)2(π*2p)2 Bond order = 1/2(10 – 6) = 2 Bond energy decreases as bond order decreases: N2– > N22+ = N22– Bond length increases as bond energy decreases, so the order of decreasing bond length will be opposite that of decreasing bond energy. Decreasing bond length: N22+ = N22–> N2– 11.4B

11-5

Solution: F22–: total electrons = 9 + 9 + 2 = 20 Configuration: (σ1s)2(σ*1s)2(σ2s)2(σ*2s)2(σ2p)2(π2p)4(π*2p)4(σ*2p)2 Bond order = 1/2(10 – 10) = 0 F2–: total electrons = 9 + 9 + 1 = 19 Configuration: (σ1s)2(σ*1s)2(σ2s)2(σ*2s)2(σ2p)2(π2p)4(π*2p)4(σ*2p)1 Bond order = 1/2(10 – 9) = 0.5 F2: total electrons = 9 + 9 = 18 Configuration: (σ1s)2(σ*1s)2(σ2s)2(σ*2s)2(σ2p)2(π2p)4(π*2p)4 Bond order = 1/2(10 – 8) = 1 +

F2 : total electrons = 9 + 9 – 1 = 17 Configuration: (σ1s)2(σ*1s)2(σ2s)2(σ*2s)2(σ2p)2(π2p)4(π*2p)3 Bond order = 1/2(10 – 7) = 1.5 2+

F2 : total electrons = 9 + 9 – 2 = 16 Configuration: (σ1s)2(σ*1s)2(σ2s)2(σ*2s)2(σ2p)2(π2p)4(π*2p)2 Bond order = 1/2(10 – 6) = 2 2–

F2 does not exist, so it has no bond energy and is not included in the list. For the other molecule and ions, bond energy increases as bond order increases: F2– < F2< F2+< F22+ Bond length decreases as bond energy increases, so the order of increasing bond length will be opposite that of increasing bond energy. Increasing bond length: F22+ < F2+ < F2< F2– F22– will not form a bond, so it has no bond length and is not included in the list. END–OF–CHAPTER PROBLEMS 11.1

Plan: Table 11.1 describes the types of hybrid orbitals that correspond to the various electron-group arrangements. The number of hybrid orbitals formed by a central atom is equal to the number of electron groups arranged around that central atom. Solution: a) trigonal planar: three electron groups - three hybrid orbitals: sp2 b) octahedral: six electron groups - six hybrid orbitals: sp3d2 c) linear: two electron groups - two hybrid orbitals: sp d) tetrahedral: four electron groups - four hybrid orbitals: sp3 e) trigonal bipyramidal: five electron groups - five hybrid orbitals: sp3d

11.2

a) sp2

11.3

Carbon and silicon have the same number of valence electrons, but the outer level of electrons is n = 2 for carbon and n = 3 for silicon. Thus, silicon has 3d orbitals in addition to 3s and 3p orbitals available for bonding in its outer level, to form up to six hybrid orbitals, whereas carbon has only 2s and 2p orbitals available in its outer level to form up to four hybrid orbitals.

11.4

Four. The same number of hybrid orbitals will form as the initial number of atomic orbitals mixed.

b) sp3

c) sp3d

d) sp3d2

11-6

11.5

Plan: The number of hybrid orbitals is the same as the number of atomic orbitals before hybridization. The type depends on the orbitals mixed. The name of the type of hybrid orbital comes from the number and type of atomic orbitals mixed. The number of each type of atomic orbital appears as a superscript in the name of the hybrid orbital. Solution: a) There are six unhybridized orbitals, and therefore six hybrid orbitals result. The type is sp3d2 since one s, three p, and two d atomic orbitals were mixed. b) Four sp3 hybrid orbitals form from three p and one s atomic orbitals.

11.6

a) two sp orbitals

11.7

b) five sp3d orbitals

Solution: a) The three electron groups (one double bond, one lone pair, and one unpaired electron) around nitrogen require three hybrid orbitals. The hybridization is sp2.

b) The nitrogen has three electron groups (one single bond, one double bond, and one unpaired electron), requiring three hybrid orbitals so the hybridization is sp2.

N O

c) The nitrogen has three electron groups (one single bond, one double bond, and one lone pair) so the hybridization is sp2. N O

11.8

a) sp2

b) sp2

11-7

c) sp

11.9

Plan: To determine hybridization, draw the Lewis structure and count the number of electron groups around the central chlorine atom. Hybridize that number of orbitals. Single, double, and triple bonds all count as one electron group. An unshared pair (lone pair) of electrons or one unshared electron also counts as one electron group. Solution: a) The Cl has four electron groups (one lone pair, one lone electron, and two double bonds) and therefore four hybrid orbitals are required; the hybridization is sp3. Note that in ClO2, the π bond is formed by the overlap of d orbitals from chlorine with p orbitals from oxygen.

Cl O

b) The Cl has four electron groups (one lone pair and three bonds) and therefore four hybrid orbitals are required; the hybridization is sp3.

c) The Cl has four electron groups (four bonds) and therefore four hybrid orbitals are required; the hybridization is sp3.

11.10

a) sp3d F

F 3

b) sp

Br O

O c) sp3d2 F F

F Br

11-8

11.11

Plan: Draw the Lewis structure and count the number of electron groups around the central atom. Hybridize that number of orbitals. Single, double, and triple bonds all count as one electron group. An unshared pair (lone pair) of electrons or one unshared electron also counts as one electron group. Once the type of hybridization is known, the types of atomic orbitals that will mix to form those hybrid orbitals are also known. Solution: a) Silicon has four electron groups (four bonds) requiring four hybrid orbitals; four sp3 hybrid orbitals are made from one s and three p atomic orbitals. H

H b) Carbon has two electron groups (two double bonds) requiring two hybrid orbitals; two sp hybrid orbitals are made from one s and one p orbital. S

c) Sulfur is surrounded by five electron groups (four bonding pairs and one lone pair), requiring five hybrid orbitals; five sp3d hybrid orbitals are formed from one s orbital, three p orbitals, and one d orbital.

F Cl Cl

S Cl

d) Nitrogen is surrounded by four electron groups (three bonding pairs and one lone pair) requiring four hybrid orbitals; four sp3 hybrid orbitals are formed from one s orbital and three p orbitals.

F 11.12

a) sp3←s + 3p O Cl

b) sp d←s + 3p + d

Cl Cl

11-9

c) sp3d←s + 3p + d F F F

P F F

d) sp3←s + 3p S

11.13

Plan: To determine hybridization, draw the Lewis structure of the reactants and products and count the number of electron groups around the central atom. Hybridize that number of orbitals. Single, double, and triple bonds all count as one electron group. An unshared pair (lone pair) of electrons or one unshared electron also counts as one electron group. Recall that sp hydrid orbitals are oriented in a linear geometry, sp2 in a trigonal planar geometry, sp3 in a tetrahedral geometry, sp3d in a trigonal bipyramidal geometry, and sp3d2 in an octahedral geometry. Solution: a) The P in PH3 has four electron groups (one lone pair and three bonds) and therefore four hybrid orbitals are required; the hybridization is sp3. The P in the product also has four electron groups (four bonds) and again four hybrid orbitals are required. The hybridization of P remains sp3. There is no change in hybridization. Illustration B best shows the hybridization of P during the reaction as sp3→sp3. b) The B in BH3 has three electron groups (three bonds) and therefore three hybrid orbitals are required; the hybridization is sp2. The B in the product has four electron groups (four bonds) and four hybrid orbitals are required. The hybridization of B is now sp3. The hybridization of B changes from sp2 to sp3; this is best shown by illustration A.

H P

H 11.14

a) The Te in TeF6 has six electron groups (six bonds) and therefore six hybrid orbitals are required; the hybridization is sp3d2. Te in TeF5– also has six electron groups (five bonds and one unshared pair) and again six hybrid orbitals are required. The hybridization of Te remains sp3d2. There is no change in hybridization. Illustration A best shows the hybridization of Te when TeF6 forms TeF5–: sp3d2→sp3d2.

Te F

11-10

b) The Te in TeF4 has five electron groups (four bonds and one unshared pair) and therefore five hybrid orbitals are required; the hybridization is sp3d. Te in TeF6 has six electron groups (six bonds) and therefore six hybrid orbitals are required; the hybridization is sp3d2. Illustration C best shows the change in hybridization of Te from sp3d to sp3d2. F

Te F

11.15

Te F

Plan: To determine hybridization, draw the Lewis structure and count the number of electron groups around the central atom. Hybridize that number of orbitals. Single, double, and triple bonds all count as one electron group. An unshared pair (lone pair) of electrons or one unshared electron also counts as one electron group. Write the electron configuration of the central atom and mix the appropriate atomic orbitals to form the hybrid orbitals. Solution: a) Germanium is the central atom in GeCl4. Its electron configuration is [Ar]4s23d104p2. Ge has four electron groups (four bonds), requiring four hybrid orbitals. Hybridization is sp3 around Ge. One of the 4s electrons is moved to a 4p orbital and the four orbitals are hybridized. Cl Cl

sp3

Isolated Ge atom Hybridized Ge atom b) Boron is the central atom in BCl3. Its electron configuration is [He]2s22p1. B has three electron groups (three bonds), requiring three hybrid orbitals. Hybridization is sp2 around B. One of the 2s electrons is moved to an empty 2p orbital and the three atomic orbitals are hybridized. One of the 2p atomic orbitals is not involved in the hybridization. Cl B Cl

s Isolated B atom

sp2

Hybridized B atom

11-11

c)) Carbon is the central atom in CH3+. Its elecctron configuraation is [He]2ss22p2. C has thrree electron grooups (three bo onds), requirin ng three hybrid orbitals. Hybridization is sp2 around C. One O of the 2s electrons is mov ved to an empty y 2p orbital; thrree orbitals aree hybridized annd one electronn is reemoved to form m the +1 ion.

H C H

H +

sp2

Issolated C atom 11.16

Hybridized C atom

a))

b))

+ 3e s

sp3

sp2

c))

11.17

Pllan: To determ mine hybridizatiion, draw the Lewis L structuree and count thee number of eleectron groups aaround the ceentral atom. Hy ybridize that nu umber of orbitaals. Single, douuble, and triplee bonds all couunt as one electtron group. An A unshared paiir (lone pair) of electrons or one o unshared eelectron also coounts as one eleectron group. W Write the ellectron configu uration of the central atom and mix the apprropriate atomicc orbitals to forrm the hybrid oorbitals. So olution: a)) In SeCl2, Se is the central attom and has fou ur electron grooups (two singlle bonds and tw wo lone pairs), requiring fo our hybrid orbitals so Se is sp p3 hybridized. The T electron coonfiguration off Se is [Ar]4s233d104p4. The 4s and 4p attomic orbitals are a hybridized.. Two sp3 hybriid orbitals are filled with lone electron pairrs and two sp3 oorbitals bo ond with the ch hlorine atoms.

sp3

Copyright © McGraw-Hill Ed ducation. This is proprietary mateerial solely for au uthorized instructtor use. Not authoorized for sale or distribution in any mann ner. This documen nt may not be cop pied, scanned, dup plicated, forward ded, distributed, oor posted on a weebsite, in whole orr part.

11-12

b) In H3O+, O is the central atom and has four electron groups (three single bonds and one lone pair), requiring four hybrid orbitals. O is sp3 hybridized. The electron configuration of O is [He]2s22p4. The 2s and 2p orbitals are hybridized. One sp3 hybrid orbital is filled with a lone electron pair and three sp3 orbitals bond with the hydrogen atoms.

H H

c) I is the central atom in IF4– with six electron groups (four single bonds and two lone pairs) surrounding it. Six hybrid orbitals are required and I has sp3d2 hybrid orbitals. The sp3d2 hybrid orbitals are composed of one s orbital, three p orbitals, and two d orbitals. Two sp3d2 orbitals are filled with a lone pair and four sp3d2 orbitals bond with the fluorine atoms.

F F

+e s

sp3d2 11.18

sp3

sp2

11-13

+e s

sp3d2

11.19 H H

sp3

sp2

11.20

Plan: A single bond is a sigma bond which is the result of two orbitals overlapping end to end; a double bond consists of one sigma bond and one pi bond; and a triple bond consists of one sigma bond and two pi bonds. A pi bond is the result of orbitals overlapping side to side. Solution: a) False, a double bond is one sigma (σ) and one pi (π) bond. b) False, a triple bond consists of one sigma (σ) and two pi (π) bonds. c) True d) True e) False, a π bond consists of one pair of electrons; it occurs after a σ bond has been previously formed. f) False, end-to-end overlap results in a bond with electron density along the bond axis.

11.21

Plan: To determine hybridization, draw the Lewis structure and count the number of electron groups around the central atom. Hybridize that number of orbitals. Single, double, and triple bonds all count as one electron group. An unshared pair (lone pair) of electrons or one unshared electron also counts as one electron group. A single bond is a sigma bond which is the result of two orbitals overlapping end to end; a double bond consists of one sigma bond and one pi bond; and a triple bond consists of one sigma bond and two pi bonds. Solution: a) Nitrogen is the central atom in NO3–. Nitrogen has three surrounding electron groups (two single bonds and one double bond), so it is sp2 hybridized. Nitrogen forms three σ bonds (one each for the N–O bonds) and one π bond (part of the N=O double bond).

b) Carbon is the central atom in CS2. Carbon has two surrounding electron groups (two double bonds), so it is sp hybridized. Carbon forms two σ bonds (one each for the C–S bonds) and two π bonds (part of the two C=S double bonds). S

11-14

c) Carbon is the central atom in CH2O. Carbon has three surrounding electron groups (two single bonds and one double bond), so it is sp2 hybridized. Carbon forms three σ bonds (one each for the two C–H bonds and one C–O bond) and one π bond (part of the C=O double bond). H

H 11.22

a) sp2

2 σ bonds and 1 π bond

O O

b) sp3d 2 σ bonds

c)sp2

3 σ bond and 1 π bond

Cl C

Cl 11.23

Plan: To determine hybridization, draw the Lewis structure and count the number of electron groups around the central nitrogen atom. Hybridize that number of orbitals. Single, double, and triple bonds all count as one electron group. An unshared pair (lone pair) of electrons or one unshared electron also counts as one electron group. A single bond is a sigma bond which is the result of two orbitals overlapping end to end; a double bond consists of one sigma bond and one pi bond; and a triple bond consists of one sigma bond and two pi bonds. Solution: a) In FNO, three electron groups (one lone pair, one single bond, and one double bond) surround the central N atom. Hybridization is sp2 around nitrogen. One sigma bond exists between F and N, and one sigma and one pi bond exist between N and O. Nitrogen participates in a total of 2 σ and 1 π bonds. F

b) In C2F4, each carbon has three electron groups (two single bonds and one double bond) with sp2 hybridization. The bonds between C and F are sigma bonds. The C–C bond consists of one sigma and one pi bond. Each carbon participates in a total of three σ and one π bonds.

F C

c) In (CN)2, each carbon has two electron groups (one single bond and one triple bond) and is sp hybridized with a sigma bond between the two carbons and a sigma and two pi bonds comprising each C–N triple bond. Each carbon participates in a total of two σ and two π bonds. N

11-15

11.24

a) sp

2 σ and 2 π bonds

b) sp3 (CH3) sp (other two C atoms) six σ and two π bonds H C

H two σ and two π bonds

c) sp2

S O 11.25

Plan: A single bond is a sigma bond which is the result of two orbitals overlapping end to end; a double bond consists of one sigma bond and one pi bond; and a triple bond consists of one sigma bond and two pi bonds. Solution: The double bond in 2-butene restricts rotation of the molecule, so that cis and trans structures result. The two structures are shown below: H H

C H

cis

H H

trans

The carbons participating in the double bond each have three surrounding groups, so they are sp2 hybridized. The =C–H σ bonds result from the head-on overlap of a C sp2 orbital and an H s orbital. The C–CH3 bonds are also σ bonds, resulting from the head-on overlap of an sp2 orbital and an sp3 orbital. The C=C bond contains 1 σ bond (head-on overlap of two sp2 orbitals) and 1 π bond (sideways overlap of unhybridized p orbitals). Finally, C–H bonds in the methyl (–CH3) groups are σ bonds resulting from the overlap of the sp3 orbital of C with the s orbital of H.

p p

1s sp2 sp2

1s sp3 sp3 1s

C sp2

p sp2 C

1s sp2 sp3

sp3 C sp3

sp2 sp3

sp3 C sp3 1s

11-16

11.26

Four molecular orbitals form from the four p atomic orbitals. In forming molecular orbitals, the total number of molecular orbitals must equal the number of atomic orbitals. Two of the four molecular orbitals formed are bonding orbitals and two are antibonding.

11.27

Two px atomic orbitals were used to form a sigma bonding MO (lower energy) and a sigma antibonding MO (higher energy). The bonding MO does not have a node separating the two halves of the orbital.

11.28

a) Bonding MOs have lower energy than antibonding MOs. The bonding MO’s lower energy, even lower than its constituent atomic orbitals, accounts for the stability of a molecule in relation to its individual atoms. However, the sum of energy of the MOs must equal the sum of energy of the AOs. b) The node is the region of an orbital where the probability of finding the electron is zero, so the nodal plane is the plane that bisects the node perpendicular to the bond axis. There is no node along the bond axis (probability is positive between the two nuclei) for the bonding MO. The antibonding MO does have a nodal plane. c) The bonding MO has higher electron density between nuclei than the antibonding MO.

11.29

A bonding MO may contain a nodal plane lying along the internuclear axis, as in π bonding. In an antibonding MO, the nodal plane is perpendicular to the bond axis, between the atoms.

11.30

Plan: Like atomic orbitals, any one MO holds a maximum of two electrons. Two atomic orbitals combine to form two molecular orbitals, a bonding and an antibonding MO. Solution: a) Two electrons are required to fill a σ-bonding molecular orbital. Each molecular orbital requires two electrons. b) Two electrons are required to fill a π-antibonding molecular orbital. There are two π-antibonding orbitals, each holding a maximum of two electrons. c) Four electrons are required to fill the two σ molecular orbitals (two electrons to fill the σ-bonding and two to fill the σ-antibonding) formed from two 1s atomic orbitals.

11.31

a) 12

11.32

Plan: Recall that a bonding MO has a region of high electron density between the nuclei while an antibonding MO has a node, or region of zero electron density between the nuclei. MOs formed from s orbitals,

b) 2

c) 4

or from p orbitals overlapping end-to-end, are called σ and MOs formed by the side-to-side overlap of p orbitals are called π. A superscript star (*) is used to designate an antibonding MO. To write the electron configuration of F2+, determine the number of valence electrons and write the sequence of MO energy levels, following the sequence order given in the text. Solution: a) A is the π*2p molecular orbital (two p orbitals overlapping side to side with a node between them); B is the σ2p molecular orbital (two p orbitals overlapping end to end with no node); C is the π2p molecular orbital (two p orbitals overlapping side to side with no node); D is the σ*2p molecular orbital (two p orbitals overlapping end to end with a node). b) F2+ has 13 valence electrons: [2 × F(7e–) – 1 (from + charge)]. The MO electron configuration is (σ2s)2(σ*2s)2(σ2p)2(π2p)2(π2p)2(π*2p)2(π*2p)1. The π*2p molecular orbital, A, σ2p molecular orbital, B, and π2p molecular orbital, C, are all occupied by at least one electron. The σ*2p molecular orbital is unoccupied. c) A π*2p molecular orbital, A, has only one electron.

11-17

11.33

a) A is the π*2p molecular orbital; B is the σ2p molecular orbital; C is the π2p molecular orbital; D is the σ*2p molecular orbital; E is the σ2s molecular orbital; F is the σ*2s molecular orbital. b) The σ*2p molecular orbital, D, is the highest in energy. c) The σ2s molecular orbital, E, is the lowest in energy. d) σ2s<σ*2s<π2p<σ2p <π*2p<σ*2p (E < F < C < B < A < D)

11.34

Plan: To write the electron configuration of Be2+, determine the number of electrons and write the sequence of MO energy levels, following the sequence order given in the text. Bond order = ½[(no. of electrons in bonding MO) – (no. of electrons in antibonding MO)]. Recall that a diamagnetic substance has no unpaired electrons. Solution: a) Be2+ has a total of seven electrons [2 × Be(4e–) – 1 (from + charge)]. The molecular orbital configuration is (σ1s)2(σ*1s)2(σ2s)2(σ*2s)1 and bond order = ½(4 – 3) = 1/2. With a bond order of 1/2 the Be2+ ion will be stable. b) No, the ion has one unpaired electron in the σ*2sMO, so it is paramagnetic, not diamagnetic. c) Valence electrons would be those in the molecular orbitals at the n = 2 level, so the valence electron configuration is (σ2s)2(σ*2s)1.

11.35

a) The molecular orbital configuration for O2– with a total of 17 electrons is (σ1s)2(σ*1s)2(σ2s)2(σ*2s)2(σ2p)2(π2p)2(π2p)2(π*2p)2(π*2p)1. Bond order = ½(10 bonding – 7 antibonding e–) = 3/2 = 1.5. O2– is stable. b) O2– is paramagnetic with an unpaired electron in the π*2p MO. c) (σ2s)2(σ*2s)2(σ2p)2(π2p)2(π2p)2(π*2p)2(π*2p)1

11.36

Plan: Write the electron configuration of each species by determining the number of electrons and writing the sequence of MO energy levels, following the sequence order given in the text. Calculate the bond order: bond order = ½[(no. of electrons in bonding MO) – (no. of electrons in antibonding MO)]. Bond energy increases as bond order increases; bond length decreases as bond order increases. Solution: C2– Total electrons = 6 + 6 + 1 = 13 MO configuration: (σ1s)2(σ*1s)2(σ2s)2(σ*2s)2(π2p)4(σ2p)1 Bond order = 1/2(9 – 4) = 2.5 C2

Total electrons = 6 + 6 = 12 MO configuration: (σ1s)2(σ*1s)2(σ2s)2(σ*2s)2(π2p)4 Bond order = 1/2(8 – 4) = 2

C2+

Total electrons = 6 + 6 – 1 = 11 MO configuration: (σ1s)2(σ*1s)2(σ2s)2(σ*2s)2(π2p)3 Bond order = 1/2(7 – 4) = 1.5

a) Bond energy increases as bond order increases: C2+< C2< C2– b) Bond length decreases as bond energy increases, so the order of increasing bond length will be opposite that of increasing bond energy. Increasing bond length: C2–< C2< C2+

11-18

11.37

Bond order +

B2 :

(σ2s) (σ*2s) (π2p)

0.5

B2:

(σ2s) (σ*2s) (π2p) (π2p)

1.0

B2–:

(σ2s)2(σ*2s)2(π2p)2(π2p)1

1.5

a) B2–> B2> B2+ b) B2+> B2> B2– 11.38 –

a) BrO3

AX3E AX4E

b) AsCl4

AX4

c) SeO4 d) BiF5

AX5E

e) SbF4

AX4 AX6

f) AlF6 g) IF4

AX4E

109.5°

none

90°

<90°

109.5°

none

90°

none

120°, 90°

<120°, <90°

octahedral sp3d2 hybrid AO

<120°, <90°

tetrahedral sp3 hybrid AO

3–

120°, 90°

square pyramidal sp3d2 hybrid AO

<109.5°

tetrahedral sp3 hybrid AO

2–

109.5°

seesaw sp3d hybrid AO

2–

Deviations

trigonal pyramidal sp3 hybrid AO

–

Ideal

seesaw sp3d hybrid AO

Lewis structures:

Cl Cl As

11-19

F F

I F F

11.39

a) There are 9 σ and 2 π bonds. Each of the six C–H bonds are sigma bonds. The C–C bond contains a sigma bond. The double bonds between the carbons consist of a pi bond in addition to the sigma bond. b) No, cis-trans structural arrangements are not possible because one of the carbons in each double bond has two hydrogens bonded to it. Cis-trans structural arrangements occur only when both carbons in the double bond are bonded to two groups that are not identical.

11.40

Plan: To determine hybridization, count the number of electron groups around each of the C, O, and N atoms. Hybridize that number of orbitals. Single, double, and triple bonds all count as one electron group. An unshared pair (lone pair) of electrons or one unshared electron also counts as one electron group. A single bond is a sigma bond which is the result of two orbitals overlapping end to end; a double bond consists of one sigma bond and one pi bond; and a triple bond consists of one sigma bond and two pi bonds. Solution: a) Each of the six C atoms in the ring has three electron groups (two single bonds and a double bond) and has sp2 hybridization; all of the other C atoms have four electron groups (four single bonds) and have sp3 hybridization; all of the O atoms have four electron groups (two single bonds and two lone pairs) and have sp3 hybridization; the N atom has four electron groups (three single bonds and a lone pair) and has sp3 hybridization. b) Each of the single bonds is a sigma bond; each of the double bonds has one sigma bond for a total of 26 sigma bonds. c) The ring has three double bonds, each of which is composed of one sigma bond and one pi bond; so there are three pi bonds each with two electrons for a total of six pi electrons.

11.41 a)

11-20

sp3d c)

sp3d2

sp2

11.42

Plan: To determine hybridization, count the number of electron groups around each C and N atom. Hybridize that number of orbitals. Single, double, and triple bonds all count as one electron group. An unshared pair (lone pair) of electrons or one unshared electron also counts as one electron group. A single bond is a sigma bond, which is the result of two orbitals overlapping end to end; a double bond consists of one sigma bond and one pi bond; and a triple bond consists of one sigma bond and two pi bonds. Solution: a) Every single bond is a sigma bond. There is one sigma bond in each double bond as well. There are 17 σ bonds in isoniazid. Every atom-to-atom connection contains a σ bond. b) All carbons have three surrounding electron groups (two single and one double bond), so their hybridization is sp2. The ring N also has three surrounding electron groups (one single bond, one double bond, and one lone pair), so its hybridization is also sp2. The other two N atoms have four surrounding electron groups (three single bonds and one lone pair) and are sp3 hybridized.

11.43

Hydrazine

Carbon disulfide

11-21

b) The electron-group arrangement around each nitrogen changes from tetrahedral to trigonal planar. The molecular shape changes from trigonal pyramidal to bent and the hybridization changes from sp3 to sp2. c) The electron-group arrangement and molecular shape around carbon change from linear to trigonal planar. The hybridization changes from sp to sp2. 11.44

Plan: To determine the hybridization in each species, count the number of electron groups around the underlined atom. Hybridize that number of orbitals. Single, double, and triple bonds all count as one electron group. An unshared pair (lone pair) of electrons or one unshared electron also counts as one electron group. Solution: a) B changes from sp2→sp3. Boron in BF3 has three electron groups with sp2 hybridization. In BF4–, four electron groups surround B with sp3 hybridization.

F F

F b) P changes from sp3→sp3d. Phosphorus in PCl3 is surrounded by four electron groups (three bonds to Cl and one lone pair) for sp3 hybridization. In PCl5, phosphorus is surrounded by five electron groups for sp3d hybridization.

Cl Cl

P Cl Cl

c) C changes from sp→sp2. Two electron groups surround C in C2H2 and three electron groups surround C in C2H4.

H H

H C

d) Si changes from sp3→sp3d2. Four electron groups surround Si in SiF4 and six electron groups surround Si in SiF62–.

F F

11-22

e) No change, S in SO2 is surrounded by three electron groups (one single bond, one double bond, and one lone pair) and in SO3 is surrounded by three electron groups (two single bonds and one double bond); both have sp2 hybridization.

O O 11.45

a) NO

Original molecules: (σ2s)2(σ*2s)2(π2p)2(π2p)2(σ2p)2(π*2p)1 2

bond order = 2.5 paramagnetic 1

(σ2s) (σ*2s) (σ2p) (π2p) (π2p) (π*2p) (π*2p)

(σ2s) (σ*2s) (π2p) (π2p) (σ2p)

bond order = 2.0 paramagnetic bond order = 3.0 diamagnetic

Ions: NO+ O2

(σ2s)2(σ*2s)2(π2p)2(π2p)2(σ2p)2 2

bond order = 3.0 diamagnetic

(σ2s) (σ*2s) (σ2p) (π2p) (π2p) (π*2p) (σ2s) (σ*2s) (π2p) (π2p) (σ2p)

b) Changing: 11.46

NO , N2

bond order = 2.5 paramagnetic bond order = 2.5 paramagnetic

Plan: To determine the molecular shape and hybridization, count the number of electron groups around the P, N, and C atoms. Hybridize that number of orbitals. Single, double, and triple bonds all count as one electron group. An unshared pair (lone pair) of electrons or one unshared electron also counts as one electron group. Solution: P (3 single bonds and 1 double bond) AX4 tetrahedral sp3 N (3 single bonds and 1 lone pair)

AX3E

trigonal pyramidal

sp3

C1 and C2 (4 single bonds)

AX4

tetrahedral

sp3

C3 (2 single bonds and 1 double bond)

AX3

trigonal planar

sp2

11.47

a)1: sp3 2: sp2 3: sp3 4: sp3 b) 28 c) a: < 109.5° b: 120° c: 120°

11.48

5: sp2

6: sp2

(σ2s)2(σ*2s)2(σ2p)2(π2p)2(π2p)2(π*2p)1(π*2p)1

bond order = 2.0 paramagnetic

(2 unpaired) O2

–

(σ2s)2(σ*2s)2(σ2p)2(π2p)2(π2p)2(π*2p)1

bond order = 2.5 paramagnetic

(1 unpaired) (σ2s)2(σ*2s)2(σ2p)2(π2p)2(π2p)2(π*2p)2(π*2p)1

bond order = 1.5 paramagnetic

(1 unpaired) O22–

(σ2s)2(σ*2s)2(σ2p)2(π2p)2(π2p)2(π*2p)2(π*2p)2

bond order = 1.0 diamagnetic

(0 unpaired) Bond length: 11.49

O2+< O2< O2–< O22–

a) All the diagrams are identical: (σ2s)2(σ*2s)2(π2p)2(π2p)2(σ2p)2 b) Nitrogen molecules are not polar, but CN– and CO are polar.

11-23

11.50

a) Yes, each one is sp2 hybridized. b) Yes, each one is sp3 hybridized. c) C–O bonds: 6 σ bonds, 1 π bond. d) No, the C=O lone pair electrons are in sp2 hybrid orbitals, while the other oxygen lone pairs occupy sp3 hybrid orbitals.

11.51

Plan: To determine the hybridization, count the number of electron groups around the atoms. Hybridize that number of orbitals. Single, double, and triple bonds all count as one electron group. An unshared pair (lone pair) of electrons or one unshared electron also counts as one electron group. Solution: a) B and D show hybrid orbitals that are present in the molecule. B shows sp3 hybrid orbitals, used by atoms that have four groups of electrons. In the molecule, the C atom in the CH3 group, the S atom, and the O atom all have four groups of electrons and would have sp3 hybrid orbitals. D shows sp2 hybrid orbitals, used by atoms that have three groups of electrons. In the molecule, the C bonded to the nitrogen atom, the C atoms involved in the C=C bond, and the nitrogen atom all have three groups of electrons and would have sp2 hybrid orbitals. b) The C atoms in the C≡C bond have only two electron groups and would have sp hybrid orbitals. These orbitals are not shown in the picture. c) There are two sets of sp hybrid orbitals, four sets of sp2 hybrid orbitals, and three sets of sp3 hybrid orbitals in the molecule.

11.52

Plan: Draw a resonance structure that places the double bond between the C and N atoms. Solution: The resonance gives the C–N bond some double bond character, which hinders rotation about the C–N bond. The C–N single bond is a σ bond; the resonance interaction exchanges a C–O π bond for a C–N π bond. O O

+ N

H 11.53

Draw the Lewis structures.

SO2

sp2

SO3

Cl Cl

11.54

sp3

Cl Cl

sp3d

SCl6

S Cl

SCl4

SO32–

sp2

sp3d2

S2Cl2

sp3

a) N has sp2 hybridization, formed from one 2s and two 2p orbitals. b) The lone pair is in a sp2 hybrid orbital. c) Hybridization of C in CH3 is sp3; C in the ring is sp2.

11-24

11.55

Plan: Write the electron configuration of H2– by determining the number of electrons and writing the sequence of MO energy levels, following the sequence order given in the text. Calculate the bond order: bond order = ½[(no. of electrons in bonding MO) – (no. of electrons in antibonding MO)]. Solution: H2– Total electrons = 1 + 1 + 1 = 3 MO configuration: (σ1s)2(σ*1s)1 Bond order = 1/2(2 – 1) = 0.5 –

H2 exists, with a bond order of 0.5. 11.56

Plan: To determine hybridization, count the number of electron groups around each C and O atom. Hybridize that number of orbitals. Single, double, and triple bonds all count as one electron group. An unshared pair (lone pair) of electrons or one unshared electron also counts as one electron group. A single bond is a sigma bond which is the result of two orbitals overlapping end to end; a double bond consists of one sigma bond and one pi bond; and a triple bond consists of one sigma bond and two pi bonds. Solution: a) The six carbons in the ring each have three surrounding electron groups (two single bonds and one double bond) with sp2 hybrid orbitals. The two carbons participating in the C=O bond are also sp2 hybridized. The single carbon in the –CH3 group has four electron groups (four single bonds) and is sp3 hybridized. The two central oxygen atoms, one in a C–O–H configuration and the other in a C–O–C configuration, each have four surrounding electron groups (two single bonds and two lone pairs) and are sp3 hybridized. The O atoms in the two C=O bonds have three electron groups (one double bond and two lone pairs) and are sp2 hybridized. Summary: C in –CH3: sp3, all other C atoms (8 total): sp2, O in C=O (2 total): sp2, O in the C–O bonds (2 total): sp3. b) The two C=O bonds are localized; the double bonds on the ring are delocalized as in benzene. c) Each carbon with three surrounding groups has sp2 hybridization and trigonal planar shape; therefore, eight carbons have this shape. Only one carbon in the CH3 group has four surrounding groups with sp3 hybridization and tetrahedral shape.

11.57

Plan: In the cis arrangement, the two H atoms are on the same side of the double bond; in the trans arrangement, the two H atoms are on different sides of the double bond. Solution: a) Four different isomeric fatty acids: trans-cis, cis-cis, cis-trans, trans-trans. b) With three double bonds, there are 2n = 23 = 8 isomers possible. cis-cis-cis trans-trans-trans cis-trans-cis trans-cis-trans cis-cis-trans trans-cis-cis cis-trans-trans trans-trans-cis

11-25

CHAPTER 12 INTERMOLECULAR FORCES: LIQUIDS, SOLIDS, AND PHASE CHANGES FOLLOW–UP PROBLEMS 12.1A

Plan: This is a three step process: warming the ice to 0.0°C, melting the ice, and warming the liquid water to 16.0°C. Use the molar heat capacities of ice and water and the relationship q = n × C × ΔT to calculate the heat use for the warming of the ice and of the water; use the heat of fusion and the relationship q = nΔHfus to calculate the heat required to melt the ice. Solution: The total heat required is the sum of three processes: 1) Warming the ice from –7.00°C to 0.00°C q1 = n × Cice × ΔT = (2.25 mol)(37.6 J/mol ∙ °C) [0.0 – (–7.00)]°C = 592.2 J = 0.592 kJ 2) Phase change of ice at 0.00°C to water at 0.00°C ⎛ 6.02 kJ ⎞⎟ q2 = nΔHfus = (2.25 mol)⎜⎜ ⎟ = 13.545 kJ = 13.5 kJ ⎝ mol ⎠⎟ 3) Warming the liquid from 0.00°C to 16.0°C q3 = n × Cwater × ΔT = (2.25 mol)(75.4 J/mol ∙ °C) [16.0 – (0.0)]°C = 2714.4 J = 2.71 kJ The three heats are positive because each process takes heat from the surroundings (endothermic). The phase change requires much more energy than the two temperature change processes. The total heat is q1 + q2 + q3 = (0.5922 kJ + 13.545 kJ + 2.7144 kJ) = 16.8516 = 16.9 kJ.

12.1B

Plan: This is a three step process: cooling the bromine vapor to 59.5°C, condensing the bromine vapor, and cooling the liquid bromine to 23.8°C. Use the molar heat capacities of bromine vapor and liquid bromine and the relationship q = n × C × ΔT to calculate the heat released by the cooling of the bromine vapor and of the liquid bromine; use the heat of vaporization and the relationship q = n(–ΔHvap) to calculate the heat released when bromine vapor condenses. Solution: ⎛ 1 mol Br ⎞⎟ 2 ⎜ ⎟⎟ = 0.3000 mol Br Amount (mol) of bromine = (47.94 g Br2) ⎜⎜ 2 ⎜⎜⎝159.8 g Br2 ⎠⎟⎟ The total heat released is the sum of three processes: 1) Cooling the bromine vapor from 73.5°C to 59.5°C q1 = n × Cgas × ΔT = (0.3000 mol)(36.0 J/mol°C) [59.5 – 73.5]°C = –151.2 J = –0.151 kJ 2) Phase change of bromine vapor at 59.5°C to liquid bromine at 59.5°C ⎛ −29.6 kJ ⎞⎟ ⎟⎟ = –8.88 kJ q2 = n(–ΔHvap) = (0.3000 mol) ⎜⎜⎜ ⎜⎝ mol ⎠⎟⎟ 3) Cooling the liquid from 59.5°C to 23.8°C q3 = n × Cliquid × ΔT = (0.3000 mol)(75.7 J/mol°C) [23.8 – 59.5]°C = –810.7 J = –0.811 kJ The three heats are negative because each process releases heat to the surroundings (exothermic). The phase change releases much more energy than the two temperature change processes. The total heat is q1 + q2 + q3 = (–0.1512 kJ + –8.88 kJ + –0.8107 kJ) = –9.8419 = –9.84 kJ.

12-1

12.2A

Plan: The variables ΔHvap, P1, T1, and T2 are given, so substitute them into the Clausius-Clapeyron equation and solve for P2. Convert both temperatures from °C to K. Convert ΔHvap to J so that the units cancel with R. Solution: T1 = 34.1 + 273.15 = 307.2 K −ΔH vap ⎛⎜ 1 P 1⎞ ln 2 = ⎜ − ⎟⎟⎟ ⎜ P1 R ⎝ T2 T1 ⎠⎟

T2 = 85.5 + 273.15 = 358.6 K

P1 = 40.1 torr

kJ −40.7 ⎛103 J ⎞⎟ ⎛ 1 1 ⎞⎟⎜ P2 ⎜ mol ⎟⎟ = 2.284105 ⎟ ⎜ − = ln ⎟⎜⎜ ⎜ ⎟⎟⎜⎜ 1 kJ ⎠⎟⎟ 8.314 J/mol • K ⎜⎝⎜ 358.6 K 307.2 K ⎠⎝ 40.1 torr P2 = 9.81689 40.1 torr

P2 = (9.81689)(40.1 torr) = 393.657 = 3.9 × 102 torr In a log term, only the numbers after the decimal point are significant. The value 2.284105 contains 3 significant figures after simplifying the expression to the right of the equal sign; this value would round to 2.28 if it was not part of an intermediate step. However, as we next perform the logarithmic function, only the 2 significant figures after the decimal are carried over through the next calculation steps. The temperature increased, so the vapor pressure should be higher. 12.2B

Plan: The variables P1, P2, T1, and T2 are given, so substitute them into the Clausius-Clapeyron equation and solve for ΔHvap. Convert T1 and T2 from °C to K. Convert ΔHvap to kJ/mol Solution: T1 = 20.2 + 273.15 = 293.35 K

P1 = 24.5 kPa

T2 = 0.95 + 273.15 = 274.10 K

P2 = 10.0 kPa

−ΔH vap ⎜⎛ 1 1⎞ P2 = ⎜⎜ − ⎟⎟⎟ R ⎝ T2 T1 ⎠⎟ P1

−ΔH vap ⎛ 24.50 kPa ⎞⎟ ⎛ 1 1 ⎞⎟ ⎜ ln ⎜⎜ = − ⎝ 10.00 kPa ⎠⎟⎟ 8.314 J / mol • K ⎜⎜⎝ 293.35K 274.10 K ⎠⎟⎟⎟

0.896088 =

−ΔHvap 8.314 J / mol • K

(−0.000239406 K−1 )

⎛ 1 kJ ⎞ ΔHvap = 31119 J/mol ⎜⎜ 3 ⎟⎟⎟ = 31.119 = 31.1 kJ/mol ⎜⎝10 J ⎠

12.3A

Plan: Refer to the phase diagram to describe the specified changes. In this problem, carbon is heated at a constant pressure. Solution: At the starting conditions (3 × 102 bar and 2000 K), the carbon sample is in the form of graphite. As the sample is heated at constant pressure, it melts at about 4500 K, and then vaporizes at around 5000 K.

12.3B

Plan: Refer to the phase diagram to describe the specified changes. In this problem, carbon is compressed at a constant temperature. Solution: At the starting conditions (4600 K and 104 bar), the carbon sample is in the form of graphite. As the sample is compressed at constant temperature, it melts at around 4 × 104 bar and then solidifies to diamond at around 2 × 105 bar.

12-2

12.4A

Plan: Hydrogen bonding occurs in compounds in which H is directly bonded to O, N, or F. In hydrogen bonding, ….. there is a –A: H–B– sequence in which A and B are O, N, or F. Solution: …..

a) The –A: H–B– sequence is present in both of the following structures:

O C

H H

H H H

H C

H …..

b) The –A: H–B– sequence can only be achieved in one arrangement.

H C

H The hydrogens attached to the carbons cannot form H bonds. c) Hydrogen bonding is not possible because there are no O–H, N–H, or F–H bonds. 12.4B

Plan: Hydrogen bonding occurs in compounds in which H is directly bonded to O, N, or F. In hydrogen bonding, ….. there is a –A: H–B– sequence in which A and B are O, N, or F. Solution: a) Hydrogen…..bonding is not possible because there are no O–H, N–H, or F–H bonds. b) The –A: H–B– sequence is present in the following structure (one of the possible hydrogen bonding structures): H

N H

…..

c) The –A: H–B– sequence is present in the following structure (one of the possible hydrogen bonding structures):

12-3

H O

H H The hydrogens attached to the carbons cannot form H bonds.

12.5A

Plan: Hydrogen bonding occurs in compounds in which hydrogen is directly bonded to O, N, or F. Dipole-dipole forces occur in compounds that are polar. Dispersion forces are the dominant forces in nonpolar compounds. Boiling point increases with increasing strength of intermolecular forces; hydrogen bonds are stronger than dipole-dipole forces, which are stronger than dispersion forces. Forces generally increase in strength as molar mass increases. Solution: a) Both CH3Br and CH3F are polar molecules that experience dipole-dipole and dispersion intermolecular interactions. Because CH3Br (ℳ = 94.9 g/mol) is ~3 times larger than CH3F (ℳ = 34.0 g/mol), dispersion forces result in a higher boiling point for CH3Br. b) CH3CH2CH2OH, n–propanol, forms hydrogen bonds and has dispersion forces whereas CH3CH2OCH3, ethyl methyl ether, has only dipole-dipole and dispersion attractions. CH3CH2CH2OH has the higher boiling point. c) Both C2H6 and C3H8 are nonpolar and experience dispersion forces only. C3H8, with the higher molar mass, experiences greater dispersion forces and has the higher boiling point.

12.5B

Plan: Hydrogen bonding occurs in compounds in which hydrogen is directly bonded to O, N, or F. Dipole-dipole forces occur in compounds that are polar. Dispersion forces are the dominant forces in nonpolar compounds. Boiling point increases with increasing strength of intermolecular forces; hydrogen bonds are stronger than dipole-dipole forces, which are stronger than dispersion forces. Forces generally increase in strength as molar mass increases. Solution: a) Both CH3CHO and CH3CH2OH are polar molecules that experience dipole-dipole and dispersion intermolecular interactions. Because of its O–H bond, CH3CH2OH also experiences hydrogen bonding. Because CH3CHO does not experience the stronger hydrogen bonds, it has a lower boiling point than CH3CH2OH. b) SO2 molecules have a bent molecular geometry; both S—O bonds are polar and the molecule is polar; SO2 has both dispersion and dipole-dipole forces. CO2 is a linear molecule and is nonpolar since the two C=O bonds balance each other so CO2 has only dispersion forces. Since CO2 has the weaker forces, it has the lower boiling point. c) Both H2NCOCH2CH3 and (CH3)2NCHO are polar molecules that experience dipole-dipole and dispersion intermolecular interactions. Because of its N–H bonds, H2NCOCH2CH3 also experiences hydrogen bonding. Because (CH3)2NCHO does not experience the stronger hydrogen bonds, it has a lower boiling point than H2NCOCH2CH3.

12.6A

12-4

Solution: a) Looking at the sulfide ions, there is one ion at each corner and one ion on each face. The total number of sulfide ions is 1/8 (8 corner ions) + 1/2 (6 face ions) = 4 S2– ions. There are also 4 Pb2+ ions due to the 1:1 ratio of S2– ions to Pb2+ ions. Each ion has 6 nearest neighbor particles, so the coordination number for both ions is 6. b) There is one atom at each corner and one atom in the center of the unit cell. The total number of atoms is 1/8 (8 corner atoms) + 1 (1 body atom) = 2 W atoms. Each atom has 8 nearest neighbor particles, so the coordination number for W is 8. c) There is one atom at each corner and one atom on each face of the unit cell. The total number of atoms is 1/8 (8 corner atoms) + 1/2 (6 face atoms) = 4 Al atoms. Each atom has 12 nearest neighbor particles, so the coordination number for Al is 12. 12.6B

Plan: To determine the number of atoms or ions in each unit cell, count the number of atoms/ions at the corners, faces, and center of the unit cell. Atoms at the eight corners are shared by eight cells for a total of 8 atoms × 1/8 atom per cell = 1 atom; atoms in the body of a cell are in that cell only; atoms at the faces are shared by two cells: 6 atoms × 1/2 atom per cell = 3 atoms. Count the number of nearest neighboring particles to obtain the coordination number. Solution: a) There is one atom at each corner and one atom in the center of the unit cell. The total number of atoms is 1/8 (8 corner atoms) + 1 (1 body atom) = 2 K atoms. Each atom has 8 nearest neighbor particles, so the coordination number for K is 8. b) There is one atom at each corner and one atom on each face of the unit cell. The total number of atoms is 1/8 (8 corner atoms) + 1/2 (6 face atoms) = 4 Pt atoms. Each atom has 12 nearest neighbor particles, so the coordination number for Pt is 12. c) Looking at the chloride ion, there is one ion in the center of the unit cell, so there is 1 Cl– ion in the unit cell. There is one Cs+ ion at each corner, so there is 1/8 (8 corner ions) = 1 Cs+ ion. Each ion has 8 nearest neighbor particles, so the coordination number for both ions is 8.

12.7A

Plan: Use the radius of the Co atom to find the volume of one Co atom. Use Avogadro’s number to find the volume of one mole of Co atoms, and convert from pm3 to cm3. Use the volume of one mole of Co atoms and the packing efficiency of the solid (0.74 for face centered cubic) to find the volume of one mole of Co metal. Calculate the density of Co metal by dividing the molar mass of Co by the volume of one mole of Co metal. Solution: V of one Co atom = 4 πr 3 = ⎛⎜ 4 ⎞⎟⎟ (π )(125 pm)3 = 8.18123 × 106 pm 3 ⎜⎝ 3 ⎠⎟ 3 V of one mole of Co atoms = 23 3 ⎛ 8.18123×106 pm3 ⎞⎛ ⎟⎟⎜⎜ 6.022 ×10 Co atoms ⎞⎛ ⎟⎟⎜⎜ 1 cm ⎞⎟⎟ ⎜⎜ = 4.9267 cm3/mole Co atoms ⎟⎟⎜ ⎟⎟⎜ 30 ⎜⎜ 3⎟ ⎟ ⎜ ⎜ 1 Co atom ⎟⎜ 1 mol Co atoms ⎠⎝ ⎟⎜10 pm ⎠⎟ ⎝⎜ ⎠⎝

V of one mole of Co metal =

4.9267 cm 3 /mol Co atoms 0.74

= 6.6578 cm3/mol Co metal

⎛ 58.93 g Co ⎞⎟⎜⎛ 1 mol Co ⎞⎟ ⎜ 3 ⎟ ⎟⎟⎜⎜ Density of Co metal = ⎜⎜ ⎟⎟⎜ 6.6578 cm 3 ⎟⎟⎟ = 8.8513 = 8.85 g/cm ⎜⎝⎜ 1 mol Co ⎠⎝ ⎜ ⎠

12.7B

Plan: Use the volume of the Fe atom to find the radius of the atom; use the radius to find the edge length of the body-centered cubic cell. Find the volume of the cell by cubing the value of the edge length. Use the fact that there are two Fe atoms in the cell. The molar mass and density of iron are then used to find the number of Fe atoms in a mole (Avogadro’s number).

12-5

Solution: 4 V = πr 3 3 r=

3(8.38 ×10 3V = 3 4π 4π

−24

cm 3 )

= 1.260042 × 10–8 cm

For a body-centered cubic unit cell, the edge length =

4 (1.260042 ×10−8 cm )

3 3 –8 = 2.90994 × 10 cm

Volume (cm3) of the cell = (edge length ) = (2.909942 ×10−8 cm) = 2.46407 × 10–23 cm3 3

⎛ ⎞⎟⎛⎜ cm3 ⎞⎟⎛ 55.85 g ⎞⎟ 2 Fe atoms ⎜ ⎟⎟⎜⎜ ⎟⎟⎜⎜ ⎟⎟ Avogadro’s number = ⎜⎜ ⎜ ⎜⎝ 2.46407 ×10−23 cm3 ⎠⎟⎟⎜⎜⎝ 7.874 g ⎠⎟⎟⎝⎜ mol ⎠⎟⎟ = 5.75711 × 1023 = 5.8 × 1023 atoms/mol 12.8A

Plan: Use the relationship between radius and edge length for a body-centered cubic structure shown in Figure 12.29. Solution: ⎛ 1 nm ⎞⎟ 4r 4(126 pm) ⎜ ⎟⎟ = 0.291 nm = (290.98454 pm) ⎜⎜ Edge length = = ⎜⎜⎝1000 pm ⎠⎟⎟ 3 3

12.8B

Plan: Use the relationship between radius and edge length for a cubic closest packed structure shown in Figure 12.29. Solution: A = 8r , so r = A/ 8 ⎛1000 pm ⎞⎟ ⎜ ⎟⎟ = 143 pm r = 0.405 nm/ 8 = (0.143 nm) ⎜⎜ ⎜⎝⎜ 1 nm ⎠⎟⎟

TOOLS OF THE LABORATORY BOXED READING PROBLEMS

B12.1

Plan: The Bragg equation gives the relationship between the angle of incoming light, θ, the wavelength of the light, λ, and the distance between layers in a crystal, d. Solution: n λ = 2d sin θ n = 1; λ = 0.709 × 10–10 m;

θ = 11.6°

–10

1(0.709 × 10 m) = 2d sin 11.6° 0.709 × 10–10 m = 0.4021558423 d d = 1.762998 × 10–10 = 1.76 × 10–10 m

B12.2

Plan: The Bragg equation gives the relationship between the angle of incoming light, θ, the wavelength of the light, λ, and the distance between layers in a crystal, d. Solution: a) d, the distance between the layers in the NaCl crystal, must be found. The edge length of the NaCl unit cell is equal to the sum of the diameters of the two ions: Edge length (pm) = 204 pm (Na+) + 362 pm (Cl–) = 566 pm. The spacing between the layers is half the distance of the edge length: 566/2 = 283 pm.

12-6

n = 1;

λ = ?;

θ = 15.9°;

d = 566 pm

nλ = 2d sin θ 2 d sinθ 2 (283 pm) sin (15.9ο ) λ= = = 155.0609178 = 155 pm n 1 λ = 155 pm; θ = ?; d = 566 pm b) n = 2; nλ = 2d sin θ

sin θ

2 (155 pm) nλ = 2d 2 (283 pm )

sin θ = 0.5477031802 θ = 33.20958 = 33.2° END–OF–CHAPTER PROBLEMS

12.1

The energy of attraction is a potential energy and denoted Ep. The energy of motion is kinetic energy and denoted Ek. The relative strength of Ep vs. Ek determines the phase of the substance. In the gas phase, Ep << Ek because the gas particles experience little attraction for one another and the particles are moving very fast. In the solid phase, Ep >> Ek because the particles are very close together and are only vibrating in place. Two properties that differ between a gas and a solid are the volume and density. The volume of a gas expands to fill the container it is in while the volume of a solid is constant, no matter what container holds the solid. Density of a gas is much less than the density of a solid. The density of a gas also varies significantly with temperature and pressure changes. The density of a solid is only slightly altered by changes in temperature and pressure. Compressibility and ability to flow are other properties that differ between gases and solids.

12.2

a) Gases are more easily compressed than liquids because the distance between particles is much greater in a gas than in a liquid. Liquids have very little free space between particles and thus can be compressed (crowded together) only very slightly. b) Liquids have a greater ability to flow because the interparticle forces are weaker in the liquid phase than in the solid phase. The stronger interparticle forces in the solid phase fix the particles in place. Liquid particles have enough kinetic energy to move around.

12.3

a) intermolecular

12.4

a) Heat of fusion refers to the change between the solid and liquid states and heat of vaporization refers to the change between liquid and gas states. In the change from solid to liquid, the kinetic energy of the molecules must increase only enough to partially offset the intermolecular attractions between molecules. In the change from liquid to gas, the kinetic energy of the molecules must increase enough to overcome the intermolecular forces. The energy to overcome the intermolecular forces for the molecules to move freely in the gaseous state is much greater than the amount of energy needed to allow the molecules to move more easily past each other but still stay very close together. b) The net force holding molecules together in the solid state is greater than that in the liquid state. Thus, to change solid molecules to gaseous molecules in sublimation requires more energy than to change liquid molecules to gaseous molecules in vaporization. c) At a given temperature and pressure, the magnitude of ΔHvap is the same as the magnitude of ΔHcond. The only difference is in the sign: ΔHvap = –ΔHcond.

12.5

Plan: Intermolecular forces (nonbonding forces) are the forces that exist between molecules that attract the molecules to each other; these forces influence the physical properties of substances. Intramolecular forces

b) intermolecular

c) intermolecular

d) intramolecular

12-7

(bonding forces) exist within a molecule and are the forces holding the atoms together in the molecule; these forces influence the chemical properties of substances. Solution: a) Intermolecular—Oil evaporates when individual oil molecules can escape the attraction of other oil molecules in the liquid phase. b) Intermolecular—The process of butter (fat) melting involves a breakdown in the rigid, solid structure of fat molecules to an amorphous, less ordered system. The attractions between the fat molecules are weakened, but the bonds within the fat molecules are not broken. c) Intramolecular—A process called oxidation tarnishes pure silver. Oxidation is a chemical change and involves the breaking of bonds and formation of new bonds. d) Intramolecular—The decomposition of O2 molecules into O atoms requires the breaking of chemical bonds, i.e., the force that holds the two O atoms together in an O2 molecule. Both (a) and (b) are physical changes, whereas (c) and (d) are chemical changes. In other words, intermolecular forces are involved in physical changes while intramolecular forces are involved in chemical changes. 12.6

a) intermolecular

b) intramolecular

12.7

a) Condensation

The water vapor in the air condenses to liquid when the temperature drops during the night. Solid ice melts to liquid water. Liquid water on clothes evaporates to water vapor.

b) Fusion (melting) c) Evaporation

c) intermolecular

d) intermolecular

12.8

a) deposition

12.9

The propane gas molecules slow down as the gas is compressed. Therefore, much of the kinetic energy lost by the propane molecules is released to the surroundings upon liquefaction.

12.10

Sublimation and deposition

12.11

The gaseous PCl3 molecules are moving faster and are farther apart than the liquid molecules. As they condense, the kinetic energy of the molecules is changed into potential energy stored in the dipole-dipole interactions between the molecules.

12.12

The two processes are the formation of solid from liquid and the formation of liquid from solid (at the macroscopic level). At the molecular level, the two processes are the removal of kinetic energy from the liquid molecules as they solidify and the overcoming of the dispersion forces between the molecules as they turn to liquid.

12.13

In closed containers, two processes, evaporation and condensation, occur simultaneously. Initially there are few molecules in the vapor phase, so more liquid molecules evaporate than gas molecules condense. Thus, the number of molecules in the gas phase increases, causing the vapor pressure of hexane to increase. Eventually, the number of molecules in the gas phase reaches a maximum where the number of liquid molecules evaporating equals the number of gas molecules condensing. In other words, the evaporation rate equals the condensation rate. At this point, there is no further change in the vapor pressure.

12.14

a) At the critical temperature, the molecules are moving so fast that they can no longer be condensed. This temperature decreases with weaker intermolecular forces because the forces are not strong enough to overcome molecular motion. Alternatively, as intermolecular forces increase, the critical temperature increases. b) As intermolecular forces increase, the boiling point increases because it becomes more difficult and takes more energy to separate molecules from the liquid phase. c) As intermolecular forces increase, the vapor pressure decreases for the same reason given in (b). At any given temperature, strong intermolecular forces prevent molecules from easily going into the vapor phase and thus vapor pressure is decreased.

b) sublimation

c) crystallization (freezing)

12-8

d) As intermolecular forces increase, the heat of vaporization increases because more energy is needed to separate molecules from the liquid phase. 12.15

Point 1 is depicted by C. This is the equilibrium between melting and freezing. Point 2 is depicted by A. This is the equilibrium between vaporization and condensation. Point 3 is depicted by D. This is the equilibrium between sublimation and deposition.

12.16

a) The final pressure will be the same, since the vapor pressure is constant as long as some liquid is present. b) The final pressure will be lower, according to Boyle’s law.

12.17

If the solid is more dense than the liquid, the solid-liquid line slopes to the right; if less dense, to the left.

12.18

When water at 100°C touches skin, the heat released is from the lowering of the temperature of the water. The specific heat of water is approximately 75 J/mol ∙ K. When steam at 100°C touches skin, the heat released is from the condensation of the gas with a heat of condensation of approximately 41 kJ/mol. Thus, the amount of heat released from gaseous water condensing will be greater than the heat from hot liquid water cooling and the burn from the steam will be worse than that from hot water.

12.19

Plan: The total heat required is the sum of three processes: warming the ice to 0.00°C, the melting point; melting the ice to liquid water; warming the water to 0.500°C. The equation q = c × mass × ΔT is used to calculate the heat involved in changing the temperature of the ice and of the water; the heat of fusion is used to calculate the heat involved in the phase change of ice to water. Solution: 1) Warming the ice from –6.00°C to 0.00°C: q1 = c × mass × ΔT = (2.09 J/g°C)(22.00 g)[0.0 – (–6.00)]°C = 275.88 J

2) Phase change of ice at 0.00°C to water at 0.00°C: ⎛ 1 mol ⎞⎟⎛ 6.02 kJ ⎞⎛103 J ⎞⎟ ο ⎟⎟⎜ q2 = n (ΔHfus ⎟ = 7349.6115 J ) = (22.0 g)⎜⎜⎜ ⎟⎟⎜⎜ ⎝18.02 g ⎠⎟⎜⎝ mol ⎠⎟⎜⎝ 1 kJ ⎠⎟⎟ 3) Warming the liquid from 0.00°C to 0.500°C: q3 = c × mass × ΔT = (4.184 J/g°C)(22.00 g)[0.500 – (0.0)]°C = 46.024 J The three heats are positive because each process takes heat from the surroundings (endothermic). The phase change requires much more energy than the two temperature change processes. The total heat is q1 + q2 + q3 = (275.88 J + 7349.6115 J + 46.024 J) = 7671.5155 = 7.67 × 103 J.

12.20

0.333 mol × 46.07 g/mol = 15.34131 g ethanol Cooling vapor to boiling point: q1 = c × mass × ΔT = (1.43 J/g°C)(15.34131 g)(78.5 – 300)°C = –4859.28 J (note ΔHcond = –ΔHvap)

Condensing vapor: q2 = n (ΔH

ο cond

) = (0.333 mol)(–40.5 kJ/mol)(103 J/kJ) = –13,486.5 J

Cooling liquid to 25.0°C: q3 = c × mass × ΔT = (2.45 J/g°C)(15.34131 g)(25.0 – 78.5)°C = –2010.86 J qtotal = q1 + q2 + q3 = (–4859.28 J) + (–13,486.5 J) + (–2010.86 J) = –20356.64 = –2.04 × 104 J

12.21

12-9

Solution: ln

ο ⎛ ⎞ −ΔH vap P2 ⎜⎜ 1 − 1 ⎟⎟ = ⎜ P1 R ⎝ T2 T1 ⎠⎟⎟

ο ΔHvap = 35.5 kJ/mol

P1 = 1.00 atm

T1 = 122°C + 273 = 395 K

P2 = ?

T2 = 113°C + 273 = 386 K

kJ 3 −35.5 ⎛ 1 1 ⎞⎟⎜⎛⎜10 J ⎞⎟⎟ P2 mol ⎜ − = ln ⎟⎟⎜⎜ 1kJ ⎟⎟⎟ = –0.2520440 8.314 J/mol • K ⎜⎝ 386 K 395 K ⎠⎜ 1.00 atm ⎝ ⎠ P2 = 0.7772105 1.00 atm

P2 = (0.7772105)(1.00 atm) = 0.7772105 = 0.777 atm

12.22

The Clausius-Clapeyron equation gives the relationship between vapor pressure and temperature. ο ⎛ ⎞ −ΔH vap P ⎜⎜ 1 − 1 ⎟⎟ ln 2 = ⎜ P1 R ⎝ T2 T1 ⎠⎟⎟

kJ 3 −29.1 1 ⎞⎟⎜⎛⎜10 J ⎞⎟⎟ ⎛ 1 P2 mol ⎜ − = ln ⎟⎟⎜⎜ 1kJ ⎟⎟⎟ = 2.2314173653 8.314 J/mol • K ⎜⎝ 368 K 298 K ⎠⎜ 0.703 atm ⎝ ⎠ P2 = 9.338762 0.703 atm

P2 = (9.338762)(0.703atm) = 6.56515 = 6.57 atm

12.23

Plan: The Clausius-Clapeyron equation gives the relationship between vapor pressure and temperature. We are ο . The pressure in torr must be given P1, P2, T1, and T2; these values are substituted into the equation to find ΔHvap converted to atm. Solution: ⎛ 1 atm ⎞⎟ P1 = (621 torr )⎜⎜ = 0.817105 atm T1 = 85.2°C + 273.2 = 358.4 K ⎝ 760 torr ⎠⎟⎟ ο ΔHvap P2 = 1 atm T2 = 95.6°C + 273.2 = 368.8 K =? ln

ο ⎛ ⎞ −ΔH vap P2 ⎜⎜ 1 − 1 ⎟⎟ = ⎜ P1 R ⎝ T2 T1 ⎠⎟⎟

ο −ΔH vap 1 atm ⎛ 1 1 ⎞⎟ ⎜ = − ⎟ ⎜ ⎝ 8.314 J /mol • K 368.8 K 358.4 K ⎠⎟ 0.817105 atm

0.2019877 = –ΔHvap(–9.463775 × 10–6)J/mol ο ΔHvap = 21,343.248 = 2 × 104 J/mol

(The significant figures in the answer are limited by the 1 atm in the problem.) 12.24

The Clausius-Clapeyron equation gives the relationship between vapor pressure and temperature. ln

ο ⎛ ⎞ −ΔH vap P2 ⎜⎜ 1 − 1 ⎟⎟ = P1 R ⎜⎝ T2 T1 ⎠⎟⎟

⎛ ⎞⎟ ⎜ 1 1 ⎟⎟ ⎜ − ⎜⎜ ln = ⎟ 8.314 J/mol • K ⎜⎜ 273 + (−100) K 1 atm 273 + (−164) K ⎠⎟⎟ ⎝ ο 3.756538 = 0.00040822 ΔHvap 42.8 atm

ο −ΔHvap

(

)

(

)

12-10

ο ΔHvap = (3.756538)/(0.00040822) = 9202.2 = 9 × 103 J/mol

(The significant figures in the answer are limited by 1 atm in the problem.) 12.25

P (atm)

50.5

L G

10−3 −200

−100 T (°C)

The pressure scale is distorted to represent the large range in pressures given in the problem, so the liquid-solid curve looks different from the one shown in the text. The important features of the graph include the distinction between the gas, liquid, and solid states, and the melting point T, which is located directly above the critical T. Solid ethylene is denser than liquid ethylene since the solid-liquid line slopes to the right with increasing pressure. 12.26

P, atm

1 30

10 T, K

Hydrogen does sublime at 0.05 atm, since 0.05 atm is below the triple point pressure.

12-11

12.27

Plan: Refer to the phase diagram to describe the specified changes. In part (a), a substance sublimes if it converts directly from a solid to a gas without passing through a liquid phase. In part (b), refer to the diagram to explain the changes that occur when sulfur is heated at a constant pressure of 1 atm. Solution: a) Rhombic sulfur will sublime when it is heated at a pressure less than 1 × 10–4 atm. b) At 90°C and 1 atm, sulfur is in the solid (rhombic form). As it is heated at constant pressure, it passes through the solid (monoclinic) phase, starting at 114°C. At about 120°C, the solid melts to form a liquid. At about 445°C, the liquid evaporates and changes to the gas state.

12.28

Plan: Refer to the phase diagram to describe the specified changes. In part (a), identify the phase at the given conditions. In part (b), refer to the diagram to explain the changes that occur when xenon is compressed at a constant temperature of –115°C. Solution: a) At room temperature and pressure, xenon is a gas. b) At –115°C and 0.5 atm, Xe is a gas. As it is compressed at constant temperature, it passes through the liquid phase (starting at about 0.6 atm). At about 0.75 atm, Xe becomes a solid (undergoes fusion).

12.29

This is a stepwise problem to calculate the total heat required. Melting SO2: ο q1 = n (ΔHfus ) = (2.500 kg)(103 g/1 kg)(1 mol SO2/64.06 g SO2)(8.619 kJ/mol)(103 J/kJ) = 336,364.3 J Warming liquid SO2: q2 = c × mass × ΔT = (0.995 J/g°C)(2.500 kg)(103 g/1 kg)[–10. – (–73)]°C = 156,712.5 J Vaporizing SO2: ο q3 = n (ΔH vap ) = (2.500 kg)(103 g/1 kg)(1 mol SO2/64.06 g SO2)(25.73 kJ/mol)(103 J/kJ) = 1,004,136.7 J Warming gaseous SO2: q4 = c × mass × ΔT = (0.622 J/g°C)(2.500 kg)(103 g/1 kg)[60 – (–10)]°C = 108,850 J qtotal = q1 + q2 + q3 + q4 = (336,364.3 J) + (156,712.5 J) + (1,004,136.7 J) + (108,850 J) = 1,606,063.5 = 1.606 × 106 J

12.30

Plan: The Clausius-Clapeyron equation gives the relationship between vapor pressure and temperature. We are ο given ΔHvap , P1, T1, and T2; these values are substituted into the equation to find P2. Convert the temperatures ο from °C to K and ΔHvap from kJ/mol to J/mol to allow cancellation with the units in R. Solution: P1 = 2.3 atm T1 = 25.0°C + 273 = 298 K ο ΔHvap P2 = ? T2 = 135°C + 273 = 408 K = 24.3 kJ/mol ο ⎛ ⎞ −Δ H P 1 vap ⎜ 1 − ⎟⎟⎟ ln 2 = ⎜ P1 R ⎜⎝ T2 T1 ⎠⎟ kJ 3 −24.3 P2 ⎛ 1 1 ⎞⎟⎛⎜10 J ⎞⎟⎟ mol ⎜ ⎜ − = ln ⎟ ⎟ 8.314 J /mol • K ⎜⎝ 408 K 298 K ⎠⎟⎜⎜⎝ 1 kJ ⎠⎟⎟ 2.3 atm ln

P2 = 2.644311 2.3 atm

P2 = 14.07374 2.3 atm

P2 = (14.07374)(2.3 atm) = 32.3696 = 32 atm

12-12

12.31

a) At 20°C and 40°C, no liquid exists, only gas. At –40°C, liquid exists. At –120°C, no liquid exists, only solid. b) No, at any pressure below the triple point pressure, the CO2(s) will sublime. c) No d) No

12.32

Intermolecular forces involve interactions of lower (partial) charges at relatively larger distances than in covalent bonds.

12.33

a) Scene A: dipole-dipole forces; Scene B: dipole-dipole forces; Scene C: ion-dipole forces; Scene D: hydrogen bonds b) dipole-dipole forces < hydrogen bonds < ion-dipole

12.34

To form hydrogen bonds, the atom bonded to hydrogen must have two characteristics: small size and high electronegativity (so that the atom has a very high electron density). With this high electron density, the attraction for a hydrogen on another molecule is very strong. Selenium is much larger than oxygen (atomic radius of 119 pm vs. 73 pm) and less electronegative than oxygen (2.4 for Se and 3.5 for O) resulting in an electron density on Se in H2Se that is too small to form hydrogen bonds.

12.35

The I–I distance within an I2 molecule is shorter than the I–I distance between adjacent molecules. This is because the I–I interaction within an I2 molecule is a true covalent bond and the I–I interaction between molecules is a dispersion force (intermolecular) which is considerably weaker.

12.36

All particles (atoms and molecules) exhibit dispersion forces, but these are the weakest of intermolecular forces. The dipole-dipole forces in polar molecules dominate the dispersion forces.

12.37

Polarity refers to a permanent imbalance in the distribution of electrons in the molecule. Polarizability refers to the ability of the electron distribution in a molecule to change temporarily. The polarity affects dipole-dipole interactions, while the polarizability affects dispersion forces.

12.38

If the electron distribution in one molecule is not symmetrical (permanent or temporary), that can induce a temporary dipole in an adjacent molecule by causing the electrons in that molecule to shift for some (often short) time.

12.39

Plan: Dispersion forces are the only forces between nonpolar substances; dipole-dipole forces exist between polar substances. Hydrogen bonds only occur in substances in which hydrogen is directly bonded to either oxygen, nitrogen, or fluorine. Solution: a) Hydrogen bonding will be the strongest force between methanol molecules since they contain O–H bonds. Dipole-dipole and dispersion forces also exist. b) Dispersion forces are the only forces between nonpolar carbon tetrachloride molecules and, thus, are the strongest forces. c) Dispersion forces are the only forces between nonpolar chlorine molecules and, thus, are the strongest forces.

12.40

a) Hydrogen bonding

12.41

Plan: Dispersion forces are the only forces between nonpolar substances; dipole-dipole forces exist between polar substances. Hydrogen bonds only occur in substances in which hydrogen is directly bonded to either oxygen, nitrogen, or fluorine. Solution: a) Dipole-dipole interactions will be the strongest forces between methyl chloride molecules because the C–Cl bond has a dipole moment. b) Dispersion forces dominate because CH3CH3 (ethane) is a symmetrical nonpolar molecule. c) Hydrogen bonding dominates because hydrogen is bonded to nitrogen, which is one of the three atoms (N, O, or F) that participate in hydrogen bonding.

b) Dipole-dipole

c) Ionic bonds

12-13

b) Dipole-dipole

c) Hydrogen bonding

12.42

a) Dispersion

12.43

Plan: Hydrogen bonds are formed when a hydrogen atom is bonded to N, O, or F. Solution: a) The presence of an OH group leads to the formation of hydrogen bonds in CH3CH(OH)CH3. There are no hydrogen bonds in CH3SCH3.

H H

H C

C H

H C

C H

H b) The presence of H attached to F in HF leads to the formation of hydrogen bonds. There are no hydrogen bonds in HBr. F

12.44

a) The presence of H directly attached to the N in (CH3)2NH leads to hydrogen bonding. More than one arrangement is possible. H H

H N

H H

C H H

b) Each of the hydrogen atoms directly attached to oxygen atoms in HOCH2CH2OH leads to hydrogen bonding. More than one arrangement is possible. In FCH2CH2F, the H atoms are bonded to C so there is no hydrogen bonding.

H O

H C

H O

C H

H O

12-14

12.45

Plan: In the vaporization process, intermolecular forces between particles in the liquid phase must be broken as the particles enter the vapor phase. In other words, the question is asking for the strongest interparticle force that must be broken to vaporize the liquid. Dispersion forces are the only forces between nonpolar substances; dipole-dipole forces exist between polar substances. Hydrogen bonds only occur in substances in which hydrogen is directly bonded to either oxygen, nitrogen, or fluorine. Solution: a) Dispersion, because hexane, C6H14, is a nonpolar molecule. b) Hydrogen bonding; hydrogen is bonded to oxygen in water. A single water molecule can engage in as many as four hydrogen bonds. c) Dispersion, although the individual Si–Cl bonds are polar, the molecule has a symmetrical, tetrahedral shape and is therefore nonpolar.

12.46

a) Dispersion

12.47

Plan: Polarizability increases down a group and decreases from left to right because as atomic size increases, polarizability increases. Solution: a) Iodide ion has greater polarizability than the bromide ion because the iodide ion is larger. The electrons can be polarized over a larger volume in a larger atom or ion. b) Ethene (CH2=CH2) has greater polarizability than ethane (CH3CH3) because the electrons involved in π bonds are more easily polarized than electrons involved in σ bonds. c) H2Se has greater polarizability than water because the selenium atom is larger than the oxygen atom.

12.48

a) Ca b) CH3CH2CH3 c) CCl4 In all cases, the larger molecule (i.e., the one with more electrons) has the higher polarizability.

12.49

Plan: Weaker attractive forces result in a higher vapor pressure because the molecules have a smaller energy barrier in order to escape the liquid and go into the gas phase. Decide which of the two substances in each pair has the weaker interparticle force. Dispersion forces are weaker than dipole-dipole forces, which are weaker than hydrogen bonds. Solution: a) C2H6 C2H6 is a smaller molecule exhibiting weaker dispersion forces than C4H10. b) CH3CH2F CH3CH2F has no H–F bonds (F is bonded to C, not to H), so it only exhibits dipole-dipole forces, which are weaker than the hydrogen bonding in CH3CH2OH. c) PH3 PH3 has weaker intermolecular forces (dipole-dipole) than NH3 (hydrogen bonding).

12.50

a) HOCH2CH2OH has a stronger intermolecular force, because there are more −OH groups to hydrogen bond. b) CH3COOH has a stronger intermolecular force, because hydrogen bonding is stronger than dipole-dipole forces. c) HF has a stronger intermolecular force, because hydrogen bonding is stronger than dipole-dipole forces.

12.51

Plan: The weaker the interparticle forces, the lower the boiling point. Decide which of the two substances in each pair has the weaker interparticle force. Dispersion forces are weaker than dipole-dipole forces, which are weaker than hydrogen bonds, which are weaker than ionic forces. Solution: a) HCl would have a lower boiling point than LiCl because the dipole-dipole intermolecular forces between hydrogen chloride molecules in the liquid phase are weaker than the significantly stronger ionic forces holding the ions in lithium chloride together.

b) Dipole-dipole

c) Hydrogen bonding

12-15

b) PH3 would have a lower boiling point than NH3 because the intermolecular forces in PH3 are weaker than those in NH3. Hydrogen bonding exists between NH3 molecules but weaker dipole-dipole forces hold PH3 molecules together. c) Xe would have a lower boiling point than iodine. Both are nonpolar with dispersion forces, but the forces between xenon atoms would be weaker than those between iodine molecules since the iodine molecules are more polarizable because of their larger size. 12.52

a) CH3CH2OH, hydrogen bonding (CH3CH2OH) vs. dispersion (CH3CH2CH3) b) NO, dipole-dipole (NO) vs. dispersion (N2) c) H2Te, the larger molecule has larger dispersion forces

12.53

Plan: The weaker the intermolecular forces, the lower the boiling point. Decide which of the two substances in each pair has the weaker intermolecular force. Dispersion forces are weaker than dipole-dipole forces, which are weaker than hydrogen bonds, which are weaker than ionic forces. Solution: a) C4H8, the cyclic molecule, cyclobutane, has less surface area exposed, so its dispersion forces are weaker than the straight chain molecule, C4H10. b) PBr3, the dipole-dipole forces of phosphorous tribromide are weaker than the ionic forces of sodium bromide. c) HBr, the dipole-dipole forces of hydrogen bromide are weaker than the hydrogen bonding forces of water.

12.54

a) CH3OH, hydrogen bonding (CH3OH) vs. dispersion forces (CH3CH3). b) FNO, greater polarity in FNO vs. ClNO c) F F C H

C H

This molecule has dipole-dipole forces since the two C–F bonds do not cancel and the molecule is polar. The other molecule has only dispersion forces since the two C–F bonds do cancel, so that the molecule is nonpolar. 12.55

The trend in both atomic size and electronegativity predicts that the trend in increasing strength of hydrogen bonds is N–H < O–H < F–H. As the atomic size decreases and electronegativity increases, the electron density of the atom increases. High electron density strengthens the attraction to a hydrogen atom on another molecule. Fluorine is the smallest of the three and the most electronegative, so its hydrogen bonds would be the strongest. Oxygen is smaller and more electronegative than nitrogen, so hydrogen bonds for water would be stronger than hydrogen bonds for ammonia.

12.56

The molecules of motor oil are long chains of CH2 units. The high molar mass results in stronger dispersions forces and leads to a high boiling point. In addition, these chains can become tangled in one another and restrict each other’s motions and ease of vaporization.

12.57

The ethylene glycol molecules have two sites (two −OH groups) which can hydrogen bond; the propanol has only one −OH group.

12.58

The molecules at the surface are attracted to one another and to those molecules in the bulk of the liquid. Since this force is directed downwards and sideways, it tends to “tighten the skin.”

12.59

The shape of the drop depends upon the competing cohesive forces (attraction of molecules within the drop itself) and adhesive forces (attraction between molecules in the drop and the molecules of the waxed floor). If the cohesive forces are strong and outweigh the adhesive forces, the drop will be as spherical as gravity will allow. If,

12-16

on the other hand, the adhesive forces are significant, the drop will spread out. Both water (hydrogen bonding) and mercury (metallic bonds) have strong cohesive forces, whereas cohesive forces in oil (dispersion) are relatively weak. Neither water nor mercury will have significant adhesive forces to the nonpolar wax molecules, so these drops will remain nearly spherical. The adhesive forces between the oil and wax can compete with the weak, cohesive forces of the oil (dispersion) and so the oil drop spreads out. 12.60

The presence of the ethanol molecules breaks up some of the hydrogen bonding interactions present between the water molecules, lowering the surface tension.

12.61

Surface tension is defined as the energy needed to increase the surface area by a given amount, so units of energy (J) per surface area (m2) describe this property.

12.62

The strength of the intermolecular forces does not change when the liquid is heated, but the molecules have greater kinetic energy and can overcome these forces more easily as they are heated. The molecules have more energy at higher temperatures, so they can break the intermolecular forces and can move more easily past their neighbors; thus, viscosity decreases.

12.63

Plan: The stronger the intermolecular force, the greater the surface tension. Decide which of the substances has the weakest intermolecular force and which has the strongest. Dispersion forces are weaker than dipole-dipole forces, which are weaker than hydrogen bonds, which are weaker than ionic forces. Solution: All three molecules exhibit hydrogen bonding (H is bonded to O), but the extent of hydrogen bonding increases with the number of O–H bonds present in each molecule. HOCH2CH(OH)CH2OH with three O–H groups can form more hydrogen bonds than HOCH2CH2OH with two O–H groups, which in turn can form more hydrogen bonds than CH3CH2CH2OH with only one O–H group. The greater the number of hydrogen bonds, the stronger the intermolecular forces, and the higher the surface tension. CH3CH2CH2OH < HOCH2CH2OH < HOCH2CH(OH)CH2OH

12.64

CH3OH > H2CO > CH3CH3 The intermolecular forces would decrease as shown (hydrogen bonding > dipole-dipole > dispersion), as would the surface tension.

12.65

Plan: Viscosity is a measure of the resistance of a liquid to flow, and is greater for molecules with stronger intermolecular forces. The stronger the force attracting the molecules to each other, the harder it is for one molecule to move past another. Thus, the substance will not flow easily if the intermolecular force is strong. Decide which of the substances has the weakest intermolecular force and which has the strongest. Dispersion forces are weaker than dipole-dipole forces, which are weaker than hydrogen bonds, which are weaker than ionic forces. Solution: The ranking of decreasing viscosity is the opposite of that for increasing surface tension (Problem 12.61). HOCH2CH(OH)CH2OH > HOCH2CH2OH > CH3CH2CH2OH The greater the number of hydrogen bonds, the stronger the intermolecular forces, and the higher the viscosity.

12.66

Viscosity and surface tension both increase with increasing strength of intermolecular forces. CH3CH3 < H2CO < CH3OH

12.67

a) The more volatile substances (volatile organic pollutants) are preferentially pulled away from the less volatile substances. b) The vapor pressure of the volatile organic pollutants increases as the temperature increases. The higher vapor pressure makes it easier to remove the vapor.

12-17

12.68

a) Calculate the energies involved using the heats of fusion. ⎛ 1 mol Hg ⎞⎛ ο ⎟⎟⎜⎜ 23.4 kJ ⎞⎟⎟ = 1.3998 = 1.40 kJ qHg = n (ΔHfus ) = (12.0 g Hg)⎜⎜⎜ ⎟⎟⎜1 mol Hg ⎠⎟⎟ ⎝ 200.6 g Hg ⎠⎝ ⎛ 1 mol CH 4 ⎞⎛ ο ⎟⎟⎜⎜ 0.94 kJ ⎞⎟⎟ = 0.70324 = 0.70 kJ qmethane = n (ΔHfus ) = (12.0 g CH 4 )⎜⎜⎜ ⎟⎟⎜1 mol CH ⎟⎟ ⎝16.04 g CH 4 ⎠⎝ 4⎠ Mercury takes more energy.

b) Calculate the energies involved using the heats of vaporization. ⎛ 1 mol Hg ⎞⎛ 59 kJ ⎞ ο qHg = n (ΔH vap ) = (12.0 g Hg)⎜⎜⎜ 200.6 g Hg ⎟⎟⎟⎟⎜⎜⎜1 mol Hg ⎟⎟⎟⎟ = 3.5294 = 3.5 kJ ⎝ ⎠⎝ ⎠ ⎛ 1 mol CH ⎞⎛ 8.9 kJ ⎞ ο qmethane = n (ΔH vap ) = (12.0 g CH 4 )⎜⎜⎜16.04 g CH4 ⎟⎟⎟⎟⎜⎜⎜1 mol CH ⎟⎟⎟⎟ = 6.65835 = 6.6 kJ ⎝ 4 ⎠⎝ 4⎠

Methane takes more energy.

c) Mercury involves metallic bonding and methane involves dispersion forces. 12.69

The pentanol has stronger intermolecular forces (hydrogen bonds) than the hexane (dispersion forces).

12.70

Water is a good solvent for polar and ionic substances and a poor solvent for nonpolar substances. Water is a polar molecule and dissolves polar substances because their intermolecular forces are of similar strength. Water is also able to dissolve ionic compounds and keep ions separated in solution through ion-dipole interactions. Nonpolar substances will not be very soluble in water since their dispersion forces are much weaker than the hydrogen bonds in water. A solute whose intermolecular attraction to a solvent molecule is less than the attraction between two solvent molecules will not dissolve because its attraction cannot replace the attraction between solvent molecules.

12.71

A single water molecule can form four hydrogen bonds. The two hydrogen atoms form a hydrogen bond each to oxygen atoms on neighboring water molecules. The two lone pairs on the oxygen atom form hydrogen bonds with hydrogen atoms on neighboring molecules.

12.72

The heat capacity of water is quite high, meaning that a large amount of heat is needed to change the temperature of a quantity of water by even a small amount.

12.73

Water has a high surface tension. The debris on the surface provides shelter and nutrients for fish, insects, etc.

12.74

Water exhibits strong capillary action, which allows it to be easily absorbed by the plant’s roots and transported to the leaves.

12.75

In ice, water molecules pack in a very specific, ordered way. When it melts, the molecular order is disrupted and the molecules pack more closely. This makes liquid water (at least below 4°C) denser than ice and allows ice to float.

12.76

As the temperature of the ice increases, the water molecules move more vigorously about their fixed positions until at some temperature, the increasing kinetic energy of the water molecules at last overcomes the attractions (hydrogen bonding) between them, allowing the water molecules to move freely through the liquid.

12.77

An amorphous solid has little order on the molecular level and has no characteristic crystal shape on the macroscopic level. An example would be rubber. A crystalline solid has a great deal of order on the molecular level and forms regularly shaped forms bounded by flat faces on the macroscopic level. An example would be NaCl.

12.78

When the unit cell is repeated infinitely in all directions, the crystal lattice is formed.

12-18

12.79

The simple, body-centered, and face-centered cubic unit cells contain one, two, and four atoms, respectively. Atoms in the body of a cell are in that cell only; atoms on faces are shared by two cells; atoms at corners are shared by eight cells. All of the cells have eight corner atoms; 8 atoms × 1/8 atom per cell = 1 atom. In addition, the body-centered cell has an atom in the center, for a total of two atoms. The face-centered cell has six atoms in the faces; 6 atoms × 1/2 atom per cell = 3 atoms, for a total of 4 in the cell (corner + face).

12.80

The unit cell is a simple cubic cell. According to the bottom row in Figure 12.27, two atomic radii (or one atomic diameter) equal the width of the cell.

12.81

The layers of a body-centered arrangement are not packed in the most efficient manner. The atoms are only in contact with four other atoms; in a face-centered cubic arrangement, they contact six other atoms. This leads to closer packing and more complete filling of the space in the face-centered arrangement.

12.82

Krypton is an atomic solid. In atomic solids, the only interparticle forces are (weak) dispersion forces. Copper is a metallic solid. In metallic solids, additional forces (metallic bonds) lead to different properties.

12.83

The energy gap is the energy difference between the highest filled energy level (valence band) and the lowest unfilled energy level (conduction band). In conductors and superconductors, the energy gap is zero because the valence band overlaps the conduction band. In semiconductors, the energy gap is small but greater than zero. In insulators, the energy gap is large and thus insulators do not conduct electricity.

12.84

a) Conductivity decreases with increasing temperature. b) Conductivity increases with increasing temperature. c) Conductivity does not change with temperature.

12.85

The density of a solid depends on the atomic mass of the element (greater mass = greater density), the atomic radius (how many atoms can fit in a given volume), and the type of unit cell, which determines the packing efficiency (how much of the volume is occupied by empty space).

12.86

Plan: The simple cubic structure unit cell contains one atom since the atoms at the eight corners are shared by eight cells for a total of 8 atoms × 1/8 atom per cell = 1 atom; the body-centered cell also has an atom in the center, for a total of two atoms; the face-centered cell has six atoms in the faces which are shared by two cells: 6 atoms × ½ atom per cell = 3 atoms plus another atom from the eight corners for a total of four atoms. Solution: a) Ni is face-centered cubic since there are four atoms/unit cell. b) Cr is body-centered cubic since there are two atoms/unit cell. c) Ca is face-centered cubic since there are four atoms/unit cell.

12.87

a) one

12.88

Plan: Use the radius of the Ca atom to find the volume of one Ca atom. Use Avogadro’s number to find the volume of one mole of Ca atoms, and convert from pm3 to cm3. Use the volume of one mole of Ca atoms and the packing efficiency of the solid (0.74 for cubic closest packing) to find the volume of one mole of Ca metal. Calculate the density of Ca metal by dividing the molar mass of Ca by the volume of one mole of Ca metal. Solution: ⎛4⎞ 4 V of one Ca atom = πr 3 = ⎜⎜ ⎟⎟ (π )(197 pm)3 = 3.2025 × 107 pm3 ⎜⎝ 3 ⎠⎟ 3

b) two

c) four

23 3 ⎛ 3.2025 ×10 7 pm 3 ⎞⎛ ⎟⎟⎜⎜ 6.022 × 10 Ca atoms ⎞⎛ ⎟⎟⎜⎜ 1 cm ⎞⎟⎟ ⎜ 3 V of one mole of Ca atoms = ⎜⎜ ⎟⎟⎜ ⎟⎟⎜ 30 ⎟ = 19.285 cm ⎜⎜ ⎜⎜ 1 mol Ca atoms ⎟⎜⎜10 pm 3 ⎟⎟ 1 Ca atom ⎟ ⎝ ⎠⎝ ⎠⎝ ⎠

12-19

V of one mole of Ca metal =

19.285 cm 3 0.74

= 26.061 cm3/mol Ca metal

⎛ 40.08 g Ca ⎞⎟⎜⎛ 1 mol Ca ⎞⎟ ⎜ ⎟⎟ = 1.5379 = 1.54 g/cm3 ⎟⎟⎜⎜ Density of Ca metal = ⎜⎜ ⎜⎝⎜ 1 mol Ca ⎠⎝ ⎟⎟⎜⎜ 26.061 cm 3 ⎠⎟⎟

12.89

Plan: Determine the volume of one mole of Cr metal by dividing the molar mass of Cr by its density. Use the volume per mol of Cr metal and the packing efficiency (0.68 for a body-centered cubic unit cell) to calculate the volume per mol of Cr atoms. Use Avogadro’s number to calculate the volume of one Cr atom; then use the formula for the volume of a sphere to calculate the radius of the Cr atom. Solution: ⎛ 1 cm 3 ⎞⎛ ⎛ 1 ⎞⎟ ⎟⎟⎜ 52.00 g Cr ⎞⎟⎟ ⎜ 3 ⎟⎟ × ℳ = ⎜⎜ Volume/mol of Cr metal = ⎜⎜⎜ ⎟⎜ ⎟ = 7.2829 cm /mol Cr ⎜⎜ 7.14 g ⎟⎟⎜⎜⎜ 1 mol Cr ⎟⎟ ⎝ density ⎠⎟ ⎠ ⎝ ⎠⎝

Volume/mol of Cr atoms: cm3/mol Cr × packing efficiency = 7.2829 cm3/mol Cr × 0.68 = 4.9524 cm3/mol Cr ⎛ 4.9524 cm 3 ⎞⎛ ⎟⎟⎜⎜ 1 mol Cr atoms ⎞⎟⎟ ⎜ –24 3 Volume of Cr atom = ⎜⎜ ⎟ ⎟ = 8.2238 × 10 cm /Cr atom ⎜⎜1 mol Cr atoms ⎟⎟⎜⎜⎜ 6.022 ×10 23 Cr atoms ⎟⎟ ⎝ ⎠⎝ ⎠

V of Cr atom =

4 3 3V πr so r 3 = 3 4π −24

r= 3

3(8.2238 ×10 3V = 3 4π 4π

cm3 )

= 1.2522 × 10–8 = 1.25 × 10–8 cm

12.90

a) There is a change in unit cell from CdO in a sodium chloride structure to CdSe in a zinc blende structure. b) Yes, the coordination number of Cd does change from six in the CdO unit cell to four in the CdSe unit cell.

12.91

a) The unit cell of Fe changes from a face-centered cubic unit cell at 1674 K to a body-centered cubic unit cell below 1181 K. b) The face-centered cubic cell has the greater packing efficiency.

12.92

Plan: Use the relationship between the edge length of the unit cell (A) and the atomic radius (r) for a bodycentered cubic unit cell to solve for the edge length of the potassium unit cell. Solution: 4 For a body-centered cubic unit cell, A = r 3 4 A= (227 pm) = 524.23 = 524 pm 3

12.93

Plan: Use the relationship between the edge length of the unit cell (A) and the atomic radius (r) for a face-centered cubic unit cell to solve for the radius of lead. Solution: A For a face-centered cubic unit cell, A = 8r so =r 8 (495 pm) r= = 175.01 = 175 pm 8

12-20

12.94

Plan: Substances composed of individual atoms are atomic solids; molecular substances composed of covalent molecules form molecular solids; ionic compounds form ionic solids; metal elements form metallic solids; certain substances that form covalent bonds between atoms or molecules form network covalent solids. Solution: a) Nickel forms a metallic solid since nickel is a metal whose atoms are held together by metallic bonds. b) Fluorine forms a molecular solid since the F2 molecules have covalent bonds and the molecules are held to each other by dispersion forces. c) Methanol forms a molecular solid since the covalently bonded CH3OH molecules are held to each other by hydrogen bonds. d) Tin forms a metallic solid since tin is a metal whose atoms are held together by metallic bonds. e) Silicon is in the same group as carbon, so it exhibits similar bonding properties. Since diamond and graphite are both network covalent solids, it makes sense that Si forms the same type of bonds. f) Xe is an atomic solid since individual atoms are held together by dispersion forces.

12.95

a) Network covalent, since this is similar to diamond. b) Ionic, since it consists of ions. c) Molecular, since this is a molecule. d) Molecular, since this is a molecule. e) Ionic, since it is an ionic compound. f) Network covalent, since this substance is isoelectronic with C (diamond).

12.96

Figure P12.90 shows the face-centered cubic array of zinc blende, ZnS. Both ZnS and ZnO have a 1:1 ion ratio, so the ZnO unit cell will also contain four Zn2+ ions.

12.97

Figure P12.91 shows the face-centered cubic array of calcium sulfide, CaS. Both CaS and NaCl have a 1:1 ion ratio, so the CaS unit cell will also contain four S2– ions.

12.98

Plan: To determine the number of Zn2+ ions and Se2– ions in each unit cell count the number of ions at the corners, faces, and center of the unit cell. Atoms at the eight corners are shared by eight cells for a total of 8 atoms × 1/8 atom per cell = 1 atom; atoms in the body of a cell are in that cell only; atoms at the faces are shared by two cells: 6 atoms × 1/2 atom per cell = 3 atoms. Add the masses of the total number of atoms in the cell to find the mass of the cell. Given the mass of one unit cell and the ratio of mass to volume (density) divide the mass, converted to grams (conversion factor is 1 amu = 1.66054 × 10–24 g), by the density to find the volume of the unit cell. Since the volume of a cube is length × width × height, the edge length is found by taking the cube root of the cell volume. Solution: a) Looking at selenide ions, there is one ion at each corner and one ion on each face. The total number of selenide ions is 1/8 (8 corner ions) + 1/2 (6 face ions) = 4 Se2– ions. There are also 4 Zn2+ ions due to the 1:1 ratio of Se ions to Zn ions. b) Mass of unit cell = (4 × mass of Zn atom) + (4 × mass of Se atom) = (4 × 65.41 amu) + (4 × 78.96 amu) = 577.48 amu ⎛1.66054 ×10−24 g ⎞⎛ ⎟⎜ cm 3 ⎞⎟⎟ ⎜ 3 –22 –22 3 ⎜ ⎟⎟⎟⎜⎜ c) Volume (cm ) = 577.48 amu ⎜ ⎟⎟ = 1.76924 × 10 = 1.77 × 10 cm ⎜⎜ ⎜ ⎟ 1 amu ⎝ ⎠⎝⎜ 5.42 g ⎠⎟

(

)

d) The volume of a cube equals (length of edge)3. Edge length (cm) = 3 1.76924 ×10−22 cm 3 = 5.6139 × 10–8 = 5.61 × 10–8 cm 12.99

a) A face-centered cubic unit cell contains four atoms. b) Volume = (4.52 × 10–8 cm)3 = 9.23454 × 10–23 = 9.23 × 10–23 cm3 c) Mass of unit cell = (1.45 g/cm3)(9.23454 × 10–23 cm3) = 1.3390 × 10–22 = 1.34 × 10–22 g

12-21

⎛1.3390 ×10−22 g ⎞⎟⎛⎜ 1 kg ⎞⎟⎛ ⎞⎟⎛⎜1 unit cell ⎞⎟ 1 amu ⎜ ⎟⎟⎜⎜ ⎟⎟⎜⎜ ⎟⎟ ⎟⎟⎜ d) Mass of atom = ⎜⎜ ⎜⎝⎜ 1 unit cell ⎠⎟⎟⎜⎜103 g ⎟⎟⎜⎝⎜1.66054 ×10−27 kg ⎠⎟⎟⎝⎜⎜⎜ 4 atoms ⎠⎟⎟ ⎝ ⎠ = 20.1592 = 20.2 amu/atom 12.100 Plan: To classify a substance according to its electrical conductivity, first locate it on the periodic table as a metal, metalloid, or nonmetal. In general, metals are conductors, metalloids are semiconductors, and nonmetals are insulators. Solution: a) Phosphorous is a nonmetal and an insulator. b) Mercury is a metal and a conductor. c) Germanium is a metalloid in Group 4A(14) and is beneath carbon and silicon in the periodic table. Pure germanium crystals are semiconductors and are used to detect gamma rays emitted by radioactive materials. Germanium can also be doped with phosphorous (similar to the doping of silicon) to form an n-type semiconductor or be doped with lithium to form a p-type semiconductor. 12.101 a) conductor

b) insulator

c) conductor

12.102 Plan: First, classify the substance as an insulator, conductor, or semiconductor. The electrical conductivity of conductors decreases with increasing temperature, whereas that of semiconductors increases with temperature. Temperature increases have little impact on the electrical conductivity of insulators. Solution: a) Antimony, Sb, is a metalloid, so it is a semiconductor. Its electrical conductivity increases as the temperature increases. b) Tellurium, Te, is a metalloid, so it is a semiconductor. Its electrical conductivity increases as temperature increases. c) Bismuth, Bi, is a metal, so it is a conductor. Its electrical conductivity decreases as temperature increases. 12.103 a) decrease (metalloid)

b) increase (metal)

c) decrease (metalloid)

12.104 Plan: Use the molar mass and the density of Po to find the volume of one mole of Po. Divide by Avogadro’s number to obtain the volume of one Po atom (and the volume of the unit cell). Since Po has a simple cubic unit cell, there is one Po atom in the cell (atoms at the eight corners are shared by eight cells for a total of 8 atoms × 1/8 atom = 1 atom per cell). Find the edge length of the cell by taking the cube root of the volume of the unit cell. The edge length of a simple cubic unit cell is twice the radius of the atom. Solution: ⎛ ⎞⎟⎛ ⎛ 209 g Po ⎞⎛ 1 mol Po ⎟⎟⎜⎜ cm3 ⎞⎟⎜ ⎜ ⎟⎟⎜⎜1 Po atom ⎞⎟⎟ ⎟⎟⎜⎜ Volume (cm3) of the unit cell = ⎜⎜ ⎟⎟⎜ 23 ⎟ ⎜ ⎜⎜⎝ 1 mol Po ⎠⎝ ⎜ ⎟⎜ 9.142 g ⎠⎝ ⎟⎜ 6.022 ×10 Po atoms ⎠⎟⎟⎜⎝1 unit cell ⎠⎟ = 3.7963332 × 10–23 cm3 Edge length (cm) of the unit cell = 3 (3.7963332 ×10−23 cm 3 ) = 3.3608937 × 10–8 cm 2r = edge length 2r = 3.3608937 × 10–8 cm r = 1.680447 × 10–8 = 1.68 × 10–8 cm

⎞⎟⎛ ⎛ 63.55 g Cu ⎞⎛ 3 ⎞⎛ 1 mol Cu ⎜ ⎟⎜ 4 Cu atoms ⎞⎟ ⎟⎟⎟⎜⎜ cm ⎟⎜ ⎟⎟⎜⎜ 12.105 Volume (cm3) of the unit cell = ⎜⎜ ⎜ ⎟⎟⎜⎜ 8.95 g ⎟⎟⎜ 6.022 ×1023 Cu atoms ⎟⎟⎟⎜⎜⎝ 1 unit cell ⎠⎟⎟ ⎜⎜⎝ 1 mol Cu ⎠⎝ ⎠⎝⎜ ⎠ = 4.71641 × 10–23 cm3 Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

12-22

Edge length (cm) of the unit cell = 3 (4.71641×10−23 cm 3 ) = 3.613022 × 10 8 cm –

Use the edge length of the cube and the Pythagorean Theorem to find the diagonal of the cell: C2 = A2 + B2 C=

2 (3.613022 ×10−8 cm) = 5.109585 × 10 cm 2

–8

C = 4r –

5.109585 × 10 8 cm = 4r –

r = 1.277396 × 10–8 = 1.28 × 10 8 cm

⎛ ⎞⎟ ⎛ 95.94 g Mo ⎞⎛ 1 mol Mo ⎟⎟⎜⎜ cm 3 ⎞⎟⎜ ⎜ ⎟⎟ 2 Mo atoms ⎟⎟⎜⎜ 12.106 a) Edge of unit cell = 3 ⎜⎜ ⎟⎟⎜ 23 ⎜⎜⎝ 1 mol Mo ⎠⎝ ⎟⎜⎜10.28 g ⎠⎝ ⎟⎟⎜⎜ 6.022 ×10 Mo atoms ⎠⎟⎟

(

–

)

–

= 3.1412218 × 10 8 = 3.141 × 10 8 cm b) The body-diagonal of a body-centered cubic unit cell is equal to four times the radius of the Mo atom. The body-diagonal is also = 4r =

3 times the length of the unit cell edge.

3 (3.1412218 ×10−8 cm) = 5.4407559 × 10 8 cm –

–

r = 1.360189 × 10 8 = 1.360 × 10 8 cm 3⎛ 3 ⎞ ⎛180.9479 g ⎞⎟⎛⎜ cm3 ⎞⎛ 10−2 m ⎞⎟ ⎛⎜ 0.01 Å ⎞⎟ ⎜⎜ 2 atoms ⎟⎟ ⎟ ⎜ ⎟ ⎜ ⎟⎜⎜ ⎟⎟ 12.107 Avogadro’s number = ⎜⎜ ⎟⎟⎜⎜ ⎟⎟ ⎜⎜ −12 ⎟⎟ ⎜⎜ ⎟⎟⎜ 3⎟ ⎟ ⎟ ⎜ ⎜ ⎜ ⎜⎝ mol ⎟ ⎟ 16.634 g 1 cm 10 m ⎟ ⎟ ⎠⎜⎝ ⎠⎝⎜ ⎠ ⎜⎝ ⎠⎟ ⎜⎝⎜ 3.3058 Å ⎠⎟

(

)

= 6.0222270 × 1023 = 6.022 × 1023 atoms/mol 12.108 A metal’s strength (as well as its other properties) depends on the number of valence electrons in the metal. An alloy of tin in copper is harder than pure copper because tin contributes additional valence electrons for the metallic bonding. 12.109 In an n-type semiconductor, an atom with more valence electrons than the host is doped in. The “extra” electrons are free to move in the conduction band. In a p-type semiconductor, an atom with fewer valence electrons than the host is doped in. This creates “holes” in the valence band, which allows valence electrons to move more readily. Either of these increases the conductivity of the host. 12.110 Liquid crystal molecules generally have a long, cylindrical shape and a structure that allows intermolecular attractions through dispersion, dipole-dipole, and/or hydrogen bonding forces, but inhibits perfect crystalline molecular packing. This allows an electric field to orient the polar molecules in approximately the same direction, so, like crystalline solids, liquid crystals may pack at the molecular level with a high degree of order. 12.111 A substance whose physical properties are the same in all directions is called isotropic; an anisotropic substance has properties that depend on direction. Liquid crystals flow like liquids but have a degree of order that gives them the anisotropic properties of a crystal. 12.112 Modern ceramics, like traditional clay ceramics, are hard and resist heat and chemical attack. Additionally, modern ceramics show superior electrical and magnetic properties. Silicon nitride is virtually inert chemically, retains its strength at high temperatures, and is an electrical insulator. Boron nitride is an electrical insulator in its graphite-like form, and is converted to an extremely hard and durable diamond-like structure at high temperature and pressure. Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

12-23

12.113 The average molar mass of a polymer sample is different from the molar mass of an individual chain because the degree of polymerization differs from chain to chain. As a result, the chain length within a sample varies, so molar masses vary. 12.114 The radius of gyration is the average distance from the center of mass of the polymer to the outside edge of the chain. A tighter random coil shape and shorter length for the polymer will give a smaller radius of gyration. 12.115 The size (molar mass), concentration of the polymer, and the strength of the intermolecular forces influence the viscosity of a polymer solution. Chain entanglement and intermolecular forces influence the viscosity of a molten polymer. Non-crystallized polymer chains form polymer glass. 12.116 Increased branching in a polymer results in less tightly packed molecules and a less rigid polymer. High-density polyethylene (HDPE) is more rigid than low-density polyethylene (LDPE). HDPE is used to make rigid food containers, such as 2-Liter soft drink bottles. LDPE is used to make less rigid food containers, such as sandwich bags. Crosslinking results in very strong polymers. Kevlar, used in bulletproof vests, is highly crosslinked. 12. 117 Plan: Germanium and silicon are elements in Group 4A with four valence electrons. If germanium or silicon is doped with an atom with more than four valence electrons, an n-type semiconductor is produced. If it is doped with an atom with fewer than four valence electrons, a p-type semiconductor is produced. Solution: a) Phosphorus has five valence electrons so an n-type semiconductor will form by doping Ge with P. b) Indium has three valence electrons so a p-type semiconductor will form by doping Si with In. 12.118 a) n-type

b) p-type

12.119 Plan: The degree of polymerization is the number of repeat units in a polymer chain and can be found by using the equation ℳpolymer = ℳ repeat × n, where n is the degree of polymerization. Solution: The molar mass of one monomer unit (ℳ repeat), C6H5CHCH2, is 104.14 g/mol and the molar mass of the polymer (ℳ polymer) is 3.5 × 105 g/mol. ℳ polymer = ℳ repeat × n 3.5×10 5 g/mol n= = 3361 monomer units. Reporting in correct significant figures, n = 3.4 × 103. 104.14 g/mol 12.120

ℳ polymer = ℳ repeat × n ℳ polymer = 62.49 g/mol × 1565 units = 9.779685 × 104 = 9.780 × 104 g/mol

12.121 Plan: Use the equation for the radius of gyration. The length of the repeat unit, l0, is given (0.252 pm). Calculate the degree of polymerization, n, by dividing the molar mass of the polymer chain (ℳ = 2.8 × 105 g/mol) by the molar mass of one monomer unit (CH3CHCH2, ℳ = 42.08 g/mol). Substitute the values l0 and n into the equation. Solution: ℳ polymer = ℳ repeat × n 2.8×10 5 g/mol n= = 6653.99 42.08 g/mol

(6653.99)(0.252 pm)

Radius of gyration = Rg = 12.122

nI 0 2 = 6

= 8.39201 = 8.4 pm

ℳ polymer = ℳ repeat × n

12-24

2.30 ×104 g/mol = 119.667 192.2 g/mol

Radius of gyration = Rg =

(

(119.667) 1.075 nm

nI 0 2 = 6

)

= 4.80087 = 4.80 nm

12.123 The vapor pressure of H2O is 4.6 torr at 0°C, 9.2 torr at 10°C, and 19.8 torr at 22°C. ⎡⎛ 44% ⎞ ⎤ ⎟⎟ 19.8 torr ⎥ 0.75 L ⎛ ⎢⎜⎜ ⎞⎛ ⎞ ⎟ ⎢⎝100% ⎠ ⎥⎦ ⎜⎜ 1 atm ⎟⎟⎜⎜18.02 g H 2 O ⎟⎟ a) Mass (g) of H2O at 22°C = PV/RT = ⎣ ⎟⎟⎜ ⎟ ⎜ L • atm ⎞⎟ ⎛ ⎜ ⎟⎜⎜ 1 mol H 2 O ⎠⎟⎟ ⎜⎜0.0821 273 + 22 K ⎝⎜ 760 torr ⎠⎝ ⎟ ⎝ mol • K ⎠ = 0.0063966877 g H2O

(

)(

((

Mass (g) of H2O at 0°C = PV/RT =

)

) )

⎛ 1 atm ⎞⎛ (4.6 torr)(0.75 L) ⎟⎟⎜⎜18.02 g H O ⎞⎟⎟ ⎜⎜ ⎟⎟⎜ ⎟ ⎜ L • atm ⎞ ⎛ ⎜⎜ 760 torr ⎠⎝ ⎟⎜⎜ 1 mol H O ⎠⎟⎟ ⎜⎝⎜0.0821 mol • K ⎠⎟⎟((273.2 + 0.0) K ) ⎝ 2

= 0.0036470057 g H2O Mass H2O condensed = (0.0063966877 g H2O) – (0.0036470057 g H2O) = 0.002749682 = 0.0027 g H2O b) Repeat the calculation at 10°C. Mass (g) of H2O at 10°C = PV/RT =

⎛ 1 atm ⎞⎛ (9.2 torr)(0.75 L) ⎟⎟⎜⎜18.02 g H O ⎞⎟⎟ ⎜⎜ ⎜ ⎟⎟⎟⎜⎜⎜ 1 mol H O ⎟⎟⎟ L • atm ⎞⎟ ⎛ ⎜ ⎠ ⎜⎜0.0821 273 + 10) K ) ⎝⎜ 760 torr ⎠⎝ ( ( ⎟ ⎝ mol • K ⎠ 2

= 0.007041427 g H2O At equilibrium, 0.0070 g of H2O could be in the vapor state. Since only 0.0063 g of H2O are actually present, no liquid would condense at 10°C. 12.124 Plan: The vapor pressure of water is temperature dependent. Table 5.3 gives the vapor pressure of water at various temperatures. Use PV = nRT to find the moles and then mass of water in 5.0 L of nitrogen at 22°C; then find the mass of water in the 2.5 L volume of nitrogen and subtract the two masses to calculate the mass of water that condenses. Solution: At 22°C the vapor pressure of water is 19.8 torr (from Table 5.3). a) Once compressed, the N2 gas would still be saturated with water. The vapor pressure depends on the temperature, which has not changed. Therefore, the partial pressure of water in the compressed gas remains the same at 19.8 torr. b) PV = nRT ⎛ 1 atm ⎞⎟ = 0.026053 atm Pressure (atm) = (19.8 torr )⎜⎜ ⎝ 760 torr ⎠⎟⎟ Moles of H2O at 22°C and 5.00 L volume:

(0.026053 atm)(5.00 L) = 0.0053785 mol H O PV = 2 RT ⎛⎜0.0821 L • atm ⎞⎟ 273 + 22 K ) ⎟⎠ ( ⎜⎝ mol • K The gas is compressed to half the volume: 5.00 L/2 = 2.50 L

(

)

⎛18.02 g H 2 O ⎞⎟ ⎟ = 0.096921 g Mass (g) of H2O at 22°C and 5.00 L volume = (0.0053785 mol H 2 O)⎜⎜ ⎜⎝ 1 mol H 2 O ⎠⎟⎟ Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

12-25

Moles of H2O at 22°C and 2.50 L volume: n=

(0.026053 atm)(2.50 L) = 0.0026892 mol H O PV = 2 RT ⎛⎜0.0821 L • atm ⎞⎟ 273 + 22 K ( ) ⎟ ⎜⎝ mol • K ⎠

(

)

⎛18.02 g H 2 O ⎞⎟ ⎟⎟ = 0.048460 g Mass (g) of H2O at 22°C and 2.50 L volume = (0.00268922 mol H 2 O)⎜⎜⎜ ⎝ 1 mol H 2 O ⎠⎟

Mass (g) of H2O condensed = (0.096921 g H2O) – (0.048460 g H2O) = 0.048461 = 0.0485 g H2O

⎛ cm 3 ⎞⎛ ⎟⎟⎜⎜ 137.3 g ⎞⎟⎟ ⎜ 3 12.125 Volume = ⎜⎜ ⎟⎟⎜ ⎟ = 37.9281768 cm /mol Ba ⎜⎜⎝ 3.62 g ⎠⎝ ⎟⎜⎜1 mol Ba ⎠⎟⎟ Volume/mol of Ba atoms = volume/mol Ba × packing efficiency The packing efficiency in the body-centered cubic unit cell is 68%. Volume/mol of Ba atoms = 37.9281768 cm3/mol Ba × 0.68 = 25.791116 cm3/mol Ba atoms ⎛ 25.791116 cm 3 ⎞⎟⎛ 1 mol Ba atoms ⎞ ⎟⎟ = 4.28282 × 10–23 cm3/atom ⎟⎜⎜ Volume of one Ba atom = ⎜⎜⎜ ⎜⎝ 1 mol Ba atoms ⎠⎟⎟⎝⎜ 6.022 ×10 23 Ba atoms ⎠⎟ Use the volume of a sphere to find the radius of the Ba atom: 4 V = πr 3 3 r= 3

3(4.28282 ×10 3V = 3 4π 4π

−23

cm 3 )

= 2.17044 × 10–8 = 2.17 × 10–8 cm

12.126 First use the Clausius-Clapeyron equation to determine the heat of vaporization of hexane. ο ⎛1 ΔH vap 1⎞ P2 ⎜⎜ − − ⎟⎟⎟ = ln R ⎜⎝ T2 T1 ⎠⎟ P1

P1 = 121 mm Hg

P2 = 760 mm Hg

T1 = 20.0°C + 273.2 = 293.2 K T2 = 68.7°C + 273.2 = 341.9 K ο −ΔH vap 760 mm Hg ⎛ 1 1 ⎞⎟ ⎜ = − ln ⎟ ⎜ ⎝ 8.314 J /mol • K 341.9 K 293.2 K ⎠⎟ 121 mm Hg

1.8375279 = –ΔHvap(–5.84327 × 10–5) J/mol ο ΔHvap = 3.1447 × 104 = 3.14 × 104 J/mol

The LFL of hexane is 1.1%. Or, the mole fraction of hexane is 0.011. According to Dalton’s law, the partial pressure of hexane at the flash point can be found: Phexane = (Xhexane)(Ptotal) Phexane = (0.011)(760 mm Hg) = 8.36 = 8.4 mm Hg Use the Clausius-Clapeyron to find the temperature of this particular vapor pressure: ΔH vap ⎛⎜ 1 1⎞ P2 − ⎟⎟⎟ =− ⎜ ⎜ R ⎝ T2 T1 ⎠⎟ P1 ο

P1 = 760 mm Hg

P2 = 8.4 mm Hg

T1 = 68.7°C + 273.2 = 341.9 K

T2 = ?

ο ΔHvap = 3.1447 × 104

12-26

4 1 ⎞⎟ 8.4 mm Hg −3.1447 ×10 J/mol ⎛⎜ 1 ⎟⎟ − ⎜⎜ = 8.314 J/mol • K ⎜⎝ T2 341.9 K ⎠⎟ 760 mm Hg

⎛1⎞ –4.5050867 = –3782.415 ⎜⎜⎜ ⎟⎟⎟ + 11.06293 ⎝ T2 ⎠⎟

T2 = 242.961 = 243.0 K – 273.2 = –30.2°C

12.127 a) I, II, III, V b) IV c) V → IV → liquid → I d) Triple point: I, II, liquid Triple point: II, IV, liquid Triple point: II, III, IV Triple point: III, IV, V Triple point: IV, V, liquid 12.128 a) Use the density of Si (d = 2.34 g/cm3) and mass of the ingot to determine its volume. Solve for the height of the cylinder using V = πr2h. Divide cylinder height by wafer thickness to determine the number of wafers possible. 3 ⎛10 3 g ⎞⎛ ⎟⎟⎜⎜ 1 cm ⎞⎟⎟ = 1709.402 cm3 Volume (cm3) of Si ingot = (4.00 kg)⎜⎜ ⎟ ⎟⎜ 2.34 g ⎠⎟⎟ ⎝⎜ 1 kg ⎠⎝ Radius, r, (cm) = ½(diameter) = ½(5.20 in) = 2.60 in = 6.604 cm ⎛1709.402 cm 3 ⎞⎟ V Height = h = 2 = ⎜⎜ ⎟ = 12.476 cm ⎜⎝ π (6.604 cm)2 ⎠⎟⎟ πr ⎛10−2 m ⎞⎟⎛ 1 wafer ⎞ ⎟⎟ = 1113.94 = 1.11 × 103 wafers ⎟⎜⎜ Number of wafers = (12.476 cm) ⎜⎜⎜ ⎜⎝ 1 cm ⎠⎟⎟⎜⎝1.12 ×10−4 m ⎠⎟

b) A single wafer is also a cylinder with dimensions h = 1.12 × 10–4 m and r = 6.604 cm. The volume of the cylinder can be converted to mass using the density (d = 2.34 g/cm3). Mass (g) = πr2hd = π(6.604 cm)2[(1.12 × 10–4 m)(1 cm/10–2 m)](2.34 g/cm3) = 3.5909 = 3.59 g c) Silicon reacts with oxygen to form silicon dioxide, SiO2(s). This oxide coating interferes with the operation of the wafer and is removed by gaseous hydrogen fluoride, HF(g). In a double displacement reaction, silicon’s four valence electrons combine with four fluorine atoms: SiO2(s) + 4HF(g) → SiF4(g) + 2H2O(g) ⎛ 1 mol Si ⎞⎟⎛ 0.750% ⎞⎛ 4 mol HF ⎞⎟ ⎜ ⎟⎟⎜ ⎟⎟ = 3.83506 × 10–3 = 3.84 × 10–3 mol HF ⎟⎟⎜⎜ d) Moles of HF = 3.5909 g Si ⎜⎜ ⎜⎜⎝ 28.09 g Si ⎠⎟⎟⎜⎜⎝ 100% ⎠⎟⎝⎜⎜⎜ 1 mol Si ⎠⎟⎟

(

)

12.129 a) The enthalpy of vaporization must be determined first. ο ⎛1 ΔH vap 1⎞ P ⎜⎜ − ⎟⎟⎟ ln 2 = − R ⎜⎝ T2 T1 ⎠⎟ P1 ln

1.00 torr 10.0 torr

⎛ ⎞⎟ ⎜⎜ 1 1 ⎟⎟ − ⎜⎜ ⎟⎟ 8.314 J/mol • K ⎜⎜ 273.2 + 54.3 K + 273.2 95.3 K ⎝ ⎠⎟ ο −ΔHvap

(

)

(

)

12-27

ο –2.302585 = –0.0000408625 ΔHvap ο ΔHvap = (–2.302585)/(–0.000040862) = 56350.28 = 5.6 × 104 J/mol

The subtraction of the 1/T terms limits the significant figures. ⎛ ⎞⎟ −56350.28 J/mol ⎜⎜ 1 1 P2 ⎟⎟ − ⎜ = ln ⎟ = –4.351324 8.314 J/mol • K ⎜⎜⎜ 273 + 25 K 10.0 torr 273.2 + 95.3 K ⎠⎟⎟ ⎝ P2 = 0.0128897 10.0 torr

(

)

(

)

P2 = (0.0128897)(10.0 torr) = 0.128897 = 0.13 torr L • atm ⎞⎟ ⎛ 1.0 mg ⎜⎜0.0821 273 + 25 K ⎛ 760 torr ⎞⎛10−3 g ⎞⎛ 1 mol ⎞ ⎟⎟⎜⎜ ⎟⎟⎜⎜ ⎟⎟ ⎝ ⎠⎟ ⎜⎜ mol • K b) V= nRT/P = ⎟⎟⎜ ⎟⎟⎜ ⎟ ⎜⎜ ⎜ ⎜ ⎟⎜ 1 mg ⎠⎝ ⎟⎜152.16 g ⎠⎟⎟ 0.128897 torr ⎝⎜ 1 atm ⎠⎝

(

((

)

(

) )

)

= 0.94805 = 0.95 L 12.130 Plan: The Clausius-Clapeyron equation gives the relationship between vapor pressure and temperature. We are given P1, P2, T1, and ΔHοvap ; these values are substituted into the equation to find T2. Solution: ο ⎛1 ΔH vap 1⎞ P2 ⎜⎜ − ⎟⎟⎟ =− ln R ⎜⎝ T2 T1 ⎠⎟ P1

P1 = 1.20 × 10–3 torr

T1 = 20.0°C + 273 = 293 K

ο ΔHvap = 59.1 kJ/mol

P2 = 5.0 × 10–5 torr

T2 = ?

−59.1 kJ/mol ⎛⎜ 1 5.0 ×10−5 torr 1 ⎞⎟⎛⎜⎜10 J ⎞⎟⎟ ⎟⎜ − = ⎜⎜ ⎟⎟⎜ −3 ⎟⎟ 8.314 J/mol • K ⎝ T2 293 K ⎠⎜ 1.20 ×10 torr ⎝ 1kJ ⎠⎟ 3

⎛1 1 ⎞⎟ ⎟ –3.178054 = –7108.492 ⎜⎜⎜ − 293 K ⎠⎟⎟ ⎝ T2 ⎛1 1 ⎞⎟ ⎟ (–3.17805)/(–7108.49) = 4.47078 × 10–4 = ⎜⎜ − ⎜⎝ T2 293 K ⎠⎟⎟

4.47078 × 10–4 + 1/293 = 1/T2 T2 = 259.064 = 259 K

12.131 a) Molar mass of fraction 1 (g/mol) = 273 units × 100.02 g/mol = 2.7305 × 104 = 2.73 × 104 g/mol Molar mass of fraction 2 (g/mol) = 330 units × 100.02 g/mol = 3.30066 × 104 = 3.30 × 104 g/mol Molar mass of fraction 3 (g/mol) = 368 units × 100.02 g/mol = 3.6807 × 104 = 3.68 × 104 g/mol Molar mass of fraction 4 (g/mol) = 483 units × 100.02 g/mol = 4.83097 × 104 = 4.83 × 104 g/mol Molar mass of fraction 5 (g/mol) = 525 units × 100.02 g/mol = 5.25105 × 104 = 5.25 × 104 g/mol Molar mass of fraction 6 (g/mol) = 575 units × 100.02 g/mol = 5.75115 × 104 = 5.75 × 104 g/mol b) ℳ n =

(100.02 g/mol)[273(0.10) + 330(0.40) + 368(1.00) + 483(0.70) + 525(0.30) + 575(0.10)] (0.10 + 0.40 + 1.00 + 0.70 + 0.30 + 0.10) mol 4

= 4.1562 × 10 = 4.2 × 104 g/mol c) ℳ w = {[(2.73 × 104 g/mol)2(0.10 mol) + (3.30 × 104 g/mol)2(0.40 mol) + (3.68 × 104 g/mol)2(1.00 mol) + (4.83 × 104 g/mol)2(0.70 mol) + (5.25 × 104 g/mol)2(0.30 mol) + (5.75 × 104 g/mol)2(0.10 mol)]}/ 108,040 g = 4.3085 × 104 = 4.3 × 104 g/mol Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

12-28

12.132 Plan: Use the volume of the liquid water and the density of water to find the moles of water present in the volume of the greenhouse. Find the pressure of this number of moles of water vapor using the ideal gas equation. Knowing that 4.20 L results in the calculated vapor pressure, the volume of water that would give a vapor pressure corresponding to 100% relative humidity can be calculated. Solution: ⎛ 1 mL ⎞⎟⎛1.00 g ⎞⎛ 1 mol H 2 O ⎞⎟ ⎜ ⎟⎟⎜⎜ ⎟⎟ = 233.0744 mol a) Moles of water = (4.20 L)⎜⎜ −3 ⎟⎟⎟⎜⎜⎜ ⎜⎝⎜10 L ⎠⎟⎜⎝ mL ⎠⎟⎟⎝⎜⎜⎜18.02 g H 2 O ⎠⎟⎟ PV = nRT

nRT P= = V

L • atm ⎞⎟ ⎛ 273 + 26 K ⎛10−3 m 3 ⎞ (233.0744 mol)⎜⎜0.0821 ⎟⎟ ⎝ ⎜⎜ mol • K ⎠⎟ ⎜⎜ ⎟⎟ 3 1 L ⎜⎝ 256 m ⎠⎟

((

(

) )

)

–2

= 2.234956 × 10 = 2.23 × 10–2 atm b) 25.2 torr is needed to saturate the air. ⎛1.00 atm ⎞⎟ ⎟⎟ = 0.0331579 atm P = (25.2 torr )⎜⎜⎜ ⎝⎜ 760 torr ⎠⎟ 4.20 L x = −2 0.0331579 atm 2.234956 ×10 atm

Volume (L) of water needed for 100% relative humidity =

( 0.0331579 atm)(4.20 L) 2.234956 ×10−2 atm

= 6.23114 = 6.23 L H2O

12.133 The formulas are TaN and TaC. 12.134 In the NaCl type lattice, there are four ions of each type. 3 ⎛ ⎞ ⎛ 4 KF ⎞⎟⎛⎜ 1 mol KF ⎞⎟⎛ 58.10 g KF ⎞⎜ unit cell ⎟⎟⎟⎛⎜ 1 Å ⎞⎟ ⎜ ⎟ ⎜⎜ ⎜ ⎟ ⎟ ⎟ ⎟⎜ ⎜ ⎟⎜ −8 ⎟ Density of KF = ⎜ ⎟⎜ ⎟⎜ ⎟ 3⎟ ⎜⎜⎝ unit cell ⎠⎟⎟⎜⎜⎜ 6.022 ×1023 KF ⎟⎟⎝⎜⎜⎜ 1 mol KF ⎠⎟⎟⎜⎜⎜ ⎟⎜10 cm ⎟⎟⎟ ⎝ ⎠ ⎠ ⎜⎝ 5.39 Å ⎠⎟⎟⎝⎜

(

)

= 2.46450 = 2.46 g/cm3 12.135 Plan: Hydrogen bonds only occur in substances in which hydrogen is directly bonded to either oxygen, nitrogen, or fluorine. Solution: a) Both furfuryl alcohol and 2-furoic acid can form hydrogen bonds since these two molecules have hydrogen directly bonded to oxygen. H O 2-furoic acid H O C H O C C O C C H C C C O C C O H H H H

12-29

H furfuryl alcohol H H C

H C

C C

H b) Both furfuryl alcohol and 2-furoic acid can form internal hydrogen bonds by forming a hydrogen bond between the O–H and the O in the ring. furfuryl alcohol H H 2-furoic acid O H O H C O O C C C C H C C C H H C C O H H H 12.136 Plan: Determine the total mass of water, with a partial pressure of 31.0 torr, contained in the air. This is done by using PV = nRT to find the moles of water at 31.0 torr and 22.0°C. Then calculate the total mass of water, with a partial pressure of 10.0 torr, contained in the air. The difference between these two values is the amount of water removed. The mass of water and the heat of condensation are necessary to find the amount of energy removed from the water. Solution: a) PV = nRT PV n= RT ⎛ 1 atm ⎞⎟ = 0.0407895 atm Convert pressure to units of atm: P (atm) = (31.0 torr )⎜⎜ ⎝ 760 torr ⎠⎟⎟ ⎛ 1L ⎞ Convert volume to units of L: V (L) = (2.4 ×106 m 3 )⎜⎜ −3 3 ⎟⎟ = 2.4 × 109 L ⎜⎝10 m ⎠⎟

(0.0407895 atm)(2.4 ×10 L) Moles of H O at 31.0 torr = = 4,039,244.229 mol ⎛ ⎞ ⎜⎜0.0821 L • atm ⎟⎟((273.2 + 22.0) K) mol • K ⎠⎟ ⎝⎜ 9

⎛ 18.02 g ⎞⎟⎛⎜ 1 kg ⎞⎛ ⎟⎜ 1 ton ⎞⎟ ⎟⎟⎜⎜ 3 ⎟⎟⎟⎜⎜ 3 ⎟⎟⎟ Mass (metric tons) of H2O at 31.0 torr = (4,039,244.229 mol H 2 O)⎜⎜⎜ ⎟⎜⎝10 g ⎠⎝ ⎟⎜⎜10 kg ⎠⎟ ⎝1 mol H 2 O ⎠⎜ = 72.7872 tons H2O ⎛ 1 atm ⎞⎟ = 0.01315789 atm Convert pressure to units of atm: P (atm) = (10.0 torr )⎜⎜ ⎝ 760 torr ⎠⎟⎟

12-30

(0.01315789 atm)(2.4 ×10 L) = 1,302,980.7 mol ⎛ ⎞ ⎜⎜0.0821 L • atm ⎟⎟((273.2 + 22.0) K) ⎜⎝ mol • K ⎠⎟ 9

Moles of H2O at 10.0 torr =

⎛ 18.02 g ⎞⎟⎜⎛ 1 kg ⎞⎛ ⎟⎜ 1 ton ⎞⎟ ⎟⎟⎜⎜ 3 ⎟⎟⎟⎜⎜ 3 ⎟⎟⎟ Mass (metric tons) of H2O at 10.0 torr = (1,302,980.7 mol H 2 O)⎜⎜ ⎟⎜⎝10 g ⎠⎝ ⎜⎝1 mol H 2 O ⎠⎜ ⎟⎜⎜10 kg ⎠⎟ = 23.4797 tons H2O Mass of water removed = (72.7872 metric tons H2O) – (23.4797 metric tons H2O) = 49.3075 = 49.3 tons H2O b) The heat of condensation for water is –40.7 kJ/mol. 3 ⎛103 kg ⎞⎛ ⎟⎟⎜⎜10 g ⎞⎛ ⎟⎟⎜⎜ 1 mol H 2 O ⎞⎛ ⎟⎟⎜⎜ −40.7 kJ ⎞⎟⎟ ⎜ Heat (kJ) = 49.3075 tons H 2 O ⎜⎜ ⎟⎟⎜ ⎟⎟⎜ ⎟⎟⎜ ⎟ ⎜⎝⎜ 1 ton ⎠⎝ ⎟⎜⎜ 1 kg ⎠⎝ ⎟⎜⎜18.02 g H 2 O ⎠⎝ ⎟⎜⎜1 mol H 2 O ⎠⎟⎟

(

)

= –1.113660 × 108 = –1.11 × 108 kJ 12.137 Plan: At the boiling point, the vapor pressure equals the atmospheric pressure, so the two boiling points can be used to find the heat of vaporization for amphetamine using the Clausius-Clapeyron equation. Then use the Clausius-Clapeyron equation to find the vapor pressure of amphetamine at a different temperature, 20.°C. Then use PV = nRT to find the concentration of amphetamine at this calculated pressure and 20.°C. Solution: ΔH vap ⎛⎜ 1 1⎞ P2 − ⎟⎟⎟ =− ⎜⎜ R ⎝ T2 T1 ⎠⎟ P1 ο

P1 = 760 torr

ο ΔHvap =? P2 = 13 torr ο 13 torr −ΔH vap ⎛ 1 1 ⎞⎟ ⎜ ln − = ⎟ ⎜ 760 torr 8.314 J /mol • K ⎝ 356 K 474 K ⎠⎟

T1 = 201°C + 273 = 474 K T2 = 83°C + 273 = 356 K

ο –4.068369076 = –0.000084109 ΔHvap ο ΔHvap = (–4.068369076)/(–0.000084109) = 48,370.199 J/mol

ΔH vap ⎛⎜ 1 1⎞ P2 − ⎟⎟⎟ =− ⎜ ⎜ R ⎝ T2 T1 ⎠⎟ P1 ο

P1 = 13 torr

T1 = 83°C + 273 = 356 K

P2 = ?

T2 = 20°C + 273 = 293 K

ΔH ln

ο vap

= 48,370.199 J/mol

P2 −48,370.199 J/mol ⎛ 1 1 ⎞⎟ ⎜ = − ⎟ = –3.5139112 8.314 J /mol • K ⎜⎝ 293 K 356 K ⎠⎟ 13 torr

P2 = 0.0297802 13 torr

P2 = (0.0297802)(13 torr) = 0.3871426 torr ⎛ 1 atm ⎞⎟ P2 (atm) = (0.3871426 torr )⎜⎜ = 5.09398 × 10–4 atm ⎝ 760 torr ⎠⎟⎟ At this pressure and a temperature of 20°C, use the ideal gas equation to calculate the concentration of amphetamine in the air. Use a volume of 1 m3. Moles from the ideal gas equation times the molar mass gives the mass in a cubic meter. (5.09398×10−4 atm)(1 m3 ) ⎛⎜⎜ 1 L ⎞⎟⎟ PV = Moles = n = ⎜ −3 3 ⎟⎟ = 0.02117612 mol amphetamine L • atm ⎞⎟ ⎛ RT ⎝⎜⎜10 m ⎠⎟ ⎜⎜0.0821 293 K ( ) ⎝ mol • K ⎠⎟ Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

12-31

⎛135.20 g amphetamine ⎞⎟ 3 Mass (g) = (0.02117612 mol)⎜⎜ ⎟⎟ = 2.8630114 = 2.9 g/m ⎝ ⎠ 1 mol

12.138 a) Unit cell edge = 3

⎛ ⎞⎟ 1 mol C cm 3 ⎛⎜⎜12.01 g C ⎞⎟⎟⎜⎜ ⎟⎟ 8 C atoms ⎟⎟⎜⎜ ⎜⎜ ⎟ 3.52 g ⎜⎝ 1 mol C ⎠⎟⎜⎝⎜ 6.022 ×10 23 C atoms ⎠⎟⎟

(

)

= 3.5654679 × 10–8 = 3.57 × 10–8 cm 3⎞ ⎛ ⎜⎜ 3.5654678 ×10−8 cm ⎟⎟⎛ 3.01 g C ⎞⎟⎜⎛ 1 mol C ⎞⎟⎜⎛ 6.022 ×1023 C atoms ⎞⎟ ⎟⎟⎜ ⎟⎟⎟⎜⎜⎜ b) C atoms/unit cell = ⎜⎜ ⎟⎟⎟ ⎟⎟⎟⎜⎜ ⎜⎜ ⎜⎜ ⎟⎟⎝⎜ cm 3 ⎠⎟⎟⎜⎝⎜⎜12.01 g C ⎠⎝ 1 unit cell 1 mol C ⎟ ⎠⎟ ⎜⎝ ⎠⎟

(

)

= 6.8409 = 6.84 C atoms/unit cell 12.139 No, filling all the available holes (8) in the face-centered cubic lattice leads to a stoichiometry of 2:1 (8 holes/4 atoms). 12.140 Gallium is not a “normal” liquid (the density of the solid is lower than that of the liquid, like water), so its liquid-solid line would slope left (negative) and the triple point temperature would be higher than the normal melting point. 12.141 a) Determine the vapor pressure of ethanol in the bottle at –11°C by applying the Clausius-Clapeyron equation. ο The boiling point of ethanol is 78.5°C at a pressure of 1 atm (760 torr). ΔHvap (40.5 kJ/mol ) is given in Figure 12.2. ο ⎛1 ΔH vap 1⎞ P ⎜⎜ − ⎟⎟⎟ P1 = ? T1 = (273 + (–11°C) = 262 K ln 2 = − ⎜ R ⎝ T2 T1 ⎠⎟ P1 ο ΔHvap = 40.5 kJ/mol

760 torr P1

P2 = 760 torr

T2 = (273.15 + 78.5°C) = 351.6 K

3 −40.5 kJ/mol ⎛ 1 1 ⎞⎟⎜⎛⎜10 J ⎞⎟⎟ ⎜ − ⎟ = 4.7380851 ⎟⎟⎜ 8.314 J/mol • K ⎜⎝ 351.6 K 262 K ⎠⎜ ⎜⎝ 1 kJ ⎠⎟⎟

= 114.2153

P1 = 6.65410 = 6.65 torr Note: The pressure should be small because not many ethanol molecules escape the liquid surface at such a cold temperature. Determine the number of moles by substituting P, V, and T into the ideal gas equation. Assume that the volume the liquid takes up in the 4.7 L space is negligible.

(6.65410 torr)(4.7 L) ⎛⎜⎜ 1 atm ⎞⎟⎟ = 0.00191306 mol C H O PV = ⎟ ⎜⎜ 2 6 L • atm ⎞⎟ ⎛ RT ⎜ 760 torr ⎠⎟⎟ ⎝ ⎜⎜0.0821 262 K ( ) ⎟ ⎝ mol • K ⎠ Convert moles of ethanol to mass of ethanol using the molar mass (ℳ = 46.07 g/mol). Mass (g) of C2H6O = (0.00191306 mol C2H6O)(46.07 g/mol C2H6O) = 0.08813467 = 8.8 × 10–2 g C2H6O n=

ΔH vap ⎛⎜ 1 1⎞ P2 − ⎟⎟⎟ =− ⎜ ⎜ R ⎝ T2 T1 ⎠⎟ P1 ο

b) ln

ο ΔHvap = 40.5 kJ/mol

P1 = ?

T1 = (273 + 20°C) = 293 K

P2 = 760 torr

T2 = (273.15 + 78.5°C) = 351.6 K

12-32

760 torr 1 1 ⎞ ⎛ 103 J ⎞ = 2.7709337 = −40.5 kJ/mol ⎛⎜ − ⎟⎜ ⎟ P1 8.314 J/mol • K ⎝ 351.6 K 293 K ⎠ ⎝ 1 kJ ⎠

760 torr = 15.97354 P1

P1 = 47.5787 = 47.6 torr

Determine the number of moles by substituting P, V, and T into the ideal gas equation. Assume that the volume the liquid takes up in the 4.7 L space is negligible.

( 47.5787 torr )( 4.7 L ) ⎛ 1 atm ⎞ PV ⎜ ⎟ = 0.01223168 mol C2H6O = L • atm ⎞ ⎛ ⎝ 760 torr ⎠ RT ⎜ 0.0821 ⎟ ( 293 K ) mol • K ⎠ ⎝ Convert moles of ethanol to mass of ethanol using the molar mass (ℳ = 46.07 g/mol). Mass (g) of C2H6O = (0.01223168 mol C2H6O)(46.07 g/mol C2H6O) = 0.5635135 = 0.56 g C2H6O The mass of ethanol present in the vapor, if excess liquid was present, is 0.56 g. Since this exceeds the 0.33 g n=

available, all of the ethanol will vaporize. c) 0.0°C = (273.2 + 0.0) = 273.2 K ln

3 −40.5 kJ/mol ⎛ 1 1 ⎞⎟⎜⎛⎜10 J ⎞⎟⎟ P2 ⎜ − = ⎟ = –3.975864 ⎟⎟⎜ 8.314 J /mol • K ⎜⎝ 273.2 K 351.6 K ⎠⎜ 760 torr ⎜⎝ 1 kJ ⎠⎟⎟

P2 = 0.0187631 760 torr

P2 = (0.018763)(760 torr) = 14.259956 torr

(14.259956 torr)(4.7 L) ⎛⎜⎜ 1 atm ⎞⎟⎟ = 0.00393168 mol C H O PV = ⎟ ⎜⎜ 2 6 L • atm ⎞⎟ ⎛ RT ⎜ 760 torr ⎠⎟⎟ ⎝ ⎜⎜0.0821 273.2 K ( ) ⎟ ⎝ mol • K ⎠ Convert moles of ethanol to mass of ethanol using the molar mass (ℳ = 46.07 g/mol). Mass C2H6O = (0.00393168 mol C2H6O)(46.07 g/mol C2H6O) = 0.18113 g C2H6O Mass (g) of ethanol in liquid = mass (g) of ethanol (total) – mass (g) of ethanol in vapor n=

Mass (g) of ethanol in liquid = 0.33 g – 0.18113 g = 0.14887 = 0.15 g C2H6O 12.142 Plan: The equation q = c × mass × ΔT is used to calculate the heat involved in changing the temperature of solid A, the temperature of liquid A after it melts, and the temperature of vapor A after the liquid boils; the heat of fusion is used to calculate the heat involved in the phase change of solid A to liquid A and the heat of vaporization is used to calculate the heat involved in the phase change of liquid A to A as a vapor. Solution: a) To heat the substance to the melting point the substance must be warmed from –40°C to –20°C: q = c × mass × ΔT = (25 g)(1.0 J/g°C)[–20 – (–40)]°C = 500 J ⎛1 min ⎞⎟ Time (min) required = (500 J )⎜⎜ ⎟ = 1.1111 = 1.1 min ⎝ 450 J ⎠⎟ b) Phase change of the substance from solid at –20°C to liquid at –20°C: ⎛180. J ⎞⎟ ο ⎟⎟ = 4500 J q = n (ΔHfus ) = (25 g)⎜⎜⎜ ⎝ 1 g ⎠⎟ ⎛1 min ⎞⎟ Time (min) required = (4500 J )⎜⎜ ⎟ = 10 min ⎝ 450 J ⎠⎟ Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

12-33

c) To heat the substance to the melting point the substance must be warmed from –20°C to 85°C: q = c × mass × ΔT = (25 g)(2.5 J/g°C)[85 – (–20)]°C = 6562.5 J ⎛1 min ⎞⎟ Time (min) required = (6562.5 J )⎜⎜ ⎟ = 14.583 = 15 min ⎝ 450 J ⎠⎟ To boil the sample at 85°C: ⎛ 500. J ⎞ ο q = n (ΔH vap ) = (25 g)⎜⎜⎜ 1 g ⎟⎟⎟⎟ = 12,500 J ⎝ ⎠ ⎛1 min ⎞⎟ Time (min) required = (12,500 J )⎜⎜ ⎟ = 27.78 = 28 min ⎝ 450 J ⎠⎟ Warming the vapor from 85°C to 100°C: q = c × mass × ΔT = (25 g)(0.5 J/g°C)[100 – 85]°C = 187.5 J ⎛1 min ⎞⎟ Time (min) required = (187.5 J )⎜⎜ ⎟ = 0.417 = 0.4 min ⎝ 450 J ⎠⎟

100 80

T, oC

60 40 20 0 −20 −40 0

t, min 12.143 Plan: Balanced chemical equations are necessary. See the section on ceramic materials for the reactions. These equations may be combined to produce an overall equation with an overall yield. Find the limiting reactant which is then used to calculate the amount of boron nitride produced. The ideal gas law is used to calculate the moles of NH3. Solution: Step 1 B(OH)3(s) + 3NH3(g) → B(NH2)3(s) + 3H2O(g) Step 2 B(NH2)3(s) → BN(s) + 2NH3(g)

Overall reaction: B(OH)3(s) + NH3(g) → BN(s) + 3 H2O(g) Yields are 85.5% for step 1 and 86.8% for step 2. Overall fractional yield = (85.5%/100%)(86.8%/100%) = 0.74214 Find the limiting reactant. Since the overall reaction has a 1:1 mole ratio, the reactant with the fewer moles will be limiting. Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

12-34

⎛ 3 ⎞⎛ 3 ⎞ ⎛ 1 mol B(OH)3 ⎞ Moles of B(OH)3 = (1.00 t B(OH)3 ) ⎜ 10 kg ⎟⎜ 10 g ⎟ ⎜ ⎟ ⎝ 1t ⎠⎝ 1 kg ⎠⎝ 61.83 g B(OH)3 ⎠ = 1.6173379 × 104 mol B(OH)3 ⎞ = 30.2985 atm 1 atm P (atm) = 3.07 × 103 kPa ⎛ ⎜ ⎟ 101.325 kPa ⎝ ⎠

(

)

⎛ ⎞ V (L) = 12.5 m3 ⎜ 1 L ⎟ = 12,500 L 3 −3 10 m ⎝ ⎠

(

)

Moles of NH3 = PV/RT =

(30.2985atm)(12,500 L) = 1.6774720 × 10 mol NH L • atm ⎞⎟ ⎛ ⎜⎜0.0821 ⎟(275 K ) ⎝ mol • K ⎠ 4

B(OH)3 is the limiting reactant. ⎛ 1 mol BN ⎞ ⎛ 24.82 g BN ⎞ ⎛ 74.214% ⎞ Mass (g) of BN = (1.6173379 × 104 mol B(OH)3 ) ⎜ ⎟⎜ ⎟⎜ ⎟ ⎝ 1 mol B(OH)3 ⎠ ⎝ 1 mol BN ⎠ ⎝ 100% ⎠ = 2.97912 × 105 = 2.98 × 105 g BN

12.144 a) A and B can form intermolecular H bonds since both have a hydrogen atom bonded to an oxygen atom. b) Highest viscosity = strongest intermolecular forces. B has the highest viscosity. ΔH vap ⎛⎜ 1 1⎞ P2 − ⎟⎟⎟ =− ⎜⎜ R ⎝ T2 T1 ⎠⎟ P1 ο

12.145 a) ln

P1 = 1.00 atm

ο ΔHvap =? P2 = 0.526 atm ο 0.526 atm −ΔH vap ⎛ ⎞⎟ 1 1 ⎜ ln = − ⎟ 1.00 atm 8.314 J /mol • K ⎜⎝ 235.4 K 249.5 K ⎠⎟

T1 = 273.2 + (–23.7°C) = 249.5 K T2 = 273.2 + (–37.8°C) = 235.4 K

ο –0.6424541 = –0.0000288787 ΔHvap ο ΔHvap = (–0.6424541/–0.0000288787)(1 kJ/103 J) = 22.2466 = 22.2 kJ/mol

b) Use Hess’s law, using the reverse of the answer in part (a). 2C(s) + 3H2(g) + 1/2O2(g) → CH3OCH3(g)

ΔHfο = –185.4 kJ/mol

CH3OCH3(g) → CH3OCH3(l)

ΔHfο = –22.2 kJ/mol

Overall: 2C(s) + 3H2(g) + 1/2O2(g) → CH3OCH3(l)

ΔHfο = –207.6 kJ/mol

12.146 Plan: A body-centered cubic unit cell has eight corner atoms; 8 atoms × 1/8 atom per cell = 1 atom. In addition, the body-centered cell has an atom in the center, for a total of two atoms. Solution: ⎛ 22.99 amu ⎞⎟ Mass (amu) of two Na atoms = (2 Na atoms)⎜⎜ = 45.98 amu. ⎝ 1 Na atom ⎠⎟⎟ 12.147 [η]solvent = K ℳ a a) [η]benzene = (9.5 × 10–3 mL/g)(104,160 g/mol)0.74 = 49 [η]cyclohexane = (8.1 × 10–2 mL/g)(104,160 g/mol)0.50 = 26 Benzene has stronger interactions with the polymer.

12-35

b) [η]benzene = (9.5 × 10–3 mL/g)(52,000 g/mol)0.74 = 29 If you have standards of known molar mass and you measure [η], K and a can be determined for that polymer and solvent pair. Then a measured [η] can be used, along with the values of K and a, to determine the molar mass of a sample. c) [η]benzene = (8.3 × 10–2 mL/g)(104,160 g/mol)0.50 = 27 [η]cyclohexane = (2.6 × 10–1 mL/g)(104,160 g/mol)0.70 = 850 Polyisobutylene has weaker interactions with benzene than polystyrene does, but much stronger interactions with the solvent cyclohexane than polystyrene does.

⎛ 1 mol H 2 ⎞⎛ ⎛ 70.8 g ⎞⎟⎛⎜ 1 L ⎞⎟ ⎟⎟⎜⎜ 22.4 L ⎞⎟⎟ 2 ⎟ 1 dm 3 ⎜⎜ ⎟⎟⎜⎜ 12.148 Volume (L) = ⎜⎜⎜ ⎟⎟⎜ ⎟⎟ = 786.6667 = 8 × 10 L ⎜ 3⎟ ⎟⎜⎝1 dm ⎠⎟⎟ ⎜⎝⎜ 2.016 g H 2 ⎠⎝ ⎜ ⎜⎝ L ⎠⎜ ⎟⎜ 1 mol ⎠⎟

(

)

12-36

CHAPTER 13 THE PROPERTIES OF MIXTURES: SOLUTIONS AND COLLOIDS FOLLOW–UP PROBLEMS 13.1A

Plan: Compare the intermolecular forces in the solutes with the intermolecular forces in the solvent. The intermolecular forces for the more soluble solute will be more similar to the intermolecular forces in the solvent than the forces in the less soluble solute. Solution: a) 1,4–Butanediol is more soluble in water than butanol. Intermolecular forces in water are primarily hydrogen bonding. The intermolecular forces in both solutes also involve hydrogen bonding with the hydroxyl groups. Compared to butanol, each 1,4–butanediol molecule will form more hydrogen bonds with water because 1,4–butanediol contains two hydroxyl groups in each molecule, whereas butanol contains only one –OH group. Since 1,4–butanediol has more hydrogen bonds to water than butanol, it will be more soluble than butanol. b) Chloroform is more soluble in water than carbon tetrachloride because chloroform is a polar molecule and carbon tetrachloride is nonpolar. Polar molecules are more soluble in water, a polar solvent.

13.1B

Plan: Compare the intermolecular forces in the solutes with the intermolecular forces in the solvent. The intermolecular forces for the solvent that dissolves more solute will be more similar to the intermolecular forces in the solute than the forces in the solvent that dissolves less solute. Solution: a) Both chloroform and chloromethane are polar molecules that experience dipole-dipole and dispersion intermolecular forces. Methanol, on the other hand, has an O–H bond and, thus, can participate in hydrogen bonding in addition to dipole-dipole and dispersion forces. Chloroform dissolves more chloromethane than methanol because of similar dipole-dipole forces. b) Pentanol has a polar O–H group that can participate in hydrogen bonding, dipole-dipole and dispersion forces. However, it has a much larger non-polar section (CH3CH2CH2CH2-) that experiences only dispersion forces. Because the nonpolar portion of the pentanol is much larger than the polar portion, the nonpolar portion determines the overall solubility of the molecule. Thus, pentanol will be more soluble in nonpolar solvents like hexane than polar solvents like water. Hexane dissolves more pentanol due to dispersion forces.

13.2A

Plan: Use the relationship ΔHsolution = ΔHlattice + ΔHhydration of the ions. Given ΔHsolution and ΔHlattice, ΔHhydration of the ions can be calculated. Solution: The two ions in potassium nitrate are K+ and NO3–. ΔHhydration of the ions = ΔHhydration(K+) + ΔHhydration(NO3–) ΔHsolution = ΔHlattice + ΔHhydration of the ions ΔHhydration of the ions = ΔHsolution – ΔHlattice = 34.89 kJ/mol – 685 kJ/mol = –650.11 = –650 kJ/mol

13.2B

Plan: Use the relationship ΔHsolution = ΔHlattice + ΔHhydration of the ions. Given ΔHsolution and ΔHlattice, ΔHhydration of the ions can be calculated. ΔHhydration of the ions and ΔHhydration (Na+) can then be used to solve for ΔHhydration(CN–). Solution: The two ions in sodium cyanide are Na+ and CN–. Due to its smaller size, Na+ should have a greater charge density and thus a larger ΔHhydr than CN–. ΔHsolution = ΔHlattice + ΔHhydration of the ions ΔHhydration of the ions = ΔHsolution – ΔHlattice = 1.21 kJ/mol – 766 kJ/mol = –764.79 = –765 kJ/mol ΔHhydration of the ions = ΔHhydration(Na+) + ΔHhydration(CN–) ΔHhydration(CN–) = ΔHhydration of the ions – ΔHhydration(Na+) = –765 kJ/mol – (–410. kJ/mol) = –355 kJ/mol

13-1

13.3A

Plan: Solubility of a gas can be found from Henry’s law: Sgas = kH × Pgas. The problem gives kH for N2 but not its partial pressure. To calculate the partial pressure, use the relationship from Chapter 5: Pgas = Xgas × Ptotal where X represents the mole fraction of the gas. Solution: To find partial pressure use the 78% N2 given for the mole fraction: Pgas = Xgas × Ptotal PN2 = 0.78 × 1 atm = 0.78 atm Use Henry’s law to find solubility at this partial pressure: Sgas = kH × Pgas SN2 = (7 × 10–4 mol/L∙atm)(0.78 atm) = 5.46 × 10–4 mol/L = 5 × 10–4 mol/L

13.3B

Plan: The Henry’s law constant of a gas can be found from Henry’s law: Sgas = kH × Pgas. The problem gives the solubility (Sgas) for N2O but not its partial pressure. To calculate the partial pressure, use Dalton’s law from Chapter 5: Pgas = Xgas × Ptotal where X represents the mole fraction of the gas. Solution: To find partial pressure use the 40.% N2O given for the mole fraction: Pgas = Xgas × Ptotal PN2 O = 0.40 × 1.2 atm = 0.48 atm Use Henry’s law to find the Henry’s law constant, kH, at 25°C: Sgas 1.2 ×10−2 mol/L –2 = = 2.5 × 10 mol/L∙atm kH = Pgas 0.48 atm

13.4A

Plan: Molality (m) is defined as amount (mol) of solute per kg of solvent. Use the molality and the mass of solvent given to calculate the amount of glucose in moles. Then convert amount (mol) of glucose to mass (g) of glucose by multiplying by its molar mass. Solution: Mass of solvent must be converted to kg. ⎛ 1 kg ⎞ Mass of solvent (kg) = (563 g)⎜⎜ 3 ⎟⎟⎟ = 0.563 kg ⎜⎝10 g ⎠⎟

⎛ 2.40 × 10−2 mol C H O ⎞⎟⎛180.16 g C H O ⎞ ⎜ 6 12 6 ⎟⎜ 6 12 6 ⎟ ⎟⎟ Mass (g) of glucose = 0.563 kg solvent ⎜⎜⎜ ⎟⎟⎟⎜⎜⎜ 1 kg solvent 1 mol C H O ⎟⎟ ⎟⎠⎜⎝ ⎜⎜⎝ 6 12 6 ⎠ = 2.4343 = 2.43 g C6H12O6

(

13.4B

)

Plan: Molality (m) is defined as amount (mol) of solute per kg of solvent. Calculate the moles of solute, I2, from the mass (g) of I2 and divide by molality to obtain mass (kg) of solvent, diethyl ether. Then convert that mass of diethyl ether to grams. Solution: ⎛ 1 mol I ⎞⎟ 2 ⎜ ⎟⎟ = 0.0598897 mol I Amount (mol) of solute = (15.20 g I2) ⎜⎜ 2 ⎜⎜⎝ 253.8 g I 2 ⎠⎟⎟ ⎛1 kg (CH CH ) O ⎟⎞⎛1000 g ⎞⎟ 3 2 2 ⎜ ⎟⎟⎜⎜ ⎟⎟ = 98.66 = 98.7 g (CH CH ) O Mass of solvent (g) = (0.05989 mol I2) ⎜⎜ 3 2 2 ⎜⎝⎜ 0.607 mol I 2 ⎟⎟⎠⎝⎜⎜⎜ 1 kg ⎠⎟⎟

13.5A

Plan: Mass percent is the mass (g) of solute per 100 g of solution. For each alcohol, divide the mass of the alcohol by the total mass. Multiply this number by 100 to obtain mass percent. To find mole fraction, first find the amount (mol) of each alcohol, then divide by the total moles.

13-2

Solution: Mass percent: Mass % propanol =

35.0 g mass of propanol × 100% ×100% = mass of propanol + mass of ethanol 35.0 + 150 g

(

)

= 18.9189 = 18.9% propanol Mass % of ethanol =

150. g mass of ethanol × 100% ×100% = mass of propanol + mass of ethanol 35.0 + 150 g

(

)

= 81.08108 = 81.1% ethanol Mole fraction: ⎛ 1 mol C 3 H 7 OH ⎞⎟ ⎜ ⎟⎟ = 0.5824596 mol propanol Moles of propanol = (35.0 g C 3 H 7 OH )⎜⎜ ⎜⎜⎝ 60.09 g C 3 H 7 OH ⎠⎟⎟ ⎛ 1 mol C 2 H 5 OH ⎞⎟ ⎜ ⎟⎟ = 3.2559149 mol ethanol Moles of ethanol = (150. g C 2 H 5 OH )⎜⎜ ⎜⎝⎜ 46.07 g C 2 H 5 OH ⎠⎟⎟ 0.5824596 mol propanol Xpropanol = = 0.151746 = 0.152 0.5824596 mol propanol + 3.2559149 mol ethanol 3.2559149 mol ethanol Xethanol = = 0.84825 = 0.848 0.5824596 mol propanol + 3.2559149 mol ethanol

13.5B

Plan: Mass percent is the mass (g) of solute per 100 g of solution. For each component of the mixture, divide the mass of the component by the total mass. Multiply this number by 100 to obtain mass percent. To find mole percent, first find the amount (mol) of each component, then divide by the total moles and multiply by 100%. Solution: Mass percent: mass of ethanol Mass % ethanol = ×100% mass of ethanol + mass of iso-octane + mass of heptane =

Mass % iso-octane =

1.87 g + 27.4 g + 4.10 g

× 100% = 5.6038 = 5.60% ethanol

mass of iso-octane mass of ethanol + mass of iso-octane + mass of heptane =

Mass % heptane =

1.87 g

27.4 g 1.87 g + 27.4 g + 4.10 g

× 100% = 82.1097 = 82.1% iso-octane

mass of heptane mass of ethanol + mass of iso-octane + mass of heptane =

4.10 g 1.87 g + 27.4 g + 4.10 g

× 100%

× 100% = 12.2865 = 12.3% heptane

Mole percent: ⎛ 1 mole ethanol ⎞⎟ ⎜ Moles of ethanol = (1.87 g ethanol) ⎜⎜ ⎟⎟ = 0.0406 mol ethanol ⎜⎜⎝ 46.07 g ethanol ⎠⎟⎟ ⎛ 1 mole iso-octane ⎞⎟ ⎜ ⎟⎟ = 0.240 mol iso-octane Moles of iso-octane = (27.4 g iso-octane) ⎜⎜ ⎜⎝⎜114.22 g iso-octane ⎠⎟⎟ Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

13-3

⎛ 1 mole heptane ⎞⎟ ⎜ ⎟⎟ = 0.0409 mol heptane Moles of heptane = (4.10 g heptane) ⎜⎜ ⎜⎜⎝100.20 g heptane ⎠⎟⎟

Mole percent ethanol =

moles of ethanol moles of ethanol + moles of iso-octane + moles of heptane =

Mole percent iso-octane =

0.0406 mol 0.0406 mol + 0.240 mol + 0.0409 mol

× 100%

× 100% = 12.6283 = 12.6% ethanol

moles of iso-octane moles of ethanol + moles of iso-octane + moles of heptane

× 100%

0.240 mol =

Mole percent heptane =

0.0406 mol + 0.240 mol + 0.0409 mol

× 100% = 74.6501 = 74.6% iso-octane

moles of heptane moles of ethanol + moles of iso-octane + moles of heptane

× 100%

0.0409 mol =

0.0406 mol + 0.240 mol + 0.0409 mol

× 100% = 12.7216 = 12.7% heptane

(Slight differences from 100% are due to rounding.) 13.6A

Plan: To find the mass percent, molality and mole fraction of HCl, the following is needed: 1) Moles of HCl in 1 L solution (from molarity) 2) Mass of HCl in 1L solution (from molarity times molar mass of HCl) 3) Mass of 1L solution (from volume times density) 4) Mass of solvent in 1L solution (by subtracting mass of solute from mass of solution) 5) Moles of solvent (by dividing the mass of solvent by molar mass of water) Mass percent is calculated by dividing mass of HCl by mass of solution and multiplying by 100. Molality is calculated by dividing moles of HCl by mass of solvent in kg. Mole fraction is calculated by dividing mol of HCl by the sum of mol of HCl plus mol of solvent. Solution: Assume the volume is exactly 1 L. ⎛11.8 mol HCl ⎞⎟ = 11.8 mol HCl 1) Mole of HCl in 1 L solution = (1.0 L) ⎜⎜ ⎝ ⎠⎟⎟ 1.0 L ⎛ 36.46 g HCl ⎞⎟ 2) Mass (g) of HCl in 1 L solution = (11.8 mol HCl)⎜⎜ = 430.228 g HCl ⎝ 1 mol HCl ⎠⎟⎟ ⎛ 1 mL ⎞⎛1.190 g ⎞⎟ = 1190 g solution 3) Mass (g) of 1 L solution = (1.0 L)⎜⎜ −3 ⎟⎟⎟⎜⎜ ⎝10 L ⎠⎝ 1 mL ⎠⎟⎟ 4) Mass (g) of solvent in 1 L solution = 1190 g – 430.228 g = 759.772 g solvent (H2O) ⎛ 1 kg ⎞ Mass (kg) of solvent in 1 L solution = (759.772 g)⎜⎜⎜ 3 ⎟⎟⎟ = 0.759772 kg solvent ⎝10 g ⎠⎟

⎛ 1 mol H O ⎞⎟ ⎜ 2 ⎟⎟ = 42.1627 mol solvent 5) Moles of solvent in 1 L solution = (759.772 g H 2 O) ⎜⎜ ⎜⎝18.02 g H 2 O ⎠⎟

430.228 g HCl mass HCl (100) = 36.1536 = 36.2% (100) = mass solution 1190 g solution mole HCl 11.8 mol HCl Molality of HCl = = = 15.530975 = 15.5 m kg solvent 0.759772 kg

Mass percent of HCl =

13-4

Mole fraction of HCl = 13.6B

mol HCl 11.8 mol = = 0.21866956 = 0.219 11.8 mol + 42.1627 mol mol HCl + mol H 2 O

Plan: To find the mass percent, molarity and mole percent of CaBr2, the following is needed: 1) Mass of CaBr2 and mass of H2O in the solution 2) Total solution volume 3) Moles of H2O in the solution Mass percent is calculated by dividing the mass of CaBr2 by the total mass of solution and multiplying by 100. Molarity is calculated by dividing the moles of CaBr2 by the volume of the solution (L). Mole percent is calculated by dividing moles of CaBr2 by the total moles in the solution and multiplying by 100. Solution: ⎛199.88 g CaBr ⎞⎟ 2 ⎜ ⎟⎟ = 1087.347 g CaBr 1) Mass of CaBr2 = (5.44 mol CaBr2) ⎜⎜ 2 ⎜⎝⎜ 1 mol CaBr2 ⎠⎟⎟ ⎛1000 g ⎞⎟ ⎜ ⎟⎟ = 1000 g H O Mass of H2O = 1 kg ⎜⎜ 2 ⎜⎝⎜ 1 kg ⎠⎟⎟ 1087.347 g CaBr2 mass of solute = × 100% = 52.09230 Mass % CaBr2 = 1087.347 g CaBr2 + 1000 g H 2 O total mass of solution = 52.1 mass % CaBr2 ⎛ 1 mL solution ⎞⎛ 1 L ⎞⎟ ⎜ ⎟⎟⎟⎜⎜ ⎟ 2) Volume(L) of solution = (1087.347 g CaBr2 + 1000 g H 2 O) ⎜⎜ ⎟⎟⎜⎜⎜1000 mL ⎟⎟⎟ ⎜⎜⎝1.70 g solution ⎠⎝ ⎠

= 1.227851 L solution 5.44 mol CaBr2 moles of solute Molarity = = = 4.43050 = 4.43 M 1.227851 L solution L of solution ⎛ 1 mol H 2 O ⎞⎟ ⎟ = 55.493896 mol H2O 3) Amount (mol) of H2O = 1000 g H 2 O ⎜⎜ ⎜⎝18.02 g H 2 O ⎠⎟⎟

Mole percent =

⎛ ⎞⎟ 5.44 mol ⎜ × 100% = ⎜⎜ ⎟⎟ × 100% moles solute + moles solvent ⎜⎜⎝ 5.44 mol + 55.493896 mol ⎠⎟⎟ moles solute

= 8.927707 = 8.93% 13.7A

Plan: Raoult’s law states that the vapor pressure of a solvent is proportional to the mole fraction of the solvent: Psolvent = Xsolvent × P°solvent. To calculate the drop in vapor pressure, a similar relationship is used with the mole fraction of the solute substituted for that of the solvent. Solution: Mole fraction of aspirin in methanol: ⎛ 1 mol aspirin ⎞⎟ ⎜ ⎟⎟ 2.00 g ⎜⎜ ⎜⎜⎝180.15 g aspirin ⎠⎟⎟ ⎛ 1 mol aspirin ⎞⎟ ⎛ 1 mol methanol ⎞⎟ ⎜ ⎟⎟ + 50.0 g ⎜⎜ ⎟⎟ 2.00 g ⎜⎜ ⎜⎜ ⎜⎜⎝180.15 g aspirin ⎠⎟⎟ ⎜⎝ 32.04 g methanol ⎠⎟⎟

(

)

(

)

Xaspirin = 7.06381 × 10–3 ΔP = Xaspirin P°methanol = (7.06381 × 10–3) (101 torr) = 0.71344 = 0.713 torr

13-5

13.7B

Plan: The drop in vapor pressure of a solvent is calculated by the following equation derived from Raoult’s law: ΔP = Xsolvent × P°solvent. Solution: ⎛ 1 mol menthol ⎞⎟ ⎜ ⎟⎟ = 0.0415333 mol menthol Amount (mol) menthol = (6.49 g menthol) ⎜⎜ ⎜⎝⎜156.26 g menthol ⎠⎟⎟ ⎛ 1 mol ethanol ⎞⎟ ⎜ ⎟⎟ = 0.542652 mol ethanol Amount (mol) ethanol = (25.0 g ethanol) ⎜⎜ ⎜⎜⎝ 46.07 g ethanol ⎠⎟⎟ ⎛ ⎞⎟ 0.542652 mol ⎜ ⎟⎟ (51.53 torr) = 47.8664 = 47.9 torr ΔP = ⎜⎜ ⎜⎜⎝ 0.0415333 mol + 0.542652 mol ⎠⎟⎟

13.8A

Plan: Find the molality of the solution by dividing the moles of P4 by the mass of the CS2 in kg. The change in freezing point is calculated from ΔTf = iKfm, where Kf is 3.83°C/m when CS2 is the solvent, i is the van’t Hoff factor, and m is the molality of particles in solution. Since P4 is a covalent compound and does not ionize in water, i = 1. Once ΔTf is calculated, the freezing point is determined by subtracting it from the freezing point of pure CS2 (–111.5°C). The boiling point of a solution is increased relative to the pure solvent by the relationship ΔTb = iKbm where Kb is 2.34°C/m when CS2 is the solvent, i is the van’t Hoff factor, and m is the molality of particles in solution. P4 is a nonelectrolyte (it is a molecular compound) so i = 1. Once ΔTb is calculated, the boiling point is determined by adding it to the boiling point of pure CS2 (46.2°C). Solution: Molality of P4 solution: ⎛ 1 mol P ⎞⎟ 4 ⎜ ⎟⎟ = 0.0681 mol Moles of P4 = (8.44 g P4) ⎜⎜ ⎜⎜⎝123.88 g P4 ⎠⎟⎟ Molality of P4 =

⎛ 0.0681 mol P ⎞⎛ 1000 g ⎞⎟ 4 ⎟⎜ ⎜ ⎟⎟⎜ ⎟ = ⎜⎜ ⎟⎟⎜⎜⎜ 1 kg ⎟⎟⎟ = 1.14 m P4 kg solvent ⎜⎜⎝ 60.0 g CS2 ⎠⎝ ⎠

moles solute

Freezing point: ΔTf = iKfm = (1)(3.83°C/m)(1.14 m) = 4.37°C The freezing point is: –111.5°C – 4.37°C = –115.9°C. Boiling point: ΔTb = iKbm = (1)(2.34°C/m)(1.14 m) = 2.67°C The boiling point is 46.2°C + 2.67°C = 48.9°C. 13.8B

Plan: The question asks for the concentration of ethylene glycol that would prevent freezing at 0.00°F. First, calculate the change in freezing point of water, which will be 0.00°F subtracted from the freezing point of pure water, 32°F. Then convert this temperature to Celsius. Use this value for ΔT in ΔTf = 1.86°C/m × molality of solution and solve for molality. Solution: Temperature conversion: ⎛ 5°C ⎞⎟ ⎛ 5°C ⎞⎟ 32°F − 0.00°F ⎜⎜ T (°C) = T (°F ) − 32°F ⎜⎜ ⎜⎝ 9°F ⎠⎟⎟ = ⎜⎝ 9°F ⎠⎟⎟ = 17.7778°C ΔTf 17.7778ο C = = 9.557956 = 9.56 m Molality of C2H6O2 = Kf 1.86ο C/m

(

)

(

)

The minimum concentration of ethylene glycol would have to be 9.56 m in order to prevent the water from freezing at 0.00°F. ⎛ 9.557956 mol C 2 H 6 O2 ⎞⎟⎛ 62.07 g C 2 H 6 O2 ⎞⎟ ⎟ = 2373.047 = 2370 g C2H6O2 ⎟⎟⎜⎜ Mass of C2H6O2 = 4.00 kg H 2 O ⎜⎜ ⎟⎜⎜ 1 mol C 2 H 6 O2 ⎠⎟⎟ ⎜⎝ 1 kg H 2 O ⎠⎝ Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

13-6

13.9A

Plan: Use the osmotic pressure, the temperature, and the gas constant to calculate the molarity of the solution with the equation Π = MRT. Multiply the molarity of the solution by the volume of the sample to calculate the number of moles in the sample. The molar mass is calculated by dividing the mass of the sample by the number of moles in the sample. Solution: Π Π = MRT, so M = RT 1 atm 8.98 torr × 760 torr M= = 4.79492 × 10–4 M ⎛ ⎞ atm i L ⎟ ⎜⎜ ⎟⎟ ((27°C + 273.15) K) ⎜⎜0.0821 mol i K ⎠⎟⎟ ⎜⎝

Amount (mol) protein = (4.79492 × 10–4 mol/L) (12.0 mL) (1 L/1000 mL) = 5.75390 × 10–6 mol ℳ=

13.9B

0.200 g 5.75390 ×10−6 mol

= 3.47590 × 104 g/mol = 3.48 × 104 g/mol

Plan: Look up the normal freezing point for benzene in the table within this section of the textbook and calculate the freezing point depression. Find the molality using ΔTf = 1.86°C/m × molality of solution. Use the given mass of the benzene to calculate the moles of naphthalene. Divide the given mass of the naphthalene by the moles of naphthalene to calculate the molar mass. Solution: ΔTf 5.56 ο C − 4.20 ο C Molality of naphthalene = = = 0.277551 m naphthalene Kf 4.90 ο C/m ⎛ 1 kg ⎞⎛ ⎟⎟⎜⎜ 0.277551 mol naphthalene ⎞⎟⎟ = .0555102 mol Amount (mol) naphthalene = 200 g benzene ⎜⎜⎜ ⎟⎟⎜ ⎟ 1 kg benzene ⎝1000 g ⎠⎝ ⎠⎟ Molar mass naphthalene =

7.01 g = 126.283 = 126 g/mol 0.0555102 mol

13.10A Plan: Write the formula of potassium phosphate, determine if the compound is soluble and, if soluble, how many cations and anions result when one unit of the compound is placed in water. Calculate the molality of the solution. The boiling point of a solution is increased relative to the pure solvent by the relationship ΔTb = iKbm where Kb is 0.512°C/m for water, i is the van’t Hoff factor (equal to the number of particles or ions produced when one unit of the compound is dissolved in the solvent), and m is the molality of particles in solution. Once ΔTb is calculated, the boiling point is determined by adding it to the boiling point of pure H2O (100.00°C). Solution: a) The formula for potassium phosphate is K3PO4. When it is placed in water, 3 K+ ions and 1 PO43– ion are formed. Scene B is the only scene that shows separate ions in a 3K+/1PO43– ratio. b) Molality of K3PO4 solution: ⎛ 1 mol K PO ⎞⎟ 3 4 ⎜ ⎟⎟ = 0.146983 mol K PO Moles of K3PO4 = (31.2 g K3PO4) ⎜⎜ 3 4 ⎜⎝⎜ 212.27 g K 3 PO 4 ⎠⎟⎟ ⎛ 0.146983 mol K PO ⎞⎛ 1000 g ⎞⎟ 3 4 ⎟⎜ ⎜ = ⎜⎜ ⎟⎟⎟⎜⎜ ⎟⎟ = 1.72921 m K3PO4 kg solvent 85.0 g H 2 O ⎟⎜⎜ 1 kg ⎠⎟⎟ ⎜⎜⎝ ⎠⎝ Because 4 ions total are formed when each unit of K3PO4 is placed in water, i = 4. ΔTb = iKbm = (4)(0.512 °C/m)(1.72921 m) = 3.5414°C The boiling point is 100.00°C + 3.5414°C = 103.5414 = 103.54°C.

Molality of K3PO4 =

moles solute

13-7

13.10B Plan: Calculate the molarity of the solution, then calculate the osmotic pressure of the magnesium chloride solution. Do not forget that MgCl2 is a strong electrolyte, and ionizes to yield three ions per formula unit. This will result in a pressure three times as great as a nonelectrolyte (glucose) solution of equal concentration. Solution: −3 ⎛ 1 mL ⎞⎛ ⎟⎟⎜10 L ⎞⎟⎟ = 0.1003499 L a) Volume (L) of solution = (100. + 0.952 g solution)⎜⎜ ⎜ ⎜⎝1.006 g ⎠⎟⎟⎜⎝ 1 mL ⎠⎟⎟ ⎛ 1 mol MgCl 2 ⎞⎟ ⎟⎟ = 9.9989 × 10–3 mol MgCl2 Moles of MgCl2 = (0.952 g MgCl 2 )⎜⎜⎜ ⎝ 95.21 g MgCl 2 ⎠⎟ mol MgCl 2 0.0099989 mol = = 0.099640 M MgCl2 Molarity of MgCl2 = volume solution 0.1003499 L ⎛ 0.099640 mol ⎞⎟⎛ L • atm ⎞⎟ ⎟⎜0.0821 Π = iMRT = 3⎜⎜⎜ ⎟((273.2 + 20.0) K ) = 7.1955 = 7.20 atm ⎟⎜ ⎜ L mol • K ⎠⎟ ⎝⎜ ⎠⎟⎝

b) The magnesium chloride solution has a higher osmotic pressure; thus, solvent will diffuse from the glucose solution into the magnesium chloride solution. The glucose side will lower and the magnesium chloride side will rise. Scene C represents this situation. CHEMICAL CONNECTIONS BOXED READING PROBLEMS

B13.1

a) The colloidal particles in water generally have negatively charged surfaces and so repel each other, slowing the settling process. Cake alum, Al2(SO4)3, is added to coagulate the colloids. The Al3+ ions neutralize the negative surface charges and allow the particles to aggregate and settle. b) Water that contains large amounts of divalent cations (such as Ca2+ and Mg2+) is called hard water. During cleaning, these ions combine with the fatty-acid anions in soaps to produce insoluble deposits. c) In reverse osmosis, a pressure greater than the osmotic pressure is applied to the solution, forcing the water back through the membrane and leaving the ions behind. d) Chlorine may give the water an unpleasant odor, and can form carcinogenic chlorinated compounds. e) The high concentration of NaCl displaces the divalent and polyvalent ions from the ion-exchange resin.

B13.2

Plan: Osmotic pressure is calculated from the molarity of particles, the gas constant, and temperature. Convert the mass of sucrose to moles using the molar mass, and then to molarity. Sucrose is a nonelectrolyte so i = 1. Solution: T = 273 + 20°C = 293 K ⎛ 1 mol sucrose ⎞⎟ ⎟ = 0.01037102 mol sucrose Moles of sucrose = (3.55 g sucrose)⎜⎜ ⎜⎝ 342.30 g sucrose ⎠⎟⎟ moles of sucrose 0.01037102 mol Molarity = = = 1.037102 × 10–2 M sucrose volume of solution 1.0 L Π = iMRT = (1)(1.037102 × 10–2 mol/L)(0.0821 L∙atm/mol∙K)(293 K) = 0.2494780 = 0.249 atm A pressure greater than 0.249 atm must be applied to obtain pure water from a 3.55 g/L solution.

END–OF–CHAPTER PROBLEMS

13.1

The composition of seawater, like all mixtures, is variable. The components of seawater (water and various ions) have not been changed and thus retain some of their properties. For example, seawater has a salty taste due to the presence of salts such as NaCl.

13.2

When a salt such as NaCl dissolves, ion-dipole forces cause the ions to separate, and many water molecules cluster around each of them in hydration shells. Ion-dipole forces hold the first shell. Additional shells are held by hydrogen bonding to inner shells.

13-8

13.3

In CH3(CH2)nCOOH, as n increases, the hydrophobic (CH) portion of the carboxylic acid increases and the hydrophilic part of the molecule stays the same, with a resulting decrease in water solubility.

13.4

Sodium stearate would be a more effective soap because the hydrocarbon chain in the stearate ion is longer than the chain in the acetate ion. A soap forms suspended particles called micelles with the polar end of the soap interacting with the water solvent molecules and the nonpolar ends forming a nonpolar environment inside the micelle. Oils dissolve in the nonpolar portion of the micelle. Thus, a better solvent for the oils in dirt is a more nonpolar substance. The long hydrocarbon chain in the stearate ion is better at dissolving oils in the micelle than the shorter hydrocarbon chain in the acetate ion.

13.5

Hexane and methanol, as gases, are free from any intermolecular forces and can simply intermix with each other. As liquids, hexane is a nonpolar molecule, whereas methanol is a polar molecule. “Like dissolves like.”

13.6

Hydrogen chloride (HCl) gas is actually reacting with the solvent (water) and thus shows a higher solubility than propane (C3H8) gas, which does not react.

13.7

Plan: A more concentrated solution will have more solute dissolved in the solvent. Determine the types of intermolecular forces in the solute and solvents. A solute tends to be more soluble in a solvent whose intermolecular forces are similar to its own. Solution: Potassium nitrate, KNO3, is an ionic compound and can form ion-dipole forces with a polar solvent like water, thus dissolving in the water. Potassium nitrate is not soluble in the nonpolar solvent CCl4. Because potassium nitrate dissolves to a greater extent in water, (a) KNO3 in H2O will result in the more concentrated solution.

13.8

b) Stearic acid in CCl4. Stearic acid will not dissolve in water. It is nonpolar while water is very polar. Stearic acid will dissolve in carbon tetrachloride, as both are nonpolar.

13.9

Plan: To identify the strongest type of intermolecular force, check the formula of the solute and identify the forces that could occur. Then look at the formula for the solvent and determine if the forces identified for the solute would occur with the solvent. Ionic forces are present in ionic compounds; dipole-dipole forces are present in polar substances, while nonpolar substances exhibit only dispersion forces. The strongest force is ion-dipole followed by dipole-dipole (including H bonds). Next in strength is ion–induced dipole force and then dipole– induced dipole force. The weakest intermolecular interactions are dispersion forces. Solution: a) Ion-dipole forces are the strongest intermolecular forces in the solution of the ionic substance cesium chloride in polar water. b) Hydrogen bonding (type of dipole-dipole force) is the strongest intermolecular force in the solution of polar propanone (or acetone) in polar water. c) Dipole–induced dipole forces are the strongest forces between the polar methanol and nonpolar carbon tetrachloride.

13.10

a) metallic bonding b) dipole-dipole c) dipole–induced dipole

13.11

13-9

Solution: a) Hydrogen bonding occurs between the H atom on water and the lone electron pair on the O atom in dimethyl ether (CH3OCH3). However, none of the hydrogen atoms on dimethyl ether participates in hydrogen bonding because the C−H bond does not have sufficient polarity. b) The dipole in water induces a dipole on the Ne(g) atom, so dipole–induced dipole interactions are the strongest intermolecular forces in this solution. c) Nitrogen gas and butane are both nonpolar substances, so dispersion forces are the principal attractive forces. 13.12

a) dispersion forces b) hydrogen bonding c) dispersion forces

13.13

Plan: CH3CH2OCH2CH3 is polar with dipole-dipole interactions as the dominant intermolecular forces. Examine the solutes to determine which has intermolecular forces more similar to those in diethyl ether. This solute is the one that would be more soluble. Solution: a) HCl would be more soluble since it is a covalent compound with dipole-dipole forces, whereas NaCl is an ionic solid. Dipole-dipole forces between HCl and diethyl ether are more similar to the dipole forces in diethyl ether than the ion-dipole forces between NaCl and diethyl ether. b) CH3CHO (acetaldehyde) would be more soluble. The dominant interactions in H2O are hydrogen bonding, a stronger type of dipole-dipole force. The dominant interactions in CH3CHO are dipole-dipole. The solute-solvent interactions between CH3CHO and diethyl ether are more similar to the solvent intermolecular forces than the forces between H2O and diethyl ether. c) CH3CH2MgBr would be more soluble. CH3CH2MgBr has a polar end (–MgBr) and a nonpolar end (CH3CH2–), whereas MgBr2 is an ionic compound. The nonpolar end of CH3CH2MgBr and diethyl ether would interact with dispersion forces, while the polar end of CH3CH2MgBr and the dipole in diethyl ether would interact with dipole-dipole forces.

13.14

a) CH3CH2-O-CH3(g), due to its smaller size (smaller molar mass). b) CH2Cl2, because it is more polar than CCl4. c) Tetrahydropyran is more water soluble due to hydrogen bonding between the oxygen atom and water molecules.

13.15

No, river water is a heterogeneous mixture, with its composition changing from one segment to another.

13.16

Plan: Determine the types of intermolecular forces present in the two compounds and in water and hexane. Substances with similar types of forces tend to be soluble while substances with different type of forces tend to be insoluble. Solution: Gluconic acid is a very polar molecule because it has –OH groups attached to every carbon. The abundance of –OH bonds allows gluconic acid to participate in extensive H bonding with water, hence its great solubility in water. On the other hand, caproic acid has a five carbon, nonpolar, hydrophobic (“water hating”) tail that does not easily dissolve in water. The dispersion forces in the nonpolar tail are more similar to the dispersion forces in hexane, hence its greater solubility in hexane.

13.17

There may be a disulfide linkage (a covalent disulfide bridge) between the ends of two cysteine side chains that bring together parts of the chain. There may be salt links between ions –COO– and –NH3+ groups. There may be hydrogen bonding between the C=O of one peptide bond and the N–H of another.

13.18

The nitrogen bases hydrogen bond to their complimentary bases. The flat, N-containing bases stack above each other, which allow extensive interaction through dispersion forces. The exterior negatively charged sugarphosphate chains form ion-dipole and hydrogen bonds to the aqueous surroundings, but this is of minor importance to the structure.

13-10

13.19

While an individual hydrogen bond is not too strong, there are very large numbers of hydrogen bonds present in DNA. The energy of so many hydrogen bonds keeps the chains together. But the hydrogen bonds are weak enough that a few are easily broken when the two chains in DNA must separate.

13.20

The more carbon and hydrogen atoms present, the more soluble the substance is in nonpolar oil droplets. Therefore, sodium propanoate is not as effective a soap as sodium stearate with the longer hydrocarbon chain.

13.21

Dispersion forces are present between the nonpolar portions of the molecules within the bilayer. Polar groups are present to hydrogen bond or to form ion-dipole interactions with the aqueous surroundings.

13.22

In soluble proteins, polar groups are found on the exterior and nonpolar groups on the interior. In proteins embedded in a membrane, the exterior of the protein that lies within the bilayer consists of nonpolar amino acid side chains, whereas the portion lying outside the bilayer has polar side chains. Amino acids with side chains that may be ionic are necessary. Two examples are lysine and glutamic acid.

13.23 13.24

The ΔHsolvent and ΔHmix components of the heat of solution combined together represent the enthalpy change during solvation, the process of surrounding a solute particle with solvent particles. When the solvent is water, solvation is called hydration.

13.25

For a general solvent, the energy changes needed to separate solvent into particles (ΔHsolvent), and that needed to mix the solvent and solute particles (ΔHmix) would be combined to obtain ΔHsolution.

13.26

a) Charge density is the ratio of the ion’s charge to its volume. An ion’s charge and size affect its charge density. b) – < + < 2– < 3+ c) The higher the charge density, the more negative the ΔHhydration. ΔHhydration increases with increasing charge and decreases with increasing size.

13.27

The solution cycle for ionic compounds in water consists of two enthalpy terms: the lattice energy, and the combined heats of hydration of the cation and anion. ΔHsolution = ΔHlattice + ΔHhydration of ions For a heat of solution to be zero (or very small) ΔHlattice ≈ ΔHhydration of ions, and they would have to have opposite signs.

13.28

a) Endothermic b) The lattice energy term is much larger than the combined ionic heats of hydration. c) The increase in entropy outweighs the increase in enthalpy, so ammonium chloride dissolves.

13.29

This compound would be very soluble in water. A large exothermic value in ΔHsolution (enthalpy of solution) means that the solution has a much lower energy state than the isolated solute and solvent particles, so the system tends to the formation of the solution. Entropy that accompanies dissolution always favors solution formation. Entropy becomes important when explaining why solids with endothermic ΔHsolution values (and higher energy states) are still soluble in water.

13.30

Plan: ΔHsolution = ΔHlattice + ΔHhydration. Lattice energy values are always positive as energy is required to separate the ions from each other. Hydration energy values are always negative as energy is released when intermolecular forces between ions and water form. Since the heat of solution for KCl is endothermic, the lattice energy must be greater than the hydration energy for an overall input of energy.

13-11

Solution:

13.31

Lattice energy values are always positive as energy is required to separate the ions from each other. Hydration energy values are always negative as energy is released when intermolecular forces between ions and water form. Since the heat of solution for NaI is exothermic, the negative hydration energy must be greater than the positive lattice energy.

13.32

Plan: Charge density is the ratio of an ion’s charge (regardless of sign) to its volume. An ion’s volume is related to its radius. For ions whose charges have the same sign (+ or –), ion size decreases as a group in the periodic table is ascended and as you proceed from left to right in the periodic table. Charge density increases with increasing charge and increases with decreasing size. Solution: a) Both ions have a +1 charge, but the volume of Na+ is smaller, so it has the greater charge density. b) Sr2+ has a greater ionic charge and a smaller size (because it has a greater Zeff), so it has the greater charge density. c) Na+ has a smaller ion volume than Cl–, so it has the greater charge density. d) O2– has a greater ionic charge and similar ion volume, so it has the greater charge density. e) OH– has a smaller ion volume than SH– (O is smaller than S), so it has the greater charge density. f) Mg2+ has the higher charge density because it has a smaller ion volume. g) Mg2+ has the higher charge density because it has both a smaller ion volume and greater charge. h) CO32– has the higher charge density because it has both a smaller ion volume and greater charge.

13-12

–

13.33

a) I has a smaller charge density (larger ion volume) than Br–. b) Ca2+ has a lower ratio than Sc3+, due to its smaller ion charge. – c) Br has a lower ratio than K+, due to its larger ion volume. – d) Cl has a lower ratio than S2–, due to its smaller ion charge. e) Sc3+ has a lower ratio than Al3+, due to its larger ion volume. f) ClO4– has a lower ratio due to its smaller ion charge. g) Fe2+ has a lower ratio due to its smaller ion charge. h) K+ has a lower ratio due to its smaller ion charge.

13.34

Plan: The ion with the greater charge density will have the larger ΔHhydration. Solution: a) Na+ would have a larger ΔHhydration than Cs+ since its charge density is greater than that of Cs+. b) Sr2+ would have a larger ΔHhydration than Rb+. c) Na+ would have a larger ΔHhydration than Cl–. d) O2– would have a larger ΔHhydration than F–. e) OH– would have a larger ΔHhydration than SH–. f) Mg2+ would have a larger ΔHhydration than Ba2+. g) Mg2+ would have a larger ΔHhydration than Na+. h) CO32– would have a larger ΔHhydration than NO3–.

13.35

a) I

13.36

Plan: Use the relationship ΔHsolution = ΔHlattice + ΔHhydration. Given ΔHsolution and ΔHlattice, ΔHhydration can be calculated. ΔHhydration increases with increasing charge density, and charge density increases with increasing charge and decreasing size. Solution: a) The two ions in potassium bromate are K+ and BrO3–. ΔHsolution = ΔHlattice + ΔHhydration ΔHhydration = ΔHsolution – ΔHlattice = 41.1 kJ/mol – 745 kJ/mol = –703.9 = –704 kJ/mol b) K+ ion contributes more to the heat of hydration because it has a smaller size and, therefore, a greater charge density.

13.37

a) ΔHhydration = ΔHsolution – ΔHlattice ΔHhydration = 17.3 kJ/mol –763 kJ/mol ΔHhydration = –745.7 = – 746 kJ/mol b) Na+ ion contributes more to the heat of hydration due to its smaller size (larger charge density).

13.38

Plan: Entropy increases as the possible states for a system increase, which is related to the freedom of motion of its particles and the number of ways they can be arranged. Solution: a) Entropy increases as the gasoline is burned. Gaseous products at a higher temperature form. b) Entropy decreases as the gold is separated from the ore. Pure gold has only the arrangement of gold atoms next to gold atoms, while the ore mixture has a greater number of possible arrangements among the components of the mixture. c) Entropy increases as a solute dissolves in the solvent.

13.39

a) Entropy increases. b) Entropy decreases. c) Entropy increases.

13.40

ΔHsolution = ΔHlattice + ΔHhydration ΔHsolution = 822 kJ/mol + (– 799 kJ/mol) ΔHsolution = 23 kJ/mol

–

b) Ca2+

–

c) Br

–

d) Cl

e) Sc3+

f) ClO4–

g) Fe2+

h) K+

13-13

13.41

Add a pinch of the solid solute to each solution. A saturated solution contains the maximum amount of dissolved solute at a particular temperature. When additional solute is added to this solution, it will remain undissolved. An unsaturated solution contains less than the maximum amount of dissolved solute and so will dissolve added solute. A supersaturated solution is unstable and addition of a “seed” crystal of solute causes the excess solute to crystallize immediately, leaving behind a saturated solution.

13.42

KMnO4(s) + H2O(l) + heat → KMnO4(aq) Prepare a mixture of more than 6.4 g KMnO4/100 g H2O and heat it until the solid completely dissolves. Then carefully cool it, without disturbing it or shaking it, back to 20°C. If no crystals form, you would then have a supersaturated solution.

13.43

An increase in temperature produces an increase in kinetic energy; the gaseous solute molecules overcome the weak intermolecular forces, which results in a decrease in solubility of any gas in water. In nearly all cases, gases dissolve exothermically (ΔHsoln < 0).

13.44

Plan: The solubility of a gas in water decreases with increasing temperature and increases with increasing pressure. Solution: a) Increasing pressure for a gas increases the solubility of the gas according to Henry’s law. b) Increasing the volume of a gas causes a decrease in its pressure (Boyle’s law), which decreases the solubility of the gas.

13.45

a) increase

13.46

Plan: Solubility for a gas is calculated from Henry’s law: Sgas = kH × Pgas. We know kH and Pgas, so Sgas can be calculated with units of mol/L. To calculate the mass of oxygen gas, convert moles of O2 to mass of O2 using the molar mass. Solution: a) Sgas = kH × Pgas ⎛ mol ⎞⎟ ⎟ 1.00 atm = 1.28 × 10–3 mol/L Sgas = ⎜⎜1.28 ×10−3 ⎜⎝ L • atm ⎠⎟

b) stay the same

(

)

⎛1.28 ×10−3 mol O ⎞⎛ ⎟⎜ 32.00 g O2 ⎞⎟⎟ 2⎟ Mass (g) of O2 = ⎜⎜⎜ ⎟⎜⎜ ⎟(2.50 L ) = 0.1024 = 0.102 g O2 ⎜⎝ L ⎠⎟⎜⎝ 1 mol O2 ⎠⎟

b) The amount of gas that will dissolve in a given volume decreases proportionately with the partial pressure of the gas. ⎛ mol ⎞⎟ ⎟ 0.209 atm = 2.6752 × 10–4 mol/L Sgas = ⎜⎜1.28 ×10−3 ⎜⎝ L • atm ⎠⎟

(

)

⎛ 2.6752 ×10−4 mol O ⎞⎛ 32.00 g O2 ⎞⎟ 2⎟ ⎟⎟⎜⎜ ⎟⎟(2.50 L ) = 0.0214016 = 0.0214 g O2 Mass (g) of O2 = ⎜⎜⎜ ⎜ ⎜⎝ L ⎠⎟⎝⎜ 1 mol O2 ⎠⎟

13.47

⎛ ⎛ 0.93% ⎞⎟ mol ⎞⎟ ⎟⎟ 1.0 atm ⎜⎜ ⎟ = 1.395 × 10–5 = 1.4 × 10–5 mol/L Solubility = ⎜⎜1.5 ×10−3 L • atm ⎠ ⎝⎜ ⎝⎜ 100% ⎠⎟

13.48

The solution is saturated.

13.49

Plan: Solubility for a gas is calculated from Henry’s law: Sgas = kH × Pgas. We know kH and Pgas, so Sgas can be calculated with units of mol/L.

(

)

13-14

Solution: Sgas = kH × Pgas = (3.7 × 10–2 mol/L∙atm)(5.5 atm) = 0.2035 = 0.20 mol/L 13.50

Solubility of gases increases with increasing partial pressure of the gas, and the goal of these devices is to increase the amount of oxygen dissolving in the bloodstream.

13.51

Molarity is defined as the number of moles of solute dissolved in one liter of solution. Molality is defined as the number of moles of solute dissolved in 1000 g (1 kg) of solvent. Molal solutions are prepared by measuring masses of solute and solvent, which are additive and not changed by temperature, so the concentration in molality does not change with temperature and is the preferred unit when the temperature of the solution may change.

13.52

Plan: Refer to the table of concentration definitions for the different methods of expressing concentration. Solution: a) Molarity and parts-by-volume (% w/v or % v/v) include the volume of the solution. b) Parts-by-mass (% w/w) include the mass of solution directly. (Others may involve the mass indirectly.) c) Molality includes the mass of the solvent.

13.53

No, 21 g solute/kg of solvent would be 21 g solute/1.021 kg solution.

13.54

Converting between molarity and molality involves conversion between volume of solution and mass of solution. Both of these quantities are given so interconversion is possible. To convert to mole fraction requires that the mass of solvent be converted to moles of solvent. Since the identity of the solvent is not given, conversion to mole fraction is not possible if the molar mass is not known.

13.55

% w/w, mole fraction, and molality are weight-to-weight relationships that are not affected by changes in temperature. % w/v and molarity are affected by changes in temperature, because the volume is temperature dependent.

13.56

Plan: The molarity is the number of moles of solute in each liter of solution: M =

mol of solute . Convert the V(L) of solution

masses to moles and the volumes to liters and divide moles by volume. Solution: ⎛ 32.3 g C12 H 22 O11 ⎞⎛ ⎟⎟⎜⎜ 1 mL ⎞⎛ ⎟⎜ 1 mol C12 H 22 O11 ⎞⎟⎟ ⎜ a) Molarity = ⎜⎜ ⎟⎟⎜ −3 ⎟⎟⎟⎜⎜ ⎟ = 0.943617 = 0.944 M C12H22O11 100. mL ⎟⎜⎜10 L ⎠⎝ ⎟⎜⎜ 342.30 g C12 H 22 O11 ⎠⎟⎟ ⎜⎜⎝ ⎠⎝ ⎛ 5.80 g LiNO3 ⎞⎟⎛ 1 mL ⎞⎛ 1 mol LiNO3 ⎞⎟ ⎜ ⎟⎟⎜⎜ ⎟⎟⎜⎜ ⎟⎟ = 0.166572 = 0.167 M LiNO b) Molarity = ⎜⎜ 3 ⎟⎜ −3 ⎜⎜⎝ 505 mL ⎠⎟⎟⎜⎝⎜10 L ⎠⎟⎝⎜⎜ 68.95 g LiNO3 ⎠⎟⎟

13.57

⎛ 0.82 g C 2 H 5 OH ⎞⎛ ⎟⎟⎜⎜ 1 mL ⎞⎛ ⎟⎜ 1 mol C 2 H 5 OH ⎞⎟⎟ ⎜ a) Molarity = ⎜⎜ ⎟⎟⎜ −3 ⎟⎟⎟⎜⎜ ⎟ = 1.69514 = 1.7 M C2H5OH ⎜⎝⎜ 10.5 mL ⎟⎜⎜10 L ⎠⎝ ⎟⎜⎜ 46.07 g C 2 H 5 OH ⎠⎟⎟ ⎠⎝ ⎛ ⎞ ⎛ ⎞ ⎜1.27 g NH 3 ⎟⎟⎜⎛ 1 mL ⎞⎟⎟⎜⎜ 1 mol NH 3 ⎟⎟ b) Molarity = ⎜⎜ ⎟⎟⎜⎜ −3 ⎟⎜ ⎟ = 2.2261 = 2.23 M NH3 ⎜⎝⎜ 33.5 mL ⎠⎟⎜⎝10 L ⎠⎟⎝⎜⎜17.03 g NH 3 ⎠⎟⎟

13.58

Plan: Dilution calculations can be done using MconcVconc = MdilVdil. Solution: a) Mconc = 0.240 M NaOH Vconc = 78.0 mL Mdil = ? Vdil = 0.250 L −3 ⎛ ⎞ ( M conc )(Vconc ) (0.240 M )(78.0 mL) ⎜⎜10 L ⎟⎟ Mdil = = ⎟ = 0.07488 = 0.0749 M ⎜ (Vdil ) (0.250 L) ⎜⎜⎝ 1 mL ⎠⎟⎟

13-15

b) Mconc = 1.2 M HNO3 Vconc = 38.5 mL Mdil = ? Vdil = 0.130 L ⎛ ⎞ − 3 (1.2 M )(38.5 mL) ⎜10 L ⎟ ( M conc )(Vconc ) ⎟⎟ = 0.355385 = 0.36 M Mdil = = ⎜⎜ (Vdil ) ⎜⎝⎜ 1 mL ⎠⎟⎟ (0.130 L) 13.59

Dilution calculations can be done using MconcVconc = MdilVdil Vconc = 25.5 mL Mdil = ? Vdil = 0.500 L a) Mconc = 6.25 M HCl −3 ⎛ ⎞⎟ (6.25 M )(25.5 mL) 10 L ( M conc )(Vconc ) ⎜⎜ ⎟⎟ = 0.31875 = 0.319 M Mdil = = ⎜ (Vdil ) ⎜⎝⎜ 1 mL ⎠⎟⎟ (0.500 L) b) Mconc = 2.00 × 10–2 M KI Vconc = 8.25 mL Mdil = ? Vdil = 12.0 mL Mdil =

13.60

(

)(

)

2.00 × 10−2 M 8.25 mL ( M conc )(Vconc ) = = 0.01375 = 0.0138 M (Vdil ) 12.0 mL

(

)

Plan: For part (a), find the number of moles of KH2PO4 needed to make 365 mL of a solution of this molarity. Convert moles to mass using the molar mass of KH2PO4. For part (b), use the relationship MconcVconc = MdilVdil to find the volume of 1.25 M NaOH needed. Solution: −2 ⎛ −3 ⎞ ⎜10 L ⎟⎟⎜⎜⎛ 8.55 ×10 mol KH 2 PO 4 ⎞⎟⎟ a) Moles of KH2PO4 = (365 mL) ⎜⎜ ⎟⎟⎜ ⎟⎟ = 0.0312075 mol L ⎜⎝⎜ 1 mL ⎠⎟⎜⎝ ⎠⎟ ⎛136.09 g KH 2 PO 4 ⎞⎟ ⎜ ⎟⎟ = 4.24703 = 4.25 g KH PO Mass (g) of KH2PO4 = (0.0312075 mol KH 2 PO 4 )⎜⎜ 2 4 ⎜⎜⎝ 1 mol KH 2 PO 4 ⎠⎟⎟ Add 4.25 g KH2PO4 to enough water to make 365 mL of aqueous solution. Mdil = 0.335 M NaOH Vdil = 465 mL b) Mconc = 1.25 M NaOH Vconc = ? ( Mdil )(Vdil ) (0.335 M )(465 mL) Vconc = = = 124.62 = 125 mL (Vconc ) (1.25 M ) Add 125 mL of 1.25 M NaOH to enough water to make 465 mL of solution.

13.61

a) Find the number of moles NaCl needed to make 2.5 L of this solution. Convert moles to mass using the molar mass of NaCl. ⎛ 0.65 mol NaCl ⎞⎟⎛⎜ 58.44 g NaCl ⎞⎟ ⎟ Mass (g) of NaCl = (2.5 L) ⎜⎜⎜ ⎟⎟⎟⎜⎜ ⎟⎟ = 94.965 = 95 g NaCl ⎜⎝ L ⎠⎟⎜ ⎝⎜ 1 mol NaCl ⎠⎟ Add 95 g NaCl to enough water to make 2.5 L of aqueous solution. b) Use the relationship MconcVconc = MdilVdil to find the volume of 2.1 M urea needed. Mconc = 2.1 M urea Vconc = ? Mdil = 0.3 M urea Vdil = 15.5 L (0.3 M )(15.5 L) ( M dil )(Vdil ) Vconc = = = 2.21429 = 2 L (2.1 M ) (Vconc ) Add 2 L of 2.1 M urea to enough water to make 15.5 L of solution. Note because of the uncertainty in the concentration of the dilute urea (0.3 M), only one significant figure is justified in the answer.

13.62

Plan: To find the mass of KBr needed in part (a), find the moles of KBr in 1.40 L of a 0.288 M solution and convert to grams using the molar mass of KBr. To find the volume of the concentrated solution that will be diluted to 255 mL in part (b), use MconcVconc = MdilVdil and solve for Vconc. Solution:

13-16

⎛ 0.288 mol KBr ⎞⎟ ⎟⎟ = 0.4032 mol a) Moles of KBr = 1.40 L ⎜⎜⎜ ⎜⎝ L ⎠⎟

(

)

⎛119.00 g KBr ⎞⎟ ⎜ ⎟⎟ = 47.9808 = 48.0 g KBr Mass (g) of KBr = 0.4032 mol ⎜⎜ ⎜⎜⎝ 1 mol KBr ⎠⎟⎟ To make the solution, weigh 48.0 g KBr and then dilute to 1.40 L with distilled water. Vconc = ? Mdil = 0.0856 M LiNO3 Vdil = 255 mL b) Mconc = 0.264 M LiNO3

(

Vconc =

)

( Mdil )(Vdil ) (0.0856 M )(255 mL) = = 82.68182 = 82.7 mL (0.264 M ) (Vconc )

To make the 0.0856 M solution, measure 82.7 mL of the 0.264 M solution and add distilled water to make a total of 255 mL. 13.63

a) To find the mass of Cr(NO3)3 needed, find the moles of Cr(NO3)3 in 57.5 mL of a 1.53 × 10–3 M solution and convert to grams using molar mass of Cr(NO3)3. −3 ⎛ −3 ⎞⎛ ⎞⎛ ⎜⎜10 L ⎟⎟⎜⎜1.53 × 10 mol Cr(NO3 )3 ⎟⎟⎜⎜ 238.03 g Cr(NO3 )3 ⎞⎟⎟ Mass (g) of Cr(NO3)3 = 57.5 mL ⎜ ⎟⎟⎜ ⎟⎟⎜ ⎟⎟ ⎜⎜ 1 mL ⎟⎜⎜ ⎜ L ⎝ ⎠⎝ ⎠⎟⎝⎜ 1 mol Cr(NO3 )3 ⎠⎟ = 0.020941 = 0.0209 g Cr(NO3)3 To make the solution, weigh 0.0209 g Cr(NO3)3 and then dilute to 57.5 mL with distilled water. b) To find the volume of the concentrated solution that will be diluted to 5.8 × 103 m3 use MconcVconc = MdilVdil and solve for Vconc. Mconc = 2.50 M NH4NO3 Vconc = ? Mdil = 1.45 M NH4NO3 Vdil = 5.8 × 103 m3

(

)

( Mdil )(Vdil ) (1.45 M )(5.8 ×10 m ) = = 3.364 × 103 = 3.4 × 103 m3 (Vconc ) (2.50 M ) 3

Vconc =

To make the 1.45 M solution, measure 3.4 × 103 m3 of the 2.50 M solution and add distilled water to make 5.8 × 103 m3. 13.64

Plan: Molality, m, =

moles of solute . Convert the mass of solute to moles and divide by the mass of solvent in kg of solvent

units of kg. Solution: ⎛ 1 mol glycine ⎞⎟ ⎜ ⎟⎟ = 1.137605 mol a) Moles of glycine = 85.4 g glycine ⎜⎜ ⎜⎜⎝ 75.07 g glycine ⎠⎟⎟ 1.137605 mol glycine m glycine = = 0.895752 = 0.896 m glycine 1.270 kg ⎛ 1 mol glycerol ⎞⎟ ⎜ ⎟⎟ = 0.093278 mol b) Moles of glycerol = 8.59 g glycerol ⎜⎜ ⎜⎜⎝ 92.09 g glycerol ⎠⎟⎟ ⎛ 1 kg ⎞⎟ ⎜ Volume (kg) of solvent = 77.0 g ⎜⎜ 3 ⎟⎟⎟ = 0.0770 kg ⎜⎝⎜10 g ⎠⎟ 0.093278 mol glycerol m glycerol = = 1.2114 = 1.21 m glycerol 0.0770 kg

13-17

13.65

Molality = moles solute/kg solvent. ⎛ 1 mol HCl ⎞⎟ ⎜ ⎟⎟ 174 g HCl ⎜⎜ ⎜⎝⎜ 36.46 g HCl ⎠⎟⎟ ⎛103 g ⎞⎟ ⎜⎜ ⎟⎟ = 6.3043 = 6.30 m HCl a) m HCl = ⎜⎜ ⎜⎝ 1 kg ⎠⎟⎟ 757 g

(

)

⎛ 1 mol naphthalene ⎞⎟ ⎜ ⎟⎟ 16.5 g naphthalene ⎜⎜ ⎜⎝⎜128.16 g naphthalene ⎠⎟⎟ ⎛103 g ⎞⎟ ⎜⎜ ⎟⎟ = 2.41548 = 2.42 m naphthalene b) m naphthalene = ⎜⎜ ⎜⎝ 1 kg ⎠⎟⎟ 53.3 g

(

13.66

moles of solute . Use the density of benzene to find the mass and then the moles of kg of solvent benzene; use the density of hexane to find the mass of hexane and convert to units of kg. Divide the moles of benzene by the mass of hexane. Solution: ⎛ 0.877 g ⎞⎟ ⎟⎟ = 38.588 g benzene Mass (g) of benzene = 44.0 mL C6 H6 ⎜⎜⎜ ⎝⎜ 1 mL ⎠⎟ ⎛ 1 mol C6 H 6 ⎞⎟ ⎜ ⎟⎟ = 0.49402 mol benzene Moles of benzene = (38.588 g C6 H 6 )⎜⎜ ⎜⎜⎝ 78.11 g C6 H 6 ⎠⎟⎟ ⎛ 0.660 g ⎞⎟⎜⎛ 1kg ⎞⎟ ⎟⎟⎜⎜ 3 ⎟⎟⎟ = 0.11022 kg hexane Mass (kg) of hexane = (167 mL C 6 H14 )⎜⎜⎜ ⎟⎜⎝10 g ⎠⎟ ⎝⎜ mL ⎠⎜

Plan: Molality, m, =

(

13.67

)

(0.49402 mol C6 H6 ) = 4.48213 = 4.48 m C H moles of solute = 6 6 kg of solvent (0.11022 kg C6 H14 )

Molality = moles solute/kg solvent. ⎛1.59 g ⎞⎛ 1 mol CCl 4 ⎞

(2.66 mL CCl )⎜⎜⎜⎜⎝ mL ⎠⎟⎟⎟⎜⎟⎜⎜⎜⎜153.81 g CCl ⎟⎟⎟⎟⎟ ⎛10 g ⎞ 4

Molality of CCl4 =

⎛1.33 g ⎞

(76.5 mL CH Cl )⎜⎜⎜⎝ mL ⎠⎟⎟⎟⎟ 2

13.68

⎝

⎠ ⎜⎜ ⎟⎟⎟ = 0.2702596 = 0.270 m CCl 4 ⎜⎜ ⎜⎝ 1 kg ⎠⎟⎟

Plan: In part (a), the total mass of the solution is 3.10 × 102 g, so masssolute + masssolvent = 3.10 × 102 g. Assume that you have 1000 g of the solvent water and find the mass of C2H6O2 needed to make a 0.125 m solution. Then a ratio can be used to find the mass of C2H6O2 needed to make 3.10 × 102 g of a 0.125 m solution. Part (b) is a dilution problem. First, determine the amount of solute in your target solution and then determine the amount of the concentrated acid solution needed to get that amount of solute. Solution: ⎛ 0.125 mol C 2 H 6 O2 ⎞⎛ ⎟⎟⎜⎜ 62.07 g C 2 H 6 O2 ⎞⎟⎟ ⎜ a) Mass (g) of C2H6O2 in 1000 g (1 kg) of H2O = 1 kg H 2 O ⎜⎜ ⎟⎟⎜ ⎟ ⎜⎝⎜ 1 kg H 2 O ⎟⎜⎜ 1 mol C 2 H 6 O2 ⎠⎟⎟ ⎠⎝

(

)

= 7.75875 g C2H6O2 in 1000 g H2O Mass (g) of this solution = 1000 g H2O + 7.75875 g C2H6O2 = 1007.75875 g ⎛ 7.75875 g C2 H6 O2 ⎞⎟ ⎟ (3.10 ×102 g solution) Mass (g) of C2H6O2 for 3.10 × 102 g of solution = ⎜⎜ ⎜⎝1007.75875 g solution ⎠⎟⎟ = 2.386695 g C2H6O2 Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

13-18

Masssolvent = 3.10 × 102 g – masssolute = 3.10 × 102 g – 2.386695 g C2H6O2 = 307.613305 = 308 g H2O Therefore, add 2.39 g C2H6O2 to 308 g of H2O to make a 0.125 m solution.

⎛ 2.20% ⎞⎟ b) Mass (kg) of HNO3 in the 2.20% solution = (1.20 kg)⎜⎜ ⎟ = 0.0264 kg HNO3 (solute) ⎝ 100% ⎠⎟ Mass % =

mass of solute (100) mass of solution

mass of solute (100) 0.0264 kg (100) = mass % 52.0% = 0.050769 = 0.0508 kg Mass of water added = mass of 2.2% solution – mass of 52.0% solution =1.20 kg – 0.050769 kg = 1.149231 = 1.15 kg Add 0.0508 kg of the 52.0% (w/w) HNO3 to 1.15 kg H2O to make 1.20 kg of 2.20% (w/w) HNO3. Mass of 52.0% solution containing 0.0264 kg HNO3 =

13.69

a) The total weight of the solution is 1.50 kg, so masssolute + masssolvent = 1.50 kg

⎛ 0.0355 mol C 2 H 5 OH ⎞⎛ ⎟⎟⎜⎜ 46.07 g C 2 H 5 OH ⎞⎟⎟ ⎜ Mass (g) of C2H5OH in 1000 g (1 kg) of H2O = 1 kg H 2 O ⎜⎜ ⎟⎟⎜ ⎟ ⎜⎜⎝ 1 kg H 2 O ⎟⎜⎜ 1 mol C 2 H 5 OH ⎠⎟⎟ ⎠⎝

(

)

= 1.635485 g C2H5OH Mass (g) of this solution = 1000 g H2O + 1.635485 g C2H5OH = 1001.635485 g

⎛ 1.635485 g C 2 H5 OH ⎞⎟ ⎟(1500 g solution) Mass (g) of C2H5OH for 1.50 kg of solution = ⎜⎜ ⎜⎝ 1001.635485 g solution ⎠⎟⎟ = 2.449222 = 2.45 g C2H5OH Masssolvent = 1500 g – masssolute = 1500 g – 2.449222 g C2H5OH = 1497.551 = 1498 g H2O Therefore, add 2.45 g C2H5OH to 1498 g of H2O to make a 0.0355 m solution. b) This is a disguised dilution problem. First, determine the amount of solute in your target solution:

⎛13.0% ⎞⎟ Mass (kg) of HCl in the 13.0% solution = (445 g)⎜⎜ ⎟ = 57.85 g HCl (solute) ⎝ 100% ⎠⎟ Then determine the amount of the concentrated acid solution needed to get 57.85 g solute: mass of solute (100) Mass % = mass of solution mass of solute (100) 57.85 g (100) = = 169.6481 = 170 g Mass of 34.1% solution containing 57.85 g HCl = mass % 34.1% Mass of water added = mass of 13.0% solution – mass of 34.1% solution = 445 g – 169.6481 g = 275.35191 = 275 g Add 170. g of the 34.1% (w/w) HCl to 275 g H2O 13.70

Plan: You know the moles of solute (C3H7OH) and the moles of solvent (H2O). Divide moles of C3H7OH by the total moles of C3H7OH and H2O to obtain mole fraction. To calculate mass percent, convert moles of solute and solvent to mass and divide the mass of solute by the total mass of solution (solute + solvent). For molality, divide the moles of C3H7OH by the mass of water expressed in units of kg. Solution: a) Mole fraction is moles of isopropanol per total moles. moles of isopropanol 0.35 mol isopropanol Xisopropanol = = = 0.2916667 = 0.29 moles of isopropanol + moles of water (0.35 + 0.85) mol (Notice that mole fractions have no units.)

13-19

mass of solute (100). From the mole amounts, find the masses mass of solution of isopropanol and water: ⎛ 60.09 g C3 H 7 OH ⎞⎟ ⎟⎟ = 21.0315 g isopropanol Mass (g) of isopropanol = (0.35 mol C3 H 7 OH)⎜⎜⎜ ⎝⎜ 1 mol C3 H 7 OH ⎠⎟ b) Mass percent =

⎛18.02 g H2 O ⎞⎟ ⎟ = 15.317 g water Mass (g) of water = (0.85 mol H2 O)⎜⎜ ⎜⎝ 1 mol H2 O ⎠⎟⎟ Mass percent =

13.71

21.0315 g isopropanol mass of solute (100) = (100) = 57.860710 = 58% mass of solution (21.0315 + 15.317) g

c) Molality of isopropanol is moles of isopropanol per kg of water. 0.35 mol isopropanol ⎛⎜103 g ⎞⎟ moles of solute ⎟⎟ = 22.85043 = 23 m isopropanol ⎜⎜ m= = kg of solvent 15.317 g water ⎜⎝⎜ 1 kg ⎠⎟⎟ a) Mole fraction is moles of NaCl per total moles. 0.100 mol NaCl XNaCl = = 0.01149425 = 0.0115 (Notice that mole fractions have no units.) (0.100 + 8.60) mol b) Mass percent is the mass of NaCl per 100 g of solution. Mass (g) of NaCl = (0.100 mol NaCl)(58.44 g/mol) = 5.844 g NaCl Mass (g) of water = (8.60 mol water)(18.02 g/mol) = 154.972 g water Mass percent NaCl =

13.72

(5.844 g NaCl) ×100%

(5.844 + 154.972) g

= 3.63396677 = 3.63% NaCl

c) Molality of NaCl is moles of NaCl per kg of solvent. 0.100 mol NaCl ⎛⎜103 g ⎞⎟ ⎟⎟ = 0.645277856 = 0.645 m NaCl ⎜⎜ Molality NaCl = 154.972 g water ⎜⎜⎝ 1 kg ⎠⎟⎟ moles of solute Plan: Molality = . Use the density of water to convert the volume of water to mass. Multiply the kg of solvent mass of water in kg by the molality to find moles of cesium bromide; convert moles to mass. To find the mole fraction, convert the masses of water and cesium bromide to moles and divide moles of cesium bromide by the total moles of cesium bromide and water. To calculate mass percent, divide the mass of cesium bromide by the total mass of solution and multiply by 100. Solution: The density of water is 1.00 g/mL. The mass of water is: ⎛ 1 mL ⎞⎛1.00 g ⎞⎟⎛⎜ 1 kg ⎞⎟ ⎟ = 0.500 kg Mass (g) of water = (0.500 L)⎜⎜ −3 ⎟⎟⎟⎜⎜ ⎜ ⎝10 L ⎠⎝ 1 mL ⎠⎟⎟⎜⎝103 g ⎠⎟⎟ moles of solute m= kg of solvent moles CsBr 0.400 m CsBr = 0.500 kg H 2 O Moles of CsBr = (0.400 m)(0.500 kg H2O) = 0.200 mol ⎛ 212.8 g CsBr ⎞⎟ Mass (g) of CsBr = (0.200 mol CsBr)⎜⎜ = 42.56 = 42.6 g CsBr ⎝ 1 mol CsBr ⎠⎟⎟ ⎛103 g ⎞⎟ ⎟ = 500 g water Mass (g) of water = (0.500 kg)⎜⎜ ⎜⎝ 1 kg ⎠⎟⎟

13-20

Mass of solution = mass (g) H2O + mass of CsBr = 500 g H2O + 42.56 g CsBr = 542.56 g ⎛ 1 mol H O ⎞⎟ 2 ⎟ = 27.74695 mol H2O Moles of H2O = (500 g H 2 O)⎜⎜⎜ ⎜⎝18.02 g H 2 O ⎠⎟⎟

mol CsBr

0.2000 mol CsBr

= 7.156 × 10–3 = 7.16 × 10–3 mol CsBr + mol H2 O (0.2000 + 27.74695) mol (42.56 g CsBr) mass of CsBr (100) = Mass percent CsBr = × 100% = 7.84429 = 7.84% CsBr mass of solution (42.56 + 500.)g XCsBr =

13.73

The density of water is 1.00 g/mL. The mass of water is: ⎛ 1 mL ⎞⎛1.00 g ⎞⎟ Mass (g) of water = (0.400 L)⎜⎜ −3 ⎟⎟⎟⎜⎜ = 4.00 102 g H2O ⎝10 L ⎠⎝ 1 mL ⎠⎟⎟ ⎛ 1 mol H O ⎞⎟ ⎜ 2 ⎟⎟ = 22.197558 mol H O Moles of H2O = (400. g H 2 O) ⎜⎜ 2 ⎜⎝18.02 g H 2 O ⎠⎟

⎛ 1 mol KI ⎞⎟ ⎟ = 1.80723 × 10–3 mol KI Moles of KI = (0.30 g KI)⎜⎜ ⎜⎝166.0 g KI ⎠⎟⎟ XKI =

1.80723 ×10−3 mol KI

= 8.14091 × 10–5 = 8.1 × 10–5

−3

(1.80723 ×10 + 22.197558)mol (0.30 g KI) × 100% = 0.07494 = 0.075% KI Mass percent KI = (0.30 + 400.)g 13.74

Plan: You are given the mass percent of the solution. Assuming 100. g of solution allows us to express the mass % as the mass of solute, NH3. To find the mass of solvent, subtract the mass of NH3 from the mass of solution and convert to units of kg. To find molality, convert mass of NH3 to moles and divide by the mass of solvent in kg. To find molarity, you will need the volume of solution. Use the density of the solution to convert the 100. g of solution to volume in liters; divide moles of NH3 by volume of solution. To find the mole fraction, convert mass of solvent to moles and divide moles of NH3 by the total moles. Solution: Determine some fundamental quantities: ⎛ 8.00% NH3 ⎞⎟ Mass (g) of NH3 = (100 g solution)⎜⎜⎜ ⎟ = 8.00 g NH3 ⎝100% solution ⎠⎟ Mass (g) H2O = mass of solution – mass NH3 = (100.00 – 8.00) g = 92.00 g H2O ⎛ 1 kg ⎞ Mass (kg) of H2O = (92.00 g H 2 O)⎜⎜ 3 ⎟⎟⎟ = 0.09200 kg H2O ⎜⎝10 g ⎠⎟

⎛ 1 mol NH3 ⎞⎟ ⎟ = 0.469759 mol NH3 Moles of NH3 = (8.00 g NH3 )⎜⎜⎜ ⎜⎝17.03 g NH3 ⎠⎟⎟ ⎛ 1 mol H2 O ⎞⎟ ⎟⎟ = 5.1054 mol H2O Moles of H2O = (92.00 g H2 O)⎜⎜⎜ ⎝18.02 g H2 O ⎠⎟ −3 ⎛ 1 mL solution ⎞⎛ ⎟⎟⎜10 L ⎞⎟⎟ = 0.103616 L Volume (L) of solution = (100.00 g solution)⎜⎜⎜ ⎜ ⎟ ⎝ 0.9651 g solution ⎠⎟⎝⎜ 1 mL ⎠⎟⎟

Using the above fundamental quantities and the definitions of the various units: ⎛ 0.469759 mol NH3 ⎞⎟ moles of solute ⎜ ⎟⎟ = 5.106076 = 5.11 m NH = ⎜⎜ Molality = m = 3 ⎜⎝⎜ 0.09200 kg H 2 O ⎠⎟⎟ kg of solvent

13-21

Molarity = M =

⎛ 0.469759 mol NH3 ⎞⎟ moles of solute ⎜ ⎟⎟ = 4.53365 = 4.53 M NH = ⎜⎜ 3 ⎟ ⎜⎜⎝ L of solution 0.103616 L ⎠⎟

Mole fraction = X = 13.75

0.469759 mol NH3 moles of NH3 = = 0.084259 = 0.0843 total moles (0.469759 + 5.1054 ) mol

The information given is 28.8 mass % FeCl3 solution with a density of 1.280 g/mL. For convenience, choose exactly 100.00 g of solution. Determine some fundamental quantities: Mass (g) of FeCl3 = (100.00 g solution)(28.8% FeCl3/100%) = 28.8 g FeCl3 Mass (g) of H2O = mass of solution – mass FeCl3 = (100.00 – 28.8) g = 71.20 g H2O Moles of FeCl3 = (28.80 g FeCl3)(1 mol FeCl3/162.20 g FeCl3) = 0.1775586 mol FeCl3 Moles of H2O = (71.20 g H2O)(1 mol H2O/18.02 g H2O) = 3.951165 mol H2O Volume of solution = (100.00 g solution)(1 mL/1.280 g)(10–3 L/1 mL) = 0.078125 L Using the above fundamental quantities and the definitions of the various units: 3 ⎛ 0.1775586 mol FeCl 3 ⎞⎛ ⎟⎟⎜⎜10 g ⎞⎟⎟ ⎜⎜ Molality = M = moles solute/kg solvent = ⎜ ⎟⎟⎜ ⎟ = 2.49380 = 2.49 m FeCl3 ⎜⎜⎝ 71.20 g H 2 O ⎟⎜⎜ 1 kg ⎠⎟⎟ ⎠⎝ 0.1775586 mol FeCl3 = 2.272750 = 2.27 M FeCl3 Molarity = m = moles solute/L solution = 0.078125 L Mole fraction = X = moles substance/total moles =

13.76

0.1775586 mol FeCl3

(0.1775586 + 3.951165) mol

= 0.043005688 = 0.0430

Plan: Use the equation for parts per million, ppm. Use the given density of solution to find the mass of solution; divide the mass of each ion by the mass of solution and multiply by 1 × 106. Solution: ⎛ 1 mL ⎞⎛1.001 g ⎞⎟ Mass (g) of solution is (100.0 L solution)⎜⎜ −3 ⎟⎟⎟⎜⎜ = 1.001 × 105 g ⎝10 L ⎠⎝ 1 mL ⎠⎟⎟ ⎛ mass solute ⎞⎟ ⎟ × 106 ppm = ⎜⎜ ⎝⎜ mass solution ⎠⎟

⎛ ⎞⎟ 0.25 g Ca 2+ ⎟ × 106 = 2.49750 = 2.5 ppm Ca2+ ppm Ca2+ = ⎜⎜⎜ 5 ⎜⎝1.001×10 g solution ⎠⎟⎟ ⎛ ⎞⎟ 0.056 g Mg 2+ ⎟ × 106 = 0.5594406 = 0.56 ppm Mg2+ ppm Mg2+ = ⎜⎜⎜ ⎜⎝1.001 ×105 g solution ⎠⎟⎟ 13.77

The information given is that ethylene glycol has a density of 1.114 g/mL and a molar mass of 62.07 g/mol. Water has a density of 1.00 g/mL. The solution has a density of 1.070 g/mL. For convenience, choose exactly 1.0000 L as the equal volumes mixed. Ethylene glycol will be designated EG. Determine some fundamental quantities: ⎛10 3 mL ⎞⎛ ⎟⎟⎜⎜1.114 g EG ⎞⎟⎟ ⎜ Mass (g) of EG = (1.0000 L EG) ⎜⎜ ⎟⎟⎜ ⎟ = 1114 g EG ⎜⎝⎜ 1 L ⎠⎝ ⎟⎜⎜ 1 mL ⎠⎟⎟ ⎛10 3 mL ⎞⎛ ⎟⎟⎜⎜1.00 g H 2 O ⎞⎟⎟ ⎜ 3 Mass (g) of H2O = (1.0000 L H2O) ⎜⎜ ⎟⎟⎜ ⎟ = 1.00 × 10 g H2O ⎜⎜⎝ 1 L ⎠⎝ ⎟⎜⎜ 1 mL ⎠⎟⎟ ⎛ 1 mol EG ⎞⎟ ⎟ = 17.94747865 mol EG Moles of EG = (1114 g EG) ⎜⎜ ⎜⎝ 62.07 g EG ⎠⎟⎟

13-22

⎛ 1 mol H 2 O ⎞⎟ ⎟⎟ = 55.49389567 mol H2O Moles of H2O = (1.00 × 103 g H2O) ⎜⎜⎜ ⎝18.02 g H 2 O ⎠⎟ ⎛ 1 mL ⎞⎛ ⎟⎟⎜⎜ 1 L ⎞⎟⎟ ⎜ Volume (L) of solution = (1114 g EG + 1.00 × 103 g H2O) ⎜⎜ ⎟⎟⎜ ⎟ = 1.97570 L ⎟⎜⎜1000 mL ⎠⎟⎟ ⎜⎜⎝1.070 g ⎠⎝ Using the above fundamental quantities and the definitions of the various units: ⎛ 1.0000 L ⎞⎟ ⎜ a) Volume percent = ⎜⎜ ⎟⎟ × 100% = 50.61497 = 50.61% v/v ⎜⎜⎝1.97570 L ⎠⎟⎟ ⎛ ⎞⎟ 1114 g EG ⎜ ⎟⎟ × 100% = 52.6963 = 52.7% w/w b) Mass percent = ⎜⎜ ⎜⎜⎝1114 g EG + 1000. g H 2 O ⎠⎟⎟

c) Molarity = moles solute/L solution =

17.94747865 mol EC 1.97570 L

= 9.08411 = 9.08 M ethylene glycol

17.94747865 mol EG ⎛⎜103 g ⎞⎟ ⎟⎟ ⎜⎜ 1.00 ×103 g H 2 O ⎜⎜⎝ 1 kg ⎠⎟⎟ = 17.94747865 = 17.9 m ethylene glycol 17.94747865 mol EG e) Mole fraction = XEG = moles substance/total moles = (17.94747865 + 55.49389567) mol

d) Molality = moles solute/kg solvent =

= 0.244378 = 0.244 13.78

Colligative properties of a solution are affected by the number of particles of solute in solution. The density of a solution would be affected by the chemical formula of the solute.

13.79

A nonvolatile nonelectrolyte is a covalently bonded molecule that does not dissociate into ions or evaporate when dissolved in a solvent. In this case, the colligative concentration is equal to the molar concentration, simplifying calculations. The “strong” in “strong electrolyte” refers to the ability of an electrolyte solution to conduct a large current. This conductivity occurs because solutes that are strong electrolytes dissociate completely into ions when dissolved in water.

13.80

13.81

Raoult’s law states that the vapor pressure of solvent above the solution equals the mole fraction of the solvent times the vapor pressure of the pure solvent. Raoult’s law is not valid for a solution of a volatile solute in solution. Both solute and solvent would evaporate based upon their respective vapor pressures.

13.82

The boiling point temperature is higher and the freezing point temperature is lower for the solution compared to the solvent because the addition of a solute lowers the freezing point and raises the boiling point of a liquid.

13.83

Yes, the vapor at the top of the fractionating column is richer in content of the more volatile component.

13.84

The boiling point of a 0.01 m KF solution is higher than that of 0.01 m glucose. KF dissociates into ions in water (K+ and F–) while the glucose does not, so the KF produces more particles.

13.85

A dilute solution of an electrolyte behaves more ideally than a concentrated one. With increasing concentration, the effective concentration deviates from the molar concentration because of ionic attractions. Thus, the more dilute 0.050 m NaF solution has a boiling point closer to its predicted value.

13-23

13.86

Univalent ions behave more ideally than divalent ions. Ionic strength (which affects “activity” concentration) is greater for divalent ions. Thus, 0.01 m NaBr has a freezing point that is closer to its predicted value.

13.87

Cyclohexane, with a freezing point depression constant of 20.1°C/m, would make calculation of molar mass of a substance easier, since ΔTf would be greater.

13.88

Plan: Strong electrolytes are substances that produce a large number of ions when dissolved in water; strong acids and bases and soluble salts are strong electrolytes. Weak electrolytes produce few ions when dissolved in water; weak acids and bases are weak electrolytes. Nonelectrolytes produce no ions when dissolved in water. Molecular compounds other than acids and bases are nonelectrolytes. Solution: a) Strong electrolyte When hydrogen chloride is bubbled through water, it dissolves and dissociates completely into H+ (or H3O+) ions and Cl– ions. HCl is a strong acid. b) Strong electrolyte Potassium nitrate is a soluble salt and dissociates into K+ and NO3– ions in water. c) Nonelectrolyte Glucose solid dissolves in water to form individual C6H12O6 molecules, but these units are not ionic and therefore do not conduct electricity. Glucose is a molecular compound. d) Weak electrolyte Ammonia gas dissolves in water, but is a weak base that forms few NH4+ and OH– ions.

13.89

a) NaMnO4 b) CH3COOH c) CH3OH d) Ca(C2H3O2)2

13.90

Plan: To count solute particles in a solution of an ionic compound, count the number of ions per mole and multiply by the number of moles in solution. For a covalent compound, the number of particles equals the number of molecules. Solution; ⎛ 0.3 mol KBr ⎞⎟⎛ 2 mol particles ⎞⎟ a) ⎜⎜ ⎟(1 L) = 0.6 mol of particles ⎟⎟⎜⎜ ⎝ ⎠⎝ L 1 mol KBr ⎠⎟

strong electrolyte weak electrolyte nonelectrolyte strong electrolyte

Each KBr forms one K+ ion and one Br– ion, two particles for each KBr. ⎛ 0.065 mol HNO3 ⎞⎟⎜⎛ 2 mol particles ⎞⎟ ⎟(1 L) = 0.13 mol of particles b) ⎜⎜⎜ ⎟⎜ ⎝ ⎠⎟⎜⎜⎝ 1 mol HNO3 ⎠⎟⎟ L HNO3 is a strong acid that forms H+(H3O+) ions and NO3– ions in aqueous solution. ⎛10−4 mol KHSO4 ⎞⎟⎜⎛ 2 mol particles ⎞⎟ –4 ⎟⎜ c) ⎜⎜ ⎟⎟(1 L) = 2 × 10 mol of particles L ⎝⎜ ⎠⎟⎟⎝⎜ 1 mol KHSO 4 ⎠⎟ Each KHSO4 forms one K+ ion and one HSO4– ion in aqueous solution, two particles for each KHSO4. ⎛ 0.06 mol C2 H 5 OH ⎞⎟⎛⎜ 1 mol particles ⎞⎟ d) ⎜⎜ ⎟⎟⎜⎝⎜1 mol C H OH ⎠⎟⎟⎟(1 L) = 0.06 mol of particles ⎝ ⎠⎜ L 2 5 Ethanol is not an ionic compound so each molecule dissolves as one particle. The number of moles of particles is the same as the number of moles of molecules, 0.06 mol in 1 L. 13.91

⎛ 0.02 mol CuSO4 ⎞⎟⎛⎜ 2 mol particles ⎞⎟⎛ 1 L ⎞⎟ –5 a) ⎜⎜ ⎟⎟⎜⎜1000 mL ⎠⎟⎟(1 mL) = 4 × 10 mol of particles ⎟⎟⎝⎜⎜ 1 mol CuSO ⎠⎟⎝ ⎝ ⎠⎜ L 4 ⎛ 0.004 mol Ba(OH)2 ⎞⎟⎜⎛ 3 mol particles ⎞⎟⎛ 1 L ⎞⎟ –5 –5 ⎜ b) ⎜⎜ ⎟⎟⎜⎜⎝1 mol Ba(OH) ⎠⎟⎟⎟⎝⎜1000 mL ⎠⎟⎟(1 mL) = 1.2 × 10 = 1 × 10 mol of particles ⎝ ⎠⎜ L 2

⎛ 0.08 mol C5 H 5 N ⎞⎟⎛⎜1 mol particles ⎞⎟⎛ 1 L ⎞⎟ –5 ⎟⎟⎜⎜⎜ c) ⎜⎜ ⎟⎟⎜⎜ ⎟⎟(1 mL) = 8 × 10 mol of particles ⎟⎝ ⎠ ⎝ ⎠⎜ L 1 mol C H N 1000 mL ⎝ ⎠ 5 5 Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

13-24

⎛ 0.05 mol (NH 4 )2 CO3 ⎞⎟⎜⎛ 3 mol particles ⎞⎟⎛ 1 L ⎞⎟ ⎟⎜ (1 mL) = 1.5 × 10–4 = 2 × 10–4 mol of particles d) ⎜⎜⎜ ⎟⎜ ⎟⎝⎜1000 mL ⎠⎟⎟ ⎝ ⎠⎟⎜⎜⎝1 mol (NH 4 )2 CO3 ⎠⎟⎜ L 13.92

Plan: The magnitude of freezing point depression is directly proportional to molality. Calculate the molality of solution by dividing the moles of solute by the mass of solvent in kg. The solution with the larger molality will have the lower freezing point. Solution: (11.0 g CH3OH) ⎛⎜⎜ 1 mol CH3OH ⎞⎛⎟⎟⎜⎜103 g ⎞⎟⎟ = 3.4332085 = 3.43 m CH OH a) Molality of CH3OH = ⎟⎜ ⎟ ⎜ 3 (100. g H 2 O) ⎜⎝⎜ 32.04 g CH3OH ⎠⎝⎟⎟⎜⎜ 1 kg ⎠⎟⎟

(22.0 g CH CH OH) ⎛⎜⎜ 1 mol CH CH OH ⎞⎛⎟⎟⎜⎜10 g ⎞⎟⎟ Molality of CH CH OH = ⎟⎜ ⎟ ⎜ (200. g H O) ⎜⎝⎜ 46.07 g CH CH OH ⎠⎝⎟⎟⎜⎜ 1 kg ⎠⎟⎟ 3

= 2.387671 = 2.39 m CH3CH2OH The molality of methanol, CH3OH, in water is 3.43 m, whereas the molality of ethanol, CH3CH2OH, in water is 2.39 m. Thus, CH3OH/H2O solution has the lower freezing point. (20.0 g H 2 O) ⎛⎜⎜ 1 mol H 2 O ⎞⎟⎟ = 1.10988 = 1.11 m H O b) Molality of H2O = ⎟ 2 ⎜ (1.00 kg CH3OH) ⎜⎜⎝18.02 g H 2 O ⎠⎟⎟ Molality of CH3CH2OH =

(20.0 g CH CH OH) ⎛⎜⎜ 1 mol CH CH OH ⎞⎟⎟ = 0.434122 = 0.434 m CH CH OH ⎟ ⎜ (1.00 kg CH OH) ⎜⎜⎝ 46.07 g CH CH OH ⎠⎟⎟ 3

The molality of H2O in CH3OH is 1.11 m, whereas CH3CH2OH in CH3OH is 0.434 m. Therefore, H2O/CH3OH solution has the lower freezing point. 13.93

The magnitude of boiling point elevation is directly proportional to molality. (38.0 g C3 H8O3 ) ⎛⎜⎜ 1 mol C3 H8 O3 ⎞⎛⎟⎟⎜⎜103 g ⎞⎟⎟ = 1.650559 = 1.65 m C H O a) Molality of C3H8O3 = ⎟⎜ ⎟ 3 8 3 ⎜ (250. g ethanol) ⎜⎜⎝ 92.09 g C3 H8 O3 ⎠⎝⎟⎟⎜⎜ 1 kg ⎠⎟⎟

(38.0 g C H O ) ⎛⎜⎜ 1 mol C H O ⎞⎛⎟⎟⎜⎜10 g ⎞⎟⎟ = 2.44885 = 2.45 m C H O ⎟⎜ ⎟ ⎜ (250. g ethanol) ⎜⎜⎝ 62.07 g C H O ⎠⎝⎟⎟⎜⎜ 1 kg ⎠⎟⎟ 3

Molality of C2H6O2 =

The molality of C2H6O2 in ethanol is 2.45 m, whereas the molality of C3H8O3 in ethanol is 1.65 m. Thus, C2H6O2/ethanol solution has the higher boiling point. (15 g C2 H6 O2 ) ⎛⎜⎜ 1 mol C2 H6 O2 ⎞⎟⎟ = 0.4833253 = 0.48 m C H O b) Molality of C2H6O2 = ⎟ ⎜ 2 6 2 (0.50 kg H 2 O) ⎜⎜⎝ 62.07 g C2 H6 O2 ⎠⎟⎟ Molality of NaCl =

(15 g NaCl) ⎛⎜⎜ 1 mol NaCl ⎞⎟⎟ = 0.513347 = 0.51 m NaCl ⎟ ⎜ (0.50 kg H O) ⎜⎜⎝ 58.44 g NaCl ⎠⎟⎟ 2

Since the NaCl is a strong electrolyte, the molality of particles would be: (2 particles/NaCl)(0.513347 mol NaCl/kg) = 1.026694 = 1.0 m particles The molality of C2H6O2 in H2O is 0.48 m, whereas NaCl in H2O is 1.0 m. Therefore, NaCl/H2O solution has the higher boiling point.

13-25

13.94

Plan: To rank the solutions in order of increasing osmotic pressure, boiling point, freezing point, and vapor pressure, convert the molality of each solute to molality of particles in the solution. The higher the molality of particles, the higher the osmotic pressure, the higher the boiling point, the lower the freezing point, and the lower the vapor pressure at a given temperature. Solution: ⎛ 2 mol particles ⎞⎟ ⎟ = 0.200 m ions (I) (0.100 m NaNO3 )⎜⎜⎜ ⎜⎝ 1 mol NaNO3 ⎠⎟⎟ NaNO3 consists of Na+ ions and NO3– ions, two particles for each NaNO3. ⎛1 mol particles ⎞⎟ ⎟ = 0.100 m molecules (II) (0.100 m glucose)⎜⎜ ⎜⎝ 1 mol glucose ⎠⎟⎟ Glucose is not an ionic compound so each molecule dissolves as one particle. The number of moles of particles is the same as the number of moles of molecules. ⎛ 3 mol particles ⎞⎟ ⎟⎟ = 0.300 m ions (III) (0.100 m CaCl2 )⎜⎜⎜ ⎝ 1 mol CaCl2 ⎠⎟ CaCl2 consists of Ca+2 ions and Cl– ions, three particles for each CaCl2. ПII < ПI < ПIII a) Osmotic pressure: b) Boiling point: bpII < bpI < bpIII c) Freezing point: fpIII < fpI < fpII d) Vapor pressure at 50°C: vpIII < vpI < vpII

13.95

I 0.04 m (H2N)2CO × 1 mol particles/1 mol (H2N)2CO = 0.04 m molecules II 0.01 m AgNO3 × 2 mol particles/1 mol AgNO3 = 0.02 m ions III 0.03 m CuSO4 × 2 mol particles/1 mol CuSO4 = 0.06 m ions ПIII > ПI > ПII a) Osmotic pressure: b) Boiling point: bpIII > bpI > bpII c) Freezing point: fpII > fpI > fpIII d) Vapor pressure at 298 K: vpII > vpI > vpIII

13.96

Plan: The mole fraction of solvent affects the vapor pressure according to the equation Psolvent = XsolventP°solvent. Convert the masses of glycerol and water to moles and find the mole fraction of water by dividing moles of water by the total number of moles. Multiply the mole fraction of water by the vapor pressure of water to find the vapor pressure of the solution. Solution: ⎛ 1 mol C3 H8 O3 ⎞⎟ ⎟ = 0.369204 mol C3H8O3 Moles of C3H8O3 = (34.0 g C3 H8 O3 )⎜⎜⎜ ⎜⎝ 92.09 g C3 H8 O3 ⎠⎟⎟

⎛ 1 mol H2 O ⎞⎟ ⎟ = 27.7469 mol H2O Moles of H2O = (500.0 g H2 O)⎜⎜ ⎜⎝18.02 g H 2 O ⎠⎟⎟

mol H2 O 27.7469 mol H2 O = = 0.9868686 mol H 2 O + mol glycerol 27.7469 mol H2 O + 0.369204 mol glycerol Psolvent = XsolventP°solvent = (0.9868686)(23.76 torr) = 23.447998 = 23.4 torr Xsolvent =

13.97

The mole fraction of solvent affects the vapor pressure according to the equation Psolvent = XsolventP°solvent. Xsolvent = (5.4 mol toluene)/[(0.39) + (5.4)] mol = 0.93264 Psolvent = XsolventP°solvent = (0.93264)(41 torr) = 38.2382 = 38 torr

13.98

Plan: The change in freezing point is calculated from ΔTf = iKfm, where Kf is 1.86°C/m for aqueous solutions, i is the van’t Hoff factor, and m is the molality of particles in solution. Since urea is a covalent compound and does not ionize in water, i = 1. Once ΔTf is calculated, the freezing point is determined by subtracting it from the freezing point of pure water (0.00°C).

13-26

Solution: ΔTf = iKfm = (1)(1.86°C/m)(0.251 m) = 0.46686°C The freezing point is 0.00°C – 0.46686°C = –0.46686 = –0.467°C. 13.99

ΔTb = iKbm = (1)(0.512°C/m)(0.200 m) = 0.1024°C The boiling point is 100.00°C + 0.1024°C = 100.1024 = 100.10°C.

13.100 Plan: The boiling point of a solution is increased relative to the pure solvent by the relationship ΔTb = iKbm. Vanillin is a nonelectrolyte (it is a molecular compound) so i = 1. To find the molality, convert mass of vanillin to moles and divide by the mass of solvent expressed in units of kg. Kb is given (1.22°C/m). Solution: ⎛ 1 mol vanillin ⎞⎟ ⎜ ⎟⎟ = 0.0420665 mol Moles of vanillin = (6.4 g vanillin) ⎜⎜ ⎜⎝⎜152.14 g vanillin ⎠⎟⎟ 0.042065 mol vanillin ⎛⎜103 g ⎞⎟ moles of vanillin ⎟⎟ ⎜⎜ Molality of vanillin = = ⎜⎜⎝ 1 kg ⎠⎟⎟ kg of solvent (ethanol) 50.0 g ethanol = 0.8413 m vanillin ΔTb = iKbm = (1)(1.22°C/m)(0.8413 m) = 1.026386°C The boiling point is 78.5°C + 1.026386°C = 79.5264 = 79.5°C. ⎛ 1 mol C H ⎞⎟ ⎜ 10 8 ⎟ 13.101 Moles of C10H8 = (5.00 g C10 H8 ) ⎜⎜ ⎟ = 0.0390137 mol C10H8 ⎜⎜⎝128.16 g C10 H8 ⎠⎟⎟

C10H8 is a nonelectrolyte so i = 1. ⎛ 1 kg ⎞⎟ ⎜ ⎟⎟ = 0.444 kg benzene Mass = (444 g benzene) ⎜⎜ ⎜⎝⎜1000 g ⎠⎟⎟ ⎛ 0.0390137 mol C10 H 8 ⎞⎟ ⎜ ⎟⎟ = 0.08786869 m Molality = ⎜⎜ ⎜⎜⎝ 0.444 kg benzene ⎠⎟⎟ ΔTf = iKfm = (1)(4.90°C/m)(0.08786869 m) = 0.43056°C Freezing point = (5.5 – 0.43056)°C = 5.06944 = 5.1°C 13.102 Plan: The molality of the solution can be determined from the relationship ΔTf = iKf m with the value 1.86°C/m inserted for Kf and i = 1 for the nonelectrolyte ethylene glycol (ethylene glycol is a covalent compound that will form one particle per molecule when dissolved). Convert the freezing point of the solution to °C and find ΔTf by subtracting the freezing point of the solvent from the freezing point of the solution. Once the molality of the solution is known, the mass of ethylene glycol needed for a solution of that molality can be found. Solution: °C = (5/9)(°F – 32.0) = (5/9)(–12.0°F – 32.0) = –24.44444°C ΔTf = Tf(solution) – Tf(solvent) = (0.00 – (–24.44444))°C = 24.44444°C ΔTf = iKf m 24.44444ο C ΔTf m= = = 13.14217 m 1.86ο C/m Kf Ethylene glycol will be abbreviated as EG. moles of EG Molality of EG = kg of solvent (water) ⎛13.14217 mol EG ⎞⎟ ⎜ ⎟⎟ 14.5 kg H O = 190.561465 mol EG Moles of EG = molality × kg of solvent = ⎜⎜ 2 ⎟ 1 kg H 2 O ⎜⎜⎝ ⎠⎟

(

)

13-27

⎛ 62.07 g EG ⎞⎟ ⎜ ⎟⎟ Mass (g) of ethylene glycol = 190.561465 mol EG ⎜⎜ ⎜⎜⎝ 1 mol EG ⎠⎟⎟ = 1.18282 × 104 = 1.18 × 104 g ethylene glycol To prevent the solution from freezing, dissolve a minimum of 1.18 × 104 g ethylene glycol in 14.5 kg water.

(

)

13.103 The molality of the solution can be determined from the relationship ΔTf = iKfm with the value 1.86°C/m inserted for Kf, i = 1 for the nonelectrolyte glycerol, and the given ΔTf of –15°C. m = ΔTf/Kf = (15°C/(1.86°C/m) = 8.06452 m Glycerol will be abbreviated as GLY. ⎛ ⎞ ⎛10−3 g ⎞⎟⎛ 1 kg ⎞⎛ 92.09 g GLY ⎞⎟ ⎜ 8.06452 mol GLY ⎟⎟ ⎜ ⎟⎜ Mass (g) of glycerol = ⎜⎜ ⎟⎟ (11.0 mg H 2 O) ⎜⎜ ⎟⎟⎜⎜⎜ 3 ⎟⎟⎜⎜ ⎟⎟ ⎟ ⎟ ⎜⎜⎝ ⎜ 1 kg H 2 O ⎟⎠ ⎜⎝ 1 mg ⎠⎟⎝⎜10 g ⎠⎝⎜⎜ 1 mol GLY ⎠⎟⎟ = 0.0081693 = 0.0082 g glycerol To prevent the solution from freezing, dissolve a minimum of 0.0082 g glycerol in 11.0 mg water. 13.104 Plan: Calculate the molarity of the protein (divide the mass of the protein in grams by the molar mass of the protein to calculate the moles of protein; then divide the moles of protein by the volume of the solution in L). Convert temperature from °C to K. Use these values and the ideal gas constant to calculate osmotic pressure. Solution: = MRT

⎛ ⎞ ⎛ 1 g ⎞⎛ ⎟⎟⎜⎜ 1 mol protien ⎞⎟⎟⎟⎟ ⎜⎜⎜ (37.5 mg protein) ⎜⎜⎜ ⎟⎟⎟⎜⎜⎜1.50 ×10 4 g protien ⎟⎟⎟⎟⎟⎟⎛ ⎜⎜ ⎜⎝⎜103 mg ⎠⎝ L i atm ⎞⎟ ⎠⎟⎟⎜⎜ ⎟⎟ ((24.0 °C + 273.15)K) = ⎜⎜ 0.0821 ⎟ ⎜ ⎟⎜⎜ ⎟ ⎛ 1 L ⎞⎟ ⎟ mol K i ⎜⎜⎜ ⎟ ⎠ ⎟⎟⎝ ⎜ ⎟⎟ ⎜⎜ (25.0 mL) ⎜⎜ 3 ⎟ ⎟ ⎟ ⎝⎜⎜10 mL ⎠⎟ ⎜⎝⎜⎜ ⎠⎟⎟⎟ = 2.43960 × 10–3 = 2.44 × 10–3 atm 13.105 Plan: Convert temperature from °C to K. Use this value, the molarity of the sucrose, and the ideal gas constant to calculate osmotic pressure using the equation = iMRT (assume i = 1). Solution: Π = iMRT = (1)(0.30 mol/L) ( 0.0821 atm⋅L mol⋅K ) (37°C + 273.15 ) = 7.63899 = 7.6 atm 13.106

Plan: Calculate the change in the freezing point then use the relationship Tf = iKfm (assume i = 1) to calculate the molality. Use the molality and the mass of the water, converted to kilograms, to calculate the moles of the unknown compound. Divide the given mass of the compound by this number of moles to calculate the molar mass. Solution: ΔTf = 0.00°C – (–1.15°C) = 1.15°C ΔTf 1.15ο C = = 0.618280 m Molality = iK f (1)1.86ο C/m Amount (mol) of solute = ⎛ 1 kg ⎞⎛ ⎟⎜ 0.618280 mol unknown compound ⎞⎟⎟ ⎜ 150. g water ⎜⎜ 3 ⎟⎟⎟⎜⎜ ⎟⎟ = 0.0927419 mol unknown compound ⎜⎝⎜10 g ⎠⎝ ⎟⎜⎜ 1 kg water ⎠⎟

molar mass =

31.7 g = 341.8087 = 342 g/mol 0.0927419 mol

13-28

13.107

Plan: Use the osmotic pressure (in atm), the ideal gas constant, and the temperature (in K) to calculate the molarity of the solution. Use the molarity and volume of the solution to calculate the number of moles of solute in the solution. Determine the molar mass (g/mol) by dividing the mass of ovalbumin in the solution by the number of moles of ovalalbumin in the solution. Solution: ⎛ 1 atm ⎞ (4.32 torr)⎜⎜⎜⎜⎜ 760 torr ⎟⎟⎟⎟⎟ ⎝ ⎠ Π Π = MRT so M = = 2.32216 × 10–4 M = ⎛ ⎞ RT ⎜ L i atm ⎟ ⎟⎟ ((25 °C + 273.15)K) ⎜⎜0.0821 ⎜⎜⎝ mol i K ⎠⎟⎟ Amount (mol) of solute = M × V =

2.32216 ×10−4 mol 1 L soln

× 0.125 L soln = 2.90270 × 10−5 mol

Molar mass of protein: ℳ=

13.108

1.31 g 2.90270 ×10−5 mol

= 4.51304 × 104 = 4.51 × 104 g/mol

Plan: Calculate the molality, m, of the solution by dividing the moles of solute by the kg of solvent. The change in freezing point is calculated from ΔTf = iKfm, where Kf is 1.86°C/m for aqueous solutions, i is the van’t Hoff factor, and m is the molality of particles in solution. Ammonium phosphate, (NH4)3PO4, is an ionic compound that dissociates in water to give 4 ions (3 ammonium ions and one phosphate ion), so i = 4. Once ΔTf is calculated, the freezing point is determined by subtracting it from the freezing point of the solvent, pure water (0.00°C). Solution: ⎛ 1 mol (NH ) PO ⎞⎟ 4 3 4 ⎟ ⎜ 13.2 g ((NH 4 )3 PO 4 ) ⎜⎜ ⎟ ⎜⎜⎝149.10 g (NH 4 )3 PO 4 ⎠⎟⎟ m= = 1.96736 m ⎛ 1 kg ⎞⎟ ⎜⎜ (45.0 g) ⎜ 3 ⎟⎟⎟ ⎜⎜10 g ⎟ ⎝ ⎠ ΔTf = iKfm = (4)(1.86°C/m)(1.96736 m) = 14.637°C The freezing point is 0.00°C – 14.637°C = –14.637 = –14.6°C.

13.109

Plan: Calculate the molality, m, of the solution by dividing the moles of solute by the kg of solvent. The change in boiling point is calculated from ΔTb = iKbm, where Kb is 0.512 °C/m for aqueous solutions, i is the van’t Hoff factor, and m is the molality of particles in solution. Calcium nitrate, Ca(NO3)2, is an ionic compound that dissociates in water to give 3 ions (1 calcium ion and 2 nitrate ions), so i = 3. Once ΔTb is calculated, the boiling point is determined by adding it to the boiling point of the solvent, pure water (100.00°C). Solution: ⎛ 1 mol (Ca(NO ) ⎞⎟ 3 2 ⎜ ⎟⎟ 32.8 g (Ca(NO3 )2 ) ⎜⎜ ⎜⎜⎝164.10 g (Ca(NO3 )2 ⎠⎟⎟ m= = 1.850723 m ⎛ 1 kg ⎞⎟ ⎜⎜ (108 g) ⎜ 3 ⎟⎟⎟ ⎜⎜10 g ⎟ ⎝ ⎠ ΔTb = iKbm = (3)(0.512°C/m)(1.850723 m) = 2.84271°C The boiling point is 100.00°C + 2.84271°C = 102.84271 = 102.84°C.

13.110 Plan: Assume 100 g of solution so that the mass of solute = mass percent. Convert the mass of solute to moles. Subtract the mass of the solute from 100 g to obtain the mass of solution. Divide moles of solute by the mass of solvent in kg to obtain molality. Use ΔT = iKfm to find the van’t Hoff factor. Kf for water = 1.86°C/m. Solution: a) Assume exactly 100 g of solution. Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

13-29

⎛ 1.00% NaCl ⎞⎟ Mass (g) of NaCl = (100.00 g solution)⎜⎜ ⎟ = 1.00 g NaCl ⎝100% solution ⎠⎟ ⎛ 1 mol NaCl ⎞⎟ ⎜ Moles of NaCl = (1.00 g NaCl)⎜⎜ ⎟⎟ = 0.0171116 mol NaCl ⎜⎜⎝ 58.44 g NaCl ⎠⎟⎟ Mass of H2O = 100.00 g solution – 1.00 g NaCl = 99.00 g H2O moles of NaCl 0.0171116 mol NaCl ⎛⎜103 g ⎞⎟ ⎟ = 0.172844 = 0.173 m NaCl Molality of NaCl = = ⎜ ⎜⎝ 1 kg ⎠⎟⎟ kg of H2 O 99.00 g H 2 O ΔTf = Tf(solution) – Tf(solvent) = 0.000°C – (–0.593)°C = 0.593°C ΔTf = iKfm ΔTf 0.593ο C i= = = 1.844537 = 1.84 ο Kf m (1.86 C/m)(0.172844 m ) The value of i should be close to two because NaCl dissociates into two particles when dissolving in water. b) For acetic acid, CH3COOH: Assume exactly 100 g of solution. ⎛ 0.500% CH3COOH ⎞⎟ Mass (g) of CH3COOH = (100.00 g solution)⎜⎜⎜ ⎟ = 0.500 g CH3COOH ⎝ 100% solution ⎠⎟ ⎛ 1 mol CH 3 COOH ⎞⎟ ⎜ ⎟⎟ = 0.0083264 mol CH COOH Moles of CH3COOH = (0.500 g CH 3 COOH )⎜⎜ 3 ⎜⎜⎝ 60.05 g CH 3 COOH ⎠⎟⎟ Mass (g) of H2O = 100.00 g solution – 0.500 g CH3COOH = 99.500 g H2O moles of CH3COOH 0.0083264 mol CH 3 COOH ⎛⎜103 g ⎞⎟ ⎟⎟ Molality of CH3COOH = = ⎜⎜ kg of H2 O 99.500 g H 2 O ⎝ 1 kg ⎠⎟

= 0.083682 = 0.0837 m CH3COOH ΔTf = Tf(solution) – Tf(solvent) = 0.000°C – (–0.159)°C = 0.159°C ΔTf = iKfm ΔTf 0.159ο C i= = = 1.02153 = 1.02 ο Kf m (1.86 C/m)(0.083682 m) Acetic acid is a weak acid and dissociates to a small extent in solution, hence a van’t Hoff factor that is close to 1. 13.111 Plan: Convert the mass % to molality and use ΔT = iKfm to find the van’t Hoff factor. Solution: a) Assume exactly 100 g of solution. Thus, the solution contains 0.500 g of KCl in 99.500 g of water. 3 ⎛ 0.500 g KCl ⎞⎛ ⎟⎟⎜⎜10 g ⎞⎛ ⎟⎟⎜⎜ 1 mol KCl ⎞⎟⎟ ⎜ Molality of KCl = ⎜⎜ ⎟⎟⎜ ⎟⎟⎜ ⎟ = 0.067406 m KCl ⎟⎜⎜ 1 kg ⎠⎝ ⎟⎜⎜ 74.55 g KCl ⎠⎟⎟ ⎜⎜⎝ 99.500 g H 2 O ⎠⎝ Calculate ΔTf = 0.000°C – (–0.234)°C = 0.234°C ΔTf = iKfm i = ΔTf/Kfm = (0.234°C)/[(1.86°C/m)(0.067406 m)] = 1.866398 = 1.87 The value of i should be close to two because KCl dissociates into two particles when dissolving in water. b) For sulfuric acid, H2SO4: Assume exactly 100 g of solution. Thus, the solution contains 1.00 g of H2SO4 in 99.00 g of water. 3 ⎛1.00 g H 2 SO 4 ⎞⎛ ⎟⎟⎜⎜10 g ⎞⎛ ⎟⎟⎜⎜ 1 mol H 2 SO 4 ⎞⎟⎟ ⎜ Molality of H2SO4 = ⎜⎜ ⎟⎟⎜ ⎟⎟⎜ ⎟ = 0.10298746 m H2SO4 ⎟⎜⎜ 1 kg ⎠⎝ ⎟⎜⎜ 98.08 g H 2 SO 4 ⎠⎟⎟ ⎜⎜⎝ 99.00 g H 2 O ⎠⎝ Calculate ΔT = 0.000°C – (–0.423)°C = 0.423°C ΔTf = iKfm i = ΔTf/Kfm = (0.423°C)/[(1.86°C/m)(0.10298746 m)] = 2.20842 = 2.21 Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

13-30

Sulfuric acid is a strong acid and dissociates to give a hydrogen ion and a hydrogen sulfate ion. The hydrogen sulfate ion may further dissociate to another hydrogen ion and a sulfate ion. If ionization in both steps were complete, the value of the van’t Hoff factor would be 3. 13.112 Plan: Use the osmotic pressure equation (Π = iMRT) to find the molarity of the solution (assuming i = 1). Solution: 0.272 atm Π M= = = 0.01111756 M L • atm ⎞⎟ ⎛ iRT (1)⎜⎜0.0821 (273 + 25) K ) ( ⎟ ⎝ mol • K ⎠ ⎛ 0.01111756 mol ⎞⎛ ⎟⎟⎜⎜ 1 L ⎞⎟⎟ ⎜ Moles = ⎜⎜ ⎟⎟⎜ ⎟(100 mL) = 0.001111756 mol 1L ⎟⎜⎜1000 mL ⎠⎟⎟ ⎜⎜⎝ ⎠⎝ 6.053 g Molar mass = = 5.4445 × 103 = 5.44 × 103 g/mol 0.01111756 mol

13.113 Plan: The mole fraction of solvent affects the vapor pressure according to the equation Psolvent = XsolventP°solvent. Find the mole fraction of each substance by dividing moles of substance by the total number of moles. Multiply the mole fraction of each compound by its vapor pressure to find the vapor pressure of the compounds above the solution. Solution: moles CH2 Cl2 1.60 mol = = 0.592593 X CH 2 Cl2 = 1.60 + 1.10 mol moles CH2 Cl2 + mol CCl 4

1.10 mol moles CCl 4 = = 0.407407 moles CH2 Cl2 + mol CCl 4 1.60 + 1.10 mol PA = XA P°A = (0.592593)(352 torr) = 208.593 = 209 torr CH2Cl2 = (0.407407)(118 torr) = 48.0740 = 48.1 torr CCl4 X CCl4 =

13.114 The fluid inside a bacterial cell is both a solution and a colloid. It is a solution of ions and small molecules, and a colloid of large molecules, proteins, and nucleic acids. 13.115 a) milk - liquid/liquid colloid b) fog - liquid/gas colloid c) shaving cream - gas/liquid colloid 13.116 Brownian motion is a characteristic movement in which the colloidal particles change speed and direction erratically by the motion of the dispersing molecules. 13.117 When light passes through a colloid, it is scattered randomly by the dispersed particles because their sizes are similar to the wavelengths of visible light. Viewed from the side, the scattered beam is visible and broader than one passing through a solution, a phenomenon known as the Tyndall effect. 13.118 Soap micelles have nonpolar “tails” pointed inward and anionic “heads” pointed outward. The charges on the “heads” on one micelle repel the “heads” on a neighboring micelle because the charges are the same. This repulsion between soap micelles keeps them from coagulating. Soap is more effective in freshwater than in seawater because the divalent cations in seawater combine with the anionic “head” to form an insoluble precipitate.

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13.119 a) Total molarity of ions: ⎛ 0.010 mol ions ⎞⎟ ⎟⎟ ⎟ 1 mL ⎞ ⎝⎜ 1 sphere ⎠⎛ ⎜ −3 ⎟⎟ = 3.2 M ⎜⎝10 L ⎠⎟ 25 mL

(8 spheres)⎜⎜⎜ Solution A: M =

⎛ 0.010 mol ions ⎞⎟ ⎟⎟ ⎟ 1 mL ⎞ ⎝⎜ 1 sphere ⎠⎛ ⎜ −3 ⎟⎟ = 2.0 M ⎜⎝10 L ⎠⎟ 50 mL

(10 spheres)⎜⎜⎜ Solution B: M =

⎛ 0.010 mol ions ⎞⎟ ⎟⎟ ⎟ 1 mL ⎞ ⎝⎜ 1 sphere ⎠⎛ ⎜ −3 ⎟⎟ = 1.2 M ⎜⎝10 L ⎠⎟ 100 mL

(12 spheres)⎜⎜⎜ Solution C: M = b) Molarity of compound:

⎛ 0.010 mol ions ⎞⎛ ⎟⎟⎜⎜1 mole compound ⎞⎟⎟ ⎜ ⎟⎟⎜ ⎟⎟ ⎜⎝⎜ 1 sphere ⎟⎜⎜ 2 moles of ions ⎠⎛ ⎟ 1 mL ⎞ ⎠⎝ ⎜⎜ −3 ⎟⎟ = 1.6 M ⎝10 L ⎠⎟ 25 mL

(8 spheres)⎜⎜ Solution A: M =

⎛ 0.010 mol ions ⎞⎛ ⎟⎟⎜⎜1 mole compound ⎞⎟⎟ ⎜ ⎟⎟⎜ ⎟⎟ ⎜⎝⎜ 1 sphere ⎟⎜⎜ 2 moles of ions ⎠⎛ ⎟ 1 mL ⎞ ⎠⎝ ⎜⎜ −3 ⎟⎟ = 1.0 M ⎝10 L ⎠⎟ 50 mL

(10 spheres)⎜⎜ Solution B: M =

⎛ 0.010 mol ions ⎞⎛ ⎟⎟⎜⎜1 mole compound ⎞⎟⎟ ⎜ ⎟⎟⎜ ⎟⎟ ⎜⎝⎜ 1 sphere ⎟⎜⎜ 3 moles of ions ⎠⎛ ⎟ 1 mL ⎞ ⎠⎝ ⎜⎜ −3 ⎟⎟ = 0.40 M ⎝10 L ⎠⎟ 100 mL

(12 spheres)⎜⎜ Solution C: M =

Solution A has the highest molarity. c) Molality of compound: ⎛ 0.010 mol ions ⎞⎛ ⎟⎟⎜⎜1 mole compound ⎞⎟⎟ ⎜ (8 spheres)⎜⎜ ⎟⎟⎜ ⎟ ⎟⎜⎜ 2 moles of ions ⎠⎟⎟ ⎛1 mL ⎞⎛103 g ⎞ ⎜⎜⎝ 1 sphere ⎠⎝ ⎟⎟⎜⎜ ⎟ ⎜⎜ Solution A: m = ⎟⎟⎜ 1 kg ⎟⎟⎟ = 1.6 m ⎜⎝ 1.0 g ⎠⎝ 25 mL ⎠

⎛ 0.010 mol ions ⎞⎛ ⎞ ⎜ ⎟⎟⎟⎜⎜1 mole compound ⎟⎟⎟ ⎟⎟⎜⎜⎜ 2 moles of ions ⎟⎟ ⎛ ⎜⎜⎝ 1 sphere ⎠⎝ ⎠ ⎜1 mL ⎞⎛ 103 g ⎞⎟ ⎟⎟⎟⎜⎜ ⎟ = 1.0 m ⎜ ⎟⎜ 1 kg ⎠⎟⎟ ⎜⎝ 1.0 g ⎠⎝ 50 mL

(10 spheres)⎜⎜ Solution B: m =

⎛ 0.010 mol ions ⎞⎛ ⎟⎟⎜⎜1 mole compound ⎞⎟⎟ ⎜ ⎟⎟⎜ ⎟ ⎟⎜⎜ 3 moles of ions ⎠⎟⎟ ⎛1 mL ⎞⎛103 g ⎞ ⎜⎜⎝ 1 sphere ⎠⎝ ⎟⎟⎜⎜ ⎟ ⎜⎜ ⎟⎟⎜ 1 kg ⎟⎟⎟ = 0.40 m ⎜⎝ 1.0 g ⎠⎝ 100 mL ⎠

(12 spheres)⎜⎜ Solution C: m =

Solution C has the lowest molality. d) Osmotic pressure: assume a temperature of 298 K Solution A: Π = iMRT = (2)(1.6 mol/L)(0.0821 L∙atm/mol∙K)(298 K) = 78.29056 = 78 atm Solution B: Π = iMRT = (2)(1.0 mol/L)(0.0821 L∙atm/mol∙K)(298 K) = 48.9316 = 49 atm Solution C: Π = iMRT = (3)(0.40 mol/L)(0.0821 L∙atm/mol∙K)(298 K) = 29.35896 = 29 atm Solution A has the highest osmotic pressure.

13.120 Assume exactly 100 g of solution. ⎛10 % glucose ⎞⎟ ⎜ ⎟⎟ = 10 g glucose Mass (g) of glucose = (100 g solution) ⎜⎜ ⎟ ⎜⎜⎝ 100 % ⎠⎟ Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

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⎛ 1 mol glucose ⎞⎟ ⎟ = 0.055506217 mol glucose Moles of glucose = (10 g glucose) ⎜⎜ ⎜⎝180.16 g glucose ⎠⎟⎟ Mass (g) of water = 100.0 g solution – 10 g glucose = 90 g of water ⎛ 1 kg ⎞⎟ ⎟ = 0.090 kg Mass (kg) of water = (90 g H2O) ⎜⎜ ⎜⎝1000 g ⎠⎟⎟

⎛ 1 mL ⎞⎛ ⎟⎟⎜⎜ 1 L ⎞⎟⎟ Volume (L) of solution = (100 g) ⎜⎜⎜ ⎟⎟⎜⎜ ⎟ = 0.096246 L ⎜⎝1.039 g ⎠⎝ 1000 mL ⎠⎟

⎛ 0.055506217 mol glucose ⎞⎟ Molarity of glucose = ⎜⎜ ⎟ = 0.57671 M glucose ⎝ ⎠⎟ 0.096246 L ⎛ 0.055506217 mol glucose ⎞⎟ Molality of glucose = ⎜⎜ ⎟⎟ = 0.6167357 m glucose ⎜⎝ 0.090 kg ⎠⎟ Glucose is a nonelectrolyte so i = 1. T = (273 + 20) = 293 K ΔTf = iKf m = (1)(1.86°C/m)(0.6167357 m) = 1.1471°C Freezing point = (0.00 – 1.1471) = –1.1471 = –1.1°C ΔTb = iKbm = (1)(0.512°C/m)(0.6167357 m) = 0.3157687°C Boiling point = (100.00 + 0.3157687) = 100.3157687 = 100.32°C Π = iMRT = (1)(0.57671 mol/L)(0.0821 L∙atm/mol∙K)(293 K) = 13.8729 = 14 atm

13.121 Plan: The density of the mixture will be the weighted average of the constituents. Thus, density of mixture = contribution from copper + contribution from zinc. The percent zinc plus the percent copper must total 100%. Zinc atoms are heavier than copper atoms, so a factor equal to the ratio of their atomic weights (65.41/63.55) must be applied to the zinc contribution. Solution: a) Density of alloy = (90.0% Cu/100%)(8.95 g/cm3) + (10.0% Zn/100%)(65.41/63.55)(8.95 g/cm3) = 8.9762 = 8.98 g/cm3 b) Density of alloy = (62.0% Cu/100%)(8.95 g/cm3) + (38.0% Zn/100%)(65.41/63.55)(8.95 g/cm3) = 9.04954 = 9.05 g/cm3 13.122 Plan: To find the volume of seawater needed, substitute the given information into the equation that describes the ppb concentration, account for extraction efficiency, and convert mass to volume using the density of seawater. Solution: 1 troy ounce = 31.1 g gold mass of gold 1.1 × 10−2 ppb = ×10 9 mass of seawater 1.1 × 10−2 ppb =

31.1 g Au mass seawater

× 109

⎡ 31.1 g ⎤ Mass (g) of seawater = ⎢ × 10 9 ⎥ = 2.827273 × 1012 g (with 100% efficiency) ⎢1.1×10−2 ⎥ ⎣ ⎦ ⎛ 100% ⎞⎟ 12 Mass (g) of seawater = (2.827273 × 1012 g) ⎜⎜ ⎜⎝ 81.5% ⎠⎟⎟ = 3.46905 × 10 g seawater (81.5% efficiency) −3 ⎛ 1 mL ⎞⎛ ⎟⎟⎜10 L ⎟⎞⎟ = 3.384439 × 10 9 = 3.4 × 109 L Volume (L) of seawater = (3.46905 × 1012 g) ⎜⎜ ⎜ ⎟ ⎝⎜1.025 g ⎠⎟⎜⎝⎜ 1 mL ⎠⎟⎟ Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

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13.123 Xe is a much larger atom than He, so it is much more polarizable. This would increase the dipole–induced dipole forces when Xe is placed in water, increasing the solubility relative to He. 13.124 C. The principal factor in the solubility of ionic compounds in water is ion-dipole forces. Virtually all of the ionic compound’s ions would become separated and surrounded by water molecules (the number depending on the sizes of the ions) interacting with the ions via H bonding or other forces. 13.125 a) Solution A has a van’t Hoff factor of 3, Solutions B and C have a van’t Hoff factor of 2, and Solution D’s van’t Hoff factor is 1. Since Solution A has the largest van’t Hoff factor, Solution A would have the highest boiling point. b) Solution A also has the lowest freezing point since it has the largest van’t Hoff factor. c) No, the solution with the highest osmotic pressure cannot be determined. Osmotic pressure is determined by the molarity, not the molality, of the solution. Since we do not know the identity of the solutes and the density of the solutions, the 0.50 m value cannot be converted to molarity. 13.126 Plan: Convert the mass of O2 dissolved to moles of O2. Use the density to convert the 1 kg mass of solution to volume in L. Divide moles of O2 by volume of solution in L to obtain molarity. Solution: 0.0°C: −3 ⎛14.5 mg O 2 ⎞⎛ ⎟⎟⎜⎜10 g ⎞⎛ ⎟⎟⎜⎜ 1 mol O 2 ⎞⎟⎟ ⎜ –4 Moles of O2 = ⎜⎜ ⎟⎟⎜ ⎟⎟⎜ ⎟ = 4.53125 × 10 mol O2 ⎜⎝⎜ 1 kg H 2 O ⎠⎝ ⎟⎜⎜ 1 mg ⎠⎝ ⎟⎜⎜ 32.00 g O 2 ⎠⎟⎟ ⎛10 3 g ⎞⎟⎛ 1 mL ⎞⎛10−3 L ⎞⎟ ⎜ ⎟⎟⎜⎜ ⎟⎟ = 1.000130017 L ⎟⎟⎜⎜ Volume (L) of solution = (1 kg)⎜⎜ ⎜⎝⎜ 1kg ⎠⎟⎟⎜⎜⎝ 0.99987 g ⎠⎟⎟⎝⎜⎜⎜ 1mL ⎠⎟⎟

moles of O2 4.53125 ×10−4 mol = = 4.53066 × 10–4 = 4.53 × 10–4 M O2 L of solution 1.000130017 L 20.0°C: −3 ⎛ 9.07 mg O 2 ⎞⎛ ⎟⎟⎜⎜10 g ⎞⎛ ⎟⎟⎜⎜ 1 mol O 2 ⎞⎟⎟ ⎜ –4 Moles of O2 = ⎜⎜ ⎟⎟⎜ ⎟⎟⎜ ⎟ = 2.834375 × 10 mol O2 ⎜⎜⎝ 1 kg H 2 O ⎠⎝ ⎟⎜⎜ 1 mg ⎠⎝ ⎟⎜⎜ 32.00 g O 2 ⎠⎟⎟ −3 ⎛103 g ⎞⎛ ⎟⎟⎜⎜ 1 mL ⎞⎛ ⎟⎟⎜⎜10 L ⎞⎟⎟ ⎜ Volume (L) of solution = (1 kg)⎜⎜ ⎟⎟⎜ ⎟⎟⎜ ⎟ = 1.001773138 L ⎜⎜⎝ 1kg ⎠⎝ ⎟⎜⎜ 0.99823 g ⎠⎝ ⎟⎜⎜ 1 mL ⎠⎟⎟

moles of O2 2.834375 ×10−4 mol = = 2.829358 × 10–4 = 2.83 × 10–4 M O2 L of solution 1.001773138 L 40.0°C: −3 ⎛ 6.44 mg O 2 ⎞⎛ ⎟⎟⎜⎜10 g ⎞⎛ ⎟⎟⎜⎜ 1 mol O 2 ⎞⎟⎟ ⎜ –4 Moles of O2 = ⎜⎜ ⎟⎟⎜ ⎟⎟⎜ ⎟ = 2.0125 × 10 mol O2 ⎟⎜⎜ 1 mg ⎠⎝ ⎟⎜⎜ 32.00 g O 2 ⎠⎟⎟ ⎜⎜⎝ 1 kg H 2 O ⎠⎝

−3 ⎛103 g ⎞⎛ ⎟⎟⎜⎜ 1 mL ⎞⎛ ⎟⎟⎜⎜10 L ⎞⎟⎟ ⎜ Volume (L) of solution = (1 kg)⎜⎜ ⎟⎟⎜ ⎟⎟⎜ ⎟ = 1.007820689 L ⎟⎜⎜ 0.99224 g ⎠⎝ ⎟⎜⎜ 1mL ⎠⎟⎟ ⎜⎜⎝ 1kg ⎠⎝

moles of O2 2.0125 ×10−4 mol = = 1.996883 × 10–4 = 2.00 × 10–4 M O2 L of solution 1.007820689 L

13.127 Pyridine has nonpolar aromatic qualities like organic solvents but also has the potential to associate with water by hydrogen bonding through its lone pair of electrons (localized on the nitrogen atom).

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13.128 Plan: First, find the molality from the freezing point depression using the relationship ΔT = iKfm and then use the molality, given mass of solute and volume of water, to calculate the molar mass of the solute compound. Assume the solute is a nonelectrolyte (i = 1). Use the mass percent data to find the empirical formula of the compound; the molar mass is used to convert the empirical formula to the molecular formula. A Lewis structure that forms hydrogen bonds must have H atoms bonded to O atoms. Solution: a) ΔTf = iKf m = 0.000 °C – (–0 .201°C) = 0.201 °C ΔT 0.201ο C m= f = = 0.1080645 m Kf i (1.86ο C/m)(1) ⎛1.00 g ⎞⎟⎛⎜ 1 kg ⎞⎟ ⎟ = 0.0250 kg water Mass (kg) of solvent = (25.0 mL )⎜⎜ ⎜ ⎝ 1 mL ⎠⎟⎟⎜⎝103 g ⎠⎟⎟ moles of solute m= kg of solvent Moles of solute = (m)(kg solvent) = (0.1080656 m)(0.0250 kg) = 0.0027016 mol 0.243 g = 89.946698 = 89.9 g/mol Molar mass = 0.0027016 mol b) Assume that 100.00 g of the compound gives 53.31 g carbon, 11.18 g hydrogen, and 100.00 – 53.31 – 11.18 = 35.51 g oxygen. ⎛ 1 mol C ⎟⎞ 4.43880 ⎟ = 4.43880 mol C; Moles C = (53.31 g C)⎜⎜ =2 ⎜⎝12.01 g C ⎟⎟⎠ 2.219375 ⎛ 1 mol H ⎞⎟ 11.09127 ⎟⎟ = 11.09127 mol H; Moles H = (11.18 g H)⎜⎜ =5 2.219375 ⎝⎜1.008 g H ⎠⎟ ⎛ 1 mol O ⎞⎟ 2.219375 ⎟ = 2.219375 mol O; =1 Moles O = (35.51 g O)⎜⎜ ⎜⎝16.00 g O ⎠⎟⎟ 2.219375 Dividing the values by the lowest amount of moles (2.219375) gives an empirical formula of C2H5O with molar mass 45.06 g/mol. Since the molar mass of the compound, 89.9 g/mol from part (a), is twice the molar mass of the empirical formula, the molecular formula is 2(C2H5O) or C4H10O2. c) There is more than one example in each case. Possible Lewis structures: Forms hydrogen bonds Does not form hydrogen bonds H H H H H H H H H O

C H

−6 ⎛ ⎞ ⎛ 60 min ⎞⎛ ⎟⎟⎜⎜11 L air ⎞⎟⎟⎜⎜ 4.0 ×10 mol CO ⎟⎟⎛⎜⎜ 28.01 g CO ⎞⎟⎟ = 0.59157 = 0.59 g CO 13.129 Mass (g) of CO = (8.0 h )⎜⎜ ⎟ ⎟⎜ 1 min ⎠⎟⎜⎜ ⎟ ⎜⎝ 1 h ⎠⎝ 1 L air ⎝ ⎠⎟⎝⎜ 1 mol CO ⎠

13.130 No, both are the same because masses are additive. moles of N 2 total moles 3 moles N 2 = 0.38 Mixture A: X N2 = 3 + 2 + 3 moles 4 moles N 2 Mixture C: X N2 = = 0.33 4 + 3 + 5 moles

13.131 a) X N2 =

Mixture B: X N2 =

4 moles N 2 = 0.40 4 + 4 + 2 moles

13-35

Mixture C has the smallest mole fraction of N2. 2 moles Ne = 0.25 b) Mixture A: X Ne = 3 + 2 + 3 moles 3 moles Ne Mixture C: X Ne = = 0.25 4 + 3 + 5 moles Mixtures A and C have the same mole fraction of Ne. 3 moles Cl 2 = 0.38 c) Mixture A: X Cl2 = 3 + 2 + 3 moles 5 moles Cl 2 Mixture C: X Cl2 = = 0.42 4 + 3 + 5 moles Mixture B < Mixture A < Mixture C

Mixture B: X Ne =

4 moles Ne = 0.40 4 + 4 + 2 moles

Mixture B: X Cl2 =

2 moles Cl 2 = 0.20 4 + 4 + 2 moles

13.132 Plan: Find the moles of NaF required to form a 4.50 × 10–5 M solution of F– with a volume of 5000. L. Then convert moles of NaF to mass. Then use the molarity to find the mass of F– in 2.0 L of solution. Solution: ⎛ 4.50 × 10−5 mol F − ⎞⎛ ⎟⎟⎜⎜1mol NaF ⎞⎟⎟ ⎜ a) Moles of NaF = (5000. L )⎜⎜ ⎟⎟⎜ ⎟ = 0.225 mol NaF ⎜⎜ ⎟⎜⎜ 1mol F − ⎠⎟⎟ L ⎝ ⎠⎝ ⎛ 41.99 g NaF ⎞⎟ ⎜ ⎟⎟ = 9.44775 = 9.45 g NaF Mass (g) of NaF = (0.225 mol)⎜⎜ ⎜⎜⎝ 1 mol NaF ⎠⎟⎟ −5 − ⎞⎛ −⎞ ⎛ ⎜ 4.50 × 10 mol F ⎟⎟⎜⎜19.00 g F ⎟⎟ – b) Mass (g) of F– = (2.0 L )⎜⎜ ⎟⎟⎜ ⎟⎟ = 0.00171 = 0.0017 g F ⎜⎜ ⎜ ⎟ ⎟ L ⎝ ⎠⎝⎜ 1 mol F ⎠

13.133 a) The solution in U tube B is the most concentrated since it has the highest osmotic pressure. b) Solution C has the smallest number of dissolved ions and thus the smallest osmotic pressure. 13.134 Calculate the individual partial pressures from Psolvent = XsolventP°solvent. Assign the “equal masses” as exactly 1 g. Liquid: ⎛ ⎞⎟ 1 g pinene ⎜⎜ ⎟⎟ ⎜⎜ ⎟ ⎝⎜136.23 g pinene /mol ⎠⎟ X (pinene) = = 0.53100 ⎛ ⎞⎟ ⎛ ⎞⎟ 1 g pinene 1 g terpineol ⎜⎜ ⎜ ⎟⎟ + ⎜⎜ ⎟⎟ ⎜⎜ ⎜⎝136.23 g pinene/mol ⎠⎟⎟ ⎝⎜⎜154.24 g terpineol/mol ⎠⎟⎟ Psolvent = XsolventP°solvent P(pinene) = (0.53100)(100.3 torr) = 53.2593 torr ⎛ ⎞⎟ 1 g terpineol ⎜⎜ ⎟⎟ ⎜⎜ ⎜⎝154.24 g terpineol/mol ⎠⎟⎟ X (terpineol) = = 0.4689985 ⎛ ⎞⎟ ⎛ ⎞⎟ 1 g pinene 1 g terpineol ⎜⎜ ⎟⎟ + ⎜⎜ ⎟⎟ ⎜⎜ ⎜ ⎜⎝136.23 g pinene/mol ⎠⎟⎟ ⎝⎜⎜154.24 g terpineol/mol ⎠⎟⎟ Psolvent = XsolventP°solvent P(terpineol) = (0.4689985)(9.8 torr) = 4.5961853 torr

13-36

Vapor:

X (pinene) =

53.2593 torr

(53.2593 + 4.5961853) torr

X (terpineol) =

= 0.9205575 = 0.921

4.5961855 torr

(53.2593 + 4.5961853) torr

= 0.0794425 = 0.079

13.135 Plan: Use the boiling point elevation of 0.45°C to calculate the molality of the solution using the relationship ΔTb = iKbm and then use the molality, given mass of solute and volume of water, to calculate the molar mass of the solute compound. If the solute is a nonelectrolyte, i = 1. If the formula is AB2 or A2B, then i = 3. For part (d), use the molar mass of CaN2O6 to calculate the molality of the compound. Then calculate i in the boiling point elevation formula. Solution: a) ΔTb = iKbm i = 1 (nonelectrolyte) ΔT = boiling point of solution – boiling point of solvent = (100.45 – 100.00)°C = 0.45°C 0.45ο C ο ΔT m = b = (0.512 C/m )(1) = 0.878906 m = 0.878906 mol/kg Kb i ⎛ 0.997 g ⎞⎟⎛⎜ 1 kg ⎞⎟ ⎟ = 0.0249250 kg water Mass (kg) of water = (25.0 mL )⎜⎜ ⎜ ⎝ 1 mL ⎠⎟⎟⎜⎝103 g ⎠⎟⎟ moles of solute m= kg of solvent Moles solute = (m)(kg solvent) = (0.878906 m)(0.0249250 kg) = 0.0219067 mol 1.50 g = 68.4722 = 68 g/mol Molar mass = 0.0219067 mol b) ΔTb = iKbm i = 3 (AB2 or A2B) 0.45ο C ΔT m= b = = 0.29296875 m = 0.29296875 mol/kg (0.512 ο C/m )(3) Kb i moles of solute m= kg of solvent Moles solute = (m)(kg solvent) = (0.29296875 m)(0.0249250 kg) = 0.00730225 mol 1.50 g = 205.416 = 2.1 × 102 g/mol Molar mass = 0.00730225 mol c) The molar mass of CaN2O6 is 164.10 g/mol. This molar mass is less than the 2.1 × 102 g/mol calculated when the compound is assumed to be a strong electrolyte and is greater than the 68 g/mol calculated when the compound is assumed to be a nonelectrolyte. Thus, the compound is an electrolyte, since it dissociates into ions in solution. However, the ions do not dissociate completely in solution. ⎛ ⎞⎟ 1 mol ⎜ ⎟⎟ = 0.0091408 mol d) Moles of CaN2O6 = (1.50 g CaN 2 O 6 )⎜⎜ ⎜⎜⎝164.10 g CaN 2 O 6 ⎠⎟⎟

moles of solute 0.00914078 mol = = 0.3667314 m kg of solvent 0.0249250 kg ΔTb = iKbm (0.45°C) ΔTb i= = = 2.39659 = 2.4 Kbm 0.512° / m 0.3667314 m m=

(

)(

)

13-37

13.136

mol C2 H5 OH (l ) mol C 2 H 5 OH (l ) ⎜⎛ 60.5 torr ⎞⎟ ⎟⎟ = ⎜⎜ (0.4801587) mol CH3OH ( g) mol CH3OH (l ) mol CH 3 OH (l ) ⎜⎝⎜126.0 torr ⎠⎟⎟ A 97:1 mass ratio gives 97 g of C2H5OH for every 1 g of CH3OH. (This limits the significant figures.) ⎛ 1 mol C 2 H 5 OH ⎞⎟ ⎜ ⎟⎟ 97 g C 2 H 5 OH ( g)⎜⎜ ⎜⎜⎝ 46.07 g C 2 H 5 OH ⎠⎟⎟ 2.10549 mol C2 H5 OH ( g) = ⎛ 1 mol CH 3 OH ⎞⎟ 0.03121 mol CH3OH ( g) ⎜ 1 g CH 3 OH ( g)⎜⎜ ⎟⎟⎟ ⎜⎜⎝ 32.04 g CH 3 OH ⎠⎟ mol C2 H5 OH ( g)

mol C2 H5 OH (l ) mol CH3OH (l )

(2.10549/0.03121) 0.4801587

= 140.4994

⎛ 46.07 g C 2 H 5 OH ⎞

(140.4994 mol C H OH)⎜⎜⎜⎜⎜ 1 mol C H OH ⎟⎟⎟⎟⎟ 2

⎝

⎛ 32.04 g CH 3 OH ⎞⎟ ⎜ ⎟⎟ 1 mol CH 3 OH ⎜⎜ ⎜⎜⎝ 1 mol CH 3 OH ⎠⎟⎟

( 13.137

⎠

= 202.0227 = 2 × 102

)

Determine the molarity of CH3Cl in 1.00 L corresponding to 100. ppb. (Assume the density of the solution is the same as for pure water, 1.00 g/mL.) ⎛ 100. g CH Cl ⎞⎛ ⎟⎟⎜⎜ 1 mol CH 3 Cl ⎞⎛ ⎟⎟⎜⎜1.00 g solution ⎞⎛ ⎟⎟⎜⎜ 1 mL ⎞⎟⎟ ⎜ 3 M = ⎜⎜ 9 ⎟⎟⎜ ⎟⎟⎜ ⎟⎟⎜ −3 ⎟⎟ ⎜⎝⎜10 g solution ⎠⎝ ⎟⎜⎜ 50.48 g CH 3 Cl ⎠⎝ ⎟⎜⎜ ⎟⎜⎜10 L ⎠⎟ mL ⎠⎝ = 1.98098 × 10–6 = 1.98 × 10–6 M CH3Cl If the density is 1.00 g/mL, then 1.00 L of solution would weigh 1.00 kg. The mass of CH3Cl is insignificant compared to 1.00 kg, thus the mass of the solution may be taken as the mass of the solvent. This makes the molarity equal to the molality, in other words: 1.98 × 10–6 m CH3Cl. Still using 1.00 L of solution: Moles of CH3Cl = (1.98098 × 10–6 mol/L)(1.00 L) = 1.98098 × 10–6 mol CH3Cl Moles of H2O = (1.00 kg)(103 g/1 kg)(1 mol H2O/18.02 g H2O) = 55.49389567 mol H2O Xchloroform = (1.98098 × 10–6 mol CH3Cl)/[(1.98098 × 10–6 + 55.49389567) mol] = 3.569726 × 10–8 = 3.57 × 10–8 Convert from ppb to pph (parts per hundred = mass percent) ⎛100. ppb ⎞⎛ ⎟⎟⎜⎜100 pph ⎞⎟⎟ = 1.00 × 10–5% ⎜⎜ ⎜⎝⎜ 10 9 ⎠⎝ ⎟⎟⎜⎜ 1 ⎠⎟⎟

13.138 a) Yes, equilibrium is a dynamic process. Solid Na2CO3 and solid Na214CO3 both dissolve in the equilibrium process. Na214CO3(s) + H2O(l) → Na214CO3(aq) b) Radioactivity would be found in all of the solid as some of the Na14 that dissolves will also precipitate back out of solution. Na214CO3(aq) + H2O(l) → Na214CO3(s) 13.139 Plan: From the osmotic pressure, the molarity of the solution can be found using the relationship Π = MRT. Convert the osmotic pressure from units of torr to atm, and the temperature from °C to K. Use the molarity of the solution to find moles of solute; divide the given mass of solute in grams by the moles of solute to obtain molar mass. To find the freezing point depression, the molarity of the solution must be converted to molality by using the density of the solution to convert volume of solution to mass of solution. Then use ΔTf = iKf m. (i = 1).

13-38

Solution: a) Π = MRT ⎛ 1 atm ⎞⎟ = 4.47368 × 10–4 atm Osmotic pressure (atm), Π = (0.340 torr)⎜⎜ ⎝ 760 torr ⎠⎟⎟ T = 25°C + 273 = 298 K Π (4.47368 × 10−4 atm) M= = = 1.828544 × 10–5 M RT L • atm ⎞⎟ ⎛ ⎜⎜0.0821 (298 K) ⎝ mol • K ⎠⎟ (M)(V) = moles ⎛10−3 L ⎞⎟ –7 Moles = (1.828544 ×10−5 M )(30.0 mL) ⎜⎜ ⎟ = 5.48563 × 10 mol ⎜⎝ 1 mL ⎠⎟⎟ ⎛10−3 g ⎞⎟ ⎟⎟ ⎝ 1 mg ⎠⎟ = 1.82294 × 104 = 1.82 × 104 g/mol Molar mass = 5.48563 ×10−7 mol ⎛ 0.997 g ⎞⎟ b) Mass (g) of solution = (30.0 mL)⎜⎜ = 29.91 g ⎝ 1 mL ⎠⎟⎟ ⎛10−3 g ⎞⎟ ⎟⎟ = 0.0100 g Mass (g) of solute = (10.0 mg)⎜⎜ ⎝⎜ 1 mg ⎠⎟

(10.0 mg)⎜⎜⎜⎜

⎛ 1 kg ⎞ Mass (kg) of solvent = mass of solution – mass of solute = 29.91 g – 0.0100 g ⎜⎜ 3 ⎟⎟⎟ = 0.0299 kg ⎜⎝10 g ⎠⎟ –7 Moles of solute = 5.48563 × 10 mol (from part a) moles of solute 5.48563 × 10−7 mol = = 1.83466 × 10–5 m Molality = kg of solvent 0.0299 kg ΔTf = iKf m = (1)(1.86°C/m)(1.83466 × 10–5 m) = 3.412 × 10–5 = 3.41 × 10–5°C (So the solution would freeze at 0 – (3.41 × 10–5°C) = –3.41 × 10–5°C.

13.140 Plan: Henry’s law expresses the relationship between gas pressure and the gas solubility (Sgas) in a given solvent. Use Henry’s law to solve for pressure of dichloroethylene (assume that the constant (kH) is given at 21°C), use the ideal gas law to find moles per unit volume, and convert moles/L to ng/L. Solution: Sgas = kHPgas ⎛ 0.65 mg C 2 H 2 Cl 2 ⎞⎟⎛⎜10−3 g ⎞⎟⎜⎛ 1 mol C 2 H 2 Cl 2 ⎞⎟ ⎟⎟ = 6.705178 × 10–6 mol/L ⎟⎜ Sgas (mol/L) = ⎜⎜ ⎟⎟⎜⎜ ⎟ ⎟ ⎟ ⎜ ⎝ ⎠ L ⎝ 1 mg ⎠⎝ 96.94 g C 2 H 2 Cl 2 ⎠

Pgas =

Sgas kH

PV = nRT

⎛ 6.705178×10−6 mol/L ⎞⎟ ⎜ –4 = ⎜⎜ ⎟⎟ = 2.031872 × 10 atm ⎜⎜⎝ 0.033 mol/L • atm ⎠⎟⎟

(

)

2.031872 × 10−4 atm n P = = 8.41794 × 10–6 mol/L = L • atm ⎞⎟ ⎛ V RT ⎜⎜0.0821 ⎟((273 + 21) K ) ⎝ mol • K ⎠ ⎛ 8.41794 ×10−6 mol C H Cl ⎞⎛ ⎟⎜ 96.94 g C 2 H 2 Cl 2 ⎞⎛ ⎟⎟⎜ 1 ng ⎞⎟⎟ 2 2 2⎟ Concentration (ng/L) = ⎜⎜⎜ ⎟⎟⎜⎜ ⎟⎜ −9 ⎟ L ⎝⎜ ⎠⎝⎜ 1 mol C 2 H 2 Cl 2 ⎠⎟⎝⎜⎜10 g ⎠⎟

= 8.16035 × 105 = 8.2 × 105 ng/L

13-39

13.141 Acircle = πr2 = π(38.6 cm/2)2 = 1.17021 × 103 cm2 ⎞⎟ 1 mol 1.17021 × 10 3 cm 2 ⎛⎜⎜ 1 mg ⎞⎟⎟⎛⎜ 283 g ⎞⎟⎟⎛⎜⎜ ⎟ Area of one molecule = ⎟⎟⎜⎜ ⎜⎜ −3 ⎟⎟⎜⎜ ⎟⎟ 23 2.50 mg ⎝⎜10 g ⎠⎟⎜⎝ mol ⎠⎟⎝⎜ 6.022 × 10 molecules ⎠⎟ = 2.19973 × 10–16 = 2.20 × 10–16 cm2/molecule 13.142 a) Looking at the data for CaCl2, K2CO3, and Na2SO4, the average conductivity is 7.0 ± 0.7 units for the 5.00 × 103 ppm solutions and 14 ± 1.7 units for the 10.00 × 103 ppm solutions. This represents a relative error of about 10% if you assume that the identity of the solute is immaterial. If your application can tolerate an error of this magnitude, then this method would be acceptable. b) This would be an unreliable estimate of the concentration for those substances which are nonelectrolytes, or weak electrolytes, as their conductivity would be much reduced in comparison to their true concentration. c) Concentration (ppm) = (14.0/16.0)(10.00 × 103 ppm) = 8.75 × 103 ppm Assume the mass of CaCl2 present is negligible relative to the mass of the solution. ⎛ 8.75 × 10 3 g CaCl ⎞⎟⎛ 1 mol CaCl ⎞⎟⎛10 3 g ⎞⎟ ⎜ 2 ⎟⎜ 2 ⎟⎟ = 0.0788430 = 0.0788 m CaCl ⎟⎟⎜⎜⎜ Molality of CaCl2 = ⎜⎜ ⎟⎟⎜⎜ 2 ⎟ 6 ⎟ ⎜⎜⎝ 10 g solution ⎠⎟⎜⎜⎝110.98 g CaCl 2 ⎠⎟⎝⎜⎜ 1 kg ⎠⎟ ⎛ 1 mol CaCl 2 ⎞⎟ ⎜ ⎟⎟ = 78.84303 mol CaCl Moles of CaCl2 = (8.75 × 103 g CaCl2) ⎜⎜ 2 ⎜⎜⎝110.98 g CaCl 2 ⎠⎟⎟ ⎛ 1 mol H 2 O ⎞⎟ ⎜ ⎟⎟ = 5.5493896 × 104 mol H O Moles of H2O = (1.00 × 106 g H2O) ⎜⎜ 2 ⎜⎜⎝18.02 g H 2 O ⎠⎟⎟ Mole fraction of CaCl2 = X =

(78.84303 mol CaCl ) 2

(78.84303 + 5.5493896 ×10 ) mol 4

= 1.4187 × 10–3 = 1.42 × 10–3

13.143 The vapor pressure of H2O above the pure water is greater than that above the sugar solution. This means that water molecules will leave the pure water and enter the sugar solution in order to make their vapor pressures closer to equal. 13.144 Plan: Assume a concentration of 1 mol/m3 for both ethanol and 2-butoxyethanol in the detergent solution. Then, from Henry’s law, the partial pressures of the two substances can be calculated. Solution: a) Sgas = kH × Pgas Sgas Pgas = kH ⎛1 mol ⎞⎛ 5 ×10−6 atm • m 3 ⎞ ⎟⎟ = 5 × 10–6 atm Pethanol = ⎜⎜⎜ 3 ⎟⎟⎟⎜⎜ mol ⎠⎟⎟ ⎝⎜ m ⎠⎟⎝⎜ ⎛1 mol ⎞⎛1.6 ×10−6 atm • m 3 ⎞⎟ ⎟ = 1.6 × 10–6 atm P2-butoxyethanol = ⎜⎜ 3 ⎟⎟⎟⎜⎜⎜ ⎟⎠⎟ mol ⎝⎜ m ⎠⎜⎝ ⎛1.6 ×10−6 atm 2-butoxyethanol ⎞⎟ ⎟⎟ = 1.6% %2-butoxyethanol = (5%)⎜⎜⎜ ⎜⎝ 5 ×10−6 atm ethanol ⎠⎟ “Down-the-drain” factor is 0.016 = 0.02 ⎛ 5 ×10−6 atm • m 3 ⎞⎟⎛ 1 L ⎞ ⎟⎟⎜⎜ −3 3 ⎟⎟ = 5 × 10–3 atm∙L/mol b) kH(ethanol) = ⎜⎜⎜ mol ⎝⎜ ⎠⎟⎜⎝10 m ⎠⎟ ⎛ 0.64 Pa • m 3 ⎞⎟⎛ ⎞⎟ 1 atm –6 –6 3 ⎟⎟⎜⎜ c) kH(ethanol) = ⎜⎜⎜ ⎟⎟ = 6.3163 × 10 = 6.3 × 10 atm∙m /mol 5 mol ⎝⎜ ⎠⎟⎝⎜1.01325 ×10 Pa ⎠

Considering the single significant figure in the measured value of 5 × 10–6, the agreement is good. Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

13-40

13.145 The fraction remaining in the water (fw) is related to the volume of water (Vw), the volume of dichloromethane (Vd), and the distribution ratio for the solubility (D = 8.35/1). fw = Vw/(Vw + DVd) Mass remaining in water = fw (original mass) a) Mass (mg) in water =

(100.0 mL)

(100.0 + 8.35 (60.0)) mL

(10.0 mg) = 1.66389 = 1.66 mg remaining

b) Perform a similar calculation to part (a), then take the result and repeat the procedure. Combine the results to get the total removed. Mass (mg) in water =

Mass (mg) in water =

(100.0 mL)

(100.0 + 8.35 (30.0)) mL

(100.0 mL) (100.0 + 8.35 (30.0)) mL

(10.0 mg) = 2.853067 = 2.85 mg remains after first extraction (2.853067 mg)

= 0.813999 = 0.814 mg remains after second extraction c) The two-step extraction extracts more of the caffeine. 13.146 Plan: Molality is defined as moles of solute per kg of solvent, so 0.150 m means 0.150 mol NaHCO3 per kg of water. The total mass of the solution would be 1 kg + (0.150 mol × molar mass of NaHCO3). Solution: ⎛ 84.01 g NaHCO3 ⎞⎟ ⎜ ⎟⎟ 0.150 mol NaHCO3 ⎜⎜ ⎟ ⎜ 0.150 mol NaHCO 3 ) ( ⎝⎜ 1 mol NaHCO3 ⎠⎟ ⎛⎜⎜ 1 kg ⎞⎟⎟ 0.150 m = = ⎜⎜ 3 ⎟⎟ 1 kg 1 kg ⎜⎝10 g ⎠⎟ = 12.6015 g NaHCO3/1000 g solvent ⎛ ⎞⎟ 12.6015 g NaHCO3 ⎜⎜ ⎟⎟ ⎜⎜ ⎟ 250 g solution = 3.111 g NaHCO3 ⎜⎜ 1000 + 12.6015 g solution ⎟⎟⎟ ⎝ ⎠ Mass (g) of H2O = 250 g – 3.111 g = 246.889 g H2O To make 250 g of a 0.150 m solution of NaHCO3, weigh 3.11 g NaHCO3 and dissolve in 247 g water.

(

)

(

)

13.147 To determine the molecular formula, both the empirical formula and the molar mass are needed. First, determine the empirical formula assuming exactly 100 g of sample, which makes the percentages equal to the mass of each element present: Moles C = 32.3 g C(1 mol C/12.01 g C) = 2.6894 mol C Moles H = 3.97 g H(1 mol H/1.008 g H) = 3.93849 mol H Moles O = (100 – 32.3 – 3.97) g O(1 mol O/16.00 g O) = 3.9831 mol O Dividing each mole value by the smallest value (moles C) gives: C = 1, H = 1.5, and O = 1.5 leading to an empirical formula of: C2H3O3. The molar mass comes from the freezing point depression: ΔTf = iKf m (Assume the compound is a nonelectrolyte, i = 1.) m = ΔTf/iKf = (1.26°C)/(1)(1.86°C/m)] = 0.677419 m m = moles of solute/kg of solvent Moles of solute = (m)(kg of solvent) = (0.677419 m)[(11.23 g)(1 kg/103 g)] = 0.007607415 mol ⎛ ⎞⎟ 0.981 g ⎜ Molar mass = ⎜⎜ ⎟⎟ = 128.953 g/mol ⎜⎜⎝ 0.007607415 mol ⎠⎟⎟ The empirical formula mass is approximately 75 g/mol.

13-41

The ratio of the molar to the empirical formula mass normally gives the conversion factor to change the empirical formula to the molecular formula. In this case, 129/75 = 1.72; this is not near a whole number. (This result is low due to dissociation of the weak acid; the assumption of i = 1 is too low. If i = 1.2, then the molar mass would increase to about 154 g/mol.) The 1.72 value implies the molecular formula is twice the empirical formula, or C4H6O6. 13.148 The range has to fall between the point where the number of moles of methanol is just greater than the number of moles of ethanol, to the point where the mass of methanol is just less than the mass of ethanol. The first point is the point at which the mole fractions are just becoming unequal. The methanol mole fraction is greater than 0.5000. Point 2: where the mass percents are just beginning to become unequal. First, find where they are equal. (1.000 g methanol/2.000 g solution) = (1.000 g ethanol/2.000 g solution) Moles of methanol = (1.000 g ethanol)(1 mol ethanol/32.04 g ethanol) = 0.031210986 mol methanol Moles of ethanol = (1.000 g ethanol)(1 mol ethanol/46.07 g ethanol) = 0.021706 mol ethanol Mole fraction of methanol = (0.031210986 mol methanol)/[(0.031210986) + (0.021706)] mol = 0.589810 Range of mole fractions of methanol: 0.5000 < Xmethanol < 0.5898 13.149 a) The molar mass comes from the boiling point elevation. ΔTb = (77.5 – 76.5) = 1.0°C ΔTb = iKbm (Assume the compound is a nonelectrolyte, i = 1.) m = ΔTb/iKb = (1.0°C)/[(1)(5.03°C/m)] = 0.198807 m m = moles of solute/kg of solvent Moles of solute = (m)(kg of solvent) = (0.198807 m)[(100.0 g)(1 kg/103 g)] = 0.0198807 mol ⎛ ⎞⎟ 5.0 g ⎜ ⎟⎟ = 251.5 = 2.5 × 102 g/mol Molar mass = ⎜⎜ ⎜⎝⎜ 0.0198807 mol ⎠⎟⎟ b) The molar mass, based on the formula, is 122.12 g/mol. The molar mass determined in part (a) is double the actual molar mass. This is because the acid dimerizes (forms pairs) in the solution. These pairs are held together by relatively strong hydrogen bonds, and give a “molecule” that is double the mass of a normal molecule. 13.150 Molarity is moles solute/L solution and molality is moles solute/kg solvent. Multiplying molality by concentration of solvent in kg solvent per liter of solution gives molarity: ⎛ mol of solute ⎞⎟⎛ kg solvent ⎞ mol of solute ⎟⎟ ⎟⎟⎜ M= = ⎜⎜⎜ L of solution ⎝ kg solvent ⎠⎟⎜⎝ L of solution ⎠⎟

⎛ kg solvent ⎞⎟ M = m ⎜⎜ ⎟ ⎝ L of solution ⎠⎟ For a very dilute solution, the assumption that mass of solvent ≅ mass of solution is valid. This equation then becomes M = m (kg solvent/L solution) = m × dsolution Thus, for very dilute solutions molality × density = molarity. In an aqueous solution, the liters of solution have approximately the same value as the kg of solvent because the density of water is close to 1 kg/L, so m = M. + ⎛ 1 mol NH 4 NO 3 ⎞⎛ ⎟⎟⎜⎜ 1 mol NH 4 ⎞⎟⎟ ⎜ –2 + 13.151 Moles = (5.66 g NH 4 NO 3 )⎜⎜ ⎟⎟⎜ ⎟ = 7.07058 × 10 mol NH4 ⎜⎝⎜ 80.05 g NH 4 NO 3 ⎠⎝ ⎟⎜⎜1 mol NH 4 NO 3 ⎠⎟⎟ + ⎛ 1 mol (NH 4 )3 PO 4 ⎞⎛ ⎞⎟ ⎟⎟⎜⎜ 3 mol NH 4 ⎜ ⎟⎟ = 8.89336 × 10–2 mol NH + Moles = (4.42 g (NH 4 )3 PO 4 )⎜⎜ ⎟⎟⎜ 4 ⎜⎝⎜149.10 g (NH 4 )3 PO 4 ⎠⎝ ⎟⎜⎜1 mol (NH 4 )3 PO 4 ⎠⎟⎟

13-42

3− ⎛ 1 mol (NH 4 )3 PO 4 ⎞⎛ ⎞⎟ ⎟⎟⎜⎜ 1 mol PO 4 ⎜ ⎟⎟ = 2.96445 × 10–2 mol PO 3– Moles = (4.42 g (NH 4 )3 PO 4 )⎜⎜ ⎟⎟⎜ 4 ⎜⎝⎜149.10 g (NH 4 )3 PO 4 ⎠⎝ ⎟⎜⎜1 mol (NH 4 )3 PO 4 ⎠⎟⎟

M NH4+ = [(7.07058 × 10–2) + (8.89336 × 10–2)] mol NH4+/20.0 L = 7.98197 × 10–3 = 7.98 × 10–3 M NH4+ M PO43– = (2.96445 × 10–2 mol PO43–)/20.0 L = 1.482225 × 10–3 = 1.48 × 10–3 M PO43– 13.152 Plan: Use Henry’s law, Sgas = kH × Pgas, to find the solubility of SO2. Solution: a) Sgas = kH × Pgas = (1.23 mol/L∙atm)(2.0 × 10–3 atm) = 2.46 × 10–3 = 2.5 × 10–3 M SO2 b) The base reacts with the sulfur dioxide to produce calcium sulfite. The reaction of sulfur dioxide makes “room” for more sulfur dioxide to dissolve. 13.153 a) Assume a 100 g sample of urea. This leads to the mass of each element being equal to the percent of that element. ⎛ 1 mol C ⎞⎟ ⎟ = 1.6736 mol C Moles C = (20.1 g C)⎜⎜ ⎜⎝12.01 g C ⎠⎟⎟ ⎛ 1 mol H ⎞⎟ ⎟ = 6.6468 mol H Moles H = (6.7 g H)⎜⎜ ⎜⎝1.008 g H ⎠⎟⎟ ⎛ 1 mol N ⎞⎟ ⎟⎟ = 3.31906 mol N Moles N = (46.5 g N)⎜⎜ ⎝⎜14.01 g N ⎠⎟ ⎛ 1 mol O ⎞⎟ ⎟ = 1.66875 mol O Moles O = ((100 − 20.1 − 6.7 − 46.5) g O)⎜⎜ ⎜⎝16.00 g O ⎠⎟⎟ Dividing all by the smallest value (1.66875 mol O) gives: C = 1, H = 4, N = 2, O = 1. Thus, the empirical formula is CH4N2O. The empirical formula weight is 60.06 g/mol. b) Use Π = MRT to solve for the molarity of the urea solution. The solution molarity is related to the concentration expressed in % w/v by using the molar mass.

M = Π/RT =

(2.04 atm)

= 0.0833817 M L • atm ⎞⎟ ⎛ ⎜⎜0.0821 + 273 25 K ( ) ( ) ⎝ mol • K ⎠⎟ Moles of urea = (M)(V) = (0.0833817 M)(1 L) = 0.0833817 mol Molar mass = 5.0 g/0.0833817 mol = 59.965 = 60 g/mol Because the molecular weight equals the empirical weight, the molecular formula is also CH4N2O. ⎛100 mL ⎞⎟⎜⎛10−3 L ⎞⎛ ⎟⎜ 0.30 mol glucose ⎞⎛ ⎟⎜180.16 g glucose ⎞⎟⎟ 13.154 a) Mass (g) of glucose = (2.5 h )⎜⎜⎜ ⎟⎟⎟⎜⎜ ⎟⎟⎟⎜⎜ ⎟ ⎟⎟⎟⎜⎜ 1L ⎜⎝ h ⎠⎟⎜⎜⎝ 1 mL ⎠⎝ ⎟⎜⎜ ⎟⎜⎜ 1 mol glucose ⎠⎟⎟ ⎠⎝ = 13.512 = 14 g glucose b) At low concentrations sodium chloride dissociates completely, forming twice as many dissolved particles per mole as glucose, so a sodium chloride solution would have to have a molarity that is one-half of glucose to be isotonic: 0.15 M. ⎛150. mL ⎞⎟⎜⎛10−3 L ⎞⎛ ⎟⎜ 0.15 mol NaCl ⎞⎛ ⎟⎟⎜⎜ 58.44 g NaCl ⎞⎟⎟ ⎟⎟⎜⎜ c) Mass (g) of NaCl = (1.5 h )⎜⎜⎜ ⎟⎟⎟⎜⎜ ⎟⎜ ⎟ h 1L ⎟⎜⎜ ⎟⎟⎜⎜ 1 mol NaCl ⎠⎟⎟ ⎝⎜ ⎠⎟⎜⎜⎝ 1 mL ⎠⎝ ⎠⎝

= 1.97235 = 2.0 g NaCl 13.155 a) There is a positive deviation since benzene molecules are held together only by weak dispersion forces while methanol molecules are held together by relatively strong hydrogen bonding. The two components will not interact with each other since the intermolecular forces are so different. Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

13-43

b) There is a positive deviation since ethyl acetate will have weaker hydrogen bonding to water than water has with itself. c) Since hexane and heptane are very similar compounds with weak dispersion forces, they will obey Raoult’s law. The behavior will be nearly ideal. d) The behavior will be nearly ideal since the hydrogen bonding in methanol and water is very similar. e) There is a negative deviation because HCl exists as ions in solution and water is in the hydration shells around the H3O+ and Cl– ions. 13.156 The enthalpy of vaporization must be determined first. ⎛ 1 bar ⎞⎛ ⎟⎟⎜ 1 atm ⎞⎟ = 0.0040464 atm P (atm) at 25°C = (4.1 mbar )⎜⎜ ⎝1000 mbar ⎠⎟⎜⎜⎝1.01325 bar ⎠⎟⎟ ο ⎛1 ΔH vap P 1⎞ ⎜⎜ ln 2 = − − ⎟⎟⎟ ⎜ R ⎝ T2 T1 ⎠⎟ P1 ⎛ ⎞⎟ ⎜⎜ 1 1 ⎟⎟ ⎜⎜ ln = − ⎟ 1 atm 8.314 J /mol • K ⎜⎜ 273 + 25 K 273.2 + 141.5 K ⎠⎟⎟⎟ ⎝ ο –5.509928 = –1.135823 × 10–4 ΔH vap ο ΔH vap = 48,510.45 J/mol

0.0040464 atm

ο −ΔH vap

(

)

(

)

⎛ ⎞⎟ P2 1 1 −48, 510.45 J/mol ⎜⎜ ⎟⎟ ⎜⎜ = − ⎟⎟ = –3.192787 1 atm 8.314 J /mol i K ⎜⎜ 273 + 65 K 273.2 + 141.5 K ⎝ ⎠⎟⎟

(

)

(

)

P2 = 0.0410573 1 atm

P2 = 0.0410573 = 0.041 atm 13.157 Plan: Use Henry’s law, Sgas = kH × Pgas, to find kH for O2. Then use that value to find the concentration of O2 at a pressure of 0.005 atm. To calculate the mole fraction of O2, assume a liter of solution. Divide the moles of O2 in 1.0 L of sample by the total moles of O2 and acrylic acid. To calculate ppm, convert moles of O2 and acrylic acid to mass; divide the mass of O2 by the total mass of the solution and multiply by 106. Solution: a) Sgas = kH × Pgas Sgas 1.64 ×10−3 mol/L kH = = = 7.8282 × 10–3 = 7.83 × 10–3 mol/L∙atm 0.2095 atm Pgas b) Sgas = kH × Pgas Sgas = (7.8282 × 10–3 mol/L∙atm)(0.005 atm) Sgas = 3.9141 × 10–5 = 4 × 10–5 M c) Assume a 1.0 L sample. Acrylic acid is 14.6 mol/L or 14.6 mol in 1.0 L. Oxygen is 4 × 10–5 mol/L or 4 × 10–5 mol in 1.0 L. moles of O 2 4 × 10−5 mol = 2.73972 × 10–6 = 3 × 10–6 X O2 = = −5 moles of O 2 + moles of acrylic acid × (4 10 ) + 14.6 mol ( ) ⎛14.6 mol acrylic acid ⎞⎟⎛⎜ 72.06 g acrylic acid ⎞⎟ ⎟ = 1052.076 g/L d) Mass (g) of acrylic acid = ⎜⎜ ⎟⎜ ⎝ ⎠⎟⎜⎝ 1 mol acrylic acid ⎠⎟⎟ L

⎛ 4 ×10−5 mol O ⎞⎛ 32.0 g O2 ⎞⎟ 2⎟ ⎟⎟⎜⎜ ⎟⎟ = 0.00128 g/L Mass of oxygen = ⎜⎜⎜ ⎜ L ⎝⎜ ⎠⎟⎝⎜ 1 mol O2 ⎠⎟ ppm =

mass of solute g (1×106 ) = 0.001280.00128 (1×106 ) = 1.2165 = 1 ppm mass of solution g + 1052.176 g

13-44

13.158 The lower the boiling point the greater the volatility. acetic acid < water < benzene < ethanol < carbon tetrachloride < chloroform < carbon disulfide < diethyl ether ΔH vap ⎛⎜ 1 P2 1⎞ − ⎟⎟⎟ =− ⎜ ⎜ R ⎝ T2 T1 ⎠⎟ P1 ο

13.159

P1 = 1.00 atm T1 = (273 + 100) K = 373 K P2 = ? T2 = (273 + 200) K = 473 K ο = 40.7 kJ/mol = 40,700 J/mol ΔH vap P −40, 700 J/mol ⎛⎜ 1 1 ⎞⎟ ln 2 = − ⎟ ⎜⎜ 1 atm 8.314 J /mol i K ⎝ 473 K 373 K ⎠⎟ ln

P2 = 2.774689665 1.00 atm

P2 = 16.03365 1.00 atm

P2 = 16.03365 = 16.0 atm 13.160 a) ΔTf = iKf m . Assume NaCl is a strong electrolyte with i = 2. ΔTf 5.0 ο C m= = = 1.344086 m NaCl iKf (2)(1.86 ο C/m ) ⎛ 58.44 g NaCl ⎞⎟ ⎛1.344086 mol NaCl ⎞⎟ 2 ⎟(5.5 kg)⎜⎜⎜ Mass (g) of NaCl = ⎜⎜⎜ ⎟⎟ = 432.016 = 4.3 × 10 g NaCl ⎟ ⎟⎠ ⎜⎜⎝ mol NaCl ⎠⎟⎟ ⎜⎝ kg

b) ΔTf = iKf m Assume CaCl2 is a strong electrolyte with i = 3. ΔTf 5.0 ο C m= = = 0.896057 m CaCl2 iKf (3)(1.86 ο C/m ) ⎛110.98 g CaCl 2 ⎞⎟ ⎛ 0.896057 mol CaCl 2 ⎞⎟ ⎟⎟ = 546.944 = 5.5 × 102 g CaCl ⎟⎟(5.5 kg)⎜⎜⎜ Mass (g) of CaCl2 = ⎜⎜⎜ 2 1 kg ⎜⎜⎝ mol CaCl 2 ⎠⎟⎟ ⎝⎜ ⎠⎟

13.161 Plan: Use the density of C6F14 to find the mass and then moles of C6F14 in 1 L. This is the molarity of C6F14 which is then used to find the molarity of O2 from its mole fraction. Then Henry’s law can be used to find kH. In part (b), use Henry’s law to find the solubility of O2 at the given temperature. Solution: ⎛ 1 mL ⎞⎛1.674 g C6 F14 ⎞⎟⎜⎛1 mol C 6 F14 ⎞⎟ a) Moles of C6F14 in 1 L (M) = (1 L)⎜⎜ −3 ⎟⎟⎟⎜⎜ ⎟⎟⎜⎜ ⎟⎟ = 4.9527 mol/L ⎝10 L ⎠⎝ ⎠⎜ 1 mL ⎝ 338 g C6 F14 ⎠⎟

mol of O2 mol of O2 + mol of C6 F14 x mol O 2 4.28 × 10–3 = 4.9527 mol Moles of O2 = 0.021198 (The moles of O2 is small enough to ignore in the denominator.) ⎛ 1 atm ⎞⎟ Pgas = (101,325 Pa )⎜⎜ ⎟ = 1 atm ⎜⎝101,325 Pa ⎠⎟

X O2 =

Sgas = kH × Pgas Sgas 0.021198 mol/L kH = = = 0.021198 = 0.0212 mol/L∙atm 1 atm Pgas

13-45

b) Sgas = kH × Pgas 1 kH = = 1.322 × 10–3 mol/L∙atm 756.7 L • atm/mol PO2 = 0.2095 × 1 atm = 0.2095 atm since O2 is 20.95% of air Sgas = (1.322 × 10–3 mol/L∙atm)(0.2095 atm) = 2.76959 × 10–4 mol/L ⎛ 2.76959×10−4 mol ⎞⎟⎛ 32.0 g O ⎞ ⎜ 2⎟ –3 –3 6 ⎟⎟⎜⎜ ⎟ ppm = ⎜⎜ ⎟⎟⎜⎜⎝ 1 mol O ⎠⎟⎟ = 8.8627 × 10 g/L = (8.8627 × 10 g/1000 g)(1 × 10 ) L ⎜⎝ ⎠ 2 = 8.8627 = 8.86 ppm c) kH: C6F14 > C6H14 > ethanol > water To dissolve oxygen in a solvent, the solvent molecules must be moved apart to make room for the gas. The stronger the intermolecular forces in the solvent, the more difficult it is to separate solvent particles and the lower the solubility of the gas. Both C6F14 and C6H14 have weak dispersion forces, with C6H14 having the weaker forces due to the electronegative fluorine atoms repelling each other. Both ethanol and water are held together by strong hydrogen bonds with those bonds being stronger in water as the boiling point indicates. 13.162 a) PN2 = (1.00 atm)(78% N2/100%) = 0.78 atm Molarity of N2 = Sgas = kH × Pgas = (6.2 × 10–4 mol/L∙atm)(0.78 atm) = 4.836 × 10–4 = 4.8 × 10–4 M N2 b) The additional pressure due to 50 ft of water must be added to 1.00 atm. Water pressure: The value, 9.80665 m/s2, is the standard acceleration of gravity from the inside back cover of the book. 2 ⎛ 1 kg ⎞⎟⎛ ⎛ 1 Pa ⎞⎟⎛ ⎞⎟ ⎛1.00 g ⎞⎟⎛⎜ 1 mL ⎞⎛ 1 atm ⎟⎟⎜⎜ 1 cm ⎞⎟⎟ ⎛⎜⎜ 2.54 cm ⎞⎛ ⎟⎟⎜⎜12 in ⎞⎟⎟ ⎜⎜ ⎟⎟⎜9.80665 m ⎞⎟⎟⎜⎜ ⎟⎟⎜⎜ ⎟⎟ ⎟⎟⎜⎜ Pwater = ⎜⎜⎜ 50.0 ft ⎟ ⎟ ⎟ ( ) ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ 3⎟ 2 3 2 2 5 − ⎟ ⎟⎟⎜⎜10 m ⎟⎟ ⎜⎜ 1 in ⎟⎟⎜⎜ 1 ft ⎟⎟ ⎟ ⎟⎜⎝1 cm ⎠⎝ ⎜⎝ mL ⎠⎜ s ⎠⎝⎜⎜1kg/m • s ⎠⎟⎟⎜⎝⎜1.013×10 Pa ⎠⎟⎟ ⎠ ⎝ ⎠⎝ ⎠ ⎝⎜⎜10 g ⎠⎟⎝ = 1.47535 atm This is the pressure due to the 50 ft of water, and it must be added to the atmospheric pressure pressing down on the surface of the water (1.00 atm). This gives a total pressure of 2.47535 atm. PN2 = (2.47535 atm)(78% N2/100%) = 1.930773 atm Molarity of N2 = Sgas = kH × Pgas = (6.2 × 10–4 mol/L∙atm)(1.903773 atm) =1.18034 × 10–3 = 1.2 × 10–3 M N2 c) Moles of N2 per liter at the surface = 4.836 × 10–4 mol N2 Moles of N2 per liter at 50 ft = 1.18034 × 10–3 mol N2 Moles N2 released per liter = (1.18034 × 10–3 – 4.836 × 10–4) mol = 6.9674 × 10–4 mol L • atm ⎞⎟ ⎛ (6.9674 ×10−4 mol)⎜⎜⎝0.0821 ⎟((273 + 25) K ) ⎛ 1 mL ⎞ mol •K⎠ ⎟⎟ ⎜⎜ PV = nRT so V = nRT/P = ⎜⎜⎝10−3 L ⎠⎟⎟ (1.00 atm) = 17.046 = 17 mL N2 13.163 a) Yes, the phases of water can still coexist at some temperature and can therefore establish equilibrium. b) The triple point would occur at a lower pressure and lower temperature because the dissolved air solute lowers the vapor pressure of the solvent. c) Yes, this is possible because the gas-solid phase boundary exists below the new triple point. d) No, the presence of the solute lowers the vapor pressure of the liquid.

13-46

13.164 a) Assuming 100 g of water, the solubilities (in g) of the indicated salts at the indicated temperatures would be: KNO3 KClO3 KCl NaCl 50°C 85 18 42 36 0°C 12 4 28 35 Difference 73 14 14 1 % recovery 86 78 33 3 (The “difference” is the number of grams of the salt, which could be recovered if a solution containing the amount of salt in the first line were cooled to 0°C. The “% recovery” is calculated by dividing the “difference” by the original amount, then multiplying by 100.) The highest percent recovery would be found for KNO3 (86%), and the lowest would be for NaCl (3%). b) If you began with 100. g of the salts given above, then the “% recovery” line above gives the number of grams which could be recovered by the process described. 13.165 Plan: Use the given percentages to find the volume of ethanol absorbed into the blood and then use the density of ethanol to convert volume of ethanol to mass and divide by the total volume of blood in mL. Use a ratio to find the volume of whiskey associated with the given blood alcohol level. Solution: ⎟⎟⎜⎜ 22% ⎞⎛ ⎟⎜ 0.789 g ⎞⎟ = 1.944096 g a) Mass (g) of ethanol dissolved in the blood = (28 mL)⎜⎜⎛ 40% ⎞⎛ ⎟⎜100% ⎠⎟⎟⎜⎝⎜ mL ⎠⎟⎟ ⎝⎜100% ⎠⎝ −3 ⎛1.944096 g ethanol ⎞⎛ ⎟⎟⎜⎜10 L ⎞⎟⎟ ⎜ –4 –4 Concentration = ⎜⎜ ⎟⎟⎜ ⎟ = 2.77728 × 10 = 2.8 × 10 g/mL ⎜⎜⎝ 7.0 L ⎟⎠⎝⎜⎜ 1 mL ⎠⎟⎟ ⎛ 28 mL whiskey ⎞⎟ ⎜ b) ⎜⎜ ⎟⎟ 8.0 × 10−4 g/mL ) = 80.6545 = 81 mL ⎜⎜ 2.77728 × 10−4 g/mL ⎟⎟( ⎝ ⎠ −2 ⎛10−3 L ⎞⎛ ⎟⎟⎜⎜ 3.3 × 10 mol ⎞⎟⎟ ⎜ 13.166 a) Moles of CO2 = (355 mL )⎜⎜ ⎟⎟⎜ ⎟ 4 atm ) = 0.04686 = 0.05 mol CO2 ⎜⎜ 1 mL ⎟⎜⎜ L i atm ⎟⎟( ⎝ ⎠⎝ ⎠ b) If it is completely flat there is no CO2 remaining or 0.00 moles CO2, however, a small amount will remain in solution: −2 ⎛10−3 L ⎞⎛ ⎟⎟⎜⎜ 3.3 × 10 mol ⎞⎟⎟ ⎜ –4 –6 –6 Moles CO2 = (355 mL) ⎜⎜ ⎟⎟⎜ ⎟⎟ (4 × 10 atm) = 4.6860 × 10 = 5 × 10 mol CO2 L i atm ⎟⎠ ⎜⎜⎝ 1 mL ⎠⎟⎜⎜⎝ c) The difference in the moles will determine the number of moles entering the gas phase. ⎡ L i atm ⎤⎥ ⎡(0.04686 − 4.6860 ×10−2 ) mol⎤ ⎢0.0821 [(273 + 25) K ] ⎣⎢ ⎦⎥ ⎢ mol i K ⎥⎥ ⎢⎣ ⎦ PV = nRT so V = nRT/P = = 1.14635 = 1 L CO2 1.00 atm

13.167 a) Scene C represents the system at the higher temperature of 298 K. At the higher temperature, the solubility of oxygen decreases so more oxygen leaves the solution to go into the vapor phase. b) Scene B represents the system when the pressure of oxygen is increased by half. The increase in pressure would result in 4 + ½(4) = 6 moles of oxygen in the vapor phase. The increased pressure results in increased solubility of oxygen in the water. Of the six moles of oxygen in the vapor phase, one mole dissolves in the water to bring the dissolved moles to three.

13-47

CHAPTER 14 PERIODIC PATTERNS IN THE MAIN-GROUP ELEMENTS END–OF–CHAPTER PROBLEMS 14.1

Ionization energy is defined as the energy required to remove the outermost electron from an atom. The further the outermost electron is from the nucleus, the less energy is required to remove it from the attractive force of the nucleus. In hydrogen, the outermost electron is in the n = 1 level and in lithium the outermost electron is in the n = 2 level. Therefore, the outermost electron in lithium requires less energy to remove, resulting in a lower ionization energy.

14.2

14.3

Plan: Recall that to form hydrogen bonds a compound must have H directly bonded to either N, O, or F. Solution: a) NH3 will hydrogen bond because H is bonded to N.

b) CH3CH2OH will hydrogen bond since H is bonded to O. CH3OCH3 has no O−H bonds, only C−H bonds. H H H H H

C H

CH3OCH3

H H CH3CH2OH

14-1

14.4

a) NH3 will form hydrogen bonds.

H b) H2O will form hydrogen bonds. H H

O H

14.5

Plan: Active metals displace hydrogen from HCl by reducing the H+ to H2. In water, H– (here in LiH) reacts as a strong base to form H2 and OH–. Solution: a) 2Al(s) + 6HCl(aq) → 2AlCl3(aq) + 3H2(g) b) LiH(s) + H2O(l) → LiOH(aq) + H2(g)

14.6

a) CaH2(s) + 2H2O(l) → Ca(OH)2(aq) + 2H2(g) b) PdCl2(aq) + H2(g) → Pd(s) + 2HCl(aq)

14.7

Plan: In metal hydrides, the oxidation state of hydrogen is –1. Solution: a) Na = +1 B = +3 H = –1 in NaBH4 Al = +3 B = +3 H = –1 in Al(BH4)3 Li = +1 Al = +3 H = –1 in LiAlH4 b) The polyatomic ion in NaBH4 is [BH4]–. There are [1 × B(3e–)] + [4 × H(1e–)] + [1e– from charge] = 8 valence electrons. All eight electrons are required to form the four bonds from the four hydrogen atoms to the boron atom. Boron is the central atom and has four surrounding electron groups; therefore, its shape is tetrahedral. H H

14.8

Since the nucleus of H contains only one proton, the electrons are not very tightly held and the H– ion will be very polarizable (i.e., its electron cloud can be very easily distorted by a neighboring ion). Stated differently, there will be different amounts of covalent character in the different compounds.

14.9

In general, the maximum oxidation number increases as you move to the right. (Max. O.N. = (old) group number) In the second period, the maximum oxidation number drops off below the group number in Groups 6A(16) and 7A(17).

14.10

For Period 2 elements in the first four groups, the number of covalent bonds equals the number of electrons in the outer level, so it increases from one covalent bond for lithium in Group 1A(1) to four covalent bonds for carbon in Group 4A(14). For the rest of Period 2 elements, the number of covalent bonds equals the difference between 8 and the number of electrons in the outer level. So for nitrogen, 8 – 5 = 3 covalent bonds; for oxygen, 8 – 6 = 2 covalent bonds; for fluorine, 8 – 7 = 1 covalent bond; and for neon, 8 – 8 = 0, no bonds. For elements in higher periods, the same pattern exists but with exceptions for Groups 3A(13) to 7A(17) when an expanded octet allows for more covalent bonds.

14-2

14.11

a) lithium fluoride, LiF beryllium fluoride, BeF2 boron trifluoride, BF3 nitrogen trifluoride, NF3 oxygen difluoride, OF2 fluorine, F2 b) ΔEN decreases left to right across the period. c) % ionic character decreases left to right across the period. d)

carbon tetrafluoride, CF4

F F

14.12

a) 32 elements, 30 metals b) between Po and At

14.13

a) E must have an oxidation of +3 to form an oxide E2O3 or fluoride EF3. E is in Group 3A(13) or 3B(3). b) If E were in Group 3B(3), the oxide and fluoride would have more ionic character because 3B elements have lower electronegativity than 3A elements. The Group 3B(3) oxides would be more basic.

14.14

Oxygen and fluorine have almost filled outer shells (2s22p4 and 2s22p5, respectively), so they both have a great ability to attract and hold bonded electrons (i.e., a large electronegativity). Neon, on the other hand, has a filled outer shell (2s22p6), so has little desire to hold additional electrons, and has essentially a zero electronegativity.

14.15

The small size of Li+ leads to a high charge density and thus to a large lattice energy for LiF, which lowers its solubility since the dissociation of LiF into ions is more difficult than for KF.

14.16

a) Alkali metals generally lose electrons (act as reducing agents) in their reactions. b) Alkali metals have relatively low ionization energies, meaning they easily lose the outermost electron. The electron configurations of alkali metals have one more electron than a noble gas configuration, so losing an electron gives a stable electron configuration. c) 2Na(s) + 2H2O(l) → 2Na+(aq) + 2OH–(aq) + H2(g) 2Na(s) + Cl2(g) → 2NaCl(s)

14.17

The large atomic radii of the Group 1A(1) elements mean that their atomic volumes are large. Since density = mass/volume, the densities will be small.

14.18

a) Density increases down a group. The increasing atomic size (volume) is not offset by the increasing size of the nucleus (mass), so m/V increases. b) Ionic size increases down a group. Electron shells are added down a group, so both atomic and ionic size increase. c) E−E bond energy decreases down a group. Shielding of the outer electron increases as the atom gets larger, so the attraction responsible for the E−E bond decreases. d) IE1 decreases down a group. Increased shielding of the outer electron is the cause of the decreasing IE1. e) ΔHhydr decreases down a group. ΔHhydr is the heat released when the metal salt dissolves in, or is hydrated by, water. Hydration energy decreases as ionic size increases. Increasing down: a and b; Decreasing down: c, d, and e

14.19

Increasing up the group: a, c, and e Decreasing up the group: b and d

14-3

14.20

Plan: Peroxides are oxides in which oxygen has a –1 oxidation state. Sodium peroxide has the formula Na2O2 and is formed from the elements Na and O2. Solution: 2Na(s) + O2(g) → Na2O2(s)

14.21

RbOH(aq) + HBr(aq) → RbBr(aq) + H2O(l)

14.22

Plan: The problem specifies that an alkali halide is the desired product. The alkali metal is K (comes from potassium carbonate, K2CO3(s)) and the halide is I (comes from hydroiodic acid, HI(aq)). Treat the reaction as a double displacement reaction. Solution: K2CO3(s) + 2HI(aq) → 2KI(aq) + H2CO3(aq) However, H2CO3(aq) is unstable and decomposes to H2O(l) and CO2(g), so the final reaction is: K2CO3(s) + 2HI(aq) → 2KI(aq) + H2O(l) + CO2(g)

14.23

a) % Li =

b) % Li =

6.941 g Li/mol 290.40 g/mol

6.941 g Li/mol 64.05 g/mol

× 100% = 2.39015 = 2.390% Li

× 100% = 10.8368 = 10.84% Li

14.24

The Group 1A(1) elements react more vigorously with water than those in Group 2A(2).

14.25

a) Li/Mg and Be/Al b) Li and Mg both form ionic nitrides and thermally unstable carbonates. Be and Al both form amphoteric oxides; their oxide coatings make both metals unreactive to water. c) The charge density (i.e., charge/radius ratio) is similar.

14.26

Metal atoms are held together by metallic bonding, a sharing of valence electrons. Alkaline earth metal atoms have one more valence electron than alkali metal atoms, so the number of electrons shared is greater. Thus, metallic bonds in alkaline earth metals are stronger than in alkali metals. Melting requires overcoming the metallic bonds. To overcome the stronger alkaline earth metal bonds requires more energy (higher temperature) than to overcome the alkali earth metal bonds. First ionization energy, density, and boiling points will be larger for alkaline earth metals than for alkali metals.

14.27

Plan: A base forms when a basic oxide, such as CaO (lime), is added to water. Alkaline earth metals reduce O2 to form the oxide. Solution: a) CaO(s) + H2O(l) → Ca(OH)2(s) b) 2Ca(s) + O2(g) → 2CaO(s)

14.28

14.29

14.30

Δ a) BaCO3(s) ⎯⎯→ BaO(s) + CO2(g) b) Mg(OH)2(s) + 2HCl(aq) → MgCl2(aq) + 2H2O(l) Δ (CaCO3 from limestone) a) CaCO3(s) ⎯⎯→ CaO(s) + CO2(g) b) Ca(OH)2(s) + SO2(g) → CaSO3(s) + H2O(l) c) 3CaO(s) + 2H3AsO4(aq) → Ca3(AsO4)2(s) + 3H2O(l) d) Na2CO3(aq) + CaO(s) + H2O(l) → CaCO3(s) + 2NaOH(aq)

Plan: The oxides of alkaline earth metals are strongly basic, but BeO is amphoteric. BeO will react with both acids and bases to form salts, but an amphoteric substance does not react with water. In part (b), each chloride ion donates a lone pair of electrons to form a covalent bond with the Be in BeCl2. Metal ions form similar covalent

14-4

bonds with ions or molecules containing a lone pair of electrons. The difference in beryllium is that the orbital involved in the bonding is a p orbital, whereas in metal ions it is usually the d orbitals that are involved. Solution: a) Here, Be does not behave like other alkaline earth metals: BeO(s) + H2O(l) →NR. b) Here, Be does behave like other alkaline earth metals: BeCl2(l) + 2 Cl–(solvated) → BeCl42–(solvated) 14.31

The pattern of ionization energies in Group 3A(13) is irregular; there is not a smooth decrease in ionization energy as you proceed down the group. This is due to the appearance of the transition metals (and the ten additional protons in the nucleus) preceding Ga, In, and Tl. The presence of the transition elements causes a contraction of the atoms and a resulting increase in ionization energy for these three elements. There is a smoother decrease in ionization energy for the elements in Group 3B(3).

14.32

Tl2O is more basic (i.e., less acidic) than Tl2O3. Acidity increases with increasing oxidation number.

14.33

The electron removed in Group 2A(2) atoms is from the outer level s orbital, whereas in Group 3A(13) atoms the electron is from the outer level p orbital. For example, the electron configuration for Be is 1s22s2 and for B is 1s22s22p1. It is easier to remove the p electron of B than the s electron of Be, because the energy of a p orbital is slightly higher than that of the s orbital from the same level. Even though the atomic size decreases from increasing Zeff, the IE decreases from Group 2A(2) to 3A(13).

14.34

a) Compounds of Group 3A(13) elements, like boron, have only six electrons in their valence shell when combined with halogens to form three bonds. Having six electrons, rather than an octet, results in an “electron deficiency.” b) As an electron deficient central atom, B is trigonal planar. Upon accepting an electron pair to form a bond, the shape changes to tetrahedral. BF3(g) + NH3(g) → F3B–NH3(g) B(OH)3(aq) + OH–(aq) → B(OH)4–(aq)

14.35

a) Boron is a metalloid, while the other elements in the group show predominately metallic behavior. It forms covalent bonds exclusively; the others at best occasionally form ions. It is also much less chemically reactive in general. b) The small size of B is responsible for these differences.

14.36

Plan: Oxide acidity increases up a group; the less metallic an element, the more acidic is its oxide. Solution: In2O3 < Ga2O3 < Al2O3

14.37

B(OH)3 < Al(OH)3 < In(OH)3

14.38

Halogens typically have a –1 oxidation state in metal-halide combinations, so the apparent oxidation state of Tl = +3. However, the anion I3– combines with Tl in the +1 oxidation state. The anion I3– has [3 × (I)7e–] + [1e– from the charge] = 22 valence electrons; four of these electrons are used to form the two single bonds between iodine atoms and sixteen electrons are used to give every atom an octet. The remaining two electrons belong to the central I atom; therefore the central iodine has five electron groups (two single bonds and three lone pairs) and has a general formula of AX2E3. The electrons are arranged in a trigonal bipyramidal with the three lone pairs in the trigonal plane. It is a linear ion with bond angles = 180°. (Tl3+) (I–)3 does not exist because of the low strength of the Tl–I bond. O.N. = +3 (apparent); = +1 (actual)

14-5

14.39

O.N. = +2 (apparent); = +1 (Ga+) and +3 (GaCl4–) (actual) class = AX4; bond angles = 109.5°; tetrahedral

Cl Cl

14.40

a) boron, B d) aluminum, Al

b) gallium, Ga e) thallium, Tl

c) boron, B

14.41

a) In: [Kr]4d105s25p1 In+: [Kr]4d105s2 In2+: [Kr]4d105s1 + 3+ 2+ b) In and In are diamagnetic while In and In are paramagnetic. c) Apparent oxidation state is 2+. d) There can be no In2+ present. Half the indium is In+ and half is In3+.

In3+: [Kr]4d10

14.42

H H

H O

O H

B(OH)3 has 120° angles around B 14.43

B(OH)4– has 109.5° angles around B

ο ο ο Plan: To calculate the enthalpy of reaction, use the relationship ΔH rxn = ∑m ΔH products – ∑n ΔH reactants .

Convert the given amount of 1.0 kg of BN to moles, find the moles of B in that amount of BN, and then find the moles and then mass of borax that provides that number of moles of B. Solution: a) B2O3(s) + 2NH3(g) → 2BN(s) + 3H2O(g) ο ο ο = ∑m ΔH products – ∑n ΔH reactants b) ΔH rxn

= {2 ΔH fο [BN(s)] + 3 ΔH fο [H2O(g)]} – {1 ΔH fο [B2O3(s)] + 2 ΔH fο [NH3(g)]} = [(2 mol)(–254 kJ/mol) + (3 mol)(–241.826 mol kJ/mol)] – [(1 mol)(–1272 kJ/mol) + (2 mol)(–45.9 kJ/mol)] = 130.322 = 1.30 × 102 kJ c) Mass (g) of borax = ⎛ 3 ⎞⎛ ⎟⎜ 1 mol B ⎞⎛ ⎟⎜1 mol Na 2 B4 O 7 i10 H 2 O ⎞⎛ ⎟⎜ 381.38 g ⎞⎟⎜ ⎟⎛100% ⎞⎟ ⎜10 g ⎟⎟⎜⎜ 1 mol BN ⎞⎛ ⎜ 1.0 kg BN ⎜ ⎟ ⎟⎟⎜ ⎟⎟⎟⎜⎜ ⎟⎟⎟⎜⎜ ⎟⎟⎟⎜⎜ ⎟⎜⎜ ⎟ ⎜⎜ 1 kg ⎟⎜⎜ 24.82 g BN ⎟⎜⎜1 mol BN ⎟⎜⎜ ⎜ 4 mol B ⎟⎜ 1 mol ⎠⎟⎟⎝ 72% ⎠⎟ ⎠⎝ ⎠⎝ ⎠⎝ ⎝ ⎠⎝

(

)

= 5.335359 × 103 = 5.3 × 103 g borax 14.44

Oxide basicity is greater for the oxide of a metal atom. Tin(IV) oxide is more basic than carbon dioxide since tin has more metallic character than carbon.

14-6

14.45

a) The increased stability of the lower oxidation state as one goes down a group. b) As the atoms become larger, the strength of the bonds to other elements becomes weaker, and insufficient energy is gained in forming the bonds to offset the additional ionization or promotion energy. c) Tl+ is more stable than Tl3+, but Al3+ is the only stable oxidation state for Al.

14.46

a) IE1 values generally decrease down a group. b) The increase in Zeff from Si to Ge is larger than the increase from C to Si because more protons have been added. Between C and Si an additional eight protons have been added, whereas between Si and Ge an additional eighteen (includes the protons for the d-block) protons have been added. The same type of change takes place when going from Sn to Pb, when the fourteen f-block protons are added. c) Group 3A(13) would show greater deviations because the single p electron receives no shielding effect offered by other p electrons.

14.47

The drop between C and Si is due to a weakening of the bonds due to increased atomic size. The drop between Ge and Sn is due to a change in bonding from covalent to metallic.

14.48

Allotropes are two forms of a chemical element which have different bonding and physical properties. C forms graphite, diamond, and buckminsterfullerene; Sn has gray (α) and white (β) forms.

14.49

Atomic size increases moving down a group. As atomic size increases, ionization energy decreases so that it is easier to form a positive ion. An atom that is easier to ionize exhibits greater metallic character.

14.50

Having four valence electrons allows all of the Group 4A(14) elements to form a large number of bonds, hence, many compounds. However, the small size of the C atom makes its bonds stronger and gives stability to a wider variety of compounds than for the heavier members of the group.

14.51

Plan: The silicate building unit is —SiO4—. There are [4 × Si(4e–)] + [12 × O(6e–)] + [8e– from charge] = 96 valence electrons in Si4O128–. Thirty-two electrons are required to form the sixteen bonds in the ion; the remaining 96 – 32 = 64 electrons are required to complete the octets of the oxygen atoms. In C2H4, there are [2 × C(4e–)] + [4 × H(1e–)] = 12 valence electrons. All 12 electrons are used to form the bonds between the atoms in the molecule. Solution: a)

There is another answer possible for C4H8.

14-7

14.52

There are numerous alternate answers for C6H12. a) O O

Si O

O O O

H C

H H

C H

C C

H H H

O Si

O O

14.53

Each alkaline earth metal ion will displace two sodium ions because of the charge difference. Determine the moles of alkaline earth metal ions: Moles of Ca2+ = (4.5 × 10–3 mol/L)(25,000 L) = 112.5 mol Ca2+ Moles of Mg2+ = (9.2 × 10–4 mol/L)(25,000 L) = 23 mol Mg2+ Total moles of M2+ = (112.5 + 23) mol = 135.5 mol M2+ Determine the moles of Na+ needed: Moles Na+ = (135.5 mol M2+)(2 mol Na+/1 mol M2+) = 271 mol Na+ Determine the molar mass of the zeolite: 12 Na(22.99 g/mol) + 12 Al(26.98 g/mol) + 12 Si(28.09 g/mol) + 54 H(1.008 g/mol) + 75 O(16.00 g/mol) = 2191.15 g/mol Determine mass of zeolite: Mass = (271 mol Na+)(1 mol zeolite/12 mol Na+)(2191.15 g zeolite/mol zeolite)(1 kg/103 g)(100%/85%) = 58.215848 = 58 kg zeolite

14.54

a) Diamond, C, a network covalent solid of carbon b) Calcium carbonate, CaCO3 (Brands that use this compound as an antacid also advertise them as an important source of calcium.) c) Carbon dioxide, CO2, is the most widely known greenhouse gas; CH4 is also implicated. d) Carbon monoxide, CO, is formed in combustion when the amount of O2 (air) is limited. e) Silicon, Si

14.55

B2H6(g) + 3O2(g) → B2O3(s) + 3H2O(g) 2Si4H10(g) + 13O2(g) → 8SiO2(s) + 10H2O(g)

14.56

All of the elements in Group 5A(15) form trihalides, but only P, As, and Sb form pentahalides. N cannot expand its octet, so it cannot form a pentahalide. The large Bi atom forms weak bonds, so it is unfavorable energetically for it to form five bonds, except with fluorine. It would also require too much energy to remove five electrons.

14.57

The bonding changes from covalent bonding in small molecules (N, P), to molecules with network covalent bonding (As, Sb), to metallic bonding in Bi. The first two elements (N, P) are nonmetals, followed by two metalloid elements (As, Sb), and then by a metallic element (Bi).

14.58

a) In Group 5A(15), all elements except bismuth have a range of oxidation states from –3 to +5. b) For nonmetals, the range of oxidation states is from the lowest at group number (A) – 8, which is 5 – 8 = –3 for Group 5A, to the highest equal to the group number (A), which is +5 for Group 5A.

14-8

14.59

In general, high oxidation states are less stable toward the bottom of the periodic table.

14.60

Bi2O3 < Sb2O3 < Sb2O5 < P4O10

14.61

Plan: Acid strength increases with increasing electronegativity of the central atom. Arsenic is less electronegative than phosphorus, which is less electronegative than nitrogen. Solution: Arsenic acid is the weakest acid and nitric acid is the strongest. Order of increasing strength: H3AsO4 < H3PO4 < HNO3

14.62

HNO3 > HNO2 > H2N2O2

14.63

Plan: With excess oxygen, arsenic will form the oxide with arsenic in its highest possible oxidation state, +5. Trihalides are formed by direct combination of the elements (except N). Metal phosphides, arsenides, and antimonides react with water to form Group 5A hydrides. Solution: a) 4 As(s) + 5O2(g) → 2As2O5(s) b) 2Bi(s) + 3F2(g) → 2BiF3(s) c) Ca3As2(s) + 6H2O(l) → 3Ca(OH)2(s) + 2AsH3(g)

14.64

a) 2Sb(s) + 3Br2(l) → 2SbBr3(s) b) 2HNO3(aq) + MgCO3(s) → Mg(NO3)2(aq) + CO2(g) + H2O(l) Δ c) 2K2HPO4(s) ⎯⎯→ K4P2O7(s) + H2O(g)

14.65

a) Aluminum is not as active a metal as Li or Mg, so heat is needed to drive this reaction. Δ N2(g) + 2Al(s) ⎯⎯→ 2AlN(s) b) The Group 5A halides react with water to form the oxoacid with the same oxidation state as the original halide. PF5(g) + 4H2O(l) → H3PO4(aq) + 5HF(g)

14.66

a) AsCl3(l) + 3H2O(l) → H3AsO3(aq) + 3HCl(g) b) Sb2O3(s) + 6NaOH(aq) → 2Na3SbO3(aq) + 3H2O(l)

14.67

Plan: There are [1 × P(5e–)] + [2 × F(7e–)] + [3 × Cl(7e–)] = 40 valence electrons in PF2Cl3. Ten electrons are required to form the five bonds between F or Cl to P; the remaining 40 – 10 = 30 electrons are required to complete the octets of the fluorine and chlorine atoms. From the Lewis structure, the phosphorus has five electron groups for a trigonal bipyramidal molecular shape. In this shape, the three groups in the equatorial plane have greater bond angles (120°) than the two groups above and below this plane (90°). The chlorine atoms would occupy the planar sites where there is more space for the larger atoms. Solution:

P Cl F

14-9

14.68

The structure would be tetrahedral at the P atoms and bent at the O atoms.

3 O

O P O

O O

O P

P O

14.69

F F

N F

b) Tetraphosphorus trisulfide, P4S3 d) Nitrogen monoxide, NO; nitrogen dioxide, NO2

14.70

a) Ammonia, NH3 c) Tetraphosphorus decaoxide, P4O10 e) Phosphoric acid, H3PO4

14.71

Plan: Set the atoms into the positions described, and then complete the Lewis structures. a) and b) N2O2 has [2 × N(5e–)] + [2 × O(6e– )] = 22 valence electrons. Six of these electrons are used to make the single bonds between the atoms, leaving 22 – 6 = 16 electrons. Since 20 electrons are needed to complete the octets of all of the atoms, two double bonds are needed. c) N2O3 has [2 × N(5e–)] + [3 × O(6e–)] = 28 valence electrons. Eight of these electrons are used to make the single bonds between the atoms, leaving 28 – 8 = 20 electrons. Since 24 electrons are needed to complete the octets of all of the atoms, two double bonds are needed. d) NO+ has [1 × N(5e–)] + [1 × O(6e–)] – [1e– (due to the + charge)] = 10 valence electrons. Two of these electrons are used to make the single bond between the atoms, leaving 10 – 2 = 8 electrons. Since 12 electrons are needed to complete the octets of both atoms, a triple bond is needed. NO3– has [1 × N(5e–)] + [3 × O(6e–) + [1 e– (due to the – charge)] = 24 valence electrons. Six of these electrons are used to make the single bond between the atoms, leaving 24 – 6 = 18 electrons. Since 20 electrons are needed to complete the octets of all of the atoms, a double bond is needed. Solution: b) a) O

N O

N N

N O

14-10

Δ NH4NO3(s) ⎯⎯→ N2O(g) +2H2O(l)

14.72 O.N.: 14.73

–3 +5

a) Thermal decomposition of KNO3 at low temperatures: Δ 2KNO3(s) ⎯⎯→ 2KNO2(s) + O2(g) b) Thermal decomposition of KNO3 at high temperatures: Δ 4KNO3(s) ⎯⎯→ 2K2O(s) + 2N2(g) + 5O2(g)

14.74

S < Se < Po; nonmetal < metalloid < metal

14.75

a) Both groups have elements that range from gas to metalloid to metal. Thus, their boiling points and conductivity vary in similar ways down a group. b) The degree of metallic character and methods of bonding vary in similar ways down a group. c) Both P and S have allotropes and both bond covalently with almost every other nonmetal. d) Both N and O are diatomic gases at normal temperatures and pressures. Both N and O have very low melting and boiling points. e) Oxygen, O2, is a reactive gas whereas nitrogen, N2, is not. Nitrogen can exist in multiple oxidation states, whereas oxygen has two oxidation states.

14.76

a) The change occurs between Periods 2 and 3. b) The H–E–H bond angle changes. c) The hybridization changes from sp3 in H2O to p (unhybridized) in the others. d) Group 5A(15) is similar.

14.77

a) To decide what type of reaction will occur, examine the reactants. Notice that sodium hydroxide is a strong base. Is the other reactant an acid? If we separate the salt, sodium hydrogen sulfate, into the two ions, Na+ and HSO4–, then it is easier to see the hydrogen sulfate ion as the acid. The sodium ions could be left out for the net ionic reaction. NaHSO4(aq) + NaOH(aq) → Na2SO4(aq) + H2O(l) b) As mentioned in the book, hexafluorides are known to exist for sulfur. These will form when excess fluorine is present. S8(s) + 24F2(g) → 8SF6(g) c) Group 6A(16) elements, except oxygen, form hydrides in the following reaction. FeS(s) + 2HCl(aq) → H2S(g) + FeCl2(aq) d) Tetraiodides, but not hexaiodides, of tellurium are known. Te(s) + 2I2(s) → TeI4(s)

14.78

a) 2H2S(g) + 3O2(g) → 2SO2(g) + 2H2O(g) b) SO3(g) + H2O(l) → H2SO4(l) c) SF4(g) + 2H2O(l) → SO2(g) + 4HF(g) d) Al2Se3(s) + 6H2O(l) → 2Al(OH)3(s) + 3H2Se(g)

14.79

Plan: The oxides of nonmetal elements are acidic, while the oxides metal elements are basic. Solution: a) Se is a nonmetal; its oxide is acidic. b) N is a nonmetal; its oxide is acidic. c) K is a metal; its oxide is basic. d) Be is an alkaline earth metal, but all of its bonds are covalent; its oxide is amphoteric. e) Ba is a metal; its oxide is basic.

14.80

a) basic

b) acidic

c) basic

d) acidic

e) amphoteric

14-11

14.81

Plan: Acid strength of binary acids increases down a group since bond energy decreases down the group. Solution: H2O < H2S < H2Te

14.82

H2SO4 > H2SO3 > HSO3–

14.83

When solid sulfur is heated, it melts at about 115°C to a mobile liquid consisting of S8 molecules. Above 150°C, the rings begin to open and become tangled, increasing the viscosity of the liquid, and causing a darkening of the liquid. At about 180°C, the dark brown mass has its highest viscosity. At higher temperatures, the chains break and untangle, decreasing the viscosity until the liquid boils at 444°C (above the end point specified). If the heated liquid (above 300°C) is poured into water, a rubbery mass (“plastic” sulfur) forms, which consists of short, tangled chains; left at room temperature for a few days, it reverts to the original crystalline solid containing S8 molecules.

14.84

a) O3, ozone b) SO3, sulfur trioxide (+6 oxidation state) c) SO2, sulfur dioxide d) H2SO4, sulfuric acid e) Na2S2O3 ∙ 5H2O, sodium thiosulfate pentahydrate

14.85

a) 0

14.86

S2F10(g) → SF4(g) + SF6(g) O.N. of S in S2F10: – (10 × –1 for F)/2 = +5 O.N. of S in SF4: – (4 × –1 for F) = +4 O.N. of S in SF6: – (6 × –1 for F) = +6

14.87

a) F2 is a pale yellow gas; Cl2 is a green gas; Br2 is a red-orange liquid; I2 is a purple-black solid. b) As the mass of the molecules increases, the strength of the dispersion forces will increase as well, and the melting and boiling points will parallel this trend by increasing with increasing molar mass.

14.88

a) Bonding with very electronegative elements: +1, +3, +5, +7. Bonding with other elements: –1 b) The electron configuration for Cl is [Ne]3s23p5. By adding one electron to form Cl–, Cl achieves an octet similar to the noble gas Ar. By forming covalent bonds, Cl completes or expands its octet by maintaining its electrons paired in bonds or lone pairs. c) Fluorine only forms the –1 oxidation state because its small size and no access to d orbitals prevent it from forming multiple covalent bonds. Fluorine’s high electronegativity also prevents it from sharing its electrons.

14.89

The halogens need one electron to complete their octets. This can be accomplished by gaining one electron (to form Cl–) or by sharing a pair of electrons to form one covalent bond (as in HCl or CCl4).

14.90

a) The Cl–Cl bond is stronger than the Br–Br bond since the chlorine atoms are smaller than the bromine atoms, so the shared electrons are held more tightly by the two nuclei. b) The Br–Br bond is stronger than the I–I bond since the bromine atoms are smaller than the iodine atoms. c) The Cl–Cl bond is stronger than the F–F bond. The fluorine atoms are smaller than the chlorine but they are so small that electron-electron repulsion of the lone pairs decreases the strength of the bond.

14.91

You would expect them to contain an odd number of atoms, so that you would have an even number of electrons.

14.92

a) A substance that disproportionates serves as both an oxidizing and reducing agent. Assume that OH– serves as the base. Write the reactants and products of the reaction, and balance like a redox reaction. 3 Br2(l) + 6OH–(aq) → 5Br–(aq) + BrO3–(aq) + 3H2O(l) b) In the presence of base, instead of water, only the oxyanion (not oxoacid) and fluoride (not hydrofluoride) form. No oxidation or reduction takes place, because Cl maintains its +5 oxidation state and F maintains its –1 oxidation state. ClF5(l) + 6OH–(aq) → 5F–(aq) + ClO3–(aq) + 3H2O(l)

b) +4

c) +6

d) –2

e) –1

f) +6

g) +2

14-12

14.93

a) 2Rb(s) + Br2(l) → 2RbBr(s) b) I2(s) + H2O(l) → HI(aq) + HIO(aq) c) Br2(l) + 2I – (aq) → I2(s) + 2Br–(aq) d) CaF2(s) + H2SO4(l) → CaSO4(s) + 2HF(g)

14.94

a) H3PO4(l) + NaI(s) → NaH2PO4(s) + HI(g) b) Cl2(g) + 2I–(aq) → 2Cl–(aq) + I2(s) c) Br2(l) + Cl–(aq) → NR d) ClF(g) + F2(g) → ClF3(g)

14.95

Plan: Acid strength increases with increasing electronegativity of the central atom and increasing number of oxygen atoms. Solution: Iodine is less electronegative than bromine, which is less electronegative than chlorine. HIO < HBrO < HClO < HClO2

14.96

HClO4 > HBrO4 > HBrO3 > HIO3

14.97

a) hydrogen fluoride, HF c) hydrofluoric acid, HF e) vinyl chloride, C2H3Cl

14.98

a) In the reaction between NaI and H2SO4 the oxidation states of iodine and sulfur change, so the reaction is an oxidation-reduction reaction. b) The reducing ability of X– increases down the group since the larger the ion the more easily it loses an electron. Therefore, I– is more easily oxidized than Cl–. c) Some acids, such as HCl, are not oxidizers, so substituting a nonoxidizing acid for H2SO4 would produce HI.

14.99

I2 < Br2 < Cl2, since Cl2 is able to oxidize Re to the +6 oxidation state, Br2 only to +5, and I2 only to +4.

b) sodium hypochlorite, NaClO d) bromine, Br2

14.100 Helium is the second most abundant element in the universe. Argon is the most abundant noble gas in Earth’s atmosphere, the third most abundant constituent after N2 and O2. 14.101 +2, +4, +6, +8 14.102 Whether a boiling point is high or low is a result of the strength of the forces between particles. Dispersion forces, the weakest of all the intermolecular forces, hold atoms of noble gases together. Only a relatively low temperature is required for the atoms to have enough kinetic energy to break away from the attractive force of other atoms and go into the gas phase. The boiling points are so low that all the noble gases are gases at room temperature. 14.103 The electrons on the larger atoms are more easily removed, transferred, or shared with another atom than those on the smaller atoms. 14.104 a) This allows the resulting molecules to have an even number of electrons. b) Xenon fluorides with an odd charge must have an odd number of fluorine atoms to maintain an even number of electrons around xenon. c) XeF3+ would be T shaped.

14.105 a) Xenon tetrafluoride, XeF4, is an AX4E2 molecule with square planar geometry. Antimony pentafluoride, SbF5, is an AX5 molecule with trigonal bipyramidal molecule geometry. XeF3+ is an AX3E2 ion with a T-shaped Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

14-13

geometry and SbF6– is an AX6 ion with octahedral molecular geometry. C best shows the molecular geometries of these substances. b) Xe in XeF4 utilizes sp3d2 hybrid orbitals; in XeF3+, xenon utilizes sp3d hybrid orbitals. 14.106 Plan: To obtain the overall reaction, reverse the first reaction and the third reaction and add these two reactions to the second reaction, canceling substances that appear on both sides of the arrow. When a reaction is reversed, the sign of its enthalpy change is reversed. Add the three enthalpy values to obtain the overall enthalpy value. Solution: H3O+(g) → H+(g) + H2O(g) ΔH = +720 kJ

Overall:

H+(g) + H2O(l) → H3O+(aq)

ΔH = –1090 kJ

H2O(g) → H2O(l)

ΔH = –40.7 kJ

H3O (g) → H3O (aq)

ΔH = –410.7 = –411 kJ

14.107 Calculation of energy from wavelength: ΔE = hc/λ. 6.626 ×10−34 J • s 2.9979 ×108 m/s ⎛⎜ 1 nm ⎞⎟ ⎟⎟ = 3.3713655 × 10–19 = 3.371 × 10–19 J ⎜⎜ ΔE = ⎜⎜10−9 m ⎟⎟ 589.2 nm ⎝ ⎠

(

)(

(

)

14.108 Be(s) + 2NaOH(aq) + 2H2O(l) → Na2Be(OH)4(aq) + H2(g) Zn(s) + 2NaOH(aq) + 2H2O(l) → Na2Zn(OH)4(aq) + H2(g) 2Al(s) + 2NaOH(aq) + 6H2O(l) → 2NaAl(OH)4(aq) + 3H2(g) 14.109 a) 5IF → IF5 + 2I2 b) Iodine pentafluoride c) This is a disproportionation redox reaction. IF acts as both the oxidizing and reducing agents. ⎛ 2.50 × 10−3 mol IF ⎞⎛ 1 mol IF5 ⎞⎟⎜⎛ 221.9 g IF5 ⎞⎟ ⎟ = 0.77665 = 0.777 g IF5 ⎟⎟⎜⎜ ⎟⎜ d) Mass (g) of IF5 = (7 IF molecules)⎜⎜⎜ ⎟ ⎜ ⎜⎝ 1 IF molecule ⎠⎟⎝ 5 mol IF ⎠⎟⎟⎜⎜⎝ 1 mol IF5 ⎠⎟⎟ ⎛ 2.50 × 10−3 mol IF ⎞⎛ 2 mol I 2 ⎞⎟⎜⎛ 253.8 g I 2 ⎞⎟ ⎟⎟ = 1.7766 = 1.78 g I2 ⎟⎟⎟⎜⎜ ⎟⎟⎜ Mass (g) of I2 = (7 IF molecules)⎜⎜⎜ ⎜ ⎝⎜ 1 IF molecule ⎠⎟⎜⎝ 5 mol IF ⎠⎟⎜⎝⎜ 1 mol I 2 ⎠⎟ ο 14.110 Plan: Examine the outer electron configuration of the alkali metals. To calculate the ΔH rxn in part (b), use Hess’s law. Solution: a) Alkali metals have an outer electron configuration of ns1. The first electron lost by the metal is the ns electron, giving the metal a noble gas configuration. Second ionization energies for alkali metals are high because the electron being removed is from the next lower energy level and electrons in a lower level are more tightly held by the nucleus. The metal would also lose its noble gas configuration. b) The reaction is 2CsF2(s) → 2CsF(s) + F2(g). You know the ΔH fο for the formation of CsF:

Cs(s) + 1/2F2(g) → CsF(s)

ΔH fο = –530 kJ/mol

You also know the ΔH fο for the formation of CsF2: ΔH fο = –125 kJ/mol Cs(s) + F2(g) → CsF2(s) To obtain the heat of reaction for the breakdown of CsF2 to CsF, combine the formation reaction of CsF with the reverse of the formation reaction of CsF2, both multiplied by 2: 2Cs(s) + F2(g) → 2CsF(s) ΔH fο = 2 × (–530 kJ) = –1060 kJ

2CsF2(s) → 2Cs(s) + 2F2(g)

ΔH fο = 2 × (+125 kJ) = 250 kJ (Note sign change)

2 CsF2(s) → 2CsF(s) + F2(g)

ο ΔH rxn = –810 kJ

14-14

810 kJ of energy are released when two moles of CsF2 convert to two moles of CsF, so heat of reaction for one mole of CsF is –810/2 or –405 kJ/mol. 14.111 4Ga(l) + P4(g) → 4GaP(s) Find the limiting reagent: ⎛ 1 mol Ga ⎞⎟⎛ 4 mol GaP ⎞⎛100.69 g GaP ⎞ ⎟⎜ ⎟ Mass (g) of GaP from Ga = (32.5 g Ga )⎜⎜ ⎟⎟⎜ ⎟⎟⎜ 1 mol GaP ⎠⎟⎟ = 46.9367 g GaP ⎝⎜ 69.72 g Ga ⎠⎟⎜⎝ 4 mol Ga ⎠⎝

Moles of P4 = n =

(195 kPa)(20.4 L) ⎛⎜⎜ 1 atm ⎞⎟⎟ = 0.928534 mol P PV = ⎟ ⎜⎜ 4 L • atm ⎞⎟ ⎛ RT ⎜⎝101.325 kPa ⎠⎟⎟ ⎜⎜0.0821 515 K ( ) ⎟ ⎝ mol • K ⎠

⎛ 4 mol GaP ⎞⎟⎛100.69 g GaP ⎞ ⎟ = 373.976 g GaP ⎟⎜ Mass (g) of GaP from P4 = (0.928534 mol P4 )⎜⎜ ⎜⎝ 1 mol P4 ⎠⎟⎟⎝⎜ 1 mol GaP ⎠⎟⎟ Since a smaller amount of product is obtained with Ga, Ga is the limiting reactant. Assuming 100% yield, 46.9367 g of GaP would be produced. Accounting for a loss of 7.2% by mass or a 100.0 – 7.2 = 92.8% yield: 46.9367 g GaP × 0.928 = 43.5573 = 43.6 g GaP

14.112 Plan: To find the molecular formula, divide the molar mass of each compound by the molar mass of the empirical formula, HNO. The result of this gives the factor by which the empirical formula is multiplied to obtain the molecular formula. Solution: a) Empirical formula HNO has a molar mass of 31.02. Hyponitrous acid has a molar mass of 62.04 g/mol, twice the mass of the empirical formula; its molecular formula is twice as large as the empirical formula, 2(HNO) = H2N2O2. The molecular formula of nitroxyl would be the same as the empirical formula, HNO, since the molar mass of nitroxyl is the same as the molar mass of the empirical formula. b) H2N2O2 has [2 × H(1e–)] + [2 × N(5e–)] + [2 × O(6e–)] = 24 valence e–. Ten electrons are used for single bonds between the atoms, leaving 24 – 10 = 14 e–. Sixteen electrons are needed to give every atom an octet; since only fourteen electrons are available, one double bond (between the N atoms) is needed. HNO has [1 × H(1e–)] + [1 × N(5e–)] + [1 × O(6e–)] = 12 valence e–. Four electrons are used for single bonds between the atoms, leaving 12 – 4 = 8e–. Ten electrons are needed to give every atom an octet; since only eight electrons are available, one double bond is needed between the N and O atoms. H

c) In both hyponitrous acid and nitroxyl, the nitrogens are surrounded by three electron groups (one single bond, one double bond, and one unshared pair), so the electron arrangement is trigonal planar and the molecular shape is bent. d) cis trans

14-15

14.113 a)

b) All have configuration (σ2s)2(σ*2s)2(π2p)2(π2p)2(σ2p)2; bond order = 3.

14.114

The three steps of the Ostwald process are given in the chapter. a) 4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g) 2NO(g) + O2(g) → 2NO2(g) 3NO2(g) + H2O(l) → 2HNO3(l) + NO(g) b) If NO is not recycled, the reaction steps proceed as written above. The molar relationships for each reaction yield NH3 consumed per mole HNO3 produced: ⎛ 3 mol NO2 ⎞⎛ mol NH 3 consumed ⎟⎟⎜⎜ 2 mol NO ⎞⎛ ⎟⎟⎜⎜ 4 mol NH 3 ⎞⎟⎟ ⎜ = ⎜⎜ ⎟⎟⎜ ⎟⎟⎜ ⎟ = 1.5 mol NH3/mol HNO3 mol HNO 3 produced ⎟⎜⎜ 2 mol NO2 ⎠⎝ ⎟⎜⎜ 4 mol NO ⎠⎟⎟ ⎜⎜⎝ 2 mol HNO3 ⎠⎝ c) The goal is to find the mass of HNO3 produced, which can be converted to volume of aqueous solution using the density and mass percent. To find mass of HNO3, determine the number of moles of NH3 present in 1 m3 of gas mixture (ideal gas law) and convert moles of NH3 to moles HNO3 using the mole ratio in part (b). Convert moles to grams using the molar mass of HNO3 and convert to volume. ⎛ 1 L ⎞⎛ 5.0 atm 1 m 3 10% NH 3 ⎞⎟ PV ⎟⎟⎜⎜ ⎜⎜ = Moles of NH3 = n = ⎟ ⎟⎟ ⎜ ⎜ ⎜10−3 m 3 ⎟⎟⎜⎜100% gas ⎟⎟ ⎛ • atm ⎞ RT L ⎜ ⎠ ⎠⎝ ⎜⎜0.0821 ⎟⎟((273 + 850) K ) ⎝ ⎜⎝ mol • K ⎠⎟

(

)(

)

= 5.42309 mol NH3 Mass (g) of HNO3 = (5.42309 mol NH3)(1 mol HNO3/1.5 mol NH3)(63.02 g HNO3/1 mol HNO3)(96%/100%) = 218.728 g HNO3 Volume (mL) of HNO3 = (218.728 g HNO3)(100%/60%)(1 mL/1.37 g) = 266.092 = 2.7 × 102 mL solution 14.115 a) Percent N = (mass N/mass NH3) × 100% = (14.01 g N/17.03 g NH3) × 100% = 82.266588 = 82.27% N b) Percent N = (mass N/mass NH4NO3) × 100% = (2 × 14.01 g N/80.05 g NH4NO3) × 100% = 35.00312 = 35.00% N c) Percent N = (mass N/mass (NH4)2HPO4) × 100% = (2 × 14.01 g N/132.06 g (NH4)2HPO4) × 100% = 21.2176 = 21.22% N Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

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14.116 Plan: Carbon monoxide and carbon dioxide would be formed from the reaction of coke (carbon) with the oxygen in the air. The nitrogen in the producer gas would come from the nitrogen already in the air. So, the calculation of mass of product is based on the mass of CO and CO2 that can be produced from 1.75 metric tons of coke. Solution: Using 100 g of sample, the percentages simply become grams. Since 5.0 g of CO2 is produced for each 25 g of CO, we can calculate a mass ratio of carbon that produces each: ⎛ 12.01 g C ⎞⎟ ⎜ ⎟⎟ 25 g CO ⎜⎜ ⎜⎝⎜ 28.01 g CO ⎠⎟⎟ = 7.85612/1 ⎛ 12.01 g C ⎞⎟ ⎜⎜ ⎟⎟ 5.0 g CO2 ⎜ ⎜⎜⎝ 44.01 g CO2 ⎠⎟⎟ Using the ratio of carbon that reacts as 7.85612:1, the total C reacting is 7.85612 + 1 = 8.85612. The mass fraction of the total carbon that produces CO is 7.85612/8.85612 and the mass fraction of the total carbon reacting that produces CO2 is 1.00/8.85612. To find the mass of CO produced from 1.75 metric tons of carbon with an 87% yield: ⎛ 7.85612 ⎞⎟ ⎛ 28.01 t CO ⎞⎟⎜⎛ 87% ⎞⎟ Mass (g) of CO = ⎜⎜ ⎟ = 3.1498656 t CO ⎟(1.75 t )⎝⎜⎜ ⎟⎜ ⎝ 8.85612 ⎠⎟ 12.01 t C ⎠⎟⎝⎜100% ⎠⎟ ⎛ 44.01 t CO2 ⎞⎟⎛ 87% ⎞⎟ ⎛ 1 ⎞⎟ ⎜ (1.75 t )⎜⎜ Mass (g) of CO2 = ⎜⎜ ⎟ ⎟ ⎟⎟⎜⎜100% ⎟⎠⎟ = 0.6299733 t CO2 ⎝ 8.85612 ⎠ ⎝ 12.01 t C ⎠⎝ The mass of CO and CO2 represent a total of 30% (100% – 70% N2) of the mass of the producer gas, so the total mass would be (3.1498656 + 0.6299733)(100%/30%) = 12.59946 = 13 metric tons.

(

)

(

)

14.117 a) 2F2(g) + 2H2O(l) → 4HF(aq) + O2(g) Oxidation states of oxygen: –2 in H2O and 0 in O2 Oxidizing agent: F2; Reducing agent: H2O 2NaOH(aq) + 2F2(g) → 2NaF(aq) + H2O(l) + OF2(g) Oxidation states of oxygen: –2 in NaOH and H2O, +2 in OF2 Oxidizing agent: F2; Reducing agent: NaOH OF2(g) + 2OH – (aq) → O2(g) + H2O(l) + 2F – (aq) Oxidation states of oxygen: +2 in OF2, 0 in O2, –2 in OH– and H2O Oxidizing agent: OF2; Reducing agent: OH– b)

F The oxygen is AX2E2, thus it is a bent molecule. 14.118 In a disproportionation reaction, a substance acts as both a reducing agent and oxidizing agent because an atom within the substance reacts to form atoms with higher and lower oxidation states. 0 –1 –1/3 a) I2(s) + KI(aq) → KI3(aq) I in I2 reduces to I in KI3. I in KI oxidizes to I in KI3. This is not a disproportionation reaction since different substances have atoms that reduce or oxidize. The reverse direction would be a disproportionation reaction because a single substance (I in KI) both oxidizes and reduces. +4 +5 +3 b) 2ClO2(g) + H2O(l) → HClO3(aq) + HClO2(aq) Yes, ClO2 disproportionates, as the chlorine reduces from +4 to +3 and oxidizes from +4 to +5. 0 –1 +1

14-17

c) Cl2(g) + 2NaOH(aq) → NaCl(aq) + NaClO(aq) + H2O(l) Yes, Cl2 disproportionates, as the chlorine reduces from 0 to –1 and oxidizes from 0 to +1. –3 +3 0 d) NH4NO2(s) → N2(g) + 2H2O(g) Yes, NH4NO2 disproportionates; the ammonium (NH4+) nitrogen oxidizes from –3 to 0, and the nitrite (NO2–) nitrogen reduces from +3 to 0. +6 +7 +4 e) 3MnO42–(aq) + 2H2O(l) → 2MnO4–(aq) + MnO2(s) + 4OH–(aq) Yes, MnO42– disproportionates; the manganese oxidizes from +6 to +7 and reduces from +6 to +4. +1 +3 0 f) 3 AuCl(s) → AuCl3(s) + 2 Au(s) Yes, AuCl disproportionates; the gold oxidizes from +1 to +3 and reduces from +1 to 0. 14.119 a) N lacks the d orbitals needed to expand its octet. b) Si has empty low-energy d orbitals which can act as a “pathway” for electron donation from the O of H2O (to form SiO2 + HCl). c) There is partial double bond character in the S—O bond. d) ClF4 would be a free radical (odd number of electrons, one electron unpaired), which would be unstable. 14.120 a) Group 5A(15) elements have five valence electrons and typically form three bonds with a lone pair to complete the octet. An example is NH3. b) Group 7A(17) elements readily gain an electron causing the other reactant to be oxidized. They form monatomic ions of formula X– and oxoanions. Examples would be Cl– and ClO–. c) Group 6A(16) elements have six valence electrons and gain a complete octet by forming two covalent bonds. An example is H2O. d) Group 1A(1) elements are the strongest reducing agents because they most easily lose an electron. As the least electronegative and most metallic of the elements, they are not likely to form covalent bonds. Group 2A(2) elements have similar characteristics. Thus, either Na or Ca could be an example. e) Group 3A(13) elements have only three valence electrons to share in covalent bonds, but with an empty orbital they can accept an electron pair from another atom. Boron would be an example of an element of this type. f) Group 8A(18), the noble gases, are the least reactive of all the elements. Xenon is an example that forms compounds, while helium does not form compounds. Δ 14.121 a) iodic acid; 2HIO3(s) ⎯⎯→ I2O5(s) + H2O(l) O

O I

I O

b) Double bond character in the terminal I–O bonds gives shorter bonds than the single bonds to the central O. c) I2O5(s) +5CO(g) → I2(s) +5CO2(g) ο 14.122 Plan: Find ΔH rxn for the reaction 2BrF(g) → Br2(g) + F2(g) by applying Hess’s law to the equations given. ο Recall that when an equation is reversed, the sign of its ΔH rxn is changed. Solution: 1) 3BrF(g) → Br2(g) + BrF3(l) ΔHrxn = –125.3 kJ 2) 5BrF(g) → 2Br2(g) + BrF5(l) ΔHrxn = –166.1 kJ 3) BrF3(l) + F2(g) → BrF5(l) ΔHrxn = –158.0 kJ Reverse equations 1 and 3, and add to equation 2: 1) Br2(g) + BrF3(l) → 3BrF(g) ΔHrxn = +125.3 kJ (note sign change) 2) 5BrF(g) → 2Br2(g) + BrF5(l) ΔHrxn = –166.1 kJ 3) BrF5(l) → BrF3(l) + F2(g) ΔHrxn = +158.0 kJ (note sign change)

Total:

2BrF(g) → Br2(g) + F2(g)

ΔHrxn = +117.2 kJ

14-18

14.123 2Ca3(PO4)2(s) + 6SiO2(s) +10C(s) → 6CaSiO3(s) + 10CO(g) + P4(g) ⎛ 2 mol Ca (PO ) ⎞⎛ 310.18 g Ca 3 (PO 4 )2 ⎞⎟⎛⎜ 1 kg ⎞⎟⎛100% ⎞ ⎜ 3 4 2⎟ ⎟ ⎟⎜ ⎟⎟⎜⎜ ⎟⎜ Mass (g) of Ca3(PO4)2 = 315 mol P4 ⎜⎜ ⎜ ⎟⎟⎜ 1 mol Ca (PO ) ⎟⎟⎟⎜⎜103 g ⎟⎟⎜⎜⎝ 90% ⎠⎟⎟ ⎜⎜ 1 mol P ⎟ ⎜ ⎜ 4 3 4 2 ⎠⎝ ⎝ ⎠⎝ ⎠ = 2.17126 × 102 = 2.2 × 102 kg Ca3(PO4)2

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)

14.124 a) E is a Group 16 element and has six valence electrons. EF5– would have [1 × E(6e–)] + [5 × F(7e–)] + [1e– from charge] = 42 valence electrons. Ten electrons are used in the single bonds between the atoms. Thirty electrons are used to complete the octets of the fluorine atoms. The remaining two electrons reside on the E atom. EF5– is thus an AX5E substance and has square pyramidal molecule geometry. b) Since element E has six regions of electron density, six hybrid orbitals are required. The hybridization is sp3d2. c) The oxidation number of E in EF5– is +4. 14.125 a)

C O b) The formal charge on carbon is –1, and the formal charge on oxygen is +1. FCO = 6 – [2 + ½(6)] = +1 FCC = 4 – [2 + ½(6)] = –1 c) The electronegativity of oxygen partially compensates for the formal charge difference. 14.126 a) Ionic size increases and charge density decreases down the column. When the charge density decreases, the ionic bond strength between the alkaline earth cation and carbonate anion will decrease. Therefore, the smaller cations release CO2 more easily at lower temperatures. b) To prepare a mixture of CaCO3 and MgO from CaCO3 and MgCO3, heat the mixture to a temperature slightly higher than 542°C, but much lower than 882°C. This should drive off CO2 from MgCO3 without significantly affecting CaCO3. 14.127 Plan: Nitrite ion, NO2–, has [1 × N(5e–)] + [2 × O(6e–)] + [1e– from charge] = 18 valence electrons. Four electrons are used in the single bonds between the atoms, leaving 18 – 4 = 14 electrons. Since sixteen electrons are required to complete the octets of the atoms, one double bond is needed. There are two resonance structures. Nitrogen dioxide, NO2, has [1 × N(5e–)] + [2 × O(6e–)] = 17 valence electrons. Four electrons are used in the single bonds between the atoms, leaving 17 – 4 = 13 electrons. Since 16 electrons are required to complete the octets of the atoms, one double bond is needed and one atom must have an unpaired electron. There are two resonance structures. The nitronium ion, NO2+, has [1 × N(5e–)] + [2 × O(6e–)] – [1e– due to + charge] = 16 valence electrons. Four electrons are used in the single bonds between the atoms, leaving 16 – 4 = 12 electrons. Since sixteen electrons are required to complete the octets of the atoms, two double bonds are needed. Solution: The Lewis structures are

The nitronium ion (NO2+) has a linear shape because the central N atom has two surrounding electron groups, which achieve maximum repulsion at 180°. Both the nitrite ion (NO2–) and nitrogen dioxide (NO2) have a central N surrounded by three electron groups. The electron-group arrangement would be trigonal planar with an ideal bond angle of 120o. The bond angle in NO2– is more compressed than that in NO2 since the lone pair of electrons Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

14-19

in NO2– takes up more space than the lone electron in NO2. Therefore the bond angle in NO2– is smaller (115°) than that of NO2 (134°) 14.128 a)

b) c)

1) CaF2(s) + H2SO4(l) → 2HF(g) + CaSO4(s) 2) NaCl(s) + H2SO4(l) → HCl(g) + NaHSO4(s) 3) FeS(s) + 2HCl(aq) → FeCl2(aq) + H2S(g) Ca3P2(s) + 6H2O(l) → 2PH3(g) + 3Ca(OH)2(s) Al4C3(s) + 12H2O(l) → 4Al(OH)3(s) + 3CH4(g)

14.129 Plan: To find the limiting reactant, find the moles of UF6 that can be produced from the given amount of uranium and then from the given amount of ClF3, use the mole ratios in the balanced equation. The density of ClF3 is used to find the mass of ClF3. The limiting reactant determines the amount of UF6 that can be produced. Solution: U(s) + 3ClF3(l) → UF6(l) + 3ClF(g) (1 metric ton = 1 t = 1000 kg) 3 ⎛103 kg ⎞⎛ ⎟⎟⎜⎜10 g ⎞⎟⎟⎛1.55% ⎞⎟⎛⎜⎜ 1 mol U ⎞⎛ ⎟⎟⎜⎜1 mol UF6 ⎞⎟⎟ ⎜ Moles of UF6 from U = 1.00 t ore ⎜⎜ ⎟⎟⎜ ⎟⎟⎜⎜⎜ ⎟⎟⎜ ⎟ ⎟ ⎜ ⎜⎜⎝ 1 t ⎠⎝ ⎟⎜⎜ 1 kg ⎠⎟⎝ 100% ⎠⎟⎝⎜⎜ 238.0 g U ⎠⎝ ⎟⎜⎜ 1 mol U ⎠⎟⎟ = 65.12605 mol UF6 ⎛ 1 mL ⎞⎜⎛1.88 g ClF3 ⎞⎛ 1 mol ClF3 ⎞⎛ 1 mol UF6 ⎞⎟ ⎟⎟⎟⎜⎜ ⎟⎟⎟⎜⎜ ⎟⎟ Moles of UF6 from ClF3 = 12.75 L ⎜⎜⎜ −3 ⎟⎟⎟⎜⎜ ⎜ ⎜ ⎟⎜⎝ 1 mL ⎠⎝ ⎟⎟⎜⎜ 92.45 g ClF3 ⎠⎝ ⎟⎟⎜⎜ 3 mol ClF3 ⎠⎟⎟ ⎝⎜10 L ⎠⎜ = 86.42509 mol UF6 Since the amount of uranium will produce less uranium hexafluoride, it is the limiting reactant. ⎛ 352.0 UF6 ⎞⎟ ⎟ = 2.2924 × 104 = 2.29 × 104 g UF6 Mass (g) of UF6 = (65.12605 mol UF6 )⎜⎜⎜ ⎜⎝ 1 mol UF6 ⎠⎟⎟

(

)

(

)

14.130 Cl2(g) + 2NaOH(aq) → NaClO(aq) + NaCl(aq) + H2O(l)

⎛ 1 mL ⎞⎟⎛1.07 g ⎞⎛ 5.25% ⎞⎛ 1 mol NaClO ⎞⎛ ⎟⎟⎜⎜ 1 mol Cl 2 ⎞⎟⎟⎛⎜ 22.4 L ⎞⎟ ⎜ ⎜⎜ ⎟⎟⎜ ⎟ ⎟⎟ Volume (L) of Cl2 = 1000. L ⎜⎜ −3 ⎟⎟⎟⎜⎜⎜ ⎟⎟⎜ ⎟⎜ ⎟ ⎜ ⎜⎜⎝10 L ⎠⎟⎝⎜ mL ⎠⎟⎟⎜⎜⎝ 100% ⎠⎟⎝⎜⎜ 74.44 g NaClO ⎠⎝ ⎟⎜⎜1 mol NaClO ⎠⎟⎟⎝⎜⎜ mol ⎠⎟

(

)

= 1.69038 × 104 = 1.69 × 104 L Cl2 14.131 Apply Hess’s law to the two-step process below. The bond energy (BE) of H2 is exothermic because heat is given off as the two H atoms at higher energy combine to form the H2 molecule at lower energy. H + H → H2 BE = –432 kJ/mol H2 + H+ → H3+

ΔH = –337 kJ/mol

Overall: H + H + H+ → H3+

ΔHrxn = –769 kJ/mol

14.132 The bond energy of H2 = –432 kJ. When two H atoms form the H2 bond, energy is released. 1) 2H(g) → H2(g) ΔH = –432.0 kJ 2)

H2(g) + 1/2O2(g) → H2O(g)

Overall: 2H(g) + 1/2O2(g) → H2O(g)

ΔH = –241.826 kJ ΔH = –673.826 = –673.8 kJ

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14.133 Plan: Determine the electron configuration of each species. Partially filled orbitals lead to paramagnetism (unpaired electrons). Solution: O+ 1s22s22p3 paramagnetic odd number of electrons – O 1s22s22p5 paramagnetic odd number of electrons O2– O2+

1s22s22p6 diamagnetic 1s22s22p2 paramagnetic

all orbitals filled (all electrons paired) Two of the 2p orbitals have one electron each. These electrons have parallel spins (Hund’s rule).

14.134 Plan: To determine mass percent, divide the mass of As in 1 mole of compound by the molar mass of the compound and multiply by 100. Find the volume of the room (length × width × height) and use the toxic concentration to find the mass of As required. The mass percent of As in CuHAsO3 is used to convert that mass of As to mass of compound. Solution: mass of As (100) a) Mass percent = mass of compound 74.92 g As % As in CuHAsO3 = (100) = 39.96160 = 39.96% As 187.48 g CuHAsO 3 % As in (CH3)3As =

74.92 g As 120.02 g (CH 3 )3 As

(

)(

(100) = 62.4229 = 62.42% As

)(

)

b) Volume (m3) of room = 12.35 m 7.52 m 2.98 m = 276.75856 m3

⎛ 0.50 mg As ⎞⎟⎜⎛10−3 g ⎞⎟ Mass (g) of As = (276.75856 m 3 )⎜⎜⎜ ⎟⎟ = 0.13838 g As ⎟⎟⎜⎜ m3 ⎝⎜ ⎠⎟⎜⎜⎝ 1 mg ⎠⎟⎟ ⎛100 g CuHAsO 3 ⎞⎟ ⎟ = 0.346282 = 0.35 g CuHAsO3 Mass (g) of CuHAsO3 = (0.13838 g As)⎜⎜ ⎜⎝ 39.96160 g As ⎠⎟⎟

14.135 a) oxidizing agent, producing H2O b) reducing agent, producing O2

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CHAPTER 15 ORGANIC COMPOUNDS AND THE ATOMIC PROPERTIES OF CARBON FOLLOW–UP PROBLEMS 15.1A

a) Plan: For compounds with seven carbons, first start with 7 C atoms in a straight chain. Then make all arrangements with 6 carbons in a straight chain and 1 carbon branched off the chain. Examine the structures to make sure they are all different. Continue in the same manner with 5 C atoms in chain and 2 branched off the chain, then 4 C atoms in chain and 3 branched off, and 3 C atoms in chain and 4 branched off. Examine all structures to guarantee there are no duplicates. Solution: Seven carbons in chain: H H H H H H H

(1) Six carbons in chain: H

H H

(2) (3) If the methyl group (—CH3) is moved to the fourth carbon from the left, the structure is the same as (3). If the methyl group is moved to the fifth carbon from the left, the structure is the same as (2). In addition, if the methyl group is moved to the sixth carbon from the left, the resulting structure has 7 carbons in a chain and is the same as (1).

15-1

Five carbons in chain: H

H H

H C

H H

H H (5)

(4) H H

H H

(6)

H H

C H

H H

H (7)

(8) Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

15-2

Starting with a methyl group on the second carbon from the left, add another methyl group in a systematic pattern. Structure (4) has the second methyl on the second carbon, (5) on the third carbon, and (6) on the 4th carbon. These are all the unique possibilities with the first methyl group on the second carbon. Any other variation will produce a structure identical to a structure already drawn. Next, move the first methyl group to the third carbon where the only unique placement for the second methyl group is on the third carbon as shown in structure (7). Which structure above would be the same as placing the first methyl group on the third carbon and the second methyl group on the fourth carbon? Are there any more arrangements with a 5-carbon chain? Think of attaching the two extra carbons as an ethyl group (—CH2CH3). If the ethyl group is attached to the second carbon from the left, the longest chain becomes 6 carbons instead of 5 and the structure is the same as (3). If the ethyl group is instead attached to the third carbon, the longest chain is still 5 carbons and the ethyl group is a side chain to give structure (8). 4 carbons in chain: H H H H H

H H

(9) H Only this arrangement of 3 methyl groups branching off the two inner carbons in a 4 C chain produces a unique structure. Attaching 4 methyl groups off a 3 C chain is impossible without making the chain longer. So, the above nine structures are all the arrangements. b) Plan: For a five-carbon compound start with 5 C atoms in a chain and place the triple bond in as many unique places as possible. Solution: H H H H H H

H H

H H (2) 4 C atoms in chain: There are two unique placements for the triple bond: one between the first and second carbons and one between the second and third carbons in the chain. (1)

The fifth carbon is added as a methyl group branched off the chain. With the triple bond between the first and second carbons, the methyl group is attached to the third carbon to give structure (3).

15-3

H H

H H (3) With the triple bond between the second and third carbons, the methyl group cannot be attached to either the second or third carbon because that would create 5 bonds to a carbon and carbon has only 4 bonds. Thus, a unique structure cannot be formed with a 4 C chain where the triple bond is between the second and third carbons. There are only 3 structures with 5 carbon atoms, one triple bond, and no rings.

15.1B

a) Plan: Start with a ring consisting of 4 atoms. Make sure to include the double bond in the ring. Then work with a ring consisting of 3 atoms and 1 carbon branched off the ring. Move the position of the double bond relative to the branch to draw the 3 remaining structures. The double bond can occur within the ring or between a ring carbon and the branch carbon. Solution: Four carbons in the ring: H H H C C

C H

H Three carbons in the ring: H

H H

C C

H C

C H

H H H H H b) Plan: For a four-carbon compound start with 4 C atoms in a chain and place the two double bonds in as many unique combinations as possible (there are 2). It will not be possible to draw a molecule with 3 C atoms in a chain, 1 branch, and 2 double bonds without a C atom having more than 4 bonds.

15-4

Solution: H

H 15.2A

Plan: Examine the structure for chain length and side groups; then use the structure to write the name. In part (d), examine the structure for carbons that are bonded to four different groups. Those are the chiral carbons. Solution: a) 3,3-diethylpentane. There are 5 single-bonded carbons in the main chain and two ethyl groups attached to carbon #3 of that chain. The end of the name, pentane, indicates a 5 C chain (pent- represents 5 C) with only single bonds between the carbons (-ane represents alkanes, only single bonds). 3,3-diethyl- means there are two ethyl groups and that each ethyl group is attached to carbon #3. CH3

CH2 H3C

CH2

CH3

CH2 CH3 b) 1-ethyl-2-methylcyclobutane. The main “chain” is a four-membered ring with an ethyl group and a methyl group attached as branches. The name cyclobutane describes the 4 C ring (cyclobut-) with only single bonds (-ane). The 1-ethyl- indicates that an ethyl group (–CH2CH3) is attached to carbon #1. The 2-methyl- indicates that a methyl group (–CH3) is attached to carbon #2. The lower number carbon is assigned to the ethyl group because it precedes methyl alphabetically. CH3

H2 C H3C

CH HC

CH2 C H2

c) trans-3-methyl-3-hexene. There are 6 carbons in the main chain, with a double bond between carbons #3 and #4 (counting from the right hand side). There is also a methyl group attached to carbon #3. The name 3-hexene describes the main chain (hex- indicates that there are 6 carbons in the chain, -ene indicates that there is a double bond in the chain, 3- indicates that the double bond starts at carbon #3). The name 3-methyl- indicates that there is a methyl group on carbon #3. Finally, because there is a double bond, we have to determine whether the groups bonded to the double bond are in a cis- (same side of the double bond) or trans- (opposite sides of the double bond) configuration. In this structure, the longer carbon chains attached to the double bond are on opposite sides of the double bond, so we add the prefix trans- to the name of the compound.

15-5

CH3 H2 C H3C

CH3

C H

C H2

d) 1-methyl-2-propylcyclopentane. The main “chain” in the molecule is a 5 C ring with only single bonds. A methyl group and a propyl group are attached to the ring. The name cyclopentane describes a 5-membered ring (cyclopent-) with all single bonds (-ane). The name 1-methyl- indicates that a methyl group is attached to the ring at the #1 position.The name 2-propyl- indicates that a propyl group is attached to the ring at the #2 position. The lower number carbon is assigned to the methyl group because it precedes propyl alphabetically. The chiral carbons (those bonded to four different groups) are marked with asterisks. H2 H2 * H3C C C

* H3C 15.2B

Plan: Analyze the name for chain length and side groups; then draw the structure. Solution: a) 3-ethyl-3-methyloctane. The end of the name, octane, indicates an 8 C chain (oct- represents 8 C) with only single bonds between the carbons (-ane represents alkanes, only single bonds). 3-ethyl means an ethyl group (–CH2CH3) attached to carbon #3 and 3-methyl means a methyl group (–CH3) attached to carbon #3. CH 3

H3C

CH2

CH3

CH2 CH3 b) 1-ethyl-3-propylcyclohexane. The hexane indicates 6 C chain (hex-) and only single bonds (-ane). The cycloindicates that the 6 carbons are in a ring. The 1-ethyl indicates that an ethyl group (–CH2CH3) is attached to carbon #1. Select any carbon atom in the ring as carbon #1 since all the carbon atoms in the ring are equivalent. The 3-propyl indicates that a propyl group (–CH2CH2CH3) is attached to carbon #3. CH2CH3

CH2CH2CH3 c) 3,3-diethyl-1-hexyne. The 1-hexyne indicates a 6 C chain with a triple bond (-yne) between C #1 and C #2. The 3,3-diethyl means two (di-) ethyl groups (–CH2CH3) both attached to C #3.

15-6

CH3 CH2 H

CH2

CH3

CH2 CH3 d) trans-3-methyl-3-heptene. The 3-heptene indicates a 7 C chain (hept-) with one double bond (-ene) between the 3rd and 4th carbons. The 3-methyl indicates a methyl group (–CH3) attached to carbon #3. The trans indicates that the arrangement around the two carbons in the double bond gives the two smaller groups on opposite sides of the double bond. Therefore, the smaller group on the third carbon (which would be the methyl group) is above the double bond while the smaller group, H, on the fourth carbon is below the double bond. H3C CH2 CH2 CH3 C H3C

CH2

C H

15.3A

Plan: In an addition reaction, atoms are added to the carbon(s) in a double bond. Atoms are removed in an elimination reaction, resulting in a product with a double bond. In a substitution reaction, an atom or a group of atoms substitutes for another one in the reactant. Solution: a) In this reaction, the Br on the carbon chain is replaced with a hydroxyl group while the carbons maintain the same number of bonds. This is a substitution reaction. b) The methyl group on the left-hand side of the structure is replaced by a hydrogen atom while the carbons maintain the same number of bonds. This is a substitution reaction. c) Water (in the form of a –H group and an –OH group) is added to the double bond, resulting in the product having three more atoms. Additionally, there is a second O bonded to the C in the product compared to the reactant. This is an addition reaction.

15.3B

Plan: a) An addition reaction involves breaking a multiple bond, in this case the double bond in 2-butene, and adding the other reactant to the carbons in the double bond. The reactant Cl2 will add –Cl to one of the carbons and –Cl to the other carbon. b) A substitution reaction involves removing one atom or group from a carbon chain and replacing it with another atom or group. For 1-bromopropane the bromine will be replaced by hydroxide. c) An elimination reaction involves removing two atoms or groups, one from each of two adjacent carbon atoms, and forming a double bond between the two carbon atoms. For 2methyl2propanol, the –OH group from the center carbon and a hydrogen from one of the terminal carbons will be removed and a double bond formed between the two carbon atoms.

15-7

Solution: (a)

CH3

Cl2

CH3

CH2

CH3

(b)

(c)

CH3 CH3

CH3

CH3 CH3

15.4A

H2O

Plan: Determine the functional group(s) of the organic reactant(s) and then examine any inorganic reactant(s) to decide on the reaction type. Solution: a) This reaction is an oxidation (elimination) reaction because Cr2O72– and H2SO4 are oxidizing agents. An alcohol group (–OH) is oxidized to a ketone group and a single bond between C and O is converted to a double bond. O

Cl b) This is a substitution reaction because the bromine atoms on the organic reactant (an alkyl halide) can be replaced with the cyanide groups from the inorganic reactant.

15.4B

H2 C

Plan: Examine any inorganic compounds and the organic product to determine the organic reactant. Look for differences between the reactants and products, if possible. Solution: a) The organic reactant must contain a single bond between the second and third carbons and the second and third carbons each have an additional group: an H atom on one carbon and a Cl atom on the other carbon.

15-8

H3C

CH3

CH3 CH3

H3C

CH2

CH3

Cl H Cl b) The third carbon from the left in the product contains the acid group. Therefore, in the reactant this carbon should have an alcohol group. H H3C

CH2

15.5A

Plan: In part (a), LiAlH4 and H2O act as reducing agents to convert a ketone or aldehyde to an alcohol. To form the product, change the carbonyl group, =O, to an –OH group. In part (b) ethyllithium, CH3CH2–Li, and H2O react to convert a ketone or aldehyde to form the alcohol and to add an ethyl group to the carbonyl carbon. To form the product, find the carbonyl carbon, add an ethyl group (from CH3CH2–Li) to the carbon and change the carbonyl group, =O, to a –OH bond. Solution: a) OH

b) H2 C

CH3

15.5B

Plan: The oxidizing agents in reaction (a) indicate that the ketone group (=O) has been oxidized from an alcohol. To form the reactant, replace the C=O with an alcohol group, (C–OH). In reaction (b) the reactants CH3CH2–Li and H2O indicate that a ketone or aldehyde reacts to form the alcohol. To form the reactant, find the carbon with the alcohol group, remove the ethyl group that came from CH3CH2–Li, and change the –OH bond to a carbonyl group, =O. Solution: a) In the reactant, a hydrogen atom is lost from the oxygen and another hydrogen is lost from the adjacent carbon. A double bond forms between the carbon and the oxygen.

15-9

OH CH3

b) In the reactant, the first carbon off the ring will have a carbonyl group (–C=O) in place of the alcohol group (–OH) and the ethyl group (–CH2CH3). HC O Check: When the reactants are combined, the products are identical to those given.

15.6A

Plan: Determine the functional group(s) of the organic reactant(s) and then examine any inorganic reactant(s) to identify the reaction type. Then draw the structure of the product(s) based on the reaction type. Solution: a) The reactant is an ester because it contains the unit: O C O R Reacting an ester with base results in a hydrolysis reaction in which an alcohol and a sodium carboxylate are formed. O

H3C

(H2C)14

O Na+

CH2

(CH2)14

CH3

b) The reactant is an amide because it contains the unit: O C

Reacting an amide with a reducing agent like LiAlH4 converts an amide to an amine. H2 C

H3C

15.6B

H N

CH3

Plan: a) The product is an ester because it contains the unit: O C

15-10

Reacting an alcohol with an acid forms an ester. Since the reactant shown is an alcohol, the other reactant must be the acid. Take the –OR group off the carbon in the ester product and replace with a hydroxyl group to get the structure of the acid. b) The product is an amide because it contains the unit: O C

An amide forms from the reaction between an ester and an amine. This is the indicated reaction because the other product is an alcohol that results when the –OR group on the ester breaks off. To draw the reactants, first take the –NH–R group off the amide and add another hydrogen to the nitrogen to form the amine. Attach the –OR group to the remaining carbonyl (–C=O) group of the amide molecule so that an ester linkage is formed. Solution: a) The –OR group attached to the oxygen in the product is –OCH3. Removing this and adding the –OH gives the acid reactant: O CH2

OH b) Take the –NHCH2CH3 group off the amide product, attach a hydrogen to the nitrogen to make the amine reactant, and attach the –OCH3 from the alcohol product in its place to make the ester reactant. O

H3C

CH2

CH3

ester 15.7A

H2N

CH2

CH3

amine

Plan: Examine the structures for known functional groups: alkenes (C=C), alkynes (C≡C), haloalkanes(C–X, where X is a halogen), alcohols (C–OH), esters (–COOR), ethers (C–O–C), amines (NR3), carboxylic acids (–COOH), amides (–CONR2), aldehydes (O=CHR), and ketones (O=CR2 where R cannot be H). Solution: a) The structure contains an aromatic ring, alkene bond, and an aldehyde group.

O CH

alkene

aldehyde

aromatic ring

15-11

b) The structure contains an amide and a haloalkane. O NH2

CH2

CH3

amide

haloalkane

15.7B

O aldehyde

N Cl 3o amine

aromatic ring haloalkane b) The structure contains an amide, a nitrile group, a ketone group, and an aromatic ring. ketone O

nitrile

H3C

H N

CH2

C H2

aromatic ring

O amide

TOOLS OF THE LABORATORY AND CHEMICAL CONNECTIONS BOXED READING PROBLEMS B15.1

Plan: Protons (hydrogen atoms) that have identical environments produce one peak, while protons in different environments produce different peaks. Examine the structure of each compound for different types (different environments) of hydrogen atoms. Solution: C4H10 C5H12

15-12

H H

In C4H10 there will be two peaks, one from the hydrogen atoms attached to the carbons labeled “a” and one from the hydrogen atoms attached to the carbons labeled “b.” In C5H12 there will be three peaks, one from the hydrogen atoms attached to the carbons labeled “a,” one from the hydrogen atoms attached to the carbons labeled “b,” and one from the hydrogen atoms attached to the carbon labeled “c.” B15.2

Plan: Convert the frequency in MHz to frequency in Hz. Use the relationship between energy and frequency, E = hν, to find the energy associated with the given frequency. Solution: ⎛1 ×106 Hz ⎞⎟ ⎟ = 2.00 × 108 Hz = 2.00 × 108 s–1 ν (s–1) = (200 MHz )⎜⎜⎜ ⎜⎝ 1 MHz ⎠⎟⎟ E (J/photon) = hν = (6.63 × 10–34 J∙s)(2.00 × 108 s–1) = 1.326 × 10–25 J/photon ⎛1.326 ×10−25 J ⎞⎛ 6.02 ×10 23 photons ⎞⎟ ⎟⎟⎟⎜⎜ ⎟⎟ = 0.0798 J/mol = 0.08 J/mol E (J/mol) = ⎜⎜⎜ ⎟⎜⎜ ⎜⎝ photon 1 mol ⎠⎝ ⎠⎟

B15.3

Plan: A compound with two peaks in the NMR spectrum must have hydrogen atoms in two different environments. Since the two peaks are in a 3:1 ratio, the ratio of hydrogen atoms in the two environments must be 3:1. Look at the structure of each isomer and identify the number of environments of hydrogen atoms in each one. Solution: Isomer A has a total of four hydrogen atoms that are each bonded to a C atom which is bonded to a Br atom and another C atom. These would produce one peak. Another four hydrogen atoms are bonded to the two C atoms in the middle of the molecule; these produce one peak. Since there are four hydrogen atoms in two different environments, two peaks are produced with a ratio of 1:1 and the two peaks would be the same size. Isomer B has six hydrogen atoms in one environment, bonded to a C atom that is bonded to another C atom that is bonded to a Br atom. These six hydrogen atoms produce one peak. Then there are two hydrogen atoms bonded to a C atom that is also bonded to a Br atom. These produce a second peak. Since there are six hydrogen atoms in one environment and two hydrogen atoms in another environment, two peaks are produced with a size ratio of 6:2 or 3:1. Isomer B is characterized by a spectrum of two peaks in a 3:1 ratio. Isomer C has hydrogen atoms in three different environments. Four hydrogen atoms are at each end of the chain, bonded to a C atom that is also bonded to a Br atom; three hydrogen atoms are bonded to a C atom which is bonded to the middle C atom; one hydrogen atom is bonded to the middle C atom. Since there are three types of hydrogen atoms, three peaks are expected in the NMR spectrum.

B15.4

Plan: When a ddNTP is incorporated into the growing DNA chain, polymerization stops since no more phosphordiester bonds can be formed. When ddATP is added, the chain stops as each base T is encountered (base A pairs with base T); when ddCTP is added, the chain stops as each base G is encountered (base C pairs with base G). Solution: With ddATP, the chain stops at each base T: TACAGGTTCAGT ddATP will give four complementary chain pieces: A, AGTCA, AAGTCA, ATGTCCAAGTCA. With ddCTP, the chain stops at each base G:

15-13

TACAGGTTCAGT ddCTP will give three complementary chain pieces: CA, CAAGTCA, CCAAGTCA B15.5

Plan: The longest complementary chain piece can be used to find the sequence of bases in the DNA fragment. Bases A and T pair and bases C and G pair. Solution: CATATG is the longest complementary chain piece; matching G for C, C for G, A for T, and T for A results in a DNA fragment of GTATAC. Electrophoresis gel: Complements Target A G C T

CATATG ATATG TATG ATG TG G END–OF–CHAPTER PROBLEMS

G T A T A C

15.1

Organic: Methane (natural gas) Inorganic: Calcium carbonate

CH4 CaCO3

15.2

a) Carbon’s electronegativity is midway between the most metallic and nonmetallic elements of Period 2. To attain a filled outer shell, carbon forms covalent bonds to other atoms in molecules (e.g., methane, CH4), network covalent solids (e.g., diamond), and polyatomic ions (e.g., carbonate, CO32–). b) Since carbon has four valence shell electrons, it forms four covalent bonds to attain an octet. c) Two noble gas configurations, He and Ne, are equally near carbon’s configuration. To reach the He configuration, the carbon atom must lose four electrons, requiring too much energy to form the C4+cation. This is confirmed by the fact that the value of the ionization energy for carbon is very high. To reach the Ne configuration, the carbon atom must gain four electrons, also requiring too much energy to form the C4– anion. The fact that a carbon anion is unlikely to form is supported by carbon’s electron affinity. The other possible ions would not have a stable noble gas configuration. d) Carbon is able to bond to itself extensively because carbon’s small size allows for closer approach and greater orbital overlap. The greater orbital overlap results in a strong, stable bond. e) The C–C bond is short enough to allow the sideways overlap of unhybridized p orbitals on neighboring C atoms. The sideways overlap of p orbitals results in double and triple bonds.

15.3

a) The elements that most frequently bond to carbon are other carbon atoms, hydrogen, oxygen, nitrogen, phosphorus, sulfur, and the halogens, F, Cl, Br, and I. b) In organic compounds, heteroatoms are defined as atoms of any element other than carbon and hydrogen. The elements O, N, P, S, F, Cl, Br, and I listed in part (a) are heteroatoms. c) Elements more electronegative than carbon are N, O, F, Cl, and Br. Elements less electronegative than carbon are H and P. Sulfur and iodine have the same electronegativity as carbon. d) The more types of atoms that can bond to carbon, the greater the variety of organic compounds that are possible.

15.4

Atomic and bonding properties produce three crucial differences between C and Si. Si is larger, forms weaker bonds, and unlike C, has d orbitals available.

15.5

Oxidation states of carbon range from –4 to +4. In carbon dioxide (CO2) carbon has a +4 oxidation state. In methane (CH4), carbon has a –4 oxidation state.

Acetic acid (in vinegar) C2H4O2 Sodium bicarbonate NaHCO3

15-14

15.6

Plan: Chemical reactivity occurs when unequal sharing of electrons in a covalent bond results in regions of high and low electron density. Solution: The C–H, C–C, and C–I bonds are unreactive because electron density is shared equally between the two atoms. The C=O bond is reactive because oxygen is more electronegative than carbon and the electron rich pi bond is above and below the C–O bond axis, making it very attractive to electron-poor atoms. The C–Li bond is also reactive because the bond polarity results in an electron-rich region around carbon and an electron-poor region around Li.

15.7

a) An alkane is an organic compound consisting of carbon and hydrogen in which there are no multiple bonds between carbons, only single bonds. A cycloalkane is an alkane in which the carbon chain is arranged in a ring. An alkene is a hydrocarbon with at least one double bond between two carbons. An alkyne is a hydrocarbon with at least one triple bond between two carbons. b) The general formula for an alkane is CnH2n+2. The general formula for a cycloalkane is CnH2n. Elimination of two hydrogen atoms is required to form the additional bond between carbons in the ring. For an alkene, assuming only one double bond, the general formula is CnH2n. When a double bond is formed in an alkane, two hydrogen atoms are removed. For an alkyne, assuming only one triple bond, the general formula is CnH2n–2. Forming a triple bond from a double bond causes the loss of two hydrogen atoms. c) For hydrocarbons, “saturated” is defined as a compound that cannot add more hydrogen. An unsaturated hydrocarbon contains multiple bonds that react with H2 to form single bonds. The alkanes and cycloalkanes are saturated hydrocarbons since they contain only single C–C bonds.

15.8

a) Constitutional isomers are those with different sequences of bonded atoms. b) Geometric isomers have different orientation of groups around a double bond or a cyclic structure. c) Optical isomers are a type of stereoisomerism that arises when a molecule and its mirror image cannot be superimposed on each other. Constitutional and geometric isomers are not stereoisomers.

15.9

Alkynes are linear about the triple bond. Aromatics can only assume one (planar) orientation in space.

15.10

Plan: An asymmetric molecule has no plane of symmetry. Solution: a) A circular clock face numbered 1 to 12 o’clock is asymmetric. Imagine that the clock is cut in half, from 12 to 6 or from 9 to 3. The one-half of the clock could never be superimposed on the other half, so the halves are not identical. Another way to visualize symmetry is to imagine cutting an object in half and holding the half up to a mirror. If the original object is “re-created” in the mirror, then the object has a plane of symmetry. b) A football is symmetric and has two planes of symmetry — one axis along the length and one axis along the fattest part of the football. c) A dime is asymmetric. Either cutting it in half or slicing it into two thin diameters results in two pieces that cannot be superimposed on one another. d) A brick, assuming that it is perfectly shaped, is symmetric and has three planes of symmetry at right angles to each other. e) A hammer is symmetric and has one plane of symmetry, slicing through the metal head and down through the handle. f) A spring is asymmetric. Every coil of the spring is identical to the one before it, so a spring can be cut in half and the two pieces can be superimposed on one another by sliding (not flipping) the second half over the first. However, if the cut spring is held up to a mirror, the resulting image is not the same as the uncut spring. Disassemble a ballpoint pen and cut the spring inside to verify this explanation.

15-15

15.11

A polarimeter is used to measure the angle that the plane of polarized light is rotated. A beam of light consists of waves moving in all planes. A polarizing filter blocks all waves except those in one plane, so the light emerging through the filter is plane-polarized. An optical isomer is said to be optically active because it rotates the plane of the polarized light. The dextrorotatory isomer (designated d or +) rotates the plane of light to the right; the levorotatory isomer (designated l or –) is the mirror image of the first and rotates the plane to the left.

15.12

Aromatic hydrocarbons have carbon in sp2 hybridization while cycloalkanes have carbon in sp3 hybridization. Aromatic hydrocarbons are planar while most cycloalkanes assume puckered ring structures.

15.13

Plan: To draw the possible skeletons, it is useful to have a systematic approach to make sure no structures are missed. Draw the chain or ring and then draw structures with branches or a double bond at different points along the chain. Solution: a) Since there are seven C atoms but only a six-carbon chain, there is one C branch off of the chain. First, draw the skeleton with the double bond between the first and second carbons and place the branched carbon in all possible positions starting with C #2. Then move the double bond to between the second and third carbons and place the branched carbon in all possible positions. Then move the double bond to between the third and fourth carbons and place the branched carbon in all possible positions. The double bond does not need to be moved further in the chain since the placement between the second and third carbons is equivalent to placement between the fourth and fifth carbons and placement between the first and second carbons is equivalent to placement between the fifth and sixth carbons. The other position to consider for the double bond is between the branched carbon and the six-carbon chain. Double bond between first and second carbons: C C C C C C C C C C C C C C

C C

Double bond between third and fourth carbons: C C C C C C C

Double bond between second and third carbons: C C C C C C C

15-16

Double bond between branched carbon and chain: C

C The total number of unique skeletons is 11. To determine if structures are the same, build a model of one skeleton and see if you can match the structure of the other skeleton by rotating the model and without breaking any bonds. If bonds must be broken to make the other skeleton, the structures are not the same. b) The same approach can be used here with placement of the double bond first between C #1 and C #2, then between C #2 and C #3. Since there are seven C atoms but only five C atoms in the chain, there are two C branches.

Double bond between first and second carbons: C C C C C C C C

C C

Double bond between second and third carbons: C C C C C C C

C C

c) Five of the carbons are in the ring and two are branched off the ring. Remember that all the carbons in the ring are equivalent and there are two groups bonded to each carbon in the ring.

15-17

C C

15.14

To draw the possible skeletons, it is useful to have a systematic approach to make sure no structures are missed. a) C C C C C C C C C C

C C

b) C

C C

C c) There are cis-trans isomers not shown because the topic has not been covered at this point. C C C C C C C

15-18

15.15

Plan: Add hydrogen atoms to make a total of four bonds to each carbon. Solution: a) C CH2 CH2 CH2 CH3 CH CH CH CH2 CH 2 2

CH3 CH2

CH2

CH3

CH2

CH3

CH2

CH3

CH2

CH3

CH3 CH

CH2

CH3

CH3 CH2

CH3

CH2

CH3

CH2

CH3

CH2

CH3 CH3

CH CH3

CH3 CH3

CH3

CH3 CH3

CH2

CH3

CH2

CH3

CH2 b) CH2

CH3

CH2

CH3

CH3 CH2

CH2

CH3

CH2

CH3

CH2

CH3

CH3 CH2

CH2

C CH3

CH3

CH2

CH3

15-19

CH3

CH2

CH3

CH3 CH3

CH3

CH3 CH3

CH3

CH3 CH3

CH3

CH2

CH3

CH2

CH3 CH CH2

CH2

CH2 CH

CH2

CH3

CH2

CH3 15.16

Add hydrogen atoms to make a total of four bonds to each carbon. a) C CH CH CH3 C CH2 CH CH CH 2

CH3

CH2

CH3

CH2

CH3

CH2

CH3

CH2

CH3

CH3 CH3

CH3

C CH3

CH2 CH2

CH2

CH3

15-20

b) CH3

CH2

CH3

CH3 CH3

CH2

CH3 c) There are cis-trans isomers not shown because the topic has not been covered at this point. CH2 CH CH2 CH3 CH CH CH 2

CH2

CH3

CH2

CH3 15.17

CH3

Plan: Remember that each C must have 4 bonds. Solution: a) The second carbon from the left in the chain is bonded to five groups. Removing one of the groups gives a correct structure. CH3 CH3

CH2

CH3

b) The first carbon in the chain has five bonds, so remove one of the hydrogen atoms on this carbon.

c) The second carbon in the chain has five bonds, so move the ethyl group from the second carbon to the third. To do this, a hydrogen atom must be removed from the third carbon atom. HC C CH CH3

CH2 CH3 d) Structure is correct. 15.18

a) Structure is correct.

15-21

CH3 C CH3 c)

CH3

CH2

CH3

CH2

CH3

d) CH3 CH3

15.19

CH2

Plan: The longest chain is named. Then we find the lowest branch numbers by counting C atoms from the end closer to a branch. Name each branch (root- + -yl) and put the names alphabetically before the chain name. Solution: a) Octane denotes an eight carbon alkane chain. A methyl group (–CH3) is located at the second and third carbon position from the left. CH3 CH3

CH2

CH3

b) Cyclohexane denotes a six-carbon ring containing only single bonds. Numbering of the carbons on the ring could start at any point, but typically, numbering starts at the top carbon atom of the ring for convenience. The ethyl group (–CH2CH3) is located at position 1 and the methyl group is located at position 3. CH2 CH3 1 6

3 5 4

CH3

c) The longest continuous chain contains seven carbon atoms, so the root name is “hept.” The molecule contains only single bonds, so the suffix is “ane.” Numbering the carbon chain from the left results in side groups (methyl groups) at positions 3 and 4. Numbering the carbon chain from the other end will result in side groups at positions 4 and 5. Since the goal is to obtain the lowest numbering position for a side group, the correct name is 3,4-dimethylheptane. Note that the prefix “di” is used to denote that two methyl side groups are present in this molecule. d) This molecule is a 4-carbon chain, with two methyl groups (dimethyl) located at the position 2 carbon. The correct name is 2,2-dimethylbutane. 15.20

a) 2-methylbutane

b) 1,3,5-trimethylcyclohexane

15-22

d) CH2

CH3

CH3 CH2

CH3

CH3 CH2

CH2

CH3

15.21

CH2

CH 3

CH2

CH3 1

Numbering from the end carbon to give the lowest value for the methyl group gives the correct name of 3-methylhexane. b) 2-ethylpentane means a five-carbon chain with an ethyl group on the second carbon: CH3

CH2 CH3

CH2CH2CH3

3 4 5 Numbering the longest chain gives the correct name, 3-methylhexane. 1 CH3

2 CH2 CH3

CH2

CH3

6 5 c) 2-methylcyclohexane means a 6 C ring with a methyl group on carbon #2: CH3

In a ring structure, whichever carbon is bonded to the methyl group is automatically assigned as carbon #1. Since this is automatic, it is not necessary to specify 1-methyl in the name. Correct name is methylcyclohexane. d) 3,3-methyl-4-ethyloctane means an 8 C chain with 2 methyl groups attached to the 3rd carbon and one ethyl group to the 4th carbon.

15-23

CH3 CH3

CH2

CH3

CH2

CH3

Numbering is good for this structure, but the fact that there are two methyl groups must be indicated by the prefix di- in addition to listing 3,3. The branch names appear in alphabetical order. Correct name is 4-ethyl-3,3dimethyloctane. 15.22

a) 3,3-dimethlybutane should be 2,2-dimethylbutane. CH3 3 4 CH 3

CH2

CH3 1

CH3

b) 1,1,1-trimethylheptane should be 2,2-dimethyloctane. CH3 8 CH 3

CH2

CH3 1

CH3

c) 1,4-diethylcyclopentane should be 1,3-diethylcyclopentane. CH2 CH3 1 2

3 4

CH3

CH2

d) 1-propylcyclohexane should be propylcyclohexane.

CH2

CH2 CH3

15-24

15.23

Plan: A carbon atom is chiral if it is attached to four different groups. Solution: The circled atoms below are chiral. (a) (b) H H

CH3

C H chiral carbon 15.24

H chiral carbon

CH3

CH2

chiral carbon

CH3

A carbon atom is chiral if it is attached to four different groups. The circled atoms below are chiral. a)

b) H H

C C

chiral carbon

H H chiral carbon

H H

H H Both can exhibit optical activity.

15.25

Plan: The longest chain is named. Then we find the lowest branch numbers by counting C atoms from the end closer to a branch. Name each branch (root- + -yl) and put the names alphabetically before the chain name. An optically active compound contains at least one chiral center, a carbon with four distinct groups bonded to it. Solution: a) This compound is a six-carbon chain with a Br on the third carbon. 3-bromohexane is optically active because carbon #3 has four distinct groups bonded to it: (1) –Br, (2) –H, (3) –CH2CH3, (4) –CH2CH2CH3. chiral carbon CH3

CH2

CH3

Br b) This compound is a five-carbon chain with a Cl and a methyl (CH3) group on the third carbon. 3-chloro-3-methylpentane is not optically active because no carbon has four distinct groups. The third carbon has three distinct groups: (1) –Cl, (2) –CH3, (3) two –CH2CH3 groups. CH3

CH3

CH2

CH3

15-25

c) This compound is a four-carbon chain with Br atoms on the first and second carbon atoms and a methyl group on the second carbon. 1,2-dibromo-2-methylbutane is optically active because the second carbon is chiral, bonded to the four groups: (1) –CH2Br, (2) –CH3, (3) –Br, (4) –CH2CH3. CH3 chiral carbon

15.26

CH2

CH3

All are optically active. a) CH2

CH2

chiral carbon CH2 CH3

b) CH3 CH3

chiral carbon

CH2

CH3

chiral carbon CH2 CH2

CH3

c) Br

CH Cl

15.27

Plan: Geometric isomers are defined as compounds with the same atom sequence but different arrangements of the atoms in space. The cis-trans geometric isomers occur when rotation is restricted around a bond, as in a double bond or a ring structure, and when two different groups are bonded to each atom in the restricted bond. Solution: a) Both carbons in the double bond are bonded to two distinct groups, so geometric isomers will occur. The double bond occurs at position 2 in a five-carbon chain. H H H CH3 C CH3CH2

C CH3

CH3CH2

C H

cis-2-pentene trans-2-pentene b) Cis-trans geometric isomerism occurs about the double bond. The ring is named as a side group (cyclohexyl) occurring at position 1 on the propene main chain. Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

15-26

H C

CH3 C

CH3

trans-1-cyclohexylpropene cis-1-cyclohexylpropene c) No geometric isomers occur because the left carbon participating in the double bond is attached to two identical methyl (–CH3) groups.

15.28

a) Cis-trans geometric isomerism occurs about the double bond. H H H C

CH3

C CH3

CH3 CH3

CH3 C

CH3

H CH3

trans-4,4-dimethyl-2-pentene cis-4,4-dimethyl-2-pentene b) No, geometric isomers occur because the right carbon participating in the double bond is attached to two identical methyl (–CH3) groups. c) Cis-trans geometric isomerism occurs about the double bond. CH CH CH 2

CH3 C

CH3

Cl trans-1-chloro-3-methyl-2-heptene 15.29

CH2 C

CH2

CH2 Cl

C CH2 CH2

CH2

CH3

cis-1-1-chloro-3-methyl-2-heptene

Plan: Geometric isomers are defined as compounds with the same atom sequence but different arrangements of the atoms in space. The cis-trans geometric isomers occur when rotation is restricted around a bond, as in a double bond or a ring structure, and when two different groups are bonded to each atom in the restricted bond. Solution: a) The structure of propene is CH2=CH–CH3. The first carbon that is involved in the double bond is bonded to two of the same type of group, hydrogen. Geometric isomers will not occur in this case. b) The structure of 3-hexene is CH3CH2CH=CHCH2CH3. Both carbons in the double bond are bonded to two distinct groups, so geometric isomers will occur.

15-27

CH3

CH2

CH2 C

CH3

CH2 C

CH2

CH3

cis-3-hexene trans-3-hexene c) The structure of 1,1-dichloroethene is CCl2=CH2. Both carbons in the double bond are bonded to two identical groups, so no geometric isomers occur. d) The structure of 1,2-dichloroethene is CHCl=CHCl. Each carbon in the double bond is bonded to two distinct groups, so geometric isomers do exist. Cl Cl Cl H C

cis-1,2-dichloroethene

15.30

trans-1,2-dichloroethene

a) The structure of 1-pentene is CH2=CH–CH2–CH2–CH3. The first carbon that is involved in the double bond is bonded to two of the same type of group, hydrogen. Geometric isomers will not occur in this case. b) CH3 CH2 CH3 CH3 H

cis-2-pentene

CH2

CH3

trans-2-pentene

c) Cl

CH3 C

H C

C H

C CH3

cis-1-chloropropene

trans-1-chloropropene d) There are no geometric isomers because the first carbon has two hydrogens attached to it.

15.31

Plan: Benzene is a planar, aromatic hydrocarbon. It is commonly depicted as a hexagon with a circle in the middle to indicate that the π bonds are delocalized around the ring and that all ring bonds are identical. With two groups attached to the ring, number the C atoms so that a group is attached to ring C-1. Alternatively, the ortho (o-), meta (m-), and para (p-) naming system is used to denote the location of attached groups in benzene compounds only, not other ring structures like the cycloalkanes.

15-28

Solution: Cl

Cl Cl 1,3-dichlorobenzene (m-dichlorobenzene)

1,2-dichlorobenzene (o-dichlorobenzene)

1,4-dichlorobenzene (p-dichlorobenzene)

15.32

CH3 CH3

CH3

1,3,5-trimethylbenzene

1,2,3-trimethylbenzene

15.33

CH3

CH3 1,2,4-trimethylbenzene

Plan: Analyzing the name gives benzene as the base structure with the following groups bonded to it: (1) on carbon #1 a hydroxy group, –OH; (2) on carbons #2 and #6 a tert-butyl group, –C(CH3)3; and (3) on carbon #4 a methyl group, –CH3. Solution: HO C(CH3)3

(H3C)3C

CH3

15-29

15.34

The compound 2-methyl-3-hexene has cis-trans isomers. CH3 CH 3

CH3 CH

CH2 C

CH3 CH

CH3

H C

CH2

CH3

trans-2-methyl-3-hexene cis-2-methyl-3-hexene The compound 2-methyl-2-hexene does not have cis-trans isomers because the #2 carbon atom is attached to two identical methyl (–CH3) groups: H3C H C

H3C

CH2CH2CH3

2-methyl-2-hexene 15.35

F Cl

Se Br

a) is optically active, because the carbon has four different groups attached. b) is not optically active, because the N does not have four different groups attached (there are two Cl atoms). c) is optically active, because the phosphorus has four different groups attached. d) is not optically active, the compound is irregular tetrahedral (seesaw), and the lone pair makes a fifth group. 15.36

Plan: Convert kJ/mol to J/bond. Use the relationship between E and wavelength, E =

hc , to find λ

the wavelength. Solution: ⎛ ⎞⎛ ⎞⎛ 1 mol ⎟⎟⎜⎜1000 J ⎞⎟⎟ = 4.152824 × 10–19 J/bond ⎜⎝ mol ⎠⎝ ⎟⎜ 6.02 × 10 23 bonds ⎠⎝ ⎟⎜ 1 kJ ⎠⎟

Energy (J/bond) = ⎜⎜ 250 kJ ⎟⎟⎟⎜⎜

hc λ

−34 8 hc (6.63×10 J • s)(3.00 ×10 m/s) λ (m) = = = 4.789512 × 10–7 m E 4.152824 ×10−19 J

λ (nm) =

(4.789512×10

−7

⎛ 1 nm ⎞ m)⎜⎜ −9 ⎟⎟ = 478.9512 = 479 nm ⎜⎝10 m ⎠⎟

15-30

b) elimination

c) substitution

15.37

a) addition

15.38

In an addition reaction, a double bond is broken to leave a single bond, and in an elimination reaction, a double bond is formed from a single bond. A double bond consists of a σ bond and a π bond. It is the π bond that breaks in the addition reaction and that forms in the elimination reaction.

15.39

Yes, an addition, elimination, or substitution reaction can be a redox reaction. Addition:

H C

(Reduction: addition of H2) Elimination: CH3 CH CH3

CH3

H + Cl

CH3

OH (Oxidation: loss of H2) Substitution: H

H Cl

H (Loss of e–)

+ H

15.40

Plan: Look for a change in the number of atoms bonded to carbon. In an addition reaction, more atoms become bonded to carbon; in an elimination reaction, fewer atoms are bonded to carbon, while in a substitution reaction, the product has the same number of atoms bonded to carbon. Solution: a) HBr is removed from the reactant so that there are fewer bonds to carbon in the product. This is an elimination reaction, and an unsaturated product is formed. b) Hydrogen is added to the double bond, resulting in the product having two more atoms bonded to carbons. This is an addition reaction, resulting in a saturated product.

15.41

a) addition reaction

15.42

Plan: In an addition reaction, atoms are added to the carbons in a double bond. Atoms are removed in an elimination reaction, resulting in a product with a double bond. In a substitution reaction, an atom or a group of atoms substitutes for another one in the reactant. Solution: a) Water (H2O or H and OH) is added to the double bond: H+ CH3CH2CH=CHCH2CH3 + H2O CH3CH2CH2CHCH2CH3

b) substitution reaction

OH Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

15-31

b) H and Br are eliminated from the molecule, resulting in a double bond: CH3CHBrCH3 + CH3CH2OK → CH3CH=CH2 + CH3CH2OH + KBr c) Two chlorine atoms are substituted for two hydrogen atoms in ethane: hυ CH3CH3 + 2Cl2 ⎯⎯ → CHCl2CH3 + 2HCl 15.43

a) CH3CHBrCH3 + KI → CH3CHICH3 + KBr b)

Cl Cl2

Cl c) Ni

O + CH3

CH3

15.44

Plan: To decide whether an organic compound is oxidized or reduced in a reaction, rely on the rules in the chapter: A C atom is oxidized when it forms more bonds to O or fewer bonds to H because of the reaction. A C atom is reduced when it forms fewer bonds to O or more bonds to H because of the reaction. Solution: a) The C atom is oxidized because it forms more bonds to O. b) The C atom is reduced because it forms more bonds to H. c) The C atom is reduced because it forms more bonds to H.

15.45

Assuming that the observed carbon is bonded to carbon: a) oxidation b) reduction c) oxidation

15.46

Plan: A C atom is oxidized when it forms more bonds to O or fewer bonds to H because of the reaction. A C atom is reduced when it forms fewer bonds to O or more bonds to H because of the reaction. Solution: a) The reaction CH3CH=CHCH2CH2CH3→CH2CH(OH)—CH(OH)CH2CH2CH3 shows the second and third carbons in the chain gaining a bond to oxygen: C–O–H. Therefore, the 2-hexene compound has been oxidized. b) The reaction shows that each carbon atom in the cyclohexane loses a bond to hydrogen to form benzene. Fewer bonds to hydrogen in the product indicates oxidation.

15.47

a) reduced

15.48

Step 1 substitution

15.49

Plan: Physical properties are determined by intermolecular forces. Decide whether the compound is polar or nonpolar and if hydrogen bonding (H bonded to O, N, or F) is present. Hydrogen bonding and polarity (dipoledipole forces) lead to higher melting and boiling points and water solubility than found in nonpolar compounds which have weaker dispersion forces.

b) oxidized Step 2 addition

15-32

Solution: a) The structures for chloroethane and methylethylamine are given below. The compound methylethylamine is more soluble due to its ability to form hydrogen bonds with water. Recall that N–H, O–H, or F–H bonds are required for H bonding. H H H H H

H H H H H H b) The compound 1–butanol is able to hydrogen bond with itself (shown below) because it contains covalent O–H bonds in the molecule. Diethyl ether molecules contain no O–H covalent bonds and experience dipole-dipole interactions instead of H bonding as intermolecular forces. Therefore, 1-butanol has a higher melting point because H bonds are stronger intermolecular forces than dipole-dipole attractions.

CH3

CH2

H O H

CH2

CH3

Hydrogen Bond H

H H3 C

CH3

diethyl ether c) Propylamine has a higher boiling point because it contains N–H bonds necessary for hydrogen bonding. Trimethylamine is a tertiary amine with no N–H bonds, and so its intermolecular forces are weaker. CH3

CH2

N H

Propylamine

15.50

CH3

CH3 Trimethylamine

Upper right result: R

CH3

Bottom result: R

CH3

15.51

Addition reactions do not occur readily with benzene due to resonance stability of the aromatic ring.

15.52

The C=C bond is nonpolar while the C=O bond is polar, since oxygen is more electronegative than carbon. Both bonds react by addition. In the case of addition to a C=O bond, an electron-rich group will bond to the carbon and an electron-poor group will bond to the oxygen, resulting in one product. In the case of addition to an alkene, the carbons are identical, or nearly so, so there will be no preference for which carbon bonds to the electron-poor group and which bonds to the electron-rich group. This may lead to two isomeric products, depending on the structure of the alkene.

15-33

When water is added to a double bond, the hydrogen is the electron-poor group and hydroxyl is the electron-rich group. For a compound with a carbonyl group, only one product results as H bonds to the O atom in the double bond and –OH bonds to the carbon atom in the double bond: CH3

CH2

CH3

CH2

CH3

H H O OH However, when water adds to a C=C, two products result since the OH can bond to either carbon in the double bond: CH3 CH2 CH CH3 O CH CH CH CH + 3

H + CH3

CH2

CH2 OH

In this reaction, very little of the second product forms.

15.53

15.54

CH3

OR'

Alcohol: R–OH + H2O → R–O– + H3O+ Carboxylic acid: O R

O OH

H2O

H3O+

The resonance-stabilized carboxylate ion allows the transfer of the proton to water. The alkoxide ion cannot show any resonance stabilization. 15.55

Esters and acid anhydrides form through dehydration-condensation reactions. Dehydration indicates that the other product is water. In the case of ester formation, condensation refers to the combination of the carboxylic acid and the alcohol. The ester forms and water is the other product.

15-34

O O CH3

CH2

CH3

Water Removed

CH2

H2O

CH3

15.56

Alcohols undergo substitution at a saturated carbon while acids undergo substitution at the carboxyl carbon.

15.57

Plan: Refer to the Table of Functional Groups in the chapter. Solution: a) Halogens, except iodine, differ from carbon in electronegativity and form a single bond with carbon. The organic compound is an alkyl halide. b) Carbon forms triple bonds with itself and nitrogen. For the bond to be polar, it must be between carbon and nitrogen. The compound is a nitrile. c) Carboxylic acids contain a double bond to oxygen and a single bond to oxygen. Carboxylic acids dissolve in water to give acidic solutions. d) Oxygen is commonly double bonded to carbon. A carbonyl group (C=O) that is at the end of a chain is found in an aldehyde.

15.58

a) amide

b) alkene

c) ketone

d) amine

15.59

O a)

b) CH3

CH2

alkene

CH2

alcohol carboxylic acid aromatic ring

haloalkane O

CH3

CH2

CH3

N O

ketone

nitrile

H amide alkene

15-35

O C

CH2

CH3

ester

15.60

b) I

CH2

haloalkane aldehyde

carboxylic acid

alkyne

c) CH2

CH2

alkene

CH3

ester

CH3

ester amide

e) CH3

haloalkane

CH2

CH3

amine alkene

15.61

Plan: Draw the longest carbon chain first and place the –OH group at different points along the chain. Then, work down to shorter chains, with the –OH group and branches at different points along the chains. Solution: For C5H12O, the longest chain is five carbons: CH2 OH

CH2

CH3

CH OH

CH2

CH3

CH2

CH3

The three structures represent all the unique positions for the alcohol group on a five-carbon chain. Next, use a four-carbon chain and attach to side groups, –OH and –CH3.

15-36

CH2

CH3

CH2

CH3

CH2

CH3

CH3 CH3

CH3

CH2

CH3

And use a three-carbon chain with three side groups: CH3 CH3

CH2

CH3

The total number of different structures is eight. 15.62

Aldehydes: O

O CH3

CH2

CH3

CH2

CH3

O CH3

CH2

CH3

Ketones:

15.63

CH3

CH2

CH3

O C

CH3

CH2

CH3

Plan: First, draw all primary amines (formula R–NH2). Next, draw all secondary amines (formula R–NH–R'). There is only one possible tertiary amine structure (formula R3–N). Eight amines with the formula C4H11N exist.

15-37

Solution: CH3 CH2

CH2

NH2

CH3

CH2

NH2

CH3 CH3 CH3

CH2

NH2

CH3

CH3 CH3

CH2

NH2

CH3 NH

CH3

CH3 CH3

CH2

CH3

CH2

CH3

15.64 CH3

CH2

CH3

CH2

CH3

CH2

CH3

15.65

CH3

Plan: With mild oxidation, an alcohol group is oxidized to a carbonyl group and an aldehyde is oxidized to a carboxylic acid. Solution: a) The product is 2-butanone. CH3 C CH2 CH3 O b) The product is 2-methylpropanoic acid. CH3 O

CH3

c) Mild oxidation of an alcohol produces a carbonyl group. The product is cyclopentanone.

15-38

15.66

a) 2-methyl-1-propanol CH3 CH3

CH2

b) 2-pentanol CH3

CH2

CH3

OH c) 3-methyl-1-butanol

CH3 CH2

CH2

CH3

15.67

Plan: These reactions are dehydration-condensation reactions, in which H and OH groups on the two reactant molecules react to form water and a new bond is formed between the two reactants. Solution: a) This reaction is a dehydration-condensation reaction to form an amide. O O

CH3

H +

H2O eliminated

H2O

b) An alcohol and a carboxylic acid undergo dehydration-condensation to form an ester. O CH3

CH3

CH2

CH CH3

H2O eliminated O CH3

CH2

CH3 O

CH CH3

15-39

c) This reaction is ester formation through dehydration-condensation. CH3 O

CH2

CH3 O

CH3

H2O eliminated

15.68

CH2

CH3

CH2

CH3

a) O CH3

CH2

H2O eliminated O CH3

CH2

CH3

b) CH3

O CH3

CH2

CH3

CH2

CH3

H2O eliminated

CH3

c) CH2

O CH3

H2O eliminated

15.69

CH3

N CH2

CH3

CH2

CH3

CH2

CH3

Plan: To break an ester apart, break the –C–O– single bond and add water (–H and –OH) as shown. O O Add H R

bond broken Add OH

15-40

Solution: (a) O CH3

(CH2)4

CH2

CH3

CH2

CH3

NH2

CH2

CH3

(b) O C

15.70

CH2

+ HO

CH2

a) O H3C

CH2

CH3

N CH3

c) O H

OH +

N H

15.71

a) Substitution of Br– occurs by the stronger base, OH–. Then a substitution reaction between the alcohol and carboxylic acid produces an ester:

15-41

O CH3

OHCH3

CH2

CH3

CH2

OH H+

O CH3

CH2

CH3

b) The strong base, CN–, substitutes for Br. The nitrile is then hydrolyzed to a carboxylic acid. Br C N CN CH3 CH2 CH CH3 CH CH CH CH 3

OH H3O+, H2O

CH3

CH2

CH3

H+, H2O

CH2

CH3

CH2

CH3

OH _

Cr2O72 , H+

15.72

CH3

CH2

OH CH3

15.73

CH3

a) b)

CH3

CH2

CH3

CH2

H2O CH3

CH2

CH3

CH2

CH3

a) The product is an ester and the given reactant is an alcohol. An alcohol reacts with a carboxylic acid to make an ester. To identify the acid, break the single bond between the carbon and oxygen in the ester group.

15-42

O CH3

CH2

CH3

portion from portion from carboxylic acid alcohol The missing reactant is propanoic acid. b) To form an amide, an amine must react with an ester to replace the –O–R group. To identify the amine that must be added, break the C–N bond in the amide. The amine is ethylamine. O CH3

CH2

a) KOH; Br2

15.75

a) CH3

CH2

CH3

carboxylic acid portion from ester

portion from amine

15.74

b) K2Cr2O7, H+; C6H5–CH2–OH; H+

CH2

CH3

CH3 1-pentanol CH3 CH3

3-methyl-1-butanol

CH2

CH3

CH2

2,2-dimethyl-1-propanol b) OH CH2

CH2

CH3

CH2

2-methyl-1-butanol OH CH3

CH3

OH CH

CH3

CH2

CH3

2-pentanol

3-methyl-2-butanol

3-pentanol

c) OH CH3

CH2

CH3

2-methyl-2-butanol Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

15-43

CH3

CH2

CH3

CH2

CH3

CH2

CH3

CH2

CH3

CH3 CH3

CH2

CH3 CH3

CH3

CH3 CH3

d) The names for the alcohols are given with their structures. 15.76

First, oxidize the methanol to formic acid. O CH3

Cr2O72 , H+

Then, produce the ester by reacting the formic acid with the ethanol. O H

+ HO

CH2

CH3

H+ H

H2O

CH2

CH3

15.77

Addition reactions and condensation reactions are the two reactions that lead to the two types of synthetic polymers that are named for the reactions that form them.

15.78

A double bond is common in the monomers resulting in addition polymers. Substituents on the alkene make the monomers different.

15.79

A free radical is an atom or a group of atoms with one unpaired electron. A free radical is highly reactive and is used to initiate a chain reaction.

15.80

Polyethylene comes in a range of strengths and flexibilities. The intermolecular dispersion forces (also called London forces) that attract the long, unbranched chains of high-density polyethylene (HDPE) are strong due to the large size of the polyethylene chains. Low-density polyethylene (LDPE) has increased branching that prevents packing and weakens intermolecular dispersion forces.

15.81

Condensation polymers are more similar chemically to biopolymers. Both are formed by the loss of water from two monomers.

15.82

Nylon is formed by the condensation reaction between an amine and a carboxylic acid resulting in an amide bond. Polyester is formed by the condensation reaction between a carboxylic acid and an alcohol to form an ester bond.

15-44

15.83

Plan: Both PVC and polypropylene are addition polymers. To draw the repeat unit, replace the double bond with a single bond, draw an additional single bond to each carbon atom, draw brackets around the molecule, and use a subscript n to denote that the monomer is repeated n times to form the polymer. Solution: a) H

Cl n

CH3 n

15.84

15.85

A carboxylic acid and an alcohol react to form an ester bond in a dehydration-condensation reaction:

15-45

n HO

CH2

+ 2n 1H2O

CH2

n The displacement reaction is: O

n H3C

H3C

CH3

n HO

CH2

H + 2n - 1CH OH 3 n

15.86 CH3 n

Si CH3

CH3 OH

+ n H2O

CH3 n

15.87

a) Amino acids form condensation polymers, called proteins. b) Alkenes form addition polymers, the simplest of which is polyethylene. c) Simple sugars form condensation polymers, called polysaccharides. d) Mononucleotides form condensation polymers, called nucleic acids.

15.88

Fibrous proteins are shaped like extended helices or sheets. The R groups come mostly from glycine, serine, alanine, and proline. Globular proteins are more compact in structure and have more complex compositions. Their R groups have more extended nonbonded interactive attraction for each other.

15-46

15.89

The amino acid sequence in a protein determines its shape and structure, which determine its function.

15.90

A phosphate ester linkage joins the nucleotides in the DNA strands.

15.91

DNA exists as two chains wrapped around each other in a double helix. The negatively charged sugar-phosphate backbone faces the aqueous surroundings, and each base in one chain pairs with a base in the other through H bonding.

15.92

The DNA base sequence contains an information template that is carried by the RNA base sequence (messenger and transfer) to create the protein amino acid sequence. In other words, the DNA sequence determines the RNA sequence, which determines the protein amino acid sequence.

15.93

Plan: Locate the specific amino acids in the Table of Amino Acids in the chapter. Solution: a) alanine b) histidine c) methionine + CH3 HN CH 3

S CH2

CH2

15.94

a) glycine

b) isoleucine

CH3

CH2

c) tyrosine OH

CH2 CH

CH3

CH2

15-47

15.95

Plan: A tripeptide contains three amino acids and two peptide (amide) bonds. Find the structures of the amino acids in the Table of Amino Acids in the chapter. Join the three acids to give the tripeptide; water is produced. Solution: O O O

H2N

CH2

O N

OH HN

NH a) aspartic acid O

H2N

H N

CH2 C

histidine O

tryptophan

CH2

O N

OH NH

b) Repeat the preceding procedure, with charges on the terminal groups as are found in cell fluid. glycine cysteine tyrosine O H O H O H3N

CH H

CH CH2

CH2

15-48

15.96

H2N

lysine

phenylalanine H O

threonine O

CH2

CH3

CH2 CH2 NH2 b)

alanine

leucine

valine

15.97

Plan: Base A always pairs with Base T; Base C always pairs with Base G. Solution: a) Complementary DNA strand is AATCGG. b) Complementary DNA strand is TCTGTA.

15.98

a) CCAATG

15.99

Plan: Uracil (U) substitutes for thymine (T) in RNA. Therefore, A pairs with U. The G-C pair and A-T pair remain unchanged. A three-base sequence constitutes a word, and each word translates into an amino acid. Solution: The RNA sequence is derived from the DNA template sequence ACAATGCCT. There are three sets of three-base sequences, so there are three words in the sequence, and three amino acids are coded in the sequence.

b) GGGCTT

15.100 CATAGTTACTTGAAC; five amino acids 15.101 Plan: Refer to the Table of Amino Acids in the chapter. The types of forces operating in proteins were discussed in Chapter 13. Disulfide bonds form between sulfur atoms, salt links form between –COO– and –NH3+ groups, and hydrogen bonding occurs between NH– and –OH groups. Nonpolar chains interact through dispersion forces.

15-49

Solution: a) Both side chains are part of the amino acid cysteine. Two cysteine R groups can form a disulfide bond (covalent bond). b) The R group (–(CH2)4–NH3+) is found in the amino acid lysine and the R group (–CH2COO–) is found in the amino acid aspartic acid. The positive charge on the amine group in lysine is attracted to the negative charge on the acid group in aspartic acid to form a salt link. c) The R group in the amino acid asparagine and the R group in the amino acid serine. The –NH– and –OH groups will hydrogen bond. d) Both the R group –CH(CH3)–CH3 from valine and the R group C6H5–CH2– from phenylalanine are nonpolar, so their interaction is through dispersion forces. ⎛ 5 × 10 5 g ⎞⎛ ⎟⎟⎜1 amino acid ⎞⎟⎟ ⎜ 3 15.102 ⎜⎜ ⎟⎜ ⎟⎟ = 5 × 10 amino acids/DNA ⎜⎜ mol DNA ⎟⎟⎜⎜⎜ 100 g ⎠⎟ ⎝ ⎠⎝

15.103 a) C4H8+ CH3OH → C5H12O b) Mass percent oxygen in MTBE = [(16.00 g O)/(88.15 g MTBE)] × 100% = 18.150879% O in MTBE ⎛ 100 g MTBE ⎞⎟ ⎟⎟ = 14.8753 = 15 g MTBE/100 g gasoline Mass (g) of MTBE = (2.7 g oxygen)⎜⎜⎜ ⎝18.150879 g oxygen ⎠⎟ −3 ⎛14.8753 g MTBE ⎞⎟⎛ 0.740 g gasoline ⎞⎛ 1 mL ⎞⎛ ⎟⎟⎜⎜ 1 mL MTBE ⎞⎛ ⎟⎟⎜10 L ⎞⎟⎟ ⎜⎜ ⎜⎜ ⎟ ⎜ ⎟ ⎜ ⎟ c) Volume (L) of MTBE = ⎜ ⎟⎜ ⎟ ⎟ ⎟⎜ −3 ⎟⎟⎜⎜ ⎟⎜ 0.740 g MTBE ⎠⎟⎟⎜⎝⎜ 1 mL ⎠⎟⎟ ⎜⎜⎝ 100. g gasoline ⎠⎟⎟⎜⎝⎜ 1 mL gasoline ⎠⎟⎝⎜⎜10 L ⎠⎝

= 0.148753 = 0.15 L MTBE/L gasoline d) 2C5H12O(l) + 15O2(g) → 10CO2(g) + 12H2O(g) ⎛ 1 mL ⎞⎛ 0.740 g MTBE ⎞⎟⎜⎛ 1 mol MTBE ⎞⎛ ⎟⎜ 15 mol O 2 ⎞⎟⎟ ⎟⎟⎜⎜ Moles of O2 = (1.00 L MTBE )⎜⎜⎜ −3 ⎟⎟⎟⎜⎜⎜ ⎟⎟⎟⎜⎜ ⎟ ⎟⎜ ⎟⎜⎝ 88.15 g MTBE ⎠⎝ mL ⎟⎜⎜ 2 mol MTBE ⎠⎟⎟ ⎝⎜10 L ⎠⎝ ⎠⎜

= 62.96086 mol O2 L atm ⎞⎟ (62.96086 mol O )⎛⎜⎜⎝0.0821 mol ⎟((273 + 24) K ) ⎛100% ⎞ K⎠ •

V = nRT/P =

•

1.00 atm

⎟⎟ ⎜⎜ ⎝ 21% ⎠⎟

= 7.310565 × 103 = 7.3 × 103 L air 15.104 a) Symbolize the monoprotic acid as HA. Then, the balanced chemical equation will be: NaOH(aq) + HA(aq) →NaA(aq) + H2O(l) Molar mass (g/mol) = ⎛1 mol NaOH ⎞⎛ ⎞⎛ 1 mL NaOH ⎞⎛ ⎞⎟ L NaOH 1 ⎟⎟ (0.2003 g HA)⎜⎜⎜⎜⎜ 1 mol HA ⎟⎟⎟⎟⎟⎜⎜⎜⎜⎜ 0.03811 mol NaOH ⎟⎟⎟⎟⎟⎜⎜⎜⎜⎜10−3 L NaOH ⎟⎟⎟⎟⎟⎜⎜⎜⎜⎜ 45.25 mL NaOH ⎠⎟⎟ ⎝ ⎠⎝ ⎠⎝ ⎠⎝ = 116.1511 = 116.2 g HA/mol b) To convert an alcohol to an acid, the alcohol loses two hydrogen atoms and gains an oxygen atom. This process must be reversed to get to the original alcohol: Molar mass (g/mol) = (116.1511 g/mol) + 2(1.008 g H/mol) – (16.00 g O/mol) = 102.1671 = 102.2 g/mol

15-50

15.105 a) CH3CHO + C6H5–MgBr →C6H5CH(OH)CH3 b) OH H2O CH3

CH2

CH3

+ CH3

CH3

MgBr

CH2

CH3

c) CH3MgBr and C6H5CHO d) HCHO (formaldehyde) e) CH3

CH3

+ CH3

CH2

CH3

CH2

CH3

MgBr

15.106 a) Perform an acid-catalyzed dehydration of the alcohol (elimination), followed by bromination of the double bond (addition of Br2): H CH3 CH2 CH2 OH CH3 CH CH2

CH3

CH2

+ Br2

CH3

CH2

Br Br b) The product is an ester, so a carboxylic acid is needed to prepare the ester. First, oxidize one mole of ethanol to acetic acid: Cr2O72 O CH3 CH2 OH CH3

H+

Then, react one mole of acetic acid with a second mole of ethanol to form the ester: O

H CH3

CH2

OH HO

CH3

CH2

CH3

H2O eliminated 15.107 a) Obtain the molar mass of A by using the ideal gas equation to find the moles of the 2.48 g sample of A. PV (1.00 atm )(1.00 L ) = = 0.02812995 mol Moles of A = n = ⎛ 0.0821 L • atm ⎞⎟ RT ⎜ ( ) 273 + 160 K ⎜⎝ ⎠⎟ mol • K 2.48 g A Molar mass of A = = 88.16226 g/mol 0.02812995 mol

15-51

Obtain the molecular formula of A from the combustion analysis: ⎛ 2 mol H ⎞⎟ ⎟⎟ = 0.0454 mol H Moles of H = (0.409 g H 2 O )⎜⎜⎜ ⎝18.02 g H 2 O ⎠⎟ ⎛1.008 g H ⎞⎟ = 0.04576 g H Mass of H = (0.0454 mol H )⎜⎜ ⎝ 1 mol H ⎠⎟⎟ ⎛ 1 mol C ⎞⎟ ⎟ = 0.0227 mol C Moles of C = (1.00 g CO 2 )⎜⎜ ⎜⎝ 44.01 g CO 2 ⎠⎟⎟ ⎛12.01 g C ⎞⎟ = 0.2726 g C Mass of C = (0.0227 mol C)⎜⎜ ⎝ 1 mol C ⎠⎟⎟ ⎛ 1 mol O ⎞⎟ ⎟ = 0.01135 mol O Moles of O = (0.500 g A − (0.04576 + 0.2726) g O)⎜⎜ ⎜⎝16.00 g O ⎠⎟⎟

0.0227 mol C =2 0.01135 mol

0.0454 mol H =4 0.01135 mol

0.01135 mol O =1 0.01135 mol

Empirical formula = C2H4O with a molar mass of 44.05 g/mol. Since the molar mass of Compound A is 88.16 g/mol, the molecular formula of Compound A is C4H8O2. Compound B is acidic so it must be a carboxylic acid. ⎛10−3 L ⎞⎟⎛1 mol COOH ⎞ ⎛ 0.5 mol NaOH ⎞⎟ ⎜ ⎟ = 0.01695 mol COOH 33.9 mL ⎟⎜ ( ) Moles of COOH = ⎜⎜ ⎜ ⎟ ⎜⎝ 1 mL ⎠⎟⎟⎜⎝ 1 mol NaOH ⎠⎟⎟ ⎝ ⎠⎟ L The 1.00 g sample of Compound B has 0.01695 moles of COOH or 1.00 g/0.01695 mol = 59 g/mol COOH. Since Compound A has a molar mass of 88 g/mol, Compound B must be a dicarboxylic acid with a molar mass of 118 g/mol and a molecular formula of C4H6O4 since there are four carbon atoms. Compound C forms when Compound B is heated and loses water so Compound C must be the anhydride of B. Compound C has the molecular formula (Compound B, C4H6O4 – H2O) = C4H4O3. The NMR of Compound C has only one peak so it has only one kind of hydrogen atom; all of the hydrogen atoms in Compound C are identical so it must be a symmetrical compound. The structures of B and C that fit are: O O C

C CH2

CH2

O CH2 C

O O

B C Compound A is not acidic so it does not have COOH groups. A is 4-hydroxybutanal.

O HOCH2CH2CH2C

15-52

b) Only one oxygen atom is added to Compound A to produce the carboxylic acid GHB, C4H8O3. The aldehyde is oxidized to the acid 4-hydroxybutanoic acid:

O HOCH2CH2CH2C

GHB 15.108 The two structures are: HN CH CH 2

CH2

H2N

NH2

CH2

cadaverine

CH2

NH2

putrescine

The addition of cyanide, [C≡N]–, to form a nitrile is a convenient way to increase the length of a carbon chain:

H + 2CN _

+ 2Br

Br Br N C C N The nitrile can then be reduced to an amine through the addition of hydrogen (NaBH4 is a hydrogen-rich reducing agent): H H reduction H2N CH2 CH2 CH2 CH2 NH2 H C C H N

15.109 Plan: Refer to the Table of Functional Groups in the chapter. Carbon atoms surrounded by four electron regions (four single bonds) are sp3 hybridized. Carbon atoms surrounded by three electron regions (two single bonds and one double bond) are sp2 hybridized. A carbon atom is chiral if it is attached to four different groups. Solution: a) Functional groups in jasmolinII: alkene CH3

CH2

CH 1 CH3

2 CH

O 3 CH

4 C

5 CH

C 6 C

C 11 CH2

O CH3

CH3

C 7 O

ester

alkene ester ketone b) Carbon 1 is sp2 hybridized. Carbon 3 is sp3 hybridized.

Carbon 2 is sp3 hybridized. Carbon 4 is sp2 hybridized.

15-53

Carbon 5 is sp3 hybridized. Carbon 6 and 7 are sp2 hybridized. c) Carbons 2, 3, and 5 are chiral centers as they are each bonded to four different groups. 15.110 a) Initial mole determination: ⎛ 1 mol CO 2 ⎞⎛ ⎟⎟⎜⎜ 1 mol C ⎞⎟⎟ ⎜ Moles of C = (1.25 g CO 2 )⎜⎜ ⎟⎟⎜ ⎟ = 0.0284026 mol C ⎜⎝⎜ 44.01 g CO 2 ⎠⎝ ⎟⎜⎜1 mol CO 2 ⎠⎟⎟ ⎛ 1 mol H 2 O ⎞⎛ ⎟⎟⎜⎜ 2 mol H ⎞⎟⎟ ⎜ Moles of H = (0.613 g H 2 O)⎜⎜ ⎟⎟⎜ ⎟ = 0.0680355 mol H ⎜⎝⎜18.02 g H 2 O ⎠⎝ ⎟⎜⎜1 mol H 2 O ⎠⎟⎟ Save the moles and now determine the masses of C and H: ⎛12.01 g C ⎞⎟ ⎜ ⎟⎟ = 0.341115 g C Mass (g) of C = (0.0284026 mol C)⎜⎜ ⎜⎝⎜ 1 mol C ⎠⎟⎟ ⎛1.008 g H ⎞⎟ ⎜ ⎟⎟ = 0.06857978 g H Mass (g) of H = (0.0680355 mol H )⎜⎜ ⎜⎝⎜ 1 mol H ⎠⎟⎟ Determine the mass and then the moles of O: Mass (g) of O =0.500 g (C, H and O) – (0.341115 g C + 0.06857978 g H) = 0.0903052 g O ⎛ 1 mol O ⎞⎟ ⎜ Moles of O = (0.0903052 g O)⎜⎜ ⎟⎟ = 0.005644075 mol O ⎜⎜⎝16.00 g O ⎠⎟⎟ Divide by the smallest number of moles: (Rounding is allowed here.) 0.0284026 mol C 0.0680355 mol H 0.005644075 mol O =5 = 12 =1 0.005644075 mol O 0.005644075 mol O 0.005644075 mol O Empirical formula A = C5H12O (empirical formula mass = 88.15 g/mol) L • atm ⎞⎟ ⎛ 0.225 g ⎜⎜0.0821 ((273 + 97) K ) ⎛ 1 mL ⎞⎛ ⎟⎟⎜⎜ 760 torr ⎞⎟⎟ ⎝ ⎜⎜ mol • K ⎠⎟ b) M = mRT/PV = ⎟ = 88.206 = 88.2 g/mol ⎜⎜ −3 ⎟⎟⎜⎜ ⎟⎜ 1 atm ⎠⎟⎟ ⎜⎝10 L ⎠⎝ 755 torr 78.0 mL

(

)

(

)(

)

This is nearly the same as the molar mass of the empirical formula so the empirical and molecular formulas are the same: C5H12O. c) Since careful oxidation yields a ketone, Compound A must be a secondary alcohol. Compound A is branched and chiral, so the structure can be: H

CH3

CH2

CH3

OH This is not the only possible correct answer. 15.111 a) The functional group in ibuprofen is the carboxylic acid group COOH. The chiral center is RC*H(CH3)COOH.

15-54

CH3

CH3 C

COOH

C H CH2

H3 C

b) React the aldehyde with methyl Grignard reagent, CH3MgBr to get the alcohol. CH3

CHO + CH3MgBr

H2O

CH OH

React the alcohol with HBr to get the brominated product. CH3

+ HBr

React the bromide with cyanide ion to produce the nitrile. CH3

+ H 2O

CH3

+ NaCN

CH3

+ NaBr

N C Then hydrolyze the nitrile with aqueous HCl to get the carboxylic acid. CH3 CH3

+ HCl(aq) N

CH COOH

15.112 Retinal is a polyunsaturated aldehyde with extensive conjugation. 15.113 The resonance structures show that the bond between carbon and nitrogen will have some double bond character that restricts rotation around the bond. O H O H

15.114 a) B2H6 + 3Cl2→ 2BCl3 + 3H2 b) 2B5H9 + 12O2→ 5B2O3 + 9H2O c) Si3H8 + 6H2O → 3SiO2 + 10H2 Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

15-55

d) Si2H6 + 6Cl2 → Si2Cl6 + 6HCl Δ → 2H2S + S8 e) 2H2S5 ⎯⎯

f) PCl5 + 4H2O → 5HCl + H3PO4 15.115 a) To prepare a polymer with a benzene containing backbone, styrene, C6H5–CH=CH2 can be used to produce polystyrene, – (–CH(C6H5) –CH2–)–n. b) p-diethenylbenzene can be used to crosslink the polymer.

CH2

15.116 Plan: Convert each mass to moles, and divide the moles by the smallest number to determine molar ratio, and thus relative numbers of the amino acids. To find the minimum molar mass, add the products of the moles of each amino acid and its molar mass to find the total mass of the amino acids and subtract the total mass of water that is eliminated during the formation of the peptide bonds. Solution: a) The hydrolysis process requires the addition of water to break the peptide bonds. b) (3.00 g gly)/(75.07 g/mol) = 0.0399627 mol glycine (0.90 g ala)/(89.10 g/mol) = 0.010101010 mol alanine (3.70 g val)/(117.15 g/mol) = 0.0315834 mol valine (6.90 g pro)/(115.13 g/mol) = 0.0599323 mol proline (7.30 g ser)/(105.10 g/mol) = 0.0694577 mol serine (86.00 g arg)/(174.21 g/mol) = 0.493657 mol arginine Divide by the smallest value (0.010101010 mol alanine), and round to a whole number. (0.0399627 mol glycine)/(0.010101010 mol) = 4 (0.010101010 mol alanine)/(0.010101010 mol) = 1 (0.0315834 mol valine)/(0.010101010 mol) = 3 (0.0599323 mol proline)/(0.010101010 mol) = 6 (0.0694577 mol serine)/(0.010101010 mol) = 7 (0.493657 mol arginine)/(0.010101010 mol) = 49 c) Mass of 70 moles of amino acids = (4 × 75.07 g/mol) + (1 × 89.09 g/mol) + (3 × 117.15 g/mol) + (6 × 115.13 g/mol) + (7 × 105.09 g/mol) + (49 × 174.20 g/mol) = 10,703.59 g To link 70 moles of amino acids, 69 peptide bonds are formed by the elimination of 69 moles of H2O. ⎛18.02 g H 2 O ⎞⎟ ⎟ = 1243.38 g Mass of H2O eliminated = (69 mol H 2 O )⎜⎜ ⎜⎝ 1 mol H 2 O ⎠⎟⎟ Minimum M of peptide = 10,703.59 g – 1243.38 g = 9,460 g/mol

15-56

15.117 When 2-butanone is reduced, equal amounts of both isomers are produced because the reaction does not favor the production of one over the other. In a 1:1 mixture of the two stereoisomers, the light rotated to the right by the other isomer cancels the light rotated to the left by one isomer. The mixture is not optically active since there is no net rotation of light. The two stereoisomers of 2-butanol are shown below. CH2CH3

CH2CH3

CH3 OH

H3C

H HO

15.118 From protein: ⎞⎛ ⎛ 8.0 g C16 H 24 O 5 N 4 ⎞⎟⎛⎜ 1 mol C16 H 24 O 5 N 4 ⎞⎛ 16 mol C ⎟⎟⎜⎜ ⎟⎟⎜⎜12.01 g C ⎞⎟⎟ ⎟⎟⎜⎜ Mass (g) of C = ⎜⎜⎜ ⎟⎟⎜ ⎟⎟⎜ ⎟ ⎟⎜⎝ 352.39 g C16 H 24 O 5 N 4 ⎠⎝ ⎜⎝ ⎟⎜⎜1 mol C16 H 24 O 5 N 4 ⎠⎝ ⎟⎜⎜ 1 mol C ⎠⎟⎟ L ⎠⎜

= 4.3624 g C/L From carbohydrate: ⎛12.0 g CH 2 O ⎞⎟⎛⎜ 1 mol CH 2 O ⎞⎛ ⎟⎟⎜⎜ 1 mol C ⎞⎛ ⎟⎟⎜⎜12.01 g C ⎞⎟⎟ ⎟⎟⎜⎜ Mass (g) of C = ⎜⎜⎜ ⎟⎟⎜ ⎟⎟⎜ ⎟ = 4.7992 g C/L ⎟⎜⎝ 30.03 g CH 2 O ⎠⎝ L ⎟⎜⎜1 mol CH 2 O ⎠⎝ ⎟⎜⎜ 1 mol C ⎠⎟⎟ ⎝⎜ ⎠⎜

From fat: ⎛ 2.0 g C 8 H16 O ⎞⎟⎜⎛ 1 mol C 8 H16 O ⎞⎛ ⎟⎟⎜⎜ 8 mol C ⎞⎛ ⎟⎟⎜⎜12.01 g C ⎞⎟⎟ ⎟⎟⎜⎜ Mass (g) of C = ⎜⎜⎜ ⎟⎟⎜ ⎟⎟⎟⎜⎜⎜ 1 mol C ⎟⎟⎟ = 1.49879 g C/L ⎟⎜⎝128.21 g C 8 H16 O ⎠⎝ L ⎟⎜⎜1 mol C 8 H16 O ⎠⎝ ⎝⎜ ⎠⎜ ⎠

TOC = (4.3624 g C/L) + (4.7992 g C/L) + (1.49879 g C/L) = 10.66039 = 10.7 g C/L

15.119 Plan: The Table of Functional Groups is used to identify the three functional groups. Aldehydes are reduced to produce alcohols while aldehydes are oxidized to produce carboxylic acids. Solution: a)

H R C O R OH aldehyde alcohol functional group functional group

O R C OH carboxylic acid functional group

Structure A is retinol, the alcohol, Structure B is retinoic acid, the carboxylic acid, and Structure C is retinal, the aldehyde. b) Converting the aldehyde retinal to retinol, the alcohol, is a reduction. Converting retinal to retinoic acid is an oxidation. c) Beta-carotene consists of two retinal molecules (minus the aldehyde functional group) joined together. The molecule is cleaved in half and oxidized to produce retinal.

15-57

CHAPTER 16 KINETICS: RATES AND MECHANISMS OF CHEMICAL REACTIONS FOLLOW–UP PROBLEMS 16.1A

Plan: Balance the equation. The rate in terms of the change in concentration with time for each substance is

1 Δ[A] , where a is the coefficient of the reactant or product A. The rate of the reactants is given a a Δt negative sign. Solution: a) The balanced equation is 4NO(g) + O2(g) → 2N2O3. Choose O2 as the reference because its coefficient is 1. Four molecules of NO (nitrogen monoxide) are consumed for every one O2 molecule, so the rate of O2 disappearance is 1/4 the rate of NO decrease. By similar reasoning, the rate of O2 disappearance is 1/2 the rate of N2O3 (dinitrogen trioxide) increase. Δ ⎡ O2 ⎤ 1 Δ[NO] 1 Δ ⎡⎣ N 2 O 3 ⎤⎦ Rate = − =− ⎣ ⎦ = + 4 Δt Δt 2 Δt b) Plan: Because NO is decreasing; its rate of concentration change is negative. Substitute the negative value into the expression and solve for Δ[O2]/Δt. Solution: expressed as

Δ ⎡ O2 ⎤ 1 −1.60 × 10−4 mol/L • s = − ⎣ ⎦ = 4.00 × 10–5 mol/L·s Δt 4 Plan: Examine the equation that expresses the rate in terms of the change in concentration with time for each Rate = −

16.1B

(

)

substance. The number in the denominator of each fraction is the coefficient for the corresponding substance in the balanced equation. The terms that are negative in the rate equation represent reactants in the balanced equation, while the terms that are positive represent products in the balanced equation. For example, the following 1 Δ[ A ] term, , describes a product (the term is positive) that has a coefficient of a in the balanced equation. In a Δt part (b), use the rate equation to compare the rate of appearance of H2O with the rate of disappearance of O2. Solution: a) The balanced equation is 4NH3 + 5O2 → 4NO + 6H2O. b) Plan: Because H2O is increasing; its rate of concentration change is positive. Substitute its rate into the expression and solve for Δ[O2]/Δt. Solution:

(

)

1 1 Δ[O 2 ] 2.52 × 10−2 mol / L • s = – 6 5 Δt Δ [O ] 5 2 = –2.10 × 10–2 mol/L·s − (2.52 × 10 –2 mol / L • s) = Δt 6 The negative value indicates that [O2] is decreasing as the reaction progresses. The rate of reaction is always expressed as a positive number, so [O2] is decreasing at a rate of 2.10 × 10–2 mol/L·s. Rate =

16.2A

16-1

Solution: a) The exponent of [I–] is 1, so the reaction is first order with respect to I–. Similarly, the reaction is first order with respect to BrO3–, and second order with respect to H+. The overall reaction order is (1 + 1 + 2) = 4, or fourth order overall. b) Rate = k[I–][BrO3–][H+]2. If [BrO3–] and [I–] are tripled and [H+] is doubled, rate = k[3 × I–][3 × BrO3–][2 × H+]2, then rate increases to 3 × 3 × 22 or 36 times its original value. The rate increases by a factor of 36. 16.2B

Plan: The reaction orders of the reactants are the exponents in the rate law. Add the individual reaction orders to obtain the overall reaction order. Use the rate law to determine how the changes listed in the problem will affect the rate. Solution: a) The exponent of [ClO2] is 2, so the reaction is second order with respect to ClO2. Similarly, the reaction is first order with respect to OH–. The overall reaction order is (1 + 2) = 3, or third order overall. b) Rate = k[ClO2]2[OH–]. If [ClO2] is halved and [OH–] is doubled, rate = k[1/2 × ClO2]2[2 × OH–], then rate increases to (1/2)2 × 2 or 1/2 its original value. The rate decreases by a factor of 1/2.

16.3A

Plan: Assume that the rate law takes the general form rate = k[H2]m[I2]n. To find how the rate varies with respect to [H2], find two experiments in which [H2] changes but [I2] remains constant. Take the ratio of rate laws for those two experiments to find m. To find how the rate varies with respect to [I2], find two experiments in which [I2] changes but [H2] remains constant. Take the ratio of rate laws for those two experiments to find n. Add m and n to obtain the overall reaction order. Use the rate law to solve for the value of k. Solution: For the reaction order with respect to [H2], compare Experiments 1 and 3: m rate 3 [H ] = 2 3m rate 1 [ H 2 ]1

9.3 × 10−23 1.9 × 10

= −23

[0.0550]3 m [ 0.0113]1

4.8947 = (4.8672566)m Therefore, m = 1 If the reaction order was more complex, an alternate method of solving for m is: log (4.8947) = m log (4.8672566); m = log (4.8947)/log (4.8672566) = 1 For the reaction order with respect to [I2], compare Experiments 2 and 4: n rate 4 [I ] = 2 4n rate 2 [ I 2 ]2

1.9 ×10−22

[0.0056]4 = n −22 [ 0.0033]2 1.1 ×10

1.72727 = (1.69697)n Therefore, n = 1 The rate law is rate = k[H2][I2] and is second order overall. Calculation of k: k = Rate/([H2][I2]) k1 = (1.9 × 10–23 mol/L∙s)/[(0.0113 mol/L)(0.0011 mol/L)] = 1.5 × 10–18 L/mol∙s k2 = (1.1 × 10–22 mol/L∙s)/[(0.0220 mol/L)(0.0033 mol/L)] = 1.5 × 10–18 L/mol∙s k3 = (9.3 × 10–23 mol/L∙s)/[(0.0550 mol/L)(0.0011 mol/L)] = 1.5 × 10–18 L/mol∙s k4 = (1.9 × 10–22 mol/L∙s)/[(0.0220 mol/L)(0.0056 mol/L)] = 1.5 × 10–18 L/mol∙s Average k = 1.5 × 10–18 L/mol∙s Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

16-2

When [H2] = [I2] = 0.0320 mol/L: Rate = k[H2][I2] = (1.5 × 10–18 L/mol∙s)(0.0320 mol/L)(0.0320 mol/L) = 1.5 × 10–21 mol/L∙s 16.3B

Plan: Assume that the rate law takes the general form rate = k[H2SeO3]m[I–]n[H+]p. To find how the rate varies with respect to [H2SeO3], find two experiments in which [H2SeO3] changes but [I–] and [H+] remain constant. Take the ratio of rate laws for those two experiments to find m. To find how the rate varies with respect to [I–], find two experiments in which [I–] changes but [H2SeO3] and [H+] remain constant. Take the ratio of rate laws for those two experiments to find n. To find how the rate varies with respect to [H+], find two experiments in which [H+] changes but [H2SeO3] and [I–] remain constant. Take the ratio of rate laws for those two experiments to find p. Add m, n, and p to obtain the overall reaction order. Use the rate law to solve for the value of k. Solution: For the reaction order with respect to [H2SeO3], compare Experiments 1 and 3: rate 3 [H 2 SeO3 ]3m = [H 2 SeO3 ]1m rate 1 [1.0 ×10 –2 mol/L]3m 3.94 ×10 –6 mol/L • s = –7 9.85 ×10 mol/L • s [2.5 ×10 –3 mol/L]1m

4 = (4)m Therefore, m = 1 For the reaction order with respect to [I–], compare Experiments 1 and 2: [I – ]n rate 2 = – 2n rate 1 [I ]1 [3.0 ×10 –2 mol/L]2n 7.88 ×10 –6 mol/L • s = 9.85 ×10 –7 mol/L • s [1.5 ×10 –2 mol/L]1n

8 = (2)n Therefore, n = 3 For the reaction order with respect to [H+], compare Experiments 2 and 4: [H+]4p rate 4 = rate 2 [H+]2p [3.0 ×10 –2 mol/L]4p 3.15 ×10 –5 mol/L • s = 7.88 ×10 –6 mol/L • s [1.5 ×10 –2 mol/L]2p 4 = (2)p Therefore, p = 2 The rate law is rate = k[H2SeO3][I–]3[H+]2 and is sixth order overall. b) Calculation of k: k = Rate/([H2SeO3][I–]3[H+]2) k1 = (9.85 × 10–7 mol/L∙s)/[(2.5 × 10–3 mol/L)(1.5 × 10–2 mol/L)3(1.5 × 10–2 mol/L)2] = 5.2 × 105 L5/mol5∙s c) When [H2SeO3] = 4.5 × 10–3 mol/L, [I–] = 2.5 × 10–2 mol/L, and [H+] = 7.0 × 10–2 mol/L: Rate = k[H2SeO3][I–]3[H+]2 = (5.2 × 105 L5/mol5 ∙ s) (4.5 × 10–3 mol/L)(2.5 × 10–2 mol/L)3(7.0 × 10–2 mol/L)2 = 1.8 × 10–4 mol/L ∙ s

16.4A

Plan: The reaction is second order in X and zero order in Y. For part (a), compare the two amounts of reactant X. For part (b), compare the two rate values. Solution: a) Since the rate law is rate = k[X]2, the reaction is zero order in Y. In Experiment 2, the amount of reactant X has not changed from Experiment 1 and the amount of Y has doubled. The rate is not affected by the doubling of Y since Y is zero order. Since the amount of X is the same, the rate has not changed. The initial rate of Experiment 2 is also 0.25 × 10–5 mol/L·s.

16-3

b)) The rate of Experiment E 3 iss four times thee rate in Experriment 1. Sincee the reaction iss second order in X, the co oncentration off X must have doubled to cau use a four-fold increase in ratte. There should be 6 black sppheres and 3 grreen spheres in n Experiment 3. 3 1.0 × 10−5 [ x ]2 = 2 −5 [3 ] 0.25 ×10

[ x ]2

9 x = 6 black b spheres

16.4B

Pllan: Examine how a change in the concentrration of the diifferent reactannts affects the rrate in order too determine th he rate law. Theen use the rate law to determiine the numberr of particles inn the scene for Experiment 4.. So olution: a)) Comparing Experiments E 1 and a 2, we can see s that the num mber of blue A spheres does not change whhile the nu umber of yellow B spheres ch hanges from 4 to 2. Althoughh the number oof yellow spherres changes, thee rate does no ot change. Theerefore, B has no n effect on thee rate, which suuggests that thee reaction is zeero order with rrespect to B. Now N that we know that the yelllow B spheress do not affect tthe rate, we caan look at the innfluence of thee blue A sp pheres. Compaaring Experimeents 2 and 3, wee can see that tthe number of bblue spheres chhanges from 4 to 2 (eexperiment 3 co oncentration off A is half of th hat of experimeent 2) while thhe rate changess from 1.6 × 100–3 mol/L∙s to –4 8..0 × 10 mol/L L∙s (the rate of experiment 3 is i half of the raate of experimeent 2). The fact that the conceentration of bllue spheres chaanges in the sam me way the ratte changes sugggests that the rreaction is firstt order with resspect to the bllue A spheres. Therefore, the rate law is: ra ate = k[A]. b)) The rate of Experiment E 4 iss twice the ratee in Experimennt 1. Since the rreaction is firstt order in A, thhe co oncentration off A must have doubled to cau use a two-fold iincrease in ratee. There shouldd be 2 × 4 partticles = 8 particles of A in the scene fo or Experiment 4.

16.5A

t decomposittion of phosphiine is a first-orrder reaction, sso use the first--order Pllan: The probleem states that the in ntegrated rate laaw. Use the iniitial amount off reactant and tthe container volume to calcuulate the initial co oncentration in n molarity of th he reactant. Sub bstitute the inittial and final cooncentrations aand the rate connstant into th he expression and a solve for tim me. So olution: ⎡ PH3 ⎤ 0.150 mol PH3 ⎦ 0 = kt ln ⎣ = 0.075 mol/L [PH3 ]0 = ⎡ PH3 ⎤ 2.0 L ⎣ ⎦t ln

0.07 75 mol/L 0.04 42 mol/L

(

)

= 0..0198 s−1 (t )

t = 29 s 16.5B

Pllan: The probleem states that the t decomposittion of hydrogeen peroxide is a first-order, reaction, so usee the firstorrder integrated rate law. Subsstitute the given n concentrationns and the timee into the expreession and solvve for the raate constant. So olution:

Copyright © McGraw-Hill Ed ducation. This is proprietary mateerial solely for au uthorized instructtor use. Not authoorized for sale or distribution in any manner.. This document may m not be copied d, scanned, dupliccated, forwarded d, distributed, or p posted on a websiite, in whole or paart.

16-4

a))

[A]0 = kt [A]t

1.28 8M = (k)(10.0 0 min) 0.85 5M

0.4094 = 0.040 094 = 0.041 min m –1 10.0 min

b)) If you start with w an initial co oncentration off hydrogen perroxide of 1.0 M ([A]0 = 1.0 M) and 25% of tthe sample deecomposes, 75% of the samplle remains ([A]t = 0.75 M). 1.00 0M ln 4 min–1)(t) = (0.04094 0.75 5M

t= 16.6A

0.2877 = 7.02736 = 7..0 min 0.0 04094 min –1

Pllan: The initiall scene contain ns 12 particles of o Substance X X. In the secondd scene, whichh occurs after 22.5 minutes haave elapsed, haalf of the particcles of Substan nce X (6 particlles) remain. Thherefore, 2.5 m min is the half-life. The haalf-life is used to find the num mber of particlees present at 5..0 min and 10.0 min. To findd the molarity oof X, moles off X is divided by b the given vo olume. So olution: a)) Since 2.5 min nutes is the halff-life, 5.0 minu utes represents two half-livess: 2.5 min 2.5 min → 6 particles → 3 paarticles of X 12 2 particles of X ⎯⎯⎯ p of X ⎯⎯⎯ After A 5.0 minutees, 3 particles of o X remain; 9 particles of X have reacted tto produce 9 paarticles of Y. Y Draw a scenee in which therre are 3 black X particles andd 9 red Y particcles.

b)) 10.0 minutes represents 4 half-lives: 2.5 min 2.5 min 2.5 min → 6 particles → 3 paarticles of X ⎯ ⎯⎯⎯ → 12 2 particles of X ⎯⎯⎯ p of X ⎯⎯⎯ 2.5 min 1..5 particles of X ⎯⎯⎯→ 0.75 particle of X ⎛ 0.20 mol ⎞⎟ ⎟⎟ = 00.15 mol Moles M of X afterr 10.0 min = (0.75 particle)⎜⎜⎜ ⎝1 particle ⎟⎠ mo ol X 0.15 mol m Molarity M = = = 0.30 M volu ume 0.50 L 16.6B

Pllan: The initial scene contains 16 particles of Substance A. Inn the second sccene, which occcurs after 24 minnutes have ellapsed, half of th he particles of Substance S A (8 particles) remaain. Therefore, 224 min is the haalf-life. The half-life is used to o find the amoun nt of time that has h passed when n only one partiicle of Substancce A remains annd to find the nuumber of paarticles of A preesent at 72 minu utes. To find thee molarity of A A, moles of A is divided by the given volume. So olution: a)) The half-life is i 24 minutes. We can use that information to determine tthe amount of ttime that has ppassed when on ne particle of Substance S A remains: 24 min

24 miin

24 minn

24 min

⎯ ⎯ → 8 A particles ⎯⎯⎯ ⎯⎯ → → 4 A par articles ⎯⎯⎯ ⎯⎯ → 2 A particlles ⎯⎯⎯⎯ → 1 A particle 16 6 A particles ⎯⎯⎯ After A 4 half-livees, only one parrticle of Substaance A remain s. 4 half-lives = 4 × (24 min/halff-life) = 96 miin Copyright © McGraw-Hill Ed ducation. This is proprietary mateerial solely for au uthorized instructtor use. Not authoorized for sale or distribution in any manner.. This document may m not be copied d, scanned, dupliccated, forwarded d, distributed, or p posted on a websiite, in whole or paart.

16-5

b) 72 min = 3 half-lives 24 min/half-life According to the scheme above in part (a), there are 2 A particles left after three half-lives.

Number of half-lives at 72 min =

16.7A

16.7B

16.8A

⎛ 0.10 mol A ⎞⎟ ⎟ = 0.20 mol A Amount (mol) of A after 72 minutes = (2 particles A) ⎜⎜ ⎜⎝1 particle A ⎠⎟⎟ 0.20 mol A M= = 0.80 M 0.25 L Plan: Rearrange the first-order half-life equation to solve for k. Solution: ln 2 ln 2 k= = 0.0529119985 = 0.0529 h–1 = t1/2 13.1 h Plan: Rearrange the first-order half-life equation to solve for half-life. Determine the number of half-lives that pass in the 40 day period described in the problem and use this number to determine the amount of pesticide remaining. Solution: a) ln2 ln2 t½ = = = 7.7016 = 8 days k 9 × 10 –2 day –1 40 days b) Number of half-lives at 40 days = = 5 half-lives 8 days/half-life After each half-life, ½ of the sample remains. 1 After five half-lives, ½ × ½ × ½ × ½ × ½ = (½)5 = of the sample remains. 32 Plan: (a) The reaction is second order, so use the integrated second-order law. Substitute the given initial concentration, the time, and the rate constant into the expression and solve for the concentration after 10 s. (b) Substitute the given initial concentration and rate constant into the second-order half-life equation. Solution: (a)

1 1 − = kt [I]t [I]0

1 1 − = (7.0 × 109 L/mol i s)(10. s) [I]t 0.055 mol/L

[I]t = 1.4×10−11 mol /L (b) t1/2 = 16.8B

1 1 = = 4.1×10−10 s k[I]0 (7.0 ×10 9 L/mol i s)(0.35 mol/L)

Plan: (a) The reaction is second order, so use the integrated second-order law. Substitute the given concentrations and the rate constant into the expression and solve for time. (b) Substitute the time and rate constant into the second-order half-life equation and solve for initial concentration. Solution: 1 1 (a) − = kt ⎡ NOBr ⎤ ⎡ NOBr ⎤ ⎣ ⎦t ⎣ ⎦0 1 1 − = (0.80 L/mol i s)(t ) [0.015 mol/L] [0.075 mol/L] t = 67 s

16-6

1 ⎡ k ⎣ NOBr ⎤⎦ 0 1 ⎡ NOBr ⎤ = 1 ⎣ ⎦ 0 (t )(k ) = (45 s)(0.80 L/mol i s) = 0.028 mol /L 1/2

(b) t1/2 =

16.9A

Plan: The activation energy, rate constant at T1, and a second temperature, T2, are given. Substitute these values into the Arrhenius equation and solve for k2, the rate constant at T2. Solution: k E ⎛1 1⎞ ln 2 = a ⎜⎜ − ⎟⎟⎟ k1 R ⎜⎝ T1 T2 ⎠⎟ k1 = 0.286 L/mol • s k2 = ? L/mol • s

T1 = 500 K Ea = 1.00 × 102 kJ/mol T2 = 490 K 3 1.00 × 10 2 kJ /mol ⎜⎛ 1 k2 ⎟⎜10 J ⎞⎟⎟ 1 ⎞⎛ ⎜⎜ ln = − ⎟⎟⎟⎜⎜ ⎟ 0.286 L/mol • s 8.314 J /mol • K ⎜⎜⎝ 500. K 490. K ⎠⎝ ⎟⎜⎜ 1 kJ ⎠⎟⎟

k2 = –0.49093 0.286 L/mol • s

k2 = 0.612057 0.286 L/mol • s k2 = (0.612057)(0.286 L/mol ∙ s) = 0.175048 = 0.175 L/mol∙s 16.9B

Plan: The activation energy and rate constant at T1 are given. We are asked to find the temperature at which the rate will be twice as fast (i.e., the temperature at which k2 = 2 × k1). Substitute the given values into the Arrhenius equation and solve for T2. Solution: k E ⎛1 1⎞ ln 2 = a ⎜⎜ − ⎟⎟⎟ k1 R ⎜⎝ T1 T2 ⎠⎟ k1 = 7.02 × 10–3 L/mol ∙ s

T1 = 500. K

–3

k2 = 2(7.02 × 10 ) L/mol ∙ s

2 (7.02 ×10 L/mol • s) –3

(7.02 ×10 L/mol • s) –3

Ea = 1.14 × 105 J/mol

T2 = unknown

1.14 ×105 J/mol ⎛⎜ 1 1⎞ – ⎟⎟⎟ ⎜⎜ 8.314 J/mol • K ⎜⎝ 500. K T2 ⎠⎟

⎛ 1 1⎞ 5.055110 × 10–5 K–1 = ⎜⎜ – ⎟⎟⎟ ⎜⎝ 500. K T2 ⎠⎟

1 = 1. 9494489 × 10–3 K–1 T2

T2 = 513 K 16.10A Plan: Begin by using Sample Problem 16.9 as a guide for labeling the diagram. Solution: The reaction energy diagram indicates that O(g) + H2O(g) → 2OH(g) is an endothermic process, because the energy of the product is higher than the energy of the reactants. The highest point on the curve indicates the transition state. In the transition state, an oxygen atom forms a bond with one of the hydrogen atoms on the H2O molecule (hashed line) and the O−H bond (dashed line) in H2O weakens. Ea(fwd) is the sum of ΔHrxn and Ea(rev).

16-7

Transition state: O

Energy

Ea(rev)= +6 kJ

Ea(fwd)= +78 kJ

DHrxn= +72 kJ

Reaction progress 16.10B Plan: Ea(fwd) is the sum of DHrxn and Ea(rev), so DHrxn can be calculated by subtracting Ea(fwd) – Ea(rev). Use Sample Problem 16.9 as a guide for drawing the diagram. Solution: DHrxn = Ea(fwd) – Ea(rev) DHrxn = 7 kJ – 72 kJ = –65 kJ The negative sign of the enthalpy indicates that the reaction is exothermic. This means that the energy of the products is lower than the energy of the reactants. The highest point on the curve indicates the transition state. In the transition state, a bond is forming between the chlorine and the hydrogen (hashed line) while the bond between the hydrogen and the bromine weakens (another hashed line): Transition state:

16.11A Plan: The overall reaction can be obtained by adding the three steps together. The molecularity of each step is the total number of reactant particles; the molecularities are used as the orders in the rate law for each step. Solution: a) (1) H2O2(aq)  2OH(aq) (2) H2O2(aq) + OH(aq)  H2O(l) + HO2(aq) (3) HO2(aq) + OH(aq)  H2O(l) + O2(g) Total: 2H2O2(aq) + 2OH(aq) + HO2(aq)  2OH(aq) + 2H2O(l) + HO2(aq) + O2(g) (overall) 2H2O2(aq)  2H2O(l) + O2(g)

16-8

(1) Unimolecular (2) Bimolecular (3) Bimolecular (1) Rate1 = k1 [H2O2] (2) Rate2 = k2 [H2O2][OH] (3) Rate3 = k3 [HO2][OH]

16.11B Plan: The overall reaction can be obtained by adding the three steps together. The molecularity of each step is the total number of reactant particles; the molecularities are used as the orders in the rate law for each step. Solution: a) (1) 2NO(g) ⇆ N2O2(g) (2) N2O2(g) + H2(g) → N2O(g) + H2O(g) (3) N2O(g) +H2(g) → N2(g) + H2O(g) Total: 2NO(g) + N2O2(g) + 2 H2(g) + N2O(g) → N2O2(g) + N2O(g) + 2H2O(g) + N2(g) (overall) 2NO(g) + 2H2(g) → 2H2O(g) + N2(g) b) (1) Bimolecular (2) Bimolecular (3) Bimolecular c) (1) Rate1 = k1 [NO]2 (2) Rate2 = k2 [N2O2][H2] (3) Rate3 = k3 [N2O][H2] 16.12A Plan: The overall reaction can be obtained by adding the three steps together. An intermediate is a substance that is formed in one step and consumed in a subsequent step. The overall rate law for the mechanism is determined from the slowest step (the rate-determining step) and can be compared to the experimental rate law. Solution: a) (1) H2O2(aq) → 2OH(aq) (2) H2O2(aq) + OH(aq) → H2O(l) + HO2(aq) (3) HO2(aq) + OH(aq) → H2O(l) + O2(g) Total: 2H2O2(aq) + 2OH(aq) + HO2(aq) → 2OH(aq) + 2H2O(l) + HO2(aq) + O2(g) (overall) 2H2O2(aq) → 2H2O(l) + O2(g) 2OH(aq) and HO2(aq) are intermediates in the given mechanism. 2OH(aq) are produced in the first step and consumed in the second and third steps; HO2(aq) is produced in the second step and consumed in the third step. Notice that the intermediates were not included in the overall reaction. b) The observed rate law is: rate = k[H2O2]. In order for the mechanism to be consistent with the rate law, the first step must be the slow step. The rate law for step one is the same as the observed rate law. 16.12B Plan: The overall reaction can be obtained by adding the three steps together. An intermediate is a substance that is formed in one step and consumed in a subsequent step. The overall rate law for the mechanism is determined from the slowest step (the rate-determining step) and can be compared to the experimental rate law. Solution: a) (1) 2NO(g) ⇆ N2O2(g) (2) N2O2(g) + H2(g) → N2O(g) + H2O(g) (3) N2O(g) +H2(g) → N2(g) + H2O(g) Total: 2NO(g) + N2O2(g) + 2 H2(g) + N2O(g) → N2O2(g) + N2O(g) + 2H2O(g) + N2(g) (overall) 2NO(g) + 2 H2(g) → 2H2O(g) + N2(g) N2O2(g) and N2O(g) are intermediates in the given mechanism. N2O2(g) is produced in the first step and consumed in the second step; N2O(g) is produced in the second step and consumed in the third step. Notice that the intermediates were not included in the overall reaction. Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

16-9

b) The observed rate law is: rate = k[NO]2[H2], and the second step is the slow, or rate-determining, step. The rate law for step two is: rate = k2[N2O2][H2]. This rate law is NOT the same as the observed rate law. However, since N2O2 is an intermediate, it must be replaced by using the first step. For an equilibrium, rateforward rxn = ratereverse rxn. For step 1 then, k1[NO]2 = k–1[N2O2]. Rearranging to solve for [N2O2] gives [N2O2] = (k1/k–1)[NO]2. Substituting this value for [N2O2] into the rate law for the second step gives the overall rate law as rate = (k2k1/k–1)[NO]2[H2] or rate = k[NO]2[H2], which is consistent with the observed rate law. CHEMICAL CONNECTIONS BOXED READING PROBLEMS B16.1

Plan: Add the two equations, canceling substances that appear on both sides of the arrow. The rate law for each step follows the format of rate = k[reactants]. An initial reactant that appears as a product in a subsequent step is a catalyst; a product that appears as a reactant in a subsequent step is an intermediate (produced in one step and consumed in a subsequent step). Solution: a) Add the two equations: (1) X(g) + O3(g) → XO(g) + O2(g) (2) XO(g) + O(g) → X(g) + O2(g) Overall O3(g) + O(g) → 2O2(g) Rate laws: Step 1 Rate1 = k1[X][O3] Step 2 Rate2 = k2[XO][O] b) X acts as a catalyst; it is a reactant in step 1 and a product in step 2. XO acts as an intermediate; it was produced in step 1 and consumed in step 2.

B16.2

Plan: Replace X in the mechanism in B16.1 with NO, the catalyst. To find the rate of ozone depletion at the given concentrations, use step (1) since it is the rate-determining (slow) step. Solution: a) (1) NO(g) + O3(g) → NO2(g) + O2(g) (2) NO2(g) + O(g) → NO(g) + O2(g) Overall O3(g) + O(g) → 2O2(g) b) Rate1 = k1[NO][O3] = (6 × 10–15 cm3/molecule ∙ s)[1.0 × 109 molecule/cm3][5 × 1012 molecule/cm3] = 3 × 107 molecule/s

B16.3

Plan: The p factor is the orientation probability factor and is related to the structural complexity of the colliding particles. The more complex the particles, the smaller the probability that collisions will occur with the correct orientation and the smaller the p factor. Solution: a) Step 1 with Cl will have the higher value for the p factor. Since Cl is a single atom, no matter how it collides with the ozone molecule, the two particles should react, if the collision has enough energy. NO is a molecule. If the O3 molecule collides with the N atom in the NO molecule, reaction can occur as a bond can form between N and O; if the O3 molecule collides with the O atom in the NO molecule, reaction will not occur as the bond between N and O cannot form. The probability of a successful collision is smaller with NO. b) The transition state would have weak bonds between the chlorine atom and an oxygen atom in ozone, and between that oxygen atom and a second oxygen atom in ozone.

O Cl

16-10

END–OF–CHAPTER PROBLEMS 16.1

Changes in concentrations of reactants (or products) as functions of time are measured to determine the reaction rate.

16.2

Rate is proportional to concentration. An increase in pressure will increase the number of gas molecules per unit volume. In other words, the gas concentration increases due to increased pressure, so the reaction rate increases. Increased pressure also causes more collisions between gas molecules.

16.3

The addition of more water will dilute the concentrations of all solutes dissolved in the reaction vessel. If any of these solutes are reactants, the rate of the reaction will decrease.

16.4

An increase in solid surface area would allow more gaseous components to react per unit time and thus would increase the reaction rate.

16.5

An increase in temperature affects the rate of a reaction by increasing the number of collisions, but more importantly, the energy of collisions increases. As the energy of collisions increases, more collisions result in reaction (i.e., reactants → products), so the rate of reaction increases.

16.6

The second experiment proceeds at the higher rate. I2 in the gaseous state would experience more collisions with gaseous H2.

16.7

The reaction rate is the change in the concentration of reactants or products per unit time. Reaction rates change with time because reactant concentrations decrease, while product concentrations increase with time.

16.8

a) For most reactions, the rate of the reaction changes as a reaction progresses. The instantaneous rate is the rate at one point, or instant, during the reaction. The average rate is the average of the instantaneous rates over a period of time. On a graph of reactant concentration vs. time of reaction, the instantaneous rate is the slope of the tangent to the curve at any one point. The average rate is the slope of the line connecting two points on the curve. The closer together the two points (shorter the time interval), the more closely the average rate agrees with the instantaneous rate. b) The initial rate is the instantaneous rate at the point on the graph where time = 0, that is when reactants are mixed.

16.9

The calculation of the overall rate is the difference between the forward and reverse rates. This complication may be avoided by measuring the initial rate, where product concentrations are negligible, so the reverse rate is negligible. Additionally, the calculations are simplified as the reactant concentrations can easily be determined from the volumes and concentrations of the solutions mixed.

16.10

At time t = 0, no product has formed, so the B(g) curve must start at the origin. Reactant concentration (A(g)) decreases with time; product concentration (B(g)) increases with time. Many correct graphs can be drawn. Two examples are shown below. The graph on the left shows a reaction that proceeds nearly to completion, i.e., [products] >> [reactants] at the end of the reaction. The graph on the right shows a reaction that does not proceed to completion, i.e., [reactants] > [products] at reaction end.

A(g)

Time

Concentration

B(g) A(g)

B(g) Time

16-11

16.11

a) Calculate the slope of the line connecting (0, [C]o) and (tf, [C]f) (final time and concentration). The negative of this slope is the average rate. b) Calculate the negative of the slope of the line tangent to the curve at t = x. c) Calculate the negative of the slope of the line tangent to the curve at t = 0. d) If you plotted [D] vs. time, you would not need to take the negative of the slopes in (a)-(c) since [D] would increase over time.

16.12

Plan: The average rate is the total change in concentration divided by the total change in time. Solution: a) The average rate from t = 0 to t = 20.0 s is proportional to the slope of the line connecting these two points: 1 Δ[AX 2 ] 1 (0.0088 mol/L − 0.0500 mol/L ) Rate = − =− = 0.00103 = 0.0010 mol/L • s 2 Δt 2 (20.0 s − 0 s) The negative of the slope is used because rate is defined as the change in product concentration with time. If a reactant is used, the rate is the negative of the change in reactant concentration. The 1/2 factor is included to account for the stoichiometric coefficient of 2 for AX2 in the reaction. b)

[AX2] vs time 0.06 0.05

[AX2]

0.04 0.03 0.02 0.01 0 0

time, s The slope of the tangent to the curve (dashed line) at t = 0 is approximately –0.004 mol/L∙s. This initial rate is greater than the average rate as calculated in part (a). The initial rate is greater than the average rate because rate decreases as reactant concentration decreases. 16.13

Plan: The average rate is the total change in concentration divided by the total change in time. Solution: a) Rate = −

1 Δ[AX 2 ] 1 (0.0088 mol/L − 0.0249 mol/L ) =− = 6.70833 × 10–4 = 6.71 × 10–4 mol/L • s 2 Δt 2 (20.0 s − 8.0 s)

b) The rate at exactly 5.0 s will be higher than the rate in part (a). The slope of the tangent to the curve at t = 5.0 s (the rate at 5.0 s) is approximately –2.8 × 10–3 mol/L • s.

16-12

0.06 0.05

[AX2]

0.04 0.03 0.02 0.01 0 0

time, s

16.14

Plan: Use Equation 16.2 to describe the rate of this reaction in terms of reactant disappearance and product appearance. A negative sign is used for the rate in terms of reactant A since A is reacting and [A] is decreasing over time. Positive signs are used for the rate in terms of products B and C since B and C are being formed and [B] and [C] increase over time. Reactant A decreases twice as fast as product C increases because two molecules of A disappear for every molecule of C that appears. Solution: Expressing the rate in terms of each component: 1 Δ[A] Δ[B] Δ[C] Rate = − = = 2 Δt Δt Δt Calculating the rate of change of [A]: 1 Δ[A] Δ[C] − = 2 Δt Δt ⎛ 2 mol A/L • s ⎞⎟ (2 mol C/L • s)⎜⎜ = –4 mol/L • s ⎝ 1 mol C/L • s ⎠⎟⎟ The negative value indicates that [A] is decreasing as the reaction progresses. The rate of reaction is always expressed as a positive number, so [A] is decreasing at a rate of 4 mol/L • s.

16.15

Plan: Use Equation 16.2 to describe the rate of this reaction in terms of reactant disappearance and product appearance. A negative sign is used for the rate in terms of reactant D since D is reacting and [D] is decreasing over time. Positive signs are used for the rate in terms of products E and F since E and F are being formed and [E] and [F] increase over time. For every 3/2 mole of product E that is formed, 5/2 mole of F is produced. Solution: Expressing the rate in terms of each component: Δ[D] 2 Δ[E] 2 Δ[F] Rate = − = = Δt 3 Δt 5 Δt Calculating the rate of change of [F]:

⎛ 5/2 mol F/L • s ⎞⎟ ⎟ = 0.416667 = 0.42 mol/L • s ⎝ 3/2 mol E/L • s ⎠⎟

(0.25 mol E/L • s)⎜⎜⎜ 16.16

Plan: Use Equation 16.2 to describe the rate of this reaction in terms of reactant disappearance and product appearance. A negative sign is used for the rate in terms of reactants A and B since A and B are reacting and [A] and [B] are decreasing over time. A positive sign is used for the rate in terms of product C since C is being

16-13

formed and [C] increases over time. The 1/2 factor is included for reactant B to account for the stoichiometric coefficient of 2 for B in the reaction. Reactant A decreases half as fast as reactant B decreases because one molecule of A disappears for every two molecules of B that disappear. Solution: Expressing the rate in terms of each component: Δ[A] 1 Δ[B] Δ[C] Rate = − =− = Δt 2 Δt Δt Calculating the rate of change of [A]: ⎛1 mol A/L • s ⎞⎟ = – 0.25 mol/L • s = – 0.2 mol/L • s (0.5 mol B/L • s)⎜⎜ ⎝ 2 mol B/L • s ⎠⎟⎟ The negative value indicates that [A] is decreasing as the reaction progresses. The rate of reaction is always expressed as a positive number, so [A] is decreasing at a rate of 0.2 mol/L • s. 16.17

Plan: Use Equation 16.2 to describe the rate of this reaction in terms of reactant disappearance and product appearance. A negative sign is used for the rate in terms of reactants D, E, and F since these substances are reacting and [D], [E], and [F] are decreasing over time. Positive signs are used for the rate in terms of products G and H since these substances are being formed and [G] and [H] increase over time. Product H increases half as fast as reactant D decreases because one molecule of H is formed for every two molecules of D that disappear. Solution: Expressing the rate in terms of each component: 1 Δ[D] 1 Δ[E] Δ[F] 1 Δ[G] Δ[H] Rate = − =− =− = = 2 Δt 3 Δt Δt 2 Δt Δt Calculating the rate of change of [H]: ⎛ 1 mol H/L • s ⎞⎟ (0.1 mol D/L • s)⎜⎜ = 0.05 mol/L • s ⎝ 2 mol D/L • s ⎠⎟⎟

16.18

Plan: A term with a negative sign is a reactant; a term with a positive sign is a product. The inverse of the fraction becomes the coefficient of the molecule. Solution: N2O5 is the reactant; NO2 and O2 are products. 2N2O5(g) → 4NO2(g) + O2(g)

16.19

Plan: A term with a negative sign is a reactant; a term with a positive sign is a product. The inverse of the fraction becomes the coefficient of the molecule. Solution: CH4 and O2 are the reactants; H2O and CO2 are products. CH4 + 2O2→ 2H2O + CO2

16.20

Plan: The average rate is the total change in concentration divided by the total change in time. The initial rate is the slope of the tangent to the curve at t = 0.0 s and the rate at 7.00 s is the slope of the tangent to the curve at t = 7.00 s Solution: Δ[NOBr] 0.0033 − 0.0100 mol/L – = 6.7 × 10–4 mol/L • s Δt 10.00 − 0.00 s 0.0055 − 0.0071 mol/L = 8.0 × 10–4 mol/L • s b) Rate = − 4.00 − 2.00 s

a) Rate = −

16-14

c))

Initial Rate R = – Δ y/Δ Δ x = – [(0.004 40 – 0.0100) m mol/L]/[(4.00 – 0.00) s] = 1.55 × 10–3 mol/L∙s d)) Rate at 7.00 s = – [(0.0030 – 0.0050) mol//L]/[(11.00 – 44.00) s] = 2.8577 × 10–4 = 2.9 × 10–4 mol/L∙s e)) Average betw ween t = 3 s and d t = 5 s is: Rate = – [(0.0050 – 0.0063) 0 mol/L]/[(5.00 – 3.00)) s] = 6.5 × 100–4 mol/L∙s –4 Rate att 4 s ≈ 6.7 × 10 0 mol/L∙s, thu us the rates aree equal at abouut 4 seconds. 16.21

16.22

Pllan: Use Equattion 16.2 to desscribe the rate of this reactionn in terms of reeactant disappeearance and prooduct ap ppearance. A negative n sign iss used for the raate in terms off the reactant O2 since it is reaacting and [O2] is deecreasing over time. A positiv ve sign is used d for the rate inn terms of the pproduct O3 sincce it is being foormed and [O O3] increases ov ver time. O3 increases 2/3 as fast as O2 decrreases because two moleculess of O3 are form med for ev very three moleecules of O2 th hat disappear. So olution: O3 ] Δ 2] 1 Δ[O 1 Δ[O = a)) Rate = − 3 Δt 2 Δt Δ b)) Use the molee ratio in the baalanced equatio on: −5 ⎛ 2.17 ol O 2 /L • s ⎞⎛ ⎟⎟⎜ 2 mol O 3 /L • s ⎞⎟⎟ = 1.446667 × 10–5 = 1.45 × 10–5 mol/L • s ⎜⎜ 2 ×10 mo ⎟⎜⎜⎜ ⎜⎜ ⎟⎠⎝ 3 mol O 2 /L • s ⎟⎟⎠ ⎝

16.23

a)) k is the rate co onstant, the pro oportionality constant c in the rrate law. k reprresents the fracction of successsful co ollisions which h includes the fraction f of colliisions with suffficient energy and the fractioon of collisionss with co orrect orientation. k is a consttant that variess with temperatture.

16-15

b) m represents the order of the reaction with respect to [A] and n represents the order of the reaction with respect to [B]. The order is the exponent in the relationship between rate and reactant concentration and defines how reactant concentration influences rate. The order of a reactant does not necessarily equal its stoichiometric coefficient in the balanced equation. If a reaction is an elementary reaction, meaning the reaction occurs in only one step, then the orders and stoichiometric coefficients are equal. However, if a reaction occurs in a series of elementary reactions, called a mechanism, then the rate law is based on the slowest elementary reaction in the mechanism. The orders of the reactants will equal the stoichiometric coefficients of the reactants in the slowest elementary reaction but may not equal the stoichiometric coefficients in the overall reaction. c) For the rate law rate = k[A] [B]2 substitute in the units: Rate (mol/L·min) = k[A]1[B]2

mol/L • min rate mol/L • min = = 1 2 1 2 mol3 [A] [B] ⎡ mol ⎤ ⎡ mol ⎤ ⎢ ⎥ ⎢ ⎥ ⎢⎣ L ⎥⎦ ⎢⎣ L ⎥⎦ L3 mol ⎛⎜ L3 ⎞⎟ ⎟ k= ⎜ L • min ⎜⎝ mol3 ⎠⎟⎟

k = L2/mol2∙ min 16.24

a) Plot either [A2] or [B2] vs. time and determine the negative of the slope of the line tangent to the curve at t = 0. b) A series of experiments at constant temperature but with different initial concentrations are run to determine different initial rates. By comparing results in which only the initial concentration of a single reactant is changed, the order of the reaction with respect to that reactant can be determined. c) When the order of each reactant is known, any one experimental set of data (reactant concentration and reaction rate) can be used to determine the reaction rate constant at that temperature.

16.25

a) The rate doubles. If rate = k[A]1 and [A] is doubled, then the rate law becomes rate = k[2 × A]1. The rate increases by 21 or 2. b) The rate decreases by a factor of four. If rate = k[B]2 and [B] is halved, then the rate law becomes rate = k[1/2 × B]2. The rate decreases to (1/2)2 or 1/4 of its original value. c) The rate increases by a factor of nine. If rate = k[C]2 and [C] is tripled, then the rate law becomes rate = k[3 × C]2. The rate increases to 32 or 9 times its original value.

16.26

Plan: The order for each reactant is the exponent on the reactant concentration in the rate law. The individual orders are added to find the overall reaction order. Solution: The orders with respect to [BrO3–] and to [Br–] are both 1 since both have an exponent of 1. The order with respect to [H+] is 2 (its exponent in the rate law is 2). The overall reaction order is 1 + 1 + 2 = 4. first order with respect to BrO3–, first order with respect to Br– , second order with respect to H+, fourth order overall

16.27

Plan: The order for each reactant is the exponent on the reactant concentration in the rate law. The individual orders are added to find the overall reaction order. Solution: The rate law may be rewritten as rate = k[O3]2[O2] –1. The order with respect to [O3] is 2 since it has an exponent of 2. The order with respect to [O2] is –1 since it has an exponent of –1. The overall reaction order is 2 + (–1) = 1. second order with respect to O3, (–1) order with respect to O2, first order overall

16.28

a) The rate is first order with respect to [BrO3–]. If [BrO3–] is doubled, rate = k[2 × BrO3–], then rate increases to 21 or 2 times its original value. The rate doubles. b) The rate is first order with respect to [Br–]. If [Br–] is halved, rate = k[1/2 × Br–], then rate decreases by a factor of (1/2)1 or 1/2 times its original value. The rate is halved.

16-16

c) The rate is second order with respect to [H+]. If [H+] is quadrupled, rate = k[4 × H+]2, then rate increases to 42 or 16 times its original value. 16.29

a) The rate is second order with respect to [O3]. If [O3] is doubled, rate = k[2 × O3]2, then rate increases to 22 or 4 times its original value. The rate increases by a factor of 4. b) [O2] has an order of –1. If [O2] is doubled, rate = k[2 × O2] –1, then rate decreases to 2–1 or 1/2 times its original value. The rate decreases by a factor of 2. c) [O2] has an order of –1. If [O2] is halved, rate = k[1/2 × O2] –1, then rate decreases by a factor of (1/2)–1 or 2 times its original value. The rate increases by a factor of 2.

16.30

16.31

Plan: The order for each reactant is the exponent on the reactant concentration in the rate law. The individual orders are added to find the overall reaction order. Solution: The rate law may be rewritten as rate = k[HNO2]4[NO] –2. The order with respect to [HNO2] is 4, and the order with respect to [NO] is –2. The overall order is: 4 + (– 2) = 2 for the overall order.

16.32

a) The rate is second order with respect to [NO2]. If [NO2] is tripled, rate = k[3 × NO2]2, then rate increases to 32 or 9 times its original value. The rate increases by a factor of 9. b) The rate is second order with respect to [NO2] and first order with respect to [Cl2]. If [NO2] and [Cl2] are doubled, rate = k[2 × NO2]2[2 × Cl2]1, then the rate increases by a factor of 22 × 21 = 8. c) The rate is first order with respect to [Cl2]. If Cl2 is halved, rate = k[1/2 × Cl2]1, then rate decreases to 1/2 times its original value. The rate is halved.

16.33

a) The rate is fourth order with respect to [HNO2]. If [HNO2] is doubled, rate = k[2 × HNO2]4, then rate increases to 24 or 16 times its original value. The rate increases by a factor of 16. b) [NO] has an order of –2. If [NO] is doubled, rate = k[2 × NO]–2, then rate increases to 2–2 or 1/(2)2 = 1/4 times its original value. The rate decreases by a factor of 4. c) The rate is fourth order with respect to [HNO2]. If [HNO2] is halved, rate = k[1/2 × HNO2]4, then rate decreases to (1/2)4 or 1/16 times its original value. The rate decreases by a factor of 16.

16.34

Plan: The rate law is rate = [A]m[B]n where m and n are the orders of the reactants. To find the order of each reactant, take the ratio of the rate laws for two experiments in which only the reactant in question changes. Once the rate law is known, any experiment can be used to find the rate constant k. Solution: a) To find the order for reactant A, first identify the reaction experiments in which [A] changes but [B] is constant. Use experiments 1 and 2 (or 3 and 4 would work) to find the order with respect to [A]. Set up a ratio of the rate laws for experiments 1 and 2 and fill in the values given for rates and concentrations and solve for m, the order with respect to [A]. m ⎛[A]exp 2 ⎟⎞ rateexp 2 ⎜ ⎟ ⎜ =⎜ ⎟ rate ⎝⎜ [A] ⎟⎟⎠ exp 1

exp 1

45.0 mol/L • min ⎛ 0.300 mol/L ⎞⎟ = ⎜⎜ ⎟ 5.00 mol/L • min ⎝ 0.100 mol/L ⎠⎟ 9.00 = (3.00)m log (9.00) = m log (3.00) m=2 Using experiments 3 and 4 also gives second order with respect to [A]. m

16-17

To find the order for reactant B, first identify the reaction experiments in which [B] changes but [A] is constant. Use experiments 1 and 3 (or 2 and 4 would work) to find the order with respect to [B]. Set up a ratio of the rate laws for experiments 1 and 3 and fill in the values given for rates and concentrations and solve for n, the order with respect to [B]. n ⎛[B]exp 3 ⎞⎟ rateexp 3 ⎜ ⎟ ⎜ =⎜ ⎟ rate ⎝⎜[B] ⎠⎟⎟ exp 1

exp 1

10.0 mol/L • min ⎛ 0.200 mol/L ⎞⎟ = ⎜⎜ ⎝ 0.100 mol/L ⎠⎟⎟ 5.00 mol/L • min 2.00 = (2.00)n log (2.00) = n log (2.00) n=1 The reaction is first order with respect to [B]. b) The rate law, without a value for k, is rate = k[A]2[B]. c) Using experiment 1 to calculate k (the data from any of the experiments can be used): Rate = k[A]2[B] 5.00 mol/L • min rate k= = = 5.00 × 103 L2/mol2 •min [0.100 mol/L] 2 [0.100 mol/L] [A]2 [B] d) When [A] = [B] = 0.400 mol/L: Rate = k[A]2[B] = k[A]2[B] = (5.00 × 103 L2/mol2•min)[0.400 mol/L]2[0.400 mol/L] n

16.35

= 320 mol/L•min Plan: The rate law is rate = k [A]m[B]n[C]p where m, n, and p are the orders of the reactants. To find the order of each reactant, take the ratio of the rate laws for two experiments in which only the reactant in question changes. Once the rate law is known, any experiment can be used to find the rate constant k. Solution: a) To find the order for reactant A, first identify the reaction experiments in which [A] changes but [B] and [C] are constant. Use experiments 1 and 2 to find the order with respect to [A]. Set up a ratio of the rate laws for experiments 1 and 2 and fill in the values given for rates and concentrations and solve for m, the order with respect to [A]. m ⎛[A]exp 2 ⎟⎞ rateexp 2 ⎟⎟ = ⎜⎜⎜ ⎜⎝ [A]exp 1 ⎟⎟⎠ rateexp 1 ⎛ 0.1000 mol/L ⎞⎟ 1.25 ×10−2 mol/L • min = ⎜⎜ ⎟ −3 ⎝⎜ 0.0500 mol/L ⎠⎟ 6.25 ×10 mol/L • min 2.00 = (2.00)m log (2.00) = m log (2.00) m=1 The order is first order with respect to A. m

To find the order for reactant B, first identify the reaction experiments in which [B] changes but [A] and [C] are constant. Use experiments 2 and 3 to find the order with respect to [B]. Set up a ratio of the rate laws for experiments 2 and 3 and fill in the values given for rates and concentrations and solve for n, the order with respect to [B]. n ⎛[B]exp 3 ⎞⎟ rateexp 3 ⎜ ⎟ ⎜ =⎜ ⎟ rate ⎝⎜[B] ⎠⎟⎟ exp 2

exp 2

⎛ 0.1000 mol/L ⎞⎟ 5.00 ×10−2 mol/L • min = ⎜⎜ ⎟ −2 ⎝⎜ 0.0500 mol/L ⎠⎟ 1.25 ×10 mol/L • min 4.00 = (2.00)n log (4.00) = n log (2.00) n=2 The reaction is second order with respect to B. n

16-18

To find the order for reactant C, first identify the reaction experiments in which [C] changes but [A] and [B] are constant. Use experiments 1 and 4 to find the order with respect to [C]. Set up a ratio of the rate laws for experiments 1 and 4 and fill in the values given for rates and concentrations and solve for p, the order with respect to [C]. p ⎛[C]exp 4 ⎞⎟ rateexp 4 ⎜ ⎟⎟ = ⎜⎜ ⎜⎝ [C]exp 1 ⎠⎟⎟ rateexp 1 ⎛ 0.0200 mol/L ⎞⎟ 6.25 ×10−3 mol/L • min = ⎜⎜ ⎟ −3 ⎜⎝ 0.0100 mol/L ⎠⎟ 6.25 ×10 mol/L • min 1.00 = (2.00)p

log (1.00) = p log (2.00) p=0

The reaction is zero order with respect to C.

b) Rate = k[A]1[B]2[C]0 Rate = k[A][B]2 c) Using the data from experiment 1 to find k: Rate = k[A][B]2 k=

rate 6.25 ×10−3 mol/L • min = = 50.0 L2/mol2 • s [A][B]2 [0.0500 mol/L][0.0500 mol/L]2

d) When [A] = 0.2000 mol/L, [B] = 0.1500 mol/L, and [C] = 0.0300 mol/L: Rate = k[A][B]2[C]0 = k[A][B]2 = (50.0 L2/mol2 • s)[0.2000 mol/L][0.1500 mol/L]2 = 2.25 × 10–1 mol/L • s 16.36

Plan: Write the appropriate rate law and enter the units for rate and concentrations to find the units of k. The units of k are dependent on the reaction orders and the unit of time. Solution: a) A first-order rate law follows the general expression, rate = k[A]. The reaction rate is expressed as a change in concentration per unit time with units of mol/L∙time. Since [A] has units of mol/L, k has units of time–1: Rate = k[A] mol mol =k L • time L mol mol L 1 × k = L • time = = = time–1 mol L • time mol time L b) A second-order rate law follows the general expression, rate = k[A]2. The reaction rate is expressed as a change in concentration per unit time with units of mol/L·time. Since [A] has units of mol2/L2, k has units of L/mol∙time: Rate = k[A]2 2 mol ⎛ mol ⎞⎟ = k ⎜⎜ ⎟ ⎝ L ⎠⎟ L • time mol L mol L2 k = L • time = = × 2 2 mol mol • time L • time mol L2

16-19

c) A third-order rate law follows the general expression, rate = k[A]3. The reaction rate is expressed as a change in concentration per unit time with units of mol/L∙time. Since [A] has units of mol3/L3, k has units of L2/mol2∙time: Rate = k[A]3 ⎛ mol ⎞⎟ mol = k ⎜⎜ ⎝ L ⎠⎟⎟ L • time mol L2 mol L3 L k = • time = = × mol 3 mol 2 • time L • time mol3 3 L d) A 5/2-order rate law follows the general expression, rate = k[A]5/2. The reaction rate is expressed as a change in concentration per unit time with units of mol/L·time. Since [A] has units of mol5/2/L5/2, k has units of L3/2/mol3/2· time: 3

5/2

mol ⎛ mol ⎞⎟ = k ⎜⎜ ⎝ L ⎠⎟⎟ L • time mol L3/2 mol L5/2 L k = • time = = × 5/2 3/2 5/2 mol mol • time L • time mol 5/2 L

16.37

Plan: Write the appropriate rate law and enter the units for rate and the rate constant to find the units of concentration. The units of concentration will give the reaction order. Solution: a) Rate = k[A]m

mol mol ⎡ mol ⎤ = ⎢ ⎥ L•s L • s ⎢⎣ L ⎥⎦ mol m ⎡ mol ⎤ ⎢ ⎥ = L•s mol ⎢⎣ L ⎥⎦ L•s m ⎡ mol ⎤ ⎢ ⎥ =1 ⎢⎣ L ⎥⎦ b) Rate = k[A]m

m must be 0. The reaction is zero order.

mol 1 ⎡ mol ⎤ = ⎢ ⎥ L • yr yr ⎣⎢ L ⎦⎥ mol m ⎡ mol ⎤ L • yr mol yr = × ⎢ ⎥ = 1 L • yr 1 ⎣⎢ L ⎦⎥ yr m

⎡ mol ⎤ ⎢ ⎥ ⎢⎣ L ⎥⎦

mol L

m must be 1. The reaction is first order.

c) Rate = k[A]m

mol mol1/ 2 ⎡ mol ⎤ = 1/ 2 ⎢ ⎥ L•s L • s ⎣⎢ L ⎦⎥

16-20

mol mol L1/ 2 • s L •s = × 1/ 2 = mol L • s mol1/2 L1 / 2 • s m ⎡ mol ⎤ mol1/2 m must be 1/2. The reaction is 1/2 order. ⎢ ⎥ = 1/2 ⎢⎣ L ⎥⎦ L ⎡ mol ⎤ ⎢ ⎥ ⎣⎢ L ⎦⎥

d) Rate = k[A]m

mol mol−5 / 2 ⎡ mol ⎤ = −5 / 2 ⎢ ⎥ L • min L • min ⎢⎣ L ⎥⎦

mol mol L−5/ 2 • min = L • min = × −5 / 2 mol L • min mol−5/2 −5 / 2 L • min m ⎡ mol ⎤ mol7/2 m must be 7/2. The reaction is 7/2 order. ⎢ ⎥ = 7/2 L ⎣⎢ L ⎦⎥ ⎡ mol ⎤ ⎢ ⎥ ⎢⎣ L ⎥⎦

16.38

Plan: The rate law is rate = k [CO]m[Cl2]n where m and n are the orders of the reactants. To find the order of each reactant, take the ratio of the rate laws for two experiments in which only the reactant in question changes. Once the rate law is known, the data in each experiment can be used to find the rate constant k. Solution: a) To find the order for CO, first identify the reaction experiments in which [CO] changes but [Cl2] is constant. Use experiments 1 and 2 to find the order with respect to [CO]. Set up a ratio of the rate laws for experiments 1 and 2 and fill in the values given for rates and concentrations and solve for m, the order with respect to [CO]. m ⎛[CO]exp 1 ⎞⎟ rateexp 1 ⎜ ⎟ = ⎜⎜ ⎟ rate ⎝⎜[CO] ⎠⎟⎟ exp 2

exp 2

⎛ 1.00 mol/L ⎞⎟ 1.29 ×10 mol/L • min = ⎜⎜ ⎟ −30 ⎜ ⎝ 0.100 mol/L ⎠⎟ 1.33 ×10 mol/L • min 9.699 = (10.0)m log (9.699) = m log (10.0) m = 0.9867 = 1 The reaction is first order with respect to [CO]. To find the order for Cl2, first identify the reaction experiments in which [Cl2] changes but [CO] is constant. Use experiments 2 and 3 to find the order with respect to [Cl2]. Set up a ratio of the rate laws for experiments 2 and 3 and fill in the values given for rates and concentrations and solve for n, the order with respect to [Cl2]. n ⎛[Cl2 ]exp 3 ⎞⎟ rateexp 3 ⎜ ⎟⎟ = ⎜⎜ ⎜⎝[Cl2 ]exp 2 ⎠⎟⎟ rateexp 2 −29

⎛ 1.00 mol/L ⎞⎟ 1.30 ×10−29 mol/L • min = ⎜⎜ ⎟ −30 ⎜⎝ 0.100 mol/L ⎠⎟ 1.33 ×10 mol/L • min 9.774 = (10.0)n log (9.774) = n log (10.0) n = 0.9901 = 1 The reaction is first order with respect to [Cl2]. Rate = k[CO][Cl2] n

16-21

b) k = rate/[CO][Cl2] Exp 1: k1 = (1.29 × 10–29 mol/L∙s)/[1.00 mol/L][0.100 mol/L] = 1.29 × 10–28 L/mol∙s Exp 2: k2 = (1.33 × 10–30 mol/L∙s)/[0.100 mol/L][0.100 mol/L] = 1.33 × 10–28 L/mol∙s Exp 3: k3 = (1.30 × 10–29 mol/L∙s)/[0.100 mol/L][1.00 mol/L] = 1.30 × 10–28 L/mol∙s Exp 4: k4 = (1.32 × 10–31 mol/L∙s)/[0.100 mol/L][0.0100 mol/L] = 1.32 × 10–28 L/mol∙s kavg = (1.29 × 10–28 + 1.33 × 10–28 + 1.30 × 10–28 + 1.32 × 10–28) L/mol∙s/4 = 1.31 × 10–28 L/mol∙s 16.39

The integrated rate law can be used to plot a graph. If the plot of [reactant] vs. time is linear, the order is zero. If the plot of ln[reactant] vs. time is linear, the order is first. If the plot of inverse concentration (1/[reactant]) vs. time is linear, the order is second. a) The reaction is first order since ln[reactant] vs. time is linear. b) The reaction is second order since 1/[reactant] vs. time is linear. c) The reaction is zero order since [reactant] vs. time is linear.

16.40

The half-life (t½) of a reaction is the time required to reach half the initial reactant concentration. For a first-order process, no molecular collisions are necessary, and the rate depends only on the fraction of the molecules having sufficient energy to initiate the reaction.

16.41

Plan: (a) To find concentration of reactant at a later time, given the initial concentration and the rate constant k for a first-order reaction, use the first-order integrated rate law. Since the time unit in k is seconds, time t must also be expressed in units of seconds. (b) To find the time required for a certain percentage of reactant to decompose, find the amount of reactant that remains and solve for t in the integrated rate law. Solution:

⎛ 60 s ⎞⎟ ⎜ ⎟⎟ = 300 s Converting t in min to s: (5.00 min) ⎜⎜ ⎜⎜⎝1 min ⎠⎟⎟

⎡ N 2 O5 ⎤ ⎦ 0 = kt ln ⎣ ⎡ N 2 O5 ⎤ ⎣ ⎦t ln

1.58 mol/L ⎡ N 2 O5 ⎤ ⎣ ⎦t

(

)

= 2.8×10−3 s−1 (300. s) = 0.84

The the antilog of each side: ln

⎡ N 2 O5 ⎤ = ⎣ ⎦t

1.58 mol/L 2.316

1.58 mol/L ⎡ N 2 O5 ⎤ ⎣ ⎦t

= e0.84 = 2.316

= 0.68221 = 0.68 mol /L

b) When 35.0% of the N2O5 decomposes, 100.0% – 35.0% = 65.0% remains at time t. [N2O5]t = 0.650 × 1.58 mol/L = 1.03 mol/L

⎡ N 2 O5 ⎤ ⎦ 0 = kt ln ⎣ ⎡ N 2 O5 ⎤ ⎣ ⎦t 1.58 mol/L = 2.8×10−3 s−1 (t ) ln 1.03 mol/L

(

)

t = 152.809 = 150 s

16-22

16.42

Plan: This is a first-order reaction so we use the first-order integrated rate law. We know t (10.0 s), [I2]0 (2.00 M), and [I2]t (0.133 M), so we solve for the rate constant, k. Solution:

⎡ I2 ⎤ ln ⎣ ⎦ 0 = kt ⎡ I2 ⎤ ⎣ ⎦t ⎡2.00⎤ ⎦ 0 = k (10.0 s) ln ⎣ ⎡0.133⎤ ⎣ ⎦t k = 0.271 s–1 16.43

Plan: The rate constant is known; use the first-order half-life equation to calculate the half-life, t1/2. Count the reactant particles in the two scenes to determine the fraction of particles remaining after the elapsed unknown time. Determine the number of half-lives that have passed and multiply by the half-life value. Solution: ln 2 ln 2 = = 11 min t1/2 = 0.063 min−1 k There are 20 A particles at t = 0 and 5 A particles at t = ?. Thus, the fraction of A remaining is ¼ of the initial amount which is two half-lives (½ × ½). Two half-lives is 2 × 11 min = 22 min.

16.44

Plan: The half-life is the time required for one-half of the reactant to be consumed. Count the reactant molecules in the three scenes to determine the number of molecules remaining after each time period. This information is used to determine the half-life. Once t1/2 is known, use the first-order half-life equation to calculate the rate constant, k. Solution: a) There are twelve molecules of reactant at t = 0, eight reactant molecules after 20 min, and three reactant molecules after 60 min. After 60 min, one-fourth (three of 12 molecules) of the initial amount of cyclopropane remains unreacted. Therefore, 60 min represents two half-lives. The half-life is 30 min. ln 2 ln 2 ln 2 = = 0.023104906 = 0.023 min–1 b) t1/2 = k= k 30 min t1/2

16.45

Plan: This is a first-order reaction so use the first-order integrated rate law. In part (a), we know t (10.5 min). Let [A]0 = 1 M and then [A]t = 50% of 1 M = 0.5 M. Solve for k. In part (b), use the value of k to find the time necessary for 75.0% of the compound to react. If 75.0% of the compound has reacted, 100–75 = 25% remains at time t. Let [A]0 = 1 M and then [A]t = 25% of 1 M = 0.25 M. Solution: a) ln [A]t = ln [A]0 – kt ln [0.5] = ln [1] – k(10.5 min) –0.693147 = 0 – k(10.5 min) 0.693147 = k(10.5 min) k = 0.0660 min–1 Alternatively, 50.0% decomposition means that one half-life has passed. Thus, the first-order half-life equation may be used: ln 2 ln 2 ln 2 t1/2 = = = 0.066014 = 0.0660 min–1 k= k 10.5 min t1/2 b) ln [A]t = ln [A]0 – kt ln[A]t − ln[A]0 =t −k ln[0.25] − ln[1] =t −0.066014 min−1

16-23

t = 21.0000 = 21.0 min If you recognize that 75.0% decomposition means that two half-lives have passed, then t = 2 (10.5 min) = 21.0 min. 16.46

Plan: This is a first-order reaction so use the first-order integrated rate law (the units of k, yr–1, indicates first order). In part (a), the first-order half-life equation may be used to solve for half-life since k is known. In part (b), use the value of k to find the time necessary for the reactant concentration to drop to 12.5% of the initial concentration. Let [A]0 = 1.00 M and then [A]t = 12.5% of 1 M = 0.125 M. Solution: ln 2 ln 2 a) t1/2 = = = 577.62 = 5.8 × 102 yr k 0.0012 yr−1 b) ln [A]t = ln [A]0 – kt [A]0 ln = kt [A]t [A]t = 0.125 M k = 0.0012 yr–1 [A]0 = 1.00 M ⎛ 1.00 M ⎞⎟ ln ⎜⎜ = (0.0012 yr–1) t ⎝ 0.125 M ⎠⎟⎟ t = 1732.86795 = 1.7 × 103 yr If the student recognizes that 12.5% remaining corresponds to three half-lives; then simply multiply the answer in part (a) by three.

16.47

Plan: The rate expression indicates that this reaction is second order overall (the order of [AB] is 2), so use the second-order integrated rate law to find time. We know k (0.2 L/mol∙s), [AB]0 (1.50 M), and [AB]t (1/3[AB]0 = 1/3(1.50 M) = 0.500 M), so we can solve for t. Solution: 1 1 = kt − [ AB]t [ AB]0 ⎛ 1 1 ⎞⎟ ⎜⎜ ⎟ − ⎜⎜⎝[ AB] [ AB] ⎠⎟⎟ t 0 t= k ⎛ 1 1 ⎞⎟⎟ ⎜⎜ − ⎟ ⎜⎜ ⎜⎝ 0.500 M 1.50 M ⎠⎟⎟ t= 0.2 L/mol • s t = 6.6667 = 7 s

16.48

Plan: The rate expression indicates that this reaction is second order overall (the order of [AB] is 2), so use the secondorder integrated rate law. We know k (0.2 L/mol∙s), [AB]0 (1.50 M), and t (10.0 s), so we can solve for [AB]t. Solution: 1 1 = kt − [ AB]t [ AB]0 1 1 = kt + [ AB]t [ AB ]0 1 1 = (0.2 L/mol ∙ s) (10.0 s) + 1.50 M [ AB]t 1 1 = 2.66667 M [ AB]t

[AB]t = 0.375 = 0.4 M Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

16-24

16.49

Plan: In a first-order reaction, ln [NH3] vs. time is a straight line with slope equal to k. The half-life can be determined using the first-order half-life equation. Solution: a) A new data table is constructed: (Note that additional significant figures are retained in the calculations.)

x-axis (time, s)

[NH3]

y-axis (ln [NH3])

4.000 M

1.38629

1.000

3.986 M

1.38279

2.000

3.974 M

1.37977

k = slope = rise/run = (y2 – y1)/(x2 – x1) k = (1.37977 – 1.38629)/(2.000 – 0) = (0.00652)/(2) = 3.260 × 10–3 s–1 = 3 × 10–3 s–1 (Note that the starting time is not exact, and hence, limits the significant figures.) ln 2 ln 2 t1/2 = = = 212.62 = 2 × 102 s k 3.260 ×10−3 s−1

16.50

The central idea of collision theory is that reactants must collide with each other in order to react. If reactants must collide to react, the rate depends on the product of the reactant concentrations.

16.51

No, collision frequency is not the only factor affecting reaction rate. The collision frequency is a count of the total number of collisions between reactant molecules. Only a small number of these collisions lead to a reaction. Other factors that influence the fraction of collisions that lead to reaction are the energy and orientation of the collision. A collision must occur with a minimum energy (activation energy) to be successful. In a collision, the orientation, that is, which ends of the reactant molecules collide, must bring the reacting atoms in the molecules together in order for the collision to lead to a reaction.

16.52

At any particular temperature, molecules have a distribution of kinetic energies, as will their collisions have a range of energies. As temperature increases, the fraction of these collisions which exceed the threshold energy, increases; thus, the reaction rate increases.

16.53

k = Ae a The Arrhenius equation indicates a negative exponential relationship between temperatures and the rate constant, k. In other words, the rate constant increases exponentially with temperature.

−E / RT

16-25

−E / RT

16.54

he Arrhenius equation, e k = Ae a , can be b used directlyy to solve for aactivation energgy at a specifieed Th teemperature if th he rate constan nt, k, and the freequency factorr, A, are knownn. However, the frequency factor is ussually not know wn. To find Ea without knowiing A, rearrangge the Arrheniuus equation to pput it in the forrm of a lin near plot: ln k = ln A – Ea/RT T where the y value is ln k andd the x value iss 1/T. Measure the rate constaant at a series off temperatures and plot ln k vs. v 1/T. The slo ope equals –Ea/R /R.

16.55

a)) The value of k increases i exponeentially with tem mperature.

b)) A plot of ln k vs. 1/T is a strraight line who ose slope is –Ea /R.

Th he activation energy e is determ mined from thee slope of the liine in the ln k vvs. 1/T graph. The slope equaals –Ea/R. 16.56

a)) As temperature increases, th he fraction of collisions c whichh exceed the aactivation energgy increases; thhus, the reeaction rate in ncreases. b)) A decrease in n activation eneergy lowers thee energy threshhold with whicch collisions muust take place tto be efffective. At a given g temperatu ure, more collisions occur wiith the lower ennergy so rate iincreases.

16.57

No N . For 4 × 10–55 moles of EF to t form, every collision must result in a reacction and no E EF molecule cann decompose baack to AB and CD. Neither condition is likeely. All collisioons will not ressult in product as some collissions will occcur with an en nergy that is lower than the acctivation energgy. In principlee, all reactions are reversible, so some EF molecules m decom mpose. Even iff all AB and CD D molecules diid combine, thhe reverse decoomposition ratee would reesult in an amo ount of EF that is less than 4 × 10–5 moles.

16.58

Collision frequeency is proportiional to the vellocity of the reeactant moleculles. At the sam me temperature, both reeaction mixturees have the sam me average kineetic energy, buut not the samee velocity. Kineetic energy equuals 1/2 mv2,

16-26

where m is mass and v velocity. The methylamine (N(CH3)3) molecule has a greater mass than the ammonia molecule, so methylamine molecules will collide less often than ammonia molecules, because of their slower velocities. Collision energy thus is less for the N(CH3)3(g) + HCl(g) reaction than for the NH3(g) + HCl(g) reaction. Therefore, the rate of the reaction between ammonia and hydrogen chloride is greater than the rate of the reaction between methylamine and hydrogen chloride. The fraction of successful collisions also differs between the two reactions. In both reactions the hydrogen from HCl is bonding to the nitrogen in NH3 or N(CH3)3. The difference between the reactions is in how easily the H can collide with the N, the correct orientation for a successful reaction. The groups (H) bonded to nitrogen in ammonia are less bulky than the groups bonded to nitrogen in trimethylamine (CH3). So, collisions with correct orientation between HCl and NH3 occur more frequently than between HCl and N(CH3)3 and the reaction NH3(g) + HCl(g) → NH4Cl(s) occurs at a higher rate than N(CH3)3(g) + HCl(g) → (CH3)3NHCl(s). Therefore, the rate of the reaction between ammonia and hydrogen chloride is greater than the rate of the reaction between methylamine and hydrogen chloride. 16.59

Each A particle can collide with three B particles, so (4 × 3) = 12 unique collisions are possible.

16.60

Plan: Use Avogadro’s number to convert moles of particles to number of particles. The number of unique collisions is the product of the number of A particles and the number of B particles. Solution: ⎛ 6.022 ×10 23 A particles ⎞⎟ ⎟⎟ = 6.08222 × 1023 particles of A Number of particles of A = (1.01 mol A )⎜⎜⎜ 1 mol A ⎝⎜ ⎠⎟ ⎛ 6.022 ×10 23 B particles ⎞⎟ ⎟⎟ = 1.279997 × 1024 particles of B Number of particles of B = (2.12 mol B)⎜⎜⎜ ⎜⎝ 1 mol B ⎠⎟

Number of collisions = (6.08222 × 1023 particles of A)(1.279997 × 1024 particles of B) = 7.76495 × 1047 = 7.76 × 1047 unique collisions 16.61

Plan: The reaction is exothermic (ΔH is negative), so the energy of the products must be lower than that of the reactants. Use the relationship ΔHrxn = Ea(fwd) – Ea(rev) to solve for Ea(rev). To draw the transition state, note that the bond between B and C will be breaking while a bond between C and D will be forming. Solution: a) Ea (fwd) 215 kJ/mol Energy

Ea(rev)

ABC + D

ΔHrxn

AB + CD

–55 kJ/mol

Reaction coordinate b) ΔHrxn = Ea(fwd) – Ea(rev) Ea(rev) = Ea(fwd) – ΔHrxn = 215 kJ/mol – (–55 kJ/mol) = 2.70 × 102 kJ/mol

16-27

c) bond forming B A

bond weakening

16.62

Plan: The forward activation energy Ea(fwd) is larger than the reverse activation energy Ea(rev) which indicates that the energy of the products must be higher than that of the reactants. Use the relationship ΔHrxn = Ea(fwd) – Ea(rev) to solve for ΔHrxn. To draw the transition state, note that the bonds in the A2 and B2 molecules will be breaking while bonds between A and B will be forming. Solution: a)

Ea(rev)

Energy

Ea(fwd)

2AB ΔHrxn

A2 + B2

Reaction coordinate b) ΔHrxn = Ea(fwd) – Ea(rev) = 125 kJ/mol – 85 kJ/mol = 40 kJ/mol c)

A.....B |

A.....B

16.63

−Ea / RT

Plan: The fraction of collisions with a specified energy is equal to the e

term in the Arrhenius equation.

Solution: −Ea / RT

f= e

T = 25°C + 273 = 298 K

Ea = 100. kJ/mol

R = 8.314 J/mol∙K = 8.314 × 10–3 kJ/mol∙K

−

100. kJ/ mol Ea =− = –40.362096 RT 8.314 ×10−3 kJ/mol • K 298 K

(

−Ea / RT

Fraction = e

)(

–40.362096

)

= 2.9577689 × 10–18 = 2.96 × 10–18

16-28

16.64

−Ea /RT

Plan: The fraction of collisions with a specified energy is equal to the e

term in the Arrhenius equation.

Solution: −Ea / RT

f= e

T = 50°C + 273 = 323 K

Ea = 100. kJ/mol

R = 8.314 J/mol∙K = 8.314 × 10–3 kJ/mol∙K

−

100. kJ/ mol Ea =− = –37.238095 RT 8.314 ×10−3 kJ/mol • K 323 K

(

)(

−Ea / RT

Fraction = e

)

= e–37.238095 = 6.725131 × 10–17

The fraction increased by (6.725131 × 10–17)/(2.9577689 × 10–18) = 22.737175 = 22.7 16.65

Plan: You are given one rate constant k1 at one temperature T1 and the activation energy Ea. Substitute these values into the Arrhenius equation and solve for k2 at the second temperature. Solution: k1 = 4.7 × 10–3 s–1 T1 = 25°C + 273 = 298 K k2 = ? T2 = 75°C + 273 = 348 K Ea = 33.6 kJ/mol = 33,600 J/mol k2 E ⎛1 1⎞ = a ⎜⎜ − ⎟⎟⎟ ln ⎜ k1 R ⎝ T1 T2 ⎠⎟ ln

k2 −3

−1

4.7 × 10 s

33,600 J /mol ⎛⎜ 1 1 ⎞⎟⎟ ⎜⎜ − ⎟ 8.314 J /mol • K ⎜⎜⎝ 298 K 348 K ⎠⎟⎟

= 1.948515 (unrounded) Raise each side to ex −3 4.7 ×10 s−1 k2 = 7.0182577 −3 4.7 ×10 s−1 k2 = (4.7 × 10–3 s–1)(7.0182577) = 0.0329858 = 0.033 s–1

16.66

Plan: You are given the rate constants, k1 and k2, at two temperatures, T1 and T2. Substitute these values into the Arrhenius equation and solve for Ea. Solution: k1 = 4.50 × 10–5 L/mol∙s T1 = 195°C + 273 = 468 K k2 = 3.20 × 10–3 L/mol∙s T2 = 258°C + 273 = 531 K Ea = ? k2 E ⎛1 1⎞ = a ⎜⎜ − ⎟⎟⎟ ln ⎜ k1 R ⎝ T1 T2 ⎠⎟ −3 ⎛ ⎞ ⎛ J ⎞⎟⎜⎜ 3.20 ×10 L /mol • s ⎟⎟ ⎜⎜8.314 ⎟⎜ln ⎟ − 5 ⎜⎝ mol • K ⎠⎟⎜⎝⎜ 4.50 ×10 L /mol • s ⎠⎟⎟ Ea = = ⎛ 1 1 ⎞⎟ ⎛ 1 1 ⎞⎟⎟ ⎜⎜ − ⎟ ⎜⎜ − ⎟ ⎜⎜ ⎜⎝ T1 T2 ⎠⎟⎟ ⎜⎝ 468 K 531 K ⎠⎟⎟

⎛ k ⎞ R ⎜⎜⎜ln 2 ⎟⎟⎟ ⎝ k1 ⎠⎟

Ea = 1.3984658 × 105 J/mol = 1.40 × 105 J/mol 16.67

Plan: You are given the rate constants, k1 and k2, at two temperatures, T1 and T2. Substitute these values into the Arrhenius equation and solve for Ea. Solution:

16-29

k1 = 0.76/s k2 = 0.87/s Ea = ? k2 E ⎛1 1⎞ = a ⎜⎜ − ⎟⎟⎟ ln ⎜ k1 R ⎝ T1 T2 ⎠⎟

T1 = 727°C + 273 = 1000. K T2 = 757°C + 273 = 1030. K

J ⎞⎟⎜⎜⎛ 0.87 / s ⎞⎟⎟ ⎛ ⎛ k ⎞ ⎜⎜8.314 ⎟ ⎟⎜ln R ⎜⎜⎜ln 2 ⎟⎟⎟ ⎝ mol • K ⎠⎜⎜⎝ 0.76 / s ⎠⎟⎟ ⎝ k1 ⎠⎟ Ea = = ⎛ 1 ⎛ 1 1 ⎞⎟ 1 ⎞⎟⎟ ⎜⎜ ⎜⎜ − ⎟ − ⎟ ⎟ ⎜⎜ ⎜⎝ T1 T2 ⎠⎟ ⎜⎝1000. K 1030. K ⎠⎟⎟ Ea = 3.8585 × 104 J/mol = 3.9 × 104 J/mol 16.68

Plan: The reaction is endothermic (ΔH is positive), so the energy of the products must be higher than that of the reactants. Use the relationship ΔHrxn = Ea(fwd) – Ea(rev) to solve for Ea(rev). To draw the transition state, note that the bond in Cl2 will be breaking while the bond between N and Cl will be forming. Solution: NOCl + Cl

Ea(rev)

Energy

ΔHrxn = +83 kJ NO + Cl2

Ea(fwd) = +86 kJ

Reaction coordinate b) ΔHrxn = Ea(fwd) – Ea(rev) Ea(rev) = Ea(fwd) – ΔHrxn = 86 kJ – 83 kJ = 3 kJ. c) To draw the transition state, look at structures of reactants and products:

N Cl

The collision must occur between one of the chlorine atoms and the nitrogen. The transition state would have weak bonds between the nitrogen and chlorine and between the two chlorine atoms. N Cl

16.69

The rate of an overall reaction depends on the slowest step. Each individual step’s reaction rate can vary widely, so the rate of the slowest step, and hence the overall reaction, will be slower than the average of the individual rates because the average contains faster rates as well as the rate-determining step.

16.70

An elementary step is a single molecular event, such as the collision of two molecules. Since an elementary step occurs in one step, its rate must be proportional to the product of the reactant concentrations. Thus, the exponents in the rate of an elementary step are identical to the coefficients in the equation for the step. Since an overall reaction is generally a series of elementary steps, it is not necessarily proportional to the product of the overall reactant concentrations.

16-30

16.71

Yes, it is often possible to devise more than one mechanism since the rate law for the slowest step determines the rate law for the overall reaction. The preferred mechanism will be the one that seems most probable, where molecules behave in their expected fashion.

16.72

Reaction intermediates have some stability, however limited, but transition states are inherently unstable. Additionally, unlike transition states, intermediates are molecules with normal bonds.

16.73

A bimolecular step (a collision between two particles) is more reasonable physically than a termolecular step (a collision involving three particles) because the likelihood that two reactant molecules will collide with the proper energy and orientation is much greater than the likelihood that three reactant molecules will collide simultaneously with the proper energy and orientation.

16.74

No, the overall rate law must contain reactants only (no intermediates) and is determined by the slow step. If the first step in a reaction mechanism is slow, the rate law for that step is the overall rate law.

16.75

If the slow step is not the first one, the faster preceding step produces intermediates that accumulate before being consumed in the slow step. Substitution of the intermediates into the rate law for the slow step will produce the overall rate law.

16.76

Plan: The overall reaction can be obtained by adding the three steps together. An intermediate is a substance that is formed in one step and consumed in a subsequent step. The molecularity of each step is the total number of reactant particles; the molecularities are used as the orders in the rate law for each step. The overall rate law for the mechanism is determined from the slowest step (the rate-determining step) and can be compared to the actual rate law. Solution: a) (1) A(g) + B(g) → X(g) fast (2) X(g) + C(g) → Y(g) slow (3) Y(g) → D(g) fast Total: A(g) + B(g) + X(g) + C(g) + Y(g) → X(g) + Y(g) + D(g) Overall: A(g) + B(g) + C(g) → D(g) b) Both X and Y are intermediates in the given mechanism. Intermediate X is produced in the first step and consumed in the second step; intermediate Y is produced in the second step and consumed in the third step. Notice that neither X nor Y were included in the overall reaction. c) Step

Molecularity

Rate law

A(g) + B(g) →X(g)

bimolecular

rate1 = k1[A][B]

X(g) + C(g) → Y(g)

bimolecular

rate2 = k2[X][C]

Y(g) → D(g)

unimolecular

rate3 = k3[Y]

d) Yes, the mechanism is consistent with the actual rate law. The slow step in the mechanism is the second step with rate law: rate = k2[X][C]. Since X is an intermediate, it must be replaced by using the first step. For an equilibrium, rateforward rxn = ratereverse rxn. For step 1 then, k1[A][B] = k–1[X]. Rearranging to solve for [X] gives [X] = (k1/k–1)[A][B]. Substituting this value for [X] into the rate law for the second step gives the overall rate law as rate = (k2k1/k–1)[A][B][C] which is identical to the actual rate law with k = k2k1/k–1. e) Yes, The one step mechanism A(g) + B(g) + C(g) → D(g) would have a rate law of rate = k[A][B][C], which is the actual rate law. 16.77

16-31

Solution: a)

(1) ClO–(aq) + H2O(l) → HClO(aq) + OH–(aq) fast (2) I–(aq) + HClO(aq) → HIO(aq) + Cl–(aq) slow (3) OH–(aq) + HIO(aq) → H2O(l) + IO–(aq) fast Total: ClO–(aq) + H2O(l) + I–(aq) + HClO(aq) + OH–(aq) + HIO(aq) → HClO(aq) + OH–(aq) + HIO(aq) + Cl–(aq) + H2O(l) + IO–(aq) – – – – (overall) ClO (aq) + I (aq) → Cl (aq) + IO (aq) b) HClO(aq), OH –(aq), and HIO(aq) are intermediates in the given mechanism. HClO(aq) is produced in the first step and consumed in the second step; OH –(aq) is produced in the first step and consumed in the third step; HIO(aq) is produced in the second step and consumed in the third step. Notice that the intermediates were not included in the overall reaction. c) (1) Bimolecular; Rate1 = k1 [ClO –] [H2O] (2) Bimolecular; Rate2 = k2 [I –][HClO] (3) Bimolecular; Rate3 = k3 [OH –][HIO] d) The slow step in the mechanism is the second step with rate law: rate = k2[I–][HClO]. Since HClO is an intermediate, it must be replaced by using the first step. For an equilibrium, rateforward rxn = ratereverse rxn. For step 1 then, leaving out the water, k1[ClO –] = k–1[HClO][ OH–]. Rearranging to solve for [HClO] gives [HClO] = (k1/k–1)[ClO–]/[OH–]. Substituting this value for [HClO] into the rate law for the second step gives the overall rate law as rate = (k2k1/k–1)[I–][ClO–]/[OH–] or rate = k[I–][ClO–]/[OH–]. This is not the observed rate law. The mechanism is not consistent with the actual rate law. 16.78

Plan: Use the rate-determining step to find the rate law for the mechanism. The concentration of the intermediate in the rate law must be expressed in terms of a true reactant which is then substituted into the rate law for the concentration of the intermediate. Solution: Nitrosyl bromide is NOBr(g). The reactions sum to the equation 2NO(g) + Br2(g) → 2NOBr(g), so criterion 1 (elementary steps must add to overall equation) is satisfied. Both elementary steps are bimolecular and chemically reasonable, so criterion 2 (steps are physically reasonable) is met. The reaction rate is determined by the slow step; however, rate expressions do not include reaction intermediates (NOBr2). The slow step in the mechanism is the second step with rate law: rate = k2[NOBr2][NO]. Since NOBr2 is an intermediate, it must be replaced by using the first step. For an equilibrium like step 1, rateforward rxn = ratereverse rxn. Solve for [NOBr2] in step 1: Rate1 (forward) = rate1 (reverse) k1[NO][Br2] = k–1[NOBr2] [NOBr2] = (k1/k–1)[NO][Br2] Rate of the slow step: Rate2 = k2[NOBr2][NO] Substitute the expression for [NOBr2] into this equation, the slow step: Rate2 = k2(k1/k–1)[NO][Br2][NO] Combine the separate constants into one constant: k = k2(k1/k–1) Rate2 = k[NO]2[Br2] The derived rate law equals the known rate law, so criterion 3 is satisfied. The proposed mechanism is valid.

16.79

Plan: Use the rate-determining step to find the rate law for each mechanism. If there is an intermediate in the rate law, the concentration of the intermediate must be expressed in terms of a true reactant which is then substituted into the rate law for the concentration of the intermediate. Solution: I. 2NO(g) + O2(g) → 2NO2(g) Rate = k[NO]2[O2] II. (1) 2NO(g) → N2O2(g) (2) N2O2(g) + O2(g) → 2NO2(g) slow N2O2 is an intermediate Rate = k2[N2O2][O2] Rate1 (forward) = rate1 (reverse)

16-32

k1[NO]2 = k–1[N2O2] [N2O2] = (k1/k–1)[NO]2 Substittute this for [N2O2] in the rate law: Rate = k[N2O2][O2] Rate = k2(k1/k–1)[NO]2[O2] Rate = k[NO]2[O2] (1) 2NO O(g) → N2(g) + O2(g) IIII. slow (2) N2(g g) + 2O2(g) → NO2(g) N2 is an intermediate Rate = k2[N2][O2]2 i Rate1 (fforward) = ratee1 (reverse) k1[NO]2 = k–1[N2][O2] [N2] = (k1/k–1)[NO]2/[O O2] = (k1/k–1)[N NO]2[O2]–1 Substittute this for [N2] in the rate laaw: Rate = k2[N2][O2]2 Rate = k2(k1/k–1)[NO]2 [O2]–1[O2]2 Rate = k[NO]2[O2] a)) All the mechaanisms are con nsistent with thee rate law. b)) The most reasonable mechaanism is II, sincce none of its eelementary stepps are more coomplicated thann being biimolecular. Meechanism I and d III have steps that are termoolecular. 16.80

Pllan: Review th he definitions of o homogeneou us vs. heterogeeneous catalystts. To draw thee reaction energgy diagrams, reecall that additiion of a catalysst to a reaction results in a low wer activation energy. So olution: a)) A heterogeneous catalyst sp peeds up a reacction that occurrs in a differentt phase. Gold iis a heterogeneeous ca atalyst since itt is a solid and is catalyzing a reaction in thee gas phase. b)) The activatio on energy for th he uncatalyzed d reaction mustt be greater thaan the activatioon for the catalyyzed reeaction.

16.81

No N , a catalyst ch hanges the mecchanism of a reeaction to one w with lower actiivation energy. Lower activaation energy means m a faster reeaction. An inccrease in tempeerature does noot influence thee activation ennergy, but instead increases th he fraction of collisions with sufficient s energ gy to react.

16-33

16.82

a) No, by definition, a catalyst is a substance that increases reaction rate without being consumed. The spark provides energy that is absorbed by the H2 and O2 molecules to achieve the threshold energy needed for reaction. b) Yes, the powdered metal acts like a heterogeneous catalyst, providing a surface upon which the reaction between O2 and H2 becomes more favorable because the activation energy is lowered.

16.83

Catalysts decrease the amount of energy required for a reaction. To carry out the reaction less energy must be generated. The generation of less energy means that fewer by-products of energy production will be released into the environment.

16.84

a) Enzymes stabilize a reaction’s transition state to a remarkable degree which lowers the activation energy and thus greatly increase the reaction rate. b) Enzymes are extremely specific and have the ability to change shape to adopt a perfect fit with the substrate.

16.85

Plan: For parts (a) and (b), find the order of the given reactant; the order of the other reactant is found by realizing that the orders m + n = 2 since the reactions are second order overall. For parts (c)-(e), use the rate laws to determine the rate changes that result from the given changes in reactant concentrations. Solution: a) The rate law is rate = [NO2]m[CO]n. When [NO2] increases by a factor of 2, the rate increases by a factor of 4: m ⎛ [NO 2 ]2 ⎞⎟ rate 2 ⎟⎟ = ⎜⎜⎜ rate1 ⎝ [NO 2 ]1 ⎠⎟ 4 = [2]m log (4.00) = m log (2.00) m=2 The reaction is second order with respect to NO2. Since the overall order is 2, m + n = 2: m+n=2 n=2–m n=2–2=0 The reaction is zero order with respect to CO. Rate = k[NO2]2[CO]0 or Rate = k[NO2]2 b) The rate law is rate = [NO]m[O3]n. When [NO] increases by a factor of 2, the rate increases by a factor of 2: m ⎛ [NO]2 ⎞⎟ rate 2 ⎟⎟ = ⎜⎜⎜ rate1 ⎝ [NO]1 ⎠⎟ 2 = [2]m log (2.00) = m log (2.00) m=1 The reaction is first order with respect to NO. Since the overall order is 2, m + n = 2: m+n=2 n=2–m n=2–1=1 The reaction is first order with respect to O3. Rate = k[NO][O3] c) The concentrations of the reactants are one-half of the original when the reaction is 50% complete. 2 k [ NO2 ] Rate initial Reaction 1: = 2 = 4 Rate 50% k ⎡⎢0.5(NO2 )⎤⎥ ⎣ ⎦ Reaction 2:

k [ NO][O3 ] Rate initial = =4 Rate 50% k ⎡⎢0.5(NO)⎤⎥ ⎡⎢0.5(O3 )⎤⎥ ⎣ ⎦⎣ ⎦

16-34

d) [NO2]initial = 2[CO]initial; when the reaction is 50% complete, [CO] = 1/2[CO]initial and [NO2] = 0.75[NO2]initial. 2 k [ NO2 ] Rate initial = 2 = 1.7778 = 1.8 Rate 50% k ⎡⎢0.75(NO2 )⎤⎥ ⎣ ⎦ e) [NO]initial = 2[O3]initial, when the reaction is 50% complete, [O3] = 1/2[O3]initial and [NO] = 0.75[NO]initial. k [ NO][O3 ] Rate initial = = 2.6667 = 2.7 Rate 50% k ⎡⎢0.75(NO)⎤⎥ ⎡⎢0.5(O3 )⎤⎥ ⎣ ⎦⎣ ⎦ 16.86

a) There are two elementary steps since there are two different values of activation energy. b) The second step is the rate-limiting step since it has the greater activation energy and would therefore be the slower of the two steps. c) The reaction is exothermic since a loss of energy results in the products having lower energy than the reactants.

16.87

Plan: An intermediate is a substance that is formed in one step and consumed in a subsequent step. The coefficients of the reactants in each elementary step are used as the orders in the rate law for the step. The overall rate law for the mechanism is determined from the slowest step (the rate-determining step) and can be compared to the actual rate law. Reactants that appear in the mechanism after the slow step do not determine the rate and therefore do not appear in the rate law. Solution: a) Water does not appear as a reactant in the rate-determining step. Note that as a solvent in the reaction, the concentration of the water is assumed not to change even though some water is used up as a reactant. This assumption is valid as long as the solute concentrations are low (∼1 M or less). So, even if water did appear as a reactant in the rate-determining step, it would not appear in the rate law. See rate law for step 2 below. b) Rate law for step (1): rate1 = k1[(CH3)3CBr] Rate law for step (2): rate2 = k2[(CH3)3C+] Rate law for step (3): rate3 = k3[(CH3)3COH2+] + c) The intermediates are (CH3)3C and (CH3)3COH2+. (CH3)3C+ is formed in step 1 and consumed in step 2; (CH3)3COH2+ is formed in step 2 and consumed in step 3. d) The rate-determining step is the slow step, (1). The rate law for this step is rate = k1[(CH3)3CBr] since the coefficient of the reactant in this slow step is 1. The rate law for this step agrees with the actual rate law with k = k1.

16.88

Plan: Radioactive decay is first order so the first-order integrated rate law is used. First use the first-order half-life equation to solve for k since the half-life is given. Since 25% of the 14C remains at time t, let [14C]0 = 100% and then [14C]t = 15%. Solution: ln 2 ln 2 ln 2 t1/2 = = = 1.20968 × 10–4 yr–1 (unrounded) k= k t1/2 5730 yr Use the first-order integrated rate law ln [14C]t = ln [14C]0 – kt or ⎡ 14 C⎤ ln ⎣ 14 ⎦ t = – kt ⎡ C⎤ ⎣ ⎦0 [15.5%]t = –(1.20968 × 10–4 yr–1) t ln [100%]0 t = 1.541176 × 104 = 1.54 × 104 yr

16-35

16.89

Plan: The activation energy can be calculated using the Arrhenius equation. Although the rate constants, k1 and k2, are not expressly stated, the relative times give an idea of the rate. The reaction rate is proportional to the rate constant. At T1 = 20°C, the rate of reaction is 1 apple/4 days while at T2 = 0°C, the rate is 1 apple/16 days. Therefore, rate1 = 1 apple/4 days and rate2 = 1 apple/16 days are substituted for k1 and k2, respectively. Solution: k1 = 1/4 T1 = 20°C + 273 = 293 K k2 = 1/16 T2 = 0°C + 273 = 273 K Ea = ? k E ⎛1 1⎞ ln 2 = a ⎜⎜⎜ − ⎟⎟⎟ k1 R ⎝ T1 T2 ⎠⎟

J ⎞⎟⎜⎜⎛ 1/16 ⎞⎟⎟ ⎛ ⎛ k ⎞ ⎜⎜8.314 ⎟ ⎟⎜ln R ⎜⎜ln 2 ⎟⎟⎟ ⎝ mol • K ⎠⎜⎜⎝ 1/4 ⎠⎟⎟ ⎜⎝ k1 ⎠⎟ Ea = = ⎛ 1 ⎛ 1 1 ⎞⎟ 1 ⎞⎟⎟ ⎜⎜ ⎜⎜ − ⎟ − ⎟ ⎜⎜ ⎜⎝ T1 T2 ⎠⎟⎟ ⎜⎝ 293 K 273 K ⎠⎟⎟ Ea = 4.6096266 × 104 J/mol = 4.61 × 104 J/mol The significant figures are based on the Kelvin temperatures. 16.90

Plan: Use the first-order integrated rate law. First use the first-order half-life equation to solve for k since the halflife is given. Since 95% of the benzoyl peroxide (BP) remains at time t, let [BP]0 = 100% and then [BP]t = 95%. Solution: ln 2 ln 2 t1/2 = = 7.07293 × 10–5 d–1 (unrounded) k= k t1/2 Use the first-order integrated rate law with BP = benzoyl peroxide: [ BP ]t ln = – kt [ BP ]0

[ 95% ]t = – (7.07293 × 10–5 d–1) t [100% ]0 t = 725.2057 = 7.3 × 102 d ln

16.91

Plan: The reaction is exothermic (ΔH is negative), so the energy of the products must be lower than that of the reactants in the reaction energy diagram. Since there are two steps in the proposed mechanism, the diagram must show two transition states. The first step in the mechanism is the slower step so its Ea value is larger than the Ea value of the second step in the mechanism. The overall rate law for the alternative mechanism is determined from the slowest step (the rate-determining step) and can be compared to the actual rate law. Solution:

Energy

a) CO + NO2

ΔH = –226kJ

CO2 + NO

Reaction coordinate Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

16-36

b) The rate of the mechanism is based on the slowest step, 2NO2(g) → N2(g) + 2O2(g). The rate law for this step is rate = k1[NO2]2 which is consistent with the actual rate law. The alternative mechanism includes an elementary reaction (step 2) that is a termolecular reaction. Thus, the original mechanism given in the text is more reasonable physically since it involves only bimolecular reactions. 16.92

Plan: The rate law is rate = [A]x[B]y[C]z where x, y, and z are the orders of the reactants. To find the order of each reactant, take the ratio of the rate laws for two experiments in which only the reactant in question changes. Once the rate law is known, any experiment can be used to find the rate constant k. For part (d), use Equation 16.2 to describe the rate of this reaction in terms of reactant disappearance and product appearance. A negative sign is used for the rate in terms of reactants A, B, and C since these substances are reacting and [A], [B], and [C] are decreasing over time. Positive signs are used for the rate in terms of products D and E since these substances are being formed and [D] and [E] increase over time. Solution: a) To find the order for reactant A, first identify the reaction experiments in which [A] changes but [B] and [C] are constant. Use experiments 1 and 2 to find the order with respect to [A]. Set up a ratio of the rate laws for experiments 1 and 2 and fill in the values given for rates and concentrations and solve for x, the order with respect to [A]. x z y k [ A ]2 [ B]2 [ C ]2 Rate 2 = x z y k [ A ]1 [ B]1 [ C ]1 Rate1 [C]1 = [C]2 [B]1 = [B]2 x k [ A ]2 Rate 2 = x k [ A ]1 Rate1

k ⎡⎢0.096 mol/L⎤⎥ ⎦2 = ⎣ x −6 ⎡ 6.0 ×10 mol/L • s k ⎢0.024 mol/L⎤⎥ ⎣ ⎦1 16 = 4x x=2 Second order with respect to A x

9.6 ×10−5 mol/L • s

In finding the order with respect to B, there are no experiments in which only [B] changes. Experiments 2 and 4 can be used as [C] is constant while [A] and [B] change. Since we already know the order with respect to A, we can determine the order with respect to B in these two experiments. Set up a ratio of the rate laws for experiments 2 and 4 and fill in the values given for rates and concentrations and solve for y, the order with respect to [B].

k [ A ]2 [ B]2 [ C ]2 Rate 2 = 2 z k [ A ]4 [ B]4y [ C ]4 Rate 4 [C]2 = [C]4 2 k [ A ]2 [ B]2y Rate 2 = 2 y k [ A ]4 [ B]4 Rate 4 2

k ⎡⎢ 0.096 mol/L ⎤⎥ ⎡⎢ 0.085 mol/L ⎤⎥ ⎦2 ⎣ ⎦2 = ⎣ 2 y −6 ⎡ ⎤ ⎡ 1.5 ×10 mol/L • s k ⎢ 0.012 mol/L ⎥ ⎢ 0.170 mol/L ⎤⎥ ⎣ ⎦4 ⎣ ⎦4 64 = (8)2 (0.5)y y=0 Zero order with respect to B

9.6 ×10−5 mol/L • s

In finding the order with respect to C, there are no experiments in which only [C] changes. Experiments 1 and 3 can be used as [A] is constant while [B] and [C] change. Since we already know the order with respect to B, we can determine the order with respect to C in these two experiments. Set up a ratio of the rate laws for experiments 1 and 3 and fill in the values given for rates and concentrations and solve for z, the order with respect to [C].

16-37

k [ A ]1 [ B]1 [ C ]1 Rate1 = 2 z 0 k [ A ]3 [ B]3 [ C ]3 Rate 3 2

[A]1 = [A]3

k [ B]1 [ C ]1 Rate1 = z 0 k [ B]3 [ C ]3 Rate 3 0

k ⎡⎢ 0.085 mol/L ⎤⎥ ⎡⎢ 0.032 mol/L ⎤⎥ ⎦1 ⎣ ⎦1 = ⎣ 0 z −5 1.5 ×10 mol/L • s k ⎡⎢ 0.034 mol/L ⎤⎥ ⎡⎢0.080 mol/L ⎤⎥ ⎣ ⎦3 ⎣ ⎦3 Since y = 0, the B term may be ignored, it is only shown here for completeness. 0.4 = 0.4z z=1 First order with respect to C b) You can use any trial to calculate k, with the rate law: Rate = k[A]2[B]0[C]1 = k[A]2[C] Using experiment 1: rate = k[A]2[C] 6.0 ×10−6 mol/L • s rate k= = = 0.32552 = 0.33 L2/mol2s [A]2 [C] [0.024 mol/L] 2 [0.032 mol/L] This value will need to be divided by the coefficient of the substance to which the initial rate refers. If the initial rate refers to the disappearance of A or B then the constant (k') is: k' = k/2 = 0.32552/2 = 0.16276 = 0.16 L2/mol2s If the initial rate refers to the disappearance of C or the formation of D then the constant (k') is: k' = k = 0.33 L2/mol2s If the initial rate refers to the formation of E then the constant (k') is: k' = k/3 = 0.32552/3 = 0.108507 = 0.11 L2/mol2s 2 c) Rate = = k[A] [C] (substitute the appropriate k value from part (b)) Δ[ C] 1 Δ[ A ] 1 Δ [ B] Δ[ D ] 1 Δ[ E ] d) Rate = − =− =− = = 2 Δt Δt 2 Δt Δt 3 Δt

6.0 ×10−6 mol/L • s

16.93

Plan: Use the given rate law, rate = k[H+][sucrose], and enter the given values. The glucose and fructose are not in the rate law, so they may be ignored. Solution: a) The rate is first order with respect to [sucrose]. The [sucrose] is changed from 1.0 M to 2.5 M, or is increased by a factor of 2.5/1.0 or 2.5. Then the rate = k[H+][2.5 × sucrose]; the rate increases by a factor of 2.5. b) The [sucrose] is changed from 1.0 M to 0.5 M, or is decreased by a factor of 0.5/1.0 or 0.5. Then the rate = k[H+][0.5 × sucrose]; the rate decreases by a factor of ½ or half the original rate. c) The rate is first order with respect to [H+]. The [H+] is changed from 0.01 M to 0.0001 M, or is decreased by a factor of 0.0001/0.01 or 0.01. Then the rate = k[0.01 × H+][sucrose]; the rate decreases by a factor of 0.01. Thus, the reaction will decrease to 1/100 the original. d) The [sucrose] decreases from 1.0 M to 0.1 M, or by a factor of (0.1 M/1.0 M) = 0.1. [H+] increases from 0.01 M to 0.1 M, or by a factor of (0.1 M/0.01 M) = 10. Then the rate will increase by k[10 × H+][0.1 × sucrose]= 1.0 times as fast. Thus, there will be no change.

16.94

Plan: The overall order is equal to the sum of the individual orders. Since the reaction is eleventh order overall, the sum of the exponents equals eleven. Add up the known orders and subtract that sum from eleven to find the unknown order. Solution: Sum of known orders = 1 + 4 + 2 + 2 = 9 Overall order – sum of known orders = 11 – 9 = 2. The reaction is second order with respect to NAD.

16-38

16.95

Plan: The reaction is A + B → products. Assume the reaction is first order with respect to A and first order with respect to B. Find the concentration of each reactant in Mixture I and use those values and the initial rate to calculate k, the rate constant, for the reaction. Knowing k, the initial rate in Mixture II can be calculated using the rate law and the reactant concentrations. Solution: Rate = k[A][B] Mixture I: ⎛ 0.010 mol A ⎞⎟ ⎟ (6 spheres A )⎜⎜⎜ ⎝ 1 sphere ⎠⎟⎟ Concentration of A = = 0.12 mol/L 0.50 L ⎛ 0.010 mol B ⎞⎟ ⎟ (5 spheres B)⎜⎜ ⎜⎝ 1 sphere ⎠⎟⎟ Concentration of B = = 0.10 mol/L 0.50 L Use the rate law to find the value of k, the rate constant: Rate = k[A][B] rate 8.3 ×10−4 mol/L • min k= = = 0.069167 L/mol·min [A][B] [0.12 mol/L][0.10 mol/L] Use this value of k to find the initial rate in Mixture II: ⎛ 0.010 mol A ⎞⎟ ⎟ (7 spheres A)⎜⎜ ⎜⎝ 1 sphere ⎠⎟⎟ Concentration of A = = 0.14 mol/L 0.50 L ⎛ 0.010 mol B ⎞⎟ ⎟ (8 spheres B)⎜⎜⎜ ⎝ 1 sphere ⎠⎟⎟ Concentration of B = = 0.16 mol/L 0.50 L Rate = k[A][B] Rate = 0.069167 L/mol·min[0.14 mol/L][0.16 mol/L] Rate = 1.5493 × 10–3 = 1.5 × 10–3 mol/L∙min

16.96

Plan: The overall rate law for the mechanism is determined from the slowest step (the rate-determining step). An overall rate law can only include reactants and products; intermediates cannot be included in the rate law. Express [intermediate] in terms of [reactant]. Compare the resulting rate law from the mechanism to the actual rate law. Solution: Initially, the slow step in the mechanism gives: Rate = k2[CHCl3][Cl] However, Cl is an intermediate, and should not be in the final rate law. For an equilibrium, rateforward rxn = ratereverse rxn. For step 1, k1[Cl2] = k–1[Cl]2. Rearranging to solve for [Cl] gives [Cl]2 = (k1/k–1)[Cl2] [Cl] = (k1/k–1)1/2[Cl2]1/2 Substituting into the rate law from the slow step: Rate = k2[CHCl3][Cl] Rate = k2(k1/k–1)1/2[CHCl3][Cl2]1/2 Combining k’s: Rate = k[CHCl3][Cl2]1/2 The rate law from the mechanism is consistent with the actual rate law.

16.97

Plan: First, find the rate constant, k, for the reaction by solving the first-order half-life equation for k. Then use the first-order integrated rate law expression to find t, the time for decay.

16-39

Solution: ln 2 ln 2 to k = k t1/2 ln 2 k= = 5.7762 × 10–2 yr–1 12 yr

Rearrange t1/2 =

Use the first-order integrated rate law: ln ln

[ DDT ]t [ DDT ]0

= – kt

[10. ppbm ]t = – (5.7762 × 10–2 yr–1) t [ 275 ppbm ]0

t = 57.3765798 = 57 yr 16.98

Plan: Solve the first-order half-life equation for k and then take the reciprocal of that value of k. Solution: ln 2 ln 2 Rearrange t1/2 = to k = k t1/2 ln 2 k= = 0.19804 min–1 (unrounded) 3.5 min The problem states that the interval t = 1/k: t = 1/(0.19804 min–1) = 5.04943 = 5.0 min

16.99

Plan: The rate law for the reaction is given. Use the given values of rate, [A]0, and [B]0 to find the value of the rate constant, k, for the reaction. The value of k can be used to calculate the rate at the second set of reactant concentrations. Solution: The rate law is: Rate = k[A]2[B] Calculate k: 0.20 mol/L • s rate k= = = 0.20 L2/mol2s 2 2 [A] [B] ⎡1.0 mol/ L ⎤ ⎡1.0 mol/L ⎤ ⎢⎣ ⎥⎦ ⎢⎣ ⎥⎦ Using the k just calculated with the rate law: Rate = k[A]2[B] Rate = (0.20 L2/mol2s) [2.0 mol/L]2[3.0 mol/L] = 2.4 mol/L∙s

16.100 Plan: The rate constant can be determined from the slope of the integrated rate law plot. To find the correct order, the data should be plotted as (1) [sucrose] vs. time – linear for zero order, (2) ln [sucrose] vs. time – linear for first order, and (3) 1/[sucrose] vs. time – linear for second order. Once the order is established, use the appropriate integrated rate law to find the time necessary for 75.0% of the sucrose to react. If 75.0% of the sucrose has reacted, 100–75 = 25% remains at time t. Let [sucrose]0 = 100% and then [sucrose]t = 25%. Solution: a) All three graphs are linear, so picking the correct order is difficult. One way to select the order is to compare correlation coefficients (R2) — you may or may not have experience with this. The best correlation coefficient is the one closest to a value of 1.00. Based on this selection criterion, the plot of ln [sucrose] vs. time for the firstorder reaction is the best. Another method when linearity is not obvious from the graphs is to examine the reaction and decide which order fits the reaction. For the reaction of one molecule of sucrose with one molecule of liquid water, the rate law would most likely include sucrose with an order of one and would not include water. The plot for a first-order reaction is described by the equation ln [A]t = –kt + ln [A]0. The slope of the plot of ln [sucrose] vs. t equals –k. The equation for the straight line in the first-order plot is y = –0.21x – 0.6936. So, k = – (–0.21 h–1) = 0.21 h–1. Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

16-40

Solve the first-order half-life equation to find t1/2: ln 2 t1/2= = 3.3007 = 3.3 h 0.21 hr−1

Integrated rate law plots

y = 0.5897x + 1.93 R2 = 0.9935

(see legend)

3 2

y = −0.077x + 0.4896

R2 = 0.9875

0 0 -1

0.5

1.5

2.5

3.5

y = −0.21x − 0.6936

-2

R2 = 0.9998

time, h

Legend: ♦ y-axis is [sucrose] ■ y-axis is ln [sucrose] ▲ y-axis is 1/[sucrose] b) If 75% of the sucrose has been reacted, 25% of the sucrose remains. Let [sucrose]0 = 100% and [sucrose]t = 25% in the first-order integrated rate law equation: [ sucrose ]t ln = – kt [ sucrose ]0 [ 25% ]t = – (0.21 h–1) t ln [100% ]0 t = 6.6014 = 6.6 h c) The reaction might be second order overall with first order in sucrose and first order in water. If the concentration of sucrose is relatively low, the concentration of water remains constant even with small changes in the amount of water. This gives an apparent zero-order reaction with respect to water. Thus, the reaction appears to be first order overall because the rate does not change with changes in the amount of water. 16.101 Plan: You are given the ratio of the rate constants, k1 and k2, at a particular temperature and you want to find the difference in Ea between the uncatalyzed and catalyzed processes. Write the Arrhenius equation twice, once for the uncatalyzed process and once for the catalyzed process and divide the two equations so that the constant A factor divides out. Solution: k2 = 2.3 × 1014 T = 37°C + 273 = 310 K k1 ΔEa = ?

16-41

k1 = Ae–Ea1/RT

and k2 = Ae–Ea2/RT

− Ea 2

( Ea1 − Ea2 ) Ae RT k2 = = e − Ea1 RT k1 Ae RT E − Ea 2 k or ln 2 = a1 RT k1

⎛ k ⎞ RT ⎜⎜ ln 2 ⎟⎟⎟ = Ea1 – Ea2 ⎜⎝ k ⎠⎟ 1

2.3 ×1014 = Ea1 – Ea2 1 Ea1 – Ea2 = 8.5230 × 104 = 8.5 × 104 J/mol (8.314 J/mol∙K) (310 K) ln

16.102 a) False, at any particular temperature, molecules have a range of kinetic energies. b) False, at reduced pressure, the number of collisions per unit time is reduced, as is the reaction rate. c) True d) False, the increase in rate depends on the activation energy for the reaction. Also, biological catalysts (enzymes) may decompose on heating, reducing their effectiveness. e) False, they also must have the correct orientation. f) False, the activation energy is unique to the mechanism of a particular reaction. g) False, since most reaction rates depend to some extent on the reactant concentrations, as these decrease during the course of the reaction, the reaction rate also decreases. h) False, see part f). i) False, a catalyst speeds up the reaction by lowering the activation energy. j) False, the speed of a reaction (kinetics) is separate from the stability of the products (thermodynamics). k) False, the frequency factor, A, is the product of the collision frequency which is affected by temperature and an orientation probability factor. l) True m) False, the catalyst changes the activation energy, not ΔH of reaction. n) True o) True p) False, bimolecular and unimolecular refer to the molecularity or the number of reactant particles involved in the reaction step. There is no direct relationship to the speed of the reaction. q) False, molecularity and molecular complexity are not related. 16.103 Plan: The overall rate law for the mechanism is determined from the slowest step (the rate-determining step). An overall rate law can only include reactants and products; intermediates cannot be included in the rate law. Express [intermediate] in terms of [reactant]. Compare the resulting rate law from the mechanism to the actual rate law. Solution: Rate law for slow step (Step 3): Rate = k3[H2I][I] Both H2I and I are intermediates and cannot be in the final rate law. For an equilibrium, rateforward rxn = ratereverse rxn. From first two steps: From step 1: k1[I2] = k–1[I]2; [I] = (k1/k–1)1/2[I2]1/2 From step 2: k2[H2][I] = k–2[H2I]; [H2I] = k2/k–2[H2][I] Substituting for [H2I] in rate = k3[H2I][I] : Rate = k3[k2/k–2[H2][I]][I] Rate = k3k2/k–2[H2][I]2 Substituting for [I] in rate = k3k2/k–2[H2][I]2: Rate = k3k2/k–2[H2][(k1/k–1)1/2[I2]1/2]2 Rate = k3k2/k–2(k1/k–1)[H2][I2] Combining k values: Rate = k[H2][I2] which is consistent with the known rate law. Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

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16.104 a) Conductometric method. The HBr that forms is a strong acid in water, so it dissociates completely into ions. As time passes, more ions form, so the conductivity of the reaction mixture increases. b) Manometric method. The reaction involves a reduction in moles of gas, so the rate can be determined from the change in pressure (at constant volume and temperature) over time. 16.105 Plan: To solve this problem, a clear picture of what is happening is useful. Initially only N2O5 is present at a pressure of 125 kPa. Then a reaction takes place that consumes the gas N2O5 and produces the gases NO2 and O2. The balanced equation gives the change in the number of moles of gas as N2O5 decomposes. Since the number of moles of gas is proportional to the pressure, this change mirrors the change in pressure. The total pressure at the end, 178 kPa, equals the sum of the partial pressures of the three gases. Solution: Balanced equation: N2O5(g) → 2NO2(g) + 1/2O2(g) Therefore, for each mole of dinitrogen pentaoxide that is consumed, 2.5 moles of gas are produced. N2O5(g) → 2NO2(g) + 1/2O2(g) Initial P (kPa) 125 0 0 total Pinitial = 125 kPa Final P (kPa) 125 – x 2x 1/2x total Pfinal = 178 kPa Solve for x: PN O + PNO + PO = (125 – x) + 2x + 1/2x = 178 2

x = 35.3333 kPa (unrounded) Partial pressure of NO2 equals 2x = 2(35.3333) = 70.667 = 71 kPa. Check: Substitute values for all partial pressures to find total final pressure: (125 – 35.3333) + (2 × 35.3333) + ((1/2) × 35.3333) = 178 kPa The result agrees with the given total final pressure. 16.106 Plan: For part (a), first use the first-order half-life equation to find the rate constant k for the reaction. Then use the first-order integrated rate law to find concentration of reactant at a later time, given the initial concentration. Since the time unit in the rate constant is minutes, t must be expressed in units of minutes. For part (b), use the first-order integrated rate law to solve for the time required for 2/3 of the pill to decompose, leaving 1/3 pill at time t. For part (c) use the Arrhenius equation to calculate Ea. Solution: ln 2 ln 2 a) Rearrange t1/2 = to k = k t1/2 ln 2 = 7.7016 × 10–3 min–1 (unrounded) k= 90 min ⎛ 60 min ⎞⎟ Converting t from h to min: (2.5 h )⎜⎜ ⎟ = 150 min ⎝ 1 h ⎠⎟ ln

[aspirin ]t = – kt or [aspirin ]0

ln [aspirin]t = ln [aspirin]0 – kt ln [aspirin]t = ln [2 mg/100 mL] – (7.7016 × 10–3 min–1)(150 min) ln [aspirin]t = – 5.06729 [aspirin]t = 6.29964 × 10–3 mg/mL or 6.29964 ×10−3 mg × 100 mL = 0.62996 mg/100 mL = 0.6 mg/100 mL mL

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b) The antibiotic pill = PILL. The pill is taken at the fever temperature, so use the fever k. [ PILL ]t ln = – kt [ PILL ]0 ⎡1/3 PILL ⎤ ⎢ ⎥⎦ t ln ⎣ = – (3.9 × 10–5 s–1) t [1 PILL ]0

⎛ 1 h ⎞⎟ t = 28169.55 s or (28169.55 s)⎜⎜ ⎟ = 7.8 h ⎝ 3600 s ⎠⎟ Pills should be taken at about eight-hour intervals. T1 = [98.6°F – 32](5/9) + 273.15 = 310.15 K k1 = 3.1 × 10–5 s–1 k2 = 3.9 × 10–5 s–1 T2 = [101.9°F – 32](5/9) + 273 = 311.98 K Ea = ? k E ⎛1 1⎞ ln 2 = a ⎜⎜ − ⎟⎟⎟ ⎜ k1 R ⎝ T1 T2 ⎠⎟

( (

) )

−5 −1 ⎞ ⎛ ⎛ ⎞⎜ 3.9×10 s ⎟⎟ ⎜⎜8.314 J ⎟⎟⎜⎜ln ⎟⎟ ⎟⎜ −5 −1 ⎟ ⎟ ⎜⎝ mol • K ⎠⎜ × 3.1 10 s ⎜⎝ ⎠⎟⎟ Ea = = ⎛ ⎞⎟ ⎛1 1⎞ 1 1 ⎜⎜ ⎜⎜ − ⎟⎟⎟ − ⎟⎟ ⎜⎜ ⎜⎝ T1 T2 ⎠⎟ ⎜⎝ 310.15 K 311.98 K ⎠⎟⎟ Ea = 1.0092 × 105 J/mol = 1 × 105 J/mol The subtraction of the 1/T terms leaves only one significant figure.

⎛ k ⎞ R ⎜⎜⎜ln 2 ⎟⎟⎟ ⎝ k1 ⎠⎟

16.107 No. The uncertainty in the pressure, P, is 5%. The reaction rate is proportional to [P]4. The relative reaction rate with 5% error would be [1.05]4 = 1.22 or 22% in error. The rate measurement has an uncertainty of 22% so a 10% change in rate is not significant. 16.108 a) The iodide ion approaches from the side opposite the relatively large chlorine. H I

H b) The “backside attack” of the I– inverts the geometry at the carbon bearing the Cl, producing this product: CH2CH3

CH3 H

c) The planar intermediate can be attached from either side, producing a racemic mixture (that is, an equal mixture of two optical isomers): CH2CH3 H3CH2C

C I and CH3 H3C H H 16.109 Plan: The activation energy can be calculated using the Arrhenius equation. Although the rate constants, k1 and k2, are not expressly stated, the relative times give an idea of the rate. The reaction rate is proportional to the rate constant. At T1 = 90.0°C, the rate of reaction is 1 egg/4.8 min while at T2 = 100.0°C, the rate is 1 egg/4.5 min. Therefore, rate1 = 1 egg/4.8 min and rate2 = 1 egg/4.5 min are substituted for k1 and k2, respectively. Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

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Solution: k1 = 1 egg/4.8 min k2 = 1 egg/4.5 min Ea = ? k E ⎛1 1⎞ ln 2 = a ⎜⎜⎜ − ⎟⎟⎟ k1 R ⎝ T1 T2 ⎠⎟

T1 = 90.0°C + 273.2 = 363.2 K T2 = 100.0°C + 273.2 = 373.2 K

The number of eggs (1) is exact, and has no bearing on the significant figures. ⎛ ⎞ J ⎞⎟⎜⎜ 1 egg/4.5min ⎟⎟⎟ ⎛ ⎛ k2 ⎞⎟ ⎜ ⎜ 8.314 ln ⎟ ⎟ ⎜⎝ R ⎜⎜ln ⎟⎟ mol • K ⎠⎟⎜⎜⎜ 1 egg/4.8min ⎟⎟⎟ ⎜⎝ k1 ⎠⎟ ⎝ ⎠ Ea = = ⎛ 1 ⎛ 1 1 ⎞⎟ 1 ⎞⎟⎟ ⎜⎜ ⎜⎜ − ⎟ − ⎟ ⎜⎜ ⎜⎝ T T ⎠⎟⎟ 1 2 ⎜⎝ 363.2 K 273.2 K ⎠⎟⎟

( (

) )

Ea = 7.2730 × 103 J/mol = 7.3 × 103 J/mol

16.110 Plan: The overall reaction can be obtained by adding the three steps together. The overall rate law for the mechanism is determined from the slowest step (the rate-determining step). An overall rate law can only include reactants and products; intermediates cannot be included in the rate law. Express [intermediate] in terms of [reactant]. Solution: (1) 2H2SO4 → H3O+ + HSO4– + SO3 [fast] (2) SO3 + C6H6 → H(C6H5+)SO3– [slow] (3) H(C6H5+)SO3– + HSO4– → C6H5SO3– + H2SO4 [fast] (4) C6H5SO3– + H3O+ → C6H5SO3H + H2O [fast] a) Add the steps together and cancel: C6H6 + H2SO4 → C6H5SO3H + H2O b) Initially: Rate = k2[SO3][C6H6] (from the slow step) SO3 is an intermediate and cannot be included in the overall rate law. SO3 is produced in step 1 and its concentration is dependent on k1 and [H2SO4]: [SO3] = k1[H2SO4]2 Substituting for [SO3] in the rate law from the slow step: Rate = k2[k1[H2SO4]2][C6H6] Rate = k[H2SO4]2[C6H6] 16.111 Plan: Starting with the fact that rate of formation of O (rate of step 1) equals the rate of consumption of O (rate of step 2), set up an equation to solve for [O] using the given values of k1, k2, [NO2], and [O2]. Solution: a) Rate1 = k1[NO2] Rate2 = k2[O][O2} Rate1 = rate2 k1[NO2] = k2[O][O2] (6.0 × 10−3 s−1 ) ⎡⎣⎢4.0 × 10−9 M ⎤⎦⎥ = 2.4 × 10–15 M k [ NO 2 ] = [O] = 1 k2 [O 2 ] (1.0 × 106 L /mol • s) ⎡⎣⎢1.0 × 10−2 M ⎤⎦⎥ b) Since the rate of the two steps is equal, either can be used to determine rate of formation of ozone. Rate2 = k2[O][O2] = (1.0 × 106 L/mol∙s)(2.4 × 10–15 M)(1.0 × 10–2 M) = 2.4 × 10–11 mol/L∙s 16.112 Plan: At time = 0.00 min, assume [A]0 = 1.00; use the given equation for % inactivation to calculate [A]t at 3.00 min. Now knowing [A]0 and [A]t, use the first-order integrated rate law to calculate the rate constant, k in part (a). For part (b), use the equation for % inactivation to calculate [A]t and then use the k value from part (a) to calculate time, t. Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

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Solution: a) Calculating [A]t at 3.00 min (99.9% inactivation): % inactivation = 100 × (1 – [A]t/[A]0) At 3.00 min: 99.9% = 100 × (1 – [A]t/1.00] 99.9% = 100 – 100[A]t [A]t = 0.001 [ A ]t ln = – kt [ A ]0 [ 0.001]t [ A ]t ln ln [1.00 ]0 [ A ]0 k= − =− = 2.302585 = 2.3 min–1 3.00 min t b) Calculating [A]t at 95% inactivation: % inactivation = 100 × (1 – [A]t/[A]0) 95% = 100 × (1 – [A]t/1.00) 95% = 100 – 100[A]t [A]t = 0.05 [ A ]t ln = – kt [ A ]0 [ A ]t [0.05]t ln ln [ A ]0 [1.00 ]0 t= − =− = 1.30103 = 1.3 min k 2.302585 min−1 16.113 Plan: The overall rate law for the mechanism is determined from the slowest step (the rate-determining step). An overall rate law can only include reactants and products; intermediates cannot be included in the rate law. Express [intermediate] in terms of [reactant]. Compare the resulting rate law from the mechanism to the actual rate law. Solution: I. Since there is only one step in the mechanism, it must be the rate-determining step: Rate = k1[N2O5]2 This rate law does not match the actual rate law so the proposed mechanism is not valid. II. The first step is the rate-determining step. From step 1: Rate = k1[N2O5]2 This rate law does not match the actual rate law so the proposed mechanism is not valid. III. The second step is the rate-determining step. From step 2: Rate = k2[NO2][N2O5] NO2 is an intermediate and cannot be in the final rate law. For an equilibrium, rateforward rxn = ratereverse rxn. From step 1: k1[N2O5] = k–1[NO3][NO2]; [NO2] = (k1/k–1)[N2O5]/[NO3] Substituting for [NO2]: Rate = k2(k1/k–1)[N2O5]2[NO3]–1 = k[N2O5]2[NO3]–1 This rate law does not match the actual rate law so the proposed mechanism is not valid. IV. The second step is the rate-determining step. From step 2: Rate = k2[N2O3][O] N2O3 and O are intermediates and cannot be in the final rate law. For an equilibrium, rateforward rxn = ratereverse rxn. From step 1: k1[N2O5]2 = k–1[NO2] 2[N2O3][O]3; [O] = (k1/k–1) 1/3[N2O5] 2/3/[NO2] 2/3[N2O3]1/3

16-46

Substituting for [O]: Rate = k2(k1/k–1)1/3[N2O5]2/3[NO2]–2/3[N2O3]2/3 This rate law does not match the actual rate law so the proposed mechanism is not valid. V. The first step is rate determining: Rate = k1[N2O5]2 This rate law does not match the actual rate law so the proposed mechanism is not valid. 16.114 Plan: This problem involves the first-order integrated rate law (ln [A]t/[A]0 = –kt). The temperature must be part of the calculation of the rate constant. The concentration of the ammonium ion is directly related to the ammonia concentration. Use the given values of [NH3]0 and [NH3]t and the calculated values of k to find time, t. Solution: a) [NH3]0 = 3.0 mol/m3 [NH3]t = 0.35 mol/m3 T = 20°C 0.095(T – 15°C) k1 = 0.47e = 0.47e0.095(20 – 15°C) = 0.75576667 d–1 [ NH 3 ]t = – kt ln [ NH 3 ]0 ln

t= −

[ NH3 ]t [ NH3 ]0

k ⎡ 0.35 mol/m 3 ⎤ ⎦t ln ⎣ ⎡3.0 mol/m 3 ⎤ ⎣ ⎦0

= 2.84272 = 2.8 d 0.75576667 d−1 b) Repeating the calculation at the different temperature: [NH3]0 = 3.0 mol/m3 [NH3]t = 0.35 mol/m3 T = 10°C 0.095(T – 15°C) k1 = 0.47e = 0.47e0.095(10 – 15°C) = 0.292285976 d–1 [ NH3 ]t ln [ NH3 ]0 t= − k ⎡ 0.35 mol/m 3 ⎤ ⎦t ln ⎣ ⎡3.0 mol/m 3 ⎤ ⎦0 t= − ⎣ 0.292285976 d−1 t = 7.35045 = 7.4 d c) For NH4+ the rate = k1[NH4+] From the balanced chemical equation: Δ[ NH 4 + ] 1 Δ[O2 ] − =− 2 Δt Δt Thus, for O2: Rate = 2 k1[NH4+] Rate = (2)(0.75576667) ⎡⎢3.0 mol/m3 ⎤⎥ = 4.5346 = 4.5 mol/m3 ⎣ ⎦ 16.115 Plan: The rate law is rate = [CS2]m where m is the order of the reactant. To find the order of the reactant, take the ratio of the rate laws for two experiments. Once the rate law is known, any experiment can be used to find the rate constant k. Solution: m ⎛[CS2 ]exp 1 ⎞⎟ Rateexp 1 ⎜ ⎟ = ⎜⎜ a) ⎟ Rateexp 4 ⎝⎜[CS2 ]exp 4 ⎠⎟⎟

16-47

2.7 ×10−7 mol/L • s ⎛⎜ 0.100 mol/L ⎞⎟ =⎜ ⎟ ⎝⎜ 0.044 mol/L ⎠⎟ 1.2 ×10−7 mol/L • s 2.25 = (2.27273)m log (2.25) = m log (2.27273) m=1 Rate = k [CS2] b) First, calculate the individual k values; then average the values. k = rate/[CS2] k1 = (2.7 × 10–7 mol/L∙s)/(0.100 mol/L) = 2.7 × 10–6 s–1 k2 = (2.2 × 10–7 mol/L∙s)/(0.080 mol/L) = 2.75 × 10–6 = 2.8 × 10–6 s–1 k3 = (1.5 × 10–7 mol/L∙s)/(0.055 mol/L) = 2.7272 × 10–6 = 2.7 × 10–6 s–1 k4 = (1.2 × 10–7 mol/L∙s)/(0.044 mol/L) = 2.7272 × 10–6 = 2.7 × 10–6 s–1 kavg = [(2.7 × 10–6 s–1) + (2.75 × 10–6) + (2.7272 × 10–6) + (2.7272 × 10–6)]/4 = 2.7261 10–6 = 2.7 × 10–6 s–1 m

16.116 a) The reaction is C=C + H2(g) ↔ C − C. Since the hydrogenation and dehydrogenation reactions are reversible, the direction of reaction is determined by the hydrogen pressure. b) Dehydrogenation will require a higher temperature. Hydrogenation, adding hydrogen to the double bond in the alkene, is exothermic. The hydrogenated product is of lower energy than the dehydrogenated reactant. The reaction pathways are the same but in reverse order so the hydrogenated material has a larger activation energy and thus a higher temperature is needed to obtain a useful reaction rate for dehydrogenation.

Energy

Ea (hydrogenation) Ea(dehydrogenation)

Alkene + H2

ΔHrxn

Hydrogenated product

Reaction coordinate c) In the hydrogenation process, when the double bond has been broken and one hydrogen atom has been added to the bond, the molecule can rotate around the resulting single bond and then lose a hydrogen atom (since hydrogenation and dehydrogenation are reversible) to restore the double bond and produce the trans fat. 16.117 Plan: Rate is the change in concentration divided by change in time. To find the average rate for each trial in part (a), the change in concentration of S2O32– is divided by the time required to produce the color. The rate law is rate = k[I–]m[S2O82–]n where m and n are the orders of the reactants. To find the order of each reactant, take the ratio of the rate laws for two experiments in which only the reactant in question changes. Once the rate law is known, any experiment can be used to find the rate constant k. Since several solutions are mixed, final concentrations of each solution must be found with dilution calculations using MiVi = MfVf in the form: Mf = MiVi/Vf. Solution: a) Mf S2O32– = [(10.0 mL)(0.0050 M)]/50.0 mL = 0.0010 M S2O32– 2− Δ[ I 2 ] 1 Δ ⎡⎣S2 O3 ⎤⎦ − =− = [1/2(0.0010 M)]/time = [0.00050 M]/time = rate 2 Δt Δt Average rates: Rate1 = (0.00050 M)/29.0 s = 1.724 × 10–5 = 1.7 × 10–5 M s−1 Rate2 = (0.00050 M)/14.5 s = 3.448 × 10–5 = 3.4 × 10–5 M s−1 Rate3 = (0.00050 M)/14.5 s = 3.448 × 10–5 = 3.4 × 10–5 M s−1 Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

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b) Mf KI = Mf I– = [(10.0 mL)(0.200 M)]/50.0 mL = 0.0400 M I– (Experiment 1) Mf I– = [(20.0 mL)(0.200 M)]/50.0 mL = 0.0800 M I– (Experiments 2 and 3) Mf Na2S2O8 = Mf S2O82– = [(20.0 mL)(0.100 M)]/50.0 mL = 0.0400 M S2O82– (Experiments 1 and 2) Mf S2O82– = [(10.0 mL)(0.100 M)]/50.0 mL = 0.0200 M S2O82– (Experiment 3) Generic rate law equation: Rate = k [I–]m[S2O82–]n To find the order for I–, use experiments 1 and 2 in which [S2O82–] is constant while [I–] changes. Set up a ratio of the rate laws for experiments 1 and 2 and fill in the values given for rates and concentrations and solve for m, the order with respect to [I–]. n m k2 [ I− ] ⎡⎣S2 O82− ⎤⎦ Rate 2 = Molarity of S2O82–is constant. n m Rate1 k1 [ I− ] ⎡⎣S2 O82− ⎤⎦ m 3.4 ×10−5 Ms−1 [ 0.0800 ] = m [ 0.0400 ] 1.7 ×10−5 Ms−1 2.0 = (2.00)m m=1 The reaction is first order with respect to I–. To find the order for S2O82–, use experiments 2 and 3 in which [I–] is constant while [S2O82–] changes. Set up a ratio of the rate laws for experiments 2 and 3 and fill in the values given for rates and concentrations and solve for n, the order with respect to [S2O82–]. n m k3 [ I− ] ⎡⎣S2 O82− ⎤⎦ Rate 3 = Molarity of I– is constant. n m Rate 2 k 2 [ I− ] ⎡⎣S2 O82− ⎤⎦ n 3.4 ×10−5 Ms−1 [ 0.0200 ] = n [ 0.0400 ] 3.4 ×10−5 Ms−1 1.0 = (0.500)n n=0 The reaction is zero order with respect to S2O82–. c) Rate = k[I–] k = rate/[I–] Using experiment 2 (unrounded rate value) k = (3.448 × 10–5 M s–1)/(0.0800 M) = 4.31 × 10–4 = 4.3 × 10–4 s–1 d) Rate = (4.3 × 10–4 s–1)[I–] 16.118 The ability of the bath to remove heat is proportional to ΔT, the difference between the bath temperature and the temperature inside the flask. Therefore when the temperature is reached such that the rate of heat increase exceeds the rate of heat loss, the reaction “runs away.” The problem with the scale-up is that the heat transfer from the flask to the cooling bath is proportional to the shared surface area of the reactant solution and the cooling bath, A, while the heat given off by the reaction is proportional to the volume of the reactants. The volume increases as the cube of the radius of the flask increases while the area increases as the square of the radius increases. Therefore the heat generation will exceed the cooling capacity at a lower temperature in the larger flask and the reaction will run away. 16.119 Plan: This is a first-order process so use the first-order integrated rate law. The increasing cell density changes the integrated rate law from –kt to +kt. In part (a), we know t(2 h), k, and [A]0 so [A]t can be found. Treat the concentration of the cells as you would molarity. Since the rate constant is expressed in units of min–1, the time interval of 2 h must be converted to a time in minutes. In part b), [A]0, [A]t, and k are known and time t is calculated.

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Solution:

⎛ 60 min ⎞⎟ a) Converting time in h to min: (2 h )⎜⎜ ⎟ = 120 min ⎝ 1 h ⎠⎟ ln [A]t = ln [A]0 + kt ln [A]t = ln [1.0 × 103] + (0.035 min–1)(120 min) ln [A]t = 11.107755 [A]t = 6.6686 × 104 = 7 × 104 cells/L b) ln [A]t = ln [A]0 + kt ln [A]t − ln [A]0 =t k ln [2 ×103 cells/L] − ln [1×103 cells/L] =t 0.035 min−1 t =19.804 = 2.0 × 101 min 16.120 a) The shape is tetrahedral and the hybridization of C1 is sp3. H C H

CH3

b) The shape is trigonal bipyramidal. Because an unhybridized p orbital is needed to overlap p orbitals on I and Br, the hybridization around C1 is sp2. H 1

Br H

I CH3

–

c) After Br is replaced with I– in the initial replacement reaction, the ethyl iodide is optically active. However, as other I– ions react with the ethyl iodide by the same mechanism the molecules change from one isomer to the other. Eventually, equal portions of each isomer exist and the ethyl iodide is optically inactive. 16.121 Plan: This is a first-order reaction so use the first-order integrated rate law. The first-order half-life equation is used to find the rate constant, k. Then, knowing k, time, and [A]0, [A]t can be calculated. The fraction remaining is [A]t/[A]0. Solution: ln 2 ln 2 Rearrange t1/2 = to k = k t1/2 ln 2 k= = 0.086212 d–1 (unrounded) 8.04 d ln [A]t = ln [A]0 – kt ln [A]t = ln [1.7 × 10–4] – (0.086212 d–1)(30. d) ln [A]t = –11.26607212 [A]t = 1.27999 × 10–5 M (unrounded) Fraction remaining = [A]t/[A]0 = (1.27999 × 10–5 M)/(1.7 × 10–4 M) = 0.0752936 = 0.075 16.122 Plan: For part (a), use the Monod equation to calculate μ for values of S between 0.0 and 1.0 kg/m3 and then graph. For parts b) and c), use the Monod equation to calculate μ at the given conditions. The value of μ is the rate constant k in the first-order integrated rate law. The increasing population density changes the integrated rate law from –kt to +kt. We know t(1 h), k, and [A]0 so [A]t can be found. Treat the density of the cells as you would molarity. Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

16-50

olution: So

1.5 × 10−4 s−1 )(0.25 kg/m 3 ) ( μmax S μ= = = 1.34 × 1 0–4 s–1 3 3 Ks + S (0.03 kg/m ) + (0.25 kg/mm ) 1.5 × 10−4 s−1 )(0.50 kg/m m3 ) ( μmax S μ= = = 1.42 × 110–4 s–1 3 3 Ks + S (0.03 kg/m ) + (0.50 kg/mm ) 1.5× 10−4 s−1 )(0.75 kg/m m3 ) ( μmax S μ= = = 1.44 × 110–4 s–1 3 3 Ks + S + 0.75 kg/m m 0.03 kg/m ( ) ( )

μ=

(1.5 ×10−4 s−1 )(1.0 kg/m 3 ) = 1.46 × 100–4 s–1 μmax S = Ks + S (0.03 kg/m 3 ) + (1.0 kg/m 3 )

b)) μ =

(1.5 ×10−4 s−1 )(0.30 kg//m 3 ) = 1.36366 × 10–4 s–1 μmax S = Ks + S (0.03 kg/mm 3 ) + (0.30 kg /m 3 )

⎛ 3600 0 s ⎞⎟ = 3600 s Converting timee in h to second ds: (1 h )⎜⎜ ⎝ 1 h ⎟⎟⎠ ln n [A]t = ln [A]0 + kt ln n [A]t = ln [5.0 × 103] + (1.36 636 × 10–4 s–1)(3600 s) ln n [A]t = 9.0080 08919 [A A]t = 8.1689 × 103 = 8.2 × 10 03 cells/m3 c)) μ =

μmax S Ks + S

(1.5 ×10 s )(0.70 kg /m ) = 1.4388356 × 10 s (0.03 kg/mm ) + (0.70 kgg/m ) −4

−1

–4

–1

ln n [A]t = ln [A]0 + kt ln n [A]t = ln [5.0 × 103] + (1.43 38356 × 10–4 s–1– )(3600 s) ln n [A]t = 9.0350 00135 [A A]t = 8.39172 × 103 = 8.4 × 10 1 3 cells/m3

16-51

16.123 Plan: The rate law is rate = k[A]m[B]n where m and n are the orders of the reactants. The initial rate for each reaction mixture is given. Calculate the concentration of A and B in each mixture. To find the order of each reactant, take the ratio of the rate laws for two experiments in which only the reactant in question changes. Once the rate law is known, Reaction Mixture I, II, or III can be used to find the rate constant k for the reactions without the solid and Reaction Mixture IV can be used to find the rate constant k for the reaction with a solid present. Solution: a) Reaction Mixture I ⎛ 0.01 mol ⎞⎟⎛ 1 ⎞ Concentration of A = (5 spheres)⎜⎜⎜ ⎟⎟ = 0.10 mol/L A ⎟⎟⎜ ⎝ 1 sphere ⎠⎟⎜⎝ 0.50 L ⎠⎟

⎛ 0.01 mol ⎞⎟⎛ 1 ⎞ ⎟⎟⎜ Concentration of B = (5 spheres)⎜⎜⎜ ⎟⎟ = 0.10 mol/L B ⎝ 1 sphere ⎠⎟⎜⎝ 0.50 L ⎠⎟ Reaction Mixture II

⎛ 0.01 mol ⎞⎟⎛ 1 ⎞ ⎟ = 0.16 mol/L A ⎟⎜ Concentration of A = (8 spheres)⎜⎜ ⎜⎝ 1 sphere ⎠⎟⎟⎜⎝ 0.50 L ⎠⎟⎟ ⎛ 0.01 mol ⎞⎟⎛ 1 ⎞ ⎟ = 0.10 mol/L B ⎟⎜ Concentration of B = (5 spheres)⎜⎜ ⎜⎝ 1 sphere ⎠⎟⎟⎜⎝ 0.50 L ⎠⎟⎟ Reaction Mixture III

⎛ 0.01 mol ⎞⎟⎛ 1 ⎞ ⎟⎟ = 0.16 mol/L A ⎟⎟⎜ Concentration of A = (8 spheres)⎜⎜⎜ ⎝ 1 sphere ⎠⎟⎜⎝ 0.50 L ⎠⎟ ⎛ 0.01 mol ⎞⎟⎛ 1 ⎞ ⎟⎟⎜ Concentration of B = (7 spheres)⎜⎜⎜ ⎟⎟ = 0.14 mol/L B ⎝ 1 sphere ⎠⎟⎜⎝ 0.50 L ⎠⎟ To find the order for A, use Mixtures I and II in which [B] is constant while [A] changes. Set up a ratio of the rate laws for Mixtures I and II and fill in the values given for rates and concentrations and solve for m, the order with respect to [A]. m Rate II k [ A ] [ B]n = 2 m n The concentration of B is constant Rate I k1 [ A ] [ B]

5.6 ×10−4 mol/L • s 3.5 ×10−4 mol/L • s

[ 0.16 ] m [ 0.10 ] m

1.6 = (1.6)m m=1 The reaction is first order with respect to A. To find the order for B, use Mixtures II and III in which [A] is constant while [B] changes. Set up a ratio of the rate laws for Mixtures II and III and fill in the values given for rates and concentrations and solve for n, the order with respect to [B].

m Rate III k [ A ] [ B]n = 2 m n Rate II k1 [ A ] [ B]

5.6 ×10−4 mol/L • s 5.6 ×10−4 mol/L • s

The concentration of A is constant. [ 0.14 ] n [ 0.10 ]

1 = (1.4)n n=0 The reaction is zero order with respect to B.

16-52

Rate law: Rate = k[A][B]0 = k[A] b) The overall reaction order is 1 + 0 = 1. c) Use Reaction Mixture I: Rate = k[A] 3.5 × 10–4 mol/L·s = k[0.10] k = 3.5 × 10–3 s–1 d) The catalyst was used in Reaction Mixture IV. ⎛ 0.01 mol ⎞⎟⎛ 1 ⎞ ⎟⎟⎜ Concentration of A = (5 spheres)⎜⎜⎜ ⎟⎟ = 0.10 mol/L A ⎝ 1 sphere ⎠⎟⎜⎝ 0.50 L ⎠⎟

⎛ 0.01 mol ⎞⎟⎛ 1 ⎞ ⎟⎟ = 0.16 mol/L B ⎟⎟⎜ Concentration of B = (8 spheres)⎜⎜⎜ ⎝ 1 sphere ⎠⎟⎜⎝ 0.50 L ⎠⎟ Rate = k[A] 4.9 × 10–4 = k[0.10] k = 4.9 × 10–3 s–1 Yes, the gray pellets had a catalytic effect. The rate of reaction and the rate constant are greater with the pellets than without.

ΔI ΔI = k Δl = k I or – Δl I where k = fraction of light removed per unit length, l, and I = light intensity. At length l = 0, I0 = intensity of light entering the solution. At some later length l, Il = intensity of light leaving the solution.

16.124 a) Rate of light intensity decrease = –

–∫

Il I0

l ΔI = k ∫ Δl =0 l I

I ln 0 = k (l − 0) Il ln I0 – ln Il = k × l ln Il – ln I0 = –k × l I ln l = − k × l = – fraction of light removed per unit length × distance (length) traveled I0 ΔS ΔS = k S or – = k Δt b) Rate of savings decline = – Δt S where k = fraction of savings lost per unit of time, t and S = savings. At time t = 0, S0 = initial value of savings. At some later time t, St = value of savings remaining. t St ΔS –∫ = k ∫ Δt t =0 S0 S S0 ln = k (t − 0) St ln S0 – ln St = k × t ln St – ln S0 = –k × t

St = − k × t = – fraction of savings lost per unit time × savings time interval S0

16.125 Plan: The figure shows the H2 molecule adsorbing to the metal surface with H2 bond breakage. The individual H atoms form bonds to the metal catalyst atoms. Ethylene also adsorbs to the metal catalyst and then the two C–H bonds form, one at a time. The resulting C2H6 leaves the metal surface. The overall reaction can be obtained by adding the steps of the mechanism together. Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

16-53

Solution: H2(g) → H2(ads) H2(ads) + 2M → 2M–H C2H4(g) → C2H4(ads) C2H4(ads) + M–H → C2H5(ads) + M C2H5(ads) + M–H → C2H6(g) + M C2H4(g) + H2(g) → C2H6(g) 16.126 Plan: Rate is proportional to the rate constant, so if the rate constant increases by a certain factor, the rate increases by the same factor. Thus, to calculate the change in rate the Arrhenius equation can be used and substitute ratecat/rateuncat = kcat/kuncat. Solution: k = Ae–Ea/RT

ΔEa = 5 kJmol = 5000 J/mol

T = 37°C + 273 = 310 K

−Ea 2

( Ea1 − Ea2 ) Ae RT k2 = = e −Ea1 RT RT k1 Ae

5000 J/mol k E − Ea 2 ln 2 = a1 = = 1.93998 J ⎞⎟ ⎛ k1 RT ⎜⎜8.314 310 K ( ) ⎝ mol • K ⎠⎟ k2 = e1.93998 = 6.9586 = 7 The rate of the enzyme-catalyzed reaction occurs at a rate 7 times faster k1 than the rate of the uncatalyzed reaction at 37°C. 16.127 Plan: This is a first-order reaction so use the first-order integrated rate law. In part (a), use the first-order half-life equation and the given value of k to find the half-life. In parts (b) and (c), solve the first-order integrated rate law to find the time necessary for 40% and 90% of the acetone to decompose. If 40% of the acetone has decomposed, 100 – 40 = 60% remains at time t. If 90% of the acetone has decomposed, 100 – 90 = 10% remains at time t. Solution:

ln 2 ln 2 = = 79.672 = 8.0 × 101 s k 8.7 ×10−3 s−1 b) ln [acetone]t = ln [acetone]0 – kt ln [acetone]t − ln [acetone]0 =t 40% of acetone has decomposed; [acetone]t = 100 – 40 = 60% −k ln [60.] − ln [100] = t = 58.71558894 = 59 s −8.7 ×10−3 s−1 c) ln [acetone]t = ln [acetone]0 – kt ln [acetone]t − ln [acetone]0 =t 90% of acetone has decomposed; [acetone]t = 100 – 90 = 10% −k ln [10.] − ln [100] = t = 264.66 = 2.6 × 102 s −3 −1 −8.7 ×10 s a) t1/2 =

16.128 Plan: The reaction begins with B as the reactant; the final product is A. Isomer C is formed during the reaction but is not a final product. Solution: a) B → C C→A B → A b) C is an intermediate. Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

16-54

CHAPTER 17 EQUILIBRIUM: THE EXTENT OF CHEMICAL REACTIONS FOLLOW–UP PROBLEMS 17.1A

Plan: First, balance the equations and then write the reaction quotient. Products appear in the numerator of the reaction quotient and reactants appear in the denominator; coefficients in the balanced reaction become exponents. Solution: a) Balanced equation: 4NH3(g) + 5O2(g) ⇆ 4NO(g) + 6H2O(g)

[ NO]4 [ H2 O]

Reaction quotient: Qc =

[ NH3 ] [O2 ] 4

b) Balanced equation: 2NO2(g) + 7H2(g) ⇆ 2NH3(g) + 4H2O(l)

[ NH3 ] Reaction quotient: Qc = 2 7 [ NO2 ] [ H2 ] 2

c) Balanced equation: 2KClO3(s) ⇆ 2KCl(s) + 3O2(g) Reaction quotient: Qc = [O2]3 17.1B

[CO]2 [ H2 ] [CH 4 ][CO2 ] 2

Reaction quotient: Qc =

b) Balanced equation: 2H2S(g) + SO2(g) ⇆ 2S(s) + 2H2O(g)

[H2 O] Reaction quotient: Qc = 2 [ H 2 S] [SO2 ] 2

c) Balanced equation: HCN(aq) + NaOH(aq) (s) ⇆ NaCN(aq) + H2O(l) Reaction quotient: Qc = 17.2A

[ NaCN ] [ HCN ][ NaOH ]

Plan: By multiplying by a factor or reversing, determine a way for Reactions 1 and 2 to sum to make the given reaction. When a reaction is multiplied by a factor, the equilibrium constant is raised to a power equal to the factor. When a reaction is reversed, the reciprocal of the equilibrium constant is used as the new equilibrium constant. Solution: (1) C(s) + CO2(g) ⇆ 2CO(g) Kc1 = 1.4 × 1012 Multiply by 2 (2) 2CO(g) + 2Cl2(g) ⇆ 2COCl2(g) Kc2 = (0.55)2 = 0.30 C(s) + CO2(g) + 2Cl2(g) ⇆ 2COCl2(g) Kc(overall) = Kc1 × Kc2 = (1.4 × 1012)(0.30) = 4.2 × 1011

17-1

17.2B

Plan: When a reaction is multiplied by a factor, the equilibrium constant is raised to a power equal to the factor. When a reaction is reversed, the reciprocal of the equilibrium constant is used as the new equilibrium constant. Solution: a) All coefficients have been multiplied by the factor 2. Additionally, the reaction has been reversed. Therefore, the reciprocal of the equilibrium constant should be raised to the 2 power. For reaction a)

æ ö÷2 1 ÷ = 5917.1598 = 5.9 × 103 Kc = çç çè1.3 ´ 10 –2 ÷ø÷

b) All coefficients have been multiplied by the factor 1/4 so the equilibrium constant should be raised to the 1/4 power. For reaction b) 17.3A

Kc = (1.3 × 10–2)1/4 = 0.33766 = 0.34

Plan: Kp and Kc for a reaction are related through the ideal gas equation as shown in KP = Kc(RT)Dn . Find Dngas, the change in the number of moles of gas between reactants and products (calculated as products minus reactants). Then, use the given Kc to solve for Kp. Solution: The total number of product moles of gas is 1 and the total number of reactant moles of gas is 2. Dn = 1 – 2 = –1 KP = Kc(RT)Dn KP = 1.67[(0.0821 atm ∙ L/mol ∙ K)(500. K)]–1 KP = 0.040682095 = 4.07 × 10–2

17.3B

Plan: Kp and Kc for a reaction are related through the ideal gas equation as shown in KP = Kc(RT)Dn . Find Dngas, the change in the number of moles of gas between reactants and products (calculated as products minus reactants). Then, use the given KP to solve for Kc. Solution: The total number of product moles of gas is 3 and the total number of reactant moles of gas is 5. Dn = 3 – 5 = –2 KP = Kc(RT)Dn KP = Kc(RT)–2 KP(RT)2 = Kc Kc = (3.0 × 10-5) [(0.0821 atm ∙ L/mol ∙ K)(1173 K)]2 = 0.27822976 = 0.28

17.4A

Plan: Write the reaction quotient for the reaction and calculate Qc for each circle. Compare Qc to Kc to determine the direction needed to reach equilibrium. If Qc > Kc, reactants are forming. If Qc < Kc, products are forming. Solution: [Y] The reaction quotient is . [X] [Y] [3] = = 0.33 Circle 1: Qc = [X] [9] Since Qc < Kc (0.33 < 1.4), the reaction will shift to the right to reach equilibrium [Y] [7] = = 1.4 Circle 2: Qc = [X] [5] Since Qc = Kc (1.4 = 1.4), there is no change in the reaction direction. The reaction is at equilibrium now.

17-2

[Y] [8] = = 2.0 [X] [4] Since Qc > Kc (2.0 > 1.4), the reaction will shift to the left to reach equilibrium

Circle 3: Qc =

17.4B

[ 5] [D] = 2 = 0.56 2 [C ] [3]

According to the problem, circle 1 is at equilibrium. Therefore, Kc = 0.56. Circle 2: Qc =

[D] [6 ] = = 0.67 [C ]2 [3]2

Since Qc > Kc (0.67 > 0.56), the reaction will shift to the left to reach equilibrium. Circle 3: Qc =

[D] [ 7] = 2 = 0.44 2 [C ] [ 4]

Since Qc < Kc (0.44 < 0.56), the reaction will shift to the right to reach equilibrium 17.5A

Plan: To decide whether CH3Cl or CH4 are forming while the reaction system moves toward equilibrium, calculate Qp and compare it to Kp. If Qp > Kp, reactants are forming. If Qp < Kp, products are forming. Solution: QP =

PCH3Cl PHCl PCH4 PCl2

(0.24 atm )(0.47atm ) = 24.7912 = 25 (0.13atm )(0.035atm )

Kp for this reaction is given as 1.6 × 104. Qp is smaller than Kp (Qp < Kp) so more products will form. CH3Cl is one of the products forming.

17.5B

Plan: To determine whether the reaction is at equilibrium or, if it is not at equilibrium, which direction to will proceed, calculate Qc and compare it to Kc. If Qc > Kc, the reaction will proceed to the left. If Qc < Kc, the reaction will proceed to the right. Solution: é 1.2 mol ù 2 2 ê ú [SO3 ] êë 2.0 L úû Qc = = = 0.166089965 = 0.17 2 2 [SO2 ] [O2 ] éê 3.4 mol ùú éê 1.5 mol ùú êë 2.0 L úû êë 2.0 L úû The system is not at equilibrium. Kc for this reaction is given as 4.2 × 10–2. Qc is larger than Kc (Qc > Kc) so the reaction will proceed to the left.

17.6A

Plan: The information given includes the balanced equation, initial pressures of both reactants, and the equilibrium pressure for one reactant. First, set up a reaction table showing initial partial pressures for reactants and 0 for product. The change to get to equilibrium is to react some of reactants to form some product. Use the equilibrium quantity for O2 and the expression for O2 at equilibrium to solve for the change. From the change find the equilibrium partial pressure for NO and NO2. Calculate Kp using the equilibrium values.

17-3

Solution: Pressures (atm)

2NO(g)

O2(g)

⇆

2NO2(g)

Initial

1.000

Change

–2x

–x

+2x

Equilibrium

1.000 – 2x 1.000 – x 2x At equilibrium PO2 = 0.506 atm = 1.000 – x; so x = 1.000 – 0.506 = 0.494 atm PNO = 1.000 – 2x = 1.000 – 2(0.494) = 0.012 atm

PNO2 = 2x = 2(0.494) = 0.988 atm Use the equilibrium pressures to calculate Kp. 2 PNO 2

KP =

17.6B

2 PNO PO2

(0.988)2 = 1.339679 × 104 = 1.3 × 104 (0.012)2 (0.506)

Plan: The information given includes the balanced equation, initial concentrations of both reactants, and the equilibrium concentration for one product. First, set up a reaction table showing initial concentrations for reactants and 0 for products. The change to get to equilibrium is to react some of reactants to form some of the products. Use the equilibrium quantity for N2O4 and the expression for N2O4 at equilibrium to solve for the change. From the change find the equilibrium concentrations for NH3, O2, and H2O. Calculate Kc using the equilibrium values. Solution: Pressures (atm)

4NH3(g) +

7O2(g)

2N2O4(g) +

6H2O(g)

Initial

2.40

Change

–4x

–7x

+2x

+6x

2.40 – 4x

2.40 – 7x

Equilibrium At equilibrium

⇆

[N2O4] = 0.134 M = 2x; so x = 0.0670 M [NH3] = 2.40 M – 4(0.0670 M) = 2.13 M [O2] = 2.40 M – 7(0.0670 M) = 1.93 M [H2O] = 6(0.0670 M) = 0.402 M

Use the equilibrium pressures to calculate Kc.

[ N2 O4 ] [ H2 O] [0.134 ]2 [0.402]6 –8 –8 = 4 7 = 3.6910 × 10 = 3.69 × 10 4 7 2.13 1.93 [ ] [ ] [ NH3 ] [O2 ] 2

Kc =

17.7A

Plan: Use the given value of Kc and the relationship Kp = Kc (RT)Δn to calculate Kp; the temperature must be in kelvins. Write the equilibrium expression for Kp, insert the given equilibrium pressures of H2 and S2 into that expression, and solve for the equilibrium pressure of H2S. Solution: Δngas = 3 − 2 = 1. æ ö atm ⋅ L K p = Kc ( RT )1 = 2.25 ´ 10-4 çç0.0821 ´ 1403 K ÷÷÷ = 0.0259 ÷ø çè mol ⋅ K

(P )(P ) K = 0.0259 = 2 H2

PH S 2

PH2 S 2

(P )(P ) = (0.370) (0.185) = 0.989 atm = 2 H2

0.0259

17-4

17.7B

Plan: Write the equilibrium expression for Kp and insert the partial pressures for PH3 and P2 as their equilibrium values. Solve for the partial pressure of H2. Solution:

(PP )(PH ) K = 2 (PPH )

( KP )( PPH3 )

PH 2 = 3

(PP )

(19.6)(0.112)2 = 1.0457 = 1.05 atm (0.215)

= 3

17.8A

Plan: Find the initial molarities of CO and H2O by dividing moles by the volume of the flask. Set up a reaction table and use the variables to find equilibrium concentrations in the equilibrium expression. Solution: 0.250 mol = 2.00 M MCO = MH O = 2 0.125 L Concentration (M)

Initial Change Equilibrium

CO(g)

2.00 –x 2.00 – x

H2O(g)

⇆

2.00 –x 2.00 – x

CO2(g)

0 +x x

H2(g)

0 +x x

é CO2 ù é H 2 ù x2 ûë û = Kc = 1.56 = ë é COù é H 2 Où (2.00 - x )2 ë ûë û Taking the square root of each side, ignoring the negative root, and solving gives x = 1.11 M. [CO] = [H2O] = 2.00 M − x = 2.00 M – 1.11 M = 0.89 M [CO2] = [H2] = x = 1.11 M 17.8B

Plan: Find the initial molarities of Cl2O and H2O by dividing moles by the volume of the flask. Set up a reaction table and use the variables to find equilibrium concentrations in the equilibrium expression. Solution: moles Cl2 O 6.15 mol = Molarity of Cl2O = = 1.23 M volume 5.00 L Molarity of H2O =

moles H 2 O 6.15 mol = = 1.23 M volume 5.00 L

Concentration (M) Cl2O(g) + Initial 1.23 Change –x Equilibrium 1.23 – x Set up equilibrium expression: Kc = 0.18 =

H2O(g) 1.23 –x 1.23 – x

⇆

2HOCl (g) 0 +2x 2x

[ HOCl ]2 [2x ]2 = [Cl2 O][ H2 O] [1.23 - x ][1.23 - x ]

Take the square root of each side.

17-5

[2x ] [1.23 - x ] 0.521844804 – 0.424264068x = 2x 0.521844804 = 2.424264068x x = 0.215259000 M [Cl2O] = [H2O] = 1.23 M – 0.215259 M = 1.014741 = 1.01 M

0.424264068 =

[HOCl] = 2(0.215259 M) = 0.430518 = 0.43 M 17.9A

Plan: Find the molarity of I2 by dividing moles of I2 by the volume. First set up the reaction table, then set up the equilibrium expression. To solve for the variable, x, first assume that x is negligible with respect to initial concentration of I2. Check the assumption by calculating the % error. If the error is greater than 5%, calculate x using the quadratic equation. The next step is to use x to determine the equilibrium concentrations of I2 and I. Solution: 0.50 mol [I2]init = = 0.20 M 2.5 L a) Equilibrium at 600 K Concentration (M) Initial Change

I2(g) 0.20 –x

Equilibrium

⇆

0.20 – x

2I(g) 0 +2x 2x

[I] = 2.94 × 10–10 [I2 ] 2

Equilibrium expression:

Kc =

[2x ]2 = 2.94 × 10–10 [0.20 - x ]

Assume x is negligible so 0.20 – x ≈ 0.20

[2x ]2 = 2.94 × 10–10 [0.20] 4x2 = (2.94 × 10–10) (0.20); x = 3.834 × 10–6 = 3.8 × 10–6 Check the assumption by calculating the % error: 3.8 ´ 10-6 (100) = 0.0019% which is smaller than 5%, so the assumption is valid. 0.20 At equilibrium [I]eq = 2x = 2(3.834 × 10–6) = 7.668 × 10–6 = 7.7 × 10–6 M and

[I2]eq = 0.20 – x = 0.20 – 3.834 × 10–6 = 0.199996 = 0.20 M b) Equilibrium at 2000 K Equilibrium expression:

[2x ]2 = 0.209 [0.20 - x ]

Kc =

[ I ]2 = 0.209 [I2 ]

Assume x is negligible so 0.20 – x is approximately 0.20

[2x ]2 = 0.209 [0.20] 4x2 = (0.209)(0.20) x = 0.102225 = 0.102 Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

17-6

Check the assumption by calculating the % error: 0.102 (100) = 51% which is larger than 5% so the assumption is not valid. Solve using quadratic 0.20 equation.

[2x ]2 = 0.209 [0.20 - x ] 4x2 + 0.209x – 0.0418 = 0 2

-0.209  (0.209) - 4 (4 )(-0.0418) 2 (4 )

= 0.0793857 or –0.1316

Choose the positive value, x = 0.0793857 At equilibrium [I]eq = 2x = 2(0.0793857) = 0.1587714 = 0.16 M and [I2]eq = 0.20 – x = 0.20 – 0.0793857 = 0.1206143 = 0.12 M 17.9B

Plan: First set up the reaction table, then set up the equilibrium expression. To solve for the variable, x, first assume that x is negligible with respect to initial partial pressure of PCl5. Check the assumption by calculating the % error. If the error is greater than 5%, calculate x using the quadratic equation. The next step is to use x to determine the equilibrium partial pressure of PCl5. Solution: a) Equilibrium at a PCl5 partial pressure of 0.18 atm: PCl5(g) ⇆ 0.18 –x 0.18 – x

Partial Pressure (atm) Initial Change Equilibrium Equilibrium expression:

KP =

PCl3(g) + 0 +x x

Cl2(g) 0 +x x

(PPCl )(PCl ) = 3.4 × 10 (PPCl ) 3

–4

(x)(x) = 3.4 × 10–4 Assume x is negligible so 0.18 – x ≈ 0.18 (0.18 - x) (x)(x) = 3.4 × 10–4 (0.18)

x2 = (3.4 × 10–4) (0.18) so x = 0.0078230428 = 7.8 × 10–3 Check the assumption by calculating the % error: (7.8 ´ 10 –3 )(100%) = 4.3% which is smaller than 5%, so the assumption is valid. (0.18) At equilibrium [PCl5]eq = 0.18 M – 7.8 × 10–3 M = 0.17 M b) Equilibrium at a PCl5 partial pressure of 0.18 atm: Partial Pressure (atm)

PCl5(g)

PCl3(g) +

Cl2(g)

Initial

0.025

Change

–x

Equilibrium

0.025 – x

⇆

17-7

Equilibrium expression:

KP =

(PPCl )(PCl ) = 3.4 × 10 (PPCl ) 3

–4

(x)(x) = 3.4 × 10–4 Assume x is negligible so 0.025 – x ≈ 0.025 (0.025 - x) (x)(x) = 3.4 × 10–4 (0.025) x2 = (3.4 × 10–4) (0.025) so x = 0.002915476 = 2.9 × 10–3

Check the assumption by calculating the % error: (2.9 ´ 10 –3 )(100%) = 12% which is larger than 5%, so the assumption is NOT valid. Solve using (0.025) quadratic equation. (x)(x) = 3.4 × 10–4 (0.025 - x) x2 + 3.4 × 10–4x – 8.5 × 10–6 = 0

+8.5 ´ 10-6  3.4 ´ 10-4 - 4(1)(-8.5 ´ 10-6 ) = 0.002750428051 or –0.003090428051 2(1) Choose the positive value, x = 0.0028 M; At equilibrium [PCl5]eq = 0.025 M – 0.0028 M = 0.022 M x=

17.10A Plan: Calculate the initial concentrations (molarity) of each substance. For part (a), calculate Qc and compare to given Kc. If Qc > Kc then the reaction proceeds to the left to make reactants from products. If Qc < Kc then the reaction proceeds to right to make products from reactants. For part (b), use the result of part (a) and the given equilibrium concentration of PCl5 to find the equilibrium concentrations of PCl3 and Cl2. Solution: 0.1050 mol Initial concentrations: [PCl5] = = 0.2100 M 0.5000 L 0.0450 mol [PCl3] = [Cl2] = = 0.0900 M 0.5000 L [ PCl3 ][Cl2 ] [0.0900][0.0900] = = 0.038571 = 0.0386 a) Qc = [0.2100 ] [ PCl5 ]

Qc, 0.0386, is less than Kc, 0.042, so the reaction will proceed to the right to make more products. b) To reach equilibrium, concentrations will increase for the products, PCl3 and Cl2, and decrease for the reactant, PCl5. Concentration (M) Initial Change Equilibrium

PCl5(g)

⇆

0.2100 –x 0.2100 – x

PCl3(g) +

Cl2(g)

0.0900

+x 0.0900 + x

+x____ 0.0900 + x

[PCl5] = 0.2065 = 0.2100 – x; x = 0.0035 M [PCl3] = [Cl2] = 0.0900 + x = 0.0900 + 0.0035 = 0.0935 M

17-8

17.10B Plan: For part (a), calculate Qc and compare to given Kc. If Qc > Kc then the reaction proceeds to the left to make reactants from products. If Qc < Kc then the reaction proceeds to right to make products from reactants. For part (b), based on this information, we determine the sign of each concentration change in the reaction table, use the balanced equation to define x in the reaction table, substitute into the Qc expression, and solve for x, from which we calculate the equilibrium concentration. Solution: M NO = M NO = 0.300 mol/2.00 L = 0.150 M ; similarly, MN O = MO = 0.260 M 2

é N 2 Où é O2 ù û ë û = (0.260)(0.260) = 3.00 Qc = ë é NO2 ù é NOù (0.150)(0.150) ë ûë û Qc > Kc, so reaction proceeds to the left.

NO(g)

⥫⥬

NO2(g)

Initial

0.150

0.260

Change

–x

0.150 + x

0.260 – x

0.150 + x

Equilibrium

N2O(g)

Concentration (M)

O2(g)

é N 2 Où é O2 ù û ë û = (0.260 - x ) Kc = 0.914 = ë 2 é NO2 ù é NOù (0.150 + x ) ë ûë û 2

Taking the square root of each side and solving gives x = 0.0596 M. [NO2] = 0.150 M + x = 0.150 M + 0.0596 M = 0.210 M 17.11A Plan: Examine each change for its impact on Qc. Then decide how the system would respond to re-establish equilibrium. Solution:

[SiF4 ][ H 2 O ]

Qc =

[ HF ]4 a) Decreasing [H2O] leads to Qc < Kc, so the reaction would shift to make more products from reactants. Therefore, the SiF4 concentration, as a product, would increase. b) Adding liquid water to this system at a temperature above the boiling point of water would result in an increase in the concentration of water vapor. The increase in [H2O] increases Qc to make it greater than Kc. To re-establish equilibrium products will be converted to reactants and the [SiF4] will decrease. c) Removing the reactant HF increases Qc, which causes the products to react to form more reactants. Thus, [SiF4] decreases. d) Removal of a solid product has no impact on the equilibrium; [SiF4] does not change. Check: Look at each change and decide which direction the equilibrium would shift using Le Châtelier’s principle to check the changes predicted above. a) Remove product, equilibrium shifts to right. b) Add product, equilibrium shifts to left. c) Remove reactant, equilibrium shifts to left. d) Remove solid reactant, equilibrium does not shift. 17.11B Plan: Examine each change for its impact on Qc. Then decide how the system would respond to re-establish equilibrium. Solution: [CO ][ H 2 ] Qc = [H2 O]

17-9

a) Adding carbon, a solid reactant, has no impact on the equilibrium. [CO] does not change. b) Removing water vapor, a reactant, increases Qc, which causes the products to react to form more reactants. Thus, [CO] decreases. c) Removing the product H2 decreases Qc, which causes the reactants to react to form more products. Thus, [CO] increases. d) Adding water vapor, a reactant, decreases Qc, which causes the reactants to react to form more products. Thus, [CO] increases. Check: Look at each change and decide which direction the equilibrium would shift using Le Châtelier’s principle to check the changes predicted above. a) Add solid reactant, equilibrium does not shift. b) Remove reactant, equilibrium shifts to the left. c) Remove product, equilibrium shifts to the right. d) Add reactant, equilibrium shifts to the right. 17.12A Plan: Changes in pressure (and volume) affect the concentration of gaseous reactants and products. A decrease in pressure, i.e., increase in volume, favors the production of more gas molecules whereas an increase in pressure favors the production of fewer gas molecules. Examine each reaction to decide whether more or fewer gas molecules will result from producing more products. If more gas molecules result, then the pressure should be increased (volume decreased) to reduce product formation. If fewer gas molecules result, then pressure should be decreased to produce more reactants. Solution: a) In 2SO2(g) + O2(g) ⇆ 2SO3(g) three molecules of gas form two molecules of gas, so there are fewer gas molecules in the product. Decreasing pressure (increasing volume) will decrease the product yield. b) In 4NH3(g) + 5O2(g) ⇆ 4NO(g) + 6H2O(g) 9 molecules of reactant gas convert to 10 molecules of product gas. Increasing pressure (decreasing volume) will favor the reaction direction that produces fewer moles of gas: towards the reactants and away from products. c) In CaC2O4(s) ⇆ CaCO3(s) + CO(g) there are no reactant gas molecules and one product gas molecule. The yield of the products will decrease when volume decreases, which corresponds to a pressure increase. 17.12B Plan: Changes in pressure (and volume) affect the concentration of gaseous reactants and products. A decrease in pressure, i.e., increase in volume, favors the production of more gas molecules whereas an increase in pressure favors the production of fewer gas molecules. Examine each reaction to determine if a decrease in pressure will shift the reaction toward the products (resulting in an increase in the yield of products) or toward the reactants (resulting in a decrease in the yield of products). Solution: a) In CH4(g) + CO2(g) ⇆ 2CO(g) + 2H2(g) two molecules of gas form four molecules of gas, so there are more gas molecules in the product. Decreasing pressure (increasing volume) will shift the reaction to the right, increasing the product yield. b) In NO(g) + CO2(g) ⇆ NO2(g) + CO(g) 2 molecules of reactant gas convert to 2 molecules of product gas. Decreasing pressure (increasing volume) will have no effect on this reaction or on the amount of product produced because the number of moles of gas does not change. c) In 2H2S(g) + SO2(g) ⇆ 3S(s) + 2H2O(g) three molecules of reactant gas convert to 2 molecules of product gas. Decreasing pressure (increasing volume) will shift the reaction toward the reactants, decreasing the product yield. 17.13A Plan: A decrease in temperature favors the exothermic direction of an equilibrium reaction. First, identify whether the forward or reverse reaction is exothermic from the given enthalpy change. DH < 0 means the forward reaction is exothermic, and DH > 0 means the reverse reaction is exothermic. If the forward reaction is exothermic then a decrease in temperature will shift the equilibrium to make more products from reactants and increase Kp. If the reverse reaction is exothermic then a decrease in temperature will shift the equilibrium to make more reactants from products and decrease Kp.

17-10

Solution: a) DH < 0 so the forward reaction is exothermic. A decrease in temperature increases the partial pressure of products and decreases the partial pressures of reactants, so PH2 decreases. With increases in product pressures and decreases in reactant pressures, Kp increases. b) DH > 0 so the reverse reaction is exothermic. A decrease in temperature decreases the partial pressure of products and increases the partial pressures of reactants, so PN2 increases. Kp decreases with decrease in product pressures and increase in reactant pressures. c) DH < 0 so the forward reaction is exothermic. Decreasing temperature increases PPCl5 and increases Kp. 17.13B Plan: A decrease in temperature favors the exothermic direction of an equilibrium reaction. First, identify whether the forward or reverse reaction is exothermic from the given enthalpy change. DH < 0 means the forward reaction is exothermic, and DH > 0 means the reverse reaction is exothermic. If the forward reaction is exothermic then a decrease in temperature will shift the equilibrium to make more products from reactants and increase Kp. If the reverse reaction is exothermic then an increase in temperature will shift the equilibrium to make more products from reactants and increase Kp. Solution: a) DH > 0 so the reverse reaction is exothermic. An increase in temperature will increase the partial pressure of products and decrease the partial pressures of reactants. Kp increases with an increase in product pressures and a decrease in reactant pressures. b) DH < 0 so the forward reaction is exothermic. A decrease in temperature increases the partial pressure of products and decreases the partial pressures of reactants. With increases in product pressures and decreases in reactant pressures, Kp increases. c) DH > 0 so the reverse reaction is exothermic. An increase in temperature increases the partial pressure of products and decreases the partial pressures of reactants. Kp increases with an increase in product pressures and a decrease in reactant pressures. 17.14A Plan: The van’t Hoff equation shows how the equilibrium constant is affected by a change in temperature. Substitute the given variables into the equation and solve for T2. Convert the given temperature to kelvins and the enthalpy change to J/mol. Solution: ln

 æ ö K2 DHrxn çç 1 - 1 ÷÷ = ÷ ç K1 R çè T1 T2 ÷ø

5.72 ´ 10 4 J/mol çæ 1 0.885 1 ÷ö çç = - ÷÷÷ ln 0.115 8.314 J/mol ⋅ K ççè 298 K T2 ÷ø

æ 1ö 2.040656 = 6879.96 K ´ ççç0.003356 K-1 - ÷÷÷ çè T2 ÷ø 0.003059 K-1 =

1 T2

T2 = 327 K 17.14B Plan: The van’t Hoff equation shows how the equilibrium constant is affected by a change in temperature. Substitute the given variables into the equation and solve for K2. Convert the given temperatures to kelvins and the enthalpy change to J/mol. Solution:

 æ ö K2 ΔHrxn çç 1 - 1 ÷÷ = ÷ K1 R ççè T1 T2 ÷ø

17-11

K2 1 ö÷÷ -1.984 ´ 105 æçç 1 = ÷ çç 0.13 8.314 J/mol ⋅ K çè1103 K 1188 K ÷÷ø

K2 = e-1.548 = 0.2127 0.13 K2 = 0.028 17.15A Plan: Given the balanced equilibrium equation, it is possible to set up the appropriate equilibrium expression (Qc). For the equation given Dn = 0 meaning that Kp = Kc. The value of K may be found for scene 1, and values for Q may be determined for the other two scenes. The reaction will shift towards the reactant side if Q > K, and the reaction will shift towards the product side if Q < K. The reaction is exothermic (DH < 0), thus, heat may be considered a product. Increasing the temperature adds a product and decreasing the temperature removes a product. Solution: a) Kp requires the equilibrium value of P for each gas. The pressure may be found from P = nRT/V æ nCD RT ö÷2 ç 2 çè V ÷÷ø PCD Kp = = æ nC2 RT öæ PC2 PD2 ÷÷ç nD2 RT ÷÷ö çç çè V ÷øèçç V ÷ø This equation may be simplified because for the sample R, T, and V are constant. Using scene 1: 2

Kp =

2 (4) nCD = =4 nC2 nD2 (2)(2) 2

b) Scene 2: Qp =

2 (6) nCD = = 36 nC2 nD2 (1)(1)

Q > K so the reaction will shift to the left (towards the reactants). 2

Scene 3: Qp =

2 (2) nCD = = 0.44 nC2 nD2 (3)(3)

Q < K so the reaction will shift to the right (towards the products). c) Moles of product will decrease. 17.15B Plan: Write the equilibrium expression for the reaction. Count the number of each type of particle in the first scene and use this information to calculate the value of K at T1. Follow a similar procedure to calculate the value of K at T2. Determine if K at T1 is larger or smaller than K at T2. Use this information to determine the sign of ΔH for the reaction. Solution: a) K =

[ AB] [ A ][ B]

Calculating K at T1:

[3] = 0.75 [2 ][2 ]

17-12

b) Going from the scene at T1 to the scene at T2, the number of product molecules decreases. This decreases the value of K. The problem states that T2 < T1, so as the temperature decreases, K also decreases. The fact that both the temperature and the value of K decreased suggests that this is an endothermic reaction, with ΔH > 0. c) Calculating K at T2: é2ù K = ë û = 0.22 é3ù é3ù ë ûë û

CHEMICAL CONNECTIONS BOXED READING PROBLEM

B17.1

Plan: To control the pathways, the first enzyme specific for a branch is inhibited by the end product of that branch. Solution: a) The enzyme that is inhibited by F is the first enzyme in that branch, which is enzyme 3. b) Enzyme 6 is inhibited by I. c) If F inhibited enzyme 1, then neither branch of the reaction would take place once enough F was produced. d) If F inhibited enzyme 6, then the second branch would not take place when enough F was made.

END–OF–CHAPTER PROBLEMS

17.1

If the rate of the forward reaction exceeds the rate of reverse reaction, products are formed faster than they are consumed. The change in reaction conditions results in more products and less reactants. A change in reaction conditions can result from a change in concentration or a change in temperature. If concentration changes, product concentration increases while reactant concentration decreases, but the Kc remains unchanged because the ratio of products and reactants remains the same. If the increase in the forward rate is due to a change in temperature, the rate of the reverse reaction also increases. The equilibrium ratio of product concentration to reactant concentration is no longer the same. Since the rate of the forward reaction increases more than the rate of the reverse reaction, Kc increases (numerator, [products], is larger and denominator, [reactants], is smaller).

Kc =

[ products] [ reactants]

17.2

The faster the rate and greater the yield, the more useful the reaction will be to the manufacturing process.

17.3

A system at equilibrium continues to be very dynamic at the molecular level. Reactant molecules continue to form products, but at the same rate that the products decompose to re-form the reactants.

17.4

If K is very large, the reaction goes nearly to completion. A large value of K means that the numerator is much larger than the denominator in the K expression. A large numerator, relative to the denominator, indicates that most of the reactants have reacted to become products. K =

[ products] [ reactants]

17.5

One cannot say with certainty whether the value of K for the phosphorus plus oxygen reaction is large or small (although it likely is large). However, it is certain that the reaction proceeds very fast.

17.6

No, the value of Q is determined by the mass action expression with arbitrary concentrations for products and reactants. Thus, its value is not constant.

17-13

17.7

The equilibrium constant expression is K = [O2] (we do not include solid substances in the equilibrium expression). If the temperature remains constant, K remains constant. If the initial amount of Li2O2 present was sufficient to reach equilibrium, the amount of O2 obtained will be constant, regardless of how much Li2O2(s) is present.

17.8

a) On the graph, the concentration of HI increases at twice the rate that H2 decreases because the stoichiometric ratio in the balanced equation is 1H2:2HI. Q for a reaction is the ratio of concentrations of products to concentrations of reactants. As the reaction progresses the concentration of reactants H2 and I2 decrease and the concentration of product HI increases, which means that Q increases as a function of time. [ HI ]2 H2(g) + I2(g) ⇆ 2HI(g) Q = [ H 2 ][ I2 ]

The value of Q increases as a function of time until it reaches the value of K. b) No, Q would still increase with time because the [I2] would decrease in exactly the same way as [H2] decreases. 17.9

A homogeneous equilibrium reaction exists when all the components of the reaction are in the same phase (i.e., gas, liquid, solid, aqueous). 2NO(g) + O2(g) ⇆ 2NO2(g) A heterogeneous equilibrium reaction exists when the components of the reaction are in different phases. Ca(HCO3)2(aq) ⇆ CaCO3(s) + H2O(l) + CO2(g)

17.10

1/2N2(g) + 1/2O2(g) ⇆ NO(g)

Qc(form) =

[ NO] 1

[ N 2 ] 2 [ O2 ] 2

NO(g) ⇆ 1/2N2(g) + 1/2O2(g) 1

Qc(decomp) =

[ N 2 ] 2 [ O2 ] 2 [ NO ]

Qc(decomp) = 1/Qc(form), so the constants do differ (they are the reciprocal of each other).

17-14

17.11

[C ]c [ D]d where A and B are [ A ]a [ B]b reactants, C and D are products, and a, b, c, and d are the stoichiometric coefficients in the balanced equation. Solution: The balanced equation for the first reaction is 3/2H2(g) + 1/2N2(g) ⇆ NH3(g) (1) The coefficient in front of NH3 is fixed at 1 mole according to the description. The reaction quotient for this [ NH3 ] reaction is Q1 = . 3 1 [H2 ] 2 [N2 ] 2

Plan: Write the reaction and then the expression for Q. Remember that Q =

In the second reaction, the coefficient in front of N2 is fixed at 1 mole. 3H2(g) + N2(g) ⇆ 2NH3(g)

(2)

[ NH3 ] 3 [H2 ] [N2 ] 2

The reaction quotient for this reaction is Q2 =

Q2 is equal to Q12. 17.12

[C ]c [ D]d where A and B are reactants, C and D are products, and a, b, c, and d are the [ A ]a [ B]b stoichiometric coefficients in the balanced equation. Solution: a) 4NO(g) + O2(g) ⇆ 2N2O3(g) Plan: Remember that Qc =

[ N 2 O3 ] Qc = [ NO ]4 [O2 ] 2

b) SF6(g) + 2SO3(g) ⇆ 3SO2F2(g)

[SO2 F2 ] 2 [SF6 ][SO3 ] 3

Qc =

c) 2SC1F5(g) + H2(g) ⇆ S2F10(g) + 2HCl(g)

Qc = 17.13

[S2 F10 ][ HCl ]2 [SClF5 ] [ H2 ] 2

a) 2C2H6(g) + 7O2(g) ⇆ 4CO2(g) + 6H2O(g)

[CO2 ] [ H 2 O] 2 7 [ C 2 H6 ] [ O2 ] 4

Qc =

b) CH4(g) + 4F2(g) ⇆ CF4(g) + 4HF(g)

Qc =

[CF4 ][ HF ]4 [CH 4 ][ F2 ]

c) 2SO3(g) ⇆ 2SO2(g) + O2(g)

[SO2 ] [O2 ] 2 [SO3 ] 2

Qc =

17-15

17.14

[C ]c [ D]d where A and B are reactants, C and D are products, and a, b, c, and d are the [ A ]a [ B]b stoichiometric coefficients in the balanced equation. Solution: a) 2NO2Cl(g) ⇆ 2NO2(g) + Cl2(g)

Plan: Remember that Qc =

[ NO2 ] [Cl2 ] 2 [ NO2 Cl ] 2

Qc =

b) 2POCl3(g) ⇆ 2PCl3(g) + O2(g)

[ PCl3 ] [O2 ] 2 [ POCl3 ] 2

Qc =

c) 4NH3(g) + 3O2(g) ⇆ 2N2(g) + 6H2O(g)

[N2 ] [H2 O] 4 3 [ NH3 ] [O2 ] 2

Qc =

17.15

a) 3O2(g) ⇆ 2O3(g)

[ O3 ] Qc = 3 [O 2 ]

b) NO(g) + O3(g) ⇆ NO2(g) + O2(g)

Qc =

[ NO2 ][O2 ] [ NO ][ O3 ]

c) N2O(g) + 4H2(g) ⇆ 2NH3(g) + H2O(g)

[ NH3 ] [ H2 O] 4 [ N2 O][ H2 ] 2

Qc =

17.16

Plan: The concentration of solids and pure liquids do not change, so their concentration terms are not written in the reaction quotient expression. Remember that stoichiometric coefficients are used as exponents in the expression for the reaction quotient. Solution: a) 2Na2O2(s) + 2CO2(g) ⇆ 2Na2CO3(s) + O2(g)

Qc =

[ O2 ] 2 [CO2 ]

b) H2O(l) ⇆ H2O(g)

Qc = [H2O(g)]

Only the gaseous water is used. The “(g)” is for emphasis.

c) NH4Cl(s) ⇆ NH3(g) + HCl(g)

Qc = [NH3][HCl] 17.17

a) H2O(l) + SO3(g) ⇆ H2SO4(aq)

Qc =

[ H2SO4 ] [SO3 ]

17-16

b) 2KNO3(s) ⇆ 2KNO2(s) + O2(g)

Qc = [O2] c) S8(s) + 24F2(g) ⇆ 8SF6(g)

[SF6 ] 24 [ F2 ]

Qc = 17.18

Qc = [CO2][H2O] b) SnO2(s) + 2H2(g) ⇆ Sn(s) + 2H2O(g)

[H2 O] 2 [H2 ]

Qc =

c) H2SO4(l) + SO3(g) ⇆ H2S2O7(l) 1 Qc = [SO3 ] 17.19

a) 2Al(s) + 2NaOH(aq) + 6H2O(l) ⇆ 2Na[Al(OH)4](aq) + 3H2(g) é Na é Al (OH ) ù ù 2 [ H ]3 2 4 û úû ê ë Qc = ë 2 [ NaOH ]

b) CO2(s) ⇆ CO2(g)

Qc = [CO2(g)]

Only the gaseous carbon dioxide is used. The “(g)” is for emphasis.

c) 2N2O5(s) ⇆ 4NO2(g) + O2(g)

Qc = [NO2]4[O2] 17.20

Plan: Compare each equation with the reference equation to see how the direction and coefficients have changed. If a reaction has been reversed, the K value is the reciprocal of the K value for the reference reaction. If the coefficients have been changed by a factor n, the K value is equal to the original K value raised to the nth power. Solution:

[ H 2 ] [S2 ] 2 [ H 2 S] 2

a) The K for the original reaction is Kc =

The given reaction 1/2S2(g) + H2(g) ⇆ H2S(g) is the reverse reaction of the original reaction and the coefficients of the original reaction have been multiplied by a factor of 1/2. The equilibrium constant for the reverse reaction is the reciprocal (1/K) of the original constant. The K value of the original reaction is raised to the 1/2 power. [ H 2 S] Kc (a) = (1/Kc)1/2 = 1 [S2 ] 2 [ H 2 ]

Kc (a) = (1/1.6 × 10–2)1/2 = 7.90569 = 7.9

17-17

b) The given reaction 5H2S(g) ⇆ 5H2(g) + 5/2S2(g) is the original reaction multiplied by 5/2. Take the original

K to the 5/2 power to find K of given reaction. 5

[ H ] [S2 ] 2 Kc (b) = (Kc) = 2 5 [ H2 S] 5

5/2

Kc (b) = (1.6 × 10–2)5/2 = 3.23817 × 10–5 = 3.2 × 10–5

[ N 2 ][ H2 O]

17.21

Kc =

[ NO ]2 [ H 2 ]

2 1

[N2 ] 2 [H2 O] a) Kc (a) = [Kc] = [ NO ][ H 2 ] 1/2

Thus, Kc (a) = [Kc]1/2 = (6.5 × 102)1/2 = 25.495 = 25

[ NO ]4 [ H 2 ]

–2

b) Kc = [Kc] =

[N2 ] [H2 O] 2

Kc = [Kc]–2 = (6.5 × 102)–2 = 2.36686 × 10–6 = 2.4 × 10–6 17.22

Write balanced chemical equations for each reaction, and then write the appropriate equilibrium expression. a) 4HCl(g) + O2(g) ⇆ 2Cl2(g) + 2H2O(g)

[Cl2 ] [ H 2 O] [ HCl ]4 [O2 ] 2

Qc =

b) 2As2O3(s) + 10F2(g) ⇆ 4AsF5(l) + 3O2(g)

[O 2 ] Qc = 10 [ F2 ] 3

c) SF4(g) + 2H2O(l) ⇆ SO2(g) + 4HF(g)

Qc =

[SO2 ][ HF ]4 [SF4 ]

d) 2MoO3(s) + 6XeF2(g) ⇆ 2MoF6(l) + 6Xe(g) + 3O2(g)

[ Xe ]6 [O2 ]

Qc =

17.23

[ XeF2 ]

Plan: Add the two equations, canceling substances that appear on both sides of the equation. Write the Qc expression for each of the steps and for the overall equation. Since the individual steps are added, their Qc’s are multiplied and common terms are canceled to obtain the overall Qc. Solution: a) The balanced equations and corresponding reaction quotients are given below. Note the second equation must be multiplied by 2 to get the appropriate overall equation.

17-18

(1) Cl2(g) + F2(g) ⇆ 2ClF(g)

Q1 =

[ClF ]2 [Cl2 ][ F2 ]

(2) 2ClF(g) + 2F2(g) ⇆ 2ClF3(g)

Q2 =

[ ClF3 ] 2 [ClF ]2 [ F2 ] 2

[ ClF3 ] Qoverall = 3 [Cl2 ][ F2 ] 2

Overall: Cl2(g) + 3F2(g) ⇆ 2ClF3(g)

b) The reaction quotient for the overall reaction, Qoverall, determined from the reaction is:

[ ClF3 ] 3 [Cl2 ][ F2 ] 2

Qoverall =

[ ClF3 ] [ ClF3 ] [ClF ]2 Qoverall = Q1Q2 = × = 2 3 2 [Cl2 ][ F2 ] [ClF ] [ F2 ] [Cl2 ][ F2 ] 2

17.24

To get the two equations to sum to the desired equation, the first equation must be reversed and doubled. This will result in squaring the reciprocal of its Kc value. The other equation does not need to be changed. Adding the two equations means the new Kc value will be the product of the individual Kc values. 2NO(g) ⇆ N2(g) + O2(g)

K1 = (Kc)–2 = 4.340 × 1018 = K2

2NO2(g) ⇆ 2NO(g) + O2(g)

K2 = Kc = 1.1 × 10–5

Overall: 2NO2(g) ⇆ N2(g) + 2O2(g)

Kc (overall) = K1K2 = 4.774 × 1013 = 4.8 × 1013 M

17.25

According to the ideal gas equation, PV = nRT. Concentration and pressure of gas are directly proportional as long as the temperature is constant: C = n/V = P/RT.

17.26

Kc and Kp are related by the equation Kp = Kc(RT)Dn, where Dn represents the change in the number of moles of gas in the reaction (moles gaseous products – moles gaseous reactants). When Dn is zero (no change in number of moles of gas), the term (RT)Dn equals 1 and Kc = Kp. When Dn is not zero, meaning that there is a change in the number of moles of gas in the reaction, then Kc ≠ Kp.

17.27

a) Kp = Kc(RT)Dn. Since Dn = number of moles gaseous products – number of moles gaseous reactants, Dn is a positive integer for this reaction. If Dn is a positive integer, then (RT)Dn is greater than 1. Thus, Kc is multiplied by a number that is greater than 1 to give Kp. Kc is smaller than Kp. b) Assuming that RT > 1 (which occurs when T > 12.2 K, because 0.0821 (R) × 12.2 = 1), Kp > Kc if the number of moles of gaseous products exceeds the number of moles of gaseous reactants. Kp < Kc when the number of moles of gaseous reactants exceeds the number of moles of gaseous product.

17.28

Plan: Dngas = moles gaseous products – moles gaseous reactants. Solution: a) Number of moles of gaseous reactants = 0; number of moles of gaseous products = 3; Dngas = 3 – 0 = 3 b) Number of moles of gaseous reactants = 1; number of moles of gaseous products = 0; Dngas = 0 – 1 = –1 c) Number of moles of gaseous reactants = 0; number of moles of gaseous products = 3; Dngas = 3 – 0 = 3

17-19

17.29

a) Dngas = 1

17.30

Plan: First, determine Dn for the reaction and then calculate Kc using Kp = Kc(RT)Dn. Solution: a) Dn = moles gaseous products – moles gaseous reactants = 1 – 2 = –1 Kp = Kc(RT)Dn

Kc =

b) Dngas = –3

c) Dngas = 1

3.9 ´ 10-2

( RT )

[(0.0821)(1000.)]-1

= Dn

= 3.2019 = 3.2

b) Dn = moles gaseous products – moles gaseous reactants = 1 – 1 = 0 Kp 28.5 Kc = = = 28.5 Dn [(0.0821)(500.)]0 ( RT ) 17.31

First, determine Dn for the reaction and then calculate Kc using Kp = Kc(RT)Dn. a) Dn = moles gaseous products – moles gaseous reactants = 2 – 2 = 0 Kp 49 Kc = = = 49 Dn [(0.0821)(730.)]0 ( RT ) b) Dn = moles gaseous products – moles gaseous reactants = 2 – 3 = –1

Kc =

2.5 ´ 1010

( RT )

[(0.0821)(500.)]-1

= Dn

= 1.02625 × 1012 = 1.0 × 1012

17.32

Plan: First, determine Dn for the reaction and then calculate Kp using Kp = Kc(RT)Dn. Solution: a) Dn = moles gaseous products – moles gaseous reactants = 2 – 1 = 1 Kp = Kc(RT)Dn = (6.1 × 10–3)[(0.0821)(298)]1 = 0.14924 = 0.15 b) Dn = moles gaseous products – moles gaseous reactants = 2 – 4 = – 2 Kp = Kc(RT)Dn = (2.4 × 10–3)[(0.0821)(1000.)]–2 = 3.5606 × 10–7 = 3.6 × 10–7

17.33

First, determine Dn for the reaction and then calculate Kp using Kp = Kc(RT)Dn. a) Dn = moles gaseous products – moles gaseous reactants = 2 – 2 = 0 Kp = Kc(RT)Dn = (0.77)[(0.0821)(1020.)]0 = 0.77 b) Dn = moles gaseous products – moles gaseous reactants = 2 – 3 = –1 Kp = Kc(RT)Dn = (1.8 × 10–56)[(0.0821)(570.)]–1 = 3.8464 × 10–58 = 3.8 × 10–58

17.34

When Q < K, the reaction proceeds to the right to form more products. The reaction quotient and equilibrium constant are determined by [products]/[reactants]. For Q to increase and reach the value of K, the concentration of products (numerator) must increase in relation to the concentration of reactants (denominator).

17.35

a) The reaction is 2D ⇆ E and Kc =

[E] . [D]2

æ 0.0100 mol ÷öæ 1 ö ÷÷ç Concentration of D = Concentration of E = (3 spheres)ççç ÷÷ = 0.0300 M è 1 sphere ÷øçè1.00 L ÷ø

[E] [0.0300] = 33.3333 = 33.3 2 = [0.0300]2 [D] b) In Scene B the concentrations of D and E are both 0.0300 mol/0.500 L = 0.0600 M [E] [0.0600] Qc = = 16.66666 = 16.7 2 = [0.0600]2 [D] Kc =

17-20

B is not at equilibrium. Since Qc < Kc, the reaction will proceed to the right. In Scene C, the concentration of D is still 0.0600 M and the concentration of E is 0.0600 mol/0.500 L = 0.120 M [E] [0.120] Qc = = = 33.3333 = 33.3 [0.0600]2 [D]2 Since Qc = Kc in Scene C, the reaction is at equilibrium. 17.36

Plan: To decide if the reaction is at equilibrium, calculate Qp and compare it to Kp. If Qp = Kp, then the reaction is at equilibrium. If Qp > Kp, then the reaction proceeds to the left to produce more reactants. If Qp < Kp, then the reaction proceeds to the right to produce more products. Solution: PH2 PBr2 (0.010)(0.010) Qp = = = 2.5 × 10–3 > Kp = 4.18 × 10–9 2 (0.20)2 PHBr Qp > Kp, thus, the reaction is not at equilibrium and will proceed to the left (towards the reactants). Thus, the numerator will decrease in size as products are consumed and the denominator will increase in size as more reactant is produced. Qp will decrease until Qp = Kp.

17.37

Qp =

2 PNO PBr2 2 PNOBr

(0.10)2 (0.10) = 0.10 < Kp = 60.6 (0.10)2

Qp < Kp. Thus, the reaction is not at equilibrium and will proceed to the right (towards the products). 17.38

There is insufficient information to calculate the partial pressures of each gas (T is not given). There is sufficient information to determine the concentrations and hence Qc. Convert the Kp given to Kc using Kp = Kc(RT)Dn. Compare the Qc to the Kc just calculated and make a prediction. Dn = moles gaseous products – moles gaseous reactants = 2 – 2 = 0 Since Dn = 0, Kp = Kc = 2.7 (Note: If Dn had any other value, we could not finish the calculation without the temperature.) [CO2 ][ H2 ] éëê0.62 /2.0ùûú éëê 0.43/2.0ùûú Qc = = = 3.662 > Kc = 2.7 é 0.13/2.0ù é 0.56/2.0ù [CO][ H2 O] êë úû êë úû

Qc > Kc. Thus, the reaction is not at equilibrium and will proceed to the left (towards the reactants). 17.39

At equilibrium, equal concentrations of CFCl3 and HCl exist, regardless of starting reactant concentrations. The equilibrium concentrations of CFCl3 and HCl would still be equal if unequal concentrations of CCl4 and HF were used. This occurs only when the two products have the same coefficients in the balanced equation. Otherwise, more of the product with the larger coefficient will be produced.

17.40

When x mol of CH4 reacts, 2x mol of H2O also reacts to form x mol of CO2 and 4x mol of H2. This is based on the 1:2:1:4 mole ratio in the reaction. The final (equilibrium) concentration of each reactant is the initial concentration minus the amount that reacts. The final (equilibrium) concentration of each product is the initial concentration plus the amount that forms.

17.41

a) The approximation applies when the change in concentration from initial to equilibrium is so small that it is insignificant. This occurs when K is small and initial concentration is large. b) This approximation will not work when the change in concentration is greater than 5%. This can occur when [reactant]initial is very small, or when [reactant]change is relatively large due to a large K.

17-21

17.42

Plan: Since all equilibrium concentrations are given in molarities and the reaction is balanced, construct an equilibrium expression and substitute the equilibrium concentrations to find Kc. Solution:

é1.87 ´ 10-3 ù 2 [ HI ]2 ëê ûú Kc = = = 50.753 = 50.8 [ H2 ][ I2 ] éê6.50 ´ 10-5 ùú éê1.06 ´ 10-3 ùú ë ûë û

[ N 2 ][ H 2 ]

[0.114 ][0.342 ]3

17.43

Kc =

17.44

Plan: Calculate the initial concentration of PCl5 from the given number of moles and the container volume; the reaction is proceeding to the right, consuming PCl5 and producing products. There is a 1:1:1 mole ratio between the reactants and products. Solution: Initial [PCl5] = 0.15 mol/2.0 L = 0.075 M Since there is a 1:1:1 mole ratio in this reaction: x = [PCl5] reacting (–x), and the amount of PCl3 and of Cl2 forming (+x). Concentration (M) PCl5(g) ⇆ PCl3(g) + Cl2(g)

[ NH3 ]

[ 0.0225]2

Initial

17.45

= 9.0077875 = 9.01

0.075

Change

–x

Equilibrium

0.075 – x

The reaction table requires that the initial [H2] and [F2] be calculated: [H2] = 0.10 mol/0.50 L = 0.20 M; [F2] = 0.050 mol/0.50 L = 0.10 M. x = [H2] = [F2] reacting (–x); 2x = [HF] forming (+2x)

17.46

Concentration (M)

H2(g)

Initial

0.20

0.10

Change

–x

+2x

Equilibrium

0.20 – x

0.10 – x

F2(g)

⇆

2HF(g)

Plan: Two of the three equilibrium pressures are known, as is Kp. Construct an equilibrium expression and solve for PNOCl. Solution:

Kp = 6.5 × 104 = 6.5 × 104 =

PNOCl =

2 PNOCl 2 PNO PCl2

P 2 NOCl (0.35)2 (0.10)

(6.5 ´10 )(0.35) (0.10) = 28.2179 = 28 atm 4

A high pressure for NOCl is expected because the large value of Kp indicates that the reaction proceeds largely to the right, i.e., to the formation of products.

17-22

17.47

C(s) + 2H2(g) ⇆ CH4(g) PCH Kp = 2 4 = 0.262 PH2

PCH 4 = K p PH22 = (0.262)(1.22)2 = 0.38996 = 0.390 atm 17.48

Plan: Use the balanced equation to write an equilibrium expression and to define x. Set up a reaction table, substitute into the Kp expression, and solve for x. Solution: NH4HS(s) ⇆ H2S(g) + NH3(g) x = [NH4HS] reacting (–x), and the amount of H2S and of NH3 forming (+x) since there is a 1:1:1 mole ratio between the reactant and products. (It is not necessary to calculate the molarity of NH4HS since, as a solid, it is not included in the equilibrium expression.) Concentration (M) NH4HS(s) ⇆ H2S(g) + NH3(g) Initial ¾ 0 0 Change

–x

Equilibrium

Kp = 0.11 = ( PH2S )( PNH3 )

(The solid NH4HS is not included.)

0.11 = (x)(x) x = PNH3 = = 0.33166 = 0.33 atm 17.49

2H2S(g) ⇆ 2H2(g) + S2(g)

[ H2S] = 0.45 mol/3.0 L = 0.15 M Concentration (M)

2H2S(g)

Initial

0.15

Change

–2x

+2x

Equilibrium

0.15 – 2x

S2(g)

2 x [x] [ H 2 ] [S2 ] ëê ûú = 2 é 0.15 - 2 x ù 2 [ H 2 S] ê ú 2

Kc = 9.30 × 10–8 =

2H2(g)

⇆

ë Assuming 0.15 M – 2x ≈ 0.15 M

é2 x ù 2 [x ] 4 x3 ê ú 9.30 × 10 = ë û 2 = 0.152 [ 0.15] –8

x = 8.0575 × 10–4 M [ H2 ] = 2x = 2 (8.0575 × 10–4 M) = 1.6115 × 10–3 = 1.6 × 10–3 M (Since (1.6 × 10–3)/(0.15) < 0.05, the assumption is OK.) 17.50

Plan: Use the balanced equation to write an equilibrium expression. Find the initial concentration of each reactant from the given amounts and container volume, use the balanced equation to define x and set up a reaction table, substitute into the equilibrium expression, and solve for x, from which the concentration of NO is calculated.

17-23

Solution: The initial concentrations of N2 and O2 are (0.20 mol/1.0 L) = 0.20 M and (0.15 mol/1.0 L) = 0.15 M, respectively. N2(g) + O2(g) ⇆ 2NO(g) There is a 1:1:2 mole ratio between reactants and products. Concentration (M)

N2(g)

Initial

0.20

0.15

O2(g)

Change

–x

+2x

Equilibrium

0.20 – x

0.15 – x

2NO(g)

⇆

(1:1:2 mole ratio)

2x 2

é2 xù [ NO ]2 ëê ûú = é N O [ 2 ][ 2 ] ê0.20 - xùú éê0.15 - x ùú ë ûë û and 0.15 M – x ≈ 0.15 M Assume 0.20 M – x ≈ 0.20 M

Kc = 4.10 × 10–4 =

4.10 × 10–4 =

4 x2

[0.20 ][ 0.15]

x = 1.753568 × 10–3 M [NO] = 2x = 2(1.753568 × 10–3 M) = 3.507136 × 10–3 = 3.5 × 10–3 M (Since (1.8 × 10–3)/(0.15) < 0.05, the assumption is OK.) 17.51

2NO2(g) ⇆ 2NO(g) + O2(g) 2NO2(g)

Pressure (atm) Initial

2NO(g)

⇆

0.75

Change

– 2x

Equilibrium

0.75 – 2x

Kp = 4.48 × 10–13 =

2 PNO PO2 2 PNO2

O2(g)

+2x

(2x ) (x )

(0.75 - 2x )

Assume 0.75 atm – 2x ≈ 0.75 atm 4.48 × 10–13 =

(4x2 )(x) (0.75)2

(4x3 ) (0.75)2

x = 3.979 × 10–5 atm = 4.0 × 10–5 atm O2

PNO = 2x = 2(3.979 × 10–5 atm) = 7.958 × 10–5 = 8.0 × 10–5 atm NO 17.52

Plan: Find the initial concentration of each reactant and product from the given amounts and container volume, use the balanced equation to define x, and set up a reaction table. The equilibrium concentration of H2 is known, so x can be calculated and used to find the other equilibrium concentrations. Solution: Initial concentrations: [HI] = (0.0244 mol)/(1.50 L) = 0.0162667 M [H2] = (0.00623 mol)/(1.50 L) = 0.0041533 M [I2] = (0.00414 mol)/(1.50 L) = 0.00276 M

17-24

2 HI(g) H2(g) + I2(g) Concentration (M)

There is a 2:1:1 mole ratio between reactants and products. 2 HI(g) ⇆ H2(g) + I2(g)

Initial

0.0162667

0.0041533

0.00276

Change

–2x

Equilibrium

0.0162667 – 2x

0.0041533 + x

0.00276 + x

2:1:1 mole ratio

[H2]eq = 0.00467 = 0.0041533 + x x = 0.0005167 M [I2]eq = 0.00276 + x = 0.00276 + 0.0005167 = 0.0032767 = 0.00328 M I2 [HI]eq = 0.0162667 – 2x = 0.0162667 – 2(0.0005167) = 0.0152333 = 0.0152 M HI 17.53

Initial concentrations: [A] = (1.75 × 10–3 mol)/(1.00 L) = 1.75 × 10–3 M [B] = (1.25 × 10–3 mol)/(1.00 L) = 1.25 × 10–3 M [C] = (6.50 × 10–4 mol)/(1.00 L) = 6.50 × 10–4 M Concentration (M) A(g) ⇆ 2B(g) + C(g) Initial 1.75 × 10–3 1.25 × 10–3 6.50 × 10–4 Change –x + 2x +x Equilibrium 1.75 × 10–3 – x 1.25 × 10–3 + 2x 6.50 × 10–4 + x [A]eq = 2.15 × 10–3 = 1.75 × 10–3 – x x = –0.00040 [B]eq = 1.25 × 10–3 + 2x = 1.25 × 10–3 + 2(–0.00040) = = 4.5 × 10–4 M [C]eq = 6.50 × 10–4 + x = 6.50 × 10–4 + (–0.00040) = 2.5 × 10–4 M

17.54

Plan: To determine in which direction the reaction is proceeding, calculate Qc and compare it to Kc. If Qc > Kc, then the reaction proceeds to the left to produce more reactants. If Qc < Kc, then the reaction proceeds to the right to produce more products. Based on that information, determine the sign of each concentration change (x, etc.) for the reaction table, substitute into the equilibrium expression, and solve for x, from which the equilibrium concentration of HF can be calculated. Solution: 2 é HF ù 2 (0.10) ë û Qc = = = 4.0 < Kc = 115 é H 2 ù é F2 ù (0.050)(0.050) ë ûë û The reaction proceeds to the right. x = [H2] = [F2] reacting (–x); 2x = [HF] forming (+2x). Concentration (M) H2(g) + F2(g) ⇆

2HF(g)

Initial

0.050

0.10

Change

–x

+2x

Equilibrium

0.050 – x

0.10 + 2x

é 0.10 + 2 x ù é 0.10 + 2 x ù 2 [HF]2 ë û û Take the square root of both sides: = = ë K c = 115 = 2 é H 2 ù é F2 ù é 0.050 - x ù é 0.050 - x ù é ù 0.050 x ë ûë û ë ûë û ë û 0.10 + 2 x 10.7238 = 0.050 - x 0.10 + 2x = 0.53619 – 10.7238x

12.7238x = 0.43649

x = 0.0343 M [HF] = 0.10 + 2x = 0.10 + 2(0.0343) = 0.169 M Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

17-25

17.55

[H2O] = [Cl2O] = [HClO] = 0.12 mol/2.0 L = 0.060 M 2 é HClOù 2 (0.060) ë û Qc = = = 1.0 > Kc = 0.090 é H2 Où éCl2 Où (0.060)(0.060) ë ûë û The reaction proceeds to the left. x = [H2O] = [Cl2O] forming (+x); 2x = [HClO] reaction (–2x). Concentration (M)

H2O(g)

Initial

0.060

Change

Equilibrium

0.060 + x

Cl2O(g)

2HClO(g)

⇆

0.060 –2x 0.060 – 2x

é 0.060 - 2 x ù 2 é 0.060 - 2 x ù 2 [HClO]2 ë û û = = ë é H 2 Où é Cl2 Où é 0.060 + x ù é 0.060 + x ù é 0.060 + x ù 2 ë ûë û ë ûë û ë û Take the square root of both sides

K c = 0.090 =

0.060 - 2 x 0.060 + x 0.060 – 2x = 0.018 + 0.30x 0.042 = 2.30x x = 0.0183 M [H2O]eq = [Cl2O]eq = 0.060 + x = 0.060 + 0.0183 = 0.078 M [HClO]eq = 0.060 – 2x = 0.060 – 2(0.0183) = 0.023 M 0.30 =

17.56

Plan: Use the balanced equation to write an equilibrium expression. Find the initial concentration of HI from the given amount and container volume, use the balanced equation to define x and set up a reaction table, substitute into the equilibrium expression, and solve for x, from which the equilibrium concentrations can be calculated. Solution: [HI]init = (2.50 mol/10.32 L) = 0.242 M Concentration (M)

2HI(g)

⥫⥬ H2(g)

Initial

0.242

Change

–2x

0.242 – 2x x éH2 ù éI2 ù [ x ][ x ] Kc = 1.26 ´ 10-3 = ë û ë 2 û = é HI ù é 0.242 - 2 x ù 2 ë û ë û

Equilibrium

1.26 ´ 10-3 =

é 0.242 - 2 x ù 2 ë û x 0.035496 = [0.242 - 2 x ]

I2(g) (2:1:1 mole ratio)

Take the square root of each side:

x = 0.008590 – 0.070992x 1.070992x = 0.008590 x = 8.02 × 10–3 [H2]eq = x = 8.02 × 10–3 M Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

17-26

17.57

Concentration (M)

SCl2(g)

Initial

0.675

2C2H4(g)

0.973

–x

Change

– 2x

0.675 – x

Equilibrium

S(CH2CH2Cl)2(g)

⇆

0.973 – 2x

[S(CH2CH2Cl)2]eq = x = 0.350 M [SCl2]eq = 0.675 – x = 0.675 – 0.350 = = 0.325 M [C2H4]eq = 0.973 – 2x = 0.973 – 2(0.350) = 0.273 M Kc =

[S(CH 2 CH 2 Cl)2 ]

[SCl 2 ][C 2 H 4 ]

[0.350] [0.325][0.273]2

= 14.4497

Kp = Kc(RT)Δn Δn = 1 mol – 3 mol = –2 Kp = (14.4497)[(0.0821)(273.2 + 20.0)]–2 = 0.0249370 = 0.0249 17.58

Plan: Use the balanced equation to write an equilibrium expression. Find the initial concentration of each reactant from the given amounts and container volume, use the balanced equation to define x, and set up a reaction table. The equilibrium concentration of N2 is known, so x can be calculated and used to find the other equilibrium concentrations. Substitute the equilibrium concentrations into the equilibrium expression to find Kc. Solution: 4NH3(g) + 3O2(g) ⇆ 2N2(g) + 6H2O(g) Initial [NH3] = Initial [O2] = (0.0150 mol)/(1.00 L) = 0.0150 M Concentration (M)

4NH3(g)

3O2(g)

Initial

0.0150

Change

–4x

–3x

+2x

+6x

Equilibrium

0.0150 – 4

0.0150 – 3x

+2x

+6x

⇆

2N2(g)

6H2O(g)

[N2] eq = 2x = 1.96 × 10–3 M x = (1.96 × 10–3 M)/2 = 9.80 × 10–4 M [H2O]eq = 6x = 6(9.80 × 10–4) = 5.8800 × 10–3 M [NH3]eq = 0.0150 – 4x = 0.0150 – 4(9.80 × 10–4 ) = 1.1080 × 10–2 M [O2]eq = 0.0150 – 3x = 0.0150 – 3(9.80 × 10–4 ) = 1.2060 × 10–2 M é

-3 ù

[ N2 ] [ H2 O] ê1.96 ´ 10 ûú ëê 5.8800 ´ 10 ûú = ë = 6.005859 × 10–6 = 6.01 × 10–6 Kc = 4 3 4 3 é1.1080 ´ 10-2 ù é1.2060 ´ 10-2 ù [ NH3 ] [O2 ] 2

17.59

Pressure (atm)

ëê FeO(s)

Initial

—

Change

ûú ëê CO(g)

⇆

1.00 –x

Equilibrium 1.00 – x PCO2 x Kp = = 0.403 = PCO 1.00 - x

ûú Fe(s)

—

CO2(g) 0 +x x

x = 0.28724 = 0.287 atm CO2 1.00 – x = 1.00 – 0.28724 = 0.71276 = 0.71 atm CO

17-27

17.60

A change in equilibrium conditions such as a change in concentration of a component, a change in pressure (volume), or a change in temperature.

17.61

Equilibrium position refers to the specific concentrations or pressures of reactants and products that exist at equilibrium, whereas equilibrium constant refers to the overall ratio of equilibrium concentrations and not to specific concentrations. Changes in reactant concentration cause changes in the specific equilibrium concentrations of reactants and products (equilibrium position), but not in the equilibrium constant.

17.62

A positive DHrxn indicates that the reaction is endothermic, and that heat is consumed in the reaction: NH4Cl(s) + heat ⇆ NH3(g) + HCl(g) a) The addition of heat (high temperature) causes the reaction to proceed to the right to counterbalance the effect of the added heat. Therefore, more products form at a higher temperature and container (B) with the largest number of product molecules best represents the mixture. b) When heat is removed (low temperature), the reaction shifts to the left to produce heat to offset that disturbance. Therefore, NH3 and HCl molecules combine to form more reactant and container (A) with the smallest number of product gas molecules best represents the mixture.

17.63

Equilibrium component concentration values may change but the mass action expression of these concentrations is a constant as long as temperature remains constant. Changes in component amounts, pressures (volumes), or addition of a catalyst will not change the value of the equilibrium constant.

17.64

a) Ratef = kf[reactants]x. An increase in reactant concentration shifts the equilibrium to the right by increasing the initial forward rate. Since Keq = kf / kr and kf and kr are not changed by changes in concentration, Keq remains constant. b) A decrease in volume causes an increase in concentrations of gases. The reaction rate for the formation of fewer moles of gases is increased to a greater extent. Again, the kf and kr values are unchanged. c) An increase in temperature increases kr to a greater extent for an exothermic reaction and thus lowers the Keq value. d) An endothermic reaction can be written as: reactants + heat products. A rise in temperature (increase in heat) favors the forward direction of the reaction, i.e., the formation of products and consumption of reactants. Since K = [products]/[reactants], the addition of heat increases the numerator and decreases the denominator, making K2 larger than K1.

17.65

XY(s) ⇆ X(g) + Y(s) Since product Y is a solid substance, addition of solid Y has no effect on the equilibrium position (as long as some Y is present). Scene A best represents the system at equilibrium after the addition of two formula units of Y. More Y is present but the amounts of X and XY do not change.

17.66

Plan: If the concentration of a substance in the reaction increases, the equilibrium position will shift to consume some of it. If the concentration of a substance in the reaction decreases, the equilibrium position will shift to produce more of it. Solution: a) Equilibrium position shifts towards products. Adding a reactant (CO) causes production of more products as the system will act to reduce the increase in reactant by proceeding toward the product side, thereby consuming additional CO. b) Equilibrium position shifts towards products. Removing a product (CO2) causes production of more products as the system acts to replace the removed product. c) Equilibrium position does not shift. The amount of a solid reactant or product does not impact the equilibrium as long as there is some solid present. d) Equilibrium position shifts towards reactants. When product is added, the system will act to reduce the increase in product by proceeding toward the reactant side, thereby consuming additional CO2; dry ice is solid carbon dioxide that sublimes to carbon dioxide gas. At very low temperatures, CO2 solid will not sublime, but since the reaction lists carbon dioxide as a gas, the assumption that sublimation takes place is reasonable.

17-28

b) no change d) shifts towards the reactants

17.67

a) no change c) shifts towards the products

17.68

Plan: An increase in container volume results in a decrease in pressure (Boyle’s law). Le Châtelier’s principle states that the equilibrium will shift in the direction that forms more moles of gas to offset the decrease in pressure. Solution: a) More F forms (two moles of gas) and less F2 (one mole of gas) is present as the reaction shifts towards the right. b) More C2H2 and H2 form (four moles of gas) and less CH4 (two moles of gas) is present as the reaction shifts towards the right.

17.69

a) less CH3OH(l); more CH3OH(g) b) less CH4 and NH3; more HCN and H2

17.70

Plan: Decreasing container volume increases the pressure (Boyle’s law). Le Châtelier’s principle states that the equilibrium will shift in the direction that forms fewer moles of gas to offset the increase in pressure. Solution: a) There are two moles of reactant gas (H2 and Cl2) and two moles of product gas (HCl). Since there is the same number of reactant and product gas moles, there is no effect on the amounts of reactants or products. b) There are three moles of reactant gases (H2 and O2) and zero moles of product gas. The reaction will shift to the right to produce fewer moles of gas to offset the increase in pressure. H2 and O2 will decrease from their initial values before the volume was changed. More H2O will form because of the shift in equilibrium position.

17.71

a) more CO2 and H2O; less C3H8 and O2 b) more NH3 and O2; less N2 and H2O

17.72

Plan: The purpose of adjusting the volume is to cause a shift in equilibrium to the right for increased product yield. Increasing the volume of the container results in a shift in the direction that forms more moles of gas, while decreasing the container volume results in a shift in the direction that forms fewer moles of gas. Solution: a) Because the number of reactant gaseous moles (4H2) equals the product gaseous moles (4H2O), a change in volume will have no effect on the yield. b) The moles of gaseous product (2CO) exceed the moles of gaseous reactant (1O2). A decrease in pressure favors the reaction direction that forms more moles of gas, so increase the reaction vessel volume.

17.73

a) increase volume

17.74

Plan: An increase in temperature (addition of heat) causes a shift in the equilibrium away from the side of o the reaction with heat. Recall that a negative value of DHrxn indicates an exothermic reaction, while a positive o value of DHrxn indicates an endothermic reaction.

b) decrease volume

Solution: o DHrxn a) CO(g) + 2H2(g) ⇆ CH3OH(g) + heat = –90.7 kJ The reaction is exothermic, so heat is written as a product. The equilibrium shifts to the left, away from heat, towards the reactants, so amount of product decreases. o DHrxn b) C(s) + H2O(g) + heat ⇆ CO(g) + H2(g) = 131 kJ The reaction is endothermic, so heat is written as a reactant. The equilibrium shifts to the right, away from heat, towards the products, so amounts of products increase. c) 2NO2(g) + heat ⇆ 2NO(g) + O2(g) The reaction is endothermic, so heat is written as a reactant. The equilibrium shifts to the right, away from heat, towards the product, so amounts of products increase.

17-29

d) 2C(s) + O2(g) ⇆ 2CO(g) + heat The reaction is exothermic, so heat is written as a product. The equilibrium shifts to the left, away from heat, towards the reactants; amount of product decreases. 17.75

a) decrease

17.76

Plan: The van’t Hoff equation shows how the equilibrium constant is affected by a change in temperature. o Substitute the given variables into the equation and solve for K2. Convert DHrxn from units of kJ/mol DH to units of J/2 mol DH. Solution: K298 = K1 = 1.80 T1 = 298 K

b) decrease

c) decrease

d) increase

K500 = K2 = ?

T2 = 500. K R = 8.314 J/mol ∙ K 3 æ10 J ö÷ æ 0.32 kJ ö÷ o ç ÷ = 6.4 × 102 J DHrxn = ççç ÷÷(2 mol DH)çç è1 mol DH ø çè 1 kJ ÷÷ø ln

o æ ö DHrxn K2 çç 1 - 1 ÷÷ =ç R çè T2 T1 ÷ø÷ K1

6.4 ´ 102 J æç 1 1 ö÷÷ K2 çç =÷ 1.80 8.314 J/mol K ççè 500. K 298 K ÷÷ø

K2 = 0.104360 1.80 K2 = 1.110 1.80 K2 = (1.80)(1.110) = 1.998 = 2.0 ln

17.77

The van’t Hoff equation shows how the equilibrium constant is affected by a change in temperature. Substitute the given variables into the equation and solve for K2. K298 = K1 = 2.25 × 104

T1 = 298 K

o DHrxn = –128 kJ/mol

K0 = K2 = ?

T2 = (273 + 0.) = 273 K

R = 8.314 J/mol ∙ K

o DHrxn = (–128 kJ/mol)(103 J/1 kJ) = –1.28 × 105 J

ln ln

o æ ö DHrxn K2 çç 1 - 1 ÷÷ =ç R çè T2 T1 ÷ø÷ K1

K2 2.25 ´ 10

-1.28 ´ 105 J æçç 1 1 ö÷÷ ÷ çç 8.314 J /mol K çè 273 K 298 K ÷ø÷

= 4.731088 2.25 ´ 104 K2 = 1.134189 × 102 2.25 ´ 104 K2 = (2.25 × 104)(1.134189 × 102) = 2.551925 × 106 = 2.55 × 106 17.78

4Fe3O4(s) + O2(g) ⇆ 6Fe2O3(s) 1 = 2.5 × 1087 a) Kp = PO2

Kp = 2.5 × 1087 at 298 K

PO2 = 4.0 × 10–88 atm Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

17-30

b) Qp =

1 = 1/(0.21) = 4.7619 PO2

Kp > Qp thus, the reaction will proceed to the right. c) Kp = Kc(RT)Dn Kc = Kp/(RT)Dn

Dn = 0 – 1 = –1

Kc = (2.5 × 10 )/[(0.0821)(298)]–1 = 6.11645 × 1088 = 6.1 × 1088 17.79

Plan: An increase in temperature (addition of heat) causes a shift in the equilibrium away from the side of the reaction with heat, while a decrease in temperature (removal of heat) causes a shift in the equilibrium towards the side with heat. Increasing the volume of the container (pressure decreases) results in a shift in the direction that forms more moles of gas, while decreasing the container volume (pressure increases) results in a shift in the direction that forms fewer moles of gas. Adding a reactant causes a shift in the direction of products. Solution: a) SO2(g) + 1/2O2(g) ⇆ SO3(g) + heat o is negative), so it is favored by lower temperatures. Lower The forward reaction is exothermic ( DHrxn temperatures will cause a shift to the right, the heat side of the reaction. There are fewer moles of gas as products (1SO3) than as reactants (1SO2(g) + 1/2O2), so products are favored by higher pressure. High pressure will cause a shift in equilibrium to the side with the fewer moles of gas. [SO3 ] , and have no impact on K. b) Addition of O2 would decrease Q since Q = [SO2 ][O2 ]1/ 2 c) To enhance yield of SO3, a low temperature is used. Reaction rates are slower at lower temperatures, so a catalyst is used to speed up the reaction.

17.80

3H2(g) + N2(g) ⇆ 2NH3(g) PNH3 = (41.49%/100%)(110. atm) = 45.639 atm 100.00% – 41.49% = 58.51% N2 + H2 PH2 + PN2 = (58.51%/100%)(110. atm) = 64.361 atm PH2 = (3/4)(64.361 atm) = 48.27075 atm PN2 = (1/4)(64.361 atm) = 16.09025 atm

(PNH ) = (45.639)2 = 1.15095 × 10 = 1.15 × 10 K = 3 3 (PH ) (PN ) (48.27075) (16.09025) 2

–3

17.81

a) 3H2(g) + N2(g) ⇆ 2NH3(g) PNH3 = 50. atm

The mole ratio H2:N2 = 3:1; at equilibrium, if N2 = x, H2 = 3x;

(PNH ) = 1.00 × 10 K = 3 (PN )(PH ) 2

–4

(50.) = 1.00 × 10–4 3 ( x )(3x ) 2

Kp =

x = 31.02016 = 31 atm N2 3x = 3(31.02016) = 93.06049 = 93 atm H2 Ptotal = Pnitrogen + Phydrogen + Pammonia = (31.02016 atm) + (93.06049 atm) + (50. atm) = 174.08065 = 174 atm total Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

17-31

b) The mole ratio H2:N2 = 6:1; at equilibrium, if N2 = x, H2 = 6x; PNH3 = 50. atm Kp =

(50.)2 = 1.00 × 10–4 3 ( x )(6x )

x = 18.445 = 18 atm N2 6x = 6(18.445) = 110.67 = 111 atm H2 Ptotal = Pnitrogen + Phydrogen + Pammonia = (18.445 atm) + (110.67 atm) + (50. atm) = 179.115 = 179 atm total This is not a valid argument. The total pressure in b) is greater than in a) to produce the same amount of NH3. 17.82

a) More CaCO3. Because the forward reaction is exothermic, decreasing the temperature will cause an increase in the amount of CaCO3 formed as the reaction shifts to the right to produce more heat. b) Less CaCO3. The only gas in the equation is a reactant. Increasing the volume (decreasing the pressure) will cause the equilibrium to shift toward the reactant side and the amount of CaCO3 formed decreases. c) More CaCO3. Increasing the partial pressure of CO2 will cause more CaCO3 to be formed as the reaction shifts to the right to consume the added CO2. d) No change. Removing half of the initial CaCO3 will have no effect on the amount of CaCO3 formed, because CaCO3 is a solid.

17.83

a) Qc =

[ XY ]2 [ X2 ][ Y2 ]

Scene A: Qc =

[0 ]2 =0 [0.4 ][0.4 ]

Scene B: Qc =

[0.4 ]2 =4 [0.2 ][0.2 ]

Scenes C–E: Qc =

[0.6 ]2 = 36 = 4 × 101 [0.1][0.1]

c) Time is progressing to the right. Frame A must be the earliest time. d) K = 4 × 101 e) Scene B. At higher temperatures, the reaction shifts to the left (forming more X2 and Y2). f) None. Volume (pressure) has no effect on the position of the equilibrium since there are two moles of gas on each side. 17.84

Plan: Use the balanced equation to write an equilibrium expression and to define x. Set up a reaction table, substitute into the Kc expression, and solve for x. Once the total concentration of the gases at equilibrium is known, the pressure can be found with PV = nRT. Solution: Concentration (M) NH2COONH4(s) ⇆ 2NH3(g) + CO2(g) Initial

7.80 g

Change

—

+2x

Equilibrium

—

The solid is irrelevant (as long as some is present) and is not included in the Kc expression. Kc = [NH3]2[CO2] Kc = 1.58 × 10–8 = (2x)2(x) x = 1.580759 × 10–3 M Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

17-32

Total concentration of gases = 2x + x = 2(1.580759 × 10–3 M) + 1.580759 × 10–3 M = 4.742277 × 10–3 M To find total pressure use the ideal gas equation: PV = nRT ænö nRT = çç ÷÷÷ RT = MRT P= èV ø V P = (4.742277 × 10–3 M)(0.0821 L ∙ atm/mol ∙ K)(273 + 250.)K = 0.203625 = 0.204 atm 17.85

(1)

2Ni3S2(s) + 7O2(g) ⇆ 6NiO(s) + 4SO2(g)

(2)

6NiO(s) + 6H2(g) ⇆ 6Ni(s) + 6H2O(g)

(3)

6Ni(s) + 24CO(g) ⇆ 6Ni(CO)4(g)

Overall: 2Ni3S2(s) + 7O2(g) + 6H2(g) + 24CO(g) ⇆ 4SO2(g) + 6H2O(g) + 6Ni(CO)4(g) b) As always, the solid is not included in the Q expression.

[SO2 ] [ H2 O] éë Ni (CO)4 ùû Qc(overall) = 7 6 [O2 ] [ H2 ] [CO]24 4

6 6 4 6 éSO2 ù 4 é H2 Où 6 éê Ni (CO) ùú [SO2 ] [ H2 O] éë Ni (CO)4 ùû 4 ë û ë û ë û ´ ´ Q1 × Q2 × Q3 = = 7 6 é O2 ù 7 é H2 ù 6 éCOù 24 [O2 ] [ H2 ] [CO]24 ë û ë û ë û

17.86

a) Since the volume is 1.00 L, the molarity equals the number of moles present. 2NH3(g)

⇆

N2(g)

3H2(g)

Initial

1.30

1.65

Change

+2x

–x

–3x

Equilibrium

2x = 0.100 M

1.30 – x

1.65 – 3x

x = 0.0500 mol [N2]eq = (1.30 – 0.0500) M = 1.25 M N2 [H2]eq = [1.65 – 3(0.0500)] M = 1.50 M H2

[ N 2 ][ H 2 ]

Kc = 1

[ NH3 ] 3

[1.25][1.50 ]3 2

[0.100 ] 1

= 421.875 = 422

é1.50ù 2 é1.25ù 2 [ N2 ] 2 [ H2 ] 2 û ë û b) Kc = = ë = 20.523177 = 20.5 é8.34 ´ 10-2 ù [ NH3 ] ê ú

17.87

ë û c) Kc in a) is the square of Kc in b). The balanced equations are different; therefore, the values of Kc are different. Plan: Write the equilibrium expression. You are given a value of Kc but the amounts of reactant and product are given in units of pressure. Convert Kc to Kp and use the equilibrium pressures of C2H5OH and H2O to obtain the equilibrium pressure of C2H4. An increase in temperature (addition of heat) causes a shift in the equilibrium away from the side of the reaction with heat, while a decrease in temperature (removal of heat) causes a shift in the equilibrium towards the side with heat. Increasing the volume of the container (pressure decreases) results in a shift in the direction that forms more moles of gas, while decreasing the container volume (pressure increases) results in a shift in the direction that forms fewer moles of gas. The van’t Hoff equation shows how the equilibrium constant is affected by a change in temperature. Substitute the given variables into the equation and solve for K at 450. K.

17-33

Solution: a) Kp = Kc(RT)Dn Dn = moles gaseous products – moles gaseous reactants = 1 – 2 = –1 (one mol of product, C2H5OH, and two mol

of reactants, C2H4 + H2O) Kp = Kc(RT)–1 = (9 × 103)[(0.0821 L ∙ atm/mol ∙ K)(600. K)]–1 = 1.82704 × 102 Substitute the given values into the equilibrium expression and solve for PC H . 2 4

PC2H5OH

200. Kp = = = 1.8270 × 102 PC2H4 PH2O PC H (400.) 2 4

PC2H4 = 2.7367 × 10–3 = 3 × 10–3 atm

o b) Since DHrxn is negative, the reaction is exothermic and heat is written as a product. To shift the reaction towards the right to yield more ethanol, heat must be removed. A low temperature favors an exothermic

reaction. The forward direction, towards the production of ethanol, produces the smaller number of moles of gas and is favored by high pressure.

æ103 J ö÷ o ÷ = –4.78 × 104 J DHrxn = (-47.8 kJ)ççç çè 1 kJ ÷÷ø T2 = 450. K R = 8.314 J/mol ∙ K o æ 1 ö÷ DHrxn ç 1 K ÷ ln 2 = çç R çè T2 T1 ÷÷ø K1

c) K1 = 9 × 103 K2 = ?

ln ln

T1 = 600. K

-4.78 ´ 104 J æç 1 1 ö÷÷ çç = ÷ 8.314 J/mol K ççè 450. K 600. K ø÷÷ 9 ´103

K2 K2

= 3.1940769 9 ´103 K2 = 24.38765 9 ´ 103 K2 = (9 × 103)(24.38765) = 2.1949 × 105 = 2 × 105 d) No, condensing the C2H5OH would not increase the yield. Ethanol has a lower boiling point (78.5°C) than water (100°C). Decreasing the temperature to condense the ethanol would also condense the water, so moles of gas from each side of the reaction are removed. The direction of equilibrium (yield) is unaffected when there is no net change in the number of moles of gas.

17.88

n/V = M = P/RT =

(2.0 atm) æ L atm ÷ö çç0.0821 ÷((273.2 + 25.0) K ) è mol K ø÷

H2(g)

CO2(g)

⇆

= 0.0816919 M each gas

H2O(g)

CO(g)

Initial

0.0816919

Change

–x

Equi:

0.0816919 – x

17-34

Kc =

[ x ][ x ] [ H2 O][CO] [ x ]2 = 0.534 = = [0.0816919 - x ][0.0816919 - x ] [ H2 ][CO2 ] [0.0816919 - x ]2

(0.534)1/2 = 0.730753 =

[x]

[0.0816919 - x ]

x = 0.03449 M M of H2 at equilibrium = 0.0816919 – x = 0.0816919 – 0.03449 = 0.0472019 mol/L æ 0.0472019 mol öæ ÷÷ç 2.016 g ÷÷ö = 0.095159 = 0.095 g H2 Mass (g) of H2 = (1.00 L )çç ÷øèç 1 mol ÷ø è L

17.89

Plan: Write the equilibrium expression. You are given a value of Kc but the amounts of reactants and product are given in units of pressure. Convert Kc to Kp and use the equilibrium pressures of SO3 and O2 to obtain the equilibrium pressure of SO2. For part b), set up a reaction table and solve for x. The equilibrium concentrations can then be used to find the Kc value at the higher temperature. The concentration of SO2 is converted to pressure using the ideal gas law, PV = nRT. Solution: a) Kp = Kc(RT)Dn Dn = moles gaseous products – moles gaseous reactants = 2 – 3 = –1 (two mol of product, SO3, and three mol of

reactants, 2 SO2 + O2) Kp = Kc(RT)Dn = Kc(RT)–1 = (1.7 × 108)[(0.0821 L ∙ atm/mol ∙ K)(600. K)]–1 = 3.451 × 106

(300.) K = 2 = 2 = 3.451 × 10 PSO2 PO2 PSO (100.) 2 2

2 PSO 3

PSO2 = 0.016149 = 0.016 atm b) Create a reaction table that describes the reaction conditions. Since the volume is 1.0 L, the moles equals the molarity. Note the 2:1:2 mole ratio between SO2:O2:SO3. Concentration (M)

2SO2(g)

Initial

0.0040

0.0028

Change

–2x

–x

+2x

Equilibrium

0.0040 – 2x

0.0028 – x

2x = 0.0020 (given)

O2(g)

⇆

2SO3(g) 0 (2:1:2 mole ratio)

x = 0.0010, therefore: [SO2] = 0.0040 – 2x = 0.0040 – 2(0.0010) = 0.0020 M [O2] = 0.0028 – x = 0.0028 – 0.0010 = 0.0018 M [SO3] = 2(0.0010) = 0.0020 M Substitute equilibrium concentrations into the equilibrium expression and solve for Kc.

[SO3 ] [0.0020 ]2 = = 555.5556 = 5.6 × 102 2 2 0.0020 0.0018 [ ] [ ] [SO2 ] [O2 ] 2

Kc =

The pressure of SO2 is estimated using the concentration of SO2 and the ideal gas law (although the ideal gas law is not well behaved at high pressures and temperatures). PV = nRT

L atm ÷ 0.0020 mol)çç0.0821 ÷(1000. K) ( ç mol K ÷ø è nRT PSO = = = 0.1642 = 0.16 atm V (1.0 L) 2

17-35

17.90

The original concentrations are: (0.350 mol/0.500 L) = 0.700 M for CO and Cl2. Concentration (M)

Cl2(g)

CO(g) +

⇆

COCl2(g)

Initial

0.700

Change

–x

Equilibrium

0.700 – x

Qc =

0.700

[x] [x] [COCl2 ] = = = 4.95 é 0.700 - x ù [0.700 - x ] é 0.490 - 1.400x + x 2 ù [CO][Cl2 ] ê ú ê ú

ë û 4.95x2 – 7.93x + 2.4255 = 0 a = 4.95

b = – 7.93

c = 2.4255

-b  b - 4ac 2a 2

-(-7.93)  (-7.93) - 4 (4.95)(2.4255) 2

2 (4.95)

x = 1.19039 or 0.41162959 (The 1.19039 value is not possible because 0.700 – x would be negative.) [CO] = [Cl2] = 0.700 – x = 0.700 – 0.41162959 = 0.28837041 = 0.288 M [COCl2] = x = 0.41162959 = 0.412 M 17.91

Plan: Set up a reaction table to find the equilibrium amount of CaCO3 after the first equilibrium is established and then the equilibrium amount after the second equilibrium is established. Solution: The equilibrium pressure of CO2 = PCO2 = 0.220 atm. CaCO3(s) Initial

0.100 mol

Change

–x

⇆

CaO(s)

CO2(g)

0.100 mol

–x

Equilibrium 0.100 – x 0.100 – x x = 0.220 atm (given) The amount of calcium carbonate solid in the container at the first equilibrium equals the original amount, 0.100 mol, minus the amount reacted to form 0.220 atm of carbon dioxide. The moles of CaCO3 reacted is equal to the number of moles of carbon dioxide produced. Use the pressure of CO2 and the ideal gas equation to calculate the moles of CO2 produced: PV = nRT PV Moles of CO2 = n = RT n=

(0.220 atm)(10.0 L) = 0.0696015 mol CO æ L atm ö÷ ç0.0821 ÷(385 K ) çè mol K ø÷

17-36

Moles of CaCO3 reacted = moles of CO2 produced = 0.0696015 mol Moles of CaCO3 remaining = initial moles – moles reacted = 0.100 mol CaCO3 – 0.0696015 mol CaCO3 = 0.0304 mol CaCO3 at first equilibrium As more carbon dioxide gas is added, the system returns to equilibrium by shifting to the left to convert the added carbon dioxide to calcium carbonate to maintain the partial pressure of carbon dioxide at 0.220 atm (Kp). Convert the added 0.300 atm of CO2 to moles using the ideal gas equation. The moles of CO2 reacted equals the moles of CaCO3 formed. PV Moles of CO2 = n = RT n=

(0.300 atm)(10.0 L) = 0.09491 mol CO æ L atm ö÷ çç0.0821 ÷(385 K ) è mol K ø÷

Moles of CaCO3 produced = moles of CO2 reacted = 0.09491 mol CaCO3 Add the moles of CaCO3 formed in the second equilibrium to the moles of CaCO3 at the first equilibrium position. Moles of CaCO3 = moles at first equilibrium + moles formed in second equilibrium = 0.0304 mol + 0.09491 = 0.12531mol CaCO3 æ100.09 g CaCO3 ö÷ ÷ = 12.542 = 12.5 g CaCO3 Mass (g) of CaCO3 = (0.12531 mol CaCO3 )ççç çè 1 mol CaCO3 ÷÷ø 17.92

a) C2H4(g) + 3O2(g) ⇆ 2CO2(g) + 2H2O(g) b) 4NO2(g) + 6H2O(g) ⇆ 4NH3(g) + 7O2(g)

17.93

C2H2(g) + H2(g) ⇆ C2H4(g) o o o - n DHreactants DHrxn = m DHproducts

= {1 DHfo [C2H4(g)]} – {1 DHfo [C2H2(g)] + 1 DHfo [H2(g)]} = [(1 mol)(52.47 kJ/mol)] – [(1 mol)(227 kJ/mol) + (1 mol)(0.0 kJ/mol)] = –174.53 kJ ln

o æ ö DHrxn K300 çç 1 - 1 ÷÷ =÷ R ççè T2 T1 ÷ø K2000

K300 2.9 ´108 K300 2.9 ´108

(-174.53 kJ /mol) çæç =-

öæ103 J ÷ö ÷ç 1 1 ÷ ÷÷çç ÷÷ç 1 kJ ÷÷÷ æ J ö÷ ççèç 300. K 2000. K ç øè ø ç8.314 ÷ çè mol K ø÷

= 59.478189

K300

= 6.77719 × 1025 2.9 ´108 K300 = (2.9 × 108)(6.77719 × 1025) = 1.9654 × 1034 = 2.0 × 1034

17-37

17.94

M2(g) + N2(g) ⇆ 2MN(g)

Kc =

[ MN ]2

[ M2 ][N 2 ]

Scene A: Concentrations: [M2] = [N2] = 0.20 M; [MN] = 0.40 M Kc =

[0.40]2

[0.20 ][0.20] Scene B: Concentration (M) Initial Change Equilibrium

= 4.0

M2(g)

0.60 –x 0.60 – x

N2(g)

⇆

2MN(g)

0.30

–x

+2x

0.30 – x

[2x ]

Kc = 4.0 =

[0.60 - x ][0.30 - x] 4x 2

4.0 =

0.18 - 0.90 x + x 2 4x = 0.72 – 3.6x + 4x2 2

3.6x = 0.72 x = 0.20 M [M2] = 0.60 – x = 0.60 – 0.20 = 0.40 M [N2] = 0.30 – x = 0.30 – 0.20 = 0.10 M [MN] = 2x = 2(0.20 M) = 0.40 M 17.95

Plan: Use the balanced reaction to write the equilibrium expression. The equilibrium concentration of S2F10 is used to write an expression for the equilibrium concentrations of SF4 and SF6. Solution: S2F10(g) ⇆ SF4(g) + SF6(g) The reaction is described by the following equilibrium expression: Kc =

[SF4 ][SF6 ] [S2 F10 ]

At the first equilibrium, [S2F10] = 0.50 M and [SF4] = [SF6] = x ([SF4]:[SF6] = 1:1). Kc =

[SF4 ][SF6 ] [ x ][ x ] = [ 0.50 ] [S2 F10 ]

x2 = 0.50Kc [SF4] = [SF6] = x =

0.50Kc

At the second equilibrium, [S2F10] = 2.5 M and [SF4] = [SF6] = x. Kc =

[SF4 ][SF6 ] [ x ][ x ] = [ 2.5 ] [S2 F10 ]

x2 = 2.5Kc [SF4] = [SF6] = x =

2.5Kc

17-38

Thus, the concentrations of SF4 and SF6 increase by a factor of:

2.5Kc 0.50 Kc 17.96

2.5 = 2.236 = 2.2 0.50

Calculate Kc. Kc =

[CO2 ][ H2 ] [0.40 ][0.10 ] = = 4.0 [0.10 ][0.10 ] [CO][ H2 O]

Calculate new concentrations. New H2 = 0.10 M + (0.60 mol/2.0 L) = 0.40 M Concentration (M)

CO(g)

H2O(g)

⇆

CO2(g) +

H2(g)

Initial

0.10

0.40

Change

–x

Equilibrium

0.10 + x

0.40 – x

[CO2 ][ H2 ] [0.40 - x ][0.40 - x ] [0.40 - x ] = = = 4.0 [0.10 + x ][0.10 + x ] [0.10 + x ]2 [CO][ H2 O] 2

Kc =

é 0.40 - x ù êë úû = 2.0 é 0.10 + x ù êë úû x = 0.066667

[CO] = [H2O] = 0.10 + x = 0.10 + 0.066667 = 0.166667 = 0.17 M [CO2] = [H2] = 0.40 – x = 0.40 – 0.066667 = 0.333333 = 0.33 M 17.97

Plan: Use the volume fraction of O2 and CO2 to find the partial pressure of each gas and substitute these pressures into the equilibrium expression to find the partial pressure of CO. Use PV = nRT to convert the partial pressure of CO to moles per liter and then convert to pg/L. Solution: a) Calculate the partial pressures of oxygen and carbon dioxide because volumes are proportional to moles of gas, so volume fraction equals mole fraction. Assume that the amount of carbon monoxide gas is small relative to the other gases, so the total volume of gases equals VCO2 + VO2 + VN2 = 10.0 + 1.00 + 50.0 = 61.0.

æ10.0 mol CO2 ö÷ ÷÷( 4.0 atm ) = 0.6557377 atm PCO2 = ççç è 61.0 mol gas ø÷ æ 1.00 mol O2 ö÷ ÷(4.0 atm ) = 0.06557377 atm PO2 = çç çè 61.0 mol gas ø÷÷ Use the partial pressures and given Kp to find PCO. 2CO2(g) ⇆ 2CO(g) + O2(g) Kp =

2 PCO PO2 2 PCO 2

2 PCO (0.06557377)

(0.6557377)

= 1.4 × 10–28

PCO = 3.0299 × 10–14 = 3.0 × 10–14 atm

17-39

b) PV = nRT

(

)

3.0299 ´ 10-14 atm nCO P = = = 4.61312 × 10–16 mol/L æ ö÷ V RT L atm çç0.0821 ÷ 800 K çè mol K ÷÷ø -16 æ mol CO ÷÷öçæ 28.01 g CO ÷öæç 1 pg ÷÷ö ç 4.61312 ´ 10 ÷ç Concentration (pg/L) of CO = çç ÷çç ÷ = 0.01292 = 0.013 pg CO/L ÷ç çè ÷øè 1 mol CO ÷÷øççè10-12 g ÷÷ø L

(

)

17.98

Although the yield is favored by low temperature, the rate of formation is not. In fact, ammonia forms so slowly at low temperatures that the process becomes uneconomical. In practice, a compromise is achieved that optimizes yield and rate (high pressure, continual removal of NH3, increasing the temperature).

17.99

Plan: Write a reaction table given that PCH 4 (init) = PCO2 (init) = 10.0 atm, substitute equilibrium values into the equilibrium expression, and solve for PH2 . Solution: a) Pressure (atm)

CH4(g)

Initial

10.0

Change

–x

Equilibrium

CO2(g)

10.0 – x 2 PCO PH22

Kp =

PCH 4 PCO2

(2x)2

(10.0 - x)

2CO(g)

(2x ) (2x )

(

)(

2H2(g)

+2x

10.0 – x 2

⇆

10.0 - x 10.0 - x

(2x )

) (10.0 - x)

= 3.548 × 106 (take square root of each side)

= 1.8836135 × 103

A quadratic is necessary: 4x2 + (1.8836135 × 103 x) – 1.8836135 × 104 = 0 a=4 x=

b = 1.8836135 × 103

-b  b2 - 4ac 2a -1.8836135 ´ 103 

c = – 1.8836135 × 104

(1.8836135 ´ 10 ) - 4 (4)(-1.8836135 ´ 10 ) 3

2 ( 4)

x = 9.796209 PH2 = 2x = 2(9.796209) = 19.592419 atm If the reaction proceeded entirely to completion, the partial pressure of H2 would be 20.0 atm (pressure is proportional to moles, and twice as many moles of H2 form for each mole of CH4 or CO2 that reacts). 19.592418 atm (100%) = 97.96209 = 98.0%. The percent yield is 20.0 atm

17-40

b) Repeat the calculations for part a) with the new Kp value. The reaction table is the same. 2 PCO PH22

Kp =

PCH 4 PCO2

(2x )2

(

10.0 - x

)

(2x )2 (2x )2

(2x )4

(10.0 - x)(10.0 - x) (10.0 - x)

= 2.626 × 107

= 5.124451 × 103

A quadratic is needed: 4x2 + (5.124451 × 103 x) – 5.124451 × 104 = 0 b = 5.124451 × 103

a=4

-5.124451 ´ 103 

c = – 5.124451 × 104

(5.124451 ´ 10 ) - 4 (4)(-5.124451 ´ 10 ) 3

2 (4)

x = 9.923144 PH2 = 2x = 2(9.923144) = 19.84629 atm If the reaction proceeded entirely to completion, the partial pressure of H2 would be 20.0 atm (pressure is proportional to moles, and twice as many moles of H2 form for each mole of CH4 or CO2 that reacts). 19.84629 atm (100%) = 99.23145 = 99.0%. The percent yield is 20.0 atm c) van’t Hoff equation: K1 = 3.548 × 106

T1 = 1200. K

K2 = 2.626 × 107

T2 = 1300. K

o DHrxn =? R = 8.314 J/mol ∙ K

o æ ö DHrxn K2 çç 1 - 1 ÷÷ =ç R çè T2 T1 ÷÷ø K1

2.626 ´ 10 7 3.548 ´ 106

æ 1 o 1 ö÷÷ DH rxn çç ÷ æ öççç1200. K J ÷è 1300. K ÷÷ø ç8.314 ÷ çè mol K ÷ø

o 2.0016628 = DHrxn (7.710195 × 10–6) o DHrxn = 2.0016628/7.710195 × 10–6 = 2.5961247 × 105 = 2.60 × 105 J/mol

(The subtraction of the 1/T terms limits the answer to three significant figures.) 17.100 a)

C3H8(g) + 3H2O(g) ⇆ 3CO(g) + 7H2(g)

K'p1 = Kp = 8.175 × 1015

3CO(g) + 3H2O(g) ⇆ 3CO2(g) + 3H2(g)

K'p2 = (Kp2)3 = (0.6944)3 = 0.33483368

(Overall): C3H8(g) + 6H2O(g) ⇆ 3CO2(g) + 10H2(g) b) Kp(overall) = K'p1 × K'p2 = (8.175 × 1015)(0.33483368) = 2.737265 × 1015 = 2.737 × 1015

(PCO ) (PH ) c) K = 6 (PC H )(PH O ) 3

3 8

The partial pressures of each reactant are proportional to the moles, and the limiting reactant may be determined from the partial pressures.

17-41

PC3H8 (initial) = (1.00/5.00) × 5.0 atm = 1.0 atm PH O(initial) = (4.00/5.00) × 5.0 atm = 4.0 atm

(limiting reactant)

PCO2(formed) = 4.0 atm H2O × (3 mol CO2/6 mol H2O) = 2.0 atm PH2(formed) = 4.0 atm H2O × (10 mol H2/6 mol H2O) = 6.6667 atm PC3H8 (remaining) = 1.0 atm C3H8 – [4.0 atm H2O × (1 mol C3H8/6 mol H2O)] = 0.3333 atm PH2O(remaining) = 0.00 atm (limiting reactant) PTotal = PCO2 + PH2 + PC3H8 + PH2O = 2.0 atm + 6.6667 atm + 0.3333 atm + 0.00 atm = 9.0 atm d) Percent C3H8(unreacted) = [0.3333 atm/1.0 atm] × 100% = 33.33% = 33% 17.101 Plan: Add the two reactions to obtain the overall reaction. Multiply the second equation by 2 to cancel the moles of CO produced in the first reaction. Kp for the second reaction is then (Kp )2. Kp for the overall reaction is equal to the product of the Kp values for the two individual reactions. Calculate Kc using Kp = Kc(RT)Dn. Solution: 2CH4(g) + O2(g) ⇆

2CO(g) + 4H2(g)

Kp = 9.34 × 1028

2CO(g) + 2H2O(g) ⇆

2CO2(g) + 2H2(g)

Kp =(1.374)2 = 1.888

2CH4(g) + O2(g) + 2H2O(g) ⇆ 2CO2(g) + 6H2(g) 28

b) Kp = (9.34 × 10 )(1.888) = 1.76339 × 1029 = 1.76 × 1029 c) Dn = moles gaseous products – moles gaseous reactants = 8 – 5 = 3 (8 moles of product gas – 5 moles of reactant gas) Kp = Kc(RT)Dn Kc =

Kp Dn

1.76339 ´ 1029

= 3.18654 × 1023 = 3.19 × 1023

( RT ) [(0.0821 atm L/mol K)(1000)] d) The initial total pressure is given as 30. atm. To find the final pressure use the relationship between pressure

and number of moles of gas: ninitial/Pinitial = nfinal/Pfinal Total mol of gas initial = 2.0 mol CH4 + 1.0 mol O2 + 2.0 mol H2O = 5.0 mol Total mol of gas final = 2.0 mol CO2 + 6.0 mol H2 = 8.0 mol

(from mole ratios)

æ 8 mol products ö÷ Pfinal = (30. atm reactants)ççç ÷ = 48 atm è 5 mol reactants ÷ø

17.102 Plan: Write an equilibrium expression. Use the balanced equation to define x and set up a reaction table, substitute into the equilibrium expression, and solve for x, from which the pressure of N or H is calculated. Convert log Kp to Kp. Convert pressures to moles using the ideal gas law, PV = nRT. Convert moles to atoms using Avogadro’s number. Solution: a) The initial pressure of N2 is 200. atm. Log Kp = –43.10; Kp = 10–43.10 = 7.94328 × 10–44 Pressure (atm)

N2(g)

⇆

2N(g)

Initial

200.

Change

–x

+2x

Equilibrium

200 – x

17-42

( PN )

Kp =

(PN )

= 7.94328 × 10–44

(2x ) = 7.94328 × 10–44 (200. - x )

Assume 200. – x ≅ 200.

(2x )2 = 7.94328 × 10–44 (200) 4x2 = 1.588656 × 10–41 x = 1.992897 × 10–21 –21

–21

PN = 2x = 2(1.992897 × 10 ) = 3.985795 × 10

= 4.0 × 10–21 atm

Kp = 10–17.30 = 5.01187 × 10–18

b) Log Kp = –17.30; Pressure (atm)

H2(g)

Initial

600.

Change

–x

+2x

Equilibrium

600 – x

( PH )

⇆

2H(g)

Kp =

(PH )

= 5.01187 × 10–18

(2x ) = 5.01187 × 10–18 (600. - x )

Assume 600. – x ≅ 600.

(2x )2 = 5.01187 × 10–18 (600) 4x2 = 3.007122 × 10–15 x = 2.741862 × 10–8 –8

–8

PH = 2x = 2(2.741862 × 10 ) = 5.48372 × 10 = 5.5 × 10

c) PV = nRT

(

)

3.985795 ´ 10-21 atm (1.00 L ) PV Moles of N atoms = = = 4.85481 × 10–15 mol æ ö RT çç0.0821 L atm ÷÷ 1000. K mol K ÷ø èç æ 6.022 ´1023 N atoms ö÷ -23 ÷÷ = 29.2356 = 29 N atoms/L mol N atoms ççç Number of N atoms = 4.85481 ´ 10 çè 1 mol N atoms ÷÷ø

(

)

(

Moles of H atoms =

)

(

)

5.48372 ´ 10-8 atm (1.00 L ) PV = = 6.67932 × 10–10 mol æ ö÷ RT L atm ç

(

)

÷ 1000.K çç0.0821 mol K ÷ø è æ 6.022 ´ 1023 H atoms ö÷ -10 ÷÷ mol H atoms ççç Number of H atoms = 6.67932 ´ 10 çè 1 mol H atoms ÷÷ø

(

)

= 4.022 × 1014 = 4.0 × 1014 H atoms/L

17-43

d) The more reasonable step is N2(g) + H(g)  NH(g) + N(g). With only twenty-nine N atoms in 1.0 L, the first reaction would produce virtually no NH(g) molecules. There are orders of magnitude more N2 molecules than N atoms, so the second reaction is the more reasonable step. 17.103 a) Scenes B and D represent equilibrium. b) C, A, B = D æ 0.025 mol ÷öæ 1 ö ÷ç c) [Y] = (4 spheres)çç ÷÷ = 0.25 M çè 1 sphere ÷÷øçè 0.40 L ø÷ æ 0.025 mol ö÷æ 1 ö ÷ç [Z] = (8 spheres)çç ÷÷ = 0.50 M çè 1 sphere ÷÷øçè 0.40 L ø÷

Kc =

[Z]2 [0.50]2 = = 1.0 [Y] [0.25]

17.104 The K is very small, thus the reaction will shift to the right to reach equilibrium. To simplify the calculations, assume the equilibrium shifts entirely to the left, and then a little material reacts to reach equilibrium. Shifting entirely to the left gives [H2S] = 0.600, and [H2] = [S2] = 0. ⇆ 2H2(g) + S2(g) 2H2S(g) Initial

0.600 M

Change

–2x

+2x

Equilibrium

0.600 – 2x

[ H 2 ] [S2 ] = 9.0 × 10–8 2 [ H 2 S] 2

Kc =

[2x ]2 [ x ] é0.600 - 2x ù 2 ëê ûú

= 9.0 × 10–8

Assume 2x is small compared to 0.600 M.

[2x ]2 [ x ] = 9.0 × 10–8 [0.600 ]2 x = 2.008 × 10–3

(assumption justified)

[H2S] = 0.600 – 2x = 0.600 – 2(2.008 × 10–3) = 0.595984 = 0.596 M H2S [H2] = 2x = 2(2.008 × 10–3) = 4.016 × 10–3 = 4.0 × 10–3 M H2 [S2] = x = 2.008 × 10–3 = 2.0 × 10–3 M S2 17.105 Plan: Write an equilibrium expression. Use the balanced equation to define x and set up a reaction table, substitute into the equilibrium expression, and solve for x, from which the equilibrium pressures of the gases are calculated. Add the equilibrium pressures of the three gases to obtain the total pressure. Use the relationship Kp = Kc(RT)Dn to find Kc. Solution: a)

Pressure (atm)

N2(g)

O2(g)

⇆

2NO(g)

Initial

0.780

0.210

Change

–x

+2x

Equilibrium

0.780 – x

0.210 – x

17-44

( PNO )

Kp =

(PN )(PO ) 2

= 4.35 × 10–31

(2 x ) = 4.35 × 10 Assume x is small because K is small. (0.780 - x)(0.210 - x) 2 (2 x ) 2

–31

(0.780)(0.210)

= 4.35 × 10–31

x = 1.33466 × 10–16 Based on the small amount of nitrogen monoxide formed, the assumption that the partial pressures of nitrogen and oxygen change to an insignificant degree holds. Pnitrogen (equilibrium) = (0.780 – 1.33466 × 10–16) atm = 0.780 atm N2 Poxygen (equilibrium) = (0.210 – 1.33466 × 10–16) atm = 0.210 atm O2 PNO (equilibrium) = 2(1.33466 × 10–16) atm = 2.66933 × 10–16 = 2.67 × 10–16 atm NO b) The total pressure is the sum of the three partial pressures: 0.780 atm + 0.210 atm + 2.67 × 10–16 atm = 0.990 atm c) Kp = Kc(RT)Dn Dn = moles gaseous products – moles gaseous reactants = 2 – 2 = 0

(two moles of product NO and two moles of reactants N2 and O2) Kp = Kc(RT) Kc = Kp = 4.35 × 10–31 because there is no net increase or decrease in the number of moles of gas in the course of the reaction. 17.106 a) Kp = Kc(RT)Dn Dn = 2 – (2 + 1) = –1 Kc = Kp/(RT)Dn

(1.3 ´ 10 ) 4

= 4.877561 × 105 = 4.9 × 105 é(0.0821)(457)ù -1 êë úû o o o b) DH rxn = å[ DHf(products) ] – å[ DHf(reactants) ] Kc =

o DH rxn = {2 DHfo [NO2(g)]} – {2 DHfo [NO(g)] +DHfo [O2(g)]}

= [(2 mol)(33.2 kJ/mol)] – [(2 mol)(90.29 kJ/mol) + (1 mol)(0.0 kJ/mol)] = –114.18 = – 114.2 kJ c) ln

o æ ö DH rxn K2 çç 1 - 1 ÷÷ =R ççè T2 T1 ÷÷ø K1

3 -114.18 kJ æç 1 1 ö÷÷æçç10 J ö÷÷ çç ÷ ÷÷ç 8.314 J/mol K çèç T2 457K ÷øèçç 1 kJ ÷÷ø 4.877561 ´ 10 æ1 1 ö÷÷ ç 9.4819931 = 13,733 çç ÷ ççè T2 457K ÷ø÷

6.4 ´ 109

T2 = 347.389 = 3.5 × 102 K Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

17-45

17.107 Plan: Use the equation Kp = Kc(RT)Dn to find Kp. The value of Kc for the formation of HI is the reciprocal of the o

o o Kc value for the decomposition of HI. Use the equation DH rxn = å[ DHf(products) ] – å[ DHf(reactants) ] to find the o o . Use the van’t Hoff equation as a second method of calculating DH rxn . value of DH rxn

Solution: a) Kp = Kc(RT)Dn Dn = moles gaseous products – moles gaseous reactants = 2 – 2 = 0

(2 mol product (1H2 + 1I2) – 2 mol reactant (HI) = 0) Kp = Kc(RT)0 = 1.26 × 10–3(RT)0 = 1.26 × 10–3

b) The equilibrium constant for the reverse reaction is the reciprocal of the equilibrium constant for the forward reaction: Kformation =

1 1 = = 793.65 = 794 Kdecomposition 1.26 ´ 10-3 o

o o c) DH rxn = å[ DHf(products) ] – å[ DHf(reactants) ] o DH rxn = {1 DHfo [H2(g)] +1 DHfo [I2(g)]} – {2 DHfo [HI(g)]} o DH rxn = [(1 mol)(0 kJ/mol) + (1 mol)(0 kJ/mol)] – [(2 mol)(25.9 kJ/mol)]

o DH rxn = – 51.8 kJ

d) ln

o æ ö K2 DH rxn çç 1 - 1 ÷÷ =ç K1 R çè T2 T1 ÷÷ø

K1 = 1.26 × 10–3; K2 = 2.0 × 10–2, T1 = 298 K; T2 = 729 K

æ 1 o DHrxn 1 ö÷÷ çç = ÷ ç 8.314 J/mol K ççè 729 K 298 K ÷÷ø 1.26 ´ 10-3

2.0 ´10-2

o 2.764621 = 2.38629 × 10–4 DH rxn o DH rxn = 1.1585 × 104 = 1.2 × 104 J/mol

17.108 C5H11OH + CH3COOH ⇆ CH3COOC5H11 + H2O Removing water should help to increase the yield of banana oil. Both isopentyl alcohol and acetic acid are more soluble in water than isopentyl acetate. Thus, removing water will increase the concentration of both reactants and cause a shift in equilibrium towards the products. 17.109 Q(g) ⇆ R(g)

[R] [Q]

For Scene A at equilibrium: [2] [R] K= = = 0.33 [6] [Q]

17-46

For Scene B: Q(g) 10

Initial

⇆

Change –x Equilibrium 10 – x [2 + x] 0.33 = [10 - x] x = 0.977 = 1

R(g) 2 +x 2+x

Q = 10 – x = 10 – 1 = 9; R = 2 + x = 2 + 1 = 3

(

17.110 a) Kp = PH2O PH 2 O =

) = 4.08 × 10 10

–25

–3 –3 4.08 ´ 10-25 = 3.6397 × 10 = 3.64 × 10 atm

b) (1) Adding more Na2SO4(s) will decrease the ratio of hydrated form/anhydrous form merely because you are increasing the value of the denominator, not because the equilibrium shifts. (2) Reducing the container size will increase the pressure (concentration) of the water vapor, which will shift the equilibrium to the reactant side. The ratio of hydrated form/anhydrous form will increase. (3) Adding more water vapor will increase the concentration of the water vapor, which will shift the equilibrium to the reactant side. The ratio of hydrated form/anhydrous form will increase. (4) Adding N2 gas will not change the partial pressure of the water vapor, so the ratio of hydrated form/anhydrous form will not change. 17.111 Plan: Use the balanced equation to write an equilibrium expression. Find the initial concentration of each reactant from the given amounts and container volume, use the balanced equation to define x, and set up a reaction table. The equilibrium concentration of CO is known, so x can be calculated and used to find the other equilibrium concentrations. Substitute the equilibrium concentrations into the equilibrium expression to find Kc. Add the molarities of all of the gases at equilibrium, use (M)(V) to find the total number of moles, and then use PV = nRT to find the total pressure. To find [CO]eq after the pressure is doubled, set up another reaction table in which the initial concentrations are equal to the final concentrations from part a) and add in the additional CO. Solution: The reaction is: CO(g) + H2O(g) ⇆ CO2(g) + H2(g) a) Initial [CO] and initial [H2O] = 0.100 mol/20.00 L = 0.00500 M. CO H2O ⇆ CO2 H2 0.00500 M –x

Initial Change

0.00500 M –x

0 +x

Equilibrium 0.00500 – x 0.00500 – x x x [CO]equilibrium = 0.00500 – x = 2.24 × 10–3 M = [H2O] (given in problem) x = 0.00276 M = [CO2] = [H2]

Kc =

[CO2 ][ H2 ] [0.00276 ][0.00276 ] = = 1.518176 = 1.52 [CO][ H2 O] [0.00224 ][0.00224 ]

17-47

b) Mtotal = [CO] + [H2O] + [CO2] + [H2] = (0.00224 M) + (0.00224 M) + (0.00276 M) + (0.00276 M) = 0.01000 M ntotal = (Mtotal)(V) = (0.01000 mol/L)(20.00 L) = 0.2000 mol total

PV = nRT

L atm ö÷ (0.2000 mol)æççè0.08206 mol ÷((273 + 900.) K) K ø÷ P = n RT/V = = 0.9625638 = 0.9626 atm (20.00 L) total

total

c) Initially, an equal number of moles must be added = 0.2000 mol CO d) Set up a table with the initial concentrations equal to the final concentrations from part a), and then add 0.2000 mol CO/20.00 L = 0.01000 M to compensate for the added CO. CO H2O ⇆ CO2 H2 Initial 0.00224 M 0.00224 M 0.00276 M 0.00276 M Added CO Change

0.01000 M –x

–x

+x__________

Equilibrium

0.01224 – x 0.00224 – x 0.00276 + x 0.00276 + x [CO2 ][ H2 ] éëê0.00276 + x ùûú éëê0.00276 + x ùûú = 1.518176 Kc = = é 0.01224 - x ù é 0.00224 - x ù [CO][ H2 O] ëê ûú ëê ûú é 7.6176 ´ 10-6 + 5.52 ´ 10-3 x + x 2 ù êë úû = 1.518176 é 2.74176 ´ 10-5 - 1.448 ´ 10-2 x + x 2 ù êë úû

7.6176 × 10–6 + 5.52 × 10–3x + x2 = (1.518176)(2.74176 × 10–5 – 1.448 × 10–2x + x2) 7.6176 × 10–6 + 5.52 × 10–3x + x2 = 4.162474 × 10–5 – 0.021983x + 1.518176x2 0.518176x2 – 0.027503x + 3.400714 × 10–5 = 0 a = 0.518176

b = – 0.027503

c = 3.400714 × 10–5

-b  b2 - 4ac

-(-0.027503) 

(-0.027503) - 4 (0.518176)(3.400714 ´ 10-5 ) 2 (0.518176) 2

x = 1.31277 × 10–3 [CO] = 0.01224 – x = 0.01224 – (1.31277 × 10–3) = 0.01092723 = 0.01093 M 17.112 a) At point A the sign of DH is negative for the reaction graphite  diamond. An increase in temperature at constant pressure will cause the formation of more graphite. Therefore, the equation must look like this: graphite  diamond + heat, and adding heat shifts the equilibrium to the reactant side. b) Diamond is denser than graphite. The slope of the diamond-graphite line is positive. An increase in pressure favors the formation of diamond.

17-48

CHAPTER 18 ACID-BASE EQUILIBRIA FOLLOW–UP PROBLEMS 18.1A

Plan: Identify the conjugate pairs by first identifying the species that donates H+ (the acid) in either reaction direction. The other reactant accepts the H+ and is the base. The acid has one more H and +1 greater charge than its conjugate base. Solution: a) CH3COOH has one more H+ than CH3COO–. H3O+ has one more H+ than H2O. Therefore, CH3COOH and H3O+ are the acids, and CH3COO– and H2O are the bases. The conjugate acid/base pairs are CH3COOH/CH3COO– and H3O+/H2O. b) H2O donates an H+ and acts as the acid. F– accepts the H+ and acts as the base. In the reverse direction, HF acts as the acid and OH– acts as the base. The conjugate acid/base pairs are H2O/OH– and HF/F –.

18.1B

Plan: To derive the formula of a conjugate base, remove one H from the acid and decrease the charge by 1 (acids donate H+). To derive the formula of a conjugate acid, add an H and increase the charge by 1 (bases accept H+). Solution: a) Adding an H+ to HSO3– gives the formula of the conjugate acid: H2SO3. b) Removing an H+ from C5H5NH+ gives the formula of the conjugate base: C5H5N c) Adding an H+ to CO32– gives the formula of the conjugate acid: HCO3–. d) Removing an H+ from HCN gives the formula of the conjugate base: CN–.

18.2A

Plan: The two possible reactions involve reacting the acid from one conjugate pair with the base from the other conjugate pair. The reaction that favors the products (Kc > 1) is the one in which the stronger acid produces the weaker acid. The reaction that favors reactants (Kc < 1) is the reaction is which the weaker acid produces the stronger acid. Solution: a) The conjugate pairs are H2SO3 (acid)/ HSO3– (base) and HCO3– (acid)/ CO32– (base). Two reactions are possible: (1) H2SO3 + CO32– ⇆ HSO3– + HCO3– and (2) HSO3– + HCO3– ⇆ H2SO3 + CO32– The first reaction is the reverse of the second. Both acids are weak. Of the two, H2SO3 is the stronger acid. Reaction (1) with the stronger acid producing the weaker acid favors products and Kc > 1. Reaction (2) with the weaker acid forming the stronger acid favors the reactants and Kc < 1. Therefore, reaction 1 is the reaction in which Kc > 1. b) The conjugate pairs are HF (acid)/F– (base) and HCN (acid)/CN– (base). Two reactions are possible: (1) HF + CN– ⇆ F– + HCN and (2) F– + HCN ⇆ HF + CN– The first reaction is the reverse of the second. Both acids are weak. Of the two, HF is the stronger acid. Reaction (1) with the stronger acid producing the weaker acid favors products and Kc > 1. Reaction (2) with the weaker acid forming the stronger acid favors the reactants and Kc < 1. Therefore, reaction 2 is the reaction in which Kc < 1.

18.2B

Plan: For a), write the reaction that shows the reaction of ammonia with water; for b), write a reaction between ammonia and HCl; for c), write the reaction between the ammonium ion and NaOH to produce ammonia. Solution: a) The following equation describes the dissolution of ammonia in water: NH3(g) + H2O(l) ⇆ NH4+(aq) + OH–(aq) weak base weak acid ← stronger acid strong base Ammonia is a known weak base, so it makes sense that it accepts an H+ from H2O. The reaction arrow indicates that the equilibrium lies to the left because the question states, “you smell ammonia” (NH4+ and OH– are odorless). NH4+ and OH– are the stronger acid and base, so the reaction proceeds to the formation of the weaker acid and base.

18-1

b) The addition of excess HCl results in the following equation: NH3(g) + H3O+(aq; from HCl) ⇆ NH4+(aq) + H2O(l) stronger base strong acid → weak acid weak base HCl is a strong acid and is much stronger than NH4+. Similarly, NH3 is a stronger base than H2O. The reaction proceeds to produce the weak acid and base, and thus the odor from NH3 disappears. c) The solution in (b) is mostly NH4+ and H2O. The addition of excess NaOH results in the following equation: NH4+(aq) + OH–(aq; from NaOH) ⇆ NH3(g) + H2O(l) stronger acid strong base → weak base weak acid NH4+ and OH– are the stronger acid and base, respectively, and drive the reaction towards the formation of the weaker base and acid, NH3(g) and H2O, respectively. The reaction direction explains the return of the ammonia odor. 18.3A

Plan: If HA is a stronger acid than HB, Kc > 1 and more HA molecules will produce HB molecules. If HB is a stronger acid than HA, Kc < 1 and more HB molecules will produce HA molecules. Solution: There are more HB molecules than there are HA molecules, so the equilibrium lies to the right and Kc > 1. HA is the stronger acid.

18.3B

Plan: Because HD is a stronger acid than HC, the reaction of HD and C– will have Kc > 1, and there should be more HC molecules than HD molecules at equilibrium. Solution: There are more green/white acid molecules in the solution than black/white acid molecules. Therefore, the green/white acid molecules represent HC, and the black/white acid molecules represent HD. The green spheres represent C–, and the black spheres represent D–. Because the reaction of the stronger acid HD with C– will have Kc > 1, the reverse reaction (HC + D–) will have Kc < 1.

18.4A

Plan: The product of [H3O+] and [OH–] remains constant at 25°C because the value of Kw is constant at a given temperature. Use Kw = [H3O+][OH–] = 1.0 × 10–14 to solve for [H3O+]. Solution: Calculating [H3O+]: Kw 1.0 ×10−14 [H3O+] = = = 1.4925×10–13 = 1.5×10–13 M − [OH ] 6.7 ×10−2 Since [OH–] > [H3O+], the solution is basic.

18.4B

Plan: The product of [H3O+] and [OH–] remains constant at 25°C because the value of Kw is constant at a given temperature. Use Kw = [H3O+][OH–] = 1.0 × 10–14 to solve for [H3O+]. Solution: Calculating [OH–]: Kw 1.0 ×10−14 [OH–] = = = 5.55555×10–5 = 5.6×10–5 M + −10 [H 3 O ] 1.8×10 Since [OH–] > [H3O+], the solution is basic.

18.5A

Plan: NaOH is a strong base that dissociates completely in water. Subtract pOH from 14.00 to find the pH, and calculate inverse logs of pOH and pH to find [OH–] and [H3O+], respectively. Alternatively, use the ion product constant of water (KW) at 25°C to calculate [H3O+] from [OH–]. Solution: pH + pOH = 14.00 pH = 14.00 – pOH = 14.00 – 4.48 = 9.52 pOH = –log [OH–] [OH–] = 10–pOH = 10–4.48 = 3.3113×10–5 = 3.3×10–5 M

18-2

18.5B

18.6A

pH = –log [H3O+] [H3O+] = 10–pH = 10–9.52 = 3.01995×10–10 = 3.0×10–10 M Kw 1.0 ×10−14 or [H3O+] = = 3.01996×10–10 = 3.0×10–10 M = ⎡ OH - ⎤ 3.3113×10−5 ⎣⎢ ⎦⎥ Plan: HCl is a strong acid that dissociates completely in water. Subtract pH from 14.00 to find the pOH, and calculate inverse logs of pH and pOH to find [H3O+] and [OH–], respectively. Solution: pH + pOH = 14.00 pOH = 14.00 – 2.28 = 11.72 pH = –log [H3O+] [H3O+] = 10–pH = 10–2.28 = 5.2481×10–3 = 5.2×10–3 M pOH = –log [OH–] [OH–] = 10–pOH = 10–11.72 = 1.9055×10–12 = 1.9×10–12 M Plan: Write a balanced equation for the dissociation of NH4+ in water. Using the given information, construct a reaction table that describes the initial and equilibrium concentrations. Construct an equilibrium expression and make assumptions where possible to simplify the calculations. Since the pH is known, [H3O+] can be found; that value can be substituted into the equilibrium expression. Solution: NH4+(aq) + H2O(l) ⇆ H3O+(aq) + NH3(g) Initial 0.2 M ——— 0 0 Change –x ——— +x +x Equilibrium 0.2 – x ——— x x The initial concentration of NH4+ = 0.2 M because each mole of NH4Cl completely dissociates to form one mole of NH4+. x = [H3O+] = [NH3] = 10–pH = 10–5.0 = 1.0×10–5 M Ka =

18.6B

[ NH 3 ][ H 3 O+ ]

[ NH +4 ]

(

x2 0.2 − x

(1.0×10 )(1.0×10 ) = 5.00250×10 = 5×10 (0.2 − 1.0×10 ) −5

)

−5

–10

−5

Plan: Write a balanced equation for the dissociation of acrylic acid in water. Using the given information, construct a reaction table that describes the initial and equilibrium concentrations. Construct an equilibrium expression. Since the pH is known, [H3O+] can be found; that value can be used to find the equilibrium concentrations of all substances, which can then be substituted into the equilibrium expression to solve for the value of Ka. Solution: H2C=CHCOOH(aq) + H2O(l) ⇆ H3O+(aq) + H2C=CHCOO– (aq) Initial 0.3 M ——— 0 0 Change –x ——— +x +x Equilibrium 0.3 – x ——— x x According to the information given in the problem, pH at equilibrium = 2.43. [H3O+]eq = 10–pH = 10–2.43 = 3.7154×10–3 = 3.7×10–3 M = x Thus, [H3O+] = [H2C=CHCOO–] = 3.7×10–3 M [H2C=CHCOOH] = (0.30 – x) = (0.30 – 3.7×10–3) M = 0.2963 M [ H CHCOO – ][H 3 O+ ] Ka = 2 [H 2 CHCOOH] Ka =

(3.7×10 –3 )(3.7×10 –3 ) = 4.6203×10–5 = 4.6×10–5 (0.2963)

18-3

18.7A

Plan: Write a balanced equation for the dissociation of HOCN in water. Using the given information, construct a table that describes the initial and equilibrium concentrations. Construct an equilibrium expression and solve the quadratic expression for x, the concentration of H3O+. Use the concentration of the hydronium ion to solve for pH. Solution: HOCN(aq) + H2O(l) ⇆ H3O+(aq) + OCN–(aq) Initial 0.10 M ——— 0 0 Change –x ——— +x +x Equilibrium 0.10 – x ——— x x − + 2 OCN H O [ ] [ ] x 3 Ka = 3.5×10–4 = = [ HOCN ] 0.10 − x

(

)

In this example, the dissociation of HOCN is not negligible in comparison to the initial concentration. Therefore, the equilibrium expression is solved using the quadratic formula. x2 = 3.5×10–4 (0.10 – x) x2 = 3.5×10–5 – 3.5×10–4 x x2 + 3.5×10–4 x – 3.5×10–5 = 0 (ax2 + bx + c = 0) –4 –5 a = 1 b = 3.5×10 c = –3.5×10 −b ± b 2 − 4ac

−3.5×10−4 ±

(3.5×10 ) − 4 (1)(−3.5×10 ) −4

−5

2 (1) –3

x = 5.7436675×10 = 5.7×10–3 M H3O+ pH = –log [H3O+] = –log [5.7436675×10–3] = 2.2408 = 2.24 18.7B

Plan: Write a balanced equation for the dissociation of C6H5COOH in water. Using the given information, construct a table that describes the initial and equilibrium concentrations. Use pKa to solve for the value of Ka. Construct an equilibrium expression, use simplifying assumptions when possible to solve for x, the concentration of H3O+. Use the concentration of the hydronium ion to solve for pH. Solution: C6H5COOH(aq) + H2O(l) ⇆ H3O+(aq) + C6H5COOH–(aq) Initial 0.25 M ——— 0 0 Change –x ——— +x +x Equilibrium 0.25 – x ——— x x Ka = 10–pKa = 10–4.20 = 6.3096×10–5 = 6.3×10–5 Ka = 6.3×10–5 =

[C6 H 5 HCOO – ][H 3 O+ ] [C6 H 5 COOH]

(x)(x) Assume x is negligible so 0.25 – x ≈ 0.25 (0.25 – x)

(x)(x) = 6.3×10–5 (0.25) x2 = (6.3×10–5) (0.25); x = 3.9686×10–3 = 4.0×10–3 Check the assumption by calculating the % error: 4.0×10 –3 (100) = 1.6% which is smaller than 5%, so the assumption is valid. 0.25 At equilibrium [H3O+]eq = 4.0×10–3 M pH = –log [H3O+] = –log [3.9686×10–3] = 2.40136 = 2.40

18-4

18.8A

Plan: Write the acid-dissociation reaction and the expression for Ka. Convert pKa to Ka. Set up a reaction table in which x = the concentration of the dissociated acid and also [H3O+]. Use the expression for Ka to solve for x, the concentration of cyanide ion at equilibrium. Then use the initial concentration of HCN and the equilibrium concentration of CN– to find % dissociation. Solution: Concentration HCN(aq) + H2O(l) ⇆ H3O+(aq) + CN–(aq) Initial 0.75 — 0 0 Change –x +x +x Equilibrium 0.75 – x x x Ka = 10−pKa = 10−9.21 = 6.2 ×10−10

Ka = 6.2×10–10 = Ka = 6.2×10–10 =

[ CN – ][H 3 O+ ] [HCN] [ x ][x]

Assume x is small compared to 0.75. [0.75 – x] [ x ][x] Ka = 6.2×10–10 = [0.75] x = 2.1564×10–5 = 2.2×10–5 M Check the assumption by calculating the % error: 2.2×10 –5 (100) = 0.0029% which is smaller than 5%, so the assumption is valid. 0.75 [HCN]dissoc (100) Percent HCN dissociated = [HCN]init Percent HCN dissociated = 18.8B

(2.1564 ×10 –5 ) (100) = .0028752 = 0.0029% 0.75

Plan: Write the acid-dissociation reaction and the expression for Ka. Percent dissociation refers to the amount of the initial concentration of the acid that dissociates into ions. Use the percent dissociation to find the concentration of acid dissociated, which also equals [H3O+]. HA will be used as the formula of the acid. Set up a reaction table in which x = the concentration of the dissociated acid and [H3O+]. Substitute [HA], [A–], and [H3O+] into the expression for Ka to find the value of Ka. Solution: HA(aq) + H2O(l) ⇆ H3O+(aq) + A–(aq) dissociated acid Percent HA = (100) initial acid 3.16% =

x 1.5 M

(100)

[Dissociated acid] = x = 0.0474 M Concentration HA(aq) + H2O(l) ⇆ H3O+(aq) + A–(aq) Initial: 1.5 0 0 Change: –x +x +x Equilibrium: 1.5 – x x x [Dissociated acid] = x = [A–] = [H3O+] = 0.0474 M [HA] = 1.5 M – 0.0474 M = 1.4526 M Solving for Ka. In the equilibrium expression, substitute the concentrations above and calculate Ka. [ H O+ ][ A− ] (0.0474)(0.0474) = 1.5467×10–3 = 1.5×10–3 Ka = 3 = [ HA ] (1.4526) Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

18-5

18.9A

Plan: Write the balanced equation and corresponding equilibrium expression for each dissociation reaction. Calculate the equilibrium concentrations of all species and convert [H3O+] to pH. Find the equilibrium constant values from Appendix C, Ka1 = 5.6×10–2 and Ka2 = 5.4×10–5. Solution: HOOC−COOH(aq) + H2O(l) ⇆ HOOC−COO– (aq) + H3O+(aq) [ HC 2 O 4− ][ H 3 O+ ] = 5.6×10–2 Ka1 = H C O [ 2 2 4] HOOC−COO– (aq) + H2O(l) ⇆ –OOC−COO– (aq) + H3O+(aq) ⎡ C 2 O 4 2− ⎤ [ H 3 O + ] ⎦ = 5.4×10–5 Ka2 = ⎣ [ HC2 O 4− ] Assumptions: 1) Since Ka1 >> Ka2, the first dissociation produces almost all of the H3O+, so [H3O+]eq = [H3O+] from C2H2O4. 2) Since Ka1 (5.6×10–2) is fairly large, solve the first equilibrium expression using the quadratic equation. HOOC−COOH(aq) + H2O(l) ⇆ H3O+(aq) + HOOC−COO– (aq) Initial 0.150 M ——— 0 0 Change –x ——— +x +x Equilibrium 0.150 – x ——— x x 2 [ HC 2 O 4− ][ H 3 O+ ] x = Ka1 = = 5.6×10–2 [H2 C2 O4 ] 0.150 − x

(

)

–2

–3

(ax2 + bx + c = 0)

x + 5.6×10 x – 8.4×10 = 0 −5.6 ×10−2 ±

(5.6×10 ) − 4 (1)(−8.4 ×10 ) −2

−3

2 (1) +

x = 0.067833 M H3O Therefore, [H3O+] = [HC2O4–] = 0.068 M and pH = –log (0.067833) = 1.16856 = 1.17. The oxalic acid concentration at equilibrium is [H2C2O4]init – [H2C2O4]dissoc = 0.150 – 0.067833 = 0.82167 = 0.082 M. Solve for [C2O42–] by rearranging the Ka2 expression: ⎡ C 2 O 4 2− ⎤ [ H 3 O + ] ⎦ Ka2 = ⎣ = 5.4×10–5 [ HC2 O 4− ] ⎡ C 2 O 4 2− ⎤ = ⎣ ⎦

18.9B

Ka2 [ HC 2 O 4− ]

[ H 3 O+ ]

(5.4 ×10 )(0.067833) −5

(0.067833)

= 5.4×10–5 M

[ H 3 O+ ][ HCO3 – ] = 4.5×10–7 [ H 2 CO3 ]

HCO3– (aq) + H2O(l) ⇆ CO32–(aq) + H3O+(aq) [ H3 O+ ] ⎡⎣CO32– ⎤⎦ Ka2 = = 4.7×10–11 HCO [ 3 ]

18-6

Assumption: 1) Since Ka1 >> Ka2, the first dissociation produces almost all of the H3O+, so [H3O+]eq = [H3O+] from H2CO3. 2) Because Ka1 (4.7×10–7) is fairly small, [H2CO3]init – x ≈ [H2CO3]init. Thus, [H2CO3] = 0.075 M – x ≈ 0.075 M Solve the first equilibrium expression making the assumption that x is small. H2O(l) ⇆ H3O+(aq) + HCO3–(aq) H2CO3(aq) + ——— 0 0 Initial 0.075 M Change –x ——— +x +x Equilibrium 0.075 – x ——— x x [ H 3O+ ][ HCO3 – ] (x)(x) (x)(x) Ka1 = = 4.5×10–7 = ≈ (0.075 – x) (0.075) [ H 2 CO3 ] x2 = (0.075)(4.5×10–7); x = 1.8371×10-4 = 1.8×10-4 M Check the assumption by calculating the % error: 1.8×10 –4 (100) = 0.24% which is smaller than 5%, so the assumption is valid. 0.075 Therefore, [H3O+] = [HCO3–] = 1.8×10−4 M and pH = –log (1.8371×10−4) = 3.73587 = 3.74. The carbonic acid concentration at equilibrium is [H2CO3]init – [H2CO3]dissoc = 0.075 – 1.8371×10−4 = 0.07482 = 0.075 M = [H2CO3]. Solve for [CO32–] by rearranging the Ka2 expression: [ H3 O+ ] ⎡⎣CO32– ⎤⎦ Ka2 = = 4.7×10–11 [ HCO3 – ] [CO32–] =

Ka2 ⎡⎣ HCO3 ⎤⎦ (4.7×10 –11 )(1.8371×10 –4 ) = 4.7×10–11 M = –4 ⎡ H O+ ⎤ × (1.8371 10 ) ⎣⎢ 3 ⎦⎥

18.10A Plan: Pyridine contains a nitrogen atom that accepts H+ from water to form OH– ions in aqueous solution. Write a balanced equation and equilibrium expression for the reaction, convert pKb to Kb, make simplifying assumptions (if valid), and solve for [OH–]. Calculate [H3O+] using [H3O+][OH–] = 1.0×10–14 and convert to pH. Solution: Kb = 10−pK b = 10–8.77 = 1.69824×10–9 C5H5NH+(aq) + OH–(aq) C5H5N(aq) + H2O(l) ⇆ ——— 0 0 Initial 0.10 M Change –x ——— +x +x Equilibrium 0.10 – x ——— x x + − [C 5 N 5 NH ][ OH ] Kb = = 1.69824×10–9 [C5 N 5 N ] Assume that 0.10 – x ≈ 0.10. [C 5 N 5 NH + ][ OH− ] x2 Kb = = = 1.69824×10–9 C N N 0.10 ( ) [ 5 5 ] x = 1.303165×10–5 = 1.3×10–5 M = [OH–] = [C5H5NH+] [OH− ] 1.303265×10−5 (100) = Since (100) = 0.01313 which < 5%, the assumption that the dissociation of 0.10 [C5 H 5 N 5 ] C5H5N5 is small is valid.

1.0 ×10−14 = 7.67362×10–10 M 1.303165×10−5 [OH− ] pH = –log (7.67362×10–10) = 9.1149995 = 9.11 (Since pyridine is a weak base, a pH > 7 is expected.) [H3O+] =

18-7

18.10B Plan: Amphetamine contains a nitrogen atom that accepts H+ from water to form OH– ions in aqueous solution. Write a balanced equation and equilibrium expression for the reaction, make simplifying assumptions (if valid), and solve for [OH–]. Calculate [H3O+] using [H3O+][OH–] = 1.0×10–14 and convert to pH. In the information below, the symbol B will be used to represent the formula of amphetamine. Solution: B(aq) + H2O(l) ⇆ BH+(aq) + OH–(aq) ——— 0 0 Initial 0.075 M Change –x ——— +x +x Equilibrium 0.075 – x ——— x x [ BH ][ OH – ] Kb = = 6.3×10–5 [ B] Assume that 0.075 – x ≈ 0.075. [ BH ][ OH – ] x2 Kb = = 6.3×10–5 = [ B] (0.075) x = 0.0021737 = 2.2×10–3 M = [OH–] = [BH+] Check the assumption by calculating the % error: 2.2×10 –3 (100) = 2.9% which is smaller than 5%, so the assumption is valid. 0.075

1.0 ×10 –14 Kw = = 4.60045×10–12 M − [OH ] 2.1737×10 –3 pH = –log (4.60045×10–12) = 11.3372 = 11.34 [H3O+] =

Since amphetamine is a weak base, a pH > 7 is expected. 18.11A Plan: The hypochlorite ion, ClO–, acts as a weak base in water. Write a balanced equation and equilibrium expression for this reaction. The Kb of ClO– is calculated from the Ka of its conjugate acid, hypochlorous acid, HClO (from Appendix C, Ka = 2.9×10–8). Make simplifying assumptions (if valid), solve for [OH–], convert to [H3O+] and calculate pH. Solution: ClO–(aq) + H2O(l) ⇆ HClO(aq) + OH–(aq) ——— 0 0 Initial 0.20 M Change –x ——— +x +x Equilibrium 0.20 – x ——— x x [ HClO ][ OH − ] Kb = [ ClO− ]

Kw 1.0×10−14 = = 3.448276×10–7 −8 Ka 2.9×10 Since Kb is very small, assume [ClO–]eq = 0.20 – x ≈ 0.2. [ HClO ][ OH − ] x2 Kb = = = 3.448276×10–7 (0.20) [ ClO− ] x = 2.6261×10–4 Therefore, [HClO] = [OH–] = 2.6×10–4 M. Kb =

2.6261×10−4 [OH− ] = 100 ( ) (100) = 0.13% which is < 5%, the assumption that the dissociation of ClO– is 0.20 [ClO− ] small is valid. Since

18-8

1.0×10−14 = 3.8079×10–11 M 2.6261×10−4 pH = –log (3.8079×10–11) = 10.4193 = 10.42 (Since hypochlorite ion is a weak base, a pH > 7 is expected.)

[H3O+] =

18.11B Plan: The nitrite ion, NO2–, acts as a weak base in water. Write a balanced equation and equilibrium expression for this reaction. The Kb of NO2– is calculated from the Ka of its conjugate acid, nitrous acid, HNO2 (from Appendix C, Ka = 7.1×10–4). Make simplifying assumptions (if valid), solve for [OH–], convert to [H3O+] and calculate pH. Solution: NO2–(aq) + H2O(l) ⇆ HNO2 (aq) + OH–(aq) ——— 0 0 Initial 0.80 M Change –x ——— +x +x Equilibrium 0.80 – x ——— x x − [ HNO2 ][ OH ] Kb = [ NO2− ]

Kw 1.0×10−14 = = 1.40845×10–11 Ka 7.1×10−4 Since Kb is very small, assume [NO2–]eq = 0.80 – x ≈ 0.8. [ HNO2 ][ OH− ] x2 Kb = = = 1.40845×10–11 (0.80) [ NO2− ] Kb =

x = 3.356725×10−6 M Check the assumption by calculating the % error: 3.3×10 –6 (100) = 0.00041% which is smaller than 5%, so the assumption is valid. 0.80 Therefore, [HNO2] = [OH–] = 3.356725×10−6 M. 1.0 ×10 –14 = 2.97909×10–9 M [H3O+] = 3.356725×10 –6 pH = –log (2.97909×10–9) = 8.5259 = 8.53 Since nitrite ion is a weak base, a pH > 7 is expected. 18.12A Plan: Examine the cations and anions in each compound. If the cation is the cation of a strong base, the cation gives a neutral solution; the cation of a weak base gives an acidic solution. An anion of a strong acid gives a neutral solution while an anion of a weak acid is basic in solution. Solution: a) The ions are K+ and ClO2–; the K+ is from the strong base KOH, and does not react with water. The ClO2– is from the weak acid HClO2, so it reacts with water to produce OH– ions. Since the base is strong and the acid is weak, the salt derived from this combination will produce a basic solution. K+ does not react with water. ClO2–(aq) + H2O(l) ⇆ HClO2(aq) + OH–(aq) b) The ions are CH3NH3+ and NO3–; CH3NH3+ is derived from the weak base methylamine, CH3NH2. Nitrate ion, NO3–, is derived from the strong acid HNO3 (nitric acid). A salt derived from a weak base and strong acid produces an acidic solution. NO3– does not react with water. CH3NH3+(aq) + H2O(l) ⇆ CH3NH2(aq) + H3O+(aq)

18-9

c) The ions are Rb+ and Br–. Rubidium ion is derived from rubidium hydroxide, RbOH, which is a strong base because Rb is a Group 1A(1) metal. Bromide ion is derived from hydrobromic acid, HBr, a strong hydrohalic acid. Since both the base and acid are strong, the salt derived from this combination will produce a neutral solution. Neither Rb+ nor Br– react with water. 18.12B Plan: Examine the cations and anions in each compound. If the cation is the cation of a strong base, the cation gives a neutral solution; the cation of a weak base gives an acidic solution. An anion of a strong acid gives a neutral solution while an anion of a weak acid is basic in solution. Solution: a) The ions are Fe3+ and Br–. The Br– is the anion of the strong acid HBr, so it does not react with water. The Fe3+ ion is small and highly charged, so the hydrated ion, Fe(H2O)63+, reacts with water to produce H3O+. Since the base is weak and the acid is strong, the salt derived from this combination will produce an acidic solution. Br– does not react with water. Fe(H2O)63+(aq) + H2O(l) ⇆ Fe(H2O)5OH2+ (aq) + H3O+ (aq) b) The ions are Ca2+ and NO2–; the Ca2+ is from the strong base Ca(OH)2, and does not react with water. The NO2– is from the weak acid HNO2, so it reacts with water to produce OH– ions. Since the base is strong and the acid is weak, the salt derived from this combination will produce a basic solution. Ca2+ does not react with water. NO2–(aq) + H2O(l) ⇆ HNO2(aq) + OH–(aq) c) The ions are C6H5NH3+ and I–; C6H5NH3+ is derived from the weak base aniline, C6H5NH2. Iodide ion, I–, is derived from the strong acid HI (hydroiodic acid). A salt derived from a weak base and strong acid produces an acidic solution. I– does not react with water. C6H5NH3+(aq) + H2O(l) ⇆ C6H5NH2(aq) + H3O+(aq) 18.13A Plan: Examine the cations and anions in each compound. If the cation is the cation of a strong base, the cation gives a neutral solution; the cation of a weak base gives an acidic solution. An anion of a strong acid gives a neutral solution while an anion of a weak acid is basic in solution. Solution: a) The two ions that comprise this salt are cupric ion, Cu2+, and acetate ion, CH3COO–. Metal ions are acidic in water. Assume that the hydrated cation is Cu(H2O)62+. The Ka is found in Appendix C. Cu(H2O)62+(aq) + H2O(l) ⇆ Cu(H2O)5OH+(aq) + H3O+(aq)

Ka = 3×10–8

Acetate ion acts likes a base in water. The Kb is calculated from the Ka of acetic acid (1.8×10–5): K 1.0×10−14 Kb = w = = 5.6×10–10 Ka 1.8×10−5 Kb = 5.6×10–10 CH3COO–(aq) + H2O(l) ⇆ CH3COOH(aq) + OH–(aq) – 2+ Cu(H2O)6 is a better proton donor than CH3COO is a proton acceptor (i.e., Ka > Kb), so a solution of Cu(CH3COO)2 is acidic. b) The two ions that comprise this salt are ammonium ion, NH4+, and fluoride ion, F–. Ammonium ion is the acid

Kw 1.0×10−14 = = 5.7×10–10. −5 Kb 1.76×10 Ka = 5.7×10–10 NH4+(aq) + H2O(l) ⇆ NH3(aq) + H3O+(aq) K 1.0×10−14 = 1.5×10–11. Fluoride ion is the base with Kb = w = −4 Ka 6.8×10 Kb = 1.5×10–11 F–(aq) + H2O(l) ⇆ HF(aq) + OH–(aq) Since Ka > Kb, a solution of NH4F is acidic. of NH3 with Ka =

18-10

c) The ions are K+ and HC6H6O6–; the K+ is from the strong base KOH, and does not react with water. The HC6H6O6– can react as an acid: Ka = 5×10–12 (from Appendix C) HC6H6O6–(aq) + H2O(l) ⇆ C6H6O62–(aq) + H3O+(aq) – HC6H6O6 can also react as a base. Its Kb value can be found by using the Ka of its conjugate acid, H2C6H6O6. Kw 1.0×10−14 Kb = = = 1.0×10–9 HC6H6O6–(aq) + H2O(l) ⇆ H2C6H6O6(aq) + OH–(aq) Ka of H2 C6 H6 O6 1.0×10−5 Since Kb > Ka, a solution of KHC6H6O6 is basic. 18.13B Plan: Examine the cations and anions in each compound. If the cation is the cation of a strong base, the cation gives a neutral solution; the cation of a weak base gives an acidic solution. An anion of a strong acid gives a neutral solution while an anion of a weak acid is basic in solution. Solution: a) The two ions that comprise this salt are sodium ion, Na+, and bicarbonate ion, HCO3–. The Na+ is from the strong base NaOH, and does not react with water. The HCO3– can react as an acid: Ka = 4.7×10–11 HCO3–(aq) + H2O(l) ⇆ CO32–(aq) + H3O+(aq) – HCO3 can also react as a base. Its Kb value can be found by using the Ka of its conjugate acid, H2CO3. K 1.0×10−14 Kb = w = = 2.2×10–8 HCO3–(aq) + H2O(l) ⇆ H2CO3(aq) + OH–(aq) Ka 4.5×10−7 Since Kb > Ka, a solution of NaHCO3 is basic. b) The two ions that comprise this salt are anilinium ion, C6H5NH3+, and nitrite ion, NO2–. K 1.0×10−14 = 2.5×10–5. Anilinium ion is the acid of C6H5NH2 with Ka = w = −10 Kb 4.0×10 Ka = 2.5×10–5 C6H5NH3+(aq) + H2O(l) ⇆ C6H5NH2(aq) + H3O+(aq) Nitrite ion is the base with Kb =

Kw 1.0×10−14 = = 1.4×10–11. Ka 7.1×10−4

NO2–(aq) + H2O(l) ⇆ HNO2(aq) + OH–(aq) Since Ka > Kb, a solution of C6H5NH3NO2 is acidic.

Kb = 1.4×10–11

c) The ions are Na+ and H2PO4–; the Na+ is from the strong base NaOH, and does not react with water. The H2PO4– can react as an acid: Ka = 6.3×10–8 H2PO4–(aq) + H2O(l) ⇆ HPO42–(aq) + H3O+(aq) – H2PO4 can also react as a base. Its Kb value can be found by using the Ka of its conjugate acid, H3PO4. K 1.0×10−14 Kb = w = = 1.4×10–12 H2PO4–(aq) + H2O(l) ⇆ H3PO4(aq) + OH–(aq) −3 Ka 7.2×10 Since Ka > Kb, a solution of NaH2PO4 is acidic. 18.14A Plan: A Lewis acid is an electron-pair acceptor while a Lewis base is an electron-pair donor. Solution: a) −

Al HO

−

HO OH

trigonal planar

OH OH

tetrahedral

18-11

Hydroxide ion, OH–, donates an electron pair and is the Lewis base; Al(OH)3 accepts the electron pair and is the Lewis acid. Note the change in geometry caused by the formation of the adduct. b) O +

S O

H O

O H

H H

Sulfur trioxide accepts the electron pair and is the Lewis acid. Water donates an electron pair and is the Lewis base. c) NH3 Co

H3N

3+ + 6 H

H H

H3N

3+ NH3 NH3

NH3

Co3+ accepts six electron pairs and is the Lewis acid. Ammonia donates an electron pair and is the Lewis base. 18.14B Plan: A Lewis acid is an electron-pair acceptor while a Lewis base is an electron-pair donor. Solution: a) B(OH)3 is the Lewis acid because it is accepting electron pairs from water, the Lewis base. b) Cd2+ accepts four electron pairs and is the Lewis acid. Each iodide ion donates an electron pair and is the Lewis base. c) Each fluoride ion donates an electron pair to form a bond with boron in SiF62–. The fluoride ion is the Lewis base and the boron tetrafluoride is the Lewis acid. END–OF–CHAPTER PROBLEMS

18.1

The Arrhenius definition classifies substances as being acids or bases by their behavior in the solvent water.

18.2

All Arrhenius acids contain hydrogen and produce hydronium ion (H3O+) in aqueous solution. All Arrhenius bases contain an OH group and produce hydroxide ion (OH–) in aqueous solution. Neutralization occurs when each H3O+ molecule combines with an OH– molecule to form two molecules of H2O. Chemists found that the ΔHrxn was independent of the combination of strong acid with strong base. In other words, the reaction of any strong base with any strong acid always produced 56 kJ/mol (ΔH = –56 kJ/mol). This was consistent with Arrhenius’s hypothesis describing neutralization, because all other counter ions (those present from the dissociation of the strong acid and base) were spectators and did not participate in the overall reaction.

18.3

The Arrhenius acid-base definition is limited by the fact that it only classifies substances as an acid or base when dissolved in the single solvent water. The anhydrous neutralization of NH3(g) and HCl(g) would not be included in the Arrhenius acid-base concept. In addition, it limits a base to a substance that contains OH in its formula. NH3 does not contain OH in its formula but produces OH– ions in H2O.

18.4

Plan: Recall that an Arrhenius acid contains hydrogen and produces hydronium ion (H3O+) in aqueous solution. Solution: a) Water, H2O, is an Arrhenius acid because it produces H3O+ ion in aqueous solution. Water is also an Arrhenius base because it produces the OH– ion as well. b) Calcium hydroxide, Ca(OH)2 is a base, not an acid.

18-12

c) Phosphorous acid, H3PO3, is a weak Arrhenius acid. It is weak because the number of O atoms equals the number of ionizable H atoms. d) Hydroiodic acid, HI, is a strong Arrhenius acid. 18.5

Only (a) NaHSO4

18.6

Plan: All Arrhenius bases contain an OH group and produce hydroxide ion (OH–) in aqueous solution. Solution: Barium hydroxide, Ba(OH)2, and potassium hydroxide, KOH, (b and d) are Arrhenius bases because they contain hydroxide ions and form OH– when dissolved in water. H3AsO4 and HClO, (a) and (c), are Arrhenius acids, not bases.

18.7

(b) H2O is a very weak Arrhenius base.

18.8

The Brønsted-Lowry theory defines acids as proton donors and bases as proton acceptors, while the Arrhenius definition looks at acids as containing ionizable H atoms and at bases as containing hydroxide ions. In both definitions, an acid produces hydronium ions and a base produces hydroxide ions when added to water. Ammonia, NH3, and carbonate ion, CO32–, are two Brønsted-Lowry bases that are not Arrhenius bases because they do not contain hydroxide ions. Brønsted-Lowry acids must contain an ionizable H atom in order to be proton donors, so a Brønsted-Lowry acid that is not an Arrhenius acid cannot be identified. (Other examples are also acceptable.)

18.9

Every acid has a conjugate base, and every base has a conjugate acid. The acid has one more H and one more positive charge than the base from which it was formed.

18.10

a) Acid-base reactions are proton transfer processes. Thus, the proton will be transferred from the stronger acid to the stronger base to form the weaker acid and weaker base. b) HB(aq) + A– (aq) → HA(aq) + B– (aq) The spontaneous direction of a Brønsted-Lowry acid-base reaction is that the stronger acid will transfer a proton to the stronger base to produce the weaker acid and base. Thus at equilibrium there should be relatively more of weaker acid and base present than there will be of the stronger acid and base. Since there is more HA and B– in sample and less HB and A–, HB must be the stronger acid and A– must be the stronger base.

18.11

An amphoteric substance can act as either an acid or a base. In the presence of a strong base (OH–), the dihydrogen phosphate ion acts like an acid by donating hydrogen: H2PO4–(aq) + OH–(aq) → H2O(aq) + HPO42–(aq) In the presence of a strong acid (HCl), the dihydrogen phosphate ion acts like a base by accepting hydrogen: H2PO4–(aq) + HCl(aq) → H3PO4(aq) + Cl–(aq)

18.12

Plan: To derive the conjugate base, remove one H from the acid and decrease the charge by 1 (acids donate H+). Since each formula in this problem is neutral, the conjugate base will have a charge of –1. Solution: a) Cl– b) HCO3– c) OH–

18.13

a) PO43–

18.14

Plan: To derive the conjugate acid, add an H and increase the charge by 1 (bases accept H+). Solution: a) NH4+ b) NH3 c) C10H14N2H+

18.15

a) OH–

b) NH3

b) HSO4–

c) S2–

c) H3O +

18-13

18.16

Plan: The acid donates the proton to form its conjugate base; the base accepts a proton to form its conjugate acid. Solution: a) HCl + H2O ⇆ Cl– + H3O+ acid base conjugate base conjugate acid Conjugate acid-base pairs: HCl/Cl– and H3O+/H2O ⇆ ClO4– + H3SO4+ b) HClO4 + H2SO4 acid base conjugate base conjugate acid Conjugate acid-base pairs: HClO4/ClO4– and H3SO4+/H2SO4 Note: Perchloric acid is able to protonate another strong acid, H2SO4, because perchloric acid is a stronger acid. (HClO4’s oxygen atoms exceed its hydrogen atoms by one more than H2SO4.) ⇆ H2PO4– + HSO4– c) HPO42– + H2SO4 base acid conjugate acid conjugate base Conjugate acid-base pairs: H2SO4/HSO4– and H2PO4–/HPO42–

18.17

HNO3 ⇆ NH4+ + NO3– acid conjugate acid conjugate base Conjugate pairs: HNO3/NO3–; NH4+ /NH3 + H2O ⇆ OH– + OH– b) O2– base acid conjugate acid conjugate base Conjugate pairs: OH–/O2–; H2O/OH– + BrO3– ⇆ NH3 + HBrO3 c) NH4+ acid base conjugate base conjugate acid Conjugate pairs: NH4+/NH3; HBrO3/BrO3–

18.18

Plan: The acid donates the proton to form its conjugate base; the base accepts a proton to form its conjugate acid. Solution: a) NH3 + H3PO4 ⇆ NH4+ + H2PO4– base acid conjugate acid conjugate base Conjugate acid-base pairs: H3PO4/H2PO4–; NH4+/NH3 + NH3 ⇆ CH3OH + NH2– b) CH3O– base acid conjugate acid conjugate base Conjugate acid-base pairs: NH3/NH2–; CH3OH/CH3O– + HSO4– ⇆ H2PO4– + SO42– c) HPO42– base acid conjugate acid conjugate base Conjugate acid-base pairs: HSO4–/SO42–; H2PO4–/HPO42–

18.19

CN– ⇆ NH3 + HCN base conjugate base conjugate acid Conjugate acid-base pairs: NH4+/NH3; HCN/CN– + HS– ⇆ OH– + H2S b) H2O acid base conjugate base conjugate acid Conjugate acid-base pairs: H2O/OH–; H2S/HS– – CH3NH2 ⇆ SO32– + CH3NH3+ c) HSO3 + acid base conjugate base conjugate acid Conjugate acid-base pairs: HSO3–/SO32–; CH3NH3+/CH3NH2

18.20

Plan: Write total ionic equations (show all soluble ionic substances as dissociated into ions) and then remove the spectator ions to write the net ionic equations. The (aq) subscript denotes that each species is soluble and dissociates in water. The acid donates the proton to form its conjugate base; the base accepts a proton to form its conjugate acid.

a) NH3 base

a) NH4+ acid

18-14

Solution: a) Na+(aq) + OH–(aq) + Na+(aq) + H2PO4–(aq) ⇆ H2O(l) + 2Na+(aq) + HPO42–(aq) Net: OH–(aq) + H2PO4–(aq) ⇆ H2O(l) + HPO42–(aq) base acid conjugate acid conjugate base Conjugate acid-base pairs: H2PO4–/HPO42– and H2O/OH– b) K+(aq) + HSO4–(aq) + 2K+(aq) + CO32–(aq) ⇆ 2K+(aq) + SO42–(aq) + K+(aq) + HCO3–(aq) Net: HSO4–(aq) + CO32–(aq) ⇆ SO42–(aq) + HCO3–(aq) acid base conjugate base conjugate acid Conjugate acid-base pairs: HSO4–/SO42– and HCO3–/CO32– 18.21

a) H3O+(aq) + CO32–(aq) ⇆ HCO3–(aq) + H2O(l) acid base conjugate acid conjugate base Conjugate acid-base pairs: H3O+/H2O; HCO3–/CO32– H2O(l) b) NH4+(aq) + OH–(aq) ⇆ NH3(aq) + acid base conjugate base conjugate acid Conjugate acid-base pairs: NH4+/NH3; H2O/OH–

18.22

18.23

Kc > 1: HNO3 + F – ⇆ NO3– + HF Kc < 1: NO3– + HF ⇆ HNO3 + F –

18.24

Plan: An acid-base reaction that favors the products (Kc > 1) is one in which the stronger acid produces the weaker acid. Use the figure to decide which of the two acids is the stronger acid. Solution: a) HCl + NH3 ⇆ NH4+ + Cl– strong acid stronger base weak acid weaker base HCl is ranked above NH4+ in the list of conjugate acid-base pair strength and is the stronger acid. NH3 is ranked above Cl– and is the stronger base. NH3 is shown as a “stronger” base because it is stronger than Cl–, but is not considered a “strong” base. The reaction proceeds toward the production of the weaker acid and base, i.e., the reaction as written proceeds to the right and Kc > 1. The stronger acid is more likely to donate a proton than the weaker acid. NH3 ⇆ HSO3– + NH4+ b) H2SO3 + stronger acid stronger base weaker base weaker acid H2SO3 is ranked above NH4+ and is the stronger acid. NH3 is a stronger base than HSO3–. The reaction proceeds toward the production of the weaker acid and base, i.e., the reaction as written proceeds to the right and Kc > 1.

18.25

Neither a or b have Kc > 1.

18.26

Plan: An acid-base reaction that favors the reactants (Kc < 1) is one in which the weaker acid produces the stronger acid. Use the figure to decide which of the two acids is the weaker acid.

18-15

Solution: a) NH4+ + HPO42– ⇆ NH3 + H2PO4– weaker acid weaker base stronger base stronger acid Kc < 1 The reaction proceeds toward the production of the weaker acid and base, i.e., the reaction as written proceeds to the left. HS– ⇆ H2SO3 + S2-– b) HSO3– + weaker base weaker acid stronger acid stronger base Kc < 1 The reaction proceeds toward the production of the weaker acid and base, i.e., the reaction as written proceeds to the left. 18.27

a) Kc < 1

b) Kc > 1

18.28

Autoionization reactions occur when a proton (or, less frequently, another ion) is transferred from one molecule of the substance to another molecule of the same substance. H2O(l) + H2O(l) ⇆ H3O+(aq) + OH– (aq) H2SO4(l) + H2SO4(l) ⇆ H3SO4+(solvated) + HSO4–(solvated)

18.29

H2O(l) + H2O(l) ⇆ H3O+(aq) + OH– (aq) [ H 3 O+ ][ OH− ] Kc = 2 [H 2 O] [H2O] is a constant and is included with the value of Kc to obtain Kw: Kw = [H2O]2 × Kc = [H3O+][OH– ]

18.30

a) pH increases by a value of 1. b) [H3O+] increases by a factor of 1000.

18.31

Plan: The lower the concentration of hydronium (H3O+) ions (the higher the concentration of hydroxide ions, OH–), the higher the pH. Solution: a) Bases produce OH– ions in solution, so the concentration of hydronium ion for a solution of a base solution is lower than that for a solution of an acid. The 0.01 M base solution has the higher pH. b) pOH equals –log [OH–]. At 25°C, the equilibrium constant for water ionization, Kw, equals 1 × 10–14 so 14 = pH + pOH. As pOH decreases, pH increases. The solution of pOH = 6.0 has the higher pH. c) Kw = 1 × 10–14 = [H3O+][OH–]. The higher the [OH–], the lower the [H3O+] and the higher the pH. The solution with [OH–] = 1 × 10–4 M has the higher pH.

18.32

Plan: The relationships are: pH = –log [H3O+] and [H3O+] = 10–pH ; pOH = –log [OH–] and [OH–] = 10–pOH ; and 14 = pH + pOH. Solution: a) [H3O+] = 10–pH = 10–9.85 = 1.4125375×10–10 = 1.4×10–10 M H3O+ pOH = 14.00 – pH = 14.00 – 9.85 = 4.15 [OH–] = 10–pOH = 10–4.15 = 7.0794578×10–5 = 7.1×10–5 M OH– b) pH = 14.00 – pOH = 14.00 – 9.43 = 4.57 [H3O+] = 10–pH = 10–4.57 = 2.691535×10–5 = 2.7×10–5 M H3O+ [OH–] = 10–pOH = 10–9.43 = 3.7153523×10–10 = 3.7×10–10 M OH–

18.33

a) [H3O+] = 10–pH = 10–3.47 = 3.38844×10–4 = 3.4×10–4 M H3O+ pOH = 14.00 – pH = 14.00 – 3.47 = 10.53 [OH–] = 10–pOH = 10–10.53 = 2.951209×10–11 = 3.0×10–11 M OH– b) pH = 14.00 – pOH = 14.00 – 4.33 = 9.67 [H3O+] = 10–pH = 10–9.67 = 2.13796×10–10 = 2.1×10–10 M H3O+ [OH–] = 10–pOH = 10–4.33 = 4.67735×10–5 = 4.7×10–5 M OH–

18-16

18.34

Plan: The relationships are: pH = –log [H3O+] and [H3O+] = 10–pH ; pOH = –log [OH–] and [OH–] = 10–pOH ; and 14 = pH + pOH. Solution: a) [H3O+] = 10–pH = 10–4.77 = 1.69824×10–5 = 1.7×10–5 M H3O+ pOH = 14.00 – pH = 14.00 – 4.77 = 9.23 [OH–] = 10–pOH = 10–9.23 = 5.8884×10–10 = 5.9×10–10 M OH– b) pH = 14.00 – pOH = 14.00 – 5.65 = 8.35 [H3O+] = 10–pH = 10–8.35 = 4.46684×10–9 = 4.5×10–9 M H3O+ [OH–] = 10–pOH = 10–5.65 = 2.23872×10–6 = 2.2×10–6 M OH–

18.35

a) [H3O+] = 10–pH = 10–8.97 = 1.071519×10–9 = 1.1×10–9 M H3O+ pOH = 14.00 – pH = 14.00 – 8.97 = 5.03 [OH–] = 10–pOH = 10–5.03 = 9.3325×10–6 = 9.3×10–6 M OH– b) pH = 14.00 – pOH = 14.00 – 11.27 = 2.73 [H3O+] = 10–pH = 10–2.73 = 1.862087×10–3 = 1.9×10–3 M H3O+ [OH–] = 10–pOH = 10–11.27 = 5.3703×10–12 = 5.4×10–12 M OH–

18.36

Plan: The pH is increasing, so the solution is becoming more basic. Therefore, OH– ion is added to increase the pH. Since 1 mole of H3O+ will react with 1 mole of OH–, the difference in [H3O+] would be equal to the [OH–] added. Use the relationship [H3O+] = 10–pH to find [H3O+] at each pH. Solution: [H3O+] = 10–pH = 10–3.15 = 7.07946×10–4 M H3O+ [H3O+] = 10–pH = 10–3.65 = 2.23872×10–4 M H3O+ Add (7.07946×10–4 M – 2.23872×10–4 M) = 4.84074×10–4 = 4.8×10–4 mol of OH– per liter.

18.37

The pH is decreasing so the solution is becoming less basic. Therefore, H3O+ ion is added to decrease the pH. 14 – 9.33 = 4.67 = pOH 14 – 9.07 = 4.93 = pOH [OH–] = 10–pOH = 10–4.67 = 2.13796×10–5 M OH– [OH–] = 10–pOH = 10–4.93 = 1.174898×10–5 M OH– Add (2.13796×10–5 M – 1.174898×10–5 M) = 9.63062×10–6 = 9.6×10–6 mol of H3O+ per liter.

18.38

Plan: The pH is increasing, so the solution is becoming more basic. Therefore, OH– ion is added to increase the pH. Since 1 mole of H3O+ will react with 1 mole of OH–, the difference in [H3O+] would be equal to the [OH–] added. Use the relationship [H3O+] = 10–pH to find [H3O+] at each pH. Solution: [H3O+] = 10–pH = 10–4.52 = 3.01995×10–5 M H3O+ [H3O+] = 10–pH = 10–5.25 = 5.623413×10–6 M H3O+ 3.01995×10–5 M – 5.623413×10–6 M = 2.4576×10–5 M OH– must be added. 2.4576 ×10−5 mol (5.6 L) = 1.3763×10–4 = 1.4×10–4 mol of OH– L

18.39

The pH is decreasing so the solution is becoming more acidic. Therefore, H3O+ ion is added to decrease the pH. 14 – 8.92 = 5.08 = pOH 14 – 8.45 = 5.55 = pOH [OH–] = 10–pOH = 10–5.08 = 8.31767 × 10–6 M OH– [OH–] = 10–pOH = 10–5.55 = 2.81838 × 10–6 M OH– Add (8.31767×10–6 M – 2.81838×10–6 M) = 5.49929×10–6 M must be added 5.49929 ×10−6 mol (0.0875 L ) = 4.81188×10–7 = 4.8×10–7 mol of H3O+ L

18-17

18.40

Scene A has a pH of 4.8. [H3O+] = 10–pH = 10–4.8 = 1.58489×10–5 M H3O+ Scene B: ⎛ 25 spheres ⎞⎟ ⎟ = 1.981125×10–4 M H3O+ [H3O+] = (1.58489×10−5 M H 3 O+ )⎜⎜ ⎜⎝ 2 spheres ⎠⎟⎟ pH = –log [H3O+] = –log [1.981125×10–4] = 3.703 = 3.7

18.41

Plan: Apply Le Châtelier’s principle in part a). In part b), given that the pH is 6.80, [H3O+] can be calculated by using the relationship [H3O+] = 10–pH. The problem specifies that the solution is neutral (pure water), meaning [H3O+] = [OH–]. A new Kw can then be calculated. Solution: a) Heat is absorbed in an endothermic process: 2H2O(l) + heat → H3O+(aq) + OH–(aq). As the temperature increases, the reaction shifts to the formation of products. Since the products are in the numerator of the Kw expression, rising temperature increases the value of Kw. b) [H3O+] = 10–pH = 10–6.80 = 1.58489×10–7 M H3O+ = 1.6×10–7 M [H3O+] = [OH–] Kw = [H3O+][OH–] = (1.58489×10–7)(1.58489×10–7) = 2.511876×10–14 = 2.5×10–14 For a neutral solution: pH = pOH = 6.80

18.42

Strong acids and bases dissociate completely into their ions when dissolved in water. For a strong acid, the concentration of hydronium ions produced by dissolving the acid is equal to the initial concentration of the undissociated acid. For a strong base, the concentration of hydroxide ions produced by dissolving the base is equal to the initial concentration of the undissociated base for MOH bases and is equal to twice the concentration of undissociated base for M(OH)2 bases. Strong acids and bases are classified as strong electrolytes because they dissociate completely to produce solutions with many ions that can strongly conduct an electric current.

18.43

The number of O atoms in strong oxoacids exceeds the number of ionizable protons by two or more.

18.44

The metals in Group 1A(1) and Ca, Sr, and Ba in Group 2A(2) form hydroxides that are strong bases.

18.45

Plan: The higher the [OH–], the lower the [H3O+] and the higher the pH. Solution: 0.25 M KOH dissociates to produce 0.25 M OH– in solution whereas 0.25 M Ca(OH)2 dissociates to produce 2 × 0.25 = 0.50 M OH– in solution. 0.25 M Ca(OH)2 has the higher pH.

18.46

Plan: Part a) can be approached two ways. Because NaOH is a strong base, the [OH–]eq = [NaOH]init. One method involves calculating [H3O+] using Kw = [H3O+][OH–], then calculating pH from the relationship pH = –log [H3O+]. The other method involves calculating pOH and then using pH + pOH = 14.00 to calculate pH. Part b) also has two acceptable methods analogous to those in part a); only one method will be shown. Solution: a) First method: Kw = [H3O+][OH–] Kw 1.0 ×10−14 = = 9.0090×10–13 M [H3O+] = − 0.0111 [OH ] pH = –log [H3O+] = –log (9.0090×10–13) = 12.04532 = 12.05 Second method: pOH = –log [OH–] = –log (0.0111) = 1.954677 pH = 14.00 – pOH = 14.00 – 1.954677 = 12.04532 = 12.05 With a pH > 7, the solution is basic. b) For a strong acid such as HCl: [H3O+] = [HCl] = 1.35×10–3 M pH = –log (1.35×10–3) = 2.869666

18-18

pOH = 14.00 – 2.869666 = 11.130334 = 11.13 With a pH < 7, the solution is acidic. 18.47

a) pH = –log (0.0333) = 1.47756 = 1.478; acidic b) pOH = –log (0.0347) = 1.45967 = 1.460; basic

18.48

Plan: HI is a strong acid, so [H3O+] = [HI] and the pH can be calculated from the relationship pH = –log [H3O+]. Ba(OH)2 is a strong base, so [OH–] = 2 × [Ba(OH)2] and pOH = –log [OH–]. Solution: a) [H3O+] = [HI] = 6.14×10–3 M. pH = –log (6.14×10–3) = 2.211832 = 2.212. Solution is acidic. – b) [OH ] = 2 × [Ba(OH)2] = 2(2.55 M) = 5.10 M pOH = –log (5.10) = –0.70757 = –0.708. Solution is basic.

18.49

a) pOH = –log (7.52×10–4) = 3.12378 pH = 14.00 – 3.12378 = 10.87622 = 10.88 basic b) pH = –log (1.59×10–3) = 2.79860 pOH = 14.00 – 2.79860 = 11.20140 = 11.20 acidic

18.50

Weak acids only partially dissociate. The characteristic property of all weak acids is that a significant number of the acid molecules are not dissociated. For a weak acid, the concentration of hydronium ions produced when the acid dissolves is less than the initial concentration of the acid.

18.51

Plan: The lower the concentration of hydronium (H3O+) ions, the higher the pH. pH increases as Ka or the molarity of acid decreases. Recall that pKa = –log Ka. Solution: a) At equal concentrations, the acid with the larger Ka will ionize to produce more hydronium ions than the acid with the smaller Ka. The solution of an acid with the smaller Ka = 4 × 10–5 has a lower [H3O+] and higher pH. b) pKa is equal to –log Ka. The smaller the Ka, the larger the pKa is. So the acid with the larger pKa, 3.5, has a lower [H3O+] and higher pH. c) Lower concentration of the same acid means lower concentration of hydronium ions produced. The 0.01 M solution has a lower [H3O+] and higher pH. d) At the same concentration, strong acids dissociate to produce more hydronium ions than weak acids. The 0.1 M solution of a weak acid has a lower [H3O+] and higher pH.

18.52

a) The concentration of a strong acid is very different before and after dissociation since a strong acid exhibits 100% dissociation. After dissociation, the concentration of the strong acid approaches 0, or [HA] ≈ 0. b) A weak acid dissociates to a very small extent (<<100%), so the acid concentration after dissociation is nearly the same as before dissociation. c) Same as b), but the percent, or extent, of dissociation is greater than in b). d) Same as a).

18.53

No, HCl and CH3COOH are never of equal strength because HCl is a strong acid with Ka > 1 and CH3COOH is a weak acid with Ka < 1. The Ka of the acid, not the concentration of H3O+ in a solution of the acid, determines the strength of the acid.

18.54

Plan: We are given the percent dissociation of the original HA solution (33%), and we know that the percent dissociation increases as the acid is diluted. Thus, we calculate the percent dissociation of each diluted sample and see which is greater than 33%. To determine percent dissociation, we use the following formula: Percent HA dissociated = ([HA]dissoc/[HA]init) × 100, with [HA]dissoc equal to the number of H3O+ (or A–) ions and [HA]init equal to the number of HA plus the number of H3O+ (or A–)

18-19

Solution: Calculating the percent dissociation of each diluted solution: Solution 1. Percent dissociation = [4/(5+4)] × 100 = 44% Solution 2. Percent dissociation = [2/(7 + 2)] × 100 = 22% Solution 3. Percent dissociation = [3/(6 + 3)] × 100 = 33% Therefore, scene 1 represents the diluted solution. 18.55

Water will add approximately 10–7 M to the H3O+ concentration. (The value will be slightly lower than for pure water.) a) CH3COOH(aq) + H2O(l) ⇆ H3O+(aq) + CH3COO–(aq) 0.10 – x x x + − [ H 3 O ][CH 3 COO ] Ka = 1.8×10–5 = [CH 3 COOH ] ( x)( x) Ka = 1.8×10–5 = Assume x is small compared to 0.1 so 0.1 – x = 0.1. 0.1 − x

(

Ka = 1.8×10–5 =

)

( x)( x)

(0.1) x = 1.3416×10 M Since the H3O+ concentration from CH3COOH is many times greater than that from H2O, [H3O+] = [CH3COO–]. b) The extremely low CH3COOH concentration means the H3O+ concentration from CH3COOH is near that from H2O. Thus [H3O+] = [CH3COO–]. c) CH3COOH(aq) + H2O(l) ⇆ H3O+(aq) + CH3COO–(aq) CH3COONa(aq) → CH3COO–(aq) + Na+(aq) –3

Ka = 1.8×10–5 =

(

( x) 0.1 + x

)

(0.1 − x)

Assume x is small compared to 0.1.

x = [H3O+] = 1.8×10–5 [CH3COO–] = 0.1 + x = 0.1 M Thus, [CH3COO–] > [H3O+] 18.56

The higher the negative charge on a species, the more difficult it is to remove a positively charged H+ ion.

18.57

Plan: Strong acids are the hydrohalic acids HCl, HBr, HI, and oxoacids in which the number of O atoms exceeds the number of ionizable protons by two or more, which include HNO3, H2SO4, and HClO4. All other acids are weak acids. Solution: a) Arsenic acid, H3AsO4, is a weak acid. The number of O atoms is 4, which exceeds the number of ionizable H atoms, 3, by one. This identifies H3AsO4 as a weak acid. b) HI is a strong acid. It is one of the strong hydrohalic acids. c) HIO is a weak acid. The number of O atoms is 1, which is equal to the number of ionizable H atoms identifying HIO as a weak acid. d) Perchloric acid, HClO4, is a strong acid. HClO4 is one example of the type of strong acid in which the number of O atoms exceeds the number of ionizable H atoms by more than 2.

18.58

a) strong acid

18.59

Plan: Ka values are listed in the Appendix. The larger the Ka value, the stronger the acid. The Ka value for hydroiodic acid, HI, is not shown because Ka approaches infinity for strong acids and is not meaningful.

b) strong acid

c) weak acid

d) weak acid

18-20

Solution: HI is the strongest acid (it is one of the six strong acids), and acetic acid, CH3COOH, is the weakest: CH3COOH < HF < HIO3 < HI 18.60

HCl > HNO2 > HClO > HCN

18.61

Plan: Ka is the equilibrium constant for an acid dissociation which has the generic equation [ H3O+ ][ A− ] HA(aq) + H2O(l) ⇆ H3O+(aq) + A–(aq). The Ka expression is . [H2O] is treated as a constant [ HA ] and omitted from the expression. Write the acid-dissociation reaction for each acid, following the generic equation, and then write the Ka expression. Solution: a) When phosphoric acid is dissolved in water, a proton is donated to the water and dihydrogen phosphate ions are generated. H3PO4(aq) + H2O(l) ⇆ H2PO4– (aq) + H3O+(aq) [ H 3 O+ ][ H 2 PO 4− ] Ka = [ H 3 PO 4 ] b) Benzoic acid is an organic acid and has only one proton to donate from the carboxylic acid group. The H atoms bonded to the benzene ring are not acidic hydrogens. C6H5COOH(aq) + H2O(l) ⇆ C6H5COO–(aq) + H3O+(aq) [ H 3 O+ ][C6 H 5 COO− ] Ka = [C6 H 5 COOH ] c) Hydrogen sulfate ion donates a proton to water and forms the sulfate ion. HSO4– (aq) + H2O(l) ⇆ SO42– (aq) + H3O+(aq) [ H 3 O+ ] ⎡⎣SO 4 2− ⎤⎦ Ka = [ HSO 4− ] d) When nitrous acid is dissolved in water, a proton is donated to the water and nitrite ions are generated. HNO2(aq) + H2O(l) ⇆ NO2– (aq) + H3O+(aq) ⎡ H O + ⎤ ⎡ NO − ⎤ 2 ⎦⎥ ⎢ 3 ⎦⎥ ⎣⎢ Ka = ⎣ ⎡ HNO 2 ⎤ ⎣ ⎦

18.62

a) Formic acid, an organic acid, has only one proton to donate from the carboxylic acid group. The remaining H atom, bonded to the carbon, is not an acidic hydrogen. HCOOH(aq) + H2O(l) ⇆ HCOO–(aq) + H3O+(aq) [ H3O+ ][ HCOO− ] Ka = [ HCOOH ] b) When chloric acid is dissolved in water, a proton is donated to the water and chlorate ions are generated. HClO3(aq) + H2O(l) ⇆ ClO3–(aq) + H3O+(aq) [ H 3 O+ ][ClO−3 ] Ka = [ HClO3 ] c) The dihydrogen arsenate ion donates a proton to water and forms the hydrogen arsenate ion. H2AsO4–(aq) + H2O(l) ⇆ HAsO42–(aq) + H3O+(aq) [ H 3 O+ ] ⎡⎣ HAsO 4 2− ⎤⎦ Ka = [ H 2 AsO 4− ]

18-21

d) When bromous acid is dissolved in water, a proton is donated to the water and bromite ions are generated. HBrO2(aq) + H2O(l) ⇆ BrO2– (aq) + H3O+(aq)

18.63

⎡ H O+ ⎤ ⎡ BrO − ⎤ 2 ⎦⎥ ⎢ 3 ⎦⎥ ⎣⎢ Ka = ⎣ ⎡ HBrO 2 ⎤ ⎣ ⎦ Plan: Write the acid-dissociation reaction and the expression for Ka. Set up a reaction table and substitute the given value of [H3O+] for x; solve for Ka. Solution: Butanoic acid dissociates according to the following equation: CH3CH2CH2COOH(aq) + H2O(l) ⇆ H3O+(aq) + CH3CH2CH2COO–(aq) 0 0 Initial: 0.15 M Change: –x +x +x Equilibrium: 0.15 – x x x According to the information given in the problem, [H3O+]eq = 1.51×10–3 M = x Thus, [H3O+] = [CH3CH2CH2COO–] = 1.51×10–3 M [CH3CH2CH2COOH] = (0.15 – x) = (0.15 – 1.51×10–3) M = 0.14849 M [ H 3 O+ ][CH 3 CH 2 CH 2 COO− ] Ka = [CH 3 CH 2 CH 2 COOH ]

(1.51×10 )(1.51×10 ) −3

Ka =

18.64

−3

(0.14849)

Any weak acid dissociates according to the following equation: HA(aq) + H2O(l) ⇆ H3O+(aq) + A–(aq) [H3O+] = 10–pH = 10–4.88 = 1.31826×10–5 M Thus, [H3O+] = [A–] = 1.31826×10–5 M, and [HA] = (0.035 – 1.31826×10–5) = 0.03499 M [ H3 O+ ][ A− ] Ka = [ HA ]

(1.31826×10 )(1.31826×10 ) −5

Ka =

18.65

= 1.53552×10–5 = 1.5×10–5

−5

(0.03499)

= 4.967×10–9 = 5.0×10–9

Plan: Write the acid-dissociation reaction and the expression for Ka. Set up a reaction table in which x = the concentration of the dissociated HNO2 and also [H3O+]. Use the expression for Ka to solve for x ([H3O+]). Solution: For a solution of a weak acid, the acid-dissociation equilibrium determines the concentrations of the weak acid, its conjugate base and H3O+. The acid-dissociation reaction for HNO2 is: + H2O(l) ⇆ H3O+(aq) + NO2–(aq) Concentration HNO2(aq) Initial 0.60 — 0 0 Change –x +x +x Equilibrium 0.60 – x x x (The H3O+ contribution from water has been neglected.) [ H 3 O+ ][ NO2− ] –4 Ka = 7.1×10 = [ HNO2 ] ( x)( x) Ka = 7.1×10–4 = Assume x is small compared to 0.60: 0.60 – x = 0.60 0.60 − x

(

)

18-22

Ka = 7.1×10–4 =

( x)( x)

(0.60) x = 0.020639767 Check assumption that x is small compared to 0.60: 0.020639767 (100) = 3.4% error, so the assumption is valid. 0.60 [H3O+ ] = [NO2– ] = 2.1×10–2 M The concentration of hydroxide ion is related to concentration of hydronium ion through the equilibrium for water: 2H2O(l) ⇆ H3O+(aq) + OH–(aq) with Kw = 1.0×10–14 Kw = 1.0×10–14 =[H3O+][OH– ] [OH–] = 1.0×10–14/0.020639767 = 4.84502×10–13 = 4.8×10–13 M OH–

18.66

For a solution of a weak acid, the acid-dissociation equilibrium determines the concentrations of the weak acid, its conjugate base and H3O+. The acid-dissociation reaction for HF is: + H2O(l) ⇆ H3O+(aq) + F–(aq) Concentration HF(aq) Initial 0.75 — 0 0 Change –x +x +x Equilibrium 0.75 – x x x (The H3O+ contribution from water has been neglected.) [ H3O+ ][ F− ] –4 Ka = 6.8×10 = [ HF ] ( x)( x) Ka = 6.8×10–4 = Assume x is small compared to 0.75. 0.75 − x

(

Ka = 6.8×10–4 =

)

( x)( x)

(0.75)

x = 0.02258 Check assumption: (0.02258/0.75) × 100% = 3% error, so the assumption is valid. [H3O+ ] = [F– ] = 2.3×10–2 M [OH–] = 1.0×10–14/0.02258 = 4.42869796×10–13 = 4.4×10–13 M OH– 18.67

Plan: Write the acid-dissociation reaction and the expression for Ka. Set up a reaction table in which x = the concentration of the dissociated acid and also [H3O+]. Use the expression for Ka to solve for x ([H3O+]). Ka is found from the pKa by using the relationship Ka = 10–pKa. Solution: Ka = 10–pKa = 10–2.87 = 1.34896×10–3 Concentration ClCH2COOH(aq) + H2O(l) ⇆ H3O+(aq) + ClCH2COO–(aq) Initial 1.25 0 0 Change –x +x +x Equilibrium 1.25 – x x x + − H O ClCH COO [ ] [ ] 3 2 Ka = 1.34896×10–3 = [ClCH2 COOH ] ( x)( x) Ka = 1.34896×10–3 = Assume x is small compared to 1.25. 1.25 − x

(

Ka = 1.34896×10–3 =

)

( x)( x)

(1.25)

18-23

x = 0.04106337 Check assumption that x is small compared to 1.25: 0.04106337 (100) = 3.3%. The assumption is good. 1.25 + [H3O ] = [ClCH2COO–] = 0.041 M [ClCH2COOH] = 1.25 – 0.04106337 = 1.20894 = 1.21 M pH = –log [H3O+] = –log (0.04106337) = 1.3865 = 1.39 18.68

Write a balanced chemical equation and equilibrium expression for the dissociation of hypochlorous acid and convert pKa to Ka. Ka = 10–pKa = 10–7.54 = 2.88403×10–8 HClO(aq) + H2O(l) ⇆ H3O+(aq) + ClO–(aq) 0.115 – x x x + H O [ ][ ClO− ] 3 Ka = 2.88403×10–8 = [ HClO ] ( x )( x ) Ka = 2.88403×10–8 = Assume x is small compared to 0.115. 0.115 − x

(

Ka = 2.88403×10–8 =

)

( x)( x)

(0.115) x = 5.75902×10–5 Check assumption: (5.75902×10–5/0.115) × 100% = 0.05%. The assumption is good. [H3O+] = [ClO–] = 5.8×10–5 M [HClO] = 0.115 – 5.75902×10–5 = 0.11494 = 0.115 M pH = –log [H3O+] = –log (5.75902×10–5) = 4.2397 = 4.24

18.69

Plan: Write the acid-dissociation reaction and the expression for Ka. Percent dissociation refers to the amount of the initial concentration of the acid that dissociates into ions. Use the percent dissociation to find the concentration of acid dissociated, which also equals [H3O+]. HA will be used as the formula of the acid. Set up a reaction table in which x = the concentration of the dissociated acid and [H3O+]. pH and [OH–] are determined from [H3O+]. Substitute [HA], [A–], and [H3O+] into the expression for Ka to find the value of Ka. Solution: a) HA(aq) + H2O(l) ⇆ H3O+(aq) + A–(aq) Percent HA =

dissociated acid initial acid

(100)

x (100) 0.20 [Dissociated acid] = x = 6.0×10–3 M Concentration HA(aq) + H2O(l) ⇆ H3O+(aq) + A–(aq) Initial: 0.20 0 0 Change: –x +x +x Equilibrium: 0.20 – x x x [Dissociated acid] = x = [A–] = [H3O+] = 6.0×10–3 M pH = –log [H3O+] = –log (6.0×10–3) = 2.22185 = 2.22 Kw = 1.0×10–14 = [H3O+][OH– ] Kw 1.0×10−14 [OH–] = = = 1.6666667×10–12 = 1.7×10–12 M + −3 [ H3O ] 6.0×10 pOH = –log [OH–] = –log (1.6666667×10–12) = 11.7782 = 11.78 3.0% =

18-24

b) In the equilibrium expression, substitute the concentrations above and calculate Ka. Ka =

18.70

[ H3 O+ ][ A− ] [ HA ]

(6.0×10 )(6.0×10 ) −3

−3

(0.20 − 6.0×10 ) −3

= 1.85567×10–4 = 1.9×10–4

Percent dissociation refers to the amount of the initial concentration of the acid that dissociates into ions. Use the percent dissociation to find the concentration of acid dissociated. HA will be used as the formula of the acid. a) The concentration of acid dissociated is equal to the equilibrium concentrations of A– and H3O+. Then, pH and [OH–] are determined from [H3O+]. dissociated acid (100) Percent HA dissociated = initial acid

x (100) 0.735 [Dissociated acid] = 9.1875×10–2 M HA(aq) + H2O(l) ⇆ H3O+(aq) + A–(aq) 0.735 – x x x [Dissociated acid] = x = [H3O+] = 9.19×10–2 M pH = –log [H3O+] = –log (9.1875×10–2) = 1.03680 = 1.037 [OH–] = Kw/[H3O+] = (1.0×10–14)/(9.1875×10–2) = 1.0884×10–13 = 1.1×10–13 M pOH = –log [OH–] = –log (1.0884×10–13) = 12.963197 = 12.963 b) In the equilibrium expression, substitute the concentrations above and calculate Ka. [ H3 O+ ][ A− ] (9.1875×10−2 )(9.1875×10−2 ) Ka = = = 1.3125×10–2 = 1.31×10–2 −2 [ HA ] 0.735 − 9.1875 × 10 ( ) 12.5% =

18.71

Plan: Write the acid-dissociation reaction and the expression for Ka. Calculate the molarity of HX by dividing moles by volume. Convert pH to [H3O+], set up a reaction table in which x = the concentration of the dissociated acid and also [H3O+], and substitute into the equilibrium expression to find Ka. ⎛ 0.250 mol ⎞⎛ 1 mL ⎞⎟ = 0.381679 M Concentration (M) of HX = ⎜⎜ ⎟⎟⎜ ⎝ 655 mL ⎠⎟⎜⎜⎝10−3 L ⎠⎟⎟ Concentration HX(aq) + H2O(l) ⇆ H3O+(aq) + X–(aq) Initial: 0.381679 0 0 Change: –x +x +x Equilibrium: 0.381679 – x x x [H3O+] = 10–pH = 10–3.54 = 2.88403×10–4 M = x Thus, [H3O+] = [X–] = 2.88403×10–4 M, and [HX] = (0.381679 – 2.88403×10–4) M Ka =

18.72

[ H3 O+ ][ X− ] [ HX ]

(2.88403×10 )(2.88403×10 ) −4

−4

(0.381679 − 2.88403×10 ) −4

= 2.18087×10–7 = 2.2×10–7

Calculate the molarity of HY by dividing moles by volume. Convert pH to [H3O+] and substitute into the equilibrium expression. Concentration of HY = (4.85×10–3 mol/0.095 L) = 0.0510526 M HY(aq) + H2O(l) ⇆ H3O+(aq) + Y–(aq) 0.0510526 – x x x [H3O+] = 10–pH = 10–2.68 = 2.089296×10–3 M = x Thus, [H3O+] = [Y–] = 2.089296×10–3 M, and [HY] = (0.0510526 – 2.089296×10–3) M Ka =

[ H3 O+ ][ Y− ] [ HY ]

(2.089296×10 )(2.089296×10 ) −3

−3

(0.0510526 − 2.089296×10 ) −3

= 8.91516×10–5 = 8.9×10–5

18-25

18.73

Plan: Write the acid-dissociation reaction and the expression for Ka. Set up a reaction table in which x = the concentration of the dissociated acid and also [H3O+]. Use the expression for Ka to solve for x ([H3O+]). OH– and then pOH can be found from [H3O+]. Solution: a) Concentration HZ(aq) + H2O(l) ⇆ H3O+(aq) + Z–(aq) Initial 0.075 — 0 0 Change –x +x +x Equilibrium 0.075 – x x x (The H3O+ contribution from water has been neglected.) [ H3O+ ][ Z− ] Ka = 2.55×10–4 = [ HZ ] ( x )( x ) Ka = 2.55×10–4 = Assume x is small compared to 0.075. 0.075 − x

(

Ka = 2.55×10–4 =

)

(x)(x)

(0.075) [H3O+] = x = 4.3732×10–3 Check assumption that x is small compared to 0.075: 4.3732×10−3 (100) = 6% error, so the assumption is not valid. 0.075 Since the error is greater than 5%, it is not acceptable to assume x is small compared to 0.075, and it is necessary to use the quadratic equation. ( x )( x ) Ka = 2.55×10–4 = 0.075 − x

(

)

–4

x + 2.55×10 x – 1.9125×10–5 = 0 a=1 b = 2.55×10–4 −b ± b 2 − 4ac x= 2a

c = –1.9125×10–5

−(2.55×10−4 ) ± (2.55×10−4 )2 − 4(1)(−1.9125×10−5 ) 2(1) x = 0.00425 or –0.004503 (The –0.004503 value is not possible.) pH = –log [H3O+] = –log (0.00425) = 2.3716 = 2.37 + H2O(l) ⇆ H3O+(aq) b) Concentration HZ(aq) Initial 0.045 — 0 Change –x +x Equilibrium 0.045 – x x (The H3O+ contribution from water has been neglected.) [H 3 O+ ][Z− ] Ka = 2.55×10–4 = [HZ] (x)(x) Ka = 2.55×10–4 = Assume x is small compared to 0.045. (0.045 − x)

Z–(aq) 0 +x x

18-26

Ka = 2.55×10–4 =

(x)(x)

(0.045) [H3O+] = x = 3.3875×10–3 Check assumption that x is small compared to 0.045: 3.3875×10−3 (100) = 7.5% error, so the assumption is not valid. 0.045 Since the error is greater than 5%, it is not acceptable to assume x is small compared to 0.045, and it is necessary to use the quadratic equation. (x)(x) Ka = 2.55×10–4 = (0.045 − x)

x2 = (2.55×10–4)(0.045 – x) = 1.1475×10–5 – 2.55×10–4 x x2 + 2.55×10–4 x – 1.1475×10–5 = 0 c = –1.1475×10–5 a = 1 b = 2.55×10–4 −b ± b 2 − 4ac

−2.55×10−4 ± (2.55×10−4 )2 − 4(1)(−1.1475×10−5 ) x=

2(1) x = 3.26238×10–3 M H3O+ Kw 1.0×10−14 = = 3.0652468×10–12 M [OH–] = + −3 [H 3 O ] 3.26238×10 pOH = –log [OH–] = –log (3.0652468×10–12) = 11.51353 = 11.51

18.74

Calculate Ka from pKa. Ka = 10–pKa = 10–4.89 = 1.28825×10–5 a) Begin with a reaction table, and then use the Ka expression as in earlier problems. Concentration HQ(aq) + H2O(l) ⇆ H3O+(aq) + Q–(aq) Initial 0.035 — 0 0 Change –x +x +x Equilibrium 0.035 – x x x (The H3O+ contribution from water has been neglected.) [H3 O+ ][Q− ] –5 Ka = 1.28825×10 = [HQ] ( x)( x) Ka = 1.28825×10–5 = Assume x is small compared to 0.035. (0.035 − x) Ka = 1.28825×10–5 =

( x )( x)

(0.035) [H3O ] = x = 6.71482×10–4 M Check assumption: (6.71482×10–4/0.035) × 100% = 2% error, so the assumption is valid. [H3O+] = 6.7×10–4 M + H2O(l) ⇆ H3O+(aq) + Q–(aq) b) Concentration HQ(aq) Initial 0.65 — 0 0 Change –x +x +x Equilibrium 0.65 – x x x +

18-27

(The H3O+ contribution from water has been neglected.) [H3 O+ ][Q− ] Ka = 1.28825×10–5 = [HQ] (x)(x) Assume x is small compared to 0.65. Ka = 1.28825×10–5 = 0.65 − x

(

Ka = 1.28825×10–5 =

)

(x)(x)

(0.65) [H3O ] = x = 2.89372×10–3 M Check assumption: (2.89372×10–3/0.65) × 100% = 0.4% error, so the assumption is valid. [OH–] = Kw/[H3O+] = (1.0×10–14)/(2.89372×10–3) = 3.4558×10–12 = 3.5×10–12 M +

18.75

Plan: Write the acid-dissociation reaction and the expression for Ka. Set up a reaction table in which x = the concentration of the dissociated acid and also [H3O+]. Use the expression for Ka to solve for x ([H3O+]). OH– and then pOH can be found from [H3O+]. Solution: a) Concentration HY(aq) + H2O(l) ⇆ H3O+(aq) + Y–(aq) Initial 0.175 — 0 0 Change –x +x +x Equilibrium 0.175 – x x x (The H3O+ contribution from water has been neglected.) [H3 O+ ][Y− ] Ka = 1.50×10–4 = [HY] (x)(x) Ka = 1.50×10–4 = Assume x is small compared to 0.175. 0.175 − x

(

Ka = 1.50×10–4 =

)

(x)(x)

(0.175) [H3O+] = x = 5.1235×10–3 M Check assumption that x is small compared to 0.175: 5.1235x10−3 (100) = 3% error, so the assumption is valid. 0.175 pH = –log [H3O+] = –log (5.1235×10–3) = 2.29043 = 2.290 b) Concentration HX(aq) + H2O(l) ⇆ H3O+(aq) + X–(aq) Initial 0.175 — 0 0 Change –x +x +x Equilibrium 0.175 – x x x (The H3O+ contribution from water has been neglected.) [H3 O+ ][X− ] Ka = 2.00×10–2 = [HX] (x)(x) Assume x is small compared to 0.175. Ka = 2.00×10–2 = 0.175 − x

(

Ka = 2.00×10–2 =

)

(x)(x)

(0.175) [H3O ] = x = 5.9161×10–2 M +

18-28

Check assumption that x is small compared to 0.175: 5.9161×10−2 (100) = 34% error, so the assumption is not valid. 0.175 Since the error is greater than 5%, it is not acceptable to assume x is small compared to 0.175, and it is necessary to use the quadratic equation. (x)(x) Ka = 2.00×10–2 = 0.175 − x

(

)

–2

x = (2.00×10 )(0.175 – x) = 0.0035 – 2.00×10–2x x2 + 2.00×10–2x – 0.0035 = 0 c = –0.0035 a = 1 b = 2.00×10–2 −b ± b 2 − 4ac

−2.00×10−2 ± (2.00×10−2 )2 − 4(1)(−0.0035) x=

2(1) x = 5.00×10–2 M H3O+ Kw 1.0×10−14 = = 2.00×10–13 M [OH–] = + [ H3O ] 5.00×10−2 pOH = –log [OH–] = –log (2.00×10–13) = 12.69897 = 12.699 18.76

a) Begin with a reaction table, then use the Ka expression as in earlier problems. + H2O(l) ⇆ H3O+(aq) + CN–(aq) Concentration HCN(aq) Initial 0.55 — 0 0 Change –x +x +x Equilibrium 0.55 – x x x (The H3O+ contribution from water has been neglected.) [ H 3 O+ ][ CN− ] –10 Ka = 6.2×10 = [ HCN ] ( x)( x) Ka = 6.2×10–10 = Assume x is small compared to 0.55. 0.55 − x

(

Ka = 6.2×10–10 =

)

( x)( x)

(0.55) [H3O ] = x = 1.84662×10–5 M Check assumption: (1.84662×10–5/0.55) × 100% = 0.0034% error, so the assumption is valid. pH = –log [H3O+] = –log (1.84662×10–5) = 4.7336 = 4.73 b) Begin this part like part a). + H2O(l) ⇆ H3O+(aq) + IO3–(aq) Concentration HIO3(aq) Initial 0.044 — 0 0 Change –x +x +x Equilibrium 0.044 – x x x (The H3O+ contribution from water has been neglected.) [ H 3 O+ ][ IO 3− ] Ka = 0.16 = [ HIO 3 ] ( x )( x ) Ka = 0.16 = Assume x is small compared to 0.044. (0.044 − x) +

18-29

Ka = 0.16 =

( x )( x )

(0.044) [H3O+] = x = 8.3905×10–2 Check assumption: (8.3905×10–2/0.044) × 100% = 191% error, so the assumption is not valid. Since the error is greater than 5%, it is not acceptable to assume x is small compared to 0.044, and it is necessary to use the quadratic equation. ( x )( x ) Ka = 0.16 = 0.044 − x

(

)

x = (0.16)(0.044 – x) = 0.00704 – 0.16 x x2 + 0.16 x – 0.00704 = 0 a=1 b = 0.16 c = – 0.00704 −b ± b 2 − 4ac

−0.16 ±

(0.16) − 4 (1)(−0.00704) 2

2 (1) +

x = 0.03593 M H3O [OH–] = Kw/[H3O+] = (1.0×10–14)/(0.03593) = 2.78×10–13 M pOH = –log [OH–] = –log (2.78×10–13) = 12.55596 = 12.56 18.77

Plan: Write the acid-dissociation reaction and the expression for Ka. Set up a reaction table in which x = the concentration of the dissociated acid and also [H3O+]. Use the expression for Ka to solve for x, the concentration of benzoate ion at equilibrium. Then use the initial concentration of benzoic acid and the equilibrium concentration of benzoate to find % dissociation. Solution: Concentration C6H5COOH(aq) + H2O(l) ⇆ H3O+(aq) + C6H5COO–(aq) Initial 0.55 — 0 0 Change –x +x +x Equilibrium 0.55 – x x x

Ka = 6.3×10–5 =

[ H 3 O+ ][C6 H 5 COO− ] [C6 H 5 COOH ]

[ x ][ x ] [ 0.55 − x ] [ x ][ x ] Ka = 6.3×10–5 = [ 0.55] x = 5.8864×10–3 M

Ka = 6.3×10–5 =

Assume x is small compared to 0.55.

Percent C6H5COOH dissociated =

[C 6 H 5 COOH ]dissociated (100) [C 6 H 5 COOH ]initial

Percent C6H5COOH dissociated =

5.8864 ×10−3 M (100) = 1.07025 = 1.1% 0.55 M

18-30

18.78

First, find the concentration of acetate ion at equilibrium. Then use the initial concentration of acetic acid and equilibrium concentration of acetate to find % dissociation. + H2O(l) ⇆ H3O+(aq) + CH3COO–(aq) Concentration CH3COOH(aq) Initial 0.050 — 0 0 Change –x +x +x Equilibrium 0.050 – x x x + − H O CH COO [ ] [ ] 3 3 Ka = 1.8×10–5 = CH COOH [ 3 ] [ x ][ x ] –5 Ka = 1.8×10 = Assume x is small compared to 0.050. [ 0.050 − x ] [ x ][ x ] Ka = 1.8×10–5 = [0.050] x = 9.48683×10–4 M [CH 3 COOH ]dissociated Percent CH3COOH dissociated = (100) [CH 3 COOH ]initial Percent CH3COOH dissociated =

18.79

9.48683×10−4 (100) = 1.897367 = 1.9% 0.050

Plan: Write balanced chemical equations and corresponding equilibrium expressions for dissociation of hydrosulfuric acid, H2S, and HS–. Since Ka1 >> Ka2, assume that almost all of the H3O+ comes from the first dissociation. Set up reaction tables in which x = the concentration of dissociated acid and [H3O+]. Solution: H2S(aq) + H2O(l) ⇆ H3O+(aq) + HS–(aq) HS–(aq) + H2O(l) ⇆ H3O+(aq) + S2–(aq) [ H 3 O+ ][ HS− ] [ H 3 O+ ] ⎡⎣S2− ⎤⎦ Ka1 = 9×10–8 = Ka2 = 1×10–17 = [ H 2 S] [ HS− ] H2S(aq) + H2O(l) ⇆ H3O+(aq) + HS–(aq) 0.10 — 0 0 –x +x +x 0.10 – x x x + − H O HS [ ] [ ] 3 Ka1 = 9×10–8 = [ H 2 S] [ x ][ x ] Ka1 = 9×10–8 = Assume x is small compared to 0.10. [ 0.10 − x ] [ x ][ x ] Ka1 = 9×10–8 = [ 0.10 ] x = 9.48683×10–5 [H3O+] = [HS–] = x = 9×10–5 M pH = –log [H3O+] = –log (9.48683×10–5) = 4.022878 = 4.0 Kw 1.0×10−14 = = 1.05409×10–10 = 1×10–10 M [OH–] = + [ H3O ] 9.48683×10−5

Concentration Initial Change Equilibrium

pOH = –log [OH–] = –log (1.05409×10–10) = 9.9771 = 10.0 [H2S] = 0.10 – x = 0.10 – 9.48683×10–5 = 0.099905 = 0.10 M Concentration is limited to one significant figure because Ka is given to only one significant figure. The pH is given to what appears to be two significant figures because the number before the decimal point (4) represents the exponent and the number after the decimal point represents the significant figures in the concentration. Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

18-31

Calculate [S2–] by using the Ka2 expression and assuming that [HS–] and [H3O+] come mostly from the first dissociation. This new calculation will have a new x value. + H2O(l) ⇆ H3O+(aq) + S2–(aq) Concentration HS–(aq) –5 –5 — 9.48683×10 0 Initial 9.48683×10 Change –x +x +x Equilibrium 9.48683×10–5 – x 9.48683×10–5 + x x [ H 3 O+ ] ⎡⎣S2− ⎤⎦ Ka2 = 1×10–17 = [ HS− ]

(9.48683×10 + x)(x) K = 1×10 = (9.48683×10 − x) −5

–17

−5

Assume x is small compared to 9.48683×10–5.

(9.48683×10 )(x) K = 1×10 = (9.48683×10 ) −5

–17

−5

x = [S2–] = 1×10–17 M The small value of x means that it is not necessary to recalculate the [H3O+] and [HS–] values. 18.80

Write balanced chemical equations and corresponding equilibrium expressions for dissociation of malonic acid (H2C3H2O4). H2C3H2O4(aq) + H2O(l) ⇆ H3O+(aq) + HC3H2O4–(aq) HC3H2O4–(aq) + H2O(l) ⇆ H3O+(aq) + C3H2O42–(aq) [ H3O+ ][ HC3 H 2 O4− ] [ H 3 O+ ] ⎡⎣C3 H 2 O 4 2− ⎤⎦ Ka1 = 1.4×10–3 = Ka2 = 2.0×10–6 = [ H 2 C3 H 2 O 4 ] [ HC3 H 2 O 4− ] Since Ka1 >> Ka2, assume that almost all of the H3O+ comes from the first dissociation. H2C3H2O4(aq) + H2O(l) ⇆ H3O+(aq) + HC3H2O4–(aq) 0.200 – x x x + − H O HC H O [ ] [ ] 3 3 2 4 Ka1 = 1.4×10–3 = [ H 2 C3 H 2 O 4 ] ( x)( x) Assume x is small compared to 0.200. Ka1 = 1.4×10–3 = 0.200 − x

(

Ka1 = 1.4×10–3 =

)

( x )( x )

(0.200)

x = 0.016733 Check assumption: (0.016733/0.200) × 100% = 8% error, so the assumption is not valid. Since the error is greater than 5%, it is not acceptable to assume x is small compared to 0.200, and it is necessary to use the quadratic equation. x2 = 2.8×10–4 – 1.4×10–3 x x2 + 1.4×10–3 x – 2.8×10–4 = 0 a = 1 b = 1.4×10–3 c = – 2.8×10–4 −b ± b 2 − 4ac

−1.4 ×10−3 ±

(1.4 ×10 ) − 4 (1)(−2.8×10 ) −3

−4

2 (1) –2

x = 1.6048×10 [H3O+] = [HC3H2O4–] = x = 1.6×10–2 M Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

18-32

pH = – log [H3O+] = – log (1.6048×10–2) = 1.79458 = 1.79 [OH–] = Kw/[H3O+] = (1.0×10–14)/(1.6048×10–2) = 6.23131×10–13 = 6.2×10–13 M pOH = – log [OH–] = – log (6.23131×10–13) = 12.2054 = 12.21 [H2C3H2O4] = (0.200 – 1.6048×10–2) M = 0.183952 = 0.18 M Concentration is limited to two significant figures because Ka is given to only two significant figures. The pH is given to what appears to be three significant figures because the number before the decimal point (1) represents the exponent and the number after the decimal point represents the significant figures in the concentration. Calculate [C3H2O42–] by using the Ka2 expression and assuming that [HC3H2O4–] and [H3O+] come mostly from the first dissociation. This new calculation will have a new x value. HC3H2O4–(aq) + H2O(l) ⇆ H3O+(aq) + C3H2O42–(aq) –2 –2 1.6048×10 – x 1.6048×10 + x x + ⎡ 2− ⎤ [H3O ] ⎣C3 H 2 O 4 ⎦ Ka2 = 2.0×10–6 = [ HC3 H 2 O 4− ] Ka2 = 2.0×10–6 =

(1.6048×10 + x)(x) (1.6048×10 − x)

Ka2 = 2.0×10–6 =

(1.6048×10 )(x) (1.6048×10 )

−2

Assume x is small compared to 1.6048×10–2.

−2

x = [C3H2O42–] = 2.0×10–5 = 2.0×10–5 M 18.81

Write balanced chemical equations and corresponding equilibrium expressions for dissociation of aspirin (HC9H7O4). HC9H7O4(aq) + H2O(l) ⇆ H3O+(aq) + C9H7O4–(aq) 0.018 – x x x + − H O C H O [ ] [ ] 3 9 7 4 Ka = 3.6×10–4 = HC H [ 9 7O4 ] ( x )( x ) Assume x is small compared to 0.018. Ka = 3.6×10–4 = 0.018 − x

(

Ka = 3.6×10–4 =

)

( x)( x)

(0.018) [H3O ] = x = 2.54558×10–3 Check assumption: (2.54558×10–3/0.018) × 100% = 14% error, so the assumption is not valid. Since the error is greater than 5%, it is not acceptable to assume x is small compared to 0.018, and it is necessary to use the quadratic equation. x2 = (3.6×10–4)(0.018 – x) = 6.48×10–6 – 3.6×10–4 x x2 + 3.6×10–4 x – 6.48×10–6 = 0 a = 1 b = 3.6×10–4 c = – 6.48×10–6 +

−b ± b 2 − 4ac

2a −3.6×10−4 ±

(3.6×10 ) − 4 (1)(−6.48×10 ) −4

−6

2 (1) –3

x = 2.37194×10 M H3O pH = –log [H3O+] = –log (2.37194×10–3) = 2.624896 = 2.62 Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

18-33

18.82

Plan: Write the acid-dissociation reaction and the expression for Ka. Set up a reaction table in which x = the concentration of the dissociated acid and also [H3O+]. Use the expression for Ka to solve for x, the concentration of formate ion at equilibrium. Then use the initial concentration of formic acid and the equilibrium concentration of formate to find % dissociation. Solution: Concentration HCOOH(aq) + H2O(l) ⇆ H3O+(aq) + HCOO–(aq) Initial 0.75 0 0 Change –x +x +x Equilibrium 0.75 – x x x + − H O HCOO [ ] [ ] 3 Ka = 1.8×10–4 = [ HCOOH ] ( x )( x) Ka = 1.8×10–4 = Assume x is small compared to 0.75. 0.75 − x

(

Ka = 1.8×10–4 =

)

( x)( x)

(0.75) x = 1.161895×10–2 Percent HCOOH dissociated =

[ HCOOH ]dissociated (100) [ HCOOH ]initial

Percent HCOOH dissociated =

1.161895×10−2 M (100) = 1.54919 = 1.5% 0.75 M

18.83

Electronegativity increases left to right across a period. As the nonmetal becomes more electronegative, the acidity of the binary hydride increases. The electronegative nonmetal attracts the electrons more strongly in the polar bond, shifting the electron density away from H+ and making the H+ more easily transferred to a surrounding water molecule to make H3O+.

18.84

As the nonmetal increases in size, its bond to hydrogen becomes longer and weaker, so that H+ is more easily lost, and a stronger acid results.

18.85

There is an inverse relationship between the strength of the bond to the acidic proton and the strength of the acid. A weak bond means the hydrogen ion is more easily lost, and hence the acid is stronger.

18.86

The two factors that explain the greater acid strength of HClO4 are: 1) Chlorine is more electronegative than iodine, so chlorine more strongly attracts the electrons in the bond with oxygen. This makes the H in HClO4 less tightly held by the oxygen than the H in HIO. 2) Perchloric acid has more oxygen atoms than HIO, which leads to a greater shift in electron density from the hydrogen atom to the oxygen atoms making the H in HClO4 more susceptible to transfer to a base.

18.87

Plan: For oxyacids, acid strength increases with increasing number of oxygen atoms and increasing electronegativity of the nonmetal in the acid. For binary acids, acid strength increases with increasing electronegativity across a row and increases with increasing size of the nonmetal down a column. Solution: a) Selenic acid, H2SeO4, is the stronger acid because it contains more oxygen atoms. b) Phosphoric acid, H3PO4, is the stronger acid because P is more electronegative than As. c) Hydrotelluric acid, H2Te, is the stronger acid because Te is larger than S and so the Te–H bond is weaker.

18.88

a) H2Se

b) H2SO4

c) H2SO3

18-34

18.89

Plan: For oxyacids, acid strength increases with increasing number of oxygen atoms and increasing electronegativity of the nonmetal in the acid. For binary acids, acid strength increases with increasing electronegativity across a row and increases with increasing size of the nonmetal in a column. Solution: a) H2Se, hydrogen selenide, is a stronger acid than H3As, arsenic hydride, because Se is more electronegative than As. b) B(OH)3, boric acid also written as H3BO3, is a stronger acid than Al(OH)3, aluminum hydroxide, because boron is more electronegative than aluminum. c) HBrO2, bromous acid, is a stronger acid than HBrO, hypobromous acid, because there are more oxygen atoms in HBrO2 than in HBrO.

18.90

a) HBr

18.91

Plan: Acidity increases as the value of Ka increases. Determine the ion formed from each salt and compare the corresponding Ka values from Appendix C. Solution: a) Copper(II) bromide, CuBr2, contains Cu2+ ion with Ka = 3×10–8. Aluminum bromide, AlBr3, contains Al3+ ion with Ka = 1×10–5. The concentrations of Cu2+ and Al3+ are equal, but the Ka of AlBr3 is almost three orders of magnitude greater. Therefore, 0.5 M AlBr3 is the stronger acid and would have the lower pH. b) Zinc chloride, ZnCl2, contains the Zn2+ ion with Ka = 1×10–9. Tin(II) chloride, SnCl2, contains the Sn2+ ion with Ka = 4×10–4. Since both solutions have the same concentration, and Ka (Sn2+) > Ka (Zn2+), 0.3 M SnCl2 is the stronger acid and would have the lower pH.

18.92

a) FeCl3

18.93

Plan: A higher pH (more basic solution) results when an acid has a smaller Ka (from the Appendix). Determine the ion formed from each salt and compare the corresponding Ka values from Appendix C. Solution: a) The Ni(NO3)2 solution has a higher pH than the Co(NO3)2 solution because Ka of Ni2+ (1×10–10) is smaller than the Ka of Co2+ (2×10–10). Note that nitrate ion is the conjugate base of a strong acid and therefore does not influence the pH of the solution. b) The Al(NO3)3 solution has a higher pH than the Cr(NO3)2 solution because Ka of Al3+ (1×10–5) is smaller than the Ka of Cr3+ (1×10–4).

18.94

a) NaCl

18.95

All Brønsted-Lowry bases contain at least one lone pair of electrons. This lone pair binds with an H+ and allows the base to act as a proton-acceptor.

18.96

The negative charge and lone pair of the anion in many cases is able to extract a proton from water forming OH– ions. Non-basic anions are from strong acids and include I–, NO3–, Cl–, ClO4–.

18.97

a) The species present are: CH3COOH(aq), CH3COO–(aq), H3O+(aq), and OH–(aq). b) CH3COOH(aq) + H2O(l) ⇆ H3O+(aq) + CH3COO–(aq) The solution is acidic because H3O+ ions are formed. CH3COO–(aq) + H2O(l) ⇆ OH–(aq) + CH3COOH(aq) The solution is basic because OH– ions are formed.

b) H3AsO4

c) HNO2

b) BeCl2

b) Co(NO3)2

18-35

18.98

Plan: Kb is the equilibrium constant for a base dissociation which has the generic equation [ BH + ][ OH− ] B(aq) + H2O(l) ⇆ BH+(aq) + OH–(aq). The Kb expression is . [H2O] is treated as a constant [ B] and omitted from the expression. Write the base-dissociation reaction for each base, showing the base accepting a proton from water, and then write the Kb expression. Solution: a) C5H5N(aq) + H2O(l) ⇆ C5H5NH+(aq) + OH–(aq) [C 5 H 5 NH + ][ OH− ] Kb = [C 5 H 5 N ] b) CO32–(aq) + H2O(l) ⇆ HCO3–(aq) + OH–(aq) [ HCO3− ][ OH− ] Kb = ⎡ CO32− ⎤ ⎣ ⎦ The bicarbonate can then also dissociate as a base, but this occurs to an insignificant amount in a solution of carbonate ions.

18.99

a) C6H5COO–(aq) + H2O(l) ⇆ OH–(aq) + C6H5COOH(aq) [C6 H 5 COOH ][ OH− ] Kb = [C6 H 5 COO− ] b) (CH3)3N(aq) + H2O(l) ⇆ OH–(aq) + (CH3)3NH+(aq) ⎡(CH 3 ) NH + ⎤ [ OH− ] ⎢ 3 ⎦⎥ Kb = ⎣ ⎡(CH 3 ) N ⎤ ⎢⎣ ⎥⎦ 3

18.100 Plan: Kb is the equilibrium constant for a base dissociation which has the generic equation [ BH + ][ OH− ] B(aq) + H2O(l) ⇆ BH+(aq) + OH–(aq). The Kb expression is . [H2O] is treated as a constant and [ B] omitted from the expression. Write the base-dissociation reaction for each base, showing the base accepting a proton from water, and then write the Kb expression. Solution: a) HONH2(aq) + H2O(l) ⇆ OH–(aq) + HONH3+(aq) [ HONH 3+ ][ OH− ] Kb = [ HONH 2 ] b) HPO42–(aq) + H2O(l) ⇆ H2PO4–(aq) + OH–(aq) [ H 2 PO 4− ][ OH− ] Kb = ⎡ HPO 4 2− ⎤ ⎣ ⎦ 18.101 a) (NH2)2C=NH(aq) + H2O(l) ⇆ OH–(aq) + (NH2)2C=NH2+(aq) [(H 2 N)2 CNH 2+ ][ OH− ] Kb = [(H 2 N)2 CNH ] b) HCC–(aq) + H2O(l) ⇆ OH–(aq) + HCCH(aq) [ HCCH ][ OH− ] Kb = [ HCC− ]

18-36

18.102 Plan: Write the balanced equation for the base reaction and the expression for Kb. Set up a reaction table in which x = the concentration of reacted base and also [OH–]. Use the expression for Kb to solve for x, [OH–], and then calculate [H3O+] and pH. Solution: The formula of dimethylamine has two methyl (CH3–) groups attached to a nitrogen: CH3

CH3

The nitrogen has a lone pair of electrons that will accept the proton from water in the base-dissociation reaction: The value for the dissociation constant is from Appendix C. Concentration (CH3)2NH(aq) + H2O(l) ⇆ OH–(aq) + (CH3)2NH2+(aq) Initial 0.070 0 0 Change –x +x +x Equilibrium 0.070 – x x x + − ⎡(CH 3 ) NH 2 ⎤ [ OH ] ⎢ 2 ⎦⎥ Kb = 5.9×10–4 = ⎣ ⎡(CH 3 ) NH ⎤ ⎢⎣ ⎥⎦ 2 [ x ][ x ] Assume 0.070 – x = 0.070 Kb = 5.9×10–4 = ⎡0.070 − x⎤ ⎣⎢ ⎦⎥ [ x ][ x ] 5.9×10–4 = [ 0.070 ] x = 6.4265×10–3 M Check assumption that x is small compared to 0.070: 6.4265×10−3 (100) = 9% error, so the assumption is not valid. 0.070 The problem will need to be solved as a quadratic. [ x ][ x ] 5.9×10–4 = ⎡0.070 − x⎤ ⎣⎢ ⎦⎥ x2 = (5.9×10–4)(0.070 – x) = 4.13×10–5 – 5.9×10–4 x x2 + 5.9×10–4 x – 4.13×10–5 = 0 a = 1 b = 5.9×10–4 c = –4.13×10–5 −b ± b 2 − 4ac

−5.9×10−4 ±

x= [H3O]+ =

(5.9×10 ) − 4 (1)(−4.13×10 ) −4

−5

2 (1)

= 6.13827×10–3 M OH–

Kw 1.0×10−14 = = 1.629124×10–12 M H3O+ [ OH− ] 6.13827×10−3

pH = –log [H3O+] = –log (1.629124×10–12) = 11.7880 = 11.79 18.103 (CH3CH2)2NH(aq) + H2O(l) ⇆ OH–(aq) + (CH3CH2)2NH2+(aq) 0.12 – x x x ⎡(CH 3 CH 2 ) NH 2 + ⎤ [ OH− ] ⎢ ⎥⎦ 2 Kb = 8.6×10–4 = ⎣ ⎡(CH 3 CH 2 ) NH ⎤ 2 ⎣⎢ ⎦⎥ Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

18-37

Kb = 8.6×10–4 = Kb = 8.6×10–4 =

[ x ][ x ]

Assume x is small compared to 0.12.

⎡ 0.12 − x ⎤ ⎢⎣ ⎥⎦ [ x ][ x ] ⎡ 0.12 ⎤ ⎣⎢ ⎦⎥

x = 0.0101587 Check assumption: (0.0101587/0.12) × 100% = 8% error, so the assumption is not valid. Since the error is greater than 5%, it is not acceptable to assume x is small compared to 0.12, and it is necessary to use the quadratic equation. x2 = (8.6×10–4)(0.12 – x) = 1.032×10–4 – 8.6×10–4 x x2 + 8.6×10–4 x – 1.032×10–4 = 0 a = 1 b = 8.6×10–4 c = –1.032×10–4 −b ± b 2 − 4ac

−8.6 ×10−4 ±

(8.6×10 ) − 4 (1)(−1.032×10 ) −4

−4

2 (1) –3

–

x = 9.7378×10 M OH [H3O]+ = Kw/[OH–] = (1.0×10–14)/(9.7378×10–3) = 1.02693×10–12 M H3O+ pH = –log [H3O+] = –log (1.02693×10–12) = 11.98846 = 11.99 18.104 Plan: Write the balanced equation for the base reaction and the expression for Kb. Set up a reaction table in which x = the concentration of reacted base and also [OH–]. Use the expression for Kb to solve for x, [OH–], and then calculate [H3O+] and pH. Solution: Concentration HOCH2CH2NH2(aq) + H2O(l) ⇆ OH–(aq) + HOCH2CH2NH3+(aq) Initial 0.25 0 0 Change –x +x +x Equilibrium 0.25 – x x x + − [ HOCH 2 CH 2 NH 3 ][ OH ] Kb = 3.2×10–5 = [ HOCH 2 CH 2 NH 2 ] [ x ][ x ] Assume x is small compared to 0.25. Kb = 3.2×10–5 = ⎡ 0.25 − x ⎤ ⎢⎣ ⎥⎦ ( x)( x) Kb = 3.2×10–5 = (0.25) x = 2.8284×10–3 M OH– Check assumption that x is small compared to 0.25: 2.8284 ×10−3 (100) = 1% error, so the assumption is valid. 0.25 Kw 1.0 ×10−14 [H3O]+ = = = 3.535568×10–12 M H3O+ − −3 [ OH ] 2.8284 ×10 pH = –log [H3O+] = –log (3.535568×10–12) = 11.4515 = 11.45 18.105 C6H5NH2(aq) + H2O(l) ⇆ OH–(aq) + C6H5NH3+(aq) 0.26 – x x x Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

18-38

Kb = 4.0×10–10 = Kb = 4.0×10–10 = Kb = 4.0×10–10 =

[C6 H 5 NH 3+ ][ OH− ] [C6 H 5 NH 2 ] [ x ][ x ]

⎡ 0.26 − x ⎤ ⎣⎢ ⎦⎥ ( x)( x)

Assume x is small compared to 0.26.

(0.26) x = 1.01980×10 M OH– Check assumption: (1.01980×10–5/0.26) × 100% = 0.004% error, so the assumption is valid. [H3O]+ = Kw/[OH–] = (1.0×10–14)/(1.01980×10–5) = 9.80584×10–10 M H3O+ pH = –log [H3O+] = –log (9.80584×10–10) = 9.008515 = 9.01 –5

18.106 Plan: The Kb of a conjugate base is related to the Ka of the conjugate acid through the equation Kw = Ka × Kb. Solution: a) Acetate ion, CH3COO–, is the conjugate base of acetic acid, CH3COOH. Kw = Ka × Kb K 1.0×10−14 Kb of CH3COO– = w = = 5.55556×10–10 = 5.6×10–10 Ka 1.8×10−5 b) Anilinium ion is the conjugate acid of the weak base aniline, C6H5NH2. K 1.0×10−14 Ka of C6H5NH3+ = w = = 2.5×10–5 −10 Kb 4.0×10 18.107 a) Benzoate ion, C6H5COO–, is the conjugate base of benzoic acid, C6H5COOH. The Kb for benzoate ion is related to the Ka for benzoic acid through the equation Kw = Ka × Kb. K 1.0×10−14 Kb of C6H5COO– = w = = 1.58730×10–10 = 1.6×10–10 Ka 6.3×10−5 b) The 2-hydroxyethylammonium ion is the conjugate acid of 2-hydroxyethylamine so the pKa for 2-hydroxyethylammonium ion is related to the pKb of 2-hydroxyethylamine by the relationship 14.00 = pKa + pKb. The Ka may be calculated from the pKa. 14.00 = pKa + pKb 14.00 = pKa + 4.49 pKa = 14.00 – 4.49 = 9.51 Ka = 10–pKa = 10–9.51 = 3.090295×10–10 = 3.1×10–10 18.108 Plan: The Kb of a conjugate base is related to the Ka of the conjugate acid through the equation Kw = Ka × Kb. Solution: a) HClO2 is the conjugate acid of chlorite ion, ClO2–. K 1.0×10−14 Kb of ClO2– = w = = 9.0909×10–13 Ka 1.1×10−2 pKb = –log (9.0909×10–13) = 12.04139 = 12.04 b) (CH3)2NH is the conjugate base of (CH3)2NH2+. K 1.0×10−14 Ka of (CH3)2NH2+ = w = = 1.694915×10–11 Kb 5.9×10−4 pKa = –log (1.694915×10–11) = 10.77085 = 10.77 18.109 a) The Ka of nitrous acid, HNO2, is reported in Appendix C. HNO2 is the conjugate acid of nitrite ion, NO2–. The Kb for nitrite ion is related to the Ka for nitrous acid through the equation Kw = Ka × Kb, and pKb = –log Kb. Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

18-39

Kw 1.0 ×10−14 = = 1.4084507×10–11 Ka 7.1×10−4 pKb = –log (1.4084507×10–11) = 10.851258 = 10.85 b) The Kb of hydrazine, H2NNH2, is reported in the problem. Hydrazine is the conjugate base of H2N–NH3+. The Ka for H2N–NH3+ is related to the Kb for H2NNH2 through the equation Kw = Ka × Kb, and pKa = –log Ka. K 1.0 ×10−14 Ka of H2N–NH3+ = w = = 1.17647×10–8 Kb 8.5×10−7 pKa = –log (1.17647×10–8) = 7.9294 = 7.93 Kb of NO2– =

18.110 Plan: In part a), potassium cyanide, when placed in water, dissociates into potassium ions, K+, and cyanide ions, CN–. Potassium ion is the conjugate acid of a strong base, KOH, so K+ does not react with water. Cyanide ion is the conjugate base of a weak acid, HCN, so it does react with a base-dissociation reaction. To find the pH first set up a reaction table and use Kb for CN– to calculate [OH–]. Find the Kb for CN– from the equation Kw = Ka × Kb. In part b), the salt triethylammonium chloride in water dissociates into two ions: (CH3CH2)3NH+ and Cl–. Chloride ion is the conjugate base of a strong acid so it will not influence the pH of the solution. Triethylammonium ion is the conjugate acid of a weak base, so an acid-dissociation reaction determines the pH of the solution. To find the pH first set up a reaction table and use Ka for (CH3CH2)3NH+ to calculate [H3O+]. Find the Ka for (CH3CH2)3NH+ from the equation Kw = Ka × Kb. Solution: a) CN–(aq) + H2O(l) ⇆ HCN(aq) + OH–(aq) Concentration (M) CN–(aq) + H2O(l) ⇆ HCN(aq) + OH–(aq) Initial 0.150 — 0 0 Change –x +x +x Equilibrium 0.150 – x x x Kw 1.0 ×10−14 – –5 Kb of CN = = = 1.612903×10 Ka 6.2 ×10−10 Kb = 1.612903×10–5 = Kb = 1.612903×10–5 = Kb = 1.612903×10–5 = –3

[ HCN ][ OH− ]

[ CN− ] [ x ][ x ]

⎡0.150 − x⎤ ⎢⎣ ⎥⎦ ( x )( x )

Assume x is small compared to 0.150.

(0.150)

–

x = 1.555×10 M OH Check assumption that x is small compared to 0.150: 1.555×10−3 (100) = 1% error, so the assumption is valid. 0.150 Kw 1.0 ×10−14 [H3O]+ = = = 6.430868×10–12 M H3O+ [ OH− ] 1.555×10−3 pH = –log [H3O+] = –log (6.430868×10–12) = 11.19173 = 11.19 b) (CH3CH2)3NH+(aq) + H2O(l) ⇆ (CH3CH2)3N(aq) + H3O+(aq) Concentration (M) Initial Change Equilibrium

(CH3CH2)3NH+(aq) + H2O(l) 0.40 — –x 0.40 – x

⇆

(CH3CH2)3N(aq) + H3O+(aq) 0 0 +x +x x x

18-40

Ka of (CH3CH2)3NH+ =

Kw 1.0 ×10−14 = = 1.9230769×10–11 Kb 5.2 ×10−4

Ka = 1.9230769×10–11 =

[ H3O+ ][(CH3CH 2 )3 N ] [(CH3CH 2 )3 NH+ ] ( x)( x)

Ka = 1.9230769×10–11 =

(0.40 − x)

Assume x is small compared to 0.40.

( x)( x)

Ka = 1.9230769×10–11 =

(0.40) [H3O ] = x = 2.7735×10 M Check assumption that x is small compared to 0.40: 2.7735×10−6 (100) = 0.0007% error, so the assumption is valid. 0.40 pH = –log [H3O+] = –log (2.7735×10–6) = 5.55697 = 5.56 +

–6

18.111 a) Sodium phenolate, when placed in water, dissociates into sodium ions, Na+, and phenolate ions, C6H5O–. Sodium ion is the conjugate acid of a strong base, NaOH, so Na+ does not react with water. Phenolate ion is the conjugate base of a weak acid, C6H5OH, so it does react with the base-dissociation reaction: C6H5O–(aq) + H2O(l) ⇆ C6H5OH(aq) + OH–(aq) To find the pH first set up a reaction table and use Kb for C6H5O– to calculate [OH–]. Concentration (M) C6H5O–(aq) + H2O(l) ⇆ C6H5OH(aq) + OH–(aq) Initial 0.100 — 0 0 Change –x +x +x Equilibrium 0.100 – x x x −14 K 1.0 × 10 Kb of C6H5O– = w = = 1.0×10–4 Ka 1.0 ×10−10

[C6 H5 OH][ OH− ] Kb = 1.0×10 = [C6 H5 O− ] –4

Kb =1.0×10–4 = Kb = 1.0×10–4 =

[ x ][ x ]

⎡0.100 − x⎤ ⎣⎢ ⎦⎥ ( x )( x )

Assume x is small compared to 0.100.

(0.100) x = 3.1622777×10–3 M OH– Check assumption: (3.1622777×10–3/0.100)x100% = 3% error, so the assumption is valid. [H3O]+ = Kw/[OH–] = (1.0×10–14)/(3.16227766×10–3) = 3.1622776×10–12 M H3O+ pH = –log [H3O+] = –log (3.16227762×10–12) = 11.50 b) The salt methylammonium bromide in water dissociates into two ions: CH3NH3+ and Br–. Bromide ion is the conjugate base of a strong acid so it will not influence the pH of the solution. Methylammonium ion is the conjugate acid of a weak base, so the acid-dissociation reaction below determines the pH of the solution. Concentration (M) CH3NH3+(aq) + H2O(l) ⇆ CH3NH2(aq) + H3O+ (aq) Initial 0.15 — 0 0 Change –x +x +x Equilibrium 0.15 – x x x −14 K 1.0 × 10 Ka of CH3NH3+ = w = = 2.272727×10–11 Kb 4.4 ×10−4 Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

18-41

Ka = 2.272727×10–11 = Ka = 2.272727×10–11 = Ka = 2.272727×10–11 =

[ H 3 O+ ][CH 3 NH 2 ] [CH 3 NH 3+ ] ( x)( x)

(0.15 − x)

Assume x is small compared to 0.15.

( x)( x)

(0.15) [H3O ] = x = 1.84637×10–6 Check assumption: (1.84637×10–6/0.15)×100% = 0.001% error, so the assumption is valid. pH = –log [H3O+] = –log (1.84637×10–6) = 5.73368 = 5.73 +

18.112 Plan: In part a), potassium formate, when placed in water, dissociates into potassium ions, K+, and formate ions, HCOO–. Potassium ion is the conjugate acid of a strong base, KOH, so K+ does not react with water. Formate ion is the conjugate base of a weak acid, HCOOH, so it does react with a base-dissociation reaction. To find the pH first set up a reaction table and use Kb for HCOO– to calculate [OH–]. Find the Kb for HCOO– from the equation Kw = Ka × Kb. In part b), the salt ammonium bromide in water dissociates into two ions: NH4+ and Br–. Bromide ion is the conjugate base of a strong acid so it will not influence the pH of the solution. Ammonium ion is the conjugate acid of the weak base NH3, so an acid-dissociation reaction determines the pH of the solution. To find the pH first set up a reaction table and use Ka for NH4+ to calculate [H3O+]. Find the Ka for NH4+ from the equation Kw = Ka × Kb. Solution: a) HCOO–(aq) + H2O(l) ⇆ HCOOH(aq) + OH–(aq) Concentration (M) HCOO–(aq) + H2O(l) ⇆ HCOOH(aq) + OH–(aq) Initial 0.65 — 0 0 Change –x +x +x Equilibrium 0.65 – x x x −14 K 1.0 × 10 Kb of HCOO– = w = = 5.55556×10–11 Ka 1.8×10−4 Kb = 5.55556×10–11 = Kb = 5.55556×10–11 = Kb = 5.55556×10–11 =

[ HCOOH ][ OH− ]

[ HCOO− ] [ x ][ x ]

⎡0.65 − x⎤ ⎣⎢ ⎦⎥ ( x)( x)

Assume x is small compared to 0.65.

(0.65) x = 6.00925×10 M OH– Check assumption that x is small compared to 0.65: 6.00925×10−6 (100) = 0.0009% error, so the assumption is valid. 0.65 Kw 1.0×10−14 [H3O]+ = = = 1.66410×10–9 M H3O+ − [ OH ] 6.00925×10−6 pH = –log [H3O+] = –log (1.66410×10–9) = 8.7788 = 8.78 + b) NH4 (aq) + H2O(l) ⇆ H3O+(aq) + NH3(aq) Concentration (M) NH4+(aq) + H2O(l) ⇆ NH3(aq) Initial 0.85 — 0 Change –x +x Equilibrium 0.85 – x x –6

H3O+(aq) 0 +x x

18-42

Ka of NH4+ =

Kw 1.0×10−14 = = 5.681818×10–10 Kb 1.76×10−5

Ka = 5.681818×10–10 = Ka = 5.681818×10–10 =

[ H 3O+ ][ NH 3 ] [ NH 4 + ] [ x ][ x ]

Assume x is small compared to 0.85. [ 0.85 − x ] [ x ][ x ] Ka = 5.681818×10–10 = [0.85] [H3O+] = x = 2.1976×10–5 M Check assumption that x is small compared to 0.85: 2.1976 ×10−5 (100) = 0.003% error, so the assumption is valid. 0.85 pH = – log [H3O+] = – log (2.1976×10–5) = 4.65805 = 4.66 18.113 a) The fluoride ion, F–, acts as the base as shown by the following equation: F–(aq) + H2O(l) ⇆ HF(aq) + OH–(aq) Because NaF is a soluble salt, [F–] = [NaF]. The sodium ion is from a strong base; therefore, it will not affect the pH, and can be ignored. Concentration (M) F–(aq) + H2O(l) ⇆ HF(aq) + OH–(aq) Initial 0.75 — 0 0 Change –x +x +x Equilibrium 0.75 – x x x Kw 1.0 ×10−14 – –11 Kb of F = = = 1.470588×10 Ka 6.8×10−4 Kb = 1.470588×10–11 = Kb = 1.470588×10–11 = Kb = 1.470588×10–11 =

[ HF ][ OH− ]

[ F− ] [ x ][ x ]

⎡0.75 − x⎤ ⎢⎣ ⎥⎦ ( x)( x)

Assume x is small compared to 0.75.

(0.75) x = 3.3210558×10 M OH– Check assumption: (3.3210558×10–6/0.75) × 100% = 0.0004% error, so the assumption is valid. [H3O]+ = Kw/[OH–] = (1.0×10–14)/(3.3210558×10–6) = 3.01109×10–9 M H3O+ pH = –log [H3O+] = –log (3.01109×10–9) = 8.521276 = 8.52 b) The pyridinium ion, C5H5NH+, acts as an acid shown by the following equation: C5H5NH+(aq) + H2O(l) ⇆ H3O+(aq) + C5H5N(aq) Because C5H5NHCl is a soluble salt, [C5H5NH+] = [C5H5NHCl]. The chloride ion is from a strong acid; therefore, it will not affect the pH, and can be ignored. Concentration (M) C5H5NH+(aq) + H2O(l) ⇆ C5H5NH(aq) + H3O+(aq) Initial 0.88 — 0 0 Change –x +x +x Equilibrium 0.88 – x x x −14 K 1.0 × 10 Ka of NH4+ = w = = 5.88235×10–6 Kb 1.7×10−9 –6

18-43

[ H3O+ ][C5 H5 N ] K = 5.88235×10 = [C5 H5 NH+ ] –6

Ka = 5.88235×10–6 = Ka = 5.88235×10–6 =

[ x ][ x ]

Assume x is small compared to 0.88.

[ 0.88 − x ] ( x)( x)

(0.88) [H3O ] = x = 2.275186×10–3 M Check assumption: (2.275185×10–3/0.88) × 100% = 0.3% error, so the assumption is valid. pH = –log [H3O+] = –log (2.275185×10–3) = 2.64298 = 2.64 +

18.114 Plan: First, calculate the initial molarity of ClO– from the mass percent. Then, set up reaction table with base dissociation of ClO–. Find the Kb for ClO– from the equation Kw = Ka × Kb, using the Ka for HClO from Appendix C. Solution: − ⎛ 1 mL solution ⎞⎛ ⎟⎟⎜⎜ 1.0 g solution ⎞⎛ ⎟⎟⎜⎜ 6.5% NaClO ⎞⎛ ⎟⎟⎜⎜ 1 mol NaClO ⎞⎛ ⎟⎟⎜⎜ 1 mol ClO ⎞⎟⎟ ⎜ Molarity of ClO– = ⎜⎜ −3 ⎟⎟⎟⎜⎜⎜1 mL solution ⎟⎟⎟⎜⎜⎜100% Solution ⎟⎟⎟⎜⎜⎜ 74.44 g NaClO ⎟⎟⎟⎜⎜⎜1 mol NaClO ⎟⎟⎟ ⎜⎝⎜10 L solution ⎠⎝ ⎠⎝ ⎠⎝ ⎠⎝ ⎠ – = 0.873186 M ClO The sodium ion is from a strong base; therefore, it will not affect the pH, and can be ignored. Concentration (M) ClO–(aq) + H2O(l) ⇆ HClO(aq) + OH–(aq) Initial 0.873186 — 0 0 Change –x +x +x Equilibrium 0.873186 – x x x −14 K 1.0 × 10 Kb of ClO– = w = = 3.448275862×10–7 Ka 2.9×10−8 Kb = 3.448275862×10–7 = Kb = 3.448275862×10–7 = Kb = 3.448275862×10–7 =

[ HClO ][ OH− ]

[ ClO− ] [ x ][ x ]

⎡0.873186 − x⎤ ⎣⎢ ⎦⎥ ( x)( x)

Assume x is small compared to 0.873186.

(0.873186) x = 5.4872×10 = 5.5×10 M OH– Check assumption that x is small compared to 0.873186: 5.4872 ×10−4 (100) = 0.006% error, so the assumption is valid. 0.873186 Kw 1.0 ×10−14 [H3O]+ = = = 1.82242×10–11 M H3O+ [ OH− ] 5.4872×10−4 pH = –log [H3O+] = –log (1.82242×10–11) = 10.73935 = 10.74 –4

–4

18.115 The cation ion, HC18H21NO3+, acts as an acid shown by the following equation: HC18H21NO3+(aq) + H2O(l) ⇆ C18H21NO3(aq) + H3O+(aq) Because HC18H21NO3Cl is a soluble salt, [HC18H21NO3+] = [HC18H21NO3Cl]. The chloride ion is from a strong acid; therefore, it will not affect the pH, and can be ignored. Concentration (M) HC18H21NO3+(aq) + H2O(l) ⇆ C18H21NO3(aq) + H3O+(aq) Initial 0.050 — 0 0 Change –x +x +x Equilibrium 0.050 – x x x Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

18-44

K b = 10 − pK b = 10 −5.80 = 1.58489 × 10−6

Ka of HC18H21NO3+ = Kw/Kb = (1.0×10–14)/(1.58489×10–6) = 6.309586×10–9 [ H3O+ ][C18 H21NO3 ] –9 Ka = 6.309586×10 = [ HC18 H21NO3+ ] ( x)( x) Ka = 6.309586×10–9 = Assume x is small compared to 0.050. 0.050 − x

(

Ka = 6.309586×10–9 =

)

( x )( x )

(0.050) [H3O+] = x = 1.7761737×10–5 M Check assumption: (1.7761737×10–5/0.050) × 100% = 0.04% error, so the assumption is valid. pH = –log [H3O+] = –log (1.7761737×10–5) = 4.75051 = 4.75 18.116 Salts that contain anions of weak acids and cations of strong bases are basic. Salts that contain cations of weak bases or small, highly charged metal cations, and anions of strong acids are acidic. Salts that contain cations of strong bases and anions of strong acids are neutral. Basic salt: KCN (K+ is the cation from the strong base KOH; CN– is the anion from the weak acid, HCN.) Acid salt: FeCl3 or NH4NO3 (Fe3+ is a small, highly charged metal cation and Cl– is the anion of the strong acid HCl; NH4+ is the cation of the weak base NH3, while NO3– is the anion of the strong acid HNO3.) Neutral salt: KNO3 (K+ is the cation of the strong base KOH, while NO3– is the anion of the strong acid HNO3.) 18.117 Sodium fluoride, NaF, contains the cation of a strong base, NaOH, and anion of a weak acid, HF. This combination yields a salt that is basic in aqueous solution as the F– ion acts as a base: F–(aq)+ H2O(l) ⇆ HF(aq) + OH–(aq) Sodium chloride, NaCl, is the salt of a strong base, NaOH, and strong acid, HCl. This combination yields a salt that is neutral in aqueous solution as neither Na+ or Cl– react in water to change the [H3O+]. 18.118 If Ka for the conjugate acid of the anion is approximately equal to Kb for the conjugate base of the cation, the solution will be close to neutral. Otherwise, the solution will be acidic or basic. In this case, the Ka for the conjugate acid (CH3COOH) is 1.8×10–5, and the Kb for the conjugate base (NH3) is 1.76×10–5. 18.119 Plan: For each salt, first break into the ions present in solution and then determine if either ion acts as a weak acid or weak base to change the pH of the solution. Cations are neutral if they are from a strong base; other cations will be weakly acidic. Anions are neutral if they are from a strong acid; other anions are weakly basic. Solution: + – O a) KBr(s) ⎯H⎯⎯ → K (aq) + Br (aq) + K is the conjugate acid of a strong base, so it does not influence pH. Br– is the conjugate base of a strong acid, so it does not influence pH. Since neither ion influences the pH of the solution, it will remain at the neutral pH of pure water. – + O b) NH4I(s) ⎯H⎯⎯ → NH4 (aq) + I (aq) + NH4 is the conjugate acid of a weak base, so it will act as a weak acid in solution and produce H3O+ as represented by the acid-dissociation reaction: NH4+(aq) + H2O(l) ⇆ NH3(aq) + H3O+(aq) – I is the conjugate base of a strong acid, so it will not influence the pH. The production of H3O+ from the ammonium ion makes the solution of NH4I acidic. + – O c) KCN(s) ⎯H⎯⎯ → K (aq) + CN (aq) + K is the conjugate acid of a strong base, so it does not influence pH. CN– is the conjugate base of a weak acid, so it will act as a weak base in solution and impact pH by the base-dissociation reaction: 2

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CN–(aq) + H2O(l) ⇆ HCN(aq) + OH–(aq) Hydroxide ions are produced in this equilibrium so solution will be basic. 18.120 a) SnCl2(s) + nH2O(l) → Sn(H2O)n2+(aq) + 2Cl–(aq) Sn(H2O)n2+(aq) + H2O(l) ⇆ Sn(H2O)n-1OH+(aq) + H3O+(aq) acidic b) NaHS(s) + H2O(l) → Na+(aq) + HS–(aq) HS–(aq) + H2O(l) ⇆ OH–(aq) + H2S(aq) basic c) Zn(CH3COO)2(s) + nH2O(l) → Zn(H2O)n2+(aq) + 2CH3COO–(aq) Zn(H2O)n2+(aq) + H2O(l) ⇆ Zn(H2O)n–1OH+(aq) + H3O+(aq) CH3COO–(aq) + H2O(l) ⇆ OH–(aq) + CH3COOH(aq) Ka (Zn(H2O)n2+) = 1×10–9 Kb (CH3COO–) = Kw/Ka = (1.0×10–14)/(1.8×10–5) = 5.5556×10–10 The two K values are similar, so the solution is close to neutral or slightly acidic. 18.121 Plan: For each salt, first break into the ions present in solution and then determine if either ion acts as a weak acid or weak base to change the pH of the solution. Cations are neutral if they are from a strong base; other cations will be weakly acidic. Anions are neutral if they are from a strong acid; other anions are weakly basic. Solution: a) The two ions that comprise sodium carbonate, Na2CO3, are sodium ion, Na+, and carbonate ion, CO32–. + 2– O Na2CO3(s) ⎯H⎯⎯ → 2Na (aq) + CO3 (aq) Sodium ion is derived from the strong base NaOH. Carbonate ion is derived from the weak acid HCO3–. A salt derived from a strong base and a weak acid produces a basic solution. Na+ does not react with water. CO32–(aq) + H2O(l) ⇆ HCO3–(aq) + OH–(aq) b) The two ions that comprise calcium chloride, CaCl2, are calcium ion, Ca2+, and chloride ion, Cl–. 2+ – O CaCl2(s) ⎯H⎯⎯ → Ca (aq) + 2Cl (aq) Calcium ion is derived from the strong base Ca(OH)2. Chloride ion is derived from the strong acid HCl. A salt derived from a strong base and strong acid produces a neutral solution. Neither Ca2+ nor Cl– reacts with water. c) The two ions that comprise cupric nitrate, Cu(NO3)2, are the cupric ion, Cu2+, and the nitrate ion, NO3–. 2+ – O Cu(NO3)2(s) ⎯H⎯⎯ → Cu (aq) + 2NO3 (aq) Small metal ions are acidic in water (assume the hydration of Cu2+ is 6): Cu(H2O)62+(aq) + H2O(l) ⇆ Cu(H2O)5OH+(aq) + H3O+(aq) Nitrate ion is derived from the strong acid HNO3. Therefore, NO3– does not react with water. A solution of cupric nitrate is acidic. 2

18.122 a) CH3NH3Cl(s) + H2O(l) → CH3NH3+(aq) + Cl–(aq) CH3NH3+(aq) + H2O(l) ⇆ H3O+(aq) + CH3NH2(aq) acidic b) LiClO4(s) + H2O(l) → Li+(aq) + ClO4–(aq) neutral c) CoF2(s) + nH2O(l) → Co(H2O)n2+(aq) + 2F–(aq) Co(H2O)n2+(aq) + H2O(l) ⇆ Co(H2O)n–1OH+(aq) + H3O+(aq) F–(aq) + H2O(l) ⇆ OH–(aq) + HF(aq) Ka (Co(H2O)n2+ ) = 2×10–10 Kb (F–) = Kw/Ka = (1.0×10–14)/(6.8×10– 4) = 1.47×10–11 The two K values are similar so the solution is close to neutral or slightly acidic. 18.123 Plan: For each salt, first break into the ions present in solution and then determine if either ion acts as a weak acid or weak base to change the pH of the solution. Cations are neutral if they are from a strong base; other cations will be weakly acidic. Anions are neutral if they are from a strong acid; other anions are weakly basic. Solution: a) A solution of strontium bromide is neutral because Sr2+ is the conjugate acid of a strong base, Sr(OH)2, and Br– is the conjugate base of a strong acid, HBr, so neither change the pH of the solution. Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

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b) A solution of barium acetate is basic because CH3COO– is the conjugate base of a weak acid and therefore forms OH– in solution whereas Ba2+ is the conjugate acid of a strong base, Ba(OH)2, and does not influence solution pH. The base-dissociation reaction of acetate ion is CH3COO–(aq) + H2O(l) ⇆ CH3COOH(aq) + OH–(aq). c) A solution of dimethylammonium bromide is acidic because (CH3)2NH2+ is the conjugate acid of a weak base and therefore forms H3O+ in solution whereas Br– is the conjugate base of a strong acid and does not influence the pH of the solution. The acid-dissociation reaction for methylammonium ion is (CH3)2NH2+(aq) + H2O(l) ⇆ (CH3)2NH(aq) + H3O+(aq). 18.124 a) Fe(HCOO)3(s) + nH2O(l) → Fe(H2O)n3+(aq) + 3HCOO–(aq) Fe(H2O)n3+(aq) + H2O(l) ⇆ Fe(H2O)n–1OH2+(aq) + H3O+(aq) HCOO–(aq) + H2O(l) ⇆ OH–(aq) + HCOOH(aq) Ka (Fe3+) = 6×10–3 Kb (HCOO–) = Kw/Ka = (1.0×10–14)(1.8×10–4) = 5.5556×10–-11 Ka (Fe3+) > Kb (HCOO–) acidic b) KHCO3(s) + H2O(l) → K+(aq) + HCO3–(aq) HCO3–(aq) + H2O(l) ⇆ H3O+(aq) + CO32–(aq) HCO3–(aq) + H2O(l) ⇆ OH–(aq) + H2CO3(aq) Kb (HCO3–) > Ka (HCO3–) basic c) K2S(s) + H2O(l) ⇆ 2K+(aq) + S2–(aq) S2–(aq) + H2O(l) ⇆ OH–(aq) + HS–(aq) basic 18.125 Plan: For each salt, first break into the ions present in solution and then determine if either ion acts as a weak acid or weak base to change the pH of the solution. Cations are neutral if they are from a strong base; other cations will be weakly acidic. Anions are neutral if they are from a strong acid; other anions are weakly basic. Solution: a) The two ions that comprise ammonium phosphate, (NH4)3PO4, are the ammonium ion, NH4+, and the phosphate ion, PO43–. NH4+(aq) + H2O(l) ⇆ NH3(aq) + H3O+(aq) Ka = Kw/Kb (NH3) = 5.7×10–10 – 3– 2– PO4 (aq) + H2O(l) ⇆ HPO4 (aq) + OH (aq) Kb = Kw/Ka3 (H3PO4) = 2.4×10–2 A comparison of Ka and Kb is necessary since both ions are derived from a weak base and weak acid. The Ka of NH4+ is determined by using the Kb of its conjugate base, NH3 (Appendix). The Kb of PO43– is determined by using the Ka of its conjugate acid, HPO42–. The Ka of HPO42– comes from Ka3 of H3PO4 (Appendix). Since Kb > Ka, a solution of (NH4)3PO4 is basic. b) The two ions that comprise sodium sulfate, Na2SO4, are sodium ion, Na+, and sulfate ion, SO42–. The sodium ion is derived from the strong base NaOH. The sulfate ion is derived from the weak acid, HSO4– . SO42–(aq) + H2O(l) ⇆ HSO4–(aq) + OH–(aq) A solution of sodium sulfate is basic. c) The two ions that comprise lithium hypochlorite, LiClO, are lithium ion, Li+, and hypochlorite ion, ClO–. Lithium ion is derived from the strong base LiOH. Hypochlorite ion is derived from the weak acid, HClO (hypochlorous acid). ClO–(aq) + H2O(l) ⇆ HClO(aq) + OH–(aq) A solution of lithium hypochlorite is basic. 18.126 a) Pb(CH3COO)2(s) + nH2O(l) → Pb(H2O)n2+(aq) + 2CH3COO–(aq) Pb(H2O)n2+(aq) + H2O(l) ⇆ Pb(H2O)n–1OH+(aq) + H3O+(aq) Ka (Pb2+) = 3×10–8 CH3COO–(aq) + H2O(l) ⇆ CH3COOH(aq) + OH–(aq) Kb (CH3COO–) = Kw/Ka = (1.0×10–14)/(1.8×10–5) = 5.5556×10–10 Ka (Pb2+) > Kb (CH3COO–) acidic b) Cr(NO2)3(s) + nH2O(l) → Cr(H2O)n3+(aq) + 3NO2–(aq) Cr(H2O)n3+(aq) + H2O(l) ⇆ Cr(H2O)n–1OH2+(aq) + H3O+(aq) Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

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NO2–(aq) + H2O(l) ⇆ OH–(aq) + HNO2(aq) Ka (Cr3+) = 1×10–4 Kb (NO2–) = Kw/Ka = (1.0×10–14)/(7.1×10–4) = 1.40845×10–11 Ka (Cr3+) > Kb (NO2–) acidic c) CsI(s) + H2O(l) → Cs+(aq) + I–(aq) neutral 18.127 Plan: For each salt, first break into the ions present in solution and then determine if either ion acts as a weak acid or weak base to change the pH of the solution. Cations are neutral if they are from a strong base; other cations will be weakly acidic. Anions are neutral if they are from a strong acid; other anions are weakly basic. Use Ka and Kb values to rank the pH; the larger the Ka value, the lower the pH and the larger the Kb value, the higher the pH. Solution: a) Order of increasing pH: Fe(NO3)2 < KNO3 < K2SO3 < K2S (assuming concentrations equivalent) Iron(II) nitrate, Fe(NO3)2, is an acidic solution because the iron ion is a small, highly charged metal ion that acts as a weak acid and nitrate ion is the conjugate base of a strong acid, so it does not influence pH. Potassium nitrate, KNO3, is a neutral solution because potassium ion is the conjugate acid of a strong base and nitrate ion is the conjugate base of a strong acid, so neither influences solution pH. Potassium sulfite, K2SO3, and potassium sulfide, K2S, are similar in that the potassium ion does not influence solution pH, but the anions do because they are conjugate bases of weak acids. Ka for HSO3– is 6.5×10–8, so Kb for SO3– is 1.5×10–7, which indicates that sulfite ion is a weak base. Ka for HS– is 1×10–17 (see the table of Ka values for polyprotic acids), so sulfide ion has a Kb equal to 1×103. Sulfide ion is thus a strong base. The solution of a strong base will have a greater concentration of hydroxide ions (and higher pH) than a solution of a weak base of equivalent concentrations. b) In order of increasing pH: NaHSO4 < NH4NO3 < NaHCO3 < Na2CO3 In solutions of ammonium nitrate, only the ammonium will influence pH by dissociating as a weak acid: NH4+(aq) + H2O(l) ⇆ NH3(aq) + H3O+(aq) with Ka = 1.0×10–14/1.8×10–5 = 5.6×10–10 Therefore, the solution of ammonium nitrate is acidic. In solutions of sodium hydrogen sulfate, only HSO4– will influence pH. The hydrogen sulfate ion is amphoteric so both the acid and base dissociations must be evaluated for influence on pH. As a base, HSO4– is the conjugate base of a strong acid, so it will not influence pH. As an acid, HSO4– is the conjugate acid of a weak base, so the acid dissociation applies: HSO4–(aq) + H2O(l) ⇆ SO42–(aq) + H3O+(aq) Ka2 = 1.2×10–2 In solutions of sodium hydrogen carbonate, only the HCO3– will influence pH and it, like HSO4–, is amphoteric: As an acid: HCO3–(aq) + H2O(l) ⇆ CO32–(aq) + H3O+(aq) Ka = 4.7×10–11, the second Ka for carbonic acid As a base: HCO3–(aq) + H2O(l) ⇆ H2CO3(aq) + OH–(aq) Kb = 1.0×10–14/4.5×10–7 = 2.2×10–8 Since Kb > Ka, a solution of sodium hydrogen carbonate is basic. In a solution of sodium carbonate, only CO32– will influence pH by acting as a weak base: CO32–(aq) + H2O(l) ⇆ HCO3–(aq) + OH–(aq) Kb = 1.0×10–14/4.7×10–11 = 2.1×10–4 Therefore, the solution of sodium carbonate is basic. Two of the solutions are acidic. Since the Ka of HSO4– is greater than that of NH4+, the solution of sodium hydrogen sulfate has a lower pH than the solution of ammonium nitrate, assuming the concentrations are relatively close. Two of the solutions are basic. Since the Kb of CO32– is greater than that of HCO3–, the solution of sodium carbonate has a higher pH than the solution of sodium hydrogen carbonate, assuming concentrations are not extremely different. 18.128 a) KClO2 > MgCl2 > FeCl2 > FeCl3 b) NaBrO2 > NaClO2 > NaBr > NH4Br Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

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18.129 Both methoxide ion and amide ion produce OH– in aqueous solution. In water, the strongest base possible is OH–. Since both bases produce OH– in water, both bases appear equally strong. CH3O–(aq) + H2O(l) → OH–(aq) + CH3OH(aq) NH2–(aq) + H2O(l) → OH–(aq) + NH3(aq) 18.130 H2SO4 is a strong acid and would be 100% dissociated in H2O and any solvent more basic than H2O (such as NH3). It would be less than 100% dissociated in solvents more acidic than H2O (such as CH3COOH). 18.131 Ammonia, NH3, is a more basic solvent than H2O. In a more basic solvent, weak acids like HF act like strong acids and are 100% dissociated. 18.132 A Lewis base must have an electron pair to donate. A Lewis acid must have a vacant orbital or the ability to rearrange its bonding to make one available. The Lewis acid-base reaction involves the donation and acceptance of an electron pair to form a new covalent bond in an adduct. 18.133 A Lewis acid is defined as an electron-pair acceptor, while a Brønsted-Lowry acid is a proton donor. If only the proton in a Brønsted-Lowry acid is considered, then every Brønsted-Lowry acid fits the definition of a Lewis acid since the proton is accepting an electron pair when it bonds with a base. There are Lewis acids that do not include a proton, so all Lewis acids are not Brønsted-Lowry acids. A Lewis base is defined as an electron-pair donor and a Brønsted-Lowry base is a proton acceptor. In this case, the two definitions are essentially the same. 18.134 a) No, a weak Brønsted-Lowry base is not necessarily a weak Lewis base. For example, the following equation shows that the weak Brønsted-Lowry base NH3 is a good Lewis base. Ni(H2O)62+(aq) + 6 NH3(aq) ⇆ Ni(NH3)62+(aq) + 6H2O(l) b) The cyanide ion has a lone pair to donate from either the C or the N, and donates an electron pair to the Cu(H2O)62+ complex. It is the Lewis base for the forward direction of this reaction. In the reverse direction, water donates one of the electron pairs on the oxygen to the Cu(CN)42– and is the Lewis base. c) Because Kc > 1, the reaction proceeds in the direction written (left to right) and is driven by the stronger Lewis base, the cyanide ion. 18.135 All three concepts can have water as the product in an acid-base neutralization reaction. It is the only product in an Arrhenius neutralization reaction. 18.136 a) NH3 can only act as a Brønsted-Lowry or Lewis base. b) AlCl3 can only act as a Lewis acid. 18.137 Plan: A Lewis acid is an electron-pair acceptor and therefore must be able to accept an electron pair. A Lewis base is an electron-pair donor and therefore must have an electron pair to donate. Solution: a) Cu2+ is a Lewis acid because it accepts electron pairs from molecules such as water. b) Cl– is a Lewis base because it has lone pairs of electrons it can donate to a Lewis acid. c) Tin(II) chloride, SnCl2, is a compound with a structure similar to carbon dioxide, so it will act as a Lewis acid to form an additional bond to the tin. d) Oxygen difluoride, OF2, is a Lewis base with a structure similar to water, where the oxygen has lone pairs of electrons that it can donate to a Lewis acid. 18.138 a) Lewis acid

b) Lewis base

c) Lewis base

d) Lewis acid

18.139 Plan: A Lewis acid is an electron-pair acceptor and therefore must be able to accept an electron pair. A Lewis base is an electron-pair donor and therefore must have an electron pair to donate. Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

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Solution: a) The boron atom in boron trifluoride, BF3, is electron deficient (has six electrons instead of eight) and can accept an electron pair; it is a Lewis acid. b) The sulfide ion, S2–, can donate any of four electron pairs and is a Lewis base. c) The Lewis dot structure for the sulfite ion, SO32–, shows lone pairs on the sulfur and on the oxygen atoms. The sulfur atom has a lone electron pair that it can donate more easily than the electronegative oxygen in the formation of an adduct. The sulfite ion is a Lewis base. d) Sulfur trioxide, SO3, acts as a Lewis acid. 18.140 a) Lewis acid

b) Lewis base

c) Lewis acid

d) Lewis acid

18.141 Plan: A Lewis acid is an electron-pair acceptor while a Lewis base is an electron-pair donor. Solution: a) Sodium ion is the Lewis acid because it is accepting electron pairs from water, the Lewis base. Na+ + 6H2O ⇆ Na(H2O)6+ Lewis acid Lewis base adduct b) The oxygen from water donates a lone pair to the carbon in carbon dioxide. Water is the Lewis base and carbon dioxide the Lewis acid. CO2 + H2O ⇆ H2CO3 Lewis acid Lewis base adduct c) Fluoride ion donates an electron pair to form a bond with boron in BF4–. The fluoride ion is the Lewis base and the boron trifluoride is the Lewis acid. F– + BF3 ⇆ BF4– Lewis base Lewis acid adduct 18.142 a) Fe3+ + Lewis acid b) H2O + Lewis acid c) 4CO + Lewis base

2H2O ⇆ Lewis base H– ⇆ Lewis base Ni ⇆ Lewis acid

FeOH2+ OH–

+ +

H3O+ H2

Ni(CO)4

18.143 Plan: In an Arrhenius acid-base reaction, H+ ions react with OH– ions to produce H2O. In a Brønsted-Lowry acid-base reaction, an acid donates H+ to a base. In a Lewis acid-base reaction, an electron pair is donated by the base and accepted by the acid. Solution: a) Since neither H+ nor OH– is involved, this is not an Arrhenius acid-base reaction. Since there is no exchange of protons, this is not a Brønsted-Lowry reaction. This reaction is only classified as Lewis acid-base reaction, where Ag+ is the acid and NH3 is the base. b) Again, no OH– is involved, so this is not an Arrhenius acid-base reaction. This is an exchange of a proton, from H2SO4 to NH3, so it is a Brønsted-Lowry acid-base reaction. Since the Lewis definition is most inclusive, anything that is classified as a Brønsted-Lowry (or Arrhenius) reaction is automatically classified as a Lewis acidbase reaction. c) This is not an acid-base reaction. d) For the same reasons listed in a), this reaction is only classified as Lewis acid-base reaction, where AlCl3 is the acid and Cl– is the base. 18.144 a) Lewis acid-base reaction c) This is not an acid-base reaction.

b) Brønsted-Lowry, Arrhenius, and Lewis acid-base reaction d) Brønsted-Lowry and Lewis acid-base reaction

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18.145 a) The hydrogen-bonded form would be: H Cl3C O ····· H O H The product of a Lewis acid-base reaction is: OH

Cl3C

OH The O atom of water (Lewis base) donates a lone pair to the C of the carbonyl group, which functions as a Lewis acid. The π bond is broken and a C–O bond is formed. The O atom of the carbonyl (Lewis base) accepts a proton from water (Lewis acid) to complete the reaction. b) Infrared spectroscopy could be used. Infrared spectroscopy is very good at identifying functional groups. C=O bonds have a characteristic range of absorption wavelengths. The hydrogen-bonded structure would have a carbonyl group and the second structure would not.

18.146 Plan: Calculate the [H3O+] using the pH values given. Determine the value of Kw from the pKw given. The [H3O+] is combined with the Kw value at 37°C to find [OH–] using Kw = [H3O+][OH–]. Solution: Kw = 10–pKw = 10–13.63 = 2.34423×10–14 Kw = [H3O+][OH–] = 2.34423×10–14 at 37°C + [H3O ] range High value (low pH) = 10–pH = 10–7.35 = 4.46684×10–8 = 4.5×10–8 M H3O+ Low value (high pH) = 10–pH = 10–7.45 = 3.54813×10–8 = 3.5×10–8 M H3O+ Range: 3.5×10–8 to 4.5×10–8 M H3O+ – [OH ] range Kw = [H3O+][OH–] = 2.34423×10–14 at 37°C Kw [OH–] = [H 3 O]+

2.34423×10−14 = 6.60695×10–7 = 6.6×10–7 M OH– 3.54813×10−8 2.34423×10−14 Low value (low pH) = = 5.24807×10–7 = 5.2×10–7 M OH– −8 4.46684 ×10 Range: 5.2×10–7 to 6.6×10–7 M OH– 18.147 a) Acids will vary in the amount they dissociate (acid strength) depending on the acid-base character of the solvent. Water and methanol have different acid-base characters. b) The Ka is the measure of an acid’s strength. A stronger acid has a smaller pKa. Therefore, phenol is a stronger acid in water than it is in methanol. In other words, water more readily accepts a proton from phenol than does methanol, i.e., methanol is a weaker base than water. c) C6H5OH(solvated) + CH3OH(l) ⇆ CH3OH2+(solvated) + C6H5O–(solvated) The term “solvated” is analogous to “aqueous.” “Aqueous” would be incorrect in this case because the reaction does not take place in water. d) In the autoionization process, one methanol molecule is the proton donor while another methanol molecule is the proton acceptor. CH3OH(l) + CH3OH(l) ⇆ CH3O–(solvated) + CH3OH2+(solvated) In this equation “(solvated)” indicates that the molecules are solvated by methanol. High value (high pH) =

18-51

The equilibrium constant for this reaction is the autoionization constant of methanol: K = [CH3O–][CH3OH2+] 18.148 a) Step 1 CO2(g) + H2O(l) ⇆ H2CO3(aq) Lewis Step 2 H2CO3(aq) + H2O(l) ⇆ HCO3–(aq) + H3O+(aq) Brønsted-Lowry and Lewis b) Molarity of CO2 = kH Pcarbon dioxide = (0.033 mol/L ∙ atm)(4×10–4 atm) = 1.320×10–5 M CO2 CO2(g) + 2H2O(l) ⇆ HCO3–(aq) + H3O+(aq) Koverall = 4.5×10–7 + − [ H3O ][ HCO3 ] Koverall = 4.5×10–7 = [CO2 ] ⎡ ⎤ ⎣ x ⎦ [x] Assume x is small compared to 1.320×10–5. [1.320 ×10 -5 − x] ⎡ x⎤ [x] Koverall = 4.5×10–7 = ⎣ ⎦ 1.320×10-5 –6 x = 2.4372×10 Check assumption that x is small compared to 1.320×10–5: 2.4372 ×10 -6 (100) = 18% error, so the assumption is not valid. 1.320 ×10 -5 Since the error is greater than 5%, it is not acceptable to assume x is small compared to 1.320×10–5, and it is necessary to use the quadratic equation. x2 = (4.5×10–7)(1.320×10–5 – x) = 5.940×10–12 – 4.5×10–7x x2 + 4.5×10–7 x – 5.94010–12 = 0 a = 1 b = 4.5×10–7 c = –5.940×10–12

Koverall = 4.5×10–7 =

−b ± b2 − 4ac x=

2a –4.5×10−7 ± (4.5 ×10−7 ) − 4(1)(–5.940 ×10−12 ) 2

2(1)

x = 2.222575×10–6 M H3O+ pH = –log [H3O+] = –log (2.222575×10–6) = 5.6531 = 5.6 c) HCO3–(aq) + H2O(l) ⇆ H3O+(aq) + CO32–(aq) [ H3O+ ] ⎡⎣CO32− ⎤⎦ –11 Ka = 4.7×10 = [ HCO3− ] Use the unrounded x from part b). ⎡2.222575×10−6 + x ⎤ [x] ⎢ ⎥⎦ –11 Ka = 4.7×10 = ⎣ −6 [2.222575×10 − x]

Assume x is small compared to 2×10–6.

⎡2.222575×10−6 ⎤ [x] ⎢ ⎦⎥ Ka = 4.7×10–11 = ⎣ [2.222575 ×10−6 ] [CO32–] = x = 4.7×10–11 = 5×10–11 M CO32– Check assumption that x is small compared to 2×10–6: 4.7 ×10−11 (100) = 0.0021% error, so the assumption is valid. 2.222575×10−6 d) New molarity of CO2 = 2kH Pcarbon dioxide = 2(0.033 mol/L ∙ atm)(4×10–4 atm) = 2.640×10–5 M CO2 Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

18-52

–7

Koverall = 4.5×10 =

[ H3O+ ][ HCO3− ] [CO2 ]

⎡ x ⎤ [x] ⎣ ⎦ Assume x is small compared to 2.640×10–5. [2.640 ×10−5 − x] ⎡ x ⎤ [x] ⎣ ⎦ Koverall = 4.5×10–7 = [2.640 ×10−5 ] x = 3.44674×10–6 Check assumption that x is small compared to 2.640×10–5: 3.44674 ×10-6 (100) = 13% error, so the assumption is not valid. 2.640 ×10-5 Since the error is greater than 5%, it is not acceptable to assume x is small compared to 2.640×10–5, and it is necessary to use the quadratic equation. x2 = (4.5×10–7)(2.640×10–5 – x) = 1.1880×10–11 – 4.5×10–7 x x2 + 4.5×10–7 x – 1.1880×10–11 = 0 a = 1 b = 4.5×10–7 c = –1.1880×10–11

Koverall = 4.5×10–7 =

−b ± b2 − 4ac x=

2a –4.5×10−7 ± (4.5×10−7 ) − 4(1)(–1.1880 ×10−11 ) 2

2(1) x = 3.229×10–6 M H3O+ pH = –log [H3O+] = –log (3.229×10–6) = 5.4909 = 5.5

18.149 At great depths, the higher pressure increases the concentration of H3O+ and the effect is to shift the dissolving reaction to the right, so seashells dissolve more rapidly. 18.150 Plan: A Lewis acid is an electron-pair acceptor and a Lewis base is an electron-pair donor. Recall that n is the main energy level and l is the orbital type. Solution: a) SnCl4 is the Lewis acid accepting an electron pair from (CH3)3N, the Lewis base. b) Tin is the element in the Lewis acid accepting the electron pair. The electron configuration of tin is [Kr]5s24d105p2. The four bonds to tin are formed by sp3 hybrid orbitals, which completely fill the 5s and 5p orbitals. The 5d orbitals are empty and available for the bond with trimethylamine. 18.151 Plan: A 10-fold dilution means that the chemist takes 1 mL of the 1.0×10–5 M solution and dilutes it to 10 mL (or dilute 10 mL to 100 mL). The chemist then dilutes the diluted solution in a 1:10 ratio, and repeats this process for the next two successive dilutions. M1V1 = M2V2 can be used to find the molarity after each dilution. After each dilution, find [H3O+] and calculate the pH. Solution: Hydrochloric acid is a strong acid that completely dissociates in water. Therefore, the concentration of H3O+ is the same as the starting acid concentration: [H3O+] = [HCl]. The original solution pH: pH = –log (1.0×10–5) = 5.00 = pH Dilution 1: M1V1 = M2V2 (1.0×10–5 M)(1.0 mL) = (x)(10. mL) [H3O+]HCl = 1.0×10–6 M H3O+ pH = –log (1.0×10–6) = 6.00 Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

18-53

Dilution 2: (1.0×10–6 M)(1.0 mL) = (x)(10. mL) [H3O+]HCl = 1.0×10–7 M H3O+ Once the concentration of strong acid is close to the concentration of H3O+ from water autoionization, the [H3O+] in the solution does not equal the initial concentration of the strong acid. The calculation of [H3O+] must be based on the water ionization equilibrium: H2O(l) + H2O(l) ⇆ H3O+(aq) + OH–(aq) with Kw = 1.0×10–14 at 25°C. The dilution gives an initial [H3O+] of 1.0×10–7 M. Assuming that the initial concentration of hydroxide ions is zero, a reaction table is set up. Concentration (M) 2H2O(l) ⇆ H3O+(aq) + OH–(aq) Initial — 1×10–7 0 Change — +x +x Equilibrium — 1×10–7 + x x + – –7 Kw = [H3O ][OH ] = (1×10 + x)(x) = 1.0×10–14 Set up as a quadratic equation: x2 + 1.0×10–7 x – 1.0×10–14 = 0 c = –1.0×10–14 a = 1 b = 1.0×10–7 −1.0 ×10−7 ±

(1.0 ×10 ) − 4 (1)(−1.0×10 ) 2

−7

−14

2 (1) –8

x = 6.18034×10 [H3O+] = (1.0×10–7 + x) M = (1.0×10–7 + 6.18034×10–8) M = 1.618034×10–7 M H3O+ pH = –log [H3O+] = –log (1.618034×10–7) = 6.79101 = 6.79 Dilution 3: (1.0×10–7 M)(1.0 mL) = (x)(10. mL) [H3O+]HCl = 1.0×10–8 M H3O+ The dilution gives an initial [H3O+] of 1.0×10–8 M. Assuming that the initial concentration of hydroxide ions is zero, a reaction table is set up. Concentration (M) 2H2O(l) ⇆ H3O+(aq) + OH–(aq) –8 Initial — 1×10 0 Change +x +x Equilibrium — 1×10–8 + x x + – –8 Kw = [H3O ][OH ] = (1×10 + x)(x) = 1.0×10–14 Set up as a quadratic equation: x2 + 1.0×10–8 x – 1.0×10–14 = 0 c = –1.0×10–14 a = 1 b = 1.0×10–8 −1.0 ×10−8 ±

(1.0×10 ) − 4 (1)(−1.0×10 ) −8

−14

2 (1) –8

x = 9.51249×10 [H3O+] = (1.0×10–8 + x) M = (1.0×10–8 + 9.51249×10–8) M = 1.051249×10–7 M H3O+ pH = –log [H3O+] = –log (1.051249×10–7) = 6.97829 = 6.98 Dilution 4: (1.0×10–8 M)(1.0 mL) = (x)(10. mL) [H3O+]HCl = 1.0×10–9 M H3O+ The dilution gives an initial [H3O+] of 1.0×10–9 M. Assuming that the initial concentration of hydroxide ions is zero, a reaction table is set up.

18-54

2H2O(l) ⇆ H3O+(aq) + OH–(aq) Concentration (M) 0 Initial — 1×10–9 Change — +x +x Equilibrium — 1×10–9 + x x + – –9 Kw = [H3O ][OH ] = (1×10 + x)(x) = 1.0×10–14 Set up as a quadratic equation: x2 + 1.0×10–9 x – 1.0×10–14 = 0 c = –1.0×10–14 a = 1 b = 1.0×10–9 −1.0 ×10−9 ±

(1.0 ×10 ) − 4 (1)(−1.0×10 ) −9

−14

2 (1) –8

x = 9.95012×10 [H3O+] = (1.0×10–9 + x) M = (1.0×10–9 + 9.95012×10–8) M = 1.005012×10–7 M H3O+ pH = –log [H3O+] = –log (1.005012×10–7) = 6.9978 = 7.00 As the HCl solution is diluted, the pH of the solution becomes closer to 7.0. Continued dilutions will not significantly change the pH from 7.0. Thus, a solution with a basic pH cannot be made by adding acid to water. 18.152 a) Steps 1, 2, and 4 are Lewis acid-base reactions. b) Step 1 Cl2 + FeCl3 ⇆ FeCl5 (or Cl+FeCl4–) Lewis acid = FeCl3 Lewis base = Cl2 Step 2 C6H6 + Cl+FeCl4– ⇆ C6H6Cl+ + FeCl4– Lewis acid = C6H6 Lewis base = Cl+FeCl4– + – Step 4 H + FeCl4 ⇆ HCl + FeCl3 Lewis acid = H+ Lewis base = FeCl4– 18.153 a) HY(aq) + H2O(l) ⇆ H3O+(aq) + Y–(aq) Ka =

[ H3 O+ ][ Y− ] [ HY ]

Concentrations in Beaker A: ⎛ 0.010 mol ⎞⎟⎛ 1 ⎞ ⎟ = 0.266667 M ⎟⎜ [HY] = (8 particles)⎜⎜ ⎜⎝ 1 particle ⎠⎟⎟⎜⎝ 0.300 L ⎠⎟⎟ ⎛ 0.010 mol ⎞⎟⎛ 1 ⎞ ⎟ = 0.133333 M ⎟⎜ [H3O+] = [Y–] = (4 particles)⎜⎜ ⎜⎝ 1 particle ⎠⎟⎟⎜⎝ 0.300 L ⎠⎟⎟

Ka =

[ H3 O+ ][ Y− ]

[ 0.133333][ 0.133333]

= 0.066667 = 0.067 [ 0.266667 ] Calculate the concentrations in Beakers B-D, then calculate Q to determine which are at equilibrium. Concentrations in Beaker B: ⎛ 0.010 mol ⎞⎟⎛ 1 ⎞ ⎟ = 0.20 M ⎟⎜ [HY] = (6 particles)⎜⎜ ⎜⎝ 1 particle ⎠⎟⎟⎜⎝ 0.300 L ⎠⎟⎟ ⎛ 0.010 mol ⎞⎟⎛ 1 ⎞ ⎟⎟ = 0.066667 M ⎟⎟⎜ [H3O+] = [Y–] = (2 particles)⎜⎜ ⎝⎜ 1 particle ⎠⎟⎜⎝ 0.300 L ⎠⎟

[ HY ]

[ H3 O+ ][ Y− ] [ HY ]

[ 0.066667 ][ 0.066667 ]

[ 0.20 ] Beaker B is not at equilibrium.

= 0.0222222 = 0.022

18-55

Concentrations in Beaker C: ⎛ 0.010 mol ⎞⎟⎛ 1 ⎞ ⎟ = 0.133333 M ⎟⎜ [HY] = (4 particles)⎜⎜ ⎜⎝ 1 particle ⎠⎟⎟⎜⎝ 0.300 L ⎠⎟⎟ ⎛ 0.010 mol ⎞⎟⎛ 1 ⎞ ⎟ = 0.066667 M ⎟⎜ [H3O+] = [Y–] = (2 particles)⎜⎜ ⎜⎝ 1 particle ⎠⎟⎟⎜⎝ 0.300 L ⎠⎟⎟

[ H3 O+ ][ Y− ]

[ 0.066667 ][ 0.066667 ]

= 0.0333345 = 0.033 [ 0.13333] Beaker C is not at equilibrium. Concentrations in Beaker D: ⎛ 0.010 mol ⎞⎟⎛ 1 ⎞ ⎟ = 0.066667 M ⎟⎜ [HY] = (2 particles)⎜⎜ ⎜⎝ 1 particle ⎠⎟⎟⎜⎝ 0.300 L ⎠⎟⎟ ⎛ 0.010 mol ⎞⎟⎛ 1 ⎞ ⎟⎟ = 0.066667 M ⎟⎟⎜ [H3O+] = [Y–] = (2 particles)⎜⎜ ⎝⎜ 1 particle ⎠⎟⎜⎝ 0.300 L ⎠⎟ Q=

[ HY ]

[ H3 O+ ][ Y− ]

[ 0.066667 ][ 0.066667 ]

= 0.066667 = 0.067 [ 0.066667 ] Beaker D is at equilibrium. b) For both beakers B and C, Q < Ka. Therefore, the reaction is proceeding to the right to produce more products. c) Yes, dilution affects the extent of dissociation of a weak acid. Dilution increases the degree of dissociation. For example, in Beaker A, 4 of 12 HY molecules have dissociated for a (4/12)100 = 33% dissociation. In Beaker D, 2 of 4 HY molecules have dissociated for a (2/4)100 = 50% dissociation. [ HY ]

18.154 a) Electrical conductivity of 0.1 M HCl is higher than that of 0.1 M CH3COOH. Conductivity is proportional to the concentration of charge in the solution. Since HCl dissociates to a greater extent than CH3COOH, the concentration of ions, and thus the charge, is greater in 0.1 M HCl than in 0.1 M CH3COOH. b) The electrical conductivity of the two solutions will be approximately the same because at low concentrations the autoionization of water is significant causing the concentration of ions, and thus the charge, to be about the same in the two solutions. In addition, the percent dissociation of a weak electrolyte such as acetic acid increases with decreasing concentration. +

18.155 In step (1), the RCOOH is the Lewis base and the H+ is the Lewis acid. In step (2), the RC(OH)2 is the Lewis acid and the R'OH is the Lewis base. 18.156 a) pH = –log [H3O+] HCl is a strong acid so [H3O+] = M HCl pH = –log (0.10) = 1.00 HClO2 and HClO are weak acids requiring a Ka from the Appendix. HClO(aq) + H2O(l) ⇆ H3O+(aq) + ClO–(aq) 0.10 – x x x + [ H O ][ ClO− ] Ka = 2.9×10–8 = 3 [ HClO ] ( x)( x) Ka = 2.9×10–8 = Assume x is small compared to 0.10. 0.10 − x

(

Ka = 2.9×10–8 =

)

( x)( x)

(0.10)

[H3O ] = x = 5.38516×10–5 M

18-56

Check assumption: (5.38516×10–5/0.10) × 100% = 0.05%. The assumption is good. pH = –log [H3O+] = –log (5.38516×10–5) = 4.2688 = 4.27 HClO2(aq) + H2O(l) ⇆ H3O+(aq) + ClO2–(aq) 0.10 – x x x + [ H 3 O ][ClO 2− ] Ka = 1.1×10–2 = [ HClO 2 ] [ x ][ x ] Ka = 1.1×10–2 = Assume x is small compared to 0.10. ⎡0.10 − x ⎤ ⎢⎣ ⎥⎦ [ x ][ x ] Ka = 1.1×10–2 = ⎡ 0.10 ⎤ ⎢⎣ ⎥⎦ x = 0.033166 Check assumption: (0.033166/0.10) × 100% = 33%. The assumption is not valid. The problem will need to be solved as a quadratic. x2 = (1.1×10–2)(0.10 – x) = 1.1×10–3 – 1.1×10–2 x x2 + 1.1×10–2 x – 1.1×10–3 = 0 a = 1 b = 1.1×10–-2 c = – 1.1×10–-3 −1.1×10−2 ±

(1.1×10 ) − 4 (1)(−1.1×10 ) −2

−3

2 (1) –2

x = 2.8119×10 M H3O pH = –log [H3O+] = –log (2.8119×10–2) = 1.550997 = 1.55 b) The lowest H3O+ concentration is from the HClO. Leave the HClO beaker alone, and dilute the other acids until they yield the same H3O+ concentration. A dilution calculation is needed to calculate the amount of water added. HCl Mi = 0.10 M Vi = 100. mL Mf = 5.38516×10–5 M Vf = ? MiVi = MfVf Vf = MiVi/Mf = [(0.10 M)(100. mL)]/(5.38516×10–5 M) = 1.85695×105 mL Volume water added = (1.85695×105 mL) – 100. mL = 1.85595×105 = 1.9×105 mL H2O added HClO2 requires the Ka for the acid with the ClO2– concentration equal to the H3O+ concentration. The final molarity of the acid will be Mf, which may be used in the dilution equation. HClO2(aq) + H2O(l) ⇆ H3O+(aq) + ClO2–(aq) –5 –5 x – 5.38516×10 5.38516×10 5.38516×10–5 + − [ H 3 O ][ClO 2 ] Ka = 1.1×10–2 = [ HClO 2 ] ⎡ 5.38516 ×10−5 ⎤ ⎡ 5.38516 ×10−5 ⎤ ⎢ ⎦⎥ ⎣⎢ ⎦⎥ Ka = 1.1×10–2 = ⎣ ⎡ x − 5.38516 ×10−5 ⎤ ⎣⎢ ⎦⎥ x = Mf = 5.41152×10–5 M Mi = 0.10 M Vi = 100. mL Mf = 5.41152×10–5 M Vf = ? MiVi = MfVf Vf = MiVi/Mf = [(0.10 M)(100. mL)]/(5.41152×10–5 M) = 1.8479×105 mL Volume water added = (1.8479×105 mL) – 100. mL = 1.8469×105 = 1.8×105 mL H2O added

18.157 Plan: Determine the hydrogen ion concentration from the pH. The molarity and the volume will give the number of moles, and with the aid of Avogadro’s number, the number of ions may be found.

18-57

Solution: M H3O+ = 10–pH = 10–6.2 = 6.30957×10–7 M −3 23 ⎛ 6.30957 ×10−7 mol H O + ⎞⎛ ⎟⎜10 L ⎞⎟⎟⎜⎛1250. mL ⎞⎟⎟⎜⎛ 7 d ⎞⎟⎟⎛⎜⎜ 6.022 ×10 H 3 O + ⎞⎟⎟ 3 ⎜⎜⎜ ⎜⎜ = 3.32467×1018 = 3×1018 H3O+ ⎟⎟⎟⎜⎜ ⎟⎟⎜⎜ ⎟ ⎟ ⎟⎟⎜⎜1 wk ⎟⎟⎜⎜⎜ 1 mol H O + ⎟⎟⎟ ⎜⎜ ⎜⎜ 1 mL ⎟⎜⎝ ⎟ L d ⎠⎝ ⎠⎝ 3 ⎝ ⎠⎝ ⎠ ⎠ The pH has only one significant figure, and limits the significant figures in the final answer.

18.158 a) NH3(l) ⇆ NH4+(am) + NH2–(am) In this equilibrium “(am)” indicated ammoniated, solvated by ammonia, instead of “(aq)” to indicate aqueous, solvated by water. [ NH 4 + ][ NH 2− ] Initially, based on the equilibrium: Kc = 2 [ NH3 ] Since NH3 is a liquid and a solvent: Kc [NH3]2 = Kam = [NH4+][NH2–] Strongest base = NH2– b) Strongest acid = NH4+ + – + NH4 > NH2–: acidic c) NH2 > NH4 : basic + HNO3(am) + NH3(l) → NH4 (am) + NO3–(am) HCOOH(am) + NH3(l) → NH4+(am) + HCOO–(am) HNO3 is a strong acid in water while HCOOH is a weak acid in water. However, both acids are equally strong (i.e., their strengths are leveled) in NH3 because they dissociate completely to form NH4+. d) Kam = [NH4+][NH2–] = 5.1×10–27 [NH4+] = [NH2–] = x Kam = [x][x] = 5.1×10–27 x = 7.1414×10–14 = 7.1×10–14 M NH4+ (sa) = solvated by sulfuric acid (sulf) e) 2H2SO4(l) ⇆ H3SO4+(sa) + HSO4–(sa) Ksulf = [H3SO4+][HSO4–] = 2.7×10–4 [H3SO4+] = [HSO4–] = x Ksulf = [x][x] = 2.7×10–4 x = 1.643×10–2 = 1.6×10–2 M HSO4– 18.159 M is the unknown molarity of the thiamine. C12H18ON4SCl2(aq) + H2O(l) ⇆ H3O+(aq) + C12H17ON4SCl2–(aq) M–x x x pH = 3.50 [H3O+] = 10–3.50 = 3.1623×10–4 M = x [ H3O+ ][C12 H17 ON 4 SCl2− ] Ka = 3.37×10–7 = [C12 H18 ON 4 SCl2 ] ( x)( x) Ka = 3.37×10–7 = M−x

(

)

(3.1623×10 )(3.1623×10 ) −4

Ka = 3.37×10–7 =

−4

( M − 3.1623×10 ) −4

M = 0.29705633 M ⎛ 337.27 g thiamine HCl ⎞⎟ ⎛ 0.29705633 mol thiamine HCl ⎞⎟⎛⎜10−3 L ⎞⎟ ⎟⎟(10.00 mL )⎜⎜ ⎟⎟ ⎟⎟⎜⎜ Mass (g) = ⎜⎜⎜ ⎜⎜ ⎜⎝ L ⎜⎝ 1 mol thiamine HCl ⎠⎟⎟ ⎠⎟⎜⎜⎝ 1 mL ⎠⎟⎟

= 1.00188 = 1.0 g thiamine hydrochloride Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

18-58

18.160 Plan: Determine Kb using the relationship Kb = 10–pKb. Write the base-dissociation equation and set up a reaction table in which x = the amount of OH– produced. Use the Kb expression to find x. From [OH–], [H3O+] and then pH can be calculated. Solution: Kb = 10pK = 10–5.91 = 1.23027×10–6 TRIS(aq) + H2O(l) ⇆ OH–(aq) + HTRIS+(aq) Initial 0.075 — 0 0 Change –x +x +x Equilibrium 0.075 – x x x + − HTRIS OH [ ] [ ] Kb = 1.23027×10–6 = [ TRIS] [ x ][ x ] Kb = 1.23027×10–6 = Assume x is small compared to 0.075. ⎡0.075 − x⎤ ⎣⎢ ⎦⎥ [ x ][ x ] Kb = 1.23027×10–6 = [0.075] – x = [OH ] = 3.03760×10–4 M OH– Check assumption that x is small compared to 0.075: 3.03760 ×10−4 (100) = 0.40% error, so the assumption is valid. 0.075 Kw 1.0×10−14 [H3O]+ = = = 3.292073×10–11 M [ OH− ] 3.03760×10−4 pH = –log [H3O+] = –log (3.292073×10–11) = 10.4825 = 10.48 18.161 Fe3+(aq) + 6H2O(l) ⇆ Fe(H2O)63+(aq) Lewis acid-base reaction Fe(H2O)63+(aq) + H2O(l) ⇆ Fe(H2O)5OH2+(aq) + H3O+(aq) Brønsted-Lowry acid-base reaction 18.162 The pH is dependent on the molar concentration of H3O+. Convert % w/v to molarity, and use the Ka of acetic acid to determine [H3O+] from the equilibrium expression. Convert % w/v to molarity using the molecular weight of acetic acid (CH3COOH): ⎛ 5.0 g CH 3 COOH ⎞⎛ ⎟⎟⎜⎜ 1 mol CH 3 COOH ⎞⎛ ⎟⎟⎜⎜ 1 mL ⎞⎟⎟ ⎜ Molarity = ⎜⎜ ⎟⎟⎜ ⎟⎟⎜ −3 ⎟⎟ = 0.832639 M CH3COOH ⎟⎜⎜ 60.05 g CH 3 COOH ⎠⎝ ⎟⎜⎜10 L ⎠⎟ ⎜⎜⎝ 100 mL solution ⎠⎝ Acetic acid dissociates in water according to the following equation and equilibrium expression: CH3COOH(aq) + H2O(l) ⇆ CH3COO–(aq) + H3O+(aq) Initial 0.832639 — 0 0 Change –x +x +x Equilibrium 0.832639 – x x x + − H O CH COO [ ] [ ] 3 3 Ka = 1.8×10–5 = [CH3 COOH ] [ x ][ x ] Ka = 1.8×10–5 = Assume x is small compared to 0.832639. ⎡0.832639 − x ⎤ ⎣⎢ ⎦⎥ [ x ][ x ] Ka = 1.8×10–5 = [ 0.832639] x = 3.8714×10–3 M = [H3O+] Check assumption: [3.871×10–3/0.832639] × 100% = 0.46%, therefore the assumption is good. pH = –log [H3O+] = –log (3.8714×10–3) = 2.412132 = 2.41 Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

18-59

18.163 Plan: When an initial acid concentration is diluted, the percent dissociation of the acid increases. Solution: When the solution is diluted, the percent dissociation of the acid HB increases. There is no single correct scene. Any scene in which the total number of HB + B– is less than in the original solution yet the relative number of HB dissociated is greater would be correct. In the original scene, 2 of 12 HB molecules, or 16.7%, have dissociated. One possible scene after dilution could show 8 HB molecules, 2 B– and 2 H3O+ ions. Then the dissociation is 2 B–/10 HB = 20% dissociation. 18.164 a) The strong acid solution would have a larger electrical conductivity. b) The strong acid solution would have a lower pH. c) The strong acid solution would bubble more vigorously. 18.165 Plan: Assuming that the pH in the specific cellular environment is equal to the optimum pH for the enzyme, the hydronium ion concentrations are [H3O+] = 10–pH. Solution: Salivary amylase, mouth: [H3O+] = 10–6.8 = 1.58489×10–7 = 2×10–7 M Pepsin, stomach: [H3O+] = 10–2.0 = 1×10–2 M Trypsin, pancreas: [H3O+] = 10–9.5 = 3.1623×10–10 = 3×10–10 M 18.166 a) CH3COOH(aq) + H2O(l) ⇆ H3O+(aq) + CH3COO–(aq) 0.240 – x x x + − [ H 3 O ][CH 3 COO ] Ka = 1.8×10–5 = [CH 3 COOH ] ( x)( x) Ka = 1.8×10–5 = Assume x is small compared to 0.240. 0.240 − x

(

Ka = 1.8×10–5 =

)

( x )( x )

(0.240) –3

x = 2.07846×10 Check assumption: [2.07846×10–3/0.240] × 100% = 0.9%, therefore the assumption is good. x = [H3O+] = 2.07846×10–3 = 2.1×10–3 M H3O+ [OH–] = Kw/[H3O+] = (1.0×10–14)/(2.07846×10–3) = 4.81125×10–12 = 4.8×10–12 M OH– pH = –log [H3O+] = –log (2.07846×10–3) = 2.682258 = 2.68 pOH = –log [OH–] = –log (4.81125×10–12) = 11.317742 = 11.32 b) NH3(aq) + H2O(l) ⇆ NH4+(aq) + OH–(aq) 0.240 – x x x + [ NH 4 ][ OH− ] Kb = 1.8×10–5 = [ NH3 ] ( x)( x) Kb = 1.8×10–5 = Assume x is small compared to 0.240. 0.240 − x

(

Kb = 1.8×10–5 =

)

( x )( x )

(0.240) –3

x = 2.07846×10 Check assumption: [2.07846×10–3 0.240] × 100% = 0.9%, therefore the assumption is good. x = [OH–] = 2.07846×10–3 = 2.1×10–3 M OH– [H3O+] = Kw/[OH–] = (1.0×10–14)/(2.07846×10–3) = 4.81125×10–12 = 4.8×10–12 M H3O+ pOH = –log [OH–] = –log (2.07846×10–3) = 2.682258 = 2.68 pH = –log [H3O+] = –log (4.81125×10–12) = 11.317742 = 11.32 Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

18-60

3− ⎛ 33 g Na 3 PO 4 ⎞⎛ ⎟⎟⎜⎜ 1 mol Na 3 PO 4 ⎞⎛ ⎟⎟⎜⎜ 1 mol PO 4 ⎞⎟⎟ ⎜ 3– 18.167 Concentration = M = ⎜⎜ ⎟⎟⎜ ⎟⎟⎜ ⎟⎟ = 0.20129 M PO4 ⎜⎝⎜ ⎜ ⎜ ⎟ ⎟ ⎟ 1L ⎠⎝⎜163.94 g Na 3 PO 4 ⎠⎝⎜1 mol Na 3 PO 4 ⎠

PO43–(aq) + H2O(l) ⇆ OH–(aq) + HPO42–(aq) (Use Ka3 for H3PO4) Kb = Kw/Ka = (1.0×10–14)/(4.2×10–13) = 0.0238095 (HPO42− )(OH− ) Kb = 0.0238095 = (PO43− ) ( x)( x)

Kb = 0.0238095 =

(0.20129 − x)

A quadratic is required. x2 + 0.0238095x – 0.00479261 = 0 a = 1 b = 0.0238095 c = – 0.00479261

−b ± b2 − 4ac x=

2a −0.0238095 ± (0.0238095) − 4 (1)(−0.00479261) 2

2 (1) x = 0.058340 = 0.058 M OH– [H3O]+ = Kw/[OH–] = (1.0×10–14)/(0.058340) = 1.7140898×10–13 M H3O+ pH = –log [H3O+] = –log (1.7140898×10–13) = 12.765966 = 12.77

18.168 a) PH3BCl3(s) ⇆ PH3(g) + BCl3(g) x = [PH3] = [BCl3] Kc = [PH3][BCl3] = (x)(x) = x2 = [8.4×10–3/3.0 L]2 = 7.84×10–6 = 7.8×10–6 b)

18.169

The freezing point depression equation is required to determine the molality of the solution. ΔT = [0.00 – (–1.93°C)] = 1.93°C = iKf m Temporarily assume i = 1. ΔT 1.93ο C m= = = 1.037634 m = 1.037634 M iKf (1)(1.86 ο C/m) This molality is the total molality of all species in the solution, and is equal to their molarity. From the equilibrium: ClCH2COOH(aq) + H2O(l) ⇆ H3O+(aq) + ClCH2COO–(aq) Initial 1.000 M x x Change –x +x +x Equilibrium 1.000 – x x x The total concentration of all species is: [ClCH2COOH] + [H3O+] + [ClCH2COO–] = 1.037634 M [1.000 – x] + [x] + [x] = 1.000 + x = 1.037634 M x = 0.037634 M

18-61

Ka = Ka =

[ H3O+ ][CH3COO− ] [CH3COOH ] (0.037634 )(0.037634 )

(1.000 − 0.037634)

= 0.0014717 = 0.00147

− ⎛ 0.42 g C17 H 35 COONa ⎞⎛ ⎟⎟⎜⎜ 1 mL ⎞⎛ ⎟⎜ 1 mol C17 H 35 COONa ⎞⎛ ⎟⎟⎜⎜ 1 mol C17 H 35 COO ⎞⎟⎟ ⎜ 18.170 Molarity = ⎜⎜ ⎟⎟⎜ −3 ⎟⎟⎟⎜⎜ ⎟⎟⎜ ⎟ 10.0 mL ⎟⎜⎜10 L ⎠⎝ ⎟⎜⎜ 306.45 g C17 H 35 COONa ⎠⎝ ⎟⎜⎜1 mol C17 H 35 COONa ⎠⎟⎟ ⎜⎜⎝ ⎠⎝

= 0.137053 M C17H35COO– Kb = Kw/Ka = (1.0×10–14)/(1.3×10–5) = 7.69231×10–10 C17H35COO–(aq) + H2O(l) ⇆ C17H35COOH(aq) + OH–(aq) 0.137053 – x x x − [C H COOH ][ OH ] Kb = 7.69231×10–10 = 17 35 [C17 H 35 COO− ] ( x)( x) Kb = 7.69231×10–10 = Assume x is small compared to 0.137053. 0.137053 − x

(

)

( x)( x)

Kb = 7.69231×10–10 =

(0.137053) x = 1.026769×10 = [OH–] Check assumption: [1.026768×10–5/0.137053] × 100% = 0.007%, therefore the assumption is good. [H3O]+ = Kw/[OH–] = (1.0×10–14)/(1.026768×10–5) = 9.739298×10–10 M H3O+ pH = – log [H3O+] = – log (9.739298×10–10) = 9.01147 = 9.01 –5

18.171 a) The two ions that comprise this salt are Ca2+ (derived from the strong base Ca(OH)2) and CH3CH2COO– (derived from the weak acid, propionic acid, CH3CH2COOH). A salt derived from a strong base and weak acid produces a basic solution. Ca2+ does not react with water. CH3CH2COO–(aq) + H2O(l) ⇆ CH3CH2COOH(aq) + OH–(aq) b) Calcium propionate is a soluble salt and dissolves in water to yield two propionate ions: Ca(CH3CH2COO)2(s) + H2O(l) → Ca2+(aq) + 2CH3CH2COO–(aq) The molarity of the solution is: − ⎛ 8.75 g Ca(CH 3 CH 2 COO)2 ⎞⎛ ⎟⎟⎜⎜ 1 mol Ca(CH 3 CH 2 COO)2 ⎞⎛ ⎟⎟⎜⎜ 2 mol CH 3 CH 2 COO ⎞⎟⎟ ⎜ Molarity = ⎜⎜ ⎟⎟⎜ ⎟⎟⎜ ⎟ ⎜⎝⎜ ⎟⎜⎜186.22 g Ca(CH 3 CH 2 COO)2 ⎠⎝ ⎟⎜⎜1 mol Ca(CH 3 CH 2 COO)2 ⎠⎟⎟ 0.500 L ⎠⎝ = 0.1879497 M CH3CH2COO– CH3CH2COO– + H2O ⇆ CH3CH2COOH + OH– Initial 0.1879497 M 0 0 Change –x +x +x Equilibrium 0.1879497 – x x x Kb = Kw/Ka = (1.0×10–14)/(1.3×10–5) = 7.69231×10–10 [CH3CH2 COOH ][ OH− ] –10 Kb = 7.69231×10 = [CH3CH2 COO− ] ( x)( x) Kb = 7.69231×10–10 = Assume x is small compared to 0.1879497. 0.1879497 − x

(

)

18-62

Kb = 7.69231×10–10 =

( x)( x)

(0.1879497) x = 1.202401×10 M = [OH–] Check assumption: [1.202401×10–5/0.1879497] × 100% = 0.006%, therefore the assumption is good. [H3O]+ = Kw/[OH–] = (1.0×10–14)/(1.202401×10–5) = 8.31669×10–10 M H3O+ pH = –log [H3O+] = –log (8.31669×10–10) = 9.0800 = 9.08 –5

18.172 a) Annual depositions: ⎛ 3.0 (NH 4 )2 SO4 ⎞⎟ 2 2 (NH4)2SO4: ⎜⎜ ⎟⎟(2.688 g/m ) = 0.8488421 = 0.85 g/m ⎝ ⎠ 9.5 total ⎛ 5.5 NH 4 HSO4 ⎞⎟ 2 2 NH4HSO4: ⎜⎜ ⎟(2.688 g/m ) = 1.55621 = 1.56 g/m ⎝ 9.5 total ⎠⎟ ⎛1.0 H2 SO4 ⎞⎟ 2 2 ⎜ H2SO4: ⎜⎝ 9.5 total ⎠⎟⎟(2.688 g/m ) = 0.282947 = 0.28 g/m (NH4)2SO4 is a weak acid; NH4HSO4 has half the acidity per mole as H2SO4 so the equivalent amount of sulfuric acid deposition would be: ⎛1.56 g NH 4 HSO 4 ⎞⎟⎛⎜ 0.50 M NH 4 HSO 4 ⎞⎛ ⎟⎟⎜⎜ 98.08 g H 2 SO 4 ⎞⎟⎟ = 0.664602 g/m2 ⎜⎜ ⎟⎟⎜ ⎟⎟⎜115.11 g NH HSO ⎟⎟ 2 ⎝ ⎠⎜⎝ 1 M H 2 SO 4 m ⎠⎝ 4 4⎠ Total as sulfuric acid = 0.66 g/m2 + 0.28 g/m2 = 0.94 g/m2 2 ⎛ ⎞ ⎛ 0.94 g H 2 SO 4 ⎞⎟ 2 ⎛1000 m ⎞ ⎟⎟ ⎜⎜ 1 kg ⎟⎟ = 9.4×103 kg ⎜ ⎟⎠(10. km )⎜⎜⎝ 2 ⎟ ⎟ ⎜⎝ m 1 km ⎠ ⎝⎜1000 g ⎠⎟⎟ b) H2SO4(aq) + CaCO3(s) → H2O(l) + CO2(g) + CaSO4(s) There is a 1:1 mole ratio between H2SO4 and CaCO3. Mass (lb) of CaCO3 = ⎛1000 g ⎞⎛ 1 mol H 2 SO 4 ⎞⎛ ⎟⎟⎜⎜1 mol CaCO3 ⎞⎟⎟⎛⎜⎜100.1 g CaCO3 ⎞⎟⎟⎛⎜ 2.205 lb ⎟⎞⎟ ⎟⎟⎜⎜ (9.4 ×103 kg H2SO4 )⎝⎜⎜⎜ 1 kg ⎠⎟⎜ ⎟⎜ ⎟⎟⎜⎜ ⎟⎟⎜⎜ ⎟⎜⎝ 98.08 g H 2 SO 4 ⎠⎝ 1 mol H 2 SO 4 ⎠⎝ 1 mol CaCO3 ⎠⎟⎝⎜ 1000 g ⎠⎟⎟ = 2.1154×104 = 2.1×104 lb CaCO3 + ⎛1000 g ⎞⎟⎜⎛ 1 mol H 2 SO 4 ⎞⎛ ⎟⎟⎜⎜ 2 mol H ⎞⎟⎟ = 1.91680×105 mol H+ ⎟⎟⎜⎜ c) Moles of H+= (9.4 ×103 kg H 2 SO 4 )⎜⎜ ⎟ ⎟⎜⎜1 mol H 2 SO 4 ⎠⎟⎟ ⎜⎝ 1 kg ⎠⎟⎜⎝ 98.08 g H 2 SO 4 ⎠⎝

⎛1000 m ⎞⎟ ⎛ 1L ⎞ (3 m)⎜⎜ −3 3 ⎟⎟⎟ = 3.0×1010 L Volume of lake = (10. km 2 )⎜⎜ ⎝ 1 km ⎠⎟⎟ ⎝10 m ⎠ 5 + 1.91680 ×10 mol H Molarity of H+ = = 6.3893×10–6 M 3.0 ×1010 L pH = –log [H+] = –log (6.3893×10–6) = 5.1945 = 5.19 2

18.173 a) 0°C Kw = [H3O+][OH–] = (x)(x) = 1.139×10–15 x = [H3O+] = 3.374907×10–8 = 3.375×10–8 M H3O+ pH = –log [H3O+] = –log (3.374907×10–8) = 7.471738 = 7.4717 50°C Kw = [H3O+][OH–] = (x)(x) = 5.474×10–14 x = [H3O+] = 2.339658×10–7 = 2.340×10–7 M H3O+ pH = –log [H3O+] = –log (2.339658×10–7) = 6.6308476 = 6.6308 b) 0°C Kw = [D3O+][OD–] = (x)(x) = 3.64×10–16 x = [D3O+] = 1.907878×10–8 = 1.91×10–8 M D3O+ pH = –log [D3O+] = –log (1.907878×10–8) = 7.719449 = 7.719 50°C Kw = [D3O+][OD–] = (x)(x) = 7.89×10–15 x = [D3O+] = 8.882567×10–8 = 8.88×10–8 M D3O+ pH = –log [D3O+] = –log (8.882567×10–8) = 7.0514615 = 7.051 Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

18-63

c) The deuterium atom has twice the mass of a normal hydrogen atom. The deuterium atom is held more strongly to the oxygen atom, so the degree of ionization is decreased. ⎛12.0 g HX ⎞⎟⎛⎜ 1 mol HX ⎞⎟ ⎟⎟ = 0.0800 M HX ⎟⎟⎜⎜ 18.174 Molarity of HX = ⎜⎜⎜ ⎟⎜⎝150. g HX ⎠⎟⎟ L ⎝⎜ ⎠⎜ ⎛ 6.00 g HY ⎞⎟⎜⎛ 1 mol HY ⎞⎟ Molarity of HY = ⎜⎜⎜ ⎟⎟⎟ = 0.120 M HY ⎟⎟⎟⎜⎜⎜ L 50.0 g HY ⎝⎜ ⎠⎜ ⎝ ⎠⎟

HX must be the stronger acid because a lower concentration of HX has the same pH (it produces the same number of H+ ions) as a higher concentration of HY.

18.175 Acid HA: HA(aq) + H2O(l) ⇆ H3O+(aq) + A–(aq) ⎡ 0.010 mol ⎤ ⎡ 0.010 mol ⎤ ⎢ 0.50 L ⎦⎥ ⎣⎢ 0.50 L ⎦⎥ = 4.0×10–3 Ka = ⎣ ⎡ 0.050 mol ⎤ ⎢⎣ 0.50 L ⎥⎦ Acid HB: HB(aq) + H2O(l) ⇆ H3O+(aq) + B–(aq) ⎡ 0.010 mol ⎤ ⎡ 0.010 mol ⎤ ⎢ 0.25 L ⎥⎦ ⎢⎣ 0.25 L ⎥⎦ = 1.0×10–2 Ka = ⎣ ⎡ 0.040 mol ⎤ 0.25 L ⎦⎥ ⎣⎢ Acid HB, with the larger Ka value, is the stronger acid. 18.176 Treat H3O+(aq) as H+(aq) because this corresponds to the listing in the Appendix. a) K+(aq) + OH−(aq) + H+(aq) + NO3−(aq) → K+(aq) + NO3−(aq) + H2O(l) Net ionic equation: OH−(aq) + H+(aq) → H2O(l) Na+(aq) + OH−(aq) + H+(aq) + Cl−(aq) → Na+(aq) + Cl−(aq) + H2O(l) Net ionic equation: OH−(aq) + H+(aq) → H2O(l) ο ο ο = ∑[ ΔHf(products) ] – ∑[ ΔHf(reactants) ] Δ H rxn + ο ο ο Δ H rxn = {1 Δ H f [H2O(l)]} – {1 Δ H f [H (aq)] + 1 Δ H fο [ OH−(aq)]} ο = [(1 mol)(−285.840 kJ/mol)] – [(1 mol)(0 kJ/mol) + (1 mol)(−229.94 kJ/mol)] Δ H rxn ο Δ H rxn = −55.90 kJ b) The neutralization reaction of a strong acid and a strong base is essentially the reaction between H+(aq) and ο OH−(aq) to form H2O(l). Therefore, Δ H rxn for KOH and HCl would be expected to be −55.90 kJ. 18.177

NH2(CH2)4NH2(aq) + H2O(l) ⇆ NH2(CH2)4NH3+(aq) + OH−(aq) 0.10 − x x x x = [OH−] = 2.1×10–3 ⎡2.1×10−3 ⎤ ⎡ 2.1×10−3 ⎤ ⎡⎢ NH 2 (CH 2 ) NH +3 ⎤⎥ [ OH− ] ⎢ ⎥⎦ ⎢⎣ ⎥⎦ 4 ⎣ ⎦ Kb = = ⎣ = 4.5045965×10−5 = 4.5×10−5 − 3 ⎡ ⎤ ⎡⎢ NH 2 (CH 2 ) NH 2 ⎤⎥ 0.10 2.1 10 − × 4 ⎣ ⎦ ⎢⎣ ⎥⎦

18.178 a) There are 20 OH− ions for every 2 H3O+ ions; in other words, [OH−] = 10 × [H3O+] Kw = 1.0×10−14 = [H3O+][OH−] 1.0×10−14 = [H3O+](10)[ H3O+] [ H3O+] = 3.162278×10−8 M pH = –log [H3O+] = –log (3.162278×10–8) = 7.4999999 = 7.5 b) For a pH of 4, [H3O+] = 10−pH = 10−4 = 1.0×10−4 M [OH−] = Kw/[H3O+] = (1.0×10–14)/(1.0×10–4) = 1.0×10–10 M OH− [H3O+]/[OH−] = 1.0×10−4/1.0×10–10 = 1.0×106 The H3O+ concentration is one million times greater than that of OH−. You would have to draw one million H3O+ ions for every one OH−. Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

18-64

18.179 a) As the pH of a water solution containing casein increases, the H+ ions from the carboxyl groups on casein will be removed. This will increase the number of charged groups, and the solubility of the casein will increase. b) As the pH of a water solution containing histones decreases, –NH2 and =NH groups will accept H+ ions from solution. This will increase the number of charged groups, and the solubility of the histones will increase. 18.180 Plan: Use Le Châtelier’s principle. Solution: a) The concentration of oxygen is higher in the lungs so the equilibrium shifts to the right. b) In an oxygen deficient environment, the equilibrium would shift to the left to release oxygen. c) A decrease in the [H3O+] concentration would shift the equilibrium to the right. More oxygen is absorbed, but it will be more difficult to remove the O2. d) An increase in the [H3O+] concentration would shift the equilibrium to the left. Less oxygen is absorbed, but it will be easier to remove the O2. 18.181 NH3(aq) + H2O(l) ⇆ NH4+(aq) + OH–(aq) Convert to a Ka relationship: NH4+(aq) + H2O(l) ⇆ NH3(aq) + H3O+(aq) Ka = Kw/Kb = (1.0×10–14)/(1.76×10–5) = 5.6818×10–10 [ NH3 ][ H 3 O+ ] Ka = [ NH +4 ]

[ NH3 ] Ka = + [ NH 4 ]+[ NH3 ] [ H3 O+ ]+ K a a) [H3O+] = 10–pH = 10–7.00 = 1.0×10–7 M H3O+ 5.6818×10−10 [ NH 3 ] = = 5.6496995×10–3 = 5.6×10–3 [ NH +4 ] + [ NH 3 ] 1.0×10−7 + 5.6818×10−10 b) [H3O+] = 10–pH = 10–10.00 = 1.0×10–10 M H3O+ [ NH 3 ] 5.6818×10−10 = = 0.8503397 = 0.85 + [ NH 4 ] + [ NH 3 ] 1.0×10−10 + 5.6818×10−10 c) Increasing the pH shifts the equilibria towards NH3. Ammonia is able to escape the solution as a gas. 18.182 Plan: The molarity of the acid is calculated by dividing moles of acid by the volume of solution. Set up a reaction table for the dissociation of the acid, in which x = the amount of propanoate ion at equilibrium. The freezing point depression is used to calculate the apparent molality and thus the apparent molarity of the solution. The total concentration of all species at equilibrium equals the apparent molarity and is used to find x. Percent dissociation is the concentration of dissociated acid divided by the initial concentration of the acid and multiplied by 100. Solution: a) Calculate the molarity of the solution (before acid dissociation). ⎛ 7.500 g CH 3 CH 2 COOH ⎞⎟⎛ 1 mL ⎞⎛ 1 mol CH 3 CH 2 COOH ⎞⎟ ⎜ ⎟⎟⎜⎜ ⎟⎟⎜⎜ ⎟⎟ = 1.012419 = 1.012 M CH CH COOH M = ⎜⎜ 3 2 ⎜⎝⎜ 100.0 mL solution ⎠⎟⎟⎜⎝⎜10−3 L ⎠⎟⎟⎝⎜⎜⎜ 74.08 g CH 3 CH 2 COOH ⎠⎟⎟ b) The freezing point depression equation is required to determine the molality of the solution. ΔT = iKf m = [0.000 – (–1.890°C)] = 1.890°C Temporarily assume i = 1. ΔT 1.890 ο C m= = = 1.016129032 m = 1.016129032 M iKf (1)(1.86 ο C/m) Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

18-65

This molality is the total molality of all species in the solution, and is equal to their molarity. From the equilibrium: CH3CH2COO–(aq) CH3CH2COOH(aq) + H2O(l) ⇆ H3O+(aq) + Initial 1.012419 M 0 0 Change –x +x +x Equilibrium 1.012419 – x x x The total concentration of all species is: [CH3CH2COOH] + [H3O+] + [CH3CH2COO–] = 1.016129032 M [1.012419 – x] + [x] + [x] = 1.012419 + x = 1.016129032 M x = 0.00371003 = 0.004 M CH3CH2COO– c) The percent dissociation is the amount dissociated (x from part b)) divided by the original concentration from part a). 0.00371003 M Percent dissociation = (100) = 0.366452 = 0.4% 1.012419 M 18.183 Plan: For parts a) and b), write the base-dissociation reaction and the Kb expression. Set up a reaction table in which x = the amount of reacted base and the concentration of OH–. Solve for x, calculate [H3O]+, and find the pH. For parts c) and d), write the acid-dissociation reaction for the conjugate acid of quinine. Find the Ka value from Kw = Ka × Kb. Set up a reaction table in which x = dissociated acid and the concentration of [H3O]+, and find the pH. Solution: Note that both pKb values only have one significant figure. This will limit the final answers. Kb (tertiary amine N) = 10–pKb = 10–5.1 = 7.94328×10–6 Kb (aromatic ring N) = 10–pKb = 10–9.7 = 1.995262×10–10 a) Ignoring the smaller Kb: C20H24N2O2(aq) + H2O(l) ⇆ OH–(aq) + HC20H24N2O2+(aq) Initial 1.6×10–3 M 0 0 Change –x +x +x Equilibrium 1.6×10–3 – x x x + − HC H N O OH [ ] [ ] 20 24 2 2 Kb = 7.94328×10–6 = [ H 2 C 20 H 24 N 2 O2 ]

⎡ x⎤ ⎡ x⎤ ⎣ ⎦⎣ ⎦ Assume x is small compared to 1.6×10–3. ⎡1.6×10−3 − x ⎤ ⎢⎣ ⎥⎦ ⎡ x⎤ ⎡ x⎤ ⎣ ⎦⎣ ⎦ Kb = 7.94328×10–6 = ⎡1.6×10−3 ⎤ ⎣⎢ ⎦⎥ x = 1.127353×10–4 Check assumption that x is small compared to 1.6×10–3: 1.127353×10−4 (100) = 7% error, so the assumption is not valid. 1.6×10−3 Since the error is greater than 5%, it is not acceptable to assume x is small compared to 1.6×10–3, and it is necessary to use the quadratic equation. x2 = (7.94328×10–6)(1.6×10–3 – x) = 1.27092×10–8 – 7.94328×10–6x x2 + 7.94328×10–6 x – 1.270925×10–8 = 0 a = 1 b = 7.94328×10–6 c = –1.27092×10–8

Kb = 7.94328×10–6 =

−b ± b2 − 4ac x=

18-66

−7.94328×10−6 ±

(7.94328×10 ) − 4 (1)(−1.270925×10 ) −6

−8

2 (1)

= 1.08834×10–4 M OH–

1.0×10−14 Kw = = 9.18830513×10–11 M H3O+ − [OH ] 1.08834 ×10−4 pH = –log [H3O+] = –log (9.18830513×10–11) = 10.03676 = 10.0 b) (Assume the aromatic N is unaffected by the tertiary amine N.) Use the Kb value for the aromatic nitrogen. C20H24N2O2(aq) + H2O(l) ⇆ OH–(aq) + HC20H24N2O2+(aq) Initial 1.6×10–3 M 0 0 Change –x +x +x Equilibrium 1.6×10–3 – x x x + − HC H N O OH [ ] [ ] 20 24 2 2 Kb = 1.995262×10–10 = [C 20 H 24 N 2 O2 ] [H3O]+ =

⎡ ⎤⎡ ⎤ ⎣ x⎦ ⎣ x⎦ ⎡1.6×10−3 − x ⎤ ⎢⎣ ⎥⎦ ⎡ x⎤ ⎡ x⎤ ⎣ ⎦⎣ ⎦ Kb = 1.995262×10–10 = ⎡1.6×10−3 ⎤ ⎣⎢ ⎦⎥ Kb = 1.995262×10–10 =

Assume x is small compared to 1.6×10–3.

x = 5.65015×10–7 M OH– The hydroxide ion from the smaller Kb is much smaller than the hydroxide ion from the larger Kb (compare the powers of ten in the concentration). c) HC20H24N2O2+(aq) + H2O(l) ⇆ H3O+(aq) + C20H24N2O2(aq) Initial 0.33 M 0 0 Change –x +x +x Equilibrium 0.33 – x x x −14 K 1.0×10 Ka = w = = 1.25893×10–9 −6 Kb 7.94328×10 Ka = 1.25893×10–9 = Ka = 1.25893×10–9 = Ka = 1.25893×10–9 =

[ H3O+ ][C20 H 24 N2 O2 ] [ HC20 H 24 N 2 O2+ ] ( x)( x)

(0.33 − x)

Assume x is small compared to 0.33.

( x)( x)

(0.33) [H3O+] = x = 2.038252×10–5 M Check assumption that x is small compared to 0.33: 2.038252×10−5 (100) = 0.006%. The assumption is good. 0.33 pH = –log [H3O+] = –log (2.038252×10–5) = 4.69074 = 4.7 d) Quinine hydrochloride will be indicated as QHCl. ⎛ 1.5% ⎞⎟⎛⎜1.0 g ⎞⎟⎜⎛⎜ 1 mL ⎞⎛ ⎟ 1 mol QHCl ⎞⎟⎟ ⎟⎟⎜ −3 ⎟⎟⎟⎜⎜⎜ M = ⎜⎜⎜ ⎟ = 0.041566 M ⎟⎟⎜⎜ ⎟⎜⎝10 L ⎠⎝ ⎝100% ⎠⎜⎝ mL ⎠⎜ ⎟⎜⎜ 360.87 g QHCl ⎠⎟⎟ HC20H24N2O2+(aq) + H2O(l) ⇆ H3O+(aq) + C20H24N2O2(aq) Initial 0.041566 M 0 0 Change –x +x +x Equilibrium 0.041566 – x x x Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

18-67

[ H3O+ ][C20 H 24 N2 O2 ] K = 1.25893×10 = [ HC20 H 24 N 2 O2+ ] –9

Ka = 1.25893×10–9 = Ka = 1.25893×10–9 =

( x)( x)

(0.041566 − x)

Assume x is small compared to 0.041566.

( x)( x)

(0.041566) [H3O+] = x = 7.233857×10–6 M Check assumption that x is small compared to 0.041566: 7.233857×10−6 (100) = 0.02%. The assumption is good. 0.041566 pH = –log [H3O+] = –log (7.233857×10–6) = 5.1406 = 5.1 18.184 a) At pH = 7.00, [H3O+] = 10–pH = 10–7.00 = 1.0×10–7 M

[ H 3 O+ ] [ HClO ] = [ HClO ]+ [ ClO− ] [ H 3 O + ] + K a ⎡ HClO ⎤ 1.0 ×10−7 ⎣ ⎦ = 0.775194 = 0.78 = − ⎡ HClO ⎤ + ⎢⎡ ClO ⎥⎤ 1.0 ×10−7 + 2.9 ×10−8 ⎣ ⎦ ⎣ ⎦

b) At pH = 10.00, [H3O+] = 10–pH = 10–10.00 = 1.0×10–10 M

[ H 3 O+ ] [ HClO ] = [ HClO ]+ [ ClO− ] [ H 3 O + ] + K a ⎡ HClO ⎤ 1.0 ×10−10 ⎣ ⎦ = 0.003436 = 0.0034 = − 10 − ⎤ ⎡ ⎤ ⎡ + 2.9 ×10−8 ⎣ HClO ⎦ + ⎢⎣ ClO ⎥⎦ 1.0 ×10

18.185

a) All scenes indicate equal initial amounts of each acid. The more H3O+ present, the stronger the acid is (greater Ka). Increasing Ka: HX < HZ < HY b) The pKa values increase in order of decreasing Ka values. Increasing pKa: HY < HZ < HX c) The order of pKb is always the reverse of pKa values: Increasing pKb: HX < HZ < HY d) Percent dissociation = (2/8) × 100% = 25% e) NaY, the weakest base, will give the highest pOH and the lowest pH.

18-68

CHAPTER 19 IONIC EQUILIBRIA IN AQUEOUS SYSTEMS FOLLOW–UP PROBLEMS 19.1A

Plan: The problems are both equilibria with the initial concentration of reactant and product given. For part (a), set up a reaction table for the dissociation of HF. Set up an equilibrium expression and solve for [H3O+], assuming that the change in [HF] and [F−] is negligible. Check this assumption after finding [H3O+]. Convert [H3O+] to pH. For part (b), first find the concentration of OH− added. Then, use the neutralization reaction to find the change in initial [HF] and [F−]. Repeat the solution in part (a) to find pH. a) Solution: Concentration (M) HF(aq) + H2O(l) ⇆ F−(aq) + H3O+(aq) Initial 0.50 — 0.45 0 Change –x — +x +x Equilibrium 0.50 – x — 0.45 + x x Assumed that x is negligible with respect to 0.50 M and 0.45 M. [ H O+ ][ F− ] Ka = 3 [ HF ] [H3O+] = K a

⎡ 0.50⎤ [ HF ] ⎦ = 7.5556×10–4 = 7.6×10–4 = (6.8×10−4 ) ⎣ − ⎡0.45⎤ [F ] ⎣ ⎦

Check assumption: Percent error in assuming x is negligible:

7.5556×10−4 (100) = 0.17%. 0.45

The error is less than 5%, so the assumption is valid. Solve for pH: pH = –log (7.5556 × 10–4) = 3.12173 = 3.12 The other method to calculate the pH of a buffer is to use the Henderson-Hasselbalch equation: [base] pH = pKa + log [acid] [F− ] pH = pKa + log pKa = –log Ka = –log(6.8×10–4) = 3.16749 [HF] [0.45] pH = 3.16749 + log [0.50] pH = 3.12173 = 3.12 Since [HF] and [F−] are similar, the pH should be close to pKa, which equals log (6.8×10–4) = 3.17. The pH should be slightly less (more acidic) than pKa because [HF] > [F–]. The calculated pH of 3.12 is slightly less than pKa of 3.17. b) Solution: What is the initial molarity of the OH– ion? − ⎛ 0.40 g NaOH ⎞⎟⎛⎜ 1mol NaOH ⎞⎛ ⎟⎜ 1mol OH ⎞⎟⎟ ⎟⎟⎜⎜ Molarity = ⎜⎜⎜ ⎟⎟⎟⎜⎜ ⎟ = 0.010 M OH− ⎟⎜⎝ 40.00 g NaOH ⎠⎝ L ⎟⎜⎜1mol NaOH ⎠⎟⎟ ⎝⎜ ⎠⎜ Set up reaction table for neutralization of 0.010 M OH− (note the quantity of water is irrelevant). HF(aq) + OH−(aq) → F−(aq) + H2O(l) Concentration (M) Before addition 0.50 — 0.45 — Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

19-1

Addition — 0.010 — — Change – 0.010 – 0.010 + 0.010________________ After addition 0.49 0 0.46 — Following the same solution path with the same assumptions as part (a): ⎡0.49⎤ [ HF ] ⎦ = 7.2434782×10–4 = 7.2×10–4 M [H3O+] = K a − = (6.8×10−4 ) ⎣ ⎡ ⎤ [F ] ⎣ 0.46⎦ Check assumption:

7.2434782×10−4 (100) = 0.16%, which is less than the 5% maximum so the assumption is valid. 0.46 Solve for pH: pH = –log (7.2434782×10–4) = 3.14005 = 3.14 or using the Henderson-Hasselbalch equation: [F− ] [0.46] pH = pKa + log = 3.16749 + log [HF] [0.49] pH = 3.14005 = 3.14 With addition of base, the pH should increase and it does, from 3.12 to 3.14. However, the pH should still be slightly less than pKa: 3.14 is still less than 3.17. 19.1B

Plan: The problems are both equilibria with the initial concentration of reactant and product given. Take the inverse log of –pKb to solve for the Kb. Then use Kb to solve for Ka: Ka × Kb = 1.0×10-14. For part (a), set up a reaction table for the dissociation of (CH3)2NH2+ (the acid component of the buffer; Cl– is a spectator ion and does not participate in the buffer reaction). Set up an equilibrium expression and solve for [H3O+], assuming that the change in [(CH3)2NH2+] and [(CH3)2NH] is negligible. Check this assumption after finding [H3O+]. Convert [H3O+] to pH. For part (b), first calculate the new initial concentrations of (CH3)2NH2+, (CH3)2NH, and H3O+ after the addition of the 50.0 mL of hydrochloric acid solution. Then, use the neutralization reaction to find the change in initial [(CH3)2NH2+] and [(CH3)2NH]. Repeat the Solution in part (a) to find pH. a) Solution: Kb = 10−pKb = 10−3.23 = 5.8884×10–4 = 5.9×10–4 K 1.0×10 –14 Ka = w = = 1.7×10–11 K b 5.9×10 –4 Concentration (M) (CH3)2NH2+(aq) + H2O(l) ⇆ (CH3)2NH(aq) + Initial 0.25 — 0.30 Change –x — +x Equilibrium 0.25 – x — 0.30 + x Assumed that x is negligible with respect to 0.25 M and 0.30 M. [(CH3 )2 NH][ H3O+ ] Ka = [(CH3 )2 NH2+ ] [H3O+] = Ka

[(CH3 )2 NH 2+ ] [(CH3 )2 NH ]

= (1.7×10–11)

H3O+(aq) 0 +x x

(0.25) = 1.4167×10–11 = 1.4×10–11 (0.30)

Check assumption: Percent error in assuming x is negligible:

1.4 ×10 –11 (100) = 5.6×10–9% 0.25

The error is less than 5%, so the assumption is valid. Solve for pH: pH = –log (1.4×10–11) = 10.8539 = 10.85 Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

19-2

The other method to calculate the pH of a buffer is to use the Henderson-Hasselbalch equation: [base] pH = pKa + log [acid] pH = pKa + log

[(CH3 )2 NH]

[(CH3 )2 NH2+ ]

pKa = –log Ka = –log(1.7×10–11) = 10.7696

(0.30) pH = 10.8488 = 10.85 (0.25) Since [(CH3)2NH2+] and [(CH3)2NH] are similar, the pH should be close to pKa, which equals log(1.7×10–11) = 10.7696. The pH should be slightly greater (more basic) than the pKa because [(CH3)2NH] > [(CH3)2NH2+]. The calculated pH of 10.85 is slightly greater than the pKa of 10.77. b) Solution: M × Vconc 0.30M×1.0L [(CH3)2NH] = conc = 0.285714 M = Mdil 1.05L pH = 10.7696 + log

[(CH3)2NH2+] =

Mconc × Vconc 0.25M ×1.0L = 0.238095 M = Mdil 1.05L

Mconc × Vconc 0.75M × 0.0500L = 0.035714 M = Mdil 1.05L Set up reaction table for neutralization of 0.035714 M H3O+. Concentration (M) (CH3)2NH(aq) + H3O+ (aq) → (CH3)2NH2+ (aq) + H2O(l) Before addition 0.285714 — 0.238095 — Addition — 0.035714 — — Change – 0.035714 – 0.035714 + 0.035714______________—______ After addition 0.250000 0 0.273809 — Following the same solution path with the same assumptions as part (a): [(CH3 )2 NH 2+ ] (0.273809) = (1.7×10–11) = 1.86190×10–11 [H3O+] = Ka (0.250000) [(CH3 )2 NH ]

[H3O+] =

Check assumption: Percent error in assuming x is negligible:

1.86190×10 –11 (100) = 7.4×10–9% 0.25

The error is less than 5%, so the assumption is valid. Solve for pH: pH = –log (1.86190×10–11) = 10.73004 = 10.73 or using the Henderson-Hasselbalch equation: [(CH3 )2 NH] pH = pKa + log [(CH3 )2 NH2+ ]

(0.250000) pH = 10.73009 = 10.73 (0.273809) With addition of acid, the pH should decrease and it does, from 10.85 to 10.73.

pH = 10.7696 + log

19.2A

Plan: For high buffer capacity, the components of a buffer should be concentrated and the concentrations of the base and acid components should be similar. Solution: a) The buffer has a much larger amount of weak acid than of the conjugate weak base. Addition of strong base would convert some HB into B− to make the ration [B−]/[HB] closer to 1 according to the reaction: HB(aq) + OH−(aq) → B−(aq) + H2O(l)

19-3

b) The buffer with the highest possible buffer capacity would have 4 HB particles and 4 B− particles. Addition of strong base would convert 3 HB particles to 3 B− particles so that there are 4 of each of the weak acid and weak base.

19.2B

Plan: The pH of a given buffer solution depends on the relative concentrations of conjugate base and acid. If [A−] > [HA], the pH of the buffer solution will be greater than its pKa. If [A−] < [HA], the pH of the buffer solution will be less than its pKa. Buffers are able to lessen the effects of added strong acid or base by neutralizing the strong acid or base. Added strong base reacts with the acid component of the buffer, and added strong acid reacts with the conjugate base component of the buffer. The most effective buffers consist of a weak acid with a pKa close to (within one pH unit of) the buffering pH and its conjugate base. Solution: a) The buffer with pH > pKa will have more conjugate base particles than weak acid particles. The buffer in scene 3 has 6 conjugate base particles to 3 weak acid particles and, thus, would have pH > pKa. b) Strong base reacts with the weak acid component of the buffer, so the more weak acid there is, the more strong base the buffer can react with. There are more weak acid particles in the buffer in scene 2 than in the other scenes. c) Buffers are most effective at buffering pH values within one unit of their pKa. The pKa of this buffer, 4.2, is more than one unit away from the buffering pH, 6.1, so this buffer would not be effective at buffering a pH of 6.1.

19.3A

Plan: Sodium benzoate is a salt so it dissolves in water to form Na+ ions and C6H5COO− ions. Only the benzoate ion is involved in the buffer system represented by the equilibrium: C6H5COOH(aq) + H2O(l) ⇆ C6H5COO−(aq) + H3O+(aq) Given in the problem are the volume and pH of the buffer and the concentration of the base, benzoate ion. The question asks for the mass of benzoic acid to add to the sodium benzoate solution. First, find the concentration of C6H5COOH needed to make a buffer with a pH of 4.25. Multiply the volume by the concentration to find moles of C6H5COOH and use the molar mass to find grams of benzoic acid. Solution: The concentration of benzoic acid is calculated from the Henderson-Hasselbalch equation: [C H COO− ] pH = pKa + log 6 5 [C6 H 5 COOH] pKa = –log Ka = –log 6.3×10–5 = 4.20066 [0.050] 4.25 = 4.20066 + log [x] [0.050] Raise each side to 10x. 0.04934 = log [x] [0.050] 1.1203146 = [x] x = 4.46303×10–2 M The number of moles of benzoic acid is determined from the concentration and volume: (4.46303×10–2 M C6H5COOH) × (5.0 L) = 0.2231515 mol C6H5COOH Mass (g) of C6H5COOH = ⎛122.12 g C6 H 5 COOH ⎞⎟ ⎟ = 27.2513 = 27 g C6H5COOH (0.2231515 mol C6 H 5 COOH)⎜⎜⎜ ⎜⎝ 1 mol C6 H 5 COOH ⎠⎟⎟

19-4

Prepare a benzoic acid/benzoate buffer by dissolving 27 g of C6H5COOH into 5.0 L of 0.050 M C6H5COONa. Using a pH meter, adjust the pH to 4.25 with strong acid or base. 19.3B

Plan: Ammonium chloride is a salt so it dissolves in water to form NH4+ ions and Cl− ions. Only the ammonium ion is involved in the buffer system represented by the equilibrium: NH4+(aq) + H2O(l) ⇆ NH3(aq) + H3O+(aq) Given in the problem are the volume and pH of the buffer and the concentration of the conjugate base, ammonia. The question asks for the mass of ammonium chloride to add to the ammonia solution. First, find the ratio of ammonia to ammonium ion needed to make a buffer with a pH of 9.18. Multiply the volume of the solution by the concentration of the ammonia to find moles of ammonia in the buffer. Then use the ratio of ammonia to ammonium as a conversion factor to calculate the moles of ammonium ion. Then use the molar mass to find grams of ammonium chloride. Solution: The ratio of ammonia to ammonium ion is calculated from the Henderson-Hasselbalch equation: [ NH3 ] pH = pKa + log [ NH 4+ ] Ka of NH4+ =

Kw 1.0×10 –14 = = 5.6818×10–10 –5 K b of NH 3 1.76×10

pKa = –log Ka = –log (5.6818×10–10) = 9.24551 9.18 = 9.24551 + log –0.0655127 = log 0.859978 =

[ NH3 ]

[ NH 4+ ]

[ NH3 ]

[ NH 4+ ]

Raise each side to 10x.

[ NH3 ]

[ NH 4+ ]

There are 0.86 moles of NH3 for every 1 mole of NH4+, or 0.86 mol NH3 = 1 mol NH4+.

⎛ 0.15 mol NH3 ⎞⎟ Amount (mol) of NH3 in buffer = (0.75 L buffer) ⎜⎜⎜ ⎟ = 0.1125 mol NH3 ⎝ ⎠⎟ 1L ⎛ 1 mol NH 4 + ⎞⎛ 1 mol NH 4 Cl ⎞⎛ ⎟⎟⎜⎜ 53.49 g NH 4 Cl ⎞⎟⎟ ⎟⎟⎟⎜⎜ Mass (g) of NH4Cl = (0.1125 mol NH3 )⎜⎜⎜ + ⎟ ⎜ ⎟⎜ 1 mol NH 4 Cl ⎠⎟⎟ ⎝⎜ 0.859978 mol NH3 ⎠⎟⎜⎝ 1 mol NH 4 ⎠⎝

= 6.99742 = 7.0 g NH4Cl 19.4A

Plan: The titration is of a weak acid, HBrO, with a strong base, NaOH. The reactions involved are: 1) Neutralization of weak acid with strong base: HBrO(aq) + OH−(aq) → H2O(l) + BrO−(aq) Note that the reaction goes to completion and produces the conjugate base, BrO−. 2) The weak acid and its conjugate base are in equilibrium based on the acid dissociation reaction: a) HBrO(aq) + H2O(l) ⇆ BrO−(aq) + H3O+(aq) or b) the base dissociation reaction: BrO–(aq) + H2O(l) ⇆ HBrO(aq) + OH–(aq) The pH of the solution is controlled by the relative concentrations of HBrO and BrO−. For each step in the titration, first think about what is present initially in the solution. Then use the two reactions to determine solution pH. It is useful in a titration problem to first determine at what volume of titrant the equivalence point occurs. For part (e), use the pH values from (a) – (d) to plot the titration curve.

19-5

Solution: a) Before any base is added, the solution contains only HBrO and water. Equilibrium reaction 2a applies and pH can be found in the same way as for a weak acid solution: Concentration (M) HBrO(aq) + H2O(aq) ⇆ BrO−(aq) + H3O+(aq) Initial 0.2000 — 0 0 Change –x — +x +x Equilibrium 0.2000 – x — x x + − [ ][ ] x x H O BrO ] [ 3 ][ Ka = = 2.3×10–9 = ⎡ 0.2000 − x ⎤ [ HBrO ] ⎣⎢ ⎦⎥ Assume 0.2000 – x = 0.2000 M [ x ][ x ] 2.3×10–9 = [ 0.2000 ] + x = [H3O ] = 2.144761×10–5 M pH = –log [H3O+] = –log (2.144761×10–5) = 4.66862 = 4.67 Check the assumption by calculating the percent error in [HBrO]eq. 2.144761×10−5 Percent error = (100) = 0.01%; this is well below the 5% maximum. 0.2000 b) When [HBrO] = [BrO−], the solution contains significant concentrations of both the weak acid and its conjugate base. Use the equilibrium expression for reaction 2a to find pH. Since [HBrO] = [BrO−], their ratio equals 1. [ H 3 O+ ][ BrO− ] Ka = [ HBrO ] [ HBrO ] [H3O+] = K a = (2.3×10−9 ) ⎡⎣1⎤⎦ = 2.3×10–9 M [ BrO− ] pH = –log [H3O+] = –log (2.3×10–9) = 8.63827 = 8.64 Note that when [HBrO] = [BrO−], the titration is at the midpoint (half the volume to the equivalence point) and pH = pKa. c) At the equivalence point, the total number of moles of HBrO present initially in solution equals the number of moles of base added. Therefore, reaction 1 goes to completion to produce that number of moles of BrO−. The solution consists of BrO− and water. Calculate the concentration of BrO−, and then find the pH using the base dissociation equilibrium, reaction 2b. First, find equivalence point volume of NaOH. ⎛1 mol NaOH ⎞⎛ ⎞⎟⎛ 1 mL ⎞ 1L ⎛ 0.2000 mol HBrO ⎞⎟⎛⎜10−3 L ⎞⎟ ⎟⎟⎜⎜ ⎟⎟ ⎟⎟ 20.00 mL ⎜⎜ ⎟⎟⎜⎜ ⎟⎟⎜⎜ Volume (mL) of NaOH = ⎜⎜⎜ ⎟ ⎜ ⎜ ⎟ −3 L ⎟⎟⎜⎜ 0.1000 mol NaOH ⎠⎟⎟⎝⎜⎜10 L ⎠⎟ ⎝⎜ ⎠⎟⎜⎜⎝ 1 mL ⎠⎟⎟ ⎜⎝⎜ 1 mol HBrO ⎠⎝ = 40.00 mL NaOH added All of the HBrO present at the beginning of the titration is neutralized and converted to BrO− at the equivalence point. Calculate the concentration of BrO−. Initial moles of HBrO: (0.2000 M)(0.02000 L) = 0.004000 mol Moles of added NaOH: (0.1000 M)(0.04000 L) = 0.004000 mol Amount (mol) HBrO(aq) + OH−(aq) → H2O(l) + BrO−(aq) Before addition 0.004000 mol — — 0 Addition — 0.004000 mol — — Change – 0.004000 mol – 0.004000 mol — +0.004000 mol After addition 0 0 — 0.004000 mol

(

)

19-6

At the equivalence point, 40.00 mL of NaOH solution has been added (see calculation above) to make the total volume of the solution (20.00 + 40.00) mL = 60.00 mL. ⎛ 0.004000 mol BrO− ⎞⎟⎛ 1 mL ⎞ ⎟⎟ = 0.06666667 M [BrO–] = ⎜⎜⎜ ⎟⎜ ⎝ ⎠⎟⎜⎝10−3 L ⎠⎟ 60.00 mL Set up reaction table with reaction 2b, since only BrO− and water are present initially: Concentration (M) BrO−(aq) + H2O(l) ⇆ HBrO(aq) + OH−(aq) Initial 0.06666667 — 0 0 Change –x — +x +x Equilibrium 0.06666667 – x — x x Kb = Kw/Ka = (1.0×10–14/2.3×10–9) = 4.347826×10–6 [ x ][ x ] [ OH− ][ HBrO ] Kb = = = 4.347826×10–6 − [ 0.06666667 − x ] [ BrO ] Assume that x is negligible, since [BrO−] >> Kb. [ x ][ x ] 4.347826×10–6 = [ 0.06666667 ] − x = [OH ] = 5.3838191×10–4 = 5.4 × 10–4 M Check the assumption by calculating the percent error in [BrO−]eq. 5.3838191×10−4 Percent error = (100) = 0.8%, which is well below the 5% maximum. 0.06666667 pOH = –log (5.3838191×10–4) = 3.26891 pH = 14 – pOH = 14 – 3.26891 = 10.73109 = 10.73 d) After the equivalence point, the concentration of excess strong base determines the pH. Find the concentration of excess base and use it to calculate the pH. Initial moles of HBrO: (0.2000 M)(0.02000 L) = 0.004000 mol Moles of added NaOH: 2 × 0.004000 mol = 0.008000 mol NaOH Amount (mol) HBrO(aq) + OH−(aq) → H2O(l) + BrO−(aq) Before addition 0.004000 mol — — 0 Addition — 0.008000 mol — — Change – 0.004000 mol – 0.004000 mol — +0.004000 mol After addition 0 0.004000 mol — 0.004000 mol Excess NaOH: 0.004000 mol ⎛ 1L ⎞⎛ ⎟⎟⎜⎜ 1 mL ⎞⎟⎟ Volume (mL) of added NaOH = (0.0080000 mol NaOH )⎜⎜ ⎟ 10−3 L ⎠⎟ ⎝ 0.1000 mol NaOH ⎠⎝ = 80.00 mL Total volume: 20.00 mL + 80.00 mL = 100.0 mL [NaOH] = 0.004000 mol NaOH/0.1000 L = 0.0400 M pOH = –log (0.0400) = 1.3979 pH = 14 – pOH = 14 – 1.3979 = 12.60 e) Plot the pH values calculated in the preceding parts of this problem as a function of the volume of titrant.

19-7

Titration of HBrO with NaOH 14 13 12 11 10 9 pH 8 7 6 5 4 3 0

10 20 30 40 50 60 70 80 90 100 Volume of 0.1000 M NaOH, mL

The plot and pH values follow the pattern for a weak acid vs. strong base titration. The pH at the midpoint of the titration does equal pKa. The equivalence point should be, and is, greater than 7. 19.4B

Plan: The titration is of a weak acid, C6H5COOH, with a strong base, NaOH. The reactions involved are: 1) Neutralization of weak acid with strong base: C6H5COOH(aq) + OH−(aq) → H2O(l) + C6H5COO−(aq) Note that the reaction goes to completion and produces the conjugate base, C6H5COO−. 2) The weak acid and its conjugate base are in equilibrium based on the acid dissociation reaction: a) C6H5COOH(aq) + H2O(l) ⇆ C6H5COO−(aq) + H3O+(aq) or b) the base dissociation reaction: C6H5COO–(aq) + H2O(l) ⇆ C6H5COOH(aq) + OH–(aq) The pH of the solution is controlled by the relative concentrations of C6H5COOH and C6H5COO−. For each step in the titration, first think about what is present initially in the solution. Then use the two reactions to determine solution pH. Solution: a) Before any base is added, the solution contains only C6H5COOH and water. Equilibrium reaction 2a applies and pH can be found in the same way as for a weak acid solution: Concentration (M) C6H5COOH(aq) + H2O(aq) Initial 0.1000 — Change –x — Equilibrium 0.1000 – x — – + [C H HCOO ][H3O ] (x)(x) Ka = 6.3×10–5 = 6 5 = (0.10 – x) [C6 H 5 COOH]

⇆ C6H5COO−(aq) 0 +x x

+ H3O+(aq) 0 +x x

Assume 0.1000 – x = 0.1000 M (x)(x) Ka = 6.3×10–5 = (0.10) x = [H3O+] = 2.50998×10–3 pH = –log [H3O+] = –log (2.50998×10–3) = 2.60033 = 2.60 Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

19-8

Check the assumption by calculating the percent error in [C6H5COOH]eq. 2.5×103 (100) = 2.5% which is smaller than 5%, so the assumption is valid. 0.10 b) Any NaOH added to the buffer reacts with the benzoic acid. Calculate the amount (mol) of NaOH (OH−) added. ⎛ 1 L ⎞⎟⎜⎛ 0.1500 mol OH – ⎞⎟ Amount (mol) of OH− added = (12.00 mL OH−) ⎜⎜ ⎟ = 0.001800 mol OH− ⎟⎜⎝ ⎝1000 mL ⎠⎟⎜ ⎠⎟ 1L The added OH− will react with the benzoic acid: Initial moles of C6H5COOH: (0.1000 M)(0.03000 L) = 0.003000 mol

Amount (mol) Before addition Addition Change After addition

C6H5COOH(aq) 0.003000 mol — – 0.001800 mol 0.001200 mol

OH−(aq) → H2O(l) + C6H5COO−(aq) — — 0 0.001800 mol — — – 0.001800 mol — +0.001800 mol 0 — 0.001800 mol

At this point in the titration, 12.00 mL of NaOH solution has been added (see calculation above) to make the total volume of the solution (12.00 + 30.00) mL = 42.00 mL (0.04200 L). Calculate the new concentrations of [C6H5COOH] and [C6H5COO−]: Molarity of C6H5COOH =

0.001200 mol = 0.02857 M 0.04200 L

Molarity of C6H5COO− =

0.001800 mol = 0.04286 M 0.04200 L

Ka = 6.3×10–5 =

[C6 H5 HCOO– ][H3O+ ] [C6 H 5 COOH]

, so [H3O+] = Ka

[C6 H 5 COOH]

[C6 H 5 HCOO – ]

[H3O+] = (6.3×10–5) (0.02857) = 4.1995×10–5 (0.04286) + pH = –log [H3O ] = –log (4.1995×10–5) = 4.3768 = 4.38 c) At the equivalence point, the total number of moles of C6H5COOH present initially in solution equals the number of moles of base added. Therefore, reaction 1 goes to completion to produce that number of moles of C6H5COO−. The solution consists of C6H5COO− and water. Calculate the concentration of C6H5COO−, and then find the pH using the base dissociation equilibrium, reaction 2b. First, find equivalence point volume of NaOH. Volume (mL) of NaOH = ⎛ 1 L ⎞⎟⎛ 0.1000 mol C6 H 5 COOH ⎞⎟⎜⎛ 1 mol NaOH ⎞⎟⎛ 1L ⎞⎟ ⎜ ⎟⎜⎜ (30.00 mL C6H5COOH) ⎜⎜ ⎟⎜⎜ ⎟ ⎟ ⎟ ⎟ ⎝1000 mL ⎠⎟⎟⎜⎝ ⎝ ⎠ ⎠⎜ 1L ⎝1 mol C6 H 5 COOH ⎠ 0.1500 mol NaOH ⎟ = 0.02000 L (20.00 mL) NaOH added All of the C6H5COOH present at the beginning of the titration is neutralized and converted to C6H5COO− at the equivalence point. Calculate the concentration of C6H5COO−. Initial moles of C6H5COOH: (0.1000 M)(0.03000 L) = 0.003000 mol Moles of added NaOH: (0.1500 M)(0.02000 L) = 0.003000 mol Amount (mol) Before addition Addition Change After addition

C6H5COOH(aq) 0.003000 mol — – 0.003000 mol 0

OH−(aq) → H2O(l) + C6H5COO−(aq) — — 0 0.003000 mol — — – 0.003000 mol — +0.003000 mol 0 — 0.003000 mol

19-9

At this point in the titration, 20.00 mL of NaOH solution has been added (see calculation above) to make the total volume of the solution (20.00 + 30.00) mL = 50.00 mL (0.05000 L). Calculate the new concentration [C6H5COO−]: Molarity of C6H5COO− =

0.003000 mol = 0.06000 M 0.05000 L

Set up reaction table with reaction 2b, since only C6H5COO− and water are present initially: Concentration (M) Initial Change Equilibrium

C6H5COO−(aq) 0.06000 –x 0.06000 – x

H2O(l) — — —

⇆

C6H5COOH (aq) + 0 +x x

OH−(aq) 0 +x x

Kb = Kw/Ka = (1.0×10–14/6.3×10–5) = 1.5873×10–10 Kb =

[C6 H 5 COOH][OH – ] [x][x] = = 1.5873×10–10 – 0.06000 – x [ ] C H HCOO [ 6 5 ]

Assume that x is negligible, since [C6H5COO−] >> Kb. [x][x] 1.6×10–10 = [ 0.06000 ] x = [OH−] = 3.086065×10–6 = 3.1×10–6 M Check the assumption by calculating the percent error in [C6H5COO−]eq. 3.1×10–6 (100) = 0.0052% which is smaller than 5%, so the assumption is valid. 0.06000 pOH = –log (3.086065×10–6) = 5.51059 pH = 14 – pOH = 14 – 5.51059 = 8.48941 = 8.49 d) After the equivalence point, the concentration of excess strong base determines the pH. Find the concentration of excess base and use it to calculate the pH. Initial moles of C6H5COOH: (0.1000 M)(0.03000 L) = 0.003000 mol Moles of added NaOH: (0.1500 M)(0.02200 L) = 0.003300 mol + OH−(aq) → H2O(l) + C6H5COO−(aq) Amount (mol) C6H5COOH(aq) Before addition 0.003000 mol — — 0 Addition — 0.003300 mol — — Change – 0.003000 mol – 0.003000 mol — +0.003000 mol After addition 0 0.000300 mol — 0.003000 mol Excess NaOH: 0.000300 mol Total volume: 22.00 mL + 30.00 mL = 52.00 mL (0.05200 L) [NaOH] = 0.000300 mol NaOH/0.05200 L = 0.00576923 M pOH = –log (0.0056923) = 2.2388 pH = 14 – pOH = 14 – 2.2388 = 11.76 (pH is shown to two decimal places) 19.5A

Plan: Write the formula of the salt and the reaction showing the equilibrium of a saturated solution. The ionproduct expression can be written from the stoichiometry of the solution reaction as the coefficients in the reaction become exponents in the ion-product expression. Solution: a) The formula of calcium sulfate is CaSO4. The equilibrium reaction is: CaSO4(s) ⇆ Ca2+(aq) + SO42−(aq) Ion–product expression: Ksp = [Ca2+][SO42−] b) Chromium(III) carbonate is Cr2(CO3)3. Cr2(CO3)3(s) ⇆ 2Cr3+(aq) + 3CO32−(aq) Ion–product expression: Ksp = [Cr3+]2[CO32−]3

19-10

c) Magnesium hydroxide is Mg(OH)2. Mg(OH)2(s) ⇆ Mg2+(aq) + 2OH−(aq) Ion–product expression: Ksp = [Mg2+][OH−]2 d) Aluminum hydroxide is Al(OH)3. Al(OH)3(s) ⇆ Al3+(aq) + 3OH−(aq) Ion-product expression: Ksp = [Al3+][OH−]3 19.5B

Plan: Examine the ion-product expressions. The exponents in the ion-product expression are the subscripts of those ions in the chemical formula. Solution: a) The compound is lead(II) chromate. Its formula is PbCrO4. b) The compound is iron(II) sulfide. Its formula is FeS. c) The compound is strontium fluoride. Its formula is SrF2. d) The compound is copper(II) phosphate. Its formula is Cu3(PO4)2.

19.6A

Plan: Calculate the solubility of CaF2 as molarity and use molar ratios to find the molarity of Ca2+ and F− dissolved in solution. Calculate Ksp from [Ca2+] and [F−] using the ion-product expression. Solution: Convert the solubility to molar solubility: ⎛1.5×10−4 g CaF ⎞⎟⎛ 1 mL ⎞⎛ 1 mol CaF ⎞ ⎜ ⎟⎜ 2 ⎟⎜ 2 ⎟ ⎟⎟ = 1.9211×10–4 M CaF ⎟⎟⎜⎜ ⎟⎜ Molarity = ⎜⎜⎜ 2 ⎟⎜ −3 ⎟ ⎟ ⎟ ⎜ ⎜ 10.0 mL ⎜⎝ ⎠⎟⎝10 L ⎠⎟⎝⎜ 78.08 g CaF2 ⎠⎟ [Ca2+] = [CaF2] = 1.9211×10–4 M because there is 1 mol of calcium ions in each mol of CaF2. [F−] = 2[CaF2] = 3.8422×10–4 M because there are 2 mol of fluoride ions in each mol of CaF2. The solubility equilibrium is: CaF2(s) ⇆ Ca2+(aq) + 2F−(aq) Ksp = [Ca2+][F−]2 Calculate Ksp using the solubility product expression from above and the saturated concentrations of calcium and fluoride ions. Ksp = [Ca2+][F−]2 = (1.9211×10–4)( 3.8422×10–4)2 = 2.836024×10–11 = 2.8×10–11 The Ksp for CaF2 is 2.8×10–11 at 18°C.

19.6B

Plan: Calculate the solubility of Ag3PO4 as molarity and use molar ratios to find the molarity of Ag+ and PO43− dissolved in solution. Calculate Ksp from [Ag+] and [PO43−] using the ion-product expression. Solution: Convert the solubility to molar solubility: ⎛ 3.2 ×10 –4 g Ag PO ⎞⎟⎛1000 mL ⎞⎛ 1 mol Ag PO ⎞⎟ 3 4 ⎟⎜ 3 4 ⎟⎟⎜⎜ ⎟⎟ = 1.5×10–5 M Ag3PO4 Molarity = ⎜⎜⎜ ⎟⎜⎜ ⎜ ⎟ ⎜ ⎟ 50. mL ⎝⎜ ⎠⎝ 1 L ⎠⎝ 418.7 g Ag3 PO 4 ⎠⎟ [Ag+] = 3[Ag3PO4] = 4.5×10–5 M because there are 3 mol of silver ions in each mol of Ag3PO4. [PO43−] = [Ag3PO4] = 1.5×10–5 M because there is 1 mol of phosphate ions in each mol of Ag3PO4. The solubility equilibrium is: ⇆ 3 Ag+(aq) + PO43−(aq) Ksp = [Ag+]3[ PO43−] Ag3PO4 (s) Calculate Ksp using the solubility product expression from above and the saturated concentrations of silver and phosphate ions. Ksp = [Ag+]3[ PO43−] = (4.5×10–5)3(1.5×10–5) = 1.3669×10–18 = 1.4×10–18 The Ksp for Ag3PO4 is 1.4×10–18 at 20°C.

19.7A

Plan: Write the solubility reaction for Mg(OH)2 and set up a reaction table, where S is the unknown molar solubility of the Mg2+ ion. Use the ion-product expression to solve for the concentration of Mg(OH)2 in a saturated solution (also called the solubility of Mg(OH)2). Solution: Concentration (M) Mg(OH)2(s) ⇆ Mg2+(aq) + 2OH−(aq) Initial — 0 0 +2S Change — +S Equilibrium — S 2S

19-11

Ksp = [Mg2+][OH−]2 = (S)(2S)2 = 4S3 = 6.3×10–10 S = 5.4004114×10–4 = 5.4×10–4 M Mg(OH)2 The solubility of Mg(OH)2 is equal to S, the concentration of magnesium ions at equilibrium, so the molar solubility of magnesium hydroxide in pure water is 5.4×10–4 M.

19.7B

Plan: Write the solubility reaction for Ca3(PO4)2 and set up a reaction table, where S is the unknown molar solubility of Ca3(PO4)2. Use the ion-product expression to solve for the concentration of Ca3(PO4)2 in a saturated solution (also called the solubility of Ca3(PO4)2). Solution: Concentration (M) Ca3(PO4)2(s) ⇆ 3Ca2+(aq) + 2PO43−(aq) Initial — 0 0 +2S Change — +3S Equilibrium — 3S 2S Ksp = [Ca2+]3[PO43−]2 = (3S)3(2S)2 = 108S5 = 1.2×10–29 S = 6.4439×10–7 = 6.4×10–7 M Ca3(PO4)2 The solubility of Ca3(PO4)2 in pure water, S, is 6.4×10–7 M.

19.8A

Plan: Write the solubility reaction of BaSO4. For part (a) set up a reaction table in which [Ba2+] = [SO42−] = S, which also equals the solubility of BaSO4. Then, solve for S using the ion-product expression. For part (b), there is an initial concentration of sulfate, so set up the reaction table including this initial [SO42−]. Solve for the solubility, S, which equals [Ba2+] at equilibrium. Solution: a) Set up reaction table. BaSO4(s) ⇆ Ba2+(aq) + SO42−(aq) Concentration (M) Initial — 0 0 +S Change — +S Equilibrium — S S Ksp = [Ba2+][SO42–] = S2 = 1.1×10–10 S = 1.0488×10–5 = 1.0×10–5 M The molar solubility of BaSO4 in pure water is 1.0×10–5 M. b) Set up another reaction table with initial [SO42−] = 0.10 M (from the Na2SO4). BaSO4(s) ⇆ Ba2+(aq) + SO42−(aq) Concentration (M) Initial — 0 0.10 +S Change — +S Equilibrium — S 0.10 + S Ksp = [Ba2+][SO42–] = S(0.10 + S) = 1.1×10–10 Assume that 0.10 + S is approximately equal to 0.10, which appears to be a good assumption based on the fact that 0.10 > 1×10–10, Ksp Ksp = S(0.10) = 1.1×10–10 S = 1.1×10–9 M Molar solubility of BaSO4 in 0.10 M Na2SO4 is 1.1×10–9 M. The solubility of BaSO4 decreases when sulfate ions are already present in the solution. The calculated decrease is from 10–5 M to 10–9 M, for a 10,000-fold decrease. This decrease is expected to be large because of the high concentration of sulfate ions.

19.8B

Plan: Write the solubility reaction of CaF2. For part (a) set up a reaction table in which [Ca2+] = S, which also equals the solubility of CaF2. Then, solve for S using the ion-product expression. For part (b), there is an initial concentration of calcium, so set up the reaction table including this initial [Ca2+]. Solve for the solubility, S, which equals [Ca2+] at equilibrium. For part (c), there is an initial concentration of fluoride, so set up the reaction table including this initial [F−]. Solve for the solubility, S, which equals [Ca2+] at equilibrium.

19-12

19.9A

Solution: a) Set up reaction table. CaF2(s) ⇆ Ca2+(aq) + 2F−(aq) Concentration (M) Initial — 0 0 +2S Change — +S Equilibrium — S 2S Ksp = [Ca2+][ F−]2 = (S)(2S)2 = 4S3 = 3.2×10–11 S = 2.0×10–4 M The molar solubility of CaF2 in pure water is 2.0×10–4 M. b) Set up another reaction table with initial [Ca2+] = 0.20 M (from the CaCl2). CaF2(s) ⇆ Ca2+(aq) + 2F−(aq) Concentration (M) Initial — 0.20 0 +2S Change — +S Equilibrium — 0.20 + S 2S Ksp = [Ca2+][ F−]2 = (0.20 + S)(2S)2 = 1.1×10–10 Assume that 0.20 + S is approximately equal to 0.20, which appears to be a good assumption based on the fact that 0.20 > 3.2×10–11, Ksp. Ksp = (0.20)(2S)2 = 3.2×10–11 S = 6.3×10–6 M Molar solubility of CaF2 in 0.20 M CaCl2 is 6.3×10–6 M. c) Set up another reaction table with initial [F−] = 0.40 M (from the NiF2; there are two fluoride ions per NiF2 unit, so an NiF2 concentration of 0.20 M gives a fluoride ion concentration of 0.40 M). CaF2(s) ⇆ Ca2+(aq) + 2F−(aq) Concentration (M) Initial — 0 0.40 +2S______ Change — +S Equilibrium — S 0.40 + 2S Ksp = [Ca2+][ F−]2 = (S)(0.40 + 2S)2 = 1.1×10–10 Assume that 0.40 + 2S is approximately equal to 0.40, which appears to be a good assumption based on the fact that 0.40 > 3.2×10–11, Ksp. Ksp = (S)(0.40)2 = 3.2×10–11 S = 2.0×10–10 M Molar solubility of CaF2 in 0.20 M NiF2 is 2.0×10–10 M. The solubility of CaF2 decreases when either calcium or fluoride ions are already present in the solution. Plan: First, write the solubility reaction for the salt. Then, check the ions produced when the salt dissolves to see if they will react with acid. Three cases are possible: 1) If OH− is produced, then addition of acid will neutralize the hydroxide ions and shift the solubility equilibrium toward the products. This causes more salt to dissolve. Write the solubility and neutralization reactions. 2) If the anion from the salt is the conjugate base of a weak acid, it will react with the added acid in a neutralization reaction. Solubility of the salt increases as the anion is neutralized. Write the solubility and neutralization reactions. 3) If the anion from the salt is the conjugate base of a strong acid, it does not react with a strong acid. The solubility of the salt is unchanged by the addition of acid. Write the solubility reaction. Solution: a) Calcium fluoride, CaF2 Solubility reaction: CaF2(s) ⇆ Ca2+(aq) + 2F−(aq) Fluoride ion is the conjugate base of HF, a weak acid. Thus, it will react with H3O+ from the strong acid, HNO3. Neutralization reaction: F−(aq) + H3O+(aq) → HF(aq) + H2O(l) The neutralization reaction decreases the concentration of fluoride ions, which causes the solubility equilibrium to shift to the right and more CaF2 dissolves. The solubility of CaF2 increases with the addition of HNO3. b) Iron(III) hydroxide, Fe(OH)3 Solubility reaction: Fe(OH)3(s) + H2O(l) ⇆ Fe3+(aq) + 3OH−(aq)

19-13

The hydroxide ion reacts with the added acid: Neutralization reaction: OH−(aq) + H3O+(aq) → 2H2O(l) The neutralization reactions decrease the concentration of hydroxide in the solubility equilibrium, which causes a shift to the right, and more Fe(OH)3 dissolves. The addition of HNO3 will increase the solubility of Fe(OH)3. c) Silver iodide, AgI Solubility reaction: AgI(s) ⇆ Ag+(aq) + I−(aq) The iodide ion is the conjugate base of a strong acid, HI. So, I− will not react with added acid. The solubility of AgI will not change with added HNO3. 19.9B

Plan: First, write the solubility reaction for the salt. Then, check the ions produced when the salt dissolves to see if they will react with acid. Three cases are possible: 1) If OH− is produced, then addition of acid will neutralize the hydroxide ions and shift the solubility equilibrium toward the products. This causes more salt to dissolve. Write the solubility and neutralization reactions. 2) If the anion from the salt is the conjugate base of a weak acid, it will react with the added acid in a neutralization reaction. Solubility of the salt increases as the anion is neutralized. Write the solubility and neutralization reactions. 3) If the anion from the salt is the conjugate base of a strong acid, it does not react with a strong acid. The solubility of the salt is unchanged by the addition of acid. Write the solubility reaction. Solution: a) Silver cyanide, AgCN Solubility reaction: AgCN(s) ⇆ Ag+(aq) + CN−(aq) Cyanide ion is the conjugate base of HCN, a weak acid. Thus, it will react with H3O+ from the strong acid, HBr. Neutralization reaction: CN−(aq) + H3O+(aq) → HCN(aq) + H2O(l) The neutralization reaction decreases the concentration of cyanide ions, which causes the solubility equilibrium to shift to the right and more AgCN dissolves. The solubility of AgCN increases with the addition of HBr. b) copper(I) chloride, CuCl Solubility reaction: CuCl (s) ⇆ Cu+(aq) + Cl−(aq) The chloride ion is the conjugate base of a strong acid, HCl. So, Cl− will not react with added acid. The solubility of CuCl will not change with added HBr. c) Magnesium phosphate, Mg3(PO4)2 Solubility reaction: Mg3(PO4)2(s) ⇆ Mg2+(aq) + 2PO43−(aq) Phosphate ion is the conjugate base of HPO42−, a weak acid. Thus, it will react with H3O+ from the strong acid, HBr. Neutralization reaction: PO43−(aq) + H3O+(aq) → HPO42−(aq) + H2O(l) HPO42−, in turn, is the conjugate base of H2PO4−, a weak acid. Thus, it will react with H3O+ from the strong acid, HBr. Neutralization reaction: HPO42−(aq) + H3O+(aq) → H2PO4−(aq) + H2O(l) H2PO4−, in turn, is the conjugate base of H3PO4, a weak acid. Thus, it will react with H3O+ from the strong acid, HBr. Neutralization reaction: H2PO4−(aq) + H3O+(aq) → H3PO4(aq) + H2O(l) Each of these neutralization reactions ultimately decreases the concentration of phosphate ions, which causes the solubility equilibrium to shift to the right and more Mg3(PO4)2 dissolves. The solubility of Mg3(PO4)2 increases with the addition of HBr.

19.10A Plan: First, write the solubility equilibrium equation and ion-product expression. Use the given concentrations of calcium and phosphate ions to calculate Qsp. Compare Qsp to Ksp. If Ksp < Qsp, precipitation occurs. If Ksp ≥ Qsp then, precipitation will not occur. Solution: Write the solubility equation: Ca3(PO4)2(s) ⇆ 3Ca2+(aq) + 2PO43−(aq) and ion-product expression: Qsp = [Ca2+]3[PO43−]2 [Ca2+] = [PO43–] = 1.0×10–9 M Qsp = [Ca2+]3[PO43−]2 = (1.0×10–9)3(1.0×10–9)2 = 1.0×10–45 Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

19-14

Compare Ksp and Qsp. Ksp = 1.2×10–29 > 1.0×10–45 = Qsp Precipitation will not occur because concentrations are below the level of a saturated solution as shown by the value of Qsp. 19.10B Plan: First, write the solubility equilibrium equation and ion-product expression. Find the concentrations of the lead and sulfide ions in the final mixture. Use these concentrations to calculate Qsp. Compare Qsp to Ksp. If Ksp < Qsp, precipitation occurs. If Ksp ≥ Qsp, then precipitation will not occur. Solution: Write the solubility equation: PbS(s) ⇆ Pb2+(aq) + S2−(aq) and ion-product expression: Qsp = [Pb2+][S2−] 25 L of a solution containing Pb2+ are mixed with 0.500 L of a solution containing S2−. The final volume is 25.5 L. ⎛ 0.015 g Pb2+ ⎞⎟⎜⎛ 1 mol Pb2+ ⎞⎟ ⎟ = 0.00180985 mol Pb2+ ⎟ Amount (mol) of Pb2+ = (25 L) ⎜⎜ ⎟⎟⎜⎜ 207.2 g Pb2+ ⎠⎟⎟ 1L ⎝⎜ ⎠⎝

0.00180985 mol Pb2+ = 7.09743×10–5 M 25.5 L ⎛ 0.10 mol S2– ⎞⎟ ⎟ = 0.050 mol S2− Amount (mol) of S2−: (0.500 L) ⎜⎜ ⎜⎝ 1L ⎠⎟⎟

Molarity of Pb2+ =

0.050 mol S2– = 2.0×10–3 M 25.5 L Qsp = [Pb2+][S2−] = (7.09743×10–5)(2.0×10–3) = 1.41949×10–7 =1.4×10–7 Compare Ksp and Qsp. Ksp = 3×10–25 < 1.4×10–7 = Qsp Precipitation will occur because concentrations are above the level of a saturated solution as shown by the value of Qsp. Molarity of S2− =

19.11A Plan: First, write the solubility equilibrium equation and ion-product expression. For b) use the given amounts of nickel(II) and hydroxide ions to calculate Qsp. Compare Qsp to Ksp. For c) check the ions produced when the salt dissolves to see if they will react with acid or if the hydroxide ion of a strong base is part of the ion-product expression, in which case, it will influence the solubility of the solid through the common ion effect. Solution: a) Write the solubility equation: Ni(OH)2(s) ⇆ Ni2+(aq) + 2OH−(aq) Scene 3 has the same relative number of ions as in the formula of Ni(OH)2. The Ni2+ and OH− ions are in a 1:2 ratio in Scene 3. b) Write the ion-product expression: Qsp = [Ni2+][OH−]2 Calculate Ksp using Scene 3: Ksp = [Ni2+][OH−]2 = [2][4]2 = 32 Calculate Qsp using Scene 1: Qsp = [Ni2+][OH−]2 = [3][4]2 = 48 Calculate Qsp using Scene 2: Qsp = [Ni2+][OH−]2 = [4][2]2 = 16 Qsp exceeds Ksp in Scene 1 (48 > 32) so additional solid will form in Scene 1. c) Hydroxide ion is one of the products of the solubility equilibrium reaction. The hydroxide ion reacts with added acid: Neutralization reaction: OH−(aq) + H3O+(aq) → 2H2O(l) The neutralization reaction decreases the concentration of OH−(aq) in the solubility equilibrium, which causes a shift to the right, and more Ni(OH)2(s) dissolves. Addition of base (OH–) shifts the equilibrium to the left due to the common-ion effect and the mass of Ni(OH)2 increases. 19.11B Plan: First, write the solubility equilibrium equation and ion-product expression. For b) use the given amounts of lead(II) and chloride ions to calculate Qsp. Compare Qsp to Ksp. For c) check the ions produced when the salt dissolves to see if they will react with acid or if the anion of the acid is part of the ion-product expression, in which case, it will influence the solubility of the solid through the common ion effect. Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

19-15

Solution: a) Write the solubility equation: PbCl2(s) ⇆ Pb2+(aq) + 2Cl−(aq) Scene 1 has the same relative number of ions as in the formula of PbCl2. The Pb2+ and Cl− ions are in a 1:2 ratio in Scene 1. b) Write the ion-product expression: Qsp = [Pb2+][Cl−]2 Calculate Qsp using Scene 1: Qsp = [Pb2+][Cl−]2 = [3][6]2 = 108 Calculate Qsp using Scene 2: Qsp = [Pb2+][Cl−]2 = [4][5]2 = 100 Calculate Ksp using Scene 3: Ksp = [Pb2+][Cl−]2 = [5][5]2 = 125 Qsp exceeds Ksp in Scene 3 (125 > 108) so additional solid will form in Scene 3. c) Chloride ion is one of the products of the solubility equilibrium reaction. Added chloride ion (from HCl) will shift the reaction to the left, due to the common ion effect. As a result of the reaction shifting to the left, the mass of solid PbCl2 will increase. 19.12A Plan: Compare the Ksp values for the two salts. Since CaSO4 is more soluble, calculate the concentration of sulfate ions in equilibrium with the Ca2+ concentration. Solution: The solubility equilibrium for CaSO4 is CaSO4(s) ⇆ Ca2+(aq) + SO42−(aq) and Ksp = [Ca2+][SO42−] = 2.4×10−5 ⎛ 2.4 ×10−5 ⎞⎟ K sp ⎜⎜ ⎡SO4 2− ⎤ = ⎟⎟ = 9.6×10−4 M = ⎣ ⎦ [Ca 2+ ] ⎜⎝⎜ 0.025 M ⎠⎟ 19.12B Plan: Compare the Ksp values for the two salts. Since BaF2 is more soluble, calculate the concentration of fluoride ions in equilibrium with the Ba2+ concentration. Solution: The solubility equilibrium for BaF2 is BaF2(s) ⇆ Ba2+(aq) + 2F−(aq) and Ksp = [Ba2+][F−]2 = 1.5×10−6 Ksp 1.5×10 –6 [F−] = = = 8.66025×10−3 = 8.7×10−3 M [Ba2+ ] (0.020) 19.13A Plan: Write the complex-ion formation equilibrium reaction. Calculate the initial concentrations of Fe3+ and CN−. The approach to complex ion equilibria problems is slightly different than for solubility equilibria because formation constants are generally large while solubility product constants are generally very small. The best mathematical approach is to assume that the equilibrium reaction goes to completion and then calculate back to find the equilibrium concentrations of reactants. So, assume that all of the Fe3+ reacts to form Fe(CN)63− and calculate the concentration of Fe(CN)63− formed from the given concentrations of Fe3+ and CN− and the concentration of the excess reactant. Then use the complex ion formation equilibrium to find the equilibrium concentration of Fe3+. Solution: Equilibrium reaction: Fe3+(aq) + 6CN−(aq) ⇆ Fe(CN)63−(aq) Initial concentrations from a simple dilution calculation: Mf = MiVi/Vf

(3.1×10

−2

[Fe ] =

M )(25.5 mL )

= 0.0130661157 M 25.5 + 35.0 mL (1.5 M )(35.0 mL) [CN−] = = 0.867768595 M 25.5 + 35.0 mL

19-16

Set up a reaction table: Fe3+(aq) + 6CN−(aq) ⇆ Fe(CN)63−(aq) Concentration (M) Initial 0.0130661157 0.867768595 0 Change −0.0130661157 + x − 6(0.013066115) +0.013066115 Equilibrium x 0.789371905 0.013066115 ⎡ Fe(CN)3− ⎤ 6 ⎦⎥ [ 0.0130661] ⎢ Kf = 4.0×1043 = ⎣ = 6 6 [ x ][ 0.789371905] ⎡ Fe3+ ⎤ ⎡CN− ⎤ ⎣⎢ ⎦⎥ ⎣⎢ ⎦⎥ x = 1.35019×10−45 = 1.4×10−45 M The concentration of Fe3+ at equilibrium is 1.4×10−45 M. This concentration is so low that it is impossible to calculate it using the initial concentrations minus a variable x. The variable would have to be so close to the initial concentration that the initial concentration of x cannot be calculated to enough significant figures (43 in this case) to get a difference in concentrations of 1×10−45 M. Thus, the approach above is the best to calculate the very low equilibrium concentration of Fe3+. 19.13B Plan: Write the complex-ion formation equilibrium reaction. Calculate the initial concentration of Al3+. The approach to complex ion equilibria problems is slightly different than for solubility equilibria because formation constants are generally large while solubility product constants are generally very small. The best mathematical approach is to assume that the equilibrium reaction goes to completion and then calculate back to find the equilibrium concentrations of reactants. So, assume that all of the Al3+ reacts to form AlF63− and calculate the concentration of AlF63− formed from the given concentrations of Al3+ and F−. Then use the complex ion formation equilibrium to find the equilibrium concentration of Al3+. Solution: Equilibrium reaction: Al3+(aq) + 6F−(aq) ⇆ AlF63−(aq) + 6H2O(l) Initial concentrations: 1 mol AlCl3 2.4 g AlCl3 × 133.33 g AlCl3 = 0.0720018 M [Al3+] = 0.250 L [F−] = 0.560 M Set up a reaction table: Al3+(aq) + 6F−(aq) ⇆ AlF63−(aq) Concentration (M) Initial 0.0720018 0.560 0 Change −0.0720018 + x − 6(0.0720018) +0.0720018 Equilibrium x 0.1279892 0.0720018 ⎡ AlF 3− ⎤ (0.0720018) 6 ⎢ ⎥⎦ Kf = 4×1019 = ⎣ = x = 4.0949×10−16 = 4×10−16 M 6 6 3+ − (x)(0.1279892) ⎡ Al ⎤ ⎡ F ⎤ ⎢⎣ ⎥⎦ ⎢⎣ ⎥⎦ 3+ The concentration of Al at equilibrium is 4×10−16 M 19.14A Plan: Write equations for the solubility equilibrium and formation of the silver-ammonia complex ion. Add the two equations to get the overall reaction. Set up a reaction table for the overall reaction with the given value for initial [NH3]. Write equilibrium expressions from the overall balanced reaction and calculate Koverall from Kf and Ksp values. Insert the equilibrium concentration values from the reaction table into the equilibrium expression and calculate solubility. Solution: Equilibria: Ksp = 5.0×10–13 AgBr(s) ⇆ Ag+(aq) + Br−(aq) + + Ag (aq) + 2NH3(aq) ⇆ Ag(NH3)2 (aq) Kf = 1.7×107 + − AgBr(s) + 2NH3(aq) ⇆ Ag(NH3)2 (aq) + Br (aq) Koverall = KspKf

19-17

Set up reaction table: AgBr(s) + 2NH3(aq) ⇆ Ag(NH3)2+(aq) + Br−(aq) Concentration (M) Initial — 1.0 0 0 +S +S Change — – 2S Equilibrium — 1.0 – 2S S S + − Ag(NH ) Br ] [ 3 2 ][ Koverall = = KspKf = (5.0×10–13)(1.7×107) = 8.5×10–6 2 [ NH3 ] Calculate the solubility of AgBr: [ S ][ S ] Koverall = = 8.5×10–6 2 ⎡1.0 − 2 S ⎤ ⎢⎣ ⎥⎦ Assume that 1.0 – 2S is approximately equal to 1.0, which appears to be a good assumption based on the fact that 1.0 >> 8.5×10–6, Koverall.

[ S ][ S ]

–6 2 = 8.5×10 [1.0 ] S = 2.9154759×10–3 = 2.9×10–3 M The solubility of AgBr in ammonia is less than its solubility in hypo (sodium thiosulfate). Since the formation constant for Ag(NH3)2+ is less than the formation constant of Ag(S2O3)23−, the addition of ammonia will increase the solubility of AgBr less than the addition of thiosulfate ion increases its solubility.

19.14B Plan: Write equations for the solubility equilibrium and formation of the lead(II)-hydroxide complex ion. Add the two equations to get the overall reaction. Set up a reaction table for the overall reaction with the given value for initial [OH−]. Write equilibrium expressions from the overall balanced reaction and calculate Koverall from Kf and Ksp values. Insert the equilibrium concentration values from the reaction table into the equilibrium expression and calculate solubility. Solution: Equilibria: Ksp = 1.7×10–5 PbCl2(s) ⇆ Pb2+(aq) + 2Cl−(aq) 2+ − − Pb (aq) + 3OH (aq) ⇆ Pb(OH)3 (aq) Kf = 8×1013 − − − PbCl2(s)+ 3OH (aq) ⇆ Pb(OH)3 (aq) + 2Cl (aq) Koverall = KspKf Set up reaction table: PbCl2 (s) + 3OH− (aq) ⇆ Pb(OH)3− (aq) + 2Cl− Concentration (M) Initial — 0.75 0 0 +S +2S Change — – 3S Equilibrium — 0.75 – 3S S 2S − − 2 [ Pb(OH)3 ][ Cl ] Koverall = = KspKf = (1.7×10–5)(8×1013) = 1.36×109 − 3 [ OH ] Calculate the solubility of PbCl2: 2 [S][ 2S] 9 Koverall = 3 = 1.36×10 [ 0.75 – 3S] 4S3 = 1.36×109 Take the cube root of both sides of the equation 3 [0.75 – 3S] 1.5874S = 1.1079×103 0.75 – 3S S = 523.4503 – 2093.801S S = 0.24988 = 0.25 M

19-18

CHEMICAL CONNECTIONS BOXED READING PROBLEMS

B19.1

Plan: Consult Figure 19.5 for the colors and pH ranges of the indicators. Solution: Litmus paper indicates the pH is below 7. The result from thymol blue, which turns yellow at a pH above 2.5, indicates that the pH is above 2.5. Bromphenol blue is the best indicator as it is green in a fairly narrow range of 3.5 < pH < 4. Methyl red turns red below a pH of 4.3. Therefore, a reasonable estimate for the rainwater pH is 3.5 to 4.

B19.2

Plan: Find the volume of the rain received by multiplying the surface area of the lake by the depth of rain. Find the volume of the lake before the rain. Express both volumes in liters. The pH of the rain is used to find the molarity of H3O+; this molarity multiplied by the volume of rain gives the moles of H3O+. The moles of H3O+ divided by the volume of the lake plus rain gives the molarity of H3O+ and the pH of the lake. Solution: a) To find the volume of rain, multiply the surface area in square inches by the depth of rain. Convert the volume in in3 to cm3 and then to L. 3 −3 ⎛ 4.840 ×103 yd 2 ⎞⎟⎛ 36 in ⎞⎟2 ⎛ 2.54 cm ⎞⎟ ⎛⎜ 1 mL ⎞⎛ ⎟⎟⎜⎜10 L ⎞⎟⎟ ⎜⎜ ⎜⎜ ⎜⎜ ⎟ ⎟ ⎟ Volume (L) of rain = 10.0 acres ⎜ ⎟⎟⎜ ⎟ ⎟⎟ ⎟⎟ 1.00 in ⎜ ⎟⎟ ⎜⎜ ⎜ 3 ⎟⎜ ⎜ ⎜ 1 acre ⎟⎜ 1 mL ⎠⎟ ⎜⎝⎜ ⎝⎜⎜ 1 in ⎠⎟ ⎝⎜1 cm ⎠⎝ ⎠⎟⎜⎝ 1 yd ⎠⎟ = 1.027902×106 L + At pH = 4.20, [H3O ] = 10–4.20 = 6.3095734×10–5 M ⎛ 6.3095734 ×10−5 mol ⎞⎟ ⎟⎟ = 64.8562 = 65 mol Moles of H3O+ = (1.027902×106 L)⎜⎜⎜ ⎜⎝ L ⎠⎟

(

)

(

)

2 3 3 2⎞ ⎛ ⎛12 in ⎞⎛ 10−3 L ⎞⎟ ⎟⎜⎜ 2.54 cm ⎞⎟⎟ ⎛⎜⎜ 1 mL ⎞⎛ ⎜ 4.840 ×10 yd ⎟⎟⎜⎜⎛ 36 in ⎞⎟⎟ ⎜ ⎟⎟⎜⎜ ⎟ ⎟ ⎜ ⎜ b) Volume (L) of the lake = 10.0 acres ⎜ ⎟⎟⎜ ⎟ ⎟⎟ 10.0 ft ⎜ ⎟⎟⎜ ⎟⎟ ⎜ ⎜ ⎟ ⎟⎟ 1 acre ⎟⎜⎜ 1 in ⎠⎟ ⎝⎜⎜1 cm 3 ⎠⎝ ⎜⎝⎜ 1 ft ⎠⎝ ⎟⎠⎝⎜⎜ 1 yd ⎠⎟ ⎟⎜⎜ 1 mL ⎠⎟ ⎜⎝⎜ = 1.23348×108 L

(

)

(

)

Total volume of lake after rain = 1.23348×108 L + 1.027902×106 L = 1.243759×108 L

mol H3O+ 64.8562 mol = = 5.214531×10–7 M L 1.243759×108 L pH = –log (5.214531×10–7) = 6.2827847 = 6.28 c) Each mol of H3O+ requires one mole of HCO3− for neutralization. − ⎛1 mol HCO3− ⎞⎛ ⎟⎟⎜⎜ 61.02 g HCO3 ⎞⎟⎟ ⎜ Mass (g) = (64.8562 mol H3O+) ⎜⎜ ⎟⎟⎟⎜⎜⎜ 1 mol HCO − ⎟⎟⎟ ⎜⎝⎜ 1 mol H 3 O+ ⎠⎝ 3 ⎠ [H3O+] =

= 3.97575×103 = 4.0×103 g HCO3– END–OF–CHAPTER PROBLEMS

19.1

The purpose of an acid-base buffer is to maintain a relatively constant pH in a solution.

19.2

The weak-acid component neutralizes added base and the weak-base component neutralizes added acid so that the pH of the buffer solution remains relatively constant. The components of a buffer do not neutralize one another when they are a conjugate acid-base pair.

19.3

The presence of an ion in common between two solutes will cause any equilibrium involving either of them to shift in accordance with Le Châtelier’s principle. For example, addition of NaF to a solution of HF will cause the equilibrium HF(aq) + H2O(l) ⇆ H3O+(aq) + F −(aq) to shift to the left, away from the excess of F−, the common ion.

19-19

19.4

A buffer with a high capacity has a great resistance to pH change. A high buffer capacity results when the weak acid and weak base are both present at high concentration. Addition of 0.01 mol of HCl to a high capacity buffer will cause a smaller change in pH than with a low capacity buffer, since the ratio [HA]/[A−] will change less.

19.5

Only the concentration of the buffer components (c) has an effect on the buffer capacity. In theory, any conjugate pair (of any pKa) can be used to make a high capacity buffer. With proper choice of components, it can be at any pH. The buffer range changes along with the buffer capacity, but does not determine it. A high capacity buffer will result when comparable quantities (i.e., buffer-component ratio < 10:1) of weak acid and weak base are dissolved so that their concentrations are relatively high.

19.6

The buffer-component ratio refers to the ratio of concentrations of the acid and base that make up the buffer. When this ratio is equal to 1, the buffer resists changes in pH with added acid to the same extent that it resists changes in pH with added base. The buffer range extends equally in both the acidic and basic direction. When the ratio shifts with higher [base] than [acid], the buffer is more effective at neutralizing added acid than base so the range extends further in the acidic than basic direction. The opposite is true for a buffer where [acid] > [base]. Buffers with a ratio equal to 1 have the greatest buffer range. The more the buffer-component ratio deviates from 1, the smaller the buffer range.

19.7

pKa (formic acid) = 3.74; pKa (acetic acid) = 4.74. Formic acid would be the better choice, since its pKa is closer to the desired pH of 3.5. If acetic acid were used, the buffer-component ratio would be far from 1:1 and the buffer’s effectiveness would be lower. The NaOH serves to partially neutralize the acid and produce its conjugate base.

19.8

Plan: Remember that the weak-acid buffer component neutralizes added base and the weak-base buffer component neutralizes added acid. Solution: a) The buffer-component ratio and pH increase with added base. The OH− reacts with HA to decrease its concentration and increase [NaA]. The ratio [NaA]/[HA] thus increases. The pH of the buffer will be more basic because the concentration of base, A−, has increased and the concentration of acid, HA, decreased. b) Buffer-component ratio and pH decrease with added acid. The H3O+ reacts with A− to decrease its concentration and increase [HA]. The ratio [NaA]/[HA] thus decreases. The pH of the buffer will be more acidic because the concentration of base, A−, has decreased and the concentration of acid, HA, increased. c) Buffer-component ratio and pH increase with the added sodium salt. The additional NaA increases the concentration of both NaA and HA, but the relative increase in [NaA] is greater. Thus, the ratio increases and the solution becomes more basic. Whenever base is added to a buffer, the pH always increases, but only slightly if the amount of base is not too large. d) Buffer-component ratio and pH decrease. The concentration of HA increases more than the concentration of NaA, so the ratio is less and the solution is more acidic.

19.9

a) pH would increase by a small amount. b) pH would decrease by a small amount. c) pH would increase by a very small amount. d) pH would increase by a large amount.

19.10

a) Buffer 3 has equal, high concentrations of both HA and A−. It has the highest buffering capacity. b) All of the buffers have the same pH range. The practical buffer range is pH = pKa ± 1, and is independent of concentration. c) Buffer 2 has the greatest amount of weak base and can therefore neutralize the greatest amount of added acid.

19.11

a) Added strong base would react with the acid component of the buffer. Buffer 3 has the largest concentration of acid and thus can react with the largest amount of added base.

19-20

(b) Buffer 1 has a [A–]/[HA] ratio of 5/3 = 1.666; buffer 2 has a [A–]/[HA] ratio of 3/3 = 1; buffer 3 has a [A–]/[HA] ratio of 4/5 = 0.8. Because it has the highest [A–]/[HA] ratio, Buffer 1 has the highest pH. (c) Buffer 2 has a [A–]/[HA] ratio of 3/3 = 1; according to the Henderson-Hasselbalch equation: [A− ] pH = pKa + log pH = pKa + log 1 pH = pKa + 0 [HA] pH = pKa in Buffer 2. 19.12

Plan: The buffer components are propanoic acid and propanoate ion, the concentrations of which are known. The sodium ion is a spectator ion and is ignored because it is not involved in the buffer. Write the propanoic acid-dissociation reaction and its Ka expression. Set up a reaction table in which x equals the amount of acid that dissociates; solving for x will result in [H3O+], from which the pH can be calculated. Alternatively, the pH can be calculated from the Henderson-Hasselbalch equation. Solution: Concentration (M) CH3CH2COOH(aq) + H2O(l) ⇆ CH3CH2COO−(aq) + H3O+(aq) Initial 0.15 — 0.35 0 Change –x — +x +x Equilibrium 0.15 – x — 0.35 + x x Assume that x is negligible with respect to both 0.15 and 0.35 since both concentrations are much larger than Ka. [ H3O+ ][CH3CH2 COO− ] [ x ][0.35 + x] [ x][0.35] Ka = 1.3×10–5 = = = [0.15] [0.15 − x ] [CH3CH2 COOH] x = [H3O+] = Ka =

⎛ 0.15 ⎞⎟ [CH3CH2 COOH] ⎟ = 5.57143×10–6 = 5.6×10–6 M = (1.3×10−5 )⎜⎜ − ⎜⎝ 0.35 ⎠⎟ [CH3CH2 COO ]

Check assumption: Percent error = (5.6×10–6/0.15)100% = 0.0037%. The assumption is valid. pH = –log [H3O+] = –log (5.57143×10–6) = 5.2540 = 5.25 Another solution path to find pH is using the Henderson-Hasselbalch equation: ⎛[base]⎞⎟ pH = pKa + log ⎜⎜ pKa = –log (1.3×10–5) = 4.886 ⎟ ⎜⎝ [acid] ⎠⎟

⎛ [CH3 CH 2 COO− ] ⎞⎟ ⎛[0.35]⎞⎟ ⎟⎟ = 4.886 + log ⎜⎜ pH = 4.886 + log ⎜⎜⎜ ⎟ ⎜⎝[0.15]⎠⎟ ⎟ ⎜⎝[CH3CH 2 COOH]⎠ pH = 5.25398 = 5.25 19.13

C6H5COOH(aq) + H2O(l) ⇆ C6H5COO–(aq) + H3O+(aq) [ H3O+ ][C6 H5COO− ] [ x ][ 0.28 + x ] [ x ][ 0.28] Ka = 6.3×10–5 = = = [ 0.33 − x ] [ 0.33] [C6 H5 COOH] x = [H3O+] = (6.3×10–5)(0.33/0.28) = 7.425×10–5 = 7.4×10–5 M Check assumption: Percent error = (7.425×10–5/0.28)100% = 0.026%. The assumption is valid. pH = –log [H3O+] = –log (7.425×10–5) = 4.1293 = 4.13

19.14

Plan: The buffer components are HNO2 and NO2−, the concentrations of which are known. The potassium ion is a spectator ion and is ignored because it is not involved in the buffer. Write the HNO2 acid-dissociation reaction and its Ka expression. Set up a reaction table in which x equals the amount of acid that dissociates; solving for x will result in [H3O+], from which the pH can be calculated. Alternatively, the pH can be calculated from the Henderson-Hasselbalch equation. Solution: Concentration (M) HNO2(aq) + H2O(l) ⇆ NO2−(aq) + H3O+(aq) Initial 0.55 — 0.75 0 Change −x — +x +x Equilibrium 0.55 − x — 0.75 + x x

19-21

Assume that x is negligible with respect to both 0.55 and 0.75 since both concentrations are much larger than Ka. [ H3O+ ][ NO2− ] [ x ] ⎡⎣⎢0.75 + x⎤⎦⎥ [ x][0.75] Ka = 7.1×10–4 = = = ⎡ 0.55 − x ⎤ [0.55] [ HNO2 ] ⎣⎢ ⎦⎥ ⎡ 0.55⎤ [ HNO2 ] ⎦ = 5.2066667×10–4 = 5.2×10–4 M x = [H3O+] = Ka = (7.1×10−4 ) ⎣ − ⎡ ⎤ [ NO2 ] ⎣ 0.75⎦ Check assumption: Percent error = (5.2066667×10–4/0.55)100% = 0.095%. The assumption is valid. pH = −log [H3O+] = −log (5.2066667×10–4) = 3.28344 = 3.28 Verify the pH using the Henderson-Hasselbalch equation. ⎛[base]⎞⎟ pH = pKa + log ⎜⎜ pKa = –log(7.1×10–4) = 3.149 ⎟ ⎝⎜ [acid] ⎠⎟

⎛ [NO2− ] ⎞⎟ ⎛[0.75]⎞⎟ ⎟ = 3.149 + log ⎜⎜ pH = 3.149 + log ⎜⎜ ⎟ ⎜⎝[0.55]⎠⎟ ⎜⎝[HNO2 ]⎠⎟⎟ pH = 3.2837 = 3.28 19.15

HF(aq) + H2O(l) ⇆ F–(aq) + H3O+(aq) [ H3O+ ][ F− ] [ x ][0.25 + x] [ x][0.25] Ka = 6.8×10−4 = = = [ 0.20 − x ] [ 0.20] [ HF ] [ HF ] [H3O+] = Ka − = (6.8×10−4)(0.20/0.25) = 5.44×10−4 = 5.4×10−4 M [F ] Check assumption: Percent error = (5.44×10−4/0.20)100% = 0.27%. The assumption is valid. pH = −log [H3O+] = −log (5.44×10−4) = 3.2644 = 3.26 Verify the pH using the Henderson-Hasselbalch equation.

19.16

Plan: The buffer components are formic acid, HCOOH, and formate ion, HCOO−, the concentrations of which are known. The sodium ion is a spectator ion and is ignored because it is not involved in the buffer. Write the HCOOH acid-dissociation reaction and its Ka expression. Set up a reaction table in which x equals the amount of acid that dissociates; solving for x will result in [H3O+], from which the pH can be calculated. Alternatively, the pH can be calculated from the Henderson-Hasselbalch equation. Solution: Ka = 10−pKa = 10−3.74 = 1.8197×10−4 HCOOH(aq) + H2O(l) ⇆ HCOO−(aq) + H3O+(aq) Concentration (M) Initial 0.45 — 0.63 0 Change −x — +x +x Equilibrium 0.45 − x — 0.63 + x x Assume that x is negligible because both concentrations are much larger than Ka. [ H3O+ ][ HCOO− ] [ x ][ 0.63 + x ] [ x ][ 0.63] Ka = 1.8197×10−4 = = = [0.45 − x] [0.45] [ HCOOH ] x = [H3O+] = Ka

⎡0.45⎤ [ HCOOH ] ⎦ = 1.29979×10−4 = 1.3×10−4 M = (1.8197×10−4 ) ⎣ − ⎡0.63⎤ [ HCOO ] ⎣ ⎦

Check assumption: Percent error = (1.29979×10−4/0.45)100% = 0.029%. The assumption is valid. pH = −log [H3O+] = −log (1.29979×10−4) = 3.886127 = 3.89 Verify the pH using the Henderson-Hasselbalch equation. ⎛[base]⎞⎟ pH = pKa + log ⎜⎜ ⎟ ⎝⎜ [acid] ⎠⎟ Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

19-22

⎛ [HCOO− ] ⎞⎟ ⎛[0.63]⎞⎟ pH = 3.74 + log ⎜⎜ ⎟⎟ = 3.74 + log ⎜⎜ ⎟ ⎜⎝[HCOOH]⎠ ⎜⎝[0.45]⎠⎟ pH = 3.8861 = 3.89 19.17

HBrO(aq) + H2O(l) ⇆ BrO−(aq) + H3O+(aq) Ka = 10−pKa = 10−8.64 = 2.2908677×10−9 −9

Ka = 2.2908677×10 =

[ H 3 O+ ][ BrO− ]

[ x ][ 0.68 + x ]

[ x ][ 0.68]

[0.95 − x] [0.95] [ HBrO ] [ HBrO ] = (2.2908677×10−9)(0.95/0.68) = 3.2004769×10−9 = 3.2×10−9 M x = [H3O+] = Ka [ BrO− ] Check assumption: Percent error = (3.2004769×10−9/0.68)100% = 0.00000047%. The assumption is valid. pH = −log [H3O+] = −log (3.2004769×10−9) = 8.4947853 = 8.49 Verify the pH using the Henderson-Hasselbalch equation. 19.18

Plan: The buffer components phenol, C6H5OH, and phenolate ion, C6H5O−, the concentrations of which are known. The sodium ion is a spectator ion and is ignored because it is not involved in the buffer. Write the C6H5OH aciddissociation reaction and its Ka expression. Set up a reaction table in which x equals the amount of acid that dissociates; solving for x will result in [H3O+], from which the pH can be calculated. Alternatively, the pH can be calculated from the Henderson-Hasselbalch equation. Solution: Ka = 10−pKa = 10–10.00 = 1.0×10–10 C6H5OH(aq) + H2O(l) ⇆ C6H5O−(aq) + H3O+(aq) Concentration (M) Initial 1.2 — 1.3 0 Change –x — +x +x Equilibrium 1.2 – x — 1.3 + x x Assume that x is negligible with respect to both 1.0 and 1.2 because both concentrations are much larger than Ka. [ H3O+ ][C6 H5O− ] [ x ][1.3 + x ] [ x ][1.3] –10 Ka = 1.0×10 = = = [1.2 − x ] [1.2 ] [C6 H5 OH ] x = [H3O+] = Ka

[C6 H5OH] ⎛1.2 ⎞ = (1.0 ×10−10 )⎜⎜ ⎟⎟ = 9.23077×10–11 M − ⎜⎝1.3 ⎠⎟ [ C6 H 5 O ]

Check assumption: Percent error = (9.23077×10–11/1.2)100% = 7.7×10–9%. The assumption is valid. pH = –log (9.23077×10–11) = 10.03476 = 10.03 Verify the pH using the Henderson-Hasselbalch equation: ⎛[base]⎞⎟ pH = pKa + log ⎜⎜ ⎟ ⎜⎝ [acid] ⎠⎟

⎛ [C H O− ] ⎞⎟ ⎛[1.3]⎞⎟ ⎟⎟ = 10.00 + log ⎜⎜ pH = 10.00 + log ⎜⎜⎜ 6 5 ⎟ ⎜⎝[1.2]⎠⎟ ⎜⎝[C6 H 5 OH]⎠⎟ pH = 10.03 19.19

H2BO3–(aq) + H3O+(aq) H3BO3(aq) + H2O(l) ⇆ –9.24 −pKa Ka = 10 = 10 = 5.7543994×10–10 [ H3O+ ][ H2 BO3− ] [ x ][ 0.82 + x ] [ x ][ 0.82] –10 Ka = 5.7543994×10 = = = [ 0.12 − x ] [ 0.12] [ H3 BO3 ]

19-23

x = [H3O+] = Ka

[ H3BO3 ] = (5.7543994×10–10)(0.12/0.82) = 8.4210723×10–11 M [ H2 BO3− ]

Check assumption: Percent error = (8.4210723×10–11/0.12)100% = 7.0×10–8%. The assumption is valid. pH = –log [H3O+] = –log (8.4210723×10–11) = 10.0746326 = 10.07 Verify the pH using the Henderson-Hasselbalch equation. 19.20

Plan: The buffer components ammonia, NH3, and ammonium ion, NH4+, the concentrations of which are known. The chloride ion is a spectator ion and is ignored because it is not involved in the buffer. Write the NH4+ aciddissociation reaction and its Ka expression. Set up a reaction table in which x equals the amount of acid that dissociates; solving for x will result in [H3O+], from which the pH can be calculated. Alternatively, the pH can be calculated from the Henderson-Hasselbalch equation. The Ka of NH4+ will have to be calculated from the pKb. Solution: 14 = pKa + pKb pKa = 14 – pKb = 14 – 4.75 = 9.25 Ka = 10−pKa = 10–9.25 = 5.62341325×10–10 NH4+(aq) + H2O(l) ⇆ NH3(aq) + H3O+(aq) Concentration (M) Initial 0.15 — 0.25 0 Change –x — +x +x Equilibrium 0.15 – x — 0.25 + x x Assume that x is negligible with respect to both 0.25 and 0.15 because both concentrations are much larger than Ka. [ NH3 ][ H3O+ ] [0.25 + x ][ H3O+ ] [0.25][ H3O+ ] Ka = .62341325×10–10 = = = [0.15 − x ] [0.15] [ NH4+ ] X = [H3O+] = Ka

[ NH 4 + ]

⎛ 0.15 ⎞⎟ ⎟⎟ = 3.374048×10–10 M = (5.62341325×10−10 )⎜⎜ ⎜ NH 0.25 ⎝ ⎠ [ 3]

Check assumption: Percent error = (3.374048×10–10/0.15)100% = 2×10–7%. The assumption is valid. pH = –log [H3O+] = –log [3.374048×10–10] = 9.4718 = 9.47 Verify the pH using the Henderson-Hasselbalch equation. ⎛[base]⎞⎟ pH = pKa + log ⎜⎜ ⎟ ⎝⎜ [acid] ⎠⎟

⎛ [NH3 ] ⎞⎟ ⎛[0.25]⎞⎟ ⎟ = 9.25 + log ⎜⎜ pH = 9.25 + log ⎜⎜ ⎟ ⎜⎝[0.15]⎠⎟ ⎜⎝[NH 4 + ]⎠⎟⎟ pH = 9.47 19.21

Kb = 10−pKb = 10–3.35 = 4.4668359×10–4 The base component is CH3NH2 and the acid component is CH3NH3+. Neglect Cl–. Assume + x and – x are negligible. ⇆ CH3NH3+(aq) + OH–(aq) CH3NH2(aq) + H2O(l) Kb = 4.4668359×10–4 =

[OH–] = Kb

[CH3 NH2 ]

[CH3 NH3+ ]

[CH3 NH3+ ][ OH− ] [CH3 NH2 ]

[ 0.60 + x ][ OH− ]

[0.50 − x ]

[ 0.60 ][ OH− ]

[0.50 ]

= (4.4668359×10–4)(0.50/0.60) = 3.7223632×10–4 M

Check assumption: Percent error = (3.7223632×10–4/0.50)100% = 0.074%. The assumption is valid. pOH = –log [OH–] = –log (3.7223632×10–4) = 3.429181254 pH = 14.00 – pOH = 14.00 – 3.429181254 = 10.57081875 = 10.57 Verify the pH using the Henderson-Hasselbalch equation. Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

19-24

19.22

Plan: The buffer components are HCO3− from the salt KHCO3 and CO32− from the salt K2CO3. Choose the Ka value that corresponds to the equilibrium with these two components. The potassium ion is a spectator ion and is ignored because it is not involved in the buffer. Write the acid-dissociation reaction and its Ka expression. Set up a reaction table in which x equals the amount of acid that dissociates; solving for x will result in [H3O+], from which the pH can be calculated. Alternatively, the pH can be calculated from the Henderson-Hasselbalch equation. Solution: a) Ka1 refers to carbonic acid, H2CO3, losing one proton to produce HCO3−. This is not the correct Ka because H2CO3 is not involved in the buffer. Ka2 is the correct Ka to choose because it is the equilibrium constant for the loss of the second proton to produce CO32− from HCO3−. b) Set up the reaction table and use Ka2 to calculate pH. HCO3−(aq) + H2O(l) ⇆ CO32−(aq) + H3O+(aq) Concentration (M) Initial 0.22 — 0.37 0 Change –x — +x +x Equilibrium 0.22 – x — 0.37 + x x Assume that x is negligible with respect to both 0.22 and 0.37 because both concentrations are much larger than Ka. [ H3O+ ] ⎡⎣CO32− ⎤⎦ [ x ] ⎡⎣⎢0.37 + x⎤⎦⎥ [ x ][ 0.37] Ka = 4.7×10−11 = = = ⎡ 0.22 − x ⎤ [ 0.22] [ HCO3− ] ⎣⎢ ⎦⎥

[ HCO3− ]

⎛ 0.22 ⎞⎟ 11 = (4.7 ×10−11 )⎜⎜ ⎟ = 2.79459×10− M ⎜ 0.37 ⎠⎟ ⎡CO32− ⎤ ⎝ ⎣ ⎦ Check assumption: Percent error = (2.79459×10−11/0.22)100% = 1.3×10-8%. The assumption is valid. pH = –log [H3O+] = –log (2.79459×10−11) = 10.5537 = 10.55 Verify the pH using the Henderson-Hasselbalch equation. ⎛[base]⎞⎟ pH = pKa + log ⎜⎜ pKa = –log (4.7×10–11) = 10.328 ⎟ ⎜⎝ [acid] ⎠⎟

[H3O+] = Ka

⎛ [CO32− ] ⎞⎟ ⎛[0.37]⎞⎟ ⎟ = 10.328 + log ⎜⎜ pH = 10.328 + log ⎜⎜⎜ ⎟ ⎜⎝[0.22]⎠⎟ ⎜⎝[HCO3− ]⎠⎟⎟ pH = 10.55 19.23

a) The conjugate acid-base pair is related by Ka2 (6.3×10–8). b) Assume that x is negligible with respect to both 0.50 and 0.40 because both concentrations are much larger than Ka. The acid component is H2PO4– and the base component is HPO42–. Neglect Na+. Assume + x and – x are negligible. ⇆ HPO42–(aq) + H3O+(aq) H2PO4–(aq) + H2O(l) [ H3O+ ] ⎡⎣ HPO4 2− ⎤⎦ [ x ] ⎡⎣⎢0.40 + x⎤⎦⎥ [ x ][ 0.40] Ka = 6.3×10–8 = = = ⎡ 0.50 − x ⎤ [0.50] [ H2 PO4− ] ⎢⎣ ⎥⎦ [H3O+] = Ka

[ H2 PO4− ]

= (6.3×10–8)(0.50/0.40) = 7.875×10–8 M ⎡ HPO4 2− ⎤ ⎣ ⎦ Check assumption: Percent error = (7.875×10–8/0.50)100% = 1.6×10–5%. The assumption is valid. pH = –log [H3O+] = –log (7.875×10–8) = 7.103749438 = 7.10 Verify the pH using the Henderson-Hasselbalch equation. 19.24

Plan: Given the pH and Ka of an acid, the buffer-component ratio can be calculated from the HendersonHasselbalch equation. Convert Ka to pKa.

19-25

Solution: pKa = –log Ka = –log (1.3×10–5) = 4.8860566 ⎛[base]⎞⎟ pH = pKa + log ⎜⎜ ⎟ ⎜⎝ [acid] ⎠⎟

⎛ [Pr− ] ⎞⎟ 5.44 = 4.8860566 + log ⎜⎜ ⎟ ⎜⎝[HPr]⎠⎟ ⎛ [Pr− ] ⎞⎟ 0.5539467 = log ⎜⎜ ⎟ ⎜⎝[HPr]⎠⎟

Raise each side to 10x.

[Pr− ] = 3.5805 = 3.6 [HPr] 19.25

pKa = –log Ka = –log (7.1×10–4) = 3.14874165

⎛[base]⎞⎟ pH = pKa + log ⎜⎜ ⎟ ⎜⎝ [acid] ⎠⎟ ⎛ [NO2− ] ⎞⎟ ⎟⎟ 2.95 = 3.14874165 + log ⎜⎜⎜ ⎝[HNO2 ]⎠⎟ ⎛ [NO2− ] ⎞⎟ ⎟ –0.19874165 = log ⎜⎜ ⎜⎝[HNO2 ]⎠⎟⎟ [NO2− ] = 0.632788 = 0.63 [HNO2 ] 19.26

Raise each side to 10x.

Plan: Given the pH and Ka of an acid, the buffer-component ratio can be calculated from the HendersonHasselbalch equation. Convert Ka to pKa. Solution: pKa = –log Ka = –log (2.3×10–9) = 8.63827 ⎛[base]⎞⎟ pH = pKa + log ⎜⎜ ⎟ ⎜⎝ [acid] ⎠⎟

⎛ [BrO− ] ⎞⎟ 7.95 = 8.63827 + log ⎜⎜ ⎟ ⎜⎝[HBrO]⎠⎟ ⎛ [BrO− ] ⎞⎟ –0.68827 = log ⎜⎜ ⎟ ⎜⎝[HBrO]⎠⎟

Raise each side to 10x.

[BrO− ] = 0.204989 = 0.20 [HBrO] 19.27

Given the pH and Ka of an acid, the buffer-component ratio can be calculated from the Henderson-Hasselbalch equation. pKa = –log Ka = –log (1.8×10–5) = 4.744727495 ⎛[base]⎞⎟ pH = pKa + log ⎜⎜ ⎟ ⎜⎝ [acid] ⎠⎟

⎛[CH3 CO O− ]⎞⎟ ⎟ 4.39 = 4.744727495 + log ⎜⎜⎜ ⎜⎝ [CH3COOH] ⎠⎟⎟ ⎛ [CH3 COO− ] ⎞⎟ ⎟ –0.35473 = log ⎜⎜⎜ ⎜⎝[CH3 COOH]⎠⎟⎟

Raise each side to 10x.

19-26

[CH 3 COO− ] = 0.441845 = 0.44 [CH 3 COOH]

19.28

Plan: Determine the pKa of the acid from the concentrations of the conjugate acid and base, and the pH of the solution. This requires the Henderson-Hasselbalch equation. Set up a reaction table that shows the stoichiometry of adding the strong base NaOH to the weak acid in the buffer. Calculate the new concentrations of the buffer components and use the Henderson-Hasselbalch equation to find the new pH. Solution: ⎛[base]⎞⎟ pH = pKa + log ⎜⎜ ⎟ ⎜⎝ [acid] ⎠⎟

⎛ [A− ] ⎞⎟ ⎛[0.1500]⎞⎟ 3.35 = pKa + log ⎜⎜ ⎟ = pKa + log ⎜⎜ ⎟ ⎜⎝[HA]⎠⎟ ⎜⎝[0.2000]⎠⎟ 3.35 = pKa – 0.1249387 pKa = 3.474939 = 3.47 Determine the moles of conjugate acid (HA) and conjugate base (A–) using (M)(V) = moles. ⎛ 0.2000 mol HA ⎞⎟ = 0.1000 mol HA ⎝ ⎠⎟⎟ 1L

Moles of HA = (0.5000 L )⎜⎜

⎛ 0.1500 mol A− ⎞⎟ – Moles of A– = (0.5000 L )⎜⎜ ⎟ = 0.07500 mol A ⎜⎝ ⎠⎟ 1L The reaction is: + NaOH(aq) → Na+(aq) + A–(aq) + H2O(l) HA(aq) Initial 0.1000 mol 0.0015 mol 0.07500 mol Change –0.0015 mol –0.0015 mol + 0.0015 mol Final 0.0985 mol 0 mol 0.0765 mol NaOH is the limiting reagent. The addition of 0.0015 mol NaOH produces an additional 0.0015 mol A– and consumes 0.0015 mol of HA. Then: 0.0765 mol A − = 0.153 M A– 0.5000 L 0.0985 mol HA [HA] = = 0.197 M HA 0.5000 L

[A–] =

⎛[base]⎞⎟ pH = pKa + log ⎜⎜ ⎟ ⎜⎝ [acid] ⎠⎟ ⎛[0.153]⎞⎟ pH = 3.474939 + log ⎜⎜ ⎟ = 3.365164 = 3.37 ⎜⎝[0.197]⎠⎟ 19.29

Determine the pKa of the acid from the concentrations of the conjugate acid and base and the pH of the solution. This requires the Henderson-Hasselbalch equation. ⎛[base]⎞⎟ pH = pKa + log ⎜⎜ ⎟ ⎜⎝ [acid] ⎠⎟

⎛ [B] ⎞⎟ ⎛[0.40]⎞⎟ 8.88 = pKa + log ⎜⎜ ⎟ = pKa + log ⎜⎜ ⎟ ⎜⎝[BH + ]⎠⎟ ⎜⎝[0.25]⎠⎟ 8.88 = pKa + 0.20411998 pKa = 8.67588 = 8.68 Determine the moles of conjugate acid (BH+) and conjugate base (B) using (M)(V) = moles. Moles BH+ = (0.25 L)(0.25 mol BH+/L) = 0.0625 mol BH+ Moles B = (0.25 L)(0.40 mol B/L) = 0.10 mol B Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

19-27

The reaction is: + HCl(aq) → BH+(aq) + Cl–(aq) + H2O(l) B(aq) Initial 0.10 mol 0.0020 mol 0.0625 mol Change –0.0020 mol –0.0020 mol +0.0020 mol Final 0.098 mol 0 mol 0.0645 mol HCl is the limiting reagent. The addition of 0.0020 mol HCl produces an additional 0.0020 mol BH+ and consumes 0.0020 mol of B. Then: [B] =

0.098 mol B = 0.392 M B 0.25 L

0.0645 mol BH+ = 0.258 M BH+ 0.25 L ⎛[base]⎞⎟ pH = pKa + log ⎜⎜ ⎟ ⎜⎝ [acid] ⎠⎟ [BH+] =

⎛[0.392]⎞⎟ pH = 8.67588 + log ⎜⎜ ⎟ = 8.857546361 = 8.86 ⎜⎝[0.258]⎠⎟ 19.30

Plan: Determine the pKa of the acid from the concentrations of the conjugate acid and base, and the pH of the solution. This requires the Henderson-Hasselbalch equation. Set up a reaction table that shows the stoichiometry of adding the strong base Ba(OH)2 to the weak acid in the buffer. Calculate the new concentrations of the buffer components and use the Henderson-Hasselbalch equation to find the new pH. Solution: ⎛[base]⎞⎟ pH = pKa + log ⎜⎜ ⎟ ⎜⎝ [acid] ⎠⎟

⎛ [Y− ] ⎞⎟ ⎛[0.220]⎞⎟ 8.77 = pKa + log ⎜⎜ ⎟ = pKa + log ⎜⎜ ⎟ ⎜⎝[HY]⎠⎟ ⎜⎝[0.110]⎠⎟ 8.77 = pKa + 0.3010299957 pKa = 8.46897 = 8.47 Determine the moles of conjugate acid (HY) and conjugate base (Y–) using (M)(V) = moles. ⎛ 0.110 mol HY ⎞⎟ = 0.0385 mol HY ⎝ ⎠⎟⎟ 1L

Moles of HY = (0.350 L )⎜⎜

⎛ 0.220 mol Y− ⎞⎟ – Moles of Y– = (0.350 L )⎜⎜ ⎟⎟ = 0.077 mol Y ⎜⎝ ⎠ 1L The reaction is: 2HY(aq) + Ba(OH)2(aq) → Ba2+(aq) + 2Y–(aq) + 2H2O(l) Initial 0.0385 mol 0.0015 mol 0.077 mol Change –0.0030 mol –0.0015 mol +0.0030 mol Final 0.0355 mol 0 mol 0.0800 mol Ba(OH)2 is the limiting reagent. The addition of 0.0015 mol Ba(OH)2 will produce 2 × 0.0015 mol Y– and consume 2 × 0.0015 mol of HY. Then: 0.0800 mol Y − = 0.228571 M Y– 0.350 L 0.0355 mol HY [HY] = = 0.101429 M HY 0.350 L

[Y–] =

⎛ [Y− ] ⎞⎟ pH = pKa + log ⎜⎜ ⎟ ⎜⎝[HY]⎠⎟ Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

19-28

⎛[0.228571]⎞⎟ pH = 8.46897 + log ⎜⎜ ⎟ = 8.82183 = 8.82 ⎜⎝[0.101429]⎠⎟ 19.31

⎛ [B] ⎞⎟ ⎛ [1.05] ⎞⎟ 9.50 = pKa + log ⎜⎜ ⎟⎟ = pKa + log ⎜⎜ ⎟ + ⎜⎝[BH ]⎠ ⎜⎝[0.750]⎠⎟ 9.50 = pKa + 0.1461280357 pKa = 9.353872 = 9.35 Determine the moles of conjugate acid (BH+) and conjugate base (B). Moles of BH+ = (0.500 L)(0.750 mol BH+/L) = 0.375 mol BH+ Moles of B = (0.500 L)(1.05 mol B/L) = 0.525 mol B The reaction is: + HCl(aq) → BH+(aq) + Cl– (aq) + H2O(l) B(aq) Initial 0.525 mol 0.0050 mol 0.375 mol Change –0.0050 mol –0.0050 mol +0.0050 mol Final 0.520 mol 0 mol 0.380 mol HCl is the limiting reagent. The addition of 0.0050 mol HCl will produce 0.0050 mol BH+ and consume 0.0050 mol of B. Then [B] =

0.520 mol B = 1.04 M B 0.500 L

[BH+] =

0.380 mol BH+ = 0.760 M BH+ 0.500 L

⎛ [B] ⎞⎟ pH = pKa + log ⎜⎜ ⎟ ⎜⎝[BH+ ]⎠⎟ ⎛ [1.04] ⎞⎟ ⎟⎟ = 9.490092 = 9.49 pH = 9.353872 + log ⎜⎜⎜ ⎜⎝ ⎡⎣0.760⎤⎦ ⎠⎟⎟ 19.32

Plan: The hydrochloric acid will react with the sodium acetate, NaC2H3O2, to form acetic acid, HC2H3O2. Calculate the number of moles of HCl and NaC2H3O2. Set up a reaction table that shows the stoichiometry of the reaction of HCl and NaC2H3O2. All of the HCl will be consumed to form HC2H3O2, and the number of moles of C2H3O2− will decrease. Find the new concentrations of NaC2H3O2 and HC2H3O2 and use the HendersonHasselbalch equation to find the pH of the buffer. Add 0.15 to find the pH of the buffer after the addition of the KOH. Use the Henderson-Hasselbalch equation to find the [base]/[acid] ratio needed to achieve that pH. Solution: ⎛ 0.452 mol HCl ⎞⎟⎛⎜10−3 L ⎞⎟ ⎟⎟ 204 mL = 0.092208 mol HCl ⎟⎟⎜⎜ a) Initial moles of HCl = ⎜⎜⎜ ⎜⎝ L ⎠⎟⎜⎜⎝ 1mL ⎠⎟⎟

(

)

⎛ 0.400 mol NaC2 H3O2 ⎞⎟ ⎟⎟ 0.500 L = 0.200 mol NaC2H3O2 Initial moles of NaC2H3O2 = ⎜⎜⎜ ⎜⎝ L ⎠⎟ HCl + NaC2H3O2 → HC2H3O2 + NaCl Initial 0.092208 mol 0.200 mol 0 mol Change –0.092208 mol –0.092208 mol +0.092208 mol Final 0 mol 0.107792 mol 0.092208 mol

(

)

19-29

Total volume = 0.500 L + (204 mL)(10–3 L/1 mL) = 0.704 L 0.092208 mol = 0.1309773 M 0.704 L 0.107792 mol = 0.1531136 M [C2H3O2−] = 0.704 L pKa = –log Ka = –log (1.8×10–5) = 4.744727495

[HC2H3O2] =

⎛ [C H O − ] ⎞ pH = pKa + log ⎜⎜⎜ 2 3 2 ⎟⎟⎟ ⎜⎝[HC 2 H3 O2 ]⎠⎟ ⎛[0.1531136]⎟⎞ pH = 4.744727495 + log ⎜⎜ ⎟ = 4.812545 = 4.81 ⎜⎝[0.1309773]⎟⎠ b) The addition of base would increase the pH, so the new pH is (4.81 + 0.15) = 4.96. The new [C2H3O2−]/[ HC2H3O2] ratio is calculated using the Henderson-Hasselbalch equation. ⎛ [C H O − ] ⎞ pH = pKa + log ⎜⎜⎜ 2 3 2 ⎟⎟⎟ ⎜⎝[HC 2 H3 O2 ]⎠⎟ ⎛ [C H O − ] ⎞ 4.96 = 4.744727495 + log ⎜⎜⎜ 2 3 2 ⎟⎟⎟ ⎜⎝[HC2 H3 O2 ]⎠⎟ ⎛ [C H O − ] ⎞ 0.215272505 = log ⎜⎜⎜ 2 3 2 ⎟⎟⎟ ⎜⎝[HC2 H3 O2 ]⎠⎟ [C 2 H 3 O2− ] = 1.64162 [HC 2 H 3 O2 ] From part a), we know that [HC2H3O2] + [C2H3O2−] = (0.1309773 M + 0.1531136 M) = 0.2840909 M. Although the ratio of [C2H3O2−] to [HC2H3O2] can change when acid or base is added, the absolute amount does not change unless acetic acid or an acetate salt is added. Given that [C2H3O2−]/[ HC2H3O2] = 1.64162 and [HC2H3O2] + [C2H3O2−] = 0.2840909 M, solve for [C2H3O2−] and substitute into the second equation. [C2H3O2−] = 1.64162[HC2H3O2] and [HC2H3O2] + 1.64162[HC2H3O2] = 0.2840909 M [HC2H3O2] = 0.1075441 M and [C2H3O2−] = 0.176547 M Moles of C2H3O2− needed = (0.176547 mol C2H3O2–/L)(0.500 L) = 0.0882735 mol Moles of C2H3O2− initially = (0.1531136 mol C2H3O2–/L)(0.500 L) = 0.0765568 mol This would require the addition of (0.0882735 mol – 0.0765568 mol) = 0.0117167 mol C2H3O2− The KOH added reacts with HC2H3O2 to produce additional C2H3O2−: HC2H3O2 + KOH → C2H3O2−+ K+ + H2O(l) To produce 0.0117167 mol C2H3O2– would require the addition of 0.0117167 mol KOH. ⎛ 56.11 g KOH ⎞⎟ Mass (g) of KOH = (0.0117167 mol KOH)⎜⎜ ⎟ = 0.657424 = 0.66 g KOH ⎝ 1 mol KOH ⎠⎟

19.33

a) The sodium hydroxide will react with the sodium bicarbonate, NaHCO3, to form carbonate ion, CO32–: NaOH + NaHCO3 → 2Na+ + CO32– + H2O Calculate the number of moles of NaOH and NaHCO3. All of the NaOH will be consumed to form CO32–, and the number of moles of NaHCO3 will decrease. The HCO3– is the important part of NaHCO3. ⎛ 0.10 mol NaOH ⎞⎟⎛⎜10−3 L ⎞⎟ ⎟⎟ 10.7 mL = 0.00107 mol NaOH ⎟⎟⎜⎜ Initial moles NaOH = ⎜⎜⎜ ⎟⎜⎝ 1 mL ⎠⎟⎟ L ⎝⎜ ⎠⎜ ⎛ 0.050 mol HCO3− ⎞⎟⎛⎜10−3 L ⎞⎟ ⎟⎟ 50.0 mL = 0.0025 mol HCO – ⎟⎟⎜⎜ Initial moles HCO3– = ⎜⎜⎜ 3 ⎟ ⎜⎝ L ⎠⎟⎜⎜⎝ 1 mL ⎠⎟

(

)

(

)

19-30

NaOH Initial 0.00107 mol Change –0.00107 mol Final 0 mol

NaHCO3 → 0.0025 mol –0.00107 mol 0.00143 mol

2Na+ + CO32– + H2O 0 mol +0.00107 mol 0.00107 mol

Total volume = (50.0 mL + 10.7 mL)(10–3 L/1 mL) = 0.0607 L 0.00143 mol = 0.023558484 M 0.0607 L 0.00107 mol = 0.017627677 M [CO32–] = 0.0607 L pKa = –log Ka = –log (4.7×10–11) = 10.32790214

[HCO3–] =

⎛ [CO32− ] ⎞⎟ ⎟ pH = pKa + log ⎜⎜⎜ ⎜⎝[HCO3− ]⎠⎟⎟ ⎛[0.017627677]⎞⎟ pH = 10.32790214 + log ⎜⎜ ⎟ = 10.2019 = 10.20 ⎜⎝[0.023558484]⎠⎟ b) The addition of acid would decrease the pH, so the new pH is (10.20 – 0.07) = 10.13. The new [CO32–]/[HCO3–] ratio is calculated using the Henderson-Hasselbalch equation. ⎛ [CO32− ] ⎞⎟ ⎟ pH = pKa + log ⎜⎜⎜ − ⎟ ⎝⎜[HCO3 ]⎠⎟

⎛ [CO32− ] ⎞⎟ ⎟ 10.13 = 10.32790214 + log ⎜⎜⎜ ⎜⎝[HCO3− ]⎠⎟⎟ ⎛ [CO 2− ] ⎞⎟ 3 − 0.19790214 = log ⎜⎜⎜ ⎟⎟ ⎜⎝[HCO3− ]⎠⎟ [CO 3 2− ] = 0.63401 [HCO 3− ] From part a), we know that [HCO3–] + [CO32–] = (0.023558484 M + 0.017627677 M) = 0.041185254 M. Although the ratio of [CO32–] to [HCO3–] can change when acid or base is added, the absolute amount does not change unless acetic acid or an acetate salt is added. Given that [CO32–]/[ HCO3–] = 0.63401 and [HCO3–] + [CO32–] = 0.041185254 M, solve for [CO32–] and substitute into the second equation. [CO32–] = 0.63401[HCO3–] and [HCO3–] + 0.63401[HCO3–] = 0.041185254 M [HCO3–] = 0.025205019 M and [CO32–] = 0.015980234 M Moles of HCO3– needed = (0.025205019 mol HCO3–/L)(10–3 L/1 mL)(25.0 mL) = 0.0006301255 mol Moles of HCO3– initially = (0.023558484 mol HCO3–/L)(10–3 L/1 mL)(25.0 mL) = 0.000588962 mol This would require the addition of (0.0006301255 mol – 0.000588962 mol) = 0.0000411635 mol HCO3– The HCl added reacts with CO32– to produce additional HCO3−: CO32– + HCl → HCO3– + Cl– To produce 0.0000411635 mol HCO3– would require the addition of 0.0000411635 mol HCl. ⎛ 36.46 g HCl ⎞⎟ Mass (g) of HCl = (0.0000411635 mol HCl )⎜⎜ = 0.0015008 = 0.0015 g HCl ⎝ 1 mol HCl ⎠⎟⎟

19.34

Plan: Select conjugate pairs with Ka values close to the desired [H3O+].

19-31

Solution: a) For pH ≈ 4.5, [H3O+] = 10–4.5 = 3.2×10–5 M. Some good selections are the HOOC(CH2)4COOH/ HOOC(CH2)4COOH− conjugate pair with Ka equal to 3.8×10–5 or C6H5CH2COOH/C6H5CH2COO− conjugate pair with Ka equal to 4.9×10–5. From the base list, the C6H5NH2/C6H5NH3+ conjugate pair comes close with Ka = Kw/Kb = 1.0×10–14/4.0×10–10 = 2.5×10–5. b) For pH ≈ 7.0, [H3O+] = 10–7.0 = 1.0×10–7 M. Two choices are the H2PO4−/HPO42− conjugate pair with Ka of 6.3×10–8 and the H2AsO4−/HAsO42− conjugate pair with Ka of 1.1×10–7. 19.35

Select conjugate pairs that have Ka or Kb values close to the desired [H3O+] or [OH–]. a) For [H3O+] ≈ 1×10–9 M, the HOBr/OBr– conjugate pair comes close with Ka equal to 2.3×10–9. From the base list, the NH3/NH4 + conjugate pair comes close with Ka = Kw/Kb = 1.0×10–14/1.76×10–5 = 5.7×10–10. b) For [OH–] ≈ 3×10–5 M, the NH3/NH4 + conjugate pair comes close; also, it is possible to choose [H3O+] = 1.0×10–14/3×10–5 = 3.3×10–10; the C6H5OH/C6H5O– comes close with Ka = 1.0×10–10.

19.36

Plan: Select conjugate pairs with pKa values close to the desired pH. Convert pH to [H3O+] for easy comparison to Ka values in the Appendix. Determine an appropriate base by [OH–] = Kw/[H3O+]. Solution: a) For pH ≈ 3.5 ([H3O+] = 10–pH = 10–3.5 = 3.2×10–4), the best selection is the HOCH2CH(OH)COOH/ HOCH2CH(OH)COOH – conjugate pair with a Ka = 2.9×10–4. The CH3COOC6H4COOH/CH3COOC6H4COO – pair, with Ka = 3.6×10–4, is also a good choice. The [OH–] = Kw/[H3O+] = 1.0×10–14/3.2×10–4 = 3.1×10–11, results in no reasonable Kb values from the Appendix. b) For pH ≈ 5.5 ([H3O+] = 10–pH = 3×10–6), no Ka1 gives an acceptable pair; the Ka2 values for adipic acid, malonic acid, and succinic acid are reasonable. The [OH–] = Kw/[H3O+] = 1.0×10–14/3×10–6 = 3×10–9; the Kb selection is C5H5N/C5H5NH+.

19.37

Select conjugate pairs that have Ka or Kb values close to the desired [H3O+] or [OH–]. a) For [OH–] ≈ 1×10–6 M, no Kb values work. The Ka values are [H3O+] = Kw/[OH–] = 1.0×10–14/1×10–6 = 1×10–8, giving the following acceptable pairs H2PO4–/HPO42– or HC6H5O72–/C6H5O73– or HOCl/OCl–. b) For [H3O+] ≈ 4×10–4 M, the HF/F– conjugate pair comes close with Ka equal to 6.8×10–4. From the base list, [OH–] = 1.0×10–14/4×10–4 = 2.5×10–11, there are no reasonable choices.

19.38

The value of the Ka from the Appendix: Ka = 2.9×10–8 pKa = –log (2.9×10–8) = 7.5376 ⎛ [ClO− ] ⎞⎟ pH = pKa + log ⎜⎜ ⎟ ⎜⎝[HClO]⎠⎟

⎛[0.100]⎞⎟ a) pH = 7.5376 + log ⎜⎜ ⎟ = 7.5376 = 7.54 ⎜⎝[0.100]⎠⎟ ⎛[0.150]⎞⎟ b) pH = 7.5376 + log ⎜⎜ ⎟ = 7.71369 = 7.71 ⎜⎝[0.100]⎠⎟ ⎛[0.100]⎞⎟ c) pH = 7.5376 + log ⎜⎜ ⎟ = 7.3615 = 7.36 ⎜⎝[0.150]⎠⎟ d) The reaction is NaOH + HClO → Na+ + ClO– + H2O. The original moles of HClO and OCl– are both = (0.100 mol/L)(1.0 L) = 0.100 mol NaOH + HClO → Na+ + ClO– + H2O Initial 0.0050 mol 0.100 mol 0.100 mol Change –0.0050 mol –0.0050 mol + 0.0050 mol Final 0 mol 0.095 mol 0.105 mol ⎛[0.105]⎞⎟ pH = 7.5376 + log ⎜⎜ ⎟ = 7.5811 = 7.58 ⎜⎝[0.095]⎠⎟ Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

19-32

19.39

Plan: Given the pH and Ka of an acid, the buffer-component ratio can be calculated from the HendersonHasselbalch equation. Convert Ka to pKa. Solution: The value of the Ka from the Appendix: Ka = 6.3×10–8 (We are using Ka2 since we are dealing with the equilibrium in which the second hydrogen ion is being lost.) pKa = –log Ka = –log (6.3×10–8) = 7.200659451 Use the Henderson-Hasselbalch equation: ⎛ [HPO 4 2− ] ⎞⎟ ⎟ pH = pKa + log ⎜⎜ ⎜⎝[H 2 PO 4− ]⎠⎟⎟

⎛ [HPO4 2− ] ⎞⎟ ⎟ 7.40 = 7.200659451 + log ⎜⎜ ⎜⎝[H 2 PO 4− ]⎠⎟⎟ ⎛ [HPO4 2− ] ⎞⎟ ⎟ 0.19934055 = log ⎜⎜⎜ − ⎟ ⎝[H 2 PO4 ]⎠⎟ [HPO4 2− ] = 1.582486 = 1.6 [H 2 PO 4− ]

19.40

a) The reactions are: OH–(aq) + H3PO4(aq) → H2PO4–(aq) + H2O(l) Ka1 = 7.2×10–3 – – 2– Ka2 = 6.3×10–8 OH (aq) + H2PO4 (aq) → HPO4 (aq) + H2O(l) The correct order is C, B, D, A. Scene C shows the solution before the addition of any NaOH. Scene B is halfway to the first equivalence point; Scene D is halfway to the second equivalence point and Scene A is at end of the titration. b) Scene B is the second scene in the correct order. This is halfway towards the first equivalence point when there are equal amounts of the acid and conjugate base, which constitutes a buffer. ⎛[H PO − ]⎞ pH = pKa + log ⎜⎜⎜ 2 4 ⎟⎟⎟ ⎜⎝ [H 3 PO 4 ] ⎠⎟ Determine the pKa using pKa = –log (7.2×10–3) = 2.142668

⎛[3]⎞ pH = 2.1426675 + log ⎜⎜ ⎟⎟⎟ = 2.1426675 = 2.14 ⎜⎝[3]⎠ c) 10.00 mL of NaOH is required to reach the first half equivalence point. Therefore, an additional 10.00 mL of NaOH is required to reach the first equivalence point, for a total of 20 mL for the first equivalence point. An additional 20.00 mL of NaOH will be required to reach the second equivalence point where only HPO42– remains. A total of 40.00 mL of NaOH is required to reach Scene A. 19.41

a) The initial pH is lowest for the flask solution of the strong acid, followed by the weak acid and then the weak base. In other words, strong acid–strong base < weak acid–strong base < strong acid–weak base in terms of initial pH. b) At the equivalence point, the moles of H3O+ equal the moles of OH–, regardless of the type of titration. However, the strong acid–strong base equivalence point occurs at pH = 7.00 because the resulting cation-anion combination does not react with water. An example is the reaction NaOH + HCl → H2O + NaCl. Neither Na+ nor Cl– ions dissociate in water. The weak acid–strong base equivalence point occurs at pH > 7, because the anion of the weak acid is weakly basic, whereas the cation of the strong base does not react with water. An example is the reaction HCOOH + NaOH → HCOO− + H2O + Na+. The conjugate base, HCOO−, reacts with water according to this reaction: HCOO− + H2O → HCOOH + OH−.

19-33

The strong acid–weak base equivalence point occurs at pH < 7, because the anion of the strong acid does not react with water, whereas the cation of the weak base is weakly acidic. An example is the reaction HCl + NH3 → NH4+ + Cl−. The conjugate acid, NH4+, dissociates slightly in water: NH4+ + H2O → NH3 + H3O+. In rank order of pH at the equivalence point, strong acid–weak base < strong acid–strong base < weak acid– strong base. 19.42

In the buffer region, comparable amounts of weak acid and its conjugate base are present. At the equivalence point, the predominant species is the conjugate base. In a strong acid–weak base titration, the weak base and its conjugate acid are the predominant species present.

19.43

At the very center of the buffer region of a weak acid–strong base titration, the concentration of the weak acid and its conjugate base are equal. If equal values for concentration are put into the Henderson-Hasselbalch equation, the [base]/[acid] ratio is 1, the log of 1 is 0, and the pH of the solution equals the pKa of the weak acid. ⎛[base]⎞⎟ pH = pKa + log ⎜⎜ ⎟ ⎜⎝ [acid] ⎠⎟

pH = pKa + log 1 19.44

You need to know the pKa value for the indicator. Its transition range is approximately pKa ± 1. If the indicator is a diprotic acid, it will have two transition ranges, one for each of the two H3O+ ions lost.

19.45

To see a distinct color in a mixture of two colors, you need one color to be about 10 times the intensity of the other. For this to take place, the concentration ratio [HIn]/[In–] needs to be greater than 10:1 or less than 1:10. This will occur when pH = pKa – 1 or pH = pKa + 1, respectively, giving a transition range of about two units.

19.46

The addition of an acid-base indicator does not affect the pH of the test solution because the concentration of indicator is very small.

19.47

The equivalence point in a titration is the point at which the number of moles of OH– equals the number of moles of H3O+ (be sure to account for stoichiometric ratios, e.g., one mol of Ca(OH)2 produces two moles of OH–). The end point is the point at which the added indicator changes color. If an appropriate indicator is selected, the end point is close to the equivalence point, but not normally the same. Using an indicator that changes color at a pH after the equivalence point means the equivalence point is reached first. However, if an indicator is selected that changes color at a pH before the equivalence point, then the end point is reached first.9.48 The titration curve for a diprotic acid has two “breaks,” i.e., two regions where the pH increases sharply. For a monoprotic acid, only one break occurs.

19.49

Plan: The reaction occurring in the titration is the neutralization of H3O+ (from HCl) by OH− (from NaOH): HCl(aq) + NaOH(aq) → H2O(l) + NaCl(aq) or, omitting spectator ions: H3O+(aq) + OH−(aq) → 2H2O(l) For the titration of a strong acid with a strong base, the pH before the equivalence point depends on the excess concentration of acid and the pH after the equivalence point depends on the excess concentration of base. At the equivalence point, there is not an excess of either acid or base so the pH is 7.0. The equivalence point occurs when 40.00 mL of base has been added. Use (M)(V) to determine the number of moles of acid and base. Note that the NaCl product is a neutral salt that does not affect the pH. Solution: The initial number of moles of HCl = (0.1000 mol HCl/L)(10–3 L/1 mL)(40.00 mL) = 4.000×10–3 mol HCl a) At 0 mL of base added, the concentration of hydronium ion equals the original concentration of HCl. pH = –log (0.1000 M) = 1.0000 b) Determine the moles of NaOH added: Moles of added NaOH = (0.1000 mol NaOH/L)(10–3 L/1 mL)(25.00 mL) = 2.500×10–3 mol NaOH + NaOH(aq) → H2O(l) + NaCl(aq) HCl(aq) 2.500×10–3 mol – 0 Initial 4.000×10–3 mol

19-34

– +2.500×10–3 mol Change –2.500×10–3 mol –2.500×10–3 mol –3 Final 1.500×10 mol 0 2.500×10–3 mol –3 The volume of the solution at this point is [(40.00 + 25.00) mL](10 L/1 mL) = 0.06500 L The molarity of the excess HCl is (1.500×10–3 mol HCl)/(0.06500 L) = 0.023077 M pH = –log (0.023077) = 1.6368 c) Determine the moles of NaOH added: Moles of added NaOH = (0.1000 mol NaOH/L)(10–3 L/1 mL)(39.00 mL) = 3.900×10–3 mol NaOH + NaOH(aq) → H2O(l) + NaCl(aq) HCl(aq) 3.900×10–3 mol – 0 Initial 4.000×10–3 mol – +3.900×10–3 mol Change –3.900×10–3 mol –3.900×10–3 mol Final 1.000×10–4 mol 0 3.900×10–3 mol –3 The volume of the solution at this point is [(40.00 + 39.00) mL](10 L/1 mL) = 0.07900 L The molarity of the excess HCl is (1.00×10–4mol HCl)/(0.07900 L) = 0.0012658 M pH = –log (0.0012658) = 2.898 d) Determine the moles of NaOH added: Moles of added NaOH = (0.1000 mol NaOH/L)(10–3 L/1 mL)(39.90 mL) = 3.990×10–3 mol NaOH + NaOH(aq) → H2O(l) + NaCl(aq) HCl(aq) 3.990×10–3 mol – 0 Initial 4.000×10–3 mol –3.990×10–3 mol – +3.990×10–3 mol Change –3.990×10–3 mol Final 1.000×10–5 mol 0 3.990×10–3 mol –3 The volume of the solution at this point is [(40.00 + 39.90) mL](10 L/1 mL) = 0.07990 L The molarity of the excess HCl is (1.0×10–5 mol HCl)/(0.07990 L) = 0.000125156 M pH = –log (0.000125156) = 3.903 e) Determine the moles of NaOH added: Moles of added NaOH = (0.1000 mol NaOH/L)(10–3 L/1 mL)(40.00 mL) = 4.000×10–3 mol NaOH + NaOH(aq) → H2O(l) + NaCl(aq) HCl(aq) 4.000×10–3 mol – 0 Initial 4.000×10–3 mol –4.000×10–3 mol – +4.000×10–3 mol Change –4.000×10–3 mol Final 0 0 4.000×10–3 mol The NaOH will react with an equal amount of the acid and 0.0 mol HCl will remain. This is the equivalence point of a strong acid–strong base titration, thus, the pH is 7.00. Only the neutral salt NaCl is in solution at the equivalence point. f) The NaOH is now in excess. It will be necessary to calculate the excess base after reacting with the HCl. The excess strong base will give the pOH, which can be converted to the pH. Determine the moles of NaOH added: Moles of added NaOH = (0.1000 mol NaOH/L)(10–3 L/1 mL)(40.10 mL) = 4.010×10–3 mol NaOH The HCl will react with an equal amount of the base, and 1.0×10–5 mol NaOH will remain. + NaOH(aq) → H2O(l) + NaCl(aq) HCl(aq) 4.010×10–3 mol – 0 Initial 4.000×10–3 mol –4.000×10–3 mol – +4.000×10–3 mol Change –4.000×10–3 mol Final 0 1.000×10–5 mol 4.000×10–3 mol –3 The volume of the solution at this point is [(40.00 + 40.10) mL](10 L/1 mL) = 0.08010 L The molarity of the excess NaOH is (1.0×10–5 mol NaOH)/(0.08010 L) = 0.00012484 M pOH = –log (0.00012484) = 3.9036 pH = 14.00 – pOH = 14.00 – 3.9036 = 10.09637 = 10.10 g) Determine the moles of NaOH added: Moles of NaOH = (0.1000 mol NaOH/L)(10–3 L/1 mL)(50.00 mL) = 5.000×10–3 mol NaOH The HCl will react with an equal amount of the base, and 1.000×10–3 mol NaOH will remain. + NaOH(aq) → H2O(l) + NaCl(aq) HCl(aq) 5.000×10–3 mol – 0 Initial 4.000×10–3 mol –4.000×10–3 mol – +4.000×10–3 mol Change –4.000×10–3 mol Final 0 1.000×10–3 mol 4.000×10–3 mol Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

19-35

The volume of the solution at this point is [(40.00 + 50.00) mL](10–3 L/ 1 mL) = 0.09000 L The molarity of the excess NaOH is (1.000×10–3 mol NaOH)/(0.09000 L) = 0.011111 M pOH = –log (0.011111) = 1.95424 pH = 14.00 – pOH = 14.00 – 1.95424 = 12.04576 = 12.05 19.50

The reaction occurring in the titration is the neutralization of OH− (from KOH) by H3O+ (from HBr): HBr(aq) + KOH(aq) → H2O(l) + KBr(aq) H3O+(aq) + OH−(aq) → 2 H2O(l) For the titration of a strong base with a strong acid, the pH before the equivalence point depends on the excess concentration of base and the pH after the equivalence point depends on the excess concentration of acid. At the equivalence point, there is not an excess of either acid or base so pH is 7.0. The equivalence point occurs when 30.00 mL of acid has been added. The initial number of moles of KOH = (0.1000 mol KOH/L)(10–3 L/1 mL)(30.00 mL) = 3.000×10–3 mol KOH a) At 0 mL of acid added, the concentration of hydroxide ion equals the original concentration of KOH. pOH = –log (0.1000 M) = 1.0000 pH = 14.00 – pOH = 14.00 – 1.0000 = 13.00 b) Determine the moles of HBr added: Moles of added HBr = (0.1000 mol HBr/L)(10–3 L/1 mL)(15.00 mL) = 1.500×10–3 mol HBr The HBr will react with an equal amount of the base, and 1.500×10–3 mol KOH will remain. The volume of the solution at this point is [(30.00 + 15.00) mL](10–3 L/1 mL) = 0.04500 L The molarity of the excess KOH is (1.500×10–3 mol KOH)/(0.04500 L) = 0.03333 M pOH = –log (0.03333) = 1.4772 pH = 14.00 – pOH = 14.00 – 1.4772 = 12.5228 = 12.52 c) Determine the moles of HBr added: Moles of added HBr = (0.1000 mol HBr/L)(10–3 L/1 mL)(29.00 mL) = 2.900×10–3 mol HBr The HBr will react with an equal amount of the base, and 1.00×10–4 mol KOH will remain. The volume of the solution at this point is [(30.00 + 29.00) mL](10–3 L/1 mL) = 0.05900 L The molarity of the excess KOH is (1.00×10–4 mol KOH)/(0.05900 L) = 0.0016949 M pOH = –log (0.0016949) = 2.7708559 pH = 14.00 – pOH = 14.00 – 2.7708559 = 11.2291441 = 11.23 d) Determine the moles of HBr added: Moles of added HBr = (0.1000 mol HBr/L)(10–3 L/1 mL)(29.90 mL) = 2.990×10–3 mol HBr The HBr will react with an equal amount of the base, and 1.0×10–5 mol KOH will remain. The volume of the solution at this point is [(30.00 + 29.90) mL](10–3 L/1 mL) = 0.05990 L The molarity of the excess KOH is (1.0×10–5 mol KOH)/(0.05990 L) = 0.000166945 M pOH = –log (0.000166945) = 3.7774266 pH = 14.00 – pOH = 14.00 – 3.7774266 = 10.2225734 = 10.2 e) Determine the moles of HBr added: Moles of added HBr = (0.1000 mol HBr/L)(10–3 L/1 mL)(30.00 mL) = 3.000×10–3 mol HBr The HBr will react with an equal amount of the base and 0.0 mol KOH will remain. This is the equivalence point of a strong acid–strong base titration; thus, the pH is 7.00. f) The HBr is now in excess. It will be necessary to calculate the excess base after reacting with the HCl. The excess strong acid will give the pH. Determine the moles of HBr added: Moles of added HBr = (0.1000 mol HBr/L)(10–3 L/1 mL)(30.10 mL) = 3.010×10–3 mol HBr The HBr will react with an equal amount of the base, and 1.0×10–5 mol HBr will remain. The volume of the solution at this point is [(30.00 + 30.10) mL](10–3 L/1 mL) = 0.06010 L The molarity of the excess HBr is (1.0×10–5 mol HBr)/(0.06010 L) = 0.000166389 M pH = –log (0.000166389) = 3.778875 = 3.8 g) Determine the moles of HBr added: Moles of added HBr = (0.1000 mol HBr/L)(10–3 L/1 mL)(40.00 mL) = 4.000×10–3 mol HBr The HBr will react with an equal amount of the base, and 1.000×10–3 mol HBr will remain.

19-36

The volume of the solution at this point is [(30.00 + 40.00) mL](10–3 L/1 mL) = 0.07000 L The molarity of the excess HBr is (1.000×10–3 mol HBr)/(0.07000 L) = 0.0142857 M pH = –log (0.0142857) = 1.845098 = 1.85 19.51

Plan: This is a titration between a weak acid and a strong base. The pH before addition of the base is dependent on the Ka of the acid (labeled HBut). Prior to reaching the equivalence point, the added base reacts with the acid to form butanoate ion (labeled But–). The equivalence point occurs when 20.00 mL of base is added to the acid because at this point, moles acid = moles base. Addition of base beyond the equivalence point is simply the addition of excess OH−. Solution: a) At 0 mL of base added, the concentration of [H3O+] is dependent on the dissociation of butanoic acid: ⇆ H3O+ + But– HBut + H2O 0 0 Initial 0.100 M Change –x +x +x Equilibrium 0.100 – x x x + − 2 H O But [ ] [ ] x x2 3 Ka = 1.54×10–5 = = = 0.1000 − x [ HBut ] 0.1000 x = [H3O+] = 1.2409674×10–3 M pH = –log [H3O+] = –log (1.2409674×10–3) = 2.9062 = 2.91 b) The initial number of moles of HBut = (M)(V) = (0.1000 mol HBut/L)(10–3 L/1 mL)(20.00 mL) = 2.000×10–3 mol HBut Determine the moles of NaOH added: Moles of added NaOH = (0.1000 mol NaOH/L)(10–3 L/1 mL)(10.00 mL) = 1.000×10–3 mol NaOH The NaOH will react with an equal amount of the acid, and 1.000×10–3 mol HBut will remain. An equal number of moles of But– will form. + NaOH(aq) → H2O(l) + But–(aq) + Na+(aq) HBut(aq) –3 –3 1.000×10 mol – 0 – Initial 2.000×10 mol –1.000×10–3 mol – +1.000×10–3 mol – Change –1.000×10–3 mol Final 1.000×10–3 mol 0 1.000×10–3 mol The volume of the solution at this point is [(20.00 + 10.00) mL](10–3 L/1 mL) = 0.03000 L The molarity of the excess HBut is (1.000×10–3 mol HBut)/(0.03000 L) = 0.03333 M The molarity of the But– formed is (1.000×10–3 mol But– /(0.03000 L) = 0.03333 M Using a reaction table for the equilibrium reaction of HBut: ⇆ H3O+ + But– HBut + H2O – 0 0.03333 M Initial 0.03333 M Change –x +x +x Equilibrium 0.03333 – x x 0.03333 + x Ka = 1.54×10–5 = +

[ H3O+ ][ But− ] [ HBut ]

(

x 0.0333 + x

0.03333 − x

) = x (0.03333) 0.03333

–5

x = [H3O ] = 1.54×10 M pH = –log [H3O+] = –log (1.54×10–5) = 4.812479 = 4.81 c) Determine the moles of NaOH added: Moles of added NaOH = (0.1000 mol NaOH/L)(10–3 L/1 mL)(15.00 mL) = 1.500×10–3 mol NaOH The NaOH will react with an equal amount of the acid, and 5.00×10–4 mol HBut will remain, and 1.500×10–3 moles of But– will form. + NaOH(aq) → H2O(l) + But–(aq) + Na+(aq) HBut(aq) –3 –3 1.500×10 mol – 0 – Initial 2.000×10 mol –1.500×10–3 mol – +1.500×10–3 mol –_____ Change –1.500×10–3 mol Final 5.000×10–4 mol 0 1.500×10–3 mol The volume of the solution at this point is [(20.00 + 15.00) mL](10–3 L/1 mL) = 0.03500 L Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

19-37

The molarity of the excess HBut is (5.00×10–4 mol HBut)/(0.03500 L) = 0.0142857 M The molarity of the But– formed is (1.500×10–3 mol But–)/(0.03500 L) = 0.042857 M Using a reaction table for the equilibrium reaction of HBut: ⇆ H3O+ + But– HBut + H2O – 0 0.042857 M Initial 0.0142857 M Change –x +x +x Equilibrium 0.0142857 – x x 0.042857 + x Ka = 1.54×10–5 =

[ H3O+ ][ But− ] [ HBut ] +

(

x 0.042857 + x

0.0142857 − x

) = x (0.042857) 0.0142857

–6

x = [H3O ] = 5.1333×10 M pH = –log [H3O+] = –log (5.1333×10–6) = 5.2896 = 5.29 d) Determine the moles of NaOH added: Moles of added NaOH = (0.1000 mol NaOH/L)(10–3 L/1 mL)(19.00 mL) = 1.900×10–3 mol NaOH The NaOH will react with an equal amount of the acid, and 1.00×10–4 mol HBut will remain, and 1.900×10–3 moles of But– will form. + NaOH(aq) → H2O(l) + But–(aq) + Na+(aq) HBut(aq) –3 –3 1.900×10 mol – 0 – Initial 2.000×10 mol –1.900×10–3 mol – +1.900×10–3 mol – Change –1.900×10–3 mol Final 1.000×10–4 mol 0 1.900×10–3 mol The volume of the solution at this point is [(20.00 + 19.00) mL](10–3 L/1 mL) = 0.03900 L The molarity of the excess HBut is (1.00×10–4 mol HBut)/(0.03900 L) = 0.0025641 M The molarity of the But– formed is (1.900×10–3 mol But–)/(0.03900 L) = 0.0487179 M Using a reaction table for the equilibrium reaction of HBut: ⇆ H3O+ + But– HBut + H2O – 0 0.0487179 M Initial 0.0025641 M Change –x – +x +x Equilibrium 0.0025641 – x +x 0.0487179 + x –5

Ka = 1.54×10 = +

[ H3O+ ][ But− ] [ HBut ]

(

x 0.0487179 + x

0.0025641 − x

) = x (0.0487179) 0.0025641

–7

x = [H3O ] = 8.1052632×10 M pH = –log [H3O+] = –log (8.1052632×10–7) = 6.09123 = 6.09 e) Determine the moles of NaOH added: Moles of added NaOH = (0.1000 mol NaOH/L)(10–3 L/1 mL)(19.95 mL) = 1.995×10–3 mol NaOH The NaOH will react with an equal amount of the acid, and 5×10–6 mol HBut will remain, and 1.995×10–3 moles of But– will form. + NaOH(aq) → H2O(l) + But–(aq) + Na+(aq) HBut(aq) –3 –3 1.995×10 mol – 0 – Initial 2.000×10 mol –1.995×10–3 mol – +1.995×10–3 mol – Change –1.995×10–3 mol Final 5.000×10–6 mol 0 1.995×10–3 mol –3 The volume of the solution at this point is [(20.00 + 19.95) mL](10 L/1 mL) = 0.03995 L The molarity of the excess HBut is (5×10–6 mol HBut)/(0.03995 L) = 0.000125156 M The molarity of the But– formed is (1.995×10–3 mol But–)/(0.03995 L) = 0.0499374 M Using a reaction table for the equilibrium reaction of HBut: ⇆ H3O+ + But– HBut + H2O – 0 0.0499374 M Initial 0.000125156 M Change –x – +x +x Equilibrium 0.000125156 – x x 0.0499374 + x

19-38

Ka = 1.54×10–5 =

[ H3O+ ][ But− ] [ HBut ]

(

x 0.0499374 + x

) = x (0.0499374 )

0.000125156 − x

0.000125156

–8

x = [H3O ] = 3.859637×10 M pH = –log [H3O+] = –log (3.859637×10–8) = 7.41345 = 7.41 f) Determine the moles of NaOH added: Moles of added NaOH = (0.1000 mol NaOH/L)(10–3 L/1 mL)(20.00 mL) = 2.000×10–3 mol NaOH The NaOH will react with an equal amount of the acid, and 0 mol HBut will remain, and 2.000×10–3 moles of But– will form. This is the equivalence point. + NaOH(aq) → H2O(l) + But–(aq) + Na+(aq) HBut(aq) –3 –3 2.000×10 mol – 0 – Initial 2.000×10 mol –2.000×10–3 mol – +2.000×10–3 mol – Change –2.000×10–3 mol Final 0 0 2.000×10–3 mol The Kb of But– is now important. The volume of the solution at this point is [(20.00 + 20.00) mL](10–3 L/1 mL) = 0.04000 L The molarity of the But– formed is (2.000×10–3 mol But–)/(0.04000 L) = 0.05000 M Kb = Kw/Ka = (1.0×10–14)/(1.54×10–5) = 6.49351×10–10 Using a reaction table for the equilibrium reaction of But–: + H2O ⇆ HBut + OH– But– – 0 0 Initial 0.05000 M Change –x – +x +x Equilibrium 0.05000 – x x x − [ x ][ x ] [ x ][ x ] [ ] HBut OH [ ] Kb = 6.49351×10–10 = = = ⎡0.05000 − x ⎤ [0.05000 ] [ But− ] ⎣⎢ ⎦⎥ – –6 [OH ] = x = 5.6980304×10 M pOH = –log (5.6980304×10–6) = 5.244275238 pH = 14.00 – pOH = 14.00 – 5.244275238 = 8.7557248 = 8.76 g) After the equivalence point, the excess strong base is the primary factor influencing the pH. Determine the moles of NaOH added: Moles of added NaOH = (0.1000 mol NaOH/L)(10–3 L/1 mL)(20.05 mL) = 2.005×10–3 mol NaOH The NaOH will react with an equal amount of the acid, 0 mol HBut will remain, and 5×10–6 moles of NaOH will be in excess. There will be 2.000×10–3 mol of But– produced, but this weak base will not affect the pH compared to the excess strong base, NaOH. + NaOH(aq) → H2O(l) + But–(aq) + Na+(aq) HBut(aq) –3 –3 2.005×10 mol – 0 – Initial 2.000×10 mol –2.000×10–3 mol – +2.000×10–3 mol – Change –2.000×10–3 mol Final 0 5.000×10–6 mol 2.000×10–3 mol The volume of the solution at this point is [(20.00 + 20.05) mL](10–3 L/1 mL) = 0.04005 L The molarity of the excess OH– is (5×10–6 mol OH–)/(0.04005 L) = 1.2484×10–4 M pOH = –log (1.2484×10–4) = 3.9036 pH = 14.00 – pOH = 14.00 – 3.9036 = 10.0964 = 10.10 h) Determine the moles of NaOH added: Moles of added NaOH = (0.1000 mol NaOH/L)(10–3 L/1 mL)(25.00 mL) = 2.500×10–3 mol NaOH The NaOH will react with an equal amount of the acid, 0 mol HBut will remain, and 5.00×10–4 moles of NaOH will be in excess. Initial Change Final

HBut(aq) 2.000×10–3 mol –2.000×10–3 mol 0

NaOH(aq) 2.500×10–3 mol –2.000×10–3 mol 5.000×10–4 mol

→

H2O(l) – –

But–(aq) + 0 +2.000×10–3 mol 2.000×10–3 mol

Na+(aq) – –

19-39

The volume of the solution at this point is [(20.00 + 25.00) mL](10–3 L/1 mL) = 0.04500 L The molarity of the excess OH– is (5.00×10–4 mol OH–)/(0.04500 L) = 1.1111×10–2 M pOH = –log (1.1111×10–2) = 1.9542 pH = 14.00 – pOH = 14.00 – 1.9542 = 12.0458 = 12.05 19.52

This is a titration between a weak base and a strong acid. The pH before addition of the acid is dependent on the Kb of the base ((CH3CH2)3N)). Prior to reaching the equivalence point, the added acid reacts with base to form (CH3CH2)3NH+ ion. The equivalence point occurs when 20.00 mL of acid is added to the base because at this point, moles acid = moles base. Addition of acid beyond the equivalence point is simply the addition of excess H3O+. The initial number of moles of (CH3CH2)3N = (0.1000 mol (CH3CH2)3N)/L)(10–3 L/1 mL)(20.00 mL) = 2.000×10–3 mol (CH3CH2)3N a) Since no acid has been added, only the weak base (Kb) is important. ⎡(CH 3 CH 2 ) NH + ⎤ [ OH− ] [ x ][ x ] [ x ][ x ] ⎢ 3 ⎦⎥ Kb = 5.2×10–4 = ⎣ = = ⎡0.1000 − x⎤ ⎡(CH 3 CH 2 ) N ⎤ [ 0.1000 ] ⎢⎣ ⎥⎦ 3 ⎣⎢ ⎦⎥ [OH–] = x = 7.2111×10–3 M pOH = –log (7.2111×10–3) = 2.141998 pH = 14.00 – pOH = 14.00 – 2.141998 = 11.8580 = 11.86 b) Determine the moles of HCl added: Moles of added HCl = (0.1000 mol HCl/L)(10–3 L/1 mL)(10.00 mL) = 1.000×10–3 mol HCl The HCl will react with an equal amount of the base, and 1.000×10–3 mol (CH3CH2)3N will remain; an equal number of moles of (CH3CH2)3NH+ will form. The volume of the solution at this point is [(20.00 + 10.00) mL](10–3 L/1 mL) = 0.03000 L The molarity of the excess (CH3CH2)3N is (1.000×10–3 mol (CH3CH2)3N)/(0.03000 L) = 0.03333 M The molarity of the (CH3CH2)3NH+ formed is (1.000×10–3 mol (CH3CH2)3NH+)/(0.03000 L) = 0.03333 M + ⎡(CH 3 CH 2 ) NH ⎤ [ OH− ] [ x ][ 0.0333] [ x ] ⎡ 0.0333 + x ⎤ ⎢⎣ ⎥⎦ ⎢⎣ ⎥⎦ –4 3 Kb = 5.2×10 = = = ⎡ 0.03333 − x ⎤ ⎡(CH 3 CH 2 ) N ⎤ [ 0.03333] ⎢⎣ ⎥⎦ 3 ⎣⎢ ⎦⎥ [OH–] = x = 5.2×10–4 M pOH = –log (5.2×10–4) = 3.283997 pH = 14.00 – pOH = 14.00 – 3.283997 = 10.7160 = 10.72 c) Determine the moles of HCl added: Moles of added HCl = (0.1000 mol HCl/L)(10–3 L/1 mL)(15.00 mL) = 1.500×10–3 mol HCl The HCl will react with an equal amount of the base, and 5.00×10–4 mol (CH3CH2)3N will remain; and 1.500×10–3 moles of (CH3CH2)3NH+ will form. The volume of the solution at this point is [(20.00 + 15.00) mL](10–3 L/1 mL) = 0.03500 L The molarity of the excess (CH3CH2)3N is (5.00×10–4 mol (CH3CH2)3N)/(0.03500 L) = 0.0142857 M The molarity of the (CH3CH2)3NH+ formed is (1.500×10–3 mol (CH3CH2)3NH+)/(0.03500 L) = 0.0428571 M ⎡(CH 3 CH 2 ) NH + ⎤ [ OH− ] [ x ][ 0.0428571] [ x ] ⎡ 0.0428571 + x ⎤ ⎢ 3 ⎦⎥ ⎣⎢ ⎦⎥ = Kb = 5.2×10–4 = ⎣ = ⎡ 0.0142857 − x ⎤ ⎡(CH 3 CH 2 ) N ⎤ [ 0.0142857] ⎢⎣ ⎥⎦ 3 ⎣⎢ ⎦⎥ [OH–] = x = 1.7333×10–4 M pOH = –log (1.7333×10–4) = 3.761126 pH = 14.00 – pOH = 14.00 – 3.761126 = 10.23887 = 10.24 d) Determine the moles of HCl added: Moles of added HCl = (0.1000 mol HCl/L)(10–3 L/1 mL)(19.00 mL) = 1.900×10–3 mol HCl

19-40

The HCl will react with an equal amount of the base, and 1.00×10–4 mol (CH3CH2)3N will remain; and 1.900×10–3 moles of (CH3CH2)3NH+ will form. The volume of the solution at this point is [(20.00 + 19.00) mL](10–3 L/1 mL) = 0.03900 L The molarity of the excess (CH3CH2)3N is (1.00×10–4 mol (CH3CH2)3N)/(0.03900 L) = 0.002564103 M The molarity of the (CH3CH2)3NH+ formed is (1.900×10–3 mol (CH3CH2)3NH+)/(0.03900 L) = 0.0487179 M ⎡(CH 3 CH 2 ) NH + ⎤ [ OH− ] [ x ][ 0.0487179] [ x ] ⎡ 0.0487179 + x ⎤ ⎢ 3 ⎦⎥ ⎣⎢ ⎦⎥ = Kb = 5.2×10–4 = ⎣ = ⎡(CH 3 CH 2 ) N ⎤ ⎡ 0.002564103 − x ⎤ [0.002564103] ⎢⎣ ⎥⎦ 3 ⎣⎢ ⎦⎥ [OH–] = x = 2.73684×10–5 M pOH = –log (2.73684×10–5) = 4.56275 pH = 14.00 – pOH = 14.00 – 4.56275 = 9.43725 = 9.44 e) Determine the moles of HCl added: Moles of added HCl = (0.1000 mol HCl/L)(10–3 L/1 mL)(19.95 mL) = 1.995×10–3 mol HCl The HCl will react with an equal amount of the base, and 5×10–6 mol (CH3CH2)3N will remain; and 1.995×10–3 moles of (CH3CH2)3NH+ will form. The volume of the solution at this point is [(20.00 + 19.95) mL](10–3 L/1 mL) = 0.03995 L The molarity of the excess (CH3CH2)3N is (5×10–6 mol (CH3CH2)3N)/(0.03995 L) = 0.000125156 M The molarity of the (CH3CH2)3NH+ formed is (1.995×10–3 mol (CH3CH2)3NH+)/(0.03995 L) = 0.0499374 M ⎡(CH 3 CH 2 ) NH + ⎤ [ OH− ] [ x ][ 0.0499374 ] [ x ] ⎡ 0.0499374 + x ⎤ ⎢ 3 ⎦⎥ ⎣⎢ ⎦⎥ = Kb = 5.2×10–4 = ⎣ = ⎡(CH 3 CH 2 ) N ⎤ ⎡ 0.000125156 − x ⎤ [0.000125156] 3 ⎣⎢ ⎦⎥ ⎢⎣ ⎥⎦ [OH–] = x = 1.303254×10–6 M pOH = –log (1.303254×10–6) = 5.88497 pH = 14.00 – pOH = 14.00 – 5.88497 = 8.11503 = 8.1 f) Determine the moles of HCl added: Moles of added HCl = (0.1000 mol HCl/L)(10–3 L/1 mL)(20.00 mL) = 2.000×10–3 mol HCl The HCl will react with an equal amount of the base, and 0 mol (CH3CH2)3N will remain; and 2.000×10–3 moles of (CH3CH2)3NH+ will form. This is the equivalence point. The volume of the solution at this point is [(20.00 + 20.00) mL](10–3 L/1 mL) = 0.04000 L The molarity of the (CH3CH2)3NH+ formed is (2.000×10–3 mol (CH3CH2)3NH+)/(0.04000 L) = 0.05000 M Ka = Kw/Kb = (1.0×10–14)/(5.2×10–4) = 1.9231×10–11 [ x ][ x ] [ x ][ x ] [ H3O+ ] ⎡⎣⎢(CH3CH 2 )3 N⎤⎦⎥ Ka = 1.9231×10–11 = = = + ⎡0.05000 − x ⎤ ⎡(CH 3 CH 2 ) NH ⎤ [0.05000 ] ⎢⎣ ⎥⎦ 3 ⎣⎢ ⎦⎥ x = [H3O+] = 9.80587×10–7 M pH = –log [H3O+] = –log (9.80587×10–7) = 6.0085 = 6.01 g) After the equivalence point, the excess strong acid is the primary factor influencing the pH. Determine the moles of HCl added: Moles of added HCl = (0.1000 mol HCl/L)(10–3 L/1 mL)(20.05 mL) = 2.005×10–3 mol HCl The HCl will react with an equal amount of the base, and 0 mol (CH3CH2)3N will remain, and 5×10–6 moles of HCl will be in excess. The volume of the solution at this point is [(20.00 + 20.05) mL](10–3 L/1 mL) = 0.04005 L The molarity of the excess H3O+ is (5×10–6 mol H3O+)/(0.04005 L) = 1.2484×10–4 M pH = –log (1.2484×10–4) = 3.9036 = 3.90 h) Determine the moles of HCl added: Moles of added HCl = (0.1000 mol HCl/L)(10–3 L/1 mL)(25.00 mL) = 2.500×10–3 mol HCl Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

19-41

The HCl will react with an equal amount of the base, and 0 mol (CH3CH2)3N will remain, and 5.00×10–4 mol of HCl will be in excess. The volume of the solution at this point is [(20.00 + 25.00) mL](10–3 L/1 mL) = 0.04500 L The molarity of the excess H3O+ is (5.00×10–4 mol H3O+)/(0.04500 L) = 1.1111×10–2 M pH = –log (1.1111×10–2) = 1.9542 = 1.95 19.53

Plan: Use (M)(V) to find the initial moles of acid and then use the mole ratio in the balanced equation to find moles of base; dividing moles of base by the molarity of the base gives the volume. At the equivalence point, the conjugate base of the weak acid is present; set up a reaction table for the base dissociation in which x = the amount of dissociated base. Use the Kb expression to solve for x from which pOH and then pH is obtained. Solution: a) The balanced chemical equation is: NaOH(aq) + CH3COOH(aq) → Na+(aq) + CH3COO–(aq) + H2O(l) The sodium ions on the product side are written as separate species because they have no effect on the pH of the solution. Calculate the volume of NaOH needed: Volume (mL) of NaOH = ⎛ 1 mol NaOH ⎞⎛ ⎞⎛ ⎛ 0.0520 mol CH 3 COOH ⎞⎟⎜⎛10−3 L ⎞⎟ ⎟⎜ ⎟⎜ 1 mL ⎞⎟ L ⎜ ⎟⎜⎜ ⎜⎜⎜ ⎟⎟⎟ 42.2 mL ⎜⎜ ⎟⎟⎟⎜⎜ ⎟⎟⎟⎜⎜ −3 ⎟⎟⎟ ⎟ L ⎟⎜⎜ 0.0372 mol NaOH ⎠⎝ ⎟⎜⎜10 L ⎠⎟ ⎝⎜ ⎠⎟⎜⎜⎝ 1 mL ⎠⎟ ⎝⎜⎜1 mol CH 3 COOH ⎠⎝ = 58.989247 = 59.0 mL NaOH Determine the moles of initially CH3COOH present: ⎛ 0.0520 mol CH 3 COOH ⎞⎟⎛⎜10−3 L ⎞⎟ ⎟⎟⎜⎜ Moles of CH3COOH = ⎜⎜⎜ ⎟⎟ 42.2 mL = 0.0021944 mol CH3COOH ⎜⎝ L ⎠⎟⎜⎜⎝ 1 mL ⎠⎟⎟

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At the equivalence point, 0.0021944 mol NaOH will be added so the moles acid = moles base. The NaOH will react with an equal amount of the acid, 0 mol CH3COOH will remain, and 0.0021944 moles of CH3COO– will be formed. + NaOH(aq) → H2O(l) + CH3COO–(aq) + Na+(aq) CH3COOH(aq) Initial 0.0021944 mol 0.0021944 mol – 0 – Change –0.0021944 mol –0.0021944 mol – +0.0021944 mol – Final 0 0 0.0021944 mol Determine the liters of solution present at the equivalence point: Volume = [(42.2 + 58.989247) mL](10–3 L/1 mL) = 0.101189247 L Concentration of CH3COO– at equivalence point: Molarity = (0.0021944 mol CH3COO–)/(0.101189247 L) = 0.0216861 M Ka CH3COOH = 1.8×10–5 Calculate Kb for CH3COO–: –14 Kb = Kw/Ka = (1.0×10 )/(1.8×10–5) = 5.556×10–10 Using a reaction table for the equilibrium reaction of CH3COO–: + H2O ⇆ CH3COOH + OH– CH3COO– – 0 0 Initial 0.0216861 M Change –x +x +x Equilibrium 0.0216861 – x x x Determine the hydroxide ion concentration from the Kb, and then determine the pH from the pOH. [ x ][ x ] [ x ][ x ] [CH3COOH ][ OH− ] Kb = 5.556×10–10 = = = − ⎡0.0216861 − x ⎤ [ 0.0216861] [CH3COO ] ⎣⎢ ⎦⎥ [OH–] = x = 3.471138×10–6 M pOH = –log (3.471138×10–6) = 5.459528 pH = 14.00 – pOH = 14.00 – 5.459528 = 8.54047 = 8.540

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b) In the titration of a diprotic acid such as H2SO3, two OH– ions are required to react with the two H+ ions of each acid molecule. Because of the large difference in Ka values, each mole of H+ is titrated separately, so H2SO3 molecules lose one H+ before any HSO3– ions do. The balanced chemical equation for the neutralization of H2SO3 at the first equivalence point is: NaOH(aq) + H2SO3(aq) → Na+(aq) + HSO3–(aq) + H2O(l) The sodium ions on the product side are written as separate species because they have no effect on the pH of the solution. Calculate the volume of NaOH needed to reach the first equivalence point: Volume (mL) of NaOH = ⎛ 1 mol NaOH ⎞⎛ ⎞⎛ ⎛ 0.0850 mol H 2 SO3 ⎞⎟⎛⎜10−3 L ⎞⎟ ⎟⎜ ⎟⎜ 1 mL ⎞⎟ L ⎜ ⎟⎜⎜ ⎜⎜⎜ ⎟⎟⎟ 28.9 mL ⎜⎜ ⎟⎟⎟⎜⎜ ⎟⎟⎟⎜⎜ −3 ⎟⎟⎟ ⎟ ⎜⎝ L ⎟⎜⎜ 0.0372 mol NaOH ⎠⎝ ⎟⎜⎜10 L ⎠⎟ ⎠⎟⎜⎜⎝ 1 mL ⎠⎟ ⎝⎜⎜1 mol H 2 SO3 ⎠⎝ = 66.034946 = 66.0 mL NaOH Determine the moles of HSO3– produced at the first equivalence point: ⎛1 mol HSO32− ⎞⎟ ⎛ 0.0850 mol H 2 SO3 ⎞⎟⎛⎜10−3 L ⎞⎟ ⎟⎟ 28.9 mL ⎜⎜ ⎟⎟ = 0.0024565 mol HSO – ⎟⎟⎜⎜ Moles of HSO3– = ⎜⎜⎜ 3 ⎜⎝⎜⎜ 1 mol H 2 SO3 ⎠⎟⎟ ⎜⎝ L ⎠⎟⎜⎜⎝ 1 mL ⎠⎟⎟

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Determine the liters of solution present at the first equivalence point: Volume = [(28.9 + 66.034946) mL](10–3 L/1 mL) = 0.094934946 L Determine the concentration of HSO3– at the first equivalence point: Molarity = (0.0024565 moles HSO3–)/(0.094934946 L) = 0.0258756 M – HSO3 is an amphoteric substance. To calculate the pH at the first equivalence point, determine whether HSO3– is a stronger acid or a stronger base. The Ka for HSO3– is 6.5 × 10-8. Ka1 for H2SO3 = 1.4×10–2 Calculate Kb for HSO3–: –14 Kb = Kw/Ka1 = (1.0×10 )/(1.4×10–2) = 7.142857×10–13 Because HSO3– is a stronger acid than it is a base (Ka is larger than Kb for HSO3–), at the first equivalence point, it will behave as an acid and donate a hydrogen ion. For the first equivalence point: Using a reaction table for the equilibrium reaction of HSO3–: + H2O ⇆ H3O+ + SO32– HSO3– 0 0 Initial 0.0258756 M Change –x +x +x Equilibrium 0.0258756 – x x x Determine the hydronium ion concentration from the Ka, and then determine the pH from the hydronium ion concentration. [ x ][ x ] [ x ][ x ] [ H3O+ ] ⎡⎣SO32- ⎤⎦ Ka = 6.5×10–8 = = = ⎡0.0258756 − x⎤ [0.0258756 ] [ HSO3 ] ⎢⎣ ⎥⎦ (assuming that x is small compared with 0.0258756 M) [H3O+] = x = 4.101114×10–5 M pH at first equivalence point = –log (4.101114×10–5) = 4.3870982 = 4.387 The balanced chemical equation for the neutralization of HSO3– at the second equivalence point is: NaOH(aq) + HSO3–(aq) → Na+(aq) + SO32–(aq) + H2O(l) A total of 66.034946 mL were required to reach the first equivalence point. It will require an equal volume of NaOH to reach the second equivalence point from the first equivalence point. (2 × 66.0 mL = 132 mL) Determine the total volume of solution present at the second equivalence point. Volume = [(28.9 + 66.034946 + 66.034946) mL](10–3 L/1 mL) = 0.160969892 L 0.0024565 mol HSO3– were present at the first equivalence point. An equal number of moles of SO32– will be present at the second equivalence point. Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

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Determine the concentration of SO32– at the second equivalence point. Molarity = (0.0024565 moles SO32–)/(0.160969892 L) = 0.0152606 M SO32– does not have a hydrogen ion to donate, so it acts as a base and accepts a hydrogen. Because it acts as a base, we must calculate its Kb. Ka HSO3– (Ka2) = 6.5×10–8 Calculate Kb for SO32–: Kb = Kw/Ka = (1.0×10–14)/(6.5×10–8) = 1.53846×10–7 For the second equivalence point: Using a reaction table for the equilibrium reaction of SO32–: + H2O ⇆ HSO3– + OH– SO32– 0 0 Initial 0.0152606 M Change –x +x +x Equilibrium 0.0152606 – x x x Determine the hydroxide ion concentration from the Kb, and then determine the pH from the pOH. [ x ][ x ] [ x ][ x ] [ HSO3− ][ OH− ] Kb = 1.53846×10–7 = = = 2 − ⎡SO3 ⎤ ⎡0.0152606 − x ⎤ [0.0152606 ] ⎣ ⎦ ⎣⎢ ⎦⎥ (assuming x is small compared with 0.0152606 M) [OH–] = x = 4.84539×10–5 M pOH = –log (4.84539×10–5) = 4.31467 pH = 14.00 – pOH = 14.00 – 4.31467 = 9.68533 = 9.685 19.54

a) The balanced chemical equation is: K+(aq) + OH–(aq) + HNO2(aq) → K+(aq) + NO2–(aq) + H2O(l) The potassium ions on the product side are written as separate species because they have no effect on the pH of the solution. Calculate the volume of NaOH needed: Volume (mL) of KOH = ⎛ 1 mol KOH ⎞⎛ ⎞⎛ ⎛ 0.0390 mol HNO 2 ⎞⎟⎜⎛10−3 L ⎞⎟ ⎟⎜ ⎟⎜ 1 mL ⎞⎟ L ⎜ ⎟⎟⎜⎜ ⎜⎜⎜ ⎟⎟⎟ 23.4 mL ⎜⎜ ⎟⎟⎟⎜⎜ ⎟⎟⎟⎜⎜ −3 ⎟⎟⎟ ⎟⎜⎝ 1 mL ⎠⎟ ⎜⎝⎜1 mol HNO 2 ⎠⎝ L ⎟⎜⎜ 0.0588 mol KOH ⎠⎝ ⎟⎜⎜10 L ⎠⎟ ⎝⎜ ⎠⎜

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= 15.5204 = 15.5 mL KOH Determine the moles of HNO2 present: ⎛ 0.0390 mol HNO 2 ⎞⎟⎛⎜10−3 L ⎞⎟ Moles of HNO2 = ⎜⎜⎜ ⎟⎟⎟ 23.4 mL = 0.0009126 mol HNO2 ⎟⎟⎟⎜⎜⎜ L 1 mL ⎝⎜ ⎠⎜ ⎝ ⎠⎟

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At the equivalence point, 0.0009126 mol KOH will be added so the moles acid = moles base. The KOH will react with an equal amount of the acid, 0 mol HNO2 will remain, and 0.0009126 moles of NO2– will be formed. + KOH(aq) → H2O(l) + NO2–(aq) + K+(aq) HNO2(aq) Initial 0.0009126 mol 0.0009126 mol – 0 – Change –0.0009126 mol –0.0009126 mol – +0.0009126 mol – Final 0 0 0.0009126mol Determine the liters of solution present at the equivalence point: Volume = [(23.4 + 15.5204) mL] (10–3 L/1 mL) = 0.0389204 L Concentration of NO2– at equivalence point: Molarity = (0.0009126 mol NO2–)/(0.0389204 L) = 0.023447858 M Ka HNO2 = 7.1×10–4 Calculate Kb for NO2–: –14 Kb = Kw/Ka = (1.0×10 )/(7.1×10–4) = 1.40845×10–11 Using a reaction table for the equilibrium reaction of NO2–: + H2O ⇆ HNO2 + OH– NO2– – 0 0 Initial 0.023447858 M Change –x +x +x Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

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Equilibrium 0.023447858 – x x x Determine the hydroxide ion concentration from the Kb, and then determine the pH from the pOH. [ x ][ x ] [ x ][ x ] [ HNO2 ][ OH− ] –11 Kb = 1.40845×10 = = = ⎡0.023447858 − x⎤ [0.023447858] [ NO2− ] ⎢⎣ ⎥⎦ [OH–] = x = 5.7468×10–7 M pOH = –log (5.7468×10–7) = 6.240574 pH = 14.00 – pOH = 14.00 – 6.240574 = 7.759426 = 7.76 b) The balanced chemical equations are: KOH(aq) + H2CO3(aq) → K+(aq) + HCO3–(aq) + H2O(l) KOH(aq) + HCO3-(aq) → K+(aq) + CO32–(aq) + H2O(l) The potassium ions on the product side are written as separate species because they have no effect on the pH of the solution. Calculate the volume of KOH needed: Volume (mL) of KOH = ⎛ 1 mol KOH ⎞⎛ ⎞⎛ ⎛ 0.130 mol H 2 CO 3 ⎞⎟⎛⎜10−3 L ⎞⎟ ⎟⎟⎜⎜ ⎟⎟⎜⎜ 1 mL ⎞⎟⎟ L ⎜⎜ ⎟⎟ 17.3 mL ⎜⎜ ⎟⎜⎜ ⎟ ⎜ ⎜ ⎟ ⎟⎟⎜⎜ 0.0588 mol KOH ⎟⎟⎟⎜⎜⎜10−3 L ⎟⎟⎟ ⎟⎜⎝ 1 mL ⎠⎟⎟ ⎜⎝⎜1 mol H 2 CO3 ⎠⎝ ⎜⎜⎝ L ⎠⎜ ⎠⎝ ⎠ = 38.248299 = 38.2 mL KOH It will require an equal volume to reach the second equivalence point (76.4 mL). Determine the moles of HCO3– produced:

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⎛1 mol HCO3− ⎞⎟ ⎛ 0.130 mol H 2 CO3 ⎞⎟⎛⎜10−3 L ⎞⎟ ⎟⎟ 17.3 mL ⎜⎜ ⎟⎟ = 0.002249 mol HCO – ⎟⎟⎜⎜ Moles = ⎜⎜⎜ 3 ⎟⎜⎝ 1 mL ⎠⎟⎟ ⎜⎝⎜⎜1 mol H 2 CO3 ⎠⎟⎟ L ⎝⎜ ⎠⎜

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An equal number of moles of CO32– will be present at the second equivalence point. Determine the liters of solution present at the first equivalence point: Volume = [(17.3 + 38.248299) mL](10–3 L/1 mL) = 0.055548 L Determine the liters of solution present at the second equivalence point: Volume = [(17.3 + 38.248299 + 38.248299) mL](10–3 L/1 mL) = 0.0937966 L Concentration of HCO3– at equivalence point: Molarity = (0.002249 mol HCO3–)/(0.055548 L) = 0.0404875 M Concentration of CO32– at equivalence point: Molarity = (0.002249 mol CO32–)/(0.0937966 L) = 0.023977 M Ka H2CO3 = 4.5×10–7 Calculate Kb for HCO3–: –14 Kb = Kw/Ka = (1.0×10 )/(4.5×10–7) = 2.222×10–8 Ka HCO3– = 4.7×10–11 Calculate Kb for CO32–: –14 Kb = Kw/Ka = (1.0×10 )/(4.7×10–11) = 2.1276596×10–4 Determine the hydroxide ion concentration from the Kb, and then determine the pH from the pOH. For the first equivalence point: [ x ][ x ] [ x ][ x ] [ H2 CO3 ][ OH− ] –8 Kb = 2.222×10 = = = − ⎡0.0404875 − x ⎤ [0.0404875] [ HCO3 ] ⎣⎢ ⎦⎥ – –5 [OH ] = x = 2.999387×10 M pOH = – log (2.999387×10–5) = 4.522967495 pH = 14.00 – pOH = 14.00 – 4.522967495 = 9.4770 = 9.48 For the second equivalence point: [ x ][ x ] [ x ][ x ] [ HCO3− ][ OH− ] Kb = 2.1276596×10–4 = = = 2 − ⎡CO3 ⎤ ⎡0.023977 − x ⎤ [ 0.023977 ] ⎣ ⎦ ⎣⎢ ⎦⎥ [OH–] = x = 2.2586477×10–3 M Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

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pOH = – log (2.2586477×10–3) = 2.6461515 pH = 14.00 – pOH = 14.00 – 2.6461515 = 11.3538 = 11.35 19.55

Plan: Use (M)(V) to find the initial moles of base and then use the mole ratio in the balanced equation to find moles of acid; dividing moles of acid by the molarity of the acid gives the volume. At the equivalence point, the conjugate acid of the weak base is present; set up a reaction table for the acid dissociation in which x = the amount of dissociated acid. Use the Ka expression to solve for x from which pH is obtained. Solution: a) The balanced chemical equation is: HCl(aq) + NH3(aq) → NH4+(aq) + Cl–(aq) The chloride ions on the product side are written as separate species because they have no effect on the pH of the solution. Calculate the volume of HCl needed: Volume (mL) of HCl = ⎛ 1 mol HCl ⎞⎛ ⎞⎛ ⎛ 0.234 mol NH 3 ⎞⎟⎛⎜10−3 L ⎞⎟ ⎟⎟⎜⎜ ⎟⎟⎜⎜ 1 mL ⎞⎟⎟ L ⎜⎜ ⎟⎟ 65.5 mL ⎜⎜ ⎟⎟⎜⎜ ⎟ ⎜ ⎜ ⎟⎟⎜⎜ 0.125 mol HCl ⎟⎟⎟⎜⎜⎜10−3 L ⎟⎟⎟ = 122.616 = 123 mL HCl ⎜⎝⎜ ⎟⎜⎝ 1 mL ⎠⎟⎟ ⎜⎝⎜1 mol NH 3 ⎠⎝ L ⎠⎜ ⎠⎝ ⎠

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Determine the moles of NH3 present: ⎛ 0.234 mol NH 3 ⎞⎟⎛⎜10−3 L ⎞⎟ ⎜ Moles = ⎜⎜ ⎟⎟⎟ 65.5 mL = 0.015327 mol NH3 ⎟⎟⎟⎜⎜⎜ ⎜⎝ L 1 mL ⎠⎜ ⎝ ⎠⎟ At the equivalence point, 0.015327 mol HCl will be added so the moles acid = moles base. The HCl will react with an equal amount of the base, 0 mol NH3 will remain, and 0.015327 moles of NH4+ will be formed. + NH3(aq) → NH4+(aq) + Cl–(aq) HCl(aq) Initial 0.015327 mol 0.015327 mol 0 – Change –0.015327 mol –0.015327 mol +0.015327 mol – Final 0 0 0.015327 mol Determine the liters of solution present at the equivalence point: Volume = [(65.5 + 122.616) mL](10–3 L/1 mL) = 0.188116 L

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Concentration of NH4+ at equivalence point: Molarity = (0.015327 mol NH4+)/(0.188116 L) = 0.081476 M Kb NH3 = 1.76×10–5 Calculate Ka for NH4+: –14 Ka = Kw/Kb = (1.0×10 )/(1.76×10–5) = 5.6818×10–10 Using a reaction table for the equilibrium reaction of NH4+: + H2O ⇆ NH3 + H3O+ NH4+ – 0 0 Initial 0.081476 M Change –x +x +x Equilibrium 0.081476 – x x x Determine the hydrogen ion concentration from the Ka, and then determine the pH. [ x ][ x ] [ x ][ x ] [ H3O+ ][ NH3 ] –10 Ka= 5.6818×10 = = = + ⎡ 0.081476 − x⎤ [ 0.081476 ] [ NH 4 ] ⎢⎣ ⎥⎦ x = [H3O+] = 6.803898×10–6 M pH = –log [H3O+] = –log (6.803898×10–6) = 5.1672 = 5.17 b) The balanced chemical equation is: HCl(aq) + CH3NH2(aq) → CH3NH3+(aq) + Cl–(aq) The chloride ions on the product side are written as separate species because they have no effect on the pH of the solution. Calculate the volume of HCl needed: Volume (mL) of HCl =

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⎛ 1 mol HCl ⎞⎛ ⎞⎟⎛ 1 mL ⎞ ⎛1.11 mol CH 3 NH 2 ⎞⎟⎛⎜10−3 L ⎞⎟ ⎟⎟⎜⎜ L ⎟ ⎜⎜ ⎟⎟ 21.8 mL ⎜⎜ ⎟⎜ ⎟⎟⎜⎜ ⎟ ⎜ ⎜ ⎟⎟⎜⎜ 0.125 mol HCl ⎟⎟⎟⎝⎜⎜⎜10−3 L ⎠⎟⎟⎟ ⎜⎝⎜ ⎜⎝⎜1 mol CH 3 NH 2 ⎠⎝ L ⎠⎟⎜⎜⎝ 1 mL ⎠⎟⎟ ⎠ = 193.584 = 194 mL HCl Determine the moles of CH3NH2 present: ⎛1.11 mol CH 3 NH 2 ⎞⎟⎛⎜10−3 L ⎞⎟ ⎟⎟ 21.8 mL = 0.024198 mol CH NH ⎟⎟⎜⎜ Moles = ⎜⎜⎜ 3 2 ⎟⎜⎝ 1 mL ⎠⎟⎟ L ⎝⎜ ⎠⎜ At the equivalence point, 0.024198 mol HCl will be added so the moles acid = moles base. The HCl will react with an equal amount of the base, 0 mol CH3NH2 will remain, and 0.024198 moles of CH3NH3+ will be formed. + CH3NH2(aq) → CH3NH3+(aq) + Cl–(aq) HCl(aq) Initial 0.024198 mol 0.024198 mol 0 – Change –0.024198 mol –0.024198 mol +0.024198 mol – Final 0 0 0.024198 mol Determine the liters of solution present at the equivalence point: Volume = [(21.8 + 193.584) mL](10–3 L/1 mL) = 0.215384 L

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Concentration of CH3NH3+ at equivalence point: Molarity = (0.024198 mol CH3NH3+)/(0.215384 L) = 0.1123482 M Kb CH3NH2 = 4.4×10–4 Calculate Ka for CH3NH3+: –14 Ka = Kw/Kb = (1.0×10 )/(4.4×10–4) = 2.2727×10–11 Using a reaction table for the equilibrium reaction of CH3NH3+: + H2O ⇆ CH3NH2 + H3O+ CH3NH3+ – 0 0 Initial 0.1123482 M Change –x +x +x Equilibrium 0.1123482 – x x x Determine the hydrogen ion concentration from the Ka, and then determine the pH. [ x ][ x ] [ x ][ x ] [ H3O+ ][CH3 NH 2 ] Ka = 2.2727×10–11 = = = + ⎡ 0.1123482 − x ⎤ [ 0.1123482 ] [CH3 NH3 ] ⎣⎢ ⎦⎥ x = [H3O+] = 1.5979×10–6 M pH = –log [H3O+] = –log (1.5979×10–6) = 5.7964 = 5.80 19.56

a) The balanced chemical equation is: HNO3(aq) + C5H5N(aq) → C5H5NH+(aq) + NO3–(aq) The nitrate ions on the product side are written as separate species because they have no effect on the pH of the solution. Calculate the volume of HNO3 needed: Volume (mL) of HNO3 = ⎛ 1 mol HNO3 ⎞⎛ ⎞⎛ ⎛ 0.0750 mol C 5 H 5 N ⎞⎟ ⎟⎜ ⎟⎜ 1 mL ⎞⎟ L ⎟⎟ 2.65 L ⎜⎜⎜ ⎜⎜⎜ ⎟⎟⎟⎜⎜ ⎟⎟⎟⎜⎜ −3 ⎟⎟⎟ ⎜⎜⎝1 mol C 5 H 5 N ⎠⎝ ⎜⎝ L ⎟⎜⎜ 0.447 mol HNO3 ⎠⎝ ⎟⎜⎜10 L ⎠⎟ ⎠⎟ = 444.63087 = 445 mL HNO3 Determine the moles of C5H5NH+ produced: ⎛1 mol C 5 H 5 NH + ⎞⎟ ⎛ 0.0750 mol C 5 H 5 N ⎞⎟ ⎟⎟ = 0.19875 mol C H NH+ ⎟⎟ 2.65 L ⎜⎜⎜ Moles of C5H5NH+ = ⎜⎜⎜ 5 5 ⎜⎜⎝ 1 mol C 5 H 5 N ⎠⎟⎟ ⎜⎝ L ⎠⎟

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Determine the liters of solution present at the equivalence point: Volume = 2.65 L + (444.63087mL)(10–3 L/1 mL) = 3.09463 L Concentration of C5H5NH+ at equivalence point: Molarity = (0.19875 mol C5H5NH+)/(3.09463 L) = 0.064224 M Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

19-47

Kb C5H5N = 1.7×10–9 Calculate Ka for C5H5NH+: –14 Ka = Kw/Kb = (1.0×10 )/(1.7×10–9) = 5.88235×10–6 Determine the hydrogen ion concentration from the Ka, and then determine the pH. [ x ][ x ] [ x ][ x ] [ H3O+ ][C5 H5 N ] –6 Ka = 5.88235×10 = = = ⎡0.064224 − x ⎤ [ 0.064224 ] [C5 H 5 NH+ ] ⎢⎣ ⎥⎦ + –4 x = [H3O ] = 6.1464×10 M pH = –log [H3O+] = –log (6.1464×10–4) = 3.211379 = 3.21 b) The balanced chemical equations are: HNO3(aq) + H2NCH2CH2NH2(aq) → H2NCH2CH2NH3+(aq) + NO3–(aq) HNO3(aq) + H2NCH2CH2NH3+(aq) → H3NCH2CH2NH32+(aq) + NO3–(aq) The nitrate ions on the product side are written as separate species because they have no effect on the pH of the solution. Calculate the volume of HNO3 needed: Volume (mL) of HNO3 = ⎛ ⎞⎛ ⎞⎛ ⎛ 0.250 mol H 2 NCH 2 CH 3 NH 2 ⎞⎟ 1 mol HNO3 ⎟⎟⎜⎜ ⎟⎟⎜⎜ 1 mL ⎞⎟⎟ L ⎜⎜ ⎟⎟ 0.188 L ⎜⎜⎜ ⎟ ⎜ ⎟⎟⎜⎜ 0.447 mol HNO ⎟⎟⎟⎜⎜⎜10−3 L ⎟⎟⎟ ⎜⎝⎜1 mol H 2 NCH 2 CH 3 NH 2 ⎠⎝ ⎜⎜⎝ L ⎠⎟ 3 ⎠⎝ ⎠ = 105.1454 = 105 mL HCl It will require an equal volume to reach the second equivalence point. (210. mL) Determine the moles of H2NCH2CH2NH3+ produced: +⎞ ⎛ ⎛ 0.250 mol H 2 NCH 2 CH 3 NH 2 ⎞⎟ ⎜1 mol H 2 NCH 2 CH 3 NH 3 ⎟⎟ + ⎜ ⎜ ⎟ Moles of H2NCH2CH2NH3 = ⎜⎜ ⎟ 0.188 L ⎜⎜ ⎟⎟ L ⎝⎜ ⎠⎟ ⎝⎜ 1 mol H 2 NCH 2 CH 3 NH 2 ⎠⎟ = 0.0470 mol H2NCH2CH2NH3+ An equal number of moles of H3NCH2CH2NH32+ will be present at the second equivalence point. Determine the liters of solution present at the first equivalence point: Volume = 0.188 L + (105.1454 mL)(10–3 L/1 mL) = 0.293145 L Determine the liters of solution present at the second equivalence point: Volume = 0.188 L + 2(105.1454 mL)(10–3 L/1 mL) = 0.39829 L Concentration of H2NCH2CH2NH3+ at equivalence point: Molarity = (0.0470 mol H2NCH2CH2NH3+)/(0.293145 L) = 0.16033 M Concentration of H3NCH2CH2NH32+ at equivalence point: Molarity = (0.0470 mol H3NCH2CH2NH32+)/(0.39829 L) = 0.11800 M Kb H2NCH2CH2NH2 = 8.5×10–5 Calculate Ka for H2NCH2CH2NH3+: –14 –5 Ka = Kw/Kb = (1.0×10 )/(8.5×10 ) = 1.17647×10–10 Kb H2NCH2CH2NH3+ = 7.1×10–8 Calculate Ka for H3NCH2CH2NH32+: –14 –8 Ka = Kw/Kb = (1.0×10 )/(7.1×10 ) = 1.40845×10–7

(

)

(

)

Determine the hydrogen ion concentration from the Ka, and then determine the pH for the first equivalence point. [ x ][ x ] [ x ][ x ] [ H3O+ ][ H2 NCH2 CH3 NH2 ] Ka = 1.17647×10–10 = = = + ⎡ 0.16033 − x⎤ [ 0.16033] [ H2 NCH2 CH3 NH3 ] ⎣⎢ ⎦⎥ x = [H3O+] = 4.3430780×10–6 M pH = –log [H3O+] = –log (4.3430780×10–6) = 5.36220 = 5.36 Determine the hydrogen ion concentration from the Ka, and then determine the pH for the second equivalence point. [ x ][ x ] [ x ][ x ] [ H3 O+ ][ H 2 NCH 2 CH3 NH3+ ] –7 Ka = 1.40845×10 = = = ⎡ H 3 NCH 2 CH3 NH32+ ⎤ ⎡0.11800 − x⎤ [ 0.11800 ] ⎣ ⎦ ⎣⎢ ⎦⎥ + –4 x = [H3O ] = 1.2891745×10 M pH = –log [H3O+] = –log (1.2891745×10–4) = 3.889688 = 3.89 Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

19-48

19.57

Plan: Indicators have a pH range that is approximated by pKa ± 1. Find the pKa of the indicator by using the relationship pKa = –log Ka. Solution: The pKa of cresol red is –log (3.5×10–9) = 8.5, so the indicator changes color over an approximate range of 8.5 ± 1 or 7.5 to 9.5.

19.58

Indicators have a pH range that is approximated by pKa ± 1. The pKa of ethyl red is –log (3.8×10–6) = 5.42, so the indicator changes color over an approximate range of 4.4 to 6.4.

19.59

Plan: Choose an indicator that changes color at a pH close to the pH of the equivalence point. Solution: a) The equivalence point for a strong acid–strong base titration occurs at pH = 7.0. Bromthymol blue is an indicator that changes color around pH 7. b) The equivalence point for a weak acid–strong base is above pH 7. Estimate the pH at equivalence point from equilibrium calculations. At the equivalence point, all of the HCOOH and NaOH have been consumed; the solution is 0.050 M HCOO−. (The volume doubles because equal volumes of base and acid are required to reach the equivalence point. When the volume doubles, the concentration is halved.) The weak base HCOO– undergoes a base reaction: COOH–(aq) + H2O(l) ⇆ HCOOH(aq) + OH–(aq) Concentration, M __ 0 0 Initial 0.050 M Change –x +x +x Equilibrium 0.050 – x x x The Ka for HCOOH is 1.8×10–4, so Kb = 1.0×10–14/1.8×10–4 = 5.5556×10–11 [ x ][ x ] [ x ][ x ] [ HCOOH ][ OH− ] –11 Kb = 5.5556×10 = = = ⎡0.050 − x ⎤ [ 0.050 ] [ HCOO− ] ⎢⎣ ⎥⎦ – –6 [OH ] = x = 1.666673×10 M pOH = –log (1.666673×10–6) = 5.7781496 pH = 14.00 – pOH = 14.00 – 5.7781496 = 8.2218504 = 8.22 Choose thymol blue or phenolphthalein.

19.60

a) Determine the Ka (of the conjugate acid) from the Kb for CH3NH2. Ka = Kw/Kb = (1.0×10–14)/(4.4×10–4) = 2.2727×10–11 An acid-base titration of two components of equal concentration and at a 1:1 ratio gives a solution of the conjugates with half the concentration. In this case, the concentration of CH3NH3+ = 0.050 M. [ H3O+ ][CH3 NH 2 ] x2 x2 Ka = 2.2727×10–11 = = = + 0.050 − x 0.050 [CH3 NH3 ] x = [H3O+] = 1.0659972×10–6 M pH = –log [H3O+] = –log (1.0659972×10–6) = 5.97224 = 5.97 Either methyl red or alizarin is acceptable. b) This is a strong acid–strong base titration; thus, the equivalence point is at pH = 7.00. The best choice would be bromthymol blue; alizarin might be acceptable.

19.61

Plan: Choose an indicator that changes color at a pH close to the pH of the equivalence point. Solution: a) The equivalence point for a weak base–strong acid is below pH 7. Estimate the pH at equivalence point from equilibrium calculations. At the equivalence point, the solution is 0.25 M (CH3)2NH2+. (The volume doubles because equal volumes of base and acid are required to reach the equivalence point. When the volume doubles, the concentration is halved.) Ka = Kw/Kb = (1.0×10–14)/(5.9×10–4) = 1.69491525×10–11

19-49

(CH3)2NH2+(aq) + H2O(l) ⇆ (CH3)2NH(aq) + H3O+(aq) __ 0 0 0.25 M –x +x +x 0.25 – x x x + 2 2 H O (CH ) NH [ 3 ][ 3 2 ] x x Ka = 1.69491525×10–11 = = = + 0.25 − x 0.25 [(CH3 )2 NH2 ]

Concentration, M Initial Change Equilibrium

x = [H3O+] = 2.0584674×10–6 M pH = –log [H3O+] = –log (2.0584674×10–6) = 5.686456 = 5.69 Methyl red is an indicator that changes color around pH 5.7. b) This is a strong acid–strong base titration; thus, the equivalence point is at pH = 7.00. Bromthymol blue is an indicator that changes color around pH 7. 19.62

a) Determine the Kb (of the conjugate base) from the Ka for C6H5COOH. Kb = Kw/Ka = (1.0×10–14)/(6.3×10–5) = 1.5873×10–10 An acid-base titration of two components of equal concentration and at a 1:1 ratio gives a solution of the conjugates with half the concentration. In this case, the concentration of C6H5COO– = 0.125 M. [ x ][ x ] [ x ][ x ] [C6 H5 COOH ][ OH− ] Kb = 1.5873×10–10 = = = − ⎡0.125 − x ⎤ [ 0.125] [C6 H5 COO ] ⎣⎢ ⎦⎥ [OH–] = x = 4.4543518×10–6 M pOH = –log (4.4543518×10–6) = 5.351215485 pH = 14.00 – pOH = 14.00 – 5.351215485 = 8.64878 = 8.65 The choices are phenolphthalein or thymol blue. b) The titration will produce a 0.25 M NH3 solution at the equivalence point. Use the Kb for NH3 from the Appendix. [ x ][ x ] [ x ][ x ] [ NH 4+ ][ OH− ] –5 Kb = 1.76×10 = = = ⎡ 0.25 − x ⎤ [0.25] [ NH3 ] ⎢⎣ ⎥⎦ [OH–] = x = 2.0976177×10–3 M pOH = –log (2.0976177×10–3) = 2.67827 pH = 14.00 – pOH = 14.00 – 2.67827 = 11.32173 = 11.32 The best choice would be alizarin yellow R; alizarin might be acceptable.

19.63

M2X(s) ⇆ 2M+(aq) + X2–(aq) Ksp = [M+]2[X2–], assuming M2X is a strong electrolyte. S = molar solubility = 5×10–5 M [M+] = 2S = 1×10–4 M [X2–] = S = 5×10–5 M The actual Ksp is lower than the calculated value because the assumption that M2X is a strong electrolyte (i.e., exists as M+ + X2–) is in error to some degree. There would be some (probably significant) amount of ion pairing to form MX–(aq), M2X(aq), etc., which reduces the effective concentrations of the ions.

19.64

Fluoride ion in BaF2 is the conjugate base of the weak acid HF. The base hydrolysis reaction of fluoride ion F−(aq) + H2O(l) ⇆ HF(aq) + OH−(aq) therefore is influenced by the pH of the solution. As the pH increases, [OH–] increases and the equilibrium shifts to the left to decrease [OH–] and increase [F−]. As the pH decreases, [OH–] decreases and the equilibrium shifts to the right to increase [OH–] and decrease [F−]. The changes in [F−] influence the solubility of BaF2. Chloride ion is the conjugate base of a strong acid so it does not react with water. Thus, its concentration is not influenced by pH, and solubility of BaCl2 does not change with pH.

19-50

19.65

To use Ksp for comparing solubilities, the Ksp expressions must be of the same mathematical form. Stated differently, AgCl and AgBr are both 1:1 electrolytes, while Ag2CrO4 is a 2:1 electrolyte.

19.66

Consider the reaction AB(s) ⇆ A+(aq) + B−(aq), where Qsp = [A+][B−]. If Qsp > Ksp, then there are more ions dissolved than expected at equilibrium, and the equilibrium shifts to the left and the compound AB precipitates. The excess ions precipitate as solid from the solution.

19.67

Plan: Write an equation that describes the solid compound dissolving to produce its ions. The ion-product expression follows the equation Ksp = [Mn+]p[Xz−]q where p and q are the subscripts of the ions in the compound’s formula. Solution: a) Ag2CO3(s) ⇆ 2Ag+(aq) + CO32−(aq) Ion-product expression: Ksp = [Ag+]2[CO32−] b) BaF2(s) ⇆ Ba2+(aq) + 2F−(aq) Ion-product expression: Ksp = [Ba2+][F−]2 c) CuS(s) + H2O(l) ⇆ Cu2+(aq) + S2−(aq) Ion-product expression: Ksp = [Cu2+][S2−]

19.68

a) Fe(OH)3(s) ⇆ Fe3+(aq) + 3OH–(aq) Ion-product expression: Ksp = [Fe3+][OH–]3 b) Ba3(PO4)2(s) ⇆ 3Ba2+(aq) + 2PO43–(aq) Ion-product expression: Ksp = [Ba2+]3[PO43–]2 c) MgF2(s) + H2O(l) ⇆ Mg2+(aq) + 2F−(aq) Ion-product expression: Ksp = [Mg2+][F–]2

19.69

Plan: Write an equation that describes the solid compound dissolving to produce its ions. The ion-product expression follows the equation Ksp = [Mn+]p[Xz−]q where p and q are the subscripts of the ions in the compound’s formula. Solution: a) CaCrO4(s) ⇆ Ca2+(aq) + CrO42−(aq) Ion-product expression: Ksp = [Ca2+][CrO42−] b) AgCN(s) ⇆ Ag+(aq) + CN−(aq) Ion-product expression: Ksp = [Ag+][CN−] c) Ag3PO4(s) + H2O(l) ⇆ 3Ag+(aq) + PO43−(aq) Ion-product expression: Ksp = [Ag+]3[PO43−]

19.70

a) PbI2(s) ⇆ Pb2+(aq) + 2I−(aq) Ion-product expression: Ksp = [Pb2+][I−]2 b) SrSO4(s) ⇆ Sr2+(aq) + SO42−(aq) Ion-product expression: Ksp = [Sr2+][SO42−] c) Cr(OH)3(s) + H2O(l) ⇆ Cr3+(aq) + 3OH−(aq) Ion-product expression: Ksp = [Cr3+][OH−]3

19.71

Plan: Write an equation that describes the solid compound dissolving in water and then write the ion-product expression. Write a reaction table, where S is the molar solubility of Ag2CO3. Substitute the given solubility, S, into the ion-expression and solve for Ksp. Solution: Concentration (M) Ag2CO3(s) ⇆ 2Ag+(aq) + CO32−(aq) Initial — 0 0

19-51

+S Change — +2S Equilibrium — 2S S S = [Ag2CO3] = 0.032 M so [Ag+] = 2S = 0.064 M and [CO32−] = S = 0.032 M Ksp = [Ag+]2[CO32−] = (0.064)2(0.032) = 1.31072×10–4 = 1.3×10–4 19.72

Write a reaction table, where S is the molar solubility of ZnC2O4: ZnC2O4(s) ⇆ Zn2+(aq) + C2O42−(aq) Concentration (M) Initial — 0 0 +S Change — +S Equilibrium — S S S = [ZnC2O4] = 7.9×10–3 M so [Zn2+] = [C2O42−] = S = 7.9×10–3 M Ksp = [Zn2+][C2O42−] = (7.9×10–3)( 7.9×10–3) = 6.241×10–5 = 6.2×10–5

19.73

Plan: Write an equation that describes the solid compound dissolving in water and then write the ion-product expression. Write a reaction table, where S is the molar solubility of Ag2Cr2O7. Substitute the given solubility, S, converted from mass/volume to molarity, into the ion-expression and solve for Ksp. Solution: The solubility of Ag2Cr2O7, converted from g/100 mL to M is: ⎛ 8.3×10−3 g Ag Cr O ⎞⎛ 1 mL ⎞⎛ ⎟⎟⎜ 1 mol Ag 2 Cr2 O 7 ⎞⎟⎟ ⎜ 2 2 7⎟ ⎟⎟⎜⎜ ⎜ Molar solubility = S = ⎜⎜ ⎟ = 0.00019221862 M ⎜ ⎟ −3 ⎟⎟⎜ 100 mL ⎟⎠⎝⎜⎜10 L ⎠⎟⎝⎜⎜ 431.8 g Ag 2 Cr2 O 7 ⎠⎟⎟ ⎜⎝⎜ The equation for silver dichromate, Ag2Cr2O7, is: Ag2Cr2O7(s) ⇆ 2Ag+(aq) + Cr2O72−(aq) Concentration (M) Initial — 0 0 +S Change — +2S Equilibrium — 2S S 2S = [Ag+] = 2(0.00019221862 M) = 0.00038443724 M S = [Cr2O72–] = 0.00019221862 M Ksp = [Ag+]2[Cr2O72–] = (2S)2(S) = (0.00038443724)2(0.00019221862) = 2.8408×10–11 = 2.8×10–11

19.74

The equation and ion-product expression for calcium sulfate, CaSO4, is: Ksp = [Ca2+][SO42−] CaSO4(s) ⇆ Ca2+(aq) + SO42−(aq) The solubility of CaSO4, converted from g/100 mL to M is: ⎛ 0.209 g CaSO 4 ⎞⎛ ⎟⎟⎜⎜ 1 mL ⎞⎛ ⎟⎜ 1 mol CaSO 4 ⎞⎟⎟ ⎜ Molar solubility = S = ⎜⎜ ⎟⎟⎜ −3 ⎟⎟⎟⎜⎜ ⎟ = 0.015351844 M ⎜⎜⎝ 100 mL ⎟⎜⎜10 L ⎠⎝ ⎟⎜⎜136.14 g CaSO 4 ⎠⎟⎟ ⎠⎝ Since one mole of CaSO4 dissociates to form one mole of Ca2+, the concentration of Ca2+ is S = 0.015350716 M. The concentration of SO42− is S = 0.015350716 M because one mole of CaSO4 dissociates to form one mole of SO42−. Ksp = [Ca2+][SO42−] = (S)(S) = (0.015351844)( 0.015351844) = 2.35679×10–4 = 2.36×10–4

19.75

Plan: Write the equation that describes the solid compound dissolving in water and then write the ion-product expression. Set up a reaction table that expresses [Sr2+] and [CO32−] in terms of S, substitute into the ion-product expression, and solve for S. In part b), the [Sr2+] that comes from the dissolved Sr(NO3)2 must be included in the reaction table. Solution: a) The equation and ion-product expression for SrCO3 is: Ksp = [Sr2+][CO32−] SrCO3(s) ⇆ Sr2+(aq) + CO32−(aq) 2+ 2– The solubility, S, in pure water equals [Sr ] and [CO3 ] Write a reaction table, where S is the molar solubility of SrCO3: SrCO3(s) ⇆ Sr2+(aq) + CO32−(aq) Concentration (M) Initial — 0 0

19-52

Change — +S Equilibrium — S Ksp = 5.4×10–10 = [Sr2+][CO32−] = [S][S] = S2 S = 2.32379×10–5 = 2.3×10–5 M

+S S

b) In 0.13 M Sr(NO3)2, the initial concentration of Sr2+ is 0.13 M. Equilibrium [Sr2+] = 0.13 + S and equilibrium [CO32−] = S where S is the solubility of SrCO3. SrCO3(s) ⇆ Sr2+(aq) + CO32−(aq) Concentration (M) Initial — 0.13 0 +S Change — +S Equilibrium — 0.13 + S S Ksp = 5.4×10–10 = [Sr2+][CO32−] = (0.13 + S)S This calculation may be simplified by assuming S is small and setting 0.13 + S = 0.13. Ksp = 5.4×10–10 = (0.13)S S = 4.1538×10–9 = 4.2×10–9 M 19.76

The equation and ion-product expression for BaCrO4 is: Ksp =[Ba2+][CrO42–] BaCrO4(s) ⇆ Ba2+(aq) + CrO42–(aq) 2+ a) The solubility, S, in pure water equals [Ba ] and [CrO42–] Ksp = 2.1×10–10 = [Ba2+][CrO42–] = S2 S = 1.4491×10–5 = 1.4×10–5 M b) In 1.5×10–3 M Na2CrO4, the initial concentration of CrO42– is 1.5×10–3 M. Equilibrium [Ba2+] = S and equilibrium [CrO42–] = 1.5×10–3 + S where S is the solubility of BaCrO4. Ksp = 2.1×10–10 = [Ba2+][CrO42–] = S(1.5×10–3 + S) Assume S is small so 1.5×10–3 + S = 1.5×10–3 Ksp = 2.1×10–10 = S(1.5×10–3) S = 1.4×10–7 M

19.77

Plan: Write the equation that describes the solid compound dissolving in water and then write the ion-product expression. Set up a reaction table that expresses [Ca2+] and [IO3−] in terms of S, substitute into the ion-product expression, and solve for S. The [Ca2+] that comes from the dissolved Ca(NO3)2 and the [IO3−] that comes from NaIO3 must be included in the reaction table. Solution: a) The equilibrium is: Ca(IO3)2(s) ⇆ Ca2+(aq) + 2IO3–(aq). From the Appendix, Ksp(Ca(IO3)2) = 7.1×10–7. Write a reaction table that reflects an initial concentration of Ca2+ = 0.060 M. In this case, Ca2+ is the common ion. Ca(IO3)2(s) ⇆ Ca2+(aq) + 2IO3−(aq) Concentration (M) Initial — 0.060 0 +2S Change — +S Equilibrium — 0.060 + S 2S Assume that 0.060 + S ≈ 0.060 because the amount of compound that dissolves will be negligible in comparison to 0.060 M. Ksp = [Ca2+][IO3−]2 = (0.060)(2S)2 = 7.1×10–7 S = 1.71998×10–3 = 1.7×10–3 M Check assumption: (1.71998×10–3 M)/(0.060 M) × 100% = 2.9% < 5%, so the assumption is good. S represents both the molar solubility of Ca2+ and Ca(IO3)2, so the molar solubility of Ca(IO3)2 is 1.7×10–3 M. b) Write a reaction table that reflects an initial concentration of IO3− = 0.060 M. IO3− is the common ion. Ca(IO3)2(s) ⇆ Ca2+(aq) + 2IO3−(aq) Concentration (M) Initial — 0 0.060 +2S Change — +S Equilibrium — S 0.060 + 2S The equilibrium concentration of Ca2+ is S, and the IO3− concentration is 0.060 + 2S. Assume that 0.060 + 2S ≈ 0.060

19-53

Ksp = [Ca2+][IO3−]2 = (S)(0.060)2 = 7.1×10–7 S = 1.97222×10–4 = 2.0×10–4 M Check assumption: (1.97222×10–4 M)/(0.060 M) × 100% = 0.3% < 5%, so the assumption is good. S represents both the molar solubility of Ca2+ and Ca(IO3)2, so the molar solubility of Ca(IO3)2 is 2.0×10–4 M.

19.78

The equilibrium is: Ag2SO4(s) ⇆ 2Ag+(aq) + SO42–(aq). From the Appendix, Ksp(Ag2SO4) = 1.5×10–5. a) Write a reaction table that reflects an initial concentration of Ag+ = 0.22 M. In this case, Ag+ is the common ion. Ag2SO4(s) ⇆ 2Ag+(aq) + SO42–(aq) Concentration (M) Initial — 0.22 0 +S Change — +2S Equilibrium — 0.22 + 2S S Assume that 0.22 + 2S ≈ 0.22 because the amount of compound that dissolves will be negligible in comparison to 0.22 M. Ksp = [Ag+]2[SO42–] = (0.22)2(S) = 1.5×10–5 S = 3.099174×10–4 = 3.1×10–4 Check assumption: (3.099174×10–4 M)/(0.22 M) × 100% = 1.4% < 5%, so the assumption is good. S represents the molar solubility of Ag2SO4(s): 3.1×10–4 M. b) Write a reaction table that reflects an initial concentration of SO42– = 0.22 M. In this case, SO42– is the common ion. Ag2SO4(s) ⇆ 2Ag+(aq) + SO42–(aq) Concentration (M) Initial — 0 0.22 +S Change — +2S Equilibrium — 2S 0.22 + S The equilibrium concentration of Ag+ is 2S, and the SO42– concentration is 0.22 + S. Assume that 0.22 + S ≈ 0.22. Ksp = [Ag+]2[SO42–] = (2S)2(0.22) = 1.5×10–5 S = 4.1286×10–3 = 4.1×10–3 Check assumption: (4.1286×10–3 M)/(0.22 M) × 100% = 1.9% < 5%, so the assumption is good. S represents the molar solubility of Ag2SO4, so the molar solubility of Ag2SO4 is 4.1×10–3 M.

19.79

Plan: The larger the Ksp, the larger the molar solubility if the number of ions are equal. Solution: a) Mg(OH)2 with Ksp = 6.3×10–10 has higher molar solubility than Ni(OH)2 with Ksp = 6×10–16. b) PbS with Ksp = 3×10–25 has higher molar solubility than CuS with Ksp = 8×10–34. c) Ag2SO4 with Ksp = 1.5×10–5 has higher molar solubility than MgF2 with Ksp = 7.4×10–9.

19.80

The larger the Ksp, the larger the molar solubility if the number of ions are equal. a) SrSO4 with Ksp = 3.2×10–7 has higher molar solubility than BaCrO4 with Ksp = 2.1×10–10. b) CaCO3 with Ksp = 3.3×10–9 has higher molar solubility than CuCO3 with Ksp = 3×10–12. c) Ba(IO3)2 with Ksp = 1.5×10–9 has higher molar solubility than Ag2CrO4 with Ksp = 2.6×10–12.

19.81

Plan: The larger the Ksp, the more water soluble the compound if the number of ions are equal. Solution: a) CaSO4 with Ksp = 2.4×10–5 is more water soluble than BaSO4 with Ksp = 1.1×10–10. b) Mg3(PO4)2 with Ksp = 5.2×10–24 is more water soluble than Ca3(PO4)2 with Ksp = 1.2×10–29. c) PbSO4 with Ksp = 1.6×10–8 is more water soluble than AgCl with Ksp = 1.8×10–10.

19.82

The larger the Ksp, the more water soluble the compound if the number of ions are equal. a) Ca(IO3)2 with Ksp = 7.1×10–7 is more water soluble than Mn(OH)2 with Ksp = 1.6×10–13. b) SrCO3 with Ksp = 5.4×10–10 is more water soluble than CdS with Ksp = 1.0×10–24. c) CuI with Ksp = 1×10–12 is more water soluble than AgCN with Ksp = 2.2×10–16.

19-54

19.83

Plan: If a compound contains an anion that is the weak conjugate base of a weak acid, the concentration of that anion, and thus the solubility of the compound, is influenced by pH. Solution: a) AgCl(s) ⇆ Ag+(aq) + Cl−(aq) The chloride ion is the anion of a strong acid, so it does not react with H3O+. The solubility is not affected by pH. b) SrCO3(s) ⇆ Sr2+(aq) + CO32−(aq) The strontium ion is the cation of a strong base, so pH will not affect its solubility. The carbonate ion is the conjugate base of a weak acid and will act as a base: CO32−(aq) + H2O(l) ⇆ HCO3−(aq) + OH−(aq) and HCO3−(aq) + H2O(l) ⇆ H2CO3(aq) + OH−(aq) The H2CO3 will decompose to CO2(g) and H2O(l). The gas will escape and further shift the equilibrium. Changes in pH will change the [CO32−], so the solubility of SrCO3 is affected. Solubility increases with addition of H3O+ (decreasing pH). A decrease in pH will decrease [OH−], causing the base equilibrium to shift to the right which decreases [CO32−], causing the solubility equilibrium to shift to the right, dissolving more solid.

19.84

a) CuBr(s) ⇆ Cu+(aq) + Br−(aq) The bromide ion is the anion of a strong acid, so it does not react with H3O+. At high pH the copper ion may precipitate. Cu+(aq) + OH–(aq) ⇆ CuOH(s) b) Ca3(PO4)2(s) ⇆ 3Ca2+(aq) + 2PO43–(aq) The calcium ion is the cation of a strong base so pH will not affect its solubility. PO43– is the anion of a weak acid, so the following equilibria would be present. PO43–(aq) + nH2O(l) ⇆ HnPO4(3 – n)–(aq) + nOH–(aq) (n = 1,2,3) Since these involve OH–, the solubility will change with changing pH. Solubility increases with addition of H3O+ (decreasing pH). A decrease in pH will decrease [OH−], causing the base equilibrium to shift to the right which decreases [PO43−], causing the solubility equilibrium to shift to the right, dissolving more solid.

19.85

Plan: If a compound contains an anion that is the weak conjugate base of a weak acid, the concentration of that anion, and thus the solubility of the compound, is influenced by pH. Solution: a) Fe(OH)2(s) ⇆ Fe2+(aq) + 2OH−(aq) The hydroxide ion reacts with added H3O+: OH−(aq) + H3O+(aq) → 2H2O(l) The added H3O+ consumes the OH−, driving the equilibrium toward the right to dissolve more Fe(OH)2. Solubility increases with addition of H3O+ (decreasing pH). b) CuS(s) + H2O(l) ⇆ Cu2+(aq) + S2−(aq) S2− is the anion of a weak acid, so it reacts with added H3O+. Solubility increases with addition of H3O+ (decreasing pH).

19.86

a) PbI2(s) ⇆ Pb2+(aq) + 2I−(aq) The iodide ion is the anion of a strong acid, so it does not react with H3O+. Thus, the solubility does not increase in acid solution. At high pH the lead ion may precipitate. b) Hg2(CN)2(s) ⇆ Hg22+(aq) + 2CN−(aq) At high pH the mercury(I) ion may precipitate. CN− is the anion of a weak acid, so the equilibrium would be CN−(aq) + H2O(l) ⇆ HCN(aq) + OH−(aq) Since this involves OH−, it would shift with changing pH. Solubility increases with addition of H3O+ (decreasing pH).

19.87

Plan: Find the initial molar concentrations of Cu2+ and OH–. The molarity of the KOH is calculated by converting mass to moles and dividing by the volume. Put these concentrations in the ion-product expression, solve for Qsp, and compare Qsp with Ksp. If Qsp > Ksp, precipitate forms.

19-55

Solution: The equilibrium is: Cu(OH)2(s) ⇆ Cu2+(aq) + 2OH–(aq). The ion-product expression is Ksp = [Cu2+][OH−]2 and, from the Appendix, Ksp equals 2.2×10–20. ⎛1.0 ×10−3 mol Cu(NO ) ⎞⎛ 1 mol Cu 2+ ⎞⎟ ⎜ 3 2⎟ –3 2+ ⎟⎟⎜⎜ ⎟ [Cu2+] = ⎜⎜ ⎜ ⎟⎟⎜1 mol Cu(NO ) ⎟⎟⎟ = 1.0×10 M Cu ⎜⎜⎝ L ⎜ 3 2⎠ ⎠⎝ − ⎛ 0.075 g KOH ⎞⎛ ⎞⎛ ⎟⎟⎜⎜ 1 mol KOH ⎟⎟⎜⎜ 1 mol OH ⎞⎟⎟ ⎜ –3 – [OH–] = ⎜⎜ ⎟⎟⎜ ⎟⎟⎜ ⎟ = 1.33666×10 M OH 1.0 L ⎟⎠⎝⎜⎜ 56.11 g KOH ⎠⎝ ⎟⎜⎜1 mol KOH ⎠⎟⎟ ⎜⎜⎝ Qsp = [Cu2+][OH−]2 = (1.0×10–3)(1.33666×10–3)2 = 1.786660×10–9 Qsp is greater than Ksp (1.8×10–9 > 2.2×10–20), so Cu(OH)2 will precipitate. 19.88

The ion-product expression for PbCl2 is Ksp = [Pb2+][Cl−]2 and, from the Appendix, Ksp equals 1.7×10–5. To decide if a precipitate will form, calculate Qsp with the given quantities and compare it to Ksp. ⎛ 0.12 mol Pb(NO3 )2 ⎞⎟⎛⎜ 1 mol Pb 2 + ⎞⎟ ⎟⎟ = 0.12 M Pb2+ ⎟⎟⎜⎜ [Pb2+] = ⎜⎜⎜ ⎜⎝ L ⎠⎟⎜⎜⎝1 mol Pb(NO3 )2 ⎠⎟⎟ −3 − ⎛ 3.5 mg NaCl ⎞⎛ ⎟⎟⎜⎜10 g ⎞⎛ ⎟⎟⎜⎜ 1 mol NaCl ⎞⎛ ⎟⎟⎜⎜ 1 mol Cl ⎞⎟⎟ ⎜ –4 – [Cl–] = ⎜⎜ ⎟⎟⎟⎜⎜⎜ 1 mg ⎟⎟⎟⎜⎜⎜ 58.45 g NaCl ⎟⎟⎟⎜⎜⎜1 mol NaCl ⎟⎟⎟ = 2.3952×10 M Cl ⎜⎝⎜ 0.250 L ⎠⎝ ⎠⎝ ⎠⎝ ⎠

Qsp = [Pb2+][Cl−]2 = (0.12)(2.3952×10–4)2 = 6.8843796×10–9 Qsp is smaller than Ksp (6.9×10–9 < 1.7×10–5), so PbCl2 will not precipitate.

19.89

Plan: Find the initial molar concentrations of Ba2+ and IO3–. The molarity of the BaCl2 is calculated by converting mass to moles and dividing by the volume. Put these concentrations in the ion-product expression, solve for Qsp, and compare Qsp with Ksp. If Qsp > Ksp, precipitate forms. Solution: The equilibrium is: Ba(IO3)2(s) ⇆ Ba2+(aq) + 2IO3–(aq). The ion-product expression is Ksp = [Ba2+][IO3−]2 and, from the Appendix, Ksp equals 1.5×10–9. −3 2+ ⎛ 7.5 mg BaCl 2 ⎞⎛ ⎟⎟⎜⎜10 g ⎞⎛ ⎟⎟⎜⎜ 1 mL ⎞⎛ ⎟⎜ 1 mol BaCl 2 ⎞⎛ ⎟⎟⎜⎜ 1 mol Ba ⎞⎟⎟ ⎜ –5 2+ [Ba2+] = ⎜⎜ ⎟⎟⎜ ⎟⎟⎜ −3 ⎟⎟⎟⎜⎜ ⎟⎟⎜ ⎟ = 7.204611×10 M Ba ⎟⎜⎜ 1 mg ⎠⎝ ⎟⎜⎜10 L ⎠⎝ ⎟⎜⎜ 208.2 g BaCl 2 ⎠⎝ ⎟⎜⎜1 mol BaCl 2 ⎠⎟⎟ ⎜⎜⎝ 500. mL ⎠⎝ ⎛ 0.023 mol NaIO3 ⎞⎟⎛⎜ 1 mol IO3− ⎞⎟ ⎟⎟ = 0.023 M IO – ⎟⎟⎜⎜ [IO3–] = ⎜⎜⎜ 3 ⎟⎜⎝1 mol NaIO3 ⎠⎟⎟ L ⎝⎜ ⎠⎜

Qsp = [Ba2+][IO3−]2 = (7.204611×10–5)(0.023)2 = 3.81124×10–8 Since Qsp > Ksp (3.8×10–8 > 1.5×10–9), Ba(IO3)2 will precipitate.

19.90

The ion-product expression for Ag2CrO4 is Ksp = [Ag+]2[CrO42–] and, from the Appendix, Ksp equals 2.6×10–12. To decide if a precipitate will form, calculate Qsp with the given quantities and compare it to Ksp. −5 + ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎜ 2.7 ×10 g AgNO3 ⎟⎟⎜⎜ 1 mL ⎟⎟⎜⎜⎛ 1 mol AgNO3 ⎞⎟⎟⎜⎜ 1 mol Ag ⎟⎟ –5 + [Ag+] = ⎜⎜ ⎟⎟⎜ −3 ⎟⎟⎜ ⎟ = 1.0594467×10 M Ag ⎟ ⎜⎜ ⎜⎜10 L ⎟⎜⎜169.9 g AgNO 3 ⎟⎟⎜⎜⎜1 mol AgNO 3 ⎟⎟ ⎟ 15.0 mL ⎠⎝ ⎝ ⎠⎝ ⎠⎝ ⎠ 2− ⎞ ⎛ 4.0 ×10−4 mol K CrO ⎞⎛ ⎜⎜ ⎜⎜ 1 mol CrO 4 ⎟⎟ 2 4⎟ –4 2– ⎟ [CrO4 ] = ⎜ ⎟⎟⎜ ⎟⎟ = 4.0×10 M CrO4 ⎜⎜ ⎜ ⎟ ⎟ L ⎝ ⎠⎝⎜1 mol K 2 CrO 4 ⎠ 2–

Qsp = [Ag+]2[ CrO42–]= (1.0594467×10–5)2(4.0×10–4) = 4.4897×10–14 Since Qsp < Ksp (4.5×10–14 < 2.6×10–12), Ag2CrO4 will not precipitate.

19.91

⎛ 9.7×10−5 g Ca 2+ ⎞⎟ ⎛ 1 mol Ca 2+ ⎞⎟ ⎟⎟(104 mL)⎜⎜ ⎟ = 2.5170×10–4 mol Ca2+ Original moles of Ca2+ = ⎜⎜⎜ 2+ ⎟ mL ⎝⎜ ⎠⎟ ⎝⎜⎜ 40.08 g Ca ⎠⎟

19-56

⎛ 1mol C 2 O 4 2− ⎞⎟ ⎛ 0.1550 mol Na 2 C 2 O 4 ⎞⎟⎛⎜10−3 L ⎞⎟ ⎜ 2– ⎟⎜⎜ Moles of C2O42– added = ⎜⎜⎜ ⎟⎟⎟ 100.0 mL ⎜⎜ ⎟⎟ = 0.01550 mol C2O4 ⎟ ⎟⎜ ⎜ L ⎜⎝1mol Na 2 C 2 O 4 ⎠⎟⎟ ⎝⎜ ⎠⎜⎝ 1 mL ⎠⎟

(

)

The Ca2+ is limiting leaving 0 M, and after the reaction there will be (0.01550 – 0.00025170) = 0.0152483 mol of C2O42– left in a total volume of 104 + 100.0 mL = 204 mL. ⎛ 0.0152483 mol C 2 O ⎞⎟⎛⎜ 1 mL ⎞⎟ ⎟⎟⎜⎜ −3 ⎟⎟ = 0.0747466 M C2O42– [C2O42–] = ⎜⎜⎜ ⎜⎝ 204 mL ⎠⎟⎜⎝⎜10 L ⎠⎟⎟ Concentration (M) CaC2O4 ∙ H2O(s) ⇆ Ca2+(aq) + C2O42–(aq) + H2O(l) Initial — 0 0.0747466 — +S — Change — +S Equilibrium — S 0.0747466 + S — Assume that 0.0747466 + S ≈ 0.0747466 because the amount of compound that dissolves will be negligible in comparison to 0.0747466 M. The Ksp from the Appendix is: 2.3×10–9 Ksp = [Ca2+][C2O42–] = (S)(0.0747466) = 2.3×10–9 S = 3.07706×10–8 = 3.1×10–8 Check assumption: (3.07706×10–8 M)/(0.0747466 M) × 100% = 0.00004% < 5%, so the assumption is good. S represents both the molar solubility of Ca2+ and CaC2O4 ∙H2O(s), so the concentration of Ca2+ is 3.1×10–8 M. 19.92

Plan: When Fe(NO3)3 and Cd(NO3)2 mix with NaOH, the insoluble compounds Fe(OH)3 and Cd(OH)2 form. The compound with the smaller value of Ksp precipitates first. Calculate the initial concentrations of Fe3+ and Cd2+ from the dilution formula MconcVconc = MdilVdil. Use the ion-product expressions to find the minimum OH– concentration required to cause precipitation of each compound. Solution: a) Fe(OH)3 will precipitate first because its Ksp (1.6×10–39) is smaller than the Ksp for Cd(OH)2 at 7.2×10–15. The precipitation reactions are: Ksp = [Fe3+][ OH–]3 Fe3+(aq) + 3OH–(aq) → Fe(OH)3(s) 2+ – Ksp = [Cd2+][ OH–]2 Cd (aq) + 2OH (aq) → Cd(OH)2(s) 3+ 2+ The concentrations of Fe and Cd in the mixed solution are found from MconcVconc = MdilVdil [Fe3+] = [(0.50 M)(50.0 mL)]/[(50.0 + 125) mL] = 0.142857 M Fe3+ [Cd2+] = [(0.25 M)(125 mL)]/[(50.0 + 125) mL] = 0.178571 M Cd2+ The hydroxide ion concentration required to precipitate the metal ions comes from the metal ion concentrations and the Ksp. [OH–]Fe = 3

Ksp

⎡ Fe3+ ⎤ ⎣ ⎦

= 3

1.6×10−39 = 2.237×10–13 = 2.2×10–13 M ⎡0.142857⎤ ⎣ ⎦

7.2 ×10−15 = 2.0079864×10–7 = 2.0×10–7 M ⎡0.178571⎤ ⎡Cd 2+ ⎤ ⎣ ⎦ ⎣ ⎦ A lower hydroxide ion concentration is required to precipitate the Fe3+. b) The two ions are separated by adding just enough NaOH to precipitate the iron(III) hydroxide, but precipitating no more than 0.01% of the cadmium. The Fe3+ is found in the solid precipitate while the Cd2+ remains in the solution. c) A hydroxide concentration between the values calculated in part a) will work. The best separation would be when Qsp = Ksp for Cd(OH)2. This occurs when [OH−] = 2.0×10–7 M. [OH–]Cd =

19.93

Ksp

The metal ion can act as a Lewis acid and bond to one or more negatively charged ligands. If the total negative charge of the ligands exceeds the positive charge on the metal ion, the complex will be negative.

19-57

19.94

In the context of this equilibrium only, the increased solubility with added OH− appears to be a violation of Le Châtelier’s principle. Adding OH− should cause the equilibrium to shift towards the left, decreasing the solubility of Zn(OH)2. Before accepting this conclusion, other possible equilibria must be considered. Zinc is a metal ion and hydroxide ion is a ligand, so it is possible that a complex ion forms between the zinc ion and hydroxide ion: Zn2+(aq) + nOH−(aq) ⇆ Zn(OH)n2− n(aq) This decreases the concentration of Zn2+, shifting the solubility equilibrium to the right to dissolve more Zn(OH)2.

19.95

Plan: A hydrated metal ion in aqueous solution, Mx+, binds ligands (such as CN−) to form a complex ion. Appendix C shows that 4 CN− ligands bind to Hg2+. Solution: Hg2+(aq) + 4CN−(aq) ⇆ Hg(CN)42−(aq) The 2+ charge from the mercury ion plus the total 4– charge from the four cyanide ions gives a net charge on the complex ion of 2–.

19.96

Zn2+(aq) + 4CN−(aq) ⇆ Zn(CN)42−(aq)

19.97

Plan: A hydrated metal ion in aqueous solution, Mx+, bind ligands (such as S2O32−) to form a complex ion. Appendix C shows that 2 S2O32− ligands bind to Ag+. Solution: The 1+ charge from the silver ion plus the total 4– charge from the two thiosulfate ions gives a net charge on the complex ion of 3–. Ag+(aq) + 2S2O32−(aq) ⇆ Ag(S2O3)23−(aq).

19.98

Al3+(aq) + 6F−(aq) ⇆ AlF63−(aq)

19.99

Plan: Write the formation reaction and the Kf expression. The initial concentrations of Ag+ and S2O32− may be determined from MconcVconc = MdilVdil. Set up a reaction table and use the limiting reactant to find the amounts of species in the mixture, assuming a complete reaction. A second reaction table is then written, with x representing the amount of complex ion that dissociates. Use the Kf expression to solve for x. Solution: Ag+(aq) + 2S2O32−(aq) ⇆ Ag(S2O3)23−(aq) [Ag+] = (0.044 M)(25.0 mL)/((25.0 + 25.0) mL) = 0.022 M Ag+ [S2O32−] = (0.57 M)(25.0 mL)/((25.0 + 25.0) mL) = 0.285 M S2O32− The reaction gives: Ag+(aq) + 2S2O32−(aq) → Ag(S2O3)23−(aq) Concentration (M) Initial 0.022 0.285 0 Change −0.022 −2(0.022) +0.022 1:2:1 mole ratio Equilibrium 0 0.241 0.022 To reach equilibrium: Ag+(aq) + 2S2O32−(aq) ⇆ Ag(S2O3)23−(aq) Concentration (M) Initial 0 0.241 0.022 Change +x +2x −x Equilibrium +x 0.241 + 2x 0.022 − x Kf is large, so [Ag(S2O3)23−] ≈ 0.022 M and [S2O32−]equil ≈ 0.241 M ⎡ Ag (S O ) 3− ⎤ [ 0.022 ] 2 3 2 ⎢ ⎥⎦ 13 Kf = 4.7×10 = ⎣ = 2 + ⎡ 2− ⎤ 2 [ x ][ 0.241] [ Ag ] ⎣S2 O 3 ⎦ x = [Ag+] = 8.0591778×10−15 = 8.1×10−15 M

19-58

19.100 The reaction between SCN− and Fe3+ produces the red complex FeSCN2+. One can assume from the much larger concentration of SCN− and large Kf that all of the Fe3+ ions react to form the complex. Calculate the initial concentrations of SCN− and Fe3+ and write a reaction table in which x is the concentration of FeSCN2+ formed. 0.0015 M Fe ( NO3 )3 0.50 L ⎛⎜ 1 mol Fe3+ ⎞⎟ ⎟⎟ = 0.00075 M Fe3+ ⎜⎜ [Fe3+]initial = ⎟⎟ ⎜ 1 mol Fe NO ( ) 3 3⎠ 0.50 + 0.50 L ⎝⎜

(

)(

)

((

) ) (0.20 M KSCN)(0.50 L) ⎛⎜⎜ 1 mol SCN ⎞⎟⎟ = 0.10 M SCN = ⎟ ⎜ ((0.50 + 0.50) L) ⎜⎝⎜1 mol KSCN ⎠⎟⎟ −

−

[SCN ]initial

−

Set up a reaction table: Fe3+(aq) + SCN−(aq) ⇆ FeSCN2+ Concentration (M) −4 0.10 0 Initial 7.5×10 Change −x −x +x Equilibrium 7.5×10−4 − x 0.10 − x x It is reasonable to assume that x is much less than 0.10, so 0.10 − x ≈ 0.10. However, it is not reasonable to assume that 0.00075 − x ≈ 0.00075, because x may be significant in relation to such a small number. The equilibrium expression and the constant, from the problem, are: ⎡ x⎤ ⎡ FeSCN 2+ ⎤ [x] ⎣ ⎦ ⎦ = Kf = 8.9×102 = ⎣ 3+ = ⎡ 7.5×10−4 − x ⎤ ⎡0.10 − x ⎤ ⎡ Fe ⎤ [SCN− ] ⎡ 7.5x10−4 − x ⎤ [ 0.10 ] ⎣ ⎦ ⎢⎣ ⎥⎦ ⎢⎣ ⎥⎦ ⎢⎣ ⎥⎦ 2 −4 −4 x = (7.5×10 − x)(0.10)(8.9×10 ) = (7.5×10 − x)(89) x = 6.675×10−2 − 89x x = 7.416667×10−4 From the reaction table, [Fe3+]eq = 7.5×10−4 – x. Therefore, [Fe3+]eq = 7.5×10−4 − 7.416667×10−4 = 8.33333×10−6 = 1×10−5 M. 19.101 Plan: Write the ion-product equilibrium reaction and the complex-ion equilibrium reaction. Add the two reactions to yield an overall reaction; multiply the two constants to obtain Koverall. Write a reaction table where S = [Cr(OH)3]dissolved = [Cr(OH)4−]. Solution: Solubility-product: Cr(OH)3(s) ⇆ Cr3+(aq) + 3OH−(aq) Ksp = 6.3×10–31 3+ − − Complex-ion Cr (aq) + 4OH (aq) ⇆ Cr(OH)4 (aq) Kf = 8.0×1029 K = KspKf = 0.504 Overall: Cr(OH)3(s) + OH−(aq) ⇆ Cr(OH)4−(aq) At pH 13.0, the pOH is 1.0 and [OH−] = 10–1.0 = 0.1 M. Reaction table: Cr(OH)3(s) + OH−(aq) ⇆ Cr(OH)4−(aq) Concentration (M) Initial —— 0.1 0 +S Change —— –S Equilibrium —— 0.1 – S S Assume that 0.1 – S ≈ 0.1. ⎡Cr (OH)4− ⎤ ⎦ = [S] Koverall = 0.504 = ⎣ [ 0.1] [ OH− ]

S = [Cr(OH)4−] = 0.0504 = 0.05 M

19-59

19.102 Write the ion-product equilibrium reaction and the complex-ion equilibrium reaction. Add the two reactions to yield an overall reaction; multiply the two constants to obtain Koverall. Write a reaction table where S = [AgI]dissolved = [Ag(NH3)2+]. Ksp = 8.3×10–17 Solubility-product: AgI(s) ⇆ Ag+(aq) + I−(aq) + + Complex-ion: Ag (aq) + 2NH3(aq) ⇆ Ag(NH3)2 (aq) Kf = 1.7×107 AgI(s) + 2NH3(aq) ⇆ Ag(NH3)2+(aq) + I−(aq) Koverall = Ksp × Kf = (8.3×10–17)(1.7×107) = 1.411×10–9 Reaction table: Ag(NH3)2+(aq) + I−(aq) Concentration (M) AgI(s) + 2NH3(aq) ⇆ Initial —— 2.5 0 0 Change —— –2S +S +S Equilibrium —— 2.5 – 2S S S Assume that 2.5 – 2S ≈ 2.5 because Koverall is so small. ⎡ Ag (NH ) + ⎤ [ I− ] [ S ][ S ] [ S ][ S ] 3 2 ⎦⎥ ⎢ –9 = = S = 9.3908×10–5 = 9.4×10–5 M Koverall = 1.411×10 = ⎣ 2 2 2 [2.5] [ NH3 ] ⎡ 2.5 − 2 S ⎤ ⎣⎢ ⎦⎥ Overall:

19.103 Plan: First, calculate the initial moles of Zn2+ and CN–, then set up reaction table assuming that the reaction first goes to completion, and then calculate back to find the reactant concentrations. Solution: The complex formation equilibrium is: Kf = 4.2×1019 Zn2+(aq) + 4CN−(aq) ⇆ Zn(CN)42−(aq) 2 + ⎛ 1 mol ZnCl 2 ⎞⎛ ⎟⎟⎜⎜ 1 mol Zn ⎞⎟⎟ ⎜ 2+ Moles of Zn2+ = 0.84 g ZnCl 2 ⎜⎜ ⎟⎟⎜ ⎟⎟ = 0.0061624 mol Zn ⎜⎝⎜136.31 g ZnCl 2 ⎠⎝ ⎜ ⎟⎜1 mol ZnCl 2 ⎠⎟ ⎛ 1 mol CN− ⎞⎟ ⎛ 0.150 mol NaCN ⎞⎟⎛⎜10−3 L ⎞⎟ ⎟⎟ 245 mL ⎜⎜ ⎟⎟ = 0.03675 mol CN– ⎟⎟⎜⎜ Moles of CN– = ⎜⎜⎜ ⎜⎜ L ⎜⎝1 mol NaCN ⎠⎟⎟ ⎝⎜ ⎠⎟⎜⎜⎝ 1 mL ⎠⎟⎟

(

)

(

)

The Zn2+ is limiting because there are significantly fewer moles of this ion, thus, [Zn2+] = 0. ⎛ 4 mol CN− ⎞⎟ = 0.0246496 mol CN– Moles of CN– reacting = (0.0061624 mol Zn 2+ )⎜⎜ ⎜⎝ 1 mol Zn 2+ ⎠⎟⎟ Moles of CN– remaining are: 0.03675 – 0.0246496 = 0.0121004 mol CN–

(0.0121004 mol CN ) ⎛⎜⎜ 1 mL ⎞⎟⎟ = 0.0493894 M CN ⎟ ⎜⎜ ⎜⎝10 L ⎠⎟⎟ (245 mL) −

[CN–] =

–

−3

The Zn2+ will produce an equal number of moles of the complex with the concentration: 2− ⎛ 0.0061624 mol Zn 2 + ⎞⎛ ⎟⎟⎜⎜ 1 mL ⎞⎟⎟⎛⎜⎜1 mol Zn (CN)4 ⎞⎟⎟ ⎜⎜ 2− 2 [Zn(CN)4 ] = ⎜ ⎟⎟ = 0.025153 M Zn(CN)4 − ⎟⎟⎜ −3 ⎟⎟⎜ 2+ ⎜⎜⎝ 245 mL ⎟⎠⎝⎜⎜10 L ⎠⎝ ⎟⎜⎜ 1 mol Zn ⎟⎠ 4CN−(aq)

Zn(CN)42−(aq)

Concentration (M)

Zn2+(aq)

Initial

0.0493894

0.025153

Change Equilibrium

+x x

+4x 0.0493894 + 4x

–x 0.025153 – x

⇆

Assume the –x and the +4x do not significantly change the associated concentrations. ⎡ Zn (CN ) 2− ⎤ [0.025153 − x ] [0.025153] 4 ⎢ ⎦⎥ = Kf = 4.2×1019 = ⎣ 2 + 4 = 4 [ x ][ 0.0493894 + 4x ] [ x ][ 0.0493894 ] ⎡ Zn ⎤ [ CN− ]4 ⎣ ⎦ Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

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x = 1.006481×10–16 = 1.0×10–16 [Zn2+] = 1.0×10–16 M Zn2+ [Zn(CN)42−] = 0.025153 – x = 0.025153 – 1.0×10–16 = 0.025153 = 0.025 M Zn(CN)42– [CN−] = 0.0493894 + 4x = 0.0493894 + 4(1.0×10–16) = 0.0493894 = 0.049 M CN– 19.104 The complex formation equilibrium is: Co2+(aq) + 4OH−(aq) ⇆ Co(OH)42−(aq) Kf = 5×109 2+ − First, calculate the initial moles of Co and OH , then set up reaction table assuming that the reaction first goes to completion and then calculate back to find reactant concentrations. 2+ ⎛ 1 mol Co ( NO3 ) ⎞⎛ ⎞⎟ ⎜⎜ ⎜⎜ 1 mol Co 2+ 2 ⎟ ⎟ ⎟⎟ = 0.013118 mol Co2+ Moles of Co = 2.4 g Co ( NO3 )2 ⎜ ⎟⎟⎜ ⎜⎝⎜182.95 g Co ( NO3 ) ⎠⎝ ⎜ ⎟ 1 mol Co (NO3 )2 ⎠⎟⎟ 2 ⎜ ⎛ 1 mol OH− ⎞⎟ ⎛ 0.22 mol KOH ⎞⎟ ⎜ − ⎜ ⎟⎟ = 0.077 mol OH− Moles of OH = ⎜⎜ ⎟⎟⎟ 0.350 L ⎜⎜⎜ ⎜⎝ L ⎠ ⎜⎝1 mol KOH ⎠⎟⎟ The Co2+ is limiting, thus, [Co2+] = 0, and the moles of OH− remaining are: [0.077 − 4(0.013118)] ⎡ 0.077 − 4 (0.013118)⎤ mol OH− ⎢ ⎦⎥ = 0.07008 M OH− [OH−] = ⎣ 0.350 L

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The Co will produce an equal number of moles of the complex with the concentration: 2− ⎞ 2 + ⎞⎛ ⎛ ⎜ 0.013118 mol Co ⎟⎟⎜⎜1 mol Co (OH)4 ⎟⎟ 2 [Co(OH)42−] = ⎜⎜ ⎟ = 0.03748 M Co(OH)4 − ⎟⎟⎜ ⎟⎠⎝⎜⎜ 1 mol Co 2 + ⎠⎟⎟ 0.350 L ⎜⎜⎝ Concentration (M) Co2+(aq) + 4OH−(aq) ⇆ Co(OH)42−(aq) Initial 0 0.07008 0.03748 Change +x +4x −x Equilibrium x 0.07008 + 4x 0.03748 − x Assume the −x and the +4x do not significantly change the associated concentrations. ⎡0.03748 − x ⎤ ⎡ Co (OH) 2− ⎤ [ 0.03748] ⎢⎣ ⎥⎦ 4 ⎢ ⎦⎥ = Kf = 5×109 = ⎣ 2 + 4 4 = 4 − [ x ][ 0.07008] ⎡ Co ⎤ [ OH ] [ x ] ⎡ 0.07008 + 4x ⎤ ⎣ ⎦ ⎣⎢ ⎦⎥ x = 3.1078×10−7 = 3.1×10−7 [Co2+] = 3.1×10−7 M Co2+ [Co(OH)42−] = 0.03748 − x = 0.037479689 = 0.037 M Co(OH)42− [OH−] = 0.07008 + 4 x = 0.070078756 = 0.070 M OH− 19.105 Plan: The NaOH will react with the benzoic acid, C6H5COOH, to form the conjugate base benzoate ion, C6H5COO−. Calculate the number of moles of NaOH and C6H5COOH. Set up a reaction table that shows the stoichiometry of the reaction of NaOH and C6H5COOH. All of the NaOH will be consumed to form C6H5COO−, and the number of moles of C6H5COOH will decrease. Find the new concentrations of C6H5COOH and C6H5COO− and use the Henderson-Hasselbalch equation to find the pH of this buffer. Once the pH of the benzoic acid/benzoate buffer is known, the Henderson-Hasselbalch equation can be used to find the ratio of formate ion and formic acid that will produce a buffer of that same pH. From the ratio, the volumes of HCOOH and NaOH are calculated. Solution: The Ka for benzoic acid is 6.3×10−5 (from the Appendix). The pKa is −log (6.3×10−5) = 4.201. The reaction of benzoic acid with sodium hydroxide is: C6H5COOH(aq) + NaOH(aq) → Na+(aq) + C6H5COO−(aq) + H2O(l) ⎛ 0.200 mol C6 H 5 COOH ⎞⎟⎛⎜10−3 L ⎞⎟ ⎟ ⎟⎜⎜ Moles of C6H5COOH = ⎜⎜⎜ ⎟⎟⎜ ⎟⎟⎟ 475 mL = 0.0950 mol C6H5COOH L 1 mL ⎝⎜ ⎠⎜ ⎝ ⎠

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⎛ 2.00 mol NaOH ⎞⎟⎛⎜10−3 L ⎞⎟ ⎟ ⎟⎟⎜⎜ Moles of NaOH = ⎜⎜⎜ ⎟⎟⎟ 25 mL = 0.050 mol NaOH ⎟⎜ L 1 mL ⎝⎜ ⎠⎜ ⎝ ⎠

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NaOH is the limiting reagent: The reaction table gives: C6H5COOH(aq) + NaOH(aq) → Na+(aq) + C6H5COO−(aq) + H2O(l) Initial 0.0950 mol 0.050 mol — 0 — Reacting −0.050 mol −0.050 mol + 0.050 mol Final 0.045 mol 0 mol 0.050 mol The concentrations after the reactions are: ⎛ 0.045 mol C H COOH ⎞⎟⎛ 1 mL ⎞ ⎜ ⎟ 6 5 ⎟⎟⎜⎜ ⎜ [C6H5COOH] = ⎜⎜ ⎟⎟⎜⎜ −3 ⎟⎟⎟ = 0.090 M C6H5COOH ⎜⎜ 475 + 25 mL ⎟⎝10 L ⎠⎟ ⎝ ⎠⎟⎜

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⎛ −⎞ ⎜ 0.050 mol C6 H 5 COO ⎟⎟⎛⎜ 1 mL ⎞⎟ − ⎟⎜ [C6H5COO−] = ⎜⎜⎜ ⎟⎟⎜⎜ −3 ⎟⎟⎟ = 0.10 M C6H5COO ⎟ 10 L ⎜⎜ 475 + 25 mL ⎠ ⎝ ⎠⎟⎟⎜⎝

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Calculating the pH from the Henderson-Hasselbalch equation:

⎛ [C H CO− ] ⎞⎟ ⎛ [ 0.10 ] ⎞⎟ ⎟ = 4.201 + log ⎜⎜ pH = pKa + log ⎜⎜⎜ 6 5 ⎟ = 4.24676 = 4.2 ⎜⎝[0.090]⎠⎟ ⎜⎝[C6 H 5 COOH]⎠⎟⎟ Calculations on formic acid (HCOOH) also use the Henderson-Hasselbalch equation. The Ka for formic acid is 1.8×10−4 and the pKa = −log (1.8×10−4) = 3.7447. The formate to formic acid ratio may now be determined: ⎛ [HCOO− ] ⎞⎟ pH = pKa + log ⎜⎜ ⎟ ⎜⎝[HCOOH]⎠⎟

⎛ [HCOO− ] ⎞⎟ 4.24676 = 3.7447 + log ⎜⎜ ⎟ ⎜⎝[HCOOH]⎠⎟ ⎛ [HCOO− ] ⎞⎟ ⎟ 0.50206 = log ⎜⎜ ⎜⎝[ HCOOH ]⎠⎟⎟ ⎛ [HCOO− ] ⎞⎟ ⎜⎜ ⎟ ⎜⎝[ HCOOH ]⎠⎟⎟ = 3.177313 [HCOO−] = 3.177313 [HCOOH] Since the conjugate acid and the conjugate base are in the same volume, the mole ratio and the molarity ratios are identical. Moles HCOO− = 3.177313 mol HCOOH The total volume of the solution is (500. mL)(10−3 L/1 mL) = 0.500 L Let Va = volume of acid solution added, and Vb = volume of base added. Thus: Va + Vb = 0.500 L The reaction between the formic acid and the sodium hydroxide is: HCOOH(aq) + NaOH(aq) → HCOONa(aq) + H2O(l) The moles of NaOH added equal the moles of HCOOH reacted and the moles of HCOONa formed. Moles NaOH = (2.00 mol NaOH/L)(Vb) = 2.00Vb mol Total moles HCOOH = (0.200 mol HCOOH/L)(Va) = 0.200Va mol The stoichiometric ratios in this reaction are all 1:1. Moles HCOOH remaining after the reaction = (0.200Va − 2.00Vb) mol Moles HCOO− = moles HCOONa = moles NaOH = 2.00 Vb

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Using these moles and the mole ratio determined for the buffer gives: Moles HCOO− = 3.177313 mol HCOOH 2.00Vb mol = 3.177313(0.200Va − 2.00Vb) mol 2.00Vb = 0.6354626Va − 6.354626Vb 8.354626 Vb = 0.6354626 Va The volume relationship given above gives Va = (0.500 − Vb) L. 8.354626 Vb = 0.6354626 (0.500 − Vb) 8.354626 Vb = 0.3177313 − 0.6354626 Vb 8.9900886 Vb = 0.3177313 Vb = 0.0353424 = 0.035 L NaOH Va = 0.500 − 0.0353424 = 0.4646576 = 0.465 L HCOOH Limitations due to the significant figures lead to a solution with only an approximately correct pH. 19.106 pKa = –log Ka = – log 6.3×10–8 = 7.200659. The Ka comes from Appendix C; it is Ka2 for phosphoric acid. ⎛ [HPO 4 2− ] ⎞⎟ ⎟ pH = pKa + log ⎜⎜ ⎜⎝[H 2 PO 4− ]⎠⎟⎟

⎛ [HPO 4 2− ] ⎞⎟ ⎟ 7.00 = 7.200659 + log ⎜⎜⎜ − ⎟ ⎝[H 2 PO 4 ]⎠⎟ ⎛ [HPO4 2− ] ⎞⎟ ⎟ –0.200659 = log ⎜⎜ ⎜⎝[H 2 PO4− ]⎠⎟⎟ [HPO 4 2− ] = 0.63000 [H 2 PO 4 − ] VHPO 42− = 0.63000 Since they are equimolar, VH2 PO4−

and VHPO42– + VH2PO4– = 100. mL so (0.63000)VH2PO4– + VH2PO4– = 100. mL VH2PO4– = 61 mL and VHPO42– = 39 mL 19.107 Plan: A formate buffer contains formate (HCOO−) as the base and formic acid (HCOOH) as the acid. The Henderson-Hasselbalch equation gives the component ratio, [HCOO−]/[HCOOH]. The ratio is used to find the volumes of acid and base required to prepare the buffer. Solution: From the Appendix, the Ka for formic acid is 1.8×10–4 and the pKa = −log (1.8×10–4) = 3.7447. ⎛ [HCOO− ] ⎞⎟ a) pH = pKa + log ⎜⎜ ⎟ ⎜⎝[HCOOH]⎠⎟

⎛ [HCOO− ] ⎞⎟ 3.74 = 3.7447 + log ⎜⎜ ⎟ ⎜⎝[HCOOH]⎠⎟ ⎛ [HCOO− ] ⎞⎟ ⎟⎟ −0.0047 = log ⎜⎜⎜ ⎝[ HCOOH ]⎠⎟ ⎛ [HCOO− ] ⎞⎟ ⎜⎜ ⎟ = 0.989236 = 0.99 ⎜⎝[ HCOOH ]⎠⎟⎟ b) To prepare solutions, set up equations for concentrations of formate and formic acid with x equal to the volume, in L, of 1.0 M HCOOH added. The equations are based on the neutralization reaction between HCOOH and NaOH that produces HCOO−.

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HCOOH(aq) + NaOH(aq) → HCOO−(aq) + Na+(aq) + H2O(l) ⎛ ⎞ − ⎜⎜ 0.700 − x L NaOH ⎟⎟⎛⎜1 mol HCOO ⎞⎟ – ⎟⎟ ⎟⎜ [HCOO ] = 1.0 M NaOH ⎜⎜ ⎟⎜ ⎜⎜ 0.700 L solution ⎟⎟⎟⎜⎜⎝ 1 mol NaOH ⎠⎟⎟ ⎝ ⎠

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⎛ ⎞ − ⎛ x L HCOOH ⎞⎟ ⎜ 0.700 − x L NaOH ⎟⎟⎜⎛1 mol HCOO ⎞⎟ ⎜ ⎟⎟ ⎟⎜ [HCOOH] = 1.0 M HCOOH ⎜⎜ ⎟⎟⎟ – 1.0 M NaOH ⎜⎜⎜ ⎟⎜ ⎜⎜⎝ 0.700 L solution ⎠⎟ ⎜⎜ 0.700 L solution ⎟⎟⎟⎜⎜⎝ 1 mol NaOH ⎠⎟⎟ ⎝ ⎠

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The component ratio equals 0.99 (from part a)). Simplify the above equations and plug into ratio: ⎡⎛0.700 − x ⎤ ⎞⎟ ⎢⎜⎜ M HCOO− ⎥ ⎟ − 0.700 ⎟ 0.700 − x ⎢⎣⎝⎜ ⎥⎦ [ HCOO ] ⎠ = = = 0.989236 ⎡⎛x − 0.700 − x ⎤ ⎞⎟ 2 x − 0.700 [ HCOOH ] ( ) ⎢⎜⎜ ⎥ ⎟ ⎢⎜ 0.700⎟⎟⎟ M HCOOH ⎥⎥ ⎢⎜⎝⎜ ⎟ ⎠ ⎢⎣ ⎥⎦ Solving for x: x = 0.46751 = 0.468 L Mixing 0.468 L of 1.0 M HCOOH and 0.700 − 0.468 = 0.232 L of 1.0 M NaOH gives a buffer of pH 3.74. c) The final concentration of HCOOH from the equation in part b): − ⎛ 0.468 L HCOOH ⎞⎟ ⎛ 0.232 L NaOH ⎞⎛ ⎟⎟⎜⎜1 mole HCOO ⎞⎟⎟ ⎜ ⎟⎟ – 1.0 M NaOH ⎜⎜ [HCOOH] = 1.0 M HCOOH ⎜⎜ ⎜⎜ ⎟⎟⎟⎜⎜⎜ 1 mole NaOH ⎟⎟⎟ ⎜⎜⎝ 0.700 L solution ⎠⎟⎟ ⎝⎜ 0.700 L solution ⎠⎝ ⎠

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= 0.33714 = 0.34 M HCOOH 19.108 This is because Ka depends on temperature (like all other equilibrium constants). In this case, since the pH drops as the temperature increases, Ka must increase with temperature, indicating that the dissociation reaction is endothermic. 19.109 H2SO4 is a strong acid and will be completely ionized: H2SO4(aq) + 2H2O(l) → SO42−(aq) + 2H3O+(aq). Calculate the moles of H3O+(aq) from the H2SO4 in the 8.0×103 lb of water and then the amount of sodium acetate trihydrate (NaC2H3O2 ∙3H2O) that will be required to neutralize that amount of H3O+(aq). ⎛ 1 kg ⎞⎟⎜⎛1000 g ⎞⎟ ⎟ = 3.628118×106 g H2O ⎟ Mass (g) of water = (8.0×103 lb water)⎜⎜ ⎜⎝ 2.205 lb ⎠⎝ ⎟⎜⎜ 1 kg ⎠⎟⎟ mass of solute ppm = ×10 6 mass of solution mass of H 2 SO4 ×106 10 ppm = 6 3.628118×10 g Mass (g) of H2SO4 = 36.28118 g + ⎛ 1 mol H 2 SO 4 ⎞⎛ ⎟⎟⎜⎜ 2 mol H3 O ⎞⎟⎟ = 0.7398 mol H O+ Moles of H3O+ = (36.28118 g H 2 SO 4 )⎜⎜⎜ 3 ⎟ ⎜ ⎟ 1 mol H 2 SO4 ⎠⎟⎟ ⎝ 98.08 g H 2 SO4 ⎠⎝

The reaction between H3O+ and the base sodium acetate is: H3O+(aq) + NaC2H3O2(aq) → H2O(l) + HC2H3O2(aq) + Na+(aq) Mass (lb) of NaC2H3O2 ∙3H2O required to neutralize the H2SO4 =

⎛1 mol NaC 2 H3O2 • 3H 2 O ⎟⎞⎛136.08 g NaC 2 H3O2 • 3H 2 O ⎞⎛ ⎞⎛ 1 kg ⎟⎟⎜⎜ ⎟⎟⎜ 2.205 lb ⎞⎟⎟ ⎟⎟⎜⎜ ⎜ ⎟ + ⎜ ⎜ ⎟⎠⎝ 1 mol NaC 2 H3O2 • 3H 2 O ⎠⎝ ⎟ 1000 g NaC 2 H3O2 • 3H 2 O ⎠⎟⎟⎜⎝ 1 kg ⎠⎟⎟ 1 mol H 3O ⎝

(0.7398 mol H3O+ )⎜⎜⎜

= 0.22198 lb NaC2H3O2 ∙3H2O

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Now consider the acetic acid. Calculate the amount of acetic acid in 8.0×103 lb or 3.628118×106 g H2O. ⎛ 0.015% ⎞⎟⎛⎜ 1 mol HC2 H 3 O2 ⎞⎟ ⎟ = 9.0627 mol ⎟⎜ Moles of acetic acid = (3.628118×106 g H 2 O)⎜⎜ ⎜⎝ 100% ⎠⎟⎜ ⎟⎜⎝ 60.05 g HC H O ⎠⎟⎟ 2

Find the amount of C2H3O2– necessary to maintain a pH of 5.

⎛ [C H O − ] ⎞ pH = pKa + log ⎜⎜⎜ 2 3 2 ⎟⎟⎟ ⎜⎝[HC 2 H3 O2 ]⎠⎟ ⎛ [C H O − ] ⎞ 5.0 = 4.7447 + log ⎜⎜ 2 3 2 ⎟⎟⎟ ⎜⎝[9.0627 mol]⎠ ⎛ [C H O − ] ⎞ 0.2553 = log ⎜⎜ 2 3 2 ⎟⎟⎟ ⎜⎝[9.0627 mol]⎠ 1.800 =

[C2 H3 O2− ] [9.0627 mol]

16.31286 mol of C2H3O2− (NaC2H3O2 ∙3H2O) will be required to maintain the pH. Mass (lb) of NaC2H3O2 ∙3H2O required =

⎛136.08 g NaC2 H3O2 • 3H2 O ⎞⎛ 1 kg ⎞⎛ 2.205 lb ⎞⎟ ⎟⎟⎜⎜ ⎟⎟⎜ ⎟ = 4.89478 lb ⎟⎟⎜⎜ 1 kg ⎠⎟⎟ ⎜⎝ 1 mol NaC2 H3O2 • 3H 2 O ⎠⎟⎟⎜⎝1000 g ⎠⎝

(16.31286 mol CH3COONa • 3H2 O)⎜⎜⎜

Total amount of NaC2H3O2 ∙3H2O required = 0.22198 lb + 4.89478 lb = 5.11676 = 5.1 lb 19.110 Plan: The minimum urate ion concentration necessary to cause a deposit of sodium urate is determined by the Ksp for the salt. Convert solubility in g/100. mL to molar solubility and calculate Ksp. Substituting [Na+] and Ksp into the ion-product expression allows one to find [Ur−]. Solution: Molar solubility of NaUr: ⎛ 0.085 g NaUr ⎞⎛ ⎟⎟⎜⎜ 1 mL ⎞⎛ ⎟⎜ 1 mol NaUr ⎞⎟⎟ ⎜ –3 [NaUr] = ⎜⎜ ⎟⎟⎜ −3 ⎟⎟⎟⎜⎜ ⎟ = 4.4713309×10 M NaUr ⎜⎜⎝ 100. mL ⎠⎝ ⎟⎜⎜10 L ⎠⎝ ⎟⎜⎜190.10 mol NaUr ⎠⎟⎟ 4.4713309×10–3 M NaUr = [Na+] = [Ur−] Ksp = [Na+][Ur−] = (4.4713309×10–3) (4.4713309×10–3) = 1.99927998×10–5 M When [Na+] = 0.15 M: Ksp = 1.99927998×10–5 M = [0.15][Ur−] [Ur−] = 1.33285×10–4 The minimum urate ion concentration that will cause precipitation of sodium urate is 1.3×10–4 M. 19.111 a) K = [CO2(aq)]/[CO2(g)] = 3.1×10–2 [CO2(aq)] = K[CO2(g)] = (3.1×10–2)(4×10–4) = 1.24×10–5 M = 1×10–5 M CO2 b) K = [Ca2+][HCO3–]2/[CO2(aq)] = (x)(2x)2/(1.24×10–5 – x) = 1×10–12 Neglect –x x = 1.458×10–6 = 1×10–6 M Ca2+ c) K = [CO2(aq)]/[CO2(g)] = 3.1×10–2 [CO2(aq)] = K[CO2(aq)] = (3.1×10–2)(2×4×10–4) = 2.48×10–5 M 2+ – 2 K = [Ca ][HCO3 ] /[CO2(aq)] = (x)(2x)2/(2.48×10–5 – x) = 1×10–12 Neglect –x x = 1.837×10–6 = 2×10–6 M Ca2+

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19.112 The buffer is made by starting with phosphoric acid and neutralizing some of the acid by adding sodium hydroxide: H3PO4(aq) + OH−(aq) → H2PO4−(aq) + H2O(l) Present initially is (0.50 L)(1.0 M H3PO4) = 0.50 mol H3PO4. Adding 0.80 mol NaOH converts all the phosphoric acid to dihydrogen phosphate ions (0.50 mol) and 0.30 mol NaOH are left. The remaining OH− will react with the dihydrogen phosphate: H2PO4−(aq) + OH−(aq) → HPO42−(aq) + H2O(l) The 0.50 mol H2PO4− reacts with the 0.30 mol OH− to produce 0.30 mol HPO42−. 0.20 mol H2PO4− will remain. The pH is determined from the equilibrium involving the conjugate pair HPO42−/H2PO4−. Ka = 6.3×10−8 H2PO4−(aq) + H2O(l) ⇆ HPO42−(aq) + H3O+(aq) ⎛ 0.30 mol HPO 4 2− ⎞⎟ ⎜⎜ ⎟⎟ ⎛ [HPO 4 2− ] ⎞⎟ ⎜ 0.50 L ⎟⎟ ⎜ −8 ⎟ ⎜ pH = pKa + log ⎜⎜ = −log (6.3×10 ) + log ⎜ − ⎟⎟⎟ = 7.37675 = 7.38 ⎜⎝[H 2 PO 4− ]⎠⎟⎟ ⎜⎜ 0.20 mol H 2 PO 4 ⎟ ⎜⎜ 0.50 L ⎠⎟⎟ ⎝ 19.113 Plan: Substitute the given molar solubility of KCl into the ion-product expression to find the Ksp of KCl. Determine the total concentration of chloride ion in each beaker after the HCl has been added. This requires the moles originally present and the moles added. Determine a Qsp value to see if Ksp is exceeded. If Qsp < Ksp, nothing will precipitate. Solution: a) The solubility equilibrium for KCl is: KCl(s) ⇆ K+(aq) + Cl−(aq) The solubility of KCl is 3.7 M. Ksp = [K+][Cl−] = (3.7)(3.7) = 13.69 = 14 b) Find the moles of Cl−: Original moles from the KCl: −3 ⎛ 3.7 mol KCl ⎞⎛ ⎛1 mol Cl− ion ⎞⎟ ⎟⎟⎜⎜10 L ⎞⎟⎟ ⎜ ⎜ ⎟⎟ = 0.37 mol Cl− Moles of K+ = moles of Cl− = ⎜⎜ ⎟⎟⎜ ⎟⎟ 100. mL ⎜⎜ 1L ⎟⎠⎝⎜⎜ 1 mL ⎠⎟ ⎜⎜⎝ ⎜⎝⎜ 1 mol KCl ⎠⎟⎟

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Original moles from the 6.0 M HCl in the first beaker: −3 ⎛ 6.0 mol HCl ⎞⎛ ⎛ 1 mol Cl− ⎞⎟ ⎟⎟⎜⎜10 L ⎞⎟⎟ ⎜ ⎜ ⎟⎟ = 0.60 mol Cl− Moles of Cl− = ⎜⎜ ⎟⎟⎜ ⎟⎟ 100. mL ⎜⎜ ⎜⎜⎝ ⎜ ⎜⎝⎜1 mol HCl ⎠⎟⎟ 1L ⎟⎠⎝⎜ 1 mL ⎠⎟

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This results in (0.37 + 0.60) mol = 0.97 mol Cl−. Original moles from the 12 M HCl in the second beaker: ⎛12 mol HCl ⎞⎛ ⎛ 1 mol Cl− ⎞⎟ 10−3 L ⎞⎟ ⎜ ⎟⎟⎟⎜⎜ ⎟⎟ 100. mL ⎜⎜ ⎟⎟ = 1.2 mol Cl− Moles of Cl− = ⎜⎜ ⎜⎜ ⎟⎟⎜⎜⎜ 1 mL ⎟⎟ ⎟ 1L 1 mol HCl ⎜⎜⎝ ⎜ ⎠⎝ ⎠ ⎝ ⎠⎟ This results in (0.37 + 1.2) mol = 1.57 mol Cl− Volume of mixed solutions = (100. mL + 100. mL)(10–3 L/1 mL) = 0.200 L After the mixing: [K+] = (0.37 mol K+)/(0.200 L) = 1.85 M K+ From 6.0 M HCl in the first beaker: [Cl−] = (0.97 mol Cl−)/(0.200 L) = 4.85 M Cl− From 12 M HCl in the second beaker: [Cl−] = (1.57 mol Cl−)/(0.200 L) = 7.85 M Cl− Determine a Qsp value to see if Ksp is exceeded. If Qsp < Ksp, nothing will precipitate. From 6.0 M HCl in the first beaker: Qsp = [K+][Cl−] = (1.85)(4.85) = 8.9725 = 9.0 < 14, so no KCl will precipitate.

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From 12 M HCl in the second beaker: Qsp = [K+][Cl−] = (1.85)(7.85) = 14.5225 = 15 > 14, so KCl will precipitate. The mass of KCl that will precipitate when 12 M HCl is added: Equal amounts of K and Cl will precipitate. Let x be the molarity change. Ksp = [K+][Cl−] = (1.85 – x)(7.85 – x) = 13.69 x = 0.08659785 = 0.09 This is the change in the molarity of each of the ions. ⎛ ⎞⎛ ⎞ ⎛ 0.08659785 mol K + ⎞⎟ ⎜1 mol KCl ⎟⎟⎜⎜ 74.55 g KCl ⎟⎟ Mass (g) of KCl = ⎜⎜⎜ ⎟ ⎟ = 1.291174 = 1 g KCl ⎟⎟⎟ 0.200 L ⎜⎜⎜ ⎜ + ⎟ L ⎟⎜⎜ 1 mol KCl ⎠⎟⎟ ⎜⎝ 1 mol K ⎠⎝ ⎝⎜ ⎠⎟

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19.114 [NH3] + [NH4+] = 0.15. If [NH3] = 0.01 M, then [NH4+] = 0.14 M. Kb = 1.76×10−5 (from the Appendix) Ka = Kw/Kb = 1.0×10−14/1.76×10−5 = 5.6818×10−10 pH = pKa + log [NH3]/[NH4+] pH = −log Ka + log [NH3]/[NH4+] pH = −log (5.6818×10−10) + log [0.01]/[0.14] pH = 8.099386 = 8.10 19.115 Determine the solubility of MnS: −4 ⎛ ⎞⎛ ⎞⎛ ⎜ 4.7 ×10 g MnS ⎟⎟⎜⎜ 1 mL ⎟⎟⎜⎜ 1 mol MnS ⎞⎟⎟ –5 S = ⎜⎜ ⎟⎟⎜ −3 ⎟⎟⎜ ⎟ = 5.401677968×10 M ⎜⎜ ⎜10 L ⎟⎜⎜ 87.01 g MnS ⎟⎟ 100 mL ⎟ ⎜ ⎠ ⎝ ⎠⎝ ⎠⎝ MnS(s) + H2O(l) ⇆ Mn2+(aq) + S2−(aq) Ksp = [Mn2+][S2–] = S2 = (5.401677968×10–5)2 = 2.9178×10–9 = 2.9×10–9 19.116 Plan: Use the Henderson-Hasselbalch equation to find the ratio of [HCO3–]/[H2CO3] that will produce a buffer with a pH of 7.40 and a buffer of 7.20. Solution: a) Ka1 = 4.5×10–7 pKa = – log Ka = – log (4.5×10–7) = 6.34679 ⎛[HCO3− ]⎞⎟ ⎟ pH = pKa + log ⎜⎜⎜ ⎜⎝ [H 2 CO3 ] ⎠⎟⎟

⎛[HCO3− ]⎞⎟ ⎟ 7.40 = 6.34679 + log ⎜⎜⎜ ⎜⎝ [H 2 CO3 ]⎠⎟⎟ ⎛[HCO3− ]⎞⎟ ⎟ 1.05321 = log ⎜⎜⎜ ⎜⎝ [H 2 CO3 ] ⎠⎟⎟ [HCO3− ] = 11.30342352 [H2 CO3 ] [H 2 CO3 ] = 0.0884688 = 0.088 [HCO3− ]

⎛[HCO3− ]⎞⎟ ⎟ pH = pKa + log ⎜⎜⎜ ⎜⎝ [H 2 CO3 ] ⎠⎟⎟ ⎛[HCO3− ]⎞⎟ ⎟ 7.20 = 6.34679 + log ⎜⎜⎜ ⎜⎝ [H 2 CO3 ]⎠⎟⎟ ⎛[HCO3− ]⎞⎟ ⎟ 0.85321 = log ⎜⎜⎜ ⎜⎝ [H 2 CO3 ] ⎠⎟⎟

19-67

[HCO3− ] = 7.131978 [H2 CO3 ] [H 2 CO3 ] = 0.14021 = 0.14 [HCO3− ]

19.117 Plan: The buffer components will be TRIS, (HOCH2)3CNH2, and its conjugate acid TRISH+, (HOCH2)3CNH3+. The conjugate acid is formed from the reaction between TRIS and HCl. Since HCl is the limiting reactant in this problem, the concentration of conjugate acid will equal the starting concentration of HCl, 0.095 M. The concentration of TRIS is the initial concentration minus the amount reacted. Once the concentrations of the TRISTRISH+ acid-base pair are known, the Henderson-Hasselbalch equation can be used to find the pH. Solution: ⎛ 1 mol TRIS ⎞⎟ ⎜ ⎟⎟ = 0.354961 mol Moles of TRIS = 43.0 g TRIS ⎜⎜ ⎜⎝⎜121.14 g TRIS ⎠⎟⎟ ⎛ 0.095 mol HCl ⎞⎟ ⎟⎟ 1.00 L = 0.095 mol HCl = mol TRISH+ Moles of HCl added = ⎜⎜⎜ ⎜⎝ L ⎠⎟ (HOCH2)3CNH2(aq) + HCl(aq) ⇆ (HOCH2)3CNH3+(aq) + Cl−(aq) Initial 0.354961 mol 0.095 mol 0 0 Reacting −0.095 mol −0.095 mol +0.095 mol — Final 0.259961 mol 0 mol 0.095 mol Since there is 1.00 L of solution, the moles of TRIS and TRISH+ equal their molarities. pKa of TRISH+ = 14 – pKb = 14 – 5.91 = 8.09 ⎛ [TRIS] ⎞⎟ ⎛[0.259961]⎞⎟ pH = pKa + log ⎜⎜ ⎟ = 8.09 + log ⎜⎜ ⎟ = 8.527185 = 8.53 ⎝⎜[TRISH+ ]⎠⎟ ⎝⎜ [0.095] ⎠⎟

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19.118

19.119

a) Since Ka (−COOH) > Kb (−NH2), the proton will be transferred from the −COOH to the −NH2, producing −COO− and −NH3+. b) pKa = −log Ka = −log (4.47×10−3) = 2.34969 ⎛ [+NH3 CH 2 COO− ] ⎞⎟ ⎟ pH = pKa + log ⎜⎜⎜ ⎜⎝[+NH3 CH 2 COOH]⎠⎟⎟

19-68

⎛ [+NH3 CH 2 COO− ] ⎞⎟ ⎟ 5.5 = 2.34969 + log ⎜⎜⎜ ⎜⎝[+NH3 CH 2 COOH]⎠⎟⎟

[ + NH3CH2 COO− ] = 1.41355×10 = 1×10 [ + NH3CH 2 COOH ] 3

H3N

O OH

H3N

H2N

CH2

NH3

NH2

pH = 7

pH = 13

pH = 1 d) at pH 1: D, at pH 7: A, at pH 13: B

19.120 a) The equilibrium is: MCl2(s) ⇆ M2+(aq) + 2Cl–(aq). The ion-product expression is Ksp = [M2+][Cl−]2. Based on the picture, the ion concentrations are: ⎛1.0 ×10−6 mol ⎞⎟ ⎟⎟ (3 spheres)⎜⎜⎜⎜ ⎝ 1 sphere ⎠⎟ ⎛⎜ 1 mL ⎞⎟ 2+ −5 [M ] = ⎜⎜ −3 ⎟⎟ = 1.2×10 M 250.0 mL ⎝10 L ⎠

[Cl−] =

⎛1.0 ×10−6 mol ⎞⎟ ⎟⎟ ⎝ 1 sphere ⎠⎟ ⎛⎜ 1 mL ⎞⎟ −5 ⎜⎜ −3 ⎟⎟ = 4.0×10 M 250.0 mL ⎝10 L ⎠

(10 spheres)⎜⎜⎜⎜

Ksp = [M2+][Cl−]2 = [1.2×10−5][ 4.0×10−5]2 = 1.92×10−14 = 1.9×10−14 b) M2+ is a common ion for M(NO3)2 and MCl2. If M(NO3)2 is added to the solution, [M2+] is increased and, according to Le Châtelier’s principle, the solubility equilibrium will shift to the left, precipitating more MCl2. The number of Cl– particles decreases, the mass of MCl2 increases, and the Ksp value remains the same. 19.121 The equilibrium is: Ca5(PO4)3OH(s) ⇆ 5Ca2+(aq) + 3PO43−(aq) + OH−(aq) Ksp = 6.8×10–37 = [Ca2+]5[PO43−]3[OH−] = (5S)5(3S)3(S) = 84375S9 S = 2.7166444×10–5 = 2.7×10–5 M Solubility = (2.7166444×10–5 mol/L)(502.32 g/mol) = 0.013646248 = 0.014 g/L Ca5(PO4)3OH The equilibrium is: Ca5(PO4)3F(s) ⇆ 5Ca2+(aq) + 3PO43−(aq) + F−(aq) Ksp = 1.0×10–60 = [Ca2+]5[PO43−]3[F−] = (5S)5(3S)3(S) = 84375S9 S = 6.1090861×10–8 = 6.1×10–8 M Solubility = (6.1090861×10–8 mol/L)(504.31 g/mol) = 3.0808732×10–5 = 3.1×10–5 g/L Ca5(PO4)3F Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

19-69

19.122 Plan: An indicator changes color when the buffer-component ratio of the two forms of the indicator changes from a value greater than 1 to a value less than 1. The pH at which the ratio equals 1 is equal to pKa. The midpoint in the pH range of the indicator is a good estimate of the pKa of the indicator. Solution: pKa = (3.4 + 4.8)/2 = 4.1 Ka = 10–4.1 = 7.943×10–5 = 8×10–5 19.123

[H3 O+]

0.1

0.05

0 0

mL NaOH added

Due to the large range of [H3O+], this plot is difficult to prepare and does not easily show the end point. A logarithmic scale (pH vs. mL OH– added) shows this more clearly. 19.124 Plan: A spreadsheet will help you to quickly calculate ΔpH/ΔV and average volume for each data point. At the equivalence point, the pH changes drastically when only a small amount of base is added, therefore, ΔpH/ΔV is at a maximum at the equivalence point. Solution: a) Example calculation: For the first two lines of data: ΔpH = 1.22 – 1.00 = 0.22; ΔV = 10.00 – 0.00 = 10.00 ΔpH 0.22 = = 0.022 ΔV 10.00

V(mL) pH 0.00 10.00 20.00 30.00 35.00 39.00 39.50 39.75 39.90 39.95 39.99 40.00 40.01 40.05

1.00 1.22 1.48 1.85 2.18 2.89 3.20 3.50 3.90 4.20 4.90 7.00 9.40 9.80

Vaverage(mL) = (0.00+ 10.00)/2 = 5.00 ΔpH ΔV

Vaverage(mL)

0.022 0.026 0.037 0.066 0.18 0.62 1.2 2.67 6 18 200 200 10

5.00 15.00 25.00 32.50 37.00 39.25 39.63 39.83 39.93 39.97 40.00 40.01 40.03

19-70

40.10 40.25 40.50 41.00 45.00 50.00 60.00 70.00 80.00

10.40 10.50 10.79 11.09 11.76 12.05 12.30 12.43 12.52

10 0.67 1.2 0.60 0.17 0.058 0.025 0.013 0.009

40.08 40.18 40.38 40.75 43.00 47.50 55.00 65.00 75.00

19.125 Check to see if the concentration of Ca(OH)2 exceeds the Ksp. M Ca(OH)2 = (6.5×10–9 mol Ca(OH)2)/(10.0 L) = 6.5×10–10 M Ca(OH)2 Determine the concentration of a saturated calcium hydroxide solution from the Ksp. Ca(OH)2(s) ⇆ Ca2+(aq) + 2OH–(aq) Ksp = 6.5×10–6 = [Ca2+][OH–]2 = (S)(2S)2 = 4S3

6.5×10−6 = 0.01175667 = 0.012 M 4 Thus, the solution is less than saturated so the Ksp does not affect the concentration of Ca(OH)2. M OH– from Ca(OH)2 = (6.5×10–10 M Ca(OH)2)(2 mol OH–/1 mol Ca(OH)2) = 1.3×10–9 M OH– Pure water has 1×10–7 M OH–, thus the contribution from the Ca(OH)2 is not significant. pH of pure water = 7.0. S= 3

19.126 Use HLac to indicate lactic acid and Lac− to indicate the lactate ion. The Henderson-Hasselbalch equation gives the pH of the buffer. Determine the final concentrations of the buffer components from MconcVconc = MdilVdil. Determine the pKa of the acid from the Ka. pKa = –log Ka = –log (1.38×10−4) = 3.86012 Determine the molarity of the diluted buffer component as Mdil = MconcVconc/Vdil. [HLac] = [(0.85 M) (225 mL)]/[(225 + 435) mL] = 0.28977 M HLac [Lac−] = [(0.68 M) (435 mL)]/[(225 + 435) mL] = 0.44818 M Lac− ⎛ [Lac− ] ⎞⎟ pH = pKa + log ⎜⎜ ⎟ ⎜⎝[HLac]⎠⎟

⎛[0.44818]⎞⎟ pH = 3.86012 + log ⎜⎜ ⎟ = 4.049519 = 4.05 ⎜⎝[0.28977]⎠⎟

19-71

19.127 The ion-product equilibrium reaction is: CaF2(s) ⇆ Ca2+(aq) + 2F–(aq) – F is a weak base with the following equilibrium reaction: F−(aq) + H2O(l) ⇆ HF(aq) + OH−(aq) (I) Pure water: There is no common-ion effect and the pH is neutral. (II) 0.01M HF: Because of the common-ion effect, less CaF2 would dissolve in this solution than in pure water. (III) 0.01M NaOH: Additional OH– ions shift the base equilibrium reaction to the left, producing more F−. The additional F− shifts the ion-product equilibrium to the left so less CaF2 would dissolve. (IV) 0.01M HCl: H+ ions remove OH– ions from solution so the base equilibrium reaction shifts to the right, consuming F−. This shifts the ion-product equilibrium to the right so that more CaF2 dissolves in this solution than in pure water. (V) 0.01M Ca(OH)2: Because of the common-ion effect, less CaF2 would dissolve in this solution than in pure water. Additional OH– ions shift the base equilibrium reaction to the left, producing more F−. The additional F− shifts the ion-product equilibrium to the left so less CaF2 would dissolve. a) 0.01M HCl b) 0.01M Ca(OH)2 19.128 a) The NaOH is a strong base, so it dissociates completely. OH–(aq) NaOH(s) → Na+(aq) + 0.050 mol 0.050 mol 0.050 mol The OH– ions from NaOH will react with HClO. HClO(aq) → H2O(l) + ClO–(aq) OH–(aq) + 0.050 mol 0.13 mol 0 –0.050 mol –0.050 mol +0.050 mol 0 0.080 mol 0.050 mol The initial amount of HClO is 0.13 mol – 0.050 mol = 0.08 mol The initial amount of ClO– is 0.050 mol ClO–. The volume of the solution is (500. mL)(10–3 L/1 mL) = 0.500 L [HClO]i = (0.080 mol HClO)/(0.500 L) = 0.16 M HClO [ClO–]I = (0.050 mol ClO–)/(0.500 L) = 0.10 M OCl– H3O+(aq) + ClO–(aq) Concentration (M) HClO(aq) + H2O ⇆ Initial 0.16 — 0.10 Change –x +x +x Equilibrium 0.16 – x x 0.10 + x x is small, so [HClO] ≈ 0.16 M HClO and [ClO–] ≈ 0.10 M ClO– [ H3O+ ][ ClO− ] Ka = [ HClO ] [H3O+] =

K a [ HClO ] [ ClO− ]

(3.0×10 )(0.16) −8

[H3O+] =

(0.10)

= 4.8×10–8 M H3O+

[OH–] = Kw/[H3O+] = (1.0×10–14)/(4.8×10–8) = 2.08333×10–7 = 2.1×10–7 M OH– [Na+] = (0.050 mol)/(0.500 L) = 0.10 M Na+ b) pH = –log [H3O+] = –log (4.8×10–8 M) = 7.31875876 = 7.32 c) If 0.0050 mol HCl is added then the ClO– will react with the H+ to form HClO. + ClO–(aq) → HClO(aq) H+(aq) 0.0050 mol 0.050 mol 0.080 mol –0.0050 mol –0.0050 mol +0.0050 mol 0 0.045 mol 0.085 mol Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

19-72

[HClO]i = (0.085 mol HClO)/(0.500 L) = 0.17 M HClO [ClO–]I = (0.045 mol ClO–)/(0.500 L) = 0.090 M OCl– Concentration (M) HClO(aq) + H2O ⇆ H3O+(aq) + ClO–(aq) Initial 0.17 — 0.090 Change –x +x +x Equilibrium 0.17 – x x 0.090 + x x is small, so [HClO] ≈ 0.17 M and [ClO–] ≈ 0.090 M K [ HClO ] [H3O+] = a [ ClO− ]

(3.0×10 )(0.17) −8

[H3O+] =

(0.090)

= 5.6667×10–8 M H3O+

pH = –log [H3O+] = –log (5.6667×10–8 M) = 7.246669779 = 7.25 19.129 In both cases the equilibrium is: CaCO3(s) ⇆ Ca2+(aq) + CO32–(aq) Ksp = [Ca2+][CO32–] = S2 At 10°C Ksp = 4.4×10–9 = [Ca2+][CO32–] = S2 S = 6.6332496×10–5 = 6.6×10–5 M CaCO3 At 30°C Ksp = 3.1×10–9 = [Ca2+][CO32–] = S2 S = 5.5677644×10–5 = 5.6×10–5 M CaCO3 19.130 Hg2C2O4(s) ⇆ Hg22+(aq) + C2O42–(aq) Ksp = 1.75×10–13 = [Hg22+][C2O42–] = (0.13 + S)S ≈ (0.13) S S = 1.3461538×10–12 = 1.3×10–12 M 19.131 [H+] = 10–9.5 = 3.16227766×10–10 M H+ pOH = 14.0 – pH = 14.0 – 9.5 = 4.5 [OH–] = 10–4.5 = 3.16227766×10–5 M OH– − ⎛ 65.0 mg HCO3− ⎟⎞⎛⎜10−3 g ⎞⎛ ⎟⎟⎜⎜ 1 mol HCO3 ⎞⎟⎟ ⎟⎟⎜⎜ [HCO3–] = ⎜⎜⎜ = 1.0652245×10–3 M HCO3– ⎟ ⎟⎠⎜⎜ 1 mg ⎟⎟⎜⎜⎜ 61.02 g HCO3− ⎟⎟⎟ L ⎝⎜ ⎝ ⎠⎝ ⎠ 2 − ⎞⎛ −3 ⎞⎛ 2− ⎞ ⎛ ⎜ 26.0 mg CO3 ⎟⎟⎜⎜10 g ⎟⎟⎜⎜ 1 mol CO3 ⎟⎟ [CO32–] = ⎜⎜ = 4.3326112×10–4 M CO32– ⎟⎟⎜ ⎟⎟⎜ 2− ⎟ L ⎟⎠⎝⎜⎜ 1 mg ⎠⎝ ⎜⎝ ⎟⎜⎜ 60.01 g CO3 ⎠⎟⎟ Alkalinity = [HCO3–] + 2[CO32–] + [OH–] – [H+] Alkalinity = (1.0652245×10–3) + 2(4.3326112×10–4) + (3.16227766×10–5) – (3.16227766×10–10) Alkalinity = 1.9633692×10–3 = 1.96×10–3 M

19.132 Plan: To determine which species are present from a buffer system of a polyprotic acid, check the pKa values for the one that is closest to the pH of the buffer. The two components involved in the equilibrium associated with this Ka are the principle species in the buffer. Use the Henderson-Hasselbalch equation to find the ratio of the phosphate species that will produce a buffer with a pH of 7.4. Solution: For carbonic acid, pKa1 [−log (8×10−7) = 6.1] is closer to the pH of 7.4, so H2CO3 and HCO3− are the species present. For phosphoric acid, pKa2 [−log (2.3×10−7) = 6.6] is closest to the pH, so H2PO4− and HPO42− are the principle species present. ⇆ HPO42−(aq) + H3O+(aq) H2PO4−(aq) + H2O(l) Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

19-73

⎛ [HPO 4 2− ] ⎞⎟ ⎟ pH = pKa + log ⎜⎜ ⎜⎝[H 2 PO 4− ]⎠⎟⎟ ⎛ [HPO 4 2− ] ⎞⎟ ⎟ 7.4 = 6.6383 + log ⎜⎜ ⎜⎝[H 2 PO 4− ]⎠⎟⎟

⎡ HPO4 2− ⎤ ⎣ ⎦ = 5.77697 = 5.8 [ H2 PO4− ] 19.133 a) Combine the separate equilibria to produce the desired equilibrium. The K values are in the Appendix. K' = (Ksp)2 = (1.8×10–10)2 = 3.24×10–20 2AgCl(s) ⇆ 2Ag+(aq) + 2Cl–(aq) + 2– 2Ag (aq) + CrO4 (aq) ⇆ Ag2CrO4(s) K" = 1/Ksp = 1/(2.6×10–12) = 3.8462×1011 – 2– 2AgCl(s) + CrO4 (aq) ⇆ Ag2CrO4(s) + 2Cl (aq) K = K'K" = 1.2462×10–8 = 1.2×10–8 b) Since the above reaction has such a small K, it lies far to the left as written. c) The mixing of equal amounts of equal molar solutions would precipitate all the AgCl, thus the silver ion concentration comes entirely from the Ksp of AgCl. Ksp = 1.8×10–10 = [Ag+][Cl–] = S2 S = [Ag+] = 1.34164×10–5 M = 1.3×10–5 M Ag+ Use the Ksp for silver chromate. Ksp = 2.6×10–12 = [Ag+]2[CrO42–] [CrO42–] = (2.6×10–12)/(1.34164×10–5)2 = 0.01444 = 0.014 M If the chromate ion concentration exceeds 0.014 M, Ag2CrO4 will precipitate. 19.134 Plan: Find the moles of quinidine initially present in the sample by dividing its mass in grams by the molar mass. Use the molar ratio between quinidine and HCl to find the moles of HCl that would react with the moles of quinidine and subtract the reacted HCl from the initial moles of HCl to find the excess. Use the molar ratio between HCl and NaOH to find the volume of NaOH required to react with the excess HCl. Then use the molar ratio between NaOH and quinidine to find the volume of NaOH required to react with the quinidine. Solution: a) To find the concentration of HCl after neutralizing the quinidine, calculate the concentration of quinidine and the amount of HCl required to neutralize it, remembering that the mole ratio for the neutralization is 2 mol HCl/1 mol quinidine. ⎛10−3 g ⎞⎛ 1 mol quinidine ⎞⎟ ⎜ –4 ⎟⎟⎟⎜⎜ ⎟ Moles of quinidine = 33.85 mg quinidine ⎜⎜ ⎟⎟⎜⎜⎜ 324.41 g quinidine ⎟⎟⎟ = 1.0434327×10 mol quinidine ⎜⎜⎝ 1 mg ⎠⎝ ⎠

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⎛10−3 L ⎞⎟⎛ 0.150 mol HCl ⎞ ⎜ ⎟⎟ ⎟⎟⎜⎜ Moles of HCl excess = 6.55 mL ⎜⎜ ⎟ ⎜⎝⎜ 1 mL ⎠⎟⎜ ⎟⎜⎝ L ⎠⎟

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⎛ 2 mol HCl ⎞⎟ ⎜ ⎟⎟ = 7.7381346×10–4 mol HCl – 1.0434327×10−4 mol quinidine ⎜⎜ ⎜⎜⎝1 mol quinidine ⎠⎟⎟ ⎛1 mol NaOH ⎞⎛ ⎞⎟⎜⎛ 1 mL ⎞⎟ 1L ⎟⎟⎜⎜ ⎜ ⎟ ⎟⎟⎜⎜ Volume (mL) of NaOH needed = 7.7381346 ×10−4 mol HCl ⎜⎜ ⎟⎟⎜ ⎟⎟⎜10−3 L ⎟⎟⎟ ⎟⎜⎜ 0.0133 mol NaOH ⎠⎝ ⎜ ⎝⎜⎜ 1 mol HCl ⎠⎝ ⎠ = 58.18146 = 58.2 mL NaOH solution b) Use the moles of quinidine and the concentration of the NaOH to determine the milliliters. ⎛ 1 mol NaOH ⎞⎛ ⎞⎟⎜⎛ 1 mL ⎞⎟ 1L ⎟⎟⎜⎜ ⎜ Volume = 1.0434327×10−4 mol quinidine ⎜⎜ ⎟⎟⎜ ⎟⎟⎟⎜⎜ −3 ⎟⎟⎟ ⎜⎜⎝1 mol quinidine ⎠⎝ ⎟⎜⎜ 0.0133 mol NaOH ⎠⎝ ⎟⎜⎜10 L ⎠⎟ = 7.84536 = 7.85 mL NaOH solution

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19-74

c) When quinidine (QNN) is first acidified, it has the general form QNH+NH+. At the first equivalence point, one of the acidified nitrogen atoms has completely reacted, leaving a singly protonated form, QNNH+. This form of quinidine can react with water as either an acid or a base, so both must be considered. If the concentration of quinidine at the first equivalence point is greater than Kb1, then the [OH–] at the first equivalence point can be estimated as: [OH–] = Kb1Kb2 = (4.0×10−6 )(1.0×10−10 ) = 2.0×10–8 M [H3O+] = Kw/[OH–] = (1.0×10–14)/(2.0×10–8) = 5.0×10–7 M pH = –log [H3O+] = –log (5.0×10–7 M) = 6.3010 = 6.30 19.135

K values from the Appendix: Ka1 = 5.6×10–2 H2C2O4(aq) ⇆ H+(aq) + HC2O4–(aq) + – 2– HC2O4 (aq) ⇆ H (aq) + C2O4 (aq) Ka2 = 5.4×10–5________ + 2– H2C2O4(aq) ⇆ 2H (aq) + C2O4 (aq) K = Ka1Ka2 = 3.024×10–6 + 2 2– K = [H ] [C2O4 ]/[H2C2O4] [C2O42–] = K[H2C2O4]/[H+]2 a) At pH = 5.5: [H+] = 10–5.5 = 3.162×10–6 M [C2O42–] = K[H2C2O4]/[H+]2 = (3.024×10–6)(3.0×10–13)/(3.162×10–6)2 [C2O42–] = 9.07359×10–8 M 2+ Q = [Ca ][C2O42–] Q = (2.6×10–3)(9.07359×10–8) = 2.3591×10–10 = 2.4×10–10 < Ksp = No precipitate b) At pH = 7.0: [H+] = 10–7.0 = 1×10–7 M [C2O42–] = K[H2C2O4]/[H+]2 = (3.024×10–6)(3.0×10–13)/(1×10–7)2 [C2O42–] = 9.072×10–5 M Q = (2.6×10–3) (9.072×10–5) = 2.35872×10–7 = 2.4×10–7 > Ksp = Precipitate forms c) The higher pH would favor precipitation.

19.136 Plan: The Henderson-Hasselbalch equation demonstrates that the pH changes when the ratio of acid to base in the buffer changes (pKa is constant at a given temperature). Solution: ⎛ [A− ] ⎞⎟ pH = pKa + log ⎜⎜ ⎟ ⎜⎝[HA]⎠⎟ The pH of the A−/HA buffer cannot be calculated because the identity of “A” and, thus, the value of pKa are unknown. However, the change in pH can be described: ⎛ [A− ] ⎞⎟ ⎛ [A− ] ⎞⎟ Δ pH = log ⎜⎜ ⎟⎟ − log ⎜⎜ ⎟ ⎜⎝[HA]⎠ ⎝⎜[HA]⎠⎟ final

initial

⎛ [A− ] ⎞⎟ Since both [HA] and [A−] = 0.10 M, log ⎜⎜ ⎟ ⎜⎝[HA]⎠⎟

= 0 because [HA] = [A−], and log (1) = 0

initial

So the change in pH is equal to the concentration ratio of base to acid after the addition of H3O+. Consider the buffer prior to addition to the medium. A–(aq) → HA(aq) H3O+ (aq) + 0.0010 mol 0.10 mol 0.10 mol –0.0010 mol –0.0010 mol +0.0010 mol 0 0.099 mol 0.101 mol When 0.0010 mol H3O+ is added to 1 L of the undiluted buffer, the [A−]/[HA] ratio changes from 0.10/0.10 to (0.099)/(0.101). The change in pH is: ΔpH = log (0.099/0.101) = –0.008686 If the undiluted buffer changes 0.009 pH units with addition of 0.0010 mol H3O+, how much can the buffer be diluted and still not change by 0.05 pH units (ΔpH < 0.05)?

19-75

Let x = fraction by which the buffer can be diluted. Assume 0.0010 mol H3O+ is added to 1 L. ⎛ 0.10x − 0.0010 ⎞⎟ ⎜ [ base ] ⎟⎟ = log ⎜⎜⎜ log ⎟⎟ = –0.05 [acid ] ⎜⎜⎝ 0.10x + 0.0010 ⎠⎟⎟

( ( ⎛ 0.10x − 0.0010 ⎞⎟ ⎜⎜ ( ) ⎟⎟ = 10 ⎜⎜ ⎟ ⎜⎜(0.10x + 0.0010) ⎟⎟⎟ ⎝ ⎠

) )

–0.05

= 0.89125

0.10x – 0.0010 = 0.89125 (0.10x + 0.0010) x = 0.173908 = 0.17 The buffer concentration can be decreased by a factor of 0.17, or 170 mL of buffer can be diluted to 1 L of medium. At least this amount should be used to adequately buffer the pH change.

19.137

a) Ka = 6.8×10–4 =

[ H3O+ ][ F− ] [ HF ]

x2 x2 ≈ 0.2500 − x 0.2500

x = [H3O ] = 0.0130384 M pH = –log [H3O+] = –log (0.0130384 M) = 1.8847757 = 1.88 ⎛10−3 L ⎞⎛ ⎞⎛ 1L ⎟⎟⎜⎜ 0.2500 mol HF ⎞⎛ ⎟⎟⎜⎜1 mol NaOH ⎞⎛ ⎟⎟⎜⎜ ⎟⎟⎜⎜ 1 mL ⎞⎟⎟ ⎜ b) Volume (mL) of Na = 35.00 mL ⎜⎜ ⎟⎟⎜ ⎟⎟⎜ ⎟⎟⎜ ⎟⎟⎜ −3 ⎟⎟ ⎜⎜⎝ 1 mL ⎠⎝ 1L ⎟⎜⎜ ⎟⎜⎜ 1 mol HF ⎠⎝ ⎟⎜⎜ 0.1532 mol NaOH ⎠⎝ ⎟⎜⎜10 L ⎠⎟ ⎠⎝

(

)

= 57.11488 = 57.11 mL NaOH ⎛10−3 L ⎞⎛ ⎟⎟⎜⎜ 0.2500 mol HF ⎞⎟⎟ ⎜ –3 c) Moles of HF (initial) = 35.00 mL ⎜⎜ ⎟⎟⎜ ⎟⎟ = 8.750×10 mol HF 1L ⎟⎜⎜ ⎟⎠ ⎜⎜⎝ 1 mL ⎠⎝ ⎛10−3 L ⎞⎛ 0.1532 mol NaOH ⎞⎟ ⎜ ⎟⎟⎟⎜⎜ ⎟⎟ = 8.6734×10–3 mol NaOH Moles of NaOH added = 57.11488 − 0.50 mL ⎜⎜ ⎟⎟⎜⎜⎜ ⎟ 1L ⎜⎜⎝ 1 mL ⎠⎝ ⎠⎟ Moles of F– formed = moles NaOH Moles of HF remaining = (8.750×10–3 – 8.6734×10–3) mol = 7.66×10–5 mol HF Volume of solution = (35.00 + 57.11488 – 0.50)(10–3 L/1 mL) = 0.091615 L [HF] = (7.66×10–5 mol HF)/(0.091615 L) = 0.00083611 M HF [F–] = (8.6734×10–3 mol F–)/(0.091615 L) = 0.09467 M F– pKa = –log Ka = –log (6.8×10–4) = 3.1675 ⎛ [F− ] ⎞⎟ pH = pKa + log ⎜⎜ ⎟ ⎜⎝[HF]⎠⎟

(

)

((

) )

⎛ [0.09467] ⎞⎟ pH = 3.1675 + log ⎜⎜ ⎟ ⎜⎝[0.0008361]⎠⎟ pH = 5.2215 = 5.22 d) At this point there are 8.750×10–3 mol of F– in (35.00 + 57.11488) mL of solution. ⎛ 8.750 ×10−3 mol F− ⎞⎟⎛ 1 mL ⎞ ⎜ ⎟ ⎟⎟⎜⎜ – – ⎜ Molarity of F = ⎜⎜ ⎟⎟⎟ = 0.09499 M F ⎟⎜ −3 ⎟ ⎜ ⎜⎜ 35.00 + 57.11488 mL ⎟⎝ 10 L ⎠⎟ ⎟ ⎜ ⎝ ⎠

(

)

–14

Kb = Kw/Ka = (1.0×10 )/(6.8×10 ) = 1.470588×10–11 Kb = 1.470588×10–11 =

–4

[ HF ][ OH− ]

[ F− ]

x2 x2 ≈ 0.09499 − x 0.09499

x = [OH–] = 1.1819×10–6 M Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

19-76

[H3O+] = Kw/[OH–] = (1.0×10–14)/(1.1819×10–6) = 8.4609527×10–9 M pH = –log [H3O+] = –log (8.4609527×10–9 M) = 8.07258 = 8.07 ⎛10−3 L ⎞⎛ ⎟⎟⎜⎜ 0.1532 mol NaOH ⎞⎟⎟ ⎜ –5 e) Moles of NaOH excess = 0.50 mL ⎜⎜ ⎟⎟⎜ ⎟⎟ = 7.66×10 mol NaOH ⎜⎜⎝ 1 mL ⎠⎝ 1L ⎟⎜⎜ ⎟⎠ Volume of solution = (35.00 + 57.11488 + 0.50)(10–3 L/1 mL) = 0.092615 L [OH–] = (7.66×10–5 mol F–)/(0.092615 L) = 8.271×10–4 M OH– The excess OH– will predominate and essentially control the pH. [H3O+] = Kw/[OH–] = (1.0×10–14)/(8.271×10–4) = 1.2090436×10–11 M pH = –log [H3O+] = –log (1.2090436×10–11 M) = 10.917558 = 10.92

(

)

19.138 a) The formula is Hg2Cl2 which simplifies to the empirical formula HgCl. Ksp = 1.5×10–18 b) The equilibrium is: Hg2Cl2(s) ⇆ Hg22+(aq) + 2Cl−(aq) 2 2 3 –18 2+ Ksp = [Hg2 ][Cl−] = (S)(2S) = 4S = 1.5×10 S = 7.2112479×10–7 = 7.2×10–7 M c) [Hg22+] = Ksp/[Cl−]2 = 1.5 ×10−18 ⎡⎛ ⎤ ⎞⎛ 1 kg ⎞⎟⎛⎜103 g ⎞⎟⎛⎜ 1 mol NaCl ⎞⎟⎛⎜ 1 mol Cl− ⎞⎟⎛⎜1 gal ⎞⎛ ⎟⎟⎜⎜1.057 qt ⎞⎟⎟⎥ ⎢⎜⎜ 0.20 lb NaCl ⎟⎜ ⎟⎟⎜ ⎟⎟⎜ ⎟⎟⎜ ⎟⎟⎜ ⎟⎟⎜⎜ ⎟ ⎟ ⎢⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎟ ⎟ ⎟ ⎟ ⎟⎟⎜⎜ 1 L ⎟⎟⎥⎥ ⎟ gal ⎢⎜ ⎠⎟⎜⎝⎜ 2.205 lb ⎠⎟⎝⎜⎜ 1 kg ⎠⎟⎝⎜⎜ 58.44 g NaCl ⎠⎟⎝⎜⎜1 mol NaCl ⎠⎟⎝⎜⎜ 4 qt ⎠⎝ ⎠⎦ ⎣⎝ –18 –18 2+ = 8.9174129×10 = 8.9×10 M Hg2

d) Use the value of S for a saturated solution (see part b)). 3 ⎛ 7.2112479×10−7 mol Hg Cl ⎞⎛ ⎞⎛ 3 ⎞ ⎛ 472.1 g Hg Cl ⎞⎟ ⎜⎜ ⎜⎜ 1 L ⎟⎟⎜⎜10 m ⎟⎟ 2 2⎟ 2 2 ⎜ ⎟ ⎟⎟ Mass (g) of Hg2Cl2 = ⎜ ⎟⎟⎜ −3 3 ⎟⎟⎜ ⎟⎟ 4900 km 3 ⎜⎜ ⎟⎟ ⎜⎜ ⎜ ⎜ ⎜ L 1 km 1 mol Hg Cl 10 m ⎜ ⎟ ⎟ ⎟ ⎜ ⎜ 2 2 ⎠ ⎝ ⎝ ⎠⎝ ⎠⎝ ⎠

(

)

= 1.6681708×1012 = 1.7×1012 g Hg2Cl2 e) Use the value determined in part c). Mass (g) of Hg2Cl2 = −18 2 + ⎞⎛ ⎛ ⎞⎛ ⎞⎛ 3 ⎞ ⎛ ⎞ ⎜⎜ 8.9174129×10 mol Hg 2 ⎟⎟⎜⎜1 mol Hg2 Cl 2 ⎟⎟⎜⎜ 1 L ⎟⎟⎜⎜10 m ⎟⎟ ⎜ 472.1 g Hg2 Cl 2 ⎟⎟ ⎟ ⎟ ⎟⎟⎜ ⎟⎟ 4900 km 3 ⎜⎜ ⎟ ⎜⎜ ⎜⎜ ⎟ − 2 + ⎟⎜ 3 3 ⎜⎜⎝ 1 mol Hg2 Cl 2 ⎠⎟⎟ ⎟⎜ 1 mol Hg 2 ⎠⎝ ⎟⎜⎜10 m ⎠⎝ ⎟⎜⎜ 1 km ⎠⎟ L ⎜⎝ ⎠⎝ 3

(

)

= 20.62856 = 21 g Hg2Cl2 19.139 a) CaF2 with Ksp = 3.2×10–11 will precipitate before BaF2 with Ksp = 1.5×10–6. b) Add KF until [F –] is such that the CaF2 precipitates but just lower than the concentration required to precipitate BaF2. c) Determine the barium concentration after mixing from MconcVconc = MdilVdil. [Ba2+] = [(0.090 M)(25.0 mL)]/[(25.0 + 35.0) mL] = 0.0375 M Use the barium ion concentration and the Ksp to find the fluoride ion concentration. [F–]2 = Ksp/[Ba2+] = (1.5×10–6)/(0.0375) = 4.0×10–5 M [F –] = 6.324555×10–3 = 6.3×10–3 M or less 19.140 Plan: Use the ideal gas law to calculate the moles of CO2 dissolved in water. Use the Ka expression for H2CO3 to find the [H3O+] associated with that CO2 concentration. Solution: Carbon dioxide dissolves in water to produce H3O+ ions: CO2(g) ⇆ CO2(aq) CO2(aq) + H2O(l) ⇆ H2CO3(aq) H2CO3(aq) ⇆ H3O+(aq) + HCO3−(aq) Ka1 = 4.5×10–7 Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

19-77

The molar concentration of CO2, [CO2], depends on how much CO2(g) from the atmosphere can dissolve in pure water. Since air is not pure CO2, account for the volume fraction of air (0.040 L/100 L) when determining the moles. ⎛10-3 L ⎞⎟⎛ 0.040% ⎞ ⎟ = 3.520×10–5 L CO2 ⎟⎜ Volume (L) of CO2 = (88 mL) ⎜⎜ ⎜⎝ 1 mL ⎠⎟⎟⎜⎜⎝ 100% ⎠⎟⎟

(1 atm)(3.520 ×10−5 L) PV = = 1.438743×10–6 mol CO2 Moles of dissolved CO2 = ⎛ ⎞ RT ⎜⎜0.0821 L • atm ⎟⎟((273+25) K ) ⎜⎝ mol • K ⎠⎟ [CO2] = (1.438743×10–6 mol CO2)/[(100 mL)(10–3 L/1 mL)] = 1.438743×10–5 M CO2 –7

Ka1 = 4.5×10 =

[ H3O+ ][ HCO3− ] [ H2 CO3 ]

[ H3O+ ][ HCO3− ] [CO2 ]

Let x = [H3O+] = [HCO3−]

⎡ x⎤ ⎡x⎤ ⎣ ⎦⎣ ⎦ Assume that x is small compared to 1.438743×10–5 ⎡1.438743×10 –5 − x ⎤ ⎢⎣ ⎥⎦ ⎡x⎤ ⎡x⎤ ⎣ ⎦⎣ ⎦ 4.5×10–7 = ⎡1.438743×10 –5 ⎤ ⎢⎣ ⎥⎦ –6 x = 2.544473×10 Check assumption that x is small compared to 1.438743×10–5: 2.544473×10−6 (100) = 18% error, so the assumption is not valid. 1.438743×10−5 Since the error is greater than 5%, it is not acceptable to assume x is small compared to 1.438743×10–5, and it is necessary to use the quadratic equation. ⎡x⎤ ⎡x⎤ ⎣ ⎦⎣ ⎦ 4.5×10–7 = ⎡1.438743×10 –5 − x ⎤ ⎢⎣ ⎥⎦ 2 –7 –12 x + 4.5×10 x – 6.474344×10 = 0

4.5×10–7 =

a=1 x=

b = 4.5×10–7

c = –6.474344×10–12

−b ± b 2 − 4ac 2a

−4.5×10−7 ± (4.5×10−7 ) − 4(1)(–6.474344 ×10−12 ) 2

2(1) x = 2.329402×10–6 M = [H3O+] pH = –log (2.329402×10–6) = 5.632756 = 5.63 19.141 a) For H2CO3, pKa = –log Ka pKa1 = –log 4.5×10–7 = 6.3468 pKa2 = –log 4.7×10–11 = 10.3279 pKa1 = 6.35 and pKa2 = 10.33. Since pKa1 > pH < pKa2, the base in the first dissociation (HCO3–) and the acid in the second dissociation (also HCO3–) will predominate. b) pH = pKa + log [base]/[acid] 8.5 = 6.35 + log [HCO3–]/[H2CO3] [HCO3–]/[H2CO3] = 1.4125×102 = 1×102 M pH = pKa + log [base]/[acid] 8.5 = 10.33 + log [CO32–]/[HCO3–] [CO32–]/[HCO3–] = 1.4791×10–2 = 1×10–2 M Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

19-78

c) In deep water, animals can exist but plants, which depend on light for photosynthesis, cannot. Photosynthesis converts carbon dioxide to oxygen; animals convert oxygen to carbon dioxide. Near the surface, plants remove carbon dioxide (which, in water, can be represented as the weak acid H2CO3) and thus the pH is higher than in deep water, where higher concentrations of carbon dioxide (H2CO3) accumulate. Also, at greater depths, the pressure is higher and so is the concentration of CO2 (Henry’s law). 19.142 Initial concentrations of Pb2+ and Ca(EDTA)2− before reaction based on mixing 100. mL of 0.10 M Na2Ca(EDTA) with 1.5 L blood: −6 2+ ⎛120 μg Pb 2+ ⎞⎛ ⎟⎟⎜⎜ 1 mL ⎞⎛ ⎟⎜ 1.5 L blood ⎞⎛ ⎟⎟⎜⎜10 g ⎞⎛ ⎟⎟⎜⎜ 1 mol Pb ⎞⎟⎟ ⎜ –6 2+ [Pb2+] = ⎜⎜ ⎟⎟⎜ −3 ⎟⎟⎟⎜⎜ ⎟⎟⎜ ⎟⎟⎜ ⎟ = 5.42953668×10 M Pb ⎟⎜⎜10 L ⎠⎝ ⎟⎜⎜1.6 L mixture ⎠⎝ ⎟⎜⎜ 1 μg ⎠⎝ ⎟⎜⎜ 207.2 g Pb 2+ ⎠⎟⎟ ⎜⎝⎜ 100 mL ⎠⎝ MconcVconc = MdilVdil

[Ca(EDTA)2–] = MconcVconc/Vdil = [(0.10 M (100 mL)(10–3 L/1 mL)]/(1.6 L) = 6.25×10–3 M Set up a reaction table assuming the reaction goes to completion: Concentration (M) [Ca(EDTA)]2−(aq) + Pb2+(aq) ⇆ [Pb(EDTA)]2−(aq) + Ca2+(aq) –3 –6 Initial 6.25×10 5.42953668×10 0 0 React –5.42953668×10–6 –5.42953668×10–6 +5.42953668×10–6 +5.42953668×10–6 6.24457×10–3 0 5.42953668×10–6 5.42953668×10–6 Now set up a reaction table for the equilibrium process: Concentration (M) [Ca(EDTA)]2−(aq) + Pb2+(aq) ⇆ [Pb(EDTA)]2−(aq) + Ca2+(aq) –3 Initial 6.24457×10 0 5.42953668×10–6 5.42953668×10–6 Change +x +x –x –x Equilibrium 6.24457×10–3 + x x 5.4295366×10–6 – x 5.4295366×10–6 – x

⎡5.42953668×10−6 ⎤ ⎡5.42953668×10−6 ⎤ ⎡ Pb (EDTA)2− ⎤ ⎡ Ca 2+ ⎤ ⎢ ⎥⎦ ⎢⎣ ⎥⎦ ⎢⎣ ⎥⎦ ⎣ ⎦ Kc = 2.5×10 = = ⎣ ⎡6.24457×10−3 ⎤ ⎡ x⎤ ⎡ Ca (EDTA)2− ⎤ ⎡ Pb2+ ⎤ ⎦ ⎣⎢ ⎦⎥ ⎣ ⎣⎢ ⎦⎥ ⎣ ⎦ 2+ –16 x = [Pb ] = 1.8883522×10 M −3 ⎛1.8883522 ×10−16 mol Pb 2 + ⎞⎛ ⎛ 207.2 g Pb 2+ ⎞⎟⎛ 1 μg ⎞ ⎟⎟⎜⎜10 L ⎞⎟⎟ ⎜ ⎜ ⎟⎟ ⎟⎟⎜⎜ Mass (μg) of Pb2+ in 100 mL = ⎜⎜ ⎟⎟⎜ ⎟⎟ 100 mL ⎜⎜ 2 + ⎟⎜ −6 ⎟ ⎟ ⎜⎜ ⎜ ⎜ L ⎟⎜ 1 mL ⎠⎟ ⎝ ⎠⎝ ⎝⎜ 1 mol Pb ⎠⎟⎜⎝10 g ⎠⎟ 7

(

)

= 3.9126658×10–9 μg Pb2+ The final concentration is 3.9×10–9 μg/100 mL. 19.143 Assume that pKa is in the center of the range, and calculate the Ka from the average pKa. Average pKa (center of range) = (7.9 + 6.5)/2 = 7.2 Ka = 10–7.2 = 6.3095734×10–8 = 6×10–8 There is only one digit after the decimal point in the pKa values; thus, there is only one significant figure. 19.144 Plan: Convert the solubility of NaCl from g/L to mol/l (molarity). Use the solubility to find the Ksp value for NaCl. Find the moles of Na+ and Cl− in the original solution; find the moles of added Cl− (from the added HCl). The molarity of the Na+ and Cl− ions are then found by dividing moles of each by the total volume after mixing. Using the molarities of the two ions, determine a Q value and compare this value to the Ksp to determine if precipitation will occur. Solution: ⎛ 317 g NaCl ⎞⎟⎛⎜ 1 mol NaCl ⎞⎟ ⎟⎟ = 5.42436687 M NaCl ⎟⎟⎜⎜ Concentration (M) of NaCl = ⎜⎜⎜ ⎟⎜⎝ 58.44 g NaCl ⎠⎟⎟ L ⎝⎜ ⎠⎜

19-79

Determine the Ksp from the molarity just calculated. NaCl(s) ⇆ Na+(aq) + Cl−(aq) Ksp = [Na+][Cl−] = S2 = (5.42436687)2 = 29.42375594 = 29.4 ⎛ 1 mol Cl− ⎞⎟ ⎛ 5.42436687 mol NaCl ⎞⎟ ⎟⎟ 0.100 L ⎜⎜⎜ Moles of Cl− initially = ⎜⎜⎜ ⎟⎟ = 0.542436687 mol Cl− L ⎜⎝⎜1 mol NaCl ⎠⎟⎟ ⎝⎜ ⎠⎟

(

)

This is the same as the moles of Na+ in the solution. ⎛ 1 mol Cl− ⎞⎟ ⎛ 8.65 mol HCl ⎞⎟⎛⎜10−3 L ⎞⎟ ⎜ ⎟⎜⎜ Moles of Cl− added = ⎜⎜⎜ ⎟⎟⎟ 28.5 mL ⎜⎜ ⎟⎟ = 0.246525 mol Cl− ⎟ ⎟⎜ ⎜ ⎜⎝ L ⎜⎝1 mol HCl ⎠⎟⎟ ⎠⎜⎝ 1 mL ⎠⎟ 0.100 L of saturated solution contains 0.542 mol each Na+ and Cl−, to which you are adding 0.246525 mol of additional Cl− from HCl. Volume of mixed solutions = 0.100 L + (28.5 mL)(10–3 L/1 mL) = 0.1285 L Molarity of Cl− in mixture = [(0.542436687 + 0.246525) mol Cl−]/(0.1285 L) = 6.13978 M Cl− Molarity of Na+ in mixture = (0.542436687 mol Na+)/(0.1285 L) = 4.22130 M Na+ Determine a Q value and compare this value to the Ksp to determine if precipitation will occur. Qsp = [Na+][Cl−] = (4.22130)(6.13978) = 25.9179 = 25.9 Since Qsp < Ksp, no NaCl will precipitate.

(

)

19.145 Plan: A buffer contains a weak acid conjugate base pair. A Ka expression is used to calculate the pH of a weak acid while a Kb expression is used to calculate the pH of a weak base. The Henderson-Hasselbalch equation is used to calculate the pH when both the weak acid and conjugate base are present (a buffer). Solution: a) For the solution to be a buffer, both HA and A− must be present in the solution. This situation occurs in A and D. b) Scene A: The amounts of HA and A− are equal. ⎛ [A− ] ⎞⎟ ⎛ [A − ] ⎞⎟ ⎜⎜ pH = pKa + log ⎜⎜ ⎟⎟ ⎟ = 1 when the amounts of HA and A− are equal ⎜⎝[HA]⎠ ⎜⎝ [HA]⎠⎟

pH = pKa + log 1 pH = pKa = –log (4.5×10–5) = 4.346787 = 4.35 Scene B: Only A− is present at a concentration of 0.10 M. The Kb for A− is needed. Kb = Kw/Ka = 1.0×10–14/4.5×10–5 = 2.222×10–10 A−(aq) + H2O(l) ⇆ OH−(aq) + HA(aq) Initial: 0.10 M 0 0 Change: –x –x –x Equilibrium: 0.10 – x x x − [ HA ] OH [ ] Kb = 2.222×10–10 = − [A ] [ x ][ x ] Assume that x is small compared to 0.10 Kb = 2.222×10–10 = ⎡ 0.10 − x ⎤ ⎣⎢ ⎦⎥ ( x)( x) Kb = 2.222×10–10 = (0.10) –6 x = 4.7138095×10 M OH− Check assumption: (4.7138095×10–6/0.10) × 100% = 0.005% error, so the assumption is valid. [H3O]+ = Kw/[OH−] = (1.0×10–14)/(4.7138095×10–6) = 2.1214264×10–9 M H3O+ pH = –log [H3O+] = –log (2.1214264×10–9) = 8.67337 = 8.67 Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

19-80

Scene C: This is a 0.10 M HA solution. The hydrogen ion, and hence the pH, can be determined from the Ka. Concentration HA(aq) + H2O(l) ⇆ H3O+(aq) + A–(aq) Initial 0.10 M — 0 0 Change –x +x +x Equilibrium 0.10 – x x x (The H3O+ contribution from water has been neglected.) [ H3O+ ][ A− ] –5 Ka = 4.5×10 = [ HA ] ( )( x ) x Ka = 4.5×10–5 = Assume that x is small compared to 0.10. (0.10 − x) ( x)( x) Ka = 4.5×10–5 = (0.10) + [H3O ] = x = 2.12132×10–3 Check assumption: (2.12132×10–3/0.10) × 100% = 2% error, so the assumption is valid. pH = –log [H3O+] = –log (2.12132×10–3) = 2.67339 = 2.67 Scene D: This is a buffer with a ratio of [A–]/[HA] = 5/3. ⎛ [A− ] ⎞⎟ pH = pKa + log ⎜⎜ ⎟ ⎜⎝[HA]⎠⎟

⎡ 5⎤ pH = –log (4.5×10–5) + log ⎢ ⎥ = 4.568636 = 4.57 ⎢⎣ 3 ⎥⎦ c) The initial stage in the titration would only have HA present. The amount of HA will decrease, and the amount of A– will increase until only A– remains. The sequence will be: C, A, D, and B. d) At the equivalence point, all the HA will have reacted with the added base. This occurs in scene B. 19.146 a) The dissolution of MZ will produce equal amounts of M2+ and Z2–. The only way unequal amounts of these ions could be present would be either if one of the ions were already present or if one of the ions were removed from the solution. Distilled water will neither add nor remove ions, thus the M2+ and Z2– must be equal; as in Scene B. b) Using box B; there are 4(2.5×10–6M) = 1.0×10–5 M for each ion. Ksp = [M2+][Z2–] = (1.0×10–5)(1.0×10–5) = 1.0×10–10 c) The addition of Na2Z would increase the Z2– and shifts the equilibrium to the left, resulting in fewer ions of M2+. There will be more Z2– than M2+. This occurs in box C. d) Lowering the pH will protonate some Z2– (the weak base CO32–). This will decrease the Z2– concentration and and shift the equilibrium to the right, resulting in more M2+. This occurs in box A. 19.147 a) Ag+ ions come from the dissolution of AgCl(s). Ksp = [Ag+][Cl–] = 1.8×10–10 1.8×10−10 [Ag+] = ⎡Cl− ⎤ ⎣⎢ ⎦⎥ + AgCl(s) ⇆ Ag (aq) + Cl– (aq) Ksp = 1.8×10–10 + – – Ag (aq) + 2Cl (aq) ⇆ AgCl2 (aq) Kf = 1.8×105 AgCl(s) + Cl–(aq) ⇆ AgCl2–(aq)

K = KspKf = (1.8×10–10)(1.8×105) = 3.24×10–5

19-81

K = 3.24×10–5 =

[ AgCl2− ] [ Cl− ]

[AgCl2–] = (3.24×10–5)[Cl–] = (3.2×10–5)[Cl–] b) [Ag+] = [AgCl2–] 1.8×10−10 = (3.24×10–5)[Cl–] ⎡Cl− ⎤ ⎣⎢ ⎦⎥ 1.8×10−10 3.24 ×10−5 [Cl–] = 2.3570226×10–3 = 2.4×10–3 M Cl– c) At low Cl– ion concentration, Ag+ ions are present in solution. As Cl– ion concentration increases, more AgCl(s) is formed as the solubility equilibrium is shifted to the left and the solubility of AgCl decreases. At even higher Cl– ion concentrations, AgCl2– ions are present in solution as the formation equilibrium is shifted to the right. [Cl–]2 =

AgCl solubility

Ag+

AgCl2−

[Cl−] d) The solubility of AgCl(s) = [Ag+] + [AgCl2–] You can use either equation from part a) to calculate [Ag+] and [AgCl2–]. [Ag+] = [AgCl2–] = (3.24×10–5)(2.3570226×10–3) = 7.6367532×10–8 = 7.6×10–8 M The solubility of AgCl(s) = (7.6367532×10–8 + 7.6367532×10–8) M = 1.52735×10–7 = 1.5×10–7 M 19.148 Co2+(aq) + EDTA4−(aq) ⇆ [Co(EDTA)]2−(aq) Kf = 1016.31 = 2.0417379×1016 ⎡Co (EDTA)2− ⎤ ⎢ ⎦⎥ Kf = ⎣ 2+ ⎡Co ⎤ ⎡ EDTA 4− ⎤ ⎣ ⎦⎣ ⎦ ⎛ 0.048 mol Co 2 + ⎞⎟⎜⎛10−3 L ⎞⎟ ⎜ 2+ ⎟ ⎟⎟⎜⎜ Moles of Co2+ (initial) = ⎜⎜ ⎟⎟⎜⎜ 1 mL ⎟⎟⎟ 50.0 mL = 0.0024 mol Co L ⎜⎝ ⎠⎝ ⎠ 4 − ⎛ ⎛ 0.050 mol EDTA ⎞⎟⎜10−3 L ⎞⎟ ⎜ ⎟ ⎟⎟⎜⎜ a) Moles of EDTA added = ⎜⎜ ⎟⎟⎜⎜ 1 mL ⎟⎟⎟ 25.0 mL = 0.00125 mol EDTA L ⎜⎝ ⎠⎝ ⎠ The moles of EDTA added equals the moles of [Co(EDTA)]2− formed. The EDTA is limiting so no EDTA is left after the reaction and the remaining Co2+: Co2+ = (0.0024 – 0.00125) = 0.00115 mol Co2+ Total volume = (50.0 + 25.0) mL (10–3 L/1 mL) = 0.0750 L [Co2+] = (0.00115 mol Co2+)/(0.0750 L) = 0.015333 M Co2+ [Co(EDTA)2−] = (0.00125 mol [Co(EDTA)]2−)/(0.0750 L) = 0.016667 M

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)

(

)

19-82

To reach equilibrium the concentrations of the species involved are: [EDTA4−] = x [Co2+] = 0.015333 + x [Co(EDTA)2–] = 0.016667 – x ⎡ 0.016667 − x ⎤ ⎡Co (EDTA)2− ⎤ ⎢ ⎦⎥ = ⎣⎢ ⎦⎥ = [ 0.016667 ] Kf = 2.0417379×1016 = ⎣ 2+ ⎡Co ⎤ ⎡ EDTA 4− ⎤ ⎡ 0.01533 + x ⎤ ⎡ x ⎤ [ 0.01533][ x ] ⎣ ⎦⎣ ⎦ ⎣⎢ ⎦⎥ ⎣⎢ ⎦⎥ x = 5.324947×10–17 [EDTA] = 5.3×10–17 M [Co2+] = 0.015 M ⎛ 0.050 mol EDTA 4− ⎞⎟⎜⎛10−3 L ⎞⎟ ⎜ ⎟ ⎟⎟⎜⎜ b) Moles of EDTA added = ⎜⎜ ⎟⎟⎜⎜ 1 mL ⎟⎟⎟ 75.0 mL = 0.00375 mol EDTA L ⎜⎝ ⎠⎝ ⎠ The Co2+ is limiting so no Co2+ is left. The original moles Co2+ equals the moles of complex formed. Moles of EDTA remaining = (0.00375 – 0.0024) mol = 0.00135 mol EDTA Total volume = (50.0 + 75.0) mL (10−3 L/1 mL) = 0.1250 L [EDTA] = (0.00135 mol EDTA)/(0.1250 L) = 0.0108 M EDTA [Co(EDTA)2−] = (0.0024 mol [Co(EDTA)]2−)/(0.1250 L) = 0.0192 M To reach equilibrium the concentrations of the species involved are: [EDTA4–] = 0.0108 + x [Co2+] = x [Co(EDTA)2–] = 0.0192 – x ⎡0.0192 − x⎤ ⎡Co (EDTA)2− ⎤ [ 0.0192 ] ⎢ ⎥⎦ ⎢⎣ ⎥⎦ 16 Kf = 2.0417379×10 = ⎣ 2+ = = ⎡ x⎤ [ 0.0108] ⎡Co ⎤ ⎡ EDTA 4− ⎤ ⎡ x ⎤ ⎡0.0108 + x ⎤ ⎣ ⎦⎣ ⎦ ⎢⎣ ⎥⎦ ⎢⎣ ⎥⎦ ⎣⎢ ⎦⎥ –17 x = 8.7071792×10 [EDTA4–] = 00108 + x = 0.0108 M [Co2+] = 8.7×10–17 M

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19-83

CHAPTER 20 THERMODYNAMICS: ENTROPY, FREE ENERGY, AND THE DIRECTION OF CHEMICAL REACTIONS FOLLOW–UP PROBLEMS 20.1A

Plan: Particles with more freedom of motion have higher entropy. In general the entropy of gases is greater than that of liquids, and the entropy of liquids is greater than that of solids. Entropy increases with temperature. For substances in the same phase, entropy increases with atomic size and molecular complexity. If the entropy of the products is greater than that of the reactants, S is positive. Solution: a) PCl5(g). For substances with the same type of atoms and in the same physical state, entropy increases with increasing number of atoms per molecule because more types of molecular motion are available. 2+ – 2+ b) BaCl2(s). Entropy increases with increasing atomic size. The Ba ion and Cl ion are larger than the Ca ion – and F ion, respectively. c) Br2(g). Entropy increases from solid  liquid  gas.

20.1B

Plan: Particles with more freedom of motion have higher entropy. In general the entropy of gases is greater than that of liquids, and the entropy of liquids is greater than that of solids. Entropy increases with temperature. For substances in the same phase, entropy increases with atomic size and molecular complexity. If the entropy of the products is greater than that of the reactants, S is positive. Solution: a) LiBr(aq). For substances with the same number of atoms and in the same physical state, entropy increases with increasing atomic size. LiBr has the lower molar mass of the two substances and, therefore, the lower entropy. b) Quartz. Quartz has a crystalline structure and the particles have less freedom (and lower entropy) in that structure than in glass, an amorphous solid. c) Cyclohexane. Ethylcyclobutane has a side chain, which has more freedom of motion than the atoms in the ring. In cyclohexane, there are no side chains. Freedom of motion is restricted by the ring structure.

20.2A

o Plan: Predict the sign of Srxn by comparing the randomness of the products with the randomness of the reactants. o o o Calculate Srxn using Appendix B values and the relationship Srxn = m Sproducts – n Sreactants . o

Solution: a) 4NO (g)  N2O(g) + N2O3(g) o o The Srxn is predicted to decrease ( Srxn < 0) because four moles of random, gaseous product are transformed into two moles of random, gaseous product (the change in gas moles is –2).

Srxn = m Sproducts – n Sreactants o

Srxn = [(1 mol N2O3)(S of N2O3) + (1 mol N2O)(S of N2O)] o

– [(4 mol NO)(S of NO)] Srxn = [(1 mol N2O3)(314.7 J/mol∙K) + (1 mol N2O)(219.7 J/mol∙K)] o

= – 308.2 J/K Srxn < 0 as predicted.

 [(4 mol NO)(210.65 J/mol∙K)]

20-1

b) CH3OH(g)  CO(g) + 2H2(g) o The change in gaseous moles is +2, so the sign of Srxn is predicted to be greater than zero. o Srxn = [(1 mol CO)(S of CO) + (2 mol H2)(S of H2)]  [(1 mol CH3OH)(S of CH3OH)] o Srxn = [(1 mol CO)(197.5 J/mol∙K) + (2 mol H2)(130.6 J/mol∙K)]  [(1 mol CH3OH)(238 J/mol∙K)] = 220.7 = 221 J/K o Srxn > 0 as predicted. 20.2B

Solution: a) 2NaOH(s) + CO2(g)  Na2CO3(s) + H2O(l)

o o The Srxn is predicted to decrease ( Srxn < 0) because the more random, gaseous reactant is transformed into a

more ordered, liquid product.

Srxn = m Sproducts – n Sreactants o

Srxn = [(1 mol Na2CO3)(S of Na2CO3) + (1 mol H2O)(S of H2O)] o

– [(2 mol NaOH)(S of NaOH) + (1 mol CO2)(S of CO2)]

Srxn = [(1 mol Na2CO3)(139 J/mol∙K) + (1 mol H2O)(69.940 J/mol∙K)] o

 [(2 mol NaOH)(64.454 J/mol∙K) + (1 mol CO2)(213.7 J/mol∙K)]

= –133.668 = – 134 J/K

Srxn < 0 as predicted. o

b) 2Fe(s) + 3H2O(g)  Fe2O3(s) + 3H2(g)

o The change in gaseous moles is zero, so the sign of Srxn is difficult to predict. Iron(III) oxide has greater entropy

than Fe because it is more complex, but this is offset by the greater molecular complexity of H 2O versus H2.

Srxn = [(1 mol Fe2O3)(S of Fe2O3) + (3 mol H2)(S of H2)] o

 [(2 mol Fe)(S of Fe) + (3 mol H2O)(S of H2O)]

Srxn = [(1 mol Fe2O3)(87.400 J/mol∙K) + (3 mol H2)(130.6 J/mol∙K)] o

 [(2 mol Fe)(27.3 J/mol∙K) + (3 mol H2O)(188.72 J/mol∙K)]

= –141.56 = –141.6 J/K

o The negative Srxn shows that the greater entropy of H2O versus H2 does outweigh the greater entropy of Fe2O3

versus Fe. 20.3A

 Plan: Write the balanced equation for the reaction and calculate the Srxn using Appendix B. Determine o o the Ssurr by first finding H rxn . Add Ssurr to Srxn to verify that Suniv is positive. Solution: P4(s) + 6Cl2(g)  4PCl3(g)

Srxn = [(4 mol PCl3)(S of PCl3)] – [(1 mol P4)(S of P4) + (6 mol Cl2)(S of Cl2)] o

Srxn = [(4 mol PCl3)(312 J/mol∙K)] o

– [(1 mol P4)(41.1 J/mol∙K) + (6 mol Cl2) (223.0J/mol∙K)]

Srxn = –131J/K (The entropy change is expected to be negative because the change in gas moles is negative.) o

H rxn = m H f(products) – n  H f(reactants) o

20-2

H rxn = [(4 mol PCl3)( Hf of PCl3)] o

– [(1 mol P4)( Hf of P4) + (6 mol Cl2)( Hf of Cl2)]

H rxn = [(4 mol PCl3)(–287 kJ/mol)] o

– [(1 mol P4)(0 kJ/mol) + (6 mol Cl2)(0 kJ/mol)]

H rxn = –1148 kJ o

1148 kJ H rxn 3 = = 3.8523 kJ/K(10 J/1 kJ) = 3850 J/K 298 K T o Suniv = Srxn + Ssurr = (–131 J/K) + (3850 J/K) = 3719 J/K Ssurr = 

Because Suniv is positive, the reaction is spontaneous at 298 K. 20.3B

o Plan: Write the balanced equation for the reaction and calculate the Srxn using Appendix B. Determine o o the Ssurr by first finding H rxn . Add Ssurr to Srxn to verify that Suniv is positive.

Solution:

2FeO(s) + 1/2O2(g)  Fe2O3(s)

Srxn = [(1 mol Fe2O3)(S of Fe2O3)] – [(2 mol FeO)(S of FeO) + (1/2 mol O2)(S of O2)] o

Srxn = [(1 mol Fe2O3)(87.400 J/mol∙K)] o

– [(2 mol FeO)(60.75 J/mol∙K) + (1/2 mol O 2) (205.0J/mol∙K)]

Srxn = –136.6 J/K (entropy change is expected to be negative because gaseous reactant o

is converted to solid product).

H rxn = m H f (products) – n  H f (reactants) o

H rxn = [(1 mol Fe2O3)( Hf of Fe2O3)] o

– [(2 mol FeO)( Hfo of FeO) + (1/2 mol O2)( Hfo of O2)]

H rxn = [(1 mol Fe2O3)(–825.5 kJ/mol)] o

– [(2 mol FeO)(–272.0 kJ/mol) + (1/2 mol O 2)(0 kJ/mol)]

H rxn = –281.5 kJ o

281.5 kJ H rxn 3 = = 0.94463 kJ/K(10 J/1 kJ) = 944.63 J/K 298 K T o Suniv = Srxn + Ssurr = (–136.6 J/K) + (944.63 J/K) = 808.03 = 808 J/K Ssurr = 

Because Suniv is positive, the reaction is spontaneous at 298 K.

This process is also known as rusting. Common sense tells us that rusting occurs spontaneously. Although the entropy change of the system is negative, the increase in entropy of the surroundings is large enough to offset o Srxn . 20.4A

o o Plan: Calculate the H rxn using Hfo values from Appendix B. Calculate Srxn from tabulated S values and then o o o use the relationship Grxn = H rxn – T Srxn .

Solution:

H rxn = m H f (products) – n  H f (reactants) o

H rxn = [(2 mol NOCl)( Hf of NOCl)] – [(2 mol NO)( Hf of NO) + (1 mol Cl2)( Hf of Cl2)] H rxn = [(2 mol NOCl)(51.71 kJ/mol)] – [(2 mol NO)(90.29 kJ/mol) + (1 mol Cl2)(0 kJ/mol)] H rxn = –77.16 kJ Srxn = m Sproducts – n Sreactants o

o o

Srxn = [(2 mol NOCl)(S of NOCl)] – [(2 mol NO)(S of NO) + (1 mol Cl2)(S of Cl2)] o

20-3

Srxn = [(2 mol NOCl)(261.6 J/mol∙K)] – [(2 mol NO)(210.65 J/mol∙K) + (1 mol Cl2)(223.0 J/mol∙K)] o

Srxn = –121.1 J/K o

Grxn = H rxn – T Srxn = –77.16 kJ – [(298 K)(–121.1 J/K)(1 kJ/10 J)] = –41.0722 = –41.1 kJ o

20.4B

o o Plan: Calculate the H rxn using Hfo values from Appendix B. Calculate Srxn from tabulated S values and then o o o use the relationship Grxn = H rxn – T Srxn .

Solution:

H rxn = m H f (products) – n  H f (reactants) o

H rxn = [(4 mol NO)( Hf of NO) + (6 mol H2O)( Hf of H2O)] o

– [(4 mol NH3)( Hfo of NH3) + (5 mol O2)( Hfo of O2)]

H rxn = [(4 mol NO)(90.29 kJ/mol) + (6 mol H2O)( –241.826 kJ/mol)] o

– [(4 mol NH3)( –45.9 kJ/mol) + (5 mol O2)(0 kJ/mol)]

H rxn = –906.196 kJ o

Srxn = m Sproducts – n Sreactants o

Srxn = [(4 mol NO)(S of NO) + (6 mol H2O)(S of H2O)] o

– [(4 mol NH3)(S of NH3) + (5 mol O2)(S of O2)]

Srxn = [(4 mol NO)(210.65 J/mol∙K) + (6 mol H2O)( 188.72 J/mol∙K)] o

– [(4 mol NH3)( 193 J/mol∙K) + (5 mol O2)(205.0 J/mol∙K)]

Srxn = 177.92 J/K o

Grxn = H rxn – T Srxn = –906.196 kJ – [(298 K)(177.92 J/K)(1 kJ/10 J)] = –959.21616 = –959 kJ o

20.5A

o Plan: Use Gfo values from Appendix B to calculate Grxn using the relationship

Grxn = m Gfo (products) – n Gfo (reactants) . o

Solution:

o a) Grxn = [(2 mol NOCl)( Gfo of NOCl)] – [(2 mol NO)( Gfo of NO) + (1 mol Cl2)( Gfo of Cl2)]

Grxn = [(2 mol NOCl)(66.07 kJ/mol)] – [(2 mol NO)(86.60 kJ/mol) + (1 mol Cl2)(0 kJ/mol)] o

Grxn = –41.06 kJ o

o b) Grxn = [(2 mol Fe)( Gfo of Fe) + (3 mol H2O)( Gfo of H2O)] –

[(3 mol H2)( Gfo of H2) + (1 mol Fe2O3)( Gfo of Fe2O3)]

Grxn = [(2 mol Fe)(0 kJ/mol) + (3 mol H2O)(–228.60 kJ/mol)] – o

Grxn = 57.8 kJ

[(3 mol H2)(0 kJ/mol) + (1 mol Fe2O3)( –743.6 kJ/mol)]

20.5B

o Plan: Use Gfo values from Appendix B to calculate Grxn using the relationship

Grxn = m Gfo (products) – n Gfo (reactants) . o

Solution:

o a) Grxn = [(4 mol NO)( Gf of NO) + (6 mol H2O)( Gf of H2O)]

– [(4 mol NH3)( Gf of NH3) + (5 mol O2)( Gf of O2)]

Grxn = [(4 mol NO)(86.60 kJ/mol) + (6 mol H2O)( –228.60 kJ/mol)] o

Grxn = –961.2 = –961 kJ

– [(4 mol NH3)( –16 kJ/mol) + (5 mol O2)(0 kJ/mol)]

o b) Grxn = [(2 mol CO)( Gfo of CO)] – [(2 mol C)( Gfo of C) + (1 mol O2)( Gfo of O2)]

20-4

Grxn = [(2 mol CO)(–137.2 kJ/mol)] – [(2 mol C)(0 kJ/mol) + (1 mol O2)(0 kJ/mol)] o

Grxn = –274.4 kJ o

20.6A

o Plan: Predict the sign of Srxn by comparing the randomness of the products with the randomness of the o o o reactants. Use the relationship Grxn = H rxn – T Srxn to answer b).

Solution: a) The reaction is X2Y2(g) → X2(g) + Y2(g). Since there are more moles of gaseous product than there are of gaseous reactant, entropy increases and S > 0.

o b) The reaction is only spontaneous above 325C or in other words, at high temperatures. In the relationship Grxn o o = H rxn – T Srxn , when S > 0 so that – TS° is < 0, G° will only be negative at high T if H° > 0.

20.6B

o Plan: Predict the sign of Srxn by comparing the randomness of the products with the randomness of the reactants.

20.7A

Plan: Use the equation G° = H° – TS° to determine if the reaction is spontaneous (G° < 0). Then examine the same equation to determine the effect of raising the temperature on the spontaneity of the reaction. Solution: 3 o o o a) Grxn = H rxn – T Srxn = –192.7 kJ – [(298 K)(–308.2 J/K)(1 kJ/10 J)] = –100.8564 = –100.9 kJ

o o o Use the relationship Grxn = H rxn – T Srxn to answer b). Solution: a) A solid forms a gas and a liquid, so S > 0. A crystalline array breaks down, so H > 0. b) For the reaction to occur spontaneously (G < 0), – TS° must be greater than H, which would occur only at higher T.

Because G < 0, the reaction is spontaneous at 298 K.

b) As temperature increases, – TS° becomes more positive, so the reaction becomes less spontaneous at higher temperatures. 3 o o o c) Grxn = H rxn – T Srxn = –192.7 kJ – [(773 K)(–308.2 J/K)(1 kJ/10 J)] = 45.5386 = 45.5 kJ 20.7B

Plan: Examine the equation G° = H° – TS° and determine which combination of enthalpy and entropy will describe the given reaction. Solution: Two choices can already be eliminated: 1) When H > 0 (endothermic reaction) and S < 0 (entropy decreases), the reaction is always nonspontaneous, regardless of temperature, so this combination does not describe the reaction. 2) When H < 0 (exothermic reaction) and S > 0 (entropy increases), the reaction is always spontaneous, regardless of temperature, so this combination does not describe the reaction. Two combinations remain: 3) H° > 0 and S° > 0, or 4) H° < 0 and S° < 0. If the reaction becomes spontaneous at –40°C, this means that G° becomes negative at lower temperatures. Case 3) becomes spontaneous at higher temperatures, when the –TS° term is larger than the positive enthalpy term. By process of elimination, Case 4) describes the reaction. At a lower temperature, the negative H° becomes larger than the positive (–TS°) value, so G° becomes negative.

20.8A

Plan: To find the temperature at which the reaction becomes spontaneous, use o o o Grxn = 0 = H rxn – T Srxn and solve for temperature. Solution:

The reaction will become spontaneous when ΔG changes from being positive to being negative. This point occurs when ΔG is 0.

Grxn = H rxn – T Srxn 3 0 = –192.7 × 10 J – (T)(–308.2 J/K) T = 625.2 K – 273.15 = 352.0°C o

20-5

20.8B

Plan: To find the temperature at which the reaction becomes spontaneous, use o o o o Grxn = 0 = H rxn – T Srxn and solve for temperature. H rxn can be calculated from the individual Hfo values of the reactants and products by using the relationship o o H rxn = m Hfo (products) – n Hfo (reactants) . Srxn can be calculated from the individual S o values of the reactants o o o and products by using the relationship Srxn = m Sproducts – n Sreactants . Solution: CaO(s) + CO2(g) → CaCO3(s)

H rxn = m Hfo (products) – n Hfo (reactants) o

H rxn = [(1 mol CaCO3)( Hf of CaCO3)] o

– [(1 mol CaO)( Hfo of CaO) + (1 mol CO2)( Hfo of CO2)]

H rxn = [(1 mol CaCO3)( –1206.9 kJ/mol)] o

– [(1 mol CaO)(–635.1 kJ/mol) + (1 mol CO2)( –393.5 kJ/mol)]

H rxn = –178.3 kJ o

o Srxn = m Sproducts – n Sreactants o

Srxn = [(1 mol CaCO3)(S of CaCO3)] – [(1 mol CaO)(S of CaO) + (1 mol CO2)(S of CO2)] o

Srxn = [(1mol CaCO3)(92.9 J/mol∙K)] o

– [(1 mol CaO)(38.2 J/mol∙K) + (1 mol CO 2)(213.7 J/mol∙K)] Srxn = –159.0 J/K = – 0.159 kJ/K o

Grxn = 0 = H rxn – T Srxn o

H rxn = T Srxn o

ΔH –178.3 kJ = = 1121.384 = 1121 K ΔS –0.1590 kJ/K The reaction becomes spontaneous at temperatures < 1121 K. o

20.9A

Plan: First find G°, then calculate K from G° = –RT ln K. Calculate Go using Gfo values in the relationship

Grxn = m Gfo (products) – n Gfo (reactants) . o

Solution: 2C(graphite) + O2(g) ⇆ 2CO(g)

Grxn = m Gfo (products) – n Gfo (reactants) o



Grxn = [(2 mol CO)(–137.2 kJ/mol)] – [(2 mol C)(0 kJ/mol) + (1 mol O2)( 0 kJ/mol)] = –274.4 kJ ΔG 274.4 kJ/mol 1000 J   ln K =  =  = 110.7536 = 111 RT 8.314 J/mol  K 298 K   1 kJ  o

K=e 20.9B

111

= 1.6095 × 10 = 1.6 × 10 48

Plan: The equilibrium constant, K, is related to G° through the equation G° = –RT ln K. Solution: G° = –RT ln K = –(8.314 J/mol∙K)(298 K) ln (2.22 × 10 ) = 8.35964 × 10 J/mol = 83.6 kJ/mol –15

20.10A Plan: Write the equilibrium expression for the reaction and calculate Qc for each scene. Remember that each particle represents 0.10 mol and that the volume is 1.0 L. A reaction that is proceeding to the right will have

G° < 0 and a reaction that is proceeding to the left will have G° > 0. A reaction at equilibrium has G° = 0. Solution: a) A(g) + 3B(g) ⇆ AB3(g) Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

20-6

b) Qc =

 AB3 

 A  B

 AB3 

Mixture 1: Qc =

 A  B

 AB3 

Mixture 2: Qc =

 A  B

 AB3 

Mixture 3: Qc =

 A  B



 0.20  3 = 0.98  0.40  0.80 



 0.30  3 = 8  0.30  0.50 



 0.40  3 = 250  0.20  0.20 

Mixture 2 is at equilibrium since Qc = Kc

c) Qc < Kc for Mixture 1 and the reaction is proceeding right to reach equilibrium; thus G° < 0. Mixture 2 is at equilibrium and G° = 0. Mixture 3 proceeds to the left to reach equilibrium since Qc > Kc and G° > 0. The ranking for most positive to most negative is 3 > 2 > 1. 20.10B Plan: Write the equilibrium expression for the reaction and calculate Qc for each scene. A reaction that is proceeding to the right will have G° < 0 and a reaction that is proceeding to the left will have G° > 0. A reaction at equilibrium has G° = 0. Solution: a) X2(g) + 2Y2(g) ⇆ 2XY2(g) Qc =

[XY2 ]2 [X 2 ][Y2 ]2

[XY2 ]2 [5]2 = = 12.5 2 [X 2 ][Y2 ] [2][1]2

Mixture 1: Qc =

[XY2 ]2 [4]2 = =2 [X 2 ][Y2 ]2 [2][2]2

Mixture 2: Qc =

[XY2 ]2 [2]2 = = 0.25 2 [X 2 ][Y2 ] [4][2]2

Mixture 3: Qc =

Mixture 2 is at equilibrium since Qc = Kc

b) Qc > Kc for Mixture 1 and the reaction is proceeding left to reach equilibrium; thus G° > 0. Mixture 2 is at equilibrium and G° = 0. Mixture 3 proceeds to the right to reach equilibrium since Qc < Kc and G°< 0. The ranking for most negative to most positive is 3 < 2 < 1. c) Any reaction mixture moves spontaneously towards equilibrium so both changes have a negativeG°. 20.11A Plan: The equilibrium constant, K, is related to G° through the equation G° = –RT ln K. The free energy of the reaction under non-standard state conditions is calculated using G = G° + RT ln Q. Solution: ΔG 33.5 kJ/mol 1000 J   a) ln K =  =  = 13.5213 8.314 J/mol  K 298 K   1 kJ  RT o

K=e

13.5213

b) Q =

= 7.4512 × 10 = 7.45 × 10 5

C 2 H 5 Cl  1.5  = 3.0 C 2 H 4  HCl  0.501.0

G = G° + RT ln Q = (–33.5 kJ/mol) + (8.314 J/mol∙K) (1 kJ/1000 J) (298 K) ln (3.0) = –30.7781 = –30.8 kJ/mol

20-7

20.11B Plan: Write a balanced equation for the dissociation of hypobromous acid in water. The free energy of the reaction at standard state is calculated using G° = –RT ln K. The free energy of the reaction under non-standard state conditions is calculated using G = G° + RT ln Q. Solution: – + HBrO(aq) + H2O(l) ⇆ BrO (aq) + H3O (aq) a) G° = –RT ln K = – (8.314 J/mol∙K)(298)ln (2.3 × 10 ) = 4.927979 × 10 = 4.9 × 10 J/mol = 49 kJ/mol –9

b) Q =

 H3O  BrO   HBrO 

6.0  104   0.10     =  0.20  

6.0  104   0.10  4    G = G° + RT ln Q = (4.927979 × 10 J/mol) + (8.314 J/mol∙K) (298 K) ln  0.20   4 4 = 2.9182399 × 10 = 2.9 × 10 J/mol = 29 kJ/mol

The value of Ka is very small, so it makes sense that G° is a positive number. The natural log of a negative –4 exponent gives a negative number (ln 3.0 × 10 ), so the value of G decreases with concentrations lower than the standard state 1 M values. CHEMICAL CONNECTIONS BOXED READING PROBLEMS B20.1

Plan: Convert mass of glucose (1 g) to moles and use the ratio between moles of glucose and moles of ATP to find the moles and then molecules of ATP formed. Do the same calculation with tristearin. Solution: a) Molecules of ATP/g glucose =  1 mol glucose   36 mol ATP   6.022  10 23 molecules ATP       1 g glucose     1 mol ATP 180.16 g glucose  1 mol glucose    23 23 = 1.20333 × 10 = 1.203 × 10 molecules ATP/g glucose b) Molecules of ATP/g tristearin =  1 mol tristearin   458 mol ATP   6.022  10 23 molecules ATP       1 g tristearin     1 mol ATP  897.50 g tristearin   1 mol tristearin   

= 3.073065 × 10 = 3.073 × 10 molecules ATP/g tristearin 23

B20.2

Plan: Add the two reactions to obtain the overall process; the values of the two reactions are then added to obtain for the overall reaction. Solution: creatine phosphate  creatine + phosphate G° = – 43.1 kJ/mol ADP + phosphate  ATP G° = +30.5 kJ/mol creatine phosphate + ADP  creatine + ATP G = –43.1 kJ/mol + 30.5 kJ/mol = –12.6 kJ/mol

END–OF–CHAPTER PROBLEMS 20.1

Spontaneous processes proceed without outside intervention. The fact that a process is spontaneous does not mean that it will occur instantaneously (in an instant) or even at an observable rate. The rusting of iron is an example of a process that is spontaneous but very slow. The ignition of gasoline is an example of a process that is not spontaneous but very fast.

20-8

20.2

A spontaneous process occurs by itself (possibly requiring an initial input of energy), whereas a nonspontaneous process requires a continuous supply of energy to make it happen. It is possible to cause a nonspontaneous process to occur, but the process stops once the energy source is removed. A reaction that is found to be nonspontaneous under one set of conditions may be spontaneous under a different set of conditions (different temperature, different concentrations).

20.3

a) The energy of the universe is constant. b) Energy cannot be created or destroyed. c) Esystem = –Esurroundings The first law is concerned with balancing energy for a process but says nothing about whether the process can, in fact, occur.

20.4

Entropy is related to the freedom of movement of the particles. A system with greater freedom of movement has higher entropy. a) and b) Probability is so remote as to be virtually impossible. Both would require the simultaneous, coordinated movement of a large number of independent particles, so are very unlikely.

20.5

Vaporization is the change of a liquid substance to a gas so Svaporization = Sgas – Sliquid. Fusion is the change of a solid substance into a liquid so Sfusion = Sliquid – Ssolid. Vaporization involves a greater change in volume than fusion. Thus, the transition from liquid to gas involves a greater entropy change than the transition from solid to liquid.

20.6

In an exothermic process, the system releases heat to its surroundings. The entropy of the surroundings increases because the temperature of the surroundings increases (Ssurr > 0). In an endothermic process, the system absorbs heat from the surroundings and the surroundings become cooler. Thus, the entropy of the surroundings decreases (Ssurr < 0). A chemical cold pack for injuries is an example of a spontaneous, endothermic chemical reaction as is the melting of ice cream at room temperature.

20.7

a) According to the third law the entropy is zero. b) Entropy will increase with temperature. c) The third law states that the entropy of a pure, perfectly crystalline element or compound may be taken as zero at zero Kelvin. Since the standard state temperature is 25°C and entropy increases with temperature, S° must be greater than zero for an element in its standard state. d) Since entropy values have a reference point (0 entropy at 0 K), actual entropy values can be determined, not just entropy changes.

20.8

Plan: A spontaneous process is one that occurs by itself without a continuous input of energy. Solution: a) Spontaneous, evaporation occurs because a few of the liquid molecules have enough energy to break away from the intermolecular forces of the other liquid molecules and move spontaneously into the gas phase. b) Spontaneous, a lion spontaneously chases an antelope without added force. This assumes that the lion has not just eaten. c) Spontaneous, an unstable substance decays spontaneously to a more stable substance.

20.9

a) The movement of Earth about the Sun is spontaneous. b) The movement of a boulder against gravity is nonspontaneous. c) The reaction of an active metal (sodium) with an active nonmetal (chlorine) is spontaneous.

20.10

Plan: A spontaneous process is one that occurs by itself without a continuous input of energy. Solution: a) Spontaneous, with a small amount of energy input, methane will continue to burn without additional energy (the reaction itself provides the necessary energy) until it is used up. b) Spontaneous, the dissolved sugar molecules have more states they can occupy than the crystalline sugar, so the reaction proceeds in the direction of dissolution.

20-9

c) Not spontaneous, a cooked egg will not become raw again, no matter how long it sits or how many times it is mixed. 20.11

a) If a satellite slows sufficiently, it will fall to Earth’s surface through a spontaneous process. b) Water is a very stable compound; its decomposition at 298 K and 1 atm is not spontaneous. c) The increase in prices tends to be spontaneous.

20.12

Plan: Particles with more freedom of motion have higher entropy. Therefore, Sgas > Sliquid > Ssolid. If the products of the process have more entropy than the reactants, Ssys is positive. If the products of the process have less entropy than the reactants, Ssys is negative. Solution: a) Ssys positive, melting is the change in state from solid to liquid. The solid state of a particular substance always has lower entropy than the same substance in the liquid state. Entropy increases during melting. b) Ssys negative, the entropy of most salt solutions is greater than the entropy of the solvent and solute separately, so entropy decreases as a salt precipitates. c) Ssys negative, dew forms by the condensation of water vapor to liquid. Entropy of a substance in the gaseous state is greater than its entropy in the liquid state. Entropy decreases during condensation.

20.13

a) Ssys positive

20.14

Plan: Particles with more freedom of motion have higher entropy. Therefore, Sgas > Sliquid > Ssolid. If the products of the process have more entropy than the reactants, Ssys is positive. If the products of the process have less entropy than the reactants, Ssys is negative. Solution: a) Ssys positive, the process described is liquid alcohol becoming gaseous alcohol. The gas molecules have greater entropy than the liquid molecules. b) Ssys positive, the process described is a change from solid to gas, an increase in possible energy states for the system. c) Ssys positive, the perfume molecules have more possible locations in the larger volume of the room than inside the bottle. A system that has more possible arrangements has greater entropy.

20.15

a) Ssys negative

20.16

Plan: Ssys is the entropy of the products – the entropy of the reactants. Use the fact that Sgas > Sliquid > Ssolid; also, the greater the number of particles of a particular phase of matter, the higher the entropy. Solution: a) Ssys negative, reaction involves a gaseous reactant and no gaseous products, so entropy decreases. The number of particles also decreases, indicating a decrease in entropy. b) Ssys negative, gaseous reactants form solid product and number of particles decreases, so entropy decreases. c) Ssys positive, when a solid salt dissolves in water, entropy generally increases since the entropy of the aqueous mixture has higher entropy than the solid.

20.17

a) Ssys negative

20.18

Plan: Ssys is the entropy of the products – the entropy of the reactants. Use the fact that Sgas > Sliquid > Ssolid; also, the greater the number of particles of a particular phase of matter, the higher the entropy. Solution: a) Ssys positive, the reaction produces gaseous CO2 molecules that have greater entropy than the physical states of the reactants. b) Ssys negative, the reaction produces a net decrease in the number of gaseous molecules, so the system’s entropy decreases. c) Ssys positive, the reaction produces a gas from a solid.

20.19

a) Ssys negative

b) Ssys positive

b) Ssys negative

b) Ssys positive

c) Ssys negative

20-10

20.20

Plan: Particles with more freedom of motion have higher entropy. In general the entropy of gases is greater than that of liquids, and the entropy of liquids is greater than that of solids. Entropy increases with temperature. For substances in the same phase, entropy increases with atomic size and molecular complexity. If the entropy of the products is greater than that of the reactants, S is positive. Solution: a) Ssys positive, decreasing the pressure increases the volume available to the gas molecules so entropy of the system increases. b) Ssys negative, gaseous nitrogen molecules have greater entropy (more possible states) than dissolved nitrogen molecules. c) Ssys positive, dissolved oxygen molecules have lower entropy than gaseous oxygen molecules.

20.21

a) Ssys negative

20.22

Plan: Particles with more freedom of motion have higher entropy. In general the entropy of gases is greater than that of liquids, and the entropy of liquids is greater than that of solids. Entropy increases with temperature. For substances in the same phase, entropy increases with atomic size and molecular complexity. Solution: a) Butane has the greater molar entropy because it has two additional CH bonds that can vibrate and has greater rotational freedom around its bond. The presence of the double bond in 2-butene restricts rotation. b) Xe(g) has the greater molar entropy because entropy increases with atomic size. c) CH4(g) has the greater molar entropy because gases in general have greater entropy than liquids.

20.23

a) N2O4(g); it has greater molecular complexity. b) CH3OCH3(l); hydrogen bonding in CH3CH2OH would increase order. c) HBr(g); it has greater mass.

20.24

20.25

a) P4(g); it has greater molecular complexity. b) HNO3(aq); because S(solution) > S(pure). c) CuSO4 · 5H2O; it has greater molecular complexity.

20.26

Plan: Particles with more freedom of motion have higher entropy. In general the entropy of gases is greater than that of liquids, and the entropy of liquids is greater than that of solids. Entropy increases with temperature. For substances in the same phase, entropy increases with atomic size and molecular complexity. Solution: a) Diamond < graphite < charcoal. Diamond has an ordered, three-dimensional crystalline shape, followed by graphite with an ordered two-dimensional structure, followed by the amorphous (disordered) structure of charcoal. b) Ice < liquid water < water vapor. Entropy increases as a substance changes from solid to liquid to gas. c) O atoms < O2 < O3. Entropy increases with molecular complexity because there are more modes of movement (e.g., bond vibration) available to the complex molecules.

20.27

a) Ribose < glucose < sucrose; entropy increases with molecular complexity. b) CaCO3(s) < (CaO(s) + CO2(g)) < (Ca(s) + C(s) + 3/2O2(g)); entropy increases with moles of gas particles. c) SF4(g) < SF6(g) < S2F10(g); entropy increases with molecular complexity.

b) Ssys positive

c) Ssys negative

20-11

20.28

Plan: Particles with more freedom of motion have higher entropy. In general the entropy of gases is greater than that of liquids, and the entropy of liquids is greater than that of solids. Entropy increases with temperature. For substances in the same phase, entropy increases with atomic size and molecular complexity. Solution: – – – a) ClO4 (aq) > ClO3 (aq) > ClO2 (aq). The decreasing order of molar entropy follows the order of decreasing molecular complexity. b) NO2(g) > NO(g) > N2(g). N2 has lower molar entropy than NO because N2 consists of two of the same atoms while NO consists of two different atoms. NO2 has greater molar entropy than NO because NO2 consists of three atoms while NO consists of only two. c) Fe3O4(s) > Fe2O3(s) > Al2O3(s). Fe3O4 has greater molar entropy than Fe2O3 because Fe3O4 is more complex and more massive. Fe2O3 and Al2O3 contain the same number of atoms but Fe2O3 has greater molar entropy because iron atoms are more massive than aluminum atoms.

20.29

a) Ba(s) > Ca(s) > Mg(s); entropy decreases with lower mass. b) C6H14 > C6H12 > C6H6; entropy decreases with lower molecular complexity and lower molecular flexibility. c) PF2Cl3(g) > PF5(g) > PF3(g); entropy decreases with lower molecular complexity.

20.30

a) X2(g) + 3Y2(g) → 2XY3(g) b) ΔS < 0 since there are fewer moles of gas in the products than in the reactants. c) XY3 is the most complex molecule and thus will have the highest molar entropy.

20.31

A system at equilibrium does not spontaneously produce more products or more reactants. For either reaction direction, the entropy change of the system is exactly offset by the entropy change of the surroundings. Therefore, for system at equilibrium, Suniv = Ssys + Ssurr = 0. However, for a system moving to equilibrium, Suniv > 0, because the second law states that for any spontaneous process, the entropy of the universe increases.

20.32

Plan: Since entropy is a state function, the entropy changes can be found by summing the entropies of the products and subtracting the sum of the entropies of the reactants. Solution: o Srxn = [(2 mol HClO)(S of HClO)] – [(1 mol H2O)(S of H2O) + (1 mol Cl2O)(S of Cl2O)]

o Rearranging this expression to solve for S of Cl2O gives: S of Cl2O = 2(S of HClO) – S of H2O – Srxn

20.33

o o o Plan: To calculate the standard entropy change, use the relationship Srxn = m Sproducts – n Sreactants .

To predict the sign of entropy recall that in general Sgas > Sliquid > Ssolid, and entropy increases as the number of

particles of a particular phase of matter increases, and with increasing atomic size and molecular complexity. Solution: a) Prediction: S° negative because number of moles of (n) gas decreases. S° = [(1 mol N2O)(S° of N2O) + (1 mol NO2)(S° of NO2)] – [(3 mol NO)(S° of NO)] S° = [(1 mol)(219.7 J/mol∙K) + (1 mol)(239.9 J/mol∙K)] – [(3 mol)(210.65 J/mol∙K)] S° = –172.35 = –172.4 J/K b) Prediction: Sign difficult to predict because n = 0, but possibly S° positive because water vapor has greater complexity than H2 gas. S° = [(2 mol Fe)(S° of Fe) + (3 mol H2O)(S° of H2O)] – [(3 mol H2)(S° of H2) + (1 mol Fe2O3)(S° of Fe2O3)] S° = [(2 mol)(27.3 J/mol∙K) + (3 mol)(188.72 J/mol∙K)] – [(3 mol)(130.6 J/mol∙K) + (1 mol)(87.400 J/mol∙K)] S° = 141.56 = 141.6 J/K c) Prediction: S° negative because a gaseous reactant forms a solid product and also because the number of moles of gas (n) decreases. S° = [(1 mol P4O10)(S° of P4O10)] – [(1 mol P4)(S°of P4) + (5 mol O2)(S° of O)] S° = [(1 mol)(229 J/mol∙K)] – [(1 mol)(41.1 J/mol∙K) + (5 mol)(205.0 J/mol∙K)] S° = –837.1 = –837 J/K

20-12

20.34

a) 3NO2(g) + H2O(l)  2HNO3(l) + NO(g) S° negative S° = [(2 mol HNO3)(S° of HNO3) + (1 mol NO)(S° of NO)] – [(3 mol NO2)(S° of NO2) +( 1 mol H2O)(S° of H2O)] S° = [(2 mol)(155.6 J/K∙mol) + (1 mol)(210.65 J/K∙mol)] – [(3 mol)(239.9 J/K∙mol) + (1 mol)(69.940 J/K∙mol)] S° = –267.79 = –267.8 J/K b) N2(g) + 3F2(g)  2NF3(g) S° negative S° = [(2 mol NF3)(S° of NF3] – [(1 mol N2)(S° of N2) + (3 mol F2)(S° of F2)] S° = [(2 mol)(260.6 J/K∙mol)] – [(1 mol)(191.5 J/K∙mol) + (3 mol)(202.7 J/K∙mol)] S° = –278.4 J/K c) C6H12O6(s) + 6O2(g)  6CO2(g) + 6H2O(g) S° positive S° = [(6 mol CO2)(S° of CO2) + (6 mol H2O)(S° of H2O)] – [(1 mol C6H12O6)(S° of C6H12O6) + (6 mol O2)(S° of O2)] S° = [(6 mol)(213.7 J/K∙mol) + (6 mol H2O)(188.72 J/K∙mol)] – [(1 mol)(212.1 J/K∙mol) + (6 mol)(205.0 J/K∙mol)] S° = 972.42 = 972.4 J/K

20.35

Plan: Write the balanced equation. To calculate the standard entropy change, use the relationship o o o Srxn = m Sproducts – n Sreactants . To predict the sign of entropy recall that in general Sgas > Sliquid > Ssolid, entropy increases as the number of particles of a particular phase of matter increases, and entropy increases with increasing atomic size and molecular complexity. Solution: The balanced combustion reaction is: 2C2H6(g) + 7O2(g)  4CO2(g) + 6H2O(g) S° = [(4 mol CO2)(S° of CO2) + (6 mol H2O)(S° of H2O)] – [(2 mol C2H6)(S° of C2H6) + (7 mol O2)(S° of O2)] S° = [(4 mol)(213.7 J/mol∙K) + (6 mol)(188.72 J/mol∙K)] – [(2 mol)(229.5 J/mol∙K) + (7 mol)(205.0 J/mol∙K)] S° = 93.12 = 93.1 J/K The entropy value is not per mole of C2H6 but per two moles. Divide the calculated value by two to obtain entropy per mole of C2H6. Yes, the positive sign of S° is expected because there is a net increase in the number of gas molecules from nine moles as reactants to ten moles as products.

20.36

CH4(g) + 2O2(g)  CO2(g) + 2H2O(l) S° = [(1 mol CO2)(S° of CO2) + (2 mol H2O)(S° of H2O)] – [(1 mol CH4)(S° of CH4) + (2 mol O2)(S° of O2)] S° = [(1 mol)(213.7 J/K∙mol) + (2 mol)(69.940 J/K∙mol)] – [(1 mol)(186.1 J/K∙mol) + (2 mol)(205.0 J/K∙mol)] S° = –242.52 = –242.5 J/K Yes, a decrease in the number of moles of gas should result in a negative S° value.

20.37

Plan: Write the balanced equation. To calculate the standard entropy change, use the relationship o o o Srxn = m Sproducts – n Sreactants . To predict the sign of entropy recall that in general Sgas > Sliquid > Ssolid, entropy increases as the number of particles of a particular phase of matter increases, and entropy increases with increasing atomic size and molecular complexity. Solution: The balanced chemical equation for the described reaction is: 2NO(g) + 5H2(g)  2NH3(g) + 2H2O(g) Because the number of moles of gas decreases, i.e., n = 4 – 7 = –3, the entropy is expected to decrease. S° = [(2 mol NH3)(S° of NH3) + (2 mol H2O)(S° of H2O)] – [(2 mol NO)(S° of NO) + (5 mol H2)(S° of H2)] S° = [(2 mol)(193 J/mol∙K) + (2 mol)(188.72 J/mol∙K)] – [(2 mol)(210.65 J/mol∙K) + (5 mol)(130.6 J/mol∙K)] S° = –310.86 = –311 J/K Yes, the calculated entropy matches the predicted decrease.



20.38

4NH3(g) + 7O2(g)  4NO2(g) + 6H2O(g) S° = [(4 mol NO2)(S° of NO2) + (6 mol H2O)(S° of H2O)] – [(4 mol NH3)(S° of NH3) + (7 mol O2)(S° of O2)] S° = [(4 mol)(239.9 J/K∙mol) + (6 mol)(188.72 J/K∙mol)] – [(4 mol)(193 J/K∙mol) + (7 mol)(205.0 J/K∙mol)]

20-13

S° = –115.08 = –115 J/K Yes, a loss of one mole of a gas should result in a small negative S° value. 20.39

Plan: Write the balanced equation. To calculate the standard entropy change (part a), use the relationship o o o Srxn = m Sproducts – n Sreactants .

To calculate ΔS of the universe in part b, first calculate ΔH of the reaction using the relationship o H rxn = m Hfo (products) – n Hfo (reactants)

Use ΔH of the reaction to calculate ΔS of the surroundings. o Hrxn Ssurr =  T o Add ΔS of the surroundings and ΔS of the reaction to calculate ΔS of the universe. If ΔS of the universe is greater than zero, the reaction is spontaneous at the given temperature. o Suniv = Srxn + Ssurr Solution:

a) The reaction for forming Cu2O from copper metal and oxygen gas is: 2Cu(s) + 1/2O2(g)  Cu2O(s)

S° = [(1 mol Cu2O)(S° of Cu2O)] – [(2 mol Cu)(S° of Cu) + (1/2 mol O2)(S° of O2)] S° = [(1 mol)(93.1 J/mol∙K)] – [(2 mol)(33.1 J/mol∙K) + (1/2 mol)(205.0 J/mol∙K)] S° = –75.6 J/K

o H rxn = m Hfo (products) – n Hfo (reactants)

o H rxn = [(1 mol Cu2O)( Hfo of Cu2O)]

– [(2 mol Cu)( Hfo of Cu) + (1/2 mol O2)( Hfo of O2)]

H rxn = [(1 mol Cu2O)(–168.6 kJ/mol)] o

H rxn = –168.6 kJ

– [(2 mol Cu)(0 kJ/mol) + (1/2 mol O 2)(0 kJ/mol)]

168.6 kJ Hrxn 3 = = 0.56577 kJ/K(10 J/1 kJ) = 565.77 J/K T 298 K o Suniv = Srxn + Ssurr = (–75.6 J/K) + (565.77 J/K) = 490.17 = 490. J/K Ssurr = 

Because Suniv is positive, the reaction is spontaneous at 298 K. 20.40

Plan: Write the balanced equation. To calculate the standard entropy change (part a), use the relationship o o o Srxn = m Sproducts – n Sreactants .

To calculate ΔS of the universe in part b, first calculate ΔH of the reaction using the relationship o H rxn = m Hfo (products) – n Hfo (reactants)

a) One mole of hydrogen iodide is formed from its elements in their standard states according to the following equation: Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

20-14

1/2H2(g) + 1/2I2(s)  HI(g) S° = [(1 mol HI)(S° of HI)] – [(1/2 mol H2)(S° of H2) + (1/2 mol I2)(S° of I2)] S° = [(1 mol)(206.33 J/K∙mol)] – [(1/2 mol)(130.6 J/K∙mol) + (1/2 mol)(116.14 J/K∙mol)] S°= 82.96 = 83.0 J/K o b) H rxn = m Hfo (products) – n Hfo (reactants) o H rxn = [(1 mol HI)( Hfo of HI)]

– [(1/2 mol H2)( Hfo of H2) + (1/2 mol I2)( Hfo of I2)]

o H rxn = [(1 mol HI)(25.9 kJ/mol)]

H rxn = 25.9 kJ

– [(1/2 mol H2)(0 kJ/mol) + (1/2 mol I2)(0 kJ/mol)]

25.9 kJ Hrxn 3 = = – 0.0869128 kJ/K(10 J/1 kJ) = –86.9128 J/K T 298 K o Suniv = Srxn + Ssurr = (82.96 J/K) + (–86.9128 J/K) = –3.9528 = –4.0 J/K Ssurr = 

Because Suniv is negative, the reaction is not spontaneous at 298 K. 20.41

Plan: Write the balanced equation. To calculate the standard entropy change (part a), use the relationship o o o Srxn = m Sproducts – n Sreactants .

To calculate ΔS of the universe in part b, first calculate ΔH of the reaction using the relationship o H rxn = m Hfo (products) – n Hfo (reactants)

Solution: a) One mole of methanol is formed from its elements in their standard states according to the following equation: C(g) + 2H2(g) + 1/2O2(g)  CH3OH(l) S° = [(1 mol CH3OH)(S° of CH3OH)] – [(1 mol C)(S° of C) + (2 mol H2)(S° of H2) + (1/2 mol O2)(S° of O2)] S° = [(1 mol)(127 J/mol∙K)] – [(1 mol)(5.686 J/mol∙K) + (2 mol)(130.6 J/mol∙K) + (1/2 mol)(205.0 J/mol∙K)] S° = –242.386 = –242 J/K o H rxn = m Hfo (products) – n Hfo (reactants)

o H rxn = [(1 mol CH3OH)( Hfo of CH3OH)]

– [(1 mol C)( Hfo of C) + (2 mol H2)( Hfo of H2) + (1/2 mol O2)( Hfo of O2)]

o H rxn = [(1 mol CH3OH)(–238.6 kJ/mol)]

H rxn = –238.6 kJ

– [(1 mol C)(0 kJ/mol) + (2 mol H2)(0 kJ/mol) + (1/2 mol O2)(0 kJ/mol)]

238.6 kJ Hrxn 3 = = 0.80067 kJ/K(10 J/1 kJ) = 800.67 J/K T 298 K o Suniv = Srxn + Ssurr = (–242.386 J/K) + (800.67 J/K) = 558.284 = 558 J/K Ssurr = 

Because Suniv is positive, the reaction is spontaneous at 298 K. 20.42

Plan: Write the balanced equation. To calculate the standard entropy change (part a), use the relationship o o o Srxn = m Sproducts – n Sreactants .

20-15

To calculate ΔS of the universe in part b, first calculate ΔH of the reaction using the relationship o H rxn = m Hfo (products) – n Hfo (reactants) o Use ΔH of the reaction to calculate ΔS of the surroundings. o Hrxn Ssurr =  T o Add ΔS of the surroundings and ΔS of the reaction to calculate ΔS of the universe. If ΔS of the universe is greater than zero, the reaction is spontaneous at the given temperature. o Suniv = Srxn + Ssurr Solution: a) One mole of dinitrogen oxide is formed from its elements in their standard states according to the following equation: N2(g) + 1/2O2(g)  N2O(g) S° = [(1 mol N2O)(S° of N2O] – [(1 mol N2)(S° of N2) + (1/2 mol O2)(S° of O2)] S° = [(1 mol)(219.7 J/K∙mol)] – [(1 mol)(191.5 J/K∙mol) + (1/2 mol)(205.0 J/K∙mol)] S° = –74.3 J/K b)

o H rxn = m Hfo (products) – n Hfo (reactants)

o H rxn = [(1 mol N2O)( Hfo of N2O)] – [(1 mol N2)( Hfo of N2) + (1/2 mol O2)( Hfo of O2)] o H rxn = [(1 mol N2O)(82.05 kJ/mol)] – [(1 mol N2)(0 kJ/mol) + (1/2 mol O2)(0 kJ/mol)] o H rxn = 82.05 kJ 82.05 kJ Hrxn   0.27534 kJ / K 103 J /1 kJ   275.34 J/K Ssurr =  = T 298 K o Suniv = Srxn + Ssurr = (–74.3 J/K) + ( 275.34 J/K) = –349.64 = –349.6 J/K o

Because Suniv is negative, the reaction is not spontaneous at 298 K. 20.43

SO2(g) + Ca(OH)2(s)  CaSO3(s) + H2O(l) S° = [(1 mol CaSO3)(S° of CaSO3) + (1 mol H2O)(S° of H2O)] – [(1 mol SO2)(S° of SO2) + (1 mol Ca(OH)2)(S° of Ca(OH)2)] S° = [1 mol)(101.4 J/K∙mol) + (1 mol)(69.940 J/K∙mol)] – [(1 mol)(248.1 J/K∙mol) + (1 mol)(83.39 J/K∙mol)] S° = –160.15 = –160.2 J/K

20.44

Plan: Write the balanced equation. To calculate the standard entropy change, use the relationship o o o Srxn = m Sproducts – n Sreactants .

Solution: Complete combustion of a hydrocarbon includes oxygen as a reactant and carbon dioxide and water as the products. C2H2(g) + 5/2O2(g)  2CO2(g) + H2O(g) S° = [(2 mol CO2)(S° of CO2) + (1 mol H2O)(S° of H2O)] – [(1 mol C2H2)(S° of C2H2) + (5/2 mol O2)(S° of O2)] S° = [(2 mol)(213.7 J/mol∙K) + (1 mol)(188.72 J/mol∙K)] – [(1 mol)(200.85 J/mol∙K) + (5/2 mol)(205.0 J/mol∙K)] S° = –97.23 = –97.2 J/K 20.45

Reaction spontaneity may now be predicted from the value of only one variable (Gsys) rather than two (Ssys and Ssurr).

20.46

A spontaneous process has Suniv > 0. Since the Kelvin temperature is always positive, Gsys must be negative (Gsys < 0) for a spontaneous process.

20-16

20.47

a) G = H – TS. Since TS > H for an endothermic reaction to be spontaneous, the reaction is more likely to be spontaneous at higher temperatures. b) The change depicted is the phase change of a solid converting to a gas (sublimation). 1

Energy must be absorbed to overcome intermolecular forces to convert a substance in the solid phase to the gas phase. This is an endothermic process and H is positive. 2 Since gases have higher entropy values than solids, the process results in an increase in entropy and S is positive. 3 This is an endothermic process so the surroundings lose energy to the system. Ssurr is negative.  G = H – TS. Both H and S are positive. At low temperature, the H term will predominate and G will be positive; at high temperatures, the TS term will predominate and G will be negative.

20.48

Plan: Examine the provided diagrams. Determine whether bonds are being formed or broken in order to determine the sign of ΔH. Determine the relative number of particles before and after the reaction in order to determine the sign of ΔS. Solution: a) The sign of ΔH is negative. Bonds are being formed, so energy is released and ΔH is negative. b) The sign of ΔS is negative. There are fewer particles in the system after the reaction, so entropy decreases. o Hrxn c) The sign of ΔSsurr is positive. Ssurr =  Since ΔHrxn is negative, ΔSsurr will be positive. T d) Because both ΔH and ΔS are negative, this reaction will become more spontaneous (ΔG will become more negative) as temperature decreases.

20.49

H° is positive and S° is positive. The reaction is endothermic (H° > 0) and requires a lot of heat from its surroundings to be spontaneous. The removal of heat from the surroundings results in S° < 0. The only way an endothermic reaction can proceed spontaneously is if S° > 0, effectively offsetting the decrease in surroundings entropy. In summary, the values of H° and S° are both positive for this reaction. Melting is an example.

20.50

For a given substance, the entropy changes greatly from one phase to another, e.g., from liquid to gas. However, the entropy changes little within a phase. As long as the substance does not change phase, the value of S° is relatively unaffected by temperature.

20.51

Plan: G° can be calculated with the relationship m Gfo (products) – n Gfo (reactants) . Solution:

a) G° = [(2 mol MgO)( Gfo of MgO)] – [(2 mol Mg)( Gfo of Mg) + (1 mol O2)( Gfo of O2)]

Both Mg(s) and O2(g) are the standard-state forms of their respective elements, so their Gf values are zero. G° = [(2 mol)(–569.0 kJ/mol)] – [(2 mol)(0) + (1 mol)(0)] = –1138.0 kJ

b)G° = [(2 mol CO2)( Gfo of CO2) + (4 mol H2O)( Gfo of H2O)]

– [(2 mol CH3OH)( Gfo of CH3OH) + (3 mol O2)( Gfo of O2)]

 G° = [(2 mol)(–394.4 kJ/mol) + (4 mol)(–228.60 kJ/mol)]  [(2 mol)(–161.9 kJ/mol) + (3 mol)(0)] G° = –1379.4 kJ

c) G° = [(1 mol BaCO3)( Gfo of BaCO3)] – [(1 mol BaO)( Gfo of BaO) + (1 mol CO2)( Gfo of CO2)] G° = [(1 mol)(–1139 kJ/mol)] – [(1 mol)(–520.4 kJ/mol) + (1 mol)(–394.4 kJ/mol)] G° = –224.2 = –224 kJ 20.52

a) H2(g) + I2(s)  2HI(g)

G° = [(2 mol HI)( Gfo of HI] – [(1 mol H2)( Gfo of H2) + (1 mol I2)( Gfo of I2)] G° = [(2 mol)(1.3 kJ/mol)] – [(1 mol)(0 kJ/mol) + (1 mol)(0 kJ/mol)] G° = 2.6 kJ

20-17

b) MnO2(s) + 2CO(g)  Mn(s) + 2CO2(g) G° = [(1 mol Mn)( Gfo of Mn) + (2 mol CO2)( Gfo of CO2)] – [(1 mol MnO2)( Gfo of MnO2) + (2 mol CO)( Gfo of CO)] G° = [(1 mol)(0 kJ/mol) + (2 mol)(–394.4 kJ/mol)] – [(1 mol)(–466.1 kJ/mol) + (2 mol)(–137.2 kJ/mol)] G° = –48.3 kJ c) NH4Cl(s)  NH3(g) + HCl(g) G° = [(1 mol NH3)( Gfo of NH3) + (1 mol HCl)( Gfo of HCl)] – [(1 mol NH4Cl)( Gfo of NH4Cl)] G° = [(1 mol)(–16 kJ/mol) + (1 mol)(–95.30 kJ/mol)] – [1 mol)(–203.0 kJ/mol)] G° = 91.7 = 92 kJ 20.53

o Plan: H rxn can be calculated from the individual Hf values of the reactants and products by using the

o o relationship H rxn = m Hfo (products) – n Hfo (reactants) . Srxn can be calculated from the individual S o values

o o o o o of the reactants and products by using the relationship Srxn = m Sproducts – n Sreactants . Once H rxn and Srxn o o o o are known, G° can be calculated with the relationship Grxn = H rxn – T Srxn . Srxn values in J/K o must be converted to units of kJ/K to match the units of H rxn .

Solution:

o a) H rxn = [(2 mol MgO)( Hfo of MgO)] – [(2 mol Mg)( Hfo of Mg) + (1 mol O2)( Hfo of O2)]

 H rxn = [(2 mol)(–601.2 kJ/mol)] –[(2 mol)(0 kJ/mol) + (1 mol)(0 kJ/mol)] o

 H rxn = –1202.4 kJ o

Srxn = [(2 mol MgO)( S of MgO)] – [(2 mol Mg)( S of Mg) + (1 mol O2)( S of O2)] o

Srxn = [(2 mol)(26.9 J/mol∙K)] – [(2 mol)(32.69 J/mol∙K) + (1 mol)(205.0 J/mol∙K)] o

Srxn = –216.58 J/K o

 G rxn =  H rxn – T  S rxn = –1202.4 kJ – [(298 K)(–216.58 J/K)(1 kJ/10 J)] = –1137.859 = –1138 kJ o

b)  H rxn = [(2 mol CO2)(  H f of CO2) + (4 mol H2O)(  H f of H2O)] o

– [(2 mol CH3OH)(  H f of CH3OH) + (3 mol O2)(  H f of O2)] o

 H rxn = [(2 mol)(–393.5 kJ/mol) + (4 mol)(–241.826 kJ/mol)] – [(2 mol)(–201.2 kJ/mol) + (3 mol)(0 kJ/mol)] o

 H rxn = –1351.904 kJ o

Srxn = [(2 mol CO2)( S of CO2) + (4 mol H2O)( S of H2O)] o

– [(2 mol CH3OH)( S of CH3OH) + (3 mol O2)( S of O2)] o

Srxn = [(2 mol)(213.7 J/mol∙K) + (4 mol)(188.72 J/mol∙K)] o

– [(2 mol)(238 J/mol∙K) + (3 mol)(205.0 J/mol∙K)] = 91.28 J/K

 G rxn =  H rxn – T  S rxn = –1351.904 kJ – [(298 K)(91.28 J/K)(1 kJ/10 J)] = –1379.105 = –1379 kJ o

c)  H rxn = [(1 mol BaCO3)(  H f of BaCO3)] – [(1 mol BaO)(  H f of BaO) + (1 mol CO2)(  H f of CO2)] o

 H rxn = [(1 mol)(–1219 kJ/mol)] – [(1 mol)(–548.1 kJ/mol) + (1 mol)(–393.5 kJ/mol)] o

 H rxn = –277.4 kJ o

Srxn = [(1 mol BaCO3)( S of BaCO3)] – [(1 mol BaO)( S of BaO) + (1 mol CO2)( S of CO2)] o

Srxn = [(1 mol)(112 J/mol∙K)] – [(1 mol)(72.07 J/mol∙K) + (1 mol)(213.7 J/mol∙K)] Srxn = –173.77 J/K 3  G rxn =  H rxn – T  S rxn = –277.4 kJ – [(298 K)(–173.77 J/K)(1 kJ/10 J)] = –225.6265 = –226 kJ o o

20-18

20.54

a)  H rxn = [(2 mol HI)(25.9 kJ/mol)] – [(1 mol H2)(0 kJ/mol) + (1 mol I2)(0 kJ/mol)] o

 H rxn = 51.8 kJ o

Srxn = [(2 mol HI)(206.33 J/K∙mol)] – [(1 mol H2)(130.6 J/K∙mol) + (1 mol I2)(116.14 J/K∙mol)] o

Srxn = 165.92 J/K o

 G rxn =  H rxn – T  S rxn = 51.8 kJ – [(298 K)(165.92 J/K)(1 kJ/10 J)] = 2.3558 = 2.4 kJ o

b)  H rxn = [(1 mol Mn)(0 kJ/mol) + (2 mol CO2)(–393.5 kJ/mol)] o

– [(1 mol MnO2)(–520.9 kJ/mol) + (2 mol CO)(–110.5 kJ/mol)]

 H rxn

= –45.1 kJ

Srxn = [(1 mol Mn)(31.8 J/K∙mol) + (2 mol CO2)(213.7 J/K∙mol)] o

– [(1 mol MnO2)(53.1 J/K∙mol) + (2 mol CO)(197.5 J/K∙mol)]

Srxn

= 11.1 J/K

 G rxn =  H rxn – T  S rxn = –45.1 kJ – [(298 K (11.1 J/K)(1 kJ/10 J)] = –48.4078 = –48.4 kJ o

c)  H rxn = [(1 mol NH3)(–45.9 kJ/mol) + (1 mol HCl)(–92.3 kJ/mol)] – [(1 mol NH4Cl)(–314.4 kJ/mol)] o

 H rxn o

= 176.2 kJ

Srxn = [(1 mol NH3)(193 J/K∙mol) + (1 mol HCl)(186.79 J/K∙mol)] – [(1 mol NH4Cl)(94.6 J/K∙mol)] o

Srxn = 285.19 J/K o

 G rxn =  H rxn – T  S rxn = 176.2 kJ – [(298 K)(285.19 J/K)(1 kJ/10 J)] = 91.213 = 91.2 kJ o

20.55

Plan:  G rxn can be calculated with the relationship m Gf (products) – n Gf (reactants) . Alternatively,  G rxn o

can be calculated with the relationship  G rxn =  H rxn – T  S rxn . Entropy decreases (is negative) when there o

are fewer moles of gaseous products than there are of gaseous reactants.

Solution: a) Entropy decreases ( ΔS o negative) because the number of moles of gas decreases from reactants (1 1/2 mol) to products (1 mole). The oxidation (combustion) of CO requires initial energy input to start the reaction, but then releases energy (exothermic, ΔH o negative) which is typical of all combustion reactions. b) Method 1: Calculate  G rxn from  Gf values of products and reactants.  = m Gf(products) – n Gf(reactants)  G rxn o

 G rxn = [(1 mol CO2)(  Gf of CO2)] – [(1 mol CO)(  Gf of CO) + (1/2 mol)(  Gf of O2)] o

 G rxn = [(1 mol)(–394.4 kJ/mol)] – [(1 mol)(–137.2 kJ/mol) + (1/2 mol)(0 kJ/mol)] = –257.2 kJ o

Method 2: Calculate  G rxn from  H rxn and Srxn at 298 K (the degree superscript indicates a reaction at o

standard state, given in the Appendix at 25°C).  H rxn = m H fo (products) – n H fo (reactants) o

 H rxn = [(1 mol CO2)(  H f of CO2)] – [(1 mol CO)(  H f of CO) + (1/2 mol)(  H f of O2)] o

 H rxn = [(1 mol)(–393.5 kJ/mol)] – [(1 mol)(–110.5 kJ/mol) + (1/2 mol)(0 kJ/mol)] = –283.0 kJ o

o – n Sreactants Srxn = m Sproducts o

Srxn = [(1 mol CO2)( S of CO2)] – [(1 mol CO)( S of CO) + (1/2 mol)( S of O2)] o

Srxn = [(1mol)(213.7 J/mol∙K)] – [(1mol)(197.5 J/mol∙K) + (1/2 mol)(205.0 J/mol∙K)] o

Srxn = –86.3 J/K o

 G rxn =  H rxn – T  S rxn = (–283.0 kJ) – [(298 K)(–86.3 J/K)(1 kJ/10 J)] = –257.2826 = –257.3 kJ o

20-19

20.56

C4H10(g) + 13/2O2(g)  4CO2(g) + 5H2O(g) a) An increase in the number of moles of gas should result in a positive S° value. The combustion of C4H10(g) will result in a release of energy or a negative H° value. b)  H rxn = m H f (products) – n H f (reactants) o

 H rxn = [(4 mol CO2)(–393.5 kJ/mol) + (5 mol H2O)(–241.826 kJ/mol)] o

– [(1 mol C4H10)(–126 kJ/mol) + (13/2 mol O2)(0 kJ/mol)]  H rxn = –2657.13 kJ o

o – n Sreactants Srxn = m Sproducts o

Srxn = [(4 mol CO2)(213.7 J/K∙mol) + (5 mol H2O)(188.72 J/K∙mol)] o

– [(1 mol C4H10)(310 J/K∙mol) + (13/2 mol O2)(205.0 J/K∙mol)] Srxn = 155.9 J/K o

 G rxn =  H rxn – T  S rxn = –2657.13 kJ – [(298 K)(155.9 J/K)(1 kJ/10 J)] = –2703.588 = –2704 kJ o

 G rxn = m Gfo (products) – n Gfo (reactants) o

 G rxn = [(4 mol CO2)(–394.4 kJ/mol) + (5 mol H2O)(–228.60 kJ/mol)] o

– [(1 mol C4H10)(–16.7 kJ/mol) + (13/2 mol O2)(0 kJ/mol)]

 G rxn = –2703.9 kJ o

20.57

Plan: Use the relationship  G rxn =  H rxn – T  S rxn to find Srxn , knowing  H rxn and  G rxn . This relationship is also used to find  G rxn at a different temperature. Solution: o

Reaction is Xe(g) + 3F2(g)  XeF6(g)

a)  G rxn =  H rxn – T  S rxn ΔH°  ΔG° 402 kJ/mol  280. kJ/mol S° = = = –0.40939597 = –0.409 kJ/mol∙K T 298 K b)  G rxn =  H rxn – T  S rxn = (–402 kJ/mol) – [(500. K)(–0.40939597 kJ/mol∙K)] = –197.302 = –197 kJ/mol 20.58

a) S° =

ΔH°  ΔG° T

220. kJ /mol  206 kJ /mol = –0.046979865 = –0.047 kJ/mol∙K 298 K

b)  G rxn =  H rxn – T  S rxn = –220. kJ/mol – (450. K)(– 0.046979865 kJ/K∙mol) = –198.859 = –199 kJ/mol o

20.59

 Plan:  H rxn can be calculated from the individual  H f values of the reactants and products by using the

  relationship  H rxn = m H f(products) – n H f(reactants) . Srxn can be calculated from the individual S  values

    of the reactants and products by using the relationship Srxn = m Sproducts – n Sreactants and . Once  H rxn      are known, G° can be calculated with the relationship  G rxn =  H rxn – T  S rxn values in J/K Srxn . Srxn  must be converted to units of kJ/K to match the units of  H rxn . The temperature at which a reaction becomes spontaneous can be calculated by setting ΔG to zero in the free energy equation (assuming the reaction is at equilibrium) and solving for T. Solution:  a)  H rxn = [(1 mol CO)(  H f of CO) + (2 mol H2)(  H f of H2)] – [(1 mol CH3OH)(  H f of CH3OH)]  = [(1 mol)(–110.5 kJ/mol) + (2 mol)(0 kJ/mol)] – [(1 mol)(–201.2 kJ/mol)]  H rxn  = 90.7 kJ  H rxn

 = [(1 mol CO)( S  of CO) + (2 mol H2)( S  of H2)] – [(1 mol CH3OH)( S  of CH3OH)] Srxn  = [(1 mol)(197.5 J/mol∙K) + (2 mol)(130.6 J/mol∙K)] – [(1 mol)(238 J/mol∙K)] Srxn  = 220.7 = 221 J/K Srxn

20-20

   b)  G rxn =  H rxn – T  S rxn 3 T1 = 28 + 273 = 301 K G° = 90.7 kJ – [(301 K)(220.7 J/K)(1 kJ/10 J)] = 24.2693 = 24.3 kJ 3 T2 = 128 + 273 = 401 K G° = 90.7 kJ – [(401 K)(220.7 J/K)(1 kJ/10 J)] = 2.1993 = 2.2 kJ 3 T3 = 228 + 273 = 501 K G° = 90.7 kJ – [(501 K)(220.7 J/K)(1 kJ/10 J)] = –19.8707 = –19.9 kJ c) For the substances in their standard states, the reaction is nonspontaneous at 28°C, near equilibrium at 128°C,   and spontaneous at 228°C. Reactions with positive values of  H rxn and Srxn become spontaneous at high temperatures. d) The reaction will become spontaneous when ΔG changes from being positive to being negative. This point occurs when ΔG is 0.    =  H rxn – T  S rxn  G rxn 3

0 = 90.7 × 10 J – (T)(220.7 J/K)

T = 411 K At temperatures above 411 K, this reaction is spontaneous. (Because both ΔH and ΔS are positive, the reaction becomes spontaneous above this temperature.) 20.60

a) N2(g) + O2(g) ⇆ 2NO(g)

 = [(2 mol NO)(90.29 kJ/mol)] – [(1 mol N2)(0 kJ/mol) + (1 mol O2)(0 kJ/mol)]  H rxn  = 180.58 kJ  H rxn

 = [(2 mol NO)(210.65 J/K∙mol)] – [(1 mol N2)(191.5 J/K∙mol) + (1 mol O2)(205.0 J/K∙mol)] Srxn  = 24.8 J/K Srxn

b) G°373 = H° – ((273 + 100.)K) (S°) 3

= 180.58 kJ – [(373 K)(24.8 J/K)(1 kJ/10 J)] = 171.3296 = 171.33 kJ G°2833 = H° – ((273 + 2560.)K) (S°) 3

= 180.58 kJ – [(2833 K)(24.8 J/K)(1 kJ/10 J)] = 110.3216 = 110.3 kJ

G°3813 = H° – ((273 + 3540.)K) (S°) 3

= 180.58 kJ – [(3813 K)(24.8 J/K)(1 kJ/10 J)] = 86.0176 = 86.0 kJ

c) The values of G became smaller at higher temperatures. The reaction is not spontaneous at any of these temperatures; however, the reaction becomes less nonspontaneous as the temperature increases. d) The reaction will become spontaneous when ΔG changes from being positive to being negative. This point occurs when ΔG is 0.    =  H rxn – T  S rxn  G rxn 3

0 = 180.58 × 10 J – (T)(24.8 J/K)

T = 7280 K At temperatures above 7280 K, this reaction is spontaneous. (Because both ΔH and ΔS are positive, the reaction becomes spontaneous above this temperature.) Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

20-21

20.61

 Plan:  H rxn can be calculated from the individual  H f values of the reactants and products by using the   relationship  H rxn = m H f(products) – n H f(reactants) . Srxn can be calculated from the individual S  values      of the reactants and products by using the relationship Srxn = m Sproducts – n Sreactants . Once  H rxn and Srxn are     known, G° can be calculated with the relationship  G rxn =  H rxn – T  S rxn . Srxn values in J/K  must be converted to units of kJ/K to match the units of  H rxn . To find the temperature at which the reaction    becomes spontaneous, use  G rxn = 0 =  H rxn – T  S rxn and solve for temperature. Solution: a) The reaction for this process is H2(g) + 1/2O2(g)  H2O(g). The coefficients are written this way (instead of 2H2(g) + O2(g)  2H2O(g)) because the problem specifies thermodynamic values “per (1) mol H2,” not per 2 mol H2.  = m H f(products) – n H f(reactants)  H rxn  = [(1 mol H2O)(  H f of H2O)] – [(1 mol H2)(  H f of H2) + (1/2 mol O2)(  H f of O2)]  H rxn  = [(1 mol H2O)(–241.826 kJ/mol)] – [(1 mol H2)(0 kJ/mol) + (1/2 mol O2)(0 kJ/mol)]  H rxn  = –241.826 kJ  H rxn

   = m Sproducts – n Sreactants Srxn

 = [(1 mol H2O)( S  of H2O)] – [(1 mol H2)( S  of H2) + (1/2 mol O2)( S  of O2)] Srxn

 = [(1 mol)(188.72 J/mol∙K)] – [(1 mol)(130.6 J/mol∙K) + (1/2 mol)(205.0 J/mol∙K)] Srxn  = –44.38 = –44.4 J/K = –0.0444 kJ/K Srxn    =  H rxn – T  S rxn  G rxn

 = –241.826 kJ – [(298 K)( –0.04438 kJ/K)]  G rxn  = –228.6008 kJ = –228.6 kJ  G rxn

b) Because H < 0 and S < 0, the reaction will become nonspontaneous at higher temperatures because the positive (–TS) term becomes larger than the negative H term.

 c) The reaction becomes spontaneous below the temperature where  G rxn =0    = 0 =  H rxn – T  S rxn  G rxn   = T  S rxn  H rxn

T= 20.62

ΔH  –241.826 kJ 3 = = 5448.986 = 5.45 × 10 K ΔS  –0.04438 kJ/K

C6H12O6(s)  2C2H5OH(l) + 2CO2(g)  = m H f(products) – n H f(reactants)  H rxn

 = [(2 mol C2H5OH)(–277.63 kJ/mol) + (2 mol CO2)(–393.5 kJ/mol)] – [1 mol C6H12O6)(–1273.3 kJ/mol)]  H rxn  = –68.96 kJ = –69.0 kJ  H rxn

   = m Sproducts – n Sreactants Srxn

 = [(2 mol C2H5OH)(161 J/K∙mol) + (2 mol CO2)(213.7 J/K∙mol)] – [(1 mol C6H12O6)(212.1 K∙mol)] Srxn  = 537.3 J/K = 537 J/K Srxn    =  H rxn – T  S rxn  G rxn

 = –68.96 kJ – [(298 K)(0.5373 kJ/K)]  G rxn

 = –229.0754 kJ/mol = –229.1 kJ/mol  G rxn

No, a reaction with a negative value for H and a positive value for S is spontaneous at all temperatures. 20.63

a) An equilibrium constant that is much less than one indicates that very little product is made to reach equilibrium. The reaction, thus, is not spontaneous in the forward direction and G° is a relatively large positive value.

20-22

b) A large negative G° indicates that the reaction is quite spontaneous and goes almost to completion. At equilibrium, much more product is present than reactant so K > 1. Q depends on initial conditions, not equilibrium conditions, so its value cannot be predicted from G°. 20.64

For a spontaneous process, G is the maximum useful work obtainable from the system. In reality, the actual amount of useful work is less due to energy lost as heat. If the process is run in a slower or more controlled fashion, the actual amount of available work approaches G.

20.65

a) Point x represents the difference between Greactants and Gproducts or G°, the standard free energy change for the reaction. b) Scene A corresponds to Point 1 on the graph. This point corresponds to the pure substances, not a mixture. c) Scene C corresponds to Point 2 on the graph. Point 2 represents equilibrium; for this reaction, products dominate at equilibrium (the minimum in the curve is close to the XY side of the graph).

20.66

The standard free energy change, G°, occurs when all components of the system are in their standard states. Standard state is defined as 1 atm for gases, 1 M for solutes, and pure solids and liquids. Standard state does not specify a temperature because standard state can occur at any temperature. G° = G when all concentrations equal 1 M and all partial pressures equal 1 atm. This occurs because the value of Q = 1 and ln Q = 0 in the equation G = G° + RT ln Q.

20.67

Plan: For each reaction, first find G°, then calculate K from G° = –RT ln K. Calculate  G  using  Gf values  in the relationship  G rxn = m Gf(products) – n Gf(reactants) .



Solution: 2+ 2– a) MgCO3(s) ⇆ Mg (aq) + CO3 (aq)  = m Gf(products) – n Gf(reactants)  G rxn

 = [(1 mol Mg )(–456.01 kJ/mol) + (1 mol CO3 )(–528.10 kJ/mol)]  G rxn 2+

2–

– [(1 mol MgCO3)(–1028 kJ/mol)] = 43.89 kJ  10 3 J  43.89 kJ /mol  ΔG   = –17.7149  ln K = =   RT   8.314 J/mol  K 298 K  1 kJ    K=e

–17.7149





= 2.0254274 × 10 = 2.0 × 10 –8



–8

b) H2(g) + O2(g) ⇆ H2O2(l)  = [(1 mol H2O2)(–120.4 kJ/mol)] – [(1 mol H2)(0 kJ/mol) + (1 mol O2)(0 kJ/mol)] = –120.4 kJ/mol  G rxn   3  120.4 kJ/mol  ΔG  10 J  = 48.59596 ln K = =    RT   8.314 J/mol  K 298 K   1 kJ   K=e 20.68

48.59596





= 1.2733777 × 10 = 1.27 × 10 21



Solution:

 a)  G rxn = [(1 mol NaCN)(  Gf of NaCN) + (1 mol H2O)(  Gf of H2O)]

– [(1 mol HCN)(  Gf of HCN) + (1 mol NaOH)(  Gf of NaOH)]

NaCN(aq) and NaOH(aq) are not listed in Appendix B.

Converting the equation to net ionic form will simplify the problem:

20-23

–

HCN(aq) + OH (aq) ⇆ CN (aq) + H2O(l)

 = [(1 mol CN )(  Gf of CN ) + (1 mol H2O)(  Gf of H2O)]  G rxn –

–

– [(1 mol HCN)(  Gf of HCN) + (1 mol OH )(  Gf of OH )] –

–

 = [(1 mol)(166 kJ/mol) + (1 mol)(–237.192 kJ/mol)]  G rxn

G

– [(1 mol)(112 kJ/mol) + (1 mol)(–157.30 kJ/mol)]

 rxn

= –25.892 kJ  10 3 J  25.892 kJ/mol  ΔG   = 10.45055  ln K = =   RT   8.314 J/mol  K 298 K  1 kJ    K=e

10.45055





= 3.4563 × 10 = 3.46 × 10 4



2–

b) SrSO4(s) ⇆ Sr (aq) + SO4 (aq)

 = [(1 mol Sr )(–557.3 kJ/mol) + (1 mol SO4 )(–741.99 kJ/mol)]  G rxn 2+

2–

– [(1 mol SrSO4)(–1334 kJ/mol)] = 34.71 kJ  10 3 J  34.71 kJ /mol  ΔG   = –14.00968  ln K = =   RT 1 kJ    8.314 J/mol  K 298 K      K=e

20.69

–14.00968





= 8.23518 × 10 = 8.2 × 10 –7

–7



Plan: The equilibrium constant, K, is related to G° through the equation G° = –RT ln K. Solution:

a) G° = −RT ln K = −(8.314 J/mol∙K)(298 K) ln (5.62 × 10 ) = –203946 J = –204 kJ 35

b) G° = –RT ln K = –(8.314 J/mol∙K)(298 K) ln (4.46 × 10 ) = 8.18680 × 10 J/mol = 8.19 × 10 J/mol 15

20.70

Plan: The equilibrium constant, K, is related to G° through the equation G° = –RT ln K. Solution: 7 4 4 a) G° = –RT ln K = –(8.314 J/mol∙K)(298 K) ln (1.58 × 10 ) = –4.10670 × 10 J/mol = –4.11 × 10 J/mol 37 5 5 b) G° = –RT ln K = –(8.314 J/mol∙K)(298 K) ln (3.25 × 10 ) = –2.13999 × 10 J/mol = –2.14 × 10 J/mol

20.71

Plan: At the normal boiling point, defined as the temperature at which the vapor pressure of the liquid equals 1 atm, the phase change from liquid to gas is at equilibrium. For a system at equilibrium, the change in Gibbs free energy is zero. Since the gas is at 1 atm and the liquid assumed to be pure, the system is at standard state and     G° = 0. The temperature at which this occurs can be found from  G rxn = 0 =  H rxn – T  S rxn .  H rxn can be  calculated from the individual  H f values of the reactants and products by using the relationship   = m H f(products) – n H f(reactants) . Srxn can be calculated from the individual S  values of the  H rxn    reactants and products by using the relationship Srxn = m Sproducts – n Sreactants . Solution: Br2(l) ⇆ Br2(g)

 = m H f(products) – n H f(reactants)  H rxn

 = [(1 mol Br2)(  H f of Br2(g))] – [(1 mol Br2)(  H f of Br2(l))]  H rxn  = [(1 mol)(30.91 kJ/mol)] – [(1 mol)(0 kJ/mol)] = 30.91 kJ  H rxn

   = m Sproducts – n Sreactants Srxn

 = [(1 mol Br2)( S  of Br2(g))] – [(1 mol Br2)( S  of Br2(l))] Srxn

 = [(1 mol)(245.38 J/K∙mol)] – [(1 mol)(152.23 J/K∙mol)] = 93.15 J/K = 0.09315 kJ/K Srxn    = 0 =  H rxn – T  S rxn  G rxn   = T  S rxn  H rxn

20-24

ΔH  30.91 kJ = = 331.830 = 331.8 K  ΔS 0.09315 kJ/K S(rhombic) ⇆ S(monoclinic)  = [1 mol S(monoclinic)(0.3 kJ/mol)] – [1 mol S(rhombic)(0 kJ/mol)] = 0.3 kJ  H rxn  Srxn = [1 mol S(monoclinic)(32.6 J/K∙mol)] – [1 mol S(rhombic)(31.9 J/K∙mol)] = 0.7 J/K = 0.0007 kJ/K    = 0 =  H rxn – T  S rxn  G rxn T=

20.72

  = T  S rxn  H rxn

ΔH  0.3 kJ 2 = = 428.571 = 4 × 10 K  ΔS 0.0007 kJ/K Plan: Write the balanced equation. First find G°, then calculate K from G° = –RT ln K. Calculate  G  using  = m Gf(products) – n Gf(reactants) .  Gf values in the relationship  G rxn Solution: The solubility reaction for Ag2S is + 2– Ag2S(s) ⇆ 2Ag (aq) + S (aq) T=

20.73

 = m Gf(products) – n Gf(reactants)  G rxn

 = [(2 mol Ag )(  Gf of Ag ) + (1 mol S )(  Gf of S )]  [(1 mol Ag2S)(  Gf of Ag2S)]  G rxn +

2–

 = [(2 mol)(77.111 kJ/mol) + (1 mol)(83.7 kJ/mol)]  [(1 mol)(–40.3 kJ/mol)]  G rxn  = 278.222 kJ  G rxn

ln K = K=e 20.74

 10 3 J  278.222 kJ /mol  ΔG   = –112.296232  =   RT   8.314 J/mol  K 298 K  1 kJ   

–112.296232



= 1.6996759 × 10 2+

–49





= 1.70 × 10

–49

–

CaF2(s) ⇆ Ca (aq) + 2F (aq) 2+ –  = [(1 mol Ca )(–553.04 kJ/mol) + (2 mol F )(–276.5 kJ/mol)] – [(1 mol CaF2)(–1162 kJ/mol)]  G rxn  = 55.96 kJ  G rxn

 10 3 J  55.96 kJ /mol  ΔG    = –22.586629  ln K = =   RT 1 kJ    8.314 J/mol  K 298 K     

K=e

20.75

–22.586629



= 1.5514995 × 10

–10





= 1.55 × 10

–10

Plan: First find G°, then calculate K from G° = –RT ln K. Calculate  G  using  Gf values in the relationship  = m Gf(products) – n Gf(reactants) . Recognize that I2(s), not I2(g), is the standard state for iodine.  G rxn

Solution:  = m Gf(products) – n Gf(reactants)  G rxn   G rxn = [(2 mol ICl)(  Gf of ICl)] – [(1 mol I2)(  Gf of I2) + (1 mol Cl2)(  Gf of Cl2)]  = [(2 mol)(–6.075 kJ/mol)] – [(1 mol)(19.38 kJ/mol) + (1 mol)(0 kJ/mol)]  G rxn  = –31.53 kJ  G rxn

ln Kp = Kp = e 20.76

 10 3 J  31.53 kJ /mol  ΔG   = 12.726169  =   RT 1 kJ    8.314 J/mol  K 298 K     

12.726169





= 3.3643794 × 10 = 3.36 × 10 5



CaCO3(s) ⇆ CaO(s) + CO2(g) Kp = PCO2

 = [(1 mol CaO)(–603.5 kJ/mol) + (1 mol CO2)(–394.4 kJ/mol)] – [(1 mol CaCO3)(–1128.8 kJ/mol)]  G rxn

20-25

 = 130.9 kJ  G rxn

ln Kp = Kp = e

 10 3 J  130.9 kJ /mol  ΔG   = –52.83398  =   RT   8.314 J/mol  K 298 K  1 kJ   

–52.83398



= 1.1336892 × 10

–23



= 1.13 × 10



–23

atm = PCO2

20.77

Plan: The equilibrium constant, K, is related to G° through the equation G° = –RT ln K. Solution: –5 4 4 G° = –RT ln K = –(8.314 J/mol∙K)(298 K) ln (1.7 × 10 ) = 2.72094 × 10 J/mol = 2.7 × 10 J/mol The large positive G° indicates that it would not be possible to prepare a solution with the concentrations of lead and chloride ions at the standard-state concentration of 1 M. A Q calculation using 1 M solutions will confirm this: 2+ PbCl2(s) ⇆ Pb (aq) + 2Cl(aq) 2+ – 2 Q = [Pb ][Cl ] 2 = (1 M)(1 M) =1 2+ – Since Q > Ksp, it is impossible to prepare a standard-state solution of Pb (aq) and Cl (aq).

20.78

G° = –RT ln K = –(8.314 J/mol∙K)(298 K) ln (3.0 × 10 ) = 8.6877 × 10 J/mol = 8.7 × 10 J/mol The large positive G° indicates that it would not be possible to prepare a solution with the concentrations of zinc and fluoride ions at the standard-state concentration of 1 M. A Q calculation using 1 M solutions will confirm this: 2+ – 2 Q = [Zn ][F ] 2 = (1 M)(1 M) =1 Since Q > Ksp, it is impossible to prepare a standard state solution of ZnF2.

20.79

Plan: The equilibrium constant, K, is related to G° through the equation G° = –RT ln K. G is found by using the relationship G = G° + RT ln Q. Solution: –6 4 4 a) G° = –RT ln K = –(8.314 J/mol∙K)(298 K) ln (9.1 × 10 ) = 2.875776 × 10 = 2.9 × 10 J/mol  b) Since  G rxn is positive, the reaction direction as written is nonspontaneous. The reverse direction, formation of reactants, is spontaneous, so the reaction proceeds to the left. c) Calculate the value for Q and then use it to find G. 2 2 2 2  Fe 2   Hg 2   0.010  0.025 –4 Q =  3 2  2 = = 1.5625 × 10 2  0.20   0.010   Fe   Hg 2      4 –4 G = G° + RT ln Q = 2.875776 × 10 J/mol + (8.314 J/mol∙K)(298) ln (1.5625 × 10 )

–2

= 7.044187 × 10 = 7.0 × 10 J/mol 3

Because G298 > 0 and Q > K, the reaction proceeds to the left to reach equilibrium. 20.80

a) G° = –RT ln K = –(8.314 J/mol∙K)((273 + 25) K) ln (5.6 × 10 ) = –4.9906841 × 10 = –5.0 × 10 J/mol  b) Since  G rxn is negative, the reaction direction as written is spontaneous. The reaction proceeds to the right. c) Calculate the value for Q and then use it to find G.  Ni(NH 3 )6 2   0.010  14  Q =  2 6 = 6 = 6.4 × 10  Ni   NH 3  0.0010 0.0050       4 14 G = G° + RT ln Q = –4.9906841 × 10 J/mol + (8.314 J/mol∙K)(298) ln (6.4 × 10 ) 4 4 = 3.4559756 × 10 = 3.5 × 10 J/mol Because G298 > 0 and Q > K, the reaction proceeds to the left to equilibrium. 8

20-26

20.81

Plan: Start by writing the balanced equation for the reaction. Use the balanced equation to write the reaction quotient expression for the reaction. For part a, count the number of each type of particle present in the individual scenes. Using the given information that each particle represents 0.10 mol and the volume of the container is 0.10 L, calculate the molarity of each species in each individual scene. Use these concentrations and the reaction quotient expression to calculate the value of the reaction quotient for each scene to determine which scene is at equilibrium. For part b, the scene that is at equilibrium has ΔG = 0. Use the values of Q and the properties of logarithms to determine the relative values of ΔG for the other scenes. Solution: a) The balanced equation for the reaction is: 2A(g) ⇆ A2(g) The corresponding reaction quotient expression is: Q=

 A2 

 A

Number of A Particles 3 5 1

Scene 1 Scene 2 Scene 3 Scene 1: Molarity of A:

Molarity of A2: Q=

 0.10 mol    particle 

3 particles

 0.10 mol    particle 

3 particles

0.10 L

Scene 2: Molarity of A:

Molarity of A2: Q=

 0.10 mol    particle 

5 particles

 0.10 mol    particle 

2 particles

Molarity of A:

0.10 L

 0.10 mol    particle 

1 particle 

0.10 L

= 3.0 M

= 5.0 M

0.10 L

2.0 M = 0.080 2 5.0 M 

Scene 3:

= 3.0 M

0.10 L

3.0 M = 0.33 2 3.0 M

Number of A2 Particles 3 2 4

= 2.0 M

= 1.0 M

20-27

Molarity of A2: Q=

 0.10 mol    particle 

 4 particles

4.0 M = 4.0 2 1.0 M

0.10 L

= 4.0 M

The reaction quotient for scene 1 is equal to the equilibrium constant (K = 0.33), so scene 1 is at equilibrium. b) G = G° + RT ln Q

Scene 1 is at equilibrium, so its ΔG = 0. Scene 2 has Q = 0.08. Since Q < 1, ln Q is negative, which makes ΔG more negative than it is for the equilibrium scene. Scene 3 has Q = 4.0. Since Q > 1, ln Q is positive and ΔG is positive for scene 3. Scene 3 has the most positive ΔG, followed by Scene 1 (ΔG = 0) and Scene 2 (ΔG is negative). 20.82

Plan: Start by writing the balanced equation for the reaction. Use the balanced equation to write the reaction quotient expression for the reaction. For part a, count the number of each type of particle present in the individual scenes. Using the given information that each particle represents 0.10 atm, calculate the partial pressure of each species in each individual scene. Use these partial pressures and the reaction quotient expression to calculate the value of the reaction quotient for each scene to determine which scene is at equilibrium. For part b, the scene that is at equilibrium has ΔG = 0. Use the values of Q and the properties of logarithms to determine the relative values of ΔG for the other scenes. Solution: a) The balanced equation for the reaction is: X(g) + Y2(g) ⇆ XY(g) + Y(g)

The corresponding reaction quotient expression is: Q=

 XY  Y   X  Y2 

Number of X Particles 1 3 2

Number of Y2 Particles 1 2 1

Number of XY Particles 4 2 3

Scene 1 Scene 2 Scene 3 Scene 1: Partial Pressure of X: (1 particle)(0.10 atm/particle) = 0.10 atm

Number of Y Particles 2 2 3

Partial Pressure of Y2: (1 particle)(0.10 atm/particle) = 0.10 atm

Partial Pressure of XY: (4 particles)(0.10 atm/particle) = 0.40 atm Partial Pressure of Y: (2 particles)(0.10 atm/particle) = 0.20 atm 0.40 atm 0.20 atm  Q= = 8.0 0.10 atm 0.10 atm  Scene 2:

Partial Pressure of X: (3 particles)(0.10 atm/particle) = 0.30 atm

Partial Pressure of Y2: (2 particles)(0.10 atm/particle) = 0.20 atm

Partial Pressure of XY: (2 particles)(0.10 atm/particle) = 0.20 atm Partial Pressure of Y: (2 particles)(0.10 atm/particle) = 0.20 atm

20-28

Q= Scene 3:

0.20 atm 0.20 atm  = 0.67 0.30 atm 0.20 atm 

Partial Pressure of X: (2 particles)(0.10 atm/particle) = 0.20 atm Partial Pressure of Y2: (1 particle)(0.10 atm/particle) = 0.10 atm Partial Pressure of XY: (3 particles)(0.10 atm/particle) = 0.30 atm Partial Pressure of Y: (3 particles)(0.10 atm/particle) = 0.30 atm 0.30 atm 0.30 atm  Q= = 4.5 0.20 atm 0.10 atm 

The reaction quotient for scene 3 is equal to the equilibrium constant (K = 4.5), so scene 3 is at equilibrium. b) G = G° + RT ln Q

Scene 3 is at equilibrium, so its ΔG = 0. Scene 2 has Q = 0.67. Since Q < 1, ln Q is negative, which makes ΔG more negative than it is for the equilibrium scene. Scene 1 has Q = 8.0. Since Q > 1, ln Q is positive and ΔG is positive for scene 1. Scene 1 has the most positive ΔG, followed by Scene 3 (ΔG = 0) and Scene 2 (ΔG is negative). 20.83

  Plan: To decide when production of ozone is favored, both the signs of  H rxn and Srxn for ozone are needed.   The values of  H f and S  can be used. Once  H rxn and Srxn are known, G° can be calculated with the

    relationship  G rxn =  H rxn – T  S rxn . Srxn values in J/K must be converted to units of kJ/K to match the  units of  H rxn . G is found by using the relationship G = G° + RT ln Q.

Solution:

a) Formation of O3 from O2 : 3O2(g) ⇆ 2O3(g) or per mole of ozone: 3/2O2(g) ⇆ O3(g).  = m H f(products) – n H f(reactants)  H rxn

 = [(1 mol O3)(  H f of O3)] – [(3/2 mol O2)(  H f of O2)]  H rxn

 = [(1 mol)(143 kJ/mol] – [(3/2 mol)(0 kJ/mol)] = 143 kJ  H rxn

   = m Sproducts – n Sreactants Srxn

 = [(1 mol O3)( S  of O3)] – [(3/2 mol O2)( S  of O2)] Srxn

 = [(1 mol)(238.82 J/mol∙K] – [(3/2 mol)(205.0 J/mol∙K)] = –68.68 J/K = –0.06868 kJ/K Srxn

  The positive sign for  H rxn and the negative sign for Srxn indicate the formation of ozone is favored at no

temperature. The reaction is nonspontaneous at all temperatures.

   b)  G rxn =  H rxn – T  S rxn

 = 143 kJ – [(298 K)(–0.06868 kJ/K)] = 163.46664 = 163 kJ for the formation of one mole of O3.  G rxn

c) Calculate the value for Q and then use to find G.

 5  10 atm   O3    –6 Q= =  = 5.195664 × 10 3 3 2  O2  0.21 atm  2 7





G = G° + RT ln Q = 163.46664 kJ/mol + (8.314 J/mol∙K)(298)(1 kJ/10 J) ln (5.195664 × 10 ) 3

–6

= 133.3203215 = 1 × 10 kJ/mol 2

20.84

2–

BaSO4(s) ⇆ Ba (aq) + SO4 (aq)

The equilibrium constant, K, is related to G° through the equation G° = –RT ln K.

20-29

 103 J  59.1 kJ /mol  ΔG   = –22.930618  =   RT   8.314 J/mol  K (273  37)K  1 kJ    –22.930618 –10 –10 K=e = 1.099915 × 10 = 1.10 × 10

ln Kp =





2–

Ksp = [Ba ][SO4 ] = 1.099915 × 10 S= 20.85

–10

1.099915  1010 = 1.0487683 × 10 = 1.05 × 10 M Ba –5

–5

a) diamond  graphite  = [(1 mol graphite)(0 kJ/mol)] – [(1 mol diamond)(1.896 kJ/mol)] = –1.896 kJ/mol  H rxn  Srxn = [(1 mol graphite)(5.686 J/K∙mol)] – [(1 mol diamond)(2.439 kJ/mol)] = 3.247 J/K  = [(1 mol graphite)(0 kJ/mol)] – [(1 mol diamond)(2.866 kJ/mol)] = –2.866 kJ/mol  G rxn b) Since G is negative, the reaction diamond  graphite is spontaneous at room temperature. However, this does not give any information about the rate of reaction, which is very slow. Therefore, diamonds are not forever, but they are for a very long time. c) graphite  diamond For this process, the signs of H and S are, like the reaction, reversed. A process with H positive and S negative is nonspontaneous at all temperatures. Thus, something other than a change in temperature is necessary. That is why diamonds also require a change in pressure. d) Graphite cannot be converted to diamond spontaneously at 1 atm. At all temperatures G > 0 (nonspontaneous).

20.86

Srxn (a) (b) (c) (d) (e) (f)

+ (+)  0 () +

Hrxn  0 + () 0 +

Grxn   (+)  + ()

Comment Spontaneous Spontaneous Not spontaneous Spontaneous Not spontaneous TS > H

a) The reaction is always spontaneous when Grxn < 0, so there is no need to look at the other values other than to check the answer. b) Because Grxn = H – TS = –TS, S must be positive for Grxn to be negative. c) The reaction is always nonspontaneous when Grxn > 0, so there is no need to look at the other values other than to check the answer. d) Because Grxn = H – TS = H, H must be negative for Grxn to be negative. e) Because Grxn = H – TS = –TS, S must be negative for Grxn to be positive. f) Because TS > H, the subtraction of a larger positive term causes Grxn to be negative. 20.87

S° > 0 (Four aqueous species become seven aqueous species. The increase in the number of species increases the entropy.) G° < 0 (The reaction has K > 1.) The reaction proceeds due to an increase in entropy.

20.88

At the freezing point, the system is at equilibrium, so G = 0. G = 0 = H – TS H = TS –3 S = H/T = (–2.39 kJ/mol)/((273.2 + 63.7)K) = 7.0940932 × 10 kJ/mol∙K –3 –3 –3 S = (0.200 mol)(7.0940932 × 10 kJ/mol∙K) = 1.4188 × 10 = 1.42 × 10 kJ/K

20-30

20.89

a) False, all spontaneous reactions occur without outside intervention. b) True, a reaction cannot be spontaneous in both directions. c) False, all exothermic processes release heat. d) True e) False, if a process increases the freedom of motion of the particles of a system, the entropy of the system increases. f ) False, the energy of the universe is constant; the entropy of the universe increases toward a maximum. g) False, all systems with their surroundings increase the freedom of motion spontaneously. h) True

20.90

Plan: Write the equilibrium expression for the reaction. The equilibrium constant, K, is related to G° through the equation G° = –RT ln K. Once K is known, the ratio of the two species is known. Solution:  Hb  CO O 2  a) For the reaction, K = since the problem states that [O2] = [CO]; the K expression simplifies to:  Hb  O2  CO  K=

 Hb  CO  Hb  O2 

 103 J  14 kJ /mol  ΔG   = 5.431956979  =   RT 1 kJ    8.314 J/mol  K (273  37)K      Hb  CO   5.431956979 2 K=e = 228.596 = 2.3 × 10 =  Hb  O2  b) By increasing the concentration of oxygen, the equilibrium can be shifted in the direction of Hb∙O 2. Administer oxygen-rich air to counteract the CO poisoning.

ln Kp =

20.91





a) MgCO3(s) ⇆ MgO(s) + CO2(g) b) Locate the thermodynamic values.  = m H f(products) – n H f(reactants)  H rxn   H rxn = [(1 mol MgO)(–601.2 kJ/mol) + (1 mol CO2)(–393.5 kJ/mol)] – [(1 mol MgCO3)(–1112 kJ/mol)] = 117.3 kJ    = m Sproducts – n Sreactants Srxn  Srxn = [(1 mol MgO)(26.9 J/K∙mol) + (1 mol CO2)(213.7 J/K∙mol)] – [(1 mol MgCO3)(65.86 J/K∙mol)] = 174.74 J/K 3    =  H rxn – T  S rxn = 117.3 kJ – (298 K)(174.74 J/K)(1 kJ/10 J) = 65.22748 = 65.2 kJ  G rxn    c)  G rxn = 0 =  H rxn – T  S rxn   = T  S rxn  H rxn

ΔH  117.3 kJ = = 671.283 = 671 K  ΔS 0.17474 kJ/K

d) Kp = PCO2 and G° = –RT ln K are necessary. ln K = K=e

 10 3 J  65.22748 kJ /mol  ΔG   = –26.327178   =   RT   8.314 J/mol  K 298 K  1 kJ   

–26.327178



= 3.6834253 × 10

–12



= 3.68 × 10



–12

atm = PCO2

e) This is similar to part d) except a new G° must be determined at 1200 K.

   =  H rxn – T  S rxn = 117.3 kJ – (1200 K)(174.74 J/K)(1 kJ/10 J) = –92.388 kJ  G rxn 3

20-31

ln K = K=e 20.92

  3  92.388 kJ /mol  ΔG  10 J  = 9.26028  =    RT   8.314 J/mol  K 1200 K  1 kJ   

9.26028







= 1.0512 × 10 = 1.05 × 10 atm = PCO2 4

   Plan: Sum the two reactions to yield an overall reaction. Use the relationship  G rxn =  H rxn – T  S rxn to    calculate  G rxn .  H rxn and Srxn will have to be calculated first.

Solution:

UO2(s) + 4HF(g)  UF4(s) + 2H2O(g) UF4(s) + F2(g)  UF6(s) UO2(s) + 4 HF(g) + F2(g)  UF6(g) + 2H2O(g) (overall process)

= m H f(products) – n H f(reactants) = [(1 mol UF6)(  H f of UF6) + (2 mol H2O)(  H f of H2O)] – [(1 mol UO2)(  H f of UO2) + (4 mol HF)(  H f of HF) + (1 mol F2)(  H f of F2)]   H rxn = [(1 mol)(–2197 kJ/mol) + (2 mol)(–241.826 kJ/mol)] – [(1 mol)(–1085 kJ/mol) + (4 mol)(–273 kJ/mol) + (1 mol)(0 kJ/mol)]  = –503.652 kJ  H rxn    = m Sproducts – n Sreactants Srxn   Srxn = [(1 mol UF6)( S of UF6) + (2 mol H2O)( S  of H2O)] – [(1 mol UO2)( S  of UO2) + (4 mol HF)( S  of HF) + (1 mol F2)( S  of F2)]  Srxn = [(1 mol)(225 J/mol∙K) + (2 mol)(188.72 J/mol∙K)] – [(1 mol)(77.0 J/mol∙K) + (4 mol)(173.67 J/mol∙K) + (1 mol)(202.7 J/mol∙K)]  = –371.94 J/K Srxn 3    =  H rxn – T  S rxn = (–503.652 kJ) – ((273 + 85)K)(–371.94 J/K)(kJ/10 J) = –370.497 = –370. kJ  G rxn H H

20.93

 rxn  rxn

CO(g) + 2H2(g)  CH3OH(l)

 a)  H rxn = [(1 mol CH3OH)(–238.6 kJ/mol)] – [(1 mol CO)(–110.5 kJ/mol) + (2 mol H2)(0 kJ/mol)]

= –128.1 kJ

 = [(1 mol CH3OH)(127 J/K∙mol)] – [(1 mol CO)(197.5 J/K∙mol) + (2 mol H2)(130.6 J/K∙mol)] Srxn

G

 rxn

= –331.7 J/K

  =  H rxn – T  S rxn = –128.1 kJ – [(298 K)(–331.7 J/K)(1 kJ/10 J)] = –29.2534 = –29.2 kJ 3

The negative value of G° indicates that the reaction is spontaneous (feasible).

b) H° negative and S° negative means the reaction is favored at low temperature. c) CH3OH(g) + 1/2O2(g)  CH2O(g) + H2O(g)

 = [(1 mol CH2O)(–116 kJ/mol) + (1 mol H2O)(–241.826 kJ/mol)]  H rxn

 [(1 mol CH3OH)(–201.2 kJ/mol) + (1/2 mol O2)(0 kJ/mol)]

S

 rxn

= –156.626 kJ

= [(1 mol CH2O)(219 J/K∙mol) + (1 mol H2O)(188.72 J/K∙mol)]  [(1 mol CH3OH)(238 J/K∙mol) + (1/2 mol O2)(205.0 J/K∙mol)] = 67.22 J/K

   =  H rxn – T  S rxn = –156.626 kJ – [((273 + 100.)K)(67.22 J/K)(1 kJ/10 J)] = –181.699 = –182 kJ  G rxn

20.94

 Plan:  G rxn can be calculated with the relationship m Gf(products) – n Gf(reactants) . G is found by using the relationship G = G° + RT ln Q.

20-32

Solution: a) 2N2O5(g) + 6F2(g)  4NF3(g) + 5O2(g)  b)  G rxn = m Gf(products) – n Gf(reactants)

  = [(4 mol NF3)(  Gf of NF3) + (5 mol O2)(  G rxn of O2)]  G rxn

  – [(2 mol N2O5)(  G rxn of N2O5)) + (6 mol F2) (  G rxn of F2)]

 = [(4 mol)(–83.3 kJ/mol)) + (5 mol)(0 kJ/mol)]  [(2 mol)(118 kJ/mol) + (6 mol)(0 kJ/mol)]  G rxn  = –569.2 = –569 kJ  G rxn

c) Calculate the value for Q and then use to find G.



 



0.25atm  0.50atm   NF3  O2      = 47.6837 Q= 2 6 2 6 =     N O F  2 5  2  0.20atm  0.20atm  4

    3 G = G° + RT ln Q = –569.2 kJ/mol + (1 kJ/10 J)(8.314 J/mol∙K)(298) ln (47.6837) = –559.625 = –5.60 × 10 kJ/mol 2

20.95

 a) Srxn = [(2 mol NO)(210.65 J/K∙mol) + (1 mol Br2)(245.38 J/K∙mol)] – [(2 mol NOBr)(272.6 J/mol∙K)]  = 121.48 = 121.5 J/K Srxn

 b)  G rxn = –RT ln K = –(8.314 J/mol∙K)(373 K) ln (0.42) = 2690.225 = 2.7 × 10 J/mol 3

   c)  G rxn =  H rxn – T  S rxn

   =  G rxn + T  S rxn = 2690.225 J/mol + (373)(121.48 J/K) = 4.8002265 × 10 = 4.80 × 10 J/mol  H rxn 4

 d)  H rxn = [(2 mol NO)(90.29 kJ/mol) + (1 mol Br2)(30.91 kJ/mol)] – [(2 mol NOBr)(  H f of NOBr)]

(2 mol NOBr)(  H f of NOBr)] = [(2 mol)(90.29 kJ) + (1 mol)(30.91 kJ)] – (4.8002265 × 10 J)(1 kJ/10 J)) 4

 H f of NOBr = 81.7438675 = 81.7 kJ/mol

   e)  G rxn =  H rxn – T  S rxn = 4.8002265 × 10 J/mol – (298 K)(121.48 J/mol∙K) 4

= 1.1801225 × 10 = 1.18 × 10 J/mol 4

 f)  G rxn = [(2 mol NO)(86.60 kJ/mol) + (1 mol Br2)(3.13 kJ/mol)] – [(2 mol NOBr(g)(  Gf of NOBr)]

(2 mol NOBr(g)(  Gf of NOBr) = [(2 mol)(86.60 kJ) + (1 mol)(3.13 kJ)] – (1.1801225 × 10 J)(1 kJ/10 J) 4

 Gf of NOBr = 82.264 = 82.3 kJ/mol

20.96

The reaction for the hydrogenation of ethene is C2H4(g) + H2(g)  C2H6(g).  = m H f(products) – n H f(reactants)  H rxn

 = [(1 mol C2H6)(  H f of C2H6)] – [(1 mol C2H4)(  H f of C2H4) + (1 mol H2)(  H f of H2)]  H rxn

= [(1 mol)(–84.667 kJ/mol)] – [(1 mol)(52.47 kJ/mol) + (1 mol)(0 kJ/mol)] = –137.137 = –137.14 kJ

   = m Sproducts – n Sreactants Srxn

 = [(1 mol C2H6)( S  of C2H6)] – [(1 mol C2H4)( S  of C2H4)) + (1 mol H2)( S  of H2)] Srxn

= [(1 mol)(229.5 J/mol∙K)] – [(1 mol)(219.22 J/mol∙K) + (1 mol)(130.6 J/mol∙K)] = –120.32 = –120.3 J/K

 = m Gf(products) – n Gf(reactants)  G rxn

20-33

 = [(1 mol C2H6)(  Gf of C2H6))] – [(1 mol C2H4)(  Gf of C2H4) + (1 mol H2)(  Gf of H2)]  G rxn

= (1 mol)(–32.89 kJ/mol)] – [(1 mol)(68.36 kJ/mol) + (1 mol)(0 kJ/mol)] = –101.25 kJ

20.97

a) C6H5CH2CH3(g) → H2(g) + C6H5CHCH2(g)

 = [(1 mol H2)(  H f of H2) + (1 mol C6H5CHCH2) ( H f of C6H5CHCH2)]  H rxn

– [(1 mol C6H5CH2CH3) ( H f of C6H5CH2CH3)]

 = [(1 mol)(0 kJ/mol) + (1 mol)(103.8 kJ/mol)] – [(1 mol)(–12.5 kJ/mol)] = 116.3 kJ  H rxn  = [(1 mol H2)(  Gf of H2) + (1 mol C6H5CHCH2)(  Gf of C6H5CHCH2)]  G rxn

G S

 rxn

– [(1 mol C6H5CH2CH3) ( Gf of C6H5CH2CH3)]

= [(1 mol)(0 kJ/mol) + (1 mol)(202.5 kJ/mol)] – [(1 mol)(119.7 kJ/mol)] = 82.8 kJ

= [(1 mol H2)( S  of H2) + (1 mol C6H5CHCH2)( S  of C6H5CHCH2)] – [(1 mol C6H5CH2CH3)( S  of C6H5CH2CH3)]

 = [(1 mol)(130.6 J/mol∙K) + (1 mol)(238 J/mol∙K)] – [(1 mol)( 255 J/mol∙K)] = 113.6 = 114 J/K Srxn

   b)  G rxn = 0 =  H rxn – T  S rxn   = T  S rxn  H rxn

ΔH  116.3 kJ = = 1023.7676= 1020 K = 750°C ΔS  0.1136 kJ/K 3    c)  G rxn =  H rxn – T  S rxn = 116.3 kJ – (600°C + 273)(114 J/K)(1 kJ/10 J) = 16.778 = 16.8 kJ/mol  10 3 J  16.778 kJ/mol  ΔG   = –2.311617  ln K = =   RT  8.314 J/mol  K 873 K  1 kJ    T=

K=e

–2.311617

= 9.9101 × 10 = 9.9 × 10 –2

–2

d) Calculate the value for Q and then use it to find G. Initial

C6H5CH2CH3(g) → H2(g) + C6H5CHCH2(g) 1.0

React

– 0.50

+0.50

+0.50 (50% conversion)

Remaining

0.50

0.50 0.50 amount of gas 0.5 Mole fraction of each gas = = = 0.076923 H 2 + styrene + ethylbenzene + steam 0.5 + 0.5 + 0.5 + 5 Partial pressure of each gas = mole fraction x total pressure = 0.076923 × 1.3 atm = 0.10 atm  H2 C6 H 5CHCH2   0.10 atm  0.10 atm  Q= = = 0.10  0.10atm  C6 H5CH2 CH2    G = G° + RT ln Q = 16.778 kJ/mol + (1 kJ/10 J)(8.314 J/mol∙K)(873) ln (0.10) 3

= 0.0655565 = 0.07 kJ/mol 20.98

a) The formation reaction for propylene is 3C(s) + 3H2(g) → CH3CH=CH2(g) Sf = [(1 mol CH3CH=CH2)( S  of CH3CH=CH2)] – [(3 mol C)( S  of C) + (3 mol H2)( S  of H2)]

= [(1 mol)(267.1 J/mol∙K)] – [(3 mol)(5.686 J/mol∙K) + (3 mol)(130.6 J/mol∙K)] = –141.758 = –141.8 J/K

b)  Gf =  H f – T  Sf

 Gf = 20.4 kJ/mol – (298 K)(–141.758 J/K)(1 kJ/10 J) = 62.643884 = 62.6 kJ/mol 3

c) The dehydrogenation reaction is CH3CH2CH3(g) → CH3CH=CH2(g) + H2(g)

 = [(1 mol CH3CH=CH2)(  H f of CH3CH=CH2) + (1 mol H2)(  H f of H2)]  H rxn

20-34

– [(1 mol CH3CH2CH3) (  H f of CH3CH2CH3)]

G

 rxn

= [(1 mol)(20.4 kJ/mol) + (1 mol)(0 kJ/mol)] – [(1 mol)( –105 kJ/mol)] = 125.4 = 125 kJ = [(1 mol CH3CH=CH2)(  Gf of CH3CH=CH2) + (1 mol H2)(  Gf of H2)] – [(1 mol CH3CH2CH3)(  Gf of CH3CH2CH3)]

= [(1 mol )(62.643884 kJ/mol) + (1 mol)(0 kJ/mol)] – [(1 mol)( –24.5 kJ/mol)] = 87.143884 = 87.1 kJ

  d) Find K from  G rxn at this elevated temperature. Srxn must be found first:  = [(1 mol CH3CH=CH2)( S  of CH3CH=CH2) + (1 mol H2)( S  of H2)] Srxn

– [(1 mol CH3CH2CH3)( S  of CH3CH2CH3)]

= [(1 mol)(267.1 J/mol∙K) + (1 mol)(130.6 J/mol∙K)] – [(1 mol)(269.9 J/mol∙K)] = 127.8 J/K

 G =  H f – T  Sf  f

 Gf = 125.4 kJ – (853 K)(127.8 J/K)(1 kJ/10 J) = 16.3866 kJ

 10 3 J  16.3866 kJ /mol  ΔG   = –2.3106267  =   RT 1 kJ    8.314 J/mol  K 853 K     

ln K = K=e K=

–2.3106267

P





= 0.099199 = 0.0992

CH 3 CH  CH 2

P



 P   H2

CH 3 CH 2 CH 3

x = pressure of CH3CH=CH2 and the pressure of H2 and 1.00 – x = pressure of CH3CH2CH3  x  x  0.099199 = 1.00  x 2

x + 0.099199x – 0.099199 = 0 Solving the quadratic equation: x= x=

b  b 2  4ac 2a

(0.099199)  0.099199  4 10.099199 2

2 1

x = 0.2692408 = 0.269 atm The theoretical yield of propylene is 27%.

e) If hydrogen could escape through the reactor walls, the reaction would be shifted to the right, improving the yield.   f)  H rxn = T  S rxn  ΔH 125.4 kJ T= = = 981.22066 = 981 K = 708°C ΔS  0.1278 kJ/K 20.99

The reactions are (from text): 4– 3– 2– + ATP + H2O ⇆ ADP + HPO4 + H 2– + – Glucose + HPO4 + H ⇆ [glucose phosphate] + H2O 4– 3–  Glucose + ATP ⇆ [glucose phosphate] + ADP G° = –RT ln K T = (273 + 25) K = 298 K  10 3 J  30.5 kJ /mol  ΔG    = 12.3104  a) ln K = =   RT 1 kJ    8.314 J/mol  K 298 K     



K=e

12.3104





= 2.2199 × 10 = 2.22 × 10 5

G° = –30.5 kJ G° = 13.8 kJ G° = –16.7 kJ

20-35

b) ln K =

 10 3 J  13.8 kJ/mol  ΔG   = –5.569969  =   RT   8.314 J/mol  K 298 K  1 kJ   



K=e

c) ln K =

–5.569969





= 3.8105985 × 10 = 3.81 × 10 –3

–3

 103 J  16.7 kJ /mol  ΔG   =   = 6.74047  RT 1 kJ    8.314 J /mol  K 298 K     



K=e

6.74047





= 8.45958 × 10 = 8.46 × 10 2

d) Repeat the calculations at the new temperature. T = (273 + 37) = 310. K  10 3 J  30.5 kJ /mol  ΔG    ln K = =   = 11.8339    RT  1 kJ    8.314 J /mol  K 310. K 



K=e

ln K =



= 1.37847 × 10 = 1.38 × 10 5

 10 3 J  13.8 kJ /mol  ΔG     =   = –5.3543576    RT   8.314 J /mol  K 310. K  1 kJ   



K=e

ln K =

11.8339



–5.3543576





= 4.7275 × 10 = 4.73 × 10 –3

–3

 10 3 J  16.7 kJ /mol  ΔG    =   = 6.4795   RT   8.314 J /mol  K 310. K  1 kJ   



K=e

6.4795





= 6.51645 × 10 = 6.52 × 10 2

20.100 a) ATP + H2O ⇆ ADP + HPO4 + H G° = –30.5 kJ  10 3 J  30.5 kJ /mol  ΔG    ln K = =    = 11.8339 RT 1 kJ   8.314 J /mol  K 310. K       4–

3–



K=e

11.8339

2–





= 1.37847 × 10 = 1.38 × 10 5

b) C6H12O6(s) + 6O2(g)  6CO2(g) + 6H2O(l)

 = [(6 mol CO2)(  Gf of CO2) + (6 mol H2O)(  Gf of H2O)]  G rxn

– [(1 mol C6H12O6 )(  Gf of C6H12O6) + (6 mol O2)(  Gf of O2)]

 = [(6 mol)(–394.4 kJ/mol) + (6 mol)(–237.192 kJ/mol)]  [(1 mol)(–910.56 kJ/mol) + (6 mol)(0 kJ/mol)]  G rxn

 = –2878.992 = –2879.0 kJ  G rxn

1 mol ATP  2878.992 kJ     = 94.39318 = 94.4 mol ATP/mol glucose c)     1 mol glucose   30.5 kJ 

d) The actual yield is 36 moles ATP (this is assumed to be an exact value). The percent yield is:  36 mol ATP     100% = 38.1383 = 38.1%   94.39318 mol ATP  20.101 SO3(g) + SCl2(l)  SOCl2(l) + SO2(g)  a)  H rxn = m H f(products) – n H f(reactants)

 = –75.2 kJ  G rxn

 = [(1 mol SOCl2)(  H f of SOCl2) + (1 mol SO2)(  H f of SO2)]  H rxn

20-36

– [(1 mol SO3)(  H f of SO3) + (1 mol SCl2 )(  H f of SCl2)]

 = [(1 mol)(–245.6 kJ/mol) + (1 mol)(–296.8 kJ/mol)] – [(1 mol)(–396 kJ/mol) + (1 mol)(–50.0 kJ/mol)]  H rxn  = –96.4 kJ  H rxn

   = –75.2 kJ/mol =  H rxn – T  S rxn  G rxn

 –75.2 kJ/mol = –96.4 kJ/mol – T  S rxn

 75.2 kJ/mol – 96.4 kJ/mol = –21.2 kJ/mol = T  S rxn

 (–21.2 kJ/mol)/(298 K) = Srxn

 = (–0.0711409 kJ/mol∙K)(10 J/1 kJ) = –71.1409 J/mol∙K Srxn 3

 = [(1 mol SOCl2)( S  of SOCl2) + (1 mol SO2)( S  of SO2)] Srxn

– [(1 mol SO3)( S  of SO3) + (1 mol SCl2)( S  of SCl2)]

–71.1409 J/mol∙K = [(1 mol)( S  of SOCl2) + (1 mol)(248.1 J/mol∙K)]

– [(1 mol)(256.7 J/mol∙K)) + (1 mol)(184 J/mol∙K)]



–71.1409 J/mol∙K = [(1 mol)( S of SOCl2) + 248.1 J/mol∙K] – [(440.7 J/mol∙K)] –71.1409 J/mol∙K – 248.1 J/mol∙K + 440.7 J/mol∙K = 1 mol( S  of SOCl2) 121.4591 J/mol∙K = 1 mol( S  of SOCl2)

S  of SOCl2 = 121 J/mol∙K

   b)  G rxn = 0 =  H rxn – T  S rxn   = T  S rxn  H rxn

ΔH  96.4 kJ 3 = = 1355.057 = 1.4 × 10 K ΔS  0.0711409 kJ/K 4Fe(s) + 3O2(g)  2Fe2O3(s)

20.102 Oxidation of iron:

The change in Gibbs free energy for this reaction will be twice  Gf of Fe2O3 since the reaction forms the oxide from its elements and is reported per mole.  Gf = [(2 mol Fe2O3)(–743.6 kJ/mol)] – [(4 mol Fe)(0 kJ/mol) + (3 mol O2)(0 kJ/mol)] = –1487.2 kJ Oxidation of aluminum:

4Al(s) + 3O2(g)  2Al2O3(s)

The change in Gibbs free energy for this reaction will be twice  Gf of Al2O3 since the reaction forms the oxide from its elements and is reported per mole.

 Gf = [(2 mol Al2O3)(–1582 kJ/mol)] – [(4 mol Al)(0 kJ/mol) + (3 mol O2)(0 kJ/mol)] = –3164 kJ

With G° < 0, both reactions are spontaneous at 25°C. 20.103 a) H2(g) + I2(g) Kc =

⇆

2HI(g)

 0.10   HI  = = 50  H 2  I 2   0.010  0.020  2

b) Kp = Kc(RT)

Kc > 1

Δngas

Kp = 50[(0.0821 Latm/molK)(733 K)]

Kp = 50 = Kc

c) G° = –RT ln K = –(8.314 J/mol∙K)(733 K) ln (50) = 2.38405 × 10 = 2.4 × 10 J/mol 4

d) Kc =

2  0.10   HI  = = 50  H 2  I 2   0.020  0.010  2

The value of Kc is 50 in this situation as in a) so G° does not change. Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

20-37

20.104 Plan: Use the relationship G̥̥° = –RT ln K to calculate G°. Use the relationship G = G° + RT ln Q to calculate G. Solution: The given equilibrium is: G6P ⇆ F6P

K = 0.510 at 298 K

a) G° = –RT ln K

= – (8.314 J/mol∙K)(298 K) ln (0.510) = 1.6682596 × 10 = 1.67 × 10 J/mol 3

b) Q = [F6P]/[G6P] = 10.0 G = G° + RT ln Q G = 1.6682596 × 10 J/mol + (8.314 J/mol∙K)(298 K) ln 10.0 3

G = 7.3730799 × 10 = 7.37 × 10 J/mol 3

c) Repeat the calculation in part b) with Q = 0.100 G = 1.6682596 × 10 J/mol + (8.314 J/mol∙K)(298 K) ln 0.100 3

G = –4.0365607 × 10 = –4.04 × 10 J/mol 3

d) G = G° + RT ln Q 3

(–2.50 kJ/mol)(10 J/1 kJ) = 1.6682596 × 10 J/mol + (8.314 J/mol∙K)(298 K) ln Q 3

(–4.1682596 × 10 J/mol) = (8.314 J/mol∙K)(298 K) ln Q 3

ln Q = (–4.1682596 × 10 J/mol)/[(8.314 J/mol∙K)(298 K)] = –1.68239696 Q = 0.18592778 = 0.19

20.105 a) Graph D depicts how Gsys changes for the chemical reaction. G decreases as the reaction proceeds from either pure reactant or pure product until it reaches the minimum at equilibrium. Beyond that in either direction the reaction is nonspontaneous. b) Graph A depicts how Gsys changes as ice melts at 1C. 1C is higher than the melting point of water, therefore the system is not at equilibrium. Melting is spontaneous at 1C and 1 atm, and the Gsys will decrease until the system reaches equilibrium. 20.106 The equilibrium may be represented as: double helix ⇆ 2 random coils. a) S° is positive. The formation of a larger number of components increases randomness. b) G = H – [TS] Energy is required to overcome the hydrogen bonds and the dispersion forces between the strands. Thus, H is positive. The Kelvin temperature is, of course, positive. (+) – [(+) (+)] G is positive when TS < H. c) G = 0 (at equilibrium) G = 0 = H – TS H = TS T = H/S 20.107 a) Respiration: C6H12O6(s) + 6O2(g)  6CO2(g) + 6H2O(l) Fermentation: C6H12O6(s)  2C2H5OH(l) + 2CO2(g) Ethanol oxidation: C2H5OH(l) + 3O2(g)  2CO2(g) + 3H2O(l)  b)  G rxn = [(6 mol CO2 )(  Gf of CO2) + (6 mol H2O)(  Gf of H2O)] – [(1 mol C6H12O6)(  Gf of C6H12O6) + (6 mol O2 )(  Gf of O2)]   G rxn = [(6 mol)(–394.4 kJ/mol) + (6 mol)(–237.192 kJ/mol)] – [(1 mol)(–910.56 kJ/mol) + (6 mol)(0 kJ/mol)]  = –2878.992 kJ for one mole of C6H12O6  G rxn

20-38

 1 mol glucose  2878.992 kJ      = –15.980195 = –16.0 kJ/g /g of glucose = 1.00 g glucose   G rxn 180.16 g glucose  1 mol glucose 





 c)  G rxn = [(2 mol CO2)(  Gf of CO2) + (2 mol C2H5OH)(  Gf of C2H5OH)]

– [(1 mol C6H12O6)(  Gf of C6H12O6)]

 = [(2 mol)(–394.4 kJ/mol) + (2 mol)(–174.8 kJ/mol)] – [(1 mol)(–910.56 kJ/mol)]  G rxn  = –227.84 kJ for one mole of C6H12O6  G rxn

 1 mol glucose  227.84 kJ      = –1.264653641 = –1.26 kJ/g /g of glucose = 1.00 g glucose   G rxn  180.16 g glucose 1 mol glucose 





 d)  G rxn = [(2 mol CO2)(  Gf of CO2) + (3 mol H2O)(  Gf of H2O)]

– [(1 mol C2H5OH)(  Gf of C2H5OH) + (3 mol O2)(  Gf of O2)]

 = [(2 mol)(–394.4 kJ/mol) + (3 mol)(–237.192 kJ/mol)] – [(1 mol)(–174.8 kJ/mol) + (3 mol)(0 kJ/mol)]  G rxn  = –1325.576 kJ for one mole of C2H5OH  G rxn

 1 mol glucose  2 mol C 2 H 5OH  1325.576 kJ      = 1.00 g glucose   G rxn    = –14.71554 = –14.7 kJ/g 180.16 g glucose  1 mol glucose 1 mol C 2 H 5OH 





ΔG  , since ln Kp = 0 when Kp = 1.00, RT  G  = 0. Use the relationship  G  = 0 =  H  – T  S  to find the temperature. For part b), calculate  G  at ΔG  the higher temperature with the relationship  G  =  H  – T  S  and then calculate K with ln K = . RT Solution:

20.108 Plan: First calculate  H  and  S  . According to the relationship ln K =

a)  H  = m H f(products) – n H f(reactants)

 H  = [(2 mol NH3)(  H f of NH3)] – [(1 mol N2)(  H f of N2) + (3 mol H2)(  H f of H2)]  H  = [(2 mol)(–45.9 kJ/mol)] – [(1 mol)(0 kJ/mol) + (3 mol)(0 kJ/mol)] = –91.8 kJ

  – n Sreactants  S  = m Sproducts

 S  = [(2 mol NH3)( S  of NH3)] – [(1 mol N2)( S  of N2) + (3 mol H2)( S  of H2)]

 S  = [(2 mol)(193 J/mol∙K)] – [(1 mol)(191.50 J/mol∙K) + (3 mol)(130.6 J/mol∙K)]  S  = –197.3 J/K

ΔG  RT Since ln Kp = 0 when Kp = 1.00,  G  = 0.

ln K =

G  = 0 = H  – T S  H  = T S 

91.8 kJ 103 J  ΔH   = 465.281 = 465 K =   197.3 J/K  1 kJ  ΔS

b)  G  =  H  – T  S  = (–91.8 kJ)(10 J/1 kJ) – (673 K)(–197.3 J/K) = 4.09829 × 10 J 3

  4.09829  10 4 J/mol  ΔG   = –7.32449 ln K = =  RT   8.314 J /mol  K 673 K    K=e

–7.32449







= 6.591958 × 10 = 6.59 × 10 –4

–4

20-39

c) The reaction rate is higher at the higher temperature. The time required (kinetics) overshadows the lower yield (thermodynamics). 20.109 a) At equilibrium, all three free energies are equal. b) Kyanite has the lowest enthalpy. Lowering the temperature will shift the equilibrium in the exothermic direction and favor the substance with the lowest enthalpy. c) Sillimanite has the highest entropy. At equilibrium, the substance with the highest enthalpy (opposite to kyanite) must have the highest entropy. d) Andalusite has the lowest density. Decreasing pressure shifts the equilibrium toward the substance with more volume or less density. 20.110 a) 2CH4(g) + 1/2O2(g) → C2H2(g) + 2H2(g) + H2O(g)

 H  = [(1 mol C2H2)(  H f of C2H2) + (2 mol H2)(  H f of H2) + (1 mol H2O)(  H f of H2O)]

– [(2 mol CH4)(  H f of CH4) + (1/2 mol O2)(  H f of O2)]

 H  = [(1 mol)(227 kJ/mol) + (2 mol)(0 kJ/mol) + (1 mol)(–241.826 kJ/mol)]  H = 134.914 = 135 kJ 

– [(2 mol)(–74.87 kJ/mol) + (1/2 mol O2)(0 kJ/mol)]

 S  = [(1 mol C2H2)( S  of C2H2) + (2 mol H2)( S  of H2) + (1 mol H2O)( S  of H2O)]

– [(2 mol CH4)( S  of CH4) + (1/2 mol O2)( S  of O2)]

 S  = [(1 mol)(200.85 J/mol∙K) + (2 mol)(130.6 J/mol∙K) + (1 mol)(188.72 J/mol∙K)]

– [(2 mol)(186.1 J/mol∙K) + (1/2 mol)(205.0 J/mol∙K)]

 S = 176.07 = 176.1 J/K 

G  = 0 = H  – T S  H  = T S 

134.914 kJ 103 J  ΔH   = 766.25206 = 766 K =  176.07 J/K  1 kJ  ΔS 

b) 2C(s) + H2(g) → C2H2(g)  H  = [(1 mol C2H2)(  H f of C2H2)] – [(2 mol C)(  H f of C) + (1 mol H2)(  H f of H2)]  H  = [(1 mol)(227 kJ/mol)] – [(2 mol)(0 kJ/mol) + (1 mol)(0 kJ/mol)]  H  = 227 kJ

 S  = [(1 mol C2H2)( S  of C2H2)] – [(2 mol C)( S  of C) + (1 mol H2)( S  of H2)]

 S  = [(1 mol)(200.85 J/mol∙K)] – [(2 mol)(5.686 J/mol∙K) + (1 mol)(130.6 J/mol∙K)]  S  = 58.878 = 58.9 J/K

G  = 0 = H  – T S  H  = T S 

227 kJ 103 J  ΔH   = 3855.43 = 3860 K =  58.878 J/K  1 kJ  ΔS  c) Considering the reverse reaction of its formation, the acetylene is produced under conditions at which acetylene is unstable and can decompose back into its elements. It must be quickly cooled to a temperature where its thermal decomposition rate is slow. T=

20-40

20.111 a) CH4(g) + H2O(g) ⇆ CO(g) + 3H2(g)

steam re-forming

 H = [(1 mol CO)(  H of CO) + (3 mol H2)(  H f of H2)] 

 f

– [(1 mol CH4)(  H f of CH4) + (1 mol H2O)(  H f of H2O)]

 H = [(1 mol)(–110.5 kJ/mol) + (3 mol)(0)] – [(1 mol)(–74.87 kJ/mol) + (1 mol)(–241.826 kJ/mol)] 

S

= 206.196 = 206.2 kJ

= [(1 mol CO)( S  of CO) + (3 mol H2) (S  of H2)] – [(1 mol CH4) (S  of CH4) + (1 mol H2O) (S  of H2O)]



 S  = [(1 mol)(197.5 J/mol∙K) + (3 mol)(130.6 J/mol∙K)]

– [(1 mol)(186.1 J/mol∙K) + (1 mol)(188.72 J/mol∙K)]

 S = 214.48 = 214.5 J/K 

 G  =  H  – T  S  = (206.196 kJ) – (1273.K)(214.48 J/K)(1 kJ/10 J) 3

 G  = –66.83704 = –66.84 kJ

CO(g) + H2O(g)

⇆ CO2(g) + H2(g)

water-gas shift reaction

 H = [(1 mol CO2)(  H of CO2) + (1 mol H2)(  H f of H2)] 

 f

– [(1 mol CO)(  H f of CH4) + (1 mol H2O)(  H f of H2O)]

 H  = [(1 mol)(–393.5 kJ/mol) + (1 mol)(0 kJ/mol)] – [(1 mol)(–110.5 kJ/mol) + (1 mol)(–241.826 kJ/mol)]

 H  = –41.174 = –41.2 kJ

 S  = [(1 mol CO2)( S  of CO2) + (1 mol H2)( S  of H2)] – [(1 mol CO)( S  of CO) + (1 mol H2O)( S  of H2O)]  S  = [(1 mol)(213.7 J/mol∙K) + (1 mol)(130.6 J/mol∙K)]  S = – 41.92 = –41.9 J/K 

– [(1 mol)(197.5 J/mol∙K) + (1 mol)(188.72 J/mol∙K)]

 G  =  H  – T  S  = (–41.174 kJ) – (1273 K)(–41.92 J/K)(1 kJ/10 J) 3

 G  = 12.19016 = 12.2 kJ

b) 2CH4(g) + 3H2O(g) ⇆ CO2(g) + CO(g) + 7H2(g) Since this reaction is two steam re-forming reactions and one water-gas shift reaction,  G  = 2(  G  for the steam re-forming reaction) +  G  for the water-gas shift reaction:  G  = 2(–66.84 kJ) + 12.1387 kJ = –121.5413 = –121.5 kJ Since  G  is negative, the reaction is spontaneous at this temperature. c) At 98 atm and 50% conversion, the partial pressures of the reactants and products are as follows (from the ratio of the coefficients: CH4, 14 atm; H2O, 21 atm; CO2, 7.0 atm; CO, 7.0 atm; H2, 49 atm). Calculate Q: Q=

7  7.07.0 49 CO2 COH 2  7 = = 1.8309 × 10 2 3 2 3 14 21 CH 4  H 2 O 7

 3 7 G = G° + RT ln Q = –121.5413 kJ/mol + (1 kJ/10 J) (8.314 J/mol∙K)(1273 K) ln (1.8309 × 10 ) = 55.4476 = 55.4 kJ/mol



The reaction is not spontaneous at this point. d) At 98 atm and 90% conversion, the partial pressures of the reactants and products are as follows: CH 4, 2.3 atm; H2O, 3.4 atm; CO2, 10.3 atm; CO, 10.3 atm; H2, 71.8 atm. Calculate Q: 7 7 10.310.371.8 CO2 COH 2  12 Q= = = 5.019415 × 10 2 3 2 3     2.3 3.4 CH H O  4  2  

20-41

G = G° + RT ln Q = –121.5413 kJ/mol + (1 kJ/10 J)(8.314 J/mol∙K)(1273 K) ln (5.019 × 10 ) 3

= 187.9726 = 188 kJ/mol The reaction is not spontaneous at this point.

20-42

CHAPTER 21 ELECTROCHEMISTRY: CHEMICAL CHANGE AND ELECTRICAL WORK FOLLOW–UP PROBLEMS 21.1A

Plan: Follow the steps for balancing a redox reaction in acidic solution: 1. Divide into half-reactions 2. For each half-reaction balance a) Atoms other than O and H, b) O atoms with H2O, + c) H atoms with H and – d) Charge with e . 3. Multiply each half-reaction by an integer that will make the number of electrons lost equal to the number of electrons gained. 4. Add the half-reactions and cancel substances appearing as both reactants and products. Then, add another step for basic solution. + 5. Add hydroxide ions to neutralize H . Cancel water. Solution: 1. Divide into half-reactions: group the reactants and products with similar atoms. – 2– MnO4 (aq)  MnO4 (aq) – – I (aq)  IO3 (aq) 2. For each half-reaction balance a) Atoms other than O and H Mn and I are balanced so no changes needed. b) O atoms with H2O – 2– MnO4 (aq)  MnO4 (aq) O already balanced – – I (aq) + 3H2O(l)  IO3 (aq) Add 3 H2O to balance oxygen. + c) H atoms with H – 2– MnO4 (aq)  MnO4 (aq) H already balanced – – + + I (aq) + 3H2O(l)  IO3 (aq) + 6H (aq) Add 6 H to balance hydrogen. – – d) Charge with e ; Total charge of reactants is –1 and of products is –2, so add 1 e to reactants to balance charge: – – 2– MnO4 (aq) + e  MnO4 (aq) – Total charge is –1 for reactants and +5 for products, so add 6 e as product: – – + – I (aq) + 3H2O(l)  IO3 (aq) + 6 H (aq) + 6e 3. Multiply each half-reaction by an integer that will make the number of electrons lost equal to the number of electrons gained. One electron is gained and 6 are lost so reduction must be multiplied by 6 for the number of electrons to be equal. – – 2– 6 {MnO4 (aq) + e  MnO4 (aq)} – – + – I (aq) + 3H2O(l)  IO3 (aq) + 6H (aq) + 6 e 4. Add half-reactions and cancel substances appearing as both reactants and products. – – 2– 6MnO4 (aq) + 6e  6MnO4 (aq) I (aq) + 3H2O(l)  IO3 (aq) + 6H (aq) + 6e – – 2– – + Overall: 6MnO4 (aq) + I (aq) + 3H2O(l)  6MnO4 (aq) + IO3 (aq) + 6H (aq) –

–

21-1

–

5. Add hydroxide ions to neutralize H . Cancel water. The 6 H are neutralized by adding 6 OH . The same number of hydroxide ions must be added to the reactants to keep the balance of O and H atoms on both sides of the reaction. – – – 2– – + – 6MnO4 (aq) + I (aq) + 3H2O(l) + 6OH (aq)  6MnO4 (aq) + IO3 (aq) + 6H (aq) +6 OH (aq) + – The neutralization reaction produces water: 6 {H + OH  H2O}. – – – 2– – 6MnO4 (aq) + I (aq) + 3H2O(l) + 6OH (aq)  6MnO4 (aq) + IO3 (aq) + 6H2O(l) Cancel water: – – – 2– – 6MnO4 (aq) + I (aq) + 3H2O(l) + 6OH (aq)  6MnO4 (aq) + IO3 (aq) + 6H2O(l) Balanced reaction is – – – 2– – 6MnO4 (aq) + I (aq) + 6OH (aq)  6MnO4 (aq) + IO3 (aq) + 3H2O(l) Balanced reaction including spectator ions is 6KMnO4(aq) + KI(aq) + 6KOH(aq)  6K2MnO4(aq) + KIO3(aq) + 3H2O(l) 21.1B

Plan: Follow the steps for balancing a redox reaction in acidic solution: 1. Divide into half-reactions 2. For each half-reaction balance a) Atoms other than O and H, b) O atoms with H2O, + c) H atoms with H and – d) Charge with e . 3. Multiply each half-reaction by an integer that will make the number of electrons lost equal to the number of electrons gained. 4. Add the half-reactions and cancel substances appearing as both reactants and products. Then, add another step for basic solution. + 5. Add hydroxide ions to neutralize H . Cancel water. Solution: 1. Divide into half-reactions: group the reactants and products with similar atoms. 2– Cr(OH)3(aq)  CrO4 (aq) – – IO3 (aq)  I (aq) 2. For each half-reaction balance a) Atoms other than O and H Cr and I are balanced so no changes needed. b) O atoms with H2O 2– H2O(l) + Cr(OH)3(aq)  CrO4 (aq) Add 1 H2O to balance oxygen. – – IO3 (aq)  I (aq) + 3H2O(l) Add 3 H2O to balance oxygen. + c) H atoms with H 2– + + H2O(l) + Cr(OH)3(aq)  CrO4 (aq) + 5H (aq) Add 5 H to balance hydrogen. + – – + 6H (aq) + IO3 (aq)  I (aq) + 3H2O(l) Add 6 H to balance hydrogen. – – d) Charge with e ; total charge of reactants is 0 and of products is +3, so add 3 e to products to balance charge: 2– + – H2O(l) + Cr(OH)3(aq)  CrO4 (aq) + 5H (aq) + 3e – Total charge is –1 for products and +5 for reactants, so add 6 e as reactant: – + – – 6e + 6H (aq) + IO3 (aq)  I (aq) + 3H2O(l) 3. Multiply each half-reaction by an integer that will make the number of electrons lost equal to the number of electrons gained. Six electrons are gained and 3 are lost so oxidation must be multiplied by 2 for the number of electrons to be equal. 2– + – 2 {H2O(l) + Cr(OH)3(aq)  CrO4 (aq) + 5H (aq) + 3e } – + – – 6e + 6H (aq) + IO3 (aq)  I (aq) + 3H2O(l) 4. Add half-reactions and cancel substances appearing as both reactants and products. 2– + – 2H2O(l) + 2Cr(OH)3(aq)  2CrO4 (aq) + 104H (aq) + 6e – + – – 6e + 6H (aq) + IO3 (aq)  I (aq) + 31H2O(l) Overall: 2Cr(OH)3(aq) + IO3 (aq)  2CrO4 (aq) + I (aq) + H2O(l) + 4H (aq) –

2–

–

21-2

–

5. Add hydroxide ions to neutralize H . Cancel water. The 4 H are neutralized by adding 4 OH . The same number of hydroxide ions must be added to the reactants to keep the balance of O and H atoms on both sides of the reaction. – – 2– – + – 2Cr(OH)3(aq) + IO3 (aq) + 4OH (aq)  2CrO4 (aq) + I (aq) + H2O(l) + 4H (aq) + 4OH (aq) + – The neutralization reaction produces water: 4 {H + OH  H2O}. – – 2– – 2Cr(OH)3(aq) + IO3 (aq) + 4OH (aq)  2CrO4 (aq) + I (aq) + 5H2O(l) Balanced reaction is – – 2– – 2Cr(OH)3(aq) + IO3 (aq) + 4OH (aq)  2CrO4 (aq) + I (aq) + 5H2O(l) Balanced reaction including spectator ions is 2Cr(OH)3(aq) + NaIO3(aq) + 4NaOH(aq)  2Na2CrO4(aq) + NaI(aq) + 5H2O(l) 21.2A

Plan: Given the solution and electrode compositions, the two half cells involve the transfer of electrons 2– 3+ 2+ 1) between chromium in Cr2O7 and Cr and 2) between Sn and Sn . The negative electrode is the anode so the tin half-cell is where oxidation occurs. The graphite electrode with the chromium ion/chromate solution is where reduction occurs. In the cell diagram, show the electrodes and the solutes involved in the half-reactions. Include the salt bridge and wire connection between electrodes. Set up the two half-reactions and balance. (Note that the 3+ 2– Cr /Cr2O7 half-cell is in acidic solution.) Write the cell notation placing the anode half-cell first, then the salt bridge, then the cathode half-cell. Solution: Cell diagram:

–

Voltmeter

e–

Salt bridge

Sn (–)

C (+) Cr , H 3+

Cr2O7

2–

Balanced equations: 2+ 2+ Anode is Sn/Sn half-cell. Oxidation of Sn produces Sn : 2+ Sn(s)  Sn (aq) All that needs to be balanced is charge: 2+ – Sn(s)  Sn (aq) + 2e 3+ 2– Cathode is the Cr /Cr2O7 half-cell. Check the oxidation number of chromium in each substance to determine 3+ 2– which is reduced. Cr oxidation number is +3 and chromium in Cr2O7 has oxidation number +6. Going from +6 2– to +3 involves gain of electrons so Cr2O7 is reduced. 2– 3+ Cr2O7 (aq)  Cr (aq) Balance Cr: 2– 3+ Cr2O7 (aq)  2Cr (aq) Balance O: 2– 3+ Cr2O7 (aq)  2Cr (aq) + 7H2O(l) Balance H: 2– + 3+ Cr2O7 (aq) + 14H (aq)  2Cr (aq) + 7H2O(l) Balance charge: 2– + – 3+ Cr2O7 (aq) + 14H (aq) + 6e  2Cr (aq) + 7H2O(l) Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

21-3

Add two half-reactions, multiplying the tin half-reaction by 3 to equalize the number of electrons transferred. 2+ – 3{Sn(s)  Sn (aq) + 2e } 2– + – 3+ Cr2O7 (aq) + 14H (aq) + 6e  2 Cr (aq) + 7H2O(l) 3Sn(s) + Cr2O7 (aq) + 14H (aq)  3Sn (aq) + 2Cr (aq) + 7H2O(l) Cell notation: 2+ + 2– 3+ Sn(s)  Sn (aq)  H (aq), Cr2O7 (aq), Cr (aq)  C(graphite) 2–

21.2B

Plan: Given the solution and electrode compositions, the two half cells involve the transfer of electrons – – 2+ 1) between Cl and ClO3 and 2) between Ni and Ni . The negative electrode is the anode so the nickel half-cell is where oxidation occurs. The graphite electrode with the chloride ion/chlorate solution is where reduction occurs. In the cell diagram, show the electrodes and the solutes involved in the half-reactions. Include the salt bridge and – – wire connection between electrodes. Set up the two half-reactions and balance. (Note that the Cl /ClO3 half-cell is in acidic solution.) Write the cell notation placing the anode half-cell first, then the salt bridge, then the cathode half-cell. Solution: Cell diagram:

Voltmeter

–

e–

Salt bridge

Ni (–)

C (+) ClO3 , –

H , Cl +

–

Balanced equations: 2+ 2+ Anode is Ni/Ni half-cell. Oxidation of Ni produces Ni : 2+ Ni(s)  Ni (aq) All that needs to be balanced is charge: 2+ – Ni(s)  Ni (aq) + 2e – – Cathode is the Cl /ClO3 half-cell. Check the oxidation number of chlorine in each substance to determine which is – – reduced. Cl oxidation number is –1 and chlorine in ClO3 has oxidation number +5. Going from +5 to –1 involves – gain of electrons so ClO3 is reduced. – – ClO3 (aq)  Cl (aq) Cl is balanced. Balance O: – – ClO3 (aq)  Cl (aq) + 3H2O(l) Balance H: + – – 6H (aq) + ClO3 (aq)  Cl (aq) + 3H2O(l) Balance charge: + – – – 6H (aq) + ClO3 (aq) + 6e  Cl (aq) + 3H2O(l) Add two half-reactions, multiplying the nickel half-reaction by 3 to equalize the number of electrons transferred. 2+ – 3{Ni(s)  Ni (aq) + 2e } + – – – 6H (aq) + ClO3 (aq) + 6e  Cl (aq) + 3H2O(l) 3Ni(s) + ClO3 (aq) + 6H (aq)  3Ni (aq) + Cl (aq) + 3H2O(l) –

–

21-4

Cell notation: 2+ + – – Ni(s)  Ni (aq)  H (aq), ClO3 (aq), Cl (aq)  C(graphite) 21.3A

ο ο ο Plan: Use the relationship Ecell = E cathode – Eanode . E ο values are found in Appendix D. Spontaneous reactions ο have Ecell > 0. Solution: Oxidation: Reduction: Overall reaction:

2{Ag(s)  Ag (aq) + e } 2+ – Cu (aq) + 2e  Cu(s) 2+ + Cu (s) + 2Ag(s)  Cu(s) + 2Ag (aq) +

–

E°(anode) = 0.80 V E°(cathode) = 0.34 V

o = 0.34 V  (0.80 V) = 0.46 V Ecell Reaction is not spontaneous under standard-state conditions because E is negative. o cell

21.3B

ο ο ο Plan: Use the relationship Ecell = E cathode – Eanode . E ο values are found in Appendix D. Spontaneous reactions o have Ecell > 0. Solution: 2+ 3+ – Oxidation: 2{Fe (aq)  Fe (aq) + e } E°(anode) = 0.77 V – – Reduction: Cl2(g) + 2e  2Cl (aq) E°(cathode) = 1.36 V 2+ –+ 3+ Overall reaction: Cl2(g) + 2Fe (aq)  2Cl (aq) + 2Fe (aq) o = 1.36 V  (0.77 V) = +0.59 V Ecell o Reaction is spontaneous under standard-state conditions because Ecell is positive.

21.4A

Plan: Divide the reaction into half-reactions showing that Br2 is reduced and V is oxidized. Use the equation

o o o o = E cathode – Eanode to solve for Eanode . Ecell Solution: Half-reactions: – – o Reduction (cathode): Br2(aq) + 2e  2 Br (aq) = 1.07 V from Appendix D E cathode 3+ 2+ + – Oxidation (anode): 2V (aq) + 2H2O(l)  2VO (aq) + 4H (aq) + 2e Overall: 3+ 2+ + – o Br2(aq) + 2V (aq) + 2H2O(l)  2VO (aq) + 4H (aq) + 2Br (aq) Ecell = 0.73 V (given) o o o Ecell = E cathode – Eanode o o o = E cathode – Ecell = 1.07 V – 0.73 V = 0.34 V Eanode

21.4B

Plan: Divide the reaction into half-reactions showing that nitrate is reduced and V is oxidized. Use the equation o o o o = E cathode – Eanode to solve for Eanode Ecell . Solution: Half-reactions: + – – Reduction (cathode): 4H (aq) + NO3 (aq) + 3e  NO(g) + 2H2O(l) 3+ 2+ + – Oxidation (anode): 3{V (aq) + H2O(l)  VO (aq) + 2H (aq) + e }

E°(anode) = 0.34 V from Follow Up Problem 21.4A

Overall: NO3 (aq) + 3V (aq) + H2O(l)  3VO (aq) + 2H (aq) + NO(g) –

o o o = E cathode – Eanode Ecell

o = 0.62 V (given) Ecell

o o o = Ecell + Eanode = 0.62 V + 0.34 V = 0.96 V E cathode

21.5A

Plan: To write a spontaneous reaction, examine two reduction half-reactions. The half-reaction with the smaller E is reversed (to become an oxidation). Reducing strength increases with decreasing E°.

21-5

Solution: o a) Combining reactions (1) and (2), reaction (1) is reversed because it has the smaller E . – – Oxidation: 3{2Ag(s) + 2OH (aq)  Ag2O(s) + H2O(l) + 2e } E°(anode) = 0.34 V – – – – Reduction: BrO3 (g) + 3H2O(l) + 6e  Br (aq) + 6OH (aq) E°(cathode) = 0.61 V – – Overall reaction: 6Ag(s) + BrO3 (g)  3Ag2O(s) + Br (aq) o = 0.61 V  (0.34 V) = +0.27 V Ecell o Combining reactions (1) and (3), reaction (3) is reversed because it has the smaller E . – – Oxidation: Zn(s) + 2OH (aq)  Zn(OH)2(s) + 2e E°(anode) = –1.25 V – – Reduction: Ag2O(s) + H2O(l) + 2e  2Ag(s) + 2OH (aq) E°(cathode) = 0.34 V Overall reaction: Zn(s) + Ag2O(s) + H2O(l)  Zn(OH)2(s) + 2Ag(s) o = 0.34 V  (–1.25 V) = +1.59 V Ecell o

Combining reactions (2) and (3), reaction (3) is reversed because it has the smaller E . – – Oxidation: 3{Zn(s) + 2OH (aq)  Zn(OH)2(s) + 2e } E°(anode) = –1.25 V – – – – Reduction: BrO3 (g) + 3H2O(l) + 6e  Br (aq) + 6OH (aq) E°(cathode) = 0.61 V – – Overall reaction: 3Zn(s) + BrO3 (g) + 3H2O(l)  3Zn(OH)2(s) + Br (aq) o = 0.61 V  (–1.25 V) = +1.86 V Ecell

b) Oxidizing agents are substances that cause another substance to be oxidized. In other words, oxidizing agents are, themselves, reduced. The more likely a substance is to be reduced (the stronger the oxidizing agent), the o greater the value of its reducing potential, E . Examining, the reduction half-reactions given in the problem statement, we can see that Zn(OH)2 has the lowest reducing potential, Ag2O has the next highest reducing potential, and BrO3 has the greatest reducing potential. Therefore, we can rank these substances by their strengths as oxidizing agents: BrO3 > Ag2O > Zn(OH)2. Reducing agents are substances that cause another substance to be reduced. In other words, reducing agents are, themselves, oxidized. The more likely a substance is to be oxidized (the stronger the reducing agent), the greater the value of its oxidizing potential. The oxidizing potentials of substances in this problem can be obtained by reversing the provided reduction half-reactions and changing the sign of the half-reaction potential. When we do this, we find that Ag has an oxidizing potential of 0.34 V, Br has an oxidizing potential of 0.61 V, and Zn has an oxidizing potential of 1.25 V. The more positive the oxidizing potential, the stronger the reducing agent. Therefore, we can rank these substances by their strengths as reducing agents: Zn > Ag > Br. 21.5B

o o Plan: To determine if the reaction is spontaneous, divide into half-reactions and calculate Ecell is negative, . If Ecell the reaction is not spontaneous, so reverse the reaction to obtain the spontaneous reaction. Reducing strength increases with decreasing E°. Solution: Divide into half-reactions and balance: 2+ – o Reduction: Fe (aq) + 2e  Fe(s) = –0.44 V E cathode o Oxidation: 2 {Fe (aq)  2Fe (aq) + e } = 0.77 V Eanode The first half-reaction is reduction, so it is the cathode half-cell. The second half-reaction is oxidation, so it is the anode half-cell. Find the half-reactions in Appendix D. o o o = E cathode – Eanode = –0.44 V – 0.77 V = –1.21 V Ecell The reaction is not spontaneous as written, so reverse the reaction: 3+ 2+ Fe(s) + 2Fe (aq)  3Fe (aq) o is now +1.21 V, so the reversed reaction is spontaneous under standard state conditions. Ecell 2+ When a substance acts as a reducing agent, it is oxidized. Both Fe and Fe can be oxidized, so they can act as 3+ reducing agents. Since Fe cannot lose more electrons, it cannot act as a reducing agent. The stronger reducing 2+ agent between Fe and Fe is the one with the smaller standard reduction potential. E° for Fe is –0.44, which is 2+ 2+ less than E° for Fe , +0.77. Therefore, Fe is a stronger reducing agent than Fe . Ranking all three in order of 2+ 3+ decreasing reducing strength gives Fe > Fe > Fe . 2+

–

21-6

21.6A

Plan: Determine the half-reactions and calculate the Ecell . Find the number of moles of electrons transferred in the balanced equation. Both K and G° can be calculated using the relationships log K =

o nEcell and 0.0592 V

ο G° = nFE cell . Solution: The balanced equation is: – – 2+ 2MnO4 (aq) + 4H2O(l) + 3Cu(s)  2MnO2(s) + 8OH (aq) + 3Cu (aq) The half reactions are: – – Reduction: 2{MnO4 (aq) + 2H2O(l) + 3e  MnO2(s) + 8OH (aq)} E°(cathode) = 0.59 V 2+ Oxidation: 3{Cu(s)  Cu (aq) + 2e } E°(anode) = 0.34 V o o o = E cathode – Eanode = 0.59 V – 0.34 V = 0.25 V Ecell In this reaction, 3 moles of copper atoms are oxidized to become 3 moles of copper(II) ions. This process requires

the loss of 3 × 2 = 6 electrons. Therefore, 6 electrons are transferred in this reaction, and n = 6. o (6)(0.25 V) nEcell log K = = = 25.3378 0.0592 V 0.0592 V K = 10

25.3378

= 2.1769 × 10 = 2.2 × 10 25

o G° = nFEcell = – (6 mol e /mol rxn )(96,485 C/mol e )(0.25 J/C) = –1.4473 × 10 = –1.4 × 10 J/mol rxn 2 = –1.4 × 10 kJ/mol rxn o Plan: Find the number of moles of electrons transferred in the balanced equation. Given ΔG°, both K and Ecell –

21.6B

–

o . Spontaneous reactions have can be calculated using the relationships ΔG° = –RT ln K and ΔG  nFEcell

o Ecell  0 and K > 1.

Solution: 3+

2[Ga (aq) + 3e → Ga(s)] 2+

3[Co(s) → Co (aq) + 3e] 3+

2Ga (aq) + 3Co(s) → 2Ga(s) + 3Co (aq) n = 6 mol e/mol rxn ΔG° = –RT ln K

ln K  

1.62 10 5 J/mol rxn ΔG   65.39 RT 8.314 J/mol rxn  K  298 K

K = e65.39  4.0 1029

ΔG  nFE ; E o cell

o cell

1.62 10 5 J/mol rxn ΔG    0.28 V nF 6 mol e /mol rxn (9.65 10 4 J/V  mol e )

Reaction is not spontaneous and will not occur as written. 21.7A

Plan: Write a balanced equation for the spontaneous reaction between Cr and Sn (we know the reaction is

spontaneous because we are told the reaction occurs in a voltaic cell). Determine the number of moles of electrons 0.0592 o o transferred, and calculate Ecell . Use the Nernst equation, Ecell = Ecell – log Q to find Ecell. n Solution: o Determining the cell reaction and Ecell : 3+ o The Cr/Cr reduction half-reaction has a smaller E . Therefore, it is reversed (and becomes the oxidation halfreaction).

21-7

Oxidation: Reduction:

2{Cr(s)  Cr (aq) + 3e } 2+ – 3{Sn (aq) + 2e  Sn(s)]} 3+

–

E° (anode) = –0.74 V E° (cathode) = –0.14 V

Overall 2Cr(s) + 3Sn (aq)  2Cr (aq) + 3Sn(s) o o o = –0.14 V – (–0.74 V) = 0.60 V Ecell = E cathode – Eanode 2+ 2+ For the reaction Q = [Cr ]/[Sn ], so the Nernst equation is: 2  Cr 3  0.0592 o   Ecell = Ecell  . log 3 n Sn 2     Substituting in values from the problem: 2+

Ecell = 0.60 V  21.7B

(6 moles of electrons transferred)

0.0592 (1.60)2 = 0.60 V  0.0592 log(320) = 0.60 V – 0.024717 = 0.575283 = 0.58 V log 3 6 (0.20) 6

o o Plan: The problem is asking for the concentration of iron ions when Ecell = Ecell + 0.25 V. 0.0592 2+ o Use the Nernst equation, Ecell = Ecell – log Q to find [Fe ]. n Solution: o Determining the cell reaction and Ecell :

Oxidation:

Fe(s)  Fe (aq) + 2e

Reduction:

Cu (aq) + 2e  Cu(s)

–

E° = –0.44 V

–

E° = 0.34 V

Overall Fe(s) + Cu (aq)  Fe (aq) + Cu(s) 2+

o o o = E cathode – Eanode = 0.34 V – (–0.44 V) = 0.78 V Ecell 2+ 2+ For the reaction Q = [Fe ]/[Cu ], so the Nernst equation is: 0.0592 [Fe 2+ ] o Ecell = Ecell  log n [Cu 2+ ] Substituting in values from the problem: o Ecell = Ecell + 0.25 V = 0.78 V + 0.25 V = 1.03 V

1.03 V = 0.78 V 

0.0592 [Fe 2+ ] log n [Cu 2+ ]

0.0592 [Fe 2+ ] log 2 [0.30] [Fe 2+ ] –8.44595 = log [0.30] [Fe 2+ ] –9 3.58141 × 10 = [0.30] 2+ –9 –9 [Fe ] = 1.074423 × 10 = 1.1 × 10 M

0.25 V = 

21.8A

Plan: Half-cell B contains a higher concentration of nickel ions, so the ions will be reduced to decrease the 2+ concentration while in half-cell A, with a lower [Ni ], nickel metal will be oxidized to increase the concentration 2+ 2+ of nickel ions. In the overall cell reaction, the lower [Ni ] appears as a product and the higher [Ni ] appears as a 2+ 2+ o reactant. This means that Q = [Ni ]lower/[Ni ]higher. Use the Nernst equation with Ecell = 0 to find Ecell. Oxidation of nickel metal occurs at the anode, half-cell A, which is negative. Solution: 0.0592 o Ecell = Ecell  log Q n

21-8

o Ecell = Ecell 

[Ni 2 ]lower 0.0592 log n [Ni 2 ]higher

–

n = 2 mol e for the reduction of Ni to Ni

0.0592 (0.015) log = 0.04221 = 0.042 V 2 (0.40) Plan: Half-cell B contains a higher concentration of gold ions, so the ions will be reduced to decrease the 3+ concentration while in half-cell A, with a lower [Au ], gold metal will be oxidized to increase the concentration 3+ 3+ of gold ions. In the overall cell reaction, the lower [Au ] appears as a product and the higher [Au ] appears as a 3+ 3+ o reactant. This means that Q = [Au ]lower/[Au ]higher. Use the Nernst equation with Ecell = 0 to find Ecell. Solution: 0.0592 o Ecell = Ecell  log Q n Ecell = 0 

21.8B

o Ecell = Ecell 

[Au3+ lower ] 0.0592 log n [Au3+ higher ]

–

n = 3 mol e for the reduction of Au to Au

0.0592 [7.0 104 ] = 0.0306427 = 0.031 V log 3 [2.5102 ] Oxidation of gold metal occurs at the anode, half-cell A, which is negative.

Ecell = 0 

21.9A

Plan: In the electrolysis, the more easily reduced metal ion will be reduced at the cathode. Compare Cs and Li as to their location on the periodic table to find which has the higher ionization energy (IE increases up and across the table). Higher ionization energy of the metal means more easily reduced. At the anode, the more easily oxidized nonmetal ion will be oxidized. For oxidation, compare the electronegativity of the two nonmetals. The ion from the less electronegative element will be more easily oxidized. Note that the standard reduction potentials cannot be used in the case of molten salts because the E° values are based on aqueous solutions of ions, not liquid salts. Solution: + + Li is above Cs, so Li is more easily reduced than Cs . The reaction at the cathode is + – Li (l) + e  Li(s) – – F is the most electronegative element, so I must be more easily oxidized than F . The reaction at the anode is – – 2 I (l)  I2(g) + 2e To add the two half-reaction, the number of electrons must be equal: + – 2 {Li (l) + e  Li(s)} 2I (l)  I2(g) + 2e –

–

Overall: 2Li (l) + 2I (l)  2Li(s) + I2(g) +

–

This is the overall reaction. Li(s) forms at the cathode, and I2(g) forms at the anode. 21.9B

Plan: In the electrolysis, the more easily reduced metal ion will be reduced at the cathode. Compare K and Al as to their location on the periodic table to find which has the higher ionization energy (IE increases up and across the table). Higher ionization energy of the metal means more easily reduced. At the anode, the more easily oxidized nonmetal ion will be oxidized. For oxidation, compare the electronegativity of the two nonmetals. The ion from the less electronegative element will be more easily oxidized. Note that the standard reduction potentials cannot be used in the case of molten salts because the E° values are based on aqueous solutions of ions, not liquid salts. Solution: 3+ + Al is above and to the right of K, so Al is more easily reduced than K . The reaction at the cathode is 3+ – Al (l) + 3e  Al(s) – – F is the most electronegative element, so Br must be more easily oxidized than F . The reaction at the anode is – – 2 Br (l)  Br2(g) + 2e

21-9

To add the two half-reaction, the number of electrons must be equal: 3+ – 2 {Al (l) + 3e  Al(s)} 3 {2Br (l)  Br2(g) + 2e } –

–

Overall: 2Al (l) + 6Br (l)  2Al(s) + 3Br2(g) This is the overall reaction. Al(s) forms at the cathode, and Br2(g) forms at the anode. 2+ – + 21.10A Plan: In aqueous Pb(NO3)2, the species present are Pb (aq), NO3 (aq), H2O, and very small amounts of H and – 2+ – OH . The possible half-reactions are reduction of either Pb or H2O and oxidation of either NO3 or H2O. Whichever reduction and oxidation half-reactions are more spontaneous will take place, with consideration of the overvoltage. Solution: The two possible reductions are 2+ – Pb (aq) + 2e  Pb(s) E° = –0.13 V – – 2H2O(l) + 2e  H2(g) + 2OH (aq) E = –0.42 V The reduction of lead ions occurs because it has a higher reduction potential (more spontaneous) than reduction of water. The two possible oxidations are – NO3 (aq)  no reaction (nitrate compounds are always soluble) + – 2H2O(l)  O2(g) + 4H (aq) + 4e E = 0.82 V Water will be oxidized at the anode since no reaction is possible for the nitrate ion. The two half-reactions that are predicted are 2+ – Cathode: Pb (aq) + 2e  Pb(s) E° = –0.13 V + – Anode: 2H2O(l)  O2(g) + 4H (aq) + 4e E = 0.82 V 2+ Thus, Pb is reduced at the cathode, and H2O is oxidized at the anode. Pb forms at the cathode, and O2 forms at the anode. 3+

–

21.10B Plan: In aqueous AuBr3, the species present are Au (aq), Br (aq), H2O, and very small amounts of H and OH . 3+ – The possible half-reactions are reduction of either Au or H2O and oxidation of either Br or H2O. Whichever reduction and oxidation half-reactions are more spontaneous will take place, with consideration of the overvoltage. Solution: The two possible reductions are 3+ – Au (aq) + 3e  Au(s) E° = 1.50 V – – 2H2O(l) + 2e  H2(g) + 2OH (aq) E = –1 V (with overvoltage) The reduction of gold ions occurs because it has a higher reduction potential (more spontaneous) than reduction of water. The two possible oxidations are – – 2Br (aq)  Br2(l) + 2e E° = 1.07 V + – 2H2O(l)  O2(g) + 4H (aq) + 4e E = 1.4 V (with overvoltage) The oxidation with the less positive reduction potential is the more spontaneous oxidation so bromide ions are oxidized at the anode. The two half-reactions that are predicted are 3+ – Cathode: Au (aq) + 3e  Au(s) E° = 1.50 V – – Anode: 2Br (aq)  Br2(l) + 2e E° = 1.07 V 3+ – Thus, Au is reduced at the cathode, and Br is oxidized at the anode. Au forms at the cathode, and Br2 forms at the anode. 21.11A Plan: To find the charge transferred, first write the balanced half-reaction and then calculate the charge from the grams of copper and moles of electrons transferred per molar mass of copper. Current is charge per time, so to find the time, divide the charge by the current. Solution: Cu (aq) + 2e  Cu(s), so 2 mol e per mole of Cu 2+

–

21-10

  1 mol Cu   2 mol e   96485 C  A  1  1 min    Time (min) = 1.50 g Cu          63.55 g Cu  1 mol Cu  1 mol e  C/s  4.75 A   60 s 





= 15.9816 = 16.0 min

21.11B Plan: Using the time and current, determine the total charge in Coulombs of the electrons transferred. Use the Faraday constant, the number of moles of electrons transferred, and the molar mass of zinc to calculate the mass of zinc deposited. Solution:  60 s  7.03 C    = 3690.75 C  Charge (C) = 8.75min  1min   s  1 mol e 1 mol Zn  65.38 g Zn      = 1.25046 = 1.25 g Zn Mass (g) of zinc = 3690.75 C   96485 C  2 mol e  1 mol Zn 

CHEMICAL CONNECTIONS BOXED READING PROBLEMS B21.1

Plan: Reduction is the gain of electrons while oxidation is the loss of electrons. Solution: 3+ – 2+ a) Reduction: Fe + e → Fe + 2+ – Oxidation: Cu → Cu + e 3+

b) Overall: Fe + Cu → Fe + Cu B21.2

o Plan: Use the relationship G° = nFEcell to calculate the free energy change. Solution: – – 5 o G° (J/mol) = nFEcell = – (2 mol e )(96485 C/mol e )(0.224 J/C) = –4.322528 × 10 J  1 kJ  G° (kJ/mol) = 4.32252810 5 J 3  = –43.22528 = –43.2 kJ/mol 10 J 

END–OF–CHAPTER PROBLEMS 21.1

Oxidation is the loss of electrons (resulting in a higher oxidation number), while reduction is the gain of electrons (resulting in a lower oxidation number). In an oxidation-reduction reaction, electrons transfer from the oxidized substance to the reduced substance. The oxidation number of the reactant being oxidized increases while the oxidation number of the reactant being reduced decreases.

21.2

An electrochemical process involves electron flow. At least one substance must lose electron(s) and one substance must gain electron(s) to produce the flow. This electron transfer is a redox process.

21.3

No, one half-reaction cannot take place independently of the other because there is always a transfer of electrons from one substance to another. If one substance loses electrons (oxidation half-reaction), another substance must gain those electrons (reduction half-reaction).

21.4

O  is too strong a base to exist in H 2O. The reaction O  + H 2O  2OH  occurs. Only species actually existing in solution can be present when balancing an equation.

21.5

Multiply each half-reaction by the appropriate integer to make e lost equal to e gained.

21.6

To remove protons from an equation, add an equal number of hydroxide ions to both sides to neutralize the H and + produce water: H (aq) + OH(aq)  H2O(l).

21-11

21.7

No, spectator ions are not used to balance the half-reactions. Add spectator ions to the balanced ionic equation to obtain the balanced molecular equation.

21.8

Spontaneous reactions, Gsys < 0, take place in voltaic cells, which are also called galvanic cells. Nonspontaneous reactions take place in electrolytic cells and result in an increase in the free energy of the cell (Gsys > 0).

21.9

a) True b) True c) True d) False, in a voltaic cell, the system does work on the surroundings. e) True f) False, the electrolyte in a cell provides a solution of mobile ions to maintain charge neutrality.

21.10

Plan: Assign oxidation numbers; the species with an atom whose oxidation number has increased is being oxidized and is the reducing agent. The species with an atom whose oxidation number has decreased is being reduced and is the oxidizing agent. Electrons flow from the reducing agent to the oxidizing agent. To write the + 2– molecular equation, pair K ions with anions and SO4 with cations in the equation to write neutral molecules. Solution: –8 +1 +

+7 –2

–1



+1 –2

16H (aq) + 2MnO4 (aq) + 10Cl (aq) → 2Mn (aq) + 5Cl2(g) + 8H2O(l) a) To decide which reactant is oxidized, look at oxidation numbers. Cl is oxidized because its oxidation number increases from –1 in Cl to 0 in Cl2. 2+ b) MnO4 is reduced because the oxidation number of Mn decreases from +7 in MnO4 to +2 in Mn . c) The oxidizing agent is the substance that causes the oxidation by accepting electrons. The oxidizing agent is the substance reduced in the reaction, so MnO4 is the oxidizing agent. d) Cl is the reducing agent because it loses the electrons that are gained in the reduction. e) From Cl, which is losing electrons, to MnO4, which is gaining electrons. f) 8H2SO4(aq) + 2KMnO4(aq) + 10KCl(aq)  2MnSO4(aq) + 5Cl2(g) + 8H2O(l) + 6K2SO4(aq) 21.11

2CrO2(aq) + 2H 2O(l) + 6ClO(aq)  2CrO4 (aq) + 3Cl2(g) + 4OH (aq) a) The CrO2 is the oxidized species because Cr increases in oxidation state from +3 to +6. b) The ClO is the reduced species because Cl decreases in oxidation state from +1 to 0. c) The oxidizing agent is ClO; the oxidizing agent is the substance reduced. d) The reducing agent is CrO2; the reducing agent is the substance oxidized. e) Electrons transfer from CrO2 to ClO. f) 2NaCrO2(aq) + 6NaClO(aq) + 2H 2O(l) 2Na2CrO4(aq) + 3Cl2(g) + 4NaOH(aq)

21.12

21-12

Balance elements other than O and H: ClO3(aq)  Cl(aq) chlorine is balanced 2I(aq)  I2(s) iodine now balanced Balance O by adding H2O: ClO3(aq)  Cl(aq) + 3H2O(l) add three waters to add three O atoms to product 2I(aq)  I2(s) no change + Balance H by adding H : + + ClO3(aq) + 6H (aq)  Cl(aq) + 3H2O(l) add six H to reactants  2I (aq)  I2(s) no change Balance charge by adding e: + – ClO3(aq) + 6H (aq) + 6e  Cl(aq) + 3 H2O(l) add 6e to reactants for a –1 charge on each side – –  2I (aq)  I2(s) + 2e add 2e to products for a –2 charge on each side Multiply each half-reaction by an integer to equalize the number of electrons: + – – ClO3(aq) + 6H (aq) + 6e  Cl(aq) + 3H2O(l) multiply by one to give 6e – –  3{2I (aq)  I2(s) + 2e } or multiply by three to give 6e –  6I (aq) 3I2(s) + 6e Add half-reactions to give balanced equation in acidic solution: + ClO3(aq) + 6H (aq) + 6I(aq)  Cl(aq) + 3H2O(l) + 3I2(s) Check balancing: Reactants: 1 Cl Products: 1 Cl 3O 3O 6H 6H 6I 6I –1 charge –1 charge Oxidizing agent is ClO3 and reducing agent is I. b) Divide into half-reactions: MnO4(aq)  MnO2(s) 2 2 SO3 (aq)  SO4 (aq) Balance elements other than O and H: MnO4(aq)  MnO2(s) Mn is balanced 2 2 SO3 (aq)  SO4 (aq) S is balanced Balance O by adding H2O: MnO4(aq)  MnO2(s) + 2H2O(l) add two H2O to products 2 2 SO3 (aq) + H2O(l)  SO4 (aq) add one H2O to reactants + Balance H by adding H : + + MnO4(aq) + 4H (aq)  MnO2(s) + 2H2O(l) add four H to reactants 2 2 + + SO3 (aq) + H2O(l)  SO4 (aq) + 2H (aq) add two H to products – Balance charge by adding e : + – – MnO4(aq) + 4H (aq) + 3e  MnO2(s) + 2H2O(l) add 3e to reactants for a 0 charge on each side 2 2 + – – SO3 (aq) + H2O(l)  SO4 (aq) + 2H (aq) + 2e add 2e to products for a –2 charge on each side Multiply each half-reaction by an integer to equalize the number of electrons: + – – 2{MnO4(aq) + 4H (aq) + 3e MnO2(s) + 2H2O(l)} or multiply by two to give 6e + –  2MnO4 (aq) + 8H (aq) + 6e  2MnO2(s) + 4H2O(l) 2 2 + – – 3{SO3 (aq) + H2O(l)  SO4 (aq) + 2H (aq) + 2e } or multiply by three to give 6e 2 2 + – 3SO3 (aq) + 3H2O(l)  3SO4 (aq) + 6H (aq) + 6e Add half-reactions and cancel substances that appear as both reactants and products: + 2 2 + 2 MnO4(aq) + 8H (aq) + 3SO3 (aq) + 3H2O(l)  2MnO2(s) + 4H2O(l) + 3SO4 (aq) + 6H (aq) The balanced equation in acidic solution is: + 2 2 2MnO4(aq) + 2H (aq) + 3SO3 (aq)  2MnO2(s) + H2O(l) + 3SO4 (aq) +  To change to basic solution, add OH to both sides of equation to neutralize H + 2 2   2MnO4 (aq) + 2H (aq) + 2OH (aq) + 3SO3 (aq)  2MnO2(s) + H2O(l) + 3SO4 (aq) + 2OH(aq) Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

21-13

2MnO4(aq) + 2H2O( l) + 3SO3 (aq)  2MnO2(s) + H2O(l) + 3SO4 (aq) + 2OH(aq) Balanced equation in basic solution: 2 2 2MnO4(aq) + H2O(l) + 3SO3 (aq)  2MnO2(s) + 3SO4 (aq) + 2OH(aq) Check balancing: Reactants: 2 Mn Products: 2 Mn 18 O 18 O 2H 2H 3S 3S –8 charge –8 charge 2 Oxidizing agent is MnO4 and reducing agent is SO3 . c) Divide into half-reactions: 2+ MnO4(aq)  Mn (aq) H2O2(aq)  O2(g) Balance elements other than O and H: 2+ MnO4(aq)  Mn (aq) Mn is balanced H2O2(aq)  O2(g) No other elements to balance Balance O by adding H2O: 2+ MnO4(aq)  Mn (aq) + 4H2O(l) add four H2O to products H2O2(aq)  O2(g) O is balanced + Balance H by adding H : + 2+ +  MnO4 (aq) + 8H (aq)  Mn (aq) + 4H2O(l) add eight H to reactants + + H2O2(aq)  O2(g) + 2H (aq) add two H to products – Balance charge by adding e : + – 2+ – MnO4(aq) + 8H (aq) + 5e  Mn (aq) + 4H2O(l) add 5e to reactants for +2 on each side + – – H2O2(aq)  O2(g) + 2H (aq) + 2e add 2e to products for 0 charge on each side Multiply each half-reaction by an integer to equalize the number of electrons: + – 2+ – 2{MnO4(aq) + 8H (aq) + 5e  Mn (aq) + 4H2O(l)} or multiply by two to give 10e + – 2+  2MnO4 (aq) + 16H (aq) + 10e 2Mn (aq) + 8H2O(l) + – – 5{H2O2(aq)  O2(g) + 2H (aq) + 2e } or multiply by five to give 10e + – 5H2O2(aq)  5O2(g) + 10H (aq) + 10e Add half-reactions and cancel substances that appear as both reactants and products: + 2+ + 2MnO4(aq) + 16H (aq) + 5H2O2(aq)  2Mn (aq) + 8H2O(l) + 5O2(g) + 10H (aq) The balanced equation in acidic solution: + 2+ 2MnO4(aq) + 6H (aq) + 5H2O2(aq)  2Mn (aq) + 8H2O(l) + 5O2(g) Check balancing: Reactants: 2 Mn Products: 2 Mn 18 O 18 O 16 H 16 H +4 charge +4 charge – Oxidizing agent is MnO4 and reducing agent is H2O2. 2

21.13

a) 3O2(g) + 4NO(g) + 2H 2O(l)  4NO3 (aq) + 4H (aq) Oxidizing agent is O2 and reducing agent is NO. 2– – b) 2CrO4 (aq) + 8H 2O(l) + 3Cu(s)  2Cr(OH)3(s) + 3Cu(OH)2(s) + 4OH (aq) 2– Oxidizing agent is CrO4 and reducing agent is Cu. 3– – – – – c) AsO4 (aq) + NO2 (aq) + H 2O(l)  AsO2 (aq) + NO3 (aq) + 2OH (aq) 3– – Oxidizing agent is AsO4 and reducing agent is NO2 .

21.14

–

21-14

–

add one OH ion to each side of the equation for every H ion present to form H2O and cancel excess H2O molecules. The substance that gains electrons is the oxidizing agent while the substance that loses electrons is the reducing agent. Solution: a) Balance the reduction half-reaction: 2 3+ Cr2O7 (aq)  2Cr (aq) balance Cr 2 3+ Cr2O7 (aq)  2Cr (aq) + 7H2O(l) balance O by adding H2O 2 + 3+ + Cr2O7 (aq) + 14H (aq)  2Cr (aq) + 7H2O(l) balance H by adding H 2 + – 3+ – Cr2O7 (aq) + 14H (aq) + 6e  2Cr (aq) + 7H2O(l) balance charge by adding 6e Balance the oxidation half-reaction: 2+ – – Zn(s)  Zn (aq) + 2e balance charge by adding 2e Add the two half-reactions multiplying the oxidation half-reaction by three to equalize the electrons: 2 + – 3+ Cr2O7 (aq) + 14H (aq) + 6e  2Cr (aq) + 7H2O(l) 2+ – 3Zn(s)  3Zn (aq) + 6e Add half-reactions and cancel substances that appear as both reactants and products: 2 + 3+ 2+ Cr2O7 (aq) + 14H (aq) + 3Zn(s)  2Cr (aq) + 7H2O(l) + 3Zn (aq) 2 Oxidizing agent is Cr2O7 and reducing agent is Zn. b) Balance the reduction half-reaction: MnO4(aq)  MnO2(s) + 2H2O(l) balance O by adding H2O + + MnO4(aq) + 4H (aq)  MnO2(s) + 2H2O(l) balance H by adding H + – –  MnO4 (aq) + 4H (aq) + 3e  MnO2(s) + 2H2O(l) balance charge by adding 3 e Balance the oxidation half-reaction: Fe(OH)2(s) + H2O(l)  Fe(OH)3(s) balance O by adding H2O + + Fe(OH)2(s) + H2O(l)  Fe(OH)3(s) + H (aq) balance H by adding H + –  Fe(OH)2(s) + H2O(l)  Fe(OH)3(s) + H (aq) + e balance charge by adding 1e Add half-reactions after multiplying oxidation half-reaction by 3: + – MnO4(aq) + 4H (aq) + 3e  MnO2(s) + 2H2O(l) + 3Fe(OH)2(s) + 3H2O(l)  3Fe(OH)3(s) + 3H (aq) + 3e Add half-reactions and cancel substances that appear as both reactants and products: + + MnO4(aq) + 4H (aq) + 3Fe(OH)2(s) + 3H2O(l)  MnO2(s) + 2H2O(l) + 3Fe(OH)3(s) + 3H (aq) +  MnO4 (aq) + H (aq) + 3Fe(OH)2(s) + H2O(l)  MnO2(s) + 3Fe(OH)3(s) – + + – Add OH to both sides to neutralize the H and convert H + OH  H2O: + – –  MnO4 (aq) + H (aq) + OH (aq) + 3Fe(OH)2(s) + H2O(l)  MnO2(s) + 3Fe(OH)3(s) + OH (aq)   MnO4 (aq) + 3Fe(OH)2(s) + 2H2O(l)  MnO2(s) + 3Fe(OH)3(s) + OH (aq) Oxidizing agent is MnO4 and reducing agent is Fe(OH)2. c) Balance the reduction half-reaction: 2NO3(aq)  N2(g) balance N 2NO3(aq)  N2(g) + 6H2O(l) balance O by adding H2O + + 2NO3(aq) + 12H (aq)  N2(g) + 6H2O(l) balance H by adding H +   2NO3 (aq) + 12H (aq) + 10e  N2(g) + 6H2O(l) balance charge by adding 10e Balance the oxidation half-reaction: 2+ Zn(s)  Zn (aq) + 2e balance charge by adding 2e Add the half-reactions after multiplying the reduction half-reaction by one and the oxidation half-reaction by five: + 2NO3(aq) + 12H (aq) + 10e  N2(g) + 6H2O(l) 2+ 5Zn(s)  5Zn (aq) + 10e Add half-reactions and cancel substances that appear as both reactants and products: + 2+ 2NO3(aq) + 12H (aq) + 5Zn(s)  N2(g) + 6H2O(l) + 5Zn (aq)  Oxidizing agent is NO3 and reducing agent is Zn. 21.15

a) 3BH 4 (aq) + 4ClO3 (aq)  3H 2BO3 (aq) + 4Cl (aq) + 3H 2O(l) – – Oxidizing agent is ClO3 and reducing agent is BH 4 . 2– + 3+ b) 2CrO4 (aq) + 3N2O(g) + 10H (aq)  2Cr (aq) + 6NO(g) + 5H 2O(l) –

–

21-15

Oxidizing agent is CrO4 and reducing agent is N2O. – – – c) 3Br2(l) + 6OH (aq)  BrO3 (aq) + 5Br (aq) + 3H 2O(l) Oxidizing agent is Br2 and reducing agent is Br2. 2–

21.16

Plan: Divide the reaction into the two half-reactions, balance elements other than oxygen and hydrogen, and + then balance oxygen by adding H2O and hydrogen by adding H . Balance the charge by adding electrons and multiply each half-reaction by an integer so that the number of electrons lost equals the number of electrons gained. Add the half-reactions together, canceling substances that appear on both sides. For basic solutions, – + add one OH ion to each side of the equation for every H ion present to form H2O and cancel excess H2O molecules. The substance that gains electrons is the oxidizing agent while the substance that loses electrons is the reducing agent. Solution: a) Balance the reduction half-reaction: NO3(aq)  NO(g) + 2H2O(l) balance O by adding H2O + + NO3(aq) + 4H (aq)  NO(g) + 2H2O(l) balance H by adding H + – –  NO3 (aq) + 4H (aq) + 3e  NO(g) + 2H2O(l) balance charge by adding 3e Balance oxidation half-reaction: 4Sb(s)  Sb4O6(s) balance Sb 4Sb(s) + 6H2O(l)  Sb4O6(s) balance O by adding H2O + + 4Sb(s) + 6H2O(l)  Sb4O6(s) + 12H (aq) balance H by adding H + – – 4Sb(s) + 6H2O(l)  Sb4O6(s) + 12H (aq) + 12e balance charge by adding 12e Multiply each half-reaction by an integer to equalize the number of electrons: + – – 4{NO3(aq) + 4H (aq) + 3e  NO(g) + 2H2O(l)} multiply by four to give 12e + – – 1{4Sb(s) + 6H2O(l)  Sb4O6(s) + 12H (aq) + 12e } multiply by one to give 12e This gives: + – 4NO3(aq) + 16H (aq) + 12e  4NO(g) + 8H2O(l) + – 4Sb(s) + 6H2O(l)  Sb4O6(s) + 12H (aq) + 12e Add half-reactions. Cancel common reactants and products: + + 4 NO3(aq) + 16H (aq) + 4Sb(s) + 6H2O(l)  4NO(g) + 8H2O(l) + Sb4O6(s) + 12H (aq) Balanced equation in acidic solution: + 4NO3(aq) + 4H (aq) + 4Sb(s)  4NO(g) + 2H2O(l) + Sb4O6(s) – Oxidizing agent is NO3 and reducing agent is Sb. b) Balance reduction half-reaction: 3+ BiO3(aq)  Bi (aq) + 3H2O(l) balance O by adding H2O + 3+ +  BiO3 (aq) + 6H (aq)  Bi (aq) + 3H2O(l) balance H by adding H + – 3+  BiO3 (aq) + 6H (aq) + 2e  Bi (aq) + 3H2O(l) balance charge to give +3 on each side Balance oxidation half-reaction: 2+ Mn (aq) + 4H2O(l)  MnO4(aq) balance O by adding H2O 2+ + + Mn (aq) + 4H2O(l)  MnO4(aq) + 8H (aq) balance H by adding H 2+ + –  Mn (aq) + 4H2O(l)  MnO4 (aq) + 8H (aq) + 5e balance charge to give +2 on each side Multiply each half-reaction by an integer to equalize the number of electrons: + – 3+ – 5{BiO3(aq) + 6H (aq) + 2e  Bi (aq) + 3H2O(l)} multiply by five to give 10e 2+ + – –  2{Mn (aq) + 4H2O(l)  MnO4 (aq) + 8H (aq) + 5e } multiply by two to give 10e This gives: + – 3+ 5BiO3(aq) + 30H (aq) + 10e  5Bi (aq) + 15H2O(l) 2+ + –  2Mn (aq) + 8H2O(l)  2MnO4 (aq) + 16H (aq) + 10e + Add half-reactions. Cancel H2O and H in reactants and products: + 2+ 3+ + 5BiO3(aq) + 30H (aq) + 2Mn (aq) + 8H2O(l)  5Bi (aq) + 15H2O(l) + 2MnO4(aq) + 16H (aq) Balanced reaction in acidic solution: + 2+ 3+ 5BiO3(aq) + 14H (aq) + 2Mn (aq)  5Bi (aq) + 7H2O(l) + 2MnO4(aq) – 2+ BiO3 is the oxidizing agent and Mn is the reducing agent.

21-16

c) Balance the reduction half-reaction: Pb(OH)3(aq)  Pb(s) + 3H2O(l) balance O by adding H2O + + Pb(OH)3(aq) + 3H (aq)  Pb(s) + 3H2O(l) balance H by adding H + –  Pb(OH)3 (aq) + 3H (aq) + 2e  Pb(s) + 3H2O(l) balance charge to give 0 on each side Balance the oxidation half-reaction: Fe(OH)2(s) + H2O(l)  Fe(OH)3(s) balance O by adding H2O + + Fe(OH)2(s) + H2O(l)  Fe(OH)3(s) + H (aq) balance H by adding H + – Fe(OH)2(s) + H2O(l)  Fe(OH)3(s) + H (aq) + e balance charge to give 0 on each side Multiply each half-reaction by an integer to equalize the number of electrons: + – – 1{Pb(OH)3(aq) + 3H (aq) + 2e  Pb(s) + 3H2O(l)} multiply by 1 to give 2e + – – 2{Fe(OH)2(s) + H2O(l)  Fe(OH)3(s) + H (aq) + e } multiply by 2 to give 2e This gives: + – Pb(OH)3(aq) + 3H (aq) + 2e  Pb(s) + 3H2O(l) + – 2Fe(OH)2(s) + 2H2O(l)  2Fe(OH)3(s) + 2H (aq) + 2e + Add the two half-reactions. Cancel H2O and H : + + Pb(OH)3(aq) + 3H (aq) + 2Fe(OH)2(s) + 2H2O(l)  Pb(s) + 3H2O(l) + 2Fe(OH)3(s) + 2H (aq) +  Pb(OH)3 (aq) + H (aq) + 2Fe(OH)2(s)  Pb(s) + H2O(l) + 2Fe(OH)3(s) – + Add one OH to both sides to neutralize H : +   Pb(OH)3 (aq) + H (aq) + OH (aq) + 2Fe(OH)2(s)  Pb(s) + H2O(l) + 2 Fe(OH)3(s) + OH(aq) Pb(OH)3(aq) + H2O(l) + 2Fe(OH)2(s)  Pb(s) + H2O(l) + 2Fe(OH)3(s) + OH(aq) Balanced reaction in basic solution: Pb(OH)3(aq) + 2Fe(OH)2(s)  Pb(s) + 2Fe(OH)3(s) + OH(aq) Pb(OH)3 is the oxidizing agent and Fe(OH)2 is the reducing agent. 21.17

a) 2NO2(g) + 2OH (aq)  NO3(aq) + NO2(aq) + H 2O(l) Oxidizing agent is NO2 and reducing agent is NO2. 2 b) 4Zn(s) + 7OH (aq) + NO3(aq) + 6H 2O(l)  4Zn(OH)4 (aq) + NH 3(aq)  Oxidizing agent is NO3 and reducing agent is Zn. + c) 24H 2S(g) + 16NO3(aq) + 16H (aq)  3S8(s) + 16NO(g) + 32H 2O(l)  Oxidizing agent is NO3 and reducing agent is H 2S.

21.18

21-17

Add the half-reactions and cancel H2O and H : + 3 2+ + 5As4O6(s) + 8MnO4(aq) + 64H (aq) + 50H2O(l)  20AsO4 (aq) + 8Mn (aq) + 32H2O(l) + 100H (aq) Balanced reaction in acidic solution: 3 2+ + 5As4O6(s) + 8MnO4(aq) + 18H2O(l)  20AsO4 (aq) + 8Mn (aq) + 36H (aq)  Oxidizing agent is MnO4 and reducing agent is As4O6. b) The reaction gives only one reactant, P4. Since both products contain phosphorus, divide the half-reactions so each includes P4 as the reactant. Balance reduction half-reaction: P4(s)  4PH3(g) balance P + + P4(s) + 12H (aq)  4PH3(g) balance H by adding H +  P4(s) + 12H (aq) + 12e  4PH3(g) balance charge by adding 12 e Balance oxidation half-reaction: 2 P4(s)  4HPO3 (aq) balance P 2 P4(s) + 12H2O(l)  4HPO3 (aq) balance O by adding H2O 2 + + P4(s) + 12H2O(l)  4HPO3 (aq) + 20H (aq) balance H by adding H 2 +  P4(s) + 12H2O(l)  4HPO3 (aq) + 20H (aq) + 12e balance charge by adding 12 e + Add two half-reactions and cancel H : + 2 + 2P4(s) + 12H (aq) + 12H2O(l)  4HPO3 (aq) + 4PH3(g) + 20H (aq) Balanced reaction in acidic solution: 2 + 2P4(s) + 12H2O(l)  4HPO3 (aq) + 4PH3(g) + 8H (aq) or 2 + P4(s) + 6H2O(l)  2HPO3 (aq) + 2PH3(g) + 4H (aq) P4 is both the oxidizing agent and reducing agent. c) Balance the reduction half-reaction: MnO4(aq)  MnO2(s) + 2H2O(l) balance O by adding H2O + + MnO4(aq) + 4H (aq)  MnO2(s) + 2H2O(l) balance H by adding H +   MnO4 (aq) + 4H (aq) + 3e  MnO2(s) + 2H2O(l) balance charge by adding 3e Balance oxidation half-reaction: CN(aq) + H2O(l)  CNO(aq) balance O by adding H2O + + CN(aq) + H2O(l)  CNO(aq)+ 2H (aq) balance H by adding H +    CN (aq) + H2O(l)  CNO (aq)+ 2H (aq) + 2e balance charge by adding 2e Multiply the oxidation half-reaction by three and reduction half-reaction by two to transfer 6e in overall reaction. + 2MnO4(aq) + 8H (aq) + 6e  2MnO2(s) + 4H2O(l) +  3CN (aq) + 3H2O(l)  3CNO(aq)+ 6H (aq) + 6e + Add the two half-reactions. Cancel the H2O and H : + + 2MnO4(aq) + 3CN(aq) + 8H (aq) + 3H2O(l)  2MnO2(s) + 3CNO (aq) + 6H (aq) + 4H2O(l) +    2MnO4 (aq) + 3CN (aq) + 2H (aq)  2MnO2(s) + 3CNO (aq) + H2O(l) + Add 2 OH to both sides to neutralize H and form H2O: +   2MnO4 (aq) + 3CN (aq) + 2H (aq) + 2OH(aq) 2MnO2(s) + 3CNO(aq) + H2O(l) + 2OH(aq) 2MnO4(aq) + 3CN(aq) + 2H2O(l)  2MnO2(s) + 3CNO(aq) + H2O(l) + 2OH(aq) Balanced reaction in basic solution: 2MnO4(aq) + 3CN(aq) + H2O(l)  2MnO2(s) + 3CNO(aq) + 2OH(aq) Oxidizing agent is MnO4 and reducing agent is CN. 21.19

a) SO3 (aq) + 2OH (aq) + Cl2(g)  SO4 (aq) + 2Cl(aq) + H 2O(l) 2 Oxidizing agent is Cl2 and reducing agent is SO3 . 3 4   b) 7Fe(CN)6 (aq) + Re(s) + 8OH (aq) + 7e  7Fe(CN)6 (aq) + ReO4(aq) + 4H 2O(l) + 7e 3 Oxidizing agent is Fe(CN)6  and reducing agent is Re. + 2+  c) 2MnO4 (aq) + 5HCOOH(aq) + 6H (aq)  2Mn (aq) + 5CO2(g) + 8H 2O(l)  Oxidizing agent is MnO4 and reducing agent is H COOH . 2

21-18

21.20

5Fe (aq) + MnO4 (aq) + 8H (aq)  Mn (aq) + 5Fe (aq) + 4H 2O(l)

21.21

a) Balance reduction half-reaction: – NO3 (aq)  NO2(g) + H2O(l)

–

balance O by adding H2O

NO3 (aq) + 2H (aq)  NO2(g) + H2O(l) –

NO3 (aq) + 2H (aq) + e  NO2(g) + H2O(l) Balance oxidation half-reaction: – – Au(s) + 4Cl (aq)  AuCl4 (aq) –

–

balance H by adding H

balance charge to give 0 on each side balance Cl

Au(s) + 4Cl (aq)  AuCl4 (aq) + 3e balance charge to –4 on each side Multiply each half-reaction by an integer to equalize the number of electrons: – + – – 3{NO3 (aq) + 2H (aq) + e  NO2(g) + H2O(l)} multiply by three to give 3e –

–

1{ Au(s) + 4Cl (aq)  AuCl4 (aq) + 3e } This gives: – + – 3NO3 (aq) + 6H (aq) + 3e  3NO2(g) + 3H2O(l) –

–

multiply by one to give 3e

–

Au(s) + 4Cl (aq)  AuCl4 (aq) + 3e Add half-reactions: – – + – Au(s) + 3NO3 (aq) + 4Cl (aq) + 6H (aq)  AuCl4 (aq) + 3NO2(g) + 3H2O(l) – b) Oxidizing agent is NO3 and reducing agent is Au. – c) The HCl provides chloride ions that combine with the unstable gold ion to form the stable ion, AuCl 4 . –

–

21.22

21.23

Unless the oxidizing and reducing agents are physically separated, the redox reaction will not generate electrical energy. This electrical energy is produced by forcing the electrons to travel through an external circuit.

21.24

The purpose of the salt bridge is to maintain charge neutrality by allowing anions to flow into the anode compartment and cations to flow into the cathode compartment.

21.25

An active electrode is a reactant or product in the cell reaction, whereas an inactive electrode is neither a reactant nor a product. An inactive electrode is present only to conduct electricity when the half-cell reaction does not include a metal. Platinum and graphite are commonly used as inactive electrodes.

21.26

a) The metal A is being oxidized to form the metal cation. To form positive ions, an atom must always lose electrons, so this half-reaction is always an oxidation. b) The metal ion B is gaining electrons to form the metal B, so it is displaced. c) The anode is the electrode at which oxidation takes place, so metal A is used as the anode.

21-19

d) Acid oxidizes metal B and metal B oxidizes metal A, so acid will oxidize metal A and bubbles will form when metal A is placed in acid. The same answer results if strength of reducing agents is considered. The fact that metal A is a better reducing agent than metal B indicates that if metal B reduces acid, then metal A will also reduce acid. 21.27

Plan: The oxidation half-cell (anode) is shown on the left while the reduction half-cell (cathode) is shown on the right. Remember that oxidation is the loss of electrons and electrons leave the oxidation half-cell and move towards the positively charged cathode. Anions from the salt bridge flow into the oxidation half-cell, while cations from the salt bridge flow into the reduction half-cell. Solution: a) Electrons flow from the anode to the cathode, so from the iron half-cell to the nickel half-cell, left to right in the figure. By convention, the anode appears on the left and the cathode on the right. b) Oxidation occurs at the anode, which is the electrode in the iron half-cell. c) Electrons enter the reduction half-cell, the nickel half-cell in this example. d) Electrons are consumed in the reduction half-reaction. Reduction takes place at the cathode, nickel electrode. e) The anode is assigned a negative charge, so the iron electrode is negatively charged. f) Metal is oxidized in the oxidation half-cell, so the iron electrode will decrease in mass. g) The solution must contain nickel ions, so any nickel salt can be added. 1 M NiSO4 is one choice. + – h) KNO3 is commonly used in salt bridges, the ions being K and NO3 . Other salts are also acceptable answers. i) Neither, because an inactive electrode could not replace either electrode since both the oxidation and the reduction half-reactions include the metal as either a reactant or a product. j) Anions will move towards the half-cell in which positive ions are being produced. The oxidation half-cell 2+ produces Fe , so salt bridge anions move from right (nickel half-cell) to left (iron half-cell). 2+ – k) Oxidation half-reaction: Fe(s)  Fe (aq) + 2e 2+ – Reduction half-reaction: Ni (aq) + 2e  Ni(s) 2+ 2+ Overall cell reaction: Fe(s) + Ni (aq)  Fe (aq) + Ni(s)

21.28

a) The electrons flow left to right. b) Reduction occurs at the electrode on the right. c) Electrons leave the cell from the left side. d) The zinc electrode generates the electrons. e) The cobalt electron has the positive charge. f) The cobalt electrode increases in mass. g) The anode electrolyte could be 1 M Zn(NO3)2. + – h) One possible pair would be K and NO3 . i) Neither electrode could be replaced because both electrodes are part of the cell reaction. j) The cations move from left to right to maintain charge neutrality. 2+ – k) Reduction: Co (aq) + 2e  Co(s) 2+ – Oxidation: Zn(s)  Zn (aq) + 2e 2+ 2+ Overall: Zn(s) + Co (aq)  Co(s) + Zn (aq)

21.29

Plan: The anode, at which the oxidation takes place, is the negative electrode. Electrons flow from the anode to the cathode. Anions from the salt bridge flow into the oxidation half-cell, while cations from the salt bridge flow into the reduction half-cell. Solution: a) If the zinc electrode is negative, it is the anode and oxidation takes place at the zinc electrode: 2+ – Zn(s)  Zn (aq) + 2e 2+ – Reduction half-reaction: Sn (aq) + 2e  Sn(s) 2+ 2+ Overall reaction: Zn(s) + Sn (aq)  Zn (aq) + Sn(s)

21-20

b) e

Voltmeter

–

Salt bridge

()

(+) 1M

1M Zn

21.30

Cation

flow

–

Salt bridge

()

21.31

–

Voltmeter

–

1M 2+ Pb

Ag (aq) + e  Ag(s) 2+ – Pb(s)  Pb (aq) + 2e + 2+ 2Ag (aq) + Pb(s)  2Ag(s) + Pb (aq)

(red half-rxn) (ox half-rxn) (overall rxn)

Anion

(+) 1M Anion

Cation

flow

Plan: The cathode, at which the reduction takes place, is the positive electrode. Electrons flow from the anode to the cathode. Anions from the salt bridge flow into the oxidation half-cell, while cations from the salt bridge flow into the reduction half-cell. Solution: a) The cathode is assigned a positive charge, so the iron electrode is the cathode. 2+ – Reduction half-reaction: Fe (aq) + 2e  Fe(s) 2+ – Oxidation half-reaction: Mn(s)  Mn (aq) + 2e 2+ 2+ Overall cell reaction: Fe (aq) + Mn(s)  Fe(s) + Mn (aq)

21-21

b) e

Voltmeter

–

Salt bridge

()

(+) 1M

1M Mn

21.32

Anion

Cation

flow

(red half-rxn) (ox half-rxn) (overall rxn)

Cu (aq) + 2e  Cu(s) 2+ – Ni(s)  Ni (aq) + 2e 2+ 2+ Cu (aq) + Ni(s)  Cu(s) + Ni (aq) 2+

–

Voltmeter

–

b) bridge

()

(+) 1M

1M Ni

21.33

Anion

Cation

flow

Plan: In cell notation, the oxidation components of the anode compartment are written on the left of the salt bridge and the reduction components of the cathode compartment are written to the right of the salt bridge. A double vertical line separates the anode from the cathode and represents the salt bridge. A single vertical line separates species of different phases. Anode || Cathode Solution: a) Al is oxidized, so it is the anode and appears first in the cell notation. There is a single vertical line separating the solid metals from their solutions. 3+ 3+ Al(s) | Al (aq) || Cr (aq) | Cr(s)

21-22

b) Cu is reduced, so Cu is the cathode and appears last in the cell notation. The oxidation of SO2 does not include a metal, so an inactive electrode must be present. Hydrogen ion must be included in the oxidation half-cell. 2– + 2+ Pt | SO2(g) | SO4 (aq), H (aq) || Cu (aq) | Cu(s) 21.34

a) Mn(s) + Cd (aq)  Mn (aq) + Cd(s) – + 2+ b) 3Fe(s) + 2NO3 (aq) + 8H (aq)  3Fe (aq) + 2NO(g) + 4H 2O(l)

21.35

An isolated reduction or oxidation potential cannot be directly measured. However, by assigning a standard halfcell potential to a particular half-reaction, the standard potentials of other half-reactions can be determined relative to this reference value. The standard reference half-cell is a standard hydrogen electrode defined to have an E° value of 0.000 V.

21.36

o A negative Ecell indicates that the cell reaction is not spontaneous, G° > 0. The reverse reaction is spontaneous

21.37

Similar to other state functions, the sign of E° changes when a reaction is reversed. Unlike G°, H° and S°, E° is an intensive property, the ratio of energy to charge. When the coefficients in a reaction are multiplied by a factor, the values of G°, H° and S° are multiplied by the same factor. However, E° does not change because both the energy and charge are multiplied by the factor and their ratio remains unchanged.

21.38

Plan: Divide the balanced equation into reduction and oxidation half-reactions and add electrons. Add water and

o with Ecell > 0.

o o o hydroxide ion to the half-reaction that includes oxygen. Use the relationship Ecell = E cathode – Eanode to find the unknown E° value. Solution: 2– – a) Oxidation: Se (aq)  Se(s) + 2e 2– – 2– – Reduction: 2SO3 (aq) + 3H2O(l) + 4e  S2O3 (aq) + 6OH (aq) o o o b) Ecell = E cathode – Eanode o o o = E cathode – Ecell = –0.57 V – 0.35 V = –0.92 V Eanode

21.39

a) Reduction:

O3(g) + 2H (aq) + 2e  O2(g) + H 2O(l)

Oxidation:

Mn (aq) + 2H 2O(l)  MnO2(s) + 4H (aq) + 2e

–

o o o b) Ecell = E cathode – Eanode

o o o = E cathode – Ecell = 2.07 V – 0.84 V= 1.23 V Eanode

21.40

Plan: The greater (more positive) the reduction potential, the greater the strength as an oxidizing agent. Solution: a) From Appendix D: 3+ – 2+ Fe (aq) + e  Fe (aq) E° = 0.77 V – – Br2(l) + 2e  2Br (aq) E° = 1.07 V 2+ – Cu (aq) + e  Cu(s) E° = 0.34 V 3+ 2+ When placed in order of decreasing strength as oxidizing agents: Br2 > Fe > Cu . b) From Appendix D: 2+ Ca (aq) + 2e–  Ca(s) E° = –2.87 V 2– + – 3+ Cr2O7 (aq) + 14H (aq) 6e  2Cr (aq) + 7H2O(l) E° = 1.33 V + – Ag (aq) + e  Ag(s) E° = 0.80 V 2+ + 2– When placed in order of increasing strength as oxidizing agents: Ca < Ag < Cr2O7 .

21-23

21.41

a) When placed in order of decreasing strength as reducing agents: SO2 > MnO2 > PbSO4 b) When placed in order of increasing strength as reducing agents: H g < Sn < Fe

21.42

o o o Plan: Use the relationship Ecell = E cathode – Eanode . o have Ecell > 0. Solution: a) Oxidation: Reduction: Overall reaction:

E o values are found in Appendix D. Spontaneous reactions

Co(s)  Co (aq) + 2e + – 2H (aq) + 2e  H2(g) + 2+ Co(s) + 2H (aq)  Co (aq) + H2(g) 2+

–

E° = –0.28 V E° = 0.00 V

o = 0.00 V (–0.28 V) = 0.28 V Ecell o Reaction is spontaneous under standard-state conditions because Ecell is positive. 2+ 2+ – Oxidation: Hg2 (aq)  2Hg (aq) + 2e E° = +0.92 V 2+ – Reduction: Hg2 (aq) + 2e  2Hg(l) E° = +0.79 V 2+ 2+ Overall: 2Hg2 (aq)  2Hg (aq) + 2Hg(l) 2+ 2+ o or Hg2 (aq)  Hg (aq) + Hg(l) = 0.79 V 0.92 V = –0.13 V Ecell o Negative Ecell indicates reaction is not spontaneous under standard-state conditions.

21.43

a) Mn (aq) + 2H 2O(l) + 2Co (aq)  MnO2(s) + 4H (aq) + 2Co (aq) o o o = E cathode – Eanode Ecell = 1.82 V – (1.23 V) = 0.59 V The reaction is spontaneous. – – + b) 3AgCl(s) + NO(g) + 2H 2O(l)  3Ag(s) + 3Cl (aq) + NO3 (aq) + 4H (aq) o o o Ecell = E cathode – Eanode = 0.22 V – (0.96 V) = –0.74 V The reaction is nonspontaneous.

21.44

o o o Plan: Use the relationship Ecell = E cathode – Eanode . E o values are found in Appendix D. Spontaneous reactions o have Ecell > 0. Solution: 2+ – a) Oxidation: 3{Cd(s)  Cd (aq) + 2e } E° = –0.40 V 2– + – 3+ Reduction: Cr2O7 (aq) + 14H (aq) + 6e  2Cr (aq) + 7H2O(l) E° = +1.33 V 2– + 3+ 2+ Overall: Cr2O7 (aq) + 3Cd(s) + 14H (aq)  2Cr (aq) + 3Cd (aq) + 7H2O(l) o = +1.33 V– (–0.40 V) = +1.73 V Ecell

The reaction is spontaneous.

Oxidation: Reduction: Overall:

Pb(s)  Pb (aq) + 2e 2+ – Ni (aq) + 2e  Ni(s) 2+ 2+ Pb(s) + Ni (aq)  Pb (aq) + Ni(s) 2+

–

The reaction is not spontaneous. 21.45

E° = –0.13 V E° = –0.25 V o = –0.25 V– (–0.13 V) = –0.12 V Ecell

a) 2Cu (aq) + PbO2(s) + 4H (aq) + SO4 (aq)  2Cu (aq) + PbSO4(s) + 2H 2O(l) o o o = E cathode – Eanode = 1.70 V – (0.15 V) = 1.55 V Ecell The reaction is spontaneous. 2+ + b) H 2O2(aq) + Ni (aq)  2H (aq) + O2(g) + Ni(s) o o o = –0.25 V – (0.68 V) = –0.93 V Ecell = E cathode – Eanode +

2–

The reaction is nonspontaneous.

21-24

21.46

o Plan: Spontaneous reactions have Ecell > 0. All three reactions are written as reductions. When two halfreactions are paired, one half-reaction must be reversed and written as an oxidation. Reverse the half-reaction o o o o that will result in a positive value of Ecell using the relationship Ecell = E cathode – Eanode . To balance each reaction, multiply each half-reaction by an integer so that the number of electrons lost equals the number of electrons gained and then add the half-reactions. The greater (more positive) the reduction potential, the greater the strength as an oxidizing agent. Solution: Adding (1) and (2) to give a spontaneous reaction involves converting (1) to oxidation: 3+ – Oxidation: 2{Al(s)  Al (aq) + 3e } E° = –1.66 V – – Reduction: 3{N2O4(g) + 2e  2NO2 (aq)} E° = +0.867 V – 3+ 3N2O4(g) + 2Al(s)  6NO2 (aq) + 2Al (aq) o = 0.867 V – (–1.66 V) = 2.53 V Ecell 3+ – Oxidizing agents: N2O4 > Al ; reducing agents: Al > NO2 Adding (1) and (3) to give a spontaneous reaction involves converting (1) to oxidation: 3+ – Oxidation: 2{Al(s)  Al (aq) + 3e } E° = –1.66 V 2– – 2– – Reduction: 3{SO4 (aq) + H2O(l) + 2e  SO3 (aq) + 2OH (aq)} E° = +0.93 V 2– 3+ 2– – 2Al(s) + 3SO4 (aq) + 3H2O(l)  2Al (aq) + 3SO3 (aq) + 6OH (aq) o = 0.93 V – (–1.66 V) = 2.59 V Ecell 2– 3+ 2– Oxidizing agents: SO4 > Al ; reducing agents: Al > SO3 Adding (2) and (3) to give a spontaneous reaction involves converting (2) to oxidation: – – Oxidation: 2NO2 (aq)  N2O4(g) + 2e E° = 0.867 V 2– – 2– – Reduction: SO4 (aq) + H2O(l) + 2e  SO3 (aq) + 2OH (aq) E° = 0.93 V 2– – 2– – SO4 (aq) + 2NO2 (aq) + H2O(l)  SO3 (aq) + N2O4(g) + 2OH (aq) o = 0.93 V – 0.867 V = 0.06 V Ecell 2– – 2– Oxidizing agents: SO4 > N2O4; reducing agents: NO2 > SO3 Rank oxidizing agents (substance being reduced) in order of increasing strength: 3+ 2– Al < N2O4 < SO4 Rank reducing agents (substance being oxidized) in order of increasing strength: 2– – SO3 < NO2 < Al

21.47

3N2O(g) + 6H (aq) + 2Cr(s)  3N2(g) + 3H 2O(l) + 2Cr (aq) o = 1.77 V – (–0.74 V) = 2.51 V Ecell 3+ Oxidizing agents: N2O > Cr ; reducing agents: Cr > N2 + 3+ 3Au (aq) + Cr(s)  3Au(s) + Cr (aq) o Ecell = 1.69 V – (–0.74 V) = 2.43 V + 3+ Oxidizing agents: Au > Cr ; reducing agents: Cr > Au + + N2O(g) + 2H (aq) + 2Au(s)  N2(g) + H 2O(l) + 2Au (aq) o = 1.77 V – (1.69 V) = 0.08 V Ecell + Oxidizing agents: N2O > Au ; reducing agents: Au > N2 + 3+ Oxidizing agents: N2O > Au > Cr ; reducing agents: Cr > Au > N2

21.48

21-25

o o o o will result in a positive value of Ecell using the relationship Ecell = E cathode – Eanode . To balance each reaction, multiply each half-reaction by an integer so that the number of electrons lost equals the number of electrons gained and then add the half-reactions. The greater (more positive) the reduction potential, the greater the strength as an oxidizing agent. Solution: Adding (1) and (2) to give a spontaneous reaction involves converting (2) to oxidation: 2+ – Oxidation: Pt(s)  Pt (aq) + 2e E° = +1.20 V + – Reduction: 2HClO(aq) + 2H (aq) + 2e  Cl2(g) + 2H2O(l) E° = +1.63 V + 2+ 2HClO(aq) + Pt(s) + 2H (aq)  Cl2(g) + Pt (aq) + 2H2O(l) o = 1.63 V – 1.20 V = 0.43 V Ecell 2+ Oxidizing agents: HClO > Pt ; reducing agents: Pt > Cl2 Adding (1) and (3) to give a spontaneous reaction involves converting (3) to oxidation: 2– – Oxidation: Pb(s) + SO4 (aq)  PbSO4(s) + 2e E° = –0.31 V + – Reduction: 2HClO(aq) + 2H (aq) + 2e  Cl2(g) + 2H2O(l) E° = +1.63 V 2– + 2HClO(aq) + Pb(s) + SO4 (aq) + 2H (aq)  Cl2(g) + PbSO4(s) + 2H2O(l) o = 1.63 V – (–0.31 V) = 1.94 V Ecell Oxidizing agents: HClO > PbSO4; reducing agents: Pb > Cl2 Adding (2) and (3) to give a spontaneous reaction involves converting (3) to oxidation: 2– – Oxidation: Pb(s) + SO4 (aq)  PbSO4(s) + 2e E° = –0.31 V 2+ – Reduction: Pt (aq) + 2e  Pt(s) E° = 1.20 V 2+ 2– Pt (aq) + Pb(s) + SO4 (aq)  Pt(s) + PbSO4(s) o = 1.20 V – (–0.31 V) = 1.51 V Ecell 2+ Oxidizing agents: Pt > PbSO4; reducing agents: Pb > Pt 2+ Order of increasing strength as oxidizing agent: PbSO4 < Pt < HClO 2– Order of increasing strength as reducing agent: Cl2 < Pt < (Pb + SO4 )

21.49

S2O8 (aq) + 2I (aq)  2SO4 (aq) + I 2(s) o = 2.01 V – (0.53 V) = 1.48 V Ecell 2– – 2– Oxidizing agents: S2O8 > I 2; reducing agents: I > SO4 2– + – 3+ Cr2O7 (aq) + 14H (aq) + 6I (aq)  2Cr (aq) + 7H 2O(l) + 3I 2(s) o Ecell = 1.33 V – (0.53 V) = 0.80 V 2– – 3+ Oxidizing agents: Cr2O7 > I 2; reducing agents: I > Cr 2– 3+ 2– 2– + 3S2O8 (aq) + 2Cr (aq) + 7H 2O(l)  6SO4 (aq) + Cr2O7 (aq) + 14H (aq) o = 2.01 V – (1.33 V) = 0.68 V Ecell 2– 2– 3+ 2– Oxidizing agents: S2O8 > Cr2O7 ; reducing agents: Cr > SO4 2– 2– – 3+ 2– Oxidizing agents: S2O8 > Cr2O7 > I 2; reducing agents: I > Cr > SO4

21.50

Metal A + Metal B salt  solid colored product on metal A Conclusion: Product is solid metal B. B is undergoing reduction and plating out on A. A is a better reducing agent than B. Metal B + acid  gas bubbles + Conclusion: Product is H2 gas produced as result of reduction of H . B is a better reducing agent than acid. Metal A + Metal C salt  no reaction Conclusion: C is not undergoing reduction. C must be a better reducing agent than A.

2–

–

2–

21-26

Since C is a better reducing agent than A, which is a better reducing agent than B and B reduces acid, then C would also reduce acid to form H2 bubbles. The order of strength of reducing agents is: C > A > B. 21.51

21.52

21.53

a) Copper metal is coating the iron. 2+ b) The oxidizing agent is Cu and the reducing agent is Fe. c) Yes, this reaction, being spontaneous, may be made into a voltaic cell. 2+ 2+ d) Cu (aq) + Fe(s)  Cu(s) + Fe (aq) o o o e) Ecell = E cathode – Eanode = 0.34 V – (–0.44 V) = 0.78 V 0.0592 V

Q  log   and G = –nFEcell K  n a) When Q/K < 1, the reaction is preceding to the right; Ecell > 0 and G < 0 and the reaction is spontaneous. When Q/K = 1, the reaction is at equilibrium; Ecell = 0 and G = 0. When Q/K > 1, the reaction is preceding to the left; Ecell < 0 and G > 0 and the reaction is not spontaneous. b) Only when Q/K < 1 will the reaction proceed spontaneously and be able to do work.

Ecell = –

At the negative (anode) electrode, oxidation occurs so the overall cell reaction is + + + + A(s) + B (aq)  A (aq) + B(s) with Q = [A ]/[B ]. a) The reaction proceeds to the right because with Ecell > 0 (voltaic cell), the spontaneous reaction occurs. As the + + cell operates, [A ] increases and [B ] decreases. b) Ecell decreases because the cell reaction takes place to approach equilibrium, Ecell = 0.  RT  A  o o ln  . c) Ecell and Ecell are related by the Nernst equation: Ecell = Ecell – nF  B  o Ecell = Ecell when o Ecell to equal Ecell .

  A  RT  A  ln  = 0. This occurs when ln  = 0. Recall that e0 = 1, so [A+] must equal [B+] for nF  B  B 

o d) Yes, it is possible for Ecell to be less than Ecell when [A ] > [B ].

21.54

o a) Examine the Nernst equation: Ecell = Ecell –

RT RT ln K – ln Q nF nF RT  K  RT  Q  ln  Ecell = ln  = – nF  Q  nF  K 

RT ln Q. nF

Ecell =

If Q/K < 1, Ecell will decrease with a decrease in cell temperature. If Q/K > 1, Ecell will increase (become less negative) with a decrease in cell temperature. RT [Active ion at anode] o b) Ecell = Ecell – ln nF [Active ion at cathode] Ecell will decrease as the concentration of an active ion at the anode increases. c) Ecell will increase as the concentration of an active ion at the cathode increases. d) Ecell will increase as the pressure of a gaseous reactant in the cathode compartment increases. 21.55

In a concentration cell, the overall reaction takes place to decrease the concentration of the more concentrated electrolyte. The more concentrated electrolyte is reduced, so it is in the cathode compartment.

21-27

21.56

o o nEcell nFEcell or log K = . Use E° values from the RT 0.0592 o o o o Appendix to calculate Ecell ( Ecell = E cathode – Eanode ) and then calculate K. The substances given in the problem must

Plan: The equilibrium constant can be found by using ln K =

be the reactants in the equation. Solution: 2+ – a) Oxidation: Ni(s)  Ni (aq) + 2e E° = –0.25 V + – Reduction: 2{Ag (aq) + 1e  Ag(s)} E° = +0.80 V + 2+ Ni(s) + 2Ag (aq)  Ni (aq) + 2Ag(s) o o o = E cathode – Eanode = 0.80 V – (–0.25 V) = 1.05 V; two electrons are transferred. Ecell log K =

o 2 1.05 V  nEcell = = 35.47297 0.0592 0.0592 V 35.47297

K = 10 35 35 K = 2.97146 × 10 = 3 × 10 2+ – b) Oxidation: 3{Fe(s)  Fe (aq) + 2e } E° = –0.44 V 3+ – Reduction: 2{Cr (aq) + 3e  Cr(s)} E° = –0.74 V 3+ 2+ 3Fe(s) + 2Cr (aq)  3Fe (aq) + 2Cr(s) o o o = E cathode – Eanode = –0.74 V – (–0.44 V) = –0.30 V; six electrons are transferred. Ecell log K = K = 10

o 6 0.30 V nEcell = = –30.4054 0.0592 0.0592 V

–30.4054

K = 3.936 × 10 21.57

–31

= 4 × 10

–31

a) 2Al(s) + 3Cd (aq)  2Al (aq) + 3Cd(s) 2+

o cell

= E

log K =

o cathode

o anode

–E

= –0.40 V – (–1.66 V) = 1.26 V

o 6 1.26 V  nEcell = = 127.7027 0.0592 0.0592 V

K = 5.04316 × 10

127

= 5 × 10

127

b) I2(s) + 2Br (aq)  Br2(l) + 2I (aq) –

o cell

–

= E

log K =

o cathode

o anode

–E

n=2

= 0.53 V – (1.07 V) = –0.54 V

o 2 0.54 V nEcell = = –18.24324 0.0592 0.0592 V

K = 5.7116 × 10 21.58

n=6

–19

= 6 × 10

–19

o o nEcell nFEcell or log K = . Use E° values from the RT 0.0592 o o o o Appendix to calculate Ecell ( Ecell = E cathode – Eanode ) and then calculate K. The substances given in the problem must

Plan: The equilibrium constant can be found by using ln K = be the reactants in the equation. Solution: a) Oxidation: 2{Ag(s)  Ag (aq) + 1e }

E° = +0.80 V

Reduction: Mn (aq) + 2e  Mn(s)

E° = –1.18 V

–

21-28

2Ag(s) + Mn (aq)  2Ag (aq) + Mn(s) 2+

o o o = E cathode – Eanode = –1.18 V – (0.80 V) = –1.98V; two electrons are transferred. Ecell

log K = K = 10

o 2 1.98 V nEcell = = –66.89189 0.0592 0.0592 V

–66.89189

K = 1.2826554 × 10

–67

= 1 × 10

–67

b) Oxidation: 2Br (aq)  Br2(l) + 2e –

–

E° = 1.07 V

Reduction: Cl2(g) + 2e  2Cl (aq) –

–

E° = 1.36 V

2Br (aq) + Cl2(g)  Br2(l) + 2Cl (aq) –

–

o o o = E cathode – Eanode = 1.36 V – 1.07 V = 0.29 V; two electrons transferred. Ecell

log K = K = 10

o 2 0.29 V  nEcell = = 9.797297 0.0592 0.0592 V

9.797297

K = 6.2704253 × 10 = 6 × 10 9

21.59

a) 2Cr(s) + 3Cu (aq)  2Cr (aq) + 3Cu(s) n=6 o o o = E cathode – Eanode = 0.34 V – (–0.74 V) = 1.08 V Ecell 2+

log K =

o 6 1.08 V  nEcell = = 109.4594595 0.0592 0.0592 V

K = 2.880444 × 10

109

= 3 × 10

109

b) Sn(s) + Pb (aq)  Sn (aq) + Pb(s) 2+

o cell

= E

log K =

o cathode

n=2

o anode

= –0.13 V – (–0.14 V) = 0.01V

–E

o 2 0.01 V  nEcell = = 0.33784 0.0592 0.0592 V

K = 2.1769 = 2 21.60

o Plan: Use G° = nFEcell to calculate G°. Substitute J/C for V in the unit for Ecell . Solution: – – 5 5 o a) G° = nFEcell = – (2 mol e )(96,485 C/mol e )(1.05 J/C) = –2.026185 × 10 = –2.03 × 10 J – – 5 5 o b) G° = nFEcell = – (6 mol e )(96,485 C/mol e )(–0.30 J/C) = 1.73673 × 10 = 1.7 × 10 J

21.61

o a) G° = nFEcell = – (6 mol e )(96,485 C/mol e )(1.26 J/C) = –7.294266 × 10 = –7.29 × 10 J – – 5 5 o b) G° = nFEcell = – (2 mol e )(96,485 C/mol e )(–0.54 J/C) = 1.042038 × 10 = 1.0 × 10 J

21.62

o o Plan: Use G° = nFEcell to calculate G°. Substitute J/C for V in the unit for E cell . Solution: – – 5 5 o a) G° = nFEcell = – (2 mol e )(96,485 C/mol e )(–1.98 J/C) = 3.820806 × 10 = 3.82 × 10 J – – 4 4 o b) G° = nFEcell = – (2 mol e )(96,485 C/mol e )(0.29 J/C) = –5.59613 × 10 = –5.6 × 10 J

21.63

o a) G° = nFEcell = – (6 mol e )(96,485 C/mol e )(1.08 J/C) = –6.252228 × 10 = –6.25 × 10 J

–

21-29

o b) G° = nFEcell = – (2 mol e )(96,485 C/mol e )(0.01 J/C) = –1.9297 × 10 = –2 × 10 J –

21.64

o Plan: Use Ecell =

–

0.0592 V o and then G° = –RT ln K to find G°. log K to find Ecell n

Solution:

T = (273 + 25)K = 298 K 0.0592 V 0.0592 V o = log (5.0 10 4 ) = 0.278179 = 0.28 V Ecell log K = 1 n

G° = –RT ln K = – (8.314 J/mol ∙ K)(298 K) ln (5.0 × 10 ) = –2.68067797 × 10 = –2.7 × 10 J 4

21.65

0.0592 V o and then G° = –RT ln K to find G°. log K to find Ecell n T = (273 + 25)K = 298 K 0.0592 V 0.0592 V o = log (5.0 106 ) = –0.31382 = –0.31 V Ecell log K = 1 n

o Use Ecell =

G° = –RT ln K = – (8.314 J/mol ∙ K)(298 K) ln (5.0 × 10 ) = 3.0241424 × 10 = 3.0 × 10 J –6

21.66

o Plan: Use Ecell =

0.0592 V o and then G° = –RT ln K to find G°. log K to find Ecell n

Solution: T = (273 + 25)K = 298 K 0.0592 0.0592 V o = log 65 = 0.0536622 = 0.054 V Ecell log K = 2 n 4 4 G° = –RT ln K = – (8.314 J/mol ∙ K)(298 K) ln (65) = –1.03423 × 10 = –1.0 × 10 J 21.67

0.0592 V o and then G° = –RT ln K to find G°. log K to find Ecell n T = (273 + 25)K = 298 K 0.0592 0.0592 V o = log 0.065 = –0.03513776 = –0.035 V Ecell log K = 2 n

o Use Ecell =

G° = –RT ln K = – (8.314 J/mol ∙ K)(298 K) ln (0.065) = 6.772116 × 10 = 6.8 × 10 J 3

21.68

Plan: The standard reference half-cell is the H2/H cell. Since this is a voltaic cell, a spontaneous reaction is +

occurring. For a spontaneous reaction between H2/H and Cu/Cu , Cu must be reduced and H2 must be 0.0592 o o oxidized. Write the balanced reaction and calculate Ecell . Use the Nernst equation, Ecell = Ecell – log Q, n 2+ to find [Cu ] when Ecell = 0.22 V. Solution:

Oxidation: H2(g)  2H (aq) + 2e 2+ – Reduction: Cu (aq) + 2e  Cu(s) 2+ + Cu (aq) + H2(g)  Cu(s) + 2H (aq) o o o = E cathode – Eanode = 0.34 V – 0.00 V = 0.34 V Ecell 0.0592 o Ecell = Ecell – log Q n +

–

E° = 0.00 V E° = 0.34 V

21-30

 H  0.0592 – log  Cu2+  PH n   2 2

Ecell = E

o cell

For a standard hydrogen electrode [H ] = 1.0 M and [H2] = 1.0 atm. 0.0592 1.0 0.22 V = 0.34 V – log Cu 2  1.0 2   0.22 V – 0.34 V = – –0.12 V = –

0.0592 1.0 log Cu 2  1.0 2  

0.0592 1.0 log 2   2 Cu  1.0

1.0 x Raise each side to 10 . Cu 2  1.0   1 4 1.132541 × 10 = [Cu 2+ ] 2+ –5 –5 [Cu ] = 8.82970 × 10 = 8.8 × 10 M 4.054054 = log

21.69

The cell reaction is: Pb (aq) + Mn(s)  Pb(s) + Mn (aq) 2+

o o o = E cathode – Eanode = –0.13 V – (–1.18 V) = 1.05 V Ecell 0.0592 o Ecell = Ecell – log Q n  Mn 2  0.0592 o Ecell = Ecell – log  2   Pb  n   1.4  0.0592 0.44 V = 1.05 V – log 2   2  Pb 

(0.44 V – 1.05 V)(–2/0.0592) = log

1.4  = 20.608108  Pb2   

1.4  20 = 4.056094 × 10  Pb2    2+ –21 –21 [Pb ] = 3.45160 × 10 = 3.5 × 10 M

21.70 Plan: Since this is a voltaic cell, a spontaneous reaction is occurring. For a spontaneous reaction between Ni/Ni

o and Co/Co , Ni must be reduced and Co must be oxidized. Write the balanced reaction and calculate E cell . Use 0.0592 o the Nernst equation, Ecell = Ecell – log Q, to find Ecell at the given ion concentrations. Then n 2+ the Nernst equation can be used to calculate [Ni ] at the given Ecell. To calculate equilibrium concentrations, recall 2+

that at equilibrium Ecell = 0.00. Solution:

a) Oxidation: Co(s)  Co (aq) + 2e 2+

–

Reduction: Ni (aq) + 2e  Ni(s) 2+

–

E° = –0.28 V E° = –0.25 V

Ni (aq) + Co(s)  Ni(s) + Co (aq) 2+

o o o = E cathode – Eanode = –0.25 V – (–0.28 V) = 0.03 V Ecell

21-31

0.0592 log Q n  Co2  0.0592 o Ecell = Ecell – log  2   Ni  n   0.0592  0.20  Ecell = 0.03 V – log 2  0.80  o Ecell = Ecell –

n = 2e

Ecell = 0.047820976 V = 0.05 V

–

b) From part a), notice that an increase in [Co ] leads to a decrease in cell potential. Therefore, the concentration of cobalt ion must increase further to bring the potential down to 0.03 V. Thus, the new 2+ 2+ concentrations will be [Co ] = 0.20 M + x and [Ni ] = 0.80 M – x (there is a 1:1 mole ratio).  Co2  0.0592 o Ecell = Ecell – log  2   Ni  n   0.0592  0.20 + x  0.03 V = 0.03 V – log 2  0.80  x  0.0592  0.20 + x  0=– log 2  0.80  x   0.20 + x  x 0 = log Raise each side to 10 .  0.80  x   0.20 + x  1=  0.80  x  0.20 + x = 0.80 – x x = 0.30 M 2+ [Ni ] = 0.80 – x = 0.80 – 0.30 = 0.50 M 2+ 2+ c) At equilibrium Ecell = 0.00; to decrease the cell potential to 0.00, [Co ] increases and [Ni ] decreases. 0.0592  0.20 + x  0.00 V = 0.03 V – log 2  0.80  x  0.0592  0.20 + x  –0.03 V = – log 2  0.80  x   0.20 + x  x 1.0135135 = log Raise each side to 10 .  0.80  x   0.20 + x  10.316051 =  0.80  x  x = 0.71163 [Co ] = 0.20 + 0.71163 = 0.91163 = 0.91 M 2+

[Ni ] = 0.80 – 0.71163 = 0.08837 = 0.09 M 2+

21.71

The spontaneous reaction (voltaic cell) is Cd (aq) + Mn(s)  Cd(s) + Mn (aq) with o o o = E cathode – Eanode = –0.40 V – (–1.18 V) = 0.78 V. Ecell 0.0592 o a) Ecell = Ecell – log Q n  Mn 2  0.0592 – o Ecell = Ecell – log  2  n = 2e Cd  n   2+

21-32

Ecell = 0.78 –

0.0592  0.090  log 2  0.060 

Ecell = 0.774787699 V = 0.77 V 2+

b) For the [Cd ] to decrease from 0.060 M to 0.050 M, a change of 0.010 M, the [Mn ] must increase by the same amount, from 0.090 M to 0.100 M. 0.0592  0.10  Ecell = 0.78 – log 2 0.050  Ecell = 0.771089512 V = 0.77 V c) Increase the manganese and decrease the cadmium by equal amounts. Total = 0.150 M  Mn 2  0.0592 0.055 = 0.78 – log  2  Cd  2   2   Mn    = 3.1134597 × 1024 Cd2    2+ 24 2+ [Mn ] = (3.1134597 × 10 )[Cd ] 2+ 2+ [Mn ] + [Cd ] = 0.150 M 24 2+ 2+ (3.1134597 × 10 )[Cd ] + [Cd ] = 0.150 M 2+ –26 [Cd ] = 4.81779 × 10 M 2+ –26 [Mn ] = 0.150 M – 4.81779 × 10 M = 0.150 M d) At equilibrium Ecell = 0.00.  Mn 2  0.0592 0.00 = 0.78 – log  2  Cd  2    Mn 2    = 2.2456980 × 1026 Cd2    2+ 26 2+ [Mn ] = (2.2456980 × 10 )[Cd ] 2+

[Mn ] + [Cd ] = 0.150 M 26

(2.2456980 × 10 )[Cd ] + [Cd ] = 0.150 M 2+

[Cd ] = 6.6794378 × 10

–28

= 7 × 10

–28

[Mn ] = 0.150 M – [Cd ] = 0.150 M 2+

21.72

Plan: The overall cell reaction proceeds to increase the 0.10 M H concentration and decrease the 2.0 M H o concentration. Use the Nernst equation to calculate Ecell. Ecell = 0 V for a concentration cell since the halfreactions are the same. Solution:

Half-cell A is the anode because it has the lower concentration. Oxidation: H2(g;0.95 atm)  2H (aq; 0.10 M) + 2e +

Reduction: 2H (aq; 2.0 M) + 2 e +

–

 H2(g; 0.60 atm)

2H (aq; 2.0 M) + H2(g; 0.95 atm)

o = 0.00 V Ecell 0.0592 o Ecell = Ecell – log Q n

–

E°(anode) = 0.00 V E°(cathode) = 0.00 V

 2H (aq; 0.10 M) + H2(g; 0.60 atm) +

–

n = 2e

21-33

 H+ anode PH cathode Q for the cell equals + 2  H cathode PH anode 2

0.10 0.60 Q= = 0.00157895 2 2.0 0.95 0.0592 Ecell = 0.00 V – log (0.00157895) = 0.0829283 = 0.083 V 2 2

21.73

Sn (0.87 M)  Sn (0.13 M) 2+

Half-cell B is the cathode.

0.0592 log Q n [Sn 2 ]anode 0.0592 o Ecell = Ecell – log 2 [Sn 2 ]cathode o Ecell = Ecell –

o = 0.00 V Ecell

–

n = 2e

0.0592 0.13 log 2 0.87 Ecell = 0.024437 = 0.024 V Ecell = 0.00 V –

21.74

Electrons flow from the anode, where oxidation occurs, to the cathode, where reduction occurs. The electrons always flow from the anode to the cathode, no matter what type of cell.

21.75

The electrodes are separated by an electrolyte paste, which for the ordinary dry cell battery contains ZnCl 2, NH4Cl, MnO2, starch, graphite, and water.

21.76

A D-sized battery is much larger than an AAA-sized battery, so the D-sized battery contains a greater amount of the cell components. The potential, however, is an intensive property and does not depend on the amount of the cell components. (Note that amount is different from concentration.) The total amount of charge a battery can produce does depend on the amount of cell components, so the D-sized battery produces more charge than the AAA-sized battery.

21.77

a) Alkaline batteries = (6.0 V)(1 alkaline battery/1.5 V) = 4 alkaline batteries. b) Voltage = (6 Ag batteries)(1.6 V/Ag battery) = 9.6 V c) The usual 12-V car battery consists of six 2 V cells. If two cells are shorted only four cells remain. Voltage = (4 cells)(2 V/cell) = 8 V

21.78

The Teflon spacers keep the two metals separated so the copper cannot conduct electrons that would promote the corrosion of the iron skeleton. Oxidation of the iron by oxygen causes rust to form and the metal to corrode.

21.79

Bridge supports rust more rapidly at the water line due to the presence of large concentrations of both O 2 and H2O.

21.80

Chromium, like iron, will corrode through the formation of a metal oxide. Unlike the iron oxide, which is relatively porous and easily cracks, the chromium oxide forms a protective coating, preventing further corrosion.

21.81

Plan: Sacrificial anodes are metals with E° values that are more negative than that for iron, –0.44 V, so they are more easily oxidized than iron.

21-34

Solution: a) E°(aluminum) = –1.66 V. Yes, except aluminum resists corrosion because once a coating of its oxide covers it, no more aluminum corrodes. Therefore, it would not be a good choice. b) E°(magnesium) = –2.37 V. Yes, magnesium is appropriate to act as a sacrificial anode. c) E°(sodium) = –2.71 V. Yes, except sodium reacts with water, so it would not be a good choice. d) E°(lead) = –0.13 V. No, lead is not appropriate to act as a sacrificial anode because its E° value is too high. e) E°(nickel) = –0.25 V. No, nickel is inappropriate as a sacrificial anode because its E° value is too high. f) E°(zinc) = –0.76 V. Yes, zinc is appropriate to act as a sacrificial anode. g) E°(chromium) = –0.74 V. Yes, chromium is appropriate to act as a sacrificial anode. 21.82

a) Oxidation occurs at the left electrode (anode). b) Elemental M forms at the right electrode (cathode). c) Electrons are being released by ions at the left electrode. d) Electrons are entering the cell at the right electrode.

21.83

3Cd (aq) + 2Cr(s)  3Cd(s) + 2Cr (aq) o = –0.40 V – (–0.74 V) = 0.34 V Ecell To reverse the reaction requires 0.34 V with the cell in its standard state. A 1.5 V supplies more than enough 2+ potential, so the cadmium metal oxidizes to Cd and chromium plates out.

21.84

o The Ehalf–cell values are different than the Ehalf-cell values because in pure water, the [H ] and [OH ] are –7 1.0 × 10 M rather than the standard-state value of 1 M.

21.85

The oxidation number of nitrogen in the nitrate ion, NO3 , is +5 and cannot be oxidized further since nitrogen – has only five electrons in its outer level. In the nitrite ion, NO 2 , on the other hand, the oxidation number of nitrogen is +3, so it can be oxidized at the anode to the +5 state.

21.86

Due to the phenomenon of overvoltage, the products predicted from a comparison of electrode potentials are not always the actual products. When gases (such as H2(g) and O2(g)) are produced at metal electrodes, there is an overvoltage of about 0.4 to 0.6 V more than the electrode potential indicates. Due to this, if H2 or O2 is the expected product, another species may be the true product.

21.87

Plan: Oxidation occurs at the anode, while reduction occurs at the cathode. Solution: – – a) At the anode, bromide ions are oxidized to form bromine (Br2). 2Br (l) → Br2(l) + 2e + – b) At the cathode, sodium ions are reduced to form sodium metal (Na). Na (l) + e → Na(s)

21.88

a) At the negative electrode (cathode) barium ions are reduced to form barium metal (Ba). b) At the positive electrode (anode), iodide ions are oxidized to form iodine (I2).

21.89

Plan: Oxidation occurs at the anode, while reduction occurs at the cathode. Decide which anion is more likely to be oxidized and which cation is more likely to be reduced. The less electronegative anion holds its electrons less tightly and is more likely to be oxidized; the cation with the higher ionization energy has the greater attraction for electrons and is more likely to be reduced. Solution: Either iodide ions or fluoride ions can be oxidized at the anode. The ion that more easily loses an electron will – form. Since I is less electronegative than F, I will more easily lose its electron and be oxidized at the anode. The product at the anode is I2 gas. The iodine is a gas because the temperature is high to melt the salts. Either potassium or magnesium ions can be reduced at the cathode. Magnesium has greater ionization energy than potassium because magnesium is located up and to the right of potassium on the periodic table. The greater ionization energy means that magnesium ions will more readily add an electron (be reduced) than potassium ions. The product at the cathode is magnesium (liquid).

–

21-35

21.90

Liquid strontium (Sr) forms at the negative electrode, and gaseous bromine (Br2) forms at the positive electrode.

21.91

21.92

Liquid calcium forms at the negative electrode and chlorine gas forms at the positive electrode.

21.93

Plan: Compare the electrode potentials of the species with those of water. The reduction half-reaction with the more positive electrode potential occurs at the cathode, and the oxidation half-reaction with the more negative electrode potential occurs at the anode. Solution: Possible reductions: 2+ – Cu (aq) + 2e  Cu(s) E° = +0.34 V 2+ – Ba (aq) + 2e  Ba(s) E° = –2.90 V 3+ – Al (aq) + 3e  Al(s) E° = –1.66 V – – 2H2O(l) + 2e  H2(g) + 2OH (aq) E = –1 V with overvoltage Copper can be prepared by electrolysis of its aqueous salt since its reduction half-cell potential is more positive than the potential for the reduction of water. The reduction of copper is more spontaneous than the reduction of 2+ 3+ water. Since the reduction potentials of Ba and Al are more negative and therefore less spontaneous than the reduction of water, these ions cannot be reduced in the presence of water since the water is reduced instead. Possible oxidations: – – 2Br (aq)  Br2(l) + 2e E° = +1.07 V + – 2H2O(l)  O2(g) + 4H (aq) + 4e E = 1.4 V with overvoltage Bromine can be prepared by electrolysis of its aqueous salt because its reduction half-cell potential is more negative than the potential for the oxidation of water with overvoltage. The more negative reduction potential for – Br indicates that its oxidation is more spontaneous than the oxidation of water.

21.94

Strontium is too electropositive to form from the electrolysis of an aqueous solution. The elements that electrolysis will separate from an aqueous solution are gold, tin, and chlorine.

21.95

21-36

Iodine can be prepared by electrolysis of its aqueous salt because its reduction half-cell potential is more negative – than the potential for the oxidation of water with overvoltage. The more negative reduction potential for I indicates that its oxidation is more spontaneous than the oxidation of water. 21.96

Electrolysis will separate both iron and cadmium from an aqueous solution.

21.97

Plan: Compare the electrode potentials of the species with those of water. The reduction half-reaction with the more positive electrode potential occurs at the cathode, and the oxidation half-reaction with the more negative electrode potential occurs at the anode. Solution: a) Possible oxidations: + – 2H2O(l)  O2(g) + 4H (aq) + 4e E = 1.4 V with overvoltage – – 2F  F2(g) + 2e E° = 2.87 V – Since the reduction potential of water is more negative than the reduction potential for F , the oxidation of water is – + more spontaneous than that of F . The oxidation of water produces oxygen gas (O2), and hydronium ions (H3O ) at the anode. Possible reductions: – – 2H2O(l) + 2e  H2(g) + 2OH (aq) E = –1 V with overvoltage + – Li (aq) + e  Li(s) E° = –3.05 V + Since the reduction potential of water is more positive than that of Li , the reduction of water is more spontaneous + – than the reduction of Li . The reduction of water produces H2 gas and OH at the cathode. b) Possible oxidations: + – 2H2O(l)  O2(g) + 4H (aq) + 4e E = 1.4 V with overvoltage + The oxidation of water produces oxygen gas (O2), and hydronium ions (H3O ) at the anode. 2– 2– The SO4 ion cannot oxidize as S is already in its highest oxidation state in SO4 . Possible reductions: – – 2H2O(l) + 2e  H2(g) + 2OH (aq) E = –1 V with overvoltage 2+ – Sn (aq) + 2e  Sn(s) E° = –0.14 V 2– + – SO4 (aq) + 4H (aq) + 2e  SO2(g) + 2H2O(l) E = –0.63 V (approximate) The potential for sulfate reduction is estimated from the Nernst equation using standard-state concentrations and + –7 pressures for all reactants and products except H , which in pure water is 1 × 10 M. –7 4

E = 0.20 V – (0.0592/2) log [1/(1 × 10 ) ] = –0.6288 = –0.63 V 2+ The most easily reduced ion is Sn with the most positive reduction potential, so tin metal forms at the cathode. 21.98

a) Solid zinc (Zn) forms at the cathode and liquid bromine (Br2) forms at the anode. + b) Solid copper (Cu) forms at the cathode and both oxygen gas (O2) and aqueous hydrogen ions (H ) form at the anode.

21.99

21-37

The most easily reduced ion is NO3 , with the most positive reduction potential so NO gas is formed at the cathode. b) Possible oxidations: + – 2H2O(l)  O2(g) + 4H (aq) + 4e E = 1.4 V with overvoltage –

2Cl (aq)  Cl2(g) + 2e E° = 1.36 V – The oxidation of chloride ions to produce chlorine gas occurs at the anode. Cl has a more negative reduction potential showing that it is more easily oxidized than water. Possible reductions: – – 2H2O(l) + 2e  H2(g) + 2OH (aq) E = –1 V with overvoltage –

–

Mn (aq) + 2e  Mn(s) E° = –1.18 V It is easier to reduce water than to reduce manganese ions, so hydrogen gas and hydroxide ions form at the 2+ cathode. The reduction potential of Mn is more negative than that of water showing that its reduction is less spontaneous than that of water. 2+

–

21.100 a) Solid iron (Fe) forms at the cathode, and solid iodine (I2) forms at the anode. – b) Gaseous hydrogen (H2) and aqueous hydroxide ion (OH ) form at the cathode, while gaseous oxygen (O2) and + aqueous hydrogen ions (H ) form at the anode. 2+

21.101 Plan: Write the half-reaction for the reduction of Mg . Convert mass of Mg to moles and use the mole ratio in the balanced reaction to find the number of moles of electrons required for every mole of Mg produced. The Faraday constant is used to find the charge of the electrons in coulombs. To find the current, the charge is divided by the time in seconds. Solution: 2+ – Mg + 2e  Mg   1 mol Mg   2 mol e  = 3.751542575 = 3.75 mol e– a) Moles of electrons = 45.6 g Mg 1 mol Mg   24.31 g Mg   96, 485 C  5 5 b) Charge = 3.751542575 mol e   = 3.619676 × 10 = 3.62 × 10 C  mol e   3.619676 10 5 C   1 h  A   c) Current =     = 28.727587 = 28.7 A   3600 s  3.50 h  C s   +

–

21.102 Na + 1e

 Na

 1 mol Na  1 mol e  –  a) Moles of electrons = 215 g Na   = 9.351892127 = 9.35 mol e  22.99 g Na 1 mol Na   96, 485 C  5 5 b) Charge = 9.351892127 mol e   = 9.0231731 × 10 = 9.02 × 10 C  mol e   9.023173110 5 C   1 h  A   c) Current =     = 26.383547 = 26.4 A C   9.50 h  3600 s    s  2+

21.103 Plan: Write the half-reaction for the reduction of Ra . Use the Faraday constant to convert charge in coulombs to moles of electrons provided; the balanced reduction reaction converts moles of electrons to moles of radium produced. Multiply moles of radium by its molar mass to obtain grams. Solution: 2+

–

Ra + 2e

 Ra

21-38

In the reduction of radium ions, Ra , to radium metal, the transfer of two electrons occurs.  1 mol e   1 mol Ra   226 g Ra   Mass (g) of Ra = (235 C)     = 0.275224 = 0.275 g Ra    96, 485 C   2 mol e   1 mol Ra  21.104 Al + 3e  Al 3+ In the reduction of aluminum ions, Al , to aluminum metal, the transfer of three electrons occurs.  1 mol e   1 mol Al  26.98 g Al   Mass (g) of Al = (305 C)   3 mol e  1 mol Al  = 0.028428944 = 0.0284 g Al  96, 485 C    3+

–

21.105 Plan: Write the half-reaction for the reduction of Zn . Convert mass of Zn to moles and use the mole ratio in the balanced reaction to find the number of moles of electrons required for every mole of Zn produced. The Faraday constant is used to find the charge of the electrons in coulombs. To find the time, the charge is divided by the current. Solution: 2+ – Zn + 2e  Zn   1 mol Zn   2 mol e   96, 485 C   1 1 A   3 3  Time (s) = (65.5 g Zn)       21.0 A  C  = 9.20169 × 10 = 9.20 × 10 s  65.41 g Zn  1 mol Zn   1 mol e   s  21.106 Ni + 2e  Ni 2+

–

 1 mol Ni  2 mol e  96, 485 C  1 1 A        Time (s) = (1.63 g Ni)   1 mol Ni  1 mol e 13.7 A  C  = 391.1944859 = 391 s  58.69 g Ni     s  +

2–

21.107 a) The sodium sulfate ionizes to produce Na and SO4 ions which make the water conductive; therefore the current will flow through the water to complete the circuit, increasing the rate of electrolysis. Pure water, which –7 + – contains very low (10 M) concentrations of H and OH , conducts electricity very poorly. + b) The reduction of H2O has a more positive half-potential (–1 V) than the reduction of Na (–2.71 V); the more spontaneous reduction of water will occur instead of the less spontaneous reduction of sodium ion. The 2– oxidation of H2O is the only oxidation possible because SO4 cannot be oxidized under these conditions. In other + 2– words, it is easier to reduce H2O than Na and easier to oxidize H2O than SO4 . –

–

21.108 Iodine is formed first since the oxidation potential of I is more positive than the oxidation potential of Br . 2+

21.109 Plan: Write the half-reaction for the reduction of Zn . Find the charge in coulombs by multiplying the current by the time in units of seconds. Use the Faraday constant to convert charge in coulombs to moles of electrons provided; the balanced reduction reaction converts moles of electrons to moles of zinc produced. Multiply moles of zinc by its molar mass to obtain grams. Solution: 2+

Zn + 2e

–

 Zn

  24 h  C   3600 s  1 mol e  1 mol Zn   65.41 g Zn   Mass (g) of Zn = (0.855 A)  s  2.50 day       1 mol Zn   A   1 h  96, 485 C   2 mol e  1 day  





= 62.59998 = 62.6 g Zn

21-39

21.110 Mn (aq) + 2H2O(l)  MnO2(s) + 4H (aq) + 2e  103 g   1 mol MnO 2   2 mol e   96485 C   1  A  1 h    Time (h) = (1.00 kg MnO 2 )       25.0 A  C  3600 s   1 kg   86.94 g MnO 2  1 mol MnO 2  1 mol e   s    2+

–

= 24.66196 = 24.7 h The MnO2 product forms at the anode, since the half-reaction is an oxidation. +

21.111 Plan: Write the reduction half-reaction for H . Use the ideal gas law to find the moles of H2 produced. The mole ratio in the balanced reduction reaction gives the moles of electrons required for that amount of H 2 and the Faraday constant converts moles of electrons to charge in coulombs. To convert coulombs to energy in joules, remember that 1 V equals 1 J/C; multiply the charge in coulombs by volts to obtain joules. Convert this energy to units of kilojoules and use the given conversion factor between mass of oil and energy to find the mass of oil combusted to provide the needed amount of energy. Solution: + – The half-reaction is: 2H (aq) + 2e  H2(g) a) First, find the moles of hydrogen gas. (12.0 atm)(3.5106 L) PV 6 n= = = 1.716682 × 10 mol H2 L  atm   RT 0.0821 273  25 K  mol  K  Then, find the coulombs knowing that there are two electrons transferred per mol of H 2.  2 mol e  96485 C     = 3.3126813 × 1011 = 3.3 × 1011 C Coulombs = (1.716682106 mole H 2 )    1 mol H 2  1 mol e  1.44 J  b) Energy (J) =  3.312681011 C = 4.77026 × 1011 = 4.8 × 1011 J  C   1 kJ   1 kg 4 4 c) Mass (kg) = 4.77026 1011 J  3   = 1.19257 × 10 = 1.2 × 10 kg 4 10 J  4.0 10 kJ 

21.112 a) The half-reactions are: H2O(l) + Zn(s)  ZnO(s) + 2H (aq) + 2e +

–

2H (aq) + MnO2(s) + 2e  Mn(OH)2(s) +

–

Two moles of electrons flow per mole of reaction.

 1 mol Zn 1 mol MnO  86.94 g MnO   2  2  b) Mass (g) of MnO2 = (4.50 g Zn)    = 5.981196 = 5.98 g MnO2   65.41 g Zn  1 mol Zn  1 mol MnO 2   1 mol Zn 1 mol H O  18.02 g H 2 O   2  Mass (g) of H2O = (4.50 g Zn)    = 1.23972 = 1.24 g H2O   1 mol H 2 O   65.41 g Zn  1 mol Zn 

c) Total mass of reactants = 4.50 g Zn + 5.981196 g MnO2 + 1.23972 g H2O = 11.720916 = 11.72 g

 1 mol Zn  2 mol e  96, 485 coulombs      = 1.32757 × 104 = 1.33 × 104 C d) Charge = (4.50 g Zn)   1 mol Zn   1 mol e  65.41 g Zn    e) An alkaline battery consists of more than just reactants. The case, electrolyte paste, cathode, absorbent, and unreacted reactants (less than 100% efficient) also contribute to the mass of an alkaline battery.

21.113 The reaction is: Ag (aq) + e  Ag(s) Mass (g) of Ag = 1.8016 g – 1.7854 g = 0.0162 g Ag produced +

–

21-40

 1 mol Ag  1 mol e  96, 485 coulombs      = 14.48616 = 14.5 C Coulombs = (0.0162 g Ag)   107.9 g Ag 1 mol Ag  1 mol e 

21.114 Plan: Write balanced half-reactions for the reduction of each metal ion. From the current, 65.0% of the moles of product will be copper and 35.0% zinc. Assume a current of exactly 100 C. The amount of current used to generate copper would be (65.0%/100%)(100 C) = 65.0 C, and the amount of current used to generate zinc would be (35.0%/100%)(100 C) = 35.0 C. Convert each coulomb amount to moles of electrons using the Faraday constant and use the balanced reduction reactions to convert moles of electrons to moles and then mass of each metal. Divide the mass of copper produced by the total mass of both metals produced and multiply by 100 to obtain mass percent. Solution: The half-reactions are: Cu (aq) + 2e  Cu(s) and Zn (aq) + 2e  Zn(s) 2+

–

 1 mol e 1 mol Cu  63.55 g Cu      = 0.021406177 g Cu Mass (g) of copper = (65.0 C)   96, 485 C  2 mol e  1 mol Cu 

 1 mol e 1 mol Zn   65.41 g Zn    Mass (g) of zinc = (35.0 C)    = 0.01186376 g Zn   96, 485 C  2 mol e   1 mol Zn  0.021406177 g Cu 100 = 64.340900 = 64.3% Cu Mass % of copper = 0.021406177 g Cu + 0.01186376 g Zn 3+

21.115 Plan: Write the reduction half-reaction for Au . Use the equation for the volume of a cylinder to find the volume of gold required; use the density to convert volume of gold to mass and then moles of gold. The mole ratio in the balanced reduction reaction is used to convert moles of gold to moles of electrons required and the Faraday constant is used to convert moles of electrons to coulombs. Divide the coulombs by the current to obtain time in seconds, which is converted to time in days. To obtain the cost, start by multiplying the moles of gold from part a) by four to get the moles of gold needed for the earrings. Convert this moles to grams, then to troy ounces, and finally to dollars. Solution: 3+ – The reaction is: Au (aq) + 3e  Au(s) 2 a) V = πr h 2 103 m  1 cm   4.00 cm  3   = 0.314159265 cm3 0.25 mm    V (cm ) = π    1 mm 102 m   2  19.3 g Au  1 mol Au   = 0.03077804 mol Au Moles of Au = (0.314159265 cm 3 )    1 cm 3 197.0 g Au   3 mol e  96, 485 C  A   1 h  1 day  1     Time (days) = (0.03077804 mol Au)         1 mol Au  1 mol e  C s  0.013 A  3600 s   24 h 

= 7.931675 = 8 days b) The time required doubles once for the second earring of the pair and doubles again for the second side, thus it will take four times as long as one side of one earring. Time = (4)(7.931675 days) = 31.7267 = 32 days 197.0 g Au  1 troy oz  $1210   c) Cost = (4)(0.03077804 mol Au)     = 943.60918 = $940  1 mol Au   31.10 g  troy oz 

21-41

21.116 a) The half-reaction is: 2H2O(l)  O2(g) + 4H (aq) + 4e Determine the moles of oxygen from the ideal gas equation. Use the half-reaction and the current to convert the moles of oxygen to time.     3 (99.8 kPa)(10.0 L) 1 atm  10 Pa  PV   n= =   = 0.398569697 mol O2  5  L  atm  RT 1.0132510 Pa  1 kPa   273  28 K      0.0821   mol  K  +

–

 4 mol e  96, 485 C  A  1  1 min      Time (min) = (0.398569697 mol O 2 )           1 mol O  1 mol e  C s 1.3 A   60 s  2

= 1.97210 × 10 = 2.0 × 10 min 3

b) The balanced chemical equation is 2H2O(l)  2H2(g) + O2(g). The moles of oxygen determined previously and this chemical equation leads to the mass of hydrogen.  2 mol H  2.016 g H 2   2   = 1.60703 = 1.61 g H Mass (g) of H2 = (0.398569697 mol O2 )  2   1 mol H 2   1 mol O2  21.117 Corrosion is an oxidation process. This would be favored by the metal being in contact with the moist ground. To counteract this, electrons should flow into the rails and away from the overhead wire, so the overhead wire should be connected to the positive terminal. 21.118 Plan: Write the half-reactions and cell reaction for the silver battery. Convert mass of zinc to moles of zinc, keeping in mind that only 80% of the zinc will react; from the moles of zinc, the moles of electrons required is obtained. The Faraday constant is used to convert moles of electrons to charge in coulombs which is divided by the current to obtain the time in seconds. The moles of zinc is also used to find the moles of Ag 2O consumed and the amount of Ag needed for that amount of Ag2O. Convert this mass of silver to troy ounces and then to dollars. Solution: The half-reactions and the cell reaction are: Zn(s) + 2OH (aq)  ZnO(s) + H2O(l) + 2e –

–

Ag2O(s) + H2O(l) + 2e  2Ag(s) + 2OH (aq) –

–

Zn(s) + Ag2O(s)  ZnO(s) + 2Ag(s)  80%  1 mol Zn   = 0.00917291 mol Zn Moles of Zn = 0.75 g Zn  100%  65.41 g Zn 

 2 mol e  96, 485 C  A  1 μA    1 h  1 day  1     a) Time (days) = 0.00917291 mol Zn           6  0.85 μA  3600 s  24 h  1 mol Zn  1 mol e  C s 10 A    





= 2.410262 × 10 = 2.4 × 10 days 4

1 mol Ag 2 O 100%  2 mol Ag 107.9 g Ag       b) Mass (g) of Ag = 0.00917291 mol Zn   1 mol Zn  95% 1 mol Ag 2 O  1 mol Ag 





= 2.0836989 = 2.1 g Ag

 95%  1 troy oz  $23.00     c) Cost = 2.0836989 g Ag  4    100%  31.10 g Ag  troy oz  2.410262 10 days  = 6.073818 × 10 = $6.1 × 10 /day –5

–5

21-42

21.119 Cu (aq) + 2e  Cu(s) 2+

–

 1 mol e   C  3600 s  1 mol Cu   63.55 g Cu   10 h   Theoretical amount of copper = 5.8 A  s       2 mol e  1 mol Cu   96, 485 C   A  1 h   









= 68.7632 g Cu actual yield 53.4 g Cu 100 = 100 = 77.6578 = 78% Efficiency = theoretical yield 68.7632 g Cu The final assumes that 10 h has two significant figures. 21.120

a) Molten electrolysis Cl2(g) Na(l) + Na (l) – Cl (l)

Anode product Cathode product Species reduced Species oxidized

b) Aqueous electrolysis Cl2(g) – H2(g) and OH (aq) H2O(l) – Cl (aq)

o 21.121 Plan: Since the cells are voltaic cells, the reactions occurring are spontaneous and will have a positive E cell . Write the two half-reactions. When two half-reactions are paired, one half-reaction must be reversed and written o as an oxidation. Reverse the half-reaction that will result in a positive value of Ecell using the relationship o o o o Ecell = E cathode – Eanode. E values are found in Appendix D. The oxidation occurs at the negative electrode (the anode). Use the Nernst equation to find cell potential at concentrations other than 1 M. Solution: 2+ a) Cell with SHE and Pb/Pb : 2+ – Oxidation: Pb(s) → Pb (aq) + 2e E o = – 0.13 V + – Reduction: 2H (aq) + 2e → H2(g) E o = 0.0 V o o o Ecell = E cathode – Eanode = 0.0 V – (– 0.13 V) = 0.13 V 2+

Cell with SHE and Cu/Cu :

Oxidation: H2(g) → 2 H (aq) + 2e 2+

E o = 0.0 V

–

E o = 0.34 V

–

Reduction: Cu (aq) + 2e → Cu(s) E

o cell

= E

o cathode

o anode

–E

= 0.34 V– 0.00 V = 0.34 V

b) The anode (negative electrode) in cell with SHE and Pb/Pb is Pb. 2+

The anode in cell with SHE and Cu/Cu is platinum in the SHE. 2+

c) The precipitation of PbS decreases [Pb ]. Use Nernst equation to see how this affects potential. Cell reaction is: + 2+ Pb(s) + 2H (aq)  Pb (aq) + H2(g) 0.0592 o Ecell = Ecell – log Q n [Pb2+ ]PH2 0.0592 – o Ecell = Ecell – log n = 2e n [H+ ]2

[Pb2+ ]PH2 0.0592 log 2 [H+ ]2 Decreasing the concentration of lead ions gives a negative value for the term: o Ecell = Ecell –

[Pb2+ ]PH2 0.0592 log 2 [H+ ]2

When this negative value is subtracted from Ecell , cell potential increases. +

d) The [H ] = 1.0 M and the H2 = 1 atm in the SHE.

Cell reaction: Cu (aq) + H2(g)  Cu(s) + 2H (aq) 2+

21-43

0.0592 log Q n [H + ]2 0.0592 o Ecell = Ecell – log n [Cu 2+ ]PH2 o Ecell = Ecell –

–

n = 2e

(1) 0.0592 log 16 (110 )(1 atm) 2 Ecell = –0.1336 = –0.13 V Ecell = 0.34 V –

 1 cm   3 3 21.122 a) Volume (cm ) = (2.0 cm 2 )(7.5106 m)  2  = 0.0015 cm 10 m 

b) Mass = (0.0015 cm )(10.5 g Ag/1 cm ) = 0.01575 = 0.016 g 3

 1 mol Ag   1 mol e  96, 485 C  A  1 mA  1 min  1          c) Time = (0.01575 g Ag)           3       1 mol Ag   1 mol e  C s 10 A  12.0 mA   60 s  107.9 g Ag 

= 19.56079 = 20. min

 1 troy oz  $23.00 100 cents     = 1.164791 = 1.2 ¢ d) Cost = (0.01575 g Ag)      $1    31.10 g Ag 1 troy oz  –

21.123 The reduction of H2O to H2 and OH is easier than the reduction of Al to Al. 21.124 The three steps equivalent to the overall reaction M (aq) + e  M(s) are: + + 1) M (aq)  M (g) Energy is –Hhydration + – 2) M (g) + e  M(g) Energy is –IE or –Hionization 3) M(g)  M(s) Energy is –Hdeposition The energy for step 3 is similar for all three elements, so the difference in the energy for the overall reaction + depends on the values for –Hhydration and –IE. The lithium ion has a more negative hydration energy than Na and + K because it is a smaller ion with large charge density that holds the water molecules more tightly. The amount of energy required to remove the waters surrounding the lithium ion offsets the lower ionization energy to make the overall energy for the reduction of lithium larger than expected. +

–

21.125 The key factor is that the table deals with electrode potentials in aqueous solution. The very high and low standard electrode potentials involve extremely reactive substances, such as F2 (a powerful oxidant), and Li (a powerful reductant). These substances react directly with water, rather than according to the desired half-reactions. An alternative (essentially equivalent) explanation is that any aqueous cell with a voltage of more than 1.23 V has the ability to electrolyze water into hydrogen and oxygen. When two electrodes with 6 V across them are placed in water, electrolysis of water will occur. 21.126 Wherever the tin surface is scratched to expose the iron, the iron corrosion occurs more rapidly because iron is a better reducing agent (compare E° values). An electrochemical cell is set up where iron becomes the anode and tin the cathode. A coating on the inside of the can separates the tin from the normally acidic contents, which would react with the tin and add an unwanted “metallic” taste. 3+

21.127 Plan: Write the half-reaction for the reduction of Al . Convert mass of Al to moles and use the mole ratio in the balanced reaction to find the number of moles of electrons required for every mole of Al produced. The Faraday constant is used to find the charge of the electrons in coulombs. To find the time, the charge is divided by the current. To calculate the electrical power, multiply the time by the current and voltage, remembering that Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

21-44

1 A = 1 C/s (thus, 100,000 A is 100,000 C/s) and 1 V = 1 J/C (thus, 5.0 V = 5.0 J/C). Change units of J to kW ∙ h. To find the cost of the electricity, convert the mass of aluminum from lb to kg and use the kW ∙ h per 1000 kg of aluminum calculated in part b) to find the kW ∙ h for that mass of aluminum, keeping in mind the 90.% efficiency. Solution: 3+ – a) Aluminum half-reaction: Al (aq) + 3 e  Al(s), so n = 3. Remember that 1 A = 1 C/s.  103 g    1 mol Al   3 mol e  96, 485 C  A  1   Time (s) = 1000 kg Al             1 kg   26.98 g Al   1 mol Al  1 mol e  C s 100,000 A 





= 1.0728503 × 10 = 1.073 × 10 s 5

The molar mass of aluminum limits the significant figures.

100, 000 C  5.0 J  1 kJ  1 kW  h    = 1.4900699 × 104 = 1.5 × 104 kW ∙ h   Power = (1.072850310 5 s)   C 103 J  3.6 103 kJ   s  1 kg 1.4900699 10 4 kW  h  0.123 cents 100%       = 0.923551 = 0.92 ¢/lb Al c) Cost = 1 lb Al     2.205 lb   1 kW h  90.%  1000 kg Al

21.128 a) Electrons flow from magnesium bar to the iron pipe since magnesium is more easily oxidized than iron. 2+ – b) The magnesium half-reaction is: Mg(s)  Mg (aq) + 2e Current is charge per time. The mass of magnesium can give the total charge. Convert the mass of magnesium to moles of magnesium and multiply by two moles of electrons produced for each mole of magnesium and by the Faraday constant to convert the moles of electrons to coulombs of charge. Time must be in seconds, so convert the 8.5 yr to s. 103 g  1 mol Mg  2 mol e   96, 485 C     (12 kg Mg)       mol e     1 kg  24.31 g Mg 1 mol Mg  charge   A  Current = =   = 0.35511 = 0.36 A      C s  time  365.25 days  24 h  3600 s  (8.5 yr)     1 yr 1 days  1 h   21.129 Plan: When considering two substances, the stronger reducing agent will reduce the other substance. Solution: Statement: Metal D + hot water  reaction Conclusion: D reduces water to produce H2(g). D is a stronger + reducing agent than H . Statement: D + E salt  no reaction Conclusion: D does not reduce E salt, so E reduces D salt. E is better reducing agent than D. Statement: D + F salt  reaction Conclusion: D reduces F salt. D is better reducing agent than F. If E metal and F salt are mixed, the salt F would be reduced producing F metal because E has the greatest reducing strength of the three metals (E is stronger than D and D is stronger than F). The ranking of increasing reducing strength is F < D < E. 21.130 3Pt(s) + 16H (aq) + 4NO3 (aq) + 18Cl (aq)  3PtCl6 (aq) + 4NO(g) + 8H2O(l) +

–

2–

21.131 Plan: Examine the change in oxidation numbers in the equations to find n, the moles of electrons transferred. Use G = nFE to calculate G. Substitute J/C for V in the unit for E . Convert G to units of kJ and divide by the total mass of reactants to obtain the ratio. Solution: a) Cell I: Oxidation number (O.N.) of H changes from 0 to +1, so one electron is lost from each of four hydrogen atoms for a total of four electrons. O.N. of oxygen changes from 0 to –2, indicating that two electrons are gained by each of the two oxygen atoms for a total of four electrons. There is a transfer of four electrons in the reaction. Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

21-45

G = –nFE = – (4 mol e )(96,485 C/mol e )(1.23 J/C) = –4.747062 × 10 = –4.75 × 10 J Cell II: In Pb(s)  PbSO4, O.N. of Pb changes from 0 to +2 and in PbO2  PbSO4, O.N. of Pb changes from +4 to +2. There is a transfer of two electrons in the reaction. – – 5 5 G = –nFE = – (2 mol e )(96,485 C/mol e )(2.04 J/C) = –3.936588 × 10 = –3.94 × 10 J Cell III: O.N. of each of two Na atoms changes from 0 to +1 and O.N. of Fe changes from +2 to 0. There is a transfer of two electrons in the reaction. – – 5 5 G = –nFE = – (2 mol e )(96,485 C/mol e )(2.35 J/C) = –4.534795 × 10 = –4.53 × 10 J –

–

 2.016 g H   32.00 g O    2 2 b) Cell I: Mass of reactants = (2 mol H 2 )    (1 mol O 2 )   = 36.032 g  1 mol H 2   1 mol O 2   4.74706210 5 J  1 kJ  wmax    =  10 3 J  = –13.17457 = –13.2 kJ/g reactant mass 36.032 g   

Cell II: Mass of reactants =  207.2 g Pb   239.2 g PbO      98.09 g H 2 SO 4  2   1 mol PbO  (1 mol Pb)   2 mol H SO   2  2 4   1 mol Pb   1 mol PbO 2   1 mol H 2 SO 4 









= 642.58 g  3.93658810 5 J  1 kJ  wmax   = –0.612622 = –0.613 kJ/g  =    10 3 J  reactant mass  642.58 g  22.99 g Na  126.75 g FeCl   2   (1 mol FeCl )  Cell III: Mass of reactants = (2 mol Na)   = 172.73 g 2   1 mol Na   1 mol FeCl  2  4.534795 10 5 J  1 kJ  wmax    =  10 3 J  = –2.625366 = –2.63 kJ/g reactant mass  172.73 g   Cell I has the highest ratio (most energy released per gram) because the reactants have very low mass while Cell II has the lowest ratio because the reactants are very massive.

21.132 The current traveling through both cells is the same, so the amount of silver is proportional to the amount of zinc based on their reduction half-reactions: 2+ – + – Zn(s)  Zn (aq) + 2e and Ag (aq) + e  Ag(s)  1 mol Zn  2 mol e 1 mol Ag 107.9 g Ag       = 3.9590 = 4.0 g Ag Mass (g) of Ag = (1.2 g Zn)   65.41 g Zn 1 mol Zn  1 mol e  1 mol Ag 

21.133 A standard thermodynamic table would allow the calculation of the equilibrium constant by the equation G° = –RT ln K. G° could be calculated from Gfo values, or Hfo and S° values. A table of standard electrode potentials would allow the calculation of the equilibrium constant by the following equation: E° = (0.0592 V/n) log K 21.134 2Fe(s) + O2(g) + 4H (aq)  2H2O(l) + 2Fe (aq) An increase in temperature increases the cell potential for the above reaction. An increase in the temperature will cause the reaction rate to increase. Finally, Kw for water increases with temperature, thus at a higher temperature there will be a higher hydrogen ion concentration. +

21.135 Plan: Write the balanced equation. Multiply the current and time to calculate total charge in coulombs. Remember that the unit 1 A is 1 C/s, so the time must be converted to seconds. From the total charge, the number of electrons transferred to form copper is calculated by dividing total charge by the Faraday constant. Each mole of copper deposited requires two moles of electrons, so divide moles of electrons by two to get moles of copper. Then Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

21-46

convert to grams of copper. The initial concentration of Cu is 1.00 M (standard condition) and initial volume is 345 mL. Use this to calculate the initial moles of copper ions, then subtract the number of moles of copper ions 2+ converted to copper metal and divide by the cell volume to find the remaining [Cu ]. Solution: 2+ a) Since the cell is a voltaic cell, write a spontaneous reaction. The reduction of Cu is more spontaneous 2+ 2+ 2+ than the reduction of Sn : Cu (aq) + Sn(s)  Cu(s) + Sn (aq)  1 mol e 1 mol Cu   C  3600 s   63.55 g Cu   (48.0 h)    Mass (g) of Cu = (0.17 A)  s        96, 485 C  2 mol e   1 mol Cu   A  1 h  = 9.674275 = 9.7 g Cu

103 L   mol Cu 2   2+  = 0.345 mol Cu2+  345 mL  b) Initial moles of Cu = 1.00    L 1 mL     





 1 mol Cu   2+  = 0.1522309205 mol Cu2+ Moles of Cu reduced = 9.674275 g Cu   63.55 g Cu 





Remaining moles of Cu = initial moles – moles reduced = 0.345 mol – 0.1522309205 = 0.1927691 mol Cu 2+

M Cu =

0.1927691 mol Cu 2  2+ = 0.558751 = 0.56 M Cu 103 L   (345 mL)    1 mL 

21.136 Plan: Write the balanced equation and determine the half reactions in the anode and cathode compartments. Use +

the Nernst equation to calculate the H ion concentration within the value of Q. Calculate the pH from this concentration. Solution: Oxidation:

H2(g)  2H (aq) + 2e

Reduction:

2[Ag (aq) + e  Ag(s)]

–

E°(anode) = 0.00 V

–

E°(cathode) = 0.80 V

2Ag (aq) + H2(g)  2Ag(s) + 2H (aq) o o o = 0.80 V – 0.00 V = 0.80 V Ecell = E cathode – Eanode 2 H  0.0592 0.0592 o o Ecell = Ecell – log Q = Ecell – log 2 n n PH2  Ag   +

0.915 V = 0.80 V –

0.0592 V

H 

n = 2 mol e

–

log

1.00 0.100 

–3.885135 = log

0.0100 take the inverse log of both sides

H 

0.000130276 =

0.0100 + 2 -6 [H ] = 1.30276 × 10 +

[H ] = 1.141385 × 10

-3

pH = –log[H ] = –log[1.141385 × 10 ] = 2.942568 = 2.94 +

-3

21-47

21.137 Plan: Examine each reaction to determine which reactant is the oxidizing agent; the oxidizing agent is the reactant that gains electrons in the reaction, resulting in a decrease in its oxidation number. Solution: From reaction between U + Cr  Cr + U , find that Cr oxidizes U . 2+ 2+ 2+ From reaction between Fe + Sn  Sn + Fe , find that Sn oxidizes Fe. 4+ 2+ 3+ From the fact that there is no reaction that occurs between Fe and U , find that Fe oxidizes U . 3+ 2+ 2+ 3+ From reaction between Cr + Fe  Cr + Fe , find that Cr oxidizes Fe. 2+ 2+ 3+ 2+ 2+ From reaction between Cr + Sn  Sn + Cr , find that Sn oxidizes Cr . 2+ 3+ 2+ 3+ Notice that nothing oxidizes Sn, so Sn must be the strongest oxidizing agent. Both Cr and Fe oxidize U , so 4+ 3+ 3+ 2+ U must be the weakest oxidizing agent. Cr oxidizes iron so Cr is a stronger oxidizing agent than Fe . The half-reactions in order from strongest to weakest oxidizing agent: 2+ – Sn (aq) + 2e  Sn(s) 3+ – 2+ Cr (aq) + e  Cr (aq) 2+ – Fe (aq) + 2e  Fe(s) 4+ – 3+ U (aq) + e  U (aq) 3+

21.138 The given half-reactions are: 3+ – 2+ (1) Fe (aq) + e ⇆ Fe (aq) 2+

–

(2) Fe (aq) + 2e ⇆ Fe(s) 3+ – (3) Fe (aq) + 3e ⇆ Fe(s) a) G° = –nFE° – – 4 4 Reaction 1: G° = – (1 mol e )(96,485 C/mol e )(0.77 J/C) = –7.429345 × 10 = –7.4 × 10 J – – 4 4 Reaction 2: G° = – (2 mol e )(96,485 C/mol e )(–0.44 J/C) = 8.49068 × 10 = 8.5 × 10 J 4 4 4 4 b) G3° = G1° + G2° = –7.429345 × 10 J + 8.49068 × 10 J = 1.061335 × 10 = 1.1 × 10 J 4 – – c) E° = –G°/nF = – (1.061335 × 10 J)/(3 mol e )(96,485 C/mol e )[V/(J/C)] = –0.036667 = –0.037 V 21.139 6e + 14H + Cr2O7  2Cr + 7H2O (red half-reaction) + – CH3CH2OH + H2O  CH3COOH + 4H + 4e (ox half-reaction) – + 2– 3+ 2{6e + 14H + Cr2O7  2Cr + 7H2O} + – 3{CH3CH2OH + H2O  CH3COOH + 4H + 4e } – + 2– 3+ 12e + 28H + 2Cr2O7  4Cr + 14H2O + – 3CH3CH2OH + 3H2O  3CH3COOH + 12H + 12e –

2–

Overall: 3CH3CH2OH + 2Cr2O7 + 16H

 3CH3COOH + 4Cr + 11H2O 3+

21.140 At STP hydrogen gas occupies 22.4 L/mol. The reduction of zinc and of hydrogen both required two moles of electrons per mole of product, thus, the current percentages are equal to the mole percentages produced. 103 g   1 mol Zn   8.50 mol H 2  22.4 L H 2   Volume (L) of H2 = 1 kg Zn     = 31.81278578 = 31.8 L H2  1 kg   65.41 g Zn   91.50 mol Zn  1 mol H 2  o o 21.141 Plan: Write a balanced equation that gives a positive Ecell for a spontaneous reaction. Calculate the Ecell and

use the Nernst equation to find the silver ion concentration that results in the given Ecell. Solution: a) The calomel half-cell is the anode and the silver half-cell is the cathode. The overall reaction is: 2Ag (aq) + 2Hg(l) + 2Cl (aq)  2Ag(s) + Hg2Cl2(s) +

–

o o o = E cathode – Eanode = 0.80 V – 0.24 V = 0.56 V with n = 2. Ecell

21-48

Use the Nernst equation to find [Ag ] when Ecell = 0.060 V. 0.0592 o Ecell = Ecell – log Q n 0.0592 V o 1 Ecell = Ecell – log  2 2 [Ag ] [Cl ]2 0.0592V 1 0.060 V = 0.56 V – log  2 2 [Ag ] [Cl ]2 –

The problem suggests assuming that [Cl ] is constant. Assume it is 1.00 M. 0.0592V 1 – 0.50 V = – log 2 [Ag  ]2 [1.00]2 16.89189 = log 10

16.89189

1 [Ag  ]2 [1.00]2

1 [Ag ]2 16

7.7963232 × 10 = 16

1 [Ag ]2 + 2

7.7963232 × 10 [Ag ] = 1 + 2

[Ag ] = 1.282656 × 10

–17

[Ag ] = 3.581419 × 10 = 3.6 × 10 M +

–9

–

b) Again use the Nernst equation and assume [Cl ] = 1.00 M. 0.0592 V o 1 Ecell = Ecell – log  2 2 [Ag ] [Cl ]2 0.0592 V 1 0.53 V = 0.56 V – log  2 2 [Ag ] [Cl ]2 0.0592 V 1 – 0.03 V = – log 2 [Ag  ]2 [Cl ]2 1 1.0135135 = log [Ag  ]2 [1.00]2 1.0135135 1 10 = [Ag ]2 1 10.31605146 = [Ag ]2 + 2

10.31605146[Ag ] = 1 + 2

[Ag ] = 0.0969363

[Ag ] = 0.311346 = 0.3 M +

21.142 Oxidation: Ag(s)  Ag (aq) + e– +

Reduction: AgCl(s) + e–  Ag(s) + Cl (aq) –

Overall:

o = +0.80 V Eanode o = + 0.22 V E cathode

AgCl(s)  Ag (aq) + Cl (aq) +

–

21-49

o o o = E cathode – Eanode = 0.22 V – 0.80 V = –0.58V Ecell o 1( 0.58 V) nEcell log Ksp = = = –9.797297297 0.0592 0.0592 V

K = 1.594788 × 10

–10

= 1.6 × 10

–10

o 21.143 Plan: Use the Nernst equation to write the relationship between Ecell and the cell potential for both the waste

stream and for the silver standard. Solution: + – a) The reaction is Ag (aq) → Ag(s) + 1e 0.0592 o Ecell = Ecell – log Q n Nonstandard cell: Standard cell: +

b) To find [Ag ]waste:

 0.0592  + o Ewaste = Ecell –   log [Ag ]waste  1 e 

 0.0592  + o Estandard = Ecell –   log [Ag ]standard  1 e 

 0.0592   0.0592  + + o = Estandard +  log [Ag ]standard = Ewaste +  Ecell  log [Ag ]waste     1e   1 e 

[Ag+ ]waste  0.0592  Ewaste – Estandard = –  log   1 e  [Ag + ]standard +

Estandard – Ewaste = (0.0592 V)(log [Ag ]waste – log [Ag ]standard) Estandard  Ewaste + + = (log [Ag ]waste – log [Ag ]standard) 0.0592 V +

log [Ag ]waste =

Estandard  Ewaste 0.0592 V

+ log [Ag ]standard

 E   standard  E waste  +  [Ag ]waste = antilog   ([Ag ]standard )    0.0592 V     + + c) Convert M to ng/L for both [Ag ]waste and [Ag ]standard:

[Ag+ ]waste [Ag + ]standard If both silver ion concentrations are in the same units, in this case ng/L, the “conversions” cancel and the equation derived in part b) applies if the standard concentration is in ng/L.   Estandard  E waste   +  Conc.  Ag   Conc. (Ag )waste = antilog   standard  0.0592 V     d) Plug the values into the answer for part c).

Ewaste – Estandard = (–0.0592 V) log





  0.003   +  (1000. ng/ L) = 889.8654 = 900 ng/L [Ag ]waste = antilog      0.0592 V   e) Temperature is included in the RT/nF term, which equals 0.0592 V/n at 25°C. To account for different temperatures, insert the RT/nF term in place of 0.0592 V/n.

21-50

 2.303RT   2.303RT  + + Estandard +  log [Ag ]standard = Ewaste +  log [Ag ]waste  nF   nF   2.303R  + + Estandard – Ewaste =  (T log [Ag ]waste – Tstandard log [Ag ]standard)  nF  waste  nF  + + (Estandard – Ewaste)  = Twaste log [Ag ]waste – Tstandard log [Ag ]standard  2.303R   nF  + + (Estandard – Ewaste)  + Tstandard log [Ag ]standard = Twaste log [Ag ]waste  2.303R   (E +  Ewaste )(nF / 2.303 R)+ Tstandard log[Ag ]standard  log [Ag ]waste =  standard   Twaste  

 (E +  Ewaste )(nF / 2.303 R)+ Tstandard log[Ag ]standard  [Ag ]waste = anti log  standard    Twaste  21.144 Reduction: Ag (aq) + e–  Ag(s) +

Oxidation: Ag(s) + 2NH3(aq)  Ag(NH3)2 (aq) + e– +

o = 0.80 V E cathode o = 0.37 V Eanode

Overall: Ag (aq) + 2NH3(aq)  Ag(NH3)2 (aq) +

o o o = E cathode – Eanode = 0.80 V  0.37 V = 0.43 V Ecell o 1(0.43 V) nEcell log Ksp = = = 7.26351 0.0592 0.0592 V

K = 1.834467 × 10 = 1.8 × 10 7

21.145 Plan: Multiply the current in amperes by the time in seconds to obtain coulombs. Convert coulombs to moles of electrons with the Faraday constant and use the mole ratio in the balanced half-reactions to convert moles of electrons to moles and then mass of reactants. Divide the total mass of reactants by the mass of the battery to find the mass percentage that consists of reactants. Solution: a) Determine the total charge the cell can produce. 103 A  3600 s  1 C  3    Capacity (C) = 300. mA  h   1 A  s  = 1.08 × 10 C  1 mA  1 h  b) The half-reactions are: 0 2+ – – – Cd  Cd + 2e and NiO(OH) + H2O(l) + e  Ni(OH)2 + OH Assume 100% conversion of reactants.  1 mol e 1 mol Cd 112.4 g Cd    Mass (g) of Cd = 1080 C   1 mol Cd  = 0.62907 = 0.629 g Cd  96,485 C  2 mol e 

 1 mol e 1 mol NiO(OH)  91.70 g NiO(OH)   Mass (g) of NiO(OH) = 1080 C   1 mol NiO(OH)   96,485 C  1 mol e = 1.026439 = 1.03 g NiO(OH)

 1 mol e 1 mol H2 O 18.02 g H 2 O   Mass (g) of H2O = 1080 C   1 mol H O  = 0.20170596 = 0.202 g H2O  96,485 C  1 mol e  2 Total mass of reactants = 0.62907 g Cd + 1.026439 g NiO(OH) + 0.20170596 g H 2O = 1.857215 = 1.86 g reactants

21-51

c) Mass % reactants = 21.146 a) 2Zn  2Zn

1.85721 g 18.3 g

100 = 10.14872 = 10.1%

+ 4e and 4e + O2  2O Four mol e flow per mole of reaction. –

–

2–

–

Mass (g) of Zn = (1/10)(0.275 g battery) = 0.0275 g Zn (assuming the 1/10 is exact.) – – Coulombs = (0.0275 g Zn)(1 mol Zn/65.41 g Zn)(4 mol e /2 mol Zn)(96,485 C/1 mol e ) = 81.1294 = 81.1 C 2 b) Free energy (J) = (volts)(coulombs) = (1.3 V)(81.1294 C) = 105.46822 = 1.1 × 10 J 21.147 Plan: For a list of decreasing reducing strength, place the elements in order of increasing (more positive) E°. Metals with potentials lower than that of water (–0.83 V) can displace hydrogen from water by reducing the hydrogen in water. Metals with potentials lower than that of hydrogen (0.00 V) can displace hydrogen from acids + by reducing the H in acid. Metals with potentials above that of hydrogen (0.00 V) cannot displace (reduce) hydrogen. Solution: Reducing agent strength: Li > Ba > Na > Al > Mn > Zn > Cr > Fe > Ni > Sn > Pb > Cu > Ag > Hg > Au These can displace H2 from water: Li, Ba, Na, Al, and Mn. These can displace H2 from acid: Li, Ba, Na, Al, Mn, Zn, Cr, Fe, Ni, Sn, and Pb. These cannot displace H2: Cu, Ag, Hg, and Au. o 21.148 Determine the Ecell for the reaction given from the free energy: o G° = –nF Ecell

o = –G°/nF = – [(–298 kJ/mol)(10 J/1 kJ)]/(2 mol e )(96485 C/mol e )[V/(J/C)] = 1.54428 V Ecell 3

–

o Using E° for Cu + 2 e  Cu (cathode) and the Ecell just found: 2+

–

o o o = E cathode – Eanode = 1.54428 V Ecell

o 1.54428 V = 0.34 V – Eanode

o = 0.34 V – 1.54428 V = –1.20428 = –1.20 V Eanode

The standard reduction potential of V is –1.20 V. 2+

o For the Ti/V cell, Ecell is 0.43 V. Vanadium is the anode o o o = 0.43 V Ecell = E cathode – Eanode o E cathode –(–1.20428 V) = 0.43 V o = 0.43 V – 1.20428 V = –0.77428 = –0.77 V E cathode 2+ The standard reduction potential of Ti is –0.77 V

21.149 a) The iodine goes from –1 to 0, so it was oxidized. 2– Iodine was oxidized, so S4O6 is the oxidizing agent. – Iodine was oxidized, so I is the reducing agent. b) G° = –nFE° n = 2 o = –G°/nF = – [(87.8 kJ/mol)(10 J/1 kJ)]/(2 mol e )(96,485 C/mol e )[V/(J/C)] = – 0.454993 = – 0.455 V Ecell 3

–

c) S4O6 (aq) + 2e  2S2O3 (aq) 2–

–

2–

Oxygen remains –2 throughout. 2–

Sulfur is +2.5 in S4O6 (2.5 is an average). 2–

Sulfur is +2 in S2O3 .

The potential for the iodine half-reaction comes from the Appendix. Since the iodine was oxidized, it is the anode. o o o = E cathode – Eanode = – 0.454993 V Ecell o – (0.53 V) = –0.454993 V E cathode

21-52

o = 0.53 V – 0.454993 V = 0.075007 = 0.08 V E cathode

21.150 a) The reference half-reaction is: Cu (aq) + 2e  Cu(s) 2+

–

E° = 0.34 V

Before the addition of the ammonia, Ecell = 0. The addition of ammonia lowers the concentration of copper ions 2+

through the formation of the complex Cu(NH3)4 . The original copper ion concentration is [Cu ]original, and the 2+

copper ion concentration in the solution containing ammonia is [Cu ]ammonia.

The Nernst equation is used to determine the copper ion concentration in the cell containing ammonia. 2+

The reaction is Cu initial(aq) + Cu(s) → Cu(s) + Cu ammonia(aq).

The half-cell with the larger concentration of copper ion (no ammonia added) is the reduction and the half-cell

with the lower concentration of copper ion due to the addition of ammonia and formation of the complex is the oxidation.

0.0592 log Q n 0.0592 V [Cu2 ]ammonia 0.129 V = 0.00 V – log 2 [Cu2 ]original o Ecell = Ecell –

0.129 V = –

0.0592 V 2

–4.358108108 = log –5

4.3842155 × 10 = 2+

log

[Cu 2 ]ammonia

[0.0100]original

[Cu 2 ]ammonia

[0.0100]original

[Cu 2 ]ammonia

[0.0100]original –7

[Cu ]ammonia = 4.3842155 × 10 M This is the concentration of the copper ion that is not in the complex. The concentration of the complex and of the uncomplexed ammonia must be determined before Kf may be calculated. The original number of moles of copper and the original number of moles of ammonia are found from the original volumes and molarities:  0.0100 mol Cu(NO 3 )2  1 mol Cu 2  103 L    (90.0 mL) Original moles of copper =     L 1 mol Cu(NO 3 )2  1 mL  –4

= 9.00 × 10 mol Cu

 0.500 mol NH 3 103 L   (10.0 mL) = 5.00 × 10–3 mol NH  Original moles of ammonia =  3   1 mL   L

Determine the moles of copper still remaining uncomplexed.

 4.3842155107 mol Cu 2  103 L  –8   Remaining moles of copper =   1 mL  100.0 mL = 4.3842155 × 10 mol Cu L   





The difference between the original moles of copper and the copper ion remaining in solution is the copper in the complex (= moles of complex). The molarity of the complex may now be found. Moles copper in complex = (9.00 × 10  4.3842155 × 10 ) mol Cu = 8.9995616 × 10 mol Cu 2  8.9995616104 mol Cu 2   1 mol Cu(NH 3 )4   1 mL   Molarity of complex =    3  2 100.0 mL  1 mol Cu 10 L     –4

–3

= 8.9995616 × 10 M Cu(NH3)4

–8

–4

21-53

The concentration of the remaining ammonia is found as follows:

  4 mol NH    3  3 4 2   (5.00 10 mol NH3 )  (8.9995616 10 mol Cu )  2    1 mol Cu  1 mL  Molarity of ammonia =   3  10 L  100.0 mL         = 0.014001754 M ammonia The Kf equilibrium is: 2+ 2+ Cu (aq) + 4NH3(aq) ⇆ Cu(NH3)4 (aq)

Cu(NH3 )4 2  [8.9995616 103 ] –11  = Kf =  2 = 5.34072 × 1011 = 5.3 × 10 4 7 4 Cu   NH3  [4.3842155 10 ][0.014001754]  

b) The Kf will be used to determine the new concentration of free copper ions.

Moles uncomplexed ammonia before the addition of new ammonia = –3

(0.014001754 mol NH3/L)(10 L/1 mL)(100.0 mL) = 0.0014001754 mol NH3 –3

Moles ammonia added = 5.00 × 10 mol NH3 (same as original moles of ammonia)

From the stoichiometry:

Cu (aq)

+ –8

Initial moles



4NH3(aq)

4.3842155 × 10 mol

Cu(NH3)4 (aq) –4

0.001400175 mol

8.9995616 × 10 mol

–3

Added moles

5.00 × 10 mol

–8

Cu is limiting

–8

–(4.3842155 × 10 mol) –4(4.3842155 × 10 mol) +(4.3842155 × 10 mol)

After the reaction

–4

0.006400 mol

9.00000 × 10 mol

Determine concentrations before equilibrium: 2+

[Cu ] = 0

–3

[NH3] = (0.006400 mol NH3/110.0 mL)(1 mL/10 L) = 0.0581818 M NH3 2+

–4

–3

[Cu(NH3)4 ] = (9.00000 × 10 mol Cu(NH3)4 /110.0 mL)(1 mL/10 L) = 0.008181818 M Cu(NH3)4

Now allow the system to come to equilibrium: 2+

Cu (aq)

4NH3(aq)

⇆

Cu(NH3)4 (aq)

Initial molarity

0.0581818

0.008181818

Change

+4x

–x

Equilibrium

0.0581818 + 4 x

0.008181818 – x

 0.008181818  x  Cu(NH3 )4 2   =   = 5.34072 × 1011 Kf =  2 4 Cu   NH3  4   x  0.0581818  4x      Assume – x and + 4x are negligible when compared to their associated numbers:  0.008181818 11 Kf = 5.34072 × 10 = 4  x  0.0581818 2+

–9

x = [Cu ] = 1.3369 × 10 M Cu

21-54

Use the Nernst equation to determine the new cell potential: 0.0592 V [Cu2 ]ammonia E = 0.00 V – log 2 [Cu2 ]original E=–

0.0592 V 2

[1.3369 109 ]

log

E = 0.203467 = 0.20 V

[0.0100]

c) The first step will be to do a stoichiometry calculation of the reaction between copper ions and hydroxide ions. 3  0.500 mol NaOH  1 mol OH  10 L  – –3 –  Moles of OH =    10.0 mL = 5.00 × 10 mol OH    L  1 mL   1 mol NaOH  –4 2+ The initial moles of copper ions were determined earlier: 9.00 × 10 mol Cu The reaction: 2+ – Cu (aq) + 2OH (aq)  Cu(OH)2(s) –4 –3 Initial moles 9.00 × 10 mol 5.00 × 10 mol 2+ –4 –4 Cu is limiting –(9.00 × 10 mol) –2(9.00 × 10 mol)



After the reaction 0 0.0032 mol Determine concentrations before equilibrium: 2+ [Cu ] = 0 –

–3



[NH3] = (0.0032 mol OH /100.0 mL)(1 mL/10 L) = 0.032 M OH Now allow the system to come to equilibrium: 2+ – Cu(OH)2(s) ⇆ Cu (aq) + 2OH (aq) Initial molarity 0.0 0.032 Change +x +2x Equilibrium

Ksp = 2.2 × 10

–20

Ksp = 2.2 × 10

–20

Ksp = 2.2 × 10

–20

–

0.032 + 2 x – 2

= [Cu ][OH ]

= [x][0.032 + 2x]

Assume 2x is negligible compared to 0.032 M. = [x][0.032]

x = [Cu ] = 2.1484375 × 10

–17

= 2.1 × 10

–17

Use the Nernst equation to determine the new cell potential: 0.0592 V [Cu 2 ]hydroxide E = 0.00 V – log 2 [Cu 2  ]original

0.0592 V

[2.14843751017 ] 2 [0.0100] E = 0.434169 = 0.43 V E=–

log

d) Use the Nernst equation to determine the copper ion concentration in the half-cell containing the hydroxide ion. 0.0592 V [Cu 2 ]hydroxide E = 0.00 V – log 2 [Cu 2  ]original 0.340 = –

0.0592 V 2

log

[Cu 2 ]hydroxide [0.0100]

21-55

–11.486486 = log –12

3.2622257 × 10

[Cu 2 ]hydroxide [0.0100]

[Cu 2 ]hydroxide

[0.0100] –14 [Cu ]hydroxide = 3.2622257 × 10 M 2+

Now use the Ksp relationship: 2+

– 2

Ksp = [Cu ][OH ] = 2.2 × 10 Ksp = 2.2 × 10

–20

–20 –14

– 2

= [3.2622257 × 10 ][OH ]

– 2

[OH ] = 6.743862 × 10

–7

[OH ] = 8.2121 × 10 = 8.2 × 10 M OH = 8.2 × 10 M NaOH –

–4

–

–4

21.151 a) The half-reactions found in the Appendix are: 3+ – Au (aq) + 3e ⇆ Au(s) E° = 1.50 V 3+

–

E° = –0.74 V

–

E° = –0.28 V

–

E° = –0.76 V

Cr (aq) + 3e ⇆ Cr(s) Co (aq) + 2e ⇆ Co(s) Zn (aq) + 2e ⇆ Zn(s) o o o Calculate Ecell = E cathode – Eanode Au/Cr

o = 1.50 V – (–0.74 V) = 2.24 V Ecell

Co/Zn

o = –0.28 V – (–0.76 V) = 0.48 V Ecell

b) Connecting the cells as two voltaic cells in series will add the voltages. o o o = EAu/Cr + ECo/Zn = 2.24 V + 0.48 V = 2.72 V Eseries

c) Connecting the cells as one voltaic cell (Au/Cr) and one electrolytic (C /Zn) in series will result in the difference in the voltages. o o o = EAu/Cr – ECo/Zn = 2.24 V – 0.48 V = 1.76 V Eseries 3+

d) In parts a-c), Au is reduced in the Au/Cr cell. 2+ In parts a-b), Co is reduced in the Co/Zn cell. 2+ The connection in part c), forces the Co/Zn cell to operate in reverse, thus, Zn is reduced.   1 mol Au   3 mol e  1 mol Zn   65.41 g Zn   e) Mass (g) of Zn = 2.00 g Au      1 mol Zn  = 0.99609 = 0.996 g Zn 197.0 g Au  1 mol Au   2 mol e  





21.152 a) The half-reactions found in the Appendix are: Oxidation:

Cd(s)  Cd (aq) + 2e

Reduction:

Cu (aq) + 2e  Cu(s)

Overall:

–

E° = – 0.40 V

–

E° = 0.34 V

Cd(s) + Cu (aq)  Cd (aq) + Cu(s) 2+

o o o = E cathode – Eanode = 0.34 V – (–0.40 V) = 0.74 V Ecell 2+

Note: Cd is a better reducing agent than Cu so Cu reduces while Cd oxidizes. o G° = nFEcell – – 5 5 G° = – (2 mol e )(96,485 C/mol e )(0.74 J/C) = –1.427978 × 10 = –1.4 × 10 J o nEcell log K = 0.0592 2 0.74  log K = = 25 0.0592 25 K = 1 × 10 Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

21-56

b) The cell reaction is: Cu (aq) + Cd(s)  Cu(s) + Cd (aq) 2+

Use the Nernst equation: 0.0592 o Ecell = Ecell – log Q n 0.0592 V 2 E = 0.74 V – log [Cd ] 2 2 [Cu  ] An increase in the cadmium concentration by 0.95 M requires an equal decrease in the copper 2+ concentration since the mole ratios are 1:1. Thus, when [Cd ] = 1.95 M, 2+ [Cu ] = (1.00 – 0.95) M = 0.05 M. 0.0592 V [1.95] E = 0.74 V – log 2 [0.05] E = 0.69290 = 0.69 V c) At equilibrium, Ecell = 0, and G = 0 2+ The Nernst equation is necessary to determine the [Cu ]. 2+ 2+ Let the copper ion completely react to give [Cu ] = 0.00 M and [Cd ] = 2.00 M. The system can now go to 2+ 2+ equilibrium giving [Cu ] = +x M and [Cd ] = (2.00 – x ) M. 0.0592 o Ecell = Ecell – log Q n [2.00  x] 0.0592 V 0.00 V = 0.74 V – log [x] 2 Assume x is negligible compared to 2.00. [2.00] 25.0 = log [x] [2.00] 25 1 × 10 = [x] –25 2+ x = 2.0 × 10 M Cu 21.153 a) The chemical equation for the combustion of octane is: 2C8H18(l) + 25O2(g)  16CO2(g) + 18H2O(g) The heat of reaction may be determined from heats of formation. o = m Hfo(products)  n H fo(reactants) H rxn

o = [(16 mol CO2)( Hfo of CO2) + (18 mol H2O)( Hfo of H2O)] H rxn

H

o rxn

– [(2 mol C8H18)( Hfo of C8H18) + (25 mol O2)( Hfo of O2)]

= [(16 mol)(–393.5 kJ/mol) + (18 mol)(–241.826 kJ/mol)]

– [(2 mol)(–250.1 kJ/mol) + (25 mol)(0 kJ/mol)]

H = –10148.868 = –10148.9 kJ o rxn

The energy from 1.00 gal of gasoline is:  4 qt  1 L  1 mL  0.7028 g  1 mol C H   10148.9 kJ   8 18      Energy (kJ) = (1.00 gal)        3      2 mol C 8 H18  1 gal 1.057 q t 10 L  mL 114.22 g C 8 H18  = –1.18158 × 10 = –1.18 × 10 kJ 5

21-57

b) The energy from the combustion of hydrogen must be found using the balanced chemical equation and the heats of formation. H2(g) + 1/2O2(g)  H2O(g) With the reaction written this way, the heat of reaction is simply the heat of formation of water vapor, and no additional calculations are necessary. H° = –241.826 kJ The moles of hydrogen needed to produce the energy from part a) are:  1 mol H   2  = 488.6075 mol Moles of H2 = (1.1815810 5 kJ)   241.826 kJ  Finally, use the ideal gas equation to determine the volume.

 L  atm  (488.6075 mol H2 ) 0.0821 (273  25)K nRT 4 4  mol  K  V= = = 1.195417 × 10 = 1.20 × 10 L P (1.00 atm)





c) This part of the problem requires the half-reaction for the electrolysis of water to produce hydrogen gas. 2H2O(l) + 2e  H2(g) + 2OH (aq) Use 1 A = 1 C/s  2 mol e  96, 485 C   s    4 4  Time (s) = (488.6075 mol H 2 )  = 9.42866 × 10 = 9.43 × 10 seconds     1 mol H  1 mol e 1.00 103 C  2 –

–

d) Find the coulombs involved in the electrolysis of 488.6075 moles of H2.  2 mol e  96485 C     = 94,286,589 C Coulombs = (488.6075 mol H 2 )    1 mol H 2 1 mol e  Joules = C × V = 94,286,589 C × 6.00 V = 565,719,534 J

Power (kW ∙ h) = 565, 719, 534 J  1 kW  h  = 157.144 = 157 kW ∙ h  3.6 10 6 J 

e) The process is only 88.0% efficient, additional electricity is necessary to produce sufficient hydrogen. This is the purpose of the (100%/88.0%) factor. 100%   0.123 cents  = 21.964445 = 22.0 ¢ Cost = 157.144 kW  h   88%  1 kW  h 

21-58

CHAPTER 22 THE ELEMENTS IN NATURE AND INDUSTRY END–OF–CHAPTER PROBLEMS 22.1

Hydrogen is abundant in the universe since its simple atoms were the first “created” after the “Big Bang.” H2 has the lowest density of any material, so it is only weakly held by the gravitational field of Earth.

22.2

Iron forms iron(III) oxide, Fe2O3 (commonly known as hematite). Calcium forms calcium carbonate, CaCO3, (commonly known as limestone). Sodium is commonly found in sodium chloride (halite), NaCl. Zinc is commonly found in zinc sulfide (sphalerite), ZnS.

22.3

a) Differentiation refers to the processes involved in the formation of Earth into regions (core, mantle, and crust) of differing composition. Substances separated according to their densities, with the more dense material in the core and less dense in the crust. b) The four most abundant elements are oxygen, silicon, aluminum, and iron in order of decreasing abundance. c) Oxygen is the most abundant element in the crust and mantle but is not found in the core. Silicon, aluminum, calcium, sodium, potassium, and magnesium are also present in the crust and mantle but not in the core.

22.4

The left half of the transition metals are usually found as oxides, while the right half of the transition metals and most of the p-block metals are found as sulfides. The more electronegative metals tend to form sulfides, the less electronegative ones, oxides.

22.5

a) Bauxite (impure Al(OH)3) e) Seawater (NaCl)

22.6

Aluminosilicates are thermodynamically very stable. Great amounts of energy are needed for the production of a metal from a silicate, making the process economically unfavorable.

22.7

Plants produced O2, slowly increasing the oxygen concentration in the atmosphere and creating an oxidative environment for metals. Evidence of Fe(II) deposits pre-dating Fe(III) deposits suggests this hypothesis is true. The decay of plant material and its incorporation into the crust increased the concentration of carbon in the crust and created large fossil-fuel deposits.

22.8

Carbon atoms form a large number of bonds, which allows for the formation of a wide variety of complex molecules necessary for life. Since C atoms are small, they form strong bonds with one another, giving organic molecules stability, particularly with respect to reaction with O2 and H2O.

22.9

Fixation refers to the process of converting a substance in the atmosphere into a form more readily usable by organisms. Examples are the fixation of carbon by plants in the form of carbon dioxide and of nitrogen by bacteria in the form of nitrogen gas. Fixation of carbon dioxide gas by plants converts the CO2 into carbohydrates during photosynthesis. Fixation of nitrogen gas by nitrogen-fixing bacteria involves the conversion of N2 to ammonia and ammonium ions.

22.10

If too much CO2 enters the atmosphere from human activities (primarily forest clearing, decomposition of limestone, and burning of carbon containing fuels), the possibility exists for an unnatural warming of Earth due to trapping of heat by the “extra” CO2.

22.11

The labeled category in the figure is “Human activity and combustion.” Two other processes are the production of compounds of Mg and Ca from dolomite, and the production of steel and aluminum.

b) Air (N2)

c) Seawater (NaCl)

d) Limestone (CaCO3)

22-1

22.12

Atmospheric nitrogen is utilized by three fixation pathways: atmospheric, industrial, and biological. Atmospheric fixation requires high-temperature reactions (e.g., lightning) to convert N2 into NO and other oxidized species. Industrial fixation involves mainly the formation of ammonia, NH3, from N2 and H2. Biological fixation occurs in nitrogen-fixing bacteria that live in the roots of legumes. “Human activity” is referred to as “industrial fixation.” Human activity is a significant factor, contributing about 17% of the nitrogen removed.

22.13

Because N–N single bonds are weak, cyclic and chain compounds containing these bonds are not very stable.

22.14

a) No gaseous phosphorus compounds are involved in the phosphorus cycle, so the atmosphere is not included in the phosphorus cycle. b) Two roles of organisms in phosphorus cycle: 1) Plants excrete acid from their roots to convert PO43– ions into more soluble H2PO4– ions, which the plant can absorb. 2) Through excretion and decay after death, organisms return soluble phosphate compounds to the cycle.

22.15

a) Atmospheric fixation requires energy from lightning or fires; industrial fixation requires high temperature to make the process proceed at a reasonable rate. b) ΔHfo for NO and NO2 are both endothermic, so their formation requires energy. In contrast, ΔHfo for NH3 is exothermic. The reaction is slow, however, and requires energy to increase its rate. c) The energy comes from a widespread low-grade energy source: sunlight. d) There would be less N2 in the atmosphere and probably a greater biomass on Earth as a result of a low activation energy for nitrogen fixation.

22.16

a) N2(g) + 2H 2O(l) → 2NO(g) + 4H +(aq) + 4e– b) 3H 2O(l) + N2O(g) → 2NO2(g) + 6H +(aq) + 6e– c) NH 3(aq) + 2H 2O(l) → NO2–(aq) + 7H +(aq) + 6e– d) NO3–(aq) + 2H +(aq) + 2e– → NO2–(aq) + H 2O(l) e) N2(g) + 6H 2O(l) → 2NO3–(aq) + 12H +(aq) + 10e–

22.17

Plan: First determine the moles of F present in 100. kg of fluorapatite, using the molar mass of Ca5(PO4)3F (504.31 g/mol); take 15% of this amount. Convert mol F to mol SiF4, and use the ideal gas law to determine the volume of the SiF4 gas. For part b), convert moles of SiF4 to moles of Na2SiF6 using the mole ratio in the given equation and convert moles of Na2SiF6 to moles and then mass of F. The definition of ppm (“parts per million”) states that 1 ppm F– = (1 g F–/(106 g H2O)). Find the volume of water that can be fluoridated with the calculated mass of F. Solution: a) Moles of F = ⎛103 g Ca 5 (PO 4 )3 F ⎞⎛ ⎞⎟⎛ 15% ⎞ 1 mol F ⎟⎟⎜⎜ 1 mol Ca 5 (PO 4 )3 F ⎞⎛ ⎟⎟⎜⎜ ⎜ ⎟⎟⎜ ⎟⎟ 100. kg Ca 5 (PO 4 )3 F ⎜⎜ ⎟⎟⎜ ⎟⎟⎜ ⎜ ⎜⎝⎜ 1 kg Ca 5 (PO 4 )3 F ⎠⎝ ⎟⎜⎜ 504.31 g Ca 5 (PO 4 )3 F ⎠⎝ ⎟⎜⎜1 mol Ca 5 (PO 4 )3 F ⎠⎟⎟⎝⎜100% ⎠⎟

(

oxidation oxidation oxidation reduction oxidation

)

= 29.74361 mol F ⎛1 mol SiF4 ⎞⎟ ⎜ ⎟⎟ = 7.43590 mol SiF Moles of SiF4 = 29.74361 mol F ⎜⎜ 4 ⎜⎜⎝ 4 mol F ⎠⎟⎟ PV = nRT L • atm ⎞⎟ ⎛ 7.43590 mol SiF4 ⎜⎜0.0821 ⎟((273 +1450.) K ) ⎝ nRT mol • K ⎠ V (L) = = = 1051.86977 = 1.1 × 103 L P 1.00 atm

(

)

(

)

⎛1 mol Na 2 SiF6 ⎞⎟ ⎟ = 3.71795 mol Na2SiF6 b) Moles of Na2SiF6 = (7.43590 mol SiF4 )⎜⎜ ⎜⎝ 2 mol SiF4 ⎠⎟⎟ Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

22-2

⎛ 6 mol F− ⎞⎟⎛19.00 g F− ⎞ ⎟⎟ = 423.8463 = 4.2 × 102 g F– ⎟⎜ Mass (g) of F– = (3.71795 mol Na 2 SiF6 )⎜⎜⎜ ⎜⎝1 mol Na 2 SiF6 ⎠⎟⎟⎜⎝⎜ 1 mol F− ⎠⎟ If 1 g F– will fluoridate 106 g H2O to a level of 1 ppm, how many grams (converted to mL using density, then converted from mL to L to m3) of water can 423.8463 g F– fluoridate to the 1 ppm level? Necessary conversion factors: 1 m3 = 1000 L, density of water = 1.00 g/mL. 3 ⎛10 6 g H 2 O ⎞⎛ mL H 2 O ⎞⎛ 1 cm 3 ⎞⎛ 10−2 m ⎞⎟ ⎟ ⎟ ⎟ ⎜ ⎜ ⎜ 3 2 3 − ⎜ Volume (m ) = (423.8463 g F )⎜⎜ ⎟⎟⎟⎜⎜ ⎟⎟⎟⎜⎜ ⎟⎟⎟⎜⎜ ⎟⎟ = 423.8463 = 4.2 × 10 m − ⎟⎜⎜1.00 g H 2 O ⎠⎝ ⎟⎜⎜ 1 mL ⎠⎝ ⎟⎜⎜ 1 cm ⎠⎟⎟ ⎜⎜⎝ 1 g F ⎠⎝ 22.18

Plan: Since the 50. t contains 2.0% Fe2O3 by mass, 100 – 2.0 = 98% by mass of the 50. t contains Ca3(PO4)2. Find the moles of Ca3(PO4)2 in the sample and convert moles of Ca3(PO4)2 to moles and then mass of P4, remembering that the production of P4 has a 90% yield. Solution: a) The iron ions form an insoluble salt, Fe3(PO4)2, that decreases the yield of phosphorus. This salt is of limited value. b) Use the conversion factor 1 metric ton (t) = 1 × 103 kg. 3 3 ⎛ 98% ⎞⎟⎜⎛⎜10 kg ⎞⎛ ⎟⎟⎜⎜10 g ⎞⎛ ⎟⎟⎜⎜ 1 mol Ca 3 (PO 4 )2 ⎞⎟⎟ Moles of Ca3(PO4)2 = (50. t )⎜⎜⎜ ⎟⎜ ⎟⎜ ⎟ ⎟⎟⎟⎜⎜⎜ 310.18 g Ca (PO ) ⎟⎟⎟ = 157972.79 mol Ca3(PO4)2 ⎟ ⎝100% ⎠⎜⎝⎜ 1 t ⎠⎝ ⎟⎜⎜ 1 kg ⎠⎝ 3 4 2⎠ Mass (metric tons) of P4 = ⎛ ⎞⎛ 1 mol P4 123.88 g P4 ⎞⎛ 1 kg ⎞⎛ 1 t ⎞⎟⎛ 90.% ⎞ ⎜ ⎟⎟⎟⎜⎜ ⎟⎟⎟⎜⎜ ⎟⎟⎜⎜ ⎟⎟⎜ ⎟⎟ ⎜ ⎜ ⎜ ⎜ 3 ⎟ 3 ⎟ ⎟ ⎟ ⎟⎜⎜ 1 mol P4 ⎠⎝ ⎟⎜⎜10 g ⎠⎝ ⎟⎜⎜10 kg ⎠⎟⎟⎜⎝100% ⎠⎟ ⎜⎜⎝ 2 mol Ca 3 (PO 4 )2 ⎠⎝

(157972.79 mol Ca 3 (PO 4 )2 )⎜⎜

= 8.80635 = 8.8 t P4 22.19

a) An ore is a naturally occurring mixture of gangue and mineral from which an element can be profitably extracted. b) A mineral is a naturally occurring, homogeneous, crystalline solid, with a well-defined composition. c) Gangue is the debris, such as sand rock and clay, associated with an ore. d) Brine is a concentrated salt solution used as a source of Na, Cl, etc.

22.20

a) Roasting involves heating the mineral in air (O2) at high temperatures to convert the mineral to the oxide. b) Smelting is the reduction of the metal oxide to the free metal using heat and a reducing agent such as coke. c) Flotation is a separation process in which the ore is removed from the gangue by exploiting the different abilities of the two to interact with detergent. The gangue sinks to the bottom and the lighter ore-detergent mix is skimmed off the top. d) Refining is the final step in the purification process to yield the pure metal with no impurities.

22.21

Factors considered include cost, reducing strength needed, and by-products formed.

22.22

a) In general, the more active metal will be the one with the lower ionization energy and/or the greater enthalpy of hydration. In this way, net energy will be given off when the more active element loses electrons and the less active one gains them, with the formation of ions of the more active element and removal from solution of the ions of the less active element. An example is: Zn(s) + Cu2+(aq) → Cu(s) + Zn2+(aq) b) Th is is similar to a) except th at h ydration enthalpies are not involved. An example is: Δ Ca(l) + 2RbCl(l) ⎯⎯→ CaCl2(l) + 2Rb(g)

22.23

In general, nonmetals are obtained by oxidation and metals are obtained by reduction.

22.24

a) 6

b) 4

c) 2

d) 1

22-3

22.25

a) Slag is a waste product of iron metallurgy formed by the reaction: CaO(s) + SiO2(s) → CaSiO3(l) In other words, slag is a by-product of steelmaking and contains the impurity SiO2. b) Pig iron, used to make cast iron products, is the impure product of iron metallurgy (containing 3–4% C) that is purified to steel. c) Steel refers to the products of iron metallurgy, specifically alloys of iron containing small amounts of other elements including 1–1.5% carbon. d) Basic-oxygen process refers to the process used to purify pig iron to form steel. The pig iron is melted and oxygen gas under high pressure is passed through the liquid metal. The oxygen oxidizes impurities to their oxides, which then react with calcium oxide to form a liquid that is decanted. Molten steel is left after the basic-oxygen process.

22.26

Pyrometallurgy uses heat to obtain the metal, electrometallurgy uses electrical energy, and hydrometallurgy uses reactions in aqueous solution. a) Fe is produced by a high-temperature process using C as the reductant. b) Na is produced (with Cl2 and H2) electrochemically in a Downs cell. c) Au is produced in a hydrometallurgical process using CN– as a complexing agent. d) Al is produced electrochemically in the Hall-Heroult process after a hydrometallurgical step to dissolve the ore.

22.27

Iron and nickel are more easily oxidized than copper, so they are separated from the copper in the roasting step and conversion to slag. In the electrorefining process, all three metals are oxidized into solution, but only Cu2+ ions are reduced at the cathode to form Cu(s).

22.28

Molten cryolite is a good solvent for Al2O3.

22.29

a) Molecules containing different isotopes of a given atom react at different rates (heavier atoms react more slowly) if a bond is broken to the atom in question. b) H compounds exhibit a relatively large effect, since the mass ratio of the isotopes is larger for H than for any other atom. c) It would be smaller for C, since the mass ratio of the isotopes is smaller.

22.30

The potassium in the molten potassium salt is reduced by molten sodium. Ordinarily, sodium would not be a good reducing agent for the more active potassium. But the reduction is carried out above the boiling point of potassium, producing gaseous potassium: Na(l) + K+(l) ⇆ Na+(l) + K(g) The potassium gas is removed as it is produced. Le Châtelier’s principle states that the system shifts toward formation of more K as the gaseous metal leaves the cell.

22.31

a) Aqueous salt solutions are mixtures of ions and water. When two half-reactions are possible at an electrode, the one with the more positive electrode potential occurs. In this case, the two half-reactions are: M+ + e– → M0 E°red = –3.05 V, –2.93 V, and –2.71 V for Li+, K+, and Na+, respectively – – 2H2O + 2e → H2 + 2OH Ered = –0.42 V, with an overvoltage of about –1 V In all of these cases, it is energetically more favorable to reduce H2O to H2 than to reduce M+ to M. b) The question asks if Ca could chemically reduce RbX, i.e., convert Rb+ to Rb0. In order for this to occur, Ca0 loses electrons (Ca0 → Ca2+ + 2e–) and each Rb+ gains an electron (2Rb+ + 2e– → 2Rb0). The reaction is written as follows: 2RbX + Ca → CaX2 + 2Rb where ΔH = IE1(Ca) + IE2(Ca) – 2IE1(Rb) = 590 + 1145 – 2(403) = +929 kJ/mol. Recall that Ca0 acts as a reducing agent for the Rb+ ion because it oxidizes. The energy required to remove an electron is the ionization energy. It requires more energy to ionize calcium’s electrons, so it seems unlikely that

22-4

Ca0 could reduce Rb+. Based on values of IE and a positive ΔH for the forward reaction, it seems more reasonable that Rb0 would reduce Ca2+. c) If the reaction is carried out at a temperature greater than 688°C (the boiling point of rubidium), the product mixture will contain gaseous Rb. This can be removed from the reaction vessel, causing a shift in equilibrium to form more Rb product. If the reaction is carried out between 688°C and 1484°C (bp for Ca), then Ca remains in the molten phase and remains separated from gaseous Rb. d) The reaction of calcium with molten CsX is written as follows: 2CsX + Ca → CaX2 + 2Cs where ΔH = IE1(Ca) + IE2(Ca) – 2IE1(Cs) = 590 + 1145 – 2(376) = +983 kJ/mol. This reaction is more unfavorable than for Rb, but Cs has a lower boiling point of 671°C. If the reaction is carried out between 671°C and 1484°C, then calcium can be used to separate gaseous Cs from molten CsX. 22.32

Plan: Write the balanced equation for the process. For every two moles of Na metal produced, one mole of Cl2 is produced. Calculate the amount of chlorine gas from the reaction stoichiometry, then use the ideal gas law to find the volume of chlorine gas. For part b), write the balanced equation for the oxidation of Cl– to Cl2; convert moles of Cl2 produced to moles of electrons required and then convert to coulombs with the conversion factor 96,485 C = 1 mole of e–. Convert coulombs to time with the conversion factor 1 C = 1 A∙s. Solution: a) 2NaCl(l) → 2Na(l) + Cl2(g) ⎛103 g ⎞⎛ ⎟⎟⎜⎜ 1 mol Na ⎞⎛ ⎟⎟⎜⎜1 mol Cl 2 ⎞⎟⎟ ⎜ Moles of Cl2 = (31.0 kg Na )⎜⎜ ⎟⎟⎜ ⎟⎟⎜ ⎟ = 674.2062 mol Cl2 ⎜⎝⎜ 1 kg ⎠⎝ ⎟⎜⎜ 22.99 g Na ⎠⎝ ⎟⎜⎜ 2 mol Na ⎠⎟⎟ PV = nRT L • atm ⎞⎟ ⎛ 674.2062 mol Cl 2 ⎜⎜0.0821 ⎟((273 + 540.) K ) ⎝ nRT mol • K ⎠ = = 4.50014 × 104 = 4.5 × 104 L V (L) = P 1.0 atm

(

)

(

)

b) Two moles of electrons are passed through the cell for each mole of Cl2 produced: 2Cl–(l) → Cl2(g) + 2e– ⎛ 2 mol e− ⎞⎟⎛ 96, 485 C ⎞ ⎟⎟ = 1.30101570 × 108 = 1.30 × 108 C ⎟⎜ Coulombs = (674.2062 mol Cl 2 )⎜⎜ ⎜⎝1 mol Cl 2 ⎠⎟⎟⎝⎜ 1 mol e− ⎠⎟

⎛ A • s ⎞⎛ 1 ⎞⎟ 6 6 c) Time (s) = (1.30101570 × 108 C)⎜⎜ ⎟⎟⎟⎜⎜ ⎟ = 1.689631 × 10 = 1.69 × 10 s ⎝⎜ 1 C ⎠⎝⎜ 77.0 A ⎠⎟ 22.33

22.34

22.35

a) C (coke) b) Fuel (increasing temperature) c) CO d) 3Fe2O3(s) + CO(g) → 2Fe3O4(s) + CO2(g) Fe3O4(s) + CO(g) → 3FeO(s) + CO2(g) FeO(s) + CO(g) → Fe(l) + CO2(g)

Δ a) CaCO3(s) ⎯⎯→ CaO(s) + CO2(g) b) The lime (CaO) reacts with SiO2 present in the iron ore to produce CaSiO3 slag. The word “flux” comes from the Latin word meaning “to flow.” The liquid slag flows down to the bottom of the furnace and floats on the more dense iron. Δ c) CaO(s) + SiO2(s) ⎯⎯→ CaSiO3(l) a) Mg2+ is more difficult to reduce than H2O, so H2(g) would be produced instead of Mg metal. Cl2(g) forms at the anode due to overvoltage. b) The ΔHfo for MgCl2(s) is –641.6 kJ/mol: Mg(s) + Cl2(g) → MgCl2(s) + heat.

22-5

High temperature favors the reverse reaction (which is endothermic), so high temperatures favor the formation of magnesium metal and chlorine gas. 22.36

a) Yes, iodine is near the bottom of the periodic table, thus it has a low electronegativity which allows it to adopt positive oxidation states. b) 2IO3–(aq) + 5HSO3–(aq) → 3HSO4–(aq) + 2SO42–(aq) + H2O(l) + I2(s) Oxidizing agent: IO3– Reducing agent: HSO3– c) 0.78 mol% NaIO3 leaves (100.00 – 0.78) mol% NaNO3 = 99.22 mol% NaNO3 Mass (g) of I2 = 3 ⎛ 1 kg ⎞⎛ 1 mol NaNO ⎞⎛ 0.78 mol% NaIO ⎞⎛ 1 mol I ⎞⎛ 253.8 g I ⎞ ⎟⎟⎜⎜10 g ⎞⎛ ⎜⎜ 2000. lb ( )⎜⎜⎜ 2.205 lb ⎟⎟⎟⎜⎜⎜ 1 kg ⎟⎟⎟⎟⎟⎜⎜⎜⎜⎜ 85.00 g NaNO3 ⎟⎟⎟⎟⎟⎜⎜⎜⎜⎜(99.22) mol% NaNO3 ⎟⎟⎟⎟⎟⎜⎜⎜⎜⎜ 2 mol IO2 − ⎟⎟⎟⎟⎟⎜⎜⎜⎜⎜ 1 mol I 2 ⎟⎟⎟⎟⎟ 3 ⎠⎝ 3 ⎠⎝ 3 ⎠⎝ 2 ⎠ ⎝ ⎠⎝ ⎠⎝ = 1.0645339 × 104 = 1.1 × 104 g I2

22.37

a) Sulfur dioxide is the reducing agent and is oxidized to the +6 state (as sulfate ion, SO42–). b) The sulfate ion formed reacts as a base in the presence of acid to form the hydrogen sulfate ion. SO42–(aq) + H+(aq) → HSO4–(aq) c) Skeleton redox equation: SO2(g) + H2SeO3(aq) → Se(s) + HSO4–(aq) Reduction half-reaction: H2SeO3(aq) → Se(s) balance O and H H2SeO3(aq) + 4H+(aq) → Se(s) + 3H2O(l) balance charge H2SeO3(aq) + 4H+(aq) + 4e– → Se(s) + 3H2O(l) Oxidation half-reaction: SO2(g) → HSO4–(aq) balance O and H SO2(g) + 2H2O(l) → HSO4–(aq) + 3H+(aq) balance charge SO2(g) + 2H2O(l) → HSO4–(aq) + 3H+(aq) + 2e– Multiply oxidation half-reaction by 2 2SO2(g) + 4H2O(l) → 2HSO4–(aq) + 6H+(aq) + 4e– Add the two half-reactions: H2SeO3(aq) + 4H+(aq) + 4e– → Se(s) + 3H2O(l) 2SO2(g) + 4H2O(l) → 2HSO4–(aq) + 6H+(aq) + 4e– H2SeO3(aq) + 2SO2(g) + H2O(l) → Se(s) + 2HSO4–(aq) + 2H+(aq)

22.38

The only halogen capable of oxidizing Cl– is F2, which is expensive and dangerous to handle. F2 also reacts with water, wasting the reagent.

22.39

3K2SiF6(s) + 4Al(s) → 6KF(s) + 3Si(s) + 4AlF3(s)

22.40

Mass % Fe =

2 (55.85 g/mol) (100%) = 69.94364 = 69.94% Fe 159.70 g/mol

Mass % Fe =

3(55.85 g/mol) (100%) = 72.36018 = 72.36% Fe 231.55 g/mol

mass % Fe =

1(55.85 g/mol) (100%) = 46.545545 = 46.55% Fe 119.99 g/mol

22.41

4P(s) + 5O2(g) → P4O10(s) P4O10(s) + 6CaO(s) → 2Ca3(PO4)2(s)

22.42

a) The oxidation state of copper in Cu2S is +1 (sulfur is –2). The oxidation state of copper in Cu2O is +1 (oxygen is –2). The oxidation state of copper in Cu is 0 (oxidation state is always 0 in the elemental form). b) The reducing agent is the species that is oxidized. Sulfur changes oxidation state from –2 in Cu2S to +4 in SO2. Therefore, Cu2S is the reducing agent, leaving Cu2O as the oxidizing agent.

22-6

22.43

Al2O3 is amphoteric and will dissolve in base: Al2O3(s) + 2NaOH (aq) + 3H 2O(l) → 2NaAl(OH)4(aq) Fe2O3 and TiO2 h ave no acidic properties, so do not dissolve and can be removed.

22.44

a) Use the surface area and thickness of the film to calculate its volume. Multiply the volume by the density to find the mass of the aluminum oxide. Convert mass to moles of aluminum oxide. The oxidation of aluminum involves the loss of three electrons for each aluminum atom or six electrons for each Al2O3 compound formed. From this, find the number of electrons produced and multiply by Faraday’s constant to find coulombs. 3 ⎛ 1 cm ⎞⎟ ⎛ 3.97 g Al O ⎞⎛ 1 mol Al O ⎞⎟ ⎜⎜ 2 3 ⎟⎜ 2 3 ⎟ 2 −6 ⎟⎟⎜⎜ Moles of Al2O3 = (2.5 m )(23 × 10 m )⎜ −2 ⎟⎟⎟ ⎜⎜⎜ 3 ⎟⎟⎜⎜101.96 g Al O ⎟⎟⎟ ⎜⎜10 m ⎟ ⎜⎝ cm ⎠⎝ 2 3⎠ ⎝ ⎠ = 2.2388682 mol Al2O3 ⎛ 6 mol e− ⎞⎟⎛ 96, 485 C ⎞ ⎟ = 1.296103 × 106 = 1.3 × 106 C ⎟⎟⎜ Coulombs = (2.2388682 mol Al 2 O3 )⎜⎜⎜ − ⎟ ⎝⎜1 mol Al 2 O3 ⎠⎟⎝⎜ 1 mol e ⎠⎟ b) Current in amps can be found by dividing charge in coulombs by the time in seconds. [(1.296103 × 106 C)/[15 min (60 s/min)] × (1 A/(C/s)) = 1200 = 1.2 × 103 A

22.45

a) Work = ΔG° = –nFE° Work = – (2 mol e–)(96,485 C/mol e–)(1.24 V)[(J/C)/V] = –2.392828 × 105 = –2.39 × 105 J (The 2 in 2 mol e– is an exact number, and has no bearing on the significant figures.) b) Efficiency = [(useful work done)/(energy input)] × 100% Efficiency = [(2.392828 × 105 J)/(400 kJ(103 J/1 kJ) )] × 100% = 59.8207 = 59.8% ⎛ ⎞⎛ ⎞⎛ ⎞ ⎜ 400 kJ ⎟⎟⎜⎜1kW • s ⎟⎟⎜⎜ 1 h ⎟⎟⎛⎜ $0.06 ⎞⎟ c) Cost = (500. mol H 2 )⎜⎜ ⎟⎟⎜ ⎟⎟⎜ ⎟⎜ ⎟ = 3.333333 = $3.33 ⎜⎝⎜1 mol H 2 ⎠⎝ ⎟⎜⎜ 1 kJ ⎠⎝ ⎟⎜⎜ 3600 s ⎠⎟⎟⎝⎜ kW • h ⎠⎟

22.46

a) H 2 and CO2 b) The zeolite acts as a molecular filter to remove molecules larger than H 2.

22.47

Plan: Free energy, ΔG ο , is the measure of the ability of a reaction to proceed spontaneously. ΔG° can be calculated with the relationship ∑m ΔGfο(products) – ∑n ΔGfο(reactants) . Solution: The direct reduction of ZnS follows the reaction: 2ZnS(s) + C(graphite) → 2Zn(s) + CS2(g) ο ΔGrxn = [(2 mol Zn) (ΔGfo of Zn) + (1 mol CS2) (ΔGfο of CS2)] – [(2 mol ZnS) (ΔGfο of ZnS) + (1 mol C) (ΔGfο of C)] ο ΔGrxn = [(2 mol)(0 kJ/mol) + (1 mol)(66.9 kJ/mol)] – [(2 mol)(–198 kJ/mol) + (1 mol)(0 kJ/mol)] ο ΔGrxn = 462.9 = +463 kJ o Since ΔGrxn is positive, this reaction is not spontaneous at standard-state conditions. The stepwise reduction involves the conversion of ZnS to ZnO, followed by the final reduction step: 2ZnO(s) + C(s) → 2Zn(s) + CO2(g) o ΔGrxn = [(2 mol Zn) (ΔGfo of Zn) + (1 mol CO2) (ΔGfo of CO2)] – [(2 mol ZnO) (ΔGfo of ZnO) + (1 mol C) (ΔGfo of C)] o ΔGrxn = [(2 mol)(0 kJ/mol) + (1 mol)(–394.4 kJ/mol)] – [(2 mol)(–318.2 kJ/mol) + (1 mol)(0 kJ/mol)] o ΔGrxn = +242.0 kJ This reaction is also not spontaneous, but the oxide reaction is less unfavorable.

22.48

The formation of sulfur trioxide is very slow at ordinary temperatures. Increasing the temperature can speed up the reaction, but the reaction is exothermic, so increasing the temperature decreases the yield. Recall that yield, or the extent to which a reaction proceeds, is controlled by the thermodynamics of the reaction. Adding a catalyst

22-7

increases the rate of the formation reaction but does not impact the thermodynamics, so a lower temperature can be used to enhance the yield. Catalysts are used in such a reaction to allow control of both rate and yield of the reaction. 22.49

a) SO3(g) + H2SO4(l) → H2S2O7(l) H2S2O7(l) + H2O(l) → 2H2SO4(l) b) At high temperature, H2O(g) catalyzes the polymerization of SO3 to (SO3)x, which forms a smoke which makes poor contact with the H2O.

22.50

All the steps in its production are exothermic, and the excess heat (in the form of steam) can be sold profitably.

22.51

a) The chlor-alkali process yields Cl2, H2, and NaOH. b) The mercury-cell process yields higher purity NaOH, but produces Hg-polluted waters that are discharged into the environment.

22.52

Plan: ΔG° at 25°C can be calculated with the relationship ∑m ΔGfo(products) – ∑n ΔGfo(reactants) . Calculate ΔG° o o o o o at 500.°C with the relationship ΔGrxn = ΔH rxn – T ΔSrxn . ΔH rxn and ΔSrxn must be calculated first. The

equilibrium constant K is calculated by using ΔG° = –RT ln K. Finally, to find the temperature at which the o o o = 0 = ΔH rxn – T ΔSrxn and solve for temperature. reaction becomes spontaneous, use ΔGrxn

Solution: a) The balanced equation for the oxidation of SO2 to SO3 is 2SO2(g) + O2(g) → 2SO3(g). o ΔGrxn = ∑m ΔGfo(products) – ∑n ΔGfo(reactants)

o ΔGrxn = [(2 mol SO3)( ΔGfo of SO3)] – [(2 mol SO2)( ΔGfo of SO2) + (1 mol O2)( ΔGfo of O2)] o ΔGrxn = [(2 mol)(–371 kJ/mol)] – [(2 mol)(–300.2 kJ/mol) + (1 mol O2)(0 kJ/mol)] o ΔGrxn = –141.6 = –142 kJ o Since ΔGrxn is negative, the reaction is spontaneous at 25°C. b) The rate of the reaction is very slow at 25°C, so the reaction does not produce significant amounts of SO3 at room temperature. o c) ΔH rxn = ∑m ΔHfo(products) – ∑n ΔH fo(reactants) o ΔH rxn = [(2 mol SO3)( ΔHfo of SO3)] – [(2 mol SO2)( ΔHfo of SO2) + (1 mol O2)( ΔHfo of O2)] o ΔH rxn = [(2 mol)(–396 kJ/mol)] – [(2 mol)(–296.8 kJ/mol) + (1 mol)(0 kJ/mol)] o ΔH rxn = –198.4 = –198 kJ o o o ΔSrxn = ∑m Sproducts – ∑n Sreactants o ΔSrxn = [(2 mol SO3)( S o of SO3)] – [(2 mol SO2)( S o of SO2) + (1 mol O2)( S o of O2)] o ΔSrxn = [(2 mol)(256.66 J/mol∙K)] – [(2 mol)(248.1 J/mol∙K) + (1 mol)(205.0 J/mol∙K)] o ΔSrxn = –187.88 = –187.9 J/K o o o ΔG500 = ΔH rxn – T ΔSrxn = –198.4 kJ – ((273 + 500.) K)(–187.88 J/K)(1 kJ/103 J) = –53.16876 = –53 kJ The reaction is spontaneous at 500°C since free energy is negative. d) The equilibrium constant at 500°C is smaller than the equilibrium constant at 25°C because the free energy at 500°C indicates the reaction does not go as far to completion as it does at 25°C. Equilibrium constants can be calculated from ΔG° = –RT ln K.

22-8

⎛103 J ⎞⎟ ⎜⎜ 142 kJ − ( ) ⎟ ⎜⎝ 1 kJ ⎟⎠ ΔG o =– = 57.31417694 At 25°C: ln K = (8.314 J/mol • K) ((273 + 25)K) −RT K25 = e57.31417694 = 7.784501 × 1024 = 7.8 × 1024 ⎛103 J ⎞⎟ ⎜⎜ 53 kJ − ( ) ⎟ ⎜⎝ 1 kJ ⎟⎠ ΔG o =– = 8.246817 At 500°C: ln K = (8.314 J/mol • K) ((273 + 500)K) −RT K500 = e8.246817 = 3.815 × 103 = 3.8 × 103 e) Temperature below which the reaction is spontaneous at standard state can be calculated by setting ΔG° equal to zero and using the enthalpy and entropy values at 25°C to calculate the temperature. o o o ΔGrxn = 0 = ΔH rxn – T ΔSrxn o o ΔH rxn = T ΔSrxn

ΔH o –198 kJ = = 1.05375 × 103 = 1.05 × 103 K o ⎛ 1 kJ ⎞⎟ ΔS –187.9 J/K ⎜⎜ 3 ⎟⎟ ⎝10 J ⎠

22.53

Plan: This problem deals with the stoichiometry of electrolysis. The balanced oxidation half-reaction for the chlor-alkali process is given in the chapter: 2Cl–(aq) → Cl2(g) + 2e– Use the Faraday constant, F, (1 F = 9.6485 × 104 C/mol e–) and the fact that one mol of Cl2 produces two mol e– or 2 F, to convert coulombs to moles of Cl2. Solution: ⎛ ⎞⎟⎛ ⎛ 1 mol e− ⎞⎛ ⎜ C ⎟⎜ 3600 s ⎞⎟ ⎟⎟⎜⎜1 mol Cl 2 ⎞⎟⎟⎛⎜ 70.90 g Cl 2 ⎞⎟⎟⎛⎜⎜ 1 kg ⎞⎟⎟⎛⎜ 2.205 lb ⎞⎟⎟ ⎟⎟(8 h)⎜⎜⎜ Mass (lb) of Cl2 = (3 × 10 4 A)⎜⎜⎜ s ⎟⎟⎟⎜⎜ ⎟ ⎟⎜ ⎟⎜ ⎟ ⎟ ⎟ ⎜⎜ 96, 485 C ⎟⎟⎜⎜⎜ 2 mol e− ⎟⎟⎜⎜⎜ 1 mol Cl 2 ⎟⎟⎜⎜⎜103 g ⎟⎟⎜⎜⎜ 1 kg ⎟⎟ ⎠⎝ ⎠ ⎜⎜⎝ A ⎠⎟⎟⎜⎜⎝ 1 h ⎠⎟ ⎝ ⎠⎝ ⎠⎝ ⎠⎝ = 699.9689 = 7 × 102 lb Cl2

22.54

a) The products are kept separate because Cl2 reacts with NaOH(aq) to produce NaOCl(aq) and NaCl(aq). b) The temperature at which the reaction is run determines the product formed. c) Cl2(g) + 2OH –(aq) → Cl–(aq) + OCl–(aq) + H 2O(l) 3Cl2(g) + 6OH –(aq) → 5Cl–(aq) + ClO3–(aq) + 3H 2O(l) Cl2/ 2OH – is the mole ratio in both cases.

22.55

2SO2(g) + O2(g) → 2SO3(g)

(P ) a) K = 3.18 = (P ) (P ) 2

SO3

PSO2 = PSO3 (1:1 mole ratio)

SO2

3.18 =

(PO2 )

PO2 = 0.3144654 = 0.314 atm O2

2 (P ) (95) = b) K = 3.18 = 2 (P ) (P ) (5) (PO ) 2

SO3 2

SO2

PO2 = 113.522 = 1 × 10 atm O2

22-9

22.56

Plan: Write the balanced equation. Use the stoichiometry in the equation to find the moles of H3PO4 produced by the reaction of the given amount of P4O10. Divide moles of H3PO4 by the volume to obtain its molarity. Since H3PO4 is a weak acid, use its equilibrium expression to find the [H3O+] and then pH. Solution: a) Because P4O10 is a drying agent, water is incorporated in the formation of phosphoric acid, H3PO4. P4O10(s) + 6H2O(l) → 4H3PO4(l) ⎛ 1 mol P4 O10 ⎞⎛ ⎟⎟⎜⎜ 4 mol H 3 PO 4 ⎟⎟⎞ ⎜ b) Moles of H3PO4 = (8.5 g P4 O10 )⎜⎜ ⎟⎟⎜ ⎟ = 0.11976892 mol H3PO4 ⎜⎝⎜ 283.88 g P4 O10 ⎠⎝ ⎟⎜⎜ 1 mol P4 O10 ⎟⎟⎠ ⎛ 0.11976892 mol H 3 PO 4 ⎞⎟ ⎜ ⎟⎟ = 0.1596919 M H PO Molarity of H3PO4 = ⎜⎜ 3 4 ⎟ ⎜⎝⎜ 0.750 L ⎠⎟ Phosphoric acid is a weak acid and only partly dissociates to form H3O+ in water, based on its Ka. Since Ka1 >> Ka2 >> Ka3, assume that the H3O+ contributed by the second and third dissociation is negligible in comparison to the H3O+ contributed by the first dissociation. H3PO4(l) + H2O(l) → H3O+(aq) + H2PO4–(aq) –3

Ka = 7.2 × 10 = Ka = 7.2 × 10–3 = Ka = 7.2 × 10–3 =

[ H3O+ ][ H2 PO4− ] [ H3 PO4 ] [ x ][ x ]

(0.1596919 − x ) [ x ][ x ]

Assume x is small compared to 0.1596919.

(0.1596919)

x = 0.0339084 M Check assumption that x is small compared to 0.1596919: 0.0339084 (100) = 21% error, so the assumption is not valid. 0.1596919 The problem will need to be solved as a quadratic. x2 = (7.2 × 10–3)(0.1596919 – x) = 1.14978 × 10–3 – 7.2 × 10–3x x2 + 7.2 × 10–3x – 1.14978 × 10–3 = 0 c = –1.14978 × 10–3 a = 1 b = 7.2 × 10–3 −7.2 × 10−3 ±

(7.2 × 10 ) − 4 (1)(−1.14978 × 10 ) −3

−3

2 (1) –2

x = 3.049897 × 10 M H3O pH = –log [H3O+] = –log (3.049897 × 10–2) = 1.51571 = 1.52 22.57

a) D+(aq), OD–(aq) and traces of CH 3O–(aq) and H +(aq). Eventually, HDO, H 2O, and CH 3OD would form. b) D2O(l) → D+(aq) + OD–(aq) CH 3OH(aq) + OD–(aq) → CH 3O–(aq) + HDO(l) CH 3O–(aq) + D2O(l) → CH 3OD(l) + OD–(aq) HDO(l) → D+(aq) + OH –(aq) HDO(l) + OH –(aq) → H 2O(l) + OD–(aq)

22.58

Plan: Write a balanced equation for the production of iron and use the reaction stoichiometry to calculate the amount of CO2 produced when the given amount of Fe is produced. Then write a balanced equation for the combustion of gasoline (C8H18.) Find the total volume of gasoline, use the density to convert volume to mass of gasoline, and use the reaction stoichiometry to calculate the amount of CO2 produced.

22-10

Solution: a) The reaction taking place in the blast furnace to produce iron is: Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g) 3 ⎛ 2000 lb ⎞⎛ ⎟⎟⎜⎜ 1 kg ⎞⎛ ⎟⎟⎜⎜10 g ⎞⎛ ⎟⎟⎜⎜ 1 mol Fe ⎞⎛ ⎟⎟⎜⎜ 3 mol CO 2 ⎞⎛ ⎟⎟⎜⎜ 44.01 g CO 2 ⎞⎟⎟ ⎜⎜ Mass (g) of CO2 = (8400. t Fe)⎜ ⎟⎟⎜ ⎟⎟⎜ ⎟⎟⎜ ⎟⎟⎜ ⎟⎟⎜ ⎟ ⎜⎝⎜ 1 t ⎠⎝ ⎟⎜⎜ 2.205 lb ⎠⎝ ⎟⎜⎜ 1 kg ⎠⎝ ⎟⎜⎜ 55.85 g Fe ⎠⎝ ⎟⎜⎜ 2 mol Fe ⎠⎝ ⎟⎜⎜ 1 mol CO 2 ⎠⎟⎟ = 9.005755 × 109 = 9.006 × 109 g CO2 b) Combustion reaction for octane: 2C8H18(l) + 25O2(g) → 16CO2(g) + 18H2O(l) Mass (g) of CO2 = ⎛ 5.0 gal ⎞⎛ 4 qt ⎞⎛ 1 L ⎞⎛ 1 mL ⎞⎟⎛ 0.74 g ⎞⎛ 1 mol C H ⎞⎛ 16 mol CO ⎞⎛ 44.01 g CO ⎞ (1.0 × 106 autos)⎜⎜⎜⎜⎜ 1 auto ⎟⎟⎟⎟⎟⎜⎜⎜⎜⎜1 gal ⎟⎟⎟⎟⎟⎜⎜⎜⎜⎜1.057 qt ⎟⎟⎟⎟⎟⎜⎜⎜⎜⎜10−3 L ⎟⎟⎟⎟⎜⎜⎜⎜ mL ⎟⎟⎟⎟⎟⎜⎜⎜⎜⎜114.22 g C8 H18 ⎟⎟⎟⎟⎟⎜⎜⎜⎜⎜ 2 mol C H 2 ⎟⎟⎟⎟⎟⎜⎜⎜⎜⎜ 1 mol CO 2 ⎟⎟⎟⎟⎟ ⎠⎝ 8 18 ⎠⎝ 8 18 ⎠ ⎝ 2 ⎠ ⎝ ⎠⎝ ⎠⎝ ⎠⎝ ⎠⎝ 10 10 = 4.31604 × 10 = 4.3 × 10 g CO2 Th e CO2 production by automobiles is much greater than that from steel production. 22.59

a) Step 1 Cl2(g) + CH2=CH2(g) → CH2ClCH2Cl(g) Step 2 CH2ClCH2Cl(g) → CH2=CHCl(g) + HCl(g) b) Cl2(g) + CH2=CH2(g) → CH2=CHCl(g) + HCl(g) c) Addition reaction d) Elimination reaction ⎛1 mol CH 2 =CHCl ⎞⎛ ⎟⎟⎜⎜ 62.49 g CH2 =CHCl ⎞⎟⎟ = 28.1205 = 28 g CH =CHCl e) (0.45 mol Cl2 )⎜⎜ 2 ⎟⎟⎜ 1 mol CH =CHCl ⎟⎟ ⎜⎝ 1 mol Cl 2 ⎠⎝ ⎠ 2

22.60

Plan: Step 2 of “Isolation of Magnesium” describes this reaction. Sufficient OH– must be added to precipitate Mg(OH)2. The necessary amount of OH– can be determined from the Ksp of Mg(OH)2. For part b), use the Ksp of Ca(OH)2 to determine the amount of OH– that a saturated solution of Ca(OH)2 provides. Substitute this amount into the Ksp expression for Mg(OH)2 to determine how much Mg2+ remains in solution with this amount of OH–. Calculate the fraction of [Mg2+] remaining and subtract from 1 to obtain the fraction of [Mg2+] that precipitated. Solution: a) Mg(OH)2(s) → Mg2+(aq) + 2OH–(aq) Ksp = 6.3 × 10–10 = [Mg2+][OH –]2 [OH –] =

6.3 × 10−10 = [Mg 2+ ]

6.3 × 10−10 = 1.100699 × 10–4 = 1.1 × 10–4 M 0.052

Thus, if [OH –] > 1.1 × 10–4 M (i.e., if pH > 10.04), Mg(OH)2 will precipitate. b) Ca(OH)2(s) → Ca2+(aq) + 2OH–(aq) Ksp = 6.5 × 10–6 = [Ca2+][OH –]2 = (x)(2x)2 = 4x3 x = 0.011756673; [OH –] = 2x = 0.023513347 M Ksp (Mg(OH)2) = 6.3 × 10–10 = [Mg2+][OH –]2 6.3 × 10−10 [Mg2+] = = 1.1394930 × 10–6 M 2 (0.023513347)

This concentration is the amount of the original 0.052 M Mg2+ that was not precipitated. The percent precipitated is the difference between the remaining [Mg2+] and the initial [Mg2+] divided by the initial concentration. Fraction Mg2+ remaining = (1.1394930 × 10–6 M)/(0.052 M) = 2.19134 × 10–5 Mg2+ precipitated = 1 – 2.19134 × 10–5 = 0.9999781 = 1 (To the limit of the significant figures, all the magnesium has precipitated.)

22-11

22.61

o o o o o Plan: The equation ΔGrxn = ΔH rxn – T ΔSrxn will be used to calculate ΔGrxn at the two temperatures. ΔH rxn o and ΔSrxn will have to be calculated first. Then use ΔG° = –RT ln K to determine K. Subscripts indicate the

temperature, and (1) or (2) indicate the reaction. Solution: For the first reaction (1): o ΔH rxn = ∑m ΔHfo(products) – ∑n ΔH fo(reactants) o ΔH rxn = [(4 mol NO)( ΔHfo of NO) + (6 mol H2O)( ΔHfo of H2O)]

– [(4 mol NH3)( ΔHfo of NH3) + (5 mol O2)( ΔHfo of O2)] o ΔH rxn = [(4 mol)(90.29 kJ/mol) + (6 mol)(–241.826 kJ/mol)] – [(4 mol)(–45.9 kJ/mol) + (5 mol)(0 kJ/mol)] o ΔH rxn = –906.196 kJ o ΔSrxn = [(4 mol NO)( So of NO) + (6 mol H2O)( So of H2O)]

– [(4 mol NH3)( So of NH3) + (5 mol O2)( So of O2)] o ΔSrxn = [(4 mol)(210.65 J/K∙mol) + (6 mol)(188.72 J/K∙mol)]

– [(4 mol)(193 J/K∙mol) + (5 mol O2)(205.0 J/K∙mol)](1 kJ/103 J) o ΔSrxn = 0.17792 kJ/K o o o ΔGrxn = ΔH rxn – T ΔSrxn o ΔG25 = –906.196 kJ – [(273 + 25)K]( 0.17792 kJ/K) = –959.216 kJ o ΔG900 = –906.196 kJ – [(273 + 900.)K]( 0.17792 kJ/K) = –1114.896 kJ

For th e second reaction (2): o ΔH rxn = [(2 mol N2)( ΔHfo of N2) + (6 mol H2O)( ΔHfo of H2O)]

– [(4 mol NH3)( ΔHfo of NH3) + (3 mol O2)( ΔHfo of O2)] o ΔH rxn = [(2 mol)(0 kJ/mol) + (6 mol)(–241.826 kJ/mol)] – [(4 mol)(–45.9 kJ/mol) + (3 mol)(0 kJ/mol)] o ΔH rxn = –1267.356 kJ o ΔSrxn = [(2 mol N2)( So of N2) + (6 mol H2O)( So of H2O)]

– [(4 mol NH3)( So of NH3) + (3 mol O2)( So of O2)] o ΔSrxn = [(2 mol)(191.5 J/K∙mol) + (6 mol)(188.72 J/K∙mol)]

– [(4 mol)(193 J/K∙mol) + (3 mol)(205.0 J/K∙mol)](1 kJ/103 J) o ΔSrxn = 0.12832 J/K o o o ΔGrxn = ΔH rxn – T ΔSrxn o ΔG25 = –1267.356 kJ – [(273 + 25)K](0.12832 kJ/K) = –1305.595 kJ o ΔG900 = –1267.356 kJ – [(273 + 900.)K](0.12832 kJ/K) = –1417.875 kJ ⎛103 J ⎞⎟ (−959.216 kJ/mol ) ΔG o ⎜⎜ ⎟⎟ = 387.15977 =− a) ln K25(1) = ⎜⎜ ⎟ −RT 1 kJ ⎜ 8.314 J /mol • K ((273 + 25) K ) ⎝ ⎠⎟

(

)

168

K25(1) = 1.38 × 10 = 1 × 10168

22-12

ln K25(2) =

(

)

⎛103 J ⎞⎟ −1305.595 kJ/mol ΔG o ⎜ =− ⎟⎟⎟ = 526.9655 ⎜⎜⎜ 1 kJ −RT ⎜ 8.314 J /mol • K ((273 + 25) K ) ⎝ ⎠⎟

(

)

228

K25(2) = 7.2146 × 10 = 7 × 10228

b) ln K900(1) =

⎛103 J ⎞⎟ (−1114.896 kJ/mol) ΔG o ⎜⎜ ⎟⎟ = 114.3210817 =− ⎜ ⎟ −RT 8.314 J /mol • K ((273 + 900) K ) ⎜⎜⎝ 1 kJ ⎠⎟

(

)

K900(1) = 4.4567 × 10 = 4.6 × 1049

ln K900(2) =

(

)

⎛103 J ⎞⎟ −1417.875 kJ/mol ΔG o ⎜⎜ ⎟⎟ = 145.388 =− ⎜⎜ ⎟ 1 kJ −RT ⎜ 8.314 J /mol • K ((273 + 900) K ) ⎝ ⎠⎟

(

)

K900(2) = 1.38422 × 10 = 1.4 × 1063 ⎞⎛ ⎛ ⎞⎛ −3 ⎞⎛ ⎞⎛ ⎞ ⎜175 mg Pt ⎟⎟⎜⎜10 g ⎟⎟⎜⎜ 1 kg ⎟⎟⎜⎜ 32.15 Troy Oz ⎟⎟⎜⎜ $1557 ⎟⎟ c) Cost ($) Pt = (1.01 × 10 7 t HNO 3 )⎜⎜ ⎟⎟⎜ ⎟⎟⎜ 3 ⎟⎟⎜ ⎟⎟⎜ ⎟ 1 kg ⎟⎜⎜ 1 mg ⎠⎟⎜⎝⎜10 g ⎠⎟⎝⎜⎜ ⎟⎜⎜1 troy oz ⎠⎟⎟ ⎝⎜⎜ 1 t HNO 3 ⎠⎝ ⎠⎝ = 8.847672 × 107 = $8.8 × 107 ⎛175 mg Pt ⎞⎟⎛ 72% ⎞⎛10−3 g ⎞⎛ ⎟⎟⎜⎜ 1 kg ⎞⎛ ⎟⎟⎜ 32.15 troy oz ⎞⎛ ⎟⎜⎜ $1557 ⎞⎟⎟ ⎟⎜ ⎟⎜⎜ ⎟ ⎟⎜ d) Cost ($) Pt = (1.01 × 10 7 t HNO3 )⎜⎜⎜ ⎜ ⎟ ⎟ ⎟ ⎟ ⎟ 3 ⎟⎜1 troy oz ⎠⎟⎟ ⎜⎝ 1 t HNO3 ⎠⎟⎜⎝100% ⎠⎜⎝⎜ 1 mg ⎠⎟⎜⎜⎝10 g ⎠⎟⎜⎝ 1 kg ⎠⎝

= 6.3703238 × 107 = $6.4 × 107 22.62

2FeTiO3(s) + 7Cl2(g) + 6C(s) → 2TiCl4(g) + 2FeCl3(l) + 6CO(g) TiCl4(g) + 2Mg(l) → Ti(s) + 2MgCl2(l) ⎛ 3 ⎞⎛ 3 ⎞⎛ ⎞⎛ ⎞⎛ ⎞ ⎜10 kg ⎟⎟⎜⎜10 g ⎟⎟⎜⎜ 1 mol FeTiO 3 ⎟⎟⎜⎜ 1 mol Ti ⎟⎟⎜⎜ 47.88 g Ti ⎟⎟⎜⎛ 84% ⎞⎛ ⎟⎜ 93% ⎞⎟ Mass Ti = (21.5 t FeTiO 3 )⎜⎜ ⎟⎟⎜ ⎟⎟⎜ ⎟⎟⎜ ⎟⎟⎜ ⎟⎜ ⎟⎟⎜⎜100% ⎠⎟⎟ ⎟⎜⎜ 1 kg ⎠⎝ ⎟⎜⎜151.73 g FeTiO 3 ⎠⎝ ⎟⎜⎜1 mol FeTiO 3 ⎠⎝ ⎟⎜⎜ 1 mol Ti ⎠⎟⎟⎜⎝100% ⎠⎝ ⎜⎝⎜ 1 t ⎠⎝

Balanced reactions:

= 5.300092 × 106 = 5.3 × 106 g Ti 22.63

a) Balance the two steps and then add them. (1) 16H2S(g) + 16O2(g) → S8(g) + 8SO2(g) + 16H2O(g) (2) 16H2S(g) + 8SO2(g) → 3S8(g) + 16H2O(g) 32H2S(g) + 16O2(g) → 4S8(g) + 32H2O(g) The coefficients in the overall equation can be reduced since all are divisible by 4. 8H2S(g) + 4O2(g) → S8(g) + 8H2O(g) b) Replace oxygen with chlorine: 8H2S(g) + 8Cl2(g) → S8(g) + 16HCl(g) Calculate ΔG° to determine if this reaction is thermodynamically possible. o ΔGrxn = ∑m ΔGfo(products) – ∑n ΔGfo(reactants) o ΔGrxn = [(1 mol S8)( ΔGfo of S8) + (16 mol HCl)( ΔGfo of HCl)]

– [(8 mol H2S)( ΔGfo of H2S) + (8 mol Cl2)( ΔGfo of Cl2)] o ΔGrxn = [(1 mol)(49.1 kJ/mol) + (16 mol)(–95.30 kJ/mol)] – [(8 mol)(–33 kJ/mol) + (8 mol)(0 kJ/mol)] o ΔGrxn = –1211.7 = –1212 kJ The reaction is spontaneous. c) Oxygen is readily available from the air, so chlorine is a more expensive reactant. In addition, water is less corrosive than HCl as a product.

22.64

Plan: Balance each reaction. Acidity is a measure of the concentration of H3O+ (H+), so any reaction that produces H+ will increase the acidity.

22-13

Solution: a) 2H2O(l) + 2FeS2(s) + 7O2(g) → 2Fe2+(aq) + 4SO42–(aq) + 4H+(aq) b) 4H+(aq) + 4Fe2+(aq) + O2(g) → 4Fe3+(aq) + 2H2O(l) c) Fe3+(aq) + 3H2O(l) → Fe(OH)3(s) + 3H+(aq) d) 8H2O(l) + FeS2(s) + 14Fe3+(aq) → 15Fe2+(aq) + 2SO42–(aq) + 16H+(aq) 22.65

Increases acidity. Increases acidity. Increases acidity.

Plan: The carbon fits into interstitial positions. The cell size will increase slightly to accommodate the carbons atoms. The increase in size is assumed to be negligible. The mass of the carbon added depends upon the carbon in the unit cell. Solution: Ferrite: ⎛ 7.86 g ⎞⎟ ⎛ 7.86 g ⎞⎟⎛ 0.0218% ⎞ 3 ⎜ ⎟ ⎟ ⎜ ⎟⎟⎜⎜ ⎜⎜⎝⎜ cm3 ⎠⎟⎟ + ⎝⎜⎜⎜ cm3 ⎠⎟⎝ ⎜ 100% ⎠⎟⎟ = 7.86171 = 7.86 g/cm Austenite: ⎛ 7.40 g ⎞⎟ ⎛ 7.40 g ⎞⎟⎛ 2.08% ⎞ 3 ⎟ + ⎜⎜ ⎟⎜ ⎜⎜⎜ ⎟⎟ = 7.55392 = 7.55 g/cm ⎜⎝ cm 3 ⎠⎟⎟ ⎝⎜⎜ cm3 ⎠⎟⎟⎝⎜⎜ 100% ⎠⎟

22.66

o ΔGrxn = ∑m ΔGfo(products) – ∑n ΔGfo(reactants)

a) N2(g) + 2O2(g) → 2NO2(g) o ΔGrxn = [(2 mol)(51 kJ/mol)] – [(1 mol)(0 kJ/mol) + (2 mol)(0 kJ/mol)] = 102 kJ (unfavorable)

b) 3NO2(g) + H2O(l) → 2HNO3(aq) + NO(g) o ΔGrxn = [(2 mol)(–110.5 kJ/mol) + (1 mol)(86.60 kJ/mol)] – [(3 mol)(51 kJ/mol) + (1 mol)(–237.192 kJ/mol)] o ΔGrxn = –50.208 = –50. kJ (favorable)

c) 2NO(g) + O2(g) → 2NO2(g) o ΔGrxn = [(2 mol)(51 kJ/mol)] – [(2 mol)(86.60 kJ/mol) + (1 mol)(0 kJ/mol)] = –71.2 = –71 kJ (favorable)

(overall) 3N2(g) + 6O2(g) + 2H2O(l) → 4HNO3(aq) + 2NO(g) o ΔGrxn = [(4 mol)(–110.5 kJ/mol) + (2 mol)(86.60 kJ/mol)]

– [(3 mol)(0 kJ/mol) + (6 mol)(0 kJ/mol) + (2 mol)(–237.192 kJ/mol)] ΔG = 205.584 = 205.6 kJ Overall, the reaction is thermodynamically unfavorable, due to the unfavorability of th e first step. o rxn

22.67

Plan: Decide what is reduced and what is oxidized of the two ions present in molten NaOH. Then, write halfreactions and balance them. Then write the overall equation for the electrolysis of NaOH and also for the reaction of Na with water. Examine the molar ratios involving water and Na in these equations. Solution: a) At the cathode, sodium ions are reduced: Na+(l) + e– → Na(l) At the anode, hydroxide ions are oxidized: 4OH–(l) → O2(g) + 2H2O(g) + 4e– b) The overall cell reaction is 4Na+(l) + 4OH–(l) → 4Na(l) + O2(g) + 2H2O(g) and the reaction between sodium metal and water is 2Na(l) + 2H2O(g) → 2NaOH(l) + H2(g). For each mole of water produced, two moles of sodium are produced, but for each mole of water that reacts only one mole of sodium reacts. So, the maximum amount of sodium that reacts with water is half of the sodium produced. In the balanced cell reaction, four moles of electrons are transferred for four moles of sodium. Since 1/2 of the Na reacts with H 2O, th e maximum efficiency is 1/2 mol Na/mol e–, or 50%.

22-14

22.68

a) 4Au(s) + O2(g) + 2H 2O(l) → 4Au+(aq) + 4OH –(aq) pH = 13.55, pOH = 14.00 – pH = 14.00 – 13.55 = 0.45 [OH–] = 10–0.45 = 0.3548 M E° = E°O2 – E°Au = 0.40 V – 1.68 V = –1.28 V ⎛[ Au+ ]4 [ OH− ]4 ⎞⎟ 0.0592 V 0.0592 ⎜ ⎟⎟ E = E° – log Q = E° – log ⎜⎜ ⎟ ⎜⎜⎝ n n PO2 ⎠⎟

⎛[0.50 ]4 [ 0.3548]4 ⎞⎟ ⎟⎟ = –1.245569 = –1.25 V log ⎜⎜⎜ ⎜⎝ 0.21 4 ⎠⎟ This reaction is not spontaneous (the voltage is negative). b) Qualitatively, the addition of CN– will lower the concentration of free Au+ by forming Au(CN)2–. This will make [Au+] much lower than 1 M, making the second term in the Nernst equation larger in magnitude. If [Au+] is lowered sufficiently, the second term offsets the first and Ecell becomes favorable. E = –1.28 V –

22.69

0.0592 V

a) Nitric oxide destroys ozone by forming nitrogen dioxide and oxygen. NO(g) + O3(g) ⇆ NO2(g) + O2(g) NO2(g) + O2(g) ⇆ NO(g) + O3(g) (reverse) b) The order translates to the exponent in the rate law, and the rate is dependent on the concentration of the reactants. Ratef = kf[NO][O3] Rater = kr[NO2][O2] o o o c) ΔGrxn = ΔH rxn – T ΔSrxn because the reaction does not take place at 298 K. o ΔH rxn = ∑m ΔHfo(products) – ∑n ΔH fo(reactants) o ΔH rxn = [(1 mol NO2)(33.2 kJ/mol) + (1 mol O2)(0 kJ/mol)]

– [(1 mol NO)(90.29 kJ/mol) + (1 mol O3)(143 kJ/mol)] ΔH

o rxn

= –200.09 kJ

o ΔSrxn = [(1 mol NO2)(239.9 J/mol∙K) + (1 mol O2)(205.0 J/mol∙K)]

– [(1 mol NO)(210.65 J/mol∙K) + (1 mol O3)(238.82 J/mol∙K)] o ΔSrxn = –4.57 J/K o o o ΔGrxn = ΔH rxn – T ΔSrxn = (–200.09 kJ) – (280. K)(–4.57 J/K)(1 kJ/103 J) = –198.810 = –199 kJ d) Assume that the reaction reaches equilibrium at 280. K. At equilibrium, the forward rate equals the reverse rate and their ratio equals the equilibrium constant: kf/kr = Keq. Both ΔG and Keq express the extent to which a reaction proceeds and are related by the equation ΔG = –RT ln Keq. ⎛ ⎞⎟⎛103 J ⎞ −198.810 kJ/mol ⎜ ⎟⎟ ΔG o ⎟⎜ = ⎜⎜⎜ ln K = ⎟⎟ = 85.402591 ⎟⎟⎜ ⎜⎜ ⎟ −RT ⎟⎝ 1 kJ ⎠⎟ ⎜⎝⎜ − 8.314 J/mol • K 280.K ⎠⎟⎜

(

)(

)

K = 1.22991 × 10 = 1.2 × 10 Therefore, the ratio of rate constants, kf/kr, is 1.2 × 1037 and the forward rate is much faster than the reverse rate. 22.70

Plan: Convert mass of CO2 to moles of O2 using the reaction stoichiometry in the balanced equation. PV = nRT is used to convert moles of O2 to volume. Temperature must be in units of Kelvin. Solution: a) The reaction for the fixation of carbon dioxide includes CO2 and H2O as reactants and (CH2O)n and O2 as products. The balanced equation is: nCO2(g) + nH2O(l) → (CH2O)n(s) + nO2(g)

22-15

b) The moles of carbon dioxide fixed equal the moles of O2 produced. ⎛ 48 g CO 2 ⎞⎟⎛⎜ 1 mol CO 2 ⎞⎛ ⎟⎟⎜⎜ 1 mol O 2 ⎞⎟⎟ ⎟⎜⎜ Mole O2/day = ⎜⎜⎜ ⎟⎟⎜ ⎟ = 1.090661 mol O2/day ⎟ ⎟⎜⎝ 44.01 g CO 2 ⎠⎝ ⎜⎝ day ⎠⎜ ⎟⎜⎜1 mol CO 2 ⎠⎟⎟ To find the volume of oxygen, use the ideal gas equation. Temperature must be converted to K. 5 5 T (in K) = ⎡⎢T (°F)− 32⎤⎥ + 273.15 = ⎡⎢T (78.°)−32⎤⎥ + 273.15 = 298.7056 K ⎦ ⎣ ⎦ 9 9⎣ PV = nRT V =

(

)(

)

1.090661 mol O2 0.0821 L • atm/mol • K 298.7056 K nRT = = 26.74708 = 27 L P 1.0 atm

(

)

c) The moles of air containing 1.090661 mol of CO2 is: ⎛ 48 g CO 2 ⎞⎟⎜⎛ 1 mol CO 2 ⎞⎛ ⎟⎜ 100 mol % air ⎞⎟⎟ 3 ⎟⎟⎜⎜ Mole CO2/day = ⎜⎜⎜ ⎟⎟⎟⎜⎜ ⎟ = 2.7266530 × 10 mol air/day ⎟⎜⎝ 44.01 g CO 2 ⎠⎝ ⎟⎜⎜ 0.040 mol % CO 2 ⎠⎟⎟ ⎝⎜ day ⎠⎜

(

)(

)

2.7266530 ×103 mol air 0.0821 L • atm /mol • K 298.7056 K nRT = = 6.6867 × 104 = 6.7 × 104 L V = P 1.0 atm

(

)

22.71

This problem requires a series of conversion steps: Concentration = − −3 3 3 ⎛ 210. kg (NH 4 )2 SO 4 ⎞⎛ ⎞⎟⎛ 37% ⎞⎛ 62.01 g NO3− ⎞⎛ ⎟⎟⎜⎜10 g ⎞⎛ ⎟⎟⎜⎜ 1 mol (NH 4 )2 SO 4 ⎞⎛ ⎟⎟⎜⎜ 2 mol NO3 ⎟⎟⎜⎜ 1 mg ⎞⎛ ⎟⎟⎜⎜10 m ⎞⎟⎟ ⎜⎜ ⎜⎜ ⎟⎟⎜ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎜⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎟⎟⎜⎜ 1 kg ⎟⎟⎜⎜132.15 g (NH ) SO ⎟⎟⎜⎜1 mol (NH ) SO ⎟⎟⎜⎜⎝100% ⎠⎟⎜⎜ 1 mol NO − ⎟⎟⎜⎜10−3 g ⎟⎟⎜⎜ 1 L ⎟⎟⎟ 1000. m 3 ⎜⎝ 4 2 4 ⎠⎝ 4 2 4⎠ 3 ⎠⎝ ⎠⎝ ⎝ ⎠⎝ ⎠⎝ ⎠ = 72.9198 = 73 mg/L This assumes the plants in the field absorb none of the fertilizer.

22.72

Plan: Write a balanced equation. Determine which one of the three reactants is the limiting reactant by determining the amount of cryolite produced by each reactant. The amount of Al(OH)3 requires conversion of kg to g; the amount of NaOH requires conversion from amount of solution to mass in grams by using the density; the amount of HF requires conversion from volume to moles using the ideal gas law. Solution: The balanced chemical equation for this reaction is: 6HF(g) + Al(OH)3(s) + 3NaOH(aq) → Na3AlF6(aq) + 6H2O(l) Assuming Al(OH)3 is the limiting reactant: ⎛10 3 g ⎞⎛ ⎟⎟⎜⎜ 1 mol Al(OH)3 ⎞⎛ ⎟⎟⎜⎜ 1 mol Na 3 AlF6 ⎞⎟⎟ ⎜ Moles of Na3AlF6 = (365 kg Al(OH)3 )⎜⎜ ⎟⎟⎜ ⎟⎟⎜ ⎟ ⎟⎜⎜ 78.00 g Al(OH)3 ⎠⎝ ⎟⎜⎜1 mol Al(OH)3 ⎠⎟⎟ ⎜⎜⎝ 1 kg ⎠⎝ = 4.6795 × 103 mol Na3AlF6 Assuming NaOH is the limiting reactant: Moles of Na3AlF6 = ⎛ 1 L ⎞⎛ 1 mL ⎞⎛1.53 g ⎞⎛ 50.0% NaOH ⎞⎛ 1 mol NaOH ⎞⎛1 mol Na 3 AlF6 ⎞⎟ ⎟ (1.20 m 3 )⎜⎜⎜⎜⎜10−3 m 3 ⎟⎟⎟⎟⎟⎜⎜⎜⎜⎜10−3 L ⎟⎟⎟⎟⎟⎜⎝⎜⎜⎜ mL ⎠⎝⎟⎟⎟⎟⎜⎜⎜⎜ 100% ⎠⎟⎟⎟⎟⎜⎜⎜⎜⎜ 40.00 g NaOH ⎟⎟⎟⎟⎟⎜⎜⎜⎜⎜ 3 mol NaOH ⎟⎟⎟ ⎝ ⎠⎝ ⎠ ⎝ ⎠⎝ ⎠ = 7.650 × 103 mol Na3AlF6 Assuming HF is the limiting reactant: PV = nRT ⎛103 Pa ⎞⎛ ⎞⎛ 305 kPa 265 m 3 1 atm ⎟⎟⎜⎜ ⎟⎟⎜⎜ 1 L ⎞⎟⎟ PV ⎜⎜ Moles of HF = = ⎟⎜ ⎜⎜ ⎟ ⎟⎟⎟⎜⎜10−3 m 3 ⎟⎟⎟ 5 ⎟⎜⎜1.01325× 10 Pa ⎠⎝ RT ⎜ 0.0821 L • atm/mol • K ((273.2 + 91.5) K ) ⎝⎜ 1 kPa ⎠⎝ ⎠

(

)(

)

= 26,640.979 mol HF Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

22-16

⎛1 mol Na 3 AlF6 ⎞⎟ ⎜ ⎟⎟ = 4.4402 × 103 mol Na AlF Moles of Na3AlF6 = (26,640.979 mol HF )⎜⎜ 3 6 ⎜⎜⎝ 6 mol HF ⎠⎟⎟ Because the 265 m3 of HF produces the smallest amount of Na3AlF6, it is the limiting reactant. Convert moles Na3AlF6 to g Na3AlF6, using the molar mass, convert to kg, and multiply by 95.6% to determine the final yield. ⎛ 209.95 g Na AlF ⎞⎟⎛ 1 kg ⎞⎛ 95.6% ⎞ ⎟⎟⎜ 3 6 ⎜ ⎜ ⎟⎟ ⎟⎟⎜ Mass (g) of cryolite = (4.4402 × 10 3 mol Na 3 AlF6 )⎜⎜ ⎟⎜ ⎜⎝⎜ 1 mol Na 3 AlF6 ⎠⎟⎟⎜⎜⎝10 3 g ⎠⎟⎟⎜⎝ 100% ⎠⎟ = 891.195 = 891 kg Na3AlF6

22.73

Plan: The rate of effusion is inversely proportional to the square root of the molar mass of the molecule (Graham’s law). Solution: 4.00 g/mol Rate H molar massD a) = = = 1.40859 2.016 g/mol molar massH Rate D The time for D2 to effuse is 1.41 times greater than that for H2. Time D2 = (1.40859)(16.5 min) = 23.2417 = 23.2 min b) Set x equal to the number of effusion steps. The ratio of mol H2 to mol D2 is 99:1. Set up the equation to solve for x: 99/1 = (1.40859)x. When solving for an exponent, take the log of both sides. log (99) = log (1.40859)x Remember that log (ab) = b log (a) log (99) = x log (1.40859) x = 13.4129 To separate H2 and D2 to 99% purity requires 14 effusion steps.

22.74

a) Th e equilibrium constant is large, the rate constant is small, and Ea is large. b) 2CO(g) ⇆ C(graphite) + CO2(g) o c) ΔGrxn = ∑m ΔGfo(products) – ∑n ΔGfo(reactants) o ΔGrxn = [(1 mol C)(0 kJ/mol) + (1 mol CO2)(–394.4 kJ/mol)] – [(2 mol CO)(–137.2 kJ/mol]) = –120.0 kJ o ΔGrxn = – RT ln Kc

ln Kc =

(

) )(

⎛ 3 ⎞ −120.0 kJ ΔG o ⎜⎜10 J ⎟⎟ =− ⎟⎟ = 48.4345 ⎜ −RT 8.314 K /mol • K 298 K ⎜⎜⎝ 1 kJ ⎠⎟

(

)

Kc = 1.0835 × 1021 = 1.1 × 1021 d) Kp = Kc(RT)Δngas Kp = 1.0835 × 1021 [(0.0821 L∙atm/mol∙K)(298)]–1 = 4.428631 × 1019 = 4.4 × 1019 22.75

Plan: Use the stoichiometric relationships found in the balanced chemical equation to find the mass of Al2O3, mass of graphite, and moles of CO2. Use the ideal gas law to convert moles of CO2 into volume. Solution: a) 2Al2O3(in Na3AlF6) + 3C(gr) → 4Al(l) + 3CO2(g) ⎛10 3 kg ⎞⎛ 10 3 g ⎞⎛ 1 mol Al ⎞⎛ 2 mol Al 2 O 3 ⎞⎛ 101.96 g Al 2 O 3 ⎞⎛ 1 kg ⎞⎛ 1 t ⎞⎟ ⎜ ⎟⎟⎟⎜⎜ ⎟⎟⎟⎜⎜ ⎟⎟⎟⎜⎜ ⎟⎟⎟⎜⎜ ⎟⎟⎟⎜⎜ ⎟⎟⎟⎜⎜ ⎟ Mass (t) of Al2O3 = (1 t Al)⎜⎜ ⎟⎟⎜⎜⎜ 1 kg ⎟⎟⎜⎜⎜ 26.98 g Al ⎟⎟⎜⎜⎜ 4 mol Al ⎟⎟⎜⎜⎜ 1 mol Al O ⎟⎟⎜⎜⎜10 3 g ⎟⎟⎜⎜⎜10 3 kg ⎟⎟⎟ ⎜⎜⎝ 1 t ⎠⎝ 2 3 ⎠⎝ ⎠⎝ ⎠⎝ ⎠⎝ ⎠⎝ ⎠ = 1.8895478 = 1.890 t Al2O3 Therefore, 1.890 t of Al2O3 are consumed in the production of 1 t of pure Al.

22-17

3 ⎛10 3 kg ⎞⎛ ⎟⎟⎜⎜10 g ⎞⎛ ⎟⎟⎜⎜ 1 mol Al ⎞⎛ ⎟⎟⎜⎜ 3 mol C ⎞⎛ ⎟⎟⎜⎜12.01 g C ⎞⎛ ⎟⎟⎜⎜ 1 kg ⎞⎛ ⎟⎜ 1 t ⎞⎟ ⎜⎜ b) Mass (t) of C = 1 t Al ⎜ ⎟⎟⎜ ⎟⎟⎜ ⎟⎟⎜ ⎟⎟⎜ ⎟⎟⎜ 3 ⎟⎟⎟⎜⎜ 3 ⎟⎟⎟ ⎜⎝⎜ 1 t ⎠⎝ ⎟⎜⎜ 1 kg ⎠⎝ ⎟⎜⎜ 26.98 g Al ⎠⎝ ⎟⎜⎜ 4 mol Al ⎠⎝ ⎟⎜⎜ 1 mol C ⎠⎝ ⎟⎜⎜10 g ⎠⎝ ⎟⎜⎜10 kg ⎠⎟

(

)

= 0.3338584 = 0.3339 t C Therefore, 0.3339 t of C is consumed in the production of 1 t of pure Al, assuming 100% efficiency. c) The percent yield with respect to Al2O3 is 100% because the actual plant requirement of 1.89 t Al2O3 equals the theoretical amount calculated in part a). d) The amount of graphite used in reality to produce 1 t of Al is greater than the amount calculated in part b). In other words, a 100% efficient reaction takes only 0.3339 t of graphite to produce a ton of Al, whereas real production requires more graphite and is less than 100% efficient. Calculate the efficiency using a simple ratio: (0.45 t)(x) = (0.3338584 t)(100%) x = 74.19076 = 74% e) For every four moles of Al produced, three moles of CO2 are produced. 3 ⎛103 kg ⎞⎛ ⎟⎟⎜⎜10 g ⎞⎛ ⎟⎟⎜⎜ 1 mol Al ⎞⎛ ⎟⎟⎜⎜ 3 mol CO 2 ⎞⎟⎟ ⎜⎜ 4 Moles of C = (1 t Al)⎜ ⎟⎟⎜ ⎟⎟⎜ ⎟⎟⎜ ⎟⎟ = 2.7798 × 10 mol CO2 ⎜⎝⎜ 1 t ⎠⎝ ⎜ ⎜ ⎜ ⎟⎜ 1 kg ⎠⎝ ⎟⎜ 26.98 g Al ⎠⎝ ⎟⎜ 4 mol Al ⎠⎟ The problem states that 1 atm is exact. Use the ideal gas law to calculate volume, given moles, temperature, and pressure. PV = nRT 4 ⎛ ⎞⎛ −3 3 ⎞ nRT ⎜⎜ 2.7798 × 10 mol CO2 0.08206 L • atm/mol • K ((273 + 960.) K )⎟⎟⎟⎜⎜10 m ⎟⎟ V = = ⎜⎜ ⎟ ⎟⎜ ⎟⎟⎜⎜ 1 L ⎟⎟ 1 atm P ⎠ ⎜⎝⎜ ⎠⎟⎝

(

)(

)

= 2.812601 × 103 = 2.813 × 103 m3 22.76

a) The reactions are: 4FeCr2O4(s) + 8Na2CO3(aq) + 7O2(g) → 8Na2CrO4(aq) + 2Fe2O3(s) + 8CO2(g) 2Na2CrO4(aq) + 2H3O+(aq) → Na2Cr2O7(s) + 3H2O(l) + 2Na+(aq) Na2Cr2O7(s) + 2C(s) → Cr2O3(s) + Na2CO3(s) + CO(g) Cr2O3(s) + 2Al(s) → 2Cr(s) + Al2O3(s) b) Mass (kg) of Cr = 3 ⎛10 3 kg ⎞⎛ ⎟⎟⎜⎜10 g ⎞⎟⎟⎛⎜ 1 mol FeCr2 O 4 ⎞⎛ ⎟⎟⎜⎜ 2 mol Cr ⎞⎛ ⎟⎟⎜⎜ 52.00 g Cr ⎞⎟⎟⎛⎜⎜ 1 kg ⎞⎟⎟ ⎜⎜ 7 1.97 10 t FeCr O × ⎟ ⎟⎟⎜⎜ ⎟ ⎟⎜ ⎟ ⎟ ( ⎜ 2 4 )⎜ ⎟ ⎟ ⎜⎜ 1 t ⎟⎜⎜ 1 kg ⎟⎜⎜ 223.85 g FeCr2 O 4 ⎟⎜⎜1 mol FeCr2 O 4 ⎟⎟⎜⎜⎜ 1 mol Cr ⎟⎟⎜⎜⎜10 3 g ⎟⎟ ⎠⎝ ⎠⎝ ⎠⎝ ⎝ ⎠⎝ ⎠⎝ ⎠ = 9.152558 × 109 = 9.2 × 109 kg Cr

22.77

a) HDO + H2 ⇆ HD + H2O b) At equilibrium, there are more reactants than products. This is a reactant-favored reaction so the equilibrium constant K must be less than 1. [ H 2 O ][ HD ] [ 0.20 ][ 0.20 ] = 0.44 c) K = = [ HDO ][ H 2 ] [ 0.30 ][ 0.30 ]

22.78

Ksp (NiS) = 1.1 × 10–18 = [Ni2+][S2–]= (0.10) [S2–], so if [S2–] < 1.1 × 10–17, Ksp will not be exceeded and NiS will not precipitate. Ksp (CuS) = 8 × 10–34 = [Cu2+][S2–] = (0.10) [S2–], so if [S2–] > 8 × 10–33, Ksp will be exceeded and CuS will precipitate. [ HS− ][ H3 O+ ] Ka1 = 9 × 10–8 = [ H2 S] Ka2 = 1 × 10–17 =

[ H3O+ ] ⎡⎣S2− ⎤⎦ [ HS− ]

22-18

[ HS− ][ H3 O+ ] [ H3O+ ] ⎡⎣S2− ⎤⎦ [ H3O+ ] ⎡⎣S2− ⎤⎦ Ka1 × Ka2 = × = = (9 × 10–8) × (1 × 10–17) = 9 × 10–25 [ H2S] [ H2 S] [ HS− ] 2

[ H3O+ ] ⎡⎣S2− ⎤⎦ 2

–25

9 × 10 =

[ H2S]

[ H 3 O+ ] ⎡⎣S2− ⎤⎦ 2

0.10

So, maintaining the pH between –3.5 and 4.0 would allow CuS to precipitate but prevent precipitation of NiS.

22.79

22.80

⎛ ⎞⎟⎛ 100% ⎞ ⎜ 1 kg ⎟⎜ ⎟⎟ = 3.6281 × 107 = 3.6 × 107 kg ore a) Mass (kg) of ore = (2.0 × 10 5 lb Cu)⎜⎜ ⎟⎜ ⎟ ⎜⎜⎝ 2.205 lb ⎠⎟⎟⎟⎝⎜ 0.25% ⎠ ⎛ 0.25% Cu ⎟⎞⎛⎜183.54 g FeCuS2 ⎞⎟ ⎟⎜⎜ b) Mass % chalcopyrite = ⎜⎜⎜ ⎟⎟(100%) = 0.7220 = 0.72% chalcopyrite ⎟⎠⎜⎝ 63.55 g Cu ⎟⎠⎟ ⎜⎝ 100% ⎟⎟⎜

Plan: Solubility of a salt can be calculated from its Ksp. Ksp for Ca3(PO4)2 is 1.2 × 10–29. Phosphate is derived from a weak acid, so the pH of the solution impacts exactly where the acid-base equilibrium lies. Phosphate can gain one H+ to form HPO42–. Gaining another H+ gives H2PO4– and a last H+ added gives H3PO4. The Ka values for phosphoric acid are Ka1 = 7.2 × 10–3, Ka2 = 6.3 × 10–8, Ka3 = 4.2 × 10–13. To find the ratios of the various forms of phosphate, use the equilibrium expressions and the concentration of H+. Solution: a) Ca3(PO4)2(s) ⇆ Ca2+(aq) + 2PO43–(aq) ___ 0 0 Initial Change –x +3x +2x Equilibrium __ 3x 2x Ksp = 1.2 × 10–29 = [Ca2+]3[PO43–]2 = (3x)3(2x)2 = 108x5 x = 6.4439 × 10–7 = 6.4 × 10–7 M Ca3(PO4)2 b) [H3O+] = 10–4.5 = 3.162 × 10–5 M H3PO4(aq) + H2O(l) ⇆ H2PO4–(aq) + H3O+(aq) [ H2 PO4− ][ H3O+ ] Ka1 = [ H3 PO4 ]

[ H 2 PO 4− ] K a1 7.2 × 10−3 = = = 227.70 [ H 3 PO 4 ] [ H 3 O+ ] 3.162 × 10−5 H2PO4–(aq) + H2O(l) ⇆ HPO42–(aq) + H3O+(aq) ⎡ HPO4 2− ⎤ [ H3O+ ] ⎦ Ka2 = ⎣ [ H2 PO4− ] −8 ⎡ HPO4 2− ⎤ ⎣ ⎦ = K a 2 = 6.3 × 10 = 1.9924 × 10–3 − + [ H2 PO4 ] [ H 3 O ] 3.162 × 10−5

HPO42–(aq) + H2O(l) ⇆ PO43–(aq) + H3O+(aq) ⎡ PO4 3− ⎤ [ H3O+ ] ⎦ Ka3 = ⎣ ⎡ HPO4 2− ⎤ ⎣ ⎦ −13 ⎡ PO4 3− ⎤ ⎣ ⎦ = K a3 = 4.2 × 10 = 1.32827 × 10–8 + −5 ⎡ HPO4 2− ⎤ H O × 3.162 10 [ ] 3 ⎣ ⎦ From the ratios, the concentration of H2PO4– is at least 100 times more than the concentration of any other species, so assume that dihydrogen phosphate is the dominant species and find the value for [PO43–]. Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

22-19

⎡ PO4 3− ⎤ ⎣ ⎦ = 1.32827 × 10–8 ⎡ HPO4 2− ⎤ ⎣ ⎦

⎡ HPO4 2− ⎤ ⎣ ⎦ = 1.9924 × 10–3 − [ H2 PO4 ]

and

[PO43–] = (1.32827 × 10–8)[HPO42–] and [HPO42–] = (1.9924 × 10–3)[H2PO4–] [PO43–] = (1.32827 × 10–8)[HPO42–] = (1.32827 × 10–8)(1.9924 × 10–3)[H2PO4–] [PO43–] = (2.6464 × 10–11)[H2PO4–] Substituting this into the Ksp expression gives Ksp = [Ca2+]3[PO43–]2 = 1.2 × 10–29 Ksp = [Ca2+]3{(2.6464 × 10–11)[H2PO4–]}2 Rearranging gives Ksp/(2.6464 × 10–11)2 = [Ca2+]3[H2PO4–]2 The concentration of calcium ions is still represented as 3x and the concentration of dihydrogen phosphate ion as 2x, since each H2PO4– comes from one PO43–. Ksp/(2.6464 × 10–11)2 = (3x)3(2x)2 (1.2 × 10–29)/(2.6464 × 10–11)2 = 108x5 x = 1.0967 × 10–2 = 1.1 × 10–2 M Acid rain increases the leaching of PO43– into the groundwater, due to the protonation of PO43– to form HPO42– and H2PO4–. As shown in calculations a) and b), solubility increases from 6.4 × 10–7 M (in pure water) to 1.1 × 10–2 M (in acidic rainwater). 22.81

a) The mole % of oxygen missing may be estimated from the discrepancy between the actual and ideal formula. ⎛ 2.000 − 1.98 mol O ⎞⎟ ⎜ ⎟⎟ Mol % O missing = ⎜⎜⎜ ⎟⎟ × 100% = 1.00% ⎜⎜ 2.000 mol O ⎝ ⎠⎟⎟

(

)

b) The molar mass is determined like a normal molar mass except that the quantity of oxygen is not an integer value. Molar mass = 1 mol Pb (207.2 g/mol) + 1.98 mol O (16.00 g/mol) = 238.88 = 238.9 g/mol 22.82

Using ΔG values: ΔG = +237.192 kJ/mol H 2O(l) → H 2(g) + 1/2O2(g) ΔG = +33 kJ/mol H 2S(g) → H 2(g) + 1/8S8(s) Both these processes require energy to proceed, but the second requires less energy and would be more favorable (or less unfavorable) based on this criterion.

22.83

Plan: Convert the unit cell edge length to cm and cube that length to obtain the volume of the cell. A facecentered cubic structure contains four atoms of silver. Divide that number by Avogadro’s number to find the moles of silver in the cell and multiply by the molar mass to obtain the mass of silver in the cell. Density is obtained by dividing the mass of silver by the volume of the cell. The calculation will now be repeated replacing the atomic mass of silver with the “atomic mass” of sterling silver. The “atomic mass” of sterling silver is simply a weighted average of the masses of the atoms present (Ag and Cu). Solution: ⎛10−12 m ⎞⎟⎛ 1 cm ⎞ ⎟ = 4.086 × 10–8 cm ⎟⎜ Edge length (cm) = (408.6 pm)⎜⎜ ⎜⎝ 1 pm ⎠⎟⎟⎜⎝10−2 m ⎠⎟⎟ Volume (cm3) of cell = (4.086 × 10–8 cm)3 = 6.82174 × 10–23 cm3 ⎞⎟⎛107.9 g Ag ⎞⎟ ⎛ 4 Ag atoms ⎞⎟⎛⎜ 1 mol Ag ⎜ ⎟⎟⎜⎜ ⎟⎟⎜⎜ ⎟⎟ = 7.1671 × 10–22 g Mass (g) of silver = ⎜⎜ ⎜⎜⎝ unit cell ⎠⎟⎟⎜⎜ 6.022 × 10 23 Ag atoms ⎟⎟⎝⎜⎜⎜ 1 mol Ag ⎠⎟⎟ ⎝ ⎠ Density of silver =

7.1671 × 10−22 g = 10.506264 = 10.51 g/cm3 −23 3 6.82174 × 10 cm

22-20

(

)

⎛107.9 g ⎞⎟⎜⎜⎛ 100.0 − 7.5 % ⎞⎟⎟ ⎛ 63.55 g ⎞⎟⎛ 7.5% ⎞ ⎜ ⎟ = 104.57375 g/mol ⎟⎟ +⎜⎜ ⎟⎜ ⎟⎜ Atomic mass of sterling silver = ⎜⎜ ⎟⎟ ⎜⎜⎝ mol ⎠⎟⎜ ⎟⎝⎜100% ⎟⎠⎟ ⎜⎝ mol ⎠⎟⎟⎜⎜⎜ 100% ⎟ ⎝ ⎠ ⎞⎟⎛104.57375 g Ag ⎞⎟ ⎛ 4 Ag atoms ⎞⎟⎜⎛ 1 mol Ag ⎜ ⎜ ⎟⎟⎜⎜ ⎟⎟ = 6.946 × 10–22 g Mass (g) of sterling silver = ⎜⎜ ⎟⎟⎜ 23 ⎟ ⎜⎝⎜ unit cell ⎠⎟⎜⎜ 6.022 × 10 Ag atoms ⎟⎟⎝⎜⎜⎜ 1 mol Ag ⎠⎟⎟ ⎝ ⎠

Density of sterling silver =

22.84

6.9461 × 10−22 g = 10.1823 = 10.2 g/cm3 6.82174 × 10−23 cm 3

⎛ 3 ⎞ ⎛ ⎞ ⎜10 g ⎟⎟⎜⎛ 0.05% ⎞⎟⎜⎜ 1 mol Ti ⎟⎟ 22 22 a) Moles of Ti = (5.98 × 10 24 kg)⎜⎜ ⎟⎜ ⎟⎟⎜⎜ ⎟ = 6.24 × 10 = 6 × 10 mol Ti ⎜⎝⎜ 1 kg ⎠⎟⎝ 100% ⎠⎟⎝⎜⎜ 47.88 g Ti ⎠⎟⎟

b) Mass (g) of ilmenite = ⎛ 3 ⎞ ⎟ 151.73 g FeTiO ⎞⎟ ⎜10 g ⎟⎟⎜⎛ 0.05% ⎞⎟⎜⎜⎛ 1 mol Ti ⎞⎟⎟⎛⎜ 50% ⎞⎟⎛⎜⎜1 mol FeTiO3 ⎞⎛ 24 ⎜ (5.98 × 10 kg)⎜⎜ 1 kg ⎟⎟⎟⎜⎝⎜ 100% ⎠⎟⎟⎜⎜ 47.88 g Ti ⎟⎟⎟⎝⎜⎜100% ⎠⎟⎟⎜⎜ 1 mol Ti ⎟⎟⎟⎟⎜⎜⎜⎜ 1 mol FeTiO 3 ⎟⎟⎟⎟ ⎜⎝ 3 ⎠ ⎠ ⎝⎜ ⎠⎝⎜ ⎝⎜ ⎠ = 4.73760 × 1024 = 5 × 1024 g FeTiO3 ⎞⎟ 1 ton ⎞⎟⎛⎜ ⎛ 0.05% ⎞⎟⎛⎜ 2.205 lbs ⎞⎛ yr ⎟⎟ = 3.296 × 1013 = 3 × 1013 yr ⎟⎟⎟⎜⎜ ⎟⎟⎜⎜ c) Yr = 5.98 × 10 24 kg ⎜⎜ ⎟⎟⎜⎜ ⎜ ⎟ ⎟⎟⎜⎜1.00 × 10 5 tons ⎟⎟ ⎝⎜ 100% ⎠⎜⎜⎝ 1 kg ⎠⎝ ⎟⎟⎜⎜ 2000 lb ⎠⎟⎜ ⎝ ⎠

(

22.85

)

2NaCl(s) + H2SO4(aq) → Na2SO4(aq) + 2HCl(g) Na2SO4(s) + 2C(s) → Na2S(s) + 2CO2(g)

Na2S(s) + CaCO3(s) → Na2CO3(s) + CaS(s) Total: CaCO3(s) + Na2SO4(s) + 2C(s) + 2NaCl(s) + H2SO4(aq) → Na2SO4(aq) + 2HCl(g) + 2CO2(g) + Na2CO3(s) + CaS(s) b) Notice that Na2SO4(aq) and Na2SO4(s) are different substances, and, as such, will not cancel in this Hess’s law calculation. The values for Na2SO4(aq) and Na2SO4(s) are not in the Appendix so they must be found from another source. The other enthalpies of formation are either given in the problem or are given in the Appendix. ο ΔH rxn = ∑m ΔHfο(products) – ∑n ΔH fο(reactants) ο ΔH rxn = [(1 mol Na2SO4(aq)) (ΔHfο of Na2SO4(aq)) + (2 mol HCl) (ΔHfο of HCl) + (2 mol CO2) (ΔHfο of CO2) + (1 mol Na2CO3) (ΔHfο of Na2CO3) + (1 mol CaS) (ΔHfο of CaS)] – [(1 mol CaCO3) (ΔHfο of CaCO3) + (1 mol Na2SO4(s)) (ΔHfο of Na2SO4(s)) + (2 mol C) (ΔHfο of C) + (2 mol NaCl) (ΔHfο of NaCl) + (1 mol H2SO4) (ΔHfο of H2SO4)] 351.8 kJ/mol = [(1 mol)(–1389.5 kJ/mol) + (2 mol)(–92.31 kJ/mol) + (2 mol)(–393.5 kJ/mol) + (1 mol)(–1130.8 kJ/mol) + (1 mol) (ΔHfο of CaS)] – [(1 mol)(–1206.9 kJ/mol) + (1 mol)(–1387.1 kJ/mol) + (2 mo)(0 kJ/mol) + (2 mol)(–411.1 kJ/mol) + (1 mol)(–907.51 kJ/mol)] ΔHfο of CaS(s) = –479.99 = –480.0 kJ c) The reaction is very endothermic, therefore it is probably not spontaneous even though the entropy change may be expected to be positive. It will take a ΔG determination to know for sure. ⎛ 1 mol NaCl ⎞⎛ ⎟⎟⎜⎜1 mol Na 2 CO 3 ⎞⎛ ⎟⎟⎜⎜105.99 g Na 2 CO 3 ⎞⎟⎟⎛ 73% ⎞⎟ ⎜ d) Mass (g) of Na2CO3 = (250. g NaCl)⎜⎜ ⎟⎟⎜ ⎟⎟⎜ ⎟⎜⎜ ⎟ ⎜⎝⎜ 58.44 g NaCl ⎠⎝ ⎟⎜⎜ 2 mol NaCl ⎠⎝ ⎟⎜⎜ 1 mol Na 2 CO 3 ⎠⎟⎟⎝⎜100% ⎠⎟ = 165.4960 = 165 g Na2CO3

22-21

22.86

a) For CaCO3: CaCO3(s) → CaO(s) + CO2(g) ο ΔH rxn = ∑m ΔHfο(products) – ∑n ΔH fο(reactants) ο ΔH rxn = [(1 mol)( ΔHfο of CaO) + (1 mol)( ΔHfο of CO2)] – [(1 mol)( ΔHfο of CaCO3)] ο ΔH rxn = [(1 mol)(–635.1 kJ/mol) + (1 mol)(–393.5 kJ/mol)] – [(1 mol)(–1206.9 kJ/mol)] ο ΔH rxn = 178.3 kJ ο ο o ΔSrxn = ∑m Sproducts – ∑n Sreactants o ΔSrxn = [(1 mol)( S o of CaO) + (1 mol)( S o of CO2)] – [( 1 mol)( S o of CaCO3)] o ΔSrxn = [(1 mol)(38.2 J/mol∙K) + (1 mol)(213.7 J/mol∙K)] – [(1 mol)(92.9 J/mol∙K)] o ΔSrxn = 159.0 J/K o o o ΔGrxn = 0 = ΔH rxn – T ΔSrxn o o ΔH rxn = T ΔSrxn 3 ΔH o ⎛⎜ 178.3 kJ ⎞⎛ ⎟⎟⎜⎜10 J ⎞⎟⎟ ⎜ T= = ⎟ ⎜ ⎜ ⎟⎟⎜⎜ 1 kJ ⎟⎟⎟ = 1121.3836 = 1121 K for CaCO3 ⎜⎜⎝159.0 J /K ⎠⎝ ΔS o ⎠

For MgCO3: MgCO3(s) → MgO(s) + CO2(g) o ΔH rxn = ∑m ΔHfo(products) – ∑n ΔH fo(reactants) o ΔH rxn = [(1 mol)( ΔHfo of (MgO) + (1 mol)( ΔHfo of (CO2)] – [(1 mol)( ΔHfo of (MgCO3)] o ΔH rxn = [(1 mol)(–601.2 kJ/mol) + (1 mol)(–393.5 kJ/mol)] – [(1 mol)(–1112 kJ/mol)] o ΔH rxn = 117.3 kJ

o o o ΔSrxn = ∑m Sproducts – ∑n Sreactants o ΔSrxn = [(1 mol)( S o of MgO)) + (1 mol)( S o of CO2)] – [(1 mol)( S o of MgCO3)] o ΔSrxn = [(1 mol)(26.9 J/mol∙K) + (1 mol)(213.7 J/mol∙K)] – [(1 mol)(65.86 J/mol∙K)] o ΔSrxn = 174.74 J/K o o o ΔGrxn = 0 = ΔH rxn – T ΔSrxn o o ΔH rxn = T ΔSrxn 3 ΔH o ⎛⎜ 117.3 kJ ⎞⎛ ⎟⎟⎜⎜10 J ⎞⎟⎟ ⎜ T= = ⎟ ⎜⎜ ⎟⎟⎜⎜⎜ 1 kJ ⎟⎟⎟ = 671.283 = 671 K for CaCO3 ΔS o ⎜⎝174.74 J /K ⎠⎝ ⎠ b) CaO(s) + SiO2(s) → CaSiO3(s) Mass (kg) of CaCO3 = ⎛ 2.236 × 10 6 tons ⎞⎟⎛ 84% ⎞⎛ 52 weeks ⎞⎜⎛ 50 kg slag ⎞⎟⎜⎛10 3 g ⎞⎛ 1 mol slag ⎞⎛ 1 mol CaCO 3 ⎞⎟⎛⎜100.09 g CaCO 3 ⎞⎟⎜⎛ 1 kg ⎞⎟ ⎟ ⎟⎟⎜⎜ ⎟⎟⎜⎜ ⎟⎟⎜⎜ ⎟⎜⎜ ⎟⎟⎜⎜ ⎟⎟⎟⎜⎜ ⎟⎟⎜⎜⎜ ⎟ ⎜⎜⎜ ⎟ ⎟⎟⎜⎜ ⎟ ⎜ ⎜ ⎟⎟⎜10 3 g ⎟⎟⎟ ⎟⎜⎝ week yr ton ⎟⎠⎟⎜⎝⎜ 1 kg ⎠⎟⎟⎜⎜⎝116.17 g slag ⎠⎝ ⎠⎜ ⎟⎟⎜⎜ 1 mol slag ⎠⎟⎟⎝⎜⎜ 1 mol CaCO 3 ⎠⎝ ⎝⎜ ⎠⎟⎝⎜100% ⎠⎟⎝ ⎜ ⎠ = 4.2075 × 109 = 4.2 × 109 kg CaCO3

22-22

CHAPTER 23 TRANSITION ELEMENTS AND THEIR COORDINATION COMPOUNDS FOLLOW–UP PROBLEMS 23.1A

Plan: Locate the element on the periodic table and use its position and atomic number to write a partial electron configuration. Add or subtract electrons to obtain the configuration for the ion. Partial electron configurations do not include the noble gas configuration and any filled f sublevels, but do include outer level electrons (n-level) and the n–1 level d orbitals. Remember that ns electrons are removed before n–1 electrons. Solution: a) Ag is in the fifth row of the periodic table, so n = 5, and it is in group 1B(11). The partial electron configuration of the element is 5s14d10. For the ion, Ag+, remove one electron from the 5s orbital. The partial electron configuration of Ag+ is 4d10. b) Cd is in the fifth row and group 2B(12), so the partial electron configuration of the element is 5s24d10. Remove two electrons from the 5s orbital for the ion. The partial electron configuration of Cd2+ is 4d10. c) Ir is in the sixth row and group 8B(9). The partial electron configuration of the element is 6s25d7. Remove two electrons from the 6s orbital and one from 5d for ion Ir3+. The partial electron configuration of Ir3+ is 5d6.

23.1B

Plan: Locate the element on the periodic table and use its position and atomic number to write a condensed electron configuration for its atom. Compare the partial electron configurations of the atom and the ion (provided in the problem statement) to determine how many electrons have been removed from the atom to produce the ion. The number of electrons removed is the charge on the ion. Because electrons are being removed, the charge will be positive. Solution: a) Ta has Z = 73. The condensed electron configuration of its atom is: [Xe] 6s24f 145d3. Three electrons were removed to produce its ion, which has a condensed electron configuration of [Xe] 4f 145d2. Therefore, the charge on the Ta ion is +3, Ta3+. b) Mn has Z = 25. The condensed electron configuration of its atom is: [Ar] 4s23d5. Four electrons were removed to produce its ion, which has a condensed electron configuration of [Ar] 3d3. Therefore, the charge on the Mn ion is +4, Mn4+. c) Os has Z = 76. The condensed electron configuration of its atom is: [Xe] 6s24f 145d6. Three electrons were removed to produce its ion, which has a condensed electron configuration of [Xe] 4f 145d5. Therefore, the charge on the Os ion is +3, Os3+.

23.2A

Plan: Find Er on the periodic table and write its electron configuration. Remove 3 electrons for the electron configuration of Er3+. Place the outer level electrons in an electron diagram, making sure to follow the Aufbau principle and Hund’s rule. Count the number of unpaired electrons. Solution: Er is a lanthanide element in row 6 and with atomic number 68. The electron configuration of the element is [Xe]6s25d14f 11. Remove three electrons: 2 from the 6s and 1 from 5d for the Er3+ ion, [Xe]4f 11.

The number of unpaired electrons is three, all in the 4f sublevel. 23.2B

Plan: Find Dy on the periodic table and write its electron configuration. Remove 3 electrons for the electron configuration of Dy3+. Place the outer level electrons in an electron diagram, making sure to follow the Aufbau principle and Hund’s rule. Count the number of unpaired electrons.

23-1

Solution: Dy is a lanthanide element in row 6 and with atomic number 66. The electron configuration of the element is [Xe]6s25d14f 9. Remove three electrons: 2 from the 6s and 1 from 5d for the Dy3+ ion, [Xe]4f 9.

The number of unpaired electrons is five, all in the 4f sublevel. 23.3A

Plan: Use the charges on the counter ions and the ligands to calculate the charge on the central metal ion, remembering that all the charges in a neutral compound must add to zero and that the charges for an ion must add to the overall charge on that ion. The coordination number, or number of ligand atoms bonded to the metal ion, is found by examining the bonded entities inside the square brackets to determine if they are monodentate, bidentate, or polydentate. Solution: a) The counter ion is NO3−, so the complex ion is [Co(en)2Br2]+. Each bromo ligand has a –1 charge and ethylenediamine is neutral, so cobalt has a +3 charge: +3 + 2(0) + 2(–1) = +1. Each ethylenediamine ligand is bidentate and each bromo ligand is monodentate, so the coordination number is 6. b) The counter ion is Mg2+, so the complex ion is [CrCl5Br]3−. Each bromo ligand and each chloro ligand has a –1 charge, so Cr has a +3 charge: +3 + 5(–1) + 1(–1) = –3. Each chloro ligand and each bromo ligand is monodentate, so the coordination number is 6.

23.3B

Plan: Examine the ligands. Determine their charges and whether they are mono-, bi-, or poly-dentate. Determine the number of ligands that are necessary for the given coordination number. Write the formula of the complex ion and determine its charge. Balance the charge on the complex ion with the counter ion. Solution: a) The ligand in this case is the cyano ligand. It is monodentate and has a charge of –1. Because cyano is monodentate, 6 cyano ligands are needed to fulfill the coordination number of 6. Therefore, the formula of the complex ion is [Mn(CN)6]4–. The charge on the complex ion is –4: (1 Mg)(+2) + (6 CN)( –1) = –4. Four potassium ions, K+, are needed to balance the charge: K4[Mn(CN)6]. b) The ligand in this case is the ethylenediamine ligand. It is bidentate and is neutral. Because ethylenediamine is bidentate, 2 ethylenediamine ligands are needed to fulfill the coordination number of 4. Therefore, the formula of the complex ion is [Cu(en)2]2+. The charge on the complex ion is +2: (1 Cu)(+2) + (2 en)(0) = +2. One sulfate ion, SO42–, is needed to balance the charge: [Cu(en)2]SO4.

23.4A

Plan: Name the compound by following rules outlined in the chapter. Use Table 23.7 for the names of ligands. Write the formula from the name by breaking down the name into pieces that follow the naming rules. Solution: a) The compound consists of the cation [Cr(H2O)5Br]2– and anion Cl–. The metal ion in the cation is chromium and its charge is found from the charges on the ligands and the total charge on the ion. total charge = (charge on Cr) + 5(charge on H2O) + (charge on Br–) –2 = (charge on Cr) + 5(0) + (–1) charge on Cr = +3 The name of the cation will end with chromium(III) to indicate the oxidation state of the chromium ion since there is more than one possible oxidation state for chromium. The water ligand is named aqua. There are five water ligands, so pentaaqua describes the (H2O)5 ligands. The bromide ligand is named bromo. The ligands are named in alphabetical order, so aqua comes before bromo. The name of the cation is pentaaquabromochromium(III). Add the chloride for the anion to complete the name: pentaaquabromochromium(III) chloride. b) The compound consists of a cation, barium ion (Ba2+), and the anion hexacyanocobaltate(III). The formula of the anion consists of six (from hexa) cyanide (from cyano) ligands and cobalt (from cobaltate) in the +3 oxidation state (from (III)). Putting the formula together gives [Co(CN)6]n+. To find the charge on the complex ion, calculate from charges on the ligands and the metal ion:

23-2

total charge = 6(charge on CN–) + (charge on cobalt ion) = 6(–1) + (+3) = –3 The formula of complex ion is [Co(CN)6]3–. Combining this with the cation, Ba2+, gives the formula for the compound: Ba3[Co(CN)6]2. The three barium ions give a +6 charge and two anions give a –6 charge, so the net result is a neutral salt. 23.4B

Plan: Name the compound by following rules outlined in the chapter. Use Table 23.7 for the names of ligands. Write the formula from the name by breaking down the name into pieces that follow the naming rules. Solution: a) The compound consists of an anion, sulfate ion (SO42–), and the cation diaquabis(ethylenediamine)cobalt(III). The formula of the cation consists of two (from di) water (from aqua) ligands, two (from bis) ethylenediamine ligands, and cobalt in the +3 oxidation state (from (III)). Putting the formula together gives [Co(H2O)2(en)2]n+. To find the charge on the complex ion, calculate from charges on the ligands and the metal ion: total charge = 2(charge on H2O) + 2(charge on en) + (charge on cobalt ion) = 2(0) + 2(0) + (+3) = +3 The formula of complex ion is [Co(H2O)2(en)2]3+. Combining this with the anion, SO42–, gives the formula for the compound: [Co(H2O)2(en)2]2(SO4)3. The three sulfate ions give a –6 charge and two cations give a +6 charge, so the net result is a neutral salt. b) The compound consists of the cation Mg2+ and anion [Cr(NH3)2Cl4]–. The metal ion in the complex ion is chromium and its charge is found from the charges on the ligands and the total charge on the ion. total charge = (charge on Cr) + 2(charge on NH3) + 4(charge on Cl–) –1 = (charge on Cr) + 2(0) + 4(–1) charge on Cr = +3 The name of the anion will end with chromate(III) to indicate the oxidation state of the chromium ion since there is more than one possible oxidation state for chromium. The –ate ending indicates that the complex ion is an anion. The ammonia ligand is named ammine. There are two ammonia ligands, so diammine describes the (NH3)2 ligands. The chloride ligand is named chloro. There are four chlorine ligands, so tetrachloro describes the Cl4 ligands. The ligands are named in alphabetical order, so ammine comes before chloro. The name of the anion is diamminetetrachlorochromate(III). Add the magnesium for the cation to complete the name. The cation is named before the anion: magnesium diamminetetrachlorochromate(III).

23.5A

Plan: The given complex ion has a coordination number of 6 (en is bidentate), so it will have an octahedral arrangement of ligands. Stereoisomers can be either geometric or optical. For geometric isomers, check if the ligands can be arranged either next to (cis) or opposite (trans) each other. Then, check if the mirror image of any of the geometric isomers is not superimposable. If mirror image is not superimposable, the structure is an optical isomer. Solution: The complex ion contains two NH3 ligands and two chloride ligands, both of which can be arranged in either the cis or trans geometry. The possible combinations are 1) cis-NH3 and cis-Cl, 2) trans-NH3 and cis-Cl, and 3) cis-NH3 and trans-Cl. Both NH3 and Cl trans is not possible because the two bonds to ethylenediamine can be arranged only in this cis-position, which leaves only one set of trans positions.

NH3 H2N

NH3 Co

H2N

NH3 H2N

Cl Cl

H2N

Co Cl

Cl cis-NH3 cis-Cl

H2N

NH3 Co

Cl NH3

trans-NH3 cis-Cl

H2N

NH3 Cl

cis-NH3 trans-Cl

23-3

The mirror images of the second two structures are superimposable since two of the ligands, either ammonia or chloride ion, are arranged in the trans position. When both types of ligands are in the cis arrangement, the mirror image is not a superimposable optical isomer:

NH3 H2N

NH3

H3N

NH2 Co

H2N

NH2 Cl

cis-NH3 cis-Cl

cis-NH3 cis-Cl +

There are four stereoisomers of [Co(NH3)2(en)Cl2] . 23.5B

Plan: The given complex ion has a coordination number of 6, so it will have an octahedral arrangement of ligands. Stereoisomers can be either geometric or optical. For geometric isomers, check if the ligands can be arranged either next to (cis) or opposite (trans) each other. Then, check if the mirror image of any of the geometric isomers is not superimposable. If mirror image is not superimposable, the structure is an optical isomer. Solution: The complex ion contains two H2O ligands, two NH3 ligands and two chloride ligands, all of which can be arranged in either the cis or trans geometry. The possible combinations are drawn below: Cl NH3

Cl OH2

Ru OH2

NH3

OH2

Cl Cl

OH2

Cl NH3

Ru NH3

H2O

OH2

Cl Cl

OH2 Ru

OH2

NH3

Ru NH3

NH3

Cl OH2

OH2 Ru

NH3

OH2

H3N

NH3

mirror optical isomers 23.6A

Plan: Compare the two ions for oxidation state of the metal ion and the relative ability of ligands to split the d-orbital energies. The splitting strength of the ligands is obtained from the spectrochemical series. The greater the splitting strength of the ligands, the greater the energy of absorbed light. Solution: The oxidation number of vanadium in both ions is the same, so compare the two ligands. Ammonia is a stronger field ligand than water, so the complex ion [V(NH3)6]3+ absorbs visible light of higher energy than [V(H2O)6]3+ absorbs.

23.6B

Plan: Each of these ions includes cobalt in the +3 oxidation state. In order to distinguish between them, compare the relative ability of ligands to split the d-orbital energies. The splitting strength of the ligands is obtained from the spectrochemical series. The greater the splitting strength of the ligands, the greater the energy of absorbed

23-4

lig ght. Place the ligands l in ordeer from lowest energy of absoorbed light (abssorbs closer to red light) to hiighest en nergy of absorb bed light (abso orbs closer to violet light). Thhen use an artisst’s wheel to deetermine the coolor of ob bserved light, which w is compllementary to th he color of the absorbed lightt. So olution: Fo our different ligands are present in the comp plex ions: en, N NH3, Cl–, and H2O. Putting thhese ions in ordder from – weaker w field ligaands to strongeer field ligandss gives: Cl < H 2O < NH3 < een. Putting the complex ions iin order frrom those with weaker field ligands (which absorb lower eenergy light) too those with strronger field liggands (which ab bsorb higher en nergy light) giv ves: (absorbs closer to red ligght) [Co(NH3)4C Cl2]+ < [Co(NH H3)5Cl]2+ < 3+ 3+ [C Co(NH3)5(H2O))] < [Co(en)3] (absorbs closser to violet ligght). Th he possible obsserved colors of o the complex xes are: red, greeen, purple, annd yellow. Corrrelate the obserrved colors with w absorbed colors (see Figu ure 23.15): Obseerved Color

d Color Absorbed

Red

Greeen

Green

Red

Purple

Yelloow

Yellow

Purpple

Pu utting the possible absorbed colors c in orderr from lowest eenergy (highestt wavelength) tto highest enerrgy (lowest wavelength) w giv ves: red < yello ow < green < purple. p Correlaate this order off absorbed coloors with the orrder of co omplex ions lissted above. Absorbed Color Red

[Co(N NH3)4Cl2]

Yellow Green Purple 23.7A

Complexx Ion [Co(N NH3)5Cl]

[Co(een)3]

Green

[Co(N NH3)5(H2O)]

Obsserved Color Purple

Red Yellow

Pllan: Determinee the charge on n the manganesse ion in the coomplex ion andd the number off electrons in itts d orbitals. Siince it is an octtahedral ligand d, the d orbitalss split into threee lower energyy orbitals and ttwo higher eneergy orbitals. Check the spectrrochemical serries to see if CN N– is a strong oor weak field liigand. If it is a strong field liggand, fill the th hree lower enerrgy orbitals beffore placing an ny electrons in the higher eneergy d orbitals. If it is a weak field ligand, pllace one electro on in each of th he five d orbitaals, low energyy before high eenergy, before ppairing any eleectrons. After filling orbitals, count c the numb ber of unpaired d electrons. Thee complex ion is low-spin if the ligand is a strong field lig gand and high--spin if the ligaand is weak fieeld. So olution: Fiind the charge on manganesee: total ch harge = (chargee on Mn) + 6(ccharge on CN– ) ch harge on Mn = –3 – [6(–1)] = +3 Th he electron con nfiguration of Mn M is [Ar]4s23d 3 5; the electroon configuratioon of Mn3+ is [A Ar]3d4. – Th he ligand is CN N , which is a strong s field ligand, so the fouur d electrons w will fill the low wer energy d orrbitals before an ny are placed in n the higher en nergy d orbitalss: Higherr energy d orbittals

Lower energy d orbitals Copyright © McGraw-Hill Education. E This is proprietary matterial solely for au uthorized instrucctor use. Not auth horized for sale orr distribution in any mann ner. This documen nt may not be cop pied, scanned, dup plicated, forward ded, distributed, oor posted on a weebsite, in whole orr part.

23-5

Two T electrons are a unpaired in n [Mn(CN)6]3–. The complex iis low spin since the cyanide ligand is a stroong field lig gand. The splittting energy is greater than th he electron pairring energy. 23.7B

Pllan: Determinee the charge on n the cobalt ion n in the compleex ion and the nnumber of elecctrons in its d oorbitals. Siince it is an octtahedral ligand d, the d orbitalss split into threee lower energyy orbitals and ttwo higher eneergy orbitals. Check the spectrrochemical serries to see if F– is a strong or w weak field ligaand. If it is a sttrong field ligannd, fill the th hree lower enerrgy orbitals beffore placing an ny electrons in the higher eneergy d orbitals. If it is a weak field ligand, pllace one electro on in each of th he five d orbitaals, low energyy before high eenergy, before ppairing any eleectrons. After filling orbitals, count c the numb ber of unpaired d electrons. Thee complex ion is low-spin if the ligand is a strong field lig gand and high--spin if the ligaand is weak fieeld. So olution: Fiind the charge on cobalt: total ch harge = (chargee on Co) + 6(ccharge on F–) ch harge on Co = –3 – [6(–1)] = +3 Th he electron con nfiguration of Co C is [Ar]4s23d d7; the electronn configurationn of Co3+ is [Arr]3d6. – Th he ligand is F , which is a weeak field ligand d, so the six d eelectrons will eenter each of thhe 5 orbitals beefore pairing up p. Higheer energy d orb bitals

Lowerr energy d orbiitals Four electrons are unpaired in n [Co(F)6]3–. Th he complex is h high spin sincee the fluoride lligand is a weaak field lig gand. The splittting energy is smaller than th he electron paiiring energy.

CHEMICAL CONNEC CTIONS BOX XED READING G PROBLEM MS B23.1

a)) The central metal m ion in chlo orophyll is Mg g2+ (oxidation sstate = +2). Thhe central metaal ion in heme iis Fe2+ (o oxidation state = +2). The central metal ion n in vitamin B122 is Co3+ (oxidaation state = +3). b)) All of these compounds c hav ve the central metal m in the cennter of a large pplanar ring conntaining four N donor attoms.

B23.2

Zinc ions preferr a tetrahedral environment. e The T other ions llisted prefer otther environmeents (Ni2+ formss square 2+ 2+ pllanar complex ions while Fe and Mn form m octahedral ccomplex ions). If these ions aare placed in a ttetrahedral en nvironment in place p of the zin nc, they cannott function as w well as zinc sincce the enzyme catalyst wouldd have a diifferent shape.

END–OF– –CHAPTER PROBLEMS P 23.1

Th he n value of th he d sublevel is always one less l than the peeriod number.

23.2

a)) All transition elements in Peeriod 5 will hav ve a “base” co nfiguration of [Kr]5s2, and w will differ in thee number of d electrons (x) th hat the configu uration contains. Therefore, thhe general elecctron configuraation is 1ss22s22p63s23p64s 4 23d104p65s24dx. b)) A general eleectron configurration for Perio od 6 transition eelements incluudes f sublevel electrons, whicch are lower in n energy than th he d sublevel. The T configurattion is 1s22s22pp63s23p64s23d1044p65s24d105p66ss24f145dx.

Copyright © McGraw-Hill Education. E This is proprietary matterial solely for au uthorized instrucctor use. Not auth horized for sale orr distribution in any mann ner. This documen nt may not be cop pied, scanned, dup plicated, forward ded, distributed, oor posted on a weebsite, in whole orr part.

23-6

23.3

Electrons with the highest n value are removed first, so the ns electrons are removed before the (n – 1)d. V is [Ar]4s23d 3. V3+ is [Ar]3d 2, with the 4s electrons and one of the 3d electrons removed. The ionization energies and the magnetic properties are used to study the electron configurations of atoms.

23.4

The maximum number of unpaired d electrons is five since there are five d orbitals. An example of an atom with five unpaired d electrons is Mn with electron configuration [Ar]4s23d 5. An ion with five unpaired electrons is Mn2+ with electron configuration [Ar]3d 5.

23.5

For the main-group elements, size decreases as you move to the right. For the transition elements, the size decreases at first and then is fairly constant since inner orbitals are being filled.

23.6

a) One would expect that the elements would increase in size as they increase in mass from Period 5 to 6. Because there are fourteen inner transition elements in Period 6, the effective nuclear charge increases significantly. As effective charge increases, the atomic size decreases or “contracts.” This effect is significant enough that Zr4+ and Hf4+ are almost the same size but differ greatly in atomic mass. b) The size increases from Period 4 to 5, but stays fairly constant from Period 5 to 6. c) Atomic mass increases significantly from Period 5 to 6, but atomic radius (and thus volume) hardly increases, so Period 6 elements are very dense.

23.7

a) 1.3–1.9 b) 0.8–2.8 c) The range is smaller for the transition elements because the electrons are occupying d orbitals. Since these are inner orbitals, they shield (screen) the nuclear charge, making electrons shared between the transition metal and some bonded atom feel a fairly constant effective nuclear charge.

23.8

a) While transition elements commonly show multiple oxidation states, main-group elements show multiple oxidation states less frequently. Since the outermost s and d electrons in transition elements are so close in energy, all of these electrons can become involved in the bonding. b) The +2 oxidation state is common because the two ns electrons are more easily removed than the (n–1)d electrons. c) Valence-state electronegativity is the effective electronegativity of a metal ion. Since a metal in a high oxidation state is more positively charged than it is as a free metal, its attraction for electrons increases and therefore its electronegativity increases. The higher the oxidation state for a particular metal, the higher its electronegativity. The electronegativity of Cr increases with the oxidation state of Cr. So Cr2+ in CrO would have a lower electronegativity than Cr3+ in Cr2O3, which has a lower electronegativity than Cr6+ in CrO3.

23.9

a) A paramagnetic substance is attracted to a magnetic field, while a diamagnetic substance is slightly repelled by one. b) Ions of transition elements often have unfilled d orbitals whose unpaired electrons make the ions paramagnetic. Ions of main-group elements usually have a noble gas configuration with no partially filled levels. When orbitals are filled, electrons are paired and the ion is diamagnetic. c) The d orbitals in the transition element ions are not filled, which allows an electron from a lower energy d orbital to move to a higher energy d orbital. The energy required for this transition is relatively small and falls in the visible wavelength range. All orbitals are filled in a main-group element ion, so enough energy would have to be added to move an electron to a higher energy level, not just another orbital within the same energy level. This amount of energy is relatively large and outside the visible range of wavelengths.

23.10

23-7

b) Yttrium is in Period 5 and Group 3B(3). Electron configuration of Y is 1s22s22p63s23p64s23d104p65s24d1 or [Kr]5s24d1. c) Mercury is in Period 6 and Group 2B(12). Electron configuration of Hg is [Xe]6s24f145d10. 23.11

a) [Kr]5s24d6

23.12

Plan: The transition elements in Periods 4 and 5 have a general electron configuration of [noble gas] ns2 (n – 1)dx. Transition elements in Periods 6 and 7 have a general electron configuration of [noble gas] ns2 (n – 2)f14 (n – 1)dx. Solution: a) Osmium is in Period 6 and Group 8B(8). Electron configuration of Os is [Xe]6s24f 145d6. b) Cobalt is in Period 4 and Group 8B(9). Electron configuration of Co is [Ar]4s23d7. c) Silver is in Period 5 and Group 1B(11). Electron configuration of Ag is [Kr]5s14d10. Note that the filled d orbital is the preferred arrangement, so the configuration is not 5s24d9.

23.13

a) [Ar]4s23d10

23.14

Plan: Write the electron configuration of the atom and then remove the electrons as indicated by the charge of the metal ion. Transition metals lose their ns orbital electrons first in forming cations. After losing the ns electrons, additional electrons may be lost from the (n – 1)d orbitals. Solution: a) The two 4s electrons and one 3d electron are removed to form Sc3+: Sc: [Ar]4s23d1; Sc3+: [Ar] or 1s22s22p63s23p6. There are no unpaired electrons. b) The single 4s electron and one 3d electron are removed to form Cu2+: Cu: [Ar]4s13d10; Cu2+: [Ar]3d9. There is one unpaired electron. c) The two 4s electrons and one 3d electron are removed to form Fe3+: Fe: [Ar]4s23d6; Fe3+: [Ar]3d5. There are five unpaired electrons since each of the five d electrons occupies its own orbital. d) The two 5s electrons and one 4d electron are removed to form Nb3+: Nb: [Kr]5s24d 3; Nb3+: [Kr]4d2. There are two unpaired electrons.

23.15

a) [Ar]3d3; three unpaired electrons b) [Ar]; 0 unpaired electrons c) [Ar]3d6; four unpaired electrons d) [Xe]4f 145d 3; three unpaired electrons

23.16

Plan: For Groups 3B(3) to 7B(7), the highest oxidation state is equal to the group number. The highest oxidation state occurs when both ns electrons and all (n–1)d electrons have been removed. Solution: a) Tantalum, Ta, is in Group 5B(5), so the highest oxidation state is +5. The electron configuration of Ta is [Xe]6s24f145d3 with a total of five electrons in the 6s and 5d orbitals. b) Zirconium, Zr, is in Group 4B(4), so the highest oxidation state is +4. The electron configuration of Zr is [Kr]5s24d 2 with a total of four electrons in the 5s and 4d orbitals. c) Manganese, Mn, is in Group 7B(7), so the highest oxidation state is +7. The electron configuration of Mn is [Ar]4s23d 5 with a total of seven electrons in the 4s and 3d orbitals.

23.17

a) +5

23.18

Plan: For Groups 3B(3) to 7B(7), the highest oxidation state is equal to the group number. The highest oxidation state occurs when both ns electrons and all (n–1)d electrons have been removed. Solution: The elements in Group 6B(6) exhibit an oxidation state of +6. These elements have a total of six electrons in the outermost s orbital and d orbital: ns1(n–1)d5. These elements include Cr, Mo, and W. Sg (Seaborgium) is also

b) +3

b) [Ar]4s13d10

b) [Ar]4s23d5

c) [Ar]4s23d8

c) [Xe]6s24f 145d5

c) +7

23-8

in Group 6B(6), but its lifetime is so short that chemical properties, like oxidation states within compounds, are impossible to measure. 23.19

Ti, Zr, and Hf

23.20

Plan: Transition elements in their lower oxidation states act more like metals. Solution: The oxidation state of chromium in CrF2 is +2 and in CrF6 is +6 (use –1 oxidation state of fluorine to find oxidation state of Cr). CrF2 exhibits greater metallic behavior than CrF6 because the chromium is in a lower oxidation state in CrF2 than in CrF6.

23.21

Due to its lower oxidation state of +3, VF3 would have greater ionic character, and would be more likely to have higher melting and boiling points than VF5 in which V has an oxidation state of +5.

23.22

While atomic size increases slightly down a group of transition elements, the nuclear charge increases much more, so the first ionization energy generally increases. The reduction potential for Mo is lower, so it is more difficult to oxidize Mo than Cr. In addition, the ionization energy of Mo is higher than that of Cr, so it is more difficult to remove electrons from, i.e., oxidize, Mo.

23.23

Higher oxidation states are more stable for lower transition metals like Re, so ReO4– would be less likely to reduce. MnO4– is the stronger oxidizing agent.

23.24

Plan: Oxides of transition metals become less basic (or more acidic) as oxidation state increases. Solution: The oxidation state of chromium in CrO3 is +6 and in CrO is +2, based on the –2 oxidation state of oxygen. The oxide of the higher oxidation state, CrO3, produces a more acidic solution.

23.25

Mn2O3 is more basic, because lower oxidation states form more basic oxides.

23.26

The abnormally small size of Au atoms (due to the lanthanide contraction) means that its electrons are tightly held, giving it low reactivity. The Group 1A(1) elements do not show the lanthanide contraction.

23.27

The “last electron in” is in an f orbital, which is very close to the nucleus. This makes all of the lanthanides very similar from the “outside.”

23.28

a) The f block contains seven orbitals, so if one electron occupied each orbital, a maximum of seven electrons would be unpaired. b) The maximum number of unpaired electrons corresponds to a half-filled f subshell.

23.29

All actinides are radioactive.

23.30

Plan: The inner transition elements have a general electron configuration of [noble gas] ns2 (n – 2)f x (n – 1)d 0. Inner transition metals lose their ns orbital electrons first in forming cations. After losing the ns electrons, additional electrons may be lost from the (n – 1)d orbitals. Solution: a) Lanthanum is a transition element in Period 6 with atomic number 57. La: [Xe]6s25d1 b) Cerium is in the lanthanide series in Period 6 with atomic number 58. Ce: [Xe]6s24f15d1, so Ce3+: [Xe]4f 1. Note that cerium is one of the three lanthanide elements that have one electron in a 5d orbital. c) Einsteinium is in the actinide series in Period 7 with atomic number 99. Es: [Rn]7s25f 11 d) Uranium is in the actinide series in Period 7 with atomic number 92. U: [Rn]7s25f 36d 1. Removing four electrons gives U4+ with configuration [Rn]5f 2.

23-9

23.31

a) [Xe]6s24f 5

b) [Xe]4f 14

23.32

Plan: Write the electron configuration of the atom and then remove electrons as indicated by the charge of the metal ion. Electrons are removed first from the 6s orbital and then from the 4f orbital. Solution: a) Europium is in the lanthanide series with atomic number 63. The configuration of Eu is [Xe]6s24f 7. The stability of the half-filled f sublevel explains why the configuration is not [Xe]6s24f 65d1. The two 6s electrons are removed to form the Eu2+ ion, followed by electron removal in the f block to form the other two ions: Eu2+: [Xe]4f 7 Eu3+: [Xe]4f 6 Eu4+: [Xe]4f 5 The stability of the half-filled f sublevel makes Eu2+ most stable. b) Terbium is in the lanthanide series with atomic number 65. The configuration of Tb is [Xe]6s24f 9. The two 6s electrons are removed to form the Tb2+ ion, followed by electron removal in the f block to form the other two ions: Tb2+: [Xe]4f 9 Tb3+: [Xe]4f 8 Tb4+: [Xe]4f 7 Tb would demonstrate a +4 oxidation state because it has the half-filled sublevel.

23.33

a) Ce2+: [Xe]5d14f 1 Ce3+: [Xe]4f 1 Ce4+: [Xe] 2+ 14 3+ 13 b) Yb : [Xe]4f Yb : [Xe]4f Yb4+: [Xe]4f 12 4+ 2+ c) Ce has a noble gas configuration and Yb has a filled f subshell.

23.34

The lanthanide element gadolinium, Gd, has electron configuration [Xe]6s24f 75d1 with eight unpaired electrons. The ion Gd3+ has seven unpaired electrons: [Xe]4f 7.

23.35

A complex ion forms as a result of a metal ion acting as a Lewis acid by accepting one or more pairs of electrons from the ligand(s), which act(s) as a Lewis base. If the ligands are neutral (or, if negative, there are too few to neutralize the positive charge on the metal), the complex ion will be positive. If there are more than enough negative ligands to cancel the charge on the metal, the complex ion will be negative. The complex ion, either positive or negative, will associate with enough ions of opposite charge to form a neutral coordination compound.

23.36

Donor atoms must have a pair of electrons which can be donated to another atom.

23.37

The coordination number indicates the number of ligand atoms bonded to the central metal ion. The oxidation number represents the number of electrons lost to form the ion. The coordination number is unrelated to the oxidation number.

23.38

Chelates contain two or more donor atoms which can simultaneously bond to a single metal ion.

23.39

Coordination number of two indicates linear geometry. Coordination number of four indicates either tetrahedral or square planar geometry. Coordination number of six indicates octahedral geometry.

23.40

Co(III): 6

23.41

The metal ion acts as a Lewis acid, accepting one or more electron pairs from the ligand(s), which are acting as Lewis bases.

23.42

The -ate ending signifies that the complex ion has a negative charge.

Pt(II): 4

c) [Rn]7s26d 2

d) [Rn]5f 11

Pt(IV): 6

23-10

23.43

The ligands are listed first (alphabetically), followed by the metal.

23.44

A linkage isomer is a constitutional isomer, since the atoms are connected in a different pattern.

23.45

Plan: Use the naming rules for coordination compounds given in the text. Solution: a) The oxidation state of nickel is found from the total charge on the ion (+2 because two Cl– charges equals –2) and the charge on ligands: charge on nickel = +2 – 6(0 charge on water) = +2 Name nickel as nickel(II) to indicate oxidation state. Ligands are six (hexa-) waters (aqua). Put together with chloride anions to give hexaaquanickel(II) chloride. b) The cation is [Cr(en)3]n+ and the anion is ClO4–, the perchlorate ion. The charge on the cation is +3 to make a neutral salt in combination with the –3 charge of the three perchlorate ions. The ligand is ethylenediamine, which has 0 charge. The charge of the cation equals the charge on chromium ion, so chromium(III) is included in the name. The three ethylenediamine ligands, abbreviated en, are indicated by the prefix tris- because the name of the ligand includes a numerical indicator, di-. The complete name is tris(ethylenediamine)chromium(III) perchlorate. c) The cation is K+ and the anion is [Mn(CN)6]4–. The charge of 4– is deduced from the four potassium +1 ions in the formula. The oxidation state of Mn is –4 – {6(–1)} = +2. The name of CN– ligand is cyano and six ligands are represented by the prefix hexa-. The name of manganese anion is manganate(II). The -ate suffix on the complex ion is used to indicate that it is an anion. The full name of compound is potassium hexacyanomanganate(II).

23.46

a) tetraamminedinitrocobalt(III) chloride b) hexaamminechromium(III) hexacyanochromate(III) c) potassium tetrachlorocuprate(II)

23.47

Plan: The charge of the central metal atom was determined in Problem 23.45 because the Roman numeral indicating oxidation state is part of the name. The coordination number, or number of ligand atoms bonded to the metal ion, is found by examining the bonded entities inside the square brackets to determine if they are unidentate, bidentate, or polydentate. Solution: a) The Roman numeral “II” indicates a +2 oxidation state. There are six water molecules bonded to Ni and each ligand is unidentate, so the coordination number is 6. b) The Roman numeral “III” indicates a +3 oxidation state. There are three ethylenediamine molecules bonded to Cr, but each ethylenediamine molecule contains two donor N atoms (bidentate). Therefore, the coordination number is 6. c) The Roman numeral “II” indicates a +2 oxidation state. There are six unidentate cyano molecules bonded to Mn, so the coordination number is 6.

23.48

a) +3, 6

23.49

Plan: Use the naming rules for coordination compounds given in the text. Solution: a) The cation is K+, potassium. The anion is [Ag(CN)2]− with the name dicyanoargentate(I) ion for the two cyanide ligands and the name of silver in an anion, argentate(I). The Roman numeral (I) indicates the oxidation number on Ag. O.N. for Ag = –1 – {2(–1)} = +1 since the complex ion has a charge of –1 and the cyanide ligands are also –1. The complete name is potassium dicyanoargentate(I). b) The cation is Na+, sodium. Since there are two +1 sodium ions, the anion is [CdCl4]2– with a charge of 2–. The anion is the tetrachlorocadmate(II) ion. With four –1 chloride ligands, the oxidation state of cadmium is +2 and the name of cadmium in an anion is cadmate. The complete name is sodium tetrachlorocadmate(II). c) The cation is [Co(NH3)4(H2O)Br]2+. The 2+ charge is deduced from the two Br– ions. The cation has the name tetraamineaquabromocobalt(III) ion, with four ammonia ligands (tetraammine), one water ligand (aqua) and one

b) +3, 6; +3, 6

c) +2, 4

23-11

bromide ligand (bromo). The oxidation state of cobalt is +3: 2 – {4(0) + 1(0) + 1(–1)}. The oxidation state is indicated by (III), following cobalt in the name. The anion is Br–, bromide. The complete name is tetraammineaquabromocobalt(III) bromide. 23.50

a) potassium amminepentachloroplatinate(IV) b) diammine(ethylenediamine)copper(II) tetrachloro(ethylenediamine)cobaltate(II) c) dibromobis(ethylenediamine)platinum(IV) perchlorate

23.51

Plan: The charge of the central metal atom was determined in Problem 23.49 because the Roman numeral indicating oxidation state is part of the name. The coordination number, or number of ligand atoms bonded to the metal ion, is found by examining the bonded entities inside the square brackets to determine if they are unidentate, bidentate, or polydentate. Solution: a) The counter ion is K+, so the complex ion is [Ag(CN)2]−. Each cyano ligand has a –1 charge, so silver has a +1 charge: +1 + 2(–1) = –1. Each cyano ligand is unidentate, so the coordination number is 2. b) The counter ion is Na+, so the complex ion is [CdCl4]2−. Each chloride ligand has a –1 charge, so Cd has a +2 charge: +2 + 4(–1) = –2. Each chloride ligand is unidentate, so the coordination number is 4. c) The counter ion is Br−, so the complex ion is [Co(NH3)4(H2O)Br]2+. Both the ammine and aqua ligands are neutral. The bromide ligand has a –1 charge, so Co has a +3 charge: +3 + 4(0) + 0 + (–1) = +2. Each ligand is unidentate, so the coordination number is 6.

23.52

a) +4, 6

23.53

Plan: Use the rules given in the chapter. Solution: a) The cation is tetramminezinc ion. The tetraammine indicates four NH3 ligands. Zinc has an oxidation state of +2, so the charge on the cation is +2. The anion is SO42–. Only one sulfate is needed to make a neutral salt. The formula of the compound is [Zn(NH3)4]SO4. b) The cation is pentaamminechlorochromium(III) ion. The ligands are five NH3 from pentaammine, and one chloride from chloro. The chromium ion has a charge of +3, so the complex ion has a charge equal to +3 from chromium, plus 0 from ammonia, plus –1 from chloride for a total of +2. The anion is chloride, Cl–. Two chloride ions are needed to make a neutral salt. The formula of compound is [Cr(NH3)5Cl]Cl2. c) The anion is bis(thiosulfato)argentate(I). Argentate(I) indicates silver in the +1 oxidation state, and bis(thiosulfato) indicates two thiosulfate ligands, S2O32–. The total charge on the anion is +1 plus 2(–2) to equal –3. The cation is sodium, Na+. Three sodium ions are needed to make a neutral salt. The formula of compound is Na3[Ag(S2O3)2].

23.54

a) [Co(en)2 Br2]2SO4

23.55

Plan: The coordination number, or number of ligand atoms bonded to the metal ion, is found by examining the bonded entities inside the square brackets to determine if they are unidentate, bidentate, or polydentate. Coordination compounds act like electrolytes, i.e., they dissolve in water to yield charged species, the counter ion, and the complex ion. However, the complex ion itself does not dissociate. The “number of individual ions per formula unit” refers to the number of ions that would form per coordination compound upon dissolution in water. Solution: a) The counter ion is SO42–, so the complex ion is [Zn(NH3)4]2+. Each ammine ligand is unidentate, so the coordination number is 4. Each molecule dissolves in water to form one SO42– ion and one [Zn(NH3)4]2+ ion, so two ions form per formula unit. b) The counter ion is Cl–, so the complex ion is [Cr(NH3)5Cl]2+. Each ligand is unidentate, so the coordination number is 6. Each molecule dissolves in water to form two Cl– ions and one [Cr(NH3)5Cl]2+ ion, so three ions form per formula unit.

b) +2, 4; +2, 6

c) +4, 6

b) [Cr(NH3)6]2[CuCl4 ]3

c) K4[Fe(CN)6]

23-12

c) The counter ion is Na+, so the complex ion is [Ag(S2O3)2]3–. Assuming that the thiosulfate ligand is unidentate, the coordination number is 2. Each molecule dissolves in water to form three Na+ ions and one [Ag(S2O3)2]3– ion, so four ions form per formula unit. 23.56

a) 6; 3 ions

23.57

Plan: Follow the naming rules given in the chapter. Solution: a) The cation is hexaaquachromium(III) with the formula [Cr(H2O)6]3+. The total charge of the ion equals the charge on chromium because water is a neutral ligand. Six water ligands are indicated by a hexa prefix to aqua. The anion is SO42–. To make the neutral salt requires three sulfate ions for two cations. The compound formula is [Cr(H2O)6]2(SO4)3. b) The anion is tetrabromoferrate(III) with the formula [FeBr4] –. The total charge on the ion equals +3 charge on iron plus 4 x –1 charge on each bromide ligand for –1 overall. The cation is barium, Ba2+. Two anions are needed for each barium ion to make a neutral salt. The compound formula is Ba[FeBr4]2. c) The anion is bis(ethylenediamine)platinum(II) ion. Charge on platinum is +2 and this equals the total charge on the complex ion because ethylenediamine is a neutral ligand. The bis preceding ethylenediamine indicates two ethylenediamine ligands, which are represented by the abbreviation “en.” The cation is [Pt(en)2]2+. The anion is CO32–. One carbonate combines with one cation to produce a neutral salt. The compound formula is [Pt(en)2]CO3.

23.58

a) K3[Cr(C2O4)3]

23.59

Plan: The coordination number, or number of ligand atoms bonded to the metal ion, is found by examining the bonded entities inside the square brackets to determine if they are unidentate, bidentate, or polydentate. Coordination compounds act like electrolytes, i.e., they dissolve in water to yield charged species, the counter ions, and the complex ion. However, the complex ion itself does not dissociate. The “number of individual ions per formula unit” refers to the number of ions that would form per coordination compound upon dissolution in water. Solution: a) The counter ion is SO42–, so the complex ion is [Cr(H2O)6]3+. Each aqua ligand is unidentate, so the coordination number is 6. Each molecule dissolves in water to form three SO42– ions and two [Cr(H2O)6]3+ ions, so five ions form per formula unit. b) The counter ion is Ba2+, so the complex ion is [FeBr4] –. Each bromo ligand is unidentate, so the coordination number is 4. Each molecule dissolves in water to form one Ba2+ ion and two [FeBr4] – ions, so three ions form per formula unit. c) The counter ion is CO32–, so the complex ion is [Pt(en)2]2+. Each ethylenediamine ligand is bidentate, so the coordination number is 4. Each molecule dissolves in water to form one CO32– ion and one [Pt(en)2]2+ ion, so two ions form per formula unit.

23.60

a) 6; 4 ions

23.61

Plan: Ligands that form linkage isomers have two different possible donor atoms. Solution: a) The nitrite ion forms linkage isomers because it can bind to the metal ion through the lone pair on the N atom or any lone pair on either O atom. Resonance Lewis structures are: O

b) 6 (Cr) and 4(Cu); 5 ions

c) 6; 5 ions

b) [Co(en)3]4[Mn(CN)5I]3 c) [Al(H2O)2(NH3 )2BrCl]NO3

b) 6 and 6; 7 ions

c) 6; 2 ions

b) Sulfur dioxide molecules form linkage isomers because the lone pair on the S atom or any lone pair on either O atom can bind the central metal ion.

23-13

c) Nitrate ions have an N atom with no lone pair and three O atoms, all with lone pairs than can bond to the metal ion. But all of the O atoms are equivalent so these ions do not form linkage isomers.

23.62

a) The thiocyanate ion can form linkage isomers because both the sulfur and nitrogen ions have lone pairs.

Only one of the resonance forms is shown. b) The thiosulfate ion can form linkage isomers because either one of the oxygen atoms or the non-central sulfur may serve as a donor atom.

c) The hydrogen sulfide ion cannot form linkage isomers, because the hydrogen does not have a lone pair.

23.63

Plan: Types of isomers for coordination compounds are coordination isomers with different arrangements of ligands and counter ions, linkage isomers with different donor atoms from the same ligand bound to the metal ion, geometric isomers with differences in ligand arrangement relative to other ligands, and optical isomers with mirror images that are not superimposable. Solution: a) Platinum ion, Pt2+, is a d 8 ion so the ligand arrangement is square planar. A cis and trans geometric isomer exist for this complex ion: H H H H CH3

CH3

Pt CH3 H

N H

Br Pt

N CH3

H H

cis isomer trans isomer No optical isomers exist because the mirror images of both compounds are superimposable on the original molecules. In general, a square planar molecule is superimposable on its mirror image.

23-14

b) A cis and trans geometric isomer exist for this complex ion. No optical isomers exist because the mirror images of both compounds are superimposable on the original molecules. H H H H

Pt H

F Pt

H H H cis isomer trans isomer c) Three geometric isomers exist for this molecule, although they are not named cis or trans because all the ligands are different. A different naming system is used to indicate the relation of one ligand to another. H H H H H H H

Pt H

F Pt

H 23.64

a) no isomers b) optical isomers:

H H

Zn Cl

Zn F

c) geometric isomers

23.65

23-15

Solution: a) Platinum ion, Pt2+, is a d8 ion, so the ligand arrangement is square planar. The ligands are two Cl– and two Br–, so the arrangement can be either both ligands trans or both ligands cis to form geometric isomers.

b) The complex ion can form linkage isomers with the NO2 ligand. Either the N or an O may be the donor.

c) In the octahedral arrangement, the two iodide ligands can be either trans to each other, 180° apart, or cis to each other, 90° apart.

23.66

a) Coordination isomers:

23-16

b) Coordination isomers:

H H

CH3

N H

CH3

Pt H

H H

N H

CH3

H Cl

Pt H

CH3

c) Geometric isomers:

2+ NH3 H2O Fe H2O NH3

2+ NH3

OH2 OH2

trans

H2O Fe H2O OH2

NH3 OH2

cis

23.67

The traditional formula does not correctly indicate that at least some of the chloride ions serve as counter ions. The NH3 molecules serve as ligands, so n could equal 6 to satisfy the coordination number requirement. This complex has the formula [Cr(NH3)6]Cl3. This is a correct formula because the name “chromium(III)” means that the chromium has a +3 charge, and this balances with the –3 charge provided by the three chloride counter ions. However, when this compound dissociates in water, it produces four ions ([Cr(NH3)6]3+ and three Cl–, whereas NaCl only produces two ions. Other possible compounds are n = 5, CrCl3∙5NH3, with formula [Cr(NH3)5Cl]Cl2; n = 4, CrCl3∙4NH3, with formula [Cr(NH3)4Cl2]Cl; n = 3, CrCl3∙3NH3, with formula [Cr(NH3)3Cl3]. The compound [Cr(NH3)4Cl2]Cl has a coordination number equal to 6 and produces two ions when dissociated in water, so it has an electrical conductivity similar to an equimolar solution of NaCl.

23.68

4 (corresponding to [M(NH3)2Cl2]Cl2

23.69

Plan: First find the charge on the palladium ion, then arrange ligands and counter ions to form the complex. Solution: a) Charge on palladium = – [(+1 on K+) + (0 on NH3) + 3(–1 on Cl–)] = +2 Palladium(II) forms four-coordinate complexes. The four ligands in the formula are one NH3 and three Cl– ions. The formula of the complex ion is [Pd(NH3)Cl3] –. Combined with the potassium cation, the compound formula is K[Pd(NH3)Cl3]. b) Charge on palladium = – [2(–1 on Cl–) + 2(0 on NH3)] = +2 Palladium(II) forms four-coordinate complexes. The four ligands are two chloride ions and two ammonia molecules. The formula is [PdCl2(NH3)2]. c) Charge on palladium = – [2(+1 on K+) + 6(–1 on Cl–)] = +4 Palladium(IV) forms six-coordinate complexes. The six ligands are the six chloride ions. The formula is K2[PdCl6]. d) Charge on palladium = – [4(0 on NH3) + 4(–1 on Cl–)] = +4

23-17

Palladium(IV) forms six-coordinate complexes. The ammonia molecules have to be ligands. The other two ligand bonds are formed with two of the chloride ions. The remaining two chloride ions are the counter ions. The formula is [Pd(NH3)4Cl2]Cl2. 23.70

a) A coordinate covalent bond is a bond formed when both electrons came from one atom. b) Yes. H2O molecules act as donors to Fe3+, forming Fe(H2O)63+. c) Yes. H2O molecules act as donors to H +, forming H3O+.

23.71

a) Four empty orbitals of equal energy are “created” to receive the donated electron pairs from four ligands. The four orbitals are hybridized from an s, two p, and one d orbital from the previous n level to form four dsp2 orbitals. b) One s and three p orbitals become four sp3 hybrid orbitals.

23.72

6; octahedral

23.73

red

23.74

Absorption of orange or yellow light gives a blue solution.

23.75

a) The crystal field splitting energy is the energy difference between the two sets of d orbitals that result from the bonding of ligands to a central transition metal atom. b) In an octahedral field of ligands, the ligands approach along the x, y, and z axes. The dx2 −y2 and dz2 orbitals are located along the x, y, and z axes, so ligand interaction is higher in energy than the other orbital-ligand interactions. The other orbital-ligand interactions are lower in energy because the dxy, dyz, and dxz orbitals are located between the x, y, and z axes. c) In a tetrahedral field of ligands, the ligands do not approach along the x, y, and z axes. The ligand interaction is greater for the dxy, dyz, and dxz orbitals and lesser for the dx2 −y2 and dz2 orbitals. The crystal field splitting is reversed, and the dxy, dyz, and dxz orbitals are higher in energy than the dx2 −y2 and dz2 orbitals.

23.76

A strong-field ligand causes a greater crystal field splitting (a greater Δ) than a weak-field ligand. CN – is a strong-field ligand, and F – is a weak-field ligand.

23.77

high-spin

23.78

If Δ is greater than Epairing, electrons will preferentially pair spins in the lower energy d orbitals before adding as unpaired electrons to the higher energy d orbitals. If Δ is less than Epairing, electrons will preferentially add as unpaired electrons to the higher d orbitals before pairing spins in the lower energy d orbitals. The first case gives a complex that is low-spin and less paramagnetic than the high-spin complex formed in the latter case.

23.79

The Δ values for tetrahedral complexes are smaller than for octahedral, so Δ is always < Epairing and electrons will always preferentially add as unpaired electrons to the higher d orbitals.

23.80

23-18

b) Electron configuration of Au: [Xe]6s14f145d10 Charge on Au: The complex ion has a –1 charge ([AuCl4]–) since K has a +1 charge. Each chloride ligand has a –1 charge, so Au has a +3 charge {+3 + 4(–1)} = 1– ion. The 6s electron and two d electrons are removed. Electron configuration of Au3+: [Xe]4f145d8 Au3+ has eight d electrons. c) Electron configuration of Rh: [Kr]5s24d 7 Charge on Rh: Each chloride ligand has a –1 charge, so Rh has a +3 charge {+3 + 6(–1)} = 3– ion. The 5s electrons and one 4d electron are removed. Electron configuration of Rh3+: [Kr]4d6 Rh3+ has six d electrons. 23.81

a) 4 d electrons

23.82

Plan: To determine the number of d electrons in a central metal ion, first write the electron configuration for the metal atom. Examine the formula of the complex to determine the charge on the central metal ion, and then write the ion’s configuration by removing the correct number of electrons, beginning with the ns electrons and then the (n – 1)d electrons. Solution: a) [(+2 on Ca2+) + 6(–1 on F–) + Ir = 0] Charge on iridium = – [(+2 on Ca2+) + 6(–1 on F–)] = +4 Configuration of Ir is [Xe]6s24f145d 7. Configuration of Ir4+ is [Xe]4f 145d 5. Five d electrons in Ir4+. b) [Hg + 4(–1 on I–)] = –2 Charge on mercury = – [4(–1 on I–)] – 2 = +2 Configuration of Hg is [Xe]6s24f145d10. Configuration of Hg2+ is [Xe]4f145d10. Ten d electrons in Hg2+. c) [Co + (–4 on EDTA)] = – 2 Charge on cobalt = – [–4 on EDTA] – 2 = +2 Configuration of Co is [Ar]4s23d 7. Configuration of Co2+ is [Ar]3d 7. Seven d electrons in Co2+.

23.83

a) 5

b) 4

b) 3 d electrons

c) 6 d electrons

c) 6

23.84

23-19

The unequal shading of the lobes of the d orbitals is a consequence of the quantum mechanical derivation of these orbitals, and does not affect the current discussion. In an octahedral field of ligands, the ligands approach along the x, y, and z axes. The dx2 − y2 orbital is located along the x and y axes, so ligand interaction is greater. The dxy orbital is offset from the x and y axes by 90°, so ligand interaction is less. The greater interaction of the dx2 − y2 orbital results in its higher energy. 23.85

The unequal shading of the lobes of the d orbitals is a consequence of the quantum mechanical derivation of these orbitals, and does not affect the current discussion. In an octahedral complex, the ligands “point” directly at the electrons in both of these orbitals. In a square planar complex, there are no ligands along the z-axis, so the dz2 is a very favorable (i.e., low-energy) place for the electrons. 23.86

Plan: Determine the electron configuration of the ion, which gives the number of d electrons. If there are only 1, 2, or 3 d electrons, the complex is always high-spin since there are not enough electrons to pair. If there are 8 or 9 d electrons, the complex is always high-spin since the higher energy d orbitals will always contain two (d 8) or one (d 9) unpaired electrons. Solution: a) Ti: [Ar]4s23d2. The electron configuration of Ti3+ is [Ar]3d1. With only one electron in the d orbitals, the titanium(III) ion cannot form high- and low-spin complexes—all complexes will contain one unpaired electron and have the same spin. b) Co: [Ar]4s23d7. The electron configuration of Co2+ is [Ar]3d7 and will form high- and low-spin complexes with seven electrons in the d orbital. c) Fe: [Ar]4s23d6. The electron configuration of Fe2+ is [Ar]3d6 and will form high- and low-spin complexes with six electrons in the d orbital. d) Cu: [Ar]4s13d10. The electron configuration of Cu2+ is [Ar]3d 9, so in complexes with both strong- and weakfield ligands, one electron will be unpaired and the spin in both types of complexes is identical. Cu2+ cannot form high- and low-spin complexes.

23-20

high-spin

low-spin

high-spin

low-spin

23.87

b and d

23.88

Plan: To draw the orbital-energy splitting diagram, first determine the number of d electrons in the transition metal ion. Examine the formula of the complex ion to determine the electron configuration of the metal ion, remembering that the ns electrons are lost first. Determine the coordination number from the number of ligands, recognizing that six ligands result in an octahedral arrangement and four ligands result in a tetrahedral or square planar arrangement. Weak-field ligands give the maximum number of unpaired electrons (high-spin) while strong-field ligands lead to electron pairing (low-spin). Solution: a) Electron configuration of Cr: [Ar]4s13d 5 Charge on Cr: The aqua ligands are neutral, so the charge on Cr is +3. Electron configuration of Cr3+: [Ar]3d 3 Six ligands indicate an octahedral arrangement. Using Hund’s rule, fill the lower energy t2g orbitals first, filling empty orbitals before pairing electrons within an orbital. b) Electron configuration of Cu: [Ar]4s13d10 Charge on Cu: The aqua ligands are neutral, so Cu has a +2 charge. Electron configuration of Cu2+: [Ar]3d 9 Four ligands and a d 9 configuration indicate a square planar geometry (only filled d sublevel ions exhibit tetrahedral geometry). Use Hund’s rule to fill in the nine d electrons. Therefore, the correct orbital-energy splitting diagram shows one unpaired electron. c) Electron configuration of Fe: [Ar]4s23d 6 Charge on Fe: Each fluoride ligand has a –1 charge for a total charge of –6, so Fe has a +3 charge to make the overall complex charge equal to –3. Electron configuration of Fe3+: [Ar]3d 5 Six ligands indicate an octahedral arrangement. Use Hund’s rule to fill the orbitals. F– is a weak-field ligand, so the splitting energy, Δ, is not large enough to overcome the resistance to electron pairing. The electrons remain unpaired, and the complex is called high-spin.

23-21

(b)

(a)

(c)

23.89

23.90

Plan: To draw the orbital-energy splitting diagram, first determine the number of d electrons in the transition metal ion. Examine the formula of the complex ion to determine the electron configuration of the metal ion, remembering that the ns electrons are lost first. Determine the coordination number from the number of ligands, recognizing that six ligands result in an octahedral arrangement and four ligands result in a tetrahedral or square planar arrangement. Weak-field ligands give the maximum number of unpaired electrons (high-spin) while strong-field ligands lead to electron pairing (low-spin). Solution: a) Electron configuration of Mo: [Kr]5s14d 5 Charge on Mo: Each chloride ligand has a –1 charge for a total charge of –6, so Mo has a +3 charge to make the overall complex charge equal to –3. Electron configuration of Mo3+: [Kr]4d 3 Six ligands indicate an octahedral arrangement. Using Hund’s rule, fill the lower energy t2g orbitals first, filling empty orbitals before pairing electrons within an orbital. b) Electron configuration of Ni: [Ar]4s23d 8 Charge on Ni: The aqua ligands are neutral, so the charge on Ni is +2. Electron configuration of Ni2+: [Ar]3d 8 Six ligands indicate an octahedral arrangement. Use Hund’s rule to fill the orbitals. H2O is a weak-field ligand, so the splitting energy, Δ, is not large enough to overcome the resistance to electron pairing. One electron occupies each of the five d orbitals before pairing in the t2g orbitals, and the complex is called high-spin. c) Electron configuration of Ni: [Ar]4s23d 8 Charge on Ni: Each cyanide ligand has a –1 charge for a total charge of –4, so Ni has a +2 charge to make the overall complex charge equal to –2. Electron configuration of Ni2+: [Ar]3d 8 The coordination number is 4 and the complex is square planar. The complex is low-spin because CN– is a strong-field ligand. Electrons pair in one set of orbitals before occupying orbitals of higher energy.

23-22

23.91

23.92

Plan: The spectrochemical series describes the spectrum of splitting energy, Δ. The greater the crystal field strength of the ligand, the greater the crystal field splitting energy, Δ, and the greater the energy of light absorbed. Solution: NO2– is a stronger field ligand than NH3, which is a stronger field ligand than H2O. NO2– produces the largest Δ, followed by NH3, and then H2O with the smallest Δ value. The energy of light absorbed increases as Δ increases since more energy is required to excite an electron from a lower energy orbital to a higher energy orbital when Δ is very large. [Cr(H2O)6]3+ < [Cr(NH3)6]3+ < [Cr(NO2)6]3–

23.93

[Cr(CN)6]3– > [Cr(en)3]3+ > [CrCl6]3–

23.94

Plan: A weaker field ligand will result in a smaller Δ in the complex and lower energy light absorbed. When a particular color of light is absorbed, the complementary color is seen. Solution: A violet complex absorbs yellow-green light. The light absorbed by a complex with a weaker ligand would be at a lower energy and longer wavelength. Light of lower energy than yellow-green light is yellow, orange, or red light. The color observed would be blue or green.

23.95

A violet complex is absorbing yellow-green light, and a green complex is absorbing red light. The green complex is absorbing lower energy light, so it must contain a weaker ligand than H2O. Thus, L could not be CN – (which is stronger than H2O), but could be Cl–.

23.96

In an octahedral d 8 complex, two electrons occupy the two eg orbitals and will be unpaired. In a square planar d 8 complex, the highest energy (dx2 – y2) orbital is unoccupied and all other levels are full, making the complex diamagnetic.

23.97

Plan: A weaker field ligand will result in a smaller Δ in the complex and lower energy light absorbed. When a particular color of light is absorbed, the complementary color is seen. Solution: The aqua ligand is weaker than the ammine ligand. The weaker ligand results in a lower splitting energy and absorbs a lower energy of visible light. The green hexaaqua complex appears green because it absorbs red light

23-23

(opposite side of the color wheel). The hexaammine complex appears violet because it absorbs yellow light, which is higher in energy (shorter λ) than red light. 23.98

NH3 > H2O > F – in ligand field strength, so [Co(NH3)6]3+ would absorb light of highest energy (shortest wavelength), followed by [Co(H2O)6]3+, and then [CoF6]3–. Yellow-orange, green, and blue complexes would absorb blue-green, red, and orange light, respectively. Thus, [Co(NH3)6]3+ absorbs blue-green and is yelloworange in color, [Co(H2O)6]3+ absorbs orange light and is blue in color, and [CoF6]3– absorbs red light and is green in color.

23.99

a) The outermost electrons in these elements are in f orbitals. b) If Np and Pu were in Groups 7B(7) and 8B(8), their maximum oxidation state would be +7 and +8, respectively (analogous to Re and Os). Since the highest fluoride of Np and of Pu is only the hexafluoride, the highest oxidation state of Np apparently is +6 and not +7 as it would be in Group 7B(7) and the highest oxidation state of Pu is apparently +6 and not +8 as it would be in Group 8B(8).

23.100 There are ten general formulas. Assuming a minimum of one of each ligand in each complex about the metal M they are: MA3BCD, MAB3CD, MABC3D, MABCD3, MA2B2CD, MA2BC2D, MA2BCD2, MAB2C2D, MAB2CD2, and MABC2D2. Neglecting optical isomers, the structures for MA3BCD are: A A A A D C D D A M C M M A D A B M A C A B C B B A A A The first of these has an optical isomer. Analogous structures arise for MAB3CD, MABC3D, and MABCD3. Each of the formulas from the following group has five isomers: MA3BCD, MAB3CD, MABC3D, and MABCD3. Neglecting optical isomers, the structures for MA2B2CD are: C B B B A D A B M M B C M A C A A A D D B

D B A

A B

C There are no optical isomers. Analogous structures arise for MA2BC2D, MA2BCD2, MAB2C2D, MAB2CD2, and MABC2D2. Each of the formulas from the following group has four isomers: MA2B2CD, MA2BC2D, MA2BCD2, MAB2C2D, MAB2CD2, and MABC2D2. 23.101 The electron configuration of Hg is [Xe]6s24f145d10 and that of Hg+ is [Xe]6s14f145d10. The electron configuration of Cu is [Ar]4s13d10 and that of Cu+ is [Ar]3d10. In the mercury(I) ion, there is one electron in the 6s orbital that can form a covalent bond with the electron in the 6s orbital of another Hg+ ion. In the copper(I) ion, there are no electrons in the s orbital to bond with another copper(I) ion. 23.102 Plan: The coordination number, or number of ligand atoms bonded to the metal ion, is found by examining the bonded entities inside the square brackets to determine if they are unidentate, bidentate, or polydentate. The

23-24

oxidation of the central metal ion is found by determining the charges of the ligands and insuring that the charges of the metal ion, ligands, and counter ion add to zero. Coordination compounds act like electrolytes, i.e., they dissolve in water to yield charged species, the counter ions and the complex ion. However, the complex ion itself does not dissociate. The “number of individual ions per formula unit” refers to the number of ions that would form per coordination compound upon dissolution in water. Solution: a) The coordination number of cobalt is 6. The two Cl– ligands are unidentate and the two ethylenediamine ligands are bidentate (each en ligand forms two bonds to the metal), so a total of six ligand atoms are connected to the central metal ion. b) The counter ion is Cl–, so the complex ion is [Co(en)2Cl2]+. Each chloride ligand has a –1 charge and each en ligand is neutral, so cobalt has a +3 charge: +3 + 2(0) + 2(−1) = +1. c) One mole of complex dissolves in water to yield one mole of [Co(en)2Cl2]+ ions and one mole of Cl– ions. Therefore, each formula unit yields two individual ions. d) One mole of compound dissolves to form one mole of Cl– ions, which reacts with the Ag+ ion (from AgNO3) to form one mole of AgCl precipitate. 23.103

23.104 a) Any reduction with a potential which, when added to 0.38 V, gives a positive total will plate out. Of the metals listed, the ones that will plate out are: Co2+, Ni2+, and Cu2+. b) Equal amounts of electricity will yield equal moles of metal. (The values would not be equal if the charges were not all the same.) The metal with the lowest molar mass, Co, will yield the least mass. 23.105 Tetrahedral complexes are never low-spin, no matter how strong the ligand. Also, certain metal ions have less tendency to go to a low-spin configuration. The statement could be revised to read: “Strong-field ligands usually give rise to low-spin octahedral complexes.” 23.106 a) The longest distance will be for two of the long bonds trans to each other. Longest distance = 2(262 pm) = 524 pm b) The shortest distance will be for two of the short bonds trans to each other. Shortest distance = 2(207 pm) = 414 pm 23.107 a) 2Fe3+(aq) + 6OCl–(aq) + 4OH–(aq) → 2FeO42–(aq) + 6Cl–(aq) + 2H2O(l) b) K4[Mn(CN)6](s) + 2K(s) → K6[Mn(CN)6](s) c) 12NaO2(s) + Co3O4(s) → 3Na4CoO4(s) + 8O2(g) d) VCl3(s) + 4Na(s) + 6CO(g) → Na[V(CO)6](s) + 3NaCl(s) e) BaO2(s) + Ni2+(aq) + 2OH–(aq) → BaNiO3(s) + H2O(l) f) 11CO(g) + 12OH–(aq) + 2Co2+(aq) → 2[Co(CO)4] –(aq) + 3CO32–(aq) + 6H2O(l) g) Cs2[CuF4](s) + F2(g) → Cs2CuF6(s) h) 6TaCl5(s) + 15Na(s) → 15NaCl(s) + Ta6Cl15(s) i) 4K2[Ni(CN)4](s) + N2H4(aq) + 4OH–(aq) → 2K4[Ni2(CN)6](s) + N2(g) + 4H2O(l) + 4CN–(aq)

Fe6+ Mn0 Co4+ V1– Ni4+ Co1– Cu4+ Ta3+ Ni1+

23-25

23.108

O Mn

23.109 Plan: Types of isomers for coordination compounds are coordination isomers with different arrangements of ligands and counter ions, linkage isomers with different donor atoms from the same ligand bound to the metal ion, geometric isomers with differences in ligand arrangement relative to other ligands, and optical isomers with mirror images that are not superimposable. Solution: This compound exhibits geometric (cis-trans) and linkage isomerism. The SCN– ligand can bond to the metal through either the S atom or the N atom. NCS NH NCS NH 3

Pt NCS

NH3

cis-diamminedithiocyanatoplatinum(II)

SCN

NH3

H3N

trans-diamminedithiocyanatoplatinum(II)

SCN

NH3

cis-diamminediisothiocyanatoplatinum(II)

SCN

NH3

NH3 Pt

Pt SCN

SCN

H3 N

NCS

trans-diamminediisothiocyanatoplatinum(II) SCN NH3

Pt NCS

NH3

H3N

SCN

cis-diamminethiocyanatoisothiocyanatoplatinum(II) trans-diamminethiocyanatoisothiocyanatoplatinum(II) 23.110 Plan: Types of isomers for coordination compounds are coordination isomers with different arrangements of ligands and counter ions, linkage isomers with different donor atoms from the same ligand bound to the metal ion, geometric isomers with differences in ligand arrangement relative to other ligands, and optical isomers with mirror images that are not superimposable. Solution: [Co(NH3)4(H2O)Cl]2+ tetraammineaquachlorocobalt(III) ion 2 geometric isomers

23-26

Cl H3N

NH3

H3N

Co H3N

Cl OH2 Co

NH3

H3N

NH3

H2O

NH3

trans Cl and H2O

cis Cl and NH3

[Cr(H2O)3Br2Cl] triaquadibromochlorochromium(III) 3 geometric isomers

Cl H2O

H2O

OH2

Cr Br

OH2

H2O

Br’s trans

Br’s cis

H2O’s meridional H2O’s facial (Unfortunately meridional and facial isomers are not covered in the text. Facial (fac) isomers have three adjacent corners of the octahedron occupied by similar groups. Meridional (mer) isomers have three similar groups around the outside of the complex.) [Cr(NH3)2(H2O)2Br2]+ diamminediaquadibromochromium(III) ion 6 isomers (5 geometric) NH3 NH3 + + H2O Br H2O Br Cr Cr Br OH2 H2O Br NH3 NH 3

All pairs are trans NH3 H2O NH3 Cr Br OH2 Br Only H2O’s are trans NH3 H2 O NH3 Cr H2 O Br Br

Only NH3’s are trans

NH3

+ H2O

+ Br

Cr Br

NH3 OH2

Only Br’s are trans NH3 H3N OH2 Cr Br OH2 Br

All pairs are cis. These are optical isomers of each other. Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

23-27

23.111 The ligand field strength of NH3 is greater than that of H2O. Therefore, the crystal field splitting energy (Δ) is larger, and higher energy (shorter wavelength) light is required to excite the electrons of [Cr(NH3)6]3+. When [Cr(NH3)6]3+ absorbs higher energy blue-violet light, yellow-orange light (the compliment of blue-violet light) is observed. When [Cr(H2O)6]3+ absorbs lower energy red light, blue-gray light is observed. 23.112 a) The coordination number of uranium is 8. b) The uranium has an oxidation number of +5 (because Na is +1 and F is –1). c)

23.113 a) [Co(NH3)6][Cr(CN)6] b) [Co(NH3)5CN][CrNH3(CN)5] 23.114

hexaamminecobalt(III) hexacyanochromate(III) pentaamminecyanocobalt(III) amminepentacyanochromate(III)

Structures A, B, and E are geometric isomers. Structures C, D, and F are identical to the original structure.

23.115 Plan: Sketch the structure of each complex ion and look for a plane of symmetry. A complex ion with a plane of symmetry does not have optical isomers. Solution: a) A plane that includes both ammonia ligands and the zinc ion is a plane of symmetry. The complex does not have optical isomers because it has a plane of symmetry. b) The Pt2+ ion is d8, so the complex is square planar. Any square planar complex has a plane of symmetry, so the complex does not have optical isomers. c) The trans octahedral complex has the two chloride ions opposite each other. A plane of symmetry can be passed through the two chlorides and platinum ion, so the trans complex does not have optical isomers. d) No optical isomers for same reason as in part c). e) The cis isomer does not have a plane of symmetry, so it does have optical isomers. 23.116 Plan: Assume a 100-g sample (making the percentages of each element equal to the mass present in grams) and convert the mass of each element to moles using the element’s molar mass. Divide each mole amount by the smallest mole amount to determine whole-number ratios of the elements to determine the empirical formula. Once the empirical formula is known, use the formula of the triethylphosphine to deduce the molecular formula. Solution: ⎛ 1 mol Pt ⎞⎟ 0.198872 ⎜ ⎟⎟ = 0.198872 mol Pt; =1 Moles of Pt = (38.8 g Pt )⎜⎜ ⎜⎜⎝195.1 g Pt ⎠⎟⎟ 0.198872 ⎛ 1 mol Cl ⎞⎟ ⎜ ⎟⎟ = 0.397743 mol Cl; Moles of C1 = (14.1 g Cl)⎜⎜ ⎜⎜⎝ 35.45 g Cl ⎠⎟⎟ ⎛ 1 mol C ⎞⎟ ⎜ ⎟⎟ = 2.389675 mol C; Moles of C = (28.7 g C)⎜⎜ ⎜⎝⎜12.01 g C ⎠⎟⎟ ⎛ 1 mol P ⎞⎟ ⎜ ⎟⎟ = 0.400387 mol P; Moles of P = (12.4 g P )⎜⎜ ⎜⎝⎜ 30.97 g P ⎠⎟⎟ ⎛ 1 mol H ⎞⎟ ⎜ ⎟⎟ = 5.972222 mol H; Moles of H = (6.02 g H)⎜⎜ ⎜⎝⎜1.008 g H ⎠⎟⎟

0.397743 =2 0.198872 2.389675 = 12 0.198872 0.400387 =2 0.198872 5.972222 = 30 0.198872

23-28

The empirical formula for the compound is PtCl2C12P2H30. Each triethylphosphine ligand, P(C2H5)3 accounts for one phosphorus atom, six carbon atoms, and fifteen hydrogen atoms. According to the empirical formula, there are two triethylphosphine ligands: 2 P(C2H5)3 “equals” C12P2H30. The compound must have two P(C2H5)3 ligands and two chloro ligands per Pt ion. The formula is [Pt[P(C2H5)3]2Cl2]. The central Pt ion has four ligands and is square planar, existing as either a cis or trans compound. C2H5 C2H5 P

C2H5 P

C2H5 C2H5

C2H5 P

Pt Cl

Cl Pt

C2H5

cis-dichlorobis(triethylphosphine)platinum(II)

trans-dichlorobis(triethylphosphine)platinum(II)

23.117 a) There are no isomers.

Cl Pt Cl

b) There is an optical isomer.

N Fe3+ N

23-29

F N

Co N

trans cis trans and cis are geometric isomers; the cis form exists as two optical isomers. d) There are no optical isomers but there are two geometric isomers.

H3N

N Cl

Co3+

H3N NH3

NH3

H3N NH3

23.118 Plan: A reaction is favored in terms of entropy if there is an increase in entropy. Entropy increases if there are more moles of product than of reactant. Solution: a) The first reaction shows no change in the number of particles. In the second reaction, the number of reactant particles is greater than the number of product particles. A decrease in the number of particles means a decrease in entropy, while no change in number of particles indicates little change in entropy. Based on entropy change only, the first reaction is favored. b) The ethylenediamine complex is more stable with respect to ligand exchange with water because the entropy change is unfavorable. 23.119 a) The immediate reaction with AgNO3 to yield a red precipitate (Ag2CrO4) indicates that the chromate is not directly coordinated to M. The potassium will not be coordinated to M. Silver nitrate does not easily form a white precipitate thus the chloride must be coordinated to M as part of an inert complex. The ammonia is probably coordinated to M. The coordination number of M comes from two chlorides plus four ammonia to give a total of 6. b) The ammonia and chloride ligands are covalently bonded to M, and the potassium and chromate are ionically bonded. Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

23-30

c) The ability of oxalate to displace chloride from only one form is the clue. Oxalate is bidentate, and will only displace chlorides in a cis arrangement, not a trans arrangement. Form A is the cis complex, and form B is the trans complex. 23.120 a) [Cr(H2O)6]3+

562 nm

(6.626 × 10 E = hc/λ =

−34

)(

)

J • s 3.00 × 108 m /s ⎛⎜ 1 nm ⎞⎟⎛ 6.022 × 10 23 ⎞⎛ 1 kJ ⎞ ⎜⎜ −9 ⎟⎟⎜⎜ ⎟⎟⎟⎜⎜ 3 ⎟⎟ ⎜⎝10 m ⎠⎟⎟⎜⎜⎝ 562 nm mol ⎠⎟⎜⎝10 J ⎠⎟

= 212.99878 = 213 kJ/mol [Cr(CN)6]

3–

381 nm

(6.626 × 10 E = hc/λ =

−34

)(

)

J • s 3.00 × 108 m /s ⎛⎜ 1 nm ⎞⎟⎛ 6.022 × 10 23 ⎞⎛ 1 kJ ⎞ ⎟⎜ ⎟ ⎟⎜ ⎜⎜⎜ −9 ⎟⎟⎜⎜⎜ ⎟⎟⎜ 3 ⎟ 381 nm mol ⎠⎟⎜⎝10 J ⎠⎟ ⎜⎝10 m ⎠⎟⎝

= 314.187= 314 kJ/mol [CrCl6]

3–

735 nm

E = hc/λ =

(6.626 × 10

−34

)(

)

J • s 3.00 × 108 m /s ⎛⎜ 1 nm ⎞⎟⎛ 6.022 × 10 23 ⎞⎛ 1 kJ ⎞ ⎟⎜ ⎟ ⎟⎜ ⎜⎜⎜ −9 ⎟⎟⎜⎜⎜ ⎟⎟⎟⎜⎜ 3 ⎟⎟ ⎟ 735 nm mol ⎝ ⎠ 10 m 10 J ⎠ ⎝⎜ ⎠⎝

= 162.864 = 163 kJ/mol [Cr(NH3)6]

462 nm

(6.626 × 10 E = hc/λ =

−34

)(

)

J • s 3.00 × 108 m /s ⎛⎜ 1 nm ⎞⎟⎛ 6.022 × 10 23 ⎞⎛ 1 kJ ⎞ ⎟⎜ ⎟ ⎟⎜ ⎜⎜⎜ −9 ⎟⎟⎜⎜⎜ ⎟⎟⎜ 3 ⎟ 462 nm mol ⎠⎟⎜⎝10 J ⎠⎟ ⎜⎝10 m ⎠⎟⎝

= 259.1024 = 259 kJ/mol 3+

[Ir(NH3)6]

244 nm

E = hc/λ =

(6.626 × 10

−34

)(

)

J • s 3.00 × 108 m /s ⎛⎜ 1 nm ⎞⎟⎛ 6.022 × 10 23 ⎞⎛ 1 kJ ⎞ ⎟⎜ ⎟ ⎟⎜ ⎜⎜⎜ −9 ⎟⎟⎜⎜⎜ ⎟⎟⎟⎜⎜ 3 ⎟⎟ ⎟ 244 nm mol ⎝ ⎠ 10 m 10 J ⎝ ⎠ ⎜⎝ ⎠

= 490.59555 = 491 kJ/mol [Fe(H2O)6]

966 nm

E = hc/λ =

(6.626×10

−34

)(

)

8 J • s 3.00 ×10 m /s ⎛⎜ 1 nm ⎞⎟⎛ 6.022 ×10 23 ⎞⎛ 1 kJ ⎞ ⎟⎟⎜ ⎟ ⎟⎜ ⎜⎜⎜ −9 ⎟⎟⎜⎜⎜ ⎟⎜ 3 ⎟ 966 nm mol ⎠⎟⎜⎝10 J ⎠⎟ ⎜⎝10 m ⎠⎟⎝

= 123.9185 = 124 kJ/mol [Fe(H2O)6]3+

730 nm

E = hc/λ =

(6.626 × 10

−34

)(

)

J • s 3.00 × 108 m /s ⎜⎛ 1 nm ⎞⎟⎛ 6.022 × 10 23 ⎞⎛ 1 kJ ⎞ ⎟⎜ ⎜ ⎟⎟⎜ ⎟⎟ ⎟⎜ ⎜⎜⎜10−9 m ⎟⎟⎜⎜⎝ ⎟⎠⎟⎝⎜⎜103 J ⎠⎟ 730 nm mol ⎝ ⎠

= 163.97988 = 164 kJ/mol [Co(NH3)6]

405 nm

(6.626 × 10 E = hc/λ =

−34

)(

)

J • s 3.00 × 108 m /s ⎜⎛ 1 nm ⎞⎟⎛ 6.022 × 10 23 ⎞⎛ 1 kJ ⎞ ⎟⎟ ⎟⎟⎜⎜ ⎜ ⎟⎟⎜ ⎜⎜⎜10−9 m ⎟⎟⎜⎜⎝ ⎟⎟⎜⎝⎜103 J ⎠⎟ 405 nm mol ⎠ ⎝ ⎠

= 295.56868 = 296 kJ/mol

23-31

[Rh(NH3)6]3+

295 nm

(6.626 × 10 E = hc/λ =

−34

)(

)

J • s 3.00 × 108 m /s ⎜⎛ 1 nm ⎞⎟⎛ 6.022 × 10 23 ⎞⎛ 1 kJ ⎞ ⎟⎟ ⎟⎟⎜⎜ ⎜ ⎟⎟⎜ ⎜⎜⎜10−9 m ⎟⎟⎜⎜⎝ ⎟⎟⎜⎝⎜103 J ⎠⎟ 295 nm mol ⎠ ⎝ ⎠

= 405.7807= 406 kJ/mol b) Comparing the energies of the chromium complexes, we find the energy increases in the order: Cl– < H2O < NH3 < CN– This is an abbreviated spectrochemical series. c) The higher the oxidation number (Fe3+ vs. Fe2+) the greater the crystal field splitting (Δ). d) When moving down a column on the periodic table, in this case from Co to Rh to Ir, the greater the energy required, the greater the crystal field splitting (Δ). 23.121 a) V is +5 in VO3– and VO2Cl2–; Al is +3 in AlCl4– and AlOCl2–. b) No elements change oxidation state, so no redox occurs in either reaction. The balanced equations are: V2O5 + 2HCl → VO2Cl2– + VO3– + 2H+ V2O5 + 6HCl → 2VOCl3 + 3H2O Neither reaction is a redox process. − − ⎛ 1 mol V2 O 5 ⎞⎛ ⎟⎟⎜⎜1 mol VO 2 Cl 2 ⎞⎛ ⎟⎟⎜⎜153.84 g VO 2 Cl 2 ⎞⎟⎟ ⎜ c) Mass (g) of VO2Cl2– = (12.5 g V2 O 5 )⎜⎜ ⎟⎟⎜ ⎟⎟⎜ − ⎟ ⎜⎜⎝181.88 g V2 O 5 ⎠⎝ ⎟⎜⎜ 1 mol V2 O 5 ⎠⎝ ⎟⎜⎜ 1 mol VO 2 Cl 2 ⎠⎟⎟ = 10.572905 = 10.6 g VO2Cl2– ⎛ 1 mol V2 O 5 ⎞⎛ ⎟⎟⎜⎜ 2 mol VOCl 3 ⎞⎛ ⎟⎟⎜⎜173.29 g VOCl 3 ⎞⎟⎟ ⎜ Mass (g) of VOCl3 = (12.5 g V2 O 5 )⎜⎜ ⎟⎟⎜ ⎟⎟⎜ ⎟ ⎟⎜⎜ 1 mol V2 O 5 ⎠⎝ ⎟⎜⎜ 1 mol VOCl 3 ⎠⎟⎟ ⎜⎜⎝181.88 g V2 O 5 ⎠⎝

= 23.8193 = 23.8 g VOCl3 23.122 a) In [Co(CN)6]3–, Co has a +3 charge and an electron configuration of [Ar]3d 6. Since CN– is a strong-field ligand, the six electrons will pair in the t2g orbitals. Diagram A is the correct diagram. b) Diagram D shows seven electrons. The Co ion that is d 7 has a charge of +2. Since the six F– ligands each have a charge of –1, the complex ion must have a charge of –4. n = –4. c) The Ni ion in both complexes has a +2 charge and configuration of [Ar]3d8. Since [Ni(CN)4]2–, is diamagnetic, Diagram E with eight paired electrons is the correct diagram. The paramagnetic complex [Ni(Cl)4]2–, is tetrahedral. Diagram B is correct. d) It is not possible to determine if the ligands in the VL6 complex are strong-field or weak-field. The V2+ ion is a d 3 ion and the arrangement of three electrons in the d orbitals is the same in both the strong-field and weak-field cases.

23-32

CHAPTER 24 NUCLEAR REACTIONS AND THEIR APPLICATIONS FOLLOW–UP PROBLEMS 24.1A

Plan: Write a skeleton equation that shows fluorine-20 undergoing beta decay. Conserve mass and atomic number by ensuring the superscripts and subscripts equal one another on both sides of the equation. Determine the identity of the daughter nuclide by using the periodic table. Solution: Fluorine-20 has Z = 9. Its symbol is 209 F. When it undergoes beta decay, a beta particle, − 01 β, and a daughter nuclide, AZ X are produced. 20 F → AZ X + −01 β 9 To conserve atomic number, Z must equal 10. Element is neon. To conserve mass number, A must equal 20. The identity of ZA X is 20 10 Ne. The balanced equation is:

24.1B

20 9

0 F → 20 10 Ne + −1 β

Plan: Write a skeleton equation that shows an unknown nuclide, ZA X, undergoing positron decay, 01 β , to form tellurium-124,

124 52

Te. Conserve mass and atomic number by ensuring the superscripts and subscripts equal one

another on both sides of the equation. Determine the identity of X by using the periodic table to identify the element with atomic number equal to Z. Solution: The unknown nuclide yields tellurium-124 and a β+ particle: A 124 0 Z X → 52Te + 1 β To conserve atomic number, Z must equal 53. Element is iodine. To conserve mass number, A must equal 124. 124 The identity of ZA X is 53 I. The balanced equation is:

124 53

I → 12452Te + 01 β

Check: A = 124 = 124 + 0 and Z = 53 = 52 + 1. 24.2A

Plan: Look at the N/Z ratio, the ratio of the number of neutrons to the number of protons. If the N/Z ratio falls in the band of stability, the nuclide is predicted to be stable. For stable nuclides of elements with atomic number greater than 20, the ratio of number of neutrons to number of protons (N/Z) is greater than one. In addition, the ratio increases gradually as atomic number increases. Also check for exceptionally stable numbers of neutrons and/or protons—the “magic” numbers of 2, 8, 20, 28, 50, 82, and (N = 126). Also, even numbers of protons and or neutrons are related to stability whereas odd numbers are related to instability. Solution: a) 58 V appears unstable/radioactive because its N/Z ratio (58 − 23)/23 = 1.52 is too high and is above the band of 23 stability. Additionally, this nuclide has both odd N(35) and Z(23). b) 60 Ni appears stable because it has both even N (32) and even Z (28) and 28 is a magic number. N/Z = 1.14 and 28 falls in the band of stability.

24-1

c) 94 Ag appears radioactive because it has both odd N (47) and odd Z (47) and N/Z = 1, which is too low for this 47 region of the band of stability. 24.2B

Plan: Nuclear stability is found in nuclides with an N/Z ratio that falls within the band of stability. Nuclides with an even N and Z, especially those nuclides that have magic numbers, are exceptionally stable. Examine the two nuclides to see which of these criteria can explain the difference in stability. Solution: Phosphorus-31 has 16 neutrons and 15 protons, with an N/Z ratio of 1.07. Phosphorus-30 has 15 neutrons and 15 protons, with an N/Z ratio of 1.00. 31P has an even N while 30P has both an odd Z and an odd N. 31P has a slightly higher N/Z ratio that is closer to the band of stability.

24.3A

Plan: Examine the N/Z ratio and determine which mode of decay will yield a product nucleus that is closer to the band. Nuclides whose N/Z ratios are too high generally decay by beta emission while nuclides whose N/Z ratios are too low decay by positron emission or electron capture. Nuclides with Z > 83 decay by alpha particle emission. Solution: a) Iron-61 has an N/Z ratio of (61 – 26)/26 = 1.35, which is too high for this region of the band. Iron-61 will undergo β– decay. 61 61 0 new N/Z = (61 – 27)/27 = 1.26 26 Fe → 27 Co + −1 β Additionally, iron has an atomic mass of 55.85 amu. The A value of 61 is higher, suggesting beta decay. b) Americium-241 has Z > 83, so it undergoes α decay. 241 237 4 95 Am → 93 Np + 2 He

24.3B

Plan: Examine the N/Z ratio and determine which mode of decay will yield a product nucleus that is closer to the band. Nuclides whose N/Z ratios are too high generally decay by beta emission while nuclides whose N/Z ratios are too low decay by positron emission or electron capture. Nuclides with Z > 83 decay by alpha particle emission. Solution: a) Titanium-40 has an N/Z ratio of (40 – 22)/22 = 0.81, which is too low for this region of the band. Titanium-40 will undergo positron decay or electron capture. Additionally, titanium’s atomic mass is 47.87 amu, which is much higher than the A value of 40, also suggesting positron decay or electron capture. b) Cobalt-65 has an N/Z ratio of (65 – 27)/27 = 1.40, which is too high for this region of the band. Cobalt-65 will undergo beta decay. Additionally, cobalt’s atomic mass is 58.93 amu, which is much lower than the A value of 65, also suggesting beta decay.

24.4A

Plan: Specific activity of a radioactive sample is its decay rate per gram. Find the mass of the sample. Calculate the specific activity by dividing the number of particles emitted per second (disintegrations per second = dps) by the mass of the sample. Convert disintegrations per second to Ci by using the conversion factor between the two units: 1 Ci = 3.70 × 1010dps. Convert Ci to Bq by using the conversion factor between the two units: 1 Ci = 3.70 × 1010Bq. Solution: ⎛ 76 g As ⎞⎟ ⎜ ⎟⎟ = 2.5840 × 10–6 = 2.6 × 10–6 g As a) Mass (g) of As = (3.4 × 10–8mol As) ⎜⎜ ⎜⎜⎝1 mol As ⎠⎟⎟ ⎛ 1.53×1011 dps ⎞⎛ ⎞⎟ 1 Ci ⎟⎟⎜⎜ ⎜ ⎟⎟ = 1.60028 × 106 = 1.6 × 106 Ci/g Specific activity (Ci/g) = ⎜⎜ ⎟⎜ −6 ⎟ 10 ⎜⎝⎜ 2.5840 ×10 g ⎠⎝ ⎟⎜⎜ 3.70 ×10 dps ⎠⎟⎟ 10 ⎛1.60028×106 Ci ⎞⎛ ⎟⎟⎜⎜ 3.70×10 Bq ⎞⎟⎟ 16 16 b) Specific activity (Bq/g) = ⎜⎜⎜ ⎟⎟⎜ ⎟⎟ = 5.92105 × 10 = 5.9 × 10 Bq/g ⎜⎝ g Ci ⎟⎠⎝⎜ ⎟⎠

24-2

24.4B

24.5A

A 9.97×1012 nuclei/h = 2.5471 × 10–10 = 2.55 × 10–10 h–1 = −2 23 N (6.50×10 mol)(6.022×10 nuclei/h )

Plan: Use the half-life of 24Na to find k. Substitute the value of k, initial activity (A0), and time of decay (4 days) into the integrated first-order rate equation to solve for activity at a later time (At). Solution: ln 2 ln 2 k= = = 0.0462098 h–1 t1 / 2 15 h ln At = ln A0 – kt ln At = ln (2.5 × 109) – (0.0462098 h–1)(4.0 days)(24 h/day) ln At = 17.203416 At = 2.960388 × 107 = 3.0 × 107 d/s

24.5B

Plan: Use the half-life of 59Fe to find k. Substitute the value of k and time of decay (17 days) into the integrated first-order rate equation. Assume that A0 is 1.0 (100% of the original sample) and solve for At, the fraction of the sample remaining after 17 days. Subtract the fraction remaining from the original sample (1.0) to calculate the fraction that has decayed. Solution: ln 2 ln 2 k= = = 1.5576 × 10–2days–1 t1 / 2 44.5 days ln At = ln A0 – kt ln At = ln (1.0) – (1.5576 × 10–2days–1)(17 days) ln At = –0.2648 At = 0.7674 = fraction of iron-59 remaining

Fraction of iron-59 decayed = 1.0 – 0.7674 = 0.2326 = 0.23 = fraction of iron-59 that has decayed 24.6A

Plan: The wood from the case came from a living organism, so A0 equals 15.3 d/min∙g. Substitute the current activity of the case (At), A0, and k into the first-order rate expression and solve for t. Find k from the half-life of carbon (5730 yr). Solution: ln 2 ln 2 k= = = 1.209680943 × 10–4yr–1 t1 / 2 5730 yr ln At = ln A0 – kt ln [9.41 d/min∙g] = ln [15.3 d/min∙g] – (1.209680943 × 10–4yr–1)(t) –0.486079875 = – (1.209681 × 10–4yr–1)(t) t = 4018.2 = 4.02 × 103 years

24-3

24.6B

Plan: The woolen tunic came from a living organism, so A0 equals 15.3 d/min∙g. Substitute the current activity of the tunic (At), A0, and k into the first-order rate expression and solve for t. Find k from the half-life of carbon (5730 yr). Solution: k=

ln 2 ln 2 = = 1.209680943 × 10–4 yr–1 t1 / 2 5730 yr

ln At = ln A0 – kt ln [12.87 d/min∙g] = ln [15.3 d/min∙g] – (1.209680943 × 10–4 yr–1)(t) –0.172953806 = – (1.209681 × 10–4yr–1)(t) t = 1429.7473 = 1430 years

24.7A

Plan: In a balanced nuclear equation, the total of mass numbers and the total of charges on the left side and the right side must be equal. Solution: 209 83

Bi + 54 Cr → 01 n + 262 Bh 24 107

Cr. The bombarding nuclide is 54 24 24.7B

24.8A

55 Mn + 11 p → 01 n + 26 Fe

Plan: Nickel-58 has 28 protons and 30 neutrons in its nucleus. Calculate the change in mass (Δm) in one 58 Ni atom, convert to MeV and divide by 58 to obtain binding energy/nucleon. Solution: Δm = [(28 × mass H atom) + (30 × mass neutron)] – mass 58Ni atom Δm = [(28 × 1.007825 amu) + (30 × 1.008665)] – 57.935346 amu Δm = 0.543704 amu ⎛ 931.5 MeV ⎞⎟ ⎜ ⎟⎟ = 506.460276 MeV Binding energy (MeV) = (0.543704 amu) ⎜⎜ ⎜⎜⎝ 1 amu ⎠⎟⎟

Binding Energy/nucleon =

506.460276 MeV 58 nucleons

= 8.7321 = 8.732 MeV/nucleon

The BE/nucleon of 56Fe is 8.790 MeV/nucleon. The energy per nucleon holding the 58Ni nucleus together is less than that for 56Fe (8.732<8.790), so 58Ni is less stable than 56Fe.

24-4

24.8B

Plan: Uranium-235 has 92 protons and 143 neutrons in its nucleus. Calculate the change in mass (Δm) in one 235 U atom, convert to MeV and divide by 235 to obtain binding energy/nucleon. Solution: Δm = [(92 × mass H atom) + (143 × mass neutron)] – mass 235U atom Δm = [(92 × 1.007825 amu) + (143 × 1.008665)] – 235.043924 amu Δm = 1.915071 amu

⎛ 931.5 MeV ⎞⎟ Binding energy (MeV) = (1.915071 amu)⎜⎜ ⎟ = 1783.888637 MeV ⎝ 1 amu ⎠⎟ 1783.888637 MeV = 7.591015 = 7.591 MeV/nucleon Binding Energy/nucleon = 235 nucleons 12 The BE/nucleon of C is 7.680 MeV/nucleon. The energy per nucleon holding the 235U nucleus together is less than that for 12C (7.591 < 7.680), so 235U is less stable than 12C. CHEMICAL CONNECTIONS BOXED READING PROBLEMS

B24.1

In the s-process, a nucleus captures a neutron sometime over a long period of time. Then the nucleus emits a beta particle to form another element. The stable isotopes of most heavy elements up to 209Bi form by the s-process. The r-process very quickly forms less stable isotopes and those with A greater than 230 by multiple neutron captures, followed by multiple beta decays.

B24.2

Plan: Find the change in mass of the reaction by subtracting the mass of the products from the mass of the reactants and convert the change in mass to energy with the conversion factor between amu and MeV. Convert the energy per atom to energy per mole by multiplying by Avogadro’s number. Solution: Δm = mass of reactants – mass of products = [(4)(1.007825)] – [4.00260 + (2)(5.48580 × 10–4)] = 4.031300 – 4.003697 = 0.02760 amu/4He atom = 0.02760284 g/mol4He ⎛ 0.02760284 amu 4 He ⎞⎛ ⎟⎟⎜⎜ 931.5 MeV ⎞⎟⎟ ⎜ Energy (MeV/atom) = ⎜⎜ ⎟⎟⎜ ⎟ = 25.7120 = 25.71 MeV/atom ⎜⎝⎜ 1 atom ⎟⎜⎜ 1 amu ⎠⎟⎟ ⎠⎝ Convert atoms to moles using Avogadro’s number. ⎛ 25.7120 MeV ⎞⎟⎛⎜ 6.022 ×10 23 atoms ⎞⎟ 25 25 Energy = ⎜⎜ ⎟ = 1.54838 × 10 = 1.548 × 10 MeV/mol ⎟⎜ ⎟⎠⎟ ⎝ ⎠⎟⎜⎝ atom 1 mol

B24.3

The simultaneous fusion of three nuclei is a termolecular process. Termolecular processes have a very low probability of occurring. The bimolecular fusion of 8Be with 4He is more likely.

B24.4

Plan: In a balanced nuclear equation, the total of mass numbers and the total of charges on the left side and the right side must be equal. Solution: 210 210 0 210 83 Bi → 84 Po + − 1 β 84 Po is Nuclide A 210 84

Po → 206 82 Pb +

206 82

Pb + 3 n →

209 82

206 82

Pb is Nuclide B

209 82

Pb is Nuclide C

0 Pb → 210 83 Bi + − 1 β

210 83

Bi is Nuclide D

1 0

209 82

4 2

24-5

END–OF–CHAPTER PROBLEMS

24.1

a) Chemical reactions are accompanied by relatively small changes in energy while nuclear reactions are accompanied by relatively large changes in energy. b) Increasing temperature increases the rate of a chemical reaction but has no effect on a nuclear reaction. c) Both chemical and nuclear reaction rates increase with higher reactant concentrations. d) If the reactant is limiting in a chemical reaction, then more reactant produces more product and the yield increases in a chemical reaction. The presence of more radioactive reagent results in more decay product, so a higher reactant concentration increases the yield in a nuclear reaction.

24.2

a) The percentage of sulfur atoms that are sulfur-32 is 95.02%, the same as the relative abundance of 32S. b) The atomic mass is larger than the isotopic mass of 32S. Sulfur-32 is the lightest isotope, as stated in the problem, so the other 5% of sulfur atoms are heavier than 31.972070 amu. The average mass of all the sulfur atoms will therefore be greater than the mass of a sulfur-32 atom.

24.3

a) She found that the intensity of emitted radiation is directly proportional to the concentration of the element in the various samples, not to the nature of the compound in which the element occurs. b) She found that certain uranium minerals were more radioactive than pure uranium, which implied that they contained traces of one or more as yet unknown, highly radioactive elements. Pitchblende is the principal ore of uranium.

24.4

Plan: Radioactive decay that produces a different element requires a change in atomic number (Z, number of protons). Solution: A A = mass number (protons + neutrons) ZX Z = number of protons (positive charge)

X = symbol for the particle N = A – Z (number of neutrons) a) Alpha decay produces an atom of a different element, i.e., a daughter with two less protons and two less neutrons. A Z

X → AZ−−42Y + 42 He

2 fewer protons, 2 fewer neutrons

b) Beta decay produces an atom of a different element, i.e., a daughter with one more proton and one less neutron. A neutron is converted to a proton and β particle in this type of decay. A Z

X → Z +A1Y + −01 β

1 more proton, 1 less neutron

c) Gamma decay does not produce an atom of a different element and Z and N remain unchanged. A Z

X * → ZA X + 00 γ

( ZA X * = energy rich state), no change in number of protons or neutrons.

d) Positron emission produces an atom of a different element, i.e., a daughter with one less proton and one more neutron. A proton is converted into a neutron and positron in this type of decay. A Z

X → Z −A1Y + +01 β

1 less proton, 1 more neutron

e) Electron capture produces an atom of a different element, i.e., a daughter with one less proton and one more neutron. The net result of electron capture is the same as positron emission, but the two processes are different. A 0 A 1 less proton, 1 more neutron Z X + −1 e → Z −1Y A different element is produced in all cases except (c).

24-6

24.5

The key factor that determines the stability of a nuclide is the ratio of the number of neutrons to the number of protons, the N/Z ratio. If the N/Z ratio is either too high or not high enough, the nuclide is unstable and decays. 3 N/Z = 1/2 2 He 2 2

24.6

N/Z = 0/2, thus it is more unstable.

A neutron-rich nuclide decays to convert neutrons to protons while a neutron-poor nuclide decays to convert protons to neutrons. The conversion of neutrons to protons occurs by beta decay: 1 1 0 0 n → 1 p + −1 β The conversion of protons to neutrons occurs by either positron decay: 1 1 0 1p → 0 n + 1β or electron capture: 1 0 1 1 p + −1 e → 0 n Neutron-rich nuclides, with a high N/Z, undergo β decay. Neutron-poor nuclides, with a low N/Z, undergo positron decay or electron capture.

24.7

Both positron emission and electron capture increase the number of neutrons and decrease the number of protons. The products of both processes are the same. Positron emission is more common than electron capture among lighter nuclei; electron capture becomes increasingly common as nuclear charge increases. For Z < 20, β+ emission is more common; for Z > 80, electron capture is more common.

24.8

Plan: In a balanced nuclear equation, the total of mass numbers and the total of charges on the left side and the right side must be equal. Solution: 4 230 a) 234 Mass: 234 = 4 + 230; Charge: 92 = 2 + 90 92 U → 2 He + 90Th 0 232 b) 232 93 Np + −1 e → 92 U

c) 24.9

12 7

N → β+ 0 1

12 6

Mass: 232 + 0 = 232; Charge: 93 + (–1) = 92 Mass: 12 = 0 + 12;

Charge: 7 = 1 + 6

0 26 a) 26 11 Na → − 1 β + 12 Mg 0 223 b) 223 87 Fr → − 1 β + 88 Ra 4 208 c) 212 83 Bi → 2 α + 81Tl

24.10

Plan: In a balanced nuclear equation, the total of mass numbers and the total of charges on the left side and the right side must be equal. Solution: a) The process converts a neutron to a proton, so the mass number is the same, but the atomic number increases by one. 27 0 27 Mass: 27 = 0 + 27; Charge: 12 = –1 + 13 12 Mg → − 1 β + 13 Al b) Positron emission decreases atomic number by one, but not mass number. 23 0 23 Mass: 23 = 0 + 23; Charge: 12 = 1 + 11 12 Mg → 1 β + 11 Na c) The electron captured by the nucleus combines with a proton to form a neutron, so mass number is constant, but atomic number decreases by one. 103 0 103 Mass: 103 + 0 = 103; Charge: 46 + (–1) = 45 46 Pd + −1 e → 45 Rh

24-7

24.11

32 a) 14 Si → −01 β + 32 15 P 4 214 b) 218 84 Po → 2 α + 82 Pb 0 110 c) 110 49 In + −1 e → 48 Cd

24.12

Plan: In a balanced nuclear equation, the total of mass numbers and the total of charges on the left side and the right side must be equal. Solution: a) In other words, an unknown nuclide decays to give Ti-48 and a positron. 48 48 0 Mass: 48 = 48 + 0; Charge: 23 = 22 + 1 23V → 22Ti + 1 β b) In other words, an unknown nuclide captures an electron to form Ag-107. 107 48

Cd + −01 e → 107 47 Ag

Mass: 107 + 0 = 107;

Charge: 48 + (–1) = 47

c) In other words, an unknown nuclide decays to give Po-206 and an alpha particle. 210 86

24.13

4 Rn → 206 84 Po + 2 He

Mass: 210 = 206 + 4;

Charge: 86 = 84 + 2

241 0 a) 241 94 Pu → 95 Am + − 1 β 228 0 b) 228 88 Ra → 89 Ac + − 1 β 203 4 c) 207 85 At → 83 Bi + 2 α

24.14

Plan: In a balanced nuclear equation, the total of mass numbers and the total of charges on the left side and the right side must be equal. Solution: a) In other words, an unknown nuclide captures an electron to form Ir-186. 186 0 186 Mass: 186 + 0 = 186; Charge: 78 + (–1) = 77 78 Pt + −1 e → 77 Ir b) In other words, an unknown nuclide decays to give Fr-221 and an alpha particle. 225 89

4 Ac → 221 87 Fr + 2 He

Mass: 225 = 221 + 4;

Charge: 89 = 87 + 2

c) In other words, an unknown nuclide decays to give I-129 and a beta particle. 129 52

24.15

0 Te → 129 53 I + − 1 β

Mass: 129 = 129 + 0;

Charge: 52 = 53 + (–1)

52 0 a) 52 26 Fe → 25 Mn + 1 β 215 4 b) 219 86 Rn → 84 Po + 2 α 81 81 c) 37 Rb + −01 e → 36 Kr

24.16

24-8

b) 59 27 Co might look unstable because its Z value is an odd number, but its N/Z ratio (59 − 27)/27 = 1.19 is in the band of stability, so 59 27 Co appears stable. c) 93 Li appears unstable because its N/Z ratio (9 − 3)/3 = 2.00 is too high and is above the band of stability. 24.17

a) 146 60 Nd

N/Z = 86/60 = 1.4

Stable, N/Z ok

b) 114 48 Cd

N/Z = 66/48 = 1.4

Stable, N/Z ok

88 42

N/Z = 46/42 = 1.1

Unstable, N/Z too small for this region of the band

24.18

Plan: Look at the N/Z ratio, the ratio of the number of neutrons to the number of protons. If the N/Z ratio falls in the band of stability, the nuclide is predicted to be stable. For stable nuclides of elements with atomic number greater than 20, the ratio of number of neutrons to number of protons (N/Z) is greater than one. In addition, the ratio increases gradually as atomic number increases. Also check for exceptionally stable numbers of neutrons and/or protons – the “magic” numbers of 2, 8, 20, 28, 50, 82, and (N = 126). Also, even numbers of protons and or neutrons are related to stability whereas odd numbers are related to instability. Solution: a) For the element iodine Z = 53. For iodine-127, N = 127 − 53 = 74. The N/Z ratio for 127I is 74/53 = 1.4. Of the examples of stable nuclides given in the book, 107Ag has the closest atomic number to iodine. The N/Z ratio for 107 Ag is 1.3. Thus, it is likely that iodine with six additional protons is stable with an N/Z ratio of 1.4. b) Tin is element number 50 (Z = 50). The N/Z ratio for 106Sn is (106 − 50)/50 = 1.1. The nuclide 106Sn is unstable with an N/Z ratio that is too low. c) For 68As, Z = 33 and N = 68 − 33 = 35 and N/Z = 1.1. The ratio is within the range of stability, but the nuclide is most likely unstable because there is an odd number of both protons and neutrons.

24.19

48 a) 19 K N/Z = 29/19 = 1.5

24.20

Unstable, N/Z too large for this region of the band

79 35

Br N/Z = 44/35 = 1.3

Stable, N/Z okay

33 18

Ar N/Z = 14/18 = 0.78

Unstable, N/Z too small

low and Z is too high. To become more stable, the nucleus decays by converting a proton to a neutron, which is positron decay. Alternatively, a nucleus can capture an electron and convert a proton into a neutron through electron capture. c) The N/Z ratio for 50 25 Mn is (50 – 25)/25 = 1.00. This number is below the band of stability, so the nuclide undergoes positron decay or electron capture. 24.21

a) 111 47 Ag

beta decay

N/Z = 1.4 which is too high

41 b) 17 Cl

beta decay

N/Z = 1.4 which is too high

110 44

beta decay

N/Z = 1.5 which is too high

24-9

24.22

Plan: Calculate the N/Z ratio for each nuclide. A neutron-rich nuclide decays to convert neutrons to protons while a neutron-poor nuclide decays to convert protons to neutrons. Neutron-rich nuclides, with a high N/Z, undergo β decay. Neutron-poor nuclides, with a low N/Z, undergo positron decay or electron capture. For Z < 20, β+ emission is more common; for Z > 80, e– capture is more common. Alpha decay is the most common means of decay for a heavy, unstable nucleus (Z > 83). Solution: a) For carbon-15, N/Z = 9/6 = 1.5, so the nuclide is neutron-rich. To decrease the number of neutrons and increase the number of protons, carbon-15 decays by beta decay. b) The N/Z ratio for 120Xe is 66/54 = 1.2. Around atomic number 50, the ratio for stable nuclides is larger than 1.2, so 120Xe is proton-rich. To decrease the number of protons and increase the number of neutrons, the xenon-120 nucleus either undergoes positron emission or electron capture. c) Thorium-224 has an N/Z ratio of 134/90 = 1.5. All nuclides of elements above atomic number 83 are unstable and decay to decrease the number of both protons and neutrons. Alpha decay by thorium-224 is the most likely mode of decay.

24.23

a) 106 49 In

positron decay or electron capture

N/Z = 1.2

b) 141 63 Eu

positron decay or electron capture

N/Z = 1.2

alpha decay

N/Z = 1.5

241 95

c) 24.24

Plan: Stability results from a favorable N/Z ratio, even numbers of N and/or Z, and the occurrence of magic numbers. Solution: 52 The N/Z ratio of 24 Cr is (52 − 24)/24 = 1.17, which is within the band of stability. The fact that Z is even does 52 not account for the variation in stability because all isotopes of chromium have the same Z. However, 24 Cr has

28 neutrons, so N is both an even number and a magic number for this isotope only. 40 20

24.25

N/Z = 20/20 = 1.0 It lies in the band of stability, and N and Z are both even and magic.

24.26

237 93

Np → 42 α + 233 91 Pa

233 91

Pa → −01 β + 233 92 U

233 92

U → 42 α + 22990Th

Th → 42 α + 225 88 Ra

229 90

24.27

Alpha emission produces helium ions which readily pick up electrons to form stable helium atoms.

24.28

207 0 4 The equation for the nuclear reaction is 235 92 U → 82 Pb + __ − 1 β + __ 2 He To determine the coefficients, notice that the beta particles will not impact the mass number. Subtracting the mass number for lead from the mass number for uranium will give the total mass number for the alpha particles released, 235 − 207 = 28. Each alpha particle is a helium nucleus with mass number 4. The number of helium atoms is determined by dividing the total mass number change by 4, 28/4 = 7 helium atoms or seven alpha particles. The equation is now 235 207 0 4 92 U → 82 Pb + __ − 1 β + 7 2 He

24-10

To find the number of beta particles released, examine the difference in number of protons (atomic number) between the reactant and products. Uranium, the reactant, has 92 protons. The atomic number in the products, lead atom and 7 helium nuclei, total 96. To balance the atomic numbers, four electrons (beta particles) must be emitted to give the total atomic number for the products as 96 − 4 = 92, the same as the reactant. In summary, seven alpha particles and four beta particles are emitted in the decay of uranium-235 to lead-207. 235 207 0 4 92 U → 82 Pb + − 1 β + 7 2 He 24.29

a) In a scintillation counter, radioactive emissions are detected by their ability to excite atoms and cause them to emit light. b) In a Geiger-Müller counter, radioactive emissions produce ionization of a gas that conducts a current to a recording device.

24.30

Since the decay rate depends only on the number of radioactive nuclei, radioactive decay is a first-order process.

24.31

No, it is not valid to conclude that t1/2 equals 1 min because the number of nuclei is so small (six nuclei). Decay rate is an average rate and is only meaningful when the sample is macroscopic and contains a large number of nuclei, as in the second case. Because the second sample contains 6 × 1012 nuclei, the conclusion that t1/2 = 1 min is valid.

24.32

High-energy neutrons in cosmic rays enter the upper atmosphere and keep the amount of 14C nearly constant through bombardment of ordinary 14N atoms. This 14 C is absorbed by living organisms, so its proportion stays relatively constant there also. 14 1 14 1 7 N + 0n → 6C + 1H

24.33

Plan: Specific activity of a radioactive sample is its decay rate per gram. Calculate the specific activity by dividing the number of particles emitted per second (disintegrations per second = dps) by the mass of the sample. Convert disintegrations per second to Ci by using the conversion factor between the two units. Solution: 1 Ci = 3.70 × 1010dps ⎛1.56 ×106 dps ⎞⎛ ⎞⎟ 1 Ci ⎟⎟⎜⎜ 1 mg ⎞⎛ ⎟⎜ ⎜ ⎟⎟ = 2.55528 × 10–2 = 2.56 × 10–2 Ci/g Specific activity (Ci/g) = ⎜⎜ ⎟⎟⎜ −3 ⎟⎟⎟⎜⎜ 10 ⎟⎜⎜10 g ⎠⎝ ⎟⎜⎜ 3.70 ×10 dps ⎠⎟⎟ ⎜⎜⎝ 1.65 mg ⎠⎝

24.34

24.35

⎛⎛ 4.13×108 d ⎞⎛ 1 h ⎞⎞⎟ ⎟⎟⎟ ⎟⎜ ⎜⎜⎜⎜ ⎟⎟⎟⎜⎜ ⎟⎟⎟⎟ ⎜⎜⎜⎝⎜⎜ ⎞⎟ h 1 Ci ⎠⎟⎝⎜⎜ 3600 s ⎠⎟ ⎟⎟⎟⎜⎛ ⎜ ⎟⎟ = 1.1925 × 10–6 = 1.2 × 10–6 Ci/g Specific activity (Ci/g) = ⎜⎜ ⎟⎟⎜⎜ ⎟⎟⎜⎝⎜ 3.70 ×1010 dps ⎟⎠⎟ 2.6 g ⎜⎜ ⎟⎟ ⎜⎜ ⎜⎜⎝ ⎟⎟⎟ ⎠ ⎟ ⎜

24-11

Solution: A becquerel is a disintegration per second (dps). ⎛⎛ 7.4 ×10 4 d ⎞⎛1 min ⎞⎞⎟ ⎟⎟⎟ ⎟⎜ ⎜⎜⎜⎜⎜ ⎟⎜ ⎟ ⎜⎜⎜⎜⎝ min ⎠⎟⎟⎟⎜⎜⎜ 60 s ⎟⎟⎟⎟⎟⎟⎛ 1 Bq ⎞ ⎝ ⎠ ⎟⎟ 8 8 ⎟⎟⎜⎜ Specific activity (Bq/g) = ⎜⎜⎜ ⎟⎟ = 1.43745 × 10 = 1.4 × 10 Bq/g ⎜⎜ ⎟ −6 ⎞ ⎛ ⎜⎜ 10 g ⎟ ⎟⎟⎝⎜1 dps ⎠⎟ ⎟⎟ ⎟⎟ ⎜⎜ 8.58 μg ⎜⎜⎜ ⎜⎝⎜ 1 μg ⎠⎟⎟ ⎟⎟ ⎜⎜⎝ ⎠⎟⎟ ⎜

24.36

24.37

⎛⎛ ⎞ ⎜⎜⎜ 3.77×10 7 d ⎞⎟⎛⎜⎜1 min ⎞⎟⎟⎟⎟ ⎟ ⎟⎜ ⎟ ⎜⎜⎜⎜ min ⎠⎟⎝⎜⎜ 60 s ⎠⎟⎟⎟⎟⎟⎜⎛ 1 Bq ⎞⎟ ⎜⎝ ⎜ ⎟ ⎟⎟ = 587.2274 = 587 Bq/g Specific activity (Bq/g) = ⎜⎜ ⎟⎜ ⎛103 g ⎞⎟ ⎟⎟⎟⎜⎜⎜⎝1 dps ⎟⎠⎟ ⎜⎜ ⎟⎟ ⎟⎟ ⎜⎜ 1.07 kg ⎜⎜⎜ ⎜⎜⎝ 1 kg ⎠⎟⎟ ⎟⎟ ⎜⎜⎝ ⎠⎟⎟ ⎜

Plan: The decay constant is the rate constant for the first-order reaction. Solution: ΔN Decay rate = − = kN Δt −1 atom − = k(1 × 1012 atom) day k = 1 × 10−12 d−1

24.38

Decay rate = −

ΔN = kN Δt

− (−2.8 × 10−12 atom/1.0 yr) = k(1 atom) k = 2.8 × 10−12yr−1

24.39

Plan: The rate constant, k, relates the number of radioactive nuclei to their decay rate through the equation A = kN. The number of radioactive nuclei is calculated by converting moles to atoms using Avogadro’s number. The decay rate is 1.39 × 105atoms/yr or more simply, 1.39 × 105yr–1 (the disintegrations are assumed). Solution: ΔN Decay rate = A = – = kN Δt ⎛1.00 ×10−12 mol ⎞⎛ 6.022 ×10 23 atoms ⎞⎟ −1.39×10 5 atoms = k ⎜⎜ − ⎟⎟⎟⎜⎜ ⎟ ⎟⎜ ⎜⎝ 1.00 yr 1 mol ⎠⎝ ⎠⎟⎟ 1.39 × 105 atom/yr = k(6.022 × 1011 atom) k = (1.39 × 105 atom/yr)/6.022 × 1011 atom k = 2.30820 × 10–7 = 2.31 × 10–7yr–1

24-12

24.40

ΔN = kN Δt – (–1.07 × 1015 atom/1.00 h) = k[(6.40 × 10–9mol)(6.022 × 1023 atom/mol)]

Decay rate = A = –

(1.07 × 1015 atom/1.00 h) = k (3.85408 × 1015 atom) k = [(1.07 × 1015 atom/1.00 h)]/(3.85408 × 1015 atom) k = 0.2776 = 0.278 h–1

24.41

Plan: Radioactive decay is a first-order process, so the integrated rate law is ln Nt = ln N0 – kt First find the value of k from the half-life and use the integrated rate law to find Nt. The time unit in the time and the k value must agree. Solution: t1/2 = 1.01 yr t = 3.75 × 103 h ln 2 ln 2 t1/2 = or k = t1 / 2 k ln 2 k= = 0.686284 yr–1 1.01 yr ln Nt = ln N0 – kt

⎛ 1 d ⎞⎛ ⎟⎟⎜ 1 yr ⎞⎟ ln Nt = ln [2.00 mg] – (0.686284 yr–1) (3.75×103 h) ⎜⎜ ⎝ 24 h ⎠⎟⎜⎜⎝ 365 d ⎠⎟⎟ ln Nt = 0.399361 Nt = e0.399361 Nt = 1.49087 = 1.49 mg

24.42

t1/2 = 1.60 × 103yr t=?h ln 2 ln 2 k= = = 0.000433217 yr–1 3 t1 / 2 1.60×10 yr

ln [0.185 g]= ln [2.50 g] – (0.000433217 yr–1)(t) t = 6010.129 yr ⎛ 365 d ⎞⎟⎛ 24 h ⎞ ⎟ = 5.264873 × 107 = 5.26 × 107 h ⎟⎜ t = (6010.129 yr) ⎜⎜ ⎜⎝ 1 yr ⎠⎟⎟⎝⎜ 1 d ⎠⎟⎟ 24.43

Plan: Lead-206 is a stable daughter of 238U. Since all of the 206Pb came from 238U, the starting amount of 238U was (270 μmol + 110 μmol) = 380 μmol = N0. The amount of 238U at time t (current) is 270 μmol = Nt. Find k from the first-order rate expression for half-life, and then substitute the values into the integrated rate law and solve for t. Solution: ln 2 ln 2 t1/2 = or k = t1 / 2 k ln 2 k= = 1.540327 × 10–10yr–1 4.5×10 9 yr N ln 0 = kt ln Nt = ln N0 – kt or Nt

24-13

380 μmol 270 μmol

= (1.540327 × 10–10yr–1)(t)

0.3417492937 = (1.540327 × 10–10yr–1)(t) t = 2.21868 × 109 = 2.2 × 109yr

24.44

The ratio (0.735) equals Nt / N0 so N0/Nt = 1.360544218 ln 2 ln 2 k= = = 1.2096809 × 10–4yr–1 t1 / 2 5730 yr N ln 0 = kt Nt ln 1.360544218 = (1.2096809 × 10–4yr–1)(t) 0.30788478 = (1.2096809 × 10–4yr–1)(t) t = 2.54517 × 103 = 2.54 × 103yr

24.45

Plan: The specific activity of the potassium-40 is the decay rate per mL of milk. Use the conversion factor 1 Ci = 3.70 × 1010 disintegrations per second (dps) to find the disintegrations per mL per s; convert the time unit to min and change the volume to 8 oz. Solution: 10 ⎛ 6 ×10−11 mCi ⎞⎟⎜⎛10−3 Ci ⎞⎛ ⎟⎟⎜⎜ 3.70 ×10 dps ⎞⎛ ⎟⎟⎜⎜ 60 s ⎞⎛ ⎟⎟⎜⎜1000 mL ⎞⎟⎟⎛⎜ 1 qt ⎞⎟⎛1 cup ⎞⎟ ⎜⎜ ⎟ ⎜ ⎜ Activity = ⎜ ⎟⎟⎜ ⎟⎟⎜ ⎟⎟⎜ ⎟⎟⎟⎜⎜⎜ 1.057 qt ⎟⎟⎟⎜⎜⎝ 4 cups ⎠⎟⎟⎟⎝⎜ 8 oz ⎠⎟⎟(8 oz) ⎜⎝ ⎟⎜⎜ ⎟⎜⎜1 min ⎠⎝ mL 1 Ci ⎟⎜⎜ 1 mCi ⎠⎝ ⎠⎝ ⎠⎝ ⎠ = 31.50426 = 30 dpm

24.46

Plutonium-239 (t1/2 = 2.41 × 104yr) Time = 7(t1/2) = 7(2.41 × 104yr) = 1.6870 × 105 = 1.69 × 105yr

24.47

Plan: Both Nt and N0 are given: the number of nuclei present currently, Nt, is found from the moles of 232Th. Each fission track represents one nucleus that disintegrated, so the number of nuclei disintegrated is added to the number of nuclei currently present to determine the initial number of nuclei, N0. The rate constant, k, is calculated from the half-life. All values are substituted into the first-order decay equation to find t. Solution: ln 2 ln 2 t1/2 = or k = t1 / 2 k ln 2 k= = 4.95105129 × 10–11yr–1 10 1.4 ×10 yr ⎛ 6.022 ×10 23 Th atoms ⎞⎟ 9 Nt = (3.1×10−15 mol Th) ⎜⎜ ⎟⎟ = 1.86682 × 10 atoms Th 1 mol Th ⎝⎜ ⎠⎟ N0 = 1.86682 × 109 atoms + 9.5 × 104 atoms = 1.866915 × 109 atoms N ln 0 = kt or ln Nt = ln N0 – kt Nt

1.866915×109 atoms 1.86682 ×109 atoms

= (4.95105129 × 10–11yr–1)(t)

24-14

5.088738 × 10–5 = (4.95105129 × 10–11yr–1)(t) t = 1.027809 × 106 = 1.0 × 106yr

24.48

The mole relationship between 40K and 40Ar is 1:1. Thus, 1.14 mmol 40Ar = 1.14 mmol 40K decayed. ln 2 ln 2 k= = = 5.5451774 × 10–10yr–1 t1 / 2 1.25×109 yr N ln 0 = kt Nt (1.38 + 1.14) mmol ln = (5.5451774 × 10–10yr–1)(t) 1.38 mmol 0.6021754 = (5.5451774 × 10–10yr–1)(t) t = 1.08594 × 109 = 1.09 × 109yr

24.49

27 13

1 Al + 42 He → 30 15 P + 0 n

They experimentally confirmed the existence of neutrons, and were the first to produce an artificial radioisotope. 24.50

Both gamma radiation and neutron beams have no charge, so neither is deflected by electric or magnetic fields. Neutron beams differ from gamma radiation in that a neutron has mass approximately equal to that of a proton. Researchers observed that a neutron beam could induce the emission of protons from a substance. Gamma rays do not cause such emissions.

24.51

A proton, for example, exits the first tube just when it becomes positive and the next tube becomes negative. Pushed by the first tube and pulled by the second, the proton accelerates across the gap between them.

24.52

Protons are repelled from the target nuclei due to the interaction of like (positive) charges. Higher energy is required to overcome the repulsion.

24.53

Plan: In a balanced nuclear equation, the total of mass numbers and the total of charges on the left side and the right side must be equal. In the shorthand notation, the nuclide to the left of the parentheses is the reactant while the nuclide written to the right of the parentheses is the product. The first particle inside the parentheses is the projectile particle while the second substance in the parentheses is the ejected particle. Solution: a) An alpha particle is a reactant with 10B and a neutron is one product. The mass number for the reactants is 10 + 4 = 14. So, the missing product must have a mass number of 14 – 1 = 13. The total atomic number for the reactants is 5 + 2 = 7, so the atomic number for the missing product is 7. 10 4 1 13 5 B + 2 He → 0 n + 7 N 2 b) A deuteron ( H) is a reactant with 28Si and 29P is one product. For the reactants, the mass number is 28 + 2 = 30 and the atomic number is 14 + 1 = 15. The given product has mass number 29 and atomic number 15, so the missing product particle has mass number 1 and atomic number 0. The particle is thus a neutron. 28 2 1 29 14 Si + 1 H → 0 n + 15 P c) The products are two neutrons and 244Cf with a total mass number of 2 + 244 = 246, and an atomic number of 98. The given reactant particle is an alpha particle with mass number 4 and atomic number 2. The missing reactant must have mass number of 246 – 4 = 242 and atomic number 98 – 2 = 96. Element 96 is Cm. 242 4 1 244 96 Cm + 2 He → 2 0 n + 98 Cf

24-15

24.54

31 29 a) 15 P + γ → 11 H + 01 n + 14 Si 31

P (γ, p, n) 29Si

10 1 257 b) 252 98 Cf + 5 B → 5 0 n + 103 Lr 252

Cf (10B, 5n) 257Lr

4 1 239 c) 238 92 U + 2 He → 3 0 n + 94 Pu 238

U (α, 3n) 239Pu

24.55

12 257 1 a) 249 98 Cf + 6 C → 104 Rf + 4 0 n 249 98

1 Cf + 157 N → 260 105 Db + 4 0 n

249 98

263 Cf + 188 O → 106 Sg + 4 01 n

249

b) Cf (12C, 4n) 257Rf 249 Cf (15N, 4n) 260Db 249

Cf (18O, 4n) 263Sg

24.56

Gamma radiation has no mass or charge while alpha particles are massive and highly charged. These differences account for the different effect on matter that these two types of radiation have. Alpha particles interact with matter more strongly than gamma particles due to their mass and charge. Therefore alpha particles penetrate matter very little. Gamma rays interact very little with matter due to the lack of mass and charge. Therefore gamma rays penetrate matter more extensively.

24.57

In the process of ionization, collision of matter with radiation dislodges an electron. The free electron and the positive ion that result are referred to as an ion-pair.

24.58

Ionizing radiation is more dangerous to children because their rapidly dividing cells are more susceptible to radiation than an adult’s slowly dividing cells.

24.59

The hydroxyl free radical forms more free radicals which go on to attack and change surrounding biomolecules, whose bonding and structure are delicately connected with their function. These changes are irreversible, as opposed to the reversible changes produced by OH–.

24.60

Plan: The rad is the amount of radiation energy absorbed in J per body mass in kg: 1 rad = 0.01 J/kg. Change the mass unit from pounds to kilograms. The conversion factor between rad and gray is 1 rad = 0.01 Gy. Solution: ⎛ 3.3×10−7 J ⎞⎛ ⎞⎟ 1 rad ⎟⎟⎜⎜ 2.205 lb ⎞⎛ ⎟⎟⎜⎜ –7 –7 a) Dose (rad) = ⎜⎜⎜ ⎟⎟⎜ ⎟⎟⎜ ⎟⎟ = 5.39 × 10 = 5.4 × 10 rad −2 ⎜⎝⎜ 135 lb ⎠⎝ ⎟⎜⎜ 1 kg ⎠⎝ ⎟⎜⎜1×10 J/ kg ⎠⎟⎟ ⎛ 0.01 Gy ⎞⎟ –9 –9 b) Gray (rad) = (5.39 ×10−7 rad) ⎜⎜ ⎟ = 5.39 × 10 = 5.4 × 10 Gy ⎜⎝ 1 rad ⎠⎟

24.61

⎛ 1 rad ⎞⎟ ⎜ ⎟⎟ = 0.0892 rad a) Dose (rad) = (8.92 × 10–4Gy) ⎜⎜ ⎜⎜⎝ 0.01 Gy ⎠⎟⎟ ⎛ 0.01 J / kg ⎞⎟ ⎜ –3 –3 b) Energy (J) = (0.0892 rad) ⎜⎜ ⎟⎟ (3.6 kg) = 3.2112 × 10 = 3.2 × 10 J ⎜⎜⎝ 1 rad ⎠⎟⎟

24-16

24.62

Plan: Multiply the number of particles by the energy of one particle to obtain the total energy absorbed. Convert the energy to dose in grays with the conversion factor 1 rad = 0.01 J/kg = 0.01 Gy. To find the millirems, convert grays to rads and multiply rads by RBE to find rems. Convert rems to mrems. Convert the dose to sieverts with the conversion factor 1 rem = 0.01 Sv. Solution: a) Energy (J) absorbed = (6.0×105 β)(8.74×10−14 J/ β) = 5.244 × 10–8 J Dose (Gy) =

⎛ ⎞ 5.244 ×10−8 J ⎜⎜ 1 rad ⎟⎟⎟⎜⎛⎜ 0.01 Gy ⎞⎟⎟ –10 –10 ⎜⎜ ⎟ = 7.4914 × 10 = 7.5 × 10 Gy ⎟⎜ 70. kg ⎜⎝⎜ 0.01 J kg ⎠⎟⎟⎟⎝⎜⎜ 1 rad ⎠⎟⎟

⎛ 1 rad ⎞⎟ ⎛ 1 mrem ⎞⎟ ⎜ ⎟⎟(1.0)⎜⎜ ⎟⎟ = 7.4914 × 10–5 = 7.5 × 10–5 mrem b) rems = rads × RBE = (7.4914 ×10−10 Gy) ⎜⎜ ⎜⎜ −3 ⎜⎜⎝ 0.01 Gy ⎠⎟⎟ ⎜⎝10 rem ⎠⎟⎟ ⎛10−3 rem ⎞⎛ ⎟⎟⎜⎜ 0.01 Sv ⎞⎟⎟ ⎜ –10 –10 Sv = (7.4914 ×10−5 mrem) ⎜⎜ ⎟⎟⎜ ⎟ = 7.4914 × 10 = 7.5 × 10 Sv ⎟⎜⎜ 1 rem ⎠⎟⎟ ⎜⎜⎝ 1 mrem ⎠⎝

24.63

(1.77×1010 β)(2.20×10−13 J / β) ⎜⎛103 g ⎞⎟⎜⎛ 1 rad ⎞⎟⎟ ⎟⎜⎜ a) Dose = ⎟⎟⎜ ⎟⎟ = 1.46943 = 1.47 rad ⎜⎜⎜ 265 g ⎟⎜⎝⎜ 0.01 J kg ⎠⎟⎟⎟ ⎜⎝ 1 kg ⎠⎜ b) Dose = (1.46943 rad)(0.01 Gy/1 rad) = 1.46943 × 10–2 = 1.47 × 10–2 Gy c) Dose = (1.46943 rad)(0.75 rem/rad)(0.01 Sv/rem) = 1.10207 × 10–2 = 1.10 × 10–2 Sv

24.64

−12 10 −13 ⎛ ⎞ ⎛ 2.50 pCi ⎞⎛ ⎛ 3600 s ⎞⎛ ⎟⎟⎜⎜1×10 Ci ⎞⎛ ⎟⎟⎜⎜ 3.70 ×10 dps ⎞⎟⎟ ⎟⎟⎜⎜ 8.25×10 J ⎞⎟⎟⎜⎜ 1 rad ⎟⎟ ⎜ ⎜ ⎟⎟ Dose = ⎜⎜ ⎟⎟⎜ ⎟⎟⎜ ⎟⎟ 65 h ⎜⎜ ⎟⎟⎜ ⎟⎟⎜⎜ 1 Ci ⎟⎜⎜ 1 pCi ⎠⎝ ⎟⎜⎜ ⎟⎜⎜ disint. ⎠⎜ ⎟⎜⎝⎜ 0.01 J kg ⎠⎟⎟⎟ ⎜⎜⎝ 95 kg ⎠⎝ ⎠⎟ ⎝⎜⎜ 1 h ⎠⎝

(

)

= 1.8796974 × 10–8 = 1.9 × 10–8 rad Dose = (1.8796974 × 10–8 rad)(0.01 Gy/1 rad) = 1.8796974 × 10–10 = 1.9 × 10–10 Gy 24.65

Use the time and disintegrations per second (Bq) to find the number of 60Co atoms that disintegrate, which equals the number of β particles emitted. The dose in rads is calculated as energy absorbed per body mass. −14 ⎞ ⎛ ⎞⎛ 3 ⎞⎛ ⎞⎛ ⎛ 60 s ⎞⎟⎛⎜ 1 rad ⎞⎟ ⎜ 475 Bq ⎟⎟⎜⎜10 g ⎟⎟⎜⎜1 dps ⎟⎟⎜⎜ 5.05×10 J ⎟⎟ ⎜ ⎟⎟ ⎟⎟⎜⎜ Dose = ⎜⎜ ⎟⎟⎜ ⎟⎟⎜ ⎟⎟⎜ ⎟⎟ (24.0 min) ⎜⎜ ⎟ ⎜⎝⎜1.858 g ⎠⎝ ⎜⎝⎜1 min ⎠⎜ ⎟⎜⎜ 1 kg ⎠⎝ ⎟⎜⎜ 1 Bq ⎠⎝ ⎟⎜⎜ 1 disint. ⎠⎟ ⎟⎟⎜⎜⎝⎜ 0.01 J kg ⎠⎟⎟⎟ = 1.8591 × 10–3 = 1.86 × 10–3 rad

24.66

A healthy thyroid gland incorporates dietary I – into I-containing hormones at a known rate. To assess thyroid function, the patient drinks a solution containing a trace amount of Na131I, and a scanning monitor follows the uptake of 131I into the thyroid. Technetium-99 is often used for imaging the heart, lungs, and liver.

24.67

NAA does not destroy the sample while chemical analysis does. Neutrons bombard a non-radioactive sample, “activating” or energizing individual atoms within the sample to create radioisotopes. The radioisotopes decay back to their original state (thus, the sample is not destroyed) by emitting radiation that is different for each isotope.

24.68

In positron-emission tomography (PET), the isotope emits positrons, each of which annihilates a nearby electron. In the process, two γ photons are emitted simultaneously, 180° apart from each other. Detectors locate the sites and the image is analyzed by computer.

24.69

The concentration of 59 Fe in the steel sample and the volume of oil would be needed.

24-17

24.70

The oxygen in formaldehyde comes from methanol because the oxygen isotope in the methanol reactant appears in the formaldehyde product. The oxygen isotope in the chromic acid reactant appears in the water product, not the formaldehyde product. The isotope traces the oxygen in methanol to the oxygen in formaldehyde.

24.71

The mass change in a chemical reaction was considered too minute to be significant and too small to measure with even the most sophisticated equipment.

24.72

When a nucleus forms from its nucleons, there is a decrease in mass called the mass defect. This decrease in mass is due to mass being converted to energy to hold the nucleus together. This energy is called the binding energy.

24.73

Energy is released when a nuclide forms from nucleons. The nuclear binding energy is the amount of energy holding the nucleus together. Energy is absorbed to break the nucleus into nucleons and is released when nucleons “come together.”

24.74

The binding energy per nucleon is the average amount of energy per each component (proton and neutron) part of the nuclide. The binding energies per nucleon are helpful in comparing the stabilities of different combinations and to provide information on the potential processes a nuclide can undergo to become more stable. The binding energy per nucleon varies considerably. The greater the binding energy per nucleon, the more strongly the nucleons are held together and the more stable the nuclide.

24.75

Plan: The conversion factors are: 1 MeV = 106 eV and 1 eV = 1.602 × 10−19 J. Solution: ⎛10 6 eV ⎞⎟ 4 a) Energy (eV) = (0.01861 MeV)⎜⎜ ⎟ = 1.861 × 10 eV ⎜⎝ 1 MeV ⎠⎟⎟ −19 ⎛106 eV ⎞⎛ ⎟⎟⎜⎜1.602 ×10 J ⎞⎟⎟ = 2.981322 × 10−15 = 2.981 × 10−15 J b) Energy (J) = (0.01861 MeV)⎜⎜ ⎟ ⎟⎜ ⎜⎝ 1 MeV ⎠⎝ 1 eV ⎠⎟⎟

24.76

24.77

24.78

⎛1000 J ⎞⎛ 1 eV ⎞⎟ 6 6 ⎟⎟⎜⎜ a) Energy (eV) = (1.57 ×10−15 kJ) ⎜⎜ ⎟ 1.602 ×10−19 J ⎠⎟⎟ = 9.8002 × 10 = 9.80 × 10 eV ⎝ 1 kJ ⎠⎝ ⎛ 1 MeV ⎞⎟ b) Energy (MeV) = (9.8002 ×10 6 eV) ⎜⎜ 6 9.8002 = 9.80 MeV ⎝10 eV ⎠⎟⎟

Plan: Convert moles of 239Pu to atoms of 239Pu using Avogadro’s number. Multiply the number of atoms by the energy per atom (nucleus) and convert the MeV to J using the conversion 1 eV = 1.602 × 10–19 J. Solution: ⎛ 6.022 ×10 23 atoms ⎞⎟ ⎜ ⎟⎟ = 9.033 × 1023 atoms Number of atoms = (1.5 mol 239 Pu) ⎜⎜ ⎟ mol ⎜⎝ ⎠⎟ −19 ⎞ ⎛ ⎞⎛ 6 ⎞⎛ ⎜ 5.243 MeV ⎟⎟⎜⎜ 10 eV ⎟⎟⎜⎜1.602 ×10 J ⎟⎟ 11 11 Energy (J) = (9.033×10 23 atoms) ⎜⎜ ⎟⎟⎜ ⎟⎟⎜ ⎟⎟ = 7.587075 × 10 = 7.6 × 10 J ⎜⎜⎝ 1 atom ⎠⎝ 1 eV ⎟⎜⎜1 MeV ⎠⎝ ⎟⎜⎜ ⎟⎠ 3 ⎛ 8.11×10 5 kJ ⎞⎛ ⎞⎛ ⎞⎟ 1 eV 1 mol 49 Cr ⎟⎟⎜⎜10 J ⎞⎛ ⎟⎟⎜⎜ ⎟⎟⎜⎜1 MeV ⎞⎛ ⎟⎟⎜⎜ ⎟ Energy (MeV) = ⎜⎜⎜ ⎟ ⎟ ⎟ ⎟ −3 49 ⎟⎟⎜⎜⎜ 1 kJ ⎟⎟⎜⎜⎜1.602 ×10−19 J ⎟⎟⎜⎜⎜ 106 eV ⎟⎟⎜⎜⎜ 6.022 ×10 23 nuclei ⎟⎟⎟ ⎜⎜⎝ 3.2 ×10 mol Cr ⎠⎝ ⎠⎝ ⎠⎝ ⎠⎝ ⎠ = 2.6270 = 2.6 MeV

24.79

Plan: Oxygen-16 has eight protons and eight neutrons. First find the Δm for the nucleus by subtracting the given mass of one oxygen atom from the sum of the masses of eight 1H atoms and eight neutrons. Use the conversion

24-18

factor 1 amu = 931.5 MeV to convert Δm to binding energy in MeV and divide the binding energy by the total number of nucleons (protons and neutrons) in the oxygen nuclide to obtain binding energy per nucleon. Convert Δm of one oxygen atom to MeV using the conversion factor for binding energy/atom. To obtain binding energy per mole of oxygen, use the relationship ΔE = Δmc2. Δm must be converted to units of kg/mol. Solution: Mass of 8 1H atoms = 8 × 1.007825 = 8.062600 amu Mass of 8 neutrons = 8 × 1.008665 = 8.069320 amu Total mass =16.131920 amu Δm = 16.131920 − 15.994915 = 0.137005 amu/16O = 0.137005 g/mol16O ⎛ 0.137005 amu 16 O ⎞⎛ ⎟⎟⎜⎜ 931.5 MeV ⎞⎟⎟ ⎜ a) Binding energy (MeV/nucleon) = ⎜⎜ ⎟⎟⎜ ⎟ = 7.976259844 = 7.976 MeV/nucleon ⎜⎝⎜ 16 nucleons ⎠⎝ ⎟⎜⎜ 1 amu ⎠⎟⎟ ⎛ 0.137005 amu 16 O ⎞⎛ ⎟⎟⎜⎜ 931.5 MeV ⎞⎟⎟ ⎜ b) Binding energy (MeV/atom) = ⎜⎜ ⎟⎟⎜ ⎟ = 127.6201575 = 127.6 MeV/atom ⎜⎝⎜ ⎟⎜⎜ 1 amu ⎠⎟⎟ 1 atom ⎠⎝ c) ΔE = Δmc2

⎛ ⎞ ⎜ 1 J ⎟⎟⎛ 1 kJ ⎞⎟ ⎛ 0.137005 g 16 O ⎞⎟⎜⎛ 1 kg ⎞⎟ 2⎜ ⎜⎜ 8 ⎜ ⎟ ⎟ ⎟⎜ ⎟⎟⎜⎜ Binding energy (kJ/mol) = ⎜ ⎟⎟⎟⎜⎜⎜103 g ⎟⎟⎟ 2.99792 ×10 m/s ⎜⎜⎜ kg • m 2 ⎟⎟⎟⎜⎜⎜103 J ⎟⎟⎟ ⎜⎝ mol ⎠⎝ ⎠ ⎝ ⎠ ⎝⎜ s2 ⎠⎟⎟

(

)

= 1.23133577 × 1010 = 1.23134 × 1010 kJ/mol 24.80

Δm is calculated from the mass of 82 protons (1H) and 124 neutrons vs. the mass of the lead nuclide. Mass of 82 1H atoms = 82 × 1.007825 = 82.641650 amu Mass of 124 neutrons = 124 × 1.008665 = 125.074460 amu Total mass = 207.716110 amu Δm = 207.716110 − 205.974440 = 1.741670 amu/206Pb = 1.741670 g/mol 206Pb ⎛1.741670 amu 206 Pb ⎞⎛ ⎟⎟⎜⎜ 931.5 MeV ⎞⎟⎟ ⎜ a) Binding energy (MeV/nucleon) = ⎜⎜ ⎟⎟⎟⎜⎜⎜ 1 amu ⎟⎟⎟ = 7.8755612 = 7.876 MeV/nucleon ⎜⎝⎜ 206 nucleons ⎠⎝ ⎠ ⎛1.741670 amu 206 Pb ⎞⎛ ⎟⎟⎜⎜ 931.5 MeV ⎞⎟⎟ ⎜ b) Binding energy (MeV/atom) = ⎜⎜ ⎟⎟⎜ ⎟ = 1622.3656 = 1622 MeV/atom ⎜⎝⎜ 1 atom ⎟⎜⎜ 1 amu ⎠⎟⎟ ⎠⎝

⎛ ⎞⎟ ⎜⎜ 1J ⎛1.741670 g 206 Pb ⎞⎟⎜⎛ 1 kg ⎞⎟ ⎟⎛ 1kJ ⎞ ⎜⎜ 8 2 ⎜ ⎟ ⎟⎟⎜⎜ ⎟⎟⎟⎜⎜ 3 ⎟⎟⎟ (2.99792×10 m/s) ⎜⎜ c) Binding energy (kJ/mol) = ⎜ 2 3 ⎟ ⎟ ⎟ ⎜⎝ mol ⎟⎜⎜10 g ⎠⎟ ⎜⎜ kg • m 2 ⎟⎟⎟⎜⎜⎝10 J ⎠⎟ ⎠⎝ ⎝ s ⎠⎟ = 1.5653301 × 1011 = 1.56533 × 1011 kJ/mol 24.81

Plan: Cobalt-59 has 27 protons and 32 neutrons. First find the Δm for the nucleus by subtracting the given mass of one cobalt atom from the sum of the masses of 27 1H atoms and 32 neutrons. Use the conversion factor 1 amu = 931.5 MeV to convert Δm to binding energy in MeV and divide the binding energy by the total number of nucleons (protons and neutrons) in the cobalt nuclide to obtain binding energy per nucleon. Convert Δm of one cobalt atom to MeV using the conversion factor for binding energy/atom. To obtain binding energy per mole of cobalt, use the relationship ΔE = Δmc2. Δm must be converted to units of kg/mol. Solution: Mass of 27 1H atoms = 27 × 1.007825 = 27.211275 amu Mass of 32 neutrons = 32 × 1.008665 = 32.27728 amu

24-19

Total mass = 59.488555 amu Δm = 59.488555 – 58.933198 = 0.555357 amu/59Co = 0.555357 g/mol59Co ⎛ 0.555357 amu 59 Co ⎞⎛ ⎟⎟⎜⎜ 931.5 MeV ⎞⎟⎟ ⎜ a) Binding energy (MeV/nucleon) = ⎜⎜ ⎟⎟⎜ ⎟ = 8.768051619 = 8.768 MeV/nucleon ⎜⎝⎜ 59 nucleons ⎟⎜⎜ 1 amu ⎠⎟⎟ ⎠⎝ 59 ⎛ ⎞⎛ ⎞ ⎜ 0.555357 amu Co ⎟⎟⎜⎜ 931.5 MeV ⎟⎟ b) Binding energy (MeV/atom) = ⎜⎜ ⎟⎟⎜ ⎟ = 517.3150 = 517.3 MeV/atom 1 atom ⎟⎜⎜ 1 amu ⎠⎟⎟ ⎜⎜⎝ ⎠⎝ c) Use ΔE = Δmc2 ⎛ ⎞ ⎜⎜ 1 J ⎟⎟⎛ 1 kJ ⎞⎟ ⎛ 0.555357 g 59 Co ⎞⎟⎜⎛ 1 kg ⎞⎟ ⎜⎜ ⎟⎟⎜⎜ 8 2 ⎜ ⎟ ⎟ ⎟⎜ Binding energy (kJ/mol) = ⎜ ⎟⎟⎟⎜⎜⎜103 g ⎟⎟⎟ (2.99792 ×10 m/s) ⎜⎜⎜ kg • m 2 ⎟⎟⎟⎜⎜⎜103 J ⎟⎟⎟ ⎜⎝ mol ⎠⎝ ⎠ ⎝ ⎠ ⎜⎝ s2 ⎠⎟⎟ 10 10 = 4.9912845 × 10 = 4.99128 × 10 kJ/mol 24.82

Δm is calculated from the mass of 53 protons (1H) and 78 neutrons vs. the mass of the iodine nuclide. Mass of 53 1H atoms = 53 × 1.007825 = 53.414725 amu Mass of 78 neutrons = 78 × 1.008665 = 78.675870 amu Total mass = 132.090595 amu Δm = 132.090595 – 130.906114 = 1.184481 amu/131I = 1.184481 g/mol131I ⎛1.184481 amu 131 I ⎞⎛ ⎟⎟⎜⎜ 931.5 MeV ⎞⎟⎟ ⎜ a) Binding energy (MeV/nucleon) = ⎜⎜ ⎟⎟⎜ ⎟ = 8.422473676 = 8.422 MeV/nucleon ⎟⎜⎜ 1 amu ⎠⎟⎟ ⎜⎜⎝ 131 nucleons ⎠⎝ ⎛1.184481 amu 131 I ⎞⎛ ⎟⎟⎜⎜ 931.5 MeV ⎞⎟⎟ ⎜ b) Binding energy (MeV/atom) = ⎜⎜ ⎟⎟⎜ ⎟ = 1103.34405 = 1103 MeV/atom 1 atom ⎟⎜⎜ 1 amu ⎠⎟⎟ ⎜⎜⎝ ⎠⎝ c) ΔE = Δmc2

⎛ ⎞ ⎜⎜ 1 J ⎟⎟⎛ 1 kJ ⎞ ⎛1.184481 g 131 I ⎞⎟⎜⎛ 1 kg ⎞⎟ ⎜⎜ 8 2 ⎜ ⎟ ⎟ ⎟⎟⎟⎜⎜ 3 ⎟⎟⎟ Binding energy (kJ/mol) = ⎜ ⎟⎟⎜⎜ 3 ⎟⎟ (2.99792×10 m/s) ⎜⎜ 2 ⎜⎝ mol ⎟⎠⎝⎜⎜10 g ⎠⎟ ⎜⎜ kg • m 2 ⎟⎟⎟⎜⎝⎜10 J ⎠⎟ ⎝ s ⎠⎟ = 1.06455518 × 1011 = 1.06456 × 1011 kJ/mol 24.83

0 80 a) 80 35 Br → − 1 β + 36 Kr (reaction 1) 80 35

80 Br + −01 e → 34 Se (reaction 2)

b) Reaction 1:

Δm = 79.918528 – 79.916380 = 0.002148 amu

Reaction 2:

Δm = 79.918528 – 79.916520 = 0.002008 amu

Since ΔE = (Δm)c2, the greater mass change (reaction 1) will release more energy. 24.84

The minimum number of neutrons from each fission event that must be absorbed by the nuclei to sustain the chain reaction is one. In reality, due to neutrons lost from the fissionable material, two to three neutrons are generally needed to continue a self-sustaining chain reaction.

24-20

24.85

24.86

In both radioactive decay and fission, radioactive particles are emitted, but the process leading to the emission is different. Radioactive decay is a spontaneous process in which unstable nuclei emit radioactive particles and energy. Fission occurs as the result of high-energy bombardment of nuclei with small particles that cause the nuclides to break into smaller nuclides, radioactive particles, and energy. In a chain reaction, all fission events are not the same. The collision between the small particle emitted in the fission and the large nucleus can lead to splitting of the large nuclei in a number of ways to produce several different products. Enriched fissionable fuel is needed in the fuel rods to ensure a sustained chain reaction. Naturally occurring U is only present in a concentration of 0.7%. This is consistently extracted and separated until its concentration is between 3-4%.

235

24.87

a) Control rods are movable rods of cadmium or boron which are efficient neutron absorbers. In doing so, they regulate the flux of neutrons to keep the reaction chain self-sustaining which prevents the core from overheating. b) The moderator is the substance flowing around the fuel and control rods that slows the neutrons, making them better at causing fission. c) The reflector is usually a beryllium alloy around the fuel-rod assembly that provides a surface for neutrons that leave the assembly to collide with and therefore, return to the fuel rods.

24.88

The water serves to slow the neutrons so that they are better able to cause a fission reaction. Heavy water ( 21 H 2 O or D2O) is a better moderator because it does not absorb neutrons as well as light water ( 11 H 2 O ) does, so more neutrons are available to initiate the fission process. However, D2O does not occur naturally in great abundance, so production of D2O adds to the cost of a heavy water reactor. In addition, if heavy water does absorb a neutron, it becomes tritiated, i.e., it contains the isotope tritium, 31 H , which is radioactive.

24.89

The advantages of fusion over fission are the simpler starting materials (deuterium and tritium), and no long-lived toxic radionuclide by-products.

24.90

Virtually all the elements heavier than helium, up to and including iron, are produced by nuclear fusion reactions in successively deeper and hotter layers of massive stars. Iron is the point at which fusion reactions cease to be energy producers. Elements heavier than iron are produced by a variety of processes, primarily during a supernova event, which distribute the Sun’s ash into the cosmos to form next generation suns and planets. Thus, the high cosmic and Earth abundance of iron is consistent with it being the most stable of all nuclei.

24.91

Mass of reactants: 3.01605 + 2.0140 = 5.03005 amu Mass of products: 4.00260 + 1.008665 = 5.011265 amu Δm = mass of reactants – mass of products = 5.03005 – 5.011265 = 0.018785 amu = 0.018785 g/mol ΔE = Δmc2

⎛ ⎞ ⎜ 1 J ⎟⎟⎛ 1 kJ ⎞⎟ ⎛ 0.018785 g ⎞⎟⎛⎜ 1 kg ⎞⎟ ⎟⎜ 8 2⎜ ⎜ ⎜ ⎟ ⎟ ⎜ ⎟ Energy (kJ/mol) = ⎜⎜ ⎟⎜ 3 ⎟⎟ (2.99792×10 m/s) ⎜⎜ kg • m 2 ⎟⎟⎟⎜⎜⎜ 3 ⎟⎟ ⎟⎜ ⎟ ⎜⎝ mol 10 J ⎠⎟ ⎜ ⎠⎝⎜10 g ⎠⎟ ⎜ ⎟ ⎝ ⎜⎝ s2 ⎠⎟⎟ = 1.6883064 × 109 = 1.69 × 109 kJ/mol 24.92

243 95

Am → 42 He + 239 93 Np

239 93

Np → −01 β + 239 94 Pu

239 94

Pu → 42 He + 235 92 U

239 93

235 Np, 239 94 Pu, and 92 U were present as products in the decay of Am-243.

24-21

24.93

Plan: Use the masses given in the problem to calculate the mass change (reactant – products) for the reaction. The conversion factor between amu and kg is 1 amu = 1.66054 × 10–27 kg. Use the relationship E = Δmc2 to convert the mass change to energy. Solution: 239 4 a) 243 96 Cm → 94 Pu + 2 He Δm (amu) = 243.0614 amu − (4.0026 + 239.0522) amu = 0.0066 amu ⎛1.661×10−24 g ⎞⎟⎛ 1 kg ⎞ ⎜ ⎟⎟ = 1.09626 × 10–29 = 1.1 × 10–29 kg ⎟⎟⎜⎜ Δm (kg) = (0.0066 amu)⎜⎜ ⎟⎜ 1 amu ⎟⎠⎝⎜103 g ⎠⎟⎟ ⎜⎝ ⎛ ⎞ ⎜ 1 J ⎟⎟ ⎟⎟ = 9.85266 × 10–13 = 9.9 × 10–13 J 2 −29 8 2⎜ b) E = Δmc = (1.09626 ×10 kg)(2.99792 ×10 m /s) ⎜⎜⎜ ⎟⎟ 2 kg m • ⎜ ⎟ 2⎟ ⎜⎝ s ⎠⎟ ⎛ 9.85266 ×10−13 J ⎞⎛ 6.022 ×10 23 reactions ⎞⎟⎛ 1 kJ ⎞⎟ 8 8 c) E released = ⎜⎜ ⎟⎟⎜⎜ ⎟⎜ ⎟ = 5.93317 × 10 = 5.9 × 10 kJ/mol ⎟ ⎟⎜ reaction mol ⎝⎜ ⎠⎝ ⎠⎟⎟⎜⎝103 J ⎠⎟ This is approximately one million times larger than a typical heat of reaction.

24.94

a) First, determine the amount of activity released by the 239Pu for the duration spent in the body (16 h) using the relationship A = kN. The rate constant is derived from the half-life and N is calculated using the molar mass and Avogadro’s number. ⎛ ⎞⎛ ⎞⎛ ln 2 ln 2 ⎟⎟⎜⎜ 1 yr ⎟⎟⎜⎜1 day ⎞⎛ ⎟⎟⎜⎜ 1 h ⎞⎟⎟ ⎜ –13 –1 k= = ⎜⎜ ⎟ ⎟ 4 ⎟⎟⎜⎜⎜ 365.25 day ⎟⎟⎜⎜⎜ 24 h ⎟⎟⎟⎜⎜⎜ 3600 s ⎟⎟⎟ = 9.113904 × 10 s t1 / 2 ⎜⎜⎝ 2.41×10 yr ⎠⎝ ⎠⎝ ⎠⎝ ⎠ 23 ⎛10−6 g ⎞⎛ ⎞⎛ 1 mol Pu ⎟⎜ 6.022 ×10 atoms Pu ⎞⎟ ⎜ ⎟⎟⎟⎜⎜ ⎟⎜ ⎟⎟ = 2.5196653 × 1015 atoms Pu N = (1.00 μg Pu) ⎜⎜ ⎟⎟⎜⎜⎜ 239 g Pu ⎟⎟⎟⎜⎜⎜ ⎟ 1 mol Pu ⎜⎜⎝ 1 μg ⎠⎝ ⎠⎝ ⎠⎟ ⎛ 1 disint. ⎞⎟ ⎜ 3 A = kN = (9.113904 × 10–13 s–1)( 2.5196653 × 1015 atoms Pu) ⎜⎜ ⎟⎟ = 2.2963988 × 10 d/s ⎜⎜⎝1 atom Pu ⎠⎟⎟ Each disintegration releases 5.15 MeV, so d/s can be converted to MeV. Convert MeV to J (using 1.602 × 10–13 J = 1 MeV) and J to rad (using 0.01 J/kg = 1 rad). ⎛ 2.2963988×103 d/s ⎞⎟⎛ 5.15 MeV ⎞⎛1.602 ×10−13 J ⎞⎟⎛⎜ 1 rad ⎞⎟⎛ 3600 s ⎞⎟ ⎜ ⎟⎜⎜ ⎜ ⎟⎟⎟⎜⎜⎜ ⎟⎟⎜ Energy = ⎜ ⎟⎟⎟⎜⎜⎜ ⎟⎟⎟ (16 h) ⎟⎟⎟⎜⎜⎜ J ⎟ ⎜⎝⎜ ⎟⎜ ⎜ 85 kg MeV ⎠⎟⎜⎝⎜ 0.01 kg ⎠⎟⎟⎟⎝⎜⎜ 1 h ⎠⎟ ⎠⎟⎝ disint. ⎠⎝⎜ = 1.2838686 × 10–4 = 1.28 × 10–4rads b) Since 0.01 Gy = 1 rad, the worker receives: Dose = (1.2838686 × 10–4 rad)(0.01 Gy/rad) = 1.2838686 × 10–6 = 1.28 × 10–6Gy

24.95

Plan: Determine k for 14C using the half-life (5730 yr). Determine the mass of carbon in 4.58 g of CaCO3. Divide the given activity of the C in d/min by the mass of carbon to obtain the activity in d/min∙g; this is At and is compared to the activity of a living organism (A0 = 15.3 d/min∙g) in the integrated rate law, solving for t. Solution: ln 2 ln 2 k= = = 1.2096809 × 10–4yr–1 t1 / 2 5730 yr ⎛ 1 mol CaCO ⎞⎛ ⎞⎛ ⎞ ⎜ ⎜ 1 mol C ⎟⎟⎜⎜12.01 g C ⎟⎟ 3 ⎟ Mass (g) of C = (4.58 g CaCO3 ) ⎜⎜ ⎟⎟⎟⎜⎜ ⎟⎟⎜ ⎟ = 0.5495634 g C ⎟⎜⎜1 mol CaCO3 ⎠⎝ ⎟⎜⎜ 1 mol C ⎠⎟⎟ ⎜⎜⎝100.09 g CaCO 3 ⎠⎝

24-22

At =

3.2 d/min = 5.8228 d/min∙g 0.5495634 g

Using the integrated rate law:

⎛A ⎞ ln ⎜⎜⎜ t ⎟⎟⎟ = –kt A0 = 15.3 d/min∙g (the ratio of 12 C:14 C in living organisms) ⎜⎝ A0 ⎠⎟ ⎛ 5.8228 d/min • g ⎞⎟ ⎜ –4 –1 ln ⎜⎜ ⎟⎟ = – (1.2096809 × 10 yr )(t) ⎜⎜⎝ 15.3 d/min • g ⎠⎟⎟ t = 7986.17 = 8.0 × 103yr

24.96

Find the rate constant from the rate of decay and the initial number of atoms. Use rate constant to calculate halflife. Initial number of atoms: 23 ⎛10−6 g ⎞⎛ ⎟⎟⎜⎜ 1 mol RaCl 2 ⎞⎛ ⎟⎟⎜⎜ 1 mol Ra ⎞⎛ ⎟⎟⎜⎜ 6.022 ×10 Ra atoms ⎞⎟⎟ ⎜ Ra atoms = (5.4 μg RaCl 2 ) ⎜⎜ ⎟⎟⎟⎜⎜⎜ 297 g RaCl ⎟⎟⎟⎜⎜⎜1 mol RaCl ⎟⎟⎟⎜⎜⎜ ⎟⎟ ⎜⎝⎜ 1 μg ⎠⎝ 1 mol Ra ⎠⎟ 2 ⎠⎝ 2 ⎠⎝ = 1.09490909 × 1016 Ra atoms A A = kN or k = N ⎛ 1.5×10 5 Bq ⎟⎟⎞⎛⎜1 d s ⎞⎟⎟ ⎜ –11 –1 k = ⎜⎜ ⎟⎟⎜⎜⎜ Bq ⎟⎟ = 1.36997675 × 10 s ⎜⎜⎝1.09490909×1016 Ra atoms ⎟⎟⎜ ⎝ ⎠ ⎠ ln 2 ln 2 t1/2 = = = 5.05955 × 1010 = 5.1 × 1010 s k 1.36997675×10−11 s−1

24.97

ln 2 ln 2 = = 1.2096809 × 10–4yr–1 t1 / 2 5730 yr ⎛1 mol 14 C ⎞⎛ 6.022 ×10 23 atoms 14 C ⎞⎟ ⎟⎟⎟⎜⎜ ⎟⎟ = 4.3014 × 1014 atoms 14C Number of atoms = (10−8 g 14 C) ⎜⎜⎜ 14 ⎟⎟⎜⎜⎜ ⎟ 1 mol 14 C ⎜⎜⎝ 14 g C ⎠⎝ ⎠⎟ k=

A = kN = [(1.2096809 × 10–4yr–1)(4.3014 × 1014 atoms 14C)](1 disintegration/1 atom) = 5.2033 × 1010dpyr ⎛ 5.2033×1010 dpyr ⎞⎟ ⎛ 0.156 MeV ⎞⎟⎛⎜1.602 ×10−13 J ⎞⎟⎛⎜ 1 rad ⎞⎟⎟ ⎜⎜ –3 –3 ⎟ ⎟⎟⎜⎜ Dose = ⎜ ⎟⎟(yr)⎜⎜⎜ ⎟⎟ = 2.001 × 10 = 10 rad ⎟⎟⎜⎜⎜ ⎟⎜ J ⎟ ⎜⎜⎝ ⎟ ⎜ 65 kg ⎟⎜⎝⎜ 0.01 kg ⎠⎟⎟ ⎝ disint. ⎠⎝⎜ 1 MeV ⎠⎜ ⎠⎟

24.98

Plan: Determine how many grams of AgCl are dissolved in 1 mL of solution. The activity of the radioactive Ag+ indicates how much AgCl dissolved, given a starting sample with a specific activity (175 nCi/g). Convert g/mL to mol/L (molar solubility) using the molar mass of AgCl. Solution: ⎞⎛ ⎛1.25×10−2 Bq ⎞⎟⎜⎛1 dps ⎞⎛ 1 Ci ⎟⎟⎜⎜ ⎟⎟⎜⎜ 1 nCi ⎞⎛ ⎟⎟⎜⎜1 g AgCl ⎞⎟⎟ ⎜ –6 ⎟⎟⎜⎜ Concentration = ⎜⎜ ⎟ ⎟ ⎜ ⎜ − 10 9 ⎟ ⎟ ⎟⎟⎜⎜ 1 Bq ⎟⎜⎜ 3.70×10 dps ⎟⎜⎜10 Ci ⎟⎟⎟⎜⎜⎜ 175 nCi ⎟⎟⎟ = 1.93050 × 10 g AgCl/mL ⎜⎝ mL ⎠⎝ ⎠⎝ ⎠⎝ ⎠⎝ ⎠ − 6 ⎛1.93050 ×10 g AgCl ⎞⎟⎜⎛ 1 mol AgCl ⎞⎟⎛ 1 mL ⎞ ⎟ = 1.34623 × 10–5 = 1.35 × 10–5M AgCl ⎟⎜ Molarity = ⎜⎜ ⎟ ⎟⎟⎜⎜143.4 g AgCl ⎠⎟⎟⎝⎜10−3 L ⎠⎟⎟ ⎜⎝ mL ⎠⎝

24.99

a) The process shown is fission in which a neutron bombards a large nucleus, splitting that nucleus into two nuclei of intermediate mass.

24-23

144 90 1 b) 01 n + 235 92 U → 55 Cs + 37 Rb + 2 0 n

c) 144 55 Cs, with 55 protons and 89 neutrons, has an n/p ratio of 1.6. This ratio places this isotope above the band of stability and decay by beta particle emission is expected. 24.100 Plan: Determine the value of k from the half-life. Then determine the fraction from the integrated rate law. Solution: ln 2 ln 2 k= = = 9.90210 × 10–10 yr–1 8 t1 / 2 7.0 ×10 yr N0 ln = kt = (9.90210 × 10–10 yr–1)(2.8 × 109 yr) = 2.772588 Nt N0 = 15.99998844 Nt Nt = 0.062500 = 6.2 × 10–2 N0 24.101 Determine the value of k from the half-life. Then determine the age from the integrated rate law. ln 2 ln 2 k= = = 1.540327 × 10–10 yr–1 t1 / 2 4.5×10 9 yr N ln 0 = kt Nt ⎛ 6 + 9 ⎞⎟ = (1.540327 × 10–10 yr–1)(t) ln ⎜⎜ ⎝ 9 ⎠⎟⎟ 0.5108256 = (1.540327 × 10–10 yr–1)(t) t = 3.316345 × 109 = 3.3 × 109 yr

24.102 Plan: Find the rate constant, k, using any two data pairs (the greater the time between the data points, the greater the reliability of the calculation). Calculate t1/2 using k. Once k is known, use the integrated rate law to find the percentage lost after 2 h. The percentage of isotope remaining is the fraction remaining after 2.0 h (Nt where t = 2.0 h) divided by the initial amount (N0), i.e., fraction remaining is Nt / N0. Solve the first-order rate expression for Nt / N0, and then subtract from 100% to get fraction lost. Solution: N a) ln t = –kt N0 ⎛ 495 photons/s ⎞⎟ ⎜ ⎟⎟ = –k(20 h) ln ⎜⎜ ⎜⎝⎜ 5000 photons/s ⎠⎟⎟ –2.312635 = –k(20 h) k = 0.11563 h–1 ln 2 ln 2 t1/2 = = = 5.9945 = 5.99 h (Assuming the times are exact, and the emissions have three k 0.11563 h−1 significant figures.) N b) ln t = –kt N0 N ln t = – (0.11563 h–1)(2.0 h) = –0.23126 N0 Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

24-24

Nt Nt = 0.793533 × 100% = 79.3533% N0 N0 The fraction lost upon preparation is 100% – 79.3533% = 20.6467% = 21%. 24.103 k =

ln 2 ln 2 = = 5.5451774 × 10–10yr–1 t1 / 2 1.25×109 yr

A = kN where A = dps

N = number of atoms Number of atoms = (1.0 mol40K)(6.022 × 1023 atoms/mol) = 6.022 × 1023 atoms 40K ⎛ 1 yr ⎞⎛ ⎛ 5.5451774 ×10−10 ⎞⎟ ⎟⎟⎜⎜1 day ⎞⎛ ⎟⎟⎜⎜ 1 h ⎞⎛ ⎟⎟⎜⎜1 disint. ⎞⎟⎟ ⎜ 7 ⎟⎟ (6.022 ×10 23 atoms) ⎜⎜ A = ⎜⎜ ⎟⎟⎜ ⎟⎟⎜ ⎟⎟⎜ ⎟ = 1.05816 × 10 dps ⎜⎝ ⎜⎜⎝ 365.25 day ⎠⎝ yr ⎠⎟ ⎟⎜⎜ 24 h ⎠⎝ ⎟⎜⎜ 3600 s ⎠⎝ ⎟⎜⎜ 1 atom ⎠⎟⎟

Dose (Ci) = (1.05816 × 107dps)(1 Ci/3.70 × 1010dps) = 2.85990 × 10–4 = 2.9 × 10–4 Ci Dose = (1.05816 × 107dps)(1 Bq/1 dps) = 1.05816 × 107 = 1.1 × 107 Bq 24.104 Plan: Use the given relationship for the fraction remaining after time t, where t = 10.0 yr, 10.0 × 103 yr, and 10.0 × 104 yr. Solution: a) Fraction remaining after 10.0 yr =

() 1 2

10.0

⎛ ⎞

= ⎜⎜⎜⎝ 1 ⎠⎟⎟⎟

5730

10.0×103

⎛ ⎞

b) Fraction remaining after 10.0 × 103 yr = ⎜⎜⎜⎝ 1 ⎠⎟⎟⎟

5730

10.0×10 4

⎛ ⎞

c) Fraction remaining after 10.0 × 104 yr = ⎜⎜⎜⎝ 1 ⎠⎟⎟⎟

5730

= 0.998791 = 0.999

= 0.298292 = 0.298 = 5.5772795 × 10–6 = 5.58 × 10–6

d) Radiocarbon dating is more reliable for b) because a significant quantity of 14C has decayed and a significant quantity remains. Therefore, a change in the amount of 14C would be noticeable. For the fraction in a), very little 14 C has decayed and for c) very little 14C remains. In either case, it will be more difficult to measure the change so the error will be relatively large. 24.105

210 86

Rn + −01 e → 210 85 At

Mass = (2.368 MeV)(1 amu/931.5 MeV) = 0.0025421363 amu Mass 210At = mass 210Rn + electron mass – mass equivalent of energy emitted. = (209.989669 + 0.000549 – 0.0025421363) amu = 209.9876759 = 209.98768 amu 24.106 Plan: At one half-life, the fraction of sample is 0.500. Find n for which (0.900)n = 0.500. Solution: (0.900)n = 0.500 n ln (0.900) = ln (0.500) n = (ln 0.500)/(ln 0.900) = 6.578813 = 6.58 h

24.107 a) β decay by vanadium-52 produces chromium-52. 51 1 52 52 0 23V + 0 n → 23V → 24 Cr + − 1 β 51 23

52 V (n,β) 24 Cr

b) Positron emission by copper-64 produces nickel-64. 63 1 64 64 0 29 Cu + 0 n → 29 Cu → 28 Ni + 1 β 63 29

Cu (n,β ) 64 28 Ni

24-25

c) β decay by aluminum-28 produces silicon-28. 27 13

28 0 Al + 01 n → 28 13 Al → 14 Si + − 1 β

27 13

28 Al (n,β) 14 Si

24.108 Determine k for 90Sr. ln 2 ln 2 k= = = 0.023902 yr–1 t1 / 2 29 yr N a) ln t = –kt N0 Nt ln = – (0.023902 yr–1)(10 yr) = – 0.23902 0.0500 g Nt = 0.787399 0.0500 g (0.787399)(0.0500 g) = 0.03936995 = 0.039 g 90Sr b) ln

Nt = –kt N0

(100 − 99.9)%

= – (0.023902 yr–1)(t)

100% t = 289.003 = 3 × 102 yr (The calculation 100 – 99.9 limits the answer to one significant figure.) 24.109 a) 126 C + 42 He → 168 O ⎛ 931.5 MeV ⎞⎟⎛⎜1.602 ×10−13 J ⎞⎟⎛ 1 kJ ⎞⎟ –14 –14 b) (7.7×10−2 amu) ⎜⎜ ⎟⎜ ⎟⎜ ⎟ = 1.1490425 × 10 = 1.1 × 10 kJ ⎝ 1 amu ⎠⎟⎜⎝ 1 MeV ⎠⎟⎟⎝⎜103 J ⎠⎟

24.110 Plan: The production rate of radon gas (volume/hour) is also the decay rate of 226Ra. The decay rate, or activity, is proportional to the number of radioactive nuclei decaying, or the number of atoms in 1.000 g of 226Ra, using the relationship A = kN. Calculate the number of atoms in the sample, and find k from the half-life. Convert the activity in units of nuclei/time (also disintegrations per unit time) to volume/time using the ideal gas law. Solution: 226 4 222 88 Ra → 2 He + 86 Rn ln 2 ln 2 k= = = 4.33879178 × 10–4 yr–1(1 yr/8766 h) = 4.94510515 × 10–8 h–1 t1 / 2 1599 yr The mass of 226Ra is 226.025402 amu/atom or 226.025402 g/mol. 23 ⎛ ⎞⎛ 1 mol Ra ⎟⎟⎜⎜ 6.022 ×10 Ra atoms ⎞⎟⎟ ⎜ 21 N = (1.000 g Ra) ⎜⎜ ⎟⎟⎜ ⎟⎟ = 2.6643023 × 10 Ra atoms ⎜⎜⎝ 226.025402 g Ra ⎠⎝ 1 mol Ra ⎟⎜⎜ ⎟⎠ A = kN = (4.94510515 × 10–8 h–1)(2.6643023 × 1021 Ra atoms) = 1.3175255 × 1014 Ra atoms/h

This result means that 1.318 × 1014 226Ra nuclei are decaying into 222Rn nuclei every hour. Convert atoms of 222Rn into volume of gas using the ideal gas law. ⎛1.3175255×1014 Ra atoms ⎞⎟⎛1 atom Rn ⎞⎛ 1 mol Rn ⎞⎟ ⎟⎟⎜⎜ Moles of Rn/h = ⎜⎜ ⎟⎟⎜⎜ ⎟ 23 ⎟ ⎟ ⎜⎝ ⎝ ⎠⎝ h ⎠ 1 atom Ra 6.022 ×10 Rn atoms ⎠⎟ = 2.1878537 × 10–10 mol Rn/h Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

24-26

nRT V= = P

L • atm ⎞⎟ ⎛ (2.1878537×10−10 mol Rn/h) ⎜⎜0.08206 ⎟ (273.15 K) ⎝ mol • K ⎠ 1 atm

= 4.904006 × 10–9 = 4.904 × 10–9 L/h Therefore, radon gas is produced at a rate of 4.904 × 10–9 L/h. Note: Activity could have been calculated as decay in moles/time, removing Avogadro’s number as a multiplication and division factor in the calculation. 24.111 Determine k: ln 2 ln 2 k= = = 0.0216608 s–1 t1 / 2 32 s Nt ln = –kt N0 90% = – (0.0216608 s–1)(t) ln 100% t = 4.86411 = 4.9 s 24.112 a) 133 55 Cs

The N/Z ratio for 140Cs is too high.

b) 79 35 Br

It has an even number of neutrons compared with 78Br.

c) 24 12 Mg

The N/Z ratio equals 1.

14 7

The N/Z ratio equals 1.

24.113 Plan: Determine k from the half-life and then use the integrated rate law, solving for time. Solution: ln 2 ln 2 k= = = 0.0239016 yr–1 t1 / 2 29 yr N ln t = –kt N0 ⎛ 1.0 ×10 4 particles ⎞⎟ ⎟⎟ = – (0.0239016 yr–1)(t) ln ⎜⎜⎜ ⎝ 7.0 ×10 4 particles ⎠⎟ –1.945910 = – (0.0239016 yr–1)(t) t = 81.413378 = 81 yr

24.114 a) 63 Li + 63 Li → 126 C (dilithium) b) Δm = 2(mass 63 Li ) – mass 126 C = 2(6.015121 amu) – 12.000000 amu = 0.030242 amu/atom 126 C (dilithium) Δm = (0.030242 amu/atom)(1.66054 × 10–27 kg/amu) =5.02180507 × 10–29 kg/atom ⎛ ⎞ ⎜ 1 J ⎟⎟ ⎟ 2 –12 −29 8 2⎜ ⎜ E = Δmc = (5.02180507×10 kg/atom)(2.99792 ×10 m/s) ⎜⎜ ⎟⎟⎟ = 4.5133595 × 10 J/atom 2 ⎜ kg • m 2 ⎟⎟ ⎜⎝ s ⎠⎟ − 12 ⎛ ⎞⎛ ⎞ ⎛ 4.5133595×10 J ⎞⎟⎜ 1 atom 1 amu ⎟⎟⎜⎜ ⎟⎟ ⎜ ⎟⎟⎜⎜ E = ⎜⎜ ⎟⎟⎜ ⎟ −27 ⎟ atom ⎟⎠⎝⎜⎜12.000000 amu ⎠⎝ ⎜⎝ ⎟⎜⎜1.66054 ×10 kg ⎠⎟⎟ = 2.2650059 × 1014 = 2.2650 × 1014 J/kg dilithium c) 4 11 H → 42 He + 2 01 β

2 positrons are released.

24-27

d) For dilithium ( 126 C ): Mass = (5.02180507 × 10–29 kg/atom)(1 atom/12.000000 amu) (1 amu/1.66054 × 10–27 kg 12C) = 2.5201667 × 10–3 = 2.5202 × 10–3 kg/kg 12C For 42 He (The mass of a positron is the same as the mass of an electron.) Δm = 4 (mass 11 H ) – [mass 42 He + 2 mass −01 e ] = 4(1.007825 amu) – [4.00260 amu + 2(5.48580 × 10–4amu)] = 0.02760 amu/atom = (0.02760 amu/atom)(1.66054 × 10–27 kg/amu) = 4.58309 × 10–29 kg/atom Mass = (4.58309 × 10–29 kg/atom)(1 atom/4.00260 amu)(1 amu/1.66054 × 10–27 kg 4He) = 6.895517 × 10–3 = 6.896 × 10–3 kg/kg 4He e) 21 H + 31 H → 42 He + 01 n Δm = [2.0140 amu + 3.01605 amu] – [4.00260 amu + 1.008665 amu] = 5.0300 amu – 5.01126 amu = 0.0188 amu = (0.0188 amu/atom)(1.66054 × 10–27 kg/amu) = 3.1218152 × 10–29 kg/atom Mass = (3.1218152 × 10–29 kg/atom)(1 atom/4.00260 amu)(1 amu/1.66054 × 10–27 kg 4He) = 4.696947 × 10–3 = 4.70 × 10–3 kg/kg 4He f) 63 Li + 01 n → 42 He + 31 H 3 6 3 6 3 ⎛ 1 mol 3 H ⎞⎛ ⎟⎟⎜⎜ 3.01605g H ⎞⎛ ⎟⎟⎜⎜ 1 mol Li ⎞⎛ ⎟⎟⎜⎜10 g Li ⎞⎛ ⎟⎟⎜⎜ 1 kg H ⎞⎟⎟ ⎜ Mass 31 H / kg 6Li = ⎜⎜ ⎟ ⎟ ⎟ ⎟ 6 ⎟⎟⎜⎜⎜ 1 mol 3 H ⎟⎟⎜⎜⎜ 6.015121 g 6 Li ⎟⎟⎜⎜⎜ 1 kg 6 Li ⎟⎟⎜⎜⎜10 3 g 3 H ⎟⎟⎟ ⎜⎜⎝1 mol Li ⎠⎝ ⎠⎝ ⎠⎝ ⎠⎝ ⎠

= 0.5014113598 = 0.501411 kg 3H/kg 6Li 3 23 3 −29 ⎛ 3 ⎞⎛ ⎞⎛ ⎞ ⎜10 g ⎟⎟⎜⎜ 1 mol H ⎟⎟⎜⎜ 6.022 ×10 atom H ⎟⎟⎜⎛⎜ 3.121852 ×10 kg ⎞⎟⎟ Mass = (0.5014113598 kg 3H) ⎜⎜ ⎟⎟⎜ ⎟ ⎟ ⎟⎟ ⎟⎟⎜⎜⎜ ⎟⎜ ⎜⎜⎝ 1 kg ⎠⎝ ⎟⎜⎜ 3.01605 g 3 H ⎠⎝ mol 3 H atom ⎠⎟ ⎠⎟⎜⎝ = 3.1254222 × 10–3 = 3.125 × 10–3 kg Change in mass for dilithium reaction: Mass = (5.02180507 × 10–29 kg/atom 12C)(1 atom 12C/2 atoms 6Li)(6.022 × 1023 atoms 6Li/mol) (1 mol 6Li/6.015121 g 6Li)(103 g/kg 6Li) = 2.513774 × 10–3 = 2.514 × 10–3 kg The change in mass for the dilithium reaction is slightly less than that for the fusion of tritium with deuterium. 24.115 Plan: Convert pCi to Bq using the conversion factors 1 Ci = 3.70 × 1010 Bq and 1 pCi = 10–12 Ci. For part b), use the first-order integrated rate law to find the activity at the later time (t = 9.5 days). You will first need to calculate k from the half-life expression. For part c), solve for the time at which Nt = the EPA recommended level. Solution: 10 ⎛ 4.0 pCi ⎞⎟⎛⎜10−12 Ci ⎞⎛ ⎟⎟⎜⎜ 3.70 ×10 Bq ⎞⎟⎟ ⎜ ⎜ ⎟ a) Activity (Bq/L) = ⎜⎜ ⎟⎟⎜ ⎟⎟ = 0.148 = 0.15 Bq/L ⎜⎝ L ⎠⎟⎟⎟⎜⎜⎝⎜ 1 pCi ⎠⎝ 1 Ci ⎟⎜⎜ ⎠⎟ The safe level is 0.15 Bq/L. b) k =

ln 2 ln 2 = = 0.181452 d–1 t1 / 2 3.82 d

Nt = – kt N0

24-28

Nt = – (0.181452 d–1)(9.5 d) = –1.723794 41.5 pCi/L

Nt = 0.17838806 41.5 pCi/L

Nt = 7.403104 pCi/L 10 ⎛ 7.403104 pCi ⎞⎟⎜⎛10−12 Ci ⎞⎛ ⎟⎜ 3.70×10 Bq ⎞⎟⎟ ⎟⎜⎜ Activity (Bq/L) = ⎜⎜⎜ ⎟⎟⎟⎜⎜ ⎟ ⎟⎟ = 0.2739148 = 0.27 Bq/L L 1 Ci ⎟⎜⎜ ⎝⎜ ⎠⎟⎟⎜⎜⎝ 1 pCi ⎠⎝ ⎠⎟

c) The desired activity is 0.15 Bq/L, however, the room air currently contains 0.27 Bq/L. N ln t = – kt N0 0.15 Bq/L ln = – (0.181452 d–1)(t) 0.2739148 Bq/L – 0.602182 = – (0.181452 d–1)(t) t = 3.318685 = 3.3 d

It takes 3.3 d more to reach the recommended EPA level. A total of 12.8 (3.3 + 9.5) d is required to reach recommended levels when the room was initially measured at 41.5 pCi/L.

ln 2 ln 2 = = 0.05653729 yr–1 t1 / 2 12.26 yr Nt ln = – kt N0 N ln t = – (0.05653729 yr–1)(5.50 yr) = –0.310955097 N0 Nt = 0.732746777 N0

24.116 k =

fraction lost = 1 – 0.732746777 = 0.267253222 = 0.267 24.117

239 92

239 0 4 235 4 231 U → −01 β + 239 93 Np → 94 Pu + − 1 β → 2 α + 92 U → 2 α + 90Th

This could begin the 235 92 U decay series. 24.118 Plan: Convert mCi to Ci to disintegrations per second; multiply the dps by the energy of each disintegration in MeV and convert to energy in J. Recall that 1 rad = 0.01 J/kg. The mass of the child must be converted from lb to kg. Solution: ⎛10−3 Ci ⎞⎛ 3.70 ×1010 dps ⎞⎛ 5.59 MeV ⎞⎛ 1.602 ×10−13 J ⎞⎟ ⎜ ⎟⎟⎟⎜⎜ ⎟⎟⎟⎜⎜ ⎟⎟⎟⎜⎜ ⎟⎟ = 3.3134166 × 10–5 J/s Energy (J/s) = (1.0 mCi) ⎜⎜ ⎟⎟⎜⎜⎜ ⎟⎟⎜⎜⎜ 1 disint. ⎟⎟⎜⎜⎜ 1 MeV ⎟ 1 Ci ⎜⎜⎝ 1 mCi ⎠⎝ ⎠⎝ ⎠⎝ ⎠⎟ Convert the mass of the child to kg:

⎛ 1 kg ⎞⎟ (54 lb)⎜⎜ ⎟ = 24.4897960 kg ⎝ 2.205 lb ⎠⎟ Time for 1.0 mrad to be absorbed: ⎛10−3 rad ⎞⎛ ⎟⎟⎜⎜ 0.01 J / kg ⎞⎛ ⎟⎟⎜⎜ 24.4897960 kg ⎞⎟⎟ ⎜ Time (s) = (1.0 mrad) ⎜⎜ ⎟⎟⎜ ⎟⎟⎜ ⎟ = 7.3911008 = 7.4 s −5 ⎟⎜⎜ 1 rad ⎠⎝ ⎟⎜⎜ 3.3134166 ×10 J /s ⎠⎟⎟ ⎜⎜⎝ 1 mrad ⎠⎝ Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

24-29

ln 2 ln 2 = = 1.2096809 × 10–4 yr–1 t1 / 2 5730 yr Nt ln = – kt N0 ⎛12.9 d /min • g ⎞⎟ ⎜ –4 –1 ln ⎜⎜ ⎟⎟ = – (1.2096809 × 10 yr )(t) ⎜⎜⎝ 15.3 d /min • g ⎠⎟⎟

24.119 k =

– 0.170625517 = – (1.2096809 × 10–4 yr–1)(t) t = 1410.500216 = 1.41 × 103 yr

The earthquake occurred 1410 years ago. 24.120 Assuming our atmosphere present today was not fully developed and the fact that much of the cosmic ionizing radiation is absorbed by the atmosphere, organisms would have been exposed to more cosmic radiation a billion years ago. In addition, the number of radioactive nuclei such as uranium would have been larger. This would also have the effect of exposing organisms to greater ionizing radiation from Earth. 24.121 Because the 1941 wine has a little over twice as much tritium in it, just over one half-life has passed between the two wines. Therefore, the older wine was produced before 1929 (1941– 12.26) but not much earlier than that. To find the number of years back in time, use the first-order rate expression, where N0 = 2.32 N, Nt = N and t = years transpired between the manufacture date and 1941. ln 2 ln 2 k= = = 0.05653729 yr–1 t1 / 2 12.26 yr N ln t = – kt N0 N = – (0.05653729 yr–1)(t) ln 2.32 N – 0.841567 = – (0.05653729 yr–1)(t) t = 14.92857 = 14.9 yr The wine was produced in (1941 – 15) = 1926.

24.122 If 99% of Pu-239 decays, 1% remains. ln 2 ln 2 k= = = 2.87612938 × 10-5 yr–1 t1 / 2 2.41×104 yr N ln t = – kt N0 1% = – (2.87612938 × 10–5yr–1)(t) ln 100% –4.60517 = – (2.8761293 × 10–5yr–1)(t) t = 1.6011693 × 105 = 2 × 105yr

(The 1% limits the answer to one significant figure.)

ln 2 ln 2 = = 1.2096809 × 10–4 yr–1 t1 / 2 5730 yr N ln t = – kt N0

24.123 k =

24-30

0.61 pCi/g 6.89 pCi/g

= – (1.2096809 × 10–4 yr–1)(t)

–2.4243674 = – (1.2096809 × 10–4 yr–1(t) t = 2.0041 × 104 = 2.0 × 104 yr

24.124 Plan: Determine the change in mass for the reaction by subtracting the masses of the products from the masses of the reactants. Use conversion factors to convert the mass change in amu to energy in eV and then to J. Solution: Δm = mass of reactants – mass of products = (14.003074 + 1.008665) – (14.003241 + 1.007825) = 0.000673 amu ⎛ 931.5 MeV ⎞⎟⎜⎛106 eV ⎞⎟ ⎟ = 6.268995 × 105 = 6.27 × 105 eV Energy (eV) = (0.000673 amu)⎜⎜ ⎟⎜ ⎝ 1 amu ⎠⎟⎜⎝ 1 MeV ⎠⎟⎟

⎛1.602 ×10−19 J ⎞⎟⎛ 1 kJ ⎞⎛ 6.022 ×1023 ⎞⎟ ⎟⎟⎜⎜ 3 ⎟⎟⎟⎜⎜ ⎟ Energy (kJ/mol) = (6.268995×105 eV) ⎜⎜ 1 eV ⎝⎜ ⎠⎟⎝10 J ⎠⎝⎜ 1 mol ⎠⎟⎟ = 6.0478524 × 107 = 6.05 × 107 kJ/mol 24.125 Calculate Δm. Mass of 3 1H atoms = 3 × 1.007825 = 3.023475 amu Mass of 4 neutrons = 4 × 1.008665 = 4.034660 amu Total mass = 7.058135 amu Δm = 7.058135 – 7.016003 = 0.042132 amu/7Li = 0.042132 g/mol 7Li ⎛ 0.042132 amu 7 Li ⎞⎟⎛ 931.5 MeV ⎞⎛ 106 eV ⎞ ⎜ ⎟⎟⎜⎜ ⎟⎟ 7 7 ⎟⎟⎜⎜ Binding energy = ⎜⎜ ⎟⎟⎜ ⎟ = 3.9245958 × 10 = 3.925 10 eV/nucleus ⎜ ⎟ ⎜⎜ ⎜⎝ 1 amu ⎠⎝ ⎜1 MeV ⎠⎟⎟ ⎟ 1 atom ⎟ ⎜ ⎜ ⎟ ⎝ ⎠

Convert MeV to kJ.

⎛ ⎞⎟ ⎜ 1J ⎟⎟⎛⎜ 1 kJ ⎞⎟ ⎛ 0.042132 g 7 Li ⎞⎟⎛⎜ 1 kg ⎞⎟⎟ 8 2⎜ ⎜ ⎜ ⎜ ⎟⎜ 3 ⎟⎟ (2.99792×10 m/s) ⎜⎜ Binding energy = ⎜ ⎟⎟⎟⎜⎜⎜ 3 ⎟⎟⎟ 2 ⎟ ⎜⎝ ⎜ mol ⎠⎜⎝10 g ⎠⎟ ⎜ kg • m 2 ⎟⎟⎝10 J ⎠ ⎜⎝ s ⎠⎟ 9 9 = 3.7866237 × 10 = 3.7866 × 10 kJ/mol 24.126

146 64

146 0 4 142 Gd + −01 e → 146 63 Eu → 62 Sm + +1 β → 2 He + 60 Nd

3⎛ R ⎞ 24.127 a) Kinetic energy = 1/2 mv2 = ⎜⎜ ⎟⎟⎟ T 2 ⎜⎝ NA ⎠⎟ ⎛ 3 ⎞ (8.314 J/mol • K)(1.00×10 K) = 2.07090668 × 10–17 = 2.07 × 10–17 J/atom 11 H Energy = ⎜ ⎟⎟⎟ 23 ⎜⎝ 2 ⎠ 6.022×10 atom/mol 6

b) A kilogram of 1H will annihilate a kilogram of anti-H; thus, two kilograms will be converted to energy: Energy = mc2 = (2.00 kg)(2.99792 × 108 m/s)2(J/(kg∙m2/s2) = 1.7975048 × 1017 J ⎛1.7975048×1017 J ⎞⎛ ⎞⎛ ⎞⎟ 1 kg ⎞⎛ 1.0078 g H ⎞⎛ 1 mol H 1 H atom ⎜ ⎟⎟⎟⎜⎜ ⎟⎟⎟⎜⎜ ⎟⎟⎟⎜⎜ ⎟⎟⎟⎜⎜ ⎟ Number of atoms = ⎜⎜ ⎟⎟⎜⎜⎜10 3 g ⎟⎟⎜⎜⎜ 1 mol H ⎟⎟⎜⎜⎜ 6.022 ×10 23 atoms H ⎟⎟⎜⎜⎜ 2.0709066 ×10−17 J ⎟⎟⎟ 1 kg H ⎜⎜⎝ ⎠⎝ ⎠⎝ ⎠⎝ ⎠⎝ ⎠

= 1.4525903 × 107 = 1.45 × 107 H atoms

24-31

c) 4 11 H → 42 He + 2 01 β

(Positrons have the same mass as electrons.)

Δm = [4(1.007825 amu)] – [4.00260 amu + 2(0.000549 amu)] = 0.027602 amu / 42 He ⎛ 0.027602 g ⎞⎟⎛ 1 kg ⎞⎛1 mol He ⎞⎛ ⎞⎟ 1 mol H ⎟⎟⎜⎜ ⎜ ⎟⎜ 7 Δm = ⎜⎜ ⎟⎟⎟⎜⎜⎜ 3 ⎟⎟⎜⎜ ⎟⎟⎜ ⎟⎟ (1.4525903×10 H atoms) 23 ⎜⎜⎝ mol He ⎠⎟⎜⎝10 g ⎠⎟⎝⎜⎜ 4 mol H ⎠⎝ ⎟⎜⎜ 6.022 ×10 H atoms ⎠⎟⎟ = 1.6644967 × 10–22 kg Energy = (Δm)c2 = (1.6644967 × 10–22 kg) (2.99792 × 108 m/s)2(J/(kg∙m2/s2) = 1.4959705 × 10–5 = 1.4960 × 10–5 J d) Calculate the energy generated in part b): ⎛ ⎞⎛ ⎞⎛ ⎞⎟ 17 ⎞⎛ 1 mol H ⎜1.7975048×10 J ⎟⎟⎜⎜ 1 kg ⎟⎟⎜⎜1.0078 g H ⎟⎟⎜⎜ ⎟⎟ = 3.0081789 × 10–10 J Energy = ⎜⎜ ⎟⎟⎜ 3 ⎟⎟⎜ ⎟⎟⎜ 23 ⎜⎝⎜ ⎟⎠⎝⎜⎜10 g ⎠⎝ ⎟⎜⎜ 1 mol H ⎠⎝ ⎟⎜⎜ 6.022 ×10 atoms H ⎠⎟⎟ 1 kg H Energy increase = (1.4959705 × 10–5 – 3.0081789 × 10–10 ) J = 1.4959404 × 10–5 = 1.4959 × 10–5 J e) 3 11 H → 23 He + 101 β Δm = [3(1.007825 amu)] – [3.01603 amu + 0.000549 amu] = 0.006896 amu/ 23 He = 0.006896 g/mol 23 He ⎛ 0.006896 g ⎞⎛ ⎞⎟ 1 mol H ⎟⎜ 1 kg ⎞⎛ ⎟⎜1 mol He ⎞⎛ ⎟⎜ ⎜ 7 Δm = ⎜⎜ ⎟⎟⎟⎜⎜ 3 ⎟⎟⎟⎜⎜ ⎟⎟⎟⎜⎜ ⎟⎟ (1.4525903×10 H atoms) 23 ⎟⎜⎜10 g ⎠⎝ ⎟⎜⎜ 3 mol H ⎠⎝ ⎟⎜⎜ 6.022 ×10 H atoms ⎠⎟⎟ ⎜⎜⎝ mol He ⎠⎝ = 5.54470426 × 10–23 kg Energy = (Δm)c2 = (5.54470426 × 10–23 kg)(2.99792 × 108 m/s)2(J/(kg∙m2/s2) = 4.9833164 × 10–6 = 4.983 × 10–6 J No, the Chief Engineer should advise the Captain to keep the current technology. 24.128 Multiply the equation E = hc/λ by Avogadro’s number (N) to get the energy change per mole. 23 −34 8 ⎛ 1pm ⎞⎟ 10 E = Nhc/λ = (6.022 ×10 /mol)(6.626 ×10 J • s)(2.99792 ×10 m/s) ⎜⎜ ⎟⎟ = 1.3702442 × 10 J/mol −12 ⎜ ⎜⎝10 m ⎠⎟ (8.73 pm) Determine the mass through a series of conversions from the chapter and the inside back cover: ⎞⎟⎛ 1 amu ⎞⎟⎛1.66054 ×10−27 kg ⎞⎟ ⎛1.3702442 ×1010 J ⎞⎟⎛⎜ 1 MeV ⎟⎟⎜⎜ ⎟⎟ ⎟⎟⎜⎜⎜ ⎟⎜ Mass (kg) = ⎜⎜⎜ ⎟⎟⎜⎜ ⎟ −13 ⎟⎜ ⎟ ⎜ mol 1 amu ⎝⎜ ⎠⎟⎜⎝1.602 ×10 J ⎠⎟⎜⎝ 931.5 MeV ⎠⎟⎝⎜⎜ ⎠⎟ = 1.52476 × 10–7 = 1.52 × 10–7 kg/mol

ln 2 ln 2 = = 0.131526979 yr–1 t1 / 2 5.27 yr N ln t = – kt N0 70.% = – (0.131526979 yr–1)(t) ln 100% – 0.3566749439 = – (0.131526979 yr–1)(t) t = 2.71180 yr The source must be replaced after the time calculated above. Two years would be March 1, 2009. Then add 0.7 yr which is (365.25 × 0.7) = 256 d (actually the significant figures limit this to 3 × 102 d). The final date is November 12, 2009.

24.129 k =

24.130 Plan: The difference in the energies of the two α particles gives the energy of the γ ray released to get from excited state I to the ground state. Use E = hc/λ to determine the wavelength. The energy of the gamma ray must be converted from MeV to J. The energy of the 2% α particle is equal to the highest energy α particle Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

24-32

minus the energy of the two γ rays (the rays are from excited state II to excited state I, and from excited state I to the ground state). Solution: a) Energy of the γ ray = (4.816 – 4.773) MeV = 0.043 MeV λ=

−34 8 hc (6.626×10 J • s)(2.99792×10 m/s) ⎛⎜ 1 MeV ⎞⎟ ⎟ = 2.8836 × 10–11 = 2.9 × 10–11 m ⎜⎜ = −13 ⎟ E ⎜ (0.043 MeV) ⎜⎝1.602×10 J ⎠⎟⎟

b) (4.816 – 0.043 – 0.060) MeV = 4.713 MeV 24.131 a) Student 2 is correct. The fraction of uranium-238 remaining is equal to the amount of uranium-238 remaining divided by the initial amount. The initial amount is equal to the sum of the amount of uranium-238 remaining plus the amount of lead-206 that resulted from the decay of uranium-238. b) Amount U-238 = 2(Pb-206) 2x x 238 92

U 2x 2x 2 = = = 206 2x + x 3x 3 92 U + 82 Pb ln 2 ln 2 k= = = 1.540327 × 10–10yr–1 t1 / 2 4.5×109 yr N ln t = – kt N0

Fraction U-238 = 238

⎛ 2 ⎞⎟ ⎜ –10 –1 ln ⎜⎜ 3 ⎟⎟⎟ = – (1.540327 × 0 yr )(t) ⎜⎜⎝ 1 ⎠⎟ – 0.405465108 = – (1.540327 × 10–10yr–1)(t) t = 2.63233 × 109 = 2.6 × 109 years old

24.132

232 90

4 Th → 228 88 Ra + 2 α

228 88

0 Ra → 228 89 Ac + − 1 β

228 89

Ac → 22890Th + −01 β

228 90

4 Th → 224 88 Ra + 2 α

224 88

4 Ra → 220 86 Rn + 2 α

220 86

4 Rn → 216 84 Po + 2 α

216 84

4 Po → 212 82 Pb + 2 α

212 82

0 Pb → 212 83 Bi + − 1 β

212 83

0 Bi → 212 84 Po + − 1 β

212 84

4 Po → 208 82 Pb + 2 α

24.133 The age of Egyptian mummies is on the order of a few thousand years. An isotope with a half-life of a few thousand years would be the best choice. Carbon-14, with a half-life of 5730 yr, would be the best choice. 24.134 Plan: Multiply each of the half-lives by 20 (the number of half-lives is considered to be exact). Solution: a) 242Cm 20(163 d) = 3.26 × 103 d b) 214Po

20(1.6 × 10–4 s) = 3.2 × 10–3 s

c) 232Th

20(1.39 × 1010yr) = 2.78 × 1011yr

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24.135 If the blade was made in 100 AD it is about 1.9 × 103 years old. ln 2 ln 2 k= = = 1.2096809 × 10–4 yr–1 t1 / 2 5730 yr At ln = – kt A0 Handle: ⎛10.1 d /min • g ⎞⎟ ⎜ ⎟⎟ = – (1.2096809 × 10–4 yr–1)(t) ln ⎜⎜ ⎜⎜⎝15.3 d /min • g ⎠⎟⎟

– 0.415317404 = – (1.2096809 × 10–4 yr–1)(t) t = 3.43328 × 103 = 3.43 × 103 yr

Inlaid: ⎛13.8 d /min • g ⎞⎟ ⎜ ⎟⎟ = – (1.2096809 × 10–4 yr–1)(t) ln ⎜⎜ ⎜⎜⎝15.3 d /min • g ⎠⎟⎟

– 0.103184236 = – (1.2096809 × 10–4 yr–1)(t) t = 8.52987 × 102 = 8.53 × 102yr

Ribbon: ⎛12.1 d /min • g ⎞⎟ ⎜ ⎟⎟ = – (1.2096809 × 10–4 yr–1)(t) ln ⎜⎜ ⎜⎜⎝15.3 d /min • g ⎠⎟⎟

– 0.2346473758 = – (1.2096809 × 10–4 yr–1)(t) t = 1.939746 × 103 = 1.94 × 103 yr

Sheath: ⎛15.0 d /min • g ⎞⎟ ⎜ ⎟⎟ = – (1.2096809 × 10–4 yr–1)(t) ln ⎜⎜ ⎜⎜⎝ 15.3 d/min • g ⎠⎟⎟

– 0.0198026273 = – (1.2096809 × 10–4 yr–1)(t) t = 1.63701 × 102 = 1.64 × 102yr

The ribbon is nearest in age to the blade. 24.136 a) When 1.00 kg of antimatter annihilates 1.00 kg of matter, the change in mass is: Δm = 0 – 2.00 kg = –2.00 kg. The energy released is calculated from ΔE = Δmc2. ΔE = (–2.00 kg)(2.99792 × 108 m/s)2(J/(kg ∙ m2/s2) = –1.7975049 × 1017 = –1.80 × 1017 J The negative value indicates the energy is released. b) Assuming that four hydrogen atoms fuse to form the two protons and two neutrons in one helium atom and release two positrons, the energy released can be calculated from the binding energy of helium-4. 4 11 H → 42 He + 2 +01 β Δm = [4(1.007825 amu)] – [4.00260 amu + 2(0.000549 amu)] = 0.02760 amu per 42 He formed ⎛ 0.02760 amu ⎞⎟⎛⎜ 1 4 He ⎞⎟ ⎟⎟ = 6.90 × 102amu ⎟⎜ Total Δm = (1.00 ×10 H atoms) ⎜⎜⎜ ⎟⎟⎜⎜ ⎟ 4 ⎜⎝ He ⎠⎟⎝⎜ 4 H atoms ⎠⎟ 5

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−16 ⎛ ⎞⎛ ⎞ ⎜⎜ 931.5 MeV ⎟⎟⎜⎜1.602 ×10 kJ ⎟⎟ –10 Energy = (6.90 ×10 amu) ⎜ ⎟⎟⎜ ⎟⎟ = 1.02966 × 10 kJ per antiH collision ⎜⎝⎜ 1 amu ⎠⎝ ⎜ 1 MeV ⎟⎜ ⎠⎟ 2

23 ⎛ 3 ⎞⎛ ⎞⎛ ⎞ ⎜10 g ⎟⎟⎜⎜ 1 mol antiH ⎟⎟⎜⎜ 6.022 ×10 antiH ⎟⎟ 26 AntiH atoms = (1.00 kg) ⎜⎜ ⎟⎟⎜ ⎟⎟⎜ ⎟ = 5.9742 × 10 antiH ⎜⎜⎝ 1 kg ⎠⎝ ⎟⎜⎜1.008 g antiH ⎠⎝ ⎟⎜⎜ 1 mol antiH ⎠⎟⎟

⎛1.02966 ×10−10 kJ ⎞⎟ 16 16 Energy released = ⎜⎜ ⎟ (5.9742×1026 antiH) = 6.15139 × 10 = 6.15 × 10 kJ ⎜⎝ antiH ⎠⎟⎟ c) From the above calculations, the procedure in part b) with excess hydrogen produces more energy per kilogram of antihydrogen. 24.137 Plan: Einstein’s equation is E = mc2, which is modified to E = Δmc2 to reflect a mass difference. The speed of light, c, is 2.99792 × 108 m/s. The mass of exactly one amu is 1.66054 × 10–27 kg (inside back cover of text). When the quantities are multiplied together, the unit will be kg∙m2/s2, which is also the unit of joules. Convert J to MeV using the conversion factor 1.602 × 10–13 J = 1 MeV. Solution: ΔE = (Δm)c2 ⎛ ⎞⎟ −27 ⎡ ⎜ ⎛ ⎞⎤ ⎞⎟ ⎟⎟⎛⎜ 1 MeV J ⎜⎜1.66054 ×10 kg ⎟⎟⎥ ⎢ 8 2⎜ ⎟⎜ ΔE = ⎢(1amu) ⎜ ⎟⎟⎥ (2.99792×10 m/s) ⎜⎜⎜ ⎟ ⎟⎟ 2 − 13 ⎜⎜ 1 amu ⎟⎥ ⎜⎜ kg • m 2 ⎟⎟⎟⎜⎝⎜1.602×10 J ⎠⎟ ⎢ ⎝ ⎠ ⎣ ⎦ ⎝ s ⎠⎟ 2 2 = 9.3159448 × 10 = 9.316 × 10 MeV 24.138 The original amount of uranium is the current amount plus the amount converted to lead. We need to determine the amount of decayed uranium from the lead in the sample: 238 238 ⎛ 1 mol 206 Pb ⎞⎛ ⎟⎟⎜⎜ 1 mol U ⎞⎛ ⎟⎟⎜⎜ 238 g U ⎞⎟⎟ ⎜ 238 Mass (g) of 238U = (0.023 g 206 Pb) ⎜⎜ ⎟ ⎟ 206 ⎟⎟⎜⎜⎜1 mol 206 Pb ⎟⎟⎜⎜⎜ 1 mol 238 U ⎟⎟⎟ = 0.0265728 g U ⎜⎜⎝ 206 g Pb ⎠⎝ ⎠⎝ ⎠ Original mass of 238U = (0.065 + 0.0265728) g 238U = 0.0915728 g 238U ln 2 ln 2 k= = = 1.540327 × 10–10yr–1 t1 / 2 4.5×109 yr N ln t = – kt N0 ⎛ 0.065 g ⎞⎟ ⎜ ⎟⎟ = – (1.540327 × 10–10yr–1)(t) ln ⎜⎜ ⎜⎜⎝ 0.0915728 g ⎠⎟⎟

– 0.3427470145 = – (1.540327 × 10–10yr–1)(t) t = 2.2251575 × 109 = 2.2 × 109 years old 238 4 24.139 a) 242 94 Pu → 92 U * + 2 α 238 238 92 U * → 92 U + γ b) Determine the MeV of the γ ray by using E = hc/λ to determine the energy. −34 8 hc (6.626×10 J • s)(2.99792×10 m/s) ⎜⎛ 1 nm ⎞⎛ ⎟⎟⎜⎜ 1 MeV ⎞⎟⎟ = 0.044975 MeV ⎜⎜ E= = ⎟⎟⎜ −13 ⎟ λ ⎜⎜⎝10−9 m ⎠⎝ (0.02757 nm) ⎟⎜⎜1.602 ×10 J ⎠⎟⎟ Adding the two energies together gives the energy required to go directly from plutonium-242 to the lowest energy uranium-238: Total energy = (4.853 + 0.044975) MeV = 4.897975 = 4.898 MeV

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24.140

249 98

263 Cf + 188 O → 106 Sg + 4 01 n

263 106

259 Sg → 104 Rf + 42 α

259 104

255 Rf → 102 No + 42 α

255 102

251 No → 100 Fm + 42 α

24.141 Plan: The rate of formation of plutonium-239 depends on the rate of decay of neptunium-239 with a half-life of 2.35 d. Calculate k from the half-life equation and use the integrated rate law to find the time necessary to react 90% of the neptunium-239 (10% left). Solution: ln 2 ln 2 k= = = 0.294956 d–1 t1/ 2 2.35 d N ln t = – kt N0

(

)

⎛ 1.00 kg 10.% ⎜ ( 100%)⎞⎟⎟⎟ –1 ln ⎜⎜⎜ ⎟⎟ = – (0.294956 d )(t) ⎜⎜ 1.00 kg ⎝ ⎠⎟⎟

– 2.3025851 = – (0.294956 d–1)(t) t = 7.806538 = 7.81 d

24.142 a) Half of the atoms do not remain after each half-life. Decay is a random process. On average half of the atoms remain after one half-life, but individual simulations may vary. b) Increasing the number of atoms is more realistic. The number of remaining atoms varies less. Also, a real radioactive sample consists of many atoms. 24.143 a) Ideally 128 atoms remain after the first half-life, 64 atoms remain after the second, 32 atoms remain after the third, 16 atoms remain after the fourth, and 8 remain after the fifth. b) To make the simulation more realistic, make the size of the sample (# of atoms) much larger. 24.144 a) Nucleus 1 is 49 Be; Nucleus 2 is 104 Be; Nucleus 3 is 47 Be b) The n/p ratio for the stable Nucleus 1 = 5/4 = 1.25; the n/p ratio for Nucleus 2 = 6/4 = 1.5. Nucleus 2 has an n/p ratio that is too high and the most likely mode of decay is beta particle emission. The n/p ratio of Nucleus 3 = 3/4 = 0.75. This ratio is too low and the expected modes of decay are electron capture and/or positron emission.

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