SOLUTION MANUAL FOR CHEMISTRY THE MOLECULAR NATURE OF MATTER AND CHANGE 3CE MARTIN SILBERBERG, PATRI by StudyGuide

SOLUTION MANUAL FOR CHEMISTRY THE MOLECULAR NATURE OF MATTER AND CHANGE 3CE MARTIN SILBERBERG, PATRICIA AMATEIS, RASHMI VENKATESWARAN, LYDIA CHEN.

Solution Manual For Chemistry The Molecular Nature of Matter and Change 3CE Martin Silberberg, Patricia Amateis, Rashmi Venkateswaran, Lydia Chen Chapter 1-25

CHAPTER 1 KEYS TO THE STUDY OF CHEMISTRY 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.

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.

1.10

Alchemical: chemical methods – distillation, extraction; chemical apparatus Medical: mineral drugs Technological: metallurgy, pottery, glass

1.11

Combustion released the otherwise undetectable phlogiston. The more phlogiston a substance contained; the more easily it burned. Once all the phlogiston was gone, the substance was no longer combustible.

1.12

The mass of the reactants and products are easily observable quantities. The explanation of combustion must include an explanation of all observable quantities. Their explanation of the mass gain required phlogiston to have a negative mass.

1.13

Lavoisier measured the total mass of the reactants and products, not just the mass of the solids. The total mass of the reactants and products remained constant. His measurements showed that a gas was involved in the reaction. He called this gas oxygen (one of his key discoveries).

1.14

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.

1.15

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.

1.16

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.

1.17

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.

1.18

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

1.19

1 m 2

100 cm 2 1000 m 2 ; to convert from m2 to cm2, use 100 cm 2 b) To convert from km2 to m2, use 1 km 2 1 m 2 c) This problem requires two conversion factors: one for distance (km to m) and one for time (h to s). It does not matter which conversion is done first and alternate methods may be used. To convert distance, km to m, use:  1000 m  3   = 10 m/km  1 km  To convert time, h to s, use: 1h  1 h  1 min      60 min  60 s  3600 s Therefore, the complete conversion factor is: 1 mh  1000 m  1 h      km  3600 s  3.6 km  s

Do the units cancel when you start with units of km/h? d) To convert from kg/m3 to g/cm3 requires two conversion factors: To convert mass, kg to g:  1000 g  3 g    10 1 kg kg   

To convert volume from cm3 to m3 use, 

1 m 3

 100 cm  

3   =106 m . cm3 

3 

 3 g   6 m3  g  m3 3 10 10  10 The complete conversion is:    cm3  kg  cm3  kg  Do the units cancel when you start with units of kg/m3? 1.20

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. To convert distance, cm to mm, use:  10 mm   1 cm 

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

The complete conversion is:  10 mm   1 min   1 mm  min  1 cm   60 s 

6 cm  s

100 cm  1 m 3

b) To convert from m3 to cm3, use

c) This problem requires two conversion factors: one for distance and one for time squared. It does not matter which conversion is done first. Alternate methods may be used. To convert distance, m to km, use:  1 km  -3   = 10 km/m 1000 m   To convert time, s2 to h2, use: 2

 60 min   60 s  = 1.296 x 107 s2/1 h2      1 h   1 min 

 103 km  1.296 107 s 2  1.296  1010 km  s 2  Therefore, the complete conversion factor is  .     1 h2 m  h2  1 m   Do the units cancel when you start with a measurement of m/s2? 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. To convert volume, mL to L, use: 

1L    1000 mL 

To convert time, s to min, use:  60 s   1 min  1 L  60 s  6 L s  1 min   100 mL  min 1000 mL   

The complete conversion factor is: 

1.21

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.

1.22

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. mass Plan: Density = . An increase in mass or a decrease in volume will increase the density. A volume decrease 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.

1.23

1.24

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. Convert 65°C to K: T (in K) = T (in °C) + 273.15 = (65°C) + 273.15 = 338 K A temperature of 65°C is 338 K. Heat will flow from the hot water (65°C or 338 K) to the cooler water (65 K). The 65°C water contains more heat than the cooler water.

1.25

When we have a set of ratios, the units will cancel out as they are multiplicative. For example, m=d/V; if we express mass in kg and density in kg/m3, it would give us the identical volume (but in different units) than if we used mass in g and density in g/cm3. In the first case, the volume will have units of m3 and in the second it will have units of cm3; however, the actual volume will be the same (if you convert one to the other). When temperature is one of the variables, however, the conversion between Celsius and Kelvin is an additive conversion, not multiplicative. Hence, in the equation PV=nRT, we must have the units for T match the units for R (which contains the temperature unit to be cancelled). This temperature, with only one or two exceptions, will be a temperature in kelvin.

1.26

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

1.27

1.28

1.29

1.30

Plan: Use conversion factors from the inside back cover: 10 –12 m = 1 pm; 10-9 m = 1 nm. Solution:  1 pm  2.22x1010 m  12   2.22  10 2 pm  10 m  Radius (Å) =  1 nm  2.22x1010 m  9   0.222 nm  10 m 









Plan: Use conversion factors: 1 m = 10-9 nm Solution:  109 nm  11 Length (nm) = 100. m     10 nm 1 m   Plan: Use the conversion factor 1 km = 106 mm to convert km to height in mm. Solution:  106 mm  Height (mm) =  0.00196 km    =1960 mm  1 km  Plan: Use conversion factors (1 cm)2 = (0.01 m)2; (1000 m)2 = (1 km)2 to express the area in km2. To calculate the cost of the patch, use the conversion factor: (2.54 cm) 2 = (1 in)2. Solution:  0.01 m 2  

a) Area (km2) = 20.7 cm 2  





1 km



2



–9



   101 cmmm    1$3.25  = $6.73  10 mm 



2 b) Cost = 20.7 cm 

1.31



   2.07x10 km =  1 cm    1000 m 





Plan: Use conversion factors (1 mm)2 = (10–3 m)2; (0.01 m)2 = (1 cm)2; Solution: 2   103 m   2  7.903  103 m 2 a) Area (m2) = 7903 mm  2   1 mm    









45 s  2  3 3 b) Time (s) = 7903 mm  2   = 2.634333x10 = 2.6x10 s 135 mm  



1.32

Plan: Use conversion factor 1 g = 1 paper clip. Solution: total mass (g) =  mass per clip  number of clips  Number of Clips =

1.33

Total mass  g  mass per clip



3.56×103 g  3.56×103 clips 1 g/clip

Plan: Use conversion factor 1000 kg = 1 metric ton. Solution:  1 kg   1 T  21 15 Mass (T) = 2.36 x10 g    3  = 2.36x10 T 1000 g 10 kg   



1.34



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 mg

in part b) with the conversion factors 1000 mg = 1 g. Volume in cm3 is converted to m3 with the conversion factor (1 cm)3 = (0.01 m)3 and to mm3 with the conversion factors (10 mm)3 = (1 cm)3. The conversions may be performed in any order. Solution: 

  1 kg  3 3   = 5.52x10 kg/m  cm   0.01 m    1000 g   1 cm 3   1000 mg  3  b) Density (mg/mm3) =  5.52 g     = 5.52 mg/mm  cm3   10 mm 3   1 g 

a) Density (kg/m3) =  5.52 g   

1 cm 

1.35

3 

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 x108 m   60 s   60 min   1 km  9 9 a) Velocity (km/h) =       3  = 1.07928x10 = 1.079x10 km/h  1s    1 min   1 h   10 m   2.998 x108 m   60 s  100 cm  12 b) Velocity (cm/min) =      = 1.799x10 cm/min 1s    1 min  1m  1.36

Plan: Use the conversion factors (1 μm)3 = (1x10–6 m)3; (1x10–3 m)3 = (1 mm)3 to convert to mm3. To convert to L, use the conversion factors (1 μm) 3 = (1x10–6 m)3; (1x10–2 m)3 = (1 cm)3; 1 cm3 = 1 mL; 1 mL = 1x10–3 L. Solution:





 1x106 m 3    3   1 mm   = 2.56x10–9 mm3/cell 3 3      cell   1 μm    1x103 m     3    6 3  1x10 m   1 cm 3   1 mL   1x103 L   b) Volume (L) = 105 cells  2.56 μm    3 3  3    cell       1 μm    1x102 m   1 cm   1 mL       

3 

a) Volume (mm3) =  2.56 μm  



= 2.56 x 10 1.37





–10

L = 1 x 10

–10









L (because of sig figs in question)

Plan: For part a), convert from mL to L (1 mL = 1x10–3 L) and then to m3 (1 L = 10–3 m3). For part b), convert from mm3 to L using the following conversion factors: (10 mm)3 = (1 cm)3, 1 mL = 1 cm3 and 1 mL = 10–3 L. Solution:  103 L  103 m3  –4 3 a) Volume (m3) = 946.4 mL    1 L  = 9.464x10 m 1 mL   

 (1 cm)3   1mL   103 L  3 4 b) Volume (L) = 835 mm  3    1 cm3   1 mL   8.35x10 L (10 mm)    



1.38



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 cm3  3 Volume (cm3) of mercury = volume of vial = 130.24 g   = 9.626016 = 9.626 cm 13.53 g   b) Volume (cm3) of water = volume of vial = 9.626016 cm3



 1 cm 

Mass (g) of water = 9.626016 cm3  0.997 g  = 9.59714 g water 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 1.39

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 cm3  3 Volume (cm3) of water = volume of flask = 247.8 g   = 247.8 = 248 cm  1.00 g  b) Volume (cm3) of chloroform = volume of flask = 247.8 cm3



 cm

Mass (g) of chloroform = 247.8 cm3  1.48 g  = 366.744 g chloroform 3 

Mass (g) of flask and chloroform = mass of flask + mass of chloroform = 241.3 g + 366.744 g = 608.044 g = 608 g 1.40

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

1.41

Plan: Use the relationship c = 2πr to find the radius of the sphere and the relationship V = 4/3πr3 to find the volume of the sphere. The volume in mm3 must be converted to cm3. Divide the mass of the sphere by the volume to find density. Solution: c = 2πr c 32.5 mm = Radius (mm) = = 5.17254 mm 2 2 4 Volume (mm3) =  r 3 =  4   (5.17254 mm)3 = 579.6958 mm3 3 3 3



Volume (cm ) = 579.6958 mm

Density (g/cm3 ) 

 103 m   1 cm 3 3    2  = 0.5796958 cm   1 mm   10 m 



mass 4.20 g  = 7.24518 = 7.25 g/cm3 volume 0.5796958 cm3

1.42

Plan: Use the equations given in the text for converting between the three temperature scales. Solution: a) T (in K) = T (in °C) + 273.15 = 18°C + 273.15 = 291.15 = 291 K b) T (in K) = T (in °C) + 273.15 = –164°C + 273.15 = 109.15 = 109 K c) T (in °C) = T (in K) – 273.15 = 0 K – 273.15 = –273.15 = –273°C

1.43

Plan: Use the equations given in the text for converting between the three temperature scales. Solution: a) T (in K) = T (in °C) + 273.15 = 37C + 273.15 = 310.15 = 310 K b) T (in K) = T (in °C) + 273.15 = 3410°C + 273 = 3683 K c) T (in C) = T (in K) –273.15 = 6.1x103 K – 273 = 5.827x103 = 5.8 x 103°C

1.44

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.

1.45

Plan: Use 1 nm = 10–9 m to convert wavelength in nm to m. To convert wavelength in pm to nm, use 1000 pm = 1 nm. Solution:  109 m  –7 a) Wavelength (m) = 247 nm   = 2.47x10 m  1 nm 

 1 nm   = 6.76 nm  1000 pm 

b) Wavelength (nm) =  6760 pm   1.46

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.

1.47

Plan: Calculate the volume of the cylinder in cm3 by using the equation for the volume of a cylinder. The diameter of the cylinder must be halved to find the radius. Convert the volume in cm 3 to dm3 by using the conversion factors (1 cm)3 = (10–2 m)3 and (10–1 m)3 = (1 dm)3. Solution: Radius = diameter/2 = 0.85 cm/2 = 0.425 cm Volume (cm3) = πr2h = π(0.425 cm)2(9.5 cm) = 5.3907766 cm3 3

 102 m   1 dm  Volume (dm3) = 5.3907766 cm3  = 5.39078x10–3 = 5.4x10–3 dm3  1 cm   101 m     



1.48



Plan: Use the percent of copper in the ore to find the mass of copper in 5.01 kg of ore. Convert the mass in kg 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: Mass (kg) of copper =  5.01 kg Covellite   66%  = 3.3066 kg copper  100% 

 1000 g  3  = 3.3066 x10 g 1 kg  

Mass (g) of copper =  3.3066 kg  

 cm3 Cu  3 3 Volume (cm3) of copper = 3.3066x10 g Cu   = 369.453 cm Cu 8.95 g Cu   2 V = r h





1 cm  –3 Radius (cm) =  0.1601 mm    = 8.005x10 cm 

 10 mm 

Height (length) in cm =

 r2

369.453 cm3

   8.005 x10 cm  3

= 1.835 x 106 cm

 102 m  6 4 4 Length (m) = 1.835x10 cm   = 1.835x10 = 1.84x10 m 1 cm  





1.49

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.

1.50

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.

1.51

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. c) A value of 15,000 to two significant figures is 1.5x10 4. Values would range from 14,501 to 15,499. Both of these values round to 1.5x10 4.

1.52

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) d) 4.0100x104 (zeros between nonzero digits are significant; terminal zeros to the right of the decimal point are significant)

1.53

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) c) 5.080x103 (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)

1.54

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

1.55

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 last digit kept remains unchanged: 140,000 (or 1.4x105)

1.56

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.

 20 x 160 x 8    = 568.89 = 6x102 3 x 3 x 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 using the original numbers.) 1.57

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 an odd number, so that digit kept is increased by 1.

 11 x 6.2 x 2.38    = 0.1691 = 0.2  24 x 2 x 20  Since there is a number in the calculation with only one significant figure, the answer can be reported only to Silberberg, Amateis, Venkateswaran, Chen, Chemistry: The Molecular Nature of Matter and Change, 3rd Canadian Edition Page 1-11 Instructor‘s Solution Manual © Copyright 2021 McGraw-Hill Ryerson Ltd.

one significant figure. (Note that the answer is 0.19 with original number of significant figures.) 1.58

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  310 m  = 133.71 = 130 m (maximum of 2 significant figures allowed since 310 has 2 sf) 6.48 m

b) V =  4   17.282 mm 3 = 21,620.74 = 21,621 mm3 (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) 1.59

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

2.420 g  15.6 g = 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 = 1.0274 = 1.0 (After the subtraction, the denominator has 2 significant figures; only one 16.1 mL  8.44 mL

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.) c) V = π(6.23 cm)2(4.630 cm) = 564.556 = 565 cm3 (maximum of 3 significant figures allowed since one of the original numbers in the calculation has only 3 significant figures) 1.60

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: a) 1.310000x105 (Note that all zeros are significant.) b) 4.7x10–4 (No zeros are significant.) c) 2.10006x105 d) 2.1605x103

1.61

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: a) 2.820x102 (Note that the zero is significant.) b) 3.80x10–2 (Note the one significant zero.) c) 4.2708x103 d) 5.82009x104

1.62

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

1.63

Solution: a) 6500. b) 0.0000346 c) 750 d) 188.56

(Use terminal decimal point since the final zero is significant.) (Do not use terminal decimal point since the zero is not significant.)

1.64

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: a) 8.025x104 (The decimal point must be moved an additional 2 places to the left: 10 2 x 102 = 104) –3 b) 1.0098x10 (The decimal point must be moved an additional 3 places to the left: 10 3 x 10–6 = 10–3) –11 c) 7.7x10 (The decimal point must be moved an additional 2 places to the right: 10–2 x 10–9 = 10–11)

1.65

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. Remember: when we multiply powers of ten, we add the exponents. Solution: a) 1.43x102 (The decimal point must be moved an additional 1 place to the left: 10 1 x 101 = 102) b) 8.51 (The decimal point must be moved an additional 2 places to the left: 102 x 10–2 = 100) c) 7.5 (The decimal point must be moved an additional 3 places to the left: 10 3 x 10–3 = 100)

1.66

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: a)

6.626 x10

34

  2.9979 x10 m/s  = 4.062185x10 J 8

–19

489 x109 m 4.06x10–19 J (489x10–9 m limits the answer to 3 significant figures; units of m and s cancel)

6.022 x 10 molecules/mol 1.23 x 10 g  = 1.6078x10 molecules b) 23

46.07 g/mol 1.61x1024 molecules (1.23x102 g limits answer to 3 significant figures; units of mol and g cancel)





 2

c) 6.022 x 1023 atoms/mol 2.18 x 1018 J/atom  1  1  = 1.82333x105 J/mol 2 2 5

1.82x10 J/mol (2.18x10 1.67

–18

3 

J/atom limits answer to 3 significant figures; unit of atoms cancels)

4.32 x107 g

= 1.3909 = 1.39 g/cm3 3 4 2  3.1416  1.95x10 cm 3 (4.32x107 g limits the answer to 3 significant figures) a)





1.84x10 g  44.7 m/s  = 1.8382x10 = 1.84x10 g·m /s b) 2

(1.84x102 g limits the answer to 3 significant figures)

1.07 x10 mol / L   3.8 x 10 mol / L  = 1.9248 x 10 = 1.9 x 10 L /mol c) 8.35 x10 mol / L  1.48 x 10 mol / L  4

3

5

2

(3.8x10–3 mol/L limits the answer to 2 significant figures; mol3/L3 in the numerator cancels mol5/L5 in the denominator to leave mol2/L2 in the denominator or units of L2/mol2) 1.68

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 Horseshoe 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 students in a classroom is an exact number. We cannot have ½ a student. d) The number of millimeters in a meter is a definition of the prefix ―milli–.‖ This is an exact number.

1.69

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 provinces and territories is a counted value. These are exact numbers.

1.70

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 half, the uncertainty is one-half the separation between the marks. Thus, since the end of the metal strip falls between 7.4 and 7.6 we can report its length as 7.5 ± 0. 1 cm.

1.71

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.

1.72

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).

1.74

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

Potential Energy

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. 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.

Potential Energy

1.73

+ + +

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.

1.75

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.

1.76

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

1.77

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 cm must be converted to units of m before calculating the volume. The density of water is used to find the mass of this volume of water. Solution:  102 m  a) Depth of water (m) = 146cm    = 1.46 m  1 cm  Volume (m3) = length x width x depth =  50.0 m  25.0 m 1.46 m  = 1825 m3 b) Using the density of water = 1.0 g/mL.  (100 cm)3   1 mL  1.0 g   1 kg  3 6 6 Mass (kg) = 1825m      1000 g  = 1.825x10 = 1.8x10 kg 3 3  mL 1 m 1 cm     



1.78



Plan: In each case, calculate the overall density of the ball and contents and compare to the density of air. The volume of the ball in cm3 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 L 3  1 mL   10 Volume of ball (L) = 560 cm    = 0.560 = 0.56 L 3    1 cm  1 mL  mass 0.12 g  Density of evacuated ball = = 0.21 g/L volume 0.560 L The evacuated ball will float because its density is less than that of air. b) density of CO2 = 1.830 g/L; mass of CO2 = dV = (1.830 g/L)(0.56 L) = 1.02 g plus mass of ball gives total mass = 1.02 g + 0.12 g = 1.14 g; d = m/V = 1.14 g/0.56 L = 2.04 g/L; more than air; will sink c) density of H2 = 0.0899 g/L; mass of H2 = dV =(0.0899 g/L)(0.56 L) = 0.050 g plus mass of ball gives total mass = 0.050 g + 0.12 g = 0.17 g; d = m/V = 0.17 g/0.56 L = 0.30 g/L; less than air; will float d) using a similar calculation, density of ball + O2 = 1.54 g/L; more than air; will sink. e) using a similar calculation, density of ball + N2 = 1.38 g/L; more than air; will sink





 0.560 L   1.189 g  f) To sink, the total mass of the ball and gas must be    1 L   0.66584 g    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:

Mass of hydrogen in the ball = 0.56 L  0.0899 g   0.0503 g 



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. 1.79

Plan: Convert the cross-sectional area of 1.0 μm2 to mm2 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:









 1x106 m 2    2   1 mm   = 1.0x10–6 mm2 2 2    3  1 μm    1x10 m    

Cross-sectional area (mm2) = 1.0 μm2  





Calculate the mass that can be held up by grunerite with a cross-sectional area of 1.0x10–6 mm2:  3.5x102 kg  1x106 mm2   3.5 104 kg  1 mm 2    Calculate the area of aluminum required to match a mass of 3.5x10 –4 kg:

 



2  1 cm 2   10 mm      1.9444x10 5 = 1.9x10–5 mm2 3.5 x104 kg   2   1.8x103 kg      1 cm  



Calculate the area of steel required to match a mass of 3.5 x 10 –4 kg:

 1.80

2  1 cm 2   10 mm      1.0000x105 = 1.0x10–5 mm2 3.5x104 kg   2   3.5x103 kg   1 cm     



Plan: Convert the surface area to m2 and then use the surface area and the depth to determine the volume of the oceans (area x depth = volume) in m3. The volume is then converted from cubic metres to litres, 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  14 2  = 3.63x10 m  1 km 2   

a) Area of ocean (m2) = 3.63x108 km 2  





Volume of ocean (m3) = (area)(depth) = (3.63x1014 m2)(3800 m) = 1.3794x1018 m3 1 L   5.8 x 109 g  18 3  12 12 Mass of gold (g) = 1.3794 x 10 m  3 3    = 8.00052x10 = 8.0x10 g L 10 m    b) Use the density of gold to convert mass of gold to volume of gold:





3 



3 

 19.3 g  1 cm   = 4.14535x10 = 4.1x10 m 



Volume of gold (m3) = 8.00052 x 1012 g  1 cm     



0.01 m



tr. oz.  $1611.46    131.1   = 4.14551x10 = $4.1x10 g  1 tr. oz. 

12 c) Value of gold = 8.00052 x 10 g 

1.81

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  100% yellow zinc 

a) Mass of zinc in the 34% zinc sample = 185 g yellow zinc 



 37% zinc  = 68.45 g Zn  100% yellow zinc 

Mass of zinc in the 37% zinc sample = 185 g 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  66% copper 

Mass of zinc = 46.5 g copper 

 37% zinc   = 27.31 = 27 g  63% copper 

Mass of zinc = 46.5 g copper  24 to 27 g zinc 1.82

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



 1L   = 5.0514 = 5.05 L  809 g 

Volume of liquid N2 = 4086.57 g  1.83

Plan: For part a), convert km to m and h to s. For part b), time is converted from h to min and length stays in km. For part c), use the average speed in km/h to find the time necessary to cover the given distance. Solution:

 9.4 km  1000 m   1 h   = 2.611 = 2.6 m/s    h  1 km   3600 s 

a) Speed (m/s) = 

b) Distance (km) =  98 min   1 h

 9.4 km  = 15.353 = 15 km    60 min  h 

c) Time (h) = 14.5 km   1 h

 = 1.5426 = 1.5 h   9.4 km 

If she starts running at 11:15 am, 1.5 hours later the time is 12:45 pm. 1.84

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. 1.85

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 (80.1oC – 5.5oC) on the °C scale. So °X =  50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. So °X =  50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. So °X =  50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. fpwater °X =  50X  (0.00°C – 5.5°C) = –3.68°X = –3.7°X  74.6C 

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

1.86

Plan: Determine the total mass of Earth‘s crust in metric tonnes (t) by finding the volume of crust (surface area x depth) in km3 and then in cm3 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:





Volume of crust (km3) = area x depth = 5.10x108 km 2  35 km  = 1.785x1010 km3  1000 m 3  



Mass of crust (t) = 1.785 x 10





1 cm



 2.8 g   1 kg  1 t  19 cm3    = 4.998x10 t   1 cm3   1000 g  1000 kg 





Mass of oxygen (g) = 4.998 x 10

Mass of silicon (g) = 4.998 x 10

 4.55 x 105 g oxygen  t   = 2.2741x1025 = 2.3x1025 g oxygen  1 t  



 2.72 x 105 g silicon  25 25 t   = 1.3595x10 = 1.4x10 g silicon  1 t  



 1 x 104 g element  t    1t   = 4.998x1015 = 5x1015 g each of ruthenium and rhodium Plan: Find the total number of washers from the fact that each washer has an outer diameter of 1 cm and the sheet is 24.0 cm x 12.0 cm. Calculate the area of the outer circle and subtract the area of the inner circle to get the surface area of the washer. Multiply the surface area of the washer by the thickness to get the volume of the washer. The



Mass of ruthenium = mass of rhodium = 4.998 x 10

1.87



   = 1.785x10 cm   1 km     0.01 m 



Volume of crust (cm3) = 1.785 x 1010 km3 



total volume of the washers is the number of washers times the volume per washer. We can find the total volume of the sheet of zinc by multiplying the length x width x height. The volume of the zinc left is the total volume of zinc minus the total volume of the washers. Once we know the volume of zinc left, the mass of zinc left is the volume times the density. Solution: We can get 24 washers along the edge that is 24.0 cm and 12 washers along the edge that is 12.0 cm. No.TOT = 24 x 12 = 288 washers

Aouter circle   r 2  (3.141)(0.5 cm) 2  0.79 cm 2

Ainner circle   r 2  (3.141)(0.25 cm) 2  0.20 cm 2 Awasher  Aouter circle  Ainner circle  0.79 cm 2  0.20 cm 2  0.59 cm 2

Vwasher  A  h  (0.59 cm 2 )(0.5 cm)=0.29 cm3 VTot, washers  Vwasher  No. of washers=(0.29 cm3 )(288)  84.8 cm3 VZn sheet  l  w  h  (24.0 cm)(12.0 cm)(0.5 cm)  144.0 cm3 VZn left over  VZn sheet  VTot, washers  144.0 cm3  84.8 cm3  59.2 cm3 m Zn left over  VZn left over  d Zn  (59.2 cm3 )(7.049

g )  417 g cm3

The volume of Zn remaining is 59.2 cm3 and the mass remaining is 417 g. 1.88

Plan: Use the inner diameter of the tube and the length to find the volume of the inside of the tube. Use the density of mercury and volume to find the mass of mercury in the tube. Use the mass of mercury, set equal to the mass of ethanol, and the density of ethanol to find the volume of ethanol. Use the volume of ethanol, the inner diameter of the new tube, and the equation for the volume of a cylinder to find the length of the tube. Solution:

1.03 cm 2 Vtube interior   r 2l  (3.141)( ) (25 cm) = 20.8 cm3 2 kg 1000 g 1 m3 mHg  Vtube interior  d Hg  (20.8 cm3 )(13534 3 )( )( )  281.9 g  280 g (2sf) m 1 kg 102 cm 3

mHg  methanol Vethanol 

ltube 

methanol  d ethanol

282 g  357 cm3 3 kg 1000 g 1m (789.00 3 )( )( 6 ) m 1 kg 10 cm 3

Vethanol 357 cm3   640 cm 0.84 cm 2  r2 (3.141)( ) 2

The length of the tube would need to be 640 cm. NOTE: in part a), the answer was given to 2 sf, but in part b), the more precise answer was used to continue the calculation. Rounding too soon will lead to large rounding errors. 1.89 Plan: We will use what we know of sand and gold to determine the answer to part a. In part b), we can use the density and the volume of gold to find its mass. In part c), we use the fact that the masses have to be the same to

keep the trap from springing and then the density and mass to find the volume of sand needed. We will use all these answers to respond to part d). Solution: a)

Sand is lighter than gold. In this question just looking at the numbers also tells us that sand is much less dense than gold. That means an object that is the same approximate size (volume), cannot have the same mass. Therefore, the premise (the idea) is not scientifically reasonable.

mAu  d Au  VAu  (19.3

g )(53.5 cm3 )  1.03 103 g =1.03 kg 3 cm

The mass of the gold statue would be 1.03 kg. c)

msand  dsand

VAu  Vsand 

1.03 kg  6.45 102 cm3  645 mL 3 kg 1m (1600 3 )( 6 ) m 10 cm3

The volume of sand having a mass equal to that of the gold statue would be 645 mL. d) This is consistent. There is no way that a bag of sand the same approximate size as the idol would have had a mass that was sufficient to keep from setting off the booby trap. (Incidentally, it did not work in the movie either!) 1.90

Plan: Find the volume of 2.59 g of Pb using the density. Then find the mass of Cu using V and d. Solution: VPb = VPb 

mPb 2.5109g   0.228cm3 d Pb 11.342 g cm3

VCu = VPb = 0.228 cm3

mCu = dCu VCu = (8.96 g/cm3)(0.228 cm3) = 2.05 g 1.91

Plan: Find the volume of 4.16 g of Hg using the density. Then find the mass of ethanol using V and d. Solution:

VHg 

mHg d Hg



4.16g  0.307cm 3 g 13.56 cm 3

Vethanol = VHg = 0.307 cm3

Methanol = dethanol Vethanol = (0.7892 g/cm3)(0.307 cm3) = 0.242 g 1.92

Plan: Use the diameter of the shot to find the radius; use the radius to find the volume; use density of Al and volume to find the mass of each shot; find the number of shot using mass of shot and the mass of the soldier. Convert units along the way. Solution: rshot = diameter/2 =

(

)(

) = 0.236 cm

(

Vshot =

1.93

)

mshot = d V = (2.7 )( mkg = (mshot ) (number of shot)

10-2 cm3) = 0.15 g

number of shot =

= 1.2 x 106 shot

accept 1 million as nearest whole number

Plan: Use mass and density to find volume; use volume of the sphere to find radius; use radius to find diameter Solution: -

√

√ (

)

= 8.5 x 10-2 cm

Diameter = 2 r = 2 (8.5 x 10-2 cm) = 0.17 cm

CHAPTER 2 THE COMPONENTS OF MATTER TOOL OF THE LABORATORY BOXED READING PROBLEMS B2.1

Plan: There is one peak for each type of Cl atom and peaks for the Cl 2 molecule. The m/e ratio equals the mass divided by 1+. Solution: a) There is one peak for the 35Cl atom and another peak for the 37Cl atom. There are three peaks for the three possible Cl2 molecules: 35Cl35Cl (both atoms are mass 35), 37Cl37Cl (both atoms are mass 37), and 35Cl37Cl (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

B2.2

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

B2.3

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.

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.

2.3

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).

2.4

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 O2, N2, S8, P4, C60 etc. which are also known as elementary substances, occur as molecules. Metals often form networks or arrays.

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). 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 glass bulb 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 Cl 2O, 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 u/58.44 u), whether the sample is 0.5000 g or 50.00 g. 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 colourless 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 colourless gas = 0.64 g/0.36 g = 1.78 Experiment 2: mass white compound/mass colourless 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 = = 0.48551 = 0.486 mass fluorite 2.76 g fluorite c) Mass percent of Ca = 0.51449 x 100 %= 51.449 %= 51.4% Mass percent of F = 0.48551 x 100 %= 48.551 %= 48.6% Mass fraction of F =

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. Solution: a) Mass (g) of sulfur = mass of galena – mass of lead = 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 = = 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% Mass fraction of S =

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  b) Mass (g) of magnesium =  534 g MgO    = 322.109 g = 322 g magnesium  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  b) Mass (kg) of zinc =  3.82 kg ZnS    = 2.5567 kg= 2.56 kg zinc  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  Mass fraction of copper =   = 0.664586  133.00 g compound 

 103 g compound   0.664586 g copper  Mass (g) of copper =  5264 kg compound    1 kg compound   1 g compound     = 3.49838 x 106 g= 3.498 x 106 g copper  44.61 g sulfur  Mass fraction of sulfur =   = 0.335414  133.00 g compound 

 103 g compound   0.335414 g sulfur  Mass (g) of sulfur =  5264 kg compound    1 kg compound   1 g compound     = 1.76562 x 106 g= 1.766 x 106 g sulfur 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  Mass fraction of cesium =   = 0.51152  125.00 g compound 

 0.51152 g cesium  Mass (g) of cesium =  38.77 g compound    = 19.83163 g= 19.83 g cesium  1 g compound   61.06 g iodine  Mass fraction of iodine =   = 0.48848  125.00 g compound   0.48848 g iodine  Mass (g) of iodine =  38.77 g compound    = 18.9384 g= 18.94 g iodine  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: Compound 1:

47.5 mass % S = 0.90476 = 0.905 52.5 mass % Cl

Compound 2:

31.1 mass % S = 0.451379 = 0.451 68.9 mass % Cl

0.905 = 2.0067 = 2.00:1.00 0.451 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

63.3 mass % Xe = 1.7248 = 1.72 36.7 mass % F

3.46 = 2.0116 = 2.01:1.00 1.72 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:

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: Mass percent calcium =

1.70 g calcium x 100% = 21.767 %= 21.8% Ca 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  Mass % in Coal A =   100%  = 2.9894 %= 2.99% S (by mass)  378 g sample 

 19.0 g sulfur  Mass % in Coal B =   100%  = 3.8384 %= 3.84% S (by mass)  495 g sample   20.6 g sulfur  Mass % in Coal C =   100%  = 3.0519 %= 3.05% S (by mass)  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 charge on one electron, Millikan calculated the mass of an electron (charge/(charge/mass)) to be 9.109x10 –28 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: –3.204x10–19 C/2 = –1.602x10–19 C –4.806x10–19 C/3 = –1.602x10–19 C –8.010x10–19 C/5 = –1.602x10–19 C –1.442x10–18 C/9 = –1.602x10–19 C The value –1.602x10–19 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

Solution: Isotope 36 Ar 38 Ar 40 Ar

Mass Number 36 38 40

# of Protons 18 18 18

# of Neutrons 18 20 22

# of Electrons 18 18 18

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 35 Cl 35 17 18 17 37 Cl 37 17 20 17

2.41

are isotopes of oxygen, and 168 O has 16 – 8 = 8 neutrons whereas 178 O has 17 – 8 = 9 neutrons. Same Z value 41 b) 40 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 Co and 60 c) 27 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 2.42

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: a) ) 31 H and 23 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 b) 146 C and 157 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) 199 F and 189 F have the same number of protons and electrons (9), but different numbers of neutrons. 19 18 19 18 9 F and 9 F are isotopes of oxygen, and 9 F has 19 – 9 = 10 neutrons whereas 9 F has 18 – 9 = 9 neutrons.

Same Z value 2.43

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: a) A = 18 + 20 = 38; Z = 18; 38 18 Ar b) A = 25 + 30 = 55; Z = 25; 55 25 Mn c) A = 47 + 62 = 109; Z = 47; 109 47 Ag

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: a) A = 6 + 7 = 13; Z = 6; 136 C b) A = 40 + 50 = 90; Z = 40; 90 40 Zr 61 Ni c) A = 28 + 33 = 61; Z = 28; 28

2.45

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: a) 49 b) 79 c) 115 B 34 Se 22Ti 22 protons 34 protons 5 protons 22 electrons 34 electrons 5 electrons 49 – 22 = 27 neutrons 79 – 34 = 45 neutrons 11 – 5 = 6 neutrons

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: a) 207 b) 94 Be c) 75 82 Pb 33 As 82 protons 4 protons 33 protons 82 electrons 4 electrons 33 electrons 207 – 82 = 125 neutrons 9 – 4 = 5 neutrons 75 – 33 = 42 neutrons

2.47

82e

4e

33e

82p+ 125n0

4p+ 5n0

33p+ 42n0

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 x fractional abundance) + (isotopic mass of isotope 2 x fractional abundance). Solution:  60.11%   39.89%    70.9247 u   Atomic mass of gallium =  68.9256 u     = 69.7230 u= 69.72 u  100%   100% 

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 x fractional abundance) + (isotopic mass of isotope 2 x fractional abundance) + (isotopic mass of isotope 3 x fractional abundance). Solution:  78.99%   10.00%   11.01%  Atomic mass of Mg =  23.9850 u      24.9858 u   100%    25.9826 u   100%  100%       = 24.3050 u= 24.31 u

2.49

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

2.50

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

2.51

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.

2.52

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.

2.53

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

2.54

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.

2.55

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

2.56

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:

Element Name Germanium Phosphorus Helium Lithium Molybdenum

2.57

Element Symbol Ge P He Li Mo

Group Number 14 15 18 1 6

Element Type Metalloid non-metal non-metal metal metal

Element Name Arsenic Calcium Bromine Potassium aluminum

Element Symbol As Ca Br K Al

Group Number 15 2 17 1 13

Element Type Metalloid metal non-metal metal metal

2.58

Plan: Review the section in the chapter on the periodic table. Remember that alkaline earth metals are in Group 2, the halogens are in Group 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 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 u. d) The symbol and atomic mass of the halogen in Period 4 are Br and 79.90 u.

2.59

Plan: Review the section in the chapter on the periodic table. Remember that the noble gases are in Group 18, the alkali metals are in Group 1, and the transition elements are the groups of elements located between Groups 2 and 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 3. c) The symbol and atomic number of the first group 16 element displaying a metallic nature are Po and 84. d) The symbol and number of protons of the Period 4 alkali metal atom are K and 19.

2.60

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.

2.61

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. 2.62

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

2.63

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 x –2 = –4) is greater than the product of charges in LiF (+1 x –1 = –1). Thus, MgO has stronger ionic bonding.

2.64

There are no molecules; BaF2 is an ionic compound consisting of Ba2+ and F– ions.

2.65

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

2.66

Plan: Locate these groups on the periodic table and assign charges to the ions that would form. Solution: The monatomic ions of Group 1 have a +1 charge (e.g., Li+, Na+, and K+) whereas the monatomic ions of Group 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.

2.67

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). Magnesium atoms transfer electrons to chlorine atoms. Each magnesium atom loses two electrons to form a Mg 2+ ion and the same number of electrons (10) as the noble gas neon. Each chlorine atom gains one electron to form a Cl– ion and the same number of electrons (18) as the noble gas argon. The Mg 2+ and Cl– ions attract each other to form an ionic compound with the ratio of one Mg2+ 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.

2.68

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 charged ions, K+ and S2–, attract each other to form an ionic compound with the ratio of two K + ions to one S2– ion. The total number of electrons lost by the potassium atoms equals the total number of electrons gained by the sulfur atoms.

2.69

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+.

2.70

Plan: Locate these elements on the periodic table and predict what ions they will form. For cations (metals), ion charge = group number; for anions (nonmetals), ion charge = group number minus 18 . Solution: Potassium (K) is in Group 1 and forms the K+ ion. Iodine (I) is in Group17 and forms the I– ion (17 –18 = –1).

2.71

Plan: Locate these elements on the periodic table and predict what ions they will form. For cations (metals), ion charge = group number; for anions (nonmetals), ion charge = group number minus 18 .

Solution: Barium in Group 2 forms a +2 ion: Ba2+. Selenium in Group 16 forms a –2 ion: Se2– (16 –18 = –2). 2.72

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 16 Period 2 b) Fluorine (atomic number = 9) mass number = 9p + 10n = 19 Group 17 Period 2 c) Calcium (atomic number = 20) mass number = 20p + 20n = 40 Group 2 Period 4

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) Bromine (atomic number = 35) mass number = 35p + 44n = 79 Group 17 Period 4 b) Nitrogen (atomic number = 7) mass number = 7p + 8n = 15 Group 15 Period 2 c) Rubidium (atomic number = 37) mass number = 37p + 48n = 85 Group 1 Period 5

2.74

Plan: Determine the charges of the ions based on their position on the periodic table. For cations (metals), ion charge = group number; for anions (nonmetals), ion charge = group number minus 18. Next, determine the ratio of the charges to get the ratio of the ions. Solution: Lithium [Group 1] forms the Li + ion; oxygen [Group 16] forms the O2– ion (16 -18 = –2). The ionic compound that forms from the combination of these two ions must be electrically neutral, so two Li + ions combine with one O2– ion to form the compound Li2O. There are twice as many Li+ ions as O2– ions in a sample of Li2O.  1 O2 ion  Number of O2– ions = (8.4x1021 Li  ions)  = 4.2x1021 O2– ions  2 Li  ions   

2.75

Plan: Determine the charges of the ions based on their position on the periodic table. For cations (metals), ion charge = group number; for anions (nonmetals), ion charge = group number minus 18. Next, determine the ratio of the charges to get the ratio of the ions. Solution: Ca [Group 2] forms Ca2+ and I [Group 17] forms I– ions (17 -18 = –1). The ionic compound that forms from the combination of these two ions must be electrically neutral, so one Ca 2+ ion combines with two I– ions to form the compound CaI2. There are twice as many I– ions as Ca2+ ions in a sample of CaI2.  2 I  ions  Number of I– ions = (7.4x1021 Ca 2 ions)  = 1.48x1022 ions= 1.5x1022 I– ions  1 Ca 2 ion   

2.76

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 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).

2.77

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 x –2 = –4) is greater than the product of charges in Li2O (+1 x –2 = –2). Thus, MgO has stronger ionic bonding. 2.78

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 subscripts indicate that there are two F– ions for every one Mg2+ ion. Using this information and the mass of each element, we could calculate the percent mass of each element.

2.79

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).

2.80

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: H 2O2 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.

2.81

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).

2.82

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

2.83

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

2.84

Plan: Locate each of the individual elements on the periodic table, and assign charges to each of the ions. For main group cations (metals), ion charge = group number; for anions (nonmetals), ion charge = group number minus 18. 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 1) ion and nitrogen is a nonmetal that forms a –3 ion (Group 15, 15-18 = –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 16, 16 -18 = –2) and strontium is a metal that forms a +2 ion (Group 2). +2 –2 Sr O The compound is SrO, strontium oxide. c) Aluminum is a metal that forms a +3 ion (Group 3) and chlorine is a nonmetal that forms a –1 ion (Group 17, 17-18 = –1). +3 –3 +3 –1 +3 –1 Al Cl AlCl 3 The compound is AlCl3, aluminum chloride.

2.85

Plan: Locate each of the individual elements on the periodic table, and assign charges to each of the ions.

For main group cations (metals), ion charge = group number; for anions (nonmetals), ion charge = group number minus 18. 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 1) ion and bromine is a nonmetal that forms a –1 ion (Group 17, 17 -18 = –1). +1 –1 Cs Br The compound is CsBr, cesium bromide. b) Sulfur is a nonmetal that forms a –2 ion (Group 16, 16-18 = –2) and barium is a metal that forms a +2 ion (Group 2). +2 –2 Ba S The compound is BaS, barium sulfide. c) Fluorine is a nonmetal that forms a –1 ion (Group 17, 17-18 = –1) and calcium is a metal that forms a +2 ion (Group 2). –2 +2 –1 +2 –1 Ca F CaF2 The compound is CaF2, calcium fluoride. 2.86

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 cations (metals), ion charge = group number; for anions (nonmetals), ion charge = group number minus 18. 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) 12L is the element Mg (Z = 12). Magnesium [Group 2] forms the Mg2+ ion. 9M is the element F (Z = 9). Fluorine [Group 17] forms the F– ion (17-18 = –1). The compound formed by the combination of these two elements is MgF2, magnesium fluoride. b) 30L is the element Zn (Z = 30). Zinc forms the Zn2+ ion (see Table 2.3). 16M is the element S (Z = 16). Sulfur [Group 16] will form the S2– ion (16-18 = –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 17] forms the Cl– ion (17-18 = –1). 38M is the element Sr (Z = 38). Strontium [Group 2] forms the Sr 2+ ion. The compound formed by the combination of these two elements is SrCl2, strontium chloride.

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 (metals), ion charge = group number; for anions (nonmetals), ion charge = group number minus 18. 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 1] forms the Rb + ion. 35R is the element Br (Z = 35). Bromine [Group 17] forms the Br– ion (17-18 = –1). The compound formed by the combination of these two elements is RbBr, rubidium bromide. b) 8Q is the O (Z = 8). Oxygen [Group 16] will form the O2– ion (16 –-18 = –2). 13R is the element Al (Z = 13). Aluminum [Group13) forms the Al3+ ion. The compound formed by the combination of these two elements is Al2O3, aluminum oxide. c) 20Q is the element Ca (Z = 20). Calcium [Group 2] forms the Ca2+ ion. 53R is the element I (Z = 53). Iodine [Group 17] forms the I– ion (17 –18 = –1). The compound formed by the combination of these two elements is CaI2, calcium iodide.

2.88

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). 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 +6 –6 2.89 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: a) Na2HPO4 = sodium hydrogen phosphate Sodium [Group 1] forms the Na+ ion; HPO42– is the hydrogen phosphate ion. b) potassium carbonate dihydrate = K2CO3•2H2O Potassium [Group 1] forms the K+ ion; carbonate is the CO32– ion. Two K+ ions are required to match the –2 charge of the carbonate ion. Dihydrate indicates two water molecules (―waters of hydration‖) that are written after a centered dot. c) NaNO2 = sodium nitrite Sodium [Group 1] forms the Na + ion; NO2– is the nitrite polyatomic ion. d) 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. 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. 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: a) cobalt(II) oxide Cobalt forms more than one monatomic ion so the ionic charge must be indicated with a Roman numeral. Since the Co is paired with one O2– ion, the charge of Co is +2. b) Hg2Cl2 The Roman numeral I indicates that mercury has an ionic charge of +1; mercury is an unusual case in which the +1 ion formed is Hg22+, not Hg+. c) lead(II) acetate trihydrate The C2H3O2– ion has a –1 charge (see Table 2.5); since there are two of these ions, the lead ion has a +2 charge which must be indicated with the Roman numeral II. The •3H2O indicates a hydrate in which the number of H2O molecules is indicated by the prefix tri-. +3 –2 d) Cr2O3 ―chromic‖ denotes a +3 charge (see Table 2.4), oxygen has a –2 charge: CrO → Cr2O3 +6 –6

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. 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) tin(IV) sulfite Tin forms more than one monatomic ion so the ionic charge must be indicated with a Roman numeral. Each SO32– polyatomic ion has a charge of –2, so the ionic charge of tin is +4. b) K2Cr2O7 Dichromate is the polyatomic ion Cr2O72–; two K+ ions are required for a neutral compound. c) iron(II) carbonate Iron forms more than one monatomic ion so the ionic charge must be indicated with a Roman numeral. The CO32– polyatomic ion has a charge of –2, so the ionic charge of iron is +2. d) Cu(NO3)2 The Roman numeral II indicates that copper has an ionic charge of +2; two NO 3– polyatomic ions are required for a neutral compound.

2.92

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) Barium [Group 2] forms Ba2+ and oxygen [Group 16] forms O2– (16 -18 = –2) so the neutral compound forms from one Ba2+ ion and one O2– ion. Correct formula is BaO. b) Iron(II) indicates Fe2+ 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. Sulfide is the S2– ion and sulfite is the SO32– ion. 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: 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. c) magnesium dichromate Mg forms Mg2+ and Cr2O72– is named dichromate ion.

2.94

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 sulfate is HSO4–, so its source acid is H2SO4. Name of acid is sulfuric acid (-ate becomes -ic acid). b) HIO3, iodic acid IO3– is the iodate ion: -ate becomes -ic acid. c) Cyanide is CN– ; its source acid is HCN hydrocyanic acid (binary acid). d) H2S, hydrosulfuric acid (binary acid).

2.95

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) Perchlorate is ClO4–, so the source acid is HClO4. Name of acid is perchloric acid (-ate becomes -ic acid). b) nitric acid, HNO3 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) hydrofluoric acid, HF (binary acid)

2.96

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 Sulfide = S2–; sulfite = SO32–; sulfate = SO42–. 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.

2.97

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 Lead(II) indicates Pb2+ while lead(IV) indicates Pb4+. b) lithium nitride = Li3N lithium nitrite = LiNO2 lithium nitrate = LiNO3 Nitride = N3–; 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

2.98

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

S2F4

Di- indicates two S atoms and tetra- indicates four F atoms.

2.99

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.

2.100

Plan: Review the nomenclature rules 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. Binary acids (H plus one other nonmetal) are named hydro- + nonmetal root + -ic acid. Solution: a) Calcium(II) dichloride, CaCl2: The name becomes calcium chloride because calcium does not require ―(II)‖ since it only forms +2 ions. Prefixes like di- are only used in naming covalent compounds between nonmetal elements. b) Copper(II) oxide, Cu2O: The charge on the oxide ion is O2–, which makes each copper a Cu+. The name becomes copper(I) oxide to match the charge on the copper. c) Stannous fluoride, SnF4: Stannous refers to Sn2+, but the tin in this compound is Sn4+ due to the charge on the fluoride ion. The tin(IV) ion is the stannic ion; this gives the name stannic fluoride or tin(IV) fluoride. d) Hydrogen chloride acid, HCl: Binary acids consist of the root name of the nonmetal (chlor in this case) with a hydro- prefix and an -ic suffix. The word acid is also needed. This gives the name hydrochloric acid.

2.101

Plan: Review the nomenclature rules 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. 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) Iron(III) oxide, Fe3O4: Iron(III) is Fe3+, which combines with O2– to give Fe2O3. b) Chloric acid, HCl: HCl is hydrochloric acid. Chloric acid includes oxygen, and has the formula HClO3. c) Mercuric oxide, Hg2O: The compound shown is mercurous oxide. Mercuric oxide contains Hg 2+, which combines with O2– to give HgO. d) Dichlorine heptoxide, Cl2O6: Heptoxide refers to seven, not six, oxygen atoms. The formula should be Cl2O7.

2.102

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 u) = 53.96 u S = 3(32.07 u) = 96.21 u O = 12(16.00 u) = 192.0 u 342.2 u b) There are 9 atoms of hydrogen in (NH4)2HPO4. The molecular mass is: N = 2(14.01 u) = 28.02 u H = 9(1.008 u) = 9.072 u P = 1(30.97u) = 30.97 u O = 4(16.00 u) = 64.00 u 132.06 u c) There are 8 atoms of oxygen in Cu3(OH)2(CO3)2. The molecular mass is: Cu = 3(63.55 u) = 190.6 u O = 8(16.00 u) = 128.0 u H = 2(1.008 u) = 2.016 u C = 2(12.01 u) = 24.02 u

2.103

344.6 u 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 9 atoms of hydrogen in C6H5COONH4. The molecular mass is: C = 7(12.01 u) = 84.07 u H = 9(1.008 u) = 9.072 u O = 2(16.00 u) = 32.00 u N = 1(14.01 u) = 14.01 u 139.15 u b) There are 2 atoms of nitrogen in N2H6SO4. The molecular mass is: N = 2(14.01 u) = 28.02 u H = 6(1.008 u) = 6.048 u S = 1(32.07 u) = 32.07 u O = 4(16.00 u) = 64.00 u 130.14 u c) There are 12 atoms of oxygen in Pb4SO4(CO3)2(OH)2. The molecular mass is: Pb = 4(207.2 u) = 828.8 u S = 1(32.07 u) = 32.07 u O = 12(16.00 u) = 192.00 u C = 2(12.01 u) = 24.02 u H = 2(1.008 u) = 2.016 u 1078.9 u

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: a) (NH4)2SO4 ammonium is NH4+ and sulfate is SO42– N = 2(14.01 u) = 28.02 u H = 8(1.008 u) = 8.064 u S = 1(32.07 u) = 32.07 u O = 4(16.00 u) = 64.00 u 132.15 u b) NaH2PO4 sodium is Na+ and dihydrogen phosphate is H2PO4– Na = 1(22.99 u) = 22.99 u H = 2(1.008 u) = 2.016 u P = 1(30.97 u) = 30.97 u O = 4(16.00 u) = 64.00 u 119.98 u c) KHCO3 potassium is K+ and bicarbonate is HCO3– K = 1(39.10 u) = 39.10 u H = 1(1.008 u) = 1.008 u C = 1(12.01 u) = 12.01 u O = 3(16.00 u) = 48.00 u 100.12 u

2.105

b) NH4ClO4 N H Cl O

= = = =

c) Mg(NO2)2•3H2O Mg = N = H = O =

ammonium is NH4+ and perchlorate is ClO4– 1(14.01 u) = 14.01 u 4(1.008 u) = 4.032 u 1(35.45 u) = 35.45 u 4(16.00 u) = 64.00 u 117.49 u magnesium is Mg2+, nitrite is NO2–, and trihydrate is 3H2O 1(24.31 u) = 24.31 u 2(14.01 u) = 28.02 u 6(1.008 u) = 6.048 u 7(16.00 u) = 112.00 u 170.38 u

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: a) dinitrogen pentoxide N2O5 (di- = 2 and penta- = 5) N = 2(14.01 u) = 28.02 u O = 5(16.00 u) = 80.00 u 108.02 u b) lead(II) nitrate Pb(NO 3)2 (lead(II) is Pb2+ and nitrate is NO3–) Pb = 1(207.2 u) = 207.2 u N = 2(14.01 u) = 28.02 u O = 6(16.00 u) = 96.00 u 331.2 u c) calcium peroxide CaO2 (calcium is Ca2+ and peroxide is O22–) Ca = 1(40.08 u) = 40.08 u O = 2(16.00 u) = 32.00 u 72.08 u

2.107

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) iron(II) acetate tetrahydrate Fe(C2H3O2)2 •4H2O (iron(II) is Fe2+, acetate is C2H3O2–, and tetrahydrate is 4H2O) Fe = 1(55.85 u) = 55.85 u C = 4(12.01 u) = 48.04 u H = 14(1.008 u) = 14.112 u O = 8(16.00 u) = 128.00 u 246.00 u b) sulfur tetrachloride SCl4 (tetra- = 4) S = 1 (32.07 u) = 32.07 u Cl = 4(35.45 u) = 141.80 u 173.87 u c) potassium permanganate KMnO4 (potassium is K+ and permanganate is MnO4–) K = 1(39.10 u) = 39.10 u Mn = 1(54.94 u) = 54.94 u O = 4(16.00 u) = 64.00 u 158.04 u

2.108

Solution: a) Formula is SO3. Name is sulfur trioxide (the prefix triindicates 3 oxygen atoms). S = 1(32.07 u) = 32.07 u O = 3(16.00 u) = 48.00 u 80.07 u b) Formula is C3H8. Since it contains only carbon and hydrogen it is a hydrocarbon and with three carbons its name is propane. C = 3(12.01 u) = 36.03 u H = 8(1.008 u) = 8.064 u 44.09 u 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 N2O. Name is dinitrogen monoxide (the prefix diindicates 2 nitrogen atoms and mono- indicates 1 oxygen atom). N = 2(14.01 u) = 28.02 u O = 1(16.00 u) = 16.00 u 44.02 u b) Formula is C2H6. Since it contains only carbon and hydrogen it is a hydrocarbon and with three carbons its name is ethane. C = 2(12.01 u) = 24.02 u H = 6(1.008 u) = 6.048 u 30.07 u

2.110

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 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 SO42– = 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 SO42– is the sulfate ion; since this is an acid, -ate becomes -ic acid. d) washing soda Na2CO3 sodium carbonate CO32– 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 SO42– = sulfate; hepta- is used to indicate the 7 waters of hydration. g) chalk CaCO3 calcium carbonate CO32– 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.

2.111

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 u) = 28.02u O = 1(16.00 u) = 16.00 u 44.02 u 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 u) = 70.90 u O = 1(16.00 u) = 16.00 u 86.90 u

2.112

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).

2.113

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, colour, state, etc.) from one part of the mixture to another.

2.114

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

2.115

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.

2.116

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) 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.

2.117

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 coloured 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.

2.118

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.

2.119

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 a) Volume (m3) of nucleus =  r 3 =  2.5x1015 m = 6.54498x10–44 m3 3 3 3 4 4 3 3 Volume (m ) of atom =  r =  3.1x1011 m = 1.24788x10–31 m3 3 3





volume of Nucleus 6.54498x1044 m3 = = 5.2449x10–13 = 5.2x10–13 volume of Atom 1.24788x1031 m3 b) Mass of nucleus = mass of atom – mass of electrons = 6.64648x10–24 g – 2(9.10939x10–28 g) = 6.64466x10–24 g Fraction of volume =

Fraction of mass =

 

 

6.64466 x1024 g mass of Nucleus = = 0.99972617 = 0.999726 mass of Atom 6.64648 x1024 g

As expected, the volume of the nucleus relative to the volume of the atom is small while its relative mass is large. 2.120

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: Strongest ionic bonding: MgO. Mg2+, Ba2+, and O2– have the largest charges. Attraction increases as distance decreases, so the positive ion with the smaller radius, Mg2+, will form a stronger ionic bond than the larger ion Ba2+. 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.

2.121

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 u) = 79.90 u F = 3(19.00 u) = 57.00 u 136.90 u

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(32.07 u) = 32.07 u Cl = 2(35.45 u) = 70.90 u 102.97 u 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 u) = 30.97 u Cl = 3(35.45 u) = 106.35 u 137.32 u 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 u) = 28.02 u O = 5(16.00 u) = 80.00 u 108.02 u 2.122

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: a) SeO42– selenate ion from SO42– = sulfate ion 3– b) AsO4 arsenate ion from PO43– = phosphate ion – c) BrO2 bromite ion from ClO2– = chlorite ion d) HSeO4– hydrogen selenate ion from HSO4– = hydrogen sulfate ion 2– e) TeO3 tellurite ion from SO32– = sulfite ion

2.123

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 u) = 14.01 u H = 6(1.008 u) = 6.048 u P = 1(30.97 u) = 30.97 u O = 4(16.00 u) = 64.00 u 115.03 u 14.01 u N Mass percent of N = 100  = 12.18% N 115.03 u compound Mass percent of P =

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

  17.03 u NH3 b) Mass (g) of ammonia (NH3) = 100. g NH 4 H 2 PO 4    = 14.80 g NH3  115.03 u NH 4 H 2 PO 4  2.124

Plan: Determine the percent oxygen in each oxide by subtracting the percent nitrogen from 100%. Express the percentage in u 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.

Solution: a) I

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

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

3.3326 N 3.3319 O = 1.0002 N = 1.0000 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 u N   63.15 u O    = 2.6303 N   = 3.9469 O  14.01 u N   16.00 u O 

III

2.6303 N 3.9469 O = 1.0000 mol N = 1.5001 O 2.6303 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 u N   74.06 u O    = 4.6288 O   = 1.8515 N 14.01u N  16.00 u O   

1.8515 N 4.6288 O = 1.0000 N = 2.5000 O 1.8515 1.8515 The simplest whole-number ratio is 1:2.5 N:O = 2:5 N:O.  53.31 u O    = 1.1418 = 1.14 g O  46.69 u N 

 63.15 u O    = 1.7137 = 1.71 g O  36.85 u N 

III

 74.06 u O    = 2.8550 = 2.86 g O  25.94 u N 

2.125

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.

2.126

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 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

2.127

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.

2.128

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  6 a) Mass (mg) of seawater = 1 kg     = 1x10 mg 1 kg 1 g     mass of substance  Mass % =   100%   mass of seawater 

 18,980 mg Cl  100%  = 1.898% Cl– Mass % Cl– =   1x106 mg seawater    

 10,560 mg Na   100%  = 1.056% Na+ Mass % Na+ =   1x106 mg seawater      2650 mg SO42   100%  = 0.265% SO42– Mass % SO42– =   1x106 mg seawater    

 1270 mg Mg2   100%  = 0.127% Mg2+ Mass % Mg2+ =   1x106 mg seawater      400 mg Ca 2   100%  = 0.04% Ca2+ Mass % Ca2+ =   1x106 mg seawater    

  380 mg K  100%  = 0.038% K+ Mass % K+ =   1x106 mg seawater      140 mg HCO3  100%  = 0.014% HCO3– Mass % HCO3– =   1x106 mg seawater     The mass percents do not add to 100% since the majority of seawater is H 2O.

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 +  % Na + =  100  = 30.71553 = 30.72%  34,380 mg total ions     c) Alkaline earth metal ions are Mg2+ and Ca2+ (Group 2 ions). Total mass % = 0.127% Mg2+ + 0.04% Ca2+ = 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. d) Anions are Cl–, SO42–, and HCO3–. Total mass % = 1.898% Cl– + 0.265% SO42– + 0.014% HCO3– = 2.177% anions Cations are Na+, Mg2+, Ca2+, and K+. Total mass % = 1.056% Na+ + 0.127% Mg2+ + 0.04% Ca2+ + 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. 2.129

2.130

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. . 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 C4H6O4. C = 4(12.01 u) = 48.04 u H = 6(1.008 u) = 6.048 u O = 4 (16.00 u) = 64.00 u 118.09 u  48.04 u C  %C=  100% = 40.6815 = 40.68% C  118.088 u 

 6.048 u H  %H=  100% = 5.1216 = 5.122% H  118.088 u 

 64.00 u O  %O=  100% = 54.1969 = 54.20% O  118.088 u  Check: Total = (40.68 + 5.122 + 54.20)% = 100.00% The answer checks. 2.131

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 g of F– to reach the toxic level.  1 mg F  Mass (mg) of fluoride for a toxic level = 0.2 g F  = 200 mg F–  0.001 g F   





 1 L water  Volume (L) of water = 200 mg  = 200 L= 2x102 L water  1 mg F    7 Volume (L) of reservoir = = 8.50 x 10 L The molecular mass of NaF = 22.99 u Na + 19.00 u F = 41.99 u. There are 19.00 mg of F – in every 41.99 mg of NaF.  1 mg F   41.99 mg NaF   103 g   1 kg NaF  Mass (kg) of NaF = 8.50x107 L   1 L H O   19.00 mg F   1 mg   103 g NaF  2      = 187.85 kg= 188 kg NaF



2.132



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 x fractional abundance) + (isotopic mass of the second isotope x fractional abundance). Solution: a) Antimony is element 51so Z = 51. Isotope of mass 120.904 u has a mass number of 121: 121 51 Sb Isotope of mass 122.904 u has a mass number of 123: 123 51 Sb b) Let x = fractional abundance of antimony-121. This makes the fractional abundance of antimony-123 = 1 – x x(120.904 u) + (1 – x) (122.904 u) = 121.8 u 120.904 u(x) + 122.904 u – 122.904 u(x) = 121.8 u 2x = 1.104 x = 0.552 = 0.55 fraction of antimony-121 1 – x = 1 – 0.552 = 0.45 fraction of antimony-123

2.133

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 (u) 15 N218O 2(15 u N) + 18 u O = 48 least common 15 N216O 2(15 u N) + 16u O = 46 14 N218O 2(14 u N) + 18 u O = 46 14 N216O 2(14 u N) + 16 u O = 44 most common 15 14 18 N N O 1(15 u N) + 1(14 u N) + 18 u O = 47 15 14 16 N N O 1(15 u N) + 1(14 u N) + 16 u O = 45

2.134

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 b) 56 26 Fe

56 – 26 = 30

30/26 = 1.2

c) 20 10 Ne

20 – 10 = 10

10/10 = 1.0

d) 107 47 Ag

107 – 47 = 60

60/47 = 1.3

2.135

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 ionic charges of the metals and nonmetal must be equal in magnitude like Na + and Cl– or Ba2+ and O2–: 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 ionic charges of the metals and nonmetal must be equal in magnitude like Na + and Cl– or Ba2+ and O2–: Brown and green or blue and red i) Metals react with nonmetals to form ionic compounds. For a compound with a formula of M 2X, the ionic charge of the nonmetal must be twice as large as that of the metal like Na+ and O2– or Ba2+ and C4–: Brown and red or blue and black j) Metals react with nonmetals to form ionic compounds. For a compound with a formula of MX 2, the ionic charge of the metal must be twice as large as that of the nonmetal like Ba2+ and Cl–: Blue and green k) Most Group 18 elements are unreactive: Purple l) Different compounds often exist between the same two nonmetal elements. Since oxygen exists as O 2– or O22–, 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

2.136

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 x fractional abundance) + (isotopic mass of isotope 2 x fractional abundance). Solution: Average atomic mass of K = (isotopic mass of 39K x fractional abundance) + (isotopic mass of 41K x fractional abundance)

 93.258%   6.730%    (40.9618 u)   = 39.093 u  100%   100% 

Average atomic mass of K = (38.9637 u) 

The formula for potassium fluoride is KF, so its molecular mass is (39.093 + 18.9984)u = 58.091 u 2.137

Plan: List all possible combinations of the isotopes. BF3 contains either 10B or 11B. 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 u B + 3(19 u F) = 67 u 10 19 B F2 = 10 u B + 2(19 u F) = 48 u

B19F B 11 19 B F3 11 19 B F2 11 19 B F 11 B 10

= 10 u B + 19 u F = 29 u = 10 u B = 10. u = 11 u B + 3(19 u F) = 68 u = 11 u B + 2(19 u F) = 49 u = 11 u B + 19 u F = 30 u = 11 u B = 11 u

2.138

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) u = 30.01 u Nitroglycerin: C3H5N3O9 = 3(12.01 u C) + 5(1.008 u H) + 3(14.01 u N) + 9(16.00 u O) = 227.10 u In C3H5N3O9 (molecular mass = 227.10 u), 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 u. total mass of NO 3(30.01 u) Mass percent of NO = 100%   100%  = 39.6433% = 39.64% mass of compound 227.10 u Isoamyl nitrate: C5H11NO3 = 5(12.01 u C) + 11(1.008 u H) + 1(14.01 u N) + 3(16.00 u O) = 133.15 u In (CH3)2CHCH2CH2ONO2 (molecular mass = 133.15 u), 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 u) Mass percent of NO = 100%   100%  = 22.5385 %= 22.54% mass of compound 133.15 u

2.139

Plan: First, count each type of atom present to produce a molecular formula. Determine the mass fraction of each total mass of the element . The mass of TNT multiplied by the mass fraction of each element. Mass fraction = molecular mass of TNT element gives the mass of that element. Solution: The molecular formula for TNT is C7H5O6N3. The molecular mass of TNT is: C = 7(12.01 u) = 84.07 u H = 5(1.008 u) = 5.040 u O = 6(16.00 u) = 96.00 u N = 3(14.01 u) = 42.03 u 227.14 u The mass fraction of each element is: 84.07 u 5.040 u C= = 0.3701 C H= = 0.02219 H 227.14 u 227.14 u

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

b) 238 92 U

neutrons 238 – 92 = 146

protons 92

electrons 92

234 92 U

234 – 92 = 142

214 82 Pb

214 – 82 = 132

2.140

210 82 Pb

210 – 82 = 128

206 82 Pb

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 u) = 195.1 u N = 2 (14.01 u) = 28.02 u H = 6(1.008 u) = 6.048 u Cl = 2(35.45 u) = 70.90 u 300.1 u mass of Pt 195.1 u Mass % Pt = 100%  = 100%  = 65.012 %= 65.01% Pt molecular mass of compound 300.1 u





 1 g Pt  b) Mass (g) of Pt = $1.00 x 106   = 31,250 g Pt  $32 

 100 g platinol  4 4 Mass (g) of platinol =  31, 250 g Pt    = 4.8070x10 g= 4.8x10 g platinol  65.01 g Pt  2.141

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 u 1 1 Carbon C 6 12.01 u 2 14 Nitrogen N 7 14.01 u 2 15 Oxygen O 8 16.00 u 2 16 b) Macronutrients: Sodium Na 11 22.99 u 3 1 Magnesium Mg 12 24.31 u 3 2 Potassium K 19 39.10 u 4 1 Calcium Ca 20 40.08 u 4 2 Phosphorus P 15 30.97 u 3 15 Sulfur S 16 32.07 u 3 16 Chlorine Cl 17 35.45 u 3 17

2.142

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.

2.143

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.

CHAPTER 3 STOICHIOMETRY OF FORMULAE

AND EQUATIONS END–OF–CHAPTER PROBLEMS 3.1 Plan: The atomic mass of an element expressed in u is numerically the same as the mass of 1 mole of the element expressed in grams. We know the amount (mol) of each element and have to find the mass (in g). To convert amount (mol) of element to mass of element, multiply the amount by the molar mass of the element. Solution: Al 26.98 u  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 u  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 amount (mol) of sucrose by 12 to obtain amount (mol) of carbon atoms; multiply the amount (mol) of carbon atoms by Avogadro‘s number to convert from amount (mol) to atoms. Solution:   12 mol C a) Amount (mol) of C atoms = 1 mol C12 H 22 O11   = 12 mol C  1 mol C12 H 22 O11     6.022 x1023 C atoms  12 mol C 25 b) C atoms =  2 mol C12 H 22O11    = 1.445x10 C atoms   1 mol C H O 1 mol C 12 22 11    

3.3

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, Cl 2. 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.

3.4

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 u 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 taking its mass. 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 amount (mol) 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 P 4 molecules (4:1) to convert amount (mol) of phosphorus atoms to amount (mol) of phosphorus molecules. Finally, multiply amount (mol) of P4 molecules by Avogadro‘s number to find the number of molecules.

Solution: Roadmap Mass (g) of Ca3(PO4)2 Divide by M (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 Multiply by 6.022x1023 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 atoms are necessary to counterbalance the mass of 6 yellow atoms. Since the yellow atoms are lighter, their atomic mass is lower, and therefore there will be more atoms per gram on the right and fewer atoms per gram on the left. Since one mole of any compound contains the same number of particles (6.022x10 23) neither side has more atoms per mole. b) This figure requires more thought because the number of red and blue atoms is unequal and their masses are unequal. If each pan only contained 3 atoms, then the red atoms would be still be lighter. The presence of 6 red atoms means that they are much lighter. Because the red atoms are lighter, more red atoms are required to make 1 g. Therefore the higher molar mass compound is on the right (blue), the left has more atoms per gram and the right has fewer atoms per gram. Since one mole of any compound contains the same number of particles (6.022x1023) neither side has more atoms per mole. c) The element on the left (orange) has a higher molar mass since the 5 atoms weigh more than the 5 purple atoms. The purples atoms (right) are lighter so there are more atoms per gram and the orange atoms (left) would have fewer atoms per gram. Since one mole of any compound contains the same number of particles (6.022x10 23) neither side has more atoms per mole. d) We can see that it takes more red atoms to exactly counter the weight of the black atoms, therefore the black atoms (left) are heavier and have a higher molar mass. The red atoms (right) are lighter and there will be more atoms per gram versus the heavier black atoms (left) which will have fewer atoms per gram. Since one mole of any compound contains the same number of particles (6.022x10 23) neither side has more atoms per mole.

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) M = (1 •M of Sr) + (2 •M of O) + (2 • M 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) M = (2 •M of N) + (3 • M of O) = (2 •14.01 g/mol N) + (3 • 16.00 g/mol O) = 76.02 g/mol of N2O3 c) M = (1 • M of Na) + (1 •M of Cl) + (3 • M 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 d) M = (2 •M of Cr) + (3 • M 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) M = (3 • M of N) + (12 •M of H) + (1 •M of P) + (4 •M 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) M = (1 •M of C) + (2 •M of H) + (2 •M 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) M = (1 •M of Cu) + (1 •M of S) + (9 •M of O) + (10 •M of H) = (1 •63.55 g/mol Cu) + (1 •32.07 g/mol S) + (9 •16.00 g/mol O) + (10 •1.008 g/mol H) = 249.70 g/mol of CuSO4•5H2O d) M = (1 •M of Br) + (3 •M 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) M = (1 •M of Sn) + (1 •M of O) = (1 •118.7 g/mol Sn) + (1 •16.00 g/mol O) = 134.7 g/mol of SnO b) M = (1 •M of Ba) + (2 •M of F) = (1 •137.3 g/mol Ba) + (2 •19.00 g/mol F) = 175.3 g/mol of BaF2 c) M = (2 •M of Al) + (3 •M of S) + (12 •M of O) = (2 •26.98 g/mol Al) + (3 •32.07 g/mol S) + (12 •16.00 g/mol O) = 342.17 g/mol of Al2(SO4)3 d) M = (1 •M of Mn) + (2 •M of Cl) = (1 •54.94 g/mol Mn) + (2 • 35.45 g/mol Cl) = 125.84 g/mol of MnCl2

3.11

a) M = (2 •M of N) + (4 •M of O) = (2 •14.01 g/mol N) + (4 •16.00 g/mol O) = 92.02 g/mol of N2O4 b) M = (4 •M of C) + (10 •M of H) + (1 •M 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) M = (1 •M of Mg) + (1 •M of S) + (11 x M of O) + (14 •M of H) = (1 •24.31 g/mol Mg) + (1 •32.07 g/mol S) + (11 •16.00 g/mol O) + (14 •1.008 g/mol H) = 246.49 g/mol of MgSO4•7H2O d) M = (1 •M of Ca) + (4 •M of C) + (6 •M of H) + (4 •M 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 3.12

Plan: Determine the molar mass of each substance, then perform the appropriate molar conversions. To find the mass in part a), multiply the amount (mol) by the molar mass of the substance. In part b), first convert mass of compound to amount (mol) 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 amount (mol) of compound to amount (mol) of oxygen atoms. In part c), convert mass of compound to amount (mol) of compound by dividing by the molar mass of the compound. Since 1 mole of compound contains 6 moles of oxygen atoms, multiply the amount (mol) of compound by 6 to obtain amount (mol) of oxygen atoms; then multiply by Avogadro‘s number to obtain the number of oxygen atoms. Solution: a) M of KMnO4 = (1 •M of K) + (1 •M of Mn) + (4 •M 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 KMnO4  2 Mass of KMnO4 = 0.68 mol KMnO 4   = 107.467 g= 1.1x10 g KMnO4  1 mol KMnO 4  b) M of Ba(NO3)2 = (1 •M of Ba) + (2 •M of N) + (6 •M 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  1 mol Ba(NO3 ) 2  Amount (mol) of Ba(NO3)2 = 8.18 g Ba(NO3 ) 2   = 0.031305 mol Ba(NO3)2  261.3 g Ba(NO3 ) 2   6 mol O atoms  Amount (mol) of O atoms = 0.031305 mol Ba(NO3 ) 2    1 mol Ba(NO3 ) 2  = 0.18783 mol= 0.188 mol O atoms c) M of CaSO4•2H2O = (1 •of Ca) + (1 •M of S) + (6 •M of O) + (4 •M of H) = (1 •40.08 g/mol Ca) + (1 •32.07 g/mol S) + (6 •16.00 g/mol O) + (4 •1.008 g/mol H) = 172.18 g/mol (Note that the waters of hydration are included in the molar mass.)  1 mol CaSO4 2H 2O  Amount (mol) of CaSO4•2H2O = 7.3x103 g CaSO4 2H 2O    172.18 g CaSO4 2H 2O 





= 4.239749x10–5 mol

 6 mol O atoms  Amount (mol) of O atoms = 4.239749 x105 mol CaSO 4 2H 2O    1 mol CaSO4 2H 2O  = 2.54385x10–4 mol O atoms  6.022 x1023 O atoms  Number of O atoms = 2.54385x104 mol O atoms   1 mol O atoms    = 1.5319x1020 atoms= 1.5x1020 O atoms



3.13



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 amount (mol) 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 amount (mol) 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 amount (mol) of compound to amount (mol) of chlorine atoms. In part c), convert mass of compound to amount (mol) of compound by dividing by the molar mass of the compound. Since 1 mole of compound contains 2 moles of H– ions, multiply the amount (mol) of compound by 2 to obtain amount (mol) of H– ions; then multiply by Avogadro‘s number to obtain the number of H – ions. Solution: a) M of NO2 = (1 x M of N) + (2 x M of O) = (1 x 14.01 g/mol N) + (2 x 16.00 g/mol O) = 46.01g/mol of NO 2   1 mol NO2 Amount (mol) of NO2 = 4.6x1021 molecules NO2  = 7.63866x10–3 mol NO2 23  6.022 x10 molecules NO  2  





 46.01 g NO2   1 kg  –4 –4 Mass (kg) of NO2 = 7.63866x10 3 mol NO2    3  = 3.51455x10 kg= 3.5x10 kg NO2 1 mol NO 10 g 2    b) M of C2H4Cl2 = (2 •M of C) + (4 •M of H) + (2 x M 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 C2 H 4 Cl2  –4 Amount (mol) of C2H4Cl2 = 0.0615 g C2 H 4 Cl2   = 6.21526x10 mol C2H4Cl2 98.95 g C H Cl 2 4 2  





 2 mol Cl atoms  –3 Amount (mol) of Cl atoms = 6.21526x10 4 mol C2 H 4Cl 2   = 1.2431x10 1 mol C H Cl 2 4 2   = 1.24x10–3 mol Cl atoms c) M of SrH2 = (1 •M of Sr) + (2 •M of H) = (1 •87.62 g/mol Sr) + (2 •1.008 g/mol H) = 89.64 g/mol of SrH2  1 mol SrH 2  Amount (mol) of SrH2 = 5.82 g SrH 2   = 0.0649264 mol SrH2  89.64 g SrH 2 





 2 mol H   Amount (mol) of H– ions = 0.0649264 mol SrH 2  = 0.1298528 mol H– ions  1 mol SrH  2  

 6.022x1023 H  ions  22 22 – Number of H– ions = 0.1298528 mol H  ions   = 7.81974x10 = 7.82x10 H ions   1 mol H  



3.14







b) M of Fe(ClO4)3 = (1 •M of Fe) + (3 •M of Cl) + (12 •M 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  Mass (g) of Fe(ClO4)3 = 15.8 kg Fe(ClO4 )3  = 1.58 x 104 kg Fe(ClO4)3  1 kg   

 1 mol Fe(ClO 4 )3  Amount (mol) of Fe(ClO4)3 = 1.58x104 g Fe(ClO4 )3    354.20 g Fe(ClO4 )3  = 44.6076 mol= 44.6 mol Fe(ClO4)3 c) M of NH4NO2 = (2 •M of N) + (4 •M of H) + (2 •M 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  Mass (g) of NH4NO2 =  92.6 mg NH 4 NO2   = 0.0926 g NH4NO2  1 mg   





 1 mol NH 4 NO 2  –3 Amount (mol) of NH4NO2 =  0.0926 g NH 4 NO2    = 1.44575x10 mol NH4NO2 64.05 g NH NO 4 2    2 mol N atoms  –3 Amount (mol) of N atoms = 1.44575x103 mol NH 4 NO2   = 2.8915x10 mol N atoms  1 mol NH 4 NO2 





 6.022 x1023 N atoms  Number of N atoms = 2.8915x103 mol N atoms   1 mol N atoms    = 1.74126 x 1021 atoms= 1.74 x 1021 N atoms



3.15



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 amount (mol) of compound. Since 1 mole of compound contains 3 moles of ions (1 mole of Sr 2+ and 2 moles of F–), multiply the amount (mol) of compound by 3 to obtain amount (mol) of ions and then multiply by Avogadro‘s number to obtain the number of ions. In part b), multiply the amount (mol) 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 amount (mol); multiply the amount (mol) by the molar mass to obtain the mass in g and then convert to mg. Solution: a) M of SrF2 = (1 •M of Sr) + (2 •M of F) = (1 •87.62 g/mol Sr) + (2 •19.00 g/mol F) = 125.62 g/mol of SrF 2  1 mol SrF2  Amount (mol) of SrF2 = 38.1 g SrF2   = 0.303296 mol SrF2  125.62 g SrF2   3 mol ions  Amount (mol) of ions = 0.303296 mol SrF2   = 0.909888 mol ions  1 mol SrF2 

 6.022 x1023 ions  Number of ions =  0.909888 mol ions   = 5.47935x1023 = 5.48x1023 ions  1 mol ions    b) M of CuCl2•2H2O = (1 •M of Cu) + (2 •M of Cl) + (4 •M of H) + (2 •M 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 CuCl 2 •2H 2O  Mass (g) of CuCl2•2H2O =  3.58 mol CuCl 2 •2H 2O    = 610.32 g CuCl2•2H2O  1 mol CuCl 2 •2H 2O   1 kg  Mass (kg) of CuCl2•2H2O = 610.32 g CuCl2 2H 2 O  3  = 0.61032 = 0.610 kg CuCl2•2H2O  10 g    c) M of Bi(NO3)3•5H2O = (1 •M of Bi) + (3 •M of N) + (10 •M of H) + (14 •M 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(NO 3)3•5H2O (Note that the waters of hydration are included in the molar mass.)   1 mol Amount (mol) of Bi(NO3)3•5H2O = 2.88 x1022 FU   = 0.047825 mol Bi(NO3)3•5H2O 23  6.022 x10 FU 





 485.1 g Bi(NO3 )3 5H 2 O  Mass (g) of Bi(NO3)3•5H2O = 0.047825 mol Bi(NO3 )3 5H 2 O   = 23.1999 g  1 mol Bi(NO3 )3 5H 2 O 

 1 mg  Mass (mg) of Bi(NO3)3•5H2O = 23.1999 g Bi(NO3 )3 5H 2 O  3   10 g    4 = 23199.9 mg= 2.32x10 mg Bi(NO3)3•5H2O

3.16

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 amount (mol) 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 amount (mol); multiply the amount (mol) by the molar mass to obtain the mass. In part c), divide the mass by the molar mass to find amount (mol) of compound and multiply amount (mol) by Avogadro‘s number to find the number of formula units. In part d), use the fact that each formula unit contains 1 Na ion, 1 perchlorate ion, and that each perchlorate ion contains 1 Cl atom, and 4 O atoms. Solution: a) Carbonate is a polyatomic anion with the formula, CO32–. Copper(I) indicates Cu+. The correct formula for this ionic compound is Cu2CO3. M of Cu2CO3 = (2 •M of Cu) + (1 •M of C) + (3 •M 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 Cu 2 CO3  3 Mass (g) of Cu2CO3 = 8.35 mol Cu 2 CO3   = 1562.4 = 1.56x10 g Cu2CO3 1 mol Cu CO 2 3   b) Dinitrogen pentaoxide has the formula N2O5. Di- indicates 2 N atoms and penta- indicates 5 O atoms. M of N2O5 = (2 •M of N) + (5 •M of O) = (2 •14.01 g/mol N) + (5 •16.00 g/mol O) = 108.02 g/mol of N2O5   1 mol N 2O5 Amount (mol) of N2O5 = 4.04 x1020 N 2O5 molecules  = 6.7087x10–4 mol N2O5  6.022x1023 N O molecules  2 5  





 108.02 g N 2O5  Mass (g) of N2O5 = 6.7087x104 mol N 2O5   = 0.072467 g= 0.0725 g N2O5  1 mol N 2O5  c) The correct formula for this ionic compound is NaClO 4; Na has a charge of +1 (Group 1 ion) and the perchlorate ion is ClO4– . M of NaClO4 = (1• M of Na) + (1 •M of Cl) + (4 •M 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  Amount (mol) of NaClO4 = 78.9 g NaClO 4   = 0.644397 mol= 0.644 mol NaClO4  122.44 g NaClO 4 





 6.022 x1023 formula units NaClO 4  Formula units of NaClO4 =  0.644397 mol NaClO 4     1 mol NaClO 4   = 3.88056x1023 formula units= 3.88x1023 formula unit NaClO4   1 Na  ion d) Number of Na+ ions = 3.88056x1023 formula units NaClO4   1 formula unit NaClO  4   = 3.88x1023 Na+ ions   1 ClO4 ion Number of ClO4– ions = 3.88056x1023 formula units NaClO4   1 formula unit NaClO  4   = 3.88x1023 ClO4– ions   1 Cl atom Number of Cl atoms = 3.88056x1023 formula units NaClO4    1 formula unit NaClO4 













= 3.88x1023 Cl atoms Silberberg, Amateis, Venkateswaran, Chen, Chemistry: The Molecular Nature of Matter and Change, 3rd Canadian Edition Page 3-61 Instructor‘s Solution Manual © Copyright 2021 McGraw-Hill Ryerson Ltd.

  4 O atoms Number of O atoms = 3.88056x1023 formula units NaClO 4    1 formula unit NaClO 4  = 1.55x1024 O atoms



3.17



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 amount (mol) 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 amount (mol); multiply the amount (mol) by the molar mass to obtain the mass. In part c), divide the mass by the molar mass to find amount (mol) of compound and multiply amount (mol) 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: a) Sulfate is a polyatomic anion with the formula, SO42–. Chromium(III) indicates Cr3+. Decahydrate indicates 10 water molecules (―waters of hydration‖). The correct formula for this ionic compound is Cr 2(SO4)3•10H2O. M of Cr2(SO4)3•10H2O = (2 •M of Cr) + (3 •M of S) + (22 •M of O) + (20 •M of H) = (2 •52.00 g/mol Cr) + (3 •32.07 g/mol S) + (22 •16.00 g/mol O) + (20 •1.008 g/mol H) = 572.4 g/mol of Cr2(SO4)3•10H2O  572.4 g  Mass (g) of Cr2(SO4)3•10H2O = 8.42 mol Cr2 (SO4 )3 10H 2O     mol  = 4819.608 g= 4.82x103 g Cr2(SO4)3•10H2O b) Dichlorine heptoxide has the formula Cl2O7. Di- indicates 2 Cl atoms and hepta- indicates 7 O atoms. M of Cl2O7 = (2 •M of Cl) + (7 •M of O) = (2 •35.45 g/mol Cl) + (7 •16.00 g/mol O) = 182.9 g/mol of Cl2O7   1 mol Amount (mol) of Cl2O7 = 1.83x1024 molecules Cl 2O7   = 3.038858 mol Cl2O7 23  6.022 x10 molecules 

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

 182.9 g Cl2 O7  2 Mass (g) of Cl2O7 = 3.038858 mol Cl2 O7   = 555.807 g= 5.56x10 g Cl2O7 1 mol   c) The correct formula for this ionic compound is Li2SO4; Li has a charge of +1 (Group 1 ion) and the sulfate ion is SO42– . M of Li2SO4 = (2 •M of Li) + (1 •M of S) + (4 •M of O) = (2 •6.941 g/mol Li) + (1 •32.07 g/mol S) + (4 •16.00 g/mol O) = 109.95 g/mol of Li2SO4  1 mol Li 2SO 4  Amount (mol) of Li2SO4 = 6.2 g Li 2SO 4   = 0.056389mol = 0.056 mol Li2SO4  109.95 g Li 2SO 4 

 6.022 x1023 FU  formula unit of Li2SO4 =  0.056389 mol Li 2SO4    1 mol Li SO  2 4   22 22 = 3.3957x10 formula unit = 3.4x10 formula unit Li2SO4   2 Li  ions d) Number of Li+ ions = 3.3957x1022 formula units Li 2SO4   1 formula unit Li SO  2 4   = 6.7914x1022 ions= 6.8x1022 Li+ ions   1 SO42  ion Number of SO42– ions = 3.3957x1022 formula units Li 2SO4   1 formula unit Li SO  2 4   22 22 2– = 3.3957x10 ions= 3.4x10 SO4 ions   1 S atom Number of S atoms = 3.3957x1022 formula units Li 2SO 4    1 formula unit Li 2SO 4 









= 3.3957x1022 atoms= 3.4x1022 S atoms

  4 O atoms Number of O atoms = 3.3957x1022 formula units Li 2SO 4    1 formula unit Li 2SO 4  = 1.3583x1023 atoms= 1.4x1023 O atoms

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3.18



Plan: Determine the formula and the molar mass of each compound. The formula gives the relative number of amount (mol) of each element present. Multiply the amount (mol) of each element by its molar mass to find the total mass of element total mass of element in 1 mole of compound. Mass percent = 100 . 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. M of NH4HCO3 = (1 •M of N) + (5 •M of H) + (1 •M of C) + (3 •M 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 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. M of NaH2PO4•7H2O = (1 •M of Na) + (16 •M of H) + (1 •M of P) + (11 •M 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 

Mass percent =

3.19

total mass O 176.00 g O 100%  = 100%  molar mass of compound 246.09 g NaH 2 PO 4 7H 2O = 71.51855 %= 71.52% O

Plan: Determine the formula and the molar mass of each compound. The formula gives the relative amount (mol) of each element present. Multiply the amount (mol) of each element by its molar mass to find the total mass of total mass of element element in 1 mole of compound. Mass percent = 100 . molar mass of compound Solution: a) Strontium periodate is an ionic compound consisting of strontium ions, Sr 2+ and periodate ions, IO4–. The formula of the compound is Sr(IO4)2. M of Sr(IO4)2 = (1 •M of Sr) + (2 •Mof I) + (8 •M 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.  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 molar mass of compound 469.4 g Sr(IO 4 ) 2 b) Potassium permanganate is an ionic compound consisting of potassium ions, K + and permanganate ions, MnO4–. The formula of the compound is KMnO4. M of KMnO4 = (1 •M of K) + (1 •M of Mn) + (4 •M of O)

Mass percent =

= (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 =

3.20

total mass Mn 54.94 g Mn 100%  = 100%  molar mass of compound 158.04 g KMnO 4 = 34.76335% = 34.76% Mn

Plan: Determine the formula and the molar mass of each compound. The formula gives the relative amount (mol) of each element present. Multiply the amount (mol) of each element by its molar mass to find the total mass of total mass of element element in 1 mole of compound. Mass fraction = . molar mass of compound Solution: 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. M of CsC2H3O2 = (1 •M of Cs) + (2 •M of C) + (3 •M of H) + (2 •M 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 = = 0.125169 = 0.1252 mass fraction C molar mass of compound 191.9 g CsC2 H3O 2 b) Uranyl sulfate trihydrate is is a salt that consists of uranyl ions, UO 22+, sulfate ions, SO42–, and three waters of hydration. The formula is UO2SO4•3H2O. Note that the waters of hydration are included in the molar mass. M of UO2SO4•3H2O = (1 •M of U) + (9 •M of O) + (1 •M of S) + (6 •M of H) = (1 •238.0 g/mol U) + (9 •16.00 g/mol O) + (1 •32.07 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 

Mass fraction =

Mass fraction = 3.21

total mass O 144.0 g O = = 0.3427755 = 0.3428 mass fraction O molar mass of compound 420.1 g UO 2SO 4 3H 2 O

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. M of N2O3 = (2 •M of N) + (3 •M 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 

Mass fraction =

Mass fraction = 3.22

total mass N 28.02 g N = = 0.368587 = 0.3686 mass fraction N molar mass of compound 76.02 g N 2 O3

Plan: Divide the mass given by the molar mass of O2 to find amount (mol). Since 1 mole of oxygen molecules contains 2 moles of oxygen atoms, multiply the amount (mol) by 2 to obtain amount (mol) of atoms and then multiply by Avogadro‘s number to obtain the number of atoms. Solution:  1 mol O 2  Amount (mol) of O2 = 38.0 g O 2   = 1.1875 mol O2  32.00 g O 2   2 mol O atoms  Amount (mol) of O atoms = 1.1875 mol O 2   = 2.375 mol O atoms  1 mol O 2 

 6.022x1023 O atoms  Number of O atoms =  2.375 mol O atoms   = 1.430225x1024 = 1.43x1024 O atoms  1 mol O atoms    3.23

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 amount (mol) of cisplatin. Since 1 mole of cisplatin contains 6 moles of hydrogen atoms, multiply the amount (mol) given by 6 to obtain amount (mol) 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. M of Pt(Cl)2(NH3) 2 = (1 •M of Pt) + (2 •M of Cl) + (2 •M of N) + (6 •M 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  1 mol cisplatin  a) Amount (mol) of cisplatin = 285.3 g cisplatin   = 0.9506831 mol= 0.9507 mol cisplatin  300.1 g cisplatin   6 mol H  b) Amount (mol) of H atoms = 0.98 mol cisplatin   = 5.88 mol H atoms 1 mol cisplatin  

 6.022 x1023 H atoms  Number of H atoms =  5.88 mol H atoms   = 3.540936x1024 atoms= 3.5x1024 H atoms  1 mol H atoms   

3.24

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 amount (mol) of compound. Since 1 mole of compound contains 6 moles of carbon atoms, multiply the amount (mol) of compound by 6 to obtain amount (mol) of carbon and then multiply by Avogadro‘s number to obtain the number of atoms. Solution: The formula, from the figure, is (C3H5)2S. M of (C3H5)2S = (6 •M of C) + (10 •M of H) + (1 •M of S) = (6 •12.01 g/mol C) + (10 •1.008 g/mol H) + (1 •32.07 g/mol N) = 114.21 g/mol of (C3H5)2S

 114.21 g allyl sulfide  a) Mass (g) of allyl sulfide =  2.63 mol allyl sulfide    = 300.3723 g= 300. g allyl sulfide  1 mol allyl sulfide   1 mol (C3 H5 ) 2 S  b) Amount (mol) of allyl sulfide =  35.7 g (C3 H 5 ) 2 S    = 0.312582 mol allyl sulfide  114.21 g (C3 H 5 ) 2 S    6 mol C Amount (mol) of C atoms =  0.312582 mol (C3H 5 ) 2 S    = 1.8755 mol C atoms  1 mol (C3 H 5 ) 2 S 

 6.022 x1023 C atoms  Number of C atoms = 1.8755 mol C atoms   = 1.129426x1024 atoms= 1.13x1024 C atoms  1 mol C atoms   

3.25

Plan: Determine the molar mass of rust. Convert mass in kg to mass in g and divide by the molar mass to find the amount (mol) of rust. Since each mole of rust contains 1 mole of Fe2O3, multiply the amount (mol) of rust by 1 to obtain amount (mol) of Fe2O3. Multiply the amount (mol) of Fe2O3 by 2 to obtain amount (mol) of Fe (1:2 Fe2O3:Fe mole ratio) and multiply by the molar mass of Fe to convert to mass. Solution: a) M of Fe2O3•4H2O = (2 •M of Fe) + (7 •M of O) + (8 •M 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  Mass (g) of rust =  45.2 kg rust   = 4.52x104 g  1 kg     1 mol rust  Amount (mol) of rust = 4.52x104 g rust   = 195.029 mol= 195 mol 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.  2 mol Fe  c) Amount (mol) of iron = 195.029 mol Fe 2 O3    = 390.058 mol Fe  1 mol Fe 2 O3 





 55.85 g Fe  4 Mass (g) of iron =  390.058 mol Fe    = 21784.74 g= 2.18x10 g Fe  1 mol Fe 

3.26

Plan: Determine the molar mass of propane. Divide the given mass by the molar mass to find the amount (mol). Since each mole of propane contains 3 moles of carbon, multiply the amount (mol) of propane by 3 to obtain amount (mol) of C atoms. Multiply the amount (mol) of C by its molar mass to obtain mass of carbon. Solution: a) The formula of propane is C3H8. M of C3H8 = (3 •M of C) + (8 •M of H) = (3 •12.01 g/mol C) + (8•1.008 g/mol H) = 44.09 g/mol  1 mol C3 H8  Amount (mol) of C3H8 =  85.5 g C3 H8    = 1.939215 mol= 1.94 mol C3H8  44.09 g C3 H8   3 mol C  b) Amount (mol) of C = 1.939215 mol C3H8    = 5.817645 mol C  1 mol C3 H8   12.01 g C  Mass (g) of C =  5.817645 mol C    = 69.86992 g= 69.9 g C  1 mol C 

3.27

Plan: Determine the formula and the molar mass of each compound. The formula gives the relative amount (mol) of nitrogen present. Multiply the amount (mol) 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 by 100 to  mol N  molar mass N  determine mass percent. Mass percent = 100%  . Then rank the values in order of molar mass of compound decreasing mass percent N. Solution:

Name Potassium nitrate Ammonium nitrate Ammonium sulfate Urea

Formula Molar Mass (g/mol) KNO3 101.11 NH4NO3 80.05 (NH4)2SO4 132.15 CO(NH2)2 60.06 1 mol N  14.01 g/mol N  x 100% = 13.856196 %= 13.86% N Mass % N in potassium nitrate = 101.11 g/mol Mass % N in ammonium nitrate = Mass % N in ammonium sulfate =

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

 2 mol N  14.01 g/mol N  132.15 g/mol

 2 mol N 14.01 g/mol N 

Mass % N in urea =

x 100% = 21.20318 %= 21.20% N

x 100% = 46.6533 %= 46.65% N

60.06 g/mol Rank is CO(NH2)2 > NH4NO3 > (NH4)2SO4 > KNO3 3.28

x 100% = 35.003123% = 35.00% N

Plan: The volume must be converted from cubic metres to cubic centimetres. The volume and the density will give the mass of galena which is then divided by molar mass to obtain amount (mol). Part b) requires a conversion from cubic decimetres to cubic centimetres. The density allows a change from volume in cubic centimetres to mass which is then divided by the molar mass to obtain amount (mol); 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: Lead(II) sulfide is composed of Pb2+ and S2– ions and has a formula of PbS. M of PbS = (1 •M of Pb) + (1 •M of S) = (1 •207.2 g/mol Pb) + (1 •32.07 g/mol S) = 239.3 g/mol 3

 100 cm  3 a) Volume (cm3) = 1.00 m3 PbS   = 1,000,000 cm  1m 









 7.46 g PbS  6 Mass (g) of PbS = 1x106 cm3 PbS   = 7.46x10 g PbS  1 cm3   1 mol PbS  4 Amount (mol) of PbS = 7.46x106 g PbS   = 31,174.26 mol = 3.12x10 mol PbS  239.3 g PbS 





  0.1 m 3   1 cm 3   b) Volume (cm ) = 1.00 dm PbS  = 1.00x103 cm3  1 dm 3   102 m 3     7.46 g PbS   Mass (g) of PbS = 1.00x103 cm3 PbS   = 7460 g PbS 3  1 cm   1 mol PbS  Amount (mol) of PbS =  7460 g PbS   = 31.17426 mol PbS  239.3 g PbS  3









 1 mol Pb  Amount (mol) of Pb =  31.17426 mol PbS   = 31.17426 mol Pb  1 mol PbS  Number of lead atoms =  6.022x1023 Pb atoms  25 25  31.17426 mol Pb    = 1.87731x10 atoms= 1.88x10 Pb atoms 1 mol Pb   3.29

Plan: If the molecular formula for hemoglobin (Hb) were known, the number of Fe 2+ 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 amount (mol) of Fe in 1 mole of hemoglobin which is also the number of ions in 1 molecule. Solution:  0.33% Fe   6.8 x104 g  Mass of Fe =   = 224.4 g Fe    100% Hb   mol   1 mol Fe  2+ Amount (mol) of Fe =  224.4 g Fe    = 4.0179 mol= 4.0 mol Fe /mol Hb 55.85 g Fe   Thus, there are 4 Fe2+/molecule Hb.

3.30

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 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.

3.31

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.

3.32

Plan: Remember that the molecular formula tells the actual amount (mol) 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 f amount (mol) 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 amount (mol) using molar mass. 2) Identify the element with the lowest f amount (mol) and use this number to divide into the number of amount (mol) 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 M (g/mol) Amount (mol) of each element

Use amount 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 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 3.33

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

3.34

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.

3.35

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(32.07 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.

3.36

The compound has 2 sulfur atoms and 2 chlorine atoms and a molecular formula of S2Cl2. 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

M of S2Cl2 = (2 •M of S) + (2 •M of Cl) = (2 •32.07 g/mol S) + (2 •35.45 g/mol Cl) = 135.04 g/mol 3.37

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 hexoxide. Phosphorus is named first since it has the lower group number. The prefix tetra- is for phosphorus since there are 4 Patoms and hexa- is used for oxygen since there are 6 O atoms. The empirical formula is P4 O 6 or P2O3. 2

M of P4O6 = (4 •M of P) + (6 •M of O) = (4 •30.97 g/mol P) + (6 •16.00 g/mol O) = 219.88 g/mol 3.38

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) 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 Whole-number multiple = =   =3 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  32.05 g/mol  molar mass of compound Whole-number multiple = =  = 2 empirical formula mass  16.03 g/mol  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  92.02 g/mol  molar mass of compound Whole-number multiple = =  =2 empirical formula mass  46.01 g/mol  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 Whole-number multiple = =  =5 empirical formula mass  27.03 g/mol  Multiplying the subscripts in CHN by 5 gives C5H5N5.

3.39

Whole-number multiple =

 78.11 g/mol  molar mass of compound =  =6 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  74.08 g/mol  molar mass of compound Whole-number multiple = =  =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  472.1 g/mol  molar mass of compound Whole-number multiple = =  =2 empirical formula mass  236.0 g/mol  Multiplying the subscripts in HgCl by 2 gives Hg2Cl2. 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  240.20 g/mol  molar mass of compound Whole-number multiple = =  =2 empirical formula mass  120.10 g/mol  Multiplying the subscripts in C7H4O2 by 2 gives C14H8O4. 3.40

Plan: The empirical formula is the smallest whole-number ratio of the atoms or moles in a formula. All data must be converted to amount (mol) 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 Cl 0.063O0.22 Converting to integer subscripts (dividing all by the smallest subscript): Cl 0.063 O 0.22 → Cl1O3.5 0.063

0.063

The formula is Cl1O3.5, which in whole numbers (x 2) is Cl2O7. b) Find amount (mol) of elements by dividing by molar mass:  1 mol Si  Amount (mol) of Si =  2.45 g Si    = 0.08722 mol Si  28.09 g Si   1 mol Cl  Amount (mol) of Cl = 12.4 g Cl    = 0.349788 mol 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 → Si1Cl4 0.08722

0.349788

The empirical formula is SiCl4. c) Assume a 100 g sample and convert the masses to amount (mol) by dividing by the molar mass:  27.3 parts C by mass   1 mol C  Amount (mol) 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  Amount (mol) of O = 100 g    = 4.5438 mol O   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 → C1O2 2.2731

2.2731

The empirical formula is CO2. 3.41

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 Fe 0.039O0.052 Converting to integer subscripts (dividing all by the smallest subscript): Fe 0.039 O 0.052 → Fe1O1.33 0.039

0.039

The formula is Fe1O1.33, which in whole numbers (x 3) is Fe3O4. b) Find amount (mol) of elements by dividing by molar mass:  1 mol P  Amount (mol) of P =  0.903 g P    = 0.029157 mol P  30.97 g P   1 mol Br  Amount (mol) of Br =  6.99 g Br    = 0.087484 mol 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 → P1Br3 0.029157

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  Amount (mol) of C = 100 g    = 6.6528 mol C   100 parts by mass   12.01 g C   20.1 parts H by mass   1 mol H  Amount (mol) of H = 100 g    = 19.940 mol H   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 → C1H3 6.6528

6.6528

The empirical formula is CH3. 3.42

Plan: The percent oxygen is 100% minus the percent nitrogen. Assume 100 grams of sample, and then the amount (mol) 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% − % N = 100% − 30.45% N = 69.55% O 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  Amount (mol) of N = 100 g    = 2.1734 mol N   100 parts by mass   14.01 g N   69.55 parts O by mass   1 mol O  Amount (mol) of O = 100 g    = 4.3469 mol O   100 parts by mass   16.00 g O  Preliminary formula is N2.1734O4.3469 Converting to integer subscripts (dividing all by the smallest subscript): N 2.1734 O 4.3469 → N1O2 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 Whole-number multiple = =   =2 empirical formula mass  46.01 g/mol  Silberberg, Amateis, Venkateswaran, Chen, Chemistry: The Molecular Nature of Matter and Change, 3rd Canadian Edition Page 3-73 Instructor‘s Solution Manual © Copyright 2021 McGraw-Hill Ryerson Ltd.

Multiplying the subscripts in NO2 by 2 gives N2O4 as the molecular formula. Note: Only an approximate value of the molar mass is needed. 3.43

Plan: The percent silicon is 100% minus the percent chorine. Assume 100 grams of sample, and then the amount (mol) of each element may be found by dividing the mass of each element by its molar mass. Divide each of the amount (mol) 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% − % Cl = 100% − 79.1% Cl = 20.9% Si 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  Amount (mol) of Si = 100 g    = 0.7440 mol Si   100 parts by mass   28.09 g Si   79.1 parts Cl by mass   1 mol Cl  Amount (mol) of Cl = 100 g    = 2.2313 mol Cl   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 → Si1Cl3 0.7440

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  269 g/mol  molar mass of compound Whole-number multiple = =   =2 empirical formula mass  134.44 g/mol  Multiplying the subscripts in SiCl3 by 2 gives Si2Cl6 as the molecular formula. 3.44

Plan: The amount (mol) of the metal is known, and the amount (mol) of fluorine atoms may be found in part a) from the M:F mole ratio in the compound formula. In part b), convert amount (mol) 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 amount (mol) of M to determine the molar mass of M which can be used to identify the element. Solution: a) Determine the amount (mol) of fluorine.  2 mol F  Amount (mol) of F = 0.600 mol M   = 1.20 mol F  1 mol M  b) Determine the mass of M.  19.00 g F  Mass of F = 1.20 mol F   = 22.8 g 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.

3.45

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

 3 mol O  Amount (mol) of O = 0.370 mol M 2 O3   = 1.11 mol O  1 mol M 2 O3  b) Determine the mass of M.  16.00 g O  Mass of O = 1.11 mol O   = 17.76 g 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  Amount (mol) M = 0.370 mol M 2 O3   = 0.740 mol M  1 mol M 2 O3  The molar mass is needed to identify the element. 37.6 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. 3.46

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 N1.23 → C5H7N1 1.23

1.23

The empirical formula is C5H7N. 3.47

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  Amount (mol) of C =  69.6 g C    = 5.7952 mol C  12.01 g C   1 mol H  Amount (mol) of H =  8.34 g H    = 8.2738 mol H  1.008 g H   1 mol O  Amount (mol) of O =  22.1 g O    = 1.38125 mol O  16.00 g O  Preliminary formula is C5.7952H8.2738O1.38125 Converting to integer subscripts (dividing all by the smallest subscript): C 5.7952 H 8.2738 O1.38125 → C4.2H6O1 1.38125

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 five: 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  362.47 g/mol  molar mass of compound Whole-number multiple = =   =1 empirical formula mass  362.45 g/mol  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. Silberberg, Amateis, Venkateswaran, Chen, Chemistry: The Molecular Nature of Matter and Change, 3rd Canadian Edition Page 3-75 Instructor‘s Solution Manual © Copyright 2021 McGraw-Hill Ryerson Ltd.

3.48

Plan: Determine the molecular formula from the figure, and the molar mass from the molecular formula. The formula gives the relative amount (mol) of each element present. Multiply the amount (mol) of each element by its molar mass to find the total mass of element in 1 mole of compound. total mass of element Mass percent = 100 . 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  Mass percent =

total mass C 96.08 g C 100%  = 100%  molar mass of compound 151.16 g C8 H 9 NO 2

= 63.5618% = 63.56% C 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% molar mass of compound 151.16 g C8 H9 NO2 = 6.00159% = 6.002% H 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 

Mass percent =

total mass N 14.01 g N 100%  = 100%  molar mass of compound 151.16 g C8 H 9 NO 2 = 9.2683% = 9.268% N 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 

Mass percent =

total mass O 32.00 g O 100%  = 100%  molar mass of compound 151.16 g C8 H 9 NO 2 = 21.1696% = 21.17% O

Mass percent =

3.49

Plan: In combustion analysis, finding the amount (mol) of carbon and hydrogen is relatively simple because all of the carbon present in the sample is found in the carbon of CO 2, 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 amount (mol) and mass of C present. Do the same to find the amount (mol) and mass of H from H 2O. The amount (mol) of oxygen are more difficult to find, because additional O 2 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 CO 2   1 mol C  Amount (mol) of C =  0.449 g CO 2     = 0.010202 mol C  44.01 g CO 2   1 mol CO 2 

 12.01 g C  Mass (g) of C =  0.010202 mol C    = 0.122526 g C  1 mol C   1 mol H 2 O  2 mol H  Amount (mol) of H =  0.184 g H 2 O     = 0.020422 mol H  18.02 g H 2 O  1 mol H 2 O   1.008 g H  Mass (g) of H =  0.020422 mol H    = 0.020585 g 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  Amount (mol) of O =  0.016389 g O    = 0.0010243 mol 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 → C10H20O1 0.0010243

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. 3.50

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.

3.51

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.

3.52

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.

3.53

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.

3.54

Balance the H next, because H is present in only one reactant and only one product. The 12 H atoms in 4H 3PO4 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 Na 3BO3 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) 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 H 2O 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 O 2 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) 3.55

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 H 2O 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 H2O 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) Silberberg, Amateis, Venkateswaran, Chen, Chemistry: The Molecular Nature of Matter and Change, 3rd Canadian Edition Page 3-78 Instructor‘s Solution Manual © Copyright 2021 McGraw-Hill Ryerson Ltd.

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 NH 3 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 H 2O 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) 3.56

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) __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 H 2O 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 H 2O 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 H 2O 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) 3.57

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 CaSiO 3 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 H 2O 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) Note there are 48 H atoms on both sides, so the equation is balanced.

3.58

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) Gallium (a solid) and oxygen (a gas) are reactants and solid gallium(III) oxide is the only product: __Ga(s) + __O2(g)  __Ga2O3(s) A coefficient of 2 in front of Ga on the left is needed to balance the 2 Ga atoms in Ga 2O3: 2Ga(s) + __O2(g)  __Ga2O3(s) The 3 O atoms in Ga2O3 on the right require a coefficient of 3/2 in front of O 2 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 CO 2 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 H 2O 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 CaCl 2 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 Na 3PO4 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)

3.59

The 4 O atoms in 2SiO2 on the right require a coefficient of 4 in front of H 2O 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 H 2): 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 HNO 3 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 NO 2 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) 3.60

3.61

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.

3.62

Plan: Write a balanced chemical reaction to obtain the mole ratio between the reactants. Compare the number of particles of each reactant with the mole ratio to find the limiting reactant. Use the limiting reactant to calculate the number of product molecules that will form. Solution: a) The reaction is A2 + B2 → AB3 or, correctly balanced, A2 + 3B2 → 2AB3. The mole ratio between A2 and B2 is 1:3. Three times as many B2 molecules are required as you have of A2 molecules. With 3 A2 molecules present, 3 x 3 = 9 B2 molecules would be required. Since you have only 6 B 2 molecules, B2 is the limiting reagent. b) The balanced equation shows that 2AB3 molecules are produced for every 3 B2 molecules that react. Use the 3:2 mole ratio between the limiting reactant, B2, and AB3:  2 AB3 molecules  Number of molecules of product = 6 B2 molecules   = 4 AB3 molecules  3 B2 molecules 

3.63

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.

3.64

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 amount (mol) of A. Use the molar ratio from the balanced equation to find the amount (mol) of C. Multiply amount (mol) of C by its molar mass to obtain mass of C. Roadmap: Mass (g) of A

Divide by M (g/mol) Amount (mol) of A

Molar ratio between A and C Amount (moles) of C

Multiply by M (g/mol) Mass (g) of C 3.65

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

Mass (g) of E

Divide by M (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

Amount (moles) of F Choose lower number of moles of F and multiply by M (g/mol)

Mass (g) of F 3.66

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 amount (mol) of Cl 2 produced. The equation shows that 1 mole of Cl2 is produced for every 4 moles of HCl that react. Multiply the amount (mol) of Cl2 produced by the molar mass to convert to mass in grams. Solution: 4HCl(aq) + MnO2(s)  MnCl2(aq) + 2H2O(g) + Cl2(g)  1 mol Cl2  a) Amount (mol) of Cl2 = 1.82 mol HCl   = 0.455 mol Cl2  4 mol HCl   70.90 g Cl2  b) Mass (g) of Cl2 = 0.455 mol Cl2   = 32.2595 g= 32.3 g Cl2  1 mol Cl2 

3.67

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 amount (mol) of reactant. Use the mole ratio from the balanced chemical equation to determine the amount (mol) 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)  1 mol Bi 2 O3  a) Amount (mol) of Bi2O3 = 283 g Bi 2 O3   = 0.607296 mol= 0.607 mol Bi2O3  466.0 g Bi 2 O3   1 mol Bi 2 O3   2 mol Bi  b) Amount (mol) of Bi = 283 g Bi 2 O3    = 1.21459 mol= 1.21 mol Bi  466.0 g Bi 2 O3   1 mol Bi 2 O3 

3.68

Plan: Convert the mass of oxygen from kilograms to grams and then convert mass (g) to amount (mol) of oxygen by dividing by its molar mass. Use the amount (mol) of oxygen and the mole ratio from the balanced chemical equation to determine the amount (mol) of KNO3 required. Multiply the amount (mol) of KNO3 by its molar mass to obtain the mass in grams. Solution:  103 g  a) Mass (g) of O2 = 56.6 kg O2  = 5.66x104 g O2  1 kg     1 mol O2  3 Amount (mol) of O2 = 5.66x104 g O2   = 1.76875x10 mol O2 32.00 g O 2    4 mol KNO3  3 Amount (mol) of KNO3 = 1.76875 mol O 2   = 1415 mol= 1.42x10 mol KNO3 5 mol O 2  





 101.11 g KNO3  5 b) Mass (g) of KNO3 = 1415 mol KNO3   = 143070.65 g= 1.43x10 g KNO3 1 mol KNO 3   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 O  5 mol O   1 mol KNO  2  2 3     5 = 143070.65 g= 1.43x10 g KNO3

3.69

Plan: Convert mass of Cr2S3 to amount (mol) by dividing by its molar mass. Use the mole ratio between Cr 2S3 and Cr2O3 from the balanced chemical equation to determine the amount (mol) of Cr 2O3 required. Multiply the amount (mol) of Cr2O3 by its molar mass to obtain the mass in grams. Solution:

 1 mol Cr2S3  a) Amount (mol) of Cr2S3 = 421 g Cr2S3   = 2.102792 mol Cr2S3  200.21 g Cr2S3   1 mol Cr2 O3  Amount (mol) of Cr2O3 = 2.102792 mol Cr2S3   = 2.102792 mol= 2.10 mol Cr2O3  1 mol Cr2S3   152.00 g Cr2 O3  2 b) Mass (g) of Cr2O3 = 2.102792 mol Cr2 O3   = 319.624 g= 3.20x10 g Cr2O3 1 mol Cr O 2 3   Combining all steps gives:  1 mol Cr2S3  1 mol Cr2 O3   152.00 g Cr2 O3  Mass (g) of Cr2O3 = 421 g Cr2S3      200.21 g Cr2S3  1 mol Cr2S3   1 mol Cr2 O3 

= 319.624 g= 3.20x102 g Cr2O3 3.70

Plan: First, balance the equation. Convert the mass (g) of diborane to amount (mol) of diborane by dividing by its molar mass. Use mole ratios from the balanced chemical equation to determine the amount (mol) 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 H 6  Amount (mol) of B2H6 = 43.82 g B2 H 6   = 1.583665 mol B2H6  27.67 g B2 H 6   2 mol H 3 BO3  Amount (mol) of H3BO3 = 1.583665 mol B2 H 6   = 3.16733 mol H3BO3  1 mol B2 H 6   61.83 g H3 BO3  Mass (g) of H3BO3 = 3.16733 mol H3 BO3   = 195.83597 g= 195.8 g H3BO3  1 mol H3 BO3 

Combining all steps gives:  1 mol B2 H 6   2 mol H3 BO3   61.83 g H3 BO3  Mass (g) of H3BO3 = 43.82 g B2 H 6      27.67 g B2 H 6   1 mol B2 H 6   1 mol H3 BO3  = 195.83597 g= 195.8 g H3BO3  6 mol H 2  Amount (mol) of H2 = 1.583665 mol B2 H 6   = 9.50199 mol H2  1 mol B2 H 6   2.016 g H 2  Mass (g) of H2 = 9.50199 mol H 2   = 19.15901 g H2 = 19.16 g H2  1 mol H 2  Combining all steps gives:  1 mol B2 H 6   6 mol H 2   2.016 g H 2  Mass (g) of H2 = 43.82 g B2 H 6     = 19.15601 g= 19.16 g H2  27.67 g B2 H 6   1 mol B2 H 6   1 mol H 2 

3.71

Plan: First, balance the equation. Convert the mass (g) of silver sulfide to amount (mol) of silver sulfide by dividing by its molar mass. Use mole ratios from the balanced chemical equation to determine the amount (mol) of the products. Multiply the mole amount of each product by its molar mass to obtain mass in grams. Solution: First, balance the equation: Ag2S(s) + 2 HCl(aq)  2 AgCl(s) + H2S(g)  1 mol Ag 2S  Amount (mol) of Ag2S = 174 g Ag 2S  = 0.7018959 mol Ag2S  247.9 g Ag 2S   2 mol AgCl  Amount (mol) of AgCl = 0.7018959 mol Ag 2S  = 1.403792 mol AgCl  1 mol Ag 2S 

 143.4 g AgCl  Mass (g) of AgCl = 1.403792 mol Ag 2S  = 201.304 g= 201 g AgCl  1 mol AgCl  Combining all steps gives:  1 mol Ag 2S   2 mol AgCl   143.4 g AgCl  Mass (g) AgCl = 174 g Ag 2S    = 201.304 = 201 g AgCl  247.9 g Ag 2S   1 mol Ag 2S   1 mol AgCl   1 mol H 2S  Amount (mol) of H2S = 0.7018959 mol Ag 2S  = 0.7018959 mol H2S  1 mol Ag 2S   34.09 g H 2S  Mass (g) of H2S = 0.7018959 mol H 2S   = 23.9276 g= 23.9 g H2S  1 mol H 2S 

Combining all steps gives:

3.72

 1 mol Ag 2S  1 mol H 2S   34.09 g H 2S  Mass (g) of H2S = 174 g Ag 2S     = 23.9276 g= 23.9 g H2S  247.9 g Ag 2S  1 mol Ag 2S   1 mol H 2S  Plan: Write the balanced equation by first writing the formulas for the reactants and products. Convert the mass of phosphorus to amount (mol) by dividing by the molar mass, use the mole ratio between phosphorus and chlorine from the balanced chemical equation to obtain amount (mol) of chlorine, and finally divide the amount (mol) of chlorine by its molar mass to obtain amount in grams. Solution: Reactants: formula for phosphorus is given as P 4 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 Balancing the equation: P4 + 10Cl2  4PCl5  1 mol P4  Amount (mol) of P4 = 455 g P4   = 3.67291 mol P4  123.88 g P4   10 mol Cl2  Amount (mol) of Cl2 = 3.67291 mol P4   = 36.7291 mol Cl2  1 mol P4   70.90 g Cl2  3 Mass (g) of Cl2 = 36.7291 mol Cl2   = 2604.09 g= 2.60x10 g Cl2 1 mol Cl 2  

Combining all steps gives:  1 mol P4   10 mol Cl2   70.90 g Cl2  3 Mass (g) of Cl2 = 455 g P4     = 2604.09267 g= 2.60x10 g Cl2 123.88 g P 1 mol P 1 mol Cl 4  4  2  

3.73

Plan: Write the balanced equation by first writing the formulas for the reactants and products. Convert the mass of sulfur to amount (mol) by dividing by the molar mass, use the mole ratio between sulfur and fluorine from the balanced chemical equation to obtain amount (mol) of fluorine, and finally divide the amount (mol) 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  Amount (mol) of S8 = 17.8 g S8   = 0.0693795 mol S8  256.56 g S8   24 mol F2  Amount (mol) of F2 = 0.0693795 mol S8   = 1.665108 mol F2  1 mol S8 

 38.00 g F2  Mass (g) of F2 = 1.665108 mol F2   = 63.274 g= 63.3 g F2  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     = 63.27409 g= 63.3 g F2  256.56 g S8   1 mol S8   1 mol F2 

3.74

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 amount (mol) by dividing by the molar mass and use the mole ratio between iodine and product to find the amount (mol) of iodine. Multiply amount (mol) by the molar mass of iodine to obtain mass of iodine. Solution: a) Step 1 I2(s) + Cl2(g)  2ICl(s) 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. I2 (s) + Cl2 (g )   2ICl(s)

2ICl(s) + 2Cl2 ( g )   2ICl3 (s) I2(s) + Cl2(g) + 2ICl(s) + 2Cl2(g)  2ICl(s) + 2ICl3(s) Overall equation: I2(s) + 3Cl2(g)  2ICl3(s)  103 g  c) Mass (g) of ICl3 = 2.45 kg ICl3  = 2450 g ICl3  1 kg     1 mol ICl3  Amount (mol) of ICl3 = 2450 g ICl3   = 10.506 mol ICl3  233.2 g ICl3   1 mol I 2  Amount (mol) of I2 = 10.506 mol ICl3   = 5.253 mol I2  2 mol ICl3   253.8 g I 2  3 Mass (g) of I2 = 5.253 mol I 2   = 1333.211 g= 1.33x10 g I2 1 mol I 2  

Combining all steps gives:  103 g   1 mol ICl3  1 mol I2  253.8 g I2  Mass (g) of I2 = 2.45 kg ICl3  = 1333.211 g= 1.33x103 g I2  1 kg   233.2 g ICl  2 mol ICl  1 mol I  3  3  2   

3.75

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 mass of lead from metric tons to grams to amount (mol) by dividing by the molar mass and use the mole ratio between lead and sulfur dioxide to find the amount (mol) of sulfur dioxide. Multiply amount (mol) 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 ) + SO2 (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) or  PbS(s) + O2(g)  Pb(l) + SO2(g)

c) 1 metric tonne = 1000 kg  103 kg  103 g  Mass (g) of Pb = ton Pb  = 1.000x106 g Pb  1 ton   1 kg      1 mol Pb  Amount (mol) of Pb = 1x106 g Pb   = 4826.255 mol Pb  207.2 g Pb 





 1 mol SO 2  Amount (mol) of SO2 = 4826.255 mol Pb   = 4826.255 mol SO2  1 mol Pb 

 64.07 g SO2   1 kg  1 mt  Metric tonne SO2 =  4826.255 mol SO2   = 0.309218 mt = 0.3092 mt SO2   3   3   1 mol SO2   10 g  10 kg  Combining all steps gives:  103 kg  103 g   1 mol Pb  1 mol SO2   64.07 g SO2   1 kg  1 mt  Metric tonne SO2 =  mt Pb    103 kg   1 mt   1 kg   207.2 g Pb  1 mol Pb   1 mol SO   103 g    2       = 0.309218 mt= 0.3092 mt SO2

3.76

Plan: Convert the given mass of each reactant to amount (mol) by dividing by the molar mass of that reactant. Use the mole ratio from the balanced chemical equation to find the amount (mol) 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 amount (mol) of CaO obtained from the limiting reactant to grams using the molar mass. Solution: 2Ca(s) + O2(g)  2CaO(s)  1 mol Ca  a) Amount (mol) of Ca = 4.20 g Ca   = 0.104790 mol Ca  40.08 g Ca   2 mol CaO  Amount (mol) of CaO from Ca = 0.104790 mol Ca   = 0.104790 mol= 0.105 mol CaO  2 mol Ca   1 mol O 2  b) Amount (mol) of O2 = 2.80 g O 2   = 0.0875 mol O2  32.00 g O 2   2 mol CaO  Amount (mol) of CaO from O2 = 0.0875 mol O 2   = 0.17500 mol= 0.175 mol CaO  1 mol O 2 

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  Mass (g) of CaO = 0.104790 mol CaO   = 5.8766 g= 5.88 g 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      = 5.8766 g= 5.88 g CaO  40.08 g Ca   2 mol Ca   1 mol CaO 

3.77

Solution: SrH2(s) + 2H2O(l)  Sr(OH)2(s) + 2H2(g)  1 mol SrH 2  a) Amount (mol) of SrH2 =  5.70 g SrH 2    = 0.0635877 mol SrH2  89.64 g SrH 2   2 mol H 2  Amount (mol) of H2 from SrH2 =  0.0635877 mol SrH 2    = 0.127175 mol= 0.127 mol H2  1 mol SrH 2   1 mol H 2 O  b) Mass (g) of H2O =  4.75 g H 2 O    = 0.263596 mol H2O  18.02 g H 2 O   2 mol H 2  Amount (mol) of H2 from H2O =  0.263596 mol H 2 O    = 0.263596 mol= 0.264 mol H2  2 mol H 2 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  Mass (g) of H2 =  0.127175 mol H 2    = 0.256385 g= 0.256 g H2  1 mol H 2  Combining all steps gives:  1 mol SrH 2   2 mol H 2   2.016 g H2  Mass (g) of H2 =  5.70 g SrH 2      = 0.256385 g= 0.256 g H2  89.64 g SrH 2   1 mol SrH 2   1 mol H2 

3.78

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(s) + 3H2O (l)  ICl(g) + HIO3(aq) + 5HCl(g) 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 amount (mol) of HIO3 from the amount (mol) of ICl3 (if H2O is limiting):  1 mol ICl3  Amount (mol) of ICl3 =  635 g ICl3    = 2.722985 mol ICl3  233.2 g ICl3   1 mol HIO3  Amount (mol) of HIO3 from ICl3 =  2.722985 mol ICl3    = 1.361492 mol= 1.36 mol HIO3  2 mol ICl3  Finding the amount (mol) of HIO3 from the amount (mol) of H2O (if ICl3 is limiting):  1 mol H 2 O  Amount (mol) of H2O = 118.5 g H 2 O    = 6.57603 mol H2O  18.02 g H 2 O   1 mol HIO3  Amount (mol) HIO3 from H2O = 6.57603 mol H 2 O   = 2.19201 mol= 2.19 mol HIO3  3 mol H 2 O  ICl3 is the limiting reagent and will produce 1.36 mol HIO3.  175.9 g HIO3  Mass (g) of HIO3 = 1.361492 mol HIO3   = 239.486 g= 239 g HIO3  1 mol HIO3 

Combining all steps gives:  1 mol ICl3  1 mol HIO3   175.9 g HIO3  Mass (g) of HIO3 = 635 g ICl3     = 239.486 g= 239 g HIO3  233.2 g ICl3  2 mol ICl3   1 mol HIO3 

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  Amount (mol) of H2O required to react with 635 g ICl3 = 2.722985 mol ICl3    2 mol ICl3  = 4.0844775 mol H2O  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 g= 73.6 g H2O reacted Mass (g) of remaining H2O = 118.5 g – 73.6 g = 44.9 g H2O 3.79

Plan: First, balance the chemical equation. To determine which reactant is limiting, calculate the amount of H 2S 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 (s) + 6H2O (l)  2Al(OH)3 (aq) + 3H2S (g) Finding the amount (mol) of H2S from the amount (mol) of Al2S3 (if H2O is limiting):  1 mol Al2S3  Amount (mol) of Al2S3 = 158 g Al2S3   = 1.05214 mol Al2S3  150.17 g Al2S3   3 mol H 2S  Amount (mol) of H2S from Al2S3 = 1.05214 mol Al2S3   = 3.15642 mol= 3.16 mol H2S  1 mol Al2S3  Finding the amount (mol) of H2S from the amount (mol) of H2O (if Al2S3 is limiting):  1 mol H 2 O  Amount (mol) of H2O = 131 g H 2 O   = 7.26970 mol H2O  18.02 g H 2 O   3 mol H 2S  Amount (mol) of H2S from H2O = 7.26970 mol H 2 O   = 3.63485mol = 3.63 mol H2S  6 mol H 2 O  Al2S3 is the limiting reagent and 3.16 mol of H 2S will form.  34.09 g H 2S  Mass (g) of H2S = 3.15642 mol H 2S  = 107.602 g= 108 g H2S  1 mol H 2S 

Combining all steps gives:  1 mol Al 2S3   3 mol H 2S   34.09 g H 2S  m (H2S) = 158 g Al2S3      = 107.602 g= 108 g H2S  150.17 g Al2S3   1 mol Al 2S3   1 mol H 2S  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  Amount (mol) of H2O required to react with 158 g of Al2S3 = 1.05214 mol Al2S3     1 mol Al2S3  = 6.31284 mol H2O  18.02 g H 2 O  Mass (g) of H2O required to react with 158 g of Al2S3 =  6.31284 mol H2 O    = 113.757 g H2O  1 mol H 2 O  Mass (g) of remaining H2O = 131 g H2O – 113.757 g H2O = 17.243 g= 17 g H2O

3.80

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 CO 2 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 CO 2 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: C(s) + O2(g)  CO2(g) Finding the amount (mol) of CO2 from the amount (mol) of carbon (if O2 is limiting):  1 mol CO 2  Amount (mol) of CO2 from C =  0.100 mol C    = 0.100 mol CO2  1 mol C  Finding the amount (mol) of CO2 from the amount (mol) of oxygen (if C is limiting):  1 mol O2  Amount (mol) of O2 =  8.00 g O2    = 0.250 mol O2  32.00 g O2   1 mol CO2  Amount (mol) of CO2 from O2 =  0.250 mol O2    = 0.25000 mol= 0.250 mol CO2  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 g= 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 O 2  Amount (mol) of O2 required to react with 0.100 mol of C =  0.100 mol C    = 0.100 mol O2  1 mol C   32.00 g O 2  Mass (g) of O2 required to react with 0.100 mol of C =  0.100 mol O 2    = 3.20 g O2  1 mol O 2  Mass (g) of remaining O2 = 8.00 g – 3.20 g = 4.80 g O2

3.81

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 H 2O 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 H 2O 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 amount (mol) of H2O from the amount (mol) of hydrogen (if O 2 is limiting):  1 mol H 2  Amount (mol) of H2 = 0.0375 g H 2   = 0.01860 mol H2  2.016 g H 2   2 mol H 2 O  Amount (mol) of H2O from H2 = 0.01860 mol H 2   = 0.01860 mol= 0.0186 mol H2O  2 mol H 2  Finding the amount (mol) of H2O from the amount (mol) of oxygen (if H2 is limiting):  2 mol H 2 O  Mole of H2O from O2 = 0.0185 mol O 2   = 0.0370 mol H2O  1 mol O 2 

The hydrogen is the limiting reactant, and will be used to determine the amount of water that will form.  18.02 g H 2 O  Mass (g) of H2O = 0.01860 mol H 2 O   = 0.335172 g= 0.335 g H2O  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 O 2  Amount (mol) of O2 required to react with 0.0375 g of H2 = 0.01860 mol H 2   = 0.00930 mol O2  2 mol H 2   32.00 g O 2  Mass (g) of O2 required to react with 0.0375 g of H2 = 0.00930 mol O 2   = 0.2976 g O2  1 mol O 2   32.00 mol O 2  Mass of O2 supplied = 0.0185 mol O 2   = 0.5920 g O2  1 mol O 2 

Mass (g) of remaining O2 = 0.5920 g – 0.2976 g = 0.2944 g= 0.294 g O2 3.82

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 amount (mol) of AlCl 3 produced to determine which is limiting. Finding the amount (mol) of AlCl3 from the amount (mol) of Al(NO2)3 (if NH4Cl is limiting):  1 mol Al(NO 2 )3  Amount (mol) of Al(NO2)3 = 72.5 g Al(NO 2 )3   = 0.43937 mol Al(NO2)3  165.01 g Al(NO 2 )3   1 mol AlCl3  Amount (mol) of AlCl3 from Al(NO2)3 =  0.43937 mol Al(NO 2 )3    = 0.43937 mol= 0.439 mol  1 mol Al(NO 2 )3  AlCl3 Finding the amount (mol) of AlCl3 from the amount (mol) of NH4Cl (if Al(NO2)3 is limiting):  1 mol NH 4 Cl  Amount (mol) of NH4Cl = 58.6 g NH 4 Cl   = 1.09553 mol NH4Cl  53.49 g NH 4 Cl   1 mol AlCl3  Amount (mol) of AlCl3 from NH4Cl = 1.09553 mol NH 4 Cl   = 0.365177 mol= 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 NH 4Cl =  1 mol Al(NO 2 )3   165.01 g Al(NO 2 )3  1.09553 mol NH 4 Cl   = 60.2579 g= 60.3 g Al(NO2)3   3 mol NH 4 Cl   1 mol Al(NO 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  Mass (g) of AlCl3 = 0.365177 mol AlCl3   = 48.689 g= 48.7 g AlCl3  1 mol AlCl3  N 2:

 3 mol N 2   28.02 g N 2  Mass (g) of N2 = 1.09553 mol NH 4 Cl    = 30.697 g= 30.7 g N2  3 mol NH 4 Cl   1 mol N 2  H2O:  6 mol H 2 O   18.02 g H 2 O  Mass (g) of H2O = 1.09553 mol NH 4 Cl    = 39.483 g= 39.5 g H2O  3 mol NH 4 Cl   1 mol H 2 O 

3.83

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, CaF 2 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 amount (mol) of CaF 2 produced to determine which is limiting. Finding the amount (mol) of CaF2 from the amount (mol) of Ca(NO3)2 (if NH4F is limiting):  1 mol Ca(NO3 ) 2  Amount (mol) of Ca(NO3)2 = 16.8 g Ca(NO3 ) 2   = 0.1023766 mol Ca(NO3)2  164.10 g Ca(NO3 ) 2   1 mol CaF2  Amount (mol) of CaF2 from Ca(NO3)2 = 0.1023766 mol Ca(NO3 ) 2   1 mol Ca(NO ) 3 2   = 0.1023766 mol= 0.102 mol CaF2 Finding the amount (mol) of CaF2 from the amount (mol) of NH4F (if Ca(NO3)2 is limiting):  1 mol NH 4 F  Amount (mol) of NH4F = 17.50 g NH 4 F   = 0.47246 mol NH4F  37.04 g NH 4 F   1 mol CaF2  Amount (mol) of CaF2 from NH4F = 0.47246 mol NH 4 F   = 0.23623 mol = 0.236 mol CaF2  2 mol NH 4 F  Calcium nitrate is the limiting reactant, and it is used for all subsequent calculations 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(NO 3)2 =  2 mol NH 4 F   37.04 g NH 4 F  0.1023766 mol Ca(NO3 )2    = 7.58406 g NH4F  1 mol Ca(NO3 ) 2   1 mol NH 4 F 

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 g= 7.99 g CaF2  1 mol Ca(NO3 ) 2   1 mol CaF2  N2O:  2 mol N 2 O  44.02 g N 2 O  Mass (g) of N2O = 0.1023766 mol Ca(NO3 ) 2    = 9.0132 g= 9.01 g N2O  1 mol Ca(NO3 ) 2   1 mol N 2 O  H2O:  4 mol H 2 O   18.02 g H 2 O  Mass (g) of H2O = 0.1023766 mol Ca(NO3 ) 2    = 7.3793 =g 7.38 g H2O  1 mol Ca(NO3 ) 2   1 mol H 2 O  Silberberg, Amateis, Venkateswaran, Chen, Chemistry: The Molecular Nature of Matter and Change, 3rd Canadian Edition Page 3-94 Instructor‘s Solution Manual © Copyright 2021 McGraw-Hill Ryerson Ltd.

3.84

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 x 0.68) x 100% = 49.64% = 50.%

3.85

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 x 0.73) x 100 %= 35.04% = 35%

3.86

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 amount (mol) by dividing by its molar mass and then use the mole ratio between tungsten(VI) oxide and water to determine the amount (mol) 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  Amount (mol) of WO3 = 45.5 g WO3   = 0.1962053 mol WO3  231.9 g WO3   3 mol H 2 O   18.02 g H 2 O  Mass (g) of H2O (theoretical yield) = 0.1962053 mol WO3    = 10.60686 g H2O  1 mol WO3   1 mol H 2 O  Determining the actual yield of H2O:  1.00 g H 2 O  Mass (g) of H2O (actual yield) = 9.60 mL H 2 O   = 9.60 g H2O  1 mL H 2 O   9.60 g H 2 O   actual Yield  % yield =   x 100% = 90.5075% = 90.5%  x 100% =   theoretical Yield   10.60686 g H 2 O 

3.87

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 amount (mol) by dividing by its molar mass and then use the mole ratio between phosphorus trichloride and HCl to determine the amount (mol) 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  Amount (mol) of PCl3 = 200. g PCl3   = 1.456452 mol PCl3  137.32 g PCl3   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  Actual yield (g) of HCl is given as 128 g HCl. Calculate the percent yield:  128 g HCl   actual Yield  % yield =   x 100% = 80.3481586 %= 80.3%  x 100% =  theoretical Yield  159.3067 g HCl   

3.88

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, CH 3Cl 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 amount (mol) of CH3Cl from the amount (mol) of CH4 (if Cl2 is limiting):  1 mol CH 4  Amount (mol) of CH4 = 20.5 g CH 4   = 1.278055 mol CH4  16.04 g CH 4   1 mol CH3Cl  Amount (mol) of CH3Cl from CH4 = 1.278055 mol CH 4   = 1.278055 mol CH3Cl  1 mol CH 4  Finding the amount (mol) of CH3Cl from the amount (mol) of Cl2 (if CH4 is limiting):  1 mol Cl2  Amount (mol) of Cl2 = 45.0 g Cl2   = 0.634697 mol Cl2  70.90 g Cl2   1 mol CH3Cl  Amount (mol) of CH3Cl from Cl2 = 0.634697 mol Cl2   = 0.634697 mol CH3Cl  1 mol Cl2  Chlorine is the limiting reactant and is used to determine the theoretical yield of CH 3Cl:  50.48 g CH3Cl  Mass (g) of CH3Cl (theoretical yield) = 0.634697 mol CH3Cl   = 32.0395 g CH3Cl  1 mol CH3Cl 

 actual yield  % yield =   x 100%  theoretical yield  % yield 75% Actual yield (g) of CH3Cl = theoretical yield  = 32.0395 g CH3Cl  100% 100% = 24.02962 g= 24.0 g CH3Cl

3.89

 1 mol Ca 3 N 2  Amount (mol) of Ca3N2 from N2 = 1.08851 mol N 2   = 1.08851 mol Ca3N2  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  Mass (g) of Ca3N2 (theoretical yield) = 0.470725 mol Ca 3 N 2   = 69.7897 g Ca3N2  1 mol Ca 3 N 2   actual yield  % yield =   x 100%  theoretical yield 

Actual yield (g) of Ca3N2 = 3.90

% yield 93% theoretical yield  = 69.7897 g Ca 3 N2  = 64.9044 g= 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 CF 4 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 amount (mol) of CF4 from the amount (mol) of (CN)2 (if F2 is limiting):  1 mol (CN) 2   2 mol CF4  Amount (mol) of CF4 from (CN)2 = 60.0 g (CN) 2    = 2.30592 mol CF4  52.04 g (CN) 2   1 mol (CN) 2  Finding the amount (mol) of CF4 from the amount (mol) of F2 (if (CN)2 is limiting):  1 mol F2   2 mol CF4  Amount (mol) of CF4 from F2 = 60.0 g F2    = 0.4511278 mol CF4  38.00 g F2   7 mol F2  F2 is the limiting reactant, and will be used to calculate the amount of CF 4 produced.  1 mol F2   2 mol CF4   88.01 g CF4  Mass (g) of CF4 = 60.0 g F2     = 39.70376 g= 39.7 g CF4  38.00 g F2   7 mol F2   1 mol CF4 

3.91 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 amount (mol) 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. Amount (mol) of Cl2O =  2 O atoms   0.050 mol O atoms   1 mol O 2 molecules   2 mol Cl2 O  3 O2 molecules     2 mol O atoms   1 mol O  1 O atom   2   1 O 2 molecule   = 0.30 mol Cl2O  6.022x1023 Cl 2O molecules  Molecules of Cl2O =  0.30 mol Cl 2O     1 mol Cl 2O   = 1.8066x1023 molecules= 1.8x1023 Cl2O molecules

3.92

Solution: The balanced chemical equation is: 4NH3(g) + 5O2(g)  4NO(g) + 6H2O(g) Determining the limiting reactant: Finding the amount (mol) of NO from the amount of NH 3 (if O2 is limiting):  1 mol NH3   4 mol NO  Amount (mol) of NO from NH3 = 485 g NH3    = 28.47915 mol NO  17.03 g NH3   4 mol NH3  Finding the amount (mol) of NO from the amount of O 2 (if NH3 is limiting):  1 mol O 2   4 mol NO  Amount (mol) of NO from O2 = 792 g O 2    = 19.8 mol NO  32.00 g O 2   5 mol O 2  O2 is the limiting reactant, and will be used to calculate the amount of NO formed:  30.01 g NO  Mass (g) of NO = 19.8 mol NO   = 594.198 g= 594 g NO  1 mol NO  Combining all of the steps gives:  1 mol O 2   4 mol NO   30.01 g NO  Mass (g) of NO = 792 g O 2     = 594.198 g= 594 g NO  32.00 g O 2   5 mol O 2   1 mol NO  3.93

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 amount (mol). Use the mole ratio between butane and oxygen to find the amount (mol) and then mass of oxygen required for the reaction. The mole ratio between butane and water is used to find the amount (mol) of water produced and the mole ratio between butane and carbon dioxide is used to find the amount (mol) of carbon dioxide produced. The total amount (mol) 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 C4 H10   1 mol C4 H10  a) Amount (mol) of C4H10 = 5.50 mL C4 H10    = 0.054792 mol C4H10  1 mL C4 H10   58.12 g C4 H10   13 mol O 2   32.00 g O 2  Mass (g) of O2 = 0.054792 mol C4 H10    = 11.3967 g= 11.4 g O2  2 mol C4 H10   1 mol O 2   10 mol H 2 O  b) Amount (mol) of H2O = 0.054792 mol C4 H10   = 0.27396 mol= 0.274 mol H2O  2 mol C4 H10   8 mol CO 2  c) Amount (mol) of CO2 = 0.054792 mol C4 H10   = 0.219168 mol CO2  2 mol C4 H10  Total amount (mol) = 0.27396 mol H2O + 0.219168 mol CO2 = 0.493128 mol  6.022 x1023 molecules  23 23 Total molecules =  0.493128 mol    = 2.96962x10 molecules= 2.97x10 molecules  1 mol  

3.94

Plan: Write a balanced equation for the reaction. Convert the given mass of each reactant to amount (mol) by dividing by the molar mass of that reactant. Use the mole ratio from the balanced chemical equation to find the amount (mol) 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 amount (mol) of NaBH 4 obtained from the limiting reactant to grams using the molar mass. This is the theoretical yield of NaBH 4. Since there is a yield of 88.5%, the amount of NaBH4 actually obtained will be 88.5% of the theoretical yield. Solution: The balanced chemical equation is: 2NaH(s) + B2H6(g)  2NaBH4(s) Determining the limiting reactant:

Finding the amount (mol) of NaBH4 from the amount of NaH (if B2H6 is limiting):  1 mol NaH   2 mol NaBH 4  Amount (mol) of NaBH4 from NaH = 7.98 g NaH    = 0.3325 mol NaBH4  24.00 g NaH   2 mol NaH  Finding the amount (mol) of NaBH4 from the amount of B2H6 (if NaH is limiting):  1 mol B2 H 6   2 mol NaBH 4  Amount (mol) of NaBH4 from B2H6 = 8.16 g B2 H 6    = 0.58981 mol NaBH4  27.67 g B2 H 6   1 mol B2 H 6  NaH is the limiting reactant, and will be used to calculate the theoretical yield of NaBH 4.  37.83 g NaBH 4  Mass (g) of NaBH4 = 0.3325 mol NaBH 4   = 12.5785 g NaBH4  1 mol NaBH 4   actual Yield  % yield =   x 100%  theoretical Yield 

88.5%   % yield  Mass (g) of NaBH4 =  theoretical yield  =    12.5785 g NaHB4   100%   100%  = 11.13197 g= 11.1 g NaBH4 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       24.00 g NaH  2 mol NaH  1 mol NaBH 4   100%  = 11.13197 g= 11.1 g NaBH4 3.95

Plan: The spheres represent particles of solute and the amount of solute per given volume of solution determines its concentration. Concentration(mol/L) = amount (mol) of solute/volume (L) of solution. Solution: a) Box C has more solute added because it contains 2 more spheres than Box A contains. b) Box B has more solvent because solvent molecules have displaced two solute molecules. c) Box C has a higher concentration, because it has a greater amount (mol) of solute per volume of solution. d) Box B has a lower concentration, because it has a smaller amount (mol) of solute per volume of solution.

3.96

Plan: Recall that concentration (mol/L) = amount (mol) of solute/volume (L) of solution. Solution: a) cdil = concentration (mol/L) of the diluted solution cconc = concentration (mol/L) 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. c x V = amount (mol); cdil x Vdil = cconc x Vconc amount (mol)dil = amount (mol)conc Amount (mol) solute b) Concentration (mol/L) = volume of solution Amount (mol) CaCl2 = concentration (mol/L) · volume of solution; Mass CaCl 2 = concentration (mol/L) · volume of solution · molar mass of CaCl2.

3.97

Plan: Remember that concentration (mol/L)is amount (mol) 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 mol/L solution and add water until the total volume is 1000. mL.‖

3.98

3.99

Plan: Recall that concentration (mol/L) = amount (mol) of solute/volume (L) of solution. Here you can use the number of particles in place of amount (mol) of solute. Solution: a) Solution B has the highest concentration (mol/L) 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 concentration (mol/L). Solutions C, D, and E all have 4 particles in a volume of 50 mL and thus have the same concentration (mol/L). c) Mixing Solutions A and C results in 8 + 4 = 12 particles in a volume of 100 mL. That is a lower concentration (mol/L) 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 concentration (mol/L) 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 concentration (mol/L) 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.  moles solute  Plan: In all cases, use the known quantities and the definition of concentration (mol/L)  c   V of solution (L)   to find the unknown quantity. Volume must be expressed in litres. The molar mass is used to convert amount (mol) to mass (g). The chemical formulas must be written to determine the molar mass. (a) You will need to convert volume from millilitres to litres, multiply by the concentration (mol/L) to find amount (mol), and convert amount (mol) to mass in grams. (b) Convert mass of solute to amount (mol) and volume from mL to litres. Divide the amount (mol) by the volume. (c) Multiply the concentration (mol/L) by the volume. Solution: a) Calculating amount (mol) of solute in solution:  103 L   0.267 mol Ca(C2 H3O2 )2  Amount (mol) of Ca(C2H3O2)2 = 185.8 mL    1 mL   1L    = 0.0496086 mol Ca(C2H3O2)2 Converting from amount (mol) of solute to mass:  158.17 g Ca(C2 H3O 2 ) 2  Mass (g) of Ca(C2H3O2)2 = 0.0496086 mol Ca(C2 H3O 2 ) 2    1 mol Ca(C2 H3O 2 ) 2  = 7.84659 g= 7.85 g Ca(C2H3O2)2 b) Converting mass of solute to amount (mol):  1 mol KI  Amount (mol) of KI =  21.1 g KI    = 0.127108 moles KI  166.0 g KI   103 L  Volume (L) = 500. mL  = 0.500 L  1 mL    0.127108 mol KI concentration (mol/L) of KI = = 0.254216 mol/L= 0.254 mol/L KI 0.500 L  0.850 mol NaCN  c) Amount (mol) of NaCN = 145.6 L   = 123.76 mol= 124 mol NaCN 1L  

3.100

solute. Use Avogadro‘s number to determine the number of ions present. (c) Divide amount (millimoles) by volume (millilitres); concentration (mol/L) may not only be expressed as moles/L, but also as mmoles/mL. Solution: a) Converting mass of solute to amount (mol):  1 mol KOH  Amount (mol) of KOH = 8.42 g KOH   = 0.15006 mol KOH  56.11 g KOH 

 1L  Volume (L) of KOH solution = 0.15006 mol KOH   = 0.066398 L KOH solution  2.26 mol   1L  Volume (mL) of KOH solution = 0.066398 L KOH  3  = 66.39823 mL= 66.4 mL KOH solution  10 mL 

 2.3 mol CuCl2  b) Amount (mol) of CuCl2 = 52 L   = 119.6 mol CuCl2 L    1 mol Cu 2   Amount (mol) of Cu2+ ions = 119.6 mol CuCl2  = 119.6 mol Cu2+ ions  1 mol CuCl  2   Converting amount (mol) of ions to number of ions:  6.022 x1023 Cu 2  ions  Number of Cu2+ ions = 119.6 mol Cu 2  ions  = 7.2023x1025 ions= 7.2x1025 Cu2+ ions  1 mol Cu 2  ions   

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 135 mmol glucose  c) c glucose =   = 0.490909 mol/L= 0.491mol/L glucose 275 mL   Note: Since 1 mmol is 10–3 mol and 1 mL is 10–3 L, we can use these units instead of converting to mol and L since concentration (mol/L) is a ratio of mol/L. Concentration (mol/L) may not only be expressed as moles/L, but also as mmoles/mL.

3.101

 moles solute  Plan: In all cases, use the known quantities and the definition of concentration (mol/L)  c   V of solution (L)   to find the unknown quantity. Volume must be expressed in litres. The molar mass is used to convert amount (mol) to mass (g). The chemical formulas must be written to determine the molar mass. (a) Convert volume in millilitres to litres, multiply the volume by the concentration (mol/L) to obtain amount (mol) of solute, and convert amount (mol) to mass in grams. (b) The simplest way will be to convert the mass (milligrams) to amount (millimoles). concentration (mol/L) may not only be expressed as moles/L, but also as mmoles/mL. (c) Convert the volume from millilitres to litres and find the amount (mol) of solute and amount (mol) of ions by multiplying the volume and concentration (mol/L). Use Avogadro‘s number to determine the number of ions present. Solution: a) Calculating amount (mol) of solute in solution:  103 L   5.62 x102 mol K 2SO4  Amount (mol) of K2SO4 =  475 mL    = 0.026695 mol K2SO4  1 mL   L    Converting amount (mol) of solute to mass:  174.27 g K 2SO 4  Mass (g) of K2SO4 = 0.026695 mol K 2SO 4   = 4.6521 g= 4.65 g K2SO4  1 mol K 2SO 4 

b) Calculating amount (mmol) of solute:  7.25 mg CaCl2   1 mmol CaCl2  amount (mmol)of CaCl2 =   = 0.065327 mmol CaCl2  1 mL    110.98 mg CaCl2  Calculating concentration (mol/L):  0.065327 mmol CaCl2  concentration (mol/L) of CaCl2 =   = 0.065327 mol/L= 0.0653 mol/L CaCl2 1 mL   If you believe that concentration (mol/L) must be moles/litres then the calculation becomes: Silberberg, Amateis, Venkateswaran, Chen, Chemistry: The Molecular Nature of Matter and Change, 3rd Canadian Edition Page 3-101 Instructor‘s Solution Manual © Copyright 2021 McGraw-Hill Ryerson Ltd.

 7.25 mg CaCl2   103 g  1 mL   1 mol CaCl2  concentration (mol/L) of CaCl2 =        3   1 mL    1 mg   10 L   110.98 g CaCl2  = 0.065327 mol/L= 0.0653 mol/L CaCl2 Notice that the two central terms cancel each other. c) Converting volume in L to mL:  103 L  Volume (L) = 1 mL  = 0.001 L  1 mL   

Calculating amount (mol) of solute and amount (mol) of ions:  0.184 mol MgBr2  –4 Amount (mol) of MgBr2 = 0.001 L   = 1.84x10 mol MgBr2 1L    1 mol Mg 2   Amount (mol) of Mg2+ ions = 1.84 x 104 mol MgBr2  = 1.84x10–4 mol Mg2+ ions  1 mol MgBr  2  

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 6.022 x1023 Mg 2  ions  Number of Mg2+ ions = 1.84x104 Mg 2  ions   1 mol Mg 2  ions    = 1.1080x1020 ions= 1.11x1020 Mg2+ ions

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3.102

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moles solute   Plan: In all cases, use the known quantities and the definition of concentration (mol/L)  c  to L of solution   find the unknown quantity. Volume must be expressed in litres. The molar mass is used to convert amount (mol) to grams. The chemical formulas must be written to determine the molar mass. (a) Convert mass of solute to amount (mol) and volume from mL to litres. Divide the amount (mol) by the volume. (b) You will need to convert mass of solute to amount (mol) and divide by the concentration (mol/L) to obtain volume in litres. (c) Divide the amount (mmol) of solute by the concentration (mol/L) to obtain volume in mL. Solution: a) Calculating amount (mol) of solute:  1 mol AgNO3  Amount (mol) of AgNO3 = 46.0 g AgNO3   = 0.2707475 mol AgNO3  169.9 g AgNO3   103 L  Volume (L) = 335 mL  = 0.335 L  1 mL    Calculating the concentration (mol/L):  0.2707574 mol AgNO3  concentration (mol/L) of AgNO3 =   = 0.80823 mol/L= 0.808 mol/L AgNO3 0.335 L   b) Calculating amount (mol) of solute:  1 mol MnSO 4  Amount (mol) of MnSO4 = 63.0 g MnSO 4   = 0.417191 mol MnSO4  151.01 g MnSO 4 

Calculating volume of solution:   1L Volume (L) of solution = 0.417191 mol MnSO 4   = 1.08361 L= 1.08 L MnSO4 solution  0.385 mol MnSO 4    1mL c) Volume (mL) of ATP solution = 1.68 mmol ATP    2  6.44 x10 mmol ATP  = 26.087 mL= 26.1 mL ATP solution

3.103

litres; it only requires that the volume units match. In part c), it is necessary to find the amount (mol) of sodium ions in each separate solution, add these two mole amounts, and divide by the total volume of the two solutions. Solution: a) c1 = 0.250 mol/L KCl V1 = 37.00 mL c2 = ? V2 = 150.00 mL c1V1= c2V2 c x V1  0.250 mol / L  37.00 mL  c2 = 1 = = 0.061667 mol/L= 0.0617 mol/L KCl V2 150.0 mL b) c1 = 0.0706 mol/L (NH4)2SO4 V1 = 25.71 mL c2 = ? V2 = 500.00 mL c1V1= c2V2 c x V1  0.0706 mol / L  25.71 mL  c2 = 1 = = 0.003630 mol/L= 0.00363 mol/L (NH4)2SO4 V2 500.0 mL  103 L   0.348 mol NaCl   1 mol Na   c) Amount (mol) of Na+ from NaCl solution = 3.58 mL      1 mL   1L   1 mol NaCl    = 0.00124584 mol Na+  103 L   6.81x102 mol Na 2SO4   2 mol Na   Amount (mol) of Na+ from Na2SO4 solution =  500. mL       1 mL   1L     1 mol Na 2SO4  = 0.0681 mol Na+ + Total amount (mol) of Na ions = 0.00124584 mol Na+ ions + 0.0681 mol Na+ ions = 0.06934584 mol Na+ ions Total volume = 3.58 mL + 500. mL = 503.58 mL = 0.50358 L

total moles Na  ions 0.06934584 mol Na  ions = total volume 0.50358 L = 0.1377057 mol/L= 0.138 mol/L Na+ ions

concentration (mol/L) of Na+ =

3.104

Plan: These are dilution problems. Dilution problems can be solved by converting to amount (mol) and using the new volume; however, it is much easier to use c1V1 = c2V2. The dilution equation does not require a volume in litres; it only requires that the volume units match. Solution: a) c1 = 2.050 mol/L Cu(NO3)2 V1 = ? c2 = 0.8543 mol/L Cu(NO3)2 V2 = 750.0 mL c1V1= c2V2  0.8543 mol / L  750.0 mL c x V2 V1 = 2 = = 312.5488 mL= 312.5 mL c1 2.050 mol / L b) c1 = 1.63 mol/L CaCl2

 1.63 mol CaCl2   2 mol Cl  – c1 Cl– =   = 3.26 mol/L Cl ions   1L    1 mol CaCl2 

c1 = 3.26 mol/L Cl– V1 = ? c2 = 2.86x10–2 mol/L Cl– ions V2 = 350. mL c1V1= c2V2

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2.86x102 mol/L  350. mL  c2 x V2 = = 3.07055 mL= 3.07 mL 3.26 mol/L c1 c) c1 = 0.155 mol/L Li2CO3 V1 = 18.0 mL c2 = 0.0700mol/L Li2CO3 c1V1= c2V2 c x V1  0.155 mol/L 18.0 mL = 39.8571 mL= 39.9 mL V2 = 1 = c2  0.0700 mol/L 

V1 =

3.105

V2 = ?

Plan: Use the density of the solution to find the mass of 1 L of solution. Volume in litres 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 amount (mol) to obtain amount (mol)/L, concentration (mol/L). Solution:  1 mL   1.41 g solution  a) Mass (g) of 1 L of solution = 1 L solution  3    = 1410 g solution  10 L  1 mL     70.0 g HNO3  Mass (g) of HNO3 in 1 L of solution = 1410 g solution   = 987 g HNO3 in 1 L  100 g solution   1 mol HNO3  b) Amount (mol) of HNO3 = 987 g HNO3   = 15.6617 mol HNO3  63.02 g HNO3 

 15.6617 mol HNO3  concentration (mol/L) of HNO3 =   = 15.6617 mol/L= 15.7 mol/L HNO3 1 L solution   3.106

Plan: Use the concentration (mol/L) of the solution to find the amount (mol) of H2SO4 in 1 mL. Convert amount (mol) 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:  18.3 mol H 2SO4   103 L  –2 a) Amount (mol) of H2SO4 in 1 mL =   = 1.83x10 mol H2SO4/mL   1 L 1 mL     98.09 g H 2SO4  b) Mass of H2SO4 in 1 mL = 1.83x102 mol H 2SO4   = 1.79505 g H2SO4  1 mol H 2SO4 

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 1.84 g  Mass of 1 mL of solution = 1 mL   = 1.84 g solution  1 mL  1.79505 g H 2SO 4 mass of H 2SO4 Mass percent = 100%  = 97.5571 %= 97.6% H2SO4 by mass 100%  = 1.84 g solution mass of solution 3.107

Plan: Convert the mass of calcium carbonate to amount (mol), and use the mole ratio in the balanced chemical equation to find the amount (mol) of hydrochloric acid required to react with this amount (mol) of calcium carbonate. Use the concentration (mol/L) of HCl to find the volume that contains this amount (mol). Solution: 2HCl(aq) + CaCO3(s)  CaCl2(aq) + CO2(g) + H2O(l) Converting from mass of CaCO3 to amount (mol):  1 mol CaCO3  Amount (mol) of CaCO3 = 16.2 g CaCO3   = 0.161854 mol CaCO3  100.09 g CaCO3  Converting from amount (mol) of CaCO3 to amount (mol) of HCl:  2 mol HCl  Amount (mol) of HCl = 0.161854 mol CaCO3   = 0.323708 mol HCl  1 mol CaCO3  Converting from amount (mol) of HCl to volume:    1 mL  1L Volume (mL) of HCl = 0.323708 mol HCl    3  = 845.1906 mL= 845 mL HCl solution  0.383 mol HCl   10 L 

3.108

Plan: Convert the volume of NaOH solution to litres and multiply by the concentration (mol/L) of the solution to obtain amount (mol) of NaOH. Use the mole ratio in the balanced chemical equation to find the amount (mol) of NaH2PO4 required to react with this amount (mol) of NaOH. Finally, convert amount (mol) of NaH 2PO4 to mass. Solution: NaH2PO4(s) + 2NaOH(aq)  Na3PO4(aq) + 2H2O(l)

 103 L  Volume (L) = 43.74 mL  = 0.04374 mL  1 mL    Finding amount (mol) of NaOH:  0.285 mol NaOH  Amount (mol) of NaOH = 0.04374 L   = 0.0124659 mol NaOH 1L  

Converting from amount (mol) of NaOH to amount (mol) of NaH 2PO4:  1 mol NaH 2 PO 4  Amount (mol) of NaH2PO4 = 0.0124659 mol NaOH   = 0.00623295 mol NaH2PO4  2 mol NaOH  Converting from amount (mol) of NaH2PO4 to mass:

3.109

 119.98 g NaH 2 PO 4  Mass (g) of NaH2PO4 = 0.00623295 mol NaH 2 PO 4   = 0.747829 g= 0.748 g NaH2PO4  1 mol NaH 2 PO 4  Plan: The first step is to write and balance the chemical equation for the reaction. Multiply the concentration (mol/L) and volume of each of the reactants to determine the amount (mol) 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 BaCl2  Amount (mol) of BaCl2 = 35.0 mL   = 0.00560 mol BaCl2  1 mL   1L    Finding the amount (mol) of BaSO4 from the amount (mol) of BaCl2 (if Na2SO4 is limiting):  1 mol BaSO 4  Amount (mol) of BaSO4 from BaCl2 = 0.00560 moL BaCl2   = 0.00560 mol BaSO4  1 mol BaCl2 

 103 L   0.065 mol Na 2SO4  Amount (mol) of Na2SO4 = 58.0 mL   = 0.00377 mol Na2SO4  1 mL   1L    Finding the amount (mol) of BaSO4 from the amount (mol) of Na2SO4 (if BaCl2 is limiting):  1 mol BaSO 4  Amount (mol) BaSO4 from Na2SO4 = 0.00377 moL Na 2SO 4   = 0.00377 mol BaSO4  1 mol Na 2SO 4  Sodium sulfate is the limiting reactant. Converting from amount (mol) of BaSO4 to mass:  233.4 g BaSO 4  Mass (g) of BaSO4 = 0.0377 moL BaSO 4   = 0.879918 g= 0.88 g BaSO4  1 mol BaSO 4  3.110

Plan: The first step is to write and balance the chemical equation for the reaction. Use the concentration (mol/L) and volume of each of the reactants to determine the amount (mol) of each. To determine which reactant is limiting, calculate the amount of either 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 reagent and the mole ratio from the balanced chemical equation to determine 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: H2SO4(aq) + 2NaOH(aq)  Na2SO4(aq) + 2H2O(l) We can use either product to determine the limiting reactant. We will use sodium sulfate.  103 L   0.210 mol H 2SO4  Amount (mol) of H2SO4 = 350.0 mL   = 0.0735 mol H2SO4  1 mL   1L   

Finding the amount (mol) of Na2SO4 from the amount (mol) of H2SO4 (if NaOH is limiting):  1 mol Na 2SO 4  Amount (mol) of Na2SO4 from H2SO4 = 0.0735 moL H 2SO 4   = 0.0735 mol Na2SO4  1 mol H 2SO 4   0.196 mol NaOH  Amount (mol) of NaOH = 0.500 L   = 0.0980 mol NaOH 1L   Finding the amount (mol) of Na2SO4 from the amount (mol) of NaOH (if H2SO4 is limiting):  1 mol Na 2SO 4  Amount (mol) of Na2SO4 from NaOH = 0.0980 mol NaOH   = 0.0490 mol Na2SO4  2 mol NaOH 

NaOH is the limiting reactant and will be used in the remainder of the calculations.  1 mol H 2SO 4  Amount (mol) of H2SO4 that react with NaOH = 0.0980 mol NaOH   = 0.0490 mol H2SO4  2 mol NaOH  Amount (mol) of H2SO4 remaining = initial amount (mol) – amount (mol) reacting with NaOH = 0.0735 mol – 0.0490 mol = 0.0245 mol H2SO4 3.111

 amount (mol) solute  Plan: Recall the definition of concentration (mol/L)  c   . Convert mass of solute to volume (L) of solution   amount (mol) and volume from mL to litres. Divide the amount (mol) by the volume. Solution:  1 mol NaClO  Amount (mol) of NaClO = 20.5 g NaClO   = 0.2785896 mol NaClO  74.44 g NaClO   103 L  Volume (L) = 375 mL  = 0.375 L  1 mL     0.2753896 mol NaClO  concentration (mol/L) of NaClO =   = 0.73437 mol/L= 0.734 mol/L NaClO 0.375 L  

3.112

Plan: The first part of the problem is a simple dilution problem (c1V1 = c2V2). The volume in units of litres can be used. In part b), convert mass of HCl to amount (mol) and use the concentration (mol/L) to find the volume that contains that amount (mol). Solution: a) c1 = 11.7 mol/L V1 = ? c2 = 3.5 mol/L V2 = 3.0 L  3.5 mol/L  3.0 L c x V2 V1 = 2 = = 0.897436 L c1 11.7 mol/L Instructions: Be sure to wear goggles to protect your eyes! Pour approximately 2.0 L of water into the container. Add slowly and with mixing 0.90 L of 11.7 mol/L HCl into the water. Dilute to 3.0 L with water. b) Converting from mass of HCl to amount (mol) of HCl:  1 mol HCl  Amount (mol) of HCl = 9.66 g HCl   = 0.264948 mol HCl  36.46 g HCl  Converting from amount (mol) of HCl to volume:    1 mL  1L Volume (mL) of solution = 0.264948 mol HCl    3   11.7 mol HCl   10 L  = 22.64513 mL= 22.6 mL muriatic acid solution

3.113

Plan: Use the volume and concentration (mol/L) of HCl to find the amount (mol) of HCl added to the metal. That amount (mol) minus the amount (mol) of HCl that remain after reaction gives the amount (mol) of HCl that actually reacted. Use the amount (mol) of reacted HCl and the mole ratio in the balanced chemical equation to find the amount (mol) of Mg that reacted. Convert amount (mol) 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  Amount (mol) of HCl added =   0.100 L  = 0.0750 mol HCl 1L   Amount (mol) of HCl reacting with Mg = amount (mol) of added HCl – amount (mol) of HCl remaining = 0.0750 mol – 0.0125 mol HCl = 0.0625 mol HCl  1 mol Mg  Amount (mol) of Mg reacting = 0.0625 mol HCl   = 0.03125 mol Mg  2 mol HCl   24.31 g Mg  Mass (g) of Mg = 0.03125 mol Mg   = 0.7596875 g 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

3.114

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.

3.115

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.

3.116

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.

3.117

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.

3.118

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: a) CaCl2 dissociates to produce one Ca2+ ion for every two Cl– ions. Scene B contains four 2+ ions and twice that number of 1– ions. b) Li2SO4 dissociates to produce two Li+ ions for every one SO42– 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.

3.119

Plan: Write the formula for magnesium nitrate and note the ratio of magnesium ions to nitrate ions. Solution: Upon dissolving the salt in water, magnesium nitrate, Mg(NO 3)2, would dissociate to form one Mg2+ 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).

3.120

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. 3.121

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.

3.122

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.

3.123

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) 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.

3.124

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) Pentane (C5H12) has no bonds of significant polarity, so it would be expected to be insoluble in the polar solvent water. c) 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. d) Ethylene glycol (HOCH2CH2OH) molecules contain polar O–H bonds, similar to water, so it would be expected to be soluble.

3.125

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.

3.126

Plan: Substances whose aqueous solutions conduct an electric current are electrolytes such as ionic compounds, acids, and bases. Solution: a) Potassium sulfate, K2SO4, is an ionic compound that is soluble in water, producing K+ and SO42– 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.

3.127

b) Each mole of Ba(OH)2•8H2O forms 1 mole of Ba2+ ions and 2 moles of OH– ions, or a total of 3 moles of ions: Ba(OH)2•8H2O(s) → Ba2+(aq) + 2OH–(aq). The waters of hydration become part of the larger bulk of water. Convert mass to amount (mol) using the molar mass.  1 mol Ba(OH) 2 •8H 2O    3 mol ions Amount (mol) of ions =  25.4 g Ba(OH) 2 •8H 2O      315.4 g Ba(OH) 2 •8H 2O   1 mol Ba(OH) 2 •8H 2O  = 0.2415980 mol= 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): LiCl(s) → Li+(aq) + Cl–(aq). Recall that a mole contains 6.022x1023 entities, so a mole of LiCl contains 6.022x1023 units of LiCl, more easily expressed as formula units.    2 mol ions  1 mol LiCl Amount (mol) of ions = 3.55x1019 FU LiCl    23  6.022 x10 FU LiCl   1 mol LiCl 





= 1.17901x10–4 mol= 1.18x10–4 mol of ions

3.128





= 7.03839x10–5 = 7.04x10–5 mol of ions c) Each mole of Sr(HCO3)2 produces 3 moles of ions (1 mole of Sr2+ ions and 2 moles of HCO3– ions): Sr(HCO3)2(s) → Sr2+(aq) + 2HCO3–(aq). Recall that a mole contains 6.022x1023 entities, so a mole of Sr(HCO3)2 contains 6.022x1023 units of Sr(HCO3)2, more easily expressed as formula units.    1 mol Sr(HCO3 )2 3 mol ions Amount (mol) of ions = 4.03x1019 FU Sr(HCO3 ) 2   6.022 x1023 FU Sr(HCO )   1 mol Sr(HCO )  3 2  3 2   = 2.0076x10–4 = 2.01x10–4 mol of ions



3.129



b) Each mole of NiBr2•3H2O forms 1mole of Ni2+ ions and 2 moles of Br– ions, or a total of 3 moles of ions: NiBr2•3H2O(s)  Ni2+(aq) + 2Br–(aq). The waters of hydration become part of the larger bulk of water. Convert mass to amount (mol) using the molar mass.  1 mol NiBr2 •3H 2O    3 mol ions Amount (mol) of ions = 6.88 x 103 g NiBr2 •3H 2O     272.54 g NiBr2 •3H 2O   1 mol NiBr2 •3H 2O 





= 7.5732x10–5 mol= 7.57x10–5 mol of ions c) Each mole of FeCl3 forms 1mole of Fe3+ ions and 3 moles of Cl– ions, or a total of 4 moles of ions: FeCl3(s) Fe3+(aq) + 3Cl–(aq). Recall that a mole contains 6.022x1023 entities, so a mole of FeCl3 contains 6.022x1023 units of FeCl3, more easily expressed as formula units.    4 mol ions  1 mol FeCl3 Amount (mol) of ions = 2.23x1022 FU FeCl3   6.022x1023 FU FeCl   1 mol FeCl  3 3   = 0.148124 mol= 0.148 mol of ions



3.130



Plan: To determine the total amount (mol) of ions released, write an equation that shows the compound dissociating into ions with the correct molar ratios. Convert mass and formula units to amount (mol) of compound and use the molar ratio to convert amount (mol) of compound to amount (mol) of ions. Solution: a) Each mole of Na2HPO4 forms 2 moles of Na+ ions and 1 mole of HPO42– ions, or a total of 3 moles of ions: Na2HPO4(s)  2Na+(aq) + HPO42–(aq).  3 mol ions  Amount (mol) of ions =  0.734 mol Na 2 HPO 4    = 2.202mol = 2.20 mol of ions  1 mol Na 2 HPO 4  b) Each mole of CuSO4•5H2O forms 1 mole of Cu2+ ions and 1 mole of SO42– ions, or a total of 2 moles of ions: CuSO4•5H2O(s)  Cu2+(aq) + SO42–(aq). The waters of hydration become part of the larger bulk of water. Convert mass to amount (mol) using the molar mass.  1 mol CuSO4 •5H 2O    2 mol ions Amount (mol) of ions =  3.86 g CuSO4 •5H 2O       249.70 g CuSO 4 •5H 2O   1 mol CuSO 4 •5H 2O  = 3.0907x10–2 mol= 3.09x10–2 mol of ions c) Each mole of NiCl2 forms 1mole of Ni2+ ions and 2 moles of Cl– ions, or a total of 3 moles of ions: NiCl2(s)  Ni2+(aq) + 2Cl–(aq). Recall that a mole contains 6.022x1023 entities, so a mole of NiCl2 contains 6.022x1023 units of NiCl2, more easily expressed as formula units.    3 mol ions  1 mol NiCl2 Amount (mol) of ions = 8.66x1020 FU NiCl 2   6.022x1023 FU NiCl   1 mol NiCl  2  2   –3 –3 = 4.31418x10 = 4.31x10 mol of ions



3.131



Plan: To determine the total amount (mol) of ions released, write an equation that shows the compound dissociating into ions with the correct molar ratios. Convert the information given to amount (mol) of compound and use the molar ratio to convert amount (mol) of compound to amount (mol) of ions. Avogadro‘s number is used to convert amount (mol) of ions to numbers of ions. Solution: a) Each mole of AlCl3 forms 1mole of Al3+ ions and 3 moles of Cl– ions: AlCl3(s)  Al3+(aq) + 3Cl–(aq). Concentration (mol/L) and volume must be converted to amount (mol) of AlCl 3.  103 L   0.45 mol AlCl3  Amount (mol) of AlCl3 = 130. mL    = 0.0585 mol AlCl3  1 mL   L   

 1 mol Al3   Amount (mol) of Al3+ =  0.0585 mol AlCl3   = 0.0585 mol = 0.058 mol Al3+  1 mol AlCl  3    6.022x1023 Al3  22 22 3+ Number of Al3+ ions = 0.0585 mol Al 3   1 mol Al3  = 3.52287x 10 ions= 3.5x10 Al ions  





 3 mol Cl  Amount (mol) of Cl– =  0.0585 mol AlCl3   = 0.1755 mol= 0.18 mol Cl–  1 mol AlCl  3   

 

Cl  = 1.05686x10 ions= 1.1x10 Cl ions 0.1755 mol Cl   6.022x10 1 mol Cl 

Number of Cl– ions =

–



  b) Each mole of Li2SO4 forms 2 moles of Li+ ions and 1 mole of SO42– ions: Li2SO4(s) 2Li+(aq) + SO42–(aq).  103 L   2.59 g Li 2SO4   1 mol Li 2SO 4  Amount (mol) of Li2SO4 =  9.80 mL      1 mL   1L  109.95 g Li 2SO 4    = 2.3085x10–4 mol Li2SO4  2 mol Li   Amount (mol) of Li+ = 2.3085x104 mol Li2SO4  = 4.6170x10–4 mol= 4.62x10–4 mol Li+  1 mol Li SO  2 4  







 

x 10 Li  = 2.7804x10 ions= 2.78x10 Li ions  4.6170x10 mol Li   6.022 1 mol Li 4

Number of Li+ ions =







 1 mol SO42   Amount (mol) of SO42– = 2.3085x104 mol Li2SO4  = 2.3085x10–4 mol= 2.31x10–4 mol SO42–  1 mol Li SO  2 4  





 6.022 x 1023 SO42   2.3085x104 mol SO4 2    1 mol SO 2   4   = 1.39018x1020 = 1.39x1020 SO42– 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 x1022 FU KBr    1 mol KBr Amount (mol) of KBr =  245 mL      = 0.01497 mol KBr 23  1 mL   L     6.022x10 FU KBr  Number of SO42– ions =





 1 mol K   Amount (mol) of K+ =  0.01497 mol KBr   = 0.01497 mol= 1.50x10–2 mol K+  1 mol KBr   

 6.022 x1023 K   Number of K+ ions = 0.01497 mol K   = 9.016x1021 ions= 9.02x1021 K+ ions  1 mol K    





 1 mol Br   Amount (mol) of Br– =  0.01497 mol KBr   = 0.01497 mol= 1.50x10–2 mol Br–  1 mol KBr    Number of Br– ions = 3.132



 

x10 Br  = 9.016x10 ions= 9.02x10 Br ions 0.01497 mol Br   6.022 1 mol Br 

–



  Plan: To determine the total amount (mol) of ions released, write an equation that shows the compound dissociating into ions with the correct molar ratios. Convert the information given to amount (mol) of compound and use the molar ratio to convert amount (mol) of compound to amount (mol) of ions. Avogadro‘s number is used to convert amount (mol) of ions to numbers of ions. Solution: a) Each mole of MgCl2 forms 1 mole of Mg2+ ions and 2 moles of Cl– ions: MgCl2(s)  Mg2+(aq) + 2Cl–(aq).  103 L   1.75 mol MgCl2  Amount (mol) of MgCl2 =  88.mL    = 0.154 mol MgCl2  1 mL   L     1 mol Mg 2   Amount (mol) of Mg2+ =  0.154 mol MgCl2   = 0.154 mol = 0.15 mol Mg2+  1 mol MgCl  2  



2 

Mg  = 9.27388x10 ions= 9.3x10 Mg ions 0.154 mol Mg   6.022x10 1 mol Mg 2

Number of Mg2+ ions =

2





 2 mol Cl  Amount (mol) of Cl– =  0.154 mol MgCl2   = 0.308mol = 0.31 mol Cl–  1 mol MgCl  2    6.022x1023 Cl   0.308 mol Cl   = 1.854776x1023 ions= 1.9x1023 Cl– ions  1 mol Cl     b) Each mole of Al2(SO4)3 forms 2 moles of Al3+ ions and 3 moles of SO42– ions: Al2(SO4)3(s)  2Al3+(aq) + 3SO42–(aq).  103 L   0.22 g Al2 (SO4 )3   1 mol Al2 (SO4 )3  Amount (mol) of Al2(SO4)3 =  321 mL      1 mL   1L  342.17 g Al2 (SO4 )3    = 2.06389x10–4 mol Al2(SO4)3  2 mol Al3  Amount (mol) of Al3+ = 2.06389x104 mol Al2 (SO4 )3   1 mol Al (SO )  2 4 3  = 4.12777x10–4 mol= 4.1x10–4 mol Al3+  6.022x1023 Al3  Number of Al3+ ions = 4.12777x104 mol Al3  = 2.4857x1020 ions= 2.5x1020 Al3+ ions  1 mol Al3    Number of Cl– ions =













 3 mol SO4 2   Amount (mol) of SO42– = 2.06389x104 mol Al2 (SO4 )3   1 mol Al (SO )  2 4 3  = 6.191659x10–4 mol= 6.2x10–4 mol SO42 6.022 x1023 SO42   Number of SO42– ions = 6.191659 x104 mol SO42    1 mol SO 2   4   20 20 2– = 3.7286x10 ions= 3.7x10 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)  8.83x1021 FU CsNO3    1 mol CsNO3 Amount (mol) of CsNO3 = 1.65 L   = 0.024194 mol CsNO3   23    6.022x10 FU CsNO  L 3  









 1 mol Cs  Amount (mol) of Cs+ =  0.024194 mol CsNO3   = 0.024194 mol= 0.0242 mol Cs+  1 mol CsNO  3    6.022x1023 Cs  Number of Cs+ ions = 0.024194 mol Cs  = 1.45695x1022 ions= 1.46x1022 Cs+ ions  1 mol Cs   





 1 mol NO3  Amount (mol) of NO3– =  0.024194 mol CsNO3   = 0.024194 mol= 0.0242 mol NO3–  1 mol CsNO  3    6.022x1023 NO3  Number of NO3– ions = 0.024194 mol NO3  = 1.45695x1022 ions= 1.46x1022 NO3– ions  1 mol NO   3  



3.133



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 amount (mol) H+ = amount (mol) acid. Calculate the amount (mol) of acid by multiplying the concentration (mol/L) by the volume in litres. Solution: a) HClO4(aq) → H+(aq) + ClO4–(aq)

 0.25 mol  + Amount (mol) H+ = mol HClO4 = 1.40 L   = 0.35 mol H  1L  b) HNO3(aq) → H+(aq) + NO3–(aq)  103 L   0.92 mol  Amount (mol) H+ = mol HNO3 =  6.8 mL   = 6.256x10–3 mol= 6.3x10–3 mol H+  1 mL   1 L    c) HCl(aq) → H+(aq) + Cl–(aq)  0.085 mol  + Amount (mol) H+ = mol HCl =  2.6 L    = 0.221 mol= 0.22 mol H 1 L   3.134

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 HNO 3 each produce one mole of H+ upon dissociation, so amount (mol) H+ = amount (mol) acid. Calculate the amount (mol) of acid by multiplying the concentration (mol/L) by the volume in litres. Solution: a) HBr(aq) → H+(aq) + Br–(aq)  103 L   0.75 mol  Amount (mol) H+ = mol HBr = 1.4 mL   = 1.05x10–3 mol= 1.0x10–3 mol H+  1 mL   1 L    + – b) HI(aq) → H (aq) + I (aq)  103 L   1.98 mol  Amount (mol) H+ = mol HI =  2.47 mL   = 4.8906x10–3 mol= 4.89x10–3 mol H+  1 mL   1 L    c) HNO3(aq) → H+(aq) + NO3–(aq)  103 L   0.270 mol  + Amount (mol) H+ = mol HNO3 =  395 mL    = 0.10665 mol= 0.107 mol H  1 mL   1 L   

3.135

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 amount (mol) of compound and then use the molar ratio in the dissociation of the compound to find the amount (mol) of each ion. The concentration (mol/L) of each ion is the amount (mol) of ion divided by the volume of the seawater. To find the total concentration (mol/L) of the alkali metal ions [Group 1], add the amount (mol) of the alkali metal ions and divide by the volume of the seawater. Perform the same calculation to find the total concentration (mol/L) of the alkaline earth metal ions [Group 2] and the anions (the negatively charged ions). Solution: a) The volume of the seawater is needed.  103 g  cm3   1 mL   103 L  Volume (L) of seawater = 1.00 kg   = 0.97560976 L  1 kg   1.025 g   1 cm3   1 mL       The amount (mol) of each ion are needed. If an ion comes from more than one source, the total amount (mol) 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  Amount (mol) of NaCl =  26.5 g NaCl    = 0.4534565 mol NaCl  58.44 g NaCl 

 1 mol Na   Amount (mol) of Na+ =  0.4534565 mol NaCl  = 0.4534565 mol Na+  1 mol NaCl     1 mol Cl  Amount (mol) of Cl– =  0.4534565 mol NaCl  = 0.4534565 mol Cl–  1 mol NaCl   

MgCl2: Each mole of MgCl2 forms 1 mole of Mg2+ ions and 2 moles of Cl– ions: MgCl2(s)  Mg2+(aq) + 2Cl–(aq)  1 mol MgCl 2  Amount (mol) of MgCl2 =  2.40 g MgCl2    = 0.025207 mol MgCl2  95.21 g MgCl2 

 1 mol Mg 2   Amount (mol) of Mg2+ =  0.025207 mol MgCl2   = 0.025207 mol Mg2+  1 mol MgCl  2    2 mol Cl  Amount (mol) of Cl– =  0.025207 mol MgCl2   = 0.050415 mol Cl–  1 mol MgCl  2   MgSO4: Each mole of MgSO4 forms 1 mole of Mg2+ ions and 1 mole of SO42– ions: MgSO4(s)  Mg2+(aq) + SO42–(aq)  1 mol MgSO4  Amount (mol) of MgSO4 =  3.35 g MgSO4    = 0.0278285 mol MgSO4  120.38 g MgSO4 

 1 mol Mg 2   Amount (mol) of Mg2+ =  0.0278285 mol MgSO4   = 0.0278285 mol Mg2+  1 mol MgSO  4    1 mol SO42   Amount (mol) of SO42– =  0.0278285 mol MgSO4   = 0.0278285 mol SO42–  1 mol MgSO  4   CaCl2: Each mole of CaCl2 forms 1 mole of Ca2+ ions and 2 moles of Cl– ions: CaCl2(s)  Ca2+(aq) + 2Cl–(aq)  1 mol CaCl2   1 mol Ca 2   Amount (mol) of CaCl2 = 1.20 g CaCl2    = 0.0108128 mol CaCl2    110.98 g CaCl2   1 mol CaCl2   1 mol Ca 2   Amount (mol) of Ca2+ =  0.0108128 mol CaCl2   = 0.0108128 mol Ca2+  1 mol CaCl  2  

 2 mol Cl  Amount (mol) of Cl– =  0.0108128 mol CaCl2   = 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)  1 mol KCl  Amount (mol) of KCl = 1.05 g KCl    = 0.0140845 mol KCl  74.55 g KCl 

 1 mol K   Amount (mol) of K+ =  0.0140845 mol KCl  = 0.0140845 mol K+  1 mol KCl     1 mol Cl  Amount (mol) of Cl– =  0.0140845 mol KCl  = 0.0140845 mol Cl–  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 NaHCO3  Amount (mol) of NaHCO3 =  0.315 g NaHCO3    = 0.00374955 mol NaHCO3  84.01 g NaHCO3 

 1 mol Na   Amount (mol) of Na+ =  0.00374955 mol NaHCO3   = 0.00374955 mol Na+  1 mol NaHCO  3   Silberberg, Amateis, Venkateswaran, Chen, Chemistry: The Molecular Nature of Matter and Change, 3rd Canadian Edition Page 3-114 Instructor‘s Solution Manual © Copyright 2021 McGraw-Hill Ryerson Ltd.

 1 mol HCO3  Amount (mol) of HCO3– =  0.00374955 mol NaHCO3   = 0.00374955 mol HCO3–  1 mol NaHCO  3   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  Amount (mol) of NaBr =  0.098 g NaBr    = 0.0009524735 mol NaBr  102.89 g NaBr 

 1 mol Na   Amount (mol) of Na+ =  0.0009524735 mol NaBr   = 0.0009524735 mol Na+  1 mol NaBr     1 mol Br   Amount (mol) of Br– =  0.0009524735 mol NaBr   = 0.0009524735 mol Br–  1 mol NaBr    Total amount (mol) of each ion: Cl–: 0.4534565 mol + 0.050415 mol + 0.0216255 mol + 0.0140845 mol = 0.5395815 mol Cl – + Na : 0.4534565 mol+ 0.00374955 mol + 0.0009524735 mol = 0.458158523 mol Na+ Mg2+: 0.025207 mol+ 0.0278285 mol= 0.0530355 mol Mg2+ SO42–: 0.0278285 mol SO42– Ca2+: 0.0108128 mol Ca2+ + K: 0.0140845 mol K+ – HCO3 : 0.00374955 mol HCO3– Br–: 0.0009524735 mol Br– Dividing each amount (mol) by the volume (0.97560976 L) and rounding to the proper number of significant figures gives the concentration (mol/L) mol c= L 0.5395815 mol Cl c Cl– = = 0.55307 mol/L = 0.553 mol/L Cl– 0.97560976 L 0.45815823 mol Na  = 0.469612 mol/L = 0.470 mol/L Na+ 0.97560976 L

c Na+ =

c Mg2+ =

0.0530355 mol Mg2  = 0.054361 mol/L = 0.0544 mol/L Mg2+ 0.97560976 L

c SO42– =

0.0278285 mol SO4 2  = 0.028524 mol/L = 0.0285 mol/L SO42– 0.97560976 L

c Ca2+ =

0.0108128 mol Ca 2  = 0.011083 mol/L = 0.0111 mol/L Ca2+ 0.97560976 L

c K+ =

0.0140845 mol K 0.97560976 L

c HCO3– =

= 0.014437 mol/L = 0.0144 mol/L K+

0.00374955 mol HCO3 = 0.003843 mol/L = 0.00384 mol/L HCO3– 0.97560976 L

0.0009524735 mol Br  = 0.0009763 mol/L = 0.00098 mol/L Br–. 0.97560976 L b) The alkali metal cations are Na+ and K+. Add the concentrations (mol/L) of the individual ions. 0.469612 mol/L Na+ + 0.014437 mol/L K+ = 0.484049 mol/L = 0.484 mol/L total for alkali metal cations c) The alkaline earth metal cations are Mg2+ and Ca2+. Add the concentrations (mol/L) of the individual ions. 0.054361 mol/L Mg2+ + 0.011083 mol/L Ca2+ = 0.065444 mol/L = 0.0654 mol/L total for alkaline earth cations d) The anions are Cl–, SO42–, HCO3–, and Br–. Add the concentrations (mol/L) of the individual ions. 0.55307 mol/L Cl– + 0.028524 mol/L SO42– + 0.003843 mol/L HCO3– + 0.0009763 mol/L Br–

c Br– =

= 0.5864133 mol/L = 0.586 mol/L total for anions 3.136

Plan: Use the concentration (mol/L) and volume of the ions to find the amount (mol) of each ion. Multiply the amount (mol) of each ion by that ion‘s charge to find the total amount (mol) of charge. Since sodium ions have a +1 charge, the total amount (mol) of charge equals the amount (mol) of sodium ions. Solution:  0.015 mol Ca 2   2+ Amount (mol) of Ca2+ = 1.0 x103 L   = 15 mol Ca  L  





 2 mol charge  Amount (mol) of charge from Ca2+ = 15 mol Ca 2   = 30. mol charge from Ca2+  1 mol Ca 2    





 0.0010 mol Fe3  3+ Amount (mol) of Fe3+ = 1.0x103 L   = 1.0 mol Fe  L  





 3 mol charge  Amount (mol) of charge from Fe3+ = 1.0 mol Fe3   = 3.0 mol charge from Fe3+  1 mol Fe3     Total amount (mol) of charge = 30. mol + 3.0 mol = 33 mol charge  1 mol Na   Amount (mol) Na+ =  33 mol charge   = 33 mol Na+  1 mol charge   





3.137

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.

3.138

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: Total ionic equation: Ba2+(aq) + 2NO3–(aq) + 2Na+(aq) + CO32–(aq)  BaCO3(s) + 2Na+(aq) + 2NO3–(aq) The spectator ions are underlined and are omitted: Net ionic equation: Ba2+(aq) + CO32–(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) Total ionic: Ba2+(aq) + 2Cl–(aq) + 2K+(aq) + CO32–(aq)  BaCO3(s) + 2K+(aq) + 2Cl–(aq) Molecular: BaBr2(aq) + (NH4)2CO3(aq)  BaCO3(s) + 2NH4Br(aq) Total ionic: Ba2+(aq) + 2Br–(aq) + 2NH4+(aq) + CO32–(aq)  BaCO3(s) + 2NH4+(aq) + 2Br–(aq)

3.139

Plan: Write the balanced chemical equation. Change masses of CO2 and O2 to amount (mol). Use stoichiometry and limiting reagents to find the amount (mol) of CO 2 formed. Use stoichiometry to determine the reagent in excess and its amount. Use the actual amount and the amount found above of CO2 to determine the percent yield.

Solution: a) 2 CO (g) + O2 (g)  2 CO2 (g) b)

We need 2 mol of CO for each mol of O2; therefore 0.625 mol of O 2 would need 1.25 mol of CO, which we do not have. So CO is the limiting reagent (LR). Therefore, 0.714 mol of CO 2 is formed. Silberberg, Amateis, Venkateswaran, Chen, Chemistry: The Molecular Nature of Matter and Change, 3rd Canadian Edition Page 3-116 Instructor‘s Solution Manual © Copyright 2021 McGraw-Hill Ryerson Ltd.

( )( ) c) Since all the CO is used, the amount of O2 used is ½ n(CO) or 0.714 mol/2 = 0.357 mol. Since we started with 0.625 mol of O2, we are left with 0.625 – 0.357 = 0.268 mol of O2, which corresponds to ( )( ) d) If the actual amount of CO2 formed is 12.7 g, the percent yield is

3.140

Plan: First determine the empirical formula. Convert the mass of each element to amount (mol) by dividing the mass of each element by its molar mass. Divide each of the amount (mol) 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  Amount (mol) of S = 2.288 g S   = 0.0713439 mol S  32.07 g S   1 mol N  Amount (mol) of N = 1.000 g N   = 0.0713776 mol N  14.01 g N  Preliminary formula is S0.0713439N0.0713776 Converting to integer subscripts (dividing all by the smallest subscript): S 0.0713439 N 0.0713776 → S1N1 0.0713439

0.0713439

The empirical formula is SN. Formula mass of empirical formula = 32.07 g/mol S + 14.01 g/mol N = 46.08 g/mol  184.27 g/mol  molar mass of compound Whole-number multiple = =   =4 empirical formula mass  46.08 g/mol  Multiplying the subscripts in SN by 4 gives S4N4 as the molecular formula. 3.141

Plan: The first step is to write and balance the chemical equation for the reaction. Convert the mass of each reactant to amount (mol) 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 amount (mol) of Ca5(PO4)3(OH) from the amount (mol) of Ca(OH)2 (if H3PO4 is limiting):  1 mol Ca(OH) 2  Amount (mol) of Ca(OH)2 = 100. g Ca(OH) 2   = 1.349528 mol Ca(OH)2  74.10 g Ca(OH) 2   1 mol Ca 5 (PO 4 )3 (OH)  Amount (mol) 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 amount (mol) of Ca5(PO4)3(OH) from the amount (mol) of H3PO4 (if Ca(OH)2 is limiting):    1 mol H3 PO 4  85 g H3 PO 4 Amount (mol) of H3PO4 = 100. g H3 PO 4 solution     100. g H3 PO 4 solution   97.99 g H3 PO 4 

= 0.867435 mol H3PO4

 1 mol Ca 5 (PO 4 )3 (OH)  Amount (mol) of Ca5(PO4)3(OH) from H3PO4 = 0.867435 mol H3 PO 4 solution   3 mol H3 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 Ca 5(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 (PO 4 )3 (OH)    1 mol Ca 5 (PO 4 )3 (OH) 

= 135.57898 g= 140 g Ca5(PO4)3(OH) 3.142

Plan: The amount (mol) of narceine and the amount (mol) of water are required. We can assume any mass of narceine hydrate (we will use 100 g), and use this mass to determine the amount (mol) of hydrate. The amount (mol) of water in the hydrate is obtained by taking 10.8% of the 100 g mass of hydrate and converting the mass to amount (mol) of water. Divide the amount (mol) of water by the amount (mol) of hydrate to find the value of x. Solution: Assuming a 100 g sample of narceine hydrate:  1 mol narceine hydrate  Amount (mol) of narceine hydrate = 100 g narceine hydrate    499.52 g narceine hydrate  = 0.20019 mol narceine hydrate   10.8% H 2 O Mass (g) of H2O = 100 g narceine hydrate   = 10.8 g H2O  100% narceine hydrate   1 mol H 2 O  Amount (mol) of H2O = 10.8 g H 2 O   = 0.59933 mol H2O  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.

3.143

C2H6

30.07

6 mol(1.008 g/mol) 100%  = 20.11% H 30.07 g

Propane

C3H8

44.09

8 mol(1.008 g/mol) 100%  = 18.29% H 44.09 g

Benzene

C6H6

78.11

6 mol(1.008 g/mol) 100%  = 7.743% H 78.11 g

Ethanol

C2H5OH

46.07

6 mol(1.008 g/mol) 100%  = 13.13% H 46.07 g

Cetyl palmitate

C32H64O2

480.83

64 mol(1.008 g/mol) 100%  = 13.42% H 480.83 g

The hydrogen percentage decreases in the following order: Ethane > Propane > Cetyl palmitate > Ethanol > Benzene

3.144

Plan: First determine the empirical formula. Assume 100 grams of sample, and then the amount (mol) of each element may be found by dividing the mass of each element by its molar mass. Divide each of the amount (mol) 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 amount (mol) by dividing by the molar mass:  68.2 parts C by mass   1 mol C  Amount (mol) of C = 100 g   = 5.6786 mol C   100 parts by mass   12.01 g C   6.86 parts H by mass   1 mol H  Amount (mol) of H = 100 g   = 6.8056 mol H   100 parts by mass   1.008 g H   15.9 parts N by mass   1 mol N  Amount (mol) of N = 100 g   = 1.1349 mol N   100 parts by mass   14.01 g N   9.08 parts O by mass   1 mol O  Amount (mol) of O = 100 g   = 0.5675 mol O   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 empirical formula mass  176.22 g/mol  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.

3.145

a) Before 1961 the u 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.

3.146

 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 2SO2 and 2 in 2H2O) require a total of 6 O or 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 vapour 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 H 2O 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 CO 2 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 H 2O 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) 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)

3.147

 1 mol CO 2   1 mol C  Amount (mol) of C = 2.657 g CO 2    = 0.06037 mol C  44.01 g CO 2   1 mol CO 2   1 mol H 2 O   2 mol H  Amount (mol) 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. 3.148

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 amount (mol) of iron by dividing by the molar mass, and the mole ratio from the balanced equation is used to find the amount (mol) and then the mass of CO required to produce that amount (mol) of iron. Solution: a) In the first step, ferric oxide (ferric denotes Fe3+) 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 tonne is equal to 1000 kg. Converting 45.0 metric tons of Fe to mass in grams:  103 kg  103 g  Mass (g) of Fe = 45.0 ton Fe  = 4.50x107 g Fe  1 ton   1 kg      1 mol Fe  5 Amount (mol) of Fe = 4.50x107 g Fe   = 8.05730x10 mol Fe 55.85 g Fe  



 3 mol CO   28.01 g CO  7 7 Mass (g) of CO = 8.05730x105 mol Fe    = 3.38527x10 g= 3.39x10 g CO 2 mol Fe 1 mol CO    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 amount (mol). Use the mole ratio between toluene and oxygen to find the amount (mol) and then mass of oxygen required for the reaction. The mole ratio between toluene and the gaseous products are used to find the amount (mol) of product produced. The amount (mol) 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 C7 H8   1 mol C7 H8  a) Amount (mol) of C7H8 = 20.0 mL C7 H8    = 0.1882123 mol C7H8  1 mL C7 H8   92.13 g C7 H8 



3.149



 9 mol O 2   32.00 g O 2  Mass (g) oxygen = 0.1882123 mol C7 H8    = 54.20514 g= 54.2 g O2  1 mol C7 H8   1 mol O 2   11 mol product gas  b) Total amount (mol) of gas = 0.1882123 mol C7 H8   = 2.07034 mol= 2.07 mol of gas 1 mol C7 H8  

The 11 mol of gas is an exact, not measured, number, so it does not affect the significant figures.  4 mol H 2 O  c) Amount (mol) of H2O = 0.1882123 mol C7 H8   = 0.7528492 mol H2O  1 mol C7 H8   6.022 x1023 H 2O molecules  Molecules of H2O =  0.7528492 mol H 2O     1 mol H 2O   = 4.53366x1023 molecules= 4.53x1023 molecules H2O

3.150

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 amount (mol) of N2 from the amount of N2O4 (if N2H4 is limiting):  1 mol N 2 O 4   3 mol N 2  Amount (mol) of N2 from N2O4 = 100.0 g N 2 O 4    = 3.26016 mol N2  92.02 g N 2 O 4   1 mol N 2 O 4  Finding the amount (mol) of N2 from the amount of N2H4 (if N2O4 is limiting):  1 mol N 2 H 4   3 mol N 2  N2 from N2H4 = 100.0 g N 2 H 4    = 4.68019 mol N2  32.05 g N 2 H 4   2 mol N 2 H 4  N2O4 is the limiting reactant.  1 mol N 2 O 4   3 mol N 2   28.02 g N 2  Theoretical yield of N2 = 100.0 g N 2 O 4     = 91.3497 g N2  92.02 g N 2 O 4   1 mol N 2 O 4   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 O 4  Mass (g) of N2O4 used = 10.0 g NO      30.01 g NO   6 mol NO   1 mol N 2 O 4 

= 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       92.02 g N 2 O 4  1 mol N 2 O 4  1 mol N 2  = 82.01285 g N2  actual yield   82.01285  Theoretical yield =   100%  =   100%  = 89.7790 %= 89.8% theoretical yield  91.3497    3.151

Plan: Write a balanced chemical equation, using X to represent the halogen element. Convert the mass of strontium sulfate to amount (mol) and use the mole ratio in the balanced equation to find the amount (mol) of strontium halide required to produce that number of amount (mol) of strontium sulfate. Divide the mass of

strontium halide by the amount (mol) of strontium halide to determine its molar mass. Subtract out the molar mass of strontium to obtain the molar mass of X 2. 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 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 SrSO 4  Amount (mol) of SrSO4 = 0.755 g SrSO 4   = 0.0041102 mol SrSO4  183.69 g SrSO4   1 mol SrX 2  Amount (mol) of SrX2 = 0.0041102 mol SrSO 4   = 0.0041102 mol SrX2  1 mol SrSO 4  The 0.652 g sample of SrX2 = 0.004110186 mol 0.652 g SrX2 = = 158.63 g/mol = molar mass 0.0041102 mol mass of X2 = mass of SrX2 – mass of Sr = 158.63 g – 87.62 g = 71.01 g = X2 Molar mass of X = 71.01 g/2 mol = 35.505 g/mol= 35.5 g/mol = Cl The original halide formula is SrCl2.

3.152

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 amount (mol) of excess reactant required to react with the limiting reactant. The difference between the initial amount (mol) of excess reactant and the amount (mol) required for reaction is the amount (mol) 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 amount (mol) of AB3 from the amount (mol) of AB2 (if B2 is limiting):  2 mol AB3  Amount (mol) of AB3 from AB2 = 5.0 mol AB2   = 5.0 mol AB3  2 mol AB2  Finding the amount (mol) of AB3 from the amount (mol) of B2 (if AB2 is limiting):  2 mol AB3  Amount (mol) of AB3 from B2 = 3.0 mol B2   = 6.0 mol AB3  1 mol B2  AB2 is the limiting reagent and 5.0 mol of AB3 is formed.  1 mol B2  d) Amount (mol) of B2 that react with 5.0 mol AB2 = 5.0 mol AB2   = 2.5 mol B2  2 mol AB2  The unreacted B2 is 3.0 mol – 2.5 mol = 0.5 mol B2.

3.153

Plan: These are dilution problems. Dilution problems can be solved by converting to amount (mol) and using the new volume; however, it is much easier to use c1V1 = c2V2. The dilution equation does not require a volume in litres; it only requires that the volume units match. Solution: a) c1 = 18.0 mol/L V1 = ? c2 = 0.429 mol/L V2 = 2.00L c1V1= c2V2  0.429 mol / L  2.00 L c x V2 V1 = 2 = = 0.047667 L= 0.0477 L c1 18.0 mol/L b) c1 = 0.225 mol/L

 103 L  V1 = 80.6 mL  = 0.0806 L  1 mL   

c2 = ?

V2 = 0.250 L

c1V1= c2V2  0.225 mol / L  0.0806 L c x V1 c2 = 1 = = 0.07254 mol/L = 0.0725 mol/L V2 0.250 L c) c1 = 0.0372 mol/L V1 = 0.130 L c2 = 0.0100 mol/L c1V1= c2V2 c x V1  0.0372 mol / L  0.130 L V2 = 1 = = 0.4836 L of solution c2  0.0100 mol / L 

V2 = ?

Volume (L) of water added = final volume  initial volume = 0.4836 L  0.130 L = 0.3536 L= 0.354 L of water  103 L  d) c1 = 0.745 mol/L V1 =  64.0 mL   = 0.0640 L c2 = ? V2 = 0.100 L  1 mL    c1V1= c2V2 c x V1  0.745 mol / L  0.0640 L  c2 = 1 = = 0.4768 mol/L V2  0.100 L  The concentration in concentration (mol/L) must now be converted to grams per millilitre.  0.4768 mol Ca(NO3 )2   103 L   164.10 g Ca(NO3 )2    = 0.07824 g/mL= 0.0782 g Ca(NO3)2/mL    L    1 mL   1 mol Ca(NO3 )2  3.154

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 x 12.01 g/mol C) + (4 x 1.008 g/mol H) + (1 x 16.00 g/mol O) = 44.05 g/mol.

3.155

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. 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 amount (mol) of product from the amount of ZrOCl 2•8H2O (if H2C2O4•2H2O is limiting): Amount (mol) of product from ZrOCl2•8H2O =  1 mol ZrOCl 2 •8H 2O    1 mol product 1.68 g ZrOCl2 •8H 2O     = 0.00521334 mol product 322.25 g ZrOCl •8H O 1 mol ZrOCl •8H O 2 2  2 2   Finding the amount (mol) of product from the amount of H2C2O4•2H2O (if ZrOCl2•8H2O is limiting): Amount (mol) of product from H2C2O4•2H2O =  1 mol H 2C2O4 •H 2O    1 mol product  5.20 g H 2C2O4 •2H 2O     = 0.0103117 mol product  126.07 g H 2C2O 4 •2H 2O   4 mol H 2C2O 4 •2H 2O  It is not necessary to find the amount (mol) 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  Mass (g) of product =  0.00521334 mol product    = 2.82318 g product  1 mol product  Calculating the percent yield:

 1.25 g   actual Yield  Percent yield =   x 100% = 44.276 %= 44.3% yield  x100% =   theoretical Yield   2.82318 g 

3.156

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 concentration (mol/L) in part c), use the mass of NaCl in 100 g of seawater; convert mass of NaCl to amount (mol) and mass of seawater to volume in litres, using the density. concentration (mol/L) = Amount (mol) of NaCl / V(L) of seawater. Solution:  85% NaCl  a)  4.0% ions    = 3.4% NaCl  100% ions   22.99 g Na   b) % Na+ ions =  3.4% NaCl   = 1.3375 %= 1.3% Na+ ions  58.44 g NaCl   

 35.45 g Cl  % Cl– ions =  3.4% NaCl   = 2.062 %= 2.1% Cl– ions  58.44 g NaCl    c) Since the mass % of NaCl is 3.4%, there are 3.4 g of NaCl in 100 g of seawater.  1 mol NaCl  Amount (mol) of NaCl =  3.4 g NaCl    = 0.0581793mol NaCl  58.44 g NaCl     103 L  1 mL Volume (L) of 100 g of seawater = 100 g seawater    = 0.097561 L    1.025 g seawater   1 mL 

c NaCl =

Amount (mol) NaCl 0.0581793 mol = = 0.596338 mol/L= 0.60 mol/L NaCl Volume (L) seawater 0.097561 L

3.157

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) True e) True

3.158

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.

3.159

Plan: To convert mass to amount (mol), divide the mass by the molar mass of the substance. To convert amount (mol) to mass, divide by the molar mass. To obtain number of particles, multiply amount (mol) by Avogadro‘s number. Divide a number of particles by Avogadro‘s number to obtain amount (mol). Solution: a) Since 1 mole of any substance contains Avogadro‘s number of entities, equal amounts in moles of various substances contain equal numbers of entities. The number of entities (O 3 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(16.0 g/mol O) = 48.0 g/mol; O has a molar mass of 1(16.0 g/mol O) = 16.0 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 O 4  c) Amount (mol) of N2O4 =  4.0 g N 2 O 4    = 0.043 mol N2O4  92.02 g N 2 O 4   1 mol SO2  Amount (mol) of SO2 =  3.3 g SO2    = 0.052 mol SO2  64.07 g SO2 

SO2 is the larger quantity in terms of amount (mol).  28.05 g C2 H 4  d) Mass (g) of C2H4 =  0.6 mol C2 H 4    = 17 g C2H4  1 mol C2 H 4   38.00 g F2  Mass (g) of F2 =  0.6 mol F2    = 23 g F2  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  e) Total amount (mol) of ions in 2.3 mol NaClO3 =  2.3 mol NaClO3    = 4.6 mol ions  1 mol NaClO3 

 3 mol ions  Total amount (mol) of ions in 2.2 mol MgCl2 =  2.2 mol MgCl2    = 6.6 mol ions  1 mol MgCl2  MgCl2 is the greater quantity in terms of total amount (mol) of ions. f) The compound with the lower molar mass will have more molecules in a given mass. H 2O (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) Amount (mol) of NaBr =  0.500 L NaBr    = 0.250 mol NaBr 1L  

 1 mol Na   Amount (mol) of Na+ =  0.250 mol NaBr   = 0.250 mol Na+  1 mol NaBr     103 g   1 mol NaCl  Amount (mol) of NaCl =  0.0146 kg NaCl   = 0.250 mol NaCl  1 kg   58.44 g NaCl      2 mol ions  + Amount (mol) of Na+ =  0.250 mol NaCl    = 0.250 mol Na  1 mol NaCl  The two quantities are equal. h) The heavier atoms, 238U, will give a greater total mass since there is an equal number of particles of both.

3.160

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) 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 O 2 to produce 283.88 g of P4O10 and 3(64.07 g/mol SO2) = 192.21 g of SO2.

3.161

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  % yield =   x 100%  theoretical Yield 

Theoretical yield (g) of H2O =

actual yield 105 kg 100  = 100  = 106.2753 kg H2O % yield 98.8%

 103 g   1 mol H 2 O  2 mol H2  2.016 g H2  Mass (g) of H2 = 106.2753 kg H2 O    1 kg   18.02 g H O  2 mol H O  1 mol H  2  2  2    = 1.18896x104 g= 1.19x104 g H2

3.162

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 N 2 in grams by its molar mass to obtain amount (mol). Solution:  3.0 kg   60 s   60 min   24 h   365.25 d  9 a) Mass of atmosphere =    1min   1 h   1 d   1 y  50 x 10 y s      





= 4.73364x1018 kg= 4.7x1018 kg atmosphere    103 g  75.5% 21 b) Mass (g) of N2 = 4.73364 x1018 kg atmosphere   = 3.573898x10 g N2    100% atmosphere   1 kg 





 1 mol N 2  20 20 Amount (mol) of N2 = 3.573898 x1021 g N 2   = 1.27548x10 mol= 1.3x10 mol N2 28.02 g N 2  





3.163

Plan: Divide the given mass of a substance by its molar mass to obtain amount (mol); multiply the given amount (mol) 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  a) Amount (mol) of NH4Br =  0.588 g NH 4 Br    = 0.0060037 mol= 0.00600 mol NH4Br  97.94 g NH 4 Br   1 mol KNO3  b) Amount (mol) of KNO3 =  88.5 g KNO3    = 0.875284 mol KNO3  101.11 g KNO3 

 1 mol K   6.022 x 1023 K  ions  Number of K+ ions =  0.875284 mol KNO3     1 mol KNO   1 mol K  3    = 5.27096x1023 ions= 5.27x1023 K+ ions  92.09 g C3 H8 O3  c) Mass (g) of C3H8O3 =  5.85 mol C3 H8 O3    = 538.7265 g= 539 g C3H8O3  1 mol C3 H8 O3   119.37 g CHCl3  d) Mass (g) of CHCl3 =  2.85 mol CHCl3    = 340.2045 g CHCl3  1 mol CHCl3    mL Volume (mL) of CHCl3 =  340.2045 g CHCl3    = 229.868 = 230. mL CHCl3  1.48 g CHCl3 

 2 mol Na   e) Amount (mol) of Na+ =  2.11 mol Na 2 CO3   = 4.22 mol Na+  1 mol Na CO  2 3    6.022 x1023 Na  ions  24 24 + Number of Na+ = 4.22 mol Na    = 2.54128x10 ions= 2.54x10 Na ions   1 mol Na  

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

 103 g   1 mol Cd  f) Amount (mol) of Cd atoms =  25.0 mg Cd   = 2.224199x10–4 mol Cd atoms  1 mg   112.4 g Cd     

 6.022 x1023 Cd atoms  Number of Cd atoms = 2.224199x104 mol Cd    1 mol Cd   = 1.3394126x1020 atoms= 1.34x1020 Cd atoms  2 mol F   6.022 x1023 F atoms  g) Number of F atoms =  0.0015 mol F2      1 mol F  1 mol F2    = 1.8066x1021 atoms= 1.8x1021 F atoms

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3.164

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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 amount (mol), and through the balanced chemical equation the amount of CO 2 produced by each gas may be found. Summing the amounts of CO 2 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 CO 2   44.01 g CO 2  Methane:  200.g Mixture      = 137.188 g CO2   100%   16.04 g CH 4   1 mol CH 4   1 mol CO 2   75.0%   1 mol C3 H8   3 mol CO 2   44.01 g CO 2  Propane:  200.g Mixture      = 449.183 g CO2   100%   44.09 g C3 H8   1 mol C3 H8   1 mol CO 2  Total CO2 = 137.188 g + 449.183 g = 586.371 = 586 g CO2 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  The total amount (mol) of CO2 produced =  748 g CO 2    = 16.996 mol CO2  44.01 g CO 2 

16.996 mol CO2 =  1 mol C3 H8   3 mol CO2   1 mol CH 4   1 mol CO2   x g CH 4       +  252  x g C3 H8   16.04 g CH 1 mol CH 4  4    44.09 g C3 H8   1 mol C3 H8  16.996 mol CO2 =

x 3(252  x) mol CO2 + mol CO2 16.04 44.09

x 756  3x mol CO2 + mol CO2 16.04 44.09 16.996 mol CO2 = 0.06234x mol CO2 + (17.147 – 0.06804x mol CO2) 16.996 = 17.147 – 0.0057x x = 26.49 g CH4 252 – x = 252 g – 26.49 g = 225.51 g C3H8 16.996 mol CO2 =

3.165

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 H8 225.51 g C3 H8 100%  = 100% = 89.5% C3H8 mass of mixture 252 g mixture

Plan: Find the amount (mol) of HCl in the final solution and amount (mol) of HCl in the solution

whose concentration (mol/L) is known. Subtract the amount (mol) of HCl in the known solution from the number of amount (mol) of HCl in the final solution to find the moles of HCl in the solution of unknown concentration (mol/L). Divide that number of amount (mol) of HCl by the volume of that solution to find its concentration (mol/L). Solution: Total volume (L) of the final solution = 1.35 L + 3.57 L = 4.92 L  0.893 mol HCl  Total amount (mol) of HCl in final solution =  4.92 L    = 4.39356 mol HCl 1L    0.325 mol HCl  Amount (mol) of HCl in solution of known concentration = 1.35 L    = 0.43875 mol HCl 1L   Amount (mol) of HCl added from solution with unknown concentration = Total amount (mol)  amount (mol) from known solution = (4.39356  0.43875) mol HCl = 3.95481 mol HCl moles solute 3.95481 mol HCl Concentration (mol/L) of unknown solution = = L of solution 3.57 L = 1.10779 mol/L= 1.11 mol/L HCl

3.166

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.

3.167

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 amount (mol) of substance, and then to amount (mol) of each element. To get the desired x:y:1.0 ratio, divide the amount (mol) of each element by the amount (mol) of potassium. Solution: A 100-gram sample of 30:10:10 fertilizer contains 30 g N, 10 g P 2O5, and 10 g K2O.  1 mol N  Amount (mol) of N =  30 g N    = 2.1413 mol N  14.01 g N   1 mol P2 O5  2 mol P  Amount (mol) of P = 10 g P2 O5     = 0.14090 mol P  141.94 g P2 O5  1 mol P2 O5   1 mol K 2 O  2 mol K  Amount (mol) of K = 10 g K 2 O     = 0.21231 mol K  94.20 g K 2 O  1 mol K 2 O 

This gives a N:P:K ratio of 2.1413:0.14090:0.21231 The ratio must be divided by the amount (mol) of K and rounded. 2.1413 mol N 0.14090 mol P = 10.086 = 0.66365 0.21231 0.21231 10.086:0.66365:1.000 or 10:0.66:1.0 3.168

0.21231 mol K =1 0.21231

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 amount (mol) of substance, and then to amount (mol) 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 P 2O5 and 10 g K2O.  1 mol N  Amount (mol) of N = 10 g N    = 0.713776 mol N  14.01 g N 

 1 mol P2 O5  2 mol P  Amount (mol) of P = 10 g P2 O5     = 0.14090 mol P  141.94 g P2 O5  1 mol P2 O5   1 mol K 2 O  2 mol K  Amount (mol) of K = 10 g K 2 O     = 0.21231 mol K  94.20 g K 2 O  1 mol K 2 O 

Since N comes from 2 sources, we will first determine how much (NH 4)2HPO4 we need to have the correct amount (mol) of P. Then we will find out the amount (mol) of N contributed from that mass of (NH4)2HPO4. The balance amount of N comes from the (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   0.14090 mol P     1 mol (NH ) HP O  = 18.6073 g (NH4)2HPO4 1 mol P   4 2 4  mass of (NH4 )2 HPO4 18.6073 g (NH4 )2 HPO4 100%  = 100% mass of mixture 100 g mixture = 18.6073% = 18.6% (NH4)2HPO4

Mass % (NH4)2HPO4 =

In 18.6073 g (NH4)2HPO4, we must now determine the amount (mol) of N

 1 mol (NH4 )2 HPO4   2 mol N  0.2818 mol N Amount (mol) N = 18.6073 g  NH4 2 HPO4   132.06 g (NH ) HPO   1 mol (NH ) SO  4 2 4  4 2 4   The amount (mol) of N from (NH4)2SO4 = 0.713776 mol N – 0.28180 mol N = 0.43198 mol N To obtain 0.43198 mol N from (NH4)2SO4:  1 mol (NH 4 ) 2 SO 4   132.15 g (NH 4 ) 2 SO 4   0.43198 mol N     1 mol (NH ) SO  = 28.5428 g (NH4)2SO4 2 mol N   4 2 4  mass of (NH 4 )2 SO 4 28.5428 g (NH 4 ) 2 SO 4 100%  = 100%  mass of mixture 100 g mixture = 28.5428% = 28.5% (NH4)2SO4 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 % (NH4)2SO4 =

Mass % KCl = 3.169

mass of KCl 15.8277 g KCl 100%  = 100%  = 15.8277% = 15.8% KCl mass of mixture 100 g mixture

Plan: In combustion analysis, finding the amount (mol) of carbon and hydrogen is relatively simple because all of the carbon present in the sample is found in the carbon of CO 2, and all of the hydrogen present in the sample is found in the hydrogen of H2O. Convert the mass of CO2 to amount (mol) and use the ratio between CO2 and C to find the amount (mol) and mass of C present. Do the same to find the amount (mol) and mass of H from H 2O. Subtracting the masses of C and H from the mass of the sample gives the mass of Fe. Convert the mass of Fe to amount (mol) of Fe. Take the amount (mol) of C, H, and Fe and divide by the smallest value to convert to whole numbers to get the empirical formula. Solution: Ferrocene + ?O2(g)  CO2 + H2O 0.9437 g 2.233 g 0.457 g  1 mol CO 2  1 mol C  Amount (mol) of C =  2.233 g CO 2     = 0.050738 mol C  44.01 g CO 2  1 mol CO 2 

 12.01 g C  Mass (g) of C =  0.050738 mol C    = 0.60936 g C  1 mol C   1 mol H 2 O  2 mol H  Amount (mol) of H =  0.457 g H 2 O     = 0.050721 mol H  18.02 g H 2 O  1 mol H 2 O   1.008 g H  Mass (g) of H =  0.050721 mol H    = 0.051127 g H  1 mol H  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  Amount (mol) of Fe =  0.283213 g Fe    = 0.005071 mol Fe  55.85 g Fe 

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 → C10H10Fe1 0.005071

0.005071

Empirical formula = C10H10Fe 3.170

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 CO 2 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  28.01 g CO   1 mol CO   1 mol C   1 mol CO 2   1 mol C   12.01 g C  Mass (g) of C from CO2 =  65.0 g CO2      = 17.738 g C  44.01 g CO 2   1 mol CO 2   1 mol C  Total mass (g) of C = 15.007 g + 17.738 g = 32.745 g C mass of C 32.745 g C Mass % C = 100%  = 100%  = 32.745 %= 32.7% C mass of mixture 100 g mixture

3.171

Plan: Write a balanced equation for the reaction. Count the molecules of each reactant to obtain the amount (mol) 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 → SiF4 + N2 + 2H2.  1.25x102 mol   3 SiH 4 molecules    1 molecule  = 0.0375 mol SiH Amount (mol) of SiH4 = 4  1.25x102 mol   3 N 2 F4 molecules     1 molecule  = 0.0375 mol N F Amount (mol) of N2F4 = 2 4 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 SiF 4 produced.  1 mol SiF4  104.09 g SiF4   0.0375 mol SiH 4      1 mol SiH 4  1 mol SiF4  = 3.903375 g SiF Mass (g) of SiF4 = 4  actual Yield  % yield =   x 100%  theoretical Yield  Actual yield (g) of SiF4 =

% yield 87%  theoretical yield  =  3.903375 g SiF4  = 3.3959 g= 3.4 g SiF4 100% 100%

3.172

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 L 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 amount (mol). 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 x 12.01 g/mol C) + (8 x 1.008 g/mol H) + (7 x 16.00 g/mol O) = 192.12 g/mol b) Converting volume of lemon juice in L to mL:  1 mL  Volume (mL) of lemon juice = 1.50 L   3  = 1.50 x 103 mL  10 L    Converting volume to mass in grams:  1.09 g  Mass (g) of lemon juice = 1.50  103 mL   = 1635 g lemon juice  mL 

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

 6.82% C6 H8 O7  Mass (g) of C6H8O7 = 1635 g lemon juice   = 111.507 g C6H8O7  100% lemon juice   1 mol C6 H8 O7  Amount (mol) of C6H8O7 = 111.507 g C6 H8 O7    = 0.580403 mol= 0.580 mol C6H8O7  192.12 g C6 H8 O7 

3.173

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 amount (mol) 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) 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 x (N2(g) + O2(g)  2NO(g)) = 2N 2(g) + 2O2(g)  4NO(g) 3 x (2NO(g) + O2(g)  2NO2(g)) = 6NO(g) + 3O2(g)  6NO2(g) 2 x (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 NO 2 on each side canceling. Adding the equations gives: 2N2(g) + 5O2(g) + 2H2O(g)  4HNO3(aq)  103 kg  103 g  c) Mass (g) of N2 = 1350 t N 2   = 1.35x109 g N2  1t   1 kg      1 mol N 2  7 Moles of N2 = 1.35x109 g N 2   = 4.817987x10 mol N2  28.02 g N 2 

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 4 mol HNO3   63.02 g HNO3  9 Mass (g) of HNO3 = 4.817987x107 mol N 2   = 6.072591x10 g HNO3   2 mol N 2   1 mol HNO3  Mass (metric tons) HNO3 =  1 kg  1t  = 6.072591x103 t= 6.07x103 t HNO3 6.072591x109 g HNO3  3  3  10 g  10 kg    

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3.174

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) → 2NO2(g) 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 NO 2 will be produced. Circle A shows the formation of 4 molecules of NO2. Circle B also shows the formation of 4 molecules of NO 2 but also has 2 unreacted molecules of NO and 1 unreacted molecule of O 2. Since neither reactant is limiting, there will be no unreacted reactant remaining after the reaction is over.

3.175

Plan: For parts a) and b), convert the masses to amount (mol). Take the amount (mol) 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  a) Amount (mol) of Pt =  0.327 g Pt    = 0.001676 mol 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  Amount (mol) of F =  0.192 g F    = 0.010105 mol F  19.00 g F  Preliminary formula = Pt0.001676F0.010105 Converting to integer subscripts (dividing all by the smallest subscript): Pt 0.001676 F0.010105 → Pt1F6 0.001676 0.001676

Empirical formula = PtF6  1 mol PtF6  b) Amount (mol) of PtF6 =  0.265 g PtF6    = 0.0008576 mol PtF6  309.1 g PtF6  Mass of Xe = mass of product – mass of Xe = 0.378 g – 0.265 g = 0.113 g Xe  1 mol Xe  Amount (mol) of Xe =  0.113 g Xe    = 0.0008606 mol 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 → Xe1(PtF6)1 0.0008576

0.0008576

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 Amount (mol) of Xe consumed = 1.85x10–4 mol present – 9.00x10–6 mol excess = 1.76x10–4 mol Xe Then x + y = 1.76x10–4 mol Xe consumed 2x + 3y = 5.00x10–4 mol F2 consumed Solve for x using the first equation and substitute the value of x into the second equation: x = 1.76x10–4 – y 2(1.76x10–4 – y) + 3y = 5.00x10–4 (3.52x10–4) – 2y + 3y = 5.00x10–4 y = (5.00x10–4) – (3.52x10–4) = 1.48x10–4 mol XeF6 x = (1.76x10–4) – (1.48x10–4) =2.8x10–5 mol XeF4 Converting amount (mol) of each product to grams using the molar masses:  207.3 g XeF4  –3 Mass (g) of XeF4 = 2.8x105 mol XeF4   = 5.8044x10 g XeF4  1 mol XeF4 

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

 245.3 g XeF6  –2 Mass (g) of XeF6 = 1.48x104 mol XeF6   = 3.63044x10 g XeF6 1 mol XeF 6   Calculate the percent of each compound using the total weight of the products: (5.8044 x 10–3 + 3.63044x10–2) g = 0.0421088 g

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3.176



Mass % XeF4 =

mass of XeF4 5.8044x103 g XeF4 100%  = 100%  = 13.784 %= 14% XeF4 total mass 0.0421088 g

Mass % XeF6 =

mass of XeF6 3.63044x102 g XeF6 100%  = 100%  = 86.2157 %= 86.2% XeF6 total mass 0.0421088 g

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 amount (mol). 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 a) Mass (g) of heme =  0.65 g hemoglobin    = 0.039 g heme 100% hemoglobin    1 mol heme  –5 –5 b) Amount (mol) of heme =  0.039 g heme    = 6.32614x10 mol= 6.3x10 mol heme 616.49 g heme    1 mol Fe   55.85 g Fe  c) Mass (g) of Fe = 6.32614x105 mol heme     1 mol heme   1 mol Fe  –3 –3 = 3.5331x10 g= 3.5x10 g Fe  1 mol hemin   651.94 g hemin  d) Mass (g) of hemin = 6.32614x105 mol heme     1 mol heme   1 mol hemin  = 4.1243x10–2 g= 4.1x10–2 g hemin







3.177



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 amount (mol) 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 part b), the amount (mol) of Mn 3+ come from the Mn2O3 and the amount (mol) of Mn2+ 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 Mass % braunite = 100%  = 100%  mass of braunite + manganosite 140.338 + 7.88221 g = 94.6821% mass of manganosite 7.88221 g Mass % manganosite = 100%  = 100%  mass of braunite + manganosite 140.338 + 7.88221 g = 5.3179% In the 542.3 g sample:  94.6821 braunite  Mass (g) of braunite =  542.3 g sample    = 513.461 g= 513 g braunite  100% sample   5.3179% manganosite  Mass (g) of manganosite =  542.3 g sample    = 28.839 g= 28.8 g manganosite 100% sample   b) Each mole of braunite, Mn2O3, contains 2 moles of Mn3+ while each mole of manganosite, MnO, contains 1 mole of Mn2+. Amount (mol) of Mn3+ = 2(0.888889 mol braunite) = 1.777778 mol Mn 3+ Amount (mol) of Mn2+ = 1(0.111111 mol manganosite) = 0.111111 mol Mn2+

Mn3+:Mn2+ = 3.178

1.777778 mol Mn 3  0.111111 mol Mn 2 

= 16.000 = 16.0

Plan: Determine the formula and the molar mass of each compound. The formula gives the relative number of amount (mol) of nitrogen present. Multiply the amount (mol) of nitrogen by its molar mass to find the total mass total mass of element of nitrogen in 1 mole of compound. Mass percent = 100%  . For part b), convert mass molar mass of compound of ornithine to amount (mol), use the mole ratio between ornithine and urea to find the moles of urea, and then use the ratio between amount (mol) of urea and nitrogen to find the amount (mol) and mass of nitrogen produced. Solution: a) Urea: CH4N2O, M = 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 % molar mass of compound 60.06 g CH 4 N 2 O = 46.65% N in urea Arginine: C6H15N4O2, M = 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 

Mass percent =

total mass N 56.04 g N 100%  = 100%  molar mass of compound 175.22 g C6 H15 N 4 O 2 = 31.98265 %= 31.98% N in arginine Ornithine: C5H13N2O2, M = 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 

Mass percent =

total mass N 28.02 g N 100%  = 100%  molar mass of compound 133.17 g C5 H13 N 2 O 2 = 21.04077 %= 21.04% N in ornithine

Mass percent =

 1 mol C5 H13 N 2 O 2   1 mol CH 4 N 2 O  b) Amount (mol) of urea = 135.2 g C5 H13 N 2 O2      133.17 g C5 H13 N 2 O2   1 mol C5 H13 N 2 O 2  = 1.015244 mol urea    14.01 g N  2 mol N Mass (g) of nitrogen = 1.015244 mol CH 4 N 2 O     = 28.447 g= 28.45 g N 1 mol CH N O 4 2   1 mol N  

3.179

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. Solution: a) Finding the amount (mol) of aspirin from the amount (mol) of C 7H6O3 (if (CH3CO)2O is limiting):  1 mol C7 H6 O3   1 mol C9 H8 O4  Amount (mol) of aspirin from C7H6O3 =  3.077 g C7 H6 O3      138.12 g C7 H6 O3   1 mol C7 H6 O3  = 0.0222777 mol C9H8O4 Finding the amount (mol) of aspirin from the amount (mol) of C 4H6O3 (if C7H6O3 is limiting):  1.080 g  Mass (g) of (CH3CO)2O =  5.50 mL (CH3CO)2 O    = 5.94 g (CH3CO)2O  1 mL  Amount (mol) of aspirin from (CH3CO)2O  1 mol (CH 3CO) 2 O  1 mol C9 H 8O 4  =  5.94 g (CH 3CO) 2 O      102.09 g (CH 3CO) 2 O  1 mol (CH 3CO) 2 O 

= 0.058183955 mol C9H8O4 The limiting reactant is C7H6O3. b) First, calculate the theoretical yield from the limiting reagent:  1 mol C7 H6 O3   1 mol C9 H8 O4   180.15 g C9 H8 O4  Mass (g) of C9H8O4 =  3.077 g C7 H6 O3       138.12 g C7 H6 O3   1 mol C7 H6 O3   1 mol C9 H8 O4  = 4.01333 g C9H8O4  3.281 g   actual Yield  Percent yield =   x 100% = 81.7526 %= 81.75% yield  x 100% =  theoretical Yield    4.01333 g  c) % atom economy = % atom economy =

no. of moles x molar mass of desired products x 100% sum of  no. of moles x molar mass  for all products

1mol 180.15 g / mol  x 100% = 75.00% atom economy 1 mol 180.15 g/mol  + 1 mol  60.05 g/mol 

3.180

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 x molar mass of desired products % atom economy = x 100% sum of  no. of moles x molar mass  for all products =

1 mol  molar mass of N 2 H 4  1 mol  molar mass of N 2 H 4  + 1 mol  molar mass of NaCl  + 1 mol  molar mass of H 2O 

1mol  32.05 g / mol x 100% 1 mol  32.05 g/mol + 1 mol 58.44 g/mol +  1 mol 18.02 g/mol

x 100%

= 29.5364 %= 29.54% atom economy 3.181

Plan: Use the mass percent to obtain the mass of lead(II) chromate needed. Use the mole ratio between PbCrO 4 and K2CrO4 and then the mole ratio between K2CrO4 and FeCr2O4 to find the amount (mol) and then mass of FeCr2O4 needed to produce that mass of PbCrO4. Solution:  103 g   0.511%  Mass (g) of PbCrO4 in 1 kg of paint = 1 kg paint   = 5.11 g PbCrO4  kg   100%     1 mol PbCrO 4   1 mol K 2 CrO 4  Amount (mol) of K2CrO4 =  5.11 g PbCrO4     = 0.015811 mol K2CrO4  323.2 g PbCrO 4   1 mol PbCrO 4   4 mol FeCr2 O 4   223.85 g FeCr2O 4  Mass of FeCr2O4 =  0.015811 mol K 2CrO 4      8 mol K 2 CrO 4   1 mol FeCr2 O 4  = 1.769606 g= 1.77 g FeCr2O4

3.182

Plan: Convert the mass of ethanol to amount (mol), 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 CH 3CH 2 OH  1 mol CH 3CH 2 OCH 2 CH 3  74.12 g CH 3CH 2OCH 2CH 3   50.0 g CH3CH 2OH      2 mol CH 3CH 2 OH  46.07 g CH 3CH 2 OH   1 mol CH 3CH 2OCH 2CH 3  = 40.2214 g diethyl ether Determining the percent yield:  actual yield   35.9 g  Percent yield =   x 100% =   x 100% = 89.2560 %= 89.3% 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. 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 CH3CH 2 OCH 2 CH3   46.07 g CH 3CH 2 OH  2 mol CH3CH 2 OH  4.3214 g CH3CH 2OCH 2CH3      74.12 g CH CH OCH CH 1 mol CH CH OCH CH 3 2 2 3  3 2 2 3  1 mol CH 3CH 2 OH  

= 5.37202 g ethanol  45.0%  Mass of ethanol producing ethylene =  5.37202 g CH3CH 2 OH    = 2.417409 g ethanol  100% 

 1 mol CH3CH 2 OH  1 mol C2 H 4  28.05 g C2 H 4  Mass (g) of ethylene =  2.417409 g CH3CH 2 OH       46.07 g CH3CH 2 OH  1 mol CH3CH 2 OH   1 mol C 2 H 4  = 1.47185 g= 1.47 g ethylene 3.183

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 amount (mol) of S8 in the reactant and the amount (mol) of S8 in the product ZnS. The difference between these two amounts is the amount (mol) of S 8 in SO2. Solution: a) The balanced equation is 8Zn(s) + S8(s)  8ZnS(s). Finding the limiting reagent: Finding the amount (mol) of ZnS from the amount (mol) of Zn (if S 8 is limiting):  1 mol Zn   8 mol ZnS  Amount (mol) of ZnS from Zn =  83.2 g Zn     = 1.27198 mol ZnS  65.41 g Zn   8 mol Zn  Finding the amount (mol) of ZnS from the amount (mol) of S8 (if Zn is limiting):  1 mol S8  8 mol ZnS  Amount (mol) of ZnS from S8 =  52.4 g S8     = 1.6339 mol ZnS  256.56 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  97.48 g ZnS  Theoretical yield (g) of ZnS =  83.2 g Zn       65.41 g Zn   8 mol Zn  1 mol ZnS  = 123.9923 g ZnS (unrounded)  104.4 g   actual Yield  Percent yield =   x 100% = 84.1988 %= 84.2% yield  x 100% =   123.9923 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   81.41 g ZnO  Mass (g) of ZnO = 13.1456 g Zn      = 16.3612 g= 16.4 g ZnO  65.41 g Zn   2 mol Zn   1 mol ZnO  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  Amount (mol) of S8 in original S8 reactant =  52.4 g S8    = 0.204241 mol S8  256.56 g S8   1 mol ZnS   1 mol S8  Amount (mol) of S8 in ZnS product = 104.4 g ZnS    = 0.133874 mol S8  97.48 g ZnS   8 mol ZnS  Silberberg, Amateis, Venkateswaran, Chen, Chemistry: The Molecular Nature of Matter and Change, 3rd Canadian Edition Page 3-138 Instructor‘s Solution Manual © Copyright 2021 McGraw-Hill Ryerson Ltd.

Amount (mol) of S8 in SO2 = 0.204241 – 0.133874 ml = 0.070367 mol S8  8 mol SO 2   64.07 g SO 2  Mass (g) of SO2 =  0.070367 mol S8     = 36.0673 g= 36.1 g SO2  1 mol S8   1 mol SO 2  3.184 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 amount (mol) of salt and then to amount (mol) 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 H 2 O   2.50 kg salt   103 g  a) Mass (g) of dissolved salt =  50.0 mL H 2 O   = 125 g salt  1 mL H O   1 L H O   1 kg  2 2      1 mol salt  b) Amount (mol) of dissolved salt = 125 g salt    = 0.367853 mol 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 Volume (L) of water needed to dissolve the cocaine = 111.588 g cocaine    = 65.64 L  1.70 g cocaine  Additional water needed = total volume needed – original volume of water = 65.64 L – 0.0500 L = 65.59 L= 65.6 L H2O

3.185

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 Cu 3O7  Amount (mol) of product from BaCO3 =  x g BaCO3      197.3 g BaCO3   4 mol BaCO3  = 0.002534x mol product  1 mol CuO   2 mol YBa 2 Cu 3O7  Amount (mol) of product from CuO =  x g CuO     6 mol CuO  79.55 g CuO    = 0.004190x mol product  1 mol Y2 O3   2 mol YBa 2 Cu 3O 7  Amount (mol) of product from Y2O3 =  x g Y2 O3     1 mol Y2 O3  225.82 g Y2 O3    = 0.008857x mol product BaCO3 is the limiting reactant. c) These calculations are based on the limiting reactant. BaCO3 remaining = 0% (limiting reagent)  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  Percent CuO =   x 100% = 39.521 %= 39.52% CuO remaining 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 Y2O3  = 0.713862x g Y2O3  0.713862 x g  Percent Y2O3 =   x 100% = 71.3862 %= 71.39% Y2O3 remaining xg  

3.186

Plan : We will use the mass of carbon dioxide produced to find the amount in mol and the mass of carbon. Similarly, we will use the mass of water to find the amount in mol of water, then hydrogen and the mass of hydrogen. We will use the mass of the compound and the masses of carbon and hydrogen to find the mass of oxygen by difference, and then amount in mol of oxygen. We will use the mol amounts of C, H, and O to find the empirical formula of sorbitol. We will finally use the molar mass and the mass of the empirical unit to find the number of units and thus the molecular formula of sorbitol. Solution:

nC  nCO2 

mCO2 M CO2



0.7871 g  1.7885×10-2 mol C 44.01 g/mol

mC  nC  M C  (1.7885 102 mol C)(12.01 g/mol)= 0.2148 g nH2O 

mH2O M H2O



0.3756 g  2.0855×10-2 mol H 2O 18.01 g/mol

nH  2nH 2O  2  2.0855×10-2 mol H 2O = 4.1710×10-2 mol H mH  nH  M H  (4.1710×10-2 mol H)(1.01 g/mol)= 4.213×10-2 g mO  msorbitol  mC  mH  0.5431g  0.2148g  4.213 10 2 g= 0.2862g nO 

mO 0.2862g   1.7886 10 2 mol M O 16.00g/mol

Making the amount (mol) of each element the subscript in the formula gives C1.7885102 H 4.1710102 O1.7886102 Dividing by the smallest number gives

C1.7885102 H 4.1710102 O1.7886102 1.7886102

1.7886102

C1H 2.33O1 Multiplying through by 3 (since 0.33  1/3) C3 H 7 O 3 The empirical formula of sorbitol is C3 H 7O3 . The molecular formula will be some multiple of the empirical formula. To find this multiple, we simply divide the molar mass of sorbitol by the mass of the empirical unit.

M empirical unit  3  M C  7  M H  3  M O

 3(12.01 g/mol)+7(1.01 g/mol)+3(16.00 g/mol)=91.1 g/mol

Number of units 

M sorbitol 182.17 g/mol  2 M empirical unit 91.1 g/mol

Therefore, the molecular formula of sorbitol is found by multiplying each of the subscripts in the empirical formula by the number of units. The molecular formula of sorbitol is C6 H 14O 6 . 3.187

Plan: Lead and sulfur are already balanced, however, we have 3 O atoms on the right and only 2 on the left. Doubling the reaction will give us whole number coefficients. Once the reaction is balanced, we check to see which reactant gives us the least amount of product. That will be the limiting reagent. Once we know the limiting reagent, we can determine what amount (mol) of PbO will be produced and can convert it to mass. Finally, we can use the actual amount of PbO produced and the theoretical amount of PbO that should be produced to find the percent yield. Solution: a)

2PbS( s) + 3O2 ( g )  2PbO( s) + 2SO2 ( g) M PbS  207.2 g/mol + 32.01 g/mol=239.21 g/mol M O2  2(16.00) g/mol=32.00 g/mol

nPbS  nO2 

mPbS 125 103 g   522.6 mol M PbS 239.21 g/mol mO2 M O2



19.0  103 g  593.8 mol 32.00 g/mol

From the reaction stoichiometry, 2 mol of PbS produces 2 mol of PbO. Therefore, 522.6 mol of PbS would produce 522.6 mol of PbO. From the reaction stoichiometry, 3 mol of O2 produces 2 mol of PbO. Therefore, 593.8 mol of O 2 would produce 2/3 (593.8)=395.9 mol of PbO. Since O2 produces LESS product than PbS, O2 is the limiting reagent. c)

nPbO produced =395.9 mol

mPbO produced  nPbO  M PbO  (395.9 mol)(223.2 g/mol)=8.84  104 g = 88.4 kg d) 3.188

% yield=

actual yield 75.3 kg 100%  100%  85.2% theoretical yield 88.4 kg

Plan: Carbon is already balanced. Next we balance oxygen and finally hydrogen. Once the reaction is balanced, we check to see which reactant gives us the least amount of product. That will be the limiting reagent. Once we know the limiting reagent, we can determine what amount (mol) of H 2 (g) will be produced and can convert it to mass. Finally, we can use the percent yield and the theoretical amount of H 2 (g) that should be produced to find the actual amount of H2 (g) produced. Solution: a) b)

CH4 ( g ) + 2H2O( g )  4H 2 ( g ) + CO2 ( g )

M CH4  12.01 g/mol + 4(1.01 g/mol)=16.05 g/mol M H2O  2(1.01) g/mol+16.00 g/mol=18.02 g/mol nCH4 

mCH4 M CH4



137 103 g  8.54 103 mol 16.05 g/mol

137 103 g nH2O    7.60  103 mol M H2O 18.02 g/mol mH2O

From the reaction stoichiometry, 1 mol of CH4 produces 4 mol of H2 (g). Therefore, 8.54x103 mol of CH4 would produce 3.42x104 mol of H2 (g). From the reaction stoichiometry, 2 mol of H2O produces 4 mol of H2 (g). Therefore, 7.60x103 mol of H2O would produce 2(7.60x103)=1.52x104 mol of H2 (g). Since CH4 produces MORE product than H2O, H2O is the limiting reagent. c)

nH2 (g) produced  1.52 104 mol

mH2 (g) produced  nH2 (g)  M H2 (g)  (1.52 104 mol)(2.02 g/mol)=3.07  104 g = 30.7 kg d)

% yield=

actual yield theoretical yield

actual yield  3.189

 theoretical yield  % yield   (30.7 kg)(73.2%)  22.5 kg 100

100

Plan : We will use the mass of carbon dioxide produced to find the amount in mol and the mass of carbon. Similarly, we will use the mass of water to find the amount in mol of water, then hydrogen and the mass of hydrogen. We will use the mass of the compound and the masses of carbon and hydrogen to find the mass of oxygen by difference, and then amount in mol of oxygen. We will use the mol amounts of C, H, and O to find the empirical formula of citronellal. We will finally use the molar mass and the mass of the empirical unit to find the number of units and thus the molecular formula of citronellal. Solution:

nC  nCO2 

mCO2 M CO2



5.2469 g  0.11922 mol C 44.01 g/mol

mC  nC  M C  (0.11922 mol C)(12.01 g/mol)=1.4318 g

nH2O 

mH2O M H2O



1.9318 g  0.10726 mol H 2O 18.01 g/mol

nH  2nH2O  2  0.10726 mol H 2 O = 0.21453 mol H mH  nH  M H  (0.21453 mol H)(1.01 g/mol)=0.21667 g

mO  mcitronellal  mC  mH  1.8392g  1.4318g  0.21667g= 0.1907g nO 

mO 0.1907g   1.192 102 mol M O 16.00g/mol

Making the amount (mol) of each element the subscript in the formula gives C0.11922 H 0.21453O1.192102 Dividing by the smallest number gives

C 0.11922 H 0.21453 O1.192102 1.192102

1.192102

C10 H18O The empirical formula of citronellal is C10 H 18O . The molecular formula will be some multiple of the empirical formula. To find this multiple, we simply divide the molar mass of citronellal by the mass of the empirical unit.

M empirical unit  10  M C  18  M H  M O

 10(12.01 g/mol)+18(1.01 g/mol)+(16.00 g/mol)=154.28 g/mol M citronellal 154.25 g/mol Number of units   1 M empirical unit 154.28 g/mol Therefore, the molecular formula of citronellal is the same as the empirical formula. The molecular formula of citronellal is C10 H 18O . 3.190

Plan: Write the known reactants and products and find out what is missing. Write the balanced reaction. Use stoichiometry to find the mass of the missing product. Solution: (NH4)2Cr2O7 (s)  Cr2O3(s) + 4 H2O (g) + ? We know that one of the products is water vapour. How did we get the 4 stoichiometric coefficient? If we just look at the reaction and remove 2 Cr and 3 O from the reactant, we are left with (NH 4)2, and 4 O. 4 O and 8 H give us exactly 4 H2O. We are left with 2 N. So the missing product must be N 2. a)

(NH4)2Cr2O7 (s)  Cr2O3(s) + 4 H2O (g) + N2 (g) mreactant = 575.0 g MMreactant = 252.092 g/mol

= 2.281 mol ( 3.191

)(

)

Plan: Write the balanced chemical equation. Determine the amount (mol) of both reagents. Determine the limiting reagent (LR). Calculate the mass of the product. Use the theoretical mass and actual mass to calculate the percentage yield. Solution: a) b)

2 NH3 (g) + 3 CuO (s)  3 Cu (s) + 3 H2O (g) + N2 (g)

LR Thus CuO is the limiting reagent and will be used to determine amount of product and percent yield. 3 CuO : 1 N2 1.16 : x 3 x = 1.16

x = 1.16/3 = 0.387 mol N2 (

3.192

)(

)

If the actual amount of N2 formed is 6.32 g, the percent yield is

Plan: Use the mass of CO2 formed to find amount (mol) of CO 2 and then amount (mol) C, then mass C. Use the mass of H2O to find amount (mol) of H2O and then with stoichiometry, amount (mol) H and mass H. Find mass O by difference, using masses of C, H and the sample. Find amount (mol) O. Use the amount (mol) of C, H, and O to find the empirical formula by placing the amount (mol) as subscripts and dividing by the smallest one. Calculate the mass of the empirical unit. Calculate the MM of the compound by dividing the mass of the sample by the amount (mol) of the sample. Use the mass of the empirical unit and the molar mass of the compound to find the molecular formula. Solution:

(

)( (

) )

(

)(

) O O

C6.5H9O  multiplying by 2  C13H18O2

Mempformula = 206 g/mol

The molar mass can be found using the mass of the compound and the corresponding amount in moles.

Therefore, since the M of the empirical formula and the MM of the compound are the same, the empirical formula and the molecular formula are both C 13H18O2.

CHAPTER 4 GASES AND THE KINETICMOLECULAR THEORY CHEMICAL CONNECTIONS BOXED READING PROBLEMS B4.1

Plan: Examine the change in density of the atmosphere as altitude changes. 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.

B4.2

Plan: The conditions that result in deviations from ideal behaviour 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 behaviour on Saturn.

B4.3

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 x potal. The mole fraction of argon is given in Table B4.1. Solution: Volume percent = mole fraction x 100 = 0.00934 x 100% = 0.934 % The total pressure at sea level is 101.3 kPa pAr = XAr x ptotal = 0.00934 x 101.3 kPa = 0.946142 .kPa = 946 Pa

B4.4

Plan: To find the amount (mol) of gas, convert the mass of the atmosphere from mt to g and divide by the molar mass of air. Knowing the amount in 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) Amount (mol) of gas = 5.14x1015 t      1 t   1 kg  28.8 g 





= 1.78472x1020 mol= 1.78x1020 mol b) pV = nRT

L•kPa  1.78472 x10 mol   8.31446 mol•K    273  25  K   20

nRT V= = P

101.3 kPa 

= 4.36527x1021 L= 4.4 x 1021 L

END–OF–CHAPTER PROBLEMS 4.1

Plan: Review the behaviour 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.

4.2

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 pushed closer together and thus 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.

4.3

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.

4.4

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.

4.5

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.

4.6

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: hH 2O d Hg  hHg d H2O

hH2O  4.7

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: hH 2O d Hg  hHg d H2O

hH2O  4.8

 103 m  1 cm   13.5 g/mL  x hHg =  = 985.5 cm= 990 cm H2O    730 mmHg     2  d H2O  1.00 g/mL   1 mm   10 m  d Hg

d Hg d H2O

 13.5 g/mL  4 x hHg =    755 mmHg  = 10,192.5 mm= 1.02x10 mm H2O 1.00 g/mL  

Plan: Use the conversion factors between pressure units: 1 atm = 760 mmHg = 760 Torr = 101.325 kPa = 1.01325 bar = 1.01325x10 5 Pa Solution:  101.325 kPa  a) Converting from atm to kPa: p(kPa) =  0.745 atm    = 75.4871 kPa = 75.5 kPa 1 atm  

 1.01325 bar  b) Converting from Torr to bar: p(bar) =  992 Torr    = 1.32256 bar= 1.32 bar  760 Torr   1000 Pa  4 4 c) Converting from kPa to Pa: p(Pa) =  36.5 kPa    = 3.6500 x10 Pa = 3.65x10 Pa  1 kPa 

 101.325 kPa  d) Converting from mmHg to kPa: p(kPa) =  804 mmHg    = 107.191 kPa= 107 kPa  760 mmHg  4.9

Plan: Use the conversion factors between pressure units: 1 atm = 760 mmHg = 760 Torr = 101.325 kPa = 1.01325 bar = 1.01325x105 Pa Solution: a) Converting from mmHg to Pa:  1.01325x105 Pa  p(Pa) =  768 mmHg   = 1.0239105 Pa= 1.02105 Pa  760 mmHg   

 101.325 kPa  3 3 b) Converting from atm to kPa: p(kPa) =  27.5 atm    = 2.786x10 kPa= 2.79x10 kPa 1 atm    1.01325 bar  c) Converting from atm to bar: p(bar) =  6.50 atm    = 6.5861 bar= 6.59 bar  1 atm   1.01325 bar  d) Converting from kPa to bar: p(bar) =  0.937 kPa    = 9.37x10 ³ bar = 9.37 mbar 101.325 kPa   4.10

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 pressure corresponding to the height difference (Δh) 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 bar. Solution:  10 mm   1 Torr   2.35 cm      = 23.5 Torr  1 cm   1 mmHg  738.5 Torr – 23.5 Torr = 715.0 Torr  1.01325 bar  p(bar) =  715.0 torr    = 0.9532540789 bar= 0.953 bar  760 torr 

4.11

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 corresponding to the height difference (Δh) 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 mm  1.30 cm    = 13.0 mmHg  1 cm  765.2 mmHg – 13.0 mmHg = 752.2 mmHg  101.325 kPa  p(kPa) =  752.2 torr    = 100.285 kPa= 100.3 kPa  760 torr 

4.12

Plan: This is a closed-end manometer. The difference in the height of the Hg (Δh) equals the gas pressure. The height difference is given in units of m and must be converted to mmHg and then to bar. Solution:  1 mmHg   1.01325 bar  p(bar) =  0.734 mHg   3 = 0.97859 bar = 0.979 bar  10 mHg   760 mmHg    

4.13

Plan: This is a closed-end manometer. The difference in the height of the Hg (Δh) 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   1.01325x105 Pa  p(Pa) =  3.56 cm   = 4746.276 Pa= 4.75x103 Pa  1 cmHg   103 mHg   760 mmHg     

4.14

Plan: Use the conversion factors between pressure units: 1 atm = 760 mmHg = 760 Torr = 1.01325x10 5 Pa = 1.01325 bar Solution:  1.01325x105 Pa  a) Converting from mmHg to Pa: p(Pa) = 2.75x102 mmHg  = 36,663.65 Pa = 3.67 x 104 Pa  760 mmHg   





 1.01325x105 Pa  b) Converting from bar to Pa: p(Pa) = 1.14 bar   = 114,000 Pa = 1.14x 105 Pa  1.01325 bar     1000 Pa  6 c) Converting from kPa to Pa: p(Pa) = 9.15x103 kPa   = 9,150,000 Pa = 9.15x10 Pa  1 kPa 





 1.01325x105 Pa  6 d) Converting from Torr to Pa: p(Pa) = 2.54x104 Torr   = 3,386,388 Pa = 3.39x10 Pa  760 Torr  



4.15



Plan: 1 atm = 1.01325x105 Pa = 1.01325x105 N/m2. So the force on 1 m2 of ocean is 1.01325x105 N where kg•m 1 N = 1 2 . Use F = Mg to find the mass of the atmosphere in kg/m2 for part a). For part b), convert this mass s to g/cm2 and use the density of osmium to find the height of this mass of osmium. Solution: a) F = Mg 1.01325x105 N = Mg kg•m 1.01325 x 105 2 = (Mass) (9.81 m/s2) s Mass = 1.03287x104 kg= 1.03x104 kg 2

kg   103 g   10 2 m   3 2 b)  1.03287x104 2      = 1.03287x10 g/cm (unrounded)  1 kg 1 cm m    

g   1 mL   1 cm3   Height =  1.03287x103  = 45.702 cm= 45.7 cm Os   cm2   22.6 g   1 mL   4.16

The statement is incomplete with respect to temperature and mass of sample. The correct statement is: At constant temperature and amount (mol) of gas, the volume of gas is inversely proportional to the pressure.

4.17

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 amount (mol) b) Avogadro‘s law: At fixed temperature and pressure, the volume occupied by a gas is directly proportional to the amount (mol) of gas. Variable: volume and amount (mol); 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 amount (mol)

4.18

Plan: Examine the ideal gas law; volume and temperature are constant and pressure and amount in moles are variable. Solution: RT pV = nRT R, T, and V are constant p= n V p = n x constant At constant temperature and volume, the pressure of the gas is directly proportional to the amount of gas in moles.

4.19

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: p 1V1 p 2V2 V1 V = 2 = a) p is fixed; both V and T double: 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. p1V1 pV p1V1 pV = 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. p 1V1 p 2V2 V1 V = = 2 or n1T1 n2T2 T1 T2 V and T are directly proportional so as V is doubled, T is doubled. 4.20

Plan: Use the relationship

p1V1 pV pV n T = 2 2 or V2 = 1 1 2 2 . n1T1 n2T2 p2 n1T1

Solution: a) As the pressure on a fixed amount of gas (n is fixed, so n1=n2) increases at constant temperature (T is fixed so T1=T2), the molecules move closer together, decreasing the volume. When the pressure is tripled, the volume decreases to one-third of the original volume at constant temperature (Boyle‘s law). V2 =

(p1 )(V1 ) n2 T2 p1V1n2T2 = p2 n1T1 (3 p1 ) n1 T1

V2 = ⅓V1

b) As the temperature of a fixed amount of gas (n is fixed, so n1=n2) increases at constant pressure (p is fixed, so p1=p2), the gas molecules gain kinetic energy. With higher energy, the gas molecules collide with the walls of the container with greater force, which increases the size (volume) of the container. If the temperature is increased by a factor of 3.0 (at constant pressure) then the volume will increase by a factor of 3.0 (Charles‘s law). V2 =

p1 (V1 ) n2 (3T1 ) p1V1n2T2 = p2 n1T1 p2 n1 (T1 )

V2 = 3V1

c) As the number of molecules of gas increases at constant pressure and temperature (p and T are fixed, p1=p2 and T1=T2), the force they exert on the container increases. This results in an increase in the volume of the container. Adding 3 moles of gas to 1 mole increases the amount in moles by a factor of 4, thus the volume increases by a factor of 4 (Avogadro‘s law). V2 =

p1 (V1 )(4n1 ) T2 p1V1n2T2 = p2 n1T1 p2 (n1 ) T1

V2 = 4V1

4.21

p1V1 pV = 2 2 T1 T2

Plan: Use the relationship

p1V1T2 . R and n are fixed. p2T1

or V2 =

Solution: a) 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 (T1=T2) (Boyle‘s law). V2 =

(p1 )(V1 ) T2 p1V1T2 = p2T1 (1/ 4 p1 ) T1

V2 =4V1

b) 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 ½. As the temperature of a fixed amount of gas (n is fixed) decreases by a factor of ½ (from 310 K to 155 K) at constant pressure, the volume decreases by a factor of ½. The changes in pressure and temperature combine to decrease the volume by a factor of 4. p1 = 1.01 bar = 101 kPa T1 = 37°C + 273 = 310 K V2 =

p1V1T2 (101 kPa)(V1 )(155 K) = p2T1 (202 kPa)(310 K)

V2 = 1 4 V1

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 2, the volume increases by a factor of 2 at constant temperature (Boyle‘s law). T2 = 32°C + 273 = 305 K p2 = 10 1kPa = 1 bar (to 1 sf) V2 =

4.22

p1V1T2 (2 bar)(V1 )(305 K) = p2T1 (1 bar)(305 K)

p1V1 pV = 2 2 T1 T2

Plan: Use the relationship

V2 = 2V1

p1V1T2 . R and n are fixed (n1=n2). p2T1

or V2 =

Solution: a) The temperature is decreased by a factor of 2, so the volume is decreased by a factor of 2 (Charles‘s law) when p is constant (p1=p2). V2 =

p1 (V1 )(400 K) p1V1T2 = p2T1 p2 (800 K)

V 2 = ½ V1

b) T1 = 250°C + 273 = 523 K T2 = 500°C + 273 = 773 K The temperature increases by a factor of 773/523 = 1.48, so the volume is increased by a factor of 1.48 (Charles‘s law).

V2 =

p1 (V1 )(773 K) p1V1T2 = p2T1 p2 (523 K)

V2 = 1.48V1

c) The pressure is increased by a factor of 3, so the volume decreases by a factor of 3 (Boyle‘s law). V2 =

4.23

(2 bar)(V1 ) T2 p1V1T2 = p2T1 (6 bar) T1

p1V1 pV = 2 2 n1T1 n2T2

Plan: Use the relationship

V2 = ⅓V1

or V2 =

p1V1n2T2 . p2 n1T1

Solution: a) Since the amount in moles of gas is decreased by a factor of 2, the volume would be decreased by a factor of 2 (Avogadro‘s law). 1

p1 (V1 )( n1 ) T2 pV n T 2 V2 = 1 1 2 2 = p2 n1T1 p2 (n1 ) T1 b)

V2 = ½V1

p1 = 72.2 kPa

 100 kPa   =95.0 kPa  1 bar 

p2 =  0.950 bar  

T1 = 90.°C +273K = 363 K

T2 = 273 K

n1 = n2 V2 =

p1V1n2T2 (72.2 kPa)(V1 )(n1 )(273 K ) = 0.5715 V1 = 0.572 V1 = p2 n1T1 (95.0 kPa)(n1 )(363 K )

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. V2 =

4.24

(p1 )(V1 ) n2 ( 1 4 T1 ) p1V1n2T2 = p2 n1T1 ( 1 4 p1 ) n1 (T1 )

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 Converting T from °C to K: T1 = 198°C + 273 = 471K Arranging the ideal gas law and solving for T2: p 1V1 p 2V2 V1 V or =  2 n1T1 n2T2 T1 T2 T2  T1

4.25

V2 = 471 K  2.50 L  = 129.396 K – 273 = –143.604 °C= –144°C V1  9.10 L 

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 kelvin. 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 = 418 K T2 = –22°C + 273 = 251 K Arranging the ideal gas law and solving for V2: p 1V1 p 2V2 V1 V or =  2 n1T1 n2T2 T1 T2 V2  V1

4.26

V2 =V1

T2 = 93 L  251 K  = 55.844 L = 56 L T1  418 K 

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 bar Solution: p1 = 153.3 kPa p2 = 1 bar = 1bar (100 kPa/1 bar) = 100 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: p1V1 pV p1V1 pV or = 2 2 = 2 2 n1T1 n 2T2 T1 T2

 T2  p1   273 K  153.3 kPa    =  25.5 L     = 35.81201 L = 35.8 L  298 K  100 kPa   T1  p2 

V2 = V1 

4.27

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 kelvin. Solution: p1 = 74.5 kPa p2 = 36.7 kPa V1 = 3.65 L V2 = unknown T1 = 298 K T2 = –14°C + 273 = 259 K n remains constant Arranging the ideal gas law and solving for V2: p1V1 pV p1V1 pV or = 2 2 = 2 2 n1T1 n 2T2 T1 T2

 T2  p1   259 K   74.5 kPa    =  3.65 L      = 6.4397 L= 6.44 L  298 K   36.7 kPa   T1  p2 

V2 = V1  4.28

Plan: Given the volume, pressure, and temperature of a gas, the amount in moles of the gas can be calculated using the ideal gas law, solving for n. The gas constant, R = 8.31446 L•kPa/mol•K, gives pressure in kPa and temperature in kelvin. The given temperature must be converted to kelvin. Solution: p = 32.8 kPa V = 5.0 L T = 37°C n = unknown Converting T from °C to K: T = 37°C + 273 = 310 K pV = nRT Solving for n: pV (32.8 kPa)(5.0 L) n=  = 0.144521 mol = 0.064 mol chlorine L•kPa  RT   8.31446 mol•K  (310 K)  

4.29

Plan: Given the volume, amount in 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 = 0.08314L•bar/mol•K, gives volume in litres and temperature in kelvin. The given volume in mL must be converted to L and the temperature converted to kelvin. Solution: V = 75.0 mL T = 26°C n = 1.47 x 10–3 mol p = unknown  103 L  Converting V from mL to L: V =  75.0 mL    = 0.0750 L  1 mL  Converting T from °C to K: T = 26°C + 273 = 299 K pV = nRT Solving for p: L•bar   1.47x103 mol  0.08314  299 K  mol•K  nRT  p= = = 0.487234 bar = 0.487 bar 0.0750 L V





4.30

Plan: Solve the ideal gas law for amount (mol) and convert to mass using the molar mass of ClF 3. The gas constant, R = 8.31446 L•kPa/mol•K, gives volume in litres, pressure in kPa, and temperature in kelvin so volume must be converted to L and temperature to K. Solution: V = 357 mL T = 45°C p = 69.9 kPa n = unknown  103 L  Converting V from mL to L: V =  357 mL    1 mL  = 0.357 L   Converting T from °C to K: T = 45°C + 273 = 318 K pV = nRT Solving for n:  69.9 kPa  0.357 L  = 0.009438 mol ClF pV n=  3 L•kPa  RT  8.31446 318 K    mol•K  

 92.45 g ClF3   = 0.87255 g = 0.873 g ClF3  1 mol ClF3 

Mass ClF3 =  0.009438 mol ClF3   4.31

Plan: Solve the ideal gas law for pressure; convert mass to amount (mol) using the molar mass of N 2O. The gas constant, R 0.08314 L•bar/mol•K, gives temperature in kelvin so the temperature must be converted to units of kelvin. Solution: V = 3.1 L T = 115°C n = 75.0 g (convert to amount (mol)) p = unknown Converting T from °C to K: T = 115°C + 273 = 388 K Converting from mass of N2O to amount (mol):

 1 mol N 2 O   = 1.70377 mol N2O  44.02 g N 2 O 

n =  75.0 g N 2 O  

pV = nRT Solving for p: p=

4.32

nRT  V

1.70377 mol  0.08314 

 3.1 L 

L•bar   388 K  mol•K 

= 17.7293 bar = 18 bar N2O

Plan: Solve the ideal gas law for amount (mol). The gas constant, R = 0.08314 L•bar/mol•K, gives pressure in bar, and temperature in kelvin, so temperature must be converted to K. Solution: V = 1.5 L T = 23°C p = 5.9 bar + 1.013 bar = 6.913 bar n = unknown Converting T from °C to K: T = 23°C + 273 = 296 K pV = nRT Solving for n:  6.913 bar 1.5 L  = 0.4213627 mol = 0.42 mol SO pV n=  2 L•bar  RT   0.08314 mol•K   296 K   

4.33

Plan: Assuming that while rising in the atmosphere the balloon will neither gain nor lose gas molecules, the amount in moles of gas calculated at sea level will be the same as the amount in 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 kelvin and pressure in bar must be converted to kPa for unit agreement. Solution: p1 = 0.993 barr p2 = 6.9 kPa V1 = 65 L V2 = unknown T1 = 25°C + 273 = 298 K T2 = –5°C + 273 = 268 K n remains constant Converting p from bar to kPa:

p =  0.993 bar   101.3 kPa  = 99.3 kPa  1.013 bar 

Arranging the ideal gas law and solving for V2: p1V1 pV p1V1 pV or = 2 2 = 2 2 n1T1 n 2T2 T1 T2

 T2  p1  268 K  99.3 kPa    =  65 L     = 841.26 L = 840 L T p  298 K  6.9 kPa   1  2 

V2 = V1 

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 99.3/6.9 = 14.4. If we label the initial volume V1, then the resulting volume is 14.4V1. 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. 4.34

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.

4.35

The molar mass of H2 is less than the average molar mass of air (mostly N 2, 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 H 2. 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.

4.36

Gases mix to form a solution and each gas in the solution behaves as if it were the only gas present.

4.37

pA = XA pT The partial pressure of a gas (pA) in a mixture is directly proportional to its mole fraction (XA).

4.38

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: nB n 4 A particles 3 B particles a) X A = A = = 0.25 XB = = = 0.1875 n ntotal 16 total particles 16 total particles total nD2 nC 4 D2 particles 5 C particles XC = = = 0.3125 = = = 0.25 X D2 = ntotal 16 total particles 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 x 0.75 bar = 0.1875 bar= 0.19 bar 4.39

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 100 kPa. Do not forget that the pressure at STP is exact and will not affect the significant figures. Solution: p = 100 kPa T = 273 K M of Xe = 131.3 g/mol d = unknown pV = nRT Rearranging to solve for density: 100 kPa 131.3 g/mol  = 5.7845 g/L = 5.78 g/L pM d=  L•kPa  RT   8.31446 mol•K   273 K   

4.40 Plan: Rearrange the ideal gas law to calculate the density of CFCl3 from its molar mass. Temperature must be converted to kelvin Solution: p = 1.5 bar T = 120°C + 273 = 393 K M of CFCl3 = 137.4 g/mol d = unknown pV = nRT Rearranging to solve for density: 1.5 bar 137.4 g/mol  = 6.307765 g/L = 6.3 g/L pM d=  L•bar  RT   0.08314 mol•K   393 K    4.41

Plan: Solve the ideal gas law for amount in moles. Convert amount in 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 100 kPa. Do not forget that the pressure at STP is exact and will not affect the significant figures. Solution: V = 0.0400 L T = 0°C + 273 = 273 K p = 100 kPa n = unknown M of AsH3 = 77.94 g/mol pV = nRT Solving for n: pV 100 kPa  0.0400 L  n=   = 1.76223x10–3 mol= 1.76x10–3 mol AsH3 RT  L•kPa   8.31446 mol•K   273 K    Converting amount in moles of AsH3 to mass of AsH3:





3 

g AsH  = 0.1373 g AsH   77.94 1 mol AsH 

3 Mass (g) of AsH3 = 1.76223x10 mol AsH3 

 0.1373 g  mass = = 3.4337 g = 3.43 g/L volume  0.0400 L 

4.42

Plan: Solve the density form of the ideal gas law for molar mass. Temperature must be converted to kelvin, 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 bar T = 0°C + 273 = 273 K d = 2.71 g/L M = unknown

pM RT

Rearranging to solve for molar mass: L•bar   2.71 g/L   0.08314  273 K  dRT mol•K   = 20.5032 g/mol = 20.5 g/mol M = = p  3.00 bar  Therefore, the gas is Ne. 4.43

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. Solution: V = 0.206 μL T = 45°C + 273 = 318 K p = 0.517 bar m = 206 ng M = unknown kPa  106 L  –7 Converting V from μL to L: V =  0.206 L    = 2.06x10 L 1  L   Converting m from ng to g:

 109 g  –7 m =  206 ng    = 2.06x10 g 1 ng  

 m  pV =   RT M 

Solving for molar mass, M:

L•bar   2.06x10 g   0.08314 mol•K   318 K   = 51.1383 g/mol = 51.1 g/mol  0.517 bar   2.06x10 L  7

mRT M =  pV 4.44

7

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 litres, pressure to kPa, and temperature to kelvin. Solution: V = 63.8 mL T = 22°C + 273 = 295 K p = 9.96x104 Pa m = 0.103 g M = unknown



  1000 Pa 

Converting p from Pa to kPa:

p = 9.96x104 Pa  1 kPa  = 99.6 kPa

Converting V from mL to L:

 103 L  V =  63.8 mL    = 0.0638 L  1 mL 

 m  pV =   RT M 

Solving for molar mass, M: mRT M =  PV

 0.103 g   8.31446

L•kPa   295 K  mol•K   = 39.7570 g/mol = 39.8g/mol  99.6 kPa  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. 4.45

Plan: Use the ideal gas law to determine the amount in moles of Ar and of O 2. The gases are combined (ntotal = nAr + n O2 ) 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. Volume must be in units of L and temperature in K. Solution: For Ar: V = 0.600 L T = 227°C + 273 = 500. K p = 1.22 bar n = unknown pV = nRT Solving for n: 1.22 bar  0.600 L  = 0.017608852 mol Ar pV n=  L•bar  RT   0.08314 mol•K   500. K    For O2: V = 0.200 L T = 127°C + 273 = 400. K p = 66.8 kPa n = unknown pV = nRT Solving for n:  66.8 kPa  0.200 L  = 0.004017097 mol O pV n=  2 L•kPa  RT  8.31446 400. K    mol•K   ntotal = nAr + n O2 = 0.017608852 mol + 0.004017097 mol = 0.021625931 mol For the mixture of Ar and O2: V = 400 mL T = 27°C + 273 = 300. K p = unknownn n = 0.021554265 mol Converting V from mL to L:

 103 L  V =  400 mL    = 0.400 L  1 mL 

pV = nRT Solving for p: pmixture = 4.46

nRT  V

 0.021624651 mol  8.31446

L•kPa   300 K  mol•K   = 134.848 kPa = 135 kPa = 1.35 bar  0.400 L 

Plan: Use the ideal gas law, solving for n to find the total amount in moles of gas. Convert the mass of Ne to amount in moles and subtract amount in moles of Ne from the total amount in moles to find amount in moles of Ar. Volume must be in units of litre, pressure in units of bar and temperature in kelvin. Solution: V = 355 mL T = 35°C + 273 = 308 K p = 0.835 bar ntotal = unknown 83.5 kPa

 103 L  V =  355 mL    = 0.355 L  1 mL 

Converting V from mL to L:

pV = nRT Solving for ntotal:  0.835 bar  0.355 L  = 0.011575882 mol Ne + mol Ar pV ntotal =  L•bar  RT   0.08314 mol•K   308 K   

 1 mol Ne   = 0.007234886 mol Ne  20.18 g Ne 

Amount (mol) Ne =  0.146 g Ne  

Amount (mol) Ar = ntotal – nNe = (0.011575882 – 0.007234886) mol = 0.004340996 = 0.0043 mol Ar 4.47

Plan: Use the ideal gas law, solving for n to find the amount (mol) of O2. Use the molar ratio from the balanced equation to determine the amount in moles (and then mass) of phosphorus that will react with the oxygen. Standard temperature is 0°C (273 K) and standard pressure is 100 kPa. Solution: V = 35.5 L T = 0°C + 273 = 273 K p = 100 kPa n = unknown pV = nRT Solving for n: 100 kPa  35.5 L  = 1.563982 mol O pV n=  2 L•kPa  RT  8.31446 273 K    mol•K   P4(s) + 5O2(g)  P4O10(s)  1 mol P4  123.88 g P4  Mass P4 = 1.563982 mol O 2     = 38.7492 g = 38.7 g P4  5 mol O 2  1 mol P4 

4.48

Plan: Use the ideal gas law, solving for n to find the amount (mol) of O2 produced. Volume must be in units of litres, pressure in bar, and temperature in kelvin. Use the molar ratio from the balanced equation to determine the amount in moles (and then mass) of potassium chlorate that reacts. Solution: V = 638 mL T = 128°C + 273 = 401 K p = 1.00 bar n = unknown  103 L  Converting V from mL to L: V =  638 mL   = 0.638 L  1 mL    pV = nRT Solving for n: 1.00 bar  0.638 L  pV = 0.0191367 mol O2 n=  L•bar  RT  0.08314 401 K    mol•K   2KClO3(s)  2KCl(s) + 3O2(g)

 2 mol KClO3   122.55 g KClO3  Mass (g) of KClO3 =  0.0191367 mol O 2    = 1.5635 g = 1.56 g KClO3   3 mol O 2  1 mol KClO3 

4.49

Plan: Since the amounts of two reactants are given, this is a limiting reactant problem. To find the mass of PH 3, write the balanced equation and use molar ratios to find the amount in moles of PH 3 produced by each reactant. The smaller amount in moles of product indicates the limiting reagent. Solve for amount in moles of H 2 using the ideal gas law. Solution: Amount (mol) of hydrogen: V = 83.0 L T = 0°C + 273 = 273 K p = 100 kPa n = unknown pV = nRT Solving for n: 100 kPa 83.0 L  = 3.656633 mol H pV n=  2 L•kPa  RT  8.31446 273 K    mol•K   P4(s) + 6H2(g)  4PH3(g)

 4 mol PH3  PH3 from H2 =  3.656633 mol H 2    = 2.43775546 mol PH3  6 mol H 2   1 mol P4  4 mol PH3  PH3 from P4 =  37.5 g P4     = 1.21085 mol PH3  123.88 g P4  1 mol P4  P4 is the limiting reactant because it forms less PH3.  1 mol P4  4 mol PH 3   33.99 g PH 3  Mass PH3 =  37.5 g P4    = 41.15676 g= 41.2 g PH3    123.88 g P4  1 mol P4  1 mol PH 3  4.50

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 amount in moles of NO produced by each reactant. Since the amount in moles of gas are directly proportional to the volumes of the gases at the same temperature and pressure, 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 amount in moles and then to mass. Solution: 4NH3(g) + 5O2(g)  4NO(g) + 6H2O(l)

 4 L NO  Mol NO from NH3 =  35.6 L NH3    = 35.6 L NO  4 L NH3   4 L NO  Mol NO from O2 =  40.5 L O2    = 32.4 L NO  5 L O2  O2 is the limiting reactant since it forms less NO. Converting volume of NO to amount (mol) and then mass: V = 32.4 L T = 0°C + 273 = 273 K p = 100 kPa n = unknown pV = nRT Solving for n: 100 kPa  32.4 L  pV = 1.42741 mol NO n=  L•kPa  RT   8.31446 mol•K   273 K     30.01 g NO  Mass (g) of NO = 1.42741 mol NO    = 42.8365 g = 42.8 g NO  1 mol NO 

4.51

Plan: First, write the balanced equation. The amount (mol) 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 vapour must be subtracted from the overall pressure given. Table 4.2 reports the pressure at 27°C as 3.5681 kPa. Volume must be in units of litres, pressure in kPa, and temperature in kelvin. Once the amount in moles of hydrogen produced are known, the molar ratio from the balanced equation is used to determine the amount in moles of aluminum that reacted. Solution: 2Al(s) + 6HCl(aq)  2AlCl3(aq) + 3H2(g) V = 35.8 mL = 0.0358 L T = 27°C + 273 = 300 K ptotal = 1.00 bar = 100. kPa n = unknown pwater vapour = 3.5681 kPa phydrogen = ptotal – pwater vapour = 100. kPa – 3.5681 kPa = 96.4319 kPa Converting V from mL to L:

 103 L  V =  35.8 mL   = 0.0358 L  1 mL   

pV = nRT Solving for n:  96.4319 kPa  0.0358 L  = 0.00138404 mol H pV n=  2 L•kPa  RT   8.31446 mol•K   300. K   

 2 mol Al  26.98 g Al  Mass (g) of Al =  0.00138404 mol H 2     = 0.0248943 g = 0.025 g Al  3 mol H 2   1 mol Al  4.52

Plan: First, write the balanced equation. Convert mass of lithium to amount in moles and use the molar ratio from the balanced equation to find the amount in 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 vapour must be subtracted from the overall pressure given. Table 4.2 reports the vapour pressure of water at 18°C (2.0647 kPa). Pressure must be in units of kPa and temperature in kelvin. Solution: 2Li(s) + 2H2O(l)  2LiOH(aq) + H2(g)  1 mol Li   1 mol H2  Amount (mol) H2 =  0.84 g Li     = 0.0605100 mol H2  6.941 g Li   2 mol Li  Finding the volume of H2: V = unknown T = 18°C + 273 = 291 K ptotal = 1.00 bar = 100 kPa n = 0.0605100 mol pwater vapour = 2.0647 kPa phydrogen = ptotal – pwater vapour = 100 kPa – 2.0647 kPa = 97.9353 kPa pV = nRT Solving for V: L•kPa   0.0605100 mol   8.31446  291 K  nRT mol•K   V= = 1.4949 L = 1.5 L H2 = p  97.9353 kPa 

4.53

Plan: Rearrange the ideal gas law to calculate the density of the air from its molar mass. Temperature must be converted to kelvin and pressure to kPa. Solution: p = 9.92x104 Pa T = 17°C + 273 = 290 K or T = 60°C + 273 = 333 K M of air = 28.8 g/mol d = unknown  1 kPa  Converting p from Pa to kPa p = 9.92x104 Pa   = 99.2 kPa  1000 Pa  pV = nRT Rearranging to solve for density: At 17°C  99.2 kPa  28.8 g/mol  = 1.18487 g/L = 1.18 g/L pM d=  L•kPa  RT   8.31446 mol•K   290 K    At 60.0°C  99.2 kPa  28.8 g/mol  = 1.03185 g/L = 1.03 g/L pM d=  L•kPa  RT   8.31446 mol•K  (333 K)  





4.54

Plan: Solve the ideal gas law for molar volume, n/V. Temperature must be converted to K. Solution: p = 86.7 kPa T = –25°C + 273 = 248 K n/V = unknown pV = nRT Solving for n/V: 86.7 kPa  n p = 0.042047 mol/L = 0.0420 mol/L =  L•kPa  V RT  8.31446 248 K    mol•K  

4.55

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 bar. 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 = 374 K p = 1.02 bar m = 0.482 g M = unknown  m  pV =   RT M  Solving for molar mass, M: L•bar   0.482 g   0.08314   374 K mRT mol•K   = 72.0275 g/mol (unrounded) M =  pV 1.02 bar  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.

4.56

Plan: Solve the ideal gas law for amount (mol) of air. Temperature must be in units of kelvin. Use Avogadro‘s number to convert amount in 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 = 298 K p = 1.01 bar n = unknown pV = nRT Solving for n: 1.01 bar 1.00 L  pV  amount in moles of air = n = = 0.0407657 mol RT  L•bar   0.08314 mol•K   298 K    Converting amount (mol) of air to molecules of air:  6.022x1023 molecules  22 Molecules of air =  0.0407657 mol    = 2.45491x10 molecules  1 mol  









 78.08% N 2 molecules  Molecules of N2 = 2.45491x1022 air molecules   100% air   22 22 = 1.9148319x10 molecules= 1.91x10 molecules N2  20.94% O2 molecules  Molecules of O2 = 2.45491x1022 air molecules   100% air   = 5.14058x1021 molecules= 5.14x1021 molecules O2  0.05% CO2 molecules  Molecules of CO2 = 2.45491x1022 air molecules   100% air   = 1.22746x1019 molecules= 1x1019 molecules CO2  0.93% Ar molecules  Molecules of Ar = 2.45491x1022 air molecules   100% air   = 2.283066x1020 molecules= 2.3x1020 molecules Ar



4.57



Plan: Since you have the pressure, volume, and temperature, use the ideal gas law to solve for n, the total amount (mol) of gas. Pressure must be in units of bar and temperature in units of kelvin. 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 = 318 K p = 1.13 bar n = unknown pV = nRT 1.13 bar  21 L  pV Amount (mol) of gas = n = = 0.89755 mol = 0.90 mol gas  L•bar  RT   0.08314 mol•K   318 K    b) The equation pSO2 = X SO2 x ptotal can be used to find partial pressure. The information given in ppm is a way of expressing the proportion, or fraction, of SO 2 present in the mixture. Since n is directly proportional to V, the volume fraction can be used in place of the mole fraction, X SO2 . There are 7.95x103 parts SO2 in a million parts of mixture, so volume fraction = (7.95x10 3/1x106) = 7.95x10–3. pSO2 = volume fraction x ptotal = (7.95x10–3) (1.13 bar) = 0.0089835 bar = 0.00898 bar

4.58

Plan: First, write the balanced equation. Convert mass of P 4S3 to amount (mol) and use the molar ratio from the balanced equation to find the amount in moles of SO2 gas produced. Use the ideal gas law to find the volume of that amount of SO2. Temperature must be in units of kelvin. Solution: P4S3(s) + 8O2(g)  P4O10(s) + 3SO2(g)  1 mol P4S3  3 mol SO 2  amount in moles SO2 =  0.800 g P4S3     = 0.010905 mol SO2  220.09 g P4S3  1 mol P4S3  Finding the volume of SO2: V = unknown T = 32°C + 273 = 305 K p = 96.7 kPa n = 0.010905 mol pV = nRT Solving for V: L•atm   0.010905 mol   8.31446   305 K  nRT mol•K   V= = 0.285978 L = p  96.7 kPa  Converting V from L to mL:  1 mL  V =  0.285978 L   3  = 285.99 = 286 mL SO2  10 L   

4.59

Plan: The amount in moles of Freon-12 produced can be calculated from the ideal gas law. Volume must be in units of L, pressure in bar, and temperature in kelvin. Then, write the balanced equation. Once the amount in moles of Freon-12 produced is known, the molar ratio from the balanced equation is used to determine the amount in moles and then mass of CCl4 that reacted. Solution: V = 16.0 dm3 T = 27°C + 273 = 300 K ptotal = 1.22 bar n = unknown  1L  Converting V from dm3 to L: V = 16.0 dm3  = 16.0 L 3  1 dm  pV = nRT Solving for n: 1.22 bar 16.0 L  pV amount in moles of Freon-12 = n = = 0.782616 mol Freon-12  L•bar  RT   0.08314 mol•K   300. K   





CCl4(g) + 2HF(g)  CF2Cl2(g) + 2HCl(g)

 1 mol CCl4  153.81 g CCl 4  Mass of CCl4 =  0.782616 mol CF2 Cl 2      1 mol CF2 Cl 2  1 mol CCl 4  = 120.374 = 1.20x102 g CCl4

4.60

Plan: First, write the balanced equation. Given the amount of xenon hexafluoride that reacts, we can find the amount in moles of silicon tetrafluoride gas formed by using the molar ratio in the balanced equation. Then, using the ideal gas law with the amount in moles of gas, the temperature, and the volume, we can calculate the pressure of the silicon tetrafluoride gas. Temperature must be in units of kelvin. Solution: 2XeF6(s) + SiO2(s)  2XeOF4(l) + SiF4(g)  1 mol XeF6  1 mol SiF4  amount in moles SiF4 = n =  2.00 g XeF6     = 0.0040766 mol SiF4  245.3 g XeF6  2 mol XeF6  Finding the pressure of SiF4: V = 1.00 L p = unknown pV = nRT Solving for p: Pressure SiF4 = p =

nRT = V

T = 25°C + 273 = 298 K n = 0.0040766 mol

 0.0040766 mol SiF4   8.31446  1.00 L

L•kPa   298 K  mol•K 

= 10.1006 kPa = 10.1 kPa SiF4

4.61

Plan: Use the ideal gas law with T and p constant; then volume is directly proportional to amount in moles. Solution: pV = nRT. At constant T and p, V α n. Since the volume of the products has been decreased to ½ the original volume, the amount in moles (and molecules) must have been decreased by a factor of ½ as well. Cylinder A best represents the products as there are 2 product molecules (there were 4 reactant molecules).

4.62

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 amount in moles of SO 2 produced by each reactant. The smaller amount in moles of product indicates the limiting reagent. Solve for amount in moles of SO2 using the ideal gas law. Solution: 2PbS(s) + 3O2(g)  2PbO(s) + 2SO2(g) Amount (mol) of oxygen: V = 228 L p = 202.7 kPa pV = nRT Solving for n:

T = 220°C + 273 = 493 K n = unknown

 202.7 kPa  228 L  pV = 11.275 mol O2  L•kPa  RT   8.31446 mol•K   493 K     2 mol SO2  amount in moles SO2 from O2 = 11.275 mol O2    = 7.5167mol SO2  3 mol O2  amount in moles of O2 = n =

 103 g   1 mol PbS   2 mol SO2  amount in moles SO2 from PbS =  3.75 kg PbS  = 15.6707 mol SO2  1 kg   239.3 g PbS   2 mol PbS     (unrounded) O2 is the limiting reagent because it forms less SO2. Finding the volume of SO2: V = unknown T = 0°C + 273 = 273 K ptotal = 100 kPa n = 7.5107 mol pV = nRT Solving for V:

V = nRT = p 4.63

 7.5167 mol   8.31446

L•kPa   273 K  mol • K   = 170.62 L = 1.71x102 L SO2 100 kPa 

Plan: First, write the balanced equation. Given the amount of HgO that reacts (20.0% of the given amount), we can find the amount in moles of oxygen gas formed by using the molar ratio in the balanced equation. Then, using the ideal gas law with the amount in moles of gas, the temperature, and the volume, we can calculate the pressure of the oxygen gas. Temperature must be in units of kelvin. Solution: 2HgO(s)  2Hg(l) + O2(g)  20.0%   1 mol HgO  1 mol O2  amount in mole O2 = n =  40.0 g HgO     = 0.01846722 mol O2   100%   216.6 g HgO   2 mol HgO  Finding the pressure of O2: V = 502 mL p = unknown Converting V from mL to L:

T = 25°C + 273 = 298 K n = 0.01846722 mol  103 L  V =  502 mL   = 0.502 L  1 mL   

pV = nRT Solving for p: Pressure O2 = p =

nRT = V

 0.01846722 mol O2   8.31446  0.502 L

L•kPa   298 K  mol•K 

= 91.15 kPa = 91.1 kPa O2 = 91,100 Pa

4.64

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.

4.65

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.

4.66

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.

4.67

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 amount in moles of O2 has a greater mass than the same amount in moles of H2. mass O2 > mass H2 pM The pressure and temperature are identical and density is directly proportional to molar b) d = RT mass, M. Since the molar mass of O2 (32.0 g/mol) is greater than the molar mass of H 2 (2.02 g/mol), the density of O2 is greater than that of H2. dO > d H 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.

e) Kinetic energy = ½mass x speed2. O2 and H2 have the same average kinetic energy at the same temperature and mass and speed are inversely proportional. The lighter H 2 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 O 2

molecules with a larger molar mass. effusion time H2 < effusion time O2 4.68

Plan: The molar masses of the three gases are 2.016 for H 2 (Flask A), 4.003 for He (Flask B), and 16.04 for CH 4 (Flask C). Since hydrogen has the smallest molar mass of the three gases, 4 g of H 2 will contain more gas molecules (about 2 mole‘s worth) than 4 g of He or 4 g of CH 4. 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. 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.

4.69

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 H 2 M of UF6 352.0 g/mol = = = 13.2137 = 13.21 Rate UF6 M of H 2 2.016 g/mol

4.70

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 O 2 M of Kr 83.80 g/mol = = = 1.618255 = 1.618 Rate Kr M of O 2 32.00 g/mol

4.71

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 behaviour 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 behaviour of F2.

4.72

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.

4.73

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

M of F2 = M of He

Rate He = Rate F2

38.00 g/mol = 3.08105 (unrounded) 4.003 g/mol

time F2 Rate He = Rate F2 time He 4.74

3.08105 time F2 = 1.00 4.85 min He

Time F2 = 14.9431 min = 14.9 min

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 H 2 = 2.42 min Time of effusion of unknown = 11.1 min

time unknown rate H 2 = = time H 2 rate unknown 11.1 min = 2.42 min

M of unknown 2.016 g/mol

4.586777 =

M of unknown 2.016 g/mol

21.03852196 =

M of unknown M of H 2

M of unknown 2.016 g/mol

M ( unknown) = 42.41366 g/mol= 42.4 g/mol 4.75

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 M of Ne = 0.404 = Rate Ne M of Px 0.404 =

20.18 g/mol M of Px

(0.404)2 =

20.18 g/mol M of Px

0.163216 =

20.18 g/mol M of Px

M ( 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 Px   30.97 g P  Thus, 4 atoms per molecule, so Px = P4. 4.76

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 kelvin. Solution:  4.003 g He   1 kg  a) 0°C = 273 K 30°C + 273 = 303 K M of He =    3  = 0.004003 kg/mol mol    10 g  1 J = kg•m2/s2

R = 8.314 J/mol•K

3RT

urms =

urmsHe (at 0°C) =

urmsHe (at 30°C) =

J   3  8.314  273 K   kg•m2 /s2  mol•K  3 3    = 1.3042x10 m/s= 1.30x10 m/s 0.004003 kg/mol J   J   3  8.314   303 K   kg•m 2 /s2  mol•K 3 3     = 1.3740x10 m/s= 1.37x10 m /s 0.004003 kg/mol J  

 131.3 g Xe   1 kg    3  = 0.1313 kg/mol mol    10 g 

b) 30°C + 273 = 303 K

M of Xe = 

R = 8.314 J/mol•K

1 J = kg•m2/s2

3RT

urms =

urmsXe (at 30°C) =

J   3  8.314   303 K   kg•m 2 /s2  mol•K     = 239.913 m/s (unrounded) 0.1313 kg/mol J  

Rate He/Rate Xe = (1.3740x103 m/s)/(239.913 m/s) = 5.727076 = 5.73 He molecules travel at almost 6 times the speed of Xe molecules. c) Ek =

1 mu 2 2

EHe =

1 2

 0.004003 kg/mol  1.3740x103 m/s  1 J/kg•m2 /s2  = 3778.58 J/mol= 3.78x103 J/mol

EXe =

1 2

 0.1313 kg/mol   239.913 m/s 2 1 J/kg•m2 /s2 = 3778.70 J/mol= 3.78x103 J/mol





1 mol   3778.58 J   –21 –21 d)    = 6.2746x10 J/atom= 6.27x10 J/He atom 23 mol    6.022x10 atoms 

4.77

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 = 102.14 g/mol; M of N2F4 = 104.02 g/mol; M of SF4 = 108.07 g/mol SF4 has the largest molar mass and S2F2 has the smallest molar mass: rateSF4 < rateN2F4 < rateS 2 F2 b)

RateS2 F2 Rate N 2 F4

M of N 2 F4 104.02 g/mol = 1.009161 = 1.0092:1 = M of S2 F2 102.14 g/mol

Rate X = 0.935 = Rate SF4

M of SF4 M of X

108.07 g/mol M of X 108.07 g/mol (0.935)2 = M of X 108.07 g/mol 0.874225 = M of X M of X = 123.61806 g/mol= 124 g/mol 0.935 =

4.78

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 4.3, a N 2 = 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. 4.79

Particle volume causes a positive deviation from ideal behaviour. Thus, VReal Gases > VIdeal Gases. The particle volume is related to the constant b. According to Table 4.4, bH = 0.02651 L/mol, bO = 0.03186 L/mol, 2

and bCl = 0.05422 L/mol. Therefore, the order is H2 < O2 < Cl2. 2

4.80

Nitrogen gas behaves more ideally at 1 bar than at 500 bar 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.

4.81

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.

4.82

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. 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) = 2.575x103 cm2 p = 1.013 bar = 1.013 bar (100 kPa/1 bar) = 101.3 kPa  1 N / m²  Conversion to N/m²: 101.3 kPa = (101.3 kPa)   = 101.3 N/m²  1 kPa  2

 1 m   101.3 N  Total force (N)= 2.575x103 cm 2     = 26.08475 N= 26.1 N 2  100 cm   1 m 





4.83

Plan: Use the ideal gas law to find the amount in moles of O 2. amount in moles of O2 is divided by 4 to find amount in moles of Hb since O2 combines with Hb in a 4:1 ratio. Divide the given mass of Hb by the amount in moles of Hb to obtain molar mass, g/mol. Temperature must be in units of kelvin, , and volume in L. Solution: V = 1.53 mL T = 37°C + 273 = 310 K p = 97.8 kPa n = unknown  103 L  Converting V from mL to L: V = 1.53 mL   = 1.53x10–3 L  1 mL    pV = nRT Solving for n:





 97.8 kPa  1.53x103 L pV amount in moles of O2 = n = = 5.80543x10–5 mol O2  L•kPa  RT   8.31446 mol•K   310 K     1 mol Hb  –5 amount in moles Hb = 5.80543x105 mol O2   = 1.45136x10 mol Hb (unrounded) 4 mol O 2   1.00 g Hb Molar mass hemoglobin = = 6.890099x104 g/mol= 6.89x104 g/mol 1.45136x105 Hb



4.84



Plan: Convert mass of NaHCO3 to amount in moles and use the molar ratio from each balanced equation to find the amount in moles of CO2 gas produced. Use the ideal gas law to find the volume of that amount of CO 2. Temperature must be in kelvin. Solution: Reaction 1: 2NaHCO3(s)  Na2CO3(s) + H2O(l) + CO2(g)  1 mol NaHCO3  1 mol CO2  –3 amount in moles CO2 = 1.00 g NaHCO3     = 5.95167x10 mol CO2  84.01 g NaHCO3  2 mol NaHCO3  Finding the volume of CO2: V = unknown T = 200.°C + 273 = 473 K p = 0.988 bar n = 5.95167x10–3 mol pV = nRT Solving for V: L•bar   5.95167x103 mol  0.08314  473 K  nRT mol•K   Volume of CO2 = V = = 0.23689359 L = p  0.988 bar 





Converting V from L to mL:  1 mL  V =  0.2368935 L   3  = 236.8935 mL= 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 CO 2  –2 amount in moles CO2 = 1.00 g NaHCO3     = 1.1903x10 mol CO2 84.01 g NaHCO 1 mol NaHCO 3  3   Finding the volume of CO2: V = unknown p = 0.988 bar pV = nRT Solving for V:

T = 200.°C + 273 = 473 K n = 1.1903x10–2 mol

L•bar  1.1903x10 mol  0.08314 mol•K   473 K   2

nRT Volume of CO2 = V = = p

 0.988 bar 

= 0.473773 L

Converting V from L to mL:  1 mL  V =  0.473773 L   3  = 473.773 mL = 474 mL CO2 in Reaction 2  10 L   

p1V1 pV pV n T = 2 2 or V2 = 1 1 2 2 . Only the initial and final conditions need to be n1T1 n2T2 p2 n1T1 considered. p is fixed while V and T change. n2 is 0.75n1 since one-fourth of the gas leaks out. Solution: p1 = 1.02 bar p2 = 1.02 bar (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 L= 369 L n1T1  n1  305 K 

4.85

Plan: Use the relationship

4.86

Plan: Convert the mass of Cl2 to amount in moles and use the ideal gas law and van der Waals equation to find the pressure of the gas. Solution:  103 g   1 mol Cl2  a) amount in moles Cl2:  0.5950 kg Cl2   = 8.3921016 mol  1 kg   70.90 g Cl  2    V = 15.50 L n = 8.3921016 mol Ideal gas law: PV = nRT Solving for P: pIGL =

nRT = V

T = 225°C + 273 = 498 K P = unknown

8.3921016 mol   8.31446  15.50 L

L•kPa   498 K  mol•K 

= 2241.829 kPa = 2.24x10³ kPa

 n2 a  b) van der Waals equation:  p  2  V  nb   nRT  V   Solving for p: nRT n2 a bar•L2  2 From Table 4.4: a = 6.343 ; V  nb V mol2 n = 8.3921016 mol from part a)

pVDW =

b = 0.05422

L mol

bar•L2   100 kPa  2 L•kPa  8.3921016 mol Cl 6.343     498 K  2    mol2   1bar  mol•K     pVDW = 2 L   15.50 L   15.50 L   8.3921016 mol Cl2   0.05422 mol   = 2123.69 kPa = 2124 kPa

8.3921016 mol Cl2   8.31446

4.87

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 amount in moles by using the molar mass of each element involved. Divide all amounts in moles by the lowest number 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 = 0.7500 L T = 70.00°C + 273.15 = 343.15 K m = 0.1000 g (each liquid) p = 6.030x10³ Pa (I); 7.138x10³ Pa (II); 5.843x10³ Pa (III) M = unknown Converting V from mL to L:

 103 L  V =  750.0 mL   = 0.7500 L  1 mL   

 m  pV =   RT M  Solving for molar mass, M: Molar mass I = M =

Molar mass II = M =

mRT  pV mRT  pV

mRT Molar mass III = M =  pV

 0.1000 g   8.31446

L•kPa   343.15 K  mol•K   = 63.0869 g/mol = 63.09 g I/mol  6.030 kPa  0.7500 L 

 0.1000 g   8.31446

L•kPa   343.15 K  mol•K   = 53.294 g/mol= 53.29 g II/mol  7.138 kPa  0.7500 L 

 0.1000 g   8.31446

L•kPa   343.15 K  mol•K   = 65.10598 g/mol = 65.11 g III/mol  5.843 kPa  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  amount in moles B =  85.63 g B    = 7.921369 mol B (unrounded)  10.81 g B 

 1 mol H  amount in moles H = 14.37 g H    = 14.25595 mol H (unrounded)  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 x 5) = 5 for B and (1.7997 x 5) = 9 for H. The empirical formula is B 5H9, 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  amount in moles B =  81.10 g B    = 7.50231 mol B (unrounded)  10.81 g B   1 mol H  amount in moles H = 18.90 g H    = 18.750 mol H (unrounded)  1.008 g H  Silberberg, Amateis, Venkateswaran, Chen, Chemistry: The Molecular Nature of Matter and Change, 3rd Canadian Edition Page 3-172 Instructor‘s Solution Manual © Copyright 2021 McGraw-Hill Ryerson Ltd.

 7.50231 mol B   18.750 mol H    = 1.00   = 2.4992 7.50231 mol B    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 x 2) = 2 for B and (2.4992 x 2) = 5 for H. The empirical formula is B 2H5, 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. III

 1 mol B  amount in moles B =  82.98 g B    = 7.6762 mol B (unrounded)  10.81 g B   1 mol H  amount in moles H = 17.02 g H    = 16.8849 mol H (unrounded)  1.008 g H 

 7.6762 mol B   16.8849 mol H    = 2.2   = 1.00  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 x 5) = 5 for B and (2.2 x 5) = 11 for H. The empirical formula is B 5H11, 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 SO 2 M of IV = Rate IV M of SO2  250.0 mL     13.04 min  = 0.657318 =  350.0 mL     12.00 min 

M of IV 64.07 g/mol

M of IV 64.07 g/mol M of IV = 27.6825 g/mol= 27.68 g/mol

0.6573182 =

% H in IV = 100% – 78.14% = 21.86% H  1 mol B  Amount (mol) B =  78.14 g B    = 7.22849 mol B (unrounded)  10.81 g B 

 1 mol H  Amount (mol) H =  21.86 g H    = 21.6865 mol H (unrounded)  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.

4.88

3 A particles 4 A particles 5 A particles = 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. I. XA =

4.89

Plan: Use the relationship

p1V1 pV PV T = 2 2 or V2 = 1 1 2 . R and n are fixed. Temperatures must be converted T1 T2 P2T1

to kelvin. Solution: a) T1 = 200°C + 273 = 473 K; T2 = 100°C + 273 = 373 K pVT (2 bar)(V1 )(373 K) V2 = 1 1 2 = = 1.577 V1 Increase p2T1 (1 bar)(473 K) b) T1 = 100°C + 273 = 373 K; T2 = 300°C + 273 = 573 K p1V1T2 (1 bar)(V1 )(573 K) V2 = = = 0.51206 V1 Decrease p2T1 (3 bar)(373 K) c) T1 = –73°C + 273 = 200 K; T2 = 127°C + 273 = 400 K p1V1T2 (3 bar)(V1 )(400 K) V2 = = = V1 Unchanged p2T1 (6 bar)(200 K) d) T1 = 300°C + 273 = 573 K; T2 = 150°C + 273 = 423 K p1V1T2 (0.2 bar)(V1 )(423 K) V2 = = = 0.3691 V1 Decrease p2T1 (0.4 bar)(573 K) 4.90 Plan: Partial pressures and mole fractions are calculated from Dalton‘s law of partial pressures: PA = XA x Ptotal. Solve the ideal gas law for amount (mol) 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 = 1.01 bar = 101 kPa pNitrogen = XNitrogen x ptotal = 0.786 x 101 kpa = 79.39 kpa = 79.4 kpa N2 pOxygen = XOxygen x ptotal = 0.209 x 101 kpa = 21.11 kpa = 21.1 kpa O2 pCarbon Dioxide = XCarbon Dioxide x ptotal = 0.0004 x 101 kpa = 0.0404 kpa = 0.04 kpa CO2 pWater = XWater x ptotal = 0.0046 x 101 kPa = 0.4646 kPa = 0.46 kPa H2O b) Mole fractions can be calculated by rearranging Dalton‘s law of partial pressures: pA XA = and multiply by 100 to express mole fraction as percent ptotal ptotal = 1.01 bar = 101 kPa 74.9 kPa XNitrogen = x 100% = 74.1584 % = 74.2 mol% N2 101 kPa 13.7 kPa XOxygen = x 100% = 13.5644 % = 13.6 mol% O2 101 kPa

5.3 kPa x 100% = 5.2475 % = 5.2 mol% CO2 101 kPa 6.2 kPa XWater = x 100% = 6.1386 % = 6.1 mol% H2O 101 kPa c) V = 0.50 L T = 37°C + 273 = 310 K p = 13.7 kPa (from table) n = unknown pV = nRT Solving for n: 13.7 kPa  0.50 L  = 0.0026576 mol O pV n=  2 L•kPa  RT  8.31446 310 K    mol•K    6.022x1023 molecules O2  21 21 Molecules of O2 =  0.0026576 mol O2    = 1.6004x10 = 1.6x10  1 mol O 2   molecules O2

X Carbon Dioxide =

4.91 of

Plan: Convert the mass of Ra to amount (mol) and then atoms using Avogadro‘s number. Convert from number Ra atoms to Rn atoms produced per second and then to Rn atoms produced per day. The number of Rn atoms is converted to amount in moles and then the ideal gas law is used to find the volume of this amount of Rn. Solution:  1 mol Ra   6.022 x1023 Ra atoms  21 Atoms Ra = 1.0 g Ra    = 2.664602x10 Ra atoms   226 g Ra 1 mol Ra   

 1.373x104 Rn atoms  Atoms Rn produced/s = 2.664602x1021 Ra atoms  = 3.65849855x1010 Rn atoms/s  1.0x1015 Ra atoms   





 3.65849855x1010 Rn atoms   3600 s   24 h   1 mol Rn  amount in moles Rn produced/day =        23  s    h   day   6.022x10 Rn atoms  = 5.248992x10–9 mole Rn/day pV = nRT Solving for V (at STP): L•kPa   5.248992x109 mol  8.31446   273 K  nRT mol•K   Volume of Rn = V = = p 100 kPa 





= 1.19144x10–7 L= 1.2x10–7 L Rn 4.92

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 amount in moles of air and then amount in moles of N 2. Solution: a) p1 = 1.933 bar p2 = 1 bar V1 = 208 mL V2 = unknown T1 = 286 K T2 = 298 K Arranging the ideal gas law and solving for V2: p1V1 pV = 2 2 T1 T2

 T  p   298 K   1.933 bar  2 V2 = V1  2  1  =  208 mL     1 bar  = 418.934 mL = 4x10 mL T p 286 K      1  2 

b) V = 208 mL p = 1.933 bar n = unknown Converting V from mL to L:

T = 286 K

 103 L  V =  208 mL   = 0.208 L  1 mL   

pV = nRT Solving for n:

1.933 bar  0.208 L  = 0.016909 mol air pV  L•bar  RT   0.08314 mol•K   286 K     77% N 2  amount in mole of N2 =n=  0.016909 mol    = 0.01302 = 0.013 mol N2  100% 

amount in moles of air = n =

4.93

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 amount in moles of NO 2 produced by each reactant. The smaller amount in moles of product indicates the limiting reagent. Solve for volume of NO 2 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 NO 2  amount in moles NO2 from Cu = n = 4.95 cm3    = 1.394256 mol NO2   cm3   63.55 g Cu  1 mol Cu 





amount in moles NO2 from HNO3 = n =  68.0% HNO3   1 cm3   1.42 g   1 mol HNO3   2 mol NO2   230.0 mL        3  100%    1 mL   cm   63.02 g  4 mol HNO3  = 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 amount in 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 + 273.2 = 301.4 K p = 0.980 bar n = 1.394256 mol NO2 pV = nRT Solving for V: L•bar  1.394256 mol   0.08314   301.4 K  nRT mol•K   = 36.71539 L= 36.7 L NO2 V=  p  0.980 bar  4.94

Plan: Solve the ideal gas law for amount in moles of air and then convert to molecules using Avogadro‘s number. Volume must be converted to litres and temperature to kelvin. Solution: a) V = 1200 mL T = 37°C + 273 = 310 K p = 1.0 bar n = unknown  3  10 L  Converting V from mL to L: V = 1200 mL   = 1.2 L  1 mL    pV = nRT Solving for n:

amount in moles of air = n = pV  RT

1.0 bar 1.2 L 

L•bar    0.08314 mol•K   310 K   

= 0.0465596 mol = 0.047 mol air

 6.022 x1023 molecules  22 22 b) Molecules of air =  0.0465596 mol air    = 2.80382099x10 = 2.8x10 molecules air  1 mol air   4.95

Plan: The amounts of two reactants are given, so the first step is to identify the limiting reactant. Use the molar ratios from the balanced equation to find the amount in moles of Br 2 produced by each reactant. The smaller amount of product indicates the limiting reagent. Solve for volume of Br 2 using the ideal gas law. Solution: 5NaBr(aq) + NaBrO3(aq) + 3H2SO4(aq)  3Br2(g) + 3Na2SO4(aq) + 3H2O(g)  1 mol NaBr  3 mol Br2  amount in 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  Amount (mol) Br2 from NaBrO3 = 175.6 g NaBrO3     = 3.491285 mol Br2  150.89 g NaBrO3  1 mol NaBrO3  The NaBr is limiting since it produces a smaller amount of Br 2. Use the ideal gas law to find the volume of Br 2: V = unknown T = 300°C + 273 = 573 K p = 86.6 kPa n = 1.60365 mol Br2 pV = nRT Solving for V: L•kPa   1.60365 mol   8.31446   573 K  nRT mol•K   Volume (L) of Br2 = V = = 88.223 L = 88.2 L Br2  p 86.6 kPa  4.96

Plan: First, write the balanced equation. Convert mass of NaN3 to amount (mol) and use the molar ratio from the balanced equation to find the amount in 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 vapour must be subtracted from the overall pressure given. Table 4.2 reports the vapour pressure of water at 26°C (3.3639 kPa). Pressure must be in units of kPa and temperature in kelvin. Solution: 2NaN3(s)  2Na(s) + 3N2(g)  1 mol NaN3  3 mol N 2  amount in moles N2 =  50.0 g NaN3     = 1.15349 mol N2  65.02 g NaN3  2 mol NaN3  Finding the volume of N2: V = unknown T = 26°C + 273 = 299 K ptotal = 9.94x104 Pa = 99.4 kPa n = 1.15319 mol pwater vapour = 3.3639 kPa pnitrogen = ptotal – pwater vapour = 99.4 kPa – 3.3639 kPa = 96.0361 kPa pV = nRT Solving for V:

nRT V= = p 4.97

1.15349 mol   8.31446

L•kPa    299 K  mol•K   = 29.8709 L = 29.9 L N2  96.0361 kPa 

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 amount in moles by using the molar mass of each element involved. Divide all amounts in moles by the lowest number 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 C  amount in moles C =  64.81 g C    = 5.39634 mol C  12.01 g C   1 mol H  amount in moles H = 13.60 g H    = 13.49206 mol H  1.008 g H   1 mol O  amount in moles O =  21.59 g O    = 1.349375 mol O  16.00 g O   5.39634 mol C   13.749206 mol H    =4   = 10 1.349375 mol O    1.349375 mol O  Empirical formula = C4H10O (empirical formula mass = 74.12 g/mol) Molecular formula: V = 2.00 mL T = 25°C + 273 = 298 K m = 2.57 g p = 0.426 bar M = unknown  m  pV =   RT M  Solving for molar mass, M: L•   2.57 g   0.08314bar  298 K  mRT mol•K   = 74.73 g/mol M =  pV  0.426 bar  2.00 L 

 1.349375 mol O    = 1.00  1.349375 mol O 

Since the molar mass (74.73 g/mol ) and the empirical formula mass (74.12 g/mol) are similar, the empirical and molecular formulas must both be: C4H10O 4.98

Plan: The empirical formula for aluminum chloride is AlCl 3 (Al3+ 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 M of He = 0.122 = Rate He M of unknown 0.122 =

4.003 g/mol M of unknown

4.003 g/mol M of unknown M of unknown = 268.9465 g/mol The whole-number multiple is 268.9465/133.33, which is about 2. Therefore, the molecular formula of the gaseous species is 2 x (AlCl3) = Al2Cl6.

0.014884 =

4.99

Plan: First, write the balanced equation. Convert mass of C8H18 to amount (mol) and use the molar ratio from the balanced equation to find the total amount in 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 amount in moles of oxygen that react with the C8H18. Use the composition of air to calculate the amount of N 2 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) amount in 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 ptotal = 0.980 bar pV = nRT Solving for V:

nRT Volume gas = V = = p

T = 350°C + 273 = 623 K n = 14.883558 mol

14.883558 mol   0.08314 

L•bar   623 K  mol•K 

 0.980 bar 

= 786.645 L = 787 L gas

 1 mol C8 H18  25 mol O2  b) amount in moles O2 = 100. g C8 H18     = 10.94379 mol O2  114.22 g C8 H18  2 mol C8 H18   78% + 1% N 2 and Ar  amount in moles other gases = 10.94379 mol O 2    = 41.1695 mol Ar + N2 21% O 2   Finding the volume of Ar + N 2: V = unknown T = 350°C + 273 = 623 K ptotal = 0.980 bar n = 41.1695 mol pV = nRT Solving for V: L•bar   41.1695 mol   0.08314   623 K  nRT mol•K   Volume gas = V = = 2175.94 L residual air = p  0.980 bar  Total volume of gaseous exhaust = 786.645 L + 2175.94 L = 2964.02 L = 2.96x103 L 4.100

Plan: First, write the balanced equation for the reaction: 2SO 2 + O2  2SO3. The total amount in 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 amount in moles of gas before and after the reaction using the ideal gas law. For each mole of SO3 formed, the total amount in moles of gas decreases by 1/2 mole. Thus, twice the decrease in amount in moles of gas equals the amount in moles of SO 3 formed. Solution: amount in moles of gas before and after reaction: V = 2.00 L T = 800. K ptotal = 192.5 kPa n = unknown pV = nRT 192.5 kPa  2.00 L pV Initial amount in moles = n = = = 0.05788109 mol RT L•kPa   8.31446 800. K    mol•K   167.2 kPa  2.00 L pV Final amount in moles = n = = = 0.05027356 mol RT L•kPa    8.31446 mol•K   800. K    amount in moles of SO3 produced = 2 x decrease in the total amount in moles = 2 x (0.05788109 mol – 0.05027356 mol) = 0.01521506 mol = 1.52x10–2 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. 4.101

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 amount in moles of gas remains the same). Solution:  28.05 g C2 H 4  1 mL   103 L  a) 1 mole C2 H 4    = 0.130465 L = 0.130 L     1 mole C2 H 4   0.215 g   1 mL  V = 0.130 L T = 10°C + 273 = 283 K + 950 K = 1233 K ptotal = unknown n = 1.00 mol bar•L2 L From Table 4.4 for CH4: a = 2.300 ; b = 0.04301 2 mol mol

 n2 a   p  2  V  nb   nRT V   Pressure of CH4 = pVDW =

nRT n2 a  2 V  nb V

bar•L2   100kPa  2 L•kPa   1.00 mol 2.300      1.00 mol  8.31446 1233 K   mol2   1bar  mol•K     pVDW = = 104,240.05 kPa 0.130 L  1.00 mol  0.04301 L/mol  0.130 L 2 = 1.04x105 kPa





1.04x105 kPa  0.130 L  pV b) = 1.32 = L•kPa  RT  8.31446 1233 K    mol•K   This value is smaller than that shown in Figure 4.23 for CH 4. 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.

4.102

Plan: First, write the balanced equation. Convert mass of ammonium nitrate to amount in moles and use the molar ratio from the balanced equation to find the amount in moles of gas produced. Use the ideal gas law to find the volume of that amount of gas. Solution: 2NH4NO3(s)  2N2(g) + O2(g) + 4H2O (g)

 103 g   1 mol NH 4 NO3  7 mol gas  amount in moles gas = 15.0 kg NH 4 NO3    1 kg   80.05 g NH NO  2 mol NH NO  4 3  4 3    = 655.840 mol gas Finding the volume of gas: V = unknown T = 307°C + 273 = 580 K p = 1.01 bar n = 655.840 mol pV = nRT Solving for V:

4.103

nRT = p

 655.840 mol   0.08314 

L•bar   580 K  mol•K 

1.01 bar 

= 3.13123x104 L= 3.13x104 L

Plan: First, balance the equations given. Multiply concentration (mol/L) and volume of the I 2 solution to find initial amount in moles of I2 added. Multiply concentration (mol/L) and volume of the S2O32– solution to obtain amount in moles of that solution and use the molar ratio in the balanced equation to find amount in moles of excess I2. Subtract the excess I2 from the initial amount in moles of added I2 to find the amount in moles of I2 reacted with the SO2; the molar ratio betweeen SO2 and I2 gives the amount in moles of SO2 present. Use the ideal gas law to find the amount in moles of air which is compared to the amount in moles of SO 2 present. Solution: The balanced equation is: I2(aq) + 2S2O32(aq)  2I(aq) + S4O62(aq).  103 L   0.01017 mol I2  –4 Initial amount in moles of I2 =  20.00 mL    = 2.034x10 mol I2 initial  1 mL   L   

 103 L   0.0105 mol S2 O32   1 mol I2  amount in moles I2 reacting with S2O32– = 11.37 mL     1 mL   2 mol S O 2   L   2 3   = 5.96925x10–5 mol I2 reacting with S2O32– (not reacting with SO2) amount in moles I2 reacting with SO2 = 2.034x10–4 mol – 5.96925x10–5 mol = 1.437075x10–4 mol I2 reacted with SO2 The balanced equation is: SO2(aq) + I2(aq) + 2H2O(l)  HSO4(aq) + 2I(aq) + 3H+(aq).  1 mol SO2  –4 amount in moles SO2 = 1.437075x10 4 mol I 2   = 1.437075x10 mol SO2 1 mol I 2  





Amount (mol) of air: V = 500 mL = 0.500 L ptotal = 0.933 bar pV = nRT amount in moles air = n =

T = 38°C + 273 = 311 K n = unknown

 0.933 bar  0.500 L = 0.018042 mol air (unrounded) pV = RT L•bar    0.08314 mol•K   311 K   

Volume % SO2 = mol % SO2 = 4.104

1.437075x104 mol SO2 100%  = 0.796523 % = 0.797% 0.018042 mol air

Plan: First, write the balanced equation. The amount in 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 kPa, and temperature in kelvin. Once the amount in 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 CO. For part b), assume the volume is 1 m3. Use the ideal gas law to solve for amount in moles of Ni(CO)4, which equals the amount in moles of Ni, and convert amount in moles to mass using the molar mass. For part c), the mass of Ni obtained from 1 m3 (part b)) can be used to calculate the amount of CO released. Use the ideal gas law to calculate the volume of CO. The vapour pressure of water at 35°C (5.629 kPa) must be subtracted from the overall pressure (see Table 4.2). Solution: a) Ni(s) + 4CO(g)  Ni(CO)4(g) V = 3.55 m3 T = 50°C + 273 = 323 K p = 100.7 kPa n = unknown  1L  Converting V from m3 to L: V = 3.55 m3  3 3  = 3550 L  10 m  pV = nRT





Solving for n: amount in moles of CO = n =

100.7 kPa  3550 L  = 133.1132 mol CO pV  L•kPa  RT   8.31446 mol•K   323 K   

 1 mol Ni   58.69 g Ni  3 Mass Ni = 133.1132 mol CO     = 1953.10 g = 1.95x10 g Ni 4 mol CO 1mol Ni    b) Ni(s) + 4 CO(g)  Ni(CO)4(g) V = 1 m3 T = 155°C + 273 = 428 K p = 21 bar n = unknown  1L  Converting V from m3 to L: V = 1 m3  3 3  = 1000 L  10 m  pV = nRT Solving for n:  21 bar 1000 L pV amount in moles of Ni(CO)4 = n = = 590.1542 mol Ni(CO)4  L•bar  RT  0.08314 428 K    mol•K    1 mol Ni   58.69 g Ni  Mass Ni =  590.1542 mol Ni(CO)4      1 mol Ni(CO) 4   1 mol Ni 





= 3.4636x104 = 3.5x104 g Ni The pressure limits the significant figures.

 1 mol Ni   4 mol CO  c) amount in moles CO = 3.4636x104 g Ni    = 2360.60658 mol CO  58.69 g Ni   1 mol Ni  Finding the volume of CO: V = unknown T = 35°C + 273 = 308 K ptotal = 1.03 bar =103 kPa n = 2390.51968 mol pwater vapour = 5.629 kPa pCO = ptotal – pwater vapour = 103 kPa – 5.629 kPa = 97.371 kPa pV = nRT Solving for V: L•kPa   2360.60658 mol   8.31446   308 K  nRT mol•K   V= = 62083.87 L CO = p  97.371 kPa 





 103 m3  V =  62083.87 L   = 62.08387 m³ = 62 m3 CO  1 L    The answer is limited to two significant figures because the mass of Ni comes from part b). Converting V from L to m3:

4.105

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 amount in moles by using the molar mass of each element involved. Divide all amount (mol) by the lowest amount in 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 masses of the elements.

 1 mol Si  amount in moles Si =  33.01 g Si    = 1.17515 mol Si  28.09 g Si   1 mol F  amount in moles F =  66.99 g F    = 3.525789 mol F  19.00 g F   1.17515 mol Si   3.525789 mol F    =1   =3  1.17515 mol Si   1.17515 mol Si  Empirical formula = SiF3 (empirical formula mass = 85.1 g/mol) Molecular formula: V = 0.250 L T = 27°C + 273 = 300 K m = 2.60 g p = 1.52 bar M = unknown  m  pV =   RT M  Solving for molar mass, M: L•bar   2.60 g   0.08314   300 K  mRT mol•K   Molar mass = M = = 170.656 g/mol  pV 1.52 bar  0.250 L  The molar mass (170.656 g/mol ) is twice the empirical formula mass (85.1 g/mol), so the molecular formula must be twice the empirical formula, or 2 x SiF3 = Si2F6. 4.106

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 amount in moles of the gas we can equate volume and amount in 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.

4.107

a) There is a total of 6x106 blue particles and 6x106 black particles. When equilibrium is reached after opening the stopcocks, the particles will be evenly distributed among the three containers. Therefore, container B will have 2x106 blue particles and 2x106 black particles. b) The particles are evenly distributed so container A has 2x106 blue particles and 2x106 black particles. c) There are 2x106 blue particles and 2x106 black particles in C for a total of 4x106 particles.   1.00 bar Final pressure in C = 4x106 particles   = 0.7 bar  6x106 particles    d) There are 2x106 blue particles and 2x106 black particles in B for a total of 4x106 particles.   1.00 bar Final pressure in B = 4x106 particles   = 0.7 bar  6x106 particles   

4.108









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 amount in moles of n-hexane. Convert amount in moles of n-hexane to mass and then to volume using the density.

Solution: a) 2C6H14(l) + 19O2(g) →12CO2(g) + 14H2O(g) For a 1.00 L sample of air at STP:  20.9 L O 2   2 L C6 H14  Volume of C6H14 vapour needed = 1.00 L air    = 0.0220 L C6H14   100 L air   19 L O 2  Volume % of C6H14 =

C6 H14 volume 0.0220 L C 6H 14 100%  = 100%  = 2.20% C6H14 air volume 1.00 L air

LFL = 0.5(2.2%) = 1.10% C6H14





 1 L   1.1% C6 H14  b) Volume of C6H14 vapour = 1.000 m3 air  3 3    = 11.0 L C6H14  10 m  100% air  V = 11.0 L T = 0°C + 273 = 273 K p = 100 kPa n = unknown pV = nRT Solving for n: 100 kPa 11.0 L pV amount in moles of C6H14 = n = = 0.484614 mol C6H14  L•kPa  RT   8.31446 mol•K   273 K     86.17 g C6 H14   1 mL Volume of C6H14 liquid =  0.484614 mol C6 H14     = 63.2715 mL 1 mol C H 0.660 g C H 6 14  6 14   = 63.3 mL C6H14 4.109

Plan: To find the factor by which a diver‘s lungs would expand, find the factor by which P changes from 38.1 m to the surface, and apply Boyle‘s law. To find that factor, calculate Pseawater at 38.1 m by converting the given depth from m-seawater to mmHg to kPa and adding the surface pressure (101.3 kPa). Solution:  1 mm  p(H2O) =  38.1m   3  = 3.81x104 mmH2O  10 m 

p(Hg):

hH2O hHg

d Hg d H2O

3.81x104 mmH 2O 13.5 g/mL = hHg 1.04 g/mL

hHg = 2935.1111 mmHg

 101.3 kPa  p(Hg) =  2935.11111 mmHg    = 391.2194 kPa (unrounded)  760 mm Hg  ptotal = (101.3 kPa) + (391.2194 kPa) = 492.5194 kPa (unrounded) Use Boyle‘s law to find the volume change of the diver‘s lungs: p1V1 = p2V2 V2 p V2 492.5194 kPa = 1 = = 4.86 V1 p2 V1 101.3 kPa To find the depth to which the diver could ascend safely, use the given safe expansion factor (1.5) and the pressure at 38.1 m, P38.1, to find the safest ascended pressure, Psafe. p38.1/psafe = 1.5 psafe = p38.1/1.5 = (492.5194 kPa)/1.5 = 328.3463 kPa (unrounded) Convert the pressure in kPa to pressure in m 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):  492.5194  328.3463  kPa   = 1231.7034 mmHg  101.3 kPa 

hH2O hHg

d Hg

hH2O

d H2O

1231.7034 mmHg

13.5 g/mL 1.04 g/mL

hH2O = 15988.458 mmH2O

 103 m   = 15.9885 m  1 mm  Therefore, the diver can safely ascend 16.0 m to a depth of (38.1- 15.9885)m = 22.1015 m = 22 m.

15988.458 mmH 2 O  

4.110

Plan: Write a balanced equation. Convert mass of CaF2 to amount in moles and use the molar ratio from the balanced equation to find the amount in 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  amount in moles HF gas = 15.0 g CaF2     = 0.3842213 mol HF  78.08 g CaF2  1 mol CaF2  Finding the temperature: V = 8.63 L T = unknown p = 1.17 bar n = 0.3842213 mol pV = nRT Solving for T: 1.17 bar  8.63 L pV = = 316.0860 K T = L•bar  nR   0.3842213 mol HF  0.08314 mol•K   The gas must be heated to 316 K.

4.111

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. A 0.100 L solution will produce (20 x 0.100 L) = 2.00 L O 2 gas. Use the ideal gas law to convert this volume of O2 gas to amount in moles of O2 gas and convert to amount in 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 = 100 kPa n = unknown pV = nRT Solving for n: 100 kPa  2.00 L pV amount in moles of O2 = n = = 8.81116x10–2 mol O2  L•kPa  RT   8.31446 mol•K   273 K     2 mol H2 O2  34.02 g H2 O2  Mass H2O2 = 8.81116x102 mol O2    = 5.995116 g = 6.00 g H2O2  1 mol O2  1 mol H2 O2 



4.112



Plan: The amount in moles of gas may be found using the ideal gas law. Multiply amount in moles of gas by Avogadro‘s number to obtain the number of molecules. Solution: V = 1 mL = 0.001 L T = 500 K p = 10–6Pa n = unknown pV = nRT

Solving for n:

1kPa  10 Pa   0.001 L   1000  Pa  6

pV amount in moles of gas = n =  RT

L•kPa    8.31446 mol•K   500 K   

= 2.405448x10–16 mol gas

 6.022 x1023 molecules  8 8 Molecules = 2.405448x1016 mol   = 1.44856x10 = 10 molecules  1 mol   (The 10–6 Pa limits the significant figures.)



4.113



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  a) 0°C = 273 K M of O2 =    3  = 0.03200 kg/mol mol    10 g  R = 8.314 J/mol•K 1 J = kg•m2/s2 urms =

urms =

3RT

M J   3  8.314  273 K   kg•m2 /s2  mol•K     = 461.2878 m/s= 461 m/s 0.03200 kg/mol J  

b) Collision frequency = 4.114

urms 461.2878 m/s = = 7.2873x109 s-1= 7.29x109 s–1 8 mean free path 6.33x10 m

Plan: Convert the volume of the tubes from m³ to L. Use the ideal gas law to find the amount in moles of gas that will occupy that volume at the given conditions of pressure and temperature. From the mole fraction of propene and the total amount in moles of gas, the amount in moles of propene can be obtained. Convert amount in moles to mass in grams and then kilograms using the molar mass and scale the amount added in 1.8 s to the amount in 1 h. Solution:  1L  Volume of tubes = 2.8 m3  3 3  = 2800 L  10 m  V = 2800 L T = 330°C + 273 = 603 K p = 2.5 bar n = unknown pV = nRT  2.5 bar  2800 L  pV amount in moles of gas = n = = 139.6274 mol gas = L•bar  RT  0.08314 603 K    mol•K   moles of propene Xpropene = moles of mixture x moles of propene 0.07 = 139.627 moles of mixture amount in moles of propene = 9.7739 moles  42.08 g C3 H 6  1 kg   1  3600 s  Mass of propene per hour =  9.7739 moles C3 H 6        1 mole C3 H 6   1000 g   1.8 s  1 hr 





= 822.573 kg = 820 kg C3H6 Silberberg, Amateis, Venkateswaran, Chen, Chemistry: The Molecular Nature of Matter and Change, 3rd Canadian Edition Page 3-186 Instructor‘s Solution Manual © Copyright 2021 McGraw-Hill Ryerson Ltd.

4.115

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 T = 730. K p = 91 bar n = 1.00 mol pV = nRT Solving for V: L•bar   1.00 mol   0.08314   730. K  nRT mol•K   V= = 0.66695 L/mol = 0.67 L/mol = p  91 bar 

4.116

Plan: The diagram below describes the two Hg height levels within the barometer. First, find the pressure of the N2. The PN 2 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 amount in moles which is converted to mass using the molar mass of nitrogen. vacuum (74.0 cm)

with N2 (64.0 cm)

Solution:

 102 m   1 mm   1 atm   101.3kPa  Pressure of the nitrogen =  74.0 cm  64.0 cm   = 13.3289 kPa  1 cm   103 m   760 mmHg   1atm       3  1 mL   10 L  Volume of the nitrogen = 1.00 x 102 cm  64.0 cm 1.20 cm 2    = 0.0432 L   1 cm3   1 mL  V = 0.0432 L T = 24°C + 273 = 297 K p = 13.3289 kPa n = unknown pV = nRT Solving for n: 13.3289 kPa  0.0432 L  pV amount in moles of N2 = n = = 2.3317799x10–4 mol N2 = L•kPa  RT   8.31446 mol•K   297 K   







 28.02 g N2  –3 –3 Mass of N2 = 2.3317799x104 mol N2   = 6.5336x10 = 6.53x10 g N2 1 mol N 2   4.117

Plan: Use the ideal gas law to find the amount in moles of ammonia gas in 10.0 L at this pressure and temperature. Concentration (mol/L) is amount (mol) per litre. Use the amount in moles of ammonia and the final volume of solution (0.750 L) to get the concentration (mol/L). Solution: V = 10.0 L T = 33°C + 273 = 306 K p = 96.7 kPa n = unknown pV = nRT

Solving for n:

 96.7 kPa 10.0 L  = 0.380076 mol pV  L•kPa  RT   8.31446 mol•K   306 K    amount(mol) ammonia 0.380076 mol = concentration (mol/L) = c = = 0.50677 mol/L = 0.507 mol/L V(L) of solution 0.750 L amount in moles of ammonia = n =

4.118

Plan: Use the ideal gas law to determine the total amount in moles of gas produced. The total amount in moles multiplied by the fraction of each gas gives the amount in moles of that gas which may be converted to metric tonnes. Solution: V = 1.5x103 m3 T = 298 K p = 1.01 bar n = unknown  1L  Converting V from m3 to L: V = 1.5x103 m3  3 3  = 1.5x106 L  10 m  pV = nRT Solving for n:









1.01 bar  1.5x106 L pV amount in moles of gas/day = n = = 6.1149x104 mol/day  L•bar  RT   0.08314 mol•K   298 K     6.1149x104 mol   365.25 day  7 Amount (mol) of gas/yr =     = 2.23345x10 mol/yr  day 1 yr     2.23345x107 mol   44.01 g CO2   1 kg  1 t  2 Mass CO2 =  0.4896      = 481.248 t/yr = 4.81x10 t CO2/yr   3  3   yr 1 mol CO 2   10 g  10 kg   

 2.23345x107 mol   28.01 g CO   1 kg  1 t  Mass CO =  0.0146      = 9.1336 t/yr= 9.16 t CO/yr   3   3  yr    1 mol CO   10 g  10 kg   2.23345x107 mol   18.02 g H 2 O   1 kg  1 t  2 Mass H2O =  0.3710       3   103 kg  =149.31558 t/yr= 1.49x10 t H2O/yr  yr 1 mol H O 10 g 2     

 2.23345x107 mol   64.07 g SO2   1 kg  1 t  2 Mass SO2 =  0.1185       3   103 kg  = 169.570 t/yr = 1.70x10 t SO2/yr  yr 1 mol SO 10 g 2       2.23345x107 mol   64.14 g S2   1 kg  1 t  –1 Mass S2 =  0.0003      3   103 kg  = 0.429760 t/yr = 4.3x10 t S2/yr  yr 1 mol S 10 g 2       

 2.23345x107 mol   2.016 g H 2   1 kg  1 t  –1 Mass H2 =  0.0047      = 0.21162 t/yr = 2.1x10 t H2/yr   3  3   yr 1 mol H 2   10 g  10 kg     2.23345x107 mol   36.46 g HCl   1 kg  1 t  –1 Mass HCl =  0.0008     3  = 0.6514538 t/yr = 6.5x10 t HCl/yr   3    yr 1 mol HCl 10 g 10 kg     

 2.23345x107 mol   34.09 g H 2S   1 kg  1 t  –1 Mass H2S =  0.0003      3   103 kg  = 0.228415 t/yr = 2x10 t H2S/yr  yr 1 mol H S 10 g 2     

4.119

Plan: Use the molar ratio from the balanced equation to find the amount in 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 amount in f moles of each gas. Solution: a) The balanced chemical equation is: 2H2(g) + O2(g)  2H2O(l)  2 mol H 2  amount in moles H2 =  28.0 mol H2 O    = 28.0 mol H2  2 mol H 2 O 

 1 mol O2  amount in moles O2 =  28.0 mol H2 O    = 14.0 mol O2  2 mol H 2 O  V = 20.0 L T = 23.8°C + 273.2 =297.0 K p = unknown n = 28.0 mol H2; 14.0 mol O2 pV = nRT Solving for p: L•kPa   28.0 mol  8.31446   297.0 K  nRT mol•K   pIGL of H2 = = = 3457.1525 kPa = 3.46x10³ kPa H2 20.0 L V L•kPa  14.0 mol   8.31446  297.0 K  nRT mol•K   pIGL of O2 = = = 1728.58 kPa = 1.73x10³ kPa O2 20.0 L V b) V = 20.0 L T =23.8°C + 273.2 =297.0 K p = unknown n = 28.0 mol H2; 14.0 mol O2 Van der Waals constants from Table 4.4: H2: a = 0.2453 O2: a = 1.382

bar•L2

mol bar•L2 mol

;

b = 0.02651

L mol

b = 0.03186

L mol

;

 n a  p  2  V  nb   nRT V   2

pVDW =

nRT n2 a  2 V  nb V

bar•L2   100 kPa  2 L•atm   28.0 mol 0.2453       28.0 mol   8.31446  297.0 K   mol2   1 bar  mol•K     pVDW of H2 = 20.0 L  28.0 mol  0.02651 L/mol  20.0 L 2 = 3542.328 kPa = 3.54x103 kPa H2

bar•L2   100 kPa  2 L•kPa   14.0 mol 1.382      14.0 mol   8.31446  297.0 K   mol2   1 bar  mol•K     PVDW of O2 = 20.0 L  28.0 mol  0.03186 L/mol  20.0 L 2 = 1741.559 kPa = 1.74x10³ kPa O2 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.

4.120

Plan: Deviations from ideal gas behaviour 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 behaviour. Solution: a) Xenon would show greater deviation from ideal behaviour 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 vapour would show greater deviation from ideal behaviour 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 vapour would show greater deviation from ideal behaviour 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 behaviour to a greater extent than methane molecules, even though the water is a vapour in this question.

4.121

Plan: Use the concentration (mol/L) and volume of the solution to find the amount in moles of HBr needed to make the solution. Then use the ideal gas law to find the volume of that amount in moles of HBr gas at the given conditions. Solution:  1.20 mol HBr  amount in moles of HBr in the hydrobromic acid:    3.50 L  = 4.20 mol HBr L   V = unknown T = 29°C + 273 =302 K p = 0.978 bar n = 4.20 mol pV = nRT Solving for V: L•bar   4.20 mol  0.08314   302 K  nRT mol•K   V= = 107.827 L = 108 L HBr = p  0.978 bar 

4.122

Plan: Convert the mass of each gas to amount in moles using the molar mass. Calculate the mole fraction of CO. Use the relationship between partial pressure and mole fraction (pA = XA x ptotal) to calculate the partial pressure of CO. Solution:  1 mol CO  Amount (mol) CO =  7.0 g CO    = 0.24991 mol CO  28.01 g CO 

 1 mol SO2  Amount (mol) SO2 = 10.0 g SO 2    = 0.156079 mol SO2  64.07 g SO 2  mol CO 0.24991 mol CO XCO = = = 0.615559 mol CO + mol SO2 0.24991 mol CO + 0.156079 mol SO2 pCO = XCO x ptotal = 0.615559 x 0.33 bar = 0.20313 bar= 0.20 bar CO 4.123

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 O 2 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 SO 2 that is produced. Since pressure is directly proportional to the amount (mol) of gas, the molar ratio may be expressed as a pressure ratio. Solution: 4FeS2(s) + 11O2(g)  8SO2(g) + 2Fe2O3(s) Pressure of N2 = 1.06 bar – 0.65 bar O2 = 0.41 bar N2

Pressure of unreacted O2 = (0.65 bar O2) [(100 – 85)%/100%] = 0.098 bar O2 Pressure of reacted O2= (0.65 bar O2) [85%/100%]=0.5525 bar O2  8 atm SO 2  Pressure of SO2 produced =  0.5525 bar O2    = 0.4018bar = 0.40 bar SO2  11 atm O 2  Total Pressure = (0.41 bar) + (0.098 bar) + (0.4018bar) = 0.9098 bar = 0.91 bar total 4.124

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 x 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 = 0.717 bar The NaOH absorbed the CO2 leaving the Kr, thus Pkrypton = 0.256bar. pcarbon dioxide = ptotal – pkrypton = 0.717 bar – 0.256 bar= 0.461 bar Determining mole fractions: pA = XA x ptotal pCO2 0.461 bar Carbon dioxide: X = = = 0.64296 ptotal 0.717 bar Krypton: X =

pKr 0.256 bar = = 0.35704 ptotal 0.717 bar

  83.80 g Kr     0.35704     mol    = 1.05737 Relative mass fraction =    44.01 g CO2      0.64296   mol    35.0 g = x g CO2 + (1.05737 x) g Kr 35.0 g = 2.05737 x mass of CO2 = x = (35.0 g)/(2.057378) = 17.01201 g = 17.0 g CO2 mass of Kr = 35.0 g – 17.01201 g CO2 = 17.98799 g = 18.0 g Kr 4.125

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.

4.126

Plan: Convert molecules of •OH to amount (mol) and use the ideal gas law to find the pressure of this amount in moles of •OH in 1 m3 of air. The mole percent is the same as the pressure percent as long as the other conditions are the same. Solution:  2.5x1012 molecules     103 m3  1 mol –15 amount in moles of •OH =    = 4.151445x10 mol/L    3 23   m    6.022 x10 molecules   1 L  V = 1.00 L p = unknown pV = nRT Solving for p:

T = 22°C + 273 =295 K n = 4.151445x10–15 mol

 4.151445x10



L•kPa   mol  8.31446  295 K  mol•K  nRT  Pressure of •OH = p = = 1.00 L V =1.018252x10–11 kPa= 1.02x10–11 kPa •OH Mole percent =

15

1.018252x1011 kPa 100%  = 1.008170x10–11 = 1.01x10–11% 100 kPa 1.01 bar 1 bar





4.127

Plan: Write the balanced equations. Use the ideal gas law to find the amount in moles of SO 2 gas and then use the molar ratio between SO2 and NaOH to find amount in moles and then concentration (mol/L) 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 T = 19°C + 273 =292 K p = 99.3 kPa n = unknown pV = nRT Solving for n:  99.3 kPa  0.200 L = 8.18017x10–3 mol SO pV amount in moles of SO2 = n =  2 L•kPa  RT   8.31446 mol•K   292 K     2 mol NaOH  amount in moles of NaOH = 8.18017x103 mol SO 2   = 0.01636034 mol NaOH  1 mol SO 2 



c NaOH =



amount(mol) NaOH 0.01636034 mol NaOH  1 mL  =  3  = 1.636034 mol/L = 1.64 mol/L NaOH volume of NaOH 10.0 mL  10 L 

4.128

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 kPa, volume in litres, and temperature in kelvin. Solution: Mass of water = mass of flask filled with water – mass of flask = 327.4 g – 65.347 g = 262.053 g    103 L  1 mL Volume of flask = volume of water =  262.053 g water    = 0.2628415 L    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 V = 0.2628415 L T = 99.8°C + 273.2 = 373.0 K p = 101.2 kPa m = 0.392 g M = unknown  m  pV =   RT M  Solving for molar mass, M: L•kPa   0.392 g   8.31446 373.0 K  mRT mol•K   = 45.7048 g/mol = 45.7 g/mol M=  pV 101.2 kPa  0.2628415 L 

4.129

Plan: Write the balanced chemical equations. Convert mass of hydride to amount in moles and use the molar ratio from the balanced equation to find the amount in 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 kPa and temperature in kelvin. 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 mol LiH   1 mol H 2  Amount (mol) H2 =  454 g     = 57.1357 mol H2  7.946 g LiH   1 mol LiH  Finding the volume of H2: V = unknown T = 27°C + 273 = 300 K p = 1.00x105 Pa = 100 kPa n = 57.0746 mol pV = nRT Solving for V:

 57.1357 mol   8.31446 

nRT V= = p

L•kPa   300 K  mol•K 

100 kPa 

= 1425.1575 L = 1430 L H2 from LiH

Magnesium hydride, MgH2

 1 mol MgH 2  2 mol H 2  amount in moles H2 =  454 g MgH 2     = 34.48538 mol H2  26.33 g MgH 2  1 mol MgH 2  Finding the volume of H2: V = unknown T = 27°C + 273 = 300 K p = 1.00x105 Pa =100 kPa n = 34.44852 mol pV = nRT Solving for V: L•kPa   34.48538 mol  8.31446  300 K  nRT mol•K   V= = 860.1819 L = 8.60x102 L H2 from MgH2 = p 100 kPa   4.130

Plan: Use the equation for root mean speed (urms). Molar mass values must be in units of kg/mol and temperature in kelvin. Solution: urms =

4.131

3 RT

urms Ne =

3 8.314 J/mol•K  370 K   103 g   kg•m2 /s2      = 676.24788 m/s= 676 m/s Ne J  20.18 g/mol   kg   

urms Ar =

3 8.314 J/mol•K  370 K   103 g   kg•m2 /s2      = 480.6269 m/s= 481 m/s Ar J  39.95 g/mol   kg   

urms He =

3 8.314 J/mol•K  370 K   103 g   kg•m2 /s2  3     = 1518.356 m/s= 1.52x10 m/s He kg J  4.003 g/mol    

Plan: Use the ideal gas law to find the amount in moles of CO2 and H2O in part a). The molar mass is then used to convert amount (mol) to mass. Temperature must be in units of kelvin, pressure in kPa, and volume in L. For part b), use the molar ratio in the balanced equation to find the amount in moles and then mass of C 6H12O6 that produces the amount in moles of CO2 exhaled during 8 h. Solution: a) V = 300 L T = 37.0°C + 273.2 = 310.2 K p = 4.00x10³ Pa =4.00 kPa n = unknown pV = nRT Solving for n:  4.00 kPa  300 L  pV amount in moles of CO2 = amount (mol) of H2O = n = = 0.46527 mol  L•kPa  RT  8.31446 310.2 K    mol•K  

 44.01 g CO2  Mass (g) of CO2 =  0.46527 mol CO2    = 20.4765 g = 20.5 g CO2  1 mol CO 2   18.02 g H 2 O  Mass (g) of H2O =  0.46527 mol H 2 O    = 8.3842 g = 8.38 g H2O  1 mol H 2 O  b) C6H12O6(s) + 6 O2(g)  6 CO2(g) + 6 H2O(g)  0.46527 mol CO2  amount in moles of CO2 exhaled in 8 h =   8 h  = 3.72216 mol CO2 h    1 mol C6 H12 O6   180.16 g C6 H12 O6  Mass (g) of C6H12O6 =  3.72216 mol CO2      6 mol CO2  1 mol C6 H12 O6  = 111.764 g = 1x102 g C6H12O6 (= body mass lost) (This assumes the significant figures are limited by the 8 h.) 4.132

Plan: For part a), the amount in moles of water vapour can be found using the ideal gas law. Convert amount in 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 vapour at the stated set of condition. Solution:  75% H 2 O  –4 a) Volume of water in kernel =  0.25 mL kernel    = 0.1875 mL = 1.875x10 L  100 % kernel  V = 1.875x10–4 L T = 170°C + 273.2 = 443 K p = 9.1 bar n = unknown pV = nRT Solving for n:





 9.1bar  1.875 104 L pV = 4.632644x10–5 mol  L•bar  RT   0.08314 mol•K   443 K     18.02 g H 2 O   100%  Mass (g) of = 4.632644 x10 5 mol H 2O    = 0.052175 g = 0.052 g  1 mol H 2 O   1.6%  b) V = unknown T = 25°C + 273 = 298 K p = 1.01 bar n = 4.632644x10–5 mol pV = nRT Solving for V: L•bar   4.632644x105 mol  0.08314  298 K  nRT mol•K   V= = 0.0011364 L = 1.1364 mL = 1.1 mL = p 1.01 bar  Amount (mol) of H2O = n =



4.133



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 amount in moles of SO 2 in that volume, assuming a temperature of 25 °C. The molar ratio in the balanced equation is used to find the amount in moles and then mass of CaSO 4 produced. For part b), use the molar ratio in the balanced equation to calculate the amount in 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 O 2 and adjust for the mole fraction of O2 in air. Solution: a) Mole fraction SO2 = 1000(2x10–10) = 2x10–7  109 L   95%  2x107  Volume of SO2 removed =  4 GL    = 760 L  1 GL   100%   



V = 760 L



T = 25°C + 273 = 298 K

p = 1.01 bar pV = nRT Solving for n:

n = unknown

amount in moles of SO2 = n =

1.01 bar  760 L  pV = 30.9819 mol  L•bar  RT  0.08314 298 K    mol•K  

CaCO3(s) + SO2(g)  CaSO3(s) + CO2(g) 2CaSO3(s) + O2(g)  2CaSO4(s)  1 mol CaSO4  136.15 g CaSO 4   1 kg  Mass (kg) of CaSO4 =  30.9819 mol SO2.      3  = 4.2182 kg= 4 kg CaSO4  1 mol SO2  1 mol CaSO 4   10 g  b) amount in moles of O2 =  103 g   1 mol CaSO4  1 mol O2   4.2182 kg CaSO4       = 15.4910 mol O2  1 kg   136.15 g CaSO4  2 mol CaSO4  V = unknown T = 25°C + 273 = 298 K The balanced chemical equations are:

p = 1.01 bar pV = nRT Solving for V:

n = 15.4910 mol

Volume of O2 = V =

nRT = p

15.4910 mol   0.08314 

L•bar   298 K  mol•K 

1.01 bar 

= 380.00 L O2

 1 L Ar  Volume of air =  380.00 L O 2    = 1818.1818 L = 2000 L air  0.209 L O 2  4.134

Plan: Use the ideal gas law to find the amount in moles of gas occupying the tank at 80% of the 86.1 bar ranking. Then use van der Waals equation to find the pressure of this amount in f moles of gas. Solution: a) V = 850. L T = 298 K 80%   p = 86.1 bar    = 68.88 bar n = unknown  100%  pV = nRT Solving for n:  68.88 bar 850. L  = 2.3631x103 mol= 2.36x103 mol Cl pV amount in moles of Cl2 = n =  2 L•bar  RT  0.08314 298 K    mol•K   b) V = 850. L T =298 K p = unknown n = 2.3631x103 mol Cl2 Van der Waals constants from Table 4.4: a = 6.343

bar•L2 mol

;

b = 0.05422

L mol

 n2 a   p  2  V  nb   nRT V   pVDW =

nRT n2 a  2 V  nb V

2 bar•L2  3 L•bar   2.3614x10 mol Cl 6.343   2.3631x103 mol Cl2  0.08314 298 K 2    mol2  mol•K     pVDW = L   850. L 2 850. L  2.36248x103 mol Cl 2  0.05422  mol   = 32.1464 bar = 32.2 bar c) The engineer did not completely fill the tank. She should have filled it to (80.0%/100%)(86.1 bar) = 68.88 bar but only filled it to 32.2 bar.



4.135





Plan: Solve the density form of the ideal gas law for molar mass. Temperature must be converted to K. Solution: p = 102.5 kPa T = 10.0°C + 273.2 = 283.2 K d = 1.26 g/L M = unknown pM RT Rearranging to solve for molar mass: L•kPa   1.26 g/L   8.31446   283.2 K  dRT mol•K   = 28.9450 g/mol= 29.0 g/mol M = = p 102.5 kPa  d=

4.136

Plan: Use the relationship

p1V1 pV pV n T = 2 2 or V2 = 1 1 2 2 . R is fixed. n1T1 n2T2 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). pVT (p )(V )(1) V2 = 1 1 2 = 1 1 V2 = ½V1 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). pVT (1)(V1 )(200 K) V2 = 1 1 2 = V 2 = ½ V1 p2T1 (1)(400 K) Cylinder B has half the volume of the original cylinder. c) T1 = 100°C + 273 = 373 K T2 = 200°C + 273 = 473 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). pVT (1)(V1 )(473 K) V2 = 1 1 2 = V2 = 1.27V1 p2T1 (1)(373 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 amount in 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 amount in 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 ½. 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 V 2 = V1 p2 n1T1 (2 p1 )(0.1)(1) Cylinder D has the same volume as the original cylinder. 4.137

Plan: Use both the ideal gas law and van der Waals equation to solve for pressure. Convert mass of ammonia to amount (mol) and temperature to kelvin. Solution: Ideal gas law: V = 3.000 L T = 0°C + 273 =273 K or 400.°C + 273 =673 K  1 mol NH3  p = unknown n =  51.1 g NH3    = 3.0005872 mol  17.03 g NH3  pV = nRT Solving for p: L•kPa   3.0005872 mol   8.31446  273 K  mol•K  nRT  pIGL of NH3 at 0°C = = = 2270.29 kPa = 2.27x10³ kPa 3.000 L V L•kPa   3.0005872 mol   8.31446   673 K  mol•K nRT   pIGL of NH3 at 400.°C = = = 5596.73 kPa = 5.60x10³ kPa 3.000 L V van der Waals equation: V = 3.000 L T = 0°C + 273 =273 K or 400.°C + 273 =673 K  1 mol NH3  p = unknown n =  51.1 g NH3    = 3.0005872 mol  17.03 g NH3  van der Waals constants from Table 4.4: a = 4.225

bar•L2 2

;

b = 0.03713

mol 2   n a  p  2  V  nb   nRT V  

L mol

nRT n2 a  2 V  nb V pVDW of of NH3 at 0°C =

pVDW =

bar•L2   100 kPa  2 L•kPa  3.0005872 mol 4.225     273 K     mol2   1bar  mol•K     3.000 L  3.0005872 mol  0.03713 L/mol   3.000 L 2

 3.0005872 mol   8.31446

= 1935.19 kPa = 1.94x10³ kPa NH3 pVDW of of NH3 at 400.°C =

bar•L2   100 kPa  2 L•kPa   3.0005872 mol 4.225       3.0005872 mol   8.31446  673 K   mol2   1bar  mol•K     3.000 L  3.0005872 mol  0.03713 L/mol   3.000 L 2 = 5389.93 kPa = 5.39x10³ kPa NH3

4.138

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 amount in moles of gas. Avogadro‘s number is needed to convert amount in 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)(1.77 bar) = 0.708 bar CH4 V = 6.0 L T = 45°C + 273 =318 K p = 0.708 bar n = unknown pV = nRT Solving for n:  0.708 bar  6.0 L  = 0.160675 mol pV amount in moles of CH4 = n =  L•bar  RT   0.08314 mol•K   318 K   

 6.022x1023 CH 4 molecules  Molecules of CH4 =  0.160675 mol CH 4     1 mol CH 4   = 9.67583x1022 molecules= 9.7x1022 molecules CH4 4.139

Plan: For part a), convert mass of glucose to amount in moles and use the molar ratio from the balanced equation to find the amount in moles of CO2 gas produced. Use the ideal gas law to find the volume of that amount of CO 2. Pressure must be in units of atm and temperature in kelvin. For part b), use the molar ratios in the balanced equation to calculate the amount in 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  amount in moles CO2:  20.0 g C6 H12 O6     = 0.666075 mol CO2  180.16 g C6 H12 O6  1 mol C6 H12 O6  Finding the volume of CO2: V = unknown T = 37°C + 273 = 310 K p = 104 kPa n = 0.666075 mol pV = nRT Solving for V: L•kPa   0.666075 mol   8.31446   310 K  nRT mol•K   V= = 16.5077 L = 16.5 L CO2 = p 104 kPa  This solution assumes that partial pressure of O2 does not interfere with the reaction conditions.  1 mol C6 H12 O6    6 mol b) amount in moles CO2 = amount (mol) O2 = 10.0 g C6 H12 O6      180.16 g C6 H12 O6   1 mol C6 H12 O6  = 0.333037 mol CO2 =0.333037 mol O2 At 37°C, the vapour pressure of water is 6.51 kPa. No matter how much water is produced, the partial pressure of H2O will still be 6.51 kPa. The remaining pressure, 104 kPa – 6.51 kPa = 97.49 kPa 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 one-half of 97.49 kPa. pwater = 6.51 kPa (97.49 kPa)/2 = 48.75 kPa = 49 kPa = poxygen = pcarbon dioxide

4.140

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 kelvin. Solution: The average kinetic energies are the same for any gases at the same temperature: Average kinetic energy = 3/2RT = (3/2)(8.314 J/mol•K) (273 K) = 3.40458x10 3 J= 3.40x103 J for both gases rms speed  28.02 g N 2   1 kg  T = 0°C + 273 = 273 K M of N2 =    3  = 0.02802 kg/mol mol    10 g 

 2.016 g H 2   1 kg    3  = 0.002016 kg/mol mol    10 g  R = 8.314 J/mol•K 1 J = kg•m2/s2

M of H2 = 

urms =

4.141

3RT

urmsN2 =

3 8.314 J/mol•K  273 K   kg•m2 /s2  2 2   = 4.92961x10 m/s= 4.93x10 m/s N2  0.02802 kg/mol   J 

urms H2 =

3 8.314 J/mol•K  273K   kg•m2 /s2  3 3   = 1.83781x10 m/s= 1.84x10 m/s H2  0.002016 kg/mol   J 

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 amount in moles of air in 1000 L of air at these conditions and compare the amount in moles of each gas to the amount in moles of air. Mass and number of molecules must be converted to amount in moles. Solution: a) Assuming the total pressure is 101.3 kPa =101,300 Pa pA = XA ptotal pBr2 30 Pa X Br2 = = = 2.961500x10–4x(106 ppmv) = 296.1500 ppmv = 300 ppmv Unsafe ptotal 101300 Pa b) X CO2 =

pCO2 ptotal

30 Pa = 2. 961500x10–4x(106 ppmv) = 296.1500 ppmv= 300 ppmv Safe 101300 Pa

(30 Pa CO2/101300 Pa)(106 ppmv) = 296.1500 ppmv = 300 ppmv CO 2 Safe  1 mol Br2  –6 c) amount in moles Br2 =  0.0004 g Br2    = 2.5031x10 mol Br2 (unrounded)  159.80 g Br2  Finding the amount in moles of air: V = 1000 L T = 0°C + 273 =273 K p = 101.3 kpa n = unknown pV = nRT 101.3 kPa 1000 L  pV amount in moles of air = n = = 44.629 mol air (unrounded)  L•kPa  RT  8.31446 273 K    mol•K   6 Concentration of Br2 = mol Br2/mol air(10 ) = [(2.5031x10–6 mol)/(44.616 mol)] (106) = 0.056103 ppmv= 0.06 ppmv Br2 Safe   1 mol CO2 d) amount in moles CO2 = 2.8x1022 molecules CO2   6.022x1023 molecules CO  = 0.046496 mol CO2 2   Concentration of CO2 = mol CO2/mol air(106) = [(0.046496 mol)/(44.616 mol)] (10 6) = 1042.1 ppmv = 1.0x103 ppmv CO2 Safe





4.142

Plan: For part a), use the ideal gas law to find the amount in moles of NO in the flue gas. The amount in moles of NO are converted to amount in moles of NH3 using the molar ratio in the balanced equation and the amount in moles of NH3 is converted to volume using the ideal gas law. For part b), the amount in 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 amount in moles of NO to amount in moles and then mass of NH 3. Solution: a) 4NH3(g) + 4NO(g) + O2(g)  4N2(g) + 6H2O(g) Finding the amount in moles of NO in 1.00 L of flue gas: V = 1.00 L T = 365°C + 273 =638 K p = 4.6 Pa n = unknown pV = nRT Solving for n: pV  4.6 Pa 1kPa 1000Pa 1.00 L  amount in moles of NO = n = = 8.6717x10–7 mol NO  L•kPa  RT   8.31446 mol•K   638 K    amount in moles of NH3 =

NH  8.6717x10 mol NO  44 mol  = 8.6717x10 mol NH mol NO  7

Volume of NH3: V = unknown p = 1.01 bar pV = nRT Solving for V:

–7

T =365°C + 273 =638 K n = 8.6717x10–7 mol

L•bar  8.6717x10 mol  0.08314 mol•K   638 K   7

nRT V= = p

1.01 bar 

= 4.5542x10–5 L= 4.6x10–5 L NH3

b) Finding the amount in moles of NO in 1.00 kL of flue gas: V = 1.00 kL = 1000 L T = 365°C + 273 =638 K p = 4.6 Pa n = unknown pV = nRT Solving for n: pV  4.6 Pa 1 kPa 1000 Pa 1000 L  amount in moles of NO = n = = 8.6717x10–4 mol NO  L•kPa  RT   8.31446 mol•K   638 K    amount in moles of NH3 =

NH  8.6717x10 mol NO  44 mol  = 8.6717x10 mol NH mol NO  4

–4

 1 mol NH3  Mass of NH3 = 8.6717x104 mol NH3   = 0.0509x10 ³ g= 0.051 mg NH3  17.03 g NH3 



4.143



Plan: Use Graham‘s law to compare effusion rates. Solution: Rate Ne = Rate Xe

Thus XNe =

M of Xe = M of Ne

2.55077 131.3 g/mol = enrichment factor (unrounded) 1 20.18 g/mol

moles of Ne 2.55077 mol = 0.71837 = 0.7184 = moles of Ne + moles of Xe 2.55077 mol + 1 mol

4.144

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 M of 238 UF6 352.04 g/mol UF6 = = 235 349.03 g/mol Rate 238 M of UF6 UF 6

= 1.004302694 enrichment factor Therefore, the abundance of 235UF6 after one membrane is 0.72% x 1.004302694 Abundance of 235UF6 after ―N‖ membranes = 0.72% x (1.004302694)N Desired abundance of 235UF6 = 3.0% = 0.72% x (1.004302694)N Solving for N: 3.0% = 0.72% x (1.004302694)N 4.16667 = (1.004302694)N ln 4.16667 = ln (1.004302694)N ln 4.16667 = N x ln (1.004302694) N = (ln 4.16667)/(ln 1.004302694) N = 1.4271164/0.004293464 = 332.39277 = 332 steps 4.145

Plan: Use van der Waals equation to calculate the pressure of the gas at the given conditions. Solution: V = 22.711 L T = 273.15 K p = unknown n = 1.000 mol Van der Waals constants from Table 4.4: a = 1.472

bar•L2 mol2

;

b = 0.03948

L mol

 n2 a   p  2  V  nb   nRT V   pVDW =

nRT n2 a  2 V  nb V

bar•L2   100 kPa  2 L•kPa  1.000 mol CO 1.472     273.15 K     mol2   1bar  mol•K     pVDW = L    22.711 L 2 22.711 L  1.000 mol CO   0.03948  mol   = 99.8885 kPa = 99.89 kPa 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: Rate O 2 M of H 2 2.016 g/mol rate O2 = = = Rate H 2 35 M of O 2 32.00 g/mol

1.000 mol CO   8.31446

4.146

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

0.250998008 =

4.147

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 31 35 35 35 136 31 37 35 35 138 31 37 37 35 140 31 37 37 37 142 b) The fraction abundances are 35Cl = 75%/100% = 0.75, and 37Cl = 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 P37 Cl3 35

Rate P Cl3

M of P 35 Cl3 M of P Cl3 37

136 g/mol 142 g/mol

= 0.978645 = 0.979 4.148

Plan: Use the combined gas law for parts a) and b). For part c), use the ideal gas law to find the amount in moles of air represented by the pressure decrease and convert amount in moles to mass using molar mass. Solution: pV pV a) 1 1 = 2 2 T1 T2

 241 kPa  318 K  p1 V1T2 pT = 1 2 = = 259.79 kPa = 260 kPa V2T1 T1 295 K b)New tire volume = V2 = V1(102%/100%) = 1.02V1 p1V1 pV = 2 2 T1 T2 p2 =

p2 =

 241 kPa V1  318 K  p1V1T2 pT = 1 2 = = 254.70 kPa = 255 kPa V2T1 T1 1.02V1  295 K 

c) Pressure decrease = 254.70 kPA – 241 kPa = 13.70 kPa; V = 218 L T = 295 K p = 13.70 kPa n = unknown pV = nRT Solving for n: 13.70 kPa  218 L  pV amount in moles of air lost = n = = 1.217646 mol  L•kPa  RT   8.31446 mol•K   295 K     28.8 g  1 min  Time = 1.217646 mol     = 23.379 min= 23 min  1 mol  1.5 

4.149

Plan: Rearrange the ideal gas law to calculate the density of O2 and O3 from their molar masses. Temperature must be converted to kelvin and pressure to kPa. Solution: p = 1.01 bar 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: 1.01 bar  32.00 g/mol  = 1.42396 g = 1.42 g O /L pM d of O2 =  2 L•bar  RT  0.08314 273 K    mol•K   1.01 bar  48.00 g/mol  = 2.13594 g= 2.14 g O /L pM d of O3 =  3 L•bar  RT  0.08314 273 K    mol•K   b) dozone/doxygen = (2.13594 g)/(1.42396 g) = 1.5 The ratio of the density is the same as the ratio of the molar masses.

4.150

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 amount in moles of IF7 produced by each reactant. The mass of I 2 is converted to amount in moles using its molar mass and the amount in moles of F 2 is found using the ideal gas law. The smaller amount in moles of product indicates the limiting reagent. Determine the amount in moles of excess reactant gas and the amount in moles of product gas and use the ideal gas law to solve for the total pressure. Solution: amount in moles of F2: V = 2.50 L T = 250. K p = 46.7 kPa n = unknown pV = nRT Solving for n:  46.7 kPa  2.50 L  = 0.0561672 mol F pV n=  2 L•kPa  RT   8.31446 mol•K   250. K    7F2(g) + I2(s)  2IF7(g  2 mol IF7  amount in moles IF7 from F2 =  0.0561672 mol F2    = 0.01604777 mol IF7 (unrounded)  7 mol F2 

 1 mol I 2  2 mol IF7  amount in moles IF7 from I2 =  2.50 g I 2     = 0.019700551 mol IF7 (unrounded)  253.8 g I2  1 mol I2  F2 is limiting. All of the F2 is consumed. amount in mole I2 remaining = original amount of moles of I2 – amount of I2 (mol) reacting with F2  1 mol I 2   1 mol I 2  –3 amount in mole I2 remaining =  2.50 g I 2    –  0.0561672 mol F2    = 1.82639x10 mol I2  253.8 g I 2   7 mol F2  Total amount in moles of gas = (0 mol F2) + (0.01604777 mol IF7) + (1.82639x10–3 mol I2) = 0.01787416 mol gas V = 2.50 L T = 550. K p = unknown n = 0.01787416 mol pV = nRT

Solving for P: p (kpa) =

nRT = V

 0.01787416 mol   8.31446  2.50 L

L•kPa   550. K  mol•K 

= 32.6950 kPa

 1 bar  p (bar) =  32.6950 kPa    = 0.32695 bar = 0.327 bar  100 kPa  piodine (bar) = Xiodine ptotal = [(1.82639x10–3 mol I2)/(0.01787416 mol)] (0.32695 bar) = 0.033408 bar = 33.4x10-3 bar 4.151

Plan : We will use the mass of carbon dioxide produced to find the amount in mol and the mass of carbon. Similarly, we will use the mass of water to find the amount in mol of water, then hydrogen and the mass of hydrogen. We will use the mass of the compound and the masses of carbon and hydrogen to find the mass of oxygen by difference, and then amount in mol of oxygen. We will use the mol amounts of C, H, and O to find the empirical formula of MTBE. We will use the mass, volume, temperature and pressure of the MTBE sample to find its actual molar mass, and finally the molar mass and the mass of the empirical unit to find the number of units and thus the molecular formula of MTBE. Solution: nC  nCO2 

mCO2 M CO2



8.2145 g  0.18665 mol C 44.01 g/mol

mC  nC  M C  (0.18665 mol C)(12.01 g/mol)= 2.2417 g

nH2 O 

mH2 O M H2 O



4.0326 g  0.22391 mol H 2 O 18.01 g/mol

nH  2nH2 O  2  0.22391 mol H 2 O = 0.44782 mol H mH  nH  M H  (0.44782 mol H)(1.01 g/mol)=0.45230 g

mO  mMTBE  mC  mH  3.2914g  2.2417g  0.45230g=0.5974g nO 

mO 0.5974g   3.7338  102 mol M O 16.00g/mol

Making the amount (mol) of each element the subscript in the formula gives C0.18665 H 0.44782 O3.7338102 Dividing by the smallest number gives C 0.18665 H 0.44782 O 3.7338102 3.7338102

3.7338102

C5 H12 O1 The empirical formula of MTBE is C5 H12O . To find the molar mass of MTBE, we can use the ideal gas law variation. bar  L (0.317 g)(0.08314 )(297.85 K) mRT mol K M   88.18 g / mol pV (0.3561 bar)(0.2500 L) The molecular formula will be some multiple of the empirical formula. To find this multiple, we simply divide the molar mass of MTBE by the mass of the empirical unit. M empirical unit  5  M C  12  M H  M O  5(12.01 g/mol)+12(1.01 g/mol)+(16.00 g/mol)=88.17g/mol M MTBE 88.18 g/mol Number of units   1 M empirical unit 88.17 g/mol Therefore, the molecular formula of MTBE is the same as the empirical formula.

4.152

The molecular formula of MTBE is C5 H12O . Plan: We balance the equation for hydrogen and oxygen since the carbon and calcium are already balanced. CaC2 (s)  2 H2 O (l )  Ca(OH)2 (s) + C2 H2 (g ) We use Dalton‘s law of partial pressures to determine the pressure of the ethyne gas, and then the ideal gas law to determine the amount in mol of ethyne collected. We then use stoichiometry to determine the amount of calcium carbide that should have reacted in mol and then gram.

Ptot  Patm  Pethyne  Pwater vapour Pethyne  Patm  Pwater vapour  (0.971 bar)  (3.3639 kPa)(

1 bar )  0.9374 bar 100 kPa

pV (0.9374 bar)(0.325 L)   1.225×10-2 mol bar  L RT (0.08314 )(299.15 K) mol  K Since the stoichiometry of the balanced reaction shows that 1 mol of CaC 2 is needed to produce 1 mol of C2H2, we can calculate the mass of CaC2 needed directly. nCaC2  nC2 H2 n

mCaC2  nCaC2  MM CaC2  (1.225 102 mol)(64.098 g/mol))=0.785 g

4.153

Plan : We will use the mass of carbon dioxide produced to find the amount in mol and the mass of carbon. By difference, we can find the mass and amount (mol) of chlorine. We will use the mol amounts of C and Cl to find the empirical formula of TCE. We will use the mass of water needed to fill the flask and the density of water to find the volume of the flask. We will use the mass, volume, temperature and pressure of the TCE sample to find its actual molar mass, and finally the molar mass and the mass of the empirical unit to find the number of units and thus the molecular formula of TCE. Solution: a)

nC  nCO2 

mCO2 M CO2



6.4913 g  0.14750 mol C 44.01 g/mol

mC  nC  M C  (0.14750 mol C)(12.01 g/mol)=1.7714 g

mCl  mTCE  mC  12.2324g  1.7714g=10.4610g nCl 

mCl 10.4610g   0.2951mol M Cl 35.45g/mol

Making the amount (mol) of each element the subscript in the formula gives C0.1475 Cl0.2951 Dividing by the smallest number gives C 0.1475 Cl 0.2951 0.1475

0.1475

CCl2 b)

The empirical formula of TCE is CCl 2 . The mass of the water in the flask is: mwater  mflask +water  mflask

 291.75 g  43.8879 g  247.86 g Since the density of water is 1.00 g/mL, the volume of the flask is:

Vflask 

mwater d water

247.86 g 1.00 g/mL  247.86 mL To find the molar mass of TCE, we can use the ideal gas law variation. bar  L (2.2323 g)(0.08314 )(303.85 K) mRT mol  K M   165.8 g / mol pV (1.372 bar)(0.24786 L) The molecular formula will be some multiple of the empirical formula. To find this multiple, we simply divide the molar mass of TCE by the mass of the empirical unit. M empirical unit  M C  2  M Cl  (12.01 g/mol)+2(35.45g/mol)=82.9g/mol 

4.154

Therefore, the molecular formula of TCE is twice that of the empirical formula. The molecular formula of TCE is C2Cl 4 . Plan : We will use the temperature, density and pressure of the ethoxyethane sample to find its actual molar mass. We will use the mass of carbon dioxide produced to find the amount in mol and the mass of carbon. Similarly, we will use the mass of water to find the amount in mol of water, then hydrogen and the mass of hydrogen. We will use the mass of the compound and the masses of carbon and hydrogen to find the mass of oxygen by difference, and then amount in mol of oxygen. We will use the mol amounts of C, H, and O to find the empirical formula of ethoxyethane. Finally, we will use the molar mass and the mass of the empirical unit to find the number of units and thus the molecular formula of ethoxyethane. Solution: a) To find the molar mass of ethoxyethane, we can use the ideal gas law variation. bar  L (1.78 g/L)(0.08314 )(293.15 K) dRT mol K M   74.2 g / mol p (0.585 bar) b) We find the empirical formula as follows: mCO2 6.076 g nC  nCO2    0.13806 mol C M CO2 44.01 g/mol mC  nC  M C  (0.13806 mol C)(12.01 g/mol)=1.65809 g

nH2 O 

mH2 O M H2 O



3.107 g  0.1725 mol H 2 O 18.01 g/mol

nH  2nH2 O  2  0.1725 mol H 2 O = 0.3450 mol H mH  nH  M H  (0.3450 mol H)(1.01 g/mol)= 0.3485 g mO  methoxyethane  mC  mH  2.56g  1.65809g  0.3485g=0.553g

nO 

mO 0.553g   3.46  102 mol M O 16.00g/mol

Making the amount (mol) of each element the subscript in the formula gives C0.13806 H 0.3450 O3.46102

Dividing by the smallest number gives C 0.13806 H 0.3450 O 3.46102 3.46102

3.46102

C4 H10 O1 The empirical formula of ethoxyethane is.

The molecular formula will be some multiple of the empirical formula. To find this multiple, we simply divide the molar mass of ethoxyethane by the mass of the empirical unit. M empirical unit  4  M C  10  M H  M O

 4(12.01 g/mol)+10(1.01 g/mol)+(16.00 g/mol)=74.14g/mol M ethoxyethane 74.2 g/mol Number of units   1 M empirical unit 74.14 g/mol

4.155

Therefore, the molecular formula of ethoxyethane is the same as the empirical formula. The molecular formula of ethoxyethane is C4 H10O . Plan: We will balance the reaction for oxygen since Na and Cl are already balanced and double it to remove the fractional coefficient. a) 2NaClO3 (s)  2NaCl (s) + 3O2 (g ) b) We will use the ideal gas law to determine the amount (mol) of oxygen gas and then use stoichiometry to determine the amount (mol) of NaClO3 and then its mass. pV (1.041 bar)(1355 L) n   55.69 mol RT (0.08314 bar  L )(304.65 K) mol  K Since the stoichiometry of the balanced reaction shows that 2 mol of NaClO 3 is needed to produce 3 mol of O2, we can calculate the mass of NaClO3 needed. 2 2 nNaClO3  nO2  (55.69 mol)= 37.13 mol 3 3 mNaClO3  nNaClO3  M NaClO3  (37.13 mol)(64.098 g/mol))= 3.95  103 g = 3.95 kg c)

4.156

We can calculate the number of candles by finding how much oxygen would be provided by one candle. Then we determine how much oxygen is needed for the 10 miners for 3 days, and then we connect the two pieces of information.  1 kg of the candle mix provides oxygen for an adult to breathe for 6.5 h. 1 candle is 3.5 kg. 3.5 kg 6.5 h h   22.75 Therefore, 1 candle will provide oxygen for candle kg candle  They are trapped for 3 days so the total time is 3 days x 24 h/day = 72 h  There are 10 miners so they need enough oxygen for 10 miners x 72h/miner = 720 h 720 h No. of candles=  32 candles  22.75 h/candle

Plan: Use the mass of CO2 formed to find amount (mol) of CO 2 and then amount (mol) C, then mass C. Use the mass of H2O to find amount (mol) of H2O and then with stoichiometry, amount (mol) H and mass H. Find mass O by difference, using masses of C, H and the sample. Find amount (mol) O. Use the amount (mol) of C, H, and O to find the empirical formula by placing the amount (mol) as subscripts and dividing by the smallest one. Calculate the mass of the empirical unit. Calculate the MM of the compound using the effusion data. Use the mass of the empirical unit and the molar mass of the compound to find the molecular formula. Solution:

(

)( (

(

)

) )(

)

C4.5H5O  multiplying by 2  C9H10O2

√

)(

(

Mempformula = 150. g/mol

)

Therefore, since the M of the empirical formula and the MM of the compound are the same, the empirical formula and the molecular formula are both C 9H10O2. 4.157

Plan: Use the mass of CO2 formed to find amount (mol) of CO 2 and then amount (mol) C, then mass C. Use the mass of H2O to find amount (mol) of H2O and then with stoichiometry, amount (mol) H and mass H. Find mass O by difference, using masses of C, H and the sample. Find amount (mol) O. Use the amount (mol) of C, H, and O to find the empirical formula by placing the amount (mol) as subscripts and dividing by the smallest one. Calculate the mass of the empirical unit. Calculate the MM of the compound using a variation of the ideal gas law. Use the mass of the empirical unit and the molar mass of the compound to find the molecular formula. Solution:

(

)( (

( -

) )

)( -

) -

C6H10O

Mempformula = 98. g/mol

(

)( (

)( )(

) )

Therefore, the molecular formula is twice the empirical formula. The empirical formula is C6H10O and the molecular formula is C12H20O2. 4.158

Plan: a) Convert mass of FeO to amount (mol) and volume of CO to amount (mol) using T and p information. Determine the limiting reagent (LR) using stoichiometry. Use LR to determine amount of Fe produced (mol) and then molar mass of Fe to find mass. b) Convert mass of iron to amount (mol). Use stoichiometry to determine amount (mol) of CO2 needed. Then find volume using ideal gas law. Remember to watch units throughout.

Solution: a) (

)(

(

)

)(

since the reaction is 1:1 stoichiometry, CO is LR

)

Therefore, nFe = 6.25 mol (

)(

)

since it is a 1:1 stoichiometry (

4.159

)(

)

Plan: Find the pressure of dry hydrogen gas. Use the pressure of dry hydrogen gas and the volume of gas collected at the temperature of the water to calculate amount (mol) of hydrogen gas. Use mass/mol conversion and stoichiometry of reaction of metal with acid to express amount (mol) of hydrogen gas in terms of mass of each metal. Use the mass of the sample to connect the two metal masses. Solve the system of equations to find the mass of one of the metals and use that mass to find the mass of the other metal. Use the mass of the sample and the mass of each metal to find the percent composition of the sample. Solution: ( ( (

)(

)

) )(

)

The total amount of hydrogen gas is the sum of the amount of hydrogen that comes from the aluminum in the alloy and the amount of hydrogen that comes from the zinc in the alloy

The amount of hydrogen that comes from Al can be determined from the stoichiometry of the reaction of Al and HCl. Al (s) + 3 HCl (aq)  AlCl3 (aq) + 3/2 H2 (g) From this stoichiometry, we can see that for every mol of Al, 1.5 mol of hydrogen gas are produced. Thus, the amount (mol) of hydrogen gas produced from Al is equal to 1.5 times the amount (mol) of Al.

The amount of hydrogen that comes from Zn can be determined from the stoichiometry of the reaction of Zn and HCl. Zn (s) + 2 HCl (aq)  ZnCl2 (aq) + H2 (g) From this stoichiometry, we can see that for every mol of Zn, 1 mol of hydrogen gas is produced. Thus, the amount (mol) of hydrogen gas produced from Zn is equal to the amount (mol) of Al. Thus,

Substituting m/MM for n, we can write,

Here, we have an equation in 2 unknowns, the mAl and the mZn. We need another equation, which we obtain from the mass of the sample.

At this point we can solve the system using simultaneous equations, matrices, or more simply, substitution.

mAl = 8.10 x 10-2 g mZn = 3.47 x 10-2 g (note that while properly rounded numbers are written in the calculation above, the full numbers were entered at all steps into the calculator)

CHAPTER 5 THERMOCHEMISTRY: ENERGY FLOW AND CHEMICAL CHANGE CHEMICAL CONNECTIONS BOXED READING PROBLEMS B5.1

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:  103 g   1 mol Na 2SO4 •10H 2 O    354 kJ Heat (kJ)=q =  500.0 kg Na 2SO4 •10H 2 O    1 kg   322.21 g Na SO •10H O   1 mol Na SO •10H O  2 4 2  2 4 2    5 5 = –5.4933x10 kJ= – 5.49x10 kJ

B5.2

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 r H values for the three reactions to get the r H for the overall gasification reaction of 2 moles of coal. Use the relationship r H = m  f (products) H – n  f (reactants) H to find the heat of combustion of 1 mole of methane. Then find the r H for the gasification of 1.00 kg of coal and r H for the combustion of the methane produced from 1.00 kg of coal and sum these values. Solution: a) 1) 2C(s, coal) + 2H2O(g)  2CO(g) + 2H2(g) 2) CO(g) + H2O(g)  CO2(g) + H2(g)

r H = 2(129.7 kJ/mol)

r H = – 41 kJ/mol

r H = –206 kJ/mol 3) CO(g) + 3H2(g)  CH4(g) + H2O(g) 2C(s, coal) + 2H2O(g)  CH4(g) + CO2(g) Silberberg, Amateis, Venkateswaran, Chen, Chemistry: The Molecular Nature of Matter and Change, 3rd Canadian Edition Page 3-210 Instructor‘s Solution Manual © Copyright 2021 McGraw-Hill Ryerson Ltd.

b) The total may be determined by doubling the value for equation 1) and adding to the other two values.

r H = 2(129.7 kJ/mol) + (–41 kJ/mol) + (–206 kJ/mol) = 12.4 kJ/mol= 12 kJ/mol c) Calculating the heat of combustion of CH4: CH4(g) + 2O2(g) → CO2(g) + 2H2O(g) r H = m  f (products) H – n f (reactants) H r H = [(  f H of CO2) + 2(  f H of H2O)] – [(  f H of CH4) + 2(  f H of O2)]

r H = [1(–393.5 kJ/mol) + 2(–241.826 kJ/mol)] – [1(–74.87kJ/mol) + 2(0.0 kJ/mol)]

r H = –802.282 kJ/mol Total heat for gasification of 1.00 kg coal:  103 g   1 mol coal  12.4 kJ  qtot = 1.00 kg coal  = 516.667 kJ  1 kg   12.00 g coal  2 mol coal      Total heat from burning the methane formed from 1.00 kg of coal: qtot = (

)(

)= –33428.42 kJ

Total heat = qtot= 516.667 kJ + (–33511.75 kJ) = –32995.083 kJ= –3.30x104 kJ

END–OF–CHAPTER PROBLEMS 5.1

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.

5.2

No, an increase in temperature means that heat has been transferred to the surroundings, which makes q negative.

5.3

U = q + w = w, since q = 0. Thus, the change in work equals the change in internal energy.

5.4

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 one person‘s warm hands on the cold hands of another. The body can lose energy if it performs work, like pushing a lawnmower, and can lose energy by losing heat to a cold room.

5.5

a) electric heater e) battery (voltaic cell)

5.6

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.

5.7

Heat energy; sound energy  Kinetic energy  Potential energy  Mechanical energy  Chemical energy

b) sound amplifier

c) light bulb

d) automobile alternator

(impact) (falling text) (raised text) (raising of text) (biological process to move muscles)

5.8

Plan: The change in a system‘s energy is U = 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: U = q + w U = (+425 J/mol) + (–425 J/mol) = 0 J/mol

5.9

q + w = –255 cal/mol + (–428 cal/mol) = –683 cal/mol

5.10

Plan: The change in a system‘s energy is U = 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/mol) =  530 cal/mol    = 2217.52 J/mol  1 cal  U = q + w = –675 J/mol + 2217.52 J/mol = 1542.52 J/mol = 1.54x103 J/mol

5.11

5.12

 103 J    103 cal   4.184 J   U = q + w =  0.615 kJ / mol   +   0.247 kcal / mol     1 kcal   1 cal    1 kJ         = 1648.4 J/mol = 1.65x103 J/mol Plan: Convert 6.6x1010 J to the other units using conversion factors. Solution: C(s) + O2(g)  CO2(g) + 6.6x1010 J (2.0 tonnes)  1 kJ  a) U (kJ) = (6.6 x 1010 J)  3  = 6.6x107 kJ  10 J   

 1 cal   1 kcal  7 7 b) U (kcal) = (6.6 x 1010 J)   = 1.577x10 kcal= 1.6x10 kcal   3 4.184 J 10 cal    5.13

CaCO3(s) + 9.0x106 kJ  CaO(s) + CO2(g) (5.0 tonnes)  103 J  a) U (J) = (9.0x106 kJ)  = 9.0x109 J  1 kJ   

 103 J   1 cal  b) U (cal) = (9.0x106 kJ)  = 2.15105x109 cal= 2.2x109 cal  1 kJ   4.184 J    

5.14

 103 cal   4.184 J  U(J) = (4.1x103 Calorie)  = 1.7154x107 = 1.7x107 J = 1.7x104 kJ  1 Calorie   1 cal    

5.15

Plan: 454 g of body fat is equivalent to about 4.1x10 3 Calories. Convert Calories to kJ with the appropriate conversion factors. Solution:  4.1x103 Cal  103 cal   4.184 J   1 kJ   h  Time =  454g   = 8.79713 h= 8.8 h  454 g   1 Cal   1 cal   103 J   1950 kJ       

5.16

The system does work and thus its internal energy is decreased. This means the sign will be negative.

5.17

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 U requires measurement of heat and PV work.

5.18

The hot pack is releasing (producing) heat, thus H is negative, and the process is exothermic.

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.

5.20

The internal energy of a substance is the sum of kinetic (UK) and potential (UP) terms. UK (total) = UK (translational) + UK (rotational) + UK (vibrational) UP = UP (atom) + UP (bonds) UP (atom) has nuclear, electronic, positional, magnetic, electrical, etc., components.

5.21

H = U + pV (constant p) a) H < U, pV is negative since Vfinal < Vinitial. b) H = U, a fixed volume means pV = 0. c) H > U, pV is positive for the transformation of solid to gas.

5.22

Plan: An exothermic reaction releases heat, so the reactants have greater H (Hinitial) than the products (Hfinal). H = Hfinal – Hinitial < 0. Solution: Increasing, H

5.19

Reactants H = (), (exothermic) Products

Increasing, H

5.23 Products H = (+), (endothermic) Reactants

5.24

Plan: Combustion of hydrocarbons and related compounds require oxygen (and a heat catalyst) to yield carbon dioxide gas, water vapour, 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).

Increasing, H

Solution: a) Combustion of ethane: 2C2H6(g) + 7O2(g)  4CO2(g) + 6H2O(g) + heat 2C2H6 + 7O2 (initial)

H = (), (exothermic) 4CO2 + 6H2O (final)

Increasing, H

b) Freezing of water: H2O(l)  H2O(s) + heat H2O(l) (initial)

H2O(s) (final)

a) Na(s) + 1/2Cl2(g)  NaCl(s) + heat Na(s) + 1/2Cl2(g)

Increasing, H

5.25

H = (), (exothermic)

exothermic NaCl(s)

Increasing, H

b) C6H6(l) + heat  C6H6(g) C6H6(g)

C6H6(l)

Plan: Combustion of hydrocarbons and related compounds require oxygen (and a heat catalyst) to yield carbon dioxide gas, water vapour, 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 2CH3OH + 3O2 (initial)

Increasing, H

5.26

H = (+), (endothermic)

H = (), (exothermic) 2CO2 + 4H2O (final)

Increasing, H

b) Nitrogen dioxide, NO2, forms from N2 and O2. 1/2N2(g) + O2(g) + heat  NO2(g) NO2 (final)

1/2N2 + O2 (initial)

a) CO2(s) + heat  CO2(g) CO2(g)

Increasing, H

5.27

H = (+), (endothermic)

H = (+), (endothermic) CO2(s)

Increasing, H

b) SO2(g) + 1/2O2(g)  SO3(g) + heat SO2(g) + 1/2O2(g)

H = (), (exothermic) SO3(g)

5.28

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 U = q + w, the work must be considered in addition to qsys to find ΔUsys. 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 (+). 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 is negative. ΔUsys = q + w. Since q is positive and w is negative, the sign of ΔUsys cannot be predicted. It will be positive if q > w and negative if q < w. c) ΔUuniv = 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.

5.29

a) There is a volume decrease; Vfinal < Vinitial so ΔV is negative. Since wsys = –PΔV, w is positive, +. b) ∆Hsys is – as heat has been removed from the system to liquefy the gas. c) ∆Usys = q + w. Since q is negative and w is positive, the sign of ΔUsys and ΔUsurr cannot be predicted. ΔUsys will be positive and ΔUsurr will be negative if w > q and ΔUsys will be negative and ΔUsurr will be positive if w < q.

5.30

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.

5.31

To determine the specific heat capacity of a substance, you need its mass, the heat added (or lost), and the change in temperature.

5.32

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.

5.33

Heat capacity is the quantity of heat required for a 1 K temperature change as applied to an object instead of a specified amount of a substance. Specific heat capacity is the quantity of heat required to raise 1g of a substance by 1 K. Molar heat capacity is the quantity of heat required to raise 1 mole of substance by 1 K. 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.

5.34

In a coffee-cup calorimeter, reactions occur at constant pressure. qp = H. In a bomb calorimeter, reactions occur at constant volume. qv = U.

5.35

Plan: The heat required to raise the temperature of water is found by using the equation q = c x mass x T. The specific heat capacity, cwater, is found in Table 5.2. Because 1°C=1K, T = 100°C – 25°C = 75°C = 75 K. Solution:  J  3 q (J) = c x mass x T =  4.184  22.0 g 75 K  = 6903.6 J= 6.9x10 J g K  

5.36

5.37

 J  q (J) = c x mass x T =  2.087  0.10 g  75  10.K = –17.7395 J= –18 J g K 





Plan: Use the relationship q = c x mass x 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. Solution: q (J) = c x mass x T Tinitial = 13.00°C Tfinal = ? mass = 295 g c = 0.900 J/g•K 3  10 J  q = 75.0 kJ  = 7.50x104 J  1 kJ    7.50x104 J = (0.900 J/g•K)(295 g)(T)

7.50x10 J  4

T =

 0.900 J    gK 

 295 g  

T = 282.4859 K = 282.4859°C

(Because 1°C=1K, T is the same in either temperature unit.)

T = Tfinal – Tinitial Tfinal T + Tinitial Tfinal = 282.4859°C + 13.00°C = 295.49 °C= 295°C 5.38

q (J) = c x mass x T –688 J = (2.42 J/g•K)(27.7 g)(T) 688 J  (T) = = –10.26345 K = –10.26345°C  2.42 J  27.7 g    gK  T = Tfinal – Tinitial Tinitial Tfinal – T Tinitial = 32.5°C – (–10.26345°C) = 42.76345°C = 42.8°C

5.39

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:  T1 + T2   100.C + 55C    =   = 77.5°C 2 2    

5.40

Plan: Use the fact that no energy is gained or lost, only transferred. Therefore, heat lost equals heat gained. Set equal, substitute terms and solve for the final T. Solution: –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

5.41

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 c x mass x T (85°C water)= c x mass x 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°C = 45°C

5.42

Plan: Use the fact that no energy is gained or lost, only transferred. Therefore, heat lost equals heat gained. Set equal, substitute terms and solve for the V. Solution: –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 – (24.4g)(23.5 – 35.0) = (mass)(23.5 – 18.2) – (24.4g)(–11.5) = (mass)(5.3) 280.6g = (mass)(5.3) 52.943 g = mass  1 mL  Volume (mL) = 52.943 g   = 52.943 mL= 53 mL  1.00 g 

5.43

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 unit 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 5.44

Plan: Use the fact that no energy is gained or lost, only transferred. Therefore, heat lost equals heat gained. Set equal, substitute terms and solve for the calloy. Solution: –qlost = qgained = qwater + qcalorimeter – (30.5 g alloy)(calloy)(31.1 – 93.0)°C = (50.0 g H2O)(4.184 J/g·K)(31.1 – 22.0)°C + (9.2 J/K)(31.1 – 22.0)°C – (30.5 g)(calloy)(–61.9K) = (50.0 g)(4.184 J/g·K)(9.1K) + (9.2 J/K)(9.1K) 1887.95 g K(calloy) = 1903.72 J + 83.72 J= 1987.44 J calloy = 1987.44J/1887.95 g·K= 1.052697 J/g·K= 1.1 J/g·K

5.45

Benzoic acid is C6H5COOH, and will be symbolized as HBz. –qreaction = qwater + qcalorimeter  1 mol HBz   3227 kJ   103 J  4 –qreaction = – 1.221 g HBz     1 kJ  = 3.226472x10 J 122.12 g HBz 1 mol HBz     qwater = c x mass x T = 4.184 J/g·K x 1200 g x T qcalorimeter = C x T = 1365 J/°C x T –qreaction = qwater + qcalorimeter 3.226472x104 J = 4.184 J/g·K x 1200 g x T + 1365 J/K x T 3.226472x104 J = 5020.8J/K(T) + 1365J/K(T) 3.226472x104 J = 6385.8J/K(T) T = 3.226472 x104 J/6385.8J/K = 5.052573K = 5.053°C

5.46

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) – (7.741/°C)(Tfinal – 100.0°C) = (13.5 1/°C)(Tfinal – 0.0°C) 774 – 7.74 1/°C Tfinal = 13.5 1/°C Tfinal 774 = (13.5 + 7.74)1/°C Tfinal = 21.24 1/°C Tfinal Tfinal = 774 °C/21.24 = 36.44068 °C= 36.4°C

5.47

Plan: Use the fact that no energy is gained or lost, only transferred. Therefore, heat lost by hydrocarbon equals heat gained by water and calorimeter. Set equal, substitute terms and solve for the final T. Solution: –qhydrocarbon = qwater + qcalorimeter –qhydrocarbon = (2.550 L H2O)(1mL /10–3L)(1.00g/mL)(4.184 J/g°C)(23.55 – 20.00)°C + (403 J/°C)(23.55 – 20.00)°C –qhydrocarbon = (2550. g)(4.184 J/g°C)(3.55°C) + (403 J/°C)(3.55°C) –qhydrocarbon = (37875.66 J) + (1430.65 J) = 39306.31 J qhydrocarbon = –3.930631x104 J qhydrocarbon/g = (–3.930631x104 J)/1.520 g = –2.5859x104 J/g= –2.59x104 J/g

5.48

Plan: Solve for q using the information given. Calculate the amount (mol) of sulfuric acid and potassium hydroxide. Use stoichiometry to determine the limiting reagent. Determine the amount (mol) of the limiting reagent. Calculate the enthalpy per mole of the limiting reagent. Solution: The reaction is: 2KOH(aq) + H2SO4(aq)  K2SO4(aq) + 2H2O(l) q (kJ) = (25.0 + 25.0) mL(1.00 g/mL)(4.184 J/g°C)(30.17 – 23.50)°C(1 kJ/103 J) = 1.395364 kJ

(The temperature increased so the heat of reaction is exothermic.) Amount (moles) of H2SO4 = (25.0 mL)(0.500 mol H2SO4/L)(10–3 L/1 mL) = 0.0125 mol H 2SO4 Amount (moles) of KOH = (25.0 mL)(1.00 mol KOH/L)(10 –3 L/1 mL) = 0.0250 mol KOH The moles show that both H2SO4 and KOH are limiting. The enthalpy change could be calculated in any of the following ways: H = –1.395364 kJ/0.0125 mol H2SO4 = – 111.62912 kJ/mol= – 112 kJ/mol H2SO4 H = –1.395364 kJ/0.0250 mol KOH = –55.81456 kJ/mol= –55.8 kJ/mol KOH (Per mole of K2SO4 gives the same value as per mole of H2SO4, and per mole of H2O gives the same value as per mole of KOH.) 5.49

Plan: Determine if the reaction has heat as reactant or product. If reaction is exothermic, heat is a product. If reaction is endothermic, heat is a reactant. Solution: Reactants  Products + Energy rH = (–) Thus, energy is a product.

5.50

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: The reaction has a positive rH, because this reaction requires the input of energy to break the oxygen-oxygen bond in O2: O2(g) + energy  2O(g)

5.51

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 vapourization of 2 mol of water would be twice the value of H for the condensation of 1 mol of water vapour but would have an opposite sign (+H). H2O(g)  H2O(l) + Energy 2H2O(l) + Energy  2H2O(g) condensationH = (–) vapourizationH = (+)2[condensationH] 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 vapour.

5.52

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 rH 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 per 1/8 mol of S8 produced.  20.2 kJ  c) q = 2.6 mol S8  = –420.16 kJ= –4.2x102 kJ  1/ 8 mol S  8   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  q= 25.0 g S8   = –15.7468 kJ= –15.7 kJ    256.56 g S8   1/ 8  mol S8 

5.53

MgCO3(s)  MgO(s) + CO2(g) a) Absorbed

rH = 117.3 kJ/mol

b) rH (reverse) = –117.3 kJ/mol  117.3 kJ  c) q = 5.35 mol CO 2   = –627.555 kJ= –628 kJ  1 mol CO 2 

 1 mol CO2  117.3 kJ  d) q= 35.5 g CO 2    = –94.618 kJ= –94.6 kJ  44.01 g CO 2  1 mol CO 2  5.54

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)  (g) H = 90.29 kJ/mol  1 mol NO  90.29 kJ  b) q= 3.50 g NO    = –10.5303 kJ= –10.5 kJ  30.01 g NO  1 mol NO  5.55 a) KBr(s)  K(s) + 1/2Br2(l) rH = 394 kJ/mol 3  10 g   1 mol KBr  394 kJ  b) q= 10.0 kg KBr  = –3.3109x104 kJ= –3.31x104 kJ  1 kg   119.00 g KBr  1 mol KBr     

5.56

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) rH = –196.1 kJ/mol 3  10 g   1 mol H 2 O2  196.1 kJ  Heat (kJ) = q = 652 kg H 2 O2  = –1.87915x106 kJ= –1.88x106 kJ  1 kg   34.02 g H O  2 mol H O  2 2  2 2   

5.57

Plan: For the reaction written, 1 mole of B2H6 releases 755.4 kJ of energy upon reaction. Use this ratio to convert between the given amount of reactant and the amount of heat released. Solution: B2H6(g) + 6Cl2(g)  2BCl3(g) + 6HCl(g) rH = –755.4 kJ/mol 3  10 g   1 mol B2 H6  755.4 kJ  Heat (kJ) = q = 1 kg  = –2.73003x104 kJ/kg= –2.730x104 kJ/kg  1 kg   27.67 g B H  1 mol B H  2 6  2 6   

5.58

4Fe(s) + 3O2(g)  2Fe2O3(s)

rH = –1.65x103 kJ/mol  103 g   1 mol Fe   1.65x103 kJ  a) Heat (kJ) = q =  0.250 kg Fe    1 kg   55.85 g Fe   4 mol Fe  = –1846.46 kJ= –1850 kJ    

 2 mol Fe2O3   159.70 g Fe 2O3  b) Mass (g) of Fe2O3 = 4.85x103 kJ   = 938.84 g= 939 g Fe2O3  3  1.65x10 kJ   1 mol Fe2O3 



5.59



2HgO(s)  2Hg(l) + O2(g)

rH = 181.6 kJ/mol  1 mol HgO  181.6 kJ  a) Heat (kJ) = q =  555 g HgO     = 232.659kJ = 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 g= 608 g Hg  181.6 kJ  1 mol Hg 

5.60

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(g) rH = –1411 kJ/mol  1 mol C2 H 4   28.05 g C 2 H 4  b) Mass (g) of C2H4 =  70.0 kJ    = 1.39157 g= 1.39 g C2H4   1411 kJ  1 mol C2 H 4 

5.61

a) C12H22O11(s) + 12O2(g)  12CO2(g) + 11H2O(g) rH = –5.64x103 kJ/mol  1 mol C12 H 22O11   5.64x103 kJ  b) Heat (kJ) = q = 1 g C12 H 22O11    = –16.47677 kJ/g= –16.5 kJ/g    342.30 g C12 H 22 O11   1 mol C12 H 22O11 

5.62

Hess‘s law: rH is independent of the number of steps or the path of the reaction.

5.63

Hess‘s law provides a useful way of calculating energy changes for reactions which are difficult or impossible to measure directly.

5.64

Plan: Two chemical equations can be written based on the description given: C(s, graphite) + O2(g)  CO2(g) H (1) CO(g) + 1/2O2(g)  CO2(g) H (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, graphite) + O2(g)  CO2(g) H CO2(g)  CO(g) + 1/2O2(g) –H (reaction is reversed) Total C(s, graphite) + 1/2O2(g)  CO(g) rH = H + – (H) How are the H values for each reaction determined? The H1 can be found by using the heats of formation in Appendix B: H = [fH (CO2)] – [fH (C) + fH (O2)] = [–393.5 kJ/mol] – [0 + 0] = –393.5 kJ/mol. The 2H can be found by using the heats of formation in Appendix B: H = [fH (CO2)] – [fH (CO) + 1/2fH (O2)] = [–393.5 kJ/mol] – [–110.5 kJ/mol + 0)] = –283 kJ/mol. rH = H+ – (H) = –393.5 kJ/mol + – (–283.0 kJ/mol) = –110.5 kJ/mol

5.65

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/mol CaO(s) + CO2(g)  CaCO3(s) H = –178.3 kJ/mol (reaction is reversed) Ca(s) + 1/2O2(g) + CO2(g) CaCO3(s) H = –813.4 kJ/mol

5.66

2NOCl(g)  2NO(g) + Cl2(g) 2NO(g)  N2(g) + O2(g) 2NOCl(g)  N2(g) + O2(g) + Cl2(g)

5.67

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. Solution: 1) N2(g) + O2(g)  2NO(g) H = 180.6 kJ/mol 2) 2NO(g) + O2(g)  2NO2(g) H = –114.2 kJ/mol

H = –2(–38.6 kJ/mol) H = –2(90.3 kJ/mol) H = 77.2 kJ + (– 180.6 kJ/mol) = –103.4 kJ/mol

3) N2(g) + 2O2(g)  2NO2(g) rH = +66.4 kJ/mol In Figure P5.67, 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. 5.68

1) 2)

P4(s) + 6Cl2(g)  4PCl3(g) 4PCl3(g) + 4Cl2(g)  4PCl5(g)

1H = –1148 kJ/mol 2H = – 460 kJ/mol

3) P4(s) + 10Cl2(g)  4PCl5(g)  rH= –1608 kJ/mol In Figure P5.68, Equation 1) = B, equation 2) = C, equation 3) = A 5.69

Plan: Vapourization is the change in state from a liquid to a gas: H 2O(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 vapourization. Solution: 1) Formation of H2O(g): H2(g) + 1/2O2(g)  H2O(g) H = –241.8 kJ/mol 2) Formation of H2O(l): H2(g) + 1/2O2(g)  H2O(l) H = –285.8 kJ/mol Reverse reaction 2 (change the sign of H) and add the two reactions: H2(g) + 1/2O2(g)  H2O(g) H = –241.8 kJ/mol H2O(l)  H2(g) + 1/2O2(g) H = +285.8 kJ/mol H2O(l)  H2O(g)

5.70

vapH = 44.0 kJ/mol

C(s) + 1/4S8(s)  CS2(l) CS2(l)  CS2(g)

H = +89.7 kJ/mol H = +27.7 kJ/mol

C(s) + 1/4S8(s)  CS2(g)

H = +117.4 kJ/mol

5.71

C (diamond) + O2(g)  CO2(g) CO2(g)  C(graphite) + O2(g) C(diamond)  C(graphite)

H = –395.4 kJ/mol H = –(–393.5 kJ/mol) H = – 1.9 kJ/mol

5.72

The standard heat of reaction, r H , is the enthalpy change for any reaction where all substances are in their standard states. The standard heat of formation,  f H , 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 1 bar for gases,1 mol/L 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.

5.73

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.

r H = m  f (products) H – n f (reactants) H 5.74

Plan:  f H is for the reaction that shows the formation of one mole of compound from its elements in their standard states. Solution: 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

5.75

Plan: Formation equations show the formation of one mole of compound from its elements. The elements must be in their most stable states (  f H = 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)

5.76

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)

5.77

r H = m  f (products) H – n f (reactants) H a) r H = {2  f H [SO2(g)] + 2  f H [H2O(g)]} – {2  f H [H2S(g)] + 3  f H [O2(g)]} = [2(–296.8 kJ/mol) + 2(–241.826 kJ/mol)] – [2(–20.2 kJ/mol) + 3(0.0 kJ/mol)] = –1036.9 kJ/mol b) The balanced equation is CH4(g) + 4Cl2(g)  CCl4(l) + 4HCl(g)

r H = {1  f H [CCl4(l)] + 4  f H [HCl(g)]} – {1  f H [CH4(g)] + 4  f H [Cl2(g)]} r H = [(–139 kJ/mol) + 4(–92.31 kJ/mol)] – [(–74.87 kJ/mol) + 4(0 kJ/mol)] = –433 kJ/mol 5.78

r H = m  f (products) H – n f (reactants) H a) r H = {1  f H [SiF4(g)] + 2  f H [H2O(l)]} – {1  f H [SiO2(s)] + 4  f H [HF(g)]} = [(–1614.9 kJ/mol) + 2(–285.840 kJ/mol)] – [(–910.9 kJ/mol) + 4(–273 kJ/mol)] = –184 kJ/mol b) 2C2H6(g) + 7O2(g)  4CO2(g) + 6H2O(g)

r H = {4  f H [CO2(g)] + 6  f H [H2O(g)]} – {2  f H [C2H6(g)] + 7  f H [ O2 (g)]} = [4(–393.5 kJ/mol) + 6(–241.826 kJ/mol)] – [2(–84.667 kJ/mol) + 7(0 kJ/mol)] = –2855.6 kJ/mol (or –1427.8 kJ for reaction of 1 mol of C2H6) 5.79

 f H of CuO must be calculated. Solution:

r H = m  f (products) H – n f (reactants) H Cu2O(s) + 1/2O2(g)  2CuO(s)

 rxn H = –146.0 kJ/mol

r H = {2  f H [CuO(s)]} – {1  f H [Cu2O(s)] + 1/2  f H [O2(g)]} Silberberg, Amateis, Venkateswaran, Chen, Chemistry: The Molecular Nature of Matter and Change, 3rd Canadian Edition Page 3-224 Instructor‘s Solution Manual © Copyright 2021 McGraw-Hill Ryerson Ltd.

–146.0 kJ/mol = {2  f H [CuO(s)]} – {(–168.6 kJ/mol) + 1/2(0 kJ/mol)} –146.0 kJ/mol = 2  f H [CuO(s)] + 168.6 kJ/mol

 f H [CuO(s)] =  5.80

314.6 kJ / mol = –157.3 kJ/mol 2

r H = m  f (products) H – n f (reactants) H  rxn H = –1255.8 kJ/mol

C2H2(g) + 5/2O2(g)  2CO2(g) + H2O(g)

r H = {2  f H [CO2(g)] + 1  f H [H2O(g)]} – {1  f H [C2H2(g)] + 5/2  f H [O2(g)]} –1255.8 kJ/mol = [2(–393.5 kJ/mol) + (–241.826 kJ/mol)] – [  f H [C2H2(g)] + 5/2(0.0 kJ/mol)] –1255.8 kJ/mol = –787.0 kJ/mol – 241.8 kJ/mol –  f H [C2H2(g)]

 f H [C2H2(g)] = 5.81

227.0 kJ/ mol = 227.0 kJ/mol 1

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  f H 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 r H ) and multiply the coefficients (and r H ) of the second reaction by 2. Solution: 2PbSO4(s) + 2H2O(l)  Pb(s) + PbO2(s) + 2H2SO4(l)

r H = m  f (products) H – n f (reactants) H a) r H = {1  f H [Pb(s)] + 1  f H [PbO2(s)] + 2  f H [H2SO4(l)]} – {2  f H [PbSO4(s)] + 2  f H [H2O(l)]} = [(0 kJ/mol) + (–276.6 kJmol) + 2(–813.989 kJ/mol)] – [2(–918.39 kJ/mol) + 2(–285.840 kJ/mol)] = 503.9 kJ/mol b) Use Hess‘s law:

5.82

PbSO4(s)  Pb(s) + PbO2(s) + 2SO3(g)

r H = –(–768 kJ/mol) Equation has been reversed.

2SO3(g) + 2H2O (l)  2H2SO4(l)

r H = 2(–132 kJ/mol)

2PbSO4(s) + 2H2O(l)  Pb(s) + PbO2(s) + 2H2SO4(l)

r H = 504 kJ/mol

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  f H 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 r H to find the heat involved for this amount of acid. For 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) r H = m  f (products) H – n  f (reactants) H

r H = {18  f H [CO2(g)] + 18  f H [H2O(g)]} – {1  f H [C18H36O2(s)] + 26  f H [O2(g)]} Silberberg, Amateis, Venkateswaran, Chen, Chemistry: The Molecular Nature of Matter and Change, 3rd Canadian Edition Page 3-225 Instructor‘s Solution Manual © Copyright 2021 McGraw-Hill Ryerson Ltd.

= [18(–393.5 kJ/mol) + 18(–241.826 kJ/mol)] – [(–948 kJ/mol) + 26(0 kJ/mol)] = –10,487.868 kJ/mol= –10,488 kJ/mol  1 mol C18 H36 O 2  10, 487.868 kJ  c) q (kJ) = 1.00 g C18 H36 O2     = –36.8681 kJ= –36.9 kJ  284.47 C18 H36 O2  1 mol C18 H36 O 2 

 1 kcal  q (kcal) =  36.8681 kJ    = –8.811688 kcal= –8.81 kcal  4.184 kJ 

 8.811688 kcal  d) q (kcal) = 11.0 g fat    = 96.9286 kcal= 96.9 kcal  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. 5.83

a) H2SO4(l)  H2SO4(aq)

r H = {1  f H [H2SO4(aq)]} – {1  f H [H2SO4(l)]} = [(–907.51 kJ/mol)] – [(–813.989 kJ/mol)] = –93.52 kJ/mol b) q (J) = c x mass x T  3.50 J   1.060 g  93.52 kJ x 103 J/kJ =   x 1000. mL    x Tfinal  25.0C   1 mL   g•C 

 3.50 J  9.352x104 J =   x 1060. g  x Tfinal  25.0C   g•C  9.352 x 104 J = (Tfinal)3710 J/ºC – 9.2750x104 J Tfinal = 50.1995 °C= 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. 5.84

Plan: Use the ideal gas law, pV = nRT, to calculate the volume of one mole of helium at each temperature. Then use the given equation for ΔE to find the change in internal energy. The equation for work, w = –pΔV, is needed for part c), and qP = ΔU + pΔV is used for part d). For part e), recall that ΔH = qP. Solution: nRT a) pV = nRT or V = p T = 273 + 15 = 288 K and T = 273 + 30 = 303 K L•bar    0.08314 mol•K   288 K  nRT   Initial volume (L) = V = = = 23.7072 L/mol= 23.7 L/mol p 1.01 bar 

L•bar    0.08314 mol • K   303 K  nRT   Final volume (L) = V = = = 24.9420 L/mol = 24.9 L/mol 1.01 bar p   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, U = 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. U = 3/2nRT = (3/2)(1.00 mol) (8.314 J/mol•K)(303 – 288)K = 187.065 J= 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•bar, so a conversion factor is needed to convert work in units of L•bar to joules.

 100 J  2 w = –pV =  1.01 bar  (24.9434  23.7085) L   = –124.72 J= –1.2x10 J  1 L•bar  d) qP = U + pV = (187.065 J) + (124.72 J) = 311.785 J = 3.1x102 J e) H = qP = 310 J. Explanation: 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 = U + pV = U – w = (q + w) – w = qP 5.85

a) Respiration: C6H12O6(s) + 6O2(g)  6CO2(g) + 6H2O(g)

r H = m  f (products) H – n f (reactants) H = {6  f H [CO2(g)] + 6  f H [H2O(g)]} – {1  f H [C6H12O6(s)] + 6  f H [O2(g)]} = [6(–393.5 kJ/mol) + 6(–241.826 kJ/mol)] – [(–1273.3 kJ/mol) + 6(0.0 kJ/mol)] = –2538.656 kJ/mol= – 2538.7 kJ/mol Fermentation: C6H12O6(s)  2CO2(g) + 2CH3CH2OH(l)

r H = {2  f H [CO2(g)] + 2  f H [CH3CH2OH(l)]} – [1  f H [C6H12O6(s)]} = [2(–393.5 kJ/mol) + 2(–277.63 kJ/mol)] – [(–1273.3 kJ/mol)] = –68.96 kJ/mol= – 69.0 kJ/mol b) Combustion of ethanol: CH3CH2OH(l) + 3O2(g)  2CO2(g) + 3H2O(g)

r H = {2  f H [CO2(g)] + 3  f H [H2O(g)]} – {1  f H [CH3CH2OH(l)] + 3  f H [O2(g)]} r H = [2(–393.5 kJ/mol) + 3(–241.826 kJ/mol)] – [(–277.63 kJ/mol) + 3(0.0 kJ/mol)] = –1234.848 kJ/mol= – 1234.8 kJ/mol Heats of combustion/mol C:  2538.656 kJ   1 mol C6 H12 O6  Sugar:    = –423.1093 kJ/mol C= –423.11 kJ/mol C 6 mol C   1 mol C6 H12 O6  

 1234.848 kJ   1 mol CH3CH 2 OH  Ethanol:    = –617.424 kJ/mol C= –617.42 kJ/mol C 2 mol C   1 mol CH3CH 2 OH   Ethanol has a higher value. 5.86

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) r H = {21  f H [CO2(g)] + 22  f H [H2O(g)]} – {[1  f H [C21H44(s)] + 32  f H [O2(g)]} = [21(–393.5 kJ/mol) + 22(–241.826 kJ/mol)] – [(–476 kJ/mol) + 32(0.0 kJ/mol)] = –13,107.672 kJ/mol= –13,108 kJ/mol 2) r H = {21  f H [CO(g)] + 22  f H [H2O(g)]} – {1  f H [C21H44(s)] + 43/2  f H [O2(g)]} = [21(–110.5 kJ/mol) + 22(–241.826 kJ/mol)] – [(–476 kJ/mol) + 43/2(0.0 kJ/mol)] = –7164.672 kJ/mol= –7165 kJ/mol 3) r H = {21 H f [C(s)] + 22 H f [H2O(g)]} – {1 H f [C21H44(s)] + 11 H f [O2(g)]} = [21(0.0 kJ/mol) +22(–241.826 kJ/mol)] – [(–476 kJ/mol) + 11(0.0 kJ/mol)] = –4844.172 kJ/mol= –4844 kJ/mol

 1 mol C21H 44  13107.672 kJ  4 4 b) q (kJ) =  254 g C21H 44     = –1.12266x10 kJ= –1.12x10 kJ 296.56 g C H 1 mol C H 21 44  21 44   c) The amount in 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. Amount in 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) + (0.0500)(0.856488 mol)(–4844.172 kJ/mol)] = –1.04655x104 kJ= –1.05x104 kJ 5.87

 vap H = 569.4 J/g(44.05 g/mol)(1 kJ/1000 J) = 25.08 kJ/mol

a) EE(l) → EE(g)

 vap H = {1  f H [EE(g)]} – {1  f H [EE(l)]} 25.08 kJ/mol = {  f H [EE(g)]} – [(–77.4 kJ/mol)]

 f H [EE(g)] = –52.32 kJ/mol EE(g) → CH4(g) + CO(g)

r H = {1  f H [CH4(g)] + 1  f H [CO(g)]} – {1  f H [EE(g)]}

r H = [(–74.87 kJ/mol) + (–110.5 kJ/mol)] – [(–52.32 kJ/mol)] r H = –133.0 kJ/mol b) Assume that you have 1.00 mole of EE(g). 1.00 mole of EE(g) produces 1.00 mole or 16.04 g of CH 4(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 x mass x T 1000 J  133.0 kJ    q  1 kJ  ΔT = = c x mass  2.5 J/gC  44.05 g  ΔT = 1207.7ºC ΔT = Tfinal – Tinitial 1207.7ºC = Tfinal – 93ºC Tfinal = 1300.72°C = 1301°C 5.88

a) 3N2O5(g) + 3NO(g) → 9NO2(g)

r H = {9  f H [NO2(g)]} – {3  f H [N2O5(g)] + 3  f H [NO(g)]} = [9(33.2 kJ/mol)] – [3(11 kJ/mol) + 3(90.29 kJ/mol)]  = –5.07 kJ/mol= –5 kJ/mol  1.50x102 mol     103 J  5.07 kJ b)  9 molecules product   = –76.05 J= –76.0 J   1 molecule product   9 moles product   1 kJ     

5.89

 1 mL   0.692 g   1 mol C8 H18   5.44x103 kJ  a) Heat (kJ) =  77.2 L   3    10 L   mL   114.22 g   1 mol C H    8 18    6 6 = –2.54437x10 kJ= –2.54x10 kJ 1h   104 km  3 3 b) Distance (km) = 2.54437x106 kJ    = 4.8112x10 km= 4.8x10 km 4  5.5x10 kJ  1 h  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.



5.90



q = c x mass x 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. J. So, 450. J α (c x ΔT). c, specific heat capacity, and ΔT are inversely proportional. The higher the ΔT, the lower the value of specific heat capacity: ΔT: B > D > C > A Specific heat capacity: B < D < C < A 5.91

5.92

ClF(g) + 1/2O2(g)  1/2Cl2O(g) + 1/2OF2(g)

r H = 1/2(167.5 kJ/mol) =

F2(g) + 1/2O2(g)  OF2(g)

r H = 1/2(–43.5 kJ/mol) = –21.75 kJ/mol

1/2Cl2O(g) + 3/2OF2(g)  ClF3(l) + O2(g)

r H = –1/2(394.1 kJ/mol) = –197.05 kJ/mol

ClF(g) + F2(g)  ClF3(l)

r H =

83.75 kJ/mol

–135.1 kJ/mol

a) AgNO3(aq) + NaI(aq)  AgI(s) + NaNO3(aq)  103 L   5.0 g AgNO3   1 mol AgNO3  Amount in moles of AgNO3 =  50.0 mL      1 mL   1L  169.9 g AgNO3    = 1.47145x10–3 mol AgNO3  103 L   5.0 g NaI  1 mol NaI  Amount in moles of NaI =  50.0 mL      1 mL   1 L  149.9 g NaI    = 1.6677785x10–3 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 = 1.47145x103 mol AgNO3     1 mol AgNO3   1 mol AgI  = 0.345496 g= 0.35 g AgI b) Ag+(aq) + I–(aq)  AgI(s)





r H = {1  f H [AgI(s)]} – {1  f H [Ag+(aq)] + 1  f H [I–(aq)]} = [(–62.38 kJ/mol)] – [(105.9 kJ/mol) + (–55.94 kJ/mol)] = –112.3 kJ/mol c) r H = q = c x mass x T

 112.3 kJ   1 mol AgI   3   1.47145x10 mol AgNO3   3 q  mol AgI   1 mol AgNO3    10 J  T = =   1 kJ  mc  4.184 J    1.00 g      g•K   50.0  50.0  mL  mL       = 0.39494 K= 0.39 K



5.93



Plan: For part a), 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  f H o values (Appendix B) are reported as energy per one mole, use the appropriate stoichiometric coefficient to reflect the higher Amount in moles. For part c), convert the amount of water in L to mass in g and use the relationship q = c x mass x T to find the heat needed; then determine the total cost of heating the water. Solution: a) CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)

r H = {1  f H o [CO2(g)] + 2  f H o [H2O(g)]} – {1  f H o [CH4(g)] + 2  f H o [O2(g)]} = [(–393.5 kJ/mol) + 2(–241.826 kJ/mol)] – [(–74.87 kJ/mol) + 2(0.0 kJ/mol)] = –802.282 kJ/mol= –802.3 kJ/mol

 1 kJ   1 mol CH4  Amount in moles of CH4 = 1x108 J  3    10 J  802.282 kJ     2 = 124.644 mol= 1.2x10 mol CH4  $0.56   1x108 J  b) Cost =  8   = $0.004493/mol = $0.0045/mol  10 J  124.644 mol    





 1 mL  1.0 g  6 c) Mass (g) of = 1202 L   3   = 1.202x10 g mL  10 L  

 4.184 J  6 q = c x mass x T =   1.202x10 g  g C 



  42.0  15.0 C = 1.357875x10 J 8

 $0.56  Cost = 1.357875x108 J  = $0.7604 = $0.76  1x108 J   



5.94



 1018 J   1 kJ   1 mol CH 4   16.04 g CH 4   1 kg  a) Mass (kg) of =  5600 EJ   = 1.12x1014 kg CH4  1 EJ   103 J   802 kJ   1 mol CH   103 g  4      

 1 yr  b) Years =  5600 EJ    = 9.6 yr = 10 yr  585 EJ   1 mL   1.00 g   4.184 J   1 kJ   1 mol CH 4  100.0  25.0  C   3   c) Amount in moles of CH4 =  0.946L   3     10 L   mL   g C     10 J         802 kJ  = 0.370143 mol CH4  16.04 g CH 4    103 m3  1L Volume (m3) of CH4 =  0.370143 mol CH 4        1 mol CH 4  0.72 g CH 4   1 L  = 8.2460x10-3 m³ = 8.2x10-3 m³  1 kJ   1 mol CH 4   16.04 g CH 4    103 m3  1L d) Volume = 2x1013 J  3          10 J     802 kJ  1 mol CH 4  0.72 g CH 4   1 L  = 5.55556x105 m³ = 6x105 m³





5.95

The reaction is exothermic. The argon atoms in the chamber after the reaction are moving with greater kinetic energy, indicating an increase in temperature.

5.96

H2SO4(aq) + 2NaOH(aq) → Na2SO4(aq) + 2H2O(l) 2H+(aq) + 2OH–(aq) → 2H2O(l)

r H = {2  f H [H2O(l)]} – {2  f H [H+(aq)] + 2  f H [OH–(aq)]} = [2(–285.84 kJ/mol)] – [2(0 kJ/mol) + 2(–229.94 kJ/mol)] = –111.8 kJ/mol 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 2SO 4     3    0.50 mol H 2SO 4   10 L  1.00 mL  = 2060 g H2SO4 solution  40.00 g NaOH   100 g solution  Mass (g) of NaOH solution =  2 mol NaOH    = 200. g NaOH solution   1 mol NaOH   40 g NaOH  q = c x mass x T

 103 J   q  1 kJ  ΔT = = 11.82ºC = mc   2060 + 200  g   4.184 J/g C 

111.8 kJ  

31ºC + 11.82ºC = 42.82°C = 43ºC This temperature is above the temperature at which a flammable vapour could be formed so the temperature increase could cause the vapour to explode. 5.97

a) 2C12H26(l) + 37O2(g) → 24CO2(g) + 26H2O(g) b) r H = {24  f H [CO2(g)] + 26  f H [H2O(g)]} – {2  f H [C12H26(g)] + 37  f H [O2(g)]} –1.50x104 kJ/mol = [24(–393.5 kJ/mol) + 26(–241.826 kJ/mol)] – [2  f H [C12H26(g)] + 37(0.0 kJ/mol)] –1.50x10 kJ/mol = –9444.0 kJ/mol + –6287.476 kJ/mol – [2  f H [C12H26(g)] + 0.0 kJ] 4

–1.50x104 kJ/mol = –15,731.476 kJ/mol – 2  f H [C12H26(g)] –1.50x104 kJ/mol + 15,731.476 kJ/mol = –2  f H [C12H26(g)] 731.476 kJ/mol = –2  f H [C12H26(g)]

 f H [C12H26(g)] = –365.738 kJ/mol= – 3.66x102 kJ /mol  1 mL   0.749 g C12 H 26   1 mol C12 H26   1.50x104 kJ  c) Heat (kJ)= q =  0.50 L   3       10 L   mL   170.33 g C12 H26   2 mol C12 H26    = –1.64900x104 kJ= –1.6x104 kJ   0.50 L d) Volume (L) = 1320 kJ   = 0.0040024 L = 40x10-3 L 4  1.64900x10 kJ    5.98

Plan: Heat of reaction is calculated using the relationship r H = m  f (products) H – n  f (reactants) H . The heats of formation for all of the species, except SiCl4, are found in Appendix B. Use reaction 3, with its given

r H , 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)

r H = {1  f H [SiO2(s)] + 4  f H [HCl(g)]} – {1  f H [SiCl4(g)] + 2  f H [H2O(g)]} –139.5 kJ = [(–910.9 kJ/mol) + 4(–92.31 kJ/mol)] – [  f H [SiCl4(g)] + 2(–241.826 kJ/mol)] –139.5 kJ/mol = –1280.14kJ/mol – [  f H [SiCl4(g)] + (–483.652 kJ/mol)] 1140.64 kJ/mol = –  f H [SiCl4(g)] + 483.652 kJ/mol

 f H [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)

r H = {1  f H [SiCl4(g)]} – {1  f H [Si(s)] + 2  f H [Cl2(g)]} = [(–656.988 kJ/mol)] – [(0 kJ/mol) + 2(0 kJ/mol)] = –656.988 kJ/mol= –657.0 kJ/mol 2) SiO2(s) + 2C(graphite) + 2Cl2(g)  SiCl4(g) + 2CO(g)

r H = {1  f H [SiCl4(g)] + 2  f H [CO(g)]}

– {1  f H [SiO2(g)] + 2  f H [C(graphite)] +2  f H [Cl2(g)]} = [(–656.988 kJ/mol) + 2(–110.5 kJ/mol)] – [(–910.9 kJ/mol) + 2(0 kJ/mol) + 2(0 kJ/mol)] = 32.912 kJ/mol= 32.9 kJ/mol b) Adding reactions 2 and 3 yields: (2) SiO2(s) + 2C(graphite) + 2Cl2(g)  SiCl4(g) + 2CO(g)

 rxn H =

(3) SiCl4(g) + 2H2O(g)  SiO2(s) + 4HCl(g)

 rxn H

= –139.5 kJ/mol

2C(graphite) + 2Cl2(g) + 2H2O(g) → 2CO(g) + 4HCl(g)

 rxn H

= –106.588 kJ/mol = –106.6 kJ/mol

32.912 kJ/mol

Confirm this result by calculating r H using Appendix B values. 2C(graphite) + 2Cl2(g) + 2H2O(g) → 2CO(g) + 4HCl(g)

r H = {2  f H [CO(g)] + 4  f H [HCl(g)]} – {2  f H [C(graphite)] + 2  f H [Cl2(g)] + 2  f H [H2O(g)]} = [2(–110.5 kJ/mol) + 4(–92.31 kJ) – [2(0 kJ/mol) + 2(0 kJ/mol) + 2(–241.826 kJ/mol)] = –106.588 kJ/mol= –106.6 kJ/mol 5.99

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 = –pΔV can be used to calculate the work of expansion. Solution: a) pV = nRT P=1.01 bar T =0°C + 273 = 273 K L•bar  1 mol   0.08314   273 K  mol•K nRT   Vi at 273 K = = = 22.4725 L p 1.01 bar 

Final volume at 819°C + 273 = 1092 K = V =

nRT = p

1 mol   0.08314 

L•bar  1092 K  mol•K  = 89.8900L 1.01 bar 

ΔV = Vfinal – Vinitial = 89.8900 L – 22.4725 L = 67.4175 L w = –PV = –(1.01 bar) x 67.4275 L = –68.0917 bar•L  100 J  3 w (J) =  68.0917 bar•L    = –6809.17 J = –6.81x10 J 1 bar•L   b) q = c m T

 28.02 g  Mass (g) of N2 = 1 mol N 2    = 28.02 g  1 mol N 2  T =

5.100

6.80917x103 J q = = 243.011 K = 243 K = 243°C (c )(m)  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  r H o values so that these equations sum to give the target equation. Then the manipulated  r H o values will add to give the  r H o 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:  r H o = –(–39.8 kJ/mol) = 39.8 kJ/mol Reverse (1) N2O3(g)  NO(g) + NO2(g)

 r H o = 2(–57.2 kJ/mol) = –114.4 kJ/mol

Multiply (3) by 2

4NO2(g)  2N2O4(g)

(5)

 r H o = (54.1 kJ/mol) = 54.1 kJ/mol N2O5(s)  N2O5(g) 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)

 r H o = –(–39.8 kJ/mol) =

Multiply (3) by 2

4NO2(g)  2N2O4(g)

 r H = 2(–57.2 kJ/mol) = –114.4 kJ/mol

(5)

N2O5(s)  N2O5(g)

r H o =

Reverse (2)

N2O5(g)  NO(g) + NO2(g) + O2(g)

 r H o = –(–112.5 kJ/mol) = 112.5 kJ/mol

(4)

2NO(g) + O2(g)  2NO2(g)

r H o =

–114.2 kJ/mol

N2O3(g) + N2O5(s)  2N2O4(g)

r H =

–22.2 kJ/mol

Total: 5.101

39.8 kJ/mol

54.1 kJ/mol

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  f H values (Appendix B) are reported as energy per one mole, use the appropriate stoichiometric coefficient to reflect the higher amount in moles. In this case,  r H o of the second reaction is known and  f H of N2H4(aq) must be calculated. For part b), calculate  r H o for the reaction between N2H4(aq) and O2, using the value of  f H for N2H4(aq) found in part a); then determine the amount in moles of O2 present by multiplying volume and concentration (mol/L) and multiply by the  r H o for the reaction. Solution: a) 2NH3(aq) + NaOCl(aq) → N2H4(aq) + NaCl(aq) + H2O(l)

 r H o = {1  f H [N2H4(aq)] + 1  f H [NaCl(aq)] + 1  f H [H2O(l)]} – {2  f H [NH3(aq)] + 1  f H [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

 f H [Na+(aq)] and  f H [Cl–(aq)]. –151 kJ/mol = [  f H [N2H4(aq)] + (–239.66 kJ/mol) + (–167.46 kJ/mol) + (–285.840 kJ/mol)] – [2(–80.83 kJ/mol) + (–346 kJ/mol)] –151 kJ/mol = [  f H [N2H4(aq)] + (– 692.96 kJ/mol)] – [–507.66 kJ/mol] –151 kJ/mol = [  f H [N2H4(aq)] + (–185.3 kJ/mol)

 f H [N2H4(aq)] = 34.3 kJ/mol= 34 kJ/mol  2.50x104 mol  b) Amount in moles of O2 = 5.00x103 L   = 1.25 mol O2  1L   N2H4(aq) + O2(g) → N2(g) + 2H2O(l)





 r H o = {1  f H [N2(g)] + 2  f H [H2O(l)]} – {1  f H [N2H4(aq)] + 1  f H [O2(g)]} = [(0 kJ/mol) + 2(–285.840 kJ/mol)] – [(34.3 kJ/mol) + (0 kJ/mol] = –605.98 kJ  605.98 kJ  Heat (kJ)= q = 1.25 mol O2    = –757.475 kJ= –757 kJ  1 mol O2 

5.102

Plan: For the given reaction, calculate the enthalpy. Determine the limiting reagent. Use the stoichiometry of the reaction and the enthalpy to calculate the heat generated. Solution: CO(g) + 2H2(g) → CH3OH(l)

 r H o = {1  f H [CH3OH(l)]} – {1  f H [CO(g)] + 2  f H [H2(g)]} = [–238.6 kJ/mol)] – [(–110.5 kJ mol) + 2(0.0 kJ/mol)] = –128.1 kJ/mol Find the limiting reactant: 112 kPa 15.0 L  PV Amount in moles of CO = = = 0.5644068 mol CO L•kPa  RT  8.31446 273  85 K     mol•K    1 mol CH3OH  Amount in moles of CH3OH from CO =  0.5644078 mol CO    = 0.5644068 mol CH3OH  1 mol CO 

 0.992 bar 18.5 L

 100 kPa    = 0.6342641 mol H2 L•kPa  1 bar     8.31446 mol•K    273  75 K     1 mol CH3 OH  Amount in moles of CH3OH from H2 =  0.6342641 mol H2    = 0.317132 mol CH3OH  2 mol H 2  H2 is limiting.  128.1 kJ  Heat (kJ) = q=  0.6342641 mol H2    = –40.6246 kJ = –40.6 kJ  2 mol H 2  Amount in moles of H2 =

5.103

PV RT

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  f H values (Appendix B) are reported as energy per one mole, use the appropriate stoichiometric coefficient to reflect the higher amount in moles. Convert the mass of methane to amount in moles and multiply that amount in moles by the heat of combustion. Solution: a) The balanced chemical equation for this reaction is: CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)

r H = {1  f H [CO2(g)] + 2  f H [H2O(g)]} – {1  f H [CH4(g)] + 2  f H [O2(g)]} = [(–393.5 kJ/mol) + 2(–241.826 kJ/mol)] – [(–74.87 kJ/mol) + 2(0.0 kJ/mol)] = –802.282 kJ/mol   1 mol Amount in moles of CH4 =  25.0 g CH 4    = 1.5586 mol CH4 16.04 g CH 4  

 802.282 kJ  3 Heat (kJ)= q = 1.5586 mol CH4    = –1250.4 = –1.25x10 kJ 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 x mass x T. The problem specifies heat capacities on a molar basis, so we modify the equation to use amount in 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 CO 2  Amount in moles of CO2(g) = 1.5586 mol CH 4    = 1.5586 mol CO2(g)  1 mol CH 4 

 2 mol H 2 O  Amount in moles of H2O(g) = 1.5586 mol CH 4    = 3.1172 mol H2O(g)  1 mol CH 4   2 mol O 2  Amount in moles of O2(g) reacted = 1.5586 mol CH 4    = 3.1172 mol O2(g)  1 mol CH 4  Amount in mole fraction N2 = (79%/100%) = 0.79 Amount in mole fraction O2 = (21%/100%) = 0.21  0.79 mol N 2  Amount in moles of N2(g) =  3.1172 mol O 2 reacted    = 11.72661 mol N2  0.21 mol O2  q = (molar heat capacity)(amount, mol)( T)  103 J  q = 1250.4 kJ   = 1.2504x106 J  1 kJ    1.2504x106 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 1.2504x106 J = 89.15192 J/°C(Tfinal) + 112.2192 J/°C(Tfinal) + 357.6616 J/°C(Tfinal) 1.2504x106 J = (559.03272 J/°C)Tfinal Tfinal = (1.2504x106 J)/(559.0324 J/°C) = 2236.72°C = 2.24x103°C 5.104

Plan: We can calculate the enthalpy of the reaction by using the table of thermochemical data in Appendix B. We use the ideal gas law to determine the amounts of the reactants and then determine which reactant is the limiting reagent. We use the limiting reagent to determine the amount of product formed and the enthalpy of reaction to find the amount of heat released. We use the ideal gas law to determine the amount of product to be formed in the third part and then use stoichiometry and the enthalpy of reaction to determine the amount of heat released. Solution:  r    m f H (products)   n f H (reactants)

  4 f H (NO)  6 f H (H2 O)    4 f H (NH3 )  5 f H (O2 )  a)

  4(90.29 kJ/mol) + 6(-241.826 kJ/mol)   4(45.9 kJ/mol) + 5(0 kJ/mol)  (1089.80 kJ/mol)  (183.6 kJ/mol)

 -906.2 kJ / mol pV (2.45 bar)(25.12 L) nNH3    2.239 mol RT (0.08314 bar  L )(330.55 K) mol  K pV (139 kPa)(37.54 L) nO2    1.744 mol RT (8.314 kPa  L )(359.85 K) mol  K From the stoichiometry, 4 mol of NH3 forms 4 mol of NO and so 2.239 mol of NO would be formed, whereas 5 mol of O2 forms 4 mol of NO, so 1.395 mol of NO would be formed. Thus O2 is the limiting reagent as it forms a smaller amount of NO.

mNO  nNO  MNO  (1.395 mol)(30.01 g/mol)=41.9 g 4 mol of NO (g) releases 906.2 kJ of energy (from the enthalpy per MOLE of reaction). (1.395 mol)(-906.2 kJ/mol)  -316.1 kJ Thus 1.395 mol of NO (g) formed would release q  4 316.1 kJ of heat would be released. Silberberg, Amateis, Venkateswaran, Chen, Chemistry: The Molecular Nature of Matter and Change, 3rd Canadian Edition Page 3-235 Instructor‘s Solution Manual © Copyright 2021 McGraw-Hill Ryerson Ltd.

1.789 bar  57.9 L  pV   3.22 mol NO bar  L  RT   0.08314   386.7 K  mol  K   4 mol NO (g) releases 906.2 kJ energy 3.22 mol NO (g) releases x kJ energy

nNO 

4x   3.22  906.2  x  730. kJ of energy released 5.105

Plan: We use the consumption of water per day and the number of days to find the volume of water, We use the density of water to find the mass of water. We use the specific heat capacity of water, the initial temperature of the water and the fact that water boils at 100 °C to find the amount of heat needed to boil this volume of water. The same amount of heat has to be released by the combustion of the camping fuel, after taking into account that only 18% of the heat generated goes towards boiling the water. We use the amount of heat needed to boil the water and the enthalpy of combustion of the camping fuel (calculated using Hess‘ Law) to find the amount of fuel (mol) needed. We use the formula of the fuel component to find the molar mass of the fuel which we then use to find the mass of fuel needed. Finally, we use the density and the mass to find the volume of the fuel needed. Solution:

Vwater  No. of days × Volume of water per day=2 days  3.0 L/day=6.0 L 1.0 g 1000 mL mH2 O  d H2 O  VH2 O  ( )( )(6.0 L)= 6.0×103 g mL 1L qH2 O  mH2 O cH2 O TH2 O  (6.0 103 g)(4.184

J )(100 C  22.5 C) g C

 -1.95×106 J Since only 18% of the heat generated goes to boil the water, 18% of qfuel  qH2 O  1.95 106 J 1.95  106 J  1.08  107 J 0.18 C7 H16 (l ) + 11 O 2 (g )  7 CO 2 (g ) + 8 H 2 O (l ) qfuel 

 r    m f H (products)   n f H (reactants)  7  f H (CO2 )  8 f H (H2 O)     f H (C7 H16 )  11 f H (O2 ) 

  7(393.5 kJ/mol) + 8(-285.840 kJ/mol)   ( 187.9 kJ/mol) + 11(0 kJ/mol)   -4853.32 kJ / mol nfuel 

qfuel 1.08  107 J   2.23 mol r H 4853.32  103 J/mol

M fuel  7  M C  16  M H  7(12.01 g/mol)+16(1.01 g/mol)=100.23 g / mol mfuel  nfuel  M fuel  (2.23 mol)(100.23 g/mol)= 224 g Vfuel  5.106

mfuel 224 g   320 mL d fuel 0.7 g/mL

Plan: We can determine the energy content per chocolate cherry by dividing the serving size by the number of pieces and dividing the energy content of one serving by the same amount. We determine the amount of energy needed for the class by multiplying the energy expended per hour by the length of the class. Then we find the number of pieces (to the closest whole number) of chocolate cherries that would provide the same amount of

energy. For part b), we write the combustion (metabolism) reaction for the sugar and then calculate the enthalpy of combustion. We know that this enthalpy is per MOLE of glucose reacted and we use the stoichiometry of the reaction to determine what amount of sugar (mol) we need to give the energy needed for the class. Then we convert amount (mol) to mass. For part c), we compare the masses of sugar in the number of chocolate pieces we determined from part a) and the value from part b) and draw a conclusion. Solution: a) If two pieces of chocolate cherry provides 586 kJ of energy, one chocolate cherry provides 293 kJ of energy.

Eneeded  E / t  t  (2092 kJ/hr)(0.75 hr)=  1569 kJ No. of pieces =

Eneeded 1569 kJ   5.4 pieces=6 pieces Eper piece 293 kJ/piece

b) The balanced chemical reaction is: C6 H12 O6 (s) + 6 O2 (g )  6 CO2 (g ) + 6 H2 O (l ) The enthalpy of combustion is:  r    m f H (products)   n f H (reactants)

 6 f H (CO2 )  6 f H (H2 O)     f H (C6 H12 O6 )  6 f H (O2 ) 

  6(393.5 kJ/mol) + 6(-285.840 kJ/mol)   (1273.3 kJ/mol) + 6(0 kJ/mol)  -2802.74 kJ / mol

E 1569 kJ   0.560 mol E / mol 2802.74 kJ/mol To get an energy of 1569 kJ, we would need: mglucose  nglucose  M glucose  (0.560 mol)(180.18 g/mol)=101 g nglucose 

5.107

2 pieces of chocolate cherry contain 20 g of sugar and in part a) we determined she would need to eat 6 pieces. Therefore, 6 pieces of chocolate cherries contains 60 g of sugar, as compared to the 101 g of glucose she would need to consume. We can thus conclude that the Ganong chocolate cherries contain ingredients other than glucose that would provide the energy she needs.

Plan: We can find the volume of the metal from the dimensions given and then the mass of metal from the density. We know the initial and final temperatures of the metal. We know the mass, specific heat capacity and initial and final temperatures of the water. Finally we know the heat capacity of the calorimeter and that it will have the same initial and final temperatures as the water. From this information, we can find the specific heat capacity of the metal. Solution:

VRh  lRh  wRh  hRh  (8.57 cm)(0.932 cm)(2.26 cm)  18.05 cm3 mRh  VRh  d Rh  (18.05 cm3 )(12.41 g/cm3 )  224.0 g qRh  qcalorimeter  qwater  0 qRh  qcalorimeter  qwater (mcT ) Rh  (C T )calorimeter  (mcT ) water (224.0 g)(c)(32.93 C  141.3 C)  (52.94

J )(32.93 C  27.35 C) C

 (239.8 mL)(1.00 g/mL)(4.184 c Rh = 0.243

J )(32.93 C  27.35 C) g C

J g o C

5.108

Plan: Set the heat gained by the water equal to the heat lost by the iridium. Solve for the mass of iridium. Use the number of spheres and total mass to find the mass of one sphere. Use the density to find the volume of one sphere. Use the volume of a sphere to find the radius and then diameter of one sphere. Solution: -( -

)

(

)

)(

)

(

)(

)

r = 0.42 cm diameter = 2r = 0.84 cm 5.109

Plan: Use the area and height of the house to find its volume. Use the volume of the house and density of air to find the mass of air in the house. Use the mass of air, its specific heat capacity and the temperature change to find the heat required. Use the enthalpy and the heat to find the amount (mol) of methane. Finally, use the ideal gas law to find the volume of methane. Solution: (

)(

)

(

)(

) )

)(

( (

)(

)

(

)(

)( (

5.110

)

Plan: Write the balanced equation for the combustion reaction. Calculate the enthalpy of reaction using values from Appendix B. Use stoichiometry and the enthalpy of reaction to determine the mass of sugar equivalent to the sugar in the chocolate bar. Solution: a) b)

C6H12O6 (s) + 6 O2 (g)  6 CO2 (g) + 6 H2O (l) [ *

( )] - [

( )

)

( )]

( )

)+ - *-

( )+

remember that 1 Cal = 1 kcal 4 pieces of chocolate : - 210 kcal 10 pieces of chocolate: x kcal 4x = - 2100 kcal

x = - 525 kcal = (-

)(

)

(

)

( 5.111

)(

)

Plan: Balance the chemical equation and calculate the enthalpy of reaction. Then, convert masses to amount (mol) and use the stoichiometry to determine the limiting reagent (LR). Then use the LR to determine the value of q. Finally, use the given value of q and the other conditions to determine the volume of chlorine gas using the ideal gas law and stoichiometry. Solution: a)

CH4 (g) + 4 Cl2 (g)  CCl4 (l) + 4 HCl (g) [ () ( )] - [ ( ) * (-

)

)+ - *-

( )]

( )+

b) Even if we take 4 times this amount, we have methane left so LR 4 mol Cl2 : 1 mol CCl4 3.17 mol Cl2 : x mol CCl4 4x = 3.17 mol

x = 0.793 mol

(

)(

)

1 mol CH4 releases 433.33 kJ of heat 0.793 mol CH4 releases x kJ of heat x = (0.793)(-433.33 kJ) = -344 kJ c)

Therefore, 344 kJ of heat are released.

433.33 kJ of heat is released for 4 mol of Cl2 reacted 284 kJ of heat is released for n mol of Cl2 reacted 433.33 x = 4 (284 kJ) (

n = 2.62 mol of Cl2 )(

)( (

)

CHAPTER 6 QUANTUM THEORY AND ATOMIC STRUCTURE The value for the speed of light will be 3.00x108 m/s except when more significant figures are necessary, in which cases, 2.9979x108 m/s will be used. TOOLS OF THE LABORATORY BOXED READING PROBLEMS 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 concentration (mol/L) of the original solution. Solution: a) Absorbance vs. Concentration:

Absorbance

B6.1

0.45 0.4 0.35 0.3 0.25 0.2 0.15 0.1 0.05 0 0

0.00001

0.00002

0.00003

0.00004

Concentration (mol/L)

This is a linear plot, thus, using the first and last points given:  0.396  0.131 y  y1 m= 2 = = 13,250 /mol/L= 1.3x104 L /mol 5 5 x2  x1 3.0x10  1.0x10 mol / L





Using the slope just calculated and any of the data points, the value of the intercept may be found. b = y – mx = 0.396 – (13,250/mol/L)(3.0x10–5 mol/L) = –0.0015 = 0.00 (absorbance has no units) b) Use the equation just determined: y = (1.3x104/mol/L) x + 0.00. x = (y – 0.00)/(1.3x104/mol/L) = (0.236/1.3x104) mol/L = 1.81538x10–5 mol/L = 1.8x10–5 mol/L This value is cf in a dilution problem (ciVi)= (cfVf) with Vi = 20.0 mL and Vf = 150. mL.

 cf Vf  1.81538x10 mol / L 150. mL Mi = = = 1.361538x10–4 mol/L = 1.4x10–4 mol/L  20.0 mL  Vi  5

B6.2

Plan: The colour of light associated with each wavelength can be found from Figure 6.3. The c frequency of each wavelength can be determined from the relationship c = or = . The



wavelength in nm must be converted to metres. Solution: 3.00x108 m/s  1 nm  14 -1 14 –1 a) red =   = 4.4709x10 s = 4.47x10 s 671 nm  109 m  3.00x108 m/s  1 nm  14 -1 14 –1 b) blue =   = 6.6225x10 s = 6.62x10 s 453 nm  109 m  3.00x108 m/s  1 nm  14 -1 14 –1 c) yellow-orange  =   = 5.0933786x10 s = 5.09x10 s 589 nm  109 m  END–OF–CHAPTER PROBLEMS 6.1

All types of electromagnetic radiation travel as waves at the same speed. They differ in their frequency, wavelength, and energy.

6.2

Plan: Recall that the shorter the wavelength, the higher the frequency and the greater the energy. Figure 6.3 describes the electromagnetic spectrum by wavelength and frequency. Solution: a) Wavelength increases from left (10–2 nm) to right (1012 nm) in Figure 6.3. 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 = λν, 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 = hν. Therefore, the trend in increasing energy matches the trend in increasing frequency: radio wave < microwave < infrared < visible < ultraviolet < x-ray.

6.3

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 colours (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.

6.4

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.

6.5

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.

6.6

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 = hν. 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.

6.7

Plan: Wavelength is related to frequency through the equation c = ν. Recall that a Hz is a reciprocal second, or 1/s = s–1. Assume that the number ―950‖ has three significant figures. Solution: c = ν 3.00x108 m/s = 315.789 m= 316 m   103 Hz   s 1   950. kHz       1 kHz   Hz  c  1 nm  11 11  (nm) = =  315.789 m   9  = 3.15789x10 m= 3.16x10 nm   10 m 

(m) =

6.8

Wavelength and frequency relate through the equation c = . Recall that a Hz is a reciprocal second, or 1/s = s–1. 3.00x108 m/s = 3.208556 m= 3.21 m   106 Hz   s 1   93.5 MHz       1 MHz   Hz  c  1 nm  9 9  (nm) = =  3.208556 m   9  = 3.208556x10 m= 3.21x10 nm   10 m 

(m) =

6.9

Plan: Frequency is related to energy through the equation E = h. Note that 1 Hz = 1 s–1. Solution: E = h E = (6.626x10–34 J•s)(3.8x1010 s–1) = 2.51788x10–23 J= 2.5x10–23 J

 6.626x10 =

34





J•s 3.00x108 m/s  1 nm  –17 –17  9  = 1.5291x10 J= 1.5x10 J 13 nm 10 m  

6.10

6.11

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. 6.12

Since energy is directly proportional to frequency (E = h): UV ( = 8.0x1015 s–1) > IR ( = 6.5x1013 s–1) > microwave ( = 9.8x1011 s–1) or UV > IR > microwave.

6.13

Plan: Wavelength is related to frequency through the equation c = ν. Recall that a Hz is a reciprocal second, or 1/s = s–1. Solution:  109 Hz  s 1   = (s–1) =  22.235 GHz   = 2.2235x1010 s–1   Hz  1 GHz   

 (nm) =



2.9979x108 m/s  1 nm  7 7   = 1.3482797x10 nm= 1.3483x10 nm 2.2235x1010 s1  109 m 

6.14

Frequency and wavelength can be calculated using the speed of light: c = ν. 3.00x108 m/s  1 m  c 13 -1 13 –1 a) = =  6  = 3.125x10 s = 3.1x10 s 9.6  m   10 m   1 m  2.9979x108 m/s c b) (m) = =   = 3.464979 µm= 3.465 m  1   s   106 m  13 8.652x10 Hz    Hz 



6.15



Frequency and energy are related by E = h, and wavelength and energy are related by E = hc/.  106 eV   1.602x1019 J  1.33 MeV     1 MeV   1 eV E    Hz  = 3.2156x1020 Hz= 3.22x1020 Hz (Hz) = =  1  34 h 6.626x10 J•s s 

 (m) =

hc = E

 6.626x10

34



J•s 3.00x108 m/s



= 9.32950x10–13 m= 9.33x10–13 m  10 eV   1.602x10 J  1.33 MeV      1 eV  1 MeV    The wavelength can also be found using the frequency calculated in the equation c =  6.16

19

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 (220 nm). Use the relationship c = to find the frequency of the photons and hc relationship E = to find the energy.



Solution: a) c = 

=



b) =

=



3.00x108 m/s  1 nm  15 -1 15 –1   = 1.239669x10 s = 1.24x10 s 242 nm  109 m 

 6.626x10 = =

34





J•s 3.00x108 m/s  1 nm  –19 –19  9  = 8.2140x10 J= 8.21x10 J 242 nm  10 m 

3.00x108 m/s  1 nm  15 -1 15 –1   = 1.3636x10 s = 1.4x10 s 220 nm  109 m 

 6.626x10 =

34





J•s 3.00x108 m/s  1 nm  –19 –19  9  = 9.03545x10 J= 9.0x10 J 220 nm 10 m  

6.17

―n‖ in the Rydberg equation is equal to a Bohr orbit of quantum number ―n‖ where n = 1, 2, 3, ...

6.18

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.

6.19

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 coloured lines against a dark background. Bohr worked with emission spectra.

6.20

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

6.21

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.

6.22

The Bohr model has successfully predicted the line spectra for the H atom and Be 3+ ion since both are one-

 

 Z 2 2.18x1018 J electron species. The energies could be predicted from En =

 where Z is the atomic number

n for the atom or ion. The line spectra for H would not match the line spectra for Be 3+ since the H nucleus contains one proton while the Be3+ 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. 6.23

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. Solution:  1 1 1   R 2  2  R = 1.096776x107 m–1    n n 2   1 n1 = 2 n2 = 5  1 1 1 1   1  R  2  2  = 1.096776x107 m 1  2  2  = 2,303,229.6 m–1    5  n2  2  n1    1 nm  1  (nm) =   = 434.1729544 nm= 434.17 nm 1   9  2,303, 229.6 m   10 m 



6.24



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  = 1.096776x107 m 1  2  2  = 9,749,120 m–1 n   3  n2  1  1





   1nm  1m 2  (nm) =    9  = 1.0257x10 nm 9, 749,120    10 m  6.25

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   1  R  2  2  = 1.096776x107 m 1  2  2  = 533,155 m–1 n   4  n2  3  1





   1 nm  1  (nm) =   = 1875.627 nm= 1875.6 nm 1   9  533,155 m   10 m  6.26

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  = 1.096776x107 m 1  2  2  = 1,523,300 m–1 n   3  n2  2  1    1 nm  1  (nm) =   = 656.4695 nm= 656.47 nm 1   9  1,523,300 m   10 m 



6.27



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   2 E = 2.18x1018 J  2  n ninitial   final 1  1 18 E = 2.18x10 J  2  2  = –4.578x10–19 J/photon 2 5  









 4.578x1019 J   6.022x1023 photons  5 5 E =     = –2.75687x10 J/mol= –2.76x10 J/mol  photon 1 mol    The energy has a negative value since this electron transition to a lower n value is an emission of energy. 6.28

To find the transition energy, use the equation for the energy of an electron transition and multiply by Avogadro‘s number.  1 1   2 E = 2.18x1018 J  2  n ninitial   final 1  1 18 E = 2.18x10 J  2  2  = 1.93778x10–18 J/photon 1  3









 1.93778x1018 J   6.022x1023 photons  6 6 E =     = 1.1669x10 J/mol= 1.17x10 J/mol  photon 1 mol    6.29

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. 6.30

b>c>a>d

6.31

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 = ?  = 9.720x10 m 1 nm  

 1 1  = 1.096776x107 m 1  2  2    n2   n1 1





1 1 1  = 1.096776x107 m 1  2  2  8 1 9.72x10 m n2  





1 1  0.93803 =  2  2  1 n2   1 = 1 – 0.93803 = 0.06197 n22

n22 = 16.14 n2 = 4 6.32

 109 m  –6  = 1281 nm    = 1.281x10 m 1 nm    1 1  1 = 1.096776x107 m 1  2  2    n2   n1





 1 1 1  = 1.096776x107 m 1  2  2  6 n 1.281x10 m 5   1





 1 1  0.07118 =  2  2  n 5   1

1 n12

= 0.07118 + 0.04000 = 0.11118

n12 = 8.9944 n1 = 3 The n = 5 energy level is higher in energy than the n = 3 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. hc

 6.626x10 =

34





J•s 3.00x108 m/s  1 nm  –19 –19  9  = 4.55917x10 J= 4.56x10 J  436 nm  10 m  

6.33

6.34

a) Absorptions: A, C, D; Emissions: B, E, F b) Energy of emissions: E < F < B c) Wavelength of absorption: D < A < C

6.35

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).

6.36

De Broglie‘s concept is supported by the diffraction properties of electrons demonstrated in an electron microscope.

6.37

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.

6.38

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.

6.39

Plan: Use the de Broglie equation. Velocity in km/h must be converted to m/s because a joule is equivalent to kg•m2/s2. Solution: a) 3  19.8 km   10 m   1 h  Velocity (m/s) =     = 5.50 m/s  h    1 km   3600 s 

=







6.626x1034 J•s  kg•m 2 /s 2  h –37 –37 =   = 7.08633x10 m= 7.09x10 m m   J mu   170g   5.50  s  

3  0.1 km   10 m   1 h  b) Uncertainty in velocity (m/s) =     = 0.02778 m/s   h   1 km   3600 s 

x•mv 

h 4

h x   4  m v

 6.626x10 J•s  34

 kg•m 2 /s 2  –35 –35    1.807674x10 m 2x10 m J  0.02778 m    4 105 kg    s  

6.40

a)  =

h = mu

 6.626x10 J•s   kg•m /s  10 g      6.6x10 g   3.4x10 ms   J  1 kg  34

24

= 2.95276x10–15 m= 3.0x10–15 m h b) x•mv  4

 6.626x10 J•s   kg•m /s  10 g   1 kg  J  0.1x10 m     4  6.6x10 g    s 34

h x   4  m v

24

   7.98910x10–15 m 8x10–15 m

6.41 Plan: Use the de Broglie equation. Mass in g must be converted to kg and wavelength in nm must be converted to m because a joule is equivalent to kg•m2/s2. Solution:  1 kg  Mass (kg) =  56.5 g   3  = 0.0565 kg  10 g   

 109 m  Wavelength (m) =  540 nm   = 5.4x10–7 m  1 nm    h = mu





6.626x1034 J•s  kg•m2 /s2  h –26 –26 u= =   = 2.1717x10 m/s= 2.2x10 m/s  7 J m  0.0565 kg  5.4x10 m  

6.42

=





h mu





6.626x1034 J•s  kg•m2 /s2   103 g   1 pm  h –23 –23 u= =      12  = 4.666197x10 m/s= 4.67x10 m/s 142 g 100. pm J 1 kg m       10 m  6.43

6.44

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 electromagnetic energy that have no mass. A light photon‘s velocity is the speed of light, 3.00x10 8 m/s. Wavelength in nm must be converted to m. Solution:  109 m  Wavelength (m) =  589 nm   = 5.89x10–7 m  1 nm    h = mu





6.626x1034 J•s  kg•m2 /s2  h –36 –36 =   = 3.7499x10 kg/photon= 3.75x10 kg/photon J u 5.89x107 m 3.00 x108 m/s  

=

h mu









34



6.626x10 J•s  kg•m2 /s2   1 nm  h –36 =    9  = 3.2916x10 kg/photon 8 J u 10 m  671 nm  3.00x10 m/s   





 3.2916x1036 kg   6.022x1023 photons  –12 –12     = 1.9822x10 kg/photon= 1.98x10 kg/mol photon mol    6.45

The quantity 2 expresses the probability of finding an electron within a specified tiny region of space.

6.46

Since 2 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.

6.47

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 52.9 pm. Thus, the probability of finding an electron at 52.9 pm is much greater for the 1s orbital than for the 2s.

6.48

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.

6.49

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).

6.50

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).

6.51

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

6.52

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

6.53

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

The variations in colouring of the p orbital are a consequence of the quantum mechanical derivation of atomic orbitals that are beyond the scope of this course. 6.54

a) pz: 2 lobes along the z-axis

b) dxy: 4 lobes

The variations in colouring of the p and d orbitals are a consequence of the phases of the orbital. 6.55

Plan: The following letter designations for the various subshell (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: subshell 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

6.56

subshell a) s (l = 0) b) d (l = 2) c) p (l = 1)

6.57

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.

allowable ml 0 –2, –1, 0, +1, +2 –1, 0, +1

# of possible orbitals 1 5 3

c) For the 4f subshell, n = 4 and l = 3. Since ml = –3, –2, –1, 0, +1, +2, +3, there are seven orbitals. 6.58

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)

6.59

Plan: Allowed values of quantum numbers: n = positive integers; l = integers from 0 to n – 1; ml = integers from – l through 0 to + l. One set of possible answers is shown. 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. 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.

6.60

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

6.61

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 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 either be read directly from the calibration curve or calculated from the slope (A = kC, where k = slope of the line and C = concentration).

6.62

Plan: For Part a, use the values of the constants h, π, 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



part c). Remember that a joule is equivalent to kg•m2/s2. Solution: a) h = 6.626x10–34 J•s; me = 9.1094x10–31 kg; a0 = 52.92x10–12 m E= 

E= 

h2  1   = 2 2 2 2 2  2  8 me a0 n 8 me a0  n  h2

6.626x10

34

J•s



 kg•m2 /s2   1    2  2   J  n  8 2 9.1094x1031 kg 52.92x1012 m 







 1   1  = –(2.17963x10–18 J)  2  = –(2.180x10–18 J)  2  n  n  This is identical with the result from Bohr‘s theory. For the H atom, Z = 1 and Bohr‘s constant = –2.18x10–18 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. Silberberg, Amateis, Venkateswaran, Chen, Chemistry: The Molecular Nature of Matter and Change, 3rd Canadian Edition Page 3-252 Instructor‘s Solution Manual © Copyright 2021 McGraw-Hill Ryerson Ltd.

1   1 E = –(2.180x10–18 J)  2  2  = 3.027778x10–19 J = 3.028x10–19 J 2  3 hc c) E =









6.626x1034 J•s 2.9979x108 m/s hc  (m) = = = 6.56061x10–7 m= 6.561x10–7 m E 3.027778x1019 J





 1 nm   (nm) = 6.56061x107 m  9  = 656.061 nm= 656.1 nm  10 m 





This is the wavelength for the observed red line in the hydrogen spectrum. 6.63

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. At that wavelength, electrons will also be ejected from K but will have a greater kinetic energy. 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.

6.64

a) E =

 6.626x10



34

J•s)(3.00x108 m/s  1 nm  –19  9  = 2.8397x10 J 700. nm  10 m   This is the value for each photon, that is, J/photon.  1 photon  17 Number of photons = 2.0x10 J  = 70.430 photons= 70. photons 19  2.8397x10 J   hc





 6.626x10 J•s)(3.00x10 m/s  1nm  = 4.18484x10 J 34

–19

 9  475. nm  10 m  This is the value for each photon, that is, J/photon. 1 photon   17 Number of photons = 2.0x10 J  = 47.7916 photons= 48 photons 19   4.18484x10 J  b) E =





6.65



Determine the wavelength:  = 1/(1953 cm–1) = 5.1203277x10–4 cm  102 m   1 nm  3  (nm) = 5.1203277x104 cm    9  = 5120.3277 = 5.120x10 nm 1 cm    10 m 





= c/= 6.66

2.9979x108 m/s  1 cm   1 Hz  13 13  2   1  = 5.8548987x10 Hz= 5.855x10 Hz 4 5.1203277x10 cm  10 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 specifies a metal ion, assume that the possible choices are Li 2+ or Be3+. Use the relationship E = h to convert the  Z2  frequency to energy and then solve Bohr‘s equation E = 2.18x10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 = (6.626x10–34 J•s) (2.961x1016 Hz) (s–1/Hz) = 1.9619586x10–17 J  Z2  E = 2.18x1018 J  2  Z = charge of the nucleus n   





En 2 1.9619586x1017 (12 ) = = 8.99998 2.18x1018 J 2.18x10 18 J Then Z2 = 9 and Z = 3. Therefore, the ion is Li2+ with an atomic number of 3. Z2 =

6.67

 6.626 x10 J•s   kg•m /s  = 2.139214x10 m= 2.1x10 m   9.11x10 kg   3.4x10 ms   J   6.626x10 J•s   kg•m /s  = 1.16696x10 m= 1.2x10 m h Proton:  = =   mu 1.67x10 kg   3.4x10 ms   J  34

h a) Electron:  = = mu

31

27

–10

34

b) E = 1/2mu2

–13

therefore u2 = 2E/m

2E m



 

2 2.7x1015 J  kg•m2 /s2  7   = 7.6991x10 m/s 31 J 9.11x10 kg  

 2  2.7x10 J   kg•m /s  Proton: u =   = 1.7982x10 m/s 1.67x10 kg   J   kg•m /s  6.626x10 J•s  h Electron:  = =   = 9.44698x10 m= 9.4x10 m m   J mu   9.11x10 kg 7.6991x10    s   kg•m /s  6.626x10 J•s  h Proton:  = =   = 2.20646 x 10 m= 2.2 x 10 m m   J mu   1.67x10 kg 1.7982x10    s  Electron: u =

15

27

34

31

–12

–13

34

27

6.68

Plan: The electromagnetic spectrum shows that the visible region goes from 400 to 750 nm. 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 nm to units of m. Solution: n = ?  n = 1;  = 121 nm (shortest  corresponds to the largest E)  109 m  –7  (m) = 121 nm    = 1.21x10 m 1 nm  

 1 1  = 1.096776x107 m 1  2  2    n2   n1 1





1  1   7 1  1   = 1.096776x10 m  2  2  7 n2   1.21x10 m  1





1 1  0.7535233 =  2  2  1 n2  

 1   2  = 1 – 0.7535233  n2   1   2  = 0.2464767  n2  n2 = 2 for line (a) (n = 2  n = 1) n = ?  n = 3;  = 1094 nm (longest  corresponds to the smallest E)  109 m  –6  (m) = 1094 nm    = 1.094x10 m 1 nm    1 1  = 1.096776x107 m 1  2  2    n2   n1 1





1  1   7 1  1   = 1.096776x10 m  2  2  6 n2   1.094x10 m  3





 1 1  0.083342158 =  2  2  3 n2   0.083342158 = 0.111111111 

1 n22

 1   2  = 0.11111111 – 0.083342158  n2 

 1   2  = 0.0277689535  n2  n22 =36.01143 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.

6.69



thus  =



hc E

















a)  (nm) =

6.626x1034 J•s 3.00x108 m/s  1 nm  hc =  9  = 432.130 nm= 432 nm E 4.60x1019 J  10 m 

b)  (nm) =

6.626x1034 J•s 3.00x108 m/s  1 nm  hc =  9  = 286.4265 nm= 286 nm E 6.94x1019 J  10 m 

6.626x1034 J•s 3.00x108 m/s  1 nm  hc c)  (nm) = =  9  = 450.748 nm= 451 nm E 4.41x1019 J  10 m  6.70

Index of refraction = c/v; v = c/(index of refraction) a) Water v = c/(index of refraction) = (3.00x108 m/s)/(1.33) = 2.2556x108 m/s= 2.26x108 m/s b) Diamond v = c/(index of refraction) = (3.00x108 m/s)/(2.42) = 1.239669x108 m/s= 1.24x108 m/s

6.71

Extra significant figures are necessary because of the data presented in the problem. He–Ne  = 632.8 nm Ar  = 6.148x1014 s–1 Ar–Kr E = 3.499x10–19 J Dye  = 663.7 nm Calculating missing  values: Ar  = c/ = (2.9979x108 m/s)/(6.148x1014 s–1) = 4.8762199x 10–7 m= 4.876x10–7 m Ar–Kr  = hc/E = (6.626x10–34 J•s) (2.9979x108 m/s)/(3.499x10–19 J) = 5.67707x10–7 m= 5.677x10–7 m Calculating missing  values: He–Ne  = c/ = (2.9979x108 m/s)/[632.8 nm (10–9 m/nm)] = 4.7375 x 1014 s-1= 4.738 x 1014 s–1 Ar–Kr  = E/h = (3.499x10–19 J)/(6.626x10–34 J•s) = 5.28071x1014 s-1= 5.281x1014 s–1 Dye  = c/ = (2.9979x108 m/s)/[663.7 nm (10–9 m/nm)] = 4.51695x1014 s~= 4.517x1014 s–1 Calculating missing E values: He–Ne E = hc/ = [(6.626x10–34 J•s)(2.9979x108 m/s)]/[632.8 nm (10–9 m/nm)] = 3.13907797x10–19 J= 3.139x10–19 J Ar E = h = (6.626x10–34 J•s)(6.148x1014 s–1) = 4.0736648x10–19 J= 4.074x10–19 J Dye E = hc/ = [(6.626x10–34 J•s)(2.9979x108 m/s)]/[663.7 nm (10–9 m/nm)] = 2.99293x10–19 J= 2.993x10–19 J The colours may be predicted from Figure 6.3 and the frequencies. He–Ne  = 4.738x1014 s–1 Orange Ar  = 6.148x1014 s–1 Green Ar–Kr  = 5.281x1014 s–1 Yellow Dye  = 4.517x1014 s–1 Red

6.72

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. 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.

6.73

 1 1  = 1.096776x107 m 1  2  2    n2   n1





1 1  1  1 nm  = 1.096776x107 m 1  2  2    9 1 94.91 nm  10 m  n2  





1 1  0.9606608 =  2  2  1 n2   n2 = 5  1 nm  1 1  7 1  1 b)  9  = 1.096776x10 m  2  2  1281 nm  10 m  5   n1





 1 1  0.071175894 =  2  2  n 5   1 n1 = 3 1 1 7 1  1 c) = 1.096776x10 m  2  2   1 3  1 = 9.74912x106 m–1











 1 nm   = 1.02573x107 m  9  = 102.573nm = 102.6 nm  10 m  6.74

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 hc to determine the ionization energy of B4+ and to find the energies of the transitions listed. Use E = to convert



energy to wavelength. Solution:

 Z2  a) E = 2.18x1018 J  2  n   





Z = atomic number

 1 1   2 Z2 E = 2.18x1018 J  2 n ninitial   final





 1 1  2  6.022x1023  E = 2.18x1018 J  2  2 Z    ninitial   1 mol   = (1.312796x106) Z2 for n = 1 b) In the ground state n = 1, the initial energy level for the single electron in B 4+. Once ionized, n =  is the final energy level. Z = 5 for B4+. E = IE = (1.312796x106) Z2 = (1.312796x106 J/mol)(52) = 3.28199x107 J/mol= 3.28x107 J/mol c) nfinal = , ninitial = 3, and Z = 2 for He+.  1 1  2 1  2 18  1 –19 E = 2.18x1018 J  2  2  Z = 2.18x10 J  2  2  2 = 9.68889x10 J n n  3 final initial    





















6.626x1034 J•s 3.00x108 m/s hc = = 2.051628x10–7 m E 9.68889x1019 J  1 nm   (nm) = 2.051628x107 m  9  = 205.1628 nm= 205 nm  10 m  d) nfinal = , ninitial = 2, and Z = 4 for Be3+.  1 1  2 1  2 18  1 –18 E = 2.18x1018 J  2  2  Z = 2.18x10 J  2  2  4 = 8.72x10 J  n initial   2  

 (m) =









hc E

 (m) =



6.626x10

34





J•s 3.00x108 m/s

8.72x10

18

 = 2.279587x10 m –8

 1 nm   (nm) = 2.279587x108 m  9  = 22.79587 nm= 22.8 nm  10 m 





6.75

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.

6.76

Plan: Use the values and the equation given in the problem to calculate the appropriate values. Solution: a) rn =

n2 h2  0

 me e 2

2 C2  12 6.626x1034 J•s  8.854x1012  J•m   kg•m 2 /s2   –11 –11 r1 =   = 5.2929377x10 m= 5.293x10 m 2 31 19 J    9.109x10 kg 1.602x10 C











2 C2  102 6.626x1034 J•s  8.854x1012  J•m   kg•m2 /s2   –9 –9 b) r10 =   = 5.2929377x10 m= 5.293x10 m 2 31 19 J    9.109x10 kg 1.602x10 C







6.77

a) rn =





n2 h2  0

 me e 2

2 C2  32 6.626x1034 J•s  8.854x1012  J•m   kg•m 2 /s2   –10 –10 r3 =   = 4.76364x10 m= 4.764x10 m 2 31 19 J    9.109x10 kg 1.602x10 C











b) Z = 1 for an H atom  Z2   12  En = 2.18x1018 J  2  = 2.18x1018 J  2  = – 2.42222x10–19 J= – 2.42x10–19 J n  3      c) Z = 3 for a Li atom









 Z2   32  En = 2.18x1018 J  2  = 2.18x1018 J  2  = – 2.18x10–18 J 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.



6.78







Plan: Refer to Chapter 5 for the calculation of the amount of heat energy absorbed by a substance from its specific heat capacity and temperature change (q = c x mass x 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 x mass x T q = (4.184 J/g°C)(252 g)(98 – 20.)°C = 8.22407x104 J E=



 6.626x10

34



J•s 3.00x108 m/s

1.55x10



2

 = 1.28245x10 J/photon –23



1 photon   4 Number of photons = 8.22407x10 J  = 6.41278x1027 photons= 6.4x1027 photons 23   1.28245x10 J  6.79

One sample calculation will be done using the equation in the book: 3

r  2  r a0  1   1  2  r a0  1   1 =  =  = 1.465532x10–3 e a0  e   e      a0      52.92 pm  For r = 50 pm: r

50

 = 1.465532x10–3 e a0 = 1.465532x10–3 e 52.92 = 5.69724x10–4 2 = (5.69724x10–4)2 = 3.24585x10–7 4r22 = 4(50)2(3.24585x10–7) = 1.0197x10–2 r (pm)  (pm–3/2) 2 (pm–3) 4r22 (pm–1) –3 –6 0 1.47x10 2.15x10 0 50 0.570x10–3 0.325x10–6 1.02x10–2 –3 –6 100 0.221x10 0.0491x10 0.616x10–2 –3 –6 200 0.0335x10 0.00112x10 0.0563x10–2

The plots are similar to Figure 6.18A in the text.

6.80

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  1 1  1 level and the n =  level. Use the relationship = R  2  2  to calculate the wavelengths. n  n2   1 Solution:  1  1 1  1  1 = R  2  2  = 1.096776x107 m 1  2  2      n2  n2   n1  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 7 1  1 = 1.096776x10 m  2  2  = 8,225,820 m–1  2  1 1 = = 1.215684272x10–7 m= 1.215684x10–7 m 8,225,820 m1 Shortest wavelength in the n1 = 2 series: 1  1 7 1  1 = 1.096776x10 m  2  = 2,741,940 m–1 2     2 1 = = 3.647052817x10–7 m= 3.647053x10–7 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. 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 7 1  1 = 1.096776x10 m  2  2  = 533,155 m–1  4  3 1 = = 1.875627163x10–6 m= 1.875627x10–6 m 533,155 m1 Shortest wavelength in n1 = 4 series has n2 = . 1  1 7 1  1 –1 = 1.096776x10 m  2   = 685,485 m  2  4 1 = = 1.458821127x10–6 m= 1.458821x10–6 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 7 1  1 –1 = 1.096776x10 m  2   = 438,710.4 m  2  5 1 = = 2.27940801x10–6 m= 2.279408x10–6 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 7 1  1 = 1.096776x10 m  2  2  = 246,774.6 m–1  5  4





























1 = 4.052281x10–6 m 246,774.6 m1 For n1 = 4, n2 = 6: 1  1 7 1  1 = 1.096776x10 m  2  2  = 380,825 m–1  6  4 1 = = = 2.625878x10–6 m 1 380,825 m For n1 = 4, n2 = 7: 1  1 7 1  1 = 1.096776x10 m  2  2  = 461,653.2 m–1  4 7   1 –6 = = 2.166128x10 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.

=

6.81









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 = (– 15x10–19 J) – (– 20x10–19 J) = 5x10–19 J  = E/h = (5x10–19 J)/(6.626x10–34 J•s) = 7.54603x1014 s-1= 8x1014 s–1  = c/ = (3.00x108 m/s)/(7.54603x1014 s–1) = 3.97560x10–7 m= 4x10–7 m b) The ionization energy (IE) is the same as the reverse of E1. Thus, the value of the IE is 20x10–19 J/atom. IE = (20x10–19 J/atom)(1kJ/103 J)(6.022x1023 atoms/mol) = 1204.4 kJ/mol= 1.2x103 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). E = E6 – E4 = hc/ = (– 2x10–19 J) – (– 11x10–19 J) = 9x10–19 J

 = hc/E =

6.82

6.626x10



34





J•s 3.00x108 m/s  1 nm  2  9  = 220.867 nm= 2x10 nm 19 10 m 9x10 J  



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). E32 = E31 – E21 = (4.854x10–17 J) – (4.098x10–17 J) = 7.56x10–18 J

=







6.626x1034 J•s 3.00x108 m/s hc = = 2.629365x10–8 m= 2.63x10–8 m 18 E 7.56x10 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). E41 = E42 + E21 = (1.024x10–17 J) + (4.098x10–17 J) = 5.122x10–17 J

=







6.626x1034 J•s 3.00x108 m/s hc = = 3.88091x10–9 m= 3.881x10–9 m 17 E 5.122x10 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 = (5.242x10–17 J) – (5.122x10–17 J) = 1.2x10–18 J

=







6.626x1034 J•s 3.00x108 m/s hc = = 1.6565x10–7 m= 1.66x10–7 m 18 E 1.2x10 J





6.83

a) A dark green colour 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 colour (and vice versa).

6.84

Plan: For part a), use the equation for kinetic energy, Ek = ½mu2. 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. 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 9.109x10–31 kg. 2  J 1 1 –19 –19 Ek = mu2 = 2 9.109x1031 kg 6.40x105 m/s   = 1.86552x10 J= 1.87x10 J 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  –7  (m) =  358.1 nm    = 3.581x10 m 1nm  





6.626x10 J•s 2.9979x10 m/s = 5.447078x10 J 3.581x10 m 34

E = hc/ =



–19

7

 = h – Ek = (5.447078x10–19 J) – (1.86552x10–19 J) = 3.581558x10–19 J= 3.58x10–19 J  kg•m /s   6.626x10 J•s    = 1.322568x10 m m   J    9.109x10 kg   5.5x10 s  34

6.85

a)  = h/mu =

31

–8

Smallest object = /2 = (1.322568x10–8 m)/2 = 6.61284x10–9 m= 6.6x10–9 m

 kg•m /s   6.626x10 J•s  b)  = h/mu =   = 2.424708x10 m m   J   9.109x10 kg 3.0x10    s  34

31

–11

Smallest object = /2 = (2.424708x10–11 m)/2 = 1.212354x10–11 m= 1.2x10–11 m 6.86

Plan: Examine Figure 7.3 and match the given wavelengths to their colours. 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  1 mol SrCl 2   6.022x1023 photons  22 Number of photons =  5.00 g SrCl 2    = 1.8994449x10 photons   158.52 g SrCl 1 mol SrCl 2  2  

 109 m  –7  (m) =  641 nm    = 6.41x10 m 1nm   Ephoton =



=

 6.626x10

34





J•s 3.00x108 m/s  1 kJ  –22  3  = 3.10109x10 kJ/photon 7 6.41x10 m  10 J 

 3.10109x1022 kJ  Etotal = 1.8994449x1022 photons   = 5.89035 kJ= 5.89 kJ  1 photon   BaCl2  1 mol BaCl2   6.022x1023 photons  22 Number of photons =  5.00 g BaCl 2    = 1.44620557x10 photons   1 mol BaCl 2  208.2 g BaCl 2   





 109 m  –7  (m) =  493 nm    = 4.93x10 m  1nm  Ephoton =



=

 6.626x10

34





J•s 3.00x108 m/s  1 kJ  –22  3  = 4.0320487x10 kJ/photon 4.93x107 m 10 J  

 4.0320487x1022 kJ  Etotal = 1.44620557x1022 photons   = 5.83117 kJ= 5.83 kJ  1 photon  



6.87



a) The highest energy line corresponds to the shortest wavelength. The shortest wavelength line is given by  1  1 1 1  1   R  2  2  = 1.096776x107 m 1  2  2  n n  n2  n2   1  1





 1 nm  1 1  7 1  1  9  = 1.096776x10 m  2   3282 nm  10 m  2   n1 304,692 m–1 = (1.096776x107 m–1) (1/n2) 1/n2 = 0.0277807 n=6 b) The lowest energy line corresponds to the longest wavelength. The longest wavelength line is given by  1  1 1  = 1.096776x107 m 1  2   n1   n1  12     1 1  1 nm  7 1  1  = 1.096776x10 m    2   n12 7460 nm  109 m  n  1   1    1  1  134,048 m–1 = 1.096776x107 m 1  2  2  n1  n  1   1    1  1  0.0122220 =  2  2   n1 n  1   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!) 6.88

Plan: Examine Figure 6.3 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. c) A known quantity of vitamin A (1.67x10–3 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.  1.67x103 g  Concentration (C1, g/mL) of Vitamin A =  = 6.68x10–6 g/mL 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  6.68x10 AC C2 = 2 1 = A1 1.018

6

g/mL

 = 4.7508x10 g/mL Vitamin A –6

 4.7508x106 g Vitamin A  –3 Mass (g) of Vitamin A in oil sample =  500. mL oil    = 2.3754x10 g Vitamin A  1 mL oil  

 2.3754x10 g  = 1.92808x10 g= 1.93x10 g Vitamin A/g oil Concentration of Vitamin A in oil sample = 3

–2

 0.1232 g Oil 

6.626x10 J•s 3.00x10 m/s   1 nm  = 261.897 nm= 262 nm  = hc/E =    10 m  7.59x10 J  34

6.89

–2

9

19

Silver is not a good choice for a photocell that uses visible light because 262 nm is in the ultraviolet region. 6.90

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.

6.91

Ek = v=

1 mv2 2

Ek 1 m 2

4.71x1015 J

 kg•m 2 /s2  8   = 1.01692775x10 m/s 1 31 J 9.109x10 kg   2





 kg•m /s   6.626x10 J•s    = 7.15304x10 m= 7.15x10 m m   J   9.109x10 kg 1.01692775x10    s  34

 = h/mv =

31

–12

6.92

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    = 5.50x10 m 1 nm   E=



=

6.626x10

34



J•s 3.00x108 m/s 7

5.50x10 m

 = 3.614182x10 J/photon –19

 1 J/s  Amount of power from the bulb =  75 W    = 75 J/s 1W   5%  Amount of power converted to light =  75 J/s    = 3.75 Js  100%   10%  Amount of light shining on book =  3.75 J/s    = 0.375 J/s  100%  1 photon  0.375 J    Number of photons:  = 1.0376x1018 photons/s= 1.0x1018 photons/s   19   s   3.614182x10 J  6.93

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.

6.94

a) 6CO2(g) + 6H2O(l)  C6H12O6(s) + 6O2(g) rH = {  f H C6H12O6] + 6  f H O2]} – {6  f H CO2] + 6  f H H2O]} rH = [(–1273.3 kJ/mol) + 6(0.0 kJ/mol)] – [6(–393.5 kJ/mol) + 6(–285.840 kJ/mol)] = 2802.74 kJ/mol= 2802.7 kJ/mol 6CO2(g) + 6H2O(l)  C6H12O6(s) + 6O2(g) rH = 2802.7 kJ per 1.00 mol C6H12O6

 6.626x10 b) E = hc/=

34





J•s 3.00x108 m/s  1 nm  –19  9  = 2.9232353x10 J/photon 680. nm 10 m  

 103 J    1 photon 24 24 Number of photons =  2802.7 kJ    1 kJ   2.9232353x1019 J  = 9.5877x10 photons= 9.59x10 photons    6.95

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 3d → 2p 4s → 2p 3p → 2s

6.96

0.9

Absorbance

0.7 0.5 0.3 0.1 0 0

10 20 Concentration, mol/L

x 105

A = kC k = 4.5x103 (mol/L)–1 = slope b) A/k = C = (0.55)/(4.5x103 (mol/L)–1) = 1.2222x10–4 mol/L= 1.2x10–4 mol/L 6.97

Plan: We can find the energy required to melt the ice by converting the mass of ice to amount (mol) and then multiplying by the enthalpy of fusion. We can determine the energy per photon using the wavelength of the light. We use the two energies to find the number of photons. In the second part of the question, we know the power output of the microwave, and the total amount of energy needed, so we can find the time required.

nice 

mice 675 g   37.48 mol M ice 18.02 g/mol

Eneeded  nice   fusion H ice  (37.48 mol)(6.02 kJ/mol)= 225.6 kJ Solution:

(6.626  10 34 J  s)(3.00 108 m/s)  1.500×10-24 J / photon  0.1325 m E 225.6 103 J No. of photons = needed   1.50×1029 photons Ephoton 1.500  1024 J/photon Ephoton  h 



E 225.6 kJ   150.4 s = 2.5 min P 1.5 kW 1.50 x 1029 photons would be required and it will take 2.5 min. t

6.98

Plan: We must find the binding energy per electron by dividing the given work function by Avogadro‘s number. When the energy of the incident light JUST equals the binding energy, the light is of maximum wavelength to allow the photoelectric effect to occur. We can find the kinetic energy of the ejected electron by subtracting the binding energy from the energy of the incident light in part b). In part c), we use the fact that the ejected electron has kinetic energy equal to ½mu2. Solution: a)

260.5 103 J/mol  4.326×10-19 J / e 6.022 1023 e  / mol For the maximum wavelength, hc  =h 





4.326 1019 J 

(6.626 1034 J  s)(3.00 108 m/s)

   4.59 107 m = 459 nm

EK  h   





(6.626 1034 J  s)(3.00 108 m/s)  4.326 1019 J 325 109 m =1.79×10-19 J 

1 2 mu 2 1 1.79 1019 J  (9.109 10 31 kg)(u 2 ) 2 u = 6.27 ×105 m / s Comment: Remember to use the mass of the electron in units of kilogram in order for the units to work! EK 

6.99

Plan: We can find the binding energy per electron by dividing the given work function by Avogadro‘s number. We know the velocity of the ejected electron so we can determine its kinetic energy. We then use the equation for the photoelectric effect to find the energy of the incident light and thus, its wavelength.

337.7 103 J/mol  5.608×10-19 J / e6.022 1023 e / mol 1 EK  mu 2 2 1  (9.109 1031 kg)(8.76 105 m/s) 2 2 EK = 3.495×10-19 J

Solution:  

EK  h    hc





 EK    3.495  1019 J + 5.608 1019 J

=9.103×10-19 J hc  E hc  9.103  1019 J (6.626 1034 J  s)(3.00  108 m/s)  9.103  1019 J -7 = 2.18×10 m = 218 nm

6.100

Plan: We can find the amount of heat required to raise the temperature of the water using E=mcT (since we know the volume and density of water and its specific heat capacity as well as the temperature change). We can find the energy per photon using the wavelength of the light. We can then find the total number of photons needed using the relationship between the two energies. For part b), we can find the energy per photon using the wavelength and we know the number of photons from part a). From these, we can find the total energy. This total energy has to equal the power multiplied by the time the lamp must be on and again multiplied by the energy efficiency (since not all the power produces the light!). From this relationship, we can find the time. Solution: a)

Eneeded  mH O  cH O TH O 2

 (137 mL)(1.00 g/mL)(4.184

J )(1 C) g C

=573.2 J Ephoton  h 



(6.626 1034 J  s)(3.00 108 m/s)  0.1447 m -24  1.374×10 J / photon No. of photons =

Eneeded 573.2 J   4.17 ×1026 photons Ephoton 1.374  1024 J/photon

Ephoton  h 



(6.626  1034 J  s)(3.00  108 m/s)  550  109 m  3.61×10-19 J / photon ETOT = Ephoton  No. of photons  (3.61 1019 J/photon)(4.17×1026 photons) = 1.51 108 J E t E (1.51108 J)  1min  1h  1day  1 year  t   1.03  108 s      = 3.25 years P (0.098)(15 W)  60s  60 min  24h  364  The lamp would have to be on for 3 and a quarter years (constantly) in order to emit the same number of photons as a microwave does to heat up roughly half a cup of water by JUST 1 °C!! P

6.101

Plan: Set the kinetic energy (KE) to zero and solve for the wavelength using the equation for the photoelectric effect. Solve for KE using the photoelectric effect. Use the relationship between KE and mass/velocity to solve for velocity. Solution: a)  kJ  1000J   1 mol J  16 φ   3.0x105    4.98x10   23 mol  kJ   6.022x10 photon  photon 

φ  hν 

hc λ

 6.626x10 Jgs   3.00x10 ms  -34

4.98x10

-16

J  photon

 6.626x10 Jgs   3.00x10 ms  -34

λ

4.98x10-16

J photon

 3.99 x10-10 m  0.399 nm b)

KE  hν  φ 

m  J s  3.00x108  J s    4.98x1016 photon 2.69x10 10 m

 6.626x10

34



 2.41x1016 J c) 1 mv 2 2 2KE v m

KE 



2(2.41x1016 J) 9.11x1031 kg

m s 6.102 Plan: Set up the equation for the photoelectric effect for each wavelength. Set the relation between the two values for the KE. Solve for the binding energy. Convert the units from J/photon to kJ/mol.  2.30x107

Solution:

KE 1  hν  φ 

m  J s  3.00x108  s   φ 275 x10 9 m

 6.626x10

34



 7.23x1019 J  φ KE 2  hν  φ 

m  J s  3.00x108  s   φ 262.7 x10 9 m

 6.626x10

34



 7.57x1019 J  φ

KE 2  2KE 1



J  φ  2 7.23x10

7.57x10

19

J  φ  14.46x10

19

J  2φ

φ  14.46x10

19

J  7.57x10 19 J

 6.89x10 19



  6.89x10

19



7.57x10

19

J photon

19



Jφ

  6.022x1023 photon   1kJ     photon   mol   1000J  J

 415 kJ

6.103

Plan: Use the dimensions of the ice to find the volume of ice. Use the density and volume to find the mass of ice. Convert the mass of ice to amount (mol) of ice. Use the relationship between heat (q) and enthalpy to determine q. Use the wavelength of the light to find the energy per photon. Apply the fact that only 20% of the light reaches the surface to calculate the net energy per photon. Use the relationship between the number of photons, the energy per photon and the total energy to calculate the number of photons. Solution: Vice   2.5cm  2.5cm  4.0cm   25cm 3 g   m ice  25cm 3  0.9168  23g 3  cm   m 23g n ice    1.3 mol g MM 18.02 mol q  Δ fus H n





J     6.02x103 1.3 mol  mol    7.7x103 J

E photon 



hc λ m  J s  3.00x108  s   9 550x10 m

 6.626x10

34



 3.6x1019 J But, only 20% of the light reaches the surface of the ice. Therefore, E photon   0.2  3.6x1019 J  7x10 20 J





E tot   E photon  # of photons  # of photons 

E tot  E photon

7.66x103 J  1x1023 photons J 20 7.23x10 photon

6.104 Plan: Use the density, volume, specific heat capacity and change in temperature to find q of the cocoa. Use the wavelength of the light to find the energy per photon. Use the connection between the total enegy and the energy per photon to find the number of photons. Solution:

E cocoa  q cocoa   mcΔT cocoa   dVcΔT cocoa  g  J     1.03  99.5 C  5.3 C   245mL   3.9 mL g C   





 9.3x104 J

E photon 



hc λ m  J s  3.00x108  s   2 13.6x10 m

 6.626x10

34



 1.46x1024 J # of photons 

E tot  E photon

9.3x104 J 1.46x10

24

J photon

 6.3x1028 photons

CHAPTER 7 ELECTRON CONFIGURATION AND CHEMICAL PERIODICITY END–OF–CHAPTER PROBLEMS 7.1

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.

7.2

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.

7.3

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 u  85.47 u = = 54.23 u (actual value = 39.10 u) 2 2 b) Predicted melting point (Br2) = Cl2  I 2 101.0C  113.6C = = 6.3°C (actual value = –7.2C) 2 2

7.4

a) Predicted boiling point (HBr) = HCl  HI 84.9C  (35.4C) = = –60.15°C = –60.2°C 2 2 b) Predicted boiling point (AsH3) = PH 3  SbH 3 87.4C  (17.1C) = = –52.25°C = –52.2°C 2 2

7.5

(actual value = –67.0C)

(actual value = –55C)

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

7.6

The quantum number ms relates to just the electron; all the others describe the orbital.

7.7

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.

7.8

In a one-electron system, all subshells of a particular level (such as 2s and 2p) have the same energy. In many electron systems, the principal energy levels are split into subshells of differing energies. This splitting is due to electron-electron repulsions. Be3+ would be more like H since both have only one 1s electron.

7.9

Shielding occurs when core electrons protect or shield valence electrons from the full nuclear attractive force. The effective nuclear charge is the nuclear charge an electron actually experiences. As the number of core electrons increases, shielding increases and the effective nuclear charge decreases.

7.10

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 valence shell electrons, since they spend most of their time at distances farther from the nucleus and are shielded from the nuclear charge by the core 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.

7.11

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 hold 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.

7.12

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. 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.

7.13

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).

7.14

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.

7.15

Properties recur periodically due to similarities in electron configurations recurring periodically. Na: 1s22s22p63s1 K: 1s22s22p63s23p64s1 The properties of Na and K are similar due to a similarity in their valence shell electron configuration; both have one electron in a valence shell s orbital.

7.16

Hund‘s rule states that electrons will fill empty orbitals in the same subshell 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

7.17

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 valence electron configurations when orbital filling is in order of increasing energy.

7.18

For elements in the same group (vertical column in periodic table), the electron configuration of the valence electrons is 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 valence level with the exception of certain of the d block and f block elements.

7.19

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.

7.20

The total electron capacity for an energy level is 2n2, so the n = 4 energy level holds a maximum of 2(4 2) = 32 electrons. A filled n = 4 energy level would have the following configuration: 4s24p64d104f14.

7.21

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. By convention, ml = –1 for the px orbital, ml = 0 for the py orbital, and ml = +1 for the pz orbital. Also, 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: a) Rb: [Kr]5s1. 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 could be n = 5, l = 0, ml = 0, and ms = +1/2 or -1/2. b) The S– ion would have the configuration [Ne]3s23p5. The electron added would go into a different orbital than the first electron paired electron and would be the second electron in the orbital it enters. Quantum numbers could be n = 3, l = 1, ml = +1, 0 or -1 (depending on where the 4th electron has been placed), and ms = –1/2 or +1/2 depending on the spin of the electron already in the orbital that the added electron enters. c) Ag atoms have the configuration [Kr]5s14d10. 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, depending on which electron was lost. d) The F atom has the configuration [He]2s22p5. The electron gained would go into the only 2p orbital with a single electron and would be the second electron in that orbital. Quantum numbers could be n = 2, l = 1, ml = +1, and ms = –1/2 or +1/2 depending on the spin of the electron already in that orbital.

7.22

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 (depending on the electrons already present) 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 (depending on the electrons already present)

7.23

7.24

a) Br: 1s22s22p63s23p64s23d104p5 b) Mg: 1s22s22p63s2 c) Se: 1s22s22p63s23p64s23d104p4

7.25

Solution: a) Cl: 1s22s22p63s23p5 b) Si: 1s22s22p63s23p2 c) Sr: 1s22s22p63s23p64s23d104p65s2 7.26

a) S+: b) Kr: c) Cs:

1s22s22p63s23p3 1s22s22p63s23p64s23d104p6 1s22s22p63s23p64s23d104p65s24d105p66s1

7.27

Plan: The atomic number gives the number of electrons and the periodic table shows the order for filling subshells. 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: a) Ti (Z = 22); [Ar]4s23d2 (note that the 2 3d electrons can be placed in any of the 5 boxes as long as they have the same spin, which could be positive or negative).

b) Cl (Z = 17); [Ne]3s23p5 (note that the unpaired electron could have positive or negative spin).

c) V (Z = 23); [Ar]4s23d3 (note that the 3 3d electrons can be placed in any of the 5 boxes as long as they have the same spin, which could be positive or negative).

7.28

a) Ba:

[Xe]6s2

b) Co:

[Ar]4s23d7 (note that the 3 unpaired electrons could have positive or negative spin).

c) Ag:

[Kr]5s14d10 (note that the unpaired s electron could have positive or negative spin).

7.29

Plan: The atomic number gives the number of electrons and the periodic table shows the order for filling subshells. 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: a) Mn (Z = 25); [Ar]4s23d5 (note that the 3d electrons could also ALL have negative spin).

b) P (Z = 15); [Ne]3s23p3 (note that the 3p electrons could also ALL have negative spin).

c) Fe (Z = 26); [Ar]4s23d6 (note that the unpaired electrons could also all have negative spins).

7.30

7.31

a) Ga: [Ar]4s23d104p1 (note that the single 4p electron could be in any box of the three boxes and have positive or negative spin).

b) Zn:

[Ar]4s23d10

c) Sc:

[Ar]4s23d1 (note that the 3d electron could be in any box and have positive or negative spin).

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 16, Period 2. (Note that the unpaired 2p electrons could both have negative spin.)

b) There are 15 electrons in the configuration; the element is P, Group 15, Period 3. (Note that the 3p electrons could all have negative spins as well.)

7.32

a) Cd; Group 12; Period = 5

b) Ni; Group 10; Period = 4 (note that the two unpaired electrons could both have negative spin).

7.33

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 17; Period 3. Note that the single electron could have negative spin.

b) There are 33 electrons in the configuration; the element is As; Group 15; Period 4. Note that the 4p electrons could all have negative spin.

7.34

a) Mn; Group 7; Period = 4 (note that the 3d electrons could all have negative spins).

b) Zr; Group 4; Period = 5 (note that the two 4d electrons could be in any of the 5 boxes and could both have negative spin as well).

7.35

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 1s22s22p63s23p64s23d104p1 or [Ar]4s23d104p1. One electron in the p level indicates the element is in Group 13. The element is Ga. b) The orbital diagram shows the 2s and 2p orbitals filled which would represent the last element in Period 2, Ne. The configuration is 1s22s22p6 or [He]2s22p6. Filled s and p orbitals indicate Group 18.

7.36

a) [Kr]5s14d4 Nb; 5 b) [He]2s22p3 N; 15

7.37

Plan: Core electrons are those seen in the previous noble gas and completed transition series (d orbitals). Valence electrons are those in the highest energy level (highest n value). 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: a) O (Z = 8); [He]2s22p4. There are 2 core electrons (represented by [He]) and 4 valence electrons. b) Sn (Z = 50); [Kr]5s24d105p2. There are 36 (from [Kr]) + 10 (from the filled 4d set) = 46 core electrons. The highest energy level is n = 5 so there are 4 valence electrons. c) Ca (Z = 20); [Ar]4s2. There are 2 valence electrons (the 4s electrons), and 18 core electrons (from [Ar]). d) Fe (Z = 26); [Ar]4s23d6. There are 8 valence electrons (2 from n = 4 level and the d orbital electrons count in this case because the subshell is not full) , and 18 core electrons (from [Ar]). e) Se (Z = 34); [Ar]4s23d104p4. There are (2 + 4 in the n = 4 level), 6 valence electrons (filled d subshells count as core electrons), and 28 core electrons (18 from [Ar] and 10 from the filled 3d set).

7.38 a) Br b) Cs c) Cr d) Sr e) F

core electrons 28 54 18 36 2

valence electrons 7 1 6 2 7

7.39

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: a) The electron configuration [He]2s22p1 has a total of 5 electrons (3 + 2 from He configuration) which is element boron with symbol B. Boron is in Group 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 Group16. Other elements in Group 16 are O, Se, Te, and Po. c) Electrons total 4 + 54 (from xenon) = 58. Element 58 is hafnium, Hf, in Group 4. Other elements in this group are Ti, Zr, and Rf.

7.40

a) Se; other members O, S, Te, Po b) Hf; other members Ti, Zr, Rf c) Mn; other members Tc, Re, Bh

7.41

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: a) The electron configuration [He]2s22p2 has a total of 6 electrons (4 + 2 from He configuration) which is element carbon with symbol C; other Group 14 elements include Si, Ge, Sn, and Pb. b) Electrons total 5 + 18 (from argon) = 23 which is vanadium; other Group 5 elements include Nb, Ta, and Db. 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 15 elements include N, As, Sb, and Bi.

7.42

a) Ge; other members C, Si, Sn, Pb b) Co; other members Rh, Ir, Mt c) Tc; other members Mn, Re, Bh

7.43

Plan: Write the ground-state electron configuration of sodium; for the excited state, move the outermost electron to the next orbital. Solution: The ground-state configuration of Na is 1s22s22p63s1. Upon excitation, the 3s1 electron is promoted to the 3p level, with configuration 1s22s22p63p1. (Note that the electron in 3p can be in any of the 3 boxes and have positive or negative spin.)

7.44

a) Mg: b) Cl: c) Mn: d) Ne:

7.45

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.

7.46

a) A = silicon; B = fluorine; C = strontium; D = sulfur b) F < S < Si < Sr c) Sr < Si < S < F

7.47

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.

7.48

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 the number of valence electrons. For instance, the electron configuration for potassium is [Ar]4s1. 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.

7.49

The first drop occurs because the 3p subshell is higher in energy than the 3s, so the 3p electron of Al is pulled off more easily than a 3s electron of Mg. The second drop occurs because the 3p4 electron occupies the same orbital as another 3p electron. The resulting electron-electron repulsion raises the orbital energy and thus it is easier to remove an electron from S (3p4) than P (3p3).

7.50

A high, endothermic IE1 means it is very difficult to remove the first valence electron. This value would exclude any metal, because metals lose a valence electron easily. A very negative, exothermic EA1 suggests that this element easily gains one electron. These values indicate that the element belongs to the halogens, Group 17, which form –1 ions.

[Ne]3s2 [Ne]3s23p5 [Ar]4s23d5 [He]2s22p6

7.51

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.

7.52

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.

7.53

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 valence 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.

7.54

a) Pb > Sn > Ge b) Sr > Sn > Te c) Na > F > Ne

7.55

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 valence 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.

7.56

a) Li > Na > K

7.57

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 1s22s22p1 which represents boron, B.

7.58

The successive ionization energies show a significant jump between the second and third IEs, indicating that the element has only two valence electrons. The configuration is 1s22s22p63s2, Mg.

b) F > C > Be

d) Na > Mg > Be

c) Ar > Cl > Na d) Cl > Br > Se

7.59

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 IE 2 represents removing an electron from a full orbital, then the IE2 will be very large. In addition, for atoms with the same valence 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: First ionization: Na ([Ne]3s1)  Na+ ([Ne]) + e– (low IE) Second ionization: Na+ ([Ne])  Na+2 ([He]2s22p5) + 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.

7.60

a) Al

7.61

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.

7.62

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.

7.63

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 14, Sn and Pb, have atomic electron configurations that look like ns2(n – 1)d10np2. 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 (n – 1)d orbital filled for the 4+ ion. The Sn2+ and Pb2+ ions form by losing the two p electrons: Sn ([Kr]5s24d105p2)  Sn2+ ([Kr]5s24d10) + 2 e– Pb ([Xe]6s24f145d106p2)  Pb2+ ([Xe]6s24f145d10) + 2 e– The Sn4+ and Pb4+ ions form by losing the two p and two s electrons: Sn ([Kr]5s24d105p2)  Sn4+ ([Kr]4d10) + 4 e– Pb ([Xe]6s24f145d106p2)  Pb4+ ([Xe] 4f145d10) + 4 e– Possible ions for tin and lead have +2 and +4 charges.

7.64

An (n – 1)d10ns0np0 configuration is called a pseudo-noble gas configuration. In3+: [Kr]4d10

7.65

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. Cu1+: [Ar]3d10 thus, diamagnetic Cu2+: [Ar]3d 9 thus, paramagnetic, with one unpaired electron

7.66

In order of increasing size, the ions are: 3+ < 2+ < 1+ < 0 < 1– < 2– < 3–

b) Sc

c) Al

7.67

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 2. c) I is more metallic because it lies below Br in Group 17.

7.68

a) S

7.69

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.

7.70

a) Rn

7.71

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: a) Cl: 1s22s22p63s23p5; chlorine atoms are one electron short of a noble gas configuration, so a –1 ion will form by adding an electron to have the same electron configuration as an argon atom: Cl –, 1s22s22p63s23p6. b) Na: 1s22s22p63s1; sodium atoms contain one more electron than the noble gas configuration of neon. Thus, a sodium atom loses one electron to form a +1 ion: Na+, 1s22s22p6. c) Ca: 1s22s22p63s23p64s2; calcium atoms contain two more electrons than the noble gas configuration of argon. Thus, a calcium atom loses two electrons to form a +2 ion: Ca2+, 1s22s22p63s23p6.

7.72

a) Rb+: 1s22s22p63s23p64s23d104p6 b) N3–: 1s22s22p6 c) Br–: 1s22s22p63s23p64s23d104p6

7.73

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: a) Al: 1s22s22p63s23p1; aluminum atoms contain three more electrons than the noble gas configuration of Ne. Thus, an aluminum atom loses its 3 valence shell electrons to form a +3 ion: Al3+, 1s22s22p6. b) S: 1s22s22p63s23p4; sulfur atoms are two electrons short of the noble gas configuration of argon. Thus, a sulfur atom gains two electrons to form a –2 ion: S2–, 1s22s22p63s23p6. c) Sr: 1s22s22p63s23p64s23d104p65s2; strontium atoms contain two more electrons than the noble gas configuration of krypton. Thus, a strontium atom loses two electrons to form a +2 ion: Sr2+, 1s22s22p63s23p64s23d104p6.

7.74

a) P3– 1s22s22p63s23p6 b) Mg2+ 1s22s22p6 c) Se2– 1s22s22p63s23p64s23d104p6

7.75

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.

b) In

b) Te

c) As

c) Se

+1 –3 –1

Solution: a) Configuration of 2 group elements: [noble gas]ns2, no unpaired electrons. The electrons in the ns orbital are paired. b) Configuration of 15 group elements: [noble gas]ns2npx1npy1npz1. Three unpaired electrons, one each in px, py, and pz. Spins can be all positive or all negative. c) Configuration of 18 group elements: noble gas configuration ns2np6 with no half-filled orbitals, no unpaired electrons. d) Configuration of 13 group elements: [noble gas]ns2np1. There is one unpaired electron in one of the p orbitals. The unpaired electron can be in any of the p orbitals and can have positive or negative spin.

7.76

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. a) Configuration of 14 group elements: [noble gas]ns2npx1npy1npz0. Two unpaired electrons. b) Configuration of 17 group elements: [noble gas]ns2npx2npy2npz1. One unpaired electron. c) Configuration of 1 group elements: [noble gas]ns1. One unpaired electron. d) Configuration of 16 group elements: [noble gas]ns2npx2npy1npz1. Two unpaired electrons.

7.77

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: a) Ga (Z = 31) = [Ar]4s23d104p1. The s and d subshells are filled, so all electrons are paired. The lone p electron is unpaired, so this element is paramagnetic. b) Si (Z = 14) = [Ne]3s23px13py13pz0. This element is paramagnetic with two unpaired electrons.

Correct Incorrect c) Be (Z = 4) = [He]2s2. The two s electrons are paired so Be is not paramagnetic. d) Te (Z = 52) = [Kr]5s24d105px25py15pz1 is paramagnetic with two unpaired electrons in the 5p set. 7.78

a) Ti2+: [Ar]3d 2 b) Zn2+: [Ar]3d10 c) Ca2+: [Ar] d) Sn2+: [Kr]5s24d10

7.79

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.

paramagnetic (Note: Electrons are removed from the highest n Value first, i.e., 4s.) diamagnetic diamagnetic diamagnetic

Solution: a) V: [Ar]4s23d3; V3+: [Ar]3d2 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. Note that the two d electrons can be in any of the boxes and can both have either positive or negative spins). Paramagnetic

b) Cd: [Kr]5s24d10; Cd2+: [Kr]4d10 Cadmium atoms lose two electrons from the 4s orbital to form the +2 ion. Diamagnetic

c) Co: [Ar]4s23d7; Co3+: [Ar]3d6 Cobalt atoms lose two 4s electrons and one 3d electron to form the +3 ion. Note that all the unpaired electrons can have either positive or negative spins. Paramagnetic

d) Ag: [Kr]5s14d10; Ag+: [Kr]4d10 Silver atoms lose the one electron in the 5s orbital to form the +1 ion. Diamagnetic

7.80

a) Mo3+: [Kr]4d3 b) Au+: [Xe]4f 145d10 c) Mn2+: [Ar]3d5 d) Hf 2+: [Xe]4f 145d2

7.81

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: You might first write the condensed electron configuration for Pd as [Kr]5s24d8. However, the partial orbital diagram is not consistent with diamagnetism.

paramagnetic diamagnetic paramagnetic paramagnetic

Promoting an s electron into the d subshell (as in (c) [Kr]5s14d9) still leaves two electrons unpaired.

The only configuration that supports diamagnetism is (b) [Kr]4d10.

7.82

The expected electron configuration for Group 5 elements is ns2(n–1)d3. Nb (expected): [Kr]5s24d 3 3 unpaired e–

Nb (actual): [Kr]5s14d 4 5 unpaired e–

7.83

Plan: The size of ions increases down a group. For ions that are isoelectronic (have the same electron configuration) size decreases with increasing atomic number. Solution: a) Increasing size: Li+ < Na+ < K+, size increases down Group 1. b) Increasing size: Rb+ < Br– < Se2–, these three ions are isoelectronic with the same electron configuration as krypton. Size decreases with increasing atomic number in an isoelectronic series. c) Increasing size: F– < O2– < N3–, the three ions are isoelectronic with an electron configuration identical to neon. Size decreases with increasing atomic number in an isoelectronic series.

7.84

a) Se2– > S2– > O2–, size increases down a group. b) Te2– > I – > Cs+, size decreases with increasing atomic number in an isoelectronic series. c) Cs+ > Ba2+ > Sr 2+, both reasons as in parts a) and b).

7.85

a) oxygen g) krypton m) manganese

7.86

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: Ce: [Xe]6s24f 15d1 Eu: [Xe]6s24f 7 Ce4+: [Xe] Eu2+: [Xe]4f 7 4+ 2+ Ce has a noble gas configuration and Eu has a half-filled f subshell.

7.87

a) The distance, d, between the electron and the nucleus would be smaller for He than for H because as we go across a row, the number of protons increases in the same shell thus pulling the electrons closer; 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, IE 1 (He) < 2 IE1(H).

7.88

b) cesium h) silicon n) lutetium

 6.626x10  = hc/E =

34

c) aluminum i) ruthenium o) sulfur



J•s 3.00x108 m/s 19

d) carbon j) vanadium p) strontium

e) rubidium k) indium

f) bismuth l) scandium

 = 4.59564x10 m= 4.6x10 m –7

–7

 1.602x10 J   2.7 eV    1 eV   The absorption of light of this wavelength (blue) leads to the complimentary colour (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‖ colour.

7.89

Plan: Remember that isoelectronic species have the same electron configuration. Atomic radius decreases up a group and left to right across a period. Solution: a) A chemically unreactive Period 4 element would be Kr in Group 18. Both the Sr 2+ ion and Br– ion are isoelectronic with Kr. Their combination results in SrBr2, strontium bromide. b) Ar is the Period 3 noble gas. Ca2+ and S2– are isoelectronic with Ar. The resulting compound is CaS, calcium sulfide. c) The smallest filled d subshell is the 3d shell, so the element must be in Period 4. Zn forms the Zn2+ ion by losing its two s subshell electrons to achieve a pseudo–noble gas configuration ([Ar]3d10). 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.

7.90

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– E = 1681 kJ/mol F(g) + e–  F–(g) E = –328 kJ/mol Reverse the electron affinity reaction to give: F–(g)  F(g) + e– E = +328 kJ/mol Summing the ionization energy reaction with the reversed electron affinity reaction (Hess‘s law): F(g)  F+(g) + e– E = 1681 kJ/mol F– (g)  F(g) + e– E = +328 kJ/mol F–(g)  F+(g) + 2 e– E = 2009 kJ/mol b) Na: ionization energy = 496 kJ/mol electron affinity = –52.9 kJ/mol Na(g)  Na+(g) + e– E = 496 kJ/mol Na(g) + e–  Na–(g) E = –52.9 kJ/mol Reverse the ionization reaction to give: Na+(g) + e–  Na(g) E = –496 kJ/mol Summing the electron affinity reaction with the reversed ionization reaction (Hess‘s law): Na(g) + e–  Na–(g) E = –52.9 kJ/mol Na+(g) + e–  Na(g) E = –496 kJ/mol Na+(g) + 2 e–  Na-(g) E = –548.9 kJ/mol= –549 kJ/mol

7.91

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: Fe [Ar]4s23d6 partially filled 3d = paramagnetic number of unpaired electrons = 4 + Fe [Ar]4s13d6 partially filled 3d = paramagnetic number of unpaired electrons = 5 Fe2+ [Ar]3d6 partially filled 3d = paramagnetic number of unpaired electrons = 4 Fe3+ [Ar]3d5 partially filled 3d = paramagnetic number of unpaired electrons = 5 Fe4+ [Ar]3d4 partially filled 3d = paramagnetic number of unpaired electrons = 4 Fe5+ [Ar]3d3 partially filled 3d = paramagnetic number of unpaired electrons = 3 Fe6+ [Ar]3d2 partially filled 3d = paramagnetic number of unpaired electrons = 2 Fe7+ [Ar]3d1 partially filled 3d = paramagnetic number of unpaired electrons = 1 Fe8+ [Ar] filled orbitals = diamagnetic number of unpaired electrons = 0 Fe9+ [Ne]3s23p5 partially filled 3p = paramagnetic number of unpaired electrons = 1 Fe10+ [Ne]3s23p4 partially filled 3p = paramagnetic number of unpaired electrons = 2 11+ Fe [Ne]3s23p3 partially filled 3p = paramagnetic number of unpaired electrons = 3 Fe12+ [Ne]3s23p2 partially filled 3p = paramagnetic number of unpaired electrons = 2 Fe13+ [Ne]3s23p1 partially filled 3p = paramagnetic number of unpaired electrons = 1 Fe14+ [Ne]3s2 filled orbitals = diamagnetic number of unpaired electrons = 0 Fe+ and Fe3+ would both be most attracted to a magnetic field. They each have 5 unpaired electrons.

7.92

a) Ga [Ar]4s23d104p1 Ge [Ar]4s23d104p2 In each case, an electron is removed from a 4p orbital. Because the Zeff of Ge is greater, IE of Ge will be higher. b) Ga+ [Ar]4s23d10 Ge+ [Ar]4s23d104p1 + 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. c) Ga2+ [Ar]4s13d10 Ge2+ [Ar]4s23d10 In each case, an electron is removed from a 4s orbital. Because the Zeff of Ge2+ is greater, IE3 of Ge2+ will be higher. d) Ga3+ [Ar]3d10 Ge2+ [Ar]4s13d10 3+ IE4 would be higher for Ga , because an electron must be removed from a full 3d orbital which forms part of the core electron group. For Ge3+ the electron is removed from the 4s orbital.

7.93

a) The energy required would be 376 kJ/mol (Cs).

6.626x10  = hc/E =

34





34





J•s 3.00x108 m/s  1 kJ  –7 –7  3  = 3.18365x10 m= 3.18x10 m kJ mol     10 J   376 mol   23     6.022x10  b) The energy required would be 503 kJ/mol (Ba).

6.626x10  = hc/E =

J•s 3.00x108 m/s  1 kJ  –7 –7  3  = 2.37983x10 m= 2.38x10 m kJ   mol    10 J   503 mol   23     6.022x10  c) No other element, other than the alkali metals, in Figure 7.16 has a lower ionization energy than Ba so the radiation, 2.39x10–7 m, cannot ionize other elements. d) Both of these photons are in the ultraviolet region of the EM spectrum.

7.94

a) Rubidium atoms form +1 ions, Rb+; bromine atoms form –1 ions, Br–. b) Rb: [Kr]5s1; Rb+: [Kr]; Rb+ is a diamagnetic ion that is isoelectronic with Kr. Br: [Ar]4s23d104p5; Br–: [Ar]4s23d104p6 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.

7.95

A: X2+ = [Kr]4d8; X = [Kr]5s24d8. The element is palladium and the oxide is PdO. B: X3+ = [Ar]3d6; X = [Ar]4s23d7. The element is cobalt and the oxide is Co2O3. C: X+ = [Kr]4d10; X = [Kr]5s14d10. The element is silver and the oxide is Ag2O. D: X+4 = [Ar]3d3; X = [Ar]4s23d5. The element is manganese and the oxide is MnO2.

7.96

balloonium inertium allotropium brinium canium fertilium liquidium utilium crimsonium

= = = = = = = = =

helium neon sulfur sodium tin nitrogen bromine aluminum strontium

CHAPTER 8 MODELS OF CHEMICAL BONDING END–OF–CHAPTER PROBLEMS

8.1

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.

8.2

A has covalent bonding, B has ionic bonding, and C has metallic bonding.

8.3

The tendency of main-group elements to form cations decreases from Group 1 to 14, and the tendency to form anions increases from Group 14 to 17. Group 1 and 2 elements form mono- and divalent cations, respectively, while Group 16 and 17 elements form di- and monovalent anions, respectively.

8.4

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 15 than N.

8.5

a) O

8.6

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.

8.7

a) covalent

8.8

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.

8.9

a) metallic

8.10

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 elements in Groups 1-2 and the group number – 10 of the main-group elements in Groups 13-18 gives the number of valence electrons. Rb is Group 1, Si is a Group 14 element, and I is a Group 17 element. Solution: a)

8.11

b) Be

c) Se

b) covalent

c) ionic

b) covalent

c) ionic

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 elements in Groups 1-2 and the group number – 10 of the main-group elements in Groups 13-18 gives the number of valence electrons. Ba is a Group 2 element, Kr is a Group 18 element, and Br is a Group 17 element. Solution: a) Ba

c) Br

8.12

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 elements in Groups 1-2 and the group number – 10 of the main-group elements in Groups 13-18 gives the number of valence electrons. Sr is a Group 1 element, P is a Group 15 element, and S is a Group 16 element. Solution: a)

b) Se

8.13

c) Ga

8.14

Plan: Assuming X is a main-group element, the number of dots (valence electrons) equals the group number if this number is 1 or 2. If the number of dots (valence electrons) is 3 or more, then we add 10 to calculate 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 Group 16 element. Its general electron configuration is [noble gas]ns2np4, 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 Group 13 element with general electron configuration [noble gas]ns2np1.

8.15

a) 15; ns2np3 b) 14; ns2np2

8.16

Energy is generally 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.

8.17

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

8.18

The lattice energy releases even more energy when the gas is converted to the solid.

8.19

The lattice energy drives the energetically unfavorable electron transfer resulting in solid formation.

8.20

Plan: Write condensed electron configurations and draw the Lewis electron-dot symbols for the atoms. The group number of the elements in Groups 1-2 and the group number – 10 of the main-group elements in Groups 13-18 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: Ba ([Xe]6s2)  Ba2+ ([Xe]) + 2e– 2+ Ba

+ 2 e

Chlorine is a nonmetal and gains 1 electron to achieve a noble gas configuration: Cl ([Ne]3s23p5) + 1e–  Cl– ([Ne]3s23p6)  Cl

+ 1 e

Two Cl atoms gain the 2 electrons lost by Ba. The ionic compound formed is BaCl2.

b) Strontium is a metal and loses 2 electrons to achieve a noble gas configuration: Sr ([Kr]5s2)  Sr2+ ([Kr]) + 2e– Oxygen is a nonmetal and gains 2 electrons to achieve a noble gas configuration: O ([He]2s22p4) + 2e–  O2– ([He]2s22p6) 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: Al ([Ne]3s23p1)  Al3+ ([Ne]) + 3e– Fluorine is a nonmetal and gains 1 electron to achieve a noble gas configuration: F ([He]2s22p5) + 1e–  F– ([He]2s22p6) 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: Rb ([Kr]5s1)  Rb+ ([Kr]) + 1e– Oxygen is a nonmetal and gains 2 electrons to achieve a noble gas configuration: O ([He]2s22p4) + 2e–  O2– ([He]2s22p6) One O atom gains the two electrons lost by two Rb atoms. The ionic compound formed is Rb2O.

8.21

a) Cesium loses 1 electron to achieve a noble gas configuration: Cs ([Xe]6s1)  Cs+ ([Xe]) + 1e– Sulfur gains 2 electrons to achieve a noble gas configuration: S ([Ne]3s23p4) + 2e–  S2– ([Ne]3s23p6) One S atom gains the two electrons lost by two Cs atoms. The ionic compound formed is Cs2S.

b) Gallium loses 3 electrons to achieve a noble gas configuration: Ga ([Ar]3d104s24p1)  Ga3+ ([Ar]3d10) + 3e– Oxygen gains 2 electrons to achieve a noble gas configuration: O ([He]2s22p4) + 2e–  O2– ([He]2s22p6) Three O atoms gain the six electrons lost by two Ga atoms. The ionic compound formed is Ga2O3.

3 O

+ 2

2 Ga

2 + 3

c) Magnesium loses 2 electrons to achieve a noble gas configuration: Mg ([Ne]3s2)  Mg2+ ([Ne]) + 2e– Nitrogen gains 3 electrons to achieve a noble gas configuration: N ([He]2s22p3) + 3e–  N3– ([He]2s22p6) Two N atoms gain the six electrons lost by three Mg atoms. The ionic compound formed is 3 2+ 3 Mg N 3 Mg + 2 + 2 N

Mg3N2.

d) Lithium loses 1 electron to achieve a noble gas configuration: Li ([He]2s1)  Li+ ([He]) + 1e– Bromine gains 1 electron to achieve a noble gas configuration: Br ([Ar]3d104s24p5) + 1e–  Br– ([Ar]3d104s24p6) One Br atoms gains the one electron lost by one Li atom. The ionic compound formed is LiBr. + Li Br Li Br + + 8.22

Plan: Find the charge of the known atom and use that charge to find the ionic charge of element X. For cations of groups 1 and 2, ion charge = the group number; for cations of groups 13 and higher, ion charge = the group number – 10. For anions, ion charge = the group number – 18. 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 2 metals form +2 ions. b) X in MgX is an anion with – 2 charge since Mg2+ is the cation. Elements in Group 16 form –2 ions (16 – 18 = –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 1.

8.23

a) 1

8.24

Plan: Find the charge of the known atom and use that charge to find the ionic charge of element X. For cations of groups 1 and 2, ion charge = the group number; for cations of groups 13 and higher, ion charge = the group number – 10. For anions, ion charge = the group number – 18. 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 13 form +3 ions. b) The carbonate ion, CO32–, has a –2 charge, so X has a +2 charge. Group 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 16 elements gain 2 electrons to form –2 ions with a noble gas configuration.

8.25

a) 17

8.26

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) 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.

8.27

a) CaO; O has a smaller radius than S. Therefore, CaO has a greater lattice energy than CaS.

b) 13

b) 16

c) 2

c) 13

b) SrO; Sr has a smaller radius than Ba. Therefore, SrO has a greater lattice energy than BaO. 8.28

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: a) BaS has the lower lattice energy because the ionic radius of Ba2+ is larger than Ca2+. A larger ionic radius results in a greater distance between ions. The lattice energy decreases with increasing distance between ions. b) NaF has the lower lattice energy since the charge on each ion (+1, –1) is half the charge on the Mg2+ and O2– ions. Lattice energy increases with increasing ion charge.

8.29

a) NaCl; Cl has a larger radius than F. b) K2S; S has a larger radius than O.

8.30

Plan: The lattice energy of NaCl is represented by the equation NaCl(s) → Na+(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 ∆Hº); you will also need to multiply the second equation (∆Hº) by ½. Solution: ∆Hº Na(s) → Na(g) 109 kJ/mol 1/2Cl2(g) → Cl(g) (1/2)(243 kJ/mol) = 121.5 kJ/mol Na(g) → Na+(g) + e– 496 kJ/mol Cl(g) + e– → Cl–(g) –349 kJ/mol NaCl(s) → Na(s) + 1/2Cl2 (g) 411 kJ/mol (Reaction is reversed; sign of ∆Hº changed.) NaCl(s) → Na+(g) + Cl–(g) 788.5 kJ/mol = 788 kJ/mol 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.

8.31

Lattice energy: MgF2(s) → Mg2+(g) + 2 F–(g) Use Hess‘s law: ∆Hº Mg(s) → Mg(g) 148 kJ/mol F2(g) → 2F(g) 159 kJ/mol Mg(g) → Mg+(g) + e– 738 kJ/mol Mg+(g) → Mg2+(g) + e– 1450 kJ/mol 2F(g) + 2e– → 2F–(g) 2(–328 kJ/mol) = – 656 kJ/mol MgF2(s) → Mg(s) + F2(g) 1123 kJ/mol (Reaction is reversed and the sign of ∆Hº changed.) MgF2(s) → Mg2+(g) + 2F–(g) 2962 kJ /mol 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.

8.32

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 predicted to be high for Al2O3 since the ions involved, Al3+ and O2–, have fairly large charges and are relatively small ions.

8.33

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 ∆Hº); you will also need to multiply the third equation (∆Hº) by ½. Solve for EA. Solution: An analogous Born-Haber cycle has been described in Figure 8.7 for LiF. Use Hess‘s law and solve for the EA of fluorine:  Hº

K(s)  K(g) 90 kJ/mol K(g)  K+(g) + e– 419 kJ/mol 1/2F2(g)  F(g) (1/2)(159 kJ/mol) = 79.5 kJ/mol F(g) + e–  F–(g) ? = EA KF(s)  K(s) + 1/2F2(g) 569 kJ/mol (Reverse the reaction and change the sign of Hº.) KF(s)  K+(g) + F–(g)

821 kJ/mol (Reverse the reaction and change the sign of Hº.)

821 kJ/mol = 90 kJ/mol + 419 kJ/mol + 79.5 kJ/mol + EA + 569 kJ/mol EA = 821 kJ/mol – (90 + 419 + 79.5 + 569) kJ/mol EA = –336.5 kJ/mol= –336 kJ/mol 8.34

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.

8.35

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  bond breaking H is positive. The same amount of energy needed to break the bond is released upon its formation, so  bond forming H has the same magnitude as  bond breaking H , but opposite in sign (always exothermic and negative).

8.36

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.

8.37

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.

8.38

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.

8.39

Plan: Bond strength increases as the atomic radii of atoms in the bond decrease; bond strength also increases as bond order increases. Solution: 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 shared between atoms in the bond increases. The triple bond is the strongest and the single bond is the weakest.

8.40

a) H–F < H–Cl < H–I c) N–H < N–O < N–S

8.41

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.

b) C=O < C–O < C–S

8.42

H2(g) + O2(g)  H–O–O–H(g)  r H = bonds broken H − bonds formed H r H

(

)

8.43

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.

8.44

Bond energies are average values for a particular bond in a variety of compounds. Heats of formation are specific for a compound.

8.45

Plan: Write the combustion reactions of methane and of methanal. 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 methanal. Since C–H bonds take less energy to break than C=O bonds, more energy is released in the combustion of methane than of methanal.

8.46

Methanol has the greater heat of reaction per mole since 3 C–H bonds must be broken in methanol while 5 C–H bonds are broken in the combustion of ethanol. Less energy is required to break the bonds in the reactant molecules in the combustion of methanol.

8.47

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 8.2. Solution: Reactant bonds broken: 1 x C=C =1(614 kJ/mol) = 614 kJ/mol 4 x C–H =4(413 kJ/mol) = 1652 kJ/mol 1 x Cl–Cl = (243 kJ/mol) = 243 kJ/mol

bonds broken H = 2509 kJ/mol Product bonds formed: 1 x C–C = (–347 kJ/mol) = –347 kJ/mol 4 x C–H = 4(–413 kJ/mol) = –1652 kJ/mol 2 x C–Cl = 2(–339 kJ/mol = –678 kJ/mol

bonds formed H = –2677 kJ/mol  r H = bonds broken H + bonds formed H = 2509 kJ/mol + (–2677 kJ/mol) = –168 kJ/mol

8.48

CO2 + 2NH3  (H2N)2CO + H2O



 r H = bonds broken H + bonds formed H  r H = [(2BEC=O + 6BEN–H] + [4(BEN–H) + (BEC=O) + 2(BEC–N) + 2(BEO–H)] = [2(799 kJ/mol) + 6(391 kJ/mol)] + = [4(–391 kJ/mol) + (–745 kJ/mol) + 2(–305 kJ/mol) + 2(–467 kJ/mol)] = 3944 kJ/mol + (– 3853 kJ/mol) = 91 kJ /mol

8.49

H Reactant bonds broken: 1 x C–O = (358 kJ/mol) = 358 kJ/mol 3 x C–H = 3(413 kJ/mol) = 1239 kJ/mol 1 x O–H = (467 kJ/mol) = 467 kJ/mol 1 x C  O = (1070 kJ/mol) = 1070 kJ/mol

bonds broken H = 3134 kJ/mol Product bonds formed: 3 x C–H = 3(–413 kJ/mol) = –1239 kJ/mol 1 x C–C = (–347 kJ/mol) = –347 kJ/mol 1 x C=O = (–745 kJ/mol) = –745 kJ/mol 1 x C–O = )(–358 kJ/mol) = –358 kJ/mol 1 x O–H = (–467 kJ/mol) = –467 kJ/mol

bonds formed H = –3156 kJ/mol 

 r H = bonds broken H + bonds formed H = 3134 kJ/mol + (–3156 kJ/mol) = –22 kJ/mol

8.50

H H Reactant bonds broken: 1 x C=C = (614 kJ/mol) = 614 kJ/mol 4 x C–H = 4(413 kJ/mol) = 1652 kJ/mol 1 x H–Br = (363 kJ/mol) = 363 kJ/mol

bonds broken H = 2629 kJ/mol

Product bonds formed: 5 x C–H = 5(–413 kJ/mol) = –2065 kJ/mol 1 x C–C = (–347 kJ/mol) = –347 kJ/mol 1 x C–Br = (–276 kJ/mol) = –276 kJ/mol

bonds formed H = –2688 kJ/mol  r H = bonds broken H + bonds formed H = 2629 kJ/mol + (–2688 kJ/mol) = –59 kJ/mol

8.51

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.

8.52

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.

8.53

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. 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.

8.54

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.

8.55

The difference in χ 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 χ difference, the more covalent the bond and the lower the partial ionic character.

8.56

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.

8.57

a) I < Br < N b) Ca < H < F

8.58

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 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.

8.59

a) Cl > Br > P b) F > O > I

8.60

Plan: The polar arrow points toward the less 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:

8.61

The more electronegative element is partially negative (δ–) and the less electronegative element is partially positive (δ+). δ+ δ– δ– δ+ δ+ δ– a) Br Cl b) F Cl c) H O

δ– Se

δ+ H

δ+ e) As

δ– H

δ+ f) S

δ– N

8.62

Plan: The more polar bond will have a greater difference in electronegativity, χ. Solution: a) N−B N: χ = 3.0; B: χ = 2.0; Δaχ = 3.0 – 2.0 = 1.0 N−O N: χ = 3.0; O: χ = 3.5; Δbχ = 3.5 – 3.0 = 0.5 N−B has greater bond polarity. b) C−S C: χ = 2.5; S: χ = 2.5; Δcχ = 2.5 – 2.5 = 0 S−O S: χ = 2.5; O: χ = 3.5; Δdχ= 3.5 – 2.5 = 1.0 S−O has greater bond polarity. c) N−H N: χ = 3.0; H: χ = 2.1; Δeχ = 3.0 – 2.1 =0.9 Cl−O Cl: χ = 3.0; O: χ = 3.5; Δfχ = 3.5 – 3.0 = 0.5 N−H has greater bond polarity.

8.63

b) F–Cl is more polar; χ is 1.0 for F–Cl and 0.2 for Br–Cl c) H–O is more polar; χ is 1.4 for H–O and 0.3 for Se–H f) S–N is more polar; χ is 0.5 for S–N and 0.1 for As–H

8.64

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 χ, 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. χ = 0. b) Bonds in RbCl are ionic because Rb is a metal and Cl is a nonmetal. χ 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 units 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 χ = 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

8.65

a) KCl ionic d) SO2 polar covalent NO2 < SO2 < BF3

8.66

Plan: Increasing ionic character occurs with increasing χ. 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 less electronegative atom. Solution: a) H: χ = 2.1; Cl: χ = 3.0; Br: χ = 2.8; I: χ = 2.5 ΔHBrχ = 2.8 – 2.1 = 0.7; ΔHClχ = 3.0 – 2.1 = 0.9; ΔHIχ = 2.5 – 2.1 = 0.4 b) H: χ = 2.1; O: χ = 3.5; C: χ = 2.5; F: χ = 4.0 ΔHOχ = 3.5 – 2.1 = 1.4; ΔCHχ = 2.5 – 2.1 = 0.4; ΔHFχ = 4.0 – 2.1 = 1.9 c) Cl: χ = 3.0; S: χ = 2.5; P: χ = 2.1; Si: χ = 1.8 ΔSClχ = 3.0 – 2.5 = 0.5; ΔPClχ = 3.0 – 2.1 = 0.9; ΔSiClχ = 3.0 – 1.8 = 1.2

8.67

Increasing ionic character occurs with increasing χ. a) ΔPClχ = 0.9, ΔPBrχ = 0.7, ΔPFχ = 1.9 P–F > P–Cl > P–Br + – + – + – b) ΔBFχ = 2.0, ΔNFχ = 1.0, ΔCFχ = 1.5 B–F > C–F > N–F + – + – + – c) ΔSeFχ = 1.6, ΔTeFχ = 1.9, ΔBrFχ  = 1.2 Te–F > Se–F > Br–F + – + – + –

8.68

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.

8.69

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.

8.70

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 NO 2: O

b) P4 e) Br2

nonpolar covalent nonpolar covalent

c) BF3 polar covalent f) NO2 polar covalent

The average of all of these structures gives equivalent N–O bonds with a bond length that is between N–O and N=O.

8.71

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.

8.72

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.

8.73

Plan: Count the valence electrons and draw Lewis structures. Solution: Total valence electrons: SiF4: [1 x Si(4e–] + [4 x F(7e–)] = 32; SeCl2: [1 x Se(6e–)] + [2 x Cl(7e–)] = 20; COF2: [1 x C(4e–)] + [1 x O(6e–)] + [2 x 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

F F

C O

8.74

Plan: Count the valence electrons and draw Lewis structures. Solution: 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 PH 4+, 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)

H H

(c)

F C

H 8.75

Plan: Count the valence electrons and draw Lewis structures. Solution: a) PF3: [1 x P(5 e–)] + [3 x F(7e–)] = 26 valence electrons. P is the central atom. Draw single bonds from P to the three F atoms, using 2e– x 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 x H(1e–)] + [1 x C(4e–)] + 3 x 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– x 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 x C(4e–)] + [2 x S(6e–)] = 16 valence electrons. C is the central atom. Draw single bonds from C to the two S atoms, using 2e– x 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–)

O H

c) CS2 (16 valence e–)

S 8.76

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 b) S2Cl2 c) CHCl3 H H H

Cl Cl

H Cl

8.77

Plan: The problem asks for resonance structures, so there must be more than one answer for each part. Solution: a) NO2+ has [1 x N(5e–)] + [2 x O(6e–)] – 1e– (+ charge) = 16 valence electrons. Draw a single bond from N to each O, using 2e– x 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.

b) NO2F has [1 x N(5e–)] + [2 x O(6e–)] + [1 x F(7e–)] = 24 valence electrons. Draw a single bond from N to each surrounding atom, using 2e– x 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:

8.78

N O

8.79

Plan: Count the valence electrons and draw Lewis structures. Additional structures are needed to show resonance. Solution: a) N3– has [3 x N(5e–)] + [1 e–(from charge)] = 16 valence electrons. Place a single bond between the nitrogen atoms. This uses 2e– x 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:

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; (3) the other end N atom could form the triple bond instead. N

b) NO2– has [1 x N(5e–)] + [2 x 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 – x 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:

8.80

a) HCO2– has 18 valence electrons.

 H

b) HBrO4 has 32 valence electrons.

8.81

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 + ½ no. of shared valence electrons]. Solution: a) IF5 has [1 x I(7e–)] + [5 x 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 ten electrons; thus, this is an exception to the octet rule. The five I– F bonds use 2e– x 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 fifteen of the sixteen pairs, so there is one pair left for the central I. This gives:

F F

F I

Calculating formal charges: FC = no. of valence electrons – [no. of unshared valence electrons + ½ no. of shared valence electrons]. For iodine: FCI = 7 – [2 + ½(10)] = 0 For each fluorine: FCF = 7 – [6 + ½(2)] = 0 Total formal charge = 0 = charge on the compound. b) AlH4– has [1 x Al(3e–)] + [4 x H(1e–)] + [1e– (from charge)] = 8 valence electrons. The four Al–H bonds use all the electrons and Al has an octet. H

H FC = no. of valence electrons – [no. of unshared valence electrons + ½ no. of shared valence electrons]. For aluminum: FCAl = 3 – [0 + ½(8)] = –1 For each hydrogen: FCH = 1 – [0 + ½(2)] = 0 8.82

a) OCS has sixteen valence electrons. S

FCS = 6 – [4 + ½(4)] = 0 FCC = 4 – [0 + ½(8)] = 0 FCO = 6 – [4 + ½(4)] = 0 b) NO (has eleven valence electrons); the odd number means there will be an exception to the octet rule. O

FCO = 6 – [4 + ½(4)] = 0 FCO = 6 – [3 + ½(4)] = +1 FCN = 5 – [3 + ½(4)] = 0 FCN = 5 – [4 + ½(4)] = –1 The first resonance structure has a more even distribution of formal charges. Therefore, the first resonance structure contributed more to the actual structure of NO. 8.83

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 + ½ no. of shared valence electrons]. Solution: a) CN–: [1 x C(4e–)] + [1 x N(5e–)] + [1 e– from charge] = 10 valence electrons. Place a single bond between the carbon and nitrogen atoms. This uses 2e– x 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 + ½ no. of shared valence electrons].

FCC = 4 – [2 + ½(6)] = –1; FCN = 5 – [2 + ½(6)] = 0 Check: The total formal charge equals the charge on the ion (–1). b) ClO–: [1 x Cl(7e–)] + [1 x O(6e–)] + [1e– from charge] = 14 valence electrons. Place a single bond between the chlorine and oxygen atoms. This uses 2e– x 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 + ½ no. of shared valence electrons]. FCCl = 7 – [6 + ½(2)] = 0 FCO = 6 – [6 + ½(2)] = –1 Check: The total formal charge equals the charge on the ion (–1). 8.84

a) BF4– has thirty-two valence electrons.

F F

F FCF = 7 – [6 + ½(2)] = 0 FCB = 3 – [0 + ½(8)] = –1 b) ClNO has eighteen valence electrons. Cl

FCCl = 7 – [6 + ½(2] = 0; 8.85

FCN = 5 – [2 + ½(6)] = 0;

FCO = 6 – [4 + ½(4)] = 0

Plan: The general procedure is similar to the preceding problems, plus the oxidation number determination. Solution: a) BrO3– has [1 x Br(7e–)] + 3 x O(6e–)] + [1e– (from charge)] = 26 valence electrons. Placing the O atoms around the central Br and forming three Br–O bonds uses 2e– x 3 bonds = 6 electrons and leaves 26 – 6 = 20 electrons (ten pairs). Placing three pairs on each O (3 x 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 + ½(6)] = +2 FCO = 6 – [6 + ½(2)] = –1 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 + ½(8)] = +1 FCO = 6 – [6 + ½(2)] = –1 FCO = 6 – [4 + ½(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 + ½(10)] = 0 FCO = 6 – [6 + ½(2)] = –1 FCO = 6 – [4 + ½(4)] = 0 (Double bonded O atoms) Structure III has the most reasonable distribution of formal charges.

Oxidation number = (No. of valence e-) – (no. of shared e- + no. of unshared e-)

(Note: when there are electrons in a bond, they are transferred completely to the more electronegative atom) The oxidation numbers are: oxidation number of Br = +5 and oxidation number of O = –2. +5 –2 Check: The total formal charge equals the charge on the ion (–1). BrO3– 2– – – – b) SO3 has [1 x S(6e )] + [3 x O(6e )] + [2e (from charge)] = 26 valence electrons. Placing the O atoms around the central S and forming three S–O bonds uses 2e– x 3 bonds = 6 electrons and leaves 26 – 6 = 20 electrons (ten pairs). Placing three pairs on each O (3 x 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 + ½(6)] = +1; FCO = 6 – [6 + ½(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 + ½(8)] = 0 FCO = 6 – [6 + ½(2)] = –1 FCO = 6 – [4 + ½(4)] = 0 (Double bonded O) 2 2 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). 8.86

+4 –2 SO32–

a) AsO43– has 32 valence electrons. See structure I. FCAs = 5 – [0 + ½(8)] = +1 FCO = 6 – [6 + ½(2)] = –1 Net formal charge (+1 – 4) = –3 The octet rule is followed by all atoms. 3 O O

For more reasonable formal charges, move a lone pair from an O to a bonded pair on As (structure below): 3 O

O FCAs = 5 – [0 + ½(10)] = 0 FCO(single bond) = 6 – [6 + ½(2)] = –1 FCO(double bond) = 6 – [4 + ½(4)] = 0 Net formal charge: (0 + 3(–1)) + 0 = –3 Improved formal charge distribution O.N.: O –2 each x 4 = –8 total; As +5 b) ClO2– has 20 valence electrons.

For structure I in which all atoms have an octet: FCCl = 7 – [4 + ½(4)] = +1 FC O = 6 – [6 + ½(2)] = –1 For more reasonable formal charges, see structures II and III:

Formal charges in structures II and III: FCCl = 7 – [4 + ½(6)] = 0 FCO(double bond) = 6 – [4 + ½(4)] = 0 O.N.: O –2 each x 2 = –4 total; Cl +3 8.87

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 x B(3e–)] + [3 x 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 x As(5e–)] +[4 x 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 x Se(6e–)] + 4 x 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 ten electrons.

B H

Cl (c)

(b)

a) PF6– has 48 valence electrons.

As F

(a)

8.88

FCO (single bond) = 6 – [6 – ½(2)] = –1

P has an expanded octet of 12 e–.

P F

F F

b) ClO3 has twenty-five 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 twenty-six 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 8.89

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 x Br(7e–)] + [3 x F(7e–)] = 28 valence electrons. Placing a single bond between Br and each F uses 2e– x 3 bonds = 6e–, leaving 28 – 6 = 22 electrons (eleven 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 x I(7e–)] + [2 x Cl(7e–)] + [1e– (from charge)] = 22 valence electrons. Placing a single bond between I and each Cl uses 2e– x 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 x Be(2e–)] + [2 x F(7e–)] = 16 valence electrons. Placing a single bond between Be and each of the F atoms uses 2e– x 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) 8.90

a) O3– has nineteen 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); hence, it is a radical. O

b) XeF2 has twenty-two valence electrons. F

Sb F

8.91

Xe has an expanded octet of 10e–.

c) SbF4– has thirty-four valence electrons.

Sb has an expanded octet of 10e–.

F F

Plan: Draw Lewis structures for the reactants and products. Solution:

Beryllium chloride has the formula BeCl2. BeCl2 has [1 x Be(2e–)] + [2 x 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 x Be(2e–)] + [4 x 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 (twelve pairs). These twelve pairs complete the octet of the Cl atoms (Be already has an octet). 2 Cl Cl Cl

8.92

Draw a Lewis structure. If the formal charges are not ideal, a second structure may be needed. BrO4– has thirty-two valence electrons.

O O

In the structure on the left, all atoms have octets. The formal charges are: FCBr = 7 – [0 + ½(8)] = +3 FCO = 6 – [6 + ½(2)] = –1 The structure on the right expands the valence shell of the Br to give more favorable formal charges. FCBr = 7 – [0 + ½(14)] = 0 FCO(single bonded) = 6 – [6 + ½(2)] = –1 FCO (double bonded) = 6 – [4 + ½(4)] = 0 8.93

Count the total valence electrons and draw a Lewis structure. AlF63– has forty-eight valence electrons. 3

F F

F Al

F F

8.94

Plan: Use the structures in the text to determine the formal charges. Formal charge (FC) = no. of valence electrons – [no. of unshared valence electrons + ½ no. of shared valence electrons]. Solution: Structure A: FCC = 4 – [0 + ½(8)] = 0; FCO = 6 – [4 + ½(4)] = 0; FCCl = 7 – [6 + ½(2)] = 0 Total FC = 0 Structure B: FCC = 4 – [0 + ½(8)] = 0; FCO = 6 – [6 + ½(2)] = –1; FCCl(double bonded) = 7 – [4 + ½(4)] = +1; FCCl(single bonded) = 7 – [6 + ½(2)] = 0 Total FC = 0 Structure C: FCC = 4 – [0 + ½(8)] = 0; FCO = 6 – [6 + ½(2)] = –1; FCCl(double bonded) = 7 – [4 + ½(4)] = +1; FCCl(single bonded) = 7 – [6 + ½(2)] = 0

Total FC = 0 Structure A has the most reasonable set of formal charges. 8.95

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. Metals in the gas phase show the two metal atoms can even share their valence electrons to form gaseous, diatomic molecules, such as Na 2.

8.96

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.

8.97

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 MgF 2 is deformed, the ions are displaced so that repulsive forces between neighboring ions of like charge cause shattering of the crystals.

8.98

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.

8.99

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 8.2. 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:  r H = –1259 kJ/mol a) C2H2 + 5/2O2  2CO2 + H2O H–CC–H + 5/2O=O  2O=C=O + H–O–H  r H = bonds broken H + bonds formed H  r H = [2BEC–H + BECC +5/2BEO=O] + [4(–BEC=O) + 2(–BEO–H)]

–1259 kJ/mol = [2(413 kJ/mol) + BECC + 5/2(498 kJ/mol)] + [4(–799 kJ/mol) + 2(–467 kJ/mol)] –1259 kJ/mol = [826 kJ/mol + BECC + 1245 kJ/mol] + [–4130kJ/mol)] –1259 kJ/mol = –2059 kJ/mol + BECC BECC = 800. kJ/mol Table 8.2 lists the value as 839 kJ/mol.  1 mol C2 H 2  1259 kJ  b) Heat (kJ) = q= 500.0 g C2 H 2     26.04 g C2 H 2  1 mol C2 H 2  = –2.4174347x104 kJ= –2.417x104 kJ  1 mol C2 H 2   2 mol CO 2   44.01 g CO 2  c) Mass (g) of CO2 = 500.0 g C2 H 2      26.04 g C2 H 2   1 mol C2 H 2   1 mol CO 2  = 1690.092 g= 1690. g CO2  1 mol C2 H 2   (5/2) mol O 2  d) Amount (mol) of O2 = 500.0 g C2 H 2     26.04 g C2 H 2   1 mol C2 H 2  = 48.0030722 mol O2 pV = nRT

 48.0030722 mol O2   8.3144  1823 kPa

nRT = p = 65.2423 L = 65.2 L O2

Volume (L) of O2 =

8.100

L • kPa   298 K  mol • K 

Br F

8.101

Al3+

Plan: The heat of formation of MgCl is represented by the equation Mg(s) + 1/2Cl2(g) → MgCl(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 ½; you will need to reverse the equation for the lattice energy [MgCl(s) → Mg+(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)

1H = 148 kJ/mol  2 H = 1/2(243 kJ/mol) = 121.5 kJ/mol

2) 1/2Cl2(g)  Cl(g) 3) Mg(g)  Mg (g) + e +

–

4) Cl(g) + e–  Cl–(g) –

5) Mg (g) + Cl (g)  MgCl(s) +

Mg(s) + 1/2Cl2(g)  MgCl(s)

3 H = 738 kJ/mol  4 H = –349 kJ/mol 5 H = –783.5 kJ/mol (= –  lattice H (MgCl))  f H (MgCl) = ?

 f H (MgCl) = 1H +  2 H + 3 H +  4 H + 5 H = 148 kJ/mol + 121.5 kJ/mol + 738 kJ/mol + (–349 kJ/mol) + (–783.5 kJ/mol) = –125 kJ/mol

b) Yes, since  f H for MgCl is negative, MgCl(s) is stable relative to its elements. c) 2MgCl(s)  MgCl2(s) + Mg(s)

r H = m f(products) H – n f(reactants) H r H = {1  f H [MgCl2(s)] + 1  f H [Mg(s)]} – {2  f H [MgCl(s)]} r H = [(–641.6 kJ/mol) + (0 kJ/mol)] – [2(–125 kJ/mol)] r H = –391.6 kJ/mol = –392 kJ/mol d) No,  f H for MgCl2 is much more negative than that for MgCl. This makes the r H value for the above reaction very negative, and the formation of MgCl2 would be favored. 8.102

a) r H = bonds broken H + bonds formed H = [(BEH-H) + (BECl-Cl)] + [2(BEH-Cl)] = [(432 kJ/mol) + (243 kJ/mol)] + [2(–427 kJ/mol)] = –179 kJ/mol b) r H = bonds broken H + bonds formed H = [(BEH-H) + (BEI-I)] + [2(BEH-I)] = [(432 kJ/mol) +(151 kJ/mol)] + [2(–295 kJ/mol)]

= –7 kJ /mol c) r H = bonds broken H + bonds formed H = [2(BEH-H) + (BEO=O)] + [4(BEH-O)] = [2(432 kJ/mol) +(498 kJ/mol)] + [4(–467 kJ/mol)] = –506 kJ /mol Reactions (a) and (c) are strongly exothermic and are a potential explosive hazard. Reaction (c) should occur most explosively. 8.103

Plan: Find the bond energy for an H–I bond from Table 8.2. For part a), calculate the wavelength with this energy using the relationship from Chapter 6: 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. For part c), speed can be calculated from the excess energy since Ek = 1/2mu2. Solution: a) Bond energy for H–I is 295 kJ/mol (Table 8.2). 3  1 mol  295 kJ   10 J   –19 Bond energy (J/photon) =      = 4.898705x10 J/photon  23  mol   1 kJ   6.022x10 photons  E = hc/

6.626x10 J•s 3.00x10 m/s  = 4.057807x10 m  4.898705x10 J   1 nm   (nm) =  4.057807x10 m    = 405.7807 nm = 406 nm  10 m  34

 (m) = hc/E =

–7

19

7

9

b) E (HI) = 4.898705x10

–19

 6.626x10 E (254 nm) = hc/ =

34





J•s 3.00x108 m/s  1 nm  –19  9  = 7.82598x10 J 254 nm  10 m  Excess energy = 7.82598x10–19 J – 4.898705x10–19 J = 2.92728x10–19 J = 2.93x10–19 J

mol  1.008 g H     1 kg  c) Mass (kg) of H =  = 1.67386x10–27 kg   23   3  mol 6.022x10 10 g     Ek = 1/2mu2 thus, u =

2E m

2(2.92728x1019 J)  kg•m 2 /s 2  4 4   = 1.8701965x10 m/s= 1.87x10 m/s J 1.67386x1027 kg  

8.104

―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). ΔPHχ = 0, ΔCSχ = 0, ΔBrClχ = 0.2, ΔBHχ = 0.1, ΔSeSeχ = 0

8.105

Rb ([Kr]5s1) has one valence electron, so the metallic bonding would be fairly weak, resulting in a soft, lowmelting material. Cd ([Kr]5s24d10) has two valence electrons so the metallic bonding is stronger. V ([Ar]4s23d3) has five valence electrons, so its metallic bonding is the strongest, that is, its hardness, melting point, and other metallic properties would be greatest.

8.106

Plan: Find the appropriate bond energies in Table 8.2. Calculate the wavelengths using E = hc/. Solution: C–Cl bond energy = 339 kJ/mol 3  1 mol  339 kJ   10 J   –19 Bond energy (J/photon) =      = 5.62936x10 J/photon  23   mol   1 kJ   6.022x10 photons  E = hc/

6.626x10 J•s 3.00x10 m/s  = 3.5311296x10 m = 3.53x10 m 5.62936x10 J  34

 (m) = hc/E = 

–7

19

O2 bond energy = 498 kJ/mol 3  1 mol  498 kJ   10 J   –19 Bond energy (J/photon) =      = 8.269678x10 J/photon  23   mol   1 kJ   6.022x10 photons 

E = hc/

6.626x10 J•s 3.00x10 m/s  = 2.40372x10 m = 2.40x10 m  (m) = hc/E = 8.269678x10 J  34

–7

19

8.107

Plan: Write balanced chemical equations for the formation of each of the compounds. Obtain the bond energy of fluorine from Table 8.2 (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:

r H = bonds broken H + bonds formed H XeF2 Xe(g) + F2(g)  XeF2(g)

XeF4

r H = –105 kJ/mol = [(159 kJ/mol)] + [2(–Xe–F)] –264 kJ/mol = 2(–Xe–F) Xe–F = 132 kJ/mol Xe(g) + 2F2(g)  XeF4(g) r H = –284 kJ/mol = [2(159 kJ/mol)] + [4(–Xe–F)]

XeF6

–602 kJ/mol = 4(–Xe–F) Xe–F = 150.5 kJ/mol = 150. kJ/mol Xe(g) + 3F2(g)  XeF6(g)

r H = –402 kJ/mol = [3(159 kJ/mol)] + [6(–Xe–F)] –879 kJ/mol = 6(–Xe–F) Xe–F = 146.5 kJ/mol = 146 kJ/mol 8.108

The difference in electronegativity produces a greater than expected overlap of orbitals, which shortens the bond. As χ becomes smaller (i.e., as you proceed from HF to HI), this effect lessens and the bond lengths become more predictable.

8.109

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 O 2 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 kJ/mol = 1420 kJ/mol It would require 1420 kJ to heterolytically cleave 1 mol of O 2.

8.110

The bond energies are needed from Table 8.2. N2 = 945 kJ/mol; O2 = 498 kJ/mol; F2 = 159 kJ/mol N 2:



O 2: 



 6.626x10  hc/E =

34



J•s 3.00x108 m/s

 = 1.26672x10 m = 1.27x10 m –7

–7

kJ   10 J   mol    945 mol   1 kJ   23      6.022x10  3

 6.626x10  = hc/E =

34



J•s 3.00x108 m/s

 = 2.40372x10 m = 2.40x10 m –7

–7

kJ   10 J   mol    498 mol   1 kJ   23  6.022x10     

F2:

 = hc/E =

 6.626x10

34



J•s 3.00x108 m/s

 = 7.528636x10 7 m = 7.53x10 m –

–7



8.111

a) To compare the two energies, the ionization energy must be converted to the energy to remove an electron from an atom. The energy needed to remove an electron from a single gaseous Ag atom (J) = 3 mol  731 kJ   10 J    = 1.21388x10–18 J = 1.21x10–18 J > 7.59x10–19 J  mol   1 kJ   23      6.022x10  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).

8.112

Plan: The heat of formation of SiO2 is represented by the equation Si(s) + O2(g) → SiO2(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) → Si4+(g) + 2O2–(g). You will need to reverse the lattice energy equation (and change the sign of ∆Hº); you will also need to multiply the fourth given equation by 2.

 kJ   10 J   mol   159 mol   1 kJ   23      6.022x10  3

Solution: Use Hess‘ law. H fo of SiO2 is found in Appendix B. 1)

Si(s)  Si(g)

Si(g)  Si (g) + 4 e

O2(g)  2O(g)

1H = 454 kJ/mol

–

 2 H = 9949 kJ/mol 3 H = 498 kJ/mol

2O(g) + 4e  2O (g)

 4 H = 2(737) kJ/mol

Si4+(g) + 2O2– (g)  SiO2(s)

5 H = –  lattice H (SiO2) = ?

Si(s) + O2(g)  SiO2(s)

 f H o (SiO2) = –910.9 kJ/mol

2–

 f H o = [ 1H +  2 H + 3 H +  4 H + (–  lattice H )] –910.9 kJ/mol = [454 kJ/mol + 9949 kJ/mol + 498 kJ/mol + 2(737) kJ/mol + (–  lattice H )] –  lattice H = –13,285.9 kJ/mol  lattice H = 13,286 kJ/mol

8.113

r H = bonds broken H + bonds formed H For ethane: r H = [(BEC – C) + 6(BEC – H) + (BEH – H)] + [8(BEC – H)] –65.07 kJ/mol = [(347 kJ/mol) + 6(BEC – H) + (432 kJ/mol)] + [8(–415 kJ/mol)]  65.07  347  432  3320 kJ / mol BEC – H = = 412.655 kJ/mol = 413 kJ/mol 6 For ethene: r H = [ (BEC =C) + 4 (BEC – H) + 2 (BEH – H)] + [8 (BEC – H)] –202.21 kJ/mol = [(614 kJ/mol) + 4 (BE C – H) + 2(432 kJ/mol)] + [8 (–415 kJ/mol)]  202.21  614  864  3320 kJ / mol  BEC – H = = 409.9475 kJ/mol = 410. kJ/mol 4 For ethyne: r H = [(BECC) + 2 (BEC – H) + 3 (BEH – H)] + [8 (BEC – H)] –376.74 kJ/mol = [(839 kJ/mol) + 2(BE C – H) + 3 (432 kJ/mol)] + [8(– 415kJ/mol)]  376.74  839  1296  3320 kJ / mol BEC – H = = 404.13 kJ/mol = 404 kJ/mol 2

8.114

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: 3  1 mol  347 kJ   10 J   –19 Bond energy (J/photon) =      = 5.762205x10 J/photon  23  mol   1 kJ   6.022x10 photons  E E = h or  = h

=

5.762205x1019 J E = = 8.6963553x1014 s-1= 8.70x1014 s–1 h 6.626 x1034 J•s

E = hc/or  = hc/E

 (m) = hc/E =

 6.626x10

34



J•s 3.00 x108 m/s 19

 = 3.44972x10 m = 3.45x10 m –7

–7

5.762205x10 J This is in the ultraviolet region of the electromagnetic spectrum.

8.115

=

E = h

 kJ   103 J   mol   467 mol   1 kJ   23      6.022 x10  6.626 x 10

 6.626x10  (m) = hc/E =

34

J•s

= 1.170374x1015 s-1= 1.17x1015 s–1



J•s 3.00x108 m/s

 = 2.56328x10 m = 2.56x10 m –7

–7

kJ   10 J   mol    467 mol   1 kJ   23      6.022x10  3

 kJ   1 mol  –22 –22 Ephoton =  467   = 7.7548987x10 kJ/photon = 7.75x10 kJ/photon  23 mol   6.022x10 photons   8.116

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)

r H = bonds broken H + bonds formed H r H = [8 x (BEC–H) + 1 x (BEO=O)] + [6 x (BEC–H) + 2 x (BEC–O) + 2 x (BEO–H)] r H = [8 (413 kJ/mol) + (498 kJ/mol)] + [6(–413 kJ/mol) + 2(–358 kJ/mol) + 2 (–467 kJ/mol)] r H = –326 kJ/mol 2CH4(g) + O2(g)  CH3CH2OH(g) + H2O(g) r H = bonds broken H + bonds formed H r H = [8 x (BEC–H) + 1 x (BEO=O)] + [5 x ( BEC–H) + 1 x (BEC–C) + 1 x (BEC–O) + 3 x ( BEO–H)] r H = [8(413 kJ/mol) + (498 kJ/mol)] + [5(–413 kJ/mol) + (–347 kJ/mol) + (–358 kJ/mol) + 3(–467 kJ/mol)] r H = –369 kJ /mol 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)

r H = –(–369 kJ) = 369 kJ/mol

2CH4(g) + O2(g)  CH3OCH3(g) + H2O(g)

r H = –326 kJ/mol

CH3CH2OH(g)  CH3OCH3(g)

r H = –326 kJ/mol + 369 kJ/mol = 43 kJ/mol

8.117

a) CH2=CH2(g) + H2O(g)  CH3CH2OH(g) Using bond energies: r H = bonds broken H + bonds formed H

r H = [4 x (BEC–H) + 1 x (BEC=C) + 2 x (BEO–H)] + [5 x (BEC–H) + 1 x (BEC–C) + 1 x (BEC–O) + 1 x (BEO–H)] r H = [4(413 kJ/mol) + (614 kJ/mol) + 2(467 kJ/mol)] + [5(–413 kJ/mol) + (–347 kJ/mol) + (–358 kJ/mol) + (–467 kJ/mol)] r H = –37 kJ/mol Using heats of formation: r H = m  f (products) H – n  f (reactants) H

r H = [1( H f of CH3CH2OH(g))] – [( H f of CH2=CH2(g)) + ( H f of H2O (g))]

r H = [–235.1 kJ/mol] – [52.47 kJ/mol + –241.826 kJ/mol] r H = –45.744 kJ/mol = –45.7 kJ /mol b) C2H4O(l) + H2O(l)  C2H6O2(l) r H = bonds broken H + bonds formed H r H = [4 x (BEC–H) + 1 x (BEC-C) + 2 x (BEC–O) + 2 x (BEO–H)] + [4 x (BEC–H) + 1 x (BEC–C) + 2 x (BEC–O) + 2 x (BEO–H)] r H = [4 (413 kJ/mol) + (347 kJ/mol) + 2(358 kJ/mol) + 2(467 kJ/mol)] + [4(–413 kJ/mol) +(–347 kJ/mol) + 2(–358 kJ/mol) + 2(–467 kJ/mol)] r H = 0 kJ/mol c) In the hydrolysis in part b), the r H appears to be 0 kJ/mol 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. 8.118

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 8.2. Solution: a) The H atoms cannot be central, and they are evenly distributed on the N atoms. N2H4 has [2 x N(5e–)] + [4 x H(1e–)] = fourteen valence electrons, ten 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 x N(5e–)] + (2 x H(1e–)] = twelve 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 x N(5 e–)] = ten 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.

Hydrazine

Diazene

Nitrogen

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 x N(5e–)] + [4 x H(1e–)] = twenty-four valence electrons, fourteen 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 +

Reactant bonds broken: 4 N–H = 4 (391 kJ/mol) = 1564 kJ /mol 2 N–N = 2 (160 kJ/mol) = 320 kJ/mol 1 N=N = (418 kJ/mol) = 418 kJ/mol

Product bonds formed: 4 N–H = 4 (–391 kJ/mol) = –1564 kJ/mol 1 N–N = (–160 kJ/mol) = –160 kJ/mol 1 N≡N = (–945 kJ/mol) = –945 kJ/mol

bonds broken H = 2302 kJ/mol

bonds formed H = –2669 kJ/mol

r H   bonds broken H  bonds formed H = 2302 kJ/mol + (–2669 kJ/mol) = –367 kJ/mol 8.119 F F

F P

(a)

(b)

(c)

Cl (d)

H (e)

(f)

8.120

Bond order (avg.) C

3.0

b)  O

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) 2.0 H C H O

1.0

H H

H e) O

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 remains1.0 Bond length a < c < e < b < d ignoring O attached to H in part e) Bond strength d<b<e<c<a 8.121

The reaction is balanced as usual: ClCH2CH2SCH2CH2Cl + 2H2O  HOCH2CH2SCH2CH2OH + 2HCl Most of the molecule remains the same. Reactant bonds broken: 2 x C–C = (2)(347 kJ/mol) = 694 kJ/mol 8 x C–H = (8)(413 kJ/mol) = 3304 kJ/mol 2 x C–Cl = (2)(339 kJ/mol) = 678 kJ/mol 2 x C–S = (2 )(259 kJ/mol) = 518 kJ/mol 4 x O–H = (4 )(467 kJ/mol) = 1868 kJ/mol

bonds broken H = 7062 kJ/mol Product bonds formed: 2 x C–C = (2)(–347 kJ/mol) = –694 kJ/mol 8 x C–H = (8 )(–413 kJ/mol) = –3304 kJ/mol 2 x H–Cl = (2 )(–427 kJ/mol) = –854 kJ/mol 2 x C–S = (2 )(–259 kJ/mol) = –518 kJ/mol 2 x C–O = (2 )(–358 kJ/mol) = –716 kJ/mol 2 x O–H = (2 )(–467 kJ/mol) = –934 kJ/mol

bonds formed H = –7020 kJ/mol

r H   bonds broken H  bonds formed H = 7062 kJ/mol + (–7020 kJ/mol) = 42 kJ/mol 8.122

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 8.2. 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). Solution: a) CH3CH2OH(g) + 3O2(g) → 2CO2(g) + 3H2O(g)

+ 3

2 O

+ 3

Reactant bonds broken: 1 x C–C = (347 kJ/mol) = 347 kJ/mol 5 x C–H = (5)(413 kJ/mol) = 2065 kJ/mol 1 x C–O = (358 kJ/mol) = 358 kJ/mol 1 x O–H = (467 kJ/mol) = 467 kJ/mol 3 x O=O = (3)(498 kJ/mol) = 1494 kJ/mol

bonds broken H = 4731 kJ/mol Product bonds formed: 4 x C=O = (4)(–799 kJ/mol) = –3196 kJ/mol 6 x O–H = (6)(–467 kJ/mol) = –2802 kJ/mol

bonds formed H = –5998 kJ/mol r H   bonds broken H  bonds formed H = 4731 kJ/mol + (–5998 kJ/mol) = –1267 kJ/mol (for each mole of ethanol burned). b) If it takes 40.5 kJ/mol to vapourize 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   = –1226.5 kJ/mol = –1226 kJ/mol (of liquid ethanol  1 mol 

combustion(liquid) H = –1267 kJ/mol + (1) 

burned) c)  r H   m f(products) H  n f(reactants) H

r H  = {2  f H  [CO2(g)] + 3  f H  [H2O(g)]} – {1  f H  [C2H5OH(l)] + 3  f H  [O2(g)]} = [(2)(–393.5 kJ/mol) + (3)(–241.826 kJ/mol)] – [(–277.63 kJ/mol) + 3 (0 kJ/mol)] = –1234.848 kJ/mol= –1234.8 kJ/mol The two answers differ by less than 10 kJ/mol. 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) → CH3CH2OH(g) The Lewis structures for the reaction are: H

H C

C H

O H

Reactant bonds broken: 1 x C=C = )(614 kJ/mol) = 614 kJ/mol 4 x C–H = (4 )(413 kJ/mol) = 1652 kJ/mol 2 x O–H = (2 )(467 kJ/mol) = 934 kJ/mol

bonds broken H = 3200 kJ/mol Product bonds formed: 1 x C–C = (–347 kJ/mol) = –347 kJ/mol 5 x C–H = (5 )(–413 kJ/mol) = –2065 kJ/mol 1 x C–O = (–358 kJ/mol) = –358 kJ/mol 1 x O–H = (–467 kJ/mol) = –467 kJ/mol

bonds formed H = –3237 kJ/mol r H   bonds broken H  bonds formed H 8.123

= 3200 kJ/mol + (–3237 kJ/mol) = –37 kJ/mol

a) C3H4: 16 valence electrons.

b) C3H6: 18 valence electrons. H H

C H

Bond Order 1 (all) H

H H c) C4H6: 22 valence electrons. Resonance exists

d) C4H4: 20 valence electrons. Resonance exists. H H H

H C

C Bond Order 1.5 (all)

H H H e) C6H6: 30 valence electrons. Resonance exists.

C H

H H

Bond Order 1.5 (all) H

H 8.124

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 amount of acidic H atoms in the formula of the acid. Finally, combine this information to construct the Lewis structure. Solution:  1 mol  Moles of H = 2.24 g H   = 2.222 mol H  1.008 g H 

 1 mol  Moles of C = 26.7 g C   = 2.223 mol C  12.01 g C   1 mol  Moles of O = 71.1 g O   = 4.444 mol 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. To determine the molecular formula, calculate the amount of NaOH required for the titration:  0.040 mol NaOH  1000 mmol   1 L  mmoles of NaOH =  50.0 mL      1000 mL  = 2.0 mmol NaOH L mol     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 x H(1e–)] + [2 x C(4e–)] + [4 x 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 ten pairs of electrons. When these ten 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 8.125

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.  1 mol  Moles of C = 24.8 g C   = 2.065 mol C  12.01 g C 

 1 mol  Moles of H = 2.08 g H   = 2.063 mol H  1.008 g H   1 mol  Moles of Cl = 73.1 g Cl   = 2.062 mol 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: pM d= RT Rearranging to solve for molar mass: 8.3144 L•kPa   4.3 g/L     273 K  dRT mol•K   = 97.6027 g/mol = 98 g/M M = = p 100 kPa  The molar mass (98 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 twenty-four valence electrons. A variety of structures may be drawn: H

Cl C

There are thirty-two 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 due to resonance.) O O

O O

FCCl = 7 – [0 + ½(8)] FCCl = +3

O O

FCCl = 7 – [0 + ½(10)] FCCl = +2

Average bond order of the last structure: 8.127

8.126

FCCl = 7 – [0 + ½(12)] FCCl = +1

FCCl = 7 – [0 + ½(14)] FCCl = 0 (most important)

(3 x 2)  (1 x 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 8.2. Divide the heat of reaction by the amount 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

+ 2

H Reactant bonds broken: 4 x C–H = (4)(413 kJ/mol) = 1652 kJ/mol 2 x O=O = (2)(498 kJ/mol) = 996 kJ/mol

H bonds broken = 2648 kJ/mol Product bonds formed: 2 x C=O = (2)(–799 kJ/mol) = –1598 kJ/mol 4 x O–H = (4)(–467 kJ/mol) = –1868 kJ/mol

H bonds formed = –3466 kJ/mol

r H   bonds broken H  bonds formed H = 2648 kJ/mol + (–3466 kJ/mol) = –818 kJ/mol Per mole of O2 = –818 kJ/mol /2 = –409 kJ/mol O2 2H2S(g) + 3O2(g)  2SO2(g) + 2H2O(g)

H Reactant bonds broken: 4 x S–H = (4)(347 kJ/mol) = 1388 kJ/mol 3 x O=O = (3)(498 kJ/mol) = 1494 kJ/mol

H bonds broken = 2882 kJ/mol Product bonds formed: 4 x S=O = (4 )(–552 kJ/mol) = –2208 kJ/mol 4 x O–H = (4 )(–467 kJ/mol) = –1868 kJ/mol

H bonds formed = –4076 kJ/mol

r H   bonds broken H  bonds formed H = 2882 kJ/mol + (–4076 kJ/mol) = –1194 kJ/per mole of O2 = –(1194 kJ/mol)/3 = –398 kJ/mol O2

8.128

F F

F (a)

(b)

(c)

F S

S F

(d) Stable: a, c, and e Unstable radicals: b and d 8.129

F F

(e)

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 8.2. Solution: a) The OH molecule has [1 x O(6e–)] + [1 x 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:

b) The formation reaction is: 1/2O2(g) + 1/2H2(g)  OH(g). The heat of reaction is:

r H   bonds broken H  bonds formed H = 39.0 kJ [½ (BEO=O) + ½ (BEH–H)] + [BEO–H] = 39.0 kJ/mol [(½ )(498 kJ/mol) + (½ )(432 kJ/mol)] + [BE O–H] = 39.0 kJ/mol 465 kJ/mol + [BEO–H] = 39.0 kJ/mol BEO–H = –426 kJ/mol or 426 kJ/mol 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 x 467 kJ/mol = 934 kJ/mol. The answer to part b) accounts for 426 kJ/mol of this, leaving: 934 kJ/mol – 426 kJ/mol = 508 kJ/mol 8.130

Both N3– and HN3 have sixteen valence electrons. Azide ion: N

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. 8.131

N2O has sixteen1 valence electrons; there are three resonance structures.

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. 8.132

Plan: Count the valence electrons and draw Lewis structures for the resonance forms. Solution: The H2C2O4 molecule has [2 x H(1e–)] + [2 x C(4e–)] + [4 x 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 electrons. If these twenty 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 x H(1e–)] + [2 x C(4e–)] + [4 x 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 twenty-two 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 x C(4e–)] + [4 x 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 twenty-four 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–:

C2O42–: O

2

O C

O C O

2

2 O

O C

2

C O

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. 8.133

The molecule has forty-two valence electrons. Thirty electrons are already accounted for in the skeleton structure in the bonds. 42 – 30 = 12 valence electrons remain. If these twelve 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)

8.134

N2H4(g) + O2(g)  N2(g) + 2H2O(g)

Reactant bonds broken: 4 x N–H = (4 )(391 kJ/mol) = 1564 kJ/mol 1 x N–N = (160 kJ/mol) = 160 kJ/mol 1 x O=O = (498 kJ/mol) = 498 kJ/mol

bonds broken H = 2222 kJ/mol

Product bonds formed: 1 x N≡N = (–945 kJ/mol) = –945 kJ/mol 4 x O–H = (4 )(–467 kJ/mol) = –1868 kJ/mol

bonds formed H

= –2813 kJ/mol

r H   bonds broken H  bonds formed H = 2222 kJ/mol + (–2813 kJ/mol) = –591 kJ/mol 8.135

Plan: Draw the Lewis structures. Calculate the heat of reaction using the bond energies in Table 8.2. Solution: SO3(g) + H2SO4(l)  H2S2O7(l) O S

O O

O H

O O

O H

Reactant bonds broken: 5 x S=O = (5)(552 kJ/mol) = 2760 kJ/mol 2 x S–O = (2)(265 kJ/mol) = 530 kJ/mol 2 x O–H = (2 )(467 kJ/mol) = 934 kJ/mol

bonds broken H = 4224 kJ/mol Product bonds formed: 4 x S=O = (4)(–552 kJ/mol) = –2208 kJ/mol 4 x S–O = (4 )(–265 kJ/mol) = –1060 kJ/mol 2 x O–H = (2 )(–467 kJ/mol) = –934 kJ/mol

bonds formed H = –4202 kJ/mol r H   bonds broken H  bonds formed H = 4224 kJ/mol + (–4202 kJ/mol) = 22 kJ/mol 8.136

The simplified Lewis structures for the reaction are: H H

N + 2H

Reactant bonds broken: 1 x C–H = (413 kJ/mol) = 413 kJ/mol 1 x CN = (891 kJ/mol) = 891 kJ/mol 2 x H–H = (2 )(432 kJ/mol) = 864 kJ/mol

bonds broken H = 2168 kJ/mol Product bonds formed: 3 x C–H = (3 )(–413 kJ/mol) = –1239 kJ/mol 1 x C–N = (–305 kJ/mol) = –305 kJ/mol 2 x N–H = (2 )(–391 kJ/mol) = –782 kJ/mol

bonds formed H = –2326 kJ/mol r H   bonds broken H  bonds formed H = 2168 kJ/mol + (–2326 kJ/mol) = –158 kJ/mol

CHAPTER 9 THE SHAPES OF MOLECULES CHEMICAL CONNECTIONS BOXED READING PROBLEMS B9.1

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 bond 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.

B9.2

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 9.1

Determine the total number of valence electrons present. Next, draw a Lewis structure. Finally, use VSEPR or valence bond theory to predict the shape.

9.2

The molecular shape and the electron-group arrangement are the same when there are no lone pairs on the central atom.

9.3

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 AX2E1 or AX2E2. Examples are: NO2– and H2O.

N 120°, AX2E1

109.5°, AX2E2

9.4

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 AX3E1 Bent or V shaped AX2E2

9.5

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.

9.6

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 (AX 2E1) or two (AX2E2) lone electron pairs. b) A trigonal planar molecule follows the formula AX 3 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 AX3E1 with three bonding pairs and one lone pair. f) A square pyramidal molecule shape follows the formula AX 5E1 with five bonding pairs and one lone pair.

9.7

Determine the geometry from the lone pairs and the number of groups attached to the central atom. a) AX3E1 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) AX4E1 trigonal bipyramidal 180°, 120°, 90° smaller

9.8

Plan: First, draw a Lewis structure, and then apply VSEPR. Solution: a) O3: The molecule has [3 x 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:

or 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 (AX2E1), a bent molecular shape, and an ideal bond angle of 120°. b) H3O+: This ion has [3 x H(1e–)] + [1 x 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:

There are four electron groups around the O, one of which is a lone pair. This gives a tetrahedral electron-group arrangement (AX3E1), a trigonal pyramidal molecular shape, and an ideal bond angle of 109.5°. c) NF3: The molecule has [1 x N(5e–)] + [3 x F(7e–)] = 26 valence electrons. Six electrons are used to place a single bond between N and each F, leaving 26 – 6 = 20 e– (ten pairs). These ten 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:

There are four electron groups around the N, one of which is a lone pair. This gives a tetrahedral electron-group arrangement (AX3E1), a trigonal pyramidal molecular shape, and an ideal bond angle of 109.5°. 9.9

Lewis structure (a)

Electron-group arrangement Tetrahedral

(b)

Trigonal planar

Tetrahedral

Ideal bond angle 109.5°

In addition, there are other resonance forms. Bent 120°

O O

In addition, there are other resonance forms. (c) Tetrahedral

9.10

Molecular shape

Trigonal pyramidal

109.5°

Plan: First, draw a Lewis structure, and then apply VSEPR. Solution: (a) CO32–: This ion has [1 x C(4e–)] + [3 x 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:

2

2 O

O C

O O 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 x S(6e–)] + [2 x O(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:

There are three groups of electrons around the S, one of which is a lone pair. This gives a trigonal planar electron-group arrangement (AX2E1), a bent (V-shaped) molecular shape, and an ideal bond angle of 120°. (c) CF4: This molecule has [1 x C(4e–)] + [4 x F(7e–)] = 32 valence electrons. Eight electrons are used to place a single bond between C and each F, leaving 32 – 8 = 24 e– (twelve pairs). Use these twelve pairs to complete the octets of the F atoms (C already has an octet). This gives the following Lewis structure:

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°. 9.11

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

9.12

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 AX3E1, 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°.

9.13

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 AX5E1, 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°.

9.14 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 x Cl(7e–)] + [2 x 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 x P(5 e–)] + [5 x 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– (fifteen pairs). The fifteen 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 x Se(6e–)] + [4 x F(7e–)] = 34 valence electrons. Eight electrons are used to place single bonds between Se and each F atom, leaving 34 – 8 = 26e– (thirteen 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 (AX 4E1). 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. d) The KrF2 molecule has [1 x Kr(8e–)] + [2 x 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 x 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

a) 9.15

a) The ClO3– ion has twenty-six valence electrons. The Cl is the central atom. There are three bonds (to the O atoms) and one lone pair on the Cl (AX3E1). The shape is trigonal pyramidal. The structure is based on a tetrahedral electron-group 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 thirty-six 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 twenty-six valence electrons. The Se is the central atom. There are three bonds to the Se which also has a lone pair (AX3E1). 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 forty-two valence electrons. The Te is the central atom. There are five bonds to the Te which also has one lone pair (AX5E1). 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

9.16

I F

F F c)

F d)

Plan: The Lewis structures must be drawn, and VSEPR applied to the structures. Solution: a) CH3OH: This molecule has [1 x C(4e–)] + [4 x H(1e–)] + [1 x O(6e–)] = fourteen 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 (AX 2E2), so it is V shaped or bent with the angles less than the ideal angle of 109.5°.

b) N2O4: This molecule has [2 x N(5e–)] + [4 x 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 N 2O4 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

9.17

O N

N O

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 (AX 2E2), so the arrangement is V shaped or bent with angles less than the ideal value of 109.5°.

b) In the CH3OCH2CH3 molecule, all atoms except the hydrogen atoms serve as central atoms. All the carbon atoms 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 (AX 2E2), so the arrangement is V shaped or bent with angles less than the ideal value of 109.5°.

9.18

Plan: The Lewis structures must be drawn, and VSEPR applied to the structures. Solution:

a) CH3COOH has [2 x C(4e–)] + [4 x H(1e–)] + [2 x O(6e–)] = twenty-four 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 CH 3COOH molecule, the carbon atoms and the O with H attached serve as central atoms. The carbon bonded to the H atoms has four groups and no lone pairs (AX 4), 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°.

b) H2O2 has [2 x H(1e–)] + [2 x O(6e–)] = fourteen 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 H 2O2 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°.

9.19

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 (AX3E1), 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 (AX 2E2), so the arrangement is V shaped or bent with an angle less than the ideal value of 109.5°.

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 (AX2E1), so it is V shaped or bent with angles less than the ideal value of 120°. The nitrogen labeled NB has three bonds and no lone pairs (AX3), so it is trigonal planar with no deviation from the ideal angle of 120°.

9.20 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:

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: AX 4, AX3E1, and AX2E2, respectively. The ideal tetrahedral bond angle is 109.5°, which is present in CF 4. The one lone pair in NF3 decreases the angle a little. The two lone pairs in OF2 decrease the angle even more. 9.21

<<109.5o <<<109.5o <109.5o 109.5o 90o 2– Bond angles: SiCl4 > PCl3 > SCl2 > OCl2 > SiCl6 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 AX3E1 species such as PCl3. A greater reduction in the ideal tetrahedral bond angle is present in AX 2E2 species such as SCl2 and OCl2 with two lone pairs. The angle is reduced less around the larger S atom.

9.22

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. 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 pairs, and reduce the angle.

9.23

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.

9.24

a) Type: AX2E1 Ideal angle: 120°

Shape: bent Actual angle: <120 (because of the lone pair)

9.25

b) Type: AX3E1 Ideal angle: 109.5°

Shape: trigonal pyramidal Actual angle: <109.5 (because of the lone pair)

c) Type: AX4 Ideal angle: 109.5°

Shape: tetrahedral Actual angle: 109.5 (there are no lone pairs)

d) Type: AX5 Ideal angles: 120° and 90°

Shape: trigonal pyramidal Actual angle: 120° and 90° (there are no lone pairs)

e) Type: AX6 Ideal angles: 90°

Shape: octahedral Actual angle: 90° (there are no lone pairs)

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. PCl5: PCl4+: PCl6–:

Cl Cl

Cl P

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 PCl 4+ (tetrahedral, 109.5°), and the other half become octahedral PCl6– (octahedral, 90°). 9.26

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.

9.27

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

9.28

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.

9.29

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 towards the more electronegative chlorine atoms.

nonpolar

polar

9.30

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 Cl = 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, AX 3E1. 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, AX4E1. SF6 is nonpolar because it is octahedral, AX6, and the bonds are arranged so they cancel.

9.31

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, AX2E1, so the polar bonds result in electron density being pulled towards one side of the molecule. SO2 has a greater dipole moment because it is the only one of the pair that is polar.

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 AX4E1, so it has a see-saw shape, where the bond dipoles do not cancel. SF4 has the greater dipole moment.

d) H2O and H2S have the same basic structure. They are both bent molecules, AX 2E2, 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. 9.32

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. 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 AX3E1, so it is trigonal pyramidal and polar. AsF3 has a greater dipole moment.

9.33 Plan: Draw Lewis structures, and then apply VSEPR. A molecule has a dipole moment if polar bonds do not cancel. Solution: C2H2Cl2 has [2 x C(4e–)] + [2 x H(1e–)] + [2 x 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 ten 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 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 H 2 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.

9.34

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. 9.35

a) AX5 = trigonal bipyramidal

b) AX4 = tetrahedral

c) AX3E1 = trigonal pyramidal

d) AX 3E2 = T shaped

e) AX2 = linear H Be

f) AX2E2 = bent

H P

H g) AX4 = tetrahedral

h) AX3E1 = trigonal pyramidal

i) AX3 = trigonal planar

j) AX4E1 = seesaw

Cl B Cl

k) AX3E1 = trigonal pyramidal

l) AX4E1 = seesaw

9.36

a) SiF4 with its thirty-two valence electrons is an AX4 molecule and has a tetrahedral molecular shape. SiF5– with its forty 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.

9.37

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 + ½ no. of shared valence electrons]. Solution: a) Formal charges for Al2Cl6: FCAl = 3 – [0 + ½(8)] = –1 FCCl, ends = 7 – [6 + ½(2)] = 0 FCCl, bridging = 7 – [4 + ½(4)] = +1 (Check: Formal charges add to zero, the charge on the compound.) Formal charges for I2Cl6: FCI = 7 – [4 + ½(8)] = –1 FCCl, ends = 7 – [6 + ½(2)] = 0 FCCl, bridging = 7 – [4 + ½(4)] = +1 (Check: Formal charges add to zero, the charge on the compound.) b) The aluminum atoms have no lone pairs and are AX 4, 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.

9.38

The Lewis structure for each is required. Compound Lewis structure XeF2 F

Molecular geometry Linear (AX2E3)

XeF4

Square planar (AX4E2) F F

XeF6

Distorted octahedral (AX6E1)

9.39

a) SO3 is an AX3 molecule and has a trigonal planar shape. SO32– is an AX3E1 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.

9.40

From the Lewis structures, both are AX2E1 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.

9.41

Xe(g) + 3F2(g)  XeF6(g)

r H = bonds broken H + bonds formed H The three F–F bonds must be broken, and six Xe–F bonds are formed. r H = 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 kJ/mol = 146 kJ/mol = BEXe–F 9.42

Plan: Draw the Lewis structures, and then use VSEPR to describe epoxypropane (propylene oxide). Solution: a) H H

C H3C

CH2

H2O2

H3C

CH2

H2O

In epoxypropane , 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 epoxypropane are 109.5° and the molecular shape around each carbon atom is tetrahedral. b) In epoxypropane , 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 carbon atoms in the rings are reduced from the ideal 109.5° to 60°. 9.43

O Cl

+ 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). 9.44

CBr4 < CH2Br2 < CH2Cl2 < CF2Cl2 < CF2Br2 < CH2F2

9.45

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.

Polar

Nonpolar No dipole moment

Polar

Nonpolar No dipole moment

9.46

a) Yes, the black sphere can represent selenium. SeF4 has thirty-four valence electrons. Eight of these electrons are used in the four Se-F single bonds and twenty-four electrons are used to complete the octets of the F atoms. The remaining electron pair goes to selenium and the molecule is AX 4E1. 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 thirty-four valence electrons and would have the seesaw molecular shape as an AX 4E1 species. There would be cation present to balance the negative 1 formal charge for nitrogen. c) For BrF4 to have the thirty-four valence electrons needed for this seesaw molecular geometry, the charge of the species must be +1. BrF4+ would have [1 x Br(7e–)] + [4 x F(7e–)] – [1e– from + charge] = 34 valence electrons.

9.47

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:

a) b) c) three groups four groups five groups (AX3) (AX3E1) (AX3E2) trigonal planar trigonal pyramidal T shaped Trigonal planar molecules, such as a), are nonpolar, so it cannot be AY 3. Trigonal pyramidal molecules b) and Tshaped molecules c) are polar, so either could represent AY3. 9.48

a) Shape A is T shaped (AX3E2); Shape B is trigonal planar (AX3); Shape C is trigonal pyramid (AX3E1). XeF3+, with twenty-eight valence electrons, has two unshared pairs on Xe and is AX 3E2 and is the T-shaped molecular shape in A. SbBr3, with twenty-six valence electrons, has one unshared pair on Sb; thus it is AX 3E1 and is the trigonal pyramidal molecular shape in C. GaCl3, with twenty-four 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 (ten) around the central atom. 9.49 a)

H C

F C

C F

All the carbon atoms 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 tetrafluoroethene are not reduced as much.

9.50

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 x C(4e–)] + [2 x 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 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 x S(6e–)] + [6 x 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 thirty-six 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 pm = 223 pm. c) PF5 has [1 x P(5e–)] + [5 x 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 thirty 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: a 2 = b2 + c2 – 2bc (cos A). a2 = (156 pm)2 + (156 pm)2 – 2(156 pm)(156 pm)cos 120o. a = 270.1999 pm = 270 pm.

PCl5(l) + SO2(g)  POCl3(l) + SOCl2(l)

9.51

O O

Cl Cl

+ Cl

Cl PCl5: SO2: POCl3: SOCl2:

AX5 AX2E1 AX4 AX3E1

Cl Cl

S Cl

trigonal bipyramidal bent tetrahedral trigonal pyramidal

The F−O−F bond angle would be expected to have a greater deviation from the ideal bond angle than the H−O−H bond angle. There are 3 lone pairs electrons on each fluorine atom while hydrogen atom has no lone pair electron. Therefore, the greater electron density on fluorine atoms will repel each other, resulting in a smaller F−O−F bond angle than a H−O−H bond angle.

9.52

9.53

O atom: bent N atom: trigonal pyramidal b) Bond angle labelled a: < 109.5° Bond angle labelled b: < 120° 9.54 a)

No, CO32− not a polar compound. It is a non-polar compound. The resonance hybrid indicates the three C−O bonds are equivalent, each with a bond order of 1 . b) The O−C−O bond angle is =120°. CO32− is trigonal planar.

CHAPTER 10 THEORIES OF COVALENT BONDING END–OF–CHAPTER PROBLEMS 10.1

Plan: Table 10.1 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

10.2

a) sp2

b) sp3

c) sp3d

d) sp3d2

10.3

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.

10.4

Four. The same number of hybrid orbitals will form as the initial number of atomic orbitals mixed.

10.5

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.

10.6

a) two sp orbitals

10.7

b) five sp3d orbitals

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.

c) The nitrogen has three electron groups (one single bond, one double bond, and one lone pair) so the hybridization is sp2.

10.8

a) sp2

b) sp2

c) sp

10.9

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  bond is formed by the overlap of d orbitals from chlorine with p orbitals from oxygen.

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.

10.10

a) sp3d

b) sp3

c) sp3d2

10.11

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.

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. 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.

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.

10.12

a) sp3  s + 3p

b) sp3d  s + 3p + d

c) sp3d  s + 3p + d

d) sp3  s + 3p

10.13

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.

10.14

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.

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.

10.15

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.

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.

Isolated B atom

Hybridized B atom

c) Carbon is the central atom in CH3+. Its electron configuration is [He]2s22p2. C has three electron groups (three bonds), requiring three hybrid orbitals. Hybridization is sp2 around C. One of the 2s electrons is moved to an empty 2p orbital; three orbitals are hybridized and one electron is removed to form the +1 ion.

Isolated C atom 10.16

Hybridized C atom

10.17

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) In SeCl2, Se is the central atom and has four electron groups (two single bonds and two lone pairs), requiring four hybrid orbitals so Se is sp3 hybridized. The electron configuration of Se is [Ar]4s23d104p4. The 4s and 4p atomic orbitals are hybridized. Two sp3 hybrid orbitals are filled with lone electron pairs and two sp3 orbitals bond with the chlorine atoms.

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.

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.

10.18

10.19

sp3

sp2

10.20

Plan: A single bond is a  bond which is the result of two orbitals overlapping end to end; a double bond consists of one  bond and one  bond; and a triple bond consists of one  bond and two bonds. A  bond is the result of orbitals overlapping side to side. Solution: a) False, a double bond is one  and one  bond. b) False, a triple bond consists of one  and two  bonds. c) True d) True e) False, a  bond consists of one pair of electrons; it occurs after a  bond has been previously formed. f) False, end-to-end overlap results in a bond with electron density along the bond axis.

10.21

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  bond which is the result of two orbitals overlapping end to end; a double bond consists of one  bond and one  bond; and a triple bond consists of one  bond and two  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  bonds (one each for the N–O bonds) and one  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  bonds (one each for the C–S bonds) and two  bonds (part of the two C=S double bonds). 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  bonds (one each for the two C–H bonds and one C–O bond) and one  bond (part of the C=O double bond).

10.22 a) Oxygen is the central atom in O3. Oxygen has two surrounding electron groups (one single bond and one double bond), so it is sp2 hybridized. Oxygen forms two  bonds and one  bond.

b) Iodine is the central atom in I3–. Iodine has five surrounding electron groups (two single bonds and three electron pairs), so it is sp3d hybridized. Iodine forms two  bonds.

10.23

Carbon is the central atom in COCl2. Carbon has three surrounding electron groups (two single bonds and one double bond), so it is sp2 hybridized. Carbon forms three  bonds and one  bond.

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  bond which is the result of two orbitals overlapping end to end; a double bond consists of one  bond and one  bond; and a triple bond consists of one  bond and two  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  bond exists between F and N, and one  and one  bond exist between N and O. Nitrogen participates in a total of 2  and 1  bonds. 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  bonds. The C–C bond consists of one  and one  bond. Each carbon participates in a total of three  and one  bonds.

c) In (CN)2, each carbon has two electron groups (one single bond and one triple bond) and is sp hybridized with a  bond between the two carbon atoms and a  and two  bonds comprising each C–N triple bond. Each carbon participates in a total of two  and two  bonds. N C C N 10.24

a) sp3d three  bonds

b) sp3 (CH3) sp (other two C atoms) H

H two  and two  bonds

c) sp2

10.25

six  and two  bonds

Plan: A single bond is a  bond which is the result of two orbitals overlapping end to end; a double bond consists of one  bond and one  bond; and a triple bond consists of one  bond and two  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

H C

C H

H H

C H

H H

cis trans The carbon atoms participating in the double bond each have three surrounding groups, so they are sp2 hybridized. The =C–H  bonds result from the head-on overlap of a C sp2 orbital and an H s orbital. The C–CH3 bonds are also  bonds, resulting from the head-on overlap of an sp2 orbital and an sp3 orbital. The C=C bond contains 1 ζ bond (head on overlap of two sp2 orbitals) and 1  bond (sideways overlap of unhybridized p orbitals). Finally, C–H bonds in the methyl (–CH3) groups are  bonds resulting from the overlap of the sp3 orbital of C with the s orbital of H.

10.26

10.27

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.

10.28

Two px atomic orbitals were used to form a  bonding MO (lower energy) and a  antibonding MO (higher energy). The bonding MO does not have a node separating the two halves of the orbital.

10.29

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.

10.30

A bonding MO may contain a nodal plane lying along the internuclear axis, as in  bonding. In an antibonding MO, the nodal plane is perpendicular to the bond axis, between the atoms.

10.31

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 -bonding molecular orbital. Each molecular orbital requires two electrons. b) Two electrons are required to fill a -antibonding molecular orbital. There are two -antibonding orbitals, each holding a maximum of two electrons. c) Four electrons are required to fill the two  molecular orbitals (two electrons to fill the -bonding and two to fill the -antibonding) formed from two 1s atomic orbitals.

10.32

a) twelve

10.33

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, or from p orbitals overlapping end to end, are called and MOs formed by the side-to-side overlap of p orbitals are called . A superscript star (*) is used to designate an antibonding MO. To write the electron configuration of F 2+, determine the number of valence electrons and write the sequence of MO energy levels, following the sequence order given in the text.

b) two

c) four

Solution: a) A is the *2p molecular orbital (two p orbitals overlapping side to side with a node between them); B is the 2p molecular orbital (two p orbitals overlapping end to end with no node); C is the 2p molecular orbital (two p orbitals overlapping side to side with no node); D is the *2p molecular orbital (two p orbitals overlapping end to end with a node). b) F2+ has thirteen valence electrons: [2 x F(7e–) – 1 (from + charge)]. The MO electron configuration is (2s)2(*2s)2(2p)2(2p)2(2p)2(*2p)2(*2p)1. The *2p molecular orbital, A, 2p molecular orbital, B, and 2p molecular orbital, C, are all occupied by at least one electron. The *2p molecular orbital is unoccupied. c) A *2p molecular orbital, A, has only one electron. 10.34

a) A is the *2p molecular orbital; B is the 2p molecular orbital; C is the 2p molecular orbital; D is the *2p molecular orbital; E is the 2s molecular orbital; F is the *2s molecular orbital. b) The *2p molecular orbital, D, is the highest in energy. c) The 2s molecular orbital, E, is the lowest in energy. d) 2s < *2s < 2p < 2p < *2p < *2p (E < F < C < B < A < D)

10.35

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 x Be(4e–) – 1 (from + charge)]. The molecular orbital configuration is (1s)2(*1s)2(2s)2(*2s)1 and bond order = ½(4 – 3) = ½. With a bond order of 1/2 the Be2+ ion will be stable. b) No, the ion has one unpaired electron in the *2s MO, 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 (2s)2(*2s)1.

10.36

a) The molecular orbital configuration for O2– with a total of seventeen electrons is (1s)2(*1s)2(2s)2(*2s)2(2p)2(2p)2(2p)2(*2p)2(*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 *2p MO. c) ( 2s)2(*2s)2(2p)2(2p)2(2p)2(*2p)2(*2p)1

10.37

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: (1s)2(*1s)2(2s)2(*2s )2(2p)4(2p)1 Bond order = 1/2(9 – 4) = 2.5 C2 Total electrons = 6 + 6 = 12 MO configuration: (1s)2(*1s)2(2s)2(*2s)2(2p)4 Bond order = 1/2(8 – 4) = 2 C2+ Total electrons = 6 + 6 – 1 = 11 MO configuration: (1s)2(*1s)2(2s)2(*2s)2(2p)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+ 10.38 B2+: B2: B2–:

(2s)2(*2s)2(2p)1 (2s)2(*2s)2(2p)1(2p)1 (2s)2(*2s)2(2p)2(2p)1

a) B2– > B2 > B2+ 10.39

Bond order 0.5 1.0 1.5 b) B2+ > B2 > B2–

Bond order = ½ (8 – 7) = ½ Yes, Ne2+ does exists. d) Yes, Ne2+ would be attracted by an external field because Ne2+ is paramagnetic. e) 10.40

(ζ2s)2 (ζ*2s)2(ζ2p)2(2p)4(*2p)4(ζ*2p)1

a) BrO3–

AX3E

trigonal pyramidal sp3 hybrid AO

Ideal

Deviations

109.5

<109.5

b) AsCl4–

AX4E

c) SeO42–

AX4

d) BiF52–

AX5E

e) SbF4+

AX4

f) AlF63–

AX6

g) IF4+

AX4E

seesaw sp3d hybrid AO tetrahedral sp3 hybrid AO square pyramidal sp3d 2 hybrid AO tetrahedral sp3 hybrid AO octahedral sp3d 2 hybrid AO seesaw sp3d hybrid AO

120, 90

<120, <90

109.5

none

90

<90

109.5

none

90

none

120, 90

<120, <90

Lewis structures:

10.41

a) There are 9  and 2  bonds. Each of the six C–H bonds are  bonds. The C–C bond contains a  bond. The double bonds between the carbon atoms consist of a bond in addition to the bond.b) No, cis-trans structural arrangements are not possible because one of the carbon atoms in each double bond has two hydrogen atoms bonded to it. Cis-trans structural arrangements only occur when both carbon atoms in the double bond are bonded to two groups that are not identical.

10.42

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  bond which is the result of two orbitals overlapping end to end; a double bond consists of one  bond and one  bond; and a triple bond consists of one  bond and two  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  bond; each of the double bonds has one  bond for a total of 26  bonds. c) The ring has three double bonds each of which is composed of one  bond and one  bond; so there are three  bonds each with two electrons for a total of six  electrons. 10.43

10.44

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  bond which is the result of two orbitals overlapping end to end; a double bond consists of one  bond and one  bond; and a triple bond consists of one  bond and two  bonds. Solution: a) Every single bond is a  bond. There is one  bond in each double bond as well. There are 17  bonds in isoniazid. Every atom-to-atom connection contains a  bond. b) All carbon atoms 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.

10.45

Hydrazine

Carbon disulfide

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. 10.46

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.

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.

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

C H

d) Si changes from sp3  sp3d2. Four electron groups surround Si in SiF4 and six electron groups surround Si in SiF62–.

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.

10.47

a) Original molecules: NO (2s)2(*2s)2(2p)2(2p)2(2p)2(*2p)1 O2 (2s)2(*2s)2(2p)2(2p)2(2p)2(*2p)1(*2p)1 N2 (2s)2(*2s)2(2p)2(2p)2(2p)2 Ions: NO+ (2s)2(*2s)2(2p)2(2p)2(2p)2 + O2 (2s)2(*2s)2(2p)2(2p)2(2p)2(*2p)1 + N2 (2s)2(*2s)2(2p)2(2p)2(2p)1 b) Changing: NO+, N2+

bond order = 2.5 paramagnetic bond order = 2.0 paramagnetic bond order = 3.0 diamagnetic bond order = 3.0 diamagnetic bond order = 2.5 paramagnetic bond order = 2.5 paramagnetic

10.48

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

10.49

a) 1: sp3 2: sp2 3: sp3 b) 28 c) a: < 109.5° b: 120° c: 120°

10.50

( 2s)2(*2s)2(2p)2(2p)2(2p)2(*2p)1(*2p)1 2 unpaired electrons

bond order = 2.0 paramagnetic

O 2+

( 2s)2(*2s)2(2p)2(2p)2(2p)2(*2p)1 1 unpaired electron

bond order = 2.5 paramagnetic

O 2–

( 2s)2(*2s)2(2p)2(2p)2(2p)2(*2p)2(*2p)1 1 unpaired electron

bond order = 1.5 paramagnetic

O22–

( 2s)2(*2s)2(2p)2(2p)2(2p)2(*2p)2(*2p)2 0 unpaired electron

bond order = 1.0 diamagnetic

Bond length:

4: sp3

5: sp2

6: sp2

O2+ < O2 < O2– < O22–

10.51

a) All the diagrams are identical: ( 2s)2(*2s)2(2p)2(2p)2(2p)2 b) Nitrogen molecules are not polar, but CN– and CO are polar.

10.52

a) Yes, each one is sp2 hybridized. b) Yes, each one is sp3 hybridized. c) C–O bonds: 6  bonds, 1  bond. d) No, the C=O lone pair electrons are in sp2 hybrid orbitals, while the other oxygen lone pairs occupy sp3 hybrid orbitals.

10.53

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 CH 3 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.

10.54

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  bond; the resonance interaction exchanges a C–O  bond for a C–N  bond.

10.55

Draw the Lewis structures.

10.56

SO2

sp2

SO3

sp2

SO32–

sp3

SCl4

sp3d

SCl6

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.

10.57

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  bond which is the result of two orbitals overlapping end to end; a double bond consists of one  bond and one  bond; and a triple bond consists of one  bond and two  bonds. Solution: a) The six carbon atoms in the ring each have three surrounding electron groups (two single bonds and one double bond) with sp2 hybrid orbitals. The two carbon atoms 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 carbon atoms have this shape. Only one carbon in the CH3 group has four surrounding groups with sp3 hybridization and tetrahedral shape.

10.58

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

10.59

a) b) c) d)

24 ζ bonds 3  bonds Carbon 1‘s unhybridized p orbital overlaps with carbon 2‘s unhybridized p orbital to form a  bond. Carbon 3‘s sp3 orbital overlaps with nitrogen 4‘s sp3 orbital to form a ζ bond.

CHAPTER 11 INTERMOLECULAR FORCES: LIQUIDS, SOLIDS, AND PHASE CHANGES TOOLS OF THE LABORATORY BOXED READING PROBLEMS B11.1

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 = 0.709x10–10 m ; θ = 11.6° 1(0.709x10–10 m) = 2d sin 11.6° 0.709x10–10 m = 0.4021558423 d d = 1.762998x10–10 m = 1.76x10–10 m

B11.2

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 pm/2 = 283 pm. n = 1; λ = ?; θ = 15.9°; d = 283 pm n λ = 2d sin θ

 

2 283 pm sin 15.9 2d sin  = = 155.0609178 pm = 155 pm 1 n b) n = 2; λ = 155 pm; θ = ?; d = 566 pm n λ = 2d sin θ λ=

sin θ =

2 155 pm 

2 283 pm 

sin θ = 0.5477031802 θ = 33.20958° = 33.2°

END–OF–CHAPTER PROBLEMS

11.1

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.

11.2

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. 11.3

a) intermolecular

11.4

a) Heat of fusion refers to the change between the solid and the liquid states and heat of vapourization 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 vapourization. c) At a given temperature and pressure, the magnitude of vap is the same as the magnitude of cond. The only difference is in the sign: vap = –cond.

11.5

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 (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 O 2 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.

11.6

a) Intermolecular

b) Intramolecular

11.7

a) Condensation

The water vapour 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 vapour.

b) Fusion (melting) c) Evaporation

b) intermolecular

b) sublimation

c) intermolecular

c) Intermolecular

d) intramolecular

d) Intermolecular

11.8

a) deposition

c) crystallization (freezing)

11.9

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.

11.10

Sublimation and deposition

11.11

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.

11.12

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.

11.13

In closed containers, two processes, evaporation and condensation, occur simultaneously. Initially there are few molecules in the vapour phase, so more liquid molecules evaporate than gas molecules condense. Thus, the amount of molecules in the gas phase increases, causing the vapour pressure of hexane to increase. Eventually, the amount of molecules in the gas phase reaches a maximum where the amount of liquid molecules evaporating equals the amount of gas molecules condensing. In other words, the evaporation rate equals the condensation rate. At this point, there is no further change in the vapour pressure.

11.14

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 vapour pressure decreases for the same reason given in b). At any given temperature, strong intermolecular forces prevent molecules from easily going into the vapour phase and thus vapour pressure is decreased. d) As intermolecular forces increase, the heat of vapourization increases because more energy is needed to separate molecules from the liquid phase.

11.15

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 vapourization and condensation. Point 3 is depicted by D. This is the equilibrium between sublimation and deposition.

11.16

a) The final pressure will be the same, since the vapour pressure is constant as long as some liquid is present. b) The final pressure will be lower, according to Boyle‘s law.

11.17

The positive slope of the solid-liquid line for substance A indicates that solid A has higher density that liquid A. Therefore, solid A will sink in liquid A. The negative slope of the solid-liquid line for substance B indicates that liquid B has higher density that solid B. Therefore, solid B will float on liquid B.

11.18

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.

11.19

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 x mass x T 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 x mass x T = (2.09 J/g•K)(22.00 g)[(0.0+273.2)K – (–6.00+273.2)K] = 275.88 J 2) Phase change of ice at 0.00°C to water at 0.00°C:



q2 = n  fus H



 1 mol   6.02 kJ   103 J   = 7349.6115 J    18.02 g   mol   1 kJ 

=  22.0 g  

3) Warming the liquid from 0.00°C to 0.500°C: q3 = c x mass x T = (4.21 J/g•K)(22.00 g)[(0.500+273.2)K – (0.0+273.2)K] = 46.31 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 + 46.31 J) = 7671.8015 J = 7.67x103 J.

11.20

0.333 mol x 46.07 g/mol = 15.34131 g ethanol Cooling vapour to boiling point: q1 = c x mass x T = (1.43 J/g•K) (15.34131 g) ((273.2+78.5)K –(273+ 300)K)) = –4859.28 J Condensing vapour: (note condH = – vapH)



 = (0.333 mol)(–40.5 kJ/mol)(10 J/kJ) = –13,486.5 J 3

q2 = n  cond H

Cooling liquid to 25.0°C: q3 = c x mass x T = (2.45 J/g•K)(15.34131 g)((273.2+25.0)K –(273.2+ 78.5)K) = –2010.86 J qtotal = q1 + q2 + q3 = (–4859.28 J) + (–13,486.5 J) + (–2010.86 J) = –20356.64 J = – 2.04x104 J 11.21

Plan: The Clausius-Clapeyron equation gives the relationship between vapour pressure and temperature. We aregiven  vap H , p1, T1, and T2; these values are substituted into the equation to find the p2, the vapour pressure. Solution: p1 = 101.3 kPa

T1 = 122°C + 273 = 395 K

p2 = ?

T2 = 113°C + 273 = 386 K

 vap H = 35.5 kJ/mol

 vap H  1 p2 1 =    p1 R T T 1  2 kJ 35.5 1   103 J   1 p2 mol  = ln   = –0.2520440 8.314 J/mol•K  386 K 395 K   1kJ  101.3 kPa

e^ ( ln

p2 ) = e^ –0.2520440 101.3 kPa

p2 = 0.7772105 101.3 kPa

p2 = (0.7772105)(101.3 kPa)= 78.731424 kPa = 78.7 kPa 11.22

The Clausius-Clapeyron equation gives the relationship between vapour pressure and temperature.

 vap H  1 p2 1 =    p1 R T1   T2 kJ 29.1 1   103 J  p2 mol  1  = ln   = 2.2314173653  298 K   1kJ  71.226 kPa 8.314 J/mol•K  368 K ln

p2 = 9.338762 71.226 kPa

p2 = (9.338762)(71.226 kPa) = 665.163 kPa = 665 kPa 11.23

Plan: The Clausius-Clapeyron equation gives the relationship between vapour pressure and temperature. We are given p1, p2, T1, and T2; these values are substituted into the equation to find  vap H . Solution: p1 =82.8 kPa

T1 = 85.2°C + 273.2 = 358.4 K

p2 = 101.3 kPa

T2 = 95.6°C + 273.2 = 368.8 K

 vap H = ?

 vap H  1 p2 1 =    p1 R T1   T2

 vap H 1 1   101.3 kPa  =  8.314 J/mol•K  368.8 K 358.4 K  82.8 kPa

0.2016583 = –  vap H (–9.463775x10–6) mol/J

 vap H = 21,308.447 kJ = 21.3 kJ/mol (The significant figures in the answer are limited by the 82.8 kPa in the problem.) 11.24

The Clausius-Clapeyron equation gives the relationship between vapour pressure and temperature.

 vap H  1 p2 1 =    p1 R T T 1  2

 vap H   1 1 4336 kPa =    8.314 J/mol•K   273  ( 100)  K 101.3 kPa  273  ( 164)  K 

3.756621 = 0.00040822 mol/J  vap H

 vap H = (3.756621)/(0.00040822 mol/J) = 9202.4 J/mol = 9.20x10³ J/mol

11.25

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 ethene is denser than liquid ethene since the solid-liquid line slopes to the right with increasing pressure.

11.26

Hydrogen does sublime at 0.05 bar, since 0.05 bar is below the triple point pressure. 11.27

This is a stepwise problem to calculate the total heat required. Melting SO2:



q1 = n  fus H

 = (2.500 kg)(10 g/1 kg)(1 mol SO /64.07 g SO )(8.619 kJ/mol)(10 J/kJ) 3

= 336,311.85 J Warming liquid SO2: q2 = c x mass x T = (0.995 J/g•K)(2.500 kg)(103 g/1 kg)[(273.+(–10.))K – (273+(–73))K)] = 156,712.5J Vapourizing SO2:



q3 = n  vap H

 = (2.500 kg)(10 g/1 kg)(1 mol SO /64.07 g SO )(25.73 kJ/mol)(10 J/kJ) 3

= 1,003,980.0 J Warming gaseous SO2: q4 = c x mass x T = (0.622 J/g•K)(2.500 kg)(103 g/1 kg)[(273+60.)K –(273+ (–10.))K] = 108,850 J qtotal = q1 + q2 + q3 + q4 = (336,311.8 J) + (156,712.5 J) + (1,003,980.0 J) + (108,850 J) = 1,605,854.35 J = 1.605x106 J 11.28

Plan: The Clausius-Clapeyron equation gives the relationship between vapour pressure and temperature. We are given  vap H , p1, T1, and T2; these values are substituted into the equation to find p2. Convert the temperatures from °C to K and  vap H from kJ/mol to J/mol to allow cancellation with the units in R. Solution: p1 = 233 kPa

T1 = 25.0°C + 273 = 298 K

p2 = ?

T2 = 135°C + 273 = 408 K

 vap H = 24.3 kJ/mol

 vap H  1 p2 1 =    p1 R T1   T2

24.3

1   10 J  p2 mol  1  = ln    8.314 J/mol•K  408 K 298 K   1 kJ  233 kPa ln

p2 = 2.644311 233 kPa

p2 = 14.07374 233 kPa

p2 = (14.07374)(233 kPa) = 3279.18 kPa = 3280 kPa 11.29

a) At 20ºC and 40ºC, no liquid exists, only gas. At –40ºC, liquid exists. At –120ºC, no liquid exists, only solid. b) No, at any pressure below the triple point pressure, the CO 2(s) will sublime. c) No d) No

11.30

Intermolecular forces involve interactions of lower (partial) charges at relatively larger distances than in covalent bonds.

11.31

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

11.32

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.

11.33

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.

11.34

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.

11.35

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.

11.36

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.

11.37

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.

11.38

a) Hydrogen bonding

11.39

Plan: Dispersion forces are the only forces between nonpolar substances; dipole-dipole forces exist between polar

b) Dipole-dipole

c) Ionic bonds

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 chloromethane 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. 11.40

a) Dispersion forces

b) Dipole-dipole

c) Hydrogen bonding

11.41

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.

b) The presence of H attached to F in HF leads to the formation of hydrogen bonds. There are no hydrogen bonds in HBr.

11.42

a) The presence of H directly attached to the N in (CH3)2NH leads to hydrogen bonding. More than one arrangement is possible.

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.

11.43

Plan: In the vapourization process, intermolecular forces between particles in the liquid phase must be broken as the particles enter the vapour phase. In other words, the question is asking for the strongest interparticle force that must be broken to vapourize 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 oxygen, nitrogen, or fluorine. Solution: a) Dispersion forces, 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 forces, although the individual Si–Cl bonds are polar, the molecule has a symmetrical, tetrahedral shape and is therefore nonpolar.

11.44

a) Dispersion

11.45

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  bonds are more easily polarized than electrons involved in  bonds. c) H2Se has greater polarizability than water because the selenium atom is larger than the oxygen atom.

11.46

a) Ca b) CH3CH2CH3 c) CCl4 In all cases, the larger molecule (i.e., the one with more electrons) has the higher polarizability.

11.47

Plan: Weaker attractive forces result in a higher vapour 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).

11.48

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.

11.49

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) PH3 would have a lower boiling point than NH 3 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.

b) Dipole-dipole

c) Hydrogen bonding

11.50

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

11.51

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.

11.52

a) CH3OH, hydrogen bonding (CH3OH) vs. dispersion forces (CH3CH3). b) FNO, greater polarity in FNO vs. ClNO c) F

F C

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. 11.53

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.

11.54

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 vapourization.

11.55

The ethylene glycol molecules have two sites (two OH groups) which can hydrogen bond; the propanol has only one OH group.

11.56

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.‖

11.57

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, 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.

11.58

The presence of the ethanol molecules breaks up some of the hydrogen bonding interactions present between the water molecules, lowering the surface tension.

11.59

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.

11.60

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.

11.61

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 CH3CH2CH2OHwith 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

11.62

CH3OH > H2CO > CH3CH3 The intermolecular forces would decrease as shown (hydrogen bonding > dipole-dipole > dispersion), as would the surface tension.

11.63

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 11.61). HOCH2CH(OH)CH2OH > HOCH2CH2OH > CH3CH2CH2OH The greater the number of hydrogen bonds, the stronger the intermolecular forces, and the higher the viscosity.

11.64

Viscosity and surface tension both increase with increasing strength of intermolecular forces. CH3CH3 < H2CO < CH3OH

11.65

a) The more volatile substances (volatile organic pollutants) are preferentially pulled away from the less volatile substances. b) The vapour pressure of the volatile organic pollutants increases as the temperature increases. The higher vapour pressure makes it easier to remove the vapour.

11.66

a) Calculate the energies involved using the heats of fusion.



qHg = n  fus H

 1 mol Hg  23.4 kJ 

 = 12.0 g Hg  200.6 g Hg  1 mol Hg  = 1.3998 kJ = 1.40 kJ



qmethane = n  fus H

 1 mol CH 4  0.94 kJ 

 = 12.0 g CH  16.04 g CH  1 mol CH  = 0.70324 kJ = 0.70 kJ 4

Mercury takes more energy. b) Calculate the energies involved using the heats of vapourization.



qHg = n  vap H



 1 mol Hg  59 kJ 

 = 12.0 g Hg  200.6 g Hg  1 mol Hg  = 3.5294 kJ = 3.5 kJ

qmethane = n  vap H

 1 mol CH 4 



 = 12.0 g CH  16.04 g CH  1 mol CH  = 6.65835 kJ = 6.6 kJ 8.9 kJ

Methane takes more energy.

c) Mercury involves metallic bonding and methane involves dispersion forces.

11.67

The pentanol has stronger intermolecular forces (hydrogen bonds) than the hexane (dispersion forces).

11.68

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.

11.69

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.

11.70

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.

11.71

Water has a high surface tension. The debris on the surface provides shelter and nutrients for fish, insects, etc.

11.72

Water exhibits strong capillary action, which allows it to be easily absorbed by the plant‘s roots and transported to the leaves.

11.73

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.

11.74

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.

11.75

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.

11.76

When the unit cells are translated in all directions, the crystal structure is generated.

11.77

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 x 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 x 1/2 atom per cell = 3 atoms, for a total of 4 in the cell (corner + face).

11.78

The unit cell is a simple cubic cell. According to the bottom row in Figure 11.29 two atomic radii (or one atomic diameter) equal the width of the cell.

11.79

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.

11.80

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.

11.81

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.

11.82

a) Conductivity decreases with increasing temperature. b) Conductivity increases with increasing temperature. c) Conductivity does not change with temperature.

11.83

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, which determines the packing efficiency (how much of the volume is occupied by empty space).

11.84

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 x 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 x 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.

11.85

a) one

11.86

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. In the CdO unit cell, each Cd2+ ion is surrounded by six O2- ions. The CdSe unit cell can be pictured as two facecentered cubic arrays, one of Cd2+ ions and the other O2- ions, penetrating each other such that each ion is tetrahedrally surrounded by four of the other ions.

11.87

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.

11.88

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.

11.89

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).

11.90

Figure P11.90 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.

11.91

Figure P11.91 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.

b) two

c) four

11.92

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 x 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 x 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 u = 1.66054x10 –24 g), by the density to find the volume of the unit cell. Since the volume of a cube is length x width x 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 x mass of Zn atom) + (4 x mass of Se atom) = (4 x 65.41 u) + (4 x 78.96 u) = 577.48 u  1.66054x1024 g  cm3  –22 –22 3 c)Volume (cm3) =  577.48 u     5.42 g  = 1.76924x10 cm³ = 1.77x10 cm 1 u    d) The volume of a cube equals (length of edge) 3. Edge length (cm) = 3 1.76924 x1022 cm3 = 5.6139x10–8 cm = 5.61x10–8 cm

11.93

a) A face-centered cubic unit cell contains four atoms. b) Volume = (4.52x10–8 cm)3 = 9.23454x10–23 cm³ = 9.23x10–23 cm3 c) Mass of unit cell = (1.45 g/cm3)(9.23454x10–23 cm3) = 1.3390x10–22 g = 1.34x10–22 g  1.3390x1022 g  1 kg    1 unit cell  1u d) Mass of atom =      3  27      1 unit cell   10 g   1.66054x10 kg   4 atoms  = 20.1592 u/atom = 20.2 u/atom

11.94

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 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.

11.95

a) conductor

11.96

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.

11.97

a) decrease (metalloid)

b) insulator

c) conductor

b) increase (metal)

c) decrease (metalloid)

11.98

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 x 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:

  1 Po atom   209 g Po   cm3   1 mol Po Volume (cm3) of the unit cell =        23  1 mol Po   9.142 g   6.022x10 Po atoms   1 unit cell  = 3.7963332x10–23 cm3





Edge length (cm) of the unit cell = 3 3.7963332x10 23 cm 3 = 3.3608937x10–8 cm 2r = edge length 2r = 3.3608937x10–8 cm r = 1.680447x10–8 cm = 1.68x10–8 cm

11.99

  4 Cu atoms   63.55 g Cu   cm3   1 mol Cu Volume (cm3) of the unit cell =        23  1 mol Cu   8.95 g   6.022x10 Cu atoms   1 unit cell  = 4.71641x10–23 cm3



Edge length (cm) of the unit cell = 3 4.71641x1023 cm 3

 = 3.613022x10 cm –8

Use the edge length of the cube and the Pythagorean Theorem to find the diagonal of the cell: C2 = A2 + B2 C=



2 3.613022x108 cm

 = 5.109585x10 cm 2

–8

C = 4r 5.109585x10–8 cm = 4r r = 1.277396x10–8 cm = 1.28x10–8 cm   95.94 g Mo   cm3   1 mol Mo 11.100 a) Edge of unit cell = 3      2 Mo atoms    23 1 mol Mo 10.28 g     6.022 x10 Mo atoms  = 3.1412218x10-8 cm = 3.141x10-8 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 = 3 times the length of the unit cell edge.

4r =





3 3.1412218x108 cm = 5.4407559x10–8 cm –8

r = 1.360189x10 cm = 1.360x10–8 cm

 3 3  102 m   2 atoms   180.9479 g   cm 11.101 Avogadro‘s Number =       3   mol    16.634 g  1 cm   3.3058 x1010 m    = 6.0222270x1023 atoms/mol = 6.022x1023 atoms/mol





11.102 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. 11.103 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.

11.104 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. 11.105 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. 11.106 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. 11. 107 Plan: Germanium and silicon are elements in Group 14 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. 11.108 a) n-type

b) p-type

11.109 The vapour pressure of H2O is 0.61115 kPa. at 0ºC, 1.2282 kPa at 10ºC, and 2.6453 kPa at 22ºC.  44%    100%   2.6453 kPa    0.75 L   18.02 g H 2 O     a) Mass (g) of H2O at 22°C = pV/RT =    L•kPA    1 mol H 2 O  8.31446 273  22 K      mol•K   = 0.006413389 g H2O  18.02 g H 2 O   0.61115 kPa  0.75 L  Mass (g) of H2O at 0°C = pV/RT =   L•kPa    1 mol H 2 O   8.31446 mol•K    273.2  0.0 K    = 0.003636211 g H2O Mass H2O condensed = (0.006413389 g H2O) – (0.003636211470057 g H2O) = 0.002777178 g = 0.0028 g H2O b) Repeat the calculation at 10°C.

1.2282 kPa  0.75 L 

 18.02 g H 2 O    L•kPa    1 mol H 2 O  8.31446 273  10 K      mol•K   = 0.007054474 g H2O At equilibrium, 0.0070 g of H2O could be in the vapour state. Since only 0.0063 g of H 2O are actually present, no liquid would condense at 10ºC.

Mass (g) of H2O at 10°C = pV/RT =

11.110 Plan: The vapour pressure of water is temperature dependent. Table 4.2 gives the vapour 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 vapour pressure of water is 2.6453 kPa (from Table 4.2). a) Once compressed, the N2 gas would still be saturated with water. The vapour pressure depends on the temperature, which has not changed. Therefore, the partial pressure of water in the compressed gas remains the same at 2.6453 kPa. b) pV = nRT

Moles of H2O at 22°C and 5.00 L volume: n=

pV = RT (

(

)( •kPa

) ) )

)((

= 0.0053925 mol H2O

The gas is compressed to half the volume: 5.00 L/2 = 2.50 L

 18.02 g H 2 O   = 0.097173 g  1 mol H 2 O 

Mass (g) of H2O at 22°C and 5.00 L volume =  0.0053925 mol H 2 O   Moles of H2O at 22°C and 2.50 L volume: n=

pV = RT (

(

)( •kPa

)((

) ) )

= 0.0026962 mol H2O

 18.02 g H 2 O   = 0.04859 g  1 mol H 2 O 

Mass (g) of H2O at 22°C and 2.50 L volume =  0.0026962 mol H 2 O  

Mass (g) of H2O condensed = (0.097173 g H2O) – (0.04859 g H2O) = 0.0485867 g = 0.0486 g H2O 11.111 First use the Clausius-Clapeyron equation to determine the heat of vapourization of hexane. ln

 vap H  1 p2 =  R  T2 p1

 vap H 1 1   101.3 kPa  = 8.314 J/mol•K  341.9 K 293.2 K  16.1 kPa

1  T1  p1 = 16.1 kPa p2 = 101.3 kPa T1 = 20.0°C + 273.2 = 293.2 K T2 = 68.7°C + 273.2 = 341.9 K 

1.8392671 = –  vap H (–5.84327x10–5) mol/J

 vap H = 3.1477x104 J/mol = 3.14x104 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)(101.3 kPa ) = 1.1143 kPa = 1.1 kPa Use the Clausius-Clapeyron to find the temperature of this particular vapour pressure: ln

 vap H  1 p2 =  R  T2 p1

p1 = 101.3 kPa

1  T1  p2 = 1.1 kPa 

T1 = 68.7°C + 273.2 = 341.9 K

T2 = ?

 vap H = 3.1447x104 J/mol

3.1447x10 J/mol  1 1  1.1 kPa  =   8.314 J/mol•K  T2 341.9 K  101.3 kPa 4

 1   + 11.06293  T2 

–4.5227762 = –3782.415 K 

T2 = 242.685 K = (243 – 273)°C = –30oC

11.112 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 11.113 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 = r2h. Divide cylinder height by wafer thickness to determine the number of wafers possible.  103 g  1 cm3  3 Volume (cm3) of Si ingot = 4.00 kg    = 1709.402 cm   1 kg  2.34 g  Radius, r, (cm) = ½(diameter) = ½(13.2 cm) = 6.604 cm Height = h = V/r2 = (1709.402 cm3)/[(6.604 cm)2] = 12.476 cm Number of wafers = [(12.476 cm)(10 –2 m/cm)]/(1.12x10–4 m/wafer) = 1113.94 wafers = 1.11x103 wafers b) A single wafer is also a cylinder with dimensions h = 1.12x10–4 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) = r2hd = (6.604 cm)2[(1.12x10–4 m)(1 cm/10–2 m)](2.34 g/cm3) = 3.5909 g = 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 –3    1 mol Si  = 3.83506x10 mol = 3.84x10 mol HF 28.09 g Si 100%      

d) Moles of HF = 3.5909 g Si 

11.114 a) The enthalpy of vapourization must be determined first. ln

 vap H  1 p2 =   R  T2 p1

 vap H  1 133 Pa  =  1333 Pa 8.314 J/mol•K   273.2  54.3 K

1  T1 



 1   273.2  95.3 K 

–2.304838 = –  vap H (0.0000408625mol/J)

 vap H = (–2.304838)/(–0.000040862 mol/J) = 56405.42 J/mol = 5.64x104 J/mol The subtraction of the 1/T terms limits the significant figures.  56405.42 J/mol  1 1 p2  =   = –4.355582 ln 8.314 J/mol•K   273  25  K 1333 Pa  273.2  95.3 K  p2 = 0.0128350 1333 Pa p2 = (0.0128350)(1333 Pa) = 17.109011 Pa = 17 Pa b) V= nRT/p =

(

•

)(

)((

(

)

) )

(

)(

)

= 0.95176 L = 0.95 L

11.115 Plan: The Clausius-Clapeyron equation gives the relationship between vapour pressure and temperature. We are given p1, p2, T1, and  vap H ; these values are substituted into the equation to find T2. Solution: ln

H vap  1 p2 =   R  T2 p1



1  T1 

p1 = 0.160 Pa

T1 = 20.0°C + 273 = 293 K

p2 = 6.67x10-3 Pa

T2 = ?

 vap H = 59.1 kJ/mol

59.1 kJ/mol  1 1   10 J  6.67x103 Pa  =    8.314 J/mol•K  T2 293 K   1kJ  0.160 Pa 3

 1

–3.177554 = –7108.492 K 

 T2



1   293 K  1

(–3.177554)/(–7108.49 K) = 4.47008x10–4 (1/K)= 

 T2



1   293 K 

–4

4.47008x10 (1/K) + 1/293 K = 1/T2 T2 = 259.069 K = 259 K 11.116 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 vapour using the ideal gas equation. Knowing that 4.20 L results in the calculated vapour pressure, the volume of water that would give a vapour pressure corresponding to 100% relative humidity can be calculated. Solution:  1 mL   1.00 g   1 mol H 2 O  a) Moles of water = 4.20 L  3    = 233.0744 mol   10 L   mL   18.02 g H 2 O  pV = nRT L•kPa   233.0744 mol   8.31446   273  26  K  103 m3  mol•K nRT   p= =   V 256 m3  1L 





= 2.26338 kPa = 2.26 kPa b) 3.3639 kPa is needed to saturate the air (at 26°C). 4.20 L x = 2.26338 kPa 3.3639 kPa

Volume (L) of water needed for 100% relative humidity =

 3.3639 kPa  4.20 L  = 6.24216 L = 6.24 L H O 2.26338 kPa

11.117 The formulas are a) TaN and b)TaC. 11.118 In the NaCl type lattice, there are four ions of each type.

   4 KF   1 mol KF   58.10 g KF   unit cell  Density of KF =       23 3   unit cell   6.022x10 KF   1 mol KF   5.39x108 cm    = 2.46450 g/cm³ = 2.46 g/cm3





11.119 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.

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 C

C C

C H

O C

2-furoic acid

C H H

11.120 Plan: Determine the total mass of water, with a partial pressure of 4.133 kPa, contained in the air. This is done by using pV= nRT to find the moles of water at 4.133 kPa and 22.0°C. Then calculate the total mass of water, with a partial pressure of 1.333 kPa, 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 n=

pV RT





 1L  Convert volume to units of L: V (L) = 2.4x106 m3  3 3  = 2.4x109 L  10 m 

Moles of H2O at 4.133 kPa=

 4.133 kPa   2.4 x109 L  L•kPa    8.31446 mol•K    273.2  22.0  K   

= 4,041,348 mol

 18.02 g   1 kg  1 t  Mass (metric tons) of H2O at 44.133 kPa =  4, 041,348 mol H 2 O     3   3   1 mol H 2 O   10 g  10 kg  = 73.552533 t H2O Moles of H2O at 1.333 kPa =

1.333 kPa   2.4x109 L 

L•kPa    8.31446 mol•K    273.2  22.0  K   

= 1,303,440 mol

 18.02 g   1 kg  1 t  Mass (metric tons) of H2O at 1.333 kPa = 1,303,440 mol H 2 O     3   3   1 mol H 2 O   10 g  10 kg  = 23.4880 tons H2O Mass of water removed = (73.5525 metric tons H 2O) – (23.4880 metric tons H2O) = 50.0645 metric tons = 50.1 metric tons H2O b) The heat of condensation for water is –40.7 kJ/mol.  103 kg  103 g   1 mol H 2 O  40.7 kJ  Heat (kJ) =  50.0645 tons H 2 O          1 ton  1 kg   18.02 g H 2O  1 mol H 2O  = –1.119574x108 kJ= –1.12x108 kJ 11.121 Plan: At the boiling point, the vapour pressure equals the atmospheric pressure, so the two boiling points can be used to find the heat of vapourization for amphetamine using the Clausius-Clapeyron equation. Then use the Clausius-Clapeyron equation to find the vapour 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: ln

 vap H  1 p2 =  R  T2 p1



1  T1 

p1 = 101.3 kPa

T1 = 201°C + 273 = 474 K

p2 = 1.733 kPa

T2 = 83°C + 273 = 356 K

 vap H = ? ln

 vap H 1   1 1.733 kPa  =  8.314 J/mol•K  356 K 474 K  101.3 kPa

–4.045413384 = –  vap H (0.000084109 mol/J)

 vap H = (–4.045413384)/(–0.000084109 mol/J) = 48,097.271 J/mol ln

 vap H  1 p2 =   R  T2 p1



1  T1 

p1 = 1.733 kPa

T1 = 83°C + 273 = 356 K

p2 = ?

T2 = 20°C + 273 = 293 K

 vap H = 48,097.271 J/mol ln

p2 1  = 48, 097.271 J/mol  1 = –3.4940841  1.733 kPa 8.314 J/mol•K  293 K 356 K 

p2 = 0.0303766 1.733 kPa

p2 = (0.0303766)(1.733 kPa) = 0.05264 kPa

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.





 0.05264 kPa  1 m3  1 L  pV Moles = n = =  3 3  = 0.02160904 mol amphetamine L•kPa  RT   10 m   8.31446 mol•K   293 K    Mass (g) =  0.02160904 mol   135.20 g amphetamine  = 2.9215422 g/m³= 2.9 g/m3 

11.122 a) Unit cell edge = 3

1 mol



 cm3  12.01 g C   1 mol C   8 C atoms    23 3.52 g  1 mol C   6.022x10 C atoms 

= 3.5654679x10–8 cm= 3.57x10–8 cm





 3.5654678x108 cm 3  23   3.01 g C   1 mol C   6.022x10 C atoms           cm3  12.01 g C  1 unit cell 1 mol C       

b) C atoms/unit cell = 

= 6.8409 atoms/unit cell= 6.84 C atoms/unit cell 11.123 No, filling all the available holes (8) in the face-centered cubic lattice leads to a stoichiometry of 2:1 (8 holes/4 atoms). 11.124 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. 11.125 a) Determine the vapour pressure of ethanol in the bottle at –11°C by applying the Clausius-Clapeyron equation. The boiling point of ethanol is 78.5°C at a pressure of 101.3 kPa.  vap H (40.5 kJ/mol ) is given in Figure 11.2. ln

 vap H  1 p2 =  R  T2 p1



1  T1 

p1 = ?

T1 = (273 + (–11°C) = 262 K

p2 = 101.3 kPa

T2 = (273.15 + 78.5°C) = 351.6 K

 vap H = 40.5 kJ/mol 40.5 kJ/mol  1 1   10 J  101.3kPa  =   = 4.7380851  8.314 J/mol•K  351.6 K 262 K   1 kJ  p1 3

101.3 kPa = 114.2153 p1

p1 = 0.88692 kPa = 0.887 kPa Note: The pressure should be small because not many ethanol molecules escape the liquid surface at such a cold temperature. Determine the amount 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. pV  0.88692 kPa  4.7 L = 0.00191358 mol C H O n= = 2 6 RT L•kPa    8.31446 mol•K   262 K    Convert moles of ethanol to mass of ethanol using the molar mass (M = 46.07 g/mol). Mass (g) of C2H6O = (0.00191358 mol C2H6O)(46.07 g/mol C2H6O) = 0.0881587 g = 8.8x10–2 g C2H6O b) ln

 vap H  1 p2 =  R  T2 p1



1  T1 

p1 = ?

T1 = (273 + 20.°C) = 293 K

p2 = 101.3 kPa

T2 = (273.15 + 78.5°C) = 351.6 K

 vap H = 40.5 kJ/mol 40.5 kJ/mol  1 1   10 J  101.3kPa  =   = 2.7709337  8.314 J/mol•K  351.6 K 293 K   1 kJ  p1 3

101.3 kPa = 15.97354 p1

p1 = 6.3417 kPa = 6.34 kPa 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. pV  6.3417 kPa  4.7 L  = 0.01223494s mol C H O n= = 2 6 RT L•kPa    8.31446 mol•K   293 K    Convert moles of ethanol to mass of ethanol using the molar mass (M = 46.07 g/mol). Mass (g) of C2H6O = (0.01223494 mol C2H6O)(46.07 g/mol C2H6O) = 0.5636637 g = 0.56 g C2H6O The mass of ethanol present in the vapour, if excess liquid was present, is 0.56 g. Since this exceeds the 0.33 g available, all of the ethanol will vapourize. c) 0.0°C = (273.2 + 0.0) K = 273.2 K

40.5 kJ/mol  1 1   10 J  p2  =   = –3.975864  8.314 J/mol•K  273.2 K 351.6 K   1 kJ  101.3 kPa 3

p2 = 0.0187631 101.3 kPa p2 = (0.018763)(101.3 kPa) = 1.90069 kPa

pV = RT 

1.90069 kPa  4.7 L 

= 0.00393273 mol C2H6O L•kPa  8.31446 273.2 K    mol•K   Convert moles of ethanol to mass of ethanol using the molar mass (M = 46.07 g/mol). Mass C2H6O = (0.00393273 mol C2H6O)(46.07 g/mol C2H6O) = 0.18118 g C2H6O Mass (g) of ethanol in liquid = mass (g) of ethanol (total) – mass (g) of ethanol in vapour Mass (g) of ethanol in liquid = 0.33 g – 0.18118 g = 0.14882 g = 0.15 g C2H6O 11.126 Plan: The equation q = c x mass x T 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 vapour 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 vapourization is used to calculate the heat involved in the phase change of liquid A to A as a vapour. Solution: a) To heat the substance to the melting point the substance must be warmed from –40.°C to –20.°C: q = c x mass x T = (1.0 J/g•K) (25 g) [(273+(–20))K –(273+ (–40.))K] = 500 J Time (min) required =  500 J   1 min  = 1.1111 min = 1.1 min  450 J 

b) Phase change of the substance from solid at –20.°C to liquid at –20.°C:



q = n  fus H

 180. J 

 =  25 g   1 g  = 4500 J

Time (min) required =  4500 J   1 min  = 10. min  450 J 

c) To heat the substance to the melting point the substance must be warmed from –20.°C to 85°C: q = c x mass x T = (25 g)(2.5 J/g•K)[(273+85)K – (273+(–20.))K] = 6562.5 J

Time (min) required =  6562.5 J   1 min  = 14.583 min = 15 min  450 J 

To boil the sample at 85°C:



q = n  vap H

 500. J 

 =  25 g   1 g  = 12,500 J

 1 min  Time (min) required = 12,500 J    = 27.78 min = 28 min  450 J  Warming the vapour from 85°C to 100.°C: q = c x mass x T = (25 g)(0.5 J/g•K)[(273+100)K –(273+ 85)K] = 187.5 J

Time (min) required = 187.5 J   1 min  = 0.417 min = 0.4 min  450 J 

11.127 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.  103 kg  103 g   1 mol B(OH)3  Moles of B(OH)3 = 1.00 t B(OH)3         1t  1 kg   61.83 g B(OH)3  = 1.6173379x104 mol B(OH)3

 3  1L V (L) = 12.5 m  3 3  = 12,500 L  10 m 





3.07x10 kPa  12,500 L = 1.6783466x10 mol NH Moles of NH = pV/RT = 3

L•kPa    8.31446 mol•K   275 K    B(OH)3 is the limiting reactant.

1 mol BN   24.82 g BN   74.214%    1 mol  B(OH)   1 mol BN   100% 



4 Mass (g) of BN = 1.6173379x10 mol B(OH)3 

= 2.97912x105 g = 2.98x105 g BN 11.128 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.

11.129 a) ln

 vap H  1 p2 =  R  T2 p1



1  T1 

p1 = 101.3 kPa

T1 = 273.2 + (–23.7°C) = 249.5

K p2 = 53.3 kPa

T2 = 273.2 + (–37.8°C) = 235.4 K

 vap H = ? ln

 vap H 1 1   53.3 kPa  =  8.314 J/mol•K  235.4 K 249.5 K  101.3 kPa

–0.6421500 = –  vap H (0.0000288787 mol/J)

 vap H = (–0.6421500/–0.0000288787 mol/J)(1 kJ/103 J) = 22.2361 kJ/mol = 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) CH3OCH3(g)  CH3OCH3(l)

 f H = –185.4 kJ/mol

Overall: 2C(s) + 3H2(g) + 1/2O2(g)  CH3OCH3(l)

 f H = –207.6 kJ/mol

= –22.2 kJ/mol

11.130 Plan: A body-centered cubic unit cell has eight corner atoms; 8 atoms x 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: Mass (u) of two Na atoms =  2 Na atoms   22.99 u  = 45.98 u.  1 Na atom 

11.131 Volume (L) = (

)(

) = 786.6667 L = 8x102 L

Note: 22.7 L/1 mol is the volume of 1 mole of an ideal gas (equation 4.6).

CHAPTER 12 THE PROPERTIES OF MIXTURES: SOLUTIONS AND COLLOIDS CHEMICAL CONNECTIONS BOXED READING PROBLEMS B12.1

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 Ca 2+ 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. B12.2

Plan: Osmotic pressure is calculated from the concentration of particles in mol/L, the gas constant, and temperature. Convert the mass of sucrose to moles using the molar mass, and then to concentration (mol/L). Sucrose is a nonelectrolyte so i = 1. Solution: T = 273 + 20.°C = 293 K  1 mol sucrose  Moles of sucrose =  3.55 g sucrose    = 0.01037102 mol sucrose  342.30 g sucrose  moles of sucrose 0.01037102 mol = = 1.037102x10–2 mol/L sucrose volume of solution 1.0 L  = icRT = (1)(1.037102x10–2 mol/L)(8.31446 L•kPa/mol•K)(293 K) = 25.26522 kPa = 25.3 kPa A pressure greater than 25.3 kPa must be applied to obtain pure water from a 3.55 g/L solution.

Concentration (mol/L) =

END–OF–CHAPTER PROBLEMS 12.1

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.

12.2

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 outermost shell. Additional shells are held by hydrogen bonding to inner shells.

12.3

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.

12.4

Sodium octadecanoate (stearate) would be a more effective soap because the hydrocarbon chain in the stearate ion is longer than the chain in the ethanoate (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.

12.5

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.‖

12.6

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. In addition, HCl and water are both polar molecules, while C3H8 is a non-polar molecule. Since, ―like dissolves like‖, polar molecules have greater solubility in polar solvent, while non-polar molecules have greater solubility in non-polar solvent.

12.7

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 CCl 4. Because potassium nitrate dissolves to a greater extent in water, a) KNO3 in H2O will result in the more concentrated solution.

12.8

b) Octadecanoic acid (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.

12.9

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 hydrogen 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.

12.10

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 hydrogen bonds). Next in strength is ion–induced dipole force and then dipole–induced dipole force. The weakest intermolecular interactions are dispersion forces. Solution: a) metallic bonding is the strongest intermolecular force of Cu(s) in Ag(s). b) dipole-dipole is the strongest intermolecular force of CH3Cl(g) in CH3OCH3(g). c) dipole–induced dipole is the strongest intermolecular force of CH3CH3(g) in CH3CH2CH2NH2(l).

12.11

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 hydrogen bonds). Next in strength is ion–induced dipole force and then dipole–induced dipole force. The weakest intermolecular interactions are dispersion forces. Solution: a) Hydrogen bonding occurs between the H atom on water and the lone electron pair on the O atom in methoxymethane (CH3OCH3). However, none of the hydrogen atoms on methoxymethane 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.

12.12

12.13 Plan: Ethoxyethane (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 ethoxyethane. 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 ethoxyethane are more similar to the dipole forces in ethoxyethane than the ion-dipole forces between NaCl and ethoxyethane. b) CH3CHO (ethanal, also known as acetaldehyde) would be more soluble. The dominant interactions in H 2O are hydrogen bonding, a stronger type of dipole-dipole force. The dominant interactions in CH3CHO are dipoledipole. The solute-solvent interactions between CH3CHO and ethoxyethane are more similar to the solvent intermolecular forces than the forces between H2O and ethoxyethane. 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 ethoxyethane would interact with dispersion forces, while the polar end of CH 3CH2MgBr and the dipole in ethoxyethane would interact with dipole-dipole forces. 12.14

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.

12.15

No, river water is a heterogeneous mixture, with its composition changing from one segment to another.

12.16

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 hydrogen 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.

12.17

The solventH and mixH 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.

12.18

For a general solvent, the energy changes needed to separate solvent into particles (solventH), and that needed to mix the solvent and solute particles (mixH) would be combined to obtain solutionH.

12.19

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 hydrationH. hydrationH increases with increasing charge and decreases with increasing size.

12.20

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. solutionH = latticeH +hydration of ionsH For a heat of solution to be zero (or very small) latticeH hydration of ionsH, and they would have to have opposite signs.

12.21

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.

12.22

This compound would be very soluble in water. A large exothermic value in solutionH (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 solutionH values (and higher energy states) are still soluble in water.

12.23

Plan: solutionH = latticeH +hydrationH. 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. Solution:

solutionH > 0 (endothermic) 12.24

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.

12.25

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.

12.26

a) I has a smaller charge density (larger ion volume) than Br–. b) Ca2+ has a lower ratio than Sc3+, due the smaller ion charge of Ca2+. – – c) Br has a lower ratio than K+, due to the larger ion volume of Br . – – 2– d) Cl has a lower ratio than S , due to the smaller ion charge of Cl . 3+ 3+ e) Sc has a lower ratio than Al , due to the larger ion volume of Sc3+. 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 and larger ionic radius.

12.27

Plan: The ion with the greater charge density will have the larger hydrationH. Solution: a) Na+ would have a larger ΔhydrationH than Cs+ since its charge density is greater than that of Cs +. b) Sr2+ would have a larger ΔhydrationH than Rb+. c) Na+ would have a larger ΔhydrationH than Cl–. d) O2– would have a larger hydrationH than F–. e) OH– would have a larger hydrationH than SH–. f) Mg2+ would have a larger hydrationH than Ba2+. g) Mg2+ would have a larger hydrationH than Na+. h) CO32– would have a larger hydrationH than NO3–.

12.28

a) I

12.29

Plan: Use the relationship solutionH = latticeH +hydrationH. Given solutionH and latticeH, hydrationH can be calculated. hydrationH 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–. solutionH = latticeH +hydrationH hydrationH = solutionH – latticeH = 41.1 kJ/mol – 745 kJ/mol = –703.9 kJ/mol = –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.

12.30

a) hydrationH = solutionH – latticeH hydrationH = 17.3 kJ/mol – 763 kJ/mol hydrationH = – 745.7 kJ/mol = – 746 kJ/mol b) Na+ ion contributes more to the heat of hydration due to its smaller size (larger charge density).

–

b) Ca2+ c) Br

–

d) Cl

e) Sc3+

f) ClO4–

g) Fe2+

h) K+

12.31

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.

12.32

a) Entropy increases. b) Entropy decreases. c) Entropy increases.

12.33

solutionH = latticeH + hydrationH solutionH = 822 kJ/mol + (– 799 kJ/mol) solutionH = 23 kJ/mol

12.34

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.

12.35

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.

12.36

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 (solnH < 0).

12.37

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.

12.38

a) increase

12.39

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 O 2 to mass of O2 using the molar mass. Solution: a) Sgas = kH  pgas mol   –3 Sgas = 1.26x105  101.3 kPa  = 1.28x10 mol/L L•kPa  

b) stay the same

 1.28x103 mol O2   32.0 g O2  Mass (g) of O2 =      2.50 L  = 0.1024 g = 0.102 g O2  L    1 mol O2 

b) The mass of gas that will dissolve in a given volume decreases proportionately with the partial pressure of the gas. mol   Sgas = 1.26x105  21.2 kPa  = 2.6712x10–4 mol/L L•kPa  

 2.6712x104 mol O2   32.0 g O 2  Mass (g) of O2 =      2.50 L  = 0.0213696 g = 0.0214 g O2  L    1 mol O2  12.40

mol    0.93%  –5 –5 Solubility = 1.5x105  101.3 kPa   100%  = 1.413x10 mol/L = 1.4x10 mol/L L•kPa    

12.41

The solution is saturated.

12.42

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. Solution: Sgas = kH  pgas = (3.7x10–4 mol/L•kPa)(557.2 kPa) = 0.2061 mol/L = 0.21 mol/L

12.43

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.

12.44

Concentration (mol/L) is defined as the amount of solute (moles) dissolved in one litre of solution. Molality is defined as the amount of solute (moles) dissolved in 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.

12.45

Plan: Refer to the table of concentration definitions for the different methods of expressing concentration. Solution: a) Concentration (mol/L) 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.

12.46

No, 21 g solute/kg of solvent would be 21 g solute/1.021 kg solution.

12.47

Converting between concentration (mol/L) 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.

12.48

% w/w, mole fraction, and molality are weight-to-weight relationships that are not affected by changes in temperature. % w/v and concentration (mol/L) are affected by changes in temperature, because the volume is temperature dependant.

12.49

Plan: The concentration (mol/L) is the amount of solute (moles) in each litre of solution: c =

mol of solute . V(L) of solution

Convert the masses to moles and the volumes to litres and divide moles by volume. Solution:  32.3 g C12 H 22 O11   1 mL   1 mol C12 H 22 O11  a) Concentration =     3   100. mL    10 L   342.30 g C12 H 22 O11  = 0.943617 mol/L= 0.944 mol/L C12H22O11

 5.80 g LiNO3   1 mL   1 mol LiNO3  b) Concentration =   = 0.166572 mol/L = 0.167 mol/L LiNO3   3    505 mL   10 L   68.95 g LiNO3 

12.50

 0.82 g C2 H5OH   1 mL   1 mol C2 H5OH  a) Concentration =   = 1.69514 mol/L = 1.7 mol/L C2H5OH   3   10.5 mL    10 L   46.07 g C2 H5OH   1.27 g NH3   1 mL   1 mol NH 3  b) Concentration =   = 2.2261 mol/L = 2.23 mol/L NH3   3    33.5 mL   10 L   17.03 g NH3 

12.51

Plan: Dilution calculations can be done using cconcVconc = cdilVdil. Solution: a) cconc = 0.240 mol/L NaOH Vconc = 78.0 mL cdil = ? Vdil = 0.250 L 3   0.240 mol / L 78.0 mL c V   10 L  conc  conc  =  cdil =   = 0.07488 mol/L = 0.0749 mol/L  0.250 L  Vdil   1 mL  b) cconc = 1.2 mol/L HNO3 Vconc = 38.5 mL cdil = ? Vdil = 0.130 L  3  cconc Vconc  1.2 mol / L  38.5 mL  10 L  cdil = =   = 0.355385 mol/L = 0.36 mol/L Vdil   0.130 L   1 mL 

12.52

Plan: Dilution calculations can be done using cconcVconc = cdilVdil Solution: a) cconc = 6.25 mol/L HCl Vconc = 25.5 mL cdil = ? Vdil = 0.500 L 3   6.25 mol / L 25.5 mL c V   10 L  conc  conc  =  cdil =   = 0.31875 mol/L = 0.319 mol/L Vdil   0.500 L   1 mL  b) cconc = 2.00x10–2 mol/L KI Vconc = 8.25 mL cdil = ? Vdil = 12.0 mL 2  cconc Vconc   2.00x10 mol / L  8.25 mL  cdil = = = 0.01375 mol/L = 0.0138 mol/L 12.0 mL  Vdil 

12.53

Plan: For part a), find the amount of KH2PO4 (moles) needed to make 365 mL of a solution of this concentration (mol/L). Convert moles to mass using the molar mass of KH 2PO4. For part b), use the relationship cconcVconc = cdilVdil to find the volume of 1.25 mol/L NaOH needed. Solution:  103 L   8.55x102 mol KH 2 PO4  a) Moles of KH2PO4 =  365 mL    = 0.0312075 mol  1 mL   L   

 136.09 g KH 2 PO 4  Mass (g) of KH2PO4 =  0.0312075 mol KH 2 PO 4    = 4.24703 g = 4.25 g KH2PO4  1 mol KH 2 PO 4  Add enough water to 4.25 g KH2PO4 to make 365 mL of aqueous solution. b) cconc = 1.25 mol/L NaOH Vconc = ? cdil = 0.335 mol/L NaOH Vdil = 465 mL  cdil Vdil   0.335 mol / L  465 mL Vconc = = = 124.62 mL= 125 mL 1.25 mol/L   cconc  Add enough water to 125 mL of 1.25 mol/L NaOH to make 465 mL of solution. 12.54

a) Find the amount of NaCl (moles) 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     1 mol NaCl  = 94.965 g = 95 g NaCl L    Add enough water to 95 g NaCl to make 2.5 L of aqueous solution.

b) Use the relationshipcconcVconc = cdilVdil to find the volume of 2.1 mol/L urea needed. cconc = 2.1 mol/L urea Vconc = ? cdil = 0.3 mol/L urea Vdil = 15.5 L cdil Vdil   0.3 mol/L 15.5 L   Vconc = = = 2.21429 L = 2 L  2.1 mol/L   cconc  Add enough water to 2 L of 2.1 mol/L urea to make 15.5 L of solution. Note because of the uncertainty in the concentration of the dilute urea (0.3 mol/L), only one significant figure is justified in the answer. 12.55

Plan: To find the mass of KBr needed in part a), find the moles of KBr in 1.40 L of a 0.288 mol/L 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 cconcVconc = cdilVdil and solve for Vconc. Solution:  0.288 mol KBr  a) Moles of KBr = 1.40 L    = 0.4032 mol L  

 119.00 g KBr  Mass (g) of KBr =  0.4032 mol    = 47.9808 g = 48.0 g KBr  1 mol KBr  To make the solution, weigh 48.0 g KBr and then dilute to 1.40 L with distilled water. b) cconc = 0.264 mol/L LiNO3 Vconc = ? cdil = 0.0856 mol/L LiNO3 Vdil = 255 mL cdil Vdil    0.0856 mol/L  255 mL Vconc = = = 82.68182 mL = 82.7 mL  0.264 mol/L   cconc  To make the 0.0856 mol/L solution, measure 82.7 mL of the 0.264 mol/L solution and add distilled water to make a total of 255 mL. 12.56

a) To find the mass of Cr(NO3)3 needed, find the moles of Cr(NO3)3 in 57.5 mL of a 1.53x10–3 mol/L solution and convert to grams using molar mass of Cr(NO3)3.  103 L   1.53x103 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 g = 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 5.8x103 m3 use cconcVconc = cdilVdil and solve for Vconc. cconc = 2.50 mol/L NH4NO3 Vconc = ? cdil = 1.45 mol/L NH4NO3 Vdil = 5.8x103 m3

 cdil Vdil  1.45 mol/L   5.8x10 m  Vconc = = = 3.364x103 m³ = 3.4x103 m3  2.50 mol/L   cconc  3

To make the 1.45 mol/L solution, measure 3.4x103 m3 of the 2.50 mol/L solution and add distilled water to make 5.8x103 m3. 12.57

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  a) Moles of glycine = 85.4 g glycine   = 1.137605 mol  75.07 g glycine  1.137605 mol glycine m glycine = = 0.895752 mol/kg= 0.896 mol/kg glycine 1.270 kg

 1 mol glycerol  b) Moles of glycerol = 8.59 g glycerol   = 0.093278 mol  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 mol/kg = 1.21 mol/kg glycerol 0.0770 kg 12.58

Molality = moles solute/kg solvent.  1 mol HCl  174 g HCl   3  36.46 g HCl   10 g  = 6.3043 mol/kg= 6.30 mol/kg HCl a) m HCl =    757 g   1 kg   1 mol naphthalene  16.5 g naphthalene   3  128.16 g naphthalene   10 g  = 2.41548 mol/kg b) m naphthalene =    53.3 g   1 kg 

= 2.42 mol/kg naphthalene 12.59

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  Mass (g) of benzene =  44.0 mL C6 H6    = 38.588 g benzene  1 mL  Plan: Molality, m =

 1 mol C6 H 6  Moles of benzene =  38.588 g C6 H 6    = 0.49402 mol benzene  78.11 g C6 H 6 

 0.660 g   1kg  Mass (kg) of hexane = 167 mL C6 H14    = 0.11022 kg hexane   mL   103 g  m=

12.60

 0.49402 mol C6 H 6  moles of solute = = 4.48213 mol/kg = 4.48 mol/kg C6H6 kg of solvent  0.11022 kg C6 H14 

Molality = moles solute/kg solvent.

1.59 g   1 mol CCl4     mL   153.81 g CCl4   103 g  Molality of CCl4 =   1.33 g   1 kg   76.5 mL CH 2Cl2     mL  = 0.2702596 mol/kg = 0.270 mol/kg CCl4

 2.66 mL CCl4  

12.61

Plan: In part a), the total mass of the solution is 3.10x102 g, so masssolute + masssolvent = 3.10x102 g. Assume that you have 1000 g of the solvent water and find the mass of C2H6O2 needed to make a 0.125 mol/kg solution. Then a ratio can be used to find the mass of C2H6O2 needed to make 3.10x102 g of a 0.125 mol/kg 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 C2 H6 O2   62.07 g C2 H6 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 C2 H6 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 H 6O2  2 Mass (g) of C2H6O2 for 3.10x102 g of solution =   3.10x10 g solution 1007.75875 g solution   = 2.386695 g C2H6O2 Masssolvent = 3.10x102 g – masssolute = 3.10x102 g – 2.386695 g C2H6O2 = 307.613305 g = 308 g H2O Therefore, add 2.39 g C2H6O2 to 308 g of H2O to make a 0.125 mol/kg 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



0.0264 kg 100 = mass % 52.0% = 0.050769 kg = 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 kg = 1.15 kg Add 0.0508 kg of the 52.0% (w/w) HNO 3 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 HNO 3 =

12.62

mass of solute 100

a) The total weight of the solution is 1.50 kg, so masssolute + masssolvent = 1.50 kg

 0.0355 mol C2 H5 OH   46.07 g C2 H5 OH  Mass (g) of C2H5OH in 1000 g (1 kg) of H2O = 1 kg H 2 O     1 kg H 2 O   1 mol C2 H5 OH  = 1.635485 g C2H5OH Mass (g) of this solution = 1000 g H2O + 1.635485 g C2H5OH = 1001.635485 g  1.635485 g C2 H 5OH  Mass (g) of C2H5OH for 1.50 kg of solution =   1500 g solution   1001.635485 g solution  = 2.449222 g = 2.45 g C2H5OH Masssolvent = 1500 g – masssolute = 1500 g – 2.449222 g C2H5OH = 1497.551 g = 1498 g H2O Therefore, add 2.45 g C2H5OH to 1498 g of H2O to make a 0.0355 mol/kg 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 Mass % = 100 mass of solution mass of solute 100  57.85 g 100  = Mass of 34.1% solution containing 57.85 g HCl = = 169.6481g = 170. g mass % 34.1% Mass of water added = mass of 13.0% solution – mass of 34.1% solution = 445 g – 169.6481 g = 275.35191 g = 275 g Add 170. g of the 34.1% (w/w) HCl to 275 g H 2O

12.63

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 propan-2-ol (isopropanol) per total moles. moles of propan-2-ol 0.35 mol propan-2-ol Xpropan-2-ol = = = 0.2916667 = 0.29 moles of isopropanol + moles of water  0.35  0.85 mol (Notice that mole fractions have no units.) mass of solute b) Mass percent = 100  . From the mole amounts, find the masses mass of solution of propan-2-ol and water:  60.09 g C3 H 7 OH  Mass (g) of propan-2-ol =  0.35 mol C3 H 7 OH    = 21.0315 g propan-2-ol  1 mol C3 H 7 OH 

 18.02 g H 2 O  Mass (g) of water =  0.85 mol H 2 O    = 15.317 g water  1 mol H 2 O  21.0315 g propan-2-ol mass of solute Mass percent = 100  = 57.860710 % = 58% 100  = mass of solution  21.0315  15.317  g c) Molality of propan-2-ol is moles of propan-2-ol per kg of water. 0.35 mol propan-2-ol  103 g  moles of solute m= =   = 22.85043 mol/kg = 23mol/kg propan-2-ol kg of solvent 15.317 g water  1 kg  12.64

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  5.844 g NaCl  x100% = 3.63396677 % = 3.63% NaCl Mass percent NaCl =  5.844  154.972  g c) Molality of NaCl is moles of NaCl per kg of solvent. 0.100 mol NaCl  103 g  Molality NaCl =   = 0.645277856 mol/kg = 0.645mol/kg NaCl 154.972 g water  1 kg 

12.65

moles of solute . 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  Mass (g) of water =  0.500 L   3   = 0.500 kg   10 L  1 mL   103 g  moles of solute m= kg of solvent Plan: Molality =

moles CsBr 0.500 kg H 2 O Moles of CsBr = (0.400 mol/kg)(0.500 kg H2O) = 0.200 mol  212.8 g CsBr  Mass (g) of CsBr =  0.200 mol CsBr    = 42.56 g = 42.6 g CsBr  1 mol CsBr  0.400 mol/kg CsBr =

 103 g  Mass (g) of water =  0.500 kg   = 500. g water  1 kg    Mass of solution = mass (g) H2O + mass of CsBr = 500. g H2O + 42.56 g CsBr = 542.56 g  1 mol H 2 O  Moles of H2O =  500. g H 2 O    = 27.74695 mol H2O  18.02 g H 2 O 

0.2000 mol CsBr mol CsBr = = 7.156x10–3 = 7.16x10–3 mol CsBr + mol H2 O  0.2000  27.74695 mol

XCsBr =

Mass percent CsBr =

12.66

The density of water is 1.00 g/mL. The mass of water is: Mass (g) of water = (0.400 L)(1 mL/10 –3 L)(1.00 g/mL) = 4.00x102 g H2O Moles of H2O = (400. g H2O)(1 mol H2O/18.02 g H2O) = 22.197558 mol H2O Moles of KI = (0.30 g KI)(1 mol KI /166.0 g KI) = 1.80723x10 –3 mol KI XKI =



1.80723x103 mol KI



1.80723x103  22.197558 mol

Mass percent KI =

12.67

 42.56 g CsBr  mass of CsBr x 100% = 7.84429 % = 7.84% CsBr 100  = mass of solution  42.56  500. g

= 8.14091x10–5 = 8.1x10–5

 0.30 g KI  x 100% = 0.07494 % KI = 0.075% KI  0.30  400. g

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 concentration (mol/L), you will need the volume of solution. Use the density of the solution to convert the 100. g of solution to volume in litres; divide moles of NH 3 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  Moles of NH3 =  8.00 g NH3    = 0.469759 mol NH3  17.03 g NH3   1 mol H 2 O  Moles of H2O =  92.00 g H 2 O    = 5.1054 mol H2O  18.02 g H 2 O 

 1 mL solution   103 L  Volume (L) of solution = 100.00 g solution    = 0.103616 L    0.9651 g solution   1 mL  Using the above fundamental quantities and the definitions of the various units:

Molality = m =

 0.469759 mol NH3  moles of solute =  = 5.106076 mol/Kg = 5.11mol/kg NH3 kg of solvent  0.09200 kg H 2 O 

Concentration = c =

 0.469759 mol NH3  moles of solute =   = 4.53365 mol/L= 4.53 mol/L NH3 0.103616 L L of solution  

Mole fraction = X =

0.469759 mol NH3 moles of NH3 = = 0.084259 = 0.0843 0.469759  5.1054 mol total moles 

12.68

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:  0.1775586 mol FeCl3   103 g  Molality = m = moles solute/kg solvent =   = 2.49380 mol/kg   71.20 g H 2 O    1 kg  = 2.49 mol/kg FeCl3 0.1775586 mol FeCl3 Concentration = c = moles solute/L solution = = 2.272750 mol/L 0.078125 L = 2.27 mol/L FeCl3 0.1775586 mol FeCl3 Mole fraction = X = moles substance/total moles = = 0.043005688 = 0.0430  0.1775586  3.951165  mol

12.69

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 1x10 6. Solution:  1 mL  1.001 g  5 Mass (g) of solution is 100.0 L solution   3   = 1.001x10 g  10 L  1 mL 

 mass solute  x 106 ppm =    mass solution    0.25 g Ca 2+ x 106 = 2.49750 ppm = 2.5 ppm Ca2+ ppm Ca2+ =   1.001x105 g solution      0.056 g Mg 2+ x 106 = 0.5594406 ppm = 0.56 ppm Mg2+ ppm Mg2+ =   1.001x105 g solution    12.70

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. Ethane-1,2-diol, also known as ethylene glycol, will be designated EG. Determine some fundamental quantities: Mass (g) of EG = (1.0000 L EG)(1mL/10 –3 L)(1.114 g EG/mL) = 1.114 x103 g EG Mass (g) of H2O = (1.0000 L H2O)(1mL/10–3 L)(1.00 g H2O/mL) = 1.00x103 g H2O Moles of EG = (1114 g EG)(1 mol EG/62.07 g EG) = 17.94747865 mol EG Moles of H2O = (1.00x103 g H2O)(1 mol H2O/18.02 g H2O) = 55.49389567 mol H2O Volume (L) of solution = (1114 g EG + 1.00x10 3 g H2O)(1 mL/1.070 g)(10–3 L/1 mL)

= 1.97570 L Using the above fundamental quantities and the definitions of the various units: a) Volume percent = (1.0000 L EG/1.97570 L)100% = 50.61497 % = 50.61% v/v b) Mass percent = [(1114 g EG)/(1114 + 1.00x10 3) g]100% = 52.6963 % = 52.7% w/w 17.94747865 mol EG c) Concentration = moles solute/L solution = = 9.08411 mol/L = 9.08 mol/L ethylene 1.97570 L glycol 17.94747865 mol EG  103 g  d) Molality = moles solute/kg solvent =   1.00 x103 g H 2O  1 kg  = 17.94747865 mol/kg EG = 17.9 mol/kg ethylene glycol 17.94747865 mol EG e) Mole fraction = XEG = moles substance/total moles = 17.94747865  55.49389567  mol  = 0.244378 = 0.244 12.71

Colligative properties of a solution are affected by the amount of particles of solute in solution. The density of a solution would be affected by the chemical formula of the solute.

12.72

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.

12.73

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.

12.74

Raoult‘s law states that the vapour pressure of solvent above the solution equals the mole fraction of the solvent times the vapour 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 vapour pressures.

12.75

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.

12.76

Yes, the vapour at the top of the fractionating column is richer in content of the more volatile component.

12.77

The boiling point of a 0.01 mol/kg KF solution is higher than that of 0.01 mol/kg glucose. KF dissociates into ions in water (K+ and F–) while the glucose does not, so the KF produces more particles.

12.78

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 mol/kg NaF solution has a boiling point closer to its predicted value.

12.79

Univalent ions behave more ideally than divalent ions. Ionic strength (which affects ―activity‖ concentration) is greater for divalent ions. Thus, 0.01 mol/kg NaBr has a freezing point that is closer to its predicted value.

12.80

Cyclohexane, with a freezing point depression constant of 20.1°C•kg/mol, would make calculation of molar mass of a substance more accurate, since fT would be greater.

12.81

Plan: Strong electrolytes are substances that produce a large amount 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 NH 4+ and OH– ions. 12.82

a) NaMnO4 b) CH3COOH c) CH3OH d) Ca(C2H3O2)2

strong electrolyte weak electrolyte nonelectrolyte strong electrolyte

12.83

Plan: To count solute particles in a solution of an ionic compound, count the amount of ions per mole and multiply by the amount of moles in solution. For a covalent compound, the amount of particles equals the amount of molecules. Solution;  0.3 mol KBr  2 mol particles  a)   1 mol KBr  1 L  = 0.6 mol of particles L    Each KBr forms one K+ ion and one Br– ion, two particles for each KBr.  0.065 mol HNO3   2 mol particles  b)   1 L  = 0.13 mol of particles  L    1 mol HNO3  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  = 2x10 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 L  = 0.06 mol of particles  L    1 mol C2 H5 OH  Ethanol is not an ionic compound so each molecule dissolves as one particle. The amount of particles (moles) is the same as the amount of moles of molecules, 0.06 mol of particles in 1 L.

12.84

a) (0.02 mol CuSO4/L)(2 mol particles/mol CuSO4)(10–3 L/1 mL)(1 mL) = 4x10–5 mol of particles b) (0.004 mol Ba(OH)2/L)(3 mol particles/mol Ba(OH)2)(10–3 L/1 mL)(1 mL) = 1.2x10–5 mol = 1x10–5 mol of particles c) (0.08 mol C5H5N/L)(1 mol particles/mol C5H5N)(10–3 L/1 mL)(1 mL) = 8x10–5 mol of particles d) (0.05 mol (NH4)2CO3/L)(3 mol particles/mol (NH4)2CO3)(10–3 L/1 mL)(1 mL) = 1.5x10–4 mol = 2x10–4 mol of particles

12.85

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  a) Molality of CH3OH =  = 3.4332085 mol/kg   100. g H2 O   32.04 g CH3OH   1 kg  =3.43 mol/kg CH3OH

Molality of CH3CH2OH =

 22.0 g CH3CH 2OH   1 mol CH3CH 2OH   103 g      200. g H2 O   46.07 g CH3CH 2OH   1 kg 

= 2.387671 mol/kg = 2.39 mol/kg CH3CH2OH The molality of methanol, CH3OH, in water is 3.43 mol/kg whereas the molality of ethanol, CH3CH2OH, in water is 2.39 mol/kg. Thus, CH3OH/H2O solution has the lower freezing point.  20.0 g H 2 O   1 mol H 2 O  b) Molality of H2O =   = 1.10988 mol/kg H2O = 1.11 mol/kg H2O 1.00 kg CH3OH   18.02 g H 2O   20.0 g CH3CH 2OH   1 mol CH3CH 2 OH  = 0.434122 mol/kg Molality of CH3CH2OH =   1.00 kg CH3OH   46.07 g CH3CH 2OH  = 0.434 mol/kg CH3CH2OH The molality of H2O in CH3OH is 1.11 mol/kg, whereas CH 3CH2OH in CH3OH is 0.434 mol/kg. Therefore, H2O/CH3OH solution has the lower freezing point. 12.86

The magnitude of boiling point elevation is directly proportional to molality.  38.0 g C3 H8 O3   1 mol C3H8O3   103 g  a) Molality of C3H8O3 =  = 1.650559 mol/kg    250. g ethanol  92.09 g C3 H8 O3   1 kg  = 1.65 mol/kg C3H8O3  38.0 g C2 H6 O2   1 mol C2 H6 O2   103 g  Molality of C2H6O2 =  = 2.44885 mol/kg=    250. g ethanol   62.07 g C2 H6 O2   1 kg  = 2.45 mol/kg C2H6O2 The molality of C2H6O2, in ethanol is 2.45mol/kg whereas the molality of C3H8O3, in ethanol is 1.65 mol/kg. Thus, C2H6O2/ethanol solution has the higher boiling point. 15 g C2 H6 O2   1 mol C2 H6 O2  b) Molality of C2H6O2 =   = 0.4833253 mol/kg  0.50 kg H2 O   62.07 g C2 H6 O2  = 0.48mol/kg C2H6O2 15 g NaCl   1 mol NaCl  Molality of NaCl = = 0.513347 mol/kg = 0.51 mol/kg NaCl  0.50 kg H 2 O   58.44 g NaCl  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 mol/kg particles The molality of C2H6O2 in H2O is 0.48 mol/kg, whereas NaCl in H2O is 1.0 mol/kg. Therefore, NaCl/H2O solution has the higher boiling point.

12.87

Plan: To rank the solutions in order of increasing osmotic pressure, boiling point, freezing point, and vapour 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 vapour pressure at a given temperature. Solution:

 2 mol particles  (I)  0.100 mol/kg NaNO3    = 0.200 mol/kg ions  1 mol NaNO3  NaNO3 consists of Na+ ions and NO3– ions, two particles for each NaNO3.  1 mol particles  (II)  0.100 mol/kg glucose   = 0.100 mol/kg molecules  1 mol glucose  Glucose is not an ionic compound so each molecule dissolves as one particle. The amount of particles (moles) is the same as the amount of moles of molecules.  3 mol particles  (III)  0.100 mol/kg CaCl2    = 0.300 mol/kg ions  1 mol CaCl2  Silberberg, Amateis, Venkateswaran, Chen, Chemistry: The Molecular Nature of Matter and Change, 3rd Canadian Edition Page 3-412 Instructor‘s Solution Manual © Copyright 2021 McGraw-Hill Ryerson Ltd.

CaCl2 consists of Ca+2 ions and Cl– ions, three particles for each CaCl2. a) Osmotic pressure: II < I < III b) Boiling point: bpII < bpI < bpIII c) Freezing point: fpIII < fpI < fpII d) Vapour pressure at 50°C: vpIII < vpI < vpII 12.88

I 0.04 mol/kg (H2N)2CO x 1 mol particles/1 mol (H2N)2CO = 0.04 mol/kg molecules II 0.01 mol/kg AgNO3 x 2 mol particles/1 mol AgNO3 = 0.02 mol/kg ions III 0.03 mol/kg CuSO4 x 2 mol particles/1 mol CuSO4 = 0.06 mol/kg ions a) Osmotic pressure: III > I > II b) Boiling point: bpIII > bpI > bpII c) Freezing point: fpII > fpI > fpIII d) Vapour pressure at 298 K: vpII > vpI > vpIII

12.89

Plan: The mole fraction of solvent affects the vapour 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 amount of moles. Multiply the mole fraction of water by the vapour pressure of water to find the vapour pressure of the solution. Solution:  1 mol C3 H8 O3  Moles of C3H8O3 =  34.0 g C3 H8 O3    = 0.369204 mol C3H8O3  92.09 g C3 H8 O3 

 1 mol H 2 O  Moles of H2O =  500.0 g H 2 O    = 27.7469 mol H2O  18.02 g H 2 O  mol H 2 O 27.7469 mol H2 O Xsolvent = = = 0.9868686 mol H 2 O + mol glycerol 27.7469 mol H2 O + 0.369204 mol glycerol psolvent = Xsolventp°solvent = (0.9868686)(3168 Pa) = 3126.4 Pa = 3.13 kPa 12.90

The mole fraction of solvent affects the vapour 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)(5467 Pa) = 5098.74 Pa = 5.10 kPa

12.91

Plan: The change in freezing point is calculated from fT = iKfm, where Kf is 1.86°C•kg/mol 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 fT is calculated, the freezing point is determined by subtracting it from the freezing point of pure water (0.00°C). Solution: fT = iKfm = (1)(1.86°C•kg/mol)(0.251 mol/kg) = 0.46686°C The freezing point is 0.00°C – 0.46686°C = –0.46686°C = –0.467°C.

12.92

bT = iKbm = (1)(0.512°C•kg/mol)(0.200 mol/kg) = 0.1024°C The boiling point is 100.00°C + 0.1024°C = 100.1024°C = 100.°C.

12.93

Plan: The boiling point of a solution is increased relative to the pure solvent by the relationship bT = 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•kg/mol). Solution:  1 mol vanillin  Moles of vanillin =  6.4 g vanillin    = 0.0420665 mol  152.14 g vanillin  Molality of vanillin =

0.042065 mol vanillin  103 g  moles of vanillin =   kg of solvent (ethanol) 50.0 g ethanol  1 kg 

= 0.8413mol/kg vanillin Silberberg, Amateis, Venkateswaran, Chen, Chemistry: The Molecular Nature of Matter and Change, 3rd Canadian Edition Page 3-413 Instructor‘s Solution Manual © Copyright 2021 McGraw-Hill Ryerson Ltd.

bT = iKbm = (1)(1.22°C•kg/mol)(0.8413 kg/mol) = 1.026386°C The boiling point is 78.5°C + 1.026386°C = 79.5264°C = 79.5°C. 12.94

Moles of C10H8 = (5.00 g C10H8)(1 mol C10H8/128.16 g C10H8) = 0.0390137 mol C10H8 C10H8 is a nonelectrolyte so i = 1. Mass = (444 g benzene)(1 kg/103 g) = 0.444 kg benzene Molality = (0.0390137 mol C10H8)/(0.444 kg) = 0.08786869 mol/kg fT = iKfm = (1)(4.90°C•kg/mol)(0.08786869 mol/kg) = 0.43056°C Freezing point = (5.5 – 0.43056)°C = 5.06944°C = 5.1°C

12.95

Plan: The molality of the solution can be determined from the relationship fT = iKf m with the value 1.86°C•kg/mol 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 fT 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: . fT = Tf(solution) – Tf(solvent) = (0.00 – (–24.4))°C = 24.4°C fT = iKf m T m= f = = 13.11828 mol/kg Kf Ethylene glycol (ethane-1,2-diol) will be abbreviated as EG. moles of EG Molality of EG = kg of solvent (water) Moles of EG = molality x kg of solvent = (

)(

)= 190.215054 mol EG

)( Mass (g) of ethylene glycol = ( ) 4 4 = 1.18066x10 g = 1.18x10 g ethylene glycol To prevent the solution from freezing, dissolve a minimum of 1.18x10 4 g ethylene glycol in 14.5 kg water. 12.96

The molality of the solution can be determined from the relationship fT = iKfm with the value 1.86°C•kg/mol inserted for Kf, i = 1 for the nonelectrolyte glycerol, and the given fT of 0 – (–15) = 15°C. m = fT/Kf = (15°C/(1.86°C•kg/mol)) = 8.06452 mol/kg Propane-1,2,3-triol, also known as glycerol, will be abbreviated as GLY.  103 g  1 kg   92.09 g GLY   8.06452 mol GLY  11.0 mg H O Mass (g) of glycerol =        2   3   1 kg H 2 O    1 mg   10 g   1 mol GLY  = 0.0081693 g = 0.0082 g glycerol To prevent the solution from freezing, dissolve a minimum of 0.0082 g glycerol in 11.0 mg water.

12.97

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 T = iKfm to find the van‘t Hoff factor. Kf for water = 1.86°C•kg/mol. Solution: a) Assume exactly 100 g of solution.  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

Molality of NaCl =

0.0171116 mol NaCl  103 g  moles of NaCl =   = 0.172844 mol/kg = 0.173 mol/kg NaCl kg of H 2 O 99.00 g H2 O  1 kg 

fT = Tf(solution) – Tf(solvent) = 0.000°C – (–0.593)°C = 0.593°C fT = iKfm T 0.593 C i= f = = 1.844537 = 1.84 Kf m 1.86 C kg/mol  0.172844 mol/kg 





The value of i should be close to two because NaCl dissociates into two particles when dissolving in water. b) For ethanoic acid (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 CH3COOH  Moles of CH3COOH =  0.500 g CH3COOH    = 0.0083264 mol CH3COOH  60.05 g CH3COOH  Mass (g) of H2O = 100.00 g solution – 0.500 g CH3COOH = 99.500 g H2O Molality of CH3COOH =

moles of CH3COOH kg of H 2 O

0.0083264 mol CH3 COOH  103 g    99.500 g H2 O  1 kg 

= 0.083682 mol/kg = 0.0837mol/kg CH3COOH fT = Tf(solution) – Tf(solvent) = 0.000°C – (–0.159)°C = 0.159°C fT = iKfm T 0.159 C i= f = = 1.02153 = 1.02 Kf m 1.86 C kg/mol  0.083682 mol





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. 12.98

12.99

Convert the mass % to molality and use T = iKfm to find the van‘t Hoff factor. a) Assume exactly 100 g of solution. Thus, the solution contains 0.500 g of KCl in 99.500 g of water.  0.500 g KCl   103 g   1 mol KCl  Molality of KCl =       = 0.067406 mol/kg KCl  99.500 g H 2 O   1 kg   74.55 g KCl  Calculate fT = 0.000°C – (–0.234)°C = 0.234°C fT = iKfm i = fT/Kfm = (0.234°C)/[(1.86°C•kg/mol)(0.067406 mol/kg)] = 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 H 2SO4 in 99.00 g of water.  1.00 g H 2SO4   103 g   1 mol H2SO4  Molality of H2SO4 =       = 0.10297696 mol/kg H2SO4  99.00 g H2 O   1 kg   98.09 g H2 SO4  Calculate T = 0.000°C – (–0.423)°C = 0.423°C fT = iKfm i = fT/Kfm = (0.423°C)/[(1.86°C•kg/mol)(0.10297696 mol/kg)] = 2.2084 = 2.21 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. Use the osmotic pressure equation ( = icRT) to find the concentration (mol/L) of the solution (assuming i = 1).

27.6 kPa = 0.01113932 mol/L L•kPa  273  25 K 1  8.31446     mol•K   Moles = (0.01113932 mol/L)(100.0 mL)(10 –3 L/1 mL) = 0.001113932 mol Molar mass = (6.053 g)/(0.001113932 mol) = 5.4339x10 3 g/mol= 5.43x103 g/mol

c = /iRT =

12.100 Plan: The mole fraction of solvent affects the vapour pressure according to the equation psolvent = Xsolventp°solvent. Find the mole fraction of each substance by dividing moles of substance by the total amount of moles. Multiply the mole fraction of each compound by its vapour pressure to find the vapour pressure of the compounds above the solution. Solution: moles CH 2 Cl2 1.60 mol X CH 2Cl2 = = = 0.592593 moles CH 2 Cl2 + mol CCl4 1.60 + 1.10 mol

moles CCl4 1.10 mol = = 0.407407 moles CH 2 Cl2 + mol CCl4 1.60 + 1.10 mol pA = XA p°A = (0.592593)(46.9 kPa) = 27.79261 kPa = 27.8 kPa CH2Cl2 = (0.407407)(15.7 kPa) = 6.39629 kPa = 6.40 kPa CCl4

X CCl4 =

12.101 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. 12.102 a) milk - liquid/liquid colloid b) fog - liquid/gas colloid c) shaving cream - gas/liquid colloid 12.103 Brownian motion is a characteristic movement in which the colloidal particles change speed and direction erratically by the motion of the dispersing molecules. 12.104 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. 12.105 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. 12.106 a) Total concentration (mol/L) of ions:  0.010 mol ions  8 spheres     1 sphere   1 mL  Solution A: c =  3  = 3.2 mol/L 25 mL  10 L 

 0.010 mol ions    1 sphere   1 mL   3  = 2.0 mol/L 50 mL  10 L 

10 spheres   Solution B: c=

 0.010 mol ions    1 sphere   1 mL   3  = 1.2 mol/L 100 mL  10 L 

12 spheres   Solution C: c =

b) Concentration (mol/L) of compound:  0.010 mol ions   1 mole compound  8 spheres      1 sphere   2 moles of ions   1 mL  = 1.6 mol/L Solution A: c=  3  25 mL  10 L 

 0.010 mol ions   1 mole compound     1 sphere   2 moles of ions   1 mL   3  = 1.0 mol/L 50 mL  10 L 

10 spheres   Solution B: c=

 0.010 mol ions   1 mole compound     1 sphere   3 moles of ions   1 mL  = 0.40 mol/L Solution C: c=  3  100 mL  10 L  Solution A has the highest concentration (mol/L). c) Molality of compound:  0.010 mol ions   1 mole compound  8 spheres     3  1 sphere   2 moles of ions   1 mL   10 g  = 1.6 mol/kg Solution A: m =     25 mL  1.0 g   1 kg   0.010 mol ions   1 mole compound  10 spheres     3  1 sphere   2 moles of ions   1 mL   10 g  = 1.0 mol/kg Solution B: m =     50 mL  1.0 g   1 kg   0.010 mol ions   1 mole compound  12 spheres     3  1 sphere   3 moles of ions   1 mL   10 g  = 0.40 mol/kg Solution C: m =     100 mL  1.0 g   1 kg  Solution C has the lowest molality. d) Osmotic pressure: assume a temperature of 298 K Solution A:  = icRT = (2)(1.6 mol/L)(8.31446 L•kPa/mol•K)(298 K) = 7928.67 kPa= 7.9x10³ kPa Solution B:  = icRT = (2)(1.0 mol/L)(8.31446 L•kPa/mol•K)(298 K) = 4955.42 kPa = 4.9x10³ kPa Solution C:  = icRT = (3)(0.40 mol/L)(8.31446 L•kPa/mol•K)(298 K) = 2973.25 kPa = 2.9x10³ kPa Solution A has the highest osmotic pressure.

12 spheres  

12.107 Assume exactly 100 g of solution. Mass (g) of glucose = (100 g)(10% glucose/100%) = 10. g glucose Moles of glucose = (10. g glucose)(1 mol glucose/180.16 g glucose) = 0.055506217 mol glucose Mass (g) of water = 100.0 g solution – 10. g glucose = 90. g of water Mass (kg) of water = (90. g H2O)(1 kg/103 g) = 0.090 kg Volume (L) of solution = (100 g)(1 mL/1.039 g)(10 –3 L/1 mL) = 0.096246 L Concentration of glucose = (0.055506217 mol glucose)/(0.096246 L) = 0.57671 mol/L glucose Molality of glucose = (0.055506217 mol glucose)/(0.090 kg) = 0.6167357 mol/kg glucose Glucose is a nonelectrolyte so i = 1. T = (273 + 20) = 293 K fT = iKf m = (1)(1.86°C•kg/mol)(0.6167357 mol/kg) = 1.1471°C Freezing point = (0.00 – 1.1471) = –1.1471 = –1.1°C bT = iKbm = (1)(0.512°C•kg/mol)(0.6167357 mol/kg) = 0.3157687°C Boiling point = (100.00 + 0.3157687) = 100.3157687 °C= 100.32°C  = icRT = (1)(0.57671 mol/L)(8.31446 L•kPa/mol•K)(293 K) = 1404.944 kPa = 1.4x10² kPa

12.108 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. 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 g/cm³ = 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 g/cm³ = 9.05 g/cm3 12.109 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 x 109 1.1x10–2 ppb = mass of seawater 1.1x10–2 ppb =

31.1 g Au x 109 mass seawater

 31.1 g  x 109  = 2.827273x1012 g (with 100% efficiency) Mass (g) of seawater =  2  1.1x10   100%  12 Mass (g) of seawater = 2.827273x1012 g   = 3.46905x10 g seawater (81.5% efficiency) 81.5%  





 1 mL   103 L  9 9 Volume (L) of seawater = 3.46905x1012 g   = 3.384439x10 L = 3.4x10 L   1.025 g 1 mL   





12.110 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. 12.111 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 amount depending on the sizes of the ions) interacting with the ions via hydrogen bonding or other forces. The number of cations is equal to the number of anions present. In other words, the molecular-scale view in ―C‖ is charge balanced, unlike in ―A‖, which is not charge balanced. 12.112 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 concentration (mol/L), 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 mol/kg value cannot be converted to concentration (mol/L). 12.113 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 concentration (mol/L). Solution: 0.0°C:  14.5 mg O2   103 g   1 mol O 2  –4 Moles of O2 =       = 4.53125x10 mol O2 1 kg H O 1 mg 32.00 g O 2 2    

 103 g   1 mL   103 L  Volume (L) of solution = 1 kg   = 1.000130017 L  1kg   0.99987 g   1mL     

moles of O2 4.53125x104 mol = = 4.53066x10–4 mol/L= 4.53x10–4 mol/L O2 L of solution 1.000130017 L 20.0°C:  9.07 mg O2   103 g   1 mol O2  –4 Moles of O2 =       = 2.834375x10 mol O2  1 kg H 2 O   1 mg   32.00 g O2  c=

 103 g   1 mL   103 L  Volume (L) of solution = 1 kg   = 1.001773138 L  1kg   0.99823 g   1mL      c=

moles of O2 2.834375x104 mol = = 2.829358x10–4 mol/L = 2.83x10–4 mol/L O2 L of solution 1.001773138 L

40.0°C:

 6.44 mg O2   103 g   1 mol O2  –4 Moles of O2 =       = 2.0125x10 mol O2 1 kg H O 1 mg 32.00 g O 2 2    

 103 g   1 mL   103 L  Volume (L) of solution = 1 kg   = 1.007820689 L  1kg   0.99224 g   1mL      c=

moles of O2 2.0125x104 mol = = 1.996883x10–4 mol/L= 2.00x10–4 mol/L O2 L of solution 1.007820689 L

12.114 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). 12.115 Plan: First, find the molality from the freezing point depression using the relationship T = 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) fT = iKfm = 0 .0 0 0 °C – ( – 0 .2 0 1 °C) = 0 .2 0 1 °C T 0.201 C m = f = = 0.1080645 mol/kg Kf i 1.86 C kg/mol 1





 1.00 g   1 kg  Mass (kg) of solvent =  25.0 mL    = 0.0250 kg water   1 mL   103 g  moles of solute kg of solvent Moles of solute = (m)(kg solvent) = (0.1080656 mol/kg)(0.0250 kg) = 0.0027016 mol 0.243 g Molar mass = = 89.946698 g/mol = 89.9 g/mol 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 Moles C =  53.31 g C   =2  = 4.43880 mol C; 2.219375  12.01 g C  m=

 1 mol H  Moles H = 11.18 g H    = 11.09127 mol H;  1.008 g H 

11.09127 =5 2.219375

 1 mol O  2.219375 Moles O =  35.51 g O   =1  = 2.219375 mol 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

C H

6  60 min   11 L air   4.0x10 mol CO   28.01 g CO  12.116 Mass (g) of CO = 8.0 h         = 0.59157 g = 0.59 g CO 1 L air  1 h   1 min     1 mol CO 

12.117 No, both are the same because masses are additive.

moles of N 2 total moles 3 moles N 2 Mixture A: X N2 = = 0.38 3 + 2 + 3 moles

12.118 a) X N2 =

4 moles N 2 = 0.33 4 + 3 + 5 moles Mixture C has the smallest mole fraction of N2. 2 moles Ne b) Mixture A: X Ne = = 0.25 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 Cl2 c) Mixture A: X Cl2 = = 0.38 3 + 2 + 3 moles 5 moles Cl2 Mixture C: X Cl2 = = 0.42 4 + 3 + 5 moles Mixture B < Mixture A < Mixture C

Mixture B: X N2 =

4 moles N 2 = 0.40 4 + 4 + 2 moles

Mixture B: X Ne =

4 moles Ne = 0.40 4 + 4 + 2 moles

Mixture B: X Cl2 =

2 moles Cl2 = 0.20 4 + 4 + 2 moles

Mixture C: X N2 =

12.119 Plan: Find the moles of NaF required to form a 4.50x10 –5 mol/L solution of F– with a volume of 5000. L. Then convert moles of NaF to mass. Then use the concentration (mol/L) to find the mass of F – in 2.0 L of solution. Solution:  4.50x105 mol F   1mol NaF  a) Moles of NaF =  5000. L   = 0.225 mol NaF      L    1mol F 

 41.99 g NaF  Mass (g) of NaF =  0.225 mol    = 9.44775 g = 9.45 g NaF  1 mol NaF 

 4.50x105 mol F   19.00 g F  – b) Mass (g) of F– =  2.0 L      = 0.00171g = 0.0017 g F  L    1 mol F 

12.120 a) The solution in U tube B is the most concentrated since it has the highest osmotic pressure. b) Solution C has the smallest amount of dissolved ions and thus the smallest osmotic pressure. 12.121 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)(13.4 kPa) = 7.1154 kPa   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)(1.31 kPa) = 0.61439 kPa Vapour: X (pinene) =

7.1154 kPa = 0.9205165 = 0.921  7.1154  0.61439  kPa

X (terpineol) =

0.61439 kPa = 0.0794834425 = 0.079  7.1154  0.61439  kPa

12.122 Plan: Use the boiling point elevation of 0.45°C to calculate the molality of the solution using the relationship bT = 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) bT = iKbm i = 1 (nonelectrolyte) T = boiling point of solution – boiling point of solvent = (100.45 – 100.00)°C = 0.45°C  T 0.45 C m = b = = 0.878906 mol/kg = 0.878906 mol/kg Kbi 0.512 C kg/mol 1





 0.997 g   1 kg  Mass (kg) of water =  25.0 mL    = 0.0249250 kg water   1 mL   103 g  moles of solute kg of solvent Moles solute = (m)(kg solvent) = (0.878906 mol•kg/mol)(0.0249250 kg) = 0.0219067 mol 1.50 g Molar mass = = 68.4722 g/mol = 68 g/mol 0.0219067 mol b) bT = iKbm i = 3(AB2 or A2B)  T 0.45 C m = b = = 0.29296875mol/kg Kbi 0.512 C kg/mol  3 m=





moles of solute kg of solvent

Moles solute = (mol solute/kg solvent)(kg solvent) = (0.29296875 mol/kg)(0.0249250 kg) = 0.00730225 mol 1.50 g Molar mass = = 205.416 g/mol = 2.1x102 g/mol 0.00730225 mol c) The molar mass of CaN2O6 is 164.10 g/mol. This molar mass is less than the 2.1x102 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 d) Moles of CaN2O6 = 1.50 g CaN 2 O6    = 0.0091408 mol  164.10 g CaN 2 O6  moles of solute 0.00914078 mol m= = = 0.3667314 mol/kg kg of solvent 0.0249250 kg bT = iKbm  0.45C   T i= b = = 2.39659 = 2.4 Kbm  0.512C kg / mol  0.3667314 mol / kg 

12.123

mol C2 H 5 OH  g 

mol C2 H5 OH  l   8.07 kPa  mol C 2 H 5 OH  l  (0.4803571)   = mol CH3OH  l   16.8 kPa  mol CH 3OH  l  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  0.03121 mol CH3OH  g   1 mol CH 3OH  1 g CH 3OH  g     32.04 g CH 3OH  mol CH 3OH  g 

mol C 2 H 5 OH  l  mol CH 3OH  l 

 2.10549/0.03121 0.4803571

 46.07 g C 2 H 5 OH    1 mol C2 H 5 OH 

() ()

140.4414 mol C2 H5OH  

 32.04 g CH 3OH  1 mol CH3OH     1 mol CH 3OH  12.124

= 201.9393 = 2x102 C2H5OH / CH3OH

Determine the concentration (mol/L) 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 CH3Cl   1 mol CH3Cl   1.00 g solution   1 mL  c = 9   3   10 g solution   50.48 g CH Cl   mL   10 L  3    = 1.98098x10–6 mol/L= 1.98x10–6 mol/L CH3Cl If the density is 1.00 g/mL, then 1.00 L of solution would weigh 1.00 kg. The mass of CH 3Cl 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 concentration (mol/L) equal to the molality, in other words: 1.98x10–6 mol/kg CH3Cl. Still using 1.00 L of solution: Moles of CH3Cl = (1.98098x10–6 mol/L)(1.00 L) = 1.98098x10–6 mol CH3Cl Moles of H2O = (1.00 kg)(103 g/1 kg)(1 mol H2O/18.02 g H2O) = 55.49389567 mol H2O Xtrichloromethane = (1.98098x10–6 mol CH3Cl)/[(1.98098x10–6 + 55.49389567) mol] = 3.569726x10–8 = 3.57x10–8 Convert from ppb to pph (parts per hundred = mass percent)  100. ppb  100 pph  –5    = 1.00x10 % 9 1  10  

12.125 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 Na 14 that dissolves will also precipitate back out of solution. Na214CO3(aq) + H2O(l)  Na214CO3(s) 12.126 Plan: From the osmotic pressure, the concentration (mol/L) of the solution can be found using the relationship  = cRT. Convert the temperature from °C to K. Use the concentration 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 concentration 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 fT = iKf m. (i = 1). Solution: a)  = cRT Osmotic pressure ,  = 45.3 Pa T = 25°C + 273 = 298 K  45.3 Pa   c= = = 1.828302x10–5 mol/L RT L•kPa    8.31446 mol•K   298 K    (c)(V) = moles  103 L  Moles = 1.828302x105 mol / L  30.0 mL   = 5.48491x10–7 mol  1 mL   

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 103 g   1 mg   Molar mass = = 1.82318x104 g/mol = 1.82x104 g/mol 5.48491x107 mol  0.997 g  b) Mass (g) of solution =  30.0 mL    = 29.91 g  1 mL 

10.0 mg  

 103 g  Mass (g) of solute = 10.0 mg   = 0.0100 g  1 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 = 5.48563x10 mol (from part a)) Molality =

5.48563 x 107 mol moles of solute = = 1.83466x10–5 mol/kg kg of solvent 0.0299 kg

fT = iKf m = (1)(1.86°C•kg/mol)(1.83466 x 10 –5 mol/kg) = 3.412x10–5 °C= 3.41x10–5 °C (So the solution would freeze at 0 °C – (3.41x10–5 °C) = –3.41x10–5 °C). 12.127 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 1,2-dichloroethene (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 C2 H 2 Cl2   103 g   1 mol C2 H 2 Cl2  –6 Sgas (mol/L) =    1 mg   96.94 g C H Cl  = 6.705178x10 mol/L L   2 2 2   Pgas =

Sgas kH

 6.705178 x106 mol/L  = = 0.0203187 kPa  0.00033 mol/L•kPa   

pV = nRT

 0.020318721kPa  n p = = 8.31218x10–6 mol/L = L•kPa  V RT   8.31446 mol•K    273  21 K     8.31218x106 mol C2 H 2 Cl2   96.94 g C2 H 2 Cl2   1 ng  Concentration (ng/L) =      9   L    1 mol C2 H 2 Cl2   10 g  = 8.0578316035x105 = 8.1x105 ng/L 12.128 Acircle = r2 = (38.6 cm/2)2 = 1.17021x103 cm2  1.17021x103 cm 2  1 mg   283 g   1 mol Area of one molecule =  3      23 2.50 mg  10 g   mol   6.022x10 molecules  = 2.19973x10–16 cm² = 2.20x10–16 cm2/molecule 12.129 a) Looking at the data for CaCl2, K2CO3, and Na2SO4, the average conductivity is 7.0 ± 0.7 units for the 5.00x103 ppm solutions and 14 ± 1.7 units for the 10.00x10 3 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)(10.00x10 3 ppm) = 8.75x103 ppm Assume the mass of CaCl2 present is negligible relative to the mass of the solution.  8.75x103 g CaCl2   1 mol CaCl 2   103 g  Molality of CaCl2 =  = 0.0788430 mol/kg CaCl2  106 g solution   110.98 g CaCl   1 kg  2     = 0.0788 mol/kg CaCl2 Moles of CaCl2 = (8.75x103 g CaCl2)(1 mol CaCl2/110.98 g CaCl2) = 78.84303 mol CaCl2 Moles of H2O = (1.00x106 g H2O)(1 mol H2O/18.02 g H2O) = 5.5493896x104 mol H2O  78.84303 mol CaCl2  Mole fraction of CaCl2 = X = = 1.4187x10–3 = 1.42x10–3 4 78.84303  5.5493896x10 mol

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12.130 The vapour 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 vapour pressures closer to equal. 12.131 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 4 3  1 mol   5 x 10 kPa•m  –4 pethanol =  3    = 5x10 kPa mol  m    4 3  1 mol   1.6x10 kPa•m  –4 p2-butoxyethanol =  3    = 1.6x10 kPa  mol  m  

 1.6x104 kPa 2-butoxyethanol  %2-butoxyethanol =  5%    = 1.6%  5x104 kPa ethanol   ―Down-the-drain‖ factor is 0.016 = 0.02

 5x104 kPa•m3   1 L  –1 b) kH(ethanol) =    3 3  = 5x10 kPa•L/mol  mol    10 m  c) kH(ethanol) = 0.64 Pa•m³/mol = 0.64 kPa•L/mol Considering the single significant figure in the measured value of 5x10 –4, the agreement is good. 12.132 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) 100.0 mL  a) Mass (mg) in water = 10.0 mg  = 1.66389 mg = 1.66 mg remaining 100.0  8.35  60.0  mL b) Perform a similar calculation to part a), then take the result and repeat the procedure. Combine the results to get the total removed. 100.0 mL  Mass (mg) in water = 10.0 mg  = 2.853067 mg = 2.85 mg remains after first extraction 100.0  8.35 30.0  mL Mass (mg) in water =

100.0 mL   2.853067 mg  100.0  8.35  30.0  mL

= 0.813999 mg = 0.814 mg remains after second extraction c) The two-step extraction extracts more of the caffeine. 12.133 Plan: Molality is defined as moles of solute per kg of solvent, so 0.150 mol/kg means 0.150 mol NaHCO3 per kg of water. The total mass of the solution would be 1 kg + (0.150 mol x molar mass of NaHCO 3). Solution:  84.01 g NaHCO3   0.150 mol NaHCO3     1 mol NaHCO3   1 kg  0.150 mol/kg = (0.150 mol NaHCO3)/(1 kg solvent) =  3  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 mol/kg solution of NaHCO3, weigh 3.11 g NaHCO3 and dissolve in 247 g water. 12.134 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: fT = iKf m (Assume the compound is a nonelectrolyte, i = 1.) m = fT/iKf = (1.26°C)/(1)(1.86°C•kg/mol)] = 0.677419 mol/kg m = moles of solute/kg of solvent Moles of solute = (m)(kg of solvent) = (0.677419 mol/kg)[(11.23 g)(1 kg/10 3 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.

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. 12.135 The range has to fall between the point where the amount of methanol (moles) is just greater than the amount of ethanol (moles), 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 methanol)(1 mol methanol/32.04 g methanol) = 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 12.136 a) The molar mass comes from the boiling point elevation. bT = (77.5 – 76.5*) = 1.0°C bT =iKbm (Assume the compound is a nonelectrolyte, i = 1.) m = bT/iKb = (1.0°C)/[(1)(5.03°C•kg/mol)] = 0.198807 mol/kg m = moles of solute/kg of solvent Moles of solute = (m)(kg of solvent) = (0.198807 mol/kg)[(100.0 g)(1 kg/103 g)] = 0.0198807 mol   5.0 g 2 Molar mass =   = 251.5 g/mol = 2.5x10 g/mol 0.0198807 mol   * Note: tetrachloromethane normal boiling point value is taken from Table 12.5. 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. 12.137 Concentration (mol/L) is moles solute/L solution and molality is moles solute/kg solvent. Multiplying molality by concentration of solvent in kg solvent per litre of solution gives concentration (mol/L): mol of solute  mol of solute   kg solvent  c= =   L of solution  kg solvent   L of solution 

 kg solvent  c= m    L of solution  For a very dilute solution, the assumption that mass of solvent  mass of solution is valid. This equation then becomes c = m (kg solvent/L solution) = m x dsolution Thus, for very dilute solutions molality x density = concentration (mol/L). In an aqueous solution, the litres 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 = c.  1 mol NH 4 NO3   1 mol NH 4   –2 + 12.138 Moles =  5.66 g NH 4 NO3    = 7.07058x10 mol NH4   80.05 g NH NO 1 mol NH NO 4 3  4 3     1 mol (NH 4 )3 PO4   3 mol NH 4 –2 + Moles =  4.42 g (NH 4 )3 PO4    = 8.89336x10 mol NH4    149.10 g (NH 4 )3 PO4   1 mol (NH 4 )3 PO4 

 1 mol (NH 4 )3 PO4   1 mol PO43   –2 3– Moles =  4.42 g (NH 4 )3 PO4    = 2.96445x10 mol PO4    149.10 g (NH 4 )3 PO4   1 mol (NH 4 )3 PO 4  Silberberg, Amateis, Venkateswaran, Chen, Chemistry: The Molecular Nature of Matter and Change, 3rd Canadian Edition Page 3-426 Instructor‘s Solution Manual © Copyright 2021 McGraw-Hill Ryerson Ltd.

c NH4+ = [(7.07058x10–2) + (8.89336x10–2)] mol NH4+/20.0 L = 7.98197x10–3 mol/L= 7.98x10–3 mol/L NH4+ c PO43– = (2.96445x10–2 mol PO43–)/20.0 L = 1.482225x10–3 mol/L= 1.48x10–3 mol/L PO43– 12.139 Plan: Use Henry‘s law, Sgas = kH  pas, to find the solubility of SO2. Solution: a) Sgas = kH  pgas = (1.21x10-2mol/L•kPa)(0.203 kPa) = 2.46x10–3 mol/L= 2.5x10–3 mol/L 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. 12.140 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  Moles C =  20.1 g C    = 1.6736 mol C  12.01 g C 

 1 mol H  Moles H =  6.7 g H    = 6.6468 mol H  1.008 g H   1 mol N  Moles N =  46.5 g N    = 3.31906 mol N  14.01 g N   1 mol O  Moles O =  (100  20.1  6.7  46.5) g O    = 1.66875 mol 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  = cRT to solve for the concentration (mol/L) of the urea solution. The solution concentration (mol/L) is related to the concentration expressed in % w/v by using the molar mass.  206.7 kPa  c = /RT = = 0.0834238 mol/L L•kPa    8.31446 mol•K    273  25 K    Moles of urea = (c)(V) = (0.0834238 mol/L)(1 L) = 0.0834238 mol Molar mass = 5.0 g/0.0834238 mol = 59.935 g/mol = 60. g/mol Because the molecular weight equals the empirical weight, the molecular formula is also CH4N2O. 3  100. mL   10 L   0.30 mol glucose  180.16 g glucose  12.141 a) Mass (g) of glucose =  2.5h        1 mL   h 1L    1 mol glucose   = 13.512 g = 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 concentration (mol/L) that is one-half of glucose to be isotonic: 0.15 mol/L . 3  150. mL   10 L   0.15 mol NaCl  58.44 g NaCl  c) Mass (g) of NaCl = 1.5 h        h 1L    1 mL    1 mol NaCl  = 1.97235 g = 2.0 g NaCl

12.142 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. 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. Silberberg, Amateis, Venkateswaran, Chen, Chemistry: The Molecular Nature of Matter and Change, 3rd Canadian Edition Page 3-427 Instructor‘s Solution Manual © Copyright 2021 McGraw-Hill Ryerson Ltd.

12.143 The enthalpy of vapourization must be determined first.  1 bar  1 kPa  p (kPa) at 25°C =  4.1 mbar     = 0.41 kPa  1000 mbar  0.01 bar 

 vap H  1 p2 1 =     p1 R  T2 T1 

 vap H  0.41 kPa 1 =   101.3 kPa 8.314 J/mol•K   273  25  K

 1   273.2  141.5  K 

–5.509685 = –  vap H (1.135823x10–4 mol/J)  vap H = 48,508.30 J/mol

p2 48,508.30 J/mol  1  =  101.3 kPa 8.314 J/mol•K   273  65 K

 1  = –3.192645  273.2  141.5 K 

p2 = 0.0410631 101.3 kPa p2 = 4.15969 kPa = 4.2 kPa 12.144 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 0.5 kPa. To calculate the mole fraction of O 2, assume a litre of solution. Divide the moles of O 2 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 10 6. Solution: a) Sgas = kH  pgas Sgas 1.64x103 mol/L kH = = = 7.7358x10-5 mol/L•kPa = 7.74x10–5 mol/L•kPa pgas 21.2 kPa b) Sgas = kH  pgas Sgas = (7.7358x10–5 mol/L•kPa)(0.5 kPa) Sgas = 3.8679x10–5 mol/L= 4x10–5 mol/L c) Assume a 1.0 L sample. Acrylic acid is 14.6 mol/L or 14.6 mol in 1.0 L. Oxygen is 4x10–5 mol/L or 4x10–5 mol in 1.0 L. moles of O 2 4x105 mol = 2.73972x10–6 = 3x10–6 X O2 = = moles of O 2 + moles of acrylic acid (4x105 ) + 14.6 mol

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 14.6 mol acrylic acid   72.06 g acrylic acid  d) Mass (g) of acrylic acid =    1 mol acrylic acid  = 1052.076 g/L L   

 4x105 mol O2   32.0 g O2  Mass of oxygen =     = 0.00128 g/L  L    1 mol O2  mass of solute 0.00128 g ppm = 1x106 = 1x106 = 1.2165 ppm= 1 ppm mass of solution 0.00128 g + 1052.176 g

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12.145 The lower the boiling point the greater the volatility. ethanoic acid (acetic acid) < water < benzene < ethanol < tetrachloromethane < trichloromethane (chloroform) < carbon disulfide < ethoxyethane (diethyl ether)

12.146

 vap H  1 p2 1 =     p1 R  T2 T1  p1 = 101.3 kPa T1 = (273 + 100) K = 373 K p2 = ? T2 = (273 + 200.)K = 473 K ln

 vap H = 40.7 kJ/mol = 40,700 J/mol

p2 40, 700 J/mol  1 1   = 373 K  101.3 kPa 8.314 J/mol•K  473 K

p2 = 2.774689665 101.3 kPa

p2 = 16.03365 101.3 kPa p2 = 1624.208 kPa = 1.62x10³ kPa 12.147 a) fT = iKf m Assume NaCl is a strong electrolyte with i = 2. m = fT/iKf = (5.0°C)/[(2)(1.86°C•kg/mol)] = 1.344086 mol/L NaCl  58.44 g NaCl   1.344086 mol NaCl  2 Mass (g) of NaCl =   = 432.016 g= 4.3x10 g NaCl   5.5 kg   kg mol NaCl     b) fT = iKf m Assume CaCl2 is a strong electrolyte with i = 3. m = fT/iKf = (5.0°C)/[(3)(1.86°C•kg/mol)] = 0.896057 mol/kg CaCl2  110.98 g CaCl2   0.896057 mol CaCl2  2 Mass (g) of CaCl2 =   = 546.944 g= 5.5x10 g CaCl2   5.5 kg   kg mol CaCl   2   12.148 Plan: Use the density of C6F14 to find the mass and then moles of C6F14 in 1 L. This is the concentration (mol/L) of C6F14 which is then used to find the concentration (mol/L) of O 2 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 O 2 at the given temperature. Solution:  1 mL   1.674 g C6 F14   1 mol C6 F14  a) Moles of C6F14 in 1 L (c) = 1 L   3    = 4.9527 mol/L  1 mL  10 L     338 g C6 F14  mol of O 2 X O2 = mol of O 2 + mol of C6 F14

x mol O 2 4.9527 mol Moles of O2 = 0.021198 (The moles of O2 is small enough to ignore in the denominator.) pgas = = 101,325 Pa Sgas = kH  pgas Sgas 0.021198 mol/L 0.021198 mol/L kH = = = = 2.0926x10-4 mol/L•kPa = 2.09x10-4 mol/L•kPa pgas 101,325 Pa 101.325 kPa 4.28x10–3 =

b) Sgas = kH  pgas 1 kH = = 1.304x10–5 mol/L•kPa 4 7.67x10 L•kPa/mol pO2 = 0.2095 x 101.3 kPa = 21.222 kPa since O 2 is 20.95% of air Sgas = (1.304x10–5 mol/L•kPa)(21.222 kPa) = 2.76357x10–4 mol/L

 2.76357x104 mol   32.0 g O2  –3 –3 6 ppm =     = 8.8434x10 g/L = (8.8434x10 g/1000 g)(1x10 )  L    1 mol O2  = 8.8434 ppm= 8.84 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 C6F14 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. 12.149 a) pN2 = (101.3 kPa)(78% N2/100%) = 79.014 kPa Concentration (mol/L) of N2 = Sgas = kH  pgas = (6.1x10–6 mol/L•kPa)(79.014 kPa) = 4.836x10–4 = 4.8x10–4 mol/L N2 b) The additional pressure due to 15 m of water must be added to 101.3 kPa. Water pressure: The value, 9.80665 m/s2, is the standard acceleration of gravity from the inside back cover of the book. 3

m   1 Pa   1 kPa   1.00 g   1 mL  1 cm   1 kg   pwater =dH2O•h•g =     3  15m   9.80665 2      3  2  s   1kg/m•s 2   1000 Pa   mL   1 cm  10 m   10 g  

= 147.150 kPa This is the pressure due to the 15 m of water, and it must be added to the atmospheric pressure pressing down on the surface of the water (101.3 kPa). This gives a total pressure of 248.5 kPa. pN2 = (248.5 kPa)(78% N2/100%) = 193.8 kPa Concentration of N2 = Sgas = kH  pgas = (6.1x10–6 mol/L•kPa)(193.8 kPa) =1.18213x10–3 mol/L = 1.2x10–3 mol/L N2 c) Moles of N2 per litre at the surface = 4.836x10–4 mol N2 Moles of N2 per litre at 15 m = 1.18213x10 –3 mol N2 Moles N2 released per litre = (1.18213x10–3 – 4.836x10–4) mol = 6.9853x10–4 mol L•kPa   6.9853x104 mol  8.31446   273  25 K  1 mL  mol•K   pV = nRT so V = nRT/p =  3  101.3kPa   10 L  = 17.0854 = 17 mL N2





12.150 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 vapour 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 vapour pressure of the liquid. 12.151 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 amount of salt (in grams), 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 KNO 3 (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 amount of salt (in grams) which could be recovered by the process described.

12.152 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:  40%  22%  0.789 g  a) Mass (g) of ethanol dissolved in the blood =  28 mL      = 1.944096 g  100%  100%  mL 

 1.944096 g ethanol   103 L  –4 –4 Concentration =   = 2.77728x10 g/mL = 2.8x10 g/mL   7.0 L 1 mL     28 mL whiskey  4 b)   8.0x10 g/mL = 80.6545 mL = 81 mL 4  2.77728x10 g/mL 





 103 L  3.3x104 mol  12.153 a) Moles of CO2 =  355 mL     405.3 kPa  = 0.04686 mol = 0.05 mol CO2  1 mL   L•kPa    b) If it is completely flat there is no CO2 remaining or 0.00 moles CO2, however, a small amount will remain in solution:  103 L  3.3x104 mol  2 –6 –6 Moles CO2 =  355 mL    3 x 10 kPa = 3.5145x10 mol = 4x10 mol CO2  1 mL   L•kPa    c) The difference in the moles will determine the amount of moles entering the gas phase.  0.04686  3.5145x106 mol  8.31446 L•kPa    273  25 K    mol•K  pV = nRT so V = nRT/p = 101.3 kPa 





= 1.14607 L = 1 L CO2 12.154 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 vapour 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 vapour phase. The increased pressure results in increased solubility of oxygen in the water. Of the six moles of oxygen in the vapour phase, one mole dissolves in the water to bring the dissolved moles to three. 12.155 The CO2 levels in the ocean increase as the atmospheric CO2 level increase. The solubility of a gas in water increases with increasing partial pressure of the gas, according to Henry‘s Law: Sgas = kHPgas So as the atmospheric CO2 levels increase, more CO2 will dissolve in water by the following reaction. CO2(g) ⇌ CO2(aq) 12.156 When water is heated, the air trapped in the water gains kinetic energy and the probability of air escaping to the gas phase increases. Raising the temperature of the water increases the rate at which air escape from the water compared to the rate at which air in the gas phase dissolving in water. As temperature of the water increases, the solubility of air in the water decreases. Therefore, we observed more bubbles forming in the water as the temperature of the water increases.

CHAPTER 13 PERIODIC PATTERNS IN THE MAIN-GROUP ELEMENTS END–OF–CHAPTER PROBLEMS 13.1

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.

13.2

13.3

Plan: Recall that to form hydrogen bonds a compound must have H directly bonded to either N, O, or F. Solution: a) NH3 will form hydrogen bonds because H is bonded to N. F

b) CH3CH2OH will form hydrogen bonds since H is bonded to O. CH 3OCH3 has no OH bonds, only CH bonds. H H H H

C H

CH3OCH3

H H CH3CH2OH

13.4

a) NH3 will form hydrogen bonds.

H b) H2O will form hydrogen bonds. H

O H

13.5

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)

13.6

a) CaH2(s) + 2H2O(l)  Ca(OH)2(aq) + 2H2(g) b) PdCl2(aq) + H2(g)  Pd(s) + 2HCl(aq)

13.7

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 x B(3e–)] + [4 x 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

13.8

Since the nucleus of H contains only one proton, the electrons in the H - ion 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.

13.9

In general, the maximum oxidation number increases as you move to the right. In the second period, the maximum oxidation number in Groups 3-17 drops off below the group number – 10 in Groups 16 and 17.

13.10

a) lithium fluoride, LiF beryllium fluoride, BeF2 boron trifluoride, BF3 nitrogen trifluoride, NF3 oxygen difluoride, OF2 fluorine, F2 b) χ decreases left to right across the period. c) % ionic character decreases left to right across the period. d)

carbon tetrafluoride, CF4

F F

13.11

a) 32 elements, 30 metals b) Between Po and At

13.12

a) E must have an oxidation of +3 to form an oxide E2O3 or fluoride EF3. E is in Group 13 or 3. b) If E were in Group 3, the oxide and fluoride would have more ionic character because Group 3 elements have lower electronegativity than Group 13 elements. The Group 3 oxides would be more basic.

13.13

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.

13.14

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.

13.15

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)

13.16

The large atomic radii of the Group 1 elements mean that their atomic volumes are large. Since density = mass/volume, the densities will be small.

13.17

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 IE 1. e) hydrH decreases down a group. hydrH 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

13.18

Increasing up the group: a, c, and e Decreasing up the group: b and d

13.19

Plan: Peroxides are oxides in which oxygen has a –1 oxidation state. Sodium peroxide has the formula Na 2O2 and is formed from the elements Na and O2. Solution: 2Na(s) + O2(g)  Na2O2(s)

13.20

RbOH(aq) + HBr(aq)  RbBr(aq) + H2O(l)

13.21

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)

13.22

a) % Li =

6.941 g Li/mol x 100% = 2.39015 %= 2.390% Li 290.40 g/mol

b) % Li =

6.941 g Li/mol x 100% = 10.8368 %= 10.84% Li 64.05 g/mol

13.23

The Group 1 elements react more vigorously with water than those in Group 2.

13.24

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.

13.25

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.

13.26

Plan: A base forms when a basic oxide, such as CaO (lime), is added to water. Alkaline earth metals reduce O 2 to form the oxide. Solution: a) CaO(s) + H2O(l)  Ca(OH)2(s) b) 2Ca(s) + O2(g)  2CaO(s)

13.27

13.28

13.29

 a) BaCO3(s)  BaO(s) + CO2(g) b) Mg(OH)2(s) + 2HCl(aq)  MgCl2(aq) + 2H2O(l)  a) CaCO3(s)  CaO(s) + CO2(g) (CaCO3 from limestone) 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 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: – 2– BeCl2(l) + 2 Cl (solvated)  BeCl4 (solvated) 13.30

The pattern of ionization energies in Group 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 3.

13.31

Tl2O is more basic (i.e., less acidic) than Tl2O3. Acidity increases with increasing oxidation number.

13.32

The electron removed in Group 2 atoms is from the outer level s orbital, whereas in Group 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 2 to 13.

13.33

a) Compounds of Group 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)

13.34

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.

13.35

Plan: Oxide acidity increases up a group; the less metallic an element, the more acidic is its oxide. Solution: In2O3 < Ga2O3 < Al2O3

13.36

B(OH)3 < Al(OH)3 < In(OH)3

13.37

a) Halogens typically have a –1 oxidation state in metal-halide combinations, so the apparent oxidation state of Tl = +3. b) However, the anion I3– combines with Tl in the +1oxidation state. c) The anion I3– has [3 x (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 AX 2E3. 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°. (Tl 3+) (I–)3 does not exist because of the low strength of the Tl–I bond. O.N. = +3 (apparent); = +1 (actual) I

13.38

a) O.N. = +2 (apparent); b) O. N. = +1 (Ga+) and +3 (GaCl4–) (actual) c) class = AX4; bond angles = 109.5°; tetrahedral

Cl Cl

13.39

a) boron, B d) aluminum, Al

b) gallium, Ga e) thallium, Tl

c) boron, B

13.40

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

13.41

H H O

H O

O H

H B(OH)4– has 109.5° angles around B.

B(OH)3 has 120° angles around B. 13.42

Plan: To write a balanced equation, need to review how to convert chemical name to chemical formula, e.g. diboron trioxide is B2O3, ammonia is NH3…etc. To calculate the enthalpy of reaction, use the relationship r H = m  products H – n reactants H . Convert the given amount of 1.0 kg of BN to moles, find the amount (mol) of B in that amount of BN, and then find the amount (mol) and then mass of borax that provides that amount (mol) of B. Solution: a) B2O3(s) + 2NH3(g)  2BN(s) + 3H2O(g) b) r H = m  products H – n reactants H = {2  f H [BN(s)] + 3  f H [H2O(g)]} – {1  f H [B2O3(s)] + 2  f H [NH3(g)]} = [(2 )(–254 kJ/mol) + (3)(–241.826 mol kJ/mol)] – [(–1272 kJ/mol) + (2 )(–45.9 kJ/mol)] = 130.322 kJ= 1.30x102 kJ/mol c) Mass (g) of borax =

(

)(

)

= 5.335359x10 g= 5.3x10 g borax 13.43

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.

13.44

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.

13.45

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 13 would show greater deviations because the single p electron receives no shielding effect offered by other p electrons.

13.46

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.

13.47

Allotropes are two forms of a chemical element which have different bonding and physical properties. C forms graphite, diamond, and buckminsterfullerene; Sn has gray () and white () forms.

13.48

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.

13.49

Having four valence electrons allows all of the Group 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.

13.50

Plan: The silicate building unit is —SiO4—. There are [4 x Si(4e–)] + [12 x 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 C 4H8, there are [4 x C(4e–)] + [8 x H(1e–)] = 24 valence electrons. All twenty four electrons are used to form the bonds between the atoms in the molecule. Solution: a) b)

8

There is another answer possible for C4H8.

13.51

There are numerous alternate answers for C6H12.

13.52

Each alkaline earth metal ion will displace two sodium ions because of the charge difference. Determine the amount (mol) of alkaline earth metal ions: Amount of Ca2+ = (4.5x10–3 mol/L)(25,000 L) = 112.5 mol Ca2+ Amount of Mg2+ = (9.2x10–4 mol/L)(25,000 L) = 23 mol Mg2+ Total amount of M2+ = (112.5 + 23) mol = 135.5 mol M2+ Determine the amount (mol) of Na+ needed: Amount of 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/10 3 g)(100%/85%) = 58.215848 g= 58 kg zeolite

13.53

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. CO2 also binds coordinately to Fe(II) in blood in a manner similar to that of O2, but CO binds more strongly to Fe(II) in blood. The preferential binding of carbon monoxide to Fe(II)is largely responsible for the asphyxiation that results from carbon monoxide poisoning. e) Silicon, Si

13.54

B2H6(g) + 3O2(g)  B2O3(s) + 3H2O(g) 2Si4H10(g) + 13O2(g)  8SiO2(s) + 10H2O(g)

13.55

All of the elements in Group 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.

13.56

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).

13.57

a) In Group 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 – 18, which is 15 – 18 = –3 for Group 15, to the highest equal to the group number – 10, which is +5 for Group 15.

13.58

In general, high oxidation states are less stable towards the bottom of the periodic table.

13.59

Bi2O3 < Sb2O3 < Sb2O5 < P4O10

13.60

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

13.61

HNO3 > HNO2 > H2N2O2

13.62

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 15 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)

13.63

a) 2Sb(s) + 3Br2(l)  2SbBr3(s) b) 2HNO3(aq) + MgCO3(s)  Mg(NO3)2(aq) + CO2(g) + H2O(l)  c) 2K2HPO4(s)  K4P2O7(s) + H2O(g)

13.64

a) Aluminum is not as active a metal as Li or Mg, so heat is needed to drive this reaction.  N2(g) + 2Al(s)  2AlN(s) b) The Group 15 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)

13.65

a) AsCl3(l) + 3H2O(l)  H3AsO3(aq) + 3HCl(g) b) Sb2O3(s) + 6NaOH(aq)  2Na3SbO3(aq) + 3H2O(l)

13.66

Plan: There are [1 x P(5e–)] + [2 x F(7e–)] + [3 x 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:

13.67

The structure would be tetrahedral at the P atoms and bent at the O atoms within the ring.

13.68

F F

F 13.69

a) Ammonia, NH3 c) Tetraphosphorus decoxide, P4O10 e) Phosphoric acid, H3PO4

13.70

Plan: Set the atoms into the positions described, and then complete the Lewis structures. a) and b) N2O2 has [2 x N(5e–)] + [2 x 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 twenty electrons are needed to complete the octets of all of the atoms, two double bonds are needed. c) N2O3 has [2 x N(5e–)] + [3 x 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 twenty-four electrons are needed to complete the octets of all of the atoms, two double bonds are needed. d) NO+ has [1x N(5e–)] + [1 x 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 twelve electrons are needed to complete the octets of both atoms, a triple bond is needed. NO3– has [1x N(5e–)] + [3 x 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 twenty electrons are needed to complete the octets of all of the atoms, a double bond is needed. Solution: b) a)

N O

N N

b) Tetraphosphorus trisulfide, P4S3 d) Nitrogen monoxide, NO; nitrogen dioxide, NO2

N O

13.71 O.N.:

 NH4NO3(s)  N2O(g) +2H2O(l) –3 + 5 +1

13.72

a) Thermal decomposition of KNO3 at low temperatures: 

2KNO3(s)   2KNO2(s) + O2(g) b) Thermal decomposition of KNO3 at high temperatures: 

4KNO3(s)   2K2O(s) + 2N2(g) + 5O2(g) 13.73

S < Se < Po; nonmetal < metalloid < metal

13.74

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 nitrogen and oxygen are diatomic gases at normal temperature and pressure. 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 six oxidation states, whereas oxygen has two.

13.75

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 unhybridized (p) in the others. d) Group 15 is similar.

13.76

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 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)

13.77

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)

13.78

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.

13.79

a) basic

13.80

Plan: Acid strength of binary acids increases down a group since bond energy decreases down the group. Solution: H2O < H2S < H2Te

13.81

H2SO4 > H2SO3 > HSO3–

b) acidic

c) basic

d) acidic

e) amphoteric

13.82

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.

13.83

a) O3, ozone b) SO3, sulfur trioxide (+6 oxidation state) c) SO2, sulfur dioxide d) H2SO4, sulfuric acid e) Na2S2O3•5H2O, sodium thiosulfate pentahydrate

13.84

a) 0

13.85

S2F10(g)  SF4(g) + SF6(g) O.N. of S in S2F10: – (10 x –1 for F)/2 = +5 O.N. of S in SF4: – (4 x –1 for F) = +4 O.N. of S in SF6: – (6 x –1 for F) = +6

13.86

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.

13.87

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 has only 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.

13.88

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).

13.89

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.

13.90

You would expect them to contain an odd number of atoms, so that you would have an even number of electrons.

13.91

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)

13.92

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)

b) +4

c) +6

d) –2

e) –1

f) +6

g) +2

13.93

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)

13.94

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

13.95

HClO4 > HBrO4 > HBrO3 > HIO3

13.96

a) hydrogen fluoride, HF b) sodium hypochlorite, NaClO c) hydrofluoric acid, HF d) bromine, Br2 e) Chloroethene, also known as vinyl chloride, C2H3Cl

13.97

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 H 2SO4 would produce HI.

13.98

I2 < Br2 < Cl2, since Cl2 is able to oxidize Re to the +6 oxidation state, Br2 only to +5, and I2 only to +4.

13.99

a) Helium is the second most abundant element in the universe. b) Argon is the most abundant noble gas in Earth‘s atmosphere, the third most abundant constituent after N 2 and O 2.

13.100 +2, +4, +6, +8 13.101 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. 13.102 The electrons on the larger atoms are more easily removed, transferred, or shared with another atom than those on the smaller atoms. 13.103 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.

13.104 a) Xenon tetrafluoride, XeF4, is an AX4E2 molecule with square planar geometry. Antimony pentafluoride, SbF 5, is an AX5 molecule with trigonal bipyramidal molecule geometry. XeF 3+ is an AX3E2 ion with a T-shaped 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.

13.105 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) H = +720 kJ/mol

Overall:

H+(g) + H2O(l)  H3O+(aq) H2O(g)  H2O(l) H3O+(g)  H3O+(aq)

H = –1090 kJ/mol H = –40.7 kJ/mol H = –410.7 kJ/mol= –411 kJ/mol

13.106 Calculation of energy from wavelength: E = hc/. 

 6.626x10 E =

34





J•s 2.9979 x108 m/s  1 nm  –19 –19  9  = 3.3713655x10 J= 3.371x10 J  589.2 nm  10 m  

13.107 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) 13.108 a) 5IF → IF5 + 2I2 b) Iodine pentafluoride c) This is a disproportionation redox reaction. IF acts both as the oxidizing and reducing agents.  2.50x103 mol IF   1 mol IF5   221.9 g IF5  d) Mass (g) of IF5 =  7 IF molecules   = 0.77665 g= 0.777 g IF5  1 IF molecule   5 mol IF   1 mol IF  5    

 2.50x103 mol IF   2 mol I 2   253.8 g I 2  Mass (g) of I2 =  7 IF molecules   = 1.7766 g= 1.78 g I2  1 IF molecule   5 mol IF   1 mol I  2     13.109 Plan: Examine the outer electron configuration of the alkali metals. To calculate the r H 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  f H for the formation of CsF: Cs(s) + 1/2F2(g)  CsF(s)

 f H = –530 kJ/mol

You also know the  f H for the formation of CsF2: Cs(s) + F2(g)  CsF2(s)  f H = –125 kJ/mol 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)  f H = 2 x (–530 kJ/mol) = –1060 kJ/mol 2CsF2(s)  2Cs(s) + 2F2(g)

 f H = 2 x (+125 kJ/mol) = 250 kJ/mol (Note sign change)

2 CsF2(s)  2CsF(s) + F2(g) r H = –810 kJ/mol 810 kJ/mol 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 ½(–810 kJ/mol) or –405 kJ/mol. 13.110 4Ga(l) + P4(g)  4GaP(s) Silberberg, Amateis, Venkateswaran, Chen, Chemistry: The Molecular Nature of Matter and Change, 3rd Canadian Edition Page 3-445 Instructor‘s Solution Manual © Copyright 2021 McGraw-Hill Ryerson Ltd.

Find the limiting reagent:

 1 mol Ga   4 mol GaP  100.69 g GaP  Mass (g) of GaP from Ga =  32.5 g Ga      = 46.9367 g GaP  69.72 g Ga   4 mol Ga  1 mol GaP 

195 kPa  20.4 L  = 0.92900 mol P pV = 4 RT L•kPa    8.3146 mol•K   515 K     4 mol GaP   100.69 g GaP  Mass (g) of GaP from P4 =  0.92900 mol P4     = 374.1644 g GaP  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 x 0.928 = 43.5573 g= 43.6 g GaP Amount of P4 = n =

13.111 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 g/mol. 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 x H(1e–)] + [2 x N(5e–)] + [2 x 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 x H(1e–)] + [1 x N(5e–)] + [1 x 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.

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 2 2 N O

O O

13.112 a) C

 C

2

b) All have configuration (2s)2(*2s)2(2p)2(2p)2(2p)2; bond order = 3.

13.113

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 NO 2  2 mol NO  4 mol NH3  mol NH 3 consumed =     = 1.5 mol NH3/mol HNO3 mol HNO3 produced  2 mol HNO3   2 mol NO 2   4 mol NO  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 amount (mol) of NH 3 present in 1 m3 of gas mixture (ideal gas law) and convert amount (mol) of NH 3 to amount (mol) of HNO3 using the mole ratio in part b). Convert the amount (mol) to grams using the molar mass of HNO 3 and convert to volume.





 507 kPa  1 m3  1 L   10.% NH3  pV =  3 3    RT L•kPa    10 m   100% gas   8.31446 mol•K    273  850. K    = 5.42028 mol NH3 Mass (g) of HNO3 = (5.42028 mol NH3)(1 mol HNO3/1.5 mol NH3)(63.02 g HNO3/1 mol HNO3)(96%/100%) = 218.6149 g HNO3 Volume (mL) of HNO3 = (218.6149 g HNO3)(100%/60.%)(1 mL/1.37 g) = 265.9549 mL= 2.7x102 mL solution Amount of NH3 = n =

13.114 a) Percent N = (mass N/mass NH3) x 100% = (14.01 g N/17.03 g NH3) x 100% = 82.266588 %= 82.27% N b) Percent N = (mass N/mass NH4NO3) x 100% = (2 x 14.01 g N/80.05 g NH 4NO3) x 100% = 35.00312 %= 35.00% N c) Percent N = (mass N/mass (NH4)2HPO4) x 100% = (2 x 14.01 g N/132.06 g (NH4)2HPO4) x 100% = 21.2176 %= 21.22% N

13.115 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 CO 2 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 CO 2     44.01 g CO 2  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 =   1.75 t   12.01 t C  100%  = 3.1498656 t CO 8.85612      1  44.01 t CO2   87%    Mass (g) of CO2 =   1.75 t   12.01 t C  100%  = 0.6299733 t CO2  8.85612     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)t(100%/30%) = 12.59946 t= 13 metric tons. 13.116 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. 13.117 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 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. 13.118 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 H 2O (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. 13.119 a) Group 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 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 16 elements have six valence electrons and gain a complete octet by forming two covalent bonds. An example is H2O. d) Group 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 2 elements have similar characteristics. Thus, either Na or Ca could be an example. e) Group 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 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. 13.120 a) iodic acid;

 2HIO3(s)  I2O5(s) + H2O(l)

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) 13.121 Plan: Find r H for the reaction 2BrF(g)  Br2(g) + F2(g) by applying Hess‘s law to the equations given. Recall that when an equation is reversed, the sign of its r H Solution: 1) 3BrF(g)  Br2(g) + BrF3(l) 2) 5BrF(g)  2Br2(g) + BrF5(l) 3) BrF3(l) + F2(g)  BrF5(l) Reverse equations 1 and 3, and add to equation 2: 1) Br2(g) + BrF3(l)  3BrF(g) 2) 5BrF(g)  2Br2(g) + BrF5(l) 3) BrF5(l)  BrF3(l) + F2(g) Total:

2BrF(g)  Br2(g) + F2(g)

is changed. rH = –125.3 kJ/mol rH= –166.1 kJ/mol rH = –158.0 kJ/mol rH = +125.3 kJ/mol (note sign change) rH = –166.1 kJ/mol rH = +158.0 kJ/mol (note sign change) rH = +117.2 kJ/mol

13.122 2Ca3(PO4)2(s) + 6SiO2(s) +10C(s)  6CaSiO3(s) + 10CO(g) + P4(g)  2 mol Ca 3  PO 4 2   310.18 g Ca 3  PO 4 2   1 kg   100%   Mass (g) of Ca3(PO4)2 =  315 mol P4       1 mol Ca 3  PO 4    103 g   90.%  1 mol P4    2  = 2.17126x102 kg = 2.2x102 kg Ca3(PO4)2 13.123 a) E is a Group 16 element and has six valence electrons. EF 5– would have [1 x E(6e–)] + [5 x 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. 13.124 a) C O b) The formal charge on carbon is –1, and the formal charge on oxygen is +1. FCC = 4 – [2 + ½(6)] = –1 FCO = 6 – [2 + ½(6)] = +1 c) The electronegativity of oxygen partially compensates for the formal charge difference.

13.125 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 stronger the ionic bond, 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. 13.126 Plan: Nitrite ion, NO2–, has [1 x N(5e–)] + [2 x 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 x N(5e–)] + [2 x O(6e–)] = 17 valence electrons. Four electrons are used in the single bonds between the atoms, leaving 17 – 4 = 13 electrons. Since sixteen 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 x N(5e–)] + [2 x 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 (NO 2–) 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 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)

13.127 a)

b) c)

i) CaF2(s) + H2SO4(l)  2HF(g) + CaSO4(s) ii) NaCl(s) + H2SO4(l)  HCl(g) + NaHSO4(s) iii) 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)

13.128 Plan: To find the limiting reactant, find the amount (mol) of UF 6 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)  103 kg  103 g   1.55%   1 mol U  1 mol UF6  Amount 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  Amount of UF6 from ClF3 = 12.75 L   3       10 L   1 mL  92.45 g ClF3  3 mol ClF3  = 86.42509 mol UF6 Since the amount of uranium will produce less uranium hexafluoride, it is the limiting reactant.  352.0 g UF6  4 4 Mass (g) of UF6 =  65.12605 mol UF6    = 2.2924x10 g= 2.29x10 g UF6  1 mol UF6  13.129 Cl2(g) + 2NaOH(aq)  NaClO(aq) + NaCl(aq) + H2O(l)  1 mL   1.07 g  5.25%   1 mol NaClO  1 mol Cl2   22.7 L  Volume (L) of Cl2 = 1000. L   3        10 L   mL   100%   74.44 g NaClO  1 mol NaClO   mol    = 1.69038x104 L= 1.69x104 L Cl2 13.130 Apply Hess‘s law to the two-step process below. The bond energy (BE) of H 2 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+ H = –337 kJ/mol Overall: H + H + H+  H3+

rH = –769 kJ/mol

13.131 The bond energy of H2 = –432 kJ/mol. When two H atoms form the H2 bond, energy is released. 1) 2H(g)  H2(g) H = –432.0 kJ/mol 2) H2(g) + 1/2O2(g)  H2O(g) H = –241.826 kJ/mol Overall: 2H(g) + 1/2O2(g)  H2O(g) H = –673.826 kJ/mol= –673.8 kJ/mol 13.133 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– 1s22s22p6 diamagnetic all orbitals filled (all electrons paired) O2+ 1s22s22p2 paramagnetic Two of the 2p orbitals have one electron each. These electrons have parallel spins (Hund‘s rule).

13.133 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 x width x 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 a) Mass percent = 100%  mass of compound % As in CuHAsO3 =

74.92 g As 100%  = 39.96160% = 39.96% As 187.48 g CuHAsO3

% As in (CH3)3As =

74.92 g As 100%  = 62.4229 %= 62.42% As 120.02 g (CH 3 )3 As

b) Volume (m3) of room = 12.35 m  7.52 m  2.98 m  = 276.75856 m3

3  0.50 mg As   10 g  Mass (g) of As = 276.75856 m3    = 0.13838 g As  m3    1 mg 





 100 g CuHAsO3  Mass (g) of CuHAsO3 =  0.13838 g As    = 0.346282 g= 0.35 g CuHAsO3  39.96160 g As  13.134 a) oxidizing agent, producing H2O b) reducing agent, producing O2

CHAPTER 14 KINETICS: RATES AND MECHANISMS OF CHEMICAL REACTIONS CHEMICAL CONNECTIONS BOXED READING PROBLEMS B14.1

Plan: Use Eq. 16.8 to find the value of k at each temperature, then use Eq. 16.10 to find the value of Ea. 



44210 K 300 K



44210 K

k  Ae RT  (2.5 1016 s 1 ) e

 2.5 1048 s 1

k  Ae RT  (2.5 1016 s 1 ) e 2500 K  5.2 108 s 1

k1 Ea  1 1      k2 R  T2 T1 

 2.5  1048 s 1  Ea 1   1 ln      8 1  J 2500 K 300 K 5.2  10 s    8.314 mol  K Ea  3.6756  105 J/mol=4  105 J/mol

B14.2

Plan: The molecularity of a reaction describes the number of species reacting in an elementary step. If a single species dissociates (one reactant) the reaction is unimolecular. If two species react, the reaction is bimolecular, and so on. In a reaction mechanism, if a substance is present as a reactant in the first step (that is, it is initially consumed) and is reproduced in a subsequent step and it does not appear in the NET reaction, it is considered to be a catalyst. If a species is produced during a reaction and is then consumed in a subsequent step and it does not appear in the NET reaction, it is considered to be an intermediate. Solution:

unimolecular

ii) bimolecular

iii) unimolecular

iv) bimolecular

v) bimolecular

b) CH3 (produced in step i and consumed in step ii); C2H5 (produced in steps ii and iv and consumed in steps iii and v); H (produced in step iii and consumed in steps iv and v).

END–OF–CHAPTER PROBLEMS 14.1

Changes in concentrations of reactants (or products) as functions of time are measured to determine the reaction rate.

14.2

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.

14.3

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.

14.4

An increase in solid surface area would allow more gaseous components to react per unit time and thus would increase the reaction rate.

14.5

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.

14.6

The second experiment proceeds at the higher rate. I2 in the gaseous state would experience more collisions with gaseous H2.

14.7

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.

14.8

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.

14.9

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.

14.10

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)

Concentration

B(g)

A(g) B(g)

Silberberg, Amateis, Venkateswaran, Chen, Chemistry: The Molecular Nature of Matter and Change, 3rd Canadian Edition Page 3-453 Time Instructor‘s Solution Manual © Copyright 2021 McGraw-Hill Ryerson Ltd. Time

14.11

14.12

a) (i) 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. (ii) Calculate the negative of the slope of the line tangent to the curve at t = x. (iii)Calculate the negative of the slope of the line tangent to the curve at t = 0. b) If you plotted [D] vs. time, you would not need to take the negative of the slopes in i)-iii) since [D]would increase over time. 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 s to t = 20.0 s is proportional to the slope of the line connecting these two points: 1  0.0088 mol/L  0.0500 mol/L 1 [AX 2 ] Rate =  =  = 0.00103 mol/L•s= 0.0010 mol/L•s 2 2 t  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.

[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 s 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. 14.13

Plan: The average rate is the total change in concentration divided by the total change in time. Solution: 1 0.0088 mol/L  0.0249 mol/L  1 [AX 2 ] a) Rate =  =  = 6.70833x10–4 = 6.71x10–4 mol/L•s 2 2 t 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 –2.8x10–3 mol/L•s. 0.06 0.05

[AX2]

0.04 0.03 0.02 0.01 y = -0.0028x + 0.047

0 0

TIME, S

14.14

Plan: Use Equation 14.2 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: a) Expressing the rate in terms of each component: 1 [A] [B] [C] Rate =  = = 2 t t t Calculating the rate of change of [A]: 1 [A] [C]  = 2 t t 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.

14.15

Plan: Use Equation 14.2 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: a) Expressing the rate in terms of each component: [D] 2 [E] 2 [F] Rate =  = = t 3 t 5 t b) Calculating the rate of change of [F]: 5/2 mol F/L•s   0.25 mol E/L•s    = 0.416667 mol/L•s = 0.42 mol/L•s  3/2 mol E/L•s 

14.16

Plan: Use Equation 14.2 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 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: a) Expressing the rate in terms of each component: [A] 1 [B] [C] Rate =  =  = t 2 t t b) Calculating the rate of change of [A]: 1 mol A/L•s   0.5 mol B/L•s    = – 0.25 mol/L•s = – 0.2 mol/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.

14.17

Plan: Use Equation 14.2 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: a) Expressing the rate in terms of each component: 1 [D] 1 [E] [F] 1 [G] [H] =  =  = = 2 t 3 t t 2 t t 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 

Rate =  b)

14.18

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)

14.19

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

14.20

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: [NOBr] 0.0033  0.0100 mol/L a) Rate =  – = 6.7x10–4 mol/L•s t 10.00  0.00 s 0.0055  0.0071 mol/L b) Rate = – = 8.0x10–4 mol/L•s 4.00  2.00 s c)

Initial Rate = –  y/ x = – [(0.0040 – 0.0100) mol/L]/[4.00 – 0.00) s] = 1.5x10–3 mol/L•s d) Rate at 7.00 s = – [(0.0030 – 0.0050) mol/L]/[11.00 – 4.00) s] = 2.857x10–4 = 2.9x10–4 mol/L•s

e) Average between t = 3 s and t = 5 s is: Rate = – [(0.0050 – 0.0063) mol/L]/[5.00 – 3.00) s] = 6.5x10–4 mol/L•s Rate at 4 s  6.7x10–4 mol/L•s, thus the rates are equal at about 4 seconds. 14.21

Plan: Use Equation 14.2 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 N 2 and H2 since these substances are reacting and [N2] and [H2] are decreasing over time. A positive sign is used for the rate in terms of the product NH 3 since it is being formed and [NH3] increases over time. Solution: [N 2 ] 1 [H 2 ] 1 [NH3 ] Rate =  =  = t 3 t 2 t

14.22

Plan: Use Equation 14.2 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 the reactant O 2 since it is reacting and [O2] is decreasing over time. A positive sign is used for the rate in terms of the product O 3 since it is being formed and [O3] increases over time. O3 increases 2/3 as fast as O2 decreases because two molecules of O3 are formed for every three molecules of O2 that disappear. Solution: 1 [O2 ] 1 [O3 ] a) Rate =  = 3 t 2 t b) Use the mole ratio in the balanced equation:

 2.17x105 mol O2 /L•s   2 mol O3 /L•s  –5     = 1.45x10 mol/L•s 3 mol O /L•s 2     14.23

a) k is the rate constant, the proportionality constant in the rate law. k represents the fraction of successful collisions which includes the fraction of collisions with sufficient energy and the fraction of collisions with correct orientation. k is a constant that varies with temperature. 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 rate mol/L•min mol/L•min k= = = 1 2 [A]1[B]2 mol3  mol   mol   L   L  L3 k=

mol  L3    L•min  mol3 

k = L2/mol2•min 14.24

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.

14.25

a) The rate doubles. If rate = k[A]1 and [A] is doubled, then the rate law becomes rate = k[2 x 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 x 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 x C]2. The rate increases to 32 or 9 times its original value.

14.26

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

14.27

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

14.28

a) The rate is first order with respect to [BrO3–]. If [BrO3–] is doubled, rate = k[2 x 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 x Br–], then rate decreases by a factor of (1/2)1 or 1/2 times its original value. The rate is halved. c) The rate is second order with respect to [H+]. If [H+] is quadrupled, rate = k[4 x H+]2, then rate increases to 42 or 16 times its original value.

14.29

a) The rate is second order with respect to [O3]. If [O3] is doubled, rate = k[2 x 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 x O2] –1, then rate changes by 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 x O2] –1, then rate changes by a factor of (1/2)–1 or 2 times its original value. The rate increases by a factor of 2.

14.30

14.31

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.

14.32

a) The rate is second order with respect to [NO 2]. If [NO2] is tripled, rate = k[3 x 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 [NO 2] and first order with respect to [Cl2]. If [NO2] and [Cl2] are doubled, rate = k[2 x NO2]2[2 x Cl2]1, then the rate increases by a factor of 22 x 21 = 8. c) The rate is first order with respect to [Cl2]. If Cl2 is halved, rate = k[1/2 x Cl2]1, then rate decreases to 1/2 times its original value. The rate is halved.

14.33

a) The rate is fourth order with respect to [HNO 2]. If [HNO2] is doubled, rate = k[2 x HNO2]4, then rate changes by 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 x NO]–2, then rate changes by 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 x HNO2]4, then rate changes by (1/2)4 or 1/16 times its original value. The rate decreases by a factor of 16.

14.34

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].

 [A]exp 2  =   [A]exp 1  rateexp 1  

rateexp 2

45.0 mol/L•min  0.300 mol/L  =  5.00 mol/L•min  0.100 mol/L  m 9.00 = (3.00) log (9.00) = m log (3.00) m=2 Using experiments 3 and 4 also gives second order with respect to [A]. 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  =   [B]exp 1  rateexp 1   n 10.0 mol/L•min  0.200 mol/L  =  5.00 mol/L•min  0.100 mol/L  n 2.00 = (2.00) 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= = = 5.00x103 L2/mol2•min 2 2 [0.100 mol/L] [0.100 mol/L] [A] [B] rateexp 3

14.35

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].

 [A]exp 2  =   [A]exp 1  rateexp 1  

rateexp 2

1.25x102 mol/L•min  0.1000 mol/L  =  3 6.25x10 mol/L•min  0.0500 mol/L  m 2.00 = (2.00) log (2.00) = m log (2.00) m=1 The order is first order with respect to A. 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]. m

 [B]exp 3  =   [B]exp 2  rateexp 2   rateexp 3

5.00  102 mol/L•min

 0.1000 mol/L  =  2 1.25  10 mol/L•min  0.0500 mol/L  4.00 = (2.00)n log (4.00) = n log (2.00) n=2 The reaction is second order with respect to B. 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].

 [C]exp 4  =   [C]exp 1  rateexp 1  

rateexp 4

6.25  103 mol/L•min

 0.0200 mol/L  =  3 6.25  10 mol/L•min  0.0100 mol/L  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 rate 6.25 x 103 mol/L min k= = = 50.0 L2/mol2•s 2 2 [A][B] [0.0500 mol/L][0.0500 mol/L] 14.36

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•time  L  mol L mol L2 k = L•time = =  2 2 mol •time L•time mol mol 2 L 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 mol 3/L3, k has units of L2/mol2•time: Rate = k[A]3 3

mol  mol  =k  L•time  L  mol mol L3 L2  k = L•time = = L•time mol 2 •time mol3 mol3 L3 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: 5/2

mol  mol  = k  L•time  L  mol mol L5/2 L3/2  k = L•time = = L•time mol 3/2 •time mol5/2 mol5/2

L5/2 14.37

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  L  = mol   L•s

 mol   L  =1   b) Rate = k[A]m mol 1  mol  = L•yr yr  L 

m must be 0. The reaction is zero order.

 mol   L     mol   L   

mol mol yr L•yr =  = 1 L yr 1 yr

mol L

m must be 1. The reaction is first order.

c) Rate = k[A]m m

mol mol1/2  mol  = 1/2  L•s L •s  L  mol m mol L1/2 •s  mol  L•s  L  = mol1/2 = L•s  mol1/2   L1/2 •s

 mol   L   

mol1/2

m must be 1/2. The reaction is 1/2 order.

L1/2

d) Rate = k[A]m mol mol 5/2  mol  = 5/2 L• min L • min  L  mol m 5/2  mol  L• min = mol  L • min =  L  L• min mol 5/2 mol5/2   m

L5/2 • min  mol   L    14.38

mol7/2 L7/2

m must be 7/2. The reaction is 7/2 order.

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 [Cl 2] 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].

 [CO]exp 1  =   [CO]exp 2  rateexp 2   rateexp 1

1.29  1029 mol/L•min

 1.00 mol/L  =  30 1.33  10 mol/L•min  0.100 mol/L  m 9.699 = (10.0) 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 [Cl 2]. 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].

 [Cl2 ]exp 3  =   [Cl2 ]exp 2  rateexp 2   rateexp 3

1.30  1029 mol/L•min

 1.00 mol/L  =  30 1.33  10 mol/L•min  0.100 mol/L  n 9.774 = (10.0) log (9.774) = n log (10.0) n = 0.9901 = 1 The reaction is first order with respect to [Cl2]. Rate = k[CO][Cl2] b) k = rate/[CO][Cl2] Exp 1: k1 = (1.29x10–29 mol/L•s)/[1.00 mol/L][0.100 mol/L] = 1.29x10 –28 L/mol•s Exp 2: k2 = (1.33x10–30 mol/L•s)/[0.100 mol/L][0.100 mol/L] = 1.33x10 –28 L/mol•s Exp 3: k3 = (1.30x10–29 mol/L•s)/[0.100 mol/L][1.00 mol/L] = 1.30x10 –28 L/mol•s Exp 4: k4 = (1.32x10–31 mol/L•s)/[0.100 mol/L][0.0100 mol/L] = 1.32x10 –28 L/mol•s kavg = (1.29x10–28 + 1.33x10–28 + 1.30x10–28 + 1.32x10–28) L/mol•s/4 = 1.31x10–28 L/mol•s 14.39 a)

b) i) ii) iii)

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. The reaction is first order since ln[reactant] vs. time is linear. The reaction is second order since 1/[reactant] vs. time is linear. The reaction is zero order since [reactant] vs. time is linear.

14.40

The half-life (t1/2) 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.

14.41

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 mol/L) = 0.500 mol/L), 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 mol / L 1.50 mol / L  t= 0.2 L/mol•s t = 6.6667 = 7 s 14.42

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. We know k (0.2 L/mol•s), [AB]0 (1.50mol/L), and t (10.0 s), so we can solve for [AB]t. Solution: 1 1 = kt  ABt AB0 1

ABt

= kt +

AB0

1 1 = (0.2 L/mol•s) (10.0 s) + 1.50 mol / L ABt 1 1 = 2.66667 mol / L ABt

[AB]t = 0.375 mol/L= 0.4 mol/L 14.43

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 mol/L and then [A]t = 50% of 1 M = 0.5 mol/L. 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 mol/L and then [A]t = 25% of 1 mol/L = 0.25 mol/L. 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 k= = = 0.066014 min-1= 0.0660 min–1 t1/2 = t1/2 k 10.5 min b) ln [A]t = ln [A]0 – kt ln[A]t  ln[A]0 =t k ln[0.25]  ln[1] =t 0.0660 min 1 t = 21.0045 min= 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.

14.44

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 mol/L and then [A]t = 12.5% of 1 mol/L = 0.125 mol/L. Solution: ln 2 ln 2 a) t1/2 = = = 577.62 yr= 5.8x102 yr  1 k 0.0012 yr b) ln [A]t = ln [A]0 – kt [A]0 ln = kt [A]t [A]0 = 1.00 mol/L [A]t = 0.125 mol/L k = 0.0012 yr–1  1.00 mol / L  –1 ln   = (0.0012 yr ) t 0.125 mol / L   t = 1732.86795 yr= 1.7x103 yr If the student recognizes that 12.5% remaining corresponds to three half-lives; then simply multiply the answer in part a) by three.

14.45

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.)

ln[NH3]

x-axis (time, s) 0 1.000 2.000

[NH3] 4.000 mol/L 3.986 mol/L 3.974 mol/L

y-axis (ln [NH3]) 1.38629 1.38279 1.37977

1.387 1.386 1.385 1.384 1.383 1.382 1.381 1.38 1.379 0

0.5

1.5

Time, s

k = slope = rise/run = (y2 – y1)/(x2 – x1) k = (1.37977 – 1.38629)/(2.000 – 0) = (0.00652)/(2) = 3.260x10 –3 s–1 = 3x10–3 s–1 (Note that the starting time is not exact, and hence, limits the significant figures.) ln 2 ln 2 b) t1/2 = = = 212.62 s= 2x102 s k 3.260x103 s 1 14.46

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.

14.47

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.

14.48

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.

14.49

k = AeEa / RT 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.

14.50

The Arrhenius equation, k = AeEa / RT , can be used directly to solve for activation energy at a specified temperature if the rate constant, k, and the pre-exponential factor, A, are known. However, the pre-exponential factor is usually not known. To find Ea without knowing A, rearrange the Arrhenius equation to put it in the form of a linear plot: ln k = ln A – Ea/RT where the y value is ln k and the x value is 1/T. Measure the rate constant at a series of temperatures and plot ln k vs. 1/T. The slope equals –Ea/R.

14.51

a) The value of k increases exponentially with temperature. b) A plot of ln k vs. 1/T is a straight line whose slope is –Ea/R. a)

The activation energy is determined from the slope of the line in the ln k vs. 1/T graph. The slope equals –Ea/R. 14.52

a) As temperature increases, the fraction of collisions which exceed the activation energy increases; thus, the reaction rate increases. b) A decrease in activation energy lowers the energy threshold with which collisions must take place to be effective. At a given temperature, more collisions occur with the lower energy so rate increases.

14.53

No. For 4x10–5 moles of EF to form, every collision must result in a reaction and no EF molecule can decompose back to AB and CD. Neither condition is likely. All collisions will not result in product as some collisions will occur with an energy that is lower than the activation energy. In principle, all reactions are reversible, so some EF molecules decompose. Even if all AB and CD molecules did combine, the reverse decomposition rate would result in an amount of EF that is less than 4x10 –5 moles.

14.54

Collision frequency is proportional to the velocity of the reactant molecules. At the same temperature, both reaction mixtures have the same average kinetic energy, but not the same velocity. Kinetic energy equals 1/2 mv2, where m is mass and v velocity. The trimethylamine (N(CH3)3) molecule has a greater mass than the ammonia molecule, so trimethylamine 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 (CH 3). 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. 14.55

Each A particle can collide with three B particles, so (4 x 3) = 12 unique collisions are possible.

14.56

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 1023 A particles  23 Number of particles of A = 1.01 mol A    = 6.08222x10 particles of A  1 mol A  

 6.022 1023 B particles  24 Number of particles of B =  2.12 mol B    = 1.279997x10 particles of B 1 mol B   23 Number of collisions = (6.08222 x 10 particles of A)(1.279997 x 1024 particles of B) = 7.76495x1047 = 7.76x1047 unique collisions 14.57

Plan: The fraction of collisions with a specified energy is equal to the eEa / RT term in the Arrhenius equation. Solution: f = eEa / RT T = 25°C + 273 = 298 K Ea = 100. kJ/mol R = 8.314 J/mol•K = 8.314x10–3 kJ/mol•K E 100. kJ/ mol  a =  = –40.362096 RT 8.314  103 kJ/mol•K  298 K 





Fraction = eEa / RT 14.58

–40.362096

= 2.9577689x10–18 = 2.96x10–18

Plan: The fraction of collisions with a specified energy is equal to the eEa / RT term in the Arrhenius equation. Solution: f = eEa / RT T = 50.°C + 273 = 323 K Ea = 100. kJ/mol R = 8.314 J/mol•K = 8.314x10–3 kJ/mol•K E 100. kJ/ mol  a =  = –37.238095 RT 8.314  103 kJ/mol•K  323 K 





Fraction = eEa / RT = e–37.238095 = 6.725131x10–17 The fraction increased by (6.725131x10–17)/(2.9577689x10–18) = 22.737175 = 22.7 14.59

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 = 4.7x10–3 s–1 T1 = 25°C + 273 = 298 K k2 = ? T2 = 75°C + 273 = 348 K Ea = 33.6 kJ/mol = 33,600 J/mol E  1 1 k ln 2 =  a    k1 R  T2 T1  ln

3 1

= 

33,600 J/mol  1 1     8.314 J/mol•K  348 K 298 K 

4.7  10 s k2 ln = 1.948515 (unrounded) 4.7x103 s 1

Raise each side to e x

k2 = 7.0182577 4.7x103 s 1 k2 = (4.7x10–3 s–1)(7.0182577) = 0.0329858 s-1= 0.033 s–1

14.60

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 = 4.50x10–5 L/mol•s T1 = 195°C + 273 = 468 K k2 = 3.20x10–3 L/mol•s T2 = 258°C + 273 = 531 K Ea = ? k E  1 1 ln 2 =  a    k1 R  T2 T1   k  J   3.20x103 L/mol•s   R  ln 2    8.314 mol•K   ln 5 k1     4.50x10 L/mol•s  Ea =   =   1  1 1 1         531 K 468 K   T2 T1  Ea = 1.3984658x105 J/mol = 1.40x105 J/mol

14.61

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 rH = 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

 rH –55 kJ/mol

AB + CD

Reaction coordinate b) rH = Ea(fwd) – Ea(rev) Ea(rev) = Ea(fwd) – rH = 215 kJ/mol – (–55 kJ/mol) = 2.70x102 kJ/mol c) bond forming

B A

bond weakening

14.62

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 rH = Ea(fwd) – Ea(rev) to solve for rH. 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)

b) rH = Ea(fwd) – Ea(rev) = 125 kJ/mol – 85 kJ/mol = 40 kJ/mol c) A.....B | | A.....B 14.63

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 = 0.76/s T1 = 727°C + 273 = 1000. K k2 = 0.87/s T2 = 757°C + 273 = 1030. K Ea = ? E  1 1 k ln 2 =  a    k1 R  T2 T1 

 k  J   0.87 / s   R  ln 2  8.314  ln   k1  mol•K   0.76 / s   Ea =   =   1 1   1 1        1030. K 1000. K   T2 T1  Ea = 3.8585x104 J/mol = 3.9x104 J/mol

14.64

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 rH = 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: a) NOCl + Cl

Ea(rev)

Energy

r = +83 kJ

Ea(fwd) = +86 kJ NO + Cl2

Reaction coordinate b) rH = Ea(fwd) – Ea(rev) Ea(rev) = Ea(fwd) – Hrxn = 86 kJ – 83 kJ = 3 kJ. c) To draw the transition state, look at structures of reactants and products: Cl

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

14.65

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.

14.66

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.

14.67

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.

14.68

Reaction intermediates have some stability, however limited, but transition states are inherently unstable. Additionally, unlike transition states, intermediates are molecules with normal bonds.

14.69

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.

14.70

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.

14.71

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.

14.72

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 bimolecular rate1 = k1[A][B] A(g) + B(g) X(g) bimolecular rate2 = k2[X][C] X(g) + C(g)  Y(g) unimolecular rate3 = k3[Y] Y(g)  D(g) 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.

14.73

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) 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, rate forward 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. 14.74

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 (NOBr 2). 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, rate forward 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.

14.75

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 Rate = k2[N2O2][O2] N 2O2 is an intermediate Rate1 (forward) = rate1 (reverse) k1[NO]2 = k–1[N2O2] [N2O2] = (k1/k–1)[NO]2 Substitute this for [N2O2] in the rate law: Rate = k[N2O2][O2] Rate = k2(k1/k–1)[NO]2[O2] Rate = k[NO]2[O2] III. (1) 2NO(g) N2(g) + O2(g) (2) N2(g) + 2O2(g)  NO2(g) slow Rate = k2[N2][O2]2 N2 is an intermediate Rate1 (forward) = rate1 (reverse) k1[NO]2 = k–1[N2][O2] [N2] = (k1/k–1)[NO]2/[[O2] = (k1/k–1)[NO]2[O2]–1 Substitute this for [N2] in the rate law: Rate = k2[N2][O2]2 Rate = k2(k1/k–1)[NO]2 [O2]–1[O2]2 Rate = k[NO]2[O2] a) All the mechanisms are consistent with the rate law.

b) The most reasonable mechanism is II, since none of its elementary steps are more complicated than being bimolecular. Mechanism I and III have steps that are termolecular. 14.76

Plan: Review the definitions of homogeneous vs. heterogeneous catalysts. To draw the reaction energy diagrams, recall that addition of a catalyst to a reaction results in a lower activation energy. Solution: a) A heterogeneous catalyst speeds up a reaction that occurs in a different phase. Gold is a heterogeneous catalyst since it is a solid and is catalyzing a reaction in the gas phase. b) The activation energy for the uncatalyzed reaction must be greater than the activation for the catalyzed reaction.

14.77

No, a catalyst changes the mechanism of a reaction to one with lower activation energy. Lower activation energy means a faster reaction. An increase in temperature does not influence the activation energy, but instead increases the fraction of collisions with sufficient energy to react.

14.78

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.

14.79

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.

14.80

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.

14.81

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:

 [NO 2 ]2  rate2 =  rate1  [NO 2 ]1 

4 = [2]m Silberberg, Amateis, Venkateswaran, Chen, Chemistry: The Molecular Nature of Matter and Change, 3rd Canadian Edition Page 3-474 Instructor‘s Solution Manual © Copyright 2021 McGraw-Hill Ryerson Ltd.

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:

 [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.

k  NO 2  Rateinitial Reaction 1: = =4 2 Rate50% k 0.5  NO 2   k  NOO3  Rateinitial Reaction 2: = =4 Rate50% k 0.5  NO   0.5  O3   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 NO 2  Rateinitial = = 1.7778 = 1.8 2 Rate50% k  0.75 NO 2  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  Rateinitial = = 2.6667 = 2.7 Rate50% k 0.75 NO  0.5 O3  2

14.82

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.

14.83

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. 14.84

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 15.5% of the 14C remains at time t, let [14C]0 = 100% and then [14C]t = 15.5%. Solution: ln 2 ln 2 ln 2 k= = = 1.20968x10–4 yr–1 (unrounded) t1/2 = t1/2 5730 yr k Use the first-order integrated rate law ln [14C]t = ln [14C]0 – kt or  14 C   t = – kt ln  14 C   0

15.5%t = –(1.20968x10–4 yr–1) t 100%0

t = 1.541176x104 yr= 1.54x104 yr 14.85

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 T 2 = 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 = ? E  1 1 k ln 2 =  a    k1 R  T2 T1 

 k  J   1/16   R  ln 2   8.314 mol•K   ln 1/4  k   1   Ea =   =   1 1  1 1         273 K 293 K   T2 T1  Ea = 4.6096266x104 J/mol = 4.61x104 J/mol The significant figures are based on the Kelvin temperatures. 14.86

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 k= = 7.07293x10–5 d–1 (unrounded) t1/2 = t1/2 k Use the first-order integrated rate law with BP = benzoyl peroxide: BPt ln = – kt BP0

t = 725.2057 d= 7.3x102 d 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: a)

Energy

14.87

95%t = – (7.07293x10–5 d–1) t 100%0

CO + NO2

 = –226kJ/mol

CO2 + NO

Reaction coordinate b) The rate of the mechanism is based on the slowest step, 2NO 2(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. 14.88

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 14.2 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].

k  A 2  B2  C2 Rate 2 = x y z Rate1 k  A   B  C  x

[B]1 = [B]2

k  A 2 Rate 2 = x Rate1 k A

[C]1 = [C]2

k  0.096 mol/L2 9.6x10 mol/L•s = 6 x 6.0x10 mol/L•s k  0.024 mol/L 5

16 = 4x x=2 Second order with respect to A 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 y z Rate 4 k  A   B  C  2

[C]2 = [C]4

k  A 2  B2 Rate 2 = 2 y Rate 4 k  A   B 2

9.6  10 mol/L•s 1.5  106 mol/L•s

k  0.096 mol/L 2 0.085 mol/L 2 2

5

k  0.012 mol/L 4 0.170 mol/L 4 2

64 = (8)2 (0.5)y y=0 Zero order with respect to B 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].

k  A 1  B1  C1 Rate1 = 2 0 z Rate3 k  A   B  C  2

[A]1 = [A]3

k  B1  C1 Rate1 = 0 z Rate3 k  B  C  0

k  0.085 mol/L 1 0.032 mol/L 1 6.0x106 mol/L•s = 0 z 1.5x105 mol/L•s k  0.034 mol/L  0.080 mol/L  0

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] rate 6.0  106 mol/L•s 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))   C   D 1  A 1   B 1  E d) Rate =  =  =  = = 2 t 2 t t t 3 t 14.89

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 mol/L to 2.5 mol/L , or is increased by a factor of 2.5/1.0 or 2.5. Then the rate = k[H+][2.5 x sucrose]; the rate increases by a factor of 2.5.

b) The [sucrose] is changed from 1.0 mol/L to 0.5 mol/L , or is decreased by a factor of 0.5/1.0 or 0.5. Then the rate = k[H+][0.5 x 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 mol/L to 0.0001 mol/L , or is decreased by a factor of 0.0001/0.01 or 0.01. Then the rate = k[0.01 x 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 mol/L to 0.1 mol/L , or by a factor of (0.1 mol/L /1.0 mol/L ) = 0.1. [H+] increases from 0.01 mol/L to 0.1 mol/L , or by a factor of (0.1 mol/L /0.01 mol/L ) = 10. Then the rate will increase by k[10 x H+][0.1 x sucrose]= 1.0 times as fast. Thus, there will be no change. 14.90

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.

14.91

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    1 sphere 

6 spheres A  Concentration of A =

Concentration of B =

0.50 L  0.010 mol B  5 spheres B   1 sphere 

= 0.12 mol/L

= 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 8.3x10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  = 0.14 mol/L Concentration of A = 0.50 L

 0.010 mol B    1 sphere 

8 spheres B Concentration of B =

0.50 L

= 0.16 mol/L

Rate = k[A][B] Rate = 0.069167 L/mol·min[0.14 mol/L][0.16 mol/L] Rate = 1.5493x10–3 mol/L•min= 1.5x10–3 mol/L•min

14.92

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.

14.93

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. Solution: ln 2 ln 2 Rearrange t1/2 = to k = t1/2 k k=

ln 2 = 5.7762x10–2 yr–1 12 yr

Use the first-order integrated rate law: ln

 DDT t = – kt  DDT 0

10. ppbmt = – (5.7762x10–2 yr–1) t  275 ppbm0

t = 57.3765798 yr= 57 yr 14.94

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 = t1/2 k ln 2 = 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 min= 5.0 min

14.95

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: rate 0.20 mol/L•s 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) [2.0mol/L]2[3.0mol/L] = 2.4 mol/L•s

14.96

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. Solve the first-order half-life equation to find t1/2: ln 2 ln 2 = = 3.3007 h= 3.3 h t1/2 = k 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

1 0 0

0.5

-1

1.5

2.5

3.5

y = -0.21x - 0.6936 R² = 0.9998

-2 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 100%  0

t = 6.6014 h= 6.6 h Silberberg, Amateis, Venkateswaran, Chen, Chemistry: The Molecular Nature of Matter and Change, 3rd Canadian Edition Page 3-481 Instructor‘s Solution Manual © Copyright 2021 McGraw-Hill Ryerson Ltd.

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. 14.97

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 = 2.3x1014 T = 37°C + 273 = 310 K k1 ΔEa = ? k1 = Ae–Ea1/RT

and

k2 = Ae–Ea2/RT

 Ea 2

RT  Ea1  Ea2  Ae k2 = = e  Ea1 RT k1 RT Ae

k2 E  Ea2 = a1 k1 RT

 k  RT  ln 2  = Ea1 – Ea2  k1 

2.3x1014 = Ea1 – Ea2 1 Ea1 – Ea2 = 8.5230x104 J/mol= 8.5x104 J/mol

(8.314 J/mol•K) (310 K) ln

14.98

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, however it also depends on the rate law for the substance and partial order with respect to that gas. 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 pre-exponential 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 H 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.

14.99

Plan: To find concentration of reactant at a later time, given the initial concentration and the rate constant k, use an integrated rate law expression. Since the units on k are s–1, this is 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. To find the fraction of reactant that has decomposed, divide that amount of reactant that has decomposed ([N2O5]0 – [N2O5]t) by the initial concentration. Solution:

 60 s  a)Converting t in min to s:  5.00 min    = 300. s  1 min 

N 2 O5 t = – kt or N 2 O5 0

ln [N2O5]t = ln [N2O5]0 – kt ln [N2O5]t = ln [1.58 mol/L] – (2.8x10–3 s–1)(300. s) ln [N2O5]t = – 0.382575 [N2O5]t = 0.68210 mol/L= 0.68 mol/L [N 2O5 ]0  [N 2O5 ]t 1.58  0.68210 mol/L b) Fraction decomposed = = = 0.56829 = 0.57 [N 2O5 ]0 1.58 mol/L 14.100 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, rate forward 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. 14.101 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. 14.102 Plan: To solve this problem, a clear picture of what is happening is useful. Initially only N 2O5 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 N 2O5 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: PN2O5  PNO2  PO2 = (125 – x) + 2x + 1/2x = 178 x = 35.3333 kPa (unrounded) Partial pressure of NO2 equals 2x = 2(35.3333 kPa) = 70.667kPa = 71 kPa. Check: Substitute values for all partial pressures to find total final pressure: (125 – 35.3333)kPa + (2 x 35.3333 kPa) + ((1/2) x 35.3333 kPa) = 178 kPa The result agrees with the given total final pressure. Silberberg, Amateis, Venkateswaran, Chen, Chemistry: The Molecular Nature of Matter and Change, 3rd Canadian Edition Page 3-483 Instructor‘s Solution Manual © Copyright 2021 McGraw-Hill Ryerson Ltd.

14.103 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 = t1/2 k

ln 2 = 7.7016x10–3 min–1 (unrounded) 90 min  60 min  Converting t from h to min:  2.5 h    = 150 min  1h  k=

aspirin t = – kt or aspirin 0

ln [aspirin]t = ln [aspirin]0 – kt ln [aspirin]t = ln [2 mg/100 mL] – (7.7016x10–3 min–1)(150 min) ln [aspirin]t = – 5.06729 [aspirin]t = 6.29964x10–3 mg/mL or

6.29964x103 mg x 100 mL = 0.62996 mg/100 mL = 0.6 mg/100 mL mL 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 = – (3.9x10–5 s–1) t 1 PILL  0

 1h  t = 28169.55 s or  28169.55 s    = 7.8 h  3600 s  Pills should be taken at about eight hour intervals. k1 = 3.1x10–5 s–1 T1 = 37.0°C + 273.15 = 310.15 K k2 = 3.9x10–5 s–1 T2 = 38.8°C + 273 = 311.95 K Ea = ? k E  1 1 ln 2 =  a    k1 R  T2 T1 

 

 

 3.9  105 s 1  J     k2   8.314 mol•K   ln R  ln  5 1     3.1  10 s k1    Ea =   =   1 1   1 1        311.95 K 310.15 K   T2 T1 

Ea = 1.0259x105 J/mol = 1x105 J/mol The subtraction of the 1/T terms leaves only one significant figure. 14.104 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.

14.105 a) The iodide ion approaches from the side opposite the relatively large chlorine.

b) The ―backside attack‖ of the I– inverts the geometry at the carbon bearing the Cl, producing this product:

c) The planar intermediate can be attached from either side, producing a racemic mixture (that is, an equal mixture of two optical isomers):

14.106 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. Solution: k1 = 1 egg/4.8 min T1 = 90.0°C + 273.2 = 363.2 K k2 = 1 egg/4.5 min T2 = 100.0°C + 273.2 = 373.2 K Ea = ? E  1 1 k ln 2 =  a    k1 R  T2 T1  The number of eggs (1) is exact, and has no bearing on the significant figures. J   1 egg/4.5min    k   R  ln 2   8.314 mol•K   ln 1 egg/4.8 min   k1     Ea =   =     1 1 1 1        373.2 K 363.2 K   T2 T1  Ea = 7.2730x103 J/mol = 7.3x103 J/mol 14.107 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. SO 3 is produced in step 1 and its concentration is dependent on k1 and [H2SO4]: [SO3] = k1[H2SO4]2 Silberberg, Amateis, Venkateswaran, Chen, Chemistry: The Molecular Nature of Matter and Change, 3rd Canadian Edition Page 3-485 Instructor‘s Solution Manual © Copyright 2021 McGraw-Hill Ryerson Ltd.

Substituting for [SO3] in the rate law from the slow step: Rate = k2[k1[H2SO4]2][C6H6] Rate = k[H2SO4]2[C6H6] 14.108 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 mol / L  k1  NO 2    [O] = = = 2.4x10–15 mol / L 6 2 k2  O 2    1.0  10 L/mol•s 1.0  10 mol / L   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] = (1.0x106 L/mol•s)(2.4x10–15 mol/L)(1.0x10–2 mol/L) = 2.4x10–11 mol/L·s





14.109 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. Solution: a) Calculating [A]t at 3.00 min (99.9% inactivation): % inactivation = 100 x (1 – [A]t/[A]0) At 3.00 min: 99.9% = 100 x (1 – [A]t/1.00] 99.9% = 100 – 100[A]t [A]t = 0.001  A t ln = – kt  A 0

ln k= 

 A t  A 0

0.001t 1.000

= 2.302585 min-1= 2.3 min–1 3.00 min b) Calculating [A]t at 95% inactivation: % inactivation = 100 x (1 – [A]t/[A]0) 95% = 100 x (1 – [A]t/1.00) 95% = 100 – 100[A]t [A]t = 0.05  A t ln = – kt  A 0

ln t= 

 A t  A 0 k

= 

ln = 

0.05t 1.000

2.302585 min 1

= 1.30103 min= 1.3 min

14.110 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: a) 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. b) The first step is the rate-determining step. From step 1: Silberberg, Amateis, Venkateswaran, Chen, Chemistry: The Molecular Nature of Matter and Change, 3rd Canadian Edition Page 3-486 Instructor‘s Solution Manual © Copyright 2021 McGraw-Hill Ryerson Ltd.

Rate = k1[N2O5]2 This rate law does not match the actual rate law so the proposed mechanism is not valid. 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, rate forward 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. 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, rate forward 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 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. 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.

14.111 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)i) [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= 

ii)

 NH 3 t  NH 3 0

k 0.35 mol/m3   t ln 3 3.0 mol/m   0 t=  = 2.84272 d= 2.8 d 1 0.75576667 d 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  NH 3 t ln  NH 3 0 t=  k 0.35 mol/m3   t ln 3.0 mol/m3   0 t=  0.292285976 d 1 t = 7.35045 d= 7.4 d For NH4+ the rate = k1[NH4+] From the balanced chemical equation:

  NH 4    =  1  O 2  2 t t Thus, for O2: Rate = 2 k1[NH4+] 

Rate = 2 0.75576667 3.0 mol/m3  = 4.5346 = 4.5 mol/m3  

14.112 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:

 [CS2 ]exp 1  a) =   [CS2 ]exp 4  Rateexp 4   Rateexp 1

2.7x107 mol/L•s  0.100 mol/L  =  1.2x107 mol/L•s  0.044 mol/L  m 2.25 = (2.27273) 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 = (2.7x10–7 mol/L•s)/(0.100 mol/L) = 2.7x10–6 s–1 k2 = (2.2x10–7 mol/L•s)/(0.080 mol/L) = 2.75x10–6 s-1= 2.8x10–6 s–1 k3 = (1.5x10–7 mol/L•s)/(0.055 mol/L) = 2.7272x10–6 s-1= 2.7x10–6 s–1 k4 = (1.2x10–7 mol/L•s)/(0.044 mol/L) = 2.7272x10–6 s-1= 2.7x10–6 s–1 kavg = [(2.7x10–6 s–1) + (2.75x10–6 s-1) + (2.7272x10–6 s-1) + (2.7272x10–6 s-1)]/4 = 2.7261 10–6 s-1= 2.7x10–6 s–1 m

14.113 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. Ea (hydrogenation) Energy

Ea(dehydrogenation)

Alkene + H2

Hrxn

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.

14.114 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 colour. 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 ciVi = cfVf in the form: cf = ciVi/Vf. Solution: a) cf S2O32– = [(10.0 mL)(0.0050 mol/L)]/50.0 mL = 0.0010 mol/L S2O32– 2    I 2  1  S2 O3   =  = [1/2(0.0010 mol/L)]/time = [0.00050 mol/L]/time = rate t 2 t Average rates: Rate1 = (0.00050 mol/L)/29.0 s = 1.724x10–5 mol/L•s= 1.7x10–5 mol/L•s Rate2 = (0.00050 mol/L)/14.5 s = 3.448x10–5 mol/L•s = 3.4x10–5 mol/L•s Rate3 = (0.00050 mol/L)/14.5 s = 3.448x10–5 mol/L•s = 3.4x10–5 mol/L•s b) cf KI = cf I– = [(10.0 mL)(0.200 mol/L)]/50.0 mL = 0.0400 mol/L I– (Experiment 1) cf I– = [(20.0 mL)(0.200 mol/L)]/50.0 mL = 0.0800 mol/L I– (Experiments 2 and 3) cf Na2S2O8 = cf S2O82– = [(20.0 mL)(0.100 mol/L)]/50.0 mL = 0.0400 mol/L S2O82– (Experiments 1 and 2) cf S2O82– = [(10.0 mL)(0.100 mol/L)]/50.0 mL = 0.0200 mol/L 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 [S 2O82–] 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–]. m

k2  I  S2 O82   Rate 2     = m n Rate1  2  k1  I  S2 O8     

Concentration (mol/L) of S2O82–is constant.

m 0.0800  = 1.7  105 mol / L s 0.0400m

3.4  105 mol / L s

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–]. m

k3  I  S2 O82   Rate3     = m n Rate 2 k2  I  S2 O82       3.4x105 mol / L s 3.4x105 mol / L s

Concentration (mol/L) of I– is constant.

0.0200n 0.0400n

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 = (3.448x10–5 mol/L•s )/(0.0800 mol/L) = 4.31x10–4 s-1= 4.3x10–4 s–1 d) Rate = (4.3x10–4 s–1)[I–]

14.115 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. 14.116 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 twelve 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 k= b) t1/2 = = = 0.023104906 min-1= 0.023 min–1 t1/2 30 min k 14.117 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. Solution:  60 min  a) Converting time in h to min:  2 h    = 120 min  1h  ln [A]t = ln [A]0 + kt ln [A]t = ln [1.0x103] + (0.035 min–1)(120 min) ln [A]t = 11.107755 [A]t = 6.6686x104 cells/L= 7x104 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 min= 2.0x101 min 14.118 a) The shape is tetrahedral and the hybridization of C1 is sp3.

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.

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. 14.119

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 = t1/2 k k=

ln 2 = 0.086212 d–1 (unrounded) 8.04 d

ln [A]t = ln [A]0 – kt ln [A]t = ln [1.7x10–4] – (0.086212 d–1)(30. d) ln [A]t = –11.26607212 [A]t = 1.27999x10–5 mol/L (unrounded) Fraction remaining = [A]t/[A]0 = (1.27999x10– 5 mol/L)/(1.7x10–4 mol/L) = 0.0752936 = 0.075 14.120 Plan: For part a), use the Monod equation to calculate  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  at the given conditions. The value of 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. Solution:

1.5x10 s 0.25 kg/m  = 1.34x10 s = = K S 0.03 kg/m   0.25 kg/m   max S s

4 1

– 4 –1

1.5x10 s 0.50 kg/m  = 1.42x10 s = = K S 0.03 kg/m   0.50 kg/m  1.5x10 s 0.75 kg/m  = 1.44x10 s  S = = K S 0.03 kg/m   0.75 kg/m  1.5x10 s 1.0 kg/m  = 1.46x10 s  S = = K S 0.03 kg/m   1.0 kg/m  1.5x10 s 0.30 kg/m  = 1.3636x10 s  S b)  = = K S 0.03 kg/m   0.30 kg/m   max S s

4 1

– 4 –1

4 1

– 4 –1

max

4 1

– 4 –1

max

4 1

–4 –1

max

 3600 s  Converting time in h to seconds: 1 h    = 3600 s  1h  ln [A]t = ln [A]0 + kt ln [A]t = ln [5.0x103] + (1.3636x10–4 s–1)(3600 s) ln [A]t = 9.00808919 [A]t = 8.1689x103 cells/m³= 8.2x103 cells/m3

1.5x10 s 0.70 kg/m  = 1.438356x10 s c)  = = K S 0.03 kg/m   0.70 kg/m   max S s

4 1

–4 –1

ln [A]t = ln [A]0 + kt ln [A]t = ln [5.0x103] + (1.438356x10–4 s–1)(3600 s) ln [A]t = 9.03500135 [A]t = 8.39172x103 cells/m³= 8.4x103 cells/m3 14.121 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  Concentration of A = 8 spheres    = 0.16 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 III  0.01 mol   1  Concentration of A = 8 spheres    = 0.16 mol/L A  1 sphere   0.50 L   0.01 mol   1  Concentration of B = 7 spheres    = 0.14 mol/L B  1 sphere   0.50 L  Silberberg, Amateis, Venkateswaran, Chen, Chemistry: The Molecular Nature of Matter and Change, 3rd Canadian Edition Page 3-492 Instructor‘s Solution Manual © Copyright 2021 McGraw-Hill Ryerson Ltd.

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].

k2 A  B Rate II = m n RateI k A  B m

The concentration of B is constant

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].

k2 A  B Rate III = m n Rate II k1 A  B m

The concentration of A is constant.

n 0.14  = 5.6  104 mol/L•s 0.10n

5.6  104 mol/L•s

1 = (1.4)n n=0 The reaction is zero order with respect to B. 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] 3.5x10–4 mol/L·s = k[0.10] k = 3.5x10–3 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  Concentration of B = 8 spheres    = 0.16 mol/L B  1 sphere   0.50 L  Rate = k[A] 4.9x10–4 = k[0.10] k = 4.9x10–3 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. I I = k l = k I or – I l 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. I I l – I l = k l  0 l 0 I I ln 0 = k l  0  Il ln I0 – ln Il = k · l

14.122 a) Rate of light intensity decrease = –

ln Il – ln I0 = –k · l I ln l =  k  l = – fraction of light removed per unit length · distance (length) travelled I0

S S = k S or – = k t t 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. S t – SSt = k t  0 t 0 S S ln 0 = k  t  0  St ln S0 – ln St = k · t ln St – ln S0 = –k · t S ln t =  k · t = – fraction of savings lost per unit time x savings time interval S0 b) Rate of savings decline = –

14.123 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. 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) 14.124 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 E  a2

Ae RT

 k2 E =  a1 = e k1 Ae RT

 Ea1  Ea 2  RT

k2 E  Ea2 5000 J/mol = a1 = = 1.93998 J  k1 RT   8.314 mol•K   310 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.

14.125 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 s= 8.0x101 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 = 58.71558894s = 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 s= 2.6x102 s 3 1 8.7 10 s

a) t1/2 =

14.126 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. 14.127 Plan: We need to determine the pressure or concentration of N 2O5 at the beginning of the reaction. Then we need to determine the pressure or concentration of NO2 formed. (It does not matter which one; but it is important to use the SAME variable for both gases. The ratio of the pressures will equal the ratio for the concentrations. Then we can write a reaction table and calculate the pressure or concentration of N2O5 at the END of the reaction. These values will give us the initial and final amounts of N 2O5. We will use the half-life of the reaction to determine the first-order rate constant and then use the first order integrated rate law to determine the time needed for the reaction. Solution: We will use concentration (n/V) since p 

nN2O5 

nRT and T and V remain the same in the system. V

m 5.750 g   5.323 102 mol MM 108.02 g/mol

n 5.323 102 mol   7.050 102 mol/L V 0.755 L m 3.250 g nNO2    7.064  102 mol MM 46.01 g/mol

 N 2 O5  

 NO2  

n 7.064 102 mol   9.356 102 mol/L V 0.755 L 2 N2O5 (g )  4 NO2 (g ) + O2 (g )

Initial 7.050 x 10-2 mol/L Change -2x +4x +x Final 7.050 x 10-2 mol/L-2x 4x x However, we know the concentration of NO 2 at the end is 9.356 x 10-2 mol/L and so we can say 4x = 9.356 x 10-2 mol/L; therefore, x = 2.339 x 10-2 mol/L Therefore, [N2O5]f = 7.050 x 10-2 mol/L - 2x = 7.050 x 10-2 mol/L – 2(2.339 x 10-2 mol/L) = 2.372 x 10-2 mol/L Since the reaction is first order, we know that

k

ln 2 ln 2   6.189  102 min 1 t1/2 11.2 min

We can now use the first order integrated rate law: Silberberg, Amateis, Venkateswaran, Chen, Chemistry: The Molecular Nature of Matter and Change, 3rd Canadian Edition Page 3-495 Instructor‘s Solution Manual © Copyright 2021 McGraw-Hill Ryerson Ltd.

 Ao   kt  At 

 7.050 10 mol/L   6.189 10 min t Therefore, it will take 17.6 min for 3.250 g of NO to form. ln     2.372 10 mol/L  2

2

1

2

t  17.6 min 14.128 Plan: We can use the rate constant at the first temperature and the rate constant at the second temperature to find the activation energy. Then we use the activation energy and either of the two rate constants (and the correct corresponding temperature) to find the rate constant at the third temperature. Solution: T1 = 35 ºC + 273 = 308 K ; T2 = 14 ºC + 273 = 287 K ;

k1 = (3.71 min-1)(1 min/60 s) = 6.18 x 10-2 s-1 k2 = 8.32 x 10-4 s-1

E 1 1 k ln 1  a    k2 R  T2 T1  J   6.18  102 s 1  k1  8.314   ln   mol  K   8.32  104 s 1  k2  Ea    1.51105 J/mol 1  1 1  1        287K 308K   T2 T1  R ln

k3 Ea  1 1      k2 R  T2 T3 

5 k3 1    1.5110 J/mol  1 ln      4 1  J  287K 328K   8.32  10 s  8.314 mol  K k3  2.69 103 4 1 8.32  10 s k 3 = 2.24 s -1

14.129 Plan: We use the half-life of the medication to find k. We then use the integrated first order rate equation to find the time taken for the amount of medication to drop from 200 mg to 50 mg. Solution:

k

ln 2 ln 2   6.418  105 s 1 4 t1/2 1.08 10 s

 A   kt  At   200 mg   6.418 105 s 1 t ln     50 mg  ln

 1 min  1 h  t  2.16  104 s=  2.16 104 s      6h  60 s  60 min  Silberberg, Amateis, Venkateswaran, Chen, Chemistry: The Molecular Nature of Matter and Change, 3rd Canadian Edition Page 3-496 Instructor‘s Solution Manual © Copyright 2021 McGraw-Hill Ryerson Ltd.

14.130 Plan: First, we will use the Arrhenius equation to find the rate constant. Then we will use the first order integrated rate law to find the concentration after 26 min. Finally, we will use the Rate law to find the initial and final rates of reaction. Solution: 

k  Ae RT   2.9 1014 s 1  e 

2.61105 J/mol J    8.314  770.05K  molK  

 5.720  10-4 s-1

 A   kt  At  13.74 mol/L   5.720×10-4 s 1 26 min  60 s  ln      At   1 min  13.74 mol/L   2.4  At   A t  = 5.6 mol / L ln

Rate  k[cyclopropane] Rateinitial  k  cyclopropane   5.720×10-4 s 1  13.74 mol/L   7.9×10-3

mol Ls mol Ratefinal  k  cyclopropanet   5.720×10-4 s 1   5.6 mol/L   3.2×10-3 L s 14.131 Plan: Use the equation for first-order half-life to find the value of k using the half-life and ln 2. Convert units as needed. Then, use the first order integrated rate equation to find [N 2O5]t . Then use the balanced chemical equation to find the pressure of oxygen at time t. Solution:

 60min  t 1  2.81h    169min 2  1hour  k

ln2 ln2   4.11  103 min 1 t1 169min 2

ln ln

I C F

pAt pA

pAt

 kt



0.993bar p A t  0.363 bar

N2O5 (g)  0.993 bar -x 0.363 bar



  4.11  103 min 1  245min 

2 NO2 (g) + ½ O2 (g) 0 0 +2x +1/2 x +2x +1/2 x

0.363 bar = 0.993 bar – x

x = 0.630 bar

pO2 

1 1 x   0.630 bar   0.315 bar 2 2

14.132 Plan: Use the ratio of the rates for different trials to find the partial orders for reactants. Once the partial orders are known, write the rate law. Use the rate law and one of the trials to calculate the value for k with the proper units. Use the rate law and the known rate constant with the given information to find the unknown concentration. Solution: Rate = k [A]m[B]n[C]p

a)i)

Using rates for trials 1 and 2:

 

   

 

2 m n p m 1.5  102 mol / L 1.0  102 mol / L  4.5  102 mol / L  Rate2 k  A 2 B2 C 2 k 4.5  10 mol / L      2 m n p Rate1 k  A m Bn C p  1.5  10 mol / L  k 1.5  102 mol / L 1.5  102 mol / L 1.0  102 mol / L 1 1 1

1.1  101 mol / Ls  3m 3.9  103 mol / Ls 28  3m log28  m log 3 m

log28 log 3

m=3 ii) Using rates for trials 2 and 3:

 

    3

    n

 

 

 

m n p 2 2 p 3 2.5  103 mol / L Rate3 k  A 3 B3 C 3 k 3.0  10 mol / L 1.5  10 mol / L  2   2.5  103 mol / L        Rate2 k  A m Bn C p k 4.5  102 mol / L 3 1.5  102 mol / L n 1.0  102 mol / L p  3   1.0  102 mol / L  2 2 2 3

2.0  103 mol / Ls  2  p     0.25  1.1  101 mol / Ls  3  3

2 p 1.8  102     0.25  3

 0.25   6.1  102 p

1  1       4   16  p=2

iii) Using rates for trials 3 and 4:

 

2 2 2 Rate4 k  A 4 B4 C 4 k 6.0  10 mol / L 4.5  10 mol / L 5.0  10 mol / L   3 n 2 Rate3 k  A m Bn C p k 3.0  102 mol / L 1.5  10 2 mol / L 2.5  10 3 mol / L m

2 1.9  10 mol / Ls 3  4.5  10 mol / L  2  2    20  3 2 2.0  10 mol / Ls 1.5  10 mol / L   1

9500   8  3   400  n

 3  3 n

n=1

Rate=k  A  BC 

Using Experiment 1 data:

Rate  k  A  B C  3

k

Rate

 A  BC  3

mol L s  3 2  2 mol   2 mol  2 mol  1.5  10 1.5  10 1.0  10  L  s   L  s  L  s    3.9  103

 7.7×108

L5 mol 5  s

Rate  k  A  BC  3

1.4

mol L5 6  7.7  108 C  5 L s mol  s mol C =3.5×10-2 L

14.133 Plan: Convert units of k to be the same. Use the equation relating k values at different T values through the activation energy to find the activation energy. Re-use the same equation using the calculated activation energy to find k at the third T. Then use the first-order integrated rate law to find the initial concentration. Solution:

1  1h  1min  3 1 k 2  12.3    3.42  10 s h  60min   60s  1  1min  k 1  14.7  0.245s 1 min  60s  ln

k1 Ea  1 1      k 2 R  T2 T1 

 0.245s  1

ln ln



 3.42  10 s  8.314 3



k 

k3 k1



1   1     288K 318K 

J mol  K Ea =1.09×105 J/mol

1

Ea  1 1     R  T1 T3 

1.09  105 J / mol  1 1     1 J 318K 303K 0.245s   8.314 mol  K k 3 =3.21  102 s 1 3





 A t   kt  A o

mol    0.721 L     3.21  102 s 1 118s ln     A o



 A o = 31.8



mol L

14.134 Plan: Set the first step as the rate determining step (RDS). Determine what the rate law would be. Check to see if it matches the experimental rate law. If it does, the first step is the RDS. If it does not, then set the first step as a fast equilibrium and the second step as the RDS. Determine what the rate law would be. If it matches the experimental rate law, then the first step is a fast equilibrium and the second step is the RDS. If not, there is another mechanism that must be considered. Solution: If the first step is the RDS, then Rate = k1[NO]2 However, this does not match the experimental rate law. Thus, the first step is not the RDS. Set the first step as a fast equilibrium and make the second step the rate law. In a fast equilibrium, we can state that the rate of the forward reaction equals the rate of the reverse reaction. k1[NO]2 = k-1 [(NO)2] from this expression, we can find the concentration of the intermediate:  NO 2  

k 1  NO

k 1 The RDS is the second step, therefore, the rate is:

Rate  k 2  NO 2  O2   k2 

k1 2  NO O2  k 1

 k ' NO O2  , where k '  k 2  2

k1 k 1

This matches the experimental rate law and thus, this is a plausible mechanism.

CHAPTER 15 EQUILIBRIUM: THE EXTENT OF CHEMICAL REACTIONS CHEMICAL CONNECTIONS BOXED READING PROBLEM B15.1

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 15.1

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).  products Kc =  reactants 15.2

The faster the rate and greater the yield, the more useful the reaction will be to the manufacturing process.

15.3

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.

15.4

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  products most of the reactants have reacted to become products. K =  reactants

15.5

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.

15.6

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.

15.7

The equilibrium constant expression is K = pO2 (remember the activity of solids is 1). If the temperature remains constant, K remains constant. If the initial amount of Li2O2 present was sufficient to reach equilibrium, the pressure of O2 obtained will be constant, regardless of how much Li2O2(s) is present.

15.8

a) On the graph, the concentration of HI increases at twice the rate that H 2 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 H 2 and I2 decrease and the concentration of product HI increases, which means that Q increases as a function of time. 2 pHI H2(g) + I2(g) 2HI(g) Q = pH 2 pI 2

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. 15.9

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)

15.10

Plan: Write the reaction and then the expression for Q 1/2N2(g) + 1/2O2(g)  NO(g) p Q(form) = 1 NO1 pN2 pO2 2

NO(g) 1/2N2(g) + 1/2O2(g) 1

pN2 pO2 2

Q(decomp) =

pNO Q(decomp) = 1/Q(form), so the constants do differ (they are the reciprocal of each other). 15.11

Plan: Write the reaction and then the expression for Q. 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 pNH reaction is Q1 = 3 31 . pH2 pN2 2

In the second reaction, the coefficient in front of N2 is fixed at 1 mole. 3H2(g) + N2(g) 2NH3(g) (2) 2 pNH The reaction quotient for this reaction is Q2 = 3 3 pH 2 pN 2 Q2 is equal to Q12. 15.12

Plan: Remember that Q =

aCc aDd

where A and B are reactants, C and D are products, and a, b, c, and d are the aAa aBb stoichiometric coefficients in the balanced equation. Solution: a) 4NO(g) + O2(g) 2N2O3(g) pN2 O Q= 4 2 3 pNO pO2 b) SF6(g) + 2SO3(g) 3SO2F2(g) 3 pSO 2 F2 Q= 2 pSF6 pSO 3 c) 2SC1F5(g) + H2(g) S2F10(g) + 2HCl(g) 2 pS F pHCl Q = 22 10 pSClF5 pH 2

15.13

a) 2C2H6(g) + 7O2(g) 4CO2(g) + 6H2O(g) 4 pCO pH6 O Q = 2 2 72 pC2 H6 pO2 b) CH4(g) + 4F2(g) CF4(g) + 4HF(g) 4 pCF4 pHF Q= pCH 4 pF42 c) 2SO3(g) 2SO2(g) + O2(g) 2 pSO p 2 O2 Q= 2 pSO3

15.14

Plan: Remember that Q will be determined using pressures in place of activities when the species in the chemical reaction are all gases. Solution: a) 2NO2Cl(g) 2NO2(g) + Cl2(g) 2 pNO p 2 Cl2 Q= 2 pNO2Cl b) 2POCl3(g) 2PCl3(g) + O2(g) 2 pPCl p 3 O2 Q= 2 pPOCl 3 c) 4NH3(g) + 3O2(g) 2N2(g) + 6H2O(g) pN2 pH6 O Q = 4 2 32 pNH3 pO2

15.15

a) 3O2(g) 2O3(g) pO2 Q = 33 pO 2 b) NO(g) + O3(g) NO2(g) + O2(g) Q=

pNO2 pO2 pNO pO3

c) N2O(g) + 4H2(g) 2NH3(g) + H2O(g) 2 pNH p 3 H2O Q= pN 2O pH4 2 15.16

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: 2

a) The K for the original reaction is K =

pH2 pS2 pH2 2S

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. K (a) = (1/K)1/2 =

pH 2S pH 2 pS12/ 2

K (a) = (1/1.6x10–2)1/2 = 7.90569 = 7.9 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. pH5 2 pS52/ 2 K (b) = (Kc)5/2 = pH5 2S K (b) = (1.6x10–2)5/2 = 3.23817x10–5 = 3.2x10–5

15.17

pN2 pH2 2O 2 pNO pH2 2

a) K (a) = [K]1/2 =

p1N/22 pH 2O pNO pH 2

Thus, K (a) = [K]1/2 = (6.5x102)1/2 = 25.495 = 25 4 pNO pH4 b) K = [K]–2 = 2 4 2 pN2 pH 2O 15.18

K = [K]–2 = (6.5x102)–2 = 2.36686x10–6 = 2.4x10–6 Plan: The activity of pure solids and pure liquids is 1, so they do not appear 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)

pO2 2 pCO 2

b) H2O(l) H2O(g) Q = pH2 O c) NH4Cl(s) NH3(g) + HCl(g) Q = pNH3 pHCl 15.19

a) H2O(l) + SO3(g) H2SO4(aq)  H 2SO 4  Q= pSO3 b) 2KNO3(s) 2KNO2(s) + O2(g) Q = pO2 c) S8(s) + 24F2(g) 8SF6(g) 8 pSF Q = 246 pF2

15.20

Plan: The activity of pure solids and pure liquids is 1, so they do not appear in the reaction quotient expression. Remember that stoichiometric coefficients are used as exponents in the expression for the reaction quotient. Solution: a) 2NaHCO3(s) Na2CO3(s) + CO2(g) + H2O(g) Q = pCO2 pH2 O b) SnO2(s) + 2H2(g) Sn(s) + 2H2O(g) pH2 2O Q= 2 pH2 c) H2SO4(l) + SO3(g) H2S2O7(l) 1 Q= pSO3

15.21

a) 2Al(s) + 2NaOH(aq) + 6H2O(l) 2Na[Al(OH)4](aq) + 3H2(g) 2

 Na  Al  OH    pH3 4  2  Q= 

 NaOH2

b) CO2(s) CO2(g) Q = pCO2 c) 2N2O5(s) 4NO2(g) + O2(g) 4 Q = pNO p 2 O2 15.22

Write balanced chemical equations for each reaction, and then write the appropriate equilibrium expression. a) 4HCl(g) + O2(g) 2Cl2(g) + 2H2O(g) 2 pCl pH2 O Q = 42 2 pHCl pO 2 b) 2As2O3(s) + 10F2(g) 4AsF5(l) + 3O2(g)

pO3 2 p10 F2

c) SF4(g) + 2H2O(l) SO2(g) + 4HF(g) Q=

4 pSO2 pHF

pSF4

d) 2MoO3(s) + 6XeF2(g) 2MoF6(l) + 6Xe(g) + 3O2(g) 6 pXe pO3 2 Q= 6 pXeF 2 15.23

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. (1) Cl2(g) + F2(g) 2ClF(g)

Q1 =

(2) 2ClF(g) + 2F2(g) 2ClF3(g)

Q2 =

Overall: Cl2(g) + 3F2(g) 2ClF3(g)

2 pClF

pCl2 pF2 2 pClF 3 2 pClF pF22

Qoverall =

2 pClF 3

pCl2 pF32

b) The reaction quotient for the overall reaction, Qoverall, determined from the reaction is: 2 pClF 3 Qoverall = pCl2 pF32

 p2  ClF  Qoverall = Q1Q2 =   pCl pF  2 2   2

2  pClF  p2 3   = ClF3 2  pClF pF22  pCl2 pF32 

15.24

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.

15.25

Kc and K are related by the equation K = Kc(RT)n, where n represents the change in amount (mol) of gas in the reaction (amount (mol) gaseous products – amount (mol) gaseous reactants). When n is zero (no change in amount (mol) of gas), the term (RT)n equals 1 and Kc = K. When n is not zero, meaning that there is a change in the amount (mol) of gas in the reaction, then Kc  K.

15.26

a) K = Kc(RT)n. Since n = amount (mol) gaseous products – amount (mol) gaseous reactants, n is a positive integer for this reaction. If n is a positive integer, then (RT)n is greater than 1. Thus, Kc is multiplied by a number that is greater than 1 to give K. Kc is smaller than K. b) Assuming that RT > 1 (which occurs when T > 12.0 K, because 0.08314 (R) x 12.0 = 1), K > Kc if the amount (mol) of gaseous products exceeds the amount (mol) of gaseous reactants. K < Kc when the amount (mol) of gaseous reactants exceeds the amount (mol) of gaseous product. Plan: ngas = moles gaseous products – moles gaseous reactants. Solution: a) Amount (mol) of gaseous reactants = 0; amount (mol) of gaseous products = 3; ngas = 3 – 0 = 3 b) Amount (mol) of gaseous reactants = 1; amount (mol) of gaseous products = 0; ngas = 0 – 1 = –1

15.27

c) Amount (mol) of gaseous reactants = 0; amount (mol) of gaseous products = 3; ngas = 3 – 0 = 3 15.28

a) ngas = 1

15.29

Plan: First, determine n for the reaction and then calculate Kc using K = Kc(RT)n. Solution: a) n = moles gaseous products – moles gaseous reactants = 1 – 2 = –1 K = Kc(RT)n Kc =

b) ngas = –3

K ( RT )

= n

c) ngas = 1

3.9x102 [(0.08314)(1000.)]1

= 3.24246 = 3.2

b) n = moles gaseous products – moles gaseous reactants = 1 – 1 = 0 K 28.5 Kc = = = 28.5 n ( RT ) [(0.08314)(500.)]0 15.30

First, determine n for the reaction and then calculate Kc using K = Kc(RT)n. a) n = moles gaseous products – moles gaseous reactants = 2 – 2 = 0 K 49 Kc = = = 49 n ( RT ) [(0.08314)(730.)]0 b) n = moles gaseous products – moles gaseous reactants = 2 – 3 = –1 Kc =

K ( RT )

= n

2.5x1010 1

[(0.08314)(500.)]

= 1.03925x1012 = 1.0x1012

15.31

Plan: First, determine n for the reaction and then calculate K using K = Kc(RT)n. Solution: a) n = moles gaseous products – moles gaseous reactants = 2 – 1 = 1 K = Kc(RT)n = (6.1x10–3)[(0.08314)(298)]1 = 0.15113 = 0.15 b) n = moles gaseous products – moles gaseous reactants = 2 – 4 = – 2 K = Kc(RT)n = (2.4x10–3)[(0.08314)(1000.)]–2 = 3.4721x10–7 = 3.5x10–7

15.32

First, determine n for the reaction and then calculate K using K = Kc(RT)n. a) n = moles gaseous products – moles gaseous reactants = 2 – 2 = 0 K = Kc(RT)n = (0.77)[(0.08314)(1020.)]0 = 0.77 b) n = moles gaseous products – moles gaseous reactants = 2 – 3 = –1 K = Kc(RT)n = (1.8x10–56)[(0.08314) (570.)]–1 = 3.7983x10–58 = 3.8x10–58

15.33

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).

15.34

a) The reaction is 2D ↔ E and Kc =

[E] [D]2

 0.0100 mol   1  Concentration of D = Concentration of E = 3 spheres    = 0.0300 mol/L  1 sphere   1.00 L  [0.0300] [E] K= = = 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 mol/L [0.0600] [E] Q= = = 16.66666 = 16.7 2 [0.0600]2 [D] Silberberg, Amateis, Venkateswaran, Chen, Chemistry: The Molecular Nature of Matter and Change, 3rd Canadian Edition Page 3-507 Instructor‘s Solution Manual © Copyright 2021 McGraw-Hill Ryerson Ltd.

B is not at equilibrium. Since Q < K, the reaction will proceed to the right. In Scene C, the concentration of D is still 0.0600 mol/L and the concentration of E is 0.0600 mol/0.500 L = 0.120 mol/L [0.120] [E] Q= = = 33.3333 = 33.3 2 [0.0600]2 [D] Since Q = K in Scene C, the reaction is at equilibrium. 15.35

Plan: To decide if the reaction is at equilibrium, calculate Q and compare it to K. If Q = K, then the reaction is at equilibrium. If Q > K, then the reaction proceeds to the left to produce more reactants. If Q < K, then the reaction proceeds to the right to produce more products. Solution: pH2 pBr2 (0.010)(0.010) Q= = = 2.5x10–3 > K = 4.18x10–9 2 2 pHBr (0.20) Q > K, 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. Q will decrease until Q = K.

15.36

15.37

n = moles gaseous products – moles gaseous reactants = 2 – 2 = 0 Since n = 0, K = Kc = 2.7 (Note: If n had any other value, we could not finish the calculation without the temperature.) CO2 H 2  = 0.62 /2.00.43/2.0 = 3.662 > K = 2.7 Q= COH 2 O 0.13/2.00.56/2.0

2 pNO pBr2

(0.10) 2 (0.10)

= 0.10 < K = 60.6 2 (0.10) 2 pNOBr Q < K Thus, the reaction is not at equilibrium and will proceed to the right (towards the products).

Q > K Thus, the reaction is not at equilibrium and will proceed to the left (towards the reactants). 15.38

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.

15.39

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 or pressure of each reactant is the initial concentration or pressure minus the amount that reacts. The final (equilibrium) concentration or pressure of each product is the initial concentration or pressure plus the amount that forms.

15.40

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.

15.41

Plan: Since all equilibrium concentrations are given in mol/L and the reaction is balanced, construct an equilibrium expression and substitute the equilibrium concentrations to find Kc. Solution: 2

1.87x10  HI2 =   Kc = = 50.753 = 50.8 H 2 I2  6.50x105  1.06x103  3

N 2 H 2 3 = 0.1140.3423 = 9.0077875 = 9.01 2 0.02252  NH 3 

15.42

Kc =

15.43

Plan: Calculate the initial concentration of PCl5 from the given amount (mol) 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 mol/L 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) Initial 0.075 0 0 Change –x +x +x Equilibrium 0.075 – x x x

15.44

The reaction table requires that the initial [H2] and [F2] be calculated: [H2] = 0.10 mol/0.50 L = 0.20 mol/L; [F2] = 0.050 mol/0.50 L = 0.10 mol/L. x = [H2] = [F2] reacting (–x); 2x = [HF] forming (+2x) Concentration (mol/L) H2(g) + F2(g)  2HF(g) Initial 0.20 0.10 0 Change –x –x +2x Equilibrium 0.20 – x 0.10 – x 2x

15.45

Plan: Two of the three equilibrium pressures are known, as is K. Construct an equilibrium expression and solve for pNOCl. Solution: K = 6.5x104 = 6.5x104 =

PNOCl =

2 pNOCl 2 pNO pCl2

P 2 NOCl (0.35)2 (0.10)

6.5x10   0.35  0.10 = 28.2179 bar = 28 bar 4

A high pressure for NOCl is expected because the large value of K indicates that the reaction proceeds largely to the right, i.e., to the formation of products. 15.46

C(s) + 2H2(g) CH4(g) pCH K = 2 4 = 0.262 pH2

pCH4 = K  pH2 2 = (0.262)(1.22)2 = 0.38996 bar = 0.390 bar 15.47

Plan: Use the balanced equation to write an equilibrium expression and to define x. Set up a reaction table, substitute into the K 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 include the NH4HS as it is a solid with activity=1). Pressure (bar) NH4HS(s)  H2S(g) + NH3(g) Initial  0 0

Change Equilibrium

 

+x x

K = 0.11 = ( pH2S )( pNH3 )

+x x

(The solid NH4HS is not included.)

0.11 = (x)(x) x = pNH3 = 0.33166 = 0.33 bar 15.48

2H2S(g) 2H2(g) + S2(g)

H 2S = 0.45 mol/3.0 L = 0.15 mol/L Concentration (mol/L) Initial Change Equilibrium

2H2S(g) 0.15 –2x 0.15 – 2x



2H2(g) 0 +2x 2x

S2(g) 0 +x x

2 2 2 x  x  H 2   S2    Kc = 9.30x10 = = 0.15  2 x 2  H 2S2 –8

Assuming 0.15 mol/LM – 2 x  0.15 mol/LM 2 2 x  x   4 x3 9.30x10 = = 0.152 0.152 –8

x = 8.0575x10–4 mol/L H 2  = 2x = 2 (8.0575x10–4 mol/L) = 1.6115x10–3 mol/L= 1.6x10–3 mol/L (Since (1.6x10–3)/(0.15) < 0.05, the assumption is OK.) 15.49

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. Solution: The initial concentrations of N2 and O2 are (0.20 mol/1.0 L) = 0.20 mol/L and (0.15 mol/1.0 L) = 0.15 mol/L, respectively. N2(g) + O2(g)  2NO(g) There is a 1:1:2 mole ratio between reactants and products. Concentration (mol/L) N2(g) + O2(g)  2NO(g) Initial 0.20 0.15 0 Change –x –x +2x (1:1:2 mole ratio) Equilibrium 0.20 – x 0.15 – x 2x Kc = 4.10x10

–4

2 2 NO  2 x   = = N 2 O2  0.20  x 0.15  x 

Assume 0.20 mol/L – x  0.20 mol/L 4 x2 4.10x10–4 = 0.200.15

and

0.15 mol/L – x  0.15 mol/L

x = 1.753568x10–3 mol/L [NO] = 2x = 2(1.753568x10–3 mol/L) = 3.507136x10–3 mol/L= 3.5x10–3 mol/L (Since (1.8x10–3)/(0.15) < 0.05, the assumption is OK.) 15.50

2NO2(g)  2NO(g) + O2(g) There is a 2:2:1 mole ratio between reactants and products. Pressure (bar) 2NO2(g)  2NO(g) + O2(g) Initial 0.75 0 0 Change – 2x +2x +x (2:2:1 mole ratio) Equilibrium 0.75 – 2x 2x x

K = 4.48x10

–13

2 pNO pO2

2 pNO 2

2x 2 x  = 0.75  2x 2

Assume 0.75 bar – 2x  0.75 bar

4x x  = 4x  = 2

4.48x10

–13

0.75 2

0.752

x = 3.979x10–5 bar = 4.0x10–5 bar O2 (assumption is justified) pNO = 2x = 2(3.979x10–5 bar) = 7.958x10–5 bar = 8.0x10–5 bar NO 15.51

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 H 2 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 mol/L [H2] = (0.00623 mol)/(1.50 L) = 0.0041533 mol/L [I2] = (0.00414 mol)/(1.50 L) = 0.00276 mol/L Equilibrium concentration of H2 is greater than the initial, so the reaction moves in the forward direction. 2 HI(g)  H2(g) + I2(g) There is a 2:1:1 mole ratio between reactants and products. Concentration (mol/L) 2 HI(g)  H2(g) + I2(g) Initial 0.0162667 0.0041533 0.00276 Change –2x +x +x (2:1:1 mole ratio) Equilibrium 0.0162667 – 2x 0.0041533 + x 0.00276 + x [H2]eq = 0.00467 = 0.0041533 + x x = 0.0005167 mol/L [I2]eq = 0.00276 + x = 0.00276 + 0.0005167 = 0.0032767 mol/L= 0.00328 mol/L I2 [HI]eq = 0.0162667 – 2x = 0.0162667 – 2(0.0005167) = 0.0152333 mol/L= 0.0152 mol/L HI

15.52

Initial concentrations: [A] = (1.75x10–3 mol)/(1.00 L) = 1.75x10–3 mol/L [B] = (1.25x10–3 mol)/(1.00 L) = 1.25x10–3 mol/L [C] = (6.50x10–4 mol)/(1.00 L) = 6.50x10–4 mol/L Concentration (mol/L) A(g)  2B(g) Initial 1.75x10–3 1.25x10–3 Change –x + 2x Equilibrium 1.75x10–3 – x 1.25x10–3 + 2x [A]eq = 2.15x10–3 = 1.75x10–3 – x x = –0.00040 [B]eq = 1.25x10–3 + 2x = 4.5x10–4 mol/L [C]eq = 6.50x10–4 + x = 2.5x10–4 mol/L

15.53

C(g) 6.50x10–4 +x 6.50x10–4 + x

Plan: Use the balanced equation to write an equilibrium expression. Find the initial concentration of ICl 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: [ICl]init = (0.500 mol/5.00 L) = 0.100 mol/L Concentration (mol/L) 2ICl(g)  I2(g) + Cl2(g) Initial 0.100 0 0 Change –2x +x +x (2:1:1 mole ratio) Equilibrium 0.100 – 2x x x

I2 Cl2  =  x  x   0.100  2 x 2 ICl2  x 2 0.110 =  0.100  2 x 2 x 0.331662 =  0.100  2 x  Kc = 0.110 =

Take the square root of each side:

x = 0.0331662 – 0.663324x 1.663324x = 0.0331662 x = 0.0199397 [I2]eq = [Cl2]eq = x = 0.0199397 mol/L= 0.0200 mol/L [ICl]eq = 0.100 – 2x = 0.100 – 2(0.0199397) = 0.0601206 mol/L= 0.060 mol/L ICl 15.54

Concentration (mol/L) SCl2(g) Initial 0.675 Change –x Equilibrium 0.675 – x

2C2H4(g) 0.973 – 2x 0.973 – 2x

  S(CH2CH2Cl)2(g) 0 +x x

[S(CH2CH2Cl)2]eq = x = 0.350 mol/L [SCl2]eq = 0.675 – x = 0.675 – 0.350 = 0.325 mol/L [C2H4]eq = 0.973 – 2x = 0.973 – 2(0.350) = 0.273 mol/L  0.350 S(CH 2 CH 2 Cl)2  = Kc = = 14.4497 2  0.325 0.2732 SCl2 C2 H 4  K= Kc(RT)n

n = 1 mol – 3 mol = –2

K = 14.4497   0.08314 273.2  20.0 15.55

2

= 0.0237617 = 0.0238

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 mol/L Concentration (mol/L) 4NH3(g) + 3O2(g)  2N2(g) + 6H2O(g) Initial 0.0150 0.0150 0 0 Change –4x –3x +2x +6x Equilibrium 0.0150 – 4x 0.0150 – 3x +2x +6x [N2] eq = 2x = 1.96x10–3 mol/L x = (1.96x10–3 mol/L)/2 = 9.80x10–4 mol/L [H2O]eq = 6x = 6(9.80x10–4) = 5.8800x10–3 mol/L [NH3]eq = 0.0150 – 4x = 0.0150 – 4(9.80x10–4 ) = 1.1080x10–2 mol/L [O2]eq = 0.0150 – 3x = 0.0150 – 3(9.80x10–4 ) = 1.2060x10–2 mol/L

N 2 2 H 2 O6 = 1.96x10   5.8800x10  = 6.005859x10–6 = 6.01x10–6 Kc = 4 3 4 3 NH3  O 2  1.1080x102  1.2060x102  3

15.56

Pressure (bar) Initial Change Equilibrium

FeO(s) —

CO(g) 1.00 –x 1.00 – x

3



Fe(s) —

CO2(g) 0 +x x

pCO2

x = 0.403 = 1.00  x pCO x = 0.28724 = 0.287 bar CO2 1.00 – x = 1.00 – 0.28724 = 0.71276 bar = 0.71 bar CO

15.57

A change in equilibrium conditions such as a change in concentration of a component, a change in pressure (volume), or a change in temperature.

15.58

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.

15.59

A positive rH 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.

15.60

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.

15.61

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.

15.62

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.

15.63

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, CO 2 solid will not sublime, but since the reaction lists carbon dioxide as a gas, the assumption that sublimation takes place is reasonable. 15.64

a) no change c) shifts towards the products

b) no change d) shifts towards the reactants

15.65

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.

15.66

a) less CH3OH(l); more CH3OH(g) b) less CH4 and NH3; more HCN and H2

15.67

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 a smaller amount (mol) 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 amount (mol) of reactant and product gas , 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 a smaller amount (mol) 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.

15.68

a) more CO2 and H2O; less C3H8 and O2 b) more NH3 and O2; less N2 and H2O

15.69

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 a larger amount (mol) of gas, while decreasing the container volume results in a shift in the direction that forms a smaller amount (mol) of gas. Solution: a) Because the amount (mol) of reactant gas (4H2) equals the amount (mol) of product gas (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 (1O 2). A decrease in pressure favors the reaction direction that forms more moles of gas, so increase the reaction vessel volume.

15.70

a) increase volume

15.71

Plan: An increase in temperature (addition of heat) causes a shift in the equilibrium away from the side of

b) decrease volume

the reaction with heat. Recall that a negative value of r H indicates an exothermic reaction, while a positive value of r H indicates an endothermic reaction. Solution: a) CO(g) + 2H2(g) CH3OH(g) + heat r H = –90.7 kJ/mol 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. b) C(s) + H2O(g) + heat CO(g) + H2(g) r H = 131 kJ/mol 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. 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. 15.72

a) increase

b) decrease

c) decrease

d) increase

15.73

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. Solution: K298 = K1 = 1.80 T1 = 298 K K500 = K2 = ? T2 = 500. K R = 8.314 J/mol•K 3  10 J   0.32 kJ  2 r H =   2 DH    = 6.4x10 J/mol  1 kJ  1 mol DH   

K2  H  1 1 =  r    K1 R  T2 T1 

K2 6.4x102 J / mol  1 1   =    8.314 J/mol•K  500. K 298 K  1.80

K2 = 0.104360 1.80 K2 = 1.110 1.80 K2 = (1.80)(1.110) = 1.998 = 2.0 ln

15.74

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 = 2.25x104 K0 = K2 = ?

T1 = 298 K T2 = (273 + 0.) = 273 K

r H = –128 kJ/mol R = 8.314 J/mol•K

r H = (–128 kJ/mol)(103 J/1 kJ) = –1.28x105 J/mol ln

K2  H  1 1 =  r    K1 R  T2 T1 

1.28x105 J / mol  1 1  K2   =   4 8.314 J/mol•K  273 K 298 K  2.25x10

K2 = 4.731088 2.25x104 K2 = 1.134189x102 2.25x104 K2 = (2.25x104)(1.134189x102) = 2.551925x106 = 2.55x106 ln

15.75

4Fe3O4(s) + O2(g) 6Fe2O3(s) 1 a) K = = 2.5x1087 pO2

K = 2.5x1087 at 298 K

pO2 = 4.0x10–88 bar

b) Q =

1 = 1/(0.21) = 4.7619 PO2

K> Q thus, the reaction will proceed to the right. c) K = Kc(RT)n Kc = K/(RT)n n = 0 – 1 = –1 Kc = (2.5x1087)/[(0.08314)(298)]–1 = 6.19393x1088 = 6.2x1088 15.76

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 a larger amount (mol) of gas, while decreasing the container volume (pressure increases) results in a shift in the direction that forms a smaller amount (mol) of gas. Adding a reactant causes a shift in the direction of products. Solution: a) SO2(g) + 1/2O2(g) SO3(g) + heat The forward reaction is exothermic ( r H is negative), so it is favored by lower temperatures. Lower temperatures will cause a shift to the right, the heat side of the reaction. The amount (mol) of gas as products (1SO3) is smaller 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 smaller amount (mol) of gas. pSO3 b) Addition of O2 would decrease Q since Q = , and have no impact on K. 1/ 2 pSO2 pO2 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.

15.77

3H2(g) + N2(g)  2NH3(g) pNH3 = (41.49%/100%)(110. bar) = 45.639 bar 100.00% – 41.49% = 58.51% N2 + H2 pH2 + pN2 = (58.51%/100%)(110. bar) = 64.361 bar pH2 = (3/4)(64.361 bar) = 48.27075 bar pN2 = (1/4)(64.361 bar) = 16.09025 bar

p  =  45.639  K= = 1.15095x10 = 1.15x10 48.27075 16.09025     p p    2

NH3

–3

15.78

–3

a) 3H2(g) + N2(g)  2NH3(g) pNH3 = 50. bar

The mole ratio H2:N2 = 3:1; at equilibrium, if N2 = x, H2 = 3x;

 p  = 1.00x10 K=  p  p  2

NH3

–4

 50. = 1.00x10–4 3  x  3x  2

x = 31.02016 = 31 bar N2 3x = 3(31.02016) = 93.06049 = 93 bar H2 ptotal = pnitrogen + phydrogen + pammonia = (31.02016 bar) + (93.06049 bar) + (50. bar) = 174.08065 bar= 174 bar total b) The mole ratio H2:N2 = 6:1; at equilibrium, if N2 = x, H2 = 6x; pNH3 = 50. bar

 50.2 K= = 1.00x10–4  x  6x 3 x = 18.445 = 18 bar N2 6x = 6(18.445) = 110.67 = 111 bar H2 ptotal = pnitrogen + phydrogen + pammonia = (18.445 bar) + (110.67 bar) + (50. bar) = 179.115 bar= 179 bar total This is not a valid argument. The total pressure in b) is greater than in a) to produce the same amount of NH 3. 15.79

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 CaCO 3 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 with an activity of 1.

15.80

a) Q =

2 pXY ; since n=0, we can use either concentration or pressure terms. pX 2 pY2

2 0  Scene A: Q = =0 0.4 0.4 2 0.4  Scene B: Q = =4 0.20.2 2 0.6  Scenes C–E: Q = = 36 = 4x101 0.10.1

c) Time is progressing to the right. Frame A must be the earliest time. d) K = 4x101 e) Scene B. At higher temperatures, the reaction shifts to the left (forming more X 2 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. 15.81

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 (mol/L) NH2COONH4(s)  2NH3(g) + CO2(g) Initial 7.80 g 0 0 Change — +2x +x Equilibrium — 2x x The solid has an activity of 1 and, as long as some is present, is not included in the Kc expression. Kc = [NH3]2[CO2] Kc = 1.58x10–8 = (2x)2(x) x = 1.580759x10–3 mol/L Total concentration of gases = 2x + x = 2(1.580759x10 –3 mol/L) + 1.580759x10–3 mol/L = 4.742277x10–3 mol/L To find total pressure use the ideal gas equation: PV = nRT nRT n p= =   RT = cRT V V  P = (4.742277x10–3 mol/L)(0.08314 L•bar/mol•K)(273 + 250.)K = 0.206205 bar = 0.206 bar

15.82

(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. a)

SO2 4  H 2O6  Ni  CO 4  Qc(overall) = O2 7  H 2 6 CO24 SO2 4 Q1 · Q 2 · Q3 = O2 7 15.83

 H 2O6  H 2 6

 Ni  CO   4 

CO24

SO2 4  H 2O6  Ni  CO 4  = O2 7  H 2 6 CO24

a) Since the volume is 1.00 L, the concentration (mol/L) equals the amount (mol) present. Concentration (mol/L) 2NH3(g)  N2(g) + 3H2(g) Initial 0 1.30 1.65 Change +2x –x –3x Equilibrium 2x = 0.100 mol/L 1.30 – x 1.65 – 3x x = 0.0500 mol [N2]eq = (1.30 – 0.0500) mol/L = 1.25 mol/L N2 [H2]eq = [1.65 – 3(0.0500)] mol/L = 1.50 mol/L H2

 N 2  H 2 3

Kc =

NH3 

(1.25)(1.50)

(0.100)2 1

= 421.875 = 422

 N 2  2  H 2  2 = (1.50) 2 (1.25) 2 = 20.523177 = 20.5 b) K = 1

8.34x102  NH 3  c) Kc in a) is the square of Kc in b). The balanced equations are different; therefore, the values of Kc are different. 15.84

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 K 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. Solution: a) K = Kc(RT)n n = amount (mol) gaseous products – amount (mol) gaseous reactants = 1 – 2 = –1 (one mol of product, C2H5OH, and two mol of reactants, C2H4 + H2O) K = Kc(RT)–1 = (9x103)[(0.08314 L•bar/mol•K)(600. K)] –1 = 1.8042x102 Substitute the given values into the equilibrium expression and solve for pC2 H 4 . K=

pC2 H5OH pC2 H 4 pH 2O

200. = 1.8042x102 PC2 H4  400.

PC2H4 = 2.7713x10–3 = 3x10–3 bar b) Since r H 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 amount (mol) of gas and is favored by high pressure. Silberberg, Amateis, Venkateswaran, Chen, Chemistry: The Molecular Nature of Matter and Change, 3rd Canadian Edition Page 3-518 Instructor‘s Solution Manual © Copyright 2021 McGraw-Hill Ryerson Ltd.

c) K1 = 9x103

T1 = 600. K

K2 = ?

T2 = 450. K

 103 J  4 r H =  47.8 kJ/mol    = –4.78x10 J/mol 1 kJ   R = 8.314 J/mol•K 1   T1 

K2  H  1 =  r  K1 R  T2

4.78x104 J / mol  1 1  K2 =     3 8.314 J/mol•K  450. K 600. K  9x10

K2 = 3.1940769 9x103 K2 = 24.38765 9x103 K2 = (9x103)(24.38765) = 2.1949x105 = 2x105 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 amount (mol) of gas. ln

15.85

 2.0 bar  = 0.08067 mol/L each gas L•bar   0.08314 273.2  25.0 K      mol•K   H2(g) + CO2(g)  H2O(g) + CO(g) Initial 0.08067 0.08067 0 0 Change –x –x +x +x Equil: 0.08067 – x 0.08067 – x x x  H 2 OCO = 0.534 = (x)(x) (x) 2 Kc = = (0.08067  x)(0.08067  x)  H 2 CO2  (0.08067  x)2 n/V = c = p/RT =

(0.534)1/2 = 0.730753 =

15.86

15.87

(x) (0.08067  x)

x = 0.03406 mol/L c of H2 at equilibrium = 0.08067 – x = 0.08067– 0.03406 = 0.0466098 mol/L  0.0466098 mol  2.016 g  Mass (g) of H2 = 1.00 L    1 mol  = 0.093965 = 0.094 g H2 L    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 = 4.340x1018 2NO2(g)  2NO(g) + O2(g) K2 = Kc = 1.1x10–5 Overall: 2NO2(g)  N2(g) + 2O2(g)Kc (overall) = K1K2 = 4.774x1013 = 4.8x1013 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 K and use the equilibrium pressures of SO 3 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) K = Kc(RT)n n = amount (mol) of gaseous products – amount (mol) of gaseous reactants = 2 – 3 = –1 (two mol of product, SO3, and three mol of reactants, 2 SO2 + O2)

K = Kc(RT)n = Kc(RT)–1 = (1.7x108)[(0.08314 L•bar/mol•K)(600. K)] –1 = 3.4079x106 2 pSO  300.2 3 K= 2 = 2 = 3.4079x106 pSO2 100. pSO2 pO2

pSO2 = 0.016251 = 0.016 bar b) Create a reaction table that describes the reaction conditions. Since the volume is 1.0 L, the amount (mol) equals the concentration (mol/L). Note the 2:1:2 mole ratio between SO 2:O2:SO3. Concentration (mol/L) 2SO2(g) + O2(g)  2SO3(g) Initial 0.0040 0.0028 0 Change –2x –x +2x (2:1:2 mole ratio) Equilibrium 0.0040 – 2x 0.0028 – x 2x = 0.0020 (given) x = 0.0010, therefore: [SO2] = 0.0040 – 2x = 0.0040 – 2(0.0010) = 0.0020 mol/L [O2] = 0.0028 – x = 0.0028 – 0.0010 = 0.0018 mol/L [SO3] = 2(0.0010) = 0.0020 mol/L Substitute equilibrium concentrations into the equilibrium expression and solve for Kc. 2

Kc =

SO3 

SO2  O2  2

(0.0020)2 2

(0.0020) (0.0018)

= 555.5556 = 5.6x102

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•bar   0.0020 mol  0.08314 1000. K nRT mol•K   pSO2 = = = 0.1663 = 0.17 bar V 1.0 L  15.88

The original concentrations are: (0.350 mol/0.500 L) = 0.700 mol/L for CO and Cl 2. Concentration (mol/L) CO(g) + Cl2(g)  COCl2(g) Initial 0.700 0.700 0 Change –x –x +x Equilibrium 0.700 – x 0.700 – x x COCl2  x x    Qc = = = = 4.95 COCl2  0.700  x  0.700  x  0.490  1.400x  x 2  4.95x2 – 7.93x + 2.4255 = 0 a = 4.95 b = – 7.93 c = 2.4255 x=

b  b 2  4ac 2a

(7.93) 

 7.932  4  4.95  2.4255  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 mol/L [COCl2] = x = 0.41162959 = 0.412 mol/L 15.89

Plan: Set up a reaction table to find the equilibrium amount of CaCO 3 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 bar. CaCO3(s)



CaO(s)

CO2(g)

Initial 0.100 mol 0.100 mol 0 Change –x –x +x Equilibrium 0.100 – x 0.100 – x x = 0.220 bar (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 bar of carbon dioxide. The amount (mol) of CaCO 3 reacted is equal to the amount (mol) of carbon dioxide produced. Use the pressure of CO 2 and the ideal gas equation to calculate the amount (mol) of CO2 produced: pV = nRT pV Amount (mol) of CO2 = n = RT 0.220 bar 10.0 L   n= = 0.068731 mol CO2 L•bar   0.08314 385 K    mol•K   Amount (mol) of CaCO3 reacted = amount (mol) of CO2 produced = 0.068731 mol Amount (mol) of CaCO3 remaining = initial amount (mol) amount (mol) reacted = 0.100 mol CaCO 3 – 0.068731 mol CaCO3 = 0.0313 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 bar (K). Convert the added 0.300 bar of CO2 to moles using the ideal gas equation. The amount (mol) of CO 2 reacted equals the amount (mol) of CaCO3 formed. pV Amount (mol) of CO2 = n = RT 0.300 bar 10.0 L   n= = 0.09372 mol CO2 L•bar   0.08314 385 K    mol•K   Amount (mol) of CaCO3 produced = amount (mol) of CO2 reacted = 0.09372 mol CaCO3 Add the amount (mol) of CaCO3 formed in the second equilibrium to the amount (mol) of CaCO 3 at the first equilibrium position. Amount (mol) of CaCO3 = amount (mol) at first equilibrium + amount (mol) formed in second equilibrium = 0.0313 mol + 0.09372 = 0.12502mol CaCO3  100.09 g CaCO3  Mass (g) of CaCO3 =  0.12502 mol CaCO3    = 12.514 = 12.5 g CaCO3  1 mol CaCO3  15.90

a) C2H4(g) + 3O2(g)  2CO2(g) + 2H2O(g) b) 4NO2(g) + 6H2O(g) 4NH3(g) + 7O2(g)

15.91

C2H2(g) + H2(g) C2H4(g)

r H = m  products H – n reactants H = {1  f H [C2H4(g)]} – {1  f H [C2H2(g)] + 1  f H [H2(g)]} = [(52.47 kJ/mol)] – [(227 kJ/mol) + (0.0 kJ/mol)] = –174.53 kJ/mol

ln ln

K 300  H  1 1 =  r    K 2000 R  T2 T1  K 300   103 J   174.53 kJ /mol   1 1 =      8 J   300. K 2000. K   1 kJ  2.9x10   8.314 mol•K   

K 300 = 59.478189 2.9x108 K 300 = 6.77719x1025 2.9x108 K300 = (2.9x108)(6.77719x1025) = 1.9654x1034 = 2.0x1034 ln

M2(g) + N2(g)  2MN(g)



Scene A: Concentrations: [M2] = [N2] = 0.20 mol/L; [MN] = 0.40 mol/L Kc =



 MN 2  M 2 [N 2 ]

15.92

Kc =

0.402 = 4.0 0.20[0.20]

Scene B: Concentration (mol/L) M2(g) Initial 0.60 Change –x Equilibrium 0.60 – x

N2(g) 0.30 –x 0.30 – x



2MN(g) 0 +2x 2x

2 2x   Kc = 4.0 = 0.60  x [0.30  x]

4.0 =

4x 2

0.18  0.90 x + x 2 4x = 0.72 – 3.6x + 4x2 3.6x = 0.72 x = 0.20 mol/L [M2] = 0.60 – x = 0.60 – 0.20 = 0.40 mol/L [N2] = 0.30 – x = 0.30 – 0.20 = 0.10 mol/L [MN] = 2x = 2(0.20 mol/L) = 0.40 mol/L 2

15.93

Plan: Use the balanced reaction to write the equilibrium expression. The equilibrium concentration of S 2F10 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: SF4 SF6  Kc = S2 F10  At the first equilibrium, [S2F10] = 0.50 mol/L and [SF4] = [SF6] = x ([SF4]:[SF6] = 1:1). SF4 SF6  = (x)(x) Kc = (0.50) S2 F10  x2 = 0.50Kc [SF4] = [SF6] = x =

0.50Kc At the second equilibrium, [S2F10] = 2.5 mol/L and [SF4] = [SF6] = x. SF4 SF6  = (x)(x) Kc = (2.5) S2 F10  x2 = 2.5Kc [SF4] = [SF6] = x =

2.5Kc Thus, the concentrations of SF4 and SF6 increase by a factor of:

2.5Kc 0.50 Kc 15.94

Calculate Kc. Kc =

2.5

= 2.236 = 2.2

0.50

CO2  H 2  = (0.40)(0.10) = 4.0 CO H 2 O (0.10)(0.10)

Calculate new concentrations. New H2 = 0.10 mol/L + (0.60 mol/2.0 L) = 0.40 mol/L Concentration (mol/L) CO(g) + H2O(g)  CO2(g) + H2(g) Initial 0.10 0.10 0.40 0.40 Change +x +x –x –x Equilibrium 0.10 + x 0.10 + x 0.40 – x 0.40 – x 2 CO2  H 2  = (0.40  x)(0.40  x) = (0.40  x) = 4.0 (take the sq root of both sides) Kc = CO H 2 O (0.10  x)(0.10  x) (0.10  x)2 (0.40  x) = 2.0 (0.10  x) x = 0.066667 [CO] = [H2O] = 0.10 + x = 0.10 + 0.066667 = 0.166667 = 0.17 mol/L [CO2] = [H2] = 0.40 – x = 0.40 – 0.066667 = 0.333333 = 0.33 mol/L

15.95

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 the amount (mol) 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 + VO 2 + VN 2 = 10.0 + 1.00 + 50.0 = 61.0.

 10.0 mol CO2  pCO2 =    4.0 bar  = 0.6557377 bar  61.0 mol gas   1.00 mol O2  pO2 =    4.0 bar  = 0.06557377 bar  61.0 mol gas  Use the partial pressures and given K to find pCO. 2CO2(g) 2CO(g) + O2(g) 2 2 pCO pO2 pCO  0.06557377  K= = = 1.4x10–28 2 pCO2  0.6557377 2 pCO = 3.0299x10–14 = 3.0x10–14 bar b) pV = nRT





3.0299x1014 bar nCO p = = = 4.55542x10–16 mol/L L•bar  V RT   0.08314 mol•K   800 K   

 4.55572x1016 mol CO   28.01 g CO   1 pg  Concentration (pg/L) of CO =      12  = 0.01276 pg/L = 0.013 pg L    1 mol CO   10 g  CO/L

15.96

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 NH 3, increasing the temperature).

15.97

Plan: Write a reaction table given that pCH4 (init) = pCO2 (init) = 10.0 bar, substitute equilibrium values into the equilibrium expression, and solve for pH 2 . Solution: a) Pressure (bar) Initial Change Equilibrium

CH4(g) 10.0 –x 10.0 – x

CO2(g) 10.0 –x 10.0 – x



2CO(g) 0 +2x 2x

2H2(g) 0 +2x 2x

 2x 2  2x 2  2x 4 K= = = = 3.548x106 (take square root of each side) pCH4 pCO2 10.0  x 10.0  x  10.0  x 2 2 pCO pH2 2

 2x 2

= 1.8836135x103 10.0  x   A quadratic is necessary: 4x2 + (1.8836135x103 x) – 1.8836135x104 = 0 a = 4 b = 1.8836135x103 c = – 1.8836135x104 b  b 2  4ac 2a

1.8836135x103  x=

1.8836135x10   4 4   1.8836135x10  3 2

2 4

x = 9.796209 pH 2 = 2x = 2(9.796209) = 19.592419 bar If the reaction proceeded entirely to completion, the partial pressure of H 2 would be 20.0 bar (pressure is proportional to amount (mol), and twice as many moles of H 2 form for each mole of CH4 or CO2 that reacts). 19.592418 bar The percent yield is 100%  = 97.96209 %= 98.0%. 20.0 bar b) Repeat the calculations for part a) with the new K value. The reaction table is the same. K=

2 pCO pH2 2

pCH4 pCO2

 2x 2

10.0  x 

 2x 2  2x 2  2x 4 = = 2.626x107 10.0  x 10.0  x  10.0  x 2

= 5.124451x103

A quadratic is needed: 4x2 + (5.124451x103 x) – 5.124451x104 = 0 a=4 b = 5.124451x103 c = – 5.124451x104

5.124451x103  x=

5.124451x10   4  4   5.124451x10  3 2

2 4

x = 9.923144 pH 2 = 2x = 2(9.923144) = 19.84629 bar If the reaction proceeded entirely to completion, the partial pressure of H2 would be 20.0 bar (pressure is proportional to moles, and twice as many moles of H 2 form for each mole of CH4 or CO2 that reacts).

19.84629 bar 100%  = 99.23145 % = 99.0%. 20.0 bar c) van‘t Hoff equation: K1 = 3.548x106 T1 = 1200. K r H = ? K2 = 2.626x107 T2 = 1300. K R = 8.314 J/mol•K K  H  1 1 ln 2 =  r    K1 R  T2 T1  The percent yield is

2.626x107   r H 1 1 =     J   1200. K 1300. K  3.548x106   8.314 mol•K   

2.0016628 = r H (7.710195x10–6)mol/J

r H = 2.0016628/(7.710195x10–6 mol/J )= 2.5961247x105 J/mol= 2.60x105 J/mol (The subtraction of the 1/T terms limits the answer to three significant figures.)

15.98

C3H8(g) + 3H2O(g) 3CO(g) + 7H2(g) K'1 = K1 = 8.175x1015 3CO(g) + 3H2O(g) 3CO2(g) + 3H2(g) K'2 = (K2)3 =(0.6944)3 = 0.33483368 (Overall): C3H8(g) + 6H2O(g) 3CO2(g) + 10H2(g) b) K (overall) = K'1 x K'2 = (8.175x1015)(0.33483368) = 2.737265x1015 = 2.737x1015 a)

p  p  c) K =  p  p  3

CO2

C3H8

H2O

The partial pressures of each reactant are proportional to the amount (mol), and the limiting reactant may be determined from the partial pressures. pC3H8 (initial) = (1.00/5.00) x 5.0 bar = 1.0 bar pH2O(initial) = (4.00/5.00) x 5.0 bar = 4.0 bar (limiting reactant) pCO2(formed) = 4.0 bar H2O x (3 mol CO2/6 mol H2O) = 2.0 bar pH2(formed) = 4.0 bar H2O x (10 mol H2/6 mol H2O) = 6.6667 bar pC3H8 (remaining) = 1.0 bar C3H8 – [4.0 bar H2O x (1 mol C3H8/6 mol H2O)] = 0.3333 bar pH2O(remaining) = 0.00 bar (limiting reactant) pTotal = pCO2 + pH2 + pC3H8 + pH2O = 2.0 bar + 6.6667 bar + 0.3333 bar + 0.00 bar = 9.0 bar d) Percent C3H8(unreacted) = [0.3333 bar/1.0 bar] x 100% = 33.33% = 33% 15.99

Plan: Add the two reactions to obtain the overall reaction. Multiply the second equation by 2 to cancel the amount (mol) of CO produced in the first reaction. K for the second reaction is then (K )2. K for the overall reaction is equal to the product of the K values for the two individual reactions. Calculate Kc using K = Kc(RT)n. Solution: a) 2CH4(g) + O2(g) 2CO(g) + 4H2(g) K = 9.34x1028 2CO(g) + 2H2O(g) 2CO2(g) + 2H2(g) K =(1.374)2 = 1.888 2CH4(g) + O2(g) + 2H2O(g)  2CO2(g) + 6H2(g) b) K = (9.34x1028)(1.888) = 1.76339x1029 = 1.76x1029 c) n = amount (mol) of gaseous products – amount (mol) of gaseous reactants = 8 – 5 = 3 (8 moles of product gas – 5 moles of reactant gas) K = Kc(RT)n 1.76339x1029 K Kc = = = 3.01713x1023 = 3.02x1023 [(0.08314 bar•L/mol•K)(1000)]3 ( RT )n

d) The initial total pressure is given as 30. bar. To find the final pressure use the relationship between pressure and amount (mol) of gas: ninitial/Pinitial = nfinal/Pfinal Total amount (mol) of gas initial = 2.0 mol CH4 + 1.0 mol O2 + 2.0 mol H2O = 5.0 mol Total amount (mol) of gas final = 2.0 mol CO2 + 6.0 mol H2 = 8.0 mol (from mole ratios)  8 mol products  pfinal =  30. bar reactants   = 48 bar  5 mol reactants  15.100 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 K to K. Convert pressures to amount (mol) using the ideal gas law, pV = nRT. Convert amount (mol) to atoms using Avogadro‘s number. Solution: a) The initial pressure of N2 is 200. bar. log K = –43.10; K =10–43.10 = 7.94328x10–44 Pressure (bar) N2(g)  2N(g) Initial 200. 0 Change –x +2x Equilibrium 200 – x 2x 2 pN   K= = 7.94328x10–44

p  N2

 2x 2

= 7.94328x10–44

Assume 200. – x  200.

 200.  x   2x 2 = 7.94328x10–44  200 

4x2 = 1.588656x10–41 x = 1.992897x10–21 pN = 2x = 2(1.992897x10–21) = 3.985795x10–21 = 4.0x10–21 bar b) Log K = –17.30; K = 10–17.30 = 5.01187x10–18 Pressure (bar) H2(g)  2H(g) Initial 600. 0 Change –x +2x Equilibrium 600 – x 2x K=

 pH  2

p 

= 5.01187x10–18

 2x 2

 600.  x 

= 5.01187x10–18

Assume 600. – x  600.

 2x 2 = 5.01187x10–18 600   4x2 = 3.007122x10–15 x = 2.741862x10–8 pH = 2x = 2(2.741862x10–8) = 5.48372x10–8 = 5.5x10–8 bar c) pV = nRT





3.985795x1021 bar 1.00 L  pV Moles of N atoms = = = 4.794076x10–23 mol RT L•bar    0.08314 mol•K  1000.K   

 6.022x1023 N atoms  Number of N atoms = 4.794076x1023 mol N atoms   1 mol N atoms  = 28.8699 atoms/L = 29 N   atoms/L



Moles of H atoms =







5.48372x108 bar 1.00 L  pV = = 6.595766x10–10 mol RT L•bar    0.08314 mol•K  1000.K   

 6.022x1023 H atoms  Number of H atoms = 6.595766x1010 mol H atoms   1 mol H atoms    = 3.97197x1014 = 4.0x1014 H atoms/L 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.





15.101 a) Scenes B and D represent equilibrium. b) C, A, B = D  0.025 mol   1  c) [Y] =  4 spheres     = 0.25 mol/L  1 sphere   0.40 L 

 0.025 mol   1  [Z] =  8 spheres     = 0.50 mol/L  1 sphere   0.40 L  Kc =

[Z]2 [0.50]2 = = 1.0 [Y] [0.25]

15.102 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 mol/L, and [H2] = [S2] = 0 mol/L. Concentration 2H2S(g)  2H2(g) + S2(g) Initial 0.600 mol/L 0 mol/L 0 mol/L Change –2x +2x +x Equilibrium 0.600 – 2x 2x x 2 H 2   S2   Kc = = 9.0x10–8 2  H 2 S

(2x) 2 (x) (0.600  2x)

= 9.0x10–8

Assume 2x is small compared to 0.600 mol/L.

(2x) 2 (x)

= 9.0x10–8 (0.600) 2 x = 2.008x10–3 (assumption justified) [H2S] = 0.600 – 2x = 0.600 – 2(2.008x10–3) = 0.595984 = 0.596 mol/L H2S [H2] = 2x = 2(2.008x10–3) = 4.016x10–3 = 4.0x10–3 mol/L H2 [S2] = x = 2.008x10–3 = 2.0x10–3 mol/L S2 15.103 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 K = Kc(RT)n to find Kc. Solution: a) Pressure (bar) N2(g) + O2(g)  2NO(g) Silberberg, Amateis, Venkateswaran, Chen, Chemistry: The Molecular Nature of Matter and Change, 3rd Canadian Edition Page 3-527 Instructor‘s Solution Manual © Copyright 2021 McGraw-Hill Ryerson Ltd.

Initial Change Equilibrium K=

0.780 –x 0.780 – x

 pNO 2

 p  p  N2

0.210 –x 0.210 – x

0 +2x 2x

= 4.35x10–31

 2 x 2 = 4.35x10–31 Assume x is small because K is small.  0.780  x  0.210  x   2 x 2 = 4.35x10–31  0.780  0.210  x = 1.33466x10–16 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 – 1.33466x10–16) bar = 0.780 bar N2 poxygen (equilibrium) = (0.210 – 1.33466x10–16) bar = 0.210 bar O2 (assumption justified) pNO (equilibrium) = 2(1.33466x10–16) bar = 2.66933x10–16 = 2.67x10–16 bar NO b) The total pressure is the sum of the three partial pressures: 0.780 bar + 0.210 bar + 2.67x10–16 bar = 0.990 bar n c) K = Kc(RT) n = amount (mol) of gaseous products – amount (mol) of gaseous reactants = 2 – 2 = 0 (two moles of product NO and two moles of reactants N2 and O2) K = Kc(RT) Kc = K = 4.35x10–31 because there is no net increase or decrease in the amount (mol) of gas in the course of the reaction. 15.104 a) K = Kc(RT)n n = 2 – (2 + 1) = –1 Kc = K/(RT)n

1.3x10  4

Kc =

 0.08314  457  

1

= 4.939347x105 = 4.9x105

b) r H = [  f(products) H ] – [ f(reactants) H ]

r H = {2  f H [NO2(g)]} – {2  f H [NO(g)] +   f H [O2(g)]} = [(2)(33.2 kJ/mol)] – [(2 l)(90.29 kJ/mol) +(0.0 kJ/mol)] = –114.18 kJ/mol = – 114.2 kJ/mol K  H  1 1 c) ln 2 =  r    K1 R  T2 T1  ln

114.18 kJ / mol  1 1   103 J    =    8.314 J/mol•K  T2 457 K   1 kJ  4.939347x105 6.4x109

 1 1   9.4694052 = 13,733 K   457 K   T2 T2 = 347.500 K= 3.5x102 K 15.105 Plan: Use the equation K = Kc(RT)n to find K. The value of Kc for the formation of HI is the reciprocal of the Kc value for the decomposition of HI. Use the equation r H = [  f(products) H ] – [ f(reactants) H ] to find the value of r H . Use the van‘t Hoff equation as a second method of calculating r H . Solution: a)K = Kc(RT)n Silberberg, Amateis, Venkateswaran, Chen, Chemistry: The Molecular Nature of Matter and Change, 3rd Canadian Edition Page 3-528 Instructor‘s Solution Manual © Copyright 2021 McGraw-Hill Ryerson Ltd.

n = amount (mol) of gaseous products – amount (mol) of gaseous reactants = 2 – 2 = 0 (2 mol product (1H2 + 1I2) – 2 mol reactant (HI) = 0) K = Kc(RT)0 = 1.26x10–3(RT)0 = 1.26x10–3 b) The equilibrium constant for the reverse reaction is the reciprocal of the equilibrium constant for the forward reaction: 1 1 Kformation = = 793.65 = 794 = Kdecomposition 1.26x103 c) r H = [  f(products) H ] – [ f(reactants) H ]

r H = {1  f H [H2(g)] +1  f H [I2(g)]} – {2  f H [HI(g)]} r H = [(0 kJ/mol) + (62.442 kJ/mol)] – [(2)(25.9 kJ/mol)] r H = 10.6 kJ/mol K2  H  1 1 =  r    K1 R  T2 T1  K1 = 1.26x10–3; K2 = 2.0x10–2, T1 = 298 K; T2 = 729 K  1 r H 1  2.0x102 ln  =    3 8.314 J/mol•K  729 K 298 K  1.26x10 d) ln

2.764621 = (2.38629x10–4 mol/J) r H

r H = 1.1585x104 J/mol= 1.2x104 J/mol 15.106 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. Also, according to le Châtelier‘s principle, if a product is removed, the reaction will shift to produce more of the product, as observed here. [R] 15.107 Q(g)  R(g) K= [Q] For Scene A at equilibrium: [R] [2] K= = = 0.33 [Q] [6] For Scene B: Q(g)  R(g) Initial 10 2 Change –x +x Equilibrium 10 – x 2+x [2 + x] 0.33 = [10  x] x = 0.977 = 1 Q = 10 – x = 10 – 1 = 9; R = 2 + x = 2 + 1 = 3



15.108 a) K = pH 2O

 = 4.08x10 10

–25

pH 2 O = 10 4.08x10 25 = 3.6397x10–3 = 3.64x10–3 bar b) (i) 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. (ii) Reducing the container size will increase the pressure (concentration) of the water vapour, which will shift the equilibrium to the reactant side. The ratio of hydrated form/anhydrous form will increase. Silberberg, Amateis, Venkateswaran, Chen, Chemistry: The Molecular Nature of Matter and Change, 3rd Canadian Edition Page 3-529 Instructor‘s Solution Manual © Copyright 2021 McGraw-Hill Ryerson Ltd.

(iii) Adding more water vapour will increase the concentration of the water vapour, which will shift the equilibrium to the reactant side. The ratio of hydrated form/anhydrous form will increase. (iv) Adding N2 gas will not change the partial pressure of the water vapour, so the ratio of hydrated form/anhydrous form will not change. 15.109 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 (c)(V) to find the total amount (mol), 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 mol/L. concentrations CO H2O  CO2 H2 Initial 0.00500 mol/L 0.00500 mol/L 0 0 Change –x –x +x +x Equilibrium 0.00500 – x 0.00500 – x x x [CO]equilibrium = 0.00500 – x = 2.24x10–3 mol/L = [H2O] (given in problem) x = 0.00276 mol/L = [CO2] = [H2] CO2  H 2  = (0.00276)(0.00276) = 1.518176 = 1.52 Kc = CO H 2 O (0.00224)(0.00224) b) ctotal = [CO] + [H2O] + [CO2] + [H2] = (0.00224 mol/L) + (0.00224 mol/L) + (0.00276 mol/L) + (0.00276 mol/L) = 0.01000 mol/L ntotal = (ctotal)(V) = (0.01000 mol/L)(20.00 L) = 0.2000 mol total pV = nRT L•bar   0.2000 mol   0.08314    273  900. K  mol•K   ptotal = ntotalRT/V = = 0.9752322 bar = 0.975 bar  20.00 L  c) Initially, an equal amount (mol) 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 mol/L to compensate for the added CO. CO H2O CO2 H2 Initial 0.00224 mol/L 0.00224 mol/L 0.00276 mol/L 0.00276 mol/L Added CO 0.01000 mol/L Change –x –x +x +x Equilibrium 0.01224 – x 0.00224 – x 0.00276 + x 0.00276 + x CO H  2  2  = (0.00276  x)(0.00276  x) = 1.518176 Kc = CO H 2 O (0.00224  x)(0.00224  x)

7.6176x106  5.52x103 x  x 2

= 1.518176 2.74176x105  1.448x102 x  x 2 7.6176x10–6 + 5.52x10–3x + x2 = (1.518176)(2.74176x10–5 – 1.448x10–2x + x2) 7.6176x10–6 + 5.52x10–3x + x2 = 4.162474x10–5 – 0.021983x + 1.518176x2 0.518176x2 – 0.027503x + 3.400714x10–5 = 0 a = 0.518176 b = – 0.027503 c = 3.400714x10–5 x=

b  b 2  4ac 2a

( 0.027503)  x=

 0.0275032  4  0.518176   3.400714x10 5  2  0.518176 

x = 1.31277x10–3 [CO] = 0.01224 – x = 0.01224 – (1.31277x10–3) = 0.01092723 = 0.01093 mol/L 15.110 a) At point A the sign of H 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. 15.111 Plan: Prepare an ICE table. Set up the expression for K and then substitute values. Where possible, make the relevant assumptions and then calculate the equilibrium pressures. Justify and explain all assumptions. Solution: 3 H2 (g) + N2 (g) ⇌ 2 NH3 (g) Initial 1.50 bar 1.00 bar -Change -3x -x +2x Equilibrium 1.50-3x 1.00-x 2x

K

2 pNH 3

pH3 2 pN2

 2x  3 1.5  3x  1.00  x  2

3.5  10

7

Since K is fairly small, we will make an assumption that 3x << 1.5 and therefore x << 1.00. Then, we could say 1.5-3x≈1.5 and 1.00-x ≈ 1.00. The equation thus becomes:

 2x  3 1.5 1.00  2

3.5  10

7

 2 x   1.18 106 2

x = 5.4×10-4 bar pH2  1.5 bar  3 x  1.5 bar  3  5.4 10 4 bar   1.5 bar (2 sf) pN2  1.00 bar  x  1.5 bar   5.4 104 bar   1.0 bar (2 sf) pNH3  2 x  2  5.4 104 bar   1.1103 bar

We assumed that the amount dissociated was smaller (significantly!) than the starting amounts of reactants (given that K was so small). This in turn meant that we assumed that the equilibrium pressures of the starting materials would not differ significantly from their initial pressures. The small K value indicates that the reaction stays mainly on the left, meaning the reactants dissociate only slightly. The values calculated justified the approximation to the number of significant digits in the question. 15.112 Plan: We need to calculate Q and then compare it to K to determine the direction in which the reaction will proceed. Then we can set up the ICE table and find the equilibrium pressures. We can substitute them into the expression for K and then solve the resulting quadratic for the value of x. We will use x to calculate the values of all the equilibrium pressures. Finally, we will use le Chatelier‘s principle to determine the direction in which the reaction will shift due to the stresses applied. Solution: Silberberg, Amateis, Venkateswaran, Chen, Chemistry: The Molecular Nature of Matter and Change, 3rd Canadian Edition Page 3-531 Instructor‘s Solution Manual © Copyright 2021 McGraw-Hill Ryerson Ltd.

We will need to calculate the pressures of all four substances.

bar  L    900. K  nRT mol  K   pA    3.74 bar V 5.00 L bar  L   0.500 mol   0.08314    900. K  nRT mol  K   pB    7.48 bar V 5.00 L bar  L   0.750 mol   0.08314    900. K  nRT mol  K   pC    11.2 bar V 5.00 L bar  L   0.450 mol   0.08314    900. K  nRT mol  K   pD    6.73 bar V 5.00 L 11.2 bar  6.73 bar   2.70 p p Q C D  p A pB  3.74 bar  7.48 bar 

 0.250 mol   0.08314

QK Therefore, the reaction moves in the REVERSE direction, towards reactants. b) A (g)

Initial Change Equilibrium

K

+ B (g)

3.74 +x 3.74+ x

⇌

7.48 +x 7.48+ x

C (g)

11.2 -x 11.2- x

+ D (g)

6.73 -x 6.73- x

pC pD 11.2  x  6.73  x    1.83 p A pB  3.74  x  7.48  x 

Rearranging and solving for x,

11.2  x  6.73  x   1.83  3.74  x  7.48  x  11.2  x  6.73  x   1.83  3.74  x  7.48  x  75.38  17.93x  x 2  51.19  20.53x  1.83x 2 0.83x 2  38.46 x  24.19  0 The quadratic can now be solved for x,

x 

b  b 2  4ac 2a 38.46 

 38.46   4  0.83 24.19  2  0.83 2

 0.62 bar Calculating equilibrium pressures:

pA  3.74 bar  x  3.74 bar  0.62 bar  4.36 bar pB  7.48 bar  x  7.48 bar  0.62 bar  8.10 bar pC  11.2 bar  x  11.2 bar  0.62 bar  10.6 bar pD  6.73 bar  x  6.73 bar  0.62 bar  6.11 bar c)

i) if the container volume is doubled, all the pressures are halved. However, n=0, so no change would occur. ii) if more C is added, the reaction will shift to the LEFT or REVERSE so as to produce more reactants. iii) If D is removed from the system, then more reactants will react to form D, so the reaction will shift RIGHT or move FORWARD. iv) If the temperature is raised, then heat is added. The reaction is endothermic meaning heat is a reactant. Added heat will push the reaction in the direction that uses up additional heat. The reaction will shift RIGHT or move FORWARD to produce more product.

15.113 Plan: Use the relationship between Kc and K to find the value of K. Use the given enthalpies and the values in Appendix B to find the enthalpy of reaction. Use the van‘t Hoff equation to find the value of K at the second T. Calculate Q using the given pressures. Compare Q to K. Set up an ICE table and solve for equilibrium pressures. Solution: a)

K  K c  RT 

Δn

 1.3  104  0.08314  298.2  

 1.3  104 (alternately, since Δn = 0, we can say K = Kc) b)



 

o o Δ r Ho  Δ f HClNO  Δ f HoNO2  Δ f HClNO  Δ f HoNO 2



kJ kJ   kJ kJ     51.71  33.2   12.13  90.29  mol mol   mol mol   kJ  17.51 mol c)

K2  Ho  1 1   r    K1 R  T1 T2  J 1.751  104 K2 1  mol  1 ln   298.2K  1346K  4 J 1.3  10   8.314 mol  K K2  4.1  103 4 1.3  10 K 2  53 ln

Q

p ClNO p NO2 p ClNO2 p NO



15.8bar  8.90bar   21.7 < K Therefore, the reaction moves FORWARD (right).  2.70bar  2.40bar  ClNO2 (g) + NO (g) ⇌ ClNO (g) + NO2 (g)

e) Initial Change Equilibrium

K

pClNO p NO2 pClNO2 p NO

2.70 -x 2.70-x



2.40 -x 2.40- x

15.8 +x 15.8+ x

8.90 +x 8.90+ x

15.8  x 8.90  x   53  2.70  x  2.40  x 

Rearranging and solving for x,

15.8  x 8.90  x   53  2.70  x  2.40  x  15.8  x 8.90  x   53  2.70  x  2.40  x  141  24.7 x  x 2  343  270 x  53 x 2 52 x 2  295 x  203  0 The quadratic can now be solved for x,

x 

b  b 2  4ac 2a 295 

 295  4  52  203 2  52  2

 0.80 bar Calculating equilibrium pressures:

pClNO2  2.70 bar  x  2.70 bar  0.80 bar  1.90 bar pNO  2.40 bar  x  2.40 bar  0.80 bar  1.60 bar pClNO  15.8 bar  x  15.8 bar  0.80 bar  16.6 bar pNO2  8.90 bar  x  8.90 bar  0.80 bar  9.70 bar f) The total pressure is :

pTOTAL  1.90 bar  1.60 bar  16.6 bar  9.70 bar = 29.8 bar

i) ii) iii) iv)

no change as Δn = 0 reaction will go right to replace NO2 No change on addition of an inert gas the reaction will move forward as it is exothermic

15.114 Plan: Use the relation between K and Kc to find K. Use the data from Appendix B to calculate the enthalpy of reaction. Then use the enthalpy and the van‘t Hoff equation to find the value of K at the other temperature. Finally, use le Châtelier‘s principle to determine which way the reaction will shift. Solution: a)

K  K c  RT 

Δn

 2.0  0.08314  373.2  

 62 b)



 

o Δ r Ho  2Δ f HoNO  Δ f HBr  2Δ f HoNOBr 2



     kJ  kJ     2  90.29  0    2  231.0   mol  mol        kJ  642.58 mol K  Ho  1 1  ln 2  r    K1 R  T1 T2  J 6.4258  105 K2 1  mol  1 ln    J 62 523.1K   373.1K 8.314 mol  K K2  6.3  1025 62 c)

K 2  3.9  1027 i) the reaction will shift left as there is a smaller amount (mol) of gas on the left ii) the reaction will shift left in order to remove added bromine iii) the reaction will shift left as it is endothermic.

CHAPTER 16 ACID-BASE EQUILIBRIA END–OF–CHAPTER PROBLEMS 16.1

The Arrhenius definition classifies substances as being acids or bases by their behavior in the solvent water.

16.2

(a) All Arrhenius acids contain hydrogen and produce hydronium ion (H 3O+) in aqueous solution. All Arrhenius bases contain an OH group and produce hydroxide ion (OH–) in aqueous solution. (b) Neutralization occurs when each H3O+ molecule combines with an OH– molecule to form two molecules of H2O. Chemists found that the rH 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 (H = –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.

16.3

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.

16.4

(a)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. (b)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.)

16.5

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.

16.6

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 are more HA and B – in the sample and less HB and A–, HB must be the stronger acid and A– must be the stronger base.

16.7

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)

16.8

(a) Strong acids and bases dissociate completely into their ions when dissolved in water. Weak acids and bases only partially dissociate. (b) The characteristic property of all weak acids and bases is that a significant number of the molecules are not dissociated. 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 weak acid, the concentration of hydronium ions produced when the acid dissolves is less than the initial concentration of the acid. The same is true for bases. A strong base dissociates to produce the same concentration of OH- ion as that of the strong base. The concentration of OH- that results from dissociation of a weak base is much less than the concentration of the original base.

16.9

Plan: Recall that an Arrhenius acid contains hydrogen and produces hydrogen ion (H +) ( 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. 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.

16.10

Only (a) NaHSO4 contains the Arrhenius acid HSO4-

16.11

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.

16.12

(b) H2O is a very weak Arrhenius base.

16.13

 H3O  CO32      Ka =  HCO3    c) HCOOH(aq) + H2O(l) H3O+(aq) + HCOO– (aq)  H3O    HCOO    Ka =   HCOOH  16.14

a) CH3NH3+(aq) + H2O(l) H3O+(aq) + CH3NH2(aq)

 H3O  CH3 NH 2    Ka = CH3 NH3    b) HClO(aq) + H2O(l) H3O+(aq) + ClO– (aq)  H 3O    ClO     Ka =   HClO c) H2S(aq) + H2O(l) H3O+(aq) + HS– (aq)  H3O    HS    Ka =   H 2S 16.15

Plan: Ka is the equilibrium constant for an acid dissociation which has the generic equation  H 3O    A      . H O(l) has an activity of 1 and HA(aq) + H2O(l)  H3O+(aq) + A–(aq). The Ka expression is  2  HA  thus does not appear in the expression. Write the acid-dissociation reaction for each acid, following the generic equation, and then write the Ka expression. Solution: a) HNO2(aq) + H2O(l) H3O+(aq) + NO2–(aq)  H3O    NO2     Ka =   HNO2  b) CH3COOH(aq) + H2O(l) H3O+(aq) + CH3COO–(aq)  H 3 O    CH 3COO     Ka =  CH 3COOH  c) HBrO2(aq) + H2O(l) H3O+(aq) + BrO2–(aq)

 H 3O    BrO 2     Ka =   HBrO2 

16.16

a) H2PO4–(aq) + H2O(l) H3O+(aq) + HPO42–(aq)

 H3O   HPO42      Ka =   H 2 PO4    b) H3PO2(aq) + H2O(l) H3O+(aq) + H2PO2–(aq)  H3O    H 2 PO 2     Ka =  H3 PO 2  c) HSO4–(aq) + H2O(l) H3O+(aq) + SO42–(aq)

 H3O  SO42      Ka =  HSO4     16.17

Plan: Ka is the equilibrium constant for an acid dissociation which has the generic equation  H 3O    A      . H O(l) has an activity of 1 and HA(aq) + H2O(l) H3O+(aq) + A–(aq). The Ka expression is  2  HA  thus does not appear in 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 3O    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 3O    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)

 H3O  SO42      Ka =   HSO4    16.18

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)  H 3O    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 3O    ClO3    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)

 H3O   HAsO42      Ka =   H 2 AsO4    16.19

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–

16.20

a) PO43–

16.21

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+

16.22

a) OH–

16.23

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 b) HClO4 + H2SO4  ClO4– + H3SO4+ 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, H 2SO4, because perchloric acid is a stronger acid. (HClO4‘s oxygen atoms exceed its hydrogen atoms by one more than H 2SO4.) c) HPO42– + H2SO4  H2PO4– + HSO4– base acid conjugate acid conjugate base Conjugate acid-base pairs: H2SO4/HSO4– and H2PO4–/HPO42–

16.24

a) NH3 base

16.25

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 b) CH3O– + NH3  CH3OH + NH2– base acid conjugate acid conjugate base Conjugate acid-base pairs: NH3/NH2–; CH3OH/CH3O– c) HPO42– + HSO4–  H2PO4– + SO42–

b) NH3

b) HSO4–

c) S2–

c) H3O+

HNO3  NH4+ + NO3– acid conjugate acid conjugate base Conjugate acid-base pairs: HNO3/NO3–; NH4+ /NH3 b) O2– + H2O  OH– + OH– base acid conjugate acid conjugate base Conjugate acid-base pairs: OH–/O2–; H2O/OH– c) NH4+ + BrO3–  NH3 + HBrO3 acid base conjugate base conjugate acid Conjugate acid-base pairs: NH4+/NH3; HBrO3/BrO3– +

base

acid conjugate acid conjugate base Conjugate acid-base pairs: HSO4–/SO42–; H2PO4–/HPO42–

CN–  NH3 + HCN base conjugate base conjugate acid Conjugate acid-base pairs: NH4+/NH3; HCN/CN– b) H2O + HS–  OH– + H2S acid base conjugate base conjugate acid Conjugate acid-base pairs: H2O/OH–; H2S/HS– c) HSO3– + CH3NH2  SO32– + CH3NH3+ acid base conjugate base conjugate acid Conjugate acid-base pairs: HSO3–/SO32–; CH3NH3+/CH3NH2

16.26

a) NH4+ acid

16.27

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. 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–

16.28

a) H3O+(aq) + CO32–(aq)  HCO3–(aq) + H2O(l) acid base conjugate acid conjugate base Conjugate acid-base pairs: H3O+/H2O; HCO3–/CO32– b) NH4+(aq) + OH–(aq)  NH3(aq) + H2O(l) acid base conjugate base conjugate acid Conjugate acid-base pairs: NH4+/NH3; H2O/OH–

16.29

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 (K > 1) is the one in which the stronger acid produces the weaker acid. The reaction that favors reactants (K < 1) is the reaction is which the weaker acid produces the stronger acid. Solution: The conjugate pairs are H2S (acid)/HS– (base) and HCl (acid)/Cl– (base). Two reactions are possible: (1) HS– + HCl H2S + Cl– and (2) H2S + Cl– HS– + HCl The first reaction is the reverse of the second. HCl is a strong acid and H 2S a weak acid. Reaction (1) with the stronger acid producing the weaker acid favors products and K > 1. Reaction (2) with the weaker acid forming the stronger acid favors the reactants and K < 1.

16.30

K > 1: HNO3 + F –   NO3– + HF – K < 1: NO3 + HF HNO3 + F –

16.31

Plan: An acid-base reaction that favors the products (K > 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. NH 3 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 K > 1. The stronger acid is more likely to donate a proton than the weaker acid. b) H2SO3 + NH3  HSO3– + NH4+ 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 K > 1. 16.32

Neither a or b have K > 1.

16.33

Plan: An acid-base reaction that favors the reactants (K < 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. Solution: a) NH4+ + HPO42–  NH3 + H2PO4– weaker acid weaker base stronger base stronger acid K < 1 The reaction proceeds toward the production of the weaker acid and base, i.e., the reaction as written proceeds to the left. b) HSO3– + HS–  H2SO3 + S2-– weaker base weaker acid stronger acid stronger base K < 1 The reaction proceeds toward the production of the weaker acid and base, i.e., the reaction as written proceeds to the left.

16.34

a) K < 1

16.35

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. Solution: HI is the strongest acid (it is one of the six strong acids), and acetic acid, CH 3COOH, is the weakest: CH3COOH < HF < HIO3 < HI

16.36

HCl  HNO2  HClO  HCN

16.37

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; these include HNO 3, H2SO4, and HClO4. All other acids are weak acids. Strong bases are metal hydroxides (or oxides) in which the metal is a Group 1 metal or Ca, Sr, or Ba in Group 2. Weak bases are NH3 and amines. Solution: a) Arsenic acid, H3AsO4, is a weak acid. The number of O atoms is four, which exceeds the number of ionizable H atoms, three, by one. This identifies H3AsO4 as a weak acid. b) Strontium hydroxide, Sr(OH)2, is a strong base. Soluble compounds containing OH– ions are strong bases. Sr is a Group 2 metal. c) HIO is a weak acid. The number of O atoms is one, 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 two.

16.38

Plan: Strong acids include 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. All other acids are weak. Solution: a) weak base b) strong base c) strong acid d) weak acid

16.39

Plan: Strong acids are the hydrohalic acids HCl, HBr, HI, and oxoacids in which the number of O atoms

b) K > 1

exceeds the number of ionizable protons by two or more; these include HNO 3, H2SO4, and HClO4. All other acids are weak acids. Strong bases are metal hydroxides (or oxides) in which the metal is a Group 1 metal or Ca, Sr, or Ba in Group 2. Weak bases are NH3 and amines. Solution: a) Rubidium hydroxide, RbOH, is a strong base because Rb is a Group 1 metal. b) Hydrobromic acid, HBr, is a strong acid, because it is one of the listed hydrohalic acids. c) Hydrogen telluride, H2Te, is a weak acid, because H is not bonded to an oxygen or halide. d) Hypochlorous acid, HClO, is a weak acid. The number of O atoms is one, which is equal to the number of ionizable H atoms identifying HClO as a weak acid. 16.40

a) weak base

b) strong acid

c) weak acid

d) weak acid

16.41

(a)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. (b)H2O(l) + H2O(l) H3O+(aq) + OH– (aq) H2SO4(l) + H2SO4(l) H3SO4+(solvated) + HSO4–(solvated)

16.42

H2O(l) + H2O(l) H3O+(aq) + OH– (aq) Kw =  H 3 O    OH   H2O(l) has an activity of 1 and thus does not appear in the expressionfor Kw:

16.43

a) pH increases by a value of 1. b) [H3O+] increases by a factor of 1000.

16.44

Plan: The lower the concentration of hydrogen (H +) ions, the higher the pH. pH increases as Ka or the concentration (mol/L) 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 = 4x10–5 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 [H+] and higher pH. c) Lower concentration of the same acid means lower concentration of hydrogen ions produced. The 0.01 mol/L solution has a lower [H+] and higher pH. d) At the same concentration, strong acids dissociate to produce more hydrogen ions than weak acids. The 0.1 mol/l solution of a weak acid has a lower [H+] and higher pH. e) Bases produce OH– ions in solution, so the concentration of hydrogen ion for a solution of a base solution is lower than that for a solution of an acid. The 0.01 mol/L base solution has the higher pH. f) pOH equals – log [OH–]. At 25°C, the equilibrium constant for water ionization, Kw, equals 1x10–14 so 14 = pH + pOH. As pOH decreases, pH increases. The solution of pOH = 6.0 has the higher pH.

16.45

Plan: Part a) can be approached two ways. Because NaOH is a strong base, the [OH–]eq = [NaOH]init. One method involves calculating [H+] using Kw = [H+][OH–], then calculating pH from the relationship pH = –log [H+]. 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 = [H+][OH–] 1.0x1014 Kw [H+] = = = 9.0090x10–13 mol/L 0.0111 [OH  ] pH = –log [H+] = –log (9.0090x10–13) = 12.04532 = 12.045 (3 sig fig for 0.0111, so 3 digits after the decimal) Second method: pOH = –log [OH–] = –log (0.0111) = 1.954677 pH = 14.00 – pOH = 14.00 – 1.954677 = 12.04532 = 12.045 With a pH > 7, the solution is basic.

b) For a strong acid such as HCl: [H+] = [HCl] = 1.35x10–3 mol/L pH = –log (1.35x10–3) = 2.869666 pOH = 14.00 – 2.869666 = 11.130334 = 11.13

With a pH < 7, the solution is acidic.

16.46

a) pH = –log (0.0333) = 1.47756 = 1.478; acidic b) pOH = –log (0.0347) = 1.45967 = 1.460; basic

16.47

Plan: HI is a strong acid, so [H+] = [HI] and the pH can be calculated from the relationship pH = –log [H3O+]. Ba(OH)2 is a strong base, so [OH–] = 2 x [Ba(OH)2] and pOH = –log [OH–]. Solution: a) [H+] = [HI] = 6.14x10–3 mol/L. pH = –log (6.14x10–3) = 2.211832 = 2.212. Solution is acidic. b) [OH–] = 2 x [Ba(OH)2] = 2(2.55 M) = 5.10 mol/L pOH = –log (5.10) = –0.70757 = –0.708. Solution is basic.

16.48

a) pOH = –log (7.52x10–4) = 3.12378 pH = 14.00 – 3.12378 = 10.87622 = 10.88 basic b) pH = –log (1.59x10–3) = 2.79860 pOH = 14.00 – 2.79860 = 11.20140 = 11.20 acidic

16.49

Plan: The relationships are: pH = –log [H+] and [H+] = 10–pH ; pOH = –log [OH–] and [OH–] = 10–pOH ; and 14 = pH + pOH. Solution: a) [H+] = 10–pH = 10–9.85 = 1.4125375x10–10 = 1.4x10–10 mol/L H+ pOH = 14.00 – pH = 14.00 – 9.85 = 4.15 [OH–] = 10–pOH = 10–4.15 = 7.0794578x10–5 = 7.1x10–5 mol/L OH– b) pH = 14.00 – pOH = 14.00 – 9.43 = 4.57 [H+] = 10–pH = 10–4.57 = 2.691535x10–5 = 2.7x10–5 mol/L H+ [OH–] = 10–pOH = 10–9.43 = 3.7153523x10–10 = 3.7x10–10 mol/L OH–

16.50

a) [H+] = 10–pH = 10–3.47 = 3.38844x10–4 = 3.4x10–4 mol/L H+ pOH = 14.00 – pH = 14.00 – 3.47 = 10.53 [OH–] = 10–pOH = 10–10.53 = 2.951209x10–11 = 3.0x10–11 mol/L OH– b) pH = 14.00 – pOH = 14.00 – 4.33 = 9.67 [H+] = 10–pH = 10–9.67 = 2.13796x10–10 = 2.1x10–10 mol/L H+ [OH–] = 10–pOH = 10–4.33 = 4.67735x10–5 = 4.7x10–5 mol/L OH–

16.51

Plan: The relationships are: pH = –log [H+] and [H+] = 10–pH ; pOH = –log [OH–] and [OH–] = 10–pOH ; and 14 = pH + pOH. Solution: a) [H+] = 10–pH = 10–4.77 = 1.69824x10–5 = 1.7x10-5 mol/L H+ pOH = 14.00 – pH = 14.00 – 4.77 = 9.23 [OH–] = 10–pOH = 10–9.23 = 5.8884x10–10 = 5.9x10–10 mol/L OH– b) pH = 14.00 – pOH = 14.00 – 5.65 = 8.35 [H+] = 10–pH = 10–8.35 = 4.46684x10–9 = 4.5x10–9 mol/L H3O+ [OH–] = 10–pOH = 10–5.65 = 2.23872x10–6 = 2.2x10–6 mol/L OH–

16.52

a) [H3O+] = 10–pH = 10–8.97 = 1.071519x10–9 = 1.1x10–9 mol/L H+ pOH = 14.00 – pH = 14.00 – 8.97 = 5.03 [OH–] = 10–pOH = 10–5.03 = 9.3325x10–6 = 9.3x10–6 mol/L OH– b) pH = 14.00 – pOH = 14.00 – 11.27 = 2.73 [H+] = 10–pH = 10–2.73 = 1.862087x10–3 = 1.9x10–3 mol/L H+ [OH–] = 10–pOH = 10–11.27 = 5.3703x10–12 = 5.4x10–12 mol/L OH–

16.53

Plan: The pH is increasing, so the solution is becoming more basic. Therefore, OH – ion is added to increase the pH. Since one mole of H3O+ will react with one mole of OH–, the difference in [H+] would be equal to the [OH–] added. Use the relationship [H+] = 10–pH to find [H+] at each pH. Solution: [H+] = 10–pH = 10–3.15 = 7.07946x10–4 mol/L H+ [H+] = 10–pH = 10–3.65 = 2.23872x10–4 mol/L H+ Add (7.07946x10–4 mol/L – 2.23872x10–4 mol/L) = 4.84074x10–4 = 4.8x10–4 mol of OH– per litre

16.54

The pH is decreasing so the solution is becoming more acidic. Therefore, H + ion is added to decrease the pH. [H+] = 10–pH = 10–9.33 = 4.67735x10–10 mol/L H+ [H+] = 10–pH = 10–9.07 = 8.51138x10–10 mol/L H+ Add (8.51138x10–10 mol/L – 4.67735x10–10 mol/L) = 3.83403x10–10 = 3.8x10–10 mol of H+ per litre

16.55

Plan: The pH is increasing, so the solution is becoming more basic. Therefore, OH – ion is added to increase the pH. Since one mole of H+ will react with one mole of OH–, the difference in [H+] would be equal to the [OH–] added. Use the relationship [H+] = 10–pH to find [H+] at each pH. Solution: [H+] = 10–pH = 10–4.52 = 3.01995x10–5 mol/L H+ [H+] = 10–pH = 10–5.25 = 5.623413x10–6 mol/L H+ 3.01995x10–5 mol/L – 5.623413x10–6 mol/L = 2.4576x10–5 mol/L OH– must be added. 2.4576x105 mol Moles of OH– =  5.6 L  = 1.3763x10–4 mol= 1.4x10–4 mol of OH– L

16.56

The pH is decreasing so the solution is becoming more acidic. Therefore, H + ion is added to decrease the pH. [H+] = 10–pH = 10–8.92 = 1.20226x10–9 mol/L H+ [H+] = 10–pH = 10–6.33 = 4.67735x10–7 mol/L H+ Add (4.67735x10–7 mol/L – 1.20226x10–9 mol/L)(87.5 mL)(10–3 L/1 mL) = 4.08216x10–8 mol= 4.1x10–8 mol of H+

16.57

Scene A has a pH of 4.8. [H+] = 10–pH = 10–4.8 = 1.58489x10–5 mol/L H+ Scene B: [H+] = 1.58489x105 mol/L H    25 spheres  = 1.98x10–4 mol/L H3O+  2 spheres 

pH = –log [H+] = –log [1.98x10–4] = 3.7 16.58

Plan: Apply Le Chatelier‘s principle in part a). In part b), given that the pH is 6.80, [H +] can be calculated by using the relationship [H+] = 10–pH. The problem specifies that the solution is neutral (pure water), meaning [H +] = [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) [H+] = 10–pH = 10–6.80 = 1.58489x10–7 mol/L H+ = 1.6x10–7 mol/L [H+] = [OH–] Kw = [H+][OH–] = (1.58489x10–7)(1.58489x10–7) = 2.511876x10–14 = 2.5x10–14 For a neutral solution: pH = pOH = 6.80

16.59

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.

16.60

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–.

16.61

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 H+ (H3O+) ions are formed. CH3COO–(aq) + H2O(l)  OH–(aq) + CH3COOH(aq) The solution is basic because OH– ions are formed. 16.62

Plan: Kb is the equilibrium constant for a base dissociation which has the generic equation  BH   OH     . [H O] is treated as a constant B(aq) + H2O(l)  BH+(aq) + OH–(aq). The Kb expression is  2 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) C5 H5 NH   OH     Kb =  C5 H5 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. 16.63

a) C6H5COO–(aq) + H2O(l) OH–(aq) + C6H5COOH(aq) Kb =

C6 H5 COOH  OH   C6 H5 COO   

b) (CH3)3N(aq) + H2O(l) OH–(aq) + (CH3)3NH+(aq)

 CH3  NH    OH   3    Kb =  CH3  N  3  

16.64

Plan: Kb is the equilibrium constant for a base dissociation which has the generic equation  BH   OH     . [H O] is treated as a constant and B(aq) + H2O(l)  BH+(aq) + OH–(aq). The Kb expression is  2  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)  HONH3   OH     Kb =   HONH 2  b) HPO42–(aq) + H2O(l) H2PO4–(aq) + OH–(aq)

 H 2 PO4  OH      Kb =  HPO42     16.65

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) HCC–(aq) + H2O(l) OH–(aq) + HCCH(aq) Kb =

 HCCH OH   HCC   

16.66

Plan: The Kb of a conjugate base is related to the Ka of the conjugate acid through the equation Kw = Ka x Kb. Solution: a) Acetate ion, CH3COO–, is the conjugate base of acetic acid, CH3COOH. Kw = Ka x Kb K 1.0x1014 Kb of CH3COO– = w = = 5.55556x10–10 = 5.6x10–10 5 Ka 1.8x10 b) Anilinium ion is the conjugate acid of the weak base aniline, C 6H5NH2. K 1.0x1014 Ka of C6H5NH3+ = w = = 2.5x10–5 Kb 4.0x1010

16.67

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 x Kb. K 1.0x1014 Kb of C6H5COO– = w = = 1.58730x10–10 = 1.6x10–10 Ka 6.3x10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 = 3.090295x10–10 = 3.1x10–10

16.68

Plan: The Kb of a conjugate base is related to the Ka of the conjugate acid through the equation Kw = Ka x Kb. Solution: a) HClO2 is the conjugate acid of chlorite ion, ClO2–. K 1.0x1014 Kb of ClO2– = w = = 9.0909x10–13 2 Ka 1.1x10 pKb = –log (9.0909x10–13) = 12.04139 = 12.04 b) (CH3)2NH is the conjugate base of (CH3)2NH2+. K 1.0x1014 Ka of (CH3)2NH2+ = w = = 1.694915x10–11 Kb 5.9x104 pKa = –log (1.694915x10–11) = 10.77085 = 10.77

16.69

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 x Kb, and pKb = –log Kb. K 1.0x1014 Kb of NO2– = w = = 1.4084507x10–11 Ka 7.1x104 pKb = –log (1.4084507x10–11) = 10.851258 = 10.85 b) The Kb of hydrazine, H2NNH2, is reported in the problem. Hydrazine is the conjugate base of H 2N–NH3+. The Ka for H2N–NH3+ is related to the Kb for H2NNH2 through the equation Kw = Ka x Kb, and pKa = –log Ka. K 1.0x1014 Ka of H2N–NH3+ = w = = 1.17647x10–8 Kb 8.5x107 pKa = –log (1.17647x10–8) = 7.9294 = 7.93

16.70

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) 16.71

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.

16.72

Water will add approximately 10–7 mol/L to the H+ 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 3O    CH 3COO     Ka = 1.8x10–5 =  CH 3COOH  Ka = 1.8x10–5 = x x 

0.1  x  Ka = 1.8x10–5 = x x  0.1

Assume x is small compared to 0.1 so 0.1 – x = 0.1.

x = 1.3416x10–3 mol/L Since the H+ 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) Ka = 1.8x10–5 = x 0.1  x 

0.1  x 

Assume x is small compared to 0.1.

x = [H3O+] = 1.8x10–5 [CH3COO–] = 0.1 + x = 0.1 mol/L Thus, [CH3COO–] > [H3O+] 16.73

The higher the negative charge on a species, the more difficult it is to remove a positively charged H+ ion.

16.74

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) Initial: 0.15 mol/L 0 0 Change: –x +x +x Equilibrium: 0.15 – x x x According to the information given in the problem, [H 3O+]eq = 1.51x10–3 mol/L = x Thus, [H3O+] = [CH3CH2CH2COO–] = 1.51x10–3 mol/L [CH3CH2CH2COOH] = (0.15 – x) = (0.15 – 1.51x10–3) mol/L = 0.14849 mol/L  H 3O    CH 3CH 2 CH 2 COO     Ka =  CH 3CH 2 CH 2 COOH 

1.51x10 1.51x10  = 1.53552x10 = 1.5x10 3

Ka =

 0.14849 

3

–5

16.75

Any weak acid dissociates according to the following equation: HA(aq) + H2O(l) H3O+(aq) + A–(aq) [H+] = 10–pH = 10–4.88 = 1.31826x10–5 mol/L Thus, [H+] = [A–] = 1.31826x10–5 mol/L, and [HA] = (0.035 – 1.31826x10–5) = 0.03499 mol/L  H3O    A     Ka =  HA 

1.31826x10 1.31826x10  = 4.967x10 = 5.0x10 K = 5

–9

 0.03499 

16.76

5

–9

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 [H+] and pH. Solution: The formula of dimethylamine has two methyl (CH 3–) groups attached to a nitrogen:

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 (mol/L) (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 Kb = 5.9x10

–4

 CH3  NH 2   OH   2    =  CH 3  NH  2  

 x  x 

Kb = 5.9x10–4 =

0.070  x  x x 5.9x10–4 =     0.070

Assume 0.070 – x = 0.070

x = 6.4265x10–3 mol/L Check assumption that x is small compared to 0.070: 6.4265x103 100%  = 9% error, so the assumption is not valid. 0.070 The problem will need to be solved as a quadratic. 5.9x10–4 =

 x  x 

0.070  x 

x2 = (5.9x10–4)(0.070 – x) = 4.13x10–5 – 5.9x10–4 x x2 + 5.9x10–4 x – 4.13x10–5 = 0 a=1 b = 5.9x10–4 c = –4.13x10–5 2 x = b  b  4ac

5.9x104  x=

5.9x10   4 1  4.13x10  4 2

2 1

5

= 6.13827x10–3 mol/L OH–

[H+] =

1.0x1014 = 1.629124x10–12 mol/L H+ 6.13827x103

OH   pH = –log [H+] = –log (1.629124x10–12) = 11.7880 = 11.79 16.77



(CH3CH2)2NH(aq) + H2O(l) OH–(aq) + (CH3CH2)2NH2+(aq) 0.12 – x x x

 CH3CH 2  NH 2  OH   2    Kb = 8.6x10 =  CH3CH 2  NH  2   –4

 x  x   0.12  x  Kb = 8.6x10–4 =  x  x   0.12  Kb = 8.6x10–4 =

Assume x is small compared to 0.12.

x = 0.0101587 Check assumption: (0.0101587/0.12) x 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 = (8.6x10–4)(0.12 – x) = 1.032x10–4 – 8.6x10–4 x x2 + 8.6x10–4 x – 1.032x10–4 = 0 a=1 b = 8.6x10–4 c = –1.032x10–4 2 x = b  b  4ac

8.6x104  x=

8.6x10   4 1  1.032x10  4 2

4

2 1 x = 9.7378x10 mol/L OH– [H]+ = Kw/[OH–] = (1.0x10–14)/(9.7378x10–3) = 1.02693x10–12 mol/L H+ pH = –log [H+] = –log (1.02693x10–12) = 11.98846 = 11.99 –3

16.78

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 (mol/L) 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 NH3   OH     Kb = 3.2x10–5 =   HOCH 2 CH2 NH 2  Kb = 3.2x10–5 =

 x  x 

 0.25  x 

Assume x is small compared to 0.25.

Kb = 3.2x10–5 =  x  x 

 0.25 

–3

x = 2.8284x10 mol/L OH– Check assumption that x is small compared to 0.25: 2.8284x103 100%  = 1% error, so the assumption is valid. 0.25

[H]+ =

1.0x1014 = 3.535568x10–12 mol/L H+ 2.8284x103

OH   pH = –log [H+] = –log (3.535568x10–12) = 11.4515 = 11.45 16.79



C6H5NH2(aq) + H2O(l) OH–(aq) + C6H5NH3+(aq) 0.26 – x x x C6 H5 NH3   OH     Kb =4.0x10–10 =  C6 H5 NH 2 

 x  x   0.26  x  Kb = 4.0x10–10 =  x  x   0.26  Kb = 4.0x10–10 =

16.80

Assume x is small compared to 0.26.

x = 1.01980x10–5 mol/L OH– Check assumption: (1.01980x10–5/0.26) x 100% = 0.004% error, so the assumption is valid. [H]+ = Kw/[OH–] = (1.0x10–14)/(1.01980x10–5) = 9.80584x10–10 mol/L H+ pH = –log [H+] = –log (9.80584x10–10) = 9.008515 = 9.01 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 ([H+]). 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: Concentration (mol/L) HNO2(aq) + H2O(l)  H3O+(aq) + NO2–(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 3O    NO 2   –4   Ka = 7.1x10 =  HNO 2 

x x  0.60  x  Ka = 7.1x10–4 = x x  0.60  Ka = 7.1x10–4 =

Assume x is small compared to 0.60: 0.60 – 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 [H+ ] = [NO2– ] = 2.1x10–2 mol/L 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 = 1.0x10–14 Kw = 1.0x10–14 =[H+][OH– ] [OH–] = 1.0x10–14/0.020639767 = 4.84502x10–13 = 4.8x10–13 mol/L OH– 16.81

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: Concentration (mol/L) HF(aq) + H2O(l)  H3O+(aq) + F–(aq) Initial 0.75 — 0 0 Change –x +x +x Equilibrium 0.75 – x x x (The H+ contribution from water has been neglected.)

 H 3 O    F    Ka = 6.8x10–4 =  HF

x x  0.75  x  Ka = 6.8x10–4 = x x  0.75  Ka = 6.8x10–4 =

Assume x is small compared to 0.75.

x = 0.02258 Check assumption: (0.02258/0.75) x 100% = 3% error, so the assumption is valid. [H+ ] = [F– ] = 2.3x10–2 mol/L [OH–] = 1.0x10–14/0.02258 = 4.42869796x10–13 = 4.4x10–13 mol/L OH– 16.82

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 ([H+]). Ka is found from the pKa by using the relationship Ka = 10–pKa. Solution: Ka = 10–pKa = 10–2.87 = 1.34896x10–3 Concentration (mol/L) ClCH2COOH(aq) + H2O(l) H3O+(aq) + ClCH2COO–(aq) Initial 1.25 0 0 Change –x +x +x Equilibrium 1.25 – x x x  H 3O    ClCH 2 COO     Ka = 1.34896x10–3 =  ClCH 2 COOH 

x x  1.25  x  Ka = 1.34896x10–3 = x x  1.25  Ka = 1.34896x10–3 =

Assume x is small compared to 1.25.

x = 0.04106337 Check assumption that x is small compared to 1.25: 0.04106337 100%  = 3.3%. The assumption is good. 1.25 [H+] = [ClCH2COO–] = 0.041 mol/L [ClCH2COOH] = 1.25 – 0.04106337 = 1.20894 = 1.21 mol/L pH = –log [H+] = –log (0.04106337) = 1.3865 = 1.39 16.83

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 = 2.88403x10–8 HClO(aq) + H2O(l) H3O+(aq) + ClO–(aq) 0.115 – x x x  H 3O    ClO     Ka = 2.88403x10–8 =  HClO Ka = 2.88403x10–8 =

x x 

0.115  x  –8 Ka = 2.88403x10 = x x  0.115 

Assume x is small compared to 0.115.

x = 5.75902x10–5 Check assumption: (5.75902x10–5/0.115) x 100% = 0.05%. The assumption is good. [H+] = [ClO–] = 5.8x10–5 mol/L [HClO] = 0.115 – 5.75902x10–5 = 0.11494 = 0.115 mol/L Silberberg, Amateis, Venkateswaran, Chen, Chemistry: The Molecular Nature of Matter and Change, 3rd Canadian Edition Page 3-551 Instructor‘s Solution Manual © Copyright 2021 McGraw-Hill Ryerson Ltd.

pH = –log [H+] = –log (5.75902x10–5) = 4.2397 = 4.24 16.84

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 x Kb. In part b), the salt triethylammonium chloride in water dissociates into two ions: (CH 3CH2)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 [H+]. Find the Ka for (CH3CH2)3NH+ from the equation Kw = Ka x Kb. Solution: a) CN–(aq) + H2O(l) HCN(aq) + OH–(aq) Concentration (mol/L) CN–(aq) + H2O(l)  HCN(aq) + OH–(aq) Initial 0.150 — 0 0 Change –x +x +x Equilibrium 0.150 – x x x Kw 1.0x1014 – –5 Kb of CN = = = 1.612903x10 Ka 6.2x1010 –5

Kb = 1.612903x10 = Kb = 1.612903x10–5 =

 HCN OH  CN    

 x  x 

0.150  x 

Assume x is small compared to 0.150.

Kb = 1.612903x10–5 =  x  x 

 0.150 

x = 1.555x10–3 mol/L OH– Check assumption that x is small compared to 0.150: 1.555x103 100%  = 1% error, so the assumption is valid. 0.150 Kw 1.0x1014 [H]+ = = = 6.430868x10–12 mol/L H+ OH   1.555x10 3   pH = –log [H+] = –log (6.430868x10–12) = 11.19173 = 11.19 b) (CH3CH2)3NH+(aq) + H2O(l) (CH3CH2)3N(aq) + H3O+(aq) (CH3CH2)3NH+(aq) + H2O(l)  (CH3CH2)3N(aq) + H3O+(aq) 0.40 — 0 0 –x +x +x 0.40 – x x x 14 K 1.0x10 Ka of (CH3CH2)3NH+ = w = = 1.9230769x10–11 Kb 5.2x104 Concentration (mol/L) Initial Change Equilibrium

Ka = 1.9230769x10

–11

 H3O  (CH3CH2 )3 N   = (CH3CH2 )3 NH   

 x  x   0.40  x  Ka = 1.9230769x10–11 =  x  x   0.40  Ka = 1.9230769x10–11 =

Assume x is small compared to 0.40.

[H+] = x = 2.7735x10–6 mol/L Silberberg, Amateis, Venkateswaran, Chen, Chemistry: The Molecular Nature of Matter and Change, 3rd Canadian Edition Page 3-552 Instructor‘s Solution Manual © Copyright 2021 McGraw-Hill Ryerson Ltd.

Check assumption that x is small compared to 0.40: 2.7735x106 100%  = 0.0007% error, so the assumption is valid. 0.40

pH = –log [H+] = –log (2.7735x10–6) = 5.55697 = 5.56

a) Sodium phenolate, when placed in water, dissociates into sodium ions, Na +, and phenolate ions, C6H5O–. Sodium ion is a spectator ion 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 (mol/L) 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 1.0x10 Kb of C6H5O– = w = = 1.0x10–4 Ka 1.0x1010 16.85

–4

Kb = 1.0x10 =

C6 H5OH  OH    C6 H 5 O    

 x  x  0.100  x  Kb = 1.0x10–4 =  x  x   0.100  Kb =1.0x10–4 =

Assume x is small compared to 0.100.

x = 3.1622777x10–3 mol/L OH– Check assumption: (3.1622777x10–3/0.100)x100% = 3% error, so the assumption is valid. [H]+ = Kw/[OH–] = (1.0x10–14)/(3.16227766x10–3) = 3.1622776x10–12 mol/L H+ pH = –log [H+] = –log (3.16227762x10–12) = 11.50 b) The salt methylammonium bromide in water dissociates into two ions: CH 3NH3+ 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 (mol/L) 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 1.0x10 Ka of CH3NH3+ = w = = 2.272727x10–11 Kb 4.4x104 Ka = 2.272727x10

–11

 H3O  CH3 NH 2    = CH3 NH3   

 x  x  Assume x is small compared to 0.15.  0.15  x  Ka = 2.272727x10–11 =  x  x   0.15  Ka = 2.272727x10–11 =

[H+] = x = 1.84637x10–6 Check assumption: (1.84637x10–6/0.15)x100% = 0.001% error, so the assumption is valid. pH = –log [H+] = –log (1.84637x10–6) = 5.73368 = 5.73 16.86

Plan: In part a), potassium formate, when placed in water, dissociates into potassium ions, K+, and formate ions, HCOO–. Potassium ion is a spectator ion, 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 aciddissociation 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 x Kb. Solution: a) HCOO–(aq) + H2O(l)  HCOOH(aq) + OH–(aq) Concentration (mol/L) 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 1.0x10 Kb of HCOO– = w = = 5.55556x10–11 Ka 1.8x104 Kb = 5.55556x10

–11

 HCOOH OH   HCOO   

 x  x   0.65  x  Kb = 5.55556x10–11 =  x  x   0.65  Kb = 5.55556x10–11 =

Assume x is small compared to 0.65.

x = 6.00925x10–6 mol/L OH– Check assumption that x is small compared to 0.65: 6.00925x106 100%  = 0.0009% error, so the assumption is valid. 0.65 Kw 1.0x1014 [H]+ = = = 1.66410x10–9 mol/L H+ OH   6.00925x106   pH = –log [H+] = –log (1.66410x10–9) = 8.7788 = 8.78 b) NH4+(aq) + H2O(l) H3O+(aq) + NH3(aq) Concentration (mol/L) NH4+(aq) + H2O(l)  NH3(aq) Initial 0.85 — 0 Change –x +x Equilibrium 0.85 – x x 14 K 1.0x10 Ka of NH4+ = w = = 5.681818x10–10 Kb 1.76x105 Ka = 5.681818x10

–10

H3O+(aq) 0 +x x

 H3O  NH3    =  NH 4    

Ka = 5.681818x10–10 =

 x  x 

0.85  x  Ka = 5.681818x10–10 =  x  x   0.85

Assume x is small compared to 0.85.

[H+] = x = 2.1976x10–5 mol/L Check assumption that x is small compared to 0.85: 2.1976x105 100%  = 0.003% error, so the assumption is valid. 0.85

pH = – log [H+] = – log (2.1976x10–5) = 4.65805 = 4.66

16.87

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 (mol/L) F–(aq) + H2O(l)  HF(aq) + OH–(aq) Initial 0.75 — 0 0 Change –x +x +x Equilibrium 0.75 – x x x 14 K 1.0x10 Kb of F– = w = = 1.470588x10–11 Ka 6.8x104 Kb = 1.470588x10

–11

 HF OH   F   

Kb = = 1.470588x10–11 =

 x  x 

 0.75  x  x –11 Kb = 1.470588x10 =   x   0.75 

Assume x is small compared to 0.75.

x = 3.3210558x10–6 mol/L OH– Check assumption: (3.3210558x10–6/0.75) x 100% = 0.0004% error, so the assumption is valid. [H]+ = Kw/[OH–] = (1.0x10–14)/(3.3210558x10–6) = 3.01109x10–9 mol/L H+ pH = –log [H+] = –log (3.01109x10–9) = 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 (mol/L) 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 1.0x10 Ka of NH4+ = w = = 5.88235x10–6 Kb 1.7x109

 H3O  C5 H5 N    Ka = 5.88235x10 = C5 H5 NH     –6

Ka = 5.88235x10–6 =

 x  x 

0.88  x  –6 Ka = 5.88235x 10 =  x  x   0.88 

Assume x is small compared to 0.88.

[H+] = x = 2.275186x10–3 mol/L Check assumption: (2.275185x10–3/0.88) x 100% = 0.3% error, so the assumption is valid. pH = –log [H+] = –log (2.275185x10–3) = 2.64298 = 2.64 16.88

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 100%  initial acid

x 100%  0.20 [Dissociated acid] = x = 6.0x10–3 mol/L Concentration (mol/L) 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+] = 6.0x10–3 mol/L pH = –log [H+] = –log (6.0x10–3) = 2.22185 = 2.22 Kw = 1.0x10–14 = [H+][OH– ] Kw 1.0x1014 [OH–] = = = 1.6666667x10–12 = 1.7x10–12 mol/L 3 + H  6.0x10   – pOH = –log [OH ] = –log (1.6666667x10–12) = 11.7782 = 11.78 b) In the equilibrium expression, substitute the concentrations above and calculate Ka. 3.0% =



3



6.0x10 6.0x10  H3O    A     = Ka =  HA  0.20  6.0x103 16.89



3



 = 1.85567x10 = 1.9x10 –4

–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 H+. Then, pH and [OH–] are determined from [H +]. dissociated acid Percent HA dissociated = 100%  initial acid

x 100% 0.735 [Dissociated acid] = 9.1875x10–2 mol/L HA(aq) + H2O(l)   H3O+(aq) + A–(aq) 0.735 – x x x [Dissociated acid] = x = [H3O+] = 9.19x10–2 mol/L pH = –log [H+] = –log (9.1875x10–2) = 1.03680 = 1.037 [OH–] = Kw/[H+] = (1.0x10–14)/(9.1875x10–2) = 1.0884x10–13 = 1.1x10–13 mol/L pOH = –log [OH–] = –log (1.0884x10–13) = 12.963197 = 12.963 b) In the equilibrium expression, substitute the concentrations above and calculate Ka. 12.5% =





9.1875x102 9.1875x102  H3O    A       Ka = = HA  0.735  9.1875x102 16.90





 = 1.3125x10 = 1.31x10 –2

–2

Plan: Write the acid-dissociation reaction and the expression for Ka. Calculate the concentration (mol/L) 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. Concentration (mol/L ) of HX =  0.250 mol   1 mL  = 0.381679 mol/L  655 mL   103 L 

Concentration (mol/L) HX(aq) + H2O(l)   H3O+(aq) + X–(aq) Initial: 0.381679 0 0 Change: –x +x +x Equilibrium: 0.381679 – x x x [H+] = 10–pH = 10–3.54 = 2.88403x10–4 mol/L = x

Thus, [H+] = [X–] = 2.88403x10–4 mol/L, and [HX] = (0.381679 – 2.88403x10–4) mol/L=0.38139 mol/L  H3O    X   2.88403x104 2.88403x10 4     Ka = = = 2.18087x10–7 = 2.2x10–7  0.38139  HX 







Calculate the concentration (mol/L) of HY by dividing moles by volume. Convert pH to [H 3O+] and substitute into the equilibrium expression. Concentration of HY = (4.85x10–3 mol/0.095 L) = 0.0510526 mol/L HY(aq) + H2O(l)  H3O+(aq) + Y–(aq) E 0.0510526 – x x x

16.91

[H+] = 10–pH = 10–2.68 = 2.089296x10–3 mol/L = x Thus, [H+] = [Y–] = 2.089296x10–3 mol/L, and [HY] = (0.0510526 – 2.089296x10–3) mol/L





2.089296x103 2.089296x103  H3O    Y       Ka = = HY  0.0510526  2.089296x103 16.92





 = 8.91516x10 = 8.9x10 –5

–5

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 ([H+]). OH– and then pOH can be found from [H+]. Solution: a) Concentration(mol/L) HZ(aq) + H2O(l)  H3O+(aq) + Z–(aq) Initial 0.075 — 0 0 Change –x +x +x Equilibrium 0.075 – x x x (The H+ contribution from water has been neglected.)  H3O    Z    Ka = 2.55x10–4 =  HZ Ka = 2.55x10–4 =

x x 

0.075  x 

Assume x is small compared to 0.075.

Ka = 2.55x10–4 = x x 

0.075 

[H+] = x = 4.3732x10–3 Check assumption that x is small compared to 0.075: 4.3732x10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. Ka = 2.55x10–4 =

x x  0.075  x 

x2 + 2.55x10–4 x – 1.9125x10–5 = 0 a=1 b = 2.55x10–4

c = –1.9125x10–5

2 x = b  b  4ac

(2.55x104 )  x=

 2.55x10   4 1  1.9125x10  4 2

5

2 1

x = 0.00425 or –0.004503 (The –0.004503 value is not possible.) pH = –log [H+] = –log (0.00425) = 2.3716 = 2.37

b) Concentration (mol/L) HZ(aq) + H2O(l)  Initial 0.045 — Change –x Equilibrium 0.045 – x (The H+ contribution from water has been neglected.)  H3O    Z  –4   Ka = 2.55x10 =  HZ

x x  0.045  x  Ka = 2.55x10–4 = x x  0.045  Ka = 2.55x10–4 =

H3O+(aq) 0 +x x

Z–(aq) 0 +x

Assume x is small compared to 0.045.

[H+] = x = 3.3875x10–3 Check assumption that x is small compared to 0.045: 3.3875x10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. Ka = 2.55x10–4 =

x x  0.045  x 

x2 = (2.55x10–4)(0.045 – x) = 1.1475x10–5 – 2.55x10–4 x x2 + 2.55x10–4 x – 1.1475x10–5 = 0 a=1 b = 2.55x10–4 c = –1.1475x10–5 2 x =  b  b  4ac

2.55x104  x=

 2.55x10   4 1  1.1475x10  4 2

5

2 1

–3

x = 3.26238x10 mol/L H Kw 1.0x1014 [OH–] = = = 3.0652468x10–12 mol/L  H3O +  3.26238x103   pOH = –log [OH–] = –log (3.0652468x10–12) = 11.51353 = 11.51 16.93

Calculate Ka from pKa. Ka = 10–pKa = 10–4.89 = 1.28825x10–5 a) Begin with a reaction table, and then use the Ka expression as in earlier problems. Concentration (mol/L) HQ(aq) + H2O(l)  H3O+(aq) + Initial 0.035 — 0 Change –x +x Equilibrium 0.035 – x x (The H3O+ contribution from water has been neglected.)  H3O   Q     Ka = 1.28825x10–5 =  HQ   Ka = 1.28825x10–5 =

x x 

0.035  x  Ka = 1.28825x10–5 = x x  0.035 

Q–(aq) 0 +x x

Assume x is small compared to 0.035.

[H+] = x = 6.71482x10–4 mol/L Check assumption: (6.71482x10–4/0.035) x 100% = 2% error, so the assumption is valid. Silberberg, Amateis, Venkateswaran, Chen, Chemistry: The Molecular Nature of Matter and Change, 3rd Canadian Edition Page 3-558 Instructor‘s Solution Manual © Copyright 2021 McGraw-Hill Ryerson Ltd.

[H+] = 6.7x10–4 mol/L b) Concentration (mol/L) HQ(aq) + H2O(l)  H3O+(aq) Initial 0.65 — 0 Change –x +x Equilibrium 0.65 – x x (The H+ contribution from water has been neglected.)  H3O   Q   –5   Ka = 1.28825x10 =  HQ

x x  0.65  x  Ka = 1.28825x10–5 = x x  0.65  Ka = 1.28825x10–5 =

Q–(aq) 0 +x x

Assume x is small compared to 0.65.

[H+] = x = 2.89372x10–3 mol/L Check assumption: (2.89372x10–3/0.65) x 100% = 0.4% error, so the assumption is valid. [OH–] = Kw/[H+] = (1.0x10–14)/(2.89372x10–3) = 3.4558x10–12 = 3.5x10–12 mol/L 16.94

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 [H+]. Use the expression for Ka to solve for x ([H+]). OH– and then pOH can be found from [H+]. Solution: a) Concentration (mol/L) 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.)  H3O    Y   –4   Ka = 1.50x10 =  HY 

x x  0.175  x  Ka = 1.50x10–4 = x x  0.175  Ka = 1.50x10–4 =

Assume x is small compared to 0.175.

[H+] = x = 5.1235x10–3 mol/L 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 [H+] = –log (5.1235x10–3) = 2.29043 = 2.290 b) Concentration (mol/L) HX(aq) + H2O(l)  H3O+(aq) Initial 0.175 — 0 Change –x +x Equilibrium 0.175 – x x (The H3O+ contribution from water has been neglected.)  H3O    X     Ka = 2.00x10–2 =  HX   Ka = 2.00x10–2 =

 x  x 

 0.175  x  Ka = 2.00x10–2 =  x  x   0.175 

X–(aq) 0 +x x

Assume x is small compared to 0.175.

[H+] = x = 5.9161x10–2 mol/L Check assumption that x is small compared to 0.175: Silberberg, Amateis, Venkateswaran, Chen, Chemistry: The Molecular Nature of Matter and Change, 3rd Canadian Edition Page 3-559 Instructor‘s Solution Manual © Copyright 2021 McGraw-Hill Ryerson Ltd.

5.9161x10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. Ka = 2.00x10–2 =

 x  x   0.175  x 

x2 = = (2.00x10–2)(0.175 – x) = 0.0035 – 2.00x10–2x x2 + 2.00x10–2x – 0.0035 = 0 a=1 b = 2.00x10–2 c = –0.0035 2 x = b  b  4ac

2.00x102  x=

 2.00x10   4 1 0.0035 2 2

2 1

–2

x = 5.00x10 mol/L H Kw 1.0x10 14 [OH–] = = = 2.00x10–13 mol/L 2 H+  5.00x10   – pOH = –log [OH ] = –log (2.00x10–13) = 12.69897 = 12.699 16.95

a) Begin with a reaction table, then use the Ka expression as in earlier problems. Concentration (mol/L) HCN(aq) + H2O(l)  H3O+(aq) + Initial 0.55 — 0 Change –x +x Equilibrium 0.55 – x x (The H3O+ contribution from water has been neglected.)  H3O    CN   –10   Ka = 6.2x10 =   HCN 

 x  x   0.55  x  Ka = 6.2x10–10 =  x  x   0.55  Ka = 6.2x10–10 =

CN–(aq) 0 +x x

Assume x is small compared to 0.55.

[H+] = x = 1.84662x10–5 mol/L Check assumption: (1.84662x10–5/0.55) x 100% = 0.0034% error, so the assumption is valid. pH = –log [H+] = –log (1.84662x10–5) = 4.7336 = 4.73 b) Begin this part like part a). Concentration (mol/L) HIO3(aq) + H2O(l)  H3O+(aq) + IO3–(aq) Initial 0.044 — 0 0 Change –x +x +x Equilibrium 0.044 – x x x (The H+ contribution from water has been neglected.)  H3O    IO3    Ka = 0.16 =  HIO3  Ka = 0.16 =

 x  x 

 0.044  x  Ka = 0.16 =  x  x   0.044 

Assume x is small compared to 0.044.

[H+] = x = 8.3905x10–2 Silberberg, Amateis, Venkateswaran, Chen, Chemistry: The Molecular Nature of Matter and Change, 3rd Canadian Edition Page 3-560 Instructor‘s Solution Manual © Copyright 2021 McGraw-Hill Ryerson Ltd.

Check assumption: (8.3905x10–2/0.044) x 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. Ka = 0.16 =

 x  x 

 0.044  x 

x2 = (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 2 x = b  b  4ac

 0.16 2  4 1 0.00704  2 1

0.16 

x = 0.03593 mol/L H+ [OH–] = Kw/[H+] = (1.0x10–14)/(0.03593) = 2.78x10–13 mol/L pOH = –log [OH–] = –log (2.78x10–13) = 12.55596 = 12.56 16.96

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 (mol/L) C6H5COOH(aq) + H2O(l)  H3O+(aq) + C6H5COO–(aq) Initial 0.55 — 0 0 Change –x +x +x Equilibrium 0.55 – x x x  H 3 O    C6 H 5 COO     Ka = 6.3x10–5 =  C6 H 5 COOH 

Ka = 6.3x10–5 =

 x  x 

0.55  x 

Assume x is small compared to 0.55.

Ka = 6.3x10–5 =  x  x 

 0.55

–3

x = 5.8864x10 mol/L

16.97

Percent C6H5COOH dissociated =

C 6 H 5 COOH dissociated 100%  C 6 H 5COOH initial

Percent C6H5COOH dissociated =

5.8864x103 mol/L 100%  = 1.07025 = 1.1% 0.55 mol/L

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. Concentration (mol/L) CH3COOH(aq) + H2O(l)  H3O+(aq) + CH3COO–(aq) Initial 0.050 — 0 0 Change –x +x +x Equilibrium 0.050 – x x x  H3O    CH 3COO     Ka = 1.8x10–5 =  CH 3COOH 

Ka = 1.8x10–5 =

 x  x 

Assume x is small compared to 0.050.

 0.050  x 

Ka = 1.8x10–5 =  x  x 

 0.050

–4

x = 9.48683x10 mol/L

16.98

Percent CH3COOH dissociated =

CH 3COOH dissociated 100%  CH 3COOH initial

Percent CH3COOH dissociated =

9.48683x10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 K a >> K a , assume that almost all of the H3O+ comes from the first 1

dissociation. Set up reaction tables in which x = the concentration of dissociated acid and [H 3O+]. Solution: H2S(aq) + H2O(l) H3O+(aq) + HS–(aq) HS–(aq) + H2O(l) H3O+(aq) + S2–(aq)  H 3O    HS     = 9x10 = Ka  H 2 S –8

K a 2 = 1x10

Concentration (mol/L) Initial Change Equilibrium

H2S(aq) + 0.10 –x 0.10 – x  H 3O    HS  –8    K a = 9x10 =  H 2 S

H2O(l) —

–17

 H3O+(aq) 0 +x x

 H3O  S2      =  HS    +

HS–(aq) 0 +x x

 x  x  0.10  x  –8  x  x  K = 9x10 =  0.10 K a1 = 9x10

–8

Assume x is small compared to 0.10.

x = 9.48683x10–5 [H+] = [HS–] = x = 9x10–5 mol/L pH = –log [H+] = –log (9.48683x10–5) = 4.022878 = 4.0 Kw 1.0x1014 [OH–] = = = 1.05409x10–10 = 1x10–10 mol/L H+  9.48683x105   pOH = –log [OH–] = –log (1.05409x10–10) = 9.9771 = 10.0 [H2S] = 0.10 – x = 0.10 – 9.48683x10–5 = 0.099905 = 0.10 mol/L 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. Calculate [S2–] by using the K a expression and assuming that [HS–] and [H3O+] come mostly from the first 2

dissociation. This new calculation will have a new x value. Concentration (mol/L) HS–(aq) + H2O(l)  H3O+(aq) –5 Initial 9.48683x10 — 9.48683x10–5 Change –x +x Equilibrium 9.48683x10 –5 – x 9.48683x10 –5 + x K a 2 = 1x10

–17

S2–(aq) 0 +x x

 H3O  S2      =  HS   

9.48683x10  x   x  K = 1x10 = 9.48683x10  x  9.48683x10   x  K = 1x10 = 9.48683x10  5

–17

5

Assume x is small compared to 9.48683x10 –5.

5

–17

5

x = [S2–] = 1x10–17 mol/L The small value of x means that it is not necessary to recalculate the [H +] and [HS–] values. 16.99

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)  H 3O    HC3 H 2 O 4      = 1.4x10 = Ka H 2 C3 H 2 O 4  –3

 H3O  C3 H 2 O42      K a = 2.0x10 =  HC3 H 2 O4     –6

Since K a >> K a , assume that almost all of the H3O+ comes from the first dissociation. 2

H2O(l)  H3O+(aq) + HC3H2O4–(aq) x x    H 3O   HC3 H 2 O 4  –3    K a = 1.4x10 = H C H O  2 3 2 4 

H2C3H2O4(aq) 0.200 – x

K a1 = 1.4x10

–3

 x  x 

 0.200  x  –3  x  x  K a = 1.4x10 =  0.200 

Assume x is small compared to 0.200.

x = 0.016733 Check assumption: (0.016733/0.200) x 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 = 2.8x10–4 – 1.4x10–3 x x2 + 1.4x10–3 x – 2.8x10–4 = 0 a=1 b = 1.4x10–3 c = – 2.8x10–4 2 x = b  b  4ac

1.4x103  x=

1.4x10   4 1  2.8x10  3 2

4

2 1

–2

x = 1.6048x10 [H+] = [HC3H2O4–] = x = 1.6x10–2 mol/L pH = – log [H+] = – log (1.6048x10–2) = 1.79458 = 1.79 [OH–] = Kw/[H+] = (1.0x10–14)/(1.6048x10–2) = 6.23131x10–13 = 6.2x10–13 mol/L pOH = – log [OH–] = – log (6.23131x10–13) = 12.2054 = 12.21 [H2C3H2O4] = (0.200 – 1.6048x10–2) mol/L = 0.183952 = 0.18 mol/L 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 K a expression and assuming that [HC3H2O4–] and [H3O+] come mostly from 2

the first dissociation. This new calculation will have a new x value. HC3H2O4–(aq) + H2O(l)  H3O+(aq) + C3H2O42–(aq) –2 –2 1.6048x10 – x 1.6048x10 + x x Silberberg, Amateis, Venkateswaran, Chen, Chemistry: The Molecular Nature of Matter and Change, 3rd Canadian Edition Page 3-563 Instructor‘s Solution Manual © Copyright 2021 McGraw-Hill Ryerson Ltd.

 H3O  C3 H 2 O42      K a = 2.0x10 =   HC3 H 2 O4    –6

1.6048x10  x   x  K = 2.0x10 = 1.6048x10  x  1.6048x10   x  K = 2.0x10 = 1.6048x10  2

–6

2

Assume x is small compared to 1.6048x10–2.

2

–6

2

x = [C3H2O42–] = 2.0x10–6 = 2.0x10–6 mol/L 16.100 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 3 O   C9 H 7 O 4    Ka = 3.6x10–4 =  HC9 H 7 O 4 

x x 

Ka = 3.6x10–4 =

0.018  x  –4 Ka = 3.6x10 =  x  x   0.018 

Assume x is small compared to 0.018.

[H+] = x = 2.54558x10–3 Check assumption: (2.54558x10–3/0.018) x 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 = (3.6x10–4)(0.018 – x) = 6.48x10–6 – 3.6x10–4 x x2 + 3.6x10–4 x – 6.48x10–6 = 0 a=1 b = 3.6x10–4 c = – 6.48x10–6 2 x = b  b  4ac

3.6 x104  x=

3.6x10   4 1  6.48 x10  4 2

6

2 1

x = 2.37194x10–3 mol/L H+ pH = –log [H+] = –log (2.37194x10–3) = 2.624896 = 2.62 16.101 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 (mol/L) HCOOH(aq) + H2O(l) H3O+(aq) + HCOO–(aq) Initial 0.75 0 0 Change –x +x +x Equilibrium 0.75 – x x x    H 3O   HCOO    Ka = 1.8x10–4 =   HCOOH 

 x  x   0.75  x  Ka = 1.8x10–4 =  x  x   0.75  Ka = 1.8x10–4 =

Assume x is small compared to 0.75.

x = 1.161895x10–2 Percent HCOOH dissociated =

 HCOOH dissociated 100%   HCOOH initial

Percent HCOOH dissociated =

1.161895x102 mol/L 100%  = 1.54919 %= 1.5% 0.75 mol/L

16.102 Plan: First, calculate the initial concentration (mol/L) 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 x Kb, using the Ka for HClO from Appendix C. Solution: Concentration (mol/L) of ClO– =  1 mL solution   1.0 g solution  6.5% NaClO  1 mol NaClO   1 mol ClO    3        10 L solution   1 mL solution  100% Solution  74.44 g NaClO   1 mol NaClO  = 0.873186 mol/L ClO– The sodium ion is from a strong base; therefore, it will not affect the pH, and can be ignored. Concentration (mol/L) 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 1.0x10 Kb of ClO– = w = = 3.448275862x10–7 Ka 2.9x108 –7

Kb = 3.448275862x10 =

 HClO OH  ClO   

 x  x  0.873186  x  Kb = 3.448275862x10–7 =  x  x   0.873186  Kb = 3.448275862x10–7 =

Assume x is small compared to 0.873186.

x = 5.4872x10–4 = 5.5x10–4 mol/L OH– Check assumption that x is small compared to 0.873186: 5.4872x104 100%  = 0.006% error, so the assumption is valid. 0.873186

[H]+ =

1.0x1014 = 1.82242x10–11 mol/L H+ 5.4872x104

OH     pH = –log [H+] = –log (1.82242x10–11) = 10.73935 = 10.74

16.103 The cation, HC18H21NO3+, (formed when codeine reacts with HCl and abstracts the proton) 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 (mol/L) HC18H21NO3+(aq) + H2O(l) C18H21NO3(aq) + H3O+(aq) Initial 0.050 — 0 0 Change –x +x +x Silberberg, Amateis, Venkateswaran, Chen, Chemistry: The Molecular Nature of Matter and Change, 3rd Canadian Edition Page 3-565 Instructor‘s Solution Manual © Copyright 2021 McGraw-Hill Ryerson Ltd.

Equilibrium 0.050 – x Kb = 10–pKb = 10–5.80 = 1.58489x10–6 Ka of HC18H21NO3+ = Kw/Kb = (1.0x10–14)/(1.58489x10–6) = 6.309586x10–9

 H3O  C18 H 21 NO3    Ka = 6.309586x10 =  HC18 H 21 NO3    –9

Ka = 6.309586x10–9 =

 x  x 

 0.050  x  Ka = 6.309586x10–9 =  x  x   0.050 

Assume x is small compared to 0.050.

[H+] = x = 1.7761737x10–5 mol/L Check assumption: (1.7761737x10–5/0.050) x 100% = 0.04% error, so the assumption is valid. pH = –log [H+] = –log (1.7761737x10–5) = 4.75051 = 4.75 16.104 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+. 16.105 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. 16.106 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. 16.107 The two factors that explain the greater acid strength of HClO 4 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 HClO 4 more susceptible to transfer to a base. 16.108 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. 16.109 a) H2Se

b) H2SO4

c) H2SO3

16.110 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. 16.111 a) HBr

b) H3AsO4

c) HNO2

16.112 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 = 3x10–8. Aluminum bromide, AlBr3, contains Al3+ ion with Ka = 1x10–5. The concentrations of Cu2+ and Al3+ are equal, but the Ka of AlBr3 is almost three orders of magnitude greater. Therefore, 0.5 mol/L AlBr3 is the stronger acid and would have the lower pH. b) Zinc chloride, ZnCl2, contains the Zn2+ ion with Ka = 1x10–9. Tin(II) chloride, SnCl2, contains the Sn2+ ion with Ka = 4x10–4. Since both solutions have the same concentration, and Ka (Sn2+) > Ka (Zn2+), 0.3 mol/L SnCl2 is the stronger acid and would have the lower pH. 16.113 a) FeCl3

b) BeCl2

16.114 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+ (1x10–10) is smaller than the Ka of Co2+ (2x10–10). 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+ (1x10–5) is smaller than the Ka of Cr3+ (1x10–4). 16.115 a) NaCl

b) Co(NO3)2

16.116 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.) 16.117 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+]. 16.118 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 1.8x10–5, and the Kb for the conjugate base (NH3) is 1.76x10–5. 16.119 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: H2 O a) KBr(s)   K+(aq) + Br–(aq) + K has no acid/base nature, so it does not influence pH. Br– has no acid/base nature, 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. H2 O b) NH4I(s)   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 has no acid/base nature , so it will not influence the pH. The production of H3O+ from the ammonium ion makes the solution of NH4I acidic. Silberberg, Amateis, Venkateswaran, Chen, Chemistry: The Molecular Nature of Matter and Change, 3rd Canadian Edition Page 3-567 Instructor‘s Solution Manual © Copyright 2021 McGraw-Hill Ryerson Ltd.

+ – H2 O c) KCN(s)   K (aq) + CN (aq) + K has no acid/base nature, 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: CN–(aq) + H2O(l) HCN(aq) + OH–(aq) Hydroxide ions are produced in this equilibrium so solution will be basic.

16.120 a) Cr(NO3)3(s) + nH2O(l)  Cr(H2O)n3+(aq) + 3NO3–(aq) Cr(H2O)n3+(aq) + H2O(l) Cr(H2O)n–1OH2+(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+) = 1x10–9 Kb (CH3COO–) = Kw/Ka = (1.0x10–14)/(1.8x10–5) = 5.5556x10–10 The two K values are similar, so the solution is close to neutral or slightly acidic. 16.121 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, Na 2CO3, are sodium ion, Na+, and carbonate ion, CO32–. + 2– H2 O Na2CO3(s)   2Na (aq) + CO3 (aq) Sodium ion is derived from the strong base NaOH. Carbonate ion is derived from the weak acid HCO 3–. 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+ – H2 O CaCl2(s)   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+ – H2 O Cu(NO3)2(s)   Cu (aq) + 2NO3 (aq) Small metal ions are acidic in water (assume the hydration of Cu 2+ 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. 16.122 a) CH3NH3Cl(s) + H2O(l)  CH3NH3+(aq) + Cl–(aq) CH3NH3+(aq) + H2O(l) H3O+(aq) + CH3NH2(aq) acidic b) KClO4(s) + H2O(l)  K+(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+ ) = 2x10– 10 Kb (F–) = Kw/Ka = (1.0x10– 14)/(6.8x10– 4) = 1.47x10– 11 The two K values are similar so the solution is close to neutral or slightly acidic. 16.123 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+ and Br– have no acid/base nature, so neither changes the pH of the solution. 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+ has no acid/base nature and does not influence solution pH. The basedissociation 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). 16.124 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+) = 6x10–3 Kb (HCOO–) = Kw/Ka = (1.0x10–14)(1.8x10–4) = 5.5556x10–-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

16.125 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, (NH 4)3PO4, are the ammonium ion, NH4+, and the phosphate ion, PO43–. NH4+(aq) + H2O(l) NH3(aq) + H3O+(aq) Ka = Kw/Kb (NH3) = 5.7x10–10 PO43–(aq) + H2O(l) HPO42–(aq) + OH–(aq) Kb = Kw/Ka3 (H3PO4) = 2.4x10–2 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, HSO 4– . 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. 16.126 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+) = 3x10–8 CH3COO–(aq) + H2O(l) CH3COOH(aq) + OH–(aq) Kb (CH3COO–) = Kw/Ka = (1.0x10–14)/(1.8x10–5) = 5.5556x10–10 Silberberg, Amateis, Venkateswaran, Chen, Chemistry: The Molecular Nature of Matter and Change, 3rd Canadian Edition Page 3-569 Instructor‘s Solution Manual © Copyright 2021 McGraw-Hill Ryerson Ltd.

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) NO2–(aq) + H2O(l) OH–(aq) + HNO2(aq) Ka (Cr3+) = 1x10–4 Kb (NO2–) = Kw/Ka = (1.0x10–14)/(7.1x10–4) = 1.40845x10–11 Ka (Cr3+) > Kb (NO2–) acidic c) CsI(s) + H2O(l)  Cs+(aq) + I–(aq) neutral 16.127 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 Kbvalues 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 has no acid/base nature , so it does not influence pH. Potassium nitrate, KNO3, is a neutral solution because potassium ion and nitrate ion have no acid/base nature , 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 6.5x10–8, so Kb for SO3– is 1.5x10–7, which indicates that sulfite ion is a weak base. Ka for HS– is 1x10–17 (see the table of Ka values for polyprotic acids), so sulfide ion has a Kb equal to 1x103. 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 = 1.0x10–14/1.8x10–5 = 5.6x10–10 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, HSO 4– is the conjugate base of a strong acid, so it will not influence pH. As an acid, HSO 4– is the conjugate acid of a weak base, so the acid dissociation applies: HSO4–(aq) + H2O(l)  SO42–(aq) + H3O+(aq) Ka2 = 1.2x10–2 In solutions of sodium hydrogen carbonate, only the HCO 3– will influence pH and it, like HSO4–, is amphoteric: As an acid: HCO3–(aq) + H2O(l) CO32–(aq) + H3O+(aq) Ka = 4.7x10–11, the second Ka for carbonic acid As a base: HCO3–(aq) + H2O(l) H2CO3(aq) + OH–(aq) Kb = 1.0x10–14/4.5x10–7 = 2.2x10–8 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 = 1.0x10–14/4.7x10–11 = 2.1x10–4 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, since their concentrations are the same. 16.128 a) KClO2 > MgCl2 > FeCl2 > FeCl3 b) NaBrO2 > NaClO2 > NaBr > NH4Br Silberberg, Amateis, Venkateswaran, Chen, Chemistry: The Molecular Nature of Matter and Change, 3rd Canadian Edition Page 3-570 Instructor‘s Solution Manual © Copyright 2021 McGraw-Hill Ryerson Ltd.

16.129 (a)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. (b) CH3O–(aq) + H2O(l)  OH–(aq) + CH3OH(aq) NH2–(aq) + H2O(l)  OH–(aq) + NH3(aq) 16.130 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 H 2O (such as CH3COOH). 16.131 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. 16.132 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. 16.133 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. 16.134 a) No, a weak Brønsted-Lowry base is not necessarily a weak Lewis base. For example, water molecules solvate metal ions very well: Ni2+(aq) + 4H2O(l) Ni(H2O)42+(aq) Water is a very weak Brønsted-Lowry base, but forms the Zn complex fairly well and is a reasonably strong Lewis base. 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 K > 1, the reaction proceeds in the direction written (left to right) and is driven by the stronger Lewis base, the cyanide ion. 16.135 a) All three concepts can have water as the product in an acid-base neutralization reaction. b) It is the only product in an Arrhenius neutralization reaction. 16.136 a) NH3 can only act as a Brønsted-Lowry or Lewis base. b) AlCl3 can only act as a Lewis acid. 16.137 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. 16.138 a) Lewis acid

b) Lewis base

c) Lewis base

d) Lewis acid

16.139 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. 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. 16.140 a) Lewis acid

b) Lewis base

c) Lewis acid

d) Lewis acid

16.141 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 BF 4–. The fluoride ion is the Lewis base and the boron trifluoride is the Lewis acid. F– + BF3  BF4– Lewis base Lewis acid adduct 16.142 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

16.143 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. 16.144 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

16.145 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. 16.146 Plan: Calculate the [H+] using the pH values given. Determine the value of Kw from the pKw given. The is combined with the Kw value at 37°C to find [OH–] using Kw = [H+][OH–]. Solution: Kw = 10–pKw = 10–13.63 = 2.34423x10–14 Kw = [H+][OH–] = 2.34423x10–14 at 37°C + [H ] range High value (low pH) = 10–pH = 10–7.35 = 4.46684x10–8 = 4.5x10–8 mol/L H+ Low value (high pH) = 10–pH = 10–7.45 = 3.54813x10–8 = 3.5x10–8 mol/L H+ Range: 3.5x10–8 to 4.5x10–8 mol/L H+ – [OH ] range Kw = [H3O+][OH–] = 2.34423x10–14 at 37°C [OH–] =

[H+]

Kw [H3O]+

High value (high pH) =

2.34423x1014 = 6.60695x10–7 = 6.6x10–7 mol/L OH– 8 3.54813x10

Low value (low pH) =

2.34423x1014 = 5.24807x10–7 = 5.2x10–7 mol/L OH– 4.46684x108

Range: 5.2x10–7 to 6.6x10–7 mol/L OH–

16.147 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. The equilibrium constant for this reaction is the autoionization constant of methanol: K = [CH3O–][CH3OH2+]

16.148 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) Concentration (mol/L) of CO2 = kH Pcarbon dioxide = (0.033 mol/L•bar)(3.2x10–4 bar) = 1.056x10–5 mol/L CO2 CO2(g) + 2H2O(l) HCO3–(aq) + H3O+(aq) Koverall = 4.5x10–7  H3O    HCO3  –7   Koverall = 4.5x10 =  CO2  Koverall = 4.5x10–7 = Koverall = 4.5x10–7 =

 x x 

1.056 x10   x  x 

5

Assume x is small compared to 1.056x10–5.

 x 

1.056 x105   

x = 2.17991x10–6 Check assumption that x is small compared to 1.056x10 –5: 2.17991x106 100%  = 21% error, so the assumption is not valid. 1.056x105 Since the error is greater than 5%, it is not acceptable to assume x is small compared to 1.056x10 –5, and it is necessary to use the quadratic equation. x2 = (4.5x10–7)(1.056x10–5 – x) = 4.752x10–12 – 4.5x10–7x x2 + 4.5x10–7 x – 4.752x10–12 = 0 a=1 b = 4.5x10–7 c = – 4.752x10–12 2 x = b  b  4ac

 4.5x10   4 1  4.752x10  7 2

4.5x107  x=

12

2 1

–6

x = 1.966489x10 mol/L H pH = –log [H+] = –log (1.966489x10–6) = 5.7063 = 5.71 c) HCO3–(aq) + H2O(l) H3O+(aq) + CO32–(aq) Ka = 4.7x10

–11

 H3O   CO32      =   HCO3   

Use the unrounded x from part b). Koverall = 4.5x10

Koverall = 4.5x10

–11

1.966489x106  x   x    = 1.966489x106  x   

–11

1.966489x106   x    = 1.966489x106   

Assume x is small compared to 2x10–6.

[CO32–] = 4.5x10–11 mol/L CO32– d) New concentration (mol/L) of CO2 = 2kH Pcarbon dioxide = 2(0.033 mol/L•bar)(3.2x10 –4 bar) = 2.112x10–5 mol/L CO2  H3O    HCO3  –7   Koverall = 4.5x10 =  CO2  Koverall = 4.5x10–7 =

 x  x 

 2.112x105  x   

Assume x is small compared to 2.112x10–5.

Koverall = 4.5x10–7 =

 x  x   2.112 x105   

x = 3.08286x10–6 Check assumption that x is small compared to 2.112x10 –5: 3.08286x106 100%  = 15% error, so the assumption is not valid. 2.112x105 Since the error is greater than 5%, it is not acceptable to assume x is small compared to 2.112x10 –5, and it is necessary to use the quadratic equation. x2 =(4.5x10–7)(2.112x10–5 – x) = 9.504x10–12 – 4.5x10–7 x x2 + 4.5x10–7 x – 9.504x10–12 = 0 a=1 b = 4.5x10–7 c = – 9.504x10–12 2 x =  b  b  4ac

4.5x107  x=

 4.5x10   4 1  9.504 x10  7 2

12

2 1

–6

x = 2.866x10 mol/L H pH = –log [H+] = –log (2.866x10–6) = 5.5427 = 5.54 16.149 At great depths, the higher pressure increases the concentration of H+ and the effect is to shift the dissolving reaction to the right, so seashells dissolve more rapidly. 16.150 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 (CH 3)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. 16.151 Plan: A 10-fold dilution means that the chemist takes 1 mL of the 1.0x10 –5 mol/L 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. c1V1 = c2V2 can be used to find the concentration (mol/L) after each dilution. After each dilution, find [H+] and calculate the pH. Solution: Hydrochloric acid is a strong acid that completely dissociates in water. Therefore, the concentration of H + is the same as the starting acid concentration: [H +] = [HCl]. The original solution pH: pH = –log (1.0x10–5) = 5.00 = pH Dilution 1: c1V1 = c2V2 (1.0x10–5 mol/L)(1.0 mL) = (x)(10. mL) [H+]HCl = 1.0x10–6 mol/L H+ pH = –log (1.0x10–6) = 6.00 Dilution 2: (1.0x10–6 mol/L)(1.0 mL) = (x)(10. mL) [H+]HCl = 1.0x10–7 mol/L H+ Once the concentration of strong acid is close to the concentration of H 3O+ from water autoionization, the [H3O+] in the solution does not equal the initial concentration of the strong acid. The calculation of [H 3O+] must be based on the water ionization equilibrium: H2O(l) + H2O(l) H3O+(aq) + OH–(aq) with Kw = 1.0x10–14 at 25°C. The dilution gives an initial [H3O+] of 1.0x10–7 mol/L. Assuming that the initial concentration of hydroxide ions is zero, a reaction table is set up. Concentration (mol/L) 2H2O(l)  H3O+(aq) + OH–(aq) –7 Initial — 1x10 0 Silberberg, Amateis, Venkateswaran, Chen, Chemistry: The Molecular Nature of Matter and Change, 3rd Canadian Edition Page 3-575 Instructor‘s Solution Manual © Copyright 2021 McGraw-Hill Ryerson Ltd.

Change — +x +x Equilibrium — 1x10–7 + x x Kw = [H+][OH–] = (1x10–7 + x)(x) = 1.0x10–14 Set up as a quadratic equation: x2 + 1.0x10–7 x – 1.0x10–14 = 0 a=1 b = 1.0x10–7 c = –1.0x10–14

1.0x107  x=

1.0x10   4 1  1.0x10  7 2

14

2 1

–8

x = 6.18034x10 [H+] = (1.0x10–7 + x) mol/L = (1.0x10–7 + 6.18034x10–8) mol/L = 1.618034x10–7 mol/L H+ pH = –log [H+] = –log (1.618034x10–7) = 6.79101 = 6.79 Dilution 3: (1.0x10–7 mol/L)(1.0 mL) = (x)(10. mL) [H+]HCl = 1.0x10–8 mol/L H+ The dilution gives an initial [H+] of 1.0x10–8 mol/L. Assuming that the initial concentration of hydroxide ions is zero, a reaction table is set up. Concentration (mol/L) 2H2O(l)  H3O+(aq) + OH–(aq) –8 Initial — 1x10 0 Change +x +x Equilibrium — 1x10–8 + x x Kw = [H+][OH–] = (1x10–8 + x)(x) = 1.0x10–14 Set up as a quadratic equation: x2 + 1.0x10–8 x – 1.0x10–14 = 0 a=1 b = 1.0x10–8 c = –1.0x10–14

1.0x108  x=

1.0x10   4 1  1.0x10  8 2

14

2 1

–8

x = 9.51249x10 [H+] = (1.0x10–8 + x) mol/L = (1.0x10–8 + 9.51249x10–8) mol/L = 1.051249x10–7 mol/L H+ pH = –log [H+] = –log (1.051249x10–7) = 6.97829 = 6.98 Dilution 4: (1.0x10–8 mol/L)(1.0 mL) = (x)(10. mL) [H+]HCl = 1.0x10–9 mol/L H+ The dilution gives an initial [H+] of 1.0x10–9 mol/L. Assuming that the initial concentration of hydroxide ions is zero, a reaction table is set up. Concentration (mol/L) 2H2O(l)  H3O+(aq) + OH–(aq) –9 Initial — 1x10 0 Change — +x +x Equilibrium — 1x10–9 + x x + – –9 –14 Kw = [H ][OH ] = (1x10 + x)(x) = 1.0x10 Set up as a quadratic equation: x2 + 1.0x10–9 x – 1.0x10–14 = 0 a=1 b = 1.0x10–9 c = –1.0x10–14

1.0x109  x=

1.0x10   4 1  1.0x10  9 2

14

2 1

–8

x = 9.95012x10 [H+] = (1.0x10–9 + x) mol/L = (1.0x10–9 + 9.95012x10–8) mol/L = 1.005012x10–7 mol/L H+ pH = –log [H+] = –log (1.005012x10–7) = 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. 16.152 a) Steps 1, 2, and 4 are Lewis acid-base reactions. b) Step 1 Cl2 + FeCl3 FeCl5 (or Cl+FeCl4–) Silberberg, Amateis, Venkateswaran, Chen, Chemistry: The Molecular Nature of Matter and Change, 3rd Canadian Edition Page 3-576 Instructor‘s Solution Manual © Copyright 2021 McGraw-Hill Ryerson Ltd.

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– 16.153 a) HY(aq) + H2O(l) H3O+(aq) + Y–(aq)  H3O    Y     Ka =   HY  Concentrations in Beaker A: [HY] =  8 particles   0.010 mol  

1  = 0.266667 mol/L  1 particle 0.300 L   

[H3O+] = [Y–] =  4 particles   0.010 mol  

1  = 0.133333 mol/L   1 particle   0.300 L 

 H3O    Y      =  0.133333 0.133333 = 0.066667 = 0.067 Ka =  HY   0.266667

Calculate the concentrations in Beakers B-D, then calculate Q to determine which are at equilibrium. Concentrations in Beaker B: [HY] =  6 particles   0.010 mol  

1  = 0.20 mol/L  1 particle 0.300 L   

[H3O+] = [Y–] =  2 particles   0.010 mol  

1  = 0.066667 mol/L  1 particle 0.300 L   

 H3O    Y      =  0.066667  0.066667  = 0.0222222 = 0.022 Q=   HY   0.20

Beaker B is not at equilibrium. Concentrations in Beaker C: [HY] = 4 particles  0.010 mol  

1  = 0.133333 mol/L  1 particle 0.300 L   

[H3O+] = [Y–] =  2 particles   0.010 mol  

1  = 0.066667 mol/L  1 particle 0.300 L   

 H3O    Y      = 0.066667 0.066667  = 0.0333345 = 0.033 Q=   HY  0.13333

Beaker C is not at equilibrium. Concentrations in Beaker D: [HY] =  2 particles   0.010 mol  

1  = 0.066667 mol/L  1 particle 0.300 L   

[H3O+] = [Y–] =  2 particles   0.010 mol  

1  = 0.066667 mol/L 1 particle 0.300 L   

 H3O   Y      = 0.066667 0.066667  = 0.066667 = 0.067 Q=  HY  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. Silberberg, Amateis, Venkateswaran, Chen, Chemistry: The Molecular Nature of Matter and Change, 3rd Canadian Edition Page 3-577 Instructor‘s Solution Manual © Copyright 2021 McGraw-Hill Ryerson Ltd.

16.154 a) Electrical conductivity of 0.1 mol/L HCl is higher than that of 0.1 mol/L 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 mol/L HCl than in 0.1 mol/L 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. +

16.155 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. 16.156 a) pH = –log [H+] HCl is a strong acid so [H+] = mol/L 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 3O   ClO     Ka = 2.9x10–8 =  HClO Ka = 2.9x10–8 =

x x 

0.10  x  –8 Ka = 2.9x10 = x x  0.10 

Assume x is small compared to 0.10.

[H+] = x = 5.38516x10–5 mol/L Check assumption: (5.38516x10–5/0.10) x 100% = 0.05%. The assumption is good. pH = –log [H+] = –log (5.38516x10–5) = 4.2688 = 4.27 HClO2(aq) + H2O(l) H3O+(aq) + ClO2–(aq) 0.10 – x x x   H 3O  ClO 2     Ka = 1.1x10–2 =  HClO2  Ka = 1.1x10–2 =

x x 

0.10  x  –2 Ka = 1.1x10 = x x  0.10 

Assume x is small compared to 0.10.

x = 0.033166 Check assumption: (0.033166/0.10) x 100% = 33%. The assumption is not valid. The problem will need to be solved as a quadratic. x2 = (1.1x10–2)(0.10 – x) = 1.1x10–3 – 1.1x10–2 x x2 + 1.1x10–2 x – 1.1x10–3 = 0 a=1 b = 1.1x10-2 c = – 1.1x10-3

1.1x102  x=

1.1x10   4 1  1.1x10  2 2

3

2 1

x = 2.8119x10–2 mol/L H+ pH = –log [H+] = –log (2.8119x10–2) = 1.550997 = 1.55 + b) The lowest H concentration is from the HClO. Leave the HClO beaker alone, and dilute the other acids until they yield the same H+ concentration. A dilution calculation is needed to calculate the amount of water added. HCl ci = 0.10 mol/L Vi = 100. mL cf = 5.38516x10–5 mol/L Vf = ? ciVi = cfVf Silberberg, Amateis, Venkateswaran, Chen, Chemistry: The Molecular Nature of Matter and Change, 3rd Canadian Edition Page 3-578 Instructor‘s Solution Manual © Copyright 2021 McGraw-Hill Ryerson Ltd.

Vf = ciVi/cf = [(0.10 mol/L)(100. mL)]/(5.38516x10–5 mol/L) = 1.85695x105 mL Volume water added = (1.85695x105 mL) – 100. mL = 1.85595x105 = 1.9x105 mL H2O added HClO2 requires the Ka for the acid with the ClO2– concentration equal to the H3O+ concentration. The final concentration (mol/L) of the acid will be cf, which may be used in the dilution equation. HClO2(aq) + H2O(l)  H3O+(aq) + ClO2–(aq) –5 –5 x – 5.38516x10 5.38516x10 5.38516x10–5  H 3O   ClO 2   –2   Ka = 1.1x10 =  HClO2 

5.38516x105  5.38516x105     Ka = 1.1x10 =  x  5.38516x105    –2

x = Mf = 5.41152x10–5 mol/L

ci = 0.10 mol/L Vi = 100. mL cf = 5.41152x10–5 mol/L Vf = ? ciVi = cfVf Vf = ciVi/cf = [(0.10 mol/L)(100. mL)]/(5.41152x10–5 mol/L) = 1.8479x105 mL Volume water added = (1.8479x105 mL) – 100. mL = 1.8469x105 mL= 1.8x105 mL H2O added 16.157 Plan: Determine the hydrogen ion concentration from the pH. The concentration (mol/L) and the volume will give the number of moles, and with the aid of Avogadro‘s number, the number of ions may be found. Solution: c H+ = 10–pH = 10–6.2 = 6.30957x10–7 mol/L

 6.30957x107 mol H 3O   103 L   1250. mL   7 d   6.022x1023 H 3O  18 18 +     = 3.32467x10 = 3x10 H     1 wk    L 1 mL d 1 mol H O     3      The pH has only one significant figure, and limits the significant figures in the final answer. 16.158 a) 2 NH3(l)  NH4+(am) + NH2–(am) In this equilibrium ―(am)‖ indicated ammoniated, solvated by ammonia, instead of ―(aq)‖ to indicate aqueous, solvated by water. Initially, based on the equilibrium: K =

 NH 4    NH 2      2

NH3 

Since NH3 is a liquid and a solvent, its activity is 1, so: Kam = [NH4+][NH2–] b) Strongest acid = NH4+ Strongest base = NH2– + – + c) NH2  NH4 : basic NH4  NH2–: acidic + 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–] = 5.1x10–27 [NH4+] = [NH2–] = x Kam = [x][x] = 5.1x10–27 x = 7.1414x10–14 = 7.1x10–14 mol/L NH4+ e) 2H2SO4(l) H3SO4+(sa) + HSO4–(sa) (sa) = solvated by sulfuric acid (sulf) Ksulf = [H3SO4+][HSO4–] = 2.7x10–4 [H3SO4+] = [HSO4–] = x Ksulf = [x][x] = 2.7x10–4 x = 1.643x10–2 = 1.6x10–2 mol/L HSO4– 16.159 c is the unknown concentration (mol/L) of the thiamine.

HC12H17ON4SCl2(aq) + H2O(l) H3O+(aq) + C12H17ON4SCl2–(aq) c–x x x pH = 3.50 [H3O+] = 10–3.50 = 3.1623x10–4 mol/L = x  H3O    C12 H17 ON 4SCl2   –7   Ka = 3.37x10 =  C12 H18 ON 4SCl2  Ka = 3.37x10–7 =  x  x 

c  x 

3.1623x10 3.1623x10  K = 3.37x10 =  c  3.1623x10  4

4

–7

4

c = 0.29705633 mol/L 3  337.27 g thiamine HCl   0.29705633 mol thiamine HCl   10 L  Mass (g) =    10.00 mL    L 1 mL    1 mol thiamine HCl   = 1.00188 g= 1.0 g thiamine hydrochloride

16.160 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 = 10pKb = 10–5.91 = 1.23027x10–6 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 = 1.23027x10–6 =  TRIS

x x  Assume x is small compared to 0.075. 0.075  x  Kb = 1.23027x10–6 = x x  0.075 Kb = 1.23027x10–6 =

x = [OH–] = 3.03760x10–4 mol/L OH– Check assumption that x is small compared to 0.075: 3.03760x104 100%  = 0.40% error, so the assumption is valid. 0.075 Kw 1.0x1014 [H]+ = = = 3.292073x10–11 mol/L OH   3.03760x104   pH = –log [H+] = –log (3.292073x10–11) = 10.4825 = 10.48 16.161 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 16.162 The pH is dependent on the molar concentration of H3O+. Convert % w/v to concentration (mol/L), and use the Ka of acetic acid to determine [H3O+] from the equilibrium expression. Convert % w/v to concentration (mol/L)using the molecular weight of acetic acid (CH3COOH):  5.0 g CH3COOH   1 mol CH3COOH   1 mL  Concentration (mol/L) =    3  = 0.832639 mol/L CH3COOH   100 mL solution  60.05 g CH3COOH   10 L  Silberberg, Amateis, Venkateswaran, Chen, Chemistry: The Molecular Nature of Matter and Change, 3rd Canadian Edition Page 3-580 Instructor‘s Solution Manual © Copyright 2021 McGraw-Hill Ryerson Ltd.

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 3O    CH 3COO     Ka = 1.8x10–5 =  CH 3COOH 

x x  0.832639  x  Ka = 1.8x10–5 = x x  0.832639 Ka = 1.8x10–5 =

Assume x is small compared to 0.832639.

x = 3.8714x10–3 mol/L = [H+] Check assumption: [3.871x10–3/0.832639] x 100% = 0.46%, therefore the assumption is good. pH = –log [H+] = –log (3.8714x10–3) = 2.412132 = 2.41 16.163 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. 16.164 Plan: Assuming that the pH in the specific cellular environment is equal to the optimum pH for the enzyme, the hydronium ion concentrations are [H+] = 10–pH. Solution: Salivary amylase, mouth: [H+] = 10–6.8 = 1.58489x10–7 = 2x10–7 mol/L Pepsin, stomach: [H+] = 10–2.0 = 1x10–2 mol/L Trypsin, pancreas: [H+] = 10–9.5 = 3.1623x10–10 = 3x10–10 mol/L 16.165 a) CH3COOH(aq) + H2O(l) H3O+(aq) + CH3COO–(aq) 0.240 – x x x    H 3O   CH 3COO    Ka = 1.8x10–5 =  CH 3COOH  Ka = 1.8x10–5 =

x x 

0.240  x  –5 Ka = 1.8x10 = x x  0.240 

Assume x is small compared to 0.240.

x = 2.07846x10–3 Check assumption: [2.07846x10–3/0.240] x 100% = 0.9%, therefore the assumption is good. x = [H+] = 2.07846x10–3 = 2.1x10–3 mol/L H+ [OH–] = Kw/[H+] = (1.0x10–14)/(2.07846x10–3) = 4.81125x10–12 = 4.8x10–12 mol/L OH– pH = –log [H+] = –log (2.07846x10–3) = 2.682258 = 2.68 pOH = –log [OH–] = –log (4.81125x10–12) = 11.317742 = 11.32 b) NH3(aq) + H2O(l) NH4+(aq) + OH–(aq) 0.240 – x x x  NH 4    OH     Kb = 1.8x10–5 =   NH 3 

x x  0.240  x  Kb = 1.8x10–5 = x x  0.240  Kb = 1.8x10–5 =

Assume x is small compared to 0.240.

x = 2.07846x10–3

Check assumption: [2.07846x10–3 0.240] x 100% = 0.9%, therefore the assumption is good. x = [OH–] = 2.07846x10–3 = 2.1x10–3 mol/L OH– [H+] = Kw/[OH–] = (1.0x10–14)/(2.07846x10–3) = 4.81125x10–12 = 4.8x10–12 mol/L H+ pOH = –log [OH–] = –log (2.07846x10–3) = 2.682258 = 2.68 pH = –log [H+] = –log (4.81125x10–12) = 11.317742 = 11.32

 33 g Na 3 PO4   1 mol Na 3PO4   1 mol PO34  16.166 Concentration = c =   = 0.20129 mol/L PO43–    1 L 163.94 g Na PO 1 mol Na PO   3 4  3 4  PO43–(aq) + H2O(l) OH–(aq) + HPO42–(aq) (Use Ka3 for H3PO4) Kb = Kw/Ka = (1.0x10–14)/(4.2x10–13) = 0.0238095

 HPO  OH  K = 0.0238095 =  PO  2



3

Kb = 0.0238095 =

 x  x 

 0.20129  x 

A quadratic is required. x2 + 0.0238095x – 0.00479261 = 0 a=1 b = 0.0238095 c = – 0.00479261 2 x = b  b  4ac

0.0238095 

 0.02380952  4 1 0.00479261 2 1

x = 0.058340 = 0.058 mol/L OH– [H]+ = Kw/[OH–] = (1.0x10–14)/(0.058340) = 1.7140898x10–13 mol/L H+ pH = –log [H+] = –log (1.7140898x10–13) = 12.765966 = 12.77 16.167 a) PH3BCl3(s)  PH3(g) + BCl3(g) x = [PH3] = [BCl3] Kc = [PH3][BCl3] = (x)(x) = x2 = [8.4x10–3/3.0 L]2 = 7.84x10–6 = 7.8x10–6 b)

16.168

The freezing point depression equation is required to determine the molality of the solution. T = [0.00 – (–1.93°C)] = 1.93°C = iKf m Temporarily assume i = 1. T 1.93 C m= = = 1.037634 m = 1.037634 mol/L iK f (1)(1.86 C/m) This molality is the total molality of all species in the solution, and is equal to their concentration (mol/L). From the equilibrium: ClCH2COOH(aq) + H2O(l) H3O+(aq) + ClCH2COO–(aq) Initial 1.000 mol/L x x Change –x +x +x Equilibrium 1.000 – x x x The total concentration of all species is:

[ClCH2COOH] + [H3O+] + [ClCH2COO–] = 1.037634 mol/L [1.000 – x] + [x] + [x] = 1.000 + x = 1.037634 mol/L x = 0.037634 mol/L  H3O    CH 3COO     Ka =  CH 3COOH  Ka =  0.037634  0.037634  = 0.0014717 = 0.00147

1.000  0.037634 

16.169 concentration ( mol/L) =

 0.42 g C17 H35 COONa   1 mL   1 mol C17 H35 COONa   1 mol C17 H35 COO       3   10.0 mL    10 L   306.45 g C17 H35 COONa   1 mol C17 H35 COONa  = 0.137053 mol/L C17H35COO– Kb = Kw/Ka = (1.0x10–14)/(1.3x10–5) = 7.69231x10–10 C17H35COO–(aq) + H2O(l) C17H35COOH(aq) + OH–(aq) 0.137053 – x x x  C17 H35 COOH  OH  Kb = 7.69231x10–10 = C17 H35 COO    Kb = 7.69231x10–10 =

x x 

0.137053  x  –10 Kb = 7.69231x10 = x x  0.137053

Assume x is small compared to 0.137053.

x = 1.026769x10–5 = [OH–] Check assumption: [1.026768x10–5/0.137053] x 100% = 0.007%, therefore the assumption is good. [H]+ = Kw/[OH–] = (1.0x10–14)/(1.026768x10–5) = 9.739298x10–10 mol/L H3O+ pH = – log [H+] = – log (9.739298x10–10) = 9.01147 = 9.01 16.170 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, CH 3CH2COOH). 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 concentration (mol/L)of the solution is: Concentration (mol/L) =  8.75 g Ca(CH3CH 2 COO)2   1 mol Ca(CH3CH 2 COO)2   2 mol CH3CH 2 COO      0.500 L   186.22 g Ca(CH3CH 2 COO)2   1 mol Ca(CH3CH 2 COO)2  = 0.1879497 mol/L CH3CH2COO– CH3CH2COO– + H2O  CH3CH2COOH + OH– Initial 0.1879497 mol/L 0 0 Change –x +x +x Equilibrium 0.1879497 – x x x Kb = Kw/Ka = (1.0x10–14)/(1.3x10–5) = 7.69231x10–10 Kb = 7.69231x10

–10

Kb = 7.69231x10–10 =

CH3CH 2 COOH  OH   CH3CH 2 COO   

x x  0.1879497  x 

Assume x is small compared to 0.1879497.

Kb = 7.69231x10–10 =

x x  0.1879497 

x = 1.202401x10–5 mol/L = [OH–] Check assumption: [1.202401x10–5/0.1879497] x 100% = 0.006%, therefore the assumption is good. [H]+ = Kw/[OH–] = (1.0x10–14)/(1.202401x10–5) = 8.31669x10–10 mol/L H+ pH = –log [H+] = –log (8.31669x10–10) = 9.0800 = 9.08 16.171 a) Annual depositions:





(NH4)2SO4:  3.0 (NH 4 )2 SO4  2.688 g/m2 = 0.8488421 = 0.85 g/m2 



9.5 total





NH4HSO4:  5.5 NH 4 HSO4  2.688 g/m2 = 1.55621 = 1.56 g/m2 9.5 total    1.0 H 2SO4  2  9.5 total  2.688 g/m  



H2SO4:

 = 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   98.1 g H 2SO 4  = 0.664796 g/m2     1 M H 2SO 4 m2     115.1 g NH 4 HSO 4 

Total as sulfuric acid = 0.66 g/m2 + 0.28 g/m2 = 0.94 g/m2 2

 0.94 g H 2SO4  2  1000 m   1 kg  3   10. km  1 km   1000 g  = 9.4x10 kg 2 m      





b) H2SO4(aq) + CaCO3(s) → H2O(l) + CO2(g) + CaSO4(s) There is a 1:1 mole ratio between H2SO4 and CaCO3. Mass (g) of CaCO3 = 

3 



3 

1 mol CaCO 100.1 g CaCO g  1 mol H SO     9.4x10 kg H SO   1000  1 kg   98.1 g H SO   1 mol H SO   1 mol CaCO  3

= 9.5916x106 g= 2.19.6x106 g CaCO3

 1000 g   1 mol H 2SO4   2 mol H +  3 5 + c) Moles of H+= 9.4x10 kg H 2SO4   = 1.91641x10 mol H     1 kg   98.1 g H 2SO4   1 mol H 2SO4 





Volume of lake = 10. km 2  1000 m  3 m   1 km 

Concentration (mol/L) of H+ =

 10

1L 3

1.91641x10 mol H 3.0x1010 L

 = 3.0x1010 L

3 

m  +

= 6.3880x10–6 mol/L

pH = –log [H+] = –log (6.3880x10–6) = 5.1946 = 5.19 Kw = [H+][OH–] = (x)(x) = 1.139x10–15 x = [H+] = 3.374907x10–8 = 3.375x10–8 mol/L H+ pH = –log [H+] = –log (3.374907x10–8) = 7.471738 = 7.4717 50°C Kw = [H+][OH–] = (x)(x) = 5.474x10–14 x = [H+] = 2.339658x10–7 = 2.340x10–7 mol/L H+ pH = –log [H+] = –log (2.339658x10–7) = 6.6308476 = 6.6308 b) 0°C Kw = [D3O+][OD–] = (x)(x) = 3.64x10–16 x = [D3O+] = 1.907878x10–8 = 1.91x10–8 mol/L D3O+ pH = –log [D3O+] = –log (1.907878x10–8) = 7.719449 = 7.719 50°C Kw = [D3O+][OD–] = (x)(x) = 7.89x10–15 x = [D3O+] = 8.882567x10–8 = 8.88x10–8 mol/L D3O+ pH = –log [D3O+] = –log (8.882567x10–8) = 7.0514615 = 7.051 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.

16.172 a) 0°C

16.173 Concentration (Molarity) of HX =  12.0 g HX   1 mol HX  = 0.0800 mol/L HX 

  150. g HX 

Concentration (mol/L) of HY =  6.00 g HY   1 mol HY  = 0.120 mol/L HY 

  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. 16.174 Acid HA: HA(aq) + H2O(l) H3O+(aq) + A–(aq)  0.010 mol   0.010 mol  0.50 L   0.50 L   Ka = = 4.0x10–3 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   Ka = = 1.0x10–2 0.040 mol  0.25 L   Acid HB, with the larger Ka value, is the stronger acid. 16.175 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)

r H = [ f(products) H ] – [ f(reactants) H ]

r H = {1  f H [H2O(l)]} – {1  f H [H+(aq)] + 1  f H [ OH(aq)]} r H = [(285.840 kJ/mol)] – [(0 kJ/mol) + (229.94 kJ/mol)]

r H =  55.90 kJ /mol b) The neutralization reaction of a strong acid and a strong base is essentially the reaction between H+(aq) and OH(aq) to form H2O(l). Therefore,  r H for KOH and HCl would be expected to be 55.90 kJ/mol. 16.176

NH2(CH2)4NH2(aq) + H2O(l)  NH2(CH2)4NH3+(aq) 0.10  x x x = [OH] = 2.1x10–3

+ OH(aq) x

 NH 2 CH 2  NH3  OH    2.1x103   2.1x103  4       Kb = = = 4.5045965x105 = 4.5x105  NH 2 CH 2  NH 2  0.10  2.1x103  4    

16.177 a) There are 20 OH ions for every 2 H+ ions; in other words, [OH] = 10 x [H+] Kw = 1.0x1014 = [H+][OH] 1.0x1014 = [H+](10)[ H+] [ H+] = 3.162278x108 mol/L pH = –log [H+] = –log (3.162278x10–8) = 7.4999999 = 7.5 b) For a pH of 4, [H+] = 10pH = 104 = 1.0x104 mol/L [OH] = Kw/[H+] = (1.0x10–14)/(1.0x10–4) = 1.0x10–10 mol/L OH [H+]/[OH] = 1.0x104/1.0x10–10 = 1.0x106 The H+ concentration is one million times greater than that of OH . You would have to draw one million H+ ions for every one OH. 16.178 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. 16.179 Plan: Use Le Chatelier‘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. 16.180 NH3(aq) + H2O(l) NH4+(aq) + OH–(aq) Convert to a Ka relationship: NH4+(aq) + H2O(l) NH3(aq) + H3O+(aq) Ka = Kw/Kb = (1.0x10–14)/(1.76x10–5) = 5.6818x10–10 Ka =

NH3   H3O   NH 4   

Ka  NH 3  =   H3O    K a  NH 4    NH 3      a) [H+] = 10–pH = 10–7.00 = 1.0x10–7 mol/L H+  NH 3  5.6818 x1010 = = 5.6496995x10–3 = 5.6x10–3 7 10   NH 4    NH 3  1.0x10  5.6818x10   b) [H+] = 10–pH = 10–10.00 = 1.0x10–10 mol/L H+  NH 3  5.6818x1010 = = 0.8503397 = 0.85  NH 4    NH 3  1.0x1010  5.6818x1010   c) Increasing the pH shifts the equilibria towards NH3. Ammonia is able to escape the solution as a gas. 16.181 Plan: The concentration (mol/L) 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 concentration (mol/L)of the solution. The total concentration of all species at equilibrium equals the apparent concentration (mol/L)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 concentration (mol/L) of the solution (before acid dissocation). Silberberg, Amateis, Venkateswaran, Chen, Chemistry: The Molecular Nature of Matter and Change, 3rd Canadian Edition Page 3-586 Instructor‘s Solution Manual © Copyright 2021 McGraw-Hill Ryerson Ltd.

c =  7.500 g CH3CH 2 COOH   1 mL   1 mol CH3CH 2 COOH  = 1.012419 mol/L 

  10

100.0 mL solution

3

L   74.08 g CH3CH 2 COOH 

= 1.012 mol/L CH3CH2COOH b) The freezing point depression equation is required to determine the molality of the solution. T = iKf m = [0.000 – (–1.890°C)] = 1.890°C Temporarily assume i = 1. T 1.890 C m= = = 1.016129032 m = 1.016129032 mol/L iK f (1)(1.86 C/m) This molality is the total molality of all species in the solution, and is equal to their concentration. From the equilibrium: CH3CH2COOH(aq) + H2O(l)  H3O+(aq) + CH3CH2COO–(aq) Initial 1.012419 mol/L 0 0 Change –x +x +x Equilibrium 1.012419 – x x x The total concentration of all species is: [CH3CH2COOH] + [H3O+] + [CH3CH2COO–] = 1.016129032 mol/L [1.012419 – x] + [x] + [x] = 1.012419 + x = 1.016129032 mol/L x = 0.00371003 = 0.004 mol/L CH3CH2COO– c) The percent dissociation is the amount dissociated (x from part b)) divided by the original concentration from part a). 0.00371003 mol / L Percent dissociation = 100%  = 0.366452 %= 0.4% 1.012419 mol / L 16.182 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 x Kb. Set up a reaction table in which x = dissociated acid and the concentration of [H 3O]+, 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 = 7.94328x10–6 Kb (aromatic ring N) = 10–pKb = 10–9.7 = 1.995262x10–10 a) Ignoring the smaller Kb: C20H24N2O2(aq) + H2O(l) OH–(aq) + HC20H24N2O2+(aq) Initial 1.6x10–3 mol/L 0 0 Change –x +x +x Equilibrium 1.6x10–3 – x x x    HC 20 H 24 N 2 O 2   OH    Kb = 7.94328x10–6 =  C 20 H 24 N 2 O 2  Kb = 7.94328x10–6 =

x x 

1.6x10 

3

Kb = 7.94328x10–6 =

 x   x  x 

Assume x is small compared to 1.6x10–3.

1.6x103   

x = 1.127353x10–4 Check assumption that x is small compared to 1.6x10 –3: 1.127353x104 100%  = 7% error, so the assumption is not valid. 1.6 x 103 Since the error is greater than 5%, it is not acceptable to assume x is small compared to 1.6x10 –3, and it is necessary to use the quadratic equation. Silberberg, Amateis, Venkateswaran, Chen, Chemistry: The Molecular Nature of Matter and Change, 3rd Canadian Edition Page 3-587 Instructor‘s Solution Manual © Copyright 2021 McGraw-Hill Ryerson Ltd.

x2 = (7.94328x10–6)(1.6x10–3 – x) = 1.27092x10–8 – 7.94328x10–6x x2 + 7.94328x10–6 x – 1.270925x10–8 = 0 a = 1 b = 7.94328x10–6 c = –1.27092x10–8 2 x =  b  b  4ac

7.94328x106  x=

7.94328x10   4 1  1.270925x10  = 1.08834x10 mol/L OH 6 2

8

–4

2 1

[H]+ =



[OH ]

–

14

1.0x10 = 9.18830513x10–11 mol/L H+ 4 1.08834x10

pH = –log [H+] = –log (9.18830513x10–11) = 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 1.6x10–3 mol/L 0 0 Change –x +x +x Equilibrium 1.6x10–3 – x x x  HC 20 H 24 N 2 O 2    OH     Kb = 1.995262x10–10 =  C 20 H 24 N 2 O 2  Kb = 1.995262x10–10 = Kb = 1.995262x10–10 =

x x 

1.6x103  x     x  x 

Assume x is small compared to 1.6x10–3.

1.6x103    –7 – x = 5.65015x10 mol/L 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 mol/L 0 0 Change –x +x +x Equilibrium 0.33 – x x x Ka =

Kw 1.0x1014 = Kb 7.94328x106

= 1.25893x10–9

 H3O  C20 H 24 N 2 O2    Ka = 1.25893x10 =  HC20 H 24 N 2 O2    –9

x x  Assume x is small compared to 0.33. 0.33  x  Ka = 1.25893x10–9 = x x  0.33 Ka = 1.25893x10–9 =

[H+] = x = 2.038252x10–5 mol/L Check assumption that x is small compared to 0.33: 2.038252x105 100%  = 0.006%. The assumption is good. 0.33

pH = –log [H+] = –log (2.038252x10–5) = 4.69074 = 4.7 d) Quinine hydrochloride will be indicated as QHCl.

 1.5%  1.0 g   1 mL   1 mol QHCl  M=    = 0.041566 mol/L    100%  mL   103 L   360.87 g QHCl  HC20H24N2O2+(aq) + H2O(l) H3O+(aq) + C20H24N2O2(aq) Initial 0.041566 mol/L 0 0 Change –x +x +x Equilibrium 0.041566 – x x x

 H3O  C20 H 24 N 2 O2    Ka = 1.25893x10 =  HC20 H 24 N 2 O2    –9

x x  0.041566  x  Ka = 1.25893x10–9 = x x  0.041566  Ka = 1.25893x10–9 =

Assume x is small compared to 0.041566.

[H3O+] = x = 7.233857x10–6 mol/L Check assumption that x is small compared to 0.33: 7.233857x106 100%  = 0.02%. The assumption is good. 0.041566

pH = –log [H+] = –log (7.233857x10–6) = 5.1406 = 5.1

16.183 Plan: To solve this question, it is important to use the equilibrium expression for the acid dissociation of HOCl. Write the acid dissociation expression in terms of Ka, [HClO], [H3O+], and [OCl-]. Then substitute for [HClO] and [ClO-] into the expression for the fraction of HClO. Once you simplify the expression, you will get the term below. Solution: a) At pH = 7.00, [H+] = 10–pH = 10–7.00 = 1.0x10–7 mol/L





HO HClO  3  = HClO   ClO   H3O   Ka 7

1.0x10 HClO = 7 8 = 0.775194 = 0.78 HClO   ClO  1.0x10  2.9x10

b) At pH = 10.00, [H+] = 10–pH = 10–10.00 = 1.0x10–10 mol/L





HO HClO  3  = HClO   ClO   H3O   Ka 10

1.0x10 HClO -3 = 10 8 = 3.4364 x 10 = 0.0034   1.0x10  2.9x10 HClO   ClO  16.184

a) All scenes indicate equal initial amounts of each acid. The more H + 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) x 100% = 25% e) NaY, the weakest base, will give the highest pOH and the lowest pH.

16.185 Plan: Determine the [H+] and pH of the HCl solution. Set up the ICE table for acetic acid and write the Ka expression. Solve for the initial concentration of the acid. Solution: [H+] =[HCl]=0.0035 mol/L pH (HCl solution) = - log (0.0035) = 2.46

Initial Change Equilibrium

CH3COOH(aq) + H2O(l) CH3COO–(aq) + H3O+(aq) c — 0 0 –x +x +x c–x x x

But we know x = 0.0035 mol/L

 H 3O +  CH 3COO -  Ka   CH3COOH 

 0.0035 x2 1.8  10    c  x  c  0.0035 2

5

 0.0035 c  0.0035 

1.8 105 c = 0.68 mol / L

16.186 Plan: Determine the [OH-], pOH and pH of the NaOH solution. Set up the ICE table for ammonia and write the Kb expression. Solve for the initial concentration of the base. Solution: [OH-] =[NaOH]=0.0079 mol/L pOH (NaOH solution) = - log (0.0079) = 2.10 pH = 14.00 – pOH = 14.00 – 2.10 = 11.90

Initial Change Equilibrium

NH3 (aq) + H2O(l) NH4+ (aq) + OH-(aq) c — 0 0 –x +x +x c–x x x

But we know x = 0.0079 mol/L

 NH 4 +   OH -  Kb   NH3 

 0.0079  x2 1.76 10    c  x  c  0.0079 2

5

 0.0079  c  0.0079 

1.76  105 c = 3.6 mol / L

16.187 Plan: Potassium acetate is a salt which will dissociate completely, leaving K + (aq), which will not contribute to the acid/base equilibrium and CH3COO- (aq), which WILL contribute to the acid/base equilibrium via abstraction of a proton from water to form a weak acid. We will set up the ICE table using c for concentration of the potassium acetate. From the pH, we can calculate the [H+] and then [OH-], which will be x. We will look up the Silberberg, Amateis, Venkateswaran, Chen, Chemistry: The Molecular Nature of Matter and Change, 3rd Canadian Edition Page 3-590 Instructor‘s Solution Manual © Copyright 2021 McGraw-Hill Ryerson Ltd.

value for Ka for acetic acid and use Ka and Kw to calculate Kb for the acetate ion. We will solve for c and then use the molar mass of the salt to find the amount in moles of salt. Finally, we will use the value of c and the amount, n of the salt to find the volume. Solution:

CH3COOK (aq)  CH3COO  (aq) + K  (aq) CH3COO (aq)  H 2O (l ) Initial Change Equilibrium

CH 3COOH (aq)  OH  (aq)

—

c –x c–x

0 +x x

pH=7.235 pOH=14.000-pH=14.000-7.235=6.765 OH    10 pOH  106.765  1.72 107 mol/L=x

Kb 

K w 1.00 1014   5.56  1010 5 Ka 1.8 10

OH    CH 3COOH  Kb  CH 3COO   1.72 107   x2    c  x  c  1.72 107 2

5.56 10

10

1.72 10  c  1.72  10 

7 2

7

5.56 1010 c = 5.33  105 mol / L MM CH3COOK  98.15 g/mol

nCH3COOK 

m 2.86 103 g   2.91 105 mol MM 98.15 g/mol

n V n 2.91105 mol V   0.547 L = 547 mL c 5.33×10-5 mol / L c=

16.188 Plan: We know that ammonia is a weak base and expect its pH to be between 7 and 14. If we add water to ammonia, it becomes more dilute and its extent of dissociation should increase. However, the concentration of hydroxide ion will also decrease. Overall, the solution should become less basic. We will find the pH of the 0.150 mol/L ammonia solution, then set the pH 1 lower and find the new concentration. Then we will use c1V 1=c2V2 to find the volume of water to be added. Solution: NH3 (aq) + H2O(l) NH4+ (aq) + OH-(aq) Initial 0.150 mol/L — 0 0 Change –x +x +x Equilibrium 0.150 – x x x

 NH 4 +  OH -  Kb   NH3  1.76 105 

x2  0.150  x 

x 2  1.76  105 x  2.64  106  0 x 

b  b 2  4ac 2a 1.76 105 

1.76 10   4 1  2.64 10  5 2

6

2 x = 1.62×10 mol / L = [OH - ] -3

pOH=-log OH -    log(1.63 103 )  2.79 pH=14.00-pOH=14.00-2.79=11.21 The new pH will need to be 10.21 pOH=3.79 OH -   10 pH  103.79  1.62 104 mol/L (this should be 2 sf but for the moment we will keep 3) NH3 (aq) + H2O(l) NH4+ (aq) + OH-(aq) c — 0 0 –x +x +x c–x x x

Initial Change Equilibrium

But we know x = 1.62x10-4 mol/L

 NH 4 +  OH -  Kb    NH3  1.62 10 4   x2 5 1.76 10    c  x  c  1.62 104 2

1.62 10  c  1.62 10 

4 2

4

c1V1  c2V2

1.76 105 c = 1.65  103 mol / L

(0.150mol/L)(500.0mL)  (1.65 10 3 mol/L)(V2 ) V2  4.6 104 mL=46 L V2  V1  Vadded Vadded  V2  V1  4.6 104 mL  500mL=4.55 10 4 mL We would have to add 45.5 L of water. Silberberg, Amateis, Venkateswaran, Chen, Chemistry: The Molecular Nature of Matter and Change, 3rd Canadian Edition Page 3-592 Instructor‘s Solution Manual © Copyright 2021 McGraw-Hill Ryerson Ltd.

16.189 Plan: Use our knowledge of acids and bases to determine if a solution of sodium ascorbate will be acidic or basic. Once determined, use the given Ka as is or use it to determine Kb. Calculate the concentration of ascorbate ion in solution using the given mass and molar mass to find amount (mol) and this value and the volume of solution to find concentration. The sodium ion does not contribute to the acid/base nature of the solution, so the pH is only dependent on the acid/base nature of the ascorbate ion. Use an ICE table to determine the concentration of the hydroxide ion in solution from which we can find the pOH and then the pH. Solution: We must calculate the molar mass of sodium ascorbate, NaHC6H6O6 MM = [22.99 + 7(1.01) + 6(12.01) + 6(16.00)] g/mol = 198.12 g/mol

n SodiumAscorbate 

m 1.3734g   6.9322 103 mol MM 198.12 g mol

n 6.9322 103 mol mol   0.1386 V 0.05000L L For an acid and its conjugate base,

 NaHC6 H6 O6   HC6 H6 O6   pK a  pK b  14.00 4.10  pK b  14.00 pK b  14.00  4.10  9.90 K b  109.90  1.3  1010

NaHC6 H 6 O6 (aq)  Na  (aq) + HC6 H 6 O 6 (aq) HC6 H 6 O6 (aq)  H 2 O(l) I C E

0.1386 M -x 0.1386-x

Kb 

H 2 C 6 H 6O 6 (aq)  OH  (aq)

0 +x x

 H 2 C6 H6 O6  OH  

 HC6 H 6 O6   x  x  x2 1.3  1010    0.1386  x   0.1386  x  Assume that x << 0.1386 Therefore, 0.1386 – x ≈ 0.1386 x2 x2 1.3  1010    0.1386  x  0.1386 x 2   0.1386  1.3  1010 

mol  OH   so the assumption is valid L pOH = - log [OH-] = - log (4.2 x 10-6) = 5.38 x  4.2  106

pH = 14.00 – pOH = 14.00 – 5.38 = 8.62 Silberberg, Amateis, Venkateswaran, Chen, Chemistry: The Molecular Nature of Matter and Change, 3rd Canadian Edition Page 3-593 Instructor‘s Solution Manual © Copyright 2021 McGraw-Hill Ryerson Ltd.

16.190 Plan: Use our knowledge of acids and bases to determine if a solution of pyridinium chloride will be acidic or basic. Once determined, use the given Kb as is or use it to determine Ka. Calculate the concentration of pyridinium ion in solution using the given mass and molar mass to find amount (mol) and this value and the volume of solution to find concentration. The chloride ion does not contribute to the acid/base nature of the solution, so the pH is only dependent on the acid/base nature of the pyridinium ion. Use an ICE table to determine the concentration of the hydrogen ion in solution from which we can find the pH. Solution: We must calculate the molar mass of pyridinium chloride, C5H5NHCl MM = [5(12.01) + 6(1.01) + 14.01 + 35.45] g/mol = 115.57 g/mol

n PyridiniumChloride 

m 2.8923g   2.5026  102 mol MM 115.57 g mol

n 2.5026 102 mol mol   0.2503 V 0.1000L L For an acid and its conjugate base,

C5 H5 NHCl  C5 H5 NH    pK a  pK b  14.00 pK a  8.80  14.00 pK a  14.00  8.80  5.20 K a  105.20  6.3  10 6

C5 H5 NHCl (aq)  C5 H5 NH  (aq) + Cl  (aq) C5 H5 NH  (aq)  H 2 O (l) I C E

0.2503 M -x 0.2503-x

Ka 

C5 H5 N (aq)  H 3O  (aq) 0 +x x

0 +x x

C5 H5 N   H3O 

 C5 H 5 NH    x  x  x2 6.3  106    0.2503  x   0.2503  x  Assume that x << 0.2503 Therefore, 0.2503 – x ≈ 0.2503 x2 x2 6.3 106    0.2503  x  0.2503 x 2   0.2503  6.3 106 

mol   H 3O     H   so the assumption is valid. L pH = - log [H+] = - log (1.3 x 10-3) = 2.90 x  1.3  103

16.191 Plan: Find the Ka values for each acid. Those without a Ka value are extremely strong acids and [H+] is the concentration of the acid. For those with a Ka value, make an ICE table and solve for [H+], then find the pH. When we make solutions of the salts of the acids, the stronger the acid, the weaker the conjugate base. Once we calculate the pH of the solutions of the salts, we can predict the trend for the acids and the conjugate bases. Silberberg, Amateis, Venkateswaran, Chen, Chemistry: The Molecular Nature of Matter and Change, 3rd Canadian Edition Page 3-594 Instructor‘s Solution Manual © Copyright 2021 McGraw-Hill Ryerson Ltd.

Solution: Ka (HClO) = 2.9 x 10-8

Ka (HClO2) = 1.1 x 10-2

No Ka value for HClO3, HClO4

HClO (aq) + H2O (l) ⇄ ClO- (aq) + H3O+ (aq) 1.0 0 0 -x +x +x 1.0 - x x x

a) i) I C E

 H 3O +  ClO-  x2 K a =2.9×10 =  1.0  x  HClO -8

Assume x << 1.0  1.0 - x  1.0 x 2  2.9 108 mol   H 3O     H   L pH   log  H     log 1.7  104   3.77 x  1.7 104

HClO2 (aq) + H2O (l) ⇄ ClO2- (aq) + H3O+ (aq) 1.0 0 0 -x +x +x 1.0 - x x x

ii) I C E

 H 3O +  ClO 2  x2 K a =1.1×10 =  1.0  x  HClO2  -2

Cannot assume anything. Solve quadratic. x 2  1.1 102 x  1.1 102  0 x 

b  b 2  4ac 2a 1.1 102 

1.110   4  1.110  2 2

2

2 mol x  1.0  101   H 3O     H   L  pH   log  H    log 1.0 101   1.00 iii) and iv)

For HClO3 and HClO4, Ka is undefined, so we can assume 100% dissociation. So, [H+] = [HClO4] = 1.0 pH = - log [H+] = - log (1.0) = 0

NaClO (aq) ⇄ Na+ (aq) + ClO- (aq)

, where [NaClO] = [Na+] = [ClO-]

ClO- (aq) + H2O (l) ⇄ HClO (aq) + OH- (aq) I C E

1.0 -x 1.0 - x

0 +x x

 HClO OH  K w 1.0 1014 x2 -7 Kb = = = 3.4×10 =  K a 2.9 108 1.0  x ClO-  -

Assume x << 1.0  1.0 - x  1.0 x 2  3.4  107 mol  OH   L pOH= - log OH -  = - log  5.9×10-4  = 3.23 x  5.9  104

pH=14.00 - pOH=14.00 - 3.23=10.77 NaClO2 (aq) ⇄ Na+ (aq) + ClO2- (aq) , where [NaClO2] = [Na+] = [ClO2-] ClO2 (aq) + H2O (l) ⇄ HClO2 (aq) + OH- (aq)

ii)

I C E

1.0 -x 1.0 - x

0 +x x

 HClO2  OH -  K w 1.0 1014 x2 -13 Kb = = = 9.1×10 =  K a 1.1 102 1.0  x ClO 2  Assume x << 1.0  1.0 - x  1.0 x 2  9.11013 mol  OH   L pOH= - log OH -  = - log  9.5×10-7  = 6.02 x  9.5  107

pH=14.00 - pOH=14.00 - 6.02=7.98 iii) and iv)

For HClO3 and HClO4, Ka is undefined, so we can assume 100% dissociation. 



Therefore, their conjugate bases, namely ClO3 and ClO4 have practically no acid/base activity in water (much like Cl- in water). Thus, a solution of these salts in water would have a pH of 7. c)

16.192

As we add O to the anion (the conjugate base), the acid strength increases. As we add O to the anion (the conjugate base), the conjugate base strength decreases.

Plan: Use the Kb value and the pH to create an ICE table that can be used to find the value of the concentration of the conjugate acid in solution. Use the formula to find the molar mass of the compound and the mass and the molar mass to find the amount in mol of the compound. Use the amount in mol of the compound and the concentration of the conjugate acid (also the concentration of the base) to find the volume.

Solution: Kb aniline (C6H5NH2) = 3.9 x 10-10

MM (aniline) = 174.054 g/mol

m aniline 5.4675g n aniline = = = 3.1413×10-2 mol g MM aniline 174.054 mol I C E

C6H5NH3+ (aq) + H2O (l) ⇄ C6H5NH2 (aq) + H3O+ (aq) c 0 0 -x +x +x c-x x x

pH = 2.5186

let the initial [C6H5NH3+]=c

So, [H+] = 10 - 2.5186 = 3.030 x 10-3 mol/L

But, x = [H3O+] = [H+] = 3.030 x 10-3 mol/L

 C6 H5 NH 2   H3O  x 2 K w 1.0 1014 -5 Ka = = = 2.6×10 =  K b 3.9  1010 cx C6 H 5 NH 3+  +

x2  2.6×10-5 cx 2

 3 mol   3.030 10  L    2.6 ×10-5 mol   3  c  3.030  10  L   mol    3 mol  2.6×10 c   2.6×10   3.030 103    3.030  10  L   L   mol c  0.36 L n c V n 3.1413×10-2 mol V   8.69 ×10-2 L  87 mL mol c 0.36 L -5

-5

CHAPTER 17 IONIC EQUILIBRIA IN AQUEOUS SYSTEMS CHEMICAL CONNECTIONS BOXED READING PROBLEMS B17.1

Plan: Consult Figure 17.5 for the colours 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. B17.2

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 litres. The pH of the rain is used to find the concentration (mol/L) of H+; this concentration (mol/L) multiplied by the volume of rain gives the amount (mol) of H+. The amount (mol) of H+ divided by the volume of the lake plus rain gives the concentration (mol/L)of H+ and the pH of the lake. Solution: a) To find the volume of rain, multiply the surface area in square kilometers by the depth of rain. Convert the volume to cm3 and then to L using the density of water. 2

 1000 m   100 cm   1cm   1 mL   103 L  25.4 mm Volume (L) of rain =  0.040 km²            3    1 km   1m   10 mm   1 cm  1 mL  = 1.016x106 L + At pH = 4.20, [H ] = 10–4.20 = 6.3095734x10–5 mol/L  6.3095734x105 mol  amount (mol) of H+ = 1.016x106 L   = 64.105266 = 64 mol  L  





 1000 m   100 cm   100 cm   1 mL   103 L  b) Volume (L) of the lake =  0.040 km²        3.05 m     3    1 km   1 m   1 m   1 cm  1 mL  = 1.22x108 L Total volume of lake after rain = 1.22x10 8 L + 1.016x106 L = 1.23016x108 L mol H3O  64.105266 mol = [H+] = = 5.211132x10–7 mol/L 8 L 1.23016x10 L pH = –log (5.211132x10–7) = 6.28307 = 6.28 c) Each mol of H+ requires one mole of HCO3 for neutralization.  1 mol HCO3  61.02 g HCO3  Mass (g) = (64.105266 mol H +)   1 mol H    1 mol HCO   3    = 3.9117x103 = 3.9x103 g HCO3–

END–OF–CHAPTER PROBLEMS 17.1

The purpose of an acid-base buffer is to maintain a relatively constant pH in a solution.

17.2

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.

17.3

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  , t he co m m o n io n .

17.4

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.

17.5

a) 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. b) 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.

17.6

Only the concentration of the buffer components (c) has an affect 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.

17.7

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.

17.8

a) 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. b) If acetic acid were used, the buffer-component ratio would be far from 1:1 and the buffer‘s effectiveness would be lower. c) The NaOH serves to partially neutralize the acid and produce its conjugate base.

17.9

Plan: Remember that the weak-acid buffer component neutralizes added base and the weak-base buffer component neutralizes added acid. This is the purpose of the Henderson-Hasselbalch equation. 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.

17.10

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. 17.11

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 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 [H+], from which the pH can be calculated. Alternatively, the pH can be calculated from the Henderson-Hasselbalch equation. Solution: Concentration (mol/L) 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. Check assumption: 0.15/1.3x10–5 = 12 000 > 400, the assumption is justified.  H 3O   CH 3CH 2 COO    x  0.35  x   x  0.35     Ka = 1.3x10–5 = = = CH 3CH 2 COOH   0.15  x   0.15  CH 3CH 2 COOH  5  0.15  –6 –6 x = [H3O+] = Ka = = 1.3x10   = 5.57143x10 = 5.6x10 mol/L  CH3CH 2 COO   0.35    pH = –log [H+] = –log (5.57143x10–6) = 5.2540 = 5.25 Another solution path to find pH is using the Henderson-Hasselbalch equation:  [base]  pKa = –log (1.3x10–5) = 4.886 pH = pKa + log   [acid]  





 [CH3CH 2 COO ]   [0.35]  = 4.886 + log  pH = 4.886 + log    [CH CH COOH]   [0.15]  3 2   pH = 5.25398 = 5.25

 C6H5COO–(aq) + H3O+(aq)  H 3O    C6 H 5 COO   (x) (0.28) (x) (0.28  x)    –5 Ka = 6.3x10 = = = C6 H 5 COOH  (0.33) (0.33  x ) –5 Check assumption: 0.33/6.3x10 = 5200 > 400, the assumption is justified. x = [H3O+] = (6.3x10–5)(0.33/0.28) = 7.425x10–5 = 7.4x10–5 mol/L pH = –log [H+] = –log (7.425x10–5) = 4.1293 = 4.13

17.12

C6H5COOH(aq) + H2O(l)

17.13

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 HNO 2 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 (mol/L) 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 Assume that x is negligible with respect to both 0.55 and 0.75 since both concentrations are much larger than Ka.  H 3O    NO 2   (x) (0.75  x) (x) (0.75)    –4 Ka = 7.1x10 = = = (0.55) (0.55  x) HNO2 

HNO2 





(0.55) = 7.1x104 = 5.2066667x10–4 = 5.2x10–4 mol/L (0.75)  NO 2     Check assumption: Percent error = (5.2066667x10 –4/0.55)100% = 0.095%. The assumption is valid. pH = log [H+] = log (5.2066667x10–4) = 3.28344 = 3.28 x = [H3O+] = Ka

Using the Henderson-Hasselbalch equation instead:  [base]  pKa = –log(7.1x10–4) = 3.149 pH = pKa + log   [acid]    [NO2  ]   [0.75]  = 3.149 + log  pH = 3.149 + log    [HNO ]   [0.55]  2   pH = 3.2837 = 3.28

17.14

HF(aq) + H2O(l)

 F–(aq) + H3O+(aq)

Assuming x << 0.20,  H 3 O    F  (x) (0.25  x) (x) (0.25)    4 Ka = 6.8x10 = = = (0.20) (0.20  x) HF   x = (6.8x104)(0.20/0.25) = 5.44x104 = 5.4x104 mol/L Check assumption: Percent error = (5.44x104/0.20)100% = 0.27%. The assumption is valid. pH = log [H+] = log (5.44x104) = 3.2644 = 3.26 Alternatively, using the Henderson-Hasselbalch equation: pKa = –log(6.8x10–4) = 3.167  [F ]   0.25  pH = 3.167 + log   3.167 + log    3.26  [HF]   0.20    17.15

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 = 1.8197x104 Concentration (mol/L) HCOOH(aq) + H2O(l)  HCOO(aq) + H3O+(aq) 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.  H 3O    HCOO   (x) (0.63  x) (x) (0.63)    4 Ka = 1.8197x10 = = = (0.45) (0.45  x) HCOOH 

HCOOH 





4 (0.45) = 1.8197x10 = 1.29979x104 = 1.3x104 mol/L (0.63)  HCOO     Check assumption: Percent error = (1.29979x104/0.45)100% = 0.029%. The assumption is valid. pH = log [H+] = log (1.29979x104) = 3.886127 = 3.89 Alternatively, using the Henderson-Hasselbalch equation.  [base]  pH = pKa + log    [acid] 

x = [H3O+] = Ka

 [HCOO  ]   [0.63]  pH = 3.74 + log   = 3.74 + log   [HCOOH]  [0.45]    pH = 3.8861 = 3.89 17.16

HBrO(aq) + H2O(l)

 BrO(aq) + H3O+(aq)

Ka = 10 pKa = 108.64 = 2.2908677x109

 H 3 O    BrO     = (x) (0.68  x) = (x) (0.68) Ka = 2.2908677x10 =  (0.95) HBrO (0.95  x) 9

x = [H3O+] = Ka

HBrO  = (2.2908677x109)(0.95/0.68) = 3.2004769x109 = 3.2x109 mol/L

 BrO     Check assumption: Percent error = (3.2004769x10 9/0.68)100% = 0.00000047%. The assumption is valid. pH = log [H+] = log (3.2004769x109) = 8.4947853 = 8.49 Alternatively, using the Henderson-Hasselbalch equation:  0.68  pH = 8.64 + log    8.49  0.95  17.17

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 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 = 10–10.00 = 1.0x10–10 Concentration (mol/L) C6H5OH(aq) + H2O(l)   C6H5O(aq) + H3O+(aq) 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.  H 3 O    C6 H 5 O   (x) (1.3  x) (x) (1.3)    Ka = 1.0x10–10 = = = (1.2  x) (1.2) C6 H 5 OH  C6 H 5 OH  10  1.2  –11 x = [H3O+] = Ka = 1.0x10  1.3  = 9.23077x10 mol/L   C6 H 5 O      Check assumption: Percent error = (9.23077x10 –11/1.2)100% = 7.7x10–9%. The assumption is valid. pH = –log (9.23077x10–11) = 10.03476 = 10.03 Using the Henderson-Hasselbalch equation:  [base]  pH = pKa + log    [acid] 





 [C H O ]   [1.3]  = 10.00 + log  pH = 10.00 + log  6 5   [C H OH]   [1.2]   6 5  pH = 10.03

17.18

H3BO3(aq) + H2O(l)   H2BO3–(aq) + H3O+(aq) Ka = 10 pKa = 10–9.24 = 5.7543994x10–10

Ka = 5.7543994x10–10 =

 H 3O    H 2 BO3  (x) (0.82  x) (x) (0.82)    = = (0.12) (0.12  x) H 3 BO3 

H3 BO3  = (5.7543994x10–10)(0.12/0.82) = 8.4210723x10–11 mol/L  H 2 BO3    Check assumption: Percent error = (8.4210723x10 –11/0.12)100% = 7.0x10–8%. The assumption is valid. pH = –log [H+] = –log (8.4210723x10–11) = 10.0746326 = 10.07 Using the Henderson-Hasselbalch equation: x = [H3O+] = Ka

 0.82  pH = 9.24 + log    10.07  0.12  17.19

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 NH 4+ 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 = 5.62341325x10–10 Concentration (mol/L) NH4+(aq) + H2O(l)  NH3(aq) + H3O+(aq) 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.   (0.25  x)  H3 O  NH3   H3O    = (0.25)[H3 O ] –10 Ka = 5.62341325x10 = = (0.15  x) (0.15)  NH 4      NH 4     10  0.15  –10 X = [H3O+] = Ka = 5.62341325x10  0.25  = 3.374048x10 mol/L  NH 3    Check assumption: Percent error = (3.374048x10 –10/0.15)100% = 2x10–7%. The assumption is valid. pH = –log [H+] = –log [3.374048x10–10] = 9.4718 = 9.47 Using the Henderson-Hasselbalch equation:  [base]  pH = pKa + log    [acid] 





 [NH3 ]   [0.25]  = 9.25 + log  pH = 9.25 + log    [NH + ]   [0.15]  4   pH = 9.47

17.20

Kb = 10 pKb = 10–3.35 = 4.4668359x10–4 The base component is CH3NH2 and the acid component is CH3NH3+. Neglect Cl–. Assume + x and – x are negligible. CH3NH2(aq) + H2O(l)  CH3NH3+(aq) + OH–(aq) CH 3 NH 3  OH   (0.60  x)  OH   (0.60)  OH        =   Kb = 4.4668359x10–4 = = CH NH (0.50  x) (0.50)  2  3

CH 3 NH 2  = (4.4668359x10–4)(0.50/0.60) = 3.7223632x10–4 mol/L  CH 3 NH 3    Check assumption: Percent error = (3.7223632x10 –4/0.50)100% = 0.074%. The assumption is valid. pOH = –log [OH–] = –log (3.7223632x10–4) = 3.429181254 pH = 14.00 – pOH = 14.00 – 3.429181254 = 10.57081875 = 10.57  0.50  Using the Henderson-Hasselbalch equation: pH = (14.00 - 3.35) + log    10.57  0.60  [OH–] = Kb

17.21

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 [H 3O+], 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. Concentration (mol/L) HCO3(aq) + H2O(l)  CO32(aq) + H3O+(aq) 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 = 4.7x1011 =  (0.22  x) (0.22)  HCO3     HCO3   = 4.7x1011  0.22  = 2.79459x1011 mol/L + [H3O ] = Ka   0.37  2  CO3      Check assumption: Percent error = (2.79459x1011/0.22)100% = 1.3x10-8%. The assumption is valid. pH = –log [H+] = –log (2.79459x1011) = 10.5537 = 10.55 Using the Henderson-Hasselbalch equation:  [base]  pKa = –log (4.7x10–11) = 10.328 pH = pKa + log   [acid]  





 [CO32 ]   [0.37]  = 10.328 + log  pH = 10.328 + log    [HCO  ]   [0.22]  3   pH = 10.55

17.22

a) The conjugate acid-base pair is related by Ka2 (6.3x10–8). 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. H2PO4–(aq) + H2O(l)  HPO42–(aq) + H3O+(aq)  H3O   HPO42     = (x) (0.40  x) = (x) (0.40) –8 Ka = 6.3x10 =   (0.50) (0.50  x)  H 2 PO4     H 2 PO4   = (6.3x10–8)(0.50/0.40) = 7.875x10–8 mol/L [H3O+] = Ka   HPO4 2     Check assumption: Percent error = (7.875x10 –8/0.50)100% = 1.6x10–5%. The assumption is valid. pH = –log [H+] = –log (7.875x10–8) = 7.103749438 = 7.10 Using the Henderson-Hasselbalch equation:  0.40  pH = {-log(6.3 10-8 )} + log    7.20  0.0969  7.10  0.50 

17.23

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 (1.3x10–5) = 4.8860566  [base]  pH = pKa + log    [acid] 

 [EtCOO  ]  5.44 = 4.8860566 + log    [EtCOOH]   [EtCOO  ]  0.5539434 = log    [EtCOOH] 

Raise each side to 10x.

[EtCOO  ] = 3.5805 = 3.6 [EtCOOH] 17.24

pKa = –log Ka = –log (7.1x10–4) = 3.14874165  [base]  pH = pKa + log    [acid]   [NO2 ]  2.95 = 3.14874165 + log   [HNO ]  2    [NO2 ]  –0.19874165 = log   [HNO ]  2  

Raise each side to 10x.

[NO 2  ] = 0.632788 = 0.63 [HNO 2 ] 17.25

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 (2.3x10–9) = 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]

17.26

Given the pH and Ka of an acid, the buffer-component ratio can be calculated from the Henderson-Hasselbalch equation. pKa = –log Ka = –log (1.8x10–5) = 4.744727495  [base]  pH = pKa + log    [acid]   [CH3 CO O ]  4.39 = 4.744727495 + log   [CH COOH]  3     [CH3COO ]  –0.35473 = log  Raise each side to 10x.  [CH COOH]  3  

[CH3COO ] = 0.441845 = 0.44 [CH3COOH] 17.27

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]  = pKa + log  3.35 = pK a + log    [HA]   [0.2000]    3.35 = pKa – 0.1249387 pKa = 3.474939 = 3.47 Determine the amount (mol) of conjugate acid (HA) and conjugate base (A–) using (c)(V) = amount (mol) .  0.2000 mol HA  amount (mol) of HA = 0.5000 L   = 0.1000 mol HA 1L    0.1500 mol A   – amount (mol) of A– = 0.5000 L   = 0.07500 mol A  1 L   The reaction is: HA(aq) + NaOH(aq)  Na+(aq) + A–(aq) + H2O(l) 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  [A–] = = 0.153 mol/L A– 0.5000 L

0.0985 mol HA = 0.197 mol/L HA 0.5000 L  [base]  pH = pKa + log    [acid] 

[HA] =

 [0.153]  pH = 3.474939 + log   = 3.365164 = 3.37  [0.197]  Note: Since the volume is identical for the conjugate base and acid, we can substitute the amount (moles) directly into the ratio (i.e., nA-/V divided by nHA/V is the same as nA-/nHA), i.e., n   [0.0765]  pH = pK a + log  base  = 3.474939 + log    3.37 n  [0.0985]   acid 

17.28

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 = pK a + log  = pKa + log   +    [0.25]   [BH ]  8.88 = pKa + 0.20411998 pKa = 8.67588 = 8.68 Determine the amount (mol) of conjugate acid (BH+) and conjugate base (B) using (c)(V) = amount (mol) . amount (mol) BH+ = (0.25 L)(0.25 mol BH+/L) = 0.0625 mol BH+ amount (mol) B = (0.25 L)(0.40 mol B/L) = 0.10 mol B The reaction is: B(aq) + HCl(aq)  BH+(aq) + Cl–(aq) + H2O(l) 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. Since the volume is identical for the base and its conjugate acid, we can substitute the amount (moles) directly into the Henderson-Hasselbalch equation.

n  pH = pK a + log  base   n acid   [0.098]  pH = 8.67588 + log   = 8.857546361 = 8.86  [0.0645]  17.29

Determine the amount (mol) of conjugate acid (HY) and conjugate base (Y–) using (c)(V) = amount (mol) .  0.110 mol HY  amount (mol) of HY = 0.350 L   = 0.0385 mol HY 1L    0.220 mol Y   – amount (mol) of Y– = 0.350 L   = 0.077 mol Y  1 L   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 x 0.0015 mol Y– and consume 2 x 0.0015 mol of HY.

n  pH = pK a + log  base   n acid   [0.0800]  pH = 8.46897 + log   = 8.82183 = 8.82  [0.0355]  17.30

 [B]   [1.05]  9.50 = pK a + log   = pKa + log   +  [BH ]  [0.750]    9.50 = pKa + 0.1461280357 pKa = 9.353872 = 9.35 Determine the amount (mol) of conjugate acid (BH+) and conjugate base (B). amount (mol) of BH+ = (0.500 L)(0.750 mol BH+/L) = 0.375 mol BH+ amount (mol) of B = (0.500 L)(1.05 mol B/L) = 0.525 mol B The reaction is: B(aq) + HCl(aq)  BH+(aq) + Cl– (aq) + H2O(l) 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 n  pH = pK a + log  base   n acid   [0.520]  = 9.490092 = 9.49 pH = 9.353872 + log    0.380   

17.31

Plan: The hydrochloric acid will react with the sodium acetate, NaC2H3O2, to form acetic acid, HC2H3O2. Calculate the amount (mol) 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 theamount (mol) of C2H3O2 will decrease. Find the new concentrations of NaC 2H3O2 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: 3  0.452 mol HCl   10 L  a) Initial amount (mol) of HCl =     1mL  204 mL  = 0.092208 mol HCl L     0.400 mol NaC2 H3O2  Initial amount (mol) of NaC2H3O2 =   0.500 L  = 0.200 mol 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 pKa = –log Ka = –log (1.8x10–5) = 4.744727495  A   pH = pK a + log       HA    

  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 = pK a + log  2 3 2   [HC2 H3O 2 ]     [C H O  ]  4.96 = 4.744727495 + log  2 3 2   [HC H O ]  2 3 2     [C H O ]  0.215272505 = log  2 3 2   [HC H O ]  2 3 2  

[C2 H3O2 ] = 1.64162 [HC2 H3O2 ] From part a), we know that [HC2H3O2] + [C2H3O2] = (0.1309773 mol/L + 0.1531136 mol/L) = 0.2840909 mol/L. 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 mol/L, solve for [C2H3O2] and substitute into the second equation. [C2H3O2] = 1.64162[HC2H3O2] and [HC2H3O2] + 1.64162[HC2H3O2] = 0.2840909 mol/L [HC2H3O2] = 0.1075441 mol/L and [C2H3O2] = 0.176547 mol/L amount (mol) of C2H3O2 needed = (0.176547 mol C2H3O2–/L)(0.500 L) = 0.0882735 mol amount (mol) 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.

17.32

 56.11 g KOH  Mass (g) of KOH = 0.0117167 mol KOH   = 0.657424 = 0.66 g KOH  1 mol KOH  a) The sodium hydroxide will react with the sodium bicarbonate, NaHCO 3, to form carbonate ion, CO32–: NaOH + NaHCO3  2Na+ + CO32– + H2O Calculate the amount (mol) of NaOH and NaHCO3. All of the NaOH will be consumed to form CO 32–, and the amount (mol) of NaHCO3 will decrease. The HCO3– is the important part of NaHCO3. 3  0.10 mol NaOH   10 L  Initial amount (mol) NaOH =     1 mL  10.7 mL  = 0.00107 mol NaOH L     0.050 mol HCO3  103 L  Initial amount (mol) HCO3– =  50.0 mL  = 0.0025 mol HCO3–     1 mL   L    + NaOH + NaHCO3  2Na + CO32– + H2O Initial 0.00107 mol 0.0025 mol 0 mol Change –0.00107 mol –0.00107 mol +0.00107 mol Final 0 mol 0.00143 mol 0.00107 mol

Total volume = (50.0 mL + 10.7 mL)(10 –3 L/1 mL) = 0.0607 L 0.00143 mol [HCO3–] = = 0.023558484 mol/L 0.0607 L 0.00107 mol [CO32–] = = 0.017627677 mol/L 0.0607 L pKa = –log Ka = –log (4.7x10–11) = 10.32790214  [CO32 ]  pH = pK a + log   [HCO  ]  3    [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 = pK a + log   [HCO  ]  3    [CO32 ]  10.13 = 10.32790214 + log   [HCO  ]  3   2   [CO3 ]  –0.19790214 = log   [HCO  ]  3  

[CO32  ]

= 0.63401 [HCO3 ] From part a), we know that [HCO3–] + [CO32–] = (0.023558484 mol/L + 0.017627677 mol/L) = 0.041186161 mol/L. Although the ratio of [CO32–] to [HCO3–] can change when acid or base is added, the absolute amount does not change unless hydrogen carbonate or carbonate salt is added. Given that [CO32–]/[ HCO3–] = 0.63401 and [HCO3–] + [CO32–] = 0.041186161 M, solve for [CO32–] and substitute into the second equation. [CO32–] = 0.63401[HCO3–] and [HCO3–] + 0.63401[HCO3–] = 0.041186161 mol/L [HCO3–] = 0.025205575 M and [CO32–] = 0.015980586 mol/L amount (mol) of HCO3– needed = (0.025205575 mol HCO3–/L)(10–3 L/1 mL)(25.0 mL) = 0.00006301394 mol amount (mol) 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 (0.00006301394 mol – 0.000588962 mol) Silberberg, Amateis, Venkateswaran, Chen, Chemistry: The Molecular Nature of Matter and Change, 3rd Canadian Edition Page 3-610 Instructor‘s Solution Manual © Copyright 2021 McGraw-Hill Ryerson Ltd.

= 0.0000411774 mol HCO3– The HCl added reacts with CO32– to produce additional HCO3: CO32– + HCl  HCO3– + Cl– To produce 0.0000411774 mol HCO3– would require the addition of 0.0000411774 mol HCl.  36.46 g HCl  Mass (g) of HCl =  0.0000411774 mol HCl    = 0.00150133 = 0.0015 g HCl  1 mol HCl  17.33

Plan: Select conjugate pairs with Ka values close to the desired [H+]. Convert pH to [H+] for easy comparison to Ka values in the Appendix. Determine an appropriate base by [OH –] = Kw/[H+]. Solution: a) For pH  4.5, [H+] = 10–4.5 = 3.2x10–5 mol/L. Some good selections are the HOOC(CH 2)4COOH/ HOOC(CH2)4COO conjugate pair with Ka equal to 3.8x10–5 or C6H5CH2COOH/C6H5CH2COO conjugate pair with Ka equal to 4.9x10–5. From the base list, the C6H5NH2/C6H5NH3+ conjugate pair comes close with Ka = Kw/Kb = 1.0x10–14/4.0x10–10 = 2.5x10–5. b) For pH  7.0, [H+] = 10–7.0 = 1.0x10–7 mol/L. Two choices are the H2PO4/HPO42 conjugate pair with Ka of 6.3x10–8 and the H2AsO4/HAsO42 conjugate pair with Ka of 1.1x10–7.

17.34

Select conjugate pairs that have Ka or Kb values close to the desired [H+] or [OH–]. a) For [H3O+]  1x10–9 mol/L, the HOBr/OBr– conjugate pair comes close with Ka equal to 2.3 x10–9. From the base list, the NH3/NH4 + conjugate pair comes close with Ka = Kw/Kb = 1.0x10–14/1.76x10–5 = 5.7x10–10. b) For [OH–]  3x10–5 mol/L, the NH3/NH4 + conjugate pair comes close; also, it is possible to choose [H3O+] = 1.0x10–14/3x10–5 = 3.3x10–10; the C6H5OH/C6H5O– comes close with Ka = 1.0x10–10.

17.35

Plan: Select conjugate pairs with pKa values close to the desired pH. Convert pH to [H+] for easy comparison to Ka values in the Appendix. Determine an appropriate base by [OH –] = Kw/[H+]. Solution: a) For pH  3.5 ([H+] = 10–pH = 10–3.5 = 3.2x10–4), the best selection is the HOCH2CH(OH)COOH/ HOCH2CH(OH)COO – conjugate pair with a Ka = 2.9x10–4. The CH3COOC6H4COOH/CH3COOC6H4COO – pair, with Ka = 3.6x10–4, is also a good choice. The [OH–] = Kw/[H+] = 1.0x10–14/3.2x10–4 = 3.1x10–11, results in no reasonable Kb values from the Appendix. b) For pH  5.5 ([H+] = 10–pH = 3x10–6), no Ka1 gives an acceptable pair; the Ka2 values for adipic acid, malonic acid, and succinic acid are reasonable. The [OH–] = Kw/[H+] = 1.0x10–14/3x10–6 = 3x10–9; the Kb selection is C5H5N/C5H5NH+.

17.36

Select conjugate pairs that have Ka or Kb values close to the desired [H+] or [OH–]. a) For [OH–]  1x10–6 mol/L, no Kb values work. The Ka values are [H+] = Kw/[OH–] = 1.0x10–14/1x10–6 = 1x10–8, giving the following acceptable pairs H2PO4–/HPO42– or HC6H5O72–/C6H5O73– or HOCl/OCl–. b) For [H+]  4x10–4 mol/L, the HF/F– conjugate pair comes close with Ka equal to 6.8x10–4. From the base list, [OH–] = 1.0x10–14/4x10–4 = 2.5x10–11, there are no reasonable choices.

17.37

The value of the Ka from the Appendix: Ka = 2.9x10–8 pKa = –log (2.9x10–8) = 7.5376  [ClO ]  pH = pK a + 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] 

17.38

d) The reaction is NaOH + HClO  Na+ + ClO– + H2O. The original amount (mol) 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]  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 = 6.3x10–8 (We are using Ka2 since we are dealing with the equilibrium in which the second hydrogen ion is being lost.) pKa = –log Ka = –log (6.3x10–8) = 7.200659451 Use the Henderson-Hasselbalch equation:  [HPO42 ]  pH = pK a + log   [H PO  ]   2 4   [HPO42 ]  7.40 = 7.200659451 + log   [H PO  ]   2 4  2  [HPO4  ]  0.19934055 = log   [H PO  ]   2 4 

[HPO 4 2  ]

[H 2 PO 4  ]

= 1.582486 = 1.6

17.39

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).

17.40

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) 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, the major 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–].

17.41

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.

17.42

To see a distinct colour in a mixture of two colours, you need one colour 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. 17.43

The addition of an acid-base indicator does not affect the pH of the test solution because the concentration of indicator is very small.

17.44

The equivalence point in a titration is the point at which the amount (mol) of OH – equals the amount (mol) 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 colour. 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 colour at a pH after the equivalence point means the equivalence point is reached first. However, if an indicator is selected that changes colour at a pH before the equivalence point, then the end point is reached first.

17.45

a) The reactions are: OH–(aq) + H3PO4(aq)  H2PO4–(aq) + H2O(l) Ka1 = 7.2x10–3 – – 2– OH (aq) + H2PO4 (aq)  HPO4 (aq) + H2O(l) Ka2 = 6.3x10–8 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 PO ]   3 4  Determine the pKa using pKa = –log (7.2x10–3) = 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 HPO 42– remains. A total of 40.00 mL of NaOH is required to reach Scene A.

17.46

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 amount (mol) of H+ equals the amount (mol) of OH–, regardless of the type of titration. However, the strong acid–strong base equivalence point occurs at pH = 7.00 because the resulting cationanion 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. 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 + NH 3  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.

17.47

(a) In the buffer region, comparable amounts of weak acid and its conjugate base are present. (b) At the equivalence point for a weak acid/strong base titration, the predominant species is the conjugate base. (c) In a strong acid–weak base titration, the weak base and its conjugate acid are the predominant species present.

17.48

At the very centre 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 17.49

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.

17.50

Plan: Write a balanced equation. Find the amount (mol) of KOH from the concentration and volume information and use the molar ratio in the balanced equation to find the amount (mol) of acid present. Divide the amount (mol) of acid by its volume to determine the concentration. Since KOH is a strong base, this reaction will continue until all the KOH is reacted OR until no acid remains. Solution: The reaction is: KOH(aq) + CH3COOH(aq)  CH3COOK(aq) + H2O(l)  3  1 0 L 0 . 1 1 8 0 m o l K O H   2 5 . 9 8 m L    Amount (mol) of KOH =  = 0.00306564 mol KOH     1 m L L    

  1 m o l C H C O O H 3 0 . 0 0 3 0 6 5 6 4 m o l K O H    Amount (mol) of CH3COOH =  = 0.00306564 mol CH3COOH m o l K O H 1  0 . 0 0 3 0 6 5 6 4 m o l C H C O O H 1 m L     3 Concentration (mol/L) of CH3COOH =     3 5 2 . 5 0 m L 1 0 L     = 0.05839314 mol/L= 0.05839 mol/L CH3COOH 17.51

Plan: Write a balanced equation. Find the amount (mol) of NaOH from the concentration(mol/L) and volume information and use the molar ratio in the balanced equation to find the amount (mol) of acid present. Divide the amount (mol) of acid by its volume to determine the concentration (mol/L). Since NaOH is a strong base, this reaction will continue until all the NaOH is reacted OR until no acid remains. Solution: The reaction is: 2NaOH(aq) + H2SO4(aq)  Na2SO4(aq) + 2H2O(l)  103 L   0.1850 mol NaOH    Amount (mol) of NaOH =  26.25 mL    = 0.00485625 mol NaOH L   1 mL  

  1 m o l H S O 2 4 0 . 0 0 4 8 5 6 2 5 m o l N a O H    Amount (mol) of H2SO4 =  = 0.002428125 mol H2SO4 2 m o l N a O H   0 . 0 0 2 4 2 8 1 2 5 m o l H S O 1 m L     2 4 Concentration (mol/L) of H2SO4 =     3 2 5 . 0 0 m L 1 0 L     = 0.097125 mol/L= 0.09712 mol/L H2SO4 17.52

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 (3.5x10–9) = 8.5, so the indicator changes colour over an approximate range of 8.5  1 or 7.5 to 9.5.

17.53

Indicators have a pH range that is approximated by pKa  1. The pKa of ethyl red is –log (3.8x10–6) = 5. 4 2 therefore the indicator changes colour over an approximate range of 4.4 to 6.4.

17.54

Plan: Choose an indicator that changes colour 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 colour 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 mol/L 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: Concentration, mol/L COOH–(aq) + H2O(l)  HCOOH(aq) + OH–(aq) Initial 0.050 mol/L __ 0 0 Change –x +x +x Equilibrium 0.050 – x x x The Ka for HCOOH is 1.8x10–4, so Kb = 1.0x10–14/1.8x10–4 = 5.5556x10–11 HCOOH OH   = (x) (x) = (x) (x) Kb = 5.5556x10–11 = (0.050) HCOO   (0.050  x) – [OH ] = x = 1.666673x10–6 mol/L pOH = –log (1.666673x10–6) = 5.7781496 pH = 14.00 – pOH = 14.00 – 5.7781496 = 8.2218504 = 8.22 Choose thymol blue or phenolphthalein. 17.55

a) Determine the Ka (of the conjugate acid) from the Kb for CH3NH2. Ka = Kw/Kb = (1.0x10–14)/(4.4x10–4) = 2.2727x10–11 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 CH 3NH3+ = 0.050 mol/L.  H3O  CH3 NH 2  x2 x2  –11 Ka = 2.2727x10 =  = = 0.050  x 0.050 CH3 NH3    + –6 x = [H3O ] = 1.0659972x10 mol/L pH = –log [H+] = –log (1.0659972x10–6) = 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.

17.56

Plan: Choose an indicator that changes colour 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 mol/L (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 = (1.0x10–14)/(5.9x10–4) = 1.69491525x10–11 Concentration, mol/L (CH3)2NH2+(aq) + H2O(l)  (CH3)2NH(aq) + H3O+(aq) Initial 0.25 mol/L __ 0 0 Change –x +x +x Equilibrium 0.25 – x x x   H3O  (CH3 )2 NH  x2 x2  Ka = 1.69491525x10–11 =  = = 0.25  x 0.25 (CH3 )2 NH 2    x = [H3O+] = 2.0584674x10–6 mol/L pH = –log [H+] = –log (2.0584674x10–6) = 5.686456 = 5.69 Methyl red is an indicator that changes colour 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 colour around pH 7.

17.57

a) Determine the Kb (of the conjugate base) from the Ka for C6H5COOH. Kb = Kw/Ka = (1.0x10–14)/(6.3x10–5) = 1.5873x10–10 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 C 6H5COO– = 0.125 mol/L. C6 H5 COOH  OH   (x) (x) (x) (x) –10 Kb = 1.5873x10 = = =  (0.125  x) (0.125) C6 H5 COO    – –6 [OH ] = x = 4.4543518x10 mol/L pOH = –log (4.4543518x10–6) = 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 mol/L NH3 solution at the equivalence point. Use the Kb for NH3 from the Appendix.  NH 4    OH   (x) (x) (x) (x)    Kb = 1.76x10–5 = = = (0.25  x) (0.25) NH3  – –3 [OH ] = x = 2.0976177x10 mol/L pOH = –log (2.0976177x10–3) = 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.

17.58

Plan: The reaction occurring in the titration is the neutralization of H 3O+ (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. Since they are the same concentration, the equivalence point occurs when 40.00 mL of base has been added. Use (c)(V) to determine the amount (mol) of acid and base. Note that the NaCl product is a neutral salt that does not affect the pH. Solution: The initial amount (mol) of HCl = (0.1000 mol HCl/L)(10 –3 L/1 mL)(40.00 mL) = 4.000x10–3 mol HCl a) At 0 mL of base added, the concentration of hydronium ion equals the original concentration of HCl. pH = –log (0.1000 mol/L) = 1.0000 b) Determine the amount (mol) of NaOH added: amount (mol) of added NaOH = (0.1000 mol NaOH/L)(10 –3 L/1 mL)(25.00 mL) = 2.500x10–3 mol NaOH HCl(aq) + NaOH(aq)  H2O(l) + NaCl(aq) Initial 4.000x10–3 mol 2.500x10–3 mol – 0 Change –2.500x10–3 mol –2.500x10–3 mol – +2.500x10–3 mol Final 1.500x10–3 mol 0 2.500x10 –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 concentration (mol/L) of the excess HCl is (1.500x10–3 mol HCl)/(0.06500 L) = 0.023077 mol/L pH = –log (0.023077) = 1.6368 c) Determine the amount (mol) of NaOH added: amount (mol) of added NaOH = (0.1000 mol NaOH/L)(10 –3 L/1 mL)(39.00 mL) = 3.900x10–3 mol NaOH HCl(aq) + NaOH(aq)  H2O(l) + NaCl(aq) Initial 4.000x10–3 mol 4.900x10–3 mol – 0 Change –3.900x10–3 mol –3.900x10–3 mol – +3.900x10–3 mol Final 1.000x10–4 mol 0 3.900x10–3 mol –3 The volume of the solution at this point is [(40.00 + 39.00) mL](10 L/1 mL) = 0.07900 L The concentration (mol/L) of the excess HCl is (1.00x 10 –4mol HCl)/(0.07900 L) = 0.0012658 mol/L pH = –log (0.0012658) = 2.898

d) Determine the amount (mol) of NaOH added: amount (mol) of added NaOH = (0.1000 mol NaOH/L)(10 –3 L/1 mL)(39.90 mL) = 3.990x10–3 mol NaOH HCl(aq) + NaOH(aq)  H2O(l) + NaCl(aq) Initial 4.000x10–3 mol 3.990x10–3 mol – 0 Change –3.990x10–3 mol –3.990x10–3 mol – +3.990x10–3 mol –5 Final 1.000x10 mol 0 3.900x10 –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 concentration (mol/L) of the excess HCl is (1.0x10 –5 mol HCl)/(0.07990 L) = 0.000125156 mol/L pH = –log (0.000125156) = 3.903 e) Determine the amount (mol) of NaOH added: amount (mol) of added NaOH = (0.1000 mol NaOH/L)(10 –3 L/1 mL)(40.00 mL) = 4.000x10–3 mol NaOH HCl(aq) + NaOH(aq)  H2O(l) + NaCl(aq) Initial 4.000x10–3 mol 4.000x10–3 mol – 0 Change –4.000x10–3 mol –4.000x10–3 mol – +4.000x10–3 mol Final 0 0 4.000x10 –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 amount (mol) of NaOH added: amount (mol) of added NaOH = (0.1000 mol NaOH/L)(10 –3 L/1 mL)(40.10 mL) = 4.010x10–3 mol NaOH The HCl will react with an equal amount of the base, and 1.0x10 –5 mol NaOH will remain. HCl(aq) + NaOH(aq)  H2O(l) + NaCl(aq) Initial 4.000x10–3 mol 4.010x10–3 mol – 0 Change –4.000x10–3 mol –4.000x10–3 mol – +4.000x10–3 mol Final 0 1.000x10 –5 mol 4.000x10 –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 concentration (mol/L) of the excess NaOH is (1.0x10 –5 mol NaOH)/(0.08010 L) = 0.00012484 mol/L pOH = –log (0.00012484) = 3.9036 pH = 14.00 – pOH = 14.00 – 3.9036 = 10.09637 = 10.10 g) Determine the amount (mol) of NaOH added: amount (mol) of NaOH = (0.1000 mol NaOH/L)(10–3 L/1 mL)(50.00 mL) = 5.000x10 –3 mol NaOH The HCl will react with an equal amount of the base, and 1.000x10–3 mol NaOH will remain. HCl(aq) + NaOH(aq)  H2O(l) + NaCl(aq) Initial 4.000x10–3 mol 5.000x10–3 mol – 0 Change –4.000x10–3 mol –4.000x10–3 mol – +4.000x10–3 mol Final 0 1.000x10 –3 mol 4.000x10 –3 mol –3 The volume of the solution at this point is [(40.00 + 50.00) mL](10 L/ 1 mL) = 0.09000 L The concentration (mol/L) of the excess NaOH is (1.000x10 –3 mol NaOH)/(0.09000 L) = 0.011111 mol/L pOH = –log (0.011111) = 1.95424 pH = 14.00 – pOH = 14.00 – 1.95424 = 12.04576 = 12.05 17.59

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 amount (mol) of KOH = (0.1000 mol KOH/L)(10 –3 L/1 mL)(30.00 mL) = 3.000x10 –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 mol/L) = 1.0000 pH = 14.00 – pOH = 14.00 – 1.0000 = 13.00 b) Determine the amount (mol) of HBr added: amount (mol) of added HBr = (0.1000 mol HBr/L)(10 –3 L/1 mL)(15.00 mL) = 1.500x10 –3 mol HBr The HBr will react with an equal amount of the base, and 1.500x10 –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 concentration (mol/L)of the excess KOH is (1.500x10–3 mol KOH)/(0.04500 L) = 0.03333 mol/L pOH = –log (0.03333) = 1.4772 pH = 14.00 – pOH = 14.00 – 1.4772 = 12.5228 = 12.52 c) Determine the amount (mol) of HBr added: amount (mol) of added HBr = (0.1000 mol HBr/L)(10 –3 L/1 mL)(29.00 mL) = 2.900x10 –3 mol HBr The HBr will react with an equal amount of the base, and 1.00x10 –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 concentration (mol/L)of the excess KOH is (1.00x10–4 mol KOH)/(0.05900 L) = 0.0016949 mol/L pOH = –log (0.0016949) = 2.7708559 pH = 14.00 – pOH = 14.00 – 2.7708559 = 11.2291441 = 11.23 d) Determine the amount (mol) of HBr added: amount (mol) of added HBr = (0.1000 mol HBr/L)(10–3 L/1 mL)(29.90 mL) = 2.990x10 –3 mol HBr The HBr will react with an equal amount of the base, and 1.0x10 –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 concentration(mol/L) of the excess KOH is (1.0x10–5 mol KOH)/(0.05990 L) = 0.000166945 mol/L pOH = –log (0.000166945) = 3.7774266 pH = 14.00 – pOH = 14.00 – 3.7774266 = 10.2225734 = 10.2 e) Determine the amount (mol) of HBr added: amount (mol) of added HBr = (0.1000 mol HBr/L)(10 –3 L/1 mL)(30.00 mL) = 3.000x10 –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 acid after reacting with the KOH. The excess strong acid will give the pH. Determine the amount (mol) of HBr added: mount (mol) of added HBr = (0.1000 mol HBr/L)(10 –3 L/1 mL)(30.10 mL) = 3.010x10–3 mol HBr The HBr will react with an equal amount of the base, and 1.0x10 –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 concentration (mol/L) of the excess HBr is (1.0x10–5 mol HBr)/(0.06010 L) = 0.000166389 mol/L pH = –log (0.000166389) = 3.778875 = 3.8 g) Determine the amount (mol) of HBr added: amount (mol) of added HBr = (0.1000 mol HBr/L)(10 –3 L/1 mL)(40.00 mL) = 4.000x10 –3 mol HBr The HBr will react with an equal amount of the base, and 1.000x10 –3 mol HBr will remain. The volume of the solution at this point is [(30.00 + 40.00) mL](10 –3 L/1 mL) = 0.07000 L The concentration (mol/L) of the excess HBr is (1.000x10–3 mol HBr)/(0.07000 L) = 0.0142857 mol/L pH = –log (0.0142857) = 1.845098 = 1.85 17.60

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, amount (mol) acid = amount (mol) 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 [H 3O+] is dependent on the dissociation of butanoic acid: HBut + H2O  H3O+ + But– Initial 0.1000 mol/L 0 0 Change –x +x +x Equilibrium 0.1000 – x x x

Ka = 1.54x10–5 =

 H 3 O    But   x2 x2    = = 0.1000  x 0.1000  HBut 

x = [H3O+] = 1.2409674x10–3 mol/L pH = –log [H+] = –log (1.2409674x10–3) = 2.9062 = 2.91 b) The initial amount(mol) of HBut = (c)(V) = (0.1000 mol HBut/L)(10–3 L/1 mL)(20.00 mL) = 2.000x10–3 mol HBut Determine the amount (mol)of NaOH added: Amount(mol)of added NaOH = (0.1000 mol NaOH/L)(10–3 L/1 mL)(10.00 mL) = 1.000x10 –3 mol NaOH The NaOH will react with an equal amount of the acid, and 1.000x10 –3 mol HBut will remain. An equal amount (mol) of But– will form. HBut(aq) + NaOH(aq)  H2O(l) + But–(aq) + Na+(aq) –3 –3 Initial 2.000x10 mol 1.000x10 mol – 0 – Change –1.000x10–3 mol –1.000x10–3 mol – +1.000x10–3 mol – Final 1.000x10–3 mol 0 1.000x10 –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 concentration (mol/L) of the excess HBut is (1.000x10–3 mol HBut)/(0.03000 L) = 0.03333 mol/L The concentration (mol/L) of the But– formed is (1.000x10–3 mol But– /(0.03000 L) = 0.03333 mol/L Using a reaction table for the equilibrium reaction of HBut: HBut + H2O  H 3O + + But– Initial 0.03333 mol/L – 0 0.03333 mol/L Change –x +x +x Equilibrium 0.03333 – x x 0.03333 + x  H 3 O    But   x  0.0333  x  x  0.03333    –5 Ka = 1.54x10 = = = 0.03333  x 0.03333  HBut  x = [H3O+] = 1.54x10–5 mol/L pH = –log [H+] = –log (1.54x10–5) = 4.812479 = 4.81 c) Determine the amount (mol) of NaOH added: amount (mol) of added NaOH = (0.1000 mol NaOH/L)(10 –3 L/1 mL)(15.00 mL) = 1.500x10–3 mol NaOH The NaOH will react with an equal amount of the acid, and 5.00x10 –4 mol HBut will remain, and 1.500x10–3 moles of But– will form. HBut(aq) + NaOH(aq)  H2O(l) + But–(aq) + Na+(aq) –3 –3 Initial 2.000x10 mol 1.500x10 mol – 0 – Change –1.500x10–3 mol –1.500x10–3 mol – +1.500x10–3 mol – Final 5.000x10–4 mol 0 1.500x10 –3 mol –3 The volume of the solution at this point is [(20.00 + 15.00) mL](10 L/1 mL) = 0.03500 L The concentration (mol/L) of the excess HBut is (5.00x10–4 mol HBut)/(0.03500 L) = 0.0142857 mol/L The concentration (mol/L) of the But– formed is (1.500x10–3 mol But–)/(0.03500 L) = 0.042857 mol/L Using a reaction table for the equilibrium reaction of HBut: HBut + H2O  H3O+ + But– Initial 0.0142857 mol/L – 0 0.042857 mol/L Change –x +x +x Equilibrium 0.0142857 – x x 0.042857 + x  H 3 O    But   x  0.042857  x  x  0.042857     Ka = 1.54x10–5 = = = 0.0142857  x 0.0142857  HBut  x = [H3O+] = 5.1333x10–6 mol/L pH = –log [H+] = –log (5.1333x10–6) = 5.2896 = 5.29 d) Determine the amount (mol) of NaOH added: amount (mol) of added NaOH = (0.1000 mol NaOH/L)(10 –3 L/1 mL)(19.00 mL) = 1.900x10–3 mol NaOH The NaOH will react with an equal amount of the acid, and 1.00x10–4 mol HBut will remain, Silberberg, Amateis, Venkateswaran, Chen, Chemistry: The Molecular Nature of Matter and Change, 3rd Canadian Edition Page 3-619 Instructor‘s Solution Manual © Copyright 2021 McGraw-Hill Ryerson Ltd.

and 1.900x10–3 moles of But– will form. HBut(aq) + NaOH(aq)  H2O(l) + But–(aq) + Na+(aq) –3 –3 Initial 2.000x10 mol 1.900x10 mol – 0 – Change –1.900x10–3 mol –1.900x10–3 mol – +1.900x10–3 mol – Final 1.000x10–4 mol 0 1.900x10–3 mol –3 The volume of the solution at this point is [(20.00 + 19.00) mL](10 L/1 mL) = 0.03900 L The concentration (mol/L) of the excess HBut is (1.00x10–4 mol HBut)/(0.03900 L) = 0.0025641 mol/L The concentration (mol/L) of the But– formed is (1.900x10–3 mol But–)/(0.03900 L) = 0.0487179 mol/L Using a reaction table for the equilibrium reaction of HBut: HBut + H2O  H 3O + + But– Initial 0.0025641 mol/L – 0 0.0487179 mol/L Change –x – +x +x Equilibrium 0.0025641 – x +x 0.0487179 + x  H 3 O    But   x  0.0487179  x  x  0.0487179     Ka = 1.54x10–5 = = = 0.0025641  x 0.0025641  HBut  x = [H3O+] = 8.1052632x10–7 mol/L pH = –log [H+] = –log (8.1052632x10–7) = 6.09123 = 6.09 e) Determine the amount (mol) of NaOH added: amount (mol) of addedNaOH = (0.1000 mol NaOH/L)(10 –3 L/1 mL)(19.95 mL) = 1.995x10–3 mol NaOH The NaOH will react with an equal amount of the acid, and 5x10 –6 mol HBut will remain, and 1.995x10–3 moles of But– will form. HBut(aq) + NaOH(aq)  H2O(l) + But–(aq) + Na+(aq) –3 –3 Initial 2.000x10 mol 1.995x10 mol – 0 – Change –1.995x10–3 mol –1.995x10–3 mol – +1.995x10–3 mol – Final 5.000x10–6 mol 0 1.995x10 –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 concentration (mol/L) of the excess HBut is (5x10 –6 mol HBut)/(0.03995 L) = 0.000125156 mol/L The concentration (mol/L) of the But– formed is (1.995x10–3 mol But–)/(0.03995 L) = 0.0499374 mol/L Using a reaction table for the equilibrium reaction of HBut: HBut + H2O  H3O+ + But– Initial 0.000125156 mol/L – 0 0.0499374 mol/L Change –x – +x +x Equilibrium 0.000125156 – x x 0.0499374 + x  H 3 O    But   x  0.0499374  x  x  0.0499374     –5 Ka = 1.54x10 = = = 0.000125156  x 0.000125156  HBut  x = [H3O+] = 3.859637x10–8 mol/L pH = –log [H+] = –log (3.859637x10–8) = 7.41345 = 7.41 f) Determine the amount (mol) of NaOH added: amount (mol) of added NaOH = (0.1000 mol NaOH/L)(10–3 L/1 mL)(20.00 mL) = 2.000x10–3 mol NaOH The NaOH will react with an equal amount of the acid, and 0 mol HBut will remain, and 2.000x10–3 moles of But– will form. This is the equivalence point. HBut(aq) + NaOH(aq)  H2O(l) + But–(aq) + Na+(aq) –3 –3 Initial 2.000x10 mol 2.000x10 mol – 0 – Change –2.000x10–3 mol –2.000x10–3 mol – +2.000x10–3 mol – Final 0 0 2.000x10 –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 concentration (mol/L) of the But– formed is (2.000x10–3 mol But–)/(0.04000 L) = 0.05000 mol/L Kb = Kw/Ka = (1.0x10–14)/(1.54x10–5) = 6.49351x10–10 Using a reaction table for the equilibrium reaction of But –: Silberberg, Amateis, Venkateswaran, Chen, Chemistry: The Molecular Nature of Matter and Change, 3rd Canadian Edition Page 3-620 Instructor‘s Solution Manual © Copyright 2021 McGraw-Hill Ryerson Ltd.

But– + Initial 0.05000 mol/L Change –x Equilibrium 0.05000 – x Kb = 6.49351x10–10 =

H2O



HBut

–

OH–

–

 HBut  OH 

+x x =

0 +x x

(x) (x) (x) (x) = (0.05000  x) (0.05000)

 But     [OH–] = x = 5.6980304x10–6 mol/L pOH = –log (5.6980304x10–6) = 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 amount (mol) of NaOH added: amount (mol) of added NaOH = (0.1000 mol NaOH/L)(10 –3 L/1 mL)(20.05 mL) = 2.005x10–3 mol NaOH The NaOH will react with an equal amount of the acid, 0 mol HBut will remain, and 5x10–6 moles of NaOH will be in excess. There will be 2.000x10–3 mol of But– produced, but this weak base will not affect the pH compared to the excess strong base, NaOH. HBut(aq) + NaOH(aq)  H2O(l) + But–(aq) + Na+(aq) –3 –3 Initial 2.000x10 mol 2.005x10 mol – 0 – Change –2.000x10–3 mol –2.000x10–3 mol – +2.000x10–3 mol – Final 0 5.0x10 –6 mol 2.000x10 –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 concentration (mol/L) of the excess OH– is (5x10–6 mol OH–)/(0.04005 L) = 1.2484x10–4 mol/L pOH = –log (1.2484x10–4) = 3.9036 pH = 14.00 – pOH = 14.00 – 3.9036 = 10.0964 = 10.10 h) Determine the amount (mol) of NaOH added: amount (mol) of added NaOH = (0.1000 mol NaOH/L)(10 –3 L/1 mL)(25.00 mL) = 2.500x10–3 mol NaOH The NaOH will react with an equal amount of the acid, 0 mol HBut will remain, and 5.00x10–4 moles of NaOH will be in excess. HBut(aq) + NaOH(aq)  H2O(l) + But–(aq) + Na+(aq) –3 –3 Initial 2.000x10 mol 2.500x10 mol – 0 – Change –2.000x10–3 mol –2.000x10–3 mol – +2.000x10–3 mol – Final 0 5.00x10 –4 mol 2.000x10 –3 mol –3 The volume of the solution at this point is [(20.00 + 25.00) mL](10 L/1 mL) = 0.04500 L The concentration (mol/L) of the excess OH– is (5.00x10–4 mol OH–/(0.04500 L) = 1.1111x10–2 mol/L pOH = –log (1.1111x10–2) = 1.9542 pH = 14.00 – pOH = 14.00 – 1.9542 = 12.0458 = 12.05 17.61

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, amount (mol) acid = amount (mol) base. Addition of acid beyond the equivalence point is simply the addition of excess H3O+. The initial amount (mol) of (CH3CH2)3N = (0.1000 mol (CH3CH2)3N)/L)(10–3 L/1 mL)(20.00 mL) = 2.000x10–3 mol (CH3CH2)3N a) Since no acid has been added, only the weak base (Kb) is important.  CH 3CH 2  NH +  OH   (x) (x) (x) (x) 3   = Kb = 5.2x10–4 =  = (0.1000) (0.1000  x)  CH 3CH 2  N  3   No approximations can be made – solve with the quadratic. Expanding and collecting,

x 2  5.2  104 x  5.2 10 5  0 b  b 2  4ac 5.2  10   5.2  10   4  5.2 10  x   6.9558  103 mol / L 2a 2 [OH–] = x = 6.9558x10–3 mol/L pOH = –log (6.9558 x10–3) = 2.1576 pH = 14.00 – pOH = 14.00 – 2.1576 = 11.84235 = 11.84 b) Determine the amount (mol) of HCl added: amount (mol) of added HCl = (0.1000 mol HCl/L)(10 –3 L/1 mL)(10.00 mL) = 1.000x10 –3 mol HCl The HCl will react with an equal amount of the base, and 1.000x10–3 mol (CH3CH2)3N will remain; an equal amount (mol) 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 concentration (mol/L) of the excess (CH3CH2)3N is (1.000x10–3 mol (CH3CH2)3N)/(0.03000 L) = 0.03333 mol/L The concentration (mol/L) of the (CH3CH2)3NH+ formed is (1.000x10–3 mol (CH3CH2)3NH+)/(0.03000 L) = 0.03333 mol/L  CH 3CH 2  NH   OH   3 –4   = (x) (0.0333  x) = (x) (0.0333) Kb = 5.2x10 =  (0.03333) (0.03333  x)  CH 3CH 2  N  3   – –4 [OH ] = x = 5.2x10 mol/L pOH = –log (5.2x10–4) = 3.283997 pH = 14.00 – pOH = 14.00 – 3.283997 = 10.7160 = 10.72 c) Determine the amount (mol) of HCl added: amount (mol) of added HCl = (0.1000 mol HCl/L)(10 –3 L/1 mL)(15.00 mL) = 1.500x10 –3 mol HCl The HCl will react with an equal amount of the base, and 5.00x10 –4 mol (CH3CH2)3N will remain; and 1.500x10–3 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 concentration (mol/L) of the excess (CH3CH2)3N is (5.00x10–4 mol (CH3CH2)3N)/(0.03500 L) = 0.0142857 mol/L The concentration (mol/L) of the (CH3CH2)3NH+ formed is (1.500x10–3 mol (CH3CH2)3NH+)/(0.03500 L) = 0.0428571 mol/L  CH 3CH 2  NH   OH   3   = (x) (0.0428571 x) = (x) (0.0428571) Kb = 5.2x10–4 =  (0.0142857) (0.0142857  x)  CH 3CH 2  N  3   – –4 [OH ] = x = 1.7333x10 mol/L pOH = –log (1.7333x10–4) = 3.761126 pH = 14.00 – pOH = 14.00 – 3.761126 = 10.23887 = 10.24 d) Determine the amount (mol) of HCl added: amount (mol) of added HCl = (0.1000 mol HCl/L)(10 –3 L/1 mL)(19.00 mL) = 1.900x10 –3 mol HCl The HCl will react with an equal amount of the base, and 1.00x10 –4 mol (CH3CH2)3N will remain; and 1.900x10–3 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 concentration (mol/L) of the excess (CH3CH2)3N is (1.00x10–4 mol (CH3CH2)3N)/(0.03900 L) = 0.002564103 mol/L The concentration (mol/L) of the (CH3CH2)3NH+ formed is (1.900x10–3 mol (CH3CH2)3NH+)/(0.03900 L) = 0.0487179 mol/L  CH 3CH 2  NH   OH   3   = (x) (0.0487179  x) = (x) (0.0487179) Kb = 5.2x10–4 =  (0.002564103) (0.002564103  x)  CH 3CH 2  N  3   [OH–] = x = 2.73684x10–5 mol/L pOH = –log (2.73684x10–5) = 4.56275 pH = 14.00 – pOH = 14.00 – 4.56275 = 9.43725 = 9.44 e) Determine the amount (mol) of HCl added: amount (mol) of added HCl = (0.1000 mol HCl/L)(10 –3 L/1 mL)(19.95 mL) = 1.995x10 –3 mol HCl 4

4 2

5

The HCl will react with an equal amount of the base, and 5x10–6 mol (CH3CH2)3N will remain; and 1.995x10–3 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 concentration (mol/L) of the excess (CH3CH2)3N is (5x10–6 mol (CH3CH2)3N)/(0.03995 L) = 0.000125156 mol/L The concentration (mol/L) of the (CH3CH2)3NH+ formed is (1.995x10–3 mol (CH3CH2)3NH+)/(0.03995 L) = 0.0499374 mol/L  CH 3CH 2  NH   OH   3   = (x) (0.0499374  x) = (x) (0.0499374) Kb = 5.2x10–4 =  (0.000125156) (0.000125156  x)  CH 3CH 2  N  3   – –6 [OH ] = x = 1.303254x10 mol/L pOH = –log (1.303254x10–6) = 5.88497 pH = 14.00 – pOH = 14.00 – 5.88497 = 8.11503 = 8.1 f) Determine the amount (mol) of HCl added: amount (mol) of added HCl = (0.1000 mol HCl/L)(10–3 L/1 mL)(20.00 mL) = 2.000x10 –3 mol HCl The HCl will react with an equal amount of the base, and 0 mol (CH 3CH2)3N will remain; and 2.000x10–3 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 concentration (mol/L)of the (CH3CH2)3NH+ formed is (2.000x10–3 mol (CH3CH2)3NH+)/(0.04000 L) = 0.05000 mol/L Ka = Kw/Kb = (1.0x10–14)/(5.2x10–4) = 1.9231x10–11  H3O   CH3CH 2  N  (x) (x) (x) (x) 3   –11 Ka = 1.9231x10 =  = =  (0.05000  x) (0.05000)  CH3CH 2  NH  3   + –7 x = [H3O ] = 9.80587x10 mol/L pH = –log [H+] = –log (9.80587x10–7) = 6.0085 = 6.01 g) After the equivalence point, the excess strong acid is the primary factor influencing the pH. Determine the amount (mol) of HCl added: amount (mol) of added HCl = (0.1000 mol HCl/L)(10 –3 L/1 mL)(20.05 mL) = 2.005x10–3 mol HCl The HCl will react with an equal amount of the base, and 0 mol (CH 3CH2)3N will remain, and 5x10–6 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 concentration (mol/L) of the excess H+ is (5x10–6 mol H+)/(0.04005 L) = 1.2484x10–4 mol/LM pH = –log (1.2484x10–4) = 3.9036 = 3.90 h) Determine the amount (mol) of HCl added: amount (mol) of added HCl = (0.1000 mol HCl/L)(10 –3 L/1 mL)(25.00 mL) = 2.500x10 –3 mol HCl The HCl will react with an equal amount of the base, and 0 mol (CH 3CH2)3N will remain, and 5.00x10–4 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 concentration (mol/L) of the excess H+ is (5.00x10–4 mol H+)/(0.04500 L) = 1.1111x10–2 mol/L pH = –log (1.1111x10–2) = 1.9542 = 1.95 17.62

Plan: Use (c)(V) to find the initial amount (mol) of acid and then use the mole ratio in the balanced equation to find amount (mol) of base; dividing amount (mol) of base by the concentration (mol/L) 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    1 mL  L  0.0520 mol CH3 COOH   103 L    42.2 mL      3     L    1 mL   1 mol CH3 COOH   0.0372 mol NaOH   10 L  = 58.989247 mL= 59.0 mL NaOH Determine the amount (mol) of initially CH3COOH present:  0.0520 mol CH3COOH   103 L  amount (mol) of CH3COOH =    42.2 mL  = 0.0021944 mol CH3COOH   L    1 mL  At the equivalence point, 0.0021944 mol NaOH will be added so the amount (mol) acid = amount (mol) 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. CH3COOH(aq) + NaOH(aq)  H2O(l) + CH3COO–(aq) + Na+(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 volume of solution in litre 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: concentration (mol/L) = (0.0021944 mol CH3COO–)/(0.101189247 L) = 0.0216861 mol/L Calculate Kb for CH3COO–: Ka CH3COOH = 1.8x10–5 –14 Kb = Kw/Ka = (1.0x10 )/(1.8x10–5) = 5.556x10–10 Using a reaction table for the equilibrium reaction of CH 3COO–: CH3COO– + H2O  CH3COOH + OH– Initial 0.0216861 mol/L – 0 0 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. CH3COOH   OH   (x) (x) (x) (x) –10 Kb = 5.556x10 = = =  (0.0216861) (0.0216861  x) CH3COO    – –6 [OH ] = x = 3.471138x10 mol/L pOH = –log (3.471138x10–6) = 5.459528 pH = 14.00 – pOH = 14.00 – 5.459528 = 8.54047 = 8.54

b) The balanced chemical equations are: NaOH(aq) + H2SO3(aq)  Na+(aq) + HSO3–(aq) + H2O(l) NaOH(aq) + HSO3–(aq)  Na+(aq) + SO32–(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    1 mL  L  0.0850 mol H 2SO3   103 L    28.9 mL      3     L    1 mL   1 mol H 2SO3   0.0372 mol NaOH   10 L  = 66.034946 mL= 66.0 mL NaOH It will require an equal volume to reach the second equivalence point for a total of 2 x 66.034946 mL = 132.1 mL. Determine the amount (mol) of HSO3– produced:  1 mol HSO3   0.0850 mol H 2SO3   103 L  – amount (mol) of HSO3– =   = 0.0024565 mol HSO3   28.9 mL     L 1 mL 1 mol H SO   2 3    An equal amount (mol) of SO32– will be present at the second equivalence point. Determine the volume of solution in litre present at the first equivalence point: Volume = [(28.9 + 66.034946) mL](10 –3 L/1 mL) = 0.094934946 L Determine the volume of solution in litre present at the second equivalence point: Volume = [(28.9 + 66.034946 + 66.034946) mL](10 –3 L/1 mL) = 0.160969892 L Concentration of HSO3– at equivalence point: Silberberg, Amateis, Venkateswaran, Chen, Chemistry: The Molecular Nature of Matter and Change, 3rd Canadian Edition Page 3-624 Instructor‘s Solution Manual © Copyright 2021 McGraw-Hill Ryerson Ltd.

concentration (mol/L) = (0.0024565 moles HSO3–)/(0.094934946 L) = 0.0258756 mol/LM Concentration of SO32– at equivalence point: concentration (mol/L) = (0.0024565 moles SO32–)/(0.160969892 L) = 0.0152606 mol/L Calculate Kb for HSO3–: Ka H2SO3 = 1.4x10–2 –14 Kb = Kw/Ka = (1.0x10 )/(1.4x10–2) = 7.142857x10–13 Calculate Kb for SO32–: Ka HSO3– = 6.5x10–8 –14 Kb = Kw/Ka = (1.0x10 )/(6.5x10–8) = 1.53846x10–7 For the first equivalence point: Using a reaction table for the equilibrium reaction of HSO3–: HSO3– + H2O  H2SO3 + OH– Initial 0.0258756 mol/L 0 0 Change –x +x +x Equilibrium 0.0258756 – x x x Determine the hydroxide ion concentration from the Kb, and then determine the pH from the pOH.  H 2SO3  OH  (x) (x) (x) (x) –13 Kb = 7.142857x10 = = =  (0.0258756  x) (0.0258756)  HSO3    [OH–] = x = 1.359506x10–7 mol/L pOH = –log (1.359506x10–7) = 6.8666189 pH = 14.00 – pOH = 14.00 – 6.8666189 = 7.13338 = 7.13 For the second equivalence point: Using a reaction table for the equilibrium reaction of SO32–: SO32– + H2O  HSO3– + OH– Initial 0.0152606 mol/L 0 0 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.  HSO3   OH   (x) (x) (x) (x)   = Kb = 1.53846x10–7 =  = (0.0152606) (0.0152606  x) SO32     [OH–] = x = 4.84539x10–5 mol/L pOH = –log (4.84539x10–5) = 4.31467 pH = 14.00 – pOH = 14.00 – 4.31467 = 9.68533 = 9.69 17.63

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 KOH needed: Volume (mL) of KOH =  1 mol KOH    1 mL  L  0.0390 mol HNO2   103 L     23.4 mL      3    L    1 mL   1 mol HNO2   0.0588 mol KOH   10 L  = 15.5204 mL= 15.5 mL KOH Determine the amount (mol) of HNO2 present:  0.0390 mol HNO2   103 L  amount (mol) of HNO2 =    1 mL   23.4 mL  = 0.0009126 mol HNO2 L    At the equivalence point, 0.0009126 mol KOH will be added so the amount (mol) acid = amount (mol) base. The KOH will react with an equal amount of the acid, 0 mol HNO 2 will remain, and 0.0009126 moles of NO2– will be formed. HNO2(aq) + KOH(aq)  H2O(l) + NO2–(aq) + K+(aq) Initial 0.0009126 mol 0.0009126 mol – 0 – Change –0.0009126 mol –0.0009126 mol – +0.0009126 mol – Final 0 0 0.00 09126mol

Determine the volume of solution in litre 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: concentration (mol/L) = (0.0009126 mol NO2–)/(0.0389204 L) = 0.023447858 mol/L Calculate Kb for NO2–: Ka HNO2 = 7.1x10–4 –14 Kb = Kw/Ka = (1.0x10 )/(7.1x10–4) = 1.40845x10–11 Using a reaction table for the equilibrium reaction of NO 2–: NO2– + H 2O  HNO2 + OH– Initial 0.023447858 mol/L – 0 0 Change –x +x +x Equilibrium 0.023447858 – x x x Determine the hydroxide ion concentration from the Kb, and then determine the pH from the pOH.  HNO2  OH  (x) (x) (x) (x) Kb = 1.40845x10–11 = = = (0.023447858) (0.023447858  x)  NO2    [OH–] = x = 5.7468x10–7 mol/L pOH = –log (5.7468x10–7) = 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    1 mL  L  0.130 mol H 2 CO3   103 L   17.3 mL      3     L    1 mL   1 mol H 2 CO3   0.0588 mol KOH   10 L  = 38.248299 mL= 38.2 mL KOH It will require an equal volume to reach the second equivalence point (76.4 mL). Determine the amount (mol) of HCO3– produced:  1 mol HCO3   0.130 mol H 2 CO3   103 L  – amount (mol) =  17.3 mL   = 0.002249 mol HCO3       L 1 mL 1 mol H CO   2 3    An equal amount (mol) of CO32– will be present at the second equivalence point. Determine the volume of solution in litre present at the first equivalence point: Volume = [(17.3 + 38.248299) mL](10 –3 L/1 mL) = 0.055548 L Determine the volume of solution in litre 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: concentration (mol/L) = (0.002249 mol HCO3–)/(0.055548 L) = 0.0404875 mol/L Concentration of CO32– at equivalence point: concentration (mol/L) = (0.002249 mol CO32–)/(0.0937966 L) = 0.023977 mol/L Calculate Kb for HCO3–: Ka H2CO3 = 4.5x10–7 –14 Kb = Kw/Ka = (1.0x10 )/(4.5x10–7) = 2.222x10–8 Calculate Kb for CO32–: Ka HCO3– = 4.7x10–11 –14 Kb = Kw/Ka = (1.0x10 )/(4.7x10–11) = 2.1276596x10–4 Determine the hydroxide ion concentration from the Kb, and then determine the pH from the pOH. For the first equivalence point: H 2 CO3  OH   (x) (x) (x) (x) –8 Kb = 2.222x10 = = =  (0.0404875) (0.0404875  x)  HCO3    – –5 [OH ] = x = 2.999387x10 mol/L pOH = – log (2.999387x10–5) = 4.522967495 pH = 14.00 – pOH = 14.00 – 4.522967495 = 9.4770 = 9.48

For the second equivalence point:  HCO3  OH   (x) (x) (x) (x)   = Kb = 2.1276596x10–4 =  = (0.023977  x) (0.023977) CO32     No assumptions – use quadratic to solve [OH–] = x = 2.1547687x10–3 M pOH = – log (2.1547687x10–3) = 2.666599 pH = 14.00 – pOH = 14.00 – 2.666599= 11.3334 = 11.33 17.64

Plan: Use (c)(V) to find the initial amount (mol) of base and then use the mole ratio in the balanced equation to find amount (mol) of acid; dividing amount (mol) of acid by the concentration (mol/L) 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    1 mL  L  0.234 mol NH3   103 L    65.5 mL      3  = 122.616 mL= 123 mL HCl    L    1 mL   1 mol NH3   0.125 mol HCl   10 L  Determine the amount (mol) of NH3 present:  0.234 mol NH3   103 L  amount (mol) =   65.5 mL  = 0.015327 mol NH3    L    1 mL  At the equivalence point, 0.015327 mol HCl will be added so the amount (mol) acid = amount (mol) 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. HCl(aq) + NH3(aq)  NH4+(aq) + Cl–(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 volume of solution in litre present at the equivalence point: Volume = [(65.5 + 122.616) mL](10 –3 L/1 mL) = 0.188116 L Concentration of NH4+ at equivalence point: concentration (mol/L) = (0.015327 mol NH4+)/(0.188116 L) = 0.081476 mol/L Calculate Ka for NH4+: Kb NH3 = 1.76x10–5 –14 Ka = Kw/Kb = (1.0x10 )/(1.76x10–5) = 5.6818x10–10 Using a reaction table for the equilibrium reaction of NH 4+: NH4+ + H2O  NH3 + H 3 O+ Initial 0.081476 mol/L – 0 0 Change –x +x +x Equilibrium 0.081476 – x x x Determine the hydrogen ion concentration from the Ka, and then determine the pH.  H3O   NH3  (x) (x) (x) (x)  –10 Ka= 5.6818x10 =  = =  (0.081476  x) (0.081476)  NH 4    x = [H3O+] = 6.803898x10–6 mol/L pH = –log [H+] = –log (6.803898x10–6) = 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 =  1 mol HCl    1 mL  L  1.11 mol CH3 NH 2   103 L    21.8 mL      3     L    1 mL   1 mol CH3 NH 2   0.125 mol HCl   10 L  = 193.584 mL= 194 mL HCl Determine the amount (mol) of CH3NH2 present:  1.11 mol CH3 NH 2   103 L  amount (mol) =   21.8 mL  = 0.024198 mol CH3NH2    L    1 mL  At the equivalence point, 0.024198 mol HCl will be added so the amount (mol) acid = amount (mol) 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. HCl(aq) + CH3NH2(aq)  CH3NH3+(aq) + Cl–(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 volume of solution in litre present at the equivalence point: Volume = [(21.8 + 193.584) mL](10 –3 L/1 mL) = 0.215384 L Concentration of CH3NH3+ at equivalence point: concentration (mol/L) = (0.024198 mol CH3NH3+)/(0.215384 L) = 0.1123482 mol/L Calculate Ka for CH3NH3+: Kb CH3NH2 = 4.4x10–4 –14 –4 Ka = Kw/Kb = (1.0x10 )/(4.4x10 ) = 2.2727x10–11 Using a reaction table for the equilibrium reaction of CH3NH3+: CH3NH3+ + H 2O  CH3NH2 + H3O+ Initial 0.1123482 mol/L – 0 0 Change –x +x +x Equilibrium 0.1123482 – x x x Determine the hydrogen ion concentration from the Ka, and then determine the pH.  H3O  CH3 NH 2  (x) (x) (x) (x)  –11 Ka = 2.2727x10 =  = =  (0.1123482) (0.1123482  x) CH3 NH3    x = [H3O+] = 1.5979x10–6 mol/L pH = –log [H+] = –log (1.5979x10–6) = 5.7964 = 5.80 17.65

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    1 mL  L  0.0750 mol C5 H5 N     3     2.65 L   L    1 mol C5 H5 N  0.447 mol HNO3   10 L  = 444.63087 mL= 445 mL HNO3 Determine the amount (mol) of C5H5NH+ produced:  1 mol C5 H5 NH    0.0750 mol C5 H5 N  amount (mol) of C5H5NH+ =  2.65 L      L    1 mol C5 H5 N  = 0.19875 mol C5H5NH+ Determine the volume of solution in litre 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: concentration (mol/L) = (0.19875 mol C5H5NH+)/(3.09463 L) = 0.064224 mol/L Calculate Ka for C5H5NH+: Kb C5H5N = 1.7x10–9

Ka = Kw/Kb = (1.0x10–14)/(1.7x10–9) = 5.88235x10–6 Determine the hydrogen ion concentration from the Ka, and then determine the pH.  H3O  C5 H5 N  (x) (x) (x) (x)  –6 Ka = 5.88235x10 =  = =  (0.064224  x) (0.064224) C5 H5 NH    x = [H3O+] = 6.1464x10–4 mol/L pH = –log [H+] = –log (6.1464x10–4) = 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 =     1 mL  1 mol HNO3 L  0.250 mol H 2 NCH 2 CH3 NH 2     3     0.188 L   L    1 mol H 2 NCH 2 CH3 NH 2  0.447 mol HNO3   10 L  = 105.1454 mL= 105 mL HCl It will require an equal volume to reach the second equivalence point. (210. mL) Determine the amount (mol) of H2NCH2CH2NH3+ produced: amount (mol) of H2NCH2CH2NH3+ =  1 mol H2 NCH2 CH3 NH3   0.250 mol H2 NCH2 CH3 NH2  0.188 L       L    1 mol H 2 NCH2 CH3 NH2  = 0.0470 mol H2NCH2CH2NH3+ An equal amount (mol) of H3NCH2CH2NH32+ will be present at the second equivalence point. Determine the volume of solution in litre present at the first equivalence point: Volume = 0.188 L + (105.1454 mL)(10–3 L/1 mL) = 0.293145 L Determine the volume of solution in litre 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: concentration (mol/L) = (0.0470 mol H2NCH2CH2NH3+)/(0.293145 L) = 0.16033 mol/L Concentration of H3NCH2CH2NH32+ at equivalence point: concentration (mol/L) = (0.0470 mol H3NCH2CH2NH32+)/(0.39829 L) = 0.11800 mol/L Calculate Ka for H2NCH2CH2NH3+: Kb H2NCH2CH2NH2 = 8.5x10–5 –14 –5 Ka = Kw/Kb = (1.0x10 )/(8.5x10 ) = 1.17647x10–10 Calculate Ka for H3NCH2CH2NH32+: Kb H2NCH2CH2NH3+ = 7.1x10–8 –14 –8 Ka = Kw/Kb = (1.0x10 )/(7.1x10 ) = 1.40845x10–7 Determine the hydrogen ion concentration from the Ka, and then determine the pH for the first equivalence point.  H3O  H 2 NCH 2 CH3 NH 2  (x) (x) (x) (x)  –10 Ka = 1.17647x10 =  = =  (0.16033  x) (0.16033)  H 2 NCH 2 CH3 NH3    + x = [H3O ] = 4.3430780x10–6 mol/L pH = –log [H+] = –log (4.3430780x10–6) = 5.36220 = 5.36 Determine the hydrogen ion concentration from the Ka, and then determine the pH for the second equivalence point.  H3O   H 2 NCH 2 CH3 NH3  (x) (x) (x) (x)   = Ka = 1.40845x10–7 =  = (0.11800  x) (0.11800)  H3 NCH 2 CH3 NH32     x = [H3O+] = 1.2891745x10–4 mol/L pH = –log [H+] = –log (1.2891745x10–4) = 3.889688 = 3.89 17.66

Plan: Write a balanced equation. Find the amount (mol) of H 2SO4 from the concentration (mol/L) and volume information and use the molar ratio in the balanced equation to find the amount (mol) of NaHCO 3 required to

react with that amount of H2SO4. Divide the amount (mol) of NaHCO3 by its concentration (mol/L) to find the volume. Solution: The reaction is: 2NaHCO3(aq) + H2SO4(aq)  Na2SO4(aq) + 2 H2O(l) + 2CO2(g)  3  2 . 6 m o l H S O 1 0 L   2 4 8 8 m L     Amount (mol) of H2SO4 = = 0.2288 mol H2SO4     1 m L L    

  2 m o l N a H C O 3 0 . 2 2 8 8 m o l H S O    Amount (mol) of NaHCO3 =  = 0.4576 mol NaHCO3 2 4 1 m o l H S O 2 4     1 L 1 m L 0 . 4 5 7 6 m o l N a H C O      Volume (mL) of NaHCO3 =  3  3   1 . 6 m o l N a H C O 1 0 L 3     = 286 mL= 2.9 x 102 mL NaHCO3

17.67

Plan: Write a balanced equation. Find the amount (mol) of HCl added from the concentration (mol/L) and volume information and use the molar ratio in the balanced equation to find the amount (mol) of NaOH present. Find the volume of NaOH from the difference in buret readings. Divide the amount (mol) of NaOH by its volume to determine the concentration. Solution: The reaction is: NaOH(aq) + HCl(aq)  NaCl(aq) + H2O(l)  3  1 0 L 0 . 1 5 2 8 m o l H C l   2 5 . 0 0 m L     Amount (mol) of HCl = = 0.00382 mol HCl     1 m L L    

  1 m o lN a O H 0 . 0 0 3 8 2 m o lH C l    Amount (mol) of NaOH =  = 0.00382 mol NaOH 1 m o l H C l   Volume (mL) of NaOH = final buret reading – initial buret reading = 39.21 mL – 2.24 mL = 36.97 mL NaOH 0 . 0 0 3 8 2 m o lN a O H 1 m L     Concentration (mol/L) of NaOH =    3 3 6 . 9 7 m L 1 0L     = 0.103327 mol/L = 0.1033 mol/L NaOH 17.68

Plan: Write balanced equations for the reaction of NaOH with oxalic acid, benzoic acid, and HCl. Find the amount (mol) of added NaOH from the concentration (mol/L) and volume information; then use the concentration (mol/L) and volume information for HCl to find the amount (mol) of HCl required to react with the excess NaOH. Use the molar ratio in the NaOH/HCl reaction to find the amount (mol) of excess NaOH. The amount (mol) 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)  3  1 0 L 0 . 1 0 0 0 m o l N a O H   1 0 0 . 0 m L    Amount (mol) of NaOH added =  = 0.01000 mol NaOH     1 m L 1 L      3  1 0 L 0 . 2 0 0 0 m o l H C l    = 0.004000 mol HCl   1 m L L   1   

2 0 . 0 0 m L   Amount (mol) of added HCl =  

1 m o l N a O H   0 . 0 0 4 0 0 0 m o l H C l  Amount (mol) of excess NaOH =  = 0.004000 mol NaOH 1  m o l H C l  

Amount (mol) of NaOH required to titrate sample = amount (mol) NaOH added – amount (mol) excess NaOH = 0.01000 mol – 0.004000 mol = 0.006000 mol NaOH Let x = mass of HC7H5O2 and 0.3471 – x = mass of H2C2O4 Amount (mol) of NaOH required to titrate HC7H5O2 =     1 m o l H C H O 1 m o l N a O H 7 5 2 x g H C H O       = 0.008189x 7 5 2 1 2 2 . 1 2 g H C H O 1 m o l H C H O 7 5 2 7 5 2     Amount (mol) of NaOH required to titrate H2C2O4 =     1 m o l H C O 2 m o l N a O H 2 2 4 ( 0 . 3 4 7 1  x ) g H C O       = 0.007710 – 0.02221x 2 2 4 9 0 . 0 4 g H C O 1 m o l H C O 2 2 4 2 2 4     Amount (mol) of NaOH required to titrate sample = 0.006000 mol = 0.008189x + (0.007710 – 0.02221x) 0.006000 = 0.007710 – 0.014021x –0.001710 = –0.014021x 0.12196 = x = mass of HC7H5O2 m a s s o f H C H O 0 . 1 2 1 9 6 g 7 5 2 1 0 0 % = 1 0 0 %     Mass % of HC7H5O2 = m = 35.1368 %= 35.14% a s s o f s a m p l e 0 . 3 4 7 1 g 17.69

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 amount (mol) of HF required to react with the amount (mol) of UO2. Divide amount (mol) of HF by its concentration (mol/L) to calculate the volume. Solution: The reaction is: UO2(s) + 4HF(aq)  UF4(s) + 2H2O(l) 3    1 m o l U O 1 0 g 2 2 . 1 5 k g U O      Amount (mol) of UO2 =  = 7.96296 mol UO2 2   1 k g2 7 0 . 0 g U O 2    

  4 m o lH F 7 . 9 6 2 9 6 m o lU O    Amount (mol) of HF =  = 31.85184 mol HF 2 1 m o lU O 2    1 L  3 1 . 8 5 1 8 4 m o lH F    Volume (L) of HF =  = 13.2716 L= 13.3 L HF 2 . 4 0 m o lH F   17.70

Plan: Write balanced reactions between HNO3 and each of the bases. Find the amount (mol) of HNO 3 from its concentration (mol/L) 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)  3  1 . 0 0 0 m o l H N O 1 0 L   3 1 7 . 3 0 m L    Amount (mol) of HNO3 =  = 0.0173 mol HNO3     1 m L 1 L     Let x = mass of Al(OH)3 and 0.4826 – x = mass of Mg(OH)2 Amount (mol) of HNO3 required to titrate Al(OH)3 =     1 m o l A l ( O H ) 3 m o l H N O 3 3 x g A l ( O H )       3 = 0.038462x 7 8 . 0 0 g A l ( O H ) 1 m o l A l ( O H ) 3 3     Amount (mol) of HNO3 required to titrate Mg(OH)2 =     2 m o l H N O 1 m o l M g ( O H ) 3 2 ( 0 . 4 8 2 6  x ) g M g ( O H )       2 5 8 . 3 3 g M g ( O H ) 1 m o l M g ( O H ) 2 2     = 0.01655 – 0.03429x Amount (mol) of HNO3 required to titrate sample = 0.0173 mol = 0.038462x + (0.01655 – 0.03429x)

0.0173 = 0.004172x + 0.01655 0.17977 g = x = mass of Al(OH)3 m a s s o f A l ( O H ) 0 . 1 7 9 7 7 g 3 1 0 0 % = 1 0 0 %     Mass % of Al(OH)3 = m = 37.2503% = 37.25% a s s o f s a m p l e 0 . 4 8 2 6 g 17.71

M2X(s)  2M+(aq) + X2–(aq) + 2 Ksp = [M ] [X2–], assuming M2X is a strong electrolyte. S = molar solubility = 5x10–5 mol/L [M+] = 2S = 1x10–4 mol/L [X2–] = S = 5x10–5 mol/L The actual Ksp is lower than the calculated value because the assumption that M 2X 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.

17.72

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.

17.73

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.

17.74

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.

17.75

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) + HS(aq) + OH(aq) Ion-product expression: Ksp = [Cu2+][HS][OH]

17.76

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) SnS(s) + H2O(l) Sn2+(aq) + HS(aq) + OH(aq) Ion-product expression: Ksp = [Sn2+][HS–][OH–]

17.77

Ion-product expression: Ksp = [Ca2+][CrO42] b) AgCN(s) Ag+(aq) + CN(aq) Ion-product expression: Ksp = [Ag+][CN] c) NiS(s) + H2O(l) Ni2+(aq) + HS(aq) + OH(aq) Ion-product expression: Ksp = [Ni2+][HS][OH] 17.78

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) CdS(s) + H2O(l) Cd2+(aq) + HS(aq) + OH(aq) Ion-product expression: Ksp = [Cd2+][HS][OH]

17.79

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 (mol/L) Ag2CO3(s)  2Ag+(aq) + CO32(aq) Initial — 0 0 Change — +2S +S Equilibrium — 2S S S = [Ag2CO3] = 0.032 mol/L so [Ag+] = 2S = 0.064 mol/L and [CO32] = S = 0.032 mol/L Ksp = [Ag+]2[CO32] = (0.064)2(0.032) = 1.31072x10–4 = 1.3x10–4

17.80

Write a reaction table, where S is the molar solubility of ZnC2O4: Concentration (mol/L) ZnC2O4(s)  Zn2+(aq) + C2O42(aq) Initial — 0 0 Change — +S +S Equilibrium — S S –3 2+ 2 S = [ZnC2O4] = 7.9x10 mol/L so [Zn ] = [C2O4 ] = S = 7.9x10–3 mol/L Ksp = [Zn2+][C2O42] = (7.9x10–3)( 7.9x10–3) = 6.241x10–5 = 6.2x10–5

17.81

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 concentration (mol/L), into the ion-expression and solve for Ksp. Solution: The solubility of Ag2Cr2O7, converted from g/100 mL to mol/L is:  8.3x103 g Ag2 Cr2 O7   1 mL   1 mol Ag2 Cr2 O7  Molar solubility = S =    3    = 0.00019221862 mol/L  100 mL    10 L   431.8 g Ag2 Cr2 O7  The equation for silver dichromate, Ag2Cr2O7, is: Concentration (mol/L) Ag2Cr2O7(s)  2Ag+(aq) + Cr2O72(aq) Initial — 0 0 Change — +2S +S Equilibrium — 2S S 2S = [Ag+] = 2(0.00019221862 mol/L) = 0.00038443724 mol/L S = [Cr2O72–] = 0.00019221862 mol/L Ksp = [Ag+]2[Cr2O72–] = (2S)2(S) = (0.00038443724)2(0.00019221862) = 2.8408x10–11 = 2.8x10–11

17.82

The equation and ion-product expression for calcium sulfate, CaSO4, is: CaSO4(s) Ca2+(aq) + SO42(aq) Ksp = [Ca2+][SO42] The solubility of CaSO4, converted from g/100 mL to mol/L is:

 0.209 g CaSO4   1 mL   1 mol CaSO 4  Molar solubility = S =   = 0.015350716 mol/L   3   100 mL    10 L   136.15 g CaSO 4  Since one mole of CaSO4 dissociates to form one mole of Ca2+, the concentration of Ca2+ is S = 0.015350716 mol/L. The concentration of SO42 is S = 0.015350716 mol/L because one mole of CaSO4 dissociates to form one mole of SO42. Ksp = [Ca2+][SO42] = (S)(S) = (0.015350716)( 0.015350716) = 2.35644x10 –4 = 2.36x10–4

17.83

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 [Sr 2+] 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: SrCO3(s) Sr2+(aq) + CO32(aq) Ksp = [Sr2+][CO32] 2+ The solubility, S, in pure water equals [Sr ] and [CO32–] Write a reaction table, where S is the molar solubility of SrCO3: Concentration (mol/L) SrCO3(s)  Sr2+(aq) + CO32(aq) Initial — 0 0 Change — +S +S Equilibrium — S S Ksp = 5.4x10–10 = [Sr2+][CO32] = [S][S] = S2 S = 2.32379x10–5 = 2.3x10–5 mol/L b) In 0.13 mol/L Sr(NO3)2, the initial concentration of Sr2+ is 0.13 mol/L. Equilibrium [Sr2+] = 0.13 + S and equilibrium [CO32] = S where S is the solubility of SrCO3. Concentration (mol/L) SrCO3(s)  Sr2+(aq) + CO32(aq) Initial — 0.13 0 Change — +S +S Equilibrium — 0.13 + S S Ksp = 5.4x10–10 = [Sr2+][CO32] = (0.13 + S)S This calculation may be simplified by assuming S is small and setting 0.13 + S = 0.13. Ksp = 5.4x10–10 = (0.13)S S = 4.1538x10–9 = 4.2x10–9 mol/L

17.84

The equation and ion-product expression for BaCrO4 is: BaCrO4(s) Ba2+(aq) + CrO42–(aq) Ksp =[Ba2+][CrO42–] 2+ a) The solubility, S, in pure water equals [Ba ] and [CrO42–] Ksp = 2.1x10–10 = [Ba2+][CrO42–] = S2 S = 1.4491x10–5 = 1.4x10–5 mol/L b) In 1.5x10–3 mol/L Na2CrO4, the initial concentration of CrO42– is 1.5x10–3 mol/L. Equilibrium [Ba2+] = S and equilibrium [CrO42–] = 1.5x10–3 + S where S is the solubility of BaCrO4. Ksp = 2.1x10–10 = [Ba2+][CrO42–] = S(1.5x10–3 + S) Assume S is small so 1.5x10–3 + S = 1.5x10–3 Ksp = 2.1x10–10 = S(1.5x10–3) S = 1.4x10–7 mol/L

17.85

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) = 7.1x10–7. Write a reaction table that reflects an initial concentration of Ca2+ = 0.060 mol/L. In this case, Ca2+ is the common ion.

Concentration (mol/L) Ca(IO3)2(s)  Ca2+(aq) + 2IO3(aq) Initial — 0.060 0 Change — +S +2S 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 mol/L. Ksp = [Ca2+][IO3]2 = (0.060)(2S)2 = 7.1x10–7 S = 1.71998x10–3 = 1.7x10–3 mol/L Check assumption: (1.71998x10–3 mol/L)/(0.060 mol/L) x 100% = 2.9% < 5%, so the assumption is good. S represents both the molar solubility of Ca 2+ and Ca(IO3)2, so the molar solubility of Ca(IO3)2 is 1.7x10–3 mol/L. b) Write a reaction table that reflects an initial concentration of IO3 = 0.060 mol/L. IO3 is the common ion. Concentration (mol/L) Ca(IO3)2(s)  Ca2+(aq) + 2IO3(aq) Initial — 0 0.060 Change — +S +2S 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 Ksp = [Ca2+][IO3]2 = (S)(0.060)2 = 7.1x10–7 S = 1.97222x10–4 = 2.0x10–4 mol/L Check assumption: (1.97222x10–4 mol/L)/(0.060 mol/L) x 100% = 0.3% < 5%, so the assumption is good. S represents both the molar solubility of Ca 2+ and Ca(IO3)2, so the molar solubility of Ca(IO3)2 is 2.0x10–4 mol/L. 17.86

The equilibrium is: Ag2SO4(s) 2Ag+(aq) + SO42–(aq). From the Appendix, Ksp(Ag2SO4) = 1.5x10–5. a) Write a reaction table that reflects an initial concentration of Ag+ = 0.22 mol/L. In this case, Ag+ is the common ion. Concentration (mol/L) Ag2SO4(s)  2Ag+(aq) + SO42–(aq) Initial — 0.22 0 Change — +2S +S 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) = 1.5x10–5 S = 3.099174x10–4 = 3.1x10–4 Check assumption: (3.099174x10–4 mol/L)/(0.22 mol/L) x 100% = 1.4% < 5%, so the assumption is good. S represents the molar solubility of Ag2SO4(s): 3.1x10–4 mol/L. b) Write a reaction table that reflects an initial concentration of SO42– = 0.22 mol/L. In this case, SO42– is the common ion. Concentration (mol/L) Ag2SO4(s)  2Ag+(aq) + SO42–(aq) Initial — 0 0.22 Change — +2S +S 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) = 1.5x10–5 S = 4.1286x10–3 = 4.1x10–3 Check assumption: (4.1286x10–3 mol/L)/(0.22 mol/L) x 100% = 1.9% < 5%, so the assumption is good. S represents the molar solubility of Ag2SO4, so the molar solubility of Ag2SO4 is 4.1x10–3 mol/L.

17.87

Plan: The larger the Ksp, the larger the molar solubility if the number of ions are equal. Solution: a) Mg(OH)2 with Ksp = 6.3x10–10 has higher molar solubility than Ni(OH)2 with Ksp = 6x10–16. b) PbS with Ksp = 3x10–25 has higher molar solubility than CuS with Ksp = 8x10–34. c) Ag2SO4 with Ksp = 1.5x10–5 has higher molar solubility than MgF2 with Ksp = 7.4x10–9.

17.88

The larger the Ksp, the larger the molar solubility if the number of ions are equal.

a) SrSO4 with Ksp = 3.2x10–7 has higher molar solubility than BaCrO4 with Ksp = 2.1x10–10. b) CaCO3 with Ksp = 3.3x10–9 has higher molar solubility than CuCO3 with Ksp = 3x10–12. c) Ba(IO3)2 with Ksp = 1.5x10–9 has higher molar solubility than Ag2CrO4 with Ksp = 2.6x10–12. 17.89

Plan: The larger the Ksp, the more water soluble the compound if the number of ions are equal. Solution: a) CaSO4 with Ksp = 2.4x10–5 is more water soluble than BaSO4 with Ksp = 1.1x10–10. b) Mg3(PO4)2 with Ksp = 5.2x10–24 is more water soluble than Ca3(PO4)2 with Ksp = 1.2x10–29. c) PbSO4 with Ksp = 1.6x10–8 is more water soluble than AgCl with Ksp = 1.8x10–10.

17.90

The larger the Ksp, the more water soluble the compound if the number of ions are equal. a) Ca(IO3)2 with Ksp = 7.1x10–7 is more water soluble than Mn(OH)2 with Ksp = 1.6x10–13. b) SrCO3 with Ksp = 5.4x10–10 is more water soluble than CdS with Ksp = 1.0x10–24. c) CuI with Ksp = 1x10–12 is more water soluble than AgCN with Ksp = 2.2x10–16.

17.91

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.

17.92

a) CuBr(s) Cu+(aq) + Br(aq) The bromide ion is the anion of a strong acid, so it does not react with H 3O+. 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.

17.93

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) + HS(aq) + OH(aq) Both HS and OH are anions of weak acids, so both ions react with added H 3O+. Solubility increases with addition of H3O+ (decreasing pH).

17.94

a) PbI2(s) Pb2+(aq) + 2I(aq) The iodide ion is the anion of a strong acid, so it does not react with H 3O+. 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).

17.95

Plan: Find the initial molar concentrations of Cu2+ and OH–. The concentration (mol/L)of the KOH is calculated by converting mass to the amount (mol) 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: Cu(OH)2(s) Cu2+(aq) + 2OH–(aq). The ion-product expression is Ksp = [Cu2+][OH]2 and, from the Appendix, Ksp equals 2.2x10–20.  1.0x103 mol Cu(NO3 )2   1 mol Cu 2   –3 2+ [Cu2+] =     = 1.0x10 mol/L Cu  L 1 mol Cu(NO ) 3 2     0.075 g KOH  1 mol KOH   1 mol OH   –3 – [OH–] =   = 1.33666x10 mol/L OH    1.0 L 56.11 g KOH 1 mol KOH     Qsp = [Cu2+][OH]2 = (1.0x10–3)(1.33666x10–3)2 = 1.786660x10–9 Qsp is greater than Ksp (1.8x10–9 > 2.2x10–20), so Cu(OH)2 will precipitate.

17.96

The ion-product expression for PbCl2 is Ksp = [Pb2+][Cl]2 and, from the Appendix, Ksp equals 1.7x10–5. 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   2+ [Pb2+] =   = 0.12 mol/L Pb   L 1 mol Pb(NO )   3 2   3.5 mg NaCl   103 g   1 mol NaCl   1 mol Cl  –4 – [Cl–] =     = 2.3952x10 mol/L Cl     0.250 L 1 mg 58.45 g NaCl 1 mol NaCl      Qsp = [Pb2+][Cl]2 = (0.12)(2.3952x10–4)2 = 6.8843796x10–9 Qsp is smaller than Ksp (6.9x10–9 < 1.7x10–5), so PbCl2 will not precipitate.

17.97

Plan: Find the initial molar concentrations of Ba2+ and IO3–. The concentration (mol/L) of the BaCl2 is calculated by converting mass to amount (mol) 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 1.5x10–9.  7.5 mg BaCl2   103 g   1 mL   1 mol BaCl2   1 mol Ba 2   –5 2+ [Ba2+] =    3    = 7.204611x10 mol/L Ba     500. mL 1 mg 208.2 g BaCl 1 mol BaCl 10 L   2  2      0.023 mol NaIO3   1 mol IO3  – [IO3–] =   = 0.023 mol/L IO3   L    1 mol NaIO3 

Qsp = [Ba2+][IO3]2 = (7.204611x10–5)(0.023)2 = 3.81124x10–8 Since Qsp > Ksp (3.8x10–8 > 1.5x10–9), Ba(IO3)2 will precipitate. 17.98

The ion-product expression for Ag2CrO4 is Ksp = [Ag+]2[CrO42–] and, from the Appendix, Ksp equals 2.6x10–12. To decide if a precipitate will form, calculate Qsp with the given quantities and compare it to Ksp.

 2.7x105 g AgNO3   1 mL   1 mol AgNO3   1 mol Ag   –5 + [Ag+] =    3    = 1.0594467x10 mol/L Ag    15.0 mL 169.9 g AgNO 1 mol AgNO 3  3    10 L    4.0x104 mol K 2CrO4   1 mol CrO4 2   –4 – [CrO42–] =     = 4.0x10 mol/L IO3  L 1 mol K CrO 2 4    Qsp = [Ag+]2[ CrO42–]= (1.0594467x10–5)2(4.0x10–4) = 4.4897x10–14 Since Qsp < Ksp (4.5x10–14 < 2.6x10–12), Ag2CrO4 will not precipitate.

17.99

 9.7x105 g Ca 2+   1 mol Ca 2+  Original amount (mol) of Ca2+ =  = 2.5170x10–4 mol Ca2+amount 104 mL        40.08 g Ca 2+  mL      1mol C2 O 4 2    0.1550 mol Na 2 C2 O4   103 L  (mol) of C2O42– added =  100.0 mL   = 0.01550 mol       1 mL  L     1mol Na 2 C2 O4  C2O42– The Ca2+ is limiting leaving 0 mol/L, and after the reaction there will be (0.01550 – 0.00025170) mol= 0.0152483 mol of C2O42– left in a total volume of 104 mL+ 100.0 mL = 204 mL.  0.0152483 mol C2 O   1 mL  2– [C2O42–] =    3  = 0.0747466 mol/L C2O4 204 mL    10 L  Concentration (mol/L) CaC2O4•H2O(s)  Ca2+(aq) + C2O42–(aq) + H2O(l) Initial — 0 0.0747466 — Change — +S +S — Equilibrium — S 0.0747466 + S — Assume that 0.0747466 mol/L + S  0.0747466mol/L because the amount of compound that dissolves will be negligible in comparison to 0.0747466 mol/L. The Ksp from the Appendix is: 2.3x10–9 Ksp = [Ca2+][C2O42–] = (S)(0.0747466) = 2.3x10–9 S = 3.07706x10–8 = 3.1x10–8 Check assumption: (3.07706x10–8 mol/L)/(0.0747466 mol/L) x 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 3.1x10–8 mol/L.

17.100 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 cconcVconc = cdilVdil. 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 (1.6x10–39) is smaller than the Ksp for Cd(OH)2 at 7.2x10–15. The precipitation reactions are: Fe3+(aq) + 3OH–(aq)  Fe(OH)3(s) Ksp = [Fe3+][ OH–]3 2+ – Cd (aq) + 2OH (aq)  Cd(OH)2(s) Ksp = [Cd2+][ OH–]2 3+ 2+ The concentrations of Fe and Cd in the mixed solution are found from cconcVconc = cdilVdil [Fe3+] = [(0.50 mol/L)(50.0 mL)]/[(50.0 + 125) mL] = 0.142857 mol/L Fe3+ [Cd2+] = [(0.25 mol/L)(125 mL)]/[(50.0 + 125) mL] = 0.178571 mol/L Cd2+ The hydroxide ion concentration required to precipitate the metal ions comes from the metal ion concentrations and the Ksp. [OH–]Fe = 3

[OH–]Cd =

Ksp  Fe3    Ksp

= 3

1.6 x1039 = 2.237x10–13 = 2.2x10–13 mol/L 0.142857

7.2x1015 = 2.0079864x10–7 = 2.0x10–7 mol/L 0.178571

 Cd 2     A lower hydroxide ion concentration is required to precipitate the Fe 3+.

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] = 2.0x10–7 mol/L. 17.101 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.

17.102

Cd(H2O)42+(aq) + I–(aq) CdI(H2O)3+(aq) + H2O(l)

CdI  H O    2 3    Kf1 = 2  Cd  H O    I  2 4    

CdI(H2O)3+(aq) + I–(aq) CdI2(H2O)2(aq) + H2O(l)

Kf2 =

–

CdI2(H2O)2(aq) + I (aq) CdI3(H2O) (aq) + H2O(l)

CdI3(H2O)–(aq) + I–(aq) CdI42–(aq) + H2O(l)

Overall: Cd(H2O)2+4(aq) + 4I–(aq) CdI42–(aq) + 4H2O(l)

 CdI 2  H 2 O   2    CdI  H O    I   2 3    

CdI  H O    3 2  Kf3 =  CdI2  H 2 O   I  2   2  CdI4    Kf4 = CdI  H O    I   3 2     CdI4 2     Kf = 4 Cd  H O  2    I  2 4     

CdI  H O   CdI  H O    CdI42    CdI 2  H 2 O   2  3 2  3      2   Kf = x x x CdI  H O    I    CdI  H O     I   Cd  H O  2    I   CdI2  H 2 O    I  3 2 2 2 2   3  4            = Kf1 x Kf2 x Kf3 x Kf4 17.103 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 PbS. Before accepting this conclusion, other possible equilibria must be considered. Lead is a metal ion and hydroxide ion is a ligand, so it is possible that a complex ion forms between the lead ion and hydroxide ion: Pb2+(aq) + nOH(aq) Pb(OH)n2  n(aq) This decreases the concentration of Pb2+, shifting the solubility equilibrium to the right to dissolve more PbS. 17.104 Plan: In many cases, a hydrated metal complex (e.g., Hg(H2O)42+) will exchange ligands when placed in a solution of another ligand (e.g., CN). Solution: Hg(H2O)42+(aq) + 4CN(aq) Hg(CN)42(aq) + 4H2O(l) Note that both sides of the equation have the same ―overall‖ charge of 2. The mercury complex changes from +2 to 2 because water is a neutral molecular ligand, whereas cyanide is an ionic ligand. 17.105 Zn(H2O)42+(aq) + 4CN (aq) Zn(CN)42–(aq) + 4H2O(l) 17.106 Plan: In many cases, a hydrated metal complex (e.g., Ag(H 2O)4+) will exchange ligands when placed in a solution of another ligand (e.g., S2O32). Solution:

The two water ligands are replaced by two thiosulfate ion ligands. The +1 charge from the silver ion plus the –4 charge from the two thiosulfate ions gives a net charge on the complex ion of 3. Ag(H2O)2+(aq) + 2S2O32(aq) Ag(S2O3)23(aq) + 2H2O(l) 17.107 Al(H2O)63+(aq) + 6F–(aq) AlF63–(aq) + 6H2O(l) 17.108 Plan: Write the formation reaction and the Kf expression. The initial concentrations of Ag+ and S2O32 may be determined from cconcVconc = cdilVdil. 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 mol/L)(25.0 mL)/((25.0 + 25.0) mL) = 0.022 mol/L Ag+ 2 [S2O3 ] = (0.57 mol/L)(25.0 mL)/((25.0 + 25.0) mL) = 0.285 mol/L S2O32 The reaction gives: Concentration (mol/L) Ag+(aq) + 2S2O32(aq)  Ag(S2O3)23(aq) 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: Concentration (mol/L) Ag+(aq) + 2S2O32(aq)  Ag(S2O3)23(aq) 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 mol/L and [S2O32]equil  0.241 mol/L  Ag S O  3   2 3 2   13  = (0.022) Kf = 4.7x10 =  2 (x) (0.022)2  Ag   S2 O32      x = [Ag+] = 8.0591778x1015 = 8.1x1015 mol/L 17.109 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. [Fe3+]initial = [SCN]initial =

0.0015 M Fe NO  0.50 L   3 3

0.50  0.50 L 

 1 mol Fe3   = 0.00075 mol/L Fe3+  1 mol Fe NO3   3  

 0.20 mol / L KSCN 0.50 L   1 mol SCN      = 0.10 mol/L SCN 1 mol KSCN   0.50  0.50  L   

Set up a reaction table: Concentration (mol/L) Fe3+(aq) + SCN(aq)  FeSCN2+ 4 Initial 7.5x10 0.10 0 Change x x +x Equilibrium 7.5x10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:  FeSCN 2   (x) (x)  = 2 Kf = 8.9x10 =  = 4 4 3    Fe  SCN  (7.5x10  x) (0.10  x) (7.5x10  x) (0.10)    x = (7.5x104  x)(0.10)(8.9x102) = (7.5x10-4  x)(89) x = 6.675x102  89x Silberberg, Amateis, Venkateswaran, Chen, Chemistry: The Molecular Nature of Matter and Change, 3rd Canadian Edition Page 3-640 Instructor‘s Solution Manual © Copyright 2021 McGraw-Hill Ryerson Ltd.

x = 7.416667x104 From the reaction table, [Fe3+]eq = 7.5x104 – x. Therefore, [Fe3+]eq = 7.5x104  7.416667x104 = 8.33333x106 = 1x105 mol/L. 17.110 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 = 6.3x10–31 Complex-ion Cr3+(aq) + 4OH(aq) Cr(OH)4(aq) Kf = 8.0x1029 Overall: Cr(OH)3(s) + OH(aq) Cr(OH)4(aq) K = KspKf = 0.504 At pH 13.0, the pOH is 1.0 and [OH] = 10–1.0 = 0.1 mol/L. Reaction table: Concentration (mol/L) Cr(OH)3(s) + OH(aq)  Cr(OH)4(aq) Initial —— 0.1 0 Change —— –S +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 mol/L

17.111 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+]. Solubility-product: AgI(s) Ag+(aq) + I(aq) Ksp = 8.3x10–17 + + Complex-ion: Ag (aq) + 2NH3(aq) Ag(NH3)2 (aq) Kf = 1.7x107 AgI(s) + 2NH3(aq) Ag(NH3)2+(aq) + I(aq) Koverall = Ksp x Kf = (8.3x10–17)(1.7x107) = 1.411x10–9 Reaction table: Concentration (mol/L) AgI(s) + 2NH3(aq)  Ag(NH3)2+(aq) + I(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  3 2    (S ) (S ) (S ) (S )  –9 Koverall = 1.411x10 =  = = S = 9.3908x10–5 = 9.4x10–5 mol/L (2.5  2S )2 (2.5)2 NH3  2 Overall:

17.112 Plan: First, calculate the initial amount (mol) 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: Zn2+(aq) + 4CN(aq) Zn(CN)42(aq) Kf = 4.2x1019  1 mol ZnCl2   1 mol Zn 2   2+ amount (mol) of Zn2+ = 0.84 g ZnCl2   = 0.0061624 mol Zn   136.31 g ZnCl 1 mol ZnCl 2  2   3  1 mol CN    0.150 mol NaCN   10 L  – amount (mol) of CN– =  245 mL       = 0.03675 mol CN   1 mL  L 1 mol NaCN      The Zn2+ is limiting because the amount (mol) of this ion is significantly smaller, thus, [Zn2+] = 0.  4 mol CN   amount (mol) of CN– reacting = 0.0061624 mol Zn 2  = 0.0246496 mol CN–  1 mol Zn 2    amount (mol) of CN– remaining are: 0.03675 – 0.0246496 = 0.0121004 mol CN–





0.0121004 mol CN  1 mL  = 0.0493894 mol/L CN [CN ] = 

–

245 mL 

–

 103 L    The Zn2+ will produce an equal amount (mol) of the complex with the concentration: 2  0.0061624 mol Zn 2   1 mL   1 mol Zn CN 4    = 0.025153 mol/L Zn(CN)42 [Zn(CN)42] =      103 L   1 mol Zn 2   245 mL     Concentration (mol/L) Zn2+(aq) + 4CN(aq) Zn(CN)42(aq) Initial 0 0.0493894 0.025153 Change +x +4x –x Equilibrium x 0.0493894 + 4x 0.025153 – x Assume the –x and the +4x do not significantly change the associated concentrations.  Zn CN  2   4  (0.025153  x) (0.025153)  19  = Kf = 4.2x10 =  = 4 4 (x) (0.0493894  4x) (x) (0.0493894)4  Zn 2   CN      x = 1.006481x10–16 = 1.0x10–16 [Zn2+] = 1.0x10–16 mol/L Zn2+ [Zn(CN)42] = 0.025153 – x = 0.025153 – 1.0x10–16 = 0.025153 = 0.025 mol/L Zn(CN)42–

[CN] = 0.0493894 + 4x =0.0493894 + 4(1.0x10 –16) = 0.0493894 = 0.049 mol/L CN– 17.113 The complex formation equilibrium is: Co2+(aq) + 4OH(aq) Co(OH)42(aq) Kf = 5x109 2+ First, calculate the initial amount (mol) 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.  1 mol Co NO3   1 mol Co 2   2   = 0.013118 mol Co2+ amount (mol) of Co2+ = 2.4 g Co NO3 2   182.95 g Co NO3   1 mol Co NO3   2  2  





 1 mol OH    0.22 mol KOH   amount (mol) of OH =  0.350 L   = 0.077 mol OH  L 1 mol KOH    

The Co2+ is limiting, thus, [Co 2+] = 0, and the amount (mol) of OH remaining are: [0.077  4(0.013118)] = 0.025 mol OH [OH] =

(0.025 mol OH  ) = 0.07008 mol/L OH  0.350 L 

The Co2+ will produce an equal amount (mol) of the complex with the concentration: 2  0.013118 mol Co 2    1 mol Co OH 4    = 0.03748 mol/L Co(OH)42 [Co(OH)42] =   2    0.350 L    1 mol Co  Concentration (mol/L) 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. Co OH  2   4  (0.03748  x) (0.03748)   9 Kf = 5x10 =  = = 4 4 (x) (0.07008  4x) (x) (0.07008)4 Co 2   OH      x = 3.1078x107 = 3.1x107 [Co2+] = 3.1x107 mol/L Co2+ [Co(OH)42] = 0.03748  x = 0.037479689 = 0.037 mol/L Co(OH)42 [OH] = 0.07008 + 4 x = 0.070078756 = 0.070 mol/L OH 17.114 Plan: The NaOH will react with the benzoic acid, C6H5COOH, to form the conjugate base benzoate ion, C6H5COO. Calculate the amount (mol) of NaOH and C6H5COOH. Set up a reaction table that shows the stoichiometry of the reaction of NaOH and C6H5COOH. Since NaOH is a limiting reagent, all of the NaOH will be consumed to form C6H5COO, and the amount (mol) of C6H5COOH will decrease. Find the new amount (mol) 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 6.3x105 (from the Appendix). The pKa is log (6.3x10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 H5 COOH   103 L  amount (mol) of C6H5COOH =   475 mL  = 0.0950 mol C6H5COOH   L    1 mL  3  2.00 mol NaOH   10 L  amount (mol) of NaOH =     1 mL  25 mL  = 0.050 mol NaOH L   

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 Calculating the pH from the Henderson-Hasselbalch equation:  n C H COO  0.050   = 4.201 + log  = 4.24676 = 4.2 pH = pK a + log  6 5  nC6 H5COOH  0.045     Calculations on formic acid (HCOOH) also use the Henderson-Hasselbalch equation. The Ka for formic acid is 1.8x104 and the pKa = log (1.8x104) = 3.7447. The formate to formic acid ratio may now be determined: n  [HCOO ]    pH = pK a + log  = pK a + log  HCOO   [HCOOH]   n HCOOH      n   4.24676 = 3.7447 + log  HCOO   n HCOOH   

n   0.50206 = log  HCOO   n HCOOH     n HCOO    = 3.177313  n HCOOH    n  = 3.177313  n HCOOH HCOO

(eqn 1)

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 (eqn 2) The reaction between the formic acid and the sodium hydroxide is: HCOOH(aq) + NaOH(aq)  HCOO- Na+ (aq) + H2O(l) The amount (mol) of NaOH added equal the amount (mol) of HCOOH reacted and the amount (mol) of HCOONa formed. amount (mol) NaOH = (2.00 mol NaOH/L)(Vb) = 2.00Vb mol Total amount (mol) HCOOH = (0.200 mol HCOOH/L)(Va) = 0.200Va mol The stoichiometric ratios in this reaction are all 1:1. amount (mol) of HCOOH remaining after the reaction = (0.200Va  2.00Vb) mol (eqn 3) amount (mol) of HCOO = amount (mol) ofHCOONa = amount (mol) of NaOH = 2.00 Vb (eqn 4) Using these amounts (mol) and the mole ratio determined for the buffer gives: amount (mol) of HCOO = 3.177313 mol HCOOH Substituting equations 3 and 4 into 1 gives: 2.00Vb mol = 3.177313(0.200Va  2.00Vb) mol 2.00Vb = 0.6354626Va  6.354626Vb 8.354626 Vb = 0.6354626 Va (eqn 5) Rearranging equation 2 gives Va = (0.500  Vb) L (eqn 2b) Substituting equation 2b into equation 5 gives: 8.354626 Vb = 0.6354626 (0.500  Vb) Solving for Vb: 8.354626 Vb = 0.3177313  0.6354626 Vb 8.9900886 Vb = 0.3177313 Vb = 0.0353424 = 0.035 L NaOH Substituting Vb into equation 2b: Va = 0.500  0.0353424 = 0.4646576 = 0.465 L HCOOH Silberberg, Amateis, Venkateswaran, Chen, Chemistry: The Molecular Nature of Matter and Change, 3rd Canadian Edition Page 3-644 Instructor‘s Solution Manual © Copyright 2021 McGraw-Hill Ryerson Ltd.

Limitations due to the significant figures lead to a solution with only an approximately correct pH. 17.115 pKa = – log Ka = – log 6.3x10–8 = 7.200659. The Ka comes from Appendix C; it is Ka2 for phosphoric acid.  [HPO42 ]  pH = pK a + log   [H PO  ]   2 4   [HPO4 2  ]  7.00 = 7.200659 + log   [H PO  ]   2 4   [HPO 4 2  ]  –0.200659 = log   [H PO  ]   2 4 

[HPO 4 2  ]

[H 2 PO 4  ]

= 0.63000

Since they are equimolar,

VHPO 2  4

VH PO  2

= 0.63000

and VHPO42– + VH2PO4– = 100. mL so (0.63000)VH2PO4– + VH2PO4– = 100. mL VH2PO4– = 61 mL and VHPO42– = 39 mL 17.116 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 1.8x10–4 and the pKa = log (1.8x10–4) = 3.7447.  [HCOO ]  a) pH = pK a + 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 mol/L HCOOH added. The equations are based on the neutralization reaction between HCOOH and NaOH that produces HCOO. HCOOH(aq) + NaOH(aq)  HCOO(aq) + Na+(aq) + H2O(l)   0.700  x  L NaOH  1 mol HCOO  [HCOO–] = 1.0 mol / L NaOH    0.700 L solution    1 mol NaOH    

 x L HCOOH  [HCOOH] = 1.0 mol / L HCOOH    –  0.700 L solution    0.700  x  L NaOH  1 mol HCOO      0.700 L solution   1 mol NaOH  The component ratio equals 0.99 (from part a)). Simplify the above equations and plug into ratio:

1.0 mol / L NaOH  





 0.700  x  mol / L HCOO   HCOO    0.700 0.700  x     = = = 0.989236 2 x  0.700  HCOOH   x   0.700  x   mol / L HCOOH   0.700    Solving for x: x = 0.46751 = 0.468 L Mixing 0.468 L of 1.0 mol/L HCOOH and 0.700  0.468 = 0.232 L of 1.0 mol/L NaOH gives a buffer of pH 3.74. c) The final concentration of HCOOH from the equation in part b):  0.468 L HCOOH  [HCOOH] = 1.0 mol / L HCOOH   –  0.700 L solution   0.232 L NaOH   1 mole HCOO    = 0.33714mol/L = 0.34 mol/L HCOOH    0.700 L solution   1 mole NaOH 

1.0 mol / L NaOH  

17.117 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. 17.118 H2SO4 is a strong acid and will be completely ionized: H2SO4(aq) + 2H2O(l)  SO42(aq) + 2H3O+(aq). Calculate the amount (mol) of H3O+(aq) from the H2SO4 in the 3.6x103 kg of water and then the amount of sodium acetate trihydrate (NaC2H3O2•3H2O) that will be required to neutralize that amount of H3O+(aq).  1000 g  6 Mass (g) of water = 3.6x103 kg water   = 3.6x10 g H2O 1 kg   mass of solute ppm = x106 mass of solution mass of H 2SO4 10 ppm = x106 6 3.6x10 g Mass (g) of H2SO4 = 36g  1 mol H 2SO 4   2 mol H3O  + amount (mol) of H3O+ =  36. g H 2SO4    = 0.7340 mol H3O    98.09 g H 2SO 4   1 mol H 2SO 4  The reaction between H3O+ and the base sodium acetate is: H3O+(aq) + NaC2H3O2(aq) → H2O(l) + HC2H3O2(aq) + Na+(aq) Mass (kg) of NaC2H3O2•3H2O required to neutralize the H2SO4 =  1 mol NaC2 H3 O2 •3H2 O   136.08 g NaC2 H3 O2 •3H2 O   1 kg 0.7340 mol H3 O        1 mol H3 O    1 mol NaC2 H3 O2 •3H2 O  1000 g NaC2 H3 O2 •3H2 O  = 0.09989 kg NaC2H3O2•3H2O Now consider the acetic acid. Calculate the amount of acetic acid in 3.6x10 3 kg or 3.6x106 g H2O.  0.015%   1 mol HC2 H3O 2  amount (mol) of acetic acid = 3.6x106 g H 2 O   = 8.9925 mol   100%   60.05 g HC2 H 3O 2  Find the amount of C2H3O2– necessary to maintain a pH of 5.  nC H O    [C H O  ]  pH = pK a + log  2 3 2  = pK a + log  2 3 2   [HC H O ]   n HC2 H3O2  2 3 2      nC H O   2 3 2  5.0 = 4.7447 + log   8.9925 mol   









 nC H O   2 3 2  0.2553 = log   8.9925 mol   

1.800 =

nC H O  2 3 2

8.9925 mol 16.1865 mol of C2H3O2 (NaC2H3O2•3H2O) will be required to maintain the pH. Mass (kg) of NaC2H3O2•3H2O required =  136.08 g NaC2 H3O 2 •3H 2 O   1 kg  16.1865 mol CH3COONa•3H 2O     = 2.2027 kg  1 mol NaC2 H3O2 •3H 2 O   1000 g  Total amount of NaC2H3O2•3H2O required = 0.0999 kg + 2.2027 kg = 2.3026 kg = 2.3 kg 17.119 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    = 4.4713309x10 mol/L NaUr 100. mL 190.10 g NaUr 10 L     4.4713309x10–3 mol/L NaUr = [Na+] = [Ur] Ksp = [Na+][Ur] = (4.4713309x10–3) (4.4713309x10–3) = 1.99927998x10–5 mol/L When [Na+] = 0.15 mol/L: Ksp = 1.99927998x10–5 mol/L = [0.15][Ur] [Ur] = 1.33285x10–4 The minimum urate ion concentration that will cause precipitation of sodium urate is 1.3x10–4 mol/L. 17.120 a) K = [CO2(aq)]/[CO2(g)] = 3.1x10–2 [CO2(aq)] = K[CO2(g)] = (3.1x10–2)(3x10–4) = 9.3x10–6 mol/L = 9x10–6 mol/L CO2 b) K = [Ca2+][HCO3–]2/[CO2(aq)] = (x)(2x)2/(9.3x10–6 – x) = 1x10–12 Neglect –x x = 1.325x10–6 = 1x10–6 mol/L Ca2+ c) K = [CO2(aq)]/[CO2(g)] = 3.1x10–2 [CO2(aq)] = K[CO2(aq)] = (3.1x10–2)(2x3x10–4) = 1.86x10–5 mol/L 2+ – 2 K = [Ca ][HCO3 ] /[CO2(aq)] = (x)(2x)2/(1.86x10–5 – x) = 1x10–12 Neglect –x x = 1.669x10–6 = 2x10–6 mol/L Ca2+ 17.121 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 mol/L 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 HPO 42/H2PO4. H2PO4(aq) + H2O(l) HPO42(aq) + H3O+(aq) Ka = 6.3x108  0.30 mol HPO 4 2     [HPO42 ]  0.50 L  8 = log (6.3x10 ) + log  pH = pK a + log   = 7.37675 = 7.38   [H PO  ]   0.20 mol H 2 PO 4   2 4   0.50 L  

17.122 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 amount (mol) originally present and the amount (mol) 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 mol/L. Ksp = [K+][Cl] = (3.7)(3.7) = 13.69 = 14 b) Find the amount (mol) of Cl: Original amount (mol) from the KCl: amount (mol) of K+ = amount (mol) of Cl  1 mol Cl ion   3.7 mol KCl   103 L   =  100. mL   = 0.37 mol Cl   1L 1 mol KCl    1 mL    Original amount (mol) from the 6.0 mol/L HCl in the first beaker:  1 mol Cl   6.0 mol HCl   103 L   amount (mol) of Cl =   100. mL   = 0.60 mol Cl   1 L 1 mL 1 mol HCl      This results in (0.37 + 0.60) mol = 0.97 mol Cl. Original amount (mol) from the 12 mol/L HCl in the second beaker:  1 mol Cl   12 mol HCl   103 L   amount (mol) of Cl =   100. mL   = 1.2 mol Cl   1L 1 mol HCl    1 mL    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 mol/L K+ From 6.0 mol/L HCl in the first beaker: [Cl] = (0.97 mol Cl)/(0.200 L) = 4.85 mol/L Cl From 12 mol/L HCl in the second beaker: [Cl] = (1.57 mol Cl)/(0.200 L) = 7.85 mol/L Cl Determine a Qsp value to see if Ksp is exceeded. If Qsp < Ksp, nothing will precipitate. From 6.0 mol/L HCl in the first beaker: Qsp = [K+][Cl] = (1.85)(4.85) = 8.9725 = 9.0 < 14, so no KCl will precipitate. From 12 mol/L 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 mol/L HCl is added: Equal amounts of K and Cl will precipitate. Let x be the concentration (mol/L) change. Ksp = [K+][Cl] = (1.85 – x)(7.85 – x) = 13.69 x = 0.08659785 = 0.09 This is the change in the concentration (mol/L) of each of the ions.   0.08659785 mol K   1 mol KCl   74.55 g KCl  Mass (g) of KCl =    0.200 L    = 1.291174 g= 1 g KCl    L  1 mol K   1 mol KCl    17.123 [NH3] + [NH4+] = 0.15. If [NH3] = 0.01 mol/L, then [NH4+] = 0.14 mol/L. Kb = 1.76x105 (from the Appendix) Ka = Kw/Kb = 1.0x1014/1.76x105 = 5.6818x1010 pH = pKa + log [NH3]/[NH4+] pH = log Ka + log [NH3]/[NH4+] pH = log (5.6818x1010) + log [0.01]/[0.14] pH = 8.099386 = 8.10 17.124 Determine the solubility of MnS: Silberberg, Amateis, Venkateswaran, Chen, Chemistry: The Molecular Nature of Matter and Change, 3rd Canadian Edition Page 3-648 Instructor‘s Solution Manual © Copyright 2021 McGraw-Hill Ryerson Ltd.

 4.7x104 g MnS   1 mL   1 mol MnS  –5 S=    3    = 5.401677968x10 mol/L  100 mL 87.01 g MnS     10 L   2+   MnS(s) + H2O(l)  Mn (aq) + HS (aq) + OH (aq) Ksp = [Mn2+][HS–][OH–] = S3 = (5.401677968x10–5)3 = 1.5761x10–13 = 1.6x10–13

17.125 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 = 4.5x10–7 pKa = – log Ka = – log (4.5x10–7) = 6.34679  [HCO3 ]  pH = pK a + log   [H CO ]   2 3   [HCO3 ]  7.40 = 6.34679 + log   [H CO ]   2 3   [HCO3 ]  1.05321 = log   [H CO ]   2 3 

[HCO3 ] = 11.30342352 [H 2 CO3 ] [H 2 CO3 ]

[HCO3 ] b)

= 0.0884688 = 0.088

 [HCO3 ]  pH = pK a + log   [H CO ]   2 3   [HCO3 ]  7.20 = 6.34679 + log   [H CO ]   2 3   [HCO3 ]  0.85321 = log   [H CO ]   2 3 

[HCO3 ] = 7.131978 [H 2 CO3 ] [H 2 CO3 ]

[HCO3 ]

= 0.14021 = 0.14

17.126 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 mol/L. The concentration of TRIS is the initial concentration minus the amount reacted. Once the concentrations of the TRIS-TRISH+ acid-base pair are known, the Henderson-Hasselbalch equation can be used to find the pH. Solution:  1 mol TRIS  amount (mol) of TRIS = 43.0 g TRIS   = 0.354961 mol  121.14 g TRIS   0.095 mol HCl  + amount (mol) of HCl added =   1.00 L  = 0.095 mol HCl = mol TRISH 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 — Silberberg, Amateis, Venkateswaran, Chen, Chemistry: The Molecular Nature of Matter and Change, 3rd Canadian Edition Page 3-649 Instructor‘s Solution Manual © Copyright 2021 McGraw-Hill Ryerson Ltd.

Final 0.259961 mol 0 mol 0.095 mol Since there is 1.00 L of solution, the amount (mol) of TRIS and TRISH+ equal their concentrations (mol/L). pKa of TRISH+ = 14 – pKb = 14 – 5.91 = 8.09  [TRIS]   [0.259961]  pH = pK a + log  = 8.09 + log   = 8.527185 = 8.53 +    [0.095]   [TRISH ]  17.127

17.128 Zinc sulfide, ZnS, is much less soluble than manganese sulfide, MnS. Convert ZnCl 2 and MnCl2 to ZnS and MnS by saturating the solution with H2S; [H2S]sat‘d = 0.10 mol/L. Adjust the pH so that the greatest amount of ZnS will precipitate and not exceed the solubility of MnS as determined by Ksp(MnS). Ksp(MnS) = [Mn2+][HS][OH] = 3x1011 [Mn2+] = [MnCl2] = 0.020 mol/L  [HS ] is calculated using the Ka1 expression: Concentration (mol/L): H2S(aq) + H2O(l)  H3O+(aq) + HS(aq) Initial 0.10 mol/L 0 0 Reacting x +x +x Final 0.10  x x x  H 3O    HS   H 3O    HS       Ka1 = 9x108 = =  Assume 0.10 – x = 0.10. (0.10  x) H 2S [H3O+][HS] = 9x109 [HS] = 9x109/[H3O+] Substituting [Mn2+] and [HS] into the Ksp(MnS) above gives: Ksp(MnS) = [Mn2+][HS][OH] = 3x1011 Ksp(MnS) = [Mn2+](9x109/[H3O+])[OH] = 3x1011 Substituting Kw/[H3O+] for [OH]:    9x109   K w  Ksp(MnS) = 3x1011 =  Mn 2       H O+     H O+     3    3   3x1011 x [H3O+]2 =  Mn 2   9x109  K w 









 Mn 2   9x109  K w  0.020  9x109 1.0x1014    = 3x1011 3x1011 pH = log [H+] = log (2.4494897x107) = 6.610924 = 6.6 [H3O+] =

 = 2.4494897x10

7

Maintain the pH below 6.6 to separate the ions as their sulfides. 17.129 a) Since Ka (COOH)  Kb (NH2), the proton will be transferred from the COOH to the NH2, producing COO and NH3+. Thus, in general, when the Ka of the acid group is larger than the Kb of the base group, the proton will be transferred and it is unlikely the acid group (carboxyl) would be protonated and likely the base group (amine) will be protonated. b) pKa = log Ka = log (4.47x103) = 2.34969  [+NH3CH 2 COO ]  pH = pK a + log   [+NH CH COOH]  3 2    [+NH3CH 2 COO ]  5.5 = 2.34969 + log   [+NH CH COOH]  3 2     NH3CH 2 COO    = 1.41355x103 = 1x103   NH3CH 2 COOH    c) Assuming that each functional group is independent from the others, at pH 1, there will be a protonation of the base (amine) groups, while the acid (carboxyl) group remains the same. At pH 7, the ε amine group will be protonated, while the acid (carboxyl) group will be deprotonated. The α amine group is less basic. Therefore, the percentage of a structure with this group protonated in water at pH 7 is very small. At pH 13 the acid (carboxyl) group will be deprotonated, while the base (amine) groups remain the same.

pH = 1

pH = 7

pH = 13

d) at pH 1: D, at pH 7: C, at pH 13: B At pH 1 the base (amine) group will be protonated, while the acid (carboxyl) groups remain the same. At pH 7, the base group will be protonated while the more acidic (carboxyl) group will be deprotonated, and at pH 13 both acid (carboxyl) groups will be deprotonated while the base (amine) group remains the same. 17.130 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:  1.0x106 mol   3 spheres    1 sphere   1 mL  2+  [M ] = = 1.2x105 mol/L   3 250.0 mL  10 L   1.0x106 mol  10 spheres      1 sphere   1 mL  = 4.0x105 mol/L [Cl ] =  3  250.0 mL  10 L  Silberberg, Amateis, Venkateswaran, Chen, Chemistry: The Molecular Nature of Matter and Change, 3rd Canadian Edition Page 3-651 Instructor‘s Solution Manual © Copyright 2021 McGraw-Hill Ryerson Ltd.

Ksp = [M2+][Cl]2 = [1.2x105][ 4.0x105]2 = 1.92x1014 = 1.9x10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 Chatalier‘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. 17.131 The equilibrium is: Ca5(PO4)3OH(s) 5Ca2+(aq) + 3PO43(aq) + OH(aq) Ksp = 6.8x10–37 = [Ca2+]5[PO43]3[OH] = (5S)5(3S)3(S) = 84375S9 S = 2.7166444x10–5 = 2.7x10–5 mol/L Solubility = (2.7166444x10–5 mol/L)(502.32 g/mol) = 0.013646248 g/L= 0.014 g/L Ca5(PO4)3OH The equilibrium is: Ca5(PO4)3F(s) 5Ca2+(aq) + 3PO43(aq) + F(aq) Ksp = 1.0x10–60 = [Ca2+]5[PO43]3[F] = (5S)5(3S)3(S) = 84375S9 S = 6.1090861x10–8 = 6.1x10–8 mol/L Solubility = (6.1090861x10–8 mol/L)(504.31 g/mol) = 3.0808732x10 –5 g/L= 3.1x10–5 g/L Ca5(PO4)3F 17.132 Plan: An indicator changes colour 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 = 7.943x10–5 = 8x10–5 17.133

Due to the large range of [H+], 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. 17.134 Plan: A spreadsheet will help you to quickly calculate pH/V 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, pH/V is at a maximum at the equivalence point. Solution: a) Example calculation: For the first two lines of data: pH = 1.22 – 1.00 = 0.22; V = 10.00 – 0.00 = 10.00 pH 0.22 = = 0.022 Vaverage(mL) = (0.00+ 10.00)/2 = 5.00 V 10.00 pH V(mL) pH Vaverage(mL) V 0.00 1.00 10.00 1.22 0.022 5.00 20.00 1.48 0.026 15.00 30.00 1.85 0.037 25.00 35.00 2.18 0.066 32.50 39.00 2.89 0.18 37.00 39.50 3.20 0.62 39.25 Silberberg, Amateis, Venkateswaran, Chen, Chemistry: The Molecular Nature of Matter and Change, 3rd Canadian Edition Page 3-652 Instructor‘s Solution Manual © Copyright 2021 McGraw-Hill Ryerson Ltd.

39.75 39.90 39.95 39.99 40.00 40.01 40.05 40.10 40.25 40.50 41.00 45.00 50.00 60.00 70.00 80.00

3.50 3.90 4.20 4.90 7.00 9.40 9.80 10.40 10.50 10.79 11.09 11.76 12.05 12.30 12.43 12.52

1.2 2.67 6 18 200 200 10 10 0.67 1.2 0.60 0.17 0.058 0.025 0.013 0.009

39.63 39.83 39.93 39.97 40.00 40.01 40.03 40.08 40.18 40.38 40.75 43.00 47.50 55.00 65.00 75.00

17.135 Check to see if the concentration of Ca(OH)2 exceeds the Ksp. mol/L Ca(OH)2 = (6.5x10–9 mol Ca(OH)2)/(10.0 L) = 6.5x10–10 mol/L Ca(OH)2 Determine the concentration of a saturated calcium hydroxide solution from the Ksp. Ca(OH)2(s) Ca2+(aq) + 2OH–(aq) Ksp = 6.5x10–6 = [Ca2+][OH–]2 = (S)(2S)2 = 4S3

6.5x106 = 0.01175667 = 0.012 mol/L 4 Thus, the solution is less than saturated so the Ksp does not affect the concentration of Ca(OH)2. mol/L OH– from Ca(OH)2 = (6.5x10–10 mol/L Ca(OH)2)(2 mol OH–/1 mol Ca(OH)2) = 1.3x10–9 mol/L OH– Pure water has 1x10–7 mol/L OH–, thus the contribution from the Ca(OH)2 is not significant. pH of pure water = 7.0. S= 3

17.136 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 cconcVconc = cdilVdil. Determine the pKa of the acid from the Ka. pKa = –log Ka = –log (1.38x104) = 3.86012 Determine the concentration (mol/L) of the diluted buffer component as cdil = cconcVconc/Vdil. [HLac] = [(0.85 M) (225 mL)]/[(225 + 435) mL] = 0.28977 mol/L HLac [Lac] = [(0.68 M) (435 mL)]/[(225 + 435) mL] = 0.44818 mol/L Lac n    [Lac ]  pH = pKa + log  = pKa + log  Lac    [HLac]   n HLac       [0.44818]  pH = 3.86012 + log   = 4.049519 = 4.05  [0.28977]  17.137 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) 0.01mol/L HF: Because of the common-ion effect, less CaF2 would dissolve in this solution than in pure water. (III) 0.01mol/L 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 CaF 2 would dissolve. (IV) 0.01mol/L 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) 0.01mol/L 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) 0.01mol/L HCl b) 0.01mol/L Ca(OH)2 17.138 a) The NaOH is a strong base, so it dissociates completely. NaOH(s)  Na+(aq) + OH–(aq) 0.050 mol 0.050 mol 0.050 mol The OH– ions from NaOH will react with HClO. OH–(aq) + HClO(aq)  H2O(l) + 0.050 mol 0.13 mol –0.050 mol –0.050 mol 0 0.080 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–.

ClO–(aq) 0 +0.050 mol 0.050 mol

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 mol/L HClO [ClO–]I = (0.050 mol ClO–)/(0.500 L) = 0.10 mol/L OCl– Concentration (mol/L) HClO(aq) + H2O  H3O+(aq) + ClO–(aq) Initial 0.16 — 0.10 Change –x +x +x Equilibrium 0.16 – x x 0.10 + x x is small, so [HClO]  0.16 mol/L HClO and [ClO–]  0.10 mol/L ClO–  H 3O    ClO      Ka = HClO [H3O+] =

K a HClO   ClO    

 3.0x10  0.16  = 4.8x10 mol/L H O [H O ] = 8

 0.10 

–8

[OH–] = Kw/[H+] = (1.0x10–14)/(4.8x10–8) = 2.08333x10–7 = 2.1x10–7 mol/L OH– [Na+] = (0.050 mol)/(0.500 L) = 0.10 mol/L Na+ b) pH = –log [H+] = –log (4.8x10–8 mol/L) = 7.31875876 = 7.32 c) If 0.0050 mol HCl is added then the ClO– will react with the H+ to form HClO. H+(aq) + ClO–(aq)  HClO(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 This is a buffer solution, so to find the pH use the Henderson-Hasselbach equation: n    [ClO ]  pH = pK a + log  = pK a + log  ClO   [HClO]   n HClO     





 0.045  pH = -log 3.0 x108 + log   = 7.246669779 = 7.25  0.085  17.139 In both cases the equilibrium is: CaCO3(s) Ca2+(aq) + CO32–(aq) Ksp = [Ca2+][CO32–] = S2 At 10°C Ksp = 4.4x10–9 = [Ca2+][CO32–] = S2 S = 6.6332496x10–5 = 6.6x10–5 mol/L CaCO3 At 30°C Ksp = 3.1x10–9 = [Ca2+][CO32–] = S2 S = 5.5677644x10–5 = 5.6x10–5 mol/L CaCO3 17.140 Hg2C2O4(s) Hg22+(aq) + C2O42–(aq) Ksp = 1.75x10–13 = [Hg22+][C2O42–] = (0.13 + S)S  (0.13) S S = 1.3461538x10–12 = 1.3x10–12 mol/L 17.141 [H+] = 10–9.5 = 3.16227766x10–10 mol/L H+ pOH = 14.0 – pH = 14.0 – 9.5 = 4.5 [OH–] = 10–4.5 = 3.16227766x10–5 mol/L OH–  65.0 mg HCO3   103 g   1 mol HCO3  –3 – [HCO3–] =     = 1.0652245x10 mol/L HCO3      L 1 mg   61.02 g HCO3   

 26.0 mg CO32    103 g   1 mol CO32   –4 2– [CO32–] =     = 4.3326112x10 mol/L CO3   2   L 1 mg   60.01 g CO3    Alkalinity = [HCO3–] + 2[CO32–] + [OH–] – [H+] Alkalinity = (1.0652245x10–3) mol/L+ 2(4.3326112x10–4) mol/L + (3.16227766x10–5) mol/L – (3.16227766x10–10) mol/L Alkalinity = 1.9633692x10–3 mol/L= 1.96x10–3 mol/L

17.142 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 (8x107) = 6.1] is closer to the pH of 7.4, so H 2CO3 and HCO3 are the species present. For phosphoric acid, pKa2 [log (2.3x107) = 6.6] is closest to the pH, so H2PO4 and HPO42 are the principle species present. H2PO4(aq) + H2O(l)   HPO42(aq) + H3O+(aq)  [HPO42 ]  pH = pKa + log   [H PO  ]   2 4   [HPO4 2 ]  7.4 = 6.6383 + log   [H PO  ]   2 4   HPO42     = 5.77697 = 5.8  H 2 PO4     17.143 Plan: Mercury sulfide, HgS, is much less soluble than nickel sulfide, NiS. Adjust the pH so that the greatest amount of HgS will precipitate and not exceed the solubility of NiS as determined by Ksp of NiS. Determine the minimum pH needed to cause the initial precipitation of NiS. Use the Ka expression for H2S to express [HS] in terms of [H3O+]; use the Kw expression for water to express [OH] in terms of [H3O+]. Solution: NiS(s) + H2O(l) Ni2+(aq) + HS(aq) + OH(aq) Ksp (NiS) = 3x1016 = [Ni2+][HS][OH] 2+ [Ni ] = 0.15 mol/L, so [HS] and [OH] must be found. From [H2S] = 0.050 mol/L and Ka1 in the Appendix: H2S (aq) + H2O(l) HS(aq) + H3O+(aq)  H 3O    HS   H 3O    HS   H 3O    HS           = 9x108 Ka1 = = =  H S (0.050  x) (0.050)  2  [HS][H3O+] = 4.5x109 or [HS] = 4.5x109/[H3O+] [OH ] = Kw/[H3O+] Ksp (NiS) = 3x1016 = [Ni2+][HS][OH]  4.5x109   1.0x1014  Ksp (NiS) = 3x1016 =  0.15   [H O]   [H O]  3 3    

3x1016 =

 0.15  4.5x109 1.0x1014 

[H O] 

 2

[H3O+]2 = [(0.15)(4.5x109)(1.0x1014)]/(3x1016) = 2.25x108 [H3O+] = 1.5x104 mol/L pH = log [H+] = log (1.5x104 mol/L) = 3.8239 = 3.8 17.144 a) Combine the separate equilibria to produce the desired equilibrium. The K values are in the Appendix.

2AgCl(s) 2Ag+(aq) + 2Cl–(aq) 2Ag+(aq) + CrO42–(aq) Ag2CrO4(s)

K' = (Ksp)2 = (1.8x10–10)2 = 3.24x10–20 K" = 1/Ksp = 1/(2.6x10–12) = 3.8462x1011

2AgCl(s) + CrO42–(aq) Ag2CrO4(s) + 2Cl–(aq) K = K'K" = 1.2462x10–8 = 1.2x10–8 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 = 1.8x10–10 = [Ag+][Cl–] = S2 S = [Ag+] = 1.34164x10–5 mol/L = 1.3x10–5 mol/L Ag+ Use the Ksp for silver chromate. Ksp = 2.6x10–12 = [Ag+]2[CrO42–] [CrO42–] = (2.6x10–12)/(1.34164x10–5)2 = 0.01444 = 0.014 mol/L If the chromate ion concentration exceeds 0.014 mol/L, Ag2CrO4 will precipitate. 17.145 Plan: Find the amount (mol) 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 amount (mol) of HCl that would react with the amount (mol) of quinidine and subtract the reacted HCl from the initial amount (mol) 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  amount (mol) of quinidine = 33.85 mg quinidine   1 mg   324.41 g quinidine     -4 = 1.0434327x10 mol quinidine  103 L   0.150 mol HCl  amount (mol) of HCl excess = 6.55 mL    1 mL   L   

 2 mol HCl  –4 – 1.0434327 x10 4 mol quinidine   = 7.7381346x10 mol HCl  1 mol quinidine 





 1 mol NaOH     1 mL  1L Volume (mL) of NaOH needed = 7.7381346x104 mol HCl     3   1 mol HCl   0.0133 mol NaOH   10 L  = 58.18146 mL= 58.2 mL NaOH solution b) Use the amount (mol) of quinidine and the concentration of the NaOH to determine the volume in millilitre.  1 mol NaOH     1 mL  1L Volume = 1.0434327x104 mol quinidine     3   1 mol quinidine   0.0133 mol NaOH   10 L  = 7.84536 mL= 7.85 mL NaOH solution 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 =



 4.0x10 1.0x10  = 2.0x10 mol/L 6

10

–8

[H3O+] = Kw/[OH–] = (1.0x10–14)/(2.0x10–8) = 5.0x10–7 mol/L pH = –log [H+] = –log (5.0x10–7 mol/L) = 6.3010 = 6.30 17.146

K values from the Appendix:

H2C2O4(aq) H+(aq) + HC2O4–(aq) HC2O4–(aq) H+(aq) + C2O42–(aq)

Ka1 = 5.6x10–2 Ka2 = 5.4x10–5

H2C2O4(aq) 2H+(aq) + C2O42–(aq) K = Ka1Ka2 = 3.024x10–6 + 2 2– K = [H ] [C2O4 ]/[H2C2O4] [C2O42–] = K[H2C2O4]/[H+]2 a) At pH = 5.5: [H+] = 10–5.5 = 3.162x10–6 mol/L [C2O42–] = K[H2C2O4]/[H+]2 = (3.024x10–6)(3.0x10–13)/(3.162x10–6)2 [C2O42–] = 9.07359x10–8 mol/L 2+ Q = [Ca ][C2O42–] Q = (2.6x10–3)(9.07359x10–8) = 2.3591x10–10 = 2.4x10–10 < Ksp = No precipitate b) At pH = 7.0: [H+] = 10–7.0 = 1x10–7 mol/L [C2O42–] = K[H2C2O4]/[H+]2 = (3.024x10–6)(3.0x10–13)/(1x10–7)2 [C2O42–] = 9.072x10–5 mol/L Q = (2.6x10–3) (9.072x10–5) = 2.35872x10–7 = 2.4x10–7 > Ksp = Precipitate forms c) The higher pH would favor precipitation. 17.147 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 = pK a + 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  ]   log   pH = log    [HA]   [HA]   final  initial  [A  ]  Since both [HA] and [A] = 0.10 mol/L, log  = 0 because [HA] = [A], and log (1) = 0  [HA]  initial So the change in pH is equal to the concentration ratio of base to acid after the addition of H 3O+. Consider the buffer prior to addition to the medium. H3O+ (aq) + A–(aq)  HA(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: pH = log (0.099/0.101) = –0.008686 If the undiluted buffer changes 0.009 pH units with addition of 0.0010 mol H 3O+, how much can the buffer be diluted and still not change by 0.05 pH units (pH < 0.05)? Let x = fraction by which the buffer can be diluted. Assume 0.0010 mol H 3O+ is added to 1 L.  0.10x  0.0010   base log = log  = –0.05  0.10x  0.0010   acid      0.10x  0.0010   –0.05 = 0.89125   = 10 0.10x  0.0010     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.

17.148

a) Ka = 6.8x10–4 =

 H 3 O    F  x2 x2    =  0.2500  x 0.2500  HF

x = [H3O+] = 0.0130384 mol/L pH = –log [H+] = –log (0.0130384 mol/L) = 1.8847757 = 1.88  103 L   0.2500 mol HF  1 mol NaOH    1 mL  1L b) Volume (mL) of Na =  35.00 mL       3   1 mL   1L  1 mol HF  0.1532 mol NaOH   10 L    = 57.11488 mL= 57.11 mL NaOH  103 L   0.2500 mol HF  –3 c) amount (mol) of HF (initial) =  35.00 mL    = 8.750x10 mol HF  1 mL   1 L     103 L   0.1532 mol NaOH  amount (mol) of NaOH added =   57.11488  0.50  mL     1 mL   1L    –3 = 8.6734x10 mol NaOH amount (mol) of F– formed = moles NaOH amount (mol) of HF remaining = (8.750x10–3 – 8.6734x10–3) mol = 7.66x10–5 mol HF Volume of solution = (35.00 + 57.11488 – 0.50)mL(10–3 L/1 mL) = 0.091615 L [HF] = (7.66x10–5 mol HF)/(0.091615 L) = 0.00083611 mol/L HF [F–] = (8.6734x10–3 mol F–)/(0.091615 L) = 0.09467 mol/L F– pKa = –log Ka = –log (6.8x10–4) = 3.1675  [F ]  pH = pK a + log   [HF]     [0.09467]  pH = 3.1675 + log    [0.0008361]  pH = 5.2215 = 5.22 d) At this point there are 8.750x10–3 mol of F– in (35.00 + 57.11488) mL of solution.  8.750x103 mol F   1 mL  Concentration (mol/L) of F– =  = 0.09499 M F–   35.00  57.11488 mL   103 L     Kb = Kw/Ka = (1.0x10–14)/(6.8x10–4) = 1.470588x10–11  HF OH  x2 x2 –11 Kb = 1.470588x10 = =  0.09499  x 0.09499  F    x = [OH–] = 1.1819x10–6 mol/L [H3O+] = Kw/[OH–] = (1.0x10–14)/(1.1819x10–6) = 8.4609527x10–9 mol/L pH = –log [H+] = –log (8.4609527x10–9 mol/L) = 8.07258 = 8.07  103 L   0.1532 mol NaOH  –5 e) amount (mol) of NaOH excess =  0.50 mL    = 7.66x10 mol NaOH  1 mL   1 L    Volume of solution = (35.00 + 57.11488 + 0.50)(10 –3 L/1 mL) = 0.092615 L [OH–] = (7.66x10–5 mol OH–)/(0.092615 L) = 8.271x10–4 mol/L OH– The excess OH– will predominate and essentially control the pH. [H3O+] = Kw/[OH–] = (1.0x10–14)/(8.271x10–4) = 1.2090436x10–11 mol/L pH = –log [H+] = –log (1.2090436x10–11 mol/L) = 10.917558 = 10.92

17.149 a) The formula is Hg2Cl2 which simplifies to the empirical formula HgCl. b) The equilibrium is: Hg2Cl2(s) Hg22+(aq) + 2Cl(aq) Ksp = 1.5x10–18 2+  2 2 3 –18 Ksp = [Hg2 ][Cl ] = (S)(2S) = 4S = 1.5x10 S = 7.2112479x10–7 = 7.2x10–7 mol/L

c) [Hg22+] = Ksp/[Cl]2 =

1.5x1018  0.024 g NaCl   1 mol NaCl   1 mol Cl   1000 cm³      58.44 g NaCl   1 mol NaCl   1 L   cm³       = 8.8938375–18 mol/L= 8.9x10–18 mol/L Hg22+ d) Use the value of S for a saturated solution (see part b)).

 7.2112479x107 mol Hg 2Cl 2   1 L   103 m  3  472.1 g Hg 2Cl 2  Mass (g) of Hg2Cl2 =    3 3    4900 km   L 1 km  1 mol Hg 2Cl 2     10 m    = 1.6681708x1012 = 1.7x1012 g Hg2Cl2 e) Use the value determined in part c). Mass (g) of Hg2Cl2 = 3





 8.8938375x1018 mol Hg 2 2    1 mol Hg 2 Cl 2   1 L   103 m  3  472.1 g Hg 2 Cl 2       3 3    4900 km   2   L 1 km  1 mol Hg 2 Cl 2      1 mol Hg 2   10 m   = 20.5740 g = 21 g Hg2Cl2





17.150 a) CaF2 with Ksp = 3.2x10–11 will precipitate before BaF2 with Ksp = 1.5x10–6. 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 cconcVconc = cdilVdil. [Ba2+] = [(0.090 mol/L)(25.0 mL)]/[(25.0 + 35.0) mL] = 0.0375 mol/L Use the barium ion concentration and the Ksp to find the fluoride ion concentration. [F–]2 = Ksp/[Ba2+] = (1.5x10–6)/(0.0375) = 4.0x10–5 [F –] = 6.324555x10–3 = 6.3x10–3 mol/L or less 17.151 Plan: Use the ideal gas law to calculate the amount (mol) 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 = 4.5x10–7 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 (0.033 L/100 L) when determining the amount (mol).  103 L   0.033%  Volume (L) of CO2 =  88 mL   = 2.904x10–5 L CO2  1 mL   100%   





1 bar  2.904x105 L PV amount (mol) of dissolved CO2 = = = 1.172115x10–6 mol CO2 RT bar•L    0.08314 mol•K    273  25 K    [CO2] = (1.172115x10–6 mol CO2)/[(100 mL)(10–3 L/1 mL)] = 1.172115x10–5 mol/L CO2  H3O    HCO3   H3O    HCO3        –7 Ka1 = 4.5x10 = = Let x = [H3O+] = [HCO3] H 2 CO3  CO2  (x) (x) 4.5x10–7 = Check assumption: 1.17211510-5/4.5x10–7 = 26 < 400 (1.172115x105  x) The assumption is not justified and it is necessary to use the quadratic equation.

4.5x10–7 =

(x) (x)

(1.172115x105  x) x + 4.5x10–7x – 5.27452x10–12 = 0 a=1 b = 4.5x10–7 2

c = – 5.27452x10–12

b  b 2  4ac 2a (4.5x107 ) 

 4.5x10   4 1  5.27452x10  7

12

2 1 –6

x = 2.0826276x10 mol/L = [H+] pH = –log (2.0826276x10–6) = 5.68139 = 5.68 17.152 a) For H2CO3, pKa = –log Ka pKa1 = –log 4.5x10–7 = 6.3468 pKa2 = –log 4.7x10–11 = 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] = 1.4125x102 = 1x102 mol/L pH = pKa + log [base]/[acid] 8.5 = 10.33 + log [CO32–]/[HCO3–] [CO32–]/[HCO3–] = 1.4791x10–2 = 1x10–2 mol/L 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 (H 2CO3) accumulate. Also, at greater depths, the pressure is higher and so is the concentration of CO2 (Henry‘s law). 17.153 Initial concentrations of Pb2+ and Ca(EDTA)2 before reaction based on mixing 100. mL of 0.10 mol/L Na2Ca(EDTA) with 1.5 L blood:  120 g Pb2    1 mL   1.5 L blood   106 g  1 mol Pb 2   [Pb2+] =  = 5.42953668x10–6 mol/L Pb2+  100 mL   103 L   1.6 L mixture   1 g   207.2 g Pb 2         cconcVconc = cdilVdil [Ca(EDTA)2–] = cconcVconc/Vdil = [(0.10 mol/L (100 mL)(10–3 L/1 mL)]/(1.6 L) = 6.25x10–3 mol/L Set up a reaction table assuming the reaction goes to completion: Concentration (mol/L) [Ca(EDTA)]2(aq) + Pb2+(aq)  [Pb(EDTA)]2(aq) + Ca2+(aq) –3 –6 Initial 6.25x10 5.42953668x10 0 0 React –5.42953668x10–6 –5.42953668x10–6 +5.42953668x10–6 +5.42953668x10–6 6.24457x10–3 0 5.42953668x10–6 5.42953668x10–6 Now set up a reaction table for the equilibrium process: Concentration (mol/L) [Ca(EDTA)]2(aq) + Pb2+(aq)  [Pb(EDTA)]2(aq) + Ca2+(aq) –3 –6 Initial 6.24457x10 0 5.42953668x10 5.42953668x10–6 Change +x +x –x –x Equilibrium 6.24457x10–3 + x x 5.4295366x10–6 – x 5.4295366x10–6 – x Assuming that x is much smaller than all the numerical quantities and hence the change caused by x is negligible,:

 Pb  EDTA 2    Ca 2   (5.42953668x106 ) (5.42953668x106 )     Kc = 2.5x10 =  = Ca  EDTA 2    Pb2   (6.24457x103 ) (x)     x = [Pb2+] = 1.8883522x10–16 mol/L  1.8883522x1016 mol Pb2    103 L   207.2 g Pb2    1 g  Mass (g) of Pb2+ in 100 mL =  100 mL          1 mL   1 mol Pb2    106 g  L       = 3.9126658x10–9 g Pb2+ The final concentration is 3.9x10–9 g/100 mL. 7

17.154 Assume that pKa is in the centre of the range, and calculate the Ka from the average pKa. Average pKa (centre of range) = (7.9 + 6.5)/2 = 7.2 Ka = 10–7.2 = 6.3095734x10–8 = 6x10–8 There is only one digit after the decimal point in the pKa values; thus, there is only one significant figure in the Ka value . 17.155 Plan: Convert the solubility of NaCl from g/L to mol/l (concentration). Use the solubility to find the Ksp value for NaCl. Find the amount (mol) of Na+ and Cl in the original solution; find the amount (mol) of added Cl (from the added HCl). The concentration (mol/L) of the Na + and Cl ions are then found by dividing amount (mol) of each by the total volume after mixing. Using the concentrations (mol/L) 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  Concentration (mol/L) of NaCl =    58.44 g NaCl  = 5.42436687 mol/L NaCl L    Determine the Ksp from the concentration (mol/L) 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   amount (mol) of Cl initially =  0.100 L     = 0.542436687 mol Cl  L 1 mol NaCl     This is the same as the moles of Na+ in the solution. 3  1 mol Cl   8.65 mol HCl   10 L   amount (mol) of Cl added =     28.5 mL    = 0.246525 mol Cl  L    1 mL   1 mol HCl  +  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 concentration (mol/L)of Cl in mixture = [(0.542436687 + 0.246525) mol Cl ]/(0.1285 L) = 6.13978 mol/L Cl concentration (mol/L)of Na+ in mixture = (0.542436687 mol Na+)/(0.1285 L) = 4.22130 mol/L 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. 17.156 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 = pK a + log    = 1 when the amounts of HA and A are equal  [HA]  [HA]     Silberberg, Amateis, Venkateswaran, Chen, Chemistry: The Molecular Nature of Matter and Change, 3rd Canadian Edition Page 3-662 Instructor‘s Solution Manual © Copyright 2021 McGraw-Hill Ryerson Ltd.

pH = pKa + log 1 pH = pKa = –log (4.5x10–5) = 4.346787 = 4.35 Scene B:

Only A is present at a concentration of 0.10 mol/L. The Kb for A is needed. Kb = Kw/Ka = 1.0x10–14/4.5x10–5 = 2.222x10–10 A(aq) + H2O(l) OH(aq) + HA(aq) Initial: 0.10 mol/L 0 0 Change: –x +x +x Equilibrium: 0.10 – x x x   HA  OH  Kb = 2.222x10–10 = A    (x) (x) Kb = 2.222x10–10 = Assume that x is small compared to 0.10 (0.10  x) Kb = 2.222x10–10 =

 x  x   0.10 

x = 4.7138095x10–6 mol/L OH Check assumption: (4.7138095x10–6/0.10) x 100% = 0.005% error, so the assumption is valid. [H3O]+ = Kw/[OH] = (1.0x10–14)/(4.7138095x10–6) = 2.1214264x10–9 mol/L H3O+ pH = –log [H+] = –log (2.1214264x10–9) = 8.67337 = 8.67 Scene C: This is a 0.10 mol/L HA solution. The hydrogen ion, and hence the pH, can be determined from the Ka. Concentration (mol/L) HA(aq) + H2O(l)  H3O+(aq) + A–(aq) Initial 0.10 mol/L — 0 0 Change –x +x +x Equilibrium 0.10 – x x x (The H3O+ contribution from water has been neglected.)  H3O    A      Ka = 4.5x10–5 = HA   Ka = 4.5x10–5 =

 x  x 

 0.10  x   x  x  Ka = 4.5x10–5 =  0.10 

Assume that x is small compared to 0.10.

[H3O+] = x = 2.12132x10–3 Check assumption: (2.12132x10–3/0.10) x 100% = 2% error, so the assumption is valid. pH = –log [H+] = –log (2.12132x10–3) = 2.67339 = 2.67 Scene D:

This is a buffer with a ratio of [A–]/[HA] = 5/3.  [A  ]  5 = –log (4.5x10–5) + log   = 4.568636 = 4.57 pH = pK a + log   [HA]  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. 17.157 a) The dissolution of MZ will produce equal amounts of M 2+ 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 Silberberg, Amateis, Venkateswaran, Chen, Chemistry: The Molecular Nature of Matter and Change, 3rd Canadian Edition Page 3-663 Instructor‘s Solution Manual © Copyright 2021 McGraw-Hill Ryerson Ltd.

from the solution. Distilled water will neither add nor remove ions, thus the M 2+ and Z2– must be equal; as in Scene B. b) Using box B; there are 4(2.5x10–6mol/L) = 1.0x10–5 mol/L for each ion. Ksp = [M2+][Z2–] = (1.0x10–5)(1.0x10–5) = 1.0x10–10 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 M 2+. This occurs in box A. 17.158 a) Ag+ ions come from the dissolution of AgCl(s). Ksp = [Ag+][Cl–] = 1.8x10–10

1.8x10 10  Cl     AgCl(s) Ag+(aq) + Cl– (aq) Ksp = 1.8x10–10 + – – Ag (aq) + 2Cl (aq) AgCl2 (aq) Kf = 1.8x105 [Ag+] =

AgCl(s) + Cl–(aq) AgCl2–(aq) K = KspKf = (1.8x10–10)(1.8x105) = 3.24x10–5  AgCl2   –5 K = 3.24x10 =  Cl    [AgCl2–] = (3.24x10–5)[Cl–] = (3.2x10–5)[Cl–] b) [Ag+] = [AgCl2–]

1.8x10 10 = (3.24x10–5)[Cl–]  Cl     1.8x1010 3.24x10 5 [Cl–] = 2.3570226x10–3 = 2.4x10–3 mol/L 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–] = (3.24x10–5)(2.3570226x10–3) = 7.6367532x10–8 = 7.6x10–8 mol/L The solubility of AgCl(s) = (7.6367532x10–8 + 7.6367532x10–8) mol/L = 1.52735x10–7 = 1.5x10–7 mol/L 2+ 17.159 Co (aq) + EDTA4(aq) [Co(EDTA)]2(aq) Kf = 1016.31 = 2.0417379x1016

Co  EDTA 2     Kf = Co 2    EDTA 4       0.048 mol Co2   103 L  amount (mol) of Co2+ (initial) =  50.0 mL  = 0.0024 mol Co2+    1 mL   L     0.050 mol EDTA 4   103 L  a) amount (mol) of EDTA added =  25.0 mL  = 0.00125 mol EDTA    1 mL   L    The amount (mol) of EDTA added equals the amount (mol) of [Co(EDTA)]2 formed. The EDTA is limiting so no EDTA is left after the reaction and the remaining Co 2+: 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 mol/L Co2+ [Co(EDTA)2-] = (0.00125 mol)/(0.0750 L) = 0.016667 mol/L To reach equilibrium the concentrations of the species involved are: [EDTA4] = x [Co2+] = 0.015333 + x [Co(EDTA)2–] = 0.016667 – x Co  EDTA 2     (0.016667  x) (0.016667) Kf = 2.0417379x1016 = = = 2  4  (0.01533  x) (x) (0.01533) (x) Co   EDTA     x = 5.324947x10–17 [EDTA] = 5.3x10–17 mol/L [Co2+] = 0.015 mol/L  0.050 mol EDTA 4   103 L  b) amount (mol) of EDTA added =  75.0 mL  = 0.00375 mol EDTA    1 mL   L    The Co2+ is limiting so no Co2+ is left. The original amount (mol) Co2+ equals the amount (mol) of complex formed. amount (mol) 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 mol/L EDTA [Co(EDTA)2–] = (0.0024 mol)/(0.1250 L) = 0.0192 mol/L To reach equilibrium the concentrations of the species involved are: [EDTA4–] = 0.0108 + x [Co2+] = x [Co(EDTA)2–] = 0.0192 – x Co  EDTA 2   (0.0192  x) (0.0192)   16 Kf = 2.0417379x10 = = = ( x) (0.0108  x) ( x) (0.0108) Co 2    EDTA 4      x = 8.7071792x10–17 [EDTA4–] = 00108 + x = 0.0108 mol/L [Co2+] = 8.7x10–17 mol/L

17.160 Plan: First look at the weak acid dissociation in water to find the [H+] in the acid solution to find the pH. Then use the Henderson-Hasselbalch equation to find the pH at the midpoint of the titration (since pH = pK a). Calculate the volume of base needed for equivalence using c AVA=cBVB. Then react the acid and base to completion (leaving no weak acid or strong base and converting both to only the weak conjugate base). Then write the dissociation reaction for the weak conjugate base in water and calculate the [OH -], the pOH and then pH. Solution: a)

CH2ClCOOH (aq) + H2O (l) ⇄ CH2ClCOO- (aq) + H3O+ (aq)

I C E

0.125 -x 0.125-x

0 +x x

 H 3 O +   CH 2 ClCOO-  x2 K a =1.4×10 =  0.125  x CH 2 ClCOOH  -3

Cannot assume anything. Solve quadratic. x 2  1.4×10-3   0.125  x  x 2  1.4  103 x  1.8  10 4  0 x 

b  b 2  4ac 2a 1.4  103 

1.4 10   4  1.8 10  3 2

4

2 mol x  1.2  10   H 3 O     H   L pH   log  H     log 1.2  102   1.90 2

At the midpoint of a titration, since [HA] = [A-], pH = pKa pH = - log Ka = - log (1.4 x 10-3) = 2.85

cAVA=cBVB 500. mL = 0.500 L (0.125 mol/L)(0.500 L) = (0.200 mol/L) V B VB = 312.5 mL At equivalence, we would have added 312.5 mL of KOH. As we are mixing 2 solutions, we need to determine the concentration of the weak acid and the strong base in the mixture AFTER mixing but BEFORE reaction. We can use c 1V1=c2V2 to calculate these concentrations.

Let  CH 2 ClCOOH i =c2 c1V1  c2 V2 mol   0.125   0.500 L cV mol L  c2   CH 2 ClCOOH i  1 1   =7.69×10-2 V2 L 0.8125 L





Let  KOH i =c2 c1V1  c2 V2 mol   0.200  0.3125 L c1V1  mol L  c2   KOH i   =7.69×10-2 V2 L 0.8125 L





Since we are reacting a weak acid with a strong base, the reaction will proceed forward until either there is no more strong base or until the weak acid is completely consumed. In this case, the amounts are stoichiometrically equal (as we would expect at the equivalence point) and thus both are completely consumed leaving only the weak conjugate base in solution with the potassium ion (which makes no acid/base contribution to the solution). Silberberg, Amateis, Venkateswaran, Chen, Chemistry: The Molecular Nature of Matter and Change, 3rd Canadian Edition Page 3-666 Instructor‘s Solution Manual © Copyright 2021 McGraw-Hill Ryerson Ltd.

CH2ClCOOH (aq) + KOH (aq)  CH2ClCOO- (aq) + K+ (aq) + H2O (l) I C F

7.69 x 10-2 -7.69 x 10-2 0

0 +7.69 x 10-2 +7.69 x 10-2

We must now consider the dissociation of the weak conjugate base in water to determine the pH at the equivalence point. CH2ClCOO- (aq) + H2O (l) ⇄ CH2ClCOOH (aq) + OH- (aq) I C E

7.69 x 10-2 -x 7.69 x 10-2 - x

0 +x x

CH 2 ClCOOH  OH  K w 1.0  1014 x2 = = 7.1×10-12 =  3 K a 1.4  10 7.69  102  x CH 2 ClCOO -  -

Kb =

Assume x << 7.69  102

 7.69  10 2 - x  7.69 10 2

x 2  5.5  1013 mol  OH   L pOH= - log OH -  = - log  7.4  107  = 6.13 x  7.4  107

pH=14.00 - pOH=14.00 - 6.13=7.87 d)

As we are mixing 2 solutions, we need to determine the concentration of the weak acid and the strong base in the mixture AFTER mixing but BEFORE reaction. We can use c 1V1=c2V2 to calculate these concentrations.

Let  CH 2 ClCOOH i =c2 c1V1  c2 V2 mol   0.125   0.500 L cV mol L  c2   CH 2 ClCOOH i  1 1   =7.68×10-2 V2 L 0.8135 L





Let  KOH i =c2 c1V1  c2 V2 mol   0.200  0.3135 L c1V1  mol L  c2   KOH i   =7.70×10-2 V2 L 0.8135 L





CH2ClCOOH (aq) + KOH (aq)  CH2ClCOO- (aq) + K+ (aq) + H2O (l) I C F

7.68 x 10-2 -7.68 x 10-2 0

7.70 x 10-2 -7.68 x 10-2 2.4 x 10-4

0 +7.68 x 10-2 +7.68 x 10-2

At this point, the strong base is in excess and thus the weak acid is the limiting reagent. We are left with a small amount of the strong base and the weak conjugate base. Although we have more of the weak conjugate base, it plays no role in determining the pH as the strong base completely dissociates and provides the total amount of OH- ion. The presence of the strong base suppresses the dissociation of the weak conjugate base even more and thus its contribution to the pH is negligible. pOH = - log [OH-] = - log (2.4 x 10-4) = 3.6 pH = 14.0 – pOH = 14.0 – 3.6 = 10.4 17.161 Plan: First look at the weak base dissociation in water to find the [OH -] in the base solution to find the pOH and then pH. Then use the Henderson-Hasselbalch equation to find the pH at the midpoint of the titration (since pH = pKa). Calculate the volume of acid needed for equivalence using c AVA=cBVB. Then react the acid and base to completion (leaving no weak base or strong acid and converting both to only the weak conjugate acid). Then write the dissociation reaction for the weak conjugate acid in water and calculate the [H +] and then pH. Solution: C8H10N4O2 (aq) + H2O (l) ⇄ C8H10N4O2H+ (aq) + OH- (aq)

a) I C E

0.175 -x 0.175-x

0 +x x

OH -   C8 H10 N 4 O 2 H +  x2 K b =4.1×10 =  0.175  x C8 H10 N 4 O2  -4

Assume x << 0.175  0.175 - x  0.175 x 2  7.2  105 mol  OH   L pOH= - log OH -  = - log  8.5  10 3  = 2.07 x  8.5  103

pH=14.00 - pOH=14.00 - 2.07=11.93

At the midpoint of a titration, since [HA] = [A-], pH = pKa

Ka =

K w 1.0 1014 = = 2.4×10-11 K b 4.1104

pH = - log Ka = - log (2.4 x 10-11) = 10.61 c)

cAVA=cBVB 250.0 mL = 0.2500 L (0.150 mol/L) VA = (0.175 mol/L)(0.2500 L) VA = 292 mL At equivalence, we would have added 292 mL of HBr (the actual figure is 291.6 mL (with the 6 repeating infinitely). We have to round to 292 mL because of the significant figures in the question. For this reason, when doing calculations, numbers may not be exactly equal, but if you keep all the digits in your calculator when doing the calculation, you will find the numbers to be equal exactly). As we are mixing 2 solutions, we need to determine the concentration of the weak base and the strong acid in the mixture AFTER mixing but BEFORE reaction. We can use c 1V1=c2V2 to calculate these concentrations.

Let  C8 H10 N 4 O 2 i =c2 c1V1  c2 V2 mol   0.175   0.2500 L cV mol L  c2   C8 H10 N 4 O 2 i  1 1   =8.08×10-2 V2 L 0.542 L





Let  HBr i =c2 c1V1  c2 V2 mol   0.150  0.292 L c1V1  mol L  c2   HBr i   =8.08×10-2 V2 L 0.542 L





Since we are reacting a weak base with a strong acid, the reaction will proceed forward until either there is no more strong acid or until the weak base is completely consumed. In this case, the amounts are stoichiometrically equal (as we would expect at the equivalence point) and thus both are completely consumed leaving only the weak conjugate acid in solution with the bromide ion (which makes no acid/base contribution to the solution). C8H10N4O2 (aq) + HBr (aq)  C8H10N4O2H+ (aq) + Br - (aq) I C F

8.08 x 10-2 -8.08 x 10-2 0

0 +8.08 x 10-2 +8.08 x 10-2

We must now consider the dissociation of the weak conjugate acid in water to determine the pH at the equivalence point. C8H10N4O2H+ (aq) + H2O (l) ⇄ C8H10N4O2 (aq) + H3O+ (aq) I C E

8.08 x 10-2 -x 8.08 x 10-2 - x

K a = 2.4×10-11 =

0 +x x

C8 H10 N 4 O 2   H3O+ 



0 +x x

x2 8.08  102  x

C8 H10 N 4 O 2 H  Assume x << 8.08  102  8.08  10 2 - x  8.08 10 2 +

x 2  2.0  1012 mol x  1.4  106   H 3 O +  L pH= - log  H 3 O +  = - log 1.4  106  = 5.85 d)

Let  C8 H10 N 4 O 2 i =c2 c1V1  c2 V2 mol   0.175   0.2500 L cV mol L  c2   C8 H10 N 4 O 2 i  1 1   =8.06×10-2 V2 L 0.543 L





Let  HBr i =c2 c1V1  c2 V2 mol   0.150  0.293 L c1V1  mol L  c2   HBr i   =8.09×10-2 V2 L 0.543 L





C8H10N4O2 (aq) + HBr (aq)  C8H10N4O2H+ (aq) + Br - (aq) I C F

8.06 x 10-2 -8.06 x 10-2 0

8.09 x 10-2 -8.06 x 10-2 2.8 x 10-4

0 +8.06 x 10-2 +8.06 x 10-2

At this point, the strong acid is in excess and thus the weak base is the limiting reagent. We are left with a small amount of the strong acid and the weak conjugate acid. Although we have more of the weak conjugate acid, it plays no role in determining the pH as the strong acid completely dissociates and provides the total amount of H+ ion. The presence of the strong acid suppresses the dissociation of the weak conjugate acid even more and thus its contribution to the pH is negligible. pH = - log [H+] = - log (2.8 x 10-4) = 3.6 17.162 Plan: Calculate the molar mass of sodium benzoate and benzoic acid. Calculate the amount (mol) of each of the acid and salt. Use the Henderson-Hasselbalch equation to find the pH of the solution. Use the same equation (H-H) with the new pH to find the ratio of conjugate base to acid needed to obtain that pH. Assume that addition of the required substance does not change the concentration of the other one. Once you know the amount (mol) of the species needed, subtract from this amount the amount that was originally in the solution to determine the amount of the species to be added. Remembering to make it the original form (if an anion or cation form was used), calculate the mass of the species to be added by multiplying the amount (mol) by the molar mass of that species. Solution: MM (benzoic acid) = 7 MMC + 6 MMH + 2 MMO = [7 (12.01) + 6 (1.01) + 2 (16.00)] g/mol = 122.13 g/mol MM (sodium benzoate)

= 7 MMC + 5 MMH + 2 MMO + MMNa = [7 (12.01) + 6 (1.01) + 2 (16.00) + 22.999] g/mol = 144.11 g/mol

m benzoic acid 5.00g   4.09  102 mol MM benzoic acid 122.13 g mol msodium benzoate 5.00g n sodium benzoate    3.47  102 mol MM sodium benzoate 144.11 g mol n benzoic acid 

Since the volume does not change upon addition of solids, and since concentration = n/V, we can use the ratio of amount (mol) in the Henderson-Hasselbalch equation instead of the concentration.

A  pH  pK a  log    HA  n and V is the same for both concentration terms, we can write V n  pH  pK a  log A n HA

since c=

 3.47 10 mol  pH   log  6.46  10   log  4.09 10 mol  2

5

2

pH  4.12 b)

Since we want the pH to decrease, we need to add the more acidic component. Hence we will add benzoic acid. n since c= and V is the same for both concentration terms, we can write V n  pH  pK a  log A n HA

where A  represents the benzoate ion from sodium benzoate and HA represents benzoic acid 4.00   log  6.46  10 5   log log n A n HA

n A n HA

 0.190

 0.646

3.47  102 mol  5.37  10 2 mol 0.646 0.646 This is the total amount of benzoic acid that needs to be in the solution to have pH=4.00. n HA 

n A



To find the amount of benzoic added, we subtract the amount that was originally in the solution from this amount. n benzoic acid added  n HA  n original  5.37  10 2 mol  4.09 10 2 mol  1.28 10 2 mol The mass of benzoic acid added is the amount (mol) of benzoic acid multiplied by the molar mass. g   m benzoic acid  n benzoic acid MM benzoic acid  1.28  10 2 mol  122.13   1.56g mol   17.163 Plan: Write the solubility equilibrium for silver perchlorate. Use the common ion effect to explain the question being asked. Solution: The solubility equilibrium equation for silver perchlorate is:

AgClO4 (s)

Ag  (aq) + ClO4 (aq)

When AgClO4 is dissolved in water, it dissociates to the extent that the Ksp of the AgClO4 allows. AgClO4 has a molar solubility of 27 mol/L, which is actually quite high (meaning it is very soluble) thus explaining the high percentage of AgClO4 by mass in the saturated solution. When AgClO4 is dissolved in perchloric acid, HClO4, the common ion ClO4 (aq) pushes the above equilibrium to the left. Perchloric is a very strong acid meaning that its dissociation is complete and hence the concentration of ClO 4 (aq) in the perchloric acid would be very high. This high concentration would push the above reaction very far to the left, causing the solubility of AgClO4 to diminish substantially, thus explaining the decrease in the mass percent of AgClO4 in the solution of perchloric acid. 17.164 Plan: Write the balanced equation for the precipitation reaction. Use the expression for the reaction quotient along with the concentrations calculated to calculate a value for Q. Compare Q to K. If Q=K, the reaction is at equilibrium and no precipitate will form. If Q<K, then the solution is unsaturated and no precipitate will form. If Q>K, the solution has past equilibrium and a precipitate will form. Solution:

The molar mass of magnesium chloride hexahydrate is 203.325 g/mol (remember to add the mass of the six waters of hydration).

22.5  103 g  1.11 104 mol MM MgCl2 6H 2 O 203.325 g mol 4 n 1.11  10 mol mol  3.40  10 4  MgCl2    Mg 2    V 0.325L L n MgCl2 6H2 O 

m MgCl2 6H2 O



MgF2 (s)  2KCl (aq)

MgCl2 (aq) + 2 KF (aq)

This is the overall balanced equation. The potential precipitate is MgF2. We will form the expression for Q from this precipitate. Mg 2  (aq) + 2 F (aq)

MgF2 (s) Qsp   Mg 2    F 

  3.40  10 4   0.035 

 4.2  107

Ksp = 3.7 x 10-8

Qsp > Ksp

Therefore, a precipitate will form in the solution.

‗CHAPTER 18 THERMODYNAMICS: ENTROPY, GIBBS ENERGY, AND THE DIRECTION OF CHEMICAL REACTIONS CHEMICAL CONNECTIONS BOXED READING PROBLEMS

B18.1

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   6.022x1023 molecules ATP  1 g glucose       1 mol ATP  180.16 g glucos e   1 mol glucose    = 1.20333x1023 molecules= 1.203x1023 molecules ATP/g glucose b) Molecules of ATP/g tristearin =  1 mol tristearin   458 mol ATP   6.022x1023 molecules ATP  1 g tristearin       1 mol ATP  897.50 g tristearin   1 mol tristearin    = 3.073065x1023 molecules= 3.073x1023 molecules ATP/g tristearin

B18.2

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 18.1

(a) 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. (b) The ignition of gasoline is an example of a process that is not spontaneous but very fast.

18.2

(a) 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. (b) 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).

18.3

i) The energy of the universe is constant. ii) Energy cannot be created or destroyed. iii) Usystem = –Usurroundings b) The first law is concerned with balancing energy for a process but says nothing about whether the process can, in fact, occur.

18.4

(a) Entropy is related to the freedom of movement of the particles. A system with greater freedom of movement has higher entropy. b) (i) and (ii) 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.

18.5

Vapourization is the change of a liquid substance to a gas so vapS = Sgas – Sliquid. Fusion is the change of a solid substance into a liquid so fusS = Sliquid – Ssolid. Vapourization 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.

18.6

(a) (i) In an exothermic process, the system releases heat to its surroundings. The entropy of the surroundings increases because the temperature of the surroundings increases (surrS > 0). (ii) In an endothermic process, the system absorbs heat from the surroundings and the surroundings become cooler. Thus, the entropy of the surroundings decreases (surrS< 0). (b) 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.

18.7

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.

18.8

Plan: A process is thermodynamically allowed if it is a spontaneous process. A spontaneous process is one that occurs by itself without a continuous input of energy. Solution: a) Thermodynamically allowed (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) Thermodynamically allowed (spontaneous), a lion spontaneously chases an antelope without added force. This assumes that the lion has not just eaten. c) Thermodynamically allowed (spontaneous), an unstable substance decays spontaneously to a more stable substance.

18.9

a) The movement of Earth about the Sun is thermodynamically allowed (spontaneous). b) The movement of a boulder against gravity is thermodynamically not allowed (nonspontaneous). c) The reaction of an active metal (sodium) with an active nonmetal (chlorine) is thermodynamically allowed (spontaneous).

18.10

Plan: A process is thermodynamically allowed if it is a spontaneous process. A spontaneous process is one that occurs by itself without a continuous input of energy. Solution: a) Thermodynamically allowed (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) Thermodynamically allowed (spontaneous), the dissolved sugar molecules have more states they can occupy than the crystalline sugar, so the reaction proceeds in the direction of dissolution. c) Not thermodynamically allowed (spontaneous), a cooked egg will not become raw again, no matter how long it sits or how many times it is mixed.

18.11

a) If a satellite slows sufficiently, it will fall to Earth‘s surface through a thermodynamically allowed (spontaneous) process. b) Water is a very stable compound; its decomposition at 298 K and 1 bar is not spontaneous. c) The increase in prices tends to be spontaneous.

18.12

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, sysS is positive. If the products of the process have less entropy than the reactants, sysS is negative. Solution: a) sysS 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) sysS 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) sysS negative, dew forms by the condensation of water vapour to liquid. Entropy of a substance in the gaseous state is greater than its entropy in the liquid state. Entropy decreases during condensation.

18.13

a) sysS positive

18.14

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, sysS is positive. If the products of the process have less entropy than the reactants, sysS is negative. Solution: a) sysS positive, the process described is liquid alcohol becoming gaseous alcohol. The gas molecules have greater entropy than the liquid molecules. b) sysS positive, the process described is a change from solid to gas, an increase in possible energy states for the system. c) sysS 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.

18.15

a) sysS negative

18.16

Plan: sysS 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) sysS 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) sysS negative, gaseous reactants form solid product and number of particles decreases, so entropy decreases. c) sysS positive, when a solid salt dissolves in water, entropy generally increases since the entropy of the aqueous mixture has higher entropy than the solid.

18.17

a) sysS negative

18.18

Plan: sysS 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) sysS positive, the reaction produces gaseous CO2 molecules that have greater entropy than the physical states of the reactants. b) sysS negative, the reaction produces a net decrease in the number of gaseous molecules, so the system‘s entropy decreases. c) sysS positive, the reaction produces a gas from a solid.

18.19

a) sysS negative

b) sysS positive

b) sysS negative

b) sysS positive

c) sysS negative

18.20

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) sysS positive, decreasing the pressure increases the volume available to the gas molecules so entropy of the system increases. b) sysS negative, gaseous nitrogen molecules have greater entropy (more possible states) than dissolved nitrogen molecules. c) sysS positive, dissolved oxygen molecules have lower entropy than gaseous oxygen molecules.

18.21

a) sysS negative

18.22

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.

18.23

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.

18.24

18.25

a) P4(g); it has greater molecular complexity. b) HNO3(aq); because S(solution) > S(pure). c) CuSO4 · 5H2O; it has greater molecular complexity.

18.26

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 vapour. 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.

18.27

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) sysS positive

c) sysS negative

18.28

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 N 2 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.

18.29

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.

18.30

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.

18.31

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, univS = sysS + surrS= 0. However, for a system moving to equilibrium, univS > 0, because the second law states that for any spontaneous process, the entropy of the universe increases.

18.32

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:  r S = [(2)(Sº of HClO)] – [ (Sº of H2O) + (Sº of Cl2O)] Rearranging this expression to solve for Sº of Cl2O gives: Sº of Cl2O = 2(Sº of HClO) – Sº of H2O –  r S

18.33

Plan: To calculate the standard entropy change, use the relationship  r S = m Sproducts – n S reactants . 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 the amount (mol) of gas (Δn) decreases. S° = (S° of N2O) + (S° of NO2)] – [(3)(S° of NO)] S° = [ (219.7 J/mol•K) + (239.9 J/mol•K)] – [(3)(210.65 J/mol•K)] S° = –172.35 J/ mol•K = –172.4 J/ mol•K b) Prediction: Sign difficult to predict because n = 0, but possibly S° positive because water vapour has greater complexity than H2 gas. S° = [(2)(S° of Fe) + (3)(S° of H2O)] – [(3)(S° of H2) + (S° of Fe2O3)] S° = [(2)(27.3 J/mol•K) + (3)(188.72 J/mol•K)] – [(3)(130.6 J/mol•K) + (87.400 J/mol•K)] S° = 141.56 J/ mol•K = 141.6 J/ mol•K c) Prediction: S° negative because a gaseous reactant forms a solid product and also because the amount (mol) of gas (n) decreases. S° = [ (S° of P4O10)] – [ (S°of P4) + (5)(S° of O)] S° = [ (229 J/mol•K)] – [ (41.1 J/mol•K) + (5)(205.0 J/mol•K)] S° = –837.1 J/ mol•K = –837 J/ mol•K

18.34

a) 3NO2(g) + H2O(l)  2HNO3(l) + NO(g) S° negative S° = [(2)(S° of HNO3) + (S° of NO)] – [(3)(S° of NO2) + (S° of H2O)] S° = [(2)(155.6 J/K•mol) + (210.65 J/K•mol)] – [(3)(239.9 J/K•mol) + (69.940 J/K•mol)] S° = –267.79 J/ mol•K = –267.8 J/ mol•K b) N2(g) + 3F2(g)  2NF3(g) S° negative S° = [(2)(S° of NF3] – [ (S° of N2) + (3)(S° of F2)] S° = [(2)(260.6 J/K•mol)] – [ (191.5 J/K•mol) + (3)(202.7 J/K•mol)] S° = –278.4 J/ mol•K c) C6H12O6(s) + 6O2(g)  6CO2(g) + 6H2O(g) S° positive S° = [(6)(S° of CO2) + (6)(S° of H2O)] – [ (S° of C6H12O6) + (6)(S° of O2)] S° = [(6)(213.7 J/K•mol) + (6)(188.72 J/K•mol)] – [ (212.1 J/K•mol) + (6)(205.0 J/K•mol)] S° = 972.42 J/ mol•K = 972.4 J/ mol•K

18.35

Plan: Write the balanced equation. To calculate the standard entropy change, use the relationship  r S = m Sproducts – n S reactants . 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)(S° of CO2) + (6)(S° of H2O)] – [(2)(S° of C2H6) + (7)(S° of O2)] S° = [(4)(213.7 J/mol•K) + (6)(188.72 J/mol•K)] – [(2)(229.5 J/mol•K) + (7)(205.0 J/mol•K)] S° = 93.12 J/mol•K = 93.1 J/mol•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. 18.36

CH4(g) + 2O2(g)  CO2(g) + 2H2O(l) S° = [ (S° of CO2) + (2)(S° of H2O)] – [ (S° of CH4) + (2)(S° of O2)] S° = [ (213.7 J/K•mol) + (2)(69.940 J/K•mol)] – [ (186.1 J/K•mol) + (2)(205.0 J/K•mol)] S° = –242.52 J/mol•K = –242.5 J/mol•K Yes, a decrease in the amount (mol) of gas should result in a negative S° value.

18.37



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 amount (mol) of gas decreases, i.e., n = 4 – 7 = –3, the entropy is expected to decrease. S° = [(2)(S° of NH3) + (2)(S° of H2O)] – [(2)(S° of NO) + (5)(S° of H2)] S° = [(2)(193 J/mol•K) + (2)(188.72 J/mol•K)] – [(2)(210.65 J/mol•K) + (5)(130.6 J/mol•K)] S° = –310.86 J/mol•K = –311 J/mol•K Yes, the calculated entropy matches the predicted decrease.

18.38

4NH3(g) + 7O2(g)  4NO2(g) + 6H2O(g) S° = [(4)(S° of NO2) + (6)(S° of H2O)] – [(4)(S° of NH3) + (7)(S° of O2)] S° = [(4)(239.9 J/K•mol) + (6)(188.72 J/K•mol)] – [(4)(193 J/K•mol) + (7)(205.0 J/K•mol)] S° = –115.08 J/mol•K = –115 J/mol•K Yes, a loss of one mole of a gas should result in a small negative S° value.

18.39

Plan: Write the balanced equation. To calculate the standard entropy change, use the relationship  r S = m Sproducts – n S reactants .

Solution: The reaction for forming Cu2O from copper metal and oxygen gas is 2Cu(s) + 1/2O2(g)  Cu2O(s) S° = [ (S° of Cu2O)] – [(2)(S° of Cu) + (1/2)(S° of O2)] S° = [ (93.1 J/mol•K)] – [(2)(33.1 J/mol•K) + (1/2)(205.0 J/mol•K)] S° = –75.6 J/mol•K 18.40

1/2H2(g) + 1/2I2(s)  HI(g) S° = [ (S° of HI)] – [(1/2)(S° of H2) + (1/2)(S° of I2)] S° = [ (206.33 J/K•mol)] – [(1/2)(130.6 J/K•mol) + (1/2)(116.14 J/K•mol)] S°= 82.96 J/K•mol = 83.0 J/K•mol

18.41

Plan: Write the balanced equation. To calculate the standard entropy change, use the relationship  r S = m Sproducts – n S reactants .

Solution: 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° = [ (S° of CH3OH)] – [ (S° of C) + (2)(S° of H2) + (1/2)(S° of O2)] S° = [ (127 J/mol•K)] – [ (5.686 J/mol•K) + (2)(130.6 J/mol•K) + (1/2)(205.0 J/mol•K)] S° = –242.386 J/mol•K = –242 J/mol•K 18.42

1/4P4(s) + 5/2Cl2(g)  PCl5(g) S° = [ (S° of PCl5] – [(1/4)(S° of P4) + (5/2)(S° of Cl2)] S° = [ (353 J/K•mol)] – [(1/4)(41.1 J/K•mol) + (5/2)(223.0 J/K•mol)] S° = –214.775 J/mol•K = –215 J/mol•K

18.43

SO2(g) + Ca(OH)2(s)  CaSO3(s) + H2O(l) S° = [ (S° of CaSO3) + (S° of H2O)] – [ (S° of SO2) + (S° of Ca(OH)2)] S° = [ (101.4 J/K•mol) + (69.940 J/K•mol)] – [ (248.1 J/K•mol) + (83.39 J/K•mol)] S° = –160.15 J/mol•K = –160.2 J/mol•K

18.44

Plan: Write the balanced equation. To calculate the standard entropy change, use the relationship  r S = m Sproducts – n S reactants .

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)(S° of CO2) + (S° of H2O)] – [ (S° of C2H2) + (5/2)(S° of O2)] S° = [(2)(213.7 J/mol•K) + (188.72 J/mol•K)] – [ (200.85 J/mol•K) + (5/2)(205.0 J/mol•K)] S° = –97.23 J/mol•K = –97.2 J/mol•K

18.45

Reaction spontaneity may now be predicted from the value of only one variable (sysG) rather than two (sysS and surrS).

18.46

A spontaneous process has univS > 0. Since the temperature in Kelvin is always positive, sysG must be negative (sysG < 0) for a spontaneous process.

18.47

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). 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. Since gases have higher entropy values than solids, the process results in an increase in entropy and S is positive. This is an endothermic process so the surroundings lose energy to the system. surrS 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.

i. ii. iii. iv.  

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 surrS < 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.

18.49

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.

18.50

Plan: G° can be calculated with the relationship m f (products) G – n  f (reactants) G . Solution: a) G° = [(2)(  f G of MgO)] – [(2)(  f G of Mg) + (  f G of O2)] Both Mg(s) and O2(g) are the standard-state forms of their respective elements, so their  f G values are zero. G° = [(2 )(–569.0 kJ/mol)] – [(2)(0 kJ/mol) + (1 )(0 kJ/mol)] = –1138.0 kJ/mol b)G° = [(2)(  f G of CO2) + (4)(  f G of H2O)]

– [(2)(  f G of CH3OH) + (3)(  f G of O2)]  G° = [(2)(–394.4 kJ/mol) + (4)(–228.60 kJ/mol)]  [(2)(–161.9 kJ/mol) + (3)(0)] G° = –1379.4 kJ/mol c) G° = [ (  f G of BaCO3)] – [ (  f G of BaO) + (  f G of CO2)] G° = [ (–1139 kJ/mol)] – (–520.4 kJ/mol) + (–394.4 kJ/mol)] G° = –224.2 kJ/mol= –224 kJ/mol 18.51

a) H2(g) + I2(s)  2HI(g) G° = [(2)(  f G of HI] – [ (  f G of H2) + (  f G of I2)] G° = [(2)(1.3 kJ/mol)] – [ (0 kJ/mol) + (0 kJ/mol)] G° = 2.6 kJ/mol b) MnO2(s) + 2CO(g)  Mn(s) + 2CO2(g) G° = [ (  f G of Mn) + (2)(  f G of CO2)] – [ (  f G of MnO2) + (2)(  f G of CO)] G° = [ (0 kJ/mol) + (2)(–394.4 kJ/mol)] – [ (–466.1 kJ/mol) + (2)(–137.2 kJ/mol)] G° = –48.3 kJ/mol

c) NH4Cl(s)  NH3(g) + HCl(g) G° = [ (  f G of NH3) + (  f G of HCl)] – [ (  f G of NH4Cl)] G° = [ (–16 kJ/mol) + (–95.30 kJ/mol)] – [ (–203.0 kJ/mol)] G° = 91.7 kJ/mol= 92 kJ/mol 18.52

Plan: r H can be calculated from the individual  f H values of the reactants and products by using the relationship r H = m  f (products) H – n f (reactants) H .  r S can be calculated from the individual S values of the reactants and products by using the relationship  r S = m Sproducts – n S reactants . Once H rxn and  r S are known, G° can be calculated with the relationship  r G = r H – T r S .  r S values in J/K must be converted to units of kJ/K to match the units of r H . Solution: a) r H = [(2)(  f H of MgO)] – [(2)(  f H of Mg) + (  f H of O2)] r H = [(2)(–601.2 kJ/mol)] –[(2)(0 kJ/mol) + (0 kJ/mol)] r H = –1202.4 kJ/mol  r S = [(2)( S of MgO)] – [(2)( S of Mg) + ( S of O2)]

 r S = [(2)(26.9 J/mol•K)] – [(2)(32.69 J/mol•K) + (205.0 J/mol•K)]  r S = –216.58 J/mol•K

 r G = r H – T r S = –1202.4 kJ/mol – [(298 K)(–216.58 J/mol•K )(1 kJ/103 J)] = –1137.859 kJ/mol = –1138 kJ/mol

b) r H = [(2)(  f H of CO2) + (4)(  f H of H2O)] – [(2)(  f H of CH3OH) + (3)(  f H of O2)] r H = [(2)(–393.5 kJ/mol) + (4)(–241.826 kJ/mol)] – [(2)(–201.2 kJ/mol) + (3)(0 kJ/mol)]

r H = –1351.904 kJ/mol  r S = [(2)( S of CO2) + (4)( S of H2O)]

– [(2)( S of CH3OH) + (3)( S of O2)]  r S = [(2)(213.7 J/mol•K) + (4)(188.72 J/mol•K)] – [(2)(238 J/mol•K) + (3)(205.0 J/mol•K)] = 91.28 J/mol•K  r G = r H – T r S = –1351.904 kJ/mol – [(298 K)(91.28 J/mol•K )(1 kJ/103 J)] = –1379.105kJ/mol = –1379 kJ/mol

c) r H = [ (  f H of BaCO3)] – [ (  f H of BaO) + (  f H of CO2)]

r H = [ (–1219 kJ/mol)] – [ (–548.1 kJ/mol) + (–393.5 kJ/mol)] r H = –277.4 kJ/mol  r S = [ ( S of BaCO3)] – [ ( S of BaO) + ( S of CO2)]  r S = [ (112 J/mol•K)] – [ (72.07 J/mol•K) + (213.7 J/mol•K)]  r S = –173.77 J/mol•K  r G = r H – T r S = –277.4 kJ/mol – [(298 K)(–173.77 J/mol•K)(1 kJ/103 J)] = –225.6265 kJ/mol = –226 kJ/mol

18.53

a) r H = [(2 )(25.9 kJ/mol)] – [(0 kJ/mol) + (0 kJ/mol)]

r H = 51.8 kJ/mol  r S = [(2)(206.33 J/K•mol)] – [(130.6 J/K•mol) +(116.14 J/K•mol)]  r S = 165.92 J/K•mol  r G = r H – T r S = 51.8 kJ/mol – [(298 K)(165.92 J/K•mol )(1 kJ/10 3 J)] = 2.3558 kJ/mol= 2.4 kJ/mol

b) r H = [(0 kJ/mol) + (2)(–393.5 kJ/mol)] – [(–520.9 kJ/mol) + (2 )(–110.5 kJ/mol)]

r H = –45.1 kJ/mol  r S = [(31.8 J/K•mol) + (2)(213.7 J/K•mol)] – [(53.1 J/K•mol) + (2)(197.5 J/K•mol)] r S

= 11.1 J/K•mol

 r G = r H – T r S = –45.1 kJ/mol – [(298 K (11.1 J/K•mol )(1 kJ/10 3 J)] = –48.4078 kJ/mol= –48.4 kJ/mol

c) r H = [(–45.9 kJ/mol) + (–92.3 kJ/mol)] – [(–314.4 kJ/mol)]

r H

= 176.2 kJ/mol

 r S = [(193 J/K•mol) + (186.79 J/K•mol)] – [(94.6 J/K•mol)]  r S = 285.19 J/K•mol  r G = r H – T r S = 176.2 kJ/mol – [(298 K)(285.19 J/K•mol )(1 kJ/10 3 J)] = 91.213 kJ/mol= 91.2 kJ/mol

18.54

Plan:  r G can be calculated with the relationship m f (products) G – n  f (reactants) G . Alternatively,  r G can be calculated with the relationship  r G = r H – T r S . Entropy decreases (is negative) when there are fewer moles of gaseous products than there are of gaseous reactants. Solution: a) Entropy decreases ( ΔS

negative) because the amount (mol) 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 o

releases energy (exothermic, ΔH negative) which is typical of all combustion reactions. b) Method 1: Calculate  r G from  f G values of products and reactants.  r G = m f (products) G – n  f (reactants) G  r G = [( Gf of CO2)] – [( Gf of CO) + (1/2)( Gf of O2)]  r G = [(–394.4 kJ/mol)] – [(–137.2 kJ/mol) + (1/2)(0 kJ/mol)] = –257.2 kJ/mol

Method 2: Calculate  r G from r H and  r S at 298 K (the degree superscript indicates a reaction at standard state, given in the Appendix at 25°C).

r H = m  f (products) H – n f (reactants) H r H = [( H f of CO2)] – [( H f of CO) + (1/2)( H f of O2)] r H = [(–393.5 kJ/mol)] – [(–110.5 kJ/mol) + (1/2)(0 kJ/mol)] = –283.0 kJ/mol  r S = m Sproducts – n S reactants  r S = [( S of CO2)] – [( S of CO) + (1/2)( S of O2)]  r S = [(213.7 J/mol•K)] – [(197.5 J/mol•K) + (1/2 )(205.0 J/mol•K)] = –86.3 J/mol•K  r G = r H – T r S = (–283.0 kJ/mol) – [(298 K)(–86.3 J/mol•K )(1 kJ/103 J)] = –257.2826 kJ/mol = –257.3 kJ/mol

18.55

C4H10(g) + 13/2O2(g)  4CO2(g) + 5H2O(g) a) An increase in the amount (mol) 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) r H = m  f (products) H – n f (reactants) H

r H = [(4 )(–393.5 kJ/mol) + (5 )(–241.826 kJ/mol)] – [(–126 kJ/mol) + (13/2 )(0 kJ/mol)] r H = –2657.13 kJ /mol  r S = m Sproducts – n S reactants  r S = [(4)(213.7 J/K•mol) + (5 )(188.72 J/K•mol)] – [(310 J/K•mol) + (13/2)(205.0 J/K•mol)]  r S = 155.9 J/K•mol  r G = r H – T r S = –2657.13 kJ/mol – [(298 K)(155.9 J/K•mol )(1 kJ/103 J)] = –2703.588 kJ/mol = –2704 kJ/mol  r G = m f (products) G – n  f (reactants) G  r G = [(4)(–394.4 kJ/mol) + (5 )(–228.60 kJ/mol)] – [(–16.7 kJ/mol) + (13/2)(0 kJ/mol)]  r G = –2703.9 kJ/mol

18.56

Plan: Use the relationship  r G = r H – T r S to find  r S , knowing r H and  r G . This relationship is also used to find  r G at a different temperature. Solution: Reaction is Xe(g) + 3F2(g)  XeF6(g) a)  r G = r H – T r S S° =

402 kJ/mol   280. kJ/mol H  G = = –0.40939597 kJ/mol•K = –0.409 kJ/mol•K T 298 K

b)  r G = r H – T r S = (–402 kJ/mol) – [(500. K)(–0.40939597 kJ/mol•K)] = –197.302 kJ/mol = –197 kJ/mol 18.57

a) S° =

 220. kJ/mol   206 kJ/mol H  G = = –0.046979865 kJ/mol•K = –0.047 kJ/mol•K T 298 K

b)  r G = r H – T r S = –199 kJ/mol 18.58

= –220. kJ/mol – (450. K)(– 0.046979865 kJ/K•mol) = –198.859kJ/mol

Plan: r H can be calculated from the individual  f H values of the reactants and products by using the relationship r H = m  f (products) H – n f (reactants) H .  r S can be calculated from the individual S values of the reactants and products by using the relationship  r S = m Sproducts – n S reactants . Once r H and  r S are known, G° can be calculated with the relationship  r G = r H – T  rxn S .  r S values in J/mol∙K must be converted to units of kJ/mol∙K to match the units of r H . Solution: a) r H = [(  f H of CO) + (2 )(  f H of H2)] – [(  f H of CH3OH)]

r H = [(–110.5 kJ/mol) + (2)(0 kJ/mol)] – [(–201.2 kJ/mol)] r H = 90.7 kJ /mol  r S = [( S of CO) + (2)( S of H2)] – [( S of CH3OH)]  r S = [(197.5 J/mol•K) + (2)(130.6 J/mol•K)] – [(238 J/mol•K)]  r S = 220.7 J/mol•K = 221 J/mol•K

b)  r G = r H – T r S T1 = 28 + 273 = 301 K G° = 90.7 kJ/mol – [(301 K)(220.7 J/mol•K )(1 kJ/10 3 J)] = 24.2693 kJ/mol = 24.3 kJ/mol T2 = 128 + 273 = 401 K G° = 90.7 kJ/mol – [(401 K)(220.7 J/mol•K)(1 kJ/10 3 J)] = 2.1993 kJ/mol = 2.2 kJ/mol T3 = 228 + 273 = 501 K G° = 90.7 kJ/mol – [(501 K)(220.7 J/mol•K)(1 kJ/10 3 J)] = –19.8707 kJ/mol = –19.9 kJ/mol 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 r H and  r S become spontaneous at high temperatures. 18.59

a) N2(g) + O2(g) 2NO(g)

r H = [(2 )(90.29 kJ/mol)] – [(1 )(0 kJ/mol) + (1 )(0 kJ/mol)] r H = 180.58 kJ/mol  r S = [(2 )(210.65 J/K•mol)] – [(1)(191.5 J/K•mol) + (1)(205.0 J/K•mol)]  r S = 24.8 J/K•mol

b) 373G° = H° – ((273 + 100.)K) (S°) = 180.58 kJ/mol – [(373 K)(24.8 J/K•mol)(1 kJ/103 J)] = 171.3296 kJ/mol = 171.33 kJ/mol 2833G° = H° – ((273 + 2560.)K) (S°) = 180.58 kJ/mol – [(2833 K)(24.8 J/K•mol)(1 kJ/103 J)] = 110.3216 kJ/mol= 110.3 kJ/mol 3813G° = H° – ((273 + 3540.)K) (S°) = 180.58 kJ/mol – [(3813 K)(24.8 J/K•mol )(1 kJ/103 J)] = 86.0176 kJ/mol= 86.0 kJ/mol 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. 18.60

Plan: At the normal boiling point, defined as the temperature at which the vapour pressure of the liquid equals 1 bar, 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 bar 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  r G = 0 = r H – T r S . r H can be calculated from the individual  f H values of the reactants and products by using the relationship

r H = m  f (products) H – n f (reactants) H .  r S can be calculated from the individual S values of the reactants and products by using the relationship  r S = m Sproducts – n S reactants . Solution: Br2(l) Br2(g)

r H = m  f (products) H – n f (reactants) H

r H = [(  f H of Br2(g))] – [(  f H of Br2(l))] r H = [(30.91 kJ/mol)] – [(0 kJ/mol)] = 30.91 kJ/mol  r S = m Sproducts – n S reactants  r S = [( S of Br2(g))] – [( S of Br2(l))]  r S = [(245.38 J/K•mol)] – [(152.23 J/K•mol)] = 93.15 J/K•mol = 0.09315 kJ/K•mol  r G = 0 = r H – T r S

r H = T r S T =

18.61

H S

30.91 kJ/mol = 331.830 K= 331.8 K 0.09315 kJ/K mol

S(rhombic) S(monoclinic)

r H = [(0.30 kJ/mol)] – [(0 kJ/mol)] = 0.30 kJ/mol  r S = [(32.6 J/K•mol)] – [(31.9 J/K•mol)] = 0.7 J/K•mol = 0.0007 J/K•mol  r G = 0 = r H – T r S

r H = T r S T =

18.62

H S

0.30 kJ/mol = 428.571 K= 4x102 K 0.0007 kJ/K mol

Plan:  r H can be calculated from the individual  f H values of the reactants and products by using the relationship  r H

= m  f (products) H – n f (reactants) H .  r S can be calculated from the individual S

values of the reactants and products by using the relationship  r S = m Sproducts – n S reactants . Once  r H and  r S are known, G° can be calculated with the relationship  r G =  r H

– T r S .  r S values in

J/mol•K must be converted to units of kJ/mol•K to match the units of  r H . To find the temperature at which the reaction becomes spontaneous, use  r G = 0 =  r H

– T r S 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 H 2,‖ not per 2 mol H2.

 r H = m  f (products) H – n f (reactants) H  r H = [(  f H of H2O)] – [(  f H of H2) + (1/2)(  f H of O2)]  r H = [(–241.826 kJ/mol)] – [(0 kJ/mol) + (1/2 )(0 kJ/mol)]  r H = –241.826 kJ/mol  r S = m Sproducts – n S reactants  r S = [( S of H2O)] – [( S of H2) + (1/2)( S of O2)]  r S = [(188.72 J/mol•K)] – [(130.6 J/mol•K) + (1/2 )(205.0 J/mol•K)]  r S = –44.38 J/mol•K = –44.4 J/mol•K = –0.0444 kJ/mol•K

 r G = r H

– T r S

 r G = –241.826 kJ/mol – [(298 K)( –0.0444 kJ/mol•K )]  r G = –228.6 kJ/mol

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  r G = 0  r G = 0 = r H

– T r S

 r H = T r S T =

18.63

H S

–241.826 kJ/mol = 5446.53 K= 5.45x103 K –0.0444 kJ/K mol

C6H12O6(s)  2C2H5OH(l) + 2CO2(g)

r H = m  f (products) H – n f (reactants) H r H = [(2)(–277.63 kJ/mol) + (2 )(–393.5 kJ/mol)] – [(–1273.3 kJ/mol)] r H = –68.96 kJ/mol = –69.0 kJ/mol  r S = m Sproducts – n S reactants  r S = [(2 )(161 J/K•mol) + (2 )(213.7 J/K•mol)] – [(212.1 K•mol)]  r S = 537.3 J/K•mol = 537 J/K•mol  r G = r H – T r S  r G = –68.96 kJ/mol – [(298 K)( 0.5373 kJ/K•mol )]  r G = –229.0754 kJ/mol = –229.1 kJ/mol

No, a reaction with a negative value for H and a positive value for S is spontaneous at all temperatures.

18.64

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. 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°.

18.65

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.

18.66

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).

18.67

The standard free energy change, G°, occurs when all components of the system are in their standard states. Standard state is defined as 1 bar for gases, 1 mol/L 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 mol/L and all partial pressures equal 1 bar. This occurs because the value of Q = 1 and ln Q = 0 in the equation G = G° + RT ln Q.

18.68

Plan: For each reaction, first find G°, then calculate K from G° = –RT ln K. Calculate  r G using  f G values in the relationship  r G = m f (products) G – n  f (reactants) G . Solution: a)  r G = m f (products) G – n  f (reactants) G  r G = [(  f G of NO2)] – [(  f G of NO) + (1/2)(  f G of O2)]  r G = [(51 kJ/mol)] – [(86.60 kJ/mol) + (1/2)(0 kJ/mol)] = –35.6 kJ/mol

G° = –RT ln K    103 J  G 35.6 kJ/mol ln K = =  = 14.3689   8.314 J/mol•K  298 K    1 kJ   RT    K = e14.3689 = 1.7391377x106 = 1.7x106 b)  r G = [(  f G of H2) + (  f G of Cl2)] – [(2)(  f G of HCl)]  r G = [(0 kJ/mol) + (0 kJ/mol)] – [(2)(–95.30 kJ/mol)] = 190.60 kJ/mol

ln K =

   103 J  G 190.60 kJ/mol =  = –76.930   8.314 J/mol•K  298 K    1 kJ   RT   

K = e–76.930 = 3.88799x10–34 = 3.89x10–34

c)  r G = [(2)(  f G of CO)] – [(2)(  f G of C) + (  f G of O2)]  r G = [(2)(–137.2 kJ/mol)] – [(2)(0 kJ/mol) + (0 kJ/mol)] = 274.4 kJ/mol

ln K =

   103 J  G 274.4 kJ/mol =  = 110.75359   8.314 J/mol•K  298 K    1 kJ   RT   

K = e110.75359 = 1.2579778x1048 = 1.26x1048 Note: You may get a different answer depending on how you rounded in earlier calculations. 18.69

a) MgCO3(s) Mg2+(aq) + CO32–(aq)  r G = m f (products) G – n  f (reactants) G



 r G = [(–456.01 kJ/mol) + (–528.10 kJ/mol)] – [(–1028 kJ/mol)] = 43.89 kJ/mol    103 J  G 43.89 kJ/mol ln K = =  = –17.7149   8.314 J/mol•K  298 K    1 kJ   RT    K = e–17.7149 = 2.0254274x10–8 = 2.0x10–8 b) 2HCl(g) + Br2(l) 2HBr(g) + Cl2(g)  r G = [(2)(–53.5 kJ/mol) + (2)(0 kJ/mol)] – [(2)(–95.30 kJ/mol) + (0 kJ/mol)] = 83.6 kJ/mol    103 J  G 83.6 kJ/mol ln K = =  = –33.7427   8.314 J/mol•K  298 K    1 kJ   RT    K = e–33.7427 = 2.2168x10–15 = 2.22x10–15 c) H2(g) + O2(g) H2O2(l)  r G = [(–120.4 kJ/mol)] – [(0 kJ/mol) + (0 kJ/mol)] = –120.4 kJ/mol

ln K =

   103 J  G 120.4 kJ/mol =  = 48.59596   8.314 J/mol•K  298 K    1 kJ   RT   

K = e48.59596 = 1.2733777x1021 = 1.27x1021

18.70

Plan: For each reaction, first find G°, then calculate K from G° = –RT ln K. Calculate G using  f G values in the relationship  r G = m f (products) G – n  f (reactants) G . Solution: a)  r G = m f (products) G – n  f (reactants) G  r G = [(2)(  f G of H2O) + (2)(  f G of SO2)]

– [(2)(  f G of H2S) + (3)(  f G of O2)]  r G = [(2 )(–228.60 kJ/mol) + (2)(–300.2 kJ/mol)] – [(2)(–33 kJ/mol) + (3)(0 kJ/mol)]  r G = –991.6 kJ/mol

ln K =

   103 J  G 991.6 kJ/mol =  = 400.2305   8.314 J/mol•K  298 K    1 kJ   RT   

K = e400.2305 = 6.571696x10173 = 6.57x10173 Comment: Depending on how you round  r G , the value for K can vary by a factor of 2 or 3 because the inverse natural log varies greatly with small changes in  r G . Your calculator might register ―error‖ when trying to calculate e400 because it cannot calculate exponents greater than 99. In this case, divide 400.2305 by 2 (= 200.115), and calculate e200.115e200.115 = (8.1066x1086)2 = (8.1066)2 x 1086x2 = 65.71696x10172 = 6.5 x10173, which equals the first answer with rounding errors. b)  r G = [(  f G of H2O) + (  f G of SO3)] – [(  f G of H2SO4)]  r G = [(–237.192 kJ/mol) + (–371 kJ/mol)] – [(–690.059 kJ/mol)]  r G = 81.867 kJ/mol

ln K =

   103 J  G 81.867 kJ/mol =  = –33.0432   8.314 J/mol•K  298 K    1 kJ   RT   

K = e–33.0432 = 4.4619x10–15 = 4.46x10–15

c)  r G = [(  f G of NaCN) + (  f G of H2O)] – [(  f G of HCN) + (  f G of NaOH)] NaCN(aq) and NaOH(aq) are not listed in Appendix B. Converting the equation to net ionic form will simplify the problem: HCN(aq) + OH–(aq) CN–(aq) + H2O(l)  r G = [(  f G of CN–) + (  f G of H2O)]

– [(  f G of HCN) + (  f G of OH–)]  r G = [(166 kJ/mol) + (–237.192 kJ/mol)] – [(112 kJ/mol) + (–157.30 kJ/mol)]  r G = –25.892 kJ/mol

   103 J  G 25.892 kJ/mol =  = 10.45055   8.314 J/mol•K  298 K    1 kJ   RT    K = e10.45055 = 3.4563x104 = 3.46x104 ln K =

18.71

a) SrSO4(s) Sr2+(aq) + SO42–(aq)  r G = [(–557.3 kJ/mol) + (–741.99 kJ/mol)] – [(–1334 kJ/mol)] = 34.71 kJ/mol

ln K =

   103 J  G 34.71 kJ/mol =  = –14.00968   8.314 J/mol•K  298 K    1 kJ   RT   

K = e–14.00968 = 8.23518x10–7 = 8.2x10–7 b) 2NO(g) + Cl2(g) 2NOCl(g)

 r G = [(2)(66.07 kJ/mol)] – [(2)(86.60 kJ/mol) + (0 kJ/mol)]  r G = –41.06 kJ/mol

ln K =

   103 J  G 41.06 kJ/mol =  = 16.5726768   8.314 J/mol•K  298 K    1 kJ   RT   

K = e16.5726768 = 1.575513x107 = 1.58x107 c) Cu2S(s) + O2(g) 2Cu(s) + SO2(g)  r G = [(2)(0 kJ/mol) + (–300.2 kJ/mol)] – [(–86.2 kJ/mol) + (0 kJ/mol)] rG

= –214.0 kJ/mol

ln K =

   103 J  G 214.0 kJ/mol =  = 86.374886   8.314 J/mol•K  298 K    1 kJ   RT   

K = e86.374886 = 3.25189x1037 = 3.25x1037 18.72

Plan: Write the balanced equation. First find G°, then calculate K from G° = –RT ln K. Calculate G using  f G values in the relationship  r G = m f (products) G – n  f (reactants) G .

Solution: The solubility reaction for Ag2S is Ag2S(s) + H2O(l) 2Ag+(aq) + HS–(aq) + OH–(aq)  r G = m  r G – n  f (reactants) G  r G = [(2)(  f G of Ag+) + (  f G of HS–) + (  f G of OH–]

 [(  f G of Ag2S) + (  f G of H2O)]  r G = [(2)(77.111 kJ/mol) + (12.6 kJ/mol) + (–157.30 kJ/mol)]

 [(–40.3 kJ/mol) + (–237.192 kJ/mol)]  r G = 287.014 kJ/mol

ln K =

   103 J  G 287.014 kJ/mol =  = –115.8448675   8.314 J/mol•K  298 K    1 kJ   RT   

K = e–115.8448675 = 4.8889241x10–51 = 4.89x10–51 18.73

CaF2(s) Ca2+(aq) + 2F – (aq)  r G = [(–553.04 kJ/mol) + (2)(–276.5 kJ/mol)] – [(–1162 kJ/mol)]  r G = 55.96 kJ/mol

ln K =

   103 J  G 55.96 kJ/mol =  = –22.586629   8.314 J/mol•K  298 K    1 kJ   RT   

K = e–22.586629 = 1.5514995x10–10 = 1.55x10–10

18.74

Plan: First find G°, then calculate K from G° = –RT ln K. Calculate G using  f G values in the relationship  r G = m f (products) G – n  f (reactants) G . Recognize that I2(s), not I2(g), is the standard state for iodine.

Solution: I2(g) + Cl2(g) 2 IClg)

 r G = m f (products) G – n  f (reactants) G

 r G = [(2)(  f G of ICl)] – [(  f G of I2) + (  f G of Cl2)]  r G = [(2)(–6.075 kJ/mol)] – [(19.38 kJ/mol) + (0 kJ/mol)]  r G = –31.53 kJ/mol

   103 J  G 31.53 kJ/mol =  = 12.726169   8.314 J/mol•K  298 K    1 kJ   RT    K = e12.726169 = 3.3643794x105 = 3.36x105 ln K =

18.75

CaCO3(s) CaO(s) + CO2(g) K = pCO2  r G = [(–603.5 kJ/mol) + (–394.4 kJ/mol)] – [(–1128.8 kJ/mol)]  r G = 130.9 kJ/mol

   103 J  G 130.9 kJ/mol =  = –52.83398   8.314 J/mol•K  298 K    1 kJ   RT    K = e–52.83398 = 1.1336892x10–23 = 1.13x10–23 bar = pCO2 ln K =

18.76

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 (1.7x10 –5) = 2.72094x104 J/mol = 2.7x104 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 mol/L. A Q calculation using 1 mol/L solutions will confirm this: PbCl2(s)  Pb2+(aq) + 2Cl(aq) 2+ – 2 Q = [Pb ][Cl ] = (1 mol/L)(1 mol/L)2 =1 Since Q > Ksp, it is impossible to prepare a standard-state solution containing both Pb2+(aq) and Cl–(aq).

18.77

G° = –RT ln K = –(8.314 J/mol•K)(298 K) ln (3.0x10–2) = 8.6877x103 J/mol = 8.7x103 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 mol/L. A Q calculation using 1 mol/L solutions will confirm this: Q = [Zn2+][F–]2 = (1 mol/L)(1 mol/L)2 =1 Since Q > Ksp, it is impossible to prepare a standard state solution of ZnF 2.

18.78

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: a) G° = –RT ln K = –(8.314 J/mol•K)(298 K) ln (9.1x10 –6) = 2.875776x104 J/mol= 2.9x104 J/mol b) Since  r G 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.

 Fe2    Hg 2       2

0.0102 0.0252 = 1.5625x10–4 2 0.20 0.010

 Fe3    Hg 22       298G = G° + RT ln Q = 2.875776x104 J/mol + (8.314 J/mol•K)(298 K) ln (1.5625x10–4) = 7.044187x103 J/mol= 7.0x103 J/mol Because 298G > 0 and Q > K, the reaction proceeds to the left to reach equilibrium. 18.79

a) G° = –RT ln K = –(8.314 J/mol•K)((273 + 25) K) ln (5.6x10 8) = –4.9906841x104J/mol = –5.0x104 J/mol b) Since  r G 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(NH3 )6 2   0.010  = Q=  = 6.4x1014 6 6 2   Ni  NH3  0.0010 0.0050      298G = G° + RT ln Q = –4.9906841x104 J/mol + (8.314 J/mol•K)(298 K) ln (6.4x10 14) = 3.4559756x104 J/mol= 3.5x104 J/mol Because 298G > 0 and Q > K, the reaction proceeds to the left to equilibrium.

18.80

Plan: To decide when production of ozone is favored, both the signs of r H and  r S for ozone are needed. The values of  f H and S can be used. Once r H and  r S are known, G° can be calculated with the relationship  r G = r H – T r S .  r S values in J/mol•K must be converted to units of kJ/mol•K to match the units of r H . 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).

r H = m  f (products) H – n f (reactants) H r H = [(1 mol O3)(  f H of O3)] – [(3/2 mol O2)(  f H of O2)] r H = [(143 kJ/mol] – [(3/2)(0 kJ/mol)] = 143 kJ/mol  r S = m Sproducts – n S reactants  r S = [( S of O3)] – [(3/2 )( S of O2)]  r S = [(238.82 J/mol•K] – [(3/2)(205.0 J/mol•K)] = –68.68 J/mol•K = –0.06868 KJ/mol•K

The positive sign for r H and the negative sign for  r S indicates the formation of ozone is favored at no temperature. The reaction is not thermodynamically allowed at any temperature. b)  r G = r H – T r S  r G = 143 kJ/mol – [(298 K)(–0.06868 J/mol•K )] = 163.46664 kJ/mol= 163 kJ/mol for the formation of one mole of O3. c) Calculate the value for Q and then use to find G.

pO3 pO3 2/ 2

5x10 bar  = 5.195664x10 = 7

–6

 0.21 bar  2 3

G = G° + RT ln Q = 163 kJ/mol + (8.314 J/mol•K)(298 K)(1 kJ/10 3 J) ln (5.195664x10–6) = 132.85368 kJ/mol= 1x102 kJ/mol

18.81

BaSO4(s) Ba2+(aq) + SO42–(aq) The equilibrium constant, K, is related to G° through the equation G° = –RT ln K.    103 J  G 59.1 kJ/mol ln Kp = =  = –22.930618   8.314 J/mol•K  (273  37)K    1 kJ   RT    K = e–22.930618 = 1.099915x10–10 = 1.10x10–10 Ksp = [Ba2+][SO42–] = 1.099915x10–10 = S2 S=

18.82

1.099915x10 10 = 1.0487683x10–5 = 1.05x10–5 mol/L Ba2+

a) diamond  graphite

r H = [(0 kJ/mol)] – [(1.896 kJ/mol)] = –1.896 kJ/mol  r S = [(5.686 J/K•mol)] – [(2.439 J/K•mol l)] = 3.247 J/K•mol  r G = [(0 kJ/mol)] – [(2.866 kJ/mol)] = –2.866 kJ/mol

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 bar. At all temperatures G > 0 (nonspontaneous). 18.83 (a) (b) (c)

 rS + (+) 

 rH  0 +

 rG   (+)

(d) (e)

0 ()

() 0

 +

(f)

()

Comment Thermodynamically allowed Thermodynamically allowed Not thermodynamically allowed Thermodynamically allowed Not thermodynamically allowed TS > H

a) The reaction is always thermodynamically allowed when rG < 0, so there is no need to look at the other values other than to check the answer. b) Because rG = H – TS = –TS, S must be positive for rG to be negative. c) The reaction is always nonspontaneous when rG > 0, so there is no need to look at the other values other than to check the answer. d) Because rG = H – TS = H, H must be negative for rG to be negative. e) Because rG= H – TS = –TS, S must be negative for rG to be positive. f) Because TS > H, the subtraction of a larger positive term causes rG to be negative. 18.84

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.

18.85

At the freezing point, the system is at equilibrium, so G = 0. G = 0 = H – TS H = TS S = H/T = (-2.39 kJ/mol)/((273.2 + 63.7)K) = -7.0940932x10–3 kJ/mol•K S = (0.200 mol)(-7.0940932x10–3 kJ/mol•K) = -1.4188x10–3 = -1.42x10–3 kJ/K

18.86

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

18.87

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O2  . Since the problem states that [O ] = [CO], we can use the K as the a) For the reaction, K = 2 c  Hb•O2 CO thermodynamic equilibrium constant; the K expression simplifies to: K =

 Hb•CO  Hb•O2 

   103 J  G 14 kJ/mol =  = 5.431956979   8.314 J/mol•K  (273  37)K    1 kJ   RT     Hb•CO K = e5.431956979 = 228.596 = 2.3x102 =  Hb•O2  ln K =

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. 18.88

a) MgCO3(s) MgO(s) + CO2(g) b) Locate the thermodynamic values.

r H = m  f (products) H – n f (reactants) H r H = [(–601.2 kJ/mol) + (–393.5 kJ/mol)] – [(–1112 kJ/mol)] = 117.3 kJ/mol  r S = m Sproducts – n S reactants  r S = [(26.9 J/K•mol) + (213.7 J/K•mol)] – [(65.86 J/K•mol)] = 174.74 J/K•mol  r G = r H – T r S = 65.2 kJ/mol

= 117.3 kJ/mol – (298 K)(174.74 J/K•mol)(1 kJ/103 J) = 65.22748 kJ/mol

c)  r G = 0 = r H – T r S

r H = T r S T =

H S

117.3 kJ/mol = 671 K 0.17474 kJ/K mol

d) Kp = pCO2 and G° = –RT ln K are necessary. ln K =

   103 J  G 65.22748 kJ/mol =  = –26.327178   8.314 J/mol•K  298 K    1 kJ   RT   

K = e–26.327178 = 3.6834253x10–12 = 3.68x10–12 bar = pCO2 e) This is similar to part d) except a new G° must be determined at 1200 K.  r G = r H – T r S = 117.3 kJ/mol – (1200 K)(174.74 J/K•mol)(1 kJ/103 J) = –92.388 kJ/mol    103 J  G 92.388 kJ/mol ln K = =  = 9.26028   8.314 J/mol•K 1200 K    1 kJ   RT    K = e9.26028 = 1.0512x104 = 1.05x104 bar = pCO2

18.89

Plan: Sum the two reactions to yield an overall reaction. Use the relationship  r G = r H – T r S to calculate  r G . r H and  r S 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)

r H = m  f (products) H – n f (reactants) H r H = [(  f H of UF6) + (2)(  f H of H2O)] – [(  f H of UO2) + (4)(  f H of HF) + (  f H of F2)]

r H = [(–2197 kJ/mol) + (2)(–241.826 kJ/mol)] – [(–1085 kJ/mol) + (4 )(–273 kJ/mol) + (0 kJ/mol)] r H = –503.652 kJ/mol  r S = m Sproducts – n S reactants  r S = [( S of UF6) + (2)( S of H2O)]

– [( S of UO2) + (4)( S of HF) + ( S of F2)]  r S = [(225 J/mol•K) + (2)(188.72 J/mol•K)] – [(77.0 J/mol•K) + (4 )(173.67 J/mol•K) + (202.7 J/mol•K)]  r S = –371.94 J/mol•K  r G = r H – T r S = –370. kJ/mol

18.90

= (–503.652 kJ/mol) – ((273 + 85)K)(–371.94 J/mol•K )(kJ/103 J) = –370.497 kJ/mol

CO(g) + 2H2(g)  CH3OH(l) a) r H = [(–238.6 kJ/mol)] – [(–110.5 kJ/mol) + (2)(0 kJ/mol)] = –128.1 kJ/mol  r S = [(127 J/K•mol)] – [(197.5 J/K•mol) + (2)(130.6 J/K•mol)] = –331.7 J/K•mol  r G = r H – T r S = –128.1 kJ/mol – [(298 K)(–331.7 J/K•mol)(1 kJ/103 J)] = –29.2534 kJ/mol = –29.2 kJ/mol 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)

r H = [(–116 kJ/mol) + (–241.826 kJ/mol)]  [(–201.2 kJ/mol) + (1/2 )(0 kJ/mol)] = –156.626 kJ/mol  r S = [(219 J/K•mol) + (188.72 J/K•mol)]

 [(238 J/K•mol) + (1/2 )(205.0 J/K•mol)] = 67.22 J/K•mol  r G = r H – T r S = –156.626 kJ/mol – [((273 + 100.)K)(67.22 J/K•mol )(1 kJ/103 J)] = –181.699 kJ/mol= –182 kJ/mol

18.91

Plan:  r G can be calculated with the relationship m f (products) G – n  f (reactants) G . G is found by using the relationship G = G° + RT ln Q. Solution: a) 2N2O5(g) + 6F2(g)  4NF3(g) + 5O2(g) b)  r G = m f (products) G – n  f (reactants) G  r G = [(4)(  f G of NF3) + (5)(  f G of O2)]

– [(2)(  f G of N2O5)) + (6) (  f G of F2)]  r G = [(4)(–83.3 kJ/mol)) + (5)(0 kJ/mol)]  [(2)(118 kJ/mol) + (6)(0 kJ/mol)]  r G = –569.2 kJ/mol= –569 kJ/mol

c) Calculate the value for Q and then use to find G. 4 pNF pO5  0.25 bar 4  0.50 bar 5 = 47.6837 Q = 2 3 62 = pN 2O5 pF2  0.20 bar 2  0.20 bar 6 G = G° + RT ln Q = –569.2 kJ/mol + (1 kJ/103 J)(8.314 J/mol•K)(298 K) ln (47.6837) = –559.625 kJ/mol= –5.60x102 kJ/mol 18.92

a)  r S = [(2)(210.65 J/K•mol) + (245.38 J/K•mol)] – [(2)(272.6 J/mol•K)]  r S = 121.48 J/K•mol = 121.5 J/K•mol

b)  r G = –RT ln K = –(8.314 J/mol•K)(373 K) ln (0.42) = 2690.225 J/mol= 2.7x103 J/mol c)  r G = r H – T r S

r H =  r G + T r S = 2690.225 J/mol + (373 K)(121.48 J/K•mol) = 4.8002265x104 J/mol= 4.80x104 J/mol d) r H = [(2)(90.29 kJ/mol) + (30.91 kJ/mol)] – [(2)(  f H of NOBr)] (2)(  f H of NOBr)] = [(2)(90.29 kJmol) + (30.91 kJ/mol)] – (4.8002265x104 J/mol)(1 kJ/103 J))  f H of NOBr = 81.7438675 kJ/mol= 81.7 kJ/mol

e)  r G = r H – T r S = 4.8002265x104 J/mol – (298 K)(121.48 J/mol•K) = 1.1801225x104 J/mol= 1.18x104 J/mol f)  r G = [(2)(86.60 kJ/mol) + (3.13 kJ/mol)] – [(2)(  f G of NOBr)] (2)(  f G of NOBr) = [(2)(86.60 kJ/mol) + (3.13 kJ/mol)] – (1.1801225x104 J/mol)(1 kJ/103 J)  f G of NOBr = 82.264 kJ/mol= 82.3 kJ/mol

18.93

r H = m  f (products) H – n f (reactants) H r H = [(  f H of C2H6)] – [(  f H of C2H4) + (  f H of H2)] = [(–84.667 kJ/mol)] – [(52.47 kJ/mol) + (0 kJ/mol)] = –137.137 kJ/mol= –137.14 kJ/mol  r S = m Sproducts – n S reactants  r S = [( S of C2H6)] – [( S of C2H4)) + ( S of H2)] = [(229.5 J/mol•K)] – [(219.22 J/mol•K) + (130.6 J/mol•K)] = –120.32 J/mol•K = –120.3 J/mol•K  r G = m f (products) G – n  f (reactants) G  r G = [(  f G of C2H6))] – [(  f G of C2H4) + (  f G of H2)] = (–32.89 kJ/mol)] – [(68.36 kJ/mol) + (0 kJ/mol)] = –101.25 kJ/mol

18.94

a) C6H5CH2CH3(g) → H2(g) + C6H5CHCH2(g)

r H = [(  f H of H2) + (  f H of C6H5CHCH2)] – [(  f H of C6H5CH2CH3)]

r H = [(0 kJ/mol) + (103.8 kJ/mol)] – [( –12.5 kJ/mol)] = 116.3 kJ/mol  r G = [ (  f G of H2) + (  f G of C6H5CHCH2)]

– [ (  f G of C6H5CH2CH3)]  r G = [ (0 kJ/mol) + (202.5 kJ/mol)] – [( 119.7 kJ/mol)] = 82.8 kJ/mol  r S = [( S of H2) + ( S of C6H5CHCH2)]

– [( S of C6H5CH2CH3)]  r S = [(130.6 J/mol•K) + (238 J/mol•K)] – [(255 J/mol•K)] = 113.6 J/mol•K = 114 J/mol•K

b)  r G = 0 = r H – T r S

r H = T  r S o T =

H S

116.3 kJ/mol = 1020.1754 K= 1020 K = 747°C and above 0.114 kJ/K mol

c)  r G = r H – T r S = 116.3 kJ/mol – (600°C + 273)K(114 J/mol•K )(1 kJ/103 J) = 16.778 kJ/mol = 16.8 kJ/mol    103 J  G 16.778 kJ/mol ln K = =  = –2.311617   8.314 J/mol•K  873 K    1 kJ   RT    K = e–2.311617 = 9.9101x10–2 = 9.9x10–2 d) Calculate the value for Q and then use it to find G. C6H5CH2CH3(g) → H2(g) + C6H5CHCH2(g) Initial 1.0 0 0 React – 0.50 +0.50 +0.50 (50% conversion) Silberberg, Amateis, Venkateswaran, Chen, Chemistry: The Molecular Nature of Matter and Change, 3rd Canadian Edition Page 3-697 Instructor‘s Solution Manual © Copyright 2021 McGraw-Hill Ryerson Ltd.

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 x 1.3 bar = 0.10 bar pH2 pC6 H5CHCH2  0.10 bar  0.10 bar  = 0.10 Q= = pC6 H 5CH 2CH 2  0.10 bar  G = G° + RT ln Q = 16.778 kJ/mol + (1 kJ/103 J)(8.314 J/mol•K)(873 K) ln (0.10) = 0.0655565kJ/mol = 0.07 kJ/mol 18.95

a) The formation reaction for propylene is 3C(s) + 3H2(g) → CH3CH=CH2(g) Sf = [ ( S of CH3CH=CH2)] – [(3)( S of C) + (3)( S of H2)] = [(267.1 J/mol•K)] – [(3)(5.686 J/mol•K) + (3)(130.6 J/mol•K)] = –141.758 J/mol•K = –141.8 J/mol•K

b)  f G =  f H – T f S  f G = 20.4 kJ/mol – (298 K)(–141.758 J/mol•K )(1 kJ/103 J) = 62.643884 kJ/mol= 62.6 kJ/mol c) The dehydrogenation reaction is CH3CH2CH3(g) → CH3CH=CH2(g) + H2(g)

r H = [(  f H of CH3CH=CH2) + (  f H of H2)] – [ (  f H of CH3CH2CH3)] = [(20.4 kJ/mol) + (0 kJ/mol)] – [( –105 kJ/mol)] = 125.4kJ/mol = 125 kJ/mol  r G = [(  f G of CH3CH=CH2) + (  f G of H2)]

– [(  f G of CH3CH2CH3)] = [(62.643884 kJ/mol) + (0 kJ/mol)] – [( –24.5 kJ/mol)] = 87.143884 kJ/mol= 87.1 kJ/mol d) Find K from Grxn at this elevated temperature.  r S must be found first:  r S = [( S of CH3CH=CH2) + ( S of H2)]

– [( S of CH3CH2CH3)] = [(267.1 J/mol•K) + (130.6 J/mol•K)] – [(269.9 J/mol•K)] = 127.8 J/mol•K  r G =  r H – T r S  r G = 125.4 kJ/mol – (853 K)( 127.8 J/mol•K )(1 kJ/103 J) = 16.3866 kJ/mol

   103 J  G 16.3866 kJ/mol =  = –2.3106267   8.314 J/mol•K  853 K    1 kJ   RT    K = e–2.3106267 = 0.099199 = 0.0992 pCH3CH CH 2 pH 2 K= pCH3CH2CH3 ln K =



  

x= pressure of CH3CH=CH2 and the pressure of H2 and 1.00 – x = pressure of CH3CH2CH3  x  x  0.099199 = 1.00  x  x2 + 0.099199x – 0.099199 = 0 Solving the quadratic equation: x=

b  b 2  4ac 2a

(0.099199)  x=

 0.099199 2  4 1 0.099199 2 1

x = 0.2692408 = 0.269 bar Silberberg, Amateis, Venkateswaran, Chen, Chemistry: The Molecular Nature of Matter and Change, 3rd Canadian Edition Page 3-698 Instructor‘s Solution Manual © Copyright 2021 McGraw-Hill Ryerson Ltd.

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) r H = T r S T =

18.96

18.97

r H r S

125.4 kJ/mol = 981.22066 K= 981 K = 708°C 0.1278 kJ/K  mol

The reactions are (from text): ATP4– (aq) + H2O (l) ADP3– (aq) + HPO42– (aq) + H+ (aq) Glucose (aq) + HPO42– (aq) + H+ (aq) [glucose phosphate]– (aq) + H2O (l) Glucose (aq) + ATP4–(aq) [glucose phosphate]– (aq) + ADP3– (aq) G° = –RT ln K T = (273 + 25) K = 298 K    103 J  G 30.5 kJ/mol a) i) ln K = =  = 12.3104   8.314 J/mol•K  298 K    1 kJ   RT    K = e12.3104 = 2.2199x105 = 2.22x105    103 J  G 13.8 kJ/mol ii) ln K = =  = –5.569969   8.314 J/mol•K  298 K    1 kJ   RT    K = e–5.569969 = 3.8105985x10–3 = 3.81x10–3    103 J  G 16.7 kJ/mol iii) ln K = =  = 6.74047   8.314 J/mol•K  298 K    1 kJ   RT    K = e6.74047 = 8.45958x102 = 8.46x102 b) Repeat the calculations at the new temperature. T = (273 + 37)K = 310. K 3    10 J  G 30.5 kJ/mol ln K = =     = 11.8339   RT   8.314 J/mol•K  310. K    1 kJ  K = e11.8339 = 1.37847x105 = 1.38x105    103 J  G 13.8 kJ/mol ln K = =  = –5.3543576   8.314 J/mol•K  310. K    1 kJ   RT    K = e–5.3543576 = 4.7275x10–3 = 4.73x10–3    103 J  G 16.7 kJ/mol ln K = =  = 6.4795   8.314 J/mol•K  310. K    1 kJ   RT    K = e6.4795 = 6.51645x102 = 6.52x102

G° =–30.5 kJ/mol G° = 13.8 kJ/mol G° =–16.7 kJ/mol

a) ATP4– (aq) + H2O (l) ADP3– (aq) + HPO42– (aq) + H+ (aq) G°′ = –30.5 kJ/mol    103 J  30.5 kJ/mol G  ln K′ = =  = 11.8339   8.314 J/mol•K  310. K    1 kJ   RT    K′ = e11.8339 = 1.37847x105 = 1.38x105 b) We will assume here that ΔrG°´ = ΔrG° C6H12O6(s) + 6O2(g)  6CO2(g) + 6H2O(l)  r G = [(6)(  f G of CO2) + (6)(  f G of H2O)]

– [(  f G of C6H12O6) + (6)(  f G of O2)]  r G = [(6)(–394.4 kJ/mol) + (6)(–237.192 kJ/mol)]  [(–910.56 kJ/mol) + (6)(0 kJ/mol)]  r G = –2878.992 kJ/mol= –2879.0 kJ/mol

 1 mol ATP   2878.992 kJ  c)    = 94.39318 = 94.4 mol ATP/mol glucose  30.5 kJ   1 mol glucose  d) The actual yield is 36 moles ATP (this is assumed to be an exact value). The percent yield is:   36 mol ATP   x 100% = 38.1383 % = 38.1%  94.39318 mol ATP 

18.98

SO3(g) + SCl2(l)  SOCl2(l) + SO2(g)

 r G = –75.2 kJ/mol

a) r H = m  f (products) H – n f (reactants) H

r H = [(  f H of SOCl2) + (  f H of SO2)] – [(  f H of SO3) + (  f H of SCl2)]

r H = [(–245.6 kJ/mol) + (–296.8 kJ/mol)] – [(–396 kJ/mol) + (–50.0 kJ/mol)] r H = –96.4 kJ/mol  r G = –75.2 kJ/mol = r H – T r S

–75.2 kJ/mol = –96.4 kJ/mol – T r S 75.2 kJ/mol – 96.4 kJ/mol = –21.2 kJ/mol = T r S (–21.2 kJ/mol)/(298 K) =  r S  r S = (–0.0711409 kJ/mol•K)(103 J/1 kJ) = –71.1409 J/mol•K  r S = [( S of SOCl2) + ( S of SO2)]

– [( S of SO3) + ( S of SCl2)] –71.1409 J/mol•K = [( S of SOCl2) + (248.1 J/mol•K)] – [(256.7 J/mol•K)) + (184 J/mol•K)] –71.1409 J/mol•K = [( 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 = ( S of SOCl2) 121.4591 J/mol•K = ( S of SOCl2) S of SOCl2 = 121 J/mol•K

b)  r G =0 = r H – T r S

r H = T r S T =

18.99

H S

96.4 kJ/mol = 1355.057 K = 1.4x103 K 0.0711409 kJ/K mol

Oxidation of iron:

4Fe(s) + 3O2(g)  2Fe2O3(s)

The change in Gibbs free energy for this reaction will be twice  f G of Fe2O3 since the reaction forms the oxide from its elements and is reported per mole.  f G = [(2)(–743.6 kJ/mol)] – [(4)(0 kJ/mol) + (3)(0 kJ/mol)] = –1487.2 kJ/mol

Oxidation of aluminum:

4Al(s) + 3O2(g)  2Al2O3(s)

The change in Gibbs free energy for this reaction will be twice  f G of Al2O3 since the reaction forms the oxide from its elements and is reported per mole.  f G = [(2)(–1582 kJ/mol)] – [(4)(0 kJ/mol) + (3)(0 kJ/mol)] = –3164 kJ/mol

With G° < 0, both reactions are spontaneous at 25°C. Silberberg, Amateis, Venkateswaran, Chen, Chemistry: The Molecular Nature of Matter and Change, 3rd Canadian Edition Page 3-700 Instructor‘s Solution Manual © Copyright 2021 McGraw-Hill Ryerson Ltd.

18.100 a) H2(g) + I2(g)

 2HI(g)

 HI =  0.102 = 50  H 2  I2   0.010 0.020 2

Kc =

Kc > 1

b) K = Kc(RT)Δngas K = 50[(0.08314 Lbar/molK)(733 K)] 0 K = 50 = Kc c) G° = –RT ln K = –(8.314 J/mol•K)(733 K) ln (50) = 2.38405x10 4 J/mol= 2.4x104 J/mol 2 2 0.10  HI    d) Kc = = = 50  H 2  I2   0.020 0.010

The value of Kc is 50 in this situation as in a) so G° does not change. 18.101 Plan: Use the relationship G° = –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) = 1.6682596x10 3 J/mol= 1.67x103 J/mol b) Q = [F6P]/[G6P] = 10.0 G = G° + RT ln Q G = 1.6682596x103 J/mol + (8.314 J/mol•K)(298 K) ln 10.0 G = 7.3730799x103 J/mol= 7.37x103 J/mol c) Repeat the calculation in part b) with Q = 0.100 G = 1.6682596x103 J/mol + (8.314 J/mol•K)(298 K) ln 0.100 G = –4.0365607x103 J/mol= –4.04x103 J/mol d) G = G° + RT ln Q (–2.50 kJ/mol)(103 J/1 kJ) = 1.6682596x103 J/mol + (8.314 J/mol•K)(298 K) ln Q (–4.1682596x103 J/mol) = (8.314 J/mol•K)(298 K) ln Q ln Q = (–4.1682596x103 J/mol)/[(8.314 J/mol•K)(298 K)] = –1.68239696 Q = 0.18592778 = 0.19 18.102 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 bar, and the Gsys will decrease until the system reaches equilibrium. 18.103 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 temperature in Kelvin 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

18.104 a) Respiration: Fermentation: Ethanol oxidation:

C6H12O6(s) + 6O2(g)  6CO2(g) + 6H2O(l) C6H12O6(s)  2C2H5OH(l) + 2CO2(g) C2H5OH(l) + 3O2(g)  2CO2(g) + 3H2O(l)

b)  r G = [(6)(  f G of CO2) +(6)(  f G of H2O)] – [(  f G of C6H12O6) + (6)(  f G of O2)]  r G = [(6)(–394.4 kJ/mol) + (6)(–237.192 kJ/mol)] – [(–910.56 kJ/mol) + (6)(0 kJ/mol)]  r G = –2878.992 kJ/mol  1 mol glucose   2878.992 kJ   r G /g of glucose = 1.00 g glucose     = –15.980195 kJ/g = –16.0 kJ/g  180.16 g glucose   1 mol glucose 

c)  r G = [(2)(  f G of CO2) + (2)(  f G of C2H5OH)] – [(  f G of C6H12O6)]  r G = [(2)(–394.4 kJ/mol) + (2)(–174.8 kJ/mol)] – [(–910.56 kJ/mol)]  r G = –227.84 kJ/mol  1 mol glucose   227.84 kJ   r G /g of glucose = 1.00 g glucose     = –1.264653641 kJ/g= –1.26 kJ/g  180.16 g glucose   1 mol glucose 

d)  r G = [(2)(  f G of CO2) + (3)(  f G of H2O)] – [(  f G of C2H5OH) + (3)(  f G of O2)]  r G = [(2)(–394.4 kJ/mol) + (3 )(–237.192 kJ/mol)] – [(–174.8 kJ/mol) + (3)(0 kJ/mol)]  r G = –1325.576 kJ/mol  1 mol glucose   2 mol C2 H5 OH   1325.576 kJ   r G / g glucose= 1.00 g glucose       180.16 g glucose   1 mol glucose   1 mol C 2 H 5OH  = –14.71554 kJ/g= –14.7 kJ/g

G , since ln K = 0 when K = 1.00,  RT G = 0. Use the relationship G = 0 = H – T S to find the temperature. For part b), calculate G at the

18.105 Plan: First calculate H and S . According to the relationship ln K =

higher temperature with the relationship G = H – T S and then calculate K with ln K =

G .  RT

Solution: a) H = m  f (products) H – n f (reactants) H

H = [(2)(  f H of NH3)] – [(  f H of N2) + (3)(  f H of H2)] H = [(2)(–45.9 kJ/mol)] – [(0 kJ/mol) + (3)(0 kJ/mol)] = –91.8 kJ/mol S = m Sproducts – n S reactants S = [(2)( S of NH3)] – [( S of N2) + (3)( S of H2)] S = [(2)(193 J/mol•K)] – [(191.50 J/mol•K) + (3)(130.6 J/mol•K)] S = –197.3 J/mol•K G  RT Since ln K = 0 when K = 1.00, G = 0. ln K =

G = 0 = H – T S

H = T S T =

H S

91.8 kJ/mol  103 J    = 465.281 K= 465 K 197.3 J/K mol  1 kJ 

b) G = H – T S = (–91.8 kJ/mol)(103 J/1 kJ) – (673 K)(–197.3 J/mol•K ) = 4.09829x104 J/mol

  G 4.09829x104 J/mol =   = –7.32449   RT   8.314 J/mol•K  673 K   K = e–7.32449 = 6.591958x10–4 = 6.59x10–4 c) The reaction rate is higher at the higher temperature. The time required (kinetics) overshadows the lower yield (thermodynamics). ln K =

18.106 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. 18.107 a) 2CH4(g) + 1/2O2(g) → C2H2(g) + 2H2(g) + H2O(g)

H = [(  f H of C2H2) + (2)(  f H of H2) + (  f H of H2O)] – [(2)(  f H of CH4) + (1/2)(  f H of O2)]

H = [(227 kJ/mol) + (2)(0 kJ/mol) + (–241.826 kJ/mol)] – [(2)(–74.87 kJ/mol) + (1/2)(0 kJ/mol)] H = 134.914 kJ/mol= 135 kJ/mol S = [( S of C2H2) + (2)( S of H2) + ( S of H2O)] – [(2)( S of CH4) + (1/2)( S of O2)]

S = [(200.85 J/mol•K) + (2)(130.6 J/mol•K) + (188.72 J/mol•K)] – [(2)(186.1 J/mol•K) + (1/2)(205.0 J/mol•K)] S = 176.07 J/mol•K = 176.1 J/mol•K G = 0 = H – T S

H = T S T =

H

134.914 kJ/mol  103 J    = 766.25206 K= 766 K 176.07 J/K mol  1 kJ 

S b) 2C(s) + H2(g) → C2H2(g)

H = [(  f H of C2H2)] – [(2)(  f H of C) + (  f H of H2)] H = [(227 kJ/mol)] – [(2)(0 kJ/mol) + (0 kJ/mol)] H = 227 kJ/mol S = [( S of C2H2)] – [(2)( S of C) + ( S of H2)] S = [(200.85 J/mol•K)] – [(2)(5.686 J/mol•K) + (130.6 J/mol•K)]

S = 58.878 J/mol•K = 58.9 J/mol•K G = 0 = H – T S

H = T S T =

H

227 kJ/mol  103 J    = 3855.43 K= 3860 K 58.878 J/K mol  1 kJ 

S 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.

18.108 a) (i) CH4(g) + H2O(g)

 CO(g) + 3H2(g)

steam re-forming

H = [(  f H of CO) + (3)(  f H of H2)] – [(  f H of CH4) + (  f H of H2O)]

H = [(–110.5 kJ/mol) + [3(0 kJ/mol)] – [(–74.87 kJ/mol) + (–241.826 kJ/mol)] = 206.196 kJ/mol= 206.2 kJ/mol S

= [( S of CO) + (3)( S of H2)] – [( S of CH4) + ( S of H2O)]

S = [(197.5 J/mol•K) + (3 )(130.6 J/mol•K)] – [(186.1 J/mol•K) + (188.72 J/mol•K)] S = 214.48 J/mol•K = 214.5 J/mol•K G = H – T S = (206.2 kJ/mol) – (1273.K)(214.5 J/mol•K )(1kJ/10³ J) G = –66.8585 kJ/mol = –66.86 kJ/mol

(ii) CO(g) + H2O(g)

 CO2(g) + H2(g)

water-gas shift reaction

H = [(  f H of CO2) + (  f H of H2)] – [(  f H of CO) + (  f H of H2O)]

H = [(–393.5 kJ/mol) + (0 kJ/mol)] – [(–110.5 kJ/mol) + (–241.826 kJ/mol)] H = –41.174 kJ/mol= –41.2 kJ/mol S

= [( S of CO2) + ( S of H2)] – [( S of CO) + ( S of H2O)]

S = [(213.7 J/mol•K) + (130.6 J/mol•K)] – [(197.5 J/mol•K) + (188.72 J/mol•K)] S = – 41.92 J/mol•K = –41.9 J/mol•K G = H – T S = (–41.2 kJ/mol) – (1273 K)(–41.9 J/mol•K)(1 kJ/10³ J) G = 12.1387 kJ/mol = 12.1 kJ/mol

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(–66.86 kJ/mol) + 12.1 kJ/mol = –121.62 kJ/mol= –121.6 kJ/mol

(i) Since G is negative, the reaction is spontaneous at this temperature (T = 1000 °C). (ii) At 98 bar and 50% conversion, the partial pressures of the reactants and products are as follows (from the ratio of the coefficients: CH4, 14 bar; H2O, 21 bar; CO2, 7.0 bar; CO, 7.0 bar; H2, 49 bar. Calculate Q:

pCO2 pCO pH7 2 2 pCH pH3 2O 4

7 7.0  7.0  49   = = 1.8309x107 2 3 14   21

G = G° + RT ln Q = –121.6 kJ/mol + (1 kJ/103 J) (8.314 J/mol•K)(1273 K) ln (1.8309x10 7) = 55.391 kJ/mol= 55.4 kJ/mol The reaction is not spontaneous at this point. (iii) At 98 bar and 90% conversion, the partial pressures of the reactants and products are as follows: CH 4, 2.3 bar; H2O, 3.4 bar; CO2, 10.3 bar; CO, 10.3 bar; H2, 71.8 bar. Calculate Q: pCO2 pCO pH7 2 10.310.3 71.8 7 = 5.019x1012 Q= = 2 pCH pH3 2O  2.32  3.4 3 4 G = G° + RT ln Q = –121.6 kJ/mol + (1 kJ/103 J)(8.314 J/mol•K)(1273 K) ln (5.019x1012) = 187.91 kJ/mol= 188 kJ/mol The reaction is not spontaneous at this point. 18.109 Plan: We will write the balanced equation for the reaction, then obtain the relevant thermodynamic data from the appendix to find the values of rH°and rS° for the reaction. We will then use Equation 18.7 to calculate rG° for the reaction. We will use Equation 18.12 to solve for K and use Le Chatelier‘s principle to explain how we can maximize the yield. Finally, we will use Equation 18.7 again to solve for rG°at 1000. K. Solution: (a) 2SO2 (g ) + O2 (g ) 2SO3 (g ) From the data in the appendix, we can use Hess‘ law to find rH°and a similar expression to find rS° for the reaction. o o    2 f H SO  r H o   2 f H SO   f H Oo 2  3  2

  2  396 kJ/mol     2  296.8 kJ/mol   0 

(b)

 198.4 kJ/mol o o    2 f SSO  r S o   2 f SSO   f SOo 2  3  2

  2  256.66 J/mol  K     2  248.1 J/mol  K   205.0 J/mol  K   187.88 J/mol  K r Go  r H o  T r S o

 198.4 103 J/mol   700. K  187.88 J/mol  K 

(c)

 66.9 kJ/mol  r G   RT ln K o

r Go RT 66.9  103 J/mol  8.314 J/mol  K  700. K 

ln K   (d)

 11.49 (e)

K  9.80  104 We could maximize the amount of product by removing SO 3 (g) as it is formed. This would push the reaction to form more of the product. As the K value is large, the reaction is already product favoured. Lowering the temperature would also favour the formation of the products, as it would increase the value of K and because the reaction is exothermic (Le Chatelier). Adding more oxygen or sulfur dioxide to the system would also favour the formation of the products by Le Chatelier‘s principle. Finally, decreasing the volume of the reaction vessel would also favour the products as there is a smaller total number of moles of gas on the products side as compared to the reactant side.

r Go  r H o  T r S o (f)

 198.4  103 J/mol  1000. K  187.88 J/mol  K   10.5 kJ/mol

18.110 Plan: We will write the balanced equation for the reaction, then obtain the relevant thermodynamic data from the appendix to find the values of rH°and rS° for the reaction. We will then use Equation 18.7 to calculate rG° for the reaction. We will use Equation 18.12 to solve for K and use Le Chatelier‘s principle to explain how we can minimize the formation of products. Finally, we will use Equation 18.7 and 18.12 together to solve the final part of the question. Solution: (a) H2SO4 (l ) + CaCO3 (s) CaSO4 (s) + H2 O (l ) + CO2 (g ) From the data in the appendix, we can use Hess‘ law to find rH°and a similar expression to find rS° for the reaction. (b) o o o     f H Ho 2SO4   f H CaCO  r H o    f H CaSO   f H Ho 2 O   f H CO 4 2  3 

  1432.7 kJ/mol   285.84 kJ/mol    393.5 kJ/mol     813.989 kJ/mol    1206.9 kJ/mol    91.15 kJ/mol o o o      f SHo 2SO4   f SCaCO  r S o    f SCaSO   f SHo 2 O   f SCO 4 2  3 

 107 J/mol  K  69.94 J/mol  K  213.7 J/mol  K   156.9 J/mol  K  92.9 J/mol  K   140.8 J/mol  K

(c)

r Go  r H o  T r S o  91.15  103 J/mol   350. K 140.8 J/mol  K   140.4 kJ/mol (d)

 r G  RT ln K J   140.4  103 J / mol    8.314   350.K  ln K molK   K  9.00 1020 (e)

We could minimize the formation of products by doing the reaction in a small vessel. As the only gas produced is the product carbon dioxide, the reaction would be pushed back towards products. The reaction is highly favourable (from the large K value). It is both exothermic and entropically favourable. Increasing the temperature thus pushes the reaction in the reverse direction. Adding CO 2 in the system might also push the reaction backwards and prevent it from happening.

K  pCO2 9.14 1020 K  new    9.14  1015 5 1 10 o  r G   r H o  T  r S o   RT ln K (f)

T 

r H o  r S o  R ln K 91.15 103 J/mol 140.8 J/mol  K   8.314 J/mol  K  ln 9.14 1015

 553 K 18.111 Plan: Look up the values for the enthalpy and entropy for the reactants and products in Appendix B. Calculate the enthalpy of reaction and entropy of reaction. Use these values and the temperature to calculate the free energy of the reaction. Use the free energy and temperature to calculate the value of K. Recalculate the free energy at the new temperature and use it to calculate the value at the new temperature of K. Use the equation and our knowledge of thermodynamically favourable and unfavourable conditions to predict if K will increase or decrease. Solution: NaCl (s) + H2SO4 (l)  NaHSO4 (s) + HCl (aq) ΔfHo (kJ/mol) ΔfSo (J/mol·K)

-411.1 72.12

-813.989 156.90

-1125.5 113

-167.46 55.06

a) ΔrHo = [ΔfHo (NaHSO4) + ΔfHo (HCl)] – [ΔfHo (NaCl) + ΔfHo (H2SO4)] = [(-1125.5 kJ/mol) + (-167.46 kJ/mol)] – [(-411.1 kJ/mol) + (-813.989 kJ/mol)] = -67.871 kJ/mol = -68.9 kJ/mol ΔrSo = [Sfo (NaHSO4) + Sfo (HCl)] – [Sfo (NaCl) + Sfo (H2SO4)] = [(113 J/mol·K) + (55.06 J/mol·K)] – [(72.12 J/mol·K) + (156.90 J/mol·K)] = -60.96 J/mol·K =-61 J/mol·K b) ΔrGo = ΔrHo – T ΔrSo = -6.7871 x 104 J/mol - (423 K)( -60.96 J/mol·K) = -4.21 x 104 J/mol = -42 kJ/mol c) ΔrGo = - RT ln K -4.21 x 104 J/mol = (-8.314 J/ mol·K)(423 K) ln K ln K = 11.97 K = 1.6 x 105 d) ΔrGo = ΔrHo – T ΔrSo = -6.7871 x 104 J/mol - (523 K)( -60.96 J/mol·K) = -3.60 x 104 J/mol = -36 kJ/mol e) ΔrGo = - RT ln K -3.60 x 104 J/mol = (-8.314 J/ mol·K)(523 K) ln K ln K = 8.30 Silberberg, Amateis, Venkateswaran, Chen, Chemistry: The Molecular Nature of Matter and Change, 3rd Canadian Edition Page 3-707 Instructor‘s Solution Manual © Copyright 2021 McGraw-Hill Ryerson Ltd.

K = 3.9 x 103 The entropy of the reaction is strongly negative, which is generally unfavorable for a reaction. The enthalpy of the reaction is slightly negative which means heat is a product and is favorable for a reaction. Raising the temperature will cause the reaction to move backwards (towards reactants) to use the excess heat, thus causing K to diminish. Also in the expression for the free energy, a negative value for the entropy coupled with a higher temperature creates a more positive term that cannot be balanced by the only slightly negative term for the enthalpy. Thus, the free energy term will also decrease. 18.112 Plan: Use the value of K and the temperature to find the value of the free energy at 751 K. Use the van‘t Hoff equation to find the value of K at the second temperature. Use the value of this K and the second temperature to find the value of the free energy. Solution: ΔrGo = - RT ln K = (-8.314 J/ mol·K)(751 K) ln 92 = - 2.82 x 104 J/mol = - 28.2 kJ/mol

K1  r H o  1 1      K2 R  T2 T1 

J 2.12  104 K1 1   1 mol ln     92 8.314 J  751K 851K  mol  K K  137 ΔrGo = - RT ln K = (-8.314 J/ mol·K)(851 K) ln 137 = - 3.48 x 104 J/mol = - 34.8 kJ/mol 18.113 Plan: Use the volume, temperature and amount (mol) to calculate the partial pressure of each substance in the vessel. Use the given equilibrium pressure to determine the value of K using an ICE table and the balanced chemical equation. Then use the value of K and the temperature to find the value of the free energy. Use the equation for the non-standard value of the free energy, the calculated value of the standard free energy and the reaction quotient to determine value of the non-standard free energy. Use the van‘t Hoff equation to solve for the enthalpy and the values of K and the T to find the value of the standard free energy at this temperature. Solution:

p



nRT V bar  L    550.K  mol  K  5.00L

 0.547mol   0.08314 

 5.00bar CO (g) + H2O (g) ⇄ CO2 (g) + H2 (g) I 5.00 bar 5.00 bar 0 0 C -x -x +x +x E 5.00 - x 5.00 - x x x=4.40 bar At equilibrium, Silberberg, Amateis, Venkateswaran, Chen, Chemistry: The Molecular Nature of Matter and Change, 3rd Canadian Edition Page 3-708 Instructor‘s Solution Manual © Copyright 2021 McGraw-Hill Ryerson Ltd.

pCO  p H O  5.00bar  x  5.00bar  4.40bar  0.60bar 2

pCO2  p H2  4.40bar

K

p CO2 p H2 p CO p H2 O

 4.40  2  0.60 



 53 ΔrGo= - RT ln K = (-8.314 J/ mol·K)(550. K) ln 53 = - 1.8 x 104 J/mol = - 18 kJ/mol

 r G   r G o  RT ln Q c)

  0.345  0.289   J     1.8  104 J / mol   8.314   550.K  ln  2  mol  K     0.547    1.3  104 J / mol  13kJ / mol ln

K1  r H o  1 1      K2 R  T2 T1 

r Ho 53 1   1     1.44 8.314 J  1000K 550K  d) mol  K J  r H o  3.66  10 4 mol kJ  37 mol o ΔrG = - RT lnK = (-8.314 J/ mol·K)(1000. K) ln 1.44 = - 3.0 x 103 J/mol = - 3.0 kJ/mol ln

CHAPTER 19 ELECTROCHEMISTRY: CHEMICAL CHANGE AND ELECTRICAL WORK CHEMICAL CONNECTIONS BOXED READING PROBLEMS B19.1

Plan: Reduction is the gain of electrons while oxidation is the loss of electrons. Solution: a) Reduction: Fe3+ + e– → Fe2+ Oxidation: Cu+ → Cu2+ + e– b) Overall: Fe3+ + Cu+ → Fe2+ + Cu2+

B19.2

Plan: Use the relationship G° =  nFEcell to calculate the free energy change. Solution: G° (J/mol) =  nFEcell = – (2 mol e–)(96485 C/mol e–)(0.224 J/C) = –4.322528x104 J/mol





 1 kJ  G° (kJ/mol) = 4.322528x104 J/mol  3  = –43.22528 kJ/mol= –43.2 kJ/mol  10 J  END–OF–CHAPTER PROBLEMS 19.1

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.

19.2

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.

19.3

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).

19.4

O2 is too strong a base to exist in H 2O. The reaction O2 + H 2O  2OH  occurs. Only species actually existing in solution can be present when balancing an equation.

19.5

Multiply each half-reaction by the appropriate integer to make e lost equal to e gained.

19.6

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).

19.7

(a) No, spectator ions are not used to balance the half-reactions. (b) Add spectator ions to the balanced ionic equation to obtain the balanced molecular equation.

19.8

(a) Spontaneous reactions, sysG < 0, take place in voltaic cells, which are also called galvanic cells. (b) Nonspontaneous reactions take place in electrolytic cells and result in an increase in the free energy of the cell (sysG > 0).

19.9

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.

19.10

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 molecular equation, pair K+ ions with anions and SO42– ions with cations to form neutral molecules. Solution: –8 +2 +1 +7 –2 –1 +2 0 +1 –2 16H+(aq) + 2MnO4(aq) + 10Cl(aq) → 2Mn2+(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. b) MnO4 is reduced because the oxidation number of Mn decreases from +7 in MnO 4 to +2 in Mn2+. 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)

19.11

2CrO2(aq) + 2H 2O(l) + 6ClO(aq)  2CrO42(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)

19.12

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) SO32(aq)  SO42(aq) Balance elements other than O and H: MnO4(aq)  MnO2(s) Mn is balanced SO32(aq)  SO42(aq) S is balanced Balance O by adding H2O: MnO4(aq)  MnO2(s) + 2H2O(l) add two H2O to products SO32(aq) + H2O(l)  SO42(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 SO32(aq) + H2O(l)  SO42(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) 3{SO32(aq) + H2O(l)  SO42(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 MnO4(aq) + 8H+(aq) + 3SO32(aq) + 3H2O(l)  2MnO2(s) + 4H2O(l) + 3SO42(aq) + 6H+(aq) The balanced equation in acidic solution is: 2MnO4(aq) + 2H+(aq) + 3SO32(aq)  2MnO2(s) + H2O(l) + 3SO42(aq) To change to basic solution, add OH to both sides of equation to neutralize H + 2MnO4(aq) + 2H+(aq) + 2OH(aq) + 3SO32(aq)  2MnO2(s) + H2O(l) + 3SO42(aq) + 2OH(aq) 2MnO4(aq) + 2H2O( l) + 3SO32(aq)  2MnO2(s) + H2O(l) + 3SO42(aq) + 2OH(aq) Balanced equation in basic solution: 2MnO4(aq) + H2O(l) + 3SO32(aq)  2MnO2(s) + 3SO42(aq) + 2OH(aq) Check balancing: Reactants: 2 Mn Products: 2 Mn 18 O 18 O 2H 2H 3S 3S –8 charge –8 charge Oxidizing agent is MnO4 and reducing agent is SO32.

c) Divide into half-reactions: MnO4(aq)  Mn2+(aq) H2O2(aq)  O2(g) Balance elements other than O and H: MnO4(aq)  Mn2+(aq) Mn is balanced H2O2(aq)  O2(g) No other elements to balance Balance O by adding H2O: MnO4(aq)  Mn2+(aq) + 4H2O(l) add four H2O to products H2O2(aq)  O2(g) O is balanced Balance H by adding H+: MnO4(aq) + 8H+(aq)  Mn2+(aq) + 4H2O(l) add eight H+ to reactants + H2O2(aq)  O2(g) + 2H (aq) add two H+ to products – Balance charge by adding e : MnO4(aq) + 8H+(aq) + 5e–  Mn2+(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{MnO4(aq) + 8H+(aq) + 5e–  Mn2+(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: 2MnO4(aq) + 16H+(aq) + 5H2O2(aq)  2Mn2+(aq) + 8H2O(l) + 5O2(g) + 10H+(aq) The balanced equation in acidic solution: 2MnO4(aq) + 6H+(aq) + 5H2O2(aq)  2Mn2+(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. 19.13

a) 3O2(g) + 4NO(g) + 2H 2O(l)  4NO3–(aq) + 4H +(aq) Oxidizing agent is O2 and reducing agent is NO. b) 2CrO42–(aq) + 8H 2O(l) + 3Cu(s)  2Cr(OH)3(s) + 3Cu(OH)2(s) + 4OH –(aq) Oxidizing agent is CrO42– and reducing agent is Cu. 3– c) AsO4 (aq) + NO2–(aq) + H 2O(l)  AsO2–(aq) + NO3–(aq) + 2OH –(aq) Oxidizing agent is AsO43– and reducing agent is NO2–.

19.14

Add the two half-reactions multiplying the oxidation half-reaction by three to equalize the electrons: Cr2O72(aq) + 14H+(aq) + 6e–  2Cr3+(aq) + 7H2O(l) 3Zn(s)  3Zn2+(aq) + 6e– Add half-reactions and cancel substances that appear as both reactants and products: Cr2O72(aq) + 14H+(aq) + 3Zn(s)  2Cr3+(aq) + 7H2O(l) + 3Zn2+(aq) Oxidizing agent is Cr2O72 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: Zn(s)  Zn2+(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) 5Zn(s)  5Zn2+(aq) + 10e Add half-reactions and cancel substances that appear as both reactants and products: 2NO3(aq) + 12H+(aq) + 5Zn(s)  N2(g) + 6H2O(l) + 5Zn2+(aq) Oxidizing agent is NO3 and reducing agent is Zn. 19.15

a) 3BH 4–(aq) + 4ClO3–(aq)  3H 2BO3–(aq) + 4Cl–(aq) + 3H 2O(l) Oxidizing agent is ClO3– and reducing agent is BH 4–. b) 2CrO42–(aq) + 3N2O(g) + 10H +(aq)  2Cr3+(aq) + 6NO(g) + 5H 2O(l) Oxidizing agent is CrO42– 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.

19.16

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: BiO3(aq)  Bi3+(aq) + 3H2O(l) balance O by adding H2O BiO3(aq) + 6H+(aq)  Bi3+(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: Mn2+(aq) + 4H2O(l)  MnO4(aq) balance O by adding H2O Mn2+(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: 5{BiO3(aq) + 6H+(aq) + 2e–  Bi3+(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: 5BiO3(aq) + 30H+(aq) + 10e–  5Bi3+(aq) + 15H2O(l) 2Mn2+(aq) + 8H2O(l)  2MnO4(aq) + 16H+(aq) + 10e– Add half-reactions. Cancel H2O and H+ in reactants and products: 5BiO3(aq) + 30H+(aq) + 2Mn2+(aq) + 8H2O(l)  5Bi3+(aq) + 15H2O(l) + 2MnO4(aq) + 16H+(aq) Balanced reaction in acidic solution: 5BiO3(aq) + 14H+(aq) + 2Mn2+(aq)  5Bi3+(aq) + 7H2O(l) + 2MnO4(aq) BiO3– is the oxidizing agent and Mn2+ is the reducing agent. 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– Silberberg, Amateis, Venkateswaran, Chen, Chemistry: The Molecular Nature of Matter and Change, 3rd Canadian Edition Page 3-715 Instructor‘s Solution Manual © Copyright 2021 McGraw-Hill Ryerson Ltd.

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. 19.17

a) 2NO2(g) + 2OH (aq)  NO3(aq) + NO2(aq) + H 2O(l) Oxidizing agent is NO2 and reducing agent is NO2. b) 4Zn(s) + 7OH (aq) + NO3(aq) + 6H 2O(l)  4Zn(OH)42(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.

19.18

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 reduction half-reaction: MnO4(aq)  Mn2+(aq) + 4H2O(l) balance O by adding H2O MnO4(aq) + 8H+(aq)  Mn2+(aq) + 4H2O(l) balance H by adding H+  +  2+ MnO4 (aq) + 8H (aq) + 5e  Mn (aq) + 4H2O(l) balance charge by adding 5e Balance oxidation half-reaction: As4O6(s)  4AsO43(aq) balance As As4O6(s) + 10H2O(l)  4AsO43(aq) balance O by adding H2O As4O6(s) + 10H2O(l)  4AsO43(aq) + 20H+(aq) balance H by adding H+ 3 +  As4O6(s) + 10H2O(l)  4AsO4 (aq) + 20H (aq) + 8e balance charge by adding 8e Multiply reduction half-reaction by 8 and oxidation half-reaction by 5 to transfer 40 e in overall reaction. 8MnO4(aq) + 64H+(aq) + 40e  8Mn2+(aq) + 32H2O(l) 5As4O6(s) + 50H2O(l)  20AsO43(aq) + 100H+(aq) + 40e Add the half-reactions and cancel H2O and H+: 5As4O6(s) + 8MnO4(aq) + 64H+(aq) + 50H2O(l)  20AsO43(aq) + 8Mn2+(aq) + 32H2O(l) + 100H+(aq) Balanced reaction in acidic solution: 5As4O6(s) + 8MnO4(aq) + 18H2O(l)  20AsO43(aq) + 8Mn2+(aq) + 36H+(aq) Oxidizing agent is MnO4 and reducing agent is As4O6. b) The reaction gives only one reactant, P 4. 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: P4(s)  4HPO32(aq) balance P P4(s) + 12H2O(l)  4HPO32(aq) balance O by adding H2O P4(s) + 12H2O(l)  4HPO32(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+: 2P4(s) + 12H+(aq) + 12H2O(l)  4HPO32(aq) + 4PH3(g) + 20H+(aq) Balanced reaction in acidic solution: 2P4(s) + 12H2O(l)  4HPO32(aq) + 4PH3(g) + 8H+(aq) or P4(s) + 6H2O(l)  2HPO32(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-. 19.19

a) SO32(aq) + 2OH (aq) + Cl2(g)  SO42(aq) + 2Cl(aq) + H 2O(l) Oxidizing agent is Cl2 and reducing agent is SO32. 3 b) 7Fe(CN)6 (aq) + Re(s) + 8OH (aq)  7Fe(CN)64(aq) + ReO4(aq) + 4H 2O(l) Oxidizing agent is Fe(CN)63 and reducing agent is Re.  c) 2MnO4 (aq) + 5HCOOH(aq) + 6H +(aq)  2Mn2+(aq) + 5CO2(g) + 8H 2O(l) Oxidizing agent is MnO4 and reducing agent is H COOH .

19.20

5Fe2+(aq) + MnO4–(aq) + 8H +(aq)  Mn2+(aq) + 5Fe3+(aq) + 4H 2O(l)

19.21

a) Balance reduction half-reaction: NO3–(aq)  NO2(g) + H2O(l) balance O by adding H2O NO3–(aq) + 2H+(aq)  NO2(g) + H2O(l) balance H by adding H+ – + – NO3 (aq) + 2H (aq) + e  NO2(g) + H2O(l) balance charge to give 0 on each side Balance oxidation half-reaction: Au(s) + 4Cl–(aq)  AuCl4–(aq) 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 } multiply by one to give 3e– This gives: 3NO3–(aq) + 6H+(aq) + 3e–  3NO2(g) + 3H2O(l) 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–.

19.22

19.23

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.

19.24

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.

19.25

(a) An active electrode is a reactant or product in the cell reaction, whereas an inactive electrode is neither a reactant nor a product. (b) An inactive electrode is present only to conduct electricity when the half-cell reaction does not include a metal. (c) Platinum and graphite are commonly used as inactive electrodes.

19.26

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. 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.

19.27

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: Zn(s)  Zn2+(aq) + 2e– Reduction half-reaction: Sn2+(aq) + 2e–  Sn(s) Overall reaction: Zn(s) + Sn2+(aq)  Zn2+(aq) + Sn(s) b) Voltmeter e e rr

Salt bridge

()

(+) mol/L

1 mol/L Zn2+ Anion flow

Cation flow

19.28

(red half-rxn) (ox half-rxn) (overall rxn)

Ag+(aq) + e–  Ag(s) Pb(s)  Pb2+(aq) + 2e– 2Ag+(aq) + Pb(s)  2Ag(s) + Pb2+(aq)

b) e

Salt bridge

Pb () 1 mol/L Pb2+

Voltmeter

e

Ag (+)

Anion flow

1 mol/L Cation Ag+ flow

19.29

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 mol/L 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 produces Fe2+, so salt bridge anions move from right (nickel half-cell) to left (iron half-cell). k) Oxidation half-reaction: Fe(s)  Fe2+(aq) + 2e– Reduction half-reaction: Ni2+(aq) + 2e–  Ni(s) Overall cell reaction: Fe(s) + Ni2+(aq)  Fe2+(aq) + Ni(s)

19.30

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 mol/L 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. k) Reduction: Co2+(aq) + 2e–  Co(s) Oxidation: Zn(s)  Zn2+(aq) + 2e– Overall: Zn(s) + Co2+(aq)  Co(s) + Zn2+(aq)

19.31

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. Reduction half-reaction: Fe2+(aq) + 2e–  Fe(s) Oxidation half-reaction: Mn(s)  Mn2+(aq) + 2e– Overall cell reaction: Fe2+(aq) + Mn(s)  Fe(s) + Mn2+(aq) b) Voltmeter e– e– 



Salt bridge

()

(+) 1 mol/L + Fe2

1 mol/L + Mn2

19.32

Anion flow

(red half-rxn) (ox half-rxn) (overall rxn)

e–

Cation flow

Cu2+(aq) + 2e–  Cu(s) Ni(s)  Ni2+(aq) + 2e– Cu2+(aq) + Ni(s)  Cu(s) + Ni2+(aq) Voltmeter



e– 

Salt bridge

()

(+)

1 mol/L + Ni2 +

1 mol/L Cu2+

Anion flow

19.33

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. Al(s) | Al3+(aq) || Cr3+(aq) | Cr(s)

b) Cu2+ 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. Pt | SO2(g) | SO42–(aq), H+(aq) || Cu2+(aq) | Cu(s) 19.34

a) Mn(s) + Cd2+(aq)  Mn2+(aq) + Cd(s) b) 3Fe(s) + 2NO3–(aq) + 8H +(aq)  3Fe2+(aq) + 2NO(g) + 4H 2O(l)

19.35

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.

19.36

(a) A negative Ecell indicates that the cell reaction is not spontaneous, G° > 0. (b) The reverse reaction is spontaneous with Ecell > 0.

19.37

(a) Similar to other state functions, the sign of E° changes when a reaction is reversed. (b) 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.

19.38

Plan: Divide the balanced equation into reduction and oxidation half-reactions and add electrons. Add water and hydroxide ion to the half-reaction that includes oxygen. Use the relationship Ecell = Ecathode – Eanode to find the unknown E° value. Solution: a) Oxidation: Se2–(aq)  Se(s) + 2e– Reduction: 2SO32–(aq) + 3H2O(l) + 4e–  S2O32–(aq) + 6OH–(aq) b) Ecell = Ecathode – Eanode

Eanode = Ecathode – Ecell = –0.57 V – 0.35 V = –0.92 V 19.39

a) Reduction: Oxidation:

O3(g) + 2H +(aq) + 2e–  O2(g) + H 2O(l) Mn2+(aq) + 2H 2O(l)  MnO2(s) + 4H +(aq) + 2e–

b) Ecell = Ecathode – Eanode

Eanode = Ecathode – Ecell = 2.07 V – 0.84 V= 1.23 V 19.40

Plan: The greater (more positive) the reduction potential, the greater the strength as an oxidizing agent. Solution: a) From Appendix D: Fe3+(aq) + e–  Fe2+(aq) E° = 0.77 V Br2(l) + 2e–  2Br–(aq) E° = 1.07 V Cu2+(aq) + e–  Cu(s) E° = 0.34 V When placed in order of decreasing strength as oxidizing agents: Br2 > Fe3+ > Cu2+. b) From Appendix D: Ca2+(aq) + 2e–  Ca(s) E° = –2.87 V Cr2O72–(aq) + 14H+(aq) 6e–  2Cr3+(aq) + 7H2O(l) E° = 1.33 V Ag+(aq) + e–  Ag(s) E° = 0.80 V When placed in order of increasing strength as oxidizing agents: Ca2+ < Ag+ < Cr2O72–.

19.41

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

19.42

Plan: Use the relationship Ecell = Ecathode – Eanode . E values are found in Appendix D. Spontaneous reactions have Ecell > 0. Solution: a) Oxidation: Reduction: Overall reaction:

Co(s)  Co2+(aq) + 2e– 2H+(aq) + 2e–  H2(g) Co(s) + 2H+(aq)  Co2+(aq) + H2(g)

E° = –0.28 V E° = 0.00 V

Ecell = 0.00 V (–0.28 V) = 0.28 V b)

Reaction is spontaneous under standard-state conditions because Ecell is positive. Oxidation: 2{Mn2+(aq) + 4H2O(l)  MnO4–(aq) + 8H+(aq) + 5e–} E° = +1.51 V Reduction: 5{Br2(l) + 2e–  2 Br–(aq)} E° = +1.07 V Overall: 2Mn2+(aq) + 5Br2(l) + 8H2O(l)  2MnO4–(aq) + 10Br–(aq) + 16H+(aq)

Ecell = 1.07 V – 1.51 V = –0.44 V c)

Reaction is not spontaneous under standard-state conditions with Ecell < 0. Oxidation: Hg22+(aq)  2Hg2+(aq) + 2e– E° = +0.92 V Reduction: Hg22+(aq) + 2e–  2Hg(l) E° = +0.85 V Overall: 2Hg22+(aq)  2Hg2+(aq) + 2Hg(l) or Hg22+(aq)  Hg2+(aq) + Hg(l)

Ecell = 0.85 V 0.92 V = –0.07 V

Negative Ecell indicates reaction is not spontaneous under standard-state conditions. 19.43

a) Cl2(g) + 2Fe2+(aq)  2Cl–(aq) + 2Fe3+(aq)

Ecell = Ecathode – Eanode = 1.36 V – (0.77 V) = 0.59 V The reaction is spontaneous. b) Mn2+(aq) + 2H 2O(l) + 2Co3+(aq)  MnO2(s) + 4H +(aq) + 2Co2+(aq) Ecell = Ecathode – Eanode = 1.82 V – (1.23 V) = 0.59 V The reaction is spontaneous. c) 3AgCl(s) + NO(g) + 2H 2O(l)  3Ag(s) + 3Cl–(aq) + NO3–(aq) + 4H +(aq) Ecell = Ecathode – Eanode = 0.22 V – (0.96 V) = –0.74 V The reaction is nonspontaneous. 19.44

Plan: Use the relationship Ecell = Ecathode – Eanode . E values are found in Appendix D. Spontaneous reactions have Ecell > 0. Solution: a) Oxidation: 2{Ag(s)  Ag+(aq) + e–} Reduction: Cu2+(aq) + 2e–  Cu(s) Overall: 2Ag(s) + Cu2+(aq)  2Ag+(aq) + Cu(s)

E° = +0.80 V E° = +0.34 V

Ecell = +0.34 V – 0.80 V = –0.46 V b)

The reaction is not spontaneous. Oxidation: 3{Cd(s)  Cd2+(aq) + 2e–} E° = –0.40 V Reduction: Cr2O72–(aq) + 14H+(aq) + 6e–  2Cr3+(aq) + 7H2O(l) E° = +1.33 V Overall: Cr2O72–(aq) + 3Cd(s) + 14H+(aq)  2Cr3+(aq) + 3Cd2+(aq) + 7H2O(l)

Ecell = +1.33 V– (–0.40 V) = +1.73 V The reaction is spontaneous.

Oxidation: Reduction: Overall:

Pb(s)  Pb2+(aq) + 2e– Ni2+(aq) + 2e–  Ni(s) Pb(s) + Ni2+(aq)  Pb2+(aq) + Ni(s)

E° = –0.13 V E° = –0.25 V

Ecell = –0.25 V– (–0.13 V) = –0.12 V The reaction is not spontaneous. 19.45

a) 2Cu+(aq) + PbO2(s) + 4H +(aq) + SO42–(aq)  2Cu2+(aq) + PbSO4(s) + 2H 2O(l)

Ecell = Ecathode – Eanode = 1.70 V – (0.15 V) = 1.55 V The reaction is spontaneous. b) H 2O2(aq) + Ni2+(aq)  2H +(aq) + O2(g) + Ni(s) Ecell = Ecathode – Eanode = –0.25 V – (0.68 V) = –0.93 V The reaction is nonspontaneous. c) 3Ag+(aq) + MnO2(s) + 4OH –(aq)  3Ag(s) + MnO4–(aq) + 2H 2O(l) Ecell = Ecathode – Eanode = 0.80 V – (0.59 V) = 0.21 V The reaction is spontaneous. 19.46

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 that will result in a positive value of Ecell using the relationship Ecell = Ecathode – 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: Oxidation: 2{Al(s)  Al3+(aq) + 3e–} E° = –1.66 V Reduction: 3{N2O4(g) + 2e–  2NO2–(aq)} E° = +0.867 V 3N2O4(g) + 2Al(s)  6NO2–(aq) + 2Al3+(aq)

Ecell = 0.867 V – (–1.66 V) = 2.53 V Oxidizing agents: N2O4 > Al3+; reducing agents: Al > NO2– Adding (1) and (3) to give a spontaneous reaction involves converting (1) to oxidation: Oxidation: 2{Al(s)  Al3+(aq) + 3e–} E° = –1.66 V Reduction: 3{SO42–(aq) + H2O(l) + 2e– SO32–(aq) + 2OH–(aq)} E° = +0.93 V 2Al(s) + 3SO42–(aq) + 3H2O(l)  2Al3+(aq) + 3SO32–(aq) + 6OH–(aq) Ecell = 0.93 V – (–1.66 V) = 2.59 V Oxidizing agents: SO4 > Al ; reducing agents: Al > SO32– Adding (2) and (3) to give a spontaneous reaction involves converting (2) to oxidation: Oxidation: 2NO2–(aq)  N2O4(g) + 2e– E° = 0.867 V Reduction: SO42–(aq) + H2O(l) + 2e–  SO32–(aq) + 2OH–(aq) E° = 0.93 V SO42–(aq) + 2NO2–(aq) + H2O(l)  SO32–(aq) + N2O4(g) + 2OH–(aq) 2–

Ecell = 0.93 V – 0.867 V = 0.06 V Oxidizing agents: SO42– > N2O4; reducing agents: NO2– > SO32– Rank oxidizing agents (substance being reduced) in order of increasing strength: Al3+ < N2O4 < SO42– Rank reducing agents (substance being oxidized) in order of increasing strength: SO32– < NO2– < Al

19.47

3N2O(g) + 6H +(aq) + 2Cr(s)  3N2(g) + 3H 2O(l) + 2Cr3+(aq)

Ecell = 1.77 V – (–0.74 V) = 2.51 V Oxidizing agents: N2O > Cr3+; reducing agents: Cr > N2 3Au+(aq) + Cr(s)  3Au(s) + Cr3+(aq) Ecell = 1.69 V – (–0.74 V) = 2.43 V Oxidizing agents: Au+ > Cr3+; reducing agents: Cr > Au N2O(g) + 2H +(aq) + 2Au(s)  N2(g) + H 2O(l) + 2Au+(aq) Ecell = 1.77 V – (1.69 V) = 0.08 V Oxidizing agents: N2O > Au+; reducing agents: Au > N2 Oxidizing agents: N2O > Au+ > Cr3+; reducing agents: Cr > Au > N2 19.48

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 that will result in a positive value of Ecell using the relationship Ecell = Ecathode – 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: Oxidation: Pt(s)  Pt2+(aq) + 2e– E° = +1.20 V Reduction: 2HClO(aq) + 2H+(aq) + 2e– Cl2(g) + 2H2O(l) E° = +1.63 V 2HClO(aq) + Pt(s) + 2H+(aq)  Cl2(g) + Pt2+(aq) + 2H2O(l)

Ecell = 1.63 V – 1.20 V = 0.43 V Oxidizing agents: HClO > Pt2+; reducing agents: Pt > Cl2 Adding (1) and (3) to give a spontaneous reaction involves converting (3) to oxidation: Oxidation: Pb(s) + SO42–(aq)  PbSO4(s) + 2e– E° = –0.31 V Reduction: 2HClO(aq) + 2H+(aq) + 2e–  Cl2(g) + 2H2O(l) E° = +1.63 V 2HClO(aq) + Pb(s) + SO42–(aq) + 2H+(aq)  Cl2(g) + PbSO4(s) + 2H2O(l) Ecell = 1.63 V – (–0.31 V) = 1.94 V Oxidizing agents: HClO > PbSO4; reducing agents: Pb > Cl2 Adding (2) and (3) to give a spontaneous reaction involves converting (3) to oxidation: Oxidation: Pb(s) + SO42–(aq)  PbSO4(s) + 2e– E° = –0.31 V Reduction: Pt2+(aq) + 2e– Pt(s) E° = 1.20 V Pt2+(aq) + Pb(s) + SO42–(aq)  Pt(s) + PbSO4(s) Ecell = 1.20 V – (–0.31 V) = 1.51 V Oxidizing agents: Pt2+ > PbSO4; reducing agents: Pb > Pt Order of increasing strength as oxidizing agent: PbSO4 < Pt2+ < HClO Order of increasing strength as reducing agent: Cl2 < Pt < (Pb + SO42–) 19.49

S2O82–(aq) + 2I –(aq)  2SO42–(aq) + I 2(s)

Ecell = 2.01 V – (0.53 V) = 1.48 V Oxidizing agents: S2O82– > I 2; reducing agents: I – > SO42– 2– Cr2O7 (aq) + 14H +(aq) + 6I –(aq)  2Cr3+(aq) + 7H 2O(l) + 3I 2(s) Ecell = 1.33 V – (0.53 V) = 0.80 V Oxidizing agents: Cr2O72– > I 2; reducing agents: I – > Cr3+ 3S2O82–(aq) + 2Cr3+(aq) + 7H 2O(l)  6SO42–(aq) + Cr2O72–(aq) + 14H +(aq) Ecell = 2.01 V – (1.33 V) = 0.68 V Oxidizing agents: S2O82– > Cr2O72–; reducing agents: Cr3+ > SO42– Oxidizing agents: S2O82– > Cr2O72– > I 2; reducing agents: I – > Cr3+ > SO42– Silberberg, Amateis, Venkateswaran, Chen, Chemistry: The Molecular Nature of Matter and Change, 3rd Canadian Edition Page 3-724 Instructor‘s Solution Manual © Copyright 2021 McGraw-Hill Ryerson Ltd.

19.50

Metal A + Metal B salt  solid coloured 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. 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.

19.51

a) Copper metal is coating the iron. b) The oxidizing agent is Cu2+ and the reducing agent is Fe. c) Yes, this reaction, being spontaneous, may be made into a voltaic cell. d) Cu2+(aq) + Fe(s)  Cu(s) + Fe2+(aq) e) Ecell = Ecathode – Eanode = 0.34 V – (–0.44 V) = 0.78 V

RT  Q  ln and G = –zFEcell zF  K  a) When Q/K < 1, the reaction is preceding to the right; Ecell > 0 and G < 0 and the reaction is thermodynamically allowed (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.

19.52

Ecell = –

19.53

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  c) Ecell and Ecell are related by the Nernst equation: Ecell = Ecell – . ln zF  B    A  A  RT   Ecell = Ecell when = 0. This occurs when ln   = 0. Recall that e0 = 1, so [A+] must equal [B+] ln zF  B   B      for Ecell to equal

Ecell .

d) Yes, it is possible for Ecell to be less than Ecell when [A+] > [B+]. 19.54

a) Examine the Nernst equation: Ecell = Ecell –

RT ln Q. zF

RT RT ln K – ln Q zF zF RT  K  RT  Q  ln Ecell =  ln  = – zF  Q  zF  K 

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. b) Ecell = Ecell –

 Active ion at anode RT ln zF  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. 19.55

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.

19.56

Plan: The equilibrium constant can be found by using ln K =

zFEcell . Use E° values from Appendix D to RT

calculate Ecell ( Ecell = Ecathode – Eanode ) and then calculate K. The substances given in the problem must be the reactants in the equation. Solution: a) Oxidation: Ni(s)  Ni2+(aq) + 2e– E° = –0.25 V Reduction: 2{Ag+(aq) + 1e–  Ag(s)} E° = +0.80 V Ni(s) + 2Ag+(aq)  Ni2+(aq) + 2Ag(s)

Ecell = Ecathode – Eanode = 0.80 V – (–0.25 V) = 1.05 V; two electrons are transferred. 2(96485 C/mol e ) 1.05 V  zFEcell = = 81.78107 RT (8.314 J/mol  K)(298 K) K = e81.78107 K = 3.28904x1035 = 3x1035 b) Oxidation: 3{Fe(s)  Fe2+(aq) + 2e–} Reduction: 2{Cr3+(aq) + 3e–  Cr(s)} 3Fe(s) + 2Cr3+(aq)  3Fe2+(aq) + 2Cr(s) 

ln K =

E° = –0.44 V E° = –0.74 V

Ecell = Ecathode – Eanode = –0.74 V – (–0.44 V) = –0.30 V; six electrons are transferred. 6(96485 C/mol e )  0.30 V  zFEcell = = –70.09806 RT (8.314 J/mol  K)(298 K) 

ln K =

K = e–70.09806 K = 3.604x10–31 = 4x10–31 19.57

a) 2Al(s) + 3Cd2+(aq)  2Al3+(aq) + 3Cd(s)

z=6

Ecell = Ecathode – Eanode = –0.40 V – (–1.66 V) = 1.26 V 6(96485 C/mol e ) 1.26 V  zFEcell = = 294.4119 RT (8.314 J/mol  K)(298 K) K = 7.2688 x10127 = 7x10127 b) I2(s) + 2Br–(aq)  Br2(l) + 2I –(aq) z=2 

ln K =

Ecell = Ecathode – Eanode = 0.53 V – (1.07 V) = –0.54 V 2(96485 C/mol e )   0.54 V  zFEcell = = –42.0588 RT (8.314 J/mol  K)(298 K) K = 5.4210 x10–19 = 5x10–19 

ln K =

19.58

Plan: The equilibrium constant can be found by using ln K =

zFEcell . Use E° values from Appendix D to RT

calculate Ecell ( Ecell = Ecathode – Eanode ) and then calculate K. The substances given in the problem must be the reactants in the equation. Solution: a) Oxidation: 2{Ag(s)  Ag+(aq) + 1e–} E° = +0.80 V Reduction: Mn2+(aq) + 2e–  Mn(s) E° = –1.18 V 2Ag(s) + Mn2+(aq)  2Ag+(aq) + Mn(s)

Ecell = Ecathode – Eanode = –1.18 V – (0.80 V) = –1.98V; two electrons are transferred. 2(96485 C/mol e )   1.98 V  zFEcell = = –-154.2157 RT (8.314 J/mol  K)(298 K) 

ln K =

K = e–154.2157 K = 1.05914x10–67 = 1x10–67 b) Oxidation: 2Br–(aq)  Br2(l) + 2e– Reduction: Cl2(g) + 2e–  2Cl–(aq) 2Br–(aq) + Cl2(g)  Br2(l) + 2Cl–(aq)

E° = 1.07 V E° = 1.36 V

Ecell = Ecathode – Eanode = 1.36 V – 1.07 V = 0.29 V; two electrons transferred. 2(96485 C/mol e )  0.29 V  zFEcell = = 22.58715 RT (8.314 J/mol  K)(298 K) K = e22.58715 K = 6.44876x109 = 6x109 

ln K =

19.59

a) 2Cr(s) + 3Cu2+(aq)  2Cr3+(aq) + 3Cu(s)

z=6

Ecell = Ecathode – Eanode = 0.34 V – (–0.74 V) = 1.08 V 6(96485 C/mol e ) 1.08 V  zFEcell = = 252.353 RT (8.314 J/mol  K)(298 K) K = 3.940 x10109 = 4x10109 b) Sn(s) + Pb2+(aq)  Sn2+(aq) + Pb(s) z=2 

ln K =

Ecell = Ecathode – Eanode = –0.13 V – (–0.14 V) = 0.01V 2(96485 C/mol e )  0.01V  zFEcell = = 0.77887 RT (8.314 J/mol  K)(298 K) K = 2.179 = 2 

ln K =

19.60

Plan: Use G° =  nFEcell to calculate G°. Substitute J/C for V in the unit for Ecell .The units of ΔG° remain as J/mol as a reminder that the value is for a reaction as written. Solution: a) G° =  zFEcell = – (2 mol e–)(96,485 C/mol e–)(1.05 J/C) = –2.026185x105 J= –2.03x105 J/mol b) G° =  zFEcell = – (6 mol e–)(96,485 C/mol e–)(–0.30 J/C) = 1.73673x105 J= 1.7x105 J/mol

19.61

a) G° =  zFEcell = – (6 mol e–)(96,485 C/mol e–)(1.26 J/C) = –7.294266x105 J= –7.29x105 J/mol b) G° =  zFEcell = – (2 mol e–)(96,485 C/mol e–)(–0.54 J/C) = 1.042038x105 J= 1.0x105 J/mol

19.62

Plan: Use G° =  zFEcell to calculate G°. Substitute J/C for V in the unit for Ecell . The units of ΔG° remain as J/mol as a reminder that the value is for a reaction as written. Solution:

a) G° =  zFEcell = – (2 mol e–)(96,485 C/mol e–)(–1.98 J/C) = 3.820806x105 J= 3.82x105 J/mol b) G° =  zFEcell = – (2 mol e–)(96,485 C/mol e–)(0.29 J/C) = –5.59613x104 J= –5.6x104 J/mol 19.63

a) G° =  zFEcell = – (6 mol e–)(96,485 C/mol e–)(1.08 J/C) = –6.252228x105 J= –6.25x105 J/mol b) G° =  zFEcell = – (2 mol e–)(96,485 C/mol e–)(0.01 J/C) = –1.9297x103 J= –2x103 J/mol

19.64

Plan: Use Ecell =

RT ln K to find Ecell and then G° = –RT ln K to find G°. zF

Solution: T = (273 + 25)K = 298 K RT (8.314 J/mol  K)(298 K) ln 5.0x104 = 0.277834 V= 0.28 V ln K = Ecell = zF (1)(96485 C/mol)





G° = –RT ln K = – (8.314 J/mol•K)(298 K) ln (5.0x10 4) = –2.68067797x104 J/mol= –2.7x104 J/mol 19.65

RT ln K to find Ecell and then G° = –RT ln K to find G°. zF T = (273 + 25)K = 298 K RT (8.314 J/mol  K)(298 K) ln 5.0x106 = –0.31343 V= –0.31 V ln K = Ecell = zF (1)(96485 C/mol)

Use Ecell =





G° = –RT ln K = – (8.314 J/mol•K)(298 K) ln (5.0x10 –6) = 3.0241424x104 J/mol= 3.0x104 J/mol 19.66

Plan: Use Ecell =

RT ln K to find Ecell and then G° = –RT ln K to find G°. zF

Solution: T = (273 + 25)K = 298 K RT (8.314 J/mol  K)(298 K) ln  65  = 0.0535956 V= 0.054 V ln K = Ecell = zF (2)(96485 C/mol) G° = –RT ln K = – (8.314 J/mol•K)(298 K) ln (65) = –1.03423x104 J/mol= –1.0x104 J/mol 19.67

RT ln K to find Ecell and then G° = –RT ln K to find G°. zF T = (273 + 25)K = 298 K RT (8.314 J/mol  K)(298 K) ln  0.065 = –0.03509414 V= –0.035 V ln K = Ecell = zF (2)(96485 C/mol)

Use Ecell =

G° = –RT ln K = – (8.314 J/mol•K)(298 K) ln (0.065) = 6.772116x10 3 J/mol= 6.8x103 J/mol 19.68

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/Cu2+, Cu2+ must be reduced and H2 must be RT oxidized. Write the balanced reaction and calculate Ecell . Use the Nernst equation, Ecell = Ecell – ln Q , zF to find [Cu2+] when Ecell = 0.22 V. Solution: Oxidation: H2(g)  2H+(aq) + 2e– E° = 0.00 V Reduction: Cu2+(aq) + 2e–  Cu(s) E° = 0.34 V Cu2+(aq) + H2(g)  Cu(s) + 2H+(aq)

Ecell = Ecathode – Eanode = 0.34 V – 0.00 V = 0.34 V Ecell = Ecell –

RT ln Q zF

H+  RT   Ecell = Ecell – ln zF Cu 2+  pH   2 For a standard hydrogen electrode [H+] = 1.0 mol/L and pH 2 = 1.0 bar.

(8.314 J/mol  K)(298 K) 1.0 ln (2)(96485 C/mol) Cu 2  1.0   (8.314 J/mol  K)(298 K) 1.0 ln 0.22 V – 0.34 V = – (2)(96485 C/mol) Cu 2  1.0   (8.314 J/mol  K)(298 K) 1.0 ln –0.12 V = – (2)(96485 C/mol) Cu 2  1.0   1.0 9.34641 = ln Raise each side to ex . Cu 2   1.0   1 1.14576x104 = [Cu 2+ ] [Cu2+] = 8.72783x10–5 = 8.7x10–5 mol/L 0.22 V = 0.34 V –

19.69

The cell reaction is: Pb2+(aq) + Mn(s)  Pb(s) + Mn2+(aq)

Ecell = Ecathode – Eanode = –0.13 V – (–1.18 V) = 1.05 V RT ln Q zF 2 RT  Mn  Ecell = Ecell – ln zF  Pb 2     (8.314 J/mol  K)(298 K) 1.4 ln 0.44 V = 1.05 V – (2)(96485 C/mol)  Pb 2   

Ecell = Ecell –

(0.44 V – 1.05 V)[(–2)(96485 C/mol)/(8.314 J/mol.K)(298 K)]) = ln

1.4 

1.4   Pb 2    

= 47.5109

= 4.30255x1020  Pb 2     [Pb2+] = 3.25388x10–21 = 3.3x10–21 mol/L 19.70

Plan: Since this is a voltaic cell, a spontaneous reaction is occurring. For a spontaneous reaction between Ni/Ni 2+ and Co/Co2+, Ni2+ must be reduced and Co must be oxidized. Write the balanced reaction and calculate Ecell . RT ln Q, to find Ecell at the given ion concentrations. Then zF the Nernst equation can be used to calculate [Ni2+] at the given Ecell. To calculate equilibrium concentrations, recall that at equilibrium Ecell = 0.00. Solution: a) Oxidation: Co(s)  Co2+(aq) + 2e– E° = –0.28 V Reduction: Ni2+(aq) + 2e–  Ni(s) E° = –0.25 V 2+ 2+ Ni (aq) + Co(s)  Ni(s) + Co (aq)

Use the Nernst equation, Ecell = Ecell –

Ecell = Ecathode – Eanode = –0.25 V – (–0.28 V) = 0.03 V

RT ln Q zF Co 2   RT  Ecell = Ecell – ln  z = 2e– zF  Ni 2     (8.314 J/mol  K)(298 K)  0.20  ln  Ecell = 0.03 V –  (2)(96485 C/mol)  0.80  Ecell = 0.04779885 V = 0.05 V b) From part a), notice that an increase in [Co2+] 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 concentrations will be [Co2+] = 0.20 mol/L + x and [Ni2+] = 0.80 mol/L – x (there is a 1:1 mole ratio). 2 RT Co  Ecell = Ecell – ln zF  Ni 2     (8.314 J/mol  K)(298 K)  0.20 + x  ln  0.03 V = 0.03 V –  (2)(96485 C/mol)  0.80 - x 

Ecell = Ecell –

0=–

(8.314 J/mol  K)(298 K)  0.20 + x  ln   (2)(96485 C/mol)  0.80 - x 

 0.20 + x   0.80  x   0.20 + x  e0 = 1 =  0.80  x  0 = ln

Raise each side to ex .

0.20 + x = 0.80 – x x = 0.30 mol/L [Ni2+] = 0.80 – x = 0.80 – 0.30 = 0.50 mol/L c) At equilibrium Ecell = 0.00; to decrease the cell potential to 0.00, [Co 2+] increases and [Ni2+] decreases. (8.314 J/mol  K)(298 K)  0.20 + x  ln  0.00 V = 0.03 V –  (2)(96485 C/mol)  0.80 - x  –0.03 V = –

(8.314 J/mol  K)(298 K)  0.20 + x  ln   (2)(96485 C/mol)  0.80 - x 

 0.20 + x  2.3366 = ln    0.80 - x 

Raise each side to ex .

 0.20 + x  10.3460 =    0.80 - x  x = 0.71186 [Co2+] = 0.20 + 0.71186 = 0.91186 = 0.91 mol/L 2+ [Ni ] = 0.80 – 0.71186 = 0.088135 = 0.09 mol/L 19.71

The spontaneous reaction (voltaic cell) is Cd2+(aq) + Mn(s)  Cd(s) + Mn2+(aq) with

Ecell = Ecathode – Eanode = –0.40 V – (–1.18 V) = 0.78 V. RT ln Q zF 2 RT  Mn  Ecell = Ecell – ln zF Cd 2    

a) Ecell = Ecell –

z = 2e–

(8.314 J/mol  K)(298 K)  0.090  ln   (2)(96485 C/mol)  0.060  Ecell = 0.774794 V = 0.77 V Ecell = 0.78 –

b) For the [Cd2+] to decrease from 0.060 mol/L to 0.050 mol/L, a change of 0.010 mol/L, the [Mn 2+] must increase by the same amount, from 0.090 mol/L to 0.100 mol/L. (8.314 J/mol  K)(298 K)  0.10  ln  Ecell = 0.78 –  (2)(96485 C/mol)  0.050  Ecell = 0.771100575 V = 0.77 V c) Increase the manganese and decrease the cadmium by equal amounts. Total = 0.150 mol/L 0.055 = 0.78 –

(8.314 J/mol  K)(298 K) [Mn 2 ] ln (2)(96485 C/mol) [Cd 2 ]

 Mn 2     = 3.33957x1024 Cd 2     [Mn2+] = (3.33957x1024)[Cd2+] [Mn2+] + [Cd2+] = 0.150 mol/L (3.33957x1024)[Cd2+] + [Cd2+] = 0.150 mol/L [Cd2+] = 4.4916x10–26 mol/L [Mn2+] = 0.150 mol/L – 4.4916x10–26 mol/L = 0.150 mol/L

d) At equilibrium Ecell = 0.00. 0.00 = 0.78 –

(8.314 J/mol  K)(298 K) [Mn 2 ] ln (2)(96485 C/mol) [Cd 2 ]

 Mn 2     = 2.42163x1026 2  Cd    [Mn2+] = (2.42163x1026)[Cd2+] [Mn2+] + [Cd2+] = 0.150 mol/L (2.42163x1026)[Cd2+] + [Cd2+] = 0.150 mol/L [Cd2+] = 6.19416x10–28 = 6x10–28 mol/L [Mn2+] = 0.150 mol/L – [Cd2+] = 0.150 mol/L

19.72

Plan: The overall cell reaction proceeds to increase the 0.10 mol/L H + concentration and decrease the 2.0 mol/L H+ concentration. Use the Nernst equation to calculate Ecell. Ecell = 0 V for a concentration cell since the halfreactions are the same. Solution: a) Half-cell A is the anode because it has the lower concentration. b) Oxidation: H2(g;0.95 bar)  2H+(aq; 0.10 mol/L) + 2e– E° = 0.00 V Reduction: 2H+(aq; 2.0 mol/L) + 2 e–  H2(g; 0.60 bar) E° = 0.00 V 2H+(aq; 2.0 mol/L) + H2(g; 0.95 bar)  2H+(aq; 0.10 mol/L) + H2(g; 0.60 bar) z = 2e–

Ecell = 0.00 V Ecell = Ecell –

RT ln Q zF 2

Q for the cell equals

H  p   anode H2 (cathode) 2

H  p   cathode H2 (anode)

 0.10 2  0.60 = 0.00157895  2.0 2  0.95

Ecell = 0.00 V – 19.73

(8.314 J/mol  K)(298 K) ln (0.00157895) = 0.082821 V= 0.083 V (2)(96485 C/mol)

Sn2+(0.87 mol/L)  Sn2+(0.13 mol/L) Half-cell B is the cathode. RT Ecell = Ecell – ln Q Ecell = 0.00 V zF Sn 2   (8.314 J/mol  K)(298 K)   anode Ecell = Ecell – ln 2  (2)(96485 C/mol) Sn    cathode Ecell = 0.00 V –

z = 2e–

(8.314 J/mol  K)(298 K)  0.13 ln (2)(96485 C/mol)  0.87 

Ecell = 0.024407 V= 0.024 V 19.74

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.

19.75

The electrodes are separated by an electrolyte paste, which for the ordinary dry cell battery contains ZnCl 2, NH4Cl, MnO2, starch, graphite, and water.

19.76

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. (a) 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.) (b) 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.

19.77

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

19.78

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.

19.79

Bridge supports rust more rapidly at the water line due to the presence of large concentrations of both O 2 and H2O.

19.80

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.

19.81

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. 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.

19.82

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.

19.83

3Cd2+(aq) + 2Cr(s)  3Cd(s) + 2Cr3+(aq)

Ecell = –0.40 V – (–0.74 V) = 0.34 V To reverse the reaction requires 0.34 V with the cell in its standard state. A 1.5 V supplies more than enough potential, so the cadmium metal oxidizes to Cd2+ and chromium plates out. 19.84

The Ehalf–cell values are different than the Ehalf-cell values because in pure water, the [H +] and [OH –] are 1.0 x 10–7 mol/L rather than the standard-state value of 1 mol/L.

19.85

The oxidation number of nitrogen in the nitrate ion, NO 3–, 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.

19.86

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.

19.87

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)

19.88

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).

19.89

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).

19.90

Because this process is an electrolysis, liquid strontium (Sr) forms at the negative electrode and gaseous bromine (Br2) forms at the positive electrode.

19.91

19.92

Because this process is an electrolysis, liquid calcium forms at the negative electrode and chlorine gas forms at the positive electrode.

19.93

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: Cu2+(aq) + 2e–  Cu(s) E° = +0.34 V Ba2+(aq) + 2e–  Ba(s) E° = –2.90 V Al3+(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 water. Since the reduction potentials of Ba2+ and Al3+ 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.

19.94

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.

19.95

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: Li+(aq) + e–  Li(s) E° = –3.05 V Zn2+(aq) + 2e–  Zn(s) E° = –0.76 V Ag+(aq) + e–  Ag(s) E° = 0.80 V 2H2O(l) + 2e–  H2(g) + 2OH–(aq) E = –1 V with overvoltage Zinc and silver can be prepared by electrolysis of their aqueous salt solutions since their reduction half-cell potentials are more positive than the potential for the reduction of water. The reduction of zinc and silver is more spontaneous than the reduction of water. Since the reduction potential of Li+ is more negative and therefore less spontaneous than the reduction of water, this ion cannot be reduced in the presence of water since the water is reduced instead. Possible oxidations: 2I–(aq)  I2(l) + 2e– E° = +0.53 V 2H2O(l)  O2(g) + 4H+(aq) + 4e– E = 1.4 V with overvoltage 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.

19.96

Electrolysis will separate both iron and cadmium from an aqueous solution.

19.97

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. The SO42– ion cannot oxidize as S is already in its highest oxidation state in SO 42–. Possible reductions: 2H2O(l) + 2e–  H2(g) + 2OH–(aq) E = –1 V with overvoltage Sn2+(aq) + 2e–  Sn(s) E° = –0.14 V SO42–(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 pressures for all reactants and products except H+, which in pure water is 1x10–7 mol/L. E = 0.20 V – (0.0592 V/2) ln [1/(1x10–7)4] = –0.6288 V= –0.63 V The most easily reduced ion is Sn2+ with the most positive reduction potential, so tin metal forms at the cathode. 19.98

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 (see Chapter 19, p. 831, summary).

19.99

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 The oxidation of water produces oxygen gas (O2), and hydronium ions (H3O+) at the anode. NO3– cannot oxidize since N is in its highest oxidation state in NO3–. Possible reductions: 2H2O(l) + 2e–  H2(g) + 2OH–(aq) E = –1 V with overvoltage Cr3+(aq) + 3e–  Cr(s) E° = –0.74 V NO3–(aq) + 4H+(aq) + 3e–  NO(g) + 2H2O(l) E = +0.13 V (approximate) The potential for nitrate reduction is estimated from the Nernst equation using standard-state concentrations and pressures for all reactants and products except H+, which in pure water is 1x10–7 mol/L. E = 0.96 V – (0.0592 V/2) log [1/(1x10–7)4] = 0.1312 V= 0.13 V 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 Mn2+(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 cathode. The reduction potential of Mn2+ is more negative than that of water showing that its reduction is less spontaneous than that of water.

19.100 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 (see Chapter 19, p. 831, summary). 19.101 Plan: Write the half-reaction for the reduction of Mg2+. Convert mass of Mg to amount (mol) and use the mole ratio in the balanced reaction to find the amount (mol) 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: Mg2+ + 2e–  Mg  1 mol Mg   2 mol e  – a) Amount (mol) of electrons =  45.6 g Mg    = 3.751542575 mol= 3.75 mol e   24.31 g Mg 1 mol Mg     96, 485 C  b) Charge = 3.751542575 mol e   = 3.619676x105 C= 3.62x105 C    mol e   3.619676x105 C   1 h   A   = 28.727587 A = 28.7 A c) Current =      3.50 h    3600 s   C s 

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

19.102 Na+ + 1e–  Na

 1 mol Na   1 mol e  – a) Amount (mol) of electrons =  215 g Na    = 9.351892127 mol= 9.35 mol e    22.99 g Na   1 mol Na 





 96, 485 C  b) Charge = 9.351892127 mol e   = 9.0231731x105 C= 9.02x105 C    mol e   9.0231731x105 C   1 h   A   = 26.383547 A= 26.4 A c) Current =      9.50 h    3600 s   C s  19.103 Plan: Write the half-reaction for the reduction of Ra2+. Use the Faraday constant to convert charge in coulombs to amount (mol) of electrons provided; the balanced reduction reaction converts amount (mol) of electrons to amount (mol) of radium produced. Multiply amount (mol) of radium by its molar mass to obtain grams. Solution: Ra2+ + 2e–  Ra In the reduction of radium ions, Ra2+, 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 g= 0.275 g Ra  96, 485 C   2 mol e    1 mol Ra      19.104 Al3+ + 3e–  Al In the reduction of aluminum ions, Al3+, 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    96, 485 C   3 mol e    1 mol Al  = 0.028428944 g= 0.0284 g Al     19.105 Plan: Write the half-reaction for the reduction of Zn2+. Convert mass of Zn to amount (mol)s and use the mole ratio in the balanced reaction to find the amount (mol) 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: Zn2+ + 2e–  Zn

 1 mol Zn   2 mol e    96, 485 C   1   1 A    = 9.20169x103 s= 9.20x103 s Time (s) =  65.5 g Zn       1 mol e    21.0 A   C  65.41 g Zn 1 mol Zn    s   

19.106 Ni2+ + 2e–  Ni  1 mol Ni   2 mol e    96, 485 C   1   1 A   = 391.1944859 s= 391 s Time (s) = 1.63 g Ni           58.69 g Ni   1 mol Ni   1 mol e   13.7 A   C s 

19.107 a) The sodium sulfate ionizes to produce Na+ and SO42– 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 contains very low (10–7 mol/L) 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 oxidation of H2O is the only oxidation possible because SO42– cannot be oxidized under these conditions. In other words, it is easier to reduce H2O than Na+ and easier to oxidize H2O than SO42–. 19.108 Iodine is formed first since the oxidation potential of I – is more positive than the oxidation potential of Br–. 19.109 Plan: Write the half-reaction for the reduction of Zn2+. 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 amount (mol) of electrons provided; the balanced reduction reaction converts amount (mol) of electrons to amount (mol) of zinc produced. Multiply amount (mol) of zinc by its molar mass to obtain grams. Solution: Zn2+ + 2e–  Zn C   24 h   3600 s   1 mol e    1 mol Zn   65.41 g Zn  Mass (g) of Zn =  0.855 A   s   2.50 day            A   1 day   1 h   96, 485 C   2 mol e   1 mol Zn    = 62.59998 g= 62.6 g Zn 19.110 Mn2+(aq) + 2H2O(l)  MnO2(s) + 4H+(aq) + 2e–  103 g   1 mol MnO2   2 mol e  96485 C   1   A   1 h  Time (h) = 1.00 kg MnO2    C       1 kg   86.94 g MnO   1 mol MnO    3600 s  2  2   1 mol e   25.0 A      s 

= 24.66196 h= 24.7 h The MnO2 product forms at the anode, since the half-reaction is an oxidation. 19.111 Plan: Write the reduction half-reaction for H +. Use the ideal gas law to find the amount (mol) of H 2 produced. The mole ratio in the balanced reduction reaction gives the moles of electrons required for that amount of H2 and the Faraday constant converts the amount (mol) 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 amount (mol) of hydrogen gas.

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

12.0 bar  3.5x106 L PV = = 1.695208x106 mol H2 RT L•bar    0.08314 mol•K    273  25 K    Then, find the coulombs knowing that there are two electrons transferred per mol of H 2. n=

 2 mol e  96485 C  Coulombs = 1.695208x106 mole H 2  = 3.27124287760x1011 = 3.3x1011 C  1 mol H    1 mol e  2   









 1.44 J  11 11 11 b) Energy (J) =   3.27124287760x10 C = 4.71058974x10 = 4.7x10 J  C 

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 1 kJ  1 kg  4 4 c) Mass (kg) = 4.71058974x1011 J  3   = 1.177647x10 = 1.2x10 kg 4  10 J  4.0x10 kJ  19.112 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 MnO2   86.94 g MnO2  b) Mass (g) of MnO2 =  4.50 g Zn       65.41 g Zn  1 mol Zn   1 mol MnO2  = 5.981196 g= 5.98 g MnO2  1 mol Zn  1 mol H 2 O   18.02 g H 2 O  Mass (g) of H2O =  4.50 g Zn    = 1.23972 g= 1.24 g H2O    65.41 g Zn  1 mol Zn   1 mol H 2O  c) Total mass of reactants = 4.50 g Zn + 5.981196 g MnO2 + 1.23972 g H2O = 11.720916 g = 11.72 g  1 mol Zn   2 mol e    96, 485 coulombs  4 4 d) Charge =  4.50 g Zn      = 1.32757x10 C= 1.33x10 C    65.41 g Zn 1 mol Zn 1 mol e     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. 19.113 The reaction is: Ag+(aq) + e–  Ag(s) Mass (g) of Ag = 1.8016 g – 1.7854 g = 0.0162 g Ag produced  1 mol Ag   1 mol e    96, 485 coulombs  Coulombs =  0.0162 g Ag      = 14.48616 C= 14.5 C   1 mol e   107.9 g Ag   1 mol Ag    19.114 Plan: Write balanced half-reactions for the reduction of each metal ion. From the current, 65.0% of the amount (mol) 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 amount (mol) of electrons using the Faraday constant and use the balanced reduction reactions to convert the amount (mol) of electrons to amount (mol) 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: Cu2+(aq) + 2e–  Cu(s) and Zn2+(aq) + 2e–  Zn(s)  1 mol e    1 mol Cu   63.55 g Cu  Mass (g) of copper =  65.0 C    96, 485 C   2 mol e    1 mol Cu  = 0.021406177 g Cu    

 1 mol e    1 mol Zn   65.41 g Zn  Mass (g) of zinc =  35.0 C    96, 485 C   2 mol e    1 mol Zn  = 0.01186376 g Zn     0.021406177 g Cu Mass % of copper = 100%  = 64.340900 %= 64.3% Cu 0.021406177 g Cu + 0.01186376 g Zn

19.115 Plan: Write the reduction half-reaction for Au3+. 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 amount (mol) of gold. The mole ratio in the balanced reduction reaction is used to convert the amount (mol) of gold to amount (mol) of electrons required and the Faraday constant is used to convert amount (mol) 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 amount (mol) of gold from part a) by four to get the amount (mol) of gold needed for the earrings. Convert this amount (mol) to grams, then to troy ounces, and finally to dollars. Solution: The reaction is: Au3+(aq) + 3e–  Au(s) a) V = r2h 2  103 m   1 cm   4.00 cm  3 V (cm3) =    0.25 mm     2  = 0.314159265 cm  2 1 mm 10 m      

 19.3 g Au   1 mol Au  Amount (mol) of Au = 0.314159265 cm3   = 0.03077804 mol Au   1 cm3   197.0 g Au 

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 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   0.013 A   3600 s   24 h      s   

= 7.931675 days= 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 days= 32 days  197.0 g Au  1 troy oz  $1615  c) Cost =  4  0.03077804 mol Au      = $1259.45 = $1300  1 mol Au  31.10 g  troy oz  19.116 a) The half-reaction is: 2H2O(l)  O2(g) + 4H +(aq) + 4e– Determine the amount (mol) of oxygen from the ideal gas equation. Use the half-reaction and the current to convert the amount (mol) of oxygen to time.     99.8 kPa 10.0 L    PV  = 0.39879897 mol O2 n= =    L•kPa  RT   8.314 mol•K    273  28 K       4 mol e  96, 485 C   A   1  1 min    Time (min) =  0.39879897 mol O2    1 mol O    1 mol e   C   1.3 A  60 s    2  s   

= 1.973237x103 min= 2.0x103 min b) The balanced chemical equation is 2H2O(l)  2H2(g) + O2(g). The amount (mol) of oxygen determined previously and this chemical equation leads to the mass of hydrogen.  2 mol H 2  2.016 g H 2  Mass (g) of H2 =  0.39879897 mol O 2     = 1.607096 g = 1.61 g H2  1 mol O 2  1 mol H 2  19.117 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. 19.118 Plan: Write the half-reactions and cell reaction for the silver battery. Convert mass of zinc to amount (mol) of zinc, keeping in mind that only 80% of the zinc will react; from the amount (mol) of zinc, the amount (mol) of electrons required is obtained. The Faraday constant is used to convert the amount (mol) of electrons to charge in coulombs which is divided by the current to obtain the time in seconds. The amount (mol) of zinc is also used to find the amount (mol) of Ag2O consumed and the amount of Ag needed for that amount of Ag2O. Convert this mass of silver to troy ounces and then to dollars. Silberberg, Amateis, Venkateswaran, Chen, Chemistry: The Molecular Nature of Matter and Change, 3rd Canadian Edition Page 3-739 Instructor‘s Solution Manual © Copyright 2021 McGraw-Hill Ryerson Ltd.

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  Amount (mol) of Zn =  0.75 g Zn    = 0.00917291 mol Zn   100%   65.41 g Zn   2 mol e    96, 485C   A   1  A     1 h   1 day  1   a) Time (days) =  0.00917291 mol Zn    1 mol Zn   1 mol e    C   106 A   0.85  A   3600 s   24 h      s    

= 2.410262x104 days = 2.4x104 days  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 g= 2.1 g Ag

 1  95%   1 troy oz  CAD 27.48   c) Cost =  2.0836989 g Ag         4  100%   31.10 g Ag  troy oz   2.410262x10 days  = 7.25689x10–5 = CAD 7.3x10–5/day 19.119 Cu2+(aq) + 2e–  Cu(s)  C   3600 s   1 mol e    1 mol Cu   63.55 g Cu  Theoretical amount of copper =  5.8 A   s   10 h        96, 485 C   2 mol e    1 mol Cu   A  1 h     = 68.7632 g Cu actual yield 53.4 g Cu Efficiency = 100%  = 100%  = 77.6578 %= 78% theoretical yield 68.7632 g Cu The final assumes that 10 h has two significant figures.

19.120

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)

19.121 Plan: Since the cells are voltaic cells, the reactions occurring are spontaneous and will have a positive Ecell . Write the two half-reactions. When two half-reactions are paired, one half-reaction must be reversed and written as an oxidation. Reverse the half-reaction that will result in a positive value of Ecell using the relationship

Ecell = Ecathode – 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 mol/L. Solution: a) Cell with SHE and Pb/Pb2+: Oxidation: Pb(s) → Pb2+(aq) + 2e–

E = – 0.13 V

Reduction: 2H+(aq) + 2e– → H2(g)

E = 0.0 V

Ecell = Ecathode – Eanode = 0.0 V – (– 0.13 V) = 0.13 V Cell with SHE and Cu/Cu2+: Oxidation: H2(g) → 2 H+(aq) + 2e– 2+

–

Reduction: Cu (aq) + 2e → Cu(s)

E = 0.0 V E = 0.34 V

Ecell = Ecathode – Eanode = 0.34 V– 0.00 V = 0.34 V b) The anode (negative electrode) for the cell with SHE and Pb/Pb2+ is Pb. The anode for the cell with SHE and Cu/Cu2+ is platinum in the SHE. c) The precipitation of PbS decreases [Pb2+]. Use Nernst equation to see how this affects potential. Cell reaction is: (8.314 J/mol  K)(298 K) ln Q Pb(s) + 2H+(aq)  Pb2+(aq) + H2(g) Ecell = Ecell – z (96485 C/mol) Ecell = Ecell –

2+ (8.314 J/mol  K)(298 K) [Pb ]pH 2 ln (2)(96485 C/mol) [H + ]2

z = 2e–

Decreasing the concentration of lead ions makes the following term more negative: 2+ (8.314 J/mol  K)(298 K) [Pb ]pH 2 ln (2)(96485 C/mol) [H + ]2 When this more negative value is subtracted from Ecell , cell potential increases. d) The [H+] = 1.0 mol/L , [Cu2+] = 1x10-16 mol/L and the H2 = 1 bar in the SHE. Cell reaction: Cu2+(aq) + H2(g)  Cu(s) + 2H+(aq) (8.314 J/mol  K)(298 K) ln Q Ecell = Ecell – z (96485 C/mol) Ecell = Ecell –

(8.314 J/mol  K)(298 K) [H + ]2 ln (2)(96485 C/mol) [Cu 2+ ]pH

z = 2e– 2

1 (8.314 J/mol  K)(298 K) ln (2)(96485 C/mol) 1x1016 1  2

Ecell = 0.34 V –

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Ecell = –0.133012 V= –0.13 V

 1 cm  19.122 a) Volume (cm3) = 2.0 cm 2 7.5x106 m  2  = 0.0015 cm3  10 m  b) Mass = (0.0015 cm3)(10.5 g Ag/1 cm3) = 0.01575 g= 0.016 g  1 mol Ag   1 mol e    96, 485 C   A   1 mA     1 min  1   c) Time =  0.01575 g Ag            C 3  107.9 g Ag   1 mol Ag   1 mol e   s   10 A   12.0 mA   60 s  = 19.56079 min= 20. min  1 troy oz   CAD 27.48   100 cents  d) Cost =  0.01575 g Ag      = 1.391672 cents = 1.4 cents  31.10 g Ag   1 troy oz   CAD1 

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19.123 The reduction of H2O to H2 and OH – is easier than the reduction of Al3+ to Al. 19.124 The three steps equivalent to the overall reaction M+(aq) + e–  M(s) are: 1) M+(aq)  M+(g) Energy is –hydrationH 2) M+(g) + e–  M(g) Energy is –IE or –ionizationH 3) M(g)  M(s) Energy is –atomizationH

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 –hydrationH 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. 19.125 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 F 2 (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. 19.126 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 becomes 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. 19.127 Plan: Write the half-reaction for the reduction of Al3+. Convert mass of Al to amount (mol) and use the mole ratio in the balanced reaction to find the amount (mol) 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 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, use the kW • h per 1000 kg of aluminum calculated in part b) to find the power (kW • h) for the mass of aluminum, keeping in mind the 90.% efficiency. Solution: a) Aluminum half-reaction: Al3+(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   100,000 A     s     5 5 = 1.0728503x10 s= 1x10 s The mass and current limit the answer to 1 significant figure. b) 5 V = 5 J/C  100,000 C   5.0 J   1 kJ   1 kW•h  4 4 Power = 1.0728503x105 s   = 1.4900699x10 =kW•h 1.5x10 kW•h  C  3  3 s     10 J   3.6x10 kJ 

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 1 kg   1.4900699x104 kW•h   CAD 0.177   100%  c) Cost =  454 g Al         = $1.3304 = $1.3 1000 kg Al  1000 g     1 kW•h   90.%  19.128 a) Electrons flow from magnesium bar to the iron pipe since magnesium is more easily oxidized than iron. b) The magnesium half-reaction is: Mg(s)  Mg2+(aq) + 2e– Current is charge per time. The mass of magnesium can give the total charge. Convert the mass of magnesium to amount (mol) of magnesium and multiply by two moles of electrons produced for each mole of magnesium and by the Faraday constant to convert the amount (mol) 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       1 kg   24.31 g Mg   1 mol Mg   mol e  ch arg e  Current = = = 0.35511 C/s= 0.36 A time  365.25 days  24 h  3600 s  8.5 yr      1 yr   1 days  1 h 

19.129 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. 19.130 3Pt(s) + 4NO3–(aq) + 18Cl–(aq) + 16H +(aq)  3PtCl62–(aq) + 4NO(g) + 8H2O(l) 19.131 Plan: Examine the change in oxidation numbers in the equations to find n, the amount (mol) 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 mole of electrons in the reaction. G = –zFE = – (4 mol e–)(96,485 C/mol e–)(1.23 J/C) = –4.747062x105 J/mol= –4.75x105 J/mol Cell II: In Pb(s)  PbSO4, O.N. of Pb changes from 0 to +2 and in PbO 2  PbSO4, O.N. of Pb changes from +4 to +2. There is a transfer of two mole of electrons in the reaction. G = –zFE = – (2 mol e–)(96,485 C/mol e–)(2.04 J/C) = –3.936588x105 J/mol= –3.94x105 J/mol 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 mole of electrons in the reaction. G = –zFE = – (2 mol e–)(96,485 C/mol e–)(2.35 J/C) = –4.534795x105 J/mol= –4.53x105 J/mol  2.016 g H 2   32.00 g O 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 

 4.747062x105 J   1 kJ  wmax =    3  = –13.17457 kJ/g= –13.2 kJ/g  reactan t mass 36.032 g    10 J  Cell II: Mass of reactants =  239.2 g PbO 2   98.09 g H 2SO 4   207.2 g Pb  1 mol Pb      2 mol H 2SO 4      1 mol PbO 2   1 mol Pb 1 mol PbO   2    1 mol H 2SO 4  = 642.58 g  3.936588x105 J   1 kJ  wmax =    3  = –0.612622 kJ/g= –0.613 kJ/g reactan t mass  642.58 g   10 J 

 126.75 g FeCl 2   22.99 g Na  Cell III: Mass of reactants =  2 mol Na    = 172.73 g   1 mol FeCl 2   1 mol Na    1 mol FeCl 2 

 4.534795x105 J   1 kJ  wmax =    3  = –2.625366 kJ/g= –2.63 kJ/g reactan t mass  172.73 g   10 J  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.

19.132 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: Zn(s)  Zn2+(aq) + 2e– and Ag+(aq) + e–  Ag(s)  1 mol Zn   2 mol e  1 mol Ag   107.9 g Ag  Mass (g) of Ag = 1.2 g Zn       = 3.9590 g= 4.0 g Ag      65.41 g Zn   1 mol Zn   1 mol e   1 mol Ag  19.133 A standard thermodynamic table would allow the calculation of the equilibrium constant by the equation G° = – RT ln K. G° could be calculated from  f G values, or  f H and S° values. A table of standard electrode potentials would allow the calculation of the equilibrium constant by the following equation: E° = (RT/zF) ln K 19.134 2Fe(s) + O2(g) + 4H+(aq)  2H2O(l) + 2Fe2+(aq); r H = -747.5 kJ/mol;  r S = 106.3 J/mol∙K Based on thermodynamics, we can see that increasing the temperature makes the reaction more spontaneous and therefore increases the cell potential for the above reaction. Based on kinetics, an increase in the temperature will also 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. 19.135 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 the amount (mol) of electrons by two to get amount (mol) of copper. Then convert to grams of copper. The initial concentration of Cu 2+ is 1.00 mol/L (standard condition) and initial volume is 345 mL. Use this to calculate the initial amount (mol) of copper ions, then subtract the amount (mol) of copper ions converted to copper metal and divide by the cell volume to find the remaining [Cu2+]. Solution: a) Since the cell is a voltaic cell, write a spontaneous reaction. The reduction of Cu 2+ is more spontaneous than the reduction of Sn2+: Cu2+(aq) + Sn(s)  Cu(s) + Sn2+(aq)  C   3600 s   1 mol e    1 mol Cu   63.55 g Cu  Mass (g) of Cu =  0.17 A   s       48.0 h       A  1 h   96, 485 C   2 mol e   1 mol Cu    = 9.674275 g= 9.7 g Cu   103 L  mol Cu 2   345 mL   b) Initial amount (mol) of Cu2+ = 1.00 = 0.345 mol Cu2+      1 mL  L    

 1 mol Cu  2+ Amount (mol) of Cu2+ reduced =  9.674275 g Cu    = 0.1522309205 mol Cu 63.55 g Cu   Remaining amount (mol) of Cu2+ = initial amount (mol) – amount (mol) reduced = 0.345 mol – 0.1522309205 mol= 0.1927691 mol Cu2+ Concentration (mol/L) Cu2+ =

0.1927691 mol Cu 2  = 0.558751 mol/L= 0.56 mol/L Cu2+  3  10 L   345 mL     1 mL 

19.136 When H2 is in its standard state, the pressure is equal to 1 bar. The pH of the solution is neutral, so hydroxide concentration is 110-7 mol/L. 2H2O(l) + 2e–  H2(g) + 2OH–(aq) (8.314 J/mol  K)(298 K) ln Q Ecell = Ecell – (2)(96485 C/mol)

Ecell = Ecell –

(8.314 J/mol  K)(298 K) ln (2)(96485 C/mol)

 p [OH ]   2





2 (8.314 J/mol  K)(298 K)  ln 1 1.0  10 –7  (2)(96485 C/mol)   = –0.83 V + 0.4139 V= –0.4161 V= –0.42 V

Ecell = –0.83 V –

19.137 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 U3+ + Cr3+  Cr2+ + U4+, find that Cr3+ oxidizes U3+. From reaction between Fe + Sn2+  Sn + Fe2+, find that Sn2+ oxidizes Fe. From the fact that there is no reaction that occurs between Fe and U 4+, find that Fe2+ oxidizes U3+. From reaction between Cr3+ + Fe  Cr2+ + Fe2+, find that Cr3+ oxidizes Fe. From reaction between Cr2+ + Sn2+  Sn + Cr3+, find that Sn2+ oxidizes Cr2+. Notice that nothing oxidizes Sn, so Sn2+ must be the strongest oxidizing agent. Both Cr3+ and Fe2+ oxidize U3+, so U4+ must be the weakest oxidizing agent. Cr3+ oxidizes iron so Cr3+ is a stronger oxidizing agent than Fe2+. The half–reactions in order from strongest to weakest oxidizing agent: Sn2+(aq) + 2e–  Sn(s) Cr3+(aq) + e–  Cr2+(aq) Fe2+(aq) + 2e–  Fe(s) U4+(aq) + e–  U3+(aq) 19.138 The given half-reactions are: (1) Fe3+(aq) + e–  Fe2+(aq) (2) Fe2+(aq) + 2e–  Fe(s) (3) Fe3+(aq) + 3e–  Fe(s) (a) E°1 = +0.77 V E°2 = –0.44 V Adding half-reaction (1) and (2) gives half-reaction (3), thus E°3 = E°1 + E°2 = +0.77 V + (–0.44 V) = 0.33 V (b) G° = –zFE° Reaction 1: G° = – (1)(96,485 C/mol)(0.77 J/C) = –7.429345x104 J/mol= –7.4x104 J/mol Reaction 2: G° = – (2)(96,485 C/mol)(–0.44 J/C) = 8.49068x104 J/mol= 8.5x104 J/mol c) G3° = G1° + G2° = –7.429345x104 J/mol + 8.49068x104 J/mol = 1.061335x104 J/mol= 1.1x104 J/mol d) E° = –G°/zF = – (1.061335x104 J)/(3)(96,485 C/mol)[V/(J/C)] = –0.036667 V= –0.037 V e) Although the half-cells (1) and (2) add to (3) , their voltages DO NOT add to E°3. Half-cell values cannot be simply added together to arrive at a third potential because potential is a ratio between energy and charge. With differing amounts of electrons, the energy is changing. This is similar to why when we multiply the half-cell reaction by a factor, we do not change the potential, because the ratio of energy to charge stays the same. 19.139 6e– + 14H + + Cr2O72–  2Cr3+ + 7H2O (red half-reaction) CH3CH2OH + H2O  CH3COOH + 4H + + 4e– (ox half-reaction) 2{6e– + 14H + + Cr2O72–  2Cr3+ + 7H2O} 3{CH3CH2OH + H2O  CH3COOH + 4H + + 4e– } 12e– + 28H + + 2Cr2O72–  4Cr3+ + 14H2O 3CH3CH2OH + 3H2O  3CH3COOH + 12H + + 12e– Overall: 3CH3CH2OH + 2Cr2O72– + 16H +  3CH3COOH + 4Cr3+ + 11H2O 19.140 At STP hydrogen gas occupies 22.7 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   22.7 L H 2  Volume (L) of H2 = 1 kg Zn   = 32.23885 L= 32.2 L H2  1 kg   65.41 g Zn  91.50 mol Zn   1 mol H    2   

19.141 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)

Ecell = Ecathode – Eanode = 0.80 V – 0.24 V = 0.56 V with z = 2. Use the Nernst equation to find [Ag+] when Ecell = 0.060 V. (8.314 J/mol  K)(298 K) ln Q Ecell = Ecell – z (96485 C/mol) 0.060 V = 0.56 V –

(8.314 J/mol  K)(298 K) 1 ln 2 2 (2)(96485 C/mol)  Ag +  Cl-     

– The problem suggests assuming that [Cl –] is constant. Assume it is standard conc. of 1.00 mol/L. (8.314 J/mol  K)(298 K) 1 ln – 0.50 V = – 2 (2)(96485 C/mol) +  Ag  (1.00)2   1 38.94337 = ln 2  Ag   1.00 2   1 e38.94337 = 2  Ag     1 8.18258x1016 = 2  Ag     8.18258x1016[Ag+]2 = 1 [Ag+]2 = 1.22211x10–17 [Ag+] = 3.49587x10–9 mol/L= 3.5x10–9 mol/L b) Again use the Nernst equation and assume [Cl–] = 1.00 mol/L. (8.314 J/mol  K)(298 K) 1 ln Ecell = Ecell – 2 2 (2)(96485 C/mol)  Ag +  Cl-      (8.314 J/mol  K)(298 K) 1 ln 0.53 V = 0.56 V – 2 2 (2)(96485 C/mol)  Ag +  Cl-      (8.314 J/mol  K)(298 K) 1 ln – 0.03 V = – 2 2 (2)(96485 C/mol)  Ag +  Cl-      1 2.33660 = ln 2  Ag   1.00 2   1 e2.33660 = 2  Ag     1 10.3460 = 2  Ag     10.3460 [Ag+]2 = 1 [Ag+]2 = 0.0966555

[Ag+] = 0.31089 mol/L = 0.3 mol/L 19.142 Oxidation: Ag(s)  Ag+(aq) + e–

Eanode = +0.80 V –

Reduction: AgCl(s) + e–  Ag(s) + Cl (aq)

Ecathode = + 0.22 V

AgCl(s)  Ag+(aq) + Cl–(aq)

Overall:

Ecell = Ecathode – Eanode = 0.22 V – 0.80 V = –0.58V

1(96485 C/mol)  0.58 V  zFEcell = = –22.58715 RT (8.314 J/mol  K)(298 K) K = 1.550685x10–10 = 1.6x10–10 ln Ksp =

19.143 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– (8.314 J/mol  K)(298 K) ln Q Ecell = Ecell – ( z )(96485 C/mol) (8.314 J/mol  K)(298 K)    ln Ag   waste (1)(96485 C/mol)

Nonstandard cell:

Ewaste = Ecell –

Standard cell:

Estandard = Ecell –

b) To find [Ag+]waste: = Ewaste +

Ecell = Estandard +

(8.314 J/mol  K)(298 K)    ln Ag   standard (1)(96485 C/mol)

(8.314 J/mol  K)(298 K)    ln Ag   waste (1)(96485 C/mol)

Estandard – Ewaste =

(8.314 J/mol  K)(298 K) (ln [Ag+]waste – ln [Ag+]standard) (1)(96485 C/mol)

Es tan dard  Ewaste = (ln [Ag+]waste – ln [Ag+]standard) 0.0256783 E  Ewaste ln [Ag+]waste = s tan dard + ln [Ag+]standard 0.0256783   Estandard  Ewaste   + [Ag ]waste =  e 0.0256783    Ag    standard      c) Convert mol/L to ng/L for both [Ag+]waste and [Ag+]standard:



Ewaste – Estandard = –



[Ag + ]waste (8.314 J/mol  K)(298 K) ln (1)(96485 C/mol) [Ag + ]standard

Remember: ln A – ln B = ln (A/B)

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  Ewaste   + [Ag ]waste =  e 0.0256783    Ag    standard      d) Plug the values into the answer for part c).





  0.003   [Ag+]waste = e 0.0256783   1000. ng/L  = 889.7363 ng/L= 900 ng/L     e) Temperature is included in the RT/zF term, in which we have been using T = 298 K. To account for different temperatures, insert the required T value in RT/zF term.  RT   RT  Estandard +  standard  ln [Ag+]standard = Ewaste +  waste  ln [Ag+]waste zF    zF 

RTwaste RT ln [Ag+]waste – standard ln [Ag+]standard zF zF R Estandard – Ewaste = (Twaste ln [Ag+]waste – Tstandard ln [Ag+]standard) zF  zF  + + (Estandard – Ewaste)   = Twaste ln [Ag ]waste – Tstandard ln [Ag ]standard  R 

Estandard – Ewaste =

 zF  + + (Estandard – Ewaste)   + Tstandard ln [Ag ]standard = Twaste ln [Ag ]waste  R   E    standard  Ewaste   zF / R   Tstandard ln  Ag  standard  ln [Ag+]waste =    Twaste    

[Ag+]waste

 E     standard  Ewaste  zF / R   Tstandard ln  Ag standard    Twaste    = e

19.144 Reduction: Ag+(aq) + e–  Ag(s) Oxidation: Ag(s) + 2NH3(aq)  Ag(NH3)2+(aq) + e– Overall: Ag+(aq) + 2NH3(aq)  Ag(NH3)2+(aq)

Ecathode = 0.80 V

Eanode = 0.37 V

Ecell = Ecathode – Eanode = 0.80 V  0.37 V = 0.43 V

1(96485 C/mol)  0.43 V  zFEcell = = 16.74565 RT (8.314 J/mol  K)(298 K) K = 1.87302x107 = 1.9x107 ln Ksp =

19.145 Plan: Multiply the current in amperes by the time in seconds to obtain coulombs. Convert coulombs to amount (mol) of electrons with the Faraday constant and use the mole ratio in the balanced half-reactions to convert amount (mol) of electrons to amount (mol) 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  Capacity (C) =  300. mA•h   = 1.08x103 C   1 mA   1 h   1 A•s    b) The half-reactions are: Cd0  Cd2+ + 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    96,485 C   2 mol e    1 mol Cd  = 0.62907 g= 0.629 g Cd  

 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      Silberberg, Amateis, Venkateswaran, Chen, Chemistry: The Molecular Nature of Matter and Change, 3rd Canadian Edition Page 3-748 Instructor‘s Solution Manual © Copyright 2021 McGraw-Hill Ryerson Ltd.

= 1.026439 g= 1.03 g NiO(OH)  1 mol e    1 mol H 2O   18.02 g H 2O  Mass (g) of H2O = 1080 C   = 0.20170596 g= 0.202 g H2O  96,485 C   1 mol e    1 mol H O  2     Total mass of reactants = 0.62907 g Cd + 1.026439 g NiO(OH) + 0.20170596 g H 2O = 1.857215 g= 1.86 g reactants 1.85721 g c) Mass % reactants = 100%  = 10.14872 % = 10.1% 18.3 g 19.146 a) 2Zn  2Zn2+ + 4e– and 4e– + O2  2O2– Four mol e– flow per mole of reaction. 1 Mass (g) of Zn = ( )(0.275 g battery) = 0.0275 g Zn (assuming the 1/10 is exact.) 10

 1 mol Zn   4 mol e –   96, 485 C  Coulombs = (0.0275 g Zn)  = 81.1294 C= 81.1 C   –   65.41 g Zn  2 mol Zn   1 mol e  b) Free energy (J) = -zFEcell = - 4 (96485 C/mol)(1.3 V) = -5.01722105 J/mol = -5.0102 kJ/mol The amount of free energy released is 5.0102 kJ/mol. 19.147 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. 19.148 Determine the Ecell for the reaction given from the free energy: G° = –zF Ecell

Ecell = –G°/zF = – [(–298 kJ/mol)(103 J/1 kJ)]/(2 mol e–)(96,485 C/mol e–)[V/(J/C)] = 1.54428 V Using E° for Cu2+ + 2e–  Cu (cathode) and the Ecell just found:

Ecell = Ecathode – Eanode = 1.54428 V 1.54428 V = 0.34 V – Eanode

Eanode = 0.34 V – 1.54428 V = –1.20428 = –1.20 V The standard reduction potential of V2+ is –1.20 V. For the Ti/V cell, Ecell = 0.43 V

Vanadium is the anode.

Ecell = Ecathode – Eanode = 0.43 V Ecathode – (–1.20428 V) = 0.43 V Ecathode = 0.43 V – 1.20428 V = – 0.77428 V = – 0.77 V The standard reduction potential of Ti is – 0.77 V.

19.149 a) The iodide ion goes from –1 to 0, so it was oxidized. Iodide was oxidized, so S4O62– is the oxidizing agent. Iodide was oxidized, so I– is the reducing agent. b) G° = –zFE° n = 2

Ecell = –G°/zF = – [(87.8 kJ/mol)(103 J/1 kJ)]/(2 mol e–)(96,485 C/mol e–)[V/(J/C)] = – 0.454993 V= – 0.455 V Silberberg, Amateis, Venkateswaran, Chen, Chemistry: The Molecular Nature of Matter and Change, 3rd Canadian Edition Page 3-749 Instructor‘s Solution Manual © Copyright 2021 McGraw-Hill Ryerson Ltd.

c) S4O62–(aq) + 2e–  2S2O32– (aq) Oxygen remains –2 throughout. Sulfur is +2.5 in S4O62– (2.5 is an average). Sulfur is +2 in S2O32– . The potential for the iodine half-reaction comes from the Appendix. Since the iodide ion was oxidized, it is the anode.

Ecell = Ecathode – Eanode = – 0.454993 V Ecathode – (0.53 V) = –0.454993 V Ecathode = 0.53 V – 0.454993 V = 0.075007 V= 0.08 V

19.150 a) The reference half-reaction is: Cu2+(aq) + 2e–  Cu(s) E° = 0.34 V Before the addition of the ammonia, Ecell = 0. The addition of ammonia lowers the concentration of copper ions through the formation of the complex Cu(NH3)42+. The original copper ion concentration is [Cu2+]original, and the copper ion concentration in the solution containing ammonia is [Cu2+]ammonia. The Nernst equation is used to determine the copper ion concentration in the cell containing ammonia. The reaction is Cu2+initial(aq) + Cu(s) → Cu(s) + Cu2+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. (8.314 J/mol  K)(298 K) ln Q Ecell = Ecell – ( z )(96485 C/mol)  2+  (8.314 J/mol  K)(298 K) Cu  ammonia 0.129 V = 0.00 V – ln (2)(96485 C/mol) Cu 2+    original  2+  (8.314 J/mol  K)(298 K) Cu  ammonia 0.129 V = – ln (2)(96485 C/mol) Cu 2+    original

Cu 2     ammonia –10.04739 = ln  0.0100  Cu 2     ammonia 0.0100   [Cu2+]ammonia = 4.329865x10–7 mol/L 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 amount (mol) of copper and the original amount (mol) of ammonia are found from the original volumes and concentrations:  0.0100 mol Cu(NO3 )2   1 mol Cu 2   103 L  Original amount (mol) of copper =     90.0 mL    L    1 mol Cu(NO3 )2   1 mL  = 9.00x10–4 mol Cu2+  0.500 mol NH3   103 L  –3 Original amount (mol) of ammonia =   10.0 mL  = 5.00x10 mol NH3   L    1 mL  Determine the amount (mol) of copper still remaining uncomplexed.  4.329865x107 mol Cu 2   103 L  100.0 mL  Remaining amount (mol) of copper =     1 mL   L    = 4.329865x10–8 mol Cu 4.329865x10–5 =

The difference between the original amount (mol) of copper and the copper ion remaining in solution is the copper in the complex (= amount (mol) of complex). The concentration (mol/L) of the complex may now be found. Amount (mol) of copper in complex = (9.00x10 –4  4.329865x10–8) mol Cu2+ = 8.9995670x10–4 mol Cu2+  8.9995670x104 mol Cu 2    1 mol Cu(NH3 ) 24    1 mL  Concentration (mol/L) of complex =    3   2  100.0 mL   1 mol Cu   10 L  –3 2+ = 8.9995670x10 mol/L Cu(NH3)4 The concentration of the remaining ammonia is found as follows: Concentration (mol/L) of ammonia =   4 mol NH3    5.00x103 mol NH 3  8.9995670x10 4 mol Cu 2   2     1 mol Cu   1 mL     3  100.0 mL    10 L      = 0.014001732 mol/L ammonia



 



The Kf equilibrium is: Cu2+(aq) + 4NH3(aq)  Cu(NH3)42+(aq) 8.9995670x103  Cu(NH3 ) 4 2       Kf = = = 5.407795x1011 = 5.4x1011 4 2  4  7 Cu  NH3   4.329865x10   0.014001732     b) The Kf will be used to determine the new concentration of free copper ions. Amount(mol) of uncomplexed ammonia before the addition of new ammonia = (0.014001732 mol NH3/L)(10–3 L/1 mL)(100.0 mL) = 0.0014001732 mol NH 3 Amount(mol) of ammonia added = 5.00x10–3 mol NH3 (same as original amount (mol) of ammonia) From the stoichiometry: Cu2+(aq) + 4NH3(aq)  Cu(NH3)42+(aq) –8 Initial amount (mol) 4.329865x10 mol 0.0014001732 mol 8.9995670x10 –4 mol –3 Added amount (mol) 5.00x10 mol Cu2+ is limiting –(4.329865x10–8 mol) –4(4.329865x10–8 mol) +(4.329865x10–8 mol) After the reaction 0 0.006400 mol 9.00000x10–4 mol Determine concentrations before equilibrium: [Cu2+] = 0 [NH3] = (0.006400 mol NH3/110.0 mL)(1 mL/10–3 L) = 0.0581818 mol/L NH3 [Cu(NH3)42+] = (9.00000x10–4 mol Cu(NH3)42+/110.0 mL)(1 mL/10–3 L) = 0.008181818 mol/L Cu(NH3)42+ Now allow the system to come to equilibrium: Cu2+(aq) + 4NH3(aq)  Cu(NH3)42+(aq) Initial molarity 0 0.0581818 0.008181818 Change +x +4x –x Equilibrium x 0.0581818 + 4 x 0.008181818 – x 2  Cu(NH3 ) 4   =  0.008181818  x  = 5.34072x1011 Kf =  2  Cu  NH3  4  x 0.0581818  4x 4   Assume – x and + 4x are negligible when compared to their associated numbers: 0.008181818 Kf = 5.34072x1011 =  x 0.05818184 x = [Cu2+] = 1.3369x10–9 mol/L Cu2+ Use the Nernst equation to determine the new cell potential:

E = 0.00 V –

E=–

 2+  (8.314 J/mol  K)(298 K) Cu  ammonia ln (2)(96485 C/mol) Cu 2+    original

(8.314 J/mol  K)(298 K)  1.3369 109  ln   0.0100  (2)(96485 C/mol)  

E = 0.203215 V= 0.20 V 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 – Amount (mol) of OH– =      1 mol NaOH  1 mL  10.0 mL  = 5.00x10 mol OH L     The initial moles of copper ions were determined earlier: 9.00x10 –4 mol Cu2+ The reaction: Cu2+(aq) + 2OH–(aq)  Cu(OH)2(s) –4 Initial amount (mol) 9.00x10 mol 5.00x10–3 mol Cu2+ is limiting –(9.00x10–4 mol) –2(9.00x10–4 mol) After the reaction 0 0.0032 mol Determine concentrations before equilibrium: [Cu2+] = 0 [NH3] = (0.0032 mol OH–/100.0 mL)(1 mL/10–3 L) = 0.032 mol/L OH– Now allow the system to come to equilibrium: Cu(OH)2(s)  Cu2+(aq) + 2OH–(aq) Initial molarity 0.0 0.032 Change +x +2x Equilibrium x 0.032 + 2 x Ksp = 2.2x10–20 = [Cu2+][OH–]2 Ksp = 2.2x10–20 = [x][0.032 + 2x]2 Assume 2x is negligible compared to 0.032 mol/L. Ksp = 2.2x10–20 = [x][0.032]2 x = [Cu2+] = 2.1484375x10–17 = 2.1x10–17 mol/L Use the Nernst equation to determine the new cell potential: Cu 2+   hydroxide (8.314 J/mol  K)(298 K)  E = 0.00 V – ln 2+ (2)(96485 C/mol) Cu    original E=–

(8.314 J/mol  K)(298 K)  2.1484375 10 17  ln    (2)(96485 C/mol) 0.0100  

E = 0.433630 V= 0.43 V d) Use the Nernst equation to determine the copper ion concentration in the half-cell containing the hydroxide ion. Cu 2+   hydroxide (8.314 J/mol  K)(298 K)  E = 0.00 V – ln (2)(96485 C/mol) Cu 2+    original

Cu 2+   hydroxide (8.314 J/mol  K)(298 K)  ln 0.340 = – (2)(96485 C/mol) (0.0100) Cu 2     hydroxide –26.48149 = ln  0.0100  3.15671x10

–12

Cu 2     hydroxide =  0.0100 

[Cu2+]hydroxide = 3.15671x10–14 mol/L Now use the Ksp relationship: Ksp = [Cu2+][OH–]2 = 2.2x10–20 Ksp = 2.2x10–20 = [3.2622257x10–14][OH–]2 [OH–]2 = 6.96928x10–7 [OH–] = 8.3482x10–4 = 8.3x10–4 mol/L OH– = 8.3x10–4 mol/L NaOH 19.151 a) The half-reactions found in the Appendix are: Au3+(aq) + 3e–  Au(s) E° = 1.50 V Cr3+(aq) + 3e–  Cr(s) E° = –0.74 V Co2+(aq) + 2e–  Co(s) E° = –0.28 V Zn2+(aq) + 2e–  Zn(s) E° = –0.76 V Calculate Ecell = Ecathode – Eanode Au/Cr

Ecell = 1.50 V – (–0.74 V) = 2.24 V

Co/Zn

Ecell = –0.28 V – (–0.76 V) = 0.48 V

b) Connecting the cells as two voltaic cells in series will add the voltages. Eseries = EAu/Cr + ECo/Zn = 2.24 V + 0.48 V = 2.72 V c) Connecting the cells as one voltaic cell (Au/Cr) and one electrolytic (Co/Zn) in series will result in the difference in the voltages. Eseries = EAu/Cr – ECo/Zn = 2.24 V – 0.48 V = 1.76 V d) In parts a-c), Au3+ is reduced in the Au/Cr cell. In parts a-b), Co2+ is reduced in the Co/Zn cell. The connection in part c), forces the Co/Zn cell to operate in reverse, thus, Zn 2+ is reduced.  1 mol Au   3 mol e  1 mol Zn   65.41 g Zn  e) Mass (g) of Zn =  2.00 g Au       = 0.99609 g= 0.996 g Zn      197.0 g Au   1 mol Au   2 mol e   1 mol Zn 

19.152 a) The half-reactions found in the Appendix are: Oxidation: Cd(s)  Cd2+(aq) + 2e– E° = – 0.40 V Reduction: Cu2+(aq) + 2e–  Cu(s) E° = 0.34 V Overall: Cd(s) + Cu2+(aq)  Cd2+(aq) + Cu(s)

Ecell = Ecathode – Eanode = 0.34 V – (–0.40 V) = 0.74 V Note: Cd is a better reducing agent than Cu so Cu2+ reduces while Cd oxidizes. G° =  zFEcell G° = – (2)(96,485 C/mol )(0.74 J/C) = –1.427978x105 J/mol= –1.4x105 J/mol lnK = ln K =

zFEcell RT 2(96485 C/mol)  0.74 V 

= 57.636 (8.314 J/mol  K)(298 K) K = 1x1025 b) The cell reaction is: Cu2+(aq) + Cd(s) Cu(s) + Cd2+(aq) Use the Nernst equation: (8.314 J/mol  K)(298 K) ln Q Ecell = Ecell – (2)(96485 C/mol)  2+  (8.314 J/mol  K)(298 K) Cd  E = 0.74 V – ln (2)(96485 C/mol) Cu 2+   

An increase in the cadmium concentration by 0.95 mol/L requires an equal decrease in the copper concentration since the mole ratios are 1:1. Thus, when [Cd 2+] = 1.95 mol/L, [Cu2+] = (1.00 – 0.95) mol/L = 0.05 mol/L. (8.314 J/mol  K)(298 K)  1.95  ln  E = 0.74 V –  (2)(96485 C/mol)  0.05  E = 0.69296 V= 0.69 V c) At equilibrium, Ecell = 0, and G = 0 The Nernst equation is necessary to determine the [Cu2+]. Let the copper ion completely react to give [Cu2+] = 0.00 mol/L and [Cd2+] = 2.00 mol/L. The system can now go to equilibrium giving [Cu2+] = +x mol/L and [Cd2+] = (2.00 – x) mol/L. (8.314 J/mol  K)(298 K) ln Q Ecell = Ecell – (2)(96485 C/mol)

(8.314 J/mol  K)(298 K)  2.00  x  ln   (2)(96485 C/mol) x   Assume x is negligible compared to 2.00. 2.00 57.636 = ln x 2.00 1x1025 = x x = 1.9x10–25 mol/L Cu2+ 0.00 V = 0.74 V –

19.153 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.

r H = m  f (products) H – n  f (reactants) H r H = [(16)(  f H of CO2) + (18 )(  f H of H2O)] – [(2 )(  f H of C8H18) + (25)(  f H of O2)]

r H = [(16 )(–393.5 kJ/mol) + (18)(–241.826 kJ/mol)] – [(2)(–250.1 kJ/mol) + (25)(0 kJ/mol)] r H = –10148.868 kJ/mol= –10148.9 kJ per two moles of octane The energy from 4.00 L of gasoline is:  1 mL   0.7028 g   1 mol C8 H18  10148.9 kJ  Energy (kJ) =  4.00 L   3    10 L   mL   114.22 g C H  2 mol C H  8 18  8 18     5 5 = –1.2489313x10 = –1.25x10 kJ 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 vapour (since the heats of formation of the pure elements are zero). . H° = –241.826 kJ/mol The amount (mol) of hydrogen needed to produce the energy from part a) is:   1 mol H 2 Amount (mol) of H2 = 1.2489313x105 kJ / mol   = 516.45867 mol  241.826 kJ / mol 





Finally, use the ideal gas equation to determine the volume. L•bar   516.45867 mol H2   0.08314   273  25 K nRT mol•K   V= = = 1.27956x104 = 1.28x104 L p 1.00 bar 

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  Time (s) =  516.45867 mol H 2   = 9.966103x104 = 9.97x104 seconds  1 mol H    1 mol e   1.00x103 C  2    d) Find the coulombs involved in the electrolysis of 516 moles of H2.  2 mol e  96485 C  Coulombs =  516.45867 mol H 2   = 99,661,030 C     1 mol H   2   1 mol e   Joules = C x V = 99,661,030 C x 6.00 V = 597,966,177 J  1 kW•h  Power (kW • h) =  597,966,177 J    = 166.102 kW•h = 166 kW•h  3.6x106 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  Cost = 166.102 kW•h     = $23.2165 = $23.2  88%  1 kW•h  19.154 Plan: Write the half-reactions and the overall reaction. Calculate Ecell by using Ecell = Ecathode – Eanode and then use the Nernst equation to find [H+] at a cell potential of 0.915 V. pH is obtained from [H +]. Solution: The half-reactions are (from the Appendix): Oxidation: H2(g)  2H+(aq) + 2e– E° = 0.00 V Reduction: 2(Ag+(aq) + 1e–  Ag(s)) E° = 0.80 V Overall: 2Ag+(aq) + H2(g)  2Ag(s) + 2H+(aq) Ecell = 0.80 V – 0.0 V = 0.80 V The hydrogen ion concentration can now be found from the Nernst equation. (8.314 J/mol  K)(298 K) ln Q Ecell = Ecell – (2)(96485 C/mol) 2

H+  (8.314 J/mol  K)(298 K) 0.915 V = 0.80 V – ln  2 (2)(96485 C/mol)  Ag +  pH 2   2

H+  (8.314 J/mol  K)(298 K)   0.915 V – 0.80 V = – ln (2)(96485 C/mol) (0.100) 2 (1.00) 2

H+  (8.314 J/mol  K)(298 K)   0.115 V = – ln (2)(96485 C/mol) (0.100) 2 (1.00) 2

H  –8.9570 = ln    0.0100  2

H  1.28835x10 =    0.0100  –4

[H+] = 1.135x10–3 mol/L pH = –log [H+] = –log (1.1413851x10–3) = 2.94498 = 2.94 19.155 Plan: A concentration cell by definition has the same components in the two half cells, but at different concentrations. Write the balanced reaction using this information. We will then use c1V1=c2V2 to calculate the concentration of copper ion in beaker A. We will use the Nernst Equation to calculate the concentration of copper ion in

beaker B. We will write the reaction for the formation of the complex and set up and complete an ICE table assuming all of the copper ion reacts. We will then write the Kf expression and substitute the appropriate values to find Kf. Solution: (a) Cu 2+ A (aq) + Cu (s )

Cu 2+ B (aq) + Cu (s)

c1V1  c2V2

(b)

 0.10 mol/L  50.0 mL    c2 100.0 mL) 

Cu 2+  A   0.050 mol/L RT EE  ln Q zF 2+ RT   Cu B     0 ln  zF   Cu 2+  A   J    8.314   298 K    Cu 2+  B  mol  K      0.289 V=  ln   0.050 mol/L)   (c)  2  96485 C/mol    2+   Cu B    22.5 ln    0.050 mol/L)     2+ Cu B   1.68  10 10 0.050 mol/L)   12 Cu 2+  mol/L B   8.38  10

Cu 2B (aq)  4NH3 (aq)  Cu(NH3 ) 4 

2

(d) Initial Change Eqbm

0.05 mol/L -0.05 mol/L ≈0

0.5 mol/L -0.2 mol/L 0.3 mol/L

(aq)

0 0.05 mol/L 0.05 mol/L

Cu(NH3 )4  Kf  4 Cu 2    NH 3   0.05   8.38 1012   0.34 2

 7.4  1011 19.156 Plan: We will look up the Ksp value for silver sulfide and use it to calculate the concentration of the silver ion in the half cell with the silver sulfide. Then we will use the Nernst equation to calculate the potential, E. Solution:

Ag 2S (s) Initial Change Eqbm

saturated -x saturated

2 Ag + (aq) + S2- (aq) 0 2x 2x

0 x x

K sp   Ag   S2     2x x 2

8  1048  4 x3 x  1.25  1016 mol/L  Ag    2 x  2 1.25  1016 mol/L   2.5 10 16 mol/L RT EE  ln Q zF + RT   Ag B     0 ln  zF   Ag +A     J    8.314   298 K   2.5  1016 mol/L  mol  K  = ln   1 96485 C/mol   0.100 mol/L 

 0.86 V 19.157

Plan: We will use the relationship between current and time to find the charge and the relationship between charge and the Faraday constant to find the amount in mol of electrons transferred. We will use the half-reaction for Cr3+ to find the mass of chromium plated. Working backwards, we will find the amount in mol of electrons transferred to plate the same mass of silver and using the given current, find the time needed for plating. Solution: Q  It (a)

 60 s    8.75 A 15.5 min     min   8.14 103 C Q mCr   z  MM F 8.14 103 C   3 mol e  51.9961 g/mol     C   1 mol Cr    96485  mol e     13.2 g Cr mCr  mAg  Q

(b)

Q  z  MM Ag F

F  mAg z  MM Ag

C    96485  13.2 g  mol e      1 mol e     107.8682 g/mol   1 mol Ag   1.18  104 C

Q  It Q I 1.18  104 C  9.25 A

t

 1.27  103 s 

1 min 60 s

 21.2 min 19.158

Plan: In a concentration cell, the value for Eo is zero. We can use the reaction quotient and the given value of Ecell to find the concentration of silver ion in the saturated half-cell. Silver loses and gains 1 e- so z=1. Then we can use the concentration of the ions to find the value of K sp. We can draw the half cell. Solution: Ag (s) ∣ Ag+ (aq, sat‘d AgBr) ∣∣ Ag+ (aq, 0.100 mol/L) ∣ Ag (s)

E cell  0.305V Eo  0 RT ln Q zF J    8.314   298K   Ag   mol  K    ln  mol 1 mol e   96485 molC e  0.100 L mol  Ag    6.94  107   Br   L

E cell  

K sp   Ag    Br     6.94  10 7   4.8 10 13 2

19.159

Plan: Use the mass and molar mass to find the amount (mol) of Al. Use the proportion of amount (mol) of metal to amount (mol) of electrons to find the amount (mol) of electrons needed to produce this mass of metal. Find the charge using the value of z and the Faraday constant. Find the time using the relationship Q=It. Solution:

m Al 106g   3.93mol MM Al 26.98 g mol  1 mol Al : 3 mol e n Al 

3.93 mol Al : x mol e  a)

x   3.93mol  3  11.8mol C   Q  zF  11.8 mol e    96485  1.14  106 C   mol e   Q  It  1.25  105 A   t   1.14  106 C t  9.10 s n Al2 O3 

m Al2 O3 MM Al2 O3



218g 101.96

g mol

 2.14mol

1 mol Al2 O3 : 2 mol Al 2.14 mol Al : x b)

x=4.28 mol Al The actual yield is 3.93 mol actual yield  100% theoretical yield 3.93mol   100% 4.28mol  91.9%

mol % yield =

19.160

Plan: Use the current and time to find the total charge, and then the Faraday constant to find the amount in mol of electrons transferred. Use the fact that 3 mol of electrons are transferred per mol of metal to find the amount (mol) of the metal formed. Use the mass of metal formed and the amount (mol) of metal formed to calculate the molar mass of the metal and find its identity. Solution:

Q  It  60 min   60s    0.3001A  0.7500h      1h   1min   8.103  102 C Q  zF Q z  F    8.103  102 C    C   96485  mol e     8.398  103 mol e  1 mol M : 3 mol e  x

: 8.398 10 3 mol e 

x  2.800  103 mol M m 0.4036g g MM M  M   144.2 n M 2.800  103 mol mol Therefore, the metal is neodymium, Nd. 19.161

Plan: Use the current and time to find the charge transferred, then use the charge and the Faraday constant to find the amount (mol) of electrons transferred. Use the ratio of amount (mol) of metal to amount (mol) of electrons to find the amount (mol) of metal formed. Use the amount (mol) of metal and the molar mass to find the mass of metal formed, then the density of the metal to find the volume formed. The volume is the surface area times the thickness, so the thickness can be calculated. Solution:

Q  It  60s   1.3A  25 min     1min   2.0  103 C Q  zF Q z  F    2.0  103 C     96485 C  mol e     2.0  102 mol e  1 mol M : 1 mol e  x

: 2.0  10 2 mol e 

x  2.0  102 mol Ag g   m Ag  n Ag MM Ag   2.0 10 2 mol  107.87   2.2g mol   m 2.2g VAg    0.21cm3 d 10.5 g cm3 V 0.21cm3 Thickness of the silver coat =   9.9 10 3 cm  0.099mm  99  m SA 21cm 2 19.162 Plan: Write down the two half reactions and the half-cell potentials. Decide which would be the oxidation and which would be reduction half reaction based on the potentials and the requirement for a positive cell potential for a spontaneous cell. Write the balanced cell reaction. Use the equation for the cell potential. Input all the values provided for the variables and for Q to be able to calculate the silver ion concentration. Find the amount (mol) of silver using the concentration and volume of the solution, then the mass using the molar mass and amount (mol) of silver. Use the mass of silver and the mass of the sample to find the percent by mass silver in the sample. Solution: Ag+ (aq) + e-  Ag (s) 2 H+ (aq) + 2 e-  H2 (g) 2 Ag+ (aq) + H2 (g)  2 H+ (aq) + Ag (s)

Eo = 0.80 V Eo = 0.00 V (SHE) Eo = 0.80 V

E cell  E o 

RT ln Q zF 2

H  RT E cell  E  ln  2 zF  Ag   p   H2 o

J    mol   8.314   298K  1  mol  K L   0.625V  0.80V  ln  2    2 mol e   96485 molC e  Ag  1bar  1 L2  8.3  105 2 mol 2  Ag   mol  Ag    1.1 10 3 L mol   4 n Ag  cV  1.1 103   0.500L   5.5  10 mol L   g   2 m Ag  n Ag MM Ag   5.5  10 4 mol  107.87   5.9  10 g mol   2 m Ag 5.9  10 g %Ag   100%   100%  5.1% msample 1.160g

CHAPTER 4 ORGANIC COMPOUNDS AND THE

ATOMIC PROPERTIES OF CARBON END–OF–CHAPTER PROBLEMS 20.1

Organic: Inorganic:

20.2

methane (natural gas), CH4 ethanoic acid (also known as acetic acid or vinegar), C2H4O2 calcium carbonate, CaCO3 sodium bicarbonate (also known as baking soda), NaHCO3

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, CH 4), network covalent solids (e.g., diamond), and polyatomic ions (e.g., carbonate, CO32–). b) Since carbon has four valence 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 C 4+ 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 to form π-bond of unhybridised p orbitals on neighboring C atoms. The sideways overlap of p orbitals results in double and triple bonds. 20.3

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. 20.4

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.

20.5

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.

20.6

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 and the C–C bonds are unreactive because electron density is shared equally between the two atoms. The C-I bond is reactive because it is long and weak. The C=O bond is reactive because oxygen is more electronegative than carbon and the electron rich  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 electronrich region around carbon and an electron-poor region around Li.

20.7

a) An alkane is an organic compound consisting of carbon and hydrogen in which there are no multiple bonds between carbon atoms, only single bonds. A cycloalkane is an alkane in which the carbon chain is arranged in a ring. 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 carbon atoms in the ring).

20.8

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 the hydrogen gives a correct structure. CH3 CH3

CH2

CH3

b) The third carbon from the left in the chain is bonded to three groups, so attach one hydrogen atom to this carbon

c) The second carbon in the chain has five bonds, so remove the ethyl group from the second carbon. One can also move this ethyl group to the third carbon. To do this, a hydrogen atom must be removed from the third carbon atom.

CH3

CH2 CH3 d) Structure is correct.

20.9

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

CH3 b) Cyclohexane denotes a six-carbon ring containing only single bonds. Numbering of the carbon atoms 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

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. The correct name is 2,2-dimethylbutane. 20.10

a) 2-methylbutane

b) 1,3,5-trimethylcyclohexane

CH2

CH3

CH3 CH2

CH3

CH3 CH2

CH2

CH3

20.11

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:

Numbering the longest chain gives the correct name, 3-methylhexane.

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. CH3 CH3

CH2

CH3

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.

20.12

a) 3,3-dimethlybutane should be 2,2-dimethylbutane.

CH3 3 4 CH3

CH2

CH3 1

CH3 b) 1,1,1-trimethylheptane should be 2,2-dimethyloctane. CH3 8 CH3

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

20.13

All non-numbered carbon atoms are secondary (2). All other carbon atoms are identified as primary (1), tertiary (3) or quaternary (4). Note that this notation is not used for carbon atoms participating in double or triple bonds.

20.14

or similar

20.15

20.16

20.17

Configurations are unique arrangements of the same atoms, differing in the connectivity (how the atoms are joined together) of the atoms. Conformations are unique arrangements of the same atoms, differing by rotation around single bonds. Configurations require bonds to be broken to be interconverted, whereas conformations are interconvertable without the need to break bonds.

20.18

B is the most stable as it minimizes the interaction between the largest substituents (methyl groups)

20.19

B is less favoured than A due to 1,3-diaxial interactions. In B, as the ethyl and the propan-2-yl (also known as isopropyl or i-Pr) groups can rotate to place a proton over the ring, they create essentially the same instability. When X= t-Bu,no protons are left and a much greater steric repulsion is felt.

20.20 B is the least stable. A and C are equally stable.

20.21

Note the stereochemistry is unimportant here as the molecule is symmetrical. 20.22

Carbon skeleton:

Fischer projection:

20.23

Structure B is more stable as the bulky groups (1,1-dimethyl ethyl, also known as tert-butyl or t-Bu, and methyl) are in the equatorial position, which reduces the 1,3-diaxial interactions. 20.24

20.25

a) Constitutional isomers are those with different sequences of bonded atoms. b) Geometric isomers are those where the connectivity between the atoms is the same, but have a different spatial arrangement of the atoms. Geometric isomers differ in the geometric arrangement of the groups attached to the double bond or to the ring. c) Optical isomers are a type of stereoisomerism that arises when a molecule and its mirror image cannot be superimposed on each other. They rotate the plane of polarized light in opposite direction. Configurational and conformational isomers are stereoisomers. Constitutional isomers are not stereoisomers.

20.26

Alkynes are linear about the triple bond. Aromatics can only assume one (planar) orientation in space.

20.27

a) The Z configuration is unstable and will isomerize to give b) The E configuration is the most stable and so will remain mostly c) Only the middle double bond is in its least thermodynamically favorable form and therefore this will isomerize to give the major product shown below.

20.28

a) trans (with respect to longest chain) but labelled Z b) trans (with respect to longest chain) but labelled E. As you can see from above E/Z and cis-trans nomenclature becomes more complicated with double bonds having more than 2 substituents. In general, we need to specify what the reference point is when talking about cis and trans. In this case we used the longest carbon chain. Moreover, for double bonds with 2 or more substituents the E/Z nomenclature is preferred.

20.29

None. Both samples will not rotate plane polarized light. They are not chiral, and therefore no optical activity will be observed.

20.30

An asymmetric carbon is sp3 hybridised and has 4 different groups attached to it. The absolute requirement is that it must have a non-superimposable mirror image.

20.31

A racemic mixture contains equal amounts of 2 enantiomers. It will not affect plane polarized light as each enantiomer acts in an equal and opposite fashion on the light passing through the sample, thus cancelling any overall affect.

20.32

H < D < O < P < Br < I

20.33

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.

20.34

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.

20.35

Aromatic hydrocarbon atoms have carbon in sp2 hybridization while cycloalkanes have carbon in sp3 hybridization. Aromatic hydrocarbons are planar while most cycloalkanes assume puckered ring structures.

20.36

The compound 2-methylhex-3-ene has cis-trans isomers.

The compound 2-methylhex-2-ene does not have cis-trans isomers because the #2 carbon atom is attached to two identical methyl (–CH3) groups:

20.37

Plan: Sketch each structure to evaluate whether the central atom is connected to at least four different groups. To be optically active the molecules mirror image must be nonsuperimposable. Solution: a) b) c) d)

F Cl

Se Br

Species a) and c) are optically active because the central atom has four different groups (an asymmetric carbon) arranged in a tetrahedral shape where the mirror image is nonsuperimposable. Species b) is not optically active because there are two Cl atoms and the mirror image is superimposable. In the case of d), although the Se has more than four electron groups, the mirror image is nonsuperimposable, making the species optically active. 20.38

Plan: Use the Cahn-Ingold-Prelog rules to evaluate each structure. In the case of a chiral centre, orient the molecule such that the lowest priority of the four substituents is pointed away from the viewer. If the priority of the remaining three substituents decreases in a clockwise direction, the carbon is labeled R. If the priority of the substituents decreases in a counterclockwise direction, the carbon is labeled S.

For double bonds, look at the atoms directly attached to each carbon of the double bond. Rank them according to decreasing atomic number. If two atoms are identical, look at all the atoms directly attached to these atoms. If the high priority groups are on the same side, then the alkene is Z (i.e., German for “together”) and if the high priority groups are on the opposite side, then the alkene is E (i.e., German for “opposite”). If there is more than one double bond, then the location needs to be included with the locant (e.g., 2E or 4Z). a)

In the molecule above, the lowest priority of the four substituents (4) is pointed towards the viewer and the priority of the remaining three substituents decreases in a clockwise direction. However, if we invert the molecule, the lowest priority of the four substituents is pointed away from the viewer and we get exactly the opposite: the priority of the substituents decreases in a counterclockwise direction. That is why the stereogenic centre is S.

R and E

2S,3R

20.39

2S,3S

20.40

20.41

Plan: In common names, the positions of two groups are indicated by o- (ortho) for groups in positions 1 and 2, m- (meta) for groups in positions 1 and 3, and p- (para) for groups in positions 1 and 4. Solution:

20.42

A is more stable as the double bonds form an aromatic system which is stabilized by resonance.

20.43

All 6 isomers of trinitrotoluene are shown below:

20.44

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 H

CH3

C H chiral carbon 20.45

CH3

CH3 chiral carbon

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 OH H H C H H C C H C H C C H H H chiral carbon chiral carbon C C

H H 20.46

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

Cl 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

20.47

CH2

CH3

All are optically active. a)

chiral carbon CH2

CH2

CH3

b) CH3 CH3

chiral carbon

CH3

CH2

CH3

chiral carbon Br

CH2

CH3

Cl 20.48

Solution: a) Both carbon atoms 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.

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.

c) No geometric isomers occur because the left carbon participating in the double bond is attached to two identical methyl (–CH3) groups. 20.49

a) Cis-trans geometric isomerism occurs about the double bond.

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.

20.50

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 hex-3-ene is CH3CH2CH═CHCH2CH3. Both carbon atoms in the double bond are bonded to two distinct groups, so geometric isomers will occur.

c) The structure of 1,1-dichloroethene is CCl2═CH2. Both carbon atoms 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

C H

cis-1,2-dichloroethene

20.51

trans-1,2-dichloroethene

a) The structure of pent-1-ene 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)

c) Cl

CH3 C

H C

C CH3

cis-1-chloropropene trans-1-chloropropene d) There are no geometric isomers because the first carbon has two hydrogen atoms attached to it. 20.52

a) An alkene is a hydrocarbon with at least one double bond between two carbon atoms. An alkyne is a

hydrocarbon with at least one triple bond between two carbon atoms. b) 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. 20.53

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 carbon 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 carbon 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

C Double bond between second and third carbons: C C C C C C C

C C

C Double bond between third and fourth carbons: C C C C C C C

C Double bond between branched carbon and chain: C C C C C C

C The total number of unique skeletons is eleven. 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

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.

C C

C 20.54

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

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

C 20.55

Plan: Add hydrogen atoms to make a total of four bonds to each carbon. Solution: a) CH2 CH2 CH2 CH3 CH2 CH CH CH2 CH2 C CH3 CH2

CH2

CH3

CH3 CH2

CH3

CH2

CH3

CH2

CH3

CH2

CH3

CH2

CH3

CH3 CH3

CH3

CH3 CH3

CH2

CH3 CH

CH2

CH3

CH2

CH3

CH2

CH3

CH2

CH3

CH2

CH3

CH3 CH3

CH2

CH3

CH2

CH3

CH2

CH3

CH2

CH3

CH2

CH3

CH3 CH2

CH3

CH2

CH3 CH2 CH3

CH3

CH2

CH3

C CH3

CH3

CH3 CH3

CH3

CH3 CH3

CH3

CH3 CH3

CH3

CH2

CH3

CH2 CH2

CH3

CH2

CH3 CH

CH2

CH2 CH

CH2

CH3

CH2

CH3 20.56

Add hydrogen atoms to make a total of four bonds to each carbon. a) CH CH CH3 CH2 CH CH2 C CH2 C

CH2

CH3 CH

C CH2

CH2

CH3

CH3 CH

CH2

CH3

CH2

CH3

CH3 CH

CH3

CH3 C CH

CH2

CH3

CH2

CH3

CH2

CH3 C

CH3

CH3 b) CH3

CH2

CH3

CH CH3

CH3 CH3

CH2

CH3 c) There are cis-trans isomers not shown here. CH2 CH CH2 CH3 CH2 CH CH3 CH2

CH2

CH3

20.57

CH2

CH3

Plan: Remember that each C must have 4 bonds. Solution: a) The first carbon in the chain has five bonds, so remove one of the hydrogen atoms on this carbon. H2C CH CH2 CH3 b) 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 c) Structure is correct.

20.58

a) Structure is correct. b)

CH3 C CH3 c) CH3

CH2

CH3

20.59

Due to resonance structures, all the bonds in benzene are equivalent and the same, having partial double bond and partial single bond characteristic.

20.60

methylbenzene = toluene, benzenol = phenol, methylbenzenol = cresol, and dimethylbenzene = xylene.

20.61

The aromatic component is circled in red.

20.62

A) Methyl: ortho B) Hydroxy: para C) Bromo: meta D) Chloro: meta E) Fluoro: ortho

20.63

Molecules with large numbers of aromatic components might be used for dyes because of their extended  systems that can absorb light in the visible range.

20.64

Plan: Identify the parent compound based on the substituents attached to the benzene ring. Note the following common names: toluene (methylbenzene); phenol (hydroxyl group on benzene ring); aniline (NH2 group on benzene ring). Ethenylbenzene is also known as vinylbenzene or styrene; however no substitution is allowed when using styrene. Solution: a) 2-bromo-4-methylphenol b) 5-bromo-2-chloro-3-methylaniline c) 1-bromo-2,6-dichloro-3-fluoro-5-methyl-4-ethenylbenzene

20.65

A has a positively charged carbon and two double bonds. 4n+2 = 4; n = ½. The molecule is not aromatic. B has three double bonds and a carbon with a lone pair which is delocalized in the ring. 4n+2 = 8; n = 1.5. B is not aromatic. C is fully conjugated molecule. The N atoms are using their one p orbital for the electrons in the double bond and their lone pair of electrons are not  electrons; therefore, the total of  electrons is 6. Applying Hückel‘s rule, we find that n = 1. Since the molecule follows all criteria for aromaticity, we conclude that C is aromatic. In D, the top carbon is sp3 hybridised, so the molecule is not aromatic.

20.66

In A, the top carbon is sp3 hybridised and so A is not aromatic. In B, the lone pair of electrons of the N at the top of the ring is delocalized in the ring, while the lone pair of electrons for the N at the bottom is not. This means that the molecule has 6 electrons. 4n+2 = 6; n = 1. B is therefore aromatic. C has a similar situation where one of the lone pairs of electrons from O is delocalized in the ring while the second pair is not. This results in 6 electrons and n = 1. C is therefore aromatic. E also has 6 electrons and n = 1 making it also aromatic. In the case of D, all the carbon atoms are sp2 hybridised and lone pair of electrons from the N is delocalized in the ring.

Since the electrons from the carbonyl are outside the ring, the total number of  electrons within the ring is 6, resulting in n = 1. D is therefore aromatic

20.67

20.68

a) 4,4-dimethylpent-2-yne b) (2Z,5E)-hepta-2,5-diene (if counted from right to left) or (2E,5Z)-hepta-2,5-diene (if counted from left to right) c) (E)-5-bromopent-3-en-1-yne

20.69

Plan: Benzene is a planar, aromatic hydrocarbon. It is commonly depicted as a hexagon with a circle in the middle to indicate that the  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. Solution: Cl Cl Cl Cl

1,2-dichlorobenzene (o-dichlorobenzene)

1,3-dichlorobenzene (m-dichlorobenzene)

Cl 1,4-dichlorobenzene (p-dichlorobenzene)

20.70

CH3 CH3

CH3

1,3,5-trimethylbenzene

1,2,3-trimethylbenzene 20.71

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 carbon atoms #2 and #6 a 1,1-dimethyl ethyl or tert-butyl group, –C(CH3)3; and 3) on carbon #4 a methyl group, –CH3. Solution: HO C(CH ) 3 3

(H3C)3C

CH3

20.72

From lowest to highest, the predicted boiling point order is: C < B < D < A. Boiling point increases with molecular weight (therefore C is lowest, followed by B). Long chains have higher boiling points than branched species since they are able to form stronger intermolecular interactions. D is more hindered than A and therefore is less able to make hydrogen bonds (therefore D is lower than A)

20.73

Ethoxyethane cannot form intermolecular hydrogen bonds whereas butan-1-ol can. Therefore, even though they have the same mass, ethoxyethane has a lower boiling point than butan-1-ol.

20.74

C. In general para substituted halobenzenes have high melting points as they fit in crystal lattices better. In fact, A and C are both liquid at room temperature and B is solid.

20.75

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 a haloalkane. 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.

20.76

a) amide

b) alkene

c) ketone

d) amine

20.77

CH3

CH2

CH3

N ketone

O nitrile

H amide alkene e)

O C

CH2

CH3

ester

20.78

c) CH2

alkene

CH2

d) O

CH3

ester

CH3

ester amide

20.79

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 carbon atoms: CH2 CH2 CH2 CH2 CH3 CH3 CH CH2 CH2 CH3 CH3 CH2 CH CH2 CH3 OH OH OH 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. CH2 CH CH2 CH3 CH2 CH2 CH CH3

CH3

CH3 CH3 CH3

CH2

CH3

OH CH3 OH And use a three-carbon chain with three side groups: CH3 CH3

CH2

CH3 The total number of different structures is eight.

20.80

Aldehydes: O

O CH3

CH2

CH3

CH2

CH3

O CH3

CH CH3

CH2

CH3

Ketones:

20.81

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. Solution: CH3 CH2 CH2 CH2 NH2 CH3 CH2 CH NH2

CH3 CH3 CH3

CH2

NH2

CH3

CH2

NH2

CH3

CH2

CH3

CH3 CH3

CH2

CH3

CH2

CH3 20.82 CH3

CH2

CH3

CH2

CH3

CH CH3

20.83

CH2

C O

CH3

OH CH3

CH3

Plan: We can calculate the degrees of unsaturation from the following general formula: 2C  N  H  X  2 DU  2 where C = number of carbon atoms, N = number of nitrogen atoms, X = number of halogens and H = number of hydrogen atoms. Solution: 2(13)  0  16  0  2 DU  =6 2

20.84

DU 

2(8)  1  13  0  2 =3 2

20.85

DU 

2(6)  2  6  2  2 =4 2

20.86

DU 

2(6)  1  7  0  2 =4 2

20.87

20.88.

2(5)  0  8  0  2 =2 2 With a degree of unsaturation of 2, the molecule could have two double bonds, one triple bond or one ring and one double bond, as shown below. DU 

2(7)  0  6  0  2 =5 2 With a degree of unsaturation equal to 5, the molecule could have five double bonds, a triple bond and three double bonds, two triple bonds and a double bond or a ring and four double bonds, as shown below (recall that an acid reacts with a base to form a salt). DU 

20.89

20.90

2(7)  1  11  0  2 =3 2 Knowing that reduction results in the formation of an amine tells us that a nitrile group is present. This then accounts for two of the degrees of unsaturation. If no alkene is present, the other degree of unsaturation must be due to a ring. A possible structure is shown below. DU 

Plan: The Table of Functional Groups is used to identify the three functional groups. Solution:

H R

aldehyde functional group

R OH

R C OH

alcohol functional group

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. 20.91

Plan: Using the given formula draw all possible isomer taking care to have the correct number of carbons and hydrogen. From Section 20.2 we learned that, within a group of isomers the more spherical an alkane is the lower boiling point it will have compared to more linear forms. Solution: Increasing Boiling Point

2,2-dimethylpropane

2-methylbutane

pentane

Since 2,2-dimethylpropane has the most branching it is the most spherical and has the lowest point, pentane is linear and has the highest boiling point. 20.92

Plan: Knowing that a linear alkane has the formula CnH2n+2 and that there are 5 carbon atoms and two of the hydrogen atoms have been substituted by bromine, the molecule will then have the formula C5H10Br2. We can now draw all possible isomers taking care to have the correct number of carbon and hydrogen.

Solution: Br

Br Br

For the second part of the problem we first draw one molecule and then the mirror image of it, this will be our first enantiomer pair. By changing the stereochemistry of one of the bromine atoms on either of our structures and then drawing the mirror image of that molecule we will have another pair of enantiomers. Each pair of enantiomers is a diastereomer for the opposite pair of enantiomers. Following the CIP rules in section 20.4 we now label the stereoisomers with R/S notation.

20.93

Plan: The general formula CnH2n refers to alkenes so we can now draw and name all of our possible structures of C5H10 making sure that all carbon and hydrogen atoms are present. Solution:

Two of the possible isomers are geometric isomers and we can name them using E/Z nomenclature from Section 20.5

20.94

Plan: In order to name these compounds we must use our priority rules from Sections 20.4 and 20.5 paying special attention to the alkene. Solution: a) For the first compound we first rotate the molecule so that the hydrogen atom is facing away from us. The alkene group takes priority as it counts as -CClR2 vs. -CH2Cl which gets second priority and the lone methyl group is lowest priority. C C

H Cl

Cl CH3

H 3C

CH3

Cl H Cl CH3

(S)-2,4-dichloro-3-methylbut-1-ene The second molecule already has the hydrogen atom facing away from us so it‘s easier to name, this time the –CH2Cl takes precedent over the alkene due to the chlorine atom. 2 H

H 3C

(S)-4-chloro-3-methylbut-1-ene b) Moving the chlorine atom would change the chirality from S to R in both molecules, since it would change the order of priority of the groups attached to the chiral centre. It would also create two new geometric isomers which would also need to be named appropriately.

20.95

Plan: Using the formula from Section 20.6 we can determine the degree of unsaturation the formula C 7H10O2 has. We can then draw the structures using the rules from 20.6, for the aromatic structure we must check any structures we draw using Hückel's rule (Section 20.5) Solution: ( )

For DU = 3, we can have 3 rings or three double bonds or 1 triple bond and 1 double bond or 2 rings and 1 double bond or 1 ring and 2 double bonds in our final structure. Below are several possible structures containing the required functional groups and following the formula C7H10O2. The ring structure below has n = 1 and therefore satisfies the Hückel rule meaning it is aromatic. HO

aldehyde

alkyne

OH O

aromatic ring

ketone

It is not possible to create a molecule containing all of the functional groups as it would require a DU of at least 6. 20.96

Plan: We know from Section 20.3 that the most stable conformation will be the one that has the most number of groups in the equatorial position and that groups which are larger will take precedent over smaller groups. Solution: The first structure (A) has a 1,1-dimethyl ethyl or tert-butyl group equatorial and a methyl group axial, the tertbutyl group defines the most stable conformation.

The second molecule (B) has one conformation in which both groups are equatorial, therefore this is the most stable.

The third structure (C) will always have one group axial and one group equatorial, therefore both conformations are equally stable.

20.97

Plan: We will need to use Hückel's rule to determine which structures are aromatic (Section 20.5) Solution: The first molecule has 4 π electrons from the two double bonds and 2 π electrons from one of the oxygen lone pairs: 6 = 4n + 2 n = 1, therefore A is aromatic. The second compound has 6 π electrons from the three double bonds and the lone pairs from the two nitrogen atoms are not in the ring: 6 = 4n + 2 n = 1, therefore B is aromatic. The third compound has no π electrons and since none of the atoms are sp2 hybridized, therefore C is not aromatic. The fourth compound has 4 π electrons: 4 = 4n + 2

n = 0.5, therefore D is aromatic.

The fifth compound has 4 π electrons from the two double bonds and 2 π electrons from the carbanion lone pairs: 6 = 4n + 2 n = 1, therefore E is aromatic. There are no chiral centres on any of these molecules. 20.98

Plan: Section 20.6 contains the section on amides which shows that the result of an amide hydrolysis reaction is a carboxylic acid and an amine, allowing us to draw the two structures and label the stereochemistry. Solution: Cl

O H

H +

N H

Br H2N

H (R)-N-((S)-2-bromopropyl)-3-chloropentanamide

( R)-3-chloropentanoic acid

(S)-2-bromopropan-1-amine

Changing the bromide for a hydroxide group would lead to a change in the name of the complex, but not affect the stereochemistry since the priority order is maintained. H

Br H2N

(S)-2-bromopropan-1-amine

OH H2N

(S)-1-aminopropan-2-ol

20.99

Plan: Aromatic nomenclature is covered in Section 20.5 and functional group naming priority is listed in Table 20.7. We also know that the boiling point is determined by both the molar mass and the presence of polar groups, which form stronger intermolecular interactions than non-polar groups. We will examine the three molecules with both these facts in mind. Solution: The naming changes for the third molecule because a carboxylic acid has a higher naming priority than a nitrile, which is higher than a ketone. Since all three molecules have essentially the same molar mass, the boiling point is determined by the presence of the most polar groups. A carboxylic acid is much more polar than a ketone; therefore 2-cyanobenzoic acid will have the highest boiling point of the three molecules. N

N C

OH O 2-acetylbenzonitrile o-acetylbenzonitrile

2-cyanobenzoic acid 3-acetylbenzonitrile

o-cyanobenzoic acid

p-acetylbenzonitrile

20.100 Plan: We will have to use our knowledge of functional groups from Section 20.6 to determine the nature of the functional groups. Ethers are weakly polar and capable of accepting hydrogen bonds, in addition they are highly flammable and also have anesthetic properties. Amines are heteroatoms and also give compounds strong fish-like odours. Carboxylic acids have the highest naming priority of all the functional groups. Solution:

c) 5-Aminohexanoic acid has a chiral centre at C-5

CHAPTER 5 ORGANIC REACTION MECHANISMS END–OF–CHAPTER PROBLEMS 21.1

a) addition

b) elimination

c) substitution

21.2

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  bond and a  bond. It is the  bond that breaks in the addition reaction and that forms in the elimination reaction.

21.3

Yes, an addition, elimination, or substitution reaction can be a redox reaction.

Addition:

(Reduction: addition of H2) Elimination: CH3

CH3

Substitution: H

OH (Oxidation: loss of H2)

H H + Cl

+ H

H H (Carbon loses electrons - oxidation state of -4 to -2 and chlorine gains electrons – oxidation state of 0 to -1). 21.4

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.

21.5

a) addition reaction

21.6

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 group of atoms substitutes for another one in the reactant.

b) substitution reaction

Solution: a) Water (H2O or H and OH) is added to the double bond: H+ CH3CH2CH=CHCH2CH3 + H2O CH3CH2CH2CHCH2CH3 OH 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: h

CH3CH3 + 2Cl2  CHCl2CH3 + 2HCl 21.7

a) CH3CHBrCH3 + KI  CH3CHICH3 + KBr b)

Cl Cl2

Cl c)

O + CH3 21.8

CH3

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.

21.9

Assuming that the observed carbon is bonded to carbon: a) oxidation b) reduction c) oxidation

21.10

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: KMnO4 a) The reaction CH3CH=CHCH 2 CH 2 CH3   CH 2CH  OH  —CH  OH  CH 2CH 2CH3 shows the cold OH  second and third carbon atoms in the chain gaining a bond to oxygen: C–O–H. Therefore, the hex-2-ene 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.



3 H2

catalyst

21.11

a) reduced

b) oxidized

21.12

Step 1 substitution

21.13

The answer is (c). A concerted process occurs in one step (i.e., one hump) and in an exothermic reaction the products have lower energy than the reactants.

21.14

a) stepwise b) exothermic c) step A

21.15

The fastest reaction is (a). It has the lowest activation energy.

21.16

Plan:When the electrons in the bond are equally divided, we call it a homolytic cleavage, while when the bond is unequally divided, the more electronegative atom gets the electrons, and we call it a heterolytic cleavage Solution: a) heterolytic b) homolytic c) heterolytic

21.17

An electrophile is an electron poor species (which may or may not be positively charged) which will react with electron rich species.

Step 2 addition

The carbon atom in (a) is the most electrophilic. The chlorine atom is more electronegative than the bromine atom, making the carbon more electron deficient. The oxygen in (c) is negatively charged, reducing its electronegativity, making the oxygen less electronegative than the chlorine atom. 21.18

The oxygen atom in (b) is the most nucleophilic. Nucleophilicity of the same atom type roughly follow the basicity of the atom. Methanol is more basic than the mildly acidic phenol (a) or the more acidic carboxylic acid (c) or sulfonic acid (d)

21.19

(d) is the most nucleophilic. Across atoms of different types, nucleophilicity and basicity increases from right to left across the periodic table. As the species given can be thought of existing as ion pairs, the CH 3 ion is the least stable anion, making it the most basic and nucleophilic.

21.20

Like DMSO or DMF, HMPA is a polar aprotic solvent that can solvate cations (Na +) well, but cannot solvate anions (F–). By solvating the fluoride‘s counterion, HMPA makes the anion less stable, and therefore more nucleophilic.

21.21

Nucleophilicity generally increases on decending a group. As sulfur is in a lower period than oxygen, it is larger in size and therefore more polarizable, making it more nucleophilic.

21.22

Potassium carbonate is not a strong enough base to fully deprotonate methanol. The iodide anion is more nucleophilic than the methanol and reacts first to form an alkyl iodide. The newly formed alkyl iodide is more reactive as an electrophile as the iodide is a better leaving group. When the methanol substitutes the alkyl iodide, it regenerates the iodide anion which can be reused as a catalyst.

21.23

The hydroxyl (OH-–) anion is a poor leaving group. By treating the alcohol with tosyl chloride to convert it into a tosylate, a better leaving group can be obtained accelerating the rate of reaction with sodium azide. 21.24

(b). As SN1 reactions are a stepwise process, they generate an intermediate carbocation. The generation of this carbocation is the rate determining step – therefore substrates which generate a more stable carbocation will have lower activation energies for the formation of this carbocation. As methyl groups are electron releasing (due to hyperconjugation) the most substituted carbocation will be the most stable.

21.25

Protonation of the alcohol functionality generates a good leaving group, which on heterolytic bond fission generates carbocation C. The carbocation can be stabilized by resonance as shown (D). The acetate nucleophile can then attack either the hindered tertiary center generating A or the unhindered center generating B. 21.26

21.27

As the reaction goes by an SN1 mechanism, there is a planar carbocation intermediate which can be attacked from either side giving both enantiomers and resulting in a racemic mixture.

21.28

Polar aprotic solvents are best for SN2 reactions. Therefore the best solvent is (c) acetonitrile.

21.29

(b). The azide (N3-) is negatively charged (making it highly nucleophilic). In addition, as the azide is linear shape it is less sterically hindered than the other two nucleophiles.

21.30

The adjacent tertiary center provides a large amount of steric hindrance, blocking the approach of incoming nucleophiles slowing down the rate of nucleophilic displacement.

21.31

The most simple answer, based on our understanding developed in this chapter relies on steric considerations. In molecule B, the methyl groups create a large amount of steric hindrance, blocking the approach of a nucleophile.

21.32

Importantly, to maintain the stereochemical information, a series of S N2 reactions should be proposed. In the first reaction, the alcohol is deprotonated and reacted with methyl iodide via an SN2 reaction. In the second case, we get inversion of the configuration by converting the alcohol into a good leaving group, by conversion to the tosylate, followed by an SN2 reaction with sodium methoxide. Other methods for activation of the alcohol are acceptable, however acidic conditions will lead to racemization due to the competitive S N1 reaction. 21.33

21.34

21.35

21.36

21.37

21.38

(b): E1 reactions require substrates that can form a stabilized carbocation (compare with the S N1 reaction). Of the three examples given (a) would generate a primary carbocation (the least stable), (c) would generate a secondary carbocation (middle stability), and (b) would generate an allylic carbocation, which is stabilized by resonance and is the most stable.

21.39

(c). As the rate determining step for an E1 reaction is the generation of the carbocation, the ability of the leaving group is of key importance in determining the rate of an E1 reaction. The strong bonds in compounds a) and f) make them poor substrates. The negative charge on the oxygen in e) makes the oxygen a poor leaving group. In terms of the halogens, leaving group ability increases as you descend the group, making iodide the best leaving group.

21.40

The major product that would be formed is the more substituted double bond. This is called Zaitsev‘s rule. The more substituted double bond is the thermodynamic product.

21.41

(b). Steric hindrance is an important consideration in E2 reactions. Molecule (a) is highly sterically congested and is therefore an E2 type elimination would be difficult. In (c) the proton is more acidic due to the phenyl ring, but only one proton is available for deprotonation and therefore the rate of elimination would be slower than in (b) where deprotonation of any of the 6 available protons would lead to an E2 elimination. In addition, (b) is also a good E1 substrate due to the stability of the carbocation. Therefore low temperatures would help to favor an E2 type mechanism.

21.42

(c). The 2-methylpropan-2-olate (tert-butyoxide) is the strongest base. It will react more rapidly to deprotonate the system accelerating the rate of elimination.

21.43

The most basic is (b), the least basic is (d). Basicity is linked to the stability of the anion. Molecules (c) and (d) are stabilized by resonance making them the most stable. The base (d) has multiple heteroatoms to delocalize the charge, making it the most stable and therefore the least basic. Molecules (a) and (b) lack any resonance forms. As methyl groups release electron density through hyperconjugation, the anion in (b) is less stable than the one in (a).

21.44

(b). Basicity across different heteroatoms parallels the electronegativity of those atoms. As such, the more electronegative atoms are generally more able to stabilize negative charge, making them less basic. This is why oxygen is less basic than nitrogen. However, the size of the atom is also important. Larger atoms are more able to stabilize negative charge than small er ones making these larger atoms less basic. As such, iodide is much less basic than nitrogen.

21.45

A is locked in a conformation where the iodide is in the axial position and therefore antiperiplanar to the adjacent proton, the conformation required for elimination. In B, the iodide is in the equatorial position and therefore no proton is antiperiplanar making elimination more difficult.

21.46

The rate determining step (RDS) for an E1 reaction is the generation of a carbocation. A general reaction coordinate diagram is given below

Consider the following reaction schemes below for 21.47 and 21.48.

21.47

(a). In the reaction with the unhindered base (MeO –) the SN2 reaction will dominate. In (b), and (c), the hindered base favors elimination.

21.48

(c). The bulky bases favor elimination as stated in 21.47. However, elimination reactions are entropically favored over substitution reactions. As such, higher temperatures favor elimination.

21.49

The major product we would expect is the more hindered double bond as the only conformation which allows elimination (antiperiplanar) leads to this product. It is worth noting that often due to the higher temperatures that are required to favor elimination, significant equilibration to the less hindered diastereomer can occur. 21.50

21.51

The sterically hindered base will abstract the more accessible proton. 21.52

Markovnikov‘s Rule dictates the regioselectivity of addition across double bonds. Roughly speaking it is phrased as ―Them that has, gets‖ (the carbon with more hydrogen atoms gets the hydrogen).

21.53

Fill in each blank with a general formula for the type of compound formed: .

21.54

21.55

(d) CH=CH2 < (b) Me < (e) OCOR < (a) OH < (c) NH 2

21.56

B > C > A > D. Reactivity order is based on which ring is most electron rich (from the most activating effect to the most deactivating effect).

21.57

Give the products of the following reactions.

21.58

Comparing the product and reactant, one sees two hydrogen atoms added at the same face of the double bond. Therefore the reagent is H2 in the presence of a catalyst such as Pt/C.

21.59

To get the halogen attached to the carbon with more hydrogen atoms (anti-Markovnikov reaction), we would use HBr and a peroxide.

21.60

The reagents indicate the substitution of an H with a Br. Methyl groups are ortho/para directing.

21.61

21.62

21.63

21.64

21.65

(b). The double bond acts as a nucleophile, therefore the most electron rich double bond will react first.

21.66 \

21.67

21.68

Aromatic molecules behave differently than other compounds with C=C bonds. In general, aromatic rings are much less reactive (more stable) than compounds with isolated C=C double bonds because of their delocalized  electrons. The benzene ring can undergo electrophilic substitution, nucleophilic substitution, elimination, and addition reactions, while the alkenes mainly undergo addition reactions. There are reactions that alkenes undergo pretty easily, but benzene does not. For example, bromine, Br2 adds easily across alkene double bonds. Ethene (CH2=CH2) would add bromine to form dibromoethane, BrCH 2-CH2Br. As a result, the red-brown colour of bromine disappears instantly if it is added to an alkene, because the bromine is rapidly consumed. Benzene can be halogenated, but is more difficult. For example, the aromatic halogenation of benzene with bromine, chlorine, or iodine needs the help of a catalyst (an iron salt or aluminum trihalide).

21.69

Addition reactions do not occur readily with benzene due to resonance stability of the aromatic ring.

21.70

C) The reaction of the electrophile E+ with the arene is the slow step since it results in the loss of aromaticity.

21.71

The reagents indicated the substitution of an H with a nitro group. Hydroxyl groups are ortho/para directing. 21.72

The reagents indicate the substitution of an H with a Cl. The methyl group is ortho/para directing while the nitro group is meta directing. This results in two products where the Cl group is directed in the meta position relative to the nitro group as well as the ortho and para positions relative to the methyl group. 21.73

The aniline is a good ortho/para director and under weak acidic conditions (right hand arrow) dominates the course of the reaction. Under strongly acidic conditions, the aniline is protonated, making it a weak meta director, and the methyl group a nd protonated aniline give the alternative product (left hand arrow).

21.74

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. When water is added to a double bond, the hydrogen is the electron-poor group and the hydroxyl is the electronrich 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:

However, when water adds to a C=C, two products result since the OH can bond to either carbon in the double bond:

In this reaction, very little of the second product forms. 21.75 Silberberg, Amateis, Venkateswaran, Chen, Chemistry: The Molecular Nature of Matter and Change, 3rd Canadian Edition Page 5-807 Instructor‘s Solution Manual © Copyright 2021 McGraw-Hill Ryerson Ltd.

21.76

Alcohol: R–OH + H2O → R–O– + H3O+ Alcohols, like water, can show either acidic or basic properties at the –OH group. They are in general slightly weaker acids than water, but definitively less acidic than carboxylic acids. Carboxylic acid:

The resonance-stabilized carboxylate ion allows the transfer of the proton to water. The alkoxide ion cannot show any resonance stabilization. 21.77

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.

21.78

Alcohols undergo substitution at a saturated carbon while acids undergo substitution at the carboxyl carbon.

21.79

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 butan-2-one.

b) The product is 2-methylpropanoic acid.

c) The product is cyclopentanone.

21.80

a) 2-methyl-propan-1-ol

b) pentan-2-ol

c) 3-methylbutan-1-ol

21.81

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.

b) An alcohol and a carboxylic acid undergo dehydration-condensation to form an ester.

c) This reaction is ester formation through dehydration-condensation.

21.82

21.83

Plan: To break an ester apart, break the –C–O– single bond and add water (–H and –OH) as shown.

Solution: (a)

O CH3

(CH2)4

CH2

CH3

CH2

CH3

(b)

O C

CH3

(c)

O CH3

21.84

CH2

CH3

+ HO

CH2

CH3

N CH3

O H

OH +

N H

21.85

a) Substitution of Br– occurs by the stronger base, OH–. Then a substitution reaction between the alcohol and carboxylic acid produces an ester:

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 CH3 CH2 CH CH3 OH H3O+, H2O

CH3 21.86

CH2

CH3

H+, H2O CH3

CH2

CH3

CH2

CH3

OH _

Cr2O72 , H+ CH3

CH2

CH3

O b)

OH CH3

CH3

CH2

CH3

CH2

H2O CH3

CH2

CH3

CH2

CH3

21.87

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. 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

CH3

carboxylic acid portion from ester

portion from amine 21.88

a) KOH; Br2 The first step is an elimination reaction, while the second is an addition. b) K2Cr2O7, H; C6H5–CH2–OH; H+ Comparing the first reagent and the first product, oxidation takes place. For the second step, an ester is produced from the carboxylic acid, so an alcohol and acid must be added.

21.89

a) CH3

CH2

CH3

CH3 3-methylbutan-1-ol

pentan-1-ol CH3

CH3

CH2

CH3

CH2

CH3

CH3 2,2-dimethylpropan-1-ol

2-methylbutan-1-ol

OH CH3

CH2

pentan-2-ol

CH2

OH CH3

CH3

OH CH

CH3 3-methylbutan-2-ol

CH3

CH2

CH3

pentan-3-ol

OH CH3

CH2

CH3

CH3 2-methylbutan-2-ol

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. 21.90

First, oxidize the methanol to methanoic acid.

O CH3

Cr2O72 , H+

H C OH Then, produce the ester by reacting the methanoic acid with the ethanol. O O

21.91

OH + HO

CH2

CH3

H+ H

H2O

CH2

CH3

a) Symbolize the monoprotic acid as HA. Then, the balanced chemical equation will be: NaOH(aq) + HA(aq)  NaA(aq) + H2O(l)  1 mol NaOH    1 mL NaOH    1L NaOH 1 Molar mass (g / mol)   0.2003 g HA        3   1 mol HA  0.03811 mol NaOH   10 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

21.92

a) CH3CHO + C6H5–MgBr  C6H5CH(OH)CH3 b)

OH H2O CH3

CH2

CH3

+ CH3

CH3

MgBr

CH2

CH3

c) CH3MgBr and C6H5CHO d) HCHO (methanal, also known as formaldehyde) e) CH3

CH3

+ CH3

CH2

CH3

CH3 MgBr

MgBr

O 21.93

CH3

CH2

a)The functional group in ibuprofen is the carboxylic acid group COOH. The chiral center is RC*H(CH 3)COOH. CH3

CH3 C C

COOH

H CH2

H3C

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

+ NaCN

+ H 2O

CH3

+ NaBr

N C Then hydrolyze the nitrile with aqueous HCl to get the carboxylic acid.

CH3

+ HCl(aq)

CH C

21.94

CH3

COOH

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

b) The product is an ester, so a carboxylic acid is needed to prepare the ester. First, oxidize one mole of ethanol to ethanoic acid (acetic acid): Cr2O72 CH3

CH2

OH H+

CH3

Then, react one mole of ethanoic acid with a second mole of ethanol to form the ester:

21.95

When butan-2-one 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 in one direction by the other isomer cancels the light rotated to the other direction by one isomer. The mixture is not optically active since there is no net rotation of light. The two stereoisomers of butan-2-ol are shown below.

21.96

Plan: Using our knowledge of nucleophiles, leaving groups and the different type of reactions we can classify each type of reaction. Solution: a) This is a substitution reaction, where Y replaces X, which is attached to a carbon in the starting material. If X was Br and Y was I then this reaction would not proceed since I - is a weaker nucleophile than Br- or conversely is a better leaving group.

b) This is an elimination reaction, where the saturated reactant becomes an unsaturated product.

( c) This is an electrophilic addition reaction. Typically these reactions follow Markovnikov‘s rule (Section 21.5) which means the X- group should have gone to the secondary carbon. In order to perform an anti-Markovnikov addition, a peroxide reagent (ROOR) must be added to the reaction.

21.97

Plan: In order to answer these questions we must recognize which mechanisms are occurring in each case. The first reaction depicts an E2 elimination reaction which must pass through an anti-periplanar transition state. The second reaction is an electrophilic halogenation reaction which results in an anti-addition product. Finally the third reaction is an epoxidation reaction which should result in a syn- product. Solution: a) The 2-chlorobutane will adopt an anti-periplanar arrangement in which the methyl groups are in an antiarrangement (see Section 20.3 for a review). The minor product will occur when the two methyl groups are in a gauche configuration. Since this is a higher energy conformation it will be less likely to form.

b) Since the addition of Br2 can only generate the anti-product, there will be only one major product, which is shown below.

c) The epoxidation mechanism is concerted and results in a syn- product only. Furthermore, the depicted minor structure would not be physically possible since the oxygen could not exist both above and below the plane of the molecule as depicted by the stereo-bonds.

21.98

Plan: Using our knowledge of electrophilic aromatic substitutions (Section 21.6) we can predict that a phenol will be an ortho/para director for the incoming chloroethane and that AlCl 3 will be required to catalyze the reaction. Solution: Based solely on an alcohol being an ortho/para directing group we can predict that the following products will form:

The following mechanism shows that there are four resonance structures available for ortho/para substitutions and that some of the positive charge can be held on the oxygen atom.

In the case of the meta reaction pathway only 3 resonance structures are produced to stabilize the cation, which is why this product is the minor one.

21.99

Plan: Identifying the type of reaction will allow us to determine the common reagents used and the likely nature of the other molecules. The first reaction (a) produces an ester using an alcohol as one of the reagents, the second reaction (b) appears to be a nucleophilic substitution reaction and the final reaction (c) is an elimination reaction. Solution: a) As we studied in Section 21.7, esters can be formed by the reaction of carboxylic acids with alcohols in an acidic medium. Therefore if we have the final product (an ester) and the reagent (an alcohol), the starting material must be an acid. This reaction is catalyzed by acids.

b) This is a substitution reaction and since it is occurring at a 1 o carbon it must follow the SN2 mechanism. OH- is an ideal nucleophile for SN2 reactions making it a likely reagent.

c) This is an elimination reaction but it does not follow Zaitsev‘s rule (Section 21.4). In order to violate Zaitsev‘s rule a bulky base must be used so that protons are only removed from the least sterically hindered carbon.

21.100

Plan: To start this reaction sequence we must realize that the first step in the reaction is an electrophilic addition of a haloalkane. From this point we can then proceed with the rest of the reactions. Solution: a) In order to perform the first reaction we only need to add HBr which will form the Markovnikov addition product 2-bromo-2-methylbutane. The bromine is now at a tertiary position which means we will need to use an SN1 substitution pathway which would call for a polar protic solvent and a mild nucleophile, CH 3OH will fill this role in order to generate the final product.

b) To make 2-methylbut-1-ene from the tertiary alkyl halide it would require an E1 reaction mechanism. We can utilize the same solvent that we used for the S N1 substitution, but by applying heat we drive the reaction to the elimination products instead of the substitution products. Unfortunately, Zaitsev‘s rule predicts the formation of the most substituted alkene, so this would not be a good technique to generate 2-methyl-1-ene, which would be the minor product of this reaction.

21.101 Plan: Given that we know that the starting material is a 3 carbon haloalkane and that the one product we are provided has 3 carbon atoms in it we can work backwards (retrosynthesis) to figure out all the missing reagents Solution: 3-methyl-N-propylbutanamide is an amide and we know from section 21.7 that amides can be formed by the reaction of an amine and an acyl chloride. This allows us to predict the structure of the acyl chloride as being 3methylbutanoyl chloride, which means that the other reagent must be prop-1-amine. The SN2 substitution reaction of 1-bromopropane in liquid ammonia would generate the prop-1-amine

21.102 Plan: Since we know what the starting material is we can examine the reagents used in order to predict the products at each subsequent step. Solution: a) The first reaction is using a strong nucleophile in a polar aprotic solvent making it an S N2 reaction, which results in the substitution of the Br for an OH. Na2Cr2O7 and acid are used to oxidize alcohols to ketones or aldehydes. Since the product of the previous reaction is a secondary alcohol, it will be oxidized to a ketone. The final step involves a Grignard reagent (Section 21.7) which means that we can expect the –CH3 to attack the carbonyl group, while the water work-up in the second step will neutralize the alkoxide formed.

b) If the solvent was changed to CH3OH in the first step of the reaction then an E2 reaction would become more favored resulting in the formation of an alkene which would not react further.

21.103 Plan: As we know what the starting material and final products are we can work backwards (retrosynthesis) in order to identify all of the correct reagents and intermediates. Solution: This question can be challenging as we figure out where to start. A first analysis is to compare the number of carbon atoms in the starting material with those in the final product. From this we can see that there are three extra carbon atoms which must be introduced in the last step. We know that a Grignard reaction is an ideal way of

adding carbon chains to molecules and that a water work-up is the last step in a Grignard reaction. This allows us to predict that propanone (acetone) and the Grignard agent below were used. In order to generate the Grignard reagent Mg must be added to the bromoalkane produced in the first reaction. Going from the starting material (alkene) to the bromoalkane would require an anti-Markovnikov addition of HBr so we need to also add a peroxide reagent.

21.104 Plan: As we know what the starting material and final product are and all of the reagents in between, we need to figure out the product for each reaction, starting with the reaction of the bromoalkene with methyllithium. Solution: a) In the first step the methyl group replaces the bromine in a substitution reaction. The second reaction is an acid catalyzed electrophilic hydration of the alkene and generates the Markovnikov product. Reaction of p-tosyl chloride with an alcohol generates a p-toluenesulfonate which is a very good leaving group.

The polar protic solvent (CH3OH) and a good leaving group allows the reaction to follow an S N1 pathway as shown below, thus leading to the final product.

b) Doubling the concentration of CH3Li would double the reaction rate since it is an SN2 reaction. c) Halving the concentration of CH3OH in the final step would have no effect as it is a S N1 reaction. 21.105 Plan: In order to answer these questions we must review what we know about hydroboration reactions (Section 21.5) as well as all of the substitution and elimination reactions. Solution: a) The hydroboration process produces syn- products since the BH3 molecule adds to both carbon atoms of the alkene from one side of the molecule in a concerted fashion, and the oxidation step replaces BH 2 group with OH without altering the stereochemistry. The hydrogenation, like the hydroboration, is a concerted mechanism that adds two hydrogen atoms to one face of the alkene only. In contrast to this, the halogenation of alkenes yields the anti- product since the first step only adds one halogen to the alkene forming a 3-membered ring structure on one face of the molecule. The remaining negatively charged halogen ion then attacks the back-side of this ring structure producing an anti-configuration. b) SN1 and E1 reactions do not preserve stereochemistry since in the first step of the reaction a carbocation is generated which is sp2 hybridized and therefore planar. The subsequent step in the reaction then has a 50% chance of occurring on the top or the bottom face of the carbocation, thus losing stereo-control.

21.106

21.107

21.108

21.109

The two structures are: cadaverine

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 (NaBH 4 is a hydrogen-rich reducing agent): H H reduction

21.110

H2N

CH2

NH2

Plan: Refer to the Table of Functional Groups in Chapter 20. 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 jasmolin II:

alkene

CH2

2 CH

CH 1 CH3

CH3 O

3 CH

4 C

5 CH

C 6 C

CH3

C 11 CH2

O CH3

C 7 O

ester

alkene ester ketone Carbon 2 is sp3 hybridized. Carbon 4 is sp2 hybridized. Carbon 6 and 7 are sp2 hybridized.

b) Carbon 1 is sp hybridized. Carbon 3 is sp3 hybridized. Carbon 5 is sp3 hybridized.

c) Carbon atoms 2, 3, and 5 are chiral centers as they are each bonded to four different groups.

CHAPTER 6 SPECIAL TOPICS IN ORGANIC

CHEMISTRY CHEMICAL CONNECTIONS BOXED READING PROBLEMS B22.1

Plan: When a ddNTP is incorporated into the growing DNA chain, polymerization stops since no more phosphodiester 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: TACAGGTTCAGT ddCTP will give three complementary chain pieces: CA, CAAGTCA, CCAAGTCA

B22.2

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

-------

G T A T A C

END–OF–CHAPTER PROBLEMS 22.1

The glass transition temperature is the midpoint of the range of temperatures when a semicrystalline substance, like a polymer, changes from a molten or rubber-like substance into a hard brittle solid. The glass transition temperature for the polymer given is 6°C.

22.2

Dissolved polymers increase the viscosity of solutions. Therefore increasing the amount of polymer should lead to a more viscous solution.

22.3

Both branching and crosslinking are bifurcations in a linear polymer, where instead of extending in only one linear direction, now two or more chains exist. In cross-linked polymers, these branches connect two chains together, whereas in a normal branched polymer they do not connect chains together. In a branched polymer, this leads to more difficulty in packing and therefore leads to a flexible polymer with low crystallinity. In a crosslinked polymer, the addition covalent linkages between chains make the polymer more rigid and strong.

22.4

A thermoplastic polymer is one that melts at high temperature and reversibly forms a solid polymer at low temperature. A thermosetting plastic is a polymer which is a liquid or soft solid at lower temperature, but irreversible forms a solid at high temperature (also known as curing) to give a infusible, insoluble polymer (which is then known as the thermoset). In general, thermosetting polymers have a much higher number of cross-links than thermoplastic polymers affording them this rigidity.

22.5

A block polymer is a special type of copolymer consisting of a larger polymer made up from blocks of polymerized monomer units. Therefore the chain consists of a number of repeats of one monomer unit followed by a number of repeats of a second different monomer unit. An ABA block copolymer can be represented as: A-A-A-A-A-A-A-A-A-A-A-A-A-B-B-B-B-B-B-B-B-B-B-B-A-A-A-A-A-A-A-A-A-A-A-A

22. 6

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.

22. 7

A double bond is common in the monomers resulting in addition polymers. Substituents on the alkene make the monomers different.

22.8

A free radical is an atom or group of atoms with one unpaired electron. A free radical is highly reactive and is used to initiate a chain reaction.

22.9

Polyethene 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 polyethene (HDPE) are strong due to the large size of the polyethene chains. Low-density polyethene (LDPE) has increased branching that prevents packing and weakens intermolecular dispersion forces.

22.10

Condensation polymers are more similar chemically to biopolymers. Both are formed by the loss of water from two monomers.

22.11

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.

22.12

Water (H2O) is formed when two monomers are joined in condensation polymerization.

22.13

Plan: Both PVC and polypropene 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)

Cl n

CH3 n

22.14

b) H

22.15

A carboxylic acid and an alcohol react to form an ester bond in a dehydration-condensation reaction:

The displacement reaction is:

22.16

22.17

Plan: The molar mass of a polymer chain (M polymer, in g/mol) depends on the molar mass of the repeat unit (M repeat) and the degree of polymerization (n), the number of repeat units in the chain: Mpolymer = Mrepeat  n. (Use Table 22.3 to determine the repeat unit for the specified polymer). Solution: Polyethene is made up of ethene (ethylene) repeating units, whose molar mass is 28 g/mol. If an individual polyethene chain has a degree of polymerization of 6600, the molar mass of this chain is: M polymer = M repeat  n = (28 g/mol)  (6.6103) = 184800 g/mol = 1.8105 g/mol

22.18

Plan: The molar mass of a polymer chain (Mpolymer, in g/mol) depends on the molar mass of the repeat unit (Mrepeat) and the degree of polymerization (n), the number of repeat units in the chain: Mpolymer = Mrepeat  n. (Use Table 22.3 to determine the repeat unit for the specified polymer). Solution: Poly(vinyl chloride) is made up of chloroethene (vinyl chloride) repeating units, whose molar mass is 62.5 g/mol. If an individual poly(vinyl chloride) chain has a degree of polymerization of 8000, the molar mass of this chain is: Mpolymer = Mrepeat  n = (62.5 g/mol)  (8103) = 500000 g/mol= 5105 g/mol

22.19

Plan: The molar mass of a polymer chain (Mpolymer, in g/mol) depends on the molar mass of the repeat unit (Mrepeat) and the degree of polymerization (n), the number of repeat units in the chain: Mpolymer = Mrepeat  n. (Use Table 22.3 to determine the repeat unit for the specified polymer). Solution: polyphenylethene or polystyrene is made up of phenylethene (styrene) repeating units, whose molar mass is 104 g/mol. If an individual polyphenylethene chain has a degree of polymerization of 10000, the molar mass of this chain is: Mpolymer = Mrepeat  n = (104g/mol)  (1104) = 1040000 g/mol= 1106 g/mol

22.20

Plan: The number-average molar mass, Mn, is equal to the total mass of all chains divided by the amount (mol) of chains. Using this definition and the given number-average molar mass, one can calculate the total mass of 0.23 moles of polyphenylethene or polystyrene. Solution: 80000 g/mol = total mass of all chains multiplied by 0.23 mol Total mass of all chains = 18 400 g = 18.4 kg

22.21

Plan: The length of an extended backbone is simply the number of repeat units (degree of polymerization, n) times the length of each repeat unit (l0). Solution: The number of repeat units in question 22.1 is 6600 and each unit is approximately 250 pm, therefore, 250 pm  6600 = 1.65 x106 pm = 1.7 x106 pm

22.22

Plan: The length of an extended backbone is simply the number of repeat units (degree of polymerization, n) times the length of each repeat unit (l0). Solution: The number of repeat units in question 22.2 is 8000 and each unit is approximately 250 pm, therefore, 250 pm  8000 = 2 x106 pm

22.23

Plan: The length of an extended backbone is simply the number of repeat units (degree of polymerization, n) times the length of each repeat unit (l0). Solution: The number of repeat units in question 22.3 is 10000 and each unit is approximately 250 pm, therefore, 250 pm  10000 = 2.5 x106 pm

22.24

Plan: The size of the coiled chain is expressed by its radius of gyration (Rg), the average distance from the centre of mass of the molecule to the outer edge of the coil. The mathematical expression for the radius of gyration includes the length of each repeat unit and the degree of polymerization: Rg 

Solution: Rg  22.25

(6600)(250)2  8291.56 = 8292 pm 6

Rg  Solution: Rg  22.26

nlo 2 6

(8000)(250)2  9128.71 = 9129 pm 6

Rg  Solution: Rg 

22.27

nlo 2 6

(10000)(250) 2  10206.2 = 10210 pm 6

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  2.7 g oxygen     18.150879 g oxygen  = 14.8753 = 15 g MTBE/100. g gasoline Mass (g) of MTBE =

 14.8753 g MTBE   0.740 g gasoline   1 mL   1 mL MTBE   10 3 L        100. g gasoline   1 mL gasoline   103 L   0.740 g MTBE   1 mL   c) Volume (L) of MTBE = = 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 O 2  1.00 L MTBE   3    88.15 g MTBE  2 mol MTBE  mL  10 L     = 62.96086 mol O2 L•bar   62.96086 mol O2   0.08314  273  24 K 100 mol•K     V = nRT/p =  21  1.00 bar   = 7,40317177×103 = 7.4x103 L air



22.28



a) To prepare a polymer with a benzene containing backbone, phenylethene or styrene, C6H5–CH=CH2 can be used to produce polyphenylethene or polystyrene, – (–CH(C6H5) –CH2–)–n. b) p-diethenylbenzene can be used to crosslink the polymer.

22.29

22.30

22.31

Addition polymers contain a large number of relatively inert bonds (C-C, C-H) whereas condensation polymers contain a large number of ester or amide bonds. As ester and amide bonds are more susceptible to hydrolysis, we would predict condensation polymers are more biodegradable.

22.32

Isotatic has a regular structure making its solid form more regular and semicrystalline. As this makes the solid more stable it will have a higher melting point than the atatic variant.

22.33

Taking our example from earlier with PVC (question 22.29).

22.34

Polyamide condensation bonds should have a higher glass transition temperature. As amides have bond hydrogen bond donor and acceptor functionality (they have parts with lone pairs and other parts with hydrogen connected to a heteroatom) they are capable of making intermolecular hydrogen bonds. Esters on the other hand, lack the N-H (hydrogen connected to a heteroatom) and therefore are less able to make intermolecular hydrogen bonds. These hydrogen bonds stabilize the solid form and raise the glass transition temperature.

22.35

In carbohydrate chemistry, the anomeric carbon is the hemiacetal or hemiketal carbon (the carbon bonded to a hydroxyl and to an alkoxy group ( sugar,

). Therefore, on the following

22.36

a) Amino acids form condensation polymers, called proteins. b) Alkenes form addition polymers, the simplest of which is polyethene. c) Simple sugars form condensation polymers, called polysaccharides. d) Mononucleotides form condensation polymers, called nucleic acids.

22.37

A phosphate ester linkage joins the nucleotides in the DNA strands.

22.38

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 hydrogen bonding.

22.39

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.

22.40

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. 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) Minimum M= (4 × 75.07 g/mol) + (1 × 89.09 g/mol) + (3 × 117.15 g/mol) + (6 × 115.13 g/mol) + (7 × 105.10 g/mol) + (49 × 174.21 g/mol) = 10,700 g/mol Glycogen is a highly branched polymer. As the enzymes that digest glycogen remove the glucose subunits from the terminus of the chains, a branched structure allows for energy to be released faster than its more linear cellulose counterpart.

22.41

22. 42

Common sugars can exist in a variety of forms. A monosaccharide is a sugar composed of a single ring. Glucose and fructose are both monosaccharide. Monosaccharide can undergo a condensation reaction, releasing a molecule of water and a disaccharide. These can be the same monosaccharide, or two different monosaccharides as is the case with sucrose – a disaccharide made up from one molecules of glucose and one molecule of fructose.

22.43

Simple monosaccharides can be classified into two groups – ketoses and aldoses – based on their linear open chain forms. To determine whether a sugar is a ketose or aldose sugar, draw the sugar in its Fischer projection, and if the sugar has a ketone in the molecule it is a ketose – as is the case for fructose. If it has an aldehyde then it is an aldose – as is the case for glucose.

22.44

Nucleoside = sugar + nucleobase Nucleotide = sugar + nucleobase + phosphate

22.45

Plan: Locate the specific amino acids in Figure 22.8 (the common amino acids) Solution: a) alanine b) histidine c) methionine

CH3

HN NH

S CH2

CH2

22.46

CH2

Plan: Locate the specific amino acids in Figure 22.8 (the common amino acids) Solution: a) glycine b) isoleucine c) tyrosine OH CH3 H

CH2 CH

CH3

CH2

22.47

Plan: A tripeptide contains three amino acids and two peptide (amide) bonds. Find the structures of the amino acids in Figure 22.8 (the common amino acids). Join the three acids to give the tripeptide; water is produced. Solution: O O O

H2N

CH2

CH2 C

O N

OH HN

NH aspartic acid

H2N

histidine

CH2 C

tryptophan

O CH

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

22.48

H2N

CH2

CH3

CH2 CH2 NH2 b)

H3N

alanine

CH CH3

leucine O CH

CH2 CH

CH3

valine O

H N

CH3

O

CH3

CH3 22.49

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

22.50

Plan: Base A always pairs with Base T; Base C always pairs with Base G. Solution: a) Complementary DNA strand is CCAATG b) Complementary DNA strand is GGGCTT

22.51

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.

22.52

The RNA sequence is derived from the DNA template sequence CATAGTTACTTGAAC. There are five sets of three-base sequences, so there are five words in the sequence, and five amino acids are coded in the sequence.

22.53

RNA has OH groups at the 2‘ position of its sugar (ribose) whereas DNA does not (deoxyribose). The nucleophilic nature of this oxygen means it can attack intramolecularly, breaking up the phosphate backbone of the RNA chain.

22.54

There may be a disulfide linkage 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.

22.55

The nitrogen-containing bases form hydrogen bonds to their complimentary bases. The flat, N-containing bases stack above each other, which allow extensive interaction through dispersion forces. The exterior negatively charged sugar-phosphate chains form ion-dipole and hydrogen bonds to the aqueous surroundings, but this is of minor importance to the structure.

22.56

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.

22.57

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.

22.58

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.

22.59

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.

22.60

Amino acids with side chains that may be ionic are necessary. Two examples are lysine and glutamic acid.

22.61

-Pleated sheets are a type of secondary structural element.

22.62

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.

22.63

The amino acid sequence in a protein determines its shape and structure, which determine its function.

22.64

Plan: Refer to Figure 22.8 in the chapter. The types of forces operating in proteins were discussed in Section 22.4. 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. 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.

22.65

Plan: Knowing the molar mass of the protein and the average molar mass of the amino acids, we can calculate how many amino acids are on a single strand of DNA. Solution:  5x105 g   1 a min o acid  3     = 5x10 amino acids/DNA 100 g   mol DNA  

22. 66

The resonance structures show that the bond between carbon and nitrogen will have some double bond character that restricts rotation around the bond.

22.67

The key intermolecular hydrogen bonds responsible for recognition are shown in blue.

22.68

Covalent bonds (the amide linkages) restrict rotation creating a rigid backbone, polar and ionic interaction of the side chains hold the three dimensional structure together through electrostatic interactions and bonding with the hydrophobic shell, hydrogen bonds create rigidity stabilizing the secondary and tertiary structure of the protein, nonpolar side chains interact through dispersion forces within the non-aqueous protein interior, and the thiol residues within two cysteine side chains can form covalent disulfide bridges.

22.69

IR absorbs light which corresponds to the energy required to vibrate bonds, therefore it tells us how difficult it is for bonds to vibrate, the higher the wavenumber the more difficult it is for the bond to vibrate.

22.70

Due to hydrogen bonding, hydrogen atoms are constantly in flux – forming and breaking between different oxygen atoms at any time. This means that O-H bonds are particularly varied in their length and strength. As there is a wide variety in the strength of these bonds, it means they absorb at a variety of frequencies. The consequence of this is that they give a broad absorption in the IR.

22.71

As chlorine is a highly electronegative atom, it increases the positive charge on the carbon of the carbonyl compared to acetone. This in turn makes the carbonyl more polarized and increases the bond strength. As stronger bonds vibrate at larger wavenumbers, it increases the frequency of the absorption.

propan-2-one (acetone)

ethanoyl chloride(acetyl chloride)

22.72

As we are examining the carbonyl bond stretching frequency, the data suggests that the amide carbonyl has less double bond character than acetone. By looking at the resonance forms of an amide we can see that they have partial single bond character, explaining the lower shift.

22.73

NMR examines the difference in energy between the two spin states of atomic nuclei. In the case of 1H NMR we are examining the nuclei of a proton. We can infer the strength of the local magnetic field from the point at which the nuclei comes in to resonance. This we can usually interpret as the electron density surrounding the nucleus.

22.74

The standard reference is tetramethylsilane (TMS). It is a good reference because it is a) unreactive to most compounds and therefore will not be destroyed or altered, b) its soluble in most NMR solvents, c) its shift is in a region away from common NMR signals so it is difficult to misassign it in the spectrum.

22.75

In general, chromatography is used to separate compounds for either analytical or preparatory purposes. In general, it takes advantage of how different molecules interact with a solid phase (either through their size, polarity, or chirality). As some molecules move faster than others over the stationary phase depending on their physical characteristic, we should be able to separate them by carefully choosing elution conditions.

22.76

Plan: The number of peaks corresponds to the number of protons with different environments. Therefore, the first step will be to draw the structure of propan-2-one (acetone). Solution:

All six protons have identical environment, therefore we would expect to see one peak in addition to the TMS peak. 22.77

Plan: The number of peaks corresponds to the number of protons with different environments. Therefore, the first step will be to draw the structure of benzaldehyde. Solution:

The total number of sets of peaks would be four. The colours represent the protons with different environment (five types of 1H environments). 22.78

Plan: The number of peaks corresponds to the number of protons with different environments. Therefore, the first step will be to draw the structure of tetrachloromethane. Solution:

0. There are no protons! 22.79

Plan: The number of peaks corresponds to the number of protons with different environments. Therefore, the first step will be to draw the structure of DMF.. Solution:

Due to the restricted rotation around the amide bond, the two methyl groups of DMF are not equivalent. Therefore they come into resonance at different frequencies leading to three signals.

22.80

The methyl circled in red will have the highest shift because it is attached to an electronegative atom, making it electron deficient. Due to the lack of electron density it will appear at a higher ppm. These protons are the most deshielded ones.

22.81

O-H bonds are readily exchangeable in CDCl3. The protons attached to the hydroxyl groups are clearly seen. When the sample is shaken with D2O, the deuterium in the D2O can exchange with OH groups of the sample, converting them to OD groups. As deuterium atoms cannot be seen in 1H NMR, the OD groups can no longer been seen.

22.82

22.83

Plan: The number of peaks corresponds to the number of carbon atoms with different environments. Therefore, the first step will be to draw the structure of propan-2-one (acetone). Solution:

Since there are two types of carbon environments (in red and in blue in the figure above), we would expect to see two peaks. 22.84

Plan: The number of peaks corresponds to the number of carbon atoms with different environments. Therefore, the first step will be to draw the structure of propan-2-one (acetone). Solution:

Since there are five carbon types of C environments (each with a different colour in the figure above), we would expect five peaks.

22.85

Plan: After drawing the structure of the ethyl ethanoate (ethyl acetate), we identify the protons with different environments and their neighbouring spin-coupled nuclei. Solution:

There are 3 peaks (three types of 1H environments):. 1 high shifted peak (approx. 4 ppm), multiplicity is a quartet (the CH2) 1 high medium shifted peak (approx. 2 ppm), multiplicity is a single (the CH3 – next to carbonyl) 1 low shifted peak (approx. 1.2 ppm), multiplicity is a triplet (the CH 3 – next to CH2) 22.86

Plan: After drawing the structure of the N,N-diethylethanamine (triethylamine), we identify the protons with different environments and their neighbouring spin-coupled nuclei. Solution:

There are two peaks (two types of 1H environments) 1 high medium shifted peak (approx. 2.5 ppm), multiplicity is a quartet (the CH 2) 1 high low shifted peak (approx. 1.0 ppm), multiplicity is a triplet (the CH 3) 22.87

Plan: After drawing the structure of the 1,2-dichloroethane, we identify the protons with different environments and their neighbouring spin-coupled nuclei. Solution:

There is one resonance (only one type of 1H environment), appearing as a singlet (all protons are in the same environment and therefore do not couple) 22.88

B is diastereomeric to A and C. Therefore we would predict we can separate B from A and C. A and C however are enantiomers of each other and as our stationary phase (silicon dioxide) is achiral – we would predict that they are inseparable from each other.

22.89

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. 1.01 bar 1.00 L  pV Moles of A  n   = 0.028055851 mol RT  0.08314 L•bar     273 + 160  K mol•K   2.48 g A Molar mass of A  = 88.39510653 g/mol 0.028055851 mol Obtain the molecular formula of A from the combustion analysis:  2 mol H  Moles of H   0.409 g H 2 O     0.0454 mol H  18.02 g H 2 O 

 1.008 g H  Mass of H   0.0454 mol H     0.04576 g H  1 mol H 

 1 mol C  Moles of C  1.00 g CO2     0.0227 mol C  44.01 g CO 2 

 12.01 g C  Mass of C   0.0227 mol C     0.2726 g C  1 mol C   1 mol O  Moles of O   0.500 g A  (0.04576 + 0.2726) g O     0.01135 mol O  16.00 g O  0.0227 mol C 0.0454 mol H 0.01135 mol O C: 2 H: 4 O: 1 0.01135 mol 0.01135 mol 0.01135 mol

Empirical formula = C2H4O with a molar mass of 44.05 g/mol. Since the molar mass of Compound A is 88.40 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  Moles of COOH   33.9 mL         0.01695 mol COOH L    1 mL   1 mol NaOH  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

Compound A is not acidic so it does not have COOH groups. A is 4-hydroxybutanal. O HOCH2CH2CH2C

A b) Only one oxygen atom is added to Compound A to produce the carboxylic acid GHB, C 4H8O3. The aldehyde is oxidized to the acid 4-hydroxybutanoic acid:

O HOCH2CH2CH2C

GHB

22.90

Kevlar is formed from 1,4-phenylenediamine with terephthaloyl chloride.

This type of reaction would be consistent with chapter 21, where amides were described as being formed by acid chlorides and amines. 22.91

Nomex is formed from 1,3-phenylene-diamine with isophthaloyl chloride.

This type of reaction would be consistent with chapter 21, where amides were described as being formed by acid chlorides and amines. 22.92

Because the amides share a para relationship in Kevlar, there is more space around them than in Nomex. This makes it much easier to form hydrogen bonds between the chains. These hydrogen bonds are responsible for the increased rigidity, and the more ordered packing they encourage makes Kevlar more dense.

22.93

Plan: We can identify the polymer by using the information from the IR and the NMR. Figure 22.25 shows us that a peak at 2280 cm-1 corresponds to a nitrile group. Since the 1H NMR has two peaks we know that there are two different types of hydrogen atoms in the polymer backbone. The splitting pattern of a quintuplet means that one type of hydrogen is next to four other equivalent hydrogen atoms according to the n+1 rule (Section 22.5). A triplet indicates that the second type of hydrogen atoms is next to two other equivalent hydrogen atoms. The information about the molar mass and the degree of polymerization can be used to confirm our answer. Solution: From the NMR spectrum we can propose that the polymer is polyacrylonitrile. In order to aid our understanding two units of the polymer are shown below. The hydrogen atoms in green will always be next to the four red hydrogen atoms throughout the polymer backbone and will therefore appear as a quintuplet. Furthermore, the hydrogen atoms in green are next to an electron withdrawing group and so we would expect them to be shifted downfield from the red hydrogen atoms, this effect also corresponds with the peak at 2.5 ppm. This means that the peak at 1.9 ppm must correspond to the hydrogen atoms in red. Since the 1.9 ppm signal is a triplet, this makes sense as the red hydrogen atoms will always be next to two green hydrogen atoms in the polymer chain. From this data, we can determine that the monomer is acrylonitrile.

The nitrile groups (bonds in blue) have a strong stretching signal in the IR at 2280 cm -1. This type of polymerization is known as an addition polymerization. Using the molecular weight and the degree of polymerization data we can find that the Mrepeat is 53.06 g/mol which is the molecular weight of acrylonitrile which confirms our choice. Mpolymer = Mrepeat × n

77 096 g/mol = Mrepeat × 1453

Mrepeat = 53.06 g/mol

22.94

Plan: Using our understanding of DNA and proteins from Section 22.4 and how polymeric materials behave (Section 22.1) we can answer these questions. Solution: a) In order to produce a protein DNA must first be copied by mRNA in the nucleus, this process is known as transcription. The mRNA then proceeds out of the nucleus to a ribosome where it combines with tRNA molecules bearing complementary 3-base pair sequences; this begins the process of translation. The ribosome then shuttles along the mRNA strand matching each 3-base pair sequence to an incoming tRNA molecule all the while an enzyme catalyzes the formation of peptide bonds between each amino acid brought by the tRNA. b) Increasing the concentration of the protein in the solution will cause the viscosity of the solution to increase. c) There is only one type of covalent bond which forms in a protein secondary structure: a disulfide bond. In order to form disulfide bonds the protein must therefore contain at least two cysteine amino acids. d) Since we are only examining a single protein it is not possible to form a quaternary structure because this level of structure requires two or more proteins which then assemble into a larger structure.

22.95

Plan: Since we are given the DNA sequence we can form the complementary sequence knowing that A and T are complementary and G and C are also complementary and that RNA molecules do not contain U instead of T. Solution: a)The complementary sequence to GATCGACTA would be: CTAGCTGAT. The RNA sequence would be different since RNA contains U instead of T, therefore the sequence would be: CUAGCUGAU. b) It requires 3 bases in order to code for one amino acid, therefore this sequence could code for 3 amino acids. c) GC would not be a good technique to separate this nucleotide sequence from a mixture of other nucleotides since the molecular weight of the sequence and polarity mean that the nucleotide is not volatile enough. Liquid chromatography or better yet gel electrophoresis would be better separation techniques.

22.96

Plan: We are given the identities of both polymers so we can calculate their Mrepeat. The diameter of the polymer is related to the radius of gyration (Rg) which can be used to determine the degree of polymerization and then the molar mass of the polymer (Mpolymer). Solution: Mrepeat(isobutene) = 56.11 g/mol and Mrepeat(styrene) = 104.15 g.mol The diameter can be converted into radius: r = d/2 = 5.5 × 103 pm / 2 = 2.75 × 103 pm The radius of gyration (Rg) can now be used to solve for the degree of polymerization. Since we know that 60 mol% is isobutene and 40 mol% is styrene, we get the following formula:

Rg 

0.4nStyrenel0 2Styrene nl02 0.6nIsobutenel0 2Isobutene 0.6n( 240 pm )2 0.4n( 260 pm )2     6 6 6 6 6

2.75 x 103 pm = (75.89 pm)

n + (67.13 pm) n 2.75 x 10 pm = (143.03 pm) n 19.23 = n 3

n = 369.7 Since we now know the degree of polymerization and mol% of each monomer we can calculate the final Mpolymer = MIsobutene x nIsobutene + MStyrene x nStyrene

Polymer.

Mpolymer = (56.11 g/mol) × 0.6(369.7) + (104.15 g.mol) × 0.4(369.7) Mpolymer = 2.79 × 104 g/mol A thermoplastic elastomer is a material that can be shaped at high temperature and becomes an elastomer at room temperature. Silberberg, Amateis, Venkateswaran, Chen, Chemistry: The Molecular Nature of Matter and Change, 3rd Canadian Edition Page 6-843 Instructor‘s Solution Manual © Copyright 2021 McGraw-Hill Ryerson Ltd.

CH3

H CH3

H H There will be at least three distinct regions in the 1H NMR spectrum; the aromatic peaks of the styrene molecules will be the most downfield (dark blue). The backbone C-H bond attached to the styrene will be next (cyan) and finally the back-bone CH2 groups (red) and the isobutene CH3 groups of the isobutene will be the closest signals to 0 ppm. An answer of two regions would also be acceptable as the aromatic protons are quite distinct from the aliphatic CH, CH2 and CH3 protons.

22.97

Plan: Using our knowledge of polymer reactions we can draw the mechanism which will then show the type of reaction. We are given that the Tg of the material is 50 °C so we must then rationalize whether this material would be a good material for gloves knowing that it will be below the Tg in most applications (ie. room temperature = 25 °C). Solution: a) This type of polymerization is a condensation polymerization (shown below) in which the monomers lost HCl during the process.

b) Rubber gloves need to be flexible at room temperature, not hard and brittle, like glass, which is how this polymer becomes when it is cooled below its Tg (50 °C). Therefore, it will not be suitable for making rubber gloves. c) Since this is a polymer, Size Exclusion Chromatography (SEC) would likely be the best technique to analyze it, though it might also be possible to use liquid chromatography. 22.98

Plan: SEC separates polymers based on their size; therefore polymers which elute earlier have a higher molar mass than those which elute later. Since we have the diameters of the polymers from the light scattering measurements, we can use the radius of gyration (Rg) to determine the molar mass of each polymer. Solution: For the polymer which has a diameter of 16.2 nm we can find the radius: r = d/2 = 1.62 ×104 pm / 2 = 103 pm

nl02 Rg  6

→

8.1 x 10 pm =

8.1 ×

n(250 pm) 2 6

8.1 × 103 pm = (102.06 pm) n → n = 79.36 n = 6.30 × 103 We can calculate the molar mass of this polymer (Mpolymer) knowing that the molar mass of the repeat unit (MRepeat) is 28.05 g/mol Mpolymer = Repeat × n Mpolymer = 28.05 g/mol × (6.30 × 103) Mpolymer = 1.77 × 105 g/mol For the polymer which has a diameter of 5.3 nm we can find the radius: r = d/2 = 5.3 × 103 pm / 2 = 2.65 × 103 pm

nl02 Rg  6

→

2.65 × 10 pm =

n(250 pm) 2 6

2.65 × 10 pm = 102.06 pm n → n = 25.97 n = 674.2 As we did before for the other polymer, we can calculate the molar mass of this polymer (Mpolymer) knowing that the molar mass of the repeat unit (MRepeat) is 28.05 g/mol Mpolymer = Repeat × n Mpolymer = 28.05 g/mol × (674.2) Mpolymer = 1.9 × 104 g/mol Since we know that the polymer with higher molar mass elutes faster from SEC, then the peak at 6 minutes must correspond to the 16.2 nm polymer, which had a molar mass of 1.77 × 105 g/mol. The peak at 9 minutes can be assigned to the 5.3 nm polymer which has a molar mass of 1.9 × 104 g/mol.

CHAPTER 23 THE ELEMENTS IN NATURE AND INDUSTRY CHEMICAL CONNECTIONS BOXED READING PROBLEMS B23.1

a) Adding both reactions we get O3 + O  2O2 Each step is considered an elementary step. Therefore for the first step, rate = k[X][O3] and for the second step, rate = k[X][O] b) X acts as a catalyst, and XO acts as an intermediate. X is a catalyst because it is not consumed in the reaction, while XO is an intermediate because it is formed and then consumed in the reaction.

B23.2

a) (1) NO + O3  XO + O2 [slow] (2) XO + O  X + O2 [fast] b) The rate-determining step is the slow step. The rate law for that is: rate = k[NO][O3] = (6x10-15 cm3/molecule•s)(5x1012 molecule/cm3)(1.0x109 molecule/cm3) = 3x107 molecule/cm3•s

END–OF–CHAPTER PROBLEMS 23.1

Hydrogen is abundant in the universe since its simple atoms were the first ―created‖ after the ―Big Bang‖, and only a few percent of the original hydrogen in the Universe has been burned to heavier elements. H2 has the lowest density of any material, so it is only weakly held by the gravitational field of Earth.

23.2

Iron forms iron(III) oxide, Fe2O3 (commonly known as hematite). Calcium forms calcium carbonate, CaCO 3, (commonly known as limestone). Sodium is commonly found in sodium chloride (halite), NaCl. Zinc is commonly found in zinc sulfide (sphalerite), ZnS.

23.3

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.

23.4

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.

23.5

a) Bauxite (impure Al(OH)3) e) Seawater (NaCl)

23.6

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.

23.7

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.

23.8

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 O 2 and H2O.

23.9

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 CO 2 into carbohydrates during photosynthesis. Fixation of nitrogen gas by nitrogen-fixing bacteria involves the conversion of N2 to ammonia and ammonium ions.

23.10

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.

23.11

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.

23.12

Atmospheric nitrogen is utilized by three fixation pathways: atmospheric, industrial, and biological. Atmospheric fixation requires high-temperature reactions (e.g., lightning) to convert N 2 into NO and other oxidized species. Industrial fixation involves mainly the formation of ammonia, NH 3, 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.

23.13

Because N–N single bonds are weak, cyclic and chain compounds containing these bonds are not very stable.

23.14

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 PO 43– 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.

23.15

a) Atmospheric fixation requires energy from lightning or fires; industrial fixation requires high temperature to make the process proceed at a reasonable rate.

b) Air (N2)

c) Seawater (NaCl)

d) Limestone (CaCO3)

b) NO and NO2 both undergo atmospheric fixation while NH3 undergoes industrial fixation.  f H for NO and NO2 are both endothermic, so their formation requires energy. In contrast,  f H 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. 23.16

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–

23.17

Plan: First determine the moles of F present in 100. kg of fluorapatite, using the molar mass of Ca 5(PO4)3F (504.31 g/mol); take 15% of this amount. Convert mol F to mol SiF 4, 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.

oxidation oxidation oxidation reduction oxidation

Solution: a) Moles of F =

 103 g Ca 5 (PO4 )3 F   1 mol Ca 5 (PO4 )3 F    15%  1 mol F       1 kg Ca 5 (PO4 )3 F   504.31 g Ca 5 (PO4 )3 F  1 mol Ca 5 (PO4 )3 F   100%  = 29.74361 mol F  1 mol SiF4  Moles of SiF4 =  29.74361 mol F   = 7.43590 mol SiF4  4 mol F  pV = nRT L•kPa   7.43590 mol SiF4   8.31446    273  1450. K nRT mol•K   V (L) = = = 1051.58267 = 1.1x103 L p 101.3 kPa  

100. kg Ca 5 (PO4 )3 F 

 1 mol Na 2SiF6  b) Moles of Na2SiF6 =  7.43590 mol SiF4    = 3.71795 mol Na2SiF6  2 mol SiF4 

 6 mol F  19.00 g F  Mass (g) of F– =  3.71795 mol Na 2SiF6   = 423.8463 g= 4.2x102 g F–  1 mol Na SiF   1 mol F  2 6    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



Volume (m ) = 423.8463 g F

23.18





 106 g H 2 O   mL H 2 O   1 cm3   102 m  2 3       = 423.8463 m³ = 4.2x10 m    1.00 g H O 1 mL 1 cm 1 g F 2        

Plan: Since the 50. t (metric tonne) 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 Ca 3(PO4)2 to moles and then mass of P4, remembering that the production of P 4 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 tonne (t) = 1x103 kg.

3 3  98%   10 kg   10 g   1 mol Ca 3 (PO4 )2  Moles of Ca3(PO4)2 =  50. t        = 157972.79 mol Ca3(PO4)2   100%   1 t   1 kg   310.18 g Ca 3 (PO 4 ) 2 

Reaction: 2Ca3(PO4)2(s) + 10C(s) + 6SiO2(s)  6CaSiO3(s) + 10CO(g) + P4(s)

(section 23.3)

Mass (metric tonnes) of P4 =

 123.88 g P4   1 kg  1 t   90.%  1 mol P4    3    3    2 mol Ca 3 (PO4 )2   1 mol P4   10 g  10 kg   100%  

157972.79 mol Ca 3 (PO4 )2  

= 8.80635 t= 8.8 t P4 23.19

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.

23.20

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.

23.21

Factors considered include cost, reducing strength needed, and by-products formed.

23.22

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) It is similar to a) except that h ydration enthalpies are not involved. An example is:   CaCl (l) + 2Rb(g) Ca(l) + 2RbCl(l)  2

23.23

In general, nonmetals are obtained by oxidation and metals are obtained by reduction.

23.24

a) 6

23.25

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.

23.26

a) Pyrometallurgy uses heat to obtain the metal, electrometallurgy uses electrical energy, and hydrometallurgy uses reactions in aqueous solution. b) (i) Fe is produced by a high-temperature process using C as the reducing agent (ii) Na is produced (with Cl2 and H2) electrochemically in a Downs cell. (iii) Au is produced in a hydrometallurgical process using CN– as a complexing agent.

b) 4

c) 2

d) 1

(iv) Al is produced electrochemically in the Hall-Heroult process after a hydrometallurgical step to dissolve the ore. 23.27

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 Cu 2+ ions are reduced at the cathode to form Cu(s).

23.28

Molten cryolite is a good solvent for Al2O3, which creates a mixture with a lower the melting point.

23.29

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.

23.30

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.

23.31

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 H 2O 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 two electrons (Ca0  Ca2+ + 2e–) and each Rb+ gains an electron (2Rb+ + 2e–  2Rb0). The reaction is written as follows: 2RbX + Ca  CaX2 + 2Rb where H = 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 Ca0 could reduce Rb+. Based on values of IE and a positive H for the forward reaction, it seems more reasonable that Rb0 would reduce Ca2+. Please note that there is a considerable uncertainty in this argument as the ionization energies are from the atoms ionizing in their gas phase. 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, in accordance with Le Chatelier‘s principle, this will lead to a shift in the equilibrium to form more Rb product. If the reaction is carried out between 688°C and 1484°C (bpt 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 H = 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.

23.32

Plan: Write the balanced equation for the process. For every two moles of Na metal produced, one mole of Cl 2 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 coulomb with the conversion factor 96,485 C = 1 mole of e–. Convert electric charge 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 Cl2  Moles of Cl2 =  31.0 kg Na   = 674.2062 mol Cl2  1 kg   22.99 g Na  2 mol Na      pV = nRT L•kPa   674.2062 mol Cl2   8.31446    273  540. K  nRT mol•K   V (L) = = = 4.49892×104 L= 4.5x104 L p 101.3 kPa  b) Two moles of electrons are passed through the cell for each mole of Cl 2 produced: 2Cl–(l)  Cl2(g) + 2e–  2 mol e    96, 485 C  Electric charge =  674.2062 mol Cl2   = 1.30101570×108 C= 1.30×108 C  1 mol Cl   1 mol e   2    A•s   1  6 6 c) Time (s) = 1.30101570x108 C    77.0 A  = 1.689631x10 s= 1.69×10 s 1 C   





23.33

23.34

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)   CaO(s) + CO (g) a) CaCO (s)  3

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.   CaSiO (l) c) CaO(s) + SiO (s)  2

23.35

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 overpotential. b) The  f H for MgCl2(s) is –641.6 kJ/mol: Mg(s) + Cl2(g)  MgCl2(s) + heat. High temperature favors the reverse reaction (which is endothermic), so high temperatures favor the formation of magnesium metal and chlorine gas.

23.36

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  106 g   1 mol NaNO3   0.78 mol% NaIO3   1 mol I2   253.8 g I2  Mass (g) of I2 = 1000. t    l t   85.00 g NaNO    99.22  mol% NaNO  2 mol IO    1 mol I  3  3  2    3  4 4 = 1.173648677 ×10 g = 1.2×10 g I2

23.37

a) Sulfur dioxide is the reducing agent and is oxidized to the +6 state (as sulfate ion, SO 42–). 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)

23.38

The only halogen capable of oxidizing Cl– is F2, which is expensive and dangerous to handle. F2 also reacts with water, wasting the reagent.

23.39

3K2SiF6(s) + 4Al(s)  6KF(s) + 3Si(s) + 4AlF3(s)

23.40

a) Mass % Fe = b) Mass % Fe = c) Mass % Fe =

2  55.85 g/mol  159.70 g/mol 3  55.85 g/mol  231.55 g/mol 1 55.85 g/mol  119.99 g/mol

100%  = 69.94364 %= 69.94% Fe 100%  = 72.36018 %= 72.36% Fe

100%  = 46.545545 %= 46.55% Fe

23.41

4P(s) + 5O2(g)  P4O10(s) P4O10(s) + 6CaO(s)  2Ca3(PO4)2(s)

23.42

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.

23.43

Al2O3 is amphoteric and will dissolve in base: Al2O3(s) + 2NaOH(aq) + 3H 2O(l)  2NaAl(OH)4(aq) Fe2O3 and TiO2 have no acidic properties, so do not dissolve and can be removed.

23.44

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 Al 2O3 compound formed. From this, find the number of electrons produced and multiply by Faraday‘s constant to find the electric charge (coulomb). 3

 1 cm   3.97 g Al 2O3   1 mol Al 2O3  Moles of Al2O3 = 2.5 m 2 23x106 m  2     101.96 g Al O  cm3  2 3  10 m   = 2.2388682 mol Al2O3  6 mol e    96, 485 C  Electric charge (C) =  2.2388682 mol Al2O3   = 1.296103×106 C= 1.3×106 C  1 mol Al O   1 mol e   2 3  b) Current in amps can be found by dividing charge in coulombs by the time in seconds. [(1.296103×106 C)/[18 min (60 s/min)] × (1 A/(C/s)) = 1200 A= 1.2×103 A







23.45

a) Work = G° = –zFE° Work = – (2 mol e–)(96,485 C/mol e–)(1.24 V)[(J/C)/V] = –2.392828×105 J= –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 

23.46

a) H 2 and CO2 b) The zeolite acts as a molecular filter to remove molecules larger than H 2.

23.47

Plan: Free energy, G , is the measure of the ability of a reaction to proceed spontaneously. G° can be calculated with the relationship m f (products) G – n  f (reactants) G . Solution: The direct reduction of ZnS follows the reaction: 2ZnS(s) + C(graphite)  2Zn(s) + CS2(g)

 r G = 2(  f G of Zn) + (  f G of CS2)]

– [2(  f G of ZnS) + (  f G of C)]  r G = [2(0 kJ/mol) + (66.9 kJ/mol)] – [2(–198 kJ/mol) + (0 kJ/mol)]  r G = 462.9kJ/mol = +463 kJ/mol

Since  r G 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)  r G = [2(  f G of Zn) + (  f G of CO2)]

– [2(  f G of ZnO) + (  f G of C)]  r G = [2(0 kJ/mol) + (–394.4 kJ/mol)] – [2(–318.2 kJ/mol) + (0 kJ/mol)]  r G = +242.0 kJ /mol This reaction is also not spontaneous, but the oxide reaction is less unfavorable.

23.48

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 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.

23.49

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.

23.50

All the steps in its production are exothermic, and the excess heat (in the form of steam) can be sold profitably.

23.51

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.

23.52

Plan: G° at 25°C can be calculated with the relationship m f (products) G – n  f (reactants) G . Calculate G° at 500.°C with the relationship  r G = r H –  r S . r H and  r S must be calculated first. The equilibrium constant K is calculated by using G° = –RT ln K. Finally, to find the temperature at which the reaction becomes spontaneous, use  r G = 0 =  rxn H –  r S and solve for temperature. Solution: a) The balanced equation for the oxidation of SO2 to SO3 is 2SO2(g) + O2(g)  2SO3(g).  r G = m f (products) G – n  f (reactants) G  r G = [2(  f G of SO3)] – [2(  f G of SO2) + (  f G of O2)]  r G = [2(–371 kJ/mol)] – [2(–300.2 kJ/mol) + (0 kJ/mol)]  r G = –141.6 kJ/mol= –142 kJ/mol

Since  r G 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 SO 3 at room temperature. c) r H = m  f (products) H – n  f (reactants) H

r H = [2(  f H of SO3)] – [2(  f H of SO2) + (  f H of O2)] r H = [2(–396 kJ/mol)] – [2(–296.8 kJ/mol) + (0 kJ/mol)] r H = –198.4 kJ/mol= –198 kJ/mol  r S = m Sproducts – n S reactants  r S = [2( S of SO3)] – [2( S of SO2) + ( S of O2)]  r S = [2(256.66 J/mol•K)] – [2(248.1 J/mol•K) + (205.0 J/mol•K)]  r S = –187.88 J/K•mol = –187.9 J/K•mol 500o CG = r H – T r S = –198.4 kJ•mol– ((273 + 500.) K)(–187.88 J/K•mol)(1 kJ/103 J)

= –53.16876 kJ/mol= –53 kJ/mol 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 G° = –RT ln K.  103 J   142 kJ    1 kJ  G  At 25°C: ln K = =– = 57.31417694  RT (8.314 J/mol•K) ((273 + 25)K)

K 25o C = e57.31417694 = 7.784501x1024 = 7.8x1024  103 J   G  1 kJ  At 500°C: ln K = =– = 8.246817  RT (8.314 J/mol•K) ((273 + 500)K)

 53 kJ  

K500o C = e8.246817 = 3.815x103 = 3.8x103 e) Temperature below which the reaction is spontaneous at standard state can be calculated by setting G° equal to zero and using the enthalpy and entropy values at 25°C to calculate the temperature.  r G = 0 = r H – T r S

r H = T r S T =

23.53

H S

–198 kJ = 1.05375x103 K= 1.05x103 K or 780C  1 kJ  –187.9 J/K  3   10 J 

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 = 9.6485x104 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:  C   1 mol e  1 mol Cl2   70.90 g Cl2   1 kg  3600 s  Mass (kg) of Cl2 = 3x104 A  s         8 h    A  1 h  96, 485 C   2 mol e   1 mol Cl2   103 g     = 317.46 kg = 3x102 kg Cl2





23.54

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) 1 Cl2 : 2 OH – is the mole ratio in both cases.

23.55

2SO2(g) + O2(g)  2SO3(g)

p  a) K = 0.0314 kPa = p  p  2

SO3

-1

pSO2  pSO3 (1:1 mole ratio)

SO2

0.0314 kPa-1=

p  O2

pO2 = 31.85534 kPa = 31.9 kPa O2

 pSO  =  952 b) K = 0.0314 kPa = 2 2  pSO   pO   5  PO  2

-1

pO2 = 11496.82 kPa = 1×10 kPa O2 23.56

Plan: Write the balanced equation. Use the stoichiometry in the equation to find the moles of H 3PO4 produced by the reaction of the given amount of P 4O10. Divide moles of H3PO4 by the volume to obtain its concentration (mol/L). Since H3PO4 is a weak acid, use its equilibrium expression to find the [H 3O+] and then pH. Solution: a) Because P4O10 is a drying agent, water is incorporated in the formation of phosphoric acid, H 3PO4. P4O10(s) + 6H2O(l)  4H3PO4(l)  1 mol P4 O10   4 mol H3 PO4  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 H3 PO 4  Concentration of H3PO4 =   = 0.1596919 mol/L H3PO4 0.750 L   Phosphoric acid is a weak acid and only partly dissociates to form H 3O+ 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)

 H 3O    H 2 PO 4     Ka = 7.2x10 =  H 3 PO 4  –3

Ka = 7.2x10–3 =

 x  x 

 0.1596919  x   x  x  Ka = 7.2x10–3 =  0.1596919 

Assume x is small compared to 0.1596919.

x = 0.0339084 mol/L 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 = (7.2x10–3)(0.1596919 – x) = 1.14978×10–3 – 7.2×10–3x x2 + 7.2x10–3x – 1.14978x10–3 = 0

a=1

b = 7.2x10–3

c = –1.14978×10–3

7.2x10   4 1  1.14978x10  3 2

7.2x103  x=

3

2 1

–2

x = 3.049897x10 mol/L H3O+ pH = –log [H3O+] = –log (3.049897×10–2) = 1.51571 = 1.52 23.57

a) D+(aq), OD–(aq) and traces of CH 3O–(aq) and H +(aq). Eventually, HD O, 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)

23.58

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 CO 2 produced. Solution: a)The reaction taking place in the blast furnace to produce iron is: Fe2O3(s) + 3CO(g)  2Fe(s) + 3CO2(g) Mass (g) of CO2 = (

)( 9

)(

)

= 9.006881×10 g= 9.007×10 g CO2 b) Combustion reaction for octane: 2C8H18(l) + 25O2(g)  16CO2(g) + 18H2O(l) Mass (g) of CO2 =  19 L   1 mL   0.74 g   1 mol C8 H18  16 mol CO2  44.01 g CO2  1.0x106 autos       3     1 auto   10 L   mL   114.22 g C8 H18  2 mol C8 H18   1 mol CO 2  = 4.33396×1010 g = 4.3×1010 g CO2 The CO2 production by automobiles is much greater than that from steel production.





23.59

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 CH 2 =CHCl  e)  0.45 mol Cl2     = 28.1205 g= 28 g CH2=CHCl 1 mol Cl2    1 mol CH 2 =CHCl 

23.60

Plan: Step 2 of ―Isolation of Magnesium‖ in Section 23.4 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 [Mg 2+] that precipitated. Solution: a) Mg(OH)2(s)  Mg2+(aq) + 2OH–(aq) Ksp = 6.3×10–10 = [Mg2+][OH –]2 [OH –] =

6.3x1010 = [Mg2 ]

6.3x1010 = 1.100699×10–4 mol/L= 1.1×10–4 mol/L 0.052

Thus, if [OH –] > 1.1×10–4 mol/L (i.e., if pH > 10.04), Mg(OH)2 will precipitate. b) Ca(OH)2(s)  Ca2+(aq) + 2OH–(aq) Ksp = 6.5x10–6 = [Ca2+][OH –]2 = (x)(2x)2 = 4x3 x = 0.011756673; [OH –] = 2x = 0.023513347 mol/L Ksp (Mg(OH)2) = 6.3x10–10 = [Mg2+][OH –]2

6.3x1010

= 1.1394930×10–6 mol/L 2 0.023513347   This concentration is the amount of the original 0.052 mol/L 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 mol/L)/(0.052 mol/L) = 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.) [Mg2+] =

23.61

Plan: The equation  r G = r H – T r S will be used to calculate  r G at the two temperatures. r H and  r S will have to be calculated first. Then use G° = –RT ln K to determine K. Subscripts indicate the temperature, and (1) or (2) indicate the reaction. Solution: For the first reaction (1):

r H = m  f (products) H – n  f (reactants) H r H = [4(  f H of NO) + 6(  f H of H2O)] – [4(  f H of NH3) + 5(  f H of O2)]

r H = [4(90.29 kJ/mol) + 6(–241.826 kJ/mol)] – [4(–45.9 kJ/mol) + 5(0 kJ/mol)] r H = –906.196 kJ /mol  r S = [4( S of NO) + 6( S of H2O)]

– [4( S of NH3) + 5( S of O2)]  r S = [4(210.65 J/K•mol) + 6(188.72 J/K•mol)] – [4(193 J/K•mol) + 5(205.0 J/K•mol)](1 kJ/10 3 J)  r S = 0.17792 kJ/K •mol  r G = r H – T r S  25o C G = –906.196 kJ/mol – [(273 + 25)K]( 0.17792 kJ/K•mol) = –959.216 kJ 900o CG = –906.196 kJ/mol – [(273 + 900.)K]( 0.17792 kJ/K•mol) = –1114.896 kJ/mol

For the second reaction (2):

r H = [2(  f H of N2) + 6(  f H of H2O)] – [4(  f H of NH3) + 3(  f H of O2)]

r H = [2(0 kJ/mol) + 6(–241.826 kJ/mol)] – [4(–45.9 kJ/mol) + 3(0 kJ/mol)] r H = –1267.356 kJ /mol  r S = [2( S of N2) + 6( S of H2O)]

– [4( S of NH3) + 3( S of O2)]  r S = [2(191.5 J/K•mol) + 6(188.72 J/K•mol)] – [4(193 J/K•mol) + 3(205.0 J/K•mol)](1 kJ/103 J)  r S = 0.12832 J/K•mol Silberberg, Amateis, Venkateswaran, Chen, Chemistry: The Molecular Nature of Matter and Change, 3rd Canadian Edition Page 6-857 Instructor‘s Solution Manual © Copyright 2021 McGraw-Hill Ryerson Ltd.

 r G = r H –  r S  25o C G = –1267.356 kJ/mol – [(273 + 25)K](0.12832 kJ/K•mol) = –1305.595 kJ/mol 900o CG = –1267.356 kJ/mol – [(273 + 900.)K](0.12832 kJ/K•mol) = –1417.875 kJ/mol

a) ln K 25C 1 

G  RT

= 

 103 J   959.216 kJ/mol   = 387.15977 8.314 J/mol•K    273  25 K   1 kJ 

K 25C 1 = 1.38×10168 = 1×10168 ln K 25C  2  

 103 J   1305.595 kJ/mol G =    = 526.9655  RT 8.314 J/mol•K    273  25 K   1 kJ 

K 25C  2  = 7.2146×10228 = 7×10228

b) ln K900C 1 

 103 J   1114.896 kJ/mol G =    = 114.3210817  RT 8.314 J/mol•K    273  900 K   1 kJ 

K 900C 1 = 4.4567×1049 = 4.5×1049

ln K900C  2  

 103 J   1417.875 kJ/mol G =    = 145.388  RT 8.314 J/mol•K    273  900 K   1 kJ 

K900C  2  = 1.38422×1063 = 1.4×1063

 175 mg Pt   10 3 g  1 kg   32.15 oz t  $1557  c) Cost ($) Pt = 6.6x10 6 t HNO 3         3     1 oz t   1 t HNO 3   1 mg   10 g   1 kg = $5.781647 ×107= $5.8×107  175 mg Pt   72%   103 g  1 kg   32.15 oz t   $1557  d) Cost ($) Pt = 6.6x106 t HNO3          3    1 t HNO3   100%   1 mg   10 g   1 kg   1 oz t  = $4.1627856×107 = $4.2×107 23.62









2FeTiO3(s) + 7Cl2(g) + 6C(s)  2TiCl4(g) + 2FeCl3(l) + 6CO(g) TiCl4(g) + 2Mg(l)  Ti(s) + 2MgCl2(l)

Balanced reactions: Mass Ti = (

)( 6

)(

)

= 5.300092×10 g= 5.3×10 g Ti 23.63

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 G° to determine if this reaction is thermodynamically possible.  r G = m f (products) G – n  f (reactants) G  r G = [(  f G of S8) + 16(  f G of HCl)]

– [8(  f G of H2S) + 8(  f G of Cl2)]  r G = [(49.1 kJ/mol) + 16(–95.30 kJ/mol)] – [8(–33 kJ/mol) + 8(0 kJ/mol)]

23.64

 r G = –1211.7 kJ/mol= –1212 kJ/mol 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. Plan: Balance each reaction. Acidity is a measure of the concentration of H 3O+ (H+), so any reaction that produces H+ will increase the acidity. Solution: a) 2H2O(l) + 2FeS2(s) + 7O2(g)  2Fe2+(aq) + 4SO42–(aq) + 4H+(aq) Increases acidity. b) 4H+(aq) + 4Fe2+(aq) + O2(g)  4Fe3+(aq) + 2H2O(l) c) Fe3+(aq) + 3H2O(l)  Fe(OH)3(s) + 3H+(aq) Increases acidity. d) 8H2O(l) + FeS2(s) + 14Fe3+(aq)  15Fe2+(aq) + 2SO42–(aq) + 16H+(aq) Increases acidity.

23.65

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    = 7.86171 g/cm³= 7.86 g/cm 3  3  100% cm cm      Austenite:  7.40 g   7.40 g  2.08%  3     = 7.55392 g/cm³= 7.55 g/cm  cm3   cm3  100% 

23.66

 r G = m f (products) G – n  f (reactants) G

(1) N2(g) + 2O2(g) → 2NO2(g)  r G = [2(51 kJ/mol)] – [(0 kJ/mol) + 2(0 kJ/mol)] = 102 kJ/mol (unfavorable) (2) 3NO2(g) + H2O(l) → 2HNO3(aq) + NO(g)  r G = [2(–110.5 kJ/mol) + (86.60 kJ/mol)] – [3(51 kJ/mol) + (–237.192 kJ/mol)]  r G = –50.208 kJ/mol= –50. kJ/mol (favorable) (3) 2NO(g) + O2(g) → 2NO2(g)  r G = [2(51 kJ/mol)] – [2(86.60 kJ/mol) + (0 kJ/mol)] = –71.2 kJ/mol= –71 kJ/mol (favorable) (overall) 3N2(g) + 6O2(g) + 2H2O(l) → 4HNO3(aq) + 2NO(g)  r G = [4(–110.5 kJ/mol) + 2(86.60 kJ/mol)] – [3(0 kJ/mol) + 6(0 kJ/mol) + 2(–237.192 kJ/mol)]  r G = 205.584 kJ/mol= 205.6 kJ/mol Overall, the reaction is thermodynamically unfavorable, due to the unfavorability of the first step.

23.67

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. Since 1/2 of the Na reacts with H 2O, the maximum efficiency is 1/2 mol Na/mol e–, or 50%.

23.68

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 mol/L E° = E°O2 – E°Au = 0.40 V – 1.68 V = –1.28 V 4    4    0.0592 0.0592 V   Au  OH   E = E° – log Q = E° – log   poxygen n n    

  0.504  0.35484  0.0592 V  = –1.245569 = –1.25 V E = –1.28 V – log    4 0.21   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 mol/L, 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. 23.69

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] c)  r G = r H – T r S because the reaction does not take place at 298 K.

r H = m  f (products) H – n  f (reactants) H r H = [(33.2 kJ/mol) + (0 kJ/mol)] – [(90.29 kJ/mol) + (143 kJ/mol)] r H = –200.09 kJ /mol Sr = [(239.9 J/mol•K) + (205.0 J/mol•K)] – [ (210.65 J/mol•K) + (238.82 J/mol•K)]  r S = –4.57 J/K•mol  r G = r H – T r S = (–200.09 kJ) – (280. K)(–4.57 J/K•mol)(1 kJ/103 J) = –198.810 kJ/mol = –199 kJ/mol 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 G and Keq express the extent to which a reaction proceeds and are related by the equation G = –RT ln Keq.    103 J  G 198.810 kJ/mol ln K = =  = 85.402591   8.314 J/mol•K  280. K    1 kJ   RT    K = 1.22991x1037 = 1.2x1037 Therefore, the ratio of rate constants, kf/kr, is 1.2x1037 and the forward rate is much faster than the reverse rate.

23.70

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 CO 2 and H2O as reactants and (CH2O)n and O2 as products. The balanced equation is: nCO2(g) + nH2O(l)  (CH2O)n(s) + nO2(g) b) The moles of carbon dioxide fixed equal the moles of O 2 produced.  48 g CO 2   1 mol CO 2   1 mol O 2  Mole O2/day =    = 1.090661 mol O2/day  day   44.01 g CO 2   1 mol CO 2  To find the volume of oxygen, use the ideal gas equation. Temperature must be converted to K. T (in K) = 25.6 °C + 273.15 = 25.6 °C + 273.15 = 298.75 K pV = nRT nRT 1.090661 mol O 2  8.31446 L•kPa /mol•K  298.75 K  V = = = 26.7437 L= 27 L p 101.3 kPa  c) The moles of air containing 1.090661 mol of CO 2 are:  48 g CO 2   1 mol CO 2  100 mol % air  3 Mole CO2/day =    = 3.11617x10 mol air/day  day 44.01 g CO 0.035 mol % CO   2  2 

V =





3.11617x103 mol air  8.31446 L•kPa /mol•K 298.75 K  nRT = = 7.64106x104 L= 7.6x104 L p 101.3 kPa 

23.71

This problem requires a series of conversion steps: Concentration =  210. kg (NH 4 )2 SO4   103 g   1 mol (NH 4 ) 2 SO4   2 mol NO3   37%   62.01 g NO3   1 mg   103 m3                  3   1000. m3    1 kg   132.15 g (NH 4 ) 2 SO4   1 mol (NH 4 ) 2 SO4   100%   1 mol NO3   10 g   1 L  = 72.9198 mg/L= 73 mg/L This assumes the plants in the field absorb none of the fertilizer.

23.72

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) a) Assuming Al(OH)3 is the limiting reactant:  103 g   1 mol Al(OH)3  1 mol Na 3AlF6  Moles of Na3AlF6 =  365 kg Al(OH)3    1 kg   78.00 g Al(OH)  1 mol Al(OH)  3  3    = 4.6795×103 mol Na3AlF6 b) 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 m3  3 3  3     40.00 g NaOH  3 mol NaOH   10 m  10 L   mL  100%        3 = 7.650x10 mol Na3AlF6 c) Assuming HF is the limiting reactant: pV = nRT









 305 kPa  265 m3  1L  PV Moles of HF = =   RT 8.31446 L•kPa /mol•K   273.2  91.5  K  103 m3 





= 26,654.83 mol HF

 1 mol Na 3 AlF6  3 Moles of Na3AlF6 =  26, 654.83 mol HF    = 4.4425×10 mol Na3AlF6 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 3 AlF6   1 kg   95.6%  Mass (g) of cryolite = 4.4425x103 mol Na 3 AlF6    3     1 mol Na 3 AlF6  10 g   100%  = 891.658 kg= 892 kg Na3AlF6



23.73



Plan: The rate of effusion is inversely proportional to the square root of the molar mass of the molecule (Graham‘s law). Solution: Rate H 4.00 g/mol molar massD a) = = = 1.40859 Rate D molar massH 2.016 g/mol The time for D2 to effuse is 1.41 times greater than that for H2. Time D2 = (1.40859)(16.5 min) = 23.2417 min= 23.2 min b) Set x equal to the number of effusion steps. The ratio of mol H 2 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 13 effusion steps.

23.74

a) The equilibrium constant is large, the rate constant is small, and Ea is large. b) 2CO(g) → C(graphite) + CO2(g) c)  r G = m f (products) G – n  f (reactants) G  r G = [(0 kJ/mol) + (–394.4 kJ/mol)] – [2(–137.2 kJ/mol]) = –120.0 kJ/mol  r G = – RT ln K

ln K =

 120.0 kJ / mol  103 J  G =    = 48.4345  RT 8.314 K/mol•K  298 K   1 kJ 

K = 1.0835×1021 = 1.1x1021 d) Kp = K(RT)ngas Kp = 1.0835×1021 [(8.31446L•kPa/mol•K)(298)]–1 = 4.3729912×1017 = 4.4×1017 23.75

Plan: Use the stoichiometric relationships found in the balanced chemical equation to find the mass of Al 2O3, 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(graphite)  4Al(l) + 3CO2(g) Mass (t) of Al2O3 =  103 kg  10 3g   1 mol Al  2 mol Al2 O3   101.96 g Al2 O3   1 kg  1 t  1 t Al        3     3    1 t  1 kg   26.98 g Al  4 mol Al  1 mol Al 2O3   10 g  10 kg  = 1.8895478t = 1.890 t Al2O3 Therefore, 1.890 t of Al2O3 are consumed in the production of 1 t of pure Al.

 103 kg  10 3g   1 mol Al  3 mol C  12.01 g C   1 kg  1 t  b) Mass (t) of C = 1 t Al    103 kg   1 t   1 kg   26.98 g Al  4 mol Al  1 mol C   103 g          = 0.3338584 t= 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 metric tonne 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.  103 kg  103 g   1 mol Al  3 mol CO2  Moles of C = 1 t Al   = 2.7798×104 mol CO2  1 t   1 kg   26.98 g Al  4 mol Al       The problem states that 101.3 kPa is exact. Use the ideal gas law to calculate volume, given moles, temperature, and pressure. pV = nRT 4   nRT  2.7798x10 mol CO 2  8.31446 L•kPa /mol•K    273  960. K    103 m3  V = =    1 L  p 101.3 kPa    3 3 3 = 2.813204x10 m³= 2.813x10 m



23.76



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 =

(

)( 10

)(

)

= 2.0442×10 kg= 2.0×10 kg Cr 23.77

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

23.78

Ksp (NiS) = 1.1×10–18 = [Ni2+][HS–][OH –] = (0.10)[HS–][OH –], so if [H S–][OH –] < 1.1×10–17, Ksp will not be exceeded and NiS will not precipitate. Ksp (CuS) = 8x10–34 = [Cu2+][HS–][OH –] = (0.10)[HS–][OH –], so if [H S–][OH –] > 8x10–33, Ksp will be exceeded and CuS will precipitate.  HS   H 3 O    HS   H3 O       =  Ka1 = 9x10–8 =  0.10 H S  2  [HS–][H 3O+] = 9×10–9 , or [HS–] = 9×10–9 mol/L/[H 3O+] Kw = [H 3O+][OH –] or [OH –] = Kw/[H 3O+] So[HS–][OH –] =(9×10–9 mol/L/[H 3O+])( 1.0×10–14/[H 3O+]) = (9×10–23)/[H 3O+]2 [HS–][OH –] < 1.1×10–17 [H 3O+] > 3×10–3 mol/L and pH < 2.5 – [HS ][OH –] > 8×10–33 [H 3O+] < 1×105 mol/L and pH > –5.0

23.79

So, maintaining the pH between –5.0 and 2.5 would allow CuS to precipitate but prevent precipitation of NiS.  100%  7 7 a) Mass (kg) of ore = 9.1  104 kg Cu   = 3.64x10 kg= 3.6x10 kg ore  0.25% 





 0.25% Cu   183.54 g FeCuS2  b) Mass % chalcopyrite =   100%  = 0.7220 %= 0.72% chalcopyrite   100%   63.55 g Cu 

23.80

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) →3Ca2+(aq) + 2PO43–(aq) ___ Initial 0 0 Change –x +3x +2x __ Equilibrium 3x 2x –29 Ksp = 1.2×10 = [Ca2+]3[PO43–]2 = (3x)3(2x)2 = 108x5 x = 6.4439×10–7 = 6.4×10–7 mol/L Ca3(PO4)2 [H3O+] = 10–4.5 = 3.162×10–5 mol/L H3PO4(aq) + H2O(l)  H2PO4–(aq) + H3O+(aq)  H 2 PO 4    H 3O     Ka1 =  H 3 PO 4   H 2 PO 4   K a1 7.2x103   = = = 227.70  H3O   H 3 PO 4  3.162x105   – H2PO4 (aq) + H2O(l)  HPO42–(aq) + H3O+(aq)  HPO4 2   H3O    Ka2 =   H 2 PO4      HPO42   6.3x108   = Ka 2 = = 1.9924×10–3  H3O    H 2 PO4   3.162x105     HPO42–(aq) + H2O(l)  PO43–(aq) + H3O+(aq)  PO43   H3O    Ka3 =   HPO42     PO43   4.2x1013   = K a3 = = 1.32827×10–8  H3O    HPO 4 2   3.162x105     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 [PO 43–].  PO43    HPO42     = 1.32827×10–8 and   = 1.9924×10–3  HPO 4 2    H 2 PO4       [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 b)

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 mol/L Acid rain increases the leaching of PO43– into the groundwater, due to the protonation of PO 43– to form HPO42– and H2PO4–. As shown in calculations a) and b), solubility increases from 6.4×10 –7 mol/L (in pure water) to 1.1×10–2 mol/L (in acidic rainwater). 23.81

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 g/mol= 238.9 g/mol

23.82

Using G values: H 2O(l)  H 2(g) + 1/2O2(g) G = +237.192 kJ/mol H 2S(g)  H 2(g) + 1/8S8(s) G = +33 kJ/mol 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  Edge length (cm) =  408.6 pm   = 4.086×10–8 cm  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 –22 Mass (g) of silver =     = 7.1671×10 g 23 unit cell 1 mol Ag 6.022x10 Ag atoms     Density of silver =

7.1671 1022 g 6.82174  10

23

= 10.506264 g/cm³= 10.51 g/cm3

 107.9 g   100.0  7.5 %   63.55 g  7.5%  Atomic mass of sterling silver =        = 104.57375 g/mol 100%  mol     mol  100%    104.57375 g Ag   4 Ag atoms   1 mol Ag –22 Mass (g) of sterling silver =     = 6.946×10 g 23 1 mol Ag  unit cell   6.022x10 Ag atoms   

Density of sterling silver =

6.9461  1022 g 6.82174  1023 cm3

= 10.1823 g/cm³= 10.2 g/cm3

23.84

 10 3 g   0.05%   1 mol Ti  a) Moles of Ti = 5.98x1024 kg  = 6.24×1022 mol= 6×1022 mol Ti  1 kg   100%   47.88 g Ti      b) Mass (g) of ilmenite =  103 g   0.05%   1 mol Ti   50%   1 mol FeTiO3   151.73 g FeTiO3  5.98x1024 kg   1 kg   100%   47.88 g Ti   100%   1 mol Ti   1 mol FeTiO  3        = 4.73760x1024 g= 5×1024 g FeTiO3  yr  0.05%   1t   c) Yr = 5.98 x1024 kg  = 3.297×1013 yr= 3×1013 yr     4   100%   1000 kg   9.07 x 10 t 







23.85



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 given in the Appendix. a)

r H = m  f (products) H – n  f (reactants) H r H = [(  f H of Na2SO4(aq)) + 2(  f H of HCl) + 2(  f H of CO2) + (  f H of Na2CO3) + (  f H of CaS)] – [(  f H of CaCO3) + (  f H of Na2SO4(s)) + 2(  f H of C) + 2(  f H of NaCl) + (  f H of H2SO4)] 351.8 kJ/mol = [(–1389.5 kJ/mol) + 2(–92.31 kJ/mol) + 2(–393.5 kJ/mol) + (–1130.8 kJ/mol) + ( H f of CaS)] – [(–1206.9 kJ/mol) + (–1387.1 kJ/mol) + 2(0 kJ/mol) + 2(–411.1 kJ/mol) + (–907.51 kJ/mol)]  f H of CaS(s) = –479.99 kJ/mol= –480.0 kJ/mol c) The reaction is very endothermic, therefore it is probably not spontaneous even though the entropy change may be expected to be positive. G calculation can be used to determine the spontaneity of this reaction.  1 mol NaCl   1 mol Na 2 CO3   105.99 g Na 2 CO3   73%  d) Mass (g) of Na2CO3 =  250. g NaCl        58.44 g NaCl   2 mol NaCl   1 mol Na 2CO 3   100% 

= 165.4960 g= 165 g Na2CO3 23.86

a) For CaCO3: CaCO3(s)  CaO(s) + CO2(g)

r H = m  f (products) H – n  f (reactants) H r H = [(  f H of CaO) + (  f H of CO2)] – [(  f H of CaCO3)] r H = [(–635.1 kJ/mol) + (–393.5 kJ/mol)] – [(–1206.9 kJ/mol)] r H = 178.3 kJ /mol  r S = m Sproducts – n S reactants  r S = [( S of CaO) + ( S of CO2)] – [( S of CaCO3)]  r S = [(38.2 J/mol•K) + (213.7 J/mol•K)] – [(92.9 J/mol•K)]  r S = 159.0 J/mol•K

 r G = 0 = r H – T r S

r H =  r S

 178.3 kJ / mol   103 J  =   = 1121.3836 K= 1121 K for CaCO3   S  159.0 J/ mol K   1 kJ  For MgCO3: MgCO3(s)  MgO(s) + CO2(g) T =

H

r H = m  f (products) H – n  f (reactants) H r H = [(  f H of (MgO) + (  f H of (CO2)] – [(  f H of (MgCO3)] r H = [(–601.2 kJ/mol) + (–393.5 kJ/mol)] – [(–1112 kJ/mol)] r H = 117.3 kJ/mol  r S = m Sproducts – n S reactants  r S = [( S of MgO)) + ( S of CO2)] – [( S of MgCO3)]  r S = [(26.9 J/mol•K) + (213.7 J/mol•K)] – [(65.86 J/mol•K)]  r S = 174.74 J/mol•K  r G = 0 = r H – T r S

r H = T r S

 117.3 kJ / mol   103 J  =   = 671.283 K= 671 K for CaCO3   S  174.74 J/mol K   1 kJ  b) CaO(s) + SiO2(s)  CaSiO3(s) Mass (kg) of CaCO3 T =

H

)( 7

)(

)

= 4.168679×10 = 4.2×x10 kg CaCO3

CHAPTER 24 THE TRANSITION ELEMENTS AND THEIR COORDINATION COMPOUNDS END–OF–CHAPTER PROBLEMS 24.1

The n value of the d subshell is always one less than the period number.

24.2

a) All transition elements in Period 5 will have a ―base‖ configuration of [Kr]5s2, and will differ in the number of d electrons (x) that the configuration contains. Therefore, the general electron configuration is 1s22s22p63s23p64s23d104p65s24dx. b) A general electron configuration for Period 6 transition elements includes f subshell electrons, which are lower in energy than the d subshell. The configuration is 1s22s22p63s23p64s23d104p65s24d105p66s24f145dx.

24.3

(a) Electrons with the highest n value are removed first, so the ns electrons are removed before the (n – 1)d. V is [Ar]3d 34s2. V3+ is [Ar]3d 2, with the 4s electrons and one of the 3d electrons removed. (b) The ionization energies and the magnetic properties are used to study the electron configurations of atoms.

24.4

(a) The maximum number of unpaired d electrons is five since there are five d orbitals. (b) An example of an atom with five unpaired d electrons is Mn with electron configuration [Ar]3d 54s2. An ion with five unpaired electrons is Mn2+ with electron configuration [Ar]3d 5.

24.5

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.

24.6

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 Zr 4+ 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.

24.7

a) Range in electronegativity across first transition series: 1.3–1.9 b) Range in electronegativity across Period 4 of main-group elements: 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 experience a fairly constant effective nuclear charge.

24.8

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 more 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 Cr 2+ in CrO would have a lower electronegativity than Cr3+ in Cr2O3, which has a lower electronegativity than Cr 6+ in CrO3.

24.9

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. While the above statement is true for almost all 3d elements, as we move to 4d and 5d levels, the energy gaps increase and so there are exceptions. 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.

24.10

Plan: The transition elements in Periods 4 and 5 have a general electron configuration of [noble gas] (n – 1)dx ns2. Transition elements in Periods 6 and 7 have a general electron configuration of [noble gas] (n – 2)f14 (n – 1)dx ns2. Solution: a) Vanadium is in Period 4 and Group 5. Electron configuration of V is 1s22s22p63s23p64s23d3 or [Ar]3d34s2. b) Yttrium is in Period 5 and Group 3. Electron configuration of Y is 1s22s22p63s23p64s23d104p65s24d1 or [Kr]5s24d15s2. c) Mercury is in Period 6 and Group 12. Electron configuration of Hg is [Xe]4f145d106s2.

24.11

a) [Kr]4d65s2

24.12

Plan: The transition elements in Periods 4 and 5 have a general electron configuration of [noble gas] (n – 1)dx ns2. Transition elements in Periods 6 and 7 have a general electron configuration of [noble gas] (n – 2)f14 (n – 1)dx ns2. Solution: a) Osmium is in Period 6 and Group 8. Electron configuration of Os is [Xe]4f 145d66s2. b) Cobalt is in Period 4 and Group 9. Electron configuration of Co is [Ar]3d74s2. c) Silver is in Period 5 and Group 11. Electron configuration of Ag is [Kr]4d105s1 . Note that the filled d orbital is the preferred arrangement, so the configuration is not 5s24d9.

24.13

a) [Ar]3d104s2

24.14

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]3d14s2; 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]3d104s1; Cu2+: [Ar]3d9. There is one unpaired electron. c) The two 4s electrons and one 3d electron are removed to form Fe 3+: Fe: [Ar]3d64s2; 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]4d 35s2; Nb3+: [Kr]4d2. There are two unpaired electrons.

24.15

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

24.16

b) [Ar]3d104s1

b) [Ar]3d54s2

c) [Ar]3d84s2

c) [Xe]4f 145d5 6s2

[Xe]4f145d36s2 with a total of five electrons in the 6s and 5d orbitals. b) Zirconium, Zr, is in Group 4, so the highest oxidation state is +4. The electron configuration of Zr is [Kr]4d 25s2 with a total of four electrons in the 5s and 4d orbitals. c) Manganese, Mn, is in Group 7, so the highest oxidation state is +7. The electron configuration of Mn is [Ar]3d 54s2 with a total of seven electrons in the 4s and 3d orbitals. 24.17

a) +5

24.18

Plan: For Groups 3 to 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 6 exhibit an oxidation state of +6. These elements have a total of six electrons in the outermost s orbital and d orbital: (n–1)d5 ns1. These elements include Cr, Mo, and W. Sg (Seaborgium) is also in Group 6, but its lifetime is so short that chemical properties, like oxidation states within compounds, are impossible to measure.

24.19

Ti, Zr, and Hf

24.20

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.

24.21

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.

24.22

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. Higher oxidation states are more stable for lower transition metals like Re, so ReO 4– would be less likely to reduce. MnO4– is the stronger oxidizing agent.

24.23

b) +3

c) +7

24.24

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.

24.25

Mn2O3 is more basic, because lower oxidation states form more basic oxides.

24.26

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 1 elements do not show the lanthanide contraction.

24.27

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.‖

24.28

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.

24.29

All actinides are radioactive.

24.30

Plan: The inner transition elements have a general electron configuration of [noble gas] (n – 2)f x (n – 1)d 0 ns2. 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]5d16s2 b) Cerium is in the lanthanide series in Period 6 with atomic number 58. Ce: [Xe]4f15d16s2, 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]5f 117s2 d) Uranium is in the actinide series in Period 7 with atomic number 92. U: [Rn]5f 36d 17s2. Removing four electrons gives U4+ with configuration [Rn]5f 2. 24.31

a) [Xe]4f 56s2

b) [Xe]4f 14

24.32

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]4f 76s2. The stability of the half-filled f subshell explains why the configuration is not [Xe]4f 65d16s2. 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 subshell makes Eu2+ most stable. b) Terbium is in the lanthanide series with atomic number 65. The configuration of Tb is [Xe]4f 96s2. 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 subshell.

24.33

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.

24.34

The lanthanide element gadolinium, Gd, has electron configuration [Xe]4f 75d1 6s2with eight unpaired electrons. The ion Gd3+ has seven unpaired electrons: [Xe]4f 7.

24.35

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.

24.36

Donor atoms must have a pair of electrons which can be donated to another atom.

24.37

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.

24.38

Chelates contain two or more donor atoms which can simultaneously bond to a single metal ion.

24.39

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.

24.40

Co(III): 6

Pt(II): 4

c) [Rn]6d 27s2

d) [Rn]5f 11

Pt(IV): 6

24.41

The metal ion acts as a Lewis acid, accepting one or more electron pairs from the ligand(s), which are acting as Lewis bases.

24.42

The -ate ending signifies that the complex ion has a negative charge.

24.43

The ligands are listed first (alphabetically), followed by the metal.

24.44

A linkage isomer is a constitutional isomer, since the atoms are connected in a different pattern.

24.45

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).

24.46

a) tetraamminedinitrocobalt(III) chloride b) hexaamminechromium(III) hexacyanochromate(III) c) potassium tetrachlorocuprate(II)

24.47

Plan: The charge of the central metal atom was determined in Problem 24.45 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.

24.48

a) +3, 6

24.49

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 [CdCl 4]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).

b) +3, 6; +3, 6

c) +2, 4

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 tetraammineaquabromocobalt(III) ion, with four ammonia ligands (tetraammine), one water ligand (aqua) and one 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. 24.50

a) potassium amminepentachloroplatinate(IV) b) diammine(ethylenediamine)copper(II) tetrachloro(ethylenediamine)cobaltate(II) c) dibromobis(ethylenediamine)platinum(IV) perchlorate

24.51

Plan: The charge of the central metal atom was determined in Problem 24.49 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.

24.52

a) +4, 6

24.53

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 NH 3 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].

24.54

a) [Co(en)2 Br2]2SO4

24.55

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. 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.

b) +2, 4; +2, 6

c) +4, 6

b) [Cr(NH3)6]2[CuCl4 ]3

c) K4[Fe(CN)6]

24.56

a) 6; 3 ions

b) 6 (Cr) and 4(Cu); 5 ions

24.57

Plan: Follow the naming rules given in the chapter. Solution: a) The cation is hexaaquachromium(III) with the formula [Cr(H 2O)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 [FeBr 4] –. The total charge on the ion equals +3 charge on iron plus 4  –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 cation 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.

24.58

a) K3[Cr(C2O4)3]

24.59

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 SO 42– 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 Ba 2+ 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 CO 32– ion and one [Pt(en)2]2+ ion, so two ions form per formula unit.

24.60

a) 6; 4 ions

24.61

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:

b) [Co(en)3]4[Mn(CN)5I]3

b) 6 and 6; 7 ions

c) 6; 5 ions

c) [Al(NH3 )2(H2O)2BrCl]NO3

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.

. 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.

24.62

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.

O S

O c) The hydrogen sulfide ion cannot form linkage isomers, because the hydrogen does not have a lone pair.

24.63

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. Cis and trans geometric isomers exist for this complex ion: H H H H

CH3

Pt CH3

Br Pt

H H CH3 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. b) Cis and trans geometric isomers 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 N

H N

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

24.64

a) no isomers b) optical isomers: H H

H H

Zn F

c) geometric isomers

2 H

_ H

C cis

24.65

Pd H

C trans

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 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.

_ Br

Pt Cl

b) The complex ion can form linkage isomers with the NO 2 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.

cis 24.66

trans

a) Coordination isomers:

b) Coordination isomers: H H

CH3

N H

CH3

Pt H

H H

N H

H Cl

Pt H

CH3

c) Geometric isomers:

trans

cis

24.67

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(NH 3)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.

24.68

4 (corresponding to [M(NH3)2Cl2]Cl2) Because two moles of AgCl precipitate for each mole of MCl 4(NH3)2, two of the Cl ions must be counter ions and not part of the complex ion.

24.69

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 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.

24.70

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+.

24.71

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.

24.72

6 (number of orbitals: (2d) + (1s) + (3p) = 6); octahedral

24.73

red

24.74

Absorption of orange or yellow light gives a blue solution.

24.75

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 d 2 2 and d z2 orbitals x y

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 d 2 2 and d z2 orbitals. The crystal field x y splitting is reversed, and the dxy, dyz, and dxz orbitals are higher in energy than the d 2

x  y2

and d z2 orbitals.

24.76

A strong-field ligand causes a greater crystal field splitting (a greater ) than a weak-field ligand. CN – is a strong-field ligand, and F – is a weak-field ligand.

24.77

high-spin A weak-field ligand gives the maximum number of unpaired electrons and a high-spin complex, whereas a strong-field ligand gives the minimum number of unpaired electrons and a low-spin complex.

24.78

If  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  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.

24.79

The  values for tetrahedral complexes are smaller than for octahedral, so  is always < Epairing and electrons will always preferentially add as unpaired electrons to the higher d orbitals.

24.80

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) Electron configuration of Ti: [Ar]3d 24s2 Charge on Ti: Each chloride ligand has a –1 charge, so Ti has a +4 charge {+4 + 6(–1)} = -2– ion. Both of the 4s electrons and both 3d electrons are removed. Electron configuration of Ti4+: [Ar] Ti4+ has no d electrons. b) Electron configuration of Au: [Xe]4f145d106s1 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]4d 75s2 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.

24.81

a) 4 d electrons

24.82

b) 3 d electrons

c) 6 d electrons

Configuration of Ir is [Xe]4f145d 76s2. 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]4f145d106s2. 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]3d 74s2. Configuration of Co2+ is [Ar]3d 7. Seven d electrons in Co2+. 24.83

a) 5

b) 4

c) 6

24.84

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]3d24s2. 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]3d74s2. 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]3d64s2. 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]3d104s1. 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.

24.85

b and d

24.86

Solution: a) Electron configuration of Cr: [Ar]3d 54s1 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]3d104s1 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 subshell 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]3d 64s2 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, , is not large enough to overcome the resistance to electron pairing. The electrons remain unpaired, and the complex is called high-spin.

24.87

24.88

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 Mo: [Kr]4d 55s1 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. (Refer to diagram below.) b) Electron configuration of Ni: [Ar]3d 84s2 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. Silberberg, Amateis, Venkateswaran, Chen, Chemistry: The Molecular Nature of Matter and Change, 3rd Canadian Edition Page 6-881 Instructor‘s Solution Manual © Copyright 2021 McGraw-Hill Ryerson Ltd.

H2O is a weak-field ligand, so the splitting energy, , 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]3d 84s2 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 most d8 metal ions form square planar complex ions (Figure 24.7B). 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.

24.89

24.90

Plan: The spectrochemical series describes the spectrum of splitting energy, . The greater the crystal field strength of the ligand, the greater the crystal field splitting energy, 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 , followed by NH3, and then H2O with the smallest value. The energy of light absorbed increases as  increases since more energy is required to excite an electron from a lower energy orbital to a higher energy orbital when is very large. [Cr(H2O)6]3+ < [Cr(NH3)6]3+ < [Cr(NO2)6]3–

24.91

[Cr(CN)6]3– > [Cr(en)3]3+ > [CrCl6]3–

24.92

In an octahedral field of ligands, the ligands approach along the x, y, and z axes. The d 2

x  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 90º, so ligand interaction is less. The greater interaction of the d 2 2 orbital results in its higher energy. x y

24.93

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 d z 2 is a very favorable (i.e., low-energy) place for the electrons.

24.94

Plan: A weaker field ligand will result in a smaller in the complex and lower energy light absorbed. When a particular colour of light is absorbed, the complementary colour 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 colour observed would be blue or green.

24.95

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 H 2O. Thus, L could not be CN – (which is stronger than H2O), but could be Cl–.

24.96

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.

24.97

Plan: A weaker field ligand will result in a smaller in the complex and lower energy light absorbed. When a particular colour of light is absorbed, the complementary colour 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 (opposite side of the colour wheel). The hexaammine complex appears violet because it absorbs yellow light, which is higher in energy (shorter ) than red light.

24.98

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(NH 3)6]3+ absorbs blue-green and is yellow-orange in colour, [Co(H2O)6]3+ absorbs orange light and is blue in colour, and [CoF6]3– absorbs red light and is green in colour.

24.99

a) The outermost electrons in these elements are in f orbitals. b) If Np and Pu were in Groups 7 and 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 7 and the highest oxidation state of Pu is apparently +6 and not +8 as it would be in Group 8.

24.100 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:

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:

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. 24.101 The electron configuration of Hg is [Xe]4f145d106s2 and that of Hg+ is [Xe]4f145d106s1. The electron configuration of Cu is [Ar]3d104s1 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. 24.102 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 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. 24.103

24.104 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 amounts (mol) 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. 24.105 Tetrahedral complexes are never low-spin, no matter how strong the ligand. Also, certain metal ions have less of a 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.‖ 24.106 a) There are no isomers.

Cl Pt Cl

b) There is an optical isomer.

N Fe3+ N

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+

NH3

Co3+

H3N

H3N NH3

NH3

24.107 a) 2Fe3+(aq) + 3OCl–(aq) + 10OH–(aq)  2FeO42-(aq) + 3Cl–(aq) + 5H2O(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+

24.108

O O

O 24.109 Plan: Types of isomers for coordination compounds include: (i) coordination isomers with different arrangements of ligands and counter ions; (ii), linkage isomers with different donor atoms from the same ligand bound to the metal ion; (iii) geometric isomers with differences in ligand arrangement relative to other ligands; and (iv) 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 NH3 NCS NH 3

Pt NCS NH3 cis-diamminedithiocyanatoplatinum(II) SCN

NH3

H3N

trans-diamminedithiocyanatoplatinum(II) SCN

SCN

NH3 Pt

Pt SCN NH3 cis-diamminediisothiocyanatoplatinum(II)

SCN

H3N

NCS

trans-diamminediisothiocyanatoplatinum(II) SCN NH3 Pt

H3N SCN NCS NH3 cis-diamminethiocyanatoisothiocyanatoplatinum(II) trans-diamminethiocyanatoisothiocyanatoplatinum(II)

24.110 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: (a) [Co(NH3)4(H2O)Cl]2+ tetraammineaquachlorocobalt(III) ion 2 geometric isomers

trans Cl and H2O

cis Cl and NH3

(b) [Cr(H2O)3Br2Cl] triaquadibromochlorochromium(III) 3 geometric isomers

Br‘s trans

Br‘s cis Br‘s cis H2O‘s facial H2O‘s meridional (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.) ( c) [Cr(NH3)2(H2O)2Br2]+ diamminediaquadibromochromium(III) ion 6 isomers (5 geometric)

All pairs are trans

Only NH3‘s are trans

Only H2O‘s are trans

Only Br‘s are trans

All pairs are cis. These are optical isomers of each other. 24.111 The ligand field strength of NH3 is greater than that of H2O. Therefore, the crystal field splitting energy () is larger, and higher energy (shorter wavelength) light is required to excite the electrons of [Cr(NH 3)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. 24.112 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) 3 F

F F

24.113

Structures A, B, and E are geometric isomers. Structures C, D, and F are identical to the original structure.

24.114 a) [Co(NH3)6][Cr(CN)6] b) [Co(NH3)5CN][CrNH3(CN)5]

hexaamminecobalt(III) hexacyanochromate(III) pentaamminecyanocobalt(III) amminepentacyanochromate(III)

24.115 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. 24.116 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 the amount (mol) 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 =1 Amount (mol) of Pt =  38.8 g Pt    = 0.198872 mol Pt; 0.198872  195.1 g Pt 

 1 mol Cl  Amount (mol) of C1 = 14.1 g Cl    = 0.397743 mol Cl;  35.45 g Cl 

0.397743 =2 0.198872

 1 mol C  Amount (mol) of C =  28.7 g C    = 2.389675 mol C;  12.01 g C 

2.389675 = 12 0.198872

 1 mol P  Amount (mol) of P = 12.4 g P    = 0.400387 mol P;  30.97 g P 

0.400387 =2 0.198872

 1 mol H  5.972222 = 30 Amount (mol) of H =  6.02 g H    = 5.972222 mol H; 1.008 g H 0.198872   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

cis-dichlorobis(triethylphosphine)platinum(II)

C2H5

C2H5 trans-dichlorobis(triethylphosphine)platinum(II)

24.117 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 24.118 Plan: A reaction is favored in terms of entropy if there is an increase in entropy. Entropy increases if the amount (mol) of product is larger than the amount (mol ) of the 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 and because the reverse reaction is favoured for the second reaction, the first reaction is favoured. b) The ethylenediamine complex is more stable with respect to ligand exchange with water because the entropy change is unfavorable. 24.119 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. 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.

24.120 a) [Cr(H2O)6]3+

562 nm

[Cr(CN)6]3–

6.626x10

34

6.626x10 E = hc/ =

[CrCl6]3–





34





6.626x10

34





6.626x10

34





34





6.626x10

34





6.626x10

34





34





J•s 3.00x108 m/s  1 nm   6.022x1023   1 kJ    3   9   381 nm mol  10 m     10 J  = 314.187kJ/mol= 314 kJ/mol 735 nm J•s 3.00x108 m/s  1 nm   6.022x1023   1 kJ    3   9   735 nm mol  10 m     10 J  = 162.864 kJ/mol= 163 kJ/mol 462 nm J•s 3.00x108 m/s  1 nm   6.022x1023   1 kJ    3   9   462 nm mol  10 m     10 J  = 259.1024 kJ/mol= 259 kJ/mol 244 nm

E = hc/ = [Ir(NH3)6]3+

J•s 3.00x108 m/s  1 nm   6.022x1023   1 kJ    3   9   244 nm mol  10 m     10 J  = 490.59555 kJ/mol= 491 kJ/mol 966 nm

E = hc/ = [Fe(H2O)6]2+

6.626x10 E = hc/ =

[Fe(H2O)6]3+

J•s 3.00x108 m/s  1 nm   6.022x1023   1 kJ    3   9   966 nm mol  10 m     10 J  = 123.9185 kJ/mol= 124 kJ/mol 730 nm J•s 3.00x108 m/s  1 nm   6.022x1023   1 kJ    3   9   730 nm mol  10 m     10 J  = 163.97988 kJ/mol= 164 kJ/mol 405 nm

E = hc/ = [Co(NH3)6]3+

J•s 3.00x108 m/s  1 nm   6.022x1023   1 kJ    3   9   405 nm mol  10 m     10 J  = 295.56868 kJ/mol= 296 kJ/mol 295 nm

E = hc/ = [Rh(NH3)6]3+



34

6.626x10 E = hc/ =

[Cr(NH3)6]3+



J•s 3.00x108 m/s  1 nm   6.022x1023   1 kJ   9     103 J  562 nm mol  10 m    = 212.99878 kJ/mol= 213 kJ/mol 381 nm

E = hc/ =

6.626x10 E = hc/ =

J•s 3.00x108 m/s  1 nm   6.022x1023   1 kJ    3   9   295 nm mol  10 m     10 J  = 405.7807kJ/mol= 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 (). 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 ().

24.121 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 O5   1 mol VO2 Cl2   153.84 g VO2 Cl2   c) Mass (g) of VO2Cl2– = 12.5 g V2 O5           181.88 g V2 O5   1 mol V2 O5  1 mol VO2 Cl2  = 10.572905 g= 10.6 g VO2Cl2–  1 mol V2 O5  2 mol VOCl3  173.29 g VOCl3  Mass (g) of VOCl3 = 12.5 g V2 O5       181.88 g V2 O5  1 mol V2 O5  1 mol VOCl3  = 23.8193 g= 23.8 g VOCl3 24.122 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.

CHAPTER 25 NUCLEAR REACTIONS AND THEIR APPLICATIONS CHEMICAL CONNECTIONS BOXED READING PROBLEMS B25.1

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.

B25.2

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 u and MeV. Convert the energy per atom to energy per mole by multiplying by Avogadro‘s number. Solution: m = mass of reactants – mass of products = [(4)(1.007825)]u – [4.00260 + (2)(5.48580x10–4)]u = (4.031300 – 4.003697)u = 0.02760 u /4He atom = 0.02760284 g/mol 4He  0.02760284 u 4 He   931.5 MeV  Energy (MeV/atom) =     = 25.7120 MeV/atom = 25.71 MeV/atom 1 atom 1u    Convert atoms to moles using Avogadro‘s number. 23  25.7120 MeV   6.022x10 atoms  25 25 Energy =   = 1.54838x10 MeV/atom = 1.548x10 MeV/mol   atom 1 mol   

B25.3

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.

B25.4

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 84 Po is nuclide A 83 Bi  84 Po + 1 210 206 4 84 Po  82 Pb + 2  206 1 209 82 Pb + 3 0 n  82 Pb 209 210 0 82 Pb  83 Bi + 1

206 82 Pb is nuclide B 209 82 Pb is nuclide C 210 83 Bi is nuclide D

END–OF–CHAPTER PROBLEMS 25.1

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.

25.2

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 u. The average mass of all the sulfur atoms will therefore be greater than the mass of a sulfur-32 atom.

25.3

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.

25.4

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 A 4 4 2 fewer protons, 2 fewer neutrons Z X  Z  2Y + 2 He 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  particle in this type of decay. A A 0 1 more proton, 1 less neutron Z X  Z 1Y + 1 c) Gamma decay does not produce an atom of a different element and Z and N remain unchanged. A A 0 Z X*  Z X + 0

( 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 A 0 1 less proton, 1 more neutron Z X  Z 1Y + 1 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). 25.5

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 He

25.6

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 The conversion of protons to neutrons occurs by either positron decay: 1 1 0 1 p  0 n + 1 or electron capture: 1 0 1 1 p + 1 e  0 n Neutron-rich nuclides, with a high N/Z, undergo  decay. Neutron-poor nuclides, with a low N/Z, undergo positron decay or electron capture.

25.7

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, + emission is more common; for Z > 80, electron capture is more common.

25.8

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: a) The process involves the loss of an  particle from the nucleus. For each  particle emited, the mass number decreases by four and the atomic number decreases by two. 234 4 230 Mass: 234 = 4 + 230; Charge: 92 = 2 + 90 92 U  2 He + 90Th b) The electron captured by the nucleus combines with a proton to form a neutron, so mass number is constant 232 0 232 Mass: 232 + 0 = 232; Charge: 93 + (–1) = 92 93 Np + 1 e  92 U c) Positron emission decreases atomic number by one, but not mass number. 12 0 12 Mass: 12 = 0 + 12; Charge: 7 = 1 + 6 7 N  1 + 6 C

25.9

0 26 a) 26 11 Na  1 + 12 Mg 0 223 b) 223 87 Fr  1 + 88 Ra

4 208 c) 212 83 Bi  2  + 81Tl

25.10

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 + 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 + 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

25.11

0 32 a) 32 14 Si  1 + 15 P 4 214 b) 218 84 Po  2  + 82 Pb 0 110 c) 110 49 In + 1 e  48 Cd

25.12

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 b) In other words, an unknown nuclide captures an electron to form Ag-107. 107 0 107 Mass: 107 + 0 = 107; Charge: 48 + (–1) = 47 48 Cd + 1 e  47 Ag c) In other words, an unknown nuclide decays to give Po-206 and an alpha particle. 210 206 4 Mass: 210 = 206 + 4; Charge: 86 = 84 + 2 86 Rn  84 Po + 2 He

25.13

241 0 a) 241 94 Pu  95 Am + 1 228 0 b) 228 88 Ra  89 Ac + 1

203 4 c) 207 85 At  83 Bi + 2 

25.14

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. 225 221 4 Mass: 225 = 221 + 4; Charge: 89 = 87 + 2 89 Ac  87 Fr + 2 He c) In other words, an unknown nuclide decays to give I-129 and a beta particle. 129 129 0 Mass: 129 = 129 + 0; Charge: 52 = 53 + (–1) 52Te  53 I + 1

25.15

52 0 a) 52 26 Fe  25 Mn + 1 215 4 b) 219 86 Rn  84 Po + 2  81 81 c) 37 Rb + 01 e  36 Kr

25.16

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) 208 O appears stable because its Z (8) value is a magic number, but its N/Z ratio (20  8)/8 = 1.50 is too high and this nuclide is above the band of stability; 208 O is unstable. 59 b) 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 59 band of stability, so 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. 25.17

25.18

a) 146 60 Nd

N/Z = 86/60 = 1.4

Stable, N/Z ok

b) 114 48 Cd 88 c) 42 Mo

N/Z = 66/48 = 1.4

Stable, N/Z ok

N/Z = 46/42 = 1.1

Unstable, N/Z too small for this region of the band

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.

25.19

25.20

48 a) 19 K N/Z = 29/19 = 1.5

Unstable, N/Z too large for this region of the band

b) 79 35 Br 32 c) 18 Ar

N/Z = 44/35 = 1.3

Stable, N/Z okay

N/Z = 14/18 = 0.78

Unstable, N/Z too small

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  decay. Neutron-poor nuclides, with a low N/Z, undergo positron decay or electron capture. For Z < 20, + 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) 238 92 U: Nuclides with Z > 83 decay through  decay. 48 b) The N/Z ratio for 24 Cr is (48 – 24)/24 = 1.00. This number is below the band of stability because N is too 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.

25.21

a) 111 47 Ag

beta decay

41 b) 17 Cl 110 c) 44 Ru

beta decay N/Z = 1.4 which is too high

N/Z = 1.4 which is too high

beta decay N/Z = 1.5 which is too high

25.22

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  decay. Neutron-poor nuclides, with a low N/Z, undergo positron decay or electron capture. For Z < 20, + 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.

25.23

a) 106 49 In

positron decay or electron capture

N/Z = 1.2

b) 141 63 Eu 241 c) 95 Am

positron decay or electron capture

N/Z = 1.2

alpha decay

N/Z = 1.5

25.24

Plan: Stability results from a favorable N/Z ratio, even numbers of N and/or Z, and the occurrence of magic numbers. Solution: The N/Z ratio of 52 24 Cr is (52  24)/24 = 1.17, which is within the band of stability. The fact that Z is even does not account for the variation in stability because all isotopes of chromium have the same Z. However, 52 24 Cr has 28 neutrons, so N is both an even number and a magic number for this isotope only.

40 20 Ca

25.25

N/Z = 20/20 = 1.0 It lies in the band of stability, and N and Z are both even and magic.

25.26

237 233 4 93 Np  2  + 91 Pa 233 233 0 91 Pa  1 + 92 U 233 229 4 92 U  2  + 90Th 229 225 4 90Th  2  + 88 Ra

25.27

Alpha emission produces helium ions which readily pick up electrons to form stable helium atoms.

25.28

207 4 0 The equation for the nuclear reaction is 235 92 U  82 Pb + __ 1 + __ 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 4 0 92 U  82 Pb + __ 1 + 7 2 He 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-4 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 4 0 92 U  82 Pb + 4 1 + 7 2 He

25.29

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.

25.30

Since the decay rate depends only on the number of radioactive nuclei, radioactive decay is a first-order process.

25.31

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 6x10 12 nuclei, the conclusion that t1/2 = 1 min is valid.

25.32

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 7N

25.33

+ 01 n 

14 1 6 C + 1H

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 = 3.70x1010 dps  1.56 x106 dps  1 mg   1 Ci Specific activity (Ci/g) =  = 2.55528x10–2 Ci/g = 2.56x10–2 Ci/g  3.70 x1010 dps   1.65 mg    103 g     

25.34

25.35

25.36

25.37

  4.13 x108 d   1 h        h  1 Ci    3600 s    –6 –6 Specific activity (Ci/g) =    = 1.1925x10 Ci/g = 1.2x10 Ci/g  10 2.6 g 3.70 x10 dps        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 Bq by using the conversion factor between the two units. Solution: A becquerel is a disintegration per second (dps).   7.4 x104 d   1 min         min   60 s    1 Bq  8 8 Specific activity (Bq/g) =    1 dps  = 1.43745 x10 Bq/g = 1.4x10 Bq/g  6   8.58 g  10 g     1 g       

  3.77 x107 d   1 min        min    60 s    1 Bq  Specific activity (Bq/g) =    1 dps  = 587.2274 Bq/g = 587 Bq/g  103 g     1.07 kg      1 kg    Plan: The decay constant is the rate constant for the first-order reaction. Solution:



1 atom = k(1x1012 atom) day

k = 1x1012 d1

25.38

25.39

 (2.8x1012 atom/1.0 yr) = k(1 atom) k = 2.8x1012 yr1 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 1.39x105 atoms/yr or more simply, 1.39x105 yr–1 (the disintegrations are assumed). Solution:



 1.00 x1012 mol  6.022 x10 23 atoms  1.39 x105 atoms = k     1.00 yr 1 mol   

1.39x105 atom/yr = k(6.022x1011 atom) k = (1.39x105 atom/yr)/6.022x1011 atom k = 2.30820x10–7 yr-1 = 2.31x10–7 yr–1

25.40

25.41

– (–1.07x1015 atom/1.00 h) = k[(6.40x10–9 mol)(6.022x1023 atom/mol)] (1.07x1015 atom/1.00 h) = k (3.85408x1015 atom) k = [(1.07x1015 atom/1.00 h)]/(3.85408x1015 atom) k = 0.2776 h-1 = 0.278 h–1 Plan: Radioactive decay is a first-order process, so the integrated rate law is ln nt = lnn0 – 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 = 3.75x103 h ln 2 ln 2 t1/2 = or k = t1/ 2 k

ln 2 = 0.686284 yr–1 1.01 yr lnnt = lnn0 – kt k=





 1 d  1 yr  lnnt = ln [2.00 mg] – (0.686284 yr–1) 3.75 x103 h     24 h  365 d  lnnt = 0.399361 nt = e0.399361 nt = 1.49087 mg = 1.49 mg

25.42

t1/2 = 1.60x103 yr t=?h ln 2 ln 2 k= = = = 0.000433217 yr–1 t1/ 2 1.60x103 yr ln [0.185 g] = ln [2.50 g] – (0.000433217 yr–1)(t) t = 6010.129 yr  365 d   24 h  7 7 t =  6010.129 yr     = 5.264873x10 h = 5.26x10 h  1 yr   1 d 

25.43

Plan: Lead-206 is a stable daughter of 238U. Since all of the 206Pb came from 238U, the starting amount of 238U was (270 mol + 110 mol) = 380 mol = n0. The amount of 238U at time t (current) is 270 mol = 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 k=

ln 2 = 1.540327x10–10 yr–1 9 4.5x10 yr

380 mol = (1.540327x10–10 yr–1)(t) 270 mol 0.3417492937 = (1.540327x10–10 yr–1)(t) t = 2.21868x109 yr = 2.2x109 yr ln

25.44

The ratio (0.735) equals nt/n0 so n0/nt = 1.360544218 ln 2 ln 2 k= = = = 1.2096809x10–4 yr–1 5730 yr t1/ 2

ln 1.360544218 = (1.2096809x10–4 yr–1)(t) 0.30788478 = (1.2096809x10–4 yr–1)(t) t = 2.54517x103 yr = 2.54x103 yr 25.45

Plan: The specific activity of the potassium-40 is the decay rate per mL of milk. Use the conversion factor 1 Ci = 3.70x1010 disintegrations per second (dps) to find the disintegrations per mL per s; convert the time unit to min. Solution:  6 x1011 mCi  103 Ci  3.70 x1010 dps   60 s  1000 mL  Activity =        0.2 L  = 26.64 dpm = 27 dpm   1 mCi   mL 1 Ci      1 min  1 L 

25.46

Plutonium-239 (t1/2 = 2.41x104 yr) Time = 7(t1/2) = 7(2.41x104 yr) = 1.6870x105 yr = 1.69x105 yr

25.47

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 k=

ln 2 =4.95105129x10–11 yr–1 1.4x1010 yr

 6.022 x1023 Th atoms  9 nt = 3.1x1015 mol Th   = 1.86682x10 atoms Th  1 mol Th   n0 =1.86682x109 atoms + 9.5x104 atoms = 1.866915x109 atoms





1.866915 x109 atoms

= (4.95105129x10–11 yr–1)(t) 1.86682 x109 atoms 5.088738x10–5 = (4.95105129x10–11 yr–1)(t) t = 1.027809x106 yr = 1.0x106 yr

25.48

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= = = 5.5451774x10–10 yr–1 t1/ 2 1.25x109 yr

1.38  1.14 mmol

= (5.5451774x10–10 yr–1)(t) 1.38 mmol 0.6021754 = (5.5451774x10–10 yr–1)(t) t = 1.08594x109 yr = 1.09x109 yr

25.49

1 + 42 He  30 15 P + 0 n They experimentally confirmed the existence of neutrons, and were the first to produce an artificial radioisotope.

25.50

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.

25.51

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.

25.52

Protons are repelled from the target nuclei due to the interaction of like (positive) charges. Higher energy is required to overcome the repulsion.

25.53

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

25.54

31 a) 15 P +   11 H + 01 n + 29 14 Si 31 29 P (, p, n) Si 252 b) 98 Cf + 105 B  5 01 n + 257 103 Lr 252 Cf (10B, 5n) 257Lr 4 1 239 c) 238 92 U + 2 He  3 0 n + 94 Pu 238 239 U (, 3n) Pu

25.55

12 1 a) 249 98 Cf + 6 C  104 Rf + 4 0 n

27 13 Al

257

249 15 260 1 98 Cf + 7 N  105 Db + 4 0 n 249 18 263 1 98 Cf + 8 O  106 Sg + 4 0 n 249 12 257

Cf ( C, 4n) Rf Cf (15N, 4n) 260Db 249 Cf (18O, 4n) 263Sg 249

25.56

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.

25.57

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.

25.58

Ionizing radiation is more dangerous to children because their rapidly dividing cells are more susceptible to radiation than an adult‘s slowly dividing cells.

25.59

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–. 

25.60

Plan: The rad is the amount of radiation energy absorbed in J per body mass in kg: 1 rad = 0.01 J/kg. The conversion factor between rad and gray is 1 rad = 0.01 Gy. Solution:  3.3x107 J   1 rad a) Dose (rad) =  = 5.39x10–7 rad = 5.4x10–7 rad  61.2 kg    1x102 J/kg     b) Gray (rad) = (

25.61

)= 5.39x10–9 Gy = 5.4x10–9 Gy

)(

 1 rad  a) Dose (rad) = (8.92x10–4 Gy)   = 0.0892 rad  0.01 Gy 

 0.01 J/kg  –3 –3 b) Energy (J) = (0.0892 rad)   3.6 kg  = 3.2112x10 J = 3.2x10 J 1 rad   25.62

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 (relative biological effectiveness) 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 x105  8.74 x10 14 J/  = 5.244x10–8 J Dose (Gy) =

  5.244  108 J  1 rad   0.01 Gy  = 7.4914x10–10 Gy = 7.5x10–10 Gy  0.01 J   1 rad  70. kg   kg   

 1 mrem   1 rad  –5 –5 b) rems = rads x RBE = 7.4914 x1010 Gy   = 7.4914x10 mrem = 7.5x10 mrem  1.0   3 0.01 Gy 10 rem    





 103 rem   0.01 Sv  sieverts = 7.4914 x105 mrem  = 7.4914x10–10 Sv = 7.5x10–10 Sv  1 mrem   1 rem     





1.77 x10  2.20 x10 a) Dose =

25.63



J /   103 g   1 rad    = 1.46943 rad = 1.47 rad     265 g  1 kg   0.01 J kg    b) Dose = (1.46943 rad)(0.01 Gy/1 rad) = 1.46943x10 –2 Gy = 1.47x10–2 Gy c) Dose = (1.46943 rad)(0.75 rem/rad)(0.01 Sv/rem) = 1.10207x10–2 Sv = 1.10x10–2 Sv 10

13

25.64

   2.50 pCi   1x1012 Ci  3.70 x1010 dps   3600 s   8.25 x1013 J   1 rad  Dose =     65 h         1 Ci disint.   0.01 J   95 kg   1 pCi   1 h   kg   –8 –8 = 1.8796974x10 rad= 1.9x10 rad Dose = (1.8796974x10–8 rad)(0.01 Gy/1 rad) = 1.8796974x10 –10 Gy = 1.9x10–10 Gy

25.65

Use the time and disintegrations per second (Bq) to find the number of 60Co atoms that disintegrate, which equals the number of  particles emitted. The dose in rad is calculated as energy absorbed per body mass.    475 Bq   103 g   1 dps   5.05  1014 J   60 s   1 rad  Dose =      24.0 min          1.858 g   1 kg   1 Bq   1 disint.   1 min   0.01 J kg    = 1.8591x10–3 rad = 1.86x10–3 rad

25.66

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 Na 131I, and a scanning monitor follows the uptake of 131I into the thyroid. Technetium-99 is often used for imaging the heart, lungs, and liver.

25.67

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.

25.68

In positron-emission tomography (PET), the isotope emits positrons, each of which annihilates a nearby electron. In the process, two  photons are emitted simultaneously, 180° apart from each other. Detectors locate the sites and the image is analyzed by computer.

25.69

The concentration of 59 Fe in the steel sample and the volume of oil would be needed.

25.70

The oxygen in methanal (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.

25.71

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.

25.72

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.

25.73

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.‖

25.74

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.

25.75

Plan: The conversion factors are: 1 MeV = 106 eV and 1 eV = 1.602x1019 J. Solution:  106 eV  4 a) Energy (eV) = 0.01861 MeV   1 MeV  = 1.861x10 eV  

 106 eV  1.602 x1019 J  15 15 b) Energy (J) =  0.01861 MeV    = 2.981322x10 J = 2.981x10 J  1 MeV   1 eV    25.76

 1 eV  1000 J   6 6 a) Energy (eV) = 1.57 x1015 kJ    = 9.8002x10 eV = 9.80x10 eV  19 J   1 kJ   1.602 x10









 1 MeV  b) Energy (MeV) = 9.8002 x106 eV  6  9.8002 MeV = 9.80 MeV  10 eV 

25.77

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 = 1.602x10 –19 J. Solution:  6.022 x1023 atoms  23 Number of atoms = 1.5 mol239 Pu   = 9.033x10 atoms  mol  





 5.243 MeV   106 eV  1.602 x1019 J  11 11 Energy (J) = 9.033x1023 atoms    = 7.587075x10 J = 7.6x10 J    1 atom 1 MeV 1 eV    



25.78

25.79



 8.11x105 kJ  103 J     1 MeV   1 eV 1 mol 49 Cr Energy (MeV) =         3.2 x103 mol 49 Cr  1 kJ   1.602 x1019 J  106 eV   6.022 x10 23 nuclei        = 2.6270 MeV = 2.6 MeV 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 factor 1 u = 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 E = mc2. m must be converted to units of kg/mol. Solution: Mass of 8 1H atoms = 8 x 1.007825 u = 8.062600 u Mass of 8 neutrons = 8 x 1.008665 u = 8.069320 u Total mass =16.131920 u m = (16.131920  15.994915)u = 0.137005 u/16O = 0.137005 g/mol 16O  0.137005 u 16 O   931.5 MeV  a) Binding energy (MeV/nucleon) =   = 7.976259844 MeV/nucleon  16 nucleons   1u    = 7.976 MeV/nucleon  0.137005 u 16 O   931.5 MeV  b) Binding energy (MeV/atom) =     = 127.6201575 MeV/atom  1 atom 1u    = 127.6 MeV/atom c) E = mc2     1 kJ  2  0.137005 g 16 O  1 kg  1 J 2.99792 x108 m/ s  Binding energy (kJ/mol) =    3   3  2    mol  kg•m   10 J     10 g  s2   = 1.23133577x1010 kJ/mol = 1.23134x1010 kJ/mol



25.80



m 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 x 1.007825 u = 82.641650 u Mass of 124 neutrons = 124 x 1.008665 u = 125.074460 u

Total mass = 207.716110 u m = (207.716110  205.974440)u = 1.741670 u/206Pb = 1.741670 g/mol 206Pb  1.741670 u 206 Pb   931.5 MeV  a) Binding energy (MeV/nucleon) =   = 7.8755612 MeV= 7.876 MeV/nucleon  206 nucleons   1u   

 1.741670 u 206 Pb   931.5 MeV  b) Binding energy (MeV/atom) =     = 1622.3656 MeV/atom = 1622 MeV/atom  1 atom 1u   

    1kJ  2  1.741670 g 206 Pb  1 kg  1J 8 2.99792 x10 m / s c) Binding energy (kJ/mol) =    3    2   103 g  mol  kg•m   10 J     s2   = 1.5653301x1011 kJ/mol = 1.56533x1011 kJ/mol



25.81

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 1u= 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 E = mc2. m must be converted to units of kg/mol. Solution: Mass of 27 1H atoms = 27 x 1.007825 u = 27.211275 u Mass of 32 neutrons = 32 x 1.008665 u = 32.27728 u Total mass = 59.488555 u m = (59.488555 – 58.933198)u = 0.555357 u/59Co = 0.555357 g/mol 59Co  0.555357 u 59 Co   931.5 MeV  a) Binding energy (MeV/nucleon) =   = 8.768051619 MeV/nucleon  59 nucleons   1u    = 8.768 MeV/nucleon  0.555357 u 59 Co   931.5 MeV  b) Binding energy (MeV/atom) =     = 517.3150 MeV/atom= 517.3 MeV/atom  1 atom 1u    2 c) Use E = mc     1 kJ  2  0.555357 g 59 Co  1 kg  1J 8 2.99792 x10 m/s  Binding energy (kJ/mol) =    3   3  2    mol  kg•m   10 J     10 g  2  s   = 4.9912845x1010 kJ/mol = 4.99128x1010 kJ/mol



25.82



m 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 x 1.007825 u = 53.414725 u Mass of 78 neutrons = 78 x 1.008665 u = 78.675870 u Total mass = 132.090595 u m = (132.090595 – 130.906114)u = 1.184481 u/131I = 1.184481 g/mol 131I  1.184481 u 131 I   931.5 MeV  a) Binding energy (MeV/nucleon) =   = 8.422473676 MeV/nucleon  131 nucleons   1u    = 8.422 MeV/nucleon  1.184481 u 131 I   931.5 MeV  b) Binding energy (MeV/atom) =     = 1103.34405 MeV/atom  1 atom 1u    = 1103 MeV/atom c) E = mc2

  2  1.184481 g 131 I  1 kg  1 J   1 kJ  8 2.99792 x10 m/s  Binding energy (kJ/mol) =   3   2   103 g  mol  kg•m   10 J     2  s   = 1.06455518x1011 kJ/mol = 1.06456x1011 kJ/mol



25.83



0 80 a) 80 35 Br  1 + 36 Kr (reaction 1)

+ 1 e  80 34 Se (reaction 2) b) Reaction 1: m = (79.918528 – 79.916380)u = 0.002148 u Reaction 2: m = (79.918528 – 79.916520)u = 0.002008 u Since E = (m)c2, the greater mass change (reaction 1) will release more energy. 80 35 Br

25.84

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.

25.85

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.

25.86

Enriched fissionable fuel is needed in the fuel rods to ensure a sustained chain reaction. Naturally occurring 235U is only present in a concentration of 0.7%. This is consistently extracted and separated until its concentration is between 3-4%.

25.87

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.

25.88

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, D 2O 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.

25.89

The advantages of fusion over fission are the simpler starting materials (deuterium and tritium), and no long-lived toxic radionuclide by-products.

25.90

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.

25.91

Mass of reactants: 3.01605 u + 2.0140 u = 5.03005 u Mass of products: 4.00260 u + 1.008665 u = 5.011265 u Since there is always a mass decrease when nucleons form a nucleus,

m = mass of reactants – mass of products = (5.03005 – 5.011265)u = 0.018785 u = 0.018785 g/mol E = mc2   2 1 J   1 kJ   0.018785 g   1 kg  8 Energy (kJ/mol) =   3    3  2.99792x10 m/s  2 mol    10 g   kg•m 2   10 J  s   = 1.6883064x109 kJ/mol= 1.69x109 kJ/mol





25.92

243 4 239 95 Am  2 He + 93 Np 239 0 239 93 Np  1 + 94 Pu 239 4 235 94 Pu  2 He + 92 U 239 239 235 93 Np , 94 Pu , and 92 U were present as products in the decay of Am-243.

25.93

Plan: Use the masses given in the problem to calculate the mass change (reactant – products) for the reaction. The conversion factor between u and kg is 1 u = 1.66054x10 –27 kg. Use the relationship E = mc2 to convert the mass change to energy. Solution: 239 4 a) 243 96 Cm  94 Pu + 2 He m (u) = 243.0614 u  (4.0026 + 239.0522) u = 0.0066 u m (kg) = (

)(



b) E = mc2 = 1.09626 x10

)( 29

) = 1.09626x10–29 kg = 1.1x10–29 kg

  kg 2.99792 x10 m/s   = 9.85266x10–13 J = 9.9x10–13 J 2  kg•m  s2  







2

 9.85266 x1013 J  6.022 x1023 reactions   1 kJ  8 8 c) E released =     3  = 5.93327x10 kJ/mol = 5.9x10 kJ/mol   reaction mol 10 J      This is approximately one million times larger than a typical heat of reaction. 25.94

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.    1 day  1 h  ln 2 1 yr ln 2 k= =  = 9.113904x10–13 s–1  2.41  104 yr   365.25 day  24 h  3600 s  t1/ 2     

 106 g   1 mol Pu   6.022 x1023 atoms Pu  15 n = 1.00 g Pu    = 2.5196653x10 atoms Pu  1 g   239 g Pu   1 mol Pu    

25.95

 1 disint.  3 a = kn = (9.113904 x 10–13 s–1)( 2.5196653x1015 atoms Pu)   = 2.2963988x10 d/s 1 atom Pu   Each disintegration releases 5.15 MeV, so d/s can be converted to MeV. Convert MeV to J (using 1.602x10–13 J = 1 MeV) and J to rad (using 0.01 J/kg = 1 rad).    2.2963988 x103 d/s   5.15 MeV   1.602 x10 13 J   1 rad   3600 s  Energy =       16 h      85 kg MeV    disint.     0.01 J kg   1 h    = 1.2838686x10–4 rads = 1.28x10–4 rads b) Since 0.01 Gy = 1 rad, the worker receives: Dose = (1.2838686x10–4 rad)(0.01 Gy/rad) = 1.2838686x10 –6 Gy = 1.28x10–6 Gy Plan: Determine k for 14C using the half-life (5730 yr). Determine the mass of carbon in 4.58 g of CaCO 3.

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 (a 0 = 15.3 d/min•g) in the integrated rate law, solving for t. Solution: ln 2 ln 2 k= = = 1.2096809x10–4 yr–1 5730 yr t1/ 2

 1 mol CaCO3  1 mol C  12.01 g C  Mass (g) of C = 4.58 g CaCO3     = 0.5495634 g C  100.09 g CaCO3  1 mol CaCO3   1 mol C  3.2 d/min at= = 5.8228 d/min•g 0.5495634 g Using the integrated rate law:

 5.8228 d/min•g  –4 –1 ln   = – (1.2096809x10 yr )(t)  15.3 d/min•g  t = 7986.17 yr = 8.0x103 yr 25.96

Find the rate constant from the rate of decay and the initial number of atoms. Use rate constant to calculate half life. Initial number of atoms:

 106 g   1 mol RaCl2  1 mol Ra   6.022 x1023 Ra atoms  Ra atoms =  5.4 g RaCl2     1 g   297 g RaCl  1 mol RaCl   1 mol Ra 2  2     16 = 1.09490909x10 Ra atoms

   1d s  1.5 x105 Bq  = 1.36997675x10–11 s–1 k =    16   Bq 1.09490909 x10 Ra atoms    t1/2 =

25.97

ln 2 ln 2 = = 5.05955x1010 s = 5.1x1010 s 11 1 k 1.36997675x10 s

ln 2 ln 2 = = 1.2096809x10–4 yr–1 5730 yr t1/ 2

 1 mol 14 C  6.022 x1023 atoms 14 C  14 14 Number of atoms = 108 g 14 C   = 4.3014x10 atoms C 14  14 g 14 C   1 mol C    a = kn = [(1.2096809x10–4 yr–1)(4.3014x1014 atoms 14C)](1 disintegration/1 atom) = 5.2033x10 10 dpyr   13  5.2033 x1010 dpyr   0.156 MeV   1.602 x10 J   1 rad  Dose =  = 2.001x10–3 rad = 10–3 rad yr            0.01 J  65 kg disint. 1 MeV       kg  



25.98



Plan: Determine the amount of AgCl (grams) 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 x102 Bq   1 dps    1 nCi   1 g AgCl  1 Ci Concentration =  = 1.93050x10–6 g AgCl/mL     10  109 Ci   175 nCi    1 Bq  3.70 x10 dps  mL      

 1.93050 x106 g AgCl   1 mol AgCl   1 mL  Concentration (mol/L) =      3   mL    143.4 g AgCl   10 L  = 1.34623x10–5 mol/L = 1.35x10–5 mol/L AgCl 25.99

a) The process shown is fission in which a neutron bombards a large nucleus, splitting that nucleus into two nuclei of intermediate mass. 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 a n/p ratio of 1.6. This ratio places this isotope above the band of stability and decay by beta particle emission is expected.

25.100 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= = = 9.90210x10–10 yr–1 t1/ 2 7.0x108 yr

25.101 Determine the value of k from the half-life. Then determine the age from the integrated rate law. ln 2 ln 2 k= = = 1.540327x10–10 yr–1 t1/ 2 4.5x109 yr

( ) = (1.540327x10–10 yr–1)(t) 0.510826 = (1.540327x10–10 yr–1)(t) t = 3.316345x109 yr = 3.3x109 yr 25.102 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:

 495 photons/s  ln   = –k(20 h)  5000 photons/s  –2.312635 = –k(20 h) k = 0.11563 h–1 ln 2 ln 2 t1/2 = = = 5.9945 h = 5.99 h (Assuming the times are exact, and the emissions have three k 0.11563 h 1 significant figures.)

The fraction lost upon preparation is 100% – 79.3533% = 20.6467% = 21%. ln 2 ln 2 = = 5.5451774x10–10 yr–1 t1/ 2 1.25  109 yr a = kn where a = dps n = number of atoms Number of atoms = (1.0 mol 40K)(6.022x1023 atoms/mol) = 6.022x1023 atoms 40K  5.5451774 x1010    1 day  1 h  1 disint.  1 yr 23 7 a=   6.022 x10 atoms      = 1.05816x10 dps  yr 365.25 day 24 h 3600 s 1 atom        Dose (Ci) = (1.05816x107 dps)(1 Ci/3.70x1010 dps) = 2.85990x10–4 Ci= 2.9x10–4 Ci Dose = (1.05816x107 dps)(1 Bq/1 dps) = 1.05816x107 Bq = 1.1x107 Bq

25.103 k =





25.104 Plan: Use the given relationship for the fraction remaining after time t, where t = 10.0 yr, 10.0x103 yr, and 10.0x104 yr. Solution: a)

(i) Fraction remaining after 10.0 yr =

 1 2

10.0

=  1  2

10.0x103

(ii) Fraction remaining after 10.0x10

yr =  1  2

10.0x104

(iii) Fraction remaining after 10.0x10 4 yr =  1  2

5730

= 0.998791 = 0.999

= 0.298292 = 0.298 = 5.5772795x10–6 = 5.58x10–6

b) Radiocarbon dating is more reliable for (ii) 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 (i), very little 14 C has decayed and for (iii) very little 14C remains. In either case, it will be more difficult to measure the change so the error will be relatively large. 25.105

+ 01 e  210 85 At Mass = (2.368 MeV)(1 u/931.5 MeV) = 0.0025421363 u Mass 210At = mass 210Rn + electron mass – mass equivalent of energy emitted. = (209.989669 + 0.000549 – 0.0025421363) u = 209.9876759 u= 209.98768 u 210 86 Rn

25.106 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 nln (0.900) = ln (0.500) n = (ln 0.500)/(ln 0.900) = 6.578813 h = 6.579 h

25.107 a)  decay by vanadium-52 produces chromium-52. 51 1 52 52 0 23V + 0 n  23V *  24 Cr + 1 51 52 23V (n,) 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 + 64 63 29 Cu (n, ) 28 Ni

c)  decay by aluminum-28 produces silicon-28. 27 1 28 28 0 13 Al + 0 n  13 Al *  14 Si + 1 27 28 13 Al (n,) 14 Si

25.108 Determine k for 90Sr. ln 2 ln 2 k= = = 0.023902 yr–1 t1/ 2 29 yr

100  99.9 %

= – (0.023902 yr–1)(t)

100% t = 289.003 yr = 3x102 yr (The calculation 100 – 99.9 limits the answer to one significant figure.)

25.109 a) 126 C + 42 He  168 O 13  931.5 MeV   1.602  10 J   1 kJ  –14 –14 b) 7.7 x102 u     3  = 1.1490425x10 kJ= 1.1x10 kJ  1 u 1 MeV 10 J    





25.110 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 = = 4.33879178x10–4 yr–1(1 yr/8766 h) = 4.94510515x10 –8 h–1 1599 yr t1/ 2 The mass of 226Ra is 226.025402 u/atom or 226.025402 g/mol.    6.022 x1023 Ra atoms  1 mol Ra 21 n = 1.000 g Ra    = 2.6643023x10 Ra atoms   226.025402 g Ra 1 mol Ra    a = kn = (4.94510515x10–8 h–1)(2.6643023x1021 Ra atoms) = 1.3175255x1014 Ra atoms/h This result means that 1.318x1014 226Ra nuclei are decaying into 222Rn nuclei every hour. Convert atoms of 222Rn into volume of gas using the ideal gas law.

 1.3175255 x1014 Ra atoms   1 atom Rn    1 mol Rn Moles of Rn/h =       23  h    1 atom Ra   6.022 x10 Rn atoms  = 2.1878537x10–10 mol Rn/h kPa•L   2.1878537x1010 mol Rn/h  8.31446  273.15 K  mol•K  nRT  V= = p 100 kPa –9 = 4.96882305x10 L/h = 4.969x10–9 L/h Therefore, radon gas is produced at a rate of 4.904x10 –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.

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25.111 Determine k: ln 2 ln 2 k= = = 0.0216608 s–1 t1/ 2 32 s

90% = – (0.0216608 s–1)(t) 100% t = 4.86411 s = 4.9 s ln

The N/Z ratio for 140Cs is too high.

25.112 a) 133 55 Cs b) 79 35 Br

It has an even number of neutrons compared with 78Br.

c) 24 12 Mg

The N/Z ratio equals 1.

d) 147 N

The N/Z ratio equals 1.

25.113 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

 1.0x104 particles  ln  = – (0.0239016 yr–1)(t)  7.0x104 particles    –1.945910 = – (0.0239016 yr–1)(t) t = 81.413378 yr = 81 yr 25.114 a) 63 Li + 63 Li  126 C (dilithium) b) m = 2(mass 63 Li ) – mass 126 C = 2(6.015121 u) – 12.000000 u = 0.030242 u/atom 126 C (dilithium) m = (0.030242 u/atom)(1.66054x10–27 kg/u) =5.02180507x10–29 kg/atom    2 1 J 29 8 E = mc2 = 5.02180507 x10 kg/atom 2.99792 x10 m/s   = 4.5133595x10–12 J/atom 2  kg•m  s2  

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

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 4.5133595 x1012 J   1 atom    1u E=       27   12.000000 u  1.66054 x10 kg  atom     = 2.2650059x1014 J/kg = 2.2650x1014 J/kg dilithium

c) 4 11 H  42 He + 2 01

2 positrons are released.

d) For dilithium ( 126 C ):

Mass = (5.02180507x10–29 kg/atom)(1 atom/12.000000 u) (1 u/1.66054x10 –27 kg 12C) = 2.5201667x10–3 kg/kg 12C = 2.5202x10–3 kg/kg 12C 4 For 2 He (The mass of a positron is the same as the mass of an electron.) m = 4 (mass 11 H ) – [mass 42 He + 2 mass 01 e ] = 4(1.007825 u) – [4.00260 u + 2(5.48580x10–4 u)] = 0.02760 u/atom = (0.02760 u/atom)(1.66054x10–27 kg/u) = 4.58309x10–29 kg/atom Mass = (4.58309x10–29 kg/atom)(1 atom/4.00260 u)(1 u/1.66054x10 –27 kg 4He) = 6.895517x10–3 kg/kg 4He = 6.896x10–3 kg/kg 4He e) 21 H + 31 H  42 He + 01 n m = [2.0140 u + 3.01605 u] – [4.00260 u + 1.008665 u] = 5.0300 u – 5.01126 u = 0.0188 u = (0.0188 u/atom)(1.66054x10–27 kg/u) = 3.1218152x10–29 kg/atom Mass = (3.1218152x10–29 kg/atom)(1 atom/4.00260 u)(1 u/1.66054x10 –27 kg 4He) = 4.696947x10–3 kg/kg 4He = 4.70x10–3 kg/kg 4He f) 63 Li + 01 n  42 He + 31 H

 1 mol 3 H  3.01605g 3 H  1 mol 6 Li  103 g 6 Li  1 kg 3 H  Mass 31 H / kg 6Li =   1 mol 6 Li   1 mol 3 H   6.015121 g 6 Li   1 kg 6 Li   103 g 3 H        = 0.5014113598 kg 3H/kg 6Li = 0.501411 kg 3H/kg 6Li  103 g   1 mol 3 H   6.022x1023 atom 3 H   3.121852x1029 kg  Mass = (0.5014113598 kg 3H)      1 kg   3.01605 g 3 H   atom mol 3 H      = 3.1254222x10–3 kg = 3.125x10–3 kg Change in mass for dilithium reaction: Mass = (5.02180507x10–29 kg/atom 12C)(1 atom 12C/2 atoms 6Li)(6.022x1023 atoms 6Li/mol) (1 mol 6Li/6.015121 g 6Li)(103 g/kg 6Li) = 2.513774x10–3 kg = 2.514x10–3 kg The change in mass for the dilithium reaction is slightly less than that for the fusion of tritium with deuterium. 25.115 Plan: Convert pCi to Bq using the conversion factors 1 Ci = 3.70x10 10 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 level recommended by Health Canada. Solution: 12 10  4.0 pCi   10 Ci  3.70  10 Bq  a) Activity (Bq/L) =     = 0.148 Bq/L= 0.15 Bq/L  1 Ci  L   1 pCi    The safe level is 0.15 Bq/L. ln 2 ln 2 b) k = = = 0.181452 d–1 t1/ 2 3.82 d

12 10  7.403104 pCi   10 Ci   3.70 x10 Bq  Activity (Bq/L) =      = 0.2739148 Bq/L = 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.

0.15 Bq/L = – (0.181452 d–1)(t) 0.2739148 Bq/L – 0.602182 = – (0.181452 d–1)(t) t = 3.318685 d = 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

25.116 k =

ln 2 ln 2 = = 0.05653729 yr–1 t1/ 2 12.26 yr

fraction lost = 1 – 0.732746777 = 0.267253222 = 0.267 25.117

239 0 239 239 0 92 U  1 + 93 Np  94 Pu + 1 This could begin the 235 92 U decay series.

4 231  42  + 235 92 U  2  + 90Th

25.118 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. Solution:  103 Ci  3.70 x1010 dps   5.59 MeV   1.602 x1013 J  –5 Energy (J/s) = 1.0 mCi      = 3.3134166x10 J/s    1 mCi   1 Ci 1 MeV     1 disint.    Time for 1.0 mrad to be absorbed:  103 rad   0.01 J/kg    24.5 kg Time (s) = 1.0 mrad   = 7.3941804 s = 7.4 s  1 mrad   1 rad   3.3134166 x105 J/s      25.119 k =

ln 2 ln 2 = = 1.2096809x10–4 yr–1 t1/ 2 5730 yr

 12.9 d /min • g  –4 –1 ln   = – (1.2096809x10 yr )(t) 15.3 d/min • g   – 0.170625517 = – (1.2096809x10–4 yr–1)(t) t = 1410.500216 yr = 1.41x103 yr The earthquake occurred 1410 years ago.

25.120 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. 25.121 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

– 0.841567 = – (0.05653729 yr–1)(t) t = 14.92857 yr = 14.9 yr The wine was produced in (1941 – 15) = 1926. 25.122 If 99% of Pu-239 decays, 1% remains. ln 2 ln 2 k= = = 2.87612938x10-5 yr–1 t1/ 2 2.41x104 yr

1% = – (2.87612938x10–5 yr–1)(t) 100% –4.60517 = – (2.8761293x10–5 yr–1)(t) t = 1.6011693x105 yr = 2x105 yr (The 1% limits the answer to one significant figure.) ln

25.123 k =

ln 2 ln 2 = = 1.2096809x10–4 yr–1 t1/ 2 5730 yr

0.61 pCi/g = – (1.2096809x10–4 yr–1)(t) 6.89 pCi/g –2.4243674 = – (1.2096809x10–4 yr–1(t) t = 2.0041x104 yr = 2.0x104 yr ln

25.124 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 u to energy in eV and then to J. Solution: m = mass of reactants – mass of products = (14.003074 + 1.008665)u – (14.003241 + 1.007825)u = 0.000673 u 6  931.5 MeV   10 eV  5 5 Energy (eV) =  0.000673 u      1 MeV  = 6.268995x10 eV = 6.27x10 eV 1u   

 1.602 x1019 J   1 kJ   6.022 x1023  Energy (kJ/mol) = 6.268995 x105 eV    3     1 eV    10 J   1 mol  = 6.0478524x107 kJ/mol = 6.05x107 kJ/mol

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25.125 Calculate m. Mass of 3 1H atoms = 3 x 1.007825 u = 3.023475 u Mass of 4 neutrons = 4 x 1.008665 u = 4.034660 u Total mass = 7.058135 u m = (7.058135 – 7.016003)u = 0.042132 u 7Li = 0.042132 g/mol 7Li  0.042132 u 7 Li   931.5 MeV   106 eV  7 Binding energy =     = 3.9245958x10 eV/nucleus    1 atom 1 u 1 MeV     = 3.925 107 eV/nucleus Convert MeV to kJ.

    1 kJ  2  0.042132 g 7 Li   1 kg  1J 8 Binding energy =    3  2.99792x10 m/s   3  2 mol  kg•m 2   10 J     10 g  s   = 3.7866237x109 kJ/mol = 3.7866x109 kJ/mol

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25.126

146 64 Gd

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146 0 4 142 + 01 e  146 63 Eu  62 Sm + 1  2 He + 60 Nd

3 R  25.127 a) Kinetic energy = 1/2 mv2 =  T 2  NA 

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 3   8.314 J/mol•K  1.00x10 K Energy =   = 2.07090668x10–17 J/atom = 2.07x10–17 J/atom 11 H 6.022  1023 atom/mol 2 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)(2.99792x108 m/s)2(J/(kg•m2/s2) = 1.7975048x1017 J  1.7975048x1017 J   1 kg   1.0078 g H     1 mol H 1 H atom Number of atoms =           3 23  17   10 g 1 kg H   1 mol H   6.022x10 atoms H   2.0709066x10 J    = 1.4525903x107 H atoms = 1.45x107 H atoms c) 4 11 H  42 He + 2 01 (Positrons have the same mass as electrons.) 6

m = [4(1.007825 u)] – [4.00260 u + 2(0.000549 u)] = 0.027602 u / 42 He

  0.027602 g   1 kg   1 mol He   1 mol H 7 m =   1.4525903x10 H atoms  3   23 mol He 4 mol H 10 g 6.022x10 H atoms      = 1.6644967x10–22 kg Energy = (m)c2 = (1.6644967x10–22 kg) (2.99792x108 m/s)2(J/(kg•m2/s2) = 1.4959705x10–5 J = 1.4960x10–5 J d) Calculate the energy generated in part b):  1.7975048 x1017 J  1 kg   1.0078 g H    1 mol H Energy =  = 3.0081789x10–10 J         103 g  1 mol H  6.022 x1023 atoms H  1 kg H      Energy increase = (1.4959705x10–5 – 3.0081789x10–10 ) J = 1.4959404x10–5 J = 1.4959x10–5 J e) 3 11 H  23 He + 101

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m = [3(1.007825 u)] – [3.01603 u + 0.000549 u] = 0.006896 u/ 23 He = 0.006896 g/mol 23 He

  0.006896 g   1 kg   1 mol He   1 mol H 7 m =   1.4525903x10 H atoms   3     23 mol He 3 mol H    10 g     6.022 x10 H atoms  = 5.54470426x10–23 kg Energy = (m)c2 = (5.54470426x10–23 kg)(2.99792x108 m/s)2(J/(kg•m2/s2) = 4.9833164x10–6 J = 4.983x10–6 J No, the Chief Engineer should advise the Captain to keep the current technology.

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25.128 Multiply the equation E = hc/ by Avogadro‘s number (N) to get the energy change per mole.

 6.022 x10 /mol  6.626 x10 J•s  2.99792 x10 m / s   1pm  = 1.3702442x10 J/mol 34

E = Nhc/ =

 12   10 m  Determine the mass through a series of conversions from the chapter and the inside back cover:  1.3702442 x1010 J      1.66054 x10 27 kg  1 Mev 1u Mass (kg) =          1.602 x1013 J  931.5 MeV   mol 1u      = 1.52476x10–7 kg/mol = 1.52x10–7 kg/mol 25.129 k =

8.73pm 

ln 2 ln 2 = = 0.131526979 yr–1 t1/ 2 5.27 yr

70.% = – (0.131526979 yr–1)(t) 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, 2023. Then add 0.7 yr which is (365.25 x 0.7) = 256 d (actually the significant figures limit this to 3x10 2 d). The final date is November 12, 2023. ln

25.130 Plan: The difference in the energies of the two  particles gives the energy of the ray released to get from excited state I to the ground state. Use E = hc/ to determine the wavelength. The energy of the gamma ray must be converted from MeV to J. The energy of the 2%  particle is equal to the highest energy  particle minus the energy of the two  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  ray = (4.816 – 4.773) MeV = 0.043 MeV 

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6.626 x1034 J•s 2.99792 x108 m/s   hc 1 MeV –11 –11 = =   = 2.8836x10 m= 2.9x10 m  13 E  0.043 MeV   1.602 x10 J 

b) (4.816 – 0.043 – 0.060) MeV = 4.713 MeV 25.131 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 206 92 U  82 Pb

Fraction U-238 = 238 k=

2x 2x 2 = = 2x  x 3 3x

ln 2 ln 2 = = 1.540327x10–10 yr–1 t1/ 2 4.5x109 yr

2  ln  3  = – (1.540327x 0–10 yr–1)(t)  1    – 0.405465108 = – (1.540327x10–10 yr–1)(t) t = 2.63233x109 yr = 2.6x109 years old Silberberg, Amateis, Venkateswaran, Chen, Chemistry: The Molecular Nature of Matter and Change, 3rd Canadian Edition Page 6-918 Instructor‘s Solution Manual © Copyright 2021 McGraw-Hill Ryerson Ltd.

25.132

232 228 4 90Th  88 Ra + 2  228 228 0 88 Ra  89 Ac + 1 228 228 0 89 Ac  90Th + 1 228 224 4 90Th  88 Ra + 2  224 220 4 88 Ra  86 Rn + 2  220 216 4 86 Rn  84 Po + 2  216 212 4 84 Po  82 Pb + 2  212 212 0 82 Pb  83 Bi + 1 212 212 0 83 Bi  84 Po + 1 212 208 4 84 Po  82 Pb + 2 

25.133 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. 25.134 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) = 3.26x103 d 214 b) Po 20(1.6x10–4 s) = 3.2x10–3 s 232 c) Th 20(1.39x1010 yr) = 2.78x1011 yr 25.135 If the blade was made in 100 C.E. it is about 1.9x103 years old. ln 2 ln 2 k= = = 1.2096809x10–4 yr–1 t1/ 2 5730 yr

Handle:

 10.1 d/min•g  –4 –1 ln   = – (1.2096809x10 yr )(t)  15.3 d/min•g  – 0.415317404 = – (1.2096809x10–4 yr–1)(t) t = 3.43328x103 yr= 3.43x103 yr Inlaid:

 13.8 d/min•g  –4 –1 ln   = – (1.2096809x10 yr )(t)  15.3 d/min•g  – 0.103184236 = – (1.2096809x10–4 yr–1)(t) t = 8.52987x102 yr = 8.53x102 yr Ribbon:

 12.1 d/min•g  –4 –1 ln   = – (1.2096809x10 yr )(t) 15.3 d/min•g   – 0.2346473758 = – (1.2096809x10–4 yr–1)(t) t = 1.939746x103 yr = 1.94x103 yr Sheath:

 15.0 d/min•g  –4 –1 ln   = – (1.2096809x10 yr )(t) 15.3 d/min•g   – 0.0198026273 = – (1.2096809x10–4 yr–1)(t) t = 1.63701x102 yr = 1.64x102 yr The ribbon is nearest in age to the blade. Silberberg, Amateis, Venkateswaran, Chen, Chemistry: The Molecular Nature of Matter and Change, 3rd Canadian Edition Page 6-919 Instructor‘s Solution Manual © Copyright 2021 McGraw-Hill Ryerson Ltd.

25.136 a) When 1.00 kg of antimatter annihilates 1.00 kg of matter, the change in mass is: m = 0 – 2.00 kg = –2.00 kg. The energy released is calculated from E = mc2. E = (–2.00 kg)(2.99792x108 m/s)2(J/(kg•m2/s2) = –1.7975049x1017 J = –1.80x1017 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 m = [4(1.007825 u)] – [4.00260 u + 2(0.000549 u)] = 0.02760 u per 42 He formed 4  0.02760 u   1 He  2 Total m = 1.00 x105 H atoms    4 H atoms  = 6.90x10 u 4 He    

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

 931.5 MeV   1.602 x1016 kJ  –10 Energy = 6.90 x102 u   = 1.02966x10 kJ per antiH collision   1 u 1 MeV   

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 103 g   1 mol antiH   6.022 x1023 antiH  AntiH atoms = 1.00 kg   = 5.9742x1026 antiH  1 kg   1.008 g antiH   1 mol antiH       1.02966 x1010 kJ  26 16 16 Energy released =   5.9742 x10 antiH = 6.15139x10 kJ = 6.15x10 kJ  antiH   c) From the above calculations, the procedure in part b) with excess hydrogen produces more energy per kilogram of antihydrogen.

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25.137 Plan: Einstein‘s equation is E = mc2, which is modified to E = mc2 to reflect a mass difference. The speed of light, c, is 2.99792x108 m/s. The mass of exactly 1 u is 1.66054x10 –27 kg (inside back cover of text). When the quantities are multiplied together, the unit will be kg•m2/s2, which is also J. Convert J to MeV using the conversion factor 1.602x10–13 J = 1 MeV. Solution: E = (m)c2     2  1.66054 x1027 kg    J 1 MeV 8 E = 1u      2.99792 x10 m/s   2 13    1u   kg•m   1.602 x10 J     2 s   = 9.3159448x102 MeV = 9.316x102 MeV

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25.138 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:  1 mol 206 Pb  1 mol 238 U  238 g 238 U  Mass (g) of 238U = 0.023 g 206 Pb  = 0.0265728 g 238U  206 g 206 Pb   1 mol 206 Pb   1 mol 238 U      Original mass of 238U = (0.065 + 0.0265728) g 238U = 0.0915728 g 238U ln 2 ln 2 k= = = 1.540327x10–10 yr–1 t1/ 2 4.5x109 yr

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 0.065 g  –10 –1 ln   = – (1.540327x10 yr )(t) 0.0915728 g   – 0.3427470145 = – (1.540327x10–10 yr–1)(t) t = 2.2251575x109 yr = 2.2x109 years old

238 4 25.139 a) 242 94 Pu  92 U * + 2  238 238 92 U *  92 U

+ b) Determine the MeV of the  ray by using E = hc/ to determine the energy.

 6.626 x10

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J•s 2.99792 x108 m/s  1 nm   1 MeV  9   = 0.044975 MeV  13    0.02757 nm   10 m  1.602 x10 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 MeV = 4.898 MeV 18 263 1 25.140 249 98 Cf + 8 O  106 Sg + 4 0 n

263 259 4 106 Sg  104 Rf + 2  259 255 4 104 Rf  102 No + 2  255 251 4 102 No  100 Fm + 2 

25.141 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

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 1.00 kg  10.%  100%  = – (0.294956 d–1)(t) ln    1.00 kg     – 2.3025851 = – (0.294956 d–1)(t) t = 7.806538 d = 7.81 d 25.142 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. 25.143 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. 25.144 a) Nucleus 1 is 94 Be ; nucleus 2 is 104 Be ; nucleus 3 is 74 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.