Chemistry, 11e Steven Zumdahl, Susan Zumdahl, Donald DeCoste (Solutions Manual All Chapters. 100% Original Verified, A+ Grade)
CHAPTER 1 CHEMICAL FOUNDATIONS
Review Questions 1.
a. Law versus theory: A law is a concise statement or equation that summarizes observed behavior. A theory is a set of hypotheses that gives an overall explanation of some phenomenon. A law summarizes what happens; a theory (or model) attempts to explain why it happens. b. Theory versus experiment: A theory is an explanation of why things behave the way they do, while an experiment is the process of observing that behavior. Theories attempt to explain the results of experiments and are, in turn, tested by further experiments. c. Qualitative versus quantitative: A qualitative observation only describes a quality while a quantitative observation attaches a number to the observation. Some qualitative observations would be: The water was hot to the touch. Mercury was found in the drinking water. Some quantitative observations would be: The temperature of the water was 62˚C. The concentration of mercury in the drinking water was 1.5 ppm. d. Hypothesis versus theory: Both are explanations of experimental observation. A theory is a set of hypotheses that has been tested over time and found to still be valid, with (perhaps) some modifications.
2.
The fundamental steps are (1) making observations; (2) formulating hypotheses; (3) performing experiments to test the hypotheses. The key to the scientific method is performing experiments to test hypotheses. If after the test of time the hypotheses seem to account satisfactorily for some aspect of natural behavior, then the set of tested hypotheses turns into a theory (model). However, scientists continue to perform experiments to refine or replace existing theories.
3.
A qualitative observation expresses what makes something what it is; it does not involve a number; e.g., the air we breathe is a mixture of gases, ice is less dense than water, rotten milk stinks. The SI units are mass in kilograms, length in meters, and volume in the derived units of m3. The assumed uncertainty in a number is 1 in the last significant figure of the number.
4.
Volume readings are estimated to one decimal place past the markings on the glassware. The assumed uncertainty is ±1 in the estimated digit. For glassware a, the volume would be
1
2
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CHEMICAL FOUNDATIONS
estimated to the tenths place since the markings are to the ones place. A sample reading would be 4.2 with an uncertainty of ±0.1. This reading has two significant figures. For glassware b, 10.52 ±0.01 would be a sample reading and the uncertainty; this reading has four significant figures. For glassware c, 18 ±1 would be a sample reading and the uncertainty, with the reading having two significant figures. 5.
Precision: reproducibility; accuracy: the agreement of a measurement with the true value. a. Imprecise and inaccurate data: 12.32 cm, 9.63 cm, 11.98 cm, 13.34 cm b. Precise but inaccurate data: 8.76 cm, 8.79 cm, 8.72 cm, 8.75 cm c. Precise and accurate data: 10.60 cm, 10.65 cm, 10.63 cm, 10.64 cm
6.
Significant figures are the digits we associate with a number. They contain all the certain digits and the first uncertain digit (the first estimated digit). What follows is one thousand indicated to varying numbers of significant figures: 1000 or 1 × 10 3 (1 S.F.); 1.0 × 103 (2 S.F.); 1.00 × 103 (3 S.F.); 1000. or 1.000 × 103 (4 S.F.).
7.
In both sets of rules, the lease precise number determines the number of significant figures in the final result. For multiplication/division, the number of significant figures in the result is the same as the number of significant figures in the least precise number used in the calculation. For addition/subtraction, the result has the same number of decimal places as the least precise number used in the calculation (not necessarily the number with the fewest significant figures). To perform the calculation, the addition/subtraction significant figure rule is applied to 1.5 − 1.0. The result of this is the one significant figure answer of 0.5. Even though both numbers in the calculation had two significant figures, they both showed uncertainty to the tenths place. Any addition or subtraction of these numbers can at best be known to the tenths place. Next, the multiplication/division rule is applied to 0.50/0.5. A two significant figure number divided by a one significant figure number yields an answer with one significant figure (answer = 1).
8.
The two scales have different zero points and different degree sizes. In converting from one to the other, one must account for both differences. The Fahrenheit scale has the smallest change in temperature per degree, while the Celsius and Kelvin scales have the largest change in temperature per degree.
9.
Consider gold with a density of 19.32 g/cm 3. The two possible ways to express this density as a conversion factor are: 19.32 g 1 cm 3
or
1 cm3 19.32 g
Use the first conversion factor form when converting from the volume of gold in cm 3 to the mass of gold and use the second form when converting from mass of gold to volume of gold. When using conversion factors, concentrate on the units canceling each other. 10.
Solid: a state of matter that has fixed volume and shape; a solid is rigid. Liquid: a state of matter that has definite volume but no specific shape; it assumes the shape of the container.
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3
Gas: a state of matter that has no fixed volume or shape; it takes on the shape and volume of the container. Unlike the solid and liquid state where the molecules/atoms are very close together, a gas is mostly empty space. Gases are easily compressed. Pure substance: a substance with constant composition Element: a substance that cannot be decomposed into simpler substances by chemical or physical means. Compound: a substance with constant composition that can be broken down into elements by chemical processes. Homogeneous mixture: a mixture having visibly indistinguishing parts. Heterogeneous mixture: a mixture having visibly distinguishable parts. Solution: another name for a homogeneous mixture Chemical change: a change of substances into other substances through a reorganization of the atoms; a chemical reaction. Physical change: a change in the form of a substance (solid, liquid, or gas), but not in its chemical composition: chemical bonds are not broken in a physical change.
Active Learning Questions 1.
a. 1 month ×
1 yr 365 days 24 hr 60 min = 4.38 × 104 min 12 months yr day hr
b. 1 month ×
4 wk 7 days 24 hr 60 min = 4.03 × 104 min month wk day hr
c. The 4.38 × 104 min answer is best. In each calculation, one of the conversion factors is not exact. In the first calculation, 365 days per year is fine for most months except for leap year where there are 366 days per year. In the second calculation, 4 weeks per month is an inexact conversion. There are 52 months per year, which comes out to 4.33 weeks per year. There is a larger percent error in the weeks to month conversion factor than in the days to year conversion factor. Hence, the first calculation gives the better estimate of minutes in a month. 2.
The best explanation is c. The marble sinks because it is denser that water. Statement c says that given equal volumes of water and a marble, the same volume of the marble has a greater mass (it is denser). For statement a, surface tension is the resistance of a liquid to increase its surface area. Water can support a paper clip because the denser metal is spread out over a large volume. However, once the paper clip breaks through the surface, it sinks. For statement b, the mass per unit volume of marble is greater than the water, but the overall mass of water in a beaker is more than likely larger than the mass of the marble. For statement d, surface tension can support more dense items on the surface for a certain period of time, but the real reason substances sink in water is that they are denser than water. And for statement e, if the marble
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has a greater mass and a greater volume, then the marble may be less dense than water and it would float. This is an ambiguous statement that does explain why the marble sinks. 3.
a. The mass of the sugar and water combined will be 280.0 g (statement iii). No chemical reaction takes place, so no gases are lost when the sugar dissolves. Because everything is present before mixing as compared to after mixing, mass is conserved. b. The volume should be significantly less than 200.0 mL (statement v). Dry sugar in the beaker has a lot of empty space between the sugar molecules. When the sugar is dissolved in water, each sugar molecule is surrounded by water molecules. This eliminates the empty space that was present in the dry sugar. Therefore, the volume will be significantly less than 200.0 mL. However, since students probably don’t realize the significant amount of empty space in a dry sugar, answer d is also a reasonable choice for students.
4.
a. When a gas boils, water is converted from the liquid phase to the gaseous phase. So water molecules in the vapor phase are present in the bubbles. b. A physical change is the change of the form of a substance (solid or liquid or gas), but not in its chemical composition. Chemical bonds are not broken in a physical change. The boiling of water is a physical change; liquid water is converted to gaseous water. In a chemical change, a substance is converted into a different substance by breaking and making new chemical bonds.
5.
a. Pudding takes the shape of the container, but the pudding molecules are close together so there is not a lot of empty space in the pudding. This defines a liquid. However, pouring pudding can be problematic. So, from this observation, a solid designation is also applicable. b. Sand takes the shape of the bucket and it can be poured, so a liquid designation can apply. However, sand is generally considered a solid made up of silicon and oxygen atoms bonded together in an extended structure. The structure of each sand particle is very ordered, hence the solid designation. A bucketful of sand can pour because there are significant amounts of air in between the sand particles in the bucket and this allows the sand to pour.
6.
a. There is no single correct answer. Any drawing showing 2 different compounds, but having visibly distinguishable parts is correct. For example, a beaker of water (H2O) with an insoluble compound like AgCl(s) at the bottom. b. Any drawing of an element and a compound have the same composition throughout is correct. The easiest drawings would be two different gases in a container. For example, a container with CO2(g) and N2(g) would be fine. The gases would be equally distributed throughout the container.
7.
Yes, this is consistent with the scientific method. Paracelsus is instructing his students to learn by observation. Observation of facts along with experimentation of potential remedies to heal a patient are key parts of the scientific method.
8.
Experimental results are the facts we deal with. Theories are our attempt to rationalize the facts. If the experiment is done correctly and the theory can’t account for the facts, then the theory is wrong.
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9.
If the results of the measurements are all close to the true value, then the data is accurate. But if the data are all close to the true value, then they are reproduceable (they are precise).
10.
You would need to know the average miles per gallon of the car used for the trip along with the miles that must be traveled to go from New York to Chicago. You would also need to know the average price of gasoline per gallon. Any calculation involving these three quantities is correct.
11.
Volume readings are estimated to one decimal place past the markings on the glassware. The assumed uncertainty is ±1 in the estimated digit. The piece of glassware estimated to the thousandths place would have markings to hundredths place. The piece of glassware that is estimated to the ones place would have markings to tens place. Drawings illustrating these specific markings would be correct.
12.
Each sample would displace 1.0-cm3 of water. The masses of the two samples would not be equal. One would expect the 1.0 cm3 sample of lead to have a much greater mass than the 1.0 cm3 sample of glass since lead is denser than glass. Volume and mass are two distinctly different quantities.
13
The mathematician would say that the sum equal 43.4. The scientist rounds the sum to the ones place giving an answer of 43. Assuming these are measurements, 15.4 implies the measurement is somewhere between 15.3 and 15.5, while 28 implies that the measurement is somewhere between 27 and 29. The sum of these two measurements can at best be know to the ones place. The least precise measurement always determines the significant figures in a mathematical operation. With addition (or subtraction), the least precise measurement is determined by the measuring device with the fewest decimal places.
14.
The mathematician says the result equals 430.466. The scientist says 430. is the result. Assuming these numbers are measurements, one of the measurements is known to 3 significant figures while the other is known to four significant figures. The 26.2 measurement implies a value known to ±0.1, somewhere between 26.1 and 26.3. If this value has uncertainty in the third significant figure, then any multiplication or division of this measurement can at best be known to 3 significant figures (assuming the other numbers used in the multiplication and/or division are known to at least 3 significant figures).
15.
False; this is a correct statement for the multiplication and division operations, but it is false for subtraction and addition. Here, the number of decimal places designates the least precise measurement, which dictates the number of decimal places in the final result.
16.
2 H2(g) + O2(g) → 2 H2O(g); a 2:1 mixture of H2 and O2 would be a homogeneous mixture since it contains two different species randomly dispersed in the container. When reacted, a 2:1 ratio of H2 and O2 would react exactly together to form H2O. Thus the final result would be a compound since no other species is present. A mixture must have two or more substances present.
17.
a. 636.6 g ×
1 penny = 210. pennies 3.03 g
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CHAPTER 1 b. 210. pennies × c. 140. dimes ×
CHEMICAL FOUNDATIONS
2.29 g 2 dimes = 140. dimes; 140. dimes × = 320.6 g = 321 g dime 3 pennies
4 pieces candy = 280. pieces of candy 2 dimes
We would get the same answer if we started with 210. pennies, then multiplied by 4 pieces of candy per 3 pennies. 10.23 g 280. pieces of candy × = 2864.4 g = 2860 g piece of candy d. If only the number of dimes is doubled, you could still only purchase 280. pieces of candy. You need both dimes and pennies to purchase candy. Once 280. pieces of candy has been purchased, all 210. pennies have been spent. No matter how many more dimes you have, you can’t purchase anymore candy unless the number of pennies also increases. 18.
a. Wax undergoes both a physical change and a chemical change. The flame melts wax from the solid state to the liquid state which is a physical change. The liquid wax with oxygen then reacts in the flame by a chemical change. b. The wick is undergoing a chemical change. Wicks are commonly made of braided cotton. In the presence of a flame, the cotton in the wick reacts with oxygen by chemical change. c. The glass rod generally turns black. The black color is from the carbon produced in the wax chemical reaction. The glass itself is not undergoing any change; it just provides a surface for the black carbon to collect. So, the glass undergoes neither a chemical nor a physical change.
Questions 19.
A law summarizes what happens, e.g., law of conservation of mass in a chemical reaction or the ideal gas law, PV = nRT. A theory (model) is an attempt to explain why something happens. Dalton’s atomic theory explains why mass is conserved in a chemical reaction. The kinetic molecular theory explains why pressure and volume are inversely related at constant temperature and moles of gas present, as well as explaining the other mathematical relationships summarized in PV = nRT.
20.
A dynamic process is one that is active as opposed to static. In terms of the scientific method, scientists are always performing experiments to prove or disprove a hypothesis or a law or a theory. Scientists do not stop asking questions just because a given theory seems to account satisfactorily for some aspect of natural behavior. The key to the scientific method is to continually ask questions and perform experiments. Science is an active process, not a static one.
21.
No, it is useful whenever a systematic approach of observation and hypothesis testing can be used.
22.
A random error has equal probability of being too high or too low. This type of error occurs when estimating the value of the last digit of a measurement. A systematic error is one that always occurs in the same direction, either too high or too low. For example, this type of error
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7
would occur if the balance you were using weighed all objects 0.20 g too high, that is, if the balance wasn’t calibrated correctly. A random error is an indeterminate error, whereas a systematic error is a determinate error. 23.
a. No
b. Yes
c. Yes
Only statements b and c can be determined from experiment. 24.
Accuracy: how close a measurement or series of measurements are to an accepted or true value. Precision: how close a series of measurements of the same item are to each other. The results, average = 14.91 ±0.03%, are precise (are close to each other) but are not accurate (are not close to the true value).
25.
Many techniques of chemical analysis require relatively pure samples. Thus, a separation step often is necessary to remove materials that will interfere with the analytical measurement.
26.
From Figure 1.9 of the text, a change in temperature of 180°F is equal to a change in temperature of 100°C and 100 K. A degree unit on the Fahrenheit scale is a smaller unit than a degree unit on the Celsius or Kelvin scales. Therefore, a 20° change in the Celsius or Kelvin temperature would correspond to a larger temperature change than a 20° change in the Fahrenheit scale. The 20° temperature change on the Celsius and Kelvin scales are equal to each other.
27.
They are not equivalent. Doubling the temperature on the Kelvin scale is a larger increase than doubling the temperature on the Celsius scale. For example, doubling the Celsius temperature from 25ºC to 50ºC corresponds to taking the Kelvin temperature from 298 K to 333 K. However, doubling the Kelvin temperature from 298 K to 596 K corresponds with taking the Celsius temperature from 25ºC to 323ºC. Doubling the temperature are clearly different on the two scales.
28.
When performing a multiple step calculation, always carry at least one extra significant figure in intermediate answers. If you round-off at each step, each intermediate answer gets further away from the actual value of the final answer. So to avoid round-off error, carry extra significant figures through intermediate answers, then round-off to the proper number of significant figures when the calculation is complete. In this solutions manual, we rounded off intermediate answers to the show the proper number significant figures at each step; our answers to multistep calculations will more than likely differ from yours because we are introducing round-off error into our calculations.
29.
The gas phase density is much smaller than the density of a solid or a liquid. The molecules in a solid and a liquid are very close together. In the gas phase, the molecules are very far apart from one another. In fact, the molecules are so far apart that a gas is considered to be mostly empty space. Because gases are mostly empty space, their density is very small.
30.
a. coffee; saltwater; the air we breathe (N2 + O2 + others); brass (Cu + Zn) b. book; human being; tree; desk c. sodium chloride (NaCl); water (H2O); glucose (C6H12O6); carbon dioxide (CO2) d. nitrogen (N2); oxygen (O2); copper (Cu); zinc (Zn)
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e. boiling water; freezing water; melting a popsicle; dry ice subliming f.
31.
Electrolysis of molten sodium chloride to produce sodium and chlorine gas; the explosive reaction between oxygen and hydrogen to produce water; photosynthesis, which converts H2O and CO2 into C6H12O6 and O2; the combustion of gasoline in our car to produce CO 2 and H2O.
The object that sinks has a greater density than water and the object that floats has a smaller density than water. Since both objects have the same mass, the sphere that sinks must have the smaller volume which makes it denser. Therefore, the object that floats has the larger volume along with the greater diameter.
32.
gas element (monoatomic)
solid element
liquid element
atoms/molecules far apart; random order; takes volume of container
atoms/molecules close together; somewhat ordered arrangement; takes volume of container
atoms/molecules close together; ordered arrangement; has its own volume
Exercises Significant Figures and Unit Conversions 33.
a.
exact
c. exact 34.
35.
b. inexact d. inexact (π has an infinite number of decimal places.)
a. two significant figure (S.F.). The implied uncertainty is 1000 pages. b. two S.F.
c. four S.F.
d. two S.F.
e. infinite number of S.F. (exact number)
a. 6.07 × 10 −15 ; 3 S.F.
b. 0.003840; 4 S.F.
c. 17.00; 4 S.F.
d. 8 × 108; 1 S.F.
e. 463.8052; 7 S.F.
f.
g. 301; 3 S.F.
h. 300.; 3 S.F.
300; 1 S.F.
f.
one S.F.
CHAPTER 1 36.
37.
CHEMICAL FOUNDATIONS
9
a. 100; 1 S.F.
b. 1.0 × 102; 2 S.F.
c. 1.00 × 103; 3 S.F.
d. 100.; 3 S.F.
e. 0.0048; 2 S.F.
f.
g. 4.80 × 10 −3 ; 3 S.F.
h. 4.800 × 10 −3 ; 4 S.F.
0.00480; 3 S.F.
When rounding, the last significant figure stays the same if the number after this significant figure is less than 5 and increases by one if the number is greater than or equal to 5. a. 3.42 × 10 −4
b. 1.034 × 104
c. 1.7992 × 101
d. 3.37 × 105
38.
a. 2 × 10 −4
b. 1.5 × 10 −4
c. 1.51 × 10 −4
d. 1.5051 × 10 −4
39.
Volume measurements are estimated to one place past the markings on the glassware. The first graduated cylinder is labeled to 0.2 mL volume increments, so we estimate volumes to the hundredths place. Realistically, the uncertainty in this graduated cylinder is 0.05 mL. The second cylinder, with 0.02 mL volume increments, will have an uncertainty of 0.005 mL. The approximate volume in the first graduated cylinder is 2.85 mL, and the volume in the other graduated cylinder is approximately 0.280 mL. The total volume would be: 2.85 mL +0.280 mL 3.13 mL We should report the total volume to the hundredths place because the volume from the first graduated cylinder is only read to the hundredths (read to two decimal places). The first graduated cylinder is the least precise volume measurement because the uncertainty of this instrument is in the hundredths place, while the uncertainty of the second graduated cylinder is to the thousandths place. It is always the lease precise measurement that limits the precision of a calculation.
40.
a. Volumes are always estimated to one position past the marked volume increments. The estimated volume of the first beaker is 32.7 mL, the estimated volume of the middle beaker is 33 mL, and the estimated volume in the last beaker is 32.73 mL. b. Yes, all volumes could be identical to each other because the more precise volume readings can be rounded to the other volume readings. But because the volumes are in three different measuring devices, each with its own unique uncertainty, we cannot say with certainty that all three beakers contain the same amount of water. c. 32.7 mL 33 mL 32.73 mL 98.43 mL = 98 mL The volume in the middle beaker can only be estimated to the ones place, which dictates that the sum of the volume should be reported to the ones place. As is always the case, the least precise measurement determines the precision of a calculation.
41.
For addition and/or subtraction, the result has the same number of decimal places as the number in the calculation with the fewest decimal places. When the result is rounded to the correct number of significant figures, the last significant figure stays the same if the number after this
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CHEMICAL FOUNDATIONS
significant figure is less than 5 and increases by one if the number is greater than or equal to 5. The underline shows the last significant figure in the intermediate answers. a. 53.5 + 5.612 + 6 = 65.112 = 65 b. 10.67 ‒ 9.5 ‒ 0.634 = 0.536 = 0.5 c. 3.25 × 103 + 6.174 × 102 = 32.5 × 102 + 6.174 × 102 = 38.674 × 102 = 3870 d. 1.65 × 10‒2 – 9.73 × 10‒3 = 16.5 × 10‒3 – 9.73 × 10‒3 = 6.77 × 10‒3 = 0.0068 When the exponents are different, it is easiest to apply the addition/subtraction rule when all numbers are based on the same power of 10. 42.
For multiplication and/or division, the result has the same number of significant figures as the number in the calculation with the fewest significant figures. a.
0.102 0.0821 273 = 2.2635 = 2.26 1.01
b. 0.14 × 6.022 × 1023 = 8.431 × 1022 = 8.4 × 1022; since 0.14 only has two significant figures, the result should only have two significant figures. c. 4.0 × 104 × 5.021 × 10 −3 × 7.34993 × 102 = 1.476 × 105 = 1.5 × 105 d.
43.
2.00 10 6 3.00 10
−7
= 6.6667 1012 = 6.67 1012
a. For this problem, apply the multiplication/division rule first; then apply the addition/subtraction rule to arrive at the one-decimal-place answer. We will generally round off at intermediate steps in order to show the correct number of significant figures. However, you should round off at the end of all the mathematical operations in order to avoid round-off error. The best way to do calculations is to keep track of the correct number of significant figures during intermediate steps, but round off at the end. For this problem, we underlined the last significant figure in the intermediate steps. 2.526 0.470 80.705 + + = 0.8148 + 0.7544 + 186.558 = 188.1 3 .1 0.623 0.4326
b. Here, the mathematical operation requires that we apply the addition/subtraction rule first, then apply the multiplication/division rule. 6.404 2.91 6.404 2.91 = = 12 18.7 − 17.1 1.6
c. 6.071 × 10 −5 − 8.2 × 10 −6 − 0.521 × 10 −4 = 60.71 × 10 −6 − 8.2 × 10 −6 − 52.1 × 10 −6 = 0.41 × 10 −6 = 4 × 10 −7 d.
3.8 10 −12 + 4.0 10 −13 38 10 −13 + 4.0 10 −13 42 10 −13 = = = 6.3 10 −26 4 1012 + 6.3 1013 4 1012 + 63 1012 67 1012
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CHEMICAL FOUNDATIONS
11
9.5 + 4.1 + 2.8 + 3.175 19.575 = = 4.89 = 4.9 4 4
Uncertainty appears in the first decimal place. The average of several numbers can only be as precise as the least precise number. Averages can be exceptions to the significant figure rules. f. 44.
8.925 − 8.905 0.020 100 = × 100 = 0.22 8.925 8.925
a. 6.022 × 1023 × 1.05 × 102 = 6.32 × 1025 b.
6.6262 10 −34 2.998 10 8 2.54 10 −9
= 7.82 10 −17
c. 1.285 × 10 −2 + 1.24 × 10 −3 + 1.879 × 10 −1 = 0.1285 × 10 −1 + 0.0124 × 10 −1 + 1.879 × 10 −1 = 2.020 × 10 −1 When the exponents are different, it is easiest to apply the addition/subtraction rule when all numbers are based on the same power of 10. d.
e.
f.
45.
(1.00866 − 1.00728 ) 6.02205 10
23
0.00138 6.02205 10 23
9.875 10 2 − 9.795 10 2 9.875 10
2
100 =
= 2.29 10 −27
0.080 10 2 9.875 10
2
100 = 8.1 10 −1
9.42 10 2 + 8.234 10 2 + 1.625 10 3 0.942 10 3 + 0.824 10 3 + 1.625 10 3 = 3 3 = 1.130 × 103
a. 8.43 cm ×
1m 1000 mm = 84.3 mm 100 cm m
c. 294.5 nm
1m 1 10 nm 9
f.
903.3 nm
a.
1 Tg
1m 1 10 9 nm
b. 2.41 × 102 cm ×
e. 235.3 m ×
1000 mm = 2.353 × 105 mm m
1 10 6 μm = 0.9033 μm m
1 1012 g 1 kg = 1 109 kg Tg 1000 g
b. 6.50 × 102 Tm
1m = 2.41 m 100 cm
100 cm = 2.945 10 −5 cm m
1 km = 14.45 km 1000 m
d. 1.445 × 104 m ×
46.
=
1 1012 m 1 10 9 nm = 6.50 10 23 nm Tm m
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CHAPTER 1 1g
c. 25 fg
1 10
d. 8.0 dm3 ×
47.
48.
dm 3
1 kg = 25 10 −18 kg = 2.5 10 −17 kg 1000 g
= 8.0 L (1 L = 1 dm3 = 1000 cm3 = 1000 mL)
1g 1 10 6 μg
1 1012 pg = 1 10 6 pg g
453 .6 g = 680 g; 715 g is the larger mass. 1 lb
a. 1.5 lb × b. 1.90 L
1 qt = 2.01 qt; 1.90 L is the larger volume. 0.9463 L
a. 70 in
1 yd 1m = 1.78 m; The 1.80 m person is taller. 36 in 1.094 yd
b. 49.
1 μg
1L
fg
1L 1 10 6 μL = 1 10 3 μL 1 000 mL L
e. 1 mL
f.
15
CHEMICAL FOUNDATIONS
100 km 0.6214 mi = 62 mi/hr; 65 mi/hr is the faster velocity. hr km
Conversion factors are found in Appendix 6. In general, the number of significant figures we use in the conversion factors will be one more than the number of significant figures from the numbers given in the problem. This is usually sufficient to avoid round-off error. 3.91 kg ×
16 oz 1 lb = 8.62 lb; 0.62 lb × = 9.9 oz lb 0.4536 kg
Baby’s weight = 8 lb and 9.9 oz or, to the nearest ounce, 8 lb and 10. oz. 51.4 cm × 50.
1 in = 20.2 in 20 1/4 in = baby’s height 2.54 cm
V = 1 × w × h = 11.0 in × 8.5 in × 1.75 in = 164 in3 (Carrying an extra significant figure.) 3
1L 2.54 cm 3 3 164 in3 × = 2700 cm ; 2700 cm 1000 cm 3 = 2.7 L in
51.
a. 908 oz ×
1 lb 0.4536 kg = 25.7 kg 16 oz lb
b. 12.8 L ×
1 qt 1 gal = 3.38 gal 0.9463 L 4 qt
c. 125 mL ×
1L 1 qt = 0.132 qt 1000 mL 0.9463 L
CHAPTER 1 52.
CHEMICAL FOUNDATIONS
a. 3.50 gal ×
53.
4 qt 1L 1000 mL = 1.32 × 104 mL 1 gal 1 .057 qt 1L
453.6 g 1 kg = 88.5 kg 1 lb 1000 g
b. 195 lb ×
c.
13
2.998 108 m 1 km 0.62137 mi 60 s 60 min = 6.706 × 108 mi/hr s 1000 m km min hr
a. 1.25 mi ×
8 furlongs 40 rods = 10.0 furlongs; 10.0 furlongs × = 4.00 × 102 rods mi furlong
4.00 × 102 rods × 2.01 × 103 m ×
5.5 yd 36 in 2.54 cm 1m = 2.01 × 103 m rod yd in 100 cm
1 km = 2.01 km 1000 m
b. Let's assume we know this distance to ±1 yard. First, convert 26 miles to yards. 26 mi ×
5280 ft 1 yd = 45,760. yd mi 3 ft
26 mi + 385 yd = 45,760. yd + 385 yd = 46,145 yards 46,145 yard ×
1 furlong 1 rod = 8390.0 rods; 8390.0 rods × = 209.75 furlongs 5.5 yd 40 rods
46,145 yard ×
1 km 36 in 2.54 cm 1m = 42,195 m; 42,195 m × = 42.195 km 1000 m yd in 100 cm 2
54.
a. 1 ha ×
10,000 m 2 1 km = 1 10 −2 km 2 ha 1000 m 2
b. 5.5 acre ×
160 rod 2 5.5 yd 36 in 2.54 cm 1m = 2.2 × 104 m2 acre yd in 100 cm rod 2
1 km = 0.022 km2 2.2 × 10 m × = 2.2 ha; 2.2 × 10 m × 4 2 1 10 m 1000 m 4
2
1 ha
4
2
c. Area of lot = 120 ft × 75 ft = 9.0 × 10 3 ft2 2
1 yd 1 rod 1 acre $6,500 $31,000 = 9.0 × 10 ft × = 0.21 acre; 2 0.21 acre acre 5.5 yd 160 rod 3 ft 3
2
We can use our result from (b) to get the conversion factor between acres and hectares (5.5 acre = 2.2 ha.). Thus 1 ha = 2.5 acre.
14
CHAPTER 1
1 ha $6,500 $77,000 = = 0.084 ha; the price is: 2.5 acre 0.084 ha ha
0.21 acre ×
55.
a. 1 troy lb ×
12 troy oz 20 pw 24 grains 0.0648 g 1 kg = 0.373 kg troy lb troy oz pw grain 1000 g
1 troy lb = 0.373 kg ×
b. 1 troy oz ×
2.205 lb = 0.822 lb kg
20 pw 24 grains 0.0648 g = 31.1 g troy oz pw grain
1 troy oz = 31.1 g ×
1 carat = 156 carats 0.200 g
c. 1 troy lb = 0.373 kg; 0.373 kg ×
56.
CHEMICAL FOUNDATIONS
a. 1 grain ap ×
1000 g 1 cm3 = 19.3 cm3 kg 19.3 g
1 scruple 1 dram ap 3.888 g = 0.06480 g 20 grain ap 3 scruples dram ap
From the previous question, we are given that 1 grain troy = 0.0648 g = 1 grain ap. So the two are the same. b. 1 oz ap ×
8 dram ap 3.888 g 1 oz troy * = 1.00 oz troy; *see Exercise 49b. oz ap dram ap 31.1 g
c. 5.00 × 102 mg × 0.386 scruple ×
d. 1 scruple ×
1g 1 dram ap 3 scruples = 0.386 scruple 1000 mg 3.888 g dram ap 20 grains ap = 7.72 grains ap scruple
1 dram ap 3.888 g = 1.296 g 3 scruples dram ap
1 capsule = 24 capsules 0.65 g
57.
15.6 g ×
58.
1.5 teaspoons ×
80. mg acet = 240 mg acetaminophen 0.50 teaspoon
240 mg acet 1 lb = 22 mg acetaminophen/kg 24 lb 0.454 kg 240 mg acet 1 lb = 15 mg acetaminophen/kg 35 lb 0.454 kg
The range is from 15 to 22 mg acetaminophen per kg of body weight.
CHAPTER 1 59.
CHEMICAL FOUNDATIONS
15
3.00 10 8 m 1.094 yd 60 s 60 min 1 knot warp 1.71 = 5.00 s m min h 2030 yd/h = 2.91 × 109 knots 8 1 mi 60 s 60 min 5.00 3.00 10 m 1 km s 1.609 km min h 1000 m
60.
= 3.36 × 109 mi/h
100 . m 100 . m 1 km 60 s 60 min = 10.4 m/s; = 37.6 km/h 9.58 s 1000 m 9.58 s min h 100 . m 1.0936 yd 3 ft 34.2 ft 1 mi 60 s 60 min = 34.2 ft/s; = 23.3 mi/h 9.58 s s 5280 ft m min yd h
1.00 × 102 yd ×
61.
1s
1m 9.58 s = 8.76 s 1.0936 yd 100 . m
1 min 1h 65 mi 5280 ft = 95.3 ft = 100 ft 60 s 60 min h mi
If you take your eyes off the road for one second traveling at 65 mph, your car travels approximately 100 feet. 62.
63.
112 km ×
0.6214 mi 1h = 1.1 h = 1 h and 6 min km 65 mi
112 km ×
0.6214 mi 1 gal 3.785 L = 9.4 L of gasoline km 28 mi gal
180 lb ×
1 kg 8.0 mg = 650 mg antibiotic/dose 2.205 lb kg
2 wk × 64.
7 days 650 mg 2 doses = 18,000 mg = 18 g antibiotic in total wk day dose
For the gasoline car: 500. mi
1 gal $3.50 28.0 mi gal
= $62.5
For the E85 car: 500. mi
1 gal $2.85 = $63.3 22.5 mi gal
The E85 vehicle would cost slightly more to drive 500. miles as compared to the gasoline vehicle ($63.3 versus $62.5).
16
CHAPTER 1
CHEMICAL FOUNDATIONS
2
65.
5280 ft Volume of lake = 100 mi × × 20 ft = 6 × 1010 ft3 mi 2
3
2.54 cm 1 mL 0.4 μg 12 in 6 × 10 ft × = 7 × 1014 μg mercury in mL cm3 ft 10
3
7 × 1014 μg
66.
1g 1 10 μg 6
1 kg 1 103 g
= 7 × 105 kg of mercury
Volume of room = 18 ft × 12 ft × 8 ft = 1700 ft3 (carrying one extra significant figure) 3
3
3
1m 12 in 2.54 cm = 48 m 3 1700 ft 100 cm ft in 3
48 m3
400,000 μg CO m
3
1 g CO 1 10 6 μg CO
= 19 g = 20 g CO (to 1 sig. fig.)
Temperature 67.
TC = TK ‒ 273 = 292 ‒ 273 = 19°C; TC =
5 5 (TF − 32) = (72°F − 32) = 22.2°C = 22°C 9 9
292 K is the coldest temperature, then 72°F, then 24°C is the hottest temperature. 68.
340. – 273 = 67°C, 350. – 273 = 77°C; the 340. to 350. K temperature range is 67°C to 77°C. TF =
9 9 9 × TC + 32 = × 67°C + 32 = 153°F = 150 °F, TF = × 77°C + 32 = 171°F = 170 °F 5 5 5
The 340. to 350. K temperature range is 150°F to 170°F. 69.
70.
a. TC =
5 5 (TF − 32) = (−459°F − 32) = −273°C; TK = TC + 273 = −273°C + 273 = 0 K 9 9
b. TC =
5 (−40.°F − 32) = −40.°C; TK = −40.°C + 273 = 233 K 9
c. TC =
5 (68°F − 32) = 20.°C; TK = 20.°C + 273 = 293 K 9
d. TC =
5 (7 × 107°F − 32) = 4 × 107°C; TK = 4 × 107°C + 273 = 4 × 107 K 9
96.1°F ±0.2°F; first, convert 96.1°F to °C. TC =
5 5 (TF − 32) = (96.1 − 32) = 35.6°C 9 9
A change in temperature of 9°F is equal to a change in temperature of 5°C. The uncertainty is: ±0.2°F ×
5 C = ±0.1°C. Thus 96.1 ±0.2°F = 35.6 ±0.1°C. 9 F
CHAPTER 1
CHEMICAL FOUNDATIONS
71.
9 9 × TC + 32 = × 39.2°C + 32 = 102.6°F 5 5
a. TF =
17 (Note: 32 is exact.)
TK = TC + 273.2 = 39.2 + 273.2 = 312.4 K
72.
b. TF =
9 × (−25) + 32 = −13°F; TK = −25 + 273 = 248 K 5
c.
TF =
9 × (−273) + 32 = −459°F; TK = −273 + 273 = 0 K 5
d. TF =
9 × 801 + 32 = 1470°F; TK = 801 + 273 = 1074 K 5
a. TC = TK − 273 = 233 − 273 = -40.°C TF =
9 9 × TC + 32 = × (−40.) + 32 = −40.°F 5 5
b. TC = 4 − 273 = −269°C; TF =
9 × (−269) + 32 = −452°F 5
c. TC = 298 − 273 = 25°C; TF =
9 × 25 + 32 = 77°F 5
d. TC = 3680 − 273 = 3410°C; TF = 73.
TF =
9 × 3410 + 32 = 6170°F 5
9 × TC + 32; from the problem, we want the temperature where T F = 2TC. 5
Substituting: 2TC =
32 9 × TC + 32, (0.2)TC = 32, TC = = 160C 0 .2 5
TF = 2TC when the temperature in Fahrenheit is 2(160) = 320F. Because all numbers when solving the equation are exact numbers, the calculated temperatures are also exact numbers. 74.
TC =
5 5 (TF – 32) = (72 – 32) = 22C; TC = TK – 273 = 313 – 273 = 40.C 9 9
The difference in temperature between Jupiter at 313 K and Earth at 72F is 40.C – 22 C = 18C. 75.
a. A change in temperature of 140C is equal to 50X. Therefore,
140 o C
is the unit con50 o X version between a degree on the X scale to a degree on the Celsius scale. To account for the different zero points, −10 must be subtracted from the temperature on the X scale to get to the Celsius scale. The conversion between X to C is:
18
CHAPTER 1 TC = TX
140 o C
− 10C, TC = TX
50 o X
14 o C 5o X
CHEMICAL FOUNDATIONS
− 10C
The conversion between C to X would be: 5o X
TX = (TC + 10C)
b. Assuming 10C and
14 o C 5o X
14 o C
TX = (22.0C + 10C)
are exact numbers: 5o X 14 o C
= 11.4X
c. Assuming exact numbers in the temperature conversion formulas: TC = 58.0X
14 o C 5o X
− 10C = 152C
TK = 152C + 273 = 425 K TF =
76.
9o F 5o C
a.
152C + 32F = 306F
100oA
115oC
100oA
160oC
0oA
-45oC
A change in temperature of 160°C equals a change in temperature of 100°A. 160 C is our unit conversion for a 100 A degree change in temperature.
So
At the freezing point: 0°A = −45°C
Combining these two pieces of information: TA = (TC + 45°C) × b. TC = (TF − 32) × TF − 32 =
100 A 5A 8 C = (TC + 45°C) × or TC = TA × − 45°C 160 C 5A 8 C
8 5 5 ; TC = TA × − 45 = (TF − 32) × 9 5 9
72 F 9 72 8 TA − 45 = TA − 81, TF = TA × − 49F 25 A 5 25 5
3TC 8 8 − 45 and TC = TA; so TC = TC × − 45, = 45, TC = 75°C = 75°A 5 5 5 8 C 72 F d. TC = 86°A × − 45°C = 93°C; TF = 86°A × − 49°F = 199°F = 2.0 × 102°F 25 A 5A
c. TC = TA ×
e. TA = (45°C + 45°C) ×
5A = 56°A 8 C
CHAPTER 1
CHEMICAL FOUNDATIONS
19
Density mass 156 g = = 13.6 g/cm 3 ; from the table, the toxic liquid is mercury. volume 11.5 cm 3
77.
Density =
78.
The least dense substance (hydrogen) would occupy the largest volume. This volume would be: 1 cm3 = 1.8 × 105 cm3 15 g 0.000084 g The most-dense substance, gold, would occupy the smallest volume (15 g of gold occupies 0.78 cm3). 3
79.
2.54 cm 453 .6 g Mass = 350 lb = 1.6 × 105 g; V = 1.2 × 104 in3 = 2.0 × 105 cm3 lb in mass 1 10 5 g = = 0.80 g/cm3 Density = 5 3 volume 2.0 10 cm Because the material has a density less than water, it will float in water.
80.
Let d = density; dcube =
140.4 g = 5.20 g/cm3 (3.00 cm) 3
If this is correct to ±1.00% then the density is 5.20 ±0.05 g/cm3 Vsphere = (4/3)r3 = (4/3)(1.42 cm)3 = 12.0 cm3 dsphere =
61.6 g = 5.13 g/cm3 = 5.13 ±0.05 g/cm3 12.0 cm 3
Since dcube is between 5.15 and 5.25 g/cm3 and dsphere is between 5.08 and 5.18 g/cm3, the error limits overlap and we can’t decisively determine if they are built of the same material. The data are not precise enough to determine. 81.
V=
Density = 82.
3
4 3 4 1000 m 100 cm 33 3 r = 3.14 7.0 10 5 km = 1.4 10 cm 3 3 km m
mass = volume
V = l × w × h = 2.9 cm × 3.5 cm × 10.0 cm = 1.0 × 10 2 cm3 d = density =
83.
1000 g kg = 1.4 × 106 g/cm3 = 1 × 106 g/cm3 33 1.4 10 cm3
2 1036 kg
615 .0 g 1.0 10 cm
a. 5.0 carat
2
3
=
6.2 g cm3
0.200 g 1 cm3 = 0.28 cm3 carat 3.51 g
20
CHAPTER 1
CHEMICAL FOUNDATIONS
1 cm3 3.51 g 1 carat = 49 carats 3 mL 0.200 g cm
b. 2.8 mL
1 mL 3.12 g = 390. g Br2; 85.0 g = 27.2 mL Br2 3 3.12 g cm
84.
1 mL = 1 cm3; 125 cm3
85.
V = 21.6 mL − 12.7 mL = 8.9 mL; density =
86.
5.25 g
1 cm 3 10.5 g
33.42 g = 3.8 g/mL = 3.8 g/cm3 8.9 mL
= 0.500 cm3 = 0.500 mL
The volume in the cylinder will rise to 11.7 mL (11.2 mL + 0.500 mL = 11.7 mL). 87.
a. Both have the same mass of 1.0 kg. b. 1.0 mL of mercury; mercury is more dense than water. Note: 1 mL = 1 cm3. 0.998 g 13.6 g = 14 g of mercury; 1.0 mL = 1.0 g of water mL mL
1.0 mL
c. Same; both represent 19.3 g of substance. 19.3 mL
0.9982 g 19.32 g = 19.3 g of water; 1.00 mL = 19.3 g of gold mL mL
d. 1.0 L of benzene (880 g versus 670 g) 75 mL
88.
a. 1.50 qt
b. 3.5 in 89.
1000 mL 0.880 g 8.96 g = 670 g of copper; 1.0 L = 880 g of benzene L mL mL
1L 1000 mL 0.789 g = 1120 g ethanol 1.0567 qt L mL
3
3
2.54 cm 13.6 g = 780 g mercury cm3 in
a. 1.0 kg feather; feathers are less dense than lead. b. 100 g water; water is less dense than gold.
90.
c. Same; both volumes are 1.0 L.
a. H2(g): V = 25.0 g ×
1 cm3 = 3.0 × 105 cm3 [H2(g) = hydrogen gas.] 0.000084 g
b. H2O(l): V = 25.0 g ×
1 cm3 = 25.0 cm3 [H2O(l) = water.] 0.9982 g
c. Fe(s): V = 25.0 g ×
1 cm3 = 3.18 cm3 [Fe(s) = iron.] 7.87 g
CHAPTER 1
CHEMICAL FOUNDATIONS
21
Notice the huge volume of the gaseous H2 sample as compared to the liquid and solid samples. The same mass of gas occupies a volume that is over 10,000 times larger than the liquid sample. Gases are indeed mostly empty space. 91.
V = 1.00 × 103 g ×
1 cm 3 = 44.3 cm3 22.57 g
44.3 cm3 = 1 × w × h = 4.00 cm × 4.00 cm × h, h = 2.77 cm 92.
V = 22 g ×
1 cm3 = 2.5 cm3; V = πr2 × l, where l = length of the wire 8.96 g 2
1 cm 0.25 mm × l, l = 5.1 × 103 cm = 170 ft 2.5 cm = π × × 10 mm 2 2
3
Classification and Separation of Matter 93.
A gas has molecules that are very far apart from each other, whereas a solid or liquid has molecules that are very close together. An element has the same type of atom, whereas a compound contains two or more different elements. Picture i represents an element that exists as two atoms bonded together (like H 2 or O2 or N2). Picture iv represents a compound (like CO, NO, or HF). Pictures iii and iv contain representations of elements that exist as individual atoms (like Ar, Ne, or He). a. Picture iv represents a gaseous compound. Note that pictures ii and iii also contain a gaseous compound, but they also both have a gaseous element present. b. Picture vi represents a mixture of two gaseous elements. c. Picture v represents a solid element. d. Pictures ii and iii both represent a mixture of a gaseous element and a gaseous compound.
94.
2 compounds
95.
compound and element (diatomic)
Homogeneous: Having visibly indistinguishable parts (the same throughout). Heterogeneous: Having visibly distinguishable parts (not uniform throughout).
22
CHAPTER 1
CHEMICAL FOUNDATIONS
a. heterogeneous (due to hinges, handles, locks, etc.) b. homogeneous (hopefully; if you live in a heavily polluted area, air may be heterogeneous.)
96.
c. homogeneous
d. homogeneous (hopefully, if not polluted)
e. heterogeneous
f.
a. heterogeneous
b. homogeneous
c. heterogeneous
d. homogeneous (assuming no imperfections in the glass)
heterogeneous
e. heterogeneous (has visibly distinguishable parts) 97.
a.
pure
b.
mixture
f.
pure
g. mixture
c.
mixture
d.
pure
h. mixture
i.
mixture
e. mixture (copper and zinc)
Iron and uranium are elements. Water (H2O) is a compound because it is made up of two or more different elements. Table salt is usually a homogeneous mixture composed mostly of sodium chloride (NaCl), but will usually contain other substances that help absorb water vapor (an anticaking agent). 98.
Initially, a mixture is present. The magnesium and sulfur have only been placed together in the same container at this point, but no reaction has occurred. When heated, a reaction occurs. Assuming the magnesium and sulfur had been measured out in exactly the correct ratio for complete reaction, the remains after heating would be a pure compound composed of magnesium and sulfur. However, if there were an excess of either magnesium or sulfur, the remains after reaction would be a mixture of the compound produced and the excess reactant.
99.
Chalk is a compound because it loses mass when heated and appears to change into another substance with different physical properties (the hard chalk turns into a crumbly substance).
100.
Because vaporized water is still the same substance as solid water (H2O), no chemical reaction has occurred. Sublimation is a physical change.
101.
A physical change is a change in the state of a substance (solid, liquid, and gas are the three states of matter); a physical change does not change the chemical composition of the substance. A chemical change is a change in which a given substance is converted into another substance having a different formula (composition). a. Vaporization refers to a liquid converting to a gas, so this is a physical change. The formula (composition) of the moth ball does not change. b. This is a chemical change since hydrofluoric acid (HF) is reacting with glass (SiO 2) to form new compounds that wash away.
CHAPTER 1
CHEMICAL FOUNDATIONS
23
c. This is a physical change because all that is happening during the boiling process is the conversion of liquid alcohol to gaseous alcohol. The alcohol formula (C 2H5OH) does not change. d. This is a chemical change since the acid is reacting with cotton to form new compounds. 102.
a. Distillation separates components of a mixture, so the orange liquid is a mixture (has an average color of the yellow liquid and the red solid). Distillation utilizes boiling point differences to separate out the components of a mixture. Distillation is a physical change because the components of the mixture do not become different compounds or elements. b. Decomposition is a type of chemical reaction. The crystalline solid is a compound, and decomposition is a chemical change where new substances are formed. c. Tea is a mixture of tea compounds dissolved in water. The process of mixing sugar into tea is a physical change. Sugar doesn’t react with the tea compounds, it just makes the solution sweeter.
ChemWork Problems 103.
The national debt is $2.0875 × 1013 and the wealthiest 1.00% represents 3.32 × 106 people.
2.0875 1013 dollars = 6.29 × 106 dollars/person= $6.29 million dollars per richest 1%. 3.32 106 people 104.
422 mg caffeine
105.
4145 mi ×
1g 6.02 10 23 molecules = 1.31 1021 caffeine molecules 1000 mg 194 g caffeine
5280 ft 1 fathom 1 cable length = 3.648 × 104 cable lengths mi 6 ft 100 fathom
1 km 1000 m = 6.671 × 106 m 0.62137 mi km 1 nautical mile 3.648 × 104 cable lengths × = 3,648 nautical miles 10 cable lengths
4145 mi ×
106. 107.
1000 m 1.25 mi 1 km = 16.9 m/s 119.2 s 0.6214 mi 1 km
Because each pill is 4.0% Lipitor by mass, for every 100.0 g of pills, there are 4.0 g of Lipitor present. Note that 100 pills is assumed to be an exact number. 100 pills
108.
126 gal ×
2.5 g 4.0 g Lipitor 1 kg = 0.010 kg Lipitor pill 100 .0 g pills 1000 g
4 qt 1L = 477 L gal 1.057 qt
24
CHAPTER 1
CHEMICAL FOUNDATIONS
5 5 (TF − 32) = (134°F − 32) = 56.7°C; phosphorus would be a liquid. 9 9
109.
TC =
110.
At 200.0°F: TC =
5 (200.0°F − 32F) = 93.33°C; TK = 93.33 + 273.15 = 366.48 K 9
At −100.0°F: TC =
5 (−100.0°F − 32°F) = −73.33°C; TK = −73.33°C + 273.15 = 199.82 K 9
T(°C) = [93.33°C − (−73.33°C)] = 166.66°C; T(K) = (366.48 K −199.82 K) = 166.66 K The “300 Club” name only works for the Fahrenheit scale; it does not hold true for the Celsius and Kelvin scales. 111.
100 cm 100 cm Total volume = 200 . m 300 . m × 4.0 cm = 2.4 × 109 cm3 m m
Volume of topsoil covered by 1 bag = 2 2 2.54 cm 12 in 2.54 cm 10. ft 2 1.0 in = 2.4 × 104 cm3 in ft in
2.4 × 109 cm3
112.
1 bag 2.4 10 4 cm3
= 1.0 × 105 bags topsoil
a. No; if the volumes were the same, then the gold idol would have a much greater mass because gold is much more dense than sand. b. Mass = 1.0 L
1 000 cm3 1 9.32 g 1 kg = 19.32 kg (= 42.59 lb) 3 L 1000 g cm
It wouldn't be easy to play catch with the idol because it would have a mass of over 40 pounds. 113.
1 light year = 1 yr 9.6 parsecs
365 day 24 h 60 min 60 s 186,000 mi = 5.87 × 1012 miles yr day h min s
3.26 light yr 5.87 1012 mi 1.609 km 1000 m = 3.0 × 1017 m parsec light yr mi km
mass 0.384 g = = 1.2 g/cm3; from the table, the other ingredient is caffeine. volume 0.32 cm3
114.
Density =
115.
a. 0.25 lb
453 .6 g 1.0 g trytophan = 1.1 g tryptophan lb 100 .0 g turkey
b. 0.25 qt
0.9463 L 1.04 kg 1000 kg 2.0 g tryptophan = 4.9 g tryptophan qt L kg 100 .0 g milk
CHAPTER 1 116.
CHEMICAL FOUNDATIONS
25
A chemical change involves the change of one or more substances into other substances through a reorganization of the atoms. A physical change involves the change in the form of a substance, but not its chemical composition. a. physical change (Just smaller pieces of the same substance.) b. chemical change (Chemical reactions occur.) c. chemical change (Bonds are broken.) d. chemical change (Bonds are broken.) e. physical change (Water is changed from a liquid to a gas.) f.
117.
physical change (Chemical composition does not change.)
a. False; sugar is generally considered to be the pure compound sucrose, C 12H22O11. b. False; elements and compounds are pure substances. c. True; air is a mixture of mostly nitrogen and oxygen gases. d. False; gasoline has many additives, so it is a mixture. e. True; compounds are broken down to elements by chemical change.
118.
The rusting of iron is the only change listed where chemical formulas change, so it is the only chemical change. The others are physical properties. Note that the red glow of a platinum wire assumes no reaction between platinum and oxygen; the red glow is just hot Pt.
119.
5.4 L blood
120.
a. For
250 mg cholestero l 1g 1000 mL = 13.5 g = 14 g cholesterol L 100.0 mL blood 1000 mg
103 1 104 102 : maximum = = 1.04; minimum = = 1.00 100 102 101 1 103 1 So: = 1.02 ± 0.02 101 1
b. For So:
c. For So:
102 100 101 1 : maximum = = 1.04; minimum = = 1.00 100 98 99 1
101 1 = 1.02 ± 0.02 99 1 100 98 99 1 : maximum = = 1.00; minimum = = 0.96 100 102 101 1
99 1 = 0.98 ± 0.02 101 1
26
CHAPTER 1
CHEMICAL FOUNDATIONS
Considering the error limits, answers to parts a and b should be expressed to three significant figures while the part c answer should be expressed to two significant figures. Using the multiplication/division rule leads to a different result in part b; according to the rule, the part b answer should be to two significant figures. If this is the case, then the answer to part b is 1.0, which implies the answer to the calculation is somewhere between 0.95 and 1.05. The actual error limit to the answer is better than this, so we should use the more precise way of expressing the answer. The significant figure rules give general guidelines for estimating uncertainty; there are exceptions to the rules. 121.
18.5 cm ×
10.0 o F = 35.2F increase; Tfinal = 98.6 + 35.2 = 133.8F 5.25 cm
TC = 5/9 (133.8 – 32) = 56.56C 122.
Massbenzene = 58.80 g − 25.00 g = 33.80 g; Vbenzene = 33.80 g Vsolid = 50.0 cm3 − 38.4 cm3 = 11.6 cm3; density =
123.
124.
25.00 g = 2.16 g/cm3 11.6 cm3
1 10 −12 m 4 4 100 cm V = π r3 = 3.14 69 pm 3 3 pm m
Density =
3
= 1.4 10 − 24 cm 3
3.35 10 −23 g mass = 24 g/cm3 = volume 1.4 10 − 24 cm 3
22,610 kg 1000 g 1 m3 = 22.61 g/cm3 3 m kg 1 106 cm3 Volume of block = 10.0 cm 8.0 cm 9.0 cm = 720 cm3;
125.
1 cm 3 = 38.4 cm3 0.880 g
22.61 g × 720 cm3 = 1.6 × 104 g cm3
a. Volume × density = mass; the orange block is more dense. Because mass (orange) > mass (blue) and because volume (orange) < volume (blue), the density of the orange block must be greater to account for the larger mass of the orange block. b. Which block is more dense cannot be determined. Because mass (orange) > mass (blue) and because volume (orange) > volume (blue), the density of the orange block may or may not be larger than the blue block. If the blue block is more dense, its density cannot be so large that its mass is larger than the orange block’s mass. c. The blue block is more dense. Because mass (blue) = mass (orange) and because volume (blue) < volume (orange), the density of the blue block must be larger in order to equate the masses. d. The blue block is denser. Because mass (blue) > mass (orange) and because the volumes are equal, the density of the blue block must be larger in order to give the blue block the larger mass.
CHAPTER 1
CHEMICAL FOUNDATIONS
27 3
126.
4π r 3 4π c c3 = = Circumference = c = 2πr; V = 3 3 2π 6π 2
Largest density =
5.25 oz (9.00 in )
6π 2 5.00 oz Smallest density = (9.25 in ) 3 6π 2
Maximum range is:
5.25 oz
=
3
12.3 in
=
5.00 oz 13.4 in
(0.373 − 0.427 ) oz in
0.427 oz
=
3
3
in 3
=
0.73 oz in 3
or 0.40 ±0.03 oz/in3
3
Uncertainty is in 2nd decimal place. 127.
V = Vfinal − Vinitial; d = dmax =
dmax =
28.90 g 9.8 cm − 6.4 cm 3
3
=
28.90 g 3
= 8.5 g/cm3
3.4 cm
mass max ; we get Vmin from 9.7 cm3 − 6.5 cm3 = 3.2 cm3. Vmin
28.93 g 3
3.2 cm
=
9.0 g 3
cm
; dmin =
mass min 28.87 g 8.0 g = = 3 3 Vmax 9.9 cm − 6.3 cm cm3
The density is 8.5 ±0.5 g/cm3. 128.
We need to calculate the maximum and minimum values of the density, given the uncertainty in each measurement. The maximum value is: dmax =
19.625 g + 0.002 g 25.00 cm − 0.03 cm 3
3
=
19.627 g 24.97 cm3
= 0.7860 g/cm3
The minimum value of the density is: dmin =
19.625 g − 0.002 g 25.00 cm + 0.03 cm 3
3
=
19.623 g 25.03 cm3
= 0.7840 g/cm3
The density of the liquid is between 0.7840 and 0.7860 g/cm 3. These measurements are sufficiently precise to distinguish between ethanol (d = 0.789 g/cm 3) and isopropyl alcohol (d = 0.785 g/cm3).
Challenge Problems 129.
In a subtraction, the result gets smaller, but the uncertainties add. If the two numbers are very close together, the uncertainty may be larger than the result. For example, let’s assume we want to take the difference of the following two measured quantities, 999,999 ±2 and 999,996 ±2. The difference is 3 ±4. Because of the uncertainty, subtracting two similar numbers is poor practice.
28 130.
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CHEMICAL FOUNDATIONS
In general, glassware is estimated to one place past the markings. a.
128.7 mL glassware
b.
18 mL glassware
130
c. 23.45 mL glassware 24
30
129
20 128 127
10
read to tenth’s place
read to one’s place
23
read to two decimal places
Total volume = 128.7 + 18 + 23.45 = 170.15 = 170. (Due to 18, the sum would be known only to the ones place.) 131.
132.
a.
2.70 − 2.64 × 100 = 2% 2.70
c.
1.000 − 0.9981 0.002 × 100 = × 100 = 0.2% 1.000 1.000
b.
| 16.12 − 16.48 | × 100 = 2.2% 16.12
a. At some point in 1982, the composition of the metal used in minting pennies was changed because the mass changed during this year (assuming the volume of the pennies were constant). b. It should be expressed as 3.08 ±0.05 g. The uncertainty in the second decimal place will swamp any effect of the next decimal places.
133.
Heavy pennies (old): mean mass = 3.08 ±0.05 g Light pennies (new): mean mass =
( 2.467 + 2.545 + 2.518 ) = 2.51 ±0.04 g 3
Because we are assuming that volume is additive, let’s calculate the volume of 100.0 g of each type of penny, then calculate the density of the alloy. For 100.0 g of the old pennies, 95 g will be Cu (copper) and 5 g will be Zn (zinc). V = 95 g Cu ×
1 cm3 1 cm3 + 5 g Zn = 11.3 cm3 (carrying one extra sig. fig.) 8.96 g 7.14 g
Density of old pennies =
100 . g = 8.8 g/cm3 11.3 cm3
For 100.0 g of new pennies, 97.6 g will be Zn and 2.4 g will be Cu. V = 2.4 g Cu ×
1 cm3 1 cm3 + 97.6 g Zn × = 13.94 cm3 (carrying one extra sig. fig.) 8.96 g 7.14 g
CHAPTER 1
CHEMICAL FOUNDATIONS
Density of new pennies = d=
100. g 13.94 cm 3
29
= 7.17 g/cm3
mass ; because the volume of both types of pennies are assumed equal, then: volume 3
7.17 g / cm d new mass new = 0.81 = = 3 d old mass old 8.8 g / cm
The calculated average mass ratio is:
mass new 2.51 g = = 0.815 mass old 3.08 g
To the first two decimal places, the ratios are the same. If the assumptions are correct, then we can reasonably conclude that the difference in mass is accounted for by the difference in alloy used. 134.
a. At 8 a.m., approximately 57 cars pass through the intersection per hour. b. At 12 a.m. (midnight), only 1 or 2 cars pass through the intersection per hour. c. Traffic at the intersection is limited to less than 10 cars per hour from 8 p.m. to 5 a.m. Starting at 6 a.m., there is a steady increase in traffic through the intersection, peaking at 8 a.m. when approximately 57 cars pass per hour. Past 8 a.m. traffic moderates to about 40 cars through the intersection per hour until noon, and then decreases to 21 cars per hour by 3 p.m. Past 3 p.m. traffic steadily increases to a peak of 52 cars per hour at 5 p.m., and then steadily decreases to the overnight level of less than 10 cars through the intersection per hour. d. The traffic pattern through the intersection is directly related to the work schedules of the general population as well as to the store hours of the businesses in downtown. e. Run the same experiment on a Sunday, when most of the general population doesn’t work and when a significant number of downtown stores are closed in the morning.
135.
Let x = mass of copper and y = mass of silver. x y + 105.0 g = x + y and 10.12 mL = ; solving and carrying 1 extra sig. fig.: 8.96 10.5 x 105 .0 − x + 10.12 = × 8.96 × 10.5, 952.1 = (10.5)x + 940.8 − (8.96)x 8.96 10.5 7 .3 g 11.3 = (1.54)x, x = 7.3 g; mass % Cu = × 100 = 7.0% Cu 105 .0 g
136.
Straight line equation: y = mx + b, where m is the slope of the line and b is the y-intercept. For the TF vs. TC plot: TF = (9/5)TC + 32 y= m x + b The slope of the plot is 1.8 (= 9/5) and the y-intercept is 32°F.
30
CHAPTER 1
CHEMICAL FOUNDATIONS
For the TC vs. TK plot: TC = TK − 273 y= mx + b The slope of the plot is 1, and the y-intercept is −273°C. 137.
a. One possibility is that rope B is not attached to anything and rope A and rope C are connected via a pair of pulleys and/or gears. b. Try to pull rope B out of the box. Measure the distance moved by C for a given movement of A. Hold either A or C firmly while pulling on the other rope.
138.
The bubbles of gas is air in the sand that is escaping; methanol and sand are not reacting. We will assume that the mass of trapped air is insignificant. Mass of dry sand = 37.3488 g − 22.8317 g = 14.5171 g Mass of methanol = 45.2613 g − 37.3488 g = 7.9125 g Volume of sand particles (air absent) = volume of sand and methanol − volume of methanol Volume of sand particles (air absent) = 17.6 mL − 10.00 mL = 7.6 mL Density of dry sand (air present) = Density of methanol =
14.5171 g = 1.45 g/mL 10.0 mL
7.9125 g = 0.7913 g/mL 10.00 mL
Density of sand particles (air absent) =
14.5171 g = 1.9 g/mL 7.6 mL
CHAPTER 2 ATOMS, MOLECULES, AND IONS Review Questions 1.
a. Atoms have specific masses and are neither created nor destroyed by chemical reactions. Because atoms are conserved in a chemical reaction, mass cannot change in a chemical reaction. Mass is conserved. b. The composition of a substance depends on the number and kinds of atoms that form it. A certain compound always has the same number and kinds of atoms in its formula. c. Compounds of the same elements differ only in the numbers of atoms of the elements forming them, i.e., NO, N2O, NO2.
2.
Deflection of cathode rays by magnetic and electric fields led to the conclusion that cathode rays were negatively charged. The cathode ray was produced at the negative electrode and repelled by the negative pole of the applied electric field.
3.
J. J. Thomson discovered electrons. He postulated that all atoms must contain electrons, but Thomson also postulated that atoms must contain positive charge in order for the atom to be electrically neutral. Henri Becquerel discovered radioactivity. Lord Rutherford proposed the nuclear model of the atom. Dalton's original model proposed that atoms were indivisible particles (that is, atoms had no internal structure). Thomson and Becquerel discovered subatomic particles, and Rutherford's model attempted to describe the internal structure of the atom composed of these subatomic particles. In addition, the existence of isotopes, atoms of the same element but with different mass, had to be included in the model.
4.
If the plum pudding model were correct (a diffuse positive charge with electrons scattered throughout), then alpha particles should have traveled through the thin foil with very minor deflections in their path. This was not the case as a few of the alpha particles were deflected at very large angles. Rutherford reasoned that the large deflections of these alpha particles could be caused only by a center of concentrated positive charge that contains most of the atom’s mass (the nuclear model of the atom).
5.
The proton and neutron have similar mass with the mass of the neutron slightly larger than that of the proton. Each of these particles has a mass approximately 1800 times greater than that of an electron. The combination of the protons and neutrons in the nucleus makes up the bulk of the mass of an atom, but the electrons make the greatest contribution to the chemical properties of the atom.
6.
The atomic number of an element is equal to the number of protons in the nucleus of an atom of that element. The mass number is the sum of the number of protons plus neutrons in the nucleus. The atomic mass is the actual mass of a particular isotope (including electrons). As we will see in Chapter Three, the average mass of an atom is taken from a measurement made on many atoms. The average atomic mass value is listed in the periodic table.
31
32
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ATOMS, MOLECULES, AND IONS
7.
A family is a set of elements in the same vertical column. A family is also called a group. A period is a set of elements in the same horizontal row.
8.
AlCl3, aluminum chloride; CrCl3, chromium(III) chloride; ICl3, iodine trichloride; AlCl3 and CrCl3 are ionic compounds, following the rules for naming ionic compounds. The major difference is that CrCl3 contains a transition metal (Cr), which generally can have two or more stable charges when in ionic compounds. We need to indicate which charged ion we have in the compound. This is generally true whenever the metal in the ionic compound is a transition metal. ICl3 is made from only nonmetals and is a covalent compound. Predicting formulas for covalent compounds is extremely difficult. Because of this, we need to indicate the number of each nonmetal in the binary covalent compound. The exception is when there is only one of the first species present in the formula; when this is the case, mono is not used (it is assumed).
9.
When in ionic compounds, the metals in groups 1A, 2A, and aluminum form +1, +2, and +3 charged ions, respectively. The nonmetals in the groups 5A, 6A, and 7A form −3, −2, and −1 charged ions, respectively, when in ionic compounds. The correct formulas are A 2S where A is an alkali metal, B3N2 where B is an alkaline earth metal, and AlC3 where C is a halogen.
10.
The polyatomic ions and acids in this problem are not named in the text. However, they are all related to other ions and acids named in the text which contain a same group element. Since HClO4 is perchloric acid, HBrO4 is perbromic acid. Since ClO3− is the chlorate ion, KIO3 is potassium iodate. Since ClO2− is the chlorite ion, NaBrO2 is sodium bromite. And finally, since HClO is hypochlorous acid, HIO is hypoiodous acid.
Active Learning Questions 1.
A singular atom of any element is neither a solid nor a liquid nor a gas (statement e is correct). Only collections of atoms can be characterized as a solid, liquid, or gas.
2.
The data needed would be the mass of chalk before writing your name and the the mass of chalk after writing your name. In a separate experiment, weigh a sample of chalk that has a known number of “chalk molecules”, then determine the mass of one “chalk molecule”. Once the mass of a “chalk molecule” is known, one can convert the mass of chalk used to write your name into the number of “chalk molecules”. Note that chalk is composed of the ionic compound calcium carbonate, CaCO3. Calcium carbonate is an ionic compound composed of ions, not molecules. This is why chalk molecules is put in quotation marks.
3.
a. Thomson’s plum pudding model of the atom consisted of a diffuse cloud of positive charge with the negative charged electrons embedded randomly in it. Since electrons were the center piece of his model, Thomson would probably consider electrons as the most important particle. When different atoms come together to react to form a compound, it is the electrons that are shared or transferred to form new substances. b. The protons are next most important. The number of protons dictates how many electrons are required to form specific neutral atoms or charged ions, which is related to the compounds that form.
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ATOMS, MOLECULES, AND IONS
33
c. Thomson worked with applying high voltage between metal electrodes in evacuated tubes to produce a stream of negative charged particles. Any model that explains the results of the cathode ray experiments would be a possible answer to this question. 4.
An ice cube has the H2O molecules packed very closely together. Steam consists of separate H2O molecules with lots of space between them. Any drawing illustrating these differences between the two states is fine. Also illustrated should be that the number and size of the H2O molecules doesn’t change when an ice cube is converted to steam in a closed container. Because the number of water molecules doesn’t change, the mass will not change.
5.
Assuming no substances can enter or leave the glass container, then the mass before the reaction should be identical to the mass after the reaction. In a chemical reaction, some bonds are broken between the atoms in the reactant compounds and some new bonds form in the product compounds. But the number and types of atoms are the same before a reaction as compared to after a reaction. Since atoms are conserved in a chemical reaction, mass is conserved. Answer b is correct.
6.
Answer b is correct; H2O consists of 2 H atoms bound to one O atom. Because a hydrogen atom’s mass is not equivalent to the mass of an oxygen atom, answer a is false. The other answers assume an incorrect formula for H2O.
7.
When water boils, H2O(l) is converted to H2O(g). So water vapor is present in the bubbles (answer d). For H2(g) and O2(g) to form, a lot more energy must be added in order to break down water into its elements.
8.
Yes, many questions are raised from Dalton’s theory. For example, how are atoms different from one another; how are atoms of the same element identical to each other; how are atoms held together when in compounds; if atoms are particles, what is its mass; etc.
9.
Elements are made up of different isotopes; these are atoms of an element that contain the same number of protons but differ in the number of neutrons in the nucleus. So atoms of any element are not necessarily identical to each other because of isotopes.
10.
An atomic element is represented by pictures d and g where the element exists as a singular atom. A molecular element is represented by pictures c and e where the same two elements are covalently bonded to each other to form a molecule. A compound consists of two or more different elements bonded to each other. Pictures a, b, and f represent compounds.
11.
Molecule is a term used to represent covalently bonded substances. CO 2 is formed from only nonmetals and is a covalent substance, hence molecule is an appropriate term for CO 2. NaCl is an ionic compound formed from a metal and a nonmetal. Ionic compounds exist as oppositely charged ions. In the solid state, it is a vast array of positive and negatively charged ions. There is no individual unit in an ionic compound. This is not the case with covalent substances. Molecules are the individual units in covalent compounds.
12.
Dihydrogen oxide is the correct systematic name for water (H 2O). The names in parts b and c indicate that the hydroxide (OH−) anion is present in water. If this where the case, the bonding in water would be ionic, which is not the case. The name in part d would be fine if the formula of water is written as OH2. This is not the typical way the formula for water is written.
34
CHAPTER 2
ATOMS, MOLECULES, AND IONS
13.
Barium is an alkaline earth metal. All alkaline earth metals form 2+ charge cations when in ionic compounds. The charge for most transition metal ions is not easily deduced from its position in the periodic table. For most transition metal compounds, the charge of the metal ion is included in the name.
14.
Calcium dichloride follows the covalent rules of nomenclature, i.e., the use of di, tri, tetra, etc, to indicate number of atoms in a formula. When the metal calcium is in a compound, it forms an ionic compound. So we use the ionic rules, and calcium chloride is correct name. For ionic compounds, the charges of each ion can be predicted (generally) from the periodic table; if the charges of the ions are known, then the formula can be deduced.
15.
Nitrogen trihydride would be the systematic name of NH 3. This compound is made up of only nonmetals so we would use covalent nomenclature rules to name it.
16.
The number of protons identifies the element and determines how many electrons are required to balance the total positive charge from the protons. So identity and number of electrons in the neutral element can be determined. However, the number of neutrons cannot be determined from just the number of protons.
17.
The electrons are on the outside of the nucleus; these are the particles that are easiest to access. Ions form when electrons are added or removed from an atom. Immense energy is required to add or subtract protons or neutrons to or from a nucleus.
18.
For each experiment, divide the larger number by the smaller. In doing so, we get: experiment 1
X = 1.0
Y = 10.5
experiment 2
Y = 1.4
Z = 1.0
experiment 3
X = 1.0
Y = 3.5
Note that compounds 1 and 3 have different amounts of Y that combine with 1.0 g of X. Therefore, compounds 1 and 3 are different compounds (part b answer). Our assumption about formulas dictates the rest of the solution (answers for parts a, c, d, and e). For example, if we assume that the formula of the compound in experiment 1 is XY and that of experiment 2 is YZ, we get relative masses of: X = 2.0; Y = 21; Z = 15 (= 21/1.4) and a formula of X3Y for experiment 3 [three times as much X must be present in experiment 3 as compared to experiment 1 (10.5/3.5 = 3)]. However, if we assume the formula for experiment 2 is YZ and that of experiment 3 is XZ, then we get: X = 2.0; Y = 7.0; Z = 5.0 (= 7.0/1.4) and a formula of XY3 for experiment 1. Any answer that is consistent with your initial assumptions is correct.
CHAPTER 2
ATOMS, MOLECULES, AND IONS
35
The answer to part e depends on which (if any) of experiments 1 and 3 have a formula of XY in the compound. If the compound in experiment 1 has a formula of XY, then: 21 g XY ×
4 .2 g Y = 19.2 g Y (and 1.8 g X) (4.2 + 0.4) g XY
If the compound in experiment 3 has the XY formula, then: 21 g XY ×
7.0 g Y = 16.3 g Y (and 4.7 g X) (7.0 + 2.0) g XY
Note that it could be that neither experiment 1 nor experiment 3 has XY as the formula. Therefore, there is no way of knowing an absolute answer here. 19.
a. The smaller parts are electrons and the nucleus. The nucleus is broken down into protons and neutrons, which can be broken down into quarks. For our purpose, electrons, neutrons, and protons are the key smaller parts of an atom. b. All atoms of hydrogen have 1 proton in the nucleus. Different isotopes of hydrogen have 0, 1, or 2 neutrons in the nucleus. Because we are talking about atoms, this implies a neutral charge, which dictates 1 electron present for all hydrogen atoms. If charged ions were included, then different ions/atoms of H could have different numbers of electrons. c. Hydrogen atoms always have 1 proton in the nucleus, and helium atoms always have 2 protons in the nucleus. The number of neutrons can be the same for a hydrogen atom and a helium atom. Tritium (3H) and 4He both have 2 neutrons. Assuming neutral atoms, then the number of electrons will be 1 for hydrogen and 2 for helium. d. Water (H2O) is always 1 g hydrogen for every 8 g of O present, whereas H 2O2 is always 1 g hydrogen for every 16 g of O present. These are distinctly different compounds, each with its own unique relative number and types of atoms present. e. A chemical equation involves a reorganization of the atoms. Bonds are broken between atoms in the reactants, and new bonds are formed in the products. However, the number and types of atoms between reactants and products do not change. Because atoms are conserved in a chemical reaction, mass is also conserved.
Questions 20.
Some elements exist as molecular substances. That is, hydrogen normally exists as H 2 molecules, not single hydrogen atoms. The same is true for N 2, O2, F2, Cl2, etc.
21.
A compound will always contain the same numbers (and types) of atoms. A given amount of hydrogen will react only with a specific amount of oxygen. Any excess oxygen will remain unreacted.
22.
The halogens have a high affinity for electrons, and one important way they react is to form anions of the type X−. The alkali metals tend to give up electrons easily and in most of their compounds exist as M+ cations. Note: These two very reactive groups are only one electron away (in the periodic table) from the least reactive family of elements, the noble gases.
36 23.
CHAPTER 2
ATOMS, MOLECULES, AND IONS
Law of conservation of mass: Mass is neither created nor destroyed. The total mass before a chemical reaction always equals the total mass after a chemical reaction. Law of definite proportion: A given compound always contains exactly the same proportion of elements by mass. For example, water is always 1 g H for every 8 g oxygen. Law of multiple proportions: When two elements form a series of compounds, the ratios of the mass of the second element that combine with 1 g of the first element always can be reduced to small whole numbers: For CO2 and CO discussed in Section 2.2, the mass ratios of oxygen that react with 1 g carbon in each compound are in a 2 : 1 ratio.
24.
a. True;
b. False; compounds have constant composition.
c. True;
d. True;
e. True
25.
Mass is conserved in a chemical reaction because atoms are conserved. Chemical reactions involve the reorganization of atoms, so formulas change in a chemical reaction, but the number and types of atoms do not change. Because the atoms do not change in a chemical reaction, mass must not change. In this equation we have two oxygen atoms and four hydrogen atoms both before and after the reaction occurs.
26.
The plum pudding model postulated that an atom consisted of a diffuse positive charge with negative electrons embedded randomly in it. If the plum pudding model was correct, then the alpha particle used in Rutherford’s metal foil experiment should easily pass through the atom with little or no deflection. In Rutherford’s experiment, most alpha particles did pass through, but some alpha particles were deflected at large angles. Because of the severe deflections of some alpha particles, the plum pudding model could not be correct.
27.
J. J. Thomson’s study of cathode-ray tubes led him to postulate the existence of negatively charged particles that we now call electrons. Thomson also postulated that atoms must contain positive charge for the atom to be electrically neutral. Ernest Rutherford and his alpha bombardment of metal foil experiments led him to postulate the nuclear atom−an atom with a tiny dense center of positive charge (the nucleus) with electrons moving about the nucleus at relatively large distances away; the distance is so large that an atom is mostly empty space.
28.
The atom is composed of a tiny dense nucleus containing most of the mass of the atom. The nucleus itself is composed of neutrons and protons. Neutrons have a mass slightly larger than that of a proton and have no charge. Protons, on the other hand, have a 1+ relative charge as compared to the 1– charged electrons; the electrons move about the nucleus at relatively large distances. The volume of space that the electrons move about is so large, as compared to the nucleus, that we say an atom is mostly empty space.
29.
The number and arrangement of electrons in an atom determine how the atom will react with other atoms, i.e., the electrons determine the chemical properties of an atom. The number of neutrons present determines the isotope identity and the mass number.
30.
Density = mass/volume; if the volumes are assumed equal, then the much more massive proton would have a much larger density than the relatively light electron.
CHAPTER 2
ATOMS, MOLECULES, AND IONS
37
31.
For lighter, stable isotopes, the number of protons in the nucleus is about equal to the number of neutrons. When the number of protons and neutrons is equal to each other, the mass number (protons + neutrons) will be twice the atomic number (protons). Therefore, for lighter isotopes, the ratio of the mass number to the atomic number is close to 2. For example, consider 28Si, which has 14 protons and (28 – 14 =) 14 neutrons. Here, the mass number to atomic number ratio is 28/14 = 2.0. For heavier isotopes, there are more neutrons than protons in the nucleus. Therefore, the ratio of the mass number to the atomic number increases steadily upward from 2 as the isotopes get heavier and heavier. For example, 238U has 92 protons and (238 – 92 =) 146 neutrons. The ratio of the mass number to the atomic number for 238U is 238/92 = 2.6.
32.
Some properties of metals are (1) conduct heat and electricity; (2) malleable (can be hammered into sheets); (3) ductile (can be pulled into wires); (4) lustrous appearance; (5) form cations when they form ionic compounds. Nonmetals generally do not have these properties, and when they form ionic compounds, nonmetals always form anions.
33.
Carbon is a nonmetal. Silicon and germanium are called metalloids because they exhibit both metallic and nonmetallic properties. Tin and lead are metals. Thus metallic character increases as one goes down a family in the periodic table. The metallic character decreases from left to right across the periodic table.
34.
Yes, 1.0 g H would react with 37.0 g 37Cl, and 1.0 g H would react with 35.0 g 35Cl. No, the mass ratio of H/Cl would always be 1 g H/37 g Cl for 37Cl and 1 g H/35 g Cl for 35Cl. If we had pure 37Cl or pure 35Cl, the ratios would always hold. If we have a mixture (such as the natural abundance of chlorine), the ratio will also be constant if the composition of the mixture of the two isotopes does not change.
35.
In the paste, sodium chloride will dissolve to form separate Na+ and Cl− ions. With the ions present and able to move about, electrical impulses will be conducted.
36.
a. A molecule has no overall charge (an equal number of electrons and protons are present). Ions, on the other hand, have extra electrons added or removed to form anions (negatively charged ions) or cations (positively charged ions). b. The sharing of electrons between atoms is a covalent bond. An ionic bond is the force of attraction between two oppositely charged ions. c. A molecule is a collection of atoms held together by covalent bonds. A compound is composed of two or more different elements having constant composition. Covalent and/or ionic bonds can hold the atoms together in a compound. Another difference is that molecules do not necessarily have to be compounds. H 2 is two hydrogen atoms held together by a covalent bond. H2 is a molecule, but it is not a compound; H 2 is a diatomic element.
38
CHAPTER 2
ATOMS, MOLECULES, AND IONS
d. An anion is a negatively charged ion, e.g., Cl−, O2−, and SO42− are all anions. A cation is a positively charged ion, e.g., Na+, Fe3+, and NH4+ are all cations. 37.
a. This represents ionic bonding. Ionic bonding is the electrostatic attraction between anions and cations. b. This represents covalent bonding where electrons are shared between two atoms. This could be the space-filling model for H2O or SF2 or NO2, etc.
38.
Natural niacin and commercially produced niacin have the exact same formula of C6H5NO2. Therefore, both sources produce niacin having an identical nutritional value. There may be other compounds present in natural niacin that would increase the nutritional value, but the nutritional value due to just niacin is identical to the commercially produced niacin.
39.
Statement d is true. For statement a, neutrons in the nucleus are neutral in charge. For statement b, the atom consists of a tiny dense nucleus with electrons around the nucleus at relatively large distances. An atom is mostly empty space. For statement c, the nucleus contains most of the mass of the nucleus. For statement e, the number of protons in a neutral atom is equal to the number of electrons.
40.
From the sulfide formula MS, since sulfur forms 2‒ charged ions when in ionic compounds with metals, M must form a 2+ charged ion. For light atoms, the number of protons and neutrons are about the same. With a mass number of 26, let’s assume the M has 13 protons and 13 neutrons. Element 13 is aluminum, but it forms 3+ charged ions when in ionic compounds. From the periodic table, element 12 is an alkaline metal which does form 2+ charged ions when in ionic compounds. Therefore, M is element 12, which is magnesium, Mg.
41.
Statements a and b are true. Element 118, Og, is a noble gas and will presumably be a nonmetal. For statement c, hydrogen has mostly nonmetallic properties. For statement d, a family of elements is also known as a group of elements. For statement e, two items are incorrect. When a metal reacts with a nonmetal, an ionic compound is produced, and the formula of the compound would be AX2 (alkaline earth metals form 2+ ions and halo-gens form 1– ions in ionic compounds). The correct statement would be: When an alkaline earth metal, A, reacts with a halogen, X, the formula of the ionic compound formed should be AX 2.
42.
Predicting charges that transition metals form when in ionic compounds is difficult. Also, most transition metals form multiple charged ions when in compounds. For example, copper forms 1+ and 2+ charged ions when in ionic compounds. Because of this, roman numerals are used to indicate the charge when naming most transition metal ionic compounds. Silver and zinc ionic compounds are exceptions to this. Silver always forms 1+ charged ions when in ionic compounds and zinc always forms 2+ charged ions when in ionic compounds. Roman numerals are never used for silver and zinc ionic compounds. Note that cadmium is another transition metal which only forms 2+ charged ions when in ionic compounds. No roman numeral is used for cadmium ionic compounds either.
43.
Predicting formulas for covalent compounds is difficult, unlike ionic compound formulas where ion charges are used to determine formulas. Covalent compounds are not formed from ions. Also, sometimes many different formulas can form between two different nonmetals. For example, between nitrogen and oxygen, some compounds that form are NO, NO2, N2O,
CHAPTER 2
ATOMS, MOLECULES, AND IONS
39
and N2O4. Because it is difficult to predict covalent compound formulas, prefixes are necessary to indicate the formula of the compound. 44.
a. Dinitrogen monoxide is correct. N and O are both nonmetals, resulting in a covalent compound. We need to use the covalent rules of nomenclature. The other two names are for ionic compounds. b. Copper(I) oxide is correct. With a metal in a compound, we have an ionic compound. Because copper, like most transition metals, forms at least a couple of different stable charged ions in compounds, we must indicate the charge on copper in the name. Copper oxide could be CuO or Cu2O, hence why we must give the charge of most transition metal compounds. Dicopper monoxide is the name if this were a covalent compound, which it is not. c. Lithium oxide is correct. Lithium forms 1+ charged ions in stable ionic compounds. Because lithium is assumed to form 1+ ions in compounds, we do not need to indicate the charge of the metal ion in the compound. Dilithium monoxide would be the name if Li 2O were a covalent compound (a compound composed of only nonmetals).
Exercises Development of the Atomic Theory 45.
a. The composition of a substance depends on the numbers of atoms of each element making up the compound (depends on the formula of the compound) and not on the composition of the mixture from which it was formed. b. Avogadro’s hypothesis (law) implies that volume ratios are proportional to molecule ratios at constant temperature and pressure. H2(g) + Cl2(g) → 2 HCl(g). From the balanced equation, the volume of HCl produced will be twice the volume of H2 (or Cl2) reacted.
46.
Avogadro’s hypothesis (law) implies that volume ratios are equal to molecule ratios at constant temperature and pressure. Here, 1 volume of N 2 reacts with 3 volumes of H2 to produce 2 volumes of the gaseous product or in terms of molecule ratios: 1 N2 + 3 H2 → 2 product For the equation to be balanced, the product must be NH3.
47.
From the law of definite proportions, a given compound always contains exactly the same proportion of elements by mass. The first sample of chloroform has a total mass of 12.0 g C + 106.4 g Cl + 1.01 g H = 119.41 g (carrying extra significant figures). The mass percent of carbon in this sample of chloroform is: 12.0 g C × 100 = 10.05% C by mass 119 .41 g total
40
CHAPTER 2
ATOMS, MOLECULES, AND IONS
From the law of definite proportions, the second sample of chloroform must also contain 10.05% C by mass. Let x = mass of chloroform in the second sample: 30 .0 g C × 100 = 10.05, x = 299 g chloroform x
48.
A compound will always have a constant composition by mass. From the initial data given, the mass ratio of H : S : O in sulfuric acid (H 2SO4) is: 2.02 32.07 64.00 : : = 1 : 15.9 : 31.7 2.02 2.02 2.02
If we have 7.27 g H, then we will have 7.27 × 15.9 = 116 g S and 7.27 × 31.7 = 230. g O in the second sample of H2SO4. 49.
Compound 1:
21.8 g C and 58.2 g O (80.0 – 21.8 = mass O)
Compound 2:
34.3 g C and 45.7 g O (80.0 – 34.3 = mass O)
The mass of carbon that combines with 1.0 g of oxygen is: Compound 1:
21.8 g C = 0.375 g C/g O 58.2 g O
Compound 2:
34.3 g C = 0.751 g C/g O 45.7 g O
0.751 2 = ; this 0.375 1 supports the law of multiple proportions because this carbon ratio is a small whole number.
The ratio of the masses of carbon that combine with 1 g of oxygen is
50.
Let’s determine the ratios of the masses of fluorine that combine with 1 g of S: 1.188 = 1.000; 1.188
2.375 = 1.999; 1.188
3.563 = 2.999 1.188
Because the masses of the fluorine that combine with 1.000 g of sulfur in the three compounds are all in whole number multiples, this illustrates the law of multiple proportions. 51.
For CO and CO2, it is easiest to concentrate on the mass of oxygen that combines with 1 g of carbon. From the formulas (two oxygen atoms per carbon atom in CO 2 versus one oxygen atom per carbon atom in CO), CO2 will have twice the mass of oxygen that combines per gram of carbon as compared to CO. For CO2 and C3O2, it is easiest to concentrate on the mass of carbon that combines with 1 g of oxygen. From the formulas (three carbon atoms per two oxygen atoms in C3O2 versus one carbon atom per two oxygen atoms in CO 2), C3O2 will have three times the mass of carbon that combines per gram of oxygen as compared to CO2. As expected, the mass ratios are whole numbers as predicted by the law of multiple proportions.
52.
Compound I:
14.0 g R 4.67 g R = ; 3.00 g Q 1.00 g Q
compound II:
7.00 g R 1.56 g R = 4.50 g Q 1.00 g Q
CHAPTER 2
ATOMS, MOLECULES, AND IONS
The ratio of the masses of R that combine with 1.00 g Q is:
41 4.67 = 2.99 3 1.56
As expected from the law of multiple proportions, this ratio is a small whole number. Because compound I contains three times the mass of R per gram of Q as compared with compound II (RQ), the formula of compound I should be R3Q. 53.
Mass is conserved in a chemical reaction. Mass:
ethanol + oxygen → water + carbon dioxide 46.0 g 96.0 g 54.0 g ?
Mass of reactants = 46.0 + 96.0 = 142.0 g = mass of products 142.0 g = 54.0 g + mass of CO2, mass of CO2 = 142.0 – 54.0 = 88.0 g 54.
Mass is conserved in a chemical reaction.
Mass:
chromium(III) oxide + aluminum → chromium + aluminum oxide 34.0 g 12.1 g 23.3 g ?
Mass aluminum oxide produced = (34.0 + 12.1) − 23.3 = 22.8 g 55.
To get the atomic mass of H to be 1.00, we divide the mass of hydrogen that reacts with 1.00 0.126 g of oxygen by 0.126; that is, = 1.00. To get Na, Mg, and O on the same scale, we do 0.126 the same division. Na:
1.500 2.875 1.00 = 22.8; Mg: = 11.9; O: = 7.94 0.126 0.126 0.126
H
O
Na
Mg
Relative value
1.00
7.94
22.8
11.9
Accepted value
1.008
16.00
22.99
24.31
For your information, the atomic masses of O and Mg are incorrect. The atomic masses of H and Na are close to the values given in the periodic table. Something must be wrong about the assumed formulas of the compounds. It turns out the correct formulas are H 2O, Na2O, and MgO. The smaller discrepancies result from the error in the assumed atomic mass of H. 56.
If the formula is InO, then one atomic mass of In would combine with one atomic mass of O, or: A 4.784 g In = , A = atomic mass of In = 76.54 16.00 1.000 g O If the formula is In2O3, then two times the atomic mass of In will combine with three times the atomic mass of O, or:
42
CHAPTER 2
ATOMS, MOLECULES, AND IONS
2A 4.784 g In = , A = atomic mass of In = 114.8 (3)16.00 1.000 g O
The latter number is the atomic mass of In used in the modern periodic table.
The Nature of the Atom 57.
From section 2.5, the nucleus has “a diameter of about 10 −13 cm” and the electrons “move about the nucleus at an average distance of about 10−8 cm from it.” We will use these statements to help determine the densities. Density of hydrogen nucleus (contains one proton only): Vnucleus =
4 3 4 r = (3.14) (5 10 −14 cm)3 = 5 10 −40 cm3 3 3
1.67 10 −24 g
d = density =
5 10
− 40
= 3 1015 g/cm3
3
cm
Density of H atom (contains one proton and one electron): Vatom =
d= 58.
4 (3.14) (1 10 −8 cm) 3 = 4 10 − 24 cm3 3
1.67 10 −24 g + 9 10 −28 g 4 10
− 24
3
cm
= 0.4 g/cm3
Because electrons move about the nucleus at an average distance of about 1 × 10 −8 cm, the diameter of an atom will be about 2 × 10 −8 cm. Let's set up a ratio: diameter of nucleus 1 mm 1 10 −13 cm = = ; solving: diameter of atom diameter of model 2 10 −8 cm
diameter of model = 2 × 105 mm = 200 m
1 electron charge
59.
5.93 10 −18 C
60.
First, divide all charges by the smallest quantity, 6.40 × 10 −13 .
1.602 10 −19 C
2.56 10 −12 6.40 10
−13
= 4.00;
= 37 negative (electron) charges on the oil drop
7.68 3.84 = 12.0; = 6.00 0.640 0.640
Because all charges are whole-number multiples of 6.40 × 10 −13 zirkombs, the charge on one electron could be 6.40 × 10 −13 zirkombs. However, 6.40 × 10 −13 zirkombs could be the charge of two electrons (or three electrons, etc.). All one can conclude is that the charge of an electron is 6.40 × 10 −13 zirkombs or an integer fraction of 6.40 × 10 −13 zirkombs. 61.
Sn−tin; Pt−platinum; Hg−mercury; Mg−magnesium; K−potassium; Ag−silver
62.
fluorine−F; chlorine−Cl; bromine−Br; sulfur−S; oxygen−O; phosphorus−P
CHAPTER 2 63.
ATOMS, MOLECULES, AND IONS
43
a. 6; the group 2A elements are Be, Mg, Ca, Sr, Ba, and Ra. b. 6; the group 6A elements are O, S, Se, Te, Po, and Lv. c. 4; the nickel family elements are Ni, Pd, Pt, amd Ds. d. 7; the noble gas group 8A elements are He, Ne, Ar, Kr, Xe, Rn, and Og.
64.
a. 6; the halogen group 7A elements are F, Cl, Br, I, At, and Ts. b. 6; the alkali group 1A elements are Li, Na, K, Rb, Cs, and Fr. H is not considered an alkali metal. c. 14; elements Ce through Lu are the 14 lanthanide series elements. d. 40; the block of elements from Sc to Zn to Cn to Ac and back to Sc are all transition metal elements.
65.
a. Metals: Mg, Ti, Au, Bi, Ge, Eu, and Am. Nonmetals: Si, B, At, Rn, and Br. b. Si, Ge, B, and At. The elements at the boundary between the metals and the nonmetals are B, Si, Ge, As, Sb, Te, Po, and At. Aluminum has mostly properties of metals, so it is generally not classified as a metalloid.
66.
a. The noble gases are He, Ne, Ar, Kr, Xe, , Rn, and Og (helium, neon, argon, krypton, xenon, Radon, and Oganesson). Radon and oganesson only have radioactive isotopes. In the periodic table, the whole number enclosed in parentheses is the mass number of the longestlived isotope of the element. b. Promethium (Pm) has only radioactive isotopes.
67.
68.
a. transition metals
b. alkaline earth metals
d. noble gases
e. halogens
c. alkali metals
Use the periodic table to identify the elements. a. Cl; halogen
b. Be; alkaline earth metal
c. Eu; lanthanide metal
d. Hf; transition metal
e. He; noble gas
f.
U; actinide metal
g. Cs; alkali metal 69.
24 25 26 27 28 29 30 The symbols are 13 Al, 13 Al, 13 Al, 13 Al, 13 Al, 13 Al, and 13 Al.
70.
Rb is element 37, so 85Rb has 85 ‒ 37 = 48 neutrons. Sr is element 38, so 86Sr also has 48 neutrons (86 ‒ 38 = 48 neutrons). Answer e is correct. 85Kr has 49 neutrons, 87Rb has 50 neutrons, 85Sr has 47 neutrons and 86Kr has 50 neutrons.
71.
a.
Element 8 is oxygen. A = mass number = 9 + 8 = 17;
17 8O
44
72.
CHAPTER 2 Chlorine is element 17. 17 Cl
d.
Z = 26; A = 26 + 31 = 57; 26 Fe
f.
Lithium is element 3.
60
c. Cobalt is element 27. 27 Co
57
e. Iodine is element 53.
131 53 I
7 3 Li
58
a. Cobalt is element 27. A = mass number = 27 + 31 = 58; 27 Co b.
73.
37
b.
ATOMS, MOLECULES, AND IONS
10 5B
c.
23 12 Mg
d.
132 53 I
e.
47 20 Ca
f.
65 29 Cu
Z is the atomic number and is equal to the number of protons in the nucleus. A is the mass number and is equal to the number of protons plus neutrons in the nucleus. X is the symbol of the element. See the front cover of the text which has a listing of the symbols for the various elements and corresponding atomic number or see the periodic table on the cover to determine the identity of the various atoms. Because all of the atoms have equal numbers of protons and electrons, each atom is neutral in charge. a. 23 11 Na
b. 199 F
c. 168 O
74.
The atomic number for carbon is 6. 14C has 6 protons, 14 − 6 = 8 neutrons, and 6 electrons in the neutral atom. 12C has 6 protons, 12 – 6 = 6 neutrons, and 6 electrons in the neutral atom. The only difference between an atom of 14C and an atom of 12C is that 14C has two additional neutrons.
75.
a.
79 35 Br: 35 protons, 79 – 35 = 44 neutrons. Because the charge of the atom is neutral,
the number of protons = the number of electrons = 35.
76.
77.
b.
81 35 Br: 35 protons, 46 neutrons, 35 electrons
c.
239 94 Pu: 94 protons, 145 neutrons, 94 electrons
d.
133 55 Cs: 55 protons, 78 neutrons, 55 electrons
e.
3 1 H: 1 proton, 2 neutrons, 1 electron
f.
56 26 Fe: 26 protons, 30 neutrons, 26 electrons
a.
235 92 U: 92 p, 143 n, 92 e
b.
28 14 Si: 14 p, 14 n, 14 e
c.
57 26 Fe: 26 p, 31 n, 26 e
d.
208 82 Pb: 82 p, 126 n, 82 e
e.
86 37 Rb: 37 p, 49 n, 37 e
f.
41 20 Ca: 20 p, 21 n, 20 e
a. Ba is element 56. Ba2+ has 56 protons, so Ba2+ must have 54 electrons in order to have a net charge of 2+. b. Zn is element 30. Zn2+ has 30 protons and 28 electrons.
CHAPTER 2
ATOMS, MOLECULES, AND IONS
45
c. N is element 7. N3− has 7 protons and 10 electrons. d. Rb is element 37. Rb+ has 37 protons and 36 electrons. e. Co is element 27. Co3+ has 27 protons and 24 electrons. f.
Te is element 52. Te2− has 52 protons and 54 electrons.
g. Br is element 35. Br− has 35 protons and 36 electrons. 78.
79.
a.
24 Mg: 12 protons, 12 neutrons, 12 electrons 12
b.
24 Mg2+: 12 p, 12 n, 10 e 12
c.
59 Co2+: 27 p, 32 n, 25 e 27
d.
59 Co3+: 27
e.
59 Co: 27 p, 32 n, 27 e 27
f.
79 Se: 34 p, 45 n, 34 e 34
g.
79 2− Se : 34 p, 45 n, 36 e 34
h.
63 Ni: 28 p, 35 n, 28 e 28
i.
59 2+ Ni : 28 p, 31 n, 26 e 28
27 p, 32 n, 24 e
Atomic number = 63 (Eu); net charge = +63 − 60 = 3+; mass number = 63 + 88 = 151; symbol:
151 3+ 63 Eu
Atomic number = 50 (Sn); mass number = 50 + 68 = 118; net charge = +50 − 48 = 2+; symbol: 118 2+ 50 Sn
80.
Atomic number = 16 (S); net charge = +16 − 18 = 2−; mass number = 16 + 18 = 34; 34
symbol: 16 S2− Atomic number = 16 (S); net charge = +16 − 18 = 2−; mass number = 16 + 16 = 32; 32
symbol: 16 S2−
46
CHAPTER 2
ATOMS, MOLECULES, AND IONS
81. Number of protons in nucleus
Number of neutrons in nucleus
Number of electrons
Net charge
238 92 U
92
146
92
0
40 2+ 20 Ca
20
20
18
2+
51 3+ 23 V
23
28
20
3+
89 39 Y
39
50
39
0
79 − 35 Br
35
44
36
1−
31 3 − 15 P
15
16
18
3−
Symbol
Number of protons in nucleus
Number of neutrons in nucleus
Number of electrons
Net charge
53 2 + 26 Fe
26
27
24
2+
59 3+ 26 Fe
26
33
23
3+
210 − 85 At
85
125
86
1–
27 3+ 13 Al
13
14
10
3+
128 2− 52 Te
52
76
54
2–
Symbol
82.
83.
In ionic compounds, metals lose electrons to form cations, and nonmetals gain electrons to form anions. Group 1A, 2A, and 3A metals form stable 1+, 2+, and 3+ charged cations, respectively. Group 5A, 6A, and 7A nonmetals form 3−, 2−, and 1− charged anions, respectively.
CHAPTER 2
84.
ATOMS, MOLECULES, AND IONS
47
a. Lose 2 e − to form Ra2+.
b. Lose 3 e − to form In3+.
c. Gain 3 e − to form P 3− .
d. Gain 2 e − to form Te 2 − .
e. Gain 1 e − to form Br−.
f.
Lose 1 e − to form Rb+.
See Exercise 83 for a discussion of charges various elements form when in ionic compounds. a. Element 13 is Al. Al forms 3+ charged ions in ionic compounds. Al3+ b. Se2−
c. Ba2+
d. N3−
e. Fr+
f.
Br−
Nomenclature 85.
86.
87.
a. sodium bromide
b. rubidium oxide
c. calcium sulfide
d. aluminum iodide
e. SrF2
f.
g. K3N
h. Mg3P2
a. mercury(I) oxide
b. iron(III) bromide
c. cobalt(II) sulfide
d. titanium(IV) chloride
e. Sn3N2
f.
g. HgO
h. CrS3
a. cesium fluoride
b. lithium nitride
Al2Se3
CoI3
c. silver sulfide (Silver only forms stable 1+ ions in compounds, so no Roman numerals are needed.) d. manganese(IV) oxide 88.
89.
90.
91.
92.
e. titanium(IV) oxide
f.
strontium phosphide
a. ZnCl2 (Zn only forms stable +2 ions in compounds, so no Roman numerals are needed.) b. SnF4
c. Ca3N2
e. Hg2Se
f.
a. barium sulfite
b. sodium nitrite
c. potassium permanganate
d. potassium dichromate
a. Cr(OH)3
b. Mg(CN)2
c. Pb(CO3)2
d. NH4C2H3O2
a. dinitrogen tetroxide
b. iodine trichloride
c. sulfur dioxide
d. diphosphorus pentasulfide
a. B2O3
b. AsF5
d. Al2S3
AgI (Ag only forms stable +1 ions in compounds.)
c. N2O
d. SCl6
48 93.
94.
CHAPTER 2
ATOMS, MOLECULES, AND IONS
a. copper(I) iodide
b. copper(II) iodide
c. cobalt(II) iodide
d. sodium carbonate
e. sodium hydrogen carbonate or sodium bicarbonate
f.
tetrasulfur tetranitride
g. selenium tetrachloride
i.
barium chromate
j.
h. sodium hypochlorite
ammonium nitrate
a. acetic acid
b. ammonium nitrite
c. cobalt(III) sulfide
d. iodine monochloride
e. lead(II) phosphate
f.
potassium chlorate
g. sulfuric acid
h. strontium nitride
i.
aluminum sulfite
j.
k. sodium chromate
l.
hypochlorous acid
tin(IV) oxide
Note: For the compounds named as acids, we assume these are dissolved in water. 95.
In the case of sulfur, SO42− is sulfate, and SO32− is sulfite. By analogy: SeO42−: selenate; SeO32−: selenite; TeO42−: tellurate; TeO32−: tellurite
96.
From the anion names of hypochlorite (ClO −), chlorite (ClO2−), chlorate (ClO3−), and perchlorate (ClO4−), the oxyanion names for similar iodine ions would be hypoiodite (IO−), iodite (IO2−), iodate (IO3−), and periodate (IO4−). The corresponding acids would be hypoiodous acid (HIO), iodous acid (HIO2), iodic acid (HIO3), and periodic acid (HIO4).
97.
a. SF2
b. SF6
c. NaH2PO4
d. Li3N
e. Cr2(CO3)3
f.
SnF2
g. NH4C2H3O2
h. NH4HSO4
i.
Co(NO3)3
l.
NaH
j. 98.
a. CrO3
b. S2Cl2
d. K2HPO4
e. AlN
f.
99.
Hg2Cl2; mercury(I) exists as Hg22+ ions.
k. KClO3
NH3 (Nitrogen trihydride is the systematic name.) (NH4)2SO3
c. NiF2
g. MnS2
h. Na2Cr2O7
i.
j.
CI4
a. Na2O
b. Na2O2
c. KCN
d. Cu(NO3)2
e. SeBr4
f.
g. PbS2
h. CuCl
HIO2
i.
GaAs (We would predict the stable ions to be Ga3+ and As 3− .)
j.
CdSe (Cadmium only forms 2+ charged ions in compounds.)
k. ZnS (Zinc only forms 2+ charged ions in compounds.) l. 100.
HNO2
m. P2O5
a. (NH4)2HPO4
b. Hg2S
c. SiO2
d. Na2SO3
e. Al(HSO4)3
f.
NCl3
CHAPTER 2
101.
102.
ATOMS, MOLECULES, AND IONS
49
g. HBr
h. HBrO2
i.
HBrO4
j. KHS
k. CaI2
l.
CsClO4
a. nitric acid, HNO3
b. perchloric acid, HClO4
d. sulfuric acid, H2SO4
e. phosphoric acid, H3PO4
c. acetic acid, HC2H3O2
a. Iron forms 2+ and 3+ charged ions; we need to include a Roman numeral for iron. Iron(III) chloride is correct. b. This is a covalent compound, so use the covalent rules. Nitrogen dioxide is correct. c. This is an ionic compound, so use the ionic rules. Calcium oxide is correct. Calcium only forms stable 2+ ions when in ionic compounds, so no Roman numeral is needed. d. This is an ionic compound, so use the ionic rules. Aluminum sulfide is correct. e. This is an ionic compound, so use the ionic rules. Mg is magnesium. Magnesium acetate is correct. f.
Phosphide is P3−, while phosphate is PO43−. Because phosphate has a 3− charge, the charge on iron is 3+. Iron(III) phosphate is correct.
g. This is a covalent compound, so use the covalent rules. Tetraphosphorus hexoxide is correct. h. Because each sodium is 1+ charged, we have the O 22− (peroxide) ion present. Sodium peroxide is correct. Note that sodium oxide would be Na2O.
103.
i.
HNO3 is nitric acid, not nitrate acid. Nitrate acid does not exist.
j.
H2S is hydrosulfuric acid or dihydrogen sulfide or just hydrogen sulfide (common name). H2SO4 is sulfuric acid.
a. (NH4)2CrO4 is ammonium chromate. b. HCl is hydrochloric acid. HClO4 is perchloric acid. c. P2S5 is diphosphorus pentasulfide. d. Iron(II) nitrate is Fe(NO3)2.
104.
AlSO3 should be Al2(SO3)3 and is aluminum sulfite. NaPO4 should be Na3PO4 and is sodium phosphate. MgI2 is a correct formula and is magnesium iodide. AgS should be Ag 2S and is silver sulfide. ZnP should be Zn3P2 and is zinc phosphide.
ChemWork Problems 105.
Answer b only contains alkaline earth metals combined with halogens to form alkaline halide compounds.
50
CHAPTER 2
ATOMS, MOLECULES, AND IONS
106.
With 15 protons, the atom is phosphorus (not sulfur). The ion is called phosphide. The most stable ion of phosphorus has 18 electrons (not 10 electrons) for a 3‒ charge. 33P will have 3315 = 18 neutrons (not 33 neutrons). Only statement a is true.
107.
There should be no difference. The composition of insulin (the number and types of atoms) from both sources will be the same and therefore, it should have the some activity regardless of the source. As a practical note, trace contaminants in the two types of insulin may be different. These trace contaminants may be important towards the activity of insulin in the body.
108.
The formula of glucose is C6(H2O)6, or C6H12O6.
109. Number of protons in nucleus
Number of neutrons in nucleus
Symbol
9
10
19 9F
13
14
1 3 Al
53
74
127 53 I
34
45
79 34 Se
16
16
32 16 S
Symbol
Number of protons in nucleus
Number of neutrons in nucleus
4 2 He
2
2
20 10 Ne
10
10
48 22 Ti
22
26
190 76 Os
76
114
50 27 Co
27
23
27
110.
CHAPTER 2 111.
112.
ATOMS, MOLECULES, AND IONS
a.
131 53 I has 53 protons and 131 – 53 = 78 neutrons.
b.
201 81 Tl has 81 protons and 201 – 81 = 120 neutrons.
51
Carbon (C); hydrogen (H); oxygen (O); nitrogen (N); phosphorus (P); sulfur (S) For lighter elements, stable isotopes usually have equal numbers of protons and neutrons in the nucleus; these stable isotopes are usually the most abundant isotope for each element. Therefore, a predicted stable isotope for each element is 12C, 2H, 16O, 14N, 30P, and 32S. These are stable isotopes except for 30P, which is radioactive. The most stable (and most abundant) isotope of phosphorus is 31P. There are exceptions. Also, the most abundant isotope for hydrogen is 1H; this has just a proton in the nucleus. 2H (deuterium) is stable (not radioactive), but 1H is also stable as well as most abundant.
113. Number of protons in nucleus
Number of neutrons in nucleus
Number of electrons
120 50 Sn
50
70
50
25 2+ 12 Mg
12
13
10
56 2+ 26 Fe
26
30
24
79 34 Se
34
45
34
35 17 Cl
17
18
17
63 29 Cu
29
34
29
Symbol
114.
a. True b. False; this was J. J. Thomson. c. False; a proton is about 1800 times more massive than an electron. d. The nucleus contains the protons and the neutrons.
115.
53 2+ has 26 protons, 53 – 26 = 27 neutrons, and two fewer electrons than protons (24 26 Fe
electrons) to have a net charge of 2+.
52
CHAPTER 2
ATOMS, MOLECULES, AND IONS
116.
Statement c and d are false. Alkali metals form stable 1+ charged ions, and transition metals form positive charged ions when in ionic compounds. Positive charged ions are called cations.
117.
From the Na2X formula, X has a 2− charge. Because 36 electrons are present, X has 34 protons and 79 − 34 = 45 neutrons, and is selenium. a. True. Nonmetals bond together using covalent bonds and are called covalent compounds. b. False. The isotope has 34 protons. c. False. The isotope has 45 neutrons. d. False. The identity is selenium, Se.
118.
a. Fe2+: 26 protons (Fe is element 26.); protons − electrons = net charge, 26 − 2 = 24 electrons; FeO is the formula since the oxide ion has a 2− charge, and the name is iron(II) oxide. b. Fe3+: 26 protons; 23 electrons; Fe2O3; iron(III) oxide c. Ba2+: 56 protons; 54 electrons; BaO; barium oxide d. Cs+: 55 protons; 54 electrons; Cs2O; cesium oxide e. S2−: 16 protons; 18 electrons; Al2S3; aluminum sulfide f.
P3−: 15 protons; 18 electrons; AlP; aluminum phosphide
g. Br−: 35 protons; 36 electrons; AlBr3; aluminum bromide h. N3−: 7 protons; 10 electrons; AlN; aluminum nitride 119.
a. Pb(C2H3O2)2: lead(II) acetate
b. CuSO4: copper(II) sulfate
c. CaO: calcium oxide
d. MgSO4: magnesium sulfate
e. Mg(OH)2: magnesium hydroxide
f.
CaSO4: calcium sulfate
g. N2O: dinitrogen monoxide or nitrous oxide (common name) 120.
121.
Co(NO2)2, cobalt(II) nitrite;
AsF5, arsenic pentafluoride;
LiCN, lithium cyanide;
K2SO3, potassium sulfite;
Li3N, lithium nitride;
PbCrO4, lead(II) chromate
carbon tetrabromide, CBr4;
cobalt(II) phosphate, Co3(PO4)2;
magnesium chloride, MgCl2;
nickel(II) acetate, Ni(C2H3O2)2;
calcium nitrate, Ca(NO3)2
CHAPTER 2 122.
ATOMS, MOLECULES, AND IONS
a. False; magnesium is Mg.
53
b. True
c. Ga is a metal and is expected to lose electrons when forming ions. d. True e. Titanium(IV) oxide is correct for this transition metal ionic compound. 123.
K will lose 1 e− to form K+.
Cs will lose 1 e− to form Cs+.
Br will gain 1 e− to form Br−.
Sulfur will gain 2 e− to form S2 −.
Se will gain 2 e− to form Se2 −. 124.
a. This is element 52, tellurium. Te forms stable 2– charged ions in ionic compounds (like other oxygen family members). b.
Rubidium. Rb, element 37, forms stable 1+ charged ions.
c.
Argon. Ar is element 18.
d.
Astatine. At is element 85.
125.
From the X3P2 formula, the charge on X is 2+. Since X2+ has 86 electrons, the ion must have 88 protons. From the periodic table, element 88 is radium, Ra. 230 − 88 = 142 neutrons.
126.
Because this is a relatively small number of neutrons, the number of protons will be very close to the number of neutrons present. The heavier elements have significantly more neutrons than protons in their nuclei. Because this element forms anions, it is a nonmetal and will be a halogen because halogens form stable 1− charged ions in ionic compounds. From the halogens listed, chlorine, with an average atomic mass of 35.45, fits the data. The two isotopes are 35Cl and 37Cl, and the number of electrons in the 1− ion is 18. Note that because the atomic mass of chlorine listed in the periodic table is closer to 35 than 37, we can assume that 35Cl is the more abundant isotope. This is discussed in Chapter 3.
127.
a. Ca2+ and N3−: Ca3N2, calcium nitride
b. K+ and O2−: K2O, potassium oxide
c. Rb+ and F−: RbF, rubidium fluoride
d. Mg2+ and S2−: MgS, magnesium sulfide
e. Ba2+ and I−: BaI2, barium iodide f.
Al3+ and Se2−: Al2Se3, aluminum selenide
g. Cs+ and P3−: Cs3P, cesium phosphide h. In3+ and Br−: InBr3, indium(III) bromide. In also forms In + ions, but one would predict In3+ ions from its position in the periodic table. 128.
These compounds are like phosphate (PO43-− ) compounds. Na3AsO4 contains Na+ ions and AsO43− ions. The name would be sodium arsenate. H3AsO4 is analogous to phosphoric acid,
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H3PO4. H3AsO4 would be arsenic acid. Mg3(SbO4)2 contains Mg2+ ions and SbO43− ions, and the name would be magnesium antimonate. 129.
a. Element 15 is phosphorus, P. This atom has 15 protons and 31 − 15 = 16 neutrons. b. Element 53 is iodine, I. 53 protons; 74 neutrons c. Element 19 is potassium, K. 19 protons; 20 neutrons d. Element 70 is ytterbium, Yb. 70 protons; 103 neutrons
130.
A is 19F, B is 2H, C is 39K+, D is 35Cl‒, E is 3H, F is 16O2‒, and G is 207Pb. Only statement e is false. The compound between the two nonmetals F and H will be covalent (HF is the compound).
131.
The law of multiple proportions does not involve looking at the ratio of the mass of one element with the total mass of the compounds. To illustrate the law of multiple proportions, we compare the mass of carbon that combines with 1.0 g of oxygen in each compound: Compound 1:
27.2 g C and 72.8 g O (100.0 – 27.2 = mass O)
Compound 2:
42.9 g C and 57.1 g O (100.0 – 42.9 = mass O)
The mass of carbon that combines with 1.0 g of oxygen is: Compound 1:
27.2 g C = 0.374 g C/g O 72.8 g O
Compound 2:
42.9 g C = 0.751 g C/g O 57.1 g O
0.751 2 = ; because the ratio is a small whole number, this supports the law of multiple 0.374 1 proportions.
132.
a. Hydrosulfuric acid if dissolved in water or dihydrogen sulfide if a gas. b. sulfur dioxide c. sulfur hexafluoride d. sodium sulfite
133.
The systematic name of Ta2O5 is tantalum(V) oxide. Tantalum is a transition metal and requires a Roman numeral. Sulfur is in the same group as oxygen, and its most common ion is S2–. Therefore, the formula of the sulfur analogue would be Ta 2S5. Total number of protons in Ta2O5: Ta, Z = 73, so 73 protons 2 = 146 protons; O, Z = 8, so 8 protons 5 = 40 protons Total protons = 186 protons
CHAPTER 2
ATOMS, MOLECULES, AND IONS
55
Total number of protons in Ta2S5: Ta, Z = 73, so 73 protons 2 = 146 protons; S, Z = 16, so 16 protons 5 = 80 protons Total protons = 226 protons Proton difference between Ta2S5 and Ta2O5: 226 protons – 186 protons = 40 protons 134.
The cation has 51 protons and 48 electrons. The number of protons corresponds to the atomic number. Thus, this is element 51, antimony. There are 3 fewer electrons than protons. Therefore, the charge on the cation is 3+. The anion has one-third the number of protons of the cation, which corresponds to 17 protons; this is element 17, chlorine. The number of electrons in this anion of chlorine is 17 + 1 = 18 electrons. The anion must have a charge of 1−. The formula of the compound formed between Sb3+ and Cl– is SbCl3. The name of the compound is antimony(III) chloride. The Roman numeral is used to indicate the charge on Sb because the predicted charge is not obvious from the periodic table.
Challenge Problems 135.
Copper (Cu), silver (Ag), and gold (Au) make up the coinage metals.
136.
Because the gases are at the same temperature and pressure, the volumes are directly proportional to the number of molecules present. Let’s assume hydrogen and oxygen to be monatomic gases and that water has the simplest possible formula (HO). We have the equation: H + O → HO But the volume ratios are also equal to the molecule ratios, which correspond to the coefficients in the equation: 2 H + O → 2 HO Because atoms cannot be created nor destroyed in a chemical reaction, this is not possible. To correct this, we can make oxygen a diatomic molecule: 2 H + O2 → 2 HO This does not require hydrogen to be diatomic. Of course, if we know water has the formula H2O, we get: 2 H + O2 → 2 H2O The only way to balance this is to make hydrogen diatomic: 2 H2 + O2 → 2 H2O
137.
Avogadro proposed that equal volumes of gases (at constant temperature and pressure) contain equal numbers of molecules. In terms of balanced equations, Avogadro’s hypothesis (law) implies that volume ratios will be identical to molecule ratios. Assuming one molecule of octane reacting, then 1 molecule of CxHy produces 8 molecules of CO2 and 9 molecules of H2O.
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CxHy + n O2 → 8 CO2 + 9 H2O. Because all the carbon in octane ends up as carbon in CO2, octane must contain 8 atoms of C. Similarly, all hydrogen in octane ends up as hydrogen in H2O, so one molecule of octane must contain 9 × 2 = 18 atoms of H. Octane formula = C 8H18, and the ratio of C : H = 8 : 18 or 4 : 9. 138.
From Section 2.5 of the text, the average diameter of the nucleus is about 10 −13 cm, and the electrons move about the nucleus at an average distance of about 10 −8 cm . From this, the diameter of an atom is about 2 10 −8 cm . 2 10 −8 cm 1 10
−13
cm
= 2 105;
1 mi 5280 ft 63,360 in = = 1 grape 1 grape 1 grape
Because the grape needs to be 2 105 times smaller than a mile, the diameter of the grape would need to be 63,360/(2 × 105) 0.3 in. This is a reasonable size for a small grape. 139.
The alchemists were incorrect. The solid residue must have come from the flask.
140.
The equation for the reaction would be 2 Na(s) + Cl2(g) → 2 NaCl(s). The sodium reactant exists as singular sodium atoms packed together very tightly and in a very organized fashion. This type of packing of atoms represents the solid phase. The chlorine reactant exists as Cl2 molecules. In the picture of chlorine, there is a lot of empty space present. This only occurs in the gaseous phase. When sodium and chlorine react, the ionic compound NaCl forms. NaCl exists as separate Na+ and Cl− ions. Because the ions are packed very closely together and are packed in a very organized fashion, NaCl is depicted in the solid phase.
141.
a. Both compounds have C2H6O as the formula. Because they have the same formula, their mass percent composition will be identical. However, these are different compounds with different properties because the atoms are bonded together differently. These compounds are called isomers of each other. b. When wood burns, most of the solid material in wood is converted to gases, which escape. The gases produced are most likely CO2 and H2O. c. The atom is not an indivisible particle but is instead composed of other smaller particles, called electrons, neutrons, and protons. d. The two hydride samples contain different isotopes of either hydrogen and/or lithium. Although the compounds are composed of different isotopes, their properties are similar because different isotopes of the same element have similar properties (except, of course, their mass).
142.
Let Xa be the formula for the atom/molecule X, Y b be the formula for the atom/molecule Y, XcYd be the formula of compound I between X and Y, and XeYf be the formula of compound II between X and Y. Using the volume data, the following would be the balanced equations to produce the two compounds. Xa + 2 Yb → 2 XcYd; 2 Xa + Yb → 2 XeYf From the balanced equations, a = 2c = e and b = d = 2f.
CHAPTER 2
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57
Substituting into the balanced equations: X2c + 2 Y2f → 2 XcY2f ; 2 X2c + Y2f → 2 X2cYf For simplest formulas, assume that c = f = 1. Thus: X2 + 2 Y2 → 2 XY2 and 2 X2 + Y2 → 2 X2Y Compound I = XY2: If X has relative mass of 1.00,
1.00 = 0.3043, y = 1.14. 1.00 + 2 y
Compound II = X2Y: If X has relative mass of 1.00,
2.00 = 0.6364, y = 1.14. 2.00 + y
The relative mass of Y is 1.14 times that of X. Thus, if X has an atomic mass of 100, then Y will have an atomic mass of 114. 143.
Most of the mass of the atom is due to the protons and the neutrons in the nucleus, and protons and neutrons have about the same mass (1.67 10−24 g). The ratio of the mass of the molecule to the mass of a nuclear particle will give a good approximation of the number of nuclear particles (protons and neutrons) present. 7.31 10 −23 g = 43.8 44 nuclear particles 1.67 10 −24 g
Thus there are 44 protons and neutrons present. If the number of protons equals the number of neutrons, we have 22 protons in the molecule. One possibility would be the molecule CO 2 [6 + 2(8) = 22 protons]. 144.
Number of electrons in the unknown ion: 2.55 × 10 −26 g ×
1 kg 1 electron = 28 electrons 1000 g 9.11 10 −31 kg
Number of protons in the unknown ion: 5.34 × 10 −23 g ×
1 kg 1 proton = 32 protons 1000 g 1.67 10 − 27 kg
Therefore, this ion has 32 protons and 28 electrons. This is element number 32, germanium (Ge). The net charge is 4+ because four electrons have been lost from a neutral germanium atom. The number of electrons in the unknown atom: 3.92 × 10 −26 g ×
1 kg 1 electron = 43 electrons 1000 g 9.11 0 −31 kg
In a neutral atom, the number of protons and electrons is the same. Therefore, this is element 43, technetium (Tc).
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The number of neutrons in the technetium atom: 9.35 × 10 −23 g ×
1 kg 1 proton = 56 neutrons 1000 g 1.67 10 − 27 kg
The mass number is the sum of the protons and neutrons. In this atom, the mass number is 43 protons + 56 neutrons = 99. Thus this atom and its mass number is 99Tc.
Marathon Problem 145.
a.
For each set of data, divide the larger number by the smaller number to determine relative masses. 0.602 = 2.04; A = 2.04 when B = 1.00 0.295 0.401 = 2.33; C = 2.33 when B = 1.00 0.172 0.374 = 1.17; C = 1.17 when A = 1.00 0.320
To have whole numbers, multiply the results by 3. Data set 1: A = 6.1 and B = 3.0 Data set 2: C = 7.0 and B = 3.0 Data set 3: C = 3.5 and A = 3.0 or C = 7.0 and A = 6.0 Assuming 6.0 for the relative mass of A, the relative masses would be A = 6.0, B = 3.0, and C = 7.0 (if simplest formulas are assumed). b. Gas volumes are proportional to the number of molecules present. There are many possible correct answers for the balanced equations. One such solution that fits the gas volume data is: 6 A2 + B4
→ 4 A3B
B4 + 4 C3 → 4 BC3 3 A2 + 2 C3 → 6 AC In any correct set of reactions, the calculated mass data must match the mass data given initially in the problem. Here, the new table of relative masses would be: 6 ( mass A 2 ) 0.602 = ; mass A2 = 0.340(mass B4) mass B 4 0.295
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59
4 ( mass C 3 ) 0.401 = ; mass C3 = 0.583(mass B4) mass B 4 0.172 2 (mass C 3 ) 0.374 = ; mass A2 = 0.570(mass C3) 3 (mass A 2 ) 0.320
Assume some relative mass number for any of the masses. We will assume that mass B = 3.0, so mass B4 = 4(3.0) = 12. Mass C3 = 0.583(12) = 7.0, mass C = 7.0/3 Mass A2 = 0.570(7.0) = 4.0, mass A = 4.0/2 = 2.0 When we assume a relative mass for B = 3.0, then A = 2.0 and C = 7.0/3. The relative masses having all whole numbers would be A = 6.0, B = 9.0, and C = 7.0. Note that any set of balanced reactions that confirms the initial mass data is correct. This is just one possibility.
CHAPTER 3 STOICHIOMETRY Review Questions 1.
Counting by weighing utilizes the average mass of a particular unit of substance. For marbles, a large sample size will contain many different individual masses for the various marbles. However, the large sample size will have an average mass so that the marbles behave as if each individual marble has that average mass. This assumption is valid if the sample size is large. When a large sample of marbles is weighed, one divides the total mass of marbles by the average mass of a marble, and this will give a very good estimate of the number of marbles present. For atoms, because we can’t count individual atoms, we “count” the atoms by weighing; converting the sample mass in grams to the number of atoms in the sample by using the average molar mass given in the periodic table and Avogadro’s number. The mole scale of atoms is a huge number (6.022 × 1023 atoms = 1 mole), so the assumption that a weighable sample size behaves as a bunch of atoms, each with an average mass, is valid and very useful.
2.
The masses of all the isotopes are relative to a specific standard. The standard is one atom of the carbon-12 isotope weighing exactly 12.0000 u. One can determine from experiment how much heavier or lighter any specific isotope is than 12C. From this information, we assign an atomic mass value to that isotope. For example, experiment tells one that 16O is about 4/3 heavier than 12C, so a mass of 4/3(12.00) = 16.00 u is assigned to 16O.
3.
The two major isotopes of boron are 10B and 11B. The listed mass of 10.81 is the average mass of a very large number of boron atoms.
4.
There are several ways to do this. The three conversion factors to use are Avogadro’s number, the molar mass, and the chemical formula. Two ways to use these conversions to convert grams of aspirin to number of H atoms are given below: molar mass of aspirin = 9(12.01) + 8(1.008) + 4(16.00) = 180.15 g/mol 1.00 g C9H8O4 ×
1 mol C 9 H 8 O 4 8 mol H 6.022 10 23 atoms H 180.15 g C 9 H 8 O 4 mol C 9 H 8 O 4 mol H
= 2.67 × 1022 H atoms or 1.00 g C9H8O4 ×
1 mol C 9 H 8 O 4 6.022 10 23 molecules C 9 H 8 O 4 × 180.15 g C 9 H 8 O 4 mol C 9 H 8 O 4
8 atoms H = 2.67 × 1022 H atoms molecule C 9 H 8 O 4
Of course the answer is the same no matter which order of the conversion factors is used.
60
CHAPTER 3 5.
STOICHIOMETRY
61
CxHyOz + oxygen → x CO2 + y/2 H2O From the equation above, the only reactant that contains carbon is the unknown compound and the only product that contains carbon is CO2. From the mass of CO2 produced, one can calculate the mass of C present which is also the mass of C in CxHyOz. Similarly, all the hydrogen in the unknown compound ends up as hydrogen in water. From the mass of H2O produced, one can calculate the mass of H in CxHyOz. Once the mass of C and H are known, the remainder of the compound is oxygen. From the masses of C, H, and O in the compound, one can then go on to determine the empirical formula.
6.
The molecular formula tells us the actual number of atoms of each element in a molecule (or formula unit) of a compound. The empirical formula tells only the simplest whole number ratio of atoms of each element in a molecule. The molecular formula is a whole number multiple of the empirical formula. If that multiplier is one, the molecular and empirical formulas are the same. For example, both the molecular and empirical formulas of water are H2O.
7.
The product of the reaction has two A atoms bonded to a B atom for a formula of A 2B. The initial reaction mixture contains 4 A2 and 8 AB molecules and the final reaction mixture contains 8 A2B molecules. The reaction is: 8 AB(g) + 4 A2(g) → 8 A2B(g) Using the smallest whole numbers, the balanced reaction is: 2 AB(g) + A2(g) → 2 A2B(g) 2.50 mol A2 ×
2 mol A 2 B mol A 2
= 5.00 mol A2B
The atomic mass of each A atom is 40.0/2 = 20.0 u and the atomic mass of each B atom is 30.0 − 20.0 = 10.0 u. The mass of A2B = 2(20.0) + 10.0 = 50.0 u. 15.0 g AB ×
1 mol A 2 40 .0 g A 2 1 mol AB = 10.0 g A2 30 .0 g AB 2 mol AB mol A 2
From the law of conservation of mass, the mass of product is: 10.0 g A2 + 15.0 g AB = 25.0 g A2B or by stoichiometric calculation: 15.0 g AB ×
1 mol A 2 B 50 .0 g A 2 B 1 mol AB = 25.0 g A2B 30 .0 g AB mol AB mol A 2 B
or 10.0 g A2 ×
1 mol A 2 2 mol A 2 B 50 .0 g A 2 B = 25.0 g A2B 40 .0 g A 2 mol A 2 mol A 2 B
Generally, there are several ways to correctly do a stoichiometry problem. You should choose the method you like best.
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CHAPTER 3 STOICHIOMETRY
8.
A limiting reactant problem gives you initial masses of two (or more) of the reactants and then asks for the amount of product that can form. Because one doesn’t know which reactant runs out first and hence determines the mass of product formed, a limiting reactant problem is a more involved problem. The first step in solving the problem is to figure which reactant runs out first (is limiting). One of the strategies outlined in the text is to calculate the mole ratio of reactants present and compare this mole ratio to that required from the balanced reaction. Whichever ratio is larger allows one to deduce the identity of the limiting reactant. After the limiting reactant is determined, that amount of the limiting reactant is used to calculate the amount of product that can form. Another strategy outlined in the text is to assume each reactant is limiting and then calculate for each reactant the amount of product that could form. This gives two or more possible answers. The correct answer is the mass of product that is smallest. Even though there is enough of the other reactant to form more product, once the smaller amount of product is formed, the limiting reactant has run out. A third strategy is to pick one of the reactants and then calculate the mass of the other reactant required to react with it. By comparing the calculated mass to the actual mass present in the problem, one can deduce the identity of the limiting reactant and go on to solve the problem.
9.
Balanced equation: 2 SO2(g) + O2(g) → 2 SO3(g) We have 6 SO2 and 6 O2 molecules present. If all six of the SO2 molecules react, then 3 molecules of O2 will react producing 6 molecules of SO3. These numbers were determined using the balanced reaction. Since 6 molecules of O2 are present, and only 3 react when the SO2 reacts completely, SO2 is limiting. The product mixture will contain 6 – 6 = 0 SO2 molecules, 6 – 3 = 3 O2 molecules in excess, and 6 molecules of SO3 formed.
Assuming SO2 is limiting, the amount of product that could be produced is: 96.0 g SO2 ×
2 mol SO 3 80 .07 g SO 3 1 mol SO 2 = 120. g SO3 64 .07 g SO 2 2 mol SO 2 mol SO 3
Now let’s assume that O2 is limiting. The amount of product that could be produced if O2 is limiting is:
CHAPTER 3
STOICHIOMETRY
32.0 g O2 ×
1 mol O 2 32.00 g O 2
63
2 mol SO 3 1 mol O 2
80.07 g SO 3 mol SO 3
= 160. g SO3
Because SO2 produces the smaller quantity of product (120. g versus 160. g), SO2 is limiting (runs out first) and 120. g of SO3 can form. As soon as 120. g of SO3 has formed, there is no more SO2 present for the reaction to continue, and the reaction ends. 10.
Side reactions may occur. For example, in the combustion of CH4 (methane) to CO2 and H2O, some CO may also form. Also, all reactions only go part way to completion. The actual endpoint of a reaction is the state of equilibrium, where both reactants and products are present (see Ch. 13 of the text).
Active Learning Questions 1.
Answer d is the best choice. For answer a, b, c, and e, a reaction that starts with any amount of reactants can result in product formation. All that matters is that all reactants must be present for a reaction to occur. The end point of the reaction is assumed to be completion. This is when one of the reactants completely runs out. A reaction can only occur if all reactants are present. A soon as one of the reactants runs out, products cannot be made anymore.
2.
A chemical formula is a representation of a molecule in which the symbols for the elements are used to indicate the types of atoms present and subscripts are used to show the relative numbers of atoms. A chemical equation is a representation of a chemical reaction showing the relative numbers of reactant and product molecules.
3.
False; for example, H2O has a larger mass percent of oxygen compared to hydrogen. Assuming 1 mol of H2O: Mass % O =
mass O 16.00 g O 100 = 100 = 88.79% mass H 2 O 18.02 g H 2 O
Water is 88.79% O and 11.21% H. The molar mass of the atoms in the formula must also be considered when determining which element has the largest mass percent. 4.
The data needed would be the mass of the pen before writing your name and the mass of the pen after writing your name. The difference of these numbers will be the mass of ink used. Also needed is the chemical make-up of the ink. In the simplest case, there is just one ink compound. Knowing the formula for the ink compound, you can determine the molar mass of the ink compound. To calculate the number of ink molecules, divide the mass of the ink by the molar mass of the compound; this gives the moles of ink molecules. Lastly, multiply the moles of ink used by Avogadro’s number to determine the number of ink molecules.
5.
1.33 cups butter ×
3 eggs = 2.00 eggs 2 cups butter
1.33 cups butter ×
72 cookies = 47.9 cookies 2 cups butter
With 1.33 cups of butter, 2 eggs will be needed and about 48 cookies can be made.
64
CHAPTER 3 STOICHIOMETRY 62.1 g ×
2 cups butter 1 egg = 1.82 eggs; 1.82 eggs × = 1.21 cups butter 34.21 g 3 eggs
1.21 cups of butter are required to react with all the eggs. Since we 1.33 cups of butter initially, we have an excess of butter (1.33-1.21 = 0.12 cups of butter in excess). The quantity of eggs present will be limiting, i.e., when we react all the eggs, we will have excess butter remaining. So, the quantity of eggs will determine the amount of cookies produced. 1.82 eggs × 6.
72 cookies = 43.7 cookies 3 eggs
A balanced equation for this reaction is N2(g) + 3 H2(g) → 2 NH3(g). We have 6 N2 and 6 H2 molecules present initially. If all the H2 reacts (if H2 is limiting), we will need only 2 molecules of N2 to react (from the 1:3 molecule ratio between nitrogen and hydrogen in the balanced equation). The amount of NH3 produced would be: 6 molecules hydrogen ×
2 molecules NH 3 = 4 molecules NH3 3 molecules N 2
Your picture of the final mixture should contain 4 molecules of NH3, 6 − 2 = 4 molecules of excess N2 molecules, and 0 molecules of H2. 7.
Equations d and e both correctly represent this reaction. Both these reactions have the correct chemical formulas of the reactant and products, and both are balanced (have the same number and types of atoms on both sides of the equation). Equation d is often thought of as the best balanced equation because it has the lowest possible overall coefficients. But both equations are fine to do stoichiometry problems. In reaction a, N2 is not a product in this reaction. The products will have different formulas than the reactants; if they don’t have different formulas, then they are just excess of one of the reactants that couldn’t react. Reaction b has the correct chemical formulas but is not balanced. And reaction c has incorrect formulas for hydrogen and nitrogen. Both are found in nature as diatomic molecules instead of individual atoms.
8.
In addition to the masses of the reactants, one needs to know the molar masses of the reactants and products as well as the balanced equation for the reaction. In order to determine which reactant is limiting, one needs to determine how the moles of each reactant is present and one needs to know the mole relationship between each reactant and the product. Lastly, to determine the mass of product, one needs to know the molar mass of the product.
9.
The mass of gases given off will be greater than 3.0 kg. Some of the elements in the charcoal will react with oxygen gas in air to form gaseous oxides of the elements. For example, the carbon in charcoal will react with oxygen from the atmosphere to form CO2(g) and the hydrogen in charcoal will end up as H2O(g). The atoms in 3.0 kg of charcoal will be reacted, and some oxygen from the atmosphere will be added to this mass.
10.
When iron rusts, some of the iron in the bar reacts with oxygen in the atmosphere to form an iron oxide, which is called rust. So, all the iron in the original bar is present, but some oxygen atoms have been added when the rusts form. The mass will be greater than 75.0 g.
CHAPTER 3
STOICHIOMETRY
65
11.
Water is formed when the hydrogen atoms in gasoline react with oxygen in the atmosphere to form H2O. So, water is not present in gasoline initially; it is formed in the combustion process when gasoline reacts with oxygen gas.
12.
The mass of product will be exactly 120 g. Since both reactants run out at the same time (both are limiting), the product will be the combined masses of the reactants. All the number and types of atoms in the reactants will be identical to the number and type of atoms in the products. It is just the formulas changed as the reaction occurred.
13.
The product has a different formula from the reactants. The properties depend on the formula of the product. The properties could be like chemical A or like chemical B or not like either. Answers d and e are best.
14.
The formulas of the species present are different. Hydrogen is H2, oxygen is O2, and water is H2O. The homogeneous mixture of hydrogen and oxygen must react to completion to form water vapor [2 H2(g) + O2(g) → 2 H2O(g)]. So yes, they are different. Also, one is a mixture and the other is a pure compound.
15.
The molar mass listed in the periodic table is an average mass of the various isotopes that make up an element. The mass of each isotope is about equal to the mass number. So 35Cl has a mass of about 35 u while 37Cl has a mass of about 37 u. The average mass of chlorine listed in the periodic table is 35.45 u. Since the average mass value is closer to the mass of the 35Cl, a typical sample of chlorine contains more 35Cl than 37Cl.
16.
There is a zero chance of a singular atom of carbon weighing 12.011 u. This number is the average mass of the various isotopes of carbon that make up a typical carbon sample. The lightest of these isotopes is 12C which has a mass of exactly 12.0000000… u. The next lightest isotope is 13C which has a mass of 13.0034 u. There are no isotopes of carbon that have a mass between 12.0000 and 13.0034.
17.
Subscripts in a chemical formula gives us the whole number ratio of the atoms that make up a compound. We can’t have fraction of atoms, so the numbers are whole numbers. For a balanced equation, we are generally most concerned with the mole relationships in the equation. Since a mole of a substance is a huge quantity, then yes, we can have fractions in a balanced equation. If you want an equation to exactly illustrate molecule relationships, then whole numbers are best. But to perform stoichiometry problems, using fractions to balance an equation is fine. Each compound has its own unique number and types of atoms in the formula. If you change the subscripts to balance an equation, then you are changing the identities of the reactants and/or products. This gives a different reaction altogether.
18.
Reactant A is used up twice as fast as reactant B is used up from the 2:1 mol ratio in the balanced equation. Since the reaction starts with equal moles of each reactant and since A is used up faster, reactant A is limiting (runs out first and determines the amount of product formed). The amount of product formed is: 1.0 mol A ×
1 mol A 2 B = 0.50 mol A2B can form 2 mol A
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CHAPTER 3 STOICHIOMETRY
19.
In a typical reaction, there is a reactant that runs out first (is limiting) with the other reactant(s) in excess. After a reaction, some product is formed, but excess of the non-limiting reactant(s) is also present. The final mass includes the mass of the product plus the mass of any excess reactants present. If all reactants run out at the same time (all are limiting), then, in this case, the mass of product will equal the total mass of reactants.
20.
Pairs a and c have the same empirical formula. The empirical formula is the smallest whole number ratio of the atoms in a formula. In pair a, both have CH as the empirical formula and in pair c, NO2 is the empirical formula for both compounds. In pair b, C2H6 has an empirical formula of CH3 while C4H10 has an empirical formula of C2H5. In pair d, C12H10O and C6H5OH each have an empirical formula which is the same as the molecular formula.
21.
The balance equation is:
From the balanced equation, we have a 3:1 molecule ratio between the first reactant and the second reactant. Since we have equal numbers of molecules of each reactant and since the first reactant is used up faster, it will be limiting; we will have excess of the second reactant. One molecule of the second reactant will be used up in the reaction leaving two molecules of the second reactant in excess. 22.
A mole is a specific number of whatever substance you are talking about. Specifically, 1 mol of things = Avogadro’s number of units of things = 6.022 × 1023 things. For example, 1 mol of book contains 6.022 × 1023 books. We define a mole to provide a way to count atoms or molecules or compounds. The mass of singular atoms is extremely miniscule, too miniscule to measure. The mass of a mol of atoms is a measurable quantity on the order of grams. So, we define a mole so we can count atoms/molecules/compounds by weighing them.
23.
In addition to the masses of the reactants, one needs to know the molar masses of the reactants and products as well as the balanced equation for the reaction. To determine which reactant is limiting, one need to determine how many moles of each reactant are present, and one needs to know the mole relationship between each reactant and the product. Lastly, to determine the mass of product, one needs to know the molar mass of the product.
24.
2 mol B ×
3 mol A = 6 mol A 1 mol B
When all the B is reacted, we need 6 mol of A to react with it. However, we only have 4 mol A present initially. This tells us that A must be limiting. It takes more A to react with all the B present than what is available. So, b is the correct statement. For statement a, we are given moles of reactants, so we don’t need molar mass to answer the question. And we just proved that statements c, d, and e are not correct. The limiting reactant depends on how many moles of reactant are present, but it also depends on the coefficients in the balanced equation.
CHAPTER 3 25.
STOICHIOMETRY
Mol of Ag = 20.0 g ×
67
1 mol Ag = 0.1854 mol Ag (carrying 1 extra sig fig) 107.9 g
For twice the number of atoms, we would need twice the moles of Ag, which is 2(0.1854) = 0.3708 mol. This number of moles has a mass of 5.0 g. Solving for the molar mass: 10.0 g = 27.0 g 0.3708 mol
From the periodic table, Al has this molar mass. Yes, it is possible for 10.0 g of another metal to have twice the number of atoms as 20.0 g Ag. For three times as many atoms, we need three times the moles of Ag which is 3(0.1854) = 0.5562 mol. This number of moles has a mass of 5.0 g. Solving for the molar mass: 5.0 g = 9.0 g/mol 0.5562 mol
This is the molar mass of Be, so this claim is possible. 26.
Let’s consider two sample calculations before answer the question. For the first problem, let’s determine which reactant is limiting when 2.0 g H2(g) reacts with 4.0 g O2(g) to form H2O(l). 2 H2(g) + O2(g) → 2 H2O(l) 2.0 g H2 ×
1 mol H 2 1 mol O 2 = 1.0 mol H2; 4.0 g O2 × = 0.25 mol O2 32.0 g 2.0 g
In this problem, O2 is limiting. When all the O2 reacts, only 0.50 mol H2 is needed (from the 2:1 mol ratio in the balanced equation). However, we have 1.0 mol H2 present. H2 is in excess and O2 is limiting. In the balanced equation, H2 has the largest coefficient, but it is not limiting. The claim that the limiting reactant has the largest coefficient in the balanced equation is false in this problem. For the second problem, let’s determine which reactant is limiting when 2.0 g H2(g) reacts with 24.0 g O2(g) to form H2O(l). 2.0 g H2 ×
1 mol H 2 1 mol O 2 = 1.0 mol H2; 24.0 g O2 × = 0.750 mol O2 2.0 g 32.0 g
In this problem, H2 is limiting. When all the H2 reacts, only 0.50 mol O2 is needed (from the 2:1 mol ratio in the balanced equation). However, we have 0.750 mol O2 present. O2 is in excess and H2 is limiting. In this problem, notice we have fewer moles of O2 initially, but H2 is limiting. The claim that the limiting reactant has the fewest moles present is also false. Let’s check the last claim that the limiting reactant has the lowest moles available/coefficient in the balance equation ratio. Determining the mole to coefficient ratio in the first problem:
68
CHAPTER 3 STOICHIOMETRY H2:
0.25 mol O2 1.0 mol H 2 mol mol = 0.50; O2: = 0.25 = = coefficient coefficient 1 2
O2 has the lowest mol/coefficient ratio, and it is limiting in problem 1. For the second problem: H2:
1.0 mol H 2 0.750 mol O 2 mol mol = 0.50; O2: = 0.750 = = coefficient 2 coefficient 1
H2 has the lowest mol/coefficient ratio, and it is limiting in problem 2. The claim that the limiting reactant has the lowest moles available to coefficient in the balanced equation ratio is correct. These are only two examples, but it is true for all limiting reactant problems. Having a small number of moles of reactant is helpful for a substance to be limiting, but also is having a large coefficient in the balanced equation. We want the smallest mole/coefficient ratio for the limiting reactant.
Questions 27.
The atomic mass of any isotope is a relative mass to a specific standard. The standard is one atom of the carbon-12 isotope weighing exactly 12.0000 u. One can determine from experiment how much heavier or lighter any specific isotope is than 12C. From this information, we assign an atomic mass value to that isotope. For example, experiment tells one that 16O is about 4/3 heavier than 12C, so a mass of 4/3(12.00) = 16.00 u is assigned to 16O. The atomic mass listed in the periodic table is also an average mass. Most elements in nature occur as a mixture of isotopes. The atomic mass of an element is the average mass of all the isotopes that make up a specific element, weighted by abundance.
28.
Consider a sample of glucose, C6H12O6. The molar mass of glucose is 180.16 g/mol. The chemical formula allows one to convert from molecules of glucose to atoms of carbon, hydrogen, or oxygen present and vice versa. The chemical formula also gives the mole relationship in the formula. One mole of glucose contains 6 mol C, 12 mol H, and 6 mol O. Thus mole conversions between molecules and atoms are possible using the chemical formula. The molar mass allows one to convert between mass and moles of compound, and Avogadro’s number (6.022 × 1023) allows one to convert between moles of compound and number of molecules.
29.
Avogadro’s number of dollars = 6.022 × 1023 dollars/mol dollars 6.022 10 23 dollars mol dollars = 7.6 × 1013 dollars/person 7.9 109 people
1 mol dollars
1 trillion = 1,000,000,000,000 = 1 × 1012; each person would have 76 trillion dollars.
CHAPTER 3 30.
STOICHIOMETRY
69
Molar mass of CO2 = 12.01 + 2(16.00) = 44.01 g/mol One mol of CO2 contains 6.022 1023 molecules of CO2, 6.022 1023 atoms of C, and 1.204 1024 atoms of O. We could also break down 1 mol of CO2 into the number of protons and the number of electrons present (1.325 1025 protons and 1.325 1025 electrons). To determine the number of neutrons present, we would need to know the isotope abundances for carbon and oxygen. The mass of 1 mol of CO2 would be 44.01 g. From the molar mass, one mol of CO2 would contain 12.01 g C and 32.00 g O. We could also break down 1 mol of CO 2 into the mass of protons and mass of electrons present (22.16 g protons and 1.207 10−2 g electrons). This assumes no mass loss when the individual particles come together to form the atom. This is not a great assumption as will be discussed in Chapter 19 on Nuclear Chemistry.
31.
Only in b are the empirical formulas the same for both compounds illustrated. In b, general formulas of X2Y4 and XY2 are illustrated, and both have XY2 for an empirical formula. For a, general formulas of X2Y and X2Y2 are illustrated. The empirical formulas for these two compounds are the same as the molecular formulas. For c, general formulas of XY and XY 2 are illustrated; these general formulas are also the empirical formulas. For d, general formulas of XY4 and X2Y6 are illustrated. XY4 is also the molecular formula, but X2Y6 has the empirical formula of XY3.
32.
The molar mass is the mass of 1 mole of the compound. The empirical mass is the mass of 1 mole of the empirical formula. The molar mass is a whole-number multiple of the empirical mass. The masses are the same when the molecular formula = empirical formula, and the masses are different when the two formulas are different. When different, the empirical mass must be multiplied by the same whole number used to convert the empirical formula to the molecular formula. For example, C6H12O6 is the molecular formula for glucose, and CH2O is the empirical formula. The whole-number multiplier is 6. This same factor of 6 is the multiplier used to equate the empirical mass (30 g/mol) of glucose to the molar mass (180 g/mol).
33.
The mass percent of a compound is a constant no matter what amount of substance is present. Compounds always have constant composition.
34.
A balanced equation starts with the correct formulas of the reactants and products. The coefficients necessary to balance the equation give molecule relationships as well as mole relationships between reactants and products. The state (phase) of the reactants and products is also given. Finally, special reaction conditions are sometimes listed above or below the arrow. These can include special catalysts used and/or special temperatures required for a reaction to occur.
35.
The theoretical yield is the stoichiometric amount of product that should form if the limiting reactant is completely consumed, and the reaction has 100% yield.
36.
One method is to assume each quantity of reactant is limiting, then calculate the amount of product that could be produced from each reactant. This gives two possible answers (assuming two reactants). The correct answer (the amount of product that could be produced) is always the smaller number. Even though there is enough of the other reactant to form more product, once the smaller quantity is reached, the limiting reactant runs out, and the reaction cannot continue.
70
CHAPTER 3 STOICHIOMETRY A second method would be to pick one of the reactants and then calculate how much of the other reactant would be required to react with all of it. How the answer compares to the actual amount of that reactant present allows one to deduce the identity of the limiting reactant. Once the identity is known, one would take the limiting reactant and convert it to mass of product formed.
Exercises Atomic Masses and the Mass Spectrometer 37.
Let A = average atomic mass; abundance of 70.95 u isotope = 100.00 – 60.16 = 39.84% A = 0.6016(68.95 u) + 0.3984(70.95 u) = 41.48 + 28.27 = 69.75 u Note: u is an abbreviation for amu (atomic mass units).
38.
The average atomic mass of bromine from the periodic table is 79.90 u. This is about halfway between the masses of the two isotopes (78.95 u and 80.95 u). Therefore, we have about a 50/50 mixture of the two isotopes in nature. Answer c is correct. The actual abundances are 50.7% and 49.3%.
39.
Let A = average atomic mass A = 0.0140(203.973) + 0.2410(205.9745) + 0.2210(206.9759) + 0.5240(207.9766) A = 2.86 + 49.64 + 45.74 + 109.0 = 207.2 u; from the periodic table, the element is Pb. Note: u is an abbreviation for amu (atomic mass units).
40.
Average atomic mass = A = 0.0800(45.952632) + 0.0730(46.951764) + 0.7380(47.947947) + 0.0550(48.947841) + 0.0540(49.944792) = 47.88 amu This is element Ti (titanium).
41.
Let A = mass of 185Re: 186.207 = 0.6260(186.956) + 0.3740(A), 186.207 − 117.0 = 0.3740(A) A=
42.
69.2 = 185 u (A = 184.95 u without rounding to proper significant figures.) 0.3740
Abundance 28Si = 100.00 − (4.70 + 3.09) = 92.21%; from the periodic table, the average atomic mass of Si is 28.09 u. 28.09 = 0.9221(27.98) + 0.0470(atomic mass 29Si) + 0.0309(29.97) Atomic mass 29Si = 29.01 u The mass of 29Si is a little less than 29 u. There are other isotopes of silicon that are considered when determining the 28.09 u average atomic mass of Si listed in the atomic table.
CHAPTER 3 43.
STOICHIOMETRY
71
Let x = % of 151Eu and y = % of 153Eu, then x + y = 100 and y = 100 − x. 151.96 =
x(150.9196) + (100 − x)(152.9209) 100
15196 = (150.9196)x + 15292.09 − (152.9209)x, −96 = −(2.0013)x x = 48%; 48% 151Eu and 100 − 48 = 52% 153Eu 44.
If silver is 51.82% 107Ag, then the remainder is 109Ag (48.18%). Determining the atomic mass (A) of 109Ag: 107.868 =
51.82(106.905) + 48.18(A) 100
10786.8 = 5540. + (48.18)A, A = 108.9 u = atomic mass of 109Ag 45.
There are three peaks in the mass spectrum, each 2 mass units apart. This is consistent with two isotopes differing in mass by two mass units. The peak at 157.84 corresponds to a Br 2 molecule composed of two atoms of the lighter isotope. This isotope has mass equal to 157.84/2 or 78.92. This corresponds to 79Br. The second isotope is 81Br with mass equal to 161.84/2 = 80.92. The peaks in the mass spectrum correspond to 79Br2, 79Br81Br, and 81Br2 in order of increasing mass. The intensities of the highest and lowest masses tell us the two isotopes are present in about equal abundance. The actual abundance is 50.68% 79Br and 49.32% 81Br.
46.
Because we are not given the relative masses of the isotopes, we need to estimate the masses of the isotopes. A good estimate is to assume that only the protons and neutrons contribute to the overall mass of the atom and that the atomic mass of a proton and neutron are each 1.00 u. So the masses are about: 54Fe, 54.00 u; 56Fe, 56.00 u; 57Fe, 57.00 u; 58Fe, 58.00 u. Using these masses, the calculated average atomic mass would be: 0.0585(54.00) + 0.9175(56.00) + 0.0212(57.00) + 0.0028(58.00) = 55.91 u The average atomic mass listed in the periodic table is 55.85 u.
Moles and Molar Masses 47.
When more than one conversion factor is necessary to determine the answer, we will usually put all the conversion factors into one calculation instead of determining intermediate answers. This method reduces round-off error and is a time saver. 500. atoms Fe
48.
500.0 g Fe ×
1 mol Fe 6.022 10 atoms Fe 23
55.85 g Fe = 4.64 × 10‒20 g Fe mol Fe
1 mol Fe = 8.953 mol Fe 55 .85 g Fe
8.953 mol Fe ×
6.022 10 23 atoms Fe = 5.391 × 1024 atoms Fe mol Fe
72
CHAPTER 3 STOICHIOMETRY 0.200 g C 1 mol C 6.022 × 10 23 atoms C × × = 1.00 × 1022 atoms C carat 12.01 g C mol C
49.
1.00 carat ×
50.
5.00 × 1020 atoms Cr × 8.30 × 10‒4 mol Cr ×
51.
1 mol Cr 6.022 10
23
= 8.30 × 10‒4 mol Cr
atoms Cr
52.00 g Cr = 0.0432 g Cr mol Cr
The formula of the metal phosphide compound would be M3P2. Let x = molar mass of M. 238.0 = 3x + 2(30.97), x = 58.69 g/mol; from the periodic table, M = Ni.
52.
The formula of the metal sulfide compound would be M2S3. Let x = molar mass of M. 298.4 = 2x + 3(32.07), x = 101.1 g/mol; from the periodic table, M = Ru
53.
C17H18F3NO (Prozac): 17(12.01) + 18(1.008) + 3(19.00) + 14.01 + 16.00 = 309.32 g/mol C17H17Cl2N (Zoloft): 17(12.01) + 17(1.008) + 2(35.45) + 14.01 = 306.22 g/mol
54.
HFC−134a, CH2FCF3: 2(12.01) + 2(1.008) + 4(19.00) = 102.04 g/mol HCFC−124, CHClFCF3: 2(12.01) + 1(1.008) + 1(35.45) + 4(19.00) = 136.48 g/mol
55.
a.
The formula is NH3. 14.01 g/mol + 3(1.008 g/mol) = 17.03 g/mol
b. The formula is N2H4. 2(14.01) + 4(1.008) = 32.05 g/mol c. (NH4)2Cr2O7: 2(14.01) + 8(1.008) + 2(52.00) + 7(16.00) = 252.08 g/mol 56.
a. The formula is P4O6. 4(30.97 g/mol) + 6(16.00 g/mol) = 219.88 g/mol b. Ca3(PO4)2: 3(40.08) + 2(30.97) + 8(16.00) = 310.18 g/mol c. Na2HPO4: 2(22.99) + 1(1.008) + 1(30.97) + 4(16.00) = 141.96 g/mol
57.
a. 1.00 g NH3 × b. 1.00 g N2H4 ×
1 mol NH 3 = 0.0587 mol NH3 17.03 g NH 3 1 mol N 2 H 4 32.05 g N 2 H 4
c. 1.00 g (NH4)2Cr2O7 ×
58.
a. 1.00 g P4O6 ×
= 0.0312 mol N2H4
1 mol (NH 4 ) 2 Cr2 O 7 = 3.97 × 10‒3 mol (NH4)2Cr2O7 252.08 g (NH 4 ) 2 Cr2 O 7
1 mol P 4 O 6 = 4.55 × 10‒3 mol P4O6 219.88 g
CHAPTER 3
59.
STOICHIOMETRY
b. 1.00 g Ca3(PO4)2 ×
1 mol Ca 3 (PO 4 ) 2 = 3.22 × 10‒3 mol Ca3(PO4)2 310.18 g
c. 1.00 g Na2HPO4 ×
1 mol Na 2 HPO 4 = 7.04 × 10‒3 mol Na2HPO4 141.96 g
a. 5.00 mol NH3 × b. 5.00 mol N2H4 ×
17.03 g NH 3 mol NH 3
61.
a. 5.00 mol P4O6 ×
mol N 2 H 4
63.
= 160. g N2H4
252.08 g (NH 4 ) 2 Cr2 O 7 = 1260 g (NH4)2Cr2O7 1 mol (NH 4 ) 2 Cr2 O 7
219.88 g = 1.10 × 103 g P4O6 1 mol P 4 O 6
b. 5.00 mol Ca3(PO4)2 ×
310.18 g = 1.55 × 103 g Ca3(PO4)2 mol Ca 3 (PO 4 ) 2
c. 5.00 mol Na2HPO4 ×
141.96 g = 7.10 × 102 g Na2HPO4 mol Na 2 HPO 4
Chemical formulas give atom ratios as well as mole ratios. a. 5.00 mol NH3 ×
1 mol N 14.01 g N = 70.1 g N mol NH 3 mol N
b. 5.00 mol N2H4 ×
2 mol N 14.01 g N = 140. g N mol N 2 H 4 mol N
c. 5.00 mol (NH4)2Cr2O7 ×
62.
= 85.2 g NH3
32 .05 g N 2 H 4
c. 5.00 mol (NH4)2Cr2O7 ×
60.
73
a. 5.00 mol P4O6 ×
2 mol N 14.01 g N = 140. g N mol (NH 4 ) 2 Cr2 O 7 mol N
4 mol P 30.97 g P = 619 g P mol P4 O 6 mol P
b. 5.00 mol Ca3(PO4)2 ×
2 mol P 30.97 g P = 310. g P mol Ca 3 (PO 4 ) 2 mol P
c. 5.00 mol Na2HPO4 ×
1 mol P 30.97 g P = 155 g P mol Na 2 HPO 4 mol P
1 mol NH3 6.022 1023 molecules NH3 a. 1.00 g NH3 × 17.03 g NH3 mol NH3 = 3.54 × 1022 molecules NH3
74
CHAPTER 3 STOICHIOMETRY b. 1.00 g N2H4 ×
1 mol N 2 H 4 32.05 g N 2 H 4
6.022 10 23 molecules N 2 H 4 mol N 2 H 4 = 1.88 × 1022 molecules N2H4
c. 1.00 g (NH4)2Cr2O7 ×
64.
1 mol (NH 4 ) 2 Cr2 O 7 252.08 g (NH 4 ) 2 Cr2 O 7
6.022 1023 formula units (NH 4 )2 Cr2 O7 = 2.39 × 1021 formula units (NH4)2Cr2O7 mol (NH 4 ) 2 Cr2 O7
a. 1.00 g P4O6 ×
1 mol P4 O6 6.022 1023 molecules = 2.74 × 1021 molecules P4O6 219.88 g mol P4 O6
b. 1.00 g Ca3(PO4)2 ×
1 mol Ca 3 (PO 4 ) 2 6.022 1023 formula units 310.18 g mol Ca 3 (PO 4 ) 2
= 1.94 × 1021 formula units Ca3(PO4)2 c. 1.00 g Na2HPO4 ×
1 mol Na 2 HPO 4 6.022 1023 formula units 141.96 g mol Na 2 HPO 4
= 4.24 × 1021 formula units Na2HPO4 65.
Using answers from Exercise 63: a. 3.54 × 1022 molecules NH3 ×
1 atom N = 3.54 × 1022 atoms N molecule NH 3
b. 1.88 × 1022 molecules N2H4 ×
2 atoms N = 3.76 × 1022 atoms N molecule N 2 H 4
c. 2.39 × 1021 formula units (NH4)2Cr2O7 ×
2 atoms N formula unit (NH 4 ) 2 Cr2 O 7
= 4.78 × 1021 atoms N 66.
Using answers from Exercise 62: a. 2.74 × 1021 molecules P4O6 ×
67.
4 atoms P = 1.10 × 1022 atoms P molecule P4 O 6
b. 1.94 × 1021 formula units Ca3(PO4)2 ×
2 atoms P = 3.88 × 1021 atoms P formula unit Ca 3 (PO 4 ) 2
c. 4.24 × 1021 formula units Na2HPO4 ×
1 atom P = 4.24 × 1021 atoms P formula unit Na 2 HPO 4
38 g Al2S3 ×
1 mol Al2S3 3 mol S 6.02 10 23 atoms S = 4.6 × 1023 atoms S 150.17 g mol Al 2S3 mol S
CHAPTER 3
STOICHIOMETRY
75
1 mol Ba 3 (PO 4 ) 2 8 mol O 6.022 10 23 atoms O 601.8 g mol Ba 3 (PO 4 ) 2 mol O = 4.6 × 1023 atoms S
68.
3.10 g Ba3(PO4)2 ×
69.
Molar mass CH4 = 12.0 + 4(1.0) = 16.0 g/mol; from the molar mass calculation, 16.0 g of methane is 1.00 mol of CH4, contains 12.0 g C, and 4.0 g of H. So, 16.0 g CH4 represents 6.02 × 1023 molecules of methane, not 16.0(6.02 × 1023). Lastly, 16.0 g CH4 contains 4 mol of H and 4(6.02 × 1023) atoms of H. Answer d is the correct answer to the question.
70.
Molar mass C8H18 = 8(12.01) + 18(1.008) = 114.22 g/mol; from the molar mass calculation, 114.2 g of isooctane is 1.000 mol of C8H18, contains 8 mol C, 8(12.01) = 96.08 g C, 18 mol H, 18(1.008) = 18.14 g H. Lastly, 114.2 g C8H18 represents 6.022 × 1023 molecules of isooctane, and 8( 6.022 × 1023) atoms of C. Only answer c is a correct statement.
71.
2.59 × 1023 atoms H ×
1 mol H 6.022 10
23
atoms H
1 mol C 4 H10 58.12 g C 4 H10 10 mol H mol C 4 H10
= 2.50 g C4H10 72.
a. 16(12.01) + 14(1.008) + 3(19.00) + 3(14.01) + 2(16.00) + 32.07 = 369.37 g/mol b. 0.75 mol C16H14F3N3O2S
19.00 g F 3 mol F = 43 g F mol F mol C16 H14 F3 N 3 O 2S
c. 0.75 mol C16H14F3N3O2S
16 mol C 6.022 1023 atoms mol C16 H14 F3 N 3O 2S mol C
= 7.2 × 1024 atoms C d. 4.25 × 1021 molecules C16H14F3N3O2S
1 mol C16 H14 F3 N 3O 2S 369.37 g 23 6.022 10 molecules mol
= 2.61 g C16H14F3N3O2S 73.
a.
1 mol 1.50 g CO2 × = 3.41 × 10 ‒2 mol CO2 44.01 g
b. 2.72 × 1021 molecules C2H5OH ×
1 mol 6.022 10 23 molecules
= 4.52 × 10 ‒3 mol C2H5OH c. 20.0 mg C8H10N4O2 × 74.
1g 1 mol = 1.03 × 10 ‒4 mol C8H10N4O2 1000 mg 194 .20 g
a. A chemical formula gives atom ratios as well as mole ratios. We will use both ideas to Illustrate how these conversion factors can be used. Molar mass of C2H5O2N = 2(12.01) + 5(1.008) + 2(16.00) + 14.0l = 75.07 g/mol
76
CHAPTER 3 STOICHIOMETRY 5.00 g C2H5O2N
1 mol C2 H 5O 2 N 6.022 1023 molecules C2 H 5O 2 N 75.07 g C2 H 5 O 2 N mol C2 H 5O 2 N
1 atom N = 4.01 × 1022 atoms N molecule C 2 H 5 O 2 N
b. Molar mass of Mg3N2 = 3(24.31) + 2(14.01) = 100.95 g/mol 1 mol Mg 3 N 2 6.022 1023 formula units Mg 3 N 2 100.95 g Mg 3 N 2 mol Mg 3 N 2
5.00 g Mg3N2
2 atoms N = 5.97 × 1022 atoms N mol Mg 3 N 2
c. Molar mass of Ca(NO3)2 = 40.08 + 2(14.01) + 6(16.00) = 164.10 g/mol 5.00 g Ca(NO3)2
1 mol Ca(NO3 ) 2 2 mol N 6.022 1023 atoms N 164.10 g Ca(NO3 ) 2 mol Ca(NO3 ) 2 mol N
= 3.67 × 1022 atoms N d. Molar mass of N2O4 = 2(14.01) + 4(16.00) = 92.02 g/mol 5.00 g N2O4
75.
Molar mass of C6H8O6 = 6(12.01) + 8(1.008) + 6(16.00) = 176.12 g/mol 10 tablets ×
8 tablets × 76.
1 mol N 2 O 4 2 mol N 6.022 1023 atoms N 92.02 g N 2 O 4 mol N 2 O4 mol N = 6.54 × 1022 atoms N
1g 1 mol 500 .0 mg = 2.839 × 10‒2 mol C6H8O6 tablet 1000 mg 176.12 g
1 mol 0.5000 g tablet 176.12 g
6.022 10 23 molecules = 1.368 × 1022 molecules mol
a. 9(12.01) + 8(1.008) + 4(16.00) = 180.15 g/mol b. 500. mg ×
1g 1 mol × = 2.78 × 10‒3 mol C9H8O4 1000 mg 180.15 g
2.78 × 10‒3 mol × 77.
a.
2(12.01) + 3(1.008) + 3(35.45) + 2(16.00) = 165.39 g/mol
b. 500.0 g ×
c.
6.022 1023 molecules = 1.67 × 1021 molecules C9H8O4 mol
1 mol = 3.023 mol C2H3Cl3O2 165 .39 g
2.0 × 10-2 mol ×
165.39 g = 3.3 g C2H3Cl3O2 mol
CHAPTER 3
STOICHIOMETRY
d. 5.0 g C2H3Cl3O2 ×
77
1 mol 6.022 1023 molecules 3 atoms Cl 165.39 g mol molecule = 5.5 × 1022 atoms of chlorine
78.
1 mol C 2 H 3 Cl3 O 2 165.39 g C 2 H 3Cl 3O 2 1 mol Cl = 1.6 g C2H3Cl3O2 × × 35.45 g 3 mol Cl mol C 2 H 3Cl 3O 2
e.
1.0 g Cl ×
f.
500 molecules ×
1 mol 6.022 10
23
molecules
165.39 g = 1.373 × 10−19 g C2H3Cl3O2 mol
As we shall see in later chapters, the formula written as (CH3)2N2O tries to tell us something about how the atoms are attached to each other. For our purposes in this problem, we can write the formula as C2H6N2O. a. 2(12.01) + 6(1.008) + 2(14.01) + 1(16.00) = 74.09 g/mol b. 250 mg ×
1g 1 mol × = 3.4 × 10−3 mol 1000 mg 74.09 g
d. 1.0 mol C2H6N2O
c. 0.050 mol ×
74.09 g = 3.7 g mol
6.022 1023 molecules C 2 H 6 N 2 O 6 atoms of H mol C2 H 6 N 2 O molecule C 2 H 6 N 2 O
= 3.6 × 1024 atoms of hydrogen e. 1.0 × 106 molecules f.
1 mol 74.09 g = 1.2 × 10−16 g 23 6.022 10 molecules mol
1 mol 74.09 g = 1.230 × 10−22 g C2H6N2O 23 6.022 10 molecules mol
1 molecule
Percent Composition 79.
a. C3H4O2: Molar mass = 3(12.01) + 4(1.008) + 2(16.00) = 36.03 + 4.032 + 32.00 = 72.06 g/mol Mass % C =
36 .03 g C × 100 = 50.00% C 72 .06 g compound
Mass % H =
4.032 g H × 100 = 5.595% H 72 .06 g compound
Mass % O = 100.00 − (50.00 + 5.595) = 44.41% O or: %O=
32 .00 g × 100 = 44.41% O 72 .06 g
b. C4H6O2: Molar mass = 4(12.01) + 6(1.008) + 2(16.00) = 48.04 + 6.048 + 32.00 = 86.09 g/mol 48.04 g 6.048 g Mass % C = × 100 = 55.80% C; mass % H = × 100 = 7.025% H 86.09 g 86.09 g Mass % O = 100.00 − (55.80 + 7.025) = 37.18% O
78
CHAPTER 3 STOICHIOMETRY c. C3H3N: Molar mass = 3(12.01) + 3(1.008) + 1(14.01) = 36.03 + 3.024 + 14.01 = 53.06 g/mol Mass % C =
36.03 g 53.06 g
× 100 = 67.90% C; mass % H =
Mass % N =
14 .01 g 53 .06 g
× 100 = 26.40% N or % N = 100.00 − (67.90 + 5.699)
3.024 g 53 .06 g
× 100 = 5.699% H
= 26.40% N 80.
In 1 mole of YBa2Cu3O7, there are 1 mole of Y, 2 moles of Ba, 3 moles of Cu, and 7 moles of O. 88.91 g Y 137.3 g Ba + 2 mol Ba Molar mass = 1 mol Y mol Y mol Ba 63.55 g Cu 16.00 g O + 7 mol O + 3 mol Cu mol Cu mol O
Molar mass = 88.91 + 274.6 + 190.65 + 112.00 = 666.2 g/mol
81.
Mass % Y =
274.6 g 88.91 g × 100 = 13.35% Y; mass % Ba = × 100 = 41.22% Ba 666.2 g 666.2 g
Mass % Cu =
190.65 g 112.0 g × 100 = 28.62% Cu; mass % O = × 100 = 16.81% O 666.2 g 666.2 g
NO: Mass % N =
14 .01 g N 30 .01 g NO
× 100 = 46.68% N
NO2: Mass % N =
14.01 g N 46.01 g NO 2
× 100 = 30.45% N
N2O: Mass % N =
2(14.01) g N × 100 = 63.65% N 44.02 g N 2 O
From the calculated mass percent values, only NO is 46.7% N by mass, so NO could be this species. Any other compound having NO as an empirical formula could also be the compound. 82.
a.
C8H10N4O2: Molar mass = 8(12.01) + 10(1.008) + 4(14.0l) + 2(16.00) = 194.20 g/mol Mass % C =
8(12.01) g C 96.08 g × 100 = × 100 = 49.47% C 194.20 g C 8 H10 N 4 O 2 194.20 g
b. C12 H22O11: Molar mass = 12(12.01) + 22(1.008) + 11(16.00) = 342.30 g/mol Mass % C = c.
12(12.01) g C × 100 = 42.10% C 342.30 g C12 H 22 O11
C2H5OH: Molar mass = 2(12.01) + 6(1.008) + 1(16.00) = 46.07 g/mol
CHAPTER 3
STOICHIOMETRY
Mass % C =
79
2(12.01) g C × 100 = 52.14% C 46.07 g C 2 H 5 OH
The order from lowest to highest mass percentage of carbon is: sucrose (C12H22O11) < caffeine (C8H10N4O2) < ethanol (C2H5OH) 83.
Assuming 100.0 g of sarin, there are 22.8 g of O present. The moles of sarin in 100.0 g is: 22.8 g O ×
1 mol O 1 mol sarin = 0.713 mol sarin 16.00 g 2 mol O
Molar mass is a ratio between the mass of substance and moles of substance. Molar mass = 84.
mass 100.0 g = 140. g/mol = mol 0.713 mol
Assuming 100.00 g of M2O3, there are 100.00 ‒ 68.42 = 31.58 g O present. The moles of M present in 100.00 g is: 31.58 g O ×
1 mol O 2 mol M = 1.316 mol M 16.00 g 3 mol O
So, 68.42 g M represents 1.316 mol M. The molar mass of M is: mass 68.42 g = 51.99 g/mol; the atomic mass of M = 51.99 u. Note: M = Cr. = mol 1.316 mol
85.
There are 0.390 g Cu for every 100.000 g of fungal laccase. Assuming 100.00 g fungal laccase: mol fungal laccase = 0.390 g Cu ×
1 mol Cu 1 mol fungal laccase = 1.53 × 10‒3 mol 63 .55 g Cu 4 mol Cu
100.000 g x g fungal laccase = , x = molar mass = 6.54 × 104 g/mol mol fungal laccase 1.53 10 −3 mol
86.
There are 0.347 g Fe for every 100.000 g hemoglobin (Hb). Assuming 100.000 g hemoglobin: 1 mol Fe 1 mol Hb × Mol Hb = 0.347 g Fe × = 1.55 × 10−3 mol Hb 55.85 g Fe 4 mol Fe x g Hb 100.000 g Hb = , x = molar mass = 6.45 × 104 g/mol 1.55 10 −3 mol Hb mol Hb
Empirical and Molecular Formulas 87.
12.01 g C 1.008 g H + 2 mol H a. Molar mass of CH2O = 1 mol C mol C mol H 16.00 g O = 30.03 g/mol + 1 mol O mol O
80
CHAPTER 3 STOICHIOMETRY %C=
12.01 g C 2.016 g H × 100 = 39.99% C; % H = × 100 = 6.713% H 30.03 g CH 2 O 30.03 g CH 2 O
%O=
16.00 g O × 100 = 53.28% O or % O = 100.00 − (39.99 + 6.713) = 53.30% 30.03 g CH 2 O
b. Molar mass of C6H12O6 = 6(12.01) + 12(1.008) + 6(16.00) = 180.16 g/mol %C=
12(1.008) g H 6(12.01) g C × 100 = 40.00%; % H = × 100 = 6.714% 180.16 g C 6 H12 O 6 180.16 g C 6 H12 O 6
% O = 100.00 − (40.00 + 6.714) = 53.29% c. Molar mass of HC2H3O2 = 2(12.01) + 4(1.008) + 2(16.00) = 60.05 g/mol %C=
2(12.01) g C × 100 = 40.00%; 60.05 g HC 2 H 3 O 2
%H=
4(1.008) g H × 100 = 6.714% 60.05 g HC 2 H 3 O 2
% O = 100.00 − (40.00 + 6.714) = 53.29% 88.
All three compounds have the same empirical formula, CH2O, and different molecular formulas. The composition of all three in mass percent is also the same (within rounding differences). Therefore, elemental analysis will give us only the empirical formula.
89.
a. The molecular formula is N2O4. The smallest whole number ratio of the atoms (the empirical formula) is NO2. b. Molecular formula: C3H6; empirical formula: CH2 c. Molecular formula: P4O10; empirical formula: P2O5 d. Molecular formula: C6H12O6; empirical formula: CH2O
90.
a. SNH: Empirical formula mass = 32.07 + 14.01 + 1.008 = 47.09 g/mol 188.35 g = 4.000; so the molecular formula is (SNH)4 or S4N4H4. 47.09 g
b. NPCl2: Empirical formula mass = 14.01 + 30.97 + 2(35.45) = 115.88 g/mol 347.64 g = 3.0000; molecular formula is (NPCl2)3 or N3P3Cl6. 115.88 g
c. CoC4O4: 58.93 + 4(12.01) + 4(16.00) = 170.97 g/mol 341.94 g = 2.0000; molecular formula: Co2C8O8 170.97 g
d. SN: 32.07 + 14.01 = 46.08 g/mol; 91.
184.32 g = 4.000; molecular formula: S4N4 46.08 g
Mass O in compound = 6.91 g FexOy ‒ 5.00 g Fe = 1.91 g O
CHAPTER 3
STOICHIOMETRY
5.00 g Fe ×
81
1 mol Fe 1 mol O = 0.0895 mol Fe; 1.91 g O × = 0.119 mol O 55.85 g Fe 16.00 g O
The mol ratio between mol O to mol Fe is:
0.119 mol O = 1.33; this is a 4:3 mol ratio. 0.0895 mol Fe
Fe3O4 is the empirical formula. 92.
Mass O in compound = 111.2 g NxOy ‒ 70.8 g N = 40.4 g O 70.8 g N ×
1 mol N 1 mol O = 5.05 mol N; 40.4 g O × = 2.53 mol O 14.01 g N 16.00 g O
The mol ratio between mol N to mol O is: 93.
5.05 mol N = 2.00; N2O is the empirical formula. 2.53 mol O
Out of 100.00 g of compound, there are: 62.58 g C ×
1 mol C 1 mol H = 5.211 mol C; 9.63 g H × = 9.55 mol H 1.008 g H 12.01 g C
27.79 g O ×
1 mol O = 1.737 mol O 16.00 g O
Dividing each mole value by the smallest number: 9.55 1.737 5.211 = 3.000; = 5.50; = 1.000 1.737 1.737 1.737
The C : H : O mol ratio is 3 : 5.5 : 1. Because a whole number ratio is required, this is 6 : 11 : 2. So the empirical formula is C6H11O2. 94.
Assuming 100.00 g of nylon-6: 63.68 g C ×
9.80 g H ×
1 mol N 1 mol C = 5.302 mol C; 12.38 g N × = 0.8837 mol N 14.01 g N 12.01 g C
1 mol H 1 mol O = 9.72 mol H; 14.14 g O × = 0.8838 mol O 1.008 g H 16.00 g O
Dividing each mole value by the smallest number: 5.302 9.72 0.8838 = 6.000; = 11.0; = 1.000 0.8837 0.8837 0.8837
The empirical formula for nylon-6 is C6H11NO 95.
Compound I: Mass O = 0.6498 g HgxOy − 0.6018 g Hg = 0.0480 g O 0.6018 g Hg ×
0.0480 g O ×
1 mol Hg = 3.000 × 10−3 mol Hg 200.6 g Hg
1 mol O = 3.00 × 10−3 mol O 16.00 g O
82
CHAPTER 3 STOICHIOMETRY The mole ratio between Hg and O is 1 : 1, so the empirical formula of compound I is HgO. Compound II: Mass Hg = 0.4172 g HgxOy − 0.016 g O = 0.401 g Hg 0.401 g Hg ×
1 mol Hg 1 mol O = 2.00 × 10−3 mol Hg; 0.016 g O × = 1.0 × 10−3 mol O 16.00 g O 200.6 g Hg
The mole ratio between Hg and O is 2 : 1, so the empirical formula is Hg2O. 96.
1.121 g N ×
1 mol N 14.01 g N
= 8.001 × 10‒2 mol N; 0.161 g H ×
0.480 g C ×
1 mol C 12.01 g C
= 4.00 × 10‒2 mol C; 0.640 g O ×
1 mol H 1.008 g H
= 1.60 × 10‒1 mol H
1 mol O 16.00 g O
= 4.00 × 10‒2 mol O
Dividing all mole values by the smallest number: 4.00 10−2 1.60 × 10-1 = 4.00; 4.00 10−2 4.00 × 10-2
8.001 10−2 = 2.00; 4.00 10−2
= 1.00
The empirical formula is N2H4CO. 97.
Out of 100.0 g, there are: 69.6 g S ×
1 mol N 1 mol S = 2.17 mol S; 30.4 g N × = 2.17 mol N 14.01 g N 32.07 g S
The empirical formula is SN because the mole values are in a 1 : 1 mole ratio. The empirical formula mass of SN is ~46 g/mol. Because 184/46 = 4.0, the molecular formula is S4N4. 98.
Assuming 100.0 g of compound: 26.7 g P ×
1 mol P 30.97 g P
61.2 g Cl ×
1 mol Cl 35.45 g Cl
= 0.862 mol P; 12.1 g N ×
1 mol N 14.01 g N
= 0.864 mol N
= 1.73 mol Cl
1.73 = 2.01; the empirical formula is PNCl2. 0.862
The empirical formula mass is 31.0 + 14.0 + 2(35.5) = 116 g/mol. Molar mass Empirical formula mass
=
580 = 5.0; the molecular formula is (PNCl2)5 = P5N5Cl10 116
CHAPTER 3 99.
STOICHIOMETRY
83
Assuming 100.00 g of compound: 47.08 g C ×
1 mol C 1 mol H = 3.920 mol C; 6.59 g H × = 6.54 mol H 1.008 g H 12.01 g C
46.33 g Cl ×
1 mol Cl = 1.307 mol Cl 35.45 g Cl
Dividing all mole values by 1.307 gives: 1.307 3.920 6.54 = 2.999; = 5.00; = 1.000 1.307 1.307 1.307
The empirical formula is C3H5Cl. The empirical formula mass is 3(12.01) + 5(1.008) + 1(35.45) = 76.52 g/mol. Molar mass 153 = = 2.00; the molecular formula is (C3H5Cl)2 = C6H10Cl2. Empirical formula mass 76.52
100.
Assuming 100.00 g of compound (mass oxygen = 100.00 g − 41.39 g C − 3.47 g H = 55.14 g O): 1 mol C 1 mol H 41.39 g C × = 3.446 mol C; 3.47 g H × = 3.44 mol H 1.008 g H 12.01 g C 55.14 g O ×
1 mol O 16.00 g O
= 3.446 mol O
All are the same mole values, so the empirical formula is CHO. The empirical formula mass is 12.01 + 1.008 + 16.00 = 29.02 g/mol. Molar mass =
15.0 g 0.129 mol
= 116 g/mol
Molar mass 116 = = 4.00; molecular formula = (CHO)4 = C4H4O4 Empirical mass 29.02
101.
When combustion data are given, it is assumed that all the carbon in the compound ends up as carbon in CO2 and all the hydrogen in the compound ends up as hydrogen in H2O. In the sample of fructose combusted, the masses of C and H are: mass C = 2.20 g CO2 ×
1 mol CO 2 1 mol C 12.01 g C = 0.600 g C 44.01 g CO 2 1 mol CO 2 mol C
mass H = 0.900 g H2O ×
1 mol H 2 O 2 mol H 1.008 g H = 0.101 g H 18.02 g H 2 O 1 mol H 2 O mol H
Mass O = 1.50 g fructose − 0.600 g C − 0.101 g H = 0.799 g O So, in 1.50 g of the fructose, we have:
84
CHAPTER 3 STOICHIOMETRY 0.600 g C ×
1 mol C 1 mol H = 0.0500 mol C; 0.101 g H × = 0.100 mol H 1.008 g H 12.01 g C
0.799 g O ×
1 mol O = 0.0499 mol O 16.00 g O
Dividing by the smallest number: 102.
0.100 = 2.00; the empirical formula is CH2O. 0.0499
This compound contains nitrogen, and one way to determine the amount of nitrogen in the compound is to calculate composition by mass percent. We assume that all the carbon in 33.5 mg CO2 came from the 35.0 mg of compound and all the hydrogen in 41.1 mg H2O came from the 35.0 mg of compound. 3.35 × 10‒2 g CO2 × Mass % C =
1 mol C 12.01 g C = 9.14 × 10‒3 g C mol CO 2 mol C
9.14 10−3 g C × 100 = 26.1% C 3.50 10−2 g compound
4.11 × 10‒2 g H2O ×
Mass % H =
1 mol CO 2 44.01 g CO 2
1 mol H 2 O 18.02 g H 2 O
2 mol H 1.008 g H = 4.60 × 10‒3 g H mol H 2 O mol H
4.60 10 −3 g H 3.50 10 − 2 g compound
× 100 = 13.1% H
The mass percent of nitrogen is obtained by difference: Mass % N = 100.0 − (26.1 + 13.1) = 60.8% N Now perform the empirical formula determination by first assuming 100.0 g of compound. Out of 100.0 g of compound, there are: 26.1 g C ×
1 mol C 1 mol H = 2.17 mol C; 13.1 g H × = 13.0 mol H 1.008 g H 12.01 g C
60.8 g N ×
1 mol N = 4.34 mol N 14.01 g N
Dividing all mole values by 2.17 gives:
13.0 2.17 4.34 = 1.00; = 5.99; = 2.00 2.17 2.17 2.17
The empirical formula is CH6N2. 103.
a. 0.2833 g CO2 ×
1 mol CO 2 1 mol C 12.01 g C = 0.07731 g C 44.01 g CO 2 1 mol CO 2 mol C
0.1160 g H2O ×
1 mol H 2 O 2 mol H 1.008 g H = 0.01298 g H 18.02 g H 2 O 1 mol H 2 O mol H
CHAPTER 3
STOICHIOMETRY
85
Mass O = 0.1006 g compound ‒ 0.07731 g C ‒ 0.01298 g H = 0.0103 g O Mass percent O =
0.0103 g O × 100 = 10.2% O 0.1006 g compound
b. From the work in part a, 0.1006 g menthol contains 0.07731 g C, 0.01298 g H, and 0.0103 g O. Let’s use this mass breakdown to solve the empirical formula problem. 1 mol C 1 mol H = 0.006437 mol C; 0.01298 g H × = 0.01288 mol H 1.008 g H 12.01 g C
0.07731 g C ×
0.0103 g O ×
1 mol O = 6.44 × 10‒4 mol O 16.00 g O
Dividing all mole values by 6.44 × 10‒4 gives: 0.006437 6.44 × 10
−4
= 10.0;
0.01288 6.44 × 10 −4
= 20.0
The empirical formula for menthol is C10H20O. 104.
a. 1.950 g AgCl ×
Mass % Cl =
1 mol Cl 35.45 g Cl = 0.4821 g Cl mol AgCl mol Cl
0.4821 g Cl × 100 = 48.21% Cl 1.000 g compound
b. 0.3678 g H2O ×
Mass % H =
1 mol AgCl 143.4 g AgCl
1 mol H 2 O 18.02 g H 2 O
2 mol H 1.008 g H = 0.04115 g H mol H 2 O mol H
0.04115 g H × 100 = 2.743% H 1.500 g compound
Mass % C = 100.00 ‒ (48.21 + 2.743) = 49.05% C Assuming 100.00 g of compound: 49.05 g C ×
1 mol C 1 mol H = 4.084 mol C; 2.743 g H × = 2.721 mol H 12.01 g C 1.008 g H
48.21 g Cl ×
1 mol Cl = 1.360 mol Cl 35.45 g Cl
4.084 2.721 = 3.003; = 2.001; the empirical formula is C3H2Cl. 1.360 1.360
105.
The combustion data allow determination of the amount of hydrogen in cumene. One way to determine the amount of carbon in cumene is to determine the mass percent of hydrogen in the compound from the data in the problem; then determine the mass percent of carbon by difference (100.0 − mass % H = mass % C).
86
CHAPTER 3 STOICHIOMETRY 42.8 10−3 g H2O × Mass % H =
1 mol H 2 O 2 mol H 1.008 g H 1000 mg = 4.79 mg H 18.02 g H 2 O mol H 2 O mol H g
4.79 mg H × 100 = 10.1% H; mass % C = 100.0 − 10.1 = 89.9% C 47 .6 mg cumene
Now solve the empirical formula problem. Out of 100.0 g cumene, we have: 89.9 g C ×
1 mol C 1 mol H = 7.49 mol C; 10.1 g H × = 10.0 mol H 12.01 g C 1.008 g H
10.0 4 = 1.34 ; the mole H to mole C ratio is 4 : 3. The empirical formula is C3H4. 3 7.49
Empirical formula mass 3(12) + 4(1) = 40 g/mol. The molecular formula must be (C3H4)3 or C9H12 because the molar mass of this formula will be between 115 and 125 g/mol (molar mass 3 × 40 g/mol = 120 g/mol). 106.
Mass C = 2.20 g CO2 ×
1 mol CO 2 44.01 g CO 2
Mass H = 0.400 g H2O ×
1 mol H 2 O 18.02 g H 2 O
1 mol C 12.01 g C = 0.600 g C mol CO 2 mol C
2 mol H 1.008 g H = 0.0448 g H mol H 2 O mol H
Mass O = 1.000 g aspirin − 0.600 g C − 0.0448 g H = 0.355 g O In 1.000 g aspirin: 0.600 g C ×
1 mol C 1 mol H = 0.0500 mol C; 0.0448 g H × = 0.0444 mol H 1.008 g H 12.01 g C
0.335 g O ×
1 mol O = 0.0222 mol O 16.00 g O
0.0500 0.0444 = 2.25; = 2.00; (C2.25H2O)×4 = C9H8O4 is the empirical formula. 0.0222 0.0222
Empirical formula mass 9(12) + 8(1) + 4(16) = 180 g/mol; this is in the range of the molar mass of the compound, so C9H8O4 is also the molecular formula of aspirin.
Balancing Chemical Equations 107.
When balancing reactions, start with elements that appear in only one of the reactants and one of the products, and then go on to balance the remaining elements. a. C6H12O6(s) + O2(g) → CO2(g) + H2O(g) Balance C atoms: C6H12O6 + O2 → 6 CO2 + H2O Balance H atoms: C6H12O6 + O2 → 6 CO2 + 6 H2O
CHAPTER 3
STOICHIOMETRY
Lastly, balance O atoms: C6H12O6(s) + 6 O2(g) → 6 CO2(g) + 6 H2O(g) b. Fe2S3(s) + HCl(g) → FeCl3(s) + H2S(g) Balance Fe atoms: Fe2S3 + HCl → 2 FeCl3 + H2S Balance S atoms: Fe2S3 + HCl → 2 FeCl3 + 3 H2S There are 6 H and 6 Cl on right, so balance with 6 HCl on left: Fe2S3(s) + 6 HCl(g) → 2 FeCl3(s) + 3 H2S(g). c. CS2(l) + NH3(g) → H2S(g) + NH4SCN(s) C and S balanced; balance N: CS2 + 2 NH3 → H2S + NH4SCN H is also balanced. CS2(l) + 2 NH3(g) → H2S(g) + NH4SCN(s) 108.
An important part to this problem is writing out correct formulas. If the formulas are incorrect, then the balanced reaction is incorrect. a. C2H5OH(l) + 3 O2(g) → 2 CO2(g) + 3 H2O(g) b. 3 Pb(NO3)2(aq) + 2 Na3PO4(aq) → Pb3(PO4)2(s) + 6 NaNO3(aq) c. Zn(s) + 2 HCl(aq) → ZnCl2(aq) + H2(g) d. Sr(OH)2(aq) + 2 HBr(aq) → 2H2O(l) + SrBr2(aq) MnO2 catalyst
109.
2 H2O2(aq)
2 H2O(l) + O2(g)
110.
C12H22O11(aq) + H2O(l) → 4 C2H5OH(aq) + 4 CO2(g)
111.
a. 3 Ca(OH)2(aq) + 2 H3PO4(aq) → 6 H2O(l) + Ca3(PO4)2(s) b. Al(OH)3(s) + 3 HCl(aq) → AlCl3(aq) + 3 H2O(l) c. 2 AgNO3(aq) + H2SO4(aq) → Ag2SO4(s) + 2 HNO3(aq)
112.
a. 2 KO2(s) + 2 H2O(l) → 2 KOH(aq) + O2(g) + H2O2(aq) or 4 KO2(s) + 6 H2O(l) → 4 KOH(aq) + O2(g) + 4 H2O2(aq) b. Fe2O3(s) + 6 HNO3(aq) → 2 Fe(NO3)3(aq) + 3 H2O(l) c. 4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g)
87
88
CHAPTER 3 STOICHIOMETRY d. PCl5(l) + 4 H2O(l) → H3PO4(aq) + 5 HCl(g) e. 2 CaO(s) + 5 C(s) → 2 CaC2(s) + CO2(g) f.
2 MoS2(s) + 7 O2(g) → 2 MoO3(s) + 4 SO2(g)
g. FeCO3(s) + H2CO3(aq) → Fe(HCO3)2(aq) 113.
a. The formulas of the reactants and products are C6H6(l) + O2(g) → CO2(g) + H2O(g). To balance this combustion reaction, notice that all of the carbon in C6H6 has to end up as carbon in CO2 and all of the hydrogen in C6H6 has to end up as hydrogen in H2O. To balance C and H, we need 6 CO2 molecules and 3 H2O molecules for every 1 molecule of C6H6. We do oxygen last. Because we have 15 oxygen atoms in 6 CO2 molecules and 3 H2O molecules, we need 15/2 O2 molecules in order to have 15 oxygen atoms on the reactant side. 15
C6H6(l) +
2
O2(g) → 6 CO2(g) + 3 H2O(g); multiply by two to give whole numbers.
2 C6H6(l) + 15 O2(g) → 12 CO2(g) + 6 H2O(g) b. The formulas of the reactants and products are C4H10(g) + O2(g) → CO2(g) + H2O(g). C4H10(g) +
13 2
O2(g) → 4 CO2(g) + 5 H2O(g); multiply by two to give whole numbers.
2 C4H10(g) + 13 O2(g) → 8 CO2(g) + 10 H2O(g) c. C12H22O11(s) + 12 O2(g) → 12 CO2(g) + 11 H2O(g) d. 2 Fe(s) +
3 2
e. 2 FeO(s) +
O2(g) → Fe2O3(s); for whole numbers: 4 Fe(s) + 3 O2(g) → 2 Fe2O3(s) 1 2
O2(g) → Fe2O3(s); for whole numbers, multiply by two.
4 FeO(s) + O2(g) → 2 Fe2O3(s) 114.
a. 16 Cr(s) + 3 S8(s) → 8 Cr2S3(s) b. 2 NaHCO3(s) → Na2CO3(s) + CO2(g) + H2O(g) c. 2 KClO3(s) → 2 KCl(s) + 3 O2(g) d. 2 Eu(s) + 6 HF(g) → 2 EuF3(s) + 3 H2(g)
115.
a. SiO2(s) + C(s) → Si(s) + CO(g); Si is balanced. Balance oxygen atoms: SiO2 + C → Si + 2 CO Balance carbon atoms: SiO2(s) + 2 C(s) → Si(s) + 2 CO(g) b. SiCl4(l) + Mg(s) → Si(s) + MgCl2(s); Si is balanced.
CHAPTER 3
STOICHIOMETRY
89
Balance Cl atoms: SiCl4 + Mg → Si + 2 MgCl2 Balance Mg atoms: SiCl4(l) + 2 Mg(s) → Si(s) + 2 MgCl2(s) c. Na2SiF6(s) + Na(s) → Si(s) + NaF(s); Si is balanced. Balance F atoms:
Na2SiF6 + Na → Si + 6 NaF
Balance Na atoms: Na2SiF6(s) + 4 Na(s) → Si(s) + 6 NaF(s) 116.
CaSiO3(s) + 6 HF(aq) → CaF2(aq) + SiF4(g) + 3 H2O(l)
Reaction Stoichiometry 117.
6 Li(s) + N2(g) → 2 Li3N(s) 20.0 g Li ×
118.
2 mol Li 3 N 34.83 g Li 3 N 1 mol Li = 33.5 g Li3N 6.941 g Li 6 mol Li mol Li 3 N
2 Al(s) + 3 I2(s) → 2 AlI3(s) 10.0 g AlI3 ×
119.
1 mol AlI3 3 mol I 2 253.8 g I 2 = 9.34 g I2 407.7 g AlI 3 2 mol AlI 3 mol I 2
The stepwise method to solve stoichiometry problems is outlined in the text. Instead of calculating intermediate answers for each step, we will combine conversion factors into one calculation. This practice reduces round-off error and saves time. Fe2O3(s) + 2 Al(s) → 2 Fe(l) + Al2O3(s) 15.0 g Fe ×
1 mol Fe 2 mol Al 26 .98 g Al = 0.269 mol Fe; 0.269 mol Fe × 55 .85 g Fe 2 mol Fe mol Al = 7.26 g Al
0.269 mol Fe × 0.269 mol Fe × 120.
1 mol Fe 2 O 3 2 mol Fe 1 mol Al 2 O 3
159.70 g Fe 2 O 3
101.96 g Al 2 O 3
mol Fe 2 O 3 mol Al 2 O 3
= 21.5 g Fe2O3 = 13.7 g Al2O3
10 KClO3(s) + 3 P4(s) → 3 P4O10(s) + 10 KCl(s) 52.9 g KClO3 ×
121.
2 mol Fe
1.000 kg Al ×
1 mol KClO 3 3 mol P4 O10 283.88 g P4 O10 × × = 36.8 g P4O10 122.55 g KClO 3 10 mol KClO 3 mol P4 O10
3 mol NH 4 ClO 4 117.49 g NH 4 ClO 4 1000 g Al 1 mol Al × × × kg Al 26.98 g Al 3 mol Al mol NH 4 ClO 4
= 4355 g = 4.355 kg NH4ClO4
90 122.
CHAPTER 3 STOICHIOMETRY a. 2 C8H18(l) + 25 O2(g) → 16 CO2(g) + 18 H2O(g) b. 1.2 1010 gal gasoline 3.1 1013 g C8H18
123.
4 qt gal
946 mL 0.692 g = 3.1 1013 g gasoline qt mL
1 mol C8 H18 16 mol CO 2 44.01 g CO 2 = 9.6 1013 g CO2 114.22 g C8 H18 2 mol C8 H18 mol CO 2
a. 1.0 × 102 mg NaHCO3
1 mol NaHCO 3 1 mol C 6 H 8 O 7 1g × × 1000 mg 84.01 g NaHCO 3 3 mol NaHCO 3
b. 0.10 g NaHCO3 ×
192.12 g C 6 H 8 O 7 mol C 6 H 8 O 7
= 0.076 g or 76 mg C6H8O7
1 mol NaHCO 3 3 mol CO 2 44.01 g CO 2 × × 84.01 g NaHCO 3 3 mol NaHCO 3 mol CO 2
= 0.052 g or 52 mg CO2 124.
a.
1.00 × 102 g C7H6O3 ×
1 mol C 7 H 6 O 3 1 mol C 4 H 6 O 3 102.09 g C 4 H 6 O 3 138.12 g C 7 H 6 O 3 1 mol C 7 H 6 O 3 1 mol C 4 H 6 O 3
= 73.9 g C4H6O3 b.
1.00 × 102 g C7H6O3 ×
1 mol C 7 H 6 O 3 1 mol C 9 H 8 O 4 180.15 g C 9 H 8O 4 138.12 g C 7 H 6 O 3 1 mol C 7 H 6 O 3 mol C 9 H 8O 4
= 1.30 × 102 g aspirin 125.
1.0 × 104 kg waste ×
1 mol C5 H 7 O 2 N 3.0 kg NH 4 + 1 mol NH 4 + 1000 g × × × + 100 kg waste kg 18.04 g NH 4 55 mol NH 4 + 113.12 g C 5 H 7 O 2 N × = 3.4 × 104 g tissue if all NH4+ converted mol C 5 H 7 O 2 N
Because only 95% of the NH4+ ions react: mass of tissue = (0.95)(3.4 × 104 g) = 3.2 × 104 g or 32 kg bacterial tissue 126.
1.0 × 103 g phosphorite ×
75 g Ca 3 (PO 4 ) 2 1 mol Ca 3 (PO 4 ) 2 100 g phosphorite 310.18 g Ca 3 (PO 4 ) 2
×
127.
1.0 ton CuO ×
1 mol P4 123.88 g P4 × = 150 g P4 2 mol Ca 3 (PO 4 ) 2 mol P4
907 kg 1000 g 1 mol CuO 1 mol C 12.01 g C × × × × ton kg 79.55 g CuO 2 mol CuO mol C
×
100. g coke = 7.2 × 104 g or 72 kg coke 95 g C
CHAPTER 3 128.
STOICHIOMETRY
91
2 LiOH(s) + CO2(g) → Li2CO3(aq) + H2O(l) The total volume of air exhaled each minute for the 7 astronauts is 7 × 20. = 140 L/min. 25,000 g LiOH ×
1 mol CO 2 44.01 g CO 2 1 mol LiOH 100 g air 23.95 g LiOH 2 mol LiOH mol CO 2 4.0 g CO 2 ×
1 mL air 1L 1 min 1h × × × = 68 h = 2.8 days 0.0010 g air 1000 mL 140 L air 60 min
Limiting Reactants and Percent Yield 129.
The product formed in the reaction is NO2; the other species present in the product representtation is excess O2. Therefore, NO is the limiting reactant. In the pictures, 6 NO molecules react with 3 O2 molecules to form 6 NO2 molecules. 6 NO(g) + 3 O2(g) → 6 NO2(g) For smallest whole numbers, the balanced reaction is: 2 NO(g) + O2(g) → 2 NO2(g)
130.
Only one product is formed in this representation. This product has two Y atoms bonded to an X. The other substance present in the product mixture is just excess of one of the reactants (Y). The best equation has smallest whole numbers. Here, answer c would be this smallest whole number equation (X + 2 Y → XY2). Answers a and b have incorrect products listed, and for answer d, an equation only includes the reactants that go to produce the product; excess reactants are not shown in an equation.
131.
In the following table we have listed three rows of information. The “Initial” row is the number of molecules present initially, the “Change” row is the number of molecules that react to reach completion, and the “After” row is the number of molecules present after completion. To determine the limiting reactant, let’s calculate how much of one reactant is necessary to react with the other. 10 molecules O2 ×
4 molecules NH 3 = 8 molecules NH3 to react with all the O2 5 molecules O 2
Because we have 10 molecules of NH3 and only 8 molecules of NH3 are necessary to react with all the O2, O2 is limiting. Now use the 10 molecules of O2 and the molecule relationships given in the balanced equation to determine the number of molecules of each product formed, then complete the table. 4 NH3(g) Initial Change After
10 molecules −8 molecules 2 molecules
+
5 O2(g) 10 molecules −10 molecules 0
→
4 NO(g) 0 +8 molecules 8 molecules
+
6 H2O(g) 0 +12 molecules 12 molecules
The total number of molecules present after completion = 2 molecules NH3 + 0 molecules O2 + 8 molecules NO + 12 molecules H2O = 22 molecules.
92
CHAPTER 3 STOICHIOMETRY
132.
The balanced equation is 2 CH3OH(g) + 3 O2(g) → 2 CO2(g) + 4 H2O(g); because we are reacting numbers of molecules of CH3OH and O2, and because O2 has the larger coefficient in the balanced equation, O2 will be limiting. Let’s summarize the problem in a table like we did for question 131. 2 CH3OH (g) Initial Change After
+
15 molecules −10 molecules 5 molecules
3 O2(g) 15 molecules −15 molecules 0
→
2 CO2(g) 0 +10 molecules 10 molecules
+
4 H2O(g) 0 +20 molecules 20 molecules
The total number of molecules present after completion = 5 molecules CH3OH + 0 molecules O2 + 10 molecules CO2 + 20 molecules H2O = 35 molecules. 133.
a. The strategy we will generally use to solve limiting reactant problems is to assume each reactant is limiting, and then calculate the quantity of product each reactant could produce if it were limiting. The reactant that produces the smallest quantity of product is the limiting reactant (runs out first) and therefore determines the mass of product that can be produced. Assuming N2 is limiting: 1.00 × 103 g N2
2 mol NH 3 17.03 g NH 3 1 mol N 2 = 1.22 × 103 g NH3 × × 28.02 g N 2 mol N 2 mol NH 3
Assuming H2 is limiting: 5.00 × 102 g H2
2 mol NH 3 17.03 g NH 3 1 mol H 2 = 2.82 × 103 g NH3 2.016 g H 2 3 mol H 2 mol NH 3
Because N2 produces the smaller mass of product (1220 g vs. 2820 g NH3), N2 is limiting and 1220 g NH3 can be produced. As soon as 1220 g of NH3 is produced, all of the N2 has run out. Even though we have enough H2 to produce more product, there is no more N2 present as soon as 1220 g of NH3 have been produced. b. 1.00 103 g N2
1 mol N 2 3 mol H 2 2.016 g H 2 × × 28.02 g N 2 mol N 2 mol H 2
= 216 g H2 reacted
Excess H2 = 500. g H2 initially – 216 g H2 reacted = 284 g H2 in excess (unreacted) 134.
Ca3(PO4)2 + 3 H2SO4 → 3 CaSO4 + 2 H3PO4 Assuming Ca3(PO4)2 is limiting: 1.0 × 103 g Ca3(PO4)2 ×
1 mol Ca 3 (PO 4 ) 2 3 mol CaSO 4 136.15 g CaSO 4 310.18 g Ca 3 (PO 4 ) 2 mol Ca 3 (PO 4 ) 2 mol CaSO 4
= 1300 g CaSO4 Assuming concentrated H2SO4 reagent is limiting: 1.0 × 103 g conc. H2SO4 ×
98 g H 2SO 4 1 mol H 2SO 4 100 g conc. H 2SO 4 98.09 g H 2SO 4
3 mol CaSO 4 136.15 g CaSO 4 = 1400 g CaSO4 3 mol H 2SO 4 mol CaSO 4
CHAPTER 3
STOICHIOMETRY
93
Because Ca3(PO4)2 produces the smaller quantity of product, Ca3(PO4)2 is limiting and 1300 g CaSO4 can be produced. 1.0 × 103 g Ca3(PO4)2
2 mol H 3 PO 4 1 mol Ca 3 (PO 4 ) 2 97.99 g H 3 PO 4 mol Ca 3 (PO 4 ) 2 310.18 g Ca 3 (PO 4 ) 2 mol H 3 PO 4
= 630 g H3PO4 produced 135.
Assuming BaO2 is limiting: 1.50 g BaO2 ×
1 mol BaO 2 1 mol H 2 O 2 34.02 g H 2 O 2 = 0.301 g H2O2 × × 169.3 g BaO 2 mol BaO 2 mol H 2 O 2
Assuming HCl is limiting: 88.0 mL ×
1 mol H 2 O 2 34.02 g H 2 O 2 0.0272 g HCl 1 mol HCl = 1.12 g H2O2 × × × mL 36.46 g HCl 2 mol HCl mol H 2O 2
BaO2 produces the smaller amount of H2O2, so it is limiting and a mass of 0.301 g of H2O2 can be produced. Initial mol HCl present: 88.0 mL
0.0272 g HCl 1 mol HCl mL 36 .46 g HCl
= 6.57 × 10−2 mol HCl
The amount of HCl reacted: 1.50 g BaO2 ×
1 mol BaO 2 2 mol HCl = 1.77 × 10‒2 mol HCl 169.3 g BaO 2 mol BaO 2
Excess mol HCl = 6.57 × 10‒2 mol − 1.77 × 10‒2 mol = 4.80 × 10‒2 mol HCl Mass of excess HCl = 4.80 × 10−2 mol HCl × 136.
36.46 g HCl = 1.75 g HCl unreacted mol HCl
Assuming Ag2O is limiting: 25.0 g Ag2O ×
2 mol AgC10 H 9 N 4SO 2 357.18 g AgC10 H 9 N 4SO 2 1 mol Ag 2 O 231.8 g Ag 2 O mol Ag 2 O mol AgC10 H 9 N 4SO 2
= 77.0 g AgC10H9N4SO2 Assuming C10H10N4SO2 is limiting: 50.0 g C10H10N4SO2 ×
1 mol C10 H10 N 4SO 2 2 mol AgC10 H 9 N 4SO 2 250.29 g C10 H10 N 4SO 2 2 mol C10 H10 N 4SO 2
357.18 g AgC10 H 9 N 4SO 2 = 71.4 g AgC10H9N4SO2 mol AgC10 H 9 N 4SO 2
Because C10H10N4SO2 produces the smaller amount of product, it is limiting and 71.4 g of silver sulfadiazine can be produced. 137.
To solve limiting-reagent problems, we will generally assume each reactant is limiting and then calculate how much product could be produced from each reactant. The reactant that produces the smallest amount of product will run out first and is the limiting reagent. 5.00 × 106 g NH3 ×
1 mol NH 3 2 mol HCN = 2.94 × 105 mol HCN × 17.03 g NH 3 2 mol NH 3
94
CHAPTER 3 STOICHIOMETRY 5.00 × 106 g O2 ×
1 mol O 2 2 mol HCN = 1.04 × 105 mol HCN × 32.00 g O 2 3 mol O 2
5.00 × 106 g CH4 ×
1 mol CH 4 2 mol HCN = 3.12 × 105 mol HCN × 16.04 g CH 4 2 mol CH 4
O2 is limiting because it produces the smallest amount of HCN. Although more product could be produced from NH3 and CH4, only enough O2 is present to produce 1.04 × 105 mol HCN. The mass of HCN produced is: 1.04 × 105 mol HCN ×
1 mol O 2 6 mol H 2 O 18.02 g H 2 O = 5.63 × 106 g H2O × × 32.00 g O 2 3 mol O 2 1 mol H 2 O
5.00 × 106 g O2 × 138.
27 .03 g HCN = 2.81 × 106 g HCN mol HCN
If C3H6 is limiting: 15.0 g C3H6 ×
1 mol C 3 H 6 2 mol C 3 H 3 N 53.06 g C 3 H 3 N × × = 18.9 g C3H3N 42.08 g C 3 H 6 2 mol C 3 H 6 mol C 3 H 3 N
If NH3 is limiting: 5.00 g NH3 ×
1 mol NH 3 2 mol C 3 H 3 N 53.06 g C 3 H 3 N × × = 15.6 g C3H3N 17.03 g NH 3 2 mol NH 3 mol C 3 H 3 N
If O2 is limiting: 10.0 g O2 ×
2 mol C 3 H 3 N 53.06 g C 3 H 3 N 1 mol O 2 × × = 11.1 g C3H3N 32.00 g O 2 3 mol O 2 mol C 3 H 3 N
O2 produces the smallest amount of product; thus O2 is limiting, and 11.1 g C3H3N can be produced. 139.
20.0 mL Br2
2 mol AlBr3 3.10 g Br2 1 mol Br2 × × mL 159.80 g Br2 3 mol Br2 ×
266.68 g AlBr3 = 69.0 g AlBr3 mol AlBr3
The theoretical yield of the reaction is 69.0 g AlBr3. From the problem, the actual yield was 50.3 g. Percent yield = 140.
actual 50.3 g × 100 = × 100 = 72.9% 69.0 g theoretical
3 Ca(s) + N2(g) → Ca3N2(s) If Ca is limiting: 10.0 g Ca ×
1 mol Ca 3 N 2 148.26 g Ca 3 N 2 1 mol Ca × × = 12.3 g Ca3N2 40.08 g Ca 3 mol Ca mol Ca 3 N 2
If N2 is limiting: 10.0 g N2 ×
1 mol Ca 3 N 2 148.26 g Ca 3 N 2 1 mol N 2 × × = 52.9 g Ca3N2 28.02 g N 2 1 mol N 2 mol Ca 3 N 2
CHAPTER 3
STOICHIOMETRY
95
Ca produces the smallest quantity of product, thus Ca is limiting and the theoretical yield is 12.3 g Ca3N2. percent yield =
141.
10.0 g actual × 100 = × 100 = 81.3% 12.3 g theoretical
C2H6(g) + Cl2(g) → C2H5Cl(g) + HCl(g) If C2H6 is limiting: 300. g C2H6 ×
1 mol C 2 H 6 1 mol C 2 H 5 Cl 64 .51 g C 2 H 5 Cl = 644 g C2H5Cl 30 .07 g C 2 H 6 mol C 2 H 6 mol C 2 H 5 Cl
If Cl2 is limiting: 650. g Cl2 ×
1 mol C 2 H 5 Cl 64.51 g C 2 H 5 Cl 1 mol Cl 2 × × = 591 g C2H5Cl 70.90 g Cl 2 mol Cl 2 mol C 2 H 5Cl
Cl2 is limiting because it produces the smaller quantity of product. Hence, the theoretical yield for this reaction is 591 g C2H5Cl. The percent yield is: percent yield =
142.
490. g actual × 100 = × 100 = 82.9% 591 g theoretical
a. 1142 g C6H5Cl ×
1 mol C 6 H 5 Cl 1 mol C14 H 9 Cl 5 354.46 g C14 H 9 Cl 5 112.55 g C 6 H 5 Cl 2 mol C 6 H 5 Cl mol C14 H 9 Cl5
= 1798 C14H9Cl5 485 g C2HOCl3 ×
1 mol C 2 HOCl3 1 mol C14 H 9 Cl5 354.46 g C14 H 9 Cl 5 147.38 g C 2 HOCl3 mol C 2 HOCl 3 mol C14 H 9 Cl 5
= 1170 g C14H9Cl5 From the masses of product calculated, C2HOCl3 is limiting and 1170 g C14H9Cl5 can be produced. b. C2HOCl3 is limiting, and C6H5Cl is in excess. c. 485 g C2HOCl3 ×
1 mol C 2 HOCl3 2 mol C 6 H 5Cl 112.55 g C 6 H 5Cl 147.38 g C 2 HOCl3 mol C 2 HOCl 3 mol C 6 H 5Cl
1142 g − 741 g = 401 g C6H5Cl in excess d. Percent yield = 143.
= 741 g C6H5Cl reacted
200.0 g DDT × 100 = 17.1% 1170 g DDT
2 Cu(s) + S(s) → Cu2S(s); the theoretical yield of Cu2S is: 10.0 g Cu2S (actual) ×
100.0 g Cu 2S (theoretical) = 13.7 g Cu2S (theoretical) 73.0 g Cu 2S (actual)
96
CHAPTER 3 STOICHIOMETRY 13.7 g Cu2S ×
1 mol Cu 2S 2 mol Cu 63.55 g Cu = 10.9 g Cu 159.17 g Cu 2S 1 mol Cu 2S mol Cu
If 10.9 g of Cu is reacted, 13.7 g of Cu2S will be produced in theory. And with a 73.0% yield, 10.0 g of Cu2S will be the actual yield. 144.
P4(s) + 6 F2(g) → 4 PF3(g); the theoretical yield of PF3 is: 120. g PF3 (actual) × 154 g PF3 ×
100.0 g PF3 (theoretical) = 154 g PF3 (theoretical) 78.1 g PF3 (actual)
1 mol PF3 6 mol F2 38.00 g F2 × × = 99.8 g F2 87.97 g PF3 4 mol PF3 mol F2
99.8 g F2 is needed to actually produce 120. g of PF3 if the percent yield is 78.1%.
ChemWork Problems 145.
12
C21H6: 2(12.000000) + 6(1.007825) = 30.046950 u
12
C1H216O: 1(12.000000) + 2(1.007825) + 1(15.994915) = 30.010565 u
14
N16O: 1(14.003074) + 1(15.994915) = 29.997989 u
The peak results from 12C1H216O. 146.
We would see the peaks corresponding to: 10
B35Cl3 [mass 10 + 3(35) = 115 u], 10B35Cl237Cl (117), 10B35Cl37Cl2 (119),
10
B37Cl3 (121), 11B35Cl3 (116), 11B35Cl237Cl (118), 11B35Cl37Cl2 (120), 11B37Cl3 (122)
We would see a total of eight peaks at approximate masses of 115, 116, 117, 118, 119, 120, 121, and 122. 147.
1 mol Fe 6.022 10 23 atoms Fe = 113 atoms Fe 55.85 g Fe mol Fe
a. 1.05 × 10‒20 g Fe ×
b. The total number of platinum atoms is 14 × 20 = 280 atoms (exact number). The mass of these atoms is: 280 atoms Pt ×
1 mol Pt 6.022 10
c. 9.071 × 10‒20 g Ru ×
148.
119 g PH3
23
atoms Pt
195.1 g Pt mol Pt
= 9.071 × 10‒20 g Pt
1 mol Ru 6.022 10 23 atoms Ru = 540.3 = 540. atoms Ru 101.1 g Ru mol Ru
1 mol PH 3 6.022 10 23 molecules PH 3 33.99 g PH 3 mol PH 3
3 atoms H = 6.32 × 1024 atoms H mol PH 3
CHAPTER 3
STOICHIOMETRY
119 g H2O
97
1 mol H 2 O 6.022 10 23 molecules H 2O 18.02 g H 2 O mol H 2 O
119 g H2S
119 g HF
2 atoms H = 7.95 × 1024 atoms H mol H 2 O
1 mol H 2S 6.022 10 23 molecules H 2S 34.09 g H 2S mol H 2S
2 atoms H = 4.20 × 1024 atoms H mol H 2S
1 atom H = 3.58 × 1024 atoms H mol HF
1 mol HF 6.022 10 23 molecules HF 20.01 g HF mol HF
The order from least to greatest number of H atoms is HF < H2S < PH3 < H2O. 3.97 g Al2 O3 1 mol Al2 O3 2 mol Al 6.022 1023 atoms Al mL 101.96 g Al2 O3 1 mol Al2 O3 mol Al = 7.03 × 1023 atoms Al
149.
15.00 mL ×
150.
Mol caffeine = 3.01 × 1023 molecules ×
1 mol 6.022 10 23 molecules
= 0.500 mol caffeine
97 g = 194 g/mol = 190 g/mol 0.500 mol 1g 1 mol caffeine 420 mg × = 2.2 × 10‒3 mol caffeine in the energy drink. 1000 mg 190 g
Molar mass =
151.
Mol compound = 1 molecule × Molar mass =
152.
1 mol 6.022 10 23 molecules
4.65 10 −23 g 1.661 10 −24 mol
Mol enzyme = 1 molecule × Molar mass of enzyme =
= 1.661 × 10‒24 mol
= 28.0 g/mol; N2 and CO are possible identities. 1 mol
6.022 10 23 molecules
1.328 10 − g 1.661 10 −24 mol
= 1.661 × 10‒24 mol
= 7.995 × 104 g/mol
Assuming 100.0 g of enzyme: mol enzyme = 100.0 g enzyme × mol Zn2+ = 0.327 g ×
1 mol enzyme 7.995 10 4 g
= 1.251 × 10‒3 mol enzyme
1 mol Zn 2+ = 5.00 × 10‒3 mol Zn2+ 65.38 g
98
CHAPTER 3 STOICHIOMETRY mol Zn 2+ 5.00 10 −3 mol = 4.00 = mol enzyme 1.251 10 −3 mol
There are 4 Zn2+ ions in each molecule of alcohol dehydrogenase. 153.
0.368 g XeFn
Molar mass XeFn = 9.03 10
20
molecules XeFn
1 mol XeFn
= 245 g/mol
6.022 10 23 molecules
245 g = 131.3 g + n(19.00 g), n = 5.98; formula = XeF6 154.
a.
14 mol C ×
12.01 g 14.01 g 1.008 g + 18 mol H × + 2 mol N × mol C mol N mol H
+ 5 mol O ×
1 mol C14 H18 N 2 O 5 = 3.40 × 10−2 mol C14H18N2O5 294.30 g C14 H18 N 2 O 5
b. 10.0 g C14H18N2O5 × c.
1.56 mol ×
d. 5.0 mg ×
16.00 g = 294.30 g mol O
294.3g = 459 g C14H18N2O5 mol
1g 1 mol 6.022 10 23 molecules 1000 mg 294.30 g mol = 1.0 × 1019 molecules C14H18N2O5
e. The chemical formula tells us that 1 molecule of C14H18N2O5 contains 2 atoms of N. If we have 1 mole of C14H18N2O5 molecules, then 2 moles of N atoms are present. 1.2 g C14H18N2O5 ×
1 mol C14 H18 N 2 O 5 2 mol N 294.30 g C14 H18 N 2 O 5 mol C14 H18 N 2 O 5
f.
1.0 × 109 molecules ×
g. 1 molecule × 155.
1 mol 6.022 10
1 mol 6.022 × 10
23
atoms
23
×
atoms
6.022 10 23 atoms N = 4.9 × 1021 atoms N mol N
294.30 g = 4.9 × 10−13 g mol
294.30 g = 4.887 × 10−22 g C14H18N2O5 mol
Molar mass = 20(12.01) + 29(1.008) + 19.00 + 3(16.00) = 336.43 g/mol Mass % C =
20(12.01) g C × 100 = 71.40% C 336.43 g compound
Mass % H =
29(1.008) g H × 100 = 8.689% H 336.43 g compound
Mass % F =
19 .00 g F × 100 = 5.648% F 336 .43 g compound
CHAPTER 3
STOICHIOMETRY
99
Mass % O = 100.00 − (71.40 + 8.689 + 5.648) = 14.26% O or: %O=
156.
3(16 .00 ) g O × 100 = 14.27% O 336 .43 g compound
100.0 g (NH4)2CO3
1 mol (NH 4 ) 2 CO 3 3 mol O 1 mol NaOH 96.09 g (NH 4 ) 2 CO 3 1 mol (NH 4 ) 2CO 3 1 mol O
157.
40.00 g NaOH = 124.9 g NaOH mol NaOH
a. NO: Mass % N =
14.01 g N 30.01 g NO
b. N2O: Mass % N =
2(14.01) g N × 100 = 63.65% N 44.02 g N 2 O
c. NH3: Mass % N =
14.01 g N 17.03 g NH 3
× 100 = 82.27% N
d. SNH: Mass % N =
14.01 g N 47.09 g SNH
× 100 = 29.75% N
× 100 = 46.68% N
From the calculations above, the order from smallest to largest mass percent N is SNH < NO < N2O < NH3. 158.
When combustion data are given, it is assumed that all the carbon in the compound ends up as carbon in CO2 and all the hydrogen in the compound ends up as hydrogen in H2O. In the sample of p-cresol combusted, the masses of C and H are: mass C = 0.983 g CO2 ×
1 mol CO 2 44.01 g CO 2
1 mol C mol CO 2
mass H = 0.230 g H2O ×
1 mol H 2 O 18.02 g H 2 O
2 mol H 1.008 g H = 0.0257 g H mol H 2 O mol H
12.01 g C = 0.268 g C mol C
Mass O = 0.345 g p-cresol − 0.268 g C − 0.0257 g H = 0.0513 g O So, in 0.345 g of the p-cresol, we have: 0.268 g C ×
1 mol C 1 mol H = 0.0223 mol C; 0.0257 g H × = 0.0255 mol H 1.008 g H 12.01 g C
0.0513 g O ×
1 mol O = 0.00321 mol O 16.00 g O
Dividing by the smallest number: The empirical formula is C7H8O.
0.0255 0.0223 = 6.95; = 7.94 0.00321 0.00321
100 159.
CHAPTER 3 STOICHIOMETRY Out of 100.00 g of adrenaline, there are: 56.79 g C ×
1 mol C 1 mol H = 4.729 mol C; 6.56 g H × = 6.51 mol H 1.008 g H 12.01 g C
28.37 g O ×
1 mol O 1 mol N = 1.773 mol O; 8.28 g N × = 0.591 mol N 16.00 g O 14.01 g N
Dividing each mole value by the smallest number: 4.729 6.51 = 8.00; 0.591 0.591
= 11.0;
1.773 0.591
= 3.00;
0.591 = 1.00 0.591
This gives adrenaline an empirical formula of C8H11O3N. 160.
Assuming 100.00 g of compound (mass hydrogen = 100.00 g − 49.31 g C − 43.79 g O = 6.90 g H): 1 mol C 1 mol H 49.31 g C × = 4.106 mol C; 6.90 g H × = 6.85 mol H 1.008 g H 12.01 g C 43.79 g O ×
1 mol O 16.00 g O
= 2.737 mol O
Dividing all mole values by 2.737 gives: 2.737 4.106 6.85 = 1.500; = 2.50; = 1.000 2.737 2.737 2.737
Because a whole number ratio is required, the empirical formula is C3H5O2. Empirical formula mass: 3(12.01) + 5(1.008) +2(16.00) = 73.07 g/mol 146.1 Molar mass = = 1.999; molecular formula = (C3H5O2)2 = C6H10O4 Empirical formula mass 73.07
161.
There are many valid methods to solve this problem. We will assume 100.00 g of compound, and then determine from the information in the problem how many moles of compound equals 100.00 g of compound. From this information, we can determine the mass of one mole of compound (the molar mass) by setting up a ratio. Assuming 100.00 g cyanocobalamin: mol cyanocobalamin = 4.34 g Co ×
1 mol Co 1 mol cyanocobalamin × 58.93 g Co mol Co
= 7.36 × 10‒2 mol cyanocobalamin 100.00 g x g cyanocobalamin = , x = molar mass = 1.36 × 103 g/mol 1 mol cyanocobalamin 7.36×10 -2 mol
162.
2 tablets ×
0.262 g C 7 H 5 BiO 4 1 mol C 7 H 5 BiO 4 1 mol Bi 209.0 g Bi tablet 362.11 g C 7 H 5 BiO 4 1 mol C 7 H 5 BiO 4 mol Bi
= 0.302 g Bi consumed
CHAPTER 3 163.
STOICHIOMETRY
Empirical formula mass = 12.01 + 1.008 = 13.02 g/mol; because 104.14/13.02 = 7.998 8, the molecular formula for styrene is (CH)8 = C8H8. 2.00 g C8H8 ×
164.
101
1 mol C8 H8 8 mol H 6.022 ×1023 atoms H × × = 9.25 × 1022 atoms H 104.14 g C8 H8 mol C8 H8 mol H 11.46 mg 12.01 mg C = 11.46 mg C; % C = × 100 = 57.85% C 44.01 mg CO 2 19.81 mg
41.98 mg CO2 ×
6.45 mg H2O ×
0.772 mg 2.01 6 mg H = 0.722 mg H; % H = × 100 = 3.64% H 18.02 mg H 2 O 19.81 mg
% O = 100.00 − (57.85 + 3.64) = 38.51% O Out of 100.00 g terephthalic acid, there are: 57.85 g C ×
1 mol C 1 mol H = 4.817 mol C; 3.64 g H × = 3.61 mol H 1.008 g H 12.01 g C
38.51 g O ×
1 mol O = 2.407 mol O 16.00 g O
4.817 2.407
= 2.001;
2.407 = 1.000 2.407
3.61 = 1.50; 2.407
The C : H : O mole ratio is 2 : 1.5 : 1 or 4 : 3 : 2. The empirical formula is C4H3O2. Mass of C4H3O2 4(12) + 3(1) + 2(16) = 83g/mol. Molar mass =
165.
17.3 g H ×
41.5 g 166 = 166 g/mol; = 2.0; the molecular formula is C8H6O4. 0.250 mol 83
1 mol H 1 mol C = 17.2 mol H; 82.7 g C × = 6.89 mol C 1.008 g H 12.01 g C
17.2 = 2.50; the empirical formula is C2H5. 6.89
The empirical formula mass is ~29 g/mol, so two times the empirical formula would put the compound in the correct molar mass range. Molecular formula = (C2H5)2 = C4H10. 1.04 × 1023 atoms C ×
1 mol C 6.022 10
23
atoms C
1 mol C 4 H10 58.12 g C 4 H10 4 mol C mol C 4 H10
= 2.51 g C4H10 166.
Assuming 100.00 g E3H8: mol E = 8.73 g H ×
1 mol H 3 mol E × = 3.25 mol E 1.008 g H 8 mol H
xgE 91.27 g E = , x = molar mass of E = 28.1 g/mol; atomic mass of E = 28.1 u 1 mol E 3.25 mol E
Note: From the periodic table, element E is silicon, Si.
102 167.
CHAPTER 3 STOICHIOMETRY Mass of H2O = 0.755 g CuSO4•xH2O − 0.483 g CuSO4 = 0.272 g H2O 0.483 g CuSO4 ×
0.272 g H2O ×
1 mol CuSO 4 159 .62 g CuSO 4
1 mol H 2 O = 0.0151 mol H2O 18 .02 g H 2 O
0.0151 mol H 2 O 0.00303 g CuSO 4
168.
= 0.00303 mol CuSO4
4.98 mol H 2 O
=
1 mol CuSO 4
; compound formula = CuSO4•5H2O, x = 5
a. Only acrylonitrile contains nitrogen. If we have 100.00 g of polymer: 8.80 g N ×
1 mol C 3 H 3 N 53.06 g C 3 H 3 N × = 33.3 g C3H3N 14.01 g N 1 mol C 3 H 3 N
% C3H3N =
33.3 g C 3 H 3 N = 33.3% C3H3N 100.00 g polymer
Only butadiene in the polymer reacts with Br2: 0.605 g Br2 ×
% C4H6 =
1 mol C 4 H 6 54.09 g C 4 H 6 1 mol Br2 × × = 0.205 g C4H6 159.8 g Br2 mol Br2 mol C 4 H 6
0.205 g × 100 = 17.1% C4H6 1.20 g
b. If we have 100.0 g of polymer: 33.3 g C3H3N ×
1 mol C 3 H 3 N = 0.628 mol C3H3N 53.06 g
17.1 g C4H6 ×
1 mol C 4 H 6 = 0.316 mol C4H6 54.09 g C 4 H 6
49.6 g C8H8 ×
1 mol C8 H 8 = 0.476 mol C8H8 104.14 g C8 H 8
Dividing by 0.316:
0.628 = 1.99; 0.316
0.316 0.476 = 1.00; = 1.51 0.316 0.316
This is close to a mole ratio of 4 : 2 : 3. Thus there are 4 acrylonitrile to 2 butadiene to 3 styrene molecules in the polymer, or (A4B2S3)n. 169.
1.20 g CO2 ×
1 mol CO 2 1 mol C 44.01 g mol CO 2
1 mol C 24 H 30 N 3O 376.51 g 24 mol C mol C 24 H 30 N 3O
= 0.428 g C24H30N3O 0.428 g C 24 H 30 N 3 O 1.00 g sample
170.
× 100 = 42.8% C24H30N3O (LSD)
Since each molecule of cortisone contains 21 C atoms, 1 mol of cortisone will contain 21 mol of carbon. Assuming 100.00 g cortisone:
CHAPTER 3
STOICHIOMETRY
Mol cortisone = 69.98 g C ×
Molar mass of cortisone = 171.
103
1 mol C 1 mol cortisone = 0.2775 mol cortisone 12.01 g C 21 mol C
100.00 g cortisone = 360.4 g/mol 0.2775 mol cortisone
Assuming 100.00 g of tetrodotoxin: 41.38 g C ×
5.37 g H ×
1 mol C 1 mol N = 3.445 mol C; 13.16 g N × = 0.9393 mol N 14.01 g N 12.01 g C
1 mol H 1 mol O = 5.33 mol H; 40.09 g O × = 2.506 mol O 1.008 g H 16.00 g O
Divide by the smallest number: 3.445 2.506 5.33 = 3.668; = 5.67; = 2.668 0.9393 0.9393 0.9393
To get whole numbers for each element, multiply through by 3. Empirical formula: (C3.668H5.67NO2.668)3 = C11H17N3O8; the mass of the empirical formula is 319.3 g/mol. 1.59 10 −21 g Molar mass tetrodotoxin = = 319 g/mol 1 mol 3 molecules 6.022 10 23 molecules Because the empirical mass and molar mass are the same, the molecular formula is the same as the empirical formula, C11H17N3O8.
1 kg 10. μg 1 10 −6 g 1 mol 6.022 10 23 molecules 165 lb × 2.2046 lb kg μg 319.3 g 1 mol = 1.4 × 1018 molecules tetrodotoxin is the LD50 dosage 172.
0.105 g
Molar mass X2 = 8.92 × 10
20
1 mol molecules × 6.022 × 1023 molecules
The mass of X = 1/2(70.9 g/mol) = 35.5 g/mol. This is chlorine, Cl. Assuming 100.00 g of MX3 (= MCl3) compound: 54.47 g Cl ×
1 mol = 1.537 mol Cl 35.45 g
1.537 mol Cl ×
1 mol M = 0.5123 mol M 3 mol Cl
= 70.9 g/mol
104
CHAPTER 3 STOICHIOMETRY Molar mass of M =
45.53 g M = 88.87 g/mol M 0.5123 mol M
M is the element yttrium (Y), and the name of YCl3 is yttrium(III) chloride. The balanced equation is 2 Y + 3 Cl2 → 2 YCl3. Assuming Cl2 is limiting: 2 mol YCl3 195.26 g YCl 3 1 mol Cl 2 × × 1.00 g Cl2 × = 1.84 g YCl3 70.90 g Cl 2 3 mol Cl 2 1 mol YCl 3 Assuming Y is limiting: 2 mol YCl3 195.26 g YCl 3 1 mol Y × × 1.00 g Y × = 2.20 g YCl3 88.91 g Y 2 mol Y 1 mol YCl 3 Cl2 is limiting, and 1.84 g YCl3 can be produced. 173.
126 g B5H9 × 192 g O2 ×
1 mol B 5 H 9 63.12 g B 5 H 9
1 mol O 2 32.00 g O 2
9 mol H 2 O 2 mol B 5 H 9
9 mol H 2 O 12 mol O 2
18.02 g H 2 O mol H 2 O
18.02 g H 2 O mol H 2 O
= 162 g H2O
= 81.1 g H2O
Because O2 produces the smallest quantity of product, O2 is limiting and 81.1g H2O can be produced. 174.
2 NaNO3(s) → 2 NaNO2(s) + O2(g); the amount of NaNO3 in the impure sample is: 2 mol NaNO 3 85.00 g NaNO 3 1 mol NaNO 2 69.00 g NaNO 2 2 mol NaNO 2 mol NaNO 3 = 0.3528 g NaNO3 0.3528 g NaNO 3 Mass percent NaNO3 = × 100 = 83.40% 0.4230 g sample
0.2864 g NaNO2
175.
a. False; this is what is done when equations are balanced. b. False; the coefficients give molecule ratios as well as mole ratios between reactants and products. c. False; the reactants are on the left, with the products on the right. d. True e. True; for mass to be conserved, there must be the same number of atoms as well as the same type of atoms on both sides.
176.
SO2(g) + 2 NaOH(aq) → Na2SO3(s) + H2O(l) Assuming SO2 is limiting: 38.3 g SO2
1 mol Na 2SO 3 126.05 g Na 2SO 3 1 mol SO 2 = 75.4 g Na2SO3 64.07 g SO 2 mol SO 2 mol Na 2SO 3
CHAPTER 3
STOICHIOMETRY
105
Assuming NaOH is limiting: 32.8 g NaOH
1 mol Na 2SO 3 1 mol NaOH 126.05 g = 51.7 g Na2SO3 40.00 g 2 mol NaOH mol Na 2SO 3
Because NaOH produces the smaller mass of product, NaOH is limiting and 51.7 g Na2SO3 can be produced. 32.8 g NaOH 177.
453 g Fe ×
1 mol H 2 O 18.02 g H 2 O 1 mol NaOH = 7.39 g H2O produced 40.00 g NaOH 2 mol NaOH mol H 2 O
1 mol Fe 2 O 3 159.70 g Fe 2 O 3 1 mol Fe = 648 g Fe2O3 55.85 g Fe 2 mol Fe mol Fe 2 O 3
Mass percent Fe2O3 =
178.
648 g Fe 2 O 3 752 g ore
× 100 = 86.2%
a. Mass of Zn in alloy = 0.0985 g ZnCl2 ×
% Zn =
65.38g Zn = 0.0473 g Zn 136.28g ZnCl 2
0.0473g Zn × 100 = 9.34% Zn; % Cu = 100.00 − 9.34 = 90.66% Cu 0.5065 g brass
b. The Cu remains unreacted. After filtering, washing, and drying, the mass of the unreacted copper could be measured. 179.
Assuming 1 mole of vitamin A (286.4 g vitamin A): mol C = 286.4 g vitamin A ×
0.8386 g C 1 mol C = 20.00 mol C × g vitamin A 12.01 g C
mol H = 286.4 g vitamin A ×
0.1056 g H 1 mol H = 30.00 mol H × g vitamin A 1.008 g H
Because 1 mole of vitamin A contains 20 mol C and 30 mol H, the molecular formula of vitamin A is C20H30E. To determine E, let’s calculate the molar mass of E: 286.4 g = 20(12.01) + 30(1.008) + molar mass E, molar mass E = 16.0 g/mol From the periodic table, E = oxygen, and the molecular formula of vitamin A is C20H30O. 180.
a. At 40.0 g of Na added, Cl2 and Na both run out at the same time (both are limiting reactants). Past 40.0 g of Na added, Cl2 is limiting, and because the amount of Cl2 present in each experiment was the same quantity, no more NaCl can be produced. Before 40.0 g of Na added, Na was limiting. As more Na was added (up to 40.0 g Na), more NaCl was produced. b. 20.0 g Na
1 mol Na 2 mol NaCl 58.44 g NaCl × × = 50.8 g NaCl 22.99 g Na 2 mol Na mol NaCl
c. At 40.0 g Na added, both Cl2 and Na are present in stoichiometric amounts. 40.0 g Na
1 mol Cl 2 70.90 g Cl 2 1 mol Na = 61.7 g Cl2 22.99 g Na 2 mol Na mol Cl 2
106
CHAPTER 3 STOICHIOMETRY 61.7 g Cl2 was present at 40.0 g Na added, and from the problem, the same 61.7 g Cl2 was present in each experiment. d. At 50.0 g Na added, Cl2 is limiting: 61.7 g Cl2 e. 20.0 g Na
1 mol Cl 2 2 mol NaCl 58.44 g NaCl = 101.7 g = 102 g NaCl 70.90 g Cl 2 1 mol Cl 2 mol NaCl
1 mol Cl 2 70.90 g Cl 2 1 mol Na = 30.8 g Cl2 reacted 22.99 g Na 2 mol Na mol Cl 2
Excess Cl2 = 61.7 g Cl2 initially – 30.8 g Cl2 reacted = 30.9 g Cl2 in excess Note: We know that 40.0 g Na is the point where Na and the 61.7 g of Cl2 run out at the same time. So if 20.0 g of Na are reacted, one-half of the Cl2 that was present at 40.0 g Na reacted will be in excess. The previous calculation confirms this. For 50.0 g Na reacted, Cl2 is limiting and 40.0 g Na will react as determined previously. Excess Na = 50.0 g Na initially – 40.0 g Na reacted = 10.0 g Na in excess. 181.
X2Z: 40.0% X and 60.0% Z by mass;
40.0/A x ( 40 .0) A z mol X or Az = 3Ax, = 2 = = mol Z 60.0/A z (60 .0) A x
where A = molar mass. For XZ2, molar mass = Ax + 2Az = Ax + 2(3Ax) = 7Ax. Mass percent X =
Ax × 100 = 14.3% X; % Z = 100.0 − 14.3 = 85.7% Z 7A x
Challenge Problems 182.
GaAs can be either 69GaAs or 71GaAs. The mass spectrum for GaAs will have 2 peaks at 144 (= 69 + 75) and 146 (= 71 + 75) with intensities in the ratio of 60 : 40 or 3 : 2.
144
146
Ga2As2 can be 69Ga2As2, 69Ga71GaAs2, or 71Ga2As2. The mass spectrum will have 3 peaks at 288, 290, and 292 with intensities in the ratio of 36 : 48 : 16 or 9 : 12 : 4. We get this ratio from the following probability table: 69
Ga (0.60)
71
Ga (0.40)
69
0.36
0.24
71
0.24
0.16
Ga (0.60) Ga (0.40)
CHAPTER 3
STOICHIOMETRY
107
288 183.
292
290
The volume of a gas is proportional to the number of molecules of gas. Thus, the formulas are: I: NH3 ;
II: N2H4;
III: HN3
The mass ratios are: I:
4.634 g N 82 .25 g N = ; 17 .75 g H gH
II:
6 .949 g N ; gH
III:
41 .7 g N gH
If we set the atomic mass of H equal to 1.008, then the atomic mass, A, for nitrogen is: I: 14.01;
II: 14.01;
For example, for compound I:
184.
85
Rb atoms
87
Rb atoms
III. 14.0 4.634 A = , A = 14.01 3(1 .008 ) 1
= 2.591
If we had exactly 100 atoms, x = number of 85Rb atoms, and 100 − x = number of 87Rb atoms. x 259.1 = 2.591, x = 259.1 − (2.591)x, x = = 72.15; 72.15% 85Rb 100 − x 3.591 0.7215(84.9117) + 0.2785(A) = 85.4678, A = 185.
85.4678 − 61.26 = 86.92 u 0.2785
First, we will determine composition in mass percent. We assume that all the carbon in the 0.213 g CO2 came from the 0.157 g of the compound and that all the hydrogen in the 0.0310 g H2O came from the 0.157 g of the compound. 0.213 g CO2 ×
12.01 g C 44.01 g CO 2
0.0310 g H2O ×
= 0.0581 g C; % C =
0.0581 g C × 100 = 37.0% C 0.157 g compound
3.47 10 −3 g 2.016 g H = 3.47 × 10−3 g H; % H = 100 = 2.21% H 18 .02 g H 2 O 0.157 g
We get the mass percent of N from the second experiment: 0.0230 g NH3 ×
14 .01 g N = 1.89 × 10−2 g N 17 .03 g NH 3
108
CHAPTER 3 STOICHIOMETRY %N=
1.89 ×10-2 g × 100 = 18.3% N 0.103 g
The mass percent of oxygen is obtained by difference: % O = 100.00 − (37.0 + 2.21 + 18.3) = 42.5% O So, out of 100.00 g of compound, there are: 37.0 g C ×
1 mol C 1 mol H = 3.08 mol C; 2.21 g H × = 2.19 mol H 1.008 g H 12.01 g C
18.3 g N ×
1 mol N 1 mol O = 1.31 mol N; 42.5 g O × = 2.66 mol O 14.01 g N 16.00 g O
Lastly, and often the hardest part, we need to find simple whole-number ratios. Divide all mole values by the smallest number: 3.08 2.19 = 2.35; 1.31 1.31
= 1.67;
1.31 2.66 = 1.00; = 2.03 1.31 1.31
Multiplying all these ratios by 3 gives an empirical formula of C7H5N3O6. 186.
1.0 × 106 kg HNO3 ×
1000 g HNO 3 1mol HNO 3 = 1.6 × 107 mol HNO3 kg HNO 3 63.02 g HNO 3
We need to get the relationship between moles of HNO3 and moles of NH3. We have to use all three equations. 2 mol HNO 3 16 mol HNO 3 2 mol NO 2 4 mol NO = 3 mol NO 2 2 mol NO 4 mol NH 3 24 mol NH 3
Thus we can produce 16 mol HNO3 for every 24 mol NH3, we begin with: 1.6 × 107 mol HNO3 ×
24 mol NH 3 17.03 g NH 3 = 4.1 × 108 g or 4.1 × 105 kg NH3 16 mol HNO 3 mol NH 3
This is an oversimplified answer. In practice, the NO produced in the third step is recycled back continuously into the process in the second step. If this is taken into consideration, then the conversion factor between mol NH3 and mol HNO3 turns out to be 1 : 1; that is, 1 mole of NH3 produces 1 mole of HNO3. Taking into consideration that NO is recycled back gives an answer of 2.7 × 105 kg NH3 reacted. 187.
Fe(s) +
1 2
O 2 (g ) → FeO(s) ; 2 Fe(s) +
20.00 g Fe
3 2
O 2 (g) → Fe 2 O 3 (s)
1 mol Fe = 0.3581 mol 55 .85 g
(11 .20 − 3.24 ) g O 2
1 mol O 2 = 0.2488 mol O2 consumed (1 extra sig. fig.) 32 .00 g
CHAPTER 3
STOICHIOMETRY
109
Let’s assume x moles of Fe reacts to form x moles of FeO. Then 0.3581 – x, the remaining moles of Fe, reacts to form Fe2O3. Balancing the two equations in terms of x: x Fe +
1 x O 2 → x FeO 2
(0.3581 − x ) mol Fe +
3 0.3581 − x 0.3581 − x mol O 2 → mol Fe 2 O 3 2 2 2
Setting up an equation for total moles of O2 consumed: 1
x +
2
3 4
(0.3581 − x) = 0.2488 mol O 2 ,
0.079 mol FeO
71.85 g FeO = 5.7 g FeO produced mol
Mol Fe2O3 produced = 0.140 mol Fe2O3 188.
x = 0.0791 = 0.079 mol FeO
0.3581 − 0.079 = 0.140 mol Fe2O3 2
159.70 g Fe 2 O 3 = 22.4 g Fe2O3 produced mol
2 C2H6(g) + 7 O2(g) → 4 CO2(g) + 6 H2O(l); C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(l) 30.07 g/mol 44.09 g/mol Let x = mass C2H6, so 9.780 − x = mass C3H8. Use the balanced equation to set up a mathematical expression for the moles of O2 required. x 7 9.780 − x 5 + = 1.120 mol O2 30.07 2 44.09 1
Solving: x = 3.7 g C2H6; 189.
3.7 g × 100 = 38% C2H6 by mass 9.780 g
The two relevant equations are: Zn(s) + 2 HCl(aq) → ZnCl2(aq) + H2(g) and Mg(s) + 2 HCl(aq) → MgCl2(aq) + H2(g) Let x = mass Mg, so 10.00 − x = mass Zn. From the balanced equations, moles H2 produced = moles Zn reacted + moles Mg reacted. Mol H2 = 0.5171 g H2 × 0.2565 =
1 mol H 2 = 0.2565 mol H2 2.016 g H 2
10.00 − x x + ; solving: x = 4.008 g Mg 24.31 65.38
4 .008 g × 100 = 40.08% Mg 10 .00 g
110 192.
CHAPTER 3 STOICHIOMETRY a N2H4 + b NH3 + (10.00 − 4.062) O2 → c NO2 + d H2O Setting up four equations to solve for the four unknowns: 2a + b = c
(N mol balance)
2c + d = 2(10.00 − 4.062)
(O mol balance)
4a + 3b = 2d
(H mol balance)
a(32.05) + b(17.03) = 61.00
(mass balance)
Solving the simultaneous equations gives a = 1.12 = 1.1 mol N2H4. 1.1 mol N 2 H 4 × 32.05 g/mol N 2 H 4 100 = 58% N2H4 61.00 g
191.
We know that water is a product, so one of the elements in the compound is hydrogen. XaHb + O2 → H2O + ? To balance the H atoms, the mole ratio between XaHb and H2O = Mol compound =
2 . b
1.39 g 1.21 g = 0.0224 mol; mol H2O = = 0.0671 mol 18 .02 g / mol 62 .09 g / mol
2 0.0224 , b = 6; XaH6 has a molar mass of 62.09 g/mol. = b 0.0671
62.09 = a(molar mass of X) + 6(1.008), a(molar mass of X) = 56.04 Some possible identities for X could be Fe (a = 1), Si (a = 2), N (a = 4), and Li (a = 8). N fits the data best, so N4H6 is the most likely formula. 192.
The balanced equation is 2 Sc(s) + 2x HCl(aq) → 2 ScClx(aq)+ x H2(g) The mole ratio of Sc : H2 = Mol Sc = 2.25 g Sc ×
2 . x
1 mol Sc = 0.0500 mol Sc 44.96 g Sc
Mol H2 = 0.1502 g H2 ×
1 mol H 2 = 0.07450 mol H2 2.016 g H 2
2 0.0500 = , x = 3; the formula is ScCl3. x 0.07450
193.
Total mass of copper used: 10,000 boards ×
(8.0 cm 16 .0 cm 0.060 cm ) 8.96 g = 6.9 × 105 g Cu 3 board cm
CHAPTER 3
STOICHIOMETRY
111
Amount of Cu to be recovered = 0.80 × (6.9 × 105 g) = 5.5 × 105 g Cu. 1mol Cu(NH 3 ) 4 Cl 2 202.59 g Cu(NH 3 ) 4Cl 2 1mol Cu × × 63.55 g Cu mol Cu mol Cu(NH 3 ) 4 Cl 2
5.5 × 105 g Cu ×
= 1.8 × 106 g Cu(NH3)4Cl2 5.5 × 105 g Cu × 194.
4 mol NH 3 17.03g NH 3 1mol Cu = 5.9 × 105 g NH3 × × 63.55 g Cu mol Cu mol NH 3
a. From the reaction stoichiometry we would expect to produce 4 mol of acetaminophen for every 4 mol of C6H5O3N reacted. The actual yield is 3 mol of acetaminophen compared to a theoretical yield of 4 mol of acetaminophen. Solving for percent yield by mass (where M = molar mass acetaminophen): percent yield =
3 mol × M × 100 = 75% 4 mol × M
b. The product of the percent yields of the individual steps must equal the overall yield, 75%. (0.87)(0.98)(x) = 0.75, x = 0.88; step III has a percent yield of 88%. 195.
10.00 g XCl2 + excess Cl2 → 12.55 g XCl4; 2.55 g Cl reacted with XCl2 to form XCl4. XCl4 contains 2.55 g Cl and 10.00 g XCl2. From the mole ratios, 10.00 g XCl2 must also contain 2.55 g Cl; mass X in XCl2 = 10.00 − 2.55 = 7.45 g X. 2.55 g Cl ×
1 mol XCl 2 1 mol Cl 35.45 g Cl 2 mol Cl
1 mol X = 3.60 × 10 −2 mol X mol XCl 2
So 3.60 × 10 −2 mol X has a mass equal to 7.45 g X. The molar mass of X is: 7.45 g X 3.60 10 − 2 mol X
196.
= 207 g/mol X; atomic mass = 207 u, so X is Pb.
4.000 g M2S3 → 3.723 g MO2 There must be twice as many moles of MO2 as moles of M2S3 in order to balance M in the reaction. Setting up an equation for 2(mol M2S3) = mol MO2 where A = molar mass M: 4.000 g 3.723 g 8.000 3.723 2 , = = 2A + 3(32.07) A + 2(16.00) 2A + 96.21 A + 32.00
(8.000)A + 256.0 = (7.446)A + 358.2, (0.554)A = 102.2, A = 184 g/mol Atomic mass of M = 184 u Note: From the periodic table, M is tungsten, W. 197.
Consider the case of aluminum plus oxygen. Aluminum forms Al3+ ions; oxygen forms O2− anions. The simplest compound between the two elements is Al2O3. Similarly, we would expect the formula of any Group 6A element with Al to be Al2X3. Assuming this, out of 100.00 g of compound, there are 12.35 g Al and 87.65 g of the unknown element, X. Let’s use this
112
CHAPTER 3 STOICHIOMETRY information to determine the molar mass of X, which will allow us to identify X from the periodic table. 12.35 g Al ×
1 mol Al 3 mol X = 0.6866 mol X 26 .98 g Al 2 mol Al
87.65 g of X must represent 0.6866 mol of X. Molar mass of X =
87.65 g X = 127.7 g/mol X 0.6866 mol X
From the periodic table, the unknown element is tellurium, and the formula is Al2Te3. 198.
Let x = mass KCl and y = mass KNO3. Assuming 100.0 g of mixture, x + y = 100.0 g. Molar mass KCl = 74.55 g/mol; molar mass KNO3 = 101.11 g/mol x y Mol KCl = ; mol KNO3 = 74.55 101.11 Knowing that the mixture is 43.2% K, then in the 100.0 g mixture, an expression for the mass of K is: y x + 39.10 = 43.2 74.55 101.11
We have two equations and two unknowns: (0.5245)x + (0.3867)y = 43.2 and x + y = 100.0 Solving, x = 32.9 g KCl; 199.
32.9 g 100 = 32.9% KCl 100.0 g
LaH2.90 is the formula. If only La3+ is present, LaH3 would be the formula. If only La2+ is present, LaH2 would be the formula. Let x = mol La2+ and y = mol La3+: (La2+)x(La3+)yH(2x + 3y) where x + y = 1.00 and 2x + 3y = 2.90 Solving by simultaneous equations: 2x + 3y = 2.90 −2x − 2y = −2.00 y = 0.90 and x = 0.10 LaH2.90 contains
200.
1.252 g Cu ×
1 9 La2+, or 10.% La2+, and La3+, or 90.% La3+. 10 10
1 mol Cu = 1.970 × 10−2 mol Cu 63.55 g Cu
The molar mass of Cu2O is 143.10 g/mol and the molar mass of CuO is 79.55 g/mol. Note that Cu2O formula contains twice the mol of Cu as compared to CuO. Let x = mass of Cu2O and y = mass of CuO, then x + y = 1.500 and: y x 2 = 1.970 × 10−2 mol Cu or (1.112)x + y = 1.567 + 79.55 143.10
CHAPTER 3
STOICHIOMETRY
113
Solving by the method of simultaneous equations: (1.112)x + y = 1.567 −x − y = −1.500 (0.112)x
= 0.067,
Mass % Cu2O = 201.
x = 0.60 g Cu2O
0.60 g Cu 2 O × 100 = 40.% Cu2O and 60.% CuO 1.500 g
The balanced equations are: 4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g) and 4 NH3(g) + 7 O2(g) → 4 NO2(g) + 6 H2O(g) Let 4x = number of moles of NO formed, and let 4y = number of moles of NO2 formed. Then: 4x NH3 + 5x O2 → 4x NO + 6x H2O and 4y NH3 + 7y O2 → 4y NO2 + 6y H2O All the NH3 reacted, so 4x + 4y = 2.00. 10.00 − 6.75 = 3.25 mol O2 reacted, so 5x + 7y = 3.25. Solving by the method of simultaneous equations: 20x + 28y = 13.0 −20x − 20y = −10.0 8y = 3.0, y = 0.38; 4x + 4 × 0.38 = 2.00, x = 0.12 Mol NO = 4x = 4 × 0.12 = 0.48 mol NO formed
202
CxHyOz + oxygen → x CO2 + y/2 H2O 2.20 g CO 2 ×
Mass % C in aspirin =
1 mol CO 2 1 mol C 12.01 g C × × 44.01 g CO 2 mol CO 2 mol C = 60.0% C 1.00 g aspirin
0.400 g H 2 O ×
Mass % H in aspirin =
1 mol H 2 O 2 mol H 1.008 g H × × 18.02 g H 2 O mol H 2 O mol H = 4.48% H 1.00 g aspirin
Mass % O = 100.00 − (60.0 + 4.48) = 35.5% O Assuming 100.00 g aspirin: 60.0 g C ×
1 mol C 1 mol H = 5.00 mol C; 4.48 g H × = 4.44 mol H 12.01 g C 1.008 g H
35.5 g O ×
1 mol O = 2.22 mol O 16.00 g O
114
CHAPTER 3 STOICHIOMETRY Dividing by the smallest number:
5.00 4.44 = 2.25; = 2.00 2.22 2.22
Empirical formula: (C2.25 H2.00O)4 = C9H8O4. Empirical mass 9(12) + 8(1) + 4(16) = 180 g/mol; this is in the 170–190 g/mol range, so the molecular formula is also C9H8O4. Balance the aspirin synthesis reaction to determine the formula for salicylic acid. CaHbOc + C4H6O3 → C9H8O4 + C2H4O2, CaHbOc = salicylic acid = C7H6O3
Marathon Problems 203.
Let M = unknown element; mass O in oxide = 3.708 g – 2.077 g = 1.631 g O In 3.708 g of compound: 1.631 g O ×
1 mol O = 0.1019 g mol O 16.00 g O
If MO is the formula of the oxide, then M has a molar mass of
2.077 g M = 20.38 g/mol. 0.1019 mol M
This is too low for the molar mass. We must have fewer moles of M than moles O present in the formula. Some possibilities are MO2, M2O3, MO3, etc. It is a guessing game as to which to try. Let’s assume an MO2 formula. Then the molar mass of M is: 2.077 g M = 40.77 g/mol 1 mol M 0.1019 mol O 2 mol O
This is close to calcium, but calcium forms an oxide having the CaO formula, not CaO2. If MO3 is assumed to be the formula, then the molar mass of M calculates to be 61.10 g/mol, which is too large. Therefore, the mol O to mol M ratio must be between 2 and 3. Some reasonable possibilities are 2.25, 2.33, 2.5, 2.67, and 2.75 (these are reasonable because they will lead to whole number formulas). Trying a mol O to mol M ratio of 2.5 : 1 gives a molar mass of: 2.077 g M = 50.96 g/mol 1 mol M 0.1019 mol O 2.5 mol O
This is the molar mass of vanadium, and V2O5 is a reasonable formula for an oxide of vanadium. The other choices for the O : M mole ratios between 2 and 3 do not give as reasonable results. Therefore, M is most likely vanadium, and the formula is V2O5. 204.
a. i.
If the molar mass of A is greater than the molar mass of B, then we cannot determine the limiting reactant because, while we have a fewer number of moles of A, we also need fewer moles of A (from the balanced reaction).
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ii. If the molar mass of B is greater than the molar mass of A, then B is the limiting reactant because we have a fewer number of moles of B and we need more B (from the balanced reaction). b. A + 5 B → 3 CO2 + 4 H2O To conserve mass: 44.01 + 5(B) = 3(44.01) + 4(18.02); solving: B = 32.0 g/mol Because B is diatomic, the best choice for B is O2. c. We can solve this without mass percent data simply by balancing the equation: A + 5 O2 → 3 CO2 + 4 H2O A must be C3H8 (which has a similar molar mass to CO2). This is also the empirical formula. Note:
3(12.01) × 100 = 81.71% C. So this checks. 3(12.01) + 8(1.008)
CHAPTER 4 TYPES OF CHEMICAL REACTIONS AND SOLUTION STOICHIOMETRY Review Questions 1.
Soluble ionic compounds break apart into their separate ions when in solution. KBr(aq) really means K+(aq) + Br−(aq). The hydration process for ions has the partial negative end of the polar water molecules surrounding and stabilizing the cations in solution. Here, many water molecules align themselves so the oxygen end of water aligns with the K+ ions. The negative ions are stabilized in water by having the partial positive end of the polar water molecules surround the anions in solution. Here, many water molecules align themselves so the hydrogen end of water aligns with the Br− ion. All this is assumed when (aq) is placed after an ionic compound. C2H5OH is a covalent compound and does not break up into ions when dissolved in water. C2H5OH is a polar covalent compound which means it has a partial negative end and a partial positive end. The hydration process for polar covalent solutes in water is again to have the opposite charged parts of the solute and solvent align themselves. Here, the hydrogens of many water molecules align with the partial negative end of each C2H5OH molecule, and the oxygens of many water molecules align with the partial positive end of each C2H5OH molecule. This is the hydration process for polar covalent compounds and is always assumed when (aq) is listed after a covalent compound. Note: at this point, you are not able to predict the partial negative and partial positive ends for polar covalent compounds.
2.
The electrolyte designation refers to how well the dissolved solute breaks up into ions. Strong electrolytes fully break up into ions when in water, weak electrolytes only partially break up into ions (less than 5% usually), and nonelectrolytes do not break up into ions when they dissolve in water. The conductivity apparatus illustrated in Figure 4.4 of the text is one way to experimentally determine the type of electrolyte. As illustrated, a bright light indicates many charge carriers (ions) are present and the solute is a strong electrolyte. A dim light indicates few ions are present so the solute is a weak electrolyte, and no light indicates no ions are present so the solute is a nonelectrolyte.
3.
The electrolyte designation refers to what happens to a substance when it dissolves in water, i.e., does it produce a lot of ions or a few ions or no ions when the substance dissolves. A weak electrolyte is a substance that only partially dissociates in water to produce only a few ions. Solubility refers to how much substance can dissolve in a solvent. "Slightly soluble" refers to substances that dissolve only to a small extent, whether it is an electrolyte or a nonelectrolyte. A weak electrolyte may be very soluble in water, or it may be slightly soluble. Acetic acid is an example of a weak electrolyte that is very soluble in water.
116
CHAPTER 4 4.
SOLUTION STOICHIOMETRY
117
Consider a 0.25 M solution of NaCl. The two ways to write 0.25 M as conversion factors are: 0.25 mol NaCl 1L or 0.25 mol NaCl L
Use the first conversion factor when converting from volume of NaCl solution (in liters) to mol NaCl and use the second conversion factor when converting from mol NaCl to volume of NaCl solution. 5.
Dilution refers to a method used to prepare solutions. In a dilution, one starts with a certain amount of a more concentrated solution; water is then added to a specific new volume, forming a solution which has a smaller concentration (it is diluted). The quantity that is constant in a dilution is the moles of solute between the concentrated solution and the dilute solution. The difference between the two solutions is that in the new solution, we have the same number of solute particles occupying a larger volume of water; the new solution is less concentrated. Molarity (mol/L) × volume (L) gives mol of solute. M1V1 = mol of solute in the concentrated solution. M2V2 = mol of solute in the diluted solution. Since the mol of solute are constant between the two solutions, M1V1 = M2V2 for dilution problems.
6.
In the first set of beakers, Pb2+ reacts with Br− to form PbBr2(s) (from the solubility rules). The Na+ and NO3− ions are spectator ions. There are 6 Na+, 6 Br−, 3 Pb2+ and 6 NO3− ions present initially. Pb2+(aq) + 2 Br−(aq) → PbBr2(s). The 3 Pb2+ ions will react with 6 Br− ions to form 3 formula units of the PbBr2 precipitate. The ions remaining in solution will be 6 Na+ ions and 6 NO3− ions floating about in solution, and there will be three formula units of PbBr 2 settled on the bottom as the precipitate (all this would be in a correct drawing). In the second set of beakers, Al3+ reacts with OH− to form Al(OH)3(s) (from the solubility rules). There are 3 Al3+ ions, 9 Cl− ions, 6 OH− ions and 6 K+ ions present initially. Al3+(aq) + 3 OH−(aq) → Al(OH)3(s). The 6 OH− ions will react with two of the three Al3+ ions to form 2 formula units of the Al(OH)3 precipitate. One Al3+ ion is in excess. Also remaining in solution are the K+ and Cl− spectator ions. Therefore, your drawing should show 1 Al3+ ion, 9 Cl− ions, and 6 K+ ions in solution, with 2 Al(OH)3 formula units shown as the precipitate.
7.
The formula equation keeps all of the ions together in nice, neutral formulas. This is not how soluble ionic compounds are present in solution. Soluble ionic compounds (indicated with aq) exist as separate ions in solution; only the precipitate has the ions together. So in the complete ionic equation, the soluble ionic compounds are shown as separate ions and the precipitate is shown as staying together. In the net ionic equation, we get rid of the ions that did nothing but balance the charge. These ions are called spectator ions. In the net ionic equation, only the ions that come together to form the precipitate are shown. In the following balanced equations, the formula equation is written first, the complete ionic equation is second, and the net ionic equation is third. 2 NaBr(aq) + Pb(NO3)2(aq) → PbBr2(s) + 2 NaNO3(aq) 2 Na (aq) + 2 Br (aq) + Pb2+(aq) + 2 NO3−(aq) → PbBr2(s) + 2 Na+(aq) + 2 NO3−(aq) +
−
Pb2+(aq) + 2 Br−(aq) → PbBr2(s)
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AlCl3(aq) + 3 KOH(aq) → Al(OH)3(s) + 3 KCl(aq) Al3+(aq) + 3 Cl−(aq) + 3 K+(aq) + 3 OH−(aq) → Al(OH)3(s) + 3 K+(aq) + 3 Cl−(aq) Al3+(aq) + 3 OH−(aq) → Al(OH)3(s) 8.
An acid-base reaction involves the transfer of a H+ ion from an acid to a base. The H+ ion is just a proton; an electron is removed from neutral hydrogen to form H +. Acid-base reactions are commonly called proton transfer reactions. The acid is the proton donor and the base is the proton acceptor. The strong bases are (by the solubility rules) LiOH, NaOH, KOH, RbOH, CsOH, Ca(OH) 2, Sr(OH)2 and Ba(OH)2. When OH− from these strong bases react with H+ (a proton), water is formed [H+(aq) + OH−(aq) → H2O(l)]. Titration: A technique in which one solution is used to analyze another. Stoichiometric point: When exactly enough of one solution has been added to react completely with the other solution. Neutralization: A term used for acid-base reactions referring to the added OH− reacting with (neutralizing) the protons from the acid. It can be reversed; the added protons neutralizing the OH− ions from the base. Either way, the neutralization reaction is H+(aq) + OH−(aq) → H2O(l). Standardization: The experimental procedure of running a controlled acid-base reaction in order to determine the concentration of a specific solution.
9.
Oxidation: A loss of electrons. Reduction: A gain of electrons. Oxidizing agent: A reactant that accepts electrons from another reactant. Reducing agent: A reactant that donates electrons to another reactant. The best way to identify a redox reaction is to assign oxidation states to all elements in the reaction. If elements show a change in oxidation states when going from reactants to products, then the reaction is a redox reaction. No change in oxidation states indicates the reaction is not a redox reaction. Note that the element oxidized shows an increase in oxidation state and the element reduced shows a decrease in oxidation state.
10.
Half-reactions: the two parts of an oxidation-reduction reaction, one representing oxidation, the other reduction. Overall charge must be balanced in any chemical reaction. We balance the charge in the halfreactions by adding electrons to either the reactant side (reduction half-reaction) or to the product side (oxidation half-reaction). In the overall balanced equation, the number of electrons lost by the oxidation half-reaction has to exactly equal the number of electrons gained in the reduction half-reaction. Since electrons lost = electrons gained, then electrons will not appear in the overall balanced equation.
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119
See Section 4.10 of the text for flow charts summarizing the half-reaction method for balancing redox reactions in acidic or basic solution. In all cases, the redox reaction must be mass balanced as well as charge balanced. Mass balance means that we have the same number and types of atoms on both sides of the equation; charge balance means that the overall net charge on each side of the reaction is the same.
Active Learning Questions 1.
HCl is a strong acid. In solution, HCl exists as separate H+ and Cl− ions. Your picture should show a solution with equal numbers of H+ and Cl− ions randomly dispersed (see Figure 4.6). The ions, of course, would be hydrated with water. But this makes for a messy picture; the hydrating water molecules are usually omitted. Mg(s) + 2 HCl(aq) → MgCl2(aq) + H2(g) or Mg(s) + 2 H+(aq) → Mg2+(aq) + H2(g)
2.
As the solution boils, water is converted into steam [H2O(l) → H2O(g)]. As the boiling process continues, more and more solvent (water) molecules are removed from solution. We have the same amount of solute in a smaller volume of solvent. Hence the solution becomes more concentrated as it boils.
3.
a. If x is the amount of sugar in solution A, then the amount of sugar in solution B will be ¼ x. b. The initial volume of solution A will be twice the volume of solution B. c. The concentration of solution A will be twice as concentrated as solution B
4.
The formula equation is Pb(NO3)2(aq) + 2 KI(aq) → PbI2(s) + 2 KNO3(aq). This is not a good representation of the species in solution. The complete ionic equation gives the best representation. The complete ionic equation is Pb2+(aq) + 2 NO3−(aq) + 2 K+(aq) + 2 I−(aq) → PbI2(s) + 2 NO3−(aq) + 2 K+(aq). Let’s assume you start with 4 formula units of lead nitrate and four formula units of potassium iodide. Your initial picture should show the 4 Pb2+ ions, 8 NO3− ions, 4 K+ ions, and 4 I− ions as completely separate from one another. The final picture should show the 2 formula units of the PbI2 precipitate at the bottom of the beaker with 2 excess Pb2+ ions, 8 NO3− ions, and 4 K+ ions still randomly dispersed in the solution above the precipitate.
5.
HNO3: +1 + x + 3(−2) = 0, x = +5; N2O: 2x + (−2) = 0, x = +1; NO2: x + 2(−2) = 0, x = +4; NH4Cl is composed of NH4+ and Cl− ions. In NH4+, x + 4(+1) = +1, x = −3; NaNO2 is composed of Na+ and NO2− ions. In NO2−: x + 2(−2) = −1, x = +3. The order of oxidation states from lowest to highest is −3 in NH4Cl, +1 in N2O, +3 in NaNO2, +4 in NO2, and +5 in HNO3.
6.
When something gains electrons, the oxidation number of the species is reduced to a lower state. Oxidation states are a way of keeping track of electron flow in a reaction. When the oxidation state of a species is lowered, it does so by gaining negative charges (electrons). Hence why the oxidation state of a species is lowered when reduced.
120 7.
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SOLUTION STOICHIOMETRY
HCl(aq) + NaOH(aq) → H2O(l) + NaOH(aq) H2SO4(aq) + 2 NaOH(aq) → 2 H2O(l) + Na2SO4(aq) Since H2SO4 is a diprotic acid while HCl is a monoprotic acid, the H2SO4 solution of a certain concentration will require twice the amount of a NaOH solution to react completely with it (neutralize it) as compared to the HCl solution having the same concentration. This can be seen from the balanced equations for the two reactions. Two moles of NaOH are required to react per mol of H2SO4 while only one mol of NaOH is required to react per mol of HCl.
8.
Your picture of a concentrated solution will have a lot more solute molecules present than the same volume of the dilute solution. Concentrated means there are a lot of solute molecules per liter of solution; a dilute solution means there are only a few solute molecules per liter of solution.
9.
For the strong electrolyte, your drawing should show cations and anions separate from each other. Each ion should also be hydrated, showing the hydrogen ends of several water molecules surrounding the anions and the oxygen end of several water molecules surrounding the cations. See Figure 4.2. For your nonelectrolyte drawing, several polar covalent compound should be drawn showing the partial negative end and the partial positive end of the polar covalent compounds. Also included in your drawing should be the hydration process with the hydrogen end of some waters surrounding the partial negative end of the polar compound and the oxygen end of several water molecules surrounding the partial positive end of the polar compound.
10.
a. CaSO4(s) will form when Na2SO4(aq) and CaCl2(aq) are mixed. b. AgI(s) will form when NH4I(aq) and AgNO3(aq) are mixed. c. Pb3(PO4)2(s) will form when K3PO4(aq) is mixed with Pb(NO3)2(aq).
11.
0.150 L
0.10 mol NaCl 58.44 g NaCl = 0.877 g = 0.88 g NaCl L mol
Place 0.88 g NaOH in a 150-mL volumetric flask; add water to dissolve the NaOH, and fill to the mark with water, mixing several times along the way. 0.150 L
0.10 mol NaCl 1L = 0.0060 L = 6.0 mL L 2.5 mol NaCl
Add 6.0 mL of 2.5 M NaOH stock solution to a 150-mL volumetric flask; fill to the mark with water, mixing several times along the way. 12.
All these choices are strong electrolytes, so by themselves, a solution of each would produce a glowing light. However, when Ba(NO3)2 is added to the H2SO4 solution, a precipitate of BaSO4 would form. This will lower the Ba2+ and SO42− ion concentrations which will dull the light since fewer of these ions are present. Similarly, when Ca(NO3)2 is added to the H2SO4 solution, a precipitate of CaSO4 would form. This will lower the Ca2+ and SO42− ion concentrations which will dull the light since fewer of these ions are present. The other choices have ions that will not react the H2SO4 solution, so adding these will just add to the total number of ions present.
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121
13.
To determine which is more concentrated, one needs to know the mass of A and the volume of water added to make each solution. The solution which is most concentrated has the largest mass of A per liter of water added. We won’t be able to calculate the actual molarity if we don’t know the molar mass and the volume of solution, but that wasn’t asked in the question.
14.
To calculate the molarity, we need to know the number of moles of solute added. So you must know the mass of solute added along with the molar mass of the solute. The second item you need to know is the volume of solution formed. Molarity is defined as the moles of solute per liter of solution. So to calculate the molarity, one needs to know the volume of solution prepared and not the volume of water added. The volume of water added usually doesn’t equate to the total volume of solution.
15.
The equation is not charge balanced. The overall charge on the reactant side is +1, while the overall charge on the product side is +2. Reactions must be mass balanced (atoms are balanced) and charge balanced. The correct mass and charge balanced equation is 2 Ag+(aq) + Cu(s) → Cu2+(aq) + 2 Ag(s).
16.
We are balancing oxidation-reduction reactions that are performed in water. Because water is present as the solvent, H2O(l) is available if needed to balance the equation.
17.
The first determination to make is whether the compound is an ionic or covalent compound. If the compound is ionic, next determine if it is soluble or insoluble. If the ionic compound is soluble, then label it a strong electrolyte. If the ionic compound is insoluble, then it has no designation. If the compound is covalent, determine whether it is an acid or a nitrogen containing base like NH3. If the compound is an acid, then determine if it is a strong acid or a weak acid. If the acid is a strong acid, then it is a strong electrolyte. If the acid is a weak acid, then it is a weak electrolyte. The 6 strong acids to memorize are HCl, HBr, HI, HNO3, HClO4, and H2SO4. Any other acid (a covalent compound with H listed first in the formula) will be a weak acid. If the compound is a neutral nitrogen containing covalent compound like NH3, then it is a weak base that is classified as a weak electrolyte. Note that weak bases will be discussed in detail in Chapter 14. If the covalent compound is not an acid or a nitrogen containing compound like NH3, then it is assumed to be a polar covalent compound and should be labeled a nonelectrolyte. A covalent compound that is insoluble in water has no designation. Predicting whether a covalent compound is polar or nonpolar is discussed in Chapter 8.
18.
a. Cu(NO3)2(aq) + 2 KOH(aq) → Cu(OH)2(s) + 2 KNO3(aq) Solution A contains 2.00 L × 2.00 mol/L = 4.00 mol Cu(NO3)2, and solution B contains 2.00 L × 3.00 mol/L = 6.00 mol KOH. In the picture in the problem, we have 4 formula units of Cu(NO3)2 (4 Cu2+ ions and 8 NO3− ions) and 6 formula units of KOH (6 K+ ions and 6 OH− ions). With 4 Cu2+ ions and 6 OH− ions present, OH− is limiting (when all 6 molecules of OH− react, we only need 3 of the 4 Cu2+ ions to react with all the OH− present). After reaction, one Cu2+ ion remains as 3 Cu(OH)2(s) formula units form as precipitate.
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The following drawing summarizes the ions that remain in solution and the relative amount of precipitate that forms. Note that K+ and NO3− ions are spectator ions. In the drawing, V1 is the volume of solution A or B, and V2 is the volume of the combined solutions, with V2 = 2V1. The drawing exaggerates the amount of precipitate that would form. V2 NO 3
V1
K
-
+
K
NO 3
+
NO 3
+
NO 3
-
2+
NO 3 -
K
-
K
Cu
-
NO 3
+
NO 3
-
K
-
K
+
+
NO 3
-
Cu(OH)2 Cu(OH)2 Cu(OH) 2
b. The spectator ion concentrations will be one-half the original spectator ion concentrations in the individual beakers because the volume was doubled. Or using moles, M K + = 8.00 mol NO 3 6.00 mol K + = 1.50 M and M NO − = 3 4.00 L 4.00 L
−
= 2.00 M. The concentration of
OH− ions will be zero because OH− is the limiting reagent. From the drawing, the number of Cu2+ ions will decrease by a factor of four as the precipitate forms. Because the volume of solution doubled, the concentration of Cu2+ ions will decrease by a factor of eight after the two beakers are mixed: 1 M Cu + = 2.00 M = 0.250 M 8
Alternately, one could certainly use moles to solve for M Cu 2 + : Mol Cu2+ reacted = 2.00 L ×
3.00 mol OH − 1 mol Cu 2 + = 3.00 mol Cu2+ reacted L 2 mol OH −
Mol Cu2+ present initially = 2.00 L ×
2.00 mol Cu 2+ = 4.00 mol Cu2+ present initially L
Excess Cu2+ present after reaction = 4.00 mol − 3.00 mol = 1.00 mol Cu2+ excess M Cu 2+ =
1.00 mol Cu 2 + = 0.250 M 2.00 L + 2.00 L
Mass of precipitate = 6.00 mol KOH ×
1 mol Cu (OH ) 2 97 .57 g Cu (OH ) 2 2 mol KOH mol Cu (OH ) 2
= 293 g Cu(OH)2
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Questions 19.
a. Polarity is a term applied to covalent compounds. Polar covalent compounds have an unequal sharing of electrons in bonds that results in unequal charge distribution in the overall molecule. Polar molecules have a partial negative end and a partial positive end. These are not full charges as in ionic compounds but are charges much smaller in magnitude. Water is a polar molecule and dissolves other polar solutes readily. The oxygen end of water (the partial negative end of the polar water molecule) aligns with the partial positive end of the polar solute, whereas the hydrogens of water (the partial positive end of the polar water molecule) align with the partial negative end of the solute. These opposite charge attractions stabilize polar solutes in water. This process is called hydration. Nonpolar solutes do not have permanent partial negative and partial positive ends; nonpolar solutes are not stabilized in water and do not dissolve. b. KF is a soluble ionic compound, so it is a strong electrolyte. KF(aq) actually exists as separate hydrated K+ ions and hydrated F− ions in solution: C6H12O6 is a polar covalent molecule that is a nonelectrolyte. C6H12O6 is hydrated as described in part a. c. RbCl is a soluble ionic compound, so it exists as separate hydrated Rb+ ions and hydrated Cl− ions in solution. AgCl is an insoluble ionic compound, so the ions stay together in solution and fall to the bottom of the container as a precipitate. d. HNO3 is a strong acid and exists as separate hydrated H+ ions and hydrated NO3− ions in solution. CO is a polar covalent molecule and is hydrated as explained in part a.
20.
2.0 L × 3.0 mol/L = 6.0 mol HCl; the 2.0 L of solution contains 6.0 mol of the solute. HCl is a strong acid; it exists in aqueous solution as separate hydrated H+ ions and hydrated Cl− ions. So the solution will contain 6.0 mol of H+(aq) and 6.0 mol of Cl− (aq). For the acetic acid solution, HC2H3O2 is a weak acid instead of a strong acid. Only some of the 6.0 moles of HC2H3O2 molecules will dissociate into H+(aq) and C2H3O2−(aq). The 2.0 L of 3.0 M HC2H3O2 solution will contain mostly hydrated HC2H3O2 molecules but will also contain some hydrated H+ ions and hydrated C2H3O2− ions.
21.
During a dilution, only water is added. There is the same quantity of solute in a larger volume. Because of this, the concentration decreases. Answer d is correct.
22.
By the solubility rules, calcium nitrate, Ca(NO3)2, is soluble, while calcium carbonate, CaCO3, is insoluble. If a mixture of both compounds were added to water, calcium nitrate would dissolve, but calcium carbonate would not dissolve. You could filter off the solid calcium carbonate, thereby separating it from the calcium nitrate.
23.
Only statement b is true. A concentrated solution can also contain a nonelectrolyte dissolved in water, e.g., concentrated sugar water. Acids are either strong or weak electrolytes. Some ionic compounds are not soluble in water, so they are not labeled as a specific type of electrolyte.
24.
The worst conducting solution will be the one with the nonelectrolyte solute present. This will be the solution which contains the covalent compound CH3OH (answer c). When dissolved in water, CH3OH stays together, and no ions are formed. The solution with HF will weakly conduct electricity since HF is a weak electrolyte. The other three solutions contain soluble
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ionic compounds, which are strong electrolytes. Solutions of these strong electrolytes will easily conduct electricity since plenty of ions are present. 25.
Zinc Chloride is not one of the insoluble chlorides, so it is soluble. All nitrate salts are soluble, so iron(II) nitrate is soluble. The rest of the compounds, c-e, will be insoluble by the solubility rules. Pb2+ ions form an insoluble sulfate, and most sulfide and chromate salts are insoluble.
26.
One mole of NaOH dissolved in 1.00 L of solution will produce 1.00 M NaOH. First, weigh out 40.00 g of NaOH (1.000 mol). Next, add some water to a 1-L volumetric flask (an instrument that is precise to 1.000 L). Dissolve the NaOH in the flask, add some more water, mix, add more water, mix, etc. until water has been added to 1.000-L mark of the volumetric flask. The result is 1.000 L of a 1.000 M NaOH solution. Because we know the volume to four significant figures as well as the mass, the molarity will be known to four significant figures. This is good practice, if you need a three-significant-figure molarity, your measurements should be taken to four significant figures. When you need to dilute a more concentrated solution with water to prepare a solution, again make all measurements to four significant figures to ensure three significant figures in the molarity. Here, we need to cut the molarity in half from 2.00 M to 1.00 M. We would start with 1 mole of NaOH from the concentrated solution. This would be 500.0 mL of 2.00 M NaOH. Add this to a 1-L volumetric flask with addition of more water and mixing until the 1.000-L mark is reached. The resulting solution would be 1.00 M.
27.
Use the solubility rules in Table 4.1. Some soluble bromides by Rule 2 would be NaBr, KBr, and NH4Br (there are others). The insoluble bromides by Rule 3 would be AgBr, PbBr2, and Hg2Br2. Similar reasoning is used for the other parts to this problem. Sulfates: Na2SO4, K2SO4, and (NH4)2SO4 (and others) would be soluble, and BaSO4, CaSO4, and PbSO4 (or Hg2SO4) would be insoluble. Hydroxides: NaOH, KOH, Ca(OH)2 (and others) would be soluble, and Al(OH)3, Fe(OH)3, and Cu(OH)2 (and others) would be insoluble. Phosphates: Na3PO4, K3PO4, (NH4)3PO4 (and others) would be soluble, and Ag3PO4, Ca3(PO4)2, and FePO4 (and others) would be insoluble. Lead: PbCl2, PbBr2, PbI2, Pb(OH)2, PbSO4, and PbS (and others) would be insoluble. Pb(NO3)2 would be a soluble Pb2+ salt.
28.
Pb(NO3)2(aq) + 2 KI(aq) → PbI2(s) + 2 KNO3(aq)
(formula equation)
Pb (aq) + 2 NO3 (aq) + 2 K (aq) + 2 I (aq) → PbI2(s) + 2 K+(aq) + 2 NO3−(aq) (complete ionic equation) 2+
−
+
−
The 1.0 mol of Pb2+ ions would react with the 2.0 mol of I− ions to form 1.0 mol of the PbI2 precipitate. Even though the Pb2+ and I− ions are removed, the spectator ions K+ and NO3− are still present. The solution above the precipitate will conduct electricity because there are plenty of charge carriers present in solution.
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125
29.
The Brønsted-Lowry definitions are best for our purposes. An acid is a proton donor, and a base is a proton acceptor. A proton is an H+ ion. Neutral hydrogen has 1 electron and 1 proton, so an H+ ion is just a proton. An acid-base reaction is the transfer of an H+ ion (a proton) from an acid to a base.
30.
The acid present is a diprotic acid (H2A), meaning that it has two H+ ions in the formula to donate to a base. The reaction is H2A(aq) + 2 NaOH(aq) → 2 H2O(l) + Na2A(aq), where A2− is what is left over from the acid formula when the two protons (H+ ions) are reacted. For the HCl reaction, the base present reacts with two protons. The most common examples are Ca(OH)2, Sr(OH)2, and Ba(OH)2. A possible reaction would be 2 HCl(aq) + Ca(OH)2(aq) → 2 H2O(l) + CaCl2(aq).
31.
a. The species reduced is the element that gains electrons. The reducing agent causes reduction to occur by itself being oxidized. The reducing agent generally refers to the entire formula of the compound/ion that contains the element oxidized. b. The species oxidized is the element that loses electrons. The oxidizing agent causes oxidation to occur by itself being reduced. The oxidizing agent generally refers to the entire formula of the compound/ion that contains the element reduced. c. For simple binary ionic compounds, the actual charges on the ions are the same as the oxidation states. For covalent compounds, nonzero oxidation states are imaginary charges the elements would have if they were held together by ionic bonds (assuming the bond is between two different nonmetals). Nonzero oxidation states for elements in covalent compounds are not actual charges. Oxidation states for covalent compounds are a bookkeeping method to keep track of electrons in a reaction.
32.
Mass balance indicates that we have the same number and type of atoms on both sides of the equation (so that mass is conserved). Similarly, net charge must also be conserved. We cannot have a buildup of charge on one side of the reaction or the other. In redox equations, electrons are used to balance the net charge between reactants and products.
Exercises Aqueous Solutions: Strong and Weak Electrolytes 33.
a. NaBr(s) → Na+(aq) + Br-(aq)
b. MgCl2(s) → Mg2+(aq) + 2 Cl−(aq) -
Na Br
+
-
Na
+
Br
-
Na Br
Cl
-
Mg
2+
-
Cl
Cl +
-
Your drawing should show equal number of Na+ and Br- ions.
2+
Mg -
Cl
-
Cl
-
Cl
2+
Mg
Your drawing should show twice the number of Cl− ions as Mg2+ ions.
126
CHAPTER 4 c. Al(NO3)3(s) → Al3+(aq) + 3 NO3−(aq) Al3
+
NO 3
NO 3
-
NO 3
-
NO 3
-
NO 3 NO 3 Al3
-
+
NO 3 Al3
d. (NH4)2SO4(s) → 2 NH4+(aq) + SO42−(aq)
-
SO 4
+
NO 3 NO 3
SOLUTION STOICHIOMETRY
2-
NH 4
+
-
SO 4 2
-
NH 4
NH 4 NH 4
-
+
+
+
SO 4 2
NH 4 NH 4
-
+ +
For e-i, your drawings should show equal numbers of the cations and anions present because each salt is a 1 : 1 salt. The ions present are listed in the following dissolution reactions. e. NaOH(s) → Na+(aq) + OH−(aq)
f.
g. KMnO4(s) → K+(aq) + MnO4− (aq)
h. HClO4(aq) → H+(aq) + ClO4−(aq)
i. 34.
FeSO4(s) → Fe2+(aq) + SO42−(aq)
NH4C2H3O2(s) → NH4+(aq) + C2H3O2−(aq)
a. Ba(NO3)2(aq) → Ba2+(aq) + 2 NO3−(aq); present in Ba(NO3)2(aq).
picture iv represents the Ba2+ and NO3− ions
b. NaCl(aq) → Na+(aq) + Cl−(aq); picture ii represents NaCl(aq). c. K2CO3(aq) → 2 K+(aq) + CO32−(aq); picture iii represents K2CO3(aq). d. MgSO4(aq) → Mg2+(aq) + SO42−(aq); picture i represents MgSO4(aq). HNO3(aq) → H+(aq) + NO3−(aq). Picture ii best represents the strong acid HNO3. Strong acids are strong electrolytes. HC2H3O2 only partially dissociates in water; acetic acid is a weak electrolyte. None of the pictures represent weak electrolyte solutions; they all are representations of strong electrolytes. 35.
By the sulfate solubility rule, CuSO4 is soluble. It will exist in solution as hydrated Cu2+ and SO42- ions moving independently of each other. Answer d is correct.
36.
HF is a weak acid (weak electrolyte), so it only partially dissociates in water. Some HF will break up into H+ and F−, but most of the molecules will stay together in solution. Answer c is correct.
37.
CaCl2(s) → Ca2+(aq) + 2 Cl−(aq)
38.
MgSO4(s) → Mg2+(aq) + SO42−(aq); NH4NO3(s) → NH4+(aq) + NO3−(aq)
Solution Concentration: Molarity 39.
a. 5.623 g NaHCO3 ×
1 mol NaHCO 3 = 6.693 × 10−2 mol NaHCO3 84.01 g NaHCO 3
CHAPTER 4 M=
SOLUTION STOICHIOMETRY 6.693 10 −2 mol 1000 mL = 0.2677 M NaHCO3 250.0 mL L
b. 0.1846 g K2Cr2O7 ×
M=
1 mol K 2 Cr 2 O 7 = 6.275 × 10−4 mol K2Cr2O7 294.20 g K 2 Cr 2 O 7
6.275 10 −4 mol 500.0 10 −3 L
c. 0.1025 g Cu ×
M=
127
= 1.255 × 10−3 M K2Cr2O7
1 mol Cu = 1.613 × 10−3 mol Cu = 1.613 × 10−3 mol Cu2+ 63.55 g Cu
1.613 10 −3 mol Cu 2+ 1000 mL = 8.065 × 10−3 M Cu2+ 200.0 mL L
40.
75.0 mL ×
41.
a.
0.79 g 1 mol 1.3 mol = 1.3 mol C2H5OH; molarity = = 5.2 M C2H5OH mL 46.07 g 0.250 L 0.100 mol Ca(NO 3 ) 2 = 1.00 M 0.100 L
M Ca(NO3 ) 2 =
Ca(NO3)2(s) → Ca2+(aq) + 2 NO3−(aq); M Ca 2+ = 1.00 M; M NO − = 2(1.00) = 2.00 M 3
b.
2.5 mol Na 2 SO 4
M Na 2SO 4 =
1.25 L
= 2.0 M
Na2SO4(s) → 2 Na+(aq) + SO42−(aq); M Na + = 2(2.0) = 4.0 M ; M SO 2− = 2.0 M 4
c. 5.00 g NH4Cl ×
1 mol NH 4 Cl = 0.0935 mol NH4Cl 53.49 g NH 4 Cl
0.0935 mol NH 4 Cl = 0.187 M 0.5000 L
M NH 4 Cl =
NH4Cl(s) → NH4+(aq) + Cl−(aq); M NH + = M Cl − = 0.187 M 4
d. 1.00 g K3PO4 ×
M K 3 P O4 =
1 mol K 3 PO 4 212 .27 g
4.71 10 −3 mol 0.2500 L
= 4.71 × 10−3 mol K3PO4
= 0.0188 M
K3PO4(s) → 3 K+(aq) + PO43−(aq); M K + = 3(0.0188) = 0.0564 M; M P O3− = 0.0188 M 4
42.
a.
M Na3 P O4 =
0.0200 mol = 2.00 M 0.0100 L
128
CHAPTER 4
SOLUTION STOICHIOMETRY
Na3PO4(s) → 3 Na+(aq) + PO 34− (aq) ; M Na + = 3(2.00) = 6.00 M; M P O3− = 2.00 M 4
b.
M Ba ( NO3 ) 2 =
0.300 mol = 0.500 M 0.6000 L
Ba(NO3)2(s) → Ba2+(aq) + 2 NO3−(aq); M Ba 2+ = 0.500 M; M NO − = 2(0.500) = 1.00 M 3
1 mol KCl 74.55 g KCl = 0.0268 M 0.5000 L
1.00 g KCl
c.
M KCl =
KCl(s) → K+(aq) + Cl−(aq); M K + = M Cl − = 0.0268 M 1 mol ( NH 4 ) 2 SO 4 132.15 g 1.50 L
132 g ( NH 4 ) 2 SO 4
d.
M ( NH4 ) 2 SO 4 =
= 0.666 M
(NH4)2SO4(s) → 2 NH4+(aq) + SO42−(aq)
M NH + = 2(0.666) = 1.33 M; M SO2− = 0.666 M 4
4
43.
Mol solute = volume (L) × molarity
mol 3+ − ; AlCl3(s) → Al (aq) + 3 Cl (aq) L
Mol Cl− = 0.1000 L ×
0.30 mol AlCl 3 L
3 mol Cl − = 9.0 × 10−2 mol Cl− mol AlCl 3
2 mol Cl − = 6.0 × 10−2 mol Cl− mol MgCl 2
MgCl2(s) → Mg2+(aq) + 2 Cl− (aq) Mol Cl− = 0.0500 L ×
0.60 mol MgCl 2 L
NaCl(s) → Na+(aq) + Cl− (aq) Mol Cl− = 0.2000 L ×
0.40 mol NaCl 1 mol Cl − = 8.0 × 10−2 mol Cl− L mol NaCl
100.0 mL of 0.30 M AlCl3 contains the most moles of Cl− ions. 44.
NaOH(s) → Na+(aq) + OH−(aq), 2 total mol of ions (1 mol Na+ and 1 mol Cl−) per mol NaOH. 0.1000 L ×
0.100 mol NaOH 2 mol ions = 2.0 × 10−2 mol ions L mol NaOH
BaCl2(s) → Ba2+(aq) + 2 Cl−(aq), 3 total mol of ions per mol BaCl2. 0.0500 L ×
0.200 mol 3 mol ions = 3.0 × 10−2 mol ions L mol BaCl 2
CHAPTER 4
SOLUTION STOICHIOMETRY
129
Na3PO4(s) → 3 Na+(aq) + PO43−(aq), 4 total mol of ions per mol Na3PO4. 0.0750 L ×
0.150 mol Na 3 PO 4 L
4 mol ions = 4.50 × 10−2 mol ions mol Na 3 PO 4
75.0 mL of 0.150 M Na3PO4 contains the largest number of ions. 45.
Molar mass of NaOH = 22.99 + 16.00 + 1.008 = 40.00 g/mol Mass NaOH = 0.2500 L ×
0.400 mol NaOH 40 .00 g NaOH = 4.00 g NaOH L mol NaOH
1 mol AgNO 3 1L = 0.24 L = 240 mL 169.9 g 0.25 mol AgNO 3
46.
10. g AgNO3 ×
47.
0.0150 L ×
48.
0.040 g C27H46O ×
137 mmol Na 1 mol 22.99 g Na = 0.0472 g Na L 1000 mmol mol Na
0.059 g C27H46O ×
1 mol C 27 H 46 O 386 .64 g C 27 H 46 O
1 mol C 27 H 46 O 386 .64 g C 27 H 46 O
= 1.0 × 10−4 mol C27H46O = 1.5 × 10−4 mol C27H46O
1.0 10 −4 mol 1.5 10 −4 mol = 1.0 × 10−3 M; = 1.5 × 10−3 M 1L 1L 1 dL 1 dL 10 dL 10 dL
HDL levels between 40.−59 mg/dL correspond to a molarity of cholesterol between 1.0 × 10−3 and 1.5 × 10−3 M. 49.
Mol CaCl2 present = 0.230 L C12H22O11 ×
0.275 mol C12 H 22 O11 = 6.33 × 10−2 mol C12H22O11 L C12 H 22 O11
The volume of C12H22O11 solution after evaporation is: 6.33 ×10−2 mol C12H22O11 ×
1 L C12 H 22 O11 = 5.75 ×10−2 L = 57.5 mL C12H22O11 1.10 mol C12 H 22 O11
Volume H2O evaporated = 230. mL − 57.5 mL = 173 mL H2O evaporated 50.
Initial mol K+ = 0.100 L ×
0.875 mol K 2 CO 3 2 mol K + = 0.175 mol K+ L mol K 2 CO 3
Volume after evaporation = 0.175 mol K+ ×
1L 2.334 mol K +
Volume H2O evaporated = 100.0 mL – 75.0 mL = 25.0 mL
= 0.0750 L = 75.0 mL
130 51.
CHAPTER 4 a. 2.00 L ×
SOLUTION STOICHIOMETRY
0.250 mol NaOH 40 .00 g NaOH = 20.0 g NaOH L mol NaOH
Place 20.0 g NaOH in a 2-L volumetric flask; add water to dissolve the NaOH, and fill to the mark with water, mixing several times along the way. b. 2.00 L ×
0.250 mol NaOH 1 L stock = 0.500 L L 1.00 mol NaOH
Add 500. mL of 1.00 M NaOH stock solution to a 2-L volumetric flask; fill to the mark with water, mixing several times along the way. c. 2.00 L ×
0.100 mol K 2 CrO 4 L
194.20 g K 2 CrO 4 mol K 2 CrO 4
= 38.8 g K2CrO4
Like the solution made in part a, instead using 38.8 g K2CrO4. d. 2.00 L ×
0.100 mol K 2 CrO 4 L
1 L stock = 0.114 L 1.75 mol K 2 CrO 4
Like the solution made in part b, instead using 114 mL of the 1.75 M K2CrO4 stock solution. 52.
0.50 mol H 2 SO 4
a. 1.00 L solution ×
L
0.50 mol H2SO4 ×
= 0.50 mol H2SO4
1L = 2.8 × 10−2 L conc. H2SO4 or 28 mL 18 mol H 2 SO 4
Dilute 28 mL of concentrated H2SO4 to a total volume of 1.00 L with water. The resulting 1.00 L of solution will be a 0.50 M H2SO4 solution. b. We will need 0.50 mol HCl. 0.50 mol HCl ×
1L 12 mol HCl
= 4.2 × 10−2 L = 42 mL
Dilute 42 mL of concentrated HCl to a final volume of 1.00 L. c. We need 0.50 mol NiCl2. 0.50 mol NiCl2 ×
1 mol NiCl 2 • 6H 2 O mol NiCl 2
237.69 g NiCl 2 • 6H 2 O mol NiCl 2 • 6H 2 O
= 118.8 g NiCl2•6H2O 120 g Dissolve 120 g NiCl2•6H2O in water, and add water until the total volume of the solution is 1.00 L. d. 1.00 L ×
0.50 mol HNO 3 L
0.50 mol HNO3 ×
= 0.50 mol HNO3
1L 16 mol HNO 3
= 0.031 L = 31 mL
CHAPTER 4
SOLUTION STOICHIOMETRY
131
Dissolve 31 mL of concentrated reagent in water. Dilute to a total volume of 1.00 L. e. We need 0.50 mol Na2CO3. 0.50 mol Na2CO3 ×
105 .99 g Na 2 CO 3 mol
= 53 g Na2CO3
Dissolve 53 g Na2CO3 in water, dilute to 1.00 L. 53.
10.8 g (NH4)2SO4 ×
Molarity =
1 mol = 8.17 × 10−2 mol (NH4)2SO4 132 .15 g
8.17 10 −2 mol 1000 mL = 0.817 M (NH4)2SO4 100.0 mL L
Moles of (NH4)2SO4 in final solution: 10.00 × 10−3 L ×
0.817 mol = 8.17 × 10−3 mol L 8.17 10 −3 mol 1000 mL = 0.136 M (NH4)2SO4 (10.00 + 50.00) mL L
Molarity of final solution =
(NH4)2SO4(s) → 2 NH4+(aq) + SO42−(aq); M NH + = 2(0.136) = 0.272 M; M SO2− = 0.136 M 4
4
54.
Molarity =
total mol HNO 3 ; total volume = 0.05000 L + 0.10000 L = 0.15000 L total volume
Total mol HNO3 = 0.05000 L
0.100 mol HNO 3 L
+ 0.10000 L
0.200 mol HNO 3 L
Total mol HNO3 = 5.00 × 10−3 mol + 2.00 × 10−2 mol = 2.50 × 10−2 mol HNO3 Molarity =
2.50 10 −2 mol HNO 3 0.15000 L
= 0.167 M HNO3
As expected, the molarity of HNO3 is between 0.100 M and 0.200 M. 55.
3.0 mol Na 2 CO 3
Mol Na2CO3 = 0.0700 L ×
L
= 0.21 mol Na2CO3
Na2CO3(s) → 2 Na+(aq) + CO32−(aq); mol Na+ = 2(0.21 mol) = 0.42 mol Mol NaHCO3 = 0.0300 L ×
1.0 mol NaHCO 3 L
= 0.030 mol NaHCO3
NaHCO3(s) → Na+(aq) + HCO3−(aq); mol Na+ = 0.030 mol
M Na + = total mol Na
+
total volume
=
0.42 mol + 0.030 mol 0.45 mol = 4.5 M Na+ = 0.0700 L + 0.0300 L 0.1000 L
132 56.
CHAPTER 4 Mol CoCl2 = 0.0500 L ×
Mol NiCl2 = 0.0250 L ×
0.250 mol CoCl 2 L
0.350 mol NiCl 2 L
SOLUTION STOICHIOMETRY
= 0.0125 mol
= 0.00875 mol
Both CoCl2 and NiCl2 are soluble chloride salts by the solubility rules. A 0.0125-mol aqueous sample of CoCl2 is actually 0.0125 mol Co2+ and 2(0.0125 mol) = 0.0250 mol Cl−. A 0.00875mol aqueous sample of NiCl2 is actually 0.00875 mol Ni2+ and 2(0.00875) = 0.0175 mol Cl−. The total volume of solution that these ions are in is 0.0500 L + 0.0250 L = 0.0750 L.
57.
M Co 2 + =
0.0125 mol Co 2 + 0.00875 mol Ni 2 + = 0.167 M ; M Ni2 + = = 0.117 M 0.0750 L 0.0750 L
M Cl − =
0.0250 mol Cl − + 0.0175 mol Cl − = 0.567 M 0.0750 L
10.0 mg
Stock solution = −6
100.0 × 10
500.0 mL
L stock ×
=
10.0 10 −3 g 500.0 mL
1000 mL L
=
2.00 10 −5 g steroid
2.00 10 −5 g steroid mL
mL
= 2.00 × 10−6 g steroid
This is diluted to a final volume of 100.0 mL. 2.00 10 −6 g steroid 100.0 mL
58.
1000 mL
L
1 mol steroid 336.43 g steroid
= 5.94 × 10−8 M steroid
Stock solution: 1.584 g Mn2+ ×
1 mol Mn 2 +
= 2.883 × 10−2 mol Mn2+
54.94 g Mn 2 +
2.833 10 −2 mol Mn 2 + 1.000 L
Molarity =
= 2.883 × 10−2 M
Solution A: 50.00 mL ×
1L 2.833 10 −2 mol 1000 mL L
Molarity =
1.442 10 −3 mol 1000.0 mL
= 1.442 × 10−3 mol Mn2+
1000 mL = 1.442 × 10−3 M 1L
Solution B: 10.0 mL ×
1L 1.442 10 −3 mol 1000 mL L
Molarity =
1.442 10 −5 mol 0.2500 L
= 1.442 × 10−5 mol Mn2+
= 5.768 × 10−5 M
CHAPTER 4
SOLUTION STOICHIOMETRY
133
Solution C: 10.00 × 10-3 L × Molarity =
5.768 10 −5 mol L
5.768 10 −7 mol 0.5000 L
= 5.768 × 10−7 mol Mn2+
= 1.154 × 10−6 M
Precipitation Reactions 59.
The solubility rules referenced in the following answers are outlined in Table 4.1 of the text. a. Soluble: Most nitrate salts are soluble (Rule 1). b. Soluble: Most chloride salts are soluble except for Ag+, Pb2+, and Hg22+ (Rule 3). c. Soluble: Most sulfate salts are soluble except for BaSO4, PbSO4, Hg2SO4, and CaSO4 (Rule 4.) d. Insoluble: Most hydroxide salts are only slightly soluble (Rule 5). Note: We will interpret the phrase “slightly soluble” as meaning insoluble and the phrase “marginally soluble” as meaning soluble. So the marginally soluble hydroxides Ba(OH) 2, Sr(OH)2, and Ca(OH)2 will be assumed soluble unless noted otherwise. e. Insoluble: Most sulfide salts are only slightly soluble (Rule 6). Again, “slightly soluble” is interpreted as “insoluble” in problems like these. f.
Insoluble: Rule 5 (see answer d).
g. Insoluble: Most phosphate salts are only slightly soluble (Rule 6). 60.
61.
The solubility rules referenced in the following answers are from Table 4.1 of the text. The phrase “slightly soluble” is interpreted to mean insoluble, and the phrase “marginally soluble” is interpreted to mean soluble. a. Soluble (Rule 3)
b. Soluble (Rule 1)
c. Insoluble (Rule 4)
d. Soluble (Rules 2 and 3)
e. Insoluble (Rule 6)
f.
g. Insoluble (Rule 6)
h. Soluble (Rule 2)
Insoluble (Rule 5)
In these reactions, soluble ionic compounds are mixed together. To predict the precipitate, switch the anions and cations in the two reactant compounds to predict possible products; then use the solubility rules in Table 4.1 to predict if any of these possible products are insoluble (are the precipitate). Note that the phrase “slightly soluble” in Table 4.1 is interpreted to mean insoluble, and the phrase “marginally soluble” is interpreted to mean soluble. a. Possible products = FeCl2 and K2SO4; both salts are soluble, so no precipitate forms. b. Possible products = Al(OH)3 and Ba(NO3)2; precipitate = Al(OH)3(s) c. Possible products = CaSO4 and NaCl; precipitate = CaSO4(s) d. Possible products = KNO3 and NiS; precipitate = NiS(s)
134 62.
CHAPTER 4
SOLUTION STOICHIOMETRY
Use Table 4.1 to predict the solubility of the possible products. a. Possible products = Hg2SO4 and Cu(NO3)2; precipitate = Hg2SO4 b. Possible products = NiCl2 and Ca(NO3)2; both salts are soluble so no precipitate forms. c. Possible products = KI and MgCO3; precipitate = MgCO3 d. Possible products = NaBr and Al2(CrO4)3; precipitate = Al2(CrO4)3
63.
Precipitates will form in all four beakers. In beaker 1, PbCl2(s) forms, in beaker 2, Pb(OH)2(s) forms, in beaker 3, Pb3(PO4)2(s) forms, and in beaker 4, PbSO4(s) forms.
64.
No precipitate forms when Ba(NO3)2 is added to either NH4Cl or RbOH as BaCl2 and Ba(OH)2 are soluble by the solubility rules. When the solution of Ba(NO3)2 is added to the solution of Na2SO4, BaSO4(s) forms, and when the solution of Ba(NO3)2 is added to the solution of K2CO3, BaCO3(s) forms.
65.
For the following answers, the balanced formula equation is first, followed by the complete ionic equation, then the net ionic equation. a. No reaction occurs since all possible products are soluble salts. b. 2 Al(NO3)3(aq) + 3 Ba(OH)2(aq) → 2 Al(OH)3(s) + 3 Ba(NO3)2(aq) 2 Al3+(aq) + 6 NO3−(aq) + 3 Ba2+(aq) + 6 OH−(aq) → 2 Al(OH)3(s) + 3 Ba2+(aq) + 6 NO3−(aq) Al3+(aq) + 3 OH−(aq) → Al(OH)3(s) c. CaCl2(aq) + Na2SO4(aq) → CaSO4(s) + 2 NaCl(aq) Ca2+(aq) + 2 Cl−(aq) + 2 Na+(aq) + SO42−(aq) → CaSO4(s) + 2 Na+(aq) + 2 Cl−(aq) Ca2+(aq) + SO42−(aq) → CaSO4(s) d. K2S(aq) + Ni(NO3)2(aq) → 2 KNO3(aq) + NiS(s) 2 K+(aq) + S2−(aq) + Ni2+(aq) + 2 NO3−(aq) → 2 K+(aq) + 2 NO3−(aq) + NiS(s) Ni2+(aq) + S2−(aq) → NiS(s)
66.
a. Hg2(NO3)2(aq) + CuSO4(aq) → Hg2SO4(s) + Cu(NO3)2(aq) Hg22+(aq) + 2 NO3−(aq) + Cu2+(aq) + SO42−(aq) → Hg2SO4(s) + Cu2+(aq) + 2 NO3−(aq) Hg22+(aq) + SO42−(aq) → Hg2SO4(s) b. No reaction occurs since both possible products are soluble. c. K2CO3(aq) + MgI2(aq) → 2 KI(aq) + MgCO3(s) 2 K+(aq) + CO32−(aq) + Mg2+(aq) + 2I−(aq) → 2 K+(aq) + 2 I−(aq) + MgCO3(s)
CHAPTER 4
SOLUTION STOICHIOMETRY
135
Mg2+(aq) + CO32−(aq) → MgCO3(s) d. 3 Na2CrO4(aq) + 2 Al(Br)3(aq) → 6 NaBr(aq) + Al2(CrO4)3(s) 6 Na+(aq) + 3 CrO42−(aq) + 2 Al3+(aq) + 6 Br−(aq) → 6 Na+(aq) + 6 Br−(aq) + 2 Al (aq) + 3 CrO4 (aq) → Al2(CrO4)3(s) 3+
67.
Al2(CrO4)3(s)
2−
a. When CuSO4(aq) is added to Na2S(aq), the precipitate that forms is CuS(s). Therefore, Na+ (the gray spheres) and SO42− (the bluish green spheres) are the spectator ions. CuSO4(aq) + Na2S(aq) → CuS(s) + Na2SO4(aq); Cu2+(aq) + S2−(aq) → CuS(s) b. When CoCl2(aq) is added to NaOH(aq), the precipitate that forms is Co(OH)2(s). Therefore, Na+ (the gray spheres) and Cl- (the green spheres) are the spectator ions. CoCl2(aq) + 2 NaOH(aq) → Co(OH)2(s) + 2 NaCl(aq) Co2+(aq) + 2 OH−(aq) → Co(OH)2(s) c. When AgNO3(aq) is added to KI(aq), the precipitate that forms is AgI(s). Therefore, K + (the red spheres) and NO3− (the blue spheres) are the spectator ions. AgNO3(aq) + KI(aq) → AgI(s) + KNO3(aq); Ag+(aq) + I−(aq) → AgI(s)
68.
There are many acceptable choices for spectator ions. We will generally choose Na+ and NO3− as the spectator ions because sodium salts and nitrate salts are usually soluble in water. a. Fe(NO3)3(aq) + 3 NaOH(aq) → Fe(OH)3(s) + 3 NaNO3(aq) b. Hg2(NO3)2(aq) + 2 NaCl(aq) → Hg2Cl2(s) + 2 NaNO3(aq) c. Pb(NO3)2(aq) + Na2SO4(aq) → PbSO4(s) + 2 NaNO3(aq) d. BaCl2(aq) + Na2CrO4(aq) → BaCrO4(s) + 2 NaCl(aq)
69.
a. (NH4)2SO4(aq) + Ba(NO3)2(aq) → 2 NH4NO3(aq) + BaSO4(s) Ba2+(aq) + SO42−(aq) → BaSO4(s) b. Pb(NO3)2(aq) + 2 NaCl(aq) → PbCl2(s) + 2 NaNO3(aq) Pb2+(aq) + 2 Cl−(aq) → PbCl2(s) c. Potassium phosphate and sodium nitrate are both soluble in water. No reaction occurs. d. No reaction occurs because all possible products are soluble.
136
CHAPTER 4
SOLUTION STOICHIOMETRY
e. CuCl2(aq) + 2 NaOH(aq) → Cu(OH)2(s) + 2 NaCl(aq) Cu2+(aq) + 2 OH−(aq) → Cu(OH)2(s) 70.
a. CrCl3(aq) + 3 NaOH(aq) → Cr(OH)3(s) + 3 NaCl(aq) Cr3+(aq) + 3 OH−(aq) → Cr(OH)3(s) b. 2 AgNO3(aq) + (NH4)2CO3(aq) → Ag2CO3(s) + 2 NH4NO3(aq) 2 Ag+(aq) + CO32−(aq) → Ag2CO3(s) c. CuSO4(aq) + Hg2(NO3)2(aq) → Cu(NO3)2(aq) + Hg2SO4(s) Hg22+(aq) + SO42−(aq) → Hg2SO4(s) d. No reaction occurs because all possible products (SrI2 and KNO3) are soluble.
71.
For effective separation of the ions, only one precipitate can be filtered off at a time. The NaCl solution cannot be added first since AgCl(s) and PbCl2(s) both form, and the Na2S solution cannot be added first since all three cations form an insoluble sulfide. So, the Na2SO4 solution is added first removing the Pb2+ ion as PbSO4(s). After the PbSO4 precipitate is removed, the NaCl solution is added next removing the Ag+ ion as AgCl(s). After the AgCl precipitate has been removed, the Na2S solution is added last to precipitate out Ni2+ as NiS(s)
72.
Because no precipitates formed upon addition of NaCl or Na2SO4, we can conclude that Hg22+ and Ba2+ are not present in the sample because Hg2Cl2 and BaSO4 are insoluble salts. However, Mn2+ may be present since Mn2+ does not form a precipitate with either NaCl or Na2SO4. A precipitate formed with NaOH; the solution must contain Mn2+ because it forms a precipitate with OH− [Mn(OH)2(s)].
73.
2 AgNO3(aq) + Na2CrO4(aq) → Ag2CrO4(s) + 2 NaNO3(aq) 0.0750 L ×
74.
0.100 mol AgNO 3 L
1 mol Na 2 CrO 4 2 mol AgNO 3
161.98 g Na 2 CrO 4 mol Na 2 CrO 4
= 0.607 g Na2CrO4
2 Na3PO4(aq) + 3 Pb(NO3)2(aq) → Pb3(PO4)2(s) + 6 NaNO3(aq) 0.1500 L ×
0.250 mol Pb(NO 3 ) 2 L
2 mol Na 3 PO 4 3 mol Pb(NO 3 ) 2
1 L Na 3 PO 4 0.100 mol Na 3 PO 4
= 0.250 L
= 250. mL Na3PO4 75.
Fe(NO3)3(aq) + 3 NaOH(aq) → Fe(OH)3(s) + 3 NaNO3(aq) Assuming Fe(NO3)3 is limiting: 0.0750 L ×
0.105 mol Fe(NO3 ) 3 1 mol Fe(OH) 3 106.87 g Fe(OH) 3 mol Fe(NO 3 ) 3 mol Fe(OH) 3 L
= 0.842 g Fe(OH)3
CHAPTER 4
SOLUTION STOICHIOMETRY
137
Assuming NaOH is limiting: 0.125 L ×
0.150 mol NaOH 1 mol Fe(OH) 3 106.87 g Fe(OH) 3 = 0.668 g Fe(OH)3 3 mol NaOH mol Fe(OH) 3 L
Because NaOH produces the smaller mass of the Fe(OH)3 precipitate, NaOH is the limiting reagent and 0.668 g Fe(OH)3 can form. 76.
The balanced equation is 3 BaCl2(aq) + Fe2(SO4)3(aq) → 3 BaSO4(s) + 2 FeCl3(aq). 100.0 mL BaCl2 ×
0.100 mol BaCl 2 3 mol BaSO 4 233.4 g BaSO 4 1L 1000 mL L 3 mol BaCl 2 mol BaSO 4
= 2.33 g BaSO4 100.0 mL Fe2(SO4)3 ×
3 mol BaSO 4 0.100 mol Fe 2 (SO 4 ) 3 1L mol Fe 2 (SO 4 ) 3 1000 mL L
233.4 g BaSO 4 mol BaSO 4
= 7.00 g BaSO4
The BaCl2 reagent produces the smaller quantity of the BaSO4 precipitate, so BaCl2 is limiting and 2.33 g BaSO4 can form. 77.
a.
3 AgNO3(aq) + K3PO4(aq) → Ag3PO4(s) + 3 KNO3(aq) Assuming AgNO3 is limiting: 50.00 mL AgNO3
1 mol Ag 3 PO 4 1.00 mol AgNO 3 1L 1000 mL L AgNO 3 3 mol AgNO 3
= 0.0167 mol Ag3PO4 Assuming K3PO4 is limiting: 25.0 mL K3PO4 ×
1.00 mol K 3 PO 4 1 mol Ag 3 PO 4 1L × 1000 mL L K 3 PO 4 mol K 3 PO 4
= 0.0250 mol Ag3PO4 The AgNO3 reagent produces the smaller quantity of Ag3PO4, so AgNO3 is limiting and 0.0167 mol Ag3PO4 can form. b. initial mol of PO43‒ = 0.0250 L ×
1.00 mol K 3 PO 4 1 mol PO 43− = 0.0250 mol PO43‒ L K 3 PO 4 mol K 3 PO 4
mol PO43‒ in the precipitate = 0.0167 mol Ag3PO4 ×
1 mol PO 43− = 0.0167 mol PO43‒ mol Ag 3 PO 4
mol PO43‒ remaining in solution = 0.0250 mol ‒ 0.0167 mol = 0.0083 mol PO43‒ PO43‒ concentration after reaction = 78.
0.0083 mol PO 43− = 0.11 M PO43‒ 0.05000 L + 0.02500 L
2 AgNO3(aq) + CaCl2(aq) → 2 AgCl(s) + Ca(NO3)2(aq)
138
CHAPTER 4
SOLUTION STOICHIOMETRY
2 mol AgCl 143.4 g AgCl = 2.9 g AgCl 2 mol AgNO 3 mol AgCl
0.1000 L
0.20 mol AgNO 3
0.1000 L ×
2 mol AgCl 143.4 g AgCl 0.15 mol CaCl 2 = 4.3 g AgCl mol CaCl mol AgCl L 2
L
AgNO3 is limiting (it produces the smaller mass of AgCl) and 2.9 g AgCl can form. The net ionic equation is Ag+(aq) + Cl−(aq) → AgCl(s). The ions remaining in solution are the unreacted Cl− ions and the spectator ions NO3− and Ca2+ (all Ag+ is used up in forming AgCl). The moles of each ion present initially (before reaction) can be easily determined from the moles of each reactant. We have 0.1000 L(0.20 mol AgNO3/L) = 0.020 mol AgNO3, which dissolves to form 0.020 mol Ag+ and 0.020 mol NO3−. We also have 0.1000 L(0.15 mol CaCl2/L) = 0.015 mol CaCl2, which dissolves to form 0.015 mol Ca2+ and 2(0.015) = 0.030 mol Cl−. To form the 2.9 g of AgCl precipitate, 0.020 mol Ag+ will react with 0.020 mol of Cl− to form 0.020 mol AgCl (which has a mass of 2.9 g). Mol unreacted Cl− = 0.030 mol Cl− initially − 0.020 mol Cl− reacted Mol unreacted Cl− = 0.010 mol Cl− M Cl− =
0.010 mol Cl − 0.010 mol Cl − = 0.050 M Cl− = total volume 0.1000 L + 0.1000 L
The molarities of the spectator ions are: 0.020 mol NO 3 0.2000 L
79.
−
= 0.10 M NO3−;
0.015 mol Ca 2+ = 0.075 M Ca2+ 0.2000 L
a. The balanced reaction is 2 KOH(aq) + Mg(NO3)2(aq) → Mg(OH)2(s) + 2 KNO3(aq). b. The precipitate is magnesium hydroxide. c. Assuming KOH is limiting: 0.1000 L KOH
1 mol Mg(OH) 2 58.33 g Mg(OH) 2 0.200 mol KOH L KOH 2 mol KOH mol Mg(OH) 2
= 0.583 g Mg(OH)2 Assuming Mg(NO3)2 is limiting: 0.1000 L Mg(NO3)2
0.200 mol Mg(NO 3 ) 2 L Mg(NO 3 ) 2
1 mol Mg(OH) 2
58.33 g Mg(OH) 2
mol Mg(NO 3 ) 2 mol Mg(OH) 2
= 1.17 g Mg(OH)2
The KOH reagent is limiting because it produces the smaller quantity of the Mg(OH)2 precipitate. So 0.583 g Mg(OH)2 can form. d. The net ionic equation for this reaction is Mg2+(aq) + 2 OH−(aq) → Mg(OH)2(s).
CHAPTER 4
SOLUTION STOICHIOMETRY
139
Because KOH is the limiting reagent, all of the OH− is used up in the reaction. So M OH − = 0 M. Note that K+ is a spectator ion, so it is still present in solution after precipitation was complete. Also present will be the excess Mg2+ and NO3− (the other spectator ion). Total Mg2+ = 0.1000 L Mg(NO3)2
0.200 mol Mg(NO 3 ) 2 L Mg(NO 3 ) 2
1 mol Mg 2 + mol Mg(NO 3 ) 2
= 0.0200 mol Mg2+ Mol Mg2+ reacted = 0.1000 L KOH
1 mol Mg(NO 3 ) 2 0.200 mol KOH L KOH 2 mol KOH
M Mg 2+ =
1 mol Mg 2 + = 0.0100 mol Mg2+ mol Mg ( NO 3 ) 2
(0.0200 − 0.0100) mol Mg mol excess Mg 2+ = total volume 0.1000 L + 0.1000 L
2+
= 5.00 × 10−2 M Mg2+
The spectator ions are K+ and NO3−. The moles of each are: mol K+ = 0.1000 L KOH
0.200 mol KOH 1 mol K + = 0.0200 mol K+ L KOH mol KOH
mol NO3− = 0.1000 L Mg(NO3)2
0.200 mol Mg(NO 3 ) 2 L Mg(NO 3 ) 2
2 mol NO 3
−
mol Mg(NO 3 ) 2
= 0.0400 mol NO3− The concentrations are: 0.0400 mol NO 3 0.0200 mol K + = 0.100 M K+; 0.2000 L 0.2000 L
80.
−
= 0.200 M NO3−
a. 3 Ba(NO3)2(aq) + 2 Na3PO4(aq) → Ba3(PO4)2(s) + 6 NaNO3(aq) Net ionic equation: 3 Ba2+(aq) + 2 PO43‒(aq) → Ba3(PO4)2(s) Mol Ba(NO3)2 = 0.150 L Ba(NO3)2
0.300 mol Ba(NO 3 ) 2 = 0.0450 mol Ba(NO3)2 L Ba(NO 3 ) 2
Since Ba(NO3)2 is soluble, it exists in solution as 0.0450 mol Ba2+ and 2(0.0450) = 0.0900 mol NO3‒. Mol Na3PO4 = 0.120 L Na3PO4 ×
0.300 mol Na 3 PO 4 = 0.0360 mol Na3PO4 L Na 3 PO 4
Since Na3PO4 is soluble, it exists as 3(0.0360) = 0.108 mol Na+ and 0.0360 mol PO43‒ in solution. Before reaction, 0.0450 mol Ba2+, 0.0900 mol NO3‒, 0.108 mol Na+, and 0.0360 mol PO43‒ are all present in solution. From the net ionic equation, Ba2+ and PO43‒ react to form Ba3(PO4)2(s). Assuming Ba2+ is limiting:
140
CHAPTER 4 0.0450 mol Ba2+
1 mol Ba 3 (PO 4 ) 2 3 mol Ba 2+
SOLUTION STOICHIOMETRY
= 0.0150 mol Ba3(PO4)2
Assuming PO43 ‒ is limiting: 0.0360 mol PO43‒ ×
1 mol Ba 3 (PO 4 ) 2 2 mol PO 43−
= 0.0180 mol Ba3(PO4)2
Ba2+ produces the smaller quantity of precipitate, so Ba2+ is limiting and will not be present after precipitation is complete. After reaction, there is excess PO43 ‒ present in solution along with the Na+ and NO3‒ spectator ions. Answer v is correct. b.
From the previous work, Ba2+ is limiting and 0.0150 mol Ba3(PO4)2(s) can form.
c.
M Ba 2+ = 0; M Na + =
0.0900 mol NO 3 − 0.108 mol Na + = 0.400 M; M NO − = = 0.333 M 3 0.270 L 0.150 L + 0.120 L
Mol PO43 ‒ in the precipitate = 0.0150 mol Ba3(PO4)2 ×
2 mol PO 43− = 0.0300 mol mol Ba 3 (PO 4 ) 2
Excess mol PO43‒ remaining in solution = 0.0360 mol PO43‒ initial - 0.0300 mol PO43‒ reacted = 0.0060 mol PO43‒ in excess. M PO 3− = 4
81.
0.0060 mol PO 4 3− = 0.022 M 0.270 L
M2SO4(aq) + CaCl2(aq) → CaSO4(s) + 2 MCl(aq) 1.36 g CaSO4 ×
1 mol CaSO 4 136.15 g CaSO 4
1 mol M 2 SO 4 mol CaSO 4
= 9.99 × 10−3 mol M2SO4
From the problem, 1.42 g M2SO4 was reacted, so: molar mass =
1.42 g M 2 SO 4 9.99 10 −3 mol M 2 SO 4
= 142 g/mol
142 u = 2(atomic mass M) + 32.07 + 4(16.00), atomic mass M = 23 u From periodic table, M = Na (sodium). 82.
a. Na+, NO3−, Cl−, and Ag+ ions are present before any reaction occurs. The excess Ag+ added will remove all of the Cl− ions present. Therefore, Na+, NO3−, and the excess Ag+ ions will all be present after precipitation of AgCl is complete. b. Ag+(aq) + Cl−(aq) → AgCl(s) c. Mass NaCl = 0.641 g AgCl ×
1 mol AgCl 1 mol Cl − 1 mol NaCl 58.44 g − 143.4 g mol AgCl mol NaCl mol Cl
= 0.261 g NaCl Mass % NaCl =
0.261 g NaCl × 100 = 17.4% NaCl 1.50 g mixture
CHAPTER 4
SOLUTION STOICHIOMETRY
141
Acid-Base Reactions 83.
All the bases in this problem are ionic compounds containing OH-. The acids are either strong or weak electrolytes. The best way to determine if an acid is a strong or weak electrolyte is to memorize all the strong electrolytes (strong acids). Any other acid you encounter that is not a strong acid will be a weak electrolyte (a weak acid), and the formula should be left unaltered in the complete ionic and net ionic equations. The strong acids to recognize are HCl, HBr, HI, HNO3, HClO4, and H2SO4. For the following answers, the order of the equations are formula, complete ionic, and net ionic. a. 2 HClO4(aq) + Mg(OH)2(s) → 2 H2O(l) + Mg(ClO4)2(aq) 2 H+(aq) + 2 ClO4−(aq) + Mg(OH)2(s) → 2 H2O(l) + Mg2+(aq) + 2 ClO4−(aq) 2 H+(aq) + Mg(OH)2(s) → 2 H2O(l) + Mg2+(aq) b. HCN(aq) + NaOH(aq) → H2O(l) + NaCN(aq) HCN(aq) + Na+(aq) + OH−(aq) → H2O(l) + Na+(aq) + CN−(aq) HCN(aq) + OH−(aq) → H2O(l) + CN−(aq) c. HCl(aq) + NaOH(aq) → H2O(l) + NaCl(aq) H+(aq) + Cl−(aq) + Na+(aq) + OH−(aq) → H2O(l) + Na+(aq) + Cl−(aq) H+(aq) + OH−(aq) → H2O(l)
84.
a. 3 HNO3(aq) + Al(OH)3(s) → 3 H2O(l) + Al(NO3)3(aq) 3 H+(aq) + 3 NO3−(aq) + Al(OH)3(s) → 3 H2O(l) + Al3+(aq) + 3 NO3−(aq) 3 H+(aq) + Al(OH)3(s) → 3 H2O(l) + Al3+(aq) b. HC2H3O2(aq) + KOH(aq) → H2O(l) + KC2H3O2(aq) HC2H3O2(aq) + K+(aq) + OH−(aq) → H2O(l) + K+(aq) + C2H3O2−(aq) HC2H3O2(aq) + OH−(aq) → H2O(l) + C2H3O2−(aq) c. Ca(OH)2(aq) + 2 HCl(aq) → 2 H2O(l) + CaCl2(aq) Ca2+(aq) + 2 OH−(aq) + 2 H+(aq) + 2 Cl−(aq) → 2 H2O(l) + Ca2+(aq) + 2 Cl−(aq) 2 H+(aq) + 2 OH−(aq) → 2 H2O(l) or H+(aq) + OH−(aq) → H2O(l)
85.
All the acids in this problem are strong electrolytes (strong acids). The acids to recognize as strong electrolytes are HCl, HBr, HI, HNO3, HClO4, and H2SO4.
142
CHAPTER 4
SOLUTION STOICHIOMETRY
a. KOH(aq) + HNO3(aq) → H2O(l) + KNO3(aq) b. Ba(OH)2(aq) + 2 HCl(aq) → 2 H2O(l) + BaCl2(aq) c. 3 HClO4(aq) + Fe(OH)3(s) → 3 H2O(l) + Fe(ClO4)3(aq) d. AgOH(s) + HBr(aq) → AgBr(s) + H2O(l) e. Sr(OH)2(aq) + 2 HI(aq) → 2 H2O(l) + SrI2(aq) 86.
a. Perchloric acid plus potassium hydroxide is a possibility. HClO4(aq) + KOH(aq) → H2O(l) + KClO4(aq) b. Nitric acid plus cesium hydroxide is a possibility. HNO3(aq) + CsOH(aq) → H2O(l) + CsNO3(aq) c. Hydroiodic acid plus calcium hydroxide is a possibility. 2 HI(aq) + Ca(OH)2(aq) → 2 H2O(l) + CaI2(aq)
87.
If we begin with 50.00 mL of 0.200 M NaOH, then: 50.00 × 10−3 L ×
0.200 mol = 1.00 × 10−2 mol NaOH is to be neutralized L
a. NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l) 1.00 × 10−2 mol NaOH ×
1 mol HCl 1L = 0.100 L or 100. mL mol NaOH 0.100 mol
b. HNO3(aq) + NaOH(aq) → H2O(l) + NaNO3(aq) 1.00 × 10−2 mol NaOH ×
1 mol HNO 3 mol NaOH
1L = 6.67 × 10 −2 L or 66.7 mL 0.150 mol HNO 3
c. HC2H3O2(aq) + NaOH(aq) → H2O(l) + NaC2H3O2(aq) 1.00 × 10−2 mol NaOH ×
1 mol HC 2 H 3 O 2 mol NaOH
1L 0.200 mol HC 2 H 3 O 2
= 5.00 × 10−2 L = 50.0 mL
88.
We begin with 25.00 mL of 0.200 M HCl or 25.00 × 10 −3 L × 0.200 mol/L = 5.00 × 10−3 mol HCl. a. HCl(aq) + NaOH(aq) → H2O(l) + NaCl(aq) 5.00 × 10−3 mol HCl ×
1 mol HCl 1L = 5.00 × 10−2 L or 50.0 mL mol NaOH 0.100 mol NaOH
b. 2 HCl(aq) + Sr(OH)2(aq) → 2 H2O(l) + SrCl2(aq) 5.00 × 10−3 mol HCl ×
1 mol Sr(OH) 2 2 mol HCl
1L = 0.0500 L = 50.0 mL 0.0500 mol Sr(OH) 2
CHAPTER 4
SOLUTION STOICHIOMETRY
143
c. HCl(aq) + KOH(aq) → H2O(l) + KCl(aq) 1 mol KOH mol HCl
5.00 × 10−3 mol HCl ×
89.
1L = 2.00 × 10−2 L = 20.0 mL 0.250 mol KOH
Ba(OH)2(aq) + 2 HCl(aq) → BaCl2(aq) + 2 H2O(l); H+(aq) + OH−(aq) → H2O(l) 75.0 × 10−3 L ×
0.250 mol HCl L
225.0 × 10−3 L ×
= 1.88 × 10 −2 mol HCl = 1.88 × 10−2 mol H+ + 1.88 × 10−2 mol Cl−
0.0550 mol Ba (OH ) 2 = 1.24 × 10−2 mol Ba(OH)2 L = 1.24 × 10−2 mol Ba2+ + 2.48 × 10−2 mol OH−
The net ionic equation requires a 1 : 1 mole ratio between OH− and H+. The actual mole OH− to mole H+ ratio is greater than 1 : 1, so OH− is in excess. Because 1.88 × 10 −2 mol OH− will be neutralized by the H+, we have (2.48 − 1.88) × 10−2 = 0.60 × 10−2 mol OH− in excess. M OH− =
90.
mol OH − excess total volume
=
6.0 10 −3 mol OH − = 2.0 × 10 −2 M OH− 0.0750 L + 0.2250 L
HCl and HNO3 are strong acids; Ca(OH)2 and RbOH are strong bases. The net ionic equation that occurs is H+(aq) + OH−(aq) → H2O(l). Mol H+ = 0.0500 L ×
0.100 mol HCl 1 mol H + L mol HCl
0.200 mol HNO 3 1 mol H + = 0.00500 + 0.0200 = 0.0250 mol H+ L mol HNO 3
+ 0.1000 L × Mol OH− = 0.5000 L × + 0.2000 L ×
0.0100 mol Ca(OH) 2 L
2 mol OH − mol Ca(OH) 2
0.100 mol RbOH 1 mol OH − = 0.0100 + 0.0200 = 0.0300 mol OH− L mol RbOH
We have an excess of OH−, so the solution is basic (not neutral). The moles of excess OH− = 0.0300 mol OH− initially − 0.0250 mol OH− reacted (with H+) = 0.0050 mol OH− excess.
M OH− = 91.
0.0050 mol OH − 0.0050 mol = 5.9 × 10−3 M = (0.0500 + 0.1000 + 0.5000 + 0.2000) L 0.8500 L
HCl(aq) + NaOH(aq) → H2O(l) + NaCl(aq) 24.16 × 10 −3 L NaOH × Molarity of HCl =
92.
0.106 mol NaOH L NaOH
2.56 10 −3 mol 25.00 10 −3 L
1 mol HCl = 2.56 × 10−3 mol HCl mol NaOH
= 0.102 M HCl
HC2H3O2(aq) + NaOH(aq) → H2O(l) + NaC2H3O2(aq)
144
CHAPTER 4
SOLUTION STOICHIOMETRY
1 mol HC 2 H 3 O 2 0.5062 mol NaOH L soln mol NaOH
a. 16.58 × 10−3 L soln ×
= 8.393 × 10−3 mol HC2H3O2 Concentration of HC2H3O2(aq) =
8.393 10 −3 mol = 0.8393 M 0.01000 L
b. If we have 1.000 L of solution: Total mass = 1000. mL × Mass of HC2H3O2 = 0.8393 mol × Mass % acetic acid = 93.
1.006 g = 1006 g solution mL
60 .05 g = 50.40 g HC2H3O2 mol
50.40 g × 100 = 5.010% 1006 g
2 HNO3(aq) + Ca(OH)2(aq) → 2 H2O(l) + Ca(NO3)2(aq) 35.00 × 10−3 L HNO3 ×
0.0500 mol HNO 3 L HNO 3
1 mol Ca(OH) 2 2 mol HNO 3
1 L Ca(OH) 2 0.0200 mol Ca(OH) 2
= 0.0438 L = 43.8 mL Ca(OH)2 94.
Strong bases contain the hydroxide ion (OH−). The reaction that occurs is H+ + OH− → H2O. 0.0120 L ×
0.150 mol H + 1 mol OH − = 1.80 × 10−3 mol OH− + L mol H
The 30.0 mL of the unknown strong base contains 1.80 × 10−3 mol OH− . 1.80 10 −3 mol OH − = 0.0600 M OH− 0.0300 L
The unknown base concentration is one-half the concentration of OH− ions produced from the base, so the base must contain 2 OH− in each formula unit. The three soluble strong bases that have two OH− ions in the formula are Ca(OH)2, Sr(OH)2, and Ba(OH)2. These are all possible identities for the strong base. 95.
Because KHP is a monoprotic acid, the reaction is (KHP is an abbreviation for potassium hydrogen phthalate): NaOH(aq) + KHP(aq) → NaKP(aq) + H2O(l) 0.1082 g KHP ×
1 mol KHP 1 mol NaOH = 5.298 × 10−4 mol NaOH 204 .22 g KHP mol KHP
There are 5.298 × 10−4 mol of sodium hydroxide in 34.67 mL of solution. Therefore, the concentration of sodium hydroxide is: 5.298 10 −4 mol 34.67 10 −3 L
= 1.528 × 10−2 M NaOH
CHAPTER 4 96.
SOLUTION STOICHIOMETRY
145
KHP is a monoprotic acid: NaOH(aq) + KHP(aq) →H2O(l) + NaKP(aq) Mass KHP = 0.02046 L NaOH ×
0.1000 mol NaOH 1 mol KHP 204 .22 g KHP L NaOH mol NaOH mol KHP
= 0.4178 g KHP
Oxidation-Reduction Reactions 97.
Apply the rules in Table 4.2. a. KMnO4 is composed of K+ and MnO4− ions. Assign oxygen an oxidation state of −2, which gives manganese a +7 oxidation state because the sum of oxidation states for all atoms in MnO4− must equal the 1− charge on MnO4−. K, +1; O, −2; Mn, +7. b. Assign O a −2 oxidation state, which gives nickel a +4 oxidation state. Ni, +4; O, −2. c. Na4Fe(OH)6 is composed of Na+ cations and Fe(OH)64−anions. Fe(OH)64− is composed of an iron cation and 6 OH− anions. For an overall anion charge of 4−, iron must have a +2 oxidation state. As is usually the case in compounds, assign O a −2 oxidation state and H a +1 oxidation state. Na, +1; Fe, +2; O, −2; H, +1. d. (NH4)2HPO4 is made of NH4+ cations and HPO42− anions. Assign +1 as the oxidation state of H and −2 as the oxidation state of O. In NH4+, x + 4(+1) = +1, x = −3 = oxidation state of N. In HPO42−, +1 + y + 4(−2) = −2, y = +5 = oxidation state of P. e. O, −2; P, +3 f.
O, −2; 3x + 4(−2) = 0, x = +8/3 = oxidation state of Fe; this is the average oxidation state of the three iron ions in Fe3O4. In the actual formula unit, there are two Fe3+ ions and one Fe2+ ion.
g. O, −2; F, −1; Xe, +6 i. 98.
O, −2; C, +2
h. F, −1; S, +4 j.
H, +1; O, −2; C, 0
a. UO22+: O, −2; for U, x + 2(−2) = +2, x = +6 b. As2O3: O, −2; for As, 2(x) + 3(−2) = 0, x = +3 c. NaBiO3: Na, +1; O, −2; for Bi, +1 + x + 3(−2) = 0, x = +5 d. As4: As, 0 e. HAsO2: Assign H = +1 and O = −2; for As, +1 + x + 2(−2) = 0, x = +3 f.
Mg2P2O7: Composed of Mg2+ ions and P2O74− ions. Mg, +2; O, −2; P, +5
g. Na2S2O3: Composed of Na+ ions and S2O32− ions. Na, +1; O, −2; S, +2 h. Hg2Cl2: Hg, +1; Cl, −1 i.
Ca(NO3)2: Composed of Ca2+ ions and NO3− ions.
Ca, +2; O, −2; N, +5
146 99.
100.
CHAPTER 4
SOLUTION STOICHIOMETRY
a. −3
b. −3
c. 2(x) + 4(+1) = 0, x = −2
d. +2
e. +1
f.
+4
g. +3
h. +5
i.
0
a. SrCr2O7: Composed of Sr2+ and Cr2O72− ions. Sr, +2; O, −2; Cr, 2x + 7(−2) = −2, x = +6 b. Cu, +2; Cl, −1
d. H, +1; O, −1
c. O, 0;
e. Mg2+ and CO32− ions present. Mg, +2; O, −2; C, +4;
f.
g. Pb2+ and SO32− ions present. Pb, +2; O, −2; S, +4;
h. O, −2; Pb, +4
Ag, 0
i.
Na+ and C2O42− ions present. Na, +1; O, −2; C, 2x + 4(−2) = −2, x = +3
j.
O, −2; C, +4
k. Ammonium ion has a 1+ charge (NH4+), and sulfate ion has a 2− charge (SO42−). Therefore, the oxidation state of cerium must be +4 (Ce4+). H, +1; N, −3; O, −2; S, +6 l.
O, −2; Cr, +3
101.
All the statements are false. The oxygen oxidation state is −2 on both the reactant and product sides of the equation. Oxygen is neither oxidized nor reduced. Carbon goes from the +2 oxidation state in CO to the +4 oxidation state in CO2. With the increase in oxidation states, carbon is oxidized, so CO is the reducing agent. Note that nitrogen goes from the +2 oxidation state in NO to the 0 oxidation state in N2. Nitrogen is reduced and NO is the oxidizing agent.
102.
The oxidation state of carbon in H2C2O4 is +3 and the oxidation state of C in CO2 is +4. Carbon is oxidized and H2C2O4 is the reducing agent. Statements a and d are true. The oxidation state of manganese in MnO4– is +7, so statement c is true. On the product side, Mn has an oxidation state of +2. Mn is reduced and MnO4– is the oxidizing agent. Statement c is false.
103.
To determine if the reaction is an oxidation-reduction reaction, assign oxidation states. If the oxidation states change for some elements, then the reaction is a redox reaction. If the oxidation states do not change, then the reaction is not a redox reaction. In redox reactions, the species oxidized (called the reducing agent) shows an increase in oxidation states, and the species reduced (called the oxidizing agent) shows a decrease in oxidation states. Redox?
Oxidizing Agent
Reducing Agent
Substance Oxidized
Substance Reduced
a. Yes
Ag+
Cu
Cu
Ag+
b. No
−
−
−
−
c. No
−
−
−
−
d. Yes
SiCl4
Mg
Mg
SiCl4 (Si)
e. No
−
−
−
In b, c, and e, no oxidation numbers change.
−
CHAPTER 4 104.
SOLUTION STOICHIOMETRY
147
The species oxidized shows an increase in oxidation states and is called the reducing agent. The species reduced shows a decrease in oxidation states and is called the oxidizing agent. The pertinent oxidation states are listed by the substance oxidized and the substance reduced. Redox?
Oxidizing Agent
Reducing Agent
a. Yes
H2O
CH4
CH4 (C, −4 → +2)
b
AgNO3
Cu
Cu (0 → +2)
AgNO3 (Ag, +1 → 0)
HCl
Zn
Zn (0 → +2)
HCl (H, +1 → 0)
Yes
c. Yes
Substance Oxidized
Substance Reduced H2O (H, +1 → 0)
d. No; there is no change in any of the oxidation numbers. 105.
Use the method of half-reactions described in Section 4.10 of the text to balance these redox reactions. The first step always is to separate the reaction into the two half-reactions, and then to balance each half-reaction separately. a. 3 I− → I3− + 2e−
ClO− → Cl− 2e− + 2H+ + ClO− → Cl− + H2O
Adding the two balanced half-reactions so electrons cancel: 3 I−(aq) + 2 H+(aq) + ClO−(aq) → I3−(aq) + Cl−(aq) + H2O(l) b. As2O3 → H3AsO4 As2O3 → 2 H3AsO4 Left 3 − O; right 8 − O
NO3− → NO + 2 H2O 4 H+ + NO3− → NO + 2 H2O − (3 e + 4 H+ + NO3− → NO + 2 H2O) × 4
Right hand side has 5 extra O. Balance the oxygen atoms first using H2O, then balance H using H+, and finally, balance charge using electrons. This gives: (5 H2O + As2O3 → 2 H3AsO4 + 4 H+ + 4 e−) × 3 Common factor is a transfer of 12 e−. Add half-reactions so that electrons cancel. 12 e− + 16 H+ + 4 NO3− → 4 NO + 8 H2O 15 H2O + 3 As2O3 → 6 H3AsO4 + 12 H+ + 12 e− 7 H2O(l) + 4 H+(aq) + 3 As2O3(s) + 4 NO3−(aq) → 4 NO(g) + 6 H3AsO4(aq) c. (2 Br− → Br2 + 2 e−) × 5
MnO4− → Mn2+ + 4 H2O (5 e + 8 H + MnO4− →Mn2+ + 4 H2O) × 2 −
+
Common factor is a transfer of 10 e−. 10 Br− → 5 Br2 + 10 e− 10 e + 16 H + 2 MnO4− → 2 Mn2+ + 8 H2O −
+
16 H+(aq) + 2 MnO4−(aq) + 10 Br−(aq) → 5 Br2(l) + 2 Mn2+(aq) + 8 H2O(l)
148
CHAPTER 4 d.
CH3OH → CH2O (CH3OH → CH2O + 2 H+ + 2 e−) × 3
SOLUTION STOICHIOMETRY
Cr2O72− → 2 Cr3+ 14 H+ + Cr2O72− → 2 Cr3+ + 7 H2O − 6 e + 14 H+ + Cr2O72− → 2 Cr3+ + 7 H2O
Common factor is a transfer of 6 e−. 3 CH3OH → 3 CH2O + 6 H+ + 6 e− 6 e + 14 H + Cr2O72− → 2 Cr3+ + 7 H2O −
+
8 H+(aq) + 3 CH3OH(aq) + Cr2O72−(aq) → 2 Cr3+(aq) + 3 CH2O(aq) + 7 H2O(l) 106.
a. (Cu → Cu2+ + 2 e−) × 3
NO3−→ NO + 2 H2O (3 e− + 4 H+ + NO3− → NO + 2 H2O) × 2
Adding the two balanced half-reactions so that electrons cancel: 3 Cu → 3 Cu2+ + 6 e− 6 e + 8 H + 2 NO3− → 2 NO + 4 H2O −
+
3 Cu(s) + 8 H+(aq) + 2 NO3−(aq) → 3 Cu2+(aq) + 2 NO(g) + 4 H2O(l) b. (2 Cl− → Cl2 + 2 e−) × 3
Cr2O72− → 2 Cr3+ + 7 H2O 6 e + 14 H + Cr2O72− → 2 Cr3+ + 7 H2O −
+
Add the two half-reactions with six electrons transferred: 6 Cl− → 3 Cl2 + 6 e− 6 e + 14 H + Cr2O72− → 2 Cr3+ + 7 H2O −
+
14 H+(aq) + Cr2O72−(aq) + 6 Cl−(aq) → 3 Cl2(g) + 2 Cr3+(aq) + 7 H2O(l) c.
Pb → PbSO4 Pb + H2SO4 → PbSO4 + 2 H+ Pb + H2SO4 → PbSO4 + 2 H+ + 2 e−
PbO2 → PbSO4 PbO2 + H2SO4 → PbSO4 + 2 H2O 2 e− + 2 H+ + PbO2 + H2SO4 → PbSO4 + 2 H2O
Add the two half-reactions with two electrons transferred: 2 e− + 2 H+ + PbO2 + H2SO4 → PbSO4 + 2 H2O Pb + H2SO4 → PbSO4 + 2 H+ + 2 e− Pb(s) + 2 H2SO4(aq) + PbO2(s) → 2 PbSO4(s) + 2 H2O(l) This is the reaction that occurs in an automobile lead-storage battery. d.
Mn2+ → MnO4− (4 H2O + Mn2+ → MnO4− + 8 H+ + 5 e−) × 2
NaBiO3 → Bi3+ + Na+ NaBiO3 → Bi3+ + Na+ 6 H+ + NaBiO3 → Bi3+ + Na+ + 3 H2O − (2 e + 6 H+ + NaBiO3 → Bi3+ + Na+ + 3 H2O) × 5
CHAPTER 4
SOLUTION STOICHIOMETRY
149
8 H2O + 2 Mn2+ → 2 MnO4− + 16 H+ + 10 e− 10 e− + 30 H+ + 5 NaBiO3 → 5 Bi3+ + 5 Na+ + 15 H2O 8 H2O + 30 H+ + 2 Mn2+ + 5 NaBiO3 → 2 MnO4− + 5 Bi3+ + 5 Na+ + 15 H2O + 16 H+ Simplifying: 14 H+(aq) + 2 Mn2+(aq) + 5 NaBiO3(s) → 2 MnO4−(aq) + 5 Bi3+(aq) + 5 Na+(aq) + 7 H2O(l) e.
H3AsO4 →AsH3 H3AsO4 → AsH3 + 4 H2O − + 8 e + 8 H + H3AsO4 → AsH3 + 4 H2O
(Zn → Zn2+ + 2 e−) × 4
8 e− + 8 H+ + H3AsO4 → AsH3 + 4 H2O 4 Zn → 4 Zn2+ + 8 e− 8 H+(aq) + H3AsO4(aq) + 4 Zn(s) → 4 Zn2+(aq) + AsH3(g) + 4 H2O(l) 107.
Use the same method as with acidic solutions. After the final balanced equation, convert H+ to OH− as described in Section 4.10 of the text. The extra step involves converting H+ into H2O by adding equal moles of OH− to each side of the reaction. This converts the reaction to a basic solution while still keeping it balanced. a.
Al → Al(OH)4− 4 H2O + Al → Al(OH)4− + 4 H+ 4 H2O + Al → Al(OH)4− + 4 H+ + 3 e−
MnO4− → MnO2 3 e + 4 H + MnO4− → MnO2 + 2 H2O −
+
4 H2O + Al → Al(OH)4− + 4 H+ + 3 e− 3 e− + 4 H+ + MnO4− → MnO2 + 2 H2O 2 H2O(l) + Al(s) + MnO4−(aq) → Al(OH)4−(aq) + MnO2(s) H+ doesn’t appear in the final balanced reaction, so we are done. b.
Cl2 → Cl− 2 e− + Cl2 → 2 Cl−
Cl2 → OCl− 2 H2O + Cl2 → 2 OCl− + 4 H+ + 2 e−
2 e− + Cl2 → 2 Cl− 2 H2O + Cl2 → 2 OCl− + 4 H+ + 2 e−
2 H2O + 2 Cl2 → 2 Cl− + 2 OCl− + 4 H+ Now convert to a basic solution. Add 4 OH− to both sides of the equation. The 4 OH− will react with the 4 H+ on the product side to give 4 H2O. After this step, cancel identical species on both sides (2 H2O). Applying these steps gives: 4 OH− + 2 Cl2 → 2 Cl− + 2 OCl− + 2 H2O, which can be further simplified to: 2 OH−(aq) + Cl2(g) → Cl−(aq) + OCl−(aq) + H2O(l) c.
NO2− → NH3 6 e + 7 H + NO2− → NH3 + 2 H2O −
+
Al → AlO2− (2 H2O + Al → AlO2− + 4 H+ + 3 e−) × 2
150
CHAPTER 4
SOLUTION STOICHIOMETRY
Common factor is a transfer of 6 e−. 6e− + 7 H+ + NO2− → NH3 + 2 H2O 4 H2O + 2 Al → 2 AlO2− + 8 H+ + 6 e− OH− + 2 H2O + NO2− + 2 Al → NH3 + 2 AlO2− + H+ + OH− Reducing gives OH−(aq) + H2O(l) + NO2−(aq) + 2 Al(s) → NH3(g) + 2 AlO2−(aq). 108.
a.
Cr → Cr(OH)3 3 H2O + Cr → Cr(OH)3 + 3 H+ + 3 e−
CrO42− → Cr(OH)3 3 e + 5 H + CrO42− → Cr(OH)3 + H2O −
+
3 H2O + Cr → Cr(OH)3 + 3 H+ + 3 e− 3 e + 5 H+ + CrO42− → Cr(OH)3 + H2O −
2 OH− + 2 H+ + 2 H2O + Cr + CrO42− → 2 Cr(OH)3 + 2 OH− Two OH− were added above to each side to convert to a basic solution. The two OH− react with the 2 H+ on the reactant side to produce 2 H2O. The overall balanced equation is: 4 H2O(l) + Cr(s) + CrO42−(aq) → 2 Cr(OH)3(s) + 2 OH−(aq) b.
S2− → S (S2− → S + 2 e−) × 5
MnO4− → MnS MnO4− + S2− → MnS − + ( 5 e + 8 H + MnO4− + S2− → MnS + 4 H2O) × 2
Common factor is a transfer of 10 e−. 5 S2− → 5 S + 10 e− 10 e + 16 H + 2 MnO4 + 2 S2− → 2 MnS + 8 H2O −
+
−
16 OH− + 16 H+ + 7 S2− + 2 MnO4− → 5 S + 2 MnS + 8 H2O + 16 OH− 16 H2O + 7 S2− + 2 MnO4− → 5 S + 2 MnS + 8 H2O + 16 OH− Reducing gives 8 H2O(l) + 7 S2−(aq) + 2 MnO4−(aq) → 5 S(s) + 2 MnS(s) + 16 OH−(aq). c.
CN− → CNO− (H2O + CN− → CNO− + 2 H+ + 2 e−) × 3
MnO4− → MnO2 (3 e + 4 H + MnO4− → MnO2 + 2 H2O) × 2 −
+
Common factor is a transfer of 6 electrons. 3 H2O + 3 CN−→ 3 CNO− + 6 H+ + 6 e− 6 e− + 8 H+ + 2 MnO4− → 2 MnO2 + 4 H2O 2 OH− + 2 H+ + 3 CN− + 2 MnO4− → 3 CNO− + 2 MnO2 + H2O + 2 OH− Reducing gives: H2O(l) + 3 CN−(aq) + 2 MnO4−(aq) → 3 CNO−(aq) + 2 MnO2(s) + 2 OH−(aq)
CHAPTER 4 109.
SOLUTION STOICHIOMETRY
151
NaCl + H2SO4 + MnO2 → Na2SO4 + MnCl2 + Cl2 + H2O We could balance this reaction by the half-reaction method, which is generally the preferred method. However, sometimes a redox reaction is not so complicated and thus balancing by inspection is a possibility. Let’s try inspection here. To balance Cl−, we need 4 NaCl: 4 NaCl + H2SO4 + MnO2 → Na2SO4 + MnCl2 + Cl2 + H2O Balance the Na+ and SO42− ions next: 4 NaCl + 2 H2SO4 + MnO2 → 2 Na2SO4 + MnCl2 + Cl2 + H2O On the left side: 4 H and 10 O; on the right side: 8 O not counting H2O We need 2 H2O on the right side to balance H and O: 4 NaCl(aq) + 2 H2SO4(aq) + MnO2(s) → 2 Na2SO4(aq) + MnCl2(aq) + Cl2(g) + 2 H2O(l)
110.
Au + HNO3 + HCl → AuCl4−+ NO Only deal with ions that are reacting (omit H+): Au + NO3− + Cl− → AuCl4− + NO The balanced half-reactions are: Au + 4 Cl− → AuCl4− + 3 e−
3 e− + 4 H+ + NO3− → NO + 2 H2O
Adding the two balanced half-reactions: Au(s) + 4 Cl−(aq) + 4 H+(aq) + NO3−(aq) → AuCl4−(aq) + NO(g) + 2 H2O(l) 111.
(H2C2O4 → 2 CO2 + 2 H+ + 2 e−) × 5
(5 e− + 8 H+ + MnO4− → Mn2+ + 4 H2O) × 2
5 H2C2O4 → 10 CO2 + 10 H+ + 10 e− 10 e + 16 H + 2 MnO4− → 2 Mn2+ + 8 H2O −
+
6 H+(aq) + 5 H2C2O4(aq) + 2 MnO4−(aq) → 10 CO2(g) + 2 Mn2+(aq) + 8 H2O(l) 0.1058 g H2C2O4 ×
1 mol H 2 C 2 O 4 90.04 g
2 mol MnO 4
−
5 mol H 2 C 2 O 4
= 4.700 × 10−4 mol MnO4−
−
Molarity = 112.
4.700 10 −4 mol MnO 4 1000 mL = 1.622 × 10−2 M MnO4− 28.97 mL L
(Fe2+ → Fe3+ + e−) × 5 5 e + 8 H + MnO4− → Mn2+ + 4 H2O −
+
8 H+(aq) + MnO4−(aq) + 5 Fe2+(aq) → 5 Fe3+(aq) + Mn2+(aq) + 4 H2O(l) From the titration data we can get the number of moles of Fe2+. We then convert this to a mass of iron and calculate the mass percent of iron in the sample.
152
CHAPTER 4 38.37 × 10−3 L MnO4− ×
0.0198 mol MnO 4 L
−
SOLUTION STOICHIOMETRY
5 mol Fe 2 + mol MnO 4
= 3.80 × 10−3 mol Fe2+
−
= 3.80 × 10−3 mol Fe present 3.80 × 10−3 mol Fe × Mass % Fe = 113.
55 .85 g Fe = 0.212 g Fe mol Fe
0.212 g × 100 = 34.6% Fe 0.6128 g
a. (Fe2+ → Fe3+ + e−) 5
5 e− + 8 H+ + MnO4− → Mn2+ + 4 H2O
The balanced equation is: 8 H+(aq) + MnO4−(aq) + 5 Fe2+(aq) → 5 Fe3+(aq) + Mn2+(aq) + 4 H2O(l) 20.62 × 10−3 L soln × Molarity =
0.0216 mol MnO 4 L soln
2.23 10 −3 mol Fe 2 + 50.00 10 −3 L
b. (Fe2+ → Fe3+ + e−) 6
−
5 mol Fe 2 + mol MnO 4
−
= 2.23 × 10−3 mol Fe2+
= 4.46 × 10−2 M Fe2+
6 e− + 14 H+ + Cr2O72− → 2 Cr3+ + 7 H2O
The balanced equation is: 14 H+(aq) + Cr2O72−(aq) + 6 Fe2+(aq) → 6 Fe3+(aq) + 2 Cr3+(aq) + 7 H2O(l) 50.00 × 10−3 L ×
2−
1 mol Cr2 O 7 4.46 10 −2 mol Fe 2 + L 6 mol Fe 2 +
1L
0.0150 mol Cr2 O 7
2−
= 2.48 × 10−2 L or 24.8 mL 2(6 e− + 14 H+ + Cr2O72− → 2 Cr3+ + 7 H2O) 3 H2O + C2H5OH → 2 CO2 + 12 H+ + 12 e−
114.
16 H+(aq) + 2 Cr2O72−(aq) + C2H5OH(aq) → 4 Cr3+(aq) + 2 CO2(g) + 11 H2O(l) 0.0600 mol Cr2 O 7 2 − 1 mol C 2 H 5 OH 46.07 g = 0.0429 g C2H5OH 2 − 2 mol Cr O mol C H OH L 2 5 2 7
0.03105 L
0.0429 g C 2 H 5 OH 100 = 0.143% C2H5OH 30 .0 g blood
ChemWork Problems 115.
Desired uncertainty is 1% of 0.02, or ±0.0002. So we want the solution to be 0.0200 ± 0.0002 M, or the concentration should be between 0.0198 and 0.0202 M. We should use a 1-L volumetric flask to make the solution. They are good to ±0.1%. We want to weigh out between 0.0198 mol and 0.0202 mol of KIO3. Molar mass of KIO3 = 39.10 + 126.9 + 3(16.00) = 214.0 g/mol
CHAPTER 4
SOLUTION STOICHIOMETRY
0.0198 mol ×
153
214.0 g 214.0 g = 4.237 g; 0.0202 mol × = 4.323 g (carrying extra sig. figs.) mol mol
We should weigh out between 4.24 and 4.32 g of KIO3. We should weigh it to the nearest milligram, or nearest 0.1 mg. Dissolve the KIO3 in water, and dilute (with mixing along the way) to the mark in a 1-L volumetric flask. This will produce a solution whose concentration is within the limits and is known to at least the fourth decimal place. 116.
6 molecules 1.5 molecules = 4 .0 L 1 .0 L
Solution A:
4 molecules ; solution B: 1 .0 L
Solution C:
4 molecules 2 molecules 6 molecules 3 molecules = ; solution D: = 2 .0 L 1 .0 L 2 .0 L 1 .0 L
Solution A has the most molecules per unit volume so solution A is most concentrated. This is followed by solution D, then solution C. Solution B has the fewest molecules per unit volume, so solution B is least concentrated. 117.
32.0 g C12H22O11 ×
1 mol C12 H 22 O11 = 0.0935 mol C12H22O11 added to blood 342 .30 g
The blood sugar level would increase by: 0.0935 mol C 12 H 22 O 11 5.0 L
118.
0.160 g MgCl2 × Molarity =
1 mol 95.21 g
= 0.019 mol/L
= 1.68 × 10 −3 mol MgCl2
1.68 10 −3 mol 1000 mL = 0.0168 M MgCl2 100.0 mL L
MgCl2(s) → Mg2+(aq) + 2 Cl−(aq); M Mg 2 + = 0.0168 M; M Cl = 2(0.0168) = 0.0336 M −
119.
Stock solution =
6.706 10 −3 g oxalic acid 0.6706 g = 100.0 mL mL
10.00 mL stock ×
6.706 10 −3 g oxalic acid = 6.706 × 10−2 g oxalic acid mL
This is diluted to a final volume of 250.0 mL. 6.706 10 −2 g H 2 C 2 O 4 250.0 mL
1000 mL L
1 mol H 2 C 2 O 4 90.04 g
= 2.979 × 10−3 M H2C2O4
120.
CaCO3(s) + 2 HCl(aq) → CaCl2(aq) + H2O(l) + CO2(g)
121.
There are other possible correct choices for most of the following answers. We have listed only three possible reactants in each case. a. AgNO3, Pb(NO3)2, and Hg2(NO3)2 would form precipitates with the Cl− ion.
154
CHAPTER 4
SOLUTION STOICHIOMETRY
Ag+(aq) + Cl−(aq) → AgCl(s); Pb2+(aq) + 2 Cl−(aq) → PbCl2(s) Hg22+(aq) + 2 Cl−(aq) → Hg2Cl2(s) b. Na2SO4, Na2CO3, and Na3PO4 would form precipitates with the Ca2+ ion. Ca2+(aq) + SO42−(aq) → CaSO4(s); Ca2+(aq) + CO32−(aq) → CaCO3(s) 3 Ca2+(aq) + 2 PO43−(aq) → Ca3(PO4)2(s) c. NaOH, Na2S, and Na2CO3 would form precipitates with the Fe3+ ion. Fe3+(aq) + 3 OH−(aq) → Fe(OH)3(s); 2 Fe3+(aq) + 3 S2−(aq) → Fe2S3(s) 2 Fe3+(aq) + 3 CO32−(aq) → Fe2(CO3)3(s) d. BaCl2, Pb(NO3)2, and Ca(NO3)2 would form precipitates with the SO42− ion. Ba2+(aq) + SO42−(aq) → BaSO4(s); Pb2+(aq) + SO42−(aq) → PbSO4(s) Ca2+(aq) + SO42−(aq) → CaSO4(s) e. Na2SO4, NaCl, and NaI would form precipitates with the Hg22+ ion. Hg22+(aq) + SO42−(aq) → Hg2SO4(s); Hg22+(aq) + 2 Cl−(aq) → Hg2Cl2(s) Hg22+(aq) + 2 I−(aq) → Hg2I2(s) f.
NaBr, Na2CrO4, and Na3PO4 would form precipitates with the Ag+ ion. Ag+(aq) + Br-(aq) → AgBr(s); 2 Ag+(aq) + CrO42−(aq) →Ag2CrO4(s) 3 Ag+(aq) + PO43−(aq) → Ag3PO4(s)
122.
a. Assume 100.00 g of material. 42.23 g C × 2.11 g B ×
1 mol F 1 mol C = 3.516 mol C; 55.66 g F × = 2.929 mol F 19 .00 g F 12 .01 g C
1 mol B = 0.195 mol B 10 .81 g B
Dividing by the smallest number:
3.516 2.929 = 18.0; = 15.0 0.195 0.195
The empirical formula is C18F15B. 0.01267 mol = 4.396 × 10 −3 mol BARF L 2.251 g Molar mass of BARF = = 512.1 g/mol 4.396 10 − 3 mol
b. 0.3470 L ×
The empirical formula mass of BARF is 511.99 g. Therefore, the molecular formula is the same as the empirical formula, C18F15B.
CHAPTER 4 123.
SOLUTION STOICHIOMETRY
155
2 NaOH(aq) + Ni(NO3)2(aq) → Ni(OH)2(s) + 2 NaNO3(aq) 0.1500 L ×
0.249 mol Ni(NO 3 ) 2 L
2 mol NaOH 1 L NaOH = 0.747 L 1 mol Ni(NO 3 ) 2 0.100 mol NaOH
= 747 mL NaOH 124.
The balanced equation is 3 BaCl2(aq) + 2 Na3PO4(aq) → Ba3(PO4)2(s) + 6 NaCl(aq). 0.4000 L BaCl2 ×
0.289 mol BaCl 2 L
1 mol Ba 3 (PO 4 ) 2 3 mol BaCl 2
601 .8 g Ba 3 (PO 4 ) 2 mol Ba 3 (PO 4 ) 2
= 23.2 g Ba3(PO4)2 0.5000 L Na3PO4 ×
0.200 mol Na 3 PO 4 L
1 mol Ba 3 (PO 4 ) 2 2 mol Na 3 PO 4
601 .8 g Ba 3 (PO 4 ) 2 mol Ba 3 (PO 4 ) 2
= 30.1 g Ba3(PO4)2
The BaCl2 reagent produces the smaller quantity of precipitate, so BaCl2 is limiting and 23.2 g Ba3(PO4)2(s) can form. 125.
3 (NH4)2CrO4(aq) + 2 Cr(NO2)3(aq) → 6 NH4NO2(aq) + Cr2(CrO4)3(s) 0.203 L × 0.307 mol ( NH 4 ) 2 CrO 4 1 mol Cr 2 (CrO 4 ) 3 452 .00 g Cr 2 (CrO 4 ) 3 L
3 mol ( NH 4 ) 2 CrO 4
mol Cr 2 (CrO 4 ) 3
= 9.39 g Cr2(CrO4)3 0.137 L ×
0.269 mol Cr ( NO 2 ) 3 1 mol Cr 2 (CrO 4 ) 3 452 .00 g Cr 2 (CrO 4 ) 3 L 2 mol Cr ( NO 2 ) 3 mol Cr 2 (CrO 4 ) 3
= 8.33 g Cr2(CrO4)3 The Cr(NO2)3 reagent produces the smaller amount of product, so Cr(NO2)3 is limiting and the theoretical yield of Cr2(CrO4)3 is 8.33 g. 0.880 = 126.
actual yield , actual yield = (8.33 g)(0.880) = 7.33 g Cr2(CrO4)3 isolated 8.33 g
a. MgCl2(aq) + 2 AgNO3(aq) → 2 AgCl(s) + Mg(NO3)2(aq) 0.641 g AgCl × 0.213 g MgCl 2 1.50 g mixture
1 mol MgCl 2 1 mol AgCl 95 .21 g = 0.213 g MgCl2 143 .4 g AgCl 2 mol AgCl mol MgCl 2
× 100 = 14.2% MgCl2
b. 0.213 g MgCl2 ×
2 mol AgNO 3 1 mol MgCl 2 1L 1000 mL 95 .21 g mol MgCl 2 0.500 mol AgNO 3 1L
= 8.95 mL AgNO3 127.
XCl2(aq) + 2 AgNO3(aq) → 2 AgCl(s) + X(NO3)2(aq)
156
CHAPTER 4 1.38 g AgCl ×
SOLUTION STOICHIOMETRY
1 mol AgCl 1 mol XCl 2 = 4.81 × 10 −3 mol XCl2 143.4 g 2 mol AgCl
1.00 g XCl 2
= 208 g/mol; x + 2(35.45) = 208, x = 137 g/mol
4.91 10 −3 mol XCl 2
From the periodic table, the metal X is barium (Ba). 128.
2 AgNO3(aq) + CaCl2(aq) → 2 AgCl(s) + Ca(NO3)2(aq) 0.4500 L
0.257 mol AgNO 3
0.4000 L ×
0.200 mol CaCl 2
L
L
2 mol AgCl 2 mol AgNO 3
143.4 g AgCl mol AgCl
= 1.66 g AgCl
2 mol AgCl 143 .4 g AgCl = 2.29 g AgCl mol CaCl 2 mol AgCl
AgNO3 is limiting (it produces the smaller mass of AgCl) and 1.66 g AgCl(s) can form. Note that we did this calculation for your information. It is typically asked in this type of problem. The net ionic equation is Ag+(aq) + Cl−(aq) → AgCl(s). The ions remaining in solution after precipitation is complete will be the unreacted Cl− ions and the spectator ions NO3− and Ca2+ (all Ag+ is used up in forming AgCl). The moles of each ion present initially (before reaction) can be determined from the moles of each reactant. We have 0.4500 L(0.257 mol AgNO3/L) = 0.116 mol AgNO3, which dissolves to form 0.116 mol Ag+ and 0.116 mol NO3−. We also have 0.4000 L(0.200 mol CaCl2/L) = 0.0800 mol CaCl2, which dissolves to form 0.0800 mol Ca2+ and 2(0.0800) = 0.160 mol Cl−. To form the 1.66 g of AgCl precipitate, 0.116 mol Ag+ will react with 0.116 mol of Cl− to form 0.116 mol AgCl (which has a mass of 1.66 g). Mol unreacted Cl− = 0.160 mol Cl− initially − 0.116 mol Cl− reacted to form the precipitate Mol unreacted Cl− = 0.044 mol Cl− M Cl− =
129.
0.044 mol Cl − total volume
=
0.044 mol Cl − 0.4500 L + 0.4000 L
= 0.052 M Cl− in excess after reaction
Zn2P2O7: 2(65.38) + 2(30.97) + 7(16.00) = 304.70 g/mol All of the zinc in Zn2P2O7 came from the zinc in the foot powder. 0.4089 g Zn2P2O7 × Mass % Zn =
130.
1 mol Zn 2 P2 O 7 304.70 g Zn 2 P2 O 7
65.38 g Zn 2 mol Zn = 0.1755 g Zn mol Zn 2 P2 O 7 mol
0.1755 g Zn × 100 = 13.07% Zn 1.343 g foot powder
From the periodic table, use aluminum in the formulas to convert from mass of Al(OH) 3 to mass of Al2(SO4)3 in the mixture.
CHAPTER 4
SOLUTION STOICHIOMETRY
0.107 g Al(OH)3 ×
1 mol Al(OH) 3 78.00 g
157
1 mol Al 3+ mol Al(OH) 3
Mass % Al2(SO4)3 = 131.
1 mol Al 2 (SO 4 ) 3 2 mol Al 3+
342.17 g Al 2 (SO 4 ) 3 mol Al 2 (SO 4 ) 3
= 0.235 g Al2(SO4)3
0 .235 g × 100 = 16.2% 1 .45 g
All the Tl in TlI came from Tl in Tl2SO4. The conversion from TlI to Tl2SO4 uses the molar masses and formulas of each compound. 0.1824 g TlI
504 .9 g Tl 2SO 4 204 .4 g Tl = 0.1390 g Tl2SO4 331 .3 g TlI 408 .8 g Tl
Mass % Tl2SO4 =
132.
0.1390 g Tl 2SO 4 × 100 = 1.465% Tl2SO4 9.486 g pesticide
a. Fe3+(aq) + 3 OH−(aq) → Fe(OH)3(s) Fe(OH)3: 55.85 + 3(16.00) + 3(1.008) = 106.87 g/mol 0.107 g Fe(OH)3 ×
55.85 g Fe 106.87 g Fe(OH) 3
= 0.0559 g Fe
b. Fe(NO3)3: 55.85 + 3(14.01) + 9(16.00) = 241.86 g/mol 0.0559 g Fe ×
241 .86 g Fe ( NO 3 ) 3 55 .85 g Fe
c. Mass % Fe(NO3)3 = 133.
= 0.242 g Fe(NO3)3
0 .242 g × 100 = 53.1% 0 .456 g
With the ions present, the only possible precipitate is Cr(OH)3. Cr(NO3)3(aq) + 3 NaOH(aq) → Cr(OH)3(s) + 3 NaNO3(aq) Mol NaOH used to form precipitate = 2.06 g Cr(OH)3 ×
1 mol Cr (OH ) 3 3 mol NaOH = 6.00 × 10−2 mol 103 .02 g mol Cr (OH ) 3
NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l) Mol NaOH used to react with HCl = 0.1000 L × MNaOH = 134.
0.400 mol HCl 1 mol NaOH = 4.00 × 10−2 mol L mol HCl
6.00 10 total mol NaOH = volume
−2
mol + 4.00 10 −2 mol = 2.00 M NaOH 0.0500 L
a. MgO(s) + 2 HCl(aq) → MgCl2(aq) + H2O(l)
158
CHAPTER 4
SOLUTION STOICHIOMETRY
Mg(OH)2(s) + 2 HCl(aq) → MgCl2(aq) + 2 H2O(l) Al(OH)3(s) + 3 HCl(aq) → AlCl3(aq) + 3 H2O(l) b. Let's calculate the number of moles of HCl neutralized per gram of substance. We can get these directly from the balanced equations and the molar masses of the substances. 2 mol HCl mol MgO
1 mol MgO 40.31 g MgO
=
4.962 10 −2 mol HCl g MgO
2 mol HCl mol Mg (OH ) 2
1 mol Mg (OH ) 2 58.33 g Mg (OH ) 2
3 mol HCl mol Al (OH ) 3
1 mol Al (OH ) 3 78.00 g Al (OH ) 3
=
=
3.429 10 −2 mol HCl g Mg (OH ) 2
3.846 10 −2 mol HCl g Al (OH ) 3
Therefore, 1 gram of magnesium oxide would neutralize the most 0.10 M HCl. 135.
Using HA as an abbreviation for the monoprotic acid acetylsalicylic acid: HA(aq) + NaOH(aq) → H2O(l) + NaA(aq) Mol HA = 0.03517 L NaOH
0.5065 mol NaOH 1 mol HA = 1.781 × 10−2 mol HA L NaOH mol NaOH
From the problem, 3.210 g HA was reacted, so: molar mass =
136.
1.781 10 − 2 mol HA
= 180.2 g/mol
Mg(s) + 2 HCl(aq) → MgCl2(aq) + H2(g) 3.00 g Mg ×
137.
3.210 g HA
1 mol Mg 24 .31 g Mg
2 mol HCl 1L = 0.049 L = 49 mL HCl mol Mg 5.0 mol HCl
Let HA = unknown monoprotic acid; HA(aq) + NaOH(aq) → NaA(aq) + H2O(l) Mol HA present = 0.0250 L × x g HA mol HA
=
0.500 mol NaOH 1 mol HA L 1 mol NaOH
= 0.0125 mol HA
2.20 g HA , x = molar mass of HA = 176 g/mol 0.0125 mol HA
Empirical formula weight 3(12) + 4(1) + 3(16) = 88 g/mol. Because 176/88 = 2.0, the molecular formula is (C3H4O3)2 = C6H8O6. 138.
We get the empirical formula from the elemental analysis. Out of 100.00 g carminic acid, there are: 53.66 g C ×
1 mol C 1 mol H = 4.468 mol C; 4.09 g H × = 4.06 mol H 12 .01 g C 1.008 g H
CHAPTER 4
SOLUTION STOICHIOMETRY
42.25 g O ×
159
1 mol O = 2.641 mol O 16 .00 g O
Dividing the moles by the smallest number gives: 4.468 = 1.692; 2.641
4.06 = 1.54 2.641
These numbers don’t give obvious mole ratios. Let’s determine the mol C to mol H ratio: 4.468 11 = 1.10 = 4.06 10
So let's try
4.468 2.641 4.06 4.06 = 0.406 as a common factor: = 11.0; = 6.50 = 10.0; 0.406 0.406 0.406 10
Therefore, C22H20O13 is the empirical formula. We can get molar mass from the titration data. The balanced reaction is HA(aq) + OH−(aq) → H2O(l) + A−(aq), where HA is an abbreviation for carminic acid, an acid with one acidic proton (H+). 18.02 × 10−3 L soln ×
0.0406 mol NaOH 1 mol carminic acid L soln mol NaOH
= 7.32 × 10−4 mol carminic acid Molar mass =
0.3602 g 492 g = −4 mol 7.32 10 mol
The empirical formula mass of C22H20O13 22(12) + 20(1) + 13(16) = 492 g. Therefore, the molecular formula of carminic acid is also C22H20O13. 139.
2 HNO3(aq) + Ca(OH)2(aq) → 2 H2O(l) + Ca(NO3)2(aq) 34.66 × 10−3 L HNO3 × Molarity of Ca(OH)2 =
140.
0.944 mol HNO 3 L HNO 3
1.64 10 −2 mol 50.00 10 −3 L
1 mol Ca(OH) 2 2 mol HNO 3
= 1.64 × 10−2 mol Ca(OH)2
= 0.328 M Ca(OH)2
Let HX = unknown monoprotic acid; HX(aq) + KOH(aq) → KX(aq) + H2O(l) Mol HX = 0.2015 L Molar mass of HX =
1.00 mol KOH 1 mol HX = 0.202 mol HX L 1 mol KOH
mass of HX 16.3 g = 80.7 g/mol = mol of HX 0.202 mol
The common monoprotic strong acids are HCl, HBr, HI, HNO3, and HClO4. HBr has a molar mass of 1.008 + 79.90 = 80.91 g/mol. The unknown monoprotic strong acid is HBr and the precipitate with Pb(NO3)2(aq) is PbBr2.
160 141.
CHAPTER 4 0.104 g AgCl
SOLUTION STOICHIOMETRY
1 mol AgCl 35.45 g Cl − 1 mol Cl − = 2.57 × 10−2 g Cl− 143.4 g AgCl mol AgCl mol Cl −
All of the Cl− in the AgCl precipitate came from the chlorisondamine chloride compound in the medication. So we need to calculate the quantity of C14H20Cl6N2 which contains 2.57 × 10−2 g Cl−. Molar mass of C14H20Cl6N2 = 14(12.01) + 20(1.008) + 6(35.45) + 2(14.01) = 429.02 g/mol There are 6(35.45) = 212.70 g chlorine for every mole (429.02 g) of C14H20Cl6N2. 2.57 × 10−2 g Cl−
429.02 g C 14 H 20 Cl 6 N 2 212.70 g Cl −
Mass % chlorisondamine chloride = 142.
5.18 10 −2 g × 100 = 4.05% 1.28 g
All the sulfur in BaSO4 came from the saccharin. The conversion from BaSO4 to saccharin utilizes the molar masses of each compound. 0.5032 g BaSO4 ×
183 .19 g C 7 H 5 NO 3S 32 .07 g S = 0.3949 g C7H5NO3S 233 .4 g BaSO 4 32 .07 g S
Average mass
0.3949 g
Tablet
=
10 tablets
Average mass % = 143.
= 5.18 × 10−2 g C14H20Cl6N2
=
3.949 10 −2 g tablet
=
39.49 mg tablet
0.3949 g C 7 H 5 NO 3S × 100 = 67.00% saccharin by mass 0.5894 g
Pb2+ + 2 Cl− → PbCl2(s) 0.0292 L
0.100 mol Pb 2+ 2 mol Cl35.45 g Cl − = 0.207 g Cl– × × 2+ L mol Pb mol Cl
Mass % Cl = 0.207 g Cl−/0.455 g douglasite = 45.5% Cl− 144.
a. Al(s) + 3 HCl(aq) → AlCl3(aq) + 3/2 H2(g) or 2 Al(s) + 6 HCl(aq) → 2 AlCl3(aq) + 3 H2(g) Hydrogen is reduced (goes from the +1 oxidation state to the 0 oxidation state), and aluminum Al is oxidized (0 → +3). b. Balancing S is most complicated since sulfur is in both products. Balance C and H first; then worry about S. CH4(g) + 4 S(s) → CS2(l) + 2 H2S(g) Sulfur is reduced (0 → −2), and carbon is oxidized (−4 → +4). c. Balance C and H first; then balance O. C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(l)
CHAPTER 4
SOLUTION STOICHIOMETRY
161
Oxygen is reduced (0 → −2), and carbon is oxidized (−8/3 → +4). d. Although this reaction is mass balanced, it is not charge balanced. We need 2 moles of silver on each side to balance the charge. Cu(s) + 2 Ag+(aq) → 2 Ag(s) + Cu2+(aq) Silver is reduced (+1 → 0), and copper is oxidized (0 → +2). 145.
HNO3 → NO2 HNO3 → NO2 + H2O − + (e + H + HNO3 → NO2 + H2O) × 2
Mn → Mn2+ + 2 e−
Mn → Mn2+ + 2 e− 2 e + 2 H + 2 HNO3 → 2 NO2 + 2 H2O −
+
2 H+(aq) + Mn(s) + 2 HNO3(aq) → Mn2+(aq) + 2 NO2(g) + 2 H2O(l) or 4 H+(aq) + Mn(s) + 2 NO3−(aq) → Mn2+(aq) + 2 NO2(g) + 2 H2O(l) (HNO3 is a strong acid.) (4 H2O + Mn2+ → MnO4− + 8 H+ + 5 e−) × 2
(2 e− + 2 H+ + IO4− → IO3− + H2O) × 5
8 H2O + 2 Mn2+ → 2 MnO4− + 16 H+ + 10 e− 10 e + 10 H+ + 5 IO4− → 5 IO3− + 5 H2O −
3 H2O(l) + 2 Mn2+(aq) + 5 IO4−(aq) → 2 MnO4−(aq) + 5 IO3−(aq) + 6 H+(aq)
Challenge Problems 146.
Let x = mass of NaCl, and let y = mass K2SO4. So x + y = 10.00. Two reactions occur: Pb2+(aq) + 2 Cl−(aq) → PbCl2(s) and Pb2+(aq) + SO42−(aq) → PbSO4(s) Molar mass of NaCl = 58.44 g/mol; molar mass of K2SO4 = 174.27 g/mol; molar mass of PbCl2 = 278.1 g/mol; molar mass of PbSO4 = 303.3 g/mol x y = moles NaCl; = moles K2SO4 58.44 174.27
Setting up an equation for total mass of solid: mass of PbCl2
+ mass PbSO4 = total mass of solid
y x (1/2)(278.1) + (303.3) = 21.75 58.44 174.27
We have two equations: (2.379)x + (1.740)y = 21.75 and x + y = 10.00. Solving: x = 6.81 g NaCl;
6.81 g NaCl 10 .00 g mixture
100 = 68.1% NaCl
162 147.
CHAPTER 4 a. 5.0 ppb Hg in water =
SOLUTION STOICHIOMETRY
5.0 ng Hg 5.0 10 −9 g Hg = g soln mL soln
5.0 10 −9 g Hg 1 mol Hg 1000 mL = 2.5 × 10 −8 M Hg mL 200.6 g Hg L
b.
1.0 10 −9 g CHCl 3 1 mol CHCl 3 1000 mL = 8.4 × 10 −9 M CHCl3 mL 119.37 g CHCl 3 L
c. 10.0 ppm As =
10.0 μg As g soln
=
10.0 10 −6 g As mL soln
−6
10.0 10 g As 1 mol As 1000 mL = 1.33 × 10 −4 M As mL 74.92 g As L
d. 148.
0.10 10 −6 g DDT 1 mol DDT 1000 mL = 2.8 × 10 −7 M DDT mL 354.46 g DDT L
We want 100.0 mL of each standard. To make the 100. ppm standard: 100. μg Cu × 100.0 mL solution = 1.00 × 104 µg Cu needed mL
1.00 × 104 µg Cu ×
1 mL stock = 10.0 mL of stock solution 1000 .0 μg Cu
Therefore, to make 100.0 mL of 100. ppm solution, transfer 10.0 mL of the 1000.0 ppm stock solution to a 100-mL volumetric flask, and dilute to the mark. Similarly: 75.0 ppm standard, dilute 7.50 mL of the 1000.0 ppm stock to 100.0 mL. 50.0 ppm standard, dilute 5.00 mL of the 1000.0 ppm stock to 100.0 mL. 25.0 ppm standard, dilute 2.50 mL of the 1000.0 ppm stock to 100.0 mL. 10.0 ppm standard, dilute 1.00 mL of the 1000.0 ppm stock to 100.0 mL. 149.
a. 0.308 g AgCl ×
35 .45 g Cl 0.0761 g = 0.0761 g Cl; % Cl = × 100 = 29.7% Cl 143 .4 g AgCl 0.256 g
Cobalt(III) oxide, Co2O3: 2(58.93) + 3(16.00) = 165.86 g/mol 0.145 g Co2O3 ×
0 .103 g 117 .86 g Co = 0.103 g Co; % Co = × 100 = 24.8% Co 0 .416 g 165 .86 g Co 2 O 3
The remainder, 100.0 − (29.7 + 24.8) = 45.5%, is water. Assuming 100.0 g of compound:
CHAPTER 4
SOLUTION STOICHIOMETRY
163
45.5 g H2O ×
2.016 g H 5.09 g H = 5.09 g H; % H = × 100 = 5.09% H 18 .02 g H 2 O 100 .0 g compound
45.5 g H2O ×
16 .00 g O 40 .4 g O = 40.4 g O; % O = × 100 = 40.4% O 18 .02 g H 2 O 100 .0 g compound
The mass percent composition is 24.8% Co, 29.7% Cl, 5.09% H, and 40.4% O. b. Out of 100.0 g of compound, there are: 24.8 g Co × 5.09 g H ×
1 mol 1 mol = 0.421 mol Co; 29.7 g Cl × = 0.838 mol Cl 35 .45 g Cl 58 .93 g Co
1 mol 1 mol = 5.05 mol H; 40.4 g O × = 2.53 mol O 16 .00 g O 1.008 g H
Dividing all results by 0.421, we get CoCl2•6H2O for the empirical formula, which is also the actual formula given the information in the problem. The •6H2O represent six waters of hydration in the chemical formula. c. CoCl2•6H2O(aq) + 2 AgNO3(aq) → 2 AgCl(s) + Co(NO3)2(aq) + 6 H2O(l) CoCl2•6H2O(aq) + 2 NaOH(aq) → Co(OH)2(s) + 2 NaCl(aq) + 6 H2O(l) Co(OH)2 → Co2O3
This is an oxidation-reduction reaction. Thus we also need to include an oxidizing agent. The obvious choice is O2.
4 Co(OH)2(s) + O2(g) → 2 Co2O3(s) + 4 H2O(l) 150.
a. C12H10-nCln + n Ag+ → n AgCl; molar mass of AgCl = 143.4 g/mol Molar mass of PCB = 12(12.01) + (10 − n)(1.008) + n(35.45) = 154.20 + (34.44)n Because n mol AgCl is produced for every 1 mol PCB reacted, n(143.4) g of AgCl will be produced for every [154.20 + (34.44)n] g of PCB reacted. Mass of AgCl (143 .4) n or massAgCl[154.20 + (34.44)n] = massPCB(143.4)n = Mass of PCB 154 .20 + (34 .44 ) n
b. 0.4971[154.20 + (34.44)n] = 0.1947(143.4)n, 76.65 + (17.12)n = (27.92)n 76.65 = (10.80)n, n = 7.097 151.
Zn(s) + 2 AgNO2(aq) → 2 Ag(s) + Zn(NO2)2(aq) Let x = mass of Ag and y = mass of Zn after the reaction has stopped. Then x + y = 29.0 g. Because the moles of Ag produced will equal two times the moles of Zn reacted: (19.0 − y) g Zn Simplifying:
1 mol Zn 2 mol Ag 1 mol Ag = x g Ag 65 .38 g Zn 1 mol Zn 107 .9 g Ag
164
CHAPTER 4
SOLUTION STOICHIOMETRY
3.059 × 10−2(19.0 − y) = (9.268 × 10−3)x Substituting x = 29.0 − y into the equation gives: 3.059 × 10−2(19.0 − y) = 9.268 × 10−3(29.0 − y) Solving: 0.581 − (3.059 × 10−2)y = 0.269 − (9.268 × 10−3)y, (2.132 × 10−2)y = 0.312, y = 14.6 g Zn 14.6 g Zn is present, and 29.0 − 14.6 = 14.4 g Ag is also present after the reaction is stopped. 152.
Ag+(aq) + Cl−(aq) → AgCl(s); let x = mol NaCl and y = mol KCl. (22.90 × 10−3 L) × 0.1000 mol/L = 2.290 × 10−3 mol Ag+ = 2.290 × 10−3 mol Cl− total x + y = 2.290 × 10−3 mol Cl−, x = 2.290 × 10−3 − y Because the molar mass of NaCl is 58.44 g/mol and the molar mass of KCl is 74.55 g/mol: (58.44)x + (74.55)y = 0.1586 g 58.44(2.290 × 10−3 − y) + (74.55)y = 0.1586, (16.11)y = 0.0248, y = 1.54 × 10−3 mol KCl Mass % KCl =
1.54 10 −3 mol 74.55 g/mol 0.1586 g
× 100 = 72.4% KCl
% NaCl = 100.0 − 72.4 = 27.6% NaCl 2−
153.
0.298 g BaSO4 ×
2− 96.07 g SO 4 = 0.123 g SO42−; % sulfate = 0.123 g SO 4 = 60.0% 233.4 g BaSO 4 0.205 g
Assume we have 100.0 g of the mixture of Na2SO4 and K2SO4. There are: 60.0 g SO42− ×
1 mol = 0.625 mol SO42− 96 .07 g
There must be 2 × 0.625 = 1.25 mol of 1+ cations to balance the 2− charge of SO42−. Let x = number of moles of K+ and y = number of moles of Na+; then x + y = 1.25. The total mass of Na+ and K+ must be 40.0 g in the assumed 100.0 g of mixture. Setting up an equation: x mol K+ ×
39.10 g 22.99 g + y mol Na+ × = 40.0 g mol mol
So we have two equations with two unknowns: x + y = 1.25 and (39.10)x + (22.99)y = 40.0 x = 1.25 − y, so 39.10(1.25 − y) + (22.99)y = 40.0
CHAPTER 4
SOLUTION STOICHIOMETRY
165
48.9 − (39.10)y + (22.99)y = 40.0, − (16.11)y = −8.9 y = 0.55 mol Na+ and x = 1.25 − 0.55 = 0.70 mol K+ Therefore: 0.70 mol K+ ×
1 mol K 2 SO 4 2 mol K
+
= 0.35 mol K2SO4; 0.35 mol K2SO4 ×
174.27 g mol
= 61 g K2SO4 We assumed 100.0 g; therefore, the mixture is 61% K2SO4 and 39% Na2SO4. 154.
a. Let x = mass of Mg, so 10.00 − x = mass of Zn. Ag+(aq) + Cl−(aq) → AgCl(s). From the given balanced equations, there is a 2 : 1 mole ratio between mol Mg and mol Cl−. The same is true for Zn. Because mol Ag+ = mol Cl− present, one can set up an equation relating mol Cl− present to mol Ag+ added. x g Mg ×
1 mol Mg 2 mol Cl − 1 mol Zn 2 mol Cl − + (10.00 − x ) g Zn 24.31 g Mg mol Mg 65.38 g Zn mol Zn
= 0.156 L ×
3.00 mol Ag + 1 mol Cl − = 0.468 mol Cl− L mol Ag +
20 .00 − 2 x 2x 2x 2(10.00 − x) = 0.468, 24.31 × 65.38 + = 0.468 + 65 .38 24.31 65.38 24 .31
(130.8)x + 486.2 − (48.62)x = 743.8 (carrying 1 extra sig. fig.) (82.2)x = 257.6, x = 3.13 g Mg; b. 0.156 L × MHCl =
155.
% Mg =
3.13 g Mg × 100 = 31.3% Mg 10 .00 g mixture
3.00 mol Ag + 1 mol Cl − = 0.468 mol Cl− = 0.468 mol HCl added + L mol Ag
0.468 mol = 6.00 M HCl 0.0780 L
Pb2+(aq) + 2 Cl−(aq) → PbCl2(s) 3.407 g PbCl2 ×
1 mol PbCl 2 278.1 g PbCl 2
1 mol Pb 2+ = 0.01225 mol Pb2+ mol PbCl 2
0.01225 mol = 6.13 M Pb2+ = 6.13 M Pb(NO3)2 2.00 10 −3 L
This is also the Pb(NO3)2 concentration in the 80.0 mL of evaporated solution. Original concentration =
moles Pb ( NO 3 ) 2 0.0800 L 6.13 mol/L = = 4.90 M Pb(NO3)2 original volume 0.1000 L
166 156.
CHAPTER 4 Mol CuSO4 = 87.7 mL × Mol Fe = 2.00 g ×
SOLUTION STOICHIOMETRY
1L 0.500 mol = 0.0439 mol 1000 mL L
1 mol Fe = 0.0358 mol 55 .85 g
The two possible reactions are: I. CuSO4(aq) + Fe(s) → Cu(s) + FeSO4(aq) II. 3 CuSO4(aq) + 2 Fe(s) → 3 Cu(s) + Fe2(SO4)3(aq) If reaction I occurs, Fe is limiting, and we can produce: 0.0358 mol Fe ×
1 mol Cu 63 .55 g Cu = 2.28 g Cu mol Fe mol Cu
If reaction II occurs, CuSO4 is limiting, and we can produce: 0.0439 mol CuSO4 ×
3 mol Cu 63 .55 g Cu = 2.79 g Cu 3 mol CuSO 4 mol Cu
Assuming 100% yield, reaction I occurs because it fits the data best. 157.
0.2750 L × 0.300 mol/L = 0.0825 mol H+; let y = volume (L) delivered by Y and z = volume (L) delivered by Z. H+(aq) + OH−(aq) → H2O(l); y(0.150 mol/L) + z(0.250 mol/L) = 0.0825 mol H+ mol OH−
0.2750 L + y + z = 0.655 L, y + z = 0.380, z = 0.380 − y y(0.150) + (0.380 − y)(0.250) = 0.0825, solving: y = 0.125 L, z = 0.255 L Flow rate for Y =
158.
255 mL 125 mL = 2.06 mL/min; flow rate for Z = = 4.20 mL/min 60 .65 min 60 .65 min
a. H3PO4(aq) + 3 NaOH(aq) → 3 H2O(l) + Na3PO4(aq) b. 3 H2SO4(aq) + 2 Al(OH)3(s) → 6 H2O(l) + Al2(SO4)3(aq) c. H2Se(aq) + Ba(OH)2(aq) → 2 H2O(l) + BaSe(s) d. H2C2O4(aq) + 2 NaOH(aq) → 2 H2O(l) + Na2C2O4(aq)
159.
2 H3PO4(aq) + 3 Ba(OH)2(aq) → 6 H2O(l) + Ba3(PO4)2(s) 0.01420 L ×
0.141 mol H 3 PO 4 3 mol Ba (OH ) 2 1 L Ba (OH ) 2 = 0.0576 L L 2 mol H 3 PO 4 0.0521 mol Ba (OH ) 2
= 57.6 mL Ba(OH)2
CHAPTER 4 160.
SOLUTION STOICHIOMETRY
2 NaOH(aq) + H2SO4(aq) → Na2SO4(aq) + 2 H2O(l) 1 mol H 2 SO 4 0.1000 mol NaOH = 1.422 × 10 −3 mol H2SO4 L NaOH 2 mol NaOH
0.02844 L NaOH ×
Mass sulfur = 1.422 × 10 −3 mol H2SO4 × Mass % S =
161.
167
32.07 g S 1 mol S = 4.560 × 10−2 g S mol H 2 SO 4 mol S
4.560 10 −2 g S × 100 = 3.502% S 1.302 g coal
Mol OH‒ added = 0.0500 L ×
0.20 mol Ba(OH) 2 2 mol OH − = 0.020 mol OH‒ L mol Ba(OH) 2
If a monoprotic acid: HA(aq) + OH‒(aq) → A‒(aq) + H2O(l) 0.020 mol OH −
Molarity of HA =
0.1000 L
1 mol HA mol OH − = 0.20 M ; answers a-c are incorrect.
If a diprotic acid: H2A(aq) + 2 OH‒(aq) → A2‒(aq) + 2 H2O(l) 1 mol H 2 A 2 mol OH − = 0.10 M 0.1000 L
0.020 mol OH −
Molarity of H2A = Answer e is correct. 162.
35.08 mL NaOH ×
Molarity = 163.
1L 2.12 mol NaOH 1000 mL L NaOH
3.72 10 −2 mol 10.00 mL
1000 mL L
1 mol H 2SO 4 = 3.72 × 10−2 mol H2SO4 2 mol NaOH
= 3.72 M H2SO4
The pertinent equations are: 2 NaOH(aq) + H2SO4(aq) → Na2SO4(aq) + 2 H2O(l) HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) Amount of NaOH added = 0.0500 L ×
0.213 mol = 1.07 × 10−2 mol NaOH L
Amount of NaOH neutralized by HCl: 0.01321 L HCl
0.103 mol HCl 1 mol NaOH = 1.36 × 10−3 mol NaOH L HCl mol HCl
The difference, 9.3 × 10−3 mol, is the amount of NaOH neutralized by the sulfuric acid. 9.3 × 10−3 mol NaOH ×
1 mol H 2SO 4 = 4.7 × 10−3 mol H2SO4 2 mol NaOH
Concentration of H2SO4 =
4.7 10 −3 mol = 4.7 × 10−2 M H2SO4 0.1000 L
168 164.
CHAPTER 4
SOLUTION STOICHIOMETRY
Let H2A = formula for the unknown diprotic acid. H2A(aq) + 2 NaOH(aq) → 2 H2O(l) + Na2A(aq) Mol H2A = 0.1375 L × Molar mass of H2A =
165.
0.750 mol NaOH L
1 mol H 2 A 2 mol NaOH
= 0.0516 mol
6.50 g = 126 g/mol 0.0516 mol
Mol C6H8O7 = 0.250 g C6H8O7 ×
1 mol C 6 H 8 O 7 192 .12 g C 6 H 8 O 7
= 1.30 × 10−3 mol C6H8O7
Let HxA represent citric acid, where x is the number of acidic hydrogens. The balanced neutralization reaction is: HxA(aq) + x OH−(aq) → x H2O(l) + Ax−(aq) Mol OH− reacted = 0.0372 L × x=
0.105 mol OH − = 3.91 × 10−3 mol OH− L
−3 mol OH − = 3.91 10 −3 mol mol citric acid 1.30 10 mol
= 3.01
Therefore, the general acid formula for citric acid is H3A, meaning that citric acid has three acidic hydrogens per citric acid molecule (citric acid is a triprotic acid). 166.
Let HxA = glutamic acid; HxA(aq) + x KOH(aq) → KxA + x H2O(l) Mol glutamic acid = 0.250 g ×
1 mol glutamic acid = 0.00170 mol 147.13 g
0.200 mol KOH = 0.00340 mol L mol KOH 0.00340 mol x= = 2.00 = mol glutamic acid 0.00170 mol
Mol KOH added = 0.0170 L ×
Glutamic acid is a diprotic acid with a formula of H2C5H7O4N (answer b is correct). 167.
X2− contains 36 electrons, so X2− has 34 protons, which identifies X as selenium (Se). The name of H2Se would be hydroselenic acid following the conventions described in Chapter 2. H2Se(aq) + 2 OH−(aq) → Se2−(aq) + 2 H2O(l) 0.0356 L ×
168.
1 mol H 2Se 80.98 g H 2Se 0.175 mol OH − = 0.252 g H2Se L mol H 2Se 2 mol OH −
a. Flow rate = 5.00 × 104 L/s + 3.50 × 103 L/s = 5.35 × 104 L/s b. CHCl =
3.50 10 3 (65.0) 5.35 10 4
= 4.25 ppm HCl
CHAPTER 4
SOLUTION STOICHIOMETRY
169
c. 1 ppm = 1 mg/kg H2O = 1 mg/L (assuming density = 1.00 g/mL) 8.00 h
60 min 60 s 1.80 10 4 L 4.25 mg HCl 1g = 2.20 × 106 g HCl h min s L 1000 mg 1 mol HCl 1 mol CaO 56 .08 g Ca = 1.69 × 106 g CaO 36 .46 g HCl 2 mol HCl mol CaO
2.20 × 106 g HCl
d. The concentration of Ca2+ going into the second plant was: 5.00 10 4 (10.2) 5.35 10 4
= 9.53 ppm
The second plant used: 1.80 × 104 L/s × (8.00 × 60 × 60) s = 5.18 × 108 L of water. 1.69 × 106 g CaO
40.08 g Ca 2+ = 1.21 × 106 g Ca2+ was added to this water. 56.08 g CaO
C Ca 2+ (plant water) = 9.53 + 1.21 10 mg = 9.53 + 2.34 = 11.87 ppm 9
5.18 108 L
Because 90.0% of this water is returned, (1.80 × 104) × 0.900 = 1.62 × 104 L/s of water with 11.87 ppm Ca2+ is mixed with (5.35 − 1.80) × 104 = 3.55 × 104 L/s of water containing 9.53 ppm Ca2+.
C Ca 2+ (final) = (1.62 10 L / s)(11.87 ppm) + (3.55 10 L / s)(9.53 ppm) = 10.3 ppm 4
4
1.62 10 4 L / s + 3.55 10 4 L / s
169.
Mol KHP used = 0.4016 g ×
1 mol = 1.967 × 10−3 mol KHP 204 .22 g
Because 1 mole of NaOH reacts completely with 1 mole of KHP, the NaOH solution contains 1.967 × 10 −3 mol NaOH. Molarity of NaOH = Maximum molarity = Minimum molarity =
1.967 10 −3 mol 25.06 10 −3 L 1.967 10 −3 mol 25.01 10 −3 L 1.967 10 −3 mol 25.11 10 −3 L
=
7.849 10 −2 mol L
=
7.865 10 −2 mol L
=
7.834 10 −2 mol L
We can express this as 0.07849 ±0.00016 M. An alternative way is to express the molarity as 0.0785 ±0.0002 M. This second way shows the actual number of significant figures in the molarity. The advantage of the first method is that it shows that we made all our individual measurements to four significant figures. 170.
a. 16 e− + 18 H+ + 3 IO3− → I3− + 9 H2O
(3 I− → I3− + 2 e−) × 8
170
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SOLUTION STOICHIOMETRY
24 I− → 8 I3− + 16 e− 16 e + 18 H + 3 IO3− → I3− + 9 H2O −
+
18 H+ + 24 I− + 3 IO3− → 9 I3− + 9 H2O Reducing: 6 H+(aq) + 8 I−(aq) + IO3−(aq) → 3 I3−(aq) + 3 H2O(l) b. 0.6013 g KIO3 ×
1 mol KIO 3 = 2.810 × 10−3 mol KIO3 214 .0 g KIO 3
2.810 × 10−3 mol KIO3 ×
8 mol KI 166 .0 g KI = 3.732 g KI mol KIO 3 mol KI
2.810 × 10−3 mol KIO3 ×
6 mol HCl 1L = 5.62 × 10−3 L = 5.62 mL HCl mol KIO 3 3.00 mol HCl
c. I3− + 2 e− → 3 I−
2 S2O32− → S4O62− + 2 e−
Adding the balanced half-reactions gives: 2 S2O32−(aq) + I3−(aq) → 3 I−(aq) + S4O62−(aq) −
d. 25.00 × 10−3 L KIO3 ×
0.0100 mol KIO 3 3 mol I 3 2 mol Na 2S 2 O 3 = − L mol KIO 3 mol I 3
1.50 × 10−3 mol Na2S2O3 M Na2S2 O 3 =
e. 0.5000 L ×
1.50 10 −3 mol 32.04 10 −3 L
= 0.0468 M Na2S2O3
0.0100 mol KIO 3 214 .0 g KIO 3 = 1.07 g KIO3 L mol KIO 3
Place 1.07 g KIO3 in a 500-mL volumetric flask; add water to dissolve the KIO3; continue adding water to the 500.0-mL mark, with mixing along the way. 171.
7 H2O + 2 Cr3+ → Cr2O72− + 14 H+ + 6 e− (2 e- + S2O82- → 2 SO42−) × 3
a.
7 H2O(l) + 2 Cr3+(aq) + 3 S2O82−(aq) → Cr2O72−(aq) + 14 H+(aq) + 6 SO42−(aq) (Fe2+ → Fe3+ + e−) × 6 6 e + 14 H + Cr2O72− → 2 Cr3+ + 7 H2O −
+
14 H+(aq) + 6 Fe2+(aq) + Cr2O72−(aq) → 2 Cr3+(aq) + 6 Fe3+(aq) + 7 H2O(l) b. 8.58 × 10−3 L
0.0520 mol Cr2 O 72 − 6 mol Fe 2 + = 2.68 × 10−3 mol of excess Fe2+ L mol Cr2 O 72 −
Fe2+ (total) = 3.000 g Fe(NH4)2(SO4)2•6H2O ×
1 mol = 7.650 × 10−3 mol Fe2+ 392 .17 g
7.650 × 10−3 − 2.68 × 10−3 = 4.97 × 10−3 mol Fe2+ reacted with Cr2O72− generated from the Cr plating.
CHAPTER 4
SOLUTION STOICHIOMETRY
171
The Cr plating contained: 4.97 × 10−3 mol Fe2+
1 mol Cr2 O 72 − 6 mol Fe
2+
2 mol Cr 3+ mol Cr2 O 72 −
= 1.66 × 10−3 mol Cr3+ = 1.66 × 10−3 mol Cr
1.66 × 10−3 mol Cr ×
52 .00 g Cr = 8.63 × 10−2 g Cr mol Cr 3
Volume of Cr plating = 8.63 × 10−2 g × 1 cm
= 1.20 × 10−2 cm3 = area × thickness
7.19 g
Thickness of Cr plating = 172.
1.20 10 −2 cm3 40.0 cm2
= 3.00 × 10−4 cm = 300. µm
The unbalanced reaction is: VO2+ + MnO4− → V(OH)4+ + Mn2+ This is a redox reaction in acidic solution and must be balanced accordingly. The two halfreactions to balance are: VO2+ → V(OH)4+ and MnO4− → Mn2+ Balancing by the half-reaction method gives: MnO4−(aq) + 5 VO2+(aq) + 11 H2O(l) → 5 V(OH)4+(aq) + Mn2+(aq) + 2 H+(aq) −
0.02645 L × 0.581 =
0.02250 mol MnO 4 5 mol VO 2 + 1 mol V 50.94 g V = 0.1516 g V − 2+ L mol V mol VO mol MnO 4
0.1516 g V , 0.1516/0.581 = 0.261 g ore sample mass of ore sample
Marathon Problems 173.
Mol BaSO4 = 0.2327 g × 1 mol = 9.970 × 10−4 mol BaSO4 233.4 g The moles of the sulfate salt depend on the formula of the salt. The general equation is: Mx(SO4)y(aq) + y Ba2+(aq) → y BaSO4(s) + x Mz+ Depending on the value of y, the mole ratio between the unknown sulfate salt and BaSO4 varies. For example, if Pat thinks the formula is TiSO4, the equation becomes: TiSO4(aq) + Ba2+(aq) → BaSO4(s) + Ti2+(aq) Because there is a 1 : 1 mole ratio between mol BaSO4 and mol TiSO4, you need 9.970 × 10 −4 mol of TiSO4. Because 0.1472 g of salt was used, the compound would have a molar mass of (assuming the TiSO4 formula): 0.1472 g/9.970 × 10 −4 mol = 147.6 g/mol From atomic masses in the periodic table, the molar mass of TiSO4 is 143.95 g/mol. From just these data, TiSO4 seems reasonable.
172
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Chris thinks the salt is sodium sulfate, which would have the formula Na2SO4. The equation is: Na2SO4(aq) + Ba2+(aq) → BaSO4(s) + 2 Na+(aq) As with TiSO4, there is a 1:1 mole ratio between mol BaSO4 and mol Na2SO4. For sodium sulfate to be a reasonable choice, it must have a molar mass of about 147.6 g/mol. Using atomic masses, the molar mass of Na2SO4 is 142.05 g/mol. Thus Na2SO4 is also reasonable. Randy, who chose gallium, deduces that gallium should have a 3+ charge (because it is in column 3A), and the formula of the sulfate would be Ga2(SO4)3. The equation would be: Ga2(SO4)3(aq) + 3 Ba2+(aq) → 3 BaSO4(s) + 2 Ga3+(aq) The calculated molar mass of Ga2(SO4)3 would be: 0.1472 g Ga 2 (SO 4 ) 3 9.970 10
−4
mol BaSO 4
3 mol BaSO 4 = 442.9 g/mol mol Ga 2 (SO 4 ) 3
Using atomic masses, the molar mass of Ga2(SO4)3 is 427.65 g/mol. Thus Ga2(SO4)3 is also reasonable. Looking in references, sodium sulfate (Na2SO4) exists as a white solid with orthorhombic crystals, whereas gallium sulfate [Ga2(SO4)3] is a white powder. Titanium sulfate exists as a green powder, but its formula is Ti2(SO4)3. Because this has the same formula as gallium sulfate, the calculated molar mass should be around 443 g/mol. However, the molar mass of Ti2(SO4)3 is 383.97 g/mol. It is unlikely, then, that the salt is titanium sulfate. To distinguish between Na2SO4 and Ga2(SO4)3, one could dissolve the sulfate salt in water and add NaOH. Ga3+ would form a precipitate with the hydroxide, whereas Na2SO4 would not. References confirm that gallium hydroxide is insoluble in water. 174.
a. Compound A = M(NO3)x; in 100.00 g of compd.: 8.246 g N ×
48 .00 g O = 28.25 g O 14 .01 g N
Thus the mass of nitrate in the compound = 8.246 + 28.25 g = 36.50 g (if x = 1). If x = 1: mass of M = 100.00 − 36.50 g = 63.50 g Mol M = mol N =
8.246 g = 0.5886 mol 14 .01 g / mol
Molar mass of metal M =
63 .50 g = 107.9 g/mol (This is silver, Ag.) 0.5886 mol
If x = 2: mass of M = 100.00 − 2(36.50) = 27.00 g Mol M = ½ mol N =
0.5886 mol = 0.2943 mol 2
Molar mass of metal M =
27.00 g = 91.74 g/mol 0.2943 mol
CHAPTER 4
SOLUTION STOICHIOMETRY
173
This is close to Zr, but Zr does not form stable 2+ ions in solution; it forms stable 4+ ions. Because we cannot have x = 3 or more nitrates (three nitrates would have a mass greater than 100.00 g), compound A must be AgNO3. Compound B: K2CrOx is the formula. This salt is composed of K+ and CrOx2− ions. Using oxidation states, 6 + x(−2) = −2, x = 4. Compound B is K2CrO4 (potassium chromate). b. The reaction is: 2 AgNO3(aq) + K2CrO4(aq) → Ag2CrO4(s) + 2 KNO3(aq) The blood red precipitate is Ag2CrO4(s). c. 331.8 g Ag2CrO4 formed; this is equal to the molar mass of Ag2CrO4, so 1 mole of precipitate formed. From the balanced reaction, we need 2 mol AgNO3 to react with 1 mol K2CrO4 to produce 1 mol (331.8 g) of Ag2CrO4. 2.000 mol AgNO3 ×
169.9 g = 339.8 g AgNO3 mol
1.000 mol K2CrO4 ×
194.2 g = 194.2 g K2CrO4 mol
The problem says that we have equal masses of reactants. Our two choices are 339.8 g AgNO3 + 339.8 g K2CrO4 or 194.2 g AgNO3 + 194.2 g K2CrO4. If we assume the 194.2-g quantities are correct, then when 194.2 g K2CrO4 (1 mol) reacts, 339.8 g AgNO3 (2.0 mol) must be present to react with all the K2CrO4. We only have 194.2 g AgNO3 present; this cannot be correct. Instead of K2CrO4 limiting, AgNO3 must be limiting, and we have reacted 339.8 g AgNO3 and 339.8 g K2CrO4. Solution A:
2.000 mol NO 3 2.000 mol Ag + = 4.000 M Ag+; 0.5000 L 0.5000 L
Solution B: 339.8 g K2CrO4 ×
−
= 4.000 M NO3−
1 mol = 1.750 mol K2CrO4 194.2 g
1.750 mol CrO 4 2 1.750 mol K + = 7.000 M K+; 0.5000 L 0.5000 L
2−
= 3.500 M CrO42−
d. After the reaction, moles of K+ and moles of NO3− remain unchanged because they are spectator ions. Because Ag+ is limiting, its concentration will be 0 M after precipitation is complete. The following summarizes the changes that occur as the precipitate forms. 2 Ag+(aq) Initial 2.000 mol Change −2.000 mol After rxn 0 M K+ =
CrO42−(aq) → Ag2CrO4(s)
1.750 mol −1.000 mol 0.750 mol
0 +1.000 mol 1.000 mol
2 1.750 mol 2.000 mol = 3.500 M K+; M NO − = = 2.000 M NO3− 3 1.0000 L 1.0000 L
M CrO 2 − = 4
+
0.750 mol = 0.750 M CrO42−; M Ag + = 0 M (the limiting reagent) 1.0000 L
CHAPTER 5 GASES Review Questions 1.
See Figure 5.2 of the text for an illustration of a barometer. A barometer initially starts with a full column of mercury which is tipped upside down and placed in a dish of mercury. The mercury in the column drops some, then levels off. The height of the column of mercury is a measure of the atmospheric pressure. Here, there are two opposite processes occurring. The weight of the mercury in the column is producing a force downward; this results in mercury wanting to flow out of the column. However, there is an opposing force keeping mercury in the column. The opposing force is that of the atmospheric gas particles colliding with the surface of the mercury in the dish; this results in mercury being pushed up into the column. When the two opposing processes are equal in strength to each other, the level of mercury in the column stays constant. The height of mercury in the column supported by the atmosphere is then a measure of pressure of the atmosphere. See Figure 5.3 for an illustration of a simple manometer. A manometer also has two opposing forces going against each other. There is the force exerted by the gas molecules in the flask. The opposing force is on the other side of the mercury filled U tube; it is the force exerted by atmospheric gases. The difference in height of the mercury in the U tube is a measure of the difference in pressure between the gas in the flask and the atmosphere. By measuring the height difference of mercury, one can determine how much greater than or less than the gas pressure in the flask is to the atmospheric pressure.
2.
Boyle’s law: P is inversely proportional to V at constant n and T. Mathematically, PV = k = constant. The plot to make to show a linear relationship is V vs. 1/P. The resulting linear plot has positive slope equal to the value of k, and the y-intercept is the origin. Charles’s law: V is directly proportional to T at constant P and n. Mathematically, V = bT where b = constant. The plot to make to show a linear relationship is V vs. T. The slope of the line is equal to b, and the y-intercept is the origin if the temperature is in Kelvin. Avogadro’s law: V is directly proportional to n at constant P and T. Mathematically, V = an where a = constant. A plot of V vs. n gives a line with a positive slope equal to the a constant value, and the y-intercept is the origin.
3.
Boyles’ law: T and n are constant. PV = nRT = constant, PV = constant nR Charles’s law: P and n are constant. PV = nRT, V = T = (constant)T P
174
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175
RT Avogadro’s law: V and T are constant. PV = nRT, V = n = (constant)n P RT P and n relationship at constant V and T: PV = nRT, P = n = (constant)n V P is directly proportional to n at constant V and T. nR P and T relationship at constant V and n. PV = nRT, P = T = (constant)T V P is directly proportional to T at constant V and n.
4.
a. Heating the can will increase the pressure of the gas inside the can, P T when V and n are constant. As the pressure increases, it may be enough to rupture the can. b. As you draw a vacuum in your mouth, atmospheric pressure pushing on the surface of the liquid forces the liquid up the straw. c. The external atmospheric pressure pushes on the can. Since there is no opposing pressure from the air inside the can, the can collapses. d. How "hard" the tennis ball is depends on the difference between the pressure of the air inside the tennis ball and atmospheric pressure. A "sea level" ball will be much "harder" at high altitude since the external pressure is lower at high altitude. A “high altitude” ball will be "soft" at sea level.
5.
Rigid container (constant volume): As reactants are converted to products, the moles of gas particles present decrease by one-half. As n decreases, the pressure will decrease (by one-half). Density is the mass per unit volume. Mass is conserved in a chemical reaction, so the density of the gas will not change since mass and volume do not change. Flexible container (constant pressure): Pressure is constant since the container changes volume in order to keep a constant pressure. As the moles of gas particles decrease by a factor of 2, the volume of the container will decrease (by one-half). We have the same mass of gas in a smaller volume, so the gas density will increase (is doubled).
6.
Boyle's law: P 1/V at constant n and T In the kinetic molecular theory (KMT), P is proportional to the collision frequency which is proportional to 1/V. As the volume increases there will be fewer collisions per unit area with the walls of the container and pressure will decrease (Boyle's law). Charles's law: V T at constant n and P When a gas is heated to a higher temperature, the average velocity of the gas molecules increases and thus the gas molecules hit the walls of the container more often and with more force. To keep the pressure constant, the volume of the container must increase (this increases surface area which decreases the number of collisions per unit area which decreases the pressure). Therefore, volume and temperature are directly related at constant n and P (Charles’s law).
176
CHAPTER 5
GASES
Avogadro’s law: V n at constant P and T As gas is added to a container (n increases), there will be an immediate increase in the number of gas particle collisions with the walls of the container. This results in an increase in pressure in the container. However, the container is such that it wants to keep the pressure constant. In order to keep pressure constant, the volume of the container increases in order to reduce the collision frequency which reduces the pressure. V is directly related to n at constant P and T. Dalton’s law of partial pressure: Ptotal = P1 + P2 + P3 + … The KMT assumes that gas particles are volumeless and that they exert no interparticle forces on each other. Gas molecules all behave the same way. Therefore, a mixture of gases behaves as one big gas sample. You can concentrate on the partial pressures of the individual components of the mixture, or you can collectively group all of the gases together to determine the total pressure. One mole of an ideal gas behaves the same whether it is a pure gas or a mixture of gases. P vs. n relationship at constant V and T. From question 3, this is a direct relationship. As gas is added to a rigid container, there will be an increase in the collision frequency, resulting in an increase in pressure. P and n are directly related at constant V and T. P vs. T relationship at constant V and n. From question 3, this is a direct relationship. As the temperature of the gas sample increases, the gas molecules move with a faster average velocity. This increases the gas collision frequency as well as increases the force of each gas particle collision. Both result in an increase in pressure. Pressure and temperature are directly related at constant V and n. 7.
a. At constant temperature, the average kinetic energy of the He gas sample will equal the average kinetic energy of the Cl2 gas sample. In order for the average kinetic energies to be the same, the smaller He atoms must move at a faster average velocity as compared to Cl2. Therefore, plot A, with the slower average velocity, would be for the Cl2 sample, and plot B would be for the He sample. Note the average velocity in each plot is a little past the top of the peak. b. As temperature increases, the average velocity of a gas will increase. Plot A would be for O2(g) at 273 K and plot B, with the faster average velocity, would be for O2(g) at 1273 K. Because a gas behaves more ideally at higher temperatures, O2(g) at 1273 K would behave most ideally.
8.
Method 1: molar mass =
dRT P
Determine the density of a gas at a measurable temperature and pressure, then use the above equation to determine the molar mass. Method 2:
effusion rate for gas 1 = effusion rate for gas 2
( molar mass) 2 ( molar mass)1
Determine the effusion rate of the unknown gas relative to some known gas; then use Graham’s law of effusion (the above equation) to determine the molar mass of the unknown gas.
CHAPTER 5
GASES
177
9.
The pressure measured for real gases is too low as compared to ideal gases. This is due to the attractions gas particles have for each other; these attractions “hold” them back from hitting the container walls as forcefully. To make up for this slight decrease in pressure for real gases, a factor is added to the measured pressure. The measured volume is too large. A fraction of the space of the container volume is taken up by the volume of gas of the molecules themselves. Therefore, the actual volume available to real gas molecules is slightly less than the container volume. A term is subtracted from the container volume to correct for the volume taken up by real gas molecules.
10.
The kinetic molecular theory assumes that gas particles do not exert forces on each other and that gas particles are volumeless. Real gas particles do exert attractive forces for each other, and real gas particles do have volumes. A gas behaves most ideally at low pressures and high temperatures. The effect of attractive forces is minimized at high temperatures since the gas particles are, in general, moving very rapidly and so are less easily attracted to each other. At low pressure, the container volume is relatively large (P and V are inversely related), so the volume of the container taken up by the gas particles is negligible.
Active Learning Questions 1.
a. As the syringe is depressed, air particles are added to the container which will increase the pressure inside the container. Assuming the membrane does not allow gas particles to pass through it, the greater outside pressure will cause the membrane to be pushed into the test tube. b. As long as no gas particles can leave the container and no gas particles can enter the test tube through the membrane, the membrane will remain pushed into the test tube.
2.
Answer c is correct. For answers a and b, there is no air pressure inside the barometer tube. The tube is completely filled with mercury, then it is carefully inverted, and then some mercury flows out of the tube leaving empty space above the mercury. When the air pressure outside the tube counterbalances the weight inside the tube, the level of mercury is constant. The height of the column is a measure of pressure.
3.
Pressure is defined as the force of the gas particle collisions per unit area. Because pressure is a per area quantity, the width or shape of the barometer tube doesn’t matter. The force per unit area exerted by the weight of mercury in each barometer will always be counterbalanced by the pressure exerted by the gas particles outside the tube. The height of mercury supported by the air pressure will be the same in each of the different shapes of barometers.
4.
Density = mass/volume: if the mass of gas doesn’t change and the volume of the container is constant, then the density of the gas will remain constant as temperature is increased. In a flexible container (a constant pressure container), as temperature is increased, the volume of the container must increase to keep the pressure constant. In this case, the density of the gas decreases as temperature is increased.
5.
There is nothing between the gas particles (answer e). Gas samples are mostly empty space.
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GASES
6.
As you put your finger over the straw and lift it from the drink, some liquid does escape. The trapped gas particles between the liquid in the straw and your finger now occupy a bigger volume as some liquid leaks out. This results in a decrease in pressure inside the straw. Eventually the outside pressure exerted at the bottom of the straw will be able to counterbalance the weight of the liquid and the pressure of trapped gas inside the straw. When this occurs, no more liquid leaves the straw.
7.
Pressure and volume are inversely related when the moles of gas and the temperature are constant. At constant T and n, as a container size is decreased, we have the same number of gas particles trapped in a smaller volume. This results in an increase in collisions per unit are (an increase in pressure). As volume decreases, pressure increases and vice versa. When filling a tire with air, the moles of gas is not constant, but is steadily increasing. So when the moles of gas is also changing, both volume and pressure can increase.
8.
X(g) + Y(g) → XY(g); as the reaction proceeds, two moles of gas particles are converted into one mole of as particles. The total moles of gas decrease as the reaction proceeds, resulting in a smaller volume in this constant pressure container. If stoichiometric amounts of a gases X and Y are reacted, then the volume should decrease by a factor of two since the moles of gas will decrease by a factor of 2.
9.
Answer d is the best choice. Hot air balloons are constant pressure systems. The pressure of the gas inside the balloon will equal the external pressure. As air is heated inside the balloon, the gas particles move faster on average causing an increase in pressure inside the balloon. The balloon doesn't like a mismatch in pressure, so hot air molecules escape out of the balloon in order to lower the pressure inside the balloon. As gas particles escape the balloon, there are fewer gas particle per unit volume. Thus, heating the balloon results in a decrease in gas density inside the balloon causing the balloon to rise.
10.
In a static drawing, there will be no difference. The number of gas particles shouldn’t change when the gas is cooled. They just move slower on average (which is hard to show on a drawing). The pressure will decrease as temperature is lowered because the gas particle don’t collide as forcefully and as frequently with the walls of the container. The reverse is true when the gas is heated. Your drawing has the same number of gas particles and they are evenly spaced; they are just moving faster on average resulting in an increase in pressure. In the evacuated drawing, you would show exactly half the number of evenly spaced gas particles as was present initially. If the moles of gas decrease by a factor of two at constant T and V, the pressure will decrease by a factor of two.
11.
Helium gas is less dense than air, so the balloon rises. A balloon is a constant pressure container. As the balloon rises, the external pressure decreases. In order to keep the inside gas pressure equal to the external pressure, the balloon volume increases. Eventually the balloon volume will expand beyond the limits of the balloon walls and it pops.
12.
n = PV/RT; at the same P, V, and T, the containers must have equal moles of gas as calculated by the ideal gas equation. Assuming ideal behavior, gases all behave the same way, regardless of identity.
13.
Boyles’ law: T and n are constant. PV = nRT = constant, PV = constant
CHAPTER 5
GASES
179
nR Charles’s law: P and n are constant. PV = nRT, V = T = (constant)T P RT Avogadro’s law: V and T are constant. PV = nRT, V = n = (constant)n P
14.
The pressure will be less than 5.00 atm. For the larger bulb initially at 2.00 atm, the partial pressure due to the He gas in this bulb will be less than 2.00 atm when the gas is expands into the larger volume. The same is true for the smaller bulb initially at 3.00 atm; the partial pressure of He gas in this bulb will be less than 3.00 atm when the gas expands into the larger volume. The total pressure is the sum of these two partial pressures. Since each partial pressure is less than the initial pressure, the total pressure will be less than 5.00 atm as the gas expands. The actual pressure will be between 2.00 and 3.00 atm when the stopcock is opened.
15.
At constant temperature, the two gases do have the same average kinetic energy. But the smaller gas particles must be moving faster on average for the average kinetic energies to be the same (KE = mv2/2). The larger gas particles do collide with a greater force, but the collisions are not as frequent since they are moving slower. The smaller gas particles collide more frequently but with not as much force. The result is the two containers have equal pressures at the same n, T, and V.
16.
Gas particles are in constant random motion. The molecules will move from one container to the other to equalize the pressure between the flask and the balloon. There will be the same number of gas particles in each container assuming the volumes are equal. This is required to equalize the pressure.
17.
Dalton’s law of partial pressures works because all ideal gases behave the same way regardless of identity. For gases to behave the same way (ideally), the size of the gas particles must be negligible regardless of identity and the gas particles must not exert attractive forces. If the size of gas particles and the attractive forces were not negligible, then different gases would behave differently and Dalton’s Law of partial pressures would not work.
18.
d = P(molar mass)/RT; the density of a gas at constant T and P is proportional to the molar mass of the gas. NH3 has a molar mass ~17 g/mol while air has an average molar mass ~29 g/mol. NH3 is less dense than air because it has a smaller molar mass.
19.
a. KEavg = (3/2)RT; only the temperature is needed to calculate the average kinetic energy of a gas sample. b. d = P(molar mass)/RT c. Ptotal = P1 + P2 + P3 + …; P1 = 1 Ptotal where 1 is the mole fraction of gas 1 in a mixture of gases. If one knows the partial pressure of all gases in a mixture except for gas 1, then the partial pressure of gas 1 will be the total pressure minus the partial pressures of all the other gases in the mixture. Or one can determine the mole fraction of gas 1 and multiply the total pressure by the mole fraction. d. µrms = (3RT/M)1/2 where M is the molar mass of the gas in kg/mol and R = 8.3145 J/K mol.
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GASES
1/2
e.
20.
M Effusion Rate1 = 2 Effusion Rate 2 M1
where M = molar mass
a. KEavg = (3/2)RT; average kinetic energy depends only on the temperature. Since the temperature is the same, the average kinetic energy will be the same. b. The force of the collisions with the walls of the container will be directly related to the mass of the gas particle and the velocity of the gas particle. Since mass and velocity are unchanged when just the volume is changes, the force of each collision is the same. However, the collision frequency per unit area in the smaller volume container will increase. This results in an overall increase in the force of collisions per unit area (pressure increases as volume decreases). c. µrms = (3RT/M)1/2; the root mean square velocity will be the same since temperature is constant and both containers hold the same gas. d. The gas density will be 3 times larger in the smaller volume container. This container has the same mass of gas in a volume three times smaller. e. P = nRT/V: the pressure in the 1.0 L container will be 3 times larger than the pressure in the 3.0 L container since each container holds 1 mol of gas.
21.
Your picture of a solid should show the atoms/molecules in set positions, very close together. A solid is rigid with a fixed shape and volume. A liquid also has the atoms/molecules very close together, but there is more disorder with the arrangement of the atoms/molecules with each other as compared to a solid. Liquids have a volume but no fixed shape; they take the shape of the container (liquids can pour). Gases are mostly empty space and they take on the volume and shape of the container. The gas particles in your picture should be far apart from each other. Because gases are mostly empty space, they are easily compressible, unlike solids and liquids where the atoms/molecule are very close together.
22.
a. Containers ii, iv, vi, and viii have volumes twice those of containers i, iii, v, and vii. Containers iii, iv, vii, and viii have twice the number of molecules present than containers i, ii, v, and vi. The container with the lowest pressure will be the one that has the fewest moles of gas present in the largest volume (containers ii and vi both have the lowest P). The smallest container with the most moles of gas present will have the highest pressure (containers iii and vii both have the highest P). All the other containers (i, iv, v, and viii) will have the same pressure between the two extremes. The order is ii = vi < i = iv = v = viii < iii = vii. b. All have the same average kinetic energy because the temperature is the same in each container. Only temperature determines the average kinetic energy. c. The least dense gas will be in container ii because it has the fewest of the lighter Ne atoms present in the largest volume. Container vii has the most dense gas because the largest number of the heavier Ar atoms are present in the smallest volume. To determine the ordering for the other containers, we will calculate the relative density of each. In the following table, m1 equals the mass of Ne in container i, V1 equals the volume of container i, and d1 equals the density of the gas in container i.
CHAPTER 5 Container mass, volume
GASES i
181 ii
iii
m1, V1
m1, 2V1
2m1, V1
2m1, 2V1
2m1, V1
2m1, 2V1
m1
m1 1 = d1 2 V1 2
2 m1
2 m1
2 m1
2 m1
density mass volume
V1
= d1
iv
= 2d1
V1
v
= d1
2 V1
vi
= 2d1
vii
= d1
2 V1
V1
viii
4m1, V1
4m1, 2V1
4 m1
4 m1
= 4d1
V1
= 2d1
2 V1
From the table, the order of gas density is ii < i = iv = vi < iii = v = viii < vii. d. µrms = (3RT/M)1/2; the root mean square velocity only depends on the temperature and the molar mass. Because T is constant, the heavier argon molecules will have a slower root mean square velocity than the neon molecules. The order is v = vi = vii = viii < i = ii = iii = iv. 23.
2 NH3(g) → N2(g) + 3 H2(g); as reactants are converted into products, we go from 2 moles of gaseous reactants to 4 moles of gaseous products (1 mol N2 + 3 mol H2). Because the moles of gas doubles as reactants are converted into products, the volume of the gases will double (at constant P and T). RT n = (constant)n; pressure is directly related to n at constant T and V. V
PV = nRT, P =
As the reaction occurs, the moles of gas will double, so the pressure will double. Because 1 o . Owing to the 3 : mole of N2 is produced for every 2 moles of NH3 reacted, PN = (1/2)PNH 2
o NH 3
2 mole ratio in the balanced equation, PH = (3/2)P 2
3
.
Because of 3:1 mol ratio between H2 and N2 in the balanced equation, PH = 3PNo . 2
2
o o o + (1/2)PNH = 2PNH Note: Ptotal = PH + PN = (3/2)PNH . As we said earlier, the total pressure 2
2
3
3
3
will double as reactants are completely converted into products.
Questions 24.
Molecules in the condensed phases (liquids and solids) are very close together. Molecules in the gaseous phase are very far apart. A sample of gas is mostly empty space. Therefore, one would expect 1 mole of H2O(g) to occupy a huge volume as compared to 1 mole of H2O(l).
25.
The column of water would have to be 13.6 times taller than a column of mercury. When the pressure of the column of liquid standing on the surface of the liquid is equal to the pressure of air on the rest of the surface of the liquid, then the height of the column of liquid is a measure of atmospheric pressure. Because water is 13.6 times less dense than mercury, the column of water must be 13.6 times longer than that of mercury to match the force exerted by the columns of liquid standing on the surface.
182
CHAPTER 5
GASES
26.
A bag of potato chips is a constant-pressure container. The volume of the bag increases or decreases to keep the internal pressure equal to the external (atmospheric) pressure. The volume of the bag increased because the external pressure decreased. This seems reasonable as atmospheric pressure is lower at higher altitudes than at sea level. We ignored n (moles) as a possibility because the question said to concentrate on external conditions. It is possible that a chemical reaction occurred that would increase the number of gas molecules inside the bag. This would result in a larger volume for the bag of potato chips. The last factor to consider is temperature. During ski season, one would expect the temperature of Lake Tahoe to be colder than Los Angeles. A decrease in T would result in a decrease in the volume of the potato chip bag. This is the exact opposite of what happened, so apparently the temperature effect is not dominant.
27.
The P versus 1/V plot is incorrect. The plot should be linear with positive slope and a y-intercept of zero. PV = k, so P = k(1/V). This is in the form of the straight-line equation y = mx + b. The y-axis is pressure, the x-axis is 1/V, and the y-intercept is the origin.
28.
The decrease in temperature causes the balloon to contract (V and T are directly related). Because weather balloons do expand, the effect of the decrease in pressure must be dominant.
29.
Pressure is directly related to the mole fraction of the various gases present in a mixture. If the molar masses of the three gases X, Y, and Z are equal, then the mole fraction of each gas would be 1/3, and the partial pressure of each gas in the mixture equals 1/3 of the total pressure. If the molar mass of gas X is greater than the other molar masses, then fewer moles of gas X are present compared to moles of gas Y and Z. This results in a mole fraction of X less than 1/3, and a partial pressure of X less than 1/3 the total pressure (answer b).
30.
Rigid container: As temperature is increased, the gas molecules move with a faster average velocity. This results in more frequent and more forceful collisions, resulting in an increase in pressure. Density = mass/volume; the moles of gas are constant, and the volume of the container is constant, so density in this case must be temperature-independent (density is constant). Flexible container: The flexible container is a constant-pressure container. Therefore, the final internal pressure will be unaffected by an increase in temperature. The density of the gas, however, will be affected because the container volume is affected. As T increases, there is an immediate increase in P inside the container. The container expands its volume to reduce the internal pressure back to the external pressure. We have the same mass of gas in a larger volume. Gas density will decrease in the flexible container as T increases.
31.
At STP (T = 273.2 K and P = 1.000 atm), the volume of 1.000 mol of gas is: nRT V= = P
1.000 mol
0.08206 L atm K mol 1.000 atm
273.2 K
= 22.42 L
At STP, the volume of 1.000 mole of any gas is 22.42 L, assuming the gas behaves ideally. Therefore, the molar volume of He(g) and N2(g) at STP both equal 22.42 L/mol. If the temperature increases to 25.0C (298.2 K), the volume of 1.000 mole of a gas will be larger than 22.42 L/mole because molar volume is directly related to the temperature at constant pressure. If 1.000 mole of a gas is collected over water at a total pressure of 1.000 atm, the
CHAPTER 5
GASES
183
partial pressure of the collected gas will be less than 1.000 atm because water vapor is present (Ptotal = Pgas + PH 2 O ). At some partial pressure below 1.000 atm, the volume of 1.000 mole of a gas will be larger than 22.42 L/mol because molar volume is inversely related to the pressure at constant temperature. 32.
Assuming the temperature and pressure are the same for the two balloons, then equal volumes will contain equal moles of gas. Because the molar mass of argon (39.95 g/mol) is about twice the molar mass of neon (20.18 g/mol), the mass of gas in the Ar balloon will be about twice as great as the mass of gas in the neon balloon since the moles of gas are equal.
33.
a. For an ideal gas, KEavg = (3/2)RT. So as temperature increases, the average kinetic energy will increase. b. avg rms (T)1/2; as temperature increases, the average velocity of the gas molecules increase. c. At constant temperature, the lighter the gas molecules (the smaller the molar mass), the faster the average velocity. This must be true for the average kinetic energies to be the same at constant T.
34.
For the first diagram, there is a total volume of 3X after the stopcock is open. The six total gas particles will be equally distributed (on average) over the entire volume (3X). So per X volume, there will be two gas particles. Your first drawing should have four gas particles in the 2X volume flask and two gas particles in the X volume flask. Applying Boyle’s law, the pressure in the two flasks after the stopcock is opened is: P1V1 = P2V2, P2 =
P1V1 P 2X 2 = 1 = P1 V2 3 3X
The final pressure in both flasks will be two-thirds that of the initial pressure in the left flask. For the second diagram, there is a total volume of 2X after the stopcock is opened. The gas particles will be equally distributed (on average) so that your drawing should have three gas particles in each flask. The final pressure is: P2 =
P1 V1 P P X = 1 = 1 V2 2 2X
The final pressure in both flasks will be one-half that of the initial pressure in the left flask. 35.
No; at any nonzero Kelvin temperature, there is a distribution of kinetic energies. Similarly, there is a distribution of velocities at any nonzero Kelvin temperature. The reason there is a distribution of kinetic energies at any specific temperature is because there is a distribution of velocities for any gas sample at any specific temperature.
36.
Density = mass/volume; the molar mass of argon (39.95 g/mol) is about twice the molar mass of neon (20.18 g/mol). Because we have equal masses of gases in the two containers, the moles of gas in the neon container will be about 2 times greater than the moles of gas in the argon container. A balloon is a constant pressure container. To keep the pressure constant, the neon balloon with almost twice the number of gas particles will approximately double in size to keep
184
CHAPTER 5
GASES
the pressure constant. Therefore, the density of Ar in the smaller container will be about twice the density of Ne. 37.
2 NO2(g) → N2(g) + 2 O2(g); at constant V and T, pressure is directly proportional to the moles of gas present. As NO2 is converted to products, the moles of gas increase from 2 total moles to 3 total moles (a 50% increase). Because of this, the total pressure will increase by 50% as reactants are converted into products. From the stoichiometry in the balanced equation, the moles of O2 produced will equal the moles of NO reacted. Because pressure is directly related to moles of gas present, the partial pressure of O2 will equal the pressure of NO2 reacted (PO2 = 4.0 atm). Also from the stoichiometry, there are twice the number of moles of O2 produced as moles of N2. This will result in a N2 partial pressure which is half the O2 partial pressure (PN2 = 2.0 atm). Note that the total pressure of gases present is 2.0 + 4.0 = 6.0 atm as stated in the problem.
38.
Statements a, c, and e are true. For statement b, if temperature is constant, then the average kinetic energy will be constant no matter what the identity of the gas (KEave = 3/2 RT). For statement d, as T increases, the average velocity of the gas particles increases. When gas particles are moving faster, the effect of interparticle interactions is minimized. For statement f, the KMT predicts that P is directly related to T at constant V and n. As T increases, the gas molecules move faster, on average, resulting in more frequent and more forceful collisions. This leads to an increase in P.
39.
The values of a are: H2,
0.244 atm L2 mol 2
; CO2, 3.59; N2, 1.39; CH4, 2.25
Because a is a measure of intermolecular attractions, the attractions are greatest for CO2. 40.
The van der Waals constant b is a measure of the size of the molecule. Thus C3H8 should have the largest value of b because it has the largest molar mass (size).
41.
PV = nRT; Figure 5.6 is illustrating how well Boyle’s law works. Boyle’s law studies the pressure-volume relationship for a gas at constant moles of gas (n) and constant temperature (T). At constant n and T, the PV product for an ideal gas equals a constant value of nRT, no matter what the pressure of the gas. Figure 5.6 plots the PV product versus P for three different gases. The ideal value for the PV product is shown with a dotted line at about a value of 22.41 L atm. From the plot, it looks like the plot for Ne is closest to the dotted line, so we can conclude that of the three gases in the plot, Ne behaves most ideally. The O2 plot is also fairly close to the dotted line, so O2 also behaves fairly ideally. CO2, on the other hand, has a plot farthest from the ideal plot; hence CO2 behaves least ideally.
42.
Dalton’s law of partial pressures holds if the total pressure of a mixture of gases depends only on the total moles of gas particles present and not on the identity of the gases in the mixtures. If the total pressure of a mixture of gases were to depend on the identities of the gases, then each gas would behave differently at a certain set of conditions, and determining the pressure of a mixture of gases would be very difficult. All ideal gases are assumed volumeless and are assumed to exert no forces among the individual gas particles. Only in this scenario can Dalton’s law of partial pressure hold true for an ideal gas. If gas particles did have a volume and/or did exert forces among themselves, then each gas, with its own identity and size, would behave differently. This is not observed for ideal gases.
CHAPTER 5
GASES
185
Exercises Pressure 43.
760 mm Hg = 3.6 × 103 mm Hg atm
a. 4.8 atm ×
b. 3.6 × 103 mm Hg ×
1 torr mm Hg
= 3.6 × 103 torr 1.013 10 5 Pa = 4.9 × 105 Pa atm
c. 4.8 atm ×
44.
a.
2200 psi ×
b. 150 atm × c. 150 atm ×
45.
6.5 cm ×
1 atm 14 .7 psi
d. 4.8 atm ×
14.7 psi atm
= 71 psi
= 150 atm
1.013 10 5 Pa 1 MPa = 15 MPa atm 1 10 6 Pa
760 torr atm
10 mm
= 1.1 × 105 torr
= 65 mm Hg = 65 torr; 65 torr ×
cm
8.6 × 10 −2 atm ×
1 atm 760 torr
= 8.6 × 10 −2 atm
1.013 10 5 Pa = 8.7 × 103 Pa atm
1 atm 2.54 cm 10 mm = 508 mm Hg = 508 torr; 508 torr × = 0.668 atm 760 torr in cm
46.
20.0 in Hg ×
47.
If the levels of mercury in each arm of the manometer are equal, then the pressure in the flask is equal to atmospheric pressure. When they are unequal, the difference in height in millimeters will be equal to the difference in pressure in millimeters of mercury between the flask and the atmosphere. Which level is higher will tell us whether the pressure in the flask is less than or greater than atmospheric. a. Pflask < Patm; Pflask = 760. − 118 = 642 torr 642 torr ×
1 atm = 0.845 atm 760 torr
0.845 atm ×
1.013 10 5 Pa = 8.56 × 104 Pa atm
b. Pflask > Patm; Pflask = 760. torr + 215 torr = 975 torr 975 torr ×
1 atm = 1.28 atm 760 torr
186
CHAPTER 5 1.28 atm ×
1.013 10 5 Pa = 1.30 × 105 Pa atm
Pflask = 635 − 118 = 517 torr; Pflask = 635 + 215 = 850. torr
c. 48.
GASES
a. The pressure is proportional to the mass of the fluid. The mass is proportional to the volume of the column of fluid (or to the height of the column assuming the area of the column of fluid is constant). d = density =
mass ; in this case, the volume of silicon oil will be the same as the volume volume of mercury in Exercise 45.
m Hg m Hg d oil m m = oil , m oil = ; VHg = Voil; d Hg d oil d Hg d
V=
Because P is proportional to the mass of liquid: d 1.30 Poil = PHg oil = PHg = (0.0956)PHg d Hg 13 . 6
This conversion applies only to the column of silicon oil. Pflask = 760. torr − (0.0956 × 118) torr = 760. − 11.3 = 749 torr 749 torr ×
1 atm 760 torr
= 0.986 atm; 0.986 atm ×
1.013 10 5 Pa = 9.99 × 104 Pa atm
Pflask = 760. torr + (0.0956 × 215) torr = 760. + 20.6 = 781 torr 781 torr ×
1 atm 760 torr
= 1.03 atm; 1.03 atm ×
1.013 10 5 Pa = 1.04 × 105 Pa atm
b. If we are measuring the same pressure, the height of the silicon oil column would be 13.6 ÷ 1.30 = 10.5 times the height of a mercury column. The advantage of using a less dense fluid than mercury is in measuring small pressures. The height difference measured will be larger for the less dense fluid. Thus the measurement will be more precise.
Gas Laws 49.
At constant n and T, PV = nRT = constant, so P1V1 = P2V2. Solving for P2: P2 =
P1 V1 V2
=
5.20 atm 0.400 L = 0.972 atm 2.14 L
As predicted by Boyle’s law, as the volume of a gas increases, pressure decreases. 50.
The pressure exerted on the balloon is constant, and the moles of gas present is constant. From Charles’s law, V1/T1 = V2/T2 at constant P and n.
CHAPTER 5 V2 =
GASES
V1 T 2
=
T1
187
700 . mL 100 . K = 239 mL ( 273 .2 + 20 .0) K
As expected, as temperature decreases, the volume decreases. 51.
nAr = 27.1 g Ar × VAr n Ar
=
VNe n Ne
1 mol Ar = 0.678 mol; at constant T and P, Avogadro’s law holds (V n). 39.95 g
, VNe =
VAr n Ne
=
n Ar
4.21 L 1.29 mol = 8.01 L 0.678 mol
As expected, as n increases, V increases. 52.
As NO2 is converted completely into N2O4, the moles of gas present will decrease by one-half (from the 2 : 1 mole ratio in the balanced equation). Using Avogadro’s law: V1 n1
=
V2 n2
, V2 = V1 ×
n2 n1
= 25.0 mL ×
1 = 12.5 mL 2
N2O4(g) will occupy one-half the original volume of NO2(g). This is expected because the moles of gas present decrease by one-half when NO2 is converted into N2O4.
53.
a. PV = nRT, V =
b. PV = nRT, n =
c. PV = nRT, T =
d. PV = nRT, P =
54.
nRT = P
0.08206 L atm (155 + 273) K K mol = 14.0 L 5.00 atm
0.300 atm 2.00 L PV = = 4.72 × 10 −2 mol 0.08206 L atm RT 155 K K mol 4.47 atm 25 .0 L PV = = 678 K = 405°C 0.08206 L atm nR 2.01 mol K mol
nRT = V
a. P = 7.74 × 103 Pa ×
PV = nRT, n =
2.00 mol
0.08206 L atm ( 273 + 75) K K mol = 133 atm 2.25 L
10.5 mol
1 atm 1.013 105 Pa
= 0.0764 atm; T = 25 + 273 = 298 K
0.0764 atm 0.0122 L PV = = 3.81 × 10 −5 mol 0 . 08206 L atm RT 298 K K mol
nRT b. PV = nRT, P = = V
0.421 mol
0.08206 L atm 223 K K mol = 179 atm 0.0430 L
188
CHAPTER 5
nRT c. V = = P
GASES
0.08206 L atm (331 + 273) K K mol = 3.6 L 1 atm 455 torr 760 torr
4.4 10 − 2 mol
1 atm 745 mm Hg 11 .2 L 760 mm Hg PV d. T = = = 334 K = 61°C 0.08206 L atm nR 0.401 mol K mol
55.
n=
2.70 atm 200.0 L PV = = 22.2 mol 0.08206 L atm RT (273 + 24) K K mol
For He: 22.2 mol × For H2: 22.2 mol ×
56.
4.003 g He mol
= 88.9 g He
2.016 g H 2 = 44.8 g H2 mol
PV 1.00 atm 6.0 L = = 0.25 mol air 0.08206 L atm RT 298 K K mol
a.
n =
b.
n =
1.97 atm 6.0 L = 0.48 mol air 0.08206 L atm/K mol 298 K
c.
n =
0.296 atm 6.0 L = 0.11 mol air 0.08206 L atm/K mol 200. K
Air is indeed “thinner” at high elevations. 57.
PV = nRT, n =
0.0449 mol O2 ×
58.
59.
14.5 atm (75.0 10 −3 L) PV = 0.0449 mol O2 = 0.08206 L atm RT 295 K K mol
32.00 g O 2 = 1.44 g O2 mol O 2
1 mol 0.08206 L atm 0.60 g ( 273 + 22 ) K 32 .00 g K mol nRT P = = = 0.091 atm V 5 .0 L 1 atm 76.0 torr × 37.0 L PV 760 torr PV = nRT; T = = = 225 K; 225 – 273 = ‒48°C 0.08206 L atm nR 0.200 mol × K mol
CHAPTER 5 60.
0.050 mL ×
V=
61.
GASES
n =
nRT = P
PV = RT
0.018 mol
62.
n =
V=
63.
189
1 mol O 2 1.149 g mL 32 .00 g
1.8 10 −3 mol
0.08206 L atm 310. K K mol = 4.6 × 10 −2 L = 46 mL 1.0 atm
1 atm 0.45 L 760 torr = 0.018 mol air 0.08206 L atm 295 K K mol
745 torr
6.022 10 23 particles = 1.1 1022 air particles mol
10.5 atm 5.00 L PV = 2.15 mol N2O = 0.08206 L atm RT 298 K K mol
nRT = P
0.08206 L atm 298 K K mol = 53.6 L 1 atm 745 torr 760 torr
2.15 mol
For a gas at two conditions:
Because V is constant:
n2 =
= 1.8 × 10 −3 mol O2
P1V1 PV = 2 2 n 1T1 n 2 T2
nPT P1 P = 2 , n2 = 1 2 1 n 1T1 P1T2 n 2 T2
0.50 mol × 2.5 atm × 298 K = 1.5 mol 0.75 atm × 323 K
Moles of gas added = n2 – n1 = 1.5 – 0.50 = 1.0 mol 64.
PV = nRT, n is constant.
V2 = (1.040)V1,
P2 =
PV PV PV = nR = constant, 1 1 = 2 2 T T2 T 1
V1 1.000 = V2 1.040
P1 V1T2 1.000 ( 273 + 58 ) K = 75 psi × = 82 psi 1.040 ( 273 + 19 ) K V2 T1
For two-condition problems, units for P and V just need to be the same units for both conditions, not necessarily atm and L. The unit conversions from other P or V units would cancel when applied to both conditions. However, temperature must be converted to the Kelvin scale. The temperature conversions between other units and Kelvin will not cancel each other.
190 65.
CHAPTER 5 At two conditions:
GASES
P1V1 P2 V2 ; all gases are assumed to follow the ideal gas law. The = n 1T1 n 2 T2
identity of the gas in container B is unimportant as long as we know the moles of gas present. 1.0 L 2.0 mol 560 . K PB V n T = 2.0 = A B B = PA VB n A TA 2.0 L 1.0 mol 280 . K
The pressure of the gas in container B is twice the pressure of the gas in container A. 66.
For the two containers: PA VA P V V P n T 0.40 atm 1.0 mol 3500 K = B B, B = A B B = = 5.0 n A TA n B TB VA PB n A TA 0.20 atm 4.0 mol 350 K
The volume of container B is 5.0 times the volume of container A. 67.
At constant n and V,
P1 T1
=
P2 T2
, T2 =
T1 P2 2500 torr = (21 + 273 K) × = 970 K = 7.0 × 102°C P1 758 torr
The flask will burst when the temperature reaches 7.0 × 102°C. 68.
Because the container is flexible, P is assumed constant. The moles of gas present are also constant. P1V1 PV V V = 2 2 , 1 = 2 ; Vsphere = 4/3 r3 n 1T1 n 2 T2 T1 T2
V2 = r23 =
69.
361 K 280 . K
= 1.29, r2 = (1.29)1/3 = 1.09 cm = radius of sphere after heating
PV P V PV = nR = constant, 1 1 = 2 2 T1 T2 T
P2 =
70.
V1T2 4/3 π (1.00 cm) 3 361 K , 4/3 π (r2 ) 3 = T1 280. K
P1 V1 T 2 V 2 T1
PV = nRT,
= 710. torr ×
5.0 10 2 mL ( 273 + 820.) K = 5.1 × 104 torr 25 mL ( 273 + 30.) K
nT V nT n T = = constant, 1 1 = 2 2 ; moles × molar mass = mass P1 P2 P R
n 1 ( molar mass )T1 n ( molar mass )T2 mass 1 T1 mass 2 T2 = 2 , = P1 P2 P1 P2
Mass2 =
71.
mass1 T1P2 1.00 10 3 g 291 K 650. psi = = 309 g T2 P1 299 K 2050. psi
PV = nRT, n is constant.
PV P V VPT PV = nR = constant, 1 1 = 2 2 , V2 = 1 1 2 T2 P2 T1 T1 T
CHAPTER 5
GASES
V2 = 1.00 L ×
72.
191
760 .torr ( 273 − 31) K = 2.82 L; ΔV = 2.82 − 1.00 = 1.82 L 220 . torr ( 273 + 23 ) K
PV = nRT, P is constant.
nT P nT n T = = constant, 1 1 = 2 2 V1 V2 V R
294 K 4.20 103 m 3 n2 TV = 0.921 = 1 2 = n1 T2 V1 335 K 4.00 103 m 3
Gas Density, Molar Mass, and Reaction Stoichiometry 73.
STP: T = 273 K and P = 1.00 atm; at STP, the molar volume of a gas is 22.42 L. 2.00 L O2 ×
1 mol O 2 4 mol Al 26 .98 g Al = 3.21 g Al 22 .42 L 3 mol O 2 mol Al
Note: We could also solve this problem using PV = nRT, where n O 2 = PV/RT. You don’t have to memorize 22.42 L/mol at STP. 74.
CO2(s) → CO2(g); 4.00 g CO2 ×
1 mol CO 2 44.01 g CO 2
= 9.09 × 10 −2 mol CO2
At STP, the molar volume of a gas is 22.42 L. 9.09 × 10 −2 mol CO2 × 75.
2 NaN3(s) → 2 Na(s) + 3 N2(g) n N2 =
1.00 atm 70 .0 L PV = = 3.12 mol N2 needed to fill air bag. 0 . 08206 L atm RT 273 K K mol
Mass NaN3 reacted = 3.12 mol N2 × 76.
2 mol NaN 3 65 .02 g NaN 3 = 135 g NaN3 3 mol N 2 mol NaN 3
Because the solution is 50.0% H2O2 by mass, the mass of H2O2 decomposed is 125/2 = 62.5 g. 62.5 g H2O2 ×
V=
77.
22.42 L = 2.04 L mol CO 2
nRT = P
1 mol H 2 O 2 1 mol O 2 = 0.919 mol O2 34 .02 g H 2 O 2 2 mol H 2 O 2
0.08206 L atm 300. K K mol = 23.0 L O2 1 atm 746 torr 760 torr
0.919 mol
3 1L 100 cm 1.0 atm 4800 m 3 1000 cm3 m PV n H2 = = = 2.1 × 105 mol 0.08206 L atm RT 273 K K mol
192
CHAPTER 5
GASES
2.1 × 105 mol H2 is in the balloon. This is 80.% of the total amount of H2 that had to be generated: 0.80(total mol H2) = 2.1 × 105, total mol H2 = 2.6 × 105 mol 2.6 × 105 mol H2 ×
1 mol Fe 55 .85 g Fe = 1.5 × 107 g Fe mol H 2 mol Fe
2.6 × 105 mol H2 ×
1 mol H 2SO 4 98.09 g H 2SO 4 100 g reagent mol H 2 mol H 2SO 4 98 g H 2SO 4
= 2.6 × 107 g of 98% sulfuric acid 78.
5.00 g S ×
1 mol S = 0.156 mol S 32 .07 g
0.156 mol S will react with 0.156 mol O2 to produce 0.156 mol SO2. More O2 is required to convert SO2 into SO3. 0.156 mol SO2 ×
1 mol O 2 = 0.0780 mol O2 2 mol SO 2
Total mol O2 reacted = 0.156 + 0.0780 = 0.234 mol O2
V=
79.
nRT = P
0.234 mol
0.08206 L atm 623 K K mol = 2.28 L O2 5.25 atm
Kr(g) + 2 Cl2(g) → KrCl4(s); nKr =
PV 0.500 atm 15.0 L = 0.147 mol Kr = 0.08206 L atm RT 623 K K mol
We could do the same calculation for Cl2. However, the only variable that changed is the pressure. Because the partial pressure of Cl2 is triple that of Kr, moles of Cl2 = 3(0.147) = 0.441 mol Cl2. The balanced equation requires 2 moles of Cl2 to react with every mole of Kr. However, we actually have three times as many moles of Cl2 as we have of Kr. So Cl2 is in excess and Kr is the limiting reagent. 0.147 mol Kr
80.
1 mol KrCl 4 mol Kr
PV = nRT, V and T are constant.
225 .60 g KrCl 4 mol Kr
= 33.2 g KrCl4
P1 P P n = 2, 2 = 2 n1 n 2 P1 n1
Let's calculate the partial pressure of C3H3N that can be produced from each of the starting materials assuming each reactant is limiting. The reactant that produces the smallest amount of product will run out first and is the limiting reagent. PC 3 H 3 N = 0.500 MPa
2 MPa C3H 3 N = 0.500 MPa if C3H6 is limiting 2 MPa C 3H 6
PC 3 H 3 N = 0.800 MPa
2 MPa C 3 H 3 N = 0.800 MPa if NH3 is limiting 2 MPa NH 3
CHAPTER 5
GASES
193
PC 3 H 3 N = 1.500 MPa
2 MPa C 3 H 3 N = 1.000 MPa if O2 is limiting 3 MPa O 2
C3H6 is limiting. Although more product could be produced from NH3 and O2, there is only enough C3H6 to produce 0.500 MPa of C3H3N. The partial pressure of C3H3N in atmospheres after the reaction is: 0.500 × 106 Pa ×
n =
PV 4.94 atm 150 . L = = 30.3 mol C3H3N 0 . 08206 L atm RT 298 K K mol
53.06 g = 1.61 × 103 g C3H3N can be produced. mol
30.3 mol × 81.
1 atm = 4.94 atm 1.013 10 5 Pa
CH3OH + 3/2 O2 → CO2 + 2 H2O or 2 CH3OH(l) + 3 O2(g) → 2 CO2(g) + 4 H2O(g) 50.0 mL ×
n O2 =
0.850 g 1 mol = 1.33 mol CH3OH(l) available mL 32 .04 g
2.00 atm 22 .8 L PV = = 1.85 mol O2 available 0.08206 L atm RT 300 . K K mol
Assuming CH3OH is limiting: 1.33 mol CH3OH
4 mol H 2 O = 2.66 mol H2O 2 mol CH 3OH
Assuming O2 is limiting: 1.85 mol O2
4 mol H 2 O = 2.47 mol H2O 3 mol O 2
Because the O2 reactant produces the smaller quantity of H2O, O2 is limiting and 2.47 mol of H2O can be produced. 82.
For ammonia (in 1 minute): n NH3 =
PNH3 VNH3 RT
=
90. atm 500. L = 1.1 × 103 mol NH3 0.08206 L atm 496 K K mol
NH3 flows into the reactor at a rate of 1.1 × 103 mol/min. For CO2 (in 1 minute):
194
CHAPTER 5 n CO 2 =
PCO 2 VCO 2 RT
=
GASES
45 atm 600. L = 6.6 × 102 mol CO2 0.08206 L atm 496 K K mol
CO2 flows into the reactor at 6.6 × 102 mol/min. If NH3 is limiting: 1.1 10 3 mol NH 3 1 mol urea 60.06 g urea = 3.3 × 104 g urea/min min 2 mol NH 3 mol urea
If CO2 is limiting: 660 mol CO 2 1 mol urea 60.06 g urea = 4.0 × 104 g urea/min min mol CO 2 mol urea
Because the NH3 reactant produces the smaller quantity of product, NH3 is limiting and 3.3 × 104 g urea/min can be formed. 83.
a. CH4(g) + NH3(g) + O2(g) → HCN(g) + H2O(g); balancing H first, then O, gives: CH4 + NH3 +
3 2
O 2 → HCN + 3 H2O or 2 CH4(g) + 2 NH3(g) + 3 O2(g) →
2 HCN(g) + 6 H2O(g) b. PV = nRT, T and P constant;
V1 n1
=
V2 n2
,
V1 V2
=
n1 n2
The volumes are all measured at constant T and P, so the volumes of gas present are directly proportional to the moles of gas present (Avogadro’s law). Because Avogadro’s law applies, the balanced reaction gives mole relationships as well as volume relationships. If CH4 is limiting: 20.0 L CH4
2 L HCN = 20.0 L HCN 2 L CH 4
If NH3 is limiting: 20.0 L NH3
2 L HCN = 20.0 L HCN 2 L NH 3
If O2 is limiting: 20.0 L O2 ×
2 L HCN = 13.3 L HCN 3 L O2
O2 produces the smallest quantity of product, so O2 is limiting and 13.3 L HCN can be produced. 84.
From the balanced equation, ethyne reacts with hydrogen in a 1 : 2 mole ratio. From the problem, equal volumes of the two reactants are flowing into the reaction chamber, yet we need a larger amount of H2 to react with C2H2. Therefore, H2 is limiting. In 1 minute:
CHAPTER 5
GASES
n H2 =
195
PV 25.0 atm × 2000. L = = 1060 mol H2 reacted 0.08206 L atm RT × 573 K K mol
Theoretical yield =
Percent yield =
1060 mol H 2 1 mol C 2 H 6 = 530. mol C2H6/min min 2 mol H 2
480 mol/min × 100 = 91% 530 mol/min
85.
d = (molar mass)P/RT; density is directly proportional to the molar mass of a gas. Helium, with the smallest molar mass of all the noble gases, will have the smallest density at STP.
86.
d = (molar mass)P/RT; at constant P and T, density is directly proportional to molar mass. If the density of the unknown gas is 2.19 times the density of O2, then the molar mass of the unknown will be 2.19 times the molar mass of O2. Molar mass of unknown gas = 2.19(32.00 g/mol) = 70.1 g/mol The empirical mass of CH2 is ~ 12.0 + 2(1.0) = 14.0;
molar mass 70.1 = = 5.01 empirical formula mass 14.0
Molecular formula = (CH2)×5 = C5H10 87.
Molar mass =
dRT , where d = density of gas in units of g/L. P 3.164 g/L
Molar mass =
0.08206 L atm 273.2 K K mol = 70.98 g/mol 1.000 atm
The gas is diatomic, so the average atomic mass = 70.93/2 = 35.47 u. From the periodic table, this is chlorine, and the identity of the gas is Cl2. 88.
P × (molar mass) = dRT, d =
mass mass × RT , P × (molar mass) = volume V
0.08206 L atm 373 K mass RT K mol Molar mass = = 96.9 g/mol = 1 atm PV (750. torr ) 0.256 L 760 torr 0.800 g
Mass of CHCl 12.0 + 1.0 + 35.5 = 48.5 g/mol;
89.
96.9 = 2.00; molecular formula is C2H2Cl2. 48.5
1 atm 745 torr 352 .0 g/mol 760 torr P (molar mass) d UF6 = = = 12.6 g/L 0.08206 L atm RT 333 K K mol
196
90.
CHAPTER 5
GASES
1 atm 635 torr 169.89 g/mol 760 torr P (molar mass) d SiCl4 = = = 4.83 g/L 0.08206 L atm RT 358 K K mol 1 atm 635 torr 135.45 g/mol 760 torr P (molar mass) d SiHCl3 = = = 3.85 g/L 0.08206 L atm RT 358 K K mol
Partial Pressure 91.
The container has 5 He atoms, 3 Ne atoms, and 2 Ar atoms for a total of 10 atoms. The mole fractions of the various gases will be equal to the molecule fractions. He =
3 Ne atoms 5 He atoms = 0.50; Ne = = 0.30 10 total atoms 10 total atoms
Ar = 1.00 – 0.50 – 0.30 = 0.20 PHe = He Ptotal = 0.50(1.00 atm) = 0.50 atm PNe = Ne PTotal = 0.30(1.00atm) = 0.30 atm PAr = 1.00 atm – 0.50 atm – 0.30 atm = 0.20 atm 92.
a. There are 6 He atoms and 4 Ne atoms, and each flask has the same volume. The He flask has 1.5 times as many atoms of gas present as the Ne flask, so the pressure in the He flask will be 1.5 times greater (assuming a constant temperature). b. Because the flask volumes are the same, your drawing should have the various atoms equally distributed between the two flasks. So each flask should have 3 He atoms and 2 Ne atoms. c. After the stopcock is opened, each flask will have 5 total atoms and the pressures will be equal. If six atoms of He gave an initial pressure of PHe, initial, then 5 total atoms will have a pressure of 5/6 PHe, initial. Using similar reasoning, 4 atoms of Ne gave an initial pressure of PNe, initial, so 5 total atoms will have a pressure of 5/4 PNe, initial. Summarizing: 5 5 PHe, initial = PNe, initial 6 4 d. For the partial pressures, treat each gas separately. For helium, when the stopcock is opened, the six atoms of gas are now distributed over a larger volume. To solve for the final partial pressures, use Boyle’s law for each gas.
Pfinal =
For He: P2 =
PHe, initial P1 V1 X = PHe, initial = V2 2X 2
CHAPTER 5
GASES
197
The partial pressure of helium is exactly halved. The same result occurs with neon so that when the volume is doubled, the partial pressure is halved. Summarizing: PHe, final =
93.
n He = 15.2 g He ×
PHe, initial 2
; PNe, final =
PNe, initial 2
1 mol O 2 1 mol He = 3.80 mol H2; n O 2 = 30.6 g O2 × 32 .00 g O 2 4.003 g He
= 0.956 mol O2 PHe = PO 2 =
94.
n He RT = V n O 2 RT V
3.80 mol
0.08206 L atm (273 + 22) K K mol = 18.4 atm 5.00 L
= 4.63 atm; Ptotal = PHe + PO 2 = 18.4 atm + 4.63 atm = 23.0 atm
n H 2 = 1.00 g H2 ×
1 mol He 1 mol H 2 = 0.496 mol H2; n He = 1.00 g He × 2.016 g H 2 4.003 g He
= 0.250 mol He
PH 2 =
PHe = 95.
n H 2 RT V
0.496 mol
=
0.08206 L atm ( 273 + 27) K K mol = 12.2 atm 1.00 L
n He RT = 6.15 atm; Ptotal = PH 2 + PHe = 12.2 atm + 6.15 atm = 18.4 atm V
Treat each gas separately and determine how the partial pressure of each gas changes when the container volume increases. Once the partial pressures of H2 and N2 are determined, the total pressure will be the sum of these two partial pressures. At constant n and T, the relationship P1V1 = P2V2 holds for each gas. For H2: P2 =
P1 V1 2.00 L = 475 torr × = 317 torr 3.00 L V2
For N2: P2 = 0.200 atm ×
760 torr 1.00 L = 0.0667 atm; 0.0667 atm × = 50.7 torr 3.00 L atm
Ptotal = PH 2 + PN 2 = 317 + 50.7 = 368 torr 96.
For H2: P2 =
P1 V1 2.00 L = 360. torr × = 240. torr 3.00 L V2
Ptotal = PH 2 + PN 2 , PN 2 = Ptotal − PH 2 = 320. torr − 240. torr = 80. torr For N2: P1 = 97.
P2 V2 3.00 L = 80. torr × = 240 torr 1.00 L V1
P1V1 = P2V2; the total volume is 1.00 L + 1.00 L + 2.00 L = 4.00 L.
198
CHAPTER 5 For He: P2 =
P1 V1 V2
= 200. torr ×
1.00 L = 50.0 torr He 4.00 L
760 torr 1.00 L = 0.100 atm; 0.100 atm × = 76.0 torr Ne 4.00 L atm
For Ne: P2 = 0.400 atm ×
For Ar: P2 = 24.0 kPa ×
GASES
2.00 L 1 atm 760 torr = 12.0 kPa; 12.0 kPa × 101 .3 kPa 4.00 L atm
= 90.0 torr Ar Ptotal = 50.0 + 76.0 + 90.0 = 216.0 torr 98.
We can use the ideal gas law to calculate the partial pressure of each gas or to calculate the total pressure. There will be less math if we calculate the total pressure from the ideal gas law. n O 2 = 1.5 × 102 mg O2
1g 1000 mg
n NH 3 = 5.0 × 1021 molecules NH3 ×
1 mol O 2 = 4.7 × 10−3 mol O2 32.00 g O 2 1 mol NH 3
6.022 10 23 molecules NH 3
= 8.3 × 10−3 mol NH3
ntotal = n N 2 + n O 2 + n NH3 = 5.0 × 10−2 + 4.7 × 10−3 + 8.3 × 10−3 = 6.3 × 10−2 mol total
Ptotal =
n total RT = V
6.3 10 − 2 mol
PN 2 = χ N 2 Ptotal, χ N 2 = PO 2 =
4.7 10 −3 6.3 10 − 2
n N2 n total
; PN 2 =
0.08206 L atm 273 K K mol = 1.4 atm 1 .0 L
5.0 10 −2 mol 6.3 10 − 2 mol
1.4 atm = 0.10 atm; PNH3 =
Pcyclopropane
1.4 atm = 1.1 atm
8.3 10 −3 6.3 10 − 2
1.4 atm = 0.18 atm
170. torr = 0.230 170. torr + 570 . torr
99.
Mole fraction cyclopropane = χ =
100.
The molar mass of H2 (2.016 g/mol) is about one-half the molar mass of He (4.003 g/mol). If we have equal masses of H2 and He, then we will have twice as many moles of H2 as we have moles of He present. If we have three total moles of gas, 2.0 moles will be H2 and 1.0 mol will be He. The mole fraction of He will be 1/3 and the mole fraction of H2 will be 2/3.
Ptotal
=
PH 2 = χ H 2 Ptotal = 0.67 × 3.0 atm = 2.0 atm
101.
a. Mole fraction CH4 = χ CH 4 = χ O 2 = 1.000 − 0.412 = 0.588
PCH 4 Ptotal
=
0.175 atm 0.175 atm + 0.250 atm
= 0.412
CHAPTER 5
GASES
199
b. PV = nRT, ntotal =
c.
n CH 4
χ CH 4 =
n total
Ptotal V RT
=
0.425 atm 10 .5 L = 0.161 mol 0.08206 L atm 338 K K mol
, n CH 4 = χ CH 4 × ntotal = 0.412 × 0.161 mol = 6.63 × 10 −2 mol CH4
6.63 × 10 −2 mol CH4 ×
16 .04 g CH 4 mol CH 4
= 1.06 g CH4
n O 2 = 0.588 × 0.161 mol = 9.47 × 10 −2 mol O2; 9.47 × mol O2 ×
32 .00 g O 2 mol O 2
= 3.03 g O2 102.
52.5 g O2
O2 =
1 mol O 2 32.00 g O 2
n O2 n total
= 1.64 mol O2; 65.1 g CO2
1 mol CO 2 44.01 g CO 2
= 1.48 mol CO2
1.64 mol = 0.526 (1.64 + 1.48) mol
=
PO 2 = O 2 Ptotal = 0.526 9.21 atm = 4.84 atm PCO 2 = 9.21 – 4.84 = 4.37 atm
103.
To calculate the volume of gas, we can use Ptotal and ntotal (V = ntotal RT/Ptotal), or we can use PHe and nHe (V = nHeRT/PHe). Because n H 2 O is unknown, we will use PHe and nHe. PHe + PH 2 O = 1.00 atm = 760. torr, PHe + 17.5 torr = 760. torr, PHe = 743 torr nHe = 0.586 g ×
V=
104.
n H e RT = PH e
1 mol 4.003 g
= 0.146 mol He 0.08206 L atm × 293 K K mol = 3.59 L 1 atm 743 torr × 760 torr
0.146 mol ×
For the O2(g), n and T are constant, so the two-condition equation reduces to P1V1 = P2V2. Solving for P1: P1 =
P2 V2 785 torr 1.94 L = = 761 torr V1 2.00 L
Ptotal = PO2 + PH 2O , PH 2 O = 785 torr − 761 torr = 24 torr 105.
Ptotal = PH 2 + PH 2 O , 1.032 atm = PH 2 + 32 torr ×
1 atm 760 torr
, PH 2 = 1.032 − 0.042 = 0.990 atm
200
CHAPTER 5 n H2 =
PH 2 V RT
=
0.990 atm 0.240 L = 9.56 × 10 −3 mol H2 0.08206 L atm 303 K K mol
9.56 × 10 −3 mol H2 × 106.
GASES
1 mol Zn 65 .38 g Zn = 0.625 g Zn mol H 2 mol Zn
2 NaClO3(s) → 2 NaCl(s) + 3 O2(g) Ptotal = PO 2 + PH 2 O , PO 2 = Ptotal − PH 2 O = 734 torr − 19.8 torr = 714 torr 1 atm 714 torr 0.0572 L PO 2 V 760 torr n O2 = = = 2.22 × 10 −3 mol O2 0.08206 L atm RT ( 273 + 22 ) K K mol
Mass NaClO3 decomposed = 2.22 × 10 −3 mol O2 ×
2 mol NaClO 3 106.44 g NaClO 3 3 mol O 2 mol NaClO 3
= 0.158 g NaClO3 Mass % NaClO3 = 107.
0.158 g 0.8765 g
× 100 = 18.0%
10.10 atm − 7.62 atm = 2.48 atm is the pressure of the amount of F2 reacted. PV = nRT, V and T are constant.
P P P P n = constant, 1 = 2 or 1 = 1 n1 P2 n2 n2 n
2.48 atm Moles F2 reacted = = 2.00; so Xe + 2 F2 → XeF4 Moles Xe reacted 1.24 atm
108.
2 C(gr) + O2(g) → 2 CO(g); C(gr) is limiting. When 1.60 mol of C(gr) reacts, it requires 0.800 mol O2 and 1.60 mol of CO will be produced (from the stoichiometry in the balanced equation). After reaction, all of C(gr) is gone, 1.20 – 0.800 = 0.40 mol O2(g) remains in excess and 1.60 mol CO(g) is produced. The total moles of gas after reaction is 0.40 + 1.60 = 2.00 mol. 0.08206 L atm 2.00 mol × × 298 K n total ×RT K mol Ptotal = = = 16.3 atm V 3.00 L
109.
2 HN3(g) → 3 N2(g) + H2(g); at constant V and T, P is directly proportional to n. In the reaction, we go from 2 moles of gaseous reactants to 4 moles of gaseous products. Because moles doubled, the final pressure will double (Ptotal = 6.0 atm). Similarly, from the 2 : 1 mole ratio between HN3 and H2, the partial pressure of H2 will be 3.0/2 = 1.5 atm. The partial pressure of N2 will be (3/2)3.0 atm = 4.5 atm. This is from the 2 : 3 mole ratio between HN3 and N2.
110.
N2(g) + 2 O2(g) → 2 NO2(g); at constant V and T, P is directly proportional to n:
CHAPTER 5
GASES
201
P1 P n P = 2 or 2 = 2 ; mole ratios equal partial pressure ratios at constant V and T. n1 n2 n1 P1
From the 2:1 mole ratio between O2 and N2, if 2.00 atm of O2 reacts, it will require 1.00 atm of N2, and 2.00 atm of NO2 will be produced. After reaction we have 2.00 – 1.00 = 1.00 atm of excess N2(g) plus 2.00 atm of NO2(g). Ptotal = 1.00 atm + 2.00 atm = 3.00 atm. 111.
2 SO2(g) + O2(g) → 2 SO3(g); because P and T are constant, volume ratios will equal mole ratios (Vf/Vi = nf/ni). Let x = mol SO2 = mol O2 present initially. From the balanced equation, 2 mol of SO2 react for every 1 mol of O2 that reacts. Because we have equal moles of SO2 and O2 present initially, and because SO2 is used up twice as fast as O2, SO2 is the limiting reagent. Therefore, no SO2 will be present after the reaction goes to completion. However, excess O2(g) will be present as well as the SO3(g) produced. Mol O2 reacted = x mol SO2 ×
1 mol O 2 = x/2 mol O2 2 mol SO 2
Mol O2 remaining = x mol O2 initially − x/2 mol O2 reacted = x/2 mol O2 Mol SO3 produced = x mol SO2 ×
2 mol SO 3 = x mol SO3 2 mol SO 2
Total moles gas initially = x mol SO2 + x mol O2 = 2x Total moles gas after reaction = x/2 mol O2 + x mol SO3 = (3/2)x = (1.5)x nf V (1.5) x 1.5 = f = = = 0.75; Vf/Vi = 0.75 : l or 3 : 4 ni Vi 2x 2
The volume of the reaction container shrinks to 75% of the initial volume. This makes sense because the total mol of gas decreased as the reaction occurs at constant P and T. 112.
N2(g) + 3 Cl2(g) → 2 NCl3(g); because P and T are constant, volume ratios will equal mole ratios (Vf/Vi = nf/ni). Let x = mol N2 = mol Cl2 present initially. From the balanced equation, 3 mol of Cl2 react for every 1 mol of N2 present. Because we have equal moles of Cl2 and N2 present initially, and because Cl2 is used up 3 times as fast as N2, Cl2 is the limiting reagent. Therefore, no Cl2 will be present after the reaction goes to completion. However, excess N2(g) will be present as well as the NCl3(g) produced. Mol N2 reacted = x mol Cl2 ×
1 mol N 2 = x/3 mol N2 3 mol Cl 2
Mol N2 remaining = x mol N2 initially − x/3 mol N2 reacted = 2x/3 mol N2 Mol NCl3 produced = x mol Cl2 ×
2 mol NCl3 = 2x/3 mol NCl3 3 mol Cl 2
Total moles gas initially = x mol Cl2 + x mol N2 = 2x Total moles gas after reaction = 2x/3 mol N2 + 2x/3 mol NCl3 = 4x/3 mol total
202
CHAPTER 5
GASES
nf V 4x / 3 Vf = f, = , Vf = 2/3(4.0 L) = 2.67 L ni Vi 2x 4.0 L
The volume of the reaction container decreases from 4.00 L to 2.67 L after the reaction is complete. This makes sense because the moles of gas decrease as the reaction occurs. 113.
150 g (CH3)2N2H2 ×
PN 2 =
nRT = V
1 mol (CH 3 ) 2 N 2 H 2 3 mol N 2 = 7.5 mol N2 produced 60 .10 g mol (CH 3 ) 2 N 2 H 2
7.5 mol
0.08206 L atm 400. K K mol = 0.98 atm 250 L
We could do a similar calculation for PH 2O and PCO 2 and then calculate Ptotal (= PN 2 + PH 2O + PCO 2 ) . Or we can recognize that 9 total moles of gaseous products form for every mole of
(CH3)2N2H2 reacted (from the balanced equation given in the problem). This is three times the moles of N2 produced. Therefore, Ptotal will be three times larger than PN 2 . Ptotal = 3 × PN 2 = 3 × 0.98 atm = 2.9 atm. 114.
The partial pressure of CO2 that reacted is 740. − 390. = 350. torr. Thus the number of moles of CO2 that react is given by: 350 . atm 3.00 L PV 760 n= = 5.75 × 10−2 mol CO2 = 0.08206 L atm RT 293 K K mol
5.75 × 10−2 mol CO2 × Mass % MgO =
1 mol MgO 40 .31 g MgO = 2.32 g MgO 1 mol CO 2 mol MgO
2.32 g × 100 = 81.4% MgO 2.85 g
Kinetic Molecular Theory and Real Gases 115.
KEavg = (3/2)RT; the average kinetic energy depends only on temperature. At each temperature, CH4 and N2 will have the same average KE. For energy units of joules (J), use R = 8.3145 J/K•mol. To determine average KE per molecule, divide the molar KEavg by Avogadro’s number, 6.022 × 1023 molecules/mol. At 273 K: KEavg =
3 8.3145 J × 273 K = 3.40 × 103 J/mol = 5.65 × 10−21 J/molecule 2 K mol
At 546 K: KEavg =
3 8.3145 J × 546 K = 6.81 × 103 J/mol = 1.13 × 10−20 J/molecule 2 K mol
CHAPTER 5 116.
nAr =
GASES
203
n CH n CH 228 g 4 4 = 5.71 mol Ar; χ CH 4 = = 0.650 = 39.95 g/mol n CH + n Ar n CH + 5.71 4
4
0.650( n CH 4 + 5.71) = n CH 4 , 3.71 = (0.350 ) n CH 4 , n CH 4 = 10.6 mol CH4 KEavg =
3 2
RT for 1 mole of gas
KEtotal = (10.6 + 5.71) mol × 3/2 × 8.3145 J/K•mol × 298 K = 6.06 × 104 J = 60.6 kJ 1/2
117.
3 RT rms = M
, where R =
8 .3145 J and M = molar mass in kg. K mol
For CH4, M = 1.604 × 10−2 kg, and for N2, M = 2.802 × 10−2 kg. 1/2
8.3145 J 273 K 3 K mol For CH4 at 273 K: rms = 1.604 10 − 2 kg/mol
= 652 m/s
Similarly, rms for CH4 at 546 K is 921 m/s. 1/2
8.3145 J 273 K 3 K mol For N2 at 273 K: rms = 2.802 10 − 2 kg/mol Similarly, for N2 at 546 K, rms = 697 m/s.
1/ 2
118.
3 RT rms = M
;
μ UF6 μ He
=
3 RT UF6 M UF 6
3 RT He M He
= 493 m/s
1/ 2
1/ 2
M He TUF6 = M UF THe 6
1/ 2
We want the root mean square velocities to be equal, and this occurs when: M He TUF6 = M UF6 THe
The ratio of the temperatures is:
TUF6 THe
=
M UF6 M He
=
352.0 = 87.93 4.003
The heavier UF6 molecules would need a temperature 87.93 times that of the He atoms in order for the root mean square velocities to be equal. 119.
The number of gas particles is constant, so at constant moles of gas, either a temperature change or a pressure change results in the smaller volume. If the temperature is constant, an increase
204
CHAPTER 5
GASES
in the external pressure would cause the volume to decrease. Gases are mostly empty space so gases are easily compressible. If the pressure is constant, a decrease in temperature would cause the volume to decrease. As the temperature is lowered, the gas particles move with a slower average velocity and don’t collide with the container walls as frequently and as forcefully. As a result, the internal pressure decreases. To keep the pressure constant, the volume of the container must decrease in order to increase the gas particle collisions per unit area. 120.
In this situation, the volume has increased by a factor of two. One way to double the volume of a container at constant pressure and temperature is to double the number of moles of gas particles present. As gas particles are added, more collisions per unit area occur and the internal pressure increases. In order to keep the pressure constant, the container volume must increase. Another way to double the volume of a container at constant pressure and moles of gas is to double the absolute temperature. As temperature increases, the gas molecules collide more frequently with the walls of the container. In order to keep pressure constant, the container volume must increase. The last variable which can be changed is pressure. If the external pressure exerted on the container is halved, the volume will double (assuming constant temperature and moles). As the external pressure applied is reduced, the volume of the container must increase in order to equalize the higher internal pressure with the lower external applied pressure.
121.
a
b
c
d
Avg. KE
increase
decrease
same (KE T)
Avg. velocity
increase
decrease
same (
1
mv2 = KE T)
same same
2
Wall coll. freq
increase
decrease
increase
increase
Average kinetic energy and average velocity depend on T. As T increases, both average kinetic energy and average velocity increase. At constant T, both average kinetic energy and average velocity are constant. The collision frequency is proportional to the average velocity (as velocity increases, it takes less time to move to the next collision) and to the quantity n/V (as molecules per volume increase, collision frequency increases). 122.
V, T, and P are all constant, so n must be constant. Because we have equal moles of gas in each container, gas B molecules must be heavier than gas A molecules. a. Both gas samples have the same number of molecules present (n is constant). b. Because T is constant, KEavg must be the same for both gases [KEavg = (3/2)RT]. c. The lighter gas A molecules will have the faster average velocity. d. The heavier gas B molecules do collide more forcefully, but gas A molecules, with the faster average velocity, collide more frequently. The end result is that P is constant between the two containers.
CHAPTER 5 123.
GASES
205
a. They will all have the same average kinetic energy because they are all at the same temperature [KEavg = (3/2)RT]. b. Flask C; H2 has the smallest molar mass. At constant T, the lighter molecules have the faster average velocity. This must be true for the average kinetic energies to be the same.
124.
a. All the gases have the same average kinetic energy since they are all at the same temperature [KEavg = (3/2)RT]. b. At constant T, the lighter the gas molecule, the faster the average velocity [avg rms (1/M)1/2]. Xe (131.3 g/mol) < Cl2 (70.90 g/mol) < O2 (32.00 g/mol) < H2 (2.016 g/mol) slowest fastest c. At constant T, the lighter H2 molecules have a faster average velocity than the heavier O2 molecules. As temperature increases, the average velocity of the gas molecules increases. Separate samples of H2 and O2 can only have the same average velocities if the temperature of the O2 sample is greater than the temperature of the H2 sample. 1/2
125.
Graham’s law of effusion:
81 =
126.
MB , MB = 81MA; answer e is correct MA 1/2
M = unknown , M CH Rate unknown 4 RateCH 4
4=
1/2
M M Rate A = B , 9 = B ; squaring both sides: Rate B MA MA
1/2
M 8 mL/min = unknown ; squaring both sides: 4 mL/min 16.0 g/mol
M unknown , Munknown = 4(16.0) = 64.0 g/mol; the unknown gas is SO2 (M = 64.0 g/mol). 16.0 1/ 2
127.
Graham’s law of effusion:
M Rate1 = 2 Rate 2 M1
Let the unknown = gas 1 and O2 = gas 2: 32.00 31.50 = 30.50 M1
1/2
, 1.067 =
32.00 , M1 = 29.99 g/mol M1
Of the choices, NO with a molar mass of 30.01 g/mol, best fits the data. So the unknown gas is nitrogen monoxide (NO).
128.
Rate 1 Rate 2
M = 2 M1
1/ 2
; rate1 =
24.0 mL min
; rate2 =
16.04 g 47.8 mL ; M1 = ? ; M2 = min mol
206
CHAPTER 5 24.0 16.04 = 47.8 M 1
1/ 2
= 0.502, 16.04 = (0.502)2 × M1, M1 =
1/2
1/2
129.
16.04 63.7 g = 0.252 mol 1/2
M M M Rate A 1.0 L/3.0 min = B , 4.0 = B = B , 1.0 mL/12.0 min Rate B MA MA MA
16 =
GASES
; squaring both sides:
MB , MB = 16MA; the molar mass of B is 16 times the molar mass of A. MA
Answer e is correct. The molar mass of O2 (gas B) is 32.0 g/mol and the molar mass of H2 (gas A) is 2.0 g/mol. 1/2
130.
M Rate1 = 2 Rate 2 M1
, where M = molar mass; let gas (1) = He and gas (2) = Cl2.
Effusion rates in this problem are equal to the volume of gas that effuses per unit time (L/min). Let t = time in the following expression. 1 .0 L 1/ 2
4.5 min 70.90 = 1 .0 L 4.003
,
t = 4.209, t = 19 min 4.5 min
t
131.
132.
133.
A gas behaves most ideally at high temperatures and low pressure. Answer b combines the highest temperature and lowest pressure, so answer b is correct. 2 n The van der Waals equation is Pmeasured + a (Vmeasured − nb) = nRT. The measured V pressure is too low compared to the ideal pressure due to intermolecular attractions between the gas particles. And the container volume (Vmeasured) is too high compared to the ideal volume of the gas because some of the space in the container is taken up by the molecules themselves. Answer c is correct.
a. PV = nRT
P=
b.
nRT = V
0.5000 mol
0.08206 L atm ( 25.0 + 273.2) K K mol = 12.24 atm 1.0000 L
2 n P + a (V − nb) = nRT; for N2: a = 1.39 atm L2/mol2 and b = 0.0391 L/mol V
2 0.5000 P + 1.39 atm (1.0000 L − 0.5000 × 0.0391 L) = 12.24 L atm 1.0000
CHAPTER 5
GASES
207
(P + 0.348 atm)(0.9805 L) = 12.24 L atm P=
12 .24 L atm − 0.348 atm = 12.48 − 0.348 = 12.13 atm 0.9805 L
c. The ideal gas law is high by 0.11 atm, or
134.
0.11 × 100 = 0.91%. 12.13
a. PV = nRT
P=
b.
nRT = V
0.5000 mol
0.08206 L atm 298.2 K K mol = 1.224 atm 10.000 L
2 n P + a (V – nb) = nRT; for N2: a = 1.39 atm L2/mol2 and b = 0.0391 L/mol V
2 0.5000 P + 1.39 atm (10.000 L − 0.5000 × 0.0391 L) = 12.24 L atm 10.000
(P + 0.00348 atm)(10.000 L − 0.0196 L) = 12.24 L atm P + 0.00348 atm =
12 .24 L atm = 1.226 atm, P = 1.226 − 0.00348 = 1.223 atm 9.980 L
c. The results agree to ±0.001 atm (0.08%). d. In Exercise 133, the pressure is relatively high, and there is significant disagreement. In Exercise 134, the pressure is around 1 atm, and both gas laws show better agreement. The ideal gas law is valid at relatively low pressures.
Atmospheric Chemistry 135.
χHe = 5.24 × 10 −6 from Table 5.4. PHe = χHe × Ptotal = 5.24 × 10 −6 × 1.0 atm = 5.2 × 10 −6 atm 5.2 10 −6 atm n P = = 2.1 × 10 −7 mol He/L = 0 . 08206 L atm V RT 298 K K mol 2.1 10 −7 mol 1L 6.022 10 23 atoms = 1.3 × 1014 atoms He/cm3 L mol 1000 cm3
136.
At 15 km, T −60°C and P = 0.1 atm. Use
V2 =
V1 P1 T2 P2 T1
=
1.0 L 1.00 atm 213 K 0.1 atm 298 K
P1 V1 PV = 2 2 since n is constant. T2 T1
=7L
208 137.
CHAPTER 5
GASES
S(s) + O2(g) → SO2(g), combustion of coal 2 SO2(g) + O2(g) → 2 SO3(g), reaction with atmospheric O2 SO3(g) + H2O(l) → H2SO4(aq), reaction with atmospheric H2O
138.
H2SO4(aq) + CaCO3(s) → CaSO4(aq) + H2O(l) + CO2(g)
139.
a. If we have 1.0 × 106 L of air, then there are 3.0 × 102 L of CO. PCO = χCOPtotal; χCO =
b. nCO =
PCO V ; RT
VCO 3.0 10 2 because V ∝ n; PCO = × 628 torr = 0.19 torr Vtotal 1.0 10 6
assuming 1.0 m3 air, 1 m3 = 1000 L:
0.19 atm (1.0 10 3 L ) 760 nCO = = 1.1 × 10−2 mol CO 0.08206 L atm 273 K K mol
1.1 × 10−2 mol ×
c. 140.
6.02 10 23 molecules = 6.6 × 1021 CO molecules in 1.0 m3 of air mol
6.6 10 21 molecules m3
3
1m 6.6 1015 molecules CO = cm3 100 cm
For benzene: 89.6 × 10-9 g ×
Vbenzene =
1 mol = 1.15 × 10−9 mol benzene 78 .11 g
n benzeneRT = P
Mixing ratio =
Or ppbv =
0.08206 L atm 296 K K mol = 2.84 × 10−8 L 1 atm 748 torr 760 torr
1.15 10 −9 mol
2.84 10 −8 L × 106 = 9.47 × 10−3 ppmv 3.00 L
vol. of X 10 9 2.84 10 −8 L × 109 = 9.47 ppbv = total vol. 3.00 L
1.15 10 −9 mol benzene 1L 6.022 10 23 molecules 3.00 L mol 1000 cm3 = 2.31 × 1011 molecules benzene/cm3
CHAPTER 5
GASES
209
For toluene: 153 × 10−9 g C7H8 ×
1 mol = 1.66 × 10−9 mol toluene 92 .13 g
n RT Vtoluene = toluene = P
Mixing ratio =
0.08206 L atm 296 K K mol 1 atm 748 torr 760 torr
1.66 10 −9 mol
= 4.10 × 10−8 L
4.10 10 −8 L × 106 = 1.37 × 10−2 ppmv (or 13.7 ppbv) 3.00 L
1.66 10 −9 mol toluene 1L 6.022 10 23 molecules 3.00 L mol 1000 cm3 = 3.33 × 1011 molecules toluene/cm3
ChemWork Problems 141.
Processes b, c, and d will all result in a doubling of the pressure. Process b doubles the pressure because the absolute temperature is doubled (from 200. K to 400. K). Process c has the effect of halving the volume, which would double the pressure by Boyle’s law. Process d doubles the pressure because the moles of gas are doubled (28 g N2 is 1 mol of N2 and 32 g O2 is 1 mol of O2). Process a won’t double the pressure because 28 g O2 is less than one mol of gas and process e won’t double the temperature since the absolute temperature is not doubled (goes from 303 K to 333 K).
142.
a. True; moles and volume are directly proportional to each other at constant P and T. b. False; for the volume to double at constant P and n, the absolute temperature must double. Here the absolute temperature went from 298 K to 323 K; the absolute temperature did not double. c. True; a barometer can be used to measure atmospheric pressure. d. True; volume and pressure are inversely proportional at constant n and T.
143.
At constant P and T, volume is directly proportional to the moles of gas present. VHe n = 4 = He ; let’s assume we have 1 mol of Ne present. For the volume of the helium VNe n Ne balloon to be 4 times larger, we will need 4 mol of He. The mass of 1 mol of Ne is 20.18 g and the mass of 4 moles of He is 4(4.003) = 16.01 g. The He:Ne mass ratio is:
16.01 = 0.7934; this is approximately a 4:5 mass ratio between He and Ne. 20.18
210
CHAPTER 5
144.
GASES
At constant V and T, pressure and moles are directly related. Because the balloon with the unknown identity has a pressure twice as great as the pressure of the Ar balloon, we must have twice the moles of gas in the unknown balloon as compared to the Ar balloon. Moles unknown = 25 g Ar ×
Molar mass = 145.
1 mol Ar 2 mol unknown = 1.25 mol (1 extra sig fig) 39.95 g 1 mol Ar
mass 20. g = 16 g/mol; the unknown gas is CH4. = mol 1.25 mol
a. PV = nRT PV = constant
PV
b. PV = nRT nR × T = const × T V
T=
P
T
nR constant = V V 1 P = constant × V
e. P =
P
V
V
T
PV = constant
P
P × V = const × V nR
P=
V
d. PV = nRT
c. PV = nRT
f.
PV = nRT PV = nR = constant T
PV T
1/V
P
Note: The equation for a straight line is y = mx + b, where y is the y-axis and x is the x-axis. Any equation that has this form will produce a straight line with slope equal to m and a y intercept equal to b. Plots b, c, and e have this straight-line form. 146.
At constant T and P, Avogadro’s law applies; that is, equal volumes contain equal moles of molecules or V2/V1 = n2/n1. Note that mole ratios and volume ratios between the various reactants and products will be equal to each other. Using volume ratios, the balanced equation is Cl2 + 5 F2 → 2 X. Two moles of X must contain 2 moles of Cl and 10 moles of F; X must have the formula ClF5 for a balanced equation.
CHAPTER 5
GASES
211 3
147.
14.1 × 102 in Hg•in3 ×
2.54 cm 10 mm 1 atm 1L 2.54 cm in 1 cm 760 mm 1000 cm 3 in
= 0.772 atm•L Boyle’s law: PV = k, where k = nRT; from Example 5.3, the k values are around 22 atm•L. Because k = nRT, we can assume that Boyle’s data and the Example 5.3 data were taken at different temperatures and/or had different sample sizes (different moles). 148.
n2 =
n 1 P2 T1 P1 T2
34.3 mol × 149.
P1 P P R = constant, = 2 = n 1 T1 n 2 T2 nT V
PV = nRT,
39.95 g mol
(273 + 25) K 2.00 MPa = 34.3 mol 8.93 MPa (273 + 19) K
= 1370 g Ar remains
PV = nRT, n is constant. V2 = 855 L ×
150.
= 150.0 mol ×
VPT P V PV PV = nR = constant, 1 1 = 2 2 , V2 = 1 1 2 P2 T1 T2 T1 T
(273 + 15) K 730. torr = 997 L; ΔV = 997 − 855 = 142 L 605 torr (273 + 25) K
Mn(s) + x HCl(g) → MnClx(s) + n H2 =
x H2(g) 2
0.951 atm 3.22 L PV = = 0.100 mol H2 0 . 08206 L atm RT 373 K K mol
Mol Cl in compound = mol HCl = 0.100 mol H2 ×
Mol Cl = Mol Mn
x mol Cl = 0.200 mol Cl x mol H 2 2
0.200 mol Cl 0.200 mol Cl = = 4.00 1 mol Mn 0.05000 mol Mn 2.747 g Mn 54 .94 g Mn
The formula of compound is MnCl4. 151.
Assume some mass of the mixture. If we had 100.0 g of the gas, we would have 50.0 g He and 50.0 g Xe. 50.0 g n He 12 .5 mol He 4.003 g/mol χ He = = = 0.970 = 50.0 g 50.0 g n He + n Xe 12 .5 mol He + 0.381 mol Xe + 4.003 g/mol 131.3 g/mol No matter what the initial mass of mixture is assumed, the mole fraction of helium will always be 0.970. PHe = χHePtotal = 0.970 × 600. torr = 582 torr; PXe = 600. − 582 = 18 torr
212 152.
CHAPTER 5
GASES
Assuming 100.0 g of cyclopropane: 85.7 g C × 14.3 g H ×
1 mol C 12 .01 g
= 7.14 mol C
1 mol H 14.2 = 14.2 mol H; = 1.99 1.008 g 7.14
The empirical formula for cyclopropane is CH2, which has an empirical mass 12.0 + 2(1.0) = 14.0 g/mol.
P × (molar mass) = dRT, molar mass =
dRT = P
1.88 g / L
0.08206 L atm 273 K K mol 1.00 atm = 42.1 g/mol
Because 42.1/14.0 3.0, the molecular formula for cyclopropane is (CH2)× 3 = C3H6. 153.
P × (molar mass) = dRT, d =
mass mass × RT , P × (molar mass) = volume V
Mass of gas = 135.87 g – 134.66 g = 1.21 g
Molar mass =
154.
mass RT = PV
750. mL juice ×
0.08206 L atm 304 K K mol = 33.3 g/mol 0.967 atm 0.936 L
1.21 g
12 mL C 2 H 5OH = 90. mL C2H5OH present 100 mL juice
90. mL C2H5OH ×
0.79 g C 2 H 5 OH 1 mol C 2 H 5 OH 2 mol CO 2 = 1.5 mol CO2 mL C 2 H 5 OH 46.07 g C 2 H 5 OH 2 mol C 2 H 5 OH
The CO2 will occupy (825 − 750. =) 75 mL not occupied by the liquid (headspace).
PCO 2 =
n CO 2 RT V
1.5 mol =
0.08206 L atm 298 K K mol = 490 atm 75 10 −3 L
Enough CO2 will dissolve in the wine to lower the pressure of CO2 to a much more reasonable value. 155.
Ptotal = PN 2 + PH 2O , PN 2 = 726 torr – 23.8 torr = 702 torr × n N2 =
PN 2 V RT
=
0.924 atm 31.8 10 −3 L 0.08206 L atm 298 K K mol
1 atm = 0.924 atm 760 torr
= 1.20 × 10−3 mol N2
CHAPTER 5
GASES
213
Mass of N in compound = 1.20 × 10−3 mol N2 × Mass % N = 156.
28.02 g N 2 = 3.36 × 10−2 g nitrogen mol
3.36 10 −2 g × 100 = 13.3% N 0.253 g
Volume of cylinder = length × area = 7.0 ft × (πr2) = 7.0 × π × (1.5)2 = 49 ft3 Volume available to gas in the cylinder = 0.320 × 49 = 16 ft3 3
3
1L 12 in 2.54 cm 16 ft × = 450 L 1000 cm3 ft in 3
For just the O2 gas in the compressed gas canister transferred to the hyperbaric chamber, n and T are constant so P1V1 = P2V2. Here, P2 = 2.50 − 1.00 = 1.50 atm of O2 must be added to reach a final pressure of 2.50 atm in the hyperbaric chamber. V1 = 157.
P2 V2 1.50 atm 450 L = 7.1 L O2(g) added from the compressed gas canister = P1 95 atm
We will apply Boyle’s law to solve. PV = nRT = constant, P1V1 = P2V2 Let condition (1) correspond to He from the tank that can be used to fill balloons. We must leave 1.0 atm of He in the tank, so P1 = 200. − 1.00 = 199 atm and V1 = 15.0 L. Condition (2) will correspond to the filled balloons with P2 = 1.00 atm and V2 = N(2.00 L), where N is the number of filled balloons, each at a volume of 2.00 L. 199 atm × 15.0 L = 1.00 atm × N(2.00 L), N = 1492.5; we can't fill 0.5 of a balloon, so N = 1492 balloons or, to 3 significant figures, 1490 balloons.
158.
Mol of He removed =
1.00 atm 1.75 10 −3 L PV = = 7.16 × 10 −5 mol 0.08206 L atm RT 298 K K mol
In the original flask, 7.16 × 10 −5 mol of He exerted a partial pressure of 1.960 − 1.710 = 0.250 atm. 7.16 10 −5 mol 0.08206 298 K nRT V= = = 7.00 × 10 −3 L = 7.00 mL 0.250 atm P
159.
1.00 × 103 kg Mo ×
1000 g 1 mol Mo = 1.04 × 104 mol Mo kg 95 .94 g Mo
1.04 × 104 mol Mo ×
VO 2 =
n O 2 RT P
1 mol MoO 3 7/2 mol O 2 = 3.64 × 104 mol O2 mol Mo mol MoO 3
3.64 10 4 mol
=
0.08206 L atm 290. K K mol = 8.66 × 105 L of O2 1.00 atm
214
CHAPTER 5 8.66 × 105 L O2 ×
100 L air 21 L O 2
mol Mo
3.12 10 4 mol
VH 2 =
160.
= 3.12 × 104 mol H2
0.08206 L atm 290. K K mol = 7.42 × 105 L of H2 1.00 atm
1.34 g = 209 g HSiCl3 = actual yield of HSiCl3 mL
a. 156 mL × nHCl =
= 4.1 × 106 L air
3 mol H 2
1.04 × 104 mol Mo ×
GASES
10 .0 atm 15 .0 L PV = 5.93 mol HCl = 0 . 08206 L atm RT 308 K K mol
5.93 mol HCl ×
1 mol HSiCl 3 135 .45 g HSiCl 3 = 268 g HSiCl3 3 mol HCl 1 mol HSiCl 3
Percent yield =
actual yield 209 g 100 = 100 = 78.0% theoretical yield 268 g
b. 209 g HiSCl3 ×
1 mol HSiCl 3 1 mol SiH 4 = 0.386 mol SiH4 135 .45 g HSiCl 3 4 mol HSiCl 3
This is the theoretical yield. If the percent yield is 93.1%, then the actual yield is: 0.386 mol SiH4 × 0.931 = 0.359 mol SiH4 VSiH4 =
161.
nRT = P
0.359 mol
0.08206 L atm 308 K K mol 10.0 atm
= 0.907 L = 907 mL SiH4
The partial pressures can be determined by using the mole fractions. Pmethane = Ptotal × methane = 1.44 atm × 0.915 = 1.32 atm; Pethane = 1.44 – 1.32 = 0.12 atm Determining the number of moles of natural gas combusted: nnatural gas =
PV 1.44 atm 15.00 L = = 0.898 mol natural gas 0.08206 L atm RT 293 K K mol
nmethane = nnatural gas × methane = 0.898 mol × 0.915 = 0.822 mol methane nethane = 0.898 − 0.822 = 0.076 mol ethane CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l); 2 C2H6 + 7 O2(g) → 4 CO2(g) + 6 H2O(l)
CHAPTER 5
GASES
215
0.822 mol CH4 ×
2 mol H 2 O 18 .02 g H 2 O = 29.6 g H2O 1 mol CH 4 mol H 2 O
0.076 mol C2H6 ×
6 mol H 2 O 18 .02 g H 2 O = 4.1 g H2O 2 mol C 2 H 6 mol H 2 O
The total mass of H2O produced = 29.6 g + 4.1 g = 33.7 g H2O. 162.
For NH3: P2 =
2 .00 L P1V1 = 0.500 atm × = 0.333 atm 3 .00 L V2
For O2: P2 =
1.00 L P1V1 = 1.50 atm × = 0.500 atm V2 3.00 L
After the stopcock is opened, V and T will be constant, so P n. Assuming NH3 is limiting: 0.333 atm NH3 Assuming O2 is limiting: 0.500 atm O2
4 atm NO = 0.333 atm NO 4 atm NH 3
4 atm NO = 0.400 atm NO 5 atm O 2
NH3 produces the smaller amount of product, so NH3 is limiting and 0.333 atm of NO can be produced. 163.
Out of 100.00 g of compound there are: 58.51 g C 7.37 g H
1 mol C 4.872 = 2.001 = 4.872 mol C; 12.01 g C 2.435
1 mol H 7.31 = 3.00 = 7.31 mol H; 1.008 g H 2.435
34.12 g N
1 mol N 14.01 g N
= 2.435 mol N;
2.435 = 1.000 2.435
The empirical formula is C2H3N. 1/2
M Rate1 M2 = 2 ; let gas (1) = He; 3.20 = , M 2 = 41.0 g/mol Rate 2 4.003 M1 1/2
The empirical formula mass of C2H3N 2(12.0) + 3(1.0) + 1(14.0) = 41.0 g/mol. So the molecular formula is also C2H3N. 164.
If Be3+, the formula is Be(C5H7O2)3 and molar mass 13.5 + 15(12) + 21(1) + 6(16) = 311 g/mol. If Be2+, the formula is Be(C5H7O2)2 and molar mass 9.0 + 10(12) + 14(1) + 4(16) = 207 g/mol. Data set I (molar mass = dRT/P and d = mass/V):
216
CHAPTER 5
GASES
0.08206 L atm 286 K mass RT K mol molar mass = = 209 g/mol = 1 atm PV −3 (765.2 torr ) ( 22.6 10 L) 760 torr 0.2022 g
Data set II: 0.08206 L atm 290. K mass RT K mol molar mass = = 202 g/mol = 1 atm PV −3 (764.6 torr ) ( 26.0 10 L) 760 torr 0.2224 g
These results are close to the expected value of 207 g/mol for Be(C5H7O2)2. Thus we conclude from these data that beryllium is a divalent element with an atomic weight (mass) of 9.0 u. 165.
12 .01 g C
0.2766 g CO2 ×
44 .01 g CO 2
0.0991 g H2O ×
2.016 g H 18 .02 g H 2 O
PV = nRT, n N 2 =
7.548 10 −2 g × 100 = 73.78% C 0.1023 g
= 1.11 × 10 −2 g H; % H =
1.11 10 −2 g × 100 = 10.9% H 0.1023 g
1.00 atm 27.6 10 −3 L PV = = 1.23 × 10 −3 mol N2 0.08206 L atm RT 273 K K mol
1.23 × 10 −3 mol N2 × Mass % N =
= 7.548 × 10 −2 g C; % C =
28 .02 g N 2 mol N 2
3.45 10 −2 g 0.4831 g
= 3.45 × 10 −2 g nitrogen
× 100 = 7.14% N
Mass % O = 100.00 − (73.78 + 10.9 + 7.14) = 8.2% O Out of 100.00 g of compound, there are: 73.78 g C × 10.9 g H ×
1 mol 12 .01 g 1 mol
1 .008 g
= 6.143 mol C; 7.14 g N × = 10.8 mol H; 8.2 g O ×
1 mol 14 .01 g
1 mol 16 .00 g
= 0.510 mol N
= 0.51 mol O
Dividing all values by 0.51 gives an empirical formula of C12H21NO. 4.02 g 0.08206 L atm 400. K dRT L K mol Molar mass = = = 392 g/mol 1 atm P 256 torr 760 torr
Empirical formula mass of C12H21NO 195 g/mol; Thus the molecular formula is C24H42N2O2.
392 2 195
CHAPTER 5 166.
GASES
217
PV = nRT, V and T are constant.
P1 P P n = 2 or 1 = 1 n1 n2 P2 n2
When V and T are constant, then pressure is directly proportional to moles of gas present, and pressure ratios are identical to mole ratios. At 125°C: 2 H2(g) + O2(g) → 2 H2O(g); all substances are gases at 125°C. The balanced equation requires 2 mol H2 for every mol O2 reacted. The same ratio (2 : 1) holds true for pressure units. So if all 2.00 atm of H2 react, only 1.00 atm of O2 will react with it. Because we have 3.00 atm of O2 present, oxygen is in excess and hydrogen is the limiting reactant. PO 2 (reacted) = 2.00 atm H2 ×
1 atm O 2 = 1.00 atm O2 2 atm H 2
PO 2 (excess) = PO 2 (initial) − PO 2 (reacted) = 3.00 atm − 1.00 atm = 2.00 atm O2 = Ptotal
We also have the H2O(g) that was produced by the reaction. PH 2O (produced) = 2.00 atm H2 ×
2 atm H 2 O = 2.00 atm H2O 2 atm H 2
Ptotal = PO 2 (excess) + PH 2O (produced) = 2.00 atm O2 + 2.00 atm H2O = 4.00 atm = Ptotal 167.
All the gases have the same average kinetic energy since they are all at the same temperature [KEavg = (3/2)RT]. At constant T, the lighter the gas molecule, the faster the average velocity [avg rms (1/M)1/2]. The average velocity order is: F2 (38.00 g/mol) < N2 (28.02 g/mol) < He (4.003 g/mol) slowest fastest
168.
33.5 mg CO2 ×
9.14 mg 12 .01 mg C = 9.14 mg C; % C = × 100 = 26.1% C 35 .0 mg 44 .01 mg CO 2
41.1 mg H2O ×
4.60 mg 2.016 mg H = 4.60 mg H; % H = × 100 = 13.1% H 18 .02 mg H 2 O 35 .0 mg
n N2 =
PN 2 V RT
740 . atm 35 .6 10 −3 L 760 = 1.42 × 10−3 mol N2 = 0.08206 L atm 298 K K mol
1.42 × 10-3 mol N2 ×
Mass % N =
28 .02 g N 2 = 3.98 × 10−2 g nitrogen = 39.8 mg nitrogen mol N 2
39 .8 mg × 100 = 61.0% N 65 .2 mg
Or we can get % N by difference: % N = 100.0 − (26.1 + 13.1) = 60.8%
218
CHAPTER 5
GASES
Out of 100.0 g: 26.1 g C ×
1 mol = 2.17 mol C; 12 .01 g
13.1 g H ×
13.0 1 mol = 13.0 mol H; = 5.99 1.008 g 2.17
60.8 g N ×
1 mol 4.34 = 4.34 mol N; = 2.00; empirical formula is CH6N2. 14 .01 g 2.17 1/ 2
Rate1 M = Rate 2 39.95
=
2.17 = 1.00 2.17
26.4 = 1.07, M = (1.07)2 × 39.95 = 45.7 g/mol 24.6
Empirical formula mass of CH6N2 12 + 6 + 28 = 46 g/mol. Thus the molecular formula is also CH6N2. 169.
Statements a-c are false. For statement a, F2, with the smaller molar mass, will have the faster average velocity, so the F2 molecules will collide with the container walls more frequently. For statement b, both gas samples are at the same P, V, and T, so n (the moles of gas present) must be equal. For statement c, because both gas samples are at the same T, they will have the same average kinetic energy. Statement d is true because the force of collisions is proportional to the mass of the gas particles. The more massive SO2 molecules will collide with the container walls more forcefully, but with less frequency than the F2 collisions.
170.
At constant T, the lighter the gas molecules, the faster the average velocity. Therefore, the pressure will increase initially because the lighter H2 molecules will effuse into container A faster than air will escape. However, the pressures will eventually equalize once the gases have had time to mix thoroughly.
Challenge Problems 171.
BaO(s) + CO2(g) → BaCO3(s); CaO(s) + CO2(g) → CaCO3(s) 750 . atm 1.50 L 760 ni = = initial moles of CO2 = = 0.0595 mol CO2 0.08206 L atm RT 303 .2 K K mol
Pi V
230 . atm 1.50 L 760 nf = = final moles of CO2 = = 0.0182 mol CO2 0.08206 L atm RT 303 .2 K K mol
Pf V
0.0595 − 0.0182 = 0.0413 mol CO2 reacted Because each metal reacts 1 : 1 with CO2, the mixture contains a total of 0.0413 mol of BaO and CaO. The molar masses of BaO and CaO are 153.3 and 56.08 g/mol, respectively.
CHAPTER 5
GASES
219
Let x = mass of BaO and y = mass of CaO, so: x + y = 5.14 g and
x y = 0.0413 mol or x + (2.734)y = 6.33 + 153.3 56.08
Solving by simultaneous equations: x + (2.734)y = 6.33 −x −y = −5.14 (1.734)y = 1.19, y – 1.19/1.734 = 0.686 y = 0.686 g CaO and 5.14 − y = x = 4.45 g BaO Mass % BaO = 172.
4.45 g BaO × 100 = 86.6% BaO; %CaO = 100.0 − 86.6 = 13.4% CaO 5.14 g
Cr(s) + 3 HCl(aq) → CrCl3(aq) + 3/2 H2(g); Zn(s) + 2 HCl(aq) → ZnCl2(aq) + H2(g) 1 atm 750 . torr 0.225 L 760 torr PV = = 9.02 × 10−3 mol H2 Mol H2 produced = n = 0.08206 L atm RT ( 273 + 27 ) K K mol
9.02 × 10−3 mol H2 = mol H2 from Cr reaction + mol H2 from Zn reaction From the balanced equation: 9.02 × 10−3 mol H2 = mol Cr × (3/2) + mol Zn × 1 Let x = mass of Cr and y = mass of Zn, then: x + y = 0.362 g and 9.02 × 10−3 =
(1.5) x y + 52.00 65.38
We have two equations and two unknowns. Solving by simultaneous equations: 9.02 × 10−3 = (0.02885)x + (0.01530)y −0.01530 × 0.362 = −(0.01530)x − (0.01530)y 3.48 × 10−3 = (0.01355)x,
x = mass of Cr =
y = mass of Zn = 0.362 g − 0.257 g = 0.105 g Zn; mass % Zn = 173.
3.48 10 −3 = 0.257 g 0.01355
0 .105 g × 100 = 29.0% Zn 0 .362 g
Assuming 1.000 L of the hydrocarbon (CxHy), then the volume of products will be 4.000 L, and the mass of products (H2O + CO2) will be: 1.391 g/L × 4.000 L = 5.564 g products Mol CxHy = n C x H y =
0.959 atm 1.000 L PV = = 0.0392 mol 0 . 08206 L atm RT 298 K K mol
220
CHAPTER 5 Mol products = np =
GASES
1.51 atm 4.000 L PV = = 0.196 mol 0 . 08206 L atm RT 375 K K mol
CxHy + oxygen → x CO2 + y/2 H2O Setting up two equations: (0.0392)x + 0.0392(y/2) = 0.196
(moles of products)
(0.0392)x(44.01 g/mol) + 0.0392(y/2)(18.02 g/mol) = 5.564 g
(mass of products)
Solving: x = 2 and y = 6, so the formula of the hydrocarbon is C2H6. 174.
a.
Let x = moles SO2 = moles O2 and z = moles He. P • MM , where MM = molar mass RT
1.924 g/L =
1.000 atm MM , MMmixture = 43.13 g/mol 0.08206 L atm 273 .2 K K mol
Assuming 1.000 total moles of mixture is present, then: x + x + z = 1.000 and: 64.07 g/mol × x + 32.00 g/mol × x + 4.003 g/mol × z = 43.13 g 2x + z = 1.000 and (96.07)x + (4.003)z = 43.13 Solving: x = 0.4443 mol and z = 0.1114 mol Thus: χHe = 0.1114 mol/1.000 mol = 0.1114 b.
2 SO2(g) + O2(g) → 2 SO3(g) Initially, assume 0.4443 mol SO2, 0.4443 mol O2, and 0.1114 mol He. Because SO2 is limiting, we end up with 0.2222 mol O2, 0.4443 mol SO3, and 0.1114 mol He in the gaseous product mixture. This gives ninitial = 1.0000 mol and nfinal = 0.7779 mol. In a reaction, mass is constant. d =
mass 1 and V n at constant P and T, so d . V n
n initial 1.0000 d 1.0000 = = final , d final = × 1.924 g/L, dfinal = 2.473 g/L n final 0.7779 d initial 0.7779
175.
a. The reaction is CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g). PV = nRT,
PCH 4 VCH 4 P V PV = air air = RT = constant, n CH 4 n air n
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221
The balanced equation requires 2 mol O2 for every mole of CH4 that reacts. For three times as much oxygen, we would need 6 mol O2 per mole of CH4 reacted (n O 2 = 6n CH 4 ). Air is 21% mole percent O2, so n O 2 = (0.21)nair. Therefore, the moles of air we would need to deliver the excess O2 are: n air n O 2 = (0.21)nair = 6n CH 4 , nair = 29n CH 4 , = 29 n CH 4 In 1 minute: Vair = VCH 4
PCH 4 n air 1.50 atm = 200. L 29 = 8.7 × 103 L air/min n CH 4 Pair 1.00 atm
b. If x mol of CH4 were reacted, then 6x mol O2 were added, producing (0.950)x mol CO2 and (0.050)x mol of CO. In addition, 2x mol H2O must be produced to balance the hydrogens. CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g); CH4(g) + 3/2 O2(g) → CO(g) + 2 H2O(g) Amount O2 reacted: (0.950)x mol CO2 ×
2 mol O 2 = (1.90)x mol O2 mol CO 2
(0.050)x mol CO ×
1.5 mol O 2 = (0.075)x mol O2 mol CO
Amount of O2 left in reaction mixture = (6.00)x − (1.90)x − (0.075)x = (4.03)x mol O2 Amount of N2 = (6.00)x mol O2 ×
79 mol N 2 = (22.6)x 23x mol N2 21 mol O 2
The reaction mixture contains: (0.950)x mol CO2 + (0.050)x mol CO + (4.03)x mol O2 + (2.00)x mol H2O + 23x mol N2 = (30.)x mol of gas total
176.
χ CO =
(0.050 ) x ( 4.03 ) x (0.950 ) x = 0.13 = 0.0017; χ CO 2 = = 0.032; χ O 2 = (30 .) x (30 .) x (30 .) x
χ H 2O =
( 2.00 ) x = 0.067; (30 .) x
χ N2 =
23 x = 0.77 (30 .) x
The reactions are: C(s) + 1/2 O2(g) → CO(g) and C(s) + O2(g) → CO2(g) RT = n(constant) V
PV = nRT, P = n
Because the pressure has increased by 17.0%, the number of moles of gas has also increased by 17.0%.
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GASES
nfinal = (1.170)ninitial = 1.170(5.00) = 5.85 mol gas = n O 2 + n CO + n CO 2 n CO + n CO 2 = 5.00 (balancing moles of C). Solving by simultaneous equations:
n O 2 + n CO + n CO 2 = 5.85 − ( n CO + n CO 2 = 5.00)
n O2
= 0.85
If all C were converted to CO2, no O2 would be left. If all C were converted to CO, we would get 5 mol CO and 2.5 mol excess O2 in the reaction mixture. In the final mixture, moles of CO equals twice the moles of O2 present ( n CO = 2n O 2 ). n CO = 2n O 2 = 1.70 mol CO; 1.70 + n CO 2 = 5.00, n CO 2 = 3.30 mol CO2 χ CO =
177.
1.70 3.30 = 0.291; χ CO 2 = = 0.564; 5.85 5.85
a. Volume of hot air: V =
χ O2 =
0.85 = 0.145 0.15 5.85
4 3 4 πr = π( 2.50 m) 3 = 65.4 m3 3 3
(Note: Radius = diameter/2 = 5.00/2 = 2.50 m) 3
1L 10 dm 65.4 m × = 6.54 × 104 L dm 3 m 3
1 atm 745 torr 6.54 10 4 L 760 torr PV = n= = 2.31 × 103 mol air 0 . 08206 L atm RT ( 273 + 65 ) K K mol
Mass of hot air = 2.31 × 103 mol ×
29.0 g = 6.70 × 104 g mol
745 atm 6.54 10 4 L PV 760 Air displaced: n = = 2.66 × 103 mol air = 0.08206 L atm RT ( 273 + 21) K K mol
Mass of air displaced = 2.66 × 103 mol ×
29.0 g = 7.71 × 104 g mol
Lift = 7.71 × 104 g − 6.70 × 104 g = 1.01 × 104 g b. Mass of air displaced is the same, 7.71 × 104 g. Moles of He in balloon will be the same as moles of air displaced, 2.66 × 103 mol, because P, V, and T are the same. Mass of He = 2.66 × 103 mol ×
4.003 g = 1.06 × 104 g mol
CHAPTER 5
GASES
223
Lift = 7.71 × 104 g − 1.06 × 104 g = 6.65 × 104 g 630 . atm (6.54 10 4 L ) PV 760 c. Hot air: n = = 1.95 × 103 mol air = 0.08206 L atm RT 338 K K mol
29.0 g = 5.66 × 104 g of hot air mol 630 . atm (6.54 10 4 L ) PV 760 Air displaced: n = = 2.25 × 103 mol air = 0.08206 L atm RT 294 K K mol
1.95 × 103 mol ×
2.25 × 103 mol ×
29.0 g = 6.53 × 104 g of air displaced mol
Lift = 6.53 × 104 g − 5.66 × 104 g = 8.7 × 103 g 178.
a. We assumed a pressure of 1.0 atm and a temperature of 25C (298 K). 50. lb 0.454 kg/lb = 23 kg n=
PV 1.0 atm 10 . L = = 0.41 mol gas 0.08206 L atm RT 298 K K mol
The lift of one balloon is: 0.41 mol(29 g/mol − 4.003 g/mol) = 10. g. To lift 23 kg = 23,000 g, we need at least 23,000/10 = 2300 balloons. This is a lot of balloons. b. The balloon displaces air as it is filled. The displaced air has mass, as does the helium in the balloon, but the displaced air has more mass than the helium. The difference in this mass is the lift of the balloon. Because volume is constant, the difference in mass is directly related to the difference in density between air and helium. 179.
a.
Average molar mass of air = 0.790 × 28.02 g/mol + 0.210 × 32.00 g/mol = 28.9 g/mol Molar mass of helium = 4.003 g/mol A given volume of air at a given set of conditions has a larger density than helium at those conditions due to the larger average molar mass of air. We need to heat the air to a temperature greater than 25°C in order to lower the air density (by driving air molecules out of the hot air balloon) until the density is the same as that for helium (at 25°C and 1.00 atm).
b. To provide the same lift as the helium balloon (assume V = 1.00 L), the mass of air in the hot air balloon (V = 1.00 L) must be the same as that in the helium balloon. Let MM = molar mass:
224
CHAPTER 5 P•MM = dRT, mass =
Mass air = 0.164 g =
GASES
MM • PV ; solving: mass He = 0.164 g RT
28 .9 g/mol 1.00 atm 1.00 L 0.08206 L atm T K mol
T = 2150 K (a very high temperature) 180.
an 2 an 2 V an 3 b P + × (V − nb) = nRT, PV + = nRT − nbP − V2 V2 V 2
PV +
an 2 an 3 b = nRT − nbP − V V2
At low P and high T, the molar volume of a gas will be relatively large. Thus the an2/V and an3b/V2 terms become negligible at low P and high T because V is large. Because nb is the actual volume of the gas molecules themselves, nb << V and the −nbP term will be neg-ligible as compared to PV. Thus PV = nRT. 181.
d = molar mass(P/RT); at constant P and T, the density of gas is directly proportional to the molar mass of the gas. Thus the molar mass of the gas has a value which is 1.38 times that of the molar mass of O2. Molar mass = 1.38(32.00 g/mol) = 44.2 g/mol Because H2O is produced when the unknown binary compound is combusted, the unknown must contain hydrogen. Let AxHy be the formula for unknown compound. Mol AxHy = 10.0 g AxHy Mol H = 16.3 g H2O
1 mol A x H y 44.2 g
= 0.226 mol AxHy
1 mol H 2 O 2 mol H = 1.81 mol H 18 .02 g mol H 2 O
1.81 mol H = 8 mol H/mol AxHy; AxHy = AxH8 0.226 mol A x H y
The mass of the x moles of A in the AxH8 formula is: 44.2 g − 8(1.008 g) = 36.1 g From the periodic table and by trial and error, some possibilities for AxH8 are ClH8, F2H8, C3H8, and Be4H8. C3H8 and Be4H8 fit the data best, and because C3H8 (propane) is a known substance, C3H8 is the best possible identity from the data in this problem. 182.
a. Initially PN 2 = PH 2 = 1.00 atm, and the total pressure is 2.00 atm (Ptotal = PN 2 + PH 2 ). The total pressure after reaction will also be 2.00 atm because we have a constant-pressure container. Because V and T are constant before the reaction takes place, there must be
CHAPTER 5
GASES
225
equal moles of N2 and H2 present initially. Let x = mol N2 = mol H2 that are present initially. From the balanced equation, N2(g) + 3 H2(g) → 2 NH3(g), H2 will be limiting because three times as many moles of H2 are required to react as compared to moles of N2. After the reaction occurs, none of the H2 remains (it is the limiting reagent). Mol NH3 produced = x mol H2 × Mol N2 reacted = x mol H2 ×
2 mol NH 3 = 2x/3 3 mol H 2
1 mol N 2 = x/3 3 mol H 2
Mol N2 remaining = x mol N2 present initially − x/3 mol N2 reacted = 2x/3 mol N2 After the reaction goes to completion, equal moles of N2(g) and NH3(g) are present (2x/3). Because equal moles are present, the partial pressure of each gas must be equal ( PN 2 = PNH3 ). Ptotal = 2.00 atm = PN 2 + PNH3 ; solving: PN 2 = 1.00 atm = PNH3 b. V n because P and T are constant. The moles of gas present initially are: n N 2 + n H 2 = x + x = 2x mol
After reaction, the moles of gas present are: 2x 2x = 4x/3 mol + 3 3
n N 2 + n NH3 =
4 x/ 3 2 Vafter n = after = = Vinitial n initial 2x 3
The volume of the container will be two-thirds the original volume, so: V = 2/3(15.0 L) = 10.0 L 1/ 2
183.
M Rate1 = 2 Rate 2 M1
; let N2O = gas 1 and the lachrymator (tear gas) = gas 2: 1/2
Rate1 176 = Rate 2 44.02
= (4.00)1/2 = 2.00
The rate of effusion of N2O is twice the rate of effusion of the tear gas. So in a given amount of time, one would expect the N2O gas to travel twice as far as the tear gas. There are 9 rows difference between row 1 and row 10. Let x = rows N2O travels and y = rows the tear gas travels. Setting up two equations: x = 2.00 and x + y = 9 rows; solving: x = 6 and y = 3. y
So N2O travels from row 1 to row 7 and the tear gas travels from row 10 to row 7 where the two gases meet causing the row 7 students to simultaneously laugh and cry.
226
CHAPTER 5
GASES
Marathon Problem 184.
a. The formula of the compound AxBy depends on which gas is limiting, A2 or B2. We need to determine both possible products. The procedure we will use is to assume one reactant is limiting, and then determine what happens to the initial total moles of gas as it is converted into the product. Because P and T are constant, volume n. Because mass is conserved in a chemical reaction, any change in density must be due to a change in volume of the container as the reaction goes to completion. Density = d
d n 1 and V n, so: after = initial d initial n after V
Assume the molecular formula of the product is AxBy where x and y are whole numbers. First, let’s consider when A2 is limiting with x moles each of A2 and B2 in our equimolar mixture. Note that the coefficient in front of AxBy in the equation must be 2 for a balanced reaction. x A2(g) Initial Change After
+ y B2(g)
x mol −x mol 0
→
2 AxBy(g)
x mol −y mol (x − y) mol
0 mol +2 mol 2 mol
d after n 2x = 1.50 = initial = d initial n after x− y +2
(1.50)x − (1.50)y + 3.00 = 2x, 3.00 − (1.50)y = (0.50)x Because x and y are whole numbers, y must be 1 because the above equation does not allow y to be 2 or greater. When y = 1, x = 3 giving a formula of A3B if A2 is limiting. Assuming B2 is limiting with y moles in the equimolar mixture:
Initial Change After
x A2(g) y −x y−x
+ y B2(g) y −y 0
→
2 AxBy(g) 0 +2 2
density after n 2y = 1.50 = initial = density before n after y−x+2
Solving gives x = 1 and y = 3 for a molecular formula of AB3 when B2 is limiting. b. In both possible products, the equations dictated that only one mole of either A or B had to be present in the formula. Any number larger than 1 would not fit the data given in the problem. Thus the two formulas determined are both molecular formulas and not just empirical formulas.
CHAPTER 6 THERMOCHEMISTRY Review Questions 1.
Potential energy: Energy due to position or composition. Kinetic energy: Energy due to motion of an object. Path-dependent function: A property that depends on how the system gets from the initial state to the final state; a property that is path-dependent. State function: A property that is independent of the pathway. System: That part of the universe on which attention is to be focused. Surroundings: Everything in the universe surrounding a thermodynamic system.
2.
Plot a represents an exothermic reaction. In an exothermic process, the bonds in the product molecules are stronger (on average) than those in the reactant molecules. The net result is that the quantity of energy Δ(PE) is transferred to the surroundings as heat when reactants are converted to products. For an endothermic process, energy flows into the system from the surroundings as heat to increase the potential energy of the system. In an endothermic process, the products have higher potential energy (weaker bonds on average) than the reactants.
3.
First law of thermodynamics: The energy of universe is constant. A system can change its internal energy by flow of work, heat, or both (E = q + w). Whenever a property is added to the system from the surroundings, the sign is positive; whenever a property is added to the surroundings by the system, the sign is negative.
4.
As a gas expands, the system does work on the surroundings so w is negative. When a gas contracts, the surroundings do work on the system so w is positive. H 2O(l) → H2O(g); To boil water, heat must be added so q is positive. The molar volume of a gas is huge compared to the molar volume of a liquid. As a liquid converts to a gas, the system will expand its volume, performing work on the surroundings; w is negative.
5.
qP = ΔH; qV = ΔE; a coffee-cup calorimeter is at constant (atmospheric) pressure. The heat released or gained at constant pressure is ΔH. A bomb calorimeter is at constant volume. The heat released or gained at constant volume is ΔE.
6.
The specific heat capacities are 0.89 J/°C•g (Al) and 0.45 J/°C•g (Fe). Al would be the better choice. It has a higher heat capacity and a lower density than Fe. Using Al, the same amount of heat could be dissipated by a smaller mass, keeping the mass of the amplifier down.
227
228
CHAPTER 6
THERMOCHEMISTRY
7.
In calorimetry, heat flow is determined into or out of the surroundings. Because ΔEuniv = 0 by the first law of thermodynamics, ΔEsys = −ΔEsurr; what happens to the surroundings is the exact opposite of what happens to the system. To determine heat flow, we need to know the heat capacity of the surroundings, the mass of the surroundings that accepts/donates the heat, and the change in temperature. If we know these quantities, qsurr can be calculated and then equated to qsys (−qsurr = qsys). For an endothermic reaction, the surroundings (the calorimeter contents) releases heat to the system. This is accompanied by a decrease in temperature of the surroundings. For an exothermic reaction, the system releases heat to the surroundings (the calorimeter) so temperature of the calorimeter contents increases.
8.
Hess’s law: In going from a particular set of reactants to a particular set of products, the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps (ΔH is path independent). When a reaction is reversed, the sign of ΔH is also reversed but the magnitude is the same. If the coefficients in a balanced reaction are multiplied by a number, the value of ΔH is multiplied by the same number while the sign is unaffected.
9.
Standard enthalpy of formation: The change in enthalpy that accompanies the formation of one mole of a compound from its elements with all substances in their standard states. The standard state for a compound has the following conventions: a. gaseous substances are at a pressure of exactly 1 atm. b. for a pure substance in a condensed state (liquid or solid), the standard state is the pure liquid or solid. c. for a substance present in solution, the standard state is a concentration of exactly 1 M. The standard state of an element is the form in which the element exists under conditions of 1 atm and 25˚C. ΔH fo values for elements in their standard state are, by definition, equal to zero. Step 1: reactants → elements in standard states
ΔH 1 = − nr ΔH fo (reactants)
Step 2: elements in standard state → products
ΔH 2 = np ΔH fo (products)
_________________________________________________________________________________________________
reactants → products
o H reaction = H 1 + H 2
o So: H reaction = np ΔH fo (products) − nr ΔH fo (reactants)
10.
Three problems are there is only a finite amount of fossil fuels, fossil fuels can be expensive, and the combustion and exploration of fossil fuels can add pollution to the biosphere whose effects may not be reversible.
Active Learning Questions 1.
Metals are good heat conductors. When you touch a piece of metal, heat easily transfers from your hand to the metal. As heat leaves your finger, your body senses this as a temperature drop and feels cold. Plastics are heat insulators; they do not readily transfer energy hence they do not feel cold to the touch.
CHAPTER 6 2.
THERMOCHEMISTRY
229
Lower in energy generally refers to the potential energy of products as compared to reactants in a reaction (process). Reactions (processes) that release energy are generally favorable. So lower in energy generally refers to the final state of the process; if energy is released, the system becomes lower in energy which is generally considered to be more stable. Liquid water is lower in energy than a mixture of gaseous H 2 and O2. When H2 and O2 are reacted, they spontaneously form water; the reverse reaction does not occur unless energy is added to water from some outside source. So water is lower in energy and more stable than a mixture of gaseous H2 and O2. Note that when H2 and O2 react to form water, this process is exothermic and energy is released.
3.
In an exothermic reaction, some of the potential energy stored in bonds is converted to thermal energy (random kinetic energy) via heat. In an endothermic reaction, energy that flows into the system as heat is used to increase the potential energy of the system. The thermal energy from the surroundings accompanies a decrease in kinetic energy due to the slowing down of the surroundings molecules.
4.
Most of these processes are exothermic combustion reactions where chemical potential energy stored in bonds is converted into thermal energy (heat). The system is the match or the crumpled paper or the log plus any other reactants necessary for the various combustion reactions [O2(g) mainly]. The surroundings are everything else. The added thermal energy to the surroundings from the combustion reactions goes to increase the random kinetic energy of the surrounding molecules by increasing their motion. The chemistry of lighting a match is complicated. In some matches, red phosphorus on a match box is struck by the match. The friction from the strike initiates the conversion of red phosphorus to white phosphorus. The white phosphorus ignites in air, which then initiates the decomposition reaction of KClO3 in the matchstick, and the match lights. Most of these reactions are exothermic. However, the conversion of red phosphorus to white phosphorus is an endothermic reaction. Here energy from the surroundings is required convert red phosphorus to white phosphorus; this energy from the surroundings goes to increase the potential energy of the system.
5.
H2O(l) → H2O(s) is an exothermic process as heat is released in this process. To freeze water, one needs to reduce the kinetic energy of the water molecules so they can “stick” together to form ice. As ice molecules “stick” together, the potential energy of the system decreases energy. This loss in potential energy is converted to thermal energy as heat is released from the system to the surroundings.
6.
The energy released by exothermic reactions comes from the potential energy stored in bonds. In an exothermic reaction, reactant molecules with weaker bonds are converted to product molecules having stronger bonds. The difference in bond energies is then released to the surroundings as thermal energy (heat). In endothermic reaction, reactants with stronger bonds are converted to products having weaker bonds. The thermal energy added from the surroundings goes to increase the potential energy of the product molecules. In general, substances with weak bonds have a high potential energy while substances with stronger bonds have a low potential energy.
230
CHAPTER 6
THERMOCHEMISTRY
7.
The hot metal when added to cooler water will cool down, while the water heats up. When the hot metal cools down, energy is released so this is an exothermic process. The water gains energy, so this is an endothermic process for the water.
8.
a. w = −PΔV = −1.50 atm(10.0 L − 5.0 L) = −7.5 L atm b. step 1: w1 = −PΔV = −2.0 atm(7.5 L − 5.0 L) = −5.0 L atm step 2: w2 = −1.5 atm(10.0 L − 7.5 L) = 3.75 L atm = −3.8 L atm wtotal = w1 + w2 = −5.0 L atm + (−3.8 L atm) = −8.8 L atm
9.
a. From the previous problem, the expansion of a gas from 5.0 L at 3.0 atm to 10.0 L at 1.5 atm was done in two different pathways. If work was a state function, then the amount of work produced by the expansion from either pathway would be the same value. But the amount of work produced were different values for the two different pathways. Work for a process depends on the pathway to get from the initial state to the final state. Hence work is not a state function because its value depends on the path one takes to get from the initial state to the final state. b.
w = −PΔV; work depends on the pressure the expansion (or compression) occurs against. The larger the pressure an expansion occurs against, the more work that is done. Hence in a multiple step expansion, it is best to do the process in steps as the pressure decreases slowly from the higher initial pressure to the lower final pressure.
10.
If energy were not conserved, calorimetry would not work to determine enthalpy changes for reactions, Hess’s law would not hold true, and enthalpy of formation values would be impossible to determine. There would be no way to determine energy changes for a process. Our understanding of happenings in the universe would change drastically.
11.
If Hess’s law were not true, it would be possible to create (or lose) energy by first performing a reaction then reversing the reaction back to the initial state using a different series of steps. This violates the first law of thermodynamics; for a cyclic process where the initial and final states are identical, the overall change in energy must be zero (energy must be conserved).
12.
w = −PΔV; when a gas expands against some constant pressure, the expanding gas is doing work on the surroundings. This has a negative sign associated with this work. In an expansion, ΔV is a positive quantity. A minus sign must be included to make the correct overall sign for the work. Note in a compression, the surroundings perform work on the system; this is a positive value for work. In a compression, ΔV is negative, so again, the minus sign is necessary to make the correct overall sign for the work.
13.
Aluminum is a lightweight metal strong enough to hold carbonated beverages and at the same time, keep the container weight low. Also, aluminum does not easily react with outside elements or with the beverage inside the container. Other factors are that Al is a very abundant metal and cans produced from aluminum are easily recycled. Finally, thin aluminum cans allow heat transfer from the fluid inside the can to the surroundings. This is important when we are trying to cool soft drinks in a refrigerator.
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231
An insulator does not easily allow the transfer of heat. Styrofoam is a good insulator because it will keep hot drinks warm for a longer period. It does this by reducing the amount of heat transfer from the hot coffee to the colder surroundings. Cold drinks stay colder for a longer period because the Styrofoam container reduces the amount of heat transfer from the warmer surroundings to the cold drink. 14.
a. With equal quantities of hot and cold water, the expected temperature after mixing should be 50°C (ii). This of course assumes no heat transfer to the surroundings. b. With more cold water than hot water, the expected final temperature would be closer to that of the cold water (iii). c. Iron has a lower heat capacity than water. Therefore, 50.0 g of hot iron will change temperature more dramatically than 50.0 g of cold water when they are added together. The final temperature will be closer to the cold water (iii).
15.
The total energy content of the universe is conserved. Energy can be transferred between a system and the surroundings through a temperature difference (called heat) and through work (a force acting over a distance). Heat and work are not individually conserved, but the combination of the two to determine energy changes is conserved. We often refer to heat as if it is contained by a system; it is not a substance but is just a means by which system and surroundings can transfer energy. If energy is transferred between two substances due to a temperature difference and not by work, then the heat loss of one of the substances must equal the heat gain of the other.
16.
a. w = −PΔV; from the ideal gas law V = nRT/P, so ΔV = Δ(nRT/P) At constant T and P, Δ(nRT/P) = (RT/P) × (Δn). Substituting this into the work equation: w = −PΔV = ‒P[(RT/P) × Δn)] = ‒RTΔn This equation is very useful to calculate work when given a chemical reaction. In a chemical equation, Δn = moles of gaseous products ‒ moles of gaseous reactants. We generally want work in units of J, so the best R to use is R = 8.3145 J/K mol. b. ΔH° = np ΔHf , products − nr ΔHf , reactants with all elements in their standard state have
ΔH f = 0 by definition. In this problem, O2(g) is the standard state for oxygen, so O2(g) has ΔH f = 0. ΔH° = (4 mol NO2 × ΔH of , NO2 + 1 mol O2(g) × ΔH of , O2 ) − (2 mol N2O5 × ΔH of, N2O5 ) ΔH° = [4 mol(33.2 kJ/mol) + 1 mol(0)] ‒ [2 mol(11.3 kJ/mol)] = 110.2 kJ Because pressure is constant at 1.00 atm, ΔH° = q = 110.2 kJ. w = −PΔV = ‒RTΔn; from the balanced equation, Δn = (4 + 1) ‒ (2) = 3 mol. w = ‒RTΔn = ‒8.3145 J/K mol(298 K)(3 mol) = ‒7430 J = ‒7.43 kJ ΔE = q + w = 110.2 kJ + (−7.43 kJ) = 102.8 kJ
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Questions 17.
Path-dependent functions for a trip from Chicago to Denver are those quantities that depend on the route taken. One can fly directly from Chicago to Denver, or one could fly from Chicago to Atlanta to Los Angeles and then to Denver. Some path-dependent quantities are miles traveled, fuel consumption of the airplane, time traveling, airplane snacks eaten, etc. State functions are path-independent; they only depend on the initial and final states. Some state functions for an airplane trip from Chicago to Denver would be longitude change, latitude change, elevation change, and overall time zone change.
18.
Products have a lower potential energy than reactants when the bonds in the products are stronger (on average) than in the reactants. This occurs generally in exothermic processes. Products have a higher potential energy than reactants when the reactants have the stronger bonds (on average). This is typified by endothermic reactions.
19.
2 C8H18(l) + 25 O2(g) → 16 CO2(g) + 18 H2O(g); the combustion of gasoline is exothermic (as is typical of combustion reactions). For exothermic reactions, heat is released into the surroundings giving a negative q value. To determine the sign of w, concentrate on the moles of gaseous reactants versus the moles of gaseous products. In this reaction, we go from 25 moles of reactant gas molecules to 16 + 18 = 34 moles of product gas molecules. As reactants are converted to products, an expansion will occur because the moles of gas increase. When a gas expands, the system does work on the surroundings, and w is a negative value.
20.
Only statement a is true. The law of conservation of energy (often called the first law of thermodynamics) tells us that energy can be interconverted from one form to another (between potential and kinetic energy) but can be neither created nor destroyed. That is, the energy of the universe is constant. Heat and work are two ways to transfer energy, but they are not constant. Their values are dependent on path, whereas ΔE changes are independent of path.
21.
ΔE = q + w; only statement d is false. When a gas expands, the system is doing work on the surroundings. For statement a, when work is done by the system, this has a negative sign, and when heat is flowing into a system, this has a positive sign. So, when more work is done (‒) than heat flowing in (+), ΔE will decrease. For statement b, at constant volume where only PV work is allowed, w = 0, so ΔE = q. For statement c, ΔE will increase when heat and work are both added to the system.
22.
H = E + PV at constant P; from the definition of enthalpy, the difference between H and E, at constant P, is the quantity PV. Thus, when a system at constant P can do pressurevolume work, then H ≠ E. When the system cannot do PV work, then H = E at constant pressure. An important way to differentiate H from E is to concentrate on q, the heat flow; the heat flow by a system at constant pressure equals H, and the heat flow by a system at constant volume equals E.
23.
Water has a relatively large heat capacity, so it takes a lot of energy to increase the temperature of a large body of water. Because of this, the temperature fluctuations of a large body of water (oceans) are small compared to the temperatures fluctuations of air. Hence, oceans act as a heat reservoir for areas close to them which results in smaller temperature changes as compared to areas farther away from the oceans.
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233
24.
A coffee-cup calorimeter is at constant (atmospheric) pressure. The heat released or gained for a reaction at constant pressure is equal to the enthalpy change (∆H) for that reaction. A bomb calorimeter is at constant volume. The heat released or gained for a reaction at constant volume is equal to the internal energy change (∆E) for that reaction.
25.
Given: CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l) CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g)
H = −891 kJ H = −803 kJ
Using Hess’s law: H2O(l) + 1/2 CO2(g) → 1/2 CH4(g) + O2(g) 1/2 CH4(g) + O2(g) → 1/2 CO2(g) + H2O(g) H2O(l) → H2O(g)
H1 = −1/2(−891 kJ) H2 = 1/2(−803 kJ) H = H1 + H2 = 44 kJ
The enthalpy of vaporization of water is 44 kJ/mol. Note: When an equation is reversed, the sign on ΔH is reversed. When the coefficients in a balanced equation are multiplied by an integer, then the value of ΔH is multiplied by the same integer. 26.
A state function is a function whose change depends only on the initial and final states and not on how one got from the initial to the final state. An extensive property depends on the amount of substance. Enthalpy changes for a reaction are path-independent, but they do depend on the quantity of reactants consumed in the reaction. Therefore, enthalpy changes are a state function and an extensive property.
27.
The zero point for ΔH of values are elements in their standard state. All substances are measured in relationship to this zero point.
28.
a. CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l)
H = ?
Utilizing Hess’s law: Reactants → Standard State Elements H = Ha + Hb = 75 + 0 = 75 kJ Standard State Elements → Products H = Hc + Hd = –394 – 572 = –966 kJ Reactants → Products
H = 75 – 966 = –891 kJ
b. The standard enthalpy of formation for an element in its standard state is given a value of zero. To assign standard enthalpy of formation values for all other substances, there needs to be a reference point from which all enthalpy changes are determined. This reference point is the elements in their standard state which is defined as the zero point. So when using standard enthalpy values, a reaction is broken up into two steps. The first step is to calculate the enthalpy change necessary to convert the reactants to the elements in their standard state. The second step is to determine the enthalpy change that occurs when the elements in their standard state go to form the products. When these two steps are added
234
CHAPTER 6
THERMOCHEMISTRY
together, the reference point (the elements in their standard state) cancels out and we are left with the enthalpy change for the reaction. c. This overall reaction is just the reverse of all the steps in the part a answer. So H = +966 – 75 = 891 kJ. Products are first converted to the elements in their standard state which requires 966 kJ of heat. Next, the elements in the standard states go to form the original reactants [CH4(g) + 2 O2(g)] which has an enthalpy change of −75 kJ. All of the signs are reversed because the entire process is reversed. 29.
No matter how insulated your thermos bottle, some heat will always escape into the surroundings. If the temperature of the thermos bottle (the surroundings) is high, less heat initially will escape from the coffee (the system); this results in your coffee staying hotter for a longer period of time.
30.
From the photosynthesis reaction, CO2(g) is used by plants to convert water into glucose and oxygen. If the plant population is significantly reduced, not as much CO2 will be consumed in the photosynthesis reaction. As the CO2 levels of the atmosphere increase, the greenhouse effect due to excess CO2 in the atmosphere will become worse.
31.
Fossil fuels contain carbon; the incomplete combustion of fossil fuels produces CO(g) instead of CO2(g). This occurs when the amount of oxygen reacting is not sufficient to convert all the carbon to CO2. Carbon monoxide is a poisonous gas to humans.
32.
Advantages: H2 burns cleanly (less pollution) and gives a lot of energy per gram of fuel. Water as a source of hydrogen is abundant and cheap. Disadvantages: Expensive and gas storage and safety issues
Exercises Potential and Kinetic Energy 33.
1 kg m 2 1 2 KE = mv ; convert mass and velocity to SI units. 1 J = 2 s2
Mass = 5.25 oz ×
Velocity =
KE =
34.
1 lb 1 kg = 0.149 kg 16 oz 2.205 lb
1.0 10 2 mi 1h 1 min 1760 yd 1m 45 m = h 60 min 60 s mi 1.094 yd s
1 1 45 m mv2 = × 0.149 kg × 2 2 s
1 1 1.0 m KE = mv2 = × 2.0 kg × 2 2 s
2
2
= 150 J 2
1 1 2.0 m = 1.0 J; KE = mv2 = × 1.0 kg × 2 2 s = 2.0 J The 1.0-kg object with a velocity of 2.0 m/s has the greater kinetic energy.
CHAPTER 6 35.
THERMOCHEMISTRY
235
a. Potential energy is energy due to position. Initially, ball A has a higher potential energy than ball B because the position of ball A is higher than the position of ball B. In the final position, ball B has the higher position so ball B has the higher potential energy. b. As ball A rolled down the hill, some of the potential energy lost by A has been converted to random motion of the components of the hill (frictional heating). The remainder of the lost potential energy was added to B to initially increase its kinetic energy and then to increase its potential energy.
36.
Ball A: PE = mgz = 2.00 kg ×
9.81 m 196 kg m 2 × 10.0 m = = 196 J s2 s2
At point I: All this energy is transferred to ball B. All of B's energy is kinetic energy at this point. Etotal = KE = 196 J. At point II, the sum of the total energy will equal 196 J. At point II: PE = mgz = 4.00 kg ×
9.81 m × 3.00 m = 118 J s2
KE = Etotal − PE = 196 J − 118 J = 78 J
Heat and Work 37.
ΔE = q + w = 45 kJ + (−29 kJ) = 16 kJ
38.
a. ΔE = q + w = −47 kJ + 88 kJ = 41 kJ b. ΔE = 82 − 47 = 35 kJ
c. ΔE = 47 + 0 = 47 kJ
d. When the surroundings do work on the system, w > 0. This is the case for a. 39.
Step 1: ΔE1 = q + w = 72 J + 35 J = 107 J; step 2: ΔE2 = 35 J − 72 J = −37 J ΔEoverall = ΔE1 + ΔE2 = 107 J − 37 J = 70. J
40.
a. ΔE = q + w = −23 J + 100. J = 77 J b. w = −PΔV = −1.90 atm(2.80 L − 8.30 L) = 10.5 L atm ×
101 .3 J = 1060 J L atm
ΔE = q + w = 350. J + 1060 = 1410 J c. w = −PΔV = −1.00 atm(29.1 L − 11.2 L) = −17.9 L atm ×
101 .3 J = −1810 J L atm
ΔE = q + w = 1037 J − 1810 J = −770 J 41.
For step 1, heat is removed from the system and work is done on the system for a compression. ΔEstep 1 = −20 + 15 = −5 J. For step 2, heat is added to the system and work is done on the surroundings for an expansion. ΔEstep 2 = 25 + w2. For the overall process, ΔEoverall = ΔE1 + ΔE2.
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Because the initial and final states are the same in the overall process (a cyclic process), ΔEoverall = 0. Note: this is true for any state function in a cyclic process. 0 = ΔE1 + ΔE2 = −5 + (25 + w2), w2 = −20 J Statements a and c are true. For statement b, ΔE 2 = − ΔE1 = 5 J 42.
For step 1, ΔE1 = q + w = 45 J − 10J = 35 J (statement b is false). Because we have the same overall initial and final state, ΔEoverall = ΔE1 + ΔE2 = 0. So, ΔE2 = − ΔE1 = −35 J. Only statement d is true. For statement a and c, nothing was said about how step 2 was performed, only that the gas went back to the same initial state. From above, ΔE2 = q + w = −35 J. Any combination of q and w that combines to give −35 J for ΔE2 is fine. Work and heat are both path dependent quantities. Unless more information is given, values cannot be determined for work and heat in the second step.
43.
w = −PΔV; we need the final volume of the gas. Because T and n are constant, P1V1 = P2V2. V2 =
V1 P1 P2
=
10.0 L(15.0 atm) = 75.0 L 2.00 atm
w = −PΔV = −2.00 atm(75.0 L − 10.0 L) = −130. L atm ×
101 .3 J 1 kJ L atm 1000 J
= −13.2 kJ = work 44.
w = −210. J = −PΔV, −210 J = −P(25 L − 10. L), P = 14 atm
45.
In this problem, q = w = −950. J. −950. J ×
1 L atm = −9.38 L atm of work done by the gases 101 .3 J
w = −PΔV, −9.38 L atm = 46.
− 650 . atm × (Vf − 0.040 L), Vf − 0.040 = 11.0 L, Vf = 11.0 L 760
ΔE = q + w, −102.5 J = 52.5 J + w, w = −155.0 J ×
1 L atm = −1.530 L atm 101 .3 J
w = −PΔV, −1.530 L atm = −0.500 atm × ΔV, ΔV = 3.06 L ΔV = Vf – Vi, 3.06 L = 58.0 L − Vi, Vi = 54.9 L = initial volume 47.
20.8 J q = molar heat capacity × mol × ΔT = o × 39.1 mol × (38.0 − 0.0)°C = 30,900 J C mol = 30.9 kJ w = −PΔV = −1.00 atm × (998 L − 876 L) = −122 L atm × ΔE = q + w = 30.9 kJ + (−12.4 kJ) = 18.5 kJ
101 .3 J = −12,400 J = −12.4 kJ L atm
CHAPTER 6 48.
THERMOCHEMISTRY
237
H2O(g) → H2O(l); ΔE = q + w; q = −40.66 kJ; w = −PΔV Volume of 1 mol H2O(l) = 1.000 mol H2O(l) ×
18.02 g 1 cm3 = 18.1 cm3 = 18.1 mL mol 0.996 g
w = −PΔV = −1.00 atm × (0.0181 L − 30.6 L) = 30.6 L atm ×
101 .3 J = 3.10 × 103 J L atm = 3.10 kJ
ΔE = q + w = −40.66 kJ + 3.10 kJ = −37.56 kJ
Properties of Enthalpy 49.
This is an endothermic reaction, so heat must be absorbed in order to convert reactants into products. The high-temperature environment of internal combustion engines provides the heat.
50.
One should try to cool the reaction mixture or provide some means of removing heat because the reaction is very exothermic (heat is released). The H2SO4(aq) will get very hot and possibly boil unless cooling is provided.
51
a. Heat is absorbed from the water (it gets colder) as KBr dissolves, so this is an endothermic process. b. Heat is released as CH4 is burned, so this is an exothermic process. c. Heat is released to the water (it gets hot) as H2SO4 is added, so this is an exothermic process. d. Heat must be added (absorbed) to boil water, so this is an endothermic process.
52.
a. The combustion of gasoline releases heat, so this is an exothermic process. b. H2O(g) → H2O(l); heat is released when water vapor condenses, so this is an exothermic process. c. To convert a solid to a gas, heat must be absorbed, so this endothermic. d. When a bond forms between two atoms, energy is released, so this exothermic.
53.
When a gas is converted to a liquid, energy is released, so q is negative. The volume occupied by 1 mol of a liquid is very small compared to the volume of 1 mol of a gas. So, when a gas is converted into a liquid, we have a compression, which makes w positive.
54.
This process corresponds to breaking a bond. To break a bond, energy must be added. So, this is endothermic (q is positive). The moles of gas double as we break the F2 bond, so this will be an expansion and w is negative.
55.
4 Fe(s) + 3 O2(g) → 2 Fe2O3(s) ΔH = −1652 kJ; note that 1652 kJ of heat is released when 4 mol Fe reacts with 3 mol O2 to produce 2 mol Fe2O3.
238
CHAPTER 6 a. 4.00 mol Fe ×
THERMOCHEMISTRY
− 1652 kJ = −1650 kJ; 1650 kJ of heat released 4 mol Fe
b. 1.00 mol Fe2O3 ×
− 1652 kJ = −826 kJ; 826 kJ of heat released 2 mol Fe 2 O 3
c. 1.00 g Fe ×
1 mol Fe − 1652 kJ = −7.39 kJ; 7.39 kJ of heat released 55.85 g 4 mol Fe
d. 10.0 g Fe ×
1 mol Fe − 1652 kJ = −73.9 kJ 55.85 g Fe 4 mol Fe
2.00 g O2 ×
1 mol O 2 − 1652 kJ = −34.4 kJ 32.00 g O 2 3 mol O 2
Because 2.00 g O2 releases the smaller quantity of heat, O2 is the limiting reactant and 34.4 kJ of heat can be released from this mixture. 56.
a. 1.00 mol H2O ×
− 572 kJ = −286 kJ; 286 kJ of heat released 2 mol H 2 O
b. 4.03 g H2 ×
1 mol H 2 − 572 kJ = −572 kJ; 572 kJ of heat released 2.016 g H 2 2 mol H 2
c. 186 g O2 ×
1 mol O 2 − 572 kJ = −3320 kJ; 3320 kJ of heat released 32.00 g O 2 mol O 2
d.
n H2 =
1.0 atm 2.0 10 8 L PV = = 8.2 × 106 mol H2 0.08206 L atm RT 298 K K mol
8.2 × 106 mol H2 × 57.
− 572 kJ = −2.3 × 109 kJ; 2.3 × 109 kJ of heat released 2 mol H 2
From Example 6.3, q = 1.3 × 108 J. Because the heat transfer process is only 60.% 100. J efficient, the total energy required is 1.3 × 108 J × = 2.2 × 108 J. 60. J Mass C3H8 = 2.2 × 108 J ×
58.
a. 1.00 g CH4 ×
PV b. n = = RT
1 mol C3H 8 44.09 g C3H 8 = 4.4 × 103 g C3H8 3 mol C3H 8 2221 10 J
1 mol CH 4 − 891 kJ = −55.5 kJ 16.04 g CH 4 mol CH 4
1 atm 1.00 10 3 L 760 torr = 39.8 mol CH4 0.08206 L atm 298 K K mol
740 . torr
39.8 mol CH4
− 891 kJ = −3.55 × 104 kJ mol CH 4
CHAPTER 6
THERMOCHEMISTRY
239
59.
When a liquid is converted into gas, there is an increase in volume. The 2.5 kJ/mol quantity is the work done by the vaporization process in pushing back the atmosphere.
60.
H = E + PV; from this equation, H > E when V > 0, H < E when V < 0, and H = E when V = 0. Concentrate on the moles of gaseous products versus the moles of gaseous reactants to predict V for a reaction. a. There are 2 moles of gaseous reactants converting to 2 moles of gaseous products, so V = 0. For this reaction, H = E. b. There are 4 moles of gaseous reactants converting to 2 moles of gaseous products, so V < 0 and H < E. c. There are 9 moles of gaseous reactants converting to 10 moles of gaseous products, so V > 0 and H > E.
Calorimetry and Heat Capacity 61.
Specific heat capacity is defined as the amount of heat necessary to raise the temperature of one gram of substance by one degree Celsius. Therefore, H 2O(l) with the largest heat capacity value requires the largest amount of heat for this process. The amount of heat for H2O(l) is: 4.18 J energy = s × m × ΔT = o × 25.0 g × (37.0°C − 15.0°C) = 2.30 × 103 J Cg The largest temperature change when a certain amount of energy is added to a certain mass of substance will occur for the substance with the smallest specific heat capacity. This is Hg(l), and the temperature change for this process is: 10.7 kJ
ΔT =
62.
1000 J kJ = 140°C
energy = 0.14 J sm 550 . g o Cg
a. s = specific heat capacity =
0.24 J o
Cg
=
0.24 J since ΔT(K) = ΔT(°C) Kg
0.24 J Energy = s × m × ΔT = o × 150.0 g × (298 K − 273 K) = 9.0 × 102 J Cg b. Molar heat capacity =
0.24 J o
Cg
107.9 g Ag = mol Ag
26 J o
C mol
0.24 J 1250 c. 1250 J = o × m × (15.2°C − 12.0°C), m = = 1.6 × 103 g Ag 0 . 24 3 . 2 Cg 63.
s = specific heat capacity =
q 133 J = = 0.890 J/°C•g m T 5.00 g (55.1 − 25.2) o C
From Table 6.1, the substance is solid aluminum.
240 64.
CHAPTER 6 s=
585 J 125 .6 g (53.5 − 20.0) o C
THERMOCHEMISTRY
= 0.139 J/°C•g
Energy = s × m × ΔT = 0.139 J/°C•g × 10.0 g × [75.0°C − (‒ 10.0°C)] = 118 J 65.
| Heat loss by hot water | = | heat gain by cooler water | The magnitudes of heat loss and heat gain are equal in calorimetry problems. The only difference is the sign (positive or negative). To avoid sign errors, keep all quantities positive and, if necessary, deduce the correct signs at the end of the problem. This means for ΔT, put the higher temperature first. Water has a specific heat capacity = s = 4.18 J/°C•g = 4.18 J/K•g (ΔT in °C = ΔT in K). Heat loss by hot water = s × m × ΔT = Heat gain by cooler water =
4.18 J × 50.0 g × (330. K − Tf) Kg
4.18 J × 30.0 g × (Tf − 280. K); heat loss = heat gain, so: Kg
125 J 209 J × (330. K − Tf) = × (Tf − 280. K) K K
6.90 × 104 − 209Tf = 125Tf − 3.50 × 104, 334Tf = 1.040 × 105, Tf = 311 K Note that the final temperature is closer to the temperature of the more massive hot water, which is as it should be. 66.
|Heat loss by hot water | = |heat gain by cold water |; keeping all quantities positive helps to avoid sign errors:
4.18 J o
Cg
mhot = 67.
× mhot × (55.0°C − 37.0°C) =
4.18 J o
Cg
× 90.0 g × (37.0°C − 22.0°C)
90.0 g 15.0 o C = 75.0 g hot water needed 18.0 o C
|Heat gained by water | = |heat lost by steel | = s × m × ΔT, where s = specific heat capacity.
4.18 J Heat gain = o × 75.0 g × (Tf − 20.0°C) Cg A common error in calorimetry problems is sign errors. Keeping all quantities positive helps to eliminate sign errors. This means for ΔT, put the higher temperature first. Heat loss =
0.490 J × 25.0 × (85.0°C − Tf) o Cg
0.490 J 4.18 J × 75.0 g × (Tf − 20.0°C) = o × 25.0 × (85.0°C − Tf) o Cg Cg
314Tf – 6270 = 1040 ‒ 12.3Tf, 326Tf = 7310, Tf = 22.4°C
CHAPTER 6 68.
THERMOCHEMISTRY
241
|Heat gain by statue | = |heat lost by water | = s × m × ΔT, where s = specific heat capacity. Heat gain = s × 622.5 g × (91.33°C − 26.00°C) = 40670s A common error in calorimetry problems is sign errors. Keeping all quantities positive, specifically ΔT, helps to eliminate sign errors. Heat loss =
4.184 J × 1000. g × (100.00°C − 91.33°C) = 36300 o Cg
40670s = 36300, s = 0.893 J/g·°C; the statue is made of aluminum. 69.
| Heat gained by water | = | heat lost by nickel | = s × m × ΔT, where s = specific heat capacity.
4.18 J Heat gain = o × 150.0 g × (25.0°C − 23.5°C) = 940 J Cg A common error in calorimetry problems is sign errors. Keeping all quantities positive, specifically ΔT, helps to eliminate sign errors. Heat loss = 940 J = 70.
0.444 J o
× mass × (99.8 − 25.0) °C, mass =
Cg
940 = 28 g 0.444 74.8
Heat gain by water = heat loss by Cu; keeping all quantities positive helps to avoid sign errors:
4.18 J o
Cg
× mass × (24.9°C − 22.3°C) =
0.20 J o
Cg
× 110. g Cu × (82.4°C − 24.9°C)
11(mass) = 1300, mass = 120 g H2O 71.
| Heat loss by Al + heat loss by Fe | = | heat gain by water |; keeping all quantities positive to avoid sign error: 0.89 J 0.45 J × 5.00 g Al × (100.0°C − Tf) + o × 10.00 g Fe × (100.0 − Tf) o Cg Cg 4.18 J = o × 97.3 g H2O × (Tf − 22.0°C) Cg
4.5(100.0 − Tf) + 4.5(100.0 − Tf) = 407(Tf − 22.0), 450 − (4.5)Tf + 450 − (4.5)Tf = 407Tf − 8950 416Tf = 9850, Tf = 23.7°C 72.
Heat released to water = 5.0 g H2 × Heat gain by water = 1.10 × 103 J =
120. J 50. J + 10. g methane × = 1.10 × 103 J g methane g H2
4.18 J o
Cg
× 50.0 g × T
T = 5.26°C, 5.26°C = Tf − 25.0°C, Tf = 30.3°C
242 73.
CHAPTER 6
THERMOCHEMISTRY
50.0 × 10−3 L × 0.100 mol/L = 5.00 × 10−3 mol of both AgNO3 and HCl are reacted. Thus 5.00 × 10−3 mol of AgCl will be produced because there is a 1 : 1 mole ratio between reactants. | Heat lost by chemicals | = | heat gained by solution | Heat gain =
4.18 J × 100.0 g × (23.40 − 22.60)°C = 330 J o Cg
Heat loss = 330 J; this is the heat evolved (exothermic reaction) when 5.00 × 10 −3 mol of AgCl is produced. So q = −330 J and ΔH (heat per mol AgCl formed) is negative with a value of: ΔH =
− 330 J 1 kJ = −66 kJ/mol −3 5.00 10 mol 1000 J
Note: Sign errors are common with calorimetry problems. However, the correct sign for ΔH can be determined easily from the ΔT data; i.e., if ΔT of the solution increases, then the reaction is exothermic because heat was released, and if ΔT of the solution decreases, then the reaction is endothermic because the reaction absorbed heat from the water. For calorimetry problems, keep all quantities positive until the end of the calculation and then decide the sign for ΔH. This will help eliminate sign errors. 74.
NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l) We have a stoichiometric mixture. All of the NaOH and HCl will react. 0.10 L ×
1.0 mol = 0.10 mol of HCl is neutralized by 0.10 mol NaOH. L
| Heat lost by chemicals | = | heat gained by solution | Volume of solution = 100.0 + 100.0 = 200.0 mL Heat gain =
1.0 g 200.0 mL × (31.3 – 24.6)C = 5.6 × 103 J = 5.6 kJ mL Cg
4.18 J o
Heat loss = 5.6 kJ; this is the heat released by the neutralization of 0.10 mol HCl. Because the temperature increased, the sign for ΔH must be negative, i.e., the reaction is exothermic. For calorimetry problems, keep all quantities positive until the end of the calculation and then decide the sign for ΔH. This will help eliminate sign errors. ΔH = 75.
− 5.6 kJ = −56 kJ/mol 0.10 mol
NH4NO3(s) → NH4+(aq) + NO3−(aq) ΔH = ?; mass of solution = 60.00 g + 4.25 g = 64.25 g | Heat lost by solution | = | heat gained as NH 4NO3 dissolves |; to help eliminate sign errors, we will keep all quantities positive (q and ΔT) and then deduce the correct sign for ΔH at the end of the problem. Here, because temperature decreases as NH 4NO3 dissolves, heat is absorbed as NH4NO3 dissolves, so this is an endothermic process (ΔH is positive).
CHAPTER
6
THERMOCHEMISTRY
Heat lost by solution =
243
4.18 J × 64.25 g × (22.0 − 16.9)°C = 1400 J o Cg
Heat gained as NH4NO3 dissolves = 1400 J ΔH = 76.
80.05 g NH 4 NO3 1400 J 1 kJ × × = 26 kJ/mol NH4NO3 dissolving 4.25 g NH 4 NO3 mol NH 4 NO 3 1000 J
NaHCO3(s) + HCl(aq) → NaCl(aq) + CO2(g) + H2O(l) H = ? | Heat lost by solution | = | heat gained by reaction |; mass solution = 50.0 g + 2.00 g = 52.0 g Note: Sign errors are common with calorimetry problems. However, the correct sign for ΔH can easily be obtained from the ΔT data. When working calorimetry problems, keep all quantities positive (ignore signs). When finished, deduce the correct sign for ΔH. For this problem, T decreases as NaHCO3 and HCl react. ΔH is positive for this reaction.
77.
Heat lost by solution =
4.18 J × 52.0 g × (28.1°C − 24.8°C) = 720 J o Cg
ΔH in units of kJ/mol =
84.01 g NaHCO3 720 J 1 kJ = 30. kJ/mol × × 2.00 g NaHCO3 mol NaHCO3 1000 J
Because ΔH is exothermic, the temperature of the solution will increase as CaCl 2(s) dissolves. Keeping all quantities positive: heat loss as CaCl2 dissolves = 11.0 g CaCl2 heat gained by solution = 8.08 × 103 J = Tf − 25.0°C =
78.
1 mol CaCl 2 81.5 kJ = 8.08 kJ 110 .98 g CaCl 2 mol CaCl 2
4.18 J × (125 + 11.0) g × (Tf − 25.0°C) o Cg
8.08 10 3 = 14.2°C, Tf = 14.2°C + 25.0°C = 39.2°C 4.18 136
0.1000 L ×
0.500 mol HCl 118 kJ heat released = 2.95 kJ of heat released if HCl limiting L 2 mol HCl
0.3000 L ×
0.100 mol Ba(OH) 2 118 kJ heat = 3.54 kJ heat released if Ba(OH)2 limiting L mol Ba(OH) 2
Because the HCl reagent produces the smaller amount of heat released, HCl is limiting and 2.95 kJ of heat are released by this reaction. Heat gained by solution = 2.95 × 10 3 J =
4.18 J × 400.0 g × ΔT o Cg
ΔT = 1.76°C = Tf − Ti = Tf − 25.0°C, Tf = 26.8°C 79.
Heat gain by calorimeter =
1.56 kJ × 3.2°C = 5.0 kJ = heat loss by quinine o C
244
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THERMOCHEMISTRY
Heat loss = 5.0 kJ, which is the heat evolved (exothermic reaction) by the combustion of 0.1964 g of quinone. Because we are at constant volume, q V = E. ΔEcomb = 80.
− 5.0 kJ = −25 kJ/g; 0.1964 g
ΔEcomb =
− 25 kJ 108 .09 g = −2700 kJ/mol g mol
| Heat loss by butter | = | heat gain by calorimeter | = (27.3°C − 23.5°C) ×
2.67 kJ o
C
= 10. kJ
Heat loss = 10. kJ, which is the heat evolved (exothermic reaction) by the combustion of 0.30 g butter. Because we are at constant volume, q V = E. ΔEcomb =
81.
− 10. kJ = −33 kJ/g 0.30 g
a. Heat gain by calorimeter = heat loss by CH4 = 6.79 g CH4
Heat capacity of calorimeter =
1 mol CH 4 802 kJ 16.04 g mol = 340. kJ
340 . kJ = 31.5 kJ/°C 10.8 o C
b. Heat loss by C2H2 = heat gain by calorimeter = 16.9°C ×
31. 5 kJ = 532 kJ o C
A bomb calorimeter is at constant volume, so the heat released/gained = q V = E: ΔEcomb = 82.
− 532 kJ 26.04 g = −1.10 × 103 kJ/mol 12.6 g C 2 H 2 mol C 2 H 2
First, we need to get the heat capacity of the calorimeter from the combustion of benzoic acid. Heat lost by combustion = heat gained by calorimeter. Heat loss = 0.1584 g ×
26.42 kJ = 4.185 kJ g
Heat gain = 4.185 kJ = Ccal × ΔT, Ccal =
4.185 kJ 2.54 o C
= 1.65 kJ/°C
Now we can calculate the heat of combustion of vanillin. | Heat loss | = | heat gain |. Heat gain by calorimeter =
1.65 kJ o
× 3.25°C = 5.36 kJ
C
Heat loss = 5.36 kJ, which is the heat evolved by combustion of the vanillin. Ecomb =
− 5.36 kJ − 25.2 kJ 152 .14 g = −25.2 kJ/g; Ecomb = = −3830 kJ/mol 0.2130 g g mol
Hess's Law 83.
Information given: C(s) + O2(g) → CO2(g) CO(g) + 1/2 O2(g) → CO2(g)
ΔH = −393.7 kJ ΔH = −283.3 kJ
CHAPTER
6
THERMOCHEMISTRY
245
Using Hess’s law: 2 C(s) + 2 O2(g) → 2 CO2(g) 2 CO2(g) → 2 CO(g) + O2(g)
ΔH1 = 2(−393.7 kJ) ΔH2 = −2(−283.3 kJ)
2 C(s) + O2(g) → 2 CO(g)
ΔH = ΔH1 + ΔH2 = −220.8 kJ
Note: When an equation is reversed, the sign on ΔH is reversed. When the coefficients in a balanced equation are multiplied by an integer, then the value of ΔH is multiplied by the same integer. 84.
Given: C4H4(g) + 5 O2(g) → 4 CO2(g) + 2 H2O(l) C4H8(g) + 6 O2(g) → 4 CO2(g) + 4 H2O(l) H2(g) + 1/2 O2(g) → H2O(l)
ΔHcomb = −2341 kJ ΔHcomb = −2755 kJ ΔHcomb = −286 kJ
By convention, H2O(l) is produced when enthalpies of combustion are given, and because per-mole quantities are given, the combustion reaction refers to 1 mole of that quantity reacting with O2(g). Using Hess’s law to solve: C4H4(g) + 5 O2(g) → 4 CO2(g) + 2 H2O(l) 4 CO2(g) + 4 H2O(l) → C4H8(g) + 6 O2(g) 2 H2(g) + O2(g) → 2 H2O(l) C4H4(g) + 2 H2(g) → C4H8(g) 85.
2 N2(g) + 6 H2(g) → 4 NH3(g) 6 H2O(g) → 6 H2(g) + 3 O2(g) 2 N2(g) + 6 H2O(g) → 3 O2(g) + 4 NH3(g)
ΔH1 = −2341 kJ ΔH2 = −(−2755 kJ) ΔH3 = 2(−286 kJ) ΔH = ΔH1 + ΔH2 + ΔH3 = −158 kJ
ΔH = −2(92 kJ) ΔH = −3(−484 kJ) ΔH = 1268 kJ
No, because the reaction is very endothermic (requires a lot of heat to react), it would not be a practical way of making ammonia because of the high energy costs required. 86.
ClF + 1/2 O2 → 1/2 Cl2O + 1/2 F2O 1/2 Cl2O + 3/2 F2O → ClF3 + O2 F2 + 1/2 O2 → F2O ClF(g) + F2(g) → ClF3(g)
87.
C6H4(OH)2 → C6H4O2 + H2 H2O2 → H2 + O2 2 H2 + O2 → 2 H2O(g) 2 H2O(g) → 2 H2O(l) C6H4(OH)2(aq) + H2O2(aq) → C6H4O2(aq) + 2 H2O(l)
ΔH = 1/2(167.4 kJ) ΔH = −1/2(341.4 kJ) ΔH = 1/2(−43.4 kJ) ΔH = −108.7 kJ ΔH = 177.4 kJ ΔH = −(−191.2 kJ) ΔH = 2(−241.8 kJ) ΔH = 2(−43.8 kJ) ΔH = −202.6 kJ
246
CHAPTER 6 2 NH3 + 3 N2O → 4 N2 + 3 H2O 3 N2H4 + 3 H2O → 3 N2O + 9 H2 9 H2 + 9/2 O2 → 9 H2O 4 N2 + 8 H2O → 4 N2H4 + 4 O2
88.
2 NH3(g) + 1/2 O2(g) → N2H4(l) + H2O(g)
ΔH = −1010. kJ ΔH = −3(−317 kJ) ΔH = 9(−286 kJ) ΔH = −4(−623 kJ) ΔH = −141 kJ
CaC2 → Ca + 2 C CaO + H2O → Ca(OH)2 2 CO2 + H2O → C2H2 + 5/2 O2 Ca + 1/2 O2 → CaO 2 C + 2 O2 → 2 CO2
89.
CaC2(s) + 2 H2O(l) → Ca(OH)2(aq) + C2H2(g) P4O10 → P4 + 5 O2 10 PCl3 + 5 O2 → 10 Cl3PO 6 PCl5 → 6 PCl3 + 6 Cl2 P4 + 6 Cl2 → 4 PCl3
90.
THERMOCHEMISTRY
P4O10(s) + 6 PCl5(g) → 10 Cl3PO(g)
ΔH = −(−62.8 kJ) ΔH = −653.1 kJ ΔH = −(−1300. kJ) ΔH = −635.5 kJ ΔH = 2(−393.5 kJ) ΔH = −713 kJ ΔH = −(−2967.3 kJ) ΔH = 10(−285.7 kJ) ΔH = −6(−84.2 kJ) ΔH = −1225.6 ΔH = −610.1 kJ
Standard Enthalpies of Formation 91.
The change in enthalpy that accompanies the formation of 1 mole of a compound from its elements, with all substances in their standard states, is the standard enthalpy of formation for a compound. The reactions that refer to H of are: Na(s) + 1/2 Cl2(g) → NaCl(s); H2(g) + 1/2 O2(g) → H2O(l) 6 C(graphite, s) + 6 H2(g) + 3 O2(g) → C6H12O6(s) Pb(s) + S(rhombic, s) + 2 O2(g) → PbSO4(s)
92.
a. Aluminum oxide = Al2O3; 2 Al(s) + 3/2 O2(g) → Al2O3(s) b. C2H5OH(l) + 3 O2(g) → 2 CO2(g) + 3 H2O(l) c. NaOH(aq) + HCl(aq) → H2O(l) + NaCl(aq) d. 2 C(graphite, s) + 3/2 H2(g) + 1/2 Cl2(g) → C2H3Cl(g)CuO e. C6H6(l) + 15/2 O2(g) → 6 CO2(g) + 3 H2O(l) Note: ΔHcomb values assume 1 mole of compound combusted. f.
NH4Br(s) → NH4+(aq) + Br−(aq)
CHAPTER 93.
6
THERMOCHEMISTRY
The standard enthalpy of formation for a compound is the change in enthalpy that accompanies the formation of 1 mole of a compound from its elements, with all substances in their standard states. For CuO, the enthalpy of formation equation is Cu(s) + ½ O 2(g) → CuO(s). Now, rearrange the given equations to add up to the enthalpy of formation equation. Applying Hess’s law will allow determination of ΔHof, CuO . CuO(s) + Cu(s) → Cu2O(s) Cu2O(s) + ½ O2(g) → 2 CuO(s)] Cu(s) + ½ O2(g) → CuO(s)
94.
247
H = ‒(11 kJ) H = ½ (−288 kJ)
ΔHof, CuO = ‒ 155 kJ/mol
C(graphite) + 2 H2(g) + ½ O2(g) → CH3OH(l)
Hrxn = ΔHof, CH3OH
CO2(g) + 2 H2O(g) → CH3OH(l) + 3/2 O2(g) C(graphite) + O2(g) → CO2(g) 2 H2(g) + O2(g) → 2 H2O(g) C(graphite) + 2 H2(g) + ½ O2(g) → CH3OH(l) 95.
H = ‒(−638 kJ) H = −394 kJ H = 2(−242 kJ) ΔHof, CH3OH = ‒240. kJ/mol
ΔH° = np ΔHf , products − nr ΔHf , reactants ; elements in their standard state have ΔH f = 0 by definition. P4O10(s) + 6 H2O(l) → 4 H3PO4(s) ΔH° = (4 mol H3PO4 × ΔHof, H3PO4 ) − (1 mol P4O10 × ΔH of , P4O10 + 6 mol H2O × ΔHof, H2O ) ΔH° = [4(‒1279 kJ)] ‒ [‒2984 kJ + 6(‒286 kJ)] = ‒416. kJ/mol 2.00 mol H2O ×
96.
−416 kJ = ‒139 kJ; 139 kJ of heat is released for the reaction 6 mol H 2 O
2 CO(g) + 2 H2(g) → CO2(g) + CH4(g); ΔH° = np ΔHf , products − nr ΔHf , reactants −393.5 kJ −74.8 kJ −110.5 kJ ΔH° = 1 mol + 1 mol − 2 mol = ‒247.3 kJ mol mol mol
14.0 g CO × 97.
1 mol CO −247.3 kJ = ‒61.8 kJ; 61.8 kJ of heat is released. 28.01 g 2 mol CO
In general, ΔH° = np ΔHf , products − nr ΔHf , reactants , and all elements in their standard state have ΔH f = 0 by definition. a. The balanced equation is 2 NH3(g) + 3 O2(g) + 2 CH4(g) → 2 HCN(g) + 6 H2O(g). ΔH° = (2 mol HCN × ΔH f , HCN + 6 mol H2O(g) × ΔH f , H 2O ) − (2 mol NH3 × ΔH f , NH3 + 2 mol CH4 × H f , CH 4 ) ΔH° = [2(135.1) + 6(−242)] − [2(−46) + 2(−75)] = −940. kJ
248
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b. Ca3(PO4)2(s) + 3 H2SO4(l) → 3 CaSO4(s) + 2 H3PO4(l)
− 1433 kJ − 1267 kJ ΔH° = 3 mol CaSO4 (s) + 2 mol H 3PO4 (l) mol mol
− 4126 kJ − 814 kJ − 1 mol Ca 3 (PO4 ) 2 (s) + 3 mol H 2SO 4 (l) mol mol ΔH° = −6833 kJ − (−6568 kJ) = −265 kJ c. NH3(g) + HCl(g) → NH4Cl(s) ΔH° = (1 mol NH4Cl × ΔH f , NH4Cl ) − (1 mol NH3 × ΔH f , NH3 + 1 mol HCl × ΔH f , HCl ) − 314 kJ − 46 kJ − 92 kJ ΔH° = 1 mol − 1 mol + 1 mol mol mol mol
ΔH° = −314 kJ + 138 kJ = −176 kJ 98.
a. The balanced equation is C2H5OH(l) + 3 O2(g) → 2 CO2(g) + 3 H2O(g).
− 393 .5 kJ − 242 kJ − 278 kJ ΔH° = 2 mol + 3 mol − 1 mol mol mol mol ΔH° = −1513 kJ − (−278 kJ) = −1235 kJ b. SiCl4(l) + 2 H2O(l) → SiO2(s) + 4 HCl(aq) Because HCl(aq) is H+(aq) + Cl−(aq), ΔH f = 0 − 167 = −167 kJ/mol.
− 167 kJ − 911 kJ − 687 kJ − 286 kJ ΔH° = 4 mol + 1 mol − 1 mol + 2 mol mol mol mol mol ΔH° = −1579 kJ − (−1259 kJ) = −320. kJ c. MgO(s) + H2O(l) → Mg(OH)2(s)
− 925 kJ − 602 kJ − 286 kJ ΔH° = 1 mol − 1 mol + 1 mol mol mol mol ΔH° = −925 kJ − (−888 kJ) = −37 kJ 99.
a. 4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g); ΔH° = np ΔHf , products − nr ΔHf , reactants
90. kJ − 242 kJ − 46 kJ ΔH° = 4 mol + 6 mol − 4 mol = −908 kJ mol mol mol
CHAPTER
6
THERMOCHEMISTRY
249
2 NO(g) + O2(g) → 2 NO2(g) 34 kJ 90. kJ ΔH° = 2 mol − 2 mol = −112 kJ mol mol
3 NO2(g) + H2O(l) → 2 HNO3(aq) + NO(g) − 207 kJ 90. kJ 34 kJ − 286 kJ ΔH° = 2 mol + 1 mol − 3 mol + 1 mol mol mol mol mol
−140. kJ Note: All
ΔH f
values are assumed ±1 kJ.
b. 12 NH3(g) + 15 O2(g) → 12 NO(g) + 18 H2O(g) 12 NO(g) + 6 O2(g) → 12 NO2(g) 12 NO2(g) + 4 H2O(l) → 8 HNO3(aq) + 4 NO(g) 4 H2O(g) → 4 H2O(l) 12 NH3(g) + 21 O2(g) → 8 HNO3(aq) + 4 NO(g) + 14 H2O(g) The overall reaction is exothermic because each step is exothermic. 100.
4 Na(s) + O2(g) → 2 Na2O(s)
− 416 kJ ΔH° = 2 mol = −832 kJ mol
2 Na(s) + 2 H2O(l) → 2 NaOH(aq) + H2(g) − 470 . kJ − 286 kJ ΔH° = 2 mol − 2 mol = −368 kJ mol mol
2Na(s) + CO2(g) → Na2O(s) + CO(g) − 416 kJ − 110 .5 kJ − 393 .5 kJ ΔH° = 1 mol + 1 mol − 1 mol = −133 kJ mol mol mol
In Reactions 2 and 3, sodium metal reacts with the "extinguishing agent." Both reactions are exothermic, and each reaction produces a flammable gas, H 2 and CO, respectively. 101.
3 Al(s) + 3 NH4ClO4(s) → Al2O3(s) + AlCl3(s) + 3 NO(g) + 6 H2O(g)
− 242 kJ 90. kJ − 704 kJ − 1676 kJ ΔH° = 6 mol + 3 mol + 1 mol + 1 mol mol mol mol mol − 295 kJ − 3 mol = −2677 kJ mol
250
CHAPTER 6
102.
5 N2O4(l) + 4 N2H3CH3(l) → 12 H2O(g) + 9 N2(g) + 4 CO2(g)
THERMOCHEMISTRY
− 242 kJ − 393 .5 kJ ΔH° = 12 mol + 4 mol mol mol − 20. kJ 54 kJ − 5 mol + 4 mol = −4594 kJ mol mol
103.
2 ClF3(g) + 2 NH3(g) → N2(g) + 6 HF(g) + Cl2(g)
ΔH° = −1196 kJ
ΔH° = (6 ΔHof , HF ) − (2 ΔHof, ClF3 + 3 ΔHof, NH3 ) − 271 kJ − 46 kJ o −1196 kJ = 6 mol − 2 ΔHf , ClF3 − 2 mol mol mol
−1196 kJ = −1626 kJ − 2 ΔHof , ClF3 + 92 kJ, ΔHof , ClF3 = 104.
C2H4(g) + 3 O2(g) → 2 CO2(g) + 2 H2O(l)
(−1626 + 92 + 1196 ) kJ − 169 kJ = mol 2 mol
ΔH° = −1411.1 kJ
ΔH° = −1411.1 kJ = 2(−393.5) kJ + 2(−285.8) kJ − ΔH f , C2H 4 −1411.1 kJ = −1358.6 kJ − ΔHf , C2H 4 , ΔH f , C2H 4 = 52.5 kJ/mol
Energy Consumption and Sources 105.
C(s) + H2O(g) → H2(g) + CO(g) ΔH° = −110.5 kJ − (−242 kJ) = 132 kJ
106.
CO(g) + 2 H2(g) → CH3OH(l)
107.
C2H5OH(l) + 3 O2(g) → 2 CO2(g) + 3 H2O(l)
ΔH° = −239 kJ − (−110.5 kJ) = −129 kJ
ΔH° = [2(−393.5 kJ) + 3(−286 kJ)] − (−278 kJ) = −1367 kJ/mol ethanol − 1367 kJ 1 mol = −29.67 kJ/g mol 46.07 g
108.
CH3OH(l) + 3/2 O2(g) → CO2(g) + 2 H2O(l) ΔH° = [−393.5 kJ + 2(−286 kJ)] − (−239 kJ) = −727 kJ/mol CH3OH − 727 kJ 1 mol = −22.7 kJ/g versus −29.67 kJ/g for ethanol (from Exercise 107) mol 32.04 g
Ethanol has a slightly higher fuel value per gram than methanol. 109.
C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(l) ΔH° = [3(−393.5 kJ) + 4(−286 kJ)] − (−104 kJ) = −2221 kJ/mol C3H8
CHAPTER
6
THERMOCHEMISTRY
251
− 50.37 kJ − 2221 kJ 1 mol = versus −47.7 kJ/g for octane (Example 6.11) g mol 44.09 g
The fuel values are very close. An advantage of propane is that it burns more cleanly. The boiling point of propane is −42°C. Thus it is more difficult to store propane, and there are extra safety hazards associated with using high-pressure compressed-gas tanks. 110.
1 mole of C2H2(g) and 1 mole of C4H10(g) have equivalent volumes at the same T and P. Enthalpy of combustion per volume of C 2 H 2 enthalpy of combustion per mol of C 2 H 2 = enthalpy of combustion per mol of C 4 H10 Enthalpy of combustion per volume of C 4 H10 Enthalpy of combustion per volume of C 2 H 2 = Enthalpy of combustion per volume of C 4 H10
− 49.9 kJ 26.04 g C 2 H 2 g C2H 2 mol C 2 H 2 = 0.452 − 49.5 kJ 58.12 g C 4 H10 g C 4 H10 mol C 4 H10
More than twice the volume of acetylene is needed to furnish the same energy as a given volume of butane. 111.
The molar volume of a gas at STP is 22.42 L (from Chapter 5). 4.19 × 106 kJ ×
112.
1 mol CH 4 22.42 L CH 4 = 1.05 × 105 L CH4 891 kJ mol CH 4
Mass of H2O = 1.00 gal ×
3.785 L 1000 mL 1.00 g = 3790 g H2O gal L mL
Energy required (theoretical) = s × m × ΔT =
4.18 J × 3790 g × 10.0 °C = 1.58 × 105 J o Cg
For an actual (80.0% efficient) process, more than this quantity of energy is needed since heat is always lost in any transfer of energy. The energy required is: 1.58 × 105 J ×
100 . J = 1.98 × 105 J 80.0 J
Mass of C2H2 = 1.98 × 105 J ×
1 mol C 2 H 2 26.04 g C 2 H 2 = 3.97 g C2H2 3 mol C 2 H 2 1300 . 10 J
ChemWork Problems 1 mol H 2 O 18.02 g H 2 O 5500 kJ = 4900 g = 4.9 kg H2O h 40.6 kJ mol
113.
2.0 h
114.
From the problem, walking 4.0 miles consumes 400 kcal of energy. 1 lb fat
115.
454 g 7.7 kcal 4 mi 1h = 8.7 h = 9 h lb g 400 kcal 4 mi
313 g He ×
1 mol He = 78.2 mol; q = molar heat capacity × mol × ΔT 4.003 g He
252
CHAPTER 6 q=
20.8 J o
THERMOCHEMISTRY
× 78.2 mol × (−41.6°C) = −67,700 J = −67.7 kJ
C mol
w = −PΔV = −1.00 atm × (1643 L − 1910. L) = 267 L atm ×
101 .3 J = 2.70 × 104 J = 27.0 kJ L atm
ΔE = q + w = −67.7 kJ kJ + 27.0 kJ = −40.7 kJ 116.
At constant pressure, H = E + PV; H = E when PΔV = 0 and this occurs when ΔV = 0. For a chemical equation, ΔV = 0 when the mol of product gases = the mol of reactant gases (it is assumed the volume change of solids and liquids is negligible and that the volume of solids and liquids are negligible compared to gases). For reaction a, the mole of product gases = 2 and the mol of reactant gases = 1. Since we do not have the same moles of product gases as reactant gases, H ≠ E for reaction a. The other reactions have equal moes of product gases and reactant gases, so they all have H ≈ E.
117.
a. 2 SO2(g) + O2(g) → 2 SO3(g); w = −PΔV; because the volume of the piston apparatus decreased as reactants were converted to products (V < 0), w is positive (w > 0). b. COCl2(g) → CO(g) + Cl2(g); because the volume increased (V > 0), w is negative (w < 0). c. N2(g) + O2(g) → 2 NO(g); because the volume did not change (V = 0), no PV work is done (w = 0). In order to predict the sign of w for a reaction, compare the coefficients of all the product gases in the balanced equation to the coefficients of all the reactant gases. When a balanced reaction has more moles of product gases than moles of reactant gases (as in b), the reaction will expand in volume (ΔV positive), and the system does work on the surroundings. When a balanced reaction has a decrease in the moles of gas from reactants to products (as in a), the reaction will contract in volume (ΔV negative), and the surroundings will do compression work on the system. When there is no change in the moles of gas from reactants to products (as in c), ΔV = 0 and w = 0.
118.
a. N2(g) + 3 H2(g) → 2 NH3(g); from the balanced equation, 1 molecule of N 2 will react with 3 molecules of H2 to produce 2 molecules of NH3. So the picture after the reaction should only have 2 molecules of NH3 present. Another important part of your drawing will be the relative volume of the product container. The volume of a gas is directly proportional to the number of gas molecules present (at constant T and P). In this problem, 4 total molecules of gas were present initially (1 N 2 + 3 H2). After reaction, only 2 molecules are present (2 NH3). Because the number of gas molecules decreases by a factor of 2 (from 4 total to 2 total), the volume of the product gas must decrease by a factor of 2 as compared to the initial volume of the reactant gases. Summarizing, the picture of the product container should have 2 molecules of NH 3 and should be at a volume which is onehalf the original reactant container volume. b. w = −PV; here the volume decreased, so V is negative. When V is negative, w is positive. As the reactants were converted to products, a compression occurred which is associated with work flowing into the system (w is positive).
CHAPTER 119.
6
THERMOCHEMISTRY
253
w = −PΔV; Δn = moles of gaseous products − moles of gaseous reactants. Only gases can do PV work (we ignore solids and liquids). When a balanced reaction has more moles of product gases than moles of reactant gases (Δn positive), the reaction will expand in volume (ΔV positive), and the system will do work on the surroundings. For example, in reaction c, Δn = 2 − 0 = 2 moles, and this reaction would do expansion work against the surroundings. When a balanced reaction has a decrease in the moles of gas from reactants to products (Δn negative), the reaction will contract in volume (ΔV negative), and the surroundings will do compression work on the system, e.g., reaction a, where Δn = 0 − 1 = −1. When there is no change in the moles of gas from reactants to products, ΔV = 0 and w = 0, e.g., reaction b, where Δn = 2 − 2 = 0. When ΔV > 0 (Δn > 0), then w < 0, and the system does work on the surroundings (c and e). When ΔV < 0 (Δn < 0), then w > 0, and the surroundings do work on the system (a and d). When ΔV = 0 (Δn = 0), then w = 0 (b).
120.
Heat is released when water freezes and when H 2 and O2 react to form water. So reactions b and d are exothermic. Heat is released when a bond forms, so reaction e is exothermic. The two reactions (a and c) refer to breaking a bond to form the individual atoms. Energy must be added to break a bond, so these are endothermic processes.
121.
54.0 g B2H6 ×
122.
The standard enthalpy of formation for a compound is the change in enthalpy that accompanies the formation of 1 mole of a compound from its elements, with all substances in their standard states. Equations c and e all have the reactants in their standard states and the equations are balanced to produce 1 mol of product. Equations c and e have Hreaction = Hf. Note that
1 mol B 2 H 6 2035 kJ heat released = 3.97 × 103 kJ heat released 27.67 g B 2 H 6 mol B 2 H 6
equation e is to produce an element in its standard state. Since the products and reactants are the same, Hf = 0. This is true for any element in their standard state. In reaction a, H is not the standard state of hydrogen gas. In equation b, SO4(l) is not the standard state of sulfur or for oxygen gas. In equation d, the reactants are not in their standard state for barium and chlorine. 123.
The kcals in one serving size of cookies is: 4 g fat 120 kcal
124.
8 kcal 4 kcal 4 kcal + 20 g carbs + 2 g protein = 120 kcal g protein g carb g fat
4.184 kJ 1 mile = 3.0 miles kcal 170 kJ
2 K(s) + 2 H2O(l) → 2 KOH(aq) + H2(g) ΔH° = 2(−481 kJ) − 2(−286 kJ) = −390. kJ 5.00 g K ×
1 mol K − 390 . kJ = −24.9 kJ 39.10 g K 2 mol K
24.9 kJ of heat is released on reaction of 5.00 g K.
254
CHAPTER 6 24,900 J =
THERMOCHEMISTRY
4.18 J 24,900 × (1.00 × 103 g) × ΔT, ΔT = = 5.96°C o g C 4.18 1.00 10 3
Final temperature = 24.0 + 5.96 = 30.0°C 125.
HNO3(aq) + KOH(aq) → H2O(l) + KNO3(aq)
ΔH = −56 kJ
0.2000 L ×
0.400 mol HNO3 56 kJ heat released × = 4.5 kJ heat released if HNO3 limiting mol HNO 3 L
0.1500 L ×
0.500 mol KOH 56 kJ heat released × = 4.2 kJ heat released if KOH limiting L mol KOH
Because the KOH reagent produces the smaller quantity of heat released, KOH is limiting and 4.2 kJ of heat released. 126.
2 NO2 → 2 NO + 2 O 2 NO + 2 O3 → 2 NO2 + 2 O2 3 O2 → 2 O3
ΔH = 2(233 kJ) ΔH = 2(−199 kJ) ΔH = −(−427 kJ)
O2(g) → 2 O(g)
ΔH = 495 kJ
127.
|qsurr| = |qsolution + qcal|; we normally assume that qcal is zero (no heat gain/loss by the calorimeter). However, if the calorimeter has a nonzero heat capacity, then some of the heat absorbed by the endothermic reaction came from the calorimeter. If we ignore q cal, then qsurr is too small, giving a calculated H value that is less positive (smaller) than it should be.
128.
V = 10.0 m × 4.0 m × 3.0 m = 1.2 × 10 2 m3 × (100 cm/m)3 = 1.2 × 108 cm3 Assuming the density of water is 1.0 g/cm 3, the mass of water is 1.2 × 108 g. q = specific heat capacity × mass × ΔT =
4.18 J o
Cg
(1.2 × 108 g) × (24.6oC − 20.2oC)
q = 2.2 × 109 J = 2.2 × 106 kJ 129.
HCl(aq) + NaOH(aq) → H2O(l) + NaCl(aq) ΔH = −56 kJ 0.1500 L ×
0.50 mol HCl 56 kJ heat released × = 4.2 kJ heat released if HCl limiting L mol HCl
0.0500 L ×
1.00 mol NaOH 56 kJ heat released × = 2.8 kJ heat released if NaOH limiting L mol NaOH
Because the NaOH reagent produces the smaller quantity of heat released, NaOH is limiting and 4.2 kJ of heat released. q = specific heat capacity × mass × ΔT, 2800 J =
4.184 J o
Cg
200.0 g × ΔT, ΔT = 3.3oC
This is an exothermic reaction so the temperature will increase by 3.3oC. Tfinal = 48.2oC + 3.3oC = 51.5oC
CHAPTER 130.
6
THERMOCHEMISTRY
255
The specific heat of water is 4.18 J/°C•g, which is equal to 4.18 kJ/°C•kg. We have 1.00 kg of H2O, so: 1.00 kg ×
4.18 kJ = 4.18 kJ/°C o C kg
This is the portion of the heat capacity that can be attributed to H2O. Total heat capacity = Ccal + C H 2O , Ccal = 10.84 − 4.18 = 6.66 kJ/°C 131.
Heat released = 1.056 g × 26.42 kJ/g = 27.90 kJ = heat gain by water and calorimeter 4.18 kJ 6.66 kJ 0.987 kg ΔT + o ΔT Heat gain = 27.90 kJ = o C C kg
27.90 = (4.13 + 6.66)ΔT = (10.79)ΔT, ΔT = 2.586°C 2.586°C = Tf − 23.32°C, Tf = 25.91°C 132.
For Exercise 101, a mixture of 3 mol Al and 3 mol NH4ClO4 yields 2677 kJ of energy. The mass of the stoichiometric reactant mixture is: 26.98 g 117 .49 g 3 mol + 3 mol = 433.41 g mol mol
For 1.000 kg of fuel: 1.000 × 103 g ×
− 2677 kJ = −6177 kJ 433 .41 g
In Exercise 102, we get 4594 kJ of energy from 5 mol of N 2O4 and 4 mol of N2H3CH3. The 92.02 g 46.08 g mass is 5 mol + 4 mol = 644.42 kJ. mol mol For 1.000 kg of fuel: 1.000 × 103 g ×
-4594 kJ = −7129 kJ 644.42 g
Thus we get more energy per kilogram from the N 2O4/N2H3CH3 mixture. 133.
2 H2(g) + 2 F2(g) → 4 HF(g)
2 C(s) + 4 F2(g) → 2 CF4(g) C2H4(g) → 2 C(s) + 2 H2(g)
H = 2H2 H = 2H3 H = − H4
C2H4(g) + 6 F2(g) → 2 CF4(g) + 4 HF(g) H1 = 2 H2 + 2 H3 − H4 Answer e is correct. 134.
To avoid fractions, let's first calculate ΔH for the reaction: 6 FeO(s) + 6 CO(g) → 6 Fe(s) + 6 CO2(g)
256
CHAPTER 6 6 FeO + 2 CO2 → 2 Fe3O4 + 2 CO 2 Fe3O4 + CO2 → 3 Fe2O3 + CO 3 Fe2O3 + 9 CO → 6 Fe + 9 CO2
135.
THERMOCHEMISTRY
ΔH° = −2(18 kJ) ΔH° = −(−39 kJ) ΔH° = 3(−23 kJ)
6 FeO(s) + 6 CO(g) → 6 Fe(s) + 6 CO2(g)
ΔH° = −66 kJ
So for FeO(s) + CO(g) → Fe(s) + CO2(g), ΔH° =
− 66 kJ = −11 kJ. 6
We want ΔH for N2H4(l) + O2(g) → N2(g) + 2 H2O(l). It will be easier to calculate ΔH for the combustion of four moles of N2H4 because we will avoid fractions. 9 H2 + 9/2 O2 → 9 H2O 3 N2H4 + 3 H2O → 3 N2O + 9 H2 2 NH3 + 3 N2O → 4 N2 + 3 H2O N2H4 + H2O → 2 NH3 + 1/2 O2
ΔH = 9(−286 kJ) ΔH = −3(−317 kJ) ΔH = −1010. kJ ΔH = − (−143 kJ)
4 N2H4(l) + 4 O2(g) → 4 N2(g) + 8 H2O(l)
ΔH = −2490. kJ
For N2H4(l) + O2(g) → N2(g) + 2 H2O(l)
ΔH =
− 2490 . kJ = −623 kJ 4
Note: By the significant figure rules, we could report this answer to four significant figures. However, because the ΔH values given in the problem are only known to ±1 kJ, our final answer will at best be ±1 kJ. 136.
An element in its standard state has a standard enthalpy of formation equal to zero. At 25oC and 1 atm, chlorine is found as Cl2(g) and hydrogen is found as H2(g). So, these two elements (a and b) have enthalpies of formation equal to zero. The other two choices (c and d) do not have the elements in their standard state. The standard state for nitrogen is N2(g) and the standard state for chlorine is Cl2(g).
137.
Na2SO4(aq) + Ba(NO3)2(aq) → BaSO4(s) + 2 NaNO3(aq)
ΔH = ?
1.00 L
2.00 mol Na 2 SO 4 1 mol BaSO 4 = 2.00 mol BaSO4 if Na2SO4 limiting L mol Na 2 SO 4
2.00 L
0.750 mol Ba ( NO 3 ) 2 1 mol BaSO 4 = 1.50 mol BaSO4 if Ba(NO3)2 limiting L mol Ba ( NO 3 ) 2
The Ba(NO3)2 reagent produces the smaller quantity of product, so Ba(NO 3)2 is limiting and 1.50 mol BaSO4 can form. Heat gain by solution = heat loss by reaction. Mass of solution = 3.00 L × Heat gain by solution =
1000 mL 2.00 g = 6.00 × 103 g L mL
6.37 J o
× 6.00 × 103 g × (42.0 − 30.0)C = 4.59 × 105 J
Cg
Because the solution gained heat, the reaction is exothermic; q = −4.59 × 105 J for the reaction. H =
− 4.59 10 5 J = −3.06 × 105 J/mol = −306 kJ/mol 1.50 mol BaSO4
CHAPTER
6
THERMOCHEMISTRY I(g) + Cl(g) → ICl(g) 1/2 Cl2(g) → Cl(g) 1/2 I2(g) → I(g) 1/2 I2(s) → 1/2 I2(g)
138.
ΔH = −(211.3 kJ) ΔH = 1/2(242.3 kJ) ΔH = 1/2(151.0 kJ) ΔH = 1/2(62.8 kJ)
1/2 I2(s) + 1/2 Cl2(g) → ICl(g) 139.
257
ΔH = 16.8 kJ/mol = ΔH of , ICl
|Heat gain by water| = |heat loss by Al|; we will keep all quantities positive avoid sign errors.
4.18 J o
× mass × (24.9°C − 22.3°C) = 25.0 g Al ×
Cg
1 mol Al 24.03 J o × (82.4°C − 24.9°C) 26.98 g Al C mol
11(mass) = 1280, mass = 116 g = 120 g H2O 140.
A→B B→C C→X
H = H1 H = H2 H = H3
A→X
HT = H1 + H2 + H3
HT ‒ H1 ‒ H2 ‒ H3 = 0; answer a is correct. 141.
a. C2H4(g) + O3(g) → CH3CHO(g) + O2(g) ΔH° = −166 kJ − [143 kJ + 52 kJ] = −361 kJ b. O3(g) + NO(g) → NO2(g) + O2(g) ΔH° = 34 kJ − [90. kJ + 143 kJ] = −199 kJ c. SO3(g) + H2O(l) → H2SO4(aq) ΔH° = −909 kJ − [−396 kJ + (−286 kJ)] = −227 kJ d. 2 NO(g) + O2(g) → 2 NO2(g)
ΔH° = 2(34) kJ − 2(90.) kJ = −112 kJ
142.
The relationship in answer c is incorrect. Huniverse = Hsystem + Hsurroundings, so Hsurroundings = Huniverse − Hsystem. For answer a, since Euniverse = Esystem + Esurroundings = 0 (by the first law of thermodynamics), Esystem = −Esurroundings. For answer b, H = E + PV and at constant pressure, q = H, so q = E + PV. For answer d, E = q + w and at constant volume, w = 0, so E = q at constant volume. For answer e, any state function for a cyclic process equals 0. E = 0 = q + w, so w = −q for the overall process.
143.
N2(g) + 2 O2(g) → 2 NO2(g) n N2 =
H = 67.7 kJ
PV 3.50 atm 0.250 L = 2.86 × 10 −2 mol N2 = 0.08206 L atm RT 373 K K mol
PV 3.50 atm 0.450 L = 5.15 × 10 −2 mol O2 = 0.08206 L atm RT 373 K K mol 2 mol NO 2 2.86 × 10 −2 mol N2 × = 5.72 × 10 −2 mol NO2 produced if N2 is limiting. 1 mol N 2
n O2 =
258
CHAPTER 6 5.15 × 10 −2 mol O2 ×
THERMOCHEMISTRY
2 mol NO 2 = 5.15 × 10 −2 mol NO2 produced if O2 is limiting. 2 mol O 2
O2 is limiting because it produces the smaller quantity of product. The heat required is: 5.15 × 10 −2 mol NO2 × 144.
67.7 kJ = 1.74 kJ 2 mol NO 2
a. 4 CH3NO2(l) + 3 O2(g) → 4 CO2(g) + 2 N2(g) + 6 H2O(g)
ΔH orxn = −1288.5 kJ = [4 mol(−393.5 kJ/mol) + 6 mol(−242 kJ/mol)] − [4 mol (H of , CH3NO2 )] Solving: ΔHof , CH3NO2 = −434 kJ/mol b.
Ptotal = 950. torr ×
n N2 =
0.168 atm 15.0 L = 0.0823 mol N2 0.08206 L atm 373 K K mol
0.0823 mol N2 × 145.
1 atm = 1.25 atm; PN2 = Ptotal N2 = 1.25 atm × 0.134 760 torr = 0.168 atm
28.02 g N 2 = 2.31 g N2 1 mol N 2
Heat loss by U = heat gain by heavy water; volume of cube = (cube edge) 3 Mass of heavy water = 1.00 × 103 mL × Heat gain by heavy water =
4.211 J o
Heat loss by U = 1.4 × 104 J = 7.0 × 102 g U ×
1.11 g = 1110 g mL
× 1110 g × (28.5 – 25.5)C = 1.4 × 104 J
Cg 0.117 J o
× mass × (200.0 – 28.5)C, mass = 7.0 × 102 g U
Cg
1 cm3 = 37 cm3; cube edge = (37 cm3)1/3 = 3.3 cm 19.05 g
Challenge Problems 146.
Only when there is a volume change can PV work be done. In pathway 1 (steps 1 + 2), only the first step does PV work (step 2 has a constant volume of 30.0 L). In pathway 2 (steps 3 + 4), only step 4 does PV work (step 3 has a constant volume of 10.0 L). Pathway 1: w = −PΔV = −2.00 atm(30.0 L − 10.0 L) = −40.0 L atm ×
101 .3 J L atm
= −4.05 × 103 J
CHAPTER
6
THERMOCHEMISTRY
259
Pathway 2: w = −PΔV = −1.00 atm(30.0 L − 10.0 L) = −20.0 L atm ×
101 .3 J L atm
= −2.03 × 103 J Note: The sign is negative because the system is doing work on the surroundings (an expansion). We get different values of work for the two pathways; both pathways have the same initial and final states. Because w depends on the pathway, work cannot be a state function. 147.
A(l) → A(g)
ΔHvap = 30.7 kJ; at constant pressure, ΔH = qp = 30.7 kJ
Because PV = nRT, at constant pressure and temperature: w = −PΔV = −RTΔn, where: Δn = moles of gaseous products − moles of gaseous reactants = 1 − 0 = 1 w = −RTΔn = −8.3145 J/K•mol(80. + 273 K)(1 mol) = −2940 J = −2.94 kJ ΔE = q + w = 30.7 kJ + (−2.94 kJ) = 27.8 kJ 148.
H = E + PV; Because PV = nRT, at constant pressure and temperature, PΔV = RTΔn, where Δn = moles of gaseous products − moles of gaseous reactants. H = E + PV = E + RTn, n = 2 − (2 + 1) = −1 H = E + RT(−1), H = E − RT; answer b is correct.
149.
Because PV = nRT, at constant pressure and temperature, PΔV = RTΔn, where Δn = moles of gaseous products − moles of gaseous reactants. For units of joules, use R = 8.3145 J/K·mol. H = E + RTn; RTn = 8.3145 J/K·mol(273 + 110. K)(25 − 0) = 7.96 × 104 J = 79.6 kJ H = E + RTn = −4674 kJ + 79.6 kJ = −4594 kJ
150.
Energy needed =
20. 10 3 g C12H 22O11 1 mol C12H 22O11 5640 kJ = 3.3 × 105 kJ/h h 342 .3 g C12H 22O11 mol
Energy from sun = 1.0 kW/m2 = 1000 W/m2 = 10,000 m2
1.0 kJ sm
2
Percent efficiency =
1000 J 1.0 kJ = 2 sm s m2
60 s 60 min = 3.6 × 107 kJ/h min h
3.3 10 5 kJ energy used per hour × 100 = × 100 = 0.92% total energy per hour 3.6 10 7 kJ
260 151.
CHAPTER 6
THERMOCHEMISTRY
40. kJ h 3600 s = 1.4 × 105 kJ s h 10. kJ 60 s 60 min Energy from the sun in 8.0 hours = × 8.0 h = 2.9 × 104 kJ/m2 min h s m2
Energy used in 8.0 hours = 40. kWh =
Only 22% of the sunlight is converted into electricity: 0.22 × (2.9 × 104 kJ/m2) × area = 1.4 × 105 kJ, area = 22 m2 152.
a. 2 HNO3(aq) + Na2CO3(s) → 2 NaNO3(aq) + H2O(l) + CO2(g) ΔH° = [2(−467 kJ) + (−286 kJ) + (−393.5 kJ)] − [2(−207 kJ) + (−1131 kJ)] = −69 kJ 2.0 × 104 gallons ×
4 qt 946 mL 1.42 g = 1.1 × 108 g of concentrated HNO3 gal qt mL
1.1 × 108 g solution × 7.7 × 107 g HNO3 ×
70.0 g HNO 3 = 7.7 × 107 g HNO3 100.0 g solution
1 mol HNO3 63.02 g HNO3
7.7 × 107 g HNO3 ×
1 mol Na 2 CO 3 105.99 g Na 2 CO 3 2 mol HNO3 mol Na 2 CO 3 = 6.5 × 107 g Na2CO3
1 mol HNO 3 − 69 kJ = −4.2 × 107 kJ 63.02 g HNO 3 2 mol HNO 3
4.2 × 107 kJ of heat was released. b. They feared the heat generated by the neutralization reaction would vaporize the unreacted nitric acid, causing widespread airborne contamination. 153.
400 kcal ×
4.18 kJ = 1.7 × 103 kJ 2 × 103 kJ kcal
1 kg 9.81 m 2.54 cm 1m = 160 J 200 J 8 in PE = mgz = 180 lb 2 2.205 lb in 100 cm s
200 J of energy is needed to climb one step. The total number of steps to climb are: 2 × 106 J × 154.
1 step = 1 × 104 steps 200 J
H2(g) + 1/2 O2(g) → H2O(l) ΔH° = ΔH of , H 2O( l) = −285.8 kJ; we want the reverse reaction: H2O(l) → H2(g) + 1/2 O2(g) ΔH° = 285.8 kJ w = −PV; because PV = nRT, at constant T and P, PV = RTn, where n = moles of gaseous products – moles of gaseous reactants. Here, Δn = (1 mol H2 + 0.5 mol O2) – (0) = 1.50 mol.
CHAPTER
6
THERMOCHEMISTRY
261
ΔE° = ΔH° − PΔV = ΔH° − RTΔn 1 kJ 1.50 mol ΔE° = 285.8 kJ − 8.3145 J/K • mol 298 K 1000 J
ΔE° = 285.8 kJ − 3.72 kJ = 282.1 kJ 155.
There are five parts to this problem. We need to calculate: (1) q required to heat H2O(s) from −30. C to 0C; use the specific heat capacity of H2O(s) (2) q required to convert 1 mol H2O(s) at 0C into 1 mol H2O(l) at 0C; use Hfusion (3) q required to heat H2O(l) from 0C to 100.C; use the specific heat capacity of H2O(l) (4) q required to convert 1 mol H2O(l) at 100.C into 1 mol H2O(g) at 100.C; use Hvaporization (5) q required to heat H2O(g) from 100.C to 140.C; use the specific heat capacity of H2O(g) We will sum up the heat required for all five parts, and this will be the total amount of heat required to convert 1.00 mol of H2O(s) at −30.C to H2O(g) at 140.C. q1 = 2.03 J/C•g × 18.02 g × [0 – (−30.)]C = 1.1 × 103 J q2 = 1.00 mol × 6.02 × 103 J/mol = 6.02 × 103 J q3 = 4.18 J/C•g × 18.02 g × (100. – 0)C = 7.53 × 103 J q4 = 1.00 mol × 40.7 × 103 J/mol = 4.07 × 104 J q5 = 2.02 J/C•g × 18.02 g × (140. – 100.)C = 1.5 × 103 J qtotal = q1 + q2 + q3 + q4 + q5 = 5.69 × 104 J = 56.9 kJ
156.
When a mixture of ice and water exists, the temperature of the mixture remains at 0C until all the ice has melted. Because an ice-water mixture exists at the end of the process, the temperature remains at 0C. All of the energy released by the element goes to convert ice into water. The energy required to do this is related to Hfusion = 6.02 kJ/mol (from Exercise 155). Heat loss by element = heat gain by ice cubes at 0C Heat gain = 109.5 g H2O ×
1 mol H 2 O 6.02 kJ = 36.6 kJ 18.02 g mol H 2 O
Specific heat of element =
q 36,600 J = = 0.375 J/C•g mass ΔT 500 .0 g (195 − 0) o C
262 157.
CHAPTER 6 88.0 g N2O ×
THERMOCHEMISTRY
1 mol N 2 O = 2.00 mol N2O 44.02 g N 2 O
At constant pressure, qp = ΔH. ΔH = (2.00 mol)(38.7 J/°C • mol)(55°C − 165°C) = −8510 J = −8.51 kJ = qp w = −PΔV = −nRΔT = −(2.00 mol)(8.3145 J/K•mol)(−110. K) = 1830 J = 1.83 kJ ΔE = q + w = −8.51 kJ + 1.83 kJ = −6.68 kJ
Marathon Problems 158.
Pathway I: Step 1: (5.00 mol, 3.00 atm, 15.0 L) → (5.00 mol, 3.00 atm, 55.0 L) w = −PΔV = −(3.00 atm)(55.0 L − 15.0 L) = −120. L atm w = −120. L atm ×
101 .3 J 1 kJ = −12.2 kJ L atm 1000 J
Step 1 is at constant pressure. The heat released/gained at constant pressure = q p = ΔH. From the problem, ΔH = nCpΔT for an ideal gas. Using the ideal gas law, let’s substitute for ΔT. Δ(PV) = Δ(nRT) = nRΔT for a specific gas sample. So: ΔT =
Δ(PV) nR
Δ(PV) CpΔ(PV) ; Note: Δ(PV) = (P2V2 − P1V1) = nR R 5 For an ideal monatomic gas, Cp = R; substituting into the above equation: 2 5 5 Δ(PV) ΔH = R = Δ ( PV ) 2 2 R
ΔH = qp = nCpΔT = nCp
ΔH = qp =
5 5 Δ(PV) = (3.00 atm × 55.0 L − 3.00 atm × 15.0 L) = 300. L atm 2 2
ΔH = qp = 300. L atm ×
101 .3 J 1 kJ = 30.4 kJ L atm 1000 J
ΔE = q + w = 30.4 kJ − 12.2 kJ = 18.2 kJ Note: We could have used ΔE = nCvΔT to calculate the same answer (ΔE = 18.2 kJ).
CHAPTER
6
THERMOCHEMISTRY
263
Step 2: (5.00 mol, 3.00 atm, 55.0 L) → (5.00 mol, 6.00 atm, 20.0 L) In this step, neither pressure nor volume are constant. We will need to determine q in a different way. However, it will always hold for an ideal gas that ΔE = nC vΔT and ΔH = nCpΔT. 3 Δ (PV ) 3 ΔE = nCvΔT = n R = ΔPV 2 nR 2 3 (120. − 165) L atm = −67.5 L atm (Carry an extra significant figure.) 2 101 .3 J 1 kJ ΔE = −67.5 L atm × = −6.8 kJ L atm 1000 J
ΔE =
5 Δ (PV ) 5 ΔH = nCpΔT = n R = ΔPV 2 nR 2
ΔH =
5 (120. − 165) L atm = −113 L atm 2
(Carry an extra significant figure.)
w = −PΔV = −(6.00 atm)(20.0 − 55.0) L = 210. L atm w = 210. L atm ×
101 .3 J 1 kJ = 21.3 kJ L atm 1000 J
ΔE = q + w, −6.8 kJ = q + 21.3 kJ, q = −28.1 kJ Summary:
Path I
Step 1
Step 2
Total
q w ΔE ΔH
30.4 kJ −12.2 kJ 18.2 kJ 30.4 kJ
−28.1 kJ 21.3 kJ −6.8 kJ −11 kJ
2.3 kJ 9.1 kJ 11.4 kJ 19 kJ
Pathway II: Step 3: (5.00 mol, 3.00 atm, 15.0 L) → (5.00 mol, 6.00 atm, 15.0 L) Step 3 is at constant volume. The heat released/gained at constant volume = q v = ΔE. 3 Δ (PV ) 3 ΔE = nCvΔT = n R = ΔPV 2 nR 2
ΔE = qv =
3 3 Δ(PV) = (6.00 atm × 15.0 L − 3.00 atm × 15.0 L) 2 2
ΔE = qv =
3 (90.0 − 45.0) L atm = 67.5 L atm 2
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101 .3 J 1 kJ = 6.84 kJ L atm 1000 J
w = −PΔV = 0 because ΔV = 0 ΔH = ΔE + Δ(PV) = 67.5 L atm + 45.0 L atm = 112.5 L atm = 11.40 kJ Step 4: (5.00 mol, 6.00 atm, 15.0 L) → (5.00 mol, 6.00 atm, 20.0 L) Step 4 is at constant pressure so qp = ΔH. 5 Δ (PV ) 5 ΔH = qp = nCpΔT = R = ΔPV 2 nR 2
ΔH =
5 (120. − 90.0) L atm = 75 L atm 2
ΔH = qp = 75 L atm ×
101 .3 J 1 kJ = 7.6 kJ L atm 1000 J
w = −PΔV = − (6.00 atm)(20.0 − 15.0) L = −30. L atm w = −30. L atm ×
101 .3 J 1 kJ = −3.0 kJ L atm 1000 J
ΔE = q + w = 7.6 kJ − 3.0 kJ = 4.6 kJ Summary:
Path II
Step 3
Step 4
Total
q w ΔE ΔH
6.84 kJ 0 6.84 kJ 11.40 kJ
7.6 kJ −3.0 kJ 4.6 kJ 7.6 kJ
14.4 kJ −3.0 kJ 11.4 kJ 19.0 kJ
State functions are independent of the particular pathway taken between two states; path functions are dependent on the particular pathway. In this problem, the overall values of ΔH and ΔE for the two pathways are the same (see the two summaries of results); hence ΔH and ΔE are state functions. However, the overall values of q and w for the two pathways are different; hence q and w are path functions. 159.
2 x + y/2 CxHy + → x CO2 + y/2 H2O 2
[x(−393.5) + y/2 (−242)] − ΔH oC x H y = −2044.5, − (393.5)x − 121y − ΔH C x H y = −2044.5 dgas =
P • MM , where MM = average molar mass of CO 2/H2O mixture RT
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0.751 g/L =
265
1.00 atm MM , MM of CO2/H2O mixture = 29.1 g/mol 0.08206 L atm 473 K K mol
Let a = mol CO2 and 1.00 − a = mol H2O (assuming 1.00 total moles of mixture) (44.01)a + (1.00 − a) × 18.02 = 29.1; solving: a = 0.426 mol CO2 ; mol H2O = 0.574 mol y 0.574 y = 2 , 2.69 = , y = (2.69)x Thus: 0.426 x x
For whole numbers, multiply by three, which gives y = 8, x = 3. Note that y = 16, x = 6 is possible, along with other combinations. Because the hydrocarbon has a lower density than Kr, the molar mass of CxHy must be less than the molar mass of Kr (83.80 g/mol). Only C3H8 works. −2044.5 = −393.5(3) − 121(8) − ΔHoC3H8 , ΔHoC3H8 = −104 kJ/mol
CHAPTER 7 ATOMIC STRUCTURE AND PERIODICITY Review Questions 1.
Wavelength: The distance between two consecutive peaks or troughs in a wave. Frequency: The number of waves (cycles) per second that pass a given point in space. Photon energy: The discrete units by which all electromagnetic radiation transmits energy; EMR can be viewed as a stream of “particles” called photons. Each photon has a unique quantum of energy associated with it; the photon energy is determined by the frequency (or wavelength) of the specific EMR. Speed of travel: All electromagnetic radiation travels at the same speed, c, the speed of light; c = 2.9979 × 108 m/s. = c, E = h = hc/: From these equations, wavelength and frequency are inversely related, photon energy and frequency are directly related, and photon energy and wavelength are inversely related. Thus, the EMR with the longest wavelength has the lowest frequency and least energetic photons. The EMR with the shortest wavelength has the highest frequency and most energetic photons. Using Figure 7.2 of the text to determine the wavelengths, the order is: wavelength: gamma rays < ultraviolet < visible < microwaves frequency: microwaves < visible < ultraviolet < gamma rays photon energy: microwaves < visible < ultraviolet < gamma rays speed: all travel at the same speed, c, the speed of light
2.
The Bohr model assumes that the electron in hydrogen can orbit the nucleus at specific distances from the nucleus. Each orbit has a specific energy associated with it. Therefore, the electron in hydrogen can only have specific energies; not all energies are allowed. The term quantized refers to the allowed energy levels for the electron in hydrogen. The great success of the Bohr model is that it could explain the hydrogen emission spectrum. The electron in H, moves about the allowed energy levels by absorbing or emitting certain photons of energy. The photon energies absorbed or emitted must be exactly equal to the energy difference between any two allowed energy levels. Because not all energies are allowed in hydrogen (energy is quantized), not all energies of EMR are absorbed/emitted. The Bohr model predicted the exact wavelengths of light that would be emitted for a hydrogen atom. Although the Bohr model has great success for hydrogen and other 1 electron ions,
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it does not explain emission spectra for elements/ions having more than one electron. The fundamental flaw is that we cannot know the exact motion of an electron as it moves about the nucleus; therefore, well defined circular orbits are not appropriate. 3.
Planck’s discovery that heated bodies give off only certain frequencies of light and Einstein’s study of the photoelectric effect support the quantum theory of light. The wave-particle duality is summed up by saying all matter exhibits both particulate and wave properties. Electromagnetic radiation, which was thought to be a pure waveform, transmits energy as if it has particulate properties. Conversely, electrons, which were thought to be particles, have a wavelength associated with them. This is true for all matter. Some evidence supporting wave properties of matter are: 1.
Electrons can be diffracted like light.
2. The electron microscope uses electrons in a fashion similar to the way in which light is used in a light microscope. However, wave properties of matter are only important for small particles with a tiny mass, e.g., electrons. The wave properties of larger particles are not significant. 4.
Four scientists whose work was extremely important to the development of the quantum mechanical model were Niels Bohr, Louis deBroglie, Werner Heisenberg, and Erwin Schrödinger. The Bohr model of the atom presented the idea of quantized energy levels for electrons in atoms. DeBroglie came up with the relationship between mass and wavelength, supporting the idea that all matter (especially tiny particles like electrons) exhibits wave properties as well as the classic properties of matter. Heisenberg is best known for his uncertainty principle which states there is a fundamental limitation to just how precisely we can know both the position and the momentum of a particle at a given time. If we know one quantity accurately, we cannot absolutely determine the other. The uncertainty principle, when applied to electrons, forbids well-defined circular orbits for the electron in hydrogen, as presented in the Bohr model. When we talk about the location of an electron, we can only talk about the probability of where the electron is located. Schrödinger put the ideas presented by the scientists of the day into a mathematical equation. He assumed wave motion for the electron. The solutions to this complicated mathematical equation give allowed energy levels for the electrons. These solutions are called wave functions, , and the allowed energy levels are often referred to as orbitals. In addition, the square of the wave function (2) indicates the probability of finding an electron near a particular point in space. When we talk about the shape of an orbital, we are talking about a surface that encompasses where the electron is located 90% of the time. The key is we can only talk about probabilities when referencing electron location.
5.
Quantum numbers give the allowed solutions to Schrödinger equation. Each solution is an allowed energy level called a wave function or an orbital. Each wave function solution is described by three quantum numbers, n, ℓ, and mℓ. The physical significance of the quantum numbers are: n: gives the energy (it completely specifies the energy only for the H atom or ions with one electron) and the relative size of the orbitals. ℓ: gives the type (shape) of orbital. mℓ: gives information about the direction in which the orbital is pointing.
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The specific rules for assigning values to the quantum numbers n, ℓ, and mℓ are covered in Section 7.6 of the text. In Section 7.8, the spin quantum number ms is discussed. Since we cannot locate electrons, we cannot see if they are spinning. The spin is a convenient model. It refers to the ability of the two electrons that can occupy any specific orbital to produce two different oriented magnetic moments. 6.
The 2p orbitals differ from each other in the direction in which they point in space. The 2p and 3p orbitals differ from each other in their size, energy and number of nodes. A nodal surface in an atomic orbital is a surface in which the probability of finding an electron is zero. The 1p, 1d, 2d, 1f, 2f, and 3f orbitals are not allowed solutions to the Schrödinger equation. For n = 1, ℓ 1, 2, 3, etc., so 1p, 1d, and 1f orbitals are forbidden. For n = 2, ℓ 2, 3, 4, etc., so 2d and 2f orbitals are forbidden. For n = 3, ℓ 3, 4, 5, etc., so 3f orbitals are forbidden. The penetrating term refers to the fact that there is a higher probability of finding a 4s electron closer to the nucleus than a 3d electron. This leads to a lower energy for the 4s orbital relative to the 3d orbitals in polyelectronic atoms and ions.
7.
The four blocks are the s, p, d, and f blocks. The s block contains the alkali and alkaline earth metals (Groups 1A and 2A). The p block contains the elements in Groups 3A, 4A, 5A, 6A, 7A, and 8A. The d block contains the transition metals. The f block contains the inner transition metals. The energy ordering is obtained by sequentially following the atomic numbers of the elements through the periodic table while keeping track of the various blocks you are transversing. The periodic table method for determining energy ordering is illustrated in Figure 7.26 of the text. The Aufbau principle states that as protons are added one by one to the nucleus to build up the elements, electrons are similarly added to hydrogenlike orbitals. The main assumptions are that all atoms have the same types of orbitals and that the most stable electron configuration, the ground state, has the electrons occupying the lowest energy levels first. Hund’s rule refers to adding electrons to degenerate (same energy) orbitals. The rule states that the lowest energy configuration for an atom is the one having the maximum number of unpaired electrons allowed by the Pauli exclusion principle. The Pauli exclusion principle states that in a given atom, no two electrons can have the same four quantum numbers. This corresponds to having only two electrons in any one orbital, and they must have opposite “spins” if the electrons are in the same orbital. The two major exceptions to the predicted electron configurations for elements 1-36 are Cr and Cu. The expected electron configurations for each are: Cr: [Ar]4s23d4 and Cu: [Ar]4s23d9 The actual electron configurations are: Cr: [Ar]4s13d5 and Cu: [Ar]4s13d10
8.
Valence electrons are the electrons in the outermost principal quantum level of an atom (those electrons in the highest n value orbitals). The electrons in the lower n value orbitals are all inner core or just core electrons. The key is that the outer most electrons are the valence electrons. When atoms interact with each other, it will be the outermost electrons that are involved in
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these interactions. In addition, how tightly the nucleus holds these outermost electrons determines atomic size, ionization energy and other properties of atoms. Elements in the same group have similar valence electron configurations and, as a result, have similar chemical properties. 9.
Ionization energy: P(g) → P+(g) + e−; electron affinity: P(g) + e− → P−(g) Across a period, the positive charge from the nucleus increases as protons are added. The number of electrons also increase, but these outer electrons do not completely shield the increasing nuclear charge from each other. The general result is that the outer electrons are more strongly bound as one goes across a period which results in larger ionization energies (and smaller size). Aluminum is out of order because the electrons in the filled 3s orbital shield some of the nuclear charge from the 3p electron. Hence, the 3p electron is less tightly bound than a 3s electron, resulting in a lower ionization energy for aluminum as compared to magnesium. The ionization energy of sulfur is lower than phosphorus because of the extra electron-electron repulsions in the doubly occupied sulfur 3p orbital. These added repulsions, which are not present in phosphorus, make it slightly easier to remove an electron from sulfur as compared to phosphorus. As successive electrons are removed, the net positive charge on the resultant ion increases. This increase in positive charge binds the remaining electrons more firmly, and the ionization energy increases. The electron configuration for Si is 1s22s22p63s23p2. There is a large jump in ionization energy when going from the removal of valence electrons to the removal of core electrons. For silicon, this occurs when the fifth electron is removed since we go from the valence electrons in n = 3 to the core electrons in n = 2. There should be another big jump when the thirteenth electron is removed, i.e., when a 1s electron is removed.
10.
Both trends are a function of how tightly the outermost electrons are held by the positive charge in the nucleus. An atom where the outermost electrons are held tightly will have a small radius and a large ionization energy. Conversely, an atom where the outermost electrons are held weakly will have a large radius and a small ionization energy. The trends of radius and ionization energy should be opposite of each other. Electron affinity is the energy change associated with the addition of an electron to a gaseous atom. Ionization energy is the energy it takes to remove an electron from a gaseous atom. Because electrons are always attracted to the positive charge of the nucleus, energy will always have to be added to break the attraction and remove the electron from a neutral charged atom. Ionization energies are always endothermic for neutral charged atoms. Adding an electron is more complicated. The added electron will be attracted to the nucleus; this attraction results in energy being released. However, the added electron will encounter the other electrons which results in electron-electron repulsions; energy must be added to overcome these repulsions. Which of the two opposing factors dominates determines whether the overall electron affinity for an element is exothermic or endothermic.
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Active Learning Questions 1.
Wavelike properties refer to movement that is characterized by a wavelength and a frequency. Particulate properties refer to anything having a mass and a velocity along with a certain kinetic energy. Planck’s discovery that heated bodies give off only certain frequencies of light and Einstein’s study of the photoelectric effect support the quantum theory of light. The wave-particle duality is summed up by saying all matter exhibits both particulate and wave properties. Electromagnetic radiation, which was thought to be a pure waveform, transmits energy as if it has particulate properties. Conversely, electrons, which were thought to be particles, have a wavelength associated with them. This is true for all matter. Some evidence supporting wave properties of matter are: 1.
Electrons can be diffracted like light.
2. The electron microscope uses electrons in a fashion similar to the way in which light is used in a light microscope. However, wave properties of matter are only important for small particles with a tiny mass, e.g., electrons. The wave properties of larger particles are not significant. 2.
The Bohr model assumes that the electron in hydrogen can orbit the nucleus at specific distances from the nucleus. Each orbit has a specific energy associated with it. Therefore, the electron in hydrogen can only have specific energies; not all energies are allowed. The term quantized refers to the allowed energy levels for the electron in hydrogen. The great success of the Bohr model is that it could explain the hydrogen emission spectrum. The electron in H moves about the allowed energy levels by absorbing or emitting certain photons of energy. The photon energies absorbed or emitted must exactly equal the energy difference between any two allowed energy levels. Because not all energies are allowed in hydrogen (energy is quantized), not all energies of EMR are absorbed/emitted. The Bohr model predicted the exact wavelengths of light that would be emitted for a hydrogen atom. Although the Bohr model has great success for hydrogen and other 1 electron ions, it does not explain emission spectra for elements/ions having more than one electron. A fundamental flaw is that we cannot know the exact motion of an electron as it moves about the nucleus; therefore, well defined circular orbits are not appropriate. Another flaw is that it does not consider the electron-electron repulsions in atoms/ions containing more than 1 electron.
3.
Each element shows a very large energy change between the second and third ionization energies. This suggests that both elements have two valence electrons, so they are alkaline earth metals. Because all ionization energies for element Y are larger than those of element X, element Y should be above element X in the periodic table. One possible example is Y = Mg and X = Ca. Of course, there are other combinations from group 2A that would work.
4.
Consecutive ionization energies always increase. So the second ionization energy of He would be greater than the first ionization energy. For the first ionization energy, helium has 2 electrons attracted to a nucleus containing two protons. For the second ionization energy, there is just a single electron attracted to the two protons in the nucleus. The nucleus exerts a stronger
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attraction when there are fewer electrons present. Hence the second ionization energy is larger than the first ionization energy. 5.
Li: 1s22s1; Be: 1s22s2: The second ionization energy for beryllium removes an electron from the valence n = 2 level. The second ionization energy for Li removes an electron from an inner core n = 1 level. Inner core electrons are much closer to the nucleus, on average, and are significantly more attracted to the nucleus than valence electrons. Hence. the second ionization energy for lithium where an inner core electron is removed should be larger.
6.
As atomic number increases across a row, ionization energy generally increases. There are two major exceptions to this trend for elements 1-36. There is a decrease in ionization when going from a 2A to a 3A element and from a 5A to a 6A element. To explain these, let’s look at specific examples in row 3. Aluminum is out of order because the electrons in the filled 3s orbital shield some of the nuclear charge from the 3p electron. Hence, the 3p electron is less tightly bound than a 3s electron, resulting in a lower ionization energy for aluminum as compared to magnesium. The ionization energy of sulfur is lower than phosphorus because of the extra electron-electron repulsions in the doubly occupied sulfur 3p orbital. These added repulsions, which are not present in phosphorus, make it slightly easier to remove an electron from sulfur as compared to phosphorus.
7.
The second ionization energy for sodium removes an inner core electron. The second ionization energy for the remaining elements in row 3 remove valence electrons. Inner core electrons always require significantly more energy to remove as compared to valence electrons. So sodium will have the largest second ionization energy. We would expect the general trend to apply for the remaining elements in row three. That is, we would expect there to be an increase in the second ionization energy from magnesium to argon (with perhaps two exceptions to mirror the first ionization energy trend exceptions).
8.
There are several possibilities for a rational answer. One possibility is that nonmetals generally have smaller size and larger ionization energies. As one goes vertically down the periodic table, size increases while ionization energy increases. This results in a decrease in nonmetallic properties as one goes down a group. As one goes across the periodic table from left to right, size decreases and ionization energy increases. This results in nonmetallic behavior increasing left to right across a periodic table. Combining these two trends, one would expect the line dividing metals from nonmetals to be a diagonal line going downward from left to right. This is what is required to have the correct nonmetal combination of relatively small size and relatively large ionization energies.
9.
In general metals have low ionization energy as compared to nonmetals. From this, it is easier for metals to lose electrons as compared to nonmetals. The electron affinities of metals are generally less negative than the nonmetals. Therefore, it is more energetically favorable for nonmetals to gain electrons as compared to metals. These two ideas together help explain why metals tend to lose electrons and nonmetals then to gain electrons
10.
An electron has wave and particle properties, and its energy is quantized. An electron’s exact location in an atom can never be known. Instead, we talk about probabilities of where electrons are in the various allowed energy levels (orbitals). Since we cannot know specifically where the electrons are in an atom, measurements of size (atomic radius) cannot be determined exactly. In terms of the quantum mechanical model, the concept of an electron is vague, not the typical negative charged particle with a known orbit concept.
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11.
One would expect the Li 1s orbital to be smaller than the H 1s orbital because of the greater nuclear charge for Li. In terms of size, H is smaller than Li. Li has 2s electrons in addition to the 1s electrons. The 2s electron for Li will be on average further from the nucleus than the 1s electron for H, making Li larger than H.
12.
The H emission spectrum is just looking at the transitions in the visible region of the spectrum. If we were to look at the infrared or ultraviolet region, we would see a lot more transitions.
13.
The ground state is the lowest energy arrangement of the electrons in the allowed energy levels. There are an infinite number of allowed energy states for electrons. Electrons in the ground state can absorb electromagnetic radiation and go up into higher energy levels. This is called an excited state electron configuration, which are always higher in energy that the ground state configuration and less stable. An electron configuration that has electrons in allowed energy levels and follows the Pauli Exclusion Principle that is not the ground state is an excited state electron configuration. There are an infinite number of excited state (higher energy) configurations, but there is only one ground state (lowest energy) configuration.
14.
Cl(g) → Cl+(g) + e‒ ΔH = ionization energy for Cl; the electron affinity energy changes for the choices a-d are: a. Cl(g) + e‒ → Cl‒(g); b. Cl‒ + e‒ → Cl2‒; c. Cl+ + e‒ → Cl; d. F + e‒ → F‒ The electron affinity of Cl+ (answer c) is the exact opposite of the ionization energy for Cl. So, these two processes are equal in magnitude but opposite in sign. The ionization energy for Cl will be positive while the electron affinity for Cl+ will be positive.
15.
Electrons in an atom are attracted to the nucleus of that atom. To remove an electron, energy must be added to break this attraction. So, ionization energies are never negative quantities. It is correct that when K loses an electron it does obtain a noble gas electron configuration. But it does not lose an electron unless energy is added.
16.
In going down a column, the added electrons go into higher n value orbitals. These higher n value orbitals have the electrons further from the nucleus, on average, resulting in a larger atomic size and smaller ionization energies. As electrons are added going across the periodic table, the added electrons go into the same n value orbitals. However, electrons in the same n value orbitals don’t completely shield the increasing nuclear charge caused by the added protons. This results is in an increase in effective nuclear charge as one goes across the periodic table, so the electrons are held more strongly and ionization energy increases (size decreases).
17.
In our current model of the atom, we can’t know exactly where electrons are. Instead, we can only talk about probabilities of where electrons are when in the various allowed energy levels (atomic orbitals).
18.
An orbital is an allowed energy level for an electron.
19.
A probability density distribution is the square of the wave function for an orbital indicating the probability of finding an electron in that orbital at a particular point in space. This probability is greatest close to the nucleus and drops off rapidly as the distance from the nucleus increases.
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The radial probability distribution represents the total probability of finding the electron at a particular distance from the nucleus in an orbital. For the distribution in the 1s wave function, the 1s orbital is divided into successive thin spherical shells and the total probability of finding the electron in each spherical shell is plotted versus distance from the nucleus. 20.
Hydrogen has all of the various orbitals, including a 3s orbital. For hydrogen, the energy of a particular orbital is just determined by n. Thus a 3s orbital is degenerate (has the same energy) as the 3p orbitals which have the same energy as the 3d orbitals.
21.
In hydrogen, a 2s and 2p orbital have the same energy. For hydrogen, the energy of a particular orbital is just determined by n. In species with more than one electron (like helium), the 2s orbital has a lower energy that the 2p orbitals. In polyelectronic species, the ns orbitals have lower energy than the np orbitals, which are lower than the nd orbitals, etc.
22.
The ionization energy for sodium is 495 kJ/mol and the ionization energy for fluorine is 1681 kJ. Each of these values represents the energy it takes for any substance to remove an electron from that atom (in the gaseous state). Since sodium has the smaller ionization energy, then it is easier to remove an electron from sodium than from fluorine.
23.
The equations relating the terms are = c, E = h, and E = hc/. From the equations, wavelength and frequency are inversely related, photon energy and frequency are directly related, and photon energy and wavelength are inversely related. In Figure 7.2, the EMR with the shortest wavelengths are gamma rays and the EMR with the longest wavelengths are AM radio waves. From the relationships discussed previously, gamma rays will have the highest frequency and the most energetic photons. Radio waves will have the lowest frequency and least energetic photons.
24.
The atom with the lowest energy valence electron will require the most energy to remove an electron. Hence, the atom with the lowest energy valence electrons will have the highest energy ionization energy. Atom X has the larger ionization energy.
25.
The first period corresponds to n = 1 which can only have 1s orbitals. The 1s orbital could hold 3 electrons; hence the first period would have three elements. The second period corresponds to n = 2, which has 2s and 2p orbitals. These four orbitals can each hold three electrons. A total of 12 elements would be in the second period. The third period has 3s and 3p orbitals, so the third row will also have 12 elements. The first three rows of the periodic table will have 3 + 12 +12 = 27 elements instead of 18 as we have in our current periodic table. A sketch of the s and p block elements for the first three rows follows.
The first three noble gases would be 3, 15, and 27. The next noble gas would have 4s, 3d, and 4p orbitals. If each of the five 3d orbitals could hold 3 electrons, the first of the row transition metals would be 15 elements long. This makes the fourth row of the periodic table 27 elements long vs the current 18 elements long. The next noble gas after 27 will be 54, and the next noble gas after 54 will be 51 + 27 = 81. The last orbitals filled for the actinide series are 5f orbitals. With 7 degenerate 5f orbitals and 3 electrons in each orbital, the actinide series will be 21 elements long.
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The third ionization energy refers to the following process: E 2+(g) → E3+(g) + e− H = I3. The electron configurations for the 2+ charged ions of Na to Ar are: Na2+: 1s22s22p5 Mg2+: 1s22s22p6 Al2+: [Ne]3s1 Si2+: [Ne]3s2 P2+: [Ne]3s23p1 S2+: [Ne]3s23p2 Cl2+: [Ne]3s23p3 Ar2+: [Ne]3s23p4 I3 for sodium and magnesium should be extremely large compared with the others because n = 2 electrons are much more difficult to remove than n = 3 electrons. Between Na2+ and Mg2+, one would expect to have the same trend as seen with I1(F) versus I1(Ne); these neutral atoms have identical electron configurations to Na2+ and Mg2+. Therefore, the 1s22s22p5 ion (Na2+) should have a lower ionization energy than the 1s22s22p6 ion (Mg2+). The remaining 2+ ions (Al2+ to Ar2+) should follow the same trend as the neutral atoms having the same electron configurations. The general ionization energy trend predicts an increase from [Ne]3s1 to [Ne]3s23p4. The exceptions occur between [Ne]3s2 and [Ne]3s23p1 and between [Ne]3s23p3 and [Ne]3s23p4. [Ne]3s23p1 is out of order because of the small penetrating ability of the 3p electron as compared with the 3s electrons. [Ne]3s 23p4 is out of order because of the extra electron-electron repulsions present when two electrons are paired in the same orbital. Therefore, the correct ordering for Al2+ to Ar2+ should be Al2+ < P2+ < Si2+ < S2+ < Ar2+ < Cl2+, where P2+ and Ar2+ are out of line for the same reasons that Al and S are out of line in the general ionization energy trend for neutral atoms. A qualitative plot of the third ionization energies for elements Na through Ar follows.
Note: The actual numbers in Table 7.4 support most of this plot. No I3 is given for Na2+, so you cannot check this. The only deviation from our discussion is I 3 for Ar2+ which is greater than I3 for Cl2+ instead of less than.
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Questions 27.
The equations relating the terms are = c, E = h, and E = hc/. Frequency is defined as the number of wave cycles that pass a certain part per second, so statement a is true. Statement c is true because wavelength and frequency are inversely related. Statements b and c are false. All electromagnetic radiation travels at the same speed, the speed of light, c. And frequency and photon energy are directly related. As the frequency is halved, photon energy is halved.
28.
Frequency is the number of waves (cycles) of electromagnetic radiation per second that pass a given point in space. Speed refers to the distance a wave travels per unit time. All electromagnetic radiation (EMR) travels at the same speed (c, the speed of light = 2.998 × 108 m/s). However, each wavelength of EMR has its own unique frequency.
29.
The photoelectric effect refers to the phenomenon in which electrons are emitted from the surface of a metal when light strikes it. The light must have a certain minimum frequency (energy) to remove electrons from the surface of a metal. Light having a frequency below the minimum value results in no electrons being emitted, whereas light at or higher than the minimum frequency does cause electrons to be emitted. For light having a frequency higher than the minimum frequency, the excess energy is transferred into kinetic energy for the emitted electron. Albert Einstein explained the photoelectric effect by applying quantum theory.
30.
The emission of light by excited atoms has been the key interconnection between the macroscopic world we can observe and measure, and what is happening on a microscopic basis within an atom. Excited atoms emit light (which we can observe and measure) because of changes in the microscopic structure of the atom. By studying the emissions of atoms, we can trace back to what happened inside the atom. Specifically, our current model of the atom relates the energy of light emitted to electrons in the atom moving from higher allowed energy states to lower allowed energy states.
31.
Example 7.3 calculates the deBroglie wavelength of a ball and of an electron. The ball has a wavelength on the order of 10 −34 m. This is incredibly short and, as far as the wave- particle duality is concerned, the wave properties of large objects are insignificant. The electron, with its tiny mass, also has a short wavelength; on the order of 10 −10 m. However, this wavelength is significant because it is on the same order as the spacing between atoms in a typical crystal. For very tiny objects like electrons, the wave properties are important. The wave properties must be considered, along with the particle properties, when hypothesizing about the electron motion in an atom.
32. a. For hydrogen (Z = 1), the energy levels in units of joules are given by the equation E n = −2.178 10−18(1/n2). As n increases, the differences between 1/n2 for consecutive energy levels becomes smaller and smaller. Consider the difference between 1/n2 values for n = 1 and n = 2 as compared to n = 3 and n = 4. For n = 1 and n = 2:
1 1 − 2 = 1 − 0.25 = 0.75 2 1 2
For n = 3 and n = 4:
1 1 − 2 = 0.1111 – 0.0625 = 0.0486 2 3 4
Because the energy differences between 1/n2 values for consecutive energy levels decrease as n increases, the energy levels get closer together as n increases.
276
CHAPTER 7
ATOMIC STRUCTURE AND PERIODICITY
b. For a spectral transition for hydrogen, ΔE = Ef − Ei:
1 1 ΔE = −2.178 × 10 −18 J 2 − 2 ni nf where ni and nf are the levels of the initial and final states, respectively. A positive value of ΔE always corresponds to an absorption of light, and a negative value of ΔE always corresponds to an emission of light. In the diagram, the red line is for the ni = 3 to nf = 2 transition.
1 1 1 1 ΔE = −2.178 × 10 −18 J 2 − 2 = −2.178 × 10 −18 J − 3 2 4 9 ΔE = −2.178 × 10 −18 J × (0.2500 − 0.1111) = −3.025 × 10 −19 J The photon of light must have precisely this energy (3.025 × 10 −19 J). |ΔE| = Ephoton = hν =
=
hc λ
hc 6.6261 10 −34 J s 2.9979 10 8 m/s = = 6.567 × 10 −7 m = 656.7 nm −19 | ΔE | 3.025 10 J
From Figure 7.2, = 656.7 nm is red light so the diagram is correct for the red line. In the diagram, the green line is for the ni = 4 to nf = 2 transition.
1 1 ΔE = −2.178 × 10 −18 J 2 − 2 = −4.084 × 10 −19 J 4 2
=
hc 6.6261 10 −34 J s 2.9979 108 m/s = = 4.864 × 10 −7 m = 486.4 nm −19 | ΔE | 4.084 10 J
From Figure 7.2, = 486.4 nm is green-blue light. The diagram is consistent with this line. In the diagram, the blue line is for the ni = 5 to nf = 2 transition.
1 1 ΔE = −2.178 × 10 −18 J 2 − 2 = −4.574 × 10 −19 J 5 2
=
hc 6.6261 10 −34 J s 2.9979 10 8 m/s = = 4.343 × 10 −7 m = 434.3 nm | ΔE | 4.574 10 −19 J
From Figure 7.2, = 434.3 nm is blue or blue-violet light. The diagram is consistent with this line also. 33.
Reference the energy-level diagram in given in question 32. Notice that the energy levels get closer and closer together as n increases. For hydrogen (Z = 1), the energy levels in units of joules are given by the equation En = −2.178 10−18(1/n2). As n increases, the differences between 1/n2 for consecutive energy levels becomes smaller and smaller.
CHAPTER 7
ATOMIC STRUCTURE AND PERIODICITY
277
If you draw in the electronic transitions in this problem on an energy level diagram, you can see that the 3 → 1 transition (answer e) will have the largest ΔE value, which will correspond to the emission having the shortest wavelength transition (E = hc/). The other transitions that differ by 2 energy levels will have much smaller ΔE values and longer wavelength transitions. The 2 → 1 transition will have the next shortest wavelength transition. The largest differences in allowed energy levels is between allowed energy states with the lowest n value. 34.
The uncertainty principle implies that we cannot know the exact motion of an electron as it moves around the nucleus. Therefore, simple circular orbits (answer b) at an exact distance from the nucleus are not allowed by the uncertainty principle.
35.
The Bohr model was an important step in the development of the current quantum mechanical model of the atom. The idea that electrons can only occupy certain, allowed energy levels is illustrated nicely (and relatively easily). We talk about the Bohr model to present the idea of quantized energy levels.
36.
The figure on the left tells us that the probability of finding the electron in the 1s orbital at points along a line drawn outward from the nucleus in any direction. This probability is greatest close to the nucleus and drops off rapidly as the distance from the nucleus increases. The figure on the right represents the total probability of finding the electron at a particular distance from the nucleus for a 1s hydrogen orbital. For this distribution, the hydrogen 1s orbital is divided into successive thin spherical shells and the total probability of finding the electron in each spherical shell is plotted versus distance from the nucleus. This graph is called the radial probability distribution. The radial probability distribution initially shows a steady increase with distance from the nucleus, reaches a maximum, then shows a steady decrease. Even though it is likely to find an electron near the nucleus, the volume of the spherical shell close to the nucleus is tiny, resulting in a low radial probability. The maximum radial probability distribution occurs at a distance of 5.29 × 10 −2 nm from the nucleus; the electron is most likely to be found in the volume of the shell centered at this distance from the nucleus. The 5.29 × 10 −2 nm distance is the exact radius of innermost (n = 1) orbit in the Bohr model.
37.
When the p and d orbital functions are evaluated at various points in space, the results sometimes have positive values and sometimes have negative values. The term phase is often associated with the + and − signs. For example, a sine wave has alternating positive and negative phases. This is analogous to the positive and negative values (phases) in the p and d orbitals.
38.
The widths of the various blocks in the periodic table are determined by the number of electrons that can occupy the specific orbital(s). In the s block, we have one orbital (ℓ = 0, mℓ = 0) that can hold two electrons; the s block is two elements wide. For the f block, there are 7 degenerate f orbitals (ℓ = 3, mℓ = −3, −2, −1, 0, 1, 2, 3), so the f block is 14 elements wide. The g block corresponds to ℓ = 4. The number of degenerate g orbitals is 9. This comes from the 9 possible mℓ values when ℓ = 4 (mℓ = −4, −3, −2, −1, 0, 1, 2, 3, 4). With 9 orbitals, each orbital holding two electrons, the g block would be 18 elements wide. The h block has ℓ = 5, mℓ = −5, −4, −3, −2, −1, 0, 1, 2, 3, 4, 5. With 11 degenerate h orbitals, the h block would be 22 elements wide.
278 39.
CHAPTER 7
ATOMIC STRUCTURE AND PERIODICITY
a. The magnetic quantum number (mℓ) is related to the orientation in space of an orbital in relation to the other orbitals in the atom. b. The principal quantum number (n) is related to the size and the energy of the orbital. c. The electron spin quantum number (ms) is related to the two oppositely aligned magnetic moments produced by an electron. d. The angular momentum quantum number (ℓ) is related to the shape of an atomic orbital.
40.
The ground state electron configuration is the lowest energy arrangement of the electrons in an atom or ion. An excited state in any other possible arrangement of electrons in the allowed energy levels that follows the Pauli exclusion principle (in any one orbital, the electrons must have oppositely aligned spins). There is only one lowest energy arrangement for the electrons in an atom or ion, hence there is only one possible ground state electron configuration. Since there are an infinite number of allowed energy levels for any atom, there are an infinite number of ways to arrange the electrons in these orbitals. Therefore, there are an infinite number of excited state electron configurations for an atom or ion.
41.
If one more electron is added to a half-filled subshell, electron-electron repulsions will increase because two electrons must now occupy the same atomic orbital. This may slightly decrease the stability of the atom.
42.
Size decreases from left to right and increases going down the periodic table. Thus, going one element right and one element down would result in a similar size for the two elements diagonal to each other. The ionization energies will be similar for the diagonal elements since the periodic trends also oppose each other. Electron affinities are harder to predict, but atoms with similar sizes and ionization energies should also have similar electron affinities.
43.
The valence electrons are strongly attracted to the nucleus for elements with large ionization energies. One would expect these species to readily accept another electron and have very exothermic electron affinities. The noble gases are an exception; they have a large ionization energy but have an endothermic electron affinity. Noble gases have a filled valence shell of electrons. The added electron in a noble gas must go into a higher n value atomic orbital, having a significantly higher energy, and this is very unfavorable.
44.
Electron-electron repulsions become more important when we try to add electrons to an atom. From the standpoint of electron-electron repulsions, larger atoms would have more favorable (more exothermic) electron affinities. Considering only electron-nucleus attractions, smaller atoms would be expected to have the more favorable (more exothermic) electron affinities. These trends are exactly the opposite of each other. Thus, the overall variation in electron affinity is not as great as ionization energy in which attractions to the nucleus dominate.
45.
For hydrogen and hydrogen-like (one-electron ions), all atomic orbitals with the same n value have the same energy. For polyatomic atoms/ions, the energy of the atomic orbitals also depends on ℓ. Because there are more nondegenerate energy levels for polyatomic atoms/ions as compared to hydrogen, there are many more possible electronic transitions resulting in more complicated line spectra.
CHAPTER 7
ATOMIC STRUCTURE AND PERIODICITY
279
46.
Each element has a characteristic spectrum because each element has unique energy levels. Thus the presence of the characteristic spectral lines of an element confirms its presence in any particular sample.
47.
Yes, the maximum number of unpaired electrons in any configuration corresponds to a minimum in electron-electron repulsions.
48.
The electron is no longer part of that atom. The proton and electron are completely separated.
49.
a. The energy levels for a one electron atom or ion only depend on n. For example, the 3s, 3p, and 3d orbitals have the same energy for hydrogen or any other one electron species. b. The energy levels for an atom or ion with more than one electron depend on n and ℓ. For example, the 3s, 3p, and 3d orbitals have different energies for atoms or ions with more than one electron.
50.
None of the noble gases and no subatomic particles had been discovered when Mendeleev published his periodic table. Thus there was no element out of place in terms of reactivity and there was no reason to predict an entire family of elements. Mendeleev ordered his table by mass; he had no way of knowing there were gaps in atomic numbers (they hadn't been invented yet).
Exercises Light and Matter c
2.998 10 8 m/s = 4.5 × 1014 s−1 1m 660 nm 1 10 9 nm
51.
ν =
52.
99.5 MHz = 99.5 × 106 Hz = 99.5 × 106 s −1 ; =
53.
ν=
=
2.998 10 8 m/s c = = 3.01 m v 99.5 10 6 s −1
c 3.00 10 8 m/s = = 3.0 × 1010 s −1 λ 1.0 10 − 2 m
E = hν = 6.63 × 10−34 J s × 3.0 × 1010 s −1 = 2.0 × 10 −23 J/photon
2.0 10 −23 J 6.02 10 23 photons = 12 J/mol photon mol
54.
E = hν =
6.63 10 −34 J s 3.00 10 8 m/s hc = = 8.0 × 10 −18 J/photon 1m λ 25 nm 1 10 9 nm
8.0 10 −18 J 6.02 10 23 photons = 4.8 × 106 J/mol photon mol
280 55.
CHAPTER 7 280 nm: =
320 nm: =
c = λ
ATOMIC STRUCTURE AND PERIODICITY
3.00 10 8 m/s = 1.1 × 1015 s −1 1m 280 nm 1 10 9 nm
3.00 10 8 m/s 320 10 −9 nm
= 9.4 × 1014 s −1
The compounds in the sunscreen absorb ultraviolet B (UVB) electromagnetic radiation having a frequency from 9.4 × 1014 s −1 to 1.1 × 1015 s −1 . 56.
S-type cone receptors: =
2.998 10 8 m/s c = = 5.00 × 10 −7 m = 500. nm 14 −1 ν 6.00 10 s
λ =
2.998 10 8 m/s 7.49 10 s 14
−1
= 4.00 × 10 −7 m = 400. nm
S-type cone receptors detect 400-500 nm light. From Figure 7.2 in the text, this is violet to green light, respectively. M-type cone receptors: λ =
λ =
2.998 10 8 m/s 4.76 10 s 14
−1
2.998 10 8 m/s 6.62 10 s 14
−1
= 6.30 × 10 −7 m = 630. nm = 4.53 × 10 −7 m = 453 nm
M-type cone receptors detect 450-630 nm light. From Figure 7.2 in the text, this is blue to orange light, respectively. L-type cone receptors: λ =
λ =
2.998 10 8 m/s 4.28 10 s 14
−1
2.998 10 8 m/s 6.00 10 s 14
−1
= 7.00 × 10 −7 m = 700. nm = 5.00 × 10 −7 m = 500. nm
L-type cone receptors detect 500-700 nm light. respectively. 57.
This represents green to red light,
The wavelength is the distance between consecutive wave peaks. Wave a shows 4 wavelengths, and wave b shows 8 wavelengths. Wave a: λ =
1.6 10 −3 m = 4.0 × 10−4 m 4
Wave b: λ =
1.6 10 −3 m = 2.0 × 10−4 m 8
Wave a has the longer wavelength. Because frequency and photon energy are both inversely proportional to wavelength, wave b will have the higher frequency and larger photon energy since it has the shorter wavelength.
CHAPTER 7
ATOMIC STRUCTURE AND PERIODICITY
ν =
2.998 10 8 m/s c = = 1.5 × 1012 s−1 λ 2.0 10 − 4 m
E =
6.626 10 −34 J s 2.998 10 8 m/s hc = = 9.9 × 10−22 J λ 2.0 10 − 4 m
281
Because both waves are examples of electromagnetic radiation, both waves travel at the same speed, c, the speed of light. From Figure 7.2 of the text, both of these waves represent infrared electromagnetic radiation. 58.
Referencing Figure 7.2 of the text, 2.12 × 10−10 m electromagnetic radiation is X rays.
λ=
c 2.9979 10 8 m / s = = 2.799 m ν 107 .1 10 6 s −1
From the wavelength calculated above, 107.1 MHz electromagnetic radiation is FM radiowaves. hc 6.626 10 −34 J s 2.998 10 8 m / s λ= = = 5.00 × 10−7 m E 3.97 10 −19 J The 3.97 × 10−19 J/photon electromagnetic radiation is visible (green) light. The photon energy and frequency order will be the exact opposite of the wavelength ordering because E and ν are both inversely related to λ. From the previously calculated wavelengths, the order of photon energy and frequency is: FM radiowaves < visible (green) light < X rays longest λ shortest λ lowest ν highest ν smallest E largest E 59.
a. =
3.00 10 8 m/s c = = 5.0 × 10 −6 m 13 −1 ν 6.0 10 s
b. From Figure 7.2, this is infrared electromagnetic radiation. c. E = h = 6.63 × 10 −34 J s × 6.0 × 1013 s −1 = 4.0 × 10 −20 J/photon 4.0 10 −20 J 6.022 10 23 photons = 2.4 × 104 J/mol photon mol
d. Frequency and photon energy are directly related (E = hv). Because 5.4 × 10 13 s −1 electromagnetic radiation (EMR) has a lower frequency than 6.0 × 10 13 s −1 EMR, the 5.4 × 1013 s −1 EMR will have less energetic photons. 60.
Ephoton = h =
6.626 10 −34 J s 2.998 10 8 m/s hc , E photon = = 2.0 × 10−15 J λ 1.0 10 −10 m
2.0 10 −15 J 6.02 10 23 photons 1 kJ = 1.2 × 106 kJ/mol photon mol 1000 J
282
CHAPTER 7 Ephoton =
ATOMIC STRUCTURE AND PERIODICITY
6.626 10 −34 J s 2.998 10 8 m/s 1.0 10 4 m
= 2.0 × 10−29 J
2.0 10 −29 J 6.02 10 23 photons 1 kJ = 1.2 × 10−8 kJ/mol photon mol 1000 J X rays do have an energy greater than the carbon-carbon bond energy. Therefore, X rays could conceivably break carbon-carbon bonds in organic compounds and thereby disrupt the function of an organic molecule. Radio waves, however, do not have sufficient energy to break carboncarbon bonds and are therefore relatively harmless. 61.
The energy needed to remove a single electron is: 279.7 kJ 1 mol = 4.645 × 10 −22 kJ = 4.645 × 10 −19 J 23 mol 6.0221 10
E=
62.
208.4 kJ 1 mol = 3.461 × 10 −22 kJ = 3.461 × 10 −19 J to remove one electron mol 6.0221 10 23
E= 63.
6.6261 10 −34 J s 2.9979 10 8 m/s hc hc , λ= = = 4.277 × 10 −7 m = 427.7 nm −19 λ E 4.645 10 J
6.6261 10 −34 J s 2.9979 10 8 m/s hc hc , λ= = = 5.739 × 10 −7 m = 573.9 nm λ E 3.461 10 −19 J
Ionization energy = energy to remove an electron; the energy required for the removal of one electron from a sulfur atom is:
1005 kJ 1000 J 1 mol = 1.669 × 10‒18 J/atom 23 mol kJ 6.0221 10 atoms To ionize an electron from one sulfur atom requires Ephoton = 1.669 × 10‒18 J. λ=
64.
hc E photon
=
6.6261 10−34 J s 2.9979 108 m/s = 1.190 × 10−7 m = 119.0 nm 1.669 10−18 J
890.1 kJ 1 mol 1.478 10 −21 kJ 1.478 10 −18 J = = mol atom atom 6.0221 10 23 atoms = ionization energy per atom
6.626 10 −34 J s 2.9979 10 8 m/s hc hc , λ = = E= = 1.344 × 10−7 m = 134.4 nm λ E 1.478 10 −18 J No, it will take light having a wavelength of 134.4 nm or less to ionize gold. A photon of light having a wavelength of 225 nm is longer wavelength and thus lower energy than 134.4 nm light. 65.
a.
10.% of speed of light = 0.10 × 3.00 × 108 m/s = 3.0 × 107 m/s λ=
6.63 10 −34 J s h , λ = = 2.4 × 10 −11 m = 2.4 × 10 −2 nm −31 7 mv 9.11 10 kg 3.0 10 m/s
CHAPTER 7
ATOMIC STRUCTURE AND PERIODICITY
Note: For units to come out, the mass must be in kg because 1 J = b.
λ=
283 1 kg m 2 s2
.
6.63 10 −34 J s h = = 3.4 × 10 −34 m = 3.4 × 10 −25 nm mv 0.055 kg 35 m/s
This number is so small that it is insignificant. We cannot detect a wavelength this small. The meaning of this number is that we do not have to worry about the wave properties of large objects. 66.
67.
a.
λ=
h 6.626 10 −34 J s = = 1.32 × 10 −13 m − 27 8 mv 1.675 10 kg (0.0100 2.998 10 m/s)
b.
λ=
h h 6.626 10 −34 J s , v= = = 5.3 × 103 m/s mv λm 75 10 −12 m 1.675 10 − 27 kg
6.63 10 −34 J s h h , m = λ= = = 1.6 × 10 −27 kg −15 8 mv v 1.5 10 m (0.90 3.00 10 m/s) This particle is probably a proton or a neutron.
68.
λ=
h h , v= ; for λ = 0.1 nm = 1.0 × 10 −10 m: mv mλ
v=
6.63 10 −34 J s 9.11 10 −31 kg 1.0 10 −10 m
= 7 × 106 m/s
Hydrogen Atom: The Bohr Model 69.
For the H atom (Z = 1): En = −2.178 × 10−18 J/n 2; for a spectral transition, ΔE = Ef − Ei: 1 1 ΔE = −2.178 × 10 −18 J 2 − 2 ni nf
where ni and nf are the levels of the initial and final states, respectively. A positive value of ΔE always corresponds to an absorption of light, and a negative value of ΔE always corresponds to an emission of light. 1 1 1 1 a. ΔE = −2.178 × 10 −18 J 2 − 2 = −2.178 × 10 −18 J − 4 9 3 2
ΔE = −2.178 × 10 −18 J × (0.2500 − 0.1111) = −3.025 × 10 −19 J The photon of light must have precisely this energy (3.025 × 10 −19 J). |ΔE| = Ephoton = hν =
hc
, =
6.6261 10 −34 J s 2.9979 10 8 m/s hc = | ΔE | 3.025 10 −19 J
= 6.567 × 10 −7 m = 656.7 nm
284
CHAPTER 7
ATOMIC STRUCTURE AND PERIODICITY
From Figure 7.2, this is visible electromagnetic radiation (red light). 1 1 b. ΔE = −2.178 × 10 −18 J 2 − 2 = −4.084 × 10 −19 J 4 2
=
6.6261 10 −34 J s 2.9979 10 8 m/s hc = = 4.864 × 10 −7 m = 486.4 nm | ΔE | 4.084 10 −19 J
This is visible electromagnetic radiation (green-blue light). 1 1 c. ΔE = −2.178 × 10 −18 J 2 − 2 = −1.634 × 10 −18 J 2 1
6.6261 10 −34 J s 2.9979 10 8 m/s
λ=
1.634 10
−18
J
= 1.216 × 10 −7 m = 121.6 nm
This is ultraviolet electromagnetic radiation. The transitions in an energy level diagram are: a. 3 → 2
5 4 3
b. 4 → 2 a
b
E 2
c. 2 → 1 c
Energy levels are not to scale.
n=1
70.
1 1 a. ΔE = −2.178 × 10 −18 J 2 − 2 = −1.059 × 10 −19 J 4 3
6.6261 10 −34 J s 2.9979 10 8 m/s hc = = = 1.876 × 10 −6 m = 1876 nm −19 | E | 1.059 10 J
From Figure 7.2, this is infrared electromagnetic radiation. 1 1 b. ΔE = −2.178 × 10 −18 J 2 − 2 = −4.901 × 10 −20 J 5 4
=
6.6261 10 −34 J s 2.9979 10 8 m/s hc = = 4.053 × 10 −6 m | ΔE | 4.901 10 − 20 J
= 4053 nm (infrared) 1 1 c. ΔE = −2.178 × 10 −18 J 2 − 2 = −1.549 × 10 −19 J 5 3
=
6.6261 10 −34 J s 2.9979 10 8 m/s hc = = 1.282 × 10 −6 m | ΔE | 1.549 10 −19 J = 1282 nm (infrared)
CHAPTER 7 5 4 3 E 2
n=1
71.
ATOMIC STRUCTURE AND PERIODICITY a. 4 → 3
b a
285
c
b. 5 → 4 c. 5 → 3 Energy levels are not to scale.
a. There are three possible transitions for an electron in the n = 4 level (4 → 3, 4 → 2, and 4 → 1). If an electron drops to the n = 3 level, two additional transitions can occur (3 → 2, and 3 → 1), and one more transition can occur from the n = 2 level (2 → 1). This gives a total of 6 possible transitions for an electron in the n = 4 level. Because each transition corresponds to a different ΔE, there will be a total of 6 different wavelength emissions. b. The lowest energy transition (smallest E) is the n = 4 to n = 3 electronic transition. c. Because energy and wavelength are inversely related, the shortest wavelength emission will correspond to the transition with the largest energy change (largest E). This is the transition from n = 4 to n = 1.
72.
The spacing between energy levels decrease as n increases. There are 4 possible transitions for an electron in the n = 5 level (5 → 4, 5 → 3, 5 → 2, and 5 → 1). If an electron initially drops to the n = 4 level, three additional transitions can occur (4 → 3, 4 → 2, and 4 → 1). Similarly, there are two more transitions from the n = 3 level (3 → 2, 3 → 1) and one more transition for the n = 2 level (2 → 1). There are a total of 10 possible transitions for an electron in the n = 5 level for a possible total of 10 different wavelength emissions. The longest wavelength emission comes from the transition having the smallest ΔE value. Because energy levels in the Bohr model get closer and closer together as n increases, the 5 → 4 transition has the longest wavelength.
1 1 1 1 ΔE = −2.178 × 10−18 J 2 − 2 = −2.178 × 10−18 J 2 − 2 = ‒4.901 × 10−20 J = Ephoton ni 5 nf 4 λ =
hc 6.6261 10−34 J s 2.9979 108 m/s = 4.053 × 10−6 m = 4053 nm = −20 E 4.901 10 J
The shortest wavelength transition comes from the transition havening the largest ΔE value. This will be the 5 → 1 transition.
1 1 1 1 ΔE = −2.178 × 10−18 J 2 − 2 = −2.178 × 10−18 J 2 − 2 = 2.091 × 10−18 J = Ephoton 5 ni 1 nf λ =
6.6261 10 −34 J s 2.9979 10 8 m/s hc = = 9.500 × 10−8 m = 95.00 nm −18 E 2.091 10 J
286 73.
CHAPTER 7
ATOMIC STRUCTURE AND PERIODICITY
1 1 1 1 ΔE = −2.178 × 10−18 J 2 − 2 = −2.178 × 10−18 J 2 − 2 = 2.091 × 10−18 J = Ephoton 5 ni 1 nf λ =
6.6261 10 −34 J s 2.9979 10 8 m/s hc = = 9.500 × 10−8 m = 95.00 nm −18 E 2.091 10 J
Because wavelength and energy are inversely related, visible light (λ 400−700 nm) is not energetic enough to excite an electron in hydrogen from n = 1 to n = 5.
1 1 ΔE = −2.178 × 10−18 J 2 − 2 = 4.840 × 10−19 J 6 2
λ =
hc 6.6261 10 −34 J s 2.9979 10 8 m / s = = 4.104 × 10−7 m = 410.4 nm E 4.840 10 −19 J
Visible light with λ = 410.4 nm will excite an electron from the n = 2 to the n = 6 energy level. 74.
Statements a, b, c, and e are all true. For statement c, the energy levels in the Bohr model get closer and closer together as n increases. So the 2 → 3 transition will have a smaller ΔE as compared to the 1 → 2 transition, which corresponds to a longer wavelength absorption. Statement d is false. The n = energy level corresponds to the energy of an electron when it is completely removed from the atom. ΔE for 2 → transition is smaller than ΔE for the 1 → transition, so the 2 → transition requires less energy.
75.
Ionization from n = 1 corresponds to the transition ni = 1 → nf = , where E∞ = 0. 1 ΔE = E∞ − E1 = −E1 = 2.178 × 10 −18 2 = 2.178 × 10 −18 J = Ephoton 1
=
6.6261 10 −34 J s 2.9979 10 8 m/s hc = = 9.120 × 10 −8 m = 91.20 nm E 2.178 10 −18 J
1 To ionize from n = 2, ΔE = E∞ − E2 = −E2 = 2.178 × 10 −18 2 = 5.445 × 10 −19 J 2
λ =
76.
6.6261 10 −34 J s 2.9979 10 8 m/s 5.445 10
−19
J
= 3.648 × 10 −7 m = 364.8 nm
1 ΔE = E∞ − En = −En = 2.178 × 10 −18 J 2 n
Ephoton =
6.626 10 −34 J s 2.9979 10 8 m/s hc = = 1.36 × 10 −19 J λ 1460 10 −9 m
1 Ephoton = ΔE = 1.36 × 10 −19 J = 2.178 × 10 −18 2 , n2 = 16.0, n = 4 n
77.
|ΔE| = Ephoton = h =
hc 6.626 10−34 J s 2.998 108 m/s , E photon = = 4.58 × 10−19 J λ 434 10−9 m
CHAPTER 7
ATOMIC STRUCTURE AND PERIODICITY
287
ΔE = −4.58 × 10 −19 J because we have an emission. 1 1 −4.58 × 10 −19 J = En – E5 = −2.178 × 10 −18 J 2 − 2 5 n 1 1 1 = 0.210, = 0.250, n2 = 4, n = 2 − 2 25 n2 n
The electronic transition is from n = 5 to n = 2. 78.
|ΔE| = Ephoton = h = 6.6261 × 10‒34 J s × 2.739 × 1014 s‒1 = 1.815 × 10‒19 J ΔE = −1.815 × 10 −19 J because we have an emission. 1 1 −1.815 × 10 −19 J = En – E6 = −2.178 × 10 −18 J 2 − 2 6 n
1 1 = 0.08333 − 2 36 n
1 = 0.1111, n2 = 9, n = 3 n2
The electronic transition is from n = 6 to n = 3.
Quantum Mechanics, Quantum Numbers, and Orbitals 79.
a. Δp = mΔv = 9.11 × 10−31 kg × 0.100 m/s =
b.
9.11 10 −32 kg m s
ΔpΔx ≥
h h 6.626 10 −34 J s , Δx = = = 5.79 × 10−4 m 4π 4π Δp 4 3.142 (9.11 10 −32 kg m / s)
Δx =
6.626 10 −34 J s h = = 3.64 × 10−33 m 4π Δp 4 3.142 0.145 kg 0.100 m/s
c. The diameter of an H atom is roughly 10−8 cm. The uncertainty in position is much larger than the size of the atom. d. The uncertainty is insignificant compared to the size of a baseball. 80.
Units of ΔE • Δt = J × s, the same as the units of Planck's constant. Units of Δ(mv) • Δx = kg ×
m kg m 2 kg m 2 m= = s = Js s s s2
81.
n = 1, 2, 3, ... ; ℓ = 0, 1, 2, ... (n − 1); mℓ = −ℓ ... −2, −1, 0, 1, 2, ...+ℓ
82.
When n = 5, ℓ = 0, 1, 2, 3, and 4. For ℓ = 3, the mℓ values are −3, ‒2, ‒1, 0, 1, 2, and 3. If mℓ = ‒2, ℓ can equal 2, 3, 4, etc.
83.
2s: n = 2, ℓ = 0, and mℓ = 0; 4p: n = 4, ℓ = 1, and mℓ = ‒1, 0, or 1; 6dxy: n = 6, ℓ = 2, and mℓ has one of the values of ‒-2 or ‒1 or 0 or 1 or 2. Which specific mℓ designates the dxy orbital cannot be predicted. For an f orbital, ℓ = 3. Since ℓ = 0, 1, 2, ... (n − 1), the first possible f orbital is when n = 4. Hence a 3f orbital is a forbidden orbital.
288 84.
CHAPTER 7
ATOMIC STRUCTURE AND PERIODICITY
a. This general shape represents a p orbital (ℓ = 1) and because there is a node in each of the lobes, this figure represents a 3p orbital (n = 3, ℓ = 1) b. This is an s orbital (ℓ = 0). And because there is one node present, this is a 2s orbital (n = 2, ℓ = 0). c. This is the shape of a specific d-oriented orbital (ℓ = 2). This orbital is designated d z 2 . Because no additional nodes are present inside any of the boundary surfaces, this is a 3d z 2 orbital (n = 3, ℓ = 2).
85.
86.
a. allowed
b. For ℓ = 3, mℓ can range from −3 to +3; thus +4 is not allowed.
c. n cannot equal zero.
d. ℓ cannot be a negative number.
a. For n = 3, ℓ = 3 is not possible.
d. ms cannot equal −1.
e. ℓ cannot be a negative number.
f.
For ℓ = 1, mℓ cannot equal 2.
The quantum numbers in parts b and c are allowed. 87.
ψ2 gives the probability of finding the electron at that point.
88.
The diagrams of the orbitals in the text give only 90% probabilities of where the electron may reside. We can never be 100% certain of the location of the electrons due to Heisenberg’s uncertainty principle.
Polyelectronic Atoms 89.
He: 1s2; Ne: 1s22s22p6; Ar: 1s22s22p63s23p6; each peak in the diagram corresponds to a subshell with different values of n. Corresponding subshells are closer to the nucleus for heavier elements because of the increased nuclear charge.
90.
In polyelectronic atoms, the orbitals of a given principal quantum level are not degenerate. In polyelectronic atoms, the energy order of the n = 1, 2, and 3 orbitals are (not to scale):
3d 3p E
3s 2p 2s 1s
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289
In general, the lower the n value for an orbital, the closer on average the electron can be to the nucleus, and the lower the energy. Within a specific n value orbital (like 2s vs. 2p or 3s vs. 3p vs. 3d), it is generally true that Ens < Enp < End < Enf. To rationalize this order, we utilize the radial probability distributions. In the 2s and 2p distribution, notice that the 2s orbital has a small hump of electron density very near the nucleus. This indicates that an electron in the 2s orbital can be very close to the nucleus some of the time. The 2s electron penetrates to the nucleus more than a 2p electron, and with this penetration comes a lower overall energy for the 2s orbital as compared to the 2p orbital. In the n = 3 radial probability distribution, the 3s electron has two humps of electron density very close to the nucleus, and the 3p orbital has one hump very close to the nucleus. The 3s orbital electron is most penetrating, with the 3p orbital electron the next most penetrating, followed by the least penetrating 3d orbital electron. The more penetrating the electron, the lower the overall energy. Hence the 3s orbital is lower energy than the 3p orbitals which is lower energy than the 3d orbitals. 91.
5p: three orbitals
3d z 2 : one orbital
4d: five orbitals
n = 5: ℓ = 0 (1 orbital), ℓ = 1 (3 orbitals), ℓ = 2 (5 orbitals), ℓ = 3 (7 orbitals), ℓ = 4 (9 orbitals); total for n = 5 is 25 orbitals. n = 4: ℓ = 0 (1), ℓ = 1 (3), ℓ = 2 (5), ℓ = 3 (7); total for n = 4 is 16 orbitals. 92.
1p, 0 electrons (ℓ ≠ 1 when n = 1); 6d x 2 − y 2 , 2 electrons (specifies one atomic orbital); 4f, 14 electrons (7 orbitals have 4f designation); 7py, 2 electrons (specifies one atomic orbital); 2s, 2 electrons (specifies one atomic orbital); n = 3, 18 electrons (3s, 3p, and 3d orbitals are possible; there are one 3s orbital, three 3p orbitals, and five 3d orbitals).
93.
a. n = 4: ℓ can be 0, 1, 2, or 3. Thus we have s (2 e− ), p (6 e− ), d (10 e− ), and f (14 e− ) orbitals present. Total number of electrons to fill these orbitals is 32. b. n = 5, mℓ = +1: For n = 5, ℓ = 0, 1, 2, 3, 4. For ℓ = 1, 2, 3, 4, all can have mℓ = +1. There are four distinct orbitals having these quantum numbers, which can hold 8 electrons. c. n = 5, ms = +1/2: For n = 5, ℓ = 0, 1, 2, 3, 4. Number of orbitals = 1, 3, 5, 7, 9 for each value of ℓ, respectively. There are 25 orbitals with n = 5. They can hold 50 electrons, and 25 of these electrons can have ms = +1/2. d. n = 3, ℓ = 2: These quantum numbers define a set of 3d orbitals. There are 5 degenerate 3d orbitals that can hold a total of 10 electrons. e. n = 2, ℓ = 1: These define a set of 2p orbitals. There are 3 degenerate 2p orbitals that can hold a total of 6 electrons.
94.
a. It is impossible to have n = 0. Thus, no electrons can have this set of quantum numbers. b. The four quantum numbers completely specify a single electron in a 2p orbital.
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c. n = 3, ms = +1/2: 3s, 3p, and 3d orbitals all have n = 3. These nine orbitals can each hold one electron with ms = +1/2; 9 electrons can have these quantum numbers d. n = 2, ℓ = 2: this combination is not possible (ℓ ≠ 2 for n = 2). Zero electrons in an atom can have these quantum numbers. e. n = 1, ℓ = 0, mℓ = 0: these define a 1s orbital that can hold 2 electrons. 95.
a. Na: 1s22s22p63s1; Na has 1 unpaired electron.
or
1s
2s
2p
3s
3s
b. Co: 1s22s22p63s23p64s23d7; Co has 3 unpaired electrons.
1s
2s
2p
3s
3p
or 4s
3d
3d
c. Kr: 1s22s22p63s23p64s23d104p6; Kr has 0 unpaired electrons.
1s
2s
4s 96.
2p
3s
3d
3p
4p
The two exceptions are Cr and Cu. Cr: 1s22s22p63s23p64s13p5; Cr has 6 unpaired electrons.
1s
2s
2p
3p
or
or 4s
3s
4s
3d
3d
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ATOMIC STRUCTURE AND PERIODICITY
291
Cu: 1s22s22p63s23p64s13d10; Cu has 1 unpaired electron.
1s
2s
2p
3s
3p
or 4s 97.
4s
3d
Si: 1s22s22p63s23p2 or [Ne]3s23p2; Ga: 1s22s22p63s23p64s23d104p1 or [Ar]4s23d104p1 As: [Ar]4s23d104p3; Ge: [Ar]4s23d104p2; Al: [Ne]3s23p1; Cd: [Kr]5s24d10 S: [Ne]3s23p4; Se: [Ar]4s23d104p4
98.
Cl: ls22s22p63s23p5 or [Ne]3s23p5
Sb: [Kr]5s24d105p3
Sr: 1s22s22p63s23p64s23d104p65s2 or [Kr]5s2
W: [Xe]6s24f145d4
Pb: [Xe]6s24f145d106p2
Cf: [Rn]7s25f10*
*Note: Predicting electron configurations for lanthanide and actinide elements is difficult since they have 0, 1, or 2 electrons in d orbitals. This is the actual Cf electron configuration. 99.
a. Zr: [Kr]5s24d2; two 4d electrons are present. b. Cd: [Kr]5s24d10; ten 4d electrons are present. c. Ir: [Kr]5s24d105p66s24f145d7; ten 4d electrons are present along with seven 5d electrons. d. Fe: [Ar]4s23d6; zero 4d electrons are present, but there are six 3d electrons.
100.
a. Radium is element #88 and from the periodic table, the 7s orbital is the last to fill. b. Iodine is element #53 and from the periodic table, the 5p orbitals are the last to fill. c. Gold is element #79 and from the periodic table, the 5d orbitals are the last to fill. d. Uranium is element #92 and from the periodic table, the 5f orbitals are the last to fill.
101.
For an element to have only 12 p electrons, the element can be Ar through Zn. The elements from Ar through Zn that 0 unpaired electrons are Ar (3p 6, ↑↓ ↑↓ ↑↓), Ca (4s2, ↑↓), and Zn (3d10, ↑↓ ↑↓ ↑↓ ↑↓ ↑↓).
102.
The elements that have exactly 12 electrons in various s orbitals and 30 electrons in various d orbitals are Hg through Rn. Of these elements, Pb has 2 unpaired electrons (6p2, ↑ ↑ ), Bi has 3 unpaired electrons (6p3, ↑ ↑ ↑ ), and Po has 2 unpaired electrons (6p 4, ↑↓ ↑ ↑ ).
292 103.
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ATOMIC STRUCTURE AND PERIODICITY
a. Both In and I have one unpaired 5p electron, but only the nonmetal I would be expected to form a covalent compound with the nonmetal F. One would predict an ionic compound to form between the metal In and the nonmetal F. I: [Kr]5s24d105p5
↑↓ ↑↓ ↑_ 5p
b. From the periodic table, this will be element 120. Element 120: [Rn]7s25f146d107p68s2 c. Rn: [Xe]6s24f145d106p6; note that the next discovered noble gas will also have 4f electrons (as well as 5f electrons). d. This is chromium, which is an exception to the predicted filling order. Cr has 6 unpaired electrons, and the next most is 5 unpaired electrons for Mn. Cr: [Ar]4s13d5 ↑ ↑ ↑ ↑ ↑ ↑_ 4s 3d 104.
a. As: 1s22s22p63s23p64s23d104p3 b. Element 116 (Lv): [Rn]7s25f146d107p4 c. Ta: [Xe]6s24f145d3 or Ir: [Xe]6s24f145d7 d. At: [Xe]6s24f145d106p5; note that element 117 (Ts) will also have electrons in the 6p atomic orbitals (as well as electrons in the 7p orbitals).
105.
a. The complete ground state electron for this neutral atom is 1s22s22p63s23p4. This atom has 2 + 2 + 6 + 2 + 4 = 16 electrons. Because the atom is neutral, it also has 16 protons, making the atom sulfur, S. b. Complete excited state electron configuration: 1s22s12p4; this neutral atom has 2 + 1 + 4 = 7 electrons, which means it has 7 protons, which identifies it as nitrogen, N. c. Complete ground state electron configuration: 1s22s22p63s23p64s23d104p5; this 1− charged ion has 35 electrons. Because the overall charge is 1−, this ion has 34 protons which identifies it as selenium. The ion is Se−.
106.
a. This atom has 10 electrons. Ne
b. S
c. The predicted ground state configuration is [Kr]5s24d9. From the periodic table, the element is Ag. Note: [Kr]5s14d10 is the actual ground state electron configuration for Ag. d. Bi: [Xe]6s24f145d106p3; the three unpaired electrons are in the 6p orbitals. 107.
Hg: 1s22s22p63s23p64s23d104p65s24d105p66s24f145d10 a. From the electron configuration for Hg, we have 3s2, 3p6, and 3d10 electrons; 18 total electrons with n = 3.
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ATOMIC STRUCTURE AND PERIODICITY
293
b. 3d10, 4d10, 5d10; 30 electrons are in the d atomic orbitals. c. 2p6, 3p6, 4p6, 5p6; each set of np orbitals contain one pz atomic orbital. Because we have 4 sets of np orbitals and two electrons can occupy the p z orbital, there are 4(2) = 8 electrons in pz atomic orbitals. d. All the electrons are paired in Hg, so one-half of the electrons are spin up (ms = +1/2) and the other half are spin down (ms = −1/2). 40 electrons have spin up. 108.
Element 115, Mc, is in Group 5A under Bi (bismuth): Mc: 1s22s22p63s23p64s23d104p65s24d105p66s24f145d106p67s25f146d107p3 a. 5s2, 5p6, 5d10, and 5f14; 32 electrons have n = 5 as one of their quantum numbers. b. ℓ = 3 are f orbitals. 4f14 and 5f14 are the f orbitals used. They are all filled, so 28 electrons have ℓ = 3. c. p, d, and f orbitals all have one of the degenerate orbitals with mℓ = 1. There are 6 orbitals with mℓ = 1 for the various p orbitals used; there are 4 orbitals with mℓ =1 for the various d orbitals used; and there are 2 orbitals with mℓ = 1 for the various f orbitals used. We have a total of 6 + 4 + 2 = 12 orbitals with mℓ = 1. Eleven of these orbitals are filled with 2 electrons, and the 7p orbitals are only half-filled. The number of electrons with mℓ = 1 is 11 × (2 e−) + 1 × (1 e−) = 23 electrons. d. The first 112 electrons are all paired; one-half of these electrons (56 e−) will have ms = −1/2. The 3 electrons in the 7p orbitals singly occupy each of the three degenerate 7p orbitals; the three electrons are spin parallel, so the 7p electrons either have ms = +1/2 or ms = −1/2. Therefore, either 56 electrons have ms = −1/2 or 59 electrons have ms = −1/2.
109.
B: 1s22s22p1
1s 1s 2s 2s 2p*
n
ℓ
mℓ
ms
1 1 2 2 2
0 0 0 0 1
0 0 0 0 −1
+1/2 −1/2 +1/2 −1/2 +1/2
*This is only one of several possibilities for the 2p electron. The 2p electron in B could have mℓ = −1, 0 or +1 and ms = +1/2 or −1/2 for a total of six possibilities.
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N: 1s22s22p3
1s 1s 2s 2s 2p 2p 2p 110.
111.
n
ℓ
mℓ
ms
1 1 2 2 2 2 2
0 0 0 0 1 1 1
0 0 0 0 −1 0 +1
+1/2 −1/2 +1/2 −1/2 +1/2 +1/2 +1/2
(Or all 2p electrons could have ms = −1/2.)
Ti : [Ar]4s23d2 n
ℓ
mℓ
4s 4s 3d
4 4 3
0 0 2
0 +1/2 0 −1/2 −2 +1/2
3d
3
2
−1 +1/2
ms
Only one of 10 possible combinations of mℓ and ms for the first d electron. For the ground state, the second d electron should be in a different orbital with spin parallel; 4 possibilities.
Group 1A: 1 valence electron; ns1; Li: [He]2s1; 2s1 is the valence electron configuration for Li. Group 2A: 2 valence electrons; ns2; Ra: [Rn]7s2; 7s2 is the valence electron configuration for Ra. Group 3A: 3 valence electrons; ns2np1; Ga: [Ar]4s23d104p1; 4s24p1 is the valence electron configuration for Ga. Note that valence electrons for the representative elements of Groups 1A-8A are considered those electrons in the highest n value, which for Ga is n = 4. We do not include the 3d electrons as valence electrons because they are not in n = 4 level. Group 4A: 4 valence electrons; ns2np2; Si: [Ne]3s23p2; 3s23p2 is the valence electron configuration for Si. Group 5A: 5 valence electrons; ns2np3; Sb: [Kr]5s24d105p3; 5s25p3 is the valence electron configuration for Sb. Group 6A: 6 valence electrons; ns2np4; Po: [Xe]6s24f145d106p4; 6s26p4 is the valence electron configuration for Po. Group 7A: 7 valence electrons; ns2np5; Ts (element 117): [Rn]7s25f146d107p5; 7s27p5 is the valence electron configuration for Ts. Group 8A: 8 valence electrons; ns2np6; Ne: [He]2s22p6; 2s22p6 is the valence electron configuration for Ne.
112.
a.
2 valence electrons; 4s2
b. 6 valence electrons; 2s22p4
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ATOMIC STRUCTURE AND PERIODICITY
c.
7 valence electrons; 7s27p5
d. 3 valence electrons; 5s25p1
e.
8 valence electrons; 3s23p6
f.
295
5 valence electrons; 6s26p3
113.
O: 1s22s22px22py2 (↑↓ ↑↓ __ ); there are no unpaired electrons in this oxygen atom. This configuration would be an excited state, and in going to the more stable ground state (↑↓ ↑ ↑ ), energy would be released.
114.
The number of unpaired electrons is in parentheses. a. excited state of boron B ground state: 1s22s22p1 c. exited state of fluorine F ground state: 1s22s22p5 ↑↓ ↑↓ ↑ 2p
(1) (1) (3) (1)
b. ground state of neon
(0)
Ne ground state: 1s22s22p6 (0) d. excited state of iron
(6)
Fe ground state: [Ar]4s23d6 (4) ↑↓ ↑
↑ ↑ ↑_ 3d
115.
None of the s block elements have 2 unpaired electrons. In the p block, the elements with either ns2np2 or ns2np4 valence electron configurations have 2 unpaired electrons. For elements 1-36, these are elements C, Si, and Ge (with ns2np2) and elements O, S, and Se (with ns2np4). For the d block, the elements with configurations nd2 or nd8 have two unpaired electrons. For elements 1-36, these are Ti (3d2) and Ni (3d8). A total of 8 elements from the first 36 elements have two unpaired electrons in the ground state.
116.
Cr: [Ar]4s13d5, 6 unpaired electrons (Cr is an exception to the normal filling order); Mn: [Ar]4s23d5, 5 unpaired e−; Fe: [Ar]4s23d6, 4 unpaired e−; Co: [Ar]4s23d7, 3 unpaired e−; Ni: [Ar]4s23d8, 2 unpaired e−; Cu: [Ar] 4s13d10, 1 unpaired e− (Cu is also an exception to the normal filling order); Zn: [Ar]4s23d10, 0 unpaired e−.
117.
We get the number of unpaired electrons by examining the incompletely filled subshells. The paramagnetic substances have unpaired electrons, and the ones with no unpaired electrons are not paramagnetic (they are called diamagnetic). Li: 1s22s1 ↑ ; paramagnetic with 1 unpaired electron. 2s N: 1s22s22p3 ↑ ↑ ↑ ; paramagnetic with 3 unpaired electrons. 2p Ni: [Ar]4s23d8 ↑↓ ↑↓ ↑↓ ↑ ↑ ; paramagnetic with 2 unpaired electrons. 3d Te: [Kr]5s24d105p4 ↑↓ ↑ ↑ ; paramagnetic with 2 unpaired electrons. 5p
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Ba: [Xe]6s2 ↑↓ ; not paramagnetic because no unpaired electrons are present. 6s Hg: [Xe]6s24f145d10 ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ; not paramagnetic because no unpaired electrons. 5d 118.
We get the number of unpaired electrons by examining the incompletely filled subshells. O: [He]2s22p4
2p4:
↑↓ ↑
↑
two unpaired e−
O+: [He]2s22p3
2p3:
↑
↑
three unpaired e−
O−: [He]2s22p5
2p5:
↑↓ ↑↓ ↑
one unpaired e−
Os: [Xe]6s24f145d6
5d6:
↑↓ ↑
Zr: [Kr]5s24d2
4d2:
↑
S: [Ne]3s23p4
3p4:
↑↓ ↑
↑
two unpaired e−
F: [He]2s22p5
2p5:
↑↓ ↑↓ ↑
one unpaired e−
Ar: [Ne]3s23p6
3p6
↑↓ ↑↓ ↑↓
zero unpaired e−
↑
↑
↑
↑
four unpaired e− two unpaired e−
↑
The Periodic Table and Periodic Properties 119.
Size (radius) decreases left to right across the periodic table, and size increases from top to bottom of the periodic table. a. S < Se < Te
b. Br < Ni < K
c. F < Si < Ba
All follow the general radius trend. 120.
a. Be < Na < Rb
b. Ne < Se < Sr
c. O < P < Fe
All follow the general radius trend. 121.
The ionization energy trend is the opposite of the radius trend; ionization energy (IE), in general, increases left to right across the periodic table and decreases from top to bottom of the periodic table. a. Te < Se < S
b. K < Ni < Br
c. Ba < Si < F
All follow the general ionization energy trend. 122.
a. Rb < Na < Be
b. Sr < Se < Ne
All follow the general ionization energy trend.
c. Fe < P < O
CHAPTER 7 123.
ATOMIC STRUCTURE AND PERIODICITY
a. He (From the general radius trend.)
297
b. Cl
c. Element 116 (Lv) is under Po, element 119 is the next alkali metal to be discovered (under Fr), and element 120 is the next alkaline earth metal to be discovered (under Ra). From the general radius trend, element 116 (Lv) will be the smallest. d. Si e. Na+; this ion has the fewest electrons as compared to the other sodium species present. Na+ has the smallest number of electron-electron repulsions, which makes it the smallest ion with the largest ionization energy. 124.
a.
Ba (From the general ionization energy trend.)
b.
K
c. O; in general, Group 6A elements have a lower ionization energy than neighboring Group 5A elements. This is an exception to the general ionization energy trend across the periodic table. d. S2−; this ion has the most electrons compared to the other sulfur species present. S2− has the largest number of electron-electron repulsions, which leads to S2− having the largest size and smallest ionization energy. e. Cs; this follows the general ionization energy trend. 125.
The false statement is c. Element 118 (Og) has a ground state electron configuration of [Rn]7s25f146d107p6. Element 116 is Lv; its electron configuration is [Rn]7s25f146d107p4. A 7p4 configuration has 2 unpaired electrons. Element 117 (Ts) is a halogen, and all halogens have 7 valence electrons. For example, Ts has a valence electron configuration of 7s27p5. Element 119 will be an alkali metal under Fr. Alkali metals form 1+ charged ions, so X2S is correct since sulfur forms 2‒ charged anions when in ionic compounds. Element 120 will be an alkaline earth metal under Ra. Alkaline earth metals form 2+ charged ions when in ionic compounds.
126.
a. Ts will have 117 electrons. [Rn]7s25f146d107p5 b. It will be in the halogen family and will be most like astatine (At). c. Ts should form 1− charged anions like the other halogens. Like the other halogens, some possibilities are NaTs, MgTs2, CTs4, and OTs2 d. Assuming Ts is like the other halogens, some possibilities are TsO−, TsO2−, TsO3−, and TsO4−.
127.
The elements are Si (a), P (b), O (c), N (d), and As (e). From the ionization energy trend, O is closest to the upper righthand corner of the periodic table where elements with the largest ionization energies exist. However, there is an exception to the ionization energy trend between groups 5A and 6A for row 2 elements. Because of this exception, N (answer d) will have the largest ionization energy.
128.
In general, ionization energies increase left to right across a periodic table. However, there are exceptions to this trend between groups 2A and 3A and between 5A and 6A. Notice that there
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CHAPTER 7
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is a decrease in ionization energy between elements X and Y. So X and Y are either group 2A and 3A elements or 5A and 6A elements. Since the elements are row 3 elements, the two possible identities for Q, X, Y, and Z are Na, Mg, Al, and Si or Si, P, S, and Cl. 129.
As: [Ar]4s23d104p3; Se: [Ar]4s23d104p4; the general ionization energy trend predicts that Se should have a higher ionization energy than As. Se is an exception to the general ionization energy trend. There are extra electron-electron repulsions in Se because two electrons are in the same 4p orbital, resulting in a lower ionization energy for Se than predicted.
130.
Expected order from the ionization energy trend: Be < B < C < N < O B and O are exceptions to the general ionization energy trend. The ionization energy of O is lower because of the extra electron-electron repulsions present when two electrons are paired in the same orbital. This makes it slightly easier to remove an electron from O compared to N. B is an exception because of the smaller penetrating ability of the 2p electron in B compared to the 2s electrons in Be. The smaller penetrating ability makes it slightly easier to remove an electron from B compared to Be. The correct ionization energy ordering, taking into account the two exceptions, is B < Be < C < O < N.
131.
a. As we remove succeeding electrons, the electron being removed is closer to the nucleus, and there are fewer electrons left repelling it. The remaining electrons are more strongly attracted to the nucleus, and it takes more energy to remove these electrons. b. Al: 1s22s22p63s23p1; for I4, we begin removing an electron with n = 2. For I3, we remove an electron with n = 3 (the last valence electron). In going from n = 3 to n = 2, there is a big jump in ionization energy because the n = 2 electrons are closer to the nucleus on average than the n = 3 electrons. Since the n = 2 electrons are closer, on average, to the nucleus, they are held more tightly and require a much larger amount of energy to remove compared to the n = 3 electrons. In general, valence electrons are much easier to remove than inner-core electrons.
132.
The general ionization energy trend says that ionization energy increases going left to right across the periodic table. However, one of the exceptions to this trend occurs between Groups 2A and 3A. Between these two groups, Group 3A elements usually have a lower ionization energy than Group 2A elements. Therefore, Al should have the lowest first ionization energy value, followed by Mg, with Si having the largest ionization energy. Looking at the values for the first ionization energy in the graph, the green plot is Al, the blue plot is Mg, and the red plot is Si. Mg (the blue plot) is the element with the huge jump between I2 and I3. Mg has two valence electrons, so the third electron removed is an inner core electron. Inner core electrons are always much more difficult to remove than valence electrons since they are closer to the nucleus, on average, than the valence electrons.
133.
a. More favorable electron affinity: C and Br; the electron affinity trend is very erratic. Both N and Ar have positive electron affinity values (unfavorable) due to their electron configurations (see text for detailed explanation). b. Higher ionization energy: N and Ar (follows the ionization energy trend) c. Larger size: C and Br (follows the radius trend)
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299
134.
Since element Y has more protons and since X and Y are in the same family, element X is closer to the top of the periodic table. Therefore, X has the smaller atomic number, the smaller radius, and the larger ionization energy. Answer d is the false statement. Note that elements X and Y have the same number of valence electrons because they are in the same family.
135.
Al(−44), Si(−120), P(−74), S(−200.4), Cl(−348.7); based on the increasing nuclear charge, we would expect the electron affinity values to become more exothermic as we go from left to right in the period. Phosphorus is out of line. The reaction for the electron affinity of P is: P(g) + e− → P−(g) [Ne]3s23p3 [Ne]3s23p4 The additional electron in P− will have to go into an orbital that already has one electron. There will be greater repulsions between the paired electrons in P −, causing the electron affinity of P to be less favorable than predicted based solely on attractions to the nucleus.
136.
Electron affinity refers to the energy associated with the process of adding an electron to a gaseous substance. Be, N, and Ne all have endothermic (unfavorable) electron affinity values. In order to add an electron to Be, N, or Ne, energy must be added. Another way of saying this is that Be, N, and Ne become less stable (have a higher energy) when an electron is added to each. To rationalize why those three atoms have endothermic (unfavorable) electron affinity values, let’s see what happens to the electron configuration as an electron is added. Be(g) + e− → Be−(g) [He]2s2 [He]2s22p1
N(g) + e− → N−(g) [He]2s22p3 [He]2s22p4
Ne(g) + e− → Ne−(g) [He]2s22p6 [He]2s22p63s1 In each case something energetically unfavorable occurs when an electron is added. For Be, the added electron must go into a higher-energy 2p atomic orbital because the 2s orbital is full. In N, the added electron must pair up with another electron in one of the 2p atomic orbitals; this adds electron-electron repulsions. In Ne, the added electron must be added to a much higher 3s atomic orbital because the n = 2 orbitals are full. 137.
The electron affinity trend is very erratic. In general, electron affinity becomes more positive down the periodic table, and becomes more negative from left to right across the periodic table. But there are many exceptions. a.
Se < S; S is most exothermic.
b. I < Br < F < Cl; Cl is most exothermic. (F is an exception).
138.
a. N < O < F, F is most exothermic.
b. Al < P < Si; Si is most exothermic.
139.
a. Se3+(g) → Se4+(g) + e−
b. S−(g) + e− → S2−(g)
c. Fe3+(g) + e− → Fe2+(g)
d. Mg(g) → Mg+(g) + e−
300 140.
CHAPTER 7
ATOMIC STRUCTURE AND PERIODICITY
a. The electron affinity (EA) of Mg2+ is ΔH for Mg2+(g) + e− → Mg+(g); this is just the reverse of the second ionization energy (I2) for Mg. EA(Mg2+) = −I2(Mg) = −1445 kJ/mol (Table 7.4). Note that when an equation is reversed, the sign on the equation is also reversed. b. I1 of Cl− is ΔH for Cl−(g) → Cl(g) + e−; I1(Cl− ) = −EA(Cl) = 348.7 kJ/mol (Table 7.6) c. Cl+(g) + e− → Cl(g)
ΔH = − I1(Cl) = −1255 kJ/mol = EA(Cl+) (Table 7.4)
d. Mg−(g) → Mg(g) + e− ΔH = −EA(Mg) = −230 kJ/mol = I1(Mg− ) 141.
C: 1s22s22p2; in the ground state, carbon has three different orbitals that hold electrons. The electrons in each orbital will have a separate peak on the PES spectrum, so we should have 3 different peaks in the spectrum. The peak at the lowest binding energy corresponds to the peak with the easiest electrons to remove. This will be the 2p electrons. The peak at the highest energy corresponds to the 1s orbital electrons since they will have the largest binding energy, and the middle peak corresponds to the 2s electrons. The area under each peak is directly related to the number of electrons in each orbital. Since all orbitals in carbon have 2 electrons, each peak area will be equal. This is shown on the PES spectrum for carbon; 3 peaks with equal area.
142.
The peak at the highest energy corresponds to the 1s electrons; assuming a ground state electron configuration, the 1s orbital will be filled with two electrons. The next two peaks grouped closely together correspond to the 2s and 2p electrons. These are also filled, corresponding to 2s22p6. The last group of peaks at the lowest energy correspond to n = 3 orbital electrons. Because only two peaks are present, only 3s and 3p electrons are present. Also, since the area under these peaks looks to be the same as the 2s and 2p electron peaks, we can conclude that there are the same number of 3s and 3p electrons as there are 2s and 2p electrons (3s 23p6). The complete ground state configuration is 1s22s22p63s23p6, which identifies the element as neon (Ne).
Alkali Metals 143.
It should be potassium peroxide (K2O2) because K+ ions are stable in ionic compounds. K 2+ ions are not stable; the second ionization energy of K is very large compared to the first.
144.
a. Li3N; lithium nitride
145.
ν=
b. NaBr; sodium bromide
c. K2S; potassium sulfide
2.9979 10 8 m/s c = = 6.582 × 1014 s −1 −9 λ 455 .5 10 m
E = hν = 6.6261 × 10 −34 J s × 6.582 × 1014 s −1 = 4.361 × 10 −19 J 146.
2.9979 10 8 m/s c = For 589.0 nm: ν = = 5.090 × 1014 s −1 −9 λ 589.0 10 m E = hν = 6.6261 × 10 −34 J s × 5.090 × 1014 s −1 = 3.373 × 10 −19 J For 589.6 nm: ν = c/λ = 5.085 × 1014 s −1 ; E = hν = 3.369 × 10 −19 J
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ATOMIC STRUCTURE AND PERIODICITY
301
The energies in kJ/mol are: 3.373 × 10 −19 J ×
1 kJ 6.0221 10 23 = 203.1 kJ/mol 1000 J mol
3.369 × 10 −19 J ×
1 kJ 6.0221 10 23 = 202.9 kJ/mol 1000 J mol
147.
Yes; the ionization energy general trend is to decrease down a group, and the atomic radius trend is to increase down a group. The data in Table 7.7 confirm both of these general trends.
148.
It should be element 119 with the ground state electron configuration [Rn]7s25f146d107p68s1.
149.
No; lithium metal is very reactive. It will react somewhat violently with water, making it completely unsuitable for human consumption. Lithium has a low first ionization energy, so it is more likely that the lithium prescribed will be in the form of a soluble lithium salt (a soluble ionic compound with Li+ as the cation).
150.
a. Li2CO3 b. 1 L
1 10 −3 mol Li 6.941 g Li = 0.007 g Li L mol Li
151.
a. 6 Li(s) + N2(g) → 2 Li3N(s)
b. 2 Rb(s) + S(s) → Rb2S(s)
152.
a. 2 Cs(s) + 2 H2O(l) → 2 CsOH(aq) + H2(g)
b. 2 Na(s) + Cl2(g) → 2 NaCl(s)
ChemWork Problems 153.
=
c 2.998 108 m/s = 516.9 m (1 extra sig fig) = ν 580. 103 s −1
Minimum antenna height = ½ = ½(516.9 m) = 258 m 154.
The photon energy for the 670.8 nm light emitted will equal the energy difference between the 2p and 2s orbitals. E=
155.
hc 6.6261 10−34 J s 2.9979 108 m/s = = 2.961 × 10‒19 J λ 670.8 10−9 m
All oxygen family elements have ns2np4 valence electron configurations, so this nonmetal is from the oxygen family. a. 2 + 4 = 6 valence electrons. b. O, S, Se, and Te are the nonmetals from the oxygen family (Po is a metal). c. Because oxygen family nonmetals form 2− charged ions in ionic compounds, K2X would be the predicted formula, where X is the unknown nonmetal.
302
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ATOMIC STRUCTURE AND PERIODICITY
d. From the size trend, this element would have a smaller radius than barium. e. From the ionization energy trend, this element would have a smaller ionization energy than fluorine. 156.
The box diagrams for B and O are correct. For C, the electrons in the 2p orbitals need to be spin parallel. For N, the 2p electrons need to be spin parallel with one electron in each of the degenerate 2p orbitals. And for F, the electrons that are paired need to have opposite spins.
157.
E=
310 kJ 1 mol = 5.15 × 10 −22 kJ = 5.15 × 10 −19 J mol 6.022 10 23
E=
6.626 10 −34 J s 2.998 10 8 m/s hc hc = = 3.86 × 10 −7 m = 386 nm , λ= −19 E λ 5.15 10 J
158.
Energy to make water boil = s × m × ΔT =
4.18 J × 50.0 g × 75.0°C = 1.57 × 104 J C g
hc 6.626 10 −34 J s 2.998 10 8 m/s = Ephoton = = 2.04 × 10−24 J λ 9.75 10 − 2 m
1.57 × 104 J ×
1 photon 1s = 20.9 s; 1.57 × 104 J × = 7.70 × 1027 photons 750 . J 2.04 10 − 24 J
159.
60 × 106 km ×
1000 m 1s = 200 s (about 3 minutes) km 3.00 10 8 m
160.
λ=
6.626 10 −34 J s 2.998 10 8 m/s 100 cm hc = = 5.53 × 10 −7 m × −19 m E 3.59 10 J = 5.53 × 10 −5 cm
From the spectrum, λ = 5.53 × 10 −5 cm is greenish-yellow light. 161.
1 1 1 1 ΔE = −RH 2 − 2 = −2.178 × 10 −18 J 2 − 2 = −4.840 × 10 −19 J 6 ni 2 nf λ=
100 cm hc 6.6261 10 −34 J s 2.9979 10 8 m / s = = 4.104 × 10 −7 m × − 19 | ΔE | m 4.840 10 J
= 4.104 × 10 −5 cm From the spectrum, λ = 4.104 × 10 −5 cm is violet light, so the n = 6 to n = 2 visible spectrum line is violet. 162.
There are 5 possible transitions for an electron in the n = 6 level (6 → 5, 6 → 4, 6 → 3, 6 → 2, and 6 → 1). If an electron initially drops to the n = 5 level, four additional transitions can occur (5 → 4, 5 → 3, 5 → 2, and 5 → 1). Similarly, there are three more transitions from the n = 4 level (4 → 3, 4 → 2, 4 → 1), two more transitions from the n = 3 level (3 → 2, 3 → 1), and one more transition from the n = 2 level (2 → 1). There are a total of 15 possible transitions for an electron in the n = 6 level for a possible total of 15 different wavelength emissions.
CHAPTER 7 163.
ATOMIC STRUCTURE AND PERIODICITY
303
Ionization from n = 1 corresponds to the transition ni = 1 → nf = , where E∞ = 0. 1 ΔE = E∞ − E1 = −E1 = 2.178 × 10 −18 2 = 2.178 × 10 −18 J 1
2.178 10−18 J 1 To ionize from n = 3, ΔE = E∞ − E3 = −E3 = 2.178 × 10 −18 2 = 9 3
The ionization energy for an H atom is 9 times larger from the ground state as compared to the ionization energy from n = 3. 164.
|ΔE| = Ephoton =
6.6261 10 −34 J s 2.9979 10 8 m/s hc = = 5.001 × 10 −19 J −9 λ 397.2 10 m
ΔE = −5.001 × 10 −19 J because we have an emission. 1 1 −5.001 × 10 −19 J = E2 − En = −2.178 × 10 −18 J 2 − 2 n 2 1 1 1 0.2296 = − 2 , 2 = 0.0204, n = 7 4 n n
165.
a. False; an electron in a 2s orbital would be on average, closer to the nucleus than an electron in a 3s orbital. b. True; the Bohr model is fundamentally incorrect. c. True;
d. False; these two terms come from different atomic structure models.
e. True; when n = 3, ℓ can equal 0 (s orbital), 1 (p orbitals), and 2 (d orbitals). 166.
Exceptions: Cr, Cu, Nb, Mo, Tc, Ru, Rh, Pd, Ag, Pt, and Au; Tc, Ru, Rh, Pd, and Pt do not correspond to the supposed extra stability of half-filled and filled subshells as normally applied.
167.
a. True for H only.
168.
2f and 2dxy orbitals are forbidden energy levels (they do not exist), so zero electrons can occupy these. The 3p and 4p sets each contain three degenerate orbitals, so each set can hold 6 electrons, and the 5dyz represents a singular energy level which can hold 2 electrons.
169.
As: 1s22s22p63s23p64s23d104p3
b. True for all atoms.
c. True for all atoms.
ℓ = 1 are p orbitals. 2p6, 3p6, and 4p3 are the p orbitals used. So 15 electrons have ℓ = 1. The s, p, and d orbitals all have one of the orbitals with mℓ = 0. There are four orbitals with mℓ = 0 from the various s orbitals used, there are three orbitals with mℓ = 0 from the various p orbitals used, and there is one orbital with mℓ = 0 from the 3d orbitals used. We have a total of 4 + 3 + 1 = 8 orbitals with mℓ = 0. Seven of these orbitals are filled with 2 electrons, and the 4p orbitals are only half-filled. The number of electrons with mℓ = 0 is 7 × (2 e−) + 1 × (1 e−) = 15 electrons.
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The p and d orbitals all have one of the degenerate orbitals with mℓ = 1. There are 3 orbitals with mℓ = 1 from the various p orbitals used and there is one orbital with mℓ =1 from the 3d orbitals used. We have a total of 3 + 1 = 4 orbitals with mℓ = 1. Three of these orbitals are filled with 2 electrons, and the 4p orbitals are only half-filled. The number of electrons with mℓ = 1 is 3 × (2 e−) + 1 × (1 e−) = 7 electrons. 170.
a. This element has 36 + 2 + 10 + 4 = 52 electrons. This is Te. b. This element has 32 electrons; this is Ge. c. This element has 9 electrons, so it is F.
171.
The X+ ion has 2 + 2 + 6 + 2 + 6 + 2 + 8 + 2 + 2 + 4 = 36 electron. Rubidium has 37 electrons; the ion is Rb+ which has 36 electrons.
172.
O: 1s22s22p4; C: 1s22s22p2; H: 1s1; N: 1s22s22p3; Ca: [Ar]4s2; P: [Ne]3s22p3 Mg: [Ne]3s2; K: [Ar]4s1
173.
From the radii trend, the smallest-size element (excluding hydrogen) would be the one in the most upper right corner of the periodic table. This would be O. The largest-size element would be the one in the most lower left of the periodic table. Thus K would be the largest. The ionization energy trend is the exact opposite of the radii trend. So K, with the largest size, would have the smallest ionization energy. From the general ionization energy trend, O should have the largest ionization energy. However, there is an exception to the general ionization energy trend between N and O. Due to this exception, N would have the largest ionization energy of the elements examined.
174.
a. The 4+ ion contains 20 electrons. Thus the electrically neutral atom will contain 24 electrons. The atomic number is 24, which identifies it as chromium. b. The ground state electron configuration of the ion must be 1s22s22p63s23p64s03d2; there are 6 electrons in s orbitals. c. 12
d. 2
e. From the mass, this is the isotope 50 24 Cr. There are 26 neutrons in the nucleus. f.
1s22s22p63s23p64s13d5 is the ground state electron configuration for Cr. Cr is an exception to the normal filling order.
175.
More favorable electron affinity: K, Br, and Se (follows general electron affinity trend). Higher ionization energy: K, Br, and Se (follows general ionization energy trend). Larger size: Cs, Te, and Ge (follows general atomic radius trend).
176.
O(g) + e‒ → O‒(g) EA1; O‒(g) + e‒ → O2‒(g) EA2; electron affinity is complicated. When an electron is added, the electron is attracted to the nucleus, but it is repelled by the other electrons in the atom. Most first electron affinities are negative, so having EA 1 < 0 is not surprising. The electron-electron repulsions when adding the second electron must be
CHAPTER 7
ATOMIC STRUCTURE AND PERIODICITY
305
substantial since EA2 is very unfavorable. Oxygen will gain 2 electrons when coupled with forming an ionic compound because the ionic bond formed with the cation will be very favorable energetically. But O atoms by themselves will not gain two electrons easily since it is not energetically favorable. 177.
E = hc/ = hν; photon energy and wavelength are inversely related, while photon energy and frequency are directly related. Statement b is true. From the ionization energy trend, S has a larger ionization energy than Se. Therefore, S will need the more energetic EMR photons to ionize it; the more energetic photon has the shorter wavelength. For answer a, Cl has a larger ionization energy than Cs, so Cl will require shorter wavelength EMR. For c, Mg requires the more energetic photon to ionize it. For d, P has the larger ionization energy, so it needs the higher frequency EMR. For e, all EMR travels at the same speed, so this statement is false.
178.
ΔHI is equal to the ionization energy for F. ΔHII is equal to the electron affinity for F. ΔHIII is equal to the ionization energy for S. None of these statements are true. Since F has a larger ionization energy than S, ΔHI will be larger (more positive) than ΔHII.
179.
For consecutive ionization energies, valence electrons are much easier to remove than inner core electrons. For element X, there is a huge jump in ionization energies between I 5 and I6. This leads to the conclusion that element X has 5 valence electrons. When the sixth electron is removed, an inner core electron is removed. The big jump in ionization energies for element Y is between 4 and 5. Element Y has 4 valence electrons. Since the X and Y are from row 3, X = P and Y = Si.
180.
Applying the general trends in radii and ionization energy allows matching of the various values to the elements. Mg : 1s22s22p63s2
: 0.738 MJ/mol
: 160 pm
: 1s22s22p63s23p4
: 0.999 MJ/mol
: 104 pm
Ca : 1s22s22p63s23p64s2
: 0.590 MJ/mol
: 197 pm
S
181.
Cd: 1s22s22p63s23p64s23d104p65s24d10 a. Electrons with ℓ = 2 are the d electrons. Since the 3d and 4d orbitals are filled, there are 20 electrons with ℓ = 2 in cadmium. b. The n = 4 electrons are the 4s2, 4p6, and 4d10 electrons, for a total of 18 electrons. c. The various p and d orbitals in cadmium can have one of their orbitals with mℓ = −1. So each of the 2p, 3p, 3d, 4p, and 4d orbitals have a mℓ = −1 orbital. All of these various p and d orbitals are filled, so there are 2(5) = 10 electrons with mℓ = −1. d. All orbitals in cadmium are filled and one-half of the electrons in these orbitals contain an electron with ms = −1/2. So there are ½(48) = 24 electrons with ms = −1/2.
182.
a. =
E 7.52 10 −19 J = h 6.626 10 −34 J s
= 1.13 × 1015 s −1
306
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ATOMIC STRUCTURE AND PERIODICITY
c 2.998 10 8 m / s = = 2.65 × 10 −7 m = 265 nm ν 1.13 1015 s −1
b. Ephoton and are inversely related (E = hc/). Any wavelength of electromagnetic radiation less than or equal to 265 nm ( 265) will have sufficient energy to eject an electron. So, yes, 259-nm EMR will eject an electron. c. This is the electron configuration for copper, Cu, an exception to the expected filling order. 183.
Na(g) → Na+(g) + e− Cl(g) + e− → Cl−(g)
I1 = 495 kJ EA = −348.7 kJ
Na(g) + Cl(g) → Na+(g) + Cl−g)
ΔH = 146 kJ
Mg(g) → Mg+(g) + e− F(g) + e−→ F-(g)
I1 = 735 kJ EA = −327.8 kJ
Mg(g) + F(g) → Mg+(g) + F−(g)
ΔH = 407 kJ
Mg+(g) → Mg2+(g) + e− F(g) + e − → F−(g)
I2 = 1445 kJ EA = −327.8 kJ
Mg+(g) + F(g) → Mg2+(g) + F−(g)
ΔH = 1117 kJ
a.
b.
c.
d. Using parts b and c, we get:
184.
Mg(g) + F(g) → Mg+(g) + F−(g) Mg+(g) + F(g) → Mg2+(g) + F−(g)
ΔH = 407 kJ ΔH = 1117 kJ
Mg(g) + 2 F(g) → Mg2+(g) + 2 F−(g)
ΔH = 1524 kJ
K: 1s22s22p63s23p64s1 (assuming a ground state configuration); there are 6 peaks in the spectrum, corresponding to the electrons in the 6 different orbitals being removed. The highest energy peak corresponds to the two 1s electrons. The next highest pair of closely grouped peaks correspond to the 2s and 2p electrons. Since there are two 2s electrons, the area under the shorter peak will equal the area under the 1s peak. The area under the larger peak will be three times the 2s peak since there are six 2p electrons. The next pair of peaks correspond to the 3s 2 and 3p6 electrons; the area difference between these two peaks will also be 1 : 3 since there are two 3s electrons versus six 3p electrons. The lowest energy peak corresponds to the removal of the valence 4s1 electron. Because it is a valence electron, it takes the least amount of energy to remove this electron, so this peak should be at the lowest energy. The area of the peak will be one-half of the area under the 1s or 2s or 3s peaks since only 1 electron is present in the 4s orbital (versus 2 electrons in the 1s or 2s or 3s orbitals).
Challenge Problems 185.
=
h , where m = mass and v = velocity; v rms = mv
3RT , λ = m
h m
3RT m
=
h 3RTm
CHAPTER 7
ATOMIC STRUCTURE AND PERIODICITY
For one atom, R = 2.31 10−11 m =
Molar mass =
307
8.3145 J 1 mol = 1.381 10−23 J/K•atom 23 K mol 6.022 10 atoms
6.626 10 −34 J s m 3(1.381 10
− 23
, m = 5.32 10−26 kg = 5.32 10−23 g
)(373 K )
5.32 10 −23 g 6.022 10 23 atoms = 32.0 g/mol atom mol
The atom is sulfur (S). 186.
Ephoton =
hc 6.6261 10 −34 J s 2.9979 10 8 m/s = = 7.839 × 10 −19 J λ 253.4 10 −9 m
ΔE = 7.839 × 10 −19 J; the general energy equation for one-electron ions is En = −2.178 × 10 −18 J (Z2)/n2, where Z = atomic number. 1 1 ΔE = −2.178 × 10 −18 J (Z)2 2 − 2 , Z = 4 for Be3+ ni nf 1 1 ΔE = −7.839 × 10 −19 J = −2.178 × 10 −18 (4)2 2 − 2 5 nf
7.839 10 −19 2.178 10
−18
16
+
1 1 1 = 2 , 2 = 0.06249, nf = 4 25 nf nf
This emission line corresponds to the n = 5 → n = 4 electronic transition. 187.
a. Because wavelength is inversely proportional to energy, the spectral line to the right of B (at a larger wavelength) represents the lowest possible energy transition; this is n = 4 to n = 3. The B line represents the next lowest energy transition, which is n = 5 to n = 3, and the A line corresponds to the n = 6 to n = 3 electronic transition. b. Because this spectrum is for a one-electron ion, En = −2.178 × 10−18 J (Z2/n2). To determine ΔE and, in turn, the wavelength of spectral line A, we must determine Z, the atomic number of the one electron species. Use spectral line B data to determine Z.
Z2 16 Z 2 Z2 ΔE5 → 3 = −2.178 × 10−18 J 2 − 2 = −2.178 × 10−18 9 25 3 5 E =
hc 6.6261 10 −34 J s(2.9979 108 m / s) = = 1.394 × 10−18 J λ 142 .5 10 −9 m
Because an emission occurs, ΔE5 → 3 = −1.394 × 10−18 J.
16 Z 2 , Z2 = 9.001, Z = 3; the ion is Li2+. ΔE = −1.394 × 10−18 J = −2.178 × 10−18 J 9 25 Solving for the wavelength of line A:
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ATOMIC STRUCTURE AND PERIODICITY
1 1 ΔE6 → 3 = −2.178 × 10−18(3)2 2 − 2 = −1.634 × 10−18 J 6 3
6.6261 10 −34 J s(2.9979 108 m / s) hc = = = 1.216 × 10−7 m = 121.6 nm ΔE 1.634 10 −18 J 188.
1 1 For hydrogen: ΔE = −2.178 × 10-18 J 2 − 2 = −4.574 × 10 −19 J 5 2
For a similar blue light emission, He+ will need about the same ΔE value. For He+: En = −2.178 × 10 −18 J (Z2/n2), where Z = 2: 22 22 ΔE = −4.574 × 10 −19 J = −2.178 × 10-18 J 2 − 2 4 nf
0.2100 =
4 nf2
−
4 4 , 0.4600 = 2 , 16 nf
nf = 2.949
The transition from n = 4 to n = 3 for He+ should emit a similar colored blue light as the n = 5 to n = 2 hydrogen transition; both these transitions correspond to very nearly the same energy change. 189.
For one-electron species, En = − R H Z2 /n 2 . The ground state ionization energy is the energy change for the n = 1 → n = transition. So: ionization energy = E − E1 = −E1 = R H Z2 /n 2 = RHZ2
4.72 10 4 kJ 1 mol 1000 J = 2.178 10−18 J (Z2); solving: Z = 6 23 mol kJ 6.022 10 Element 6 is carbon (X = carbon), and the charge for a one-electron carbon ion is 5+ (m = 5). The one-electron ion is C5+. 190.
A node occurs when ψ = 0. ψ300 = 0 when 27 − 18σ + 2σ2 = 0. Solving using the quadratic formula: σ =
18 (18) 2 − 4(2)(27 ) 4
=
18 108 4
σ = 7.10 or σ = 1.90; because σ = r/ao, the nodes occur at r = (7.10)ao = 3.76 × 10−10 m and at r = (1.90)ao = 1.01 × 10−10 m, where r is the distance from the nucleus. 191.
For r = ao and θ = 0° (Z = 1 for H): 1 ψ 2pz = 1/2 −11 4(2π ) 5.29 10 1
3/2
(1) e −1/2 cos 0 = 1.57 × 1014; ψ2 = 2.46 × 1028
For r = ao and θ = 90°, ψ 2 p z = 0 since cos 90° = 0; ψ2 = 0; there is no probability of finding an electron in the 2pz orbital with θ = 0°. As expected, the xy plane, which corresponds to θ = 0°, is a node for the 2pz atomic orbital.
CHAPTER 7 192.
ATOMIC STRUCTURE AND PERIODICITY
309
a. Each orbital could hold 4 electrons. b. The first period corresponds to n = 1 which can only have 1s orbitals. The 1s orbital could hold 4 electrons; hence the first period would have four elements. The second period corresponds to n = 2, which has 2s and 2p orbitals. These four orbitals can each hold four electrons. A total of 16 elements would be in the second period. c. 4 × 5 = 20 elements
193 .
d. 4 × 7 = 28 elements
a. 1st period:
p = 1, q = 1, r = 0, s = ±1/2 (2 elements)
2nd period:
p = 2, q = 1, r = 0, s = ±1/2 (2 elements)
3rd period:
p = 3, q = 1, r = 0, s = ±1/2 (2 elements) p = 3, q = 3, r = −2, s = ±1/2 (2 elements) p = 3, q = 3, r = 0, s = ±1/2 (2 elements) p = 3, q = 3, r = +2, s = ±1/2 (2 elements)
4th period:
p = 4; q and r values are the same as with p = 3 (8 total elements) 1
2
3
4
5
6
7
8
9 10 11 12
13 14 15 16 17 18 19 20
b. Elements 2, 4, 12, and 20 all have filled shells and will be least reactive. c. Draw similarities to the modern periodic table. XY could be X+Y−, X2+Y2−, or X3+Y3−. Possible ions for each are: X+ could be elements 1, 3, 5, or 13; Y− could be 11 or 19. X2+ could be 6 or 14; Y2− could be 10 or 18. X3+ could be 7 or 15; Y3− could be 9 or 17. Note: X4+ and Y4− ions probably won’t form. XY2 will be X2+(Y− )2; See above for possible ions. X2Y will be (X+)2Y2− See above for possible ions. XY3 will be X3+(Y− )3; See above for possible ions. X2Y3 will be (X3+)2(Y2− )3; See above for possible ions. d. p = 4, q = 3, r = −2 , s = ±1/2 (2 electrons) p = 4, q = 3, r = 0, s = ±1/2 (2 electrons) p = 4, q = 3, r = +2, s = ±1/2 (2 electrons) A total of 6 electrons can have p = 4 and q = 3.
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ATOMIC STRUCTURE AND PERIODICITY
e.
p = 3, q = 0, r = 0; this is not allowed; q must be odd. Zero electrons can have these quantum numbers.
f.
p = 6, q = 1, r = 0, s = ±1/2 (2 electrons) p = 6, q = 3, r = −2, 0, +2; s = ±1/2 (6 electrons) p = 6, q = 5, r = −4, −2, 0, +2, +4; s = ±1/2 (10 electrons) Eighteen electrons can have p = 6.
194.
6.626 10 −34 kg m 2 /s h = = 6.68 10 −26 kg/atom m= −15 8 λv 3.31 10 m (0.0100 2.998 10 m/s) 6.68 10 −26 kg 6.022 10 23 atoms 1000 g = 40.2 g/mol atom mol 1 kg The element is calcium, Ca.
195.
The ratios for Mg, Si, P, Cl, and Ar are about the same. However, the ratios for Na, Al, and S are higher. For Na, the second ionization energy is extremely high because the electron is taken from n = 2 (the first electron is taken from n = 3). For Al, the first electron requires a bit less energy than expected by the trend due to the fact it is a 3p electron versus a 3s electron. For S, the first electron requires a bit less energy than expected by the trend due to electrons being paired in one of the p orbitals.
196.
Size also decreases going across a period. Sc and Ti along with Y and Zr are adjacent elements. There are 14 elements (the lanthanides) between La and Hf, making Hf considerably smaller.
197.
a. Assuming the Bohr model applies to the 1s electron, E 1s = −RHZ2/n2 = −RH (Zeff)2, where n = 1. Ionization energy = E − E1s = 0 – E1s = RH (Zeff)2. 2.462 10 6 kJ 1 mol 1000 J = 2.178 10−18 J (Zeff)2, Zeff = 43.33 23 mol kJ 6.0221 10
b. Silver is element 47, so Z = 47 for silver. Our calculated Zeff value is less than 47. Electrons in other orbitals can penetrate the 1s orbital. Thus a 1s electron can be slightly shielded from the nucleus by these penetrating electrons, giving a Zeff close to but less than Z. 198.
There are seven peaks representing electrons from seven different orbitals being removed. The first five peaks at the highest energy correspond to the 1s 2, 2s2, 2p6, 3s2, and 3p6 electrons. The relative areas under the peaks confirms this. For the last two peaks at the lowest energy, the relative area indicates 1 and 2 electrons present. This gives a total of 21 electrons which identifies the element as scandium (1s22s22p63s23p64s23d1). The only inconsistency is the relative placement of the two peaks. One would expect the 3d1 peak to be at a lower energy as compared to the 4s2 peak. This is not the case; the peaks are switched. All we can conclude is that the energies of the 3d and 4s orbitals must be very close together.
CHAPTER 8 BONDING: GENERAL CONCEPTS Review Questions 1.
Electronegativity is the ability of an atom in a molecule to attract electrons to itself. Electronegativity is a bonding term. Electron affinity is the energy change when an electron is added to a substance. Electron affinity deals with isolated atoms in the gas phase. A covalent bond is a sharing of electron pair(s) in a bond between two atoms. An ionic bond is a complete transfer of electrons from one atom to another to form ions. The electrostatic attraction of the oppositely charged ions is the ionic bond. A pure covalent bond is an equal sharing of shared electron pair(s) in a bond. A polar covalent bond is an unequal sharing. Ionic bonds form when there is a large difference in electronegativity between the two atoms bonding together. This usually occurs when a metal with a small electronegativity is bonded to a nonmetal having a large electronegativity. A pure covalent bond forms between atoms having identical or nearly identical eletronegativities. A polar covalent bond forms when there is an intermediate electronegativity difference. In general, nonmetals bond together by forming covalent bonds, either pure covalent or polar covalent. Ionic bonds form due to the strong electrostatic attraction between two oppositely charged ions. Covalent bonds form because the shared electrons in the bond are attracted to two different nuclei, unlike the isolated atoms where electrons are only attracted to one nuclei. The attraction to another nuclei overrides the added electron-electron repulsions.
2.
Anions are larger than the neutral atom and cations are smaller than the neutral atom. For anions, the added electrons increase the electron-electron repulsions. To counteract this, the size of the electron cloud increases, placing the electrons further apart from one another. For cations, as electrons are removed, there are fewer electron-electron repulsions and the electron cloud can be pulled closer to the nucleus. Isoelectronic: Same number of electrons. Two variables, the number of protons and the number of electrons, determine the size of an ion. Keeping the number of electrons constant, we only have to consider the number of protons to predict trends in size. The ion with the most protons attracts the same number of electrons most strongly resulting in a smaller size.
3.
Lattice energy: The change in energy that takes place when separated gaseous ions are packed together to form an ionic solid. The reason ionic compounds form is the extremely favorable lattice energy value (large and negative). Looking at Figure 8.11, there are many processes that occur when forming an ionic compound from the elements in their standard state. Most of these processes (if not all) are unfavorable (endothermic). However, the large, exothermic lattice energy value dominates and the ionic compound forms.
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The lattice energy follows Coulomb’s law (E Q1Q2/r). Because MgO has ions with +2 and −2 charges, it will have a more favorable lattice energy than NaF where the charge on the ions are −1 and +1. The reason MgO has +2 and −2 charged ions and not +1 and −1 charged ions is that lattice energy is more favorable as the charges increase. However, there is a limit to the magnitude of the charges. To form Mg3+O3−, the ionization energy would be extremely unfavorable for Mg2+ since an inner core (n = 2) electron is being removed. The same is true for the electron affinity of O2−; it would be very unfavorable as the added electron goes into the n = 3 level. The lattice energy would certainly be more favorable for Mg 3+O3−, but the unfavorable ionization energy and electron affinity would dominate making Mg 3+O3− energetically unfavorable overall. In general, ionic compounds want large charges, but only up to the point where valence electrons are removed or added. When we go beyond the valence shell, the energies become very unfavorable. 4.
When reactants are converted into products, reactant bonds are broken and product bonds are formed. Thus, H for a reaction should be the energy it takes to break the reactant bonds minus the energy released when bonds are formed. Bond energies give good estimates for gas phase reactions but give poor estimates when solids or liquids are present. This is because bond energy calculations ignore the attractive forces holding solids and liquids together. Gases have the molecules very far apart and they have minimal (assumed zero) attractive forces. This is not true for solids and liquids where the molecules are very close together. Attractive forces in substances are discussed in Chapter 10. For an exothermic reaction, stronger bonds are formed in the products as compared to the strength of the bonds broken in the reactants so energy is released. For endothermic reactions, the product bonds are weaker overall and energy must be absorbed. As the number of bonds increase, bond strength increases and bond length decreases.
5.
Nonmetals, which form covalent bonds, have valence electrons in the s and p orbitals. Since there are 4 total s and p orbitals, there is room for only 8 valence electrons (the octet rule). The valence shell for hydrogen is just the 1s orbital. This orbital can hold 2 electrons, so hydrogen follows the duet rule. Drawing Lewis structures is mostly trial and error. The first step is to sum the valence electrons available. Next, attach the bonded atoms with a single bond. This is called the skeletal structure. In general, the atom listed first in a compound is called the central atom; all other atoms listed after the first atom are attached (bonded) to this central atom. If the skeletal structure is something different, we will generally give you hints to determine how the atoms are attached. The final step in drawing Lewis structures is to arrange the remaining electrons around the various atoms to satisfy the octet rule for all atoms (duet role for H). Be and B are the usual examples for molecules that have fewer than 8 electrons. BeH2 and BH3 only have 4 and 6 total valence electrons, respectively; it is impossible to satisfy the octet rule for BeH2 and BH3 because fewer than 8 electrons are present. All row three and heavier nonmetals can have more than 8 electrons around them, but only if they must. Always satisfy the octet rule when you can; exceptions to the octet rule occur when there are no other options. Of the molecules listed in Review Question 10, KrF 2, IF3, SF4, XeF4, PF5, IF5, and SCl6 are all examples of central atoms having more than 8 electrons. In all cases, exceptions occur because they must.
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The octet rule cannot be satisfied when there is an odd number of valence electrons. There must be an unpaired electron somewhere in the molecule and molecules do not like unpaired electrons. In general, odd electron molecules are very reactive; they react to obtain an even number of valence electrons. NO2 is a good example. NO2 has 17 valence electrons; when two NO2 molecules react, N2O4, which has 34 valence electrons forms. The octet rule can be satisfied for N2O4. 6.
Resonance occurs when more than one valid Lewis structure can be drawn for a particular molecule. A common characteristic of resonance structures is a multiple bond(s) that can be drawn in different positions. We say the electrons in the multiple bond(s) are delocalized in the molecule. This helps us rationalize why the bonds in a molecule that exhibit resonance are all equivalent in length and strength. Any one of the resonance structures indicates different types of bonds within that molecule. This is not correct, hence none of the individual resonance structures are correct. We think of the actual structure as an average of all the resonance structures; again this helps explain the equivalent bonds within the molecule that experiment tells us we have.
7.
Formal charge: A made up charge assigned to an atom in a molecule or polyatomic ion derived from a specific set of rules. The equation to calculate formal charge (FC) is: FC = (number of valence electrons of the free atom) – (number of valence electrons assigned to the atom in the molecule) The assigned electrons are all the lone pair electrons plus one-half of the bonding electrons. Formal charge can be utilized when more than one nonequivalent resonance structure can be drawn for a molecule. The best structure, from a formal charge standpoint, is the structure that has the atoms in the molecule with a formal charge of zero. For organic compounds, carbon has 4 valence electrons and needs 4 more electrons to satisfy the octet rule. Carbon does this by forming 4 bonds to other atoms and by having no lone pairs of electrons. Any carbon with 4 bonds and no lone pairs has a formal charge of zero. Hydrogen needs just 1 more electron to obtain the He noble gas electron configuration. Hydrogen is always attached with a single bond to one other atom. N has 5 valence electrons. For a formal charge of zero, N will form 3 bonds to other atom(s) for 6 shared electrons, and the remaining 2 electrons are a lone pair on N. Oxygen will have a formal charge of zero when it is attached to other atom(s) with 2 bonds and has 2 lone pairs. The halogens obtain a formal charge of zero by forming 1 bond to another atom as well as having 3 lone pairs.
8.
VSEPR = Valence Shell Electron-Pair Repulsion model. The main postulate is that the structure around a given atom is determined principally by minimizing electron-pair repulsion. Electrons don’t like each other, so a molecule adopts a geometry to place the electron pairs about a central atom as far apart as possible. The five base geometries and bond angles are: Number of bonded atoms plus lone pairs about a central atom 2 3 4 5 6
Geometry linear trigonal planar tetrahedral trigonal bipyramid octahedral
Bond Angle(s) 180˚ 120˚ 109.5˚ 90˚, 120˚ 90˚
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To discuss deviations from the predicted VSEPR bond angles, let us examine CH 4, NH3, and H2O. CH4 has the true 109.5˚ bond angles, but NH3 (107.3˚) and H2O (104.5˚) do not. CH4 does not have any lone pairs of electrons about the central atom, while H 2O and NH3 do. These lone pair electrons require more room than bonding electrons, which tends to compress the angles between the bonding pairs. The bond angle for H 2O is the smallest because oxygen has two lone pairs on the central atom; the bond angle is compressed more than in NH 3 where N has only one lone pair. So, in general, lone pairs compress the bond angles to a value slightly smaller than predicted by VSEPR. 9.
The two general requirements for a polar molecule are: 1. polar bonds 2. a structure such that the bond dipoles of the polar bonds do not cancel. CF4, 4 + 4(7) = 32 valence electrons
tetrahedral, 109.5˚
XeF4, 8 + 4(7) = 36 e−
square planar, 90˚
SF4, 6 + 4(7) = 34 e− 90˚ 120˚ 90˚ see-saw, 90˚, 120˚ The arrows indicate the individual bond dipoles in the three molecules (the arrows point to the more electronegative atom in the bond which will be the partial negative end of the bond dipole). All three of these molecules have polar bonds. To determine the polarity of the overall molecule, we sum the effect of all of the individual bond dipoles. In CF 4, the bonded fluorine atoms are symmetrically arranged about the central carbon atom. The net result is for all of the individual C−F bond dipoles to cancel each other out giving a nonpolar molecule. In XeF 4, the 4 Xe−F bond dipoles are also symmetrically arranged and XeF 4 is also nonpolar. The individual bond dipoles cancel out when summed together. In SF 4, we also have 4 polar bonds. But in SF4, the bond dipoles are not symmetrically arranged, and they do not cancel each other out. SF4 is polar. It is the positioning of the lone pair that disrupts the symmetry in SF 4.
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CO2, 4 + 2(6) = 16 e−
315
COS, 4 + 6 + 6 = 16 e−
CO2 is nonpolar because the individual bond dipoles cancel each other out, but COS is polar. By replacing an O with a less electronegative S atom, the molecule is not symmetric any more. The individual bond dipoles do not cancel since the C−S bond dipole is smaller than the C−O bond dipole resulting in a polar molecule. 10.
To predict polarity, draw in the individual bond dipoles, then sum up the net effect of the bond dipoles on each other. If the net effect is to have the bond dipoles cancel each other out, then the molecule is nonpolar. If the net effect of the bond dipoles is to not cancel each other out, then the molecule will have a partial positive end and a partial negative end (the molecule is polar). This is called a dipole moment or a polar molecule. CO2, 4 + 2(6) = 16 valence electrons
SO2, 6 + 2(6) = 18 e−
linear, 180˚, nonpolar
V-shaped, 120˚, polar
KrF2, 8 + 2(7) = 22 e−
SO3, 6 + 3(6) = 24 e−
+ 2 other resonance structures
linear, 180˚, nonpolar
trigonal planar, 120˚, nonpolar
NF3, 5 + 3(7) = 26 e−
IF3, 7 + 3(7) = 28 e−
trigonal pyramid, < 109.5˚, polar
T-shaped, 90˚, polar
The bond angles will be somewhat less than 109.5˚ due to the lone pair on the central nitrogen atom needing more space.
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tetrahedral, 109.5˚, nonpolar
BONDING: GENERAL CONCEPS
SF4, 6 + 4(7) = 34 e−
see-saw, 90˚ and 120˚, polar
XeF4, 8 + 4(7) = 36 e−
PF5, 5 + 5(7) = 40 e−
square planar, 90˚, nonpolar
trigonal bipyramid, 90˚ and 120˚, nonpolar
IF5, 7 + 5(7) = 42 e−
SCl6, 6 + 6(7) = 48 e−
square pyramid, 90˚, polar
octahedral, 90˚, nonpolar
Active Learning Questions 1.
Electronegativity increases left to right across the periodic table and decreases from top to bottom. This trend is identical to the ionization energy trend and exactly opposite the atomic radii trend. Electronegativity is the ability of an atom in a molecule to attract electrons to itself. Atoms where the valence electrons are attracted more strongly by the nucleus are expected to have large ionization energies and small atomic radii. Likewise, atoms whose outer electrons are more strongly attracted by the nucleus would be expected to attract bonding electrons to itself more strongly and have large electronegativities.
2.
Formation of ionic compounds depend on the energy released when opposite charged ions are attracted to each other to form the ionic bond. For this factor, the higher the charge, the more
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energy released when the ionic bond forms. Another factor to consider is the energy it takes to form the specific ions in the ionic bond. In general, the cations form by losing the valence electrons. Losing electron beyond the valence shell requires a significant increase in energy. For the anions, they gain electrons to fill the valence shell, but no more. Adding electrons beyond the valence shell also requires a significant increase in energy. The result is that it is most energetically favorable for the ions to form up to the valence shell, but not beyond the valence shell. 3.
The predicted most stable ions would be Na+, Mg2+, Al3+, S2 −, Cl−, K+, Ca2+, and Ga3+. In general, cations are smaller than anions, so S2 − and Cl − should be largest. These two ions are isoelectronic with 18 electrons. Cl− should be smaller since it has one more proton attracting the 18 electrons and will do so more strongly. Na+, Mg2+, and Al3+ are isoelectronic with 10 electrons. Al3+, with the most protons, should be smallest followed by Mg 2+ with Na+ having the largest size of these three cations. K+ and Ca2+ are in row 3 versus row 2 for the Na+, Mg2+, and Al3+ ions. One would expect the ions in row 3 to be larger than the ions in row 2 since the electrons are in higher n values. Between K+ and Ca2+, which are isoelectronic, Ca2+ should be smaller because it has the most protons. Putting all of this together, the expected size order would be Al 3+ < Mg2+ < Na+ < Ca2+ < K+ < Cl − < S2 −. Figure 3.5 confirms this order. The radius of Ga3+ is harder to predict because the first-row transition metals are between Ca2+ and Ga3+. In general, size increases down a group so from this, Ga3+ should be larger than Al3+. But how it compares to the other cations is hard to predict. From Figure 3.5, Ga3+ is smaller than all of the other cations except for Al3+. Note that Ga3+ is smaller than Ca2+ and K+. The effect of the first-row transition metals between Ca2+ and Ga3+ has little effect. Ga3+ behaves as it was isoelectronic with Ca2+ since with more protons, Ga3+ is smaller.
4.
The Lewis structures for CH4 (8 valence electrons) and CHBr3 (26 valence electrons are:
Although we assume it doesn’t, bond strength depends on the environment of the atoms bound to the central atom. Here, we replace hydrogen atoms with the more electronegative bromine atoms. The more electronegative Br atoms can draw electrons away from C‒H bond. The net effect is to weaken the C‒H bond in CHBr3. 5.
Oxygen only gains 2 electrons when it is going to form an ionic bond; O atoms do not form O2‒ ions on there own. The second electron affinity of oxygen [O‒(g) + e‒ → O2‒(g)] is a large positive number; so large that is energetically unfavorable form O 2‒ ions to form on their own. But if a cation is available, O2‒ will form and go on to form an ionic bond with the cation. This overall process is energetically favorable.
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NO2‒ has 18 valence electrons and NO3‒ has 24 valence electrons. O N
O
O N
O
O
N O
O
O
N O
O
N O
O
O
When resonance structures can be drawn, the average of all of the resonance structures generally gives a good representation of the bonding in the molecule. For NO 2− the average bond would be a 1 1/2 bond, about halfway between a single bond and a double bond in length. For NO3− the average bond is a 1 1/3 bond; a bond length between a single and double bond, but closer to a single bond in length. Between the two ions, NO3‒ should have the longer bond. 7.
NCO − (which is the same as OCN − shown in the Lewis structures below) and CNO − each have 16 valence electrons; each also has 3 resonance structures.
Formal charge
Formal charge
O
C
N
O
C
N
O
C
N
0
0
-1
-1
0
0
+1
0
-2
C
N
O
C
N
O
C
N
O
-2
+1
0
-1
+1
-1
-3
+1
+1
The best resonance structure has formal charge values as close to zero as possible. The middle resonance structure in each set above is best. For NCO − the first two resonance structures have formal charge values closest to zero; but the second structure puts the 1− charge on oxygen, which seems more reasonable since oxygen is more electronegative than nitrogen. For CNO − the middle structure has the formal charge values closest to zero, so it is best from a formal charge standpoint. 8.
Electronegativity is the ability of an atom to attract bonding electrons to itself. Ti4+ with the large positive charge will attract bonding electrons more strongly than the other species. So Ti4+ is most electronegative, followed by Ti3+, followed by Ti2+, followed by Ti+, with Ti the least electronegative species. The positive charge on the ion will affect the electronegativity of a species.
9.
Electron affinity is a complicated process. The electron is attracted to the nucleus which is favorable, but the added electron will be repelled by the other electrons which is unfavorable. For O‒ and S‒, the added electron-electron repulsions when the second electron is added must significant. So much so that it is overall energetically unfavorable for O 2‒ and S2‒ to form on their own.
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Chemical bonds are the forces that hold atoms together in compounds. When two nonmetals form a bond, the electrons are shared between the two nonmetals to form a covalent bond. When a metal and a nonmetal form a bond, electrons are transferred from the metal to the nonmetal to form a cation and an anion. The force of attraction between the anion and cation is known as an ionic bond. Nonmetal elements are found in nature in whatever form is most stable. Of the nonmetal elements, only the noble gases are found as singular atoms; this is how they are most stable. The other nonmetals are found as molecules or compounds. This is because they are more stable bonded to another atom versus existing as singular atoms.
11.
The type of bond formed depends on the electronegativity of the two atoms forming the bond. When the difference in electronegativity is large between the two atoms forming the bond, a transfer of electrons occurs and the bond is ionic. This occurs general when metal with a low electronegativity value forms a bond with a nonmetal having a high electronegativity value. When the electronegativity difference is intermediate or zero between the two atoms forming the bond, then the two atoms share electrons to form a covalent bonds. This generally occurs when two nonmetals form a bond.
12.
In N2 and CO, the bonding is covalent in nature, with the bonding electrons shared between the atoms. In N2, the two atoms are identical, so the sharing is equal; in CO, the two atoms are different with different electronegativities, so the sharing in unequal. As a result, the bond in CO is polar covalent. Both of these bonds are in marked contrast to the situation in NaCl. NaCl is an ionic compound, where an electron has been completely transferred from sodium to chlorine, producing a sodium cation and a chloride anion.
13.
C−H bonds are assumed to be nonpolar when predicting overall molecular polarity. Since CH 4 doesn’t have any bond dipoles, it is nonpolar. CCl4 is also nonpolar. CCl4 is a tetrahedral molecule where all four C‒Cl bond dipoles cancel when added together. Let’s consider just the C and two of the Cl atoms. There will be a net dipole pointing in the direction of the middle of the two Cl atoms. C Cl
Cl
There will be an equal and opposite dipole arising from the other two Cl atoms. Combining these two overall bond dipoles gives: Cl
Cl C
Cl
Cl
The two CCl2 net dipoles are equal but point in opposite directions so they exactly cancel out each other. Thus CCl4 is a nonpolar molecule For the remaining molecules, picture the pairs of atoms each combing to form a net dipole, then add the two overall net dipoles together to get the polarity of the molecule (like was done above
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with CCl4). Since fluorine is more electronegative than the other atoms, the CF2 pair will have the largest overall net dipole between a pair of atoms. The CCl 2 pair is the next most polar pair with the CH2 pair being nonpolar. With this in mind, one would expect the CF 2H2 molecule to be most polar, the CCl2H2 molecule to be next most polar, and the CF2Cl2 to be the least polar of these three molecules. In the CF2Cl2, one would expect the CCl2 net dipole to partially offset the CF2 net dipole. However, the two overall net dipoles do not completely offset each other. So CF2CCl2 should be polar, but it should have a smaller overall dipole moment than CCl 2H2. The expected ordering from most polar to least polar is CF 2H2 > CCl2H2 > CF2Cl2 > CH4 = CCl4. 14.
When drawing Lewis structures, it is the total number of valence electrons that is important and not necessarily the specific number donated by each atom. Your most general goal for Lewis structures is to arrange the valence electrons around the atoms in a compound of ion so to satisfy the octet rule for all atoms (duet rule for hydrogen). A Lewis structure does not necessarily indicate where the electrons came from.
15.
This is generally true. The larger the atom, the smaller the electronegativity. Note that the size trend and the electronegativity trends are the exact opposite of each other. So, the biggest atoms will have the smallest electronegativity.
16.
The valence shell electron-repulsion model assumes the electrons about a central atom (bonding electrons and lone pairs) repel each other. The central atom will adopt a geometry to minimize the electron-pair repulsions; the electrons about a central atom want to be as far apart as possible so to minimize electron pair repulsions.
17.
Effective pair refers to double and triple bonded atoms when predicting molecular structures using VSEPR. When determining the electron pairs about a central atom, a triple bonded atom and a double bonded atom behave as if they are one single pair of bonding electrons. Hence the term effective pair since double and tripled bonds behave as if they are just a single bonding pair of electrons.
18.
Mg: 1s22s22p63s2; O: 1s22s22p4; Mg2+ forms by removing the two valence electrons. O2‒ forms by gaining two electrons to fill the n = 2 valence orbitals. Both processes require energy to be added to form the ions. However, the energy released by the attraction between Mg2+ and O2‒ ions is large enough to make formation of MgO favorable. To form Mg 3+ or higher charged ions requires removing inner core (n = 2) electrons. There is a huge increase in ionization energies when removing inner core electrons versus valence electrons. To form O3‒ or more negative charged ions requires the extra electrons to go into the n = 3 orbitals. This also requires a huge increase in energy to add electrons outside of the valence shell. So yes, the energy of attraction between Mg3+ and O3‒ (or higher charge ions) will be more energetically favorable, but the energy it takes to form the ions is too large; so large that it overcomes the more favorable energy of attraction, and it is overall energetically unfavorable for the compound to form with higher charged ions.
19.
Linear structure (180° bond angle)
Polar; bond dipoles do not cancel.
Nonpolar; bond dipoles cancel.
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Trigonal planar structure (120° bond angle)
+ 2 other resonance structures Polar; bond dipoles do not cancel.
Nonpolar; bond dipoles cancel.
Tetrahedral structure (109.5° bond angles)
Polar; bond dipoles do not cancel.
Nonpolar; bond dipoles cancel.
Trigonal bipyramid structure (90° and 120° bond angles)
Polar; bond dipoles do not cancel.
Nonpolar; bond dipoles cancel
Octahedral structure (90° bond angles)
Polar; bond dipoles do not cancel.
Nonpolar; bond dipoles cancel
Two example compounds having lone pairs but are nonpolar are XeF2 and XeF4. In each case the individual bond dipoles cancel out each other when summed together.
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Questions 20.
The simple answer is that atoms are more stable as compounds than they are as individual atoms. Substances in nature tend toward the lowest energy state, which is also the most stable state. When atoms come together to form bonds in compounds, they lower their energy state and become more stable.
21.
a. This diagram represents a polar covalent bond as in HCl. In a polar covalent bond, there is an electron rich region (indicated by the red color) and an electron poor region (indicated by the blue color). In HCl, the more electronegative Cl atom (on the red side of the diagram) has a slightly greater ability to attract the bonding electrons than does H (on the blue side of the diagram), which in turn produces a dipole moment. b. This diagram represents an ionic bond as in NaCl. Here, the electronegativity differences between the Na and Cl are so great that the valence electron of sodium is transferred to the chlorine atom. This results in the formation of a cation, an anion, and an ionic bond. c. This diagram represents a pure covalent bond as in H2. Both atoms attract the bonding electrons equally, so there is no bond dipole formed. This is illustrated in the electrostatic potential diagram as the various red and blue colors are equally distributed about the molecule. The diagram shows no one region that is red nor one region that is blue (there is no specific partial negative end and no specific partial positive end), so the molecule is nonpolar.
22.
In F2 the bonding is pure covalent, with the bonding electrons shared equally between the two fluorine atoms. In HF, there is also a shared pair of bonding electrons, but the shared pair is drawn more closely to the fluorine atom. This is called a polar covalent bond as opposed to the pure covalent bond in F2.
23.
Of the compounds listed, P2O5 is the only compound containing only covalent bonds. (NH4)2SO4, Ca3(PO4)2, K2O, and KCl are all compounds composed of ions, so they exhibit ionic bonding. The polyatomic ions in (NH4)2SO4 are NH4+ and SO42−. Covalent bonds exist between the N and H atoms in NH4+ and between the S and O atoms in SO42−. Therefore, (NH4)2SO4 contains both ionic and covalent bonds. The same is true for Ca 3(PO4)2. The bonding is ionic between the Ca2+ and PO43− ions and covalent between the P and O atoms in PO43−. Therefore, (NH4)2SO4 and Ca3(PO4)2 are the compounds with both ionic and covalent bonds.
24.
Ionic solids are held together by strong electrostatic forces that are omnidirectional. i.
For electrical conductivity, charged species must be free to move. In ionic solids, the charged ions are held rigidly in place. Once the forces are disrupted (melting or dissolution), the ions can move about (conduct).
ii. Melting and boiling disrupts the attractions of the ions for each other. Because these electrostatic forces are strong, it will take a lot of energy (high temperature) to accomplish this.
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iii. If we try to bend a piece of material, the ions must slide across each other. For an ionic solid the following might happen:
strong attraction
strong repulsion
Just as the layers begin to slide, there will be very strong repulsions causing the solid to snap across a fairly clean plane. iv. Polar molecules are attracted to ions and can break up the lattice. These properties and their correlation to chemical forces will be discussed in detail in Chapters 10 and 11. 25.
Electronegativity increases left to right across the periodic table and decreases from top to bottom. Hydrogen has an electronegativity value between B and C in the second row and identical to P in the third row. Going further down the periodic table, H has an electronegativity value between As and Se (row 4) and identical to Te (row 5). It is important to know where hydrogen fits into the electronegativity trend, especially for rows 2 and 3. If you know where H fits into the trend, then you can predict bond dipole directions for nonmetals bonded to hydrogen.
26.
This statement is false. Lone pairs do not make a compound polar. Lone pairs do affect the molecular structure of a compound. This usually results in the individual bond dipoles in the molecule not cancelling each other out, resulting in a polar compound. However, there are compounds with lone pairs about the central atom that are nonpolar. KrF2 and XeF4 are two examples of nonpolar compounds which have lone pairs about the central atom.
27.
For ions, concentrate on the number of protons and the number of electrons present. The species whose nucleus holds the electrons most tightly will be smallest. For example, anions are larger than the neutral atom. The anion has more electrons held by the same number of protons in the nucleus. These electrons will not be held as tightly, resulting in a bigger size for the anion as compared to the neutral atom. For isoelectronic ions, the same number of electrons are held by different numbers of protons in the various ions. The ion with the most protons holds the electrons tightest and is smallest in size.
28.
Two other factors that must be considered are the ionization energy needed to produce more positively charged ions and the electron affinity needed to produce more negatively charged ions. The favorable lattice energy more than compensates for the unfavorable ionization energy of the metal and for the unfavorable electron affinity of the nonmetal as long as electrons are added to or removed from the valence shell. Once the valence shell is empty, the ionization energy required to remove the next (inner-core) electron is extremely unfavorable; the same is true for electron affinity when an electron is added to a higher n shell. These two quantities are so unfavorable after the valence shell is complete that they overshadow the favorable lattice energy, and the higher charged ionic compounds do not form.
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29.
Fossil fuels contain a lot of carbon and hydrogen atoms. Combustion of fossil fuels (reaction with O2) produces CO2 and H2O. Both these compounds have very strong bonds. Because stronger product bonds are formed than reactant bonds broken, combustion reactions are very exothermic.
30.
Statements a and c are true. For statement a, XeF2 has 22 valence electrons, and it is impossible to satisfy the octet rule for all atoms with this number of electrons. The best Lewis structure is:
For statement c, NO+ has 10 valence electrons, whereas NO− has 12 valence electrons. The Lewis structures are:
Because a triple bond is stronger than a double bond, NO+ has a stronger bond. For statement b, SF4 has five electron pairs around the sulfur in the best Lewis structure; it is an exception to the octet rule. Because OF4 has the same number of valence electrons as SF4, OF4 would also have to be an exception to the octet rule. However, row 2 elements such as O never have more than 8 electrons around them, so OF4 does not exist. For statement d, two resonance structures can be drawn for ozone:
When resonance structures can be drawn, the actual bond lengths and strengths are all equal to each other. Even though each Lewis structure implies the two O−O bonds are different, this is not the case in real life. In real life, both of the O−O bonds are equivalent. When resonance structures can be drawn, you can think of the bonding as an average of all of the resonance structures. Carbon: FC = 4 − 2 − 1/2(6) = −1; oxygen: FC = 6 − 2 − 1/2(6) = +1
31.
Electronegativity predicts the opposite polarization. The two opposing effects seem to partially cancel to give a much less polar molecule than expected. 32.
All these molecules have polar bonds that are symmetrically arranged about the central atoms. In each molecule the individual bond dipoles cancel to give no net overall dipole moment. All these molecules are nonpolar even though they all contain polar bonds.
Exercises Chemical Bonds and Electronegativity 33.
Using the periodic table, the general trend for electronegativity is: (1) Increase as we go from left to right across a period
CHAPTER 8
BONDING: GENERAL CONCEPTS
325
(2) Decrease as we go down a group Using these trends, the expected orders are: a. C < N < O
b. Se < S < Cl
c. Sn < Ge < Si
d. Tl < Ge < S
34.
a. Rb < K < Na
b. Ga < B < O
c. Br < Cl < F
d. S < O < F
35.
The most polar bond will have the greatest difference in electronegativity between the two atoms. From positions in the periodic table, we would predict: a. Ge‒F
b. P‒Cl
c. S‒F
d. Ti‒Cl
36.
a. Sn‒H
b. Tl‒Br
c. Si‒O
d. O‒F
37.
The general trends in electronegativity used in Exercises 33 and 35 are only rules of thumb. In this exercise, we use experimental values of electronegativities and can begin to see several exceptions. The order of EN from Figure 8.3 is: a. C (2.5) < N (3.0) < O (3.5)
same as predicted
b. Se (2.4) < S (2.5) < Cl (3.0)
same
c. Si = Ge = Sn (1.8)
different
d. Tl (1.8) = Ge (1.8) < S (2.5)
different
Most polar bonds using actual EN values: a. Si‒F and Ge‒F have equal polarity (Ge‒F predicted). b. P‒Cl (same as predicted) c. S‒F (same as predicted) 38.
d. Ti‒Cl (same as predicted)
The order of EN from Figure 8.3 is: a. Rb (0.8) = K (0.8) < Na (0.9), different
b. Ga (1.6) < B (2.0) < O (3.5), same
c. Br (2.8) < Cl (3.0) < F (4.0), same
d. S (2.5) < O (3.5) < F (4.0), same
Most polar bonds using actual EN values: a. C‒H most polar (Sn‒H predicted) b. Al‒Br most polar (Tl‒Br predicted).
c. Si‒O (same as predicted).
d. Each bond has the same polarity, but the bond dipoles point in opposite directions. Oxygen is the positive end in the O‒F bond dipole, and oxygen is the negative end in the O‒Cl bond dipole (O‒F predicted).
326 39.
CHAPTER 8
Use the electronegativity trend to predict the partial negative end and the partial positive end of the bond dipole (if there is one). To do this, you need to remember that H has electronegativity between B and C and identical to P. Answers b, d, and e are incorrect. For d (Br 2), the bond between two Br atoms will be a pure covalent bond, where there is equal sharing of the bonding electrons, and no dipole moment exists. For b and e, the bond polarities are reversed. In Cl−I, the more electronegative Cl atom will be the partial negative end of the bond dipole, with I having the partial positive end. In O−P, the more electronegative oxygen will be the partial negative end of the bond dipole, with P having the partial positive end. In the following, we used arrows to indicate the bond dipole. The arrow always points to the partial negative end of a bond dipole (which always is the most electronegative atom in the bond).
Cl 40.
42.
I
O
P
See Exercise 39 for a discussion on bond dipoles. We will use arrows to indicate the bond dipoles. The arrow always points to the partial negative end of the bond dipole, which will always be to the more electronegative atom. The tail of the arrow indicates the partial positive end of the bond dipole. a.
41.
BONDING: GENERAL CONCEPS
C
O
b. P−H is a pure covalent (nonpolar) bond because P and H have identical electronegativities.
c.
H
Cl
d.
e.
Se
S
Br
Te
The actual electronegativity difference between Se and S is so small that this bond is probably best characterized as a pure covalent bond having no bond dipole.
Bonding between a metal and a nonmetal is generally ionic. Bonding between two nonmetals is covalent, and in general, the bonding between two different nonmetals is usually polar covalent. When two different nonmetals have very similar electronegativities, the bonding is pure covalent or just covalent. a. ionic
b. covalent
c. polar covalent
d. ionic
e. polar covalent
f.
covalent
The possible ionic bonds that can form are between the metal Cs and the nonmetals P, O, and H. These ionic compounds are Cs3P, Cs2O, and CsH. The bonding between the various nonmetals will be covalent. P4, O2, and H2 are all pure covalent (or just covalent) with equal sharing of the bonding electrons. P−H will also be a covalent bond because P and H have identical electronegativities. The other possible covalent bonds that can form will all be polar covalent because the nonmetals involved in the bonds all have intermediate differences in electronegativities. The possible polar covalent bonds are P−O and O−H. Note: The bonding among cesium atoms is called metallic. This type of bonding between metals will be discussed in Chapter 10.
CHAPTER 8
BONDING: GENERAL CONCEPTS
327
43.
In general bonds between a metal and a nonmetal are usually ionic and bonds between two nonmetals are covalent. MgS is ionic, and the others are covalent. The electronegativities of C and H are close to each other, so C-H bonds are usually considered nonpolar (like F2). Of the remaining covalent bonds, the electronegativity difference between As and F is larger than the electronegativity difference between Se-O (from the electronegativity trend). So As-F is the most polar covalent bond of the bonds listed.
44.
All these bonds are between nonmetals, so they are all considered covalent bonds. The most polar covalent bond has the largest electronegativity difference between the atoms in the bond. The bonds are between N, O, S, and P. O has the largest electronegativity of these atoms and P has the smallest electronegativity. So, the most polar bond will be the P‒O bond. H has an electronegativity value identical to phosphorus, so the P-H bond is the purest covalent bond.
45.
Electronegativity values increase from left to right across the periodic table. The order of electronegativities for the atoms from smallest to largest electronegativity will be H = P < C < N < O < F. The most polar bond will be F‒H since it will have the largest difference in electronegativities, and the least polar bond will be P‒H since it will have the smallest difference in electronegativities (ΔEN = 0). The order of the bonds in decreasing polarity will be F‒H > O‒H > N‒H > C‒H > P‒H.
46.
Ionic character is proportional to the difference in electronegativity values between the two elements forming the bond. Using the trend in electronegativity, the order will be: Br‒Br < N‒O < C‒F < Ca‒O < K‒F least most ionic character ionic character Note that Br‒Br, N‒O, and C‒F bonds are all covalent bonds since the elements are all nonmetals. The Ca‒O and K‒F bonds are ionic, as is generally the case when a metal forms a bond with a nonmetal.
47.
48.
A permanent dipole moment exists in a molecule if the molecule has one specific area with a partial negative end (a red end in an electrostatic potential diagram) and a different specific region with a partial positive end (a blue end in an electrostatic potential diagram). If the blue and red colors are equally distributed in the electrostatic potential diagrams, then no permanent dipole exists. a. Has a permanent dipole.
b. Has no permanent dipole.
c. Has no permanent dipole.
d. Has a permanent dipole.
e. Has no permanent dipole.
f.
Has no permanent dipole.
a. H2O; both H2O and NH3 have permanent dipole moments in part due to the polar O−H and N−H bonds. But because oxygen is more electronegative than nitrogen, one would expect H2O to have a slightly greater dipole moment. This diagram has the more intense
328
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BONDING: GENERAL CONCEPS
red color on one end and the more intense blue color at the other end indicating a larger dipole moment. b. NH3; this diagram is for a polar molecule, but the colors are not as intense as the diagram in part a. Hence, this diagram is for a molecule which is not as polar as H 2O. Since N is less electronegative than O, NH3 will not be as polar as H2O. c. CH4; this diagram has no one specific red region and has four blue regions arranged symmetrically about the molecule. This diagram is for a molecule which has no dipole moment. This is only true for CH 4. The C‒H bonds are at best, slightly polar because carbon and hydrogen have similar electronegativity values. In addition, the slightly polar C‒H bond dipoles are arranged about carbon so that they cancel each other out, making CH4 a nonpolar molecule. See Example 8.2.
Ions and Ionic Compounds 49.
Al3+: [He]2s22p6; Ba2+: [Kr]5s24d105p6; Se2−: [Ar]4s23d104p6 ; I−: [Kr]5s24d105p6
50.
Te2−: [Kr]5s24d105p6; Cl−: [Ne]3s23p6; Sr2+: [Ar]4s23d104p6; Li+: 1s2
51.
a. Li+ and N3− are the expected ions. The formula of the compound would be Li 3N (lithium nitride). b. Ga3+ and O2−; Ga2O3, gallium(III) oxide or gallium oxide
52.
53.
c. Rb+ and Cl−; RbCl, rubidium chloride
d. Ba2+ and S2−; BaS, barium sulfide
a. Al3+ and Cl−; AlCl3, aluminum chloride
b. Na+ and O2−; Na2O, sodium oxide
c. Sr2+ and F−; SrF2, strontium fluoride
d. Ca2+ and S2−; CaS, calcium sulfide
a. Mg2+: 1s22s22p6; K+: 1s22s22p63s23p6; Al3+: 1s22s22p6 b. N3−, O2−, and F−: 1s22s22p6; Te2-: [Kr]5s24d105p6
54.
a. Sr2+: [Ar]4s23d104p6; Cs+: [Kr]5s24d105p6; In+: [Kr]5s24d10; Pb2+: [Xe]6s24f145d10 b. P3− and S2−: [Ne]3s23p6; Br−: [Ar]4s23d104p6
55.
a. Sc3+: [Ar]
b. Te2−: [Xe]
c. Ce4+: [Xe] and Ti4+: [Ar]
d. Ba2+: [Xe]
All these ions have the noble gas electron configuration shown in brackets. 56.
a. Cs2S is composed of Cs+ and S2−. Cs+ has the same electron configuration as Xe, and S 2− has the same configuration as Ar. b. SrF2; Sr2+ has the Kr electron configuration, and F− has the Ne configuration. c. Ca3N2; Ca2+ has the Ar electron configuration, and N3− has the Ne configuration. d. AlBr3; Al3+ has the Ne electron configuration, and Br− has the Kr configuration.
CHAPTER 8 57.
BONDING: GENERAL CONCEPTS
329
a. Na+ has 10 electrons. F−, O2−, and N3− are some possible anions also having 10 electrons. b. Ca2+ has 18 electrons. Cl−, S2−, and P3− also have 18 electrons. c. Al3+ has 10 electrons. F−, O2−, and N3− also have 10 electrons. d. Rb+ has 36 electrons. Br−, Se2−, and As3− also have 36 electrons.
58.
a. Ne has 10 electrons. AlN, MgF2, and Na2O are some possible ionic compounds where each ion has 10 electrons. b. CaS, K3P, and KCl are some examples where each ion is isoelectronic with Ar; i.e., each ion has 18 electrons. c. Each ion in Sr3As2, SrBr2, and Rb2Se is isoelectronic with Kr. d. Each ion in BaTe and CsI is isoelectronic with Xe.
59.
Se2−, Br−, Rb+, Sr2+, Y3+, and Zr4+ are some ions that are isoelectronic with Kr (36 electrons). In terms of size, the ion with the most protons will hold the electrons tightest and will be the smallest. The size trend is: Zr4+ < Y3+ < Sr2+ < Rb+ < Br− < Se2− smallest largest
60.
All these ions have 18 e−; the smallest ion (Sc3+) has the most protons attracting the 18 e−, and the largest ion has the fewest protons (S 2−). The order in terms of increasing size is Sc3+ < Ca2+ < K+ < Cl− < S2−. In terms of the atom size indicated in the question:
61.
In each group, the smallest species is O‒, Mg2+, and Mg2+. As electrons are added to a species, size increases. Conversely, as electrons are removed from a species, size decreases. Between Mg2+ and O2‒, both have 10 electrons. Mg2+ has 12 protons compared to 8 protons for O2‒. Mg2+ with the most protons will be smallest.
62.
The stable ions in ionic compounds are N3‒, F‒, Na+, and Al3+. These are all isoelectronic (each has 10 electrons). The correct order is Al3+ < Na+ < F‒ < N3‒. Al3+ has the most protons, so it is smallest in size. The largest ion in this isoelectronic series, N3‒, has the fewest protons.
63.
a. Fe > Fe2+ > Fe3+
b. Au+ > Ag+ > Cu+
d. Br‒ > Cl‒ > F‒
e. Se2− > Rb+ > Yb3+
c. S2− > S− > S
330
CHAPTER 8
BONDING: GENERAL CONCEPS
For answer a, as electrons are removed from an atom, size decreases. Answers b and d follow the radius trend. For answer c, as electrons are added to an atom, size increases. Answer e follows the trend for an isoelectronic series; i.e., the smallest ion has the most protons. 64.
65.
66.
67.
68.
a. V > V2+ > V3+ > V5+
b. Cs+ > Rb+ > K+ > Na+
d. P3− > P2− > P− > P
e. Te2− > Se2− > S2− > O2−
Lattice energy is proportional to −Q1Q2/r, where Q is the charge of the ions and r is the distance between the centers of the ions. The more negative the lattice energy, the more stable the ionic compound. So greater charged ions as well as smaller sized ions lead to more negative lattice energy values and more stable ionic compounds. a. NaCl; Na+ is smaller than K+.
b. LiF; F− is smaller than Cl−.
c. MgO; O2− has a greater charge than OH-.
d. Fe(OH)3; Fe3+ has a greater charge than Fe2+.
e. Na2O; O2− has a greater charge than Cl−.
f. MgO; both ions are smaller in MgO.
a. LiF; Li+ is smaller than Cs+.
b. NaBr; Br- is smaller than I−.
c. BaO; O2− has a greater charge than Cl-.
d. CaSO4; Ca2+ has a greater charge than Na+.
e. K2O; O2− has a greater charge than F-.
f. Li2O; both ions are smaller in Li2O.
K(s) → K(g) K(g) → K+(g) + e− 1/2 Cl2(g) → Cl(g) Cl(g) + e− → Cl−(g) + K (g) + Cl−(g) → KCl(s)
ΔH = 90. kJ (sublimation) ΔH = 419 kJ (ionization energy) ΔH = 239/2 kJ (bond energy) ΔH = −349 kJ (electron affinity) ΔH = −690. kJ (lattice energy)
K(s) + 1/2 Cl2(g) → KCl(s)
ΔH of = −411 kJ/mol
Mg(s) → Mg(g) Mg(g) → Mg+(g) + e− Mg+(g) → Mg2+(g) + e− F2(g) → 2 F(g) 2 F(g) + 2 e− → 2 F−(g) Mg2+(g) + 2 F−(g) → MgF2(s)
ΔH = 150. kJ ΔH = 735 kJ ΔH = 1445 kJ ΔH = 154 kJ ΔH = 2(−328) kJ ΔH = −2913 kJ
Mg(s) + F2(g) → MgF2(s) 69.
c. Te2− > I− > Cs+ > Ba2+
(sublimation) (IE1) (IE2) (BE) (EA) (LE)
ΔH of = −1085 kJ/mol
From the data given, it takes less energy to produce Mg +(g) + O−(g) than to produce Mg2+(g) + O2−(g). However, the lattice energy for Mg 2+O2− will be much more exothermic than that for Mg+O− due to the greater charges in Mg2+O2−. The favorable lattice energy term dominates, and Mg2+O2− forms.
CHAPTER 8 70.
BONDING: GENERAL CONCEPTS
Na(g) → Na+(g) + e− F(g) + e− → F−(g) Na(g) + F(g) → Na+(g) + F−(g)
331
ΔH = IE = 495 kJ (Table 7.5) ΔH = EA = −327.8 kJ (Table 7.7) ΔH = 167 kJ
The described process is endothermic. What we haven’t accounted for is the extremely favorable lattice energy. Here, the lattice energy is a large negative (exothermic) value, making the overall formation of NaF a favorable exothermic process. 71.
Use Figure 8.11 as a template for this problem. Li(s) → Li(g) Li(g) → Li+(g) + e− 1/2 I2(g) → I(g) I(g) + e− → I−(g) + Li (g) + I−(g) → LiI(s) Li(s) + 1/2 I2(g) → LiI(s)
ΔHsub = ? ΔH = 520. kJ ΔH = 151/2 kJ ΔH = −295 kJ ΔH = −753 kJ ΔH = −292 kJ
ΔHsub + 520. + 151/2 − 295 − 753 = −292, ΔHsub = 161 kJ 72.
Let us look at the complete cycle for Na2S. 2 Na(s) → 2 Na(g) 2 Na(g) → 2 Na+(g) + 2 e− S(s) → S(g) S(g) + e− → S−(g) S−(g) + e− → S2−(g) + 2 Na (g) + S2−(g) → Na2S(s) 2 Na(s) + S(s) → Na2S(s)
2ΔHsub, Na = 2(109) kJ 2IE = 2(495) kJ ΔHsub, S = 277 kJ EA1 = −200. kJ EA2 = ? LE = −2203 kJ
ΔH of = −365 kJ
ΔH of = 2ΔHsub, Na + 2IE + ΔHsub, S + EA1 + EA2 + LE, −365 = −918 + EA2, EA2 = 553 kJ For each salt: ΔH of = 2ΔHsub, M + 2IE + 277 − 200. + LE + EA2 K2S: −381 = 2(90.) + 2(419) + 277 − 200. − 2052 + EA2, EA2 = 576 kJ Rb2S: −361 = 2(82) + 2(409) + 277 − 200. − 1949 + EA2, EA2 = 529 kJ Cs2S: −360. = 2(78) + 2(382) + 277 − 200. − 1850. + EA2, EA2 = 493 kJ We get EA2 values from 493 to 576 kJ. The mean value is: ±50 kJ.
553 + 576 + 529 + 493 = 538 kJ. We can represent the results as EA2 = 540 4
332 73.
CHAPTER 8
Ca2+ has a greater charge than Na+, and Se2− is smaller than Te2−. The effect of charge on the lattice energy is greater than the effect of size. We expect the trend from most exothermic lattice energy to least exothermic to be: CaSe > CaTe > Na2Se > Na2Te (−2862) (−2721) (−2130) (−2095)
74.
BONDING: GENERAL CONCEPTS
This is what we observe.
Lattice energy is proportional to the charge of the cation times the charge of the anion Q1Q2. Compound
Q1Q2
Lattice Energy
FeCl2
(+2)( −1) = −2
−2631 kJ/mol
FeCl3
(+3)( −1) = −3
−5359 kJ/mol
Fe2O3
(+3)( −2) = −6
−14,744 kJ/mol
Bond Energies 75.
When the overall reactant bond strengths are stronger than the product bond strengths, the reaction will be endothermic. This is what we have with this reaction; the bond strengths of the N2 and O2 bonds are overall stronger than the bond strengths of two NO bonds.
76.
This reaction will be exothermic. Here the bond strengths of two HF molecules must be stronger than the combined bond strengths of the F 2 and H2 molecules. This results in an exothermic reaction. Energy is released when going from weaker bonds to stronger bonds.
77.
a. Bonds broken:
Bonds formed:
1 H‒H (432 kJ/mol) 1 Cl‒Cl (239 kJ/mol)
2 H‒Cl (427 kJ/mol)
ΔH = ΣDbroken − ΣDformed, ΔH = 432 kJ + 239 kJ − 2(427) kJ = −183 kJ b.
N
N+ 3 H
H
2H
N
H
H
Bonds broken: 1 N≡N (941 kJ/mol) 3 H‒H (432 kJ/mol)
Bonds formed: 6 N‒H (391 kJ/mol)
ΔH = 941 kJ + 3(432) kJ − 6(391) kJ = −109 kJ
CHAPTER 8 78.
BONDING: GENERAL CONCEPTS
333
Sometimes some of the bonds remain the same between reactants and products. To save time, only break and form bonds that are involved in the reaction. H
H
C
N
H
H
a. H
C
N+2 H
H
H
Bonds broken:
Bonds formed:
1 C≡N (891 kJ/mol) 2 H−H (432 kJ/mol)
1 C−N (305 kJ/mol) 2 C−H (413 kJ/mol) 2 N−H (391 kJ/mol)
ΔH = 891 kJ + 2(432 kJ) − [305 kJ + 2(413 kJ) + 2(391 kJ)] = −158 kJ H
b.
H N
+2 F
N
H
4 H
F
F + N
N
H
Bonds broken:
Bonds formed:
1 N−N (160. kJ/mol) 4 N−H (391 kJ/mol) 2 F−F (154 kJ/mol)
4 H−F (565 kJ/mol) 1 N ≡ N (941 kJ/mol)
ΔH = 160. kJ + 4(391 kJ) + 2(154 kJ) − [4(565 kJ) + 941 kJ] = −1169 kJ H
79. H
C N
H C
H
H
Bonds broken: 1 C‒N (305 kJ/mol)
C C N H
Bonds formed: 1 C‒C (347 kJ/mol)
ΔH = ΣDbroken − ΣDformed, ΔH = 305 − 347 = −42 kJ Note: Sometimes some of the bonds remain the same between reactants and products. To save time, only break and form bonds that are involved in the reaction. 80.
H H
C O H + C O H
H
H
O
C
C O H
H
334
CHAPTER 8 Bonds broken:
BONDING: GENERAL CONCEPTS
Bonds formed:
1 C≡O (1072 kJ/mol) 1 C‒O (358 kJ/mol)
1 C‒C (347 kJ/mol) 1 C=O (745 kJ/mol) 1 C‒O (358 kJ/mol)
ΔH = 1072 + 358 − [347 + 745 + 358] = −20. kJ 81. Bonds broken:
Bonds formed:
4 O−H (467 kJ/mol) 2 F−F (154 kJ/mol)
4 H−F (565 kJ/mol) 1 O=O (495 kJ/mol)
H = 4(467) + 2(154) − [4(565) + 495] = −579 kJ 82.
Bonds broken:
Bonds formed:
4 C−H (413 kJ/mol) 2 O−H (467 kJ/mol)
1 CO (1072 kJ/mol) 3 H−H (432 kJ/mol)
H = 4(413) + 2(467) – [1072 + 3(432)] = 218 kJ 83.
H−C≡C−H + 5/2 O=O → 2 O=C=O + H−O−H Bonds broken:
Bonds formed:
2 C−H (413 kJ/mol) 1 C≡C (839 kJ/mol) 5/2 O=O (495 kJ/mol)
2 × 2 C=O (799 kJ/mol) 2 O−H (467 kJ/mol)
H = 2(413 kJ) + 839 kJ + 5/2 (495 kJ) – [4(799 kJ) + 2(467 kJ)] = −1228 kJ 84.
CH4
+ 2 O=O
→ O=C=O + 2 H−O−H
Bonds broken: 4 C−H (413 kJ/mol) 2 O=O (495 kJ/mol)
Bonds formed: 2 C=O (799 kJ/mol) 2 2 O−H (467 kJ/mol)
H = 4(413 kJ) + 2(495 kJ) – [2(799 kJ) + 4(467 kJ)] = −824 kJ
CHAPTER 8
BONDING: GENERAL CONCEPTS
85. H
H C
H
+ F
C
F
H
H
F
F
C
C
H
H
Bonds broken:
335
H
H = -549 kJ
Bonds formed:
1 C=C (614 kJ/mol) 1 F−F (154 kJ/mol)
1 C‒C (347 kJ/mol) 2 C‒F (DCF = C‒F bond energy)
ΔH = −549 kJ = 614 kJ + 154 kJ − [347 kJ + 2DCF], 2DCF = 970., DCF = 485 kJ/mol 86.
Let x = bond energy for A2, so 2x = bond energy for AB. H = −285 kJ = x + 432 kJ – [2(2x)], 3x = 717, x = 239 kJ/mol The bond energy for A2 is 239 kJ/mol.
87.
a. ΔH° = 2 H fo, HCl = 2 mol(−92 kJ/mol) = −184 kJ (−183 kJ from bond energies) b. ΔH° = 2 H fo, NH = 2 mol(–46 kJ/mol) = −92 kJ (−109 kJ from bond energies) 3
Comparing the values for each reaction, bond energies seem to give a reasonably good estimate for the enthalpy change of a reaction. The estimate is especially good for gas phase reactions. 88.
CH3OH(g) + CO(g) → CH3COOH(l)
ΔH° = −484 kJ − [(−201 kJ) + (−110.5 kJ)] = −173 kJ
Using bond energies, ΔH = −20. kJ. For this reaction, bond energies give a much poorer estimate for ΔH as compared with gas-phase reactions (see Exercise 87). The major reason for the large discrepancy is that not all species are gases in Exercise 80. Bond energies do not account for the energy changes that occur when liquids and solids form instead of gases. These energy changes are due to intermolecular forces and will be discussed in Chapter 10. 89.
a. Using SF4 data: SF4(g) → S(g) + 4 F(g) ΔH° = 4DSF = 278.8 + 4 (79.0) − (−775) = 1370. kJ DSF =
1370 . kJ = 342.5 kJ/mol = S−F bond energy 4 mol SF bonds
Using SF6 data: SF6(g) → S(g) + 6 F(g) ΔH° = 6DSF = 278.8 + 6 (79.0) − (−1209) = 1962 kJ DSF =
1962 kJ = 327.0 kJ/mol = S−F bond energy 6 mol
336
CHAPTER 8
BONDING: GENERAL CONCEPTS
b. The S‒F bond energy in Table 8.5 is 327 kJ/mol. The value in the table was based on the S−F bond in SF6. c. S(g) and F(g) are not the most stable forms of the elements at 25°C and 1 atm. The most stable forms are S8(s) and F2(g); H f = 0 for these two species. 90.
NH3(g) → N(g) + 3 H(g) ΔH° = 3DNH = 472.7 + 3(216.0) − (−46.1) = 1166.8 kJ DNH =
1166 .8 kJ = 388.93 kJ/mol 389 kJ/mol 3 mol NH bonds
Dcalc = 389 kJ/mol as compared with 391 kJ/mol in Table 8.5. There is good agreement. 91.
H‒H(g) + O=O(g) → H‒O‒O‒H(g) ΔHrxn = H of, H2 O2 Bonds broken:
Bonds formed:
1 H ‒H (432 kJ/mol) 1 O=O (495 kJ/mol)
2 H‒O (467 kJ/mol) 1 O‒O (146 kJ/mol)
H of, H2 O2 = 432 + 495 – [2(467) + 146] = ‒153 kJ/mol
92.
1/2 N2(g) + 1/2 O2(g) → NO(g) Bonds broken: 1/2 NN (941 kJ/mol) 1/2 O=O (495 kJ/mol)
H = 90. kJ Bonds formed: 1 NO (DNO = NO bond energy)
H = 90. kJ = 1/2(941) + 1/2(495) – (DNO), DNO = 628 kJ/mol From this data, the calculated NO bond energy is 628 kJ/mol.
Lewis Structures and Resonance 93.
Drawing Lewis structures is mostly trial and error. However, the first two steps are always the same. These steps are (1) count the valence electrons available in the molecule/ion, and (2) attach all atoms to each other with single bonds (called the skeletal structure). Unless noted otherwise, the atom listed first is assumed to be the atom in the middle, called the central atom, and all other atoms in the formula are attached to this atom. The most notable exceptions to the rule are formulas that begin with H, e.g., H2O, H2CO, etc. Hydrogen can never be a central atom since this would require H to have more than two electrons. In these compounds, the atom listed second is assumed to be the central atom. After counting valence electrons and drawing the skeletal structure, the rest is trial and error. We place the remaining electrons around the various atoms in an attempt to satisfy the octet rule (or duet rule for H). Keep in mind that practice makes perfect. After practicing, you can (and will) become very adept at drawing Lewis structures.
CHAPTER 8
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337
a. F2 has 2(7) = 14 valence electrons.
b. O2 has 2(6) = 12 valence electrons.
c. CO has 4 + 6 = 10 valence electrons.
d. CH4 has 4 + 4(1) = 8 valence electrons.
e. NH3 has 5 + 3(1) = 8 valence electrons.
f.
H2O has 2(1) + 6 = 8 valence electrons.
g. HF has 1 + 7 = 8 valence electrons.
94.
a. H2CO has 2(1) + 4 + 6 = 12 valence e−.
b. CO2 has 4 + 2(6) = 16 valence electrons.
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c. HCN has 1 + 4 + 5 = 10 valence electrons.
95.
The Lewis structure uses 14 e− valence electrons and the formula of the ion is X22‒. Let’s setup an equation for determining the total number of valence electrons in X 22‒: 14 = 2(x) + 2, x = 6 number of valence electrons of X Element X is a group 6A nonmetal. Possible identities are O, S, Se, and Te. O22‒ is most likely.
96.
C22− has 2(4) + 2 + 10 valence e−, N22− has 2(5) + 2 = 12 valence e−, O22+ has 2(6) ‒ 2 =10 valence e−, and F22+ has 2(7) ‒ 2 = 12 valence e−. The Lewis structures are:
All the ions require double or triple bonds to satisfy the octet rule. 97.
Drawing Lewis structures is mostly trial and error. However, the first two steps are always the same. These steps are (1) count the valence electrons available in the molecule/ion, and (2) attach all atoms to each other with single bonds (called the skeletal structure). Unless noted otherwise, the atom listed first is assumed to be the atom in the middle, called the central atom, and all other atoms in the formula are attached to this atom. The most notable exceptions to the rule are formulas that begin with H, e.g., H2O, H2CO, etc. Hydrogen can never be a central atom since this would require H to have more than two electrons. In these compounds, the atom listed second is assumed to be the central atom. After counting valence electrons and drawing the skeletal structure, the rest is trial and error. We place the remaining electrons around the various atoms in an attempt to satisfy the octet rule (or duet rule for H). a. CCl4 has 4 + 4(7) = 32 valence electrons.
b. NCl3 has 5 + 3(7) = 26 valence electrons.
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BONDING: GENERAL CONCEPTS
c. SeCl2 has 6 + 2(7) = 20 valence electrons.
98.
339
d. ICl has 7 + 7 = 14 valence electrons.
a. POCl3 has 5 + 6 + 3(7) = 32 valence electrons. O Cl
O
P
Cl
Cl
P
Cl
Cl
Skeletal structure
Lewis structure
Cl
Note: This structure uses all 32 e− while satisfying the octet rule for all atoms. This is a valid Lewis structure.
SO42− has 6 + 4(6) + 2 = 32 valence electrons. 2-
O O
S
Note: A negatively charged ion will have additional electrons to those that come from the valence shell of the atoms. The magnitude of the negative charge indicates the number of extra electrons to add in.
O
O
XeO4, 8 + 4(6) = 32 e−
PO43−, 5 + 4(6) + 3 = 32 e−
O O
Xe
O
O
ClO4− has 7 + 4(6) + 1 = 32 valence electrons -
O O
Cl
O
O
Note: All of these species have the same number of atoms and the same number of valence electrons. They also have the same Lewis structure.
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CHAPTER 8
SO32−, 6 + 3(6) + 2 = 26 e−
b. NF3 has 5 + 3(7) = 26 valence electrons.
F
N
F
F
N
2-
F
F
F
Skeletal structure
Lewis structure
BONDING: GENERAL CONCEPTS
O
S
O
O
PO33−, 5 + 3(6) + 3 = 26 e−
ClO3−, 7 + 3(6) + 1 = 26 e−
Note: Species with the same number of atoms and valence electrons have similar Lewis structures. c. ClO2− has 7 + 2(6) + 1 = 20 valence
O Cl
O
O Cl
SCl2, 6 + 2(7) = 20 e−
O PCl2−, 5 + 2(7) + 1 = 20 e−
Note: Species with the same number of atoms and valence electrons have similar Lewis structures. d. Molecules ions that have the same number of valence electrons and the same number of atoms will have similar Lewis structures. 99.
BeH2, 2 + 2(1) = 4 valence electrons
BH3, 3 + 3(1) = 6 valence electrons
H B H 100.
a. NO2, 5 + 2(6) = 17 e−
Plus others
H
N2O4, 2(5) + 4(6) = 34 e−
Plus other resonance structures
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b. BH3, 3 + 3(1) = 6 e−
341
NH3, 5 + 3(1) = 8 e−
H B H
H
BH3NH3, 6 + 8 = 14 e−
H H
H B
H
N
H H
In reaction a, NO2 has an odd number of electrons, so it is impossible to satisfy the octet rule. By dimerizing to form N2O4, the odd electron on two NO2 molecules can pair up, giving a species whose Lewis structure can satisfy the octet rule. In general, odd-electron species are very reactive. In reaction b, BH3 is electron-deficient. Boron has only six electrons around it. By forming BH3NH3, the boron atom satisfies the octet rule by accepting a lone pair of electrons from NH3 to form a fourth bond. 101.
PF5, 5 +5(7) = 40 valence electrons
SF4, 6 + 4(7) = 34 e−
ClF3, 7 + 3(7) = 28 e−
Br3−, 3(7) + 1 = 22 e−
Row 3 and heavier nonmetals can have more than 8 electrons around them when they have to. Row 3 and heavier elements have empty d orbitals that are close in energy to valence s and p orbitals. These empty d orbitals can accept extra electrons. For example, P in PF5 has its five valence electrons in the 3s and 3p orbitals. These s and p orbitals have room for three more electrons, and if it has to, P can use the empty 3d orbitals for any electrons above 8.
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102.
SF6, 6 + 6(7) = 48 e−
ClF5, 7 + 5(7) = 42 e−
103.
BH3 only has 6 valence electrons, so it must be an exception to the octet rule for the central B atom. NS2 has an odd number of valence electrons (17 e−), so it must be an exception to the octet rule for one of the atoms in the compound. For the other two compounds, here are the Lewis structures: BrF3, 7 + 3(7) = 28 e−
XeF4, 8 + 4(7) = 36 e−
OCl2, 6 + 2(7) = 20
BrF3 must also be an exception to the octet rule. With 28 valence electrons, it is impossible to satisfy the octet rule for all atom in the compound. OCl2 satisfies the octet rule. 104.
KrF4, 8 + 4(7) = 36 e−
ICl3, 7 + 3(7) = 28 e−
XeCl2, 8 + 2(7) = 22 e−
SCl2, 6 + 2(7) = 20 e−
PF3, 5 + 3(7) = 26
All but PF3 have at least two lone pairs about the central atom. 105.
The Lewis structure has 34 valence electrons, and the formula of the ion is EF2O2‒. Let x = the number of valence electrons of E: 34 = x + 2(7) + 2(6) + 1, x = 7 E is a halogen. In addition, E must be a row 3 or heavier element since this ion has more than eight electrons around the central E atom (row 2 elements never have more than eight electrons around them). Some possible halogens for E are Cl, Br, I, and At.
CHAPTER 8 106.
BONDING: GENERAL CONCEPTS
343
The Lewis structure has 36 valence electrons, and the formula of the ion is XF 3S3‒. Let x = the number of valence electrons of X: 36 = x + 3(7) + 6 + 3, x = 6 X is in Group 6A. In addition, X must also be a row 3 or heavier element since this ion has more than eight electrons around the central X atom (row 2 elements never have more than eight electrons around them). Some possible group 6A nonmetals for X are S, Se, and Te.
107.
a. NO2− has 5 + 2(6) + 1 = 18 valence electrons. The skeletal structure is O‒N‒O. To get an octet about the nitrogen and only use 18 e- , we must form a double bond to one of the oxygen atoms. O N
O
O N
O
Because there is no reason to have the double bond to a particular oxygen atom, we can draw two resonance structures. Each Lewis structure uses the correct number of electrons and satisfies the octet rule, so each is a valid Lewis structure. Resonance structures occur when you have multiple bonds that can be in various positions. We say the actual structure is an average of these two resonance structures. NO3− has 5 + 3(6) + 1 = 24 valence electrons. We can draw three resonance structures for NO3−, with the double bond rotating among the three oxygen atoms. O
O
N O
O
N O
O
N O
O
O
N2O4 has 2(5) + 4(6) = 34 valence electrons. We can draw four resonance structures for N2O4. O
O N
O
N
O N
N
O
O
O
O
O
O
O
O
N O
N
N O
O
N O
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b. OCN− has 6 + 4 + 5 + 1 = 16 valence electrons. We can draw three resonance structures for OCN−. O C N
O C N
O C N
SCN− has 6 + 4 + 5 + 1 = 16 valence electrons. Three resonance structures are possible. S C N
S C N
S C N
N3− has 3(5) + 1 = 16 valence electrons. As with OCN − and SCN−, three different resonance structures can be drawn. N N N
108.
N N N
N N N
Ozone: O3 has 3(6) = 18 valence electrons.
Sulfur dioxide: SO2 has 6 + 2(6) = 18 valence electrons.
Sulfur trioxide: SO3 has 6 + 3(6) = 24 valence electrons. O
O
O
S
S
S
O
109.
O
O
O
O
O
Benzene has 6(4) + 6(1) = 30 valence electrons. Two resonance structures can be drawn for benzene. The actual structure of benzene is an average of these two resonance structures; i.e., all carbon-carbon bonds are equivalent with a bond length and bond strength somewhere between a single and a double bond. H H
H
C C C
H
H
C H
H
C
C
C
C H
H
H
C C C C H
H
CHAPTER 8 110.
BONDING: GENERAL CONCEPTS
345
Borazine (B3N3H6) has 3(3) + 3(5) + 6(1) = 30 valence electrons. The possible resonance structures are like those of benzene in Exercise 109. H H
H H
B N B N
H
H
N
N
B
B H
N B N
H
H
111.
H
B
H
H
We will use a hexagon to represent the six-member carbon ring, and we will omit the four hydrogen atoms and the three lone pairs of electrons on each chlorine. If no resonance existed, we could draw four different molecules: Cl
Cl Cl
Cl
Cl Cl
Cl Cl
If the double bonds in the benzene ring exhibit resonance, then we can draw only three different dichlorobenzenes. The circle in the hexagon represents the delocalization of the three double bonds in the benzene ring (see Exercise 109). Cl
Cl
Cl
Cl
Cl Cl
With resonance, all carbon-carbon bonds are equivalent. We can’t distinguish between a single and double bond between adjacent carbons that have a chlorine attached. That only three isomers are observed supports the concept of resonance. 112.
CO32− has 4 + 3(6) + 2 = 24 valence electrons. 2-
O C O
2-
O
C
C O
O
2-
O
O
O
O
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Three resonance structures can be drawn for CO 32−. The actual structure for CO32− is an average of these three resonance structures. That is, the three C‒O bond lengths are all equivalent, with a length somewhere between a single and a double bond. The actual bond length of 136 pm is consistent with this resonance view of CO 32−. 113.
CH3NCO has 4 + 3(1) + 5 + 4 + 6 = 22 valence electrons. Three resonance structures can be drawn for methyl isocyanate. H H
C
H N
C
O
H
H
C
H N
C
O
H
H
C
N
C
O
H
114.
Amprya has 5(4) + 6(1) +2(5) = 36 valence electrons. Two resonance structures can be drawn for Amprya.
115.
The Lewis structures for the various species are: CO (10 e−):
Triple bond between C and O.
CO2 (16 e−):
Double bond between C and O. 2-
O
CO32− (24 e−):
C O
2-
O
C
C O
O
O
Average of 1 1/3 bond between C and O in CO32−.
CH3OH (14 e−):
2-
O
Single bond between C and O.
O
O
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BONDING: GENERAL CONCEPTS
347
As the number of bonds increases between two atoms, bond strength increases, and bond length decreases. With this in mind, then: Longest → shortest C – O bond: CH3OH > CO32− > CO2 > CO Weakest → strongest C – O bond: CH3OH < CO32− < CO2 < CO 116.
H2NOH (14 e−) H
N
O
H
Single bond between N and O
H N 2O (16 e-):
N
N
N
O
N
O
N
N
O
Average of a double bond between N and O N O+ (10 e-):
N
N O2- (18 e-):
O N
+
O
Triple bond between N and O -
O
O N
O
-
Average of 1 1/2 bond between N and O N O3- (24 e-):
-
O N O
-
O N
O
O
-
O N
O
O
O
Average of 1 1/3 bond between N and O
From the Lewis structures, the order from shortest → longest N‒O bond is: NO+ < N2O < NO2− < NO3− < H2NOH
Formal Charge 117.
BF3 has 3 + 3(7) = 24 valence electrons. The two Lewis structures to consider are: F
F +1
0
B -1 F F
0
0
0
B 0 F F
0
The formal charges for the various atoms are assigned in the Lewis structures. Formal charge = number of valence electrons on free atom − number of lone pair electrons on atoms − 1/2 (number of shared electrons of atom). For B in the first Lewis structure, formal charge (FC) = 3 − 0 − 1/2(8) = −1. For F in the first structure with the double bond, FC = 7 − 4 − 1/2(4) = +1. The others all have a formal charge equal to zero [FC = 7 – 6 – 1/2(2) = 0].
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The first Lewis structure obeys the octet rule but has a +1 formal charge on the most electronegative element there is, fluorine, and a negative formal charge on a much less electronegative element, boron. This is just the opposite of what we expect: negative formal charge on F and positive formal charge on B. The other Lewis structure does not obey the octet rule for B but has a zero formal charge on each element in BF 3. Because structures generally want to minimize formal charge, then BF 3 with only single bonds is best from a formal charge point of view. 118.
CO2, 4 + 2(6) = 16 valence electrons
The formal charges are shown above the atoms in the three Lewis structures. The best Lewis structure for CO2 from a formal charge standpoint is the first structure having each oxygen double bonded to carbon. This structure has a formal charge of zero on all atoms (which is preferred). The other two resonance structures have nonzero formal charges on the oxygen atoms, making them less reasonable. For CO 2, we usually ignore the last two resonance structures and think of the first structure as the true Lewis structure for CO 2. 119.
See Exercise 98 for the Lewis structures of POCl3, SO42−, ClO4− and PO43−. All these compounds/ions have similar Lewis structures to those of SO 2Cl2 and XeO4 shown below. Formal charge = [number of valence electrons on free atom] − [number of lone pair electrons on atom + 1/2(number of shared electrons of atom)]. a. POCl3: P, FC = 5 − 1/2(8) = +1
b. SO42−: S, FC = 6 − 1/2(8) = +2
c. ClO4−: Cl, FC = 7 − 1/2(8) = +3
d. PO43−: P, FC = 5 − 1/2(8) = +1
e. SO2Cl2, 6 + 2(6) + 2(7) = 32 e−
f.
XeO4, 8 + 4(6) = 32 e-
O Cl
S
O Cl
O
O
Xe
O
O
S, FC = 6 – 1/2(8) = +2
Xe, FC = 8 – 1/2(8) = +4
g. ClO3−, 7 + 3(6) + 1 = 26 e−
h. NO43−, 5 + 4(6) + 3 = 32 e− 3-
O
O Cl
O
O
Cl, FC = 7 – 2 – 1/2(6) = +2
O N
O
O
N, FC = 5 – 1/2(8) = +1
CHAPTER 8 120.
BONDING: GENERAL CONCEPTS
349
For SO42−, ClO4−, PO43− and ClO3−, only one of the possible resonance structures is drawn. a. Must have five bonds to P to minimize formal charge of P. The best choice is to form a double bond to O since this will give O a formal charge of zero, and single bonds to Cl for the same reason.
b. Must form six bonds to S to minimize formal charge of S.
Cl
P
2-
O
O
S, FC = 0
O S O
P, FC = 0
Cl
O
Cl
c. Must form seven bonds to Cl to minimize formal charge.
d. Must form five bonds to P to to minimize formal charge. 3-
O O
P
O
P, FC = 0
O
e.
f. O Cl
S
S, FC = 0 Cl, FC = 0 O, FC = 0
Cl
O
g.
O O
Xe
O
Xe, FC = 0
O
O Cl
O
Cl, FC = 0
O
h. We can’t. The following structure has a zero formal charge for N: 3-
O O N
O
O
but N does not expand its octet. We wouldn’t expect this resonance form to exist. 121.
O2F2 has 2(6) + 2(7) = 26 valence e−. The formal charge and oxidation number (state) of each atom is below the Lewis structure of O2F2. F
O
O
F
Formal Charge
0
0
0
0
Oxid. Number
-1
+1
+1
-1
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Oxidation states are more useful when accounting for the reactivity of O 2F2. We are forced to assign +1 as the oxidation state for oxygen due to the bonding to fluorine. Oxygen is very electronegative, and +1 is not a stable oxidation state for this element. 122.
OCN− has 6 + 4 + 5 + 1 = 16 valence electrons.
Formal charge
O
C
N
O
C
N
O
C
N
0
0
-1
-1
0
0
+1
0
-2
Only the first two resonance structures should be important. The third places a positive formal charge on the most electronegative atom in the ion and a –2 formal charge on N. CNO− will also have 16 valence electrons.
Formal charge
C
N
O
C
N
O
C
N
O
-2
+1
0
-1
+1
-1
-3
+1
+1
All the resonance structures for fulminate (CNO −) involve greater formal charges than in cyanate (OCN−), making fulminate more reactive (less stable). 123.
SCl, 6 + 7 = 13; the formula could be SCl (13 valence electrons), S 2Cl2 (26 valence electrons), S3Cl3 (39 valence electrons), etc. For a formal charge of zero on S, we will need each sulfur in the Lewis structure to have two bonds to it and two lone pairs [FC = 6 – 4 – 1/2(4) = 0]. Cl will need one bond and three lone pairs for a formal charge of zero [FC = 7 – 6 – 1/2(2) = 0]. Since chlorine wants only one bond to it, it will not be a central atom here. With this in mind, only S2Cl2 can have a Lewis structure with a formal charge of zero on all atoms. The structure is:
124.
CS2, 4 + 2(6) = 16 e−; C3S2, 3(4) + 2(6) = 24 e−; for carbon to have a formal charge of zero, we must draw the Lewis structures with 4 bonds to each carbon along with no lone pairs. For sulfur to have a formal charge of zero and follow the octet rule, we must draw the Lewis structures with 2 bonds to each sulfur along with two lone pairs of electrons on each sulfur. The correct Lewis structures for CS 2 and C3S2 that have formal charges of zero for all atoms are:
125.
For formal charge values of zero: (1) each carbon in the structure has 4 bonding pairs of electrons and no lone pairs; (2) each N has 3 bonding pairs of electrons and 1 lone pair of electrons;
CHAPTER 8
BONDING: GENERAL CONCEPTS
351
(3) each O has 2 bonding pairs of electrons and 2 lone pairs of electrons; (4) each H is attached by only a single bond (1 bonding pair of electrons). Following these guidelines, the Lewis structure is:
126.
For a formal charge of zero, carbon atoms in the structure will all satisfy the octet rule by forming four bonds (with no lone pairs). Oxygen atoms have a formal charge of zero by forming two bonds and having two lone pairs of electrons. Hydrogen atoms have a formal charge of zero by forming a single bond (with no lone pairs). Following these guidelines, two resonance structures can be drawn for benzoic acid.
Molecular Structure and Polarity 127.
The first step always is to draw a valid Lewis structure when predicting molecular structure. When resonance is possible, only one of the possible resonance structures is necessary to predict the correct structure because all resonance structures give the same structure. The Lewis structures are in Exercises 97 and 107. The structures and bond angles for each follow. 97:
a. CCl4: tetrahedral, 109.5°
b. NCl3: trigonal pyramid, <109.5°
c. SeCl2: V-shaped or bent, <109.5°
d. ICl:
linear, but there is no bond angle present
Note: NCl3 and SeCl2 both have lone pairs of electrons on the central atom that result in bond angles that are something less than predicted from a tetrahedral arrangement (109.5°). However, we cannot predict the exact number. For the solutions manual, we will insert a less than sign to indicate this phenomenon. For bond angles equal to 120°, the lone pair
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BONDING: GENERAL CONCEPTS
phenomenon isn’t as significant as compared to smaller bond angles. For these molecules, for example, NO2−, we will insert an approximate sign in front of the 120° to note that there may be a slight distortion from the VSEPR predicted bond angle. 107:
a. NO2−: V-shaped, ≈120°; NO3−: trigonal planar, 120° N2O4: trigonal planar, 120° about both N atoms b. OCN−, SCN−, and N3− are all linear with 180° bond angles.
128.
See Exercises 98 and 108 for the Lewis structures. 98:
a. All are tetrahedral; 109.5° b. All are trigonal pyramid; <109.5° c. All are V-shaped; <109.5°
108:
129.
O3 and SO2 are V-shaped (or bent) with a bond angle ≈120°. SO3 is trigonal planar with 120° bond angles.
From the Lewis structures (see Exercise 101), Br3− would have a linear molecular structure, ClF3 would have a T-shaped molecular structure, and SF4 would have a see-saw molecular structure. For example, consider ClF3 (28 valence electrons): The central Cl atom is surrounded by five electron pairs, which requires a trigonal bipyramid geometry. Since there are three bonded atoms and two lone pairs of electrons about Cl, we describe the molecular structure of ClF3 as T-shaped with predicted bond angles of about 90°. The actual bond angles will be slightly less than 90° due to the stronger repulsive effect of the lone-pair electrons as compared to the bonding electrons.
F Cl
F
F
130.
From the Lewis structures (see Exercise 102), XeF4 would have a square planar molecular structure, and ClF5 would have a square pyramid molecular structure.
131.
a. V-shaped or bent
b. see-saw
d. trigonal bipyramid
e. tetrahedral
132.
c. trigonal pyramid
a. For bent or V-shaped, there are two possible molecular structures. One is a structure based off the tetrahedral geometry, for which H 2O is an example. The other is a structure based off the trigonal planar geometry for which SO2 is an example. Both of these compounds have a bent or V-shaped molecular structures; the H2O bond angle will be ~109.5o and the SO2 bond angle will be ~120o.
S
O H
O
O
H
+ 1 other
CHAPTER 8
BONDING: GENERAL CONCEPTS
353
b. TeF4 is an example of a compound with see-saw molecular structure. It is based on the trigonal bipyramid geometry and TeF4 exhibits ~90o and ~120o bond angles. F F
120o
Te F F 90o
c. NF3 is an example of a compound with trigonal pyramid molecular structure. It is based on the tetrahedral geometry and NF3 exhibits ~109o bond angles. N F
F F
d. PCl5 is an example of a compound with trigonal bipyramid molecular structure. Because there are no lone pairs of electrons about the central atom, the geometry and molecular structure are the same for PCl5. This structure exhibits 90o and 120o bond angles. 90o Cl o
120
Cl P
Cl
Cl Cl
e. CCl4 is an example of a compound with tetrahedral molecular structure. Because there are no lone pairs of electrons on the central atom, the geometry and molecular structure are the same for CCl4. This structure exhibits 109o bond angles.
133.
a. SeO3, 6 + 3(6) = 24 eO 120
o
120
Se O
O 120
O
O
Se
Se
o
O
O
O
O
o
SeO3 has a trigonal planar molecular structure with all bond angles equal to 120°. Note that any one of the resonance structures could be used to predict molecular structure and bond angles.
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BONDING: GENERAL CONCEPTS
b. SeO2, 6 + 2(6) = 18 eSe
Se
O
O
120
O
O
o
SeO2 has a V-shaped molecular structure. We would expect the bond angle to be approximately 120° as expected for trigonal planar geometry. Note: Both SeO3 and SeO2 structures have three effective pairs of electrons about the central atom. All of the structures are based on a trigonal planar geometry, but only SeO 3 is described as having a trigonal planar structure. Molecular structure always describes the relative positions of the atoms. 134.
a. PCl3 has 5 + 3(7) = 26 valence electrons.
b. SCl2 has 6 + 2(7) = 20 valence electrons.
P Cl
S
Cl
Cl
Cl
Trigonal pyramid; all angles are <109.5°.
Cl
V-shaped; angle is <109.5°.
c. SiF4 has 4 + 4(7) = 32 valence electrons.
F
Tetrahedral; all angles are 109.5°.
Si F
F F
Note: In PCl3, SCl2, and SiF4, there are four pairs of electrons about the central atom in each case in this exercise. All of the structures are based on a tetrahedral geometry, but only SiF 4 has a tetrahedral structure. We consider only the relative positions of the atoms when describing the molecular structure. 135.
a. XeCl2 has 8 + 2(7) = 22 valence electrons. Cl
Xe
Cl
180o
There are five pairs of electrons about the central Xe atom. The structure will be based on a trigonal bipyramid geometry. The most stable arrangement of the atoms in XeCl2 is a linear molecular structure with a 180° bond angle.
CHAPTER 8
BONDING: GENERAL CONCEPTS
355
b. ICl3 has 7 + 3(7) = 28 valence electrons. Cl
90o
I
Cl
Cl
T-shaped; the ClICl angles are ≈90°. Since the lone pairs will take up more space, the ClICl bond angles will probably be slightly less than 90°.
90o
c. TeF4 has 6 + 4(7) = 34 valence electrons.
d. PCl5 has 5 + 5(7) = 40 valence electrons. 90o
F
Cl
F
120o
Te
Cl
o
120
F
P
Cl
Cl
F
Cl
90o
See-saw or teeter-totter or distorted tetrahedron
Trigonal bipyramid
All the species in this exercise have five pairs of electrons around the central atom. All the structures are based on a trigonal bipyramid geometry, but only in PCl5 are all the pairs, bonding pairs. Thus PCl5 is the only one for which we describe the molecular structure as trigonal bipyramid. Still, we had to begin with the trigonal bipyramid geometry to get to the structures (and bond angles) of the others. 136.
a. ICl5 , 7 + 5(7) = 42 e-
b. XeCl4 , 8 + 4(7) = 36 e-
90o
90o Cl Cl
Cl
Cl 90o
I Cl
Cl
Cl Xe
Cl
90o
Cl
90o
Square pyramid, ≈90° bond angles
Square planar, 90° bond angles
c. SeCl6 has 6 + 6(7) = 48 valence electrons. Cl Cl
Cl
Octahedral, 90° bond angles
Se Cl
Cl Cl
Note: All these species have six pairs of electrons around the central atom. All three structures are based on the octahedron, but only SeCl6 has an octahedral molecular structure.
356 137.
CHAPTER 8
BONDING: GENERAL CONCEPTS
To have the trigonal pyramid structure, the Lewis structure must be:
This structure requires 26 valence electrons and has a formula of EBr3 formula. 26 = x + 3(7), x = 5 valence electrons E has 5 valence electrons, so it is from Group 5A. Since the compound is covalent, some possible nonmetals in Group 5A for E are N, P, and As. 138.
To have the trigonal pyramid structure, the Lewis structure for ECl 4 must be:
This structure requires 34 valence electrons and has a formula of ECl 4 formula: 34 = x + 4(7), x = 6 valence electrons E has 6 valence electrons, so it is from Group 6A. E can’t be O since the central atom has 10 electrons around it. Some possible nonmetals in Group 6A for E are S, Se, and Te 139.
SeO3 and SeO2 both have polar bonds, but only SeO2 has a dipole moment. The three bond dipoles from the three polar Se‒O bonds in SeO3 will all cancel when summed together. Hence SeO3 is nonpolar since the overall molecule has no resulting dipole moment. In SeO 2, the two Se‒O bond dipoles do not cancel when summed together; hence SeO2 has a net dipole moment (is polar). Since O is more electronegative than Se, the negative end of the dipole moment is between the two O atoms, and the positive end is around the Se atom. The arrow in the following illustration represents the overall dipole moment in SeO2. Note that to predict polarity for SeO2, either of the two resonance structures can be used.
Se O
140.
O
All have polar bonds; in SiF4, the individual bond dipoles cancel when summed together, and in PCl3 and SCl2, the individual bond dipoles do not cancel. Therefore, SiF 4 has no net dipole moment (is nonpolar), and PCl3 and SCl2 have net dipole moments (are polar). For PCl3, the negative end of the dipole moment is between the more electronegative chlorine atoms, and the positive end is around P. For SCl2, the negative end is between the more electronegative Cl atoms, and the positive end of the dipole moment is around S.
CHAPTER 8 141.
BONDING: GENERAL CONCEPTS
357
All have polar bonds, but only TeF 4 and ICl3 have dipole moments. The bond dipoles from the five P‒Cl bonds in PCl5 cancel each other when summed together, so PCl5 has no net dipole moment. The bond dipoles in XeCl2 also cancel:
Because the bond dipoles from the two Xe‒Cl bonds are equal in magnitude but point in opposite directions, they cancel each other, and XeCl2 has no net dipole moment (is nonpolar). For TeF4 and ICl3, the arrangement of these molecules is such that the individual bond dipoles do not all cancel, so each has an overall net dipole moment (is polar). 142.
All have polar bonds, but only ICl5 has an overall net dipole moment. The six bond dipoles in SeCl6 all cancel each other, so SeCl6 has no net dipole moment. The same is true for XeCl4: Cl
Cl Xe
Cl
Cl
When the four bond dipoles are added together, they all cancel each other, resulting in XeCl4 having no overall dipole moment (is nonpolar). ICl5 has a structure in which the individual bond dipoles do not all cancel, hence ICl5 has a dipole moment (is polar) 143.
Molecules that have an overall dipole moment are called polar molecules, and molecules that do not have an overall dipole moment are called nonpolar molecules. a. OCl2, 6 + 2(7) = 20 e−
KrF2, 8 + 2(7) = 22 e−
V-shaped, polar; OCl2 is polar because the two O−Cl bond dipoles don’t cancel each other. The resulting dipole moment is shown in the drawing.
Linear, nonpolar; the molecule is nonpolar because the two Kr‒F bond dipoles cancel each other.
BeH2, 2 + 2(1) = 4 e−
SO2, 6 + 2(6) = 18 e−
Linear, nonpolar; Be−H bond dipoles are equal and point in opposite directions. They cancel each other. BeH2 is nonpolar.
V-shaped, polar; the S−O bond dipoles do not cancel, so SO2 is polar (has a net dipole moment). Only one resonance structure is shown.
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CHAPTER 8
BONDING: GENERAL CONCEPTS
Note: All four species contain three atoms. They have different structures because the number of lone pairs of electrons around the central atom are different in each case. b. SO3, 6 + 3(6) = 24 e−
NF3, 5 + 3(7) = 26 e-
O
N F
S O
F F
O
Trigonal planar, nonpolar; bond dipoles cancel. Only one resonance structure is shown.
Trigonal pyramid, polar; bond dipoles do not cancel.
IF3 has 7 + 3(7) = 28 valence electrons. F
F
I
T-shaped, polar; bond dipoles do not cancel.
F
Note: Each molecule has the same number of atoms but different structures because of differing numbers of lone pairs around each central atom. c. CF4, 4 + 4(7) = 32 e−
F
F
F Se
C F
SeF4, 6 + 4(7) = 34 e−
F
F
F
F
Tetrahedral, nonpolar; bond dipoles cancel.
See-saw, polar; bond dipoles do not cancel.
KrF4, 8 + 4(7) = 36 valence electrons F
F Kr
F
F
Square planar, nonpolar; bond dipoles cancel.
Note: Again, each molecule has the same number of atoms but different structures because of differing numbers of lone pairs around the central atom.
CHAPTER 8
BONDING: GENERAL CONCEPTS
d. IF5, 7 + 5(7) = 42 e−
359
AsF5, 5 + 5(7) = 40 e− F
F F
F
F
F
F
As
I F
F
Square pyramid, polar; bond dipoles do not cancel.
144.
F
Trigonal bipyramid, nonpolar; bond dipoles cancel.
Note: Yet again, the molecules have the same number of atoms but different structures because of the presence of differing numbers of lone pairs. a. b.
O
C
N
S
C
O
H Polar; the bond dipoles do not cancel.
c.
Polar; the C‒O bond is a more polar bond than the C‒S bond, so the two bond dipoles do not cancel each other. d. F
F
Xe
F
C F
Cl Cl
Nonpolar; the two Xe‒F bond dipoles cancel each other.
e.
Polar; all the bond dipoles are not equivalent, and they don’t cancel each other. f.
F F
F Se
F
F
H C
O
H
F
Nonpolar; the six Se−F bond dipoles cancel each other. 145.
Polar; the bond dipoles are not equivalent, and they don’t cancel
EO3− is the formula of the ion. The Lewis structure has 26 valence electrons. Let x = number of valence electrons of element E. 26 = x + 3(6) + 1, x = 7 valence electrons
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BONDING: GENERAL CONCEPTS
Element E is a halogen because halogens have seven valence electrons. Some possible identities are F, Cl, Br, and I. The EO3− ion has a trigonal pyramid molecular structure with bond angles of less than 109.5° (<109.5°). 146.
The formula is EF2O2−, and the Lewis structure has 28 valence electrons. 28 = x + 2(7) + 6 + 2, x = 6 valence electrons for element E Element E must belong to the Group 6A elements since E has six valence electrons. E must also be a row 3 or heavier element since this ion has more than eight electrons around the central E atom (row 2 elements never have more than eight electrons around them). Some possible identities for E are S, Se, and Te. The ion has a T-shaped molecular structure with bond angles of ~90°.
147.
N2F2: 2(5) + 2(7) = 24 e−; in the Lewis structure, the two central nitrogen atoms exhibit trigonal planar geometry with ~120o bond angles. The two possible arrangements for the atoms in the N2F2 Lewis structures are: N
F
N
F
F
N
N
F
Polar
Nonpolar
In the first structure, the N−F bond dipoles are both pointing at angles downward somewhat, so they will add together to make this structure of N 2F2 polar. In the second structure, the N2F2 bond dipoles point in opposite directions from one another; they will cancel each other out making this structure nonpolar. 148.
XeF2Cl2, 8 + 2(7) + 2(7) = 36 e−
Polar
Nonpolar
The two possible structures for XeF 2Cl2 are above. In the first structure, the F atoms are 90° apart from each other, and the Cl atoms are also 90° apart. The individual bond dipoles would not cancel in this molecule, so this molecule is polar. In the second possible structure, the F atoms are 180° apart, as are the Cl atoms. Here, the bond dipoles are symmetrically arranged so they do cancel each other out, and this molecule is nonpolar. Therefore, measurement of the dipole moment would differentiate between the two compounds. These are different compounds and not resonance structures.
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BONDING: GENERAL CONCEPTS
361
ChemWork Problems 149.
Bonding between a metal and a nonmetal is generally ionic. Bonding between two nonmetals is covalent, and in general, the bonding between two different nonmetals is usually polar covalent. When two different nonmetals have very similar electronegativities, the bonding is pure covalent or nonpolar covalent. a. nonpolar covalent
b. ionic
c. ionic
d. polar covalent
e. polar covalent
f.
g. polar covalent
h.
polar covalent
ionic
150.
Metals generally have low electronegativity values, while nonmetals generally have high electronegativity values. The electronegativity difference is large when a metal forms a bond with a nonmetal, so large that the nonmetal removes the electron away from the metal and an ionic bond forms. When two nonmetals form a bond, the electronegativity difference is not very large. Neither nonmetal can remove the bonding electron from the other, so a sharing of electrons occurs resulting in a covalent bond.
151.
The elements are identified by their electron configurations: [Ar]4s13d5 = Cr; [Ne]3s23p3 = P; [Ar]4s23d104p3 = As; [Ne]3s23p5 = Cl Following the electronegativity trend, the order is Cr < As < P < Cl.
152.
2 F(g) → F2(g); when a bond forms, energy is released. This is an exothermic process, so q is negative. In the equation, 2 moles of gas are converted to 1 mol of gas. This corresponds to a compression where the surroundings does work on the system. So, w is positive.
153.
a. Radius: N+ < N < N− ; IE: N− < N < N+ N+ has the fewest electrons held by the seven protons in the nucleus, whereas N− has the most electrons held by the seven protons. The seven protons in the nucleus will hold the electrons most tightly in N+ and least tightly in N− . Therefore, N+ has the smallest radius with the largest ionization energy (IE), and N− is the largest species with the smallest IE. b. Radius: Cl+ < Cl < Se < Se− ; IE: Se− < Se < Cl < Cl+ The general trends tell us that Cl has a smaller radius than Se and a larger IE than Se. Cl+, with fewer electron-electron repulsions than Cl, will be smaller than Cl and have a larger IE. Se−, with more electron-electron repulsions than Se, will be larger than Se and have a smaller IE. c. Radius: Ca2+ < K+ < Cl− ; IE: Cl− < K+ < Ca2+ These ions are isoelectronic. The species with the most protons (Ca2+) will hold the electrons most tightly and will have the smallest radius and largest IE. The ion with the fewest protons (Cl−) will hold the electrons least tightly and will have the largest radius and smallest IE.
362 154.
155.
CHAPTER 8 a. Na+(g) + Cl−(g) → NaCl(s)
b. NH4+(g) + Br-(g) → NH4Br(s)
c. Mg2+(g) + S2−(g) → MgS(s)
d. O2(g) → 2 O(g)
a.
b.
HF(g) → H(g) + F(g) H(g) → H+(g) + e− F(g) + e− → F−(g)
c.
ΔH = 1549 kJ
HCl(g) → H(g) + Cl(g) H(g) → H+(g) + e− Cl(g) + e−→ Cl−(g)
ΔH = 427 kJ ΔH = 1312 kJ ΔH = −348.7 kJ
HI(g) → H(g) + I(g) H(g) → H+(g) + e− I(g) + e− → I−(g) HI(g) → H+(g) + I−(g)
d.
H2O(g) → OH(g) + H(g) H(g) → H+(g) + e− OH(g) + e− → OH−(g) H2O(g) → H+(g) + OH−(g)
157.
ΔH = 565 kJ ΔH = 1312 kJ ΔH = −327.8 kJ
HF(g) → H+(g) + F−(g)
HCl(g) → H+(g) + Cl−(g)
156.
BONDING: GENERAL CONCEPTS
ΔH = 1390. kJ ΔH = 295 kJ ΔH = 1312 kJ ΔH = −295.2 kJ ΔH = 1312 kJ ΔH = 467 kJ ΔH = 1312 kJ ΔH = −180. kJ ΔH = 1599 kJ
Ba(s) → Ba(g) Ba(g) → Ba+(g) + e− Ba+(g) → Ba2+(g) + e− Br2(g) → 2 Br(g) 2 Br(g) + 2 e− → 2 Br−(g) Ba2+(g) + 2 Br−(g) → BaBr2(s)
ΔH = 178 kJ ΔH = 503 kJ ΔH = 965 kJ ΔH = 193 kJ ΔH = 2(−325) kJ ΔH = −1985 kJ
Ba(s) + Br2(g) → BaBr2(s)
ΔH of = −796 kJ/mol
(sublimation) (IE1) (IE2) (BE) (EA) (LE)
The stable species are: a. NaBr: In NaBr2, the sodium ion would have a 2+ charge, assuming that each bromine has a 1− charge. Sodium doesn’t form stable Na2+ ionic compounds. b. ClO4−: ClO4 has 31 valence electrons, so it is impossible to satisfy the octet rule for all atoms in ClO4. The extra electron from the 1− charge in ClO4- allows for complete octets for all atoms. c. XeO4: We can’t draw a Lewis structure that obeys the octet rule for SO 4 (30 electrons), unlike XeO4 (32 electrons). d. SeF4: Both compounds require the central atom to expand its octet. O is too small and doesn’t have low-energy d orbitals to expand its octet (which is true for all row 2 elements).
CHAPTER 8 158.
BONDING: GENERAL CONCEPTS
363
a. All have 24 valence electrons and the same number of atoms in the formula. All have the same resonance Lewis structures; the structures are all trigonal planar with 120 bond angles. The Lewis structures for NO3− and CO32− will be the same as the three SO3 Lewis structures shown below.
b. All have 18 valence electrons and the same number of atoms. All have the same resonance Lewis structures; the molecular structures are all V-shaped with 120 bond angles. O3 and SO2 have the same two Lewis structures as is shown for NO2−.
159.
a. XeCl4, 8 + 4(7) = 36 e−
Square planar, 90, nonpolar
XeCl2, 8 + 2(7) = 22 e−
Linear, 180, nonpolar
Both compounds have a central Xe atom with lone pairs and terminal Cl atoms, and both compounds do not satisfy the octet rule. In addition, both are nonpolar because the Xe−Cl bond dipoles and lone pairs around Xe are arranged in such a manner that they cancel each other out. The last item in common is that both have 180 bond angles. Although we haven’t emphasized this, the bond angles between the Cl atoms on the diagonal in XeCl4 are 180 apart from each other. b. All of these are polar covalent compounds. The bond dipoles do not cancel out each other when summed together. The reason the bond dipoles are not symmetrically arranged in these compounds is that they all have at least one lone pair of electrons on the central atom, which disrupts the symmetry. Note that there are molecules that have lone pairs and are nonpolar, e.g., XeCl4 and XeCl2 in the preceding problem. A lone pair on a central atom does not guarantee a polar molecule. 160.
Resonance structures can be drawn for O 3 (18 valence e–), CNO–(16 valence e–), and CO32– (24 valence e–). The resonance structures are:
364
CHAPTER 8 O C N
O C N
2-
O
O C N
2-
O
C O
BONDING: GENERAL CONCEPTS
C O
O
2-
O C
O
O
O
Resonance structures cannot be drawn for AsI3 or AsF3. The Lewis structure for AsF3 (26 valence e–) is below. AsI3 has a similar Lewis structure to AsF3.
161.
a. SeCl4 (34 e−)
b. SO2 (18 e−)
see-saw, 90˚ and 120˚, polar c. KrF4 (36 e−)
V-shaped, 120˚, polar d. CBr4 (32 e−)
square planar, 90˚, nonpolar e. IF3 (28 e−)
T-shaped, 90˚, polar
tetrahedral, 109.5˚, nonpolar f.
ClF5 (42 e−)
square pyramid, 90˚, polar
CHAPTER 8 162.
BONDING: GENERAL CONCEPTS
365
The general structure of the trihalide ion: X
X
X
Bromine and iodine are large enough and have low-energy, empty d orbitals to accommodate the expanded octet. Fluorine is small, and its valence shell contains only 2s and 2p orbitals (four orbitals) and cannot expand its octet. The lowest-energy d orbitals in F are 3d; they are too high in energy compared with 2s and 2p to be used in bonding. 163.
NO3‒ has three resonance structures with the double bond rotated to each of the three oxygen atoms. Whenever resonance structures can be drawn for a substance, the actual bonding between the atoms is an average of all the resonance structures. In NO3‒, the average bond is 1 1/3 bonds (a single in two resonance structures and a double in the third structure). The actual nitrogen-oxygen bond will all be equivalent with a bond length shorter than a single bond but longer than a double bond. With an average bond of 1 1/3, the bond length will be closer to the N-O single bond length with an estimate of 1.67 units long.
164.
The oxygen with the hydrogen bonded to it has four pairs of electrons around it, so it exhibits tetrahedral geometry. The expected bond angle will not be 90°, but something closer to 109.5°. The other bond angles as predicted by VSEPR about the C and N atoms are shown below.
The carbon and nitrogen atoms with only single bonds are predicted to have tetrahedral bond angels of 109.5°, and the carbon and nitrogen atoms having a double bond are expected to be 120°. The bond angles about the nitrogen and oxygen atoms are probably something a little smaller than predicted due to the lone pair taking up more space than bonding electrons. 165.
SiF4, 4 + 4(7) = 32 e−
SeF4, 6 + 4(7) = 34 e−
XeF4, 8 + 4(7) = 36 e−
SiF4 is tetrahedral and SeF4 is polar (the bond dipoles cancel each other in SiF4 and XeF4).
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CHAPTER 8
BONDING: GENERAL CONCEPTS
166.
To get from reactants to products, a C‒O bond and a H‒O bond must be broken. When the fragments are put back together, a C‒O bond and a H‒O bond must form to complete the products. The same bonds are broken that are reformed. So bond energies gives ΔH = 0 for this reaction. You would get the same answer if you broke all reactant bonds then reformed all product bonds. 167.
168.
The bonds are H‒H, F‒F, O=O, N≡N, and Cl‒Cl. A triple bond is the strongest bond, so N 2 should have the largest bond energy value. ½ H2 + ½ Cl2 → 1 HCl H → ½ H2 Cl → ½ Cl2
H = ½(−184 kJ) H = ‒½(432 kJ) H = ‒½(240 kJ)
H + Cl → H‒Cl
ΔH = ‒428 kJ/mol = HCl bond energy
169.
Bonds broken:
Bonds formed:
1 C=C (614 kJ/mol) 1 O‒O (146 kJ/mol)
1 C‒C (347 kJ/mol) 2 C‒O (358 kJ/mol)
H = 614 kJ + 146 kJ – [347 kJ + 2(358 kJ)] = −303 kJ Note: Sometimes some of the bonds remain the same between reactants and products. To save time, only break and form bonds that are involved in the reaction.
170. H
H
H
C
C
H
H
O
H +3 O
O
2O
C
O +3 H
O
H
CHAPTER 8
BONDING: GENERAL CONCEPTS
Bonds broken:
367
Bonds formed: 2 × 2 C=O (799 kJ/mol) 3 × 2 O‒H (467 kJ/mol)
5 C‒H (413 kJ/mol) 1 C‒C (347 kJ/mol) 1 C‒O (358 kJ/mol) 1 O‒H (467 kJ/mol) 3 O=O (495 kJ/mol)
ΔH = 5(413 kJ) + 347 kJ + 358 kJ + 467 kJ + 3(495 kJ) − [4(799 kJ) + 6(467 kJ)] = −1276 kJ 1 mol C 2 H 5 OH − 1276 kJ = −27.70 kJ/g versus −47.8 kJ/g for gasoline mol C 2 H 5 OH 46.07 g C 2 H 5 OH
From the calculated values, gasoline produces a significantly larger amount of energy per gram as compared to ethanol. 171. H 4
N H
+ 5
N
C H
O
H
N O
H
O 12 H
N
O
H + 9 N
N + 4 O
C
O
O
H
Bonds broken:
Bonds formed:
9 N‒N (160. kJ/mol) 4 N‒C (305 kJ/mol) 12 C‒H (413 kJ/mol) 12 N‒H (391 kJ/mol) 10 N=O (607 kJ/mol) 10 N‒O (201 kJ/mol)
24 O‒H (467 kJ/mol) 9 N≡N (941 kJ/mol) 8 C=O (799 kJ/mol)
ΔH = [9(160.) + 4(305) + 12(413) + 12(391) + 10(607) + 10(201)] − [24(467) + 9(941) + 8(799)] ΔH = 20,388 kJ − 26,069 kJ = −5681 kJ 172.
The nitrogen-nitrogen bond length of 112 pm is between a double (120 pm) and a triple (110 pm) bond. The nitrogen-oxygen bond length of 119 pm is between a single (147 pm) and a double bond (115 pm). The third resonance structure shown below doesn’t appear to be as important as the other two since there is no evidence from bond lengths for a nitrogen-oxygen triple bond or a nitrogen-nitrogen single bond as in the third resonance form. We can adequately describe the structure of N2O using the resonance forms: N
N
O
N
N
O
Assigning formal charges for all three resonance forms:
368
CHAPTER 8
BONDING: GENERAL CONCEPTS
N
N
O
N
N
O
N
N
O
-1
+1
0
0
+1
-1
-2
+1
+1
N
, FC = 5 - 4 - 1/2(4) = -1
For:
N
, FC = 5 - 1/2(8) = +1 , Same for
, FC = 5 - 6 - 1/2(2) = -2 ;
N O ,
FC = 6 - 4 - 1/2(4) = 0 ;
O ,
FC = 6 - 2 - 1/2(6) = +1
N
O
N
and
,
FC = 5 - 2 - 1/2(6) = 0
,
FC = 6 - 6 - 1/2(2) = -1
N
We should eliminate N‒N≡O because it has a formal charge of +1 on the most electronegative element (O). This is consistent with the observation that the N‒N bond is between a double and triple bond and that the N‒O bond is between a single and double bond. 173.
Yes, each structure has the same number of effective pairs around the central atom, giving the same predicted molecular structure for each compound/ion. (A multiple bond is counted as a single group of electrons.)
174.
a. The C‒H bonds are assumed nonpolar since the electronegativities of C and H are about equal. δ+ δ− H C‒Cl is the charge distribution for each C‒Cl bond. In CH2Cl2, the two individual C‒Cl bond dipoles C add together to give an overall dipole moment for H Cl the molecule. The overall dipole will point from C Cl (positive end) to the midpoint of the two Cl atoms (negative end). In CHCl3, the C‒H bond is essentially nonpolar. The three C‒Cl bond dipoles in CHCl3 add together to give an overall dipole moment for the molecule. The overall dipole will have the negative end at the midpoint of the three chlorines and the positive end around the carbon. H C Cl
Cl Cl
CHAPTER 8
BONDING: GENERAL CONCEPTS
369
CCl4 is nonpolar. CCl4 is a tetrahedral molecule where all four C‒Cl bond dipoles cancel when added together. Let’s consider just the C and two of the Cl atoms. There will be a net dipole pointing in the direction of the middle of the two Cl atoms. C Cl
Cl
There will be an equal and opposite dipole arising from the other two Cl atoms. Combining: Cl
Cl C
Cl
Cl
The two dipoles cancel, and CCl4 is nonpolar. b. CO2 is nonpolar. CO2 is a linear molecule with two equivalence bond dipoles that cancel. N2O, which is also a linear molecule, is polar because the nonequivalent bond dipoles do not cancel.
N
+ N
− O
c. NH3 is polar. The 3 N‒H bond dipoles add together to give a net dipole in the direction of the lone pair. We would predict PH3 to be nonpolar on the basis of electronegativitity, i.e., P‒H bonds are nonpolar. However, the presence of the lone pair makes the PH 3 molecule slightly polar. The net dipole is in the direction of the lone pair and has a magnitude about one third that of the NH3 dipole. N
δ− δ+ N‒H 175.
H
H
P H
H
H
H
TeF5− has 6 + 5(7) + 1 = 42 valence electrons.
-
F F
F Te
F
F
The lone pair of electrons around Te exerts a stronger repulsion than the bonding pairs of electrons. This pushes the four square-planar F atoms away from the lone pair and reduces the bond angles between the axial F atom and the square-planar F atoms.
370
CHAPTER 8
BONDING: GENERAL CONCEPTS
176.
C≡O (1072 kJ/mol); N≡N (941 kJ/mol); CO is polar, whereas N2 is nonpolar. This may lead to a great reactivity for the CO bond.
177.
Assuming 100.00 g of compound: 42.81 g F =
1 mol X = 2.253 mol F 19.00 g F
The number of moles of X in XF5 is: 2.53 mol F ×
1 mol X = 0.4506 mol X 5 mol F
This number of moles of X has a mass of 57.19 g (= 100.00 g – 42.81 g). The molar mass of X is: 57 .19 g X = 126.9 g/mol; this is element I and the compound is IF 5. 0.4506 mol X IF5, 7 + 5(7) = 42 e−
The molecular structure is square pyramid.
178.
If X2− has a configuration of [Ar]4s23d104p6, then X must have a configuration with two fewer electrons, [Ar]4s23d104p4. This is element Se. SeCN−, 6 + 4 + 5 + 1 = 16 e−
Challenge Problems 179.
a. There are two attractions of the form
( +1)(−1) , where r = 1 × 10−10 m = 0.1 nm. r
( +1)(−1) −18 −18 V = 2 × (2.31 × 10−19 J nm) = −4.62 × 10 J = −5 × 10 J 0 . 1 nm
b. There are four attractions of +1 and −1 charges at a distance of 0.1 nm from each other. The two negative charges and the two positive charges repel each other across the diagonal of the square. This is at a distance of 2 × 0.1 nm. (+1)(+1) ( +1)(−1) −19 V = 4 × (2.31 × 10−19) + 2.31 × 10 2 (0.1) 0 .1 (−1)(−1) + 2.31 × 10−19 2 (0.1)
V = −9.24 × 10−18 J + 1.63 × 10−18 J + 1.63 × 10−18 J = −5.98 × 10−18 J = −6 × 10−18 J Note: There is a greater net attraction in arrangement b than in a.
CHAPTER 8
BONDING: GENERAL CONCEPTS
180.
(IE − EA)
(IE − EA)/502
2006 kJ/mol 1604 kJ/mol 1463 kJ/mol 1302 kJ/mol
4.0 3.2 2.9 2.6
F Cl Br I
EN (text)
371 2006/502 = 4.0
4.0 3.0 2.8 2.5
The values calculated from ionization energies and electron affinities show the same trend as (and agree closely) with the values given in the text. 181.
The reaction is: 1/2 I2(g) + 1/2 Cl2(g) → ICl(g)
ΔH of = ?
Using Hess’s law: 1/2 I2(s) → 1/2 I2(g) 1/2 I2(g) → I (g) 1/2 Cl2(g) → Cl(g) I(g) + Cl(g) → ICl(g) 1/2 I2(s) + 1/2 Cl2(g) → ICl(g) 182.
H = 1/2(62 kJ) H = 1/2(149 kJ) H = 1/2(239 kJ) H = −208 kJ
(Appendix 4) (Table 8.5) (Table 8.5) (Table 8.5)
H = 17 kJ so ΔH of = 17 kJ/mol
2 Li+(g) + 2 Cl−(g) → 2 LiCl(s) 2 Li(g) → 2 Li+(g) + 2 e− 2 Li(s) → 2 Li(g) 2 HCl(g) → 2 H(g) + 2 Cl(g) 2 Cl(g) + 2 e− → 2 Cl−(g) 2 H(g) → H2(g)
H = 2(−829 kJ) H = 2(520. kJ) H = 2(166 kJ) H = 2(427 kJ) H = 2(−349 kJ) H = −(432 kJ)
2 Li(s) + 2 HCl(g) → 2 LiCl(s) + H2(g)
H = −562 kJ
183.
See Figure 8.11 to see the data supporting MgO as an ionic compound. Note that the lattice energy is large enough to overcome all of the other processes (removing two electrons from Mg, etc.). The bond energy for O2 (247 kJ/mol) and electron affinity (737 kJ/mol) are the same when making CO. However, ionizing carbon to form a C 2+ ion must be too large. See Figure 7.30 to see that the first ionization energy for carbon is about 350 kJ/mol greater than the first ionization energy for magnesium. If all other numbers were equal, the overall energy change would be down to ~250 kJ/mol (see Figure 8.11). It is not unreasonable to assume that the second ionization energy for carbon is more than 250 kJ/mol greater than the second ionization energy of magnesium. This would result in a positive H value for the formation of CO as an ionic compound. One wouldn’t expect CO to be ionic if the energetics were unfavorable.
184.
a. (1) Removing an electron from the metal: ionization energy, positive (H > 0) (2) Adding an electron to the nonmetal: electron affinity, often negative (H < 0) (3) Allowing the metal cation and nonmetal anion to come together: lattice energy, negative (H < 0)
372
CHAPTER 8
BONDING: GENERAL CONCEPTS
b. Often the sign of the sum of the first two processes is positive (or unfavorable). This is especially true due to the fact that we must also vaporize the metal and often break a bond on a diatomic gas. For example, the ionization energy for Na is +495 kJ/mol, and the electron affinity for F is −328 kJ/mol. Overall, the energy change is +167 kJ/mol (unfavorable). c. For an ionic compound to form, the sum must be negative (exothermic). d. The lattice energy must be favorable enough to overcome the endothermic process of forming the ions; i.e., the lattice energy must be a large negative quantity. e. While Na2Cl (or NaCl2) would have a greater lattice energy than NaCl, the energy to make a Cl2− ion (or Na2+ ion) must be larger (more unfavorable) than what would be gained by the larger lattice energy. The same argument can be made for MgO compared to MgO 2 or Mg2O. The energy to make the ions is too unfavorable or the lattice energy is not favorable enough, and the compounds do not form. 185.
As the halogen atoms get larger, it becomes more difficult to fit three halogen atoms around the small nitrogen atom, and the NX3 molecule becomes less stable.
186. a. I. H
H * O
H C
H
+
C * H O
Bonds broken (*):
H * C
C
N
C
C
H
H
H
+
H * O
* N
H
Bonds formed (*):
1 C‒O (358 kJ) 1 C‒H (413 kJ)
1 O‒H (467 kJ) 1 C‒C (347 kJ)
ΔHI = 358 kJ + 413 kJ − (467 kJ + 347 kJ) = −43 kJ II. OH H H
*
H
C * C H
H
C
N
*
Bonds broken (*): 1 C‒O (358 kJ/mol) 1 C‒H (413 kJ/mol) 1 C‒C (347 kJ/mol)
H C
H
C *
C
N
Bonds formed (*): 1 H‒O (467 kJ/mol) 1 C=C (614 kJ/mol)
ΔHII = 358 kJ + 413 kJ + 347 kJ − [467 kJ + 614 kJ] = +37 kJ ΔHoverall = ΔHI + ΔHII = −43 kJ + 37 kJ = −6 kJ
H
CHAPTER 8 b.
BONDING: GENERAL CONCEPTS H
H
H C
4
C
C
H
373
H 4
+ 6 NO
H
C C
H
+ 6 H
C
H
Bonds broken:
N O
H + N
N
H
Bonds formed:
4 × 3 C‒H (413 kJ/mol) 6 N=O (630. kJ/mol)
4 C≡N (891 kJ/mol) 6 × 2 H‒O (467 kJ/mol) 1 N≡N (941 kJ/mol)
ΔH = 12(413) + 6(630.) − [4(891) + 12(467) + 941] = −1373 kJ c. H H 2
H
C C
H
H
C
H
+ 2
H
N
H H + 3 O2
H
C C
2 H
Bonds broken: 2 × 3 C‒H (413 kJ/mol) 2 × 3 N‒H (391 kJ/mol) 3 O=O (495 kJ/mol)
N + 6
C
H
O
H
Bonds formed: 2 C≡N (891 kJ/mol) 6 × 2 O‒H (467 kJ/mol)
ΔH = 6(413) + 6(391) + 3(495) − [2(891) + 12(467)] = −1077 kJ d. Because both reactions are highly exothermic, the high temperature is not needed to provide energy. It must be necessary for some other reason. The reason is to increase the speed of the reaction. This is discussed in Chapter 12 on kinetics. 187.
a. i.
C6H6N12O12 → 6 CO + 6 N2 + 3 H2O + 3/2 O2 The NO2 groups are assumed to have one N‒O single bond and one N=O double bond, and each carbon atom has one C‒H single bond. We must break and form all bonds. Bonds broken: 3 C‒C (347 kJ/mol) 6 C‒H (413 kJ/mol) 12 C‒N (305 kJ/mol) 6 N‒N (160. kJ/mol) 6 N‒O (201 kJ/mol) 6 N=O (607 kJ/mol) ΣDbroken = 12,987 kJ
Bonds formed: 6 C≡O (1072 kJ/mol) 6 N≡N (941 kJ/mol) 6 H‒O (467 kJ/mol) 3/2 O=O (495 kJ/mol) ΣDformed = 15,623 kJ
ΔH = ΣDbroken − ΣDformed = 12,987 kJ − 15,623 kJ = −2636 kJ
H
374
CHAPTER 8
BONDING: GENERAL CONCEPTS
ii. C6H6N12O12 → 3 CO + 3 CO2 + 6 N2 + 3 H2O Note: The bonds broken will be the same for all three reactions. Bonds formed: 3 C≡O (1072 kJ/mol) 6 C=O (799 kJ/mol) 6 N≡N (941 kJ/mol) 6 H‒O (467 kJ/mol) ΣDformed = 16,458 kJ ΔH = 12,987 kJ −16,458 kJ = −3471 kJ iii. C6H6N12O12 → 6 CO2 + 6 N2 + 3 H2 Bonds formed: 12 C=O (799 kJ/mol) 6 N≡N (941 kJ/mol) 3 H‒H (432 kJ/mol) ΣDformed = 16,530. kJ ΔH = 12,987 kJ − 16,530. kJ = −3543 kJ b. Reaction iii yields the most energy per mole of CL-20, so it will yield the most energy per kilogram. − 3543 kJ 1 mol 1000 g = −8085 kJ/kg mol 438 .23 g kg
188.
We can draw resonance forms for the anion after the loss of H +, we can argue that the extra stability of the anion causes the proton to be more readily lost, i.e., makes the compound a better acid. a. -
O H
C
-
O
O
H
C
O
b. O CH3
C
-
O CH
C
CH3
O CH3
C
O CH
C
CH3 O
CH3
C
O CH
C
CH3
CHAPTER 8
BONDING: GENERAL CONCEPTS
375
c. O
O
O
O
O
In all three cases, extra resonance forms can be drawn for the anion that are not possible when the H+ is present, which leads to enhanced stability. 189.
For carbon atoms to have a formal charge of zero, each C atom must satisfy the octet rule by forming four bonds (with no lone pairs). For nitrogen atoms to have a formal charge of zero, each N atom must satisfy the octet rule by forming three bonds and have one lone pair of electrons. For oxygen atoms to have a formal charge of zero, each O atom must satisfy the octet rule by forming two bonds and have two lone pairs of electrons. With these bonding requirements in mind, then the Lewis structure of histidine, where all atoms have a formal charge of zero, is: H
2 H
C
N
C
C H N H
H
C
H
N
C
C
H
H
O
1 O
H
We would expect 120° bond angles about the carbon atom labeled 1 and ≈109.5° bond angles about the nitrogen atom labeled 2. The nitrogen bond angles should be slightly smaller than 109.5° due to the lone pair of electrons on nitrogen. 190.
This molecule has 30 valence electrons. The only C–N bond that can possibly have a doublebond character is the N bound to the C with O attached. Double bonds to the other two C–N bonds would require carbon in each case to have 10 valence electrons (which carbon never does). The resonance structures are:
376
CHAPTER 8 O H
H
C
N
C
BONDING: GENERAL CONCEPTS O
H
H
C
H N
H H
C
H
H
H
H
H
191.
C
C
H
H
a. BrFI2, 7 +7 + 2(7) = 28 e−; two possible structures exist with Br as the central atom; each has a T-shaped molecular structure.
I
F Br
I
Br
F
I
I 90° bond angles between I atoms
180° bond angles between I atoms
b. XeO2F2, 8 + 2(6) + 2(7) = 34 e−; three possible structures exist with Xe as the central atom; each has a see-saw molecular structure.
O
O
F
O
F
F
90° bond angle between O atoms
F Xe
Xe
Xe F
O
O
O
180° bond angle between O atoms
F
120° bond angle between O atoms
c. TeF2Cl3−; 6 + 2(7) + 3(7) + 1 = 42 e−; three possible structures exist with Te as the central atom; each has a square pyramid molecular structure.
Cl
F
Cl
Cl
F
F
Te
Te Cl
Cl
F
One F is 180° from the lone pair.
Cl
Cl
Cl
Te F
Both F atoms are 90° from the lone pair and 90° from each other.
Cl
F
Both F atoms are 90° from the lone pair and 180° from each other.
CHAPTER 8 192.
BONDING: GENERAL CONCEPTS
377
The skeletal structure of caffeine is: H O
H
C
H
H
H C H
C
N
N
C
C
C
C O
N H
C
H
N H
H
For a formal charge of zero on all atoms, the bonding requirements are: (1) four bonds and no lone pairs for each carbon atom; (2) three bonds and one lone pair for each nitrogen atom; (3) two bonds and two lone pairs for each oxygen atom; (4) one bond and no lone pairs for each hydrogen atom. Following these guidelines gives a Lewis structure that has a formal charge of zero for all the atoms in the molecule. The Lewis structure is:
193.
The complete Lewis structure follows. All but two of the carbon atoms exhibit 109.5 bond angles. The two carbon atoms that contain the double bond exhibit 120 bond angles (see * in the following Lewis structure).
378
CHAPTER 8 CH3 H
H C
H2C H
H C
CH3
H2C
C
C
*C
H
BONDING: GENERAL CONCEPTS CH2
CH CH3
CH2 CH2
CH3 CH
H C
CH3
C
CH2
C
CH2
H C C
H
H C
HO H
H
* C
CH2
H
No; most of the carbons are not in the same plane since a majority of carbon atoms exhibit a tetrahedral structure (109.5 bond angles). Note: HO, CH, CH2, H2C, and CH3 are shorthand for oxygen and carbon atoms singly bonded to hydrogen atoms.
Marathon Problem 194.
Compound A: This compound is a strong acid (part g). HNO 3 is a strong acid and is available in concentrated solutions of 16 M (part c). The highest possible oxidation state of nitrogen is +5, and in HNO3, the oxidation state of nitrogen is +5 (part b). Therefore, compound A is most likely HNO3. The Lewis structures for HNO3 are: O
H
O
N O
H
N O
O
O
Compound B: This compound is basic (part g) and has one nitrogen (part b). The formal charge of zero (part b) tells us that there are three bonds to the nitrogen and that the nitrogen has one lone pair. Assuming compound B is monobasic, then the data in part g tell us that the molar mass of B is 33.0 g/mol (21.98 mL of 1.000 M HCl = 0.02198 mol HCl; thus there are 0.02198 mol of B; 0.726 g/0.02198 mol = 33.0 g/mol). Because this number is rather small, it limits the possibilities. That is, there is one nitrogen, and the remainder of the atoms are O and H. Since the molar mass of B is 33.0 g/mol, then only one O oxygen atom can be present. The N and O atoms have a combined molar mass of 30.0 g/mol; the rest is made up of hydrogens (3 H atoms), giving the formula NH3O. From the list of Kb values for weak bases in Appendix 5 of the text, compound B is most likely NH2OH. The Lewis structure is: H
N H
O
H
CHAPTER 8
BONDING: GENERAL CONCEPTS
379
Compound C: From parts a and f and assuming compound A is HNO 3 , then compound C contains the nitrate ion, NO3-. Because part b tells us that there are two nitrogen atoms, the other ion needs to have one N atom and some H atoms. In addition, compound C must be a weak acid (part g), which must be due to the other ion since NO 3- has no acidic properties. Also, the nitrogen atom in the other ion must have an oxidation state of −3 (part b) and a formal charge of +1. The ammonium ion fits the data. Thus compound C is most likely NH 4NO3. A Lewis structure is: +
H H
N
-
O N
H O
H
O
Note: Two more resonance structures can be drawn for NO3-. Compound D: From part f, this compound has one less oxygen atom than compound C; thus NH4NO2 is a likely formula. Data from part e confirm this. Assuming 100.0 g of compound, we have: 43.7 g N × 1 mol/14.01 g = 3.12 mol N 50.0 g O × 1 mol/16.00 g = 3.12 mol O 6.3 g H × 1 mol/1.008 g = 6.25 mol H There is a 1 : 1 : 2 mole ratio among N to O to H. The empirical formula is NOH2, which has an empirical formula mass of 32.0 g/mol. Molar mass =
2.86 g/L( 0.08206 L atm/K • mol)(273 K) dRT = = 64.1 g/mol P 1.00 atm
For a correct molar mass, the molecular formula of compound D is N 2O2H4 or NH4NO2. A Lewis structure is:
H
N
-
+
H
N
H
O
O
H
Note: One more resonance structure for NO2− can be drawn. Compound E: A basic solution (part g) that is commercially available at 15 M (part c) is ammonium hydroxide (NH4OH). This is also consistent with the information given in parts b and d. The Lewis structure for NH4OH is: +
H H
N H
H
O
H
380 195.
CHAPTER 8
BONDING: GENERAL CONCEPTS
There are four possible ionic compounds we need to consider. They are MX, composed of either M+ and X− ions or M2+ and X2− ions, M2X composed of M+ and X2− ions; or MX2 composed of M2+ and X− ions. For each possible ionic compound, let’s calculate ΔH of , the enthalpy of formation. The compound with the most negative enthalpy of formation will be the ionic compound most likely to form. For MX composed of M2+ and X2−: M(s) → M(g) M(g) → M+(g) + e− M+(g) → M2+(g) + e− 1/2 X2(g) → X(g) X(g) + e− → X−(g) X−(g) + e− → X2−(g) 2+ M (g) + X2−(g) → MX(s)
H = 110. kJ H = 480. kJ H = 4750. kJ H = 1/2 (250. kJ) H = −175 kJ H = 920. kJ H = −4800. kJ
M(s) + 1/2 X2(g) → MX(s)
ΔH of = 1410 kJ
For MX composed of M+ and X−, M(s) + 1/2 X2(g) → MX(s):
ΔH of = 110. + 480. + 1/2 (250.) – 175 – 1200. = −660. kJ For M2X composed of M+ and X2− , 2 M(s) + 1/2 X2(g) → M2X(s):
ΔH of = 2(110.) + 2(480.) + 1/2 (250.) – 175 + 920. – 3600. = –1550. kJ For MX2 composed of M2+ and X−, M(s) + X2(g) → MX2(s):
ΔH of = 110. + 480. + 4750. + 250 + 2(−175) – 3500. = 1740. kJ Only M+X− and (M+)2X2− have exothermic enthalpies of formation, so these are both theoretically possible. Because M2X has the more negative (more favorable) ΔH of value, we would predict the M2X compound most likely to form. The charges of the ions in M2X are M+ and X2−.
CHAPTER 9 COVALENT BONDING: ORBITALS Review Questions 1.
The valence orbitals of the nonmetals are the s and p orbitals. The lobes of the p orbitals are 90˚ and 180˚ apart from each other. If the p orbitals were used to form bonds, then all bonds should be 90˚ or 180˚. This is not the case. To explain the observed geometry (bond angles) that molecules exhibit, we need to make up (hybridize) orbitals that point to where the bonded atoms and lone pairs are located. We know the geometry; we hybridize orbitals to explain the geometry. Sigma bonds have shared electrons in the area centered on a line joining the atoms. The orbitals that overlap to form the sigma bonds must overlap head-to-head or end to end. The hybrid orbitals about a central atom always are directed at the bonded atoms. Hybrid orbitals will always overlap head-to-head to form sigma bonds.
2.
geometry linear trigonal planar tetrahedral
hybridization
unhybridized p atomic orbitals
sp sp2 sp3
2 1 0
The unhybridized p atomic orbitals are used to form bonds. Two unhybridized p atomic orbitals, each from a different atom, overlap side to side, resulting in a shared electron pair occupying the space above and below the line joining the atoms (the internuclear axis).
3.
H2S, 2(1) + 6 = 8 e−
CH4, 4 + 4(1) = 8 e−
H2CO, 2(1) + 4 + 6 = 12 e−
HCN, 1 + 4 + 5 = 10 e−
381
382
CHAPTER 9
COVALENT BONDING: ORBITALS
H2S and CH4 both have four effective pairs of electrons about the central atom. Both central atoms will be sp3 hybridized. For H2S, two of the sp3 hybrid orbitals are occupied by lone pairs. The other two sp3 hybrid orbitals overlap with 1s orbitals from hydrogen to form the 2 S−H sigma bonds. For CH4, the four C−H bonds are formed by overlap of the sp3 hybrid orbitals from carbon with 1s orbitals on H. H2CO has a trigonal planar geometry, so carbon is sp2 hybridized. Two of the sp2 hybrid orbitals overlap with hydrogen 1s orbitals to form the two C−H sigma bonds. The third sp2 hybrid orbital is used to form the sigma bond in the double bond by overlapping head to head with an sp2 hybrid orbital from oxygen. The second bond in the double bond is a bond. The unhybridized p atomic orbital on carbon will overlap with a parallel p atomic orbital on O to form the bond. HCN has a linear geometry, so carbon is sp hybridized. HCN has one C−H sigma bond, one C−N sigma bond and two C−N bonds. The C−H sigma bond is formed from sp−1s orbital overlap. The C−N sigma bond is formed from a sp hybrid orbital on carbon overlapping with an sp hybrid orbital from nitrogen. The bonds are formed from the two unhybridized p atomic orbitals from carbon overlapping with two unhybridized p atomic orbitals from N. Each bond is formed from the p orbitals overlapping side to side. Because the p orbitals used must be perpendicular to each other, the bonds must be in two different planes that are perpendicular to each other and perpendicular to the internuclear axis. 4.
Molecules having trigonal bipyramid geometry have five pairs of electrons around the central atom. We need five hybrid orbitals to account for the location of these five sets of electrons. We use the valence s and the three degenerate p valence atomic orbitals for four of the five orbitals; the fifth is an empty d orbital close in energy to the valence atomic orbitals. We call this hybridization dsp3. For octahedral geometry, we need six hybrid orbitals to account for the locations of six pairs of electrons about the central atom. We use the s and three p valence atomic orbitals along with two empty d orbitals. This hybridization is called d2sp3. We mix these six atomic orbitals together and come up with six d2sp3 hybrid orbitals which point to the vertices of an octahedron. PF5 and SF4 both have five pairs of electrons about the central atoms so both exhibit dsp3 hybridization to account for the trigonal bipyramid arrangement of electron pairs. In PF 5, the five pairs of electrons are bonding electrons in the five P−F sigma bonds. Overlap of the dsp3 hybrid orbitals from phosphorus with the appropriate orbitals on each F atom go to form the sigma bonds. SF4 has four S−F bonds and a lone pair of electrons about the sulfur. Four of the sulfur dsp3 hybrid orbitals overlap with appropriate orbitals on the fluorine atoms to form the four S−F sigma bonds, the fifth dsp3 hybrid orbital holds the lone pair of electrons on the sulfur. SF6 and IF5 both have six pairs of electrons about the central atoms so both exhibit d2sp3 hybridization to account for the octahedral geometries of electron pairs. In SF 6, the six d2sp3 hybrid orbitals overlap with appropriate orbitals from F to form the six S−F sigma bonds. In IF5, five of the six d2sp3 hybrid orbitals go to form the five I−F sigma bonds with the sixth d2sp3 holding the lone pair of electrons on iodine.
5.
The electrons in sigma bonding molecular orbitals are attracted to two nuclei, which is a lower, more stable energy arrangement for the electrons than in separate atoms. In sigma antibonding
CHAPTER 9
COVALENT BONDING: ORBITALS
383
molecular orbitals, the electrons are mainly outside the space between the nuclei, which is a higher, less stable energy arrangement than in the separated atoms. 6.
See Figure 9.32 of the text for the 2s bonding and antibonding molecular orbitals. Reference Section 9.3 of the text for the 2p bonding, antibonding, bonding, and antibonding molecular orbitals.
7.
Bond energy is directly proportional to bond order. Bond length is inversely proportional to bond order. Bond energy and bond length can be measured; bond order is calculated from the molecular orbital energy diagram (bond order is the difference between the number of bonding electrons and the number of antibonding electrons divided by two). Paramagnetic: A kind of induced magnetism, associated with unpaired electrons, that causes a substance to be attracted into an inducing magnetic field. Diamagnetic: A type of induced magnetism, associated with paired electrons, that causes a substance to be repelled from the inducing magnetic field. The key is that paramagnetic substances have unpaired electrons in the molecular orbital diagram while diamagnetic substances have only paired electrons in the MO diagram. To determine the type of magnetism, measure the mass of a substance in the presence and absence of a magnetic field. A substance with unpaired electrons will be attracted by the magnetic field, giving an apparent increase in mass in the presence of the field. A greater number of unpaired electrons will give a greater attraction and a greater observed mass increase. A diamagnetic species will not be attracted by a magnetic field and will not show a mass increase (a slight mass decrease is observed for diamagnetic species).
8.
a. H2 has two valence electrons to put in the MO diagram for H2 while He2 has 4 valence electrons. H2: (1s)2 He2: (1s)2(1s*)2
Bond order = B.O. = (2−0)/2 = 1 B.O. = (2−2)/2 = 0
H2 has a nonzero bond order so MO theory predicts it will exist. The H2 molecule is stable with respect to the two free H atoms. He2 has a bond order of zero so it should not form. The He2 molecule is not more stable than the two free He atoms. b. See Figure 9.38 of the text for the MO energy-level diagrams of B2, C2, N2, O2, and F2. B2 and O2 have unpaired electrons in their electron configuration so they are predicted to be paramagnetic. C2, N2 and F2 have no unpaired electrons in the MO diagrams; they are all diamagnetic. c. From the MO energy diagram in Figure 9.38, N2 maximizes the number of electrons in the lower energy bonding orbitals and has no electrons in the antibonding 2p molecular orbitals. N2 has the highest possible bond order of three so it should be a very strong (stable) bond. d. NO+ has 5 + 6 – 1 = 10 valence electrons to place in the MO diagram and NO− has 5 + 6 + 1 = 12 valence electrons. The MO diagram for these two ions is assumed to be the same as that used for N2. The MO electron configurations are:
384
CHAPTER 9 NO+: (2s)2(2s*)2(2p)4(2p)2 NO−: (2s)2(2s*)2(2p)4(2p)2(2p*)2
COVALENT BONDING: ORBITALS B.O. = (8−2)/2 = 3 B.O. = (8−4)/2 = 2
NO+ has a larger bond order than NO− , so NO+ should be more stable than NO−. 9.
In HF, it is assumed that the hydrogen 1s atomic orbital overlaps with a fluorine 2p orbital to form the bonding molecular orbital. The specific 2p orbital used in forming the bonding MO is the p orbital on the internuclear axis. This p orbital will overlap head to head with the hydrogen 1s orbital forming a bonding and a antibonding MO. In the MO diagram, the unpaired H 1s electron and the unpaired fluorine 2p electron fill the bonding MO. No electrons are in the antibonding orbital. Therefore, HF has a bond order of (2−0)/2 = 1 and it should (and does) form. We also use the MO diagram to explain the polarity of the H−F bond. The fluorine 2p orbitals are assumed lower in energy than the hydrogen 1s orbital because F is more electro-negative. Because the bonding MO is closer in energy to the fluorine 2p atomic orbitals, we say the bonding orbital has more fluorine 2p character than hydrogen 1s character. With more fluorine 2p character, the electrons in the bonding orbital will have a greater probability of being closer to F. This leads to a partial negative charge on F and a partial positive charge on H.
10.
Molecules that exhibit resonance have delocalized bonding. This means that the electrons are not permanently stationed between two specific atoms, but instead can roam about over the surface of a molecule. We use the concept of delocalized electrons to explain why molecules that exhibit resonance have equal bonds in terms of strength. Because the electrons can roam about over the entire surface of the molecule, the electrons are shared by all of the atoms in the molecule giving rise to equal bond strengths. The classic example of delocalized electrons is benzene, C6H6. Figure 9.47 of the text shows the molecular orbital system for benzene. Each carbon in benzene is sp2 hybridized, leaving one unhybridized p atomic orbital. All six of the carbon atoms in benzene have an unhybridized p orbital pointing above and below the planar surface of the molecule. Instead of just two unhybridized p orbitals overlapping, we say all six of the unhybridized p orbitals overlap resulting in delocalized electrons roaming about above and below the entire surface of the benzene molecule. O3, 6 + 2(6) = 18 e−
Ozone has a delocalized system. Here the central atom is sp2 hybridized. The unhybridized p atomic orbital on the central oxygen will overlap with parallel p orbitals on each adjacent O atom. All three of these p orbitals overlap together resulting in the electrons moving about above and below the surface of the O3 molecule. With the delocalized electrons, the O−O bond lengths in O3 are equal (and are not different as each individual Lewis structure indicates).
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Active Learning Questions 1.
Molecular orbitals are the solutions (allowed energy levels) to the quantum mechanical treatment of molecules. Atomic orbitals are the solutions (allowed energy levels) to the quantum mechanical treatment of atoms. The shape of bonding orbitals show the electrons localized between the two nuclei. Electrons in bonding orbitals are attracted to two nuclei, which is a lower, more stable energy arrangement for the electrons than in separate atoms. The shape of antibonding orbitals show the electrons are mainly outside the space between the nuclei. This represents a higher, less stable energy arrangement of the electrons as compared to separate atoms.
2.
Sigma () orbitals have the electron density evenly distributed on a line passing through the two nuclei. Sigma bonding orbitals generally result when in-phase atomic orbitals are added head-to-head. Pi () orbitals have the electron density above and below the line between the two nuclei. Pi orbitals generally form when atomic orbitals add side-to-side. Bonding and antibonding molecular orbitals are both solutions to the quantum mechanical treatment of the molecule. Bonding orbitals form when in-phase orbitals combine to give constructive interference. This results in enhanced electron probability located between the two nuclei resulting in an energy for the bonding MO that is lower than the energy of the atomic orbitals from which it is composed. Antibonding orbitals form when out-of-phase orbitals combine. The mismatched phases produce destructive interference leading to a node of electron probability between the two nuclei. With electron distribution pushed to the outside, the energy of an antibonding orbital is higher than the energy of the atomic orbitals from which it is composed. When valence s and p orbitals overlap to form molecular orbitals in diatomic molecules, the type of molecular orbital formed depends on the orientation of the valence atomic orbitals. Valence s orbitals overlap head-to-head when added together to form a sigma molecular orbital. For valence p orbitals, the type of overlap can be head-to-head of side-to-side. Assuming the z-axis is the internuclear axis, the pz atomic orbital overlap head-to-head to form another sigma molecular orbital. The py and px atomic orbitals overlap side-to-side to form molecular orbitals. Hence when valence s and p atomic orbitals overlap with each other, two molecular orbitals and two molecular orbitals form.
3.
The calculated bond order for an H2 molecule with one electron excited to the antibonding orbital would be (1 − 1)/2 = 0. This means there is no net energy stabilization for this arrangement; 0 is also the bond order when you have two separate atoms. One only expects diatomic molecules/ions to form when there is a bond order greater than 0.
4.
Li: (1s)2; bond order = B.O. = (2−0)/2 = 1; because Li2 has a bond order greater than zero, it is energetically favorable for Li2 molecules to form. Hence, Li2 is a stable molecule (has lower energy than two separate Li atoms). However, there is a more stable form for Li. Lithium exists as a solid in nature, containing many Li atoms bound to each other. Bonding in metals will be discussed in Ch 10.
5.
HCN, 1 + 4 + 5 = 10 valence electrons
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Assuming N is hybridized, both C and N atoms are sp hybridized. Your picture for the he C‒ H bond should show that it is formed from overlap of a carbon sp hybrid orbital with a hydrogen 1s atomic orbital. The triple bond is composed of one bond and two bonds. Your picture should show that the sigma bond is formed from head-to-head overlap of the sp hybrid orbitals from the C and N atoms. Your picture should also show the two bonds in the triple bond as being formed from overlap of the two unhybridized p atomic orbitals from each C and N atom. Assuming the z-axis is the internuclear axis, side-to-side overlap of the 2px atomic orbitals from C and N form one of the bonds, and the other is formed from side-to-side overlap of the 2py atomic orbitals. 6.
The second statement is best. Molecular shape is an experimentally determined fact. Hybridization is a model made up to try to explain the bonding in molecules. In this case, CH4 is known to be a tetrahedral molecule by experiment; we hypothesize the central carbon atom is sp3 hybridized to explain the bonding in CH4.
7.
The MO model is a complex mathematical model that assumes the electrons in a molecule occupy allowed energy levels called molecular orbitals. At the heart of the localized electron model is Lewis structures. The MO model works great for diatomic molecules or ions. However the MO model is very difficult to apply to polyatomic molecules. For polyatomic molecule/ions, the localized electron model is much easier to apply, so it is more useful for polyatomic species.
8.
Bond order is directly related to bond energy and inversely related to bond length. A diatomic molecule with a larger bond order will have a stronger bond, but a shorter bond length. Bond order is determined from the MO electron configuration, not from experiment. Bond energy and bond length are determined from experiment. So bond length and bond energy are measured quantities while bond order is determined from the MO model.
9.
In N2, the 2s and 2p atomic orbitals that combine to form the molecular orbitals have the same energy. Hence, each atom’s atomic orbitals contribute equally to the bonding and antibonding orbitals. In CO, we would expect the oxygen atomic orbitals to be lower in energy than the corresponding atomic orbitals from carbon. This will create bonding and antibonding orbitals having unequal electron distribution. In general, one would expect the bonding orbitals, which are closer in energy to the oxygen atomic orbitals, to place more electron density closer to the O atom. The antibonding MOs would place more electron density closer to H and would have a greater contribution from the higher-energy hydrogen 1s atomic orbital.
10.
Lone pairs affect the arrangement of electron pairs about a central atom. We hybridize orbitals to explain the arrangement of electron pairs about the central atom. If the arrangement of electron pairs about a central atom depends on lone pairs, then the hybridization of the central atom also depends on lone pairs.
11.
Figure 9.35 is the expected energy ordering of the molecular orbitals for molecules have 2p valence electrons. We would expect the 2p to have a lower energy than the 2p orbitals since the electrons in sigma bonding orbitals are, on average, concentrated between the two nuclei, resulting in electrons closer to the nuclei. Figure 9.37 is the ordering of the molecular orbitals for B2 which is necessary to explain that B2 is paramagnetic. Note in Figure 9.37, the 2p orbitals
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are lower in energy the 2p orbital. From experiment, B2 is paramagnetic (has unpaired electrons). When Figure 9.35 is used to give the molecular orbital electron configuration for B2, the bond order is 1 (which is fine) but B2 is predicted to be diamagnetic. This goes against experiment. So, the model is modified to come up with the Figure 9.37 ordering. This ordering gives a bond order of 1 and correctly predicts that B2 is paramagnetic. The model is modified to explain experimental data. O2 would not be useful in predicting the energy ordering between the 2p and 2p orbitals. In both diagrams, all the bonding orbitals are filled. It doesn’t matter which is lower in energy since both diagrams give the same result of 6 bonding electrons. The important electrons in a molecular orbital electron configuration are the last electrons placed in the energy levels. In O2, the last electrons are placed into degenerate antibonding orbitals. Since not all electrons are paired, the model correctly predicts that O2 is paramagnetic with a bond order of 2. We get the same prediction using either Figure 9.35 or 9.37. Ionization energy is the energy to remove an electron from some species. In B 2, the electron removed comes from a 2p orbital. In B atoms, the electron removed comes from the 2p atomic orbital. Since the electrons in B2 are in bonding orbitals, the electron removed will be in a lower energy orbital as compared to the 2p atomic orbital of B atoms. Hence it will take more energy to remove the lower energy electron in B2 as compared to B. In O2, the electron removed is from a *2p orbital. Antibonding orbitals are higher in energy than the atomic orbitals that form them. For O2, it takes less energy to remove an electron from O2 as compared to O atoms. 12.
Benzoic acid (C7H6O2) has 7(4) + 6(1) + 2(6) = 46 valence electrons. The Lewis structure for benzoic acid is:
The circle in the ring indicates the delocalized bonding in the benzene ring. The two benzene resonance Lewis structures have three alternating double bonds in the ring (see Figure 9.46).
The six carbons in the ring and the carbon bonded to the ring are all sp 2 hybridized. The five C−H sigma bonds are formed from overlap of the sp2 hybridized carbon atoms with hydrogen 1s atomic orbitals. The seven C−C bonds are formed from head-to-head overlap of sp2 hybrid orbitals from each carbon. The C−O single bond is formed from overlap of a sp2 hybrid orbital on carbon with an sp3 hybrid orbital from oxygen. The C−O bond in the double bond is formed from overlap of carbon sp2 hybrid orbital with an oxygen sp2 orbital. The bond in the C−O double bond is formed from overlap of parallel p unhybridized atomic orbitals from C and O. The delocalized bonding system in the ring is formed from overlap of all six unhybridized p atomic orbitals from the six carbon atoms. Because the π bonding electrons are delocalized, all C-C bonds are equivalent. See Figure 9.48 for delocalized bonding system in the benzene ring.
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Questions 13.
In hybrid orbital theory, some or all of the valence atomic orbitals of the central atom in a molecule are mixed together to form hybrid orbitals; these hybrid orbitals point to where the bonded atoms and lone pairs are oriented. The sigma bonds are formed from the hybrid orbitals overlapping head to head with an appropriate orbital from the bonded atom. The bonds, in hybrid orbital theory, are formed from unhybridized p atomic orbitals. The p orbitals overlap side to side to form the bond, where the electrons occupy the space above and below a line joining the atoms (the internuclear axis). Assuming the z-axis is the internuclear axis, then the pz atomic orbital will always be hybridized whether the hybridization is sp, sp 2, sp3, dsp3 or d2sp3. For sp hybridization, the px and py atomic orbitals are unhybridized; they are used to form two bonds to the bonded atom(s). For sp2 hybridization, either the px or the py atomic orbital is hybridized (along with the s and pz orbitals); the other p orbital is used to form a bond to a bonded atom. For sp3 hybridization, the s and all the p orbitals are hybridized; no unhybridized p atomic orbitals are present, so no bonds form with sp3 hybridization. For dsp3 and d2sp3 hybridization, we just mix in one or two d orbitals into the hybridization process. Which specific d orbitals are used is not important to our discussion.
14.
The MO theory is a mathematical model. The allowed electron energy levels (molecular orbitals) in a molecule are solutions to the mathematical problem. The square of the solutions gives the shapes of the molecular orbitals. A sigma bond is an allowed energy level where the greatest electron probability is between the nuclei forming the bond. Valence s orbitals form sigma bonds, and if the z-axis is the internuclear axis, then valence pz orbitals also form sigma bonds. For a molecule like HF, a sigma-bonding MO results from the combination of the H 1s orbital and the F 2pz atomic orbital. For bonds, the electron density lies above and below the internuclear axis. The bonds are formed when px orbitals are combined (side-to-side overlap) and when py orbitals are combined.
15.
We use d orbitals when we have to, i.e., we use d orbitals when the central atom on a molecule has more than eight electrons around it. In hybrid orbital theory, the d orbitals are used to accommodate the electrons over eight. Row 2 elements never have more than eight electrons around them, so they never hybridize d orbitals. We rationalize this by saying there are no d orbitals close in energy to the valence 2s and 2p orbitals (2d orbitals are forbidden energy levels). However, for row 3 and heavier elements, there are 3d, 4d, 5d, etc. orbitals that will be close in energy to the valence s and p orbitals. It is row 3 and heavier nonmetals that hybridize d orbitals when they have to. For sulfur, the valence electrons are in 3s and 3p orbitals. Therefore, 3d orbitals are closest in energy and are available for hybridization. Arsenic would hybridize 4d orbitals to go with the valence 4s and 4p orbitals, whereas iodine would hybridize 5d orbitals since the valence electrons are in n = 5.
16.
Rotation occurs in a bond if the orbitals that go to form that bond still overlap when the atoms are rotating. Sigma bonds, with the head-to-head overlap, remain unaffected by rotating the atoms in the bonds. Atoms that are bonded together by only a sigma bond (single bond) exhibit this rotation phenomenon. The bonds, however, cannot be rotated. The p orbitals must be parallel to each other to form the bond. If we try to rotate the atoms in a bond, the p orbitals would no longer have the correct alignment necessary to overlap. Because bonds are present
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in double and triple bonds (a double bond is composed of 1 and 1 bond, and a triple bond is always 1 and 2 bonds), the atoms in a double or triple bond cannot rotate (unless the bond is broken). 17. The darker green orbitals about carbon are sp hybrid orbitals. The lighter green orbitals about each oxygen are sp2 hybrid orbitals, and the gold orbitals about all of the atoms are unhybridized p atomic orbitals. In each double bond in CO2, one sigma and one bond exists. The two carbon-oxygen sigma bonds are formed from overlap of sp hybrid orbitals from carbon with a sp2 hybrid orbital from each oxygen. The two carbon-oxygen bonds are formed from side-to-side overlap of the unhybridized p atomic orbitals from carbon with an unhybridized p atomic orbital from each oxygen. These two bonds are oriented perpendicular to each other as illustrated in the figure. 18.
The bonds between two S atoms and between C and S atoms are not as strong as compared to the bonds between two O atoms and between C and O atoms. This is because the atomic orbitals of sulfur are larger and do not overlap as well with each other or with carbon atoms as compared to the smaller atomic orbitals of oxygen. This leads to weaker bonds.
19.
Bonding and antibonding molecular orbitals are both solutions to the quantum mechanical treatment of the molecule. Bonding orbitals form when in-phase orbitals combine to give constructive interference. This results in enhanced electron probability located between the two nuclei. The end result is that a bonding MO is lower in energy than the atomic orbitals from which it is composed. Antibonding orbitals form when out-of-phase orbitals combine. The mismatched phases produce destructive interference leading to a node of electron probability between the two nuclei. With electron distribution pushed to the outside, the energy of an antibonding orbital is higher than the energy of the atomic orbitals from which it is composed.
20.
The driving force for molecule formation is for the molecule to have a lower energy arrangement for the bonding electrons as compared to the separate atoms. Nature tends toward the lowest energy state. If a molecule or ion has a bond order greater than 0, then the molecule or ion is a lower energy arrangement of the electrons as compared to the separate atoms.
21.
H2: (σ1s)2 H2−: (σ1s)2(σ1s*)1
Bond Order = (2−0)/2 = 1 Bond Order = (2−1)/2 = ½
The calculated bond order for H2 is two times larger than the bond order for H2‒. From the simple bond order calculation, the H2 bond is predicted to be twice as strong as the H2‒ bond. 22.
From experiment, B2 is paramagnetic. If the 2p MO is lower in energy than the two degenerate 2p MOs, the electron configuration for B2 would have all electrons paired. Experiment tells us we must have unpaired electrons. Therefore, the MO diagram is modified to have the 2p orbitals lower in energy than the 2p orbitals. This gives two unpaired electrons in the electron configuration for B2, which explains the paramagnetic properties of B2. The model allowed for s and p orbitals to mix, which shifted the energy of the 2p orbital to above that of the 2p orbitals.
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23.
The localized electron model does not deal effectively with molecules containing unpaired electrons. We can draw all of the possible structures for NO with its odd number of valence electrons but still not have a good feel for whether the bond in NO is weaker or stronger than the bond in NO−. MO theory can handle odd electron species without any modifications. From the MO electron configurations, the bond order is 2.5 for NO and 2 for NO−. Therefore, NO should have the stronger bond (and it does). In addition, hybrid orbital theory does not predict that NO− is paramagnetic. The MO theory correctly makes this prediction.
24.
Molecular orbitals are conserved. If 8 total atomic orbitals are used to construct molecular orbitals, then 8 molecular orbitals must form. If the electron probability of a molecular orbital is centered along a line passing through the two nuclei, then the orbital is a sigma bonding orbital. If the electron probability of a molecular orbital lies above and below the line between the two nuclei, then the orbital is a π bonding orbital.
25.
A localized bond has the electron probability centered between two nuclei or centered above and below two nuclei. A delocalized bond has electron probability centered over multiple nuclei; it occurs when unhybridized p atomic orbitals on 3 or more atoms in a molecule/ion all combine. We use delocalize π bonding to explain why molecules that exhibit resonance have equivalent bonds in length and strength.
26.
NO3−, 5 + 3(6) + 1 = 24 e−
When resonance structures can be drawn, it is usually due to a multiple bond that can be in different positions. This is the case for NO3−. Experiment tells us that the three N−O bonds are equivalent in length and strength. To explain this, we say the electrons are delocalized in the molecule. For NO3−, the bonding system is composed of an unhybridized p atomic orbital from all the atoms in NO3−. These p orbitals are oriented perpendicular to the plane of the atoms in NO3−. The bonding system consists of all of the perpendicular p orbitals overlapping forming a diffuse electron cloud above and below the entire surface of the NO3− ion. Instead of having the electrons situated above and below two specific nuclei, we think of the electrons in NO3− as extending over the entire surface of the molecule (hence the term delocalized). See Figure 9.49 for an illustration of the bonding system in NO3−.
Exercises The Localized Electron Model and Hybrid Orbitals 27.
H2O has 2(1) + 6 = 8 valence electrons. O H
H
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391
H2O has a tetrahedral arrangement of the electron pairs about the O atom that requires sp 3 hybridization. Two of the four sp3 hybrid orbitals are used to form bonds to the two hydrogen atoms, and the other two sp3 hybrid orbitals hold the two lone pairs on oxygen. The two O−H bonds are formed from overlap of the sp3 hybrid orbitals from oxygen with the 1s atomic orbitals from the hydrogen atoms. Each O‒H covalent bond is called a sigma (σ) bond since the shared electron pair in each bond is centered in an area on a line running between the two atoms. 28.
CCl4 has 4 + 4(7) = 32 valence electrons.
CCl4 has a tetrahedral arrangement of the electron pairs about the carbon atom that requires sp3 hybridization. The four sp3 hybrid orbitals from carbon are used to form the four bonds to chlorine. The chlorine atoms also have a tetrahedral arrangement of electron pairs, and we will assume that they are also sp3 hybridized. The C‒Cl sigma bonds are all formed from overlap of sp3 hybrid orbitals from carbon with sp3 hybrid orbitals from each chlorine atom. 29.
H2CO has 2(1) + 4 + 6 = 12 valence electrons. O C H
H
The central carbon atom has a trigonal planar arrangement of the electron pairs that requires sp2 hybridization. The two C−H sigma bonds are formed from overlap of the sp2 hybrid orbitals from carbon with the hydrogen 1s atomic orbitals. The double bond between carbon and oxygen consists of one σ and one π bond. The oxygen atom, like the carbon atom, also has a trigonal planar arrangement of the electrons that requires sp2 hybridization. The σ bond in the double bond is formed from overlap of a carbon sp2 hybrid orbital with an oxygen sp2 hybrid orbital. The π bond in the double bond is formed from overlap of the unhybridized p atomic orbitals. Carbon and oxygen each has one unhybridized p atomic orbital that is parallel with the other. When two parallel p atomic orbitals overlap, a π bond results where the shared electron pair occupies the space above and below a line joining the atoms in the bond. 30.
C2H2 has 2(4) + 2(1) = 10 valence electrons.
H
C
C
H
Each carbon atom in C2H2 is sp hybridized since each carbon atom is surrounded by two effective pairs of electrons; i.e., each carbon atom has a linear arrangement of the electrons. Since each carbon atom is sp hybridized, then each carbon atom has two unhybridized p atomic
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orbitals. The two C−H sigma bonds are formed from overlap of carbon sp hybrid orbitals with hydrogen 1s atomic orbitals. The triple bond is composed of one σ bond and two π bonds. The sigma bond between to the carbon atoms is formed from overlap of sp hybrid orbitals from each carbon atom. The two π bonds of the triple bond are formed from parallel overlap of the two unhybridized p atomic orbitals from each carbon. 31.
Ethane, C2H6, has 2(4) + 6(1) = 14 valence electrons. H H
H C
H
C
H H
The carbon atoms are sp3 hybridized. The six C‒H sigma bonds are formed from overlap of the sp3 hybrid orbitals from C with the 1s atomic orbitals from the hydrogen atoms. The carboncarbon sigma bond is formed from overlap of an sp3 hybrid orbital from each C atom. Ethanol, C2H6O has 2(4) + 6(1) + 6 = 20 e−
The two C atoms and the O atom are sp3 hybridized. All bonds are formed from overlap with these sp3 hybrid orbitals. The C‒H and O‒H sigma bonds are formed from overlap of sp3 hybrid orbitals with hydrogen 1s atomic orbitals. The C‒C and C‒O sigma bonds are formed from overlap of the sp3 hybrid orbitals from each atom. 32.
HCN, 1 + 4 + 5 = 10 valence electrons
Assuming N is hybridized, both C and N atoms are sp hybridized. The C‒H bond is formed from overlap of a carbon sp hybrid orbital with a hydrogen 1s atomic orbital. The triple bond is composed of one bond and two bonds. The sigma bond is formed from head-to-head overlap of the sp hybrid orbitals from the C and N atoms. The two bonds in the triple bond are formed from overlap of the two unhybridized p atomic orbitals from each C and N atom. COCl2, 4 + 6 + 2(7) = 24 valence electrons
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393
Assuming all atoms are hybridized, the carbon and oxygen atoms are sp2 hybridized, and the two chlorine atoms are sp3 hybridized. The two C‒Cl bonds are formed from overlap of sp2 hybrids from C with sp3 hybrid orbitals from Cl. The double bond between the carbon and oxygen atoms consists of one and one bond. The bond in the double bond is formed from head-to-head overlap of an sp2 orbital from carbon with an sp2 hybrid orbital from oxy-gen. The bond is formed from parallel overlap of the unhybridized p atomic orbitals from each atom of C and O. 33.
See Exercises 8.97 and 8.107 for the Lewis structures. To predict the hybridization, first determine the arrangement of electron pairs about each central atom using the VSEPR model; then use the information in Figure 9.24 of the text to deduce the hybridization required for that arrangement of electron pairs. 8.97
8.107
a. CCl4: C is sp3 hybridized.
b. NCl3: N is sp3 hybridized.
c. SeCl2: Se is sp3 hybridized.
d. ICl: Both I and Cl are sp3 hybridized.
a. The central N atom is sp2 hybridized in NO2- and NO3-. In N2O4, both central N atoms are sp2 hybridized. b. In OCN- and SCN-, the central carbon atoms in each ion are sp hybridized, and in N3-, the central N atom is also sp hybridized.
34.
See Exercises 8.98 and 8.108 for the Lewis structures. 8.98
a. All the central atoms are sp3 hybridized. b. All the central atoms are sp3 hybridized. c. All the central atoms are sp3 hybridized.
8.108
In O3 and in SO2, the central atoms are sp2 hybridized, and in SO3, the central sulfur atom is also sp2 hybridized.
35.
All exhibit dsp3 hybridization. All of these molecules/ions have a trigonal bipyramid arrangement of electron pairs about the central atom; all have central atoms with dsp 3 hybridization. See Exercise 8.101 for the Lewis structures.
36.
All these molecules have an octahedral arrangement of electron pairs about the central atom; all have central atoms with d2sp3 hybridization. See Exercise 8.102 for the Lewis structures.
37.
The molecules in Exercise 8.133 all have a trigonal planar arrangement of electron pairs about the central atom, so all have central atoms with sp2 hybridization. The molecules in Exercise 8.134 all have a tetrahedral arrangement of electron pairs about the central atom, so all have central atoms with sp3 hybridization. See Exercises 8.133 and 8.134 for the Lewis structures.
38.
The molecules in Exercise 8.135 all have central atoms with dsp3 hybridization because all are based on the trigonal bipyramid arrangement of electron pairs. The molecules in Exercise 8.136 all have central atoms with d2sp3 hybridization because all are based on the octahedral arrangement of electron pairs. See Exercises 8.135 and 8.136 for the Lewis structures.
394 39.
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a.
b. F
N F
F
C
F
F
F F
sp3 nonpolar
tetrahedral 109.5
sp3 polar
trigonal pyramid <109.5
The angles in NF3 should be slightly less than 109.5° because the lone pair requires more space than the bonding pairs. c.
d. F
O F
F
B F
sp3 polar
V-shaped <109.5 e. 40.
H‒Be‒H
a.
F
sp2 nonpolar
trigonal planar 120
linear, 180°, sp, nonpolar
S O
V-shaped, sp2, 120 O
Only one resonance form is shown. Resonance does not change the position of the atoms. We can predict the geometry and hybridization from any one of the resonance structures. b.
c.
O
2-
O
S O
O
S S
O O
trigonal planar, 120, sp2 (plus two other resonance structures)
tetrahedral, 109.5, sp3
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d.
2-
O O
395 e.
S
O
O
S
O
2-
S
O
O
O O
O
O
trigonal pyramid, <109.5, sp3
tetrahedral geometry about each S, 109.5, sp3 hybrids; V-shaped arrangement about peroxide O’s, 109.5, sp3 f.
2-
O
g. S
S O
F
O
F
O
V-shaped, <109.5, sp3
tetrahedral, 109.5, sp3 41.
a. The molecular structure of TeF4 is see-saw. Bond angle a is 120° and angle b is 90°. The hybridization of the central Te atom is dsp3. The individual bond dipoles in TeF4 will not all cancel each other, so it is a polar molecule. F F b a
Te
F F
b.
c.
F
F
a
As
b
F
Kr
F
F
dsp3 nonpolar
trigonal bipyramid a) 90, b) 120 d.
F
F
F
F
square planar 90°
e.
F Kr
dsp3 nonpolar
linear 180 F F
90 o
F Se
F
F
F F
d2sp3 nonpolar
octahedral 90°
d2sp3 nonpolar
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f.
g.
F
F F
F
I
F
F
I F
d2sp3 polar
square pyramid 90 42.
dsp3 polar
T-shaped 90
a.
b. 90
o
120
F
F F
F
F
S
o
F
S F
F
F
F
see-saw, 90 and 120, dsp3 c.
octahedral, 90, d2sp3 d.
F b
S F
c
a
F
S
F 120
o
a) 109.5
b) 90 c) 120 see-saw about S atom with one lone pair (dsp3); bent about S atom with two lone pairs (sp3)
F S
F
F
43.
F
+ 90 o F
F
trigonal bipyramid, 90 and 120, dsp3
With d2sp3 hybridization, we have six pairs of electrons about the central atom. There are three molecular structures that have six pairs of electrons about the central atom. They are octahedral, square pyramid, and square planar. Only the square pyramid molecular structure has 5 bonded atoms, so QF5 has a square pyramid molecular structure. IF5 is an example of a compound with a square pyramid molecular structure: F F
F
I F
F
This Lewis structure has 42 valence electrons. Since the five F atoms have 35 valence electrons, the central Q atom must have 7 valence electrons. So, Q is a halogen. But since the central atom exceeds the octet rule, Q can’t be F. Some possible identities for Q are Cl, Br, I, and At.
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397
44.
T-shaped and see-saw each have 5 electron pairs about the central atom, so they are dsp3 hybridized. Square planar and square pyramid have 6 electron pairs about the central atom, so they are d2sp3 hybridized. Only the trigonal planar molecular structure does not require d orbitals for the hybrids. A trigonal pyramid structure has 4 electron pairs about the central atom and is sp3 hybridized.
45.
ICl5 has 7 + 5(7) = 42 valence electrons.
The only statement that is false is answer e. The central atom is d2sp3 hybridized. The bond dipoles will not all cancel when added together, hence ICl5 is polar. 46.
The Lewis structures are:
−
The correct answer is b. All have 5 electron pairs about the central atom which require a pyramid bipyramidal arrangement and dsp3 hybridization. XeF2 does not have a 90° bond angle, just 180°. XeF2 and AsF5 are nonpolar. None of these have octahedral molecular structure and none of these have 109° bond angles. 47.
H
H C
C
H
H
For the p orbitals to properly line up to form the π bond, all six atoms are forced into the same plane. If the atoms are not in the same plane, then the π bond could not form since the p orbitals would no longer be parallel to each other. 48.
No, the CH2 planes are mutually perpendicular to each other. The center C atom is sp hybridized and is involved in two π bonds. The p orbitals used to form each π bond must be perpendicular to each other. This forces the two CH2 planes to be perpendicular.
H
H C
H
C
C H
398
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COVALENT BONDING: ORBITALS
49.
Statement e is false. The triple bond consists of one σ bond formed from overlap of sp hybrid orbitals from each C and two π bonds from overlap of unhybridized p atomic orbitals from each carbon.
50.
Acetic acid, HC2H3O2, has 4(1) + 2(4) + 2(6) = 24 valence electrons. The Lewis structure is:
Statement b is false. O3 has 4 electron pairs about it; this dictates a tetrahedral arrangement of electron pairs and sp3 hybridization. 51.
The two nitrogen atoms in urea both have a tetrahedral arrangement of electron pairs, so both atoms are sp3 hybridized. The carbon atom has a trigonal planar arrangement of electron pairs, so C is sp2 hybridized. O is also sp2 hybridized because it also has a trigonal planar arrangement of electron pairs. Each of the four N−H sigma bonds are formed from overlap of an sp3 hybrid orbital from nitrogen with a 1s orbital from hydrogen. Each of the two N−C sigma bonds are formed from an sp3 hybrid orbital from N with an sp2 hybrid orbital from carbon. The double bond between carbon and oxygen consists of one and one bond. The bond in the double bond is formed from overlap of a carbon sp2 hybrid orbital with an oxygen sp2 hybrid orbital. The bond in the double bond is formed from overlap of the unhybridized p atomic orbitals. Carbon and oxygen each have one unhybridized p atomic orbital, and they are assumed to be parallel to each other. When two parallel p atomic orbitals overlap side to side, a bond results.
52.
Creatine has 14 σ and 2 π bonds (single bonds are always σ bonds and double bonds are 1 σ and 1 π bond). For bond 1, oxygen has a tetrahedral arrangement of electron pairs, so it is sp3 hybridized. The single bond between oxygen and hydrogen is formed by combining a sp3 hybrid orbital from oxygen with a 1s orbital from hydrogen. For bond 2, the carbon atom has a trigonal planar arrangement of electron pairs, and the nitrogen has a tetrahedral arrangement of electron pairs. So, the bond is formed by combining a sp2 hybrid orbital from carbon with a sp3 hybrid orbital from nitrogen. The two π bonds, one in each double bond, are formed from overlap of unhybridized p atomic orbitals. These unhybridized p orbitals are parallel to each other and overlap side to side to form the bonds
53.
To complete the Lewis structures, just add lone pairs of electrons to satisfy the octet rule for the atoms with fewer than eight electrons. Biacetyl (C4H6O2) has 4(4) + 6(1) + 2(6) = 34 valence electrons.
CHAPTER 9
COVALENT BONDING: ORBITALS O H
H
C
C H
sp2
All CCO angles are 120°. The six atoms are not forced to lie in the same plane because of free rotation about the carboncarbon single (sigma) bonds. There are 11 σ and 2 π bonds in biacetyl.
O
sp3 H
C C
399
H
H
Acetoin (C4H8O2) has 4(4) + 8(1) + 2(6) = 36 valence electrons. sp2 sp3
H
H
C
C
O C
H
H C b
H
The carbon with the doubly bonded O is sp2 hybridized. The other three C atoms are sp3 hybridized. Angle a = 120° and angle b = 109.5°. There are 13 σ and 1 π bonds in acetoin.
a
O
H H sp3
H
Note: All single bonds are σ bonds, all double bonds are one σ and one π bond, and all triple bonds are one σ and two π bonds. 54.
Acrylonitrile: C3H3N has 3(4) + 3(1) + 5 = 20 valence electrons. H
a) 120
C
a
b) 120
H
b
H
C
c
c) 180
6 σ and 3 π bonds
C
sp2
N
sp
All atoms of acrylonitrile must lie in the same plane. The π bond in the double bond dictates that the C and H atoms are all in the same plane, and the triple bond dictates that N is in the same plane with the other atoms. Methyl methacrylate (C5H8O2) has 5(4) + 8(1) + 2(6) = 40 valence electrons. d) 120° e) 120°
H
109.5°
H
C C
f)
H
H
C
H sp2
d
sp3
C
O
e
O
14 σ and 2 π bonds
H C
f
H
H
400 55.
CHAPTER 9
COVALENT BONDING: ORBITALS
a. Add lone pairs to complete octets for each O and N. H a H
e
N C O
c
N b
N
O C
d
H 2C
NH2
N
C C f
g
C
O
O
Azodicarbonamide
CH 3
h
methyl cyanoacrylate
Note: NH2, CH2 (H2C), and CH3 are shorthand for nitrogen or carbon atoms singly bonded to hydrogen atoms. b. In azodicarbonamide, the two carbon atoms are sp2 hybridized, the two nitrogen atoms with hydrogens attached are sp3 hybridized, and the other two nitrogen atoms are sp2 hybridized. In methyl cyanoacrylate, the CH3 carbon is sp3 hybridized, the carbon with the triple bond is sp hybridized, and the other three carbons are sp2 hybridized. c. Azodicarbonamide contains three π bonds and methyl cyanoacrylate contains four π bonds.
56.
d. a) 109.5°
b) 120°
c) 120°
f) 120°
g) 109.5°
h) 120°
d) 120°
e) 180°
a. Piperine and capsaicin are molecules classified as organic compounds, i.e., compounds based on carbon. The majority of Lewis structures for organic compounds have all atoms with zero formal charge. Therefore, carbon atoms in organic compounds will usually form four bonds, nitrogen atoms will form three bonds and complete the octet with one lone pair of electrons, and oxygen atoms will form two bonds and complete the octet with two lone pairs of electrons. Using these guidelines, the Lewis structures are:
CHAPTER 9
COVALENT BONDING: ORBITALS
401
Note: The ring structures are all shorthand notation for rings of carbon atoms. In piperine the first ring contains six carbon atoms and the second ring contains five carbon atoms (plus nitrogen). Also notice that CH3, CH2, and CH are shorthand for carbon atoms singly bonded to hydrogen atoms. b. piperine: 0 sp, 11 sp2 and 6 sp3 carbons; capsaicin: 0 sp, 9 sp2, and 9 sp3 carbons c. The nitrogen atoms are sp3 hybridized in each molecule. d. a) 120°
57.
b) 120°
c) 120°
d) 120°
e) 109.5°
f) 109.5°
g) 120°
h) 109.5°
i)
120°
j)
k) 120°
l)
109.5°
109.5°
To complete the Lewis structure, just add lone pairs of electrons to satisfy the octet rule for the atoms that have fewer than eight electrons. H
O
C
C H H H
O
C H H
N
C
C
C N
O
H
H
H
O C
H H C C
C
N
H
H
N N
58.
a. 6
b. 4
c. The center N in ‒N=N=N group
d. 33 σ
e. 5 π bonds
f.
g. 109.5°
h. sp3
180°
Hydrogen atoms are usually omitted from ring structures. In organic compounds, the carbon atoms generally form four bonds. With this in mind, the following structure has the missing hydrogen atoms included in order to give each carbon atom the four bond requirement.
402
CHAPTER 9
H
H
COVALENT BONDING: ORBITALS
H H a b
H H
H N
N
N
H H
H
e
H
d
H
c N
H
f O
H
N H
a. The two nitrogen atoms in the ring with the double bonds are sp2 hybridized. The other three N atoms are sp3 hybridized. b. The five carbon atoms in the ring with one nitrogen are all sp3 hybridized. The four carbon atoms in the other ring with the double bonds are all sp2 hybridized. c. Angles a, b, and e: 109.5°; angles c, d, and f: 120° d. 31 sigma bonds e. 3 pi bonds
The Molecular Orbital Model 59.
a. The bonding molecular orbital is on the right and the antibonding molecular orbital is on the left. The bonding MO has the greatest electron probability between the nuclei, while the antibonding MO has greatest electron probability outside the space between the two nuclei. b. The bonding MO is lower in energy. Because the electrons in the bonding MO have the greatest probability of lying between the two nuclei, these electrons are attracted to two different nuclei, resulting in a lower energy.
60.
a.
p When p orbitals are combined head-to-head and the phases are the same sign (the orbital lobes have the same sign), a sigma bonding molecular orbital is formed.
CHAPTER 9
COVALENT BONDING: ORBITALS
403
b. p
When parallel p orbitals are combined in-phase (the signs match up), a pi bonding molecular orbital is formed. c. p* When p orbitals are combined head-to-head and the phases are opposite (the orbital lobes have opposite signs), a sigma antibonding molecular orbital is formed. d. p*
When parallel p orbitals are combined out-of-phase (the orbital lobes have opposite signs), a pi antibonding molecular orbital is formed. 61.
If we calculate a nonzero bond order for a molecule, then we predict that it can exist (is stable). H2+: (σ1s)1 H2: (σ1s)2 H2−: (σ1s)2(σ1s*)1 H22−: (σ1s)2(σ1s*)2
62.
If we calculate a nonzero bond order for a molecule, then we predict that it can exist (is stable). He22+: (σ1s)2 He2+: (σ1s)2(σ1s*)1 He2: (σ1s)2(σ1s*)2
63.
B.O. = bond order = (1−0)/2 = 1/2, stable B.O. = (2−0)/2 = 1, stable B.O. = (2−1)/2 = 1/2, stable B.O. = (2−2)/2 = 0, not stable
B.O. = (2−0)/2 = 1, stable B.O. = (2−1)/2 = 1/2, stable B.O. = (2−2)/2 = 0, not stable
If we calculate a nonzero bond order for a molecule, then we predict that it can exist (is stable). N22−: (σ2s)2(σ2s*)2(π2p)4(σ2p)2(π2p*)2 B.O. = bond order = (8−4)/2 = 2, stable 2− 2 2 2 4 4 O2 : (σ2s) (σ2s*) (σ2p) (π2p) (π2p*) B.O. = (8−6)/2 = 1, stable 2− 2 2 2 4 4 2 F2 : (σ2s) (σ2s*) (σ2p) (π2p) (π2p*) (σ2p*) B.O. = (8−8)/2 = 0, not stable
404 64.
CHAPTER 9
If we calculate a nonzero bond order for a molecule, then we predict that it can exist (is stable). Be2: (σ2s)2(σ2s*)2(π2p)2 B2: (σ2s)2(σ2s*)2(π2p)4 Ne2: (σ2s)2(σ2s*)2(σ2p)2(π2p)4(π2p*)4(σ2p*)2 Ne22+: (σ2s)2(σ2s*)2(σ2p)2(π2p)4(π2p*)4
65.
COVALENT BONDING: ORBITALS
B.O. = (4−2)/2 = 1, stable B.O. = (6−2)/2 = 2, stable B.O. = (8−8)/2 = 0, not stable B.O. = (8−6)/2 = 1, stable
The electron configurations are: (σ2s)2
B.O. = (2−0)/2 = 1, diamagnetic (0 unpaired e−)
b. C2:
(σ2s)2(σ2s*)2(π2p)4
B.O. = (6−2)/2 = 2, diamagnetic (0 unpaired e−)
c.
(σ3s)2(σ3s*)2(σ3p)2(π3p)4(π3p*)2
B.O. = (8−4)/2 = 2, paramagnetic (2 unpaired e−)
a.
Li2:
S2:
66.
There are 14 valence electrons in the MO electron configuration. Also, the valence shell is n = 3. Some possibilities from row 3 having 14 valence electrons are Cl2, SCl−, S22−, and Ar22+.
67.
N2+ and N2− each have a bond order of 2.5.
68.
N2+: (σ2s)2(σ2s*)2(π2p)4(σ2p)1
B.O. = bond order = (7−2)/2 = 2.5
N2−: (σ2s)2(σ2s*)2(π2p)4(σ2p)2(π2p*)1
B.O. = bond order = (8−3)/2 = 2.5
The electron configurations are: F2+: (σ2s)2(σ2s*)2(σ2p)2(π2p)4(π2p*)3
B.O. = (8−5)/2 = 1.5; 1 unpaired e−
F2: (σ2s)2(σ2s*)2(σ2p)2(π2p)4(π2p*)4
B.O. = (8−6)/2 = 1;
F2−: (σ2s)2(σ2s*)2(σ2p)2(π2p)4(π2p*)4(σ2p*)1
B.O. = (8−7)/2 = 0.5; 1 unpaired e−
0 unpaired e−
From the calculated bond orders, the order of bond lengths should be F2+ < F2 < F2−. 69.
O22−: (σ2s)2(σ2s*)2(σ2p)2(π2p)4(π2p*)4
B.O. = (8−6)/2 = 1;
O2−: (σ2s)2(σ2s*)2(σ2p)2(π2p)4(π2p*)3
B.O. = (8−5)/2 = 1.5;
0 unpaired e− 0 unpaired e−
Because O2− has a larger bond order than O22−, the superoxide bond length is expected to be shorter than the peroxide bond length.
CHAPTER 9 70.
COVALENT BONDING: ORBITALS
405
Considering only the 12 valence electrons in O2, the MO models would be: p*
p*
p
p
2s*
2s
O2 ground state
Arrangement of electrons consistent with the Lewis structure (double bond and no unpaired electrons).
It takes energy to pair electrons in the same orbital. Thus the structure with no unpaired electrons is at a higher energy; it is an excited state. 71.
The electron configurations are (assuming the same orbital order as that for N2): a. CO: (σ2s)2(σ2s*)2(π2p)4(σ2p)2
B.O. = (8-2)/2 = 3, diamagnetic
b. CO+: (σ2s)2(σ2s*)2(π2p)4(σ2p)1
B.O. = (7-2)/2 = 2.5, paramagnetic
c. CO2+: (σ2s)2(σ2s*)2(π2p)4
B.O. = (6-2)/2 = 2, diamagnetic
Because bond order is directly proportional to bond energy and inversely proportional to bond length: Shortest → longest bond length: CO < CO+ < CO2+ Smallest → largest bond energy: CO2+ < CO+ < CO 72.
a. CN+:
(σ2s)2(σ2s*)2(π2p)4
B.O. = (6−2)/2 = 2, diamagnetic
b. CN:
(σ2s)2(σ2s*)2(π2p)4(σ2p)1
B.O. = (7−2)/2 = 2.5, paramagnetic
c. CN−:
(σ2s)2(σ2s*)2(π2p)4(σ2p)2
B.O. = 3, diamagnetic
The bond orders are CN+, 2; CN, 2.5; CN−, 3; because bond order is directly proportional to bond energy and inversely proportional to bond length: Shortest → longest bond length: CN− < CN < CN+ Smallest → largest bond energy: CN+ < CN < CN−
406 73.
CHAPTER 9
COVALENT BONDING: ORBITALS
a. H2: (σ1s)2 b. B2: (σ2s)2(σ2s*)2(π2p)2 c. C22−: (σ2s)2(σ2s*)2(π2p)4(σ2p)2 d. OF: (σ2s)2(σ2s*)2(σ2p)2(π2p)4(π2p*)3 The bond strength will weaken if the electron removed comes from a bonding orbital. Of the molecules listed, H2, B2, and C22− would be expected to have their bond strength weaken as an electron is removed. OF has the electron removed from an antibonding orbital, so its bond strength increases.
74.
a. CN: (σ2s)2(σ2s*)2(π2p)4(σ2p)1 NO: (σ2s)2(σ2s*)2(π2p)4(σ2p)2(π2p*)1 b. O22+: (σ2s)2(σ2s*)2(σ2p)2(π2p)4 N22+: (σ2s)2(σ2s*)2(π2p)4 If the added electron goes into a bonding orbital, the bond order would increase, making the species more stable and more likely to form. Between CN and NO, CN would most likely form CN− since the bond order increases (unlike NO− , where the added electron goes into an antibonding orbital). Between O22+ and N22+, N2+ would most likely form since the bond order increases (unlike O22+ going to O2+).
75.
The two types of overlap that result in bond formation for p orbitals are in-phase side-to-side overlap (π bond) and in-phase head-to-head overlap (σ bond).
σ2p (in-phase; the signs match up)
π2p (in-phase; the signs match up) 76.
+
+
* (out-of-phase; the signs oppose each other)
+
(in-phase; the signs match up)
These molecular orbitals are sigma MOs because the electron density is cylindrically symmetric about the internuclear axis.
CHAPTER 9 77.
COVALENT BONDING: ORBITALS
407
a. The electron density would be closer to F on average. The F atom is more electronegative than the H atom, and the 2p orbital of F is lower in energy than the 1s orbital of H. b. The bonding MO would have more fluorine 2p character since it is closer in energy to the fluorine 2p atomic orbital. c. The antibonding MO would place more electron density closer to H and would have a greater contribution from the higher-energy hydrogen 1s atomic orbital.
78.
a. See the illustrations in the solution to Exercise 9.60 for the bonding and antibonding MOs in OH. b. The antibonding MO will have more hydrogen 1s character because the hydrogen 1s atomic orbital is closer in energy to the antibonding MO. c. No, the net overall overlap is zero. The px orbital does not have proper symmetry to overlap with a 1s orbital. The 2px and 2py orbitals are called nonbonding orbitals. + + -
d. H
HO
O *
1s
2p x
2p y
2p z
2p x
2p y
2s
2s
e. Bond order = (2 – 0)/2 = 1; Note: The 2s, 2px, and 2py electrons have no effect on the bond order. f.
79.
To form OH+, a nonbonding electron is removed from OH. Because the number of bonding electrons and antibonding electrons is unchanged, the bond order is still equal to one.
C22− has 10 valence electrons. The Lewis structure predicts sp hybridization for each carbon with two unhybridized p orbitals on each carbon. sp hybrid orbitals form the σ bond and the two unhybridized p atomic orbitals from each carbon form the two π bonds. MO: (σ2s)2(σ2s*)2(π2p)4(σ2p)2, B.O. = (8 − 2)/2 = 3
408
CHAPTER 9
COVALENT BONDING: ORBITALS
Both give the same picture, a triple bond composed of one σ and two π bonds. Both predict the ion will be diamagnetic. Lewis structures deal well with diamagnetic (all electrons paired) species. The Lewis model cannot really predict magnetic properties. 80.
Lewis structures:
Note: Lewis structures do not handle odd numbered electron species very well. MO model: NO+: (σ2s)2(σ2s*)2(π2p)4(σ2p)2, NO: (σ2s)2(σ2s*)2(π2p)4(σ2p)2(π2p*)1, NO−: (σ2s)2(σ2s*)2(π2p)4(σ2p)2(π2p*)2
B.O. = 3, 0 unpaired e− (diamagnetic) B.O. = 2.5, 1 unpaired e− (paramagnetic) B.O. = 2, 2 unpaired e− (paramagnetic)
The two models give the same results only for NO+ (a triple bond with no unpaired electrons). Lewis structures are not adequate for NO and NO−. The MO model gives a better representation for all three species. For NO, Lewis structures are poor for odd electron species. For NO−, both models predict a double bond, but only the MO model correctly predicts that NO− is paramagnetic. 81.
Statement b is false. All carbon‒carbon bonds in benzene are equivalent. Each carbon is sp2 hybridized, leaving 1 unhybridized p atomic orbital that is perpendicular to the plane of the molecule. All six of the unhybridized p orbitals combine to form a delocalized π bonding system resulting in equivalent carbon‒carbon bonds in length and strength.
82.
Borazine has another resonance structure. Whenever resonance structures can be drawn, the actual bonding is an average of the various resonance structures. Here all boron‒nitrogen bonds will be equivalent, with a length and strength between a single and a double bond. Delocalized π bonding is used to explain the equal bonds in the ring. The sigma bonds between boron and nitrogen are formed from overlap of sp2 hybrid orbitals from each atom in the ring. This leaves an unhybridized p atomic orbital on each boron and nitrogen that is perpendicular to the plane of the molecule. All six of these p atomic orbitals combine to form a delocalized π bonding system with electrons above and below the entire ring surface. This results in equivalent bonds in the ring.
83.
O3 and NO2− are isoelectronic, so we only need consider one of them since the same bonding ideas apply to both. The Lewis structures for O3 are: O O
O O
O
O
For each of the two resonance forms, the central O atom is sp 2 hybridized with one unhybridized p atomic orbital. The sp2 hybrid orbitals are used to form the two sigma bonds to the central atom and hold the lone pair of electrons on the central O atom. The localized electron view of the π bond uses unhybridized p atomic orbitals. The π bond resonates between the two
CHAPTER 9
COVALENT BONDING: ORBITALS
409
positions in the Lewis structures; the actual structure of O3 is an average of the two resonance structures:
In the MO picture of the π bond, all three unhybridized p orbitals overlap at the same time, resulting in π electrons that are delocalized over the entire surface of the molecule. This is represented as:
or
84.
The Lewis structures for CO32− are (24 e−): 2-
O C O
2-
O C
O
O
2-
O C
O
O
O
In the localized electron view, the central carbon atom is sp2 hybridized; the sp2 hybrid orbitals are used to form the three sigma bonds in CO32−. The central C atom also has one unhybridized p atomic orbital that overlaps with another p atomic orbital from one of the oxygen atoms to form the π bond in each resonance structure. This localized π bond moves (resonates) from one position to another. In the molecular orbital model for CO32−, all four atoms in CO32− have a p atomic orbital that is perpendicular to the plane of the ion. All four of these p orbitals overlap at the same time to form a delocalized π bonding system where the π electrons can roam above and below the entire surface of the ion. The π molecular orbital system for CO32− is analogous to that for NO3− which is shown in Figure 9.49 of the text.
410
CHAPTER 9
COVALENT BONDING: ORBITALS
ChemWork Problems 85.
Only statement c is true. Here an atom with a trigonal planar arrangement of electron pairs which is sp2 hybridized is bonded to an atom with a tetrahedral arrangement of electron pairs which is sp3 hybridized. These hybrid orbitals would combine to form a σ bond. For statement a, only valence orbitals combine to form hybrid orbitals. For statement b, the hybrid orbitals always point to where the electron pairs are located. Hybrid orbitals are used to explain the bond angles about a central atom. For statement d, hybrid orbitals always combine head-to-head resulting in σ bonds. And for statement e, the unhybridized p atomic orbital go to form the π bonds. With sp2 hybridization, only one p atomic orbital is unhybridized, so only 1 π bond can form.
86.
The Lewis structures for the compounds are: O
H
C
C
O
H
F
C
F H
H
In sigma bonds, the atoms can rotate while the bond remains intact. In pi bonds, the atoms cannot rotate unless the π bond is broken. All double and triple bonds contain π bonds. So C2H2, HCN, and CH2O all have at least one bond where the atoms cannot rotate freely. F2O contains only single (σ) bonds, so atoms in both bonds can rotate freely. 87.
V-shaped, 120˚,
SeO2 (18 e−)
polar, sp2 hybridized PCl3 (26 e−)
trigonal pyramid, < 109.5˚ polar, sp3 hybridized
NNO (16 e−)
linear, 180˚ +2 others
COS (16 e−)
linear, 180˚ +2 others
PF3 (26 e−)
polar, sp hybridized
polar, sp hybridized trigonal pyramid, < 109.5˚ polar, sp3 hybridized
CHAPTER 9
COVALENT BONDING: ORBITALS
411
All these compounds are polar. In each compound, the bond dipoles do not cancel each other out, leading to a polar compound. SeO2 is the only compound exhibiting ~120˚ bond angles. PCl3 and PF3 both have a tetrahedral arrangement of electron pairs, so these are the compounds that have central atoms that are sp3 hybridized. NNO and COS both have linear molecular structure. 88.
TeCl4 (34 e−) see-saw, 90˚ and 120˚ polar, dsp3 hybridized
ICl5 (42 e−) square pyramid, 90˚ polar, d2sp3 hybridized PCl5 (40 e−) trigonal bipyramid, 90˚ and 120˚ nonpolar, dsp3 hybridized
KrCl4 (36 e−) square planar, 90˚ nonpolar, d2sp3 hybridized
XeCl2 (22 e−)
linear, 180˚ nonpolar, dsp3 hybridized
TeCl4 and PCl5 exhibit at least one ~120˚ bond angle. ICl5 and KrCl4 exhibit d2sp3 hybridization. KrCl4 is square planar. TeCl4 and ICl5 are polar. 89.
a. XeO3, 8 + 3(6) = 26 e−
b. XeO4, 8 + 4(6) = 32 e− O
Xe O
O O
Xe O
O O
trigonal pyramid; sp3
tetrahedral; sp3
412
CHAPTER 9 c. XeOF4, 8 + 6 + 4(7) = 42 e− O
COVALENT BONDING: ORBITALS d. XeOF2, 8 + 6 + 2(7) = 28 e−
F
F
F Xe
F
F
F Xe
or F
O
F
F
O
Xe
or
F
Xe
O
F
square pyramid; d2sp3
F
T-shaped; dsp3
e. XeO3F2 has 8 + 3(6) + 2(7) = 40 valence electrons. F
F
O Xe
O
or
O
F Xe
F
or
Xe
O F
90.
O
O
O O
F3ClO, 3(7) + 7 + 6 = 34 e−
F trigonal bipyramid; dsp3
O
F2ClO2+, 2(7) + 7 + 2(6) − 1 = 32 e− +
O Cl F
see-saw, dsp3
F
O
tetrahedral, sp3
Note: Similar to Exercise 89c, d, and e, F3ClO has one additional Lewis structure that is possible, and F3ClO2 (below) has two additional Lewis structure that are possible. The predicted hybridization is unaffected. F3ClO2, 3(7) + 7 + 2(6) = 40 e− trigonal pyramid; dsp3
91.
The Lewis structure for N2 is: Each nitrogen has two effective electron pairs, so each N is sp hybridized. The sp hybrid orbitals combine to form the σ bond between the two N atoms. The two unhybridized p atomic orbitals on each N combine to form two π bonds in the triple bond. Answer c is correct.
CHAPTER 9 92.
COVALENT BONDING: ORBITALS
413
C6N4 has 6(4) + 4(5) = 44 valence electrons. The Lewis structure is:
The carbon atoms with a triple bond are all sp hybridized and the carbons with the double bond are sp2 hybridized. The four C‒C single bonds are all σ bonds formed from combining a sp hybrid orbital with a sp2 hybrid orbital. 93.
Molecule A has a tetrahedral arrangement of electron pairs because it is sp3 hybridized. Molecule B has 6 electron pairs about the central atom, so it is d2sp3 hybridized. Molecule C has two and two π bonds to the central atom, so it either has two double bonds to the central atom (as in CO2) or one triple bond and one single bond (as in HCN). Molecule C is consistent with a linear arrangement of electron pairs exhibiting sp hybridization. There are many correct possibilities for each molecule; an example of each is: Molecule A: CH4
Molecule B: XeF4
Molecule C: CO2 or HCN
H C H
H H
tetrahedral; 109.5°; sp3 94.
square planar; 90°; d2sp3
linear; 180°; sp
a. To complete the Lewis structure, add two lone pairs to each sulfur atom.
sp3 H 3C
sp C
sp C
H
H
C
C
C
C S
sp C
sp C
sp C
sp C
sp2 CH
sp2 CH 2
S
b. See the Lewis structure. The four carbon atoms in the ring are all sp2 hybridized, and the two sulfur atoms are sp3 hybridized. c. 23 σ and 9 π bonds. Note: CH3 (H3C), CH2, and CH are shorthand for carbon atoms singly bonded to hydrogen atoms. 95.
a. No, some atoms are in different places. Thus these are not resonance structures; they are different compounds.
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CHAPTER 9 b.
COVALENT BONDING: ORBITALS
For the first Lewis structure, all nitrogen atoms are sp3 hybridized and all carbon atoms are sp2 hybridized. In the second Lewis structure, all nitrogen atoms and carbon atoms are sp2 hybridized.
c. For the reaction: H O
O
N
H
O
O
N
C
C
C
C
N
N
N
N
H
C
H
C
H
O
O
H
Bonds broken: 3 C=O (745 kJ/mol) 3 C‒N (305 kJ/mol) 3 N‒H (391 kJ/mol)
Bonds formed: 3 C=N (615 kJ/mol) 3 C‒O (358 kJ/mol) 3 O‒H (467 kJ/mol)
ΔH = 3(745) + 3(305) + 3(391) ‒ [3(615) + 3(358) + 3(467)] ΔH = 4323 kJ ‒ 4320 kJ = 3 kJ The bonds are slightly stronger in the first structure with the carbon-oxygen double bonds since ΔH for the reaction is positive. However, the value of ΔH is so small that the best conclusion is that the bond strengths are comparable in the two structures. 96.
a. The V-shaped (or bent) molecular structure occurs with both a trigonal planar and a tetrahedral arrangement of electron pairs. If there is a trigonal planar arrangement, the central atom is sp2 hybridized. If there is a tetrahedral arrangement, the central atom is sp 3 hybridized. b. The see-saw structure is a trigonal bipyramid arrangement of electron pairs which requires dsp3 hybridization. c. The trigonal pyramid structure occurs when a central atom has three bonded atoms and a lone pair of electrons. Whenever a central atom has four effective pairs about the central atom (exhibits a tetrahedral arrangement of electron pairs), the central atom is sp 3 hybridized. d. A trigonal bipyramidal arrangement of electron pairs requires dsp3 hybridization. e. A tetrahedral arrangement of electron pairs requires sp3 hybridization.
CHAPTER 9 97.
COVALENT BONDING: ORBITALS
415
For carbon, nitrogen, and oxygen atoms to have formal charge values of zero, each C atom will form four bonds to other atoms and have no lone pairs of electrons, each N atom will form three bonds to other atoms and have one lone pair of electrons, and each O atom will form two bonds to other atoms and have two lone pairs of electrons. Using these bonding requirements gives the following two resonance structures for vitamin B6: O b
a
H O g H
H
C
C
C
H
f
O
C
H H
c
C d e
C
O
C
H
O
H
C N
C
H
H
H C
C
H
C
C C
H
H H
C
O
H
H
C N
H
a. 21 σ bonds; 4 π bonds (The electrons in the three π bonds in the ring are delocalized.) b. Angles a), c), and g): 109.5°; angles b), d), e), and f): 120° c. 6 sp2 carbons; the five carbon atoms in the ring are sp2 hybridized, as is the carbon with the double bond to oxygen. d. 4 sp3 atoms; the two carbons that are not sp2 hybridized are sp3 hybridized, and the oxygens marked with angles a and c are sp3 hybridized. e. Yes, the π electrons in the ring are delocalized. The atoms in the ring are all sp2 hybridized. This leaves a p orbital perpendicular to the plane of the ring from each atom. Overlap of all six of these p orbitals results in a π molecular orbital system where the electrons are delocalized above and below the plane of the ring (similar to benzene in Figure 9.48 of the text). 98.
For carbon, nitrogen, and oxygen atoms to have formal charge values of zero, each C atom will form four bonds to other atoms and have no lone pairs of electrons, each N atom will form three bonds to other atoms and have one lone pair of electrons, and each O atom will form two bonds to other atoms and have two lone pairs of electrons. Following these bonding requirements, a Lewis structure for aspartame is:
416
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Another resonance structure could be drawn having the double bonds in the benzene ring moved over one position. Atoms that have trigonal planar geometry of electron pairs are assumed to have sp2 hybridization, and atoms with tetrahedral geometry of electron pairs are assumed to have sp 3 hybridization. All the N atoms have tetrahedral geometry, so they are all sp3 hybridized (no sp2 hybridization). The oxygens double bonded to carbon atoms are sp2 hybridized; the other two oxygens with two single bonds are sp3 hybridized. For the carbon atoms, the six carbon atoms in the benzene ring are sp2 hybridized, and the three carbons double bonded to oxygen are also sp2 hybridized (tetrahedral geometry). Answering the questions: •
9 sp2 hybridized C and N atoms (9 from C’s and 0 from N’s)
•
7 sp3 hybridized C and O atoms (5 from C’s and 2 from O’s)
•
39 bonds and 6 bonds (this includes the 3 bonds in the benzene ring that are delocalized)
99. Cl
H C
H
Cl
C
C Cl
H
Cl
Cl
C
H
C Cl
C
H
H
In order to rotate about the double bond, the molecule must go through an intermediate stage where the π bond is broken and the sigma bond remains intact. Bond energies are 347 kJ/mol for C‒C and 614 kJ/mol for C=C. If we take the single bond as the strength of the σ bond, then the strength of the π bond is (614 − 347 = ) 267 kJ/mol. In theory, 267 kJ/mol must be supplied to rotate about a carbon-carbon double bond. 100.
CO, 4 + 6 = 10 e−;
CO2, 4 + 2(6) = 16 e−;
C3O2, 3(4) + 2(6) = 24 e−
There is no molecular structure for the diatomic CO molecule. The carbon in CO is sp hybridized. CO2 is a linear molecule, and the central carbon atom is sp hybridized. C3O2 is a linear molecule with all the central carbon atoms exhibiting sp hybridization. 101.
a. N2F2 has 2(5) + 2(7) = 24 valence electrons. Can also be: N
N
F
F
V-shaped about both N’s; 120° about both N’s; both N atoms: sp2
F N
N
F
polar
nonpolar
CHAPTER 9
COVALENT BONDING: ORBITALS
417
These are distinctly different molecules. b. C4H6 has 4(4) + 6(1) = 22 valence electrons. H H
C
H
C
C
H
C
H
H
All C atoms are trigonal planar with 120° bond angles and sp2 hybridization. Because C and H have similar electronegativity values, the C−H bonds are essentially nonpolar, so the molecule is nonpolar. All neutral compounds composed of only C and H atoms are nonpolar. 102.
a. Yes, both have four sets of electrons about the P. We would predict a tetrahedral structure for both. See part d for the Lewis structures. b. The hybridization is sp3 for P in each structure since both structures exhibit a tetrahedral arrangement of electron pairs. c. P uses one of its d orbitals to form the π bond since the p orbitals are all used to form the hybrid orbitals. d. Formal charge = number of valence electrons of an atom − [(number of lone pair electrons) + 1/2(number of shared electrons)]. The formal charges calculated for the O and P atoms are next to the atoms in the following Lewis structures. O +1
Cl
P
-1
0
O Cl
Cl
P
Cl
0
Cl
Cl
In both structures, the formal charges of the Cl atoms are all zeros. The structure with the P=O bond is favored on the basis of formal charge since it has a zero formal charge for all atoms. 103.
a. The Lewis structures for NNO and NON are: N
N
O
N
N
O
N
N
O
N
O
N
N
O
N
N
O
N
The NNO structure is correct. From the Lewis structures, we would predict both NNO and NON to be linear. However, we would predict NNO to be polar and NON to be nonpolar. Since experiments show N2O to be polar, NNO is the correct structure.
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b. Formal charge = number of valence electrons of atoms − [(number of lone pair electrons) + 1/2(number of shared electrons)]. N
N
O
N
N
O
N
N
O
-1
+1
0
0
+1
-1
-2
+1
+1
The formal charges for the atoms in the various resonance structures are below each atom. The central N is sp hybridized in all the resonance structures. We can probably ignore the third resonance structure on the basis of the relatively large formal charges as compared to the first two resonance structures. c. The sp hybrid orbitals from the center N overlap with atomic orbitals (or appropriate hybrid orbitals) from the other two atoms to form the two sigma bonds. The remaining two unhybridized p orbitals from the center N overlap with two p orbitals from the peripheral N to form the two π bonds. 2px sp
sp
z axis
2p y
104.
For CO, we will assume the same orbital ordering as that for N2. CO: (2s)2(2s*)2(2p)4(2p)2
B.O. = (8 – 2)/2 = 3; 0 unpaired electrons
O2: (2s)2(2s*)2(2p)2(2p)4(2p*)2
B.O. = (8 – 4)/2 = 2; 2 unpaired electrons
The most obvious differences are that CO has a larger bond order than O2 (3 versus 2) and that CO is diamagnetic, whereas O2 is paramagnetic. 105.
N2: (σ2s)2(σ2s*)2(π2p)4(σ2p)2;
N2+: (σ2s)2(σ2s*)2(π2p)4(σ2p)1
N2−: (σ2s)2(σ2s*)2(π2p)4(σ2p)2(π2p*)1 106.
The electron configurations are (assuming the same orbital order as that for N2): B2+: (σ2s)2(σ2s*)2(π2p)1
B.O. = (3-2)/2 = 0.5
B2: (σ2s)2(σ2s*)2(π2p)2
B.O. = (4-2)/2 = 1
B2−: (σ2s)2(σ2s*)2(π2p)3
B.O. = (5-2)/2 = 1.5
Because bond order is directly proportional to bond energy and inversely proportional to bond length: Shortest → longest bond length: B2− < B2 < B2+ Smallest → largest bond energy: B2+ < B2 < B2−
CHAPTER 9
COVALENT BONDING: ORBITALS
107.
O2:
(σ2s)2(σ2s*)2(σ2p)2(π2p)4(π2p*)2
B.O. = bond order = (8 – 4)/2 = 2
N2:
(σ2s)2(σ2s*)2(π2p)4(σ2p)2
B.O. = (8 – 2)/2 = 3
419
In O2, an antibonding electron is removed, which will increase the bond order to 2.5 [= (8 – 3)/2]. The bond order increases as an electron is removed, so the bond strengthens. In N2, a bonding electron is removed, which decreases the bond order to 2.5 = [(7 – 2)/2]. So the bond strength weakens as an electron is removed from N2. 108.
N2 (ground state): (σ2s)2(σ2s*)2(π2p)4(σ2p)2; B.O. = 3; diamagnetic (0 unpaired e−) N2 (1st excited state): (σ2s)2(σ2s*)2(π2p)4(σ2p)1(π2p*)1 B.O. = (7 ‒ 3)/2 = 2; paramagnetic (2 unpaired e−) The first excited state of N2 should have a weaker bond and should be paramagnetic.
109.
F2: (σ2s)2(σ2s*)2(σ2p)2(π2p)4(π2p*)4; F2 should have a lower ionization energy than F. The electron removed from F2 is in a π2p* antibonding molecular orbital that is higher in energy than the 2p atomic orbitals from which the electron in atomic fluorine is removed. Because the electron removed from F2 is higher in energy than the electron removed from F, it should be easier to remove an electron from F2 than from F.
110.
dxz
+
pz
x
x
z z
The two orbitals will overlap side to side, so when the orbitals are in phase, a π bonding molecular orbital would be expected to form. 111.
Side-to-side in-phase overlap of these d orbitals would produce a π bonding molecular orbital. There would be no probability of finding an electron on the axis joining the two nuclei, which is characteristic of π MOs.
112.
a. HF, 1 + 7 = 8 e−
linear, sp3 (if F is hybridized)
SbF5, 5 + 5(7) = 40 e−
trigonal bipyramid, dsp3
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CHAPTER 9
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H2F+, 2(1) + 7 – 1 = 8 e−
SbF6−, 5 + 6(7) + 1 = 48 e−
V-shaped, sp3
octahedral, d2sp3
b. 2.93 mL
0.975 g HF 1 mol HF 1 mol [H 2 F]+ [SbF6 ]− 256.8 g × mL 20.01 g HF 2 mol HF mol [H 2 F]+ [SbF6 ]−
= 18.3 g [H2F]+[SbF6]− 10.0 mL
3.10 g SbF5 mL
1 mol SbF5 216.8 g SbF5
1 mol [H 2 F]+ [SbF6 ] − mol SbF5
256.8 g mol [H 2 F] + [SbF6 ] −
= 36.7 g [H2F]+[SbF6]− Because HF produces the smaller amount of product, HF is limiting and 18.3 g of [H2F]+[SbF6]− can be produced. 113.
Element X has 36 protons, which identifies it as Kr. Element Y has one less electron than Y−, so the electron configuration of Y is 1s22s22p5. This is F. KrF3+, 8 + 3(7) – 1 = 28 e−
T-shaped, dsp3
114.
a. −307 kJ = (−1136 + x) – [(−254 kJ) + 3(−96 kJ)], x = ΔH of , NI3 = 287 kJ/mol b. IF2+, 7 + 2(7) – 1 = 20 e−
V-shaped; sp3
BF4−, 3 + 4(7) + 1 = 32 e−
Tetrahedral; sp3
CHAPTER 9 115.
COVALENT BONDING: ORBITALS
a. Li2: (σ 2s ) 2
421
B.O. = (2 – 0)/2 = 1
B2: (σ 2s ) 2 (σ*2s ) 2 ( π 2 p ) 2
B.O. = (4 – 2)/2 = 1; both have a bond order of 1.
b. B2 has four more electrons than Li2, so four electrons must be removed from B2 to make it isoelectronic with Li2. The isoelectronic ion is B24+. c. To form B24+, it takes 6455 kJ of energy to remove 4 mol of electrons from 1 mol of B2. 1 mol B 2 1000 g 6455 kJ 1.5 kg B2 = 4.5 × 105 kJ 1 kg 21 .62 g B 2 mol B 2 116.
Statement c is false. The electrons in the π bond(s) are the electrons that are delocalized over the entire surface of the molecule/ion. The delocalized π bonding system is formed from unhybridized p atomic orbitals. So, each atom in the molecule/ion forming the delocalized π bonding must have unhybridized p atomic orbitals which dictates sp or sp2 hybridization.
117.
SO2 (18 e−)
a. True; the central sulfur has a trigonal planar arrangement of electron pairs, hence the bond angle is ~120o and the central sulfur atom is sp2 hybridized. b. False; all the S−O bonds are equivalent in strength with a bond length value somewhere between a single and a double bond. c. True; 118.
d. True;
e. False; the two resonance structures are shown above.
Molecules with resonance exhibit delocalized π electrons. SO32‒ and XeO3 both have 3 singly bonded oxygen atoms to the central atom and a lone pair on the central atom. Neither exhibits resonance. The molecules where resonance structures are possible are CO2, NO2+, and NO3‒. The resonance structures are:
O
O
N O
O
N O
O
N O
O
O
When resonance structures are possible, the bond lengths are all equal; the actual bond is an average of all the resonance structures. The average bond in CO2 and NO2+ is a double bond; CO2 and NO2+ are not the molecule/ion described in the problem.
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This leaves NO3‒. In any of the three resonance structures, any one bond is a single bond in two of the resonance structures and a double bond in the third. The average bond in NO3‒ is a 1.33 bond. This fits the description given in the question. NO3‒ will have N‒O bonds stronger than a single bond, but significantly weaker than a double bond.
Challenge Problems 119.
The following Lewis structure has a formal charge of zero for all of the atoms in the molecule.
The three C atoms each bonded to three H atoms are sp3 hybridized (tetrahedral geometry); the other five C atoms with trigonal planar geometry are sp2 hybridized. The one N atom with the double bond is sp2 hybridized, and the other three N atoms are sp3 hybridized. The answers to the questions are:
120.
•
6 total C and N atoms exhibit 120 bond angles
•
6 total C and N atoms are sp3 hybridized
•
0 C and N atoms are sp hybridized (linear geometry)
•
25 bonds and 4 bonds
The complete Lewis structure follows on the next page. All but two of the carbon atoms are sp3 hybridized. The two carbon atoms that contain the double bond are sp2 hybridized (see * in the following Lewis structure).
CHAPTER 9
COVALENT BONDING: ORBITALS
423
CH3 H
H C
H2C H
H C
CH3
H 2C
C
C
*C
H
CH2 CH
CH2
CH3
CH2
CH
H
CH3
C
CH3
C
CH2
C
CH2
H C C
H
H * C
C
HO H
H
CH2
H
Note: HO, CH, CH2, H2C, and CH3 are shorthand for oxygen and carbon atoms singly bonded to hydrogen atoms. 121.
a. NCN2− has 5 + 4 + 5 + 2 = 16 valence electrons.
H2NCN has 2(1) + 5 + 4 + 5 = 16 valence electrons. H +1 N H
H 0
-1
C
N
0
H
N
0
0
C
N
favored by formal charge
NCNC(NH2)2 has 5 + 4 + 5 + 4 + 2(5) + 4(1) = 32 valence electrons. H
0 -1
0
+1
N
C
N
0
C
N
0
0
0
H N H
N
C
N
0
H
H
0 0
C
N H N H 0
H
favored by formal charge
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CHAPTER 9
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Melamine (C3N6H6) has 3(4) + 6(5) + 6(1) = 48 valence electrons. H
H H
N
N
N
C
C
N
N
H
H
N
N
H
C
N
N
H
H
C
C H
N
C
H
N
N H
H
b. NCN2−: C is sp hybridized. Each resonance structure predicts a different hybridization for the N atom. Depending on the resonance form, N is predicted to be sp, sp2, or sp3 hybridized. For the remaining compounds, we will give hybrids for the favored resonance structures as predicted from formal charge considerations. NH2
H N
C
H
N
N
C
N
sp3
C NH2
sp
2
sp
sp
3
sp
Melamine: N in NH2 groups are all sp3 hybridized; atoms in ring are all sp2 hybridized. c. NCN2−: 2 σ and 2 π bonds; H2NCN: 4 σ and 2 π bonds; dicyandiamide: 9 σ and 3 π bonds; melamine: 15 σ and 3 π bonds d. The π-system forces the ring to be planar, just as the benzene ring is planar (see Figure 9.48 of the text). e. The structure: N
C
N C N
H
H
N
H
H
best agrees with experiments because it has three different CN bonds. This structure is also favored on the basis of formal charge.
CHAPTER 9 122.
COVALENT BONDING: ORBITALS
425
One of the resonance structures for benzene is: H C
H
H
H
C
C
C
C H
C H
To break C6H6(g) into C(g) and H(g) requires the breaking of 6 C‒H bonds, 3 C=C bonds, and 3 C‒C bonds: C6H6(g) → 6 C(g) + 6 H(g)
ΔH = 6DC‒H + 3DC=C + 3DC‒C
ΔH = 6(413 kJ) + 3(614 kJ) + 3(347 kJ) = 5361 kJ The question asks for H f for C6H6(g), which is ΔH for the reaction: 6 C(s) + 3 H2(g) → C6H6(g)
ΔH = ΔH f , C H ( g ) 6
6
To calculate ΔH for this reaction, we will use Hess’s law along with the value ΔH f for C(g) and the bond energy value for H2 ( D H 2 = 432 kJ/mol). 6 C(g) + 6 H(g) → C6H6(g) 6 C(s) → 6 C(g) 3 H2(g) → 6 H(g)
ΔH1 = −5361 kJ ΔH2 = 6(717 kJ) ΔH3 = 3(432 kJ)
6 C(s) + 3 H2(g) → C6H6(g)
ΔH = ΔH1 + ΔH2 + ΔH3 = 237 kJ; ΔH f , C H ( g ) = 237 kJ/mol 6
6
The experimental ΔH of for C6H6(g) is more stable (lower in energy) by 154 kJ than the ΔH of calculated from bond energies (83 − 237 = −154 kJ). This extra stability is related to benzene’s ability to exhibit resonance. Two equivalent Lewis structures can be drawn for benzene. The π bonding system implied by each Lewis structure consists of three localized π bonds. This is not correct because all C‒C bonds in benzene are equivalent. We say the π electrons in benzene are delocalized over the entire surface of C6H6 (see Section 9.5 of the text). The large discrepancy between ΔH of values is due to the delocalized π electrons, whose effect was not accounted for in the calculated ΔH of value. The extra stability associated with benzene can be called resonance stabilization. In general, molecules that exhibit resonance are usually more stable than predicted using bond energies.
426 123.
CHAPTER 9 a. E =
COVALENT BONDING: ORBITALS
hc (6.626 10 −34 J s)(2.998 108 m / s) = 7.9 × 10−18 J = λ 25 10 −9 m
7.9 × 10−18 J ×
6.022 10 23 1 kJ = 4800 kJ/mol mol 1000 J
Using ΔH values from the various reactions, 25-nm light has sufficient energy to ionize N2 and N and to break the triple bond. Thus N2, N2+, N, and N+ will all be present, assuming excess N2. b. To produce atomic nitrogen but no ions, the range of energies of the light must be from 941 kJ/mol to just below 1402 kJ/mol. 941 kJ 1 mol 1000 J = 1.56 × 10−18 J/photon 23 mol 1 kJ 6.022 10
λ =
hc (6.6261 10 −34 J s)(2.998 108 m / s) = 1.27 × 10−7 m = 127 nm = E 1.56 10 −18 J
1402 kJ 1 mol 1000 J = 2.328 × 10−18 J/photon 23 mol kJ 6.0221 10
λ =
hc (6.6261 10 −34 J s)(2.9979 108 m / s) = 8.533 × 10−8 m = 85.33 nm = E 2.328 10 −18 J
Light with wavelengths in the range of 85.33 nm < λ < 127 nm will produce N but no ions. c. N2: (σ2s)2(σ2s*)2(π2p)4(σ2p)2; the electron removed from N2 is in the σ2p molecular orbital, which is lower in energy than the 2p atomic orbital from which the electron in atomic nitrogen is removed. Because the electron removed from N2 is lower in energy than the electron removed from N, the ionization energy of N2 is greater than that for N. 124.
NO43− 3-
O N O
O O
Both NO43− and PO43− have 32 valence electrons so both have similar Lewis structures. From the Lewis structure for NO43−, the central N atom has a tetrahedral arrangement of electron pairs. N is small. There is probably not enough room for all four oxygen atoms around N. P is larger; thus PO43− is stable.
PO3− O P O
O
PO3− and NO3− both have 24 valence electrons, so both have similar Lewis structures. From the Lewis structure, PO3− has a trigonal arrangement of electron pairs about the central P atom (two single bonds and one double bond). P=O bonds are not particularly stable, whereas N=O bonds are stable. Thus NO3− is stable.
CHAPTER 9 125.
COVALENT BONDING: ORBITALS
427
O=N‒Cl: The bond order of the NO bond in NOCl is 2 (a double bond). NO: From molecular orbital theory, the bond order of this NO bond is 2.5. (See Figure 9.40 of the text. ) Both reactions apparently involve only the breaking of the N‒Cl bond. However, in the reaction ONCl → NO + Cl, some energy is released in forming the stronger NO bond, lowering the value of ΔH. Therefore, the apparent N‒Cl bond energy is artificially low for this reaction. The first reaction involves only the breaking of the N‒Cl bond.
126.
The molecular orbitals for BeH2 are formed from the two hydrogen 1s orbitals and the 2s and one of the 2p orbitals from beryllium. One of the sigma bonding orbitals forms from overlap of the hydrogen 1s orbitals with a 2s orbital from beryllium. Assuming the z axis is the internuclear axis in the linear BeH2 molecule, then the 2pz orbital from beryllium has proper symmetry to overlap with the 1s orbitals from hydrogen; the 2px and 2py orbitals are nonbonding orbitals since they don’t have proper symmetry necessary to overlap with 1s orbitals. The type of bond formed from the 2pz and 1s orbitals is a sigma bond since the orbitals overlap head to head. The MO diagram for BeH2 is: Be
2H
*s *p 2px
2p 2s
2py 1s
1s
p
s Bond order = (4 − 0)/2 = 2; the MO diagram predicts BeH2 to be a stable species and also predicts that BeH2 is diamagnetic. Note: The σs MO is a mixture of the two hydrogen 1s orbitals with the 2s orbital from beryllium, and the σp MO is a mixture of the two hydrogen 1s orbitals with the 2pz orbital from beryllium. The MOs are not localized between any two atoms; instead, they extend over the entire surface of the three atoms. 127.
The ground state MO electron configuration for He2 is (1s)2(1s*)2, giving a bond order of 0. Therefore, He2 molecules are not predicted to be stable (and are not stable) in the lowest- energy ground state. However, in a high-energy environment, electron(s) from the anti-bonding orbitals in He2 can be promoted into higher-energy bonding orbitals, thus giving a nonzero
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bond order and a “reason” to form. For example, a possible excited-state MO electron configuration for He2 would be (1s)2(1s*)1(2s)1, giving a bond order of (3 – 1)/2 = 1. Thus excited He2 molecules can form, but they spontaneously break apart as the electron(s) fall back to the ground state, where the bond order equals zero. 128.
2p
2s
O2
O2−
O2+
O
The order from lowest IE to highest IE is: O2− < O2 < O2+ < O. The electrons for O2−, O2, and O2+ that are highest in energy are in the π *2 p MOs. But for O2−, these electrons are paired. O2− should have the lowest ionization energy (its paired π *2 p electron is easiest to remove). The species O2+ has an overall positive charge, making it harder to remove an electron from O2+ than from O2. The highest-energy electrons for O (in the 2p atomic orbitals) are lower in energy than the π *2 p electrons for the other species; O will have the highest ionization energy because it requires a larger quantity of energy to remove an electron from O as compared to the other species. 129.
The electron configurations are: N2: (σ2s)2(σ2s*)2(π2p)4(2p)2 O2: (σ2s)2(σ2s*)2(σ2p)2(π2p)4(π2p*)2
Note: The ordering of the 2p and 2p orbitals is not important to this question.
N22−: (σ2s)2(σ2s*)2(2p)4(σ2p)2(π2p*)2 N2−: (σ2s)2(σ2s*)2(2p)4(σ2p)2(π2p*)1 O2+: (σ2s)2(σ2s*)2(σ2p)2(π2p)4(π2p*)1 The species with the smallest ionization energy has the electron that is easiest to remove. From the MO electron configurations, O2, N22−, N2−, and O2+ all contain electrons in the same higherenergy antibonding orbitals (π*2p ) , so they should have electrons that are easier to remove as compared to N2, which has no π *2 p electrons. To differentiate which has the easiest π *2 p to remove, concentrate on the number of electrons in the orbitals attracted to the number of protons in the nucleus.
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429
N22− and N2− both have 14 protons in the two nuclei combined. Because N22− has more electrons, one would expect N22− to have more electron repulsions, which translates into having an easier electron to remove. Between O2 and O2+, the electron in O2 should be easier to remove. O2 has one more electron than O2+, and one would expect the fewer electrons in O2+ to be better attracted to the nuclei (and harder to remove). Between N22− and O2, both have 16 electrons; the difference is the number of protons in the nucleus. Because N22− has two fewer protons than O2, one would expect the N22− to have the easiest electron to remove, which translates into the smallest ionization energy. 130.
a. F2−(g) → F(g) + F−(g) H = F2− bond energy Using Hess’s law: F2−(g) → F2(g) + e− F2(g) → 2 F(g) F(g) + e− → F−(g) F2−(g) → F(g) + F−(g)
H = 290. kJ (ionization energy for F2−) H = 154 kJ (bond energy for F2 from Table 8.5) H = –327.8 kJ (electron affinity for F from Table 7.7) H = 116 kJ; bond energy for F2− = 116 kJ/mol
Note that F2− has a smaller bond energy than F2. b. F2: (σ2s)2(σ2s*)2(σ2p)2(π2p)4(π2p*)4 F2−: (σ2s)2(σ2s*)2(σ2p)2(π2p)4(π2p*)4(2p*)1
B.O. = (8 – 6)/2 = 1 B.O. = (8 – 7)/2 = 0.5
MO theory predicts that F2 should have a stronger bond than F2− because F2 has the larger bond order. As determined in part a, F2 indeed has a stronger bond because the F2 bond energy (154 kJ/mol) is greater than the F2− bond energy (116 kJ/mol). 131.
a. The CO bond is polar with the negative end at the more electronegative oxygen atom. We would expect metal cations to be attracted to and bond to the oxygen end of CO on the basis of electronegativity. b.
FC (carbon) = 4 − 2 − 1/2(6) = −1 FC (oxygen) = 6 − 2 − 1/2(6) = +1 From formal charge, we would expect metal cations to bond to the carbon (with the negative formal charge).
c. In molecular orbital theory, only orbitals with proper symmetry overlap to form bonding orbitals. The metals that form bonds to CO are usually transition metals, all of which have outer electrons in the d orbitals. The only molecular orbitals of CO that have proper symmetry to overlap with d orbitals are the π2p* orbitals, whose shape is similar to the d orbitals. Because the antibonding molecular orbitals have more carbon character (carbon is less electronegative than oxygen), one would expect the bond to form through carbon.
CHAPTER 10 LIQUIDS AND SOLIDS Review Questions 1.
Intermolecular forces are the relatively weak forces between molecules that hold the molecules together in the solid and liquid phases. Intramolecular forces are the forces within a molecule. These are the covalent bonds in a molecule. Intramolecular forces (covalent bonds) are much stronger than intermolecular forces. Dipole forces are the forces that act between polar molecules. The electrostatic attraction between the positive end of one polar molecule and the negative end of another is the dipole force. Dipole forces are generally weaker than hydrogen bonding. Both forces are due to dipole moments in molecules. Hydrogen bonding is given a separate name from dipole forces because hydrogen bonding is a particularly strong dipole force. Any neutral molecule that has a hydrogen covalently bonded to N, O, or F exhibits the relatively strong hydrogen bonding intermolecular forces. London dispersion forces are accidental-induced dipole forces. Like dipole forces, London dispersion forces are electrostatic in nature. Dipole forces are the electrostatic forces between molecules having a permanent dipole. London dispersion forces are the electrostatic forces between molecules having an accidental or induced dipole. All covalent molecules (polar and nonpolar) have London dispersion forces, but only polar molecules (those with permanent dipoles) exhibit dipole forces. As the size of a molecule increases, the strength of the London dispersion forces increases. This is because, as the electron cloud about a molecule gets larger, it is easier for the electrons to be drawn away from the nucleus. The molecule is said to be more polarizable. London dispersion (LD) < dipole-dipole < H bonding < metallic bonding, covalent network, ionic. Yes, there is considerable overlap. Consider some of the examples in Exercise 10.136 of the text. Benzene (only LD forces) has a higher boiling point than acetone (dipole-dipole forces). Also, there is even more overlap among the stronger forces (metallic, covalent, and ionic).
2.
a. Surface tension: The resistance of a liquid to an increase in its surface area. b. Viscosity: The resistance of a liquid to flow. c. Melting point: The temperature (at constant pressure) where a solid converts entirely to a liquid if heat is applied. A more detailed definition is the temperature at which the solid and liquid states have the same vapor pressure under conditions where the total pressure is constant.
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d. Boiling point: The temperature (at constant pressure) where a liquid converts entirely to a gas if heat is applied. The detailed definition is the temperature at which the vapor pressure of the liquid is exactly equal to the external pressure. e. Vapor pressure: The pressure of the vapor over a liquid at equilibrium. As the strengths of intermolecular forces increase, surface tension, viscosity, melting point and boiling point increase, while vapor pressure decreases. 3.
Solid: Rigid; has fixed volume and shape; slightly compressible Liquid: Definite volume but no specific shape; assumes shape of the container; slightly compressible Gas: No fixed volume or shape; easily compressible
4.
a. Crystalline solid:
Regular, repeating structure
Amorphous solid: Irregular arrangement of atoms or molecules b. Ionic solid: Molecular solid:
c. Molecular solid:
Made up of discrete covalently bonded molecules held together in the solid phase by weaker forces (LD, dipole, or hydrogen bonds). Discrete, individual molecules
Network solid:
No discrete molecules; A network solid is one large molecule. The forces holding the atoms together are the covalent bonds between atoms.
d. Metallic solid:
Completely delocalized electrons, conductor of electricity (cations in a sea of electrons)
Network solid: 5.
Made up of ions held together by ionic bonding
Localized electrons; Insulator or semiconductor
Lattice: A three-dimensional system of points designating the positions of the centers of the components of a solid (atoms, ions, or molecules). Unit cell: The smallest repeating unit of a lattice. A simple cubic unit cell has an atom, ion or molecule located at the eight corners of a cube. There is one net atom per simple cubic unit cell. Because the atoms in the cubic unit cell are assumed to touch along the cube edge, cube edge = d = 2r where r = radius of the atom. A body-centered cubic unit cell has an atom, ion or molecule at the eight corners of a cube and one atom, ion, or molecule located at the center of the cube. There are two net atoms per bodycentered cubic unit cell. Because the atoms in the cubic unit cell are assumed to touch along the body diagonal of the cube, body diagonal = 3 d = 4r where d = cube edge and r = radius of atom. A face-centered cubic unit cell has an atom, ion, or molecule at the eight corners of a cube and an atom, ion, or molecule located at the six faces of the cube. There are four net atoms
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per face-centered unit cell. Because the atoms in the cubic unit cell are assumed to touch along the face diagonal of the cube, face diagonal = 2 d = 4r. 6.
Closest packing: the packing of atoms (uniform, hard spheres) in a manner that most efficiently uses the available space with the least amount of empty space. The two types of closest packing are hexagonal closest packing and cubic closest packing. In both closest packed arrangements, the atoms (spheres) are packed in layers. The difference between the two closest packed arrangements is the ordering of the layers. Hexagonal closest packing has the third layer directly over the first layer forming a repeating layer pattern of abab… In cubic closest packing the layer pattern is abcabc… The unit cell for hexagonal closest packing is a hexagonal prism. See Figure 10.15 of the text for an illustration of the hexagonal prism unit cell. The unit cell for cubic closest packing is the face-centered cubic unit cell.
7.
Conductor:
The energy difference between the filled and unfilled molecular orbitals is minimal. We call this energy difference the band gap. Because the band gap is minimal, electrons can easily move into the conduction bands (the unfilled molecular orbitals).
Insulator:
Large band gap; electrons do not move from the filled molecular orbitals to the conduction bands since the energy difference is large.
Semiconductor: Small band gap; the energy difference between the filled and unfilled molecular orbitals is smaller than in insulators, so some electrons can jump into the conduction bands. The band gap, however, is not as small as with conductors, so semiconductors have intermediate conductivity. a. As the temperature is increased, more electrons in the filled molecular orbitals have sufficient kinetic energy to jump into the conduction bands (the unfilled molecular orbitals). b. A photon of light is absorbed by an electron which then has sufficient energy to jump into the conduction bands. c. An impurity either adds electrons at an energy near that of the conduction bands (n-type) or creates holes (unfilled energy levels) at energies in the previously filled molecular orbitals (p-type). Both n-type and p-type semiconductors increase conductivity by creating an easier path for electrons to jump from filled to unfilled energy levels. In conductors, electrical conductivity is inversely proportional to temperature. Increases in temperature increase the motions of the atoms, which gives rise to increased resistance (decreased conductivity). In a semiconductor, electrical conductivity is directly proportional to temperature. An increase in temperature provides more electrons with enough kinetic energy to jump from the filled molecular orbitals to the conduction bands, increasing conductivity. To produce an n-type semiconductor, dope Ge with a substance that has more than 4 valence electrons, e.g., a group 5A element. Phosphorus or arsenic are two substances which will produce n-type semiconductors when they are doped into germanium. To produce a p-type semiconductor, dope Ge with a substance that has fewer than 4 valence electrons, e.g., a group 3A element. Gallium or indium are two substances which will produce p-type semi-conductors when they are doped into germanium.
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The structures of most binary ionic solids can be explained by the closest packing of spheres. Typically, the larger ions, usually the anions, are packed in one of the closest packing arrangements, and the smaller cations fit into holes among the closest packed anions. There are different types of holes within the closest packed anions which are determined by the number of spheres that form them. Which of the three types of holes are filled usually depends on the relative size of the cation to the anion. Ionic solids will always try to maximize electrostatic attractions among oppositely charged ions and minimize the repulsions among ions with like charges. The structure of sodium chloride can be described in terms of a cubic closest packed array of Cl− ions with Na+ ions in all of the octahedral holes. An octahedral hole is formed between 6 Cl− anions. The number of octahedral holes is the same as the number of packed ions. So in the face-centered unit cell of sodium chloride, there are 4 net Cl− ions and 4 net octahedral holes. Because the stoichiometry dictates a 1:1 ratio between the number of Cl − anions and Na+ cations, all of the octahedral holes must be filled with Na+ ions. In zinc sulfide, the sulfide anions also occupy the lattice points of a cubic closest packing arrangement. But instead of having the cations in octahedral holes, the Zn2+ cations occupy tetrahedral holes. A tetrahedral hole is the empty space created when four spheres are packed together. There are twice as many tetrahedral holes as packed anions in the closest packed structure. Therefore, each face-centered unit cell of sulfide anions contains 4 net S2− ions and 8 net tetrahedral holes. For the 1:1 stoichiometry to work out, only one-half of the tetrahedral holes are filled with Zn2+ ions. This gives 4 S2− ions and 4 Zn2+ ions per unit cell for an empirical formula of ZnS.
9.
a. Evaporation: Process where liquid molecules escape the liquid’s surface to form a gas. b. Condensation: Process where gas molecules hit the surface of a liquid and convert to a liquid. c. Sublimation: Process where a solid converts directly to a gas without passing through the liquid state. d. Boiling: The temperature and pressure at which a liquid completely converts to a gas as long as heat is applied. e. Melting: Temperature and pressure at which a solid completely converts to a liquid as long as heat is applied. f.
Enthalpy of vaporization (Hvap): The enthalpy change that occurs at the boiling point when a liquid converts into a gas.
g. Enthalpy of fusion (Hfus): The enthalpy change that occurs at the melting point when a solid converts into a liquid. h. Heating curve: A plot of temperature versus time as heat is applied at a constant rate to some substance.
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Fusion refers to a solid converting to a liquid, and vaporization refers to a liquid converting to a gas. Only a fraction of the hydrogen bonds in ice are broken in going from the solid phase to the liquid phase. Most of the hydrogen bonds in water are still present in the liquid phase and must be broken during the liquid to gas phase transition. Thus, the enthalpy of vaporization is much larger than the enthalpy of fusion because more intermolecular forces are broken during the vaporization process. A volatile liquid is one that evaporates relatively easily. Volatile liquids have large vapor pressures because the intermolecular forces that prevent evaporation are relatively weak.
11.
See Figures 10.48 and 10.51 of the text for the phase diagrams of H2O and CO2. Most substances exhibit only three different phases: solid, liquid, and gas. This is true for H2O and CO2. Also typical of phase diagrams is the positive slopes for both the liquid-gas equilibrium line and the solid-gas equilibrium line. This is also true for both H2O and CO2. The solid-liquid equilibrium line also generally has a positive slope. This is true for CO2, but not for H2O. In the H2O phase diagram, the slope of the solid-liquid line is negative. The determining factor for the slope of the solid-liquid line is the relative densities of the solid and liquid phases. The solid phase is denser than the liquid phase in most substances; for these substances, the slope of the solid-liquid equilibrium line is positive. For water, the liquid phase is denser than the solid phase which corresponds to a negative sloping solid-liquid equilibrium line. Another difference between H2O and CO2 is the normal melting points and normal boiling points. The term normal just dictates a pressure of 1 atm. H2O has a normal melting point (0˚C) and a normal boiling point (100˚C), but CO2 does not. At 1 atm pressure, CO2 only sublimes (goes from the solid phase directly to the gas phase). There are no temperatures at 1 atm for CO 2 where the solid and liquid phases are in equilibrium or where the liquid and gas phases are in equilibrium. There are other differences, but those discussed above are the major ones. The relationship between melting points and pressure is determined by the slope of the solidliquid equilibrium line. For most substances (CO2 included), the positive slope of the solidliquid line shows a direct relationship between the melting point and pressure. As pressure increases, the melting point increases. Water is just the opposite since the slope of the solidliquid line in water is negative. Here the melting point of water is inversely related to the pressure. For boiling points, the positive slope of the liquid-gas equilibrium line indicates a direct relationship between the boiling point and pressure. This direct relationship is true for all substances including H2O and CO2. The critical temperature for a substance is defined as the temperature above which the vapor cannot be liquefied no matter what pressure is applied. The critical temperature, like the boiling point temperature, is directly related to the strength of the intermolecular forces. Since H2O exhibits relatively strong hydrogen bonding interactions and CO2 only exhibits London dispersion forces, one would expect a higher critical temperature for H2O than for CO2.
Active Learning Questions 1.
Water exhibits relatively strong hydrogen bonding intermolecular forces. The forces are strong enough to support a paperclip on the surface without breaking. However, soap weakens these forces and soapy water is not able to support a paperclip on the surface.
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2.
Answer d is best. The evaporation/condensation cycle is continuous. When the rate that water evaporates equals the rate that water condenses, there is a steady amount of water vapor leading to a specific vapor pressure. Even when the specific vapor pressure is reached at the designated temperature, the evaporation/condensation cycle continues, just at equal rates.
3.
In covalent compounds, the individual molecules pack together to form the substance in the condensed phases. For example, H2O molecules pack together to form the ice structure. In ionic solids, there is an extended lattice of positive and negative charged ions packed in an orderly fashion; there aren’t individual NaCl units. In diamond, each carbon is bound to four other carbon atoms to form one huge molecule. Diamond isn’t a bunch of singular carbon atoms packed together.
4.
Yes; naphthalene (C10H8), which is composed of only nonpolar bonds and exhibits only London dispersion forces, is a solid at room temperature. Whereas water, which exhibits hydrogen bonding, is a liquid at room temperature. Here the dispersion force in naphthalene are stronger than the hydrogen bonding intermolecular forces in water. When nonpolar compounds become large, the relatively weak dispersion forces can add up to be very significant in strength.
5.
The nature of the forces stays the same. It is the amount of intermolecular forces that differ in the various phases. Solids have the largest amount of intermolecular forces. To convert a solid into a liquid requires breaking about 15-20% of the intermolecular forces; but the types of forces are still the same. To go from a liquid to a gas requires breaking the remaining intermolecular forces in the liquid phase. In the gas phase, no intermolecular forces are assumed to exist.
6.
At any temperature, the molecules in a liquid have a range of kinetic energies. Some of the high energy molecules in the liquid can be on the surface. If these molecules have enough energy to break the intermolecular forces, they do so and go into the vapor phase. The molecules in a solid phase also have a range of kinetic energies at any temperature. Similarly, solids also have a vapor pressure as some the high energy molecules can be on the surface of the solid and can escape to the vapor phase. In each case, when rate of the molecules leaving the surface of the liquid or solid equal the rate that molecules return from the vapor phase, the pressure of the vapor above the surface is defined as the vapor pressure. As temperature rises, the kinetic energy distribution shifts so that the molecules have higher energy in general. So as temperature increases, more molecules now have the required energy necessary to break the intermolecular forces holding them in the liquid (or solid) phase and vapor pressure increases.
7.
The vapor pressure above a liquid is dictated by the external pressure. If the external pressure is constant, then the vapor pressure above a liquid can never get above the external pressure. So, if the vapor pressure above a water is less than the external pressure, then water will evaporate to try to reach the external pressure. In an open beaker, the vapor above a liquid is free to move away from the surface of the liquid. Hence, water in an open beaker generally all evaporates as it tries to reach a vapor pressure equal to the external pressure.
8.
Assuming 1 atm external pressure, water at 100C has a vapor pressure equal to the external pressure (1 atm) as it boils. All liquids boil when the external pressure equals the vapor pressure.
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9.
The plateaus in the heating curve occur when a phase change is occurring. In a phase change, some intermolecular forces must be broken. At the phase change temperature and pressure, all added heat goes to break the intermolecular forces required for the phase change to occur instead of increasing the temperature.
10.
The intramolecular forces (bonds) are stronger than the intermolecular forces. When water boils on a stove, the added energy from the burner breaks intermolecular forces in the liquid as water is converted into a gas. If the intramolecular forces were weaker than the intermolecular forces, then heat added to water on a stove would break the covalent O-H bonds in water. As these bonds are broken, water would convert to H2 and O2 molecules; a chemical reaction would occur instead of a physical change. This doesn’t happen.
11.
At any temperature, the molecules in a liquid have a range of kinetic energies. Some of the high energy molecules in the liquid can be on the surface. If these molecules have enough energy to break the intermolecular forces, they do so and go into the vapor phase. Hence, liquids evaporate at any temperature because some of the surface molecules have energy enough to break the intermolecular forces holding them in the liquid phase.
12.
A liquid is held together by intermolecular forces. At room temperature for N2, there is enough energy available to break all the intermolecular forces that could potentially form between N2 molecules. So, N2 is a gas at room temperature. As the temperature is lowered, the molecules move slower and have less energy. At a low enough temperature, the molecules lose enough energy that the intermolecular forces are strong enough to hold the N2 molecules together in the liquid phase.
13.
Phosphorus exists as covalent P4 molecules in nature, not as separate P atoms. Sulfur exists as S8 molecules in nature, not as separate S atoms. In diamond, the carbon atoms are covalently bonded to other carbon atoms to form one big extended structure. Diamond does not exist in nature as individual C atoms.
14.
In the gas phase, the induvial particles that make up a substance are very far apart from each other. For polar substances, the individual polar molecules are so far apart that they don’t interact with each other. A gas is assumed to have no intermolecular forces because the molecules are so far apart.
15.
The evaporation of water is the process of converting H2O(l) into H2O(g). This is an endothermic process that requires energy to break the intermolecular forces in the liquid phase. The energy for this process comes from the surroundings. As heat is removed from the surroundings, the molecules in the surroundings slow down resulting in a temperature decrease of the surroundings.
16.
a. Four layers are required. The arrangement of the four layers is:
Layer 1
Layer 2
Layer 3
Layer 4
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A total of 20 cannon balls will be needed. b. The layering alternates abcabc, which is cubic closest packing. c. tetrahedron 17.
Water boils when the vapor pressure above the water equals the external pressure. As the flask is cooled, the pressure above the water is lowered. If lowered enough, the vapor pressure equals the lowered external pressure and water boils again. In general, the boiling point of a liquid decreases as the external pressure decreases.
Questions 18.
Since the boiling point of water is less than the boiling point of C25H52, the nonpolar hydrocarbon has the stronger intermolecular forces. Because C25H52 is a nonpolar compound, it only exhibits London dispersion forces. We generally consider these forces to be weak. However, with compounds that have large molar masses such as C25H52, these London dispersion forces add up significantly and can overtake the relatively strong hydrogen-bonding interactions in water. This is the case here.
19.
Answer a is correct. Intermolecular forces are the forces between molecules that hold the substances together in the solid and liquid phases. Hydrogen bonding is a specific type of intermolecular forces. In this figure, the dotted lines represent the hydrogen bonding interactions that hold individual H2O molecules together in the solid and liquid phases. The solid lines represent the O−H covalent bonds.
20.
Hydrogen bonding occurs when hydrogen atoms are covalently bonded to highly electronegative atoms such as oxygen, nitrogen, or fluorine. Because the electronegativity difference between hydrogen and these highly electronegative atoms is relatively large, the N−H, O−H, and F−H bonds are very polar covalent bonds. This leads to strong dipole forces. Also, the small size of the hydrogen atom allows the dipoles to approach each other more closely than can occur between most polar molecules. Both of these factors make hydrogen bonding a special type of dipole interaction.
21.
The halogens are all nonpolar substances, and nonpolar substances only exhibit London dispersion forces. In general, the larger the molar mass, the stronger the LD forces. I2 has the largest molar mass, so it has the highest boiling point.
22.
Statement a is the best explanation. HCl and Ar have about the same molar mass (36.5 vs 40 g/mol), so their London dispersion forces should be about the same strength. Ar is a nonpolar substance that only exhibits LD forces. HCl is a polar compound that exhibits additional dipole forces. HCl has the stronger intermolecular forces and the higher boiling point.
23.
Ideal gas molecules are assumed to exhibit no intermolecular forces and are assumed to be volumeless. Real gases deviate from ideal gas behavior because real gases do exhibit intermolecular forces and real gas molecules do have a volume. Between two gases, the gas with the weaker intermolecular forces and the smaller size should behave more ideally at some conditions. Both N2 and CO have similar size (molar mass), but the nonpolar N2 molecules
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only exhibit London dispersion forces while the polar CO molecules exhibit additional dipole force. N2 will have the weaker intermolecular forces and should behave more ideally than CO. 24.
Doping silicon involves the replacing of a small fraction of the silicon atoms with other atoms. These other atoms (dopants) either add electrons at an energy near that of the conduction band in silicon (n-type) or creates holes (unfilled energy levels) at energies in the previously filled molecular orbitals in silicon (p-type). Both n-type and p-type semiconductors increase conductivity by creating an easier path for electrons to jump from filled to unfilled energy levels.
25.
Atoms have an approximately spherical shape (on average). It is impossible to pack spheres together without some empty space among the spheres.
26.
A simple cubic unit cell has 1 net atom per unit cell, a body-centered cubic unit cell has 2 net atoms per unit cell, and a face-centered unit cell has 4 net atoms per unit cell. Cubic closest packing is one of the most efficient ways of packing atoms; the unit cell in cubic closest packing is the face-centered unit cell. Of the three cubic cells, face-centered cubic has the smallest amount of empty space per unit cell.
27.
Atoms touch along the edge in a simple cubic unit cell, so Po has a simple cubic unit cell. Atoms touch along the face diagonal in a face-centered unit cell, so Al has a face-centered unit cell. Atoms touch along the body-diagonal in a body-centered unit cell, so Na has a bodycentered unit cell.
28.
Critical temperature: The temperature above which a liquid cannot exist, i.e., the gas cannot be liquefied by increased pressure. Critical pressure: The pressure that must be applied to a substance at its critical temperature to produce a liquid.
The kinetic energy distribution changes as one raises the temperature (T4 > Tc > T3 > T2 > T1). At the critical temperature Tc, all molecules have kinetic energies greater than the intermolecular forces F, and a liquid can't form. Note: The various temperature distributions shown in the plot are not to scale. The area under each temperature distribution should be equal (area = total number of molecules). 29.
An alloy is a substance that contains a mixture of elements and has metallic properties. In a substitutional alloy, some of the host metal atoms are replaced by other metal atoms of similar size, e.g., brass, pewter, plumber’s solder. An interstitial alloy is formed when some of the interstices (holes) in the closest packed metal structure are occupied by smaller atoms, e.g., carbon steels.
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Equilibrium: There is no change in composition; the vapor pressure is constant. Dynamic: Two processes, vapor → liquid and liquid → vapor, are both occurring but with equal rates, so the composition of the vapor is constant.
31.
a. As the strength of the intermolecular forces increase, the rate of evaporation decreases. b. As temperature increases, the rate of evaporation increases. c. As surface area increases, the rate of evaporation increases.
32.
H2O(l) → H2O(g) H = 44 kJ/mol; the vaporization of water is an endothermic process. To evaporate, water must absorb heat from the surroundings. In this case, part of the surroundings is our body. So, as water evaporates, our body supplies heat, and as a result, our body temperature can cool down. From Le Châtelier’s principle, the less water vapor in the air, the more favorable the evaporation process. Thus, the less humid the surroundings, the more favorably water converts into vapor, and the more heat that is lost by our bodies.
33.
C2H5OH(l) → C2H5OH(g) is an endothermic process. Heat is absorbed when liquid ethanol vaporizes; the internal heat from the body provides this heat, which results in the cooling of the body.
34.
The phase change H2O(g) → H2O(l) releases heat that can cause additional damage. Also, steam can be at a temperature greater than 100°C.
35.
Sublimation will occur, allowing water to escape as H2O(g).
36.
Water boils when the vapor pressure equals the external pressure. Because the external pressure is significantly lower at high altitudes, a lower temperature is required to equalize the vapor pressure of water to the external pressure. Thus, food cooked in boiling water at high elevations cooks at a lower temperature, so it takes longer.
37.
The strength of intermolecular forces determines relative boiling points. The types of intermolecular forces for covalent compounds are London dispersion forces, dipole forces, and hydrogen bonding. Because the three compounds are assumed to have similar molar mass and shape, the strength of the London dispersion forces will be about equal among the three compounds. One of the compounds will be nonpolar, so it only has London dispersion forces. The other two compounds will be polar, so they have additional dipole forces and will boil at a higher temperature than the nonpolar compound. One of the polar compounds will have an H covalently bonded to either N, O, or F. This gives rise to the strongest type of covalent intermolecular forces, hydrogen bonding. The compound that hydrogen bonds will have the highest boiling point, whereas the polar compound with no hydrogen bonding will boil at a temperature in the middle of the other compounds.
38.
a. Both forms of carbon are network solids. In diamond, each carbon atom is surrounded by a tetrahedral arrangement of other carbon atoms to form a huge molecule. Each carbon atom is covalently bonded to four other carbon atoms. The structure of graphite is based on layers of carbon atoms arranged in fused sixmembered rings. Each carbon atom in a particular layer of graphite is surrounded by three other carbons in a trigonal planar arrangement. This requires sp2 hybridization. Each carbon
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has an unhybridized p atomic orbital; all of these p orbitals in each six-membered ring overlap with each other to form a delocalized electron system. b. Silica is a network solid having an empirical formula of SiO2. The silicon atoms are singly bonded to four oxygens. Each silicon atom is at the center of a tetrahedral arrangement of oxygen atoms that are shared with other silicon atoms. The structure of silica is based on a network of SiO4 tetrahedra with shared oxygen atoms rather than discrete SiO2 molecules. Silicates closely resemble silica. The structure is based on interconnected SiO4 tetrahedra. However, in contrast to silica, where the O/Si ratio is 2 : 1, silicates have O/Si ratios greater than 2 : 1 and contain silicon-oxygen anions. To form a neutral solid silicate, metal cations are needed to balance the charge. In other words, silicates are salts containing metal cations and polyatomic silicon-oxygen anions. When silica is heated above its melting point and cooled rapidly, an amorphous (disordered) solid called glass results. Glass more closely resembles a very viscous solution than it does a crystalline solid. To affect the properties of glass, several different additives are thrown into the mixture. Some of these additives are Na2CO3, B2O3, and K2O, with each compound serving a specific purpose relating to the properties of glass. 39.
a. Both CO2 and H2O are molecular solids. Both have an ordered array of the individual molecules, with the molecular units occupying the lattice points. A difference within each solid lattice is the strength of the intermolecular forces. CO2 is nonpolar and only exhibits London dispersion forces. H2O exhibits the relatively strong hydrogen-bonding interactions. The differences in strength is evidenced by the solid-phase changes that occur at 1 atm. CO2(s) sublimes at a relatively low temperature of −78˚C. In sublimation, all of the intermolecular forces are broken. However, H2O(s) doesn’t have a phase change until 0˚C, and in this phase change from ice to water, only a fraction of the intermolecular forces are broken. The higher temperature and the fact that only a portion of the intermolecular forces are broken are attributed to the strength of the intermolecular forces in H2O(s) as compared to CO2(s). Related to the intermolecular forces are the relative densities of the solid and liquid phases for these two compounds. CO2(s) is denser than CO2(l), whereas H2O(s) is less dense than H2O(l). For CO2(s) and for most solids, the molecules pack together as close as possible; hence solids are usually more dense than the liquid phase. H2O is an exception to this. Water molecules are particularly well suited for hydrogen bonding interaction with each other because each molecule has two polar O−H bonds and two lone pairs on the oxygen. This can lead to the association of four hydrogen atoms with each oxygen atom: two by covalent bonds and two by dipoles. To keep this arrangement (which maximizes the hydrogen-bonding interactions), the H2O(s) molecules occupy positions that create empty space in the lattice. This translates into a smaller density for H2O(s) as compared to H2O(l). b. Both NaCl and CsCl are ionic compounds with the anions at the lattice points of the unit cells and the cations occupying the empty spaces created by anions (called holes). In NaCl, the Cl− anions occupy the lattice points of a face-centered unit cell, with the Na+ cations occupying the octahedral holes. Octahedral holes are the empty spaces created by six Cl− ions. CsCl has the Cl− ions at the lattice points of a simple cubic unit cell, with the Cs+ cations occupying the middle of the cube.
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40.
Because silicon carbide is made from Group 4A elements, and because it is extremely hard, one would expect SiC to form a covalent network structure similar to diamond.
41.
Chalk is composed of the ionic compound calcium carbonate (CaCO3). The electrostatic forces in ionic compounds are much stronger than the intermolecular forces in covalent compounds. Therefore, CaCO3 should have a much higher boiling point than the covalent compounds found in motor oil and in H2O. Motor oil is composed of nonpolar C−C and C−H bonds. The intermolecular forces in motor oil are therefore London dispersion forces. We generally consider these forces to be weak. However, with compounds that have large molar masses, these London dispersion forces add up significantly and can overtake the relatively strong hydrogen-bonding interactions in water.
42.
The interparticle forces in ionic solids (the ionic bonds) are much stronger than the interparticle forces in molecular solids (dipole forces, London forces, etc.). The difference in intermolecular forces is most clearly shown in the huge difference in melting points between ionic and molecular solids. Table salt and ordinary sugar are both crystalline solids at room temperature that look very similar to each other. However, sugar can be melted easily in a saucepan during the making of candy, whereas the full heat of a stove will not melt salt. When a substance melts, some interparticle forces must be broken. Ionic solids (salt) require a much larger amount of energy to break the interparticle forces as compared to the relatively weak forces in molecular solids (sugar).
43.
The mathematical equation that relates the vapor pressure of a substance to temperature is: ln Pvap = − y
ΔH vap 1 +C R T m x +b
This equation is in the form of the straight-line equation (y = mx + b) If one plots ln Pvap versus 1/T with temperature in Kelvin, the slope (m) of the straight line is −Hvap/R. Because Hvap is always positive, the slope of the straight line will be negative. 44.
The typical phase diagram for a substance shows three phases and has a positive-sloping solidliquid equilibrium line (water is atypical). A sketch of the phase diagram for I2 would look like this: s l g
P 90 torr
115o C T
Statements a and e are true. For statement a, the liquid phase is always more dense than the gaseous phase (gases are mostly empty space). For statement e, because the triple point is at 90
442
CHAPTER 10
LIQUIDS AND SOLIDS
torr, the liquid phase cannot exist at any pressure less than 90 torr, no matter what the temperature. For statements b, c, and d, examine the phase diagram to prove to yourself that they are false.
Exercises Intermolecular Forces and Physical Properties 45.
46.
Ionic compounds have ionic forces. Covalent compounds all have London dispersion (LD) forces, whereas polar covalent compounds have dipole forces and/or hydrogen bonding forces. For hydrogen-bonding (H-bonding) forces, the covalent compound must have either a N−H, O−H, or F−H bond in the molecule. a. LD only
b. dipole, LD
d. ionic
e. LD only (CH4 is a nonpolar covalent compound.)
f.
g. ionic
dipole, LD
a. ionic
c. H-bonding, LD
b. dipole, LD (LD = London dispersion)
c. LD only
d. LD only (For all practical purposes, a C ‒ H bond can be considered as a nonpolar bond.) e. ionic 47.
f.
LD only
g. H-bonding, LD
NaF, CaF2, and CrF3 are ionic compounds that exhibit extremely strong ionic forces. HF is a covalent compound which can form hydrogen-bonding interactions. This leaves CF4. The Lewis structure is:
CF4 is a nonpolar covalent compound and nonpolar compounds only exhibit London dispersion forces. Just one of the compounds (CF4) only exhibits LD forces. 48.
These compounds all can form hydrogen-bonding interaction due to the O‒H bond in each compound, so this is a constant. What differs is the molar mass of the compounds. As the number of nonpolar C‒C and C‒H bonds increase (as the molar mass increases), the boiling point increases. This is due to London dispersion forces. As the compound gets bigger, the strength of the LD forces increases, and a higher boiling point results.
49.
a. OCS; OCS is polar and has dipole-dipole forces in addition to London dispersion (LD) forces. All polar molecules have dipole forces. CO2 is nonpolar and only has LD forces. To predict polarity, draw the Lewis structure and deduce whether the individual bond dipoles cancel.
CHAPTER 10
LIQUIDS AND SOLIDS
443
b. SeO2; both SeO2 and SO2 are polar compounds, so they both have dipole forces as well as LD forces. However, SeO2 is a larger molecule, so it would have stronger LD forces. c. H2NCH2CH2NH2; more extensive hydrogen bonding (H-bonding) is possible because two NH2 groups are present. d. H2CO; H2CO is polar, whereas CH3CH3 is nonpolar. H2CO has dipole forces in addition to LD forces. CH3CH3 only has LD forces. e. CH3OH; CH3OH can form relatively strong H-bonding interactions, unlike H2CO. 50.
Only statement c is true. ICl is a polar substance that has additional dipole forces that the nonpolar Br2 does not have. KF is ionic and the rest of the compounds are covalent. Ionic intermolecular forces are always much stronger than the intermolecular forces for covalent compounds. So KF has the strongest intermolecular forces and the highest boiling point.
51.
Ar exists as individual atoms that are held together in the condensed phases by London dispersion forces. The molecule that will have a boiling point closest to Ar will be a nonpolar substance with about the same molar mass as Ar (39.95 g/mol); this same size nonpolar substance will have about equivalent strength of London dispersion forces. Of the choices, only Cl2 (70.90 g/mol) and F2 (38.00 g/mol) are nonpolar. Because F2 has a molar mass closest to that of Ar, one would expect the boiling point of F2 to be close to that of Ar.
52.
Ar is a nonpolar substance that only has London Dispersion forces. C4H12, Cl2, and Br2 are also nonpolar substances that only have LD forces. Because the molar masses of these three compounds are significantly larger than the molar mass of Ar (39.95 g/mol), their boiling points will be greater than the boiling point of Ar. HCl has about the same molar mass as Ar, but it is polar. HCl has additional dipole forces, so it has a greater boiling point than Ar. HBr has a larger molar mass than Ar and it is polar, so HBr also has a boiling point greater than Ar. All the compounds (5 total) have a boiling point greater than that of Ar.
53.
The Lewis structure for CF2H2 is:
To form hydrogen-bonding intermolecular forces, you must have either a N‒H, an O‒H, and/or a H‒F covalent bond in the compound. CF2H2 has a hydrogen and a fluorine in the formula, but the compound does not have a H-F bond. So CF2H2 does not exhibit H-bonding (statement a is false). All the other statements are true. 54.
The substance that is the solid has the strongest intermolecular forces. These are all nonpolar substances that only exhibit London dispersion forces. I2, with the largest molar mass, has the strongest London dispersion forces, so it is the solid.
444 55.
CHAPTER 10
LIQUIDS AND SOLIDS
a. Kr and HBr have similar molar masses, so the London dispersion forces in each substance should be about the same. However, HBr is a polar compound while Kr is nonpolar. HBr has additional dipole forces so it boils at a higher temperature. b. HF is capable of H-bonding; HBr is not. c. LiCl is ionic, and HF is a molecular solid with hydrogen bonding and LD forces. Ionic forces are much stronger than the forces for molecular solids. d. n-Hexane is a larger molecule, so it has stronger LD forces.
56.
The electrostatic potential diagrams indicate that ethanol and acetone are polar substances, and that propane is a nonpolar substance. Ethanol, with the O−H covalent bond, will exhibit relatively strong hydrogen bonding intermolecular forces in addition to London dispersion forces. The polar acetone will exhibit dipole forces in addition to London dispersion forces, and the nonpolar propane will only exhibit London dispersion (LD) forces. Therefore, ethanol (with the H-bonding capacity) should have the highest boiling point, with polar acetone having the next highest boiling point, and the nonpolar propane, with the weakest intermolecular forces, will have the lowest boiling point.
57.
Boiling points and freezing points are assumed directly related to the strength of the intermolecular forces, whereas vapor pressure is inversely related to the strength of the intermolecular forces. a. HBr; HBr is polar, whereas Kr and Cl2 are nonpolar. HBr has dipole forces unlike Kr and Cl2. So HBr has the stronger intermolecular forces and the highest boiling point. b. NaCl; the ionic forces in NaCl are much stronger than the intermolecular forces for molecular substances, so NaCl has the highest melting point. c. I2; all are nonpolar, so the largest molecule (I2) will have the strongest LD (London Dispersion) forces and the lowest vapor pressure. d. N2; nonpolar and smallest, so it has the weakest intermolecular forces. e. CH4; smallest, nonpolar molecule, so it has the weakest LD forces. f.
HF; HF can form relatively strong H-bonding interactions, unlike the others.
g. CH3CH2CH2OH; H-bonding, unlike the others, so it has strongest intermolecular forces. 58.
a. CBr4; largest of these nonpolar molecules, so it has the strongest LD (London Dispersion) forces. b. F2; ionic forces in LiF are much stronger than the molecular forces in F2 and HCl. HCl has dipole forces, whereas the nonpolar F2 does not exhibit these. So F2 has the weakest intermolecular forces and the lowest freezing point. c. CH3CH2OH; can form H-bonding interactions, unlike the other covalent compounds. d. H2O2; the H−O−O−H structure has twice the number of H-bonding sites as compared to HF, so H2O2 has the stronger H-bonding interactions and the greatest viscosity.
CHAPTER 10
LIQUIDS AND SOLIDS
445
e. H2CO; H2CO is polar, so it has dipole forces, unlike the other nonpolar covalent compounds, so H2CO will have the highest enthalpy of vaporization. f.
I2; I2 has only LD forces, whereas CsBr and CaO have much stronger ionic forces. I2 has the weakest intermolecular forces, so it has smallest ΔHfusion.
Properties of Liquids 59.
The attraction of H2O for glass is stronger than the H2O‒H2O attraction. The meniscus is concave to increase the area of contact between glass and H2O. The Hg‒Hg attraction is greater than the Hg‒glass attraction. The meniscus is convex to minimize the Hg‒glass contact.
60.
Water is a polar substance, and wax is a nonpolar substance; they are not attracted to each other. A molecule at the surface of a drop of water is subject to attractions only by water molecules below it and to each side. The effect of this uneven pull on the surface water molecules tends to draw them into the body of the water and causes the droplet to assume the shape that has the minimum surface area, a sphere.
61.
The structure of H2O2 is H‒O‒O‒H, which produces greater hydrogen bonding than in water. Thus the intermolecular forces are stronger in H2O2 than in H2O, resulting in a higher normal boiling point for H2O2 and a lower vapor pressure.
62.
CO2 is a gas at room temperature. As melting point and boiling point increase, the strength of the intermolecular forces also increases. Therefore, the strength of forces is CO2 < CS2 < CSe2. From a structural standpoint, this is expected. All three are linear, nonpolar molecules. Thus only London dispersion forces are present. Because the molecules increase in size from CO 2 < CS2 < CSe2, the strength of the intermolecular forces will increase in the same order.
Structures and Properties of Solids 1 154 pm nλ = = 313 pm = 3.13 × 10 −10 m 2 sin 14.22 2 sin θ
63.
nλ = 2d sin θ, d =
64.
d=
2 154 pm nλ = = 408 pm = 4.08 × 10 −10 m 2 sin 22.20 2 sin θ
65.
=
2 d sin θ 2 1.36 10 −10 m sin 15 .0 o = = 7.04 × 10 −11 m = 0.704 Å = 70.4 pm n 1
66.
nλ 1 2.63 A nλ = 2d sin θ, d = = 4.91 Å = 4.91 × 10−10 m = 491 pm = 2 sin θ 2 sin 15.55 o
o
o
nλ 2 2.63 A = sin θ = = 0.536, θ = 32.4° o 2d 2 4.91 A
67.
A cubic closest packed structure has a face-centered cubic unit cell. In a face-centered cubic unit, there are:
446
CHAPTER 10 8 corners ×
LIQUIDS AND SOLIDS
1 / 8 atom 1 / 2 atom + 6 faces = 4 atoms corner face
The atoms in a face-centered cubic unit cell touch along the face diagonal of the cubic unit cell. Using the Pythagorean formula, where l = length of the face diagonal and r = radius of the atom:
l2 + l2 = (4r)2 2l2 = 16r2 l=r 8
l = r 8 = 197 × 10 −12 m ×
8 = 5.57 × 10 −10 m = 5.57 × 10 −8 cm
Volume of a unit cell = l3 = (5.57 × 10 −8 cm)3 = 1.73 × 10 −22 cm3 Mass of a unit cell = 4 Ca atoms ×
Density = 68.
1 mol Ca 6.022 10
23
atoms
40.08 g Ca = 2.662 × 10 −22 g Ca mol Ca
mass 2.662 10 −22 g = = 1.54 g/cm3 volume 1.73 10 − 22 cm3
From the previous question, cubic closest packing has a face-centered unit cell. There are 4 atoms per unit cell and the relationship between the cube edge length and the radius of the atom is l = r 8 . l = r 8 = 143 × 10 −12 m ×
8 = 4.04 × 10 −10 m = 4.04 × 10 −8 cm
Volume of a unit cell = l3 = (4.04 × 10 −8 cm)3 = 6.59 × 10 −23 cm3 Mass of a unit cell = 4 Al atoms ×
Density = 69.
1 mol Al 26.98 g Al × = 1.792 × 10 −22 g Al 23 6.022 × 10 atoms mol Al
mass 1.792 × 10−22 g = 2.72 g/cm3 = −23 3 volume 6.59 × 10 cm
There are four Ni atoms in each unit cell. For a unit cell:
density =
mass = 6.84 g/cm3 = volume
4 Ni atoms
Solving: l = 3.85 × 10 −8 cm = cube edge length
1 mol Ni 6.022 10 l
3
23
atoms
58.69 g Ni mol Ni
CHAPTER 10
LIQUIDS AND SOLIDS
447 For a face-centered cube: (4r)2 = l 2 + l 2 = 2l 2 r 8 = l, r = l / 8 r = 3.85 × 10 −8 cm / 8 r = 1.36 × 10 −8 cm = 136 pm
70.
There are four Ir atoms in each unit cell. For a unit cell:
density =
mass = 22.6 g/cm3 = volume
4 Ir atoms ×
1 mol Ir 192.2 g Ir × 23 6.022 × 10 atoms mol Ir 3 l
Solving: l = 3.84 × 10 −8 cm = cube edge length From Exercise 67, the relationship between the cube edge length and the radius of the atom is l=r 8. r = l / 8 = 3.84 × 10 −8 cm/ 8 = 1.36 × 10 −8 cm = radius of Ir atom 71.
A face-centered cubic unit cell contains four atoms. For a unit cell: mass of X = volume × density = (4.09 × 10 −8 cm)3 × 10.5 g/cm3 = 7.18 × 10 −22 g mol X = 4 atoms X ×
Molar mass =
72.
1 mol X 6.022 10
7.18 10 −22 g X 6.642 10 − 24 mol X
23
= 6.642 × 10 −24 mol X atoms
= 108 g/mol; the metal is silver (Ag).
For a face-centered unit cell, the radius r of an atom is related to the length of a cube edge l by the equation l = r 8 (see Exercise 53). Radius = r = l / 8 = 392 × 10−12 m / 8 = 1.39 × 10−10 m = 1.39 × 10−8 cm The volume of a unit cell is l3, so the mass of the unknown metal (X) in a unit cell is: volume × density = (3.92 × 10−8 cm)3 ×
21 .45 g X cm
3
= 1.29 × 10−21 g X
Because each face-centered unit cell contains four atoms of X: mol X in unit cell = 4 atoms X
1 mol X 6.022 10
23
atoms X
= 6.642 × 10−24 mol X
448
CHAPTER 10
LIQUIDS AND SOLIDS
Therefore, each unit cell contains 1.29 × 10−21 g X, which is equal to 6.642 × 10−24 mol X. The molar mass of X is: 1.29 10 −21 g X 6.642 10 − 24 mol X
= 194 g/mol
The atomic mass would be 194 u. From the periodic table, the best choice for the metal is platinum. 73.
For a body-centered unit cell, 8 corners ×
1 / 8 Ti + Ti at body center = 2 Ti atoms. corner
All body-centered unit cells have two atoms per unit cell. For a unit cell where l = cube edge length: 1 mol Ti 47.88 g Ti 2 atoms Ti 23 mol Ti 6.022 10 atoms density = 4.50 g/cm3 = 3 l Solving: l = edge length of unit cell = 3.28 × 10 −8 cm = 328 pm Assume Ti atoms just touch along the body diagonal of the cube, so body diagonal = 4 × radius of atoms = 4r. The triangle we need to solve is:
4r
l
l 2
3.28 x 10 -8 cm
(3.28 x 10 -8cm) 2
(4r)2 = (3.28 × 10 −8 cm)2 + [(3.28 × 10 −8 cm) 2 ]2, r = 1.42 × 10 −8 cm = 142 pm For a body-centered unit cell (bcc), the radius of the atom is related to the cube edge length by: 4r = l 3 or l = 4r/ 3 . 74.
From Exercise 73: 16r2 = l2 + 2l2 4r
l 2
l
l = 4r/ 3 = 2.309 r l = 2.309(222 pm) = 513 pm = 5.13 × 10−8 cm
CHAPTER 10
LIQUIDS AND SOLIDS
449
In a body-centered cubic unit cell, there are two atoms per unit cell. For a unit cell:
density = 75.
mass = volume
2 atoms Ba
1 mol Ba 137.3 g Ba 23 mol Ba 3.38 g 6.022 10 atoms = −8 3 (5.13 10 cm) cm3
If gold has a face-centered cubic structure, then there are four atoms per unit cell, and from Exercise 67: 2l2 = 16r2 l = r 8 = (144 pm) 8 = 407 pm l = 407 × 10−12 m = 4.07 × 10−8 cm
4 atoms Au
Density =
1 mol Au 197.0 g Au 23 mol Au 6.022 10 atoms = 19.4 g/cm3 −8 3 ( 4.07 10 cm)
If gold has a body-centered cubic structure, then there are two atoms per unit cell, and from Exercise 73: 16r2 = l2 + 2l2 4r
l = 4r/ 3 = 333 pm = 333 × 10−12 m
l
l = 333 × 10−10 cm = 3.33 × 10−8 cm l 2
2 atoms Au
Density =
1 mol Au 197.0 g Au 23 mol Au 6.022 10 atoms = 17.7 g/cm3 −8 3 (3.33 10 cm)
The measured density of gold is consistent with a face-centered cubic unit cell. 76.
If face-centered cubic: l = r 8 = (137 pm) 8 = 387 pm = 3.87 × 10 −8 cm 4 atoms W
density =
If body-centered cubic:
1 mol
6.022 10
23
atoms
(3.87 10
−8
3
cm )
183 .9 g W mol
= 21.1 g/cm3
450
CHAPTER 10 l=
4r
=
4 137 pm 3
3
2 atoms W
density =
LIQUIDS AND SOLIDS
= 316 pm = 3.16 × 10 −8 cm 1 mol
6.022 10
23
atoms
(3.16 10
−8
3
cm )
183 .9 g W mol
= 19.4 g/cm3
The measured density of tungsten is consistent with a body-centered unit cell. 77.
In a face-centered unit cell (a cubic closest packed structure), the atoms touch along the face diagonal: (4r)2 = l2 + l2 l=r 8 Vcube = l 3 = (r 8 )3 = (22.63)r3
There are four atoms in a face-centered cubic cell (see Exercise 53). Each atom has a volume of (4/3)πr3 = volume of a sphere. Vatoms = 4 × So
4 3
πr3 = (16.76)r3
Vatoms (16.76) r 3 = = 0.7406, or 74.06% of the volume of each unit cell is occupied by Vcube ( 22.63) r 3
atoms. In a simple cubic unit cell, the atoms touch along the cube edge l:
2(radius) = 2r = l
2r
l
Vcube = l3 = (2r)3 = 8r3
There is one atom per simple cubic cell (8 corner atoms × 1/8 atom per corner = 1 atom/unit cell). Each atom has an assumed volume of (4/3)πr3 = volume of a sphere. Vatom =
4 3
πr3 = (4.189)r3
CHAPTER 10 So
Vatom
Vcube atoms.
LIQUIDS AND SOLIDS =
(4.189)r3
451
= 0.5236, or 52.36% of the volume of each unit cell is occupied by
8r 3
A cubic closest packed structure (face-centered cubic unit cell) packs the atoms much more efficiently than a simple cubic structure. 78.
From Exercise 73, a body-centered unit cell contains two net atoms, and the length of a cube edge l is related to the radius of the atom r by the equation l = 4r/ 3 . Volume of unit cell = l3 = (4r/ 3 )3 = (12.32)r3 Volume of atoms in unit cell = 2 ×
So
4 3
πr3 = (8.378)r3
Vatoms (8.378) r 3 = = 0.6800 = 68.00% occupied Vcube (12.32) r 3
To determine the radius of the Fe atoms, we need to determine the cube edge length l.
1 mol Fe
6.022 10 23 atoms
Volume of unit cell = 2 Fe atoms
55 .85 g Fe 1 cm 3 mol Fe 7.86 g
= 2.36 × 10 −23 cm3 Volume = l3 = 2.36 × 10 −23 cm3, l = 2.87 × 10 −8 cm l = 4r/ 3 , r = l 3 / 4 = 2.87 × 10 −8 cm ×
3 / 4 = 1.24 × 10 −8 cm = 124 pm
79.
Doping silicon with phosphorus produces an n-type semiconductor. The phosphorus adds electrons at energies near the conduction band of silicon. Electrons do not need as much energy to move from filled to unfilled energy levels, so conduction increases. Doping silicon with gallium produces a p-type semiconductor. Because gallium has fewer valence electrons than silicon, holes (unfilled energy levels) at energies in the previously filled molecular orbitals are created, which induces greater electron movement (greater conductivity).
80.
A rectifier is a device that produces a current that flows in one direction from an alternating current that flows in both directions. In a p-n junction, a p-type and an n-type semiconductor are connected. The natural flow of electrons in a p-n junction is for the excess electrons in the n-type semiconductor to move to the empty energy levels (holes) of the p-type semiconductor. Only when an external electric potential is connected so that electrons flow in this natural direction will the current flow easily (forward bias). If the external electric potential is connected in reverse of the natural flow of electrons, no current flows through the system (reverse bias). A p-n junction only transmits a current under forward bias, thus converting the alternating current to direct current.
81.
In has fewer valence electrons than Se. Thus Se doped with In would be a p-type semiconductor.
452
CHAPTER 10
LIQUIDS AND SOLIDS
82.
To make a p-type semiconductor, we need to dope the material with atoms that have fewer valence electrons. The average number of valence electrons is four when 50-50 mixtures of Group 3A and Group 5A elements are considered. We could dope with more of the Group 3A element or with atoms of Zn or Cd. Cadmium is the most common impurity used to produce ptype GaAs semiconductors. To make an n-type GaAs semiconductor, dope with an excess Group 5A element or dope with a Group 6A element such as sulfur.
83.
Egap = 2.5 eV × 1.6 × 10 −19 J/eV = 4.0 × 10 −19 J; we want Egap = Elight = hc/, so: λ =
−34 8 hc (6.63 10 J s) (3.00 10 m/s) = = 5.0 × 10 −7 m = 5.0 × 102 nm −19 E 4.0 10 J
(6.626 10 −34 J s) (2.998 10 8 m/s) hc = = 2.72 × 10 −19 J = energy of band gap −9 λ 730. 10 m
84.
E=
85.
Sodium chloride structure: 8 corners ×
12 edges ×
1 / 2 Cl − 1 / 8 Cl − + 6 faces × = 4 Cl− ions corner face
1 / 4 Na + + 1 Na+ at body center = 4 Na+ ions; NaCl is the formula. edge
Cesium chloride structure: 1 Cs+ ion at body center; 8 corners ×
1 / 8 Cl − = 1 Cl− ion corner
CsCl is the formula. Zinc sulfide structure: There are four Zn2+ ions inside the cube. 8 corners ×
1 / 2 S 2− 1 / 8 S 2− + 6 faces × = 4 S2− ions; ZnS is the formula. corner face
Titanium oxide structure: 4 faces ×
8 corners ×
1 / 8 Ti 4+ + 1 Ti4+ at body center = 2 Ti4+ ions corner
1/2 O 2− + 2 O2− inside cube = 4 O2− ions; TiO2 is the formula. face
86.
MgO has the NaCl structure containing 4 Mg2+ ions and 4 O2− ions per face-centered unit cell. The O2− ions are cubic closest packed with the Mg2+ ions occupying all the octahedral holes.
87.
There is one octahedral hole per closest packed anion in a closest packed structure. If onehalf of the octahedral holes are filled, then there is a 2 : 1 ratio of fluoride ions to cobalt ions in the crystal. The formula is CoF2, which is composed of Co2+ and F− ions.
88.
There are two tetrahedral holes per closest packed anion. Let f = fraction of tetrahedral holes filled by the cations.
CHAPTER 10
LIQUIDS AND SOLIDS
Na2O: Cation-to-anion ratio =
CdS: Cation-to-anion ratio =
ZrI4: Cation-to-anion ratio =
453
2 2f , f = 1; all the tetrahedral holes are filled by Na+ = 1 1 cations.
1 2f 1 = , f = ; one-half the tetrahedral holes are filled by 1 1 2 Cd2+ cations. 1 2f 1 = , f = ; one-eighth the tetrahedral holes are filled 4 1 8 by Zr4+ cations.
89.
In a cubic closest packed array of anions, there are twice the number of tetrahedral holes as anions present, and an equal number of octahedral holes as anions present. A cubic closest packed array of sulfide ions will have four S2− ions, eight tetrahedral holes, and four octahedral holes. In this structure we have 1/8(8) = 1 Zn2+ ion and 1/2(4) = 2 Al3+ ions present, along with the 4 S2− ions. The formula is ZnAl2S4.
90.
A repeating pattern in the two-dimensional structure is:
Assuming the anions A are the larger circles, there are four anions completely in this repeating square. The corner cations (smaller circles) are shared by four different repeating squares. Therefore, there is one cation in the middle of the square plus 1/4 (4) = 1 net cation from the corners. Each repeating square has two cations and four anions. The empirical formula is MA2. 91.
8 F− ions at corners × 1/8 F−/corner = 1 F− ion per unit cell; Because there is one cubic hole per cubic unit cell, there is a 2 : 1 ratio of F− ions to metal ions in the crystal if only half of the body centers are filled with the metal ions. The formula is MF2, where M2+ is the metal ion.
92.
Corners are shared by 8 cubic unit cells and edges are shared by 4 unit cells. Re ions at 8 corners: 8(1/8) = 1 Re ion; O ions at 12 edges: 12(1/4) = 3 O ions; the formula is ReO3. Assuming the oxygen ion is a −2 charge, then the charge on Re is +3.
93.
From Figure 10.36, MgO has the NaCl structure containing 4 Mg2+ ions and 4 O2− ions per face-centered unit cell. 4 MgO formula units ×
1 mol MgO 6.022 10
23
atoms
40.31 g MgO = 2.678 × 10 −22 g MgO 1 mol MgO
454
CHAPTER 10 Volume of unit cell = 2.678 × 10 −22 g MgO ×
LIQUIDS AND SOLIDS
1 cm3 = 7.48 × 10 −23 cm3 3.58 g
Volume of unit cell = l3, l = cube edge length; l = (7.48 × 10 −23 cm3)1/3 = 4.21 × 10 −8 cm For a face-centered unit cell, the O2− ions touch along the face diagonal: 2 l = 4 rO 2− , rO 2− =
2 4.21 10 −8 cm = 1.49 × 10 −8 cm 4
The cube edge length goes through two radii of the O2− anions and the diameter of the Mg2+ cation, so: l = 2 rO 2− + 2 rMg 2+ , 4.21 × 10 −8 cm = 2(1.49 × 10 −8 cm) + 2 rMg 2+ , rMg 2+ = 6.15 × 10 −9 cm 94. Assuming K+ and Cl− just touch along the cube edge l: l = 2(314 pm) = 628 pm = 6.28 × 10−8 cm Volume of unit cell = l3 = (6.28 × 10−8 cm)3
The unit cell contains four K+ and four Cl− ions. For a unit cell: 4 KCl formula units
density =
1 mol KCl
6.022 10
23
formula units
(6.28 10
−8
cm ) 3
74 .55 g KCl mol KCl
= 2.00 g/cm3 95.
CsCl is a simple cubic array of Cl− ions with Cs+ in the middle of each unit cell. There is one Cs+ and one Cl− ion in each unit cell. Cs+ and Cl− ions touch along the body diagonal. Body diagonal = 2 rCs + + 2 rCl− = 3 l , l = length of cube edge In each unit cell: mass = 1 CsCl formula unit ×
1 mol CsCl 6.022 10
23
formula units
168.4 g CsCl mol CsCl
= 2.796 × 10 −22 g volume = l 3 = 2.796 × 10 −22 g CsCl ×
1 cm3 = 7.04 × 10 −23 cm3 3.97 g CsCl
l 3 = 7.04 × 10 −23 cm3, l = 4.13 × 10 −8 cm = 413 pm = length of cube edge
CHAPTER 10
LIQUIDS AND SOLIDS
455
2 rCs + + 2 rCl − = 3 l = 3 (413 pm) = 715 pm
The distance between ion centers = rCs + + rCl − = 715 pm/2 = 358 pm From ionic radius: rCs + = 169 pm and rCl − = 181 pm; rCs + + rCl − = 169 + 181 = 350. pm The distance calculated from the density is 8 pm (2.3%) greater than that calculated from tables of ionic radii. 96.
a. The NaCl unit cell has a face-centered cubic arrangement of the anions with cations in the octahedral holes. There are four NaCl formula units per unit cell, and since there is a 1 : 1 ratio of cations to anions in MnO, then there would be four MnO formula units per unit cell, assuming an NaCl-type structure. The CsCl unit cell has a simple cubic structure of anions with the cations in the cubic holes. There is one CsCl formula unit per unit cell, so there would be one MnO formula unit per unit cell if a CsCl structure is observed. Formula units of MnO 5.28 g MnO 1 mol MnO = ( 4.47 10 −8 cm ) 3 3 Unit cell 70 .94 g MnO cm
6.022 10 23 formula units MnO = 4.00 formula units MnO mol MnO 4
From the calculation, MnO crystallizes in the NaCl type structure. b. From the NaCl structure and assuming the ions touch each other, then l = cube edge length = 2rMn 2+ + 2rO 2− . l = 4.47 × 10−8 cm = 2rMn 2+ + 2(1.40 × 10−8 cm), rMn 2+ = 8.35 × 10−8 cm = 84 pm 97.
98.
99.
a. CO2: molecular
b. SiO2: network
c. Si: atomic, network
d. CH4: molecular
e. Ru: atomic, metallic
f.
I2: molecular
g. KBr: ionic
h. H2O: molecular
i.
NaOH: ionic
j.
k. CaCO3: ionic
l.
PH3: molecular
U: atomic, metallic
a. diamond: atomic, network
b. PH3: molecular
c. H2: molecular
d. Mg: atomic, metallic
e. KCl: ionic
f.
quartz: network
g. NH4NO3: ionic
h. SF2: molecular
i.
Ar: atomic, group 8A
j.
k. C6H12O6: molecular
Cu: atomic, metallic
a. The unit cell consists of Ni at the cube corners and Ti at the body center or Ti at the cube corners and Ni at the body center. b. 8 × 1/8 = 1 atom from corners + 1 atom at body center; empirical formula = NiTi c. Both have a coordination number of 8 (both are surrounded by 8 atoms).
456 100.
CHAPTER 10 Al: 8 corners ×
LIQUIDS AND SOLIDS
1/8 Al 1/2 Ni = 1 Al; Ni: 6 face centers × = 3 Ni corner face center
The composition of the specific phase of the superalloy is AlNi3. 101.
Structure 1 (on left) 8 corners × 6 faces ×
1 / 8 Ca = 1 Ca atom corner
1/ 2 O = 3 O atoms face
1 Ti at body center. Formula = CaTiO3
Structure 2 (on right) 8 corners ×
1 / 8 Ti = 1 Ti atom corner
12 edges ×
1/ 4 O = 3 O atoms corner
1 Ca at body center. Formula = CaTiO3
In the extended lattice of both structures, each Ti atom is surrounded by six O atoms. 102.
With a cubic closest packed array of oxygen ions, we have 4 O2− ions per unit cell. We need to balance the total −8 charge of the anions with a +8 charge from the Al3+ and Mg2+ cations. The only combination of ions that gives a +8 charge is 2 Al3+ ions and 1 Mg2+ ion. The formula is Al2MgO4. There are an equal number of octahedral holes as anions (4) in a cubic closest packed array and twice the number of tetrahedral holes as anions in a cubic closest packed array. For the stoichiometry to work out, we need 2 Al3+ and 1 Mg2+ per unit cell. Hence one-half of the octahedral holes are filled with Al3+ ions, and one-eighth of the tetrahedral holes are filled with Mg2+ ions.
103.
a. Y: 1 Y in center; Ba: 2 Ba in center Cu: 8 corners × O: 20 edges ×
1 / 4 Cu 1 / 8 Cu = 1 Cu, 8 edges × = 2 Cu, total = 3 Cu atoms edge corner
1/ 4 O 1/ 2 O = 5 O atoms, 8 faces × = 4 O atoms, total = 9 O atoms edge face
Formula: YBa2Cu3O9 b. The structure of this superconductor material follows the second perovskite structure described in Exercise 101. The YBa2Cu3O9 structure is three of these cubic perovskite unit cells stacked on top of each other. The oxygen atoms are in the same places, Cu takes the place of Ti, two of the calcium atoms are replaced by two barium atoms, and one Ca is replaced by Y. c. Y, Ba, and Cu are the same. Some oxygen atoms are missing. 12 edges ×
1/ 4 O 1/ 2 O = 3 O, 8 faces × = 4 O, total = 7 O atoms edge face
Superconductor formula is YBa2Cu3O7.
CHAPTER 10 104.
LIQUIDS AND SOLIDS
457
a. Structure (a): 1 / 8 Tl = 1 Tl corner
Ba: 2 Ba inside unit cell; Tl: 8 corners × Cu: 4 edges ×
O: 6 faces ×
1 / 4 Cu = 1 Cu edge
1/ 4 O 1/ 2 O + 8 edges × = 5 O; Formula = TlBa2CuO5. edge face
Structure (b): Tl and Ba are the same as in structure (a). Ca: 1 Ca inside unit cell; Cu: 8 edges ×
O: 10 faces ×
1 / 4 Cu = 2 Cu edge
1/ 4 O 1/ 2 O + 8 edges × = 7 O; Formula = TlBa2CaCu2O7. edge face
Structure (c): Tl and Ba are the same, and two Ca are located inside the unit cell. Cu: 12 edges ×
1 / 4 Cu 1/ 4 O 1/ 2 O = 3 Cu; O: 14 faces × + 8 edges × =9O edge edge face
Formula = TlBa2Ca2Cu3O9. Structure (d):
Following similar calculations, formula = TlBa2Ca3Cu4O11.
b. Structure (a) has one planar sheet of Cu and O atoms, and the number increases by one for each of the remaining structures. The order of superconductivity temperature from lowest to highest temperature is (a) < (b) < (c) < (d). c. TlBa2CuO5: 3 + 2(2) + x + 5(−2) = 0, x = +3 Only Cu3+ is present in each formula unit. TlBa2CaCu2O7: 3 + 2(2) + 2 + 2(x) + 7(−2) = 0, x = +5/2 Each formula unit contains 1 Cu2+ and 1 Cu3+. TlBa2Ca2Cu3O9: 3 + 2(2) + 2(2) + 3(x) + 9(−2) = 0, x = +7/3 Each formula unit contains 2 Cu2+ and 1 Cu3+. TlBa2Ca3Cu4O11: 3 + 2(2) + 3(2) + 4(x) + 11(−2) = 0, x = +9/4 Each formula unit contains 3 Cu2+ and 1 Cu3+.
458
CHAPTER 10
LIQUIDS AND SOLIDS
d. This superconductor material achieves variable copper oxidation states by varying the numbers of Ca, Cu, and O in each unit cell. The mixtures of copper oxidation states are discussed in part c. The superconductor material in Exercise 103 achieves variable copper oxidation states by omitting oxygen at various sites in the lattice.
Phase Changes and Phase Diagrams 105.
The compound with the weakest intermolecular forces will have the highest vapor pressure. Compounds d and e are nonpolar compounds consisting of only nonpolar C‒C and C‒H bonds. The only intermolecular force that nonpolar compounds have is London dispersion forces. Compound e with the smaller molar mass will have the weakest LD forces and the highest vapor pressure. The compound with the lowest vapor pressure has the strongest intermolecular forces. Compound a-c all have dipole forces, but compound b can also form the relatively strong hydrogen-bonding forces. Compound b has the lowest vapor pressure at 20°C.
106.
Statement c is false. Ethanol can hydrogen-bond, while dimethyl ether cannot. Ethanol will have the higher normal boiling point.
107.
If we graph ln Pvap versus 1/T with temperature in Kelvin, the slope of the resulting straight line will be −ΔHvap/R. Pvap
ln Pvap
T (Li)
1 torr 10. 100. 400. 760.
0 2.3 4.61 5.99 6.63
1023 K 1163 1353 1513 1583
1/T 9.775 × 10−4 K−1 8.598 × 10−4 7.391 × 10−4 6.609 × 10−4 6.317 × 10−4
T (Mg) 893 K 1013 1173 1313 1383
1/T 11.2 × 10−4 K−1 9.872 × 10−4 8.525 × 10−4 7.616 × 10−4 7.231 × 10−4
For Li: We get the slope by taking two points (x, y) that are on the line we draw. For a line, slope = y/x, or we can determine the straight-line equation using a calculator. The general straight-line equation is y = mx + b, where m = slope and b = y intercept.
CHAPTER 10
LIQUIDS AND SOLIDS
459
The equation of the Li line is: ln Pvap = −1.90 × 104(1/T) + 18.6, slope = −1.90 × 104 K Slope = −ΔHvap/R, ΔHvap = −slope × R = 1.90 × 104 K × 8.3145 J/K•mol ΔHvap = 1.58 × 105 J/mol = 158 kJ/mol For Mg: The equation of the line is: ln Pvap = −1.67 × 104(1/T) + 18.7, slope = −1.67 × 104 K ΔHvap = −slope × R = 1.67 × 104 K × 8.3145 J/K•mol ΔHvap = 1.39 × 105 J/mol = 139 kJ/mol The bonding is stronger in Li because ΔHvap is larger for Li. 108.
We graph ln Pvap vs 1/T. The slope of the line equals − ΔHvap/R. T(K)
103/T (K−1) Pvap (torr)
ln Pvap
273 283 293 303 313 323 353
3.66 3.53 3.41 3.30 3.19 3.10 2.83
2.67 3.28 3.87 4.40 4.89 5.34 6.51
Slope =
14.4 26.6 47.9 81.3 133 208 670.
6.6 − 2.5 (2.80 10
−4600 K =
−3
− ΔH vap R
− 3.70 10 −3 ) K −1
=
− ΔH vap 8.3145 J/K • mol
= − 4600 K , ΔHvap = 38,000 J/mol = 38 kJ/mol
To determine the normal boiling point, we can use the following formula: ΔH vap 1 P 1 ln 1 = − T1 R T2 P2
At the normal boiling point, the vapor pressure equals 1.00 atm or 760. torr. At 273 K, the vapor pressure is 14.4. torr (from data in the problem). 38,000 J/mol 1 1 14 .4 , −3.97 = 4.6 × 103(1/T2 − 3.66 × 10−3) ln − = 760 . 8.3145 J/K • mol T2 273 K
−8.6 × 10−4 + 3.66 × 10−3 = 1/T2 = 2.80 × 10−3, T2 = 357 K or 84°C = normal boiling point
460 109.
CHAPTER 10
LIQUIDS AND SOLIDS
At 100.°C (373 K), the vapor pressure of H2O is 1.00 atm = 760. torr. For water, ΔHvap = 40.7 kJ/mol. P H vap 1 P H vap 1 1 1 ln 1 = − or ln 2 = − P P R T2 R T1 T1 T2 2 1
1 520. torr 1 40.7 10 3 J/mol 1 1 , −7.75 × 10−5 = = ln − − 8.3145 J/K • mol 373 K T2 T2 760. torr 373 K
−7.75 × 10−5 = 2.68 × 10−3 −
1 T2
110.
,
1 T2
= 2.76 × 10−3, T2 =
1 2.76 10 −3
= 362 K or 89°C
At 100.°C (373 K), the vapor pressure of H2O is 1.00 atm. For water, ΔHvap = 40.7 kJ/mol. H vap 1 H vap 1 P P 1 1 or ln 2 = ln 1 = − − R T2 T1 R T1 T2 P2 P1 P2 40.7 10 3 J/mol 1 1 = , ln P2 = 0.51, P2 = e0.51 = 1.7 atm ln − 1.00 atm 8.3145 J/K • mol 373 K 388 K
1 40.7 10 3 J/mol 1 1 1 3.50 , 2.56 10 − 4 = ln − − = 8.3145 J/K • mol 373 K T2 T2 1.00 373 K
2.56 × 10−4 = 2.68 × 10−3 −
111.
1 1 1 , = 2.42 × 10−3, T2 = = 413 K or 140.°C T2 T2 2.42 10 −3
ΔH vap 1 ΔH vap P 836 torr 1 1 1 = ln 1 = − , ln − P T R 213 torr 8.3145 J/K • mol 313 K 353 K T1 2 2
Solving: Hvap = 3.1 × 104 J/mol; for the normal boiling point, P = 1.00 atm = 760. torr. 760. torr 3.1 10 4 J/mol 1 1 1 1 , = ln − − = 3.4 × 10 −4 213 torr 8.3145 J/K • mol 313 K T 313 T 1 1
T1 = 350. K = 77°C; the normal boiling point of CCl4 is 77°C. 112.
Let P1 = 760. torr, T1 = 273.2 + 56.5 = 329.7 K, P2 = 630. torr, T2 = ?: 1 760. torr 3.20 10 4 J/mol 1 1 1 − − ln = 4.87 × 10 −5 , = 8.3145 J/K • mol T2 329.7 K T2 329.7 630. torr
T2 = 324 K = 51°C Let P1 = 760. torr, T1 = 273.2 + 56.5 = 329.7 K, P2 = ?, T2 = 25.0oC = 298.2 K
CHAPTER 10
LIQUIDS AND SOLIDS
461
H vap 1 H vap 1 P P 1 1 or ln 2 = ln 1 = − − R T2 T1 R T1 T2 P2 P1 3 P2 1 1 P2 32.0 10 J/mol ln = − = e−1.23, P2 = 222 torr , 760. torr 8.3145 J/K • mol 329.7 K 298.2 K 760 .
113.
Let P1 = 760. torr, T1 = 273.2 + 34.6 = 307.8 K, P2 = 400. torr, T2 = 273.2 + 17.9 = 291.1 K: ΔH vap 1 ΔH vap P 760 torr 1 1 1 = ln 1 = − , ln − P R T2 307.8 K T1 400. torr 8.3145 J/K • mol 291.1 K 2
Solving: Hvap = 2.86 × 104 J/mol = 28.6 kJ/mol
114.
H vap 1 H vap 1 P P 1 1 or ln 2 = ln 1 = − − R T2 T1 R T1 T2 P2 P1
P2 59.1 10 3 J/mol 1 1 = ln − 8.3145 J/K • mol 630. K 298 K 1.00 atm
ln P2 = −12.57, P2 = e−12.57 = 3.47 10−6 atm = mercury vapor pressure at 25oC 115.
116.
a. The plateau at the lowest temperature signifies the melting/freezing of the substance. Hence the freezing point is 20C. b. The higher temperature plateau signifies the boiling/condensation of the substance. The temperature of this plateau is 120C. c. X(s) → X(l) H = Hfusion; X(l) → X(g) H = Hvaporization The heat of fusion and the heat of vaporization terms refer to enthalpy changes for the specific phase changes illustrated in the equations above. In a heating curve, energy is
462
CHAPTER 10
LIQUIDS AND SOLIDS
applied at a steady rate. So the longer, higher temperature plateau has a larger enthalpy change associated with it as compared to the shorter plateau. The higher temperature plateau occurs when a liquid is converting to a gas, so the heat of vaporization is greater than the heat of fusion. This is always the case because significantly more intermolecular forces are broken when a substance boils than when a substance melts. 117.
a. Many more intermolecular forces must be broken to convert a liquid to a gas as compared with converting a solid to a liquid. Because more intermolecular forces must be broken, much more energy is required to vaporize a liquid than is required to melt a solid. Therefore, Hvap is much larger than Hfus. b. 1.00 g Na ×
1 mol Na 2.60 kJ = 0.113 kJ = 113 J to melt 1.00 g Na 22 .99 g mol Na
c. 1.00 g Na ×
1 mol Na 97 .0 kJ = 4.22 kJ = 4220 J to vaporize 1.00 g Na 22 .99 g mol Na
d. This is the reverse process of that described in part c, so the energy change is the same quantity but opposite in sign. Therefore, q = −4220 J; i.e., 4220 of heat will be released. 118.
Melt: 8.25 g C6H6 ×
1 mol C 6 H 6 9.92 kJ = 1.05 kJ 78 .11 g mol C 6 H 6
Vaporize: 8.25 g C6H6 ×
1 mol C 6 H 6 30 .7 kJ = 3.24 kJ 78 .11 g mol C 6 H 6
As is typical, the energy required to vaporize a certain quantity of substance is much larger than the energy required to melt the same quantity of substance. A lot more intermolecular forces must be broken to vaporize a substance as compared to melting a substance. 119.
Let’s break up the process in a series of steps. For each step, we will calculate the energy released, then sum up the energy for all the steps to determine the total amount energy removed. X(g, 100.0°C) → X(g, 75°C), 1.0 J q1 = sg × m × ΔT = o × 250. g × (‒25°C) = ‒6.3 × 103 J = ‒6.3 kJ g C H2O(g, 75°C) → H2O(l, 75°C), q2 = 250. g × H2O(l, 75°C) → H2O(l, ‒15°C), q3 =
1 mol −20. kJ × = ‒67 kJ 75.0 g mol
2.5 J × 250. g × (‒90.°C) = ‒5.6 × 104 J = ‒56 kJ g oC
H2O(l, ‒15°C) → H2O(s, ‒15°C), q4 = 250 × H2O(s, ‒15°C) → H2O(s, ‒50°C), q5 =
1 mol −5.0 kJ × = ‒17 kJ 75 g mol
3.0 J × 250. g × (‒35°C) = ‒2.6 × 104 J = ‒26 kJ g oC
qtotal = q1 + q2 + q3 + q4 + q5 = ‒6.3 + ‒67 + ‒56 + ‒17 + ‒26 = ‒172 kJ 172 kJ of energy must be removed to do the overall process.
CHAPTER 10 120.
LIQUIDS AND SOLIDS
463
The process that takes the longest is the one that takes the most amount of energy. All processes are for 18 g of water. The temperature changes for processes a, c, and e are all 100.0C. Because H2O(l) has the largest heat capacity of the three phases, process c must take more energy than processes a and e. This amount of energy is: q = s × m × ΔT =
4.18 J g oC
× 18 g × 100.0°C = 7.5 × 104 J = 7.5 kJ
For the processes of melting 18 g of ice and boiling 18 g of water, boiling the water will require the larger amount of heat because the enthalpy of vaporization is greater than the enthalpy of fusion. The amount of energy to boil 18 g of water is: q = 18 g H2O ×
1 mol 40 .7 kJ = 41 kJ 18 .02 g mol
This is more energy than the process of raising 18 g of water by 100.0C. So, the process of boiling water (process d) requires the most energy. 121.
To calculate qtotal, break up the heating process into five steps. H2O(s, −20.0°C) → H2O(s, 0.0°C), ΔT = 20.0°C; let sice = specific heat capacity of ice: q1 = ss × m × ΔT =
2.03 J g oC
× 500. g × 20.0°C = 2.03 × 104 J = 20.3 kJ
H2O(s, 0.0°C) → H2O(l, 0.0°C), q2 = 500. g H2O × H2O(l, 0.0°C) → H2O(l, 100.0°C), q3 =
4.18 J o
g C
× 500. g × 100.0°C = 2.09 × 105J = 209 kJ
H2O(l, 100.0°C) → H2O(g, 100.0°C), q4 = 500. g × H2O(g, 100.0°C) → H2O(g, 250.0°C), q5 =
1 mol 6.02 kJ = 167 kJ 18 .02 g mol
2.02 J o
g C
1 mol 40 .7 kJ = 1130 kJ 18 .02 g mol
× 500. g × 150.0°C = 1.52 × 105 J = 152 kJ
qtotal = q1 + q2 + q3 + q4 + q5 = 20.3 + 167 + 209 + 1130 + 152 = 1680 kJ 122.
To calculate qtotal, break up the heating process into five steps. 1.00 mol H2O = 18.0 g H2O(s, −30.0°C) → H2O(s, 0.0°C), ΔT = 30.0°C; let sice = specific heat capacity of ice: q1 = sice × m × ΔT =
2.03 J g oC
× 18.0 g × 30.0°C = 1.10 × 103 J = 1.10 kJ
H2O(s, 0.0°C) → H2O(l, 0.0°C), q2 = 1.00 mol × H2O(l, 0.0°C) → H2O(l, 100.0°C), q3 =
4.18 J g oC
6.02 kJ = 6.02 kJ mol
× 18.0 g × 100.0°C = 7520 J = 7.52 kJ
464
CHAPTER 10 H2O(l, 100.0°C) → H2O(g, 100.0°C), q4 = 1.00 mol × H2O(g, 100.0°C) → H2O(g, 140.0°C), q5 =
2.02 J g oC
40.7 kJ mol
LIQUIDS AND SOLIDS = 40.7 kJ
× 18.0 g × 40.0°C = 1450 J = 1.45 kJ
qtotal = q1 + q2 + q3 + q4 + q5 = 1.10 + 6.02 + 7.52 + 40.7 + 1.45 = 56.8 kJ 123.
Let x = mass of water; break the process up into three steps, then set the total energy added to the energy required for each step. Step 1 is heating water from 50.0°C to 100.0°C, step 2 is boiling the water at 100.0C, and step three is heating gaseous water at 100.0°C to 150.0 °C. q1 = q3 =
4.18 J g oC 2.02 J g oC
× x × 50.0°C = 209x; q2 = x ×
1 mol 40.7 103 J = 2260x 18.02 g mol
× x × 50.0°C = 101x
385 × 103 = q1 + q2 + q3 = 209x + 2260x + 101x = 2570x, x = 150. g 124.
Step 1 is heating the liquid ethanol from 48C to 78C, step 2 is boiling the ethanol at 78C, and step 3 is converting gaseous ethanol at 78C to 108C. q1 =
2.5 J 46.07 g × 2.00 mol C2H5OH × × 30.°C = 6900 J = 6.9 kJ o mol C 2 H 5 OH g C
q2 = ΔHvap × 2.00 mol = 2.00ΔHvap with ΔHvap units of kJ/mol. q3 =
1.5 J 46.07 g × 2.00 mol C2H5OH × × 30.°C = 4140 J = 4.1 kJ o mol C 2 H 5 OH g C
95 kJ = q1 + q2 + q3 = 6.9 + 2.00ΔHvap + 4.1, ΔHvap = 42 kJ/mol 125.
Total mass H2O = 18 cubes ×
1 mol H 2 O 30.0 g = 540. g; 540. g H2O × = 30.0 mol H2O 18 .02 g cube
Heat removed to produce ice at −5.0°C: 4.18 J 6.02 10 3 J 2.03 J 30 .0 mol + 540. g 22.0 o C + 540. g 5.0 o C mol gC gC
= 4.97 × 104 J + 1.81 × 105 J + 5.5 × 103 J = 2.36 × 105 J 2.36 × 105 J ×
126.
1 g CF2 Cl 2 = 1.49 × 103 g CF2Cl2 must be vaporized. 158 J
Heat released = 0.250 g Na ×
1 mol 368 kJ = 2.00 kJ 22 .99 g 2 mol
CHAPTER 10
LIQUIDS AND SOLIDS
465
To melt 50.0 g of ice requires: 50.0 g ice ×
1 mol H 2 O 6.02 kJ = 16.7 kJ 18 .02 g mol
The reaction doesn't release enough heat to melt all of the ice. The temperature will remain at 0°C. 127.
A: solid
B: liquid
C: vapor
D: solid + vapor
E: solid + liquid + vapor
F: liquid + vapor
G: liquid + vapor
triple point: E
critical point: G
H: vapor
Normal freezing point: Temperature at which solid-liquid line is at 1.0 atm (see following plot). Normal boiling point: Temperature at which liquid-vapor line is at 1.0 atm (see following plot ). 1.0 atm
nfp
nbp
Because the solid-liquid line equilibrium has a positive slope, the solid phase is denser than the liquid phase. 128.
a. 3 b. Triple point at 95.31°C: rhombic, monoclinic, gas Triple point at 115.18°C: monoclinic, liquid, gas Triple point at 153°C: rhombic, monoclinic, liquid c. From the phase diagram, the monoclinic solid phase is stable at T = 100.°C and P = 1 atm. d. Normal melting point = 115.21°C; normal boiling point = 444.6°C; the normal melting and boiling points occur at P = 1.0 atm. e. Rhombic is the densest phase because the rhombic-monoclinic equilibrium line has a positive slope, and because the solid-liquid equilibrium lines also have positive slopes. f.
No; P = 1.0 × 10 −5 atm is at a pressure somewhere between the 95.31 and 115.18°C triple points. At this pressure, the rhombic and gas phases are never in equilibrium with each other, so rhombic sulfur cannot sublime at P = 1.0 × 10 −5 atm. However, monoclinic sulfur can sublime at this pressure.
466
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LIQUIDS AND SOLIDS
g. From the phase diagram, we would start off with gaseous sulfur. At 100.°C and ~1 × 10−5 atm, S(g) would convert to the solid monoclinic form of sulfur. Finally at 100.°C and some large pressure less than 1420 atm, S(s, monoclinic) would convert to the solid rhombic form of sulfur. Summarizing, the phase changes are S(g) → S(monoclinic) → S(rhombic). 129.
a. two b. Higher-pressure triple point: graphite, diamond and liquid; lower-pressure triple point at 107 Pa: graphite, liquid and vapor c. It is converted to diamond (the more dense solid form). d. Diamond is more dense, which is why graphite can be converted to diamond by applying pressure.
130.
The following sketch of the Br2 phase diagram is not to scale. Because the triple point of Br2 is at a temperature below the freezing point of Br2, the slope of the solid-liquid line is positive.
100 Solid Pressure (atm)
Liquid
1 0.05
Gas
-7.3 -7.2
59
320
o
Temperature ( C) The positive slopes of all the lines indicate that Br2(s) is more dense than Br2(l), which is more dense than Br2(g). At room temperature (~22°C) and 1 atm, Br 2(l) is the stable phase. Br2(l) cannot exist at a temperature below the triple-point temperature of −7.3°C or at a temperature above the critical-point temperature of 320°C. The phase changes that occur as temperature is increased at 0.10 atm are solid → liquid → gas. 131.
Because the density of the liquid phase is greater than the density of the solid phase, the slope of the solid-liquid boundary line is negative (as in H2O). With a negative slope, the melting points increase with a decrease in pressure, so the normal melting point of X should be greater than 225°C.
CHAPTER 10
LIQUIDS AND SOLIDS
467
132. 760
s
l
P (torr)
g 280
-121
-112
-107
T(C)
From the three points given, the slope of the solid-liquid boundary line is positive, so Xe(s) is more dense than Xe(l). Also, the positive slope of this line tells us that the melting point of Xe increases as pressure increases. The same direct relationship exists for the boiling point of Xe because the liquid-gas boundary line also has a positive slope.
ChemWork Problems 133.
The covalent bonds within a molecule are much stronger than the intermolecular forces between molecules. Consider water, which exhibits a very strong type of dipole force called hydrogen bonding. When water boils at 100oC, hydrogen bonding intermolecular forces are broken in the liquid resulting in H2O(l) converting to H2O(g). At 100oC, the covalent bonds within each water molecule are not broken; if they were broken, then H and O atoms would be produced. It takes a lot more energy than what is provided at 100oC to break the covalent bonds within a water molecule; the intramolecular forces within a covalent compound are much stronger than the intermolecular forces between molecules.
134.
Neopentane is more compact than n-pentane. There is less surface-area contact among neighboring neopentane molecules. This leads to weaker London dispersion (LD) forces and a lower boiling point.
135.
a. dipole, LD (SF4 is a polar compound.)
b. LD only (CO2 is a nonpolar compound.)
c. H-bonding, LD (H-bonding due to the O−H bond) e. dipole, LD (ICl5 is a polar compound.)
d. H-bonding, LD
f. LD only (XeF4 is a nonpolar compound.)
CO2 and XeF4 are nonpolar covalent compounds, so they only exhibit London dispersion forces. CH3CH2OH and HF are covalent compounds that contain either an N−H, O−H, or F−H bond, so they exhibit hydrogen bonding forces. SF4 and ICl5 are polar covalent compounds, so they exhibit dipole-dipole forces. 136.
a.
True;
b.
False; large nonpolar compounds are liquids and solids at 25 oC.
468
CHAPTER 10
LIQUIDS AND SOLIDS
c. False; H2O has perfect symmetry to form two H−bonding interactions per molecule (two covalently boned H atoms and two lone pairs of electrons on the central atom). NH3, with only one lone pair, can only form one H−bonding interaction per molecule. Therefore, the H−bonding forces are stronger in H2O as compared to NH3. This is why the boiling point of H2O is greater than that of NH3. d. True; SO2 is a polar molecule so it exhibits dipole−dipole forces (as well as London dispersion forces). e. True; in general, the larger the molar mass of a compound, the stronger the London dispersion forces. 137.
As the strength of the intermolecular forces increase, boiling points increase while vapor pressures at a specific temperature decrease. a. False; the ionic forces in LiF are much stronger than molecular intermolecular forces found in H2S. b. True; HF is capable of H−bonding, HBr is not. c. True; the larger Cl2 molecule will have the stronger London dispersion forces. d. True; HCl is a polar compound while CCl4 is a nonpolar compound. e. False; the ionic forces in MgO are much stronger than the molecular intermolecular forces found in CH3CH2OH.
138.
All these substances are liquids at 20°C. The substance to freeze first has the freezing point closest to 20°C, which will be the compound with the highest freezing point. The substance to freeze last as the temperature is lowered has the lowest freezing point. Compound b can form hydrogen-bonding forces while the other similar size covalent compounds cannot H-bond. So compound b has the strongest intermolecular forces and will have the highest freezing point of these liquids. Compound b will freeze first as the temperature is lowered form 20°C.
139.
As the physical properties indicate, the intermolecular forces are slightly stronger in D2O than in H2O.
140.
CH3CO2H:
H-bonding + dipole forces + London dispersion (LD) forces
CH2ClCO2H:
H-bonding + larger electronegative atom replacing H (greater dipole) + LD forces
CH3CO2CH3:
Dipole forces (no H-bonding) + LD forces
From the intermolecular forces listed above, we predict CH3CO2CH3 to have the weakest intermolecular forces and CH2ClCO2H to have the strongest. The boiling points are consistent with this view. 141.
At any temperature, the plot tells us that substance A has a higher vapor pressure than substance B, with substance C having the lowest vapor pressure. Therefore, the substance with the weakest intermolecular forces is A, and the substance with the strongest intermolecular forces is C.
CHAPTER 10
LIQUIDS AND SOLIDS
469
NH3 can form hydrogen-bonding interactions, whereas the others cannot. Substance C is NH3. The other two are nonpolar compounds with only London dispersion forces. Because CH4 is smaller than SiH4, CH4 will have weaker LD forces and is substance A. Therefore, substance B is SiH4. 142.
As the electronegativity of the atoms covalently bonded to H increases, the strength of the hydrogen-bonding interaction increases. N ⋅⋅⋅ H‒N < N ⋅⋅⋅ H‒ O < O ⋅⋅⋅ H‒O < O ⋅⋅⋅ H‒F < F ⋅⋅⋅ H‒F weakest strongest
143.
X(s, −35.0°C) → X(s, −15.0°C), ΔT = 20.0°C; let ssolid = specific heat capacity of X(s): q1 = ssolid × m × ΔT =
3.00 J × 10.0 g × 20.0°C = 600. J g oC
X(s, −15.0°C) → X(l, −15.0°C), q2 = 10.0 g X × X(l, −15.0°C) → X(l, 25.0°C), q3 =
1 mol 5.00 kJ = 0.500 kJ = 500. J 100.0 g mol
2.50 J × 10.0 g × 40.0°C = 1.00 × 103 J o g C
qtotal = q1 + q2 + q3 = 600. + 500. + 1.00 × 103 = 2.10 × 103 J 2.10 × 103 J × 144.
1 min = 4.67 min 450.0 J
Step 1: X(g, 100.0°C) → X(g, 75°C), q1 = sgas × 100.0 g × −25.0°C = −(2.50 × 103)sgas Step 2: X(g, 75.0°C) → X(l, 75.0°C), q2 = 1.000 mol × Step 3: X(l, 75.0°C) → X(l, 50.0°C), q3 =
− 20.00 kJ = −20.00 kJ mol
2.50 J × 100.0 g × −25.0°C = −6.25 × 103 J g oC
qtotal = q1 + q2 + q3, −28.75 × 103 J = −(2.50 × 103)sgas − 20.00 × 103 J − 6.25 × 103 J (2.50 × 103)sgas = 2.50 × 103, sgas = 1.00 J/g•oC 145.
8 corners ×
1/ 4 F 1 / 8 Xe + 1 Xe inside cell = 2 Xe; 8 edges × + 2 F inside cell = 4 F edge corner
The empirical formula is XeF2. 146.
One B atom and one N atom together have the same number of electrons as two C atoms. The description of physical properties sounds a lot like the properties of graphite and diamond, the two solid forms of carbon. The two forms of BN have structures similar to graphite and diamond.
147.
B2H6: This compound contains only nonmetals, so it is probably a molecular solid with covalent bonding. The low boiling point confirms this. SiO2: This is the empirical formula for quartz, which is a network solid.
470
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LIQUIDS AND SOLIDS
CsI: This is a metal bonded to a nonmetal, which generally form ionic solids. The electrical conductivity in aqueous solution confirms this. W: Tungsten is a metallic solid as the conductivity data confirm. 148.
Ar is cubic closest packed. There are four Ar atoms per unit cell, and with a face-centered unit cell, the atoms touch along the face diagonal. Let l = length of cube edge. Face diagonal = 4r = l 2 , l = 4(190. pm)/ 2 = 537 pm = 5.37 × 10−8 cm
Density = 149.
mass = volume
4 atoms
1 mol 39.95 g 23 mol 6.022 10 atoms = 1.71 g/cm3 −8 3 (5.37 10 cm)
In simple cubic unit cell, there are: 8 corner atoms × 1/8 atom per corner = 1 atom/unit cell The volume of the unit cell is l3 where l = cube edge length. For a unit cell of polonium:
density =
mass = 9.2 g/cm3 = volume
1 Po atom
1 mol Po 209 g Po 23 mol Po 6.022 10 atoms 3 l
Solving: l = 3.35 × 10 −8 cm = cube edge length (carrying an extra sig. fig.) In a simple cubic unit cell, the atoms touch along the cube edge l:
2(radius) = 2r = l
2r
l
2r = 3.35 × 10 −8 cm r = 1.7 × 10 −8 cm = 170 pm
150.
There are two tetrahedral holes per closest packed anion. Let f = fraction of tetrahedral holes filled by the cations. K2O: Cation-to-anion ratio =
2 2f , f=1 = 1 1
All the tetrahedral holes are filled by K+ cations.
151.
8 corner Mn ions ×
1/8 Mn + 1 Mn ion inside cell = 2 Mn ions per unit cell corner
CHAPTER 10
LIQUIDS AND SOLIDS
4 face O ions ×
471
1/2 O + 2 O ions inside cell = 4 O ions per unit cell face
The empirical formula is MnO2. Assuming each oxygen ion has a 2− charge, then each manganese ion has a 4+ charge. 152.
153.
a. Kr: group 8A
b. SO2: molecular
c. Ni: metallic
d. SiO2: network
e. NH3: molecular
f.
24.7 g C6H6 ×
PC6 H 6 =
154.
Pt: metallic
1 mol = 0.316 mol C6H6 78 .11 g
nRT = V
0.316 mol
0.08206 L atm 293.2 K K mol = 0.0760 atm, or 57.8 torr 100.0 L
To set up an equation, we need to know what phase exists at the final temperature. To heat 20.0 g of ice from −10.0 to 0.0°C requires: q=
2 .03 J × 20.0 g × 10.0°C = 406 J g C
To convert ice to water at 0.0°C requires: q = 20.0 g ×
1 mol 6.02 kJ = 6.68 kJ = 6680 J × 18.02 mol
To chill 100.0 g of water from 80.0 to 0.0°C requires: q=
4 .18 J × 100.0 g × 80.0°C = 33,400 J of heat removed g C
From the heat values above, the liquid phase exists once the final temperature is reached (a lot more heat is lost when the 100.0 g of water is cooled to 0.0°C than the heat required to convert the ice into water). To calculate the final temperature, we will equate the heat gain by the ice to the heat loss by the water. We will keep all quantities positive in order to avoid sign errors. The heat gain by the ice will be the 406 J required to convert the ice to 0.0°C plus the 6680 J required to convert the ice at 0.0°C into water at 0.0°C plus the heat required to raise the temperature from 0.0°C to the final temperature. Heat gain by ice = 406 J + 6680 J + Heat loss by water =
4 .18 J × 20.0 g × (Tf − 0.0°C) = 7.09 × 103 + (83.6)Tf g C
4 .18 J × 100.0 g × (80.0°C − Tf) = 3.34 × 104 − 418Tf g C
Solving for the final temperature: 7.09 × 103 + (83.6)Tf = 3.34 × 104 − 418Tf, 502Tf = 2.63 × 104, Tf = 52.4°C
472 155.
CHAPTER 10
LIQUIDS AND SOLIDS
The heat released when 500.0 g of water is cooled from 23.0 to 2.0°C is: q=
4 .18 J × 500.0 g × ‒21.0°C = ‒4.39 × 104 J g C
Let n = number of ice cubes. Since each ice cube is 1 mol, the mass of ice cubes is 18.02n. The heat gained when the ice cubes at ‒5.0°C is converted to ice at 0.0°C is: q=
2 .03 J × 18.02n g × 5.0°C = 180n J g C
The heat gained to convert the ice cubes ice at 0.0°C to water at 0.0°C is: n×
6.02 kJ 1000 J = 6020n mol kJ
The heat gained when the n moles of liquid water 0.0°C is raised to 2.0°C is: 4 .18 J × 18.02n × 2.0°C = 150n J g C
Now equate the magnitude of the heat loss of the water to the heat gain of the ice cubes: 4.39 × 104 = 180n + 6020n + 150n, n = 6.9 ice cubes 7 ice cubes are necessary to cool 500.0 g of water from 23.0 to 2.0°C. 156.
Let s = specific heat capacity of the unknown element in units of J/g·°C. The heat loss of the hot element when it is cooled from 195 to 0°C is: s × 500.0 g × ‒195°C = (‒9.75 × 104)s J The heat gained when 109.5 g of ice melts at 0°C is: 109.5 g ×
1 mol 6.02 kJ = 36.6 kJ = 3.66 × 104 J × 18.02 mol
Equating the magnitude of the heat loss to the heat gain: (9.75 × 104)s = 3.66 × 104, s = 0.375 J/g·°C = specific heat capacity of unknown element 157.
If we extend the liquid-vapor line of the water phase diagram to below the freezing point, we find that supercooled water will have a higher vapor pressure than ice at −10°C (see Figure 10.44 of the text). To achieve equilibrium, there must be a constant vapor pressure. Over time, supercooled water will be transformed through the vapor into ice in an attempt to equilibrate the vapor pressure. Eventually there will only be ice at −10°C along with H2O(g) at the vapor pressure given by the solid-vapor line in the phase diagram at −10°C.
158.
1.00 lb ×
454 g = 454 g H2O; a change of 1.00°F is equal to a change of 5/9°C. lb
The amount of heat in J in 1 Btu is
4.18 J 5 454 g C = 1.05 × 103 J = 1.05 kJ. gC 9
CHAPTER 10
LIQUIDS AND SOLIDS
473
It takes 40.7 kJ to vaporize 1 mol H2O (ΔHvap). Combining these: 1.00 10 4 Bu 1.05 kJ 1 mol H 2 O = 258 mol/h; or: h Btu 40.7 kJ
258 mol 18.02 g H 2 O = 4650 g/h = 4.65 kg/h h mol
159.
60 s 750. J = 2.25 × 105 J = 225 kJ min s
Energy added to ice cubes = 5.00 min × Energy to melt ice cubes: 403 g H2O ×
1 mol H 2 O 18.02 g
6.02 kJ = 135 kJ mol H 2 O
Energy remaining to heat water = 225 kJ – 135 kJ = 90. kJ 9.0 × 104 J = swater × mass × ΔT =
4 .18 J g oC
× 403 g × ΔT, ΔT = 53.4oC
The final temperature of the water will increase from 0 oC to 53.4oC. 160.
a. False; the normal melting point is about 185 K; the normal boiling point is about 270 K. b. False;
c.
True
d. False; at 1.5 atm, the melting point is about 250 K. e. True 161.
The critical temperature is the temperature above which the vapor cannot be liquefied no matter what pressure is applied. Since N2 has a critical temperature below room temperature (~22°C), it cannot be liquefied at room temperature. NH3, with a critical temperature above room temperature, can be liquefied at room temperature.
162.
C2H5OH(l) → C2H5OH(g)
H = 42.4 kJ; ΔE = q + w
w = −PV; because PV = nRT, at constant T and P, PV = RTn, where n = moles of gaseous products – moles of gaseous reactants. Here, Δn = 1 - 0 = 1 mol. w = −PV = ‒ RTn = ‒
8.3145 J × 351 K × 1 mol = ‒2.92 × 103 J = ‒2.92 kJ K mol
At constant pressure, q = H = 42.4 kJ. ΔE = q + w = 42.4 kJ ‒ 2.92 kJ = 39.5 kJ 163.
Molar mass of XY =
19 .0 g = 144 g/mol 0.132 mol
X: [Kr] 5s24d10; this is cadmium, Cd.
474
CHAPTER 10
LIQUIDS AND SOLIDS
Molar mass Y = 144 – 112.4 = 32 g/mol; Y is sulfur, S. The semiconductor is CdS. The dopant has the electron configuration of bromine, Br. Because Br has one more valence electron than S, doping with Br will produce an n-type semiconductor. 164.
P H vap 1 1 296 J 200.6 g = 5.94 × 104 J/mol Hg ln 1 = − ; H vap = P R g mol T T 1 2 2 2.56 10 −3 torr 5.94 10 4 J/mol 1 1 = ln − P2 8.3145 J/K • mol 573 K 298.2 K 2.56 10 −3 torr = –11.5, P2 = 2.56 × 10 −3 torr/e −11.5 = 253 torr ln P2
1 atm 253 torr 15 .0 L 760 torr PV = n= = 0.106 mol Hg 0.08206 L atm RT 573 K K mol
0.106 mol Hg ×
6.022 10 23 atoms Hg = 6.38 × 1022 atoms Hg mol Hg
Challenge Problems 165.
H = qp = 30.79 kJ; E = qp + w, w = −PV w = −PV = −1.00 atm(28.90 L) = −28.9 L atm ×
101 .3 J = −2930 J L atm
E = 30.79 kJ + (−2.93 kJ) = 27.86 kJ 166.
XeCl2F2, 8 + 2(7) + 2(7) = 36 e−
Polar (bond dipoles do not cancel each other)
Nonpolar (bond dipoles cancel each other)
These are two possible square planar molecular structures for XeCl2F2. One structure has the Cl atoms 90° apart; the other has the Cl atoms 180° apart. The structure with the Cl atoms 90° apart is polar; the other structure is nonpolar. The polar structure will have additional dipole forces, so it has the stronger intermolecular forces and is the liquid. The gas form of XeCl 2F2 is the nonpolar form having the Cl atoms 180° apart.
CHAPTER 10 167.
LIQUIDS AND SOLIDS
475
A single hydrogen bond in H2O has a strength of 21 kJ/mol. Each H2O molecule forms two H bonds. Thus it should take 42 kJ/mol of energy to break all of the H bonds in water. Consider the phase transitions:
Hsub = 6.0 kJ/mol + 40.7 kJ/mol = 46.7 kJ/mol; it takes a total of 46.7 kJ/mol to convert solid H2O to vapor. This would be the amount of energy necessary to disrupt all of the intermolecular forces in ice. Thus (42 ÷ 46.7) × 100 = 90.% of the attraction in ice can be attributed to H bonding. 168.
1 gal
3785 mL 0.998 g 1 mol H 2 O = 210. mol H2O gal mL 18 .02 g
From Table 10.9, the vapor pressure of H2O at 25C is 23.756 torr. The water will evaporate until this partial pressure is reached. nRT V= = P
0.08206 L atm 298 K K mol = 1.64 105 L 1 atm 23 .756 torr 760 torr
210 . mol
Dimension of cube = (1.64 105 L 1 dm3/L)1/3 = 54.7 dm 54.7 dm
1m 1.094 yards 3 ft = 18.0 ft 10 dm m yard
The cube has dimensions of 18.0 ft 18.0 ft 18.0 ft. 169.
The structures of the two C2H6O compounds (20 valence e−) are:
exhibits relatively strong hydrogen bonding
does not exhibit hydrogen bonding
The liquid will have the stronger intermolecular forces. Therefore, the first compound (ethanol) with hydrogen bonding is the liquid and the second compound (dimethyl ether) with the weaker intermolecular forces is the gas.
476 170.
CHAPTER 10 Benzene
LIQUIDS AND SOLIDS
Naphthalene H
H
H
H
H
H
H
H
H
H
H
H
H
LD forces only
H
LD forces only
Note: London dispersion (LD) forces in molecules such as benzene and naphthalene are fairly large. The molecules are flat, and there is efficient surface-area contact between molecules. Large surface-area contact leads to stronger London dispersion forces. Cl
Carbon tetrachloride (CCl4) has polar bonds but is a nonpolar molecule. CCl4 only has LD forces.
C Cl
Cl Cl
In terms of size and shape, the order of the strength of LD forces is: CCl4 < C6H6 < C10H8 The strengths of the LD forces are proportional to size and are related to shape. Although the size of CCl4 is fairly large, the overall spherical shape gives rise to relatively weak LD forces as compared to flat molecules such as benzene and naphthalene. The physical properties given in the problem are consistent with the order listed above. Each of the physical properties will increase with an increase in intermolecular forces. Acetone
Acetic acid O
O H 3C
C
H 3C
CH 3
LD, dipole
C
O
H
LD, dipole, H-bonding
Benzoic acid O C
O
H
LD, dipole, H-bonding
We would predict the strength of interparticle forces of the last three molecules to be: acetone < acetic acid < benzoic acid polar
H-bonding
H-bonding, but large LD forces because of greater size and shape
CHAPTER 10
LIQUIDS AND SOLIDS
477
This ordering is consistent with the values given for boiling point, melting point, and ΔHvap. The overall order of the strengths of intermolecular forces based on physical properties are: acetone < CCl4 < C6H6 < acetic acid < naphthalene < benzoic acid The order seems reasonable except for acetone and naphthalene. Because acetone is polar, we would not expect it to boil at the lowest temperature. However, in terms of size and shape, acetone is the smallest molecule, and the LD forces in acetone must be very small compared to the other molecules. Naphthalene must have very strong LD forces because of its size and flat shape. 171.
NaCl, MgCl2, NaF, MgF2, and AlF3 all have very high melting points indicative of strong intermolecular forces. They are all ionic solids. SiCl4, SiF4, F2, Cl2, PF5, and SF6 are nonpolar covalent molecules. Only London dispersion (LD) forces are present. PCl3 and SCl2 are polar molecules. LD forces and dipole forces are present. In these eight molecular substances, the intermolecular forces are weak and the melting points low. AlCl3 doesn't seem to fit in as well. From the melting point, there are much stronger forces present than in the nonmetal halides, but they aren't as strong as we would expect for an ionic solid. AlCl3 illustrates a gradual transition from ionic to covalent bonding, from an ionic solid to discrete molecules.
172.
Total charge of all iron ions present in a formula unit is +2 to balance the −2 charge from the one O atom. The sum of iron ions in a formula unit is 0.950. Let x = fraction Fe2+ ions in a formula unit and y = fraction of Fe3+ ions present in a formula unit. Setting up two equations: x + y = 0.950 and 2x + 3y = 2.000 Solving: 2x + 3(0.950 − x) = 2.000, x = 0.85 and y = 0.10 0.10 = 0.11 = fraction of iron as Fe3+ ions 0.95
If all Fe2+, then 1.000 Fe2+ ion/O2− ion; 1.000 − 0.950 = 0.050 = vacant sites. 5.0% of the Fe2+ sites are vacant. 173.
Assuming 100.00 g: 1 mol 1 mol 28.31 g O × = 1.769 mol O; 71.69 g Ti × = 1.497 mol Ti 16 .00 g 47 .88 g 1.769 1.497 = 1.182; = 0.8462; the formula is TiO1.182 or Ti0.8462O. 1.497 1.769
For Ti0.8462O, let x = Ti2+ per mol O2− and y = Ti3+ per mol O2−. Setting up two equations and solving: x + y = 0.8462 (mass balance) and 2x + 3y = 2 (charge balance) 2x + 3(0.8462 − x) = 2, x = 0.539 mol Ti2+/mol O2- and y = 0.307 mol Ti3+/mol O2− 0 .539 × 100 = 63.7% of the titanium ions are Ti2+ and 36.3% are Ti3+ (a 1.75 : 1 ion ratio). 0.8462
478 174.
CHAPTER 10
LIQUIDS AND SOLIDS
First, we need to get the empirical formula of spinel. Assume 100.0 g of spinel: 37.9 g Al ×
1 mol Al = 1.40 mol Al 26 .98 g Al
17.1 g Mg ×
1 mol Mg = 0.703 mol Mg 24 .31 g Mg
45.0 g O ×
1 mol O = 2.81 mol O 16 .00 g O
The mole ratios are 2 : 1 : 4.
Empirical formula = Al2MgO4
Assume each unit cell contains an integral value (x) of Al2MgO4 formula units. Each Al2MgO4 formula unit has a mass of 24.31 + 2(26.98) + 4(16.00) = 142.27 g/mol. x formula units
Density =
1 mol 142.27 g 23 mol 3.57 g 6.022 10 formula units = −8 3 (8.09 10 cm) cm3
Solving: x = 8.00 Each unit cell has 8 formula units of Al2MgO4 or 16 Al, 8 Mg, and 32 O ions. 175.
mass Mn volume Cu volume Cu Density Mn mass Mn = = Density Cu volume Mn mass Cu mass Cu volume Mn
The type of cubic cell formed is not important; only that Cu and Mn crystallize in the same type of cubic unit cell is important. Each cubic unit cell has a specific relationship between the cube edge length l and the radius r. In all cases l r. Therefore, V l3 r3. For the mass ratio, we can use the molar masses of Mn and Cu since each unit cell must contain the same number of Mn and Cu atoms. Solving: volume Cu ( rCu ) 3 density Mn mass Mn 54.94 g/mol = = density Cu mass Cu volume Mn 63.55 g/mol (1.056 rCu ) 3 3
density Mn 1 = 0.8645 × = 0.7341 density Cu 1.056
densityMn = 0.7341 × densityCu = 0.7341 × 8.96 g/cm3 = 6.58 g/cm3 176.
Assuming 100.00 g of MO2: 23.72 g O ×
1 mol O = 1.483 mol O 16 .00 g O
1.483 mol O ×
1 mol M = 0.7415 mol M 2 mol O
100.00 g – 23.72 g = 76.28 g M
CHAPTER 10
LIQUIDS AND SOLIDS
Molar mass M =
479
76 .28 g = 102.9 g/mol 0.7415 mol
From the periodic table, element M is rhodium (Rh). The unit cell for cubic closest packing is face-centered cubic (4 atoms/unit cell). The atoms for a face-centered cubic unit cell are assumed to touch along the face diagonal of the cube, so the face diagonal = 4r. The distance between the centers of touching Rh atoms will be the distance of 2r, where r = radius of Rh atom. Face diagonal =
2 l, where l = cube edge.
Face diagonal = 4r = 2 × 269.0 × 10 −12 m = 5.380 × 10 −10 m 2 l = 4r = 5.38 × 10 −10 m, l =
5.380 10 −10 m
= 3.804 × 10 −10 m = 3.804 × 10 −8 cm
2
4 atoms Rh
Density = 177.
1 mol Rh 102.9 g Rh 23 mol Rh 6.0221 10 atoms = 12.42 g/cm3 −8 3 (3.804 10 cm) a
P c
b
As P is lowered, we go from a to b on the phase diagram. The water boils. The boiling of water is endothermic, and the water is cooled (b → c), forming some ice. If the pump is left on, the ice will sublime until none is left. This is the basis of freeze drying.
T
178.
w = −PΔV; assume a constant P of 1.00 atm.
V373 =
nRT = P
1.00 mol
0.08206 L atm 373 K K mol = 30.6 L for 1 mol of water vapor 1.00 atm
Because the density of H2O(l) is 1.00 g/cm3, 1.00 mol of H2O(l) occupies 18.0 cm3 or 0.0180 L. w = −1.00 atm(30.6 L − 0.0180 L) = −30.6 L atm w = −30.6 L atm ×
101.3 J = −3.10 × 103 J = −3.10 kJ L atm
ΔE = q + w = 40.7 kJ − 3.10 kJ = 37.6 kJ 37.6 × 100 = 92.4% of the energy goes to increase the internal energy of the water. 40.7
The remainder of the energy (7.6%) goes to do work against the atmosphere.
480 179.
CHAPTER 10
LIQUIDS AND SOLIDS
For a cube: (body diagonal)2 = (face diagonal)2 + (cube edge length)2 In a simple cubic structure, the atoms touch on cube edge, so the cube edge = 2r, where r = radius of sphere.
2r
2r 4r 2 + 4r 2 = r 8 = 2 2 r
Face diagonal =
(2r ) 2 + (2r ) 2 =
Body diagonal =
( 2 2 r ) 2 + ( 2r ) 2 =
12 r 2 = 2 3 r
The diameter of the hole = body diagonal − 2(radius of atoms at corners). Diameter = 2 3 r − 2r; thus the radius of the hole is: The volume of the hole is 180.
4 3
2 3 r − 2r = ( 3 − 1) r 2
( 3 − 1)r . 3
Using the ionic radii values given in the question, let’s calculate the density of the two structures. Normal pressure: Rb+ and Cl− touch along cube edge (form NaCl structure). Cube edge = l = 2(148 + 181) = 658 pm = 6.58 × 10−8 cm; there are four RbCl units per unit cell. 4(85 .47 ) + 4(35 .45 ) Density = d = = 2.82 g/cm3 6.022 10 23 (6.58 10 −8 ) 3 High pressure: Rb+ and Cl− touch along body diagonal (form CsCl structure). 2r− + 2r+ = 658 pm = body diagonal = l 3 , l = 658 pm/ 3 = 380. pm Each unit cell contains 1 RbCl unit: d =
85 .47 + 35 .45 6.022 10 23 (3.80 10 −8 ) 3
= 3.66 g/cm3
The high-pressure form has the higher density. The density ratio is 3.66/2.82 = 1.30. We would expect this because the effect of pressure is to push things closer together and thus increase density.
Marathon Problem 181.
q = s × m × T; heat loss by metal = heat gain by calorimeter. The change in temperature for the calorimeter is T = 25.2°C 0.2°C – 25.0°C 0.2°C. Including the error limits, T can
CHAPTER 10
LIQUIDS AND SOLIDS
481
range from 0.0 to 0.4°C. Because the temperature change can be 0.0°C, there is no way that the calculated heat capacity has any meaning. The density experiment is also not conclusive. d=
4g 0.42 cm 3
dhigh = dlow =
= 10 g/cm3 (1 significant figure)
5g 0.40 cm 3
3g 0.44 cm 3
= 12.5 g/cm3 = 10 g/cm3 to 1 sig fig = 7 g/cm3
From Table 1.5, the density of copper is 8.96 g/cm3. The results from this experiment cannot be used to distinguish between a density of 8.96 g/cm3 and 9.2 g/cm3. The crystal structure determination is more conclusive. Assuming the metal is copper: volume of unit cell = (600. pm)
Cu mass in unit cell = 4 atoms
−10
3
cm = 2.16 10 − 22 cm 3 1 pm
3 1 10
1 mol Cu 6.022 10
23
atoms
63.55 g Cu = 4.221 × 10 −22 g Cu mol Cu
−22
d=
mass 4.221 10 g = = 1.95 g/cm3 − 22 3 volume 2.16 10 cm
Because the density of Cu is 8.96 g/cm3, then one can assume this metal is not copper. If the metal is not Cu, then it must be kryptonite (as the question reads). Because we don’t know the molar mass of kryptonite, we cannot confirm that the calculated density would be close to 9.2 g/cm3. To improve the heat capacity experiment, a more precise balance is a must and a more precise temperature reading is needed. Also, a larger piece of the metal should be used so that T of the calorimeter has more significant figures. For the density experiment, we would need a more precise balance and a more precise way to determine the volume. Again, a larger piece of metal would help in order to ensure more significant figures in the volume.
CHAPTER 11 PROPERTIES OF SOLUTIONS Review Questions 1.
Mass percent: The percent by mass of the solute in the solution. Mole fraction: The ratio of the number of moles of a given component to the total number of moles of solution. Molarity: The number of moles of solute per liter of solution. Molality: The number of moles of solute per kilogram of solvent. Volume is temperature dependent, whereas mass and the number of moles are not. Only molarity has a volume term so only molarity is temperature dependent.
2.
KF(s) → K+(aq) + F−(aq) H = Hsoln; K+(g) + Cl−(g) → K+(aq) + F−(aq) H = Hhyd KF(s) → K+(g) + F−(g) H1 = −HLE K (g) + F−(g) → K+(aq) + F−(aq) H2 = Hhyd __________________________________________________________ KF(s) → K+(aq) + F−(aq) H = Hsoln = −HLE + Hhyd +
It is true that H1 and H2 have large magnitudes for their values; however, the signs are opposite (H1 is large and positive because it is the reverse of the lattice energy and H2, the hydration energy, is large and negative). These two H values basically cancel out each other giving a Hsoln value close to zero. 3.
“Like dissolves like” refers to the nature of the intermolecular forces. Polar solutes and ionic solutes dissolve in polar solvents because the types of intermolecular forces present in solute and solvent are similar. When they dissolve, the strength of the intermolecular forces in solution are about the same as in pure solute and pure solvent. The same is true for nonpolar solutes in nonpolar solvents. The strength of the intermolecular forces (London dispersion forces) are about the same in solution as in pure solute and pure solvent. In all cases of like dissolves like, the magnitude of Hsoln is either a small positive number (endothermic) or a small negative number (exothermic). For polar solutes in nonpolar solvents and vice versa, Hsoln is a very large, unfavorable value (very endothermic). Because the energetics are so unfavorable, polar solutes do not dissolve in nonpolar solvents and vice versa.
4.
Structure effects refer to solute and solvent having similar polarities in order for solution formation to occur. Hydrophobic solutes are mostly nonpolar substances that are “waterfearing.” Hydrophilic solutes are mostly polar or ionic substances that are “water-loving.” Pressure has little effect on the solubilities of solids or liquids; it does significantly affect the solubility of a gas. Henry’s law states that the amount of a gas dissolved in a solution is directly
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proportional to the pressure of the gas above the solution (C = kP). The equation for Henry’s law works best for dilute solutions of gases that do not dissociate in or react with the solvent. HCl(g) does not follow Henry’s law because it dissociates into H +(aq) and Cl−(aq) in solution (HCl is a strong acid). For O2 and N2, Henry’s law works well since these gases do not react with the water solvent. An increase in temperature can either increase or decrease the solubility of a solid solute in water. It is true that a solute dissolves more rapidly with an increase in temperature, but the amount of solid solute that dissolves to form a saturated solution can either decrease or increase with temperature. The temperature effect is difficult to predict for solid solutes. However, the temperature effect for gas solutes is easier to predict as the solubility of a gas typically decreases with increasing temperature. 5.
o Raoult’s law: Psoln = solventPsolvent ; when a solute is added to a solvent, the vapor pressure of a solution is lowered from that of the pure solvent. The quantity solvent, the mole fraction of solvent, is the fraction that the solution vapor pressure is lowered.
For the experiment illustrated in Figure 11.10 of the text, the beaker of water will stop having a net transfer of water molecules out of the beaker when the equilibrium vapor pressure, PHo2O , is reached. This can never happen. The beaker with the solution wants an equilibrium vapor pressure of PH2O , which is less than PHo2O . When the vapor pressure over the solution is above
PH2O , a net transfer of water molecules into the solution will occur to try to reduce the vapor pressure to PH2O . The two beakers can never obtain the equilibrium vapor pressure they want. The net transfer of water molecules from the beaker of water to the beaker of solution stops after all the water has evaporated. If the solute is volatile, then we can get a transfer of both the solute and solvent back and forth between the beakers. A state can be reached in this experiment where both beakers have the same solute concentration and hence the same vapor pressure. When this state is reached, no net transfer of solute or water molecules occurs between the beakers, so the levels of solution remain constant. When both substances in a solution are volatile, then Raoult’s law applies to both. The total vapor pressure above the solution is the equilibrium vapor pressure of the solvent plus the equilibrium vapor pressure of the solute. Mathematically:
PTOTAL = PA + PB = χ A PAo + χ B PBo where A is either the solute or solvent and B is the other one. 6.
An ideal liquid-liquid solution follows Raoult’s law:
PTOTAL = χ A PAo + χ B PBo A nonideal liquid-liquid solution does not follow Raoult’s law, either giving a total pressure greater than predicted by Raoult’s law (positive deviation) or less than predicted (negative deviation).
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In an ideal solution, the strength of the intermolecular forces in solution are equal to the strength of the intermolecular forces in pure solute and pure solvent. When this is true, Hsoln = 0 and Tsoln = 0. For positive deviations from Raoult’s law, the solution has weaker intermolecular forces in solution than in pure solute and pure solvent. Positive deviations have Hsoln > 0 (are endothermic) and Tsoln < 0. For negative deviations, the solution has stronger intermolecular forces in solution than in pure solute or pure solvent. Negative deviations have Hsoln < 0 (are exothermic) and Tsoln > 0. Examples of each type of solution are: ideal: positive deviations: negative deviations: 7.
benzene-toluene ethanol-hexane acetone-water
Colligative properties are properties of a solution that depend only on the number, not the identity, of the solute particles. A solution of some concentration of glucose (C6H12O6) has the same colligative properties as a solution of sucrose (C 12H22O11) having the same concentration. A substance freezes when the vapor pressure of the liquid and solid are identical to each other. Adding a solute to a substance lowers the vapor pressure of the liquid. A lower temperature is needed to reach the point where the vapor pressures of the solution and solid are identical. Hence, the freezing point is depressed when a solution forms. A substance boils when the vapor pressure of the liquid equals the external pressure. Because a solute lowers the vapor pressure of the liquid, a higher temperature is needed to reach the point where the vapor pressure of the liquid equals the external pressure. Hence, the boiling point is elevated when a solution forms. The equation to calculate the freezing point depression or boiling point elevation is: T = Km where K is the freezing point or boiling point constant for the solvent and m is the molality of the solute. Table 11.5 of the text lists the K values for several solvents. The solvent which shows the largest change in freezing point for a certain concentration of solute is camphor; it has the largest Kf value. Water, with the smallest Kb value, will show the smallest increase in boiling point for a certain concentration of solute. To calculate molar mass, you need to know the mass of the unknown solute, the mass and identity of solvent used, and the change in temperature (T) of the freezing or boiling point of the solution. Since the mass of unknown solute is known, one manipulates the freezing point data to determine the number of moles of solute present. Once the mass and moles of solute are known, one can determine the molar mass of the solute. To determine the moles of solute present, one determines the molality of the solution from the freezing point data and multiplies this by the kilograms of solvent present; this equals the moles of solute present.
8.
Osmotic pressure: the pressure that must be applied to a solution to stop osmosis; osmosis is the flow of solvent into the solution through a semipermeable membrane. The equation to calculate osmotic pressure, , is: = MRT
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where M is the molarity of the solution, R is the gas constant, and T is the Kelvin temperature. The molarity of a solution approximately equals the molality of the solution when 1 kg solvent 1 L solution. This occurs for dilute solutions of water since d H 2O = 1.00 g/cm3. With addition of salt or sugar, the osmotic pressure inside the fruit cells (and bacteria) is less than outside the cell. Water will leave the cells which will dehydrate bacteria present, causing them to die. Dialysis allows the transfer of solvent and small solute molecules through a membrane. To purify blood, the blood is passed through a cellophane tube (the semipermeable membrane); this cellophane tube is immersed in a dialyzing solution which contains the same concentrations of ions and small molecules as in blood but has none of the waste products normally removed by the kidney. As blood is passed through the dialysis machine, the unwanted waste products pass through the cellophane membrane, cleansing the blood. Desalination is the removal of dissolved salts from an aqueous solution. Here, a solution is subjected to a pressure greater than the osmotic pressure and reverse osmosis occurs, i.e., water passes from the solution through the semipermeable membrane back into pure water. Desalination plants can turn seawater with its high salt content into drinkable water. 9.
A strong electrolyte completely dissociates into ions in solution, a weak electrolyte only partially dissociates into ions in solution, and a nonelectrolyte does not dissociate into ions when dissolved in solution. Colligative properties depend on the total number of solute particles in solution. By measuring a property such as freezing point depression, boiling point elevation, or osmotic pressure, we can determine the number of solute particles present from the solute and thus characterize the solute as a strong, weak, or nonelectrolyte. The van’t Hoff factor i is the number of moles of particles (ions) produced for every mol of solute dissolved. For NaCl, i = 2 since Na+ and Cl− are produced in water; for Al(NO3)3, i = 4 since Al3+ and 3 NO3− ions are produced when Al(NO3)3 dissolves in water. In real life, the van’t Hoff factor is rarely the value predicted by the number of ions a salt dissolves into; i is generally something less than the predicted number of ions. This is due to a phenomenon called ion pairing where at any instant a small percentage of oppositely charged ions pair up and act like a single solute particle. Ion pairing occurs most when the concentration of ions is large. Therefore, dilute solutions behave most ideally; here i is close to that determined by the number of ions in a salt.
10.
A colloidal dispersion is a suspension of particles in a dispersing medium. See Table 11.7 of the text for some examples of different types of colloids. Both solutions and colloids have suspended particles in some medium. The major difference between the two is the size of the particles. A colloid is a suspension of relatively large particles as compared to a solution. Because of this, colloids will scatter light while solutions will not. The scattering of light by a colloidal suspension is called the Tyndall effect. Coagulation is the destruction of a colloid by the aggregation of many suspended particles to form a large particle that settles out of solution.
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Active Learning Questions 1.
a. Each beaker wants a specific vapor pressure above it to be at equilibrium. The beaker of water wants a vapor pressure of pure water at the specific temperature of the experiment. Let’s call this value PHo2O . The beaker with the solution needs a different vapor pressure to be at equilibrium. This vapor pressure will be lower than that of pure water by Raoult’s law, Psoln = PHo2O . The beaker of water will stop having a net transfer of water molecules out of the beaker when the equilibrium vapor pressure, PHo2O , is reached. This can never happen. The beaker with the solution wants an equilibrium vapor pressure of PH2O , which is less than PHo2O by Rauolt’s law. When the vapor pressure over the solution is greater than
PH2O , a net transfer of water molecules into the solution will occur to try to reduce the vapor pressure to PH2O . The two beakers can never obtain the equilibrium vapor pressures needed because they want two different values. The net transfer of water molecules from the beaker of water to the beaker of solution stops after all of the water has evaporated. b. Equilibrium is when the rate of two opposing processes are equal to each other. The opposing processes for vapor pressure are the rate that liquid is converted to gas and the rate that gas is converted back to liquid. When the rate of these two processes are equal, there is a constant amount of gas above the liquid which creates a certain gas pressure; this pressure is called the vapor pressure. c. Yes, all the water will end up in the second beaker. The two beakers can never obtain the equilibrium vapor pressure needed because they require two different values. The net transfer of water molecules from the beaker of water (which wants a higher vapor pressure) to the beaker of solution (which wants a lower vapor pressure) stops after all of the water has evaporated. d. In each beaker, water is evaporating, and water is condensing. These two processes keep occurring endlessly. No net change in each beaker will occur once equilibrium is reached. But as explained above, equilibrium can never be reached because two different equilibrium vapor pressures are required. Yes, water does evaporate from the solution, but overall, water is condensing into this beaker faster than water is evaporating. As the net transfer of water molecules into the beaker with the solution occurs, the mole fraction of solvent gets larger. And from Raoult’s law, a higher vapor pressure is required to get to equilibrium. So the amount of water that evaporates increases over time as the solution becomes more dilute. But the vapor pressure above the solution can never reach that of pure water, PHo2O . This is because there will always be solute molecules in the solution, and because it is a solution, the equilibrium vapor pressure is always some value less than PHo2O . 2.
Here there will be a transfer of both substances to and from each beaker. In Figure 11.9, only water could transfer back and forth, the solute could not transfer.
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As time passes, there will be a net transfer of A molecules into the liquid B beaker and vice versa. So each beaker will form solutions containing mixtures of A and B molecules. The liquid levels of the two beakers will become constant when each beaker has identical concentrations of A and B. This will occur when mole fraction of A = mole fraction of B = 0.50. Since liquid B is more volatile (has a higher pure vapor pressure), one would expect the beaker that initially held pure A to have a larger volume than the beaker that held pure B. But the concentrations of A and B in each beaker will be identical once equilibrium is reached. 3.
Freshwater plant cells when placed in salt water will have a net transfer of water molecules out of the cells to equalize the salt concentration inside and outside of the cells. The net result is that the freshwater cells will shrivel up. The saltwater cells when placed in freshwater will have a net transfer of water molecules from outside the cells to inside the cells, causing them to bloat. Again, the driving force is to equalize the concentration of solute molecules between inside and outside the cells.
4.
In an ideal solution, the strength of the intermolecular forces in solution are equal to the strength of the intermolecular forces in pure solute and pure solvent. When this is true, Hsoln = 0. For positive deviations from Raoult’s law, the solution has weaker intermolecular forces in solution than in pure solute and pure solvent. Positive deviations have Hsoln > 0 (are endothermic). For negative deviations, the solution has stronger intermolecular forces than in pure solute or pure solvent. Negative deviations have Hsoln < 0 (are exothermic).
5.
A substance freezes when the vapor pressure of the liquid and solid are identical to each other. Adding a solute to a substance lowers the vapor pressure of the liquid. A lower temperature is needed to reach the point where the vapor pressures of the solution and solid are identical. Hence, the freezing point is depressed when a solution forms. Salt does not get in the way of water freezing. In fact, when a solution freezes, pure H2O(s) is formed; the solid is not a mixture of solid salt and H2O molecules. A substance boils when the vapor pressure of the liquid equals the external pressure. Because a solute lowers the vapor pressure of the liquid, a higher temperature is needed to reach the point where the vapor pressure of the liquid equals the external pressure. Hence, the boiling point is elevated when a solution forms. The salt solute does not bond the water molecules together.
6.
The ice cube has a certain vapor pressure as does the saltwater solution at the specific temperature of the experiment. The saltwater solution has a lower vapor pressure than pure water since a solute is present. Here, the ice cube wants a higher vapor pressure to be at equilibrium while the saltwater wants a lower vapor pressure to be at equilibrium. As vapor is removed from the ice cube to reach the higher equilibrium vapor pressure of the ice, some water vapor will condense in the saltwater solution to lower the vapor pressure to its equilibrium value. This net process continues. Since a solute is always present in the salt water and assuming the temperature is constant, the ice cube and the salt water can never come to an equilibrium value that satisfies both. So the ice cube will eventually disappear; as more vapor condenses in the salt water solution to decrease its vapor pressure, more vapor is extracted from the ice cube to increase its vapor pressure. This continues until the ice cube disappears.
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7.
See Figure 10.48 for a phase diagram of water. Note that the slope of the solid-liquid line is negative. This means that as pressure is applied to ice (from a heavy car travelling on it), the freezing point of water is lowered. As water forms as the applied pressure increases, the added salt can dissolve in the water to form a solution. This solution has a lower vapor pressure than pure water as dictated by Raoult’s law, Psoln = PHo2O . In order to have the ice and solution to have equal vapor pressures so they are at equilibrium, the temperature must decrease. The result is that salt sprinkled on icy roads can cause the ice to melt (the freezing point is lowered.)
8.
The concentration of CO2 gas dissolved in solution is determined by Henry’s law, C = kP where C = concentration of dissolved gas and P = the partial pressure of the gas above the solution. When the two bottles are capped, each has CO 2(g) escape from solution into the head space above the cola. When enough CO2(g) has escaped to satisfy Henry’s law, then the bottles will be at equilibrium between the dissolved CO2 in solution and the pressure of CO2 in the head space above the cola. The bottle with the smaller head space will need less CO 2 molecules above it to satisfy Henry’s law. Hence, the friend’s bottle having more soda and a smaller head space will lose the least amount of dissolved CO 2 and will be most carbonated.
9.
Molarity = moles solute/L of solution; molality = moles solute/kg solvent; mass and moles quantities are not temperature dependent. However, volume does depend on temperature. As temperature increases, the volume of solution increases. Therefore, molarity does depend on the temperature. For freezing point depression and boiling point elevation, the temperature is changing. In the equations to calculate the temperature changes, the concentration unit used should not be temperature dependent. Hence, molality is used in the freezing point depression and the boiling point elevation equations since it is not temperature dependent, unlike molarity.
10.
Molaity = moles solute/kg solvent; the mass of the salt (a) and the molar mass of the salt (b) must be known to determine the moles of solute present. The volume of water added (c) and the density of water (d) must be known so you can determine the mass of solvent present (mass = volume × density). Mass percent = (mass of solute/mass of solution) × 100; the mass of the salt (a) must be known. To calculate the mass of solution, we have the mass of solute, so we need the mass of solvent added (mass of solution = mass of solute + mass of solvent). To calculate the mass of solvent (water) you need the volume of water added (c) and the density of water (d). Molarity = mols solute/volume solution; the mass of the salt (a) and the molar mass of the salt (b) must be known to determine the moles of solute present. The only other quantity needed os the volume of solution (e).
11.
Consider two solutions: solution A contains 9.0 g of urea (molar mass = 60 g/mol) in 1.0 L of solution and solution B contains 18 g of glucose (molar mass of 180 g/mol) in 1 L of solution. These are dilute solutions so 1 L of aqueous solution has a mass of about 1 kg of water. Between these two solutions, the glucose solution has the larger mass percent since twice the mass of glucose was used to prepare the solution. In terms of molarity, 9.0 g urea represents 0.15 mol of solute in 1.0 L solution for a 0.15 M solution, while 18 of glucose represents 0.1 mol of solute in 1.0 L of solution for a 0.10 M solution. Here, the urea solution has the larger concentration in terms of molarity. So yes, it is possible for one solution to have a greater mass percent composition while another has a greater molarity.
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12.
For an acid-base reaction, normality is the mass of a substance that can furnish or accept 1 mol of protons. Monoprotic acids like HCl, HCN, and HC2H3O2 have molarity = normality because 1 mole of these acids donate 1 mol of protons. Bases that contain one OH‒ per formula unit like KOH or NaOH also have molarity = normality because one mol of these bases have 1 mol of OH‒ than can react with 1 mol of protons. Acids that donate two protons per formula unit like H2SO4 and H2CO3 have normality = 2(molarity). Bases that donate two OH ‒ per formula unit like Ca(OH)2 and Ba(OH)2 also have normality = 2(molarity). Another example where normality and molarity are different would be H 3PO4 which can furnish 3 mol of protons per formula unit. For H3PO4, a 1 molar solution would have a normality of 3 N.
13.
For oxidation-reduction reactions, normality is the mass of substance that can accept or furnish 1 mole of electrons. Substances that lose one electron like Na when it forms Na+ have a molarity that equals the normality. Substances that gain one electron like Cu2+ when it forms Cu+ also have a molarity that equals its normality. A substance that gains or loses two moles of electrons like Mn2+ when it forms Mn will have normality = 2(molarity). For another example, a substance that gains or loses 3 moles of electrons like Fe when it forms Fe 3+ has normality = 3(molarity).
14.
In an open beaker, the net change will be the loss of water from the solution as water molecules evaporate and are swept away. As the solution loses water molecules, it becomes more concentrated in solute. This is turn will lower the vapor pressure of the saltwater solution. So the vapor pressure of the salt water solution will decrease over time.
15.
Solutions with positive deviations from Raoult’s law have weaker intermolecular forces in solution that if the solution behaved ideally. With the weaker intermolecular forces, the boiling point of the solution would be lower than if ideal.
16.
In theory, both salts break up into two ions, so the effective solute concentration of each salt is predicted to be 0.20 m. However, due to ion pairing, the observed molality of ions present is typically less than if the salt solution behaved ideally. Ion pairing is worse with MgSO 4 than with NaCl, so the effective ion concentration in the NaCl solution will be greater than in the MgSO4 solution. Thus, the NaCl solution has the higher boiling point. Note in Table 11.6 that the observed van’t Hoff factor (i) is greater for NaCl than for MgSO 4, indicating greater ion pairing in solutions of MgSO4 than for solutions of NaCl.
17.
a. H1 and H2 refer to the breaking of intermolecular forces in pure solute and in pure solvent. H3 refers to the formation of the intermolecular forces in solution between the solute and solvent. Hsoln is the sum H1 + H2 + H3. For solution i, the electrostatic potential diagram illustrates that acetone (CH3COCH3), like water, is a polar substance (each has a red end indicating the partial negative end of the dipole moment and a blue end indicating the partial positive end). For a polar solute in a polar solvent, H1 and H2 will be large and positive, while H3 will be a large negative value. As discussed in section 11.4 on nonideal solutions, acetone-water solutions exhibit negative deviations from Raoult’s law. Acetone and water can hydrogen bond with each other, which gives the solution stronger intermolecular forces as compared to the pure states of both solute and solvent. In the pure state, acetone cannot H−bond with itself. Because acetone and water show negative deviations from Raoult’s law, one would expect
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Hsoln to be slightly negative. Here H3 will be more than negative enough to overcome the large positive value from the H1 and H2 terms combined. For solution ii, these two molecules are named ethanol (CH3CH2OH) and water. Ethanolwater solutions show positive deviations from Raoult’s law. Both substances can hydrogen bond in the pure state, and they can continue this in solution. However, the solute-solvent interactions are somewhat weaker for ethanol-water solutions due to the small nonpolar part of ethanol (CH3CH2 is the nonpolar part of ethanol). This nonpolar part of ethanol slightly weakens the intermolecular forces in solution. So as in part a, when a polar solute and polar solvent are present, H1 and H2 are large and positive, while H3 is large and negative. For positive deviations from Raoult’s law, the interactions in solution are weaker than the interactions in pure solute and pure solvent. Here, Hsoln will be slightly positive because the H3 term is not negative enough to overcome the large, positive H1 and H2 terms combined. For solution iii, as the electrostatic potential diagrams indicate, both heptane (C 7H16) and hexane (C6H14) are nonpolar substances. For a nonpolar solute dissolved in a nonpolar solvent, H1 and H2 are small and positive, while the H3 term is small and negative. These three terms have small values due to the relatively weak London dispersion forces that are broken and formed for solutions consisting of a nonpolar solute in a nonpolar solvent. Because H1, H2, and H3 are all small values, the Hsoln value will be small. Here, heptane and hexane would form an ideal solution because the relative strengths of the London dispersion forces are about equal in pure solute and pure solvent as compared to those LD forces in solution. For ideal solutions, Hsoln = 0. For solution iv, this combination represents a nonpolar solute in a polar solvent. H1 will be small due to the relative weak London dispersion forces which are broken when the solute (C7H16) expands. H2 will be large and positive because of the relatively strong hydrogen bonding interactions that must be broken when the polar solvent (water) is expanded. And finally, the H3 term will be small because the nonpolar solute and polar solvent do not interact with each other. The result is that Hsoln is a large positive value. b. The first solution has negative deviations from Raoult’s law, so ΔT of solution formation will be positive (solution formation is exothermic). For solution ii, which has positive deviations from Raoult’s law, it will have a ΔT of solution formation which is negative (solution formation is endothermic). Solution iii is ideal so there should be no increase in temperature when this solution forms. And solution iv will have a decrease in temperature because of the positive deviations from Raoult’s law. 18.
a. Water boils when the vapor pressure equals the pressure above the water. In an open pan, Patm 1.0 atm. In a pressure cooker, Pinside > 1.0 atm, and water boils at a higher temperature. The higher the cooking temperature, the faster is the cooking time. b. In the CO2 phase diagram in Chapter 10, the triple point is above 1 atm, so CO 2(g) is the stable phase at 1 atm and room temperature. CO 2(l) can't exist at normal atmospheric pressures. Therefore, dry ice sublimes instead of boils. In a fire extinguisher, P > 1 atm, and CO2(l) can exist. When CO2 is released from the fire extinguisher, CO 2(g) forms, as predicted from the phase diagram.
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c. When water freezes from a solution, it freezes as pure water, leaving behind a more concentrated salt solution. Therefore, the melt of frozen sea ice is pure water. d. CaCl2 is a soluble salt. Solutions of CaCl2 will have a freezing point lower than that of pure water. So CaCl2 is spread on roads to lower the freezing point of water so the ice melts. e. NaCl is a soluble ionic compound that breaks up into two ions, while urea is a soluble nonelectrolyte. Solutions of NaCl have an effective solute concentration which is twice that of urea. With twice the solute particles, the NaCl solution has twice the boiling point increase.
Solution Review 1 mol C3 H 7 OH 60.09 g C3 H 7 OH = 9.74 M 1.00 L
585 g C3 H 7 OH
19.
0.100 mol 134 .00 g = 3.35 g Na2C2O4 L mol
20.
0.250 L
21.
1.00 L ×
22.
1.28 g CaCl2 ×
23.
Mol K2S = 2.00 L ×
0.040 mol HCl 1L = 0.040 mol HCl; 0.040 mol HCl = 0.16 L L 0.25 mol HCl = 160 mL
1 mol CaCl 2 1L 1000 mL = 19.9 mL 110 .98 g CaCl 2 0.580 mol CaCl 2 L 1.00 mol K 2S = 2.00 mol K2S L
K2S(s) → 2 K+(aq) + S2−(aq); mol K+ = 2(2.00) = 4.00 mol K+ Mol KHSO4 = 1.00 L ×
1.00 mol KHSO 4 = 1.00 mol KHSO4 L
KHSO4 (s) → K+(aq) + HSO4−(aq); mol K+ = 1.00 mol
M K+ = 24.
total mol K + 4.00 mol +1.00 mol 5.00 mol = 1.67 M K+ = = total volume 2.00 L +1.00 L 3.00 L
a. HNO3(l) → H+(aq) + NO3−(aq)
b. Na2SO4(s) → 2 Na+(aq) + SO42−(aq)
c. Al(NO3)3(s) → Al3+(aq) + 3 NO3−(aq)
d. SrBr2(s) → Sr2+(aq) + 2 Br−(aq)
e. KClO4(s) → K+(aq) + ClO4−(aq)
f.
g. NH4NO3(s) → NH4+(aq) + NO3−(aq)
h. CuSO4(s) → Cu2+(aq) + SO42−(aq)
i.
NaOH(s) → Na+(aq) + OH−(aq)
NH4Br(s) → NH4+(aq) + Br−(aq)
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Questions 25.
As the temperature increases, the gas molecules will have a greater average kinetic energy. A greater fraction of the gas molecules in solution will have a kinetic energy greater than the attractive forces between the gas molecules and the solvent molecules. More gas molecules are able to escape to the vapor phase, and the solubility of the gas decreases.
26.
Henry’s law is obeyed most accurately for dilute solutions of gases that do not dissociate in or react with the solvent. NH3 is a weak base and reacts with water by the following reaction: NH3(aq) + H2O(l) → NH4+(aq) + OH−(aq) O2 will bind to hemoglobin in the blood. Due to these reactions in the solvent, NH3(g) in water and O2(g) in blood do not follow Henry’s law.
27.
Because the solute is volatile, both the water and solute will transfer back and forth between the two beakers. The volume in each beaker will become constant when the concentrations of solute in the beakers are equal to each other. Because the solute is less volatile than water, one would expect there to be a larger net transfer of water molecules into the right beaker than the net transfer of solute molecules into the left beaker. This results in a larger solution volume in the right beaker when equilibrium is reached, i.e., when the solute concentration is identical in each beaker.
28.
Solutions of A and B have vapor pressures less than ideal (see Figure 11.14 of the text), so this plot shows negative deviations from Raoult’s law. Negative deviations occur when the intermolecular forces are stronger in solution than in pure solvent and solute. This results in an exothermic enthalpy of solution. The only statement that is false is e. A substance boils when the vapor pressure equals the external pressure. Because B = 0.6 has a lower vapor pressure at the temperature of the plot than either pure A or pure B, one would expect this solution to require the highest temperature in order for the vapor pressure to reach the external pressure. Therefore, the solution with B = 0.6 will have a higher boiling point than either pure A or pure B. (Note that because P°B > P°A, B is more volatile than A, and B will have a lower boiling point temperature than A).
29.
No, the solution is not ideal. For an ideal solution, the strengths of intermolecular forces in solution are the same as in pure solute and pure solvent. This results in ΔH soln = 0 for an ideal solution. ΔHsoln for methanol-water is not zero. Because ΔHsoln < 0 (heat is released), this solution shows a negative deviation from Raoult’s law.
30.
The micelles form so that the ionic ends of the detergent molecules, the SO4− ends, are exposed to the polar water molecules on the outside, whereas the nonpolar hydrocarbon chains from the detergent molecules are hidden from the water by pointing toward the inside of the micelle. Dirt, which is basically nonpolar, is stabilized in the nonpolar interior of the micelle and is washed away. See the illustration on the following page.
CHAPTER 11
PROPERTIES OF SOLUTIONS
493
= detergent molecule = SO 4= nonpolar hydrocarbon
= dirt
31.
The main intermolecular forces are: hexane (C6H14): London dispersion; chloroform (CHCl3): dipole-dipole, London dispersion; methanol (CH3OH): H-bonding; and H2O: H-bonding (two places) There is a gradual change in the nature of the intermolecular forces (weaker to stronger). Each preceding solvent is miscible in its predecessor because there is not a great change in the strengths of the intermolecular forces from one solvent to the next.
32.
It is true that the sodium chloride lattice must be broken in order to dissolve in water, but a lot of energy is released when the water molecules hydrate the Na + and Cl− ions. These two processes have relatively large values for the amount of energy associated with them, but they are opposite in sign. The end result is they basically cancel each other out resulting in a Hsoln 0. So energy is not the reason why ionic solids like NaCl are so soluble in water. The answer lies in nature’s tendency toward the higher probability of the mixed state. Processes, in general, are favored that result in an increase in disorder because the disordered state is the easiest (most probable) state to achieve. The tendency of processes to increase disorder will be discussed in Chapter 17 when entropy, S, is introduced.
33.
Only statement b is true. A substance freezes when the vapor pressure of the liquid and solid are the same. When a solute is added to water, the vapor pressure of the solution at 0°C is less than the vapor pressure of the solid, and the net result is for any ice present to convert to liquid in order to try to equalize the vapor pressures (which never can occur at 0°C). A lower temperature is needed to equalize the vapor pressure of water and ice, hence, the freezing point is depressed. For statement a, the vapor pressure of a solution is directly related to the mole fraction of solvent (not solute) by Raoult’s law. For statement c, colligative properties depend on the number of solute particles present and not on the identity of the solute. For statement d, the boiling point of water is increased because the sugar solute decreases the vapor pressure of the
494
CHAPTER 11
PROPERTIES OF SOLUTIONS
water; a higher temperature is required for the vapor pressure of the solution to equal the external pressure so boiling can occur. 34.
This is true if the solute will dissolve in camphor. Camphor has the largest K b and Kf constants. This means that camphor shows the largest change in boiling point and melting point as a solute is added. The larger the change in T, the more precise the measurement and the more precise the calculated molar mass. However, if the solute won’t dissolve in camphor, then camphor is no good and another solvent must be chosen which will dissolve the solute.
35.
Adding a solute to a solvent increases the boiling point and decreases the freezing point of the solvent. Thus the solvent is a liquid over a wider range of temperatures when a solute is dissolved.
36.
Table sugar (C12H22O11) and urea [(NH2)2CO] are nonelectrolytes in water, i.e., they stay together when dissolved in water. CaCl2 is a strong electrolyte which breaks up into three ions when dissolved in water. A 1 molal solution of CaCl2 has an effective particle concentration of 3 molal (assuming complete dissociation). If the three solutions have equal molal concentrations, the CaCl2 solution will have many more particles present and will have a lower freezing point; it will be more effective at melting ice on roads.
37.
Colligative properties are ones that only depend on the number of solute particles present and not on the identity of the solute. So if the boiling points of the two aqueous solutions are the same, then the two solutions have the same moles of proteins present. Other colligative properties that will be the same between the two solutions are vapor pressure, boiling point, and osmotic pressure. This assumes that molarity = molality for the solutions, which is a reasonable assumption for dilute solutions.
38.
For an acid-base reaction, normality is the mass of a substance that can furnish or accept 1 mol of protons. a. H2CO3 acid has 2 mol H+ per mol of compound, so normality = 2(molarity). b. H3PO4 has 3 mol H+ per mol of compound, so normality = 3(molarity). c. One of OH‒ reacts with 1 mol of H+ to form 1 mol of H2O. Because Ba(OH)2 has 2 mol of OH‒ per compound, normality = 2(molarity). d. KOH has 1 mol of OH‒ per compound, so normality = molarity. e. HC2H3O2 has 1mol H+ per mol of compound, so normality = molarity.
39.
For oxidation-reduction reactions, normality is the mass of substance that can accept or furnish 1 mole of electrons; this mass is called an equivalent. a. The reduction of Al3+ to Al requires 3 moles of electrons or 3 equivalents, so equivalent mass = 1/3(molar mass of Al). b. Two equivalents are required, so equivalent mass = 1/2(molar mass of Cu).
CHAPTER 11
PROPERTIES OF SOLUTIONS
495
c. Equivalent mass = molar mass of Cu because only 1 mol of electrons is required. d. The oxidation state of oxygen goes from 0 in O 2 to ‒2 in H2O. This is a gain of 2 mol electrons (2 equivalents). However, we have two moles of O in O 2, so equivalent mass = 1/4(molar mass of O2). 40.
In dialysis, the blood is passed through cellophane tube which acts as a semipermeable membrane. The tube is surrounded by a dialyzing solution which has the same ion and small particle concentrations as healthy blood. As the patient’s blood passes through the cellophane tube, waste products in the blood move from the patient’s blood into the dialyzing solution, which cleanses the blood.
41.
Isotonic solutions are those which have identical osmotic pressures. Crenation and hemolysis refer to phenomena that occur when red blood cells are bathed in solutions having a mismatch in osmotic pressures inside and outside the cell. When red blood cells are in a solution having a higher osmotic pressure than that of the cells, the cells shrivel as there is a net transfer of water out of the cells. This is called crenation. Hemolysis occurs when the red blood cells are bathed in a solution having lower osmotic pressure than that inside the cell. Here, the cells rupture as there is a net transfer of water to into the red blood cells.
42.
Ion pairing is a phenomenon that occurs in solution when oppositely charged ions aggregate and behave as a single particle. For example, when NaCl is dissolved in water, one would expect sodium chloride to exist as separate hydrated Na + ions and Cl− ions. A few ions, however, stay together as NaCl and behave as just one particle. Ion pairing increases in a solution as the ion concentration increases (as the molality increases).
43.
A colloid is a suspension of tiny particles is some medium. If the medium is water, these tiny, suspended particles scatter a bright light when shown through the colloidal dispersion. This scattering of light by the suspended particles is called the Tyndall effect. A solution will not scatter a bright light.
44.
The main factor for stabilization seems to be electrostatic repulsion. The center of a colloid particle is surrounded by a layer of same charged ions, with oppositely charged ions forming another charged layer on the outside. Overall, there are equal numbers of charged and oppositely charged ions, so the colloidal particles are electrically neutral. However, since the outer layers are the same charge, the particles repel each other and do not easily aggregate for precipitation to occur. Heating increases the velocities of the colloidal particles. This causes the particles to collide with enough energy to break the ion barriers, allowing the colloids to aggregate and eventually precipitate out. Adding an electrolyte neutralizes the adsorbed ion layers, which allows colloidal particles to aggregate and then precipitate out.
496
CHAPTER 11
PROPERTIES OF SOLUTIONS
Exercises Solution Composition 45.
0.0075 kg × Mass % =
46.
mass of NaCl 11 g 100 = × 100 = 1.9% mass of solution 500. + 75
5.0 g C6H12O6•H2O × Molality =
47.
2.5 mol NaCl 58.44 g = 11g NaCl kg mol NaCl
0.025 mol 1 mol = 0.025 mol; molarity = = 0.25 mol/L 0.1000 L 198.17 g
0.025 mol C 6 H 12 O 6 • H 2 O = 0.25 mol/kg 1.01 g 1 kg 100.0 mL mL 1000 g
Assuming 100.0 g of solution, we would have 10.0 g NaOH and 90.0 g of H 2O.
Molality =
moles solute = kg solvent
1 mol NaOH 40.00 g = 2.78 mol/kg 0.0900 kg H 2O
10.0 g NaOH
1 mol NaOH moles solute 40.00 g = Molarity = = 2.77 mol/L 1 cm3 1L L solution 100.0 g solution 1.109 g 1000 cm3 10.0 g NaOH
48.
Let’s assume 1.000 L of solution. Mol H2SO4 = 1.000 L ×
9.61 mol H 2SO 4 98.09 g = 9.61 mol; 9.61 mol × = 943 g H2SO4 L mol
Mass solution = 1000. cm3 ×
Mol H2O = 577 g × 49.
1.520 g = 1520. g; 1520. solution ‒ 943 g H2SO4 = 577 g H2O cm3
mol H 2SO 4 9.61 1 mol = = 32.0 mol; χ H2 SO4 = = 0.231 mol solution 9.61 + 32.0 18.02 g
Because the density of water is 1.00 g/mL, 100.0 mL of water has a mass of 100. g. Density =
10.0 g H 3 PO 4 + 100 . g H 2 O mass = = 1.06 g/mL = 1.06 g/cm3 volume 104 mL
Mol H3PO4 = 10.0 g × Mol H2O = 100. g ×
1 mol = 0.102 mol H3PO4 97.99 g
1 mol = 5.55 mol H2O 18.02 g
CHAPTER 11
PROPERTIES OF SOLUTIONS
Mole fraction of H3PO4 =
0.102 mol H 3 PO 4 = 0.0180 (0.102 + 5.55) mol
χ H 2O = 1.000 – 0.0180 = 0.9820
50.
Molarity =
0.102 mol H 3 PO 4 = 0.981 mol/L 0.104 L
Molality =
0.102 mol H 3 PO 4 = 1.02 mol/kg 0.100 kg
Molality =
40.0 g C 2 H 6 O 2 1000 g 1 mol C 2 H 6 O 2 = 10.7 mol/kg 60.0 g H 2 O kg 62.07 g
Molarity =
40.0 g C 2 H 6 O 2 1 mol C 2 H 6 O 2 1.05 g 1000 cm3 = 6.77 mol/L 3 100.0 g solution L 62.07 g cm
40.0 g C2H6O2 × χ EG =
51.
1 mol 1 mol = 0.644 mol C2H6O2; 60.0 g H2O × = 3.33 mol H2O 18.02 g 62 .07 g
0.644 = 0.162 = mole fraction ethylene glycol (C2H6O2) 3.33 + 0.644
Hydrochloric acid (HCl): molarity =
38 g HCl 1.19 g soln 1000 cm3 1 mol HCl = 12 mol/L 100. g soln L 36.5 g cm3 soln
molality =
38 g HCl 1000 g 1 mol HCl = 17 mol/kg 62 g solvent kg 36.5 g
38 g HCl ×
1 mol 1 mol = 1.0 mol HCl; 62 g H2O × = 3.4 mol H2O 36 .5 g 18.0 g
mole fraction of HCl = χ HCl =
1 .0 = 0.23 3 .4 + 1 .0
Nitric acid (HNO3): 70. g HNO3 1.42 g soln 1000 cm3 1 mol HNO3 = 16 mol/L 100 . g soln L 63.0 g cm3 soln
70. g HNO 3 1000 g 1 mol HNO 3 = 37 mol/kg 30. g solvent kg 63.0 g
70. g HNO3 × χ HNO3 =
1 mol 1 mol = 1.1 mol HNO3; 30. g H2O × = 1.7 mol H2O 63 .0 g 18.0 g
1 .1 = 0.39 1 .7 + 1 . 1
Sulfuric acid (H2SO4):
497
498
CHAPTER 11
PROPERTIES OF SOLUTIONS
95 g H 2SO 4 1.84 g soln 1000 cm3 1 mol H 2SO 4 = 18 mol/L 3 100 . g soln L 98.1 g H 2SO 4 cm soln
95 g H 2SO 4 1000 g 1 mol = 194 mol/kg 200 mol/kg 5 g H 2O kg 98.1 g
95 g H2SO4 × χ H 2SO 4 =
1 mol 1 mol = 0.97 mol H2SO4; 5 g H2O × = 0.3 mol H2O 18.0 g 98.1 g
0.97 = 0.76 0.97 + 0.3
Acetic acid (CH3CO2H): 99 g CH 3CO 2 H 1.05 g soln 1000 cm3 1 mol = 17 mol/L 3 100 . g soln L 60.05 g cm soln
99 g CH 3CO 2 H 1000 g 1 mol = 1600 mol/kg 2000 mol/kg 1 g H 2O kg 60.05 g
99 g CH3CO2H × χ CH 3CO 2 H =
1 mol 1 mol = 1.6 mol CH3CO2H; 1 g H2O × = 0.06 mol H2O 60.05 g 18.0 g
1 .6 = 0.96 1.6 + 0.06
Ammonia (NH3): 28 g NH3 0.90 g 1000 cm3 1 mol = 15 mol/L 3 100 . g soln L 17 .0 g cm
28 g NH 3 1000 g 1 mol = 23 mol/kg 72 g H 2 O kg 17.0 g
28 g NH3 × χ NH3 =
52.
1 mol 1 mol = 1.6 mol NH3; 72 g H2O × = 4.0 mol H2O 17.0 g 18.0 g
1 .6 = 0.29 4 .0 + 1 .6
a. If we use 100. mL (100. g) of H2O, we need: 0.100 kg H2O ×
2.0 mol KCl 74.55 g = 14.9 g = 15 g KCl kg mol KCl
Dissolve 15 g KCl in 100. mL H2O to prepare a 2.0 m KCl solution. This will give us slightly more than 100 mL, but this will be the easiest way to make the solution. Because we don’t know the density of the solution, we can’t calculate the molarity and use a volumetric flask to make exactly 100 mL of solution. b. If we took 15 g NaOH and 85 g H2O, the volume probably would be less than 100 mL. To make sure we have enough solution, let’s use 100. mL H 2O (100. g H2O). Let x = mass of NaCl.
CHAPTER 11
PROPERTIES OF SOLUTIONS
Mass % = 15 =
499
x × 100, 1500 + 15x = (100.)x, x = 17.6 g 18 g 100 . + x
Dissolve 18 g NaOH in 100. mL H2O to make a 15% NaOH solution by mass. c. In a fashion similar to part b, let’s use 100. mL CH3OH. Let x = mass of NaOH. 0.79 g = 79 g CH3OH mL x Mass % = 25 = × 100, 25(79) + 25x = (100.)x, x = 26.3 g 26 g 79 + x
100. mL CH3OH ×
Dissolve 26 g NaOH in 100. mL CH3OH. d. To make sure we have enough solution, let’s use 100. mL (100. g) of H 2O. Let x = mol C6H12O6. 100. g H2O ×
1 mol H 2 O = 5.55 mol H2O 18.02 g
χ C6H12O6 = 0.10 =
x , (0.10)x + 0.56 = x, x = 0.62 mol C6H12O6 x + 5.55
0.62 mol C6H12O6 ×
180 .2 g = 110 g C6H12O6 mol
Dissolve 110 g C6H12O6 in 100. mL of H2O to prepare a solution with χ C6H12O6 = 0.10. 53.
25 mL C5H12 ×
0.63 g 0.63 g 1 mol = 16 g C5H12; 25 mL × = 0.22 mol C5H12 mL mL 72.15 g
45 mL C6H14 ×
0.66 g 0.66 g 1 mol = 30. g C6H14; 45 mL × = 0.34 mol C6H14 mL mL 86 .17 g
Mass % pentane = χpentane =
54.
16 g mass pentane × 100 = × 100 = 35% 16 g + 30. g total mass
mol pentane 0.22 mol = = 0.39 0.22 mol + 0.34 mol total mol
Molality =
0.22 mol mol pentane = = 7.3 mol/kg 0.030 kg kg hexane
Molarity =
0.22 mol 1000 mL mol pentane = = 3.1 mol/L 25 mL + 45 mL 1L L solution
0.867 g 0.874 g = 43.4 g toluene; 125 mL benzene × = 109 g benzene mL mL mass of toluene 43.4 g Mass % toluene = × 100 = × 100 = 28.5% total mass 43.4 g + 109 g
50.0 mL toluene ×
500
55.
CHAPTER 11
PROPERTIES OF SOLUTIONS
Molarity =
43.4 g toluene 1000 mL 1 mol toluene = 2.69 mol/L 175 mL soln L 92.13 g toluene
Molality =
43.4 g toluene 1000 g 1 mol toluene = 4.32 mol/kg 109 g benzene kg 92.13 g toluene
43.4 g toluene ×
1 mol = 0.471 mol toluene 92.13 g
109 g benzene ×
1 mol benzene 0.471 = 1.40 mol benzene; toluene = = 0.252 78.11 g benzene 0.471 + 1.40
If we have 100.0 mL of wine: 1.00 g 0.789 g = 9.86 g C2H5OH and 87.5 mL H2O × = 87.5 g H2O mL mL 9.86 g Mass % ethanol = × 100 = 10.1% by mass 87 .5 g + 9.86 g
12.5 mL C2H5OH ×
Molality =
56.
9.86 g C 2 H 5 OH 1 mol = 2.45 mol/kg 0.0875 kg H 2 O 46.07 g
1.00 mol acetone 1 mol = 1.00 molal; 1.00 × 103 g C2H5OH × = 21.7 mol C2H5OH 1.00 kg ethanol 46 .07 g
χacetone =
1.00 = 0.0441 1.00 + 21.7
1 mol CH3COCH3 ×
58.08 g CH 3COCH 3 1 mL = 73.7 mL CH3COCH3 mol CH 3COCH 3 0.788 g
1.00 × 103 g ethanol × Molarity = 57.
1 mL = 1270 mL; total volume = 1270 + 73.7 = 1340 mL 0.789 g
1.00 mol = 0.746 M 1.34 L
If we have 1.00 L of solution: 192 .12 g = 263 g citric acid (H3C6H5O7) mol 1.10 g 1.00 × 103 mL solution × = 1.10 × 103 g solution mL 263 g Mass % of citric acid = × 100 = 23.9% 1.10 10 3 g
1.37 mol citric acid ×
In 1.00 L of solution, we have 263 g citric acid and (1.10 × 10 3 − 263) = 840 g of H2O. Molality =
1.37 mol citric acid = 1.6 mol/kg 0.84 kg H 2 O
CHAPTER 11
PROPERTIES OF SOLUTIONS
840 g H2O ×
501
1 mol 1.37 = 47 mol H2O; χ citric acid = = 0.028 18.02 g 47 + 1.37
Because citric acid is a triprotic acid, the number of protons citric acid can provide is three times the molarity. Therefore, normality = 3 × molarity: normality = 3 × 1.37 M = 4.11 N 58.
When expressing concentration in terms of normality, equivalents per liter are determined. For acid-base reactions, equivalents are equal to the moles of H + an acid can donate or the moles of H+ a base can accept. For monoprotic acids like HCl, the equivalents of H + furnished equals the moles of acid present. Diprotic acids like H2SO4 furnish two equivalents of H+ per mole of acid, whereas triprotic acids like H3PO4 furnish three equivalents of H+ per mole of acid. For the bases in this problem, the equivalents of H + accepted equals the number of OH − anions present in the formula (H+ + OH− → H2O). Finally, the equivalent mass of a substance is the mass of acid or base that can furnish or accept 1 mole of protons (H + ions). a. Normality =
0.250 mol HCl 1 equivalent 0.250 equivalent s = L mol HCl L
Equivalent mass = molar mass of HCl = 36.46 g 0.105 mol H 2 SO 4 2 equivalent s 0.210 equivalent s = L mol H 2 SO 4 L
b. Normality =
Equivalent mass = 1/2(molar mass of H2SO4) = 1/2(98.09) = 49.05 g c. Normality =
5.3 10 −2 mol H 3 PO 4 3 equivalent s 0.16 equivalent s = L mol H 3 PO 4 L
Equivalent mass = 1/3(molar mass of H3PO4) = 1/3(97.09) = 32.66 g d. Normality =
0.134 mol NaOH 1 equivalent 0.134 equivalent s = L mol NaOH L
Equivalent mass = molar mass of NaOH = 40.00 g e. Normality =
0.00521 mol Ca (OH ) 2 2 equivalent s 0.0104 equivalent s = L mol Ca (OH ) 2 L
Equivalent mass = 1/2[molar mass of Ca (OH ) 2 ] = 1/2(74.10) = 37.05 g
Energetics of Solutions and Solubility 59.
Using Hess’s law: NaI(s) → Na+(g) + I−(g) Na (g) + I−(g) → Na+(aq) + I−(aq) +
NaI(s) → Na+(aq) + I−(aq)
ΔH = −ΔHLE = −(−686 kJ/mol) ΔH = ΔHhyd = −694 kJ/mol ΔHsoln = −8 kJ/mol
502
CHAPTER 11
PROPERTIES OF SOLUTIONS
ΔHsoln refers to the heat released or gained when a solute dissolves in a solvent. Here, an ionic compound dissolves in water. 60.
a.
CaCl2(s) → Ca2+(g) + 2 Cl−(g) Ca (g) + 2 Cl−(g) → Ca2+(aq) + 2 Cl−(aq)
ΔH = −ΔHLE = −(−2247 kJ) ΔH = ΔHhyd
2+
CaCl2(s) → Ca2+(aq) + 2 Cl−(aq)
ΔHsoln = –46 kJ
−46 kJ = 2247 kJ + ΔHhyd, ΔHhyd −2293 kJ CaI2(s) → Ca2+(g) + 2 I−(g) Ca2+(g) + 2 I−(g) → Ca2+(aq) + 2 I−(aq)
ΔH = −ΔHLE = −(−2059 kJ) ΔH = ΔHhyd
CaI2(s) → Ca2+(aq) + 2 I−(aq)
ΔHsoln = −104 kJ
−104 kJ = 2059 kJ + ΔHhyd, ΔHhyd = –2163 kJ b. The enthalpy of hydration for CaCl2 is more exothermic than for CaI2. Any differences must be due to differences in hydration between Cl− and I−. Thus the smaller chloride ion is more strongly hydrated than the iodide ion. 61.
Both Al(OH)3 and NaOH are ionic compounds. Since the lattice energy is proportional to the charge of the ions, the lattice energy of aluminum hydroxide is greater than that of sodium hydroxide. The attraction of water molecules for Al3+ and OH− cannot overcome the larger lattice energy, and Al(OH)3 is insoluble. For NaOH, the favorable hydration energy is large enough to overcome the smaller lattice energy, and NaOH is soluble.
62.
The dissolving of an ionic solute in water can be thought of as taking place in two steps. The first step, called the lattice-energy term, refers to breaking apart the ionic compound into gaseous ions. This step, as indicated in the problem, requires a lot of energy and is unfavorable. The second step, called the hydration-energy term, refers to the energy released when the separated gaseous ions are stabilized as water molecules surround the ions. Because the interactions between water molecules and ions are strong, a lot of energy is released when ions are hydrated. Thus the dissolution process for ionic compounds can be thought of as consisting of an unfavorable and a favorable energy term. These two processes basically cancel each other out, so when ionic solids dissolve in water, the heat released or gained is minimal, and the temperature change is minimal.
63.
Water is a polar solvent and dissolves polar solutes and ionic solutes. Carbon tetrachloride (CCl4) is a nonpolar solvent and dissolves nonpolar solutes (like dissolves like). To predict the polarity of the following molecules, draw the correct Lewis structure and then determine if the individual bond dipoles cancel or not. If the bond dipoles are arranged in such a manner that they cancel each other out, then the molecule is nonpolar. If the bond dipoles do not cancel each other out, then the molecule is polar. a. KrF2, 8 + 2(7) = 22 e−
nonpolar; soluble in CCl4
b. SF2, 6 + 2(7) = 20 e−
polar; soluble in H2O
CHAPTER 11
PROPERTIES OF SOLUTIONS
c. SO2, 6 + 2(6) = 18 e−
503
d. CO2, 4 + 2(6) = 16 e− + 1 more
polar; soluble in H2O
nonpolar; soluble in CCl4
e. MgF2 is an ionic compound so it is soluble in water. f.
CH2O, 4 + 2(1) + 6 = 12 e−
polar; soluble in H2O
64.
g. C2H4, 2(4) + 4(1) = 12 e−
nonpolar (like all compounds made up of only carbon and hydrogen); soluble in CCl4
Water is a polar solvent and dissolves polar solutes and ionic solutes. Hexane (C6H14) is a nonpolar solvent and dissolves nonpolar solutes (like dissolves like). a. Water; Cu(NO3)2 is an ionic compound. b. C6H14; CS2 is a nonpolar molecule.
c. Water; CH3OH is polar.
d. C6H14; the long nonpolar hydrocarbon chain favors a nonpolar solvent (the molecule is mostly nonpolar). e. Water; HCl is polar. 65.
f.
C6H14; C6H6 is nonpolar.
Water exhibits H-bonding in the pure state and is classified as a polar solvent. Water will dissolve other polar solutes and ionic solutes. a. NH3; NH3 is capable of H-bonding, unlike PH3. b. CH3CN; CH3CN is polar, while CH3CH3 is nonpolar. c. CH3CO2H; CH3CO2H is capable of H-bonding, unlike the other compound.
66.
As the length of the hydrocarbon chain increases, the solubility decreases. The ‒OH end of the alcohols can hydrogen-bond with water. The hydrocarbon chain, however, is basically nonpolar and interacts poorly with water. As the hydrocarbon chain gets longer, a greater portion of the molecule cannot interact with the water molecules, and the solubility decreases; i.e., the effect of the ‒OH group decreases as the alcohols get larger.
504 67.
CHAPTER 11 C = kP,
8.21 10 −4 mol = k × 0.790 atm, k = 1.04 × 10 −3 mol/L•atm L
C = kP, C =
68.
PROPERTIES OF SOLUTIONS
C = kP =
1.04 10 −4 mol × 1.10 atm = 1.14 × 10 −3 mol/L L atm
1.3 10 −3 mol 1 atm × 120 torr × = 2.1 × 10−4 mol/L 760 torr L atm
Vapor Pressures of Solutions 69.
Psoln = χ C2H5OHPCo2H5OH ; χ C2H5OH = 53.6 g C3H8O3 ×
moles of C 2 H 5 OH total moles in solution
1 mol C 3 H 8 O 3 = 0.582 mol C3H8O3 92.09 g
133.7 g C2H5OH ×
1 mol C 2 H 5 OH = 2.90 mol C2H5OH 46.07 g
Total mol = 0.582 + 2.90 = 3.48 mol 113 torr = 70.
2.90 mol PCo2 H5OH , PCo2 H 5OH = 136 torr 3.48 mol
P = χP°, 19.6 torr = χ(23.8 torr), χ = 0.824 = mole fraction of water P = 0.824(71.9 torr) = 59.2 torr = vapor pressure at 45oC for glucose solution When equal moles of NaCl is substituted for glucose, we must consider that NaCl breaks up into two ions in solution. So the moles of solute particles will be double for the NaCl solution as compared to the glucose solution (assuming complete dissociation of NaCl). If we have 100.0 moles of the glucose solution, χ = 0.824 determines that we have 82.4 moles of H2O and 17.6 moles of glucose. The moles of solute particles if 17.6 moles of NaCl is substituted for the glucose will be 2(17.6) = 35.2 mol. χ=
moles of H 2 O 82.4 = = 0.701; P = 0.701(71.9 torr) = 50.4 torr total moles in solution 35.2 + 82.4
As expected the NaCl solution with more solute particles has the lower vapor pressure.
71.
The normal boiling point of a substance is the boiling point at 1 atm pressure. So for this problem, P° = 760. torr at 34.5°C (the normal boiling point of diethyl ether). P = χP°; 698 torr = χ(760. torr), χ = 0.918 = mole fraction of diethyl ether
72.
The normal boiling point of a substance is the boiling point at 1 atm pressure. So, for this problem, P° = 760.0 torr at 64.7°C (the normal boiling point of methanol). P = χP° = 0.268(760.0 torr), P = 204 torr = vapor pressure of the solution
CHAPTER 11 73.
PROPERTIES OF SOLUTIONS
505
Let x = mol octane, so mol heptane = 2x. L χ oct =
mol octane in solution x = = 1/3 = 0.3333; χ Lhept = 1.0000 – 0.3333 = 0.6667 total mol in solution x + 2x
L o = 0.3333(31.2 torr) = 10.4 torr; Phept = 0.6667(92.0 torr) = 61.3 torr Poct = χ oct Poct
Ptotal = Poct + Phept = 10.4 + 61.3 = 71.7 torr 74.
χ LA =
mol A in solution 2.0 = = 0.40; χ LB = 1.00 ‒ 0.40 = 0.60; P = χP° total mol in solution 2.0 + 3.0
Ptotal = PA + PB, 240. torr = 0.40(150. torr) + 0.60( PB ), PB = 3.0 × 102 torr 75.
a. 25 mL C5H12 × 45 mL C6H14 × L pen =
0.63 g 1 mol = 0.22 mol C5H12 mL 72.15 g 0.66 g 1 mol = 0.34 mol C6H14; total mol = 0.22 + 0.34 = 0.56 mol mL 86 .17 g
mol pentane in solution 0.22 mol = = 0.39, χ Lhex = 1.00 – 0.39 = 0.61 total mol in solution 0.56 mol
o = 0.39(511 torr) = 2.0 × 102 torr; Phex = 0.61(150. torr) = 92 torr Ppen = χ Lpen Ppen
Ptotal = Ppen + Phex = 2.0 × 102 + 92 = 292 torr = 290 torr b. From Chapter 5 on gases, the partial pressure of a gas is proportional to the number of moles of gas present (at constant volume and temperature). For the vapor phase: χ Vpen =
76.
Ppen mol pentane in vapor 2.0 10 2 torr = 0.69 = = total mol vapor Ptotal 290 torr
Note: In the Solutions Guide, we added V or L superscripts to the mole fraction symbol to emphasize for which value we are solving. If the L or V is omitted, then the liquid phase is assumed. 0.0300 mol CH 2 Cl 2 L = Ptotal = PCH 2Cl 2 + PCH 2 Br2 ; P = χ L P o ; χ CH = 0.375 2 Cl 2 0.0800 mol total Ptotal = 0.375(133 torr) + (1.000 − 0.375)(11.4 torr) = 49.9 + 7.13 = 57.0 torr V = In the vapor: χ CH 2 Cl 2
PCH 2Cl 2 Ptotal
=
49.9 torr V = 0.875; χ CH = 1.000 – 0.875 = 0.125 2 Br2 57.0 torr
Note: In the Solutions Guide, we added V or L superscripts to the mole fraction symbol to emphasize for which value we are solving. If the L or V is omitted, then the liquid phase is assumed. 77.
Ptotal = Pmeth + Pprop, 174 torr = χ Lmeth (303 torr ) + χ Lprop (44.6 torr); χ Lprop = 1.000 − χ Lmeth
506
CHAPTER 11 174 = 303 χ Lmeth + (1.000 − χ Lmeth )44.6 torr ,
PROPERTIES OF SOLUTIONS
129 = χ Lmeth = 0.500 258
χ Lprop = 1.000 − 0.500 = 0.500 78.
o o Ptol = χ LtolPtol , Ppen = χ Lben Pben ; for the vapor, χ VA = PA/Ptotal. Because the mole fractions of benzene and toluene are equal in the vapor phase, Ptol = Pben . o o o χ LtolPtol = χ Lben Pben = (1.00 − χ Ltol )Pben , χ Ltol (28 torr ) = (1.00 − χ Ltol ) 95 torr
123 χ Ltol = 95, χ Ltol = 0.77; χ Lben = 1.00 – 0.77 = 0.23 79.
Compared to H2O, solution d (methanol-water) will have the highest vapor pressure since methanol is more volatile than water (PHo 2O = 23.8 torr at 25°C). Both solution b (glucosewater) and solution c (NaCl-water) will have a lower vapor pressure than water by Raoult's law. NaCl dissolves to give Na+ ions and Cl− ions; glucose is a nonelectrolyte. Because there are more solute particles in solution c, the vapor pressure of solution c will be the lowest.
80.
Solution d (methanol-water); methanol is more volatile than water, which will increase the total vapor pressure to a value greater than the vapor pressure of pure water at this temperature.
81.
The first diagram shows positive deviation from Raoult's law. This occurs when the solutesolvent interactions are weaker than the interactions in pure solvent and pure solute. The second diagram illustrates negative deviation from Raoult's law. This occurs when the solutesolvent interactions are stronger than the interactions in pure solvent and pure solute. The third diagram illustrates an ideal solution with no deviation from Raoult's law. This occurs when the solute-solvent interactions are about equal to the pure solvent and pure solute interactions. a. These two molecules are named acetone (CH3COCH3) and water. As discussed in section 11.4 on nonideal solutions, acetone-water solutions exhibit negative deviations from Raoult’s law. Acetone and water have the ability to hydrogen bond with each other, which gives the solution stronger intermolecular forces as compared to the pure states of both solute and solvent. In the pure state, acetone cannot H−bond with itself. So the middle diagram illustrating negative deviations from Raoult’s law is the correct choice for acetonewater solutions. b. These two molecules are named ethanol (CH3CH2OH) and water. Ethanol-water solutions show positive deviations from Raoult’s law. Both substances can hydrogen bond in the pure state, and they can continue this in solution. However, the solute-solvent interactions are somewhat weaker for ethanol-water solutions due to the significant nonpolar part of ethanol (CH3−CH2 is the nonpolar part of ethanol). This nonpolar part of ethanol weakens the intermolecular forces in solution. So the first diagram illustrating positive deviations from Raoult’s law is the correct choice for ethanol-water solutions. c. These two molecules are named heptane (C7H16) and hexane (C6H14). Heptane and hexane are very similar nonpolar substances; both are composed entirely of nonpolar C −C bonds and relatively nonpolar C−H bonds, and both have a similar size and shape. Solutions of
CHAPTER 11
PROPERTIES OF SOLUTIONS
507
heptane and hexane should be ideal. So the third diagram illustrating no deviation from Raoult’s law is the correct choice for heptane-hexane solutions. d. These two molecules are named heptane (C 7H16) and water. The interactions between the nonpolar heptane molecules and the polar water molecules will certainly be weaker in solution as compared to the pure solvent and pure solute interactions. This results in positive deviations from Raoult’s law (the first diagram). 82.
a. An ideal solution would have a vapor pressure at any mole fraction of H 2O between that of pure propanol and pure water (between 74.0 and 71.9 torr). The vapor pressures of the various solutions are not between these limits, so water and propanol do not form ideal solutions. b. From the data, the vapor pressures of the various solutions are greater than if the solutions behaved ideally (positive deviation from Raoult’s law). This occurs when the intermolecular forces in solution are weaker than the intermolecular forces in pure solvent and pure solute. This gives rise to endothermic (positive) ΔH soln values. c. The interactions between propanol and water molecules are weaker than between the pure substances because the solutions exhibit a positive deviation from Raoult’s law. d. At χ H 2O = 0.54, the vapor pressure is highest as compared to the other solutions. Because a solution boils when the vapor pressure of the solution equals the external pressure, the χ H 2O = 0.54 solution should have the lowest normal boiling point; this solution will have a vapor pressure equal to 1 atm at a lower temperature as compared to the other solutions.
Colligative Properties 83.
Molality = m = ΔTb = Kbm =
27.0 g N 2 H 4 CO 1000 g 1 mol N 2 H 4 CO mol solute = = 3.00 molal kg solvent 150.0 g H 2 O kg 60.06 g N 2 H 4 CO
0.51 C × 3.00 molal = 1.5°C molal
The boiling point is raised from 100.0 to 101.5°C (assuming P = 1 atm). 84.
Molality = m =
25.6 g C10 H 8 1 mol C10 H 8 mol solute = = 0.666 molal kg solvent 0.300 kg benzene 128.16 g
ΔTf = Kfm, ΔTf =
5.12 °C × 0.666 molal = 3.41°C molal
The freezing point is lowered from 5.5°C ‒ 3.41°C = 2.1°C. 85.
ΔTb = 77.85°C − 76.50°C = 1.35°C; m = Mol biomolecule = 0.0150 kg solvent ×
ΔTb 1.35C = = 0.268 mol/kg Kb 5.03 C kg / mol
0.268 mol hydrocarbon = 4.02 × 10 −3 mol kg solvent
508
CHAPTER 11
PROPERTIES OF SOLUTIONS
From the problem, 2.00 g biomolecule was used. This mass must contain 4.02 × 10 −3 mol biomolecule. The molar mass of the biomolecule is:
2.00 g 4.02 10 −3 mol 86.
m=
= 498 g/mol
ΔTf 1.02°C = = 0.199 mol/kg Kf 5.12 °C kg/mol
Mol unknown compound = 0.250 kg solvent ×
0.199 mol unknown = 0.0498 mol unknown kg solvent
From the problem, 6.62 g of the unknown compound was used. This mass must contain 0.0498 mol unknown. The molar mass of the unknown compound is: 6.62 g = 133 g/mol 0.0498 mol
87.
ΔTf = Kfm, ΔTf = 1.50°C = 0.200 kg H2O ×
88.
1.86 C × m, m = 0.806 mol/kg molal
0.806 mol C 3 H 8 O 3 92.09 g C 3 H 8 O 3 kg H 2 O mol C 3 H 8 O 3
ΔTf = 25.50°C − 24.59°C = 0.91°C = Kfm, m = Mass H2O = 0.0100 kg t-butanol
89.
Molality = m =
= 14.8 g C3H8O3
0.91 C = 0.10 mol/kg 9.1 C / molal
0.10 mol H 2 O 18.02 g H 2 O = 0.018 g H2O kg t - butanol mol H 2 O
50.0 g C 2 H 6 O 2 1000 g 1 mol = 16.1 mol/kg 50.0 g H 2 O kg 62.07 g
ΔTf = Kfm = 1.86°C/molal × 16.1 molal = 29.9°C; Tf = 0.0°C − 29.9°C = –29.9°C ΔTb = Kbm = 0.51°C/molal × 16.1 molal = 8.2°C; Tb = 100.0°C + 8.2°C = 108.2°C 90.
m=
Tf 25.0 C = = 13.4 mol C2H6O2/kg Kf 1.86 C kg / mol
Because the density of water is 1.00 g/cm3, the moles of C2H6O2 needed are: 15.0 L H2O ×
1.00 kg H 2 O 13.4 mol C 2 H 6 O 2 = 201 mol C2H6O2 L H 2O kg H 2 O
Volume C2H6O2 = 201 mol C2H6O2 × ΔTb = Kbm =
62.07 g 1 cm3 = 11,200 cm3 = 11.2 L mol C 2 H 6 O 2 1.11 g
0.51 C × 13.4 molal = 6.8°C; Tb = 100.0°C + 6.8°C = 106.8°C molal
CHAPTER 11 91.
PROPERTIES OF SOLUTIONS
ΔTf = Kfm, m =
509
Tf 2.63o C 6.6 10 −2 mol reserpine = = Kf kg solvent 40. o C kg/mol
The moles of reserpine present is: 0.0250 kg solvent ×
6.6 10 −2 mol reserpine = 1.7 × 10−3 mol reserpine kg solvent
From the problem, 1.00 g reserpine was used, which must contain 1.7 × 10−3 mol reserpine. The molar mass of reserpine is:
1.00 g 1.7 10 −3 mol 92.
m =
= 590 g/mol (610 g/mol if no rounding of numbers)
Tb 0.55 o C = = 0.32 mol/kg Kb 1.71 o C kg/mol
Mol hydrocarbon = 0.095 kg solvent ×
0.32 mol hydrocarbon = 0.030 mol hydrocarbon kg solvent
From the problem, 3.75 g hydrocarbon was used, which must contain 0.030 mol hydrocarbon. The molar mass of the hydrocarbon is: 3.75 g = 130 g/mol (120 g/mol if no rounding of numbers) 0.030 mol
93.
a. M =
1.0 g protein 1 mol = 1.1 × 10 −5 mol/L; π = MRT 4 L 9.0 10 g
At 298 K: π =
1.1 10 −5 mol L
760 torr 0.08206 L atm × 298 K × , π = 0.20 torr atm K mol
Because d = 1.0 g/cm3, 1.0 L solution has a mass of 1.0 kg. Because only 1.0 g of protein is present per liter of solution, 1.0 kg of H2O is present to the correct number of significant figures, and molality equals molarity. ΔTf = Kfm =
1.86 C × 1.1 × 10 −5 molal = 2.0 × 10 −5 °C molal
b. Osmotic pressure is better for determining the molar mass of large molecules. A temperature change of 10 −5 °C is very difficult to measure. A change in height of a column of mercury by 0.2 mm (0.2 torr) is not as hard to measure precisely. 94.
m=
ΔT 0.406 o C = = 0.218 mol/kg Kf 1.86 o C/molal
= MRT, where M = mol/L; we must assume that molarity = molality so that we can calculate the osmotic pressure. This is a reasonable assumption for dilute solutions when 1.00 kg of water 1.00 L of solution. Assuming NaCl exists as Na+ and Cl– ions in solution, a 0.218 m solution corresponds to 6.37 g NaCl dissolved in 1.00 kg of water. The volume of solution may be a
510
CHAPTER 11
PROPERTIES OF SOLUTIONS
little larger than 1.00 L but not by much (to three sig. figs.). The assumption that molarity = molality will be good here. = (0.218 M)(0.08206 L atm/K • mol)(298 K) = 5.33 atm
0.745 torr
95.
M=
1 atm 760 torr
π = 3.98 × 10−5 mol/L = 0.08206 L atm RT 300 . K K mol
3.98 10 −5 mol = 3.98 × 10−5 mol catalase L 10.00 g Molar mass = = 2.51 × 105 g/mol 3.98 10 −5 mol
1.00 L ×
96.
π = MRT, π = 18.6 torr ×
0.08206 L atm 1 atm =M× × 298 K, M = 1.00 × 10−3 mol/L 760 torr K mol
1.00 10 −3 mol protein = 2.0 × 10−6 mol protein L 0.15 g Molar mass = = 7.5 × 104 g/mol 2.0 10 −6 mol
Mol protein = 0.0020 L ×
97.
π = MRT, M =
RT
=
15 atm = 0.62 M 0.08206 L atm 295 K K mol
0.62 mol 342.30 g = 212 g/L 210 g/L L mol C12 H 22 O11
Dissolve 210 g of sucrose in some water and dilute to 1.0 L in a volumetric flask. To get 0.62 ±0.01 mol/L, we need 212 ±3 g sucrose. 98.
M=
RT
=
15 atm = 0.62 M solute particles 0.08206 L atm 295 K K mol
This represents the total molarity of the solute particles. NaCl is a soluble ionic compound that breaks up into two ions, Na+ and Cl−. Therefore, the concentration of NaCl needed is 0.62/2 = 0.31 M; this NaCl concentration will produce a 0.62 M solute particle solution assuming complete dissociation. 1.0 L ×
0.31 mol NaCl 58.44 g NaCl = 18.1 18 g NaCl L mol NaCl
Dissolve 18 g of NaCl in some water and dilute to 1.0 L in a volumetric flask. To get 0.31 ±0.01 mol/L, we need 18.1 g ±0.6 g NaCl in 1.00 L solution.
CHAPTER 11
PROPERTIES OF SOLUTIONS
511
Properties of Electrolyte Solutions 99.
Na3PO4(s) → 3 Na+(aq) + PO43−(aq), i = 4.0; CaBr2(s) → Ca2+(aq) + 2 Br−(aq), i = 3.0 KCl(s) → K+(aq) + Cl−(aq), i = 2.0 The effective particle concentrations of the solutions are (assuming complete dissociation): 4.0(0.010 molal) = 0.040 molal for the Na3PO4 solution; 3.0(0.020 molal) = 0.060 molal for the CaBr2 solution; 2.0(0.020 molal) = 0.040 molal for the KCl solution; slightly greater than 0.020 molal for the HF solution because HF only partially dissociates in water (it is a weak acid). a. The 0.010 m Na3PO4 solution and the 0.020 m KCl solution both have effective particle concentrations of 0.040 m (assuming complete dissociation), so both of these solutions should have the same boiling point as the 0.040 m C6H12O6 solution (a nonelectrolyte). b. P = χP°; as the solute concentration decreases, the solvent’s vapor pressure increases because χ increases. Therefore, the 0.020 m HF solution will have the highest vapor pressure because it has the smallest effective particle concentration. c. ΔT = Kfm; the 0.020 m CaBr2 solution has the largest effective particle concentration, so it will have the largest freezing point depression (largest ΔT).
100.
101.
The solutions of C12H22O11, NaCl, and CaCl2 will all have lower freezing points, higher boiling points, and higher osmotic pressures than pure water. The solution with the largest particle concentration will have the lowest freezing point, the highest boiling point, and the highest osmotic pressure. The CaCl2 solution will have the largest effective particle concentration because it produces three ions per mole of compound. a. pure water
b. CaCl2 solution
d. pure water
e. CaCl2 solution
a. m =
c. CaCl2 solution
5.0 g NaCl 1 mol = 3.4 molal; NaCl(aq) → Na+(aq) + Cl−(aq), i = 2.0 0.025 kg 58.44 g
ΔTf = iKfm = 2.0 × 1.86°C/molal × 3.4 molal = 13°C; Tf = −13°C ΔTb = iKbm = 2.0 × 0.51°C/molal × 3.4 molal = 3.5°C; Tb = 103.5°C b.
m=
2.0 g Al ( NO 3 ) 3 1 mol = 0.63 mol/kg 0.015 kg 213 .01 g
Al(NO3)3(aq) → Al3+(aq) + 3 NO3−(aq), i = 4.0 ΔTf = iKfm = 4.0 × 1.86°C/molal × 0.63 molal = 4.7°C; T f = −4.7°C ΔTb = iKbm = 4.0 × 0.51°C/molal × 0.63 molal = 1.3°C; Tb = 101.3°C
512 102
CHAPTER 11
PROPERTIES OF SOLUTIONS
NaCl(s) → Na+(aq) + Cl−(aq), i = 2.0 π = iMRT = 2.0 ×
0.10 mol L
0.08206 L atm × 293 K = 4.8 atm K mol
A pressure greater than 4.8 atm should be applied to ensure purification by reverse osmosis. 103.
Na3PO4(s) → 3 Na+(aq) + PO43−(aq); if the solution is ideal, i = 4.0. ΔTf = iKfm = 4.0 × 1.86 °C kg/mol × 0.25 mol/kg = 1.9°C If ideal, the solution would have a freezing point of −1.9°C. Since the freezing point is only −1.6°C, the solution is not ideal and there is some ion pairing in solution. 0.10 mol
0.08206 L atm × 298 K, i = 1.2 K mol L The observed van’t Hoff factor is 1.2, well below 2.0 which is the expected value assuming complete dissociation of MgSO4.
104.
π = iMRT, 2.9 atm = i ×
105.
There are six cations and six anions in the illustration which indicates six solute formula units initially. There are a total of 10 solute particles in solution (a combined ion pair counts as one solute particle). So the value for the van’t Hoff factor is:
i =
moles of particles in solution moles of solute dissolved
=
10 = 1.67 6
106.
From Table 11.6, MgSO4 has an observed i value of 1.3 while the observed i value for NaCl is 1.9. Both salts have an expected i value of 2. The expected i value for a salt is determined by assuming 100% of the salt breaks up into separate cations and anions. The MgSO 4 solu-tion is furthest from the expected i value because it forms the most combined ion pairs in solution. So the figure on the left with the most combined ion pairs represents the MgSO4 solution. The figure on the right represents the NaCl solution. When NaCl is in solution, it has very few combined ion pairs and, hence, has a van’t Hoff factor very close to the expected i value.
107.
a. MgCl2(s) → Mg2+(aq) + 2 Cl−(aq), i = 3.0 mol ions/mol solute ΔTf = iKfm = 3.0 × 1.86 °C/molal × 0.050 molal = 0.28°C Assuming water freezes at 0.00°C, the freezing point would be –0.28°C. freezes at 0.00°C.) ΔTb = iKbm = 3.0 × 0.51 °C/molal × 0.050 molal = 0.077°C; T b = 100.077°C (Assuming water boils at 100.000°C.) b. FeCl3(s) → Fe3+(aq) + 3 Cl−(aq), i = 4.0 mol ions/mol solute ΔTf = iKfm = 4.0 × 1.86 °C/molal × 0.050 molal = 0.37°C; Tf = −0.37°C ΔTb = iKbm = 4.0 × 0.51 °C/molal × 0.050 molal = 0.10°C; Tb = 100.10°C
CHAPTER 11 108.
PROPERTIES OF SOLUTIONS
513
a. MgCl2, i (observed) = 2.7 ΔTf = iKfm = 2.7 × 1.86 °C/molal × 0.050 molal = 0.25°C; Tf = −0.25°C ΔTb = iKbm = 2.7 × 0.51 °C/molal × 0.050 molal = 0.069°C; Tb = 100.069°C b. FeCl3, i (observed) = 3.4 ΔTf = iKfm = 3.4 × 1.86 °C/molal × 0.050 molal = 0.32°C; Tf = –0.32°C ΔTb = iKbm = 3.4 × 0.51°C/molal × 0.050 molal = 0.087°C; T b = 100.087°C
109.
ΔTf = iKfm, i = i=
Tf 0.110 C = = 2.63 for 0.0225 m CaCl2 Kf m 1.86 C / molal 0.0225 molal
1.330 0.440 = 2.60 for 0.0910 m CaCl2; i = = 2.57 for 0.278 m CaCl2 1.86 0.0910 1.86 0.278
iave = (2.63 + 2.60 + 2.57)/3 = 2.60 Note that i is less than the ideal value of 3.0 for CaCl2. This is due to ion pairing in solution. Also note that as molality increases, i decreases. More ion pairing appears to occur as the solute concentration increases. 110.
For CaCl2: i =
Tf 0.440 o C = = 2.6 K f m 1.86 o C/molal 0.091 molal
Percent CaCl2 ionized = For CsCl: i =
2 .6 − 1 .0 × 100 = 80.%; 20.% ion association occurs. 3 .0 − 1 .0
Tf 0.320 o C = = 1.9 K f m 1.86 o C/molal 0.091 molal
Percent CsCl ionized =
1 .9 − 1 .0 × 100 = 90.%; 10% ion association occurs. 2 .0 − 1 .0
The ion association is greater in the CaCl2 solution. 111.
a. TC = 5(TF − 32)/9 = 5(−29 − 32)/9 = −34°C Assuming the solubility of CaCl2 is temperature independent, the molality of a saturated CaCl2 solution is: 74.5 g CaCl 2 1000 g 1 mol CaCl 2 6.71 mol CaCl 2 = 100 .0 g H 2 O kg 110 .98 g CaCl 2 kg H 2 O
ΔTf = iKfm = 3.00 × 1.86 °C kg/mol × 6.71 mol/kg = 37.4°C Assuming i = 3.00, a saturated solution of CaCl2 can lower the freezing point of water to −37.4°C. Assuming these conditions, a saturated CaCl2 solution should melt ice at −34°C (−29°F).
514
CHAPTER 11
PROPERTIES OF SOLUTIONS
b. From Exercise 109, i 2.6; ΔTf = iKfm = 2.6 × 1.86 × 6.71 = 32°C; Tf = −32°C. Assuming i = 2.6, a saturated CaCl2 solution will not melt ice at −34°C (−29°F). 112.
= iMRT, M =
π = iRT
2.50 atm = 5.11 × 10 −2 mol/L 0.08206 L atm 2.00 298 K K mol
Molar mass of compound =
0.500 g 5.11 10 −2 mol 0.1000 L L
= 97.8 g/mol
ChemWork Problems 113.
Benzoic acid is capable of hydrogen-bonding, but a significant part of benzoic acid is the nonpolar benzene ring. In benzene, a hydrogen-bonded dimer forms. O
O C
O
H
H
O
C
C O
H
O
The dimer is relatively nonpolar and thus more soluble in benzene than in water. Because benzoic acid forms dimers in benzene, the effective solute particle concentration will be less than 1.0 molal. Therefore, the freezing-point depression would be less than 5.12°C (Tf = Kf m). 114.
Benzoic acid (see Exercise 113) would be more soluble in a basic solution because of the reaction: C6H5CO2H + OH− → C6H5CO2− + H2O By removing the proton from benzoic acid, an anion forms, and like all anions, the species becomes more soluble in water.
115.
Mass % CsCl =
mass CsCl 5 2. 3 g × 100 = × 100 = 46.6% total mass 52.3 g + 60.0. g
Mol CsCl = 52.3 g ×
1 mol = 0.311 mol CsCl 168.4 g
Mol H2O = 60.0 g ×
1 mol = 3.33 mol H2O 18.02 g
Molarity =
0.311 mol CsCl mol CsCl = = 4.91 mol/L L solution 0.0633 L
Molality =
mol CsCl 0.311 mol CsCl = = 5.18 mol/kg kg solvent 0.0600 kg
Mole fraction = χCsCl =
mol CsCl 0.331 mol = = 0.0904 total mol 0.331 mol + 3.33 mol
CHAPTER 11 116.
PROPERTIES OF SOLUTIONS
515
Using Hess’s law: NaCl(s) → Na+(g) + Cl−(g) Na (g) + Cl−(g) → Na+(aq) + Cl−(aq)
ΔH = −ΔHLE = −(−786 kJ/mol) ΔH = ΔHhyd = −783 kJ/mol
+
NaCl(s) → Na+(aq) + Cl−(aq)
ΔHsoln = 3 kJ/mol
ΔHsoln refers to the heat released or gained when a solute dissolves in a solvent. Here, an ionic compound dissolves in water. 117.
MX(s) → M+(g) + X−(g) M (g) + X−(g) → M+(aq) + X−(aq) +
MX(s) → M+(aq) + X−(aq)
ΔH = −ΔHLattice Energy ΔH = ΔHhydration ΔHsoln = −ΔHLattice Energy + ΔHhydration
For KF: ‒15 kJ/mol = ‒(‒804 kJ/mol) + ΔHhydration, ΔHhydration = ‒819 kJ/mol For RbF: ‒24 kJ/mol = ‒(‒768 kJ/mol) + ΔHhydration, ΔHhydration = ‒792 kJ/mol The enthalpy of hydration term refers to the interactions of the ions with water. Because KF has the more negative ΔHhydration value, it forms the stronger attraction to water (answer e). 118.
Water is a polar solvent because the H2O molecule exhibits a dipole moment, that is, H2O is a molecule which has a partial negative charged end and a partial positive charged end. The electrostatic potential diagram for H2O illustrates this with colors. The partial negative end of the dipole moment is the red end and the partial positive end is around the blue end. Because water is a polar solvent, it will dissolve other polar covalent compounds. The electrostatic potential diagram for NH3 illustrates that NH3 has a dipole moment (the red end is the partial negative end and the light blue end is the partial positive end). Because NH 3 has a dipole moment (is polar), it will be soluble in water. However, the electrostatic potential diagram for CH4 doesn’t have one specific negative (red) end and has four blue regions arranged symmetrically about the molecule. CH4 does not have a dipole moment (is nonpolar) and will not be soluble in water.
119.
120.
For ionic compounds, as the charge of the ions increases and/or the size of the ions decreases, the attraction to water (hydration) increases. a. Fe3+; smaller size, higher charge
b. Be2+; smaller size
c. Cl−; smaller size
d. SO42−; higher charge
e. Mg2+; smaller size, higher charge
f.
750. mL grape juice ×
F−; smaller size
12 mL C 2 H 5OH 0.79 g C 2 H 5 OH 1 mol C 2 H 5 OH 100 . mL juice mL 46.07 g
2 mol CO 2 = 1.54 mol CO2 (carry extra significant figure) 2 mol C 2 H 5 OH
1.54 mol CO2 = total mol CO2 = mol CO2(g) + mol CO2(aq) = ng + naq
516
CHAPTER 11
PCO 2 =
n g RT V
PROPERTIES OF SOLUTIONS
0.08206 L atm (298 K ) n g mol K = = 326ng −3 75 10 L
n aq PCO 2 =
C 0.750 L = = (43.0)naq k 3.1 10 − 2 mol L atm
PCO 2 = 326ng = (43.0)naq, and from above, naq = 1.54 − ng; solving: 326ng = 43.0(1.54 − ng), 369ng = 66.2, ng = 0.18 mol
PCO 2 = 326(0.18) = 59 atm in gas phase C = kPCO2 = 121.
1.8 mol CO 2 3.1 10 −2 mol × 59 atm = (in wine) L L atm
A 92-proof ethanol solution is 46% C2H5OH by volume. Assuming 100.0 mL of solution: mol ethanol = 46 mL C2H5OH × molarity =
122.
0.79 g 1 mol C 2 H 5 OH = 0.79 mol C2H5OH mL 46.07 g
0.79 mol = 7.9 M ethanol 0.1000 L
10.0 mL blood
1.0 mg C 4 H 7 N 3 O 1 mol C 4 H 7 N 3 O 1L 10 dL 1g 1000 mL 1L 1 dL blood 1000 mg 113.13 g
= 8.8 × 10 −7 mol C4H7N3O Mass of blood = 10.0 mL Molality =
8.8 10 −7 mol = 8.5 × 10 −5 mol/kg 0.0103 kg
= MRT, M =
8.8 10 −7 mol = 8.8 × 10 −5 mol/L 0.0100 L
= 8.8 × 10 −5 mol/L 123.
1.025 g = 10.3 g mL
0.08206 L atm × 298 K = 2.2 × 10 −3 atm K mol
a. NH4NO3(s) → NH4+(aq) + NO3−(aq) ΔHsoln = ? Heat gain by dissolution process = heat loss by solution; we will keep all quantities positive to avoid sign errors. Because the temperature of the water decreased, the dissolution of NH4NO3 is endothermic (ΔH is positive). Mass of solution = 1.60 + 75.0 = 76.6 g. Heat loss by solution =
4.18 J × 76.6 g × (25.00°C − 23.34°C) = 532 J C g
CHAPTER 11
PROPERTIES OF SOLUTIONS
ΔHsoln =
517
80.05 g NH 4 NO 3 532 J = 2.66 × 104 J/mol = 26.6 kJ/mol 1.60 g NH 4 NO 3 mol NH 4 NO 3
b. We will use Hess’s law to solve for the lattice energy. The lattice-energy equation is: NH4+(g) + NO3−(g) → NH4NO3(s)
ΔH = lattice energy
NH4+(g) + NO3−(g) → NH4+(aq) + NO3−(aq) NH4+(aq) + NO3−(aq) → NH4NO3(s) NH4+(g) + NO3−(g) → NH4NO3(s)
ΔH = ΔHhyd = –630. kJ/mol ΔH = –ΔHsoln = –26.6 kJ/mol ΔH = ΔHhyd – ΔHsoln ΔH = –657 kJ/mol
124.
Answer c is false. When a nonvolatile solute is dissolved in a water, the freezing point of the solution is lowered from 0°C and the boiling point is raised above 100.°C Thus, the solution is a liquid over a wider range of temperatures than pure water.
125.
Because partial pressures are proportional to the moles of gas present, then: V CS = PCS2 /Ptotal 2
V PCS2 = CS P = 0.855(263 torr) = 225 torr 2 total L L PCS2 = CS P o , CS = 2 CS2 2
126.
PCS2 o PCS 2
=
225 torr = 0.600 375 torr
PB = B PBo , B = PB / PBo = 0.900 atm/0.930 atm = 0.968
0.968 =
mol benzene 1 mol ; mol benzene = 78.11 g C6H6 × = 1.000 mol 78.11 g total mol
Let x = mol solute, then: χB = 0.968 = Molar mass =
127.
10.0 g = 303 g/mol 3.0 × 102 g/mol 0.033 mol
50.0 g CH3COCH3 × 50.0 g CH3OH × L = χ acetone
1.000 mol , 0.968 + (0.968)x = 1.000, x = 0.033 mol 1.000 + x
1 mol = 0.861 mol acetone 58.08 g
1 mol = 1.56 mol methanol 32.04 g
0.861 L = 0.356; χ Lmethanol = 1.000 − χ acetone = 0.644 0.861 + 1.56
Ptotal = Pmethanol + Pacetone = 0.644(143 torr) + 0.356(271 torr) = 92.1 + 96.5 = 188.6 torr Because partial pressures are proportional to the moles of gas present, in the vapor phase:
518
CHAPTER 11 V acetone =
PROPERTIES OF SOLUTIONS
Pacetone 96.5 torr = = 0.512; χ Vmethanol = 1.000 − 0.512 = 0.488 Pt otal 188 .6 torr
The actual vapor pressure of the solution (161 torr) is less than the calculated pressure assuming ideal behavior (188.6 torr). Therefore, the solution exhibits negative deviations from Raoult’s law. This occurs when the solute-solvent interactions are stronger than in pure solute and pure solvent. 128.
π = MRT =
0.1 mol 0.08206 L atm 298 K = 2.45 atm 2 atm L K mol
π = 2 atm ×
760 mm Hg 2000 mm 2 m atm
The osmotic pressure would support a mercury column of approximately 2 m. The height of a fluid column in a tree will be higher because Hg is more dense than the fluid in a tree. If we assume the fluid in a tree is mostly H2O, then the fluid has a density of 1.0 g/cm3. The density of Hg is 13.6 g/cm3. Height of fluid 2 m × 13.6 30 m 129.
Molality = m = ΔTb = Kbm =
mol solute 35.0 g solute 1000 g 1 mol solute = = 1.01 molal kg solvent 600.0 g H 2 O kg 58.0 g solute
0.51 C × 1.01 molal = 0.52°C; 99.725 oC + 0.52 oC = 100.25 oC molal
The boiling point is raised from 99.725 oC to 100.25°C. 130.
π = MRT, π = 0.56 torr ×
0.08206 L atm 1 atm =M× × 298 K, M = 3.0 × 10−5 mol/L 760 torr K mol
Mol protein = 0.00025 L ×
3.0 10 −5 mol protein = 7.5 × 10−9 mol protein L
Molar mass =
131.
4.7 10 −5 g 7.5 10
−9
mol
= 6.3 × 103 g/mol
5.86 10 −2 mol thyroxine Tf 0.300 C = = ΔTf = Kfm, m = Kf 5.12 C kg/mol kg benzene The moles of thyroxine present are: 0.0100 kg benzene ×
5.86 10 −2 mol thyroxine kg benzene
= 5.86 × 10 −4 mol thyroxine
From the problem, 0.455 g thyroxine was used; this must contain 5.86 × 10 −4 mol thyroxine. The molar mass of the thyroxine is: molar mass =
0.455 g 5.86 × 10 −4 mol
= 776 g/mol
CHAPTER 11
PROPERTIES OF SOLUTIONS
=
8.00 atm = 0.327 mol/L 0.08206 L atm/K • mol 298 K
132.
π = MRT, M =
133.
Out of 100.00 g, there are:
RT
31.57 g C × 5.30 g H ×
519
2.629 1 mol C = 2.629 mol C; = 1.000 2.629 12.01 g
5.26 1 mol H = 5.26 mol H; = 2.00 2.629 1.008 g
63.13 g O ×
3.946 1 mol O = 3.946 mol O; = 1.501 2.629 16 .00 g
Empirical formula: C2H4O3; use the freezing-point data to determine the molar mass. m=
Tf 5.20 C = = 2.80 molal Kf 1.86 C/molal
Mol solute = 0.0250 kg × Molar mass =
2.80 mol solute = 0.0700 mol solute kg
10.56 g = 151 g/mol 0.0700 mol
The empirical formula mass of C2H4O3 = 76.05 g/mol. Because the molar mass is about twice the empirical mass, the molecular formula is C 4H8O6, which has a molar mass of 152.10 g/mol. Note: We use the experimental molar mass to determine the molecular formula. Knowing this, we calculate the molar mass precisely from the molecular formula using the atomic masses in the periodic table. 134.
a. As discussed in Figure 11.19 of the text, the water would migrate from left to right (to the side with the solution). Initially, the level of liquid in the left arm would go down, and the level in the right arm would go up. At some point the rate of solvent (H2O) transfer will be the same in both directions, and the levels of the liquids in the two arms will stabilize. The height difference between the two arms is a measure of the osmotic pressure of the solution. b. Initially, H2O molecules will have a net migration into the solution side. However, the solute can now migrate into the H2O side. Because solute and solvent transfer are both possible, the levels of the liquids will be equal once the rates of solute and solvent transfer are equal in both directions. At this point the concentration of solute will be equal in both chambers, and the levels of liquid will be equal.
135.
If ideal, NaCl dissociates completely, and i = 2.00. ΔTf = iKfm; assuming water freezes at 0.00°C: 1.28°C = 2 × 1.86°C kg/mol × m, m = 0.344 mol NaCl/kg H2O Assume an amount of solution that contains 1.00 kg of water (solvent).
520
CHAPTER 11
PROPERTIES OF SOLUTIONS
58.44 g = 20.1 g NaCl mol 20.1 g Mass % NaCl = × 100 = 1.97% 1.00 10 3 g + 20.1 g
0.344 mol NaCl ×
136.
The pair of compounds which have similar strength intermolecular forces will behave most ideally. This will be the pair of compounds which have the most similar structures. a. CH3(CH2)6CH3 and CH3(CH2)5CH3 have similar nonpolar structures so this pair will behave most ideally. CH3-O-CH3 will have significantly weaker intermolecular forces than H2O which exhibits extensive hydrogen bonding. b. CH3OH and CH3CH2OH have the most similar structures and will behave most ideally. HF, with its ability to hydrogen bond, will have significantly stronger intermolecular forces than CH3F which cannot hydrogen bond.
137.
T = Kfm, m =
ΔT 2.79 o C = 1.50 molal = Kf 1.86 o C/molal
a. T = Kbm, T = (0.51°C/molal)(1.50 molal) = 0.77°C, Tb = 100.77°C b.
o Psoln = χ water Pwater , χ water =
mol H 2 O mol H 2 O + mol solute
Assuming 1.00 kg of water, we have 1.50 mol solute, and: mol H2O = 1.00 × 103 g H2O × water =
1 mol H 2 O = 55.5 mol H2O 18.02 g H 2 O
55 .5 mol = 0.974; Psoln = (0.974)(23.76 mm Hg) = 23.1 mm Hg 1.50 + 55 .5
c. We assumed ideal behavior in solution formation, we assumed the solute was nonvolatile, and we assumed i = 1 (no ions formed). 138.
L = Ptotal = PCH 3OH + PCH 3CH 2CH 2OH ; P = χ L P o ; χ CH 3OH
1.000 mol CH 3 OH = 0.239 4.18 mol total
Ptotal = 0.239(303 torr) + (1.000 − 0.239)(44.6 torr) = 72.4 + 33.9 = 106.3 torr V = In the vapor: χ CH 3OH
PCH 3OH Ptotal
=
72.4 torr = 0.681 106.3 torr
V = 1.000 – 0.681 = 0.319 χ CH 3CH 2CH 2OH
Note: In the Solutions Guide, we added V or L superscripts to the mole fraction symbol to emphasize for which value we are solving. If the L or V is omitted, then the liquid phase is assumed.
CHAPTER 11 139.
PROPERTIES OF SOLUTIONS
521
0.995 g = 159 g = 0.159 kg mL 0.378 mol Mol NaDTZ = 0.159 kg = 0.0601 mol kg
Mass of H2O = 160. mL
Molar mass of NaDTZ =
38 .4 g = 639 g/mol 0.0601 mol
Psoln = χ H 2 O PHo 2O ; mol H2O = 159 g
1 mol = 8.82 mol 18.02 g
Sodium diatrizoate is a salt because there is a metal (sodium) in the compound. From the short-hand notation for sodium diatrizoate, NaDTZ, we can assume this salt breaks up into Na+ and DTZ− ions. So the moles of solute particles are 2(0.0601) = 0.120 mol solute particles.
χ H 2O = 140.
8.82 mol = 0.987; Psoln = 0.987 × 34.1 torr = 33.7 torr 0.120 mol + 8.82 mol
Water exhibits H-bonding in the pure state and is classified as a polar solvent. Water will dissolve other polar solutes and ionic solutes. a. NH3; NH3 is more polar CH3NH3. The CH3 group, like all hydrocarbon groups, is a nonpolar group. b. CH3CN; CH3CN is more polar than CH3OCH3 due to the 2 nonpolar CH3 groups in CH3OCH3. c. CH3CH2OH; CH3CH2OH is polar while CH3CH2CH3 is nonpolar. d. CH3OH; CH3OH is more polar than CH3CH2OH. e. (CH3)3CCH2OH; (CH3)3CCH2OH has a slightly smaller nonpolar part than CH3(CH2)6OH. Neither compound would be expected to be very soluble in water. f.
141.
CH3CO2H; CH3CO2H is capable of H-bonding, unlike the other compound.
If we have 1.000 kg of water, then we have 1.00 mol of solute. 1000. g H2O ×
1 mol H 2 O = 55.49 mol H2O 18.02 g
Psoln = χ H 2 O PHo 2O =
142.
T = Kfm, m =
55.49 mol × 92.5 torr = 90.9 torr 1.00 mol + 55.49 mol
ΔT 0.930o C = 0.50 molal = Kf 1.86 o C/molal
522
CHAPTER 11
PROPERTIES OF SOLUTIONS
Mol unknown = 0.250 kg ×
0.50 mol solute = 0.125 mol solute kg
Molar mass of unknown =
22.5 g = 180. g/mol 0.125 mol
The empirical formula mass of CH2O is ~30 g/mol. Because the molar mass of the unknown is six times the empirical formula mass, the molecular formula is (CH2O)×6 = C6H12O6. 143.
ΔT 2.79 o C = = 3.00 mK f 0.250 mol 1.86 o C kg 0.500 kg mol
T = imKf, i =
We have three ions in solutions, and we have twice as many anions as cations. Therefore, the formula of Q is MCl2. Assuming 100.00 g of compound: 38.68 g Cl
1 mol Cl = 1.091 mol Cl 35.45 g
mol M = 1.091 mol Cl Molar mass of M =
144.
1 mol M = 0.5455 mol M 2 mol Cl
61.32 g M = 112.4 g/mol; M is Cd, so Q = CdCl2. 0.5455 mol M
14.2 mg CO2 ×
3.88 mg 12 .01 mg C = 3.88 mg C; % C = × 100 = 80.8% C 44 .01 mg CO 2 4.80 mg
1.65 mg H2O ×
0.185 mg 2.016 mg H = 0.185 mg H; % H = × 100 = 3.85% H 18 .02 mg H 2 O 4.80 mg
Mass % O = 100.00 − (80.8 + 3.85) = 15.4% O Out of 100.00 g: 80.8 g C ×
6.73 1 mol = 6.73 mol C; = 6.99 7 0.963 12 .01 g
3.85 g H ×
3.82 1 mol = 3.82 mol H; = 3.97 4 0.963 1.008 g
15.4 g O ×
0.963 1 mol = 0.963 mol O; = 1.00 0.963 16.00 g
Therefore, the empirical formula is C7H4O. ΔTf = Kfm, m =
ΔTf 22.3 o C = = 0.56 molal Kf 40. o C / molal
Mol anthraquinone = 0.0114 kg camphor ×
0.56 mol anthraquin one = 6.4 × 10−3 mol kg camphor
CHAPTER 11
PROPERTIES OF SOLUTIONS
Molar mass =
523
1.32 g = 210 g/mol 6.4 10 −3 mol
The empirical mass of C7H4O is 7(12) + 4(1) + 16 104 g/mol. Because the molar mass is twice the empirical mass, the molecular formula is C 14H8O2.
Challenge Problems 145.
m=
ΔTf 5.23 o C = 2.81 molal = Kf 1.86 o C/molal
2.81 mol solute n , n = 0.141 mol of ions in solution = kg solvent 0.0500 kg
Since NaNO3 and Mg(NO3)2 are strong electrolytes: 0.141 mol ions = 2(x mol of NaNO3) + 3[y mol of Mg(NO3)2] In addition: 6.50 g = x mol NaNO3 ×
148.3 g 85.00 g + y mol of Mg(NO3)2 × mol mol
We have two equations: 2x + 3y = 0.141 and 85.00 x + 148.3 y = 6.50 Solving by simultaneous equations: −85.00 x – 127.5 y = −5.99 85.00 x + 148.3 y = 6.50 20.8 y = 0.51, y = 0.025 mol Mg(NO3)2 Mass of Mg(NO3)2 = 0.025 mol ×
148.3 g = 3.7 g Mg(NO3)2 mol
Mass of NaNO3 = 6.50 g – 3.7 g = 2.8 g NaNO3 Mass % NaNO3 = 146.
2 .8 g × 100 = 43% NaNO3 and 57% Mg(NO3)2 by mass 6.50 g
For the second vapor collected, χ VB, 2 = 0.714 and χ TV, 2 = 0.286. Let χ LB, 2 = mole fraction of benzene in the second solution and χ TL , 2 = mole fraction of toluene in the second solution.
χ LB, 2 + χ TL, 2 = 1.000 χ VB, 2 = 0.714 =
χ LB, 2 (750 .0 torr ) PB PB = = L Ptotal PB + PT χ B, 2 (750 .0 torr ) + (1.000 − χ LB, 2 )(300 .0 torr )
Solving: χ LB, 2 = 0.500 = χ TL , 2 This second solution came from the vapor collected from the first (initial) solution, so, χ VB , 1 = χ TV, 1 = 0.500. Let χ LB , 1 = mole fraction benzene in the first solution and χ TL , 1 = mole fraction of toluene in first solution. χ LB, 1 + χ TL, 1 = 1.000.
524
CHAPTER 11 χ VB , 1 = 0.500 =
PROPERTIES OF SOLUTIONS
χ LB, 1 (750 .0 torr ) PB PB = = L Ptotal PB + PT χ B, 1 (750 .0 torr ) + (1.000 − χ LB, 1 )(300 .0 torr )
Solving: χ LB , 1 = 0.286; the original solution had B = 0.286 and T = 0.714. 147.
For 30.% A by moles in the vapor, 30. = 0.30 =
PA × 100: PA + PB
χAx χAx , 0.30 = χAx + χBy χ A x + (1.00 − χ A ) y
χA x = 0.30(χA x) + 0.30 y – 0.30(χA y), χA x – (0.30)χA x + (0.30)χA y = 0.30 y χA(x – 0.30 x + 0.30 y) = 0.30 y, χA =
0.30 y ; χB = 1.00 – χA 0.70 x + 0.30 y
Similarly, if vapor above is 50.% A: χ A = If vapor above is 80.% A: χA =
y y ; χ B = 1.00 − x+ y x+ y
0.80 y ; χB = 1.00 − χA 0.20 x + 0.80 y
If the liquid solution is 30.% A by moles, χA = 0.30. Thus χ VA =
148.
PA 0.30 x 0.30 x and χ VB = 1.00 − = PA + PB 0.30 x + 0.70 y 0.30 x + 0.70 y
If solution is 50.% A: χ VA =
x x+ y
If solution is 80.% A: χ VA =
0.80 x and χ VB = 1.00 − χ VA 0.80 x + 0.20 y
and χ VB = 1.00 − χ VA
a. Freezing-point depression is determined using molality for the concentration units, whereas molarity units are used to determine osmotic pressure. We need to assume that the molality of the solution equals the molarity of the solution. b. Molarity =
moles solvent moles solvent ; molality = liters solution kg solvent
When the liters of solution equal the kilograms of solvent present for a solution, then molarity equals molality. This occurs for an aqueous solution when the density of the solution is equal to the density of water, 1.00 g/cm 3. The density of a solution is close to 1.00 g/cm3 when not a lot of solute is dissolved in solution. Therefore, molarity and molality values are close to each other only for dilute solutions. c. T = Kf m, m =
ΔT 0.621o C = = 0.334 mol/kg Kf 1.86 o C kg/mol
Assuming 0.334 mol/kg = 0.334 mol/L:
CHAPTER 11
PROPERTIES OF SOLUTIONS = MRT =
d. m =
525
0.334 mol 0.08206 L atm 298 K = 8.17 atm L K mol
ΔT 2.0 o C = = 3.92 mol/kg Kb 0.51 o C kg/mol
This solution is much more concentrated than the isotonic solution in part c. Here, water will leave the plant cells in order to try to equilibrate the ion concentration both inside and outside the cell. Because there is such a large concentration discrepancy, all the water will leave the plant cells, causing them to shrivel and die. 149.
m=
ΔTf 0.426 o C = 0.229 molal = Kf 1.86 o C / molal
Assuming a solution density = 1.00 g/mL, then 1.00 L contains 0.229 mol solute. NaCl → Na+ + Cl− i = 2; so: 2(mol NaCl) + mol C12H22O11 = 0.229 mol Mass NaCl + mass C12H22O11 = 20.0 g 2nNaCl + nC12H22O11 = 0.229 and 58.44(nNaCl) + 342.3(nC12H22O11 ) = 20.0 Solving: nC12H22O11 = 0.0425 mol = 14.5 g and nNaCl = 0.0932 mol = 5.45 g Mass % C12H22O11 = (14.5 g/20.0 g) × 100 = 72.5 % and 27.5% NaCl by mass
χ C12H22O11 = 150.
0.0425 mol = 0.313 0.0425 mol + 0.0932 mol
a. π = iMRT, iM =
RT
=
7.83 atm = 0.320 mol/L 0.08206 L atm/K • mol 298 K
Assuming 1.000 L of solution: total mol solute particles = mol Na+ + mol Cl− + mol NaCl = 0.320 mol mass solution = 1000. mL ×
1.071 g = 1071 g solution mL
mass NaCl in solution = 0.0100 × 1071 g = 10.7 g NaCl mol NaCl added to solution = 10.7 g ×
1 mol = 0.183 mol NaCl 58.44 g
Some of this NaCl dissociates into Na+ and Cl− (two moles of ions per mole of NaCl), and some remains undissociated. Let x = mol undissociated NaCl = mol ion pairs. Mol solute particles = 0.320 mol = 2(0.183 − x) + x 0.320 = 0.366 − x, x = 0.046 mol ion pairs Fraction of ion pairs =
0.046 = 0.25, or 25% 0.183
526
CHAPTER 11
PROPERTIES OF SOLUTIONS
b. ΔT = Kfm, where Kf = 1.86 °C kg/mol; from part a, 1.000 L of solution contains 0.320 mol of solute particles. To calculate the molality of the solution, we need the kilograms of solvent present in 1.000 L of solution. Mass of 1.000 L solution = 1071 g; mass of NaCl = 10.7 g Mass of solvent in 1.000 L solution = 1071 g − 10.7 g = 1060. g ΔT = 1.86 °C kg/mol ×
0.320 mol = 0.562°C 1.060 kg
Assuming water freezes at 0.000°C, then Tf = −0.562°C. 151.
χ Vpen = 0.15 =
Ppen Ptotal
o ; Ppen = χ Lpen Ppen ; Ptotal = Ppen + Phex = χ Lpen (511) + χ Lhex (150.)
L L L L Because hex = 1.000 − χ Lpen : Ptotal = pen (511) + (1.000 − pen )(150.) = 150. + 361 pen
χ Vpen
=
Ppen Ptotal
, 0.15 =
χ Lpen (511) 150. + 361 χ Lpen
23 + 54 χ Lpen = 511 χ Lpen , χ Lpen =
152.
a. m =
, 0.15(150 . + 361 χ Lpen ) = 511 χ Lpen
23 = 0.050 457
ΔTf 1.32 o C = = 0.258 mol/kg Kf 5.12 o C kg/mol
Mol unknown = 0.01560 kg × Molar mass of unknown =
0.258 mol unknown = 4.02 × 10−3 mol kg
1.22 g = 303 g/mol 4.02 10 −3 mol
Uncertainty in temperature =
0.04 × 100 = 3% 1.32
A 3% uncertainty in 303 g/mol = 9 g/mol. So molar mass = 303 ±9 g/mol. b. No, codeine could not be eliminated since its molar mass is in the possible range including the uncertainty. c. We would like the uncertainty to be ±1 g/mol. We need the freezing-point depression to be about 10 times what it was in this problem. Two possibilities are: (1) make the solution 10 times more concentrated (may be solubility problem) (2) use a solvent with a larger Kf value, e.g., camphor
CHAPTER 11 153.
PROPERTIES OF SOLUTIONS
ΔTf = 5.51 − 2.81 = 2.70°C; m =
527
ΔTf 2.70 o C = 0.527 molal = Kf 5.12 o C/molal
Let x = mass of naphthalene (molar mass = 128.2 g/mol). Then 1.60 − x = mass of anthracene (molar mass = 178.2 g/mol). 1.60 − x x = moles naphthalene and = moles anthracene 128.2 178.2 x 1.60 − x + 0.527 mol solute 178.2 , 1.05 10 −2 = (178.2) x + 1.60(128.2) − (128.2) x = 128.2 kg solvent 0.0200 kg solvent 128.2(178.2)
(50.0)x + 205 = 240., (50.0)x = 240. − 205, (50.0)x = 35, x = 0.70 g naphthalene So the mixture is: 0.70 g × 100 = 44% naphthalene by mass and 56% anthracene by mass 1.60 g
154.
iM =
RT
=
0.3950 atm = 0.01614 mol/L = total ion concentration 0.08206 L atm 298 .2 K K mol
0.01614 mol/L = M Mg 2+ + M Na + + M Cl − ; M Cl − = 2M Mg2+ + M Na + (charge balance) Combining: 0.01614 = 3M Mg 2 + + 2M Na+ Let x = mass MgCl2 and y = mass NaCl; then x + y = 0.5000 g. x y and M Na + = (Because V = 1.000 L.) 58.44 95.21 3x 2y + Total ion concentration = = 0.01614 mol/L 95.21 58.44 M Mg 2 + =
Rearranging: 3x + (3.258)y = 1.537 Solving by simultaneous equations: 3x –3(x
+ (3.258)y = 1.537 + y) = –3(0.5000) (0.258)y = 0.037, y = 0.14 g NaCl
Mass MgCl2 = 0.5000 g − 0.14 g = 0.36 g; mass % MgCl2 = 155.
0.36 g × 100 = 72% 0.5000 g
HCO2H → H+ + HCO2−; only 4.2% of HCO2H ionizes. The amount of H+ or HCO2− produced is 0.042 × 0.10 M = 0.0042 M. The amount of HCO2H remaining in solution after ionization is 0.10 M − 0.0042 M = 0.10 M.
528
CHAPTER 11
PROPERTIES OF SOLUTIONS
The total molarity of species present = M HCO2H + M H+ + M HCO − 2
= 0.10 + 0.0042 + 0.0042 = 0.11 M Assuming 0.11 M = 0.11 molal, and assuming ample significant figures in the freezing point and boiling point of water at P = 1 atm: ΔT = Kfm = 1.86°C/molal × 0.11 molal = 0.20°C; freezing point = –0.20°C ΔT = Kbm = 0.51°C/molal × 0.11 molal = 0.056°C; boiling point = 100.056°C 156.
Let χ LA = mole fraction A in solution, so 1.000 − χ LA = χ LB . From the problem, χ VA = 2 χ LA .
χ VA =
PA χ LA (350 .0 torr ) = L Ptotal χ A (350 .0 torr ) + (1.000 − χ LA )(100 .0 torr )
χ VA = 2 χ LA =
(350 .0)χ LA , (250 .0)χ LA = 75.0, χ LA = 0.300 (250 .0)χ LA + 100 .0
The mole fraction of A in solution is 0.300. 157.
a. Assuming MgCO3(s) does not dissociate, the solute concentration in water is: 560 μg MgCO 3 (s) 560 mg 560 10 −3 g 1 mol MgCO 3 = = mL L L 84.32 g = 6.6 × 10−3 mol MgCO3/L
An applied pressure of 8.0 atm will purify water up to a solute concentration of: M =
RT
=
8.0 atm 0.32 mol = 0.08206 L atm/K mol 300. K L
When the concentration of MgCO3(s) reaches 0.32 mol/L, the reverse osmosis unit can no longer purify the water. Let V = volume (L) of water remaining after purifying 45 L of H2O. When V + 45 L of water has been processed, the moles of solute particles will equal: 6.6 × 10−3 mol/L × (45 L + V) = 0.32 mol/L × V Solving: 0.30 = (0.32 − 0.0066) × V, V = 0.96 L The minimum total volume of water that must be processed is 45 L + 0.96 L = 46 L. Note: If MgCO3 does dissociate into Mg2+ and CO32− ions, then the solute concentration increases to 1.3 × 10−2 M, and at least 47 L of water must be processed. b. No; a reverse osmosis system that applies 8.0 atm can only purify water with a solute concentration of less than 0.32 mol/L. Salt water has a solute concentration of 2(0.60 M) = 1.2 mol/L ions. The solute concentration of salt water is much too high for this reverse osmosis unit to work.
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529
Marathon Problem 158.
a. The average values for each ion are: 300. mg Na+, 15.7 mg K+, 5.45 mg Ca2+, 388 mg Cl−, and 246 mg lactate (C3H5O3−) Note: Because we can precisely weigh to ±0.1 mg on an analytical balance, we'll carry extra significant figures and calculate results to ±0.1 mg. The only source of lactate is NaC3H5O3. 246 mg C3H5O3− ×
112.06 mg NaC3H 5O3 89.07 mg C3H 5O3
−
= 309.5 mg sodium lactate
The only source of Ca2+ is CaCl2•2H2O. 5.45 mg Ca2+ ×
147.0 mg CaCl 2 • 2H 2 O 40.08 mg Ca 2 +
= 19.99 or 20.0 mg CaCl2•2H2O
The only source of K+ is KCl. 15.7 mg K+ ×
74.55 mg KCl = 29.9 mg KCl 39.10 mg K +
From what we have used already, let's calculate the mass of Na+ added. 309.5 mg sodium lactate − 246.0 mg lactate = 63.5 mg Na+ Thus we need to add an additional 236.5 mg Na+ to get the desired 300. mg. 236.5 mg Na+ ×
58.44 mg NaCl = 601.2 mg NaCl 22.99 mg Na +
Now let's check the mass of Cl− added: 20.0 mg CaCl2•2H2O ×
70.90 mg Cl − = 9.6 mg Cl− 147.0 mg CaCl2 • 2H 2 O
20.0 mg CaCl2•2H2O = 9.6 mg Cl− 29.9 mg KCl − 15.7 mg K+ = 14.2 mg Cl− 601.2 mg NaCl − 236.5 mg Na+ = 364.7 mg Cl− Total Cl− = 388.5 mg Cl− This is the quantity of Cl− we want (the average amount of Cl−). An analytical balance can weigh to the nearest 0.1 mg. We would use 309.5 mg sodium lactate, 20.0 mg CaCl2•2H2O, 29.9 mg KCl, and 601.2 mg NaCl. b. To get the range of osmotic pressure, we need to calculate the molar concentration of each ion at its minimum and maximum values. At minimum concentrations, we have:
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285 mg Na + 1 mmol 14.1 mg K + 1 mmol = 0.124 M; = 0.00361 M 100 . mL 100 . mL 39.10 mg 22.99 mg
368 mg Cl − 4.9 mg Ca 2+ 1 mmol 1 mmol = 0.0012 M; = 0.104 M 100 . mL 100 . mL 35.45 mg 40.08 mg −
231 mg C3 H 5O 3 1 mmol = 0.0259 M (Note: molarity = mol/L = mmol/mL.) 100 . mL 89.07 mg
Total = 0.124 + 0.00361 + 0.0012 + 0.104 + 0.0259 = 0.259 M π = MRT =
0.259 mol 0.08206 L atm × 310. K = 6.59 atm L K mol
Similarly, at maximum concentrations, the concentration for each ion is: Na+: 0.137 M; K+: 0.00442 M; Ca2+: 0.0015 M; Cl−: 0.115 M; C3H5O3−: 0.0293 M The total concentration of all ions is 0.287 M. π=
0.287 mol 0.08206 L atm × 310. K = 7.30 atm L K mol
Osmotic pressure ranges from 6.59 atm to 7.30 atm.
CHAPTER 12 CHEMICAL KINETICS Review Questions 1.
The reaction rate is defined as the change in concentration of a reactant or product per unit time. Consider the general reaction: A → Products where rate =
− [ A ] t
If we graph [A] vs. t, it will usually look like the solid line in the following plot.
a
[A]
b c 0
t1
t2
time
An instantaneous rate is the slope of a tangent line to the graph of [A] vs. t. We can determine the instantaneous rate at any time during the reaction. On the plot, tangent lines at t 0 and t = t1 are drawn. The slope of these tangent lines would be the instantaneous rates at t 0 and t = t1. We call the instantaneous rate at t 0 the initial rate. The average rate is measured over a period of time. For example, the slope of the dashed line connecting points a and c is the average rate of the reaction over the entire length of time 0 to t 2 (average rate = Δ[A]/Δt). An average rate is determined over some time period, whereas an instantaneous rate is determined at one specific time. The rate that is largest is generally the initial rate. At t 0, the slope of the tangent line is greatest, which means the rate is largest at t 0. The initial rate is used by convention so that the rate of reaction only depends on the forward reaction; at t 0, the reverse reaction is insignificant because no products are present yet. 2.
The differential rate law describes the dependence of the rate on the concentration of reactants. The integrated rate law expresses reactant concentrations as a function of time. The differential rate law is generally just called the rate law. The rate constant k is a constant that allows one to
531
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equate the rate of a reaction to the concentration of reactants. The order is the exponent that the reactant concentrations are raised to in the rate equation. 3.
The method of initial rates uses the results from several experiments where each experiment is carried out at a different set of initial reactant concentrations and the initial rate is determined. The results of the experiments are compared to see how the initial rate depends on the initial concentrations. If possible, two experiments are compared where only one reactant concentration changes. For these two experiments, any change in the initial rate must be due to the change in that one reactant concentration. The results of the experiments are compared until all of the orders are determined. After the orders are determined, then one can go back to any (or all) of the experiments and set the initial rate equal to the rate law using the concentrations in that experiment. The only unknown is k, which is then solved for. The units on k depend on the orders in the rate law. Because there are many different rate laws, there are many different units for k. Rate = k[A]n; for a first-order rate law, n = 1. If [A] is tripled, then the rate is tripled. When [A] is quadrupled (increased by a factor of four), and the rate increases by a factor of 16, then A must be second order (42 = 16). For a third order reaction, as [A] is doubled, the rate will increase by a factor of 23 = 8. For a zero-order reaction, the rate is independent of the concentration of A. The only stipulation for zero order reactions is that the reactant or reactants must be present; if they are, then the rate is a constant value (rate = k).
4.
Zero order:
− d [A] = k, dt [ A ]t
[A]
[ A ]t
t
[ A ]0
0
d[A] = − kdt
t
= − kt , [A]t − [A]0 = − kt, [A]t = −kt + [A]0
[ A ]0
0
First order: [ A ]t
t
− d [A] d [A] = k[A], = − kdt dt [ A ] [A] 0
0
[A]t
ln[ A]
= − kt, ln[ A]t − ln[ A]0 = − kt, ln[A]t = −kt + ln[A]0
[A]0
Second order: [ A ]t
t
− d [A] d [A] = k[A] 2 , = − kdt 2 dt [ A ] [A ] 0
0
−
1 [A]t 1 1 1 1 = − kt, − + = − kt, = kt + [A] [A]0 [A] t [A] 0 [A] t [A] 0
CHAPTER 12 5.
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533
The integrated rate laws can be put into the equation for a straight line, y = mx + b where x and y are the x and y axes, m is the slope of the line, and b is the y-intercept. Zero order: [A] = −kt + [A] y = mx + b A plot of [A] vs. time will be linear with a negative slope equal to −k and a y-intercept equal to [A]. First order: ln[A] = −kt + ln[A]0 y = mx + b A plot of ln[A] vs. time will be linear with a negative slope equal to −k and a y-intercept equal to ln[A]0. Second order:
1 1 = kt + [A]0 [A ] y = mx + b
A plot of 1/[A] vs. time will be linear with a positive slope equal to k and a y-intercept equal to 1/[A]0. When two or more reactants are studied, only one of the reactants is allowed to change during any one experiment. This is accomplished by having a large excess of the other reactant(s) as compared to the reactant studied; so large that the concentration of the other reactant(s) stays effectively constant during the experiment. The slope of the straight-line plot equals k (or –k) multiplied by the other reactant concentrations raised to the correct orders. Once all the orders are known for a reaction, then any (or all) of the slopes can be used to determine k. 6.
At t = t1/2, [A] = 1/2[A]0; Plugging these terms into the integrated rate laws yields the following half-life expressions: zero order t1/2 =
[A]0 2k
first order t1/2 =
ln 2 k
second order t1/2 =
1 k[A]0
The first order half-life is independent of concentration, the zero-order half-life is directly related to the concentration, and the second order half-life is inversely related to concentration. For a first order reaction, if the first half-life equals 20. s, the second half-life will also be 20. s because the half-life for a first order reaction is concentration independent. The second half-life for a zero-order reaction will be 1/2(20.) = 10. s. This is because the half-life for a zero-order reaction has a direct relationship with concentration (as the concentration decreases by a factor of 2, the half-life decreases by a factor of 2). For a second order reaction which has an inverse relationship between t1/2 and [A]0, the second half-life will be 40. s (twice the first half-life value).
534 7.
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a. An elementary step (reaction) is one for which the rate law can be written from the molecularity, i.e., from coefficients in the balanced equation. b. The molecularity is the number of species that must collide to produce the reaction represented by an elementary step in a reaction mechanism. c. The mechanism of a reaction is the series of proposed elementary reactions that may occur to give the overall reaction. The sum of all the steps in the mechanism gives the balanced chemical reaction. d. An intermediate is a species that is neither a reactant nor a product but that is formed and consumed in the reaction sequence. e. The rate-determining step is the slowest elementary reaction in any given mechanism.
8.
For a mechanism to be acceptable, the sum of the elementary steps must give the overall balanced equation for the reaction, and the mechanism must give a rate law that agrees with the experimentally determined rate law. A mechanism can never be proven absolutely. We can only say it is possibly correct if it follows the two requirements described above. Most reactions occur by a series of steps. If most reactions were unimolecular, then most reactions would have a first order overall rate law, which is not the case.
9.
The premise of the collision model is that molecules must collide to react, but not all collisions between reactant molecules result in product formation. a. The larger the activation energy, the slower the rate. b. The higher the temperature, the more molecular collisions with sufficient energy to convert to products and the faster the rate. c. The greater the frequency of collisions, the greater the opportunities for molecules to react, and, hence, the greater the rate. d. For a reaction to occur, it is the reactive portion of each molecule that must be involved in a collision. Only some of all the possible collisions have the correct orientation to convert reactants to products. endothermic, E > 0
exothermic, E < 0
Reaction Progress
Reaction Progress
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535
The activation energy for the reverse reaction will be the energy difference between the products and the transition state at the top of the potential energy “hill.” For an exothermic reaction, the activation energy for the reverse reaction (Ea,r) is larger than the activation energy for the forward reaction (Ea), so the rate of the forward reaction will be greater than the rate of the reverse reaction. For an endothermic reaction, Ea,r < Ea so the rate of the forward reaction will be less than the rate of the reverse reaction (with other factors being equal). 10.
− Ea 1 + lnA R T y = m x + b
Arrhenius equation: k = Ae − Ea / RT ; ln(k) =
The data needed is the value of the rate constant as a function of temperature. One would plot lnk versus1/T to get a straight line (with temperature in kelvin). The slope of the line is equal to –Ea/R and the y-intercept is equal to ln(A). The R value is 8.3145 J/K•mol. If one knows the rate constant at two different temperatures, then the following equation allows determination of Ea: k E 1 1 ln 2 = a − R T1 T2 k1
11.
A catalyst increases the rate of a reaction by providing reactants with an alternate pathway (mechanism) to convert to products. This alternate pathway has a lower activation energy, thus increasing the rate of the reaction. A homogeneous catalyst is one that is in the same phase as the reacting molecules, and a heterogeneous catalyst is in a different phase than the reactants. The heterogeneous catalyst is usually a solid, although a catalyst in a liquid phase can act as a heterogeneous catalyst for some gas phase reactions. Since the catalyzed reaction has a different mechanism than the uncatalyzed reaction, the catalyzed reaction most likely will have a different rate law.
Active Learning Questions 1.
Kinetics refers to the speed of a reaction. Thermodynamics refers to the energetics of the reaction. A reaction can be favorable from an energetic standpoint (exothermic reactions are favorable from an energy standpoint), but it may be a very slow reaction, so it never occurs. Thermodynamics deals with the energetics of the initial vs. the final state of the reaction, kinetics deals with the pathway reactants take to get to products.
2.
One experimental method to determine rate laws is the method of initial rates. Several experiments are carried out using different initial concentrations of reactants, and the initial rate is determined for each experiment. The results are then compared to see how the initial rate depends on the initial concentrations. This allows the orders in the rate law to be determined. The value of the rate constant is determined from the experiments once the orders are known. The second experimental method utilizes the fact that the integrated rate laws can be put in the form of a straight-line equation. Concentration versus time data are collected for a reactant as a reaction is run. These data are then manipulated and plotted to see which plot gives a straight line. From the straight-line plot we get the order of the reactant, and the slope of the line is mathematically related to k, the rate constant.
536 3.
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The half-life equations are: Zero order: t1/2 =
[A]0 ln2 1 ; first order: t1/2 = ; 2nd order: t1/2 = k k[ A]0 2k
As a reaction is run over time, the concentration of reactant decreases. From the equations, the half-life for a first order reaction does not depend on concentration. So, if a reaction has a constant half-life value over time, it is probably a first order reaction. The half-life expression for a second order reaction has an inverse relationship on concentration. As the concentration decreases by a factor of two over time, the half-life for a second order reaction will double over time. The half-life for a zero-order reaction has a direct relationship with concentration. As the concentration decreases by a factor of two over time, the half-life for a zero-order reaction will decrease by a factor of two over time. 4.
In general, as temperature increases, the value of k increases exponentially. As temperature increases, more of the collisions between reactant molecules have sufficient energy to convert to products. This sufficient energy is called the activation energy.
5.
A reaction does not need to warm up. Some reactions are fast, and some are slow. There is no one general rate for a reaction. The true part to the statements is that most reactions slow down as reactants convert to products; rate does depend on concentration (except for zero order reactions).
6.
One possibility is that the rate determining step in the reaction only involves B molecules, so only B is present in the rate law. Another possibility is that this is a catalyzed reaction by a metal catalyst or an enzyme. Once the catalyst or enzyme is saturated with reactants, the rate of the reaction is a constant value, no matter what the concentration of reactants.
7.
Most reactions do not directly convert from reactants to products. Instead, most reactions occur in a series of steps called a mechanism. The rate of the slow step in the mechanism determines the rate law. The slow step does not have to include all the reactants in a balanced equation.
8.
To determine the order of a reaction (aA → bB), one needs concentration of A vs time data. One can then plot the data in a specific way to get a straight-line plot that proves the order of the reaction. In all cases, the slope of the straight line is related to k, the rate constant for the reaction. For a zero-order reaction, a plot of [A] vs time will be linear with a negative slope equal to ‒k. For a first order reaction, a plot of ln[A] vs time will be linear with a negative slope equal to ‒k. For a second order reaction, a plot of 1/[A] vs time is linear with a positive slope equal to k.
9.
k = Ae−Ea/RT; from the Arrhenius equation, the rate constant depends on temperature and activation energy. Other factors in the A term of the equation are collision frequency and an orientation of collision factor. The concentrations and orders are taken care of the rate law.
10.
For a graph of [A] vs time for a zero-order reaction, reference Figure 12.6. Note the plot is a straight line; the rate is a constant value no matter the specific concentration of reactant. Because there is no concentration dependence on rate, each successive half-life gets shorter and shorter because there is a smaller change in concentration for each successive half-life. It takes half the amount of time to go from 0.50[A]0 to 0.25[A]0 than to go from [A]0 to 0.50[A]0; all because the rate does not depend on concentration.
CHAPTER 12
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537
For a graph of [A] vs time for a first order reaction, see Figure 12.4 of the text. Notice for a first order reaction, the half-life is constant; each successive half-life is the same value. Rate depends on [A]. As the reaction proceeds, the concentration decreases so the rate of the reaction decreases. For each successive half-life, there is a smaller change in concentration, but the rate of reaction also decreases. The result is that successive half-lives for a first order reaction are constant. For a second order reaction, the concentration decreases more quickly with time as compared to the first order plot and the successive half-lives get longer. There is a smaller change in concentration for successive half-lives, but the rate of reaction decreases more severely than a first order reaction due to the squared dependence on concentration. Since the rate decreases faster than the concentration decreases, successive half-lives get longer and longer for a second order reaction. For example, consider the concentration vs time data in Example 12.5 of the text. When this is plotted, the first half-life is about 1700s while the second half-life is about 3400 s. 11.
a. CH3CH2CH2CH2‒Br + OH‒ → CH3CH2CH2CH2‒OH + Br‒
b. Rate = k[1-bromobutane]x[OH‒]y Comparing the first two experiments where [OH‒] is unchanged while [1-bromobutane] doubles, the rate doubles. Therefore, x = 1, and the reaction is first order in 1-bromobutane. Comparing the first and third experiments where [1-bromobutane] is unchanged while [OH‒] triples, the rate triples. Therefore, y = 1, and the reaction is first order in OH‒. Rate = k[1-bromobutane]1[OH‒]1 For the first experiment:
3.0 × 10−3 mol 0.0010 mol 0.010 mol 2 = k , k = 3.0 × 10 L/mol•s Ls L L The other experiments give the same value for k. Rate = k[2-bromo-2- methylpropane]x[OH‒]y Comparing the first two experiments where [OH‒] is unchanged while [2-bromo-2methylpropane] doubles, and the rate doubles. Therefore, x = 1, and the reaction is first order in 2-bromo-2-methylpropane. Comparing the first and third experiments where [2bromo-2-methylpropane] is unchanged while [OH‒] triples, the rate does not change. Therefore, y = 0, and the reaction is zero order in OH‒. Rate = k[2-bromo-2- methylpropane]1
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For the first experiment:
1.2 × 10-2 mol 0.0010 mol ‒1 = k , k = 12 s Ls L The other experiments give the same value for k. c. The one step mechanism is consistent with the data for the reaction of 1-bromobutane with OH‒. The derived rate law of the one step mechanism is first order in both reactants, which is the same rate law determined from experiment. For the reaction of 2-bromo-2methylpropane with OH‒, the two-step mechanism with the first step rate determining (the slow step) is consistent with the data. If the first step is rate determining, the derived rate law is rate = k[2-bromo-2-methylpropane], which is the same rate law determined from experiment.
Questions 12.
The average rate decreases with time because the reverse reaction occurs more frequently as the concentration of products increase. Initially, with no products present, the rate of the forward reaction is at its fastest, but as time goes on, the rate gets slower and slower since products are converting back into reactants. The instantaneous rate will also decrease with time. The only rate that is constant is the initial rate. This is the instantaneous rate taken at t 0. At this time, the amount of products is insignificant, and the rate of the reaction only depends on the rate of the forward reaction.
13.
One experimental method to determine rate laws is the method of initial rates. Several experiments are carried out using different initial concentrations of reactants, and the initial rate is determined for each experiment. The results are then compared to see how the initial rate depends on the initial concentrations. This allows the orders in the rate law to be determined. The value of the rate constant is determined from the experiments once the orders are known. The second experimental method utilizes the fact that the integrated rate laws can be put in the form of a straight-line equation. Concentration versus time data are collected for a reactant as a reaction is run. These data are then manipulated and plotted to see which manipulation gives a straight line. From the straight-line plot we get the order of the reactant, and the slope of the line is mathematically related to k, the rate constant.
14.
a. T2 > T1; as temperature increases, the distribution of collision energies shifts to the right. That is, as temperature increases, there are fewer collision energies with small energies and more collisions with large energies. b. As temperature increases, more of the collisions have the required activation energy necessary to convert reactants into products. Hence, the rate of the reaction increases with increasing temperature.
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539
15.
All these choices would affect the rate of the reaction, but only b and c affect the rate by effecting the value of the rate constant k. The value of the rate constant depends on temperature. The value of the rate constant also depends on the activation energy. A catalyst will change the value of k because the activation energy changes. Increasing the concentration (partial pressure) of either O2 or NO does not affect the value of k, but it does increase the rate of the reaction because both concentrations appear in the rate law.
16.
a. Activation energy and ΔE are independent of each other. Activation energy depends on the path reactants to take to convert to products. The overall energy change ΔE only depends on the initial and final energy states of the reactants and products. ΔE is pathindependent. b. The rate law can only be determined from experiment, not from the overall balanced reaction. c. Most reactions occur by a series of steps. The rate of the reaction is determined by the rate of the slowest step in the mechanism.
17.
In a unimolecular reaction, a single reactant molecule decomposes to products. In a bimolecular reaction, two molecules collide to give products. The probability of the simultaneous collision of three molecules with enough energy and the proper orientation is very small, making termolecular steps very unlikely.
18.
The most common method to experimentally determine the differential rate law is the method of initial rates. Once the differential rate law is determined experimentally, the integrated rate law can be derived. However, sometimes it is more convenient and more accurate to collect concentration versus time data for a reactant. When this is the case, then we do “proof” plots to determine the integrated rate law. Once the integrated rate law is determined, the differential rate law can be determined. Either experimental procedure allows determination of both the integrated and the differential rate law; and which rate law is determined by experiment and which is derived is usually decided by which data are easiest and most accurately collected.
19.
[A] Rate 2 k[A]2x = = 2 x Rate 1 k[A]1 [A]1
x
The rate doubles as the concentration quadruples: 2 = (4)x, x = 1/2 The order is 1/2 (the square root of the concentration of reactant). For a reactant that has an order of −1 and the reactant concentration is doubled: Rate 2 1 = ( 2) −1 = Rate 1 2
The rate will decrease by a factor of 1/2 when the reactant concentration is doubled for a −1 order reaction. Negative orders are seen for substances that hinder or slow down a reaction.
540 20.
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Some energy must be added to get the reaction started, that is, to overcome the activation energy barrier. Chemically what happens is: Energy + H2 → 2 H The hydrogen atoms initiate a chain reaction that proceeds very rapidly. Collisions of H 2 and O2 molecules at room temperature do not have sufficient kinetic energy to form hydrogen atoms and initiate the reaction.
21.
Two reasons are: (1) The collision must involve enough energy to produce the reaction; that is, the collision energy must be equal to or exceed the activation energy. (2) The relative orientation of the reactants when they collide must allow formation of any new bonds necessary to produce products.
22.
a. The blue plot is the catalyzed pathway. The catalyzed pathway has the lower activation. Therefore the catalyzed pathway is faster. b. E1 represents the activation energy for the uncatalyzed pathway. c. E2 represents the energy difference between the reactants and products. Note that E2 is the same for both the catalyzed and the uncatalyzed pathways. It is the activation energy that is different for a catalyzed pathway versus an uncatalyzed pathway. d. Because the products have a higher total energy as compared to reactants, this is an endothermic reaction.
23.
Enzymes are very efficient catalysts. As is true for all catalysts, enzymes speed up a reaction by providing an alternative pathway for reactants to convert to products. This alternative pathway has a smaller activation energy and hence, a faster rate. Also true is that catalysts are not used up in the overall chemical reaction. Once an enzyme encounters the correct reagent, the chemical reaction quickly occurs, and the enzyme is then free to catalyze another reaction. Because of the efficiency of the reaction step, only a relatively small amount of enzyme is needed to catalyze a specific reaction, no matter how complex the reaction.
24.
The enzyme lactase catalyzes the breakdown of the sugar lactose found in dairy products. Without sufficient lactase, individuals with lactose intolerance are unable to fully digest the lactose sugar in milk. This can result in diarrhea, gas, and bloating after eating or drinking dairy products.
25.
The Arrhenius equation is k = A exp(−Ea/RT) or, in logarithmic form, ln k = −Ea/RT + ln A. This equation is in the form of a straight-line equation, y = mx + b. A graph of ln k versus 1/T yields a straight line with a slope equal to ‒Ea/R and y-intercept of lnA. So the slope and the yintercept of the plot can be used to determine the activation energy, Ea, and the frequency factor, A, for a reaction. We can then use these values to estimate k at other temperatures.
26.
The slope of the ln k versus 1/T plot (with temperature in Kelvin) is equal to −Ea/R. Because Ea for the catalyzed reaction will be smaller than E a for the uncatalyzed reaction, the slope of the catalyzed plot should be less negative.
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541
27.
Increasing the concentration of A will have no effect on the rate of this reaction since the reaction is zero order in A. Increasing the concentration of B, raising the temperature, and lowering the activation energy will all increase the rate of the reaction.
28.
t1/2 =
[A]0 ln2 1 ; first order: t1/2 = ; 2nd order: t1/2 = k k [ A ]0 2k
Statement a and c are true. The half-life expression is directly related to the concentration. As the reaction proceeds, the concentration of reactant decreases, so the half-life decreases. From the equations above, the half-life for a first order reaction does not depend on concentration of reactant, so its half-life is constant as the reaction proceeds. And a second order reaction has an inverse relationship with concentration. For a second order reaction, the half-life increases as the reaction proceeds. 29.
Consider the energy plot below for a reaction with a positive ΔE. The activation energy for the reverse reaction is the energy that products must have to convert back to reactants. When a reaction has a positive ΔE, the activation energy for the reverse reaction is smaller than the activation energy for the forward reaction. Note that Ea in the plot below is the activation energy for the forward reaction. The activation energy for the reverse reaction will be smaller than the activation energy of the forward reaction by the ΔE quantity.
E
Ea
P E
R
RC
30.
A catalyst provides an alternative pathway for reactants to convert to products that has a lower activation energy. Thus, a catalyst lowers the activation energy for both the forward and reverse reactions. Because the activation energy is lowered, the rate of the forward reaction and the reverse reactions are increased. ΔE is unaffected by a catalyst as is the amount of product produced. A catalyst doesn’t affect the final endpoint of a reaction, you just get there quicker.
Exercises Reaction Rates 31.
The coefficients in the balanced reaction relate the rate of disappearance of reactants to the rate of production of products. From the balanced reaction, the rate of production of P4 will be 1/4 the rate of disappearance of PH3, and the rate of production of H 2 will be 6/4 the rate of disappearance of PH3. By convention, all rates are given as positive values. Rate =
− [PH 3 ] − ( −0.048 mol/2.0 L) = = 2.4 × 10 −3 mol/L•s t s
542
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Δ[P4 ] 1 Δ[ PH 3 ] =− = 2.4 × 10 −3 /4 = 6.0 × 10 −4 mol/L•s Δt 4 Δt Δ[H 2 ] 6 Δ[PH 3 ] =− = 6(2.4 × 10 −3 )/4 = 3.6 × 10 −3 mol/L•s Δt 4 Δt
32.
The coefficients in the balanced reaction relate the rate of disappearance of reactants to the rate of production of products. From the balanced reaction, the rate of production of S4O62will be 1/2 the rate of disappearance of S2O32-. Rate =
−Δ[S2 O32− ] −( − 7.05 10−3 mol/L) = 6.41 × 10‒4 mol/L•s = Δt 11.0 s
Δ[S4O62− ] 1 Δ[S2O32− ] = 6.41 × 10‒4/2 = 3.21 × 10 −4 mol/L•s =− Δt 2 Δt 33.
Using the coefficients in the balanced equation to relate the rates: [ NH 3 ] [H 2 ] [ N 2 ] [ N 2 ] =3 and = −2 t t t t [ NH 3 ] 1 [ H 2 ] 2 [ H 2 ] 1 [ NH 3 ] So : − =− = or t 3 t 3 t 2 t
Ammonia is produced at a rate equal to 2/3 of the rate of consumption of hydrogen. 34.
2 NO2(g) → 2 NO(g) + O2(g); from the coefficients in the balanced equation, the rate of production for NO will equal the rate of disappearance of NO 2, and the rate of production of O2 will be ½ the rate of disappearance. From the plot, rate of disappearance of NO 2 is -0.0026 mol/L/110 s = 2.4 × 10‒5 mol/L•s. So, the rate of appearance of NO2 is also 2.4 × 10‒5 mol/L•s, and the rate of appearance of O 2 will be 1.2 × 10‒5 mol/L•s.
35.
a. Average rate =
− [H 2O 2 ] − (0.500 M − 1.000 M) = 2.31 × 10 −5 mol/L•s = 4 t (2.16 10 s − 0)
From the coefficients in the balanced equation:
b.
Δ[O 2 ] 1 Δ[H 2 O 2 ] =− = 1.16 × 10 −5 mol/L•s Δt 2 Δt − [H 2O 2 ] − (0.250 − 0.500) M = 1.16 × 10 −5 mol/L•s = 4 4 t (4.32 10 − 2.16 10 ) s
Δ[O 2 ] = 1/2 (1.16 × 10 −5 ) = 5.80 × 10 −6 mol/L•s Δt
Notice that as time goes on in a reaction, the average rate decreases. 36.
0.0120/0.0080 = 1.5; reactant B is used up 1.5 times faster than reactant A. This corres-ponds to a 3 to 2 mole ratio between B and A in the balanced equation. 0.0160/0.0080 = 2; product C is produced twice as fast as reactant A is used up, so the coefficient for C is twice the coefficient for A. A possible balanced equation is 2A + 3B → 4C.
CHAPTER 12 37.
CHEMICAL KINETICS
543
a. The units for rate are always mol/L•s. c. Rate = k[A],
mol mol = k Ls L
b. Rate = k; k must have units of mol/L•s. d. Rate = k[A]2,
k must have units of s −1 .
mol mol = k Ls L
2
k must have units of L/mol•s.
e. L2/mol2•s 1/ 2
38.
Rate = k[Cl]1/2[CHCl3],
mol mol = k Ls L
mol ; k must have units of L1/2/mol1/2•s. L
Rate Laws from Experimental Data: Initial Rates Method 39.
a. In the first two experiments, [NO] is held constant and [Cl2] is doubled. The rate also doubled. Thus the reaction is first order with respect to Cl 2. Or mathematically, Rate = k[NO]x[Cl2]y. 0.36 k (0.10) x (0.20) y (0.20) y , 2.0 = 2.0y, y = 1 = = 0.18 k (0.10) x (0.10) y (0.10) y
We can get the dependence on NO from the second and third experiments. Here, as the NO concentration doubles (Cl2 concentration is constant), the rate increases by a factor of four. Thus the reaction is second order with respect to NO. Or mathematically: 1.45 k (0.20) x (0.20) (0.20) x , 4.0 = 2.0x, x = 2; so Rate = k[NO]2[Cl2]. = = 0.36 k (0.10) x (0.20) (0.10) x
Try to examine experiments where only one concentration changes at a time. The more variables that change, the harder it is to determine the orders. Also, these types of problems can usually be solved by inspection. In general, we will solve using a mathematical approach, but keep in mind that you probably can solve for the orders by simple inspection of the data. b. The rate constant k can be determined from the experiments. From experiment 1: 2
0.18 mol 0.10 mol 0.10 mol = k , k = 180 L2/mol2•min L min L L
From the other experiments: k = 180 L2/mol2•min (second exp.); k = 180 L2/mol2•min (third exp.) The average rate constant is kmean = 1.8 × 102 L2/mol2•min. 40.
a. Rate = k[I−]x[S2O82−]y;
12.5 10 −6 k (0.080) x (0.040) y = , 2.00 = 2.0x, x = 1 6.25 10 −6 k (0.040) x (0.040) y
k (0.080 )(0.040 ) y 12.5 10 −6 = , 2.00 = 2.0y, y = 1; Rate = k[I−][S2O82−] y −6 k (0.080 )(0.020 ) 6.25 10
544
CHAPTER 12
CHEMICAL KINETICS
b. For the first experiment: 12.5 10 −6 mol 0.080 mol 0.040 mol = k , k = 3.9 × 10−3 L/mol•s Ls L L
Each of the other experiments also gives k = 3.9 × 10 −3 L/mol•s, so kmean = 3.9 × 10−3 L/mol•s. 41.
a. Rate = k[NOCl]n; using experiments two and three: 2.66 10 4 k (2.0 1016 ) n = , 4.01 = 2.0n, n = 2; Rate = k[NOCl]2 6.64 10 3 k (1.0 1016 ) n 2
b.
3.0 1016 molecules 5.98 10 4 molecules , k = 6.6 × 10−29 cm3/molecules•s = k 3 cm3 s cm
The other three experiments give (6.7, 6.6, and 6.6) × 10−29 cm3/molecules•s, respectively. The mean value for k is 6.6 × 10 −29 cm3/molecules•s. 6.6 10 −29 cm3 1L 6.022 10 23 molecules 4.0 10 −8 L = molecules s mol mol s 1000 cm3
c. 42.
Rate = k[N2O5]x; the rate laws for the first two experiments are: 2.26 × 10−3 = k(0.190)x and 8.90 × 10−4 = k(0.0750)x Dividing the two rate laws: 2.54 = k =
(0.190 ) x = (2.53)x, x = 1; Rate = k[N2O5] x (0.0750 )
8.90 10 −4 mol/L • s Rate = 1.19 × 10−2 s−1 = [N 2 O 5 ] 0.0750 mol/L
The other experiments give similar values for k. kmean = 1.19 × 10−2 s−1 43.
a. Rate = k[I−]x[OCl−]y;
7.91 10 −2
k (0.12) x (0.18) y
3.95 10
k (0.060 ) x (0.18) y
= −2
3.95 10 −2
k (0.060 )(0.18) y
9.88 10
k (0.030 )(0.090 ) y
= −3
= 2.0x, 2.00 = 2.0x, x = 1
, 4.00 = 2.0 × 2.0y, 2.0 = 2.0y, y = 1
Rate = k[I−][OCl−] b. From the first experiment:
7.91 10 −2 mol 0.12 mol 0.18 mol = k , k = 3.7 L/mol•s Ls L L
All four experiments give the same value of k to two significant figures. c.
Rate =
3.7 L 0.15 mol 0.15 mol = 0.083 mol/L • s mol s L L
CHAPTER 12 44.
CHEMICAL KINETICS
545
Rate = k[NO]x[O2]y; comparing the first two experiments, [O2] is unchanged, [NO] is tripled, and the rate increases by a factor of nine. Therefore, the reaction is second order in NO (3 2 = 9). The order of O2 is more difficult to determine. Comparing the second and third experiments: 3.13 1017 k (2.50 1018 ) 2 (2.50 1018 ) y = 1.80 1017 k (3.00 1018 ) 2 (1.00 1018 ) y
1.74 = 0.694(2.50)y, 2.51 = 2.50y, y = 1 Rate = k[NO]2[O2]; from experiment 1: 2.00 × 1016 molecules/cm3•s = k(1.00 × 1018 molecules/cm3)2 (1.00 × 1018 molecules/cm3) k = 2.00 × 10−38 cm6/molecules2•s = kmean 2
6.21 1018 molecules 7.36 1018 molecules 2.00 10 −38 cm6 Rate = cm3 cm3 molecules 2 s
Rate = 5.68 × 1018 molecules/cm3•s 45.
a. Rate = k[Hb]x[CO]y Comparing the first two experiments, [CO] is unchanged, [Hb] doubles, and the rate doubles. Therefore, x = 1, and the reaction is first order in Hb. Comparing the second and third experiments, [Hb] is unchanged, [CO] triples, and the rate triples. Therefore, y = 1, and the reaction is first order in CO. b. Rate = k[Hb][CO] c. From the first experiment: 0.619 µmol/L•s = k(2.21 µmol/L)(1.00 µmol/L), k = 0.280 L/µmol•s The second and third experiments give similar k values, so k mean = 0.280 L/mol•s d. Rate = k[Hb][CO] =
46.
0.280 L 3.36 μmol 2.40 μmol = 2.26 µmol/L•s μmol s L L
a. Rate = k[ClO2]x[OH−]y; from the first two experiments: 2.30 × 10−1 = k(0.100)x(0.100)y and 5.75 × 10−2 = k(0.0500)x(0.100)y Dividing the two rate laws: 4.00 =
(0.100 ) x = 2.00x, x = 2 (0.0500 ) x
Comparing the second and third experiments: 2.30 × 10−1 = k(0.100)(0.100)y and 1.15 × 10−1 = k(0.100)(0.0500)y
546
CHAPTER 12 Dividing: 2.00 =
CHEMICAL KINETICS
(0.100) y = 2.0y, y = 1 (0.050) y
The rate law is Rate = k[ClO2]2[OH−]. 2.30 × 10−1 mol/L•s = k(0.100 mol/L)2(0.100 mol/L), k = 2.30 × 102 L2/mol2•s = kmean b. Rate = 47.
2.30 10 2 L2 mol 2 s
2
0.0844 mol 0.175 mol = 0.594 mol/L•s L L
Rate = k[(CH3)3CBr]x[OH‒]y; comparing the first two experiments, [OH‒] is unchanged, [(CH3)3CBr] doubles, and the rate doubles. Therefore, the reaction is first order in (CH3)3CBr. Comparing the first and third experiments, [(CH3)3CBr] is unchanged, [OH‒] triples, and the rate does not change. Therefore, the reaction is zero order in OH‒. Rate = k[(CH3)3CBr]1 For the first experiment:
1.0 × 10-3 mol 0.10 mol ‒2 ‒1 = k , k = 1.0 × 10 s = kmean Ls L Rate = k[(CH3)3CBr]1 = 1.0 × 10‒2 s‒1 (0.30 mol/L) = 3.0 × 10‒3 mol/L•s 48.
In experiments 1 and 2, doubling [MnO4−] while keeping the other concentrations constant results in a rate increase of a factor of 4. The reaction is second order in MnO4−. In experiments 2 and 3, doubling [H2C2O4] while keeping the other concentrations constant results in the rate increase of a factor of 2. The reaction is first order in H2C2O4. In experiments 3 and 4, doubling [H+] while keeping the other concentrations constant causes no change to the rate. The reaction zero order in H+. Rate = k[MnO4−]2[H2C2O4], using experiment 3 to determine k: 1.6 × 10−3 mol/L•s = k(2.0 × 10‒3 mol/L)2(2.0 × 10‒3 mol/L), k = 2.0 × 105 L2/mol2•s Rate = k[MnO4−]2[H2C2O4] = 2.0 × 105(3.5 × 10‒3)2(3.0 × 10‒3) = 7.4 × 10‒3 mol/L•s
Integrated Rate Laws 49.
The first assumption to make is that the reaction is first order. For a first order reaction, a graph of ln[H2O2] versus time will yield a straight line. If this plot is not linear, then the reaction is not first order, and we make another assumption.
CHAPTER 12
CHEMICAL KINETICS
Time (s)
[H2O2] (mol/L)
ln[H2O2]
0 120. 300. 600. 1200. 1800. 2400. 3000. 3600.
1.00 0.91 0.78 0.59 0.37 0.22 0.13 0.082 0.050
0.000 −0.094 −0.25 −0.53 −0.99 −1.51 −2.04 −2.50 −3.00
547
Note: We carried extra significant figures in some of the natural log values to reduce roundoff error. For the plots, we will do this most of the time when the natural log function is involved. The plot of ln[H2O2] versus time is linear. Thus, the reaction is first order. The rate law and integrated rate law are Rate = k[H2O2] and ln[H2O2] = −kt + ln[H2O2]0. We determine the rate constant k by determining the slope of the ln[H2O2] versus time plot (slope = −k). Using two points on the curve gives: slope = −k =
Δy 0 − (3.00 ) = = −8.3 × 10−4 s−1, k = 8.3 × 10−4 s−1 Δx 0 − 3600 .
To determine [H2O2] at 4000. s, use the integrated rate law, where [H 2O2]0 = 1.00 M. [H 2O 2 ] = −kt ln[H2O2] = −kt + ln[H2O2]0 or ln [ H 2 O 2 ]0 [H O ] ln 2 2 = −8.3 × 10−4 s−1 × 4000. s, ln[H2O2] = −3.3, [H2O2] = e−3.3 = 0.037 M 1.00
50.
a. Because the ln[A] versus time plot was linear, the reaction is first order in A. The slope of the ln[A] versus time plot equals −k. Therefore, the rate law, the integrated rate law, and the rate constant value are: Rate = k[A]; ln[A] = −kt + ln[A]0; k = 2.97 × 10−2 min−1 b. The half-life expression for a first order rate law is: t1/2 =
ln 2 0.6931 0.6931 = , t1/2 = = 23.3 min k k 2.97 10 − 2 min −1
548
CHAPTER 12
CHEMICAL KINETICS
c. 2.50 × 10−3 M is 1/8 of the original amount of A present initially, so the reaction is 87.5% complete. When a first-order reaction is 87.5% complete (or 12.5% remains), then the reaction has gone through 3 half-lives: 100% → 50.0% → 25.0% → 12.5%; t = 3 × t1/2 = 3 × 23.3 min = 69.9 min t1/2 t1/2 t1/2 Or we can use the integrated rate law:
2.50 10 −3 M [ A] −2 −1 = − kt , ln ln 2.00 10 −2 M = −(2.97 × 10 min )t [ A ] 0 t= 51.
ln( 0.125 ) = 70.0 min − 2.97 10 − 2 min −1
Assume the reaction is first order and see if the plot of ln[NO2] versus time is linear. If this isn’t linear, try the second-order plot of 1/[NO2] versus time because second-order reactions are the next most common after first-order reactions. The data and plots follow. Time (s)
[NO2] (M)
ln[NO2]
1/[NO2] ( M −1 )
0 1.20 × 103 3.00 × 103 4.50 × 103 9.00 × 103 1.80 × 104
0.500 0.444 0.381 0.340 0.250 0.174
−0.693 −0.812 −0.965 −1.079 −1.386 −1.749
2.00 2.25 2.62 2.94 4.00 5.75
The plot of 1/[NO2] versus time is linear. The reaction is second order in NO 2. The rate law and integrated rate law are: Rate = k[NO2]2 and
1 1 = kt + [ NO 2 ] [ NO 2 ]0
The slope of the plot 1/[NO2] vs. t gives the value of k. Using a couple of points on the plot:
CHAPTER 12
CHEMICAL KINETICS
slope = k =
549
y (5.75 − 2.00) M −1 = = 2.08 × 10 −4 L/mol•s x (1.80 10 4 − 0) s
To determine [NO2] at 2.70 × 104 s, use the integrated rate law, where 1/[NO2]o = 1/0.500 M = 2.00 M −1 . 1 1 1 2.08 10 −4 L × 2.70 × 104 s + 2.00 M −1 = kt + , = [ NO2 ] [ NO2 ]0 [ NO2 ] mol s
1 = 7.62, [NO2] = 0.131 M [ NO 2 ]
52.
a. Because the 1/[A] versus time plot was linear, the reaction is second order in A. The slope of the 1/[A] versus time plot equals the rate constant k. Therefore, the rate law, the integrated rate law, and the rate constant value are: Rate = k[A]2;
1 1 = kt + ; k = 3.60 × 10−2 L/mol • s [A]0 [A ]
b. The half-life expression for a second-order reaction is: t1/2 = For this reaction: t1/2 =
3.60 10
−2
1 k[A]0
1 L/mol • s 2.80 10 −3 mol/L
= 9.92 × 103 s
Note: We could have used the integrated rate law to solve for t1/2, where [A] = (2.80 × 10−3/2) mol/L. c. Because the half-life for a second-order reaction depends on concentration, we must use the integrated rate law to solve. 1 3.60 10 −2 L 1 1 1 = t + = kt + , [A]0 7.00 10 − 4 M mol s [A ] 2.80 10 −3 M
1.43 × 103 − 357 = (3.60 × 10−2)t, t = 2.98 × 104 s 53.
a. Because the [C2H5OH] versus time plot was linear, the reaction is zero order in C2H5OH. The slope of the [C2H5OH] versus time plot equals −k. Therefore, the rate law, the integrated rate law, and the rate constant value are: Rate = k[C2H5OH]0 = k; [C2H5OH] = −kt + [C2H5OH]0; k = 4.00 × 10−5 mol/L•s b. The half-life expression for a zero-order reaction is t1/2 = [A]0/2k. t1/2 =
[C 2 H 5 OH]0 1.25 10 −2 mol/L = = 156 s 2k 2 4.00 10 −5 mol/L • s
Note: We could have used the integrated rate law to solve for t1/2, where [C2H5OH] = (1.25 × 10−2/2) mol/L.
550
CHAPTER 12
CHEMICAL KINETICS
c. [C2H5OH] = −kt + [C2H5OH]0 , 0 mol/L = −(4.00 × 10−5 mol/L•s)t + t= 54.
1.25 10 4.00 10
−2
−5
mol/L
mol/L • s
1.25 × 10−2 mol/L
= 313 s
From the data, the pressure of C2H5OH decreases at a constant rate of 13 torr for every 100. s. Because the rate of disappearance of C2H5OH is not dependent on concentration, the reaction is zero order in C2H5OH. k=
13 torr 1 atm = 1.7 × 10 −4 atm/s 100 . s 760 torr
The rate law and integrated rate law are: 1 atm = −kt + 0.329 atm Rate = k = 1.7 × 10 −4 atm/s; PC2 H5OH = −kt + 250. torr 760 torr
At 900. s: PC2 H5OH = −1.7 × 10 −4 atm/s × 900. s + 0.329 atm = 0.176 atm = 0.18 atm = 130 torr 55.
The first assumption to make is that the reaction is first order. For a first-order reaction, a graph of ln[C4H6] versus t should yield a straight line. If this isn't linear, then try the second- order plot of 1/[C4H6] versus t. The data and the plots follow: Time
195
604
1246
2180
6210 s
[C4H6]
1.6 × 10−2
1.5 × 10−2
1.3 × 10−2
1.1 × 10−2
0.68 × 10−2 M
ln[C4H6] 1/[C4H6]
−4.14 62.5
−4.20 66.7
−4.34 76.9
−4.51 90.9
−4.99 147 M −1
Note: To reduce round-off error, we carried extra significant figures in the data points.
The natural log plot is not linear, so the reaction is not first order. Because the second-order plot of 1/[C4H6] versus t is linear, we can conclude that the reaction is second order in butadiene. The rate law is: Rate = k[C4H6]2
CHAPTER 12
CHEMICAL KINETICS
For a second-order reaction, the integrated rate law is
551 1 1 = kt + . [ C 4 H 6 ]0 [C 4 H 6 ]
The slope of the straight line equals the value of the rate constant. Using the points on the line at 1000. and 6000. s: k = slope = 56.
144 L/mol − 73 L/mol = 1.4 × 10−2 L/mol•s 6000. s − 1000. s
a. First, assume the reaction to be first order with respect to O. Hence a graph of ln[O] versus t would be linear if the reaction is first order. [O] (atoms/cm3)
t (s) 0 10. × 10−3 20. × 10−3 30. × 10−3
9
5.0 × 10 1.9 × 109 6.8 × 108 2.5 × 108
ln[O] 22.33 21.37 20.34 19.34
Because the graph is linear, we can conclude the reaction is first order with respect to O. b. The overall rate law is Rate = k[NO2][O]. Because NO2 was in excess, its concentration is constant. Thus, for this experiment, the rate law is Rate = k[O], where k = k[NO2]. In a typical first-order plot, the slope equals −k. For this experiment, the slope equals −k = −k[NO2]. From the graph: slope =
19.34 − 22.23 = −1.0 × 102 s−1, k = −slope = 1.0 × 102 s−1 (30. 10 −3 − 0) s
To determine k, the actual rate constant: k = k[NO2], 1.0 × 102 s−1 = k(1.0 × 1013 molecules/cm3) k = 1.0 × 10−11 cm3/molecules•s
552
CHAPTER 12
CHEMICAL KINETICS
57.
Because the 1/[A] versus time plot is linear with a positive slope, the reaction is second order with respect to A. The y intercept in the plot will equal 1/[A]0. Extending the plot, the y intercept will be about 10, so 1/10 = 0.1 M = [A]0.
58.
a. The slope of the 1/[A] versus time plot in Exercise 57 will equal k. Slope = k =
(60 − 20) L/mol = 10 L/mol•s (5 − 1) s
1 1 10 L 1 = kt + = 9s + = 100, [A] = 0.01 M [A] [A]o mol s 0.1 M
b. For a second-order reaction, the half-life does depend on concentration: t1/2 = First half-life: t1/2 =
1 10 L 0.1 mol mol s L
1 k [ A ]0
=1s
Second half-life ([A]0 is now 0.05 M): t1/2 = 1/(10 × 0.05) = 2 s Third half-life ([A]0 is now 0.025 M): t1/2 = 1/(10 × 0.025) = 4 s 59.
a. [A] = −kt + [A]0; if k = 5.0 × 10−2 mol/L•s and [A]0 = 1.00 × 10−3 M, then: [A] = −(5.0 × 10−2 mol/L•s)t + 1.00 × 10−3 mol/L b.
[ A ]0 = −(5.0 × 10−2)t1/2 + [A]0 because at t = t1/2, [A] = [A]o/2. 2
−0.50[A]0 = −(5.0 × 10−2)t1/2, t1/2 =
0.50(1.00 10 −3 ) = 1.0 × 10−2 s −2 5.0 10
Note: We could have used the t1/2 expression to solve (t1/2 =
[ A ]0 ). 2k
c. [A] = −kt + [A]0 = −(5.0 × 10−2 mol/L•s)(5.0 × 10−3 s) + 1.00 × 10−3 mol/L [A] = 7.5 × 10−4 mol/L [A]reacted = 1.00 × 10−3 mol/L − 7.5 × 10−4 mol/L = 2.5 × 10−4 mol/L [B]produced = [A]reacted = 2.5 × 10−4 M 60.
a. The integrated rate law for this zero-order reaction is [HI] = −kt + [HI]0. 1.20 10 −4 mol 25 min 60 s + 0.250 mol [HI] = −kt + [HI]0, [HI] = − Ls min L
[HI] = −0.18 mol/L + 0.250 mol/L = 0.07 M
CHAPTER 12
CHEMICAL KINETICS
b. [HI] = 0 = −kt + [HI]0, kt = [HI]0, t = t= 61.
0.250 mol/L 1.20 10 − 4 mol/L • s
553 [ HI ]0 k
= 2080 s = 34.7 min
If [A]0 = 100.0, then after 65 s, 45.0% of A has reacted, or [A] = 55.0. For first order reactions: [A] 55.0 = − kt , ln ln = −k(65 s), k = 9.2 × 10−3 s−1 [ A ] 100 . 0 0
t1/2 =
62.
ln 2 0.693 = 75 s = k 9.2 10 −3 s −1
a. When a reaction is 75.0% complete (25.0% of reactant remains), this represents two halflives (100% → 50% → 25%). The first-order half-life expression is t1/2 = (ln 2)/k. Because there is no concentration dependence for a first-order half-life, 320. s = two half-lives, t1/2 = 320./2 = 160. s. This is both the first half-life, the second half-life, etc. b. t1/2 =
ln 2 ln 2 ln 2 , k = = = 4.33 × 10 −3 s −1 k t1/ 2 160 . s
At 90.0% complete, 10.0% of the original amount of the reactant remains, so [A] = 0.100[A]0. [A] 0.100[A] 0 ln( 0.100 ) = − kt, ln ln = −(4.33 × 10 −3 s −1 )t, t = = 532 s [A] 0 − 4.33 10 −3 s −1 [A] 0
63.
For a first-order reaction, the integrated rate law is ln([A]/[A]0) = −kt. Solving for k: 0.250 mol/L = −k × 120. s, k = 0.0116 s −1 ln 1.00 mol/L
0.350 mol/L = −0.0116 s −1 × t, t = 150. s ln 2.00 mol/L
64.
[A] ln 2 0.693 = − kt ; k = ln = = 0.0124 d −1 [ A ] t 56 . 0 days 0 1/ 2 1.41 10 −7 mol/L = −(0.0124 d−1)t, t = 519 days ln 8.75 10 −5 mol/L
65.
The integrated rate law for a second order reaction is 1/[A] = kt + 1/[A] 0. Using the integrated rate law: _ (50. 10.) L/mol 1 1 , t = = 0.40 L/mol•min × t + = 1.0 102 min 0.020 mol/L 0.10 mol/L 0.40 L/mol • min
554 66.
CHAPTER 12
CHEMICAL KINETICS
a. The integrated rate law for a second order reaction is 1/[A] = kt + 1/[A]0, and the halflife expression is t1/2 = 1/k[A]0. We could use either to solve for t1/2. Using the integrated rate law: 1 1 1.11 L/mol , k = = k × 2.00 s + = 0.555 L/mol•s 0.900 mol/L (0.900/2) mol/L 2.00 s
b. 67.
8.9 L/mol 1 1 , t = = 0.555 L/mol•s × t + = 16 s 0.900 mol/L 0.100 mol/L 0.555 L/mol • s
Successive half-lives double as concentration is decreased by one-half. This is consistent with second-order reactions, so assume the reaction is second order in A. t1/2 = a.
1 1 1 , k= = = 1.0 L/mol•min k[A ]0 t1 / 2 [ A ]0 10.0 min( 0.10 M )
1 1 1 1.0 L = kt + = × 80.0 min + = 90. M −1 , [A] = 1.1 × 10 −2 M [A] 0.10 M [A ]0 mol min
b. 30.0 min = 2 half-lives, so 25% of original A is remaining. [A] = 0.25(0.10 M) = 0.025 M 68.
a. The first half-life is 20.0 min (time it takes to go from 10.0 to 5.00 mol/L). The second half-life is 40.0 mins (time it takes to go from 5.00 to 2.50 mol/L. The next half-life takes the A concentration from 2.50 to 1.25 mol/L. From the data, the third half-life should be double the previous half-life giving a value of 80.0 min. The total time it takes to go to 1.25 mol/L is 20.0 + 40.0 + 80.0 = 140.0 mins. b. The successive half-lives double as concentration is decreased by one-half. This is consistent with second-order reactions, so assume the reaction is second order in A. t1/2 =
1 1 1 , k= = = 5.00 × 10‒3 L/mol•min k[A]0 t1/2 [A]0 20.0 min(10.0 mol/L)
1 1 1 5.00 10−3 L × 150. min + = 0.850 M −1 , = kt + = 10.0 mol/L [A] [A]0 mol min
Solving: [A] = 1.18 mol/L 69.
The consecutive half-life values of 24 hours, then 12 hours, show a direct relationship with concentration; as the concentration decreases, the half-life decreases. Assuming the drug reaction is either zero, first, or second order, only a zero-order reaction shows this direct relationship between half-life and concentration. Therefore, assume the reaction is zero order in the drug. t1/2 =
70.
[A]0 [A]0 2.0 10 −3 mol/L , k = = = 4.2 10 −5 mol/L•h 2k 2t 1/2 2(24 h)
Note that the first two half-lives double as the reaction proceed. This is characteristic of a zeroorder reaction. So, assume the reaction is zero order.
CHAPTER 12 t1/2 =
CHEMICAL KINETICS
555
[A]0 [A]0 4.00 mol/L , k = = = 4.2 10−2 mol/L•min 2k 2t1/2 2(48 min)
[A] = −kt + [A]0 = ‒4.2 × 10‒2 mol/L•min(81 min) + 4.00 mol/L = 0.6 mol/L 71.
Because [B]0 >> [A]0, the concentration of B is essentially constant in this experiment. We have a pseudo-first-order reaction in A: Rate = k[A][B]2 = k[A], where k = k B0
2
[A] 2.3 10−4 M ‒1 = −kt, ln ln = −k × 3.0 min, k = 0.26 min −4 5.0 [ A 10 ] M 0
k = k B0 , 0.26 min ‒1 = k(5.0)2, k = 0.010 L2/mol2•min 2
72.
Because [B]0 >> [A]0, the B concentration is essentially constant during this experiment, so rate = k[A]2 where k = k[B]. For this experiment, the reaction is a pseudo-second-order reaction in A. 1 1 = k × 8.0 min + , k = 7.1 × 104 L/mol•min −6 5.0 10 −5 mol/L 1.7 10 mol/L k = k[B], 7.1 × 104 L/mol•min = k(0.40), k = 1.8 × 105 L2/mol2•min
73.
Because [V]0 >> [AV]0, the concentration of V is essentially constant in this experiment. We have a pseudo-first-order reaction in AV: Rate = k[AV][V] = k[AV], where k = k[V]0 The slope of the ln[AV] versus time plot is equal to −k. k = −slope = 0.32 s−1; k =
74.
k 0.32 s −1 = = 1.6 L/mol•s [V]0 0.20 mol/L
Because [B]0 >> [A]0, the B concentration is essentially constant during this experiment, so rate = k[A] where k = k[B]2. For this experiment, the reaction is a pseudo-first-order reaction in A. a.
[A] 3.8 10 −3 M −1 = −kt, ln ln 1.0 10 − 2 M = −k × 8.0 s, k = 0.12 s [ A ] 0
For the reaction: k = k[B]2, k = 0.12 s−1/(3.0 mol/L)2 = 1.3 × 10 −2 L2 mol−2 s−1 b. t1/2 = c.
ln 2 0.693 = = 5.8 s k' 0.12 s −1
[A] [ A] = −0.12 s −1 × 13.0 s, = e−0.12(13.0) = 0.21 ln −2 −2 1 . 0 10 1 . 0 10 M [A] = 2.1 × 10 −3 M
556
CHAPTER 12
CHEMICAL KINETICS
d. [A] reacted = 0.010 M − 0.0021 M = 0.008 M [C] reacted = 0.008 M ×
2 mol C = 0.016 M 0.02 M 1 mol A
[C]remaining = 2.0 M − 0.02 M = 2.0 M; as expected, the concentration of C basically remains constant during this experiment since [C]0 >> [A]0.
Reaction Mechanisms 75.
76.
For elementary reactions, the rate law can be written using the coefficients in the balanced equation to determine orders. a. Rate = k[CH3NC]
b. Rate = k[O3][NO]
c. Rate = k[O3]
d. Rate = k[O3][O]
From experiment (Exercise 49), we know the rate law is Rate = k[H2O2]. A mechanism consists of a series of elementary reactions where the rate law for each step can be determined using the coefficients in the balanced equation for each respective step. For a plausible mechanism, the rate law derived from a mechanism must agree with the rate law determined from experiment. To derive the rate law from the mechanism, the rate of the reaction is assumed to equal the rate of the slowest step in the mechanism. This mechanism will agree with the experimentally determined rate law only if step 1 is the slow step (called the rate-determining step). If step 1 is slow, then Rate = k[H2O]2 which agrees with experiment. Another important property of a mechanism is that the sum of all steps must give the overall balanced equation. Summing all steps gives: H2O2 → 2 OH H2O2 + OH → H2O + HO2 HO2 + OH → H2O + O2 2 H2O2 → 2 H2O + O2
77.
A mechanism consists of a series of elementary reactions in which the rate law for each step can be determined using the coefficients in the balanced equations. For a plausible mechanism, the rate law derived from a mechanism must agree with the rate law determined from experiment. To derive the rate law from the mechanism, the rate of the reaction is assumed to equal the rate of the slowest step in the mechanism. Because step 1 is the rate-determining step, the rate law for this mechanism is Rate = k[C4H9Br]. To get the overall reaction, we sum all the individual steps of the mechanism. Summing all steps gives: C4H9Br → C4H9+ + Br− + C4H9 + H2O → C4H9OH2+ C4H9OH2+ + H2O → C4H9OH + H3O+ C4H9Br + 2 H2O → C4H9OH + Br− + H3O+
CHAPTER 12
CHEMICAL KINETICS
557
Intermediates in a mechanism are species that are neither reactants nor products but that are formed and consumed during the reaction sequence. The intermediates for this mechanism are C4H9+ and C4H9OH2+. 78.
Because the rate of the slowest elementary step equals the rate of a reaction: Rate = rate of step 1 = k[NO2]2 The sum of all steps in a plausible mechanism must give the overall balanced reaction. Summing all steps gives: NO2 + NO2 → NO3 + NO NO3 + CO → NO2 + CO2 NO2 + CO → NO + CO2
79.
The rate law is Rate = k[NO]2[Cl2]. If we assume the first step is rate-determining, we expect the rate law to be Rate = k1[NO][Cl2]. This isn't correct. If we assume the second step is ratedetermining, then Rate = k2[NOCl2][NO]. To see if this agrees with experiment, we must substitute for the intermediate NOCl2 concentration. Assuming a fast-equilibrium first step (rate reverse = rate forward): k-1[NOCl2] = k1[NO][Cl2], [NOCl2] = Rate =
k1 [NO][Cl2]; substituting into the rate equation: k −1
k 2 k1 k k [NO]2[Cl2] = k[NO]2[Cl2] where k = 2 1 k −1 k −1
This is a possible mechanism with the second step the rate-determining step because the derived rate law agrees with the experimentally determined rate law. 80.
Let's determine the rate law for each mechanism. If the rate law derived from the mechanism is the same as the experimental rate law, then the mechanism is possible (assuming the sum of all the steps in the mechanism gives the overall balanced equation). When deriving rate laws from a mechanism, we must substitute for all intermediate concentrations. a. Rate = k1[NO][O2]; not possible b. Rate = k2[NO3][NO] and k1[NO][O2] = k−1[NO3] or [NO3] = Rate =
k 2 k1 [NO]2[O2]; k −1
k1 [NO][O2] k −1
possible
c. Rate = k1[NO]2; not possible d. Rate = k2[N2O2] and [N2O2] =
k1 k k [NO]2, Rate = 2 1 [NO]2 ; not possible k −1 k −1
Only the mechanism in b is consistent with the experimentally determined rate law, so only mechanism b is a possible mechanism for this reaction.
558 81.
CHAPTER 12
CHEMICAL KINETICS
In the first experiment where [A]0 >> [B0], a plot of ln[B] vs time is linear with a negative slope. So, the reaction is first order in B. In the second experiment where [B]0 >> [A]0, a plot of 1/[A] vs time is linear with a positive slope. So, the reaction is second order in A. The rate law from experiment is Rate = [A]2[B]. Mechanism I: Rate = k2[E][A]; from the fast equilibrium first step, k1[A][B] = k−1[E], [E] = Substituting: Rate = k2[E][A] = k2
k1 [A][B] k −1
k1 [A][B][A] = k[A]2[B]; this is a possible mechanism. k −1
Mechanism II: Rate = k2[E][B]; from the fast equilibrium first step, k1[A][B] = k−1[E], [E] = Substituting: Rate = k2[E][B] = k2
k1 [A][B] k −1
k1 [A][B][B] = k[A] [B]2; not a possible mechanism. k −1
Mechanism III: Rate = k1[A]2; not a possible mechanism Only mechanism I gives a derived rate law that agrees with experiment. So, only mechanism I is a possible mechanism for this reaction. 82.
From experiments 1 and 2, as [I‒] is doubled and the other concentrations are constant, the rate increases by a factor of ~2. Assume the reaction is first order in I ‒. From experiments 2 and 3, as [H2O2] is doubled, the rate increases by a factor of ~2. Assume the reaction is first order in H2O2. From experiments 3 and 4, as [H2O2] and [H+] are both doubled, the rate increases by a factor of 2. As determined before, doubling just the [H 2O2] will double the rate. Therefore, the reaction is zero order in [H+]. The rate law is Rate = k[I‒][H2O2]. Mechanism I is consistent with this rate law assuming the first step rate determining. With the first step rate determining, the derived rate is also k[I‒][H2O2]. The derived rate law from mechanism II does not agree with experiment, so mechanism II is not plausible.
Temperature Dependence of Rate Constants and the Collision Model 83.
In the following plot, R = reactants, P = products, E a = activation energy, and RC = reaction coordinate, which is the same as reaction progress. Note for this reaction that ΔE is positive because the products are at a higher energy than the reactants.
CHAPTER 12
CHEMICAL KINETICS
E
559
Ea
P E
R
RC
84.
When ΔE is positive, the products are at a higher energy relative to reactants, and when ΔE is negative, the products are at a lower energy relative to reactants. RC = reaction coordinate, which is the same as reaction progress.
85. 125 kJ/mol
E
R Ea, reverse
216 kJ/mol
P RC
The activation energy for the reverse reaction is: Ea, reverse = 216 kJ/mol + 125 kJ/mol = 341 kJ/mol
560
CHAPTER 12
CHEMICAL KINETICS
86.
The activation energy for the reverse reaction is E R in the diagram. ER = 167 − 28 = 139 kJ/mol 87.
The Arrhenius equation is k = A exp(−Ea/RT) or, in logarithmic form, ln k = −Ea/RT + ln A. Hence a graph of ln k versus 1/T should yield a straight line with a slope equal to −Ea/R since the logarithmic form of the Arrhenius equation is in the form of a straight-line equation, y = mx + b. Note: We carried extra significant figures in the following ln k values in order to reduce round off error.
Slope =
T (K)
1/T (K−1)
k (s−1)
ln k
338 318 298
2.96 × 10−3 3.14 × 10−3 3.36 × 10−3
4.9 × 10−3 5.0 × 10−4 3.5 × 10−5
−5.32 −7.60 −10.26
− 10.76 − (−5.85) = −1.2 × 104 K = −Ea/R 3.40 10 −3 − 3.00 10 −3
Ea = −slope × R = 1.2 × 104 K ×
8.3145 J , Ea = 1.0 × 105 J/mol = 1.0 × 102 kJ/mol K mol
CHAPTER 12 88.
CHEMICAL KINETICS
561
From the Arrhenius equation in logarithmic form (ln k = −Ea/RT + ln A), a graph of ln k versus 1/T should yield a straight line with a slope equal to −Ea/R and a y intercept equal to ln A. a. Slope = −Ea/R, Ea = 1.10 × 104 K ×
8.3145 J = 9.15 × 104 J/mol = 91.5 kJ/mol K mol
b. The units for A are the same as the units for k ( s −1 ). y intercept = ln A, A = e33.5 = 3.54 × 1014 s −1 c. ln k = −Ea/RT + ln A or k = A exp(−Ea/RT)
− 9.15 10 4 J/mol = 3.24 × 10 −2 s −1 k = 3.54 × 1014 s −1 × exp 8.3145 J/K • mol 298 K 89.
k = A exp(−Ea/RT) or ln k =
− Ea + ln A (the Arrhenius equation) RT
k E 1 1 (Assuming A is temperature independent.) − For two conditions: ln 2 = a R T1 T2 k1
Let k1 = 3.52 × 10 −7 L/mol•s, T1 = 555 K; k2 = ?, T2 = 645 K; Ea = 186 × 103 J/mol 1.86 10 5 J/mol 1 k2 1 = = 5.6 ln − 3.52 10 −7 8.3145 J/K • mol 555 K 645 K
k2 3.52 10
90.
−7
= e5.6 = 270, k2 = 270(3.52 × 10 −7 ) = 9.5 × 10 −5 L/mol•s
k E 1 1 (Assuming A is temperature independent.) − For two conditions: ln 2 = a R T1 T2 k1 100. s −1 1 Ea 1 ln = − −1 400. K 10.0 s 8.3145 J/K • mol 300. K
2.30 =
91.
Ea (8.3 × 10 −4 ), Ea = 2.3 × 104 J/mol = 23 kJ/mol 8.3145
k E 1 1 k2 ; ln 2 = a − = 7.00, T1 = 295 K, Ea = 54.0 × 103 J/mol k R T T k 2 1 1 1
ln(7.00) =
1 1 1 5.4 10 4 J/mol 1 , − − = 3.00 × 10 −4 T2 295 K T2 8.3145 J/K • mol 295 K
1 = 3.09 × 10 −3 , T2 = 324 K = 51°C T2
562
CHAPTER 12
92.
k E 1 1 ; because the rate doubles, k2 = 2k1. ln 2 = a − R T1 T2 k1
ln(2.00) =
CHEMICAL KINETICS
1 Ea 1 , Ea = 5.3 × 104 J/mol = 53 kJ/mol − 8.3145 J/K • mol 298 K 308 K
93.
H3O+(aq) + OH−(aq) → 2 H2O(l) should have the faster rate. H3O+ and OH− will be electrostatically attracted to each other; Ce4+ and Hg22+ will repel each other. The activation energy for the Ce4+ and Hg22+ reaction should be a larger quantity, making it the slower reaction.
94.
Carbon cannot form the fifth bond necessary for the transition state because of the small atomic size of carbon and because carbon doesn’t have low-energy d orbitals available to expand the octet.
Catalysts 95.
C2H4 + H+ → C2H5+ C2H5+ + H2O → C2H5OH2+ C2H5OH2+ → C2H5OH + H+ C2H4 + H2O → C2H5OH C2H4 and H2O are reactants, and C2H5OH is the product. The other species do not appear in the overall balanced equation, so they are either catalysts or intermediates. Catalysts will always appear first on the reactant side in one of the steps; intermediates always appear first on the product side of one of the steps. H+ is a catalyst; it is a reactant in step 1 but produced in step 3. It does not appear in the overall balanced reaction. C2H5+ and C2H5OH2+ are intermediates in the mechanism. These ions appear first on the product side of one of the steps but are reactants in a subsequent step.
96.
S(s) + O2(g) → SO2(g) SO2(g) + NO2(g) → SO3(g) + NO(g) NO(g) + ½ O2(g) → NO2(g) H2O(l) + SO3(g) → H2SO4(aq) S(s) + 3/2 O2(g) + H2O(l) → H2SO4(aq) S, O2, and H2O are reactants, and H2SO4 is the product. The other species do not appear in the overall balanced equation, so they are either catalysts or intermediates. Catalysts will always appear first on the reactant side in one of the steps; intermediates always appear first on the product side of one of the steps. NO2 is a catalyst; it is a reactant in step 2 but produced in step 3. It does not appear in the overall balanced reaction. NO and SO3 are intermediates in the mechanism. These ions appear first on the product side of one of the steps but are reactants in a subsequent step.
97.
a. NO is the catalyst. NO is present in the first step of the mechanism on the reactant side, but it is not a reactant. NO is regenerated in the second step and does not appear in overall balanced equation.
CHAPTER 12
CHEMICAL KINETICS
563
b. NO2 is an intermediate. Intermediates also never appear in the overall balanced equation. In a mechanism, intermediates always appear first on the product side, whereas catalysts always appear first on the reactant side. c. k = A exp(−Ea/RT);
k cat A exp[−E a (cat)/RT] E (un ) − E a (cat) = = exp a k un A exp[−E (un)/RT] RT
2100 J/mol k cat = e0.85 = 2.3 = exp 8.3145 J/K • mol 298 K k un
The catalyzed reaction is approximately 2.3 times faster than the uncatalyzed reaction at 25°C. 98.
The mechanism for the chlorine catalyzed destruction of ozone is: O3 + Cl → O2 + ClO ClO + O → O2 + Cl
(slow) (fast)
O3 + O → 2 O2 Because the chlorine atom-catalyzed reaction has a lower activation energy, the Cl-catalyzed rate is faster. Hence Cl is a more effective catalyst. Using the activation energy, we can estimate the efficiency that Cl atoms destroy ozone compared to NO molecules (see Exercise 79c). At 25°C:
(−2100 + 11,900) J/mol 3.96 k Cl E ( NO ) − E a (Cl ) = exp + a = exp = e = 52 k NO RT RT (8.3145 298) J/mol
At 25°C, the Cl-catalyzed reaction is roughly 52 times faster than the NO-catalyzed reaction, assuming the frequency factor A is the same for each reaction. 99.
The reaction at the surface of the catalyst is assumed to follow the steps:
D
D
H
H
H
C
C
DCH 2 H
D
D
D
CH 2
D
D
CH 2DCH 2D(g)
metal surface
Thus CH2D‒CH2D should be the product. If the mechanism is possible, then the reaction must be: C2H4 + D2 → CH2DCH2D If we got this product, then we could conclude that this is a possible mechanism. If we got some other product, for example, CH3CHD2, then we would conclude that the mechanism is wrong. Even though this mechanism correctly predicts the products of the reaction, we cannot say conclusively that this is the correct mechanism; we might be able to conceive of other mechanisms that would give the same products as our proposed one.
564 100.
CHAPTER 12
CHEMICAL KINETICS
a. W because it has a lower activation energy than the Os catalyst. b. kw = Aw exp[−Ea(W)/RT]; kuncat = Auncat exp[−Ea(uncat)/RT]; assume Aw = Auncat.
kw E (uncat) − E a ( W) = exp + a k uncat RT RT − 163,000 J/mol + 335,000 J/mol kw 30 = exp = 1.41 × 10 k uncat 8.3145 J/K • mol 298 K
The W-catalyzed reaction is approximately 1030 times faster than the uncatalyzed reaction. c. Because [H2] is in the denominator of the rate law, the presence of H2 decreases the rate of the reaction. For the decomposition to occur, NH 3 molecules must be adsorbed on the surface of the catalyst. If H2 is also adsorbed on the catalyst surface, then there are fewer sites for NH3 molecules to be adsorbed, and the rate decreases. 101.
The rate depends on the number of reactant molecules adsorbed on the surface of the catalyst. This quantity is proportional to the concentration of reactant. However, when all the catalyst surface sites are occupied, the rate becomes independent of the concentration of reactant.
102.
At high [S], the enzyme is completely saturated with substrate. Once the enzyme is com-pletely saturated, the rate of decomposition of ES can no longer increase, and the overall rate remains constant.
103.
Assuming the catalyzed and uncatalyzed reactions have the same form and orders, and because concentrations are assumed equal, the rates will be equal when the k values are equal. k = A exp(−Ea/RT); kcat = kun when Ea,cat/RTcat = Ea,un/RTun. 4.20 10 4 J/mol 7.00 10 4 J/mol = , Tun = 488 K = 215C 8.3145 J/K • mol 293 K 8.3145 J/K • mol Tun
104.
Rate =
− [ A ] = k[A ] x t
Assuming the catalyzed and uncatalyzed reaction have the same form and orders, and because concentrations are assumed equal, rate 1/t, where t = time. Rate cat rate cat k Δt 2400 yr = un = = cat and Rate un rate un k un Δt cat Δt cat A exp[ − E a (cat)/RT] Rate cat k − E a (cat) + E a (un) = cat = = exp Rate un k un A exp[ − E a (un)/RT] RT
− 5.90 10 4 J/mol + 1.84 10 5 J/mol k cat = 7.62 × 1010 = exp k un 8.3145 J/K • mol 600. K Δt un rate cat k = = cat , Δt cat rate un k un
2400 yr = 7.62 × 1010, tcat = 3.15 × 10 −8 yr 1 s Δt cat
CHAPTER 12
CHEMICAL KINETICS
565
ChemWork Problems 105.
Box a has 8 NO2 molecules. Box b has 4 NO2 molecules, and box c has 2 NO2 molecules. Box b represents what is present after the first half-life of the reaction, and box c represents what is present after the second half-life. a. For first order kinetics, t1/2 = 0.693/k; the half-life for a first order reaction is concentration independent. Therefore, the time for box c, the time it takes to go through two half-lives, will be 10 + 10 = 20 minutes. b. For second order kinetics, t1/2 = 1/k[A]0; the half-life for a second order reaction is inversely proportional to the initial concentration. So if the first half-life is 10 minutes, the second half-life will be 20 minutes. For a second order reaction, the time for box c will be 10 + 20 = 30 minutes. c. For zero order kinetics, t1/2 = [A]0/2k; the half-life for a zero-order reaction is directly related to the initial concentration. So if this reaction was zero order, then the second halflife would decrease from 10 min to 5 min. The time for box c will be 10 + 5 = 15 minutes if the reaction is zero order.
106.
Rate = k[H 2SeO 3 ] x [H + ] y [I − ] z ; comparing the first and second experiments: 3.33 10 −7
k ( 2.0 10 −4 ) x (2.0 10 −2 ) y (2.0 10 −2 ) z
1.66 10
k (1.0 10 − 4 ) x (2.0 10 − 2 ) y (2.0 10 − 2 ) z
= −7
, 2.01 = 2.0x, x = 1
Comparing the first and fourth experiments: 6.66 10 −7 1.66 10
= −7
k (1.0 10 −4 )(4.0 10 −2 ) y ( 2.0 10 −2 ) z −4
−2 y
−2 z
k (1.0 10 )(2.0 10 ) (2.0 10 )
, 4.01 = 2.0y, y = 2
Comparing the first and sixth experiments: 13.2 10 −7
k (1.0 10 −4 )(2.0 10 −2 ) 2 (4.0 10 −2 ) z
1.66 10
k (1.0 10 − 4 )(2.0 10 − 2 ) 2 (2.0 10 − 2 ) z
= −7
7.95 = 2.0 z , log(7.95) = z log(2.0), z =
log (7.95) = 2.99 3 log( 2.0)
Rate = k[H2SeO3][H+]2[I−]3 Experiment 1: 2
1.0 10 −4 mol 2.0 10 −2 mol 2.0 10 −2 mol 1.66 10 −7 mol = k Ls L L L
k = 5.19 × 105 L5/mol5•s = 5.2 × 105 L5/mol5•s = kmean 107.
The integrated rate law for each reaction is:
3
566
CHAPTER 12
CHEMICAL KINETICS
ln[A] = −4.50 × 10−4 s−1(t) + ln[A]0 and ln[B] = −3.70 × 10−3 s−1(t) + ln[B]0 Subtracting the second equation from the first equation (ln[A] 0 = ln[B]0): [A] = 3.25 × 10−3(t) ln[A] − ln[B] = −4.50 × 10−4(t) + 3.70 × 10−3(t), ln [ B ]
When [A] = 4.00 [B], ln(4.00) = 3.25 × 10 −3(t), t = 427 s. 108.
The pressure of a gas is directly proportional to concentration. Therefore, we can use the pressure data to solve the problem because Rate = −Δ[SO2Cl2]/Δt − ΔPSO 2Cl 2 /Δt. Assuming a first order equation, the data and plot follow. Time (hour)
0.00
1.00
2.00
4.00
8.00
16.00
PSO2Cl 2 (atm)
4.93
4.26
3.52
2.53
1.30
0.34
ln PSO2Cl 2
1.595
1.449
1.258
0.928
0.262
−1.08
Because the ln PSO2Cl 2 versus time plot is linear, the reaction is first order in SO 2Cl2. a. Slope of ln(P) versus t plot: −0.168 hour −1 = −k, k = 0.168 hour −1 = 4.67 × 10 −5 s −1 Because concentration units don’t appear in first-order rate constants, this value of k determined from the pressure data will be the same as if concentration data in molarity units were used. b. t1/2 =
ln 2 0.6931 0.6931 = 4.13 hour = = k k 0.168 h −1
CHAPTER 12 c.
CHEMICAL KINETICS
567
P PSO 2Cl 2 = −kt = −0.168 h−1(20.0 h) = −3.36, SO 2Cl 2 = e −3.36 = 3.47 × 10 −2 ln P 0 P0 Fraction remaining = 0.0347 = 3.47%
109.
From 338 K data, a plot of ln[N2O5] versus t is linear, and the slope = −4.86 × 10 −3 (plot not included). This tells us the reaction is first order in N 2O5 with k = 4.86 × 10 −3 at 338 K. From 318 K data, the slope of ln[N2O5] versus t plot is equal to −4.98 × 10 −4 , so k = 4.98 × 10 −4 at 318 K. We now have two values of k at two temperatures, so we can solve for E a.
4.86 10 −3 k E 1 1 Ea 1 1 = , ln ln 2 = a − − 4.98 10 −4 8.3145 J/K • mol 318 K R T1 T2 338 K k1 Ea = 1.0 × 105 J/mol = 1.0 × 102 kJ/mol 110.
The Arrhenius equation is k = A exp(−Ea/RT) or, in logarithmic form, ln k = −Ea/RT + ln A. Hence a graph of ln k versus 1/T should yield a straight line with a slope equal to -Ea/R since the logarithmic form of the Arrhenius equation is in the form of a straight-line equation, y = mx + b. Note: We carried one extra significant figure in the following ln k values in order to reduce round-off error. T (K)
1/T (K−1)
195 230. 260. 298 369
5.13 × 10−3 4.35 × 10−3 3.85 × 10−3 3.36 × 10−3 2.71 × 10−3
k (L/mol•s) 1.08 × 109 2.95 × 109 5.42 × 109 12.0 × 109 35.5 × 109
ln k 20.80 21.81 22.41 23.21 24.29
Using a couple of points on the plot: slope =
− Ea 20.95 − 23.65 − 2.70 = = −1.35 × 103 K = −3 −3 −3 R 5.00 10 − 3.00 10 2.00 10
Ea = 1.35 × 103 K × 8.3145 J/K•mol = 1.12 × 104 J/mol = 11.2 kJ/mol
568 111.
CHAPTER 12
CHEMICAL KINETICS
Because the ln[A] versus time plot is linear, the reaction is first order in A. The slope of the ln[A] versus time plot equals −k. Therefore, the integrated rate law and the rate constant value are: ln[A] = −kt + ln[A]0; k = 7.35 × 10−3 s−1 From the problem, 77.1% A remains: [A] = 0.771(1.00 × 10 −2 mol/L) = 7.71 × 10−3 mol/L
7.71 10 −3 M [A] = −(7.35 × 10−3 s−1)t = − kt, ln ln −2 [A]0 1.00 10 M ln(0.771) t= = 35.4 s − 7.35 10 −3 s −1 112.
Because the 1/[A] versus time plot was linear, the reaction is second order in A. The slope of the 1/[A] versus time plot equals the rate constant k. Therefore, the rate law, the integrated rate law, and the rate constant value are: Rate = k[A]2;
1 1 = kt + ; k = 6.90 × 10-2 L/ mol•s [ A ] [A ] 0
The half-life expression for a second-order reaction is: t1/2 = For this reaction: t1/2 =
6.90 10
−2
1 k[A]0
1 = 145 s L/mol • s 0.100 mol/L
Note: We could have used the integrated rate law to solve for t1/2, where [A] = (0.100/2) M. Because the half-life for a second-order reaction depends on concentration, we will use the integrated rate law to solve. For the second half-life, [A] = 0.100/4 = 0.0250 mol/L and [A]0 = 0.0500 mol/L. 1 6.90 10 −2 L 1 1 1 = kt + , , t = 290. s = 2nd half-life = t + [A]0 0.0250 M mol s 0.0500 M [A ]
113.
a.
t (s)
[C4H6] (M)
ln[C4H6]
1/[C4H6] (M −1)
0 1000. 2000. 3000.
0.01000 0.00629 0.00459 0.00361
−4.6052 −5.069 −5.384 −5.624
1.000 × 102 1.59 × 102 2.18 × 102 2.77 × 102
The plot of 1/[C4H6] versus t is linear, thus the reaction is second order in butadiene. From the plot (not included), the integrated rate law is: 1 = (5.90 × 10−2 L/mol•s)t + 100.0 M −1 [C 4 H 6 ]
b. When dimerization is 1.0% complete, 99.0% of C4H6 is left.
CHAPTER 12
CHEMICAL KINETICS
[C4H6] = 0.990(0.01000) = 0.00990 M;
569 1 = (5.90 × 10−2)t + 100.0 0.00990
101 = (5.90 × 10−2)t + 100.0, t = 17 s 20 s c. 10.0% complete, [C4H6] = 0.00900 M;
1 = (5.90 × 10−2)t + 100.0, 0.00900
t = 188 s 190 s d.
1 1 = kt + ; [C4H6]0 = 0.0200 M; at t = t1/2, [C4H6] = 0.0100 M. [C 4 H 6 ] [ C 4 H 6 ]0 1 1 = (5.90 × 10−2)t1/2 + , t1/2 = 847 s = 850 s 0.0100 0.0200 1 1 Or: t1/2 = = = 847 s −2 k[A]0 (5.90 10 L/mol • s)(2.00 10 − 2 M)
e. From Exercise 12.55, k = 1.4 × 10−2 L/mol•s at 500. K. From this problem, k = 5.90 × 10−2 L/mol•s at 620. K. k E 1 5.90 10−2 1 Ea 1 1 ln 2 = a − , ln = − −2 R T1 T2 1.4 10 8.3145 J/K • mol 500. K 620. K k1
12 = Ea(3.9 × 10−4), Ea = 3.1 × 104 J/mol = 31 kJ/mol 114.
0.200(0.1000 M) = 0.0200 mol/L of A reacted; [A] = 0.1000 – 0.0200 = 0.0800 M 1 1 1 1 = k 40.0 min + = kt + , , k = 0.063 L/ mol•s [A]0 0.0800 M 0.1000 M [A ]
115.
The integrated rate law for a zero-order reaction is [A] = −kt + [A]0. [A] = −kt + [A]0,
116.
2.50 10 −2 mol 0.021 mol 0.0800 mol t + = − , t = 2.36 s L Ls L
When the pressure decreases from 536 torr to 268 torr, this is a factor of one-half. We need to determine the half-life of this reaction at 295oC, which requires us to calculate k at 295oC. k E 1 1 (Assuming A is temperature independent.) − For two conditions: ln 2 = a R T1 T2 k1
1.7 10 −2 s −1 1 Ea 1 = ln − − 4 −1 720. K 7.2 10 s 8.3145 J/K • mol 660. K 3.2 =
Ea (1.3 × 10 −4 ), Ea = 2.0 × 105 J/mol 8.3145
Let k1 = 1.7 × 10 −2 s−1, T1 = 720. K; k2 = ?, T2 = 568 K (295oC); Ea = 2.0 × 105 J/mol
570
CHAPTER 12
CHEMICAL KINETICS
5 k2 1 = 2.0 10 J/mol 1 = −8.9 ln − − 2 1.7 10 8.3145 J/K • mol 720. K 568 K
k2 1.7 10
−2
= e−8.9, k2 = 2.3 × 10 −6 s−1
For a first order reaction, t1/2 = 117.
ln 2 0.693 ; t1/2 = = 3.0 105 s = 3.5 days k 2.3 10 − 6 s −1
To determine the rate of reaction, we need to calculate the value of the rate constant k. The activation energy data can be manipulated to determine k.
− 26.2 10 3 J/mol = 3.29 × 10−5 s−1 k = Ae − E a /RT = 0.850 s−1 × exp 8.3145 J/K • mol 310.2 K Rate = k[acetycholine receptor-toxin complex] 0.200 mol Rate = 3.29 10 −5 s −1 = 6.58 × 10−6 mol/L•s L
118.
Rate = k[DNA]x[CH3I]y; comparing the second and third experiments: 1.28 10 −3
k (0.200 ) x (0.200 ) y
6.40 10
k (0.100 ) x (0.200 ) y
= −4
, 2.00 = 2.00 x , x = 1
Comparing the first and second experiments: 6.40 10 −4
k (0.100 )(0.200 ) y
3.20 10
k (0.100 )(0.100 ) y
= −4
, 2.00 = 2.00 y , y = 1
The rate law is Rate = k[DNA][CH3I]. Mechanism I is possible because the derived rate law from the mechanism (Rate = k[DNA][CH3I]) agrees with the experimentally determined rate law. The derived rate law for Mechanism II will equal the rate of the slowest step. This is step 1 in the mechanism giving a derived rate law that is Rate = k[CH3I]. Because this rate law does not agree with experiment, Mechanism II would not be a possible mechanism for the reaction. 119.
k E 1 1 (Assuming A is temperature independent.) − For two conditions: ln 2 = a R T1 T2 k1
0.891 L/mol • s ln −4 2.45 10 L/mol • s 8.20 =
1 Ea 1 = − 8.3145 J/K • mol 575 K 781 K
Ea (4.6 × 10 −4 ), Ea = 1.5 × 105 J/mol = 150 kJ/mol 8.3145
Let k1 = 0.891 L/mol•s, T1 = 781 K; k2 = ?, T2 = 648 K; Ea = 1.5 × 105 J/mol
CHAPTER 12
CHEMICAL KINETICS
571
5 k2 = 1.5 10 J/mol 1 − 1 = −4.7 ln 0.891 8.3145 J/K • mol 781 K 648 K
k2 = e−4.7 = 9.1 × 10−3, k2 = 8.1 × 10 −3 L/mol•s 0.891
120.
The Arrhenius equation is k = A exp(−Ea/RT) or, in logarithmic form, ln k = −Ea/RT + ln A. Hence a graph of ln k versus 1/T should yield a straight line with a slope equal to -Ea/R since the logarithmic form of the Arrhenius equation is in the form of a straight-line equation, y = mx + b. Note: We carried one extra significant figure in the following ln k values in order to reduce round-off error. T (K)
1/T (K−1)
k (min −1)
ln k
298.2 293.5 290.5
3.353 × 10−3 3.407 × 10−3 3.442 × 10−3
178 126 100.
5.182 4.836 4.605
The plot of ln k vs. 1/T gives a straight line. The equation for the straight line is: ln k = −6.48 × 103(1/T) + 26.9; slope = −6.48 × 10 3 K = −Ea/R Ea = 6.48 × 103 K × 8.3145 J/mol•K = 5.39 × 104 J/mol = 53.9 kJ/mol ln k = −6.48 × 103 (1/280.7) + 26.9 = 3.8, k = e3.8 = 45 min−1 The chirping rate will be about 45 chirps per minute at 7.5oC. 121.
a. If the interval between flashes is 16.3 s, then the rate is: 1 flash/16.3 s = 6.13 × 10−2 s−1 = k Interval
k
16.3 s 13.0 s
6.13 × 10−2 s−1 7.69 × 10−2 s−1
T 21.0°C (294.2 K) 27.8°C (301.0 K)
k E 1 1 ; solving using above data: Ea = 2.5 × 104 J/mol = 25 kJ/mol ln 2 = a − k R T T 2 1 1
b.
2.5 10 4 J/mol k 1 1 = = 0.30 ln − −2 8.3145 J/K • mol 294.2 K 303.2 K 6.13 10 k = e0.30 × (6.13 × 10−2) = 8.3 × 10−2 s−1; interval = 1/k = 12 seconds.
572
CHAPTER 12 c.
T
Interval
54-2(Intervals)
21.0°C 27.8°C 30.0°C
16.3 s 13.0 s 12 s
21°C 28°C 30.°C
CHEMICAL KINETICS
This rule of thumb gives excellent agreement to two significant figures. 122.
k = A exp(−Ea/RT);
A cat exp (−E a , cat / RT) − E a , cat + E a , uncat k cat = = exp k uncat A uncat exp (−E a , uncat / RT) RT
− E a, cat + 5.00 10 4 J/mol k cat 2.50 × 10 = = exp 8.3145 J/K • mol 310. K k uncat 3
ln(2.50 × 103) × 2.58 × 103 J/mol = −Ea, cat + 5.00 × 104 J/mol
E a , cat = 5.00 × 104 J/mol − 2.02 × 104 J/mol = 2.98 × 104 J/mol = 29.8 kJ/mol 123.
Rate = k[A]x[B]y; using data from experiment 3 and experiment 4:
3.46 10 −2 4.32 10
= −3
k (0.24) x (0.090 ) y = 8.0x, 8.00 = 8.0x, x = 1 x y k (0.030 ) (0.090 )
From experiments 1 and 4:
3.46 10 −2 3.46 10
= −2
k (0.24)(0.090 ) y , 1.00 = 2.0 × (1/2)y, 1/2 = (1/2)y, y = 1 k (0.12)(0.18) y
The reaction is first order with respect to both A and B; rate = k[A][B] From the first experiment:
3.46 10 −2 mol 0.12 mol 0.18 mol = k , Ls L L
k = 1.6 L/mol•s; all four experiments give the same value of k to two significant figures. 124.
Rate = k[A]x[B]y; comparing the first and second experiments:
k (0.50) x (1.0) y 11.8 = , 3.9 = 2.5 x , taking the log of both sides gives : x y 3.0 k (0.20) (1.0) log(3.9) = x log( 2.5), x =
log(3.9) = 1.5 log(2.5)
Comparing the second and third experiments: k (2.0)1.5 (2.0) y 189.5 = , 16.1 = (4)1.5 (2.0) y , y = 1 11.8 k (0.50)1.5 (1.0) y
The rate law is Rate = k[A]1.5[B]. Rate = k[A]1.5[B], 11.8 mol/L•s = k(0.50 M)1.5(1.0 M), k = kmean = 33 L1.5/mol1.5•s
CHAPTER 12 125.
CHEMICAL KINETICS
573
a. Because [A]0 << [B]0 or [C]0, the B and C concentrations remain constant at 1.00 M for this experiment. Thus Rate = k[A]2[B][C] = k[A]2, where k = k[B][C]. For this pseudo-second-order reaction: 1 1 1 1 , = kt + = k(3.00 min) + −5 [A ]0 3.26 10 M [A ] 1.00 10 − 4 M
k = 6890 L/mol•min = 115 L/mol•s k = k[B][C], k =
115 L/mol • s k , k = = 115 L3/mol3•s [B] [C] (1.00 M )(1.00 M )
b. For this pseudo-second-order reaction: Rate = k[A]2, t1/2 = c.
1 1 = = 87.0 s k [A] 0 115 L/mol • s(1.00 10 − 4 mol/L)
1 1 1 = kt + = 115 L/mol•s × 600. s + = 7.90 × 104 L/mol [ A ]0 [A ] 1.00 10 −4 mol/L
[A] = 1/7.90 × 104 L/mol = 1.27 × 10–5 mol/L From the stoichiometry in the balanced reaction, 1 mol of B reacts with every 3 mol of A. Amount A reacted = 1.00 × 10–4 M – 1.27 × 10–5 M = 8.7 × 10–5 M Amount B reacted = 8.7 × 10–5 mol/L ×
1 mol B = 2.9 × 10–5 M 3 mol A
[B] = 1.00 M − 2.9 × 10–5 M = 1.00 M As we mentioned in part a, the concentration of B (and C) remains constant because the A concentration is so small compared to the B (or C) concentration. 126.
The activation energy represents the minimum energy that reactants must have to convert to products. Diagrams a and c both show a 40 kJ activation energy for the forward reaction. ΔE is the energy difference between products and reactants. With a minus ΔE value, the products have a lower energy than the reactants by 100 kJ. Diagrams a and e both show a ΔE = ‒100 kJ. Only plot a shows both the correct activation energy and the correct ΔE.
127.
Only the plot of ln[Ru2+] vs time is linear. This is the proof plot for a first order reaction. So, the reaction is first order with respect to Ru 2+. Answers b and c both indicate a first order reaction. The y-intercept of the ln[Ru2+] vs time plot will equal the ln[Ru2+]0. If the initial concentration of Ru2+ was 1.0 M, the y-intercept would be 0 [ln(1.0) = 0]. However, the yintercept looks to be around ‒1.0. Because ln(0.30) = ‒1.2, this confirms answer b as correct (first order with an initial concentration of 0.3 M). Note that the concentration vs time plot confirms an initial Ru2+ concentration of about 0.30 M.
574 128.
CHAPTER 12
CHEMICAL KINETICS
The logarithmic form of the Arrhenius equation is ln k = −Ea/RT + ln A. A graph of ln k versus 1/T yields a straight line with a slope equal to -Ea/R and y-intercept of ln A. For this reaction at 298 K: ln k = ‒982.7(1/T) ‒ 0.0726 = ‒982.7(1/298) ‒ 0.0726 = ‒3.37, k = e‒3.37 = 0.0344 L/mol•s Rate = k[A][B] = 0.0344 L/mol•s(0.050 mol/L)(0.075 mol/L) = 1.3 × 10 ‒4 mol/L•s
129.
8.75 h ×
3600 s ln 2 ln 2 = 3.15 × 104 s; k = = 2.20 × 10 −5 s −1 = 4 h t1/2 3.15 10 s
The partial pressure of a gas is directly related to the concentration in mol/L. So, instead of using mol/L as the concentration units in the integrated first-order rate law, we can use partial pressures of SO2Cl2. P P 3600 s = − (2.20 × 10 −5 s −1 ) × 12.5 h × ln = −kt, ln h 791 torr P0
PSO2Cl 2 = 294 torr × n=
0.387 atm 1.25 L PV = 9.94 × 10 −3 mol SO2Cl2 = 0.08206 L atm RT 593 K K mol
9.94 × 10 −3 mol ×
130.
k=
1 atm = 0.387 atm 760 torr
6.022 10 23 molecules = 5.99 × 1021 molecules SO2Cl2 mol
ln 2 ln 2 = = 1.04 × 10 −3 s −1 t1/ 2 667 s
2.38 g InCl
[In+]0 =
1 mol InCl 1 mol In + 150 .3 g mol InCl = 0.0317 mol/L 0.500 L
[In + ] [In + ] 3600 s = − (1.04 × 10 −3 s −1 ) × 1.25 h × ln + = − kt , ln h 0.0317 M [In ]0 Solving: [In+] = 2.94 × 10 −4 mol/L The balanced equation for the reaction is: 3 In +(aq) → 2 In(s) + In3+(aq) 0.0317 mol 2.94 10 −4 mol − 0.500 L L L = 1.57 × 10 −2 mol In+ 2 mol In 114 .8 g In 1.57 × 10 −2 mol In+ × = 1.20 g In + mol In 3 mol In
Mol In+ reacted = 0.500 L ×
CHAPTER 12 131.
CHEMICAL KINETICS
575
1.7 10 −2 s −1 k E 1 1 Ea 1 1 = ; ln ln 2 = a − − − 4 − 1 7.2 10 s 8.3145 J/K • mol 660. K R T1 T2 720. K k1 Solving: Ea = 2.1 × 105 J/mol For k at 325°C (598 K): 5 1.7 10 −2 s −1 1 = 2.1 10 J/mol 1 − , k = 1.3 × 10 −5 s −1 ln 8.3145 J/K • mol 598 K k 720. K
For three half-lives, we go from 100% → 50% → 25% → 12.5%. After three half-lives, 12.5% of the original amount of C2H5I remains. Partial pressures are directly related to gas concentrations in mol/L:
PC2H5I = 894 torr × 0.125 = 112 torr after 3 half-lives
Challenge Problems − d[A] d[ A ] = − k dt = k[A]3, 3 dt [ A ]0 [ A ] 0 [ A ]t
132.
n x dx =
t
1 xn +1 ; so: − n +1 2[A]2
[ A ]t
= − kt , −
[ A ]0
1 1 + = − kt 2 2[A]t 2[A]02
For the half-life equation, [A]t = 1/2[A]0: −
−
1 1 2 [A]0 2 3 2[ A ]02
2
+
4 1 1 + = − kt 1/ 2 = − kt 1/ 2 , − 2 2 2[ A ]0 2[ A ]02 2[A]0
= − kt 1/ 2 , t1/2 =
3 2[ A ]02 k
The first half-life is t1/2 = 40. s and corresponds to going from [A]0 to 1/2 [A]0. The second half-life corresponds to going from 1/2 [A]0 to 1/4 [A]0 . First half-life =
3 ; second half-life = 2[ A ]02 k
3 2
1 2 [A]0 k 2
=
6 [ A ]02 k
3 2
2[A]0 k First half - life = = 3/12 = 1/4 6 Second half - life [A]02 k
Because the first half-life is 40. s, the second half-life will be four times this, or 160 s.
576 133.
CHAPTER 12
CHEMICAL KINETICS
Rate = k[I−]x[OCl−]y[OH−]z; comparing the first and second experiments: 18.7 10 −3 k (0.0026 ) x (0.012 ) y (0.10) z , 2.0 = 2.0x, x = 1 = 9.4 10 −3 k (0.0013) x (0.012 ) y (0.10) z
Comparing the first and third experiments: 9.4 10 −3 k (0.0013)(0.012 ) y (0.10) z , 2.0 = 2.0y, y = 1 = 4.7 10 −3 k (0.0013)(0.0060 ) y (0.10) z
Comparing the first and sixth experiments: 4.8 10 −3 k (0.0013)(0.012 )(0.20) z , 1/2 = 2.0z, z = −1 = 9.4 10 −3 k (0.0013)(0.012 )(0.10) z
Rate =
k[I − ][OCl− ] ; the presence of OH− decreases the rate of the reaction. [OH− ]
For the first experiment: 9.4 10 −3 mol (0.0013 mol/L) (0.012 mol/L) , k = 60.3 s −1 = 60. s −1 =k Ls (0.10 mol/L)
For all experiments, kmean = 60. s −1 . 134.
For second order kinetics:
a.
1 1 1 − = kt and t1/2 = [ A ] [ A ]0 k[A]0
1 1 1 1 , = (0.250 L/mol•s)t + = 0.250 × 180. s + [ A ]0 [A] [A ] 1.00 10 −2 M 1 = 145 M −1 , [A] = 6.90 × 10 −3 M [A ]
Amount of A that reacted = 0.0100 − 0.00690 = 0.0031 M. [A2] =
1 2
(3.1 × 10 −3 M) = 1.6 × 10 −3 M
b. After 3.00 minutes (180. s): [A] = 3.00[B], 6.90 × 10 −3 M = 3.00[B], [B] = 2.30 × 10 −3 M 1 1 1 1 = k2t + = k 2 (180. s) + , , k2 = 2.19 L/mol•s −3 [B] [B] 0 2.30 10 M 2.50 10 − 2 M
c. t1/2 = 135.
1 1 = = 4.00 × 102 s k[A] 0 0.250 L/mol • s 1.00 10 − 2 mol/L
a. We check for first-order dependence by graphing ln[concentration] versus time for each set of data. The rate dependence on NO is determined from the first set of data because the ozone concentration is relatively large compared to the NO concentration, so [O 3] is effectively constant.
CHAPTER 12
CHEMICAL KINETICS
577
[NO] (molecules/cm3)
Time (ms)
8
0 100. 500. 700. 1000.
6.0 × 10 5.0 × 108 2.4 × 108 1.7 × 108 9.9 × 107
ln[NO] 20.21 20.03 19.30 18.95 18.41
Because ln[NO] versus t is linear, the reaction is first order with respect to NO. We follow the same procedure for ozone using the second set of data. The data and plot are: Time (ms) [O3] (molecules/cm3) ln[O3] 1.0 × 1010 8.4 × 109 7.0 × 109 4.9 × 109 3.4 × 109
0 50. 100. 200. 300.
23.03 22.85 22.67 22.31 21.95
The plot of ln[O3] versus t is linear. Hence the reaction is first order with respect to ozone. b. Rate = k[NO][O3] is the overall rate law. c. For NO experiment, Rate = k[NO] and k = − (slope from graph of ln[NO] versus t). k = −slope = −
18.41 − 20.21 (1000 . − 0) 10
−3
s
= 1.8 s −1
For ozone experiment, Rate = k[O3] and k = − (slope from ln[O3] versus t plot).
578
CHAPTER 12 k = −slope = −
(21.95 − 23.03) (300. − 0) 10
−3
CHEMICAL KINETICS
= 3.6 s −1
s
d. From the NO experiment, Rate = k[NO][O3] = k [NO] where k = k[O3]. k = 1.8 s −1 = k(1.0 × 1014 molecules/cm3), k = 1.8 × 10 −14 cm3/molecules•s We can check this from the ozone data. Rate = k [O3] = k[NO][O3], where k = k[NO]. k = 3.6 s −1 = k(2.0 × 1014 molecules/cm3), k = 1.8 × 10 −14 cm3/molecules•s Both values of k agree. 136.
Rate = k3[Br−][H2BrO3+]; we must substitute for the intermediate concentration. Because steps 1 and 2 are fast-equilibrium steps, rate forward reaction = rate reverse reaction. k2[HBrO3][H+] = k-2[H2BrO3+]; k1[BrO3−][H+] = k-1[HBrO3] [HBrO3] =
Rate = 137.
k k k k1 [BrO3−][H+]; [H2BrO3+] = 2 [HBrO3][H+] = 2 1 [BrO3−][H+]2 k − 2 k −1 k −2 k −1
k 3 k 2 k1 [Br−][BrO3−][H+]2 = k[Br−][BrO3−][H+]2 k − 2 k −1
a. Rate = k3[COCl][Cl2]; from the fast-equilibrium reactions 1 and 2: k k [COCl ] = 2 , [COCl] = 2 [CO][Cl] [Cl ][CO ] k −2 k −2 1/ 2
k k [Cl ]2 = 1 , [Cl ] = 1 [Cl 2 ] [Cl 2 ] k −1 k −1 1/ 2
k k Thus [COCl] = 2 1 k − 2 k −1
[CO][Cl2]1/2; Substituting into rate law:
1/ 2
k k Rate = k 3 2 1 k − 2 k −1
[CO][Cl2]3/2 = k[CO][Cl2]3/2
b. Cl and COCl are intermediates. 138.
On the energy profile to the right, R = reactants, P = products, Ea = activation energy, ΔE = overall energy change for the reaction, I = intermediate, and RC = reaction coordinate, which is the same as reaction progress. a-d.
See plot to the right
CHAPTER 12
CHEMICAL KINETICS
579
e. This is a two-step reaction since an intermediate plateau appears between the reactant and the products. This plateau represents the energy of the intermediate. The general reaction mechanism for this reaction is: R→I I→P R→P In a mechanism, the rate of the slowest step determines the rate of the reaction. The activation energy for the slowest step will be the largest energy barrier that the reaction must overcome. Since the second hump in the diagram is at the highest energy, the second step has the largest activation energy and will be the rate-determining step (the slow step). 139.
k E 1 rate 2 k 1 ; assuming = 2 = 40.0 : − ln 2 = a rate k1 k R T T 1 2 1 1
ln(40.0) =
1 Ea 1 , Ea = 1.55 105 J/mol = 155 kJ/mol − 8.3145 J/K • mol 308 K 328 K
(carrying an extra sig. fig.) Note that the activation energy is close to the F 2 bond energy. Therefore, the ratedetermining step probably involves breaking the F 2 bond. H2(g) + F2(g) → 2 HF(g); for every 2 moles of HF produced, only 1 mole of the reactant is used up. Therefore, to convert the data to Preactant versus time, Preactant = 1.00 atm – (1/2)PHF. Preactant
Time
1.000 atm 0.850 atm 0.700 atm 0.550 atm 0.400 atm 0.250 atm
0 min 30.0 min 65.8 min 110.4 min 169.1 min 255.9 min
The plot of ln Preactant versus time (plot not included) is linear with negative slope, so the reaction is first order with respect to the limiting reagent. For the reactant in excess, because the values of the rate constant are the same for both experiments, one can conclude that the reaction is zero order in the excess reactant. a. For a three-step reaction with the first step limiting, the energy-level diagram could be:
E R P Reaction coordinate
580
CHAPTER 12
CHEMICAL KINETICS
Note that the heights of the second and third humps must be lower than the first-step activation energy. However, the height of the third hump could be higher than the second hump. One cannot determine this absolutely from the information in the problem. b. We know the reaction has a slow first step, and the calculated activation energy indicates that the rate-determining step involves breaking the F 2 bond. The reaction is also first order in one of the reactants and zero order in the other reactant. All this points to F 2 being the limiting reagent. The reaction is first order in F 2, and the rate-determining step in the mechanism is F2 → 2 F. Possible second and third steps to complete the mechanism follow. F2 → 2 F F + H2 → HF + H H + F → HF
slow fast fast
F2 + H2 → 2 HF c. F2 was the limiting reactant. 140.
k E 1 1 − We need the value of k at 500. K; ln 2 = a R T1 T2 k1
1.11 10 5 J/mol 1 k2 1 = = 22.2 ln − −12 8.3145 J/K • mol 273 K 500 K 2.3 10 L/mol • s k2 = e22.2, k2 = 1.0 × 10−2 L/mol•s −12 2.3 10
Because the decomposition reaction is an elementary reaction, the rate law can be written using the coefficients in the balanced equation. For this reaction, Rate = k[NO 2]2. To solve for the time, we must use the integrated rate law for second-order kinetics. The major problem now is converting units so they match. Rearranging the ideal gas law gives n/V = P/RT. Substituting P/RT for concentration units in the second-order integrated rate equation: 1 1 1 1 RT RT RT P0 − P = kt + , = kt + , − = kt , t = [ NO 2 ] [ NO 2 ]0 P / RT P0 / RT P P0 k P P0
t=
141.
(0.08206 L atm/K • mol)(500. K) 1.0 10
−2
L/mol • s
2.5 atm − 1.5 atm = 1.1 × 103 s 1.5 atm 2.5 atm
a. [B] >> [A], so [B] can be considered constant over the experiments. This gives us a pseudo-order rate-law equation. b. Note that in each case the half-life doubles as time increases (in experiment 1, the first halflife is 40. s, the second half-life is 80. s; in experiment 2, the first half- life is 20. s, the second half-life is 40. s). This occurs only for a second-order reaction, so the reaction is second order in [A]. Between expt. 1 and expt. 2, we double [B] and the reaction rate doubles, thus it is first order in [B]. The overall rate-law equation is rate = k[A]2[B].
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Using t1/2 =
581
1 1 , we get k = = 0.25 L/mol•s; but this is actually k k[A]0 (40.)(10.0 10 − 2 )
where Rate = k[A]2 and k = k[B]. k= c. i.
k 0.25 = 0.050 L2/ mol2•s = [B] 5.0
This mechanism gives the wrong stoichiometry, so it can’t be correct.
ii. Rate = k[E][A] k1[A][B] = k-1[E]; [E] =
k 1 [A][B] k k1 ; Rate = [A]2 [B] k −1 k −1
This mechanism gives the correct stoichiometry and gives the correct rate law. This is a possible mechanism for this reaction. iii. Rate = k[A]2 This mechanism gives the wrong derived rate law, so it can’t be correct. Only mechanism ii is possible. 142.
a. Rate = k[A]x[B]y; looking at the data in experiment 2, notice that the concentration of A is cut in half every 10. s. Only first-order reactions have a half-life that is independent of concentration. The reaction is first order in A. In the data for experiment 1, notice that the half-life is 40. s. This indicates that in going from experiment 1 to experiment 2, where the B concentration doubled, the rate of reaction increased by a factor of four. This tells us that the reaction is second order in B. Rate = k[A][B]2 b. This reaction in each experiment is pseudo-first order in [A] because the concentration of B is so large, it is basically constant. Rate = k[B]2[A] = k [A], where k = k[B]2 For a first-order reaction, the integrated rate law is: [A] = − kt ln [ A ]0
Use any set of data you want to calculate k. For example, in experiment 1, from 0 to 20. s the concentration of A decreased from 0.010 M to 0.0071 M: 0.0071 ln = − k (20. s), k = 1.7 × 10 −2 s −1 0 . 010
k = k[B]2, 1.7 × 10 −2 s −1 = k(10.0 mol/L)2 k = 1.7 × 10−4 L2/mol2•s
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We get similar values for k using other data from either experiment 1 or experiment 2. c.
143.
[A] = − kt = − (1.7 × 10−2 L2/mol2•s) × 30. s, [A] = 6.0 × 10 −3 M ln 0 . 010 M
Rate = k[A]x[B]y[C]z; during the course of experiment 1, [A] and [C] are essentially constant, and Rate = k[B]y , where k = k[A]0x [C]0z . [B] (M)
Time (s)
ln[B]
1.0 × 10 −3 2.7 × 10 −4 1.6 × 10 −4 1.1 × 10 −4 8.5 × 10 −5 6.9 × 10 −5 5.8 × 10 −5
0 1.0 × 105 2.0 × 105 3.0 × 105 4.0 × 105 5.0 × 105 6.0 × 105
−6.91 −8.22 −8.74 −9.12 −9.37 −9.58 −9.76
1/[B] (M -1) 1.0 × 103 3.7 × 103 6.3 × 103 9.1 × 103 12 × 103 14 × 103 17 × 103
A plot of 1/[B] versus t is linear (plot not included), so the reaction is second order in B, and the integrated rate equation is: 1/[B] = (2.7 × 10−2 L/mol•s)t + 1.0 × 103 L/mol; k = 2.7 × 10−2 L/mol•s For experiment 2, [B] and [C] are essentially constant, and Rate = k[A]x, where k = k[B]0y [C]0z = k[B]02 [C]0z . [A] (M)
Time (s)
ln[A]
1/[A] (M−1)
1.0 × 10 −2 8.9 × 10 −3 5.5 × 10 −3 3.8 × 10 −3 2.9 × 10 −3 2.0 × 10 −3
0 1.0 5.0 8.0 10.0 13.0
−4.61 −4.95 −5.20 −5.57 −5.84 −6.21
1.0 × 102 140 180 260 340 5.0 × 102
A plot of ln[A] versus t is linear, so the reaction is first order in A, and the integrated rate law is: ln[A] = −(0.123 s−1)t − 4.61; k = 0.123 s −1 Note: We will carry an extra significant figure in k. Experiment 3: [A] and [B] are constant; Rate = k[C]z The plot of [C] versus t is linear. Thus z = 0. The overall rate law is Rate = k[A][B]2.
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From Experiment 1 (to determine k): k = 2.7 × 10−2 L/mol•s = k[A]0x [C]0z = k[A]0 = k(2.0 M), k = 1.4 × 10−2 L2/mol2•s From Experiment 2: k= 0.123 s −1 = k[B]02 , k =
0.123 s −1 = 1.4 × 10−2 L2/mol2•s (3.0 M ) 2
Thus Rate = k[A][B]2 and k = 1.4 × 10−2 L2/mol2•s. 144.
a. Rate = (k1 + k2[H+])[I−]m[H2O2]n In all the experiments, the concentration of H2O2 is small compared to the concentrations of I− and H+. Therefore, the concentrations of I− and H+ are effectively constant, and the rate law reduces to: Rate = kobs[H2O2]n, where kobs = (k1 + k2[H+])[I−]m Because all plots of ln[H2O2] versus time are linear, the reaction is first order with respect to H2O2 (n = 1). The slopes of the ln[H2O2] versus time plots equal −kobs, which equals −(k1 + k2[H+])[I−]m. To determine the order of I−, compare the slopes of two experiments in which I− changes and H+ is constant. Comparing the first two experiments: − [k1 + k 2 (0.0400 M )] (0.3000 M ) m slope (exp. 2) − 0.360 = = slope (exp. 1) − 0.120 − [k1 + k 2 (0.0400 M )] (0.1000 M ) m m
0.3000 M = (3.000)m, m = 1 3.00 = 0.1000 M
The reaction is also first order with respect to I-. b. The slope equation has two unknowns, k1 and k2. To solve for k1 and k2, we must have two equations. We need to take one of the first set of three experiments and one of the second set of three experiments to generate the two equations in k1 and k2. Experiment 1: Slope = −(k1 + k2[H+])[I−] −0.120 min−1 = −[k1 + k2(0.0400 M)](0.1000 M) or 1.20 = k1 + k2(0.0400) Experiment 4: −0.0760 min−1 = −[k1 + k2(0.0200 M)](0.0750 M) or 1.01 = k1 + k2(0.0200) Subtracting 4 from 1: 1.20 = k1 + k2(0.0400) −1.01 = −k1 − k2(0.0200) 0.19 =
k2(0.0200), k2 = 9.5 L2/mol2•min
1.20 = k1 + 9.5(0.0400), k1 = 0.82 L/mol•min
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c. There are two pathways, one involving H + with Rate = k2[H+][I−][H2O2] and another not involving H+ with Rate = k1[I−][H2O2]. The overall rate of reaction depends on which of these two pathways dominates, and this depends on the H + concentration.
Marathon Problem 145.
a. Rate = k[CH3X]x[Y]y; for experiment 1, [Y] is in large excess, so its concentration will be constant. Rate = k[CH3X]x, where k = k(3.0 M)y. A plot (not included) of ln[CH3X] versus t is linear (x = 1). The integrated rate law is: ln[CH3X] = −(0.93)t − 3.99; k = 0.93 h−1 For experiment 2, [Y] is again constant, with Rate = k [CH3X]x, where k = k(4.5 M)y. The natural log plot is linear again with an integrated rate law: ln[CH3X] = −(0.93)t − 5.40; k = 0.93 h−1 Dividing the rate-constant values:
k' 0.93 k (3.0) y , 1.0 = (0.67)y, y = 0 = = k" 0.93 k (4.5) y
The reaction is first order in CH3X and zero order in Y. The overall rate law is: Rate = k[CH3X], where k = 0.93 h−1 at 25°C b. t1/2 = (ln 2)/k = 0.6931/(7.88 × 108 h−1) = 8.80 × 10−10 hour
c.
7.88 10 8 k E 1 1 Ea 1 1 = , ln ln 2 = a − − R T1 T2 0.93 8.3145 J/K • mol 298 K 358 K k1 Ea = 3.0 × 105 J/mol = 3.0 × 102 kJ/mol
d. From part a, the reaction is first order in CH3X and zero order in Y. From part c, the activation energy is close to the C-X bond energy. A plausible mechanism that explains the results in parts a and c is: CH3X → CH3 + X CH3 + Y → CH3Y
(slow) (fast)
CH3X + Y → CH3Y + X Note: This is a plausible mechanism because the derived rate law is the same as the experimental rate law (and the sum of the steps gives the overall balanced equation).
CHAPTER 13 CHEMICAL EQUILIBRIUM
Review Questions 1.
a. The rates of the forward and reverse reactions are equal at equilibrium. b. There is no net change in the composition (if temperature is constant). See Figure 13.5 of the text for an illustration of the concentration vs. time plot for this reaction. In Figure 13.5, A = H2, B = N2, and C = NH3. Notice how the reactant concentrations decrease with time until they reach equilibrium where the concentrations remain constant. The product concentration increases with time until equilibrium is reached. Also note that H 2 decreases faster than N2; this is due to the 3:1 mole ratio in the balanced equation. H 2 should be used up three times faster than N2. Similarly, NH3 should increase at rate twice the rate of decrease of N2. This is shown in the plot. Reference Figure 13.4 for the reaction rate vs. time plot. As concentrations of reactants decrease, the rate of the forward reaction decreases. Also, as product concentration increases, the rate of the reverse reaction increases. Eventually they reach the point where the rate that reactants are converted into products exactly equals the rate that products are converted into reactants. This is equilibrium and the concentrations of reactants and products do not change.
2.
The law of mass action is a general description of the equilibrium condition; it defines the equilibrium constant expression. The law of mass action is based on experimental observation. K is a constant; the value (at constant temperature) does not depend on the initial conditions. Table 13.1 of the text illustrates this nicely. Three experiments were run with each experiment having a different initial condition (only reactants present initially, only products present initially, and some reactants and products present initially). In all three experiments, the value of K calculated is the same in each experiment (as it should be). Equilibrium and rates of reaction (kinetics) are independent of each other. A reaction with a large equilibrium constant value may be a fast reaction or a slow reaction. The same is true for a reaction with a small equilibrium constant value. Kinetics is discussed in detail in Chapter 12 of the text. The equilibrium constant is a number that tells us the relative concentrations (pressures) of reactants and products at equilibrium. An equilibrium position is a set of concentrations that satisfy the equilibrium constant expression. More than one equilibrium position can satisfy the same equilibrium constant expression.
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From Table 13.1, each of the three experiments have different equilibrium positions; that is, each experiment has different equilibrium concentrations. However, when these equilibrium concentrations are inserted into the equilibrium constant expression, each experiment gives the same value for K. The equilibrium position depends on the initial concentrations one starts with. Since there are an infinite number of initial conditions, there are an infinite number of equilibrium positions. However, each of these infinite equilibrium positions will always give the same value for the equilibrium constant (assuming temperature is constant). 3.
2 NOCl(g) ⇌ 2 NO(g) + Cl2(g)
K = 1.6 × 10−5
The expression for K is the product concentrations divided by the reactant concentrations. When K has a value much less than one, the product concentrations are relatively small and the reactant concentrations are relatively large. 2 NO(g) ⇌ N2(g) + O2(g)
K = 1 × 1031
When K has a value much greater than one, the product concentrations are relatively large, and the reactant concentrations are relatively small. In both cases, however, the rate of the forward reaction equals the rate of the reverse reaction at equilibrium (this is a definition of equilibrium). 4.
The difference between K and Kp are the units used to express the amounts of reactants and products present. K is calculated using units of molarity. Kp is calculated using partial pressures in units of atm (usually). Both have the same form; the difference is the units used to determine the values. Kp is only used when the equilibria involve gases; K can be used for gas phase equilibria and for solution equilibria. Kp = K(RT)n where n = moles gaseous products in the balanced equation – moles gaseous reactants in the balanced equation. K = Kp when n = 0 (when moles of gaseous products = moles of gaseous reactants). K Kp when n 0. When a balanced equation is multiplied by a factor n, Knew = (Koriginal)n. So, if a reaction is 3 tripled, Knew = K original . If a reaction is reversed, Knew = 1/Koriginal. Here, Kp, new = 1/Kp, original.
5.
When reactants and products are all in the same phase, these are homogeneous equilibria. Heterogeneous equilibria involve more than one phase. In general, for a homogeneous gas phase equilibrium, all reactants and products are included in the K expression. In heterogeneous equilibria, equilibrium does not depend on the amounts of pure solids or liquids present. The amount of solids and liquids present are not included in K expressions; they just have to be present. On the other hand, gases and solutes are always included in K expressions. Solutes have (aq) written after them.
6.
For the gas phase reaction aA + bB ⇌ cC + dD: the equilibrium constant expression is: K =
[C]c [D]d [A]a [B]b
and the reaction quotient has the same form: Q =
[C] c [D] d [A] a [B]b
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587
The difference is that in the expression for K, we use equilibrium concentrations, i.e., [A], [B], [C], and [D] are all in equilibrium with each other. Any set of concentrations can be plugged into the reaction quotient expression. Typically, we plug initial concentrations into the Q expression and then compare the value of Q to K to see if the reaction is at equilibrium. If Q = K, the reaction is at equilibrium with these concentrations. If Q K, then the reaction will have to shift either to products or to reactants to reach equilibrium. For Q > K, the net change in the reaction to get to equilibrium must be a conversion of products into reactants. We say the reaction shifts left to reach equilibrium. When Q < K, the net change in the reaction to get to equilibrium must be a conversion of reactants into products; the reaction shifts right to reach equilibrium. 7.
The steps to solve equilibrium problems are outlined at the beginning of section 13.6. The ICE table is a convenient way to summarize an equilibrium problem. We make three rows under the balanced reaction. The initials in ICE stand for initial, change, and equilibrium. The first step is to fill in the initial row, then deduce the net change that must occur to reach equilibrium. You then define x or 2x or 3x … as the change (molarity or partial pressure) that must occur to reach equilibrium. After defining your x, fill in the change column in terms of x. Finally, sum the initial and change columns together to get the last row of the ICE table, the equilibrium concentrations (or equilibrium partial pressures). The ICE table again summarizes what must occur for a reaction to reach equilibrium. This is vital in solving equilibrium problems.
8.
The assumption comes from the value of K being much less than 1. For these reactions, the equilibrium mixture will not have a lot of products present; mostly reactants are present at equilibrium. If we define the change that must occur in terms of x as the amount (molarity or partial pressure) of a reactant that must react to reach equilibrium, then x must be a small number because K is a very small number. We want to know the value of x in order to solve the problem, so we don’t assume x = 0. Instead, we concentrate on the equilibrium row in the ICE table. Those reactants (or products) have equilibrium concentrations in the form of 0.10 – x or 0.25 + x or 3.5 – 3x, etc., is where an important assumption can be made. The assumption is that because K << 1, x will be small (x << 1) and when we add x or subtract x from some initial concentration, it will make little or no difference. That is, we assume that 0.10 – x 0.10 or 0.25 + x 0.25 or 3.5 – 3x 3.5; we assume that the initial concentration of a substance is equal to the equilibrium concentration. This assumption makes the math much easier, and usually gives a value of x that is well within 5% of the true value of x (we get about the same answer with a lot less work). We check the assumptions for validity using the 5% rule. From doing a lot of these calculations, it is found that when an assumption like 0.20 – x 0.20 is made, if x is less than 5% of the number the assumption was made against, then our final answer is within acceptable error limits of the true value of x (as determined when the equation is solved exactly). For our example above (0.20 – x 0.20), if (x/0.20) × 100 5%, then our assumption is valid by the 5% rule. If the error is greater than 5%, then we must solve the equation exactly or use a math trick called the method of successive approximations. See Appendix 1 for details regarding the method of successive approximations as well as for a review in solving quadratic equations exactly.
9.
LeChâtelier’s Principle: if a change is imposed on a system at equilibrium, the position of the equilibrium will shift in the direction that tends to reduce that change.
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a. When a gaseous (or aqueous) reactant is added to a reaction at equilibrium, the reaction shifts right to use up some of the added reactant. Adding NOCl causes the equilibrium to shift to the right. b. If a gaseous (or aqueous) product is added to a system at equilibrium, the reaction shifts left to use up some of the added product. Adding NO(g) causes the equilibrium to shift left. c. Here a gaseous reactant is removed (NOCl), so the reaction shifts left to produce more of the NOCl. d. Here a gaseous product is removed (Cl2), so the reaction shifts right to produce more of the Cl2. e. In this reaction, 2 moles of gaseous reactants are converted into 3 moles of gaseous products. If the volume of the container is decreased, the reaction shifts to the side that occupies a smaller volume. Here, the reactions shift left to side with the fewer moles of reactant gases present. For all these changes, the value of K does not change. If temperature is constant, the value of K is constant. 2 H2(g) + O2(g) ⇌ 2 H2O(l); in this reaction, the amount of water present has no effect on the equilibrium; H2O(l) just has to be present whether it’s 0.0010 grams or 1.0 × 10 6 grams. The same is true for solids. When solids or liquids are in a reaction, addition or removal of these solids or liquids has no effect on the equilibrium (the reaction remains at equilibrium). Note that for this example, if the temperature is such that H2O(g) is the product, then the amount of H2O(g) present does affect the equilibrium. A change in volume will change the partial pressure of all reactants and products by the same factor. The shift in equilibrium depends on the number of gaseous particles on each side. An increase in volume will shift the equilibrium to the side with the greater number of particles in the gas phase. A decrease in volume will favor the side with fewer gas phase particles. If there are the same number of gas phase particles on each side of the reaction, a change in volume will not shift the equilibrium. When we change the pressure by adding an unreactive gas, we do not change the partial pressures (or concentrations) of any of the substances in equilibrium with each other. This is because the volume of the container did not change. If the partial pressures (and concentrations) are unchanged, the reaction is still at equilibrium. 10.
In an exothermic reaction, heat is a product. When the temperature increases, heat (a product) is added and the reaction shifts left to use up the added heat. For an exothermic reaction, the value of K decreases as temperature increases. In an endothermic reaction, heat is a reactant. Heat (a reactant) is added when the temperature increases, and the reaction shifts right to use up the added heat in order to reestablish equilibrium. The value of K increases for an endothermic reaction as temperature increases. A decrease in temperature corresponds to the removal of heat. Here, the value of K increases as T decreases. This indicates that as heat is removed, more products are produced. Heat must be a product, so this is an exothermic reaction.
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589
Active Learning Questions 1.
a. The net change would be for some, but not all, of the added A to react to reach equilibrium. So, the amount of A overall will increase. Reactant B will be smaller since some of it reacted with the added A to reach the new equilibrium. The amounts of C and D products will be greater since the reaction shifted right to reach equilibrium with the added A. b. The net change would be for the reaction to shift left to reestablish equilibrium. Some, but not all, of the added D would react by the reverse reaction. D overall would increase, C would decrease, and A and B would increase in amounts as compared to the original equilibrium mixture.
2.
Let x = number of molecules of either reactant that reacts to reach equilibrium. When the ICE table is set-up, there are 12 − x of each reactant molecules that remain at equilibrium and x amount of each product molecules that are present at equilibrium. Placing these equilibrium amounts into the K expression gives:
( x) 2 = 25 (12 − x)2 Using trial and error to solve, x = 10. So your picture at equilibrium should have 12 − 10 = 2 each of the reactant molecules and x = 10 each of the product molecules. Note: 102/22 = 25. K=
3.
Both problems give the same overall answers. You can reach equilibrium in several little steps or one larger step. If the sum of the steps for each path are stoichiometrically identical, each path will give identical answers. Prove it to yourself by assuming a K value and solving each problem. The answers are the same.
4.
Reaction ii is closest to equilibrium initially. It has three of the four species in the reaction present initially, so it will have the smallest net change. The other two situations only have reactants present initially. Reaction i is next closest to equilibrium while reaction iii is furthest away from equilibrium since it has the larger amount of reactant A present initially. The order of increasing concentration of product D is ii, i, and iii. For increasing concentration of reactant B, the order is just reversed, iii, i, and ii.
5.
H2O(g) + CO(g) ⇌ H2(g) + CO2(g); addition of the gaseous reactant H2O will cause the reaction to shift to products (shift right) to use up some of the added H2O. Note that only some of the added H2O will react, not all of it. If all the H2O reacted, then we would be at the previous equilibrium concentration for H2O, but way below the previous equilibrium concentration of CO. In this situation, the reaction would have to shift right to establish equilibrium. All the added H2O can’t react, just some of it. The concentration of H2O will be larger than the initial equilibrium concentration of H2O since we added H2O and only some of it reacted. The other reactant, CO, will decrease from its equilibrium value. The products (H2 and CO2) will both have larger concentrations as the reaction shifted right to reestablish equilibrium. Added H2 to an equilibrium mixture would cause the reaction to shift to reactants (shift left) so that some of the added H2 is used up. The overall concentration of H2 will be larger than the initial equilibrium concentration of H2 since we added H2 and only some of it reacted. The other product, CO2, will decrease from its equilibrium value. The reactants (H 2O and CO) will both have larger concentrations as the reaction shifted left to reestablish equilibrium.
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6.
If a reaction is at equilibrium and some reactant gas is added, this will give a reaction quotient value less than K. When Q < K, the reaction shifts right to reach equilibrium. This means some of the added A must react with some of the B that was previously present from the equilibrium mixture to form C and D. So, amounts of products C and D increase to reach the new equilibrium concentrations, reactant B decreases from the previous equilibrium value, and there is a net increase in the amount of A present. Only some of the added A reacts to reach equilibrium, not all of it.
7.
It is true that both sets of numbers represent possible equilibrium positions for this reaction since they give a K value of 2. However, the stoichiometry tells us that the second set of concentrations cannot be correct for this actual problem. The ICE table set-up for this problem would be: 4 A(g) + B(g) ⇌ C (g) K = [C] = =2 2(1) [A][B] Initial 2M 1+3=4M 4M x mol/L of A reacts to reach equilibrium Change −x −x → +x Equil. 2−x 4−x 4+x From the problem, [A]eq = 1 M and from the ICE table, [A] = 2 – x. Solving, x = 1 M. [C]eq = 4 + x = 4 + 1 = 5 M But the problem tells us that [C]eq = 6 M. This is impossible with this problem. So the second equilibrium position given in the problem cannot come from the described problem where 3 M B was added to the initial equilibrium mixture. The reaction stoichiometry doesn’t work out.
8.
Equilibrium is reached when the rate reactants are converting to products exactly equals the rate products turn back into reactants. The forward and reverse reaction keep occurring, so microscopically, the reaction is still ongoing. However, since the opposing rates are equal, there is no net change in the concentrations of reactants and products. Macroscopically, we have reached the endpoint of the reaction.
9.
K is a constant for a reaction if temperature is constant. However, if the temperature changes, the value of K changes. Only changes in temperature result in a K value change. Note that K does depend on which reactants and products are included in the K expression. H 2O(g) is included in the K expression, but H2O(l) or H2O(s) are not included.
10.
A closed system is one that does not allow transfer of matter in or out of the system. Equilibrium is reached when the rates of the forward and reverse reactions are equal. Once equilibrium is reached, the concentrations of reactants and products satisfy the equilibrium constant value. This can only occur if the reactants and products are not allowed to escape. If the system was open, then reactant and product molecules could escape making it impossible to satisfy the equilibrium position.
11.
Vapor pressure is defined as the pressure of a vapor over a liquid at equilibrium. The two opposing processes that are at equilibrium are evaporation and condensation. When the rate that liquid molecules convert to gaseous molecules (rate of evaporation) exactly equals the rate that gaseous molecules return to the liquid phase (rate of condensation), there is no net change in the number of gaseous molecules above the liquid; this creates a certain pressure above the liquid. Once the vapor pressure of a liquid has been reached, the system is at equilibrium.
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12.
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591
a. The reaction shifts left to reach equilibrium once 1.0 M C is added. How much C reacts depends on the value of the equilibrium constant. If K is very small, then most of the added C will react. From the stoichiometry in the balanced equation, if most of the added C reacts, then a little less than 1.0 M A will form giving a final A molarity close to 2.0 M. If K is a very large value, then not a lot of the added C will react and the amount of A present will be a little larger than 1.0 M. So the range of A is from, 1.0 to 2.0 M. b. If K is a very small value, then most of the added C will react to reach equilibrium. From the 2:1 mol ratio in the equation, if most of the added C reacts, the upper limit for B will be something a little less than 3.0 M. The lower limit for B will be 1.0 M; this occurs when K is a very large number so only a small amount of the added C reacts. c. If K is small, most of the added C reacts leaving a little more than 1.0 M C at equilibrium. If K is very large, then very little C reacts to reach equilibrium leaving an equilibrium C concentration close to 2.0 M. d. If K is a very small value, then most of the added C reacts leaving a D concentration close to zero. If K is large, not a lot of C (and D) will react to reach equilibrium leaving a D concentration a little less than 1.0 M. For this problem you must know the value of K to refine the estimates of reactants and products present. Also important to this problem is that the mole relationships in the balanced equation must always be considered when solving equilibrium problems.
13.
Adding N2 to a constant pressure container results in a larger container volume; the container volume must increase in order to keep the pressure constant as more gas is added. As the volume of the container increases, all reactant and product concentrations change. So, we have added more N2 to the container, but the volume of the container has increased which decreases the H2 and NH3 concentrations. Unless we know the new concentrations, we can’t predict which way the reaction shifts; either shifting right or shifting left to reestablish equilibrium is possible. If you have an equilibrium mixture in a constant volume container, adding an inert gas increases the total pressure, but it doesn’t change any of the concentrations of the reactants and products (unlike in a constant pressure container). Since the amounts of the reactant and products does not change and the volume of the container does not change, we are still at equilibrium with the exact same equilibrium concentrations before the inert gas was added (assuming temperature is constant).
Questions 14.
a. This experiment starts with only H2 and N2, and no NH3 present. From the initial mixture diagram, there is three times as many H 2 as N2 molecules. So the green line, at the highest initial concentration is the H2 plot, the blue line is the N2 plot, and the red line, which has an initial concentration of zero, is the NH 3 plot. b. N2(g) + 3H2(g) ⇌ 2NH3(g); when a reaction starts with only reactants present initially, the reactant concentrations decrease with time while the product concentrations increase with time. This is seen in the various plots. Also notice that the H 2 concentration initially
592
CHAPTER 13
CHEMICAL EQUILIBRIUM
decreases more rapidly as compared to the initial decrease in N2 concentration. This is due to the stoichiometry in the balanced equation, which dictates that for every 1 molecule of N2 that reacts, 3 molecules of H2 must also react. One would expect the NH3 plot to initially increase faster than the N2 plot decreases (due to the 2 : 1 mole ratio in the balanced equation), and for the H2 plot to initially decrease faster than the NH 3 plot increases (due to the 3 : 2 mole ratio). This is seen in the various plots. c. Equilibrium is reached when the rate of the forward reaction equals the rate of the reverse reaction. At this time, there is no net change in any of the reactant and product concentrations; so the various plots indicate equilibrium has been reached when their concentrations no longer change with time (when the plots reach a plateau). 15.
No, equilibrium is a dynamic process. Both reactions: H2O + CO → H2 + CO2 and H2 + CO2 → H2O + CO are occurring at equal rates. Thus 14C atoms will be distributed between CO and CO2.
16.
No, it doesn't matter from which direction the equilibrium position is reached (if temperature is constant). Both experiments will give the same equilibrium position because both experiments started with stoichiometric amounts of reactants or products.
17.
When the rate of the forward reaction is faster than the rate of the reverse reaction, there is a net gain of products. In essence, the reaction shifts right to get to equilibrium. As the reaction continues and there is a net change of reactants into products, the forward reaction will slow down while the reverse reaction rate will increase. Eventually the rates will be equal, and equilibrium is reached. This true for all reactions, whether K >> 1, K <<1, or K ≈ 1. So, statement d is false. Since Δn ≠ 0, K ≠ Kp.
18.
When the rate of the reverse reaction is greater than the rate of the forward reaction, there is a net change of products into reactants. The reaction shifts left to reach equilibrium. With the shift left, the reactant concentrations will be greater 0.50 M (1.0 mol/2.0 L = 0.50 M) and the product concentrations will be less than 0.50 M at equilibrium. Answer c is correct. Because there is a 2:1 mol ratio between H 2 and O2, the equilibrium concentration of H2 will be greater than O2 (H2 is produced faster than O2 as the reaction shifts left to reach equilibrium). As far as the value of K is concerned, there isn’t enough information in the problem to make this determination. No matter what the value of K, when the rate of the reverse reaction is faster than the rate of the forward reaction, there is a net change of products into reactants.
19.
K = [products]y/[reactants]x; When a reaction contains mostly products at equilibrium, the value of K will be much larger than 1. Only answer b has a K value much larger than 1, so it is the correct answer.
20.
K = [products]y/[reactants]x; because HC2H3O2 is a weak electrolyte, at equilibrium we have mostly HC2H3O2 present, with much smaller amounts of the product ions H+ and C2H3O2‒ present. When a reaction contains mostly reactants at equilibrium, the value of K will be much smaller than 1. Only answer e has a K value much smaller than 1, so it is the correct answer.
CHAPTER 13 21.
CHEMICAL EQUILIBRIUM
H2O(g) + CO(g) ⇌ H2(g) + CO2(g)
K=
593 [ H 2 ][CO 2 ] = 2.0 [ H 2 O ][CO ]
K is a unitless number because there are an equal number of moles of product gases as moles of reactant gases in the balanced equation. Therefore, we can use units of molecules per liter instead of moles per liter to determine K. We need to start somewhere, so let’s assume 3 molecules of CO react. If 3 molecules of CO react, then 3 molecules of H2O must react, and 3 molecules each of H 2 and CO2 are formed. We would have 6 − 3 = 3 molecules of CO, 8 − 3 = 5 molecules of H2O, 0 + 3 = 3 molecules of H2, and 0 + 3 = 3 molecules of CO2 present. This will be an equilibrium mixture if K = 2.0: 3 molecules H 2 3 molecules CO 2 L L = 3 K= 5 5 molecules H 2 O 3 molecules CO L L
Because this mixture does not give a value of K = 2.0, this is not an equilibrium mixture. Let’s try 4 molecules of CO reacting to reach equilibrium. Molecules CO remaining = 6 − 4 = 2 molecules of CO Molecules H2O remaining = 8 − 4 = 4 molecules of H2O Molecules H2 present = 0 + 4 = 4 molecules of H2 Molecules CO2 present = 0 + 4 = 4 molecules of CO2 4 molecules H 2 4 molecules CO 2 L L = 2.0 K= 4 molecules H 2 O 2 molecules CO L L
Because K = 2.0 for this reaction mixture, we are at equilibrium. 22.
When equilibrium is reached, there is no net change in the amount of reactants and products present because the rates of the forward and reverse reactions are equal to each other. The first diagram has 4 A2B molecules, 2 A2 molecules, and 1 B2 molecule present. The second diagram has 2 A2B molecules, 4 A2 molecules, and 2 B2 molecules. Therefore, the first diagram cannot represent equilibrium because there was a net change in reactants and products. Is the second diagram the equilibrium mixture? That depends on whether there is a net change between reactants and products when going from the second diagram to the third diagram. The third diagram contains the same numbers and types of molecules as the second diagram, so the second diagram is the first illustration that represents equilibrium. The reaction container initially contained only A 2B. From the first diagram, 2 A2 molecules and 1 B2 molecule are present (along with 4 A2B molecules). From the balanced reaction, these 2 A2 molecules and 1 B2 molecule were formed when 2 A2B molecules decomposed. Therefore, the initial number of A2B molecules present equals 4 + 2 = 6 molecules A 2B.
594
CHAPTER 13
CHEMICAL EQUILIBRIUM
23.
K and Kp are equilibrium constants, as determined by the law of mass action. For K, concentration units of mol/L are used, and for Kp, partial pressures in units of atm are used (generally). Q is called the reaction quotient. Q has the exact same form as K or K p, but instead of equilibrium concentrations, initial concentrations are used to calculate the Q value. The use of Q is when it is compared with the K value. When Q = K (or when Qp = Kp), the reaction is at equilibrium. When Q K, the reaction is not at equilibrium, and one can deduce the net change that must occur for the system to get to equilibrium.
24.
H2(g) + I2(g) → 2 HI(g)
K=
[HI] 2 [H 2 ][I 2 ]
H2(g) + I2(s) → 2 HI(g)
K=
[HI]2 (Solids are not included in K expressions.) [H 2 ]
Some property differences are: (1) the reactions have different K expressions. (2) for the first reaction, K = Kp (since n = 0), and for the second reaction, K Kp (since n 0). (3) a change in the container volume will have no effect on the equilibrium for reaction 1, whereas a volume change will affect the equilibrium for reaction 2 (shifts the reaction left or right depending on whether the volume is decreased or increased). 25.
CaCO3(s)
⇌ CaO(s) + CO2(g)
Kp = PCO 2 = 1.04; solids and liquids do not appear in
equilibrium expressions. Once the CO2 pressure reaches 1.04 atm, the reaction is at equilibrium. There can’t be an equilibrium CO2 pressure greater than or lesser than 1.04 atm. Answer e is correct. 26.
In all experiments, the reaction will be at equilibrium once the equilibrium constant value is reached. The K value does not depend on the initial concentrations, just the temperature. Answer c is true. Statement a is false because at equilibrium, the rates of the reverse and forward reaction are equal. Statement b is false because equilibrium can be reached in either direction. And statement d is false; K is a constant for a reaction if temperature is constant.
27.
We always try to make good assumptions that simplify the math. In some problems we can set up the problem so that the net change x that must occur to reach equilibrium is a small number. This comes in handy when you have expressions like 0.12 – x or 0.727 + 2x, etc. When x is small, we can assume that it makes little difference when subtracted from or added to some relatively big number. When this is the case, 0.12 – x 0.12 and 0.727 + 2x 0.727, etc. If the assumption holds by the 5% rule, the assumption is assumed valid. The 5% rule refers to x (or 2x or 3x, etc.) that is assumed small compared to some number. If x (or 2x or 3x, etc.) is less than 5% of the number the assumption was made against, then the assumption will be assumed valid. If the 5% rule fails to work, one can use a math procedure called the method of successive approximations to solve the quadratic or cubic equation. Of course, one could always solve the quadratic or cubic equation exactly. This is generally a last resort (and is usually not necessary unless K or Kp ≈ 1).
CHAPTER 13
CHEMICAL EQUILIBRIUM
595
28.
Only statement e is correct. Addition of a catalyst has no effect on the equilibrium position; the reaction just reaches equilibrium more quickly. Statement a is false for reactants that are either solids or liquids (adding more of these has no effect on the equilibrium). Statement b is false always. If temperature remains constant, then the value of K is constant. Statement c is false for exothermic reactions where an increase in temperature decreases the value of K. For statement d, only reactions that have more reactant gases than product gases will shift left with an increase in container volume. If the moles of gas are equal, or if there are more moles of product gases than reactant gases, the reaction will not shift left with an increase in volume.
29.
There will be a net increase in the amount of N2O present once equilibrium is reestablished. As N2O(g) is added, the reaction will shift right to use up some of the added N 2O. However, only some of the added N2O will react, not all. The net effect is for N2O to increase once equilibrium is reestablished. The amount of the NO product will also increase, but the amount of the O2 reactant will decrease. As far as the K value is concerned, as long as the temperature didn’t change, K will remain a constant value.
30.
Plot d best represents the [SO3] vs time plot for this experiment. When a reaction contains only reactants initially, there is a net change of reactants into products. As the concentration of products increases, the rate that reactants convert to products slows down. Eventually equilibrium is reached when the rate of the forward and reverse reactions are equal. At this point, there is no net change in the concentrations of either the reactants or products. Plot d best represents this process. Note that plot e would be best for one of the product concentrations in this experiment as equilibrium is reached.
31.
2 NaHCO3(s) Initial
Change Equil.
⇌ Na2CO3(s)
+ CO2(g) + H2O(g)
K = [CO2][H2O] = 0.25
0 0 Let some NaHCO3(s) decompose to form x atm each of CO2(g) and H2O(g) at equilibrium. → +x +x x x
Because solids do not appear in the equilibrium constant expression, we don’t worry about the actual amounts of solids in the ICE table. Statement d is false. The mass of solid initially reacted will not affect the equilibrium position (assuming there is enough solid present to produce the equilibrium amounts of products). As soon as some of the NaHCO3 reacts to give x amount of CO2 and x amount of H2O, the reaction is at equilibrium; it doesn’t matter how much excess solid is present. From the ICE table, [CO2] = [H2O] = x at equilibrium. For statement e, since Δn ≠ 0, K ≠ Kp. 32.
In an exothermic reaction, heat is a product. To maximize yield of products, one would want as low a temperature as possible since high temperatures would shift the reaction left (away from products). Temperature changes also change the value of K. As temperature decreases, the value of K will increase, which maximizes yield of product. The problem with lowering the temperature is the effect it will have on the speed of the reaction. These two factors must both be considered when coming up with the best temperature to run the process.
596
CHAPTER 13
CHEMICAL EQUILIBRIUM
The Equilibrium Constant 33.
34.
a. K =
[ NO]2 [ N 2 ][O 2 ]
b. K =
[ NO 2 ]2 [ N 2O 4 ]
c. K =
[SiCl 4 ][H 2 ]2 [SiH 4 ][Cl 2 ]2
d. K =
[PCl3 ]2 [Br2 ]3 [PBr3 ]2 [Cl 2 ]3
2 PNO a. Kp = PN 2 PO 2
c. Kp =
35.
b. Kp =
PSiCl 4 PH2 2
PN 2 O 4 2 3 PPCl PBr 3 2
d. Kp =
2 PSiH 4 PCl 2
K = 1.3 × 10 −2 =
2 PNO 2
2 3 PPBr PCl 3 2
[ NH3 ]2 for N2(g) + 3 H2(g) ⇋ 2 NH3(g). [ N 2 ][H 2 ]3
When a reaction is reversed, then Knew = 1/Koriginal. When a reaction is multiplied through by a value of n, then Knew = (Koriginal)n. a. 1/2 N2(g) + 3/2 H2(g) ⇌ NH3 (g) K = b. 2 NH3(g) ⇌ N2(g) + 3 H2(g) K =
[NH 3 ] [N 2 ]1/2 [H 2 ]3/2
[ N 2 ][H 2 ]3 [ NH3 ]
2
=
= K1/2 = (1.3 × 10 −2 )1/2 = 0.11
1 1 = 77 = K 1.3 10 −2 1/ 2
[ N 2 ]1/ 2 [H 2 ]3 / 2 1 c. NH3(g) ⇌ 1/2 N2(g) + 3/2 H2(g) K = = [ NH3 ] K
1/ 2
1 = −2 1 . 3 10
= 8.8 [ NH3 ]4 = (K)2 = (1.3 × 10 −2 )2 = 1.7 × 10 −4 [ N 2 ] 2 [ H 2 ]6
d. 2 N2(g) + 6 H2(g) ⇌ 4 NH3(g) K = 36.
2 PHBr = 3.5 × 104 (PH 2 ) (PBr2 )
H2(g) + Br2(g) ⇌ 2 HBr(g)
Kp =
a. HBr ⇌ 1/2 H2 + 1/2 Br2
(PH2 ) K 'p =
1/2
(PBr2 )1/2
PHBr
1/2
1 = K p
1/2
1 4 3.5 10
=
= 5.3 × 10−3 b. 2 HBr ⇌ H2 + Br2
K 'p' =
c. 1/2 H2 + 1/2 Br2 ⇌ HBr
(PH 2 )(PBr2 ) 2 PHBr
K 'p'' =
=
1 1 = 2.9 × 10−5 = Kp 3.5 10 4
PHBr 1/2
(PH 2 ) (PBr2 )
1/2
= (K p )1/2 = 190
CHAPTER 13 37.
CHEMICAL EQUILIBRIUM
2 NO(g) + 2 H2(g) ⇌ N2(g) + 2 H2O(g) K=
(5.3 10 −2 )(2.9 10 −3 ) 2 (8.1 10 −3 ) 2 (4.1 10 −5 ) 2
K=
[ N 2 ][H 2 O]2 [ NO]2 [H 2 ]2
= 4.0 × 106
(4.7 10 −4 ) 2 [NO]2 = 6.9 × 10−4 = [N 2 ][O 2 ] (0.041)(0.0078 )
38.
K=
39.
[NO] =
2.4 mol 4.5 10 −3 mol = 1.5 × 10 −3 M; [Cl2] = = 0.80 M 3 .0 L 3.0 L
[NOCl] =
40.
597
[N2O] =
1.0 mol = 0.33 M; 3. 0 L
K=
[NO]2 [Cl 2 ] [NOCl]2
=
(1.5 10 −3 ) 2 (0.80) (0.33) 2
= 1.7 × 10 −5
2.00 10 −2 mol 2.80 10 −4 mol 2.50 10 −5 mol ; [N2] = ; [O2] = 2.00 L 2.00 L 2.00 L 2
2.00 10 − 2 2.00 [ N 2 O]2 (1.00 10 − 2 ) 2 = = K= [ N 2 ]2 [O 2 ] 2.80 10 − 4 2 2.50 10 −5 (1.40 10 − 4 ) 2 (1.25 10 −5 ) 2.00 2.00
= 4.08 × 108 If the given concentrations represent equilibrium concentrations, then they should give a value of K = 4.08 × 108. (0.200 ) 2 = 4.08 × 108 (2.00 10 − 4 ) 2 (0.00245 )
Because the given concentrations when plugged into the equilibrium constant expression give a value equal to K (4.08 × 108), this set of concentrations is a system at equilibrium. 41.
42.
Kp =
Kp =
2 PNO PO 2 2 PNO 2
2 PNH 3
PN 2 PH3 2
=
=
(0.0167 ) 2 (0.525) (0.00761) 3
(6.5 10 −5 ) 2 (4.5 10 −5 ) (0.55) 2
= 6.3 × 10 −13
(3.1 10 −2 ) 2 = 3.8 × 104 (0.85)(3.1 10 −3 ) 3
= 1.21 × 103
When the given partial pressures in atmospheres are plugged into the K p expression, the value does not equal the Kp value of 3.8 × 104. Therefore, one can conclude that the given set of partial pressures does not represent a system at equilibrium.
598 43.
CHAPTER 13
CHEMICAL EQUILIBRIUM
Kp = K(RT)Δn, where Δn = sum of gaseous product coefficients − sum of gaseous reactant coefficients. For this reaction, Δn = 3 − 1 = 2. K=
[CO][H 2 ]2 (0.24)(1.1) 2 = 1.9 = (0.15) [CH 3OH]
Kp = K(RT)2 = 1.9(0.08206 L atm/K•mol × 600. K)2 = 4.6 × 103 44.
Kp = K(RT)Δn , K = K=
45.
Kp (RT) Δn
; Δn = 2 − 3 = − 1
0.25 = 23 (0.08206 L atm/K • mol 1100 K) −1
There are many potential balanced equations for this reaction; two possible balanced equations are 2 N2O5(g) ⇌ O2(g) + 4 NO2(g) or N2O5(g) ⇌ ½ O2(g) + 2 NO2(g). So, there are many possibly correct answers to this question, depending on the balanced equation used. Only answer e corresponds to a correctly balanced equation converting N 2O5 into products.
46.
47.
Xe(g) + 2 F2(g) ⇌ XeF4(s) is one possible balanced equation, but there are others. Note the product is a solid, so it will not appear in the K expression. Only answer a represents a correct K expression for a correctly balanced equation for the formation of solid XeF 4. Solids and liquids do not appear in equilibrium expressions. Only gases and dissolved solutes appear in equilibrium expressions. a. K =
PH 2O [H 2 O] ; Kp = 2 2 [ NH 3 ] [CO 2 ] PNH3 PCO 2
c. K = [O2]3; Kp = PO3 2 48.
a. Kp =
c. Kp =
1 (PO 2 )
3/ 2
PCO PH 2 PH 2O
3 b. K = [N2][Br2]3; Kp = PN2 PBr 2
d. K =
b. Kp =
d. Kp =
PH 2 O [ H 2 O] ; Kp = [H 2 ] PH 2
1 PCO 2
PO3 2 PH2 O 2
49.
Kp = K(RT)Δn, where Δn equals the difference in the sum of the coefficients between gaseous products and gaseous reactants (Δn = mol gaseous products − mol gaseous reactants). When Δn = 0, then Kp = K. In Exercise 47, only reaction d has Δn = 0, so only reaction d has Kp = K.
50.
Kp = K when Δn = 0. In Exercise 48, none of the reactions have Kp = K because none of the reactions have Δn = 0. The values of Δn for the various reactions are −1.5, −1, 1, and 1, respectively.
CHAPTER 13 51.
CHEMICAL EQUILIBRIUM
Solids and liquids do not appear in equilibrium expressions. Only gases and dissolved solutes appear in equilibrium expressions. 4 KO2(s) + 2 CO2(g) ⇌ 2 K2CO3(s) + 3 O2(g)
K=
52.
0.4 mol O 2 4.0 L
K=
[O2 ]3 [CO2 ]2
3
4 10 mol CO2 4.0 L −10
2
= 1 × 1017
Because solids do not appear in the equilibrium constant expression, K = 1/[O 2]3. [O2] =
53.
599
Kp =
1 1 1 1.0 10 −3 mol ; K= = = 8.0 × 109 = 3 3 −4 3 2.0 L [O 2 ] − 3 ( 5 . 0 10 ) 1.0 10 2.0
PH4 2 PH4 2O
; Ptotal = PH 2O + PH 2 , 36.3 torr = 15.0 torr + PH2 , PH2 = 21.3 torr 4
1 atm 21.3 torr 760 torr Because l atm = 760 torr: Kp = = 4.07 4 1 atm 15.0 torr 760 torr Note: Solids and pure liquids are not included in K expressions. 54.
Ptotal = PCO 2 + PCO , PCO = 0.572 − 0.0020 = 0.570 atm; since solids are not included: Kp =
2 PCO
PCO 2
=
(0.570 ) 2 , Kp = 160 0.0020
Equilibrium Calculations 55.
H2O(g) + Cl2O(g) → 2 HOCl(g)
K=
[HOCl]2 = 0.0900 [H 2 O][Cl2 O]
Use the reaction quotient Q to determine which way the reaction shifts to reach equilibrium. For the reaction quotient, initial concentrations given in a problem are used to calculate the value for Q. If Q < K, then the reaction shifts right to reach equilibrium. If Q > K, then the reaction shifts left to reach equilibrium. If Q = K, then the reaction does not shift in either direction because the reaction is already at equilibrium.
600
CHAPTER 13
CHEMICAL EQUILIBRIUM
2
1.0 mol 2 1.0 L [HOCl]0 a. Q = = = 1.0 × 102 [H 2 O]0 [Cl 2 O]0 0.10 mol 0.10 mol 1.0 L 1.0 L
Q > K, so the reaction shifts left to produce more reactants to reach equilibrium. 2
0.084 mol 2.0 L b. Q = = 0.090 = K; at equilibrium 0.98 mol 0.080 mol 2.0 L 2.0 L 2
0.25 mol 3.0 L c. Q = = 110 0.56 mol 0.0010 mol 3.0 L 3.0 L
Q > K, so the reaction shifts to the left to reach equilibrium. 56.
As in Exercise 55, determine Q for each reaction, compare this value to Kp (= 2.4 × 103), and then determine which direction the reaction shifts to reach equilibrium. Note that for this reaction, K = Kp because n = 0. a. Q =
PN 2 PO 2 2 PNO
=
(0.11 atm)(2.0 atm) = 1.5 × 103 (0.012 atm) 2
Q < Kp, so the reaction shifts right to reach equilibrium. b. Q =
(0.36 atm)(0.67 atm) = 4.0 × 103 (0.0078 atm) 2
Q > Kp, so the reaction shifts left to reach equilibrium. c. Q = 57.
(0.51 atm)(0.18 atm) = 2.4 × 103 = Kp; at eqilibrium (0.0062 atm) 2
CaCO3(s) ⇌ CaO(s) + CO2(g) Kp = PCO 2 = 1.04 a. Q = PCO2 ; we only need the partial pressure of CO2 to determine Q because solids do not appear in equilibrium expressions (or Q expressions). At this temperature, all CO2 will be in the gas phase. Q = 2.55, so Q > Kp; the reaction will shift to the left to reach equilibrium; the mass of CaO will decrease. b. Q = 1.04 = Kp, so the reaction is at equilibrium; mass of CaO will not change. c. Q = 1.04 = Kp, so the reaction is at equilibrium; mass of CaO will not change. d. Q = 0.211 < Kp; the reaction will shift to the right to reach equilibrium; mass of CaO will increase.
CHAPTER 13 58.
CHEMICAL EQUILIBRIUM
NH4NO3(s) ⇌ N2O(g) + 2 H2O(g)
601
K = 4.85
Q = [N2O]0[H2O]02 = 2.0 mol/L(2.0 mol/L)2 = 8.0; Q > K, so the reaction shifts left to reach equilibrium. As the reaction shifts left, the amount of reactants increase. The mass of NH4NO3 at equilibrium will be greater than 200. g. 59.
Q=
[NH3 ]02 (0.500 mol/L)2 = 4.00 = [N 2 ]0 [H 2 ]30 (0.500 mol/L)(0.500 mol/L)3
Since Q > K, the reaction shifts left to reach equilibrium. 3 H2(g) Initial Change Equil
+
N2(g)
⇌
2 NH3(g)
0.500 M 0.500 M 0.500 M Let 2x mol/L of NH3 react to reach equilibrium. +3x +x ‒2x 0.500 + 3x 0.500 + x 0.500 ‒2x
From the ICE table, the [NH3] will be less than 0.500 M and the reactant concentrations will be greater than 0.500 M. So, NH3 has the smallest equilibrium concentration. Also from the ICE table, H2 will have the largest concentration at equilibrium as 0.500 + 3x > 0.500 + x. 60.
Q = (2.0)2/2.0(2.0) = 1.0; Q > K, so reaction shifts left to reach equilibrium.
Initial Change Equil.
H2(g) + I2(g) ⇌ 2 HI(g) 2.0 M 2.0 M 2.0 M Let 2x mol/L of HI react to reach equilibrium. +x +x −2x 2.0 + x 2.0 + x 2.0 − 2x
The only true statement is a. The concentration of I2 will be larger than the concentration of HI. Note that H2 and I2 will have equal concentrations and it will be greater than 2.0 M. The HI concentration will be less than 2.0 M. 61.
K=
[H 2 ] 2 [O 2 ] [H 2 O] 2
, 2.4 10 −3 =
Moles of O2 = 2.0 L 2 PNOBr
(1.9 10 −2 ) 2 [O 2 ] (0.11) 2
0.080 mol O 2 = 0.16 mol O2 L
, 109 =
(0.0768) 2 , PNO = 0.0583 atm 2 PNO 0.0159
62.
KP =
63.
SO2(g) + NO2(g) ⇌ SO3(g) + NO(g) K =
2 PNO PBr2
, [O2] = 0.080 M
[SO 3 ][ NO ] [SO 2 ][ NO 2 ]
To determine K, we must calculate the equilibrium concentrations. The initial concentrations are:
602
CHAPTER 13 [SO3]0 = [NO]0 = 0; [SO2]0 = [NO2]0 =
CHEMICAL EQUILIBRIUM
2.00 mol = 2.00 M 1.00 L
Next, we determine the change required to reach equilibrium. At equilibrium, [NO] = 1.30 mol/1.00 L = 1.30 M. Because there was zero NO present initially, 1.30 M of SO2 and 1.30 M NO2 must have reacted to produce 1.30 M NO as well as 1.30 M SO3, all required by the balanced reaction. The equilibrium concentration for each substance is the sum of the initial concentration plus the change in concentration necessary to reach equilibrium. The equilibrium concentrations are: [SO3] = [NO] = 0 + 1.30 M = 1.30 M; [SO2] = [NO2] = 2.00 M − 1.30 M = 0.70 M We now use these equilibrium concentrations to calculate K: K=
64.
[SO 3 ][ NO ] (1.30)(1.30) = = 3.4 [SO 2 ][ NO 2 ] (0.70)(0.70)
S8(g) ⇌ 4 S2(g)
Kp =
PS42 PS8
Initially: PS8 = 1.00 atm and PS2 = 0 atm Change: Because 0.25 atm of S8 remain at equilibrium, 1.00 atm − 0.25 atm = 0.75 atm of S8 must have reacted in order to reach equilibrium. Because there is a 4 : 1 mole ratio between S 2 and S8 (from the balanced reaction), 4(0.75 atm) = 3.0 atm of S 2 must have been produced when the reaction went to equilibrium (moles and pressure are directly related at constant T and V). Equilibrium: PS8 = 0.25 atm, PS2 = 0 + 3.0 atm = 3.0 atm; solving for Kp: Kp = 65.
(3 . 0 ) 4 = 3.2 × 102 0.25
When solving equilibrium problems, a common method to summarize all the information in the problem is to set up a table. We commonly call this table an ICE table because it summarizes initial concentrations, changes that must occur to reach equilibrium, and equilibrium concentrations (the sum of the initial and change columns). For the change column, we will generally use the variable x, which will be defined as the amount of reactant (or product) that must react to reach equilibrium. In this problem, the reaction must shift right to reach equilibrium because there are no products present initially. Therefore, x is defined as the amount of reactant SO3 that reacts to reach equilibrium, and we use the coefficients in the balanced equation to relate the net change in SO3 to the net change in SO2 and O2. The general ICE table for this problem is: 2 SO3(g) Initial Change Equil.
⇌
2 SO2(g)
+
O2(g)
12.0 mol/3.0 L 0 0 Let x mol/L of SO3 react to reach equilibrium. −x → +x +x/2 4.0 − x x x/2
K=
[SO 2 ]2 [O 2 ] [SO 3 ]2
CHAPTER 13
CHEMICAL EQUILIBRIUM
603
From the problem, we are told that the equilibrium SO2 concentration is 3.0 mol/3.0 L = 1.0 M ([SO2]e = 1.0 M). From the ICE table setup, [SO2]e = x, so x = 1.0. Solving for the other equilibrium concentrations: [SO3]e = 4.0 − x = 4.0 − 1.0 = 3.0 M; [O2] = x/2 = 1.0/2 = 0.50 M. K=
(1.0) 2 (0.50) [SO 2 ]2 [O 2 ] = = 0.056 [SO 3 ]2 (3.0) 2
Alternate method: Fractions in the change column can be avoided (if you want) be defining x differently. If we were to let 2x mol/L of SO3 react to reach equilibrium, then the ICE table setup is: [SO 2 ]2 [O 2 ] 2 SO3(g) ⇌ 2 SO2(g) + O2(g) K= [SO 3 ]2 Initial 4.0 M 0 0 Let 2x mol/L of SO3 react to reach equilibrium. Change −2x → +2x +x Equil. 4.0 − 2x 2x x Solving: 2x = [SO2]e = 1.0 M, x = 0.50 M; [SO3]e = 4.0 − 2(0.50) = 3.0 M; [O2]e = x = 0.50 M These are the same equilibrium concentrations as solved for previously, thus K will be the same (as it must be). The moral of the story is to define x in a manner that is most comfortable for you. Your final answer is independent of how you define x initially. 66.
When solving equilibrium problems, a common method to summarize all the information in the problem is to set up a table. We commonly call this table the ICE table because it summarizes initial concentrations, changes that must occur to reach equilibrium, and equilibrium concentrations (the sum of the initial and change rows). For the change row, we generally use the variable x, which will be defined as the amount of reactant (or product) that must react to reach equilibrium. Note in this problem, the equilibrium concentration of C is greater than the initial concentration, so the reaction shifts right to reach equilibrium. The general ICE table for this problem is: A(aq) Initial Change Equil
+
2 B(aq)
⇌ 3 C(aq)
1.00 M 1.00 M 1.00 M Let x mol/L of A react to reach equilibrium −x −2x → +3x 1.00 − x 1.00 − 2x 1.00 + 3x
In the problem, we are told that [C]e = 1.96 M. From the set-up, [C]e = 1.00 + 3x = 1.96 M. Solving: x = 0.32 mol/L. The other concentrations are [A] = 1.00 – 0.32 = 0.68 mol/L and [B]e = 1.00 −2(0.32) = 0.36 mol/L. Calculating K: K=
[C]3 (1.96 M )3 = 85 = [A][B]2 0.68 M (0.36 M ) 2
604
CHAPTER 13
67.
3 H2(g) Initial
+
N2(g)
⇌
CHEMICAL EQUILIBRIUM
2 NH3 (g)
[H2]0 [N2]0 0 x mol/L of N2 reacts to reach equilibrium −3x −x → +2x [H2]0 − 3x [N2]0 − x 2x
Change Equil
From the problem: [NH3]e = 4.0 M = 2x, x = 2.0 M; [H2]e = 5.0 M = [H2]0 −3x; [N2]e = 8.0 M = [N2]0 − x 5.0 M = [H2]0 − 3(2.0 M), [H2]0 = 11.0 M; 8.0 M = [N2]0 − 2.0 M, [N2]0 = 10.0 M 68.
Let’s set -up the ICE table for this problem. The reaction shifts right to reach equilibrium since we have no products initially. Note that As will not be in the K expression since it is a solid. Since As is not in the K expression, we will not worry about it in the ICE table. 2 AsH3(g) Initial
⇌ 2 As(s)
+ 3 H2(g)
K=
[H2 ]3 [AsH3 ]2
[AsH3]0 0 Let 2x mol/L of AsH3 react to reach equilibrium. −2x → +3x [AsH3]0 − 2x 3x
Change Equil.
From the problem, [H2] = 3.0 mol/1.0 L = 3.0 mol/L, which equals 3x. So, x = 1.0 mol/L. [AsH3]0 − 2x = 6.0 mol/L, [AsH3]0 = 6.0 + 2(1.0) = 8.0 mol/L 69.
Q = 1.00, which is less than K. The reaction shifts to the right to reach equilibrium. Summarizing the equilibrium problem in a table: SO2(g) Initial Change Equil.
+
NO2(g)
⇌
SO3(g)
0.800 M 0.800 M 0.800 M x mol/L of SO2 reacts to reach equilibrium −x −x → +x 0.800 − x 0.800 − x 0.800 + x
+
NO(g)
K = 3.75
0.800 M +x 0.800 + x
Plug the equilibrium concentrations into the equilibrium constant expression: K=
[SO 3 ][ NO] (0.800 + x) 2 , 3.75 = ; taking the square root of both sides: [SO 2 ][ NO2 ] (0.800 − x) 2
0.800 + x = 1.94, 0.800 + x = 1.55 − (1.94)x, (2.94)x = 0.75, x = 0.26 M 0.800 − x
The equilibrium concentrations are: [SO3] = [NO] = 0.800 + x = 0.800 + 0.26 = 1.06 M; [SO2] = [NO2] = 0.800 − x = 0.54 M
CHAPTER 13 70.
CHEMICAL EQUILIBRIUM
605
Q = 1.00, which is greater than K. The reaction shifts to the left to reach equilibrium. Summarizing the problem in a table: 2 PHI H2(g) + I2(g) ⇌ 2 HI(g) Kp = 0.010 = PH 2 PI 2 Initial Change Equil. Kp = 0.010 =
0.10 =
1.00 atm 1.00 atm 1.00 atm Let 2x atm of HI react to reach equilibrium. +x +x −2x 1.00 + x 1.00 + x 1.00 − 2x 2 PHI
PH 2 PI 2
=
(1.00 − 2x) 2 , taking the square root of both sides: (1.00 + x)(1.00 + x)
1.00 − 2x , 0.10 + (0.10)x = 1.00 − 2x, (2.10)x = 0.90, x = 0.43 atm 1.00 + x
PH 2 = PI 2 = 1.00 + 0.43 = 1.43 atm; PHI = 1.00 − 2(0.43)x = 0.14 atm 71.
Because only reactants are present initially, the reaction must proceed to the right to reach equilibrium. Summarizing the problem in a table: 2 PClF Cl2(g) + F2(g) ⇌ 2 ClF(g) Kp = 16 = PCl2 PF2 Initial Change Equil.
2 PClF (2 x)2 = , taking the square root of both sides: PCl2 PF2 (3.00 − x)(3.00 − x)
Kp = 16 = 4.0 =
3.00 atm 3.00 atm 0 Let x atm of Cl2 react to reach equilibrium. −x −x → +2x 3.00 − x 3.00 − x 2x
2x , 12 − (4.0)x = 2x, (6.0)x = 12, x = 2.0 atm 3.00 − x
PCl2 = 3.00 − 2.0 = 1.0 atm 72.
Rection will shift left since no reactants are present initially. 2 HF(g) Initial Change Equil. Kp = 4.0 =
2.0 =
⇌
H2(g)
+
F2(g)
0 1.00 atm 1.00 atm Let x atm of H2 react to reach equilibrium. +2x −x −x 2x 1.00 − x 1.00 − x PH2 PF 2 2 PHF
=
(1.00 − x) 2 ; taking the square root of both sides: (2 x) 2
1.00 − x , 4.0x = 1.00 − x, 5.0x = 1.00, x = 0.20 atm 2x
PH 2 = PF2 = 1.00 − 0.20 = 0.80 atm; PHF = 2(0.20) = 0.40 atm
606 73.
CHAPTER 13
CHEMICAL EQUILIBRIUM
Because only reactants are present initially, the reaction must proceed to the right to reach equilibrium. Summarizing the problem in a table: N2(g) Initial Change Equil. Kp = 0.050 =
+
O2(g)
⇌
2 NO(g)
Kp = 0.050
0.80 atm 0.20 atm 0 x atm of N2 reacts to reach equilibrium −x −x → +2x 0.80 − x 0.20 − x 2x 2 PNO (2 x) 2 = , 0.050[0.16 − (1.00)x + x2] = 4x2 ( 0 . 80 − x )( 0 . 20 − x ) PN 2 PO2
4x2 = 8.0 × 10−3 − (0.050)x + (0.050)x2, (3.95)x2 + (0.050)x − 8.0 × 10−3 = 0 Solving using the quadratic formula (see Appendix 1 of the text): x=
− b (b 2 − 4ac)1/ 2 − 0.050 [(0.050 )]2 − 4(3.95)(−8.0 10 −3 )]1/ 2 = 2a 2(3.95)
x = 3.9 × 10−2 atm or x = −5.2 × 10−2 atm; only x = 3.9 × 10−2 atm makes sense (x cannot be negative), so the equilibrium NO concentration is: PNO = 2x = 2(3.9 × 10−2 atm) = 7.8 × 10−2 atm 74.
H2O(g) + Cl2O(g) ⇌ 2 HOCl(g)
K = 0.090 =
[HOCl] 2 [H 2 O][Cl 2 O]
a. The initial concentrations of H2O and Cl2O are: 1.0 g H 2 O 1 mol 2.0 g Cl 2 O 1 mol = 5.5 × 10 −2 mol/L; = 2.3 × 10 −2 mol/L 1 .0 L 86.90 g 1.0 L 18.02 g
H2O(g)
+
Cl2O(g)
⇌
2 HOCl(g)
5.5 × 10 −2 M 2.3 × 10 −2 M 0 x mol/L of H2O reacts to reach equilibrium Change −x −x → +2x −2 −2 Equil. 5.5 × 10 − x 2.3 × 10 − x 2x Initial
K = 0.090 =
(2 x) 2 (5.5 10 − 2 − x)(2.3 10 − 2 − x)
1.14 × 10 −4 − (7.02 × 10 −3 )x + (0.090)x2 = 4x2 (3.91)x2 + (7.02 × 10 −3 )x − 1.14 × 10 −4 = 0 (We carried extra significant figures.) Solving using the quadratic formula:
CHAPTER 13
CHEMICAL EQUILIBRIUM
607
− 7.02 10 −3 (4.93 10 −5 + 1.78 10 −3 )1 / 2 = 4.6 × 10 −3 or −6.4 × 10 −3 7.82
A negative answer makes no physical sense; we can't have less than nothing. Thus x = 4.6 × 10 −3 M. [HOCl] = 2x = 9.2 × 10 −3 M; [Cl2O] = 2.3 × 10 −2 − x = 0.023 − 0.0046 = 1.8 × 10 −2 M [H2O] = 5.5 × 10 −2 − x = 0.055 − 0.0046 = 5.0 × 10 −2 M b.
H2O(g) Initial
+
⇌
Cl2O(g)
2 HOCl(g)
0 0 1.0 mol/2.0 L = 0.50 M 2x mol/L of HOCl reacts to reach equilibrium +x +x ← −2x x x 0.50 − 2x
Change Equil. K = 0.090 =
(0.50 − 2 x) 2 [HOCl]2 = [H 2 O][Cl2 O] x2
The expression is a perfect square, so we can take the square root of each side: 0.30 =
0.50 − 2 x , (0.30)x = 0.50 − 2x, (2.30)x = 0.50 x
x = 0.217 (We carried extra significant figures.) x = [H2O] = [Cl2O] = 0.217 = 0.22 M; [HOCl] = 0.50 − 2x = 0.50 − 0.434 = 0.07 M 75.
2 SO2(g) Initial Change Equil.
+
O2(g)
⇋
2 SO3(g)
Kp = 0.25
0.50 atm 0.50 atm 0 2x atm of SO2 reacts to reach equilibrium −2x −x → +2x 0.50 − 2x 0.50 − x 2x
Kp = 0.25 =
2 PSO 3 2 PSO PO 2 2
=
( 2 x) 2 (0.50 − 2 x) 2 (0.50 − x)
This will give a cubic equation. Graphing calculators can be used to solve this expression. If you don’t have a graphing calculator, an alternative method for solving a cubic equation is to use the method of successive approximations (see Appendix 1 of the text). The first step is to guess a value for x. Because the value of K is small (K < 1), not much of the forward reaction will occur to reach equilibrium. This tells us that x is small. Let’s guess that x = 0.050 atm. Now we take this estimated value for x and substitute it into the equation everywhere that x appears except for one. For equilibrium problems, we will substitute the estimated value for x into the denominator and then solve for the numerator value of x. We continue this process until the estimated value of x and the calculated value of x converge on the same number. This is the same answer we would get if we were to solve the cubic equation exactly. Applying the method of successive approximations and carrying extra significant figures:
608
CHAPTER 13 4x2 [0.50 − 2(0.050 )]2 [0.50 − (0.050 )] 4x2 [0.50 − 2(0.067 )]2 [0.50 − (0.067 )]
=
4x2 (0.40) 2 (0.45)
=
CHEMICAL EQUILIBRIUM
= 0.25 , x = 0.067
4x2 (0.366 ) 2 (0.433)
= 0.25 , x = 0.060
4x2 4x2 = 0.25, x = 0.063; = 0.25, (0.374 ) 2 (0.437 ) (0.38) 2 (0.44)
x = 0.062
The next trial gives the same value for x = 0.062 atm. We are done except for determining the equilibrium concentrations. They are:
PSO 2 = 0.50 − 2x = 0.50 − 2(0.062) = 0.376 = 0.38 atm PO 2 = 0.50 − x = 0.438 = 0.44 atm; PSO3 = 2x = 0.124 = 0.12 atm 76.
a. The reaction must proceed to products to reach equilibrium because no product is present initially. Summarizing the problem in a table where x atm of N2O4 reacts to reach equilibrium: N2O4(g) Initial Change Equil. Kp =
4.5 atm −x 4.5 − x
2 PNO 2
PN 2O 4
=
⇌
2 NO2(g)
→
0 +2x 2x
Kp = 0.25
( 2 x) 2 = 0.25, 4x2 = 1.125 − (0.25)x, 4x2 + (0.25)x − 1.125 = 0 4.5 − x
We carried extra significant figures in this expression (as will be typical when we solve an expression using the quadratic formula). Solving using the quadratic formula (Appendix 1 of text): − 0.25 [(0.25) 2 − 4(4)(−1.125)]1/ 2 − 0.25 4.25 , x = 0.50 (Other value is = 2(4) 8 negative.) PNO2 = 2x = 1.0 atm; PN 2O 4 = 4.5 − x = 4.0 atm
x=
b. The reaction must shift to reactants (shifts left) to reach equilibrium. N2O4(g) Initial Change Equil. Kp =
0 +x x
⇌
2 NO2(g)
9.0 atm −2x 9.0 − 2x
(9.0 − 2 x) 2 = 0.25, 4x2 − (36.25)x + 81 = 0 (carrying extra significant figures) x
CHAPTER 13
CHEMICAL EQUILIBRIUM
Solving: x =
609
− (−36.25) [(−36.25) 2 − 4(4)(81)]1 / 2 , x = 4.0 atm 2(4)
The other value, 5.1, is impossible. PN 2O 4 = x = 4.0 atm; PNO2 = 9.0 − 2x = 1.0 atm c. No, we get the same equilibrium position starting with either pure N2O4 or pure NO2 in stoichiometric amounts. 77.
a. The reaction must proceed to products to reach equilibrium because only reactants are present initially. Summarizing the problem in a table: 2 NOCl(g) Initial
Change Equil.
⇌
2 NO(g)
+
Cl2(g)
K = 1.6 × 10 −5
2.0 mol = 1.0 M 0 0 2 .0 L 2x mol/L of NOCl reacts to reach equilibrium −2x → +2x +x 1.0− 2x 2x x
K = 1.6 × 10 −5 =
[NO]2 [Cl 2 ] [NOCl]2
=
(2 x) 2 ( x) (1.0 − 2 x) 2
If we assume that 1.0 − 2x 1.0 (from the small size of K, we know that the product concentrations will be small), then: 1.6 × 10 −5 =
4x 3 , x = 1.6 × 10 −2 ; now we must check the assumption. 1 .0 2
1.0 − 2x = 1.0 − 2(0.016) = 0.97 = 1.0 (to proper significant figures) Our error is about 3%; that is, 2x is 3.2% of 1.0 M. Generally, if the error we introduce by making simplifying assumptions is less than 5%, we go no further; the assumption is said to be valid. We call this the 5% rule. Solving for the equilibrium concentrations: [NO] = 2x = 0.032 M; [Cl2] = x = 0.016 M; [NOCl] = 1.0 − 2x = 0.97 M 1.0 M Note: If we were to solve this cubic equation exactly (a longer process), we get x = 0.016. This is the exact same answer we determined by making a simplifying assumption. We saved time and energy. Whenever K is a very small value (K << 1), always make the assumption that x is small. If the assumption introduces an error of less than 5%, then the answer you calculated making the assumption will be considered the correct answer. b.
2 NOCl(g) Initial Change Equil.
⇌
2 NO(g)
+
Cl2(g)
1.0 M 1.0 M 0 2x mol/L of NOCl reacts to reach equilibrium −2x → +2x +x 1.0 − 2x 1.0 + 2x x
610
CHAPTER 13 (1.0 + 2 x) 2 ( x)
1.6 × 10 −5 =
(1.0 − 2 x) 2
=
CHEMICAL EQUILIBRIUM
(1.0) 2 ( x) (assuming 2x << 1.0) (1.0) 2
x = 1.6 × 10 −5 ; assumptions are great (2x is 3.2 × 10−3% of 1.0). [Cl2] = 1.6 × 10 −5 M and [NOCl] = [NO] = 1.0 M c.
⇌
2 NOCl(g)
2 NO(g)
+
Cl2(g)
Initial
2.0 M 0 1.0 M 2x mol/L of NOCl reacts to reach equilibrium Change −2x → +2x +x Equil. 2.0 − 2x 2x 1.0 + x (2 x) 2 (1.0 + x) 4x2 = 1.6 × 10 −5 = (assuming x << 1.0) 4. 0 ( 2.0 − 2 x) 2 Solving: x = 4.0 × 10 −3 ; assumptions good (x is 0.4% of 1.0 and 2x is 0.4% of 2.0). [Cl2] = 1.0 + x = 1.0 M; [NO] = 2(4.0 × 10 −3 ) = 8.0 × 10 −3 M; [NOCl] = 2.0 M 78.
2 CO2(g) Initial Change Equil.
⇌
2 CO(g)
+
O2(g)
K=
[CO ] 2 [O 2 ] [CO 2 ] 2
= 2.0 × 10−6
2.0 mol/5.0 L 0 0 2x mol/L of CO2 reacts to reach equilibrium −2x → +2x +x 0.40 − 2x 2x x
K = 2.0 × 10−6 = 2.0 × 10-6
[CO ] 2 [O 2 ] [CO 2 ] 2
=
(2 x) 2 ( x) ; assuming 2x << 0.40: (0.40 − 2 x) 2
4x3 4x 3 −6 , 2 .0 × 10 = , x = 4.3 × 10−3 M 0.16 (0.40) 2
Checking assumption:
2(4.3 10 −3 ) × 100 = 2.2%; assumption is valid by the 5% rule. 0.40
[CO2] = 0.40 − 2x = 0.40 − 2(4.3 × 10−3) = 0.39 M [CO] = 2x = 2(4.3 × 10−3) = 8.6 × 10−3 M; [O2] = x = 4.3 × 10−3 M 79.
COCl2(g) Initial Change Equil.
⇌
CO(g)
+ Cl2(g)
Kp =
1.0 atm 0 0 x atm of COCl2 reacts to reach equilibrium −x → +x +x 1.0 − x x x
PCO PCl2 PCOCl2
= 6.8 × 10 −9
CHAPTER 13
CHEMICAL EQUILIBRIUM
6.8 × 10 −9 =
PCO PCl2 PCOCl2
=
x2 x2 1 .0 1.0 − x
611 (Assuming 1.0 − x 1.0.)
x = 8.2 × 10 −5 atm; assumption is good (x is 8.2 × 10-3% of 1.0).
PCOCl 2 = 1.0 − x = 1.0 − 8.2 × 10 −5 = 1.0 atm; PCO = PCl 2 = x = 8.2 × 10 −5 atm 80.
4 NO2(g) Initial Change Equil.
+
O2(g)
⇌
2 N2O5(g)
K=
[N 2 O5 ]2 = 6.2 × 10‒6 4 [NO2 ] [O 2 ]
4.0 mol/2.0 L 4.0 mol/2.0 L 0 Let x mol/L of O2 react to reach equilibrium −4x −x → +2x 2.0 − 4x 2.0 − x 2x
K = 6.2 × 10‒6 =
4 x2 (2 x)2 (2.0 − 4x) 4 (2.0 − x) (2.0)4 (2.0)
Since K << 1, x should be a small value, so we assumed that 2.0 − 4x 2.0 and 2.0 − x 2.0. Solving: 6.2 × 10‒6
4 x2 , x2 = 8.0(6.2 × 10‒6), x = 0.0070 mol/L (2.0)4 (2.0)
Assumptions are good. 4x is 1.4% of 2.0 and x is 0.35% of 2.0. Answering the question: [N2O5] = 2x = 2(0.0070) = 0.014 mol/L 81.
This is a typical equilibrium problem except that the reaction contains a solid. Whenever solids and liquids are present, we basically ignore them in the equilibrium problem. NH4OCONH2(s)
⇌ 2 NH3(g) + CO2(g)
Kp = 2.9 × 10 −3
Initial
0 0 Let some NH4OCONH2 decomposes to produce 2x atm of NH3 and x atm of CO2. Change → +2x +x Equil. 2x x 2 Kp = 2.9 × 10 −3 = PNH PCO2 = (2x)2(x) = 4x3 3
1/ 3
2.9 10 −3 x= 4
= 9.0 × 10 −2 atm; PNH3 = 2x = 0.18 atm; PCO 2 = x = 9.0 × 10 −2 atm
Ptotal = PNH3 + PCO2 = 0.18 atm + 0.090 atm = 0.27 atm 82.
NH4Cl(s) ⇌ NH3(g) + HCl(g)
Kp = PNH3 × PHCl
For this system to reach equilibrium, some of the NH 4Cl(s) decomposes to form equal moles of NH3(g) and HCl(g) at equilibrium. Because moles of HCl produced = moles of NH3 produced, the partial pressures of each gas must be equal.
612
CHAPTER 13
CHEMICAL EQUILIBRIUM
At equilibrium: Ptotal = PNH3 + PHCl and PNH3 = PHCl Ptotal = 4.4 atm = 2PNH3 , 2.2 atm = PNH3 = PHCl; Kp = (2.2)(2.2) = 4.8 83.
C(s) + CO2(g) ⇌ 2 CO(g)
Kp =
2 PCO
PCO 2
= 2.00
Let x = equilibrium PCO and y = equilibrium PCO 2 , then Ptotal = x + y = 6.00 and
x2 = 2.00. y
We have two equations and two unknowns. Solving: y=
x2 x2 ; 6.00 = x + , 12.0 = 2.00x + x2, x2 + 2.00x − 12.0 = 0 2.00 2.00
x=
− 2.00 [(2.00) 2 − 4(1)(−12.0)]1 / 2 , x = 2.61 2(1)
PCO = x = 2.61 atm; PCO 2 = 6.00 − 2.61 = 3.39 atm
84.
NH4SH(s)
⇌
NH3(g) + H2S(g)
Kp = 0.10
Initial
0 0 Let some NH4SH decomposes to produce x atm of NH3 and x atm of H2S. Change → +x +x Equil. x x Kp = PNH3 PH2S , 0.10 = (x)(x) = x2, x = (0.10)1/2 = 0.32 atm At equilibrium, we have partial pressures of NH 3 and H2S equal to 0.32 atm. We need to calculate the quantity of NH4SH that must react to produce these pressures. First, determine the moles of NH3 (or H2S) that must be present to produce a 0.32 atm partial pressure.
n NH3 =
PNH3 V RT
0.12 mol NH3 ×
=
0.32 atm 10.0 L = 0.12 mol NH3 0.08206 L atm 300. K K mol
1 mol NH 4 HS 51.12 g NH 4 HS = 6.1 g NH4HS mol NH 3 mol NH 4 HS
At a minimum, there must be 6.1 g of NH 4HS present in order for this reaction to be at equilibrium.
Le Châtelier's Principle 85.
a. No effect; adding more of a pure solid or pure liquid has no effect on the equilibrium position.
CHAPTER 13
CHEMICAL EQUILIBRIUM
613
b. Shifts left; HF(g) will be removed by reaction with the glass. As HF(g) is removed, the reaction will shift left to produce more HF(g). c. Shifts right; as H2O(g) is removed, the reaction will shift right to produce more H 2O(g). 86.
a. The reaction will shift left to reestablish equilibrium. Adding more NH 3 will shift the equilibrium left to use up some of the added gaseous product. b. Adding more of this solid has no effect on the reaction. Because solids are not included in the equilibrium constant expression, the amount of a solid present does not affect the equilibrium position (it just has to be present). c. The reaction shifts right to reestablish equilibrium. For this balanced reaction, there are 2 moles of gaseous products versus 0 mole of gaseous reactants. When the volume of a reaction container increases, the reaction will shift to the side of the reaction that occupies a larger volume; this will be the side with more moles of gas (in this reaction, to the product side). d. The reaction will shift left to reestablish equilibrium. Heat is a reactant in this endothermic process. As the temperature decreases, heat is removed, so the reaction shifts left to produce more heat.
87.
a. No change because there are equal numbers of reactant and product moles of gas in the balanced equation. So the equilibrium is not affected by a volume change. b. The mole fraction of products will increase with an increase in volume. The reaction shifts right (to products) because there are more moles of product gas (4) than moles of reactant gas (2). c. The mole fraction of reactants will increase with an increase in volume. In the balanced equation, there are more moles of reactant gas (2) than product gas (1); solids are ignored. When the volume increases, the reaction shifts to the side with more mol of gas.
88.
When the volume of a reaction container is increased, the reaction itself will want to increase its own volume by shifting to the side of the reaction that contains the most molecules of gas. When the molecules of gas are equal on both sides of the reaction, then the reaction will remain at equilibrium no matter what happens to the volume of the container. a. Reaction shifts left (to reactants) because the reactants contain 4 molecules of gas compared with 2 molecules of gas on the product side. b. Reaction shifts right (to products) because there are more product molecules of gas (2) than reactant molecules (1). c. No change because there are equal reactant and product molecules of gas. d. Reaction shifts right. e. Reaction shifts right to produce more CO 2(g). One can ignore the solids and only concentrate on the gases because gases occupy a relatively huge volume compared with solids. We make the same assumption when liquids are present (only worry about the gas molecules).
614 89.
CHAPTER 13 a. Right
b. Right
CHEMICAL EQUILIBRIUM
c. No effect; He(g) is neither a reactant nor a product.
d. Left; because the reaction is exothermic, heat is a product: CO(g) + H2O(g) → H2(g) + CO2(g) + heat Increasing T will add heat. The equilibrium shifts to the left to use up the added heat. e. No effect; because the moles of gaseous reactants equals the moles of gaseous products (2 mol versus 2 mol), a change in volume will have no effect on the equilibrium. 90.
a. The moles of SO3 will increase because the reaction will shift left to use up the added O2(g). b. Increase; because there are fewer reactant gas molecules than product gas molecules, the reaction shifts left with a decrease in volume. c. No effect; the partial pressures of sulfur trioxide, sulfur dioxide, and oxygen are unchanged, so the reaction is still at equilibrium. d. Increase; heat + 2 SO3 ⇌ 2 SO2 + O2; decreasing T will remove heat, shifting this endothermic reaction to the left to add heat. e. Decrease
91.
a. Left
b. Right
c. Left
d. No effect; the reactant and product concentrations/partial pressures are unchanged. e. No effect; because there are equal numbers of product and reactant gas molecules, a change in volume has no effect on this equilibrium position. f.
92.
Right; a decrease in temperature will shift the equilibrium to the right because heat is a product in this reaction (as is true in all exothermic reactions).
a. Shift to left b. Shift to right; because the reaction is endothermic (heat is a reactant), an increase in temperature will shift the equilibrium to the right. c. No effect; the reactant and product concentrations/partial pressures are unchanged. d. Shift to right e. Shift to right; because there are more gaseous product molecules than gaseous reactant molecules, the equilibrium will shift right with an increase in volume.
93.
Only increasing the volume (d) will cause this reaction to shift left. There are 4 mol of reactant gases vs 3 mol of product gases (solids and liquids are ignored when determining moles of gaseous reactants and products). An increase in volume will shift this reaction to the side with
CHAPTER 13
CHEMICAL EQUILIBRIUM
615
more moles of gas; this is the reactant side. Adding H2O(g) and increasing the temperature of this endothermic reaction will shift the reaction right. Adding Ar(g) has no effect and adding more of the solid also has no effect on the equilibrium. Solids just need to be present for a reaction to get to equilibrium; the amount of excess solid is immaterial. 94.
Only answer a is a true statement. Raising the temperature of this endothermic reaction will shift the reaction right as well as increase the value of K. For the false statements, adding more solid will have no effect; solids just need to be present; the amount of excess solid present doesn’t matter to the equilibrium. There are 3 moles of reactant gases compared to 4 moles of product gases. A decrease in volume will shift the reaction left to the side will fewer moles of gas. Removing the product gas B2H6 will shift the reaction right. And adding more of the product gas O2 does shift the reaction left, but does not change the value of K. The value of K only changes when the temperature changes.
95.
An endothermic reaction, where heat is a reactant, will shift right to products with an increase in temperature. The amount of NH3(g) will increase as the reaction shifts right, so the smell of ammonia will increase.
96.
From the data, as temperature increases, the value of K decreases. This is consistent with an exothermic reaction. In an exothermic reaction, heat is a product, and an increase in temperature shifts the equilibrium to the reactant side (as well as lowering the value of K).
ChemWork Problems [O 3 ] 2
[O 3 ] 2
[O 2 ]
(0.062) 3
, 1.8 10 −7 = 3
97.
K=
98.
2 NF3(g) ⇌ N2(g) + 3 F2(g)
K=
, [O3] = 6.5 × 10−6 M
[N 2 ][F2 ]3 = 4.0 [NF3 ]2
The set of concentrations in choice e are not equilibrium concentrations. When these concentrations are plugged into the equilibrium expression, we get a value of 2.0. Because K =4.0, these cannot be equilibrium concentrations. All the other sets of concentrations when plugged into the equilibrium expression give a value of 4.0. 99.
O(g) + NO(g) ⇌ NO2(g) NO2(g) + O2(g) ⇌ NO(g) + O3(g) O2(g) + O(g) ⇌ O3(g)
100.
a.
K = 1/(6.8 × 10−49) = 1.5 × 1048 K = 1/(5.8 × 10−34) = 1.7 × 1033 K = (1.5 × 1048)(1.7 × 1033) = 2.6 × 1081
Na2O(s) ⇌ 2 Na(l) + 1/2 O2(g) 2 Na(l) + O2(g) ⇌ Na2O2(s) Na2O(s) + 1/2 O2(g) ⇌ Na2O2(s) −25
K = (K1)(1/K3) =
2 10 = 4 × 103 5 10 − 29
K1 1/K3 K = (K1)(1/K3)
616
CHAPTER 13 NaO(g) ⇌ Na(l) + 1/2 O2(g) Na2O(s) ⇌ 2 Na(l) + 1/2 O2(g) 2 Na(l) + O2(g) ⇌ Na2O2(s)
b.
c.
CHEMICAL EQUILIBRIUM
K2 K1 1/K3
NaO(g) + Na2O(s) ⇌ Na2O2(s) + Na(l)
K = K2(K1)(1/K3) = 8 × 10−2
2 NaO(g) ⇌ 2 Na(l) + O2(g) 2 Na(l) + O2(g) ⇌ Na2O2(s)
(K2)2 1/K3
2 NaO(g) ⇌ Na2O2(s) 101.
4 NH3(g)
+
3 O2(g)
K = (K 2 ) 2 (1/K 3 ) = 8 × 1018 ⇌
2 N2(g)
+
6 H2O(g)
Initial
2.0 mol/2.0 L 2.0 mol/2.0 L 0 0 Let 4x mol/L of NH3 react to reach equilibrium Change −4x −3x → +2x +6x Equil. 1.0 − 4x 1.0 − 3x 2x 6x To come up with an equilibrium concentration of 2x for N2, 4x of NH3 had to react. From the ICE Table, the equilibrium concentrations are: [NH3] = 1.0 – 4x, [O2] = 1.0 – 3x, [H2O] = 6x 102.
a. The reaction must proceed to products to reach equilibrium because only reactants are present initially. Completing the table: 2 NOCl(g) Initial
Change Equil.
⇌
2 NO(g)
+
Cl2(g)
K = 1.6 × 10 −5
5.2 mol = 2.6 M 0 0 2.0 L Let x mol/L of Cl2 be present at equilibrium (requires 2x mol/L of NOCl). −2x → +2x +x 2.6 − 2x 2x x
b. K = 1.6 × 10 −5 =
[NO]2 [Cl 2 ] [NOCl]2
=
(2 x) 2 ( x) (2.6 − 2 x) 2
If we assume that 2.6 − 2x 2.6 (from the small size of K, we know that the product concentrations will be small), then: 1.6 × 10 −5 =
4x 3 , x = 3.0 × 10 −2 M; assumption good (2x is 2.3% of 2.6) 2.6 2
Solving for the equilibrium concentrations: [NO] = 2x = 0.060 M; [Cl2] = x = 0.030 M; [NOCl] = 2.6 − 2x = 2.5 M 103.
5.63 g C5H6O3 ×
1 mol C 5 H 6 O 3 = 0.0493 mol C5H6O3 initially 114 .10 g
CHAPTER 13
CHEMICAL EQUILIBRIUM
Total moles of gas at equilibrium = ntotal =
C5H6O3(g) Initial Change Equil.
617
PtotalV 1.63 atm 2.50 L = = 0.105 mol 0.08206 L atm RT 473 K K mol
⇌ C2H6(g) + 3 CO(g)
0.0493 mol 0 0 Let x mol C5H6O3 react to reach equilibrium. −x → +x +3x 0.0493 − x x 3x
0.105 mol total = 0.0493 – x + x + 3x = 0.0493 + 3x, x = 0.0186 mol 0.0186 mol C 2 H 6 3(0.0186) mol CO 2.50 L 2.50 L [C 2 H 6 ][CO]3 K= = [C 5 H 6 O 3 ] (0.0493 − 0.0186) mol C 5 H 6 O 3 2.50 L
104.
a. N2(g) + O2(g) ⇌ 2 NO(g)
Kp = 1 × 10−31 =
3
= 6.74 × 10 −6
2 2 PNO PNO = (0.8)(0.2) PN 2 PO2
Solving: PNO = 1 × 10−16 atm In 1.0 cm3 of air: nNO =
PV (1 10 −16 atm)(1.0 10 −3 L) = 4 × 10−21 mol NO = RT 0.08206 L atm (298 K) K mol
4 10 −21 mol NO 6.02 10 23 molecules 2 10 3 molecules NO = mol NO cm3 cm3
b. There is more NO in the atmosphere than we would expect from the value of K. The answer must lie in the rates of the reaction. At 25°C, the rates of both reactions: N2 + O2 → 2 NO and 2 NO → N2 + O2 are so slow that they are essentially zero. Very strong bonds must be broken; the activation energy is very high. Therefore, the reaction essentially doesn’t occur at low temperatures. Nitric oxide, however, can be produced in high-energy or high- temperature environments because the production of NO is endothermic. In nature, some NO is produced by lightning, and the primary manmade source is automobiles. At these high temperatures, K will increase, and the rates of the reaction will also increase, resulting in a higher production of NO. Once the NO gets into a more normal temperature environment, it doesn’t go back to N2 and O2 because of the slow rate. 105.
a.
2 AsH3(g) Initial Equil.
⇌
2 As(s)
392.0 torr 392.0 − 2x
Using Dalton’s law of partial pressure:
+ 3 H2(g) 0 3x
618
CHAPTER 13
CHEMICAL EQUILIBRIUM
Ptotal = 488.0 torr = PAsH3 + PH 2 = 392.0 − 2x + 3x, x = 96.0 torr
PH 2 = 3x = 3(96.0) = 288 torr × b.
PAsH3 = 392.0 − 2(96.0) = 200.0 torr × Kp =
106.
1 atm = 0.379 atm 760 torr
(PH 2 )3 (PAsH3 )
= 2
(0.379 )3
= 0.786
(0.2632 ) 2
NH4Cl(s) ⇌ NH3(g) + HCl(g)
1 atm = 0.2632 atm 760 torr
Kp = PNH3 × PHCl
For this system to reach equilibrium, some of the NH 4Cl(s) decomposes to form equal moles of NH3(g) and HCl(g) at equilibrium. Because moles of HCl produced = moles of NH 3 produced, the partial pressures of each gas must be equal to each other. At equilibrium: PNH3 = PHCl = 2.9 atm; Kp = (2.9)(2.9) = 8.4 107.
NH3(g)
+
H2S(g)
⇌ NH4HS(s)
K = 400. =
1 [ NH 3 ][H 2S]
2.00 mol 2.00 mol 5.00 L 5.00 L x mol/L of NH3 reacts to reach equilibrium Change −x −x Equil. 0.400 − x 0.400 − x
Initial
1 1 K = 400. = , 0.400 − x = (0.400 − x )(0.400 − x ) 400 .
Moles NH4HS(s) produced = 5.00 L
1/ 2
= 0.0500, x = 0.350 M
0.350 mol NH 3 1 mol NH 4 HS = 1.75 mol L mol NH 3
Total moles NH4HS(s) = 2.00 mol initially + 1.75 mol produced = 3.75 mol total 3.75 mol NH4HS
51.12 g NH 4 HS = 192 g NH4HS mol NH 4 HS
[H2S]e = 0.400 M – x = 0.400 M – 0.350 M = 0.050 M H2S PH 2S =
108.
n H 2S RT V
=
n H 2S V
RT =
0.050 mol 0.08206 L atm 308 .2 K = 1.3 atm L K mol
Assuming 100.00 g naphthalene: 93.71 g C × 6.29 g H ×
1 mol C = 7.803 mol C 12.01 g
1 mol H = 6.24 mol H; 1.008 g
7.803 = 1.25 6.24
CHAPTER 13
CHEMICAL EQUILIBRIUM
619
Empirical formula = (C1.25H)×4 = C5H4; molar mass =
32.8 g = 128 g/mol 0.256 mol
Because the empirical mass (64.08 g/mol) is one-half of 128, the molecular formula is C10H8. C10H8(s) Initial Equil.
⇌
C10H8(g)
K = 4.29 × 10 −6 = [C10H8]
0 Let some C10H8(s) sublime to form x mol/L of C10H8(g) at equilibrium. x
K = 4.29 × 10 −6 = [C10H8] = x Mol C10H8 sublimed = 5.00 L × 4.29 × 10 −6 mol/L = 2.15 × 10 −5 mol C10H8 sublimed Mol C10H8 initially = 3.00 g × Percent C10H8 sublimed = 109.
1 mol C10 H 8 = 2.34 × 10 −2 mol C10H8 initially 128 .16 g
2.15 10 −5 mol × 100 = 0.0919% 2.34 10 − 2 mol
There is a little trick we can use to solve this problem without having to solve a quadratic equation. Because K is very large (K >> 1), the reaction will have mostly products at equilibrium. So we will let the reaction go to completion (with Fe3+ limiting), and then solve an equilibrium problem to determine the molarity of reactants present at equilibrium (see the following set- up). Fe3+(aq) Before Change After Change Equil.
+
SCN−(aq)
⇌
FeSCN2+(aq)
K = 1.1 × 103
0.020 M 0.10 M 0 Let 0.020 mol/L Fe3+ react completely (K is large; products dominate). −0.020 −0.020 → +0.020 React completely 0 0.08 0.020 New initial x mol/L FeSCN2+ reacts to reach equilibrium +x +x −x x 0.08 + x 0.020 − x
K = 1.1 × 103 =
0.020 − x 0.020 [FeSCN 2+ ] = 3+ − ( x)(0.08 + x) (0.08) x [Fe ][SCN ]
x = 2 × 10−4 M; x is 1% of 0.020. Assumptions are good by the 5% rule. x = [Fe3+] = 2 × 10−4 M; [SCN−] = 0.08 + 2 × 10−4 = 0.08 M [FeSCN2+] = 0.020 − 2 × 10−4 = 0.020 M Note: At equilibrium, we do indeed have mostly products present. Our assumption to first let the reaction go to completion is good.
620
110.
CHAPTER 13
a.
CHEMICAL EQUILIBRIUM
2.450 g PCl5 0.08206 L atm 600. K n PCl5 RT 208.22 g/mol K mol PPCl5 = = = 1.16 atm V 0.500 L
b.
⇌
PCl5(g)
PCl3(g)
+
Cl2(g)
Kp =
PPCl3 PCl 2 PPCl5
= 11.5
Initial
1.16 atm 0 0 x atm of PCl5 reacts to reach equilibrium Change −x → +x +x Equil. 1.16 − x x x Kp =
x2 = 11.5, x2 + (11.5)x − 13.3 = 0 1.16 − x
Using the quadratic formula: x = 1.06 atm
PPCl5 = 1.16 − 1.06 = 0.10 atm c.
PPCl3 = PCl 2 = 1.06 atm; PPCl5 = 0.10 atm Ptotal = PPCl5 + PPCl3 + PCl 2 = 0.10 + 1.06 + 1.06 = 2.22 atm
d. Percent dissociation = 111. Initial Change Equil.
1.06 x × 100 = × 100 = 91.4% 1.16 1.16
SO2Cl2(g)
⇌
Cl2(g)
P0 −x P0 − x
→
0 +x x
+
SO2(g) 0 +x x
P0 = initial pressure of SO2Cl2
Ptotal = 0.900 atm = P0 − x + x + x = P0 + x x × 100 = 12.5, P0 = (8.00)x P0
Solving: 0.900 = P0 + x = (9.00)x, x = 0.100 atm x = 0.100 atm = PCl 2 = PSO 2 ; P0 − x = 0.800 − 0.100 = 0.700 atm = PSO2Cl 2 Kp =
112.
K=
PCl 2 PSO 2 PSO 2Cl 2
=
(0.100) 2 = 1.43 × 10 −2 0.700
(0.400) 2 [HF]2 = = 320.; 0.200 mol F2/5.00 L = 0.0400 M F2 added [H 2 ][F2 ] (0.0500)(0 .0100)
After F2 has been added, the concentrations of species present are [HF] = 0.400 M, [H2] = [F2] = 0.0500 M. Q = (0.400)2/(0.0500)2 = 64.0; because Q < K, the reaction will shift right to reestablish equilibrium.
CHAPTER 13
CHEMICAL EQUILIBRIUM H2(g)
Initial
+
F2(g)
621
⇋
2 HF(g)
0.0500 M 0.0500 M 0.400 M x mol/L of F2 reacts to reach equilibrium −x −x → +2x 0.0500 − x 0.0500 − x 0.400 + 2x
Change Equil. K = 320. = 17.9 =
(0.400 + 2 x) 2 ; taking the square root of each side: (0.0500 − x) 2
0.400 + 2 x , 0.895 − (17.9)x = 0.400 + 2x, (19.9)x = 0.495, x = 0.0249 mol/L 0.0500 − x
[HF] = 0.400 + 2(0.0249) = 0.450 M; [H2] = [F2] = 0.0500 − 0.0249 = 0.0251 M 113.
Add N2(g): reaction shifts right to reestablish equilibrium. [N 2] will increase overall, [H2] will decrease and [NH3] will increase. Note that only some of the added N 2 will react to get back to equilibrium. Therefore, [N2] will show a net increase overall. Remove H2(g): reaction shifts left to reestablish equilibrium. [N2] will increase, [H2] will decrease overall, and [NH3] will decrease. Add NH3(g): reaction will shift left to reestablish equilibrium. [N2] and [H2] will increase, while [NH3] will also increase overall. Add Ne(g) at constant volume: no change for [N 2], [H2], or [NH3]. Increase temperature (add heat): because this is an exothermic reaction, heat is a product in this reaction. As temperature is increased, the reaction will shift left to reestablish equilibrium. [N2] and [H2] will increase, while [NH3] will decrease. Decrease volume: because there are more moles of gaseous reactants as compared to gaseous products, a decrease in volume will cause this reaction to shift right to reestablish equilibrium. [N2] and [H2] will decrease, while [NH3] will increase. Add a catalyst: no change for [N2], [H2], or [NH3].
114.
115.
The only changes that cause the value of K to change are temperature changes. For an endothermic reaction, an increase in temperature causes the reaction to shift right which will increase the value of K. So only option a will increase the value of K. Option b will cause a decrease in K and options c-g will not change the value of K. CoCl2(s) + 6 H2O(g) ⇌ CoCl2•6H2O(s); if rain is imminent, there would be a lot of water vapor in the air. Because water vapor is a reactant gas, the reaction would shift to the right and would take on the color of CoCl2•6H2O, which is pink.
116.
a. Doubling the volume will decrease all concentrations by a factor of one-half. 1
Q=
2
[FeSCN2+ ]eq
1 1 3+ − [Fe ]eq [SCN ]eq 2 2
= 2K, Q > K
The reaction will shift to the left to reestablish equilibrium.
622
CHAPTER 13
CHEMICAL EQUILIBRIUM
b. Adding Ag+ will remove SCN− through the formation of AgSCN(s). The reaction will shift to the left to produce more SCN-. c. Removing Fe3+ as Fe(OH)3(s) will shift the reaction to the left to produce more Fe3+. d. Reaction will shift to the right as Fe3+ is added. 117.
118.
H+ + OH− → H2O; sodium hydroxide (NaOH) will react with the H+ on the product side of the reaction. This effectively removes H + from the equilibrium, which will shift the reaction to the right to produce more H+ and CrO42−. Because more CrO42− is produced, the solution turns yellow. N2(g) + 3 H2(g) ⇌ 2 NH3(g) + heat a. This reaction is exothermic, so an increase in temperature will decrease the value of K (see Table 13.3 of text.) This has the effect of lowering the amount of NH 3(g) produced at equilibrium. The temperature increase, therefore, must be for kinetics reasons. As temperature increases, the reaction reaches equilibrium much faster. At low temperatures, this reaction is very slow, too slow to be of any use. b. As NH3(g) is removed, the reaction shifts right to produce more NH3(g). c. A catalyst has no effect on the equilibrium position. The purpose of a catalyst is to speed up a reaction so it reaches equilibrium more quickly. d. When the pressure of reactants and products is high, the reaction shifts to the side that has fewer gas molecules. Since the product side contains two molecules of gas as compared to four molecules of gas on the reactant side, then the reaction shifts right to products at high pressures of reactants and products.
119.
Assume the final reaction is 2A ⇌ B2 which has K = 0.01. To get to this final reaction, the previous reaction was multiplied by 2. So, multiply the final reaction by ½ to get the previous reaction. This gives A ⇌ ½ B2. When a reaction is multiplied through by some number n, the new K value is the previous K value raised to n. So, K for this reaction is (0.01)1/2 = 0.1. Now to get the initial reaction, we need to reverse this intermediate reaction. This gives ½ B 2 ⇌ A. When a reaction is reversed, the new K value is the inverse of the old K value. So, K for the initial reaction is 1/0.1 = 10.
120.
See the hint for Exercise 99. 2 C(g) ⇌ 2 A(g) + 2 B(g) 2 A(g) + D(g) ⇌ C(g) C(g) + D(g) ⇌ 2 B(g)
K1 = (1/3.50)2 = 8.16 × 10 −2 K2 = 7.10 K = K1 × K2 = 0.579
Kp = K(RT)n, n = 2 – (1 + 1) = 0; because n = 0, Kp = K = 0.579.
CHAPTER 13
CHEMICAL EQUILIBRIUM C(g)
+
Initial 1.50 atm Equil. 1.50 – x 0.579 = K =
D(g)
⇌
623
2 B(g)
1.50 atm 1.50 – x
0 2x
(2 x) 2 (2 x) 2 = (1.50 − x)(1.50 − x) (1.50 − x) 2
2x = (0.579)1/2 = 0.761, x = 0.413 atm 1.50 − x
PB (at equilibrium) = 2x = 2(0.413) = 0.826 atm Ptotal = PC + PD + PB = 2(1.50 – 0.413) + 0.826 = 3.00 atm PB = BPtotal, B =
121.
N2(g) Initial Change Equil.
PB 0.826 atm = = 0.275 Ptotal 3.00 atm
+
3 H2(g)
⇌
2 NH3(g)
Kp =
2 PNH 3
PN 2 PH3 2
1.00 atm 2.00 atm 0 x atm of N2 reacts to reach equilibrium −x −3x → +2x 1.00 − x 2.00 − 3x 2x
From the setup: Ptotal = 2.00 atm = PN2 + PH2 + PNH3 2.00 atm = (1.00 – x) + (2.00 – 3x) + 2x = 3.00 – 2x, x = 0.500
PH 2 = 2.00 − 3x = 2.00 – 3(0.500) = 0.50 atm Kp = 122.
(1.00) 2 (2 x) 2 [2(0.500 )]2 = = 16 = (1.00 − x)(2.00 − 3x) 3 (1.00 − 0.500 )[2.00 − 3(0.500 )]3 (0.50)(0.50) 3
CaCO3(s) ⇌ CaO(s) + CO2(g)
Kp = 1.16 = PCO 2
Some of the 20.0 g of CaCO3 will react to reach equilibrium. The amount that reacts is the quantity of CaCO3 required to produce a CO2 pressure of 1.16 atm (from the Kp expression).
n CO 2 =
PCO 2 V RT
=
1.16 atm 10.0 L = 0.132 mol CO2 0.08206 L atm 1073 K K mol
Mass CaCO3 reacted = 0.132 mol CO2 Mass percent of CaCO3 reacted =
1 mol CaCO 3 100.09 g = 13.2 g CaCO3 mol CO 2 mol CaCO 3
13 .2 g × 100 = 66.0% 20 .0 g
624
CHAPTER 13
CHEMICAL EQUILIBRIUM
peptide(aq) + H2O(l) ⇌ acid group(aq) + amine group(aq) K = 3.1 × 10 −5
123. Initial
Change Equil.
1.0 mol = 1.0 M 0 1.0 L x mol/L peptide reacts to reach equilibrium. −x → +x 1.0 – x x
0
+x x
Note: Because water is not included in the K expression, the amount of water present initially and the amount of water that reacts are not needed to solve this problem. K = 3.1 × 10−5 = x=
x( x) x2 (assuming 1.0 – x 1.0) , 3.1 10 −5 1.0 − x 1.0
3.1 10 −5 = 5.6 × 10−3 M; assumption good (0.56% error).
[peptide] = 1.0 – x = 1.0 – 5.6 × 10−3 = 1.0 M [acid group] = [amine group] = x = 5.6 × 10−3 M CH3OH(aq) ⇌ H2CO(aq) + H2(aq)
124.
K = 3.7 × 10−10 =
[H 2 CO][H 2 ] [CH 3 OH]
Initial
1.24 M 0 0 x mol/L CH3OH reacts to reach equilibrium Change −x → +x +x Equil. 1.24 – x x x 3.7 × 10−10 =
x( x) x2 (assuming x << 1.24) 1.24 − x 1.24
x = 2.1 × 10−5 M; assumption good (1.7 × 10−3% error). [H2CO] = [H2] = x = 2.1 × 10−5 M; [CH3OH] = 1.24 – 2.1 × 10−5 = 1.24 M As formaldehyde is removed from the equilibrium by forming some other substance, the equilibrium shifts right to produce more formaldehyde. Hence the concentration of methanol (a reactant) decreases as formaldehyde (a product) reacts to form formic acid.
Challenge Problems 125.
P0 (for O2) = n O2 RT / V = (6.400 g × 0.08206 × 684 K)/(32.00 g/mol × 2.50 L) = 4.49 atm 2 O2(g) → CO2(g) + 2 H2O(g) −2x → +x +2x
Change
CH4(g) + −x
Change
CH4(g) + 3/2 O2(g) → CO(g) + 2 H2O(g) −y −3/2 y → +y +2y
CHAPTER 13
CHEMICAL EQUILIBRIUM
625
Amount of O2 reacted = 4.49 atm − 0.326 atm = 4.16 atm O2 2x + 3/2 y = 4.16 atm O2 and 2x + 2y = 4.45 atm H2O Solving using simultaneous equations: 2x + 2y = 4.45 −2x − (3/2)y = −4.16 (0.50)y = 0.29, y = 0.58 atm = PCO 4.45 − 1.16 = 1.65 atm = PCO 2 2
2x + 2(0.58) = 4.45, x = 126.
4.72 g CH3OH
1 mol = 0.147 mol CH3OH initially 32.04 g
Rate 1 = Rate 2
Graham’s law of effusion:
RateH 2 RateCH 3OH
=
M CH 3OH M H2
M2 M1
32.04 = 3.987 2.016
=
The effused mixture has 33.0 times as much H 2 as CH3OH. When the effusion rate ratio is multiplied by the equilibrium mole ratio of H2 to CH3OH, the effused mixture will have 33.0 times as much H2 as CH3OH. Let n H 2 and n CH 3OH equal the equilibrium moles of H2 and CH3OH, respectively. 33.0 = 3.987
Initial Change Equil.
n H2 n CH 3OH
,
n H2
= 8.28
n CH 3OH
CH3OH(g)
⇌
CO(g) + 2 H2(g)
0.147 mol −x 0.147 − x
→
0 +x x
From the ICE table, 8.28 =
n H2 n CH 3OH
0 +2x 2x
=
2x 0.147 − x
Solving: x = 0.118 mol 2
0.118 mol 2(0.118 mol ) 2 1.00 L 1.00 L [CO][H 2 ] = 0.23 K= = (0.147 − 0.118) mol [CH 3OH] 1.00 L 127.
There is a little trick we can use to solve this problem to avoid solving a cubic equation. Because K for this reaction is very small (K << 1), the reaction will contain mostly reactants at equilibrium (the equilibrium position lies far to the left). We will let the products react to
626
CHAPTER 13
CHEMICAL EQUILIBRIUM
completion by the reverse reaction, and then we will solve the forward equilibrium problem to determine the equilibrium concentrations. Summarizing these steps in a table:
⇌
2 NOCl(g) Before Change After Change Equil.
2 NO(g) + Cl2(g)
K = 1.6 × 10−5
0 2.0 M 1.0 M Let 1.0 mol/L Cl2 react completely. (K is small, reactants dominate.) +2.0 −2.0 −1.0 React completely 2.0 0 0 New initial conditions 2x mol/L of NOCl reacts to reach equilibrium −2x → +2x +x 2.0 − 2x 2x x (2 x) 2 ( x) 4x3 (2.0 − 2 x) 2 2.0 2
K = 1.6 × 10−5 =
(assuming 2.0 − 2x 2.0)
x3 = 1.6 × 10−5, x = 2.5 × 10−2 M; assumption good by the 5% rule (2x is 2.5% of 2.0). [NOCl] = 2.0 − 0.050 = 1.95 M = 2.0 M; [NO] = 0.050 M; [Cl2] = 0.025 M Note: If we do not break this problem into two parts (a stoichiometric part and an equilibrium part), then we are faced with solving a cubic equation. The setup would be: 2 NOCl
⇌
2 NO
2.0 M −2y 2.0 − 2y
Initial Change Equil.
0 +2y 2y
1.6 × 10−5 =
(2.0 − 2 y ) 2 (1.0 − y ) (2 y ) 2
+
Cl2 1.0 M −y 1.0 − y
If we say that y is small to simplify the problem, then: 1.6 × 10−5 =
2.0 2 ; we get y = 250. This is impossible! 4y2
To solve this equation, we cannot make any simplifying assumptions; we have to solve the cubic equation exactly. 128.
a.
2 NO(g) Initial Change Equil.
+
Br2(g)
⇌
2 NOBr(g)
98.4 torr 41.3 torr 0 2x torr of NO reacts to reach equilibrium −2x −x → +2x 98.4 − 2x 41.3 − x 2x
Ptotal = PNO + PBr2 + PNOBr = (98.4 − 2x) + (41.3 − x) + 2x = 139.7 − x Ptotal = 110.5 = 139.7 − x, x = 29.2 torr; PNO = 98.4 − 2(29.2) = 40.0 torr = 0.0526 atm
PBr2 = 41.3 − 29.2 = 12.1 torr = 0.0159 atm; PNOBr = 2(29.2) = 58.4 torr = 0.0768 atm
CHAPTER 13
CHEMICAL EQUILIBRIUM
Kp =
627
2 PNOBr (0.0768 atm) 2 = = 134 2 PNO PBr2 (0.0526 atm) 2 (0.0159 atm)
b.
2 NO(g) Initial Change Equil.
+
Br2(g)
⇌
2 NOBr(g)
0.30 atm 0.30 atm 0 2x atm of NO reacts to reach equilibrium −2x −x → +2x 0.30 − 2x 0.30 − x 2x
This would yield a cubic equation, which can be difficult to solve unless you have a graphing calculator. Because Kp is pretty large, let’s approach equilibrium in two steps: Assume the reaction goes to completion, and then solve the back-equilibrium problem. 2 NO Before
+
Br2
⇌
2 NOBr
0.30 atm 0.30 atm 0 Let 0.30 atm NO react completely. −0.30 −0.15 → +0.30 React completely 0 0.15 0.30 New initial 2y atm of NOBr reacts to reach equilibrium +2y +y −2y 2y 0.15 + y 0.30 − 2y
Change After Change Equil.
K p = 134 =
(0.30 − 2 y ) 2 (0.30 − 2 y ) 2 , = 134 4y2 = 536y2 (0.15 + y ) (2 y ) 2 (0.15 + y )
(0.30) 2 536y2, then y = 0.034; assumptions are poor (y is 23% of 0.15). 0.15 Use 0.034 as an approximation for y, and solve by successive approximations (see Appendix 1 in the text):
If y << 0.15:
(0.30 − 0.068) 2 = 536y2, y = 0.023; 0.15 + 0.034
(0.30 − 0.046 ) 2 = 536y2, y = 0.026 0.15 + 0.023
(0.30 − 0.052 ) 2 = 536y2, y = 0.026 atm (consistent answer) 0.15 + 0.026
So: PNO = 2y = 0.052 atm; PBr2 = 0.15 + y = 0.18 atm; PNOBr = 0.30 − 2y = 0.25 atm 129.
N2(g) Initial Change Equil.
+
3 H2(g) ⇌ 2 NH3(g)
Kp = 5.3 × 105
0 0 P0 P0 = initial pressure of NH3 2x atm of NH3 reacts to reach equilibrium +x +3x −2x x 3x P0 − 2x
From problem, P0 − 2x =
P0 , so P0 = (4.00)x 2.00
628
CHAPTER 13 Kp =
CHEMICAL EQUILIBRIUM
4.00 [(4.00) x − 2 x]2 [(2.00) x]2 (4.00) x 2 = = 5.3 × 105, x = 5.3 × 10−4 atm = = 3 3 4 2 ( x)(3x) ( x)(3x) 27 x 27 x
P0 = (4.00)x = 4.00(5.3 × 10−4 atm) = 2.1 × 10−3 atm 130.
P4(g) ⇌ 2 P2(g) Kp = 0.100 =
PP22 PP4
; PP4 + PP2 = Ptotal = 1.00 atm, PP4 = 1.00 atm − PP2
Let y = PP2 at equilibrium, then Kp =
y2 = 0.100 1.00 − y
Solving: y = 0.270 atm = PP ; PP = 1.00 − 0.270 = 0.73 atm 2
4
To solve for the fraction dissociated, we need the initial pressure of P4 (mol ∝ pressure). P4(g) Initial Change Equil.
⇌
2 P2(g)
P0 0 P0 = initial pressure of P4 in atm. x atm of P4 reacts to reach equilibrium −x → +2x P0 − x 2x
Ptotal = P0 − x + 2x = 1.00 atm = P0 + x Solving: 0.270 atm = PP2 = 2x, x = 0.135 atm; P0 = 1.00 − 0.135 = 0.87 atm Fraction dissociated =
131.
x 0.135 = = 0.16, or 16% of P4 is dissociated to reach equilibrium. P0 0.87
N2O4(g) ⇌ 2 NO2(g)
Kp =
2 PNO 2
PN 2O 4
=
(1.20) 2 = 4.2 0.34
Doubling the volume decreases each partial pressure by a factor of 2 (P = nRT/V). PNO 2 = 0.600 atm and PN O = 0.17 atm are the new partial pressures. 2
4
2
Q=
(0.600 ) = 2.1, so Q < K; equilibrium will shift to the right. 0.17
N2O4(g) Initial Change Equil. Kp = 4.2 =
⇌ 2 NO2(g)
0.17 atm 0.600 atm x atm of N2O4 reacts to reach equilibrium −x → +2x 0.17 − x 0.600 + 2x (0.600 + 2 x) 2 , 4x2 + (6.6)x − 0.354 = 0 (carrying extra significant figures) (0.17 − x)
Solving using the quadratic formula: x = 0.052 PNO 2 = 0.600 + 2(0.052) = 0.704 atm; PN O = 0.17 − 0.052 = 0.12 atm 2 4
CHAPTER 13
CHEMICAL EQUILIBRIUM
132.
2 NaHCO3(s)
a.
⇌ Na2CO3(s)
629 + CO2(g) + H2O(g)
Kp = 0.25
Initial
0 0 Let some NaHCO3(s) decompose to form x atm each of CO2(g) and H2O(g) at equilibrium. Change → +x +x Equil. x x Kp = 0.25 = PCO PH O , 0.25 = x2, x = PCO2 = PH2O = 0.50 atm 2
b.
n CO 2 =
PCO 2 V RT
=
2
(0.50 atm) (1.00 L) = 1.5 × 10 −2 mol CO2 (0.08206 L atm/K • mol) (398 K)
Mass of Na2CO3 produced: 1.5 × 10 −2 mol CO2 ×
1 mol Na 2 CO 3 105.99 g Na 2 CO 3 = 1.6 g Na2CO3 mol CO 2 mol Na 2 CO 3
Mass of NaHCO3 reacted: 1.5 × 10 −2 mol CO2 ×
2 mol NaHCO 3 84.01 g NaHCO 3 = 2.5 g NaHCO3 1 mol CO 2 mol
Mass of NaHCO3 remaining = 10.0 − 2.5 = 7.5 g c. 10.0 g NaHCO3 ×
1 mol NaHCO 3 1 mol CO 2 = 5.95 × 10 −2 mol CO2 84.01 g NaHCO 3 2 mol NaHCO 3
When all the NaHCO3 has just been consumed, we will have 5.95 × 10 −2 mol CO2 gas at a pressure of 0.50 atm (from a). V=
133.
(5.95 10 −2 mol)(0.082 06 L atm/K • mol) (398 K) nRT = 3.9 L = P (0.50 atm)
a. N2(g) + 3 H2(g) ⇌ 2 NH3(g); because the temperature is constant, the value of K will be the same for both container volumes. Since we now the volume in the final mixture, let’s calculate K using this mixture. In this final mixture, 2 N2 molecules, 2 H2 molecules, and 6 NH3 molecules are present in a 1.00 L container. Using units of molecules/L for concentrations:
K=
[NH 3 ] 2 [N 2 ][H 2 ]
= 3
6 NH 3 molecules 1.00 L
2
2 N 2 molecules 2 H 2 molecules 1.00 L 1.00 L
= 2.25 3
L2 molecules 2
For the K value in typical mol/L units for the concentrations: 2
6.022 10 23 molecules L2 L2 = 8.16 10 47 K = 2.25 mol molecules 2 mol 2
630
CHAPTER 13
CHEMICAL EQUILIBRIUM
b. Because temperature is constant, the initial mixture at the larger volume must also have K = 2.25 L2/molecules2. In the initial mixture, there are 2 NH3 molecules, 4 N2 molecules, and 8 H2 molecules in some unknown volume, V. 2 NH 3 molecules V
K = 2.25 =
V= 134.
2
4 N 2 molecules 8 H 2 molecules V V
= 3
4V 2 V2 = 4(512) 512
2.25(512 ) = 33.9 L; the volume of the initial container would be 33.9 L.
a. If the volume is increased, equilibrium will shift to the right, so the mole percent of N 2O5 decomposed will be greater than 0.50%. b.
2 N2O5(g) Initial Change Equil. Kp =
⇌
4 NO2(g)
1.000 atm −0.0050 → 0.995
+
0 +0.010 0.010
O2(g) 0 +0.0025 0.0025
(0.010 ) 4 (0.0025 ) = 2.5 × 10−11 (0.995 ) 2
The new volume is 10.0 times the old volume. Therefore, the initial partial pressure of N2O5 will decrease by a factor of 10.0.
PN 2O5 = 1.00 atm
1.00 = 0.100 atm 10 .0
2 N2O5 Initial Change Equil.
⇌ 4 NO2
0.100 atm −2x → 0.100 − 2x
2.5 × 10−11 =
0 +4x 4x
+
O2 0 +x x
( 4 x) 4 ( x) (4 x) 4 ( x) , 2x = 2.0 × 10−3 atm = PN 2O5 decomposed (0.100 − 2 x) 2 (0.100 ) 2
2.0 10 −3 × 100 = 2.0% N2O5 decomposed (moles and P are directly related) 0.100
135. Initial Change Equil.
SO3(g)
⇌
SO2(g) + 1/2 O2(g)
P0 −x P0 − x
→
0 +x x
0 +x/2 x/2
P0 = initial pressure of SO3
Average molar mass of the mixture is: molar mass =
(1.60 g/L) (0.08206 L atm/K • mol ) (873 K) dRT = = 63.7 g/mol P 1.80 atm
CHAPTER 13
CHEMICAL EQUILIBRIUM
631
The average molar mass is determined by: average molar mass =
n SO 3 (80.07 g/mol) + n SO 2 (64.07 g/mol) + n O 2 (32.00 g/mol) n total
Because χA = mol fraction of component A = nA/ntotal = PA/Ptotal: 63.7 g/mol = =
PSO 3 (80.07 ) + PSO 2 (64.07 ) + PO 2 (32.00) Ptotal
Ptotal = P0 − x + x + x/2 = P0 + x/2 = 1.80 atm, P0 = 1.80 – x/2 (P0 − x) (80.07 ) + x (64.07 ) +
63.7 =
x (32.00) 2
1.80
(1.80 − 3/2 x) (80.07) + x (64.07) +
63.7 =
x (32.00) 2
1.80
115 = 144 − (120.1)x + (64.07)x + (16.00)x, (40.0)x = 29, x = 0.73 atm
PSO3 = P0 − x = 1.80 − (3/2)x = 0.71 atm; PSO 2 = 0.73 atm; PO 2 = x/2 = 0.37 atm Kp =
136.
PSO 2 PO1/22 PSO 3
=
(0.73) (0.37 )1/ 2 = 0.63 (0.71)
The first reaction produces equal amounts of SO3 and SO2. Using the second reaction, calculate the SO3, SO2 and O2 partial pressures at equilibrium. SO3(g) Initial P0 Change −x Equil. P0 − x
⇌
SO2(g)
→
P0 +x P0 + x
+
1/2 O2(g) 0 +x/2 x/2
P0 = initial pressure of SO3 and SO2 after first reaction occurs.
Ptotal = P0 − x + P0 + x + x/2 = 2P0 + x/2 = 0.836 atm
PO 2 = x/2 = 0.0275 atm, x = 0.0550 atm 2P0 + x/2 = 0.836 atm; 2P0 = 0.836 − 0.0275 = 0.809 atm, P0 = 0.405 atm
PSO3 = P0 − x = 0.405 − 0.0550 = 0.350 atm; PSO 2 = P0 + x = 0.405 + 0.0550 = 0.460 atm For the reaction 2 FeSO4(s) ⇌ Fe2O3(s) + SO3(g) + SO2(g): Kp = PSO 2 PSO3 = (0.460)(0.350) = 0.161
632
CHAPTER 13
CHEMICAL EQUILIBRIUM
For the reaction SO3(g) ⇌ SO2(g) + 1/2 O2(g): Kp =
137.
PSO 2 PO1/22 PSO 3
=
(0.460) (0.0275) 1/2 = 0.218 0.350
When exactly 100 O2 molecules are initially present at 5000 K and 1.000 atm: O2(g) Initial Change Equil.
⇌
2 O(g)
100 −83 17
0 +166 166
Mole fraction O = χO =
166 = 0.9071 and χ O2 = 0.0929; PO 2 = χ O 2 Ptotal and PO = χO Ptotal 183
Because initially Ptotal = 1.000 atm, PO 2 = 0.0929 atm and PO = 0.9071 atm. Kp =
PO2 (0.9071) 2 = = 8.86 atm 0.0929 PO 2
At 95.0% O2 dissociated, let x = initial partial pressure of O2 and y = amount (atm) of O2 that dissociates to reach equilibrium. O2 ⇌ 2 O x 0 −y → +2y x−y 2y
Initial Change Equil.
y (2 y ) 2 = 8.86; × 100 = 95.0; we have two equations and two unknowns. x x− y
Solving: x = 0.123 atm and y = 0.117 atm; Ptotal = (x − y) + 2y = 0.240 atm 138.
a.
⇌
N2O4(g) x −(0.16)x (0.84)x
Initial Change Equil.
2 NO2(g) 0 → +(0.32)x (0.32)x (0.42) 2 = 0.16 1.1
(0.84)x + (0.32)x = 1.5 atm, x = 1.3 atm; Kp = b.
N2O4 Equil.
x
⇌ 2 NO2 ; x + y = 1.0 atm; y
2
x
= Kp = 0.16
y
We have 2 equations and 2 unknowns. Solving: x = 0.67 atm (= PN 2O 4 ) and y = 0.33 atm (= PNO2 )
CHAPTER 13
CHEMICAL EQUILIBRIUM
c.
N2O4 Initial Change Equil.
633
⇌ 2 NO2
P0 −x → 0.67 atm
0 +2x 0.33 atm
P0 = initial pressure of N2O4
2x = 0.33, x = 0.165 (using extra sig. figs.) P0 − x = 0.67, P0 = 0.67 + 0.165 = 0.84 atm; 139.
0.165 × 100 = 20.% dissociated 0.84
a. Because the density (mass/volume) decreased while the mass remained constant (mass is conserved in a chemical reaction), the volume must have increased as reactants were converted to products. The volume increased because the number of moles of gas increased (V n at constant T and P).
V n Density (initial ) 4.495 g/L = = 1.100 = equil. = equil. Density (equil.) 4.086 g/L Vinitial n initial Assuming an initial volume of 1.000 L: 4.495 g NOBr
Initial Change Equil.
n equil. n initial
=
1 mol NOBr = 0.04090 mol NOBr initially 109 .91 g
2 NOBr(g)
⇌
2 NO(g) + Br2(g)
0.04090 mol −2x 0.04090 − 2x
→
0 +2x 2x
0 +x x
0.04090 − 2 x + 2 x + x = 1.100; solving: x = 0.00409 mol 0.04090
If the initial volume is 1.000 L, then the equilibrium volume will be 1.110(1.000 L) = 1.110 L. Solving for the equilibrium concentrations: [NOBr] = [Br2] = K=
0.03272 mol 0.00818 mol = 0.02975 M ; [NO] = = 0.00744 M 1.100 L 1.100 L
0.00409 mol = 0.00372 M 1.100 L
(0.00744 ) 2 (0.00372 ) = 2.33 10−4 (0.02975 ) 2
b. The argon gas will increase the volume of the container. This is because the container is a constant-pressure system, and if the number of moles increases at constant T and P, the volume must increase. An increase in volume will dilute the concentrations of all gaseous reactants and gaseous products. Because there are more moles of product gases versus reactant gases (3 mol versus 2 mol), the dilution will decrease the numerator of K more than the denominator will decrease. This causes Q < K and the reaction shifts right to get back to equilibrium. Because temperature was unchanged, the value of K will not change. K is a constant as long as temperature is constant.
634
CHAPTER 13
140.
CCl4(g) Initial P0 Change −x Equil. P0− x
⇌
C(s)
+ 2 Cl2(g) 0 +2x 2x
→
CHEMICAL EQUILIBRIUM
Kp = 0.76 P0 = initial pressure of CCl4
Ptotal = P0 − x + 2x = P0 + x = 1.20 atm 4 x 2 + (0.76) x (2 x) 2 = 0.76, 4x2 = (0.76)P0 − (0.76)x, P0 = 0.76 P0 − x
Kp =
Substituting into P0 + x = 1.20: 4x2 + x + x = 1.20 atm, (5.3)x2 + 2x − 1.20 = 0; solving using the quadratic formula: 0.76 x =
141.
− 2 (4 + 25.4)1/ 2 = 0.32 atm; P0 + 0.32 = 1.20, P0 = 0.88 atm 2(5.3)
Initial moles VCl4 = 6.6834 g VCl4 × 1 mol VCl4/192.74 g VCl4 = 3.4676 × 10−2 mol VCl4 T 5.97 o C = Total molality of solute particles = im = = 0.200 mol/kg Kf 29.8 o C kg/mol
Because we have 0.1000 kg CCl4, the total moles of solute particles present is: 0.200 mol/kg(0.1000 kg) = 0.0200 mol 2 VCl4 Initial Equil.
⇌
V2Cl8
K=
[V2 Cl 8 ] [VCl 4 ]2
3.4676 × 10−2 mol 0 2x mol VCl4 reacts to reach equilibrium 3.4676 × 10−2 − 2x x
Total moles solute particles = 0.0200 mol = mol VCl4 + mol V2Cl8 = 3.4676 × 10−2 − 2x + x 0.0200 = 3.4676 × 10−2 − x, x = 0.0147 mol At equilibrium, we have 0.0147 mol V 2Cl8 and 0.0200 − 0.0147 = 0.0053 mol VCl4. To determine the equilibrium constant, we need the total volume of solution in order to calculate equilibrium concentrations. The total mass of solution is 100.0 g + 6.6834 g = 106.7 g. Total volume = 106.7 g × 1 cm3/1.696 g = 62.91 cm3 = 0.06291 L The equilibrium concentrations are: [V2Cl8] = K =
0.0147 mol 0.0053 mol = 0.234 mol/L; [VCl4] = = 0.084 mol/L 0.06291 L 0.06291 L
[V2 Cl 8 ] 0.234 = = 33 2 [VCl 4 ] (0.084 ) 2
CHAPTER 13
CHEMICAL EQUILIBRIUM
635
Marathon Problem 142.
2.00 g = 0.0121 mol XY (initially) 165 g/mol
(0.350)(0.0121 mol) = 4.24 10−3 mol XY dissociated XY(g) Initial Change Equil.
⇌
0.0121 mol −0.00424 → 0.0079 mol
X(g)
+
Y(g)
0 +0.00424 0.00424 mol
0 +0.00424 0.00424 mol
Total moles of gas = 0.0079 + 0.00424 + 0.00424 = 0.0164 mol For an ideal gas at constant P and T, V ∝ n. So:
Vinitial =
Vfinal n 0.0164 mol = final = = 1.36 Vinitial n initial 0.0121 mol
nRT (0.0121 mol) ( 0.008206 L atm/K • mol)(298 K) = = 0.306 L P 0.967 atm
Vfinal = 0.306 L(1.36) = 0.416 L Because mass is conserved in a chemical reaction: density (final) =
mass 2.00 g = = 4.81 g/L volume 0.416 L
0.00424 mol 0.416 L [X][Y] K= = [XY] 0.0079 mol 0.416 L
2
= 5.5 10 −3
CHAPTER 14 ACIDS AND BASES Review Questions 1.
a. Arrhenius acid: produce H+ in water b. Brønsted-Lowry acid: proton (H+) donor c. Lewis acid: electron pair acceptor The Lewis definition is most general. The Lewis definition can apply to all Arrhenius and Brønsted-Lowry acids; H+ has an empty 1s orbital and forms bonds to all bases by accepting a pair of electrons from the base. In addition, the Lewis definition incorporates other reactions not typically considered acid-base reactions, e.g., BF3(g) + NH3(g) → F3B−NH3(s). NH3 is something we usually consider a base and it is a base in this reaction using the Lewis definition; NH3 donates a pair of electrons to form the N−B bond.
2.
a. The Ka reaction always refers to an acid reacting with water to produce the conjugate base of the acid and the hydronium ion (H3O+). For a general weak acid HA, the Ka reaction is: HA(aq) + H2O(l) ⇌ A−(aq) + H3O+(aq) where A− = conjugate base of the acid HA This reaction is often abbreviated as: HA(aq) ⇌ H+(aq) + A−(aq) b. The Ka equilibrium constant is the equilibrium constant for the Ka reaction of some substance. For the general Ka reaction, the Ka expression is: Ka =
[A − ][H 3 O + ] [H + ][A − ] or K a = [HA] [HA]
(for the abbreviated Ka reaction)
c. The Kb reaction alwlays refers to a base reacting with water to produce the conjugate acid of the base and the hydroxide ion (OH−). For a general base, B, the Kb reaction is: B(aq) + H2O(l) ⇌ BH+(aq) + OH−(aq) where BH+ = conjugate acid of the base B
d. The Kb equilibrium constant for the general Kb reaction is: Kb =
[BH + ][OH − ] [B]
e. A conjugate acid-base pair consists of two substances related to each other by the donating and accepting of a single proton (H +). The conjugate bases of the acids HCl, HNO2, HC2H3O2, and H2SO4 are Cl−, NO2−, C2H3O2−, and HSO4−, respectively. The conjugate acids of the bases NH3, C5H5N, and HONH2 are NH4+, C5H5NH+, and HONH3+, respectively. Conjugate acid-base pairs only differ by H+ in their respective formulas.
636
CHAPTER 14 3.
ACIDS AND BASES
637
a. Amphoteric: A substance that can behave either as an acid or as a base. b. The Kw reaction is also called the autoionization of water reaction. The reaction always occurs when water is present as the solvent. The reaction is: H2O(l) + H2O(l) ⇌ H3O+(aq) + OH−(aq) or H2O(l) ⇌ H+(aq) + OH−(aq) c. The Kw equilibrium constant is also called the ion-product constant or the dissociation constant of water. It is the equilibrium constant for the autoionization reaction of water: Kw = [H3O+][OH−] or Kw = [H+][OH−] At typical solution temperatures of 25˚C, Kw = 1.0 × 10 −14. d. pH is a mathematical term which is equal to the –log of the H+ concentration of a solution (pH = –log[H+]). e. pOH is a mathematical term which is equal to the –log of the OH− concentration of a solution (pOH = –log[OH−]). f.
The p of any quantity is the –log of that quantity. So: pKw = –logKw. At 25˚C, pKw = –log(1.0 × 10 −14 ) = 14.00.
Neutral solution at 25˚C: Kw = 1.0 × 10 −14 = [H+][OH−] and pH + pOH = 14.00 [H+] = [OH− ] = 1.0 × 10 −7 M; pH = pOH = –log(1.0 × 10 −7 ) = 7.00 Acidic solution at 25˚C: [H+] > [OH− ]; [H+] > 1.0 × 10 −7 M; [OH− ] < 1.0 × 10 −7 M; pH < 7.00; pOH > 7.00 Basic solution at 25˚C: [OH−] > [H+]; [OH−] > 1.0 × 10 −7 M; [H+] < 1.0 × 10 −7 M; pOH < 7.00; pH > 7.00 As a solution becomes more acidic, [H+] increases, so [OH−] decreases, pH decreases, and pOH increases. As a solution becomes more basic, [OH −] increases, so [H+] decreases, pH increases, and pOH decreases. 4.
The Ka value refers to the reaction of an acid reacting with water to produce the conjugate base and H3O+. The stronger the acid, the more conjugate base and H 3O+ produced, and the larger the Ka value. Strong acids are basically 100% dissociated in water. Therefore, the strong acids have a Ka >> 1 because the equilibrium position lies far to the right. The conjugate bases of strong acids are terrible bases; much worse than water, so we can ignore their basic properties in water.
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Weak acids are only partially dissociated in water. We say that the equilibrium lies far to the left, thus giving values for Ka < 1 (weak acids have mostly reactants at equilibrium and few products present). The conjugate bases of weak acids are better bases than water. When we have a solution composed of just the conjugate base of a weak acid in water, the resulting pH is indeed basic (pH > 7.0). In general, as the acid strength increases, the conjugate base strength decreases, or as acid strength decreases, the conjugate base strength increases. They are inversely related. Base strength is directly related to the Kb value. The larger the Kb value, the more OH− produced from the Kb reaction, and the more basic the solution (the higher the pH). Weak bases have a Kb < 1 and their conjugate acids behave as weak acids in solution. As the strength of the base increases, the strength of the conjugate acid gets weaker. Another way of saying this is the weaker the base, the stronger the conjugate acid. 5.
Strong acids are assumed 100% dissociated in water, and we assume that the amount of H+ donated by water is negligible. Hence, the equilibrium [H +] of a strong acid is generally equal to the initial acid concentration ([HA]0). Note that solutions of H2SO4 can be different from this because H2SO4 is a diprotic acid. Also, when you have very dilute solutions of a strong acid, the H+ contribution from water by itself must be considered. The strong acids to memorize are HCl, HBr, HI, HNO3, HClO4, and H2SO4. Ka values for weak acids are listed in Table 14.2 and in Appendix 5 of the text. Because weak acids only partially dissociate in water, we must solve an equilibrium problem to determine how much H+ is added to water by the weak acid. We write down the Ka reaction, set-up the ICE table, then solve the equilibrium problem. The two assumptions generally made are that acids are less than 5% dissociated in water and that the H + contribution from water is negligible. The 5% rule comes from the assumptions that weak acids are less than 5% dissociated. When this is true, the mathematics of the problem are made much easier. The equilibrium expression we get for weak acids in water generally has the form (assuming an initial acid concentration of 0.10 M): Ka =
x2 x2 0.10 − x 0.10
The 5% rule refers to assuming 0.10 – x 0.10. The assumption is valid if x is less than 5% of the number the assumption was made against ([HA]0). When the 5% rule is valid, solving for x is very straight forward. When the 5% rule fails, we must solve the mathematical expression exactly using the quadratic equation (or your graphing calculator). Even if you do have a graphing calculator, reference Appendix 1 to review the quadratic equation. Appendix 1 also discusses the method of successive approximations which can also be used to solve quadratic (and cubic) equations. 6.
Strong bases are soluble ionic compounds containing the OH − anion. Strong bases increase the OH− concentration in water by just dissolving. Thus, for strong bases like LiOH, NaOH, KOH, RbOH, and CsOH, the initial concentration of the strong base equals the equilibrium [OH −] of water.
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The other strong bases to memorize have +2 charged metal cations. The soluble ones to know are Ca(OH)2, Sr(OH)2, and Ba(OH)2. These are slightly more difficult to solve because they donate 2 moles OH− for every mole of salt dissolved. Here, the [OH −] is equal to two times the initial concentration of the soluble alkaline earth hydroxide salt dissolved. Neutrally charged organic compounds containing at least one nitrogen atom generally behave as weak bases. The nitrogen atom has an unshared pair of electrons around it. This lone pair of electrons is used to form a bond to H+. Weak bases only partially react with water to produce OH −. To determine the amount of OH− produced by the weak base (and, in turn, the pH of the solution), we set-up the ICE table using the Kb reaction of the weak base. The typical weak base equilibrium expression is: Kb =
x2 x2 0.25 − x 0.25
(assuming [B]0 = 0.25 M)
Solving for x gives us the [OH−] in solution. We generally assume that weak bases are less than 5% reacted with water and that the OH − contribution from water is negligible. The 5% assumption makes the math easier. By assuming an expression like 0.25 M – x 0.25 M, the calculation is very straight forward. The 5% rule applied here is that if (x/0.25) × 100 is less than 5%, the assumption is valid. When the assumption is not valid, then we solve the equilibrium expression exactly using the quadratic equation (or by the method of successive approximations). 7.
Monoprotic acid: An acid with one acidic proton; the general formula for monoprotic acids is HA. Diprotic acid: An acid with two acidic protons (H2A). Triprotic acid: An acid with three acidic protons (H 3A). H2SO4(aq) → HSO4−(aq) + H+(aq)
K a1 >> 1; this is a strong acid.
HSO4−(aq) ⇌ SO42−(aq) + H+(aq)
K a 2 = 0.012; this is a weak acid.
When H2SO4 is dissolved in water, the first proton is assumed 100% dissociated because H 2SO4 is a strong acid. After H2SO4 dissociates, we have H+ and HSO4− present. HSO4− is a weak acid and can donate some more protons to water. To determine the amount of H+ donated by HSO4−, one must solve an equilibrium problem using the K a 2 reaction for HSO4−. H3PO4(aq) ⇌ H+(aq) + H2PO4−(aq)
K a1 = 7.5 × 10 −3
H2PO4−(aq) ⇌ H+(aq) + HPO42−(aq)
K a 2 = 6.2 × 10 −8
HPO42−(aq) ⇌ H+(aq) + PO43−(aq)
K a 3 = 4.8 × 10 −13
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When H3PO4 is added to water, the three acids that are present are H 3PO4, H2PO4−, and HPO42−. H3PO4, with the largest Ka value, is the strongest of these weak acids. The conjugate bases of the three acids are H2PO4−, HPO42−, and PO43−. Because HPO42− is the weakest acid (smallest Ka value), its conjugate base (PO43−) will have the largest Kb value and is the strongest base. See Examples 14.15-14.17 of the text on the strategies used to solve for the pH of polyprotic acids. The strategy to solve most polyprotic acid solutions is covered in Example 14.15. For typical polyprotic acids, K a1 >> K a 2 (and K a 3 if a triprotic acid). Because of this, the dominant producer of H+ in solution is just the K a1 reaction. We set-up the equilibrium problem using the K a1 reaction and solve for H+. We then assume that the H+ donated by the K a 2 (and K a 3 if triprotic) reaction is negligible, that is, the H+ donated by the K a1 reaction is assumed to be the H+ donated by the entire acid system. This assumption is great when K a1 >> K a 2 (roughly a 1000-fold difference in magnitude). Examples 14.16 and 14.17 cover strategies for the other type of polyprotic acid problems. This other type is solutions of H2SO4. As discussed previously, H2SO4 problems are both a strong acid and a weak acid problem in one. To solve for the [H +], we sometimes must worry about the H+ contribution from HSO4−. Example 14.16 is an example of an H2SO4 solution where the HSO4− contribution of H+ can be ignored. Example 14.17 illustrates an H 2SO4 problem where we can’t ignore the H+ contribution from HSO4−. 8.
a. H2O and CH3CO2− b. An acid-base reaction can be thought of as a competition between two opposing bases. Since this equilibrium lies far to the left (Ka < 1), CH3CO2− is a stronger base than H2O. c. The acetate ion is a better base than water and produces basic solutions in water. When we put acetate ion into solution as the only major basic species, the reaction is: CH3CO2− + H2O ⇌ CH3CO2H + OH− Now the competition is between CH3CO2− and OH− for the proton. Hydroxide ion is the strongest base possible in water. The equilibrium above lies far to the left, resulting in a Kb value less than one. Those species we specifically call weak bases ( 10 −14 < Kb < 1) lie between H2O and OH− in base strength. Weak bases are stronger bases than water but are weaker bases than OH−. The NH4+ ion is a weak acid because it lies between H 2O and H3O+ (H+) in terms of acid strength. Weak acids are better acids than water, thus their aqueous solutions are acidic. They are weak acids because they are not as strong as H 3O+ (H+). Weak acids only partially dissociate in water and have Ka values between 10 −14 and 1. For a strong acid HX having Ka = 1 × 106, the conjugate base, X−, has Kb = Kw/Ka = 1.0 × 10 −14 /1 × 106 = 1 × 10 −20 (a very small value).
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The conjugate bases of strong acids have extremely small values for Kb; so small that they are worse bases than water (Kb << Kw). Therefore, conjugate bases of strong acids have no basic properties in water. They are present, but they only balance charge in solution and nothing else. The conjugate bases of the six strong acids are Cl−, Br−, I−, NO3−, ClO4−, and HSO4−. Summarizing the acid-base properties of conjugates: a. The conjugate base of a weak acid is a weak base ( 10 −14 < Kb < 1) b. The conjugate acid of a weak base is a weak acid ( 10 −14 < Ka < 1) c. The conjugate base of a strong acid is a worthless base (Kb << 10 −14 ) d. The conjugate acid of a strong base is a worthless acid (Ka << 10 −14 ) Identifying/recognizing the acid-base properties of conjugates is crucial to understand the acidbase properties of salts. The salts we will give you will be salts containing the conjugates discussed above. Your job is to recognize the type of conjugate present, and then use that information to solve an equilibrium problem (if necessary). 9.
A salt is an ionic compound composed of a cation and an anion. Weak base anions: These are the conjugate bases of the weak acids having the HA general formula. Table 14.2 of the text lists several HA type acids. Some weak base anions derived from the acids in Table 14.2 are ClO2−, F−, NO2−, C2H3O2−, OCl−, and CN−. Garbage anions (those anions with no basic or acidic properties): These are the conjugate bases of the strong acids having the HA general formula. These are Cl −, NO3−, Br−, I−, and ClO4−. Weak acid cations: These are the conjugate acids of the weak bases which contain nitrogen. Table 14.3 of the text lists several nitrogen-containing bases. Some weak acid cations derived from the weak bases in Table 14.3 are NH 4+, CH3NH3+, C2H5NH3+, C6H5NH3+, and C5H5NH+. Garbage cations (those cations with no acidic properties or basic properties): The most common ones used are the cations in the strong bases. These are Li +, Na+, K+, Rb+, Cs+, Ca2+, Sr2+, and Ba2+. We mix and match the cations and anions to get what type of salt we want. For a weak base salt, we combine a weak base anion with a garbage cation. Some weak base salts are NaF, KNO2, Ca(CN)2, and RbC2H3O2. To determine the pH of a weak base salt, we write out the Kb reaction for the weak base anion and determine Kb (= Kw/Ka). We set up the ICE table under the Kb reaction, and then solve the equilibrium problem to calculate [OH−] and, in turn, pH. For a weak acid salt, we combine a weak acid cation with a garbage anion. Some weak acid salts are NH4Cl, C5H5NHNO3, CH3NH3I, and C2H5NH3ClO4. To determine the pH, we write out the Ka reaction for the weak acid cation and determine Ka (= Kw/Kb). We set up the ICE table under the Ka reaction, and then solve the equilibrium problem to calculate [H +] and, in turn, pH.
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For a neutral (pH = 7.0) salt, we combine a garbage cation with a garbage anion. Some examples are NaCl, KNO3, BaBr2, and Sr(ClO4)2. For salts that contain a weak acid cation and a weak base anion, we compare the Ka value of the weak acid cation to the Kb value for the weak base anion. When Ka > Kb, the salt produces an acidic solution (pH < 7.0). When Kb > Ka, the salt produces a basic solution. And when Ka = Kb, the salt produces a neutral solution (pH = 7.0). 10.
a. The weaker the X−H bond in an oxyacid, the stronger the acid. b. As the electronegativity of neighboring atoms increases in an oxyacid, the strength of the acid increases. c. As the number of oxygen atoms increases in an oxyacid, the strength of the acid increases. In general, the weaker the acid, the stronger the conjugate base and vice versa. a. Because acid strength increases as the X−H bond strength decreases, conjugate base strength will increase as the strength of the X−H bond increases. b. Because acid strength increases as the electronegativity of neighboring atoms increases, conjugate base strength will decrease as the electronegativity of neighboring atoms increases. c. Because acid strength increases as the number of oxygen atoms increases, conjugate base strength decreases as the number of oxygen atoms increases. Nonmetal oxides form acidic solutions when dissolved in water: SO3(g) + H2O(l) → H2SO4(aq) Metal oxides form basic solutions when dissolved in water: CaO(s) + H2O(l) → Ca(OH)2(aq)
Active Learning Questions 1.
As temperature changes from 25C, the value of Kw changes from 1.0 10−14. At a temperature different from 25C, the pH of neutral water is different from 7.00. The solutions are still neutral (neither acidic nor basic), just not a pH value of 7.00. However, at any temperature for neutral water, pH = pOH since [H+] = [OH−].
2.
HCl is always classified as a strong acid because in solution HCl exists mostly as separate H + and Cl− ions. NH3 is always classified as a weak base because in solution only a small amount of NH3 reacts with water to reach equilibrium. The terms concentrated and diluted apply to both strong and weak acids/bases. These terms both refer to the quantity of acid or base dissolved in water. The conjugate base of a weak acid is a weak base because K b for the conjugate base is between 1 and 10−14.
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3.
As shown in Interactive Example 14.10 of the text, as the initial concentration of the acid increases, the percent dissociation decreases. Your plot should show this inverse relationship. As the initial concentration of acid increases, the [H+] increases. This plot should show a direct relationship.
4.
Weak acids only partially dissociate in water. In terms of major species, HA is the way we find the weak acid most of the time; major species are HA and H 2O. HCl is a strong acid which is assumed to dissociate completely in water. H+, Cl−, and H2O are the major species for HCl. The major species for a solution with both a weak acid and HCl are HA, H +, Cl−, and H2O. To calculate the pH, we generally assume the amount of H + donated from the weak acid is negligible compared to the amount of H+ from HCl. That is we ignore the amount of H+ from the weak acid. Added NaA is listed as Na+ and A− in terms of major species. Sodium salts are all soluble in water. So, a solution that contains a weak acid HA, HCl, and NaA, the major species would be HA, H+, Cl−, Na+, A−, and H2O. When excess H+ is in the list of major species, react it first (if possible), react the best acid and best base together, and assume the reaction goes to completion. Since conjugate bases of weak acids are weak bases, the H + from the strong acid will react to completion with A− to form HA. After this reaction has gone to completion, take stock of what is left in solution as major species to determine the next step. If NaOH is then added, the major species present would be HA, H +, Cl−, Na+, A−, OH−, and H2O. The H+ and OH− in the major species list dominate. Assume these two substances react to completion to form water. What you do next depends on what major species remain in solution.
5.
A salt is an ionic compound composed of a cation and an anion. Acidic and basic salts generally contain the conjugate acids of the weak bases in Table 14.3 or the conjugate bases of the weak acids in Table 14.2. The conjugates of things weak are weak themselves. For example, the conjugate acid of the weak base NH 3 is NH4+ and it is a weak acid with Ka = 5.6 10−10. To make a solution having NH4+, we need to counterbalance the positive charge with a negative charge; we need to form a salt. The anions that have no acidic or basic properties are the conjugate bases of the strong acids having the general formula HA. These are Cl −, NO3−, Br−, I−, and ClO4−. An example of an acidic salt would be NH4Cl, which breaks up into the weak acid NH4+ and the ion Cl− which has no acidic/basic properties. Hence when NH4Cl is added to water, an acidic solution will result from this salt. Basic salts are generally composed of the conjugate bases of the weak acids in Table 14.2. For example, F− is a weak base with Kb = 7.2 10−11. To counterbalance the negative charge of F−, we use a species having a positive charge, but also a species that has no acidic or basic properties. These are the cations of the strong bases. Examples are are Li+, Na+, K+, Rb+, Cs+, Ca2+, Sr2+, and Ba2+. A basic salt example would be NaF which dissolves to Na+ (has no acidic or basic properties) and the weak base F −. The most common example of neutral salts are ones where both ions have no acidic or basic properties. NaCl is a neutral salt since neither Na+ nor Cl− have any acidic or basic properties in water. When NaCl is dissolved in water at 25C, pH = 7.00.
6.
pH = –log[H+] = –log[H3O+]; pH is a measure of how acidic a solution is. The larger the [H3O+] in a solution, the more acidic that solution is and the lower the pH. A strong acid does not have
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to have a lower pH than a weak acid. A very dilute solution of a strong acid can have a higher pH than a more concentrated weak acid solution. 7.
All solutions have an H+ concentration. It is not very large in basic solutions. The value of [H+] in any solution is equal to Kw/[OH−]. This comes from the autoionization of water equilibrium that always is present when water is the solvent.
8.
Answer c is best; all the possible reactions occur at the same time in an actual solution. For problem solving, we assume answer a is correct. Solving equilibrium problems is a difficult task. Instead of trying to satisfy all the different equilibriums at once, we order the reactions in a stepwise fashion to simplify the process.
9.
The assumptions are when excess H+ from a strong acid is present, react it with the best base present and let the reaction go to completion. So here the reaction is H+ + A− → HA. In this problem, we have 0.010 moles of H+ and 0.010 moles of A−, which react to form 0.010 moles of HA. Now, combine the 0.010 moles of HA present initially with the 0.010 moles of HA produced from the strong acid reaction to give 0.020 moles HA total. Now we have 300.0 mL of solution that contain 0.020 moles of HA, along with Na + and Cl−; these last two ions have no acidic or basic properties. So now we solve the weak acid problem to determine the pH of the solution (pH = 3.59). This is the general order that the reactions are assumed to occur. All these reactions are occurring at the same time. However, considering several processes at the same time is extremely difficult. To simplify the process, we take one reaction at a time.
10.
The pH of the solution would give you a clue as to if the anion is an acid or a base. Most likely the anion is a conjugate base of the HA type weak acids in Table 14.2. These anions have the general formula of A− and are all weak bases. Note that none of these A− anions are strong bases. An acidic anion that could be present is HSO 4−. Also, we could have the anion having amphoteric properties. These anions would be conjugates of the polyprotic acids in Table 13.4. For example, HCO3− is both a weak acid and a weak base, hence why it is called amphoteric. The pH of these amphoteric anions depends on how good a weak acid vs weak base each ion is. Some form acidic solutions, while some form basic solutions. Lastly, the anion could be the conjugate base of a strong acid (Cl−, Br−, I−, NO3−, and ClO4−), all of which have no acidic or basic properties.
11.
Consider the following Kb reaction for NH3 and the Ka reaction for its conjugate acid, NH4+. +
NH3(aq) + H2O(l) ⇌ NH4+(aq) + OH−(aq)
Kb =
[ NH4 ][OH− ] [ NH3 ]
NH4+(aq) + H2O(l) ⇌ NH3(aq) + H3O+(aq)
Ka =
[NH3 ][H3O+ ] [NH 4 + ]
Notice that these reactions are not the reverse of each other. The K b reaction produces OH− while the Ka reaction produces H3O+. The relationship between Ka and Kb for a conjugate acidbase pair is Ka × Kb = Kw. For NH4+ and NH3: Ka × Kb =
[NH3 ][H3O+ ] [ NH4 + ][OH− ] = [H3O+][OH −] = Kw [ NH3 ] [NH 4 + ]
This is true for all conjugate acid-base pairs and conjugate base-acid pairs.
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a. An acid-base reaction can be thought of as a competition between two opposing bases. If the better base is on the reactant side, then the forward reaction is dominate. Since water is assumed to be a better base and it is on the reactant side, then the equilibrium lies to the right. b. This is the general Ka reaction of an acid reacting with water. When this reaction lies to the right as it does here, the acid is strong. When the equilibrium lies to the left, we have a weak acid. c. Because the equilibrium lies to the right, Ka will be greater than 1.
13.
The pH will be between 4.0 and 5.0. A pH = 4.0 solution has [H+] = 1.0 × 10‒4 M and a pH = 6.0 has [H+] = 1.0 × 10‒6 M. If we have 1.0 L of each solution, 1.0 × 10 ‒4 mol H+ is present in the pH = 4.0 solution and 1.0 × 10 ‒6 mol H+ is present in the pH = 6.0 solution. The H+ added from the pH = 6.0 solution is very tiny compared to the H + from the pH = 4.0 solution. To 2 significant figures, we can ignore the H+ from the pH = 6.0 solution with our 1.0 L assumption of each solution. So, when we add the solutions together, we have 1.0 × 10 ‒4 mol of H+ in 2.0 L of solution. This gives [H+] = 5 × 10‒5 M and a pH = 4.30. When we add these two solutions together, the pH = 6.0 solution adds very little H+ but it does double the volume of the pH = 4.0 solution. To change the pH by 1.0 pH units, we need to dilute the solution by a factor of 10. Here we diluted the pH = 4.0 solution by just a factor of 2, so the pH will be between 4.0 and 5.0.
14.
Both are salt solutions that dissolve to form Na+ and X − or Y −. The sodium ion has no acidic or basic properties. The anions are probably the conjugate bases of the acids HX and HY. From the relationship Ka × Kb = Kw, the weaker acid between HX and HY would have the stronger conjugate base and would produce a more basic solution (higher pH). So, we would need to know the Ka values for the acids HX and HY. The acid with the smaller K a value would have the stronger conjugate base. The sodium salt of that conjugate base would have the higher pH.
15.
At 25C, Kw = 1.0 × 10 −14 . A pH = 7.00 solution has [H3O+] = 1.0 × 10−7 M, pOH = 7.00, and [OH−] = 1.0 × 10−7 M. We know that for water, [H3O+][OH−] = Kw = 1.0 × 10 −14 . Plugging in the concentration values for H3O+ and OH−: [H3O+][OH−] = 1.0 × 10−7(1.0 × 10−7) = 1.0 × 10−14 = Kw at 25C At 25C, pH of neutral water must equal 7.00 in order for the K w relationship to hold true. At a different temperature where Kw does not equal 1.0 × 10−14, the pH of neutral water does not equal 7.00.
16.
The pH of a very acidic solution can be negative. Specifically, when [H 3O+] > 1.0 M, the pH is a negative value. Any strong acid solution having a concentration larger than 1.0 M has a negative pH.
17.
The conjugate base of a weak acid is a weak base, not a strong base. For conjugate acid-base pairs, Kw = Ka × Kb = 1.0 × 10−14 (assuming 25C). A weak acid has a Ka value between 1 and 1.0 × 10−14. From the Kw = Ka × Kb relationship, the Kb value for the conjugate base of a weak acid must be between 1.0 × 10−14 and 1. Any base with a Kb value between 1 and 1.0 × 10 −14 is
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treated as a weak base when problem solving. So, the math dictates that the conjugate base of a weak acid is a weak base. It is also true that the conjugate acid of a weak base is a weak acid. For a strong acid like HCl, the Ka value is much larger than 1; let’s assume for HCl that Ka = 1 × 106. So, the conjugate base Cl− has Kb = Kw/Ka = 1.0 × 10 −14 /1 × 106 = 1 × 10 −20 (a very small value). The Kb value for conjugate bases of strong acids is smaller than K w. This indicates that conjugate bases of strong acids are worse bases than water. So, Cl − has no effect on the pH of an aqueous solution. 18.
All these statements could be true. The B+ cation could be the conjugate acid of one of the weak bases in Table 14.3. All these conjugate acids are weak acids and would produce acidic solutions. Typically, in salt solutions containing the conjugate acids of the bases in Table 14.3, the anion X − comes from a strong acid so we don’t have to worry about the basic properties of X − (conjugate bases of strong acids are terrible bases). However, we could make an acidic salt using a conjugate base of the weak acids in Table 14.2. The salt BX would be acidic if Ka for B+ > Kb for X−. Another possibility for an acidic salt is NaHSO 4. Here, Na+ does not affect the pH, but HSO4 − is a weak acid with Ka = 0.012. This salt has the BX general formula and would be acidic. So conceivably, all these statements could be true, but they don’t have to be true.
19.
a. Your picture with HCl should have all 10 molecules of HCl dissociated into 10 H + and 10 Cl− ions since HCl is a strong acid. Your picture of the weak acid should show only partial dissociating of the acid. Let’s assume 1 in 10 molecules of HA dissociates to reach equilibrium. So, this picture would have 1 H+ ion, 1 A− ion, and 9 HA molecules. b. Major species for HCl are H+, Cl−, and H2O. Major species for the weak acid solution are HA are HA and H2O. c. The Ka value for HCl is very large. From our picture, we can’t calculate a K a value since we assumed it dissociates 100% of the time. In real life, HCl is maybe only 99.99% dissociated in water. We didn’t want to make a drawing with 1000 HCl molecules to show the one HCl molecule that stays together. For a 1 M weak acid solution and assuming 10% dissociation: Ka =
[H + ][A- ] 0.1(0.1) = = 0.01 [HA] 0.9
Notice that this is a large value for a weak acid whose Ka values are generally 10−4 or less. Typical weak acids are 5% or less dissociated in water. d. A 0.1 M solution of the strong acid has [H+] = 0.1 M, giving a pH = 1.0. For the 0.1 M HA (Ka = 0.01) solution, HA is the major source of H+. But only a little of it dissociates. We use the Ka equilibrium reaction for the weak acid to determine the [H+]. HA Initial Change Equil.
⇌
H+
+
A−
0.1 M ~0 0 x mol/L HA dissociates to reach equilibrium –x → +x +x 0.1 – x x x
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Ka = 0.01 =
647
x2 , 0.001 ‒ 0.01x = x2, x2 + 0.01x ‒ 0.001 = 0 0.1 − x
Solving: x = [H+] = 0.027 = 0.03M; pH = ‒log(0.03) = 1.5 e. The correct order from strongest to weakest base is A −, H2O, and Cl−. Cl− is the conjugate base of a strong acid and as such, it is a terrible base. Cl − has virtually no affinity for H+ ions. A− is the conjugate base of a weak acid; in general, conjugate bases of weak acids are weak bases themselves with Kb values between 1 and 1 10−14. In terms of base strength, H2O is placed between weak bases and worthless bases. 20.
Let’s identify the species and determine Ka or Kb values to help in the ordering. HBr: strong acid; NaOH: strong base; HF, weak acid with K a = 7.2 × 10−4; HCN: weak acid with Ka = 6.2 × 10−10; NH3: weak base with Kb = 1.8 × 10 −5 . NaF → Na+ + F−: F− is a weak base with Kb = Kw/Ka for HF = 1.4 × 10−11, and Na+ has no effect on pH. NaCN → Na+ + CN−: CN− is a weak base with Kb = Kw/Ka for HCN = 1.6 × 10−5, and Na+ has no effect on pH. NH4F → NH4+ + F−: NH4+ is a weak acid with Ka = Kw/Kb for NH3 = 5.6 × 10−10, and F− is a weak base with Kb = Kw/Ka for HF = 1.4 × 10−11. Since the Ka for the weak acid is slightly larger than the Kb for the weak base, this solution will be slightly acidic. CH3NH3F → CH3NH3+ + F−: CH3NH3+ is a weak acid with Ka = Kw/Kb for CH3NH2 = 1.0 × 10−14/4.34 × 10−4 = 2.3 × 10−11, and F− is a weak base with Kb = Kw/Ka for HF = 1.4 × 10−11. The Ka for the weak acid is almost equal to the Kb for the weak base, so this solution will have a pH close to 7.0. It will be a little less than 7.0 since K a > Kb. The CH3NH3F solution has pH = 6.5 and the NH4F solution has pH = 6. The HBr solution has pH = 1. From the Ka values, HF is a better weak acid than HCN, so the HF solution has pH = 2 and the HCN solution has pH = 5. We have four basic pH values remaining: 8, 11, 11, and 13. The strong base solution of NaOH has pH = 13. The weakest of the remaining bases is F − in NaF. This solution will have pH = 8. The remaining weak bases NH 3 and CN− have similar Kb values on the order of 10−5, so they both will have similar basic pH values. The NH3 and NaCN solutions have pH = 11.
Questions 21.
Acids are proton (H+) donors, and bases are proton acceptors.
HCO3− as an acid: HCO3−(aq) + H2O(l) ⇌ CO32−(aq) + H3O+(aq) HCO3− as a base: HCO3−(aq) + H2O(l) ⇌ H2CO3(aq) + OH−(aq) H2PO4− as an acid: H2PO4− + H2O(l) ⇌ HPO42−(aq) + H3O+(aq) H2PO4− as a base: H2PO4− + H2O(l) ⇌ H3PO4(aq) + OH−(aq)
648
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22.
Acidic solutions (at 25C) have an [H+] > 1.0 × 10−7 M, which gives a pH < 7.0. Because [H+][OH−] = 1.0 × 10−14 and pH + pOH = 14.00 for an aqueous solution at 25C, an acidic solution must also have [OH−] < 1.0 × 10−7 M and pOH > 7.00. From these relationships, the solutions in parts a, b, and d are acidic. The solution in part c will have a pH > 7.0 (pH = 14.00 – 4.51 = 9.49) and is therefore not acidic (solution is basic).
23.
Basic solutions (at 25C) have an [OH−] > 1.0 × 10−7 M, which gives a pOH < 7.0. Because [H+][OH−] = 1.0 × 10−14 and pH + pOH = 14.00 for any aqueous solution at 25C, a basic solution must also have [H+] < 1.0 × 10−7 M and pH > 7.00. From these relationships, the solutions in parts b, c, and d are basic solutions. The solution in part a will have a pH < 7.0 (pH = 14.00 – 11.21 = 2.79) and is therefore not basic (solution is acidic).
24.
Only answer e is false. A 1 × 10‒11 M NaOH solution only donates 1 × 10‒11 M OH‒. Since this is an aqueous solution, water all by itself has 1 × 10-7 M OH‒ and 1 × 10-7 M H+. The pH of this very dilute solution of a strong base will be 7.0. In this problem, the amount of OH ‒ donated by the strong base is insignificant compared to that of water. The pH of the solution is that of neutral water.
25.
The Ka reaction always refers to an acid reacting with water to form H3O+ and the conjugate base of the acid. We often leave out water in the reaction since it isn’t included in the K expression. Only the reaction in answer d is a reaction associated with the K a equilibrium constant. Answer c is a reaction associated with the Kb equilibrium constant (a base reacting with water to produce OH‒ and the conjugate acid of the base).
26.
Mathematically, pKa = –log Ka. As with pH and [H+], there is an inverse relationship between pKa and Ka. As acid strength increases, Ka increases, and pKa decreases. So as pKa increases, the Ka value for the acid decreases, which dictates a decrease in acid strength.
27.
10.78 (4 S.F.); 6.78 (3 S.F.); 0.78 (2 S.F.); a pH value is a logarithm. The numbers to the left of the decimal point identify the power of 10 to which [H+] is expressed in scientific notation, for example, 10 −11 , 10 −7 , 10 −1 . The number of decimal places in a pH value identifies the number of significant figures in [H+]. In all three pH values, the [H+] should be expressed only to two significant figures because these pH values have only two decimal places.
28.
A Lewis acid must have an empty orbital to accept an electron pair, and a Lewis base must have an unshared pair of electrons.
29.
NH3 + NH3 ⇌ NH2− + NH4+ Acid Base Conjugate Conjugate Base Acid One of the NH3 molecules acts as a base and accepts a proton to form NH4+ . The other NH3 molecule acts as an acid and donates a proton to form NH 2−. NH4+ is the conjugate acid of the NH3 base. In the reverse reaction, NH4+ donates a proton. NH2− is the conjugate base of the NH3 acid. In the reverse reaction, NH2− accepts a proton. Conjugate acid-base pairs only differ by a H+ in the formula.
CHAPTER 14 30.
ACIDS AND BASES
649
a. The first equation is for the reaction of some generic acid, HA, with H2O. HA + H2O Acid Base
⇌
H3O+ + A− Conjugate Conjugate Acid of H2O Base of HA
HA is the proton donor (the acid) and H2O is the proton acceptor (the base). In the reverse reaction, H3O+ is the proton donor (the acid) and A − is the proton acceptor (the base). The second equation is for some generic base, B, with some generic acid, HX. Note that B has three hydrogens bonded to it. B + HX Base Acid
⇌
BH+ + X− Conjugate Conjugate Acid of B Base of HX
B is the proton acceptor (the base) and HX is the proton donor (the acid). When B accepts a proton, the central atom goes from having 3 bonded hydrogens to 4 bonded hydrogens. In the reverse reaction, BH+ is the proton donor (the acid) and X− is the proton acceptor (the base). b. Arrhenius acids produce H+ in solution. So HA in the first equation is an Arrhenius acid. However, in the second equation, H+ is not a product, so HX is not an Arrhenius acid. Both HA in the first equation and HX in the second equation are proton donors, so both are considered Brønsted-Lowry acids. All Brønsted-Lowry acids are Lewis acids, that is, all Brønsted-Lowry acids are electron pair acceptors. So, both HA and HX are Lewis acids. It is the proton (H+) that accepts the lone pair For the bases in the two equations, H2O and B, none of them produce OH− in their equations, so none of them are Arrhenius bases. Both H2O and B accept protons, so both are Brønsted-Lowry bases. As with Brønsted-Lowry acids, all Brønsted-Lowry bases are also electron-pair donors, so both H2O and B are Lewis bases. The oxygen in H2O and the central atom in B will always have a lone pair to donate to the Lewis acid (the proton). 31.
a. These are solutions of strong acids like HCl, HBr, HI, HNO3, H2SO4, and HClO4. So 0.10 M solutions of any of the acids would be examples of a strong electrolyte solution that is very acidic. b. These are solutions containing salts of the conjugate acids of the bases in Table 14.3. These conjugate acids are all weak acids, and they are cations with a 1+ charge. NH4Cl, CH3NH3NO3, and C2H5NH3Br are three examples of this type of slightly acidic salts. Note that the anions used to form these salts are conjugate bases of strong acids; this is so because they have no acidic or basic properties in water (except for HSO4−, which has weak acid properties). c. These are solutions of strong bases like LiOH, NaOH, KOH, RbOH, CsOH, Ca(OH)2, Sr(OH)2, and Ba(OH)2. All these strong bases are strong electrolytes.
650
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ACIDS AND BASES
d. These are solutions containing salts of the conjugate bases of the neutrally charged weak acids in Table 14.2. These conjugate bases are all weak bases, and they are anions with a 1− charge. Three examples of this type of slightly basic salts are NaClO2, KC2H3O2, and CaF2. The cations used to form these salts are Li+, Na+, K+, Rb+, Cs+, Ca2+, Sr2+, and Ba2+ because these cations have no acidic or basic properties in water. Notice that these are the cations of the strong bases you should memorize. e. There are two ways to make a neutral salt solutions. The easiest way is to combine a conjugate base of a strong acid (except for HSO4−) with one of the cations from a strong base. These ions have no acidic/basic properties in water, so salts of these ions are neutral. Three examples are NaCl, KNO3, and SrI2. Another type of strong electrolyte that can produce neutral solutions are salts that contain an ion with weak acid properties combined with an ion of opposite charge having weak base properties. If the K a for the weak acid ion is equal to the Kb for the weak base ion, then the salt will produce a neutral solution. The most common example of this type of salt is ammonium acetate (NH4C2H3O2). For this salt, Ka for NH4+ = Kb for C2H3O2− = 5.6 × 10 −10 . This salt at any concentration produces a neutral solution. 32.
Ka × Kb = Kw, –log(Ka × Kb) = –log Kw –log Ka – log Kb = –log Kw, pKa + pKb = pKw = 14.00 (Kw = 1.0 × 10−14 at 25°C)
33.
a. H2O(l) + H2O(l) ⇌ H3O+(aq) + OH−(aq) or H2O(l) ⇌ H+(aq) + OH−(aq)
K = Kw = [H+][OH−]
b. HF(aq) + H2O(l) ⇌ F−(aq) + H3O+(aq) or HF(aq) ⇌ H+(aq) + F−(aq)
K = Ka =
[H + ][F− ] [HF]
c. C5H5N(aq) + H2O(l) ⇌ C5H5NH+(aq) + OH−(aq) 34.
K = Kb =
[C5 H 5 NH + ][OH − ] [C 5 H 5 N ]
Aspirin is a weak acid and adrenaline is a weak base. The structural feature for aspirin that identifies it as a weak acid is the carboxylic acid group. The structure of the carboxylic acid grouping is: O C
OH
Abbreviations for this group are –COOH or –CO2H. When the carboxylic acid group is present in a structure, the compound is generally a weak acid. For adrenaline, the structural feature that identifies it as a weak base is a neutral charged nitrogen atom in the formula. If the nitrogen atom is neutral charged, then it has a lone pair of electrons which can accept a proton. See Table 14.3 for several examples of weak bases all of which have a neutral charged nitrogen in the structure. 35.
a. H2SO3
b. HClO3
c. H3PO3
CHAPTER 14
ACIDS AND BASES
651
NaOH and KOH are soluble ionic compounds composed of Na+ and K+ cations and OH− anions. All soluble ionic compounds dissolve to form the ions from which they are formed. In oxyacids, the compounds are all covalent compounds in which electrons are shared to form bonds (unlike ionic compounds). When these compounds are dissolved in water, the covalent bond between oxygen and hydrogen breaks to form H+ ions. 36.
Only statement a is true (assuming the species is not amphoteric). You cannot add a base to water and get an acidic pH (pH < 7.0). For statement b, you can have negative pH values; this just indicates an [H+] > 1.0 M. For statement c, a dilute solution of a strong acid can have a higher pH than a more concentrated weak acid solution. For statement d, the Ba(OH) 2 solution will have an [OH−] twice of the same concentration of KOH, but this does not correspond to a pOH value twice that of the same concentration of KOH (prove it to yourselves).
37.
a. This expression holds true for solutions of monoprotic strong acids having a concentration greater than 1.0 × 10 −6 M. 0.10 M HCl, 7.8 M HNO3, and 3.6 × 10 −4 M HClO4 are examples where this expression holds true. b. This expression holds true for solutions of weak acids where the two normal assumptions hold. The two assumptions are that water does not contribute enough H + to solution to make a difference, and that the acid is less than 5% dissociated in water (from the assumption that x is small compared to some number). This expression will generally hold true for solutions of weak acids having a Ka value less than 1 × 10 −4 , if there is a significant amount of weak acid present. Three example solutions are 1.5 M HC2H3O2, 0.10 M HOCl, and 0.72 M HCN. c. This expression holds true for strong bases that donate 2 OH − ions per formula unit. As long as the concentration of the base is above 5 × 10 −7 M, this expression will hold true. Three examples are 5.0 × 10 −3 M Ca(OH)2, 2.1 × 10 −4 M Sr(OH)2, and 9.1 × 10 −5 M Ba(OH)2. d. This expression holds true for solutions of weak bases where the two normal assumptions hold. The assumptions are that the OH− contribution from water is negligible and that and that the base is less than 5% ionized in water (for the 5% rule to hold). For the 5% rule to hold, you generally need bases with Kb < 1 × 10 −4 , and concentrations of weak base greater than 0.10 M. Three examples are 0.10 M NH3, 0.54 M C6H5NH2, and 1.1 M C5H5N.
38.
H2CO3 is a weak acid with K a1 = 4.3 × 10 −7 and K a 2 = 5.6 × 10 −11 . The [H+] concentration in solution will be determined from the K a1 reaction because K a1 >> K a 2 . Because K a1 << 1, the [H+] < 0.10 M; only a small percentage of the 0.10 M H2CO3 will dissociate into HCO3− and H+. So statement a best describes the 0.10 M H2CO3 solution. H2SO4 is a strong acid as well as a very good weak acid (K a1 >> 1, K a 2 = 1.2 × 10 −2 ). All the 0.10 M H2SO4 solution will dissociate into 0.10 M H+ and 0.10 M HSO4−. However, because HSO4− is a good weak acid due to the relatively large Ka value, some of the 0.10 M HSO4− will dissociate into some more H+ and SO42−. Therefore, the [H+] will be greater than 0.10 M but will not reach 0.20 M because only some of 0.10 M HSO4− will dissociate. Statement c is best for a 0.10 M H2SO4 solution.
652
CHAPTER 14
ACIDS AND BASES
39.
Answer c is best. For conjugate acid-base pairs, Ka × Kb = Kw = 1.0 × 10‒14 (at 25°C). A weak acid has a Ka value between 1 and 1.0 × 10 ‒14. Applying Ka × Kb = Kw = 1.0 × 10‒14, the Kb value for the conjugate base must be between 1.0 × 10‒14 and 1. Bases with Kb values between 1 and 1.0 × 10‒14 are classified as weak bases.
40.
Answer e is correct. Cl‒ is the conjugate base of the strong acid HCl. In general, conjugate bases of strong acids have no basic properties in water. This is because they are worse bases than water. Assuming a strong acid Ka value of 1 × 106, Kb for the conjugate base is 1 × 10‒20. This is smaller than the Kw value for water, hence the conclusion that conjugate bases of strong acids are worse bases than water.
41.
One reason HF is a weak acid is that the H−F bond is unusually strong and is difficult to break. This contributes significantly to the reluctance of the HF molecules to dissociate in water.
42.
a. Sulfur reacts with oxygen to produce SO2 and SO3. These sulfur oxides both react with water to produce H2SO3 and H2SO4, respectively. Acid rain can result when sulfur emissions are not controlled. Note that, in general, nonmetal oxides react with water to produce acidic solutions. b. CaO reacts with water to produce Ca(OH)2, a strong base. A gardener mixes lime (CaO) into soil to raise the pH of the soil. The effect of adding lime is to add Ca(OH)2. Note that, in general, metal oxides react with water to produce basic solutions.
Exercises Nature of Acids and Bases 43.
a. HClO4(aq) + H2O(l) → H3O+(aq) + ClO4−(aq). Only the forward reaction is indicated because HClO4 is a strong acid and is basically 100% dissociated in water. For acids, the dissociation reaction is commonly written without water as a reactant. The common abbreviation for this reaction is HClO4(aq) → H+(aq) + ClO4−(aq). This reaction is also called the Ka reaction because the equilibrium constant for this reaction is designated as Ka . b. Propanoic acid is a weak acid, so it is only partially dissociated in water. The dissociation reaction is: CH3CH2CO2H(aq) + H2O(l) ⇌ H3O+(aq) + CH3CH2CO2−(aq) or CH3CH2CO2H(aq) ⇌ H+(aq) + CH3CH2CO2−(aq). c. NH4+ is a weak acid. Like propanoic acid, the dissociation reaction is: NH4+(aq) + H2O(l) ⇌ H3O+(aq) + NH3(aq) or NH4+(aq) ⇌ H+(aq) + NH3(aq)
44.
The dissociation reaction (the Ka reaction) of an acid in water commonly omits water as a reactant. We will follow this practice. All dissociation reactions produce H+ and the conjugate base of the acid that is dissociated.
CHAPTER 14
45.
ACIDS AND BASES
653
a. HCN(aq) ⇌ H+(aq) + CN−(aq)
Ka =
[H + ][CN − ] [HCN]
b. HOC6H5(aq) ⇌ H+(aq) + OC6H5−(aq)
Ka =
[H + ][OC 6 H 5 ] [HOC 6 H 5 ]
c. C6H5NH3+(aq) ⇌ H+(aq) + C6H5NH2(aq)
Ka =
[H + ][C 6 H 5 NH2 ] [C 6 H 5 NH3+ ]
An acid is a proton (H+) donor, and a base is a proton acceptor. A conjugate acid-base pair differs by only a proton (H+). Acid
Base
Conjugate Base of Acid
Conjugate Acid of Base
a.
H2CO3
H2 O
HCO3−
H 3 O+
b.
C5H5NH+
H2 O
C5H5N
H 3 O+
c.
C5H5NH+
HCO3−
C5H5N
H2CO3
Acid
Base
Conjugate Base of Acid
Conjugate Acid of Base
a.
Al(H2O)63+
H2 O
Al(H2O)5(OH)2+
H3O+
b.
HONH3+
H2 O
HONH2
H3O+
c.
HOCl
C6H5NH2
OCl-
C6H5NH3+
46.
47.
−
Strong acids have a Ka >> 1, and weak acids have Ka < 1. Table 14.2 in the text lists some Ka values for weak acids. Ka values for strong acids are hard to determine, so they are not listed in the text. However, there are only a few common strong acids so, if you memorize the strong acids, then all other acids will be weak acids. The strong acids to memorize are HCl, HBr, HI, HNO3, HClO4, and H2SO4. a. HClO4 is a strong acid. b. HOCl is a weak acid (Ka = 3.5 × 10−8). c. H2SO4 is a strong acid. d. H2SO3 is a weak diprotic acid because the Ka1 and Ka2 values are much less than 1.
48.
The beaker on the left represents a strong acid in solution; the acid HA is 100% dissociated into the H+ and A− ions. The beaker on the right represents a weak acid in solution; only a little bit of the acid HB dissociates into ions, so the acid exists mostly as undissociated HB molecules in water. a. HNO2: weak acid beaker
b. HNO3: strong acid beaker
c. HCl: strong acid beaker
d. HF: weak acid beaker
e. HC2H3O2: weak acid beaker
654 49.
CHAPTER 14
ACIDS AND BASES
The Ka value is directly related to acid strength. As Ka increases, acid strength increases. For water, use Kw when comparing the acid strength of water to other species. The Ka values are: HClO4: strong acid (Ka >> 1); HClO2: Ka = 1.2 × 10 −2 NH4+: Ka = 5.6 × 10 −10 ; H2O: Ka = Kw = 1.0 × 10 −14 From the Ka values, the ordering is HClO4 > HClO2 > NH4+ > H2O.
50.
Except for water, these are the conjugate bases of the acids in the previous exercise. In general, the weaker the acid, the stronger is the conjugate base. ClO4− is the conjugate base of a strong acid; it is a terrible base (worse than water). The ordering is NH3 > ClO2− > H2O > ClO4− .
51.
a. HCl is a strong acid, and water is a very weak acid with Ka = Kw = 1.0 × 10 −14. HCl is a much stronger acid than H2O. b. H2O, Ka = Kw = 1.0 × 10 −14 ; HNO2, Ka = 4.0 × 10 −4 ; HNO2 is a stronger acid than H2O because Ka for HNO2 > Kw for H2O. c. HOC6H5 , Ka = 1.6 × 10 −10 ; HCN, Ka = 6.2 × 10 −10 ; HCN is a slightly stronger acid than HOC6H5 because Ka for HCN > Ka for HOC6H5.
52.
a. H2O; the conjugate bases of strong acids are extremely weak bases (Kb < 1 × 10 −14 ). b. NO2−; the conjugate bases of weak acids are weak bases (1 × 10 −14 < Kb < 1). c.
OC6H5−; for a conjugate acid-base pair, Ka × Kb = Kw. From this relationship, the stronger the acid, the weaker is the conjugate base (Kb decreases as Ka increases). Because HCN is a stronger acid than HOC6H5 (Ka for HCN > Ka for HOC6H5), OC6H5− will be a stronger base than CN−.
Autoionization of Water and the pH Scale 53.
At 25°C, the relationship [H+][OH−] = Kw = 1.0 × 10 −14 always holds for aqueous solutions. When [H+] is greater than 1.0 × 10 −7 M (pH < 7.0), the solution is acidic; when [H+] is less than 1.0 × 10 −7 M (pH > 7.0), the solution is basic; when [H+] = 1.0 × 10 −7 M (pH = 7.0), the solution is neutral. In terms of [OH−], an acidic solution has [OH−] < 1.0 × 10 −7 M (pOH > 7.0), a basic solution has [OH−] > 1.0 × 10 −7 M (pOH < 7.0), and a neutral solution has [OH−] = 1.0 × 10 −7 M (pOH = 7.0). At 25C, pH + pOH = 14.00. a. [OH−] =
Kw 1.0 10 −14 = = 1.0 × 10-7 M; the solution is neutral. [H + ] 1.0 10 −7
pH = −log[H+] = −log(1.0 × 10 −7 ) = 7.00; pOH = 14.00 − 7.00 = 7.00 b. [OH−] =
1.0 10 −14 = 12 M; the solution is basic. 8.3 10 −16
pH = −log(8.3 × 10 −16 ) = 15.08; pOH = 14.00 − 15.08 = −1.08
CHAPTER 14
ACIDS AND BASES
c. [OH−] =
655
1.0 10 −14 = 8.3 × 10 −16 M; the solution is acidic. 12
pH = −log(12) = −1.08; pOH = 14.00 − (−1.08) = 15.08 d. [OH−] =
1.0 10 −14 = 1.9 × 10 −10 M; the solution is acidic. 5.4 10 −5
pH = −log(5.4 × 10 −5 ) = 4.27; pOH = 14.00 − 4.27 = 9.73 Note that pH is greater than 14.00 when [OH −] is greater than 1.0 M (an extremely basic solution). Also note the the pH is negative when [H +] is greater than 1.0 M (an extremely acidic solution). 54.
[H+] =
Kw [OH − ]
=
1.0 10 −14 = 6.7 × 10 −15 M; basic 1.5
pOH = –log[OH−] = –log(1.5) = –0.18; pH = 14.00 – pOH = 14.00 – (–0.18) = 14.18 a. [H+] =
1.0 10 −14 3.6 10 −15
= 2.8 M; acidic
pOH = –log(3.6 × 10 −15 ) = 14.44; pH = 14.00 – 14.44 = –0.44 b. [H+] =
1.0 10 −14 = 1.0 × 10 −7 M; neutral −7 1.0 10
pOH = –log(1.0 × 10 −7 ) = 7.00; pH = 14.00 – 7.00 = 7.00 c. [H+] =
1.0 10 −14 = 1.4 × 10 −11 M; basic 7.3 10 − 4
pOH = –log(7.3 × 10 −4 ) = 3.14; pH = 14.00 – 3.14 = 10.86 55.
a. Because the value of the equilibrium constant increases as the temperature increases, the reaction is endothermic. In endothermic reactions, heat is a reactant, so an increase in temperature (heat) shifts the reaction to produce more products and increases K in the process. b. H2O(l) ⇌ H+(aq) + OH−(aq)
Kw = 5.47 × 10 −14 = [H+][OH−] at 50.°C
In pure water [H+] = [OH−], so 5.47 × 10 −14 = [H+]2, [H+] = 2.34 × 10 −7 M = [OH−] 56.
a. [H+] = 10−pH, [H+] = 10−6.77 = 1.7 × 10−7 M; this is also the [OH‒] since we have neutral water where [H+] = [OH‒] = 1.7 × 10−7 M. b. Kw = [H+][OH−] = (1.7 × 10−7)( 1.7 × 10−7) = 2.9 × 10−14 c. [H+] = Kw/[OH−] = (2.9 × 10 −14 )/0.10) = 2.9 × 10 −13 M; pH = –log(2.9 × 10 −13 ) = 12.54
656 57.
CHAPTER 14
ACIDS AND BASES
a. [H+] = 10−pH, [H+] = 10−7.40 = 4.0 × 10−8 M pOH = 14.00 − pH = 14.00 − 7.40 = 6.60; [OH−] = 10−pOH = 10−6.60 = 2.5 × 10−7 M or [OH−] =
Kw 1.0 10 −14 = 2.5 × 10−7 M; this solution is basic since pH > 7.00. = [H + ] 4.0 10 −8
b. [H+] = 10−15.3 = 5 × 10−16 M; pOH = 14.00 − 15.3 = −1.3; [OH−] = 10− (−1.3) = 20 M; basic c. [H+] = 10− (−1.0) = 10 M; pOH = 14.0 − (−1.0) = 15.0; [OH−] = 10-15.0 = 1 × 10−15 M; acidic d. [H+] = 10−3.20 = 6.3 × 10−4 M; pOH = 14.00 − 3.20 = 10.80; [OH−] = 10−10.80 = 1.6 × 10−11 M; acidic e. [OH−] = 10−5.0 = 1 × 10−5 M; pH = 14.0 − pOH = 14.0 − 5.0 = 9.0; [H+] = 10−9.0 = 1 × 10−9 M; basic f. 58.
[OH−] = 10−9.60 = 2.5 × 10−10 M; pH = 14.00 − 9.60 = 4.40; [H+] = 10−4.40 = 4.0 × 10−5 M; acidic
a. pOH = 14.00 – 9.63 = 4.37; [H+] = 10 −9.63 = 2.3 × 10 −10 M [OH−] = 10 −4.37 = 4.3 × 10 −5 M; basic b. [H+] =
1.0 10 −14 = 2.6 × 10 −9 M; pH = –log(2.6 × 10 −9 ) = 8.59 −6 3.9 10
pOH = 14.00 – 8.59 = 5.41; basic c. pH = –log(0.027) = 1.57; pOH = 14.00 – 1.57 = 12.43 [OH−] = 10 −12.43 = 3.7 × 10 −13 M; acidic d. pH = 14.00 – 1.22 = 12.78; [H+] = 10 −12.78 = 1.7 × 10−13 M [OH−] = 10 −1.22 = 0.060 M; basic 59.
pOH = 14.0 – pH = 14.0 – 2.1 = 11.9; [H+] = 10 − pH = 10 −2.1 = 8 × 10 −3 M (1 sig. fig.) −
[OH ] =
Kw +
[H ]
=
1.0 10 −14 8 10
−3
= 1 × 10 −12 M or [OH−] = 10 − pOH = 10 −11.9 = 1 × 10 −12 M
The sample of gastric juice is acidic because the pH is less than 7.00 at 25°C. 60.
pH = 14.0 – pOH = 14.0 – 2.3 = 11.7; [H+] = 10 − pH = 10 −11.7 = 2 × 10 −12 M [OH−] =
Kw +
[H ]
=
1.0 10 −14 2 10
−12
= 5 × 10 −3 M or [OH−] = 10 − pOH = 10 −2.3 = 5 × 10 −3 M
The solution of antacid is basic because the pH is greater than 7.00 at 25°C.
CHAPTER 14
ACIDS AND BASES
657
Solutions of Acids 61.
HI is a strong acid (100% dissociated in water) and the rest are weak acids. Statement e is false. From the acid strength listing, HNO2 is a stronger weak acid than HCN, so the Ka value for HNO2 is greater than the Ka value for HCN. A 0.10 M solution of HNO2 will be more acidic than that of HCN, which means the HNO 2 solution will have the lower pH. For statement c, the stronger the acid the weaker the conjugate base. Because HI is a strong acid while HNO2 is a weak acid, NO2‒ will be a stronger base than I‒. For statement d, the acid strength ordering indicates that HOCl is a stronger acid than HCN, so solutions of HOCl are more acidic meaning they have a higher [H+].
62.
The Ka value is directly related to acid strength. The larger the K a value, the stronger the acid. As acid strength increases, more of the acid dissociates leading to a more acidic solution. The more acidic a solution, the higher the [H+] and the lower the pH. Also, the stronger the acid, the weaker the conjugate base. All these statements are true.
63.
All the acids in this problem are strong acids that are always assumed to completely dissociate in water. The general dissociation reaction for a strong acid is HA(aq) → H+(aq) + A−(aq), where A− is the conjugate base of the strong acid HA. For 0.250 M solutions of these strong acids, 0.250 M H+ and 0.250 M A− are present when the acids completely dissociate. The amount of H+ donated from water will be insignificant in this problem since H 2O is a very weak acid. a. Major species present after dissociation = H+, ClO4−, and H2O pH = –log[H+] = −log(0.250) = 0.602 b. Major species = H+, NO3−, and H2O; pH = 0.602
64.
Both are strong acids, which are assumed to completely dissociate in water. 0.0500 L × 0.050 mol/L = 2.5 × 10 −3 mol HBr = 2.5 × 10 −3 mol H+ + 2.5 × 10 −3 mol Br− 0.1500 L × 0.10 mol/L = 1.5 × 10 −2 mol HI = 1.5 × 10 −2 mol H+ + 1.5 × 10 −2 mol I− [H+] =
65.
(2.5 10 −3 + 1.5 10 −2 ) mol 0.2000 L
= 0.088 M; pH = −log(0.088) = 1.06
Strong acids are assumed to completely dissociate in water, for example, HCl(aq) + H 2O(l) → H3O+(aq) + Cl−(aq) or HCl(aq) → H+(aq) + Cl−(aq). a. A 0.10 M HCl solution gives 0.10 M H+ and 0.10 M Cl− because HCl completely dissociates. The amount of H+ from H2O will be insignificant. pH = −log[H+] = −log(0.10) = 1.00 b. 5.0 M H+ is produced when 5.0 M HClO4 completely dissociates. The amount of H+ from H2O will be insignificant. pH = −log(5.0) = −0.70 (Negative pH values just indicate very concentrated acid solutions.)
658
CHAPTER 14
ACIDS AND BASES
c. 1.0 × 10−11 M H+ is produced when 1.0 × 10−11 M HI completely dissociates. If you take the negative log of 1.0 × 10−11, this gives pH = 11.00. This is impossible! We dissolved an acid in water and got a basic pH. What we must consider in this problem is that water by itself donates 1.0 × 10−7 M H+. We can normally ignore the small amount of H + from H2O except when we have a very dilute solution of an acid (as in the case here). Therefore, the pH is that of neutral water (pH = 7.00) because the amount of HI present is insignificant. 66.
HNO3(aq) → H+(aq) + NO3−(aq); HNO3 is a strong acid, which means it is assumed to completely dissociate in water. The initial concentration of HNO3 will equal the [H+] donated by the strong acid. a. pH = −log[H+] = −log(2.0 × 10−2) = 1.70 b. pH = −log(4.0) = −0.60 c. Because the concentration of HNO3 is so dilute, the pH will be that of neutral water (pH = 7.00). In this problem, water is the major H+ producer present. Whenever the strong acid has a concentration less than 1.0 × 10−7 M, the [H+] contribution from water must be considered.
67.
[H+] = 10−pH = 10−4.25 = 5.6 × 10−5 M. Because HBr is a strong acid, a 5.6 × 10−5 M HBr solution is necessary to produce a pH = 4.25 solution.
68.
[H+] = 10−pH = 10−5.10 = 7.9 × 10−6 M. Because HNO3 is a strong acid, a 7.9 × 10−6 M HNO3 solution will produce 7.9 × 10−6 M H+ along with 7.9 × 10−6 M NO3− , giving a pH = 5.10. Mol NO3− = 0.250 L
69.
7.9 10 −6 mol NO3− = 2.0 10−6 mol NO3− L
HCl is a strong acid. [H+] = 10 −1.50 = 3.16 × 10 −2 M (carrying one extra sig. fig.) M1V1 = M2V2, V1 =
M 2 V2 3.16 10 −2 mol/L 1.6 L = 4.2 × 10 −3 L = M1 12 mol/L
Add 4.2 mL of 12 M HCl to water with mixing; add enough water to make 1600 mL of solution. The resulting solution will have [H +] = 3.2 × 10 −2 M and pH = 1.50. 70.
50.0 mL conc. HCl soln ×
1.19 g 38 g HCl 1 mol HCl = 0.62 mol HCl mL 100 g conc. HCl soln 36.5 g
20.0 mL conc. HNO3 soln ×
70. g HNO 3 1 mol HNO 3 1.42 g = 0.32 mol HNO3 mL 100 g soln 63.0 g HNO 3
HCl(aq) → H+(aq) + Cl−(aq) and HNO3(aq) → H+(aq) + NO3−(aq) (Both are strong acids.) So we will have 0.62 + 0.32 = 0.94 mol of H+ in the final solution. [H+] =
0.94 mol = 0.94 M; pH = −log[H+] = −log(0.94) = 0.027 = 0.03 1.00 L
[OH−] =
Kw 1.0 10 −14 = = 1.1 × 10−14 M 0.94 [H + ]
CHAPTER 14 71.
ACIDS AND BASES
659
HA is a weak acid, so write out the Ka reaction and set-up the general ICE table.
⇌
HA Initial
H+
A−
+
1.0 mol/1.0 L ~0 0 Let x mol/L HA dissociate to reach equilibrium –x → +x +x 1.0 – x x x
Change Equil.
Since HA is a weak acid, the amount of the weak acid that dissociates, x, will typically be less than 5%. Assuming this, [HA] > [H+] and [A‒]. Since this is an acidic solution, [H+] > [OH‒]. And from the ICE table, [H+] = [A‒] = x. Only answer d is a correct statement. 72.
HOCl is a weak acid (Ka = 3.5 × 10−8). Let’s write out the ICE table for this weak acid problem. HOCl Initial
⇌
H+
OCl−
+
1.0 mol/2.0 L ~0 0 Let x mol/L HOCl dissociate to reach equilibrium –x → +x +x 0.50 M – x x x
Change Equil.
Because Ka for HOCl << 1, not a lot of HOCl will dissociate. Typically, weak acids are less than 5% dissociated in water, so x will be less than 5% of 0.50 M. HOCl is an acid, so the solution will be acidic where [H+] > [OH‒]. From the ICE table, [H+] = [OCl‒] = x. Answer e is correct. 73.
a. HNO2 (Ka = 4.0 × 10 −4 ) and H2O (Ka = Kw = 1.0 × 10 −14 ) are the major species. HNO2 is a much stronger acid than H2O, so it is the major source of H+. However, HNO2 is a weak acid (Ka < 1), so it only partially dissociates in water. We must solve an equilibrium problem to determine [H+]. In the Solutions Guide, we will summarize the initial, change, and equilibrium concentrations into one table called the ICE table. Solving the weak acid problem: HNO2 Initial Change Equil.
⇌
H+
+
NO2−
0.250 M ~0 0 x mol/L HNO2 dissociates to reach equilibrium –x → +x +x 0.250 – x x x −
Ka =
[ H + ][ NO 2 ] x2 = 4.0 × 10 −4 = ; if we assume x << 0.250, then: [HNO 2 ] 0.250 − x
4.0 × 10 −4
x2 , x = 0.250
We must check the assumption:
4.0 10 − 4 (0.250 ) = 0.010 M
x 0.010 100 = × 100 = 4.0% 0.250 0.250
All the assumptions are good. The H+ contribution from water (1 × 10 −7 M) is negligible, and x is small compared to 0.250 (percent error = 4.0%). If the percent error is less than
660
CHAPTER 14
ACIDS AND BASES
5% for an assumption, we will consider it a valid assumption (called the 5% rule). Finishing the problem: x = 0.010 M = [H+]; pH = –log(0.010) = 2.00 b. CH3CO2H or HC2H3O2 (Ka = 1.8 × 10 −5 ) and H2O (Ka = Kw = 1.0 × 10 −14 ) are the major species. CH3CO2H is the major source of H+. Solving the weak acid problem: CH3CO2H
⇌
H+
+
CH3CO2−
Initial
0.250 M ~0 0 x mol/L CH3CO2H dissociates to reach equilibrium Change –x → +x +x Equil. 0.250 – x x x − + 2 [H ][CH 3CO 2 ] x2 x Ka = , 1.8 × 10 −5 = (assuming x << 0.250) 0.250 0.250 − x [CH 3CO 2 H] x = 2.1 × 10-3 M; checking assumption:
2.1 10 −3 × 100 = 0.84%. Assumptions good. 0.250
[H+] = x = 2.1 × 10 −3 M; pH = −log(2.1 × 10 −3 ) = 2.68 74.
a. HOC6H5 (Ka = 1.6 × 10 −10 ) and H2O (Ka = Kw = 1.0 × 10 −14 ) are the major species. The major equilibrium is the dissociation of HOC6H5. Solving the weak acid problem: HOC6H5 Initial Change Equil.
⇌
H+
+
OC6H5−
0.250 M ~0 0 x mol/L HOC6H5 dissociates to reach equilibrium –x → +x +x 0.250 – x x x
Ka = 1.6 × 10
−10
−
[H + ][OC6 H 5 ] x2 x2 = = (assuming x << 0.250) 0.250 0.250 − x [HOC6 H 5 ]
x = [H+] = 6.3 × 10 −6 M; checking assumption: x is 2.5 × 10–3% of 0.250, so assumption is valid by the 5% rule. pH = –log(6.3 × 10 −6 ) = 5.20 b. HCN (Ka = 6.2 × 10 −10 ) and H2O are the major species. HCN is the major source of H+. HCN Initial Change Equil.
⇌
H+
+
CN−
0.250 M ~0 0 x mol/L HCN dissociates to reach equilibrium –x → +x +x 0.250 – x x x
Ka = 6.2 × 10 −10 =
[H + ][CN − ] x2 x2 = 0.250 0.250 − x [HCN]
(assuming x << 0.250)
CHAPTER 14
ACIDS AND BASES
661
x = [H+] = 1.2 × 10 −5 M; checking assumption: x is 4.8 × 10−3% of 0.250. Assumptions good. pH = –log(1.2 × 10 −5 ) = 4.92 75.
This is a weak acid in water. Solving the weak acid problem: HF Initial Change Equil.
⇌
H+
+
F−
Ka = 7.2 × 10−4
0.020 M ~0 0 x mol/L HF dissociates to reach equilibrium −x → +x +x 0.020 − x x x
Ka = 7.2 × 10−4 =
[H + ][F − ] x2 x2 = [HF] 0.020 − x 0.020
(assuming x << 0.020)
x = [H+] = 3.8 × 10−3 M; check assumptions: x 3.8 10 −3 × 100 = 19% 100 = 0.020 0.020
The assumption x << 0.020 is not good (x is more than 5% of 0.020). We must solve x2/(0.020 − x) = 7.2 × 10−4 exactly by using either the quadratic formula or the method of successive approximations (see Appendix 1 of the text). Using successive approximations, we let 0.016 M be a new approximation for [HF]. That is, in the denominator try x = 0.0038 (the value of x we calculated making the normal assumption) so that 0.020 − 0.0038 = 0.016; then solve for a new value of x in the numerator. x2 x2 = 7.2 × 10−4, x = 3.4 × 10−3 0.020 − x 0.016 We use this new value of x to further refine our estimate of [HF], that is, 0.020 − x = 0.020 − 0.0034 = 0.0166 (carrying an extra sig. fig.). x2 x2 = 7.2 × 10−4, x = 3.5 × 10−3 0.020 − x 0.0166
We repeat until we get a self-consistent answer. This would be the same answer we would get solving exactly using the quadratic equation. In this case it is, x = 3.5 × 10−3. Thus: [H+] = [F−] = x = 3.5 × 10−3 M; [OH−] = Kw/[H+] = 2.9 × 10−12 M [HF] = 0.020 − x = 0.020 − 0.0035 = 0.017 M; pH = 2.46 Note: When the 5% assumption fails, use whichever method you are most comfortable with to solve exactly. The method of successive approximations is probably fastest when the percent error is less than ~25% (unless you have a graphing calculator). 76.
Major species: HC2H2ClO2 (Ka = 1.35 × 10 −3 ) and H2O; major source of H+: HC2H2ClO2
662
CHAPTER 14 HC2H2ClO2 Initial Change Equil.
⇌
H+
ACIDS AND BASES
+ C2H2ClO2−
0.10 M ~0 0 x mol/L HC2H2ClO2 dissociates to reach equilibrium –x → +x +x 0.10 – x x x
Ka = 1.35 × 10 −3 =
x2 x2 , x = 1.2 × 10 −2 M 0.10 0.10 − x
Checking the assumptions finds that x is 12% of 0.10, which fails the 5% rule. We must solve 1.35 × 10 −3 = x2/(0.10 – x) exactly using either the method of successive approximations or the quadratic equation. Using either method gives x = [H+] = 1.1 × 10 −2 M. pH = –log[H+] = –log(1.1 × 10‒2) = 1.96. 77.
HC3H5O2 (Ka = 1.3 × 10 −5 ) and H2O (Ka = Kw = 1.0 × 10 −14 ) are the major species present. HC3H5O2 will be the dominant producer of H+ because HC3H5O2 is a stronger acid than H2O. Solving the weak acid problem: HC3H5O2 Initial Change Equil.
⇌
H+
C3H5O2-
+
0.100 M ~0 0 x mol/L HC3H5O2 dissociates to reach equilibrium –x → +x +x 0.100 – x x x
Ka = 1.3 × 10 −5 =
−
[H + ][C3 H 5O 2 ] x2 x2 = 0.100 [HC3 H 5O 2 ] 0.100 − x
x = [H+] = 1.1 × 10 −3 M; pH = –log(1.1 × 10 −3 ) = 2.96 Assumption follows the 5% rule (x is 1.1% of 0.100). [H+] = [C3H5O2−] = 1.1 × 10 −3 M; [OH−] = Kw/[H+] = 9.1 × 10 −12 M [HC3H5O2] = 0.100 – 1.1 × 10 −3 = 0.099 M Percent dissociation = 78.
[H + ] 1.1 10 −3 × 100 = × 100 = 1.1% [ HC 3 H 5 O 2 ] 0 0.100
This is a weak acid in water. We must solve a weak acid problem. Let HBz = C 6H5CO2H. 0.56 g HBz ×
1 mol HBz = 4.6 × 10−3 mol; [HBz]0 = 4.6 × 10−3 M 122 .1 g
HBz Initial Change Equil.
⇌
H+
+
Bz-
4.6 × 10−3 M ~0 0 x mol/L HBz dissociates to reach equilibrium −x → +x +x 4.6 × 10−3 − x x x
CHAPTER 14
ACIDS AND BASES
Ka = 6.4 × 10−5 =
663
[H + ][Bz − ] x2 x2 = [HBz] (4.6 10 −3 − x) 4.6 10 −3
x = [H+] = 5.4 × 10−4; check assumptions:
x 5.4 10 −4 × 100 = 12% 100 = −3 4.6 10 4.6 10 −3
Assumption is not good (x is 12% of 4.6 × 10−3). When assumption(s) fail, we must solve exactly using the quadratic formula or the method of successive approximations (see Appendix 1 of text). Using successive approximations: x2 = 6.4 × 10−5, x = 5.1 × 10−4 (4.6 10 −3 ) − (5.4 10 − 4 ) x2 = 6.4 × 10−5, x = 5.1 × 10−4 M (consistent answer) (4.6 10 −3 ) − (5.1 10 − 4 )
Thus: x = [H+] = [Bz−] = [C6H5CO2−] = 5.1 × 10−4 M [HBz] = [C6H5CO2H] = 4.6 × 10−3 − x = 4.1 × 10−3 M pH = −log(5.1 × 10−4) = 3.29; pOH = 14.00 − pH = 10.71; [OH−] = 10−10.71 = 1.9 × 10−11 M 2 tablets
79.
[HC9H7O4] = HC9H7O4 Initial Change Equil.
0.325 g HC9 H 7 O 4 1 mol HC9 H 7 O 4 tablet 180 .15 g = 0.0152 M 0.237 L
⇌
H+
+
C9H7O4−
0.0152 M ~0 0 x mol/L HC9H7O4 dissociates to reach equilibrium –x → +x +x 0.0152 – x x x
Ka = 3.3 × 10
−4
−
[ H + ] [C 9 H 7 O 4 ] x2 x2 = = , x = 2.2 × 10 −3 M 0.0152 [HC9 H 7 O 4 ] 0.0152 − x
Assumption that 0.0152 – x 0.0152 fails the 5% rule:
2.2 10 −3 × 100 = 14% 0.0152
Using successive approximations or the quadratic equation gives an exact answer of x = 2.1 × 10 −3 M. [H+] = x = 2.1 × 10 −3 M; pH = –log(2.1 × 10 −3 ) = 2.68
1 mol HC7 H 4 NSO 3 183.19 g = 1.6 M 0.34 L
100.0 g HC7 H 4 NSO 3 80.
[HC7H4NSO3] =
664
CHAPTER 14 HC7H4NSO3 Initial Change Equil.
⇌
H+
+
ACIDS AND BASES
C7H4NSO3−
1.6 M ~0 0 x mol/L HC7H4NSO3 dissociates to reach equilibrium –x → +x +x 1.6 – x x x
Ka = 2.0 × 10
−
[H + ] [C 7 H 4 NSO 3 ] x2 x2 = = , x = 1.8 × 10 −6 M [HC 7 H 4 NSO 3 ] 1 .6 1.6 − x
−12
Checking assumption: x is 1.1 × 10-4% of 1.6. Assumptions good. [H+] = 1.8 × 10 −6 M; pH = –log(1.8 × 10 −6 ) = 5.74 81.
HA Initial Change Equil.
⇌
H+
+
A−
0.10 M ~0 0 Let x mol/L HA dissociate to reach equilibrium –x → +x +x 0.10 ‒ x x x
From the ICE table, [H+] = [A‒]. Since pH = 3.0, [H+] = [A‒] = 10‒3.0 = 1 × 10‒3 M. 82.
HX Initial Change Equil.
⇌
H+
+
X−
1.0 × 10−2 M ~0 0 Let x mol/L HX dissociate to reach equilibrium –x → +x +x 1.0 × 10−2 – x x x
From the problem, [X−] = 2.5 10-6 M. This equals x which also equals the [H+]. Note that x < 1.0 × 10−2 , so HX is a weak acid (only some of it dissociates). pH = ‒log(2.5 10-6) = 5.60; statement e is false. The other statements are all true. [HX] = 1.0 × 10−2 – 2.5 10-6 = 1.0 × 10−2 . pOH = 14.00 − pH = 8.40; [OH−] = 10−8.40 = 4.0 × 10−9 M. % dissociation = (x/ 1.0 × 10−2 )100 = 2.5 10−2%. 83.
HF and HOC6H5 are both weak acids with Ka values of 7.2 × 10 −4 and 1.6 × 10 −10 , respectively. Since the Ka value for HF is much greater than the K a value for HOC6H5, HF will be the dominant producer of H+ (we can ignore the amount of H+ produced from HOC6H5 because it will be insignificant). HF Initial Change Equil.
⇌
H+
+
F−
1.0 M ~0 0 x mol/L HF dissociates to reach equilibrium –x → +x +x 1.0 – x x x
CHAPTER 14
ACIDS AND BASES
Ka = 7.2 × 10 −4 =
665
[H + ][F− ] x2 x2 = 1 .0 1.0 − x [HF]
x = [H+] = 2.7 × 10 −2 M; pH = –log(2.7 × 10 −2 ) = 1.57; assumptions good. Solving for [OC6H5−] using HOC6H5 ⇌ H+ + OC6H5- equilibrium: Ka = 1.6 × 10 −10 =
−
− (2.7 10 −2 )[OC 6 H 5 ] [H + ][OC6 H 5 ] = , [OC6H5−] = 5.9 × 10 −9 M 1.0 [HOC6 H 5 ]
Note that this answer indicates that only 5.9 × 10 −9 M HOC6H5 dissociates, which confirms that HF is truly the only significant producer of H+ in this solution. 84.
0.50 M HA, Ka = 1.0 × 10−3; 0.20 M HB, Ka = 1.0 × 10−10; 0.10 M HC, Ka = 1.0 × 10−12 Major source of H+ is HA because its Ka value is significantly larger than other Ka values. HA Initial Change Equil. Ka =
⇌
H+
+
A−
0.50 M ~0 0 x mol/L HA dissociates to reach equilibrium –x → +x +x 0.50 − x x x
0.022 x2 x2 , 1.0 × 10−3 , x = 0.022 M = [H+], × 100 = 4.4% error 0.50 0.50 0.50 − x
Assumption good. Let's check out the assumption that only HA is an important source of H+. For HB: 1.0 × 10−10 =
(0.022 ) [B − ] , [B−] = 9.1 × 10−10 M (0.20)
At most, HB will produce an additional 9.1 × 10 −10 M H+. Even less will be produced by HC. Thus our original assumption was good. [H+] = 0.022 M. 85.
In all parts of this problem, acetic acid (HC2H3O2) is the best weak acid present. We must solve a weak acid problem. a.
HC2H3O2 Initial Change Equil.
⇌
H+
+ C2H3O2−
0.50 M ~0 0 x mol/L HC2H3O2 dissociates to reach equilibrium −x → +x +x 0.50 − x x x −
Ka = 1.8 × 10−5 =
[H + ][C 2 H 3O 2 ] x2 x2 = [HC2 H 3O 2 ] 0.50 − x 0.50
x = [H+] = [C2H3O2−] = 3.0 × 10−3 M; assumptions good.
666
CHAPTER 14 Percent dissociation =
ACIDS AND BASES
3.0 10 −3 [H + ] × 100 = × 100 = 0.60% 0.50 [HC2 H 3O 2 ]0
b. The setup for solutions b and c are similar to solution a except that the final equation is different because the new concentration of HC 2H3O2 is different. Ka = 1.8 × 10−5 =
x2 x2 0.050 − x 0.050
x = [H+] = [C2H3O2−] = 9.5 × 10−4 M; assumptions good. Percent dissociation = c. Ka = 1.8 × 10−5 =
9.5 10 −4 × 100 = 1.9% 0.050
x2 x2 0.0050 − x 0.0050
x = [H+] = [C2H3O2−] = 3.0 × 10−4 M; check assumptions. Assumption that x is negligible is borderline (6.0% error). We should solve exactly. Using the method of successive approximations (see Appendix 1 of the text): 1.8 × 10−5 =
x2 x2 , x = 2.9 × 10−4 = −4 0 . 0047 0.0050 − (3.0 10 )
Next trial also gives x = 2.9 × 10−4. Percent dissociation =
2.9 10 −4 × 100 = 5.8% 5.0 10 −3
d. As we dilute a solution, all concentrations are decreased. Dilution will shift the equilibrium to the side with the greater number of particles. For example, suppose we double the volume of an equilibrium mixture of a weak acid by adding water; then: [H + ]eq [X − ]eq 2 2 = 1 K Q= a 2 [HX]eq 2
Q < Ka, so the equilibrium shifts to the right or toward a greater percent dissociation. e. [H+] depends on the initial concentration of weak acid and on how much weak acid dissociates. For solutions a-c, the initial concentration of acid decreases more rapidly than the percent dissociation increases. Thus [H+] decreases.
86.
HOBr Initial Change Equil.
⇌
H+
+
OBr−
Ka = 2.0 × 10−9
0.70 M ~0 0 x mol/L HOBr dissociates to reach equilibrium −x → +x +x 0.70 − x x x
CHAPTER 14
ACIDS AND BASES
Ka = 2.0 × 10−9 =
[H + ][OBr − ] x2 x2 , x = 3.7 × 10−5 M; assumptions good = [HOBr] 0.70 − x 0.70
Percent dissociation = 87.
667
3.7 10−5 × 100 = 5.3 × 10−3% 0.70
Let HA symbolize the weak acid. Set up the problem like a typical weak acid equilibrium problem.
⇌
HA Initial Change Equil.
H+
A−
+
0.15 M ~0 0 x mol/L HA dissociates to reach equilibrium –x → +x +x 0.15 – x x x
If the acid is 3.0% dissociated, then x = [H+] is 3.0% of 0.15: x = 0.030 × (0.15 M) = 4.5 × 10 −3 M. Now that we know the value of x, we can solve for Ka. Ka =
[H + ][A − ] x2 (4.5 10 −3 ) 2 = = = 1.4 × 10 −4 0.15 − x [HA] 0.15 − (4.5 10 −3 )
88.
⇌
HOCN Initial Change Equil.
H+
+
OCN−
0.010 M ~0 0 x mol/L HOCN dissociates to reach equilibrium –x → +x +x 0.010 – x x x
If the acid is 17% dissociated, then x = [H+] is 17% of 0.010: x = 0.17 × (0.010 M) = 1.7 × 10 −3 M. Now that we know the value of x, we can solve for Ka. Ka = 89.
(1.7 10 −3 ) 2 [H + ][OCN− ] x2 = = = 3.5 × 10 −4 0.010 − x [HOCN] 0.010 − (1.7 10 −3 )
HClO4 is a strong acid with [H+] = 0.040 M. This equals the [H+] in the trichloroacetic acid solution. Set up the problem using the Ka equilibrium reaction for CCl3CO2H. CCl3CO2H Initial Equil.
⇌
0.050 M 0.050 − x −
H+ ~0 x
+
CCl3CO2− 0 x
Ka =
[H + ][CCl3CO 2 ] x2 ; from the problem, x = [H+] = 4.0 × 10−2 M = [CCl3CO 2 H] 0.050 − x
Ka =
(4.0 10 −2 ) 2 = 0.16 0.050 − (4.0 10 − 2 )
668 90.
CHAPTER 14 Set up the problem using the Ka equilibrium reaction for HOBr.
⇌
HOBr Initial
Ka = Ka =
H+
OBr−
+
0.063 M ~0 0 x mol/L HOBr dissociates to reach equilibrium −x → +x +x 0.063 − x x x
Change Equil.
91.
ACIDS AND BASES
[H + ][OBr − ] x2 ; from pH = 4.95: x = [H+] = 10−pH = 10−4.95 = 1.1 × 10−5 M = [HOBr] 0.063 − x
(1.1 10 −5 ) 2 0.063 − 1.1 10 −5
= 1.9 × 10−9
HCl is a strong acid, so [H+] = 0.010 M. Set up the ICE table using the Ka reaction for HF. HF Initial
⇌
H+
+
F−
C ~0 0 x mol/L HF dissociates to reach equilibrium −x → +x +x C−x x x
Change Equil.
Ka = 1.8 × 10−4 = 7.2 × 10−4 =
where C = [HF]0
[H + ][F- ] x2 = , where x = [H+] = 0.010 M [HF] C−x
1.0 10−4 [H + ]2 (0.010)2 , C − 0.010 = , C = 0.15 M = C - [H + ] C − 0.010 7.2 10−4
A 0.15 M HF solution will have [H+] = 0.010 M.
92.
Major species: HC2H3O2 (acetic acid) and H2O; major source of H+: HC2H3O2 HC2H3O2 Initial
⇌
H+
+
C2H3O2−
C ~0 0 x mol/L HC2H3O2 dissociates to reach equilibrium −x → +x +x C−x x x
Change Equil.
where C = [HC2H3O2]0
−
Ka = 1.8 × 10−5 =
[H + ][C 2 H 3 O 2 ] x2 = , where x = [H+] [HC 2 H 3 O 2 ] C−x
[H + ] 2 1.8 × 10 = ; from pH = 3.0: [H+] = 10−3.0 = 1 × 10−3 M C − [H + ] −5
1.8 × 10−5 =
(1 10 −3 ) 2
1 10 −6 −3 , C − (1 10 ) = , C = 5.7 × 10−2 6 × 10−2 M C − (1 10 −3 ) 1.8 10 −5
A 6 × 10−2 M acetic acid solution will have pH = 3.0.
CHAPTER 14 93.
[HA]0 =
ACIDS AND BASES 1.0 mol = 0.50 mol/L; solve using the Ka equilibrium reaction. 2.0 L
HA Initial Equil. Ka =
669
⇌
H+
0.50 M 0.50 – x
A−
+
~0 x
0 x
[H + ][A − ] x2 = ; in this problem, [HA] = 0.45 M so: 0.50 − x [HA]
[HA] = 0.45 M = 0.50 M − x, x = 0.05 M; 94.
HX Initial Change Equil.
⇌
H+
+
Ka =
(0.05) 2 = 6 × 10 −3 0.45
X−
I ~0 0 x mol/L HX dissociates to reach equilibrium –x → +x +x I–x x x
where I = [HX]0
From the problem, x = 0.25(I) and I – x = 0.30 M. I – 0.25(I) = 0.30 M, I = 0.40 M and x = 0.25(0.40 M) = 0.10 M Ka =
[H + ][X − ] (0.10) 2 x2 = = = 0.033 I−x 0.30 [HX]
Solutions of Bases 95.
96.
97.
All Kb reactions refer to the base reacting with water to produce the conjugate acid of the base and OH−. + [ NH4 ][OH− ] a. NH3(aq) + H2O(l) ⇌ NH4+(aq) + OH−(aq) Kb = [ NH3 ] b. C5H5N(aq) + H2O(l) ⇌C5H5NH+(aq) + OH−(aq)
Kb =
[C5 H 5 NH + ][OH − ] [C 5 H 5 N ]
a. C6H5NH2(aq) + H2O(l) ⇌C6H5NH3+(aq) + OH−(aq)
Kb =
[C 6 H 5 NH3 ][OH− ] [C 6 H 5 NH2 ]
b. (CH3)2NH(aq) + H2O(l) ⇌ (CH3)2NH2+(aq) + OH−(aq)
Kb =
[(CH 3 ) 2 NH 2 ][OH − ] [(CH 3 ) 2 NH]
+
+
NO3−: Because HNO3 is a strong acid, NO3− is a terrible base (Kb << Kw). All conjugate bases of strong acids have no base strength. H2O: Kb = Kw = 1.0 × 10 −14 ; NH3: Kb = 1.8 × 10 −5 ; C5H5N: Kb = 1.7 × 10 −9 Base strength = NH3 > C5H5N > H2O > NO3- (As Kb increases, base strength increases.)
670
CHAPTER 14
ACIDS AND BASES
Excluding water, the acid list contains the conjugate acids of the bases in the initial list. In general, the stronger the base, the weaker is the conjugate acid. Note: Even though NH4+ and C5H5NH+ are conjugate acids of weak bases, they are still weak acids with K a values between Kw and 1. Prove this to yourself by calculating the K a values for NH4+ and C5H5NH+ (Ka = Kw/Kb). Acid strength = HNO3 > C5H5NH+ > NH4+ > H2O 98.
The base with the largest Kb value is the strongest base ( K b, C6 H5 NH2 = 3.8 × 10 −10 ,
K b, CH3NH2 = 4.4 × 10 −4 ). OH− is the strongest base possible in water. a. C6H5NH2
b. C6H5NH2
c. OH−
d. CH3NH2
The acid with the largest Ka value is the strongest acid. To calculate Ka values for C6H5NH3+ and CH3NH3+, use Ka = Kw/Kb, where Kb refers to the bases C6H5NH2 or CH3NH2. Because C6H5NH2 has a smaller Kb value (is a weaker base) than CH3NH2, C6H5NH3+ will be a stronger conjugate acid than CH3NH3+. e. HClO4 (a strong acid) 99.
f.
C6H5NH3+
g. C6H5NH3+
NaOH(aq) → Na+(aq) + OH−(aq); NaOH is a strong base that completely dissociates into Na+ and OH−. The initial concentration of NaOH will equal the concentration of OH − donated by NaOH. a. [OH−] = 0.10 M; pOH = –log[OH−] = –log(0.10) = 1.00 pH = 14.00 – pOH = 14.00 – 1.00 = 13.00 Note that H2O is also present, but the amount of OH− produced by H2O will be insignificant as compared to the 0.10 M OH− produced from the NaOH. b. The [OH−] concentration donated by the NaOH is 1.0 × 10 −10 M. Water by itself donates 1.0 × 10 −7 M. In this exercise, water is the major OH− contributor, and [OH−] = 1.0 × 10 −7 M. pOH = −log(1.0 × 10 −7 ) = 7.00; pH = 14.00 − 7.00 = 7.00 c. [OH−] = 2.0 M; pOH = −log(2.0) = −0.30; pH = 14.00 – (–0.30) = 14.30
100.
a. Ca(OH)2 → Ca2+ + 2 OH−; Ca(OH)2 is a strong base and dissociates completely. [OH−] = 2(0.00040) = 8.0 × 10 −4 M; pOH = –log[OH−] = 3.10 pH = 14.00 − pOH = 10.90 b.
25 g KOH 1 mol KOH = 0.45 mol KOH/L L 56.11 g KOH
CHAPTER 14
ACIDS AND BASES
671
KOH is a strong base, so [OH−] = 0.45 M; pOH = −log (0.45) = 0.35; pH = 13.65 c.
150 .0 g NaOH 1 mol = 3.750 M; NaOH is a strong base, so [OH−] = 3.750 M. L 40.00 g
pOH = −log(3.750) = −0.5740 and pH = 14.0000 − (−0.5740) = 14.5740 Although we are justified in calculating the answer to four decimal places, in reality, the pH can only be measured to 0.01 pH units. 101.
a. Major species: K+, OH−, H2O (KOH is a strong base.) [OH−] = 0.015 M, pOH = −log(0.015) = 1.82; pH = 14.00 − pOH = 12.18 b. Major species: Ba2+, OH−, H2O; Ba(OH)2(aq) → Ba2+(aq) + 2 OH−(aq); because each mole of the strong base Ba(OH)2 dissolves in water to produce two mol OH−, [OH−] = 2(0.015 M) = 0.030 M. pOH = –log(0.030) = 1.52; pH =14.00 − 1.52 = 12.48
102.
Major species: Na+, Ca2+, OH−, H2O [NaOH and Ca(OH)2 are both strong bases.] Mol OH‒ from NaOH = 0.1000 L ×
0.020 mol NaOH 1 mol OH − = 0.0020 mol OH− L mol NaOH
Mol OH‒ from Ca(OH)2 = 0.1500 L× [OH−] =
0.010 mol Ca(OH)2 2 mol OH − = 0.0030 mol OH− L mol Ca(OH)2
total mol OH − 0.0020 mol + 0.0030 mol = 0.020 M = total volume 0.1000 L + 0.1500 L
pOH = –log[0.020] = 1.70; pH = 14.00 − 1.70 = 12.30 103.
pOH = 14.00 – 11.56 = 2.44; [OH−] = [KOH] = 10 −2.44 = 3.6 × 10 −3 M 0.8000 L
104.
3.6 10 −3 mol KOH 56.11 g KOH = 0.16 g KOH L mol KOH
pH = 10.50; pOH = 14.00 – 10.50 = 3.50; [OH-] = 10-3.50 = 3.2 × 10-4 M Sr(OH)2(aq) → Sr2+(aq) + 2 OH−(aq); Sr(OH)2 donates 2 mol OH− per mol Sr(OH)2. [Sr(OH)2] =
3.2 10 −4 mol OH − 1 mol Sr (OH) 2 × = 1.6 × 10 −4 M Sr(OH)2 L 2 mol OH−
A 1.6 × 10-4 M Sr(OH)2 solution will produce a pH = 10.50 solution.
672 105.
CHAPTER 14
ACIDS AND BASES
For a weak base in water, write out the K b reaction then set-up the ICE table. B(aq) + H2O(l)
⇌
BH+(aq)
+
OH−(aq)
Initial
1.0 mol/1.0 L 0 ~0 Let x mol/L B react with H2O to reach equilibrium Change –x → +x +x Equil. 1.0 – x x x Kb =
[BH + ][OH − ] x2 = [B] 1.0 − x
In a typical weak base problem, it is assumed that 1.0 – x ≈ 1.0. Assuming this assumption is good, then x will be less than 5% of 1.0 (which is typical for most weak bases). B will have the largest concentration, the concentrations of BH + and OH– will be equal and will have the next largest equilibrium concentration, and H + will have the smallest equilibrium concentration since this is a basic solution. Only statement c is true ([BH+] = [OH–] = x). 106.
Set-up the ICE table using the Kb reaction for the weak base CH3NH2. CH3NH2 + H2O Initial Change Equil.
⇌
CH3NH3+
+
OH‒
2.0 mol/4.0 L 0 ~0 Let x mol/L of CH3NH2 react with H2O to reach equilibrium –x → +x +x 0.50 – x x x
[CH3 NH 2 ][OH − ] x2 x2 (if the 5% assumption is valid) = [CH3 NH 2 ] 0.50 − x 0.50 The only statement that is false is a. The [CH 3NH2] will be larger than the [CH3NH3+] because methylamine is a weak base and not a lot of it will react with water. From the ICE table, the conjugate acid concentration, [CH3NH3+], will be equal to [OH‒] (both equal x). And since this is a basic solution, pH > 7.0 and [OH‒] > [H+]. 107.
NH3 is a weak base with Kb = 1.8 × 10 −5 . The major species present will be NH3 and H2O (Kb = Kw = 1.0 × 10 −14 ). Because NH3 has a much larger Kb value than H2O, NH3 is the stronger base present and will be the major producer of OH−. To determine the amount of OH− produced from NH3, we must perform an equilibrium calculation using the Kb reaction for NH3. NH3(aq) + H2O(l)
⇌
NH4+(aq)
+
OH−(aq)
Initial
0.150 M 0 ~0 x mol/L NH3 reacts with H2O to reach equilibrium Change –x → +x +x Equil. 0.150 – x x x Kb = 1.8 × 10 −5 =
+
[ NH4 ][OH− ] x2 x2 = (assuming x << 0.150) 0.150 0.150 − x [ NH3 ]
CHAPTER 14
ACIDS AND BASES
673
x = [OH−] = 1.6 × 10 −3 M; check assumptions: x is 1.1% of 0.150, so the assumption 0.150 – x 0.150 is valid by the 5% rule. Also, the contribution of OH − from water will be insignificant (which will usually be the case). Finishing the problem: pOH = −log[OH−] = −log(1.6 × 10 −3 M) = 2.80; pH = 14.00 − pOH = 14.00 − 2.80 = 11.20. 108.
Major species: H2NNH2 (Kb = 3.0 × 10 −6 ) and H2O (Kb = Kw = 1.0 × 10 −14 ); the weak base H2NNH2 will dominate OH− production. We must perform a weak base equilibrium calculation. H2NNH2 + H2O
⇌
H2NNH3+
OH−
+
Kb = 3.0 × 10 −6
Initial
2.0 M 0 ~0 x mol/L H2NNH2 reacts with H2O to reach equilibrium Change –x → +x +x Equil. 2.0 – x x x + − 2 2 [H 2 NNH3 ][OH ] x x Kb = 3.0 × 10 −6 = = (assuming x << 2.0) 2 .0 [H 2 NNH2 ] 2 .0 − x x = [OH−] = 2.4 × 10 −3 M; pOH = 2.62; pH = 11.38; assumptions good (x is 0.12% of 2.0). [H2NNH3+] = 2.4 × 10 −3 M; [H2NNH2] = 2.0 M; [H+] = 10 −11.38 = 4.2 × 10 −12 M 109.
These are solutions of weak bases in water. a.
C6H5NH2 + H2O Initial Change Equil.
⇌
C6H5NH3+ +
OH−
Kb = 3.8 × 10 −10
0.40 M 0 ~0 x mol/L of C6H5NH2 reacts with H2O to reach equilibrium –x → +x +x 0.40 – x x x
3.8 × 10 −10 =
x2 x2 , x = [OH−] = 1.2 × 10 −5 M; assumptions good. 0.40 0.40 − x
[H+] = Kw/[OH−] = 8.3 × 10 −10 M; pH = 9.08 b.
CH3NH2 + H2O Initial Equil.
0.40 M 0.40 − x
Kb = 4.38 × 10 −4 =
⇌ CH3NH3+ +
OH−
0 x
~0 x
Kb = 4.38 × 10 −4
x2 x2 , x = 1.3 × 10 −2 M; assumptions good. 0.40 0.40 − x
[OH−] = 1.3 × 10 −2 M; [H+] = Kw/[OH−] = 7.7 × 10 −13 M; pH = 12.11
674 110.
CHAPTER 14
ACIDS AND BASES
This is a solution of weak base in water. We must solve an equilibrium weak base problem.
⇌
(CH3)3N + H2O
Kb = 7.4 × 10 −5
(CH3)3NH+ + OH−
Initial
0.40 M 0 ~0 x mol/L of (CH3)3N reacts with H2O to reach equilibrium Change –x → +x +x Equil. 0.40 – x x x Kb = 7.4 × 10 −5 =
[(CH3 ) 3 NH + ][OH− ] x2 x2 = , x = [OH−] = 5.4 × 10 −3 M [(CH3 ) 3 N] 0.40 0.40 − x
Assumptions good (x is 1.4% of 0.40). [OH−] = 5.4 × 10 −3 M [H+] = 111.
Kw 1.0 10 −14 = = 1.9 × 10 −12 M; pH = 11.72 − −3 [OH ] 5.4 10
This is a solution of a weak base in water. We must solve the weak base equilibrium problem. C2H5NH2 Initial Change Equil.
⇌
+ H2O
C2H5NH3+ + OH−
Kb = 5.6 × 10−4
0.20 M 0 ~0 x mol/L C2H5NH2 reacts with H2O to reach equilibrium –x → +x +x 0.20 – x x x
Kb = 5.6 × 10−4 =
+
[C 2 H 5 NH3 ][OH − ] x2 x2 = (assuming x << 0.20) 0.20 0.20 − x [C 2 H 5 NH 2 ]
x = 1.1 × 10 −2 ; checking assumption:
1.1 10 −2 × 100 = 5.5% 0.20
The assumption fails the 5% rule. We must solve exactly using either the quadratic equation or the method of successive approximations (see Appendix 1 of the text). Using successive approximations and carrying extra significant figures: x2 x2 = 5.6 × 10 −4 , x = 1.0 × 10 −2 M (consistent answer) = 0.20 − 0.011 0.189
x = [OH−] = 1.0 × 10 −2 M; [H+] = 112.
(C2H5)2NH + H2O Initial Equil.
⇌
0.050 M 0.050 – x
Kb = 1.3 × 10 −3 =
1.0 10 −14 Kw = = 1.0 × 10 −12 M; pH = 12.00 [OH − ] 1.0 10 − 2 (C2H5)2NH2+ 0 x
+
+
OH− ~0 x
[(C 2 H 5 ) 2 NH 2 ][OH − ] x2 x2 = 0.050 [(C 2 H 5 ) 2 NH] 0.050 − x
x = 8.1 × 10 −3 ; assumption is bad (x is 16% of 0.20).
Kb = 1.3 × 10 −3
CHAPTER 14
ACIDS AND BASES
675
Using successive approximations: 1.3 × 10 −3 =
x2 , x = 7.4 × 10 −3 0.050 − 0.081
1.3 × 10 −3 =
x2 , x = 7.4 × 10 −3 (consistent answer) 0.050 − 0.074
[OH−] = x = 7.4 × 10 −3 M; [H+] = Kw/[OH−] = 1.4 × 10 −12 M; pH = 11.85 113.
To solve for percent ionization, we first solve the weak base equilibrium problem. a.
NH3 + H2O
⇌
NH4+
Initial 0.10 M Equil. 0.10 − x Kb = 1.8 × 10−5 =
0 x
~0 x
⇌
NH4+
0.010 M 0.010 − x
1.8 × 10−5 =
Kb = 1.8 × 10−5
1.3 10 −3 M x 100 = 100 = 1.3% [ NH 3 ] 0 0.10 M
NH3 + H2O Initial Equil.
OH−
x2 x2 , x = [OH−] = 1.3 × 10−3 M; assumptions good. 0.10 − x 0.10
Percent ionization = b.
+
+
0 x
OH− ~0 x
x2 x2 , x = [OH−] = 4.2 × 10−4 M; assumptions good. 0.010 − x 0.010
Percent ionization =
4.2 10 −4 100 = 4.2% 0.010
Note: For the same base, the percent ionization increases as the initial concentration of base decreases. c.
CH3NH2 + H2O Initial Equil.
0.10 M 0.10 − x
⇌
CH3NH3+ 0 x
+
OH−
Kb = 4.38 × 10−4
~0 x
x2 x2 , x = 6.6 × 10−3; assumption fails the 5% rule (x is 0.10 − x 0.10 6.6% of 0.10). Using successive approximations and carrying extra significant figures:
4.38 × 10−4 =
x2 x2 = = 4.38 × 10−4, x = 6.4 × 10−3 0.10 − 0.0066 0.093
Percent ionization =
6.4 10 −3 100 = 6.4% 0.10
(consistent answer)
676
CHAPTER 14
114. Initial Equil.
C5H5N + H2O ⇌ C5H5N+ + OH− 0.10 M 0 ~0 0.10 – x x x
Kb = 1.7 × 10 −9 =
Kb = 1.7 × 10 −9
x2 x2 , x = [C5H5N] = 1.3 × 10 −5 M; assumptions good. 0.10 0.10 − x
Percent C5H5N ionized = 115.
ACIDS AND BASES
1.3 10 −5 M × 100 = 1.3 × 10 −2 % 0.10 M
Using the Kb reaction to solve where PT = p-toluidine (CH3C6H4NH2):
Initial Change Equil. Kb =
PT + H2O ⇌ PTH+ + OH− 0.016 M 0 ~0 x mol/L of PT reacts with H2O to reach equilibrium −x → +x +x 0.016 − x x x
[PTH+ ][OH − ] x2 = 0.016 − x [PT]
Because pH = 8.60: pOH = 14.00 − 8.60 = 5.40 and [OH−] = x = 10−5.40 = 4.0 × 10−6 M Kb = 116.
(4.0 10 −6 ) 2 = 1.0 × 10−9 0.016 − (4.0 10 −6 )
Let cod = codeine, C18H21NO3; using the Kb reaction to solve: cod + H2O
⇌
codH+
OH−
+
1.7 × 10 −3 M 0 ~0 x mol/L codeine reacts with H2O to reach equilibrium Change –x → +x +x −3 Equil. 1.7 × 10 – x x x Initial
Kb =
x2 ; pH = 9.59; pOH = 14.00 − 9.59 = 4.41 1.7 10 −3 − x
[OH−] = x = 10 −4.41 = 3.9 × 10 −5 M; Kb =
(3.9 10 −5 ) 2 1.7 10
−3
−5
− (3.9 10 )
= 9.2 × 10 −7
Polyprotic Acids 117.
H2SO3(aq) ⇌ HSO3−(aq) + H+(aq)
K a1 =
HSO3−(aq) ⇌ SO32−(aq) + H+(aq)
Ka2 =
−
[HSO3 ][H + ] [H 2SO 3 ]
[SO 32 − ][ H + ] −
[ HSO 3 ]
CHAPTER 14 118.
119.
ACIDS AND BASES
677 −
H3C6H5O7(aq) ⇌ H2C6H5O7 (aq) + H (aq)
[H 2 C 6 H 5O 7 ][H + ] K a1 = [ H 3C 6 H 5 O 7 ]
H2C6H5O7−(aq) ⇌ HC6H5O72−(aq) + H+(aq)
Ka2 =
HC6H5O72−(aq) ⇌ C6H5O73−(aq) + H+(aq)
K a3 =
−
+
2−
[ HC 6 H 5 O 7 ][ H + ] −
[H 2 C 6 H 5O 7 ] 3−
[C 6 H 5 O 7 ][H + ] 2−
[ HC 6 H 5 O 7 ]
For H2CO3, The relevant reactions are: H2CO3 ⇌ H+ + HCO3−
K a1 = 4.3 × 10−7; HCO3− ⇌ H+ + CO32− K a 2 = 5.6 × 10−11
H2CO3 is a weak acid, and since K a1 >> K a 2 , only the first dissociation step makes an important contribution to [H+].
⇌
H2CO3 Initial Equil.
0.10 M 0.10 − x
K a1 = 4.3 × 10−7 =
H+
+ HCO3−
~0 x
0 x
−
[H + ][HCO3 ] x2 = [H 2 CO3 ] 0.10 − x
When solving this, x = [H+] will be much less than 0.10 M since H2CO3 is a weak acid with Ka ~ 10‒7. For H2SO4, the first dissociation occurs to completion. So, a 0.10 M H2SO4 solution gives 0.10 M HSO4− and 0.10 M H+. The hydrogen sulfate ion (HSO4−) is a pretty good weak acid with K a 2 = 1.2 × 10−2. We also need to consider this equilibrium for additional H+ production: HSO4− Initial Change Equil.
⇌
H+
+
SO42−
0.10 M 0.10 M 0 − x mol/L HSO4 dissociates to reach equilibrium −x → +x +x 0.10 − x 0.10 + x x
Since H2SO4 is both a strong acid and a pretty good weak acid problem in one, x will not be insignificant compared to 0.10, so the [H+] will be greater than 0.10 M. From the set-ups above, the only true statement is c. Both the Ka values for H2SO4 are greater than the Ka values for H2CO3. Hence H2SO4 is a better acid than H2CO3, and HSO4− is a better acid than HCO3−. This means that more of HSO4− will dissociate than HCO3−, so more of the conjugate base of HSO4− will be present at equilibrium. Therefore at equilibrium, [SO42–] > [CO32–]. For statements a and d, H2SO4 is a much better acid than H2CO3, so H2SO4 solution will have the higher [H+] and lower pH. For statement b, the ICE table tell us that the concentration of H+ will be greater than 0.10 M. And for statement e, since H2SO4 is a strong acid while H2CO3 is a weak acid, the equilibrium concentration of H 2CO3 will be larger than the equilibrium concentration of H2SO4.
678 120.
CHAPTER 14
ACIDS AND BASES
For H2SO4, the first dissociation occurs to completion. So a 0.10 M H2SO4 solution gives 0.10 M HSO4− and 0.10 M H+. The hydrogen sulfate ion (HSO4−) is a pretty good weak acid with K a 2 = 1.2 × 10−2. We also need to consider this equilibrium for additional H + production:
⇌
HSO4− Initial
H+
SO42−
+
0.10 M 0.10 M 0 x mol/L HSO4− dissociates to reach equilibrium −x → +x +x 0.10 − x 0.10 + x x
Change Equil.
K a 2 = 0.012 =
(0.10 + x ) x x, x = 0.012; assumption is poor (12% error). 0.10 − x
To solve this problem, we need to use the quadratic equation. However, we do not need to solve the problem exactly to answer the question. In this problem, x will not be insignificant. A good guess for the value of x is around 0.01 M (from the calculation above). From the ICE table, the order of the concentrations will be [SO42−] < [HSO4−] < [H+] (answer e is correct). Note that H+ and H3O+ are the same species. 121.
For H2C6H6O6. K a1 = 7.9 × 10 −5 and K a 2 = 1.6 × 10 −12 . Because K a1 K a 2 , the amount of H+ produced by the K a 2 reaction will be negligible. 1 mol H 2 C 6 H 6 O 6 176 .12 g = 0.0142 M 0.2000 L
0.500 g
[H2C6H6O6]0 =
Initial Equil.
H2C6H6O6(aq)
⇌ HC6H6O6−(aq)
+ H+(aq)
0.0142 M 0.0142 – x
0 x
~0 x
K a1 = 7.9 × 10 −5 =
K a1 = 7.9 × 10 −5
x2 x2 , x = 1.1 × 10 −3 ; assumption fails the 5% rule. 0.0142 − x 0.0142
Solving by the method of successive approximations: 7.9 × 10 −5 =
x2 , x = 1.0 × 10 −3 M (consistent answer) −3 0.0142 − 1.1 10
Because H+ produced by the K a 2 reaction will be negligible, [H+] = 1.0 × 10 −3 and pH = 3.00. 122.
The reactions are: H3AsO4 ⇌ H+ + H2AsO4−
K a1 = 5.5 × 10−3
H2AsO4− ⇌ H+ + HAsO42− K a 2 = 1.7 × 10−7 HAsO42− ⇌ H+ + AsO43−
K a 3 = 5.1 × 10−12
We will deal with the reactions in order of importance, beginning with the largest K a, K a1 .
CHAPTER 14
ACIDS AND BASES
⇌
H3AsO4
H
679
+
+
H2AsO4
~0 x
−
[H + ][H 2 AsO4 ] K a1 = 5.5 × 10 = [H 3 AsO4 ]
−
−3
Initial Equil.
0.20 M 0.20 – x
0 x
5.5 × 10−3 =
x2 x2 , x = 3.3 × 10−2 M; assumption fails the 5% rule. 0.20 − x 0.20
Solving by the method of successive approximations: 5.5 × 10−3 = x2/(0.20 − 0.033), x = 3.03 × 10−2 (carrying an extra significant figure) 5.5 × 10−3 = x2/(0.20 − 0.0303), x = 3.06 × 10−2 5.5 × 10−3 = x2/(0.20 − 0.0306), x = 3.05 × 10−2 (consistent answer) [H+] = [H2AsO4−] = 3.05 × 10−2 = 3.1 × 10−2 M; [H3AsO4] = 0.20 - 0.031 = 0.17 M 2−
[ H + ][HAsO 4 ]
= 1.7 × 10−7 is much smaller than the K a1 value, very little − [ H 2 AsO 4 ] of H2AsO4− (and HAsO42−) dissociates as compared to H3AsO4. Therefore, [H+] and [H2AsO4−] will not change significantly by the K a 2 reaction. Using the previously calculated concentrations of H+ and H2AsO4− to calculate the concentration of HAsO42−: Because K a 2 =
2−
−7
1.7 × 10 =
(3.1 10 −2 )[HAsO 4 ] 3.1 10 − 2
, [HAsO42−] = 1.7 × 10−7 M
The assumption that the K a 2 reaction does not change [H+] and [H2AsO4−] is good. We repeat the process using K a 3 to get [AsO43−]. 3−
K a 3 = 5.1 × 10−12 =
[ H + ][AsO 4 ] 2−
[ HAsO 4 ]
3−
=
(3.1 10 −2 )[AsO 4 ] 1.7 10 −7
[AsO43−] = 2.8 × 10−17; assumption good. So in 0.20 M analytical concentration of H3AsO4: [H3AsO4] = 0.17 M; [H+] = [H2AsO4−] = 3.1 × 10−2 M ; [HAsO42−] = 1.7 × 10−7 M [AsO43−] = 2.8 × 10−17 M; [OH−] = Kw/[H+] = 3.2 × 10−13 M 123.
Because K a 2 for H2S is so small, we can ignore the H+ contribution from the K a 2 reaction. H2 S Initial Equil.
0.10 M 0.10 – x
⇌
H+
HS−
~0 x
0 x
K a1 = 1.0 × 10 −7
680
CHAPTER 14 K a1 = 1.0 × 10 −7 =
ACIDS AND BASES
x2 x2 , x = [H+] = 1.0 × 10 −4 ; assumptions good. 0.10 0.10 − x
pH = – log(1.0 × 10 −4 ) = 4.00 Use the K a 2 reaction to determine [S2−].
⇌
HS− Initial Equil.
S2−
+
1.0 × 10 −4 M 1.0 × 10 −4 + x
1.0 × 10 −4 M 1.0 × 10 −4 – x
K a 2 = 1.0 × 10 −19 =
H+
(1.0 10 −4 + x) x (1.0 10 − 4 − x)
0 x
(1.0 10 −4 )x 1.0 10 − 4
x = [S2−] = 1.0 × 10 −19 M; assumptions good. 124.
The relevant reactions are: H2CO3 ⇌ H+ + HCO3−
K a1 = 4.3 × 10−7; HCO3− ⇌ H+ + CO32− K a 2 = 5.6 × 10−11
Initially, we deal only with the first reaction (since K a1 >> K a 2 ), and then let those results control values of concentrations in the second reaction. H2CO3 Initial Equil.
⇌
0.010 M 0.010 − x
H+
+ HCO3−
~0 x
0 x
−
[H + ][HCO3 ] x2 x2 = K a1 = 4.3 × 10 = 0.010 − x 0.010 [H 2 CO3 ] −7
x = 6.6 × 10−5 M = [H+] = [HCO3−]; assumptions good. HCO3− Initial Equil.
6.6 × 10−5 M 6.6 × 10−5 − y
⇌
H+
+
6.6 × 10−5 M 6.6 × 10−5 + y
CO32− 0 y 2−
If y is small, then [H+] = [HCO3−], and K a 2 = 5.6 × 10−11 =
[ H + ][CO 3 ] −
[HCO 3 ]
y.
y = [CO32−] = 5.6 × 10−11 M; assumptions good. The amount of H+ from the second dissociation is 5.6 × 10−11 M or:
5.6 10 −11 6.6 10 −5
× 100 = 8.5 × 10−5%
This result justifies our treating the equilibria separately. If the second dissociation contributed a significant amount of H+, we would have to treat both equilibria simultaneously.
CHAPTER 14
ACIDS AND BASES
681
The reaction that occurs when acid is added to a solution of HCO3− is: HCO3−(aq) + H+(aq) → H2CO3(aq) → H2O(l) + CO2(g) The bubbles are CO2(g) and are formed by the breakdown of unstable H 2CO3 molecules. We should write H2O(l) + CO2(aq) or CO2(aq) for what we call carbonic acid. It is for convenience, however, that we write H2CO3(aq). 125.
The dominant H+ producer is the strong acid H2SO4. A 2.0 M H2SO4 solution produces 2.0 M HSO4- and 2.0 M H+. However, HSO4− is a weak acid that could also add H+ to the solution.
⇌
HSO4− Initial
H+
SO42−
+
2.0 M 2.0 M 0 − x mol/L HSO4 dissociates to reach equilibrium −x → +x +x 2.0 − x 2.0 + x x
Change Equil.
2−
K a 2 = 1.2 × 10−2 =
[H + ][SO 4 ] −
=
[HSO 4 ]
( 2 .0 + x ) x 2 .0 ( x ) , x = 1.2 × 10−2 M 2 .0 − x 2 .0
Because x is 0.60% of 2.0, the assumption is valid by the 5% rule. The amount of additional H+ from HSO4− is 1.2 × 10−2 M. The total amount of H+ present is: [H+] = 2.0 + (1.2 × 10−2) = 2.0 M; pH = −log(2.0) = −0.30 Note: In this problem, H+ from HSO4− could have been ignored. However, this is not usually the case in more dilute solutions of H2SO4. 126.
For H2SO4, the first dissociation occurs to completion. The hydrogen sulfate ion (HSO4−) is a weak acid with K a 2 = 1.2 × 10−2. We will consider this equilibrium for additional H + production: HSO4− Initial Change Equil.
⇌
H+
+
SO42−
0.0050 M 0.0050 M 0 x mol/L HSO4− dissociates to reach equilibrium −x → +x +x 0.0050 − x 0.0050 + x x
K a 2 = 0.012 =
(0.0050 + x) x x, x = 0.012; assumption is horrible (240% error). 0.0050 − x
Using the quadratic formula: 6.0 × 10−5 − (0.012)x = x2 + (0.0050)x, x2 + (0.017)x − 6.0 × 10−5 = 0 x=
− 0.017 (2.9 10 −4 + 2.4 10 −4 )1/ 2 − 0.017 0.023 = , x = 3.0 × 10−3 M 2 2
[H+] = 0.0050 + x = 0.0050 + 0.0030 = 0.0080 M; pH = 2.10 Note: We had to consider both H2SO4 and HSO4− for H+ production in this problem.
682
CHAPTER 14
ACIDS AND BASES
Acid-Base Properties of Salts 127.
KHSO4 → K+ + HSO4‒; K+ has no acidic or basic properties as is true for all alkali metal ions. HSO4‒ is a weak acid with Ka = 0.012. This solution will be acidic since a weak acid is present. HONH3Cl → HONH3+ + Cl‒; HONH3+ is the conjugate acid of the weak base HONH2 (Kb = 1.1 × 10−8). The conjugate acids of weak bases are weak acids. So, HONH3+ is an acidic ion. Cl‒ is the conjugate base of the strong acid HCl. Cl‒ has no basic properties. This is an acidic salt due to the HONH3+ ion. C5H5NHNO3 → C5H5H+ + NO3‒; C5H5NH+ is the conjugate acid of the weak base C5H5N (Kb = 1.7 × 10−9). The conjugate acids of weak bases are weak acids. So, C 5H5NH+ is an acidic ion. NO3‒ is the conjugate base of the strong acid HNO3. NO3‒ has no basic properties. This is an acidic salt due to the C5H5NH+ ion. H2NNH2 is a weak base with Kb = 3.0 × 10−6. The solution will be basic. NaHCO3 → Na+ + HCO3‒; Na+ has no acidic or basic properties, while HCO 3‒ is a weak acid with Ka = 5.6 × 10−11. This solution will be acidic since a weak acid is present. From the discussion above, all solutions except for H 2NNH2 will be acidic. Appendix 5 of the text has pertinent Ka and Kb values.
128.
Answer b is correct. NH3 is a weak base with Kb = 1.8 × 10−5. KCN is composed of the K+ and CN‒ ions. K+ has no acidic or basic properties. CN‒ is the conjugate base of the weak acid HCN (Ka = 6.2 × 10−10). The conjugate bases of weak acids are weak bases. So KCN will produce a basic solution due to the CN‒ ion. LiF is composed of the Li+ and F‒ ions. Li+ has no acidic or basic properties. F‒ is the conjugate base of the weak acid HF (Ka = 7.2 × 10−4). The conjugate bases of weak acids are weak bases. So LiF will produce a basic solution due to the F‒ ion. In answer a, the NH4+ ion from the NH4NO3 salt is a weak acid (it is the conjugate acid of the weak base NH3). In answer c, all the salts produce neutral solutions. All these salts are composed of a cation with no acidic properties with an anion with no basic properties. In answer d, all the salts contain an acidic ion (NH4+ and HSO4‒). In answer e, KClO4 is a neutral salt. K+ has no acidic properties and ClO4‒ has no basic properties; it is the conjugate base of the strong acid HClO4.
129.
One difficult aspect of acid-base chemistry is recognizing what types of species are present in solution, that is, whether a species is a strong acid, strong base, weak acid, weak base, or a neutral species. Below are some ideas and generalizations to keep in mind that will help in recognizing types of species present. a. Memorize the following strong acids: HCl, HBr, HI, HNO3, HClO4, and H2SO4 b. Memorize the following strong bases: LiOH, NaOH, KOH, RbOH, CsOH, Ca(OH)2, Sr(OH)2, and Ba(OH)2 c. Weak acids have a Ka value of less than 1 but greater than Kw. Some weak acids are listed in Table 14.2 of the text. Weak bases have a Kb value of less than 1 but greater than Kw. Some weak bases are listed in Table 14.3 of the text.
CHAPTER 14
ACIDS AND BASES
683
d. Conjugate bases of weak acids are weak bases; that is, all have a Kb value of less than 1 but greater than Kw. Some examples of these are the conjugate bases of the weak acids listed in Table 14.2 of the text. e. Conjugate acids of weak bases are weak acids; that is, all have a Ka value of less than 1 but greater than Kw. Some examples of these are the conjugate acids of the weak bases listed in Table 14.3 of the text. f.
Alkali metal ions (Li+, Na+, K+, Rb+, Cs+) and heavier alkaline earth metal ions (Ca2+, Sr2+, Ba2+) have no acidic or basic properties in water.
g. All conjugate bases of strong acids (Cl−, Br−, I−, NO3−, ClO4−, HSO4−) have no basic properties in water (Kb << Kw), and only HSO4− has any acidic properties in water. Let’s apply these ideas to this problem to see what type of species are present. The letters in parenthesis is(are) the generalization(s) above that identifies the species. KOH: Strong base (b) KNO3: Neutral; K+ and NO3− have no acidic/basic properties (f and g). KCN: CN− is a weak base, Kb = Kw/Ka, HCN = 1.0 × 10 −14 /6.2 × 10 −10 = 1.6 × 10 −5 (c and d). Ignore K+ (f). NH4Cl: NH4+ is a weak acid, Ka = 5.6 × 10-10 (c and e). Ignore Cl− (g). HCl:
Strong acid (a)
The most acidic solution will be the strong acid solution, with the weak acid solution less acidic. The most basic solution will be the strong base solution, with the weak base solution less basic. The KNO3 solution will be neutral at pH = 7.00. Most acidic → most basic: HCl > NH4Cl > KNO3 > KCN > KOH 130.
See Exercise 129 for some generalizations on acid-base properties of salts. The letter in parenthesis is(are) the generalization(s) listed in Exercise 129 that identifies the species. CaBr2:
Neutral; Ca2+ and Br− have no acidic/basic properties (f and g).
KNO2:
NO2− is a weak base, Kb = K w /K a, HNO2 = (1.0 × 10 −14 )/(4.0 × 10 −4 ) = 2.5 × 10 −11 (c and d). Ignore K+ (f).
HClO4:
Strong acid (a)
HNO2:
Weak acid, Ka = 4.0 × 10 −4 (c)
HONH3ClO4:
HONH3+ is a weak acid, Ka = K w /K b, HONH2 = (1.0 × 10 −14 )/(1.1 × 10 −8 ) = 9.1 × 10 −7 (c and e). Ignore ClO4− (g). Note that HNO2 has a larger Ka value than HONH3+, so HNO2 is a stronger weak acid than HONH3+.
Using the information above (identity and the Ka or Kb values), the ordering is: Most acidic → most basic: HClO4 > HNO2 > HONH3ClO4 > CaBr2 > KNO2
684 131.
CHAPTER 14
ACIDS AND BASES
Answer e is true. NH3 is a weak base with Kb = 1.8 × 10−5. NaC2H3O2 is a basic salt due to the C2H3O2‒ ion. This ion is the conjugate base of the weak acid HC2H3O2. The Kb value for C2H3O2‒ is Kw/Ka = 1.0 × 10−14/1.8 × 10−5 = 5.6 × 10−10. Because the Kb value for NH3 is larger than the Kb value for the C2H3O2‒ ion, the NH3 solution is more basic with the higher pH. In answer a, NaC2H3O2 is a basic salt while NH4Cl is an acidic salt due to the NH4+ ion. This ion is a weak acid since it is the conjugate acid a weak base. So, the NH4Cl solution has a lower pH. In answer b, NH3 is a stronger weak base than C2H3O2‒ (see above). In part c and d, Ka for HC2H3O2 is 1.8 × 10−5 and Ka for NH4+ = Kw/Kb = 1.0 × 10−14/1.8 × 10−5 = 5.6 × 10−10. HC2H3O2 has the larger Ka value, making it a better weak acid than NH4+. The HC2H3O2 solution will have the lower pH.
132.
Solutions b and d will be acidic, so they will have low pH. The other three solutions will be basic. The solution containing the strongest base with have the highest pH. NaCN is a basic salt due to the CN‒ ion (Kb = Kw/Ka for HCN = 1.0 × 10−14/6.2 × 10−10 = 1.6 × 10−5). NaC2H3O is a basic salt due to the C2H3O2‒ ion (Kb = Kw/Ka for HC2H3O2 = 1.0 × 10−14/1.8 × 10−5 = 5.6 × 10−10). HONH2 is a weak base with Kb = 1.1 × 10−8. From the Kb values, CN‒ is the best base, so the NaCN solution will have the highest pH.
133.
a. KCl is a soluble ionic compound that dissolves in water to produce K+(aq) and Cl−(aq). K+ (like the other alkali metal cations) has no acidic or basic properties. Cl− is the conjugate base of the strong acid HCl. Cl− has no basic (or acidic) properties. Therefore, a solution of KCl will be neutral because neither of the ions has any acidic or basic properties. The 1.0 M KCl solution has [H+] = [OH−] = 1.0 × 10 −7 M and pH = pOH = 7.00. b. KC2H3O2 is also a soluble ionic compound that dissolves in water to produce K+(aq) and C2H3O2−(aq). The difference between the KCl solution and the KC2H3O2 solution is that C2H3O2− does have basic properties in water, unlike Cl−. C2H3O2− is the conjugate base of the weak acid HC2H3O2, and as is true for all conjugate bases of weak acids, C2H3O2− is a weak base in water. We must solve an equilibrium problem in order to determine the amount of OH− this weak base produces in water. C2H3O2− + H2O Initial Change Equil.
⇌ HC2H3O2 + OH−
Kb =
Kw K a, C 2 H 3O 2
=
1.0 10 −14 1.8 10 −5
1.0 M 0 ~0 Kb = 5.6 × 10 −10 x mol/L of C2H3O2− reacts with H2O to reach equilibrium –x → +x +x 1.0 − x x x
Kb = 5.6 × 10 −10 =
[HC 2 H 3 O 2 ][OH − ] −
, 5.6 × 10 −10 =
[C 2 H 3 O 2 ]
x2 x2 1. 0 1.0 − x
x = [OH−] = 2.4 × 10 −5 M ; assumptions good pOH = 4.62; pH = 14.00 – 4.62 = 9.38; [H+] = 10 −9.38 = 4.2 × 10 −10 M 134.
C2H5NH3Cl → C2H5NH3+ + Cl− ; C2H5NH3+ is the conjugate acid of the weak base C2H5NH2 (Kb = 5.6 × 10 −4 ). As is true for all conjugate acids of weak bases, C 2H5NH3+ is a weak acid.
CHAPTER 14
ACIDS AND BASES
685
Cl− has no basic (or acidic) properties. Ignore Cl−. Solving the weak acid problem: C2H5NH3+ Initial Change Equil.
⇌
C2H5NH2
+
Ka = Kw/5.6 × 10 −4 = 1.8 × 10 −11
H+
0.25 M 0 ~0 x mol/L C2H5NH3+ dissociates to reach equilibrium –x → +x +x 0.25 – x x x
Ka = 1.8 × 10 −11 =
[C 2 H 5 NH 2 ][H + ] +
[C 2 H 5 NH 3 ]
=
x2 x2 (assuming x << 0.25) 0.25 − x 0.25
x = [H+] = 2.1 × 10 −6 M; pH = 5.68; assumptions good. [C2H5NH2] = [H+] = 2.1 × 10 −6 M; [C2H5NH3+] = 0.25 M; [Cl−] = 0.25 M [OH−] = Kw/[H+] = 4.8 × 10 −9 M 135.
a. CH3NH3Cl → CH3NH3+ + Cl−: CH3NH3+ is a weak acid. Cl− is the conjugate base of a strong acid. Cl− has no basic (or acidic) properties. CH3NH3+ ⇌ CH3NH2 + H+
[CH 3 NH 2 ][H + ]
Ka =
+
[CH 3 NH 3 ]
=
Kw 1.00 10 −14 = Kb 4.38 10 − 4
= 2.28 × 10 −11 CH3NH3+ Initial Change Equil.
⇌
CH3NH2
+
H+
0.10 M 0 ~0 x mol/L CH3NH3+ dissociates to reach equilibrium –x → +x +x 0.10 – x x x
Ka = 2.28 × 10 −11 =
x2 x2 0.10 − x 0.10
(assuming x << 0.10)
x = [H+] = 1.5 × 10 −6 M; pH = 5.82; assumptions good. b. NaCN → Na+ + CN−: CN− is a weak base. Na+ has no acidic (or basic) properties. CN− + H2O
⇌
Kw 1.0 10 −14 = Ka 6.2 10 −10
HCN +
OH−
Kb =
0
~0
Kb = 1.6 × 10 −5
Initial
0.050 M
Change Equil.
x mol/L CN− reacts with H2O to reach equilibrium –x → +x +x 0.050 – x x x
686
CHAPTER 14 Kb = 1.6 × 10 −5 =
ACIDS AND BASES
x2 [HCN][OH − ] x2 = 0.050 0.050 − x [CN − ]
x = [OH−] = 8.9 × 10 −4 M; pOH = 3.05; pH = 10.95; assumptions good. 136.
a. KNO2 → K+ + NO2−: NO2− is a weak base. Ignore K+. NO2− + H2O ⇌ HNO2 + OH− Initial 0.12 M Equil. 0.12 – x
0 x
Kw 1.0 10 −14 = = 2.5 × 10 −11 Ka 4.0 10 − 4
~0 x
[OH − ][HNO 2 ]
Kb = 2.5 × 10 −11 =
Kb =
−
[ NO 2 ]
=
x2 x2 0.12 − x 0.12
x = [OH−] = 1.7 × 10 −6 M; pOH = 5.77; pH = 8.23; assumptions good. b. NaOCl → Na+ + OCl−: OCl− is a weak base. Ignore Na+. OCl− + H2O
⇌ HOCl + OH−
Initial 0.45 M Equil. 0.45 – x Kb = 2.9 × 10 −7 =
0 x
Kb =
Kw 1.0 10 −14 = = 2.9 × 10 −7 Ka 3.5 10 −8
~0 x
x2 x2 [HOCl][OH− ] = 0.45 − x 0.45 [OCl− ]
x = [OH−] = 3.6 × 10 −4 M; pOH = 3.44; pH = 10.56; assumptions good. c. NH4ClO4 → NH4+ + ClO4−: NH4+ is a weak acid. ClO4− is the conjugate base of a strong acid. ClO4− has no basic (or acidic) properties. NH4+
⇌
NH3
+ H+
0 x
~0 x
Initial 0.40 M Equil. 0.40 – x Ka = 5.6 × 10 −10 =
[ NH 3 ][H + ] +
[ NH 4 ]
=
Ka =
Kw 1.0 10 −14 = = 5.6 × 10 −10 Kb 1.8 10 −5
x2 x2 0.40 − x 0.40
x = [H+] = 1.5 × 10 −5 M; pH = 4.82; assumptions good. 137.
NaN3 → Na+ + N3−; azide (N3−) is a weak base because it is the conjugate base of a weak acid. All conjugate bases of weak acids are weak bases (K w < Kb < 1). Ignore Na+.
CHAPTER 14
ACIDS AND BASES N3− + H2O
Initial Change Equil.
HN3 + OH−
Kb =
Kw 1.0 10 −14 = 5.3 × 10−10 = Ka 1.9 10 −5
0.010 M 0 ~0 x mol/L of N3- reacts with H2O to reach equilibrium −x → +x +x 0.010 − x x x
[ HN 3 ][OH − ]
Kb =
⇌
687
−
[N3 ]
x2 x2 (assuming x << 0.010) 0.010 − x 0.010
, 5.3 × 10−10 =
x = [OH−] = 2.3 × 10−6 M; [H+] =
1.0 10 −14 = 4.3 × 10−9 M; assumptions good. 2.3 10 −6
[HN3] = [OH−] = 2.3 × 10−6 M; [Na+] = 0.010 M; [N3−] = 0.010 − 2.3 × 10−6 = 0.010 M 138.
30.0 mg papH + Cl − 1000 mL 1g 1 mol papH + Cl − 1 mol papH + mL soln L 1000 mg 378.85 g mol papH + Cl −
= 0.0792 M
⇌
papH+ Initial 0.0792 M Equil. 0.0792 – x Ka = 2.5 × 10 −6 =
pap + H+ 0 x
Ka =
Kw K b , pap
−14
=
2.1 10 = 2.5 × 10 −6 −9 8.33 10
~0 x
x2 x2 , x = [H+] = 4.4 × 10 −4 M 0.0792 − x 0.0792
pH = −log(4.4 × 10 −4 ) = 3.36; assumptions good. 139.
All these salts contain Na+, which has no acidic/basic properties, and a conjugate base of a weak acid (except for NaCl, where Cl− is a neutral species). All conjugate bases of weak acids are weak bases since Kb values for these species are between Kw and 1. To identify the species, we will use the data given to determine the Kb value for the weak conjugate base. From the K b value and data in Table 14.2 of the text, we can identify the conjugate base present by calculating the Ka value for the weak acid. We will use A − as an abbreviation for the weak conjugate base. A− + Initial Change Equil. Kb =
H2O
⇌
HA
+
OH−
0.100 mol/1.00 L 0 ~0 x mol/L A− reacts with H2O to reach equilibrium −x → +x +x 0.100 − x x x
[HA][OH − ] x2 ; from the problem, pH = 8.07: = 0.100 − x [A − ]
pOH = 14.00 − 8.07 = 5.93; [OH−] = x = 10−5.93 = 1.2 × 10−6 M
688
CHAPTER 14 Kb =
ACIDS AND BASES
(1.2 10 −6 ) 2 = 1.4 × 10−11 = Kb value for the conjugate base of a weak acid. 0.100 − (1.2 10 −6 )
1.0 10 −14 = 7.1 × 10−4 1.4 10 −11
The Ka value for the weak acid equals Kw/Kb: Ka =
From Table 14.2 of the text, this Ka value is closest to HF. Therefore, the unknown salt is NaF. 140.
BHCl → BH+ + Cl−; Cl− is the conjugate base of the strong acid HCl, so Cl− has no acidic/ basic properties. BH+ is a weak acid because it is the conjugate acid of a weak base B. Determining the Ka value for BH+:
⇌
BH+ Initial Change Equil. Ka =
B
H+
+
0.10 M 0 ~0 x mol/L BH+ dissociates to reach equilibrium −x → +x +x 0.10 − x x x
[B][H + ] x2 = ; from the problem, pH = 5.82: 0.10 − x [BH+ ]
[H+] = x = 10−5.82 = 1.5 × 10−6 M; Ka =
(1.5 10 −6 ) 2 = 2.3 × 10−11 0.10 − (1.5 10 −6 )
Kb for the base B = Kw/Ka = (1.0 × 10−14)/(2.3 × 10−11) = 4.3 × 10−4. From Table 14.3 of the text, this Kb value is closest to CH3NH2, so the unknown salt is CH3NH3Cl. 141.
B− is a weak base. Use the weak base data to determine K b for B−. B− + H2O Initial Equil.
⇌
HB
0.050 M 0.050 − x
+
0 x
OH− ~0 x
From pH = 9.00: pOH = 5.00, [OH−] = 10−5.00 = 1.0 × 10−5 M = x.
Kb =
(1.0 10 −5 ) 2 [HB][OH− ] x2 = = = 2.0 10 −9 − −5 0.050 − x [B ] 0.050 − (1.0 10 )
Because B− is a weak base, HB will be a weak acid. Solve the weak acid problem. HB Initial Equil.
0.010 M 0.010 − x
⇌
H+
+ B−
~0 x
0 x
CHAPTER 14 Ka =
ACIDS AND BASES
689
Kw 1.0 10 −14 x2 x2 −6 = , 5 . 0 10 = Kb 0.010 − x 0.010 2.0 10 −9
x = [H+] = 2.2 × 10−4 M; pH = 3.66; assumptions good. 142.
From the pH, C7H4ClO2− is a weak base. Use the weak base data to determine K b for C7H4ClO2− (which we will abbreviate as CB−). CB−
+
⇌
H2O
Initial 0.20 M Equil. 0.20 − x
HCB
OH−
+
0 x
~0 x
Because pH = 8.65, pOH = 5.35 and [OH −] = 10−5.35 = 4.5 × 10−6 = x. Kb =
[HCB][OH − ] [CB − ]
=
(4.5 10 −6 ) 2 x2 = 1.0 × 10−10 = −6 0.20 − x 0.20 − (4.5 10 )
Because CB− is a weak base, HCB, chlorobenzoic acid, is a weak acid. Solving the weak acid problem: HCB Initial Equil. Ka =
⇌
0.20 M 0.20 − x
H+ ~0 x
+
CB− 0 x
Kw 1.0 10 −14 x2 x2 −4 = , 1 . 0 10 = Kb 0.20 − x 0.20 1.0 10 −10
x = [H+] = 4.5 × 10−3 M; pH = 2.35; assumptions good. 143.
Major species present: Al(H2O)63+ (Ka = 1.4 × 10−5), NO3− (neutral), and H2O (Kw = 1.0 × 10 −14 ); Al(H2O)63+ is a stronger acid than water, so it will be the dominant H+ producer. Al(H2O)63+ Initial Change Equil.
⇌ Al(H2O)5(OH)2+ +
H+
0.050 M 0 ~0 x mol/L Al(H2O)63+ dissociates to reach equilibrium –x → +x +x 0.050 – x x x
Ka = 1.4 × 10 −5 =
[Al(H 2 O) 5 (OH) 2+ ][H + ] x2 x2 = 0.050 0.050 − x [Al(H 2 O) 36+ ]
x = 8.4 × 10 −4 M = [H+]; pH = –log(8.4 × 10 −4 ) = 3.08; assumptions good. 144.
Major species: Co(H2O)63+ (Ka = 1.0 × 10 −5 ), Cl- (neutral), and H2O (Kw = 1.0 × 10 −14 ); Co(H2O)63+ will determine the pH because it is a stronger acid than water. Solving the weak acid problem in the usual manner:
690
CHAPTER 14 Co(H2O)63+ Initial Equil.
⇌
Co(H2O)5(OH)2+
0.10 M 0.10 – x
Ka = 1.0 × 10 −5 =
H+
+
0 x
ACIDS AND BASES Ka = 1.0 × 10 −5
~0 x
x2 x2 , x = [H+] = 1.0 × 10 −3 M 0.10 − x 0.10
pH = –log(1.0 × 10 −3 ) = 3.00; assumptions good. 145.
Reference Table 14.6 of the text and the solution to Exercise 129 for some generalizations on acid-base properties of salts. a. NaNO3 → Na+ + NO3− neutral; neither species has any acidic/basic properties. b. NaNO2 → Na+ + NO2− basic; NO2− is a weak base, and Na+ has no effect on pH. NO2− + H2O ⇌ HNO2 + OH−
Kb =
Kw K a , HNO2
=
1.0 10 −14 4.0 10 − 4
= 2.5 × 10 −11
c. C5H5NHClO4 → C5H5NH+ + ClO4− acidic; C5H5NH+ is a weak acid, and ClO4− has no effect on pH. 1.0 10 −14 Kw C5H5NH+ ⇌ H+ + C5H5N Ka = = = 5.9 × 10 −6 −9 K b , C5 H 5 N 1.7 10 d. NH4NO2 → NH4+ + NO2− acidic; NH4+ is a weak acid (Ka = 5.6 × 10 −10 ), and NO2− is a weak base (Kb = 2.5 × 10 −11 ). Because K a , NH+ K b, NO− , the solution is acidic. 4
2
NH4+ ⇌ H+ + NH3 Ka = 5.6 × 10 −10 ; NO2− + H2O ⇌ HNO2 + OH−
Kb = 2.5 × 10 −11
e. KOCl → K+ + OCl− basic; OCl− is a weak base, and K+ has no effect on pH. OCl− + H2O ⇌ HOCl + OH− f.
Kb =
Kw K a , HOCl
=
1.0 10 −14 = 2.9 × 10 −7 3.5 10 −8
NH4OCl → NH4+ + OCl− basic; NH4+ is a weak acid, and OCl− is a weak base. Because K b, OCl− K a , NH+ , the solution is basic. 4
NH4+ ⇌ NH3 + H+ Ka = 5.6 × 10 −10 ; OCl− + H2O ⇌ HOCl + OH− 146.
Kb = 2.9 × 10 −7
a. Sr(NO3)2 → Sr2+ + 2 NO3− neutral; Sr2+ and NO3– have no effect on pH. b. NH4C2H3O2 → NH4+ + C2H3O2− neutral; NH4+ is a weak acid, and C2H3O2− is a weak base. Because K a, NH + = K b, C H O − , the solution will be neutral (pH = 7.00). 4
NH4+ ⇌ NH3 + H+
2
Ka =
3
2
1.0 10 −14 Kw = = 5.6 × 10 −10 −5 K b, NH3 1.8 10
C2H3O2− + H2O ⇌ HC2H3O2 + OH−
Kb =
Kw K a, HC2 H3O 2
=
1.0 10 −14 = 5.6 × 10 −10 −5 1.8 10
CHAPTER 14
ACIDS AND BASES
691
c. CH3NH3Cl → CH3NH3+ + Cl− acidic; CH3NH3+ is a weak acid, and Cl− has no effect on pH. Because only a weak acid is present, the solution will be acidic. CH3NH3+ ⇌ H+ + CH3NH2
Ka =
Kw K b, CH 3 NH2
=
1.00 10 −14 4.38 10 − 4
= 2.28 × 10 −11
d. C6H5NH3ClO2 → C6H5NH3+ + ClO2− acidic; C6H5NH3+ is a weak acid, and ClO2− is a very weak base. Because K a, C H NH+ K b, ClO − , the solution is acidic. 6
5
3
2
C6H5NH3+ ⇌ H+ + C6H5NH2
Ka =
ClO2− + H2O ⇌ HClO2 + OH−
Kb =
Kw K b, C6 H 5 NH 2
Kw K a, HClO2
=
=
1.0 10 −14 3.8 10 −10
= 2.6 × 10 −5
1.0 10 −14 = 8.3 × 10 −13 −2 1.2 10
e. NH4F → NH4+ + F− acidic; NH4+ is a weak acid, and F− is a weak base. Because K a , NH+ K b, F− , the solution is acidic. 4
NH4+ ⇌ H+ + NH3 Ka = 5.6 × 10 −10 ; F− + H2O ⇌ HF + OH− Kb = 1.4 × 10 −11 f.
CH3NH3CN → CH3NH3+ + CN− basic; CH3NH3+ is a weak acid, and CN− is a weak base. Because K b, CN − K a , CH NH+ , the solution is basic. 3
3
CH3NH3+ ⇌ H+ + CH3NH2
Ka = 2.28 × 10-11
CN + H2O ⇌ HCN + OH
1.0 10 −14 Kw Kb = = = 1.6 × 10 −5 −10 K a , HCN 6.2 10
−
−
Relationships Between Structure and Strengths of Acids and Bases 147.
a. HIO3 < HBrO3; as the electronegativity of the central atom increases, acid strength increases. b. HNO2 < HNO3; as the number of oxygen atoms attached to the central nitrogen atom increases, acid strength increases. c. HOI < HOCl; same reasoning as in a. d. H3PO3 < H3PO4; same reasoning as in b.
148.
a. BrO3− < IO3−; these are the conjugate bases of the acids in Exercise 147a. Since HBrO3 is the stronger acid, the conjugate base of HBrO3 (BrO3−) will be the weaker base. IO3− will be the stronger base because HIO3 is the weaker acid. b. NO3− < NO2−; these are the conjugate bases of the acids in Exercise 147b. Conjugate base strength is inversely related to acid strength. c. OCl− < OI−; these are the conjugate bases of the acids in Exercise 147c.
692 149.
CHAPTER 14
ACIDS AND BASES
a. H2O < H2S < H2Se; as the strength of the H‒X bond decreases, acid strength increases. b. CH3CO2H < FCH2CO2H < F2CHCO2H < F3CCO2H; as the electronegativity of neighboring atoms increases, acid strength increases. c. NH4+ < HONH3+; same reason as in b. d. NH4+ < PH4+; same reason as in a.
150.
In general, the stronger the acid, the weaker is the conjugate base. a. SeH− < SH− < OH−; these are the conjugate bases of the acids in Exercise 149a. The ordering of the base strength is the opposite of the acids. b. PH3 < NH3 (See Exercise 149d.) c. HONH2 < NH3 (See Exercise149c.)
151.
In general, metal oxides form basic solutions when dissolved in water, and nonmetal oxides form acidic solutions in water. a. Basic; CaO(s) + H2O(l) → Ca(OH)2(aq); Ca(OH)2 is a strong base. b. Acidic; SO2(g) + H2O(l) → H2SO3(aq); H2SO3 is a weak diprotic acid. c. Acidic; Cl2O(g) + H2O(l) → 2 HOCl(aq); HOCl is a weak acid.
152.
a. Basic; Li2O(s) + H2O(l) → 2 LiOH(aq); LiOH is a strong base. b. Acidic; CO2(g) + H2O(l) → H2CO3(aq); H2CO3 is a weak diprotic acid. c. Basic; SrO(s) + H2O(l) → Sr(OH)2(aq); Sr(OH)2 is a strong base.
Lewis Acids and Bases 153.
A Lewis base is an electron pair donor, and a Lewis acid is an electron pair acceptor. a. B(OH)3, acid; H2O, base
b. Ag+, acid; NH3, base
c. BF3, acid; F−, base
154.
a. Fe3+, acid; H2O, base
b. H2O, acid; CN−, base
c. HgI2, acid; I−, base
155.
Al(OH)3(s) + 3 H+(aq) → Al3+(aq) + 3 H2O(l) Al(OH)3(s) + OH− (aq) → Al(OH)4−(aq)
156.
157.
(Brønsted-Lowry base, H+ acceptor)
(Lewis acid, electron pair acceptor)
Zn(OH)2(s) + 2 H+(aq) → Zn2+(aq) + 2 H2O(l)
(Brønsted-Lowry base)
Zn(OH)2(s) + 2 OH− (aq) → Zn(OH)42− (aq)
(Lewis acid)
Fe3+ should be the stronger Lewis acid. Fe3+ is smaller and has a greater positive charge. Because of this, Fe3+ will be more strongly attracted to lone pairs of electrons as compared to Fe2+.
CHAPTER 14 158.
ACIDS AND BASES
693
The Lewis structures for the reactants and products are: O O
C
O
C
O
+ H
O
H
O
H
H
In this reaction, H2O donates a pair of electrons to carbon in CO2, which is followed by a proton shift to form H2CO3. H2O is the Lewis base, and CO2 is the Lewis acid.
ChemWork Problems 159.
H2O(l) ⇌ H+(aq) + OH−(aq)
Kw = [H+][OH−]
In a neutral solution, [H+] = [OH−], so answer c is true. The value of K w depends on temperature. At 25°C, Kw = 1.0 × 10−14, which gives a pH = pOH = 7.00. However, at different temperatures, the value of Kw is different, giving different pH values for neutral solutions. For answer b, the pH of the solution when moles of base = moles of acid depends on the identity of the acid and base; it can produce a neutral solution, but it certainly doesn’t have to be neutral. 160.
Ca(OH)2 → Ca2+ + 2 OH‒; a 0.10 M Ca(OH)2 solution produces 0.20 M OH‒. This gives a pOH = −log(0.20) = 0.70 and a pH = 14.00 – 0.70 = 13.30. Answer d is false. The other answers are true for these strong acid or strong base solutions. For answer a, this is a very dilute solution of a strong acid. The highest pH any acid solution can have is pH = 7.00, which is the pH of water at 25°C. So, the pH of this very dilute solution of HBr is 7.00. Note that it is fine to have negative pH values and pH values above 14.00. When this is the case, we have solutions of a strong acid or of a strong base having a concentration greater than 1.0 M.
161.
At pH = 2.000, [H+] = 10 −2.000 = 1.00 × 10 −2 M At pH = 4.000, [H+] = 10 −4.000 = 1.00 × 10 −4 M Moles H+ present = 0.0100 L ×
0.0100 mol H + = 1.00 × 10 −4 mol H+ L
Let V = total volume of solution at pH = 4.000: 1.00 × 10 −4 mol/L =
1.00 10 −4 mol H + , V = 1.00 L V
Volume of water added = 1.00 L – 0.0100 L = 0.99 L = 990 mL 162.
Conjugate acid-base pairs differ by an H+ in the formula. Pairs in parts a, c, and d are conjugate acid-base pairs. For part b, HSO4− is the conjugate base of H2SO4. In addition, HSO4− is the conjugate acid of SO42−.
694
CHAPTER 14
163.
Kw = [H+][OH−]; pH + pOH = 14.00; HA(aq) ⇌ H+(aq) + A−(aq)
ACIDS AND BASES Ka =
[H + ][A − ] [HA]
As acid strength increase, more of the acid dissociates to reach equilibrium. The results in more H+ being produce, a larger value for Ka, and the solution becoming more acidic.
164.
a. [H+] will increase.
b. pH will decrease as solution becomes more acidic.
c. [OH –] will decrease.
d. pOH will increase.
e. Ka will increase.
a. HC3H5O3 is a weak acid, so HC3H5O3 and H2O are the major species. b.
HC3H5O3
⇌
H+
+
C3H5O3−
Initial
0.60 M ~0 0 x mol/L HC3H5O3 dissociates to reach equilibrium Change –x → +x +x Equil. 0.60 – x x x − + 2 [H ][C 3 H 5 O 3 ] x2 x c. Ka = 1.4 × 10 −4 = = [HC 3 H 5 O 3 ] 0.60 0.60 − x x = 9.2 × 10 −3 M; assumption follows the 5% rule (x is 1.5% of 0.60). x = [C3H5O3−] = 9.2 × 10 −3 M d. [H+] = x = 9.2 × 10−3 M, pH = –log[H+] = –log(0.0092) = 2.04 165.
HA is a stronger weak acid than HB since the pH of the 1.0 M acid solution is lower. This means that Ka for HA is greater than Ka for HB. In general, the weaker the acid, the stronger the conjugate base. So, B- will be a better base than A-. Answers a and d are true. For answer b, if HA was a strong acid, then the pH would be 0.00 [= ‒log(1.0)]. And for answer c, the most acidic solution has the highest [H+]; this would be the solution with HA.
166.
a. C2H5NH2 is a weak base, so the major species are C2H5NH2 and H2O. b.
C2H5NH2
+ H2O
⇌
C2H5NH3+ + OH−
Kb = 5.6 × 10 −4
Initial
0.67 M 0 ~0 x mol/L C2H5NH2 reacts with H2O to reach equilibrium Change –x → +x +x Equil. 0.67 – x x x Kb = 5.6 × 10 −4 =
+
[C 2 H 5 NH3 ][OH − ] x2 x2 = (assuming x << 0.67) 0.67 0.67 − x [C 2 H 5 NH 2 ]
x = [OH−] = 1.9 × 10 −2 M; assumption follows the 5% rule (x is 2.8% of 0.67). pOH = −log[OH−] = −log(1.9 × 10 −2 ) = 1.72; pH = 14.00 − pOH = 14.00 − 1.72 = 12.28
CHAPTER 14 167.
168.
ACIDS AND BASES
695
The light bulb is bright because a strong electrolyte is present; that is, a solute is present that dissolves to produce a lot of ions in solution. The pH meter value of 4.6 indicates that a weak acid is present. (If a strong acid were present, the pH would be close to zero.) Of the possible substances, only HCl (strong acid), NaOH (strong base), and NH4Cl are strong electrolytes. Of these three substances, only NH4Cl contains a weak acid (the HCl solution would have a pH close to zero, and the NaOH solution would have a pH close to 14.0). NH 4Cl dissociates into NH4+ and Cl- ions when dissolved in water. Cl− is the conjugate base of a strong acid, so it has no basic (or acidic properties) in water. NH4+, however, is the conjugate acid of the weak base NH3, so NH4+ is a weak acid and would produce a solution with a pH = 4.6 when the concentration is ~1.0 M. NH4Cl is the solute. CO2(aq) + H2O(l) ⇌ H2CO3(aq)
K=
[H 2 CO 3 ] [CO 2 ]
During exercise: [H2CO3] = 26.3 mM and [CO2] = 1.63 mM, so: K = At rest: K = 16.1 =
26.3 mM = 16.1 1.63 mM
24.9 mM , [CO2] = 1.55 mM [CO 2 ]
169.
When a patient’s breathing becomes too slow or weak, not enough CO 2(g) is exhaled causing a buildup of dissolved CO2 in the blood. From the equilibrium in Exercise 168, as the CO2 concentration in the blood builds up, the equilibrium is shifted to the right which results in more H+ in the blood and a lowering of the blood pH.
170.
In respiratory alkalosis, blood pH is increased over normal values. When a person hyperventilates, an excess of CO2(g) is removed from the body, which results in a decrease in dissolved CO2 in the blood. As CO2 is removed, the equilibrium described in Exercise 168 shifts left resulting in the removal of H+ from the blood. As the H+ concentration in blood decreases, blood pH increases.
171.
a. In the lungs there is a lot of O2, and the equilibrium favors Hb(O2)4. In the cells there is a lower concentration of O2, and the equilibrium favors HbH44+. b. CO2 is a weak acid, CO2 + H2O ⇌ HCO3− + H+. Removing CO2 essentially decreases H+, which causes the hemoglobin reaction to shift right. Hb(O2)4 is then favored, and O2 is not released by hemoglobin in the cells. Breathing into a paper bag increases CO2 in the blood, thus increasing [H+], which shifts the hemoglobin reaction left. c. CO2 builds up in the blood, and it becomes too acidic, driving the hemoglobin equilibrium to the left. Hemoglobin can't bind O2 as strongly in the lungs. Bicarbonate ion acts as a base in water and neutralizes the excess acidity.
172.
CaO(s) + H2O(l) → Ca(OH)2(aq); Ca(OH)2(aq) → Ca2+(aq) + 2 OH−(aq) 0.25 g CaO
[OH−] =
1 mol CaO 1 mol Ca(OH)2 2 mol OH − 56.08 g 1 mol CaO mol Ca(OH)2 = 5.9 × 10 −3 M 1.5 L
pOH = –log(5.9 × 10 −3 ) = 2.23, pH = 14.00 – 2.23 = 11.77
696
CHAPTER 14
173.
⇌
HBz Initial Change Equil. Ka =
H+
Bz−
+
ACIDS AND BASES
HBz = C6H5CO2H
C ~0 0 x mol/L HBz dissociates to reach equilibrium −x → +x +x C−x x x
C = [HBz]0 = concentration of HBz that dissolves to give saturated solution.
[H + ][Bz − ] x2 = 6.4 × 10 −5 = , where x = [H+] C−x [HBz]
6.4 × 10 −5 =
[ H + ]2 ; pH = 2.80; [H+] = 10 −2.80 = 1.6 × 10 −3 M + C − [H ]
C − (1.6 × 10 −3 ) =
(1.6 10 −3 ) 2 = 4.0 × 10 −2 −5 6.4 10
C = (4.0 × 10 −2 ) + (1.6 × 10 −3 ) = 4.2 × 10 −2 M The molar solubility of C6H5CO2H is 4.2 × 10 −2 mol/L. 174.
[H+]0 = (1.0 × 10−2) + (1.0 × 10−2) = 2.0 × 10−2 M from strong acids HCl and H2SO4. HSO4− is a good weak acid (Ka = 0.012). However, HCN is a poor weak acid (K a = 6.2 × 10−10) and can be ignored. Calculating the H+ contribution from HSO4−: HSO4− Initial Equil. Ka =
⇌
0.010 M 0.010 − x
H+
SO42−
+
0.020 M 0.020 + x
Ka = 0.012
0 x
x(0.020 + x) x(0.020 ) , 0.012 , x = 0.0060; assumption poor (60% error). 0.010 − x 0.010
Using the quadratic formula: x2 + (0.032)x − 1.2 × 10−4 = 0, x = 3.4 × 10−3 M [H+] = 0.020 + x = 0.020 + (3.4 × 10−3) = 0.023 M; pH = 1.64 175.
For this problem we will abbreviate CH2=CHCO2H as Hacr and CH2=CHCO2− as acr−. a. Solving the weak acid problem: Hacr Initial Equil.
0.10 M 0.10 − x
⇌
H+ ~0 x
+
acr−
Ka = 5.6 × 10 −5
0 x
x2 x2 = 5.6 × 10 −5 , x = [H+] = 2.4 × 10 −3 M; pH = 2.62; assumptions good. 0.10 − x 0.10
b. Percent dissociation =
2.4 10 −3 [H + ] × 100 = × 100 = 2.4% [ Hacr]0 0.10
CHAPTER 14
ACIDS AND BASES
697
c. acr− is a weak base and the major source of OH − in this solution. acr−
+ H2 O
⇌
Hacr
Initial 0.050 M Equil. 0.050 – x Kb =
+
0 x
1.0 10 −14 Kw = Ka 5.6 10 −5
OH−
Kb =
~0 x
Kb = 1.8 × 10 −10
x2 [Hacr][OH − ] x2 −10 10 , 1.8 × = 0.050 0.050 − x [acr− ]
x = [OH−] = 3.0 × 10 −6 M; pOH = 5.52; pH = 8.48; assumptions good. 176.
In deciding whether a substance is an acid or a base, strong or weak, you should keep in mind a couple of ideas: (1) There are only a few common strong acids and strong bases, all of which should be memorized. Common strong acids = HCl, HBr, HI, HNO 3, HClO4, and H2SO4. Common strong bases = LiOH, NaOH, KOH, RbOH, CsOH, Ca(OH) 2, Sr(OH)2, and Ba(OH)2. (2) All other acids and bases are weak and will have Ka and Kb values of less than 1 but greater than Kw (1.0 10−14). Reference Table 14.2 for Ka values for some weak acids and Table 14.3 for Kb values for some weak bases. There are too many weak acids and weak bases to memorize them all. Therefore, use the tables of K a and Kb values to help you identify weak acids and weak bases. Appendix 5 contains more complete tables of K a and Kb values.
177.
a. weak acid (Ka = 4.0 × 10−4)
b. strong acid
c. weak base (Kb = 4.38 × 10−4)
d. strong base
e. weak base (Kb = 1.8 × 10−5)
f.
g. weak acid (Ka = 1.8 × 10−4)
h. strong base
weak acid (Ka = 7.2 × 10−4) i.
strong acid
a. HA is a weak acid. Most of the acid is present as HA molecules; only one set of H + and A− ions is present. In a strong acid, all the acid would be dissociated into H+ and A− ions. b. This picture is the result of 1 out of 10 HA molecules dissociating. Percent dissociation = HA Initial Change Equil.
1 100 = 10% (an exact number) 10
⇌
H+
+
A−
Ka =
[H + ][A − ] [HA]
0.20 M ~0 0 x mol/L HA dissociates to reach equilibrium −x → +x +x 0.20 − x x x
698
CHAPTER 14
ACIDS AND BASES
[H+] = [A−] = x = 0.10 0.20 M = 0.020 M; [HA] = 0.20 – 0.020 = 0.18 M Ka =
178.
(0.020) 2 = 2.2 10−3 0.18
1.0 g quinine 1 mol quinine = 1.6 × 10−3 M quinine; let Q = quinine = C20H24N2O2. 1.9000 L 324 .4 g quinine
Q Initial Change Equil.
+ H2O
⇌
QH+
+
OH−
Kb = 10−5.1 = 8 × 10−6
1.6 × 10−3 M 0 ~0 x mol/L quinine reacts with H2O to reach equilibrium −x → +x +x 1.6 × 10−3 − x x x
Kb = 8 × 10−6 =
x2 x2 [QH + ][OH − ] = [Q] (1.6 10 −3 − x) 1.6 10 −3
x = 1 × 10−4; assumption fails 5% rule (x is 6% of 0.0016). Using successive approximations: x2 = 8 × 10−6, x = 1 × 10−4 M (consistent answer) (1.6 10 −3 − 1 10 − 4 ) x = [OH−] = 1 × 10−4 M; pOH = 4.0; pH = 10.0 179.
HONH2 + H2O Initial Equil.
⇌ HONH3+ + OH−
I I–x
Kb = 1.1 × 10 −8 =
0 x
~0 x
Kb = 1.1 × 10 −8 I = [HONH2]0
x2 I−x
From problem, pH = 10.00, so pOH = 4.00 and x = [OH−] = 1.0 × 10 −4 M. 1.1 × 10 −8 =
(1.0 10 −4 ) 2 , I = 0.91 M I − (1.0 10 − 4 )
Mass HONH2 = 0.2500 L 180.
0.91 mol HONH 2 33.03 g HONH 2 = 7.5 g HONH2 L mol HONH 2
Codeine = C18H21NO3; codeine sulfate = C36H44N2O10S The formula for codeine sulfate works out to (codeineH +)2SO42−, where codeineH+ = HC18H21NO3+. Two codeine molecules are protonated by H 2SO4, forming the conjugate acid of codeine. The SO42− then acts as the counter ion to give a neutral compound. Codeine sulfate is an ionic compound that is more soluble in water than codeine in the neutral covalent form. This allows the full dose of the drug to dissolve into the bloodstream.
CHAPTER 14 181.
ACIDS AND BASES Fe(H2O)63+ + H2O
a. Initial Equil. Ka =
699
⇌ Fe(H2O)5(OH)2+
0.10 M 0.10 – x
0 x
+
H3 O+ ~0 x
[Fe(H 2 O) 5 (OH) 2+ ][H 3O + ] x2 x2 −3 10 , 6.0 × = 0.10 0.10 − x [Fe(H 2 O) 36+ ]
x = 2.4 × 10 −2 ; assumption is poor (x is 24% of 0.10). Using successive approximations: x2 = 6.0 × 10 −3 , x = 0.021 0.10 − 0.024 x2 x2 = 6.0 × 10 −3 , x = 0.022; = 6.0 × 10 −3 , x = 0.022 0.10 − 0.021 0.10 − 0.022
x = [H+] = 0.022 M; pH = 1.66 b. Because of the lower charge, Fe2+(aq) will not be as strong an acid as Fe3+(aq). A solution of iron(II) nitrate will be less acidic (have a higher pH) than a solution with the same concentration of iron(III) nitrate. 182.
One difficult aspect of acid-base chemistry is recognizing what types of species are present in solution, that is, whether a species is a strong acid, strong base, weak acid, weak base, or a neutral species. Below are some ideas and generalizations to keep in mind that will help in recognizing types of species present. a.
Memorize the following strong acids: HCl, HBr, HI, HNO 3, HClO4, and H2SO4
b.
Memorize the following strong bases: LiOH, NaOH, KOH, RbOH, CsOH, Ca(OH)2, Sr(OH)2, and Ba(OH)2
c.
Weak acids have a Ka value of less than 1 but greater than Kw. Some weak acids are listed in Table 14.2 of the text. Weak bases have a K b value of less than 1 but greater than Kw. Some weak bases are listed in Table 14.3 of the text.
d.
Conjugate bases of weak acids are weak bases; that is, all have a Kb value of less than 1 but greater than Kw. Some examples of these are the conjugate bases of the weak acids listed in Table 14.2 of the text.
e.
Conjugate acids of weak bases are weak acids; that is, all have a Ka value of less than 1 but greater than Kw. Some examples of these are the conjugate acids of the weak bases listed in Table 14.3 of the text.
f.
Alkali metal ions (Li+, Na+, K+, Rb+, Cs+) and some alkaline earth metal ions (Ca2+, Sr2+, Ba2+) have no acidic or basic properties in water.
g.
Conjugate bases of strong acids (Cl−, Br-, I−, NO3−, ClO4−, HSO4−) have no basic properties in water (Kb << Kw), and only HSO4- has any acidic properties in water.
Let’s apply these ideas to this problem to see what types of species are present. a. HI:
Strong acid; HF: weak acid (Ka = 7.2 × 10−4)
700
CHAPTER 14
ACIDS AND BASES
NaF:
F− is the conjugate base of the weak acid HF, so F − is a weak base. The Kb value for F− = Kw/Ka, HF = 1.4 × 10−11. Na+ has no acidic or basic properties.
NaI:
Neutral (pH = 7.0); Na+ and I− have no acidic/basic properties. In order of increasing pH, we place the compounds from most acidic (lowest pH) to most basic (highest pH). Increasing pH: HI < HF < NaI < NaF.
b. NH4Br:
NH4+ is a weak acid (Ka = 5.6 × 10−10), and Br- is a neutral species.
HBr:
Strong acid
KBr:
Neutral; K+ and Br− have no acidic/basic properties.
NH3:
Weak base, Kb = 1.8 × 10−5
Increasing pH: HBr < NH4Br < KBr < NH3 Most Most acidic basic c. C6H5NH3NO3:
C6H5NH3+ is a weak acid (Ka = K w /K b, C6H5 NH2 = 1.0 × 10−14/3.8 × 10−10 = 2.6 × 10−5), and NO3− is a neutral species.
NaNO3:
Neutral; Na+ and NO3− have no acidic/basic properties.
NaOH:
Strong base
HOC6H5:
Weak acid (Ka = 1.6 × 10−10)
KOC6H5:
OC6H5− is a weak base (K b = K w /K a, HOC6H5 = 6.3 × 10−5), and K+ is a neutral species.
C6H5NH2:
Weak base (Kb = 3.8 × 10−10)
HNO3:
Strong acid
This is a little more difficult than the previous parts of this problem because two weak acids and two weak bases are present. Between the weak acids, C 6H5NH3+ is a stronger weak acid than HOC6H5 since the Ka value for C6H5NH3+ is larger than the Ka value for HOC6H5. Between the two weak bases, because the K b value for OC6H5- is larger than the Kb value for C6H5NH2, OC6H5− is a stronger weak base than C6H5NH2. Increasing pH: HNO3 < C6H5NH3NO3 < HOC6H5 < NaNO3 < C6H5NH2 < KOC6H5 < NaOH Most acidic Most basic 183.
a. CaBr2 is a soluble ionic compound that dissolves in water to produce Ca2+(aq) and Br−(aq). Ca2+ has no acidic or basic properties. Br− is the conjugate base of the strong acid HBr. Br− has no basic (or acidic) properties. Therefore, a solution of CaBr 2 will be neutral because neither of the ions has any acidic or basic properties. The 1.2 M CaBr2 solution has [H+] = [OH−] = 1.0 × 10 −7 M and pH = pOH = 7.00.
CHAPTER 14
ACIDS AND BASES
701
b. C6H5NH3NO3 → C6H5NH3+ + NO3− ; C6H5NH3+ is the conjugate acid of the weak base C6H5NH2 (Kb = 3.8 × 10 −10 ). As is true for all conjugate acids of weak bases, C 6H5NH3+ is a weak acid. NO3− has no basic (or acidic) properties. Ignore NO 3−. Solving the weak acid problem:
⇌
C6H5NH3+ Initial Change Equil.
C6H5NH2
+
H+
Ka = Kw/3.8 × 10 −10 = 2.6 × 10 −5
0.84 M 0 ~0 x mol/L C6H5NH3+ dissociates to reach equilibrium –x → +x +x 0.84 – x x x
Ka = 2.6 × 10 −5 =
[C 6 H 5 NH 2 ][H + ] +
[C 6 H 5 NH 3 ]
=
x2 x2 (assuming x << 0.84) 0.84 − x 0.84
x = [H+] = 4.7 × 10 −3 M; pH = –log[H+] = 2.33; assumptions good. c. KC7H5O2 is a soluble ionic compound that dissolves in water to produce K +(aq) and C7H5O2−(aq). K+ has no acidic (or basic) properties. C7H5O2− is the conjugate base of the weak acid HC7H5O2, and as is true for all conjugate bases of weak acids, C 7H5O2− is a weak base in water. We must solve an equilibrium problem in order to determine the amount of OH− this weak base produces in water.
⇌ HC7H5O2 + OH−
C7H5O2− + H2O Initial Change Equil.
Kb =
Kw K a, C7 H 5O 2
=
1.0 10 −14 6.4 10 −5
0.57 M 0 ~0 Kb = 1.6 × 10 −10 x mol/L of C7H5O2− reacts with H2O to reach equilibrium –x → +x +x 0.57 − x x x
Kb = 1.6 × 10 −10 =
[HC 7 H 5 O 2 ][OH − ] −
, 1.6 × 10 −10 =
[C 7 H 5 O 2 ]
x2 x2 0.57 − x 0.57
x = [OH−] = 9.5 × 10 −6 M ; assumptions good pOH = –log[OH–] = 5.02; pH = 14.00 – 5.02 = 8.98 184.
185.
In general, the weakest acid will have the strongest conjugate base. The acids to consider are H2CO3, HCO3‒, H2S, and HS‒. HS‒ is the weakest of the acids since it has the smallest Ka value. Therefore, the strongest base is S2‒. The solution is acidic from HSO4− ⇌ H+ + SO42−. Solving the weak acid problem: HSO4− Initial Equil.
⇌
0.10 M 0.10 – x
1.2 × 10 −2 =
H+ ~0 x
2−
[H + ][SO 4 ] −
[HSO 4 ]
=
+
SO42−
Ka = 1.2 × 10−2
0 x
x2 x2 , x = 0.035 0.10 0.10 − x
702
CHAPTER 14
ACIDS AND BASES
Assumption is not good (x is 35% of 0.10). Using successive approximations: x2 x2 = = 1.2 × 10 −2 , x = 0.028 0.10 − 0.035 0.10 − x x2 x2 = 1.2 × 10 −2 , x = 0.029; = 1.2 × 10 −2 , x = 0.029 0.10 − 0.028 0.10 − 0.029
x = [H+] = 0.029 M; pH = 1.54 186.
a. NH3 + H3O+ ⇌ NH4+ + H2O Keq =
+ 1.8 10 −5 K b for NH 3 [ NH 4 ] 1 = = = = 1.8 × 109 + + −14 K [ NH 3 ][H ] 1.0 10 K a for NH 4 w
b. NO2− + H3O+ ⇌ HNO2 + H2O Keq =
[HNO2 ] −
[NO2 ][H + ]
=
1 1 = K a for HNO2 4.0 10 − 4
= 2.5 × 103 c. NH4+ + OH− ⇌ NH3 + H2O
Keq =
1 1 = = 5.6 × 104 K b for NH 3 1.8 10 −5
d. HNO2 + OH− ⇌ H2O + NO2− − 4.0 10 −4 K a for HNO2 [ NO2 ] [H + ] Keq = = = 4.0 × 1010 = Kw [HNO2 ][OH − ] [H + ] 1.0 10 −14
187.
a. The initial concentrations are halved since equal volumes of the two solutions are mixed. Use the Ka reaction for HC2H3O2 to determine the amount of H+ produced from this weak acid. HC2H3O2 Initial Equil.
0.100 M 0.100 − x
Ka = 1.8 × 10−5 =
⇌
H+
+
5.00 × 10−4 M 5.00 × 10−4 + x
C2H3O2− 0 x
x(5.00 10 −4 + x) x(5.00 10 −4 ) 0.100 − x 0.100
x = 3.6 × 10−3; assumption is horrible. Using the quadratic formula: x2 + (5.18 × 10−4)x − 1.8 × 10−6 = 0 x = 1.1 × 10−3 M; [H+] = 5.00 × 10−4 + x = 1.6 × 10−3 M; pH = 2.80 b. x = [C2H3O2−] = 1.1 × 10−3 M
CHAPTER 14 188.
ACIDS AND BASES
703
Major species: HIO3, H2O; major source of H+: HIO3 (a weak acid, Ka = 0.17)
⇌
HIO3 Initial
H+
IO3−
+
0.010 M ~0 0 x mol/L HIO3 dissociates to reach equilibrium −x → +x +x 0.010 − x x x
Change Equil.
−
Ka = 0.17 =
[H + ][IO3 ] x2 x2 , x = 0.041; check assumption. = [HIO3 ] 0.010 − x 0.010
Assumption is horrible (x is more than 400% of 0.010). When the assumption is this poor, it is generally quickest to solve exactly using the quadratic formula (see Appendix 1 in text). Using the quadratic formula and carrying extra significant figures: 0.17 = x=
x2 , x2 = 0.17(0.010 − x), x2 + (0.17)x − 1.7 × 10−3 = 0 0.010 − x
− 0.17 [(0.17) 2 − 4(1)(−1.7 10 −3 )]1/ 2 − 0.17 0.189 , x = 9.5 × 10−3 M = 2(1) 2 (x must be positive)
x = 9.5 × 10−3 M = [H+]; pH = −log(9.5 × 10−3) = 2.02 189.
The typical ICE table set-up to solve a weak acid problem is: HA Initial Change Equil. Ka =
⇌
H+
+
A−
HA = weak acid
1.0 M ~0 0 x mol/L HA dissociates to reach equilibrium −x → +x +x 1.0 − x x x
[H + ][A − ] x2 x2 = ≈ when the 5% rule applies. 1 .0 1.0 − x [HA]
We want x to be less than or equal to 5.0% of 1.0 to follow the 5% rule: 2
(0.050) x2 = 2.5 × 10−3 = 1.0 1.0 When Ka ≤ 2.5 × 10−3, a 1.0 M weak acid solution will follow the 5% rule. x = 0.050 1.0 M = 0.050 M; this gives: Ka =
190.
Let’s assume we have an H2SO4 with an initial concentration of I. An I M H2SO4 solution produces I M HSO4- and I M H+. Now let’s consider the amount of H+ produced from the weak acid HSO4− (which we want to figure out a value for I so we can ignore this part of the calculation). HSO4− Initial Change Equil.
⇌
H+
+
SO42−
I I 0 x mol/L HSO4− dissociates to reach equilibrium −x → +x +x I−x I+x x
704
CHAPTER 14 2−
K a 2 = 1.2 × 10−2 =
[ H + ][SO 4 ] −
[ HSO 4 ]
=
ACIDS AND BASES
( I + x ) x I( x ) , x = 1.2 × 10−2 M I−x I
If the assumption that the amount of H + (= x) from HSO4− can be ignored is to hold, then x (= 0.012 M) must be equal to or less than 5% of I, the initial connetration of the H2SO4 solution. 0.050 =
0.012 x 0.012 = , I= = 0.24 M 0.050 I I
When the H2SO4 solution has a concentration of 0.24 M or greater, then the H+ concentration from HSO4− can be ignored by the 5% rule. 191.
For H3PO4, K a1 = 7.5 10 −3 , K a 2 = 6.2 10 −8 , and K a 3 = 4.8 10 −13. Because K a1 is much larger than K a 2 and K a 3 , the dominant H+ producer is H3PO4, and the H+ contributed from H2PO4− and HPO42− can be ignored. Solving the weak acid problem in the typical manner. H3PO4 Initial Equil.
⇌
H2PO4−
0.007 M 0.007 − x
K a1 = 7.5 10 −3 =
0 x
+
H+
K a1 = 7.5 10 −3
~0 x
−
[H 2 PO 4 ][H + ] x2 x2 = [H 3 PO 4 ] 0.007 − x 0.007
x = 7.2 × 10−3; assumption is horrible because x is more than 100% of 0.007. We will use the quadratic equation to solve exactly and carry extra significant figures through the calculation. 7.5 × 10−3 =
x2 , x2 = 5.25 × 10−5 − (7.5 × 10−3)x, x2 + (7.5 × 10−3)x − 5.25 × 10−5 = 0 0.007 − x
x = [H + ] =
− 7.5 10 −3 [(7.5 10 −3 ) 2 − 4(1)(−5.25 10 −5 )]1 / 2 = 4.4 × 10−3 = 4 × 10−3 M 2(1)
pH = −log(4 × 10−3) = 2.4
Challenge Problems 192.
The pH of this solution is not 8.00 because water will donate a significant amount of H + from the autoionization of water. You can’t add an acid to water and get a basic pH. The pertinent equations are: H2O ⇌ H+ + OH− Kw = [H+][OH−] = 1.0 × 10 −14 HCl → H+ + Cl−
Ka is very large, so we assume that only the forward reaction occurs.
In any solution, the overall net positive charge must equal the overall net negative charge (called the charge balance). For this problem: [positive charge] = [negative charge], so [H +] = [OH−] + [Cl−]
CHAPTER 14
ACIDS AND BASES
705
From Kw, [OH−] = Kw/[H+], and from 1.0 × 10 −8 M HCl, [Cl−] = 1.0 × 10 −8 M. Substituting into the charge balance equation: [H+] =
1.0 10 −14 +
[H ]
+ 1.0 × 10 −8 , [H+]2 – (1.0 × 10 −8 )[H+] – 1.0 × 10 −14 = 0
Using the quadratic formula to solve: [H+] =
− (−1.0 10 −8 ) [(−1.0 10 −8 ) 2 − 4(1)(−1.0 10 −14 )]1 / 2 , [H+] = 1.1 × 10 −7 M 2(1)
pH = –log(1.1 × 10 −7 ) = 6.96 193.
Because this is a very dilute solution of NaOH, we must worry about the amount of OH − donated from the autoionization of water. NaOH → Na+ + OH− H2O ⇌ H+ + OH−
Kw = [H+][OH−] = 1.0 × 10−14
This solution, like all solutions, must be charge balanced; that is, [positive charge] = [negative charge]. For this problem, the charge balance equation is: [Na+] + [H+] = [OH−], where [Na+] = 1.0 × 10−7 M and [H+] =
Kw [OH − ]
Substituting into the charge balance equation: 1.0 × 10−7 +
1.0 10 −14 = [OH−], [OH−]2 − (1.0 × 10−7)[OH−] − 1.0 × 10−14 = 0 − [OH ]
Using the quadratic formula to solve: [OH−] =
− (−1.0 10 −7 ) [(−1.0 10 −7 ) 2 − 4(1)(−1.0 10 −14 )]1/ 2 2(1)
[OH−] = 1.6 × 10−7 M; pOH = −log(1.6 × 10−7) = 6.80; pH = 7.20 194.
Ca(OH)2 (s) → Ca2+(aq) + 2 OH−(aq) This is a very dilute solution of Ca(OH)2, so we can't ignore the OH− contribution from H2O. From the dissociation of Ca(OH)2 alone, 2[Ca2+] = [OH−]. Including the H2O autoionization into H+ and OH−, the overall charge balance is: 2[Ca2+] + [H+] = [OH−] 2(3.0 × 10−7 M) + Kw/[OH−] = [OH−], [OH−]2 = (6.0 × 10−7)[OH−] + Kw [OH−]2 − (6.0 × 10−7)[OH−] − 1.0 × 10−14 = 0; using quadratic formula: [OH−] = 6.2 × 10−7 M
706
CHAPTER 14
195.
⇌
HA Initial Equil. Ka =
H+
C C − 1.00 × 10−4
A−
+
~0 1.00 × 10−4
ACIDS AND BASES
Ka = 1.00 × 10−6
0 1.00 × 10−4
C = [HA]0 , for pH = 4.000, x = [H+] = 1.00 × 10−4 M
(1.00 10 −4 ) 2 = 1.00 × 10−6; solving: C = 0.0101 M (C − 1.00 10 − 4 )
The solution initially contains 50.0 × 10 −3 L × 0.0101 mol/L = 5.05 × 10−4 mol HA. We then dilute to a total volume V in liters. The resulting pH = 5.000, so [H +] = 1.00 × 10−5. In the typical weak acid problem, x = [H+], so:
⇌
HA
H+
Initial 5.05 × 10−4 mol/V Equil. (5.05 × 10−4/V) − (1.00 × 10−5) Ka =
A−
+
~0 1.00 × 10−5
0 1.00 × 10−5
(1.00 10 −5 ) 2 = 1.00 × 10−6 (5.05 10 − 4 /V) − (1.00 10 −5 )
1.00 × 10−4 = (5.05 × 10−4/V) − 1.00 × 10−5 V = 4.59 L; 50.0 mL are present initially, so we need to add 4540 mL of water. 196.
⇌
HBrO
H+
Ka = 2 × 10 −9
BrO−
+
1.0 × 10 −6 M ~0 0 x mol/L HBrO dissociates to reach equilibrium Change –x → +x +x Equil. 1.0 × 10 −6 – x x x Initial
Ka = 2 × 10 −9 =
x2 x2 , x = [H+] = 4 × 10 −8 M; pH = 7.4 (1.0 10 −6 − x) 1.0 10 −6
Let’s check the assumptions. This answer is impossible! We can't add a small amount of an acid to water and get a basic solution. The highest possible pH for an acid in water is 7.0. In the correct solution we would have to consider the autoionization of water. 197.
Major species present are H2O, C5H5NH+ [Ka = K w /K b, C5H5 N = (1.0 × 10−14)/(1.7 × 10−9) = 5.9 × 10−6], and F− [Kb = K w /K a, HF = (1.0 × 10−14)/(7.2 × 10−4) = 1.4 × 10−11]. The reaction to consider is the best acid present (C5H5NH+) reacting with the best base present (F−). Let’s solve by first setting up an ICE table.
Initial Change Equil.
C5H5NH+(aq)
+
F−(aq)
⇌
0.200 M −x 0.200 − x
0.200 M −x 0.200 − x
→
1
= 5.9 × 10−6
K = K a , C H NH+ 5
5
K a , HF
C5H5N(aq) + HF(aq) 0 +x x
0 +x x
1 = 8.2 × 10−3 7.2 10 − 4
CHAPTER 14 K=
ACIDS AND BASES
707
[C 5 H 5 N ][HF ] x2 , 8.2 × 10−3 = ; taking the square root of both sides: + − [C 5 H 5 NH ][F ] (0.200 − x) 2
0.091 =
x , x = 0.018 − (0.091)x, x = 0.016 M 0.200 − x
From the setup to the problem, x = [C5H5N] = [HF] = 0.016 M, and 0.200 − x = 0.200 − 0.016 = 0.184 M = [C5H5NH+] = [F−]. To solve for the [H+], we can use either the Ka equilibrium for C5H5NH+ or the Ka equilibrium for HF. Using C5H5NH+ data:
Ka , C H NH+ = 5.9 × 10−6 = 5
5
[C5 H 5 N][H + ] (0.016 )[H + ] , [H+] = 6.8 × 10−5 M = + 0.184 [C5 H 5 NH ]
pH = −log(6.8 × 10−5) = 4.17 As one would expect, because the Ka for the weak acid is larger than the Kb for the weak base, a solution of this salt should be acidic. 198.
Major species: NH4+, OCl−, and H2O; Ka for NH4+ = (1.0 × 10−14)/(1.8 × 10−5) = 5.6 × 10 −10 and Kb for OCl− = (1.0 × 10−14)/(3.5 × 10−8) = 2.9 × 10 −7 . Because OCl− is a better base than NH4+ is an acid, the solution will be basic. The dominant equilibrium is the best acid (NH4+) reacting with the best base (OCl−) present. NH4+ Initial Change Equil.
0.50 M –x 0.50 – x
K = K a , NH + 4
K = 0.016 =
1 K a , HOCl
0.50 M –x → 0.50 – x
NH3
+ HOCl
0 +x x
0 +x x
= (5.6 × 10−10)/(3.5 × 10−8) = 0.016
[NH3 ][HOCl] +
⇌
OCl−
+
−
=
[NH 4 ][OCl ]
x(x) (0.50 − x)(0.50 − x )
2
x x = (0.016)1/2 = 0.13, x = 0.058 M = 0.016, 2 0 . 50 − x (0.50 − x)
To solve for the H+, use any pertinent Ka or Kb value. Using Ka for NH4+:
Ka , NH + = 5.6 × 10 −10 = 4
199.
[ NH 3 ][H + ] +
[ NH 4 ]
=
(0.058 )[H + ] , [H+] = 4.3 × 10 −9 M, pH = 8.37 0.50 − 0.058
Because NH3 is so concentrated, we need to calculate the OH − contribution from the weak base NH3. NH3 + H2O Initial 15.0 M Equil. 15.0 − x
⇌
NH4+ 0 x
+
OH−
Kb = 1.8 × 10−5
0.0100 M (Assume no volume change.) 0.0100 + x
708
CHAPTER 14
ACIDS AND BASES
x(0.0100 + x) x(0.0100 ) , x = 0.027; assumption is horrible 15.0 − x 15.0 (x is 270% of 0.0100). Using the quadratic formula:
Kb = 1.8 × 10−5 =
(1.8 × 10−5)(15.0 − x) = (0.0100)x + x2, x2 + (0.0100)x − 2.7 × 10−4 = 0 x = 1.2 × 10−2 M, [OH−] = (1.2 × 10−2) + 0.0100 = 0.022 M 200.
For 0.0010% dissociation: [NH4+] = 1.0 × 10 −5 (0.050) = 5.0 × 10 −7 M NH3 + H2O ⇌ NH4+ + OH−
Kb =
(5.0 10 −7 )[OH − ] = 1.8 × 10 −5 −7 0.050 − 5.0 10
Solving: [OH−] = 1.8 M; assuming no volume change: 1.0 L ×
201.
1.000 L
1.8 mol NaOH 40.00 g NaOH = 72 g of NaOH L mol NaOH
1.00 10 −4 mol HA = 1.00 × 10−4 mol HA L
25.0% dissociation gives: moles H+ = 0.250 (1.00 × 10−4) = 2.50 × 10−5 mol moles A− = 0.250 (1.00 × 10−4) = 2.50 × 10−5 mol moles HA = 0.750 (1.00 × 10−4) = 7.50 × 10−5 mol
2.50 10 −5 2.50 10 −5 + − V V [ H ][ A ] 1.00 × 10−4 = Ka = = − 5 [HA] 7.50 10 V 1.00 × 10−4 =
(2.50 10 −5 ) 2 (2.50 10 −5 ) 2 , V = = 0.0833 L = 83.3 mL (1.00 10 − 4 )(7.50 10 −5 ) (7.50 10 −5 )(V)
The volume goes from 1000. mL to 83.3 mL, so 917 mL of water evaporated. 202.
HC2H3O2 ⇌ H+ + Initial 1.00 M Equil. 1.00 − x 1.8 × 10 −5 =
~0 x
C2H3O2−
Ka = 1.8 × 10 −5
0 x
x2 x2 , x = [H+] = 4.24 × 10 −3 M (using one extra sig. fig.) 1.00 1.00 − x
pH = –log(4.24 × 10 −3 ) = 2.37; assumptions good.
CHAPTER 14
ACIDS AND BASES
709
We want to double the pH to 2(2.37) = 4.74 by addition of the strong base NaOH. As is true with all strong bases, they are great at accepting protons. In fact, they are so good that we can assume they accept protons 100% of the time. The best acid present will react the strong base. This is HC2H3O2. The initial reaction that occurs when the strong base is added is: HC2H3O2 + OH− → C2H3O2− + H2O Note that this reaction has the net effect of converting HC 2H3O2 into its conjugate base, C2H3O2−. For a pH = 4.74, let’s calculate the ratio of [C 2H3O2−]/[HC2H3O2] necessary to achieve this pH. − [ H + ][C 2 H 3 O 2 ] HC2H3O2 ⇌ H+ + C2H3O2− Ka = [ HC 2 H 3 O 2 ] When pH = 4.74, [H+] = 10 −4.74 = 1.8 × 10 −5. Ka = 1.8 × 10 −5 =
−
−
(1.8 10 −5 )[C 2 H 3O 2 ] [C 2 H 3O 2 ] = 1.0 , [HC2 H 3O 2 ] [HC2 H 3O 2 ]
For a solution having pH = 4.74, we need to have equal concentrations (equal moles) of C2H3O2− and HC2H3O2. Therefore, we need to add an amount of NaOH that will convert onehalf of the HC2H3O2 into C2H3O2−. This amount is 0.50 M NaOH. HC2H3O2
+ OH−
Before 1.00 M Change –0.50 After 0.50 M completion
→
0.50 M –0.50 → 0
C2H3O2− + H2O 0 +0.50 0.50 M
From the preceding stoichiometry problem, adding enough NaOH(s) to produce a 0.50 M OH− solution will convert one-half the HC2H3O2 into C2H3O2−; this results in a solution with pH = 4.74. Mass NaOH = 1.00 L 203.
0.50 mol NaOH 40.00 g NaOH = 20. g NaOH L mol
PO43− is the conjugate base of HPO42−. The Ka value for HPO42− is K a 3 = 4.8 × 10−13. PO43−(aq) + H2O(l) ⇌ HPO42−(aq) + OH−(aq)
Kb =
1.0 10 −14 Kw = = 0.021 K a3 4.8 10 −13
HPO42− is the conjugate base of H2PO4− ( K a 2 = 6.2 × 10 −8 ). HPO42−+ H2O
⇌ H2PO4 + OH −
−
1.0 10 −14 Kw Kb = = = 1.6 × 10 −7 −8 K a1 6.2 10
710
CHAPTER 14
ACIDS AND BASES
H2PO4− is the conjugate base of H3PO4 ( K a1 = 7.5 × 10 −3 ). H2PO4− + H2O ⇌ H3PO4 + OH− Kb =
−14
1.0 10 Kw = = 1.3 × 10 −12 −3 K a1 7.5 10
From the Kb values, PO43− is the strongest base. This is expected because PO43− is the conjugate base of the weakest acid (HPO 42−). 204.
Major species: Na+, PO43− (a weak base), H2O; From the Kb values calculated in Exercise 203, the dominant producer of OH− is the Kb reaction for PO43−. We can ignore the contribu-tion of OH− from the Kb reactions for HPO42− and H2PO4− . From Exercise 203, Kb for PO43− = 0.021. PO43− + H2O Initial Equil.
⇌
HPO42− + OH−
0.10 M 0.10 – x
0 x
Kb = 0.021
~0 x
x2 ; because Kb is so large, the 5% assumption will not hold. Solving 0.10 − x using the quadratic equation:
Kb = 0.021 =
x2 + (0.021)x – 0.0021 = 0, x = [OH−] = 3.7 × 10 −2 M, pOH = 1.43, pH = 12.57 205.
a. NH4(HCO3) → NH4+ + HCO3− K a , NH + = 4
Kw 1.0 10 −14 1.0 10 −14 −10 = 5.6 × 10 ; = 2.3 × 10−8 = K − = b, HCO3 K a1 1.8 10 −5 4.3 10 −7
The solution is basic because HCO3− is a stronger base than NH4+ is as an acid. The acidic properties of HCO3− were ignored because K a 2 is very small (4.8 × 10−11). b. NaH2PO4 → Na+ + H2PO4−; ignore Na+.
K a , H PO − = 6.2 × 10−8; K b, H PO − = 2
2
4
2
4
Kw 1.0 10 −14 = 1.3 × 10−12 = K a1 7.5 10 −3
Solution is acidic because Ka > Kb. c. Na2HPO4 → 2 Na+ + HPO42−; ignore Na+.
K a , HPO 2− = 4.8 × 10−13; K b, HPO 2− = 3
4
4
Kw 1.0 10 −14 = 1.6 × 10−7 = Ka2 6.2 10 −8
Solution is basic because Kb > Ka. d. NH4(H2PO4) → NH4+ + H2PO4− NH4+ is a weak acid, and H2PO4− is also acidic (see part b). Solution with both ions present will be acidic.
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711
e. NH4(HCO2) → NH4+ + HCO2−; from Appendix 5, K a , HCO2H = 1.8 × 10−4.
K a , NH + = 5.6 × 10−10; K b, HCO − = 4
2
Kw 1.0 10 −14 = 5.6 × 10−11 = Ka 1.8 10 − 4
Solution is acidic because NH4+ is a stronger acid than HCO2− is as a base. 206.
a. HCO3− + HCO3− ⇌ H2CO3 + CO32− 2−
Keq =
[H 2 CO 3 ][CO 3 ] −
−
[HCO3 ][HCO3 ]
K a2 5.6 10 −11 [H + ] = 1.3 × 10−4 = = + −7 K [H ] 4.3 10 a1
b. The reaction in part a produces equal moles of both products, so [H2CO3] = [CO32−]. c.
H2CO3 ⇌ 2 H+ + CO32−
2−
Keq =
[H + ]2 [CO3 ] = K a1 K a 2 [H 2 CO3 ]
Because [H2CO3] = [CO32−] from part b, [H+]2 = K a1 K a 2 . [H+] = (K a1 K a 2 )1/ 2 , or taking the −log of both sides: pH =
pK a1 + pK a 2 2
d. [H+] = [(4.3 × 10−7) × (5.6 × 10−11)]1/2, [H+] = 4.9 × 10−9 M; pH = 8.31 1 mol 100 .0 g = 2.00 × 10 −3 mol/kg 2.00 × 10 −3 mol/L 0.5000 kg
0.100 g
207.
Molality = m =
ΔTf = iKfm, 0.0056°C = i(1.86°C/molal)(2.00 × 10 −3 molal), i = 1.5 If i = 1.0, the percent dissociation of the acid = 0%, and if i = 2.0, the percent dissociation of the acid = 100%. Because i = 1.5, the weak acid is 50.% dissociated. HA ⇌ H+ + A−
Ka =
[H + ][A − ] [HA]
Because the weak acid is 50.% dissociated: [H+] = [A−] = [HA]0 × 0.50 = 2.00 × 10 −3 M × 0.50 = 1.0 × 10 −3 M [HA] = [HA]0 – amount HA reacted = 2.00 × 10 −3 M – 1.0 × 10 −3 M = 1.0 × 10 −3 M Ka = 208.
(1.0 10 −3 ) (1.0 10 −3 ) [H + ][A − ] = = 1.0 × 10 −3 −3 [HA] 1.0 10
a. Assuming no ion association between SO42−(aq) and Fe3+(aq), then i = 5 for Fe2(SO4)3. π = iMRT = 5(0.0500 mol/L)(0.08206 L atm/K•mol)(298 K) = 6.11 atm b. Fe2(SO4)3(aq) → 2 Fe3+(aq) + 3 SO42−(aq)
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Under ideal circumstances, 2/5 of π calculated above results from Fe 3+ and 3/5 results from SO42−. The contribution to π from SO42− is 3/5 × 6.11 atm = 3.67 atm. Because SO42− is assumed unchanged in solution, the SO42− contribution in the actual solution will also be 3.67 atm. The contribution to the actual osmotic pressure from the Fe(H2O)63+ dissociation reaction is 6.73 – 3.67 = 3.06 atm. The initial concentration of Fe(H2O)62+ is 2(0.0500) = 0.100 M. The set up for the weak acid problem is: [H + ][Fe(OH)(H 2 O) 52+ ] Fe(H2O)63+ ⇌ H+ + Fe(OH)(H2O)52+ Ka = [Fe(H 2 O) 36+ ] Initial 0.100 M ~0 0 x mol/L of Fe(H2O)63+ reacts to reach equilibrium Equil. 0.100 – x x x π = iMRT; total ion concentration = iM =
3.06 atm π = = 0.125 M RT 0.8206 L atm/K • mol(298)
0.125 M = 0.100 – x + x + x = 0.100 + x, x = 0.025 M Ka =
[H + ][Fe(OH)(H 2 O) 52+ ] (0.025) 2 (0.025) 2 x2 = = = 0.100 − x (0.100 − 0.025) 0.075 [Fe(H 2 O) 36+ ]
Ka = 8.3 × 10−3 209.
dRT = Molar mass = P
5.11 g / L
0.08206 L atm 298 K K mol = 125 g/mol 1.00 atm
1 mol 125 g = 0.120 M; pH = 1.80, [H+] = 10−1.80 = 1.6 × 10−2 M 0.100 L
1.50 g [HA]0 =
HA
⇌
Initial 0.120 M Equil. 0.120 − x Ka =
H+
+ A−
~0 x
0 x
where x = [H+] = 1.6 × 10−2 M
[H + ][A − ] (1.6 10 −2 ) 2 = 2.5 × 10−3 = [HA] 0.120 − 0.016
Marathon Problems 210.
Let’s abbreviate HCO2H with HA and CH3CH2CO2H with HB. Fom the problem, we have: 0.0500 M HCO2H (HA), Ka = 1.77 × 10−4; 0.150 M CH3CH2CO2H (HB), Ka = 1.34 × 10−5 Because two comparable weak acids are present, each will contribute to the total pH. To solve for the pH, we need to determine some relationships that must hold true in this solution. Two equations that must hold true are the K a expressions for the two acids:
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713
[H + ][A − ] [H + ][B− ] = 1.34 × 10−5 = 1.77 10 −4 ; [HA] [HB] Another relationship that must hold true recognizes that all of the HA originally dissolved (0.0500 M) must be present as either A− or HA at equilibrium. So: 0.0500 = [HA] + [A−] A similar relationship holds for HB: 0.150 = [HB] + [B −] The last relationship we need recognizes that in any solution, the net positive charge must equal the net negative charge. This is called the charge balance equation. This equation is: [H+] = [A−] + [B−] + [OH−] = [A−] + [B−] + Kw/[H+] We now have five equations and five unknowns ([HA], [HB], [A−], [B−], and [H+]). We should be able to manipulate the equations to solve for [H +]. Solving: [H+] = [A−] + [B−] + Kw/[H+]; [H+]2 = [H+][A−] + [H+][B−] + Kw [H+][A−] = (1.77 × 10−4)[HA] = (1.77 × 10−4) (0.0500 − [A−]) If [A−] << 0.0500, then [H+][A−] (1.77 × 10−4) (0.0500) = 8.85 × 10−6. Similarly, assume [H+][B−] (1.34 × 10−5)(0.150) = 2.01 × 10−6. [H+]2 = 8.85 × 10−6 + 2.01 × 10−6 + 1.00 × 10−14, [H+] = 3.30 × 10−3 mol/L Checking assumptions: [H+][A−] 8.85 × 10−6, [A−]
8.85 10 −6 2.68 × 10−3 3.30 10 −3
We assumed 0.0500 − [A−] 0.0500. This assumption is borderline (2.68 × 10 −3 is 5.4% of 0.0500). The HB assumption is good (0.4% error). Using successive approximations to refine the [H +][A−] value gives [H+][A−] = 8.39 × 10−6, which gives: [H+] = 3.22 × 10−3 M, pH = −log(3.22 × 10−3) = 2.492 Note: If we treat each acid separately: H+ from HA = 2.9 × 10−3 M H+ from HB = 1.4 × 10−3 M 4.3 × 10−3 M = [H+]total This assumes the acids did not suppress each other’s ionization. They do, and we expect the [H+] to be less than 4.3 × 10−3 M. We get such an answer ([H+] = 3.22 × 10−3 M).
714 211.
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a. Strongest acid from group I = HCl; weakest base (smallest K b) from group II = NaNO2 0.20 M HCl + 0.20 M NaNO2; major species = H+, Cl−, Na+, NO2−, and H2O; let the H+ react to completion with the NO2−; then solve the back equilibrium problem.
Before After
H+
+ NO2−
→ HNO2
0.10 M 0
0.10 M 0
0 0.10 M
(Molarities are halved due to dilution.)
HNO2 ⇌ H+ + NO2− Initial 0.10 M 0 0 Change −x → +x +x Equil. 0.10 − x x x
Ka = 4.0 × 10−4
x2 = 4.0 × 10−4; solving: x = [H+] = 6.1 × 10−3 M; pH = 2.21 0.10 − x
b. Weakest acid from group I = (C2H5)3NHCl; best base from group II = KOI; the dominant equilibrium will be the best base reacting with the best acid. OI− + (C2H5)3NH+ Initial Equil. K=
0.10 M 0.10 − x
K a , ( C H ) NH+ 2
5 3
K a , HOI
⇌
0.10 M 0.10 − x =
1.0 10 −14 4.0 10
−4
HOI
+ (C2H5)3N
0 x
0 x
1 = 1.25 (carrying extra sig. fig.) 2.0 10 −11
x x2 = 1.25, = 1.12, x = 0.053 M 2 0.10 − x (0.10 − x)
So [HOI] = 0.053 M and [OI−] = 0.10 – x = 0.047 M; using the Ka equilibrium constant for HOI to solve for [H+]: 2.0 × 10−11 =
[H + ](0.047 ) , [H+] = 2.3 × 10−11 M; pH = 10.64 (0.053)
c. Ka for (C2H5)3NH+ =
1.0 10 −14 1.0 10 −14 − −11 = 2.5 × 10 ; K for NO = = 2.5 × 10−11 b 2 4.0 10 − 4 4.0 10 − 4
Because Ka = Kb, mixing (C2H5)3NHCl with NaNO2 will result in a solution with pH = 7.00.
CHAPTER 15 ACID-BASE EQUILIBRIA Review Questions 1.
A common ion is an ion that appears in an equilibrium reaction but came from a source other than that reaction. Addition of a common ion (H + or NO2−) to the reaction HNO2 ⇌ H+ + NO2− will drive the equilibrium to the left as predicted by Le Châtelier’s principle. When a weak acid solution has some of the conjugate base added from an outside source, this solution is called a buffer. Similarly, a weak base solution with its conjugate acid added from an outside source would also be classified as a buffer.
2.
A buffer solution is one that resists a change in its pH when either hydroxide ions or protons (H+) are added. Any solution that contains a weak acid and its conjugate base or a weak base and its conjugate acid is classified as a buffer. The pH of a buffer depends on the [base]/[acid] ratio. When H+ is added to a buffer, the weak base component of the buffer reacts with the H + and forms the acid component of the buffer. Even though the concentrations of the acid and base component of the buffer change some, the ratio of [base]/[acid] does not change that much. This translates into a pH that doesn’t change much. When OH − is added to a buffer, the weak acid component is converted into the base com-ponent of the buffer. Again, the [base]/[acid] ratio does not change a lot (unless a large quantity of OH − is added), so the pH does not change much. The concentrations of weak acid and weak base do not have to be equal in a buffer. If both a weak acid and a weak base are present, then the solution will be buffered. If the concentrations are the same, the buffer will have the same capacity towards added H + and added OH−. Also, buffers with equal concentrations of weak acid and conjugate base have pH = pKa. Because both the weak acid and conjugate base are major species present, both equilibriums that refer to these species must hold true. That is, the Ka equilibrium must hold because the weak acid is present, and the Kb equilibrium for the conjugate base must hold true because the conjugate base is a major species. Both the Ka and Kb equilibrium have the acid and conjugate base concentrations in their expressions. The same equilibrium concentrations of the acid and conjugate base must satisfy both equilibriums. In addition, the [H +] and [OH−] concentrations must be related through the Kw constant. This leads to the same pH answer whether the Ka or Kb equilibrium is used. The third method to solve a buffer problem is to use the Henderson-Hasselbalch equation. The equation is: [base] pH = pKa + log [acid]
715
716
CHAPTER 15
ACID-BASE EQUILIBRIA
where the base is the conjugate base of the weak acid present, or the acid is the conjugate acid of the weak base present. The equation considers the normal assumptions made for buffers. Specifically, it is assumed that the initial concentration of the acid and base component of the buffer equal the equilibrium concentrations. For any decent buffer, this will always hold true. 3.
Whenever strong acid is added to a solution, always react the H+ from the strong acid with the best base present in solution. The best base has the largest Kb value. For a buffer, this will be the conjugate base (A−) of the acid component of the buffer. The H+ reacts with the conjugate base, A−, to produce the acid, HA. The assumption for this reaction is that because strong acids are great at what they do, they are assumed to donate the proton to the conjugate base 100% of the time. That is, the reaction is assumed to go to completion. Completion is when a reaction goes until one or both reactants run out. This reaction is assumed to be a stoichiometry problem like those we solved in Chapter 3 of the text. Whenever a strong base is added to a buffer, the OH − ions react with the best acid present. This reaction is also assumed to go to completion. In a buffer, the best acid present is the acid component of the buffer (HA). The OH − rips a proton away from the acid to produce the conjugate base of the acid (A−) and H2O. Again, we know strong bases are great at accepting protons, so we assume this reaction goes to completion. It is assumed to be a stoichiometry problem like the ones we solved in Chapter 3. Note: For BH+/B type buffers (buffers composed of a weak base B and its conjugate acid BH+), added OH− reacts to completion with BH+ to produce B and H2O, while added H+ reacts to completion with B to produce BH+. An example of a BH+/B type buffer is NH4+/NH3.
4.
When [HA] = [A−] (or [BH+] = [B]) for a buffer, the pH of the solution is equal to the pKa value for the acid component of the buffer (pH = pKa because [H+] = Ka). A best buffer has equal concentrations of the acid and base component so it can be equally efficient at absorbing added H+ or OH−. For a pH = 4.00 buffer, we would choose the acid component having a Ka close to 10 −4.00 = 1.0 × 10 −4 (pH = pKa for a best buffer). For a pH = 10.00 buffer, we would want the acid component of the buffer to have a Ka close to 10 −10.00 = 1.0 × 10 −10 . Of course, we can have a buffer solution made from a weak base (B) and its conjugate acid (BH +). For a pH = 10.00 buffer, our conjugate acid should have Ka 1.0 × 10 −10 which translates into a Kb value of the base close to 1.0 × 10 −4 (Kb = Kw/Ka for conjugate acid-base pairs). The capacity of a buffer is a measure of how much strong acid or strong base the buffer can neutralize. All the buffers listed have the same pH (= pKa = 4.74) because they all have a 1:1 concentration ratio between the weak acid and the conjugate base. The 1.0 M buffer has the greatest capacity; the 0.01 M buffer the least capacity. In general, the larger the concentrations of weak acid and conjugate base, the greater the buffer capacity, i.e., the more strong acid or strong base that can be neutralized with little pH change.
5.
Let’s review the strong acid-strong base titration using the example (case study) covered in section 15.4 of the text. The example used was the titration of 50.0 mL of 0.200 M HNO3 titrated by 0.100 M NaOH. See Figure 15.1 for the titration curve. The important points are: a. Initially, before any strong base has been added. Major species: H +, NO3−, and H2O. To determine the pH, determine the [H+] in solution after the strong acid has completely dissociated as we always do for strong acid problems.
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ACID-BASE EQUILIBRIA
717
b. After some strong base has been added, up to the equivalence point. For our example, this is from just after 0.00 mL NaOH added up to just before 100.0 mL NaOH added. Major species before any reaction: H+, NO3−, Na+, OH−, and H2O. Na+ and NO3− have no acidic or basic properties. In this region, the OH− from the strong base reacts with some of the H+ from the strong acid to produce water (H + + OH− → H2O). As is always the case when something strong reacts, we assume the reaction goes to completion. Major species after reaction: H+, NO3−, Na+, and H2O: To determine the pH of the solution, we first determine how much of the H+ is neutralized by the OH−. Then we determine the excess [H +] and take the –log of this quantity to determine pH. From 0.1 mL to 99.9 mL NaOH added, the excess H+ from the strong acid determines the pH. c. The equivalence point (100.0 mL NaOH added). Major species before reaction: H +, NO3−, Na+, OH−, and H2O. Here, we have added just enough OH− to neutralize all of the H+ from the strong acid (moles OH− added = moles H+ present). After the stoichiometry reaction (H+ + OH− → H2O), both H+ and OH− have run out (this is the definition of the equivalence point). Major species after reaction: Na+, NO3−, and H2O. All we have in solution are some ions with no acidic or basic properties (NO 3− and Na+ in H2O). The pH = 7.00 at the equivalence point of a strong acid by a strong base. d. Past the equivalence point (volume of NaOH added > 100.0 mL). Major species before reaction H+, NO3−, Na+, OH−, and H2O. After the stoichiometry reaction goes to completion (H+ + OH− → H2O), we have excess OH− present. Major species after reaction: OH−, Na+, NO3−, and H2O. We determine the excess [OH−] and convert this into the pH. After the equivalence point, the excess OH− from the strong base determines the pH. 6.
See Figure 15.2 of the text for a titration curve of a strong base by a strong acid. The stoichiometry problem is still the same, H+ + OH− → H2O, but what is in excess after this reaction goes to completion is reverse of the strong acid-strong base titration. The pH up to just before the equivalence point is determined by the excess OH − present. At the equivalence point, pH = 7.00 because we have added just enough H + from the strong acid to react with all of the OH− from the strong base (mole base present = mole acid added). Past the equivalence point, the pH is determined by the excess H+ present. As can be seen from Figures 15.1 and 15.2, both strong by strong titrations have pH = 7.00 at the equivalence point, but the curves are the reverse of each other before and after the equivalence point.
7.
In Section 15.4 of the text, the case study for the weak acid-strong base titration is the titration of 50.0 mL of 0.10 M HC2H3O2 by 0.10 M NaOH. See Figure 15.3 for the titration curve. As soon as some NaOH has been added to the weak acid, OH − reacts with the best acid present. This is the weak acid titrated (HC2H3O2 in our problem). The reaction is: OH− + HC2H3O2 → H2O + C2H3O2−. Because something strong is reacting, we assume the reaction goes to completion. This is the stoichiometry part of a titration problem. To solve for the pH, we see what is in solution after the stoichiometry problem and decide how to proceed. The various parts to the titration are: a. Initially, before any OH− has been added. The major species present is the weak acid, HC2H3O2, and water. We would use the Ka reaction for the weak acid and solve the equilibrium problem to determine the pH.
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b. Just past the start of the titration up to just before the equivalence point (0.1 mL to 49.9 mL NaOH added). In this region, the major species present after the OH − reacts to completion are HC2H3O2, C2H3O2−, Na+, and water. We have a buffer solution because both a weak acid and a conjugate base are present. We can solve the equilibrium buffer problem using the Ka reaction for HC2H3O2, the Kb reaction for C2H3O2−, or the Henderson-Hasselbalch equation. A special point in the buffer region is the halfway point to equivalence. At this point (25.0 mL of NaOH added), exactly one-half of the weak acid has been converted into its conjugate base. At this point, we have [weak acid] = [conjugate base] so that pH = pKa. For the HC2H3O2 titration, the pH at 25.0 mL NaOH added is –log (1.8 × 10 −5 ) = 4.74; in this titration the pH is acidic at the halfway point to equivalence. However, other weak acid-strong base titrations could have basic pH values at the halfway point. This just indicates that the weak acid has Ka < 1 × 10 −7 , which is fine. c. The equivalence point (50.0 mL NaOH added). Here we have added just enough OH − to convert all of the weak acid into its conjugate base. In our example, the major species present are C2H3O2−, Na+, and H2O. Because the conjugate base of a weak acid is a weak base, we will have a basic pH (pH > 7.0) at the equivalence point. To calculate the pH, we write out the Kb reaction for the conjugate base and then set-up and solve the equilibrium problem. For our example, we would write out the Kb reaction for C2H3O2−. d. Past the equivalence point (V > 50.0 mL). Here we added an excess of OH −. After the stoichiometry part of the problem, the major species are OH−, C2H3O2−, H2O, and Na+. We have two bases present, the excess OH − and the weak conjugate base. The excess OH − dominates the solution and thus determines the pH. We can ignore the OH − contribution from the weak conjugate base. See the titration curve before Figure 15.3 that compares strong acid-strong base titrations to weak acid-strong base titration. The two curves have the same pH only after the equivalence point where the excess strong base added determines the pH. The strong acid titration is acidic at every point before the equivalence point, has a pH = 7.0 at the equivalence point, and is basic at every point after the equivalence point. The weak acid titration is much more complicated because we cannot ignore the basic properties of the conjugate base of the weak acid; we could ignore the conjugate base in the strong acid titration because it had no basic properties. In the weak acid titration, we start off acidic (pH < 7.0), but where it goes from here depends on the strength of the weak acid titrated. At the halfway point where pH = pKa, the pH may be acidic or basic depending on the Ka value of the weak acid. At the equivalence, the pH must be basic. This is due to the presence of the weak conjugate base. Past the equivalence point, the strong acid and weak acid titrations both have their pH determined by the excess OH − added. 8.
The case study of a weak base-strong acid titration in Section 15.4 of the text is the titration of 100.0 mL of 0.050 M NH3 by 0.10 M HCl. The titration curve is in Figure 15.5. As HCl is added, the H+ from the strong acid reacts with the best base present, NH 3. Because something strong is reacted, we assume the reaction goes to completion. The reaction used for the stoichiometry part of the problem is: H+ + NH3 → NH4+. Note that the effect of this reaction is to convert the weak base into its conjugate acid. The various parts to a weak base-strong acid titration are:
CHAPTER 15
ACID-BASE EQUILIBRIA
719
a. Initially before any strong acid is added. We have a weak base in water. Write out the K b reaction for the weak base, set-up the ICE table, and then solve. b. From 0.1 mL HCl added to just before the equivalence point (49.9 mL HCl added). The major species present in this region are NH3, NH4+, Cl−, and water. We have a weak base and its conjugate acid present at the same time; we have a buffer. Solve the buffer problem using the Ka reaction for NH4+, the Kb reaction for NH3, or the Henderson-Hasselbalch equation. The special point in the buffer region of the titration is the halfway point to equivalence. Here, [NH3] = [NH4+], so pH = pKa = −log (5.6 × 10 −10 ) = 9.25. For this titration, the pH is basic at the halfway point. However, if the Ka for the weak acid component of the buffer has a Ka > 1 × 10 −7 (Kb for the base < 1 × 10−7), then the pH will be acidic at the halfway point. This is fine. In this review question, it is asked what the Kb value for the weak base is where the halfway point to equivalence has pH = 6.0 (which is acidic). Because pH = pKa at this point, pKa = 6.0, so pKb = 14.00 – 6.0 = 8.0; Kb = 10 −8.0 = 1.0 × 10 −8 . c. The equivalence point (50.0 mL HCl added). Here, just enough H + has been added to convert all the NH3 into NH4+. The major species present are NH4+, Cl− and H2O. The only important species present is NH4+, a weak acid. This is always the case in a weak basestrong acid titration. Because a weak acid is always present at the equivalence point, the pH is always acidic (pH < 7.0). To solve for the pH, write down the Ka reaction for the conjugate acid and then determine Ka (= Kw/Kb). Fill in the ICE table for the problem, and then solve to determine pH. d. Past the equivalence point (V > 50.0 mL HCl added). The excess strong acid added determines the pH. The major species present for our example would be H + (excess), NH4+, Cl−, and H2O. We have two acids present, but NH4+ is a weak acid. Its H+ contribution will be negligible compared to the excess H+ added from the strong acid. The pH is determined by the molarity of the excess H+. Examine Figures 15.2 and 15.5 to compare a strong base-strong acid titration with a weak basestrong acid titration. The points in common are only after the equivalence point has been reached where excess strong acid determines the pH. Leading up to and including the equivalence point, the two titrations are very different. This is because the conjugate acid of a weak base is a weak acid, whereas, when a strong base dissolves in water, only OH − is important; the cation in the strong base is garbage (has no acidic/basic properties). There is no conjugate acid to worry about when a strong base is titrated. This is not the case when a weak base is titrated. For a strong base-strong acid titration, the excess OH− from the strong base determines the pH from the initial point all the way up to the equivalence point. At the equivalence point, the added H+ has reacted with all of the OH− and pH = 7.0. For a weak base-strong acid, we initially have a weak base problem to solve in order to calculate the pH. After H + has been added, some of the weak base is converted into its conjugate acid and we have buffer solutions to solve to calculate the pH. At the equivalence point, we have a weak acid problem to solve. Here, all the weak base has been converted into its conjugate acid by the added H +.
720 9.
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An acid-base indicator marks the end point of a titration by changing color. Acid-base indicators are weak acids themselves. We abbreviate the acid form of an indicator as HIn and the conjugate base form as In −. The reason there is a color change with indicators is that the HIn form has one color associated with it while the In − form has a different color associated with it. Which form dominates in solution and dictates the color is determined by the pH of the solution. The related equilibrium is: HIn ⇌ H+ + In−. In a very acidic solution, there are lots of H+ ions present which drives the indicator equilibrium to the left. The HIn form dominates and the color of the solution is the color due to the HIn form. In a very basic solution, H + has been removed from solution. This drives the indicator equilibrium to the right and the In − form dominates. In very basic solutions, the solution takes on the color of the In − form. In between very acidic and very basic solutions, there is a range of pH values where the solution has significant amounts of both the HIn and In − forms present. This is where the color change occurs, and we want this pH to be close to the stoichiometric point of the titration. The pH that the color change occurs is determined by the Ka of the indicator. Equivalence point: When enough titrant has been added to react exactly with the substance in the solution being titrated. Endpoint: indicator changes color. We want the indicator to tell us when we have reached the equivalence point. We can detect the end point visually and assume it is the equivalence point for doing stoichiometric calculations. They don’t have to be as close as 0.01 pH units since, at the equivalence point, the pH is changing very rapidly with added titrant. The range over which an indicator changes color only needs to be close to the pH of the equivalence point.
10.
The two forms of an indicator are different colors. The HIn form has one color and the In− form has another color. To see only one color, that form must be in an approximately ten-fold excess or greater over the other form. When the ratio of the two forms is less than 10, both colors are present. To go from [HIn]/[In −] = 10 to [HIn]/[In−] = 0.1 requires a change of 2 pH units (a 100-fold decrease in [H+]) as the indicator changes from the HIn color to the In − color. From Figure 15.8 of the text, thymol blue has three colors associated with it: orange, yellow, and blue. For this to happen, thymol blue must be a diprotic acid. The H2In form has the orange color, the HIn− form has the yellow color, and the In 2− form has the blue color associated with it. Thymol blue cannot be monoprotic; monoprotic indicators only have two colors associated with them (either the HIn color or the In − color).
Active Learning Questions 1.
NaHSO4 → Na+ + HSO4−: Na+ has no acidic or basic properties while HSO4− is a weak acid with Ka = 0.012. As is true with any weak acid, the larger the concentration, the more H + produced and the lower the pH. Note that HSO 4− is also the conjugate base of the strong acid H2SO4. As is true for all conjugate bases of weak acids, HSO 4− is a terrible base. NaHCO3 → Na+ + HCO3−: Na+ has no acidic or basic properties. HCO3− is amphoteric; it is a weak acid with Ka = 5.6 10 −11, but HCO3− is also a weak base since it is the conjugate base of the weak acid H2CO3. Adding more NaHCO3 increases both the amount of weak acid and weak base present. This is like a buffer solution where both the weak acid and the conjugate
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base concentrations are increased the same amount. The pH of the solution does not change as more NaHCO3 is added. 2.
A buffer solution starts with both the weak acid and the conjugate base present in relatively large quantities. It is true that added strong base converts some of the weak acid into its conjugate base. However, the key is that the concentration ratio of weak acid to conjugate base does not change that much after the strong base has reacted. As long as we started with large concentrations of both the weak acid and conjugate base, small quantities of added strong base or strong acid doesn’t change the weak acid to conjugate base concentration ratio very much and the pH of the solution doesn’t change much.
3
HC2H3O2 + OH− → C2H3O2− + H2O; added OH− from NaOH converts HC2H3O2 into its conjugate base C2H3O2−. Solutions that contain both a weak acid and a conjugate base at the same time are buffers. So as we add NaOH, we make a buffer solution. The best buffer is when we have added just enough OH− to convert one-half of the HC2H3O2 present initially into its conjugate base C2H3O2−. Here pH = pKa for acetic acid since we have equal moles of HC2H3O2 and C2H3O2− present at the same time. The best buffers will be when we have large quantities of both HC2H3O2 and C2H3O2−. When we add just a little bit of NaOH, we don’t have a lot of C2H3O2− yet so the resulting solution is not a great buffer. When we get close to adding equal moles of HC2H3O2 and NaOH, we have mostly C2H3O2− in solution and very little HC2H3O2. This will also not make a very good buffer solution.
4.
HCl + NaOH → H2O + NaCl; when a strong acid and a strong base are added together, we don’t produce anything that has acidic or basic properties. This is because the conjugates of species that are strong generally have no acidic or basic properties. In this reaction, we produce water with Na+ and Cl− ions floating around. Neither Na+ nor Cl− have any acidic or basic properties. This is not the case when NaOH is added to a weak acid or when HCl is added to a weak base. In each of these cases, the conjugate that is produced (conjugate base of the weak acid or conjugate acid of the weak base) has acidic or basic properties and a buffer solution can form. Solutions containing a strong acid and its conjugate base are not considered buffers. This again is because the conjugate base of a strong acid is a terrible base. We need both an acid and a base present at the same time to make a buffer. Not just one or the other, but both must be present at the same time.
5.
See the titration curve before Figure 14.3 that compares strong acid-strong base titrations to weak acid-strong base titration. The two curves have the same pH only after the equivalence point where the excess strong base added determines the pH. The strong acid titration is acidic at every point before the equivalence point, has a pH = 7.0 at the equivalence point, and is basic at every point after the equivalence point. The weak acid titration is much more complicated because we cannot ignore the basic properties of the conjugate base of the weak acid; we could ignore the conjugate base in the strong acid titration because it had no basic properties. In the weak acid titration, we start off acidic (pH < 7.0), but where it goes from here depends on the strength of the weak acid titrated. At the halfway point where pH = pK a, the pH may be acidic or basic depending on the Ka value of the weak acid. At the equivalence, the pH must be basic. This is due to the presence of the weak conjugate base. Past the equivalence point, the strong acid and weak acid titrations both have their pH determined by the excess OH− added.
722 6.
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In Section 14.4 of the text, the case study for the weak acid-strong base titration is the titration of 50.0 mL of 0.10 M HC2H3O2 by 0.10 M NaOH. See Figure 14.3 for the titration curve. As soon as some NaOH has been added to the weak acid, OH− reacts with the best acid present. This is the weak acid titrated (HC2H3O2 in our problem). The reaction is: OH− + HC2H3O2 → H2O + C2H3O2−. Because something strong is reacting, we assume the reaction goes to completion. This is the stoichiometry part of a titration problem. To solve for the pH, we see what is in solution after the stoichiometry problem and decide how to proceed. The various parts to the titration are: a. Initially, before any OH− has been added. The major species present is the weak acid, HC2H3O2, and water. We would use the Ka reaction for the weak acid and solve the equilibrium problem to determine the pH. b. Just past the start of the titration up to just before the equivalence point (0.1 mL to 49.9 mL NaOH added). In this region, the major species present after the OH − reacts to completion are HC2H3O2, C2H3O2−, Na+, and water. We have a buffer solution because both a weak acid and a conjugate base are present. We can solve the equilibrium buffer problem using the Ka reaction for HC2H3O2, the Kb reaction for C2H3O2−, or the Henderson-Hasselbalch equation. A special point in the buffer region is the halfway point to equivalence. At this point (25.0 mL of NaOH added), exactly one-half of the weak acid has been converted into its conjugate base. At this point, we have [weak acid] = [conjugate base] so that pH = pKa. For the HC2H3O2 titration, the pH at 25.0 mL NaOH added is –log (1.8 × 10 −5 ) = 4.74; in this titration the pH is acidic at the halfway point to equivalence. However, other weak acid-strong base titrations could have basic pH values at the halfway point. This just indicates that the weak acid has Ka < 1 × 10 −7 , which is fine. c. The equivalence point (50.0 mL NaOH added). Here we have added just enough OH − to convert all of the weak acid into its conjugate base. In our example, the major species present are C2H3O2−, Na+, and H2O. Because the conjugate base of a weak acid is a weak base, we will have a basic pH (pH > 7.0) at the equivalence point. To calculate the pH, we write out the Kb reaction for the conjugate base and then set-up and solve the equilibrium problem. For our example, we would write out the Kb reaction for C2H3O2−. d. Past the equivalence point (V > 50.0 mL). Here we added an excess of OH −. After the stoichiometry part of the problem, the major species are OH −, C2H3O2−, H2O, and Na+. We have two bases present, the excess OH − and the weak conjugate base. The excess OH − dominates the solution and thus determines the pH. We can ignore the OH − contribution from the weak conjugate base.
7.
Major species: HA, H+, Cl−, and H2O; since HA is a weak acid, not a lot of A − will be present, hence why A− is not listed as a major species. We need to know the initial concentration of HA, the amount of H+ added from the strong acid, and we need to know the Ka for the weak acid. To solve, set-up the ICE table using the Ka reaction for HA and solve. Since we initially have some H+ from the strong acid present, this will suppress the amount of HA that reacts to reach equilibrium. The equilibrium concentration of H + will be overall higher than when just a weak acid is present, but the equilibrium A− concentration will be lower. Since the equilibrium
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H+ concentration is higher, the pH of the solution will be lower than when just the weak acid is present. 8.
Major species: HA, Na+, A−, and H2O; we need to know the initial concentration of HA, the amount of A− added from NaA, and we need to know the Ka for the weak acid. To solve, setup the ICE table using the Ka reaction for HA and solve. Since we initially have some A − from the NaA present, this will suppress the amount of HA that reacts to reach equilibrium. The equilibrium concentration of H+ will be overall lower than when just a weak acid is present. Since the equilibrium H+ concentration is lower, the pH of the solution will be higher than when just the weak acid is present.
9.
A best buffer has about equal concentrations of weak acid and its conjugate base or about equal concentrations of a weak base and its conjugate acid. When this occurs, pH ≈ pK a. For a best pH = 7.0 buffer, look for a weak acid/conjugate base or weak base/conjugate acid combination that has a pKa value close to 7.0. For a pKa of 7.0, we want Ka ≈ 1 × 10‒7. From A5.1, the acid with a Ka value closest to 1 × 10‒7 is HOCl. So, a HOCl + KOCl solution would be a good choice for a pH = 7.0 buffer. From Table A5.2, H2S has a Ka value equal to 1 × 10‒7, making a H2S + KHS solution a good choice for a pH = 7.0 buffer. Table A5.3 contains weak bases. For a pH = 7.0 buffer, we want the conjugate acid of the weak base to have a K a value close to 1 × 10‒7. The best choice would be a solution containing HONH2 + HONH3Cl. The Ka value for HONH3+ is 1.0 × 10‒14/1.1 × 10‒8 = 9.1 × 10‒7.
10.
See Figure 15.12 for a diprotic acid titration curve and Figure 15.11 for a triprotic acid titration curve. The diprotic acid curve has two titration curves in one. Let’s assume it takes 100.0 mL to reach the first equivalence point. Initially we start off with a beaker of H 2A. As a strong base like KOH is added, it converts the H2A into its conjugate base HA‒. From 1‒99 mL of KOH added, we have a solution containing both H 2A and HA‒. At 50.0 mL of KOH added, we have reached the first halfway point to equivalence where [H 2A] and [HA‒] and pH = pK a 1 . At 100.0 mL of KOH added, we have converted all the H2A into HA‒. The major species present at 100.0 mL KOH added is HA‒. From 101-200.0 mL of KOH added, HA‒ is converted into its conjugate base A2‒. From 101-199 mL of KOH added, we have a solution containing both HA‒ and A2‒. At 150.0 mL of KOH added, we have reached the second halfway point to equivalence where [HA‒] and [A2‒] and pH = pK a 2 . At 200.0 mL of KOH added, we have converted all the HA‒ into A2‒. The major species present at 200.0 mL KOH added is A 2‒. Past the second equivalence point, the major species present are excess OH ‒ from the KOH and A2‒. The triprotic acid curve has three titration curves in one. Let’s assume it takes 50.0 mL to reach the first equivalence point. Initially we start off with a beaker of H 3A. As a strong base like KOH is added, it converts the H3A into its conjugate base H2A‒. From 1-49 mL of KOH added, we have a solution containing both H3A and H2A‒. At 25.0 mL of KOH added, we have reached the first halfway point to equivalence where [H 3A] = [H2A‒] and pH = pK a 1 . At 50.0 mL of KOH added, we have converted all the H3A into H2A‒. The major species present at 50.0 mL KOH added is H2A‒. From 50.1-100.0 mL of KOH added, H2A‒ is converted into its conjugate base HA2‒. From 51-99 mL of KOH added, we have a solution containing H 2A‒ and HA2‒. At 75.0 mL of KOH added, we have reached the second halfway point to equivalence where [H2A‒] = [HA2‒] and pH = pK a 2 . At 100.0 mL of KOH added, we have converted all the H2A‒ into HA2‒. The major species present at 100.0 mL KOH added is HA2‒. From 100.1-150.0 mL of KOH added, HA2‒ is converted into its conjugate base A3‒. From 101-149 mL of KOH added,
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we have a solution containing HA2‒ and A3‒. At 125.0 mL of KOH added, we have reached the third halfway point to equivalence where [HA 2‒] = [A3‒] and pH = pK a 3 . At 150.0 mL of KOH added, we have converted all the HA2‒ into A3‒. The major species present at 150.0 mL KOH added is A3‒. Past the third equivalence point, the major species present are excess OH ‒ from the KOH and A3‒. 11.
The neutralization reactions are: HCO3‒(aq) + H+(aq) → H2CO3(aq), H2CO3(aq) + OH‒(aq) → HCO3‒(aq) + H2O(l) Added strong acid or added strong base coverts one of the buffer components into the other component. Acidosis occurs when excess acid is present resulting in an increase in H2CO3(aq) and a decrease of HCO3‒(aq). The ratio of H2CO3: HCO3− increases with the onset of acidosis. Alkalosis occurs when excess base is present resulting in an increase in HCO3‒(aq) and a decrease of H2CO3(aq). The ratio of H2CO3: HCO3− decreases with the onset of alkalosis. Breathing regulates the amount of CO2 in the equilibrium: CO2(g) + H2O(l) ⇌ H2CO3(aq) ⇌ HCO3−(aq) + H+(aq) Increased breathing removes CO2 from the body. As CO2 is removed, the above equilibrium will shift left which decreases the H+ concentration in blood. So, when the blood becomes too acidic, the body will increase breathing. The opposite occurs when the blood becomes too basic. The body decreases breathing to increase the CO 2 concentration in the body. This shifts the equilibrium to the right, producing H+. Removal of NaHCO3 (HCO3−) causes the buffer equilibrium to shift right to produce more HCO3− along with more H+. This causes the blood to become more acidic, so blood pH decreases. Treatment involves the intravenous addition of HCO 3−.
12.
Titration i is a strong acid titrated by a strong base. The pH is very acidic until just before the equivalence point; at the equivalence point, pH = 7.00, and past the equivalence the pH is very basic. Titration ii is a strong base titrated by a strong acid. Here, the pH is very basic until just before the equivalence point; at the equivalence point, pH = 7.00, and past the equivalence point the pH is very acidic. Titration iii is a weak base titrated by a strong acid. The pH starts out basic because a weak base is present. However, the pH will not be as basic as in titration ii, where a strong base is titrated. The pH drops as HCl is added; then at the halfway point to equivalence, pH = pKa. Because Kb = 4.4 × 10 −4 for CH3NH2, CH3NH3+ has Ka = Kw/Kb = 2.3 × 10 −11 and pKa = 10.64. So, at the halfway point to equivalence for this weak base-strong acid titration, pH = 10.64. The pH continues to drop as HCl is added; then at the equivalence point the pH is acidic (pH < 7.00) because the only important major species present is a weak acid (the conjugate acid of the weak base). Past the equivalence point the pH becomes more acidic as excess HCl is added. Titration iv is a weak acid titrated by a strong base. The pH starts off acidic, but not nearly as acidic as the strong acid titration (i). The pH increases as NaOH is added; then at the halfway point to equivalence, pH = pKa for HF = −log(7.2 × 10 −4 ) = 3.14. The pH continues to increase past the halfway point; then at the equivalence point the pH is basic (pH > 7.0) because the only important major species present is a weak base (the conjugate
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base of the weak acid). Past the equivalence point the pH becomes more basic as excess NaOH is added. a. All require the same volume of titrant to reach the equivalence point. At the equivalence point for all these titrations, moles acid = moles base (MAVA = MBVB). Because all the molarities and volumes are the same in the titrations, the volume of titrant will be the same (50.0 mL titrant added to reach equivalence point). b. Increasing initial pH: i < iv < iii < ii; the strong acid titration has the lowest pH, the weak acid titration is next, followed by the weak base titration, with the strong base titration having the highest pH. c. i < iv < iii < ii; the strong acid titration has the lowest pH at the halfway point to equivalence, and the strong base titration has the highest halfway point pH. For the weak acid titration, pH = pKa = 3.14, and for the weak base titration, pH = pKa = 10.64. d. Equivalence point pH: iii < ii = i < iv; the strong-by-strong titrations have pH = 7.00 at the equivalence point. The weak base titration has an acidic pH at the equivalence point, and a weak acid titration has a basic equivalence point pH. The only different answer when the weak acid and weak base are changed would be for part c. This is for the halfway point to equivalence, where pH = pK a. HOC6H5; Ka = 1.6 × 10 −10 , pKa = −log(1.6 × 10 −10 ) = 9.80 C5H5NH+, Ka =
Kw K b , C5 H 5 N
=
1.0 10 −14 = 5.9 × 10 −6 , pKa = 5.23 1.7 10 −9
From the pKa values, the correct ordering at the halfway point to equivalence would be i < iii < iv < ii. Note that for the weak base-strong acid titration using C5H5N, the pH is acidic at the halfway point to equivalence, whereas the weak acid-strong base titration using HOC6H5 is basic at the halfway point to equivalence. This is fine; this will always happen when the weak base titrated has a Kb < 1 × 10 −7 (so Ka of the conjugate acid is greater than 1 × 10−7) or when the weak acid titrated has a Ka < 1 × 10 −7 (so Kb of the conjugate base is greater than 1 × 10−7).
Questions 13.
When an acid dissociates, ions are produced. The common ion effect is observed when one of the product ions in a particular equilibrium is added from an outside source. For a weak acid dissociating to its conjugate base and H +, the common ion would be the conjugate base; this would be added by dissolving a soluble salt of the conjugate base into the acid solution. The presence of the conjugate base from an outside source shifts the equilibrium to the left so less acid dissociates.
14.
pH = pKa + log
[ base] [base] [base] < 0. ; when [acid] > [base], then < 1 and log [acid] [acid] [acid ]
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From the Henderson-Hasselbalch equation, if the log term is negative, then pH < pK a. When one has more acid than base in a buffer, the pH will be on the acidic side of the pKa value; that is, the pH is at a value lower than the pKa value. When one has more base than acid in a buffer ([conjugate base] > [weak acid]), then the log term in the Henderson-Hasselbalch equation is positive, resulting in pH > pKa. When one has more base than acid in a buffer, the pH is on the basic side of the pKa value; that is, the pH is at a value greater than the pK a value. The other scenario you can run across in a buffer is when [acid] = [base]. Here, the log term is equal to zero, and pH = pKa. 15.
The more weak acid and conjugate base present, the more H + and/or OH− that can be absorbed by the buffer without significant pH change. When the concentrations of weak acid and conjugate base are equal (so that pH = pKa), the buffer system is equally efficient at absorbing either H+ or OH−. If the buffer is overloaded with weak acid or with conjugate base, then the buffer is not equally efficient at absorbing either H + or OH−.
16.
A buffer consists of a weak acid and its weak conjugate base present at the same time or a weak base and its weak conjugate acid present at the same time. To solve for the pH, one can set-up the ICE table using either the Ka reaction for the acid component of the buffer or using the Kb reaction for the base component of the buffer. Both the weak acid and the conjugate base or weak base and conjugate acid are present, so both the K a and Kb equilibriums must hold.
17.
If a strong acid and/or strong base is/are present, react them first, react the best base with the best acid present, and assume the reaction goes to completion. After the strong acid and/or strong base has reacted to completion, take stock of what remains in solution to determine what to do next. Strong acids and strong bases are great at what they do, either donating a proton or accepting a proton. So always react the strong acid and/or strong base first.
18.
For a best buffer, we want about equal concentrations of weak acid and conjugate base. When this occurs in a buffer, pH ≈ pKa. Here we want a pH = 7.0, so we want a Ka value for the acid component of the buffer to be close to 1 × 10‒7. This makes H2PO4‒ having a Ka = 6.2 × 10−8 the best choice for the acid component of the buffer. To complete the buffer, we need the conjugate base of H2PO4‒ to also be present, which is HPO 42‒. So, the best pH = 7.0 buffer using phosphoric acid components would consist of KH 2PO4 and K2HPO4. Note that sodium salts of these ions would be fine also.
19.
Only statement e is false. At the halfway point to equivalence for a weak base-strong acid titration, a buffer solution consisting of equal amounts of the weak base and its conjugate acid are present. At the halfway point, pH = pKa for the conjugate acid. If the Ka value for the conjugate acid is greater than 1 × 10‒7, then the pH will be acidic at the halfway point to equivalence. Weak bases with Kb values less than 1 × 10‒7 have conjugate acids with Ka values greater than 1 × 10‒7; these weak bases have an acidic pH at the halfway point (Kw = Ka × Kb for conjugate acid-base pairs).
20.
a. The red plot is the pH curve for the strong acid and the blue plot is the pH curve for the weak acid. The pH at the equivalence point is 7.00 for the strong acid-strong base titration, while the pH is greater than 7.00 if a weak acid is titrated. Another point one could look at is the initial point. Because both acids have the same concentration, the strong acid curve will be at the lowest initial pH. Any point at any volume up to the equivalence point for the strong acid plot will have a lower pH than the weak acid plot (assuming equal
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concentrations and volumes). Another difference would be the pH at the halfway point to equivalence. For the weak acid titration, the pH of solution equals the pKa value for the weak acid at the halfway point to equivalence; this is not the case when a strong acid is titrated. b. A buffer is a solution that resists pH change. From this definition, both titrations have regions where the pH doesn’t change much on addition of strong base, so both could be labeled to have buffer regions. However, we don’t normally include strong acids as a component of buffer solutions. Strong acids certainly can absorb added OH − by reacting with it to form water. But when more strong acid is added, the H + concentration increases steadily; there is nothing present in a strong acid solution to react with added H+. This is not the case in the weak acid-strong base titration. After some OH− has been added, some weak acid is converted into its conjugate base. We now have a typical buffer solution because there are significant amounts of weak acid and conjugate base present at the same time. The buffer region extends from a little past the initial point in the titration up to just a little before the equivalence point. This entire region is a buffer region because both the weak acid and conjugate base are present in significant quantities in this region. c. True; HA + OH− → A− + H2O; both reactions have the same neutralization reaction. In both cases, the equivalence point is reached when enough OH − has been added to exactly react with the acid present initially. Because all acid concentrations and volumes are the same, we have equal moles of each acid which requires the same moles of OH− to reach the equivalence point. Therefore, each acid requires the same volume of 0.10 M NaOH to reach the equivalence point. d. False; the pH for the strong acid-strong base titration will be 7.00 at the equivalence point. The pH for the weak acid-strong base titration will be greater than 7.00 at the equivalence point. In both titrations, the major species present at the equivalence points are Na+, H2O, and the conjugate base of the acid titrated. Because the conjugate base of a strong acid has no basic characteristics, pH = 7.00 at the equivalence point. However, the conjugate base of a weak acid is a weak base. A weak base is present at the equivalence point of a weak acid-strong base titration, so the pH is basic (pH > 7.0). 21.
a. Let’s call the acid HA, which is a weak acid. When HA is present in the beakers, it exists in the undissociated form, making it a weak acid. A strong acid would exist as separate H+ and A− ions. b. Beaker a contains 4 HA molecules and 2 A− ions, beaker b contains 6 A− ions, beaker c contains 6 HA molecules, beaker d contains 6 A− and 6 OH− ions, and beaker e contains 3 HA molecules and 3 A− ions. HA + OH− → A− + H2O; this is the neutralization reaction that occurs when OH− is added. We start off the titration with a beaker full of weak acid (beaker c). When some OH− is added, we convert some weak acid HA into its conjugate base A− (beaker a). At the halfway point to equivalence, we have converted exactly onehalf of the initial amount of acid present into its conjugate base (beaker e). We finally reach the equivalence point when we have added just enough OH− to convert all of the acid present initially into its conjugate base (beaker b). Past the equivalence point, we have added an excess of OH−, so we have excess OH− present as well as the conjugate base of the acid produced from the neutralization reaction (beaker d). The order of the beakers from start to finish is:
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beaker c → beaker a → beaker e → beaker b → beaker d c. pH = pKa when a buffer solution is present that has equal concentrations of the weak acid and conjugate base. This is beaker e. d. The equivalence point is when just enough OH − has been added to exactly react with all the acid present initially. This is beaker b. e. Past the equivalence point, the pH is dictated by the concentration of excess OH − added from the strong base. We can ignore the amount of hydroxide added by the weak conjugate base that is also present. This is beaker d. 22.
Titration i is a strong acid-strong base titration. The pH is very acidic initially and very acidic up to just before the equivalence point. At the equivalence point, pH = 7.0. and past the equivalence point the pH is very basic (see Figure 14.1 for a typical strong acid-strong base titration curve). Titration ii is a strong base-strong acid titration. The pH is very basic initially and very basic up to just before the equivalence point. At the equivalence point, pH = 7.0. and past the equivalence point the pH is very acidic (see Figure 14.2 for a typical strong acid-strong base titration curve). Titration iii is a weak acid-strong base titration. The pH is acidic initially. Past the initial point and up to the equivalence point, buffer solutions exist. At the half-way to equivalence, pH = pKa of the weak acid titrated. At the equivalence point, pH is basic because the major species present is the conjugate base of the weak acid titrated. Past the equivalence point the pH is very basic (see Figure 14.3 for an example of a weak acid-strong base titration curve). Titration iv is a weak base-strong acid titration. The pH is basic initially. Past the initial point and up to the equivalence point, buffer solutions exist. At the half-way to equivalence, pH = pKa of the conjugate acid of the weak base titrated. At the equivalence point, pH is acidic because the major species present is the conjugate acid of the weak base titrated. Past the equivalence point the pH is very acidic (see Figure 14.5 for an example of a weak base-strong acid titration curve). a. Titrations i and iii are acidic initially, and titrations ii and iv are basic initially. b. Titration i is very acidic at the halfway point, and titration ii is very basic at the halfway point. The other two titrations have buffer solutions present at the halfway point where pH = pKa. For titration iii, pH = pKa = ‒log(3.5 × 10‒8) = 7.46; the pH is basic for titration iii at the halfway point. For titration iv, pH = pK a = ‒log(1.0 × 10‒14/1.1 × 10‒8) = 6.04; the pH is acidic at the halfway point. c. Titrations i and ii are neutral at the equivalence point (pH = 7.00). Titration iii is basic at the equivalence point and titration iv is acidic at the equivalence point.
23.
The three key points to emphasize in your sketch are the initial pH, the pH at the halfway point to equivalence, and the pH at the equivalence point. For all the weak bases titrated, pH = pKa at the halfway point to equivalence (50.0 mL HCl added) because [weak base] = [conjugate acid] at this point. Here, the weak base with Kb = 1 10 −5 has a conjugate acid with Ka = 1 10−9, so pH = 9.0 at the halfway point. The weak base with Kb = 1 10 −10 has a pH = 4.0 at the halfway point to equivalence. For the initial pH, the strong base has the highest pH (most basic), whereas the weakest base has the lowest pH (least basic). At the equivalence point
CHAPTER 15
ACID-BASE EQUILIBRIA
729
(100.0 mL HCl added), the strong base titration has pH = 7.0. The weak bases titrated have acidic pH’s because the conjugate acids of the weak bases titrated are the major species present. The weakest base has the strongest conjugate acid so its pH will be lowest (most acidic) at the equivalence point.
strong base
Kb = 10-5
pH 7.0
Kb = 10-10
Volume HCl added (mL) 24.
HIn ⇌ H+ + In−
Ka =
[H + ][In − ] [HIn]
Indicators are weak acids themselves. The special property they have is that the acid form of the indicator (HIn) has one distinct color, whereas the conjugate base form (In−) has a different distinct color. Which form dominates and thus determines the color of the solution is determined by the pH. An indicator is chosen in order to match the pH of the color change at about the pH of the equivalence point. 25.
H3AsO4 is a triprotic acid, so it has three equivalence points. The initial titration reaction is H3AsO4 + OH− → H2AsO4− + H2O. Because the concentration of the H3AsO4 and NaOH solutions are equal, it will take 100.0 mL of NaOH to reach the first equivalence point. The second equivalence point where H2AsO4− is titrated will occur at 200.0 mL of NaOH added, and the third equivalence point where HAsO 42− is titrated will occur at 300.0 mL of NaOH added. At 50.0 mL of NaOH added, this is the first halfway point to equivalence where H 3AsO4 and H2AsO4− are the major species present (along with H2O). At this point, we have a buffer solution where [H3AsO4] = [H2AsO4−] and pH = pKa1 = −log(5.5 10−3) = 2.26. At 150.0 mL of NaOH added, this is the second halfway point to equivalence where H 2AsO4− and HAsO42− are the major species present (along with H2O). At this point, we have a buffer solution where [H2AsO4−] = [HAsO42−] and pH = pKa2 = −log(1.7 10−7) = 6.77. At the third halfway point to equivalence, [HAsO42−] = [AsO43−] and pH = pKa3. This third halfway point will be at 250.0 mL of NaOH added.
730 26.
CHAPTER 15
ACID-BASE EQUILIBRIA
Salicylic acid and adipic acid are both diprotic acids. The titration curve for a diprotic acid should look like Figure 15.12 of the text where two equivalence points are shown. But there are unique issues with these two diprotic acids. For salicylic acid, we will see only the first stoichiometric point in the titration. This is because K a 2 is so small; so small that salicylic acid behaves as a monoprotic acid. For adipic acid, the Ka values are fairly close to each other. Both protons will be titrated almost simultaneously, giving us only one break. The stoichiometric points will occur when 1 mol of OH− is added per mole of salicylic acid present and when 2 mol of OH− is added per mole of adipic acid present. Thus the 25.00-mL volume corresponded to the titration of salicylic acid, and the 50.00-mL volume corresponded to the titration of adipic acid.
Exercises Buffers 27.
Only the third (lower) beaker represents a buffer solution. A weak acid and its conjugate base must both be present in large quantities to have a buffer solution. This is only the case in the third beaker. The first beaker represents a beaker full of strong acid which is 100% dissociated. The second beaker represents a weak acid solution. In a weak acid solution, only a small fraction of the acid is dissociated. In this representation, 1/10 of the weak acid has dissociated. The only B− present in this beaker is from the dissociation of the weak acid. A buffer solution has B− added from another source.
28.
A buffer solution is a solution containing a weak acid plus its conjugate base or a weak base plus its conjugate acid. Solution c contains a weak acid (HOCl) plus its conjugate base (OCl−), so it is a buffer. Solution e is also a buffer solution. It contains a weak base (H 2NNH2) plus its conjugate acid (H2NNH3+). Solution a contains a strong acid (HBr) and a weak acid (HOBr). Solution b contains a strong acid (HClO4) and a strong base (RbOH). Solution d contains a strong base (KOH) and a weak base (HONH2).
29.
When strong acid or strong base is added to a bicarbonate-carbonate mixture, the strong acid(base) is neutralized. The reaction goes to completion, resulting in the strong acid(base) being replaced with a weak acid(base), resulting in a new buffer solution. The reactions are: H+(aq) + CO32−(aq) → HCO3−(aq); OH− + HCO3−(aq) → CO32−(aq) + H2O(l)
30.
Like the HCO3−/CO32− buffer discussed in Exercise 29, the HONH3+/HONH2 buffer absorbs added OH− and H+ in the same fashion. HONH2(aq) + H+(aq) → HONH3+(aq) HONH3+(aq) + OH−(aq) → HONH2(aq) + H2O(l)
31.
a. This is a weak acid problem. Let HC3H5O2 = HOPr and C3H5O2− = OPr−.
CHAPTER 15
ACID-BASE EQUILIBRIA HOPr(aq)
Initial
⇌
731
H+(aq)
+
OPr−(aq)
Ka = 1.3 × 10−5
0.100 M ~0 0 x mol/L HOPr dissociates to reach equilibrium −x → +x +x 0.100 − x x x
Change Equil.
Ka = 1.3 × 10−5 =
[H + ][OPr− ] x2 x2 = [HOPr] 0.100 − x 0.100
x = [H+] = 1.1 × 10−3 M; pH = 2.96; assumptions good by the 5% rule. b. This is a weak base problem. OPr−(aq) + H2O(l) ⇌ Initial Change Equil.
HOPr(aq) + OH−(aq) Kb =
Kw = 7.7 × 10−10 Ka
0.100 M 0 ~0 − x mol/L OPr reacts with H2O to reach equilibrium −x → +x +x 0.100 − x x x
Kb = 7.7 × 10−10 =
[HOPr][OH− ] [OPr− ]
=
x2 x2 0.100 − x 0.100
x = [OH−] = 8.8 × 10−6 M; pOH = 5.06; pH = 8.94; assumptions good. c. Pure H2O, [H+] = [OH−] = 1.0 × 10−7 M; pH = 7.00 d. This solution contains a weak acid and its conjugate base. This is a buffer solution. We will solve for the pH through the weak acid equilibrium reaction. HOPr(aq) Initial Change Equil. 1.3 × 10−5 =
⇌
H+(aq) +
OPr−(aq)
Ka = 1.3 × 10−5
0.100 M ~0 0.100 M x mol/L HOPr dissociates to reach equilibrium −x → +x +x 0.100 − x x 0.100 + x (0.100 + x)( x) (0.100 )( x) = x = [H+] 0.100 − x 0.100
[H+] = 1.3 × 10−5 M; pH = 4.89; assumptions good. Alternately, we can use the Henderson-Hasselbalch equation to calculate the pH of buffer solutions. pH = pKa + log
[base] 0.100 = pKa + log = pKa = −log(1.3 × 10−5) = 4.89 0 . 100 [acid]
The Henderson-Hasselbalch equation will be valid when an assumption of the type 0.1 + x 0.1 that we just made in this problem is valid. From a practical standpoint, this will almost always be true for useful buffer solutions. If the assumption is not valid, the solution will have such a low buffering capacity it will not be of any use to control the pH. Note: The Henderson-Hasselbalch equation can only be used to solve for the pH of buffer solutions.
732 32.
CHAPTER 15
ACID-BASE EQUILIBRIA
a. Weak base problem: HONH2 Initial Change Equil.
⇌
+ H2O
HONH3+
+
OH−
Kb = 1.1 × 10 −8
0.100 M 0 ~0 x mol/L HONH2 reacts with H2O to reach equilibrium –x → +x +x 0.100 – x x x
Kb = 1.1 × 10 −8 =
x2 x2 0.100 0.100 − x
x = [OH−] = 3.3 × 10 −5 M; pOH = 4.48; pH = 9.52; assumptions good. b. Weak acid problem (Cl− has no acidic/basic properties): HONH3+ Initial Change Equil. Ka =
⇌
HONH2
+
H+
0.100 M 0 ~0 x mol/L HONH3+ dissociates to reach equilibrium –x → +x +x 0.100 – x x x
Kw [ HONH 2 ][H + ] x2 x2 = 9.1 × 10 −7 = = + Kb 0.100 0.100 − x [HONH 3 ]
x = [H+] = 3.0 × 10 −4 M; pH = 3.52; assumptions good. c. Pure H2O, pH = 7.00 d. Buffer solution where pKa = –log(9.1 × 10 −7 ) = 6.04. Using the Henderson-Hasselbalch equation: pH = pKa + log
33.
(0.100 ) [HONH2 ] [base] = 6.04 + log = 6.04 + log = 6.04 + (0.100 ) [acid] [HONH3 ]
0.100 M HC3H5O2: % dissociation =
1.1 10 −3 M [H + ] × 100 = × 100 = 1.1% 0.100 M [HC3 H 5 O 2 ]0
0.100 M HC3H5O2 + 0.100 M NaC3H5O2: % dissociation =
1.3 10 −5 × 100 = 1.3 × 10 −2 % 0.100
The percent dissociation of the acid decreases from 1.1% to 1.3 × 10 −2 % (a factor of 85) when C3H5O2− is present. This is known as the common ion effect. The presence of the conjugate base of the weak acid inhibits the acid dissociation reaction. 34.
0.100 M HONH2: Percent ionization
3.3 10 −5 M [OH − ] × 100 = × 100 0.100 M [HONH2 ]0
= 3.3 × 10 −2 %
CHAPTER 15
ACID-BASE EQUILIBRIA
733
0.100 M HONH2 + 0.100 M HONH3+: % ionization =
1.1 10 −8 × 100 = 1.1 × 10 −5 % 0.100
The percent ionization decreases by a factor of 3000. The presence of the conjugate acid of the weak base inhibits the weak base reaction with water. This is known as the common ion effect. 35.
a. We have a weak acid (HOPr = HC3H5O2) and a strong acid (HCl) present. The amount of H+ donated by the weak acid will be negligible. To prove it, consider the weak acid equilibrium reaction:
⇌
HOPr Initial Change Equil.
H+
OPr−
+
Ka = 1.3 × 10−5
0.100 M 0.020 M 0 x mol/L HOPr dissociates to reach equilibrium −x → +x +x 0.100 − x 0.020 + x x
[H+] = 0.020 + x 0.020 M; pH = 1.70; assumption good (x = 6.5 × 10−5 is << 0.020). Note: The H+ contribution from the weak acid HOPr was negligible. The pH of the solution can be determined by only considering the amount of strong acid present. b. Added H+ reacts completely with the best base present, OPr−. OPr− Before Change After
+
0.100 M −0.020 0.080
→
H+
0.020 M −0.020 → 0
HOPr 0 +0.020 0.020 M
Reacts completely
After reaction, a weak acid, HOPr , and its conjugate base, OPr −, are present. This is a buffer solution. Using the Henderson-Hasselbalch equation where pKa = −log (1.3 × 10−5) = 4.89: pH = pKa + log
(0.080 ) [base] = 4.89 + log = 5.49; assumptions good. (0.020 ) [acid]
c. This is a strong acid problem. [H+] = 0.020 M; pH = 1.70 d. Added H+ reacts completely with the best base present, OPr−. OPr− Before Change After
+
0.100 M −0.020 0.080
H+
→
HOPr
0.020 M −0.020 0
→
0.100 M +0.020 Reacts completely 0.120
A buffer solution results (weak acid + conjugate base). Using the HendersonHasselbalch equation: pH = pKa + log
(0.080 ) [base] = 4.89 + log = 4.71; assumptions good. (0.120 ) [acid]
734 36.
CHAPTER 15
ACID-BASE EQUILIBRIA
a. Added H+ reacts completely with HONH2 (the best base present) to form HONH3+. HONH2 Before Change After
H+
+
0.100 M –0.020 0.080
0.020 M –0.020 0
→
HONH3+
→
0 +0.020 0.020
Reacts completely
After this reaction, a buffer solution exists; that is, a weak acid (HONH3+) and its conjugate base (HONH2) are present at the same time. Using the Henderson-Hasselbalch equation to solve for the pH where pKa = –log(Kw /Kb) = 6.04: pH = pKa + log
(0.080 ) [base] = 6.04 + log = 6.04 + 0.60 = 6.64 (0.020 ) [acid]
b. We have a weak acid and a strong acid present at the same time. The H+ contribution from the weak acid, HONH3+, will be negligible. So we have to consider only the H+ from HCl. [H+] = 0.020 M; pH = 1.70 c. This is a strong acid in water. [H+] = 0.020 M; pH = 1.70 d. Major species: H2O, Cl−, HONH2, HONH3+, H+ H+ will react completely with HONH2, the best base present. HONH2
→
H+
+
HONH3+
Before 0.100 M 0.020 M 0.100 M Change –0.020 –0.020 → +0.020 Reacts completely After 0.080 0 0.120 A buffer solution results after reaction. Using the Henderson-Hasselbalch equation: pH = 6.04 + log 37.
[HONH2 ] +
[HONH3 ]
= 6.04 + log
(0.080 ) = 6.04 − 0.18 = 5.86 (0.120 )
a. OH− will react completely with the best acid present, HOPr. HOPr Before Change After
+
0.100 M −0.020 0.080
OH−
→
OPr-
0.020 M −0.020 0
→
0 +0.020 0.020
+
H2 O Reacts completely
A buffer solution results after the reaction. Using the Henderson-Hasselbalch equation: pH = pKa + log
(0.020 ) [base] = 4.89 + log = 4.29; assumptions good. (0.080 ) [acid]
b. We have a weak base and a strong base present at the same time. The amount of OH added by the weak base will be negligible. To prove it, let’s consider the weak base equilibrium:
CHAPTER 15
ACID-BASE EQUILIBRIA OPr−
Initial Change Equil.
+
H2 O
735
⇌ HOPr
OH−
+
Kb = 7.7 × 10−10
0.100 M 0 0.020 M x mol/L OPr− reacts with H2O to reach equilibrium −x → +x +x 0.100 − x x 0.020 + x
[OH−] = 0.020 + x 0.020 M; pOH = 1.70; pH = 12.30; assumption good. Note: The OH− contribution from the weak base OPr− was negligible (x = 3.9 × 10−9 M as compared to 0.020 M OH− from the strong base). The pH can be determined by only con-sidering the amount of strong base present. c. This is a strong base in water. [OH−] = 0.020 M; pOH = 1.70; pH = 12.30 d. OH− will react completely with HOPr, the best acid present. HOPr Before Change After
+
0.100 M −0.020 0.080
OH−
→
OPr−
0.020 M −0.020 → 0
+
H2O
0.100 M +0.020 0.120
Reacts completely
Using the Henderson-Hasselbalch equation to solve for the pH of the resulting buffer solution: pH = pKa + log 38.
(0.120 ) [base] = 4.89 + log = 5.07; assumptions good. (0.080 ) [acid]
a. We have a weak base and a strong base present at the same time. The OH− contribution from the weak base, HONH2, will be negligible. Consider only the added strong base as the primary source of OH−. [OH−] = 0.020 M; pOH = 1.70; pH = 12.30 b. Added strong base will react to completion with the best acid present, HONH 3+. OH− Before Change After
+
0.020 M –0.020 0
HONH3+
→
0.100 M –0.020 0.080
→ +0.020
HONH2
+
H2O
0 Reacts completely
0.020
The resulting solution is a buffer (a weak acid and its conjugate base). Using the Henderson-Hasselbalch equation: pH = 6.04 + log
(0.020 ) = 6.04 – 0.60 = 5.44 (0.080 )
c. This is a strong base in water. [OH−] = 0.020 M; pOH = 1.70; pH = 12.30
736
CHAPTER 15
ACID-BASE EQUILIBRIA
d. Major species: H2O, Cl−, Na+, HONH2, HONH3+, OH−; again, the added strong base reacts completely with the best acid present, HONH3+. HONH3+ Before Change After
→
HONH2
0.020 M –0.020 → 0
0.100 M +0.020 0.120
OH−
+
0.100 M –0.020 0.080
+ H2 O Reacts completely
A buffer solution results. Using the Henderson-Hasselbalch equation: pH = 6.04 + log 39.
[HONH2 ] +
[HONH3 ]
= 6.04 + log
(0.120 ) = 6.04 + 0.18 = 6.22 (0.080 )
Consider all the results to Exercises 31, 35, and 37: Solution
Initial pH
After Added H+
After Added OH−
a b c d
2.96 8.94 7.00 4.89
1.70 5.49 1.70 4.71
4.29 12.30 12.30 5.07
The solution in Exercise 31d is a buffer; it contains both a weak acid (HC 3H5O2) and a weak base (C3H5O2−). Solution d shows the greatest resistance to changes in pH when either a strong acid or a strong base is added, which is the primary property of buffers. 40.
Consider all the results to Exercises 32, 36, and 38. Solution
Initial pH
After Added H+
After Added OH−
a b c d
9.52 3.52 7.00 6.04
6.64 1.70 1.70 5.86
12.30 5.44 12.30 6.22
The solution in Exercise 32d is a buffer; it shows the greatest resistance to a change in pH when strong acid or base is added. The solution in Exercise 32d contains a weak acid (HONH3+) and a weak base (HONH2), which constitutes a buffer solution. 41.
Statement c is false. Added OH‒ will convert HCN into CN‒. So, the concentration of HCN will decrease and the concentration of CN ‒ will increase. For statement a, the buffer solution of HCN and NaCN will have a higher pH than just a solution of the acid HCN. For statement b, pH = pKa ([H+] = Ka) when the weak acid concentration equals the conjugate base concentration. For statement d, adding more of the base component of the buffer will increase the pH.
42.
Only statement a is true. When [H+] = Ka for a buffer, the the weak acid concentration equals the conjugate base concentration (or the weak base concentration equals the conjugate acid concentration). For statement b, adding the strong base NaOH will increase the pH. For statement c, adding the strong acid HCl will increase the [H +]. For statement d, added HCl converts HONH2 into its conjugate acid, HONH3+, so [HONH3+] < [HONH2]. For statement e,
CHAPTER 15
ACID-BASE EQUILIBRIA
737
when a buffer solution has more of the acid component than the base component, the [H +] is greater than the Ka value (or the pH is less than pK a). So, the pH of the equal molar buffer solution will have a higher pH. 43.
Major species: HNO2, NO2− and Na+. Na+ has no acidic or basic properties. One appropriate equilibrium reaction you can use is the Ka reaction of HNO2, which contains both HNO2 and NO2−. However, you could also use the Kb reaction for NO2− and come up with the same answer. Solving the equilibrium problem (called a buffer problem): HNO2 Initial Change Equil.
⇌
NO2−
H+
+
1.00 M 1.00 M ~0 x mol/L HNO2 dissociates to reach equilibrium –x → +x +x 1.00 – x 1.00 + x x
Ka = 4.0 × 10 −4 =
−
[NO 2 ] [H + ] (1.00 )( x ) (1.00 + x )( x ) = (assuming 1.00 1.00 − x [HNO 2 ]
x << 1.00)
x = 4.0 × 10 −4 M = [H+]; assumptions good (x is 4.0 × 10 −2 % of 1.00). pH = –log(4.0 × 10 −4 ) = 3.40 Note: We would get the same answer using the Henderson-Hasselbalch equation. Use whichever method you prefer. 44.
Major species: HF, F−, K+, and H2O. K+ has no acidic or basic properties. This is a solution containing a weak acid and its conjugate base. This is a buffer solution. One appropriate equilibrium reaction you can use is the Ka reaction of HF, which contains both HF and F−. However, you could also use the Kb reaction for F− and come up with the same answer. Alternately, you could use the Henderson-Hasselbalch equation to solve for the pH. For this problem, we will use the Ka reaction and set up an ICE table to solve for the pH. HF Initial Change Equil.
⇌
F−
+
H+
0.60 M 1.00 M ~0 x mol/L HF dissociates to reach equilibrium –x → +x +x 0.60 – x 1.00 + x x
Ka = 7.2 × 10 −4 =
(1.00 )( x ) (1.00 + x )( x ) [F− ][H + ] = (assuming x << 0.60) 0.60 0 . 60 − x [HF]
x = [H+] = 0.60 (7.2 × 10−4) = 4.3 × 10 −4 M; assumptions good. pH = −log(4.3 × 10−4) = 3.37 45.
Major species after NaOH added: HNO2, NO2−, Na+, and OH−. The OH− from the strong base will react with the best acid present (HNO2). Any reaction involving a strong base is assumed to go to completion. Because all species present are in the same volume of solution, we can use
738
CHAPTER 15
ACID-BASE EQUILIBRIA
molarity units to do the stoichiometry part of the problem (instead of moles). The stoichiometry problem is: OH− Before Change After
+
HNO2
0.10 mol/1.00 L –0.10 M 0
1.00 M –0.10 M 0.90
→
NO2− + H2O
→
1.00 M +0.10 M 1.10
Reacts completely
After all the OH− reacts, we are left with a solution containing a weak acid (HNO 2) and its conjugate base (NO2–). This is what we call a buffer problem. We will solve this buffer problem using the Ka equilibrium reaction. One could also use the Kb equilibrium reaction or use the Henderson-Hasselbalch equation to solve for the pH. HNO2 Initial Change Equil.
⇌
NO2−
+
H+
0.90 M 1.10 M ~0 x mol/L HNO2 dissociates to reach equilibrium –x → +x +x 0.90 – x 1.10 + x x
Ka = 4.0 × 10 −4 =
(1.10 )( x ) (1.10 + x )( x ) , x = [H+] = 3.3 × 10 −4 M; pH = 3.48; 0.90 0.90 − x assumptions good.
Note: The added NaOH to this buffer solution changes the pH only from 3.40 to 3.48. If the NaOH were added to 1.0 L of pure water, the pH would change from 7.00 to 13.00. Major species after HCl added: HNO2, NO2−, H+, Na+, Cl−; the added H+ from the strong acid will react completely with the best base present (NO 2−).
Before Change After
H+ 0.20 mol 1.00 L
+
NO2−
→
1.00 M
–0.20 M 0
–0.20 M 0.80
HNO2 1.00 M
→
+0.20 M 1.20
Reacts completely
After all the H+ has reacted, we have a buffer solution (a solution containing a weak acid and its conjugate base). Solving the buffer problem: HNO2 Initial Equil.
1.20 M 1.20 – x
Ka = 4.0 × 10 −4 =
⇌
NO2− 0.80 M 0.80 + x
+
H+ 0 +x
(0.80)( x ) (0.80 + x)( x) , x = [H+] = 6.0 × 10 −4 M; pH = 3.22; 1.20 1.20 − x
assumptions good. Note: The added HCl to this buffer solution changes the pH only from 3.40 to 3.22. If the HCl were added to 1.0 L of pure water, the pH would change from 7.00 to 0.70.
CHAPTER 15 46.
ACID-BASE EQUILIBRIA
739
Major species after NaOH added: HF, F −, K+, Na+, OH−, and H2O. The OH− from the strong base will react with the best acid present (HF). Any reaction involving a strong base is assumed to go to completion. Because all species present are in the same volume of solution, we can use molarity units to do the stoichiometry part of the problem (instead of moles). The stoichiometry problem is: OH− + HF → F− + H2O Before 0.10 mol/1.00 L Change –0.10 M After 0
0.60 M –0.10 M 0.50
→
1.00 M +0.10 M 1.10
Reacts completely
After all the OH− reacts, we are left with a solution containing a weak acid (HF) and its conjugate base (F−). This is what we call a buffer problem. We will solve this buffer problem using the Ka equilibrium reaction. One could also use the K b equilibrium reaction or use the Henderson-Hasselbalch equation to solve for the pH. HF Initial Change Equil.
⇌
F−
+
H+
0.50 M 1.10 M ~0 x mol/L HF dissociates to reach equilibrium –x → +x +x 0.50 – x 1.10 + x x
Ka = 7.2 × 10 −4 =
(1.10 )( x ) (1.10 + x )( x ) , x = [H+] = 3.3 × 10 −4 M; pH = 3.48; 0.50 0.50 − x assumptions good.
Note: The added NaOH to this buffer solution changes the pH only from 3.37 to 3.48. If the NaOH were added to 1.0 L of pure water, the pH would change from 7.00 to 13.00. Major species after HCl added: HF, F −, H+, K+, Cl−, and H2O; the added H+ from the strong acid will react completely with the best base present (F −). H+ 0.20 mol 1.00 L
Before Change After
+
F−
→
1.00 M
–0.20 M 0
–0.20 M 0.80
HF 0.60 M
→
+0.20 M 0.80
Reacts completely
After all the H+ has reacted, we have a buffer solution (a solution containing a weak acid and its conjugate base). Solving the buffer problem:
Initial Equil.
HF 0.80 M 0.80 – x
Ka = 7.2 × 10 −4 =
⇌
F− 0.80 M 0.80 + x
+
H+ 0 x
(0.80)( x ) (0.80 + x)( x) , x = [H+] = 7.2 × 10 −4 M; pH = 3.14; 0.80 0.80 − x assumptions good.
Note: The added HCl to this buffer solution changes the pH only from 3.37 to 3.14. If the HCl were added to 1.0 L of pure water, the pH would change from 7.00 to 0.70.
740 47.
CHAPTER 15
ACID-BASE EQUILIBRIA
Ka for C5H5NH+ = 1.0 × 10‒14/1.7 × 10‒9 = 5.9 × 10‒6; pKa = ‒log(5.9 × 10‒6) = 5.23 A solution with equal concentrations of weak base and conjugate acid ([C5H5N] = [C5H5NH+]) will have [H+] = Ka = 5.9 × 10‒6 M and pH = 5.23. Statement d is false. If a solution has a larger weak base concentration than conjugate acid concentration ([C5H5N] > [C5H5NH+]), then the buffer will have [H+] < Ka = 5.9 × 10‒6 M and pH > 5.23. Statement b is false. If the concentration of conjugate acid is greater than the weak base concentration ([C5H5N] < [C5H5NH+]), then the buffer will have [H+] > Ka = 5.9 × 10‒6 M and pH < 5.23. Statement c is true. Statement a is false as the solution with the larger concentrations will have the greater capacity.
48.
Ka for C2H5NH3+ = 1.0 × 10‒14/5.6 × 10‒4 = 1.79 × 10‒11; pKa = ‒log(1.79 × 10‒11) = 10.75 A solution with equal concentrations of weak base and conjugate acid ([C2H5NH2] = [C2H5NH3+]) will have [H+] = Ka = 1.8 × 10‒11 M and pH = 10.75. Statement a is false. If a solution has a larger weak base concentration than conjugate acid concentration ([C2H5NH2] > [C2H5NH3+]), then the buffer will have [H+] < Ka = 1.8 × 10‒11 M and pH > 10.75. Statement b is true, and statement d must be false. In statement d, we have more weak base than conjugate acid in the buffer, so pH must be greater than 10.75. If the concentration of conjugate acid is greater than the weak base concentration ([C2H5NH2] < [C2H5NH3+]), then the buffer will have [H+] > Ka = 1.8 × 10‒11 M and pH < 10.75. Statement c must be false. In statement c, we have more conjugate acid than weak base in the buffer, so the pH must be less than 10.75.
49.
a.
HC2H3O2 Initial Change Equil. 1.8 × 10−5 =
⇌
H+
+
C2H3O2−
Ka = 1.8 × 10−5
0.10 M ~0 0.25 M x mol/L HC2H3O2 dissociates to reach equilibrium −x → +x +x 0.10 − x x 0.25 + x x(0.25 + x) x(0.25) (assuming 0.25 + x 0.25 and 0.10 – x 0.10) (0.10 − x) 0.10
x = [H+] = 7.2 × 10−6 M; pH = 5.14; assumptions good by the 5% rule. Alternatively, we can use the Henderson-Hasselbalch equation: pH = pKa + log
[base] , where pKa = −log(1.8 × 10−5) = 4.74 [acid]
pH = 4.74 + log
(0.25) = 4.74 + 0.40 = 5.14 (0.10)
The Henderson-Hasselbalch equation will be valid when assumptions of the type, 0.10 − x 0.10, that we just made are valid. From a practical standpoint, this will almost always be true for useful buffer solutions. Note: The Henderson-Hasselbalch equation can only be used to solve for the pH of buffer solutions. b. pH = 4.74 + log
(0.10) = 4.74 + (−0.40) = 4.34 (0.25)
CHAPTER 15
50.
ACID-BASE EQUILIBRIA
c. pH = 4.74 + log
(0.20) = 4.74 + 0.40 = 5.14 (0.080 )
d. pH = 4.74 + log
(0.080 ) = 4.74 + (−0.40) = 4.34 (0.20)
741
We will use the Henderson-Hasselbalch equation to solve for the pH of these buffer solutions. a. pH = pKa + log
Ka =
[base] ; [base] = [C2H5NH2] = 0.50 M; [acid] = [C2H5NH3+] = 0.25 M [acid]
Kw 1.0 10 −14 = = 1.8 10 −11 −4 Kb 5.6 10
pH = −log(1.8 × 10−11) + log
(0.50 M ) = 10.74 + 0.30 = 11.04 (0.25 M )
b. pH = 10.74 + log
(0.25 M ) = 10.74 + (−0.30) = 10.44 (0.50 M )
c. pH = 10.74 + log
(0.50 M ) = 10.74 + 0 = 10.74 (0.50 M )
1 mol HC7 H 5 O 2 122.12 g = 0.880 M 0.2000 L
21.5 g HC7 H 5 O 2 51.
[HC7H5O2] =
−
1 mol NaC7 H 5 O 2 1 mol C 7 H 5 O 2 37.7 g NaC7 H 5 O 2 144 .10 g mol NaC7 H 5 O 2 [C7H5O2−] = = 1.31 M 0.2000 L
We have a buffer solution since we have both a weak acid and its conjugate base present at the same time. One can use the Ka reaction or the Kb reaction to solve. We will use the Ka reaction for the acid component of the buffer. HC7H5O2 Initial Change Equil.
⇌
H+
+
C7H5O2−
0.880 M ~0 1.31 M x mol/L of HC7H5O2 dissociates to reach equilibrium –x → +x +x 0.880 – x x 1.31 + x
Ka = 6.4 × 10 −5 =
x(1.31) x(1.31 + x) , x = [H+] = 4.3 × 10 −5 M 0.880 0.880 − x
pH = −log(4.3 × 10 −5 ) = 4.37; assumptions good. Alternatively, we can use the Henderson-Hasselbalch equation to calculate the pH of buffer solutions. pH = pKa + log
− [C H O ] [base] = pKa + log 7 5 2 [HC 7 H 5 O 2 ] [acid]
742
CHAPTER 15
ACID-BASE EQUILIBRIA
1.31 pH = −log(6.4 × 10 −5 ) + log = 4.19 + 0.173 = 4.36 0.880
Within round-off error, this is the same answer we calculated solving the equilibrium problem using the Ka reaction. The Henderson-Hasselbalch equation will be valid when an assumption of the type 1.31 + x 1.31 that we just made in this problem is valid. From a practical standpoint, this will almost always be true for useful buffer solutions. If the assumption is not valid, the solution will have such a low buffering capacity that it will be of no use to control the pH. Note: The HendersonHasselbalch equation can only be used to solve for the pH of buffer solutions. 52.
50.0 g NH4Cl ×
1 mol NH 4 Cl = 0.935 mol NH4Cl added to 1.00 L; [NH4+] = 0.935 M 53.49 g NH 4 Cl
Using the Henderson-Hasselbalch equation to solve for the pH of this buffer solution: pH = pKa + log
53.
[H+] added = a.
0.75 = −log(5.6 × 10 −10 ) + log = 9.25 − 0.096 = 9.15 0.935 [NH 4 ]
[NH 3 ] +
0.010 mol = 0.040 M; the added H+ reacts completely with NH3 to form NH4+. 0.2500 L
NH3 Before Change After
H+
→
NH4+
0.040 M −0.040 0
→
0.15 M +0.040 0.19
+
0.050 M −0.040 0.010
Reacts completely
A buffer solution still exists after H+ reacts completely. Using the HendersonHasselbalch equation: pH = pKa + log b.
NH3 Before Change After
0.50 M −0.040 0.46
0.010 = −log(5.6 × 10−10) + log = 9.25 + (−1.28) = 7.97 0.19 [ NH 4 ]
[ NH3 ] +
+
→
NH4+
0.040 M −0.040 → 0
1.50 M +0.040 1.54
H+
A buffer solution still exists. pH = pKa + log
Reacts completely
0.46 , 9.25 + log = 8.73 1.54 [ NH 4 ]
[ NH3 ] +
The two buffers differ in their capacity and not their initial pH (both buffers had an initial pH = 8.77). Solution b has the greatest capacity since it has the largest concentrations of weak acid and conjugate base. Buffers with greater capacities will be able to absorb more added H+ or OH−. 54.
a. Major species: H2O, Na+, OH−, HC3H5O2, C3H5O2−; OH- will react completely with HC3H5O2. Here, HC3H5O2 is the limiting reagent.
CHAPTER 15
ACID-BASE EQUILIBRIA
743 →
HC3H5O2 + OH− 0.050 M 0.15 M \ −0.050 −0.050 0 0.10
Before Change After
C3H5O2− 0.080 M +0.050 0.130
→
+
H2O Reacts completely
OH− from the strong base is in excess. The OH− contribution from the weak base C3H5O2− will be negligible. Consider only the excess strong base to determine the pH. [OH−] = 0.10 M; pOH = 1.00; pH = 13.00 −
Note: Original pH of buffer is: pH = pKa + log
[C 3 H 5 O 2 ] 0.080 = 4.89 + log = 5.09 [HC 3 H 5 O 2 ] 0.050
b. OH− will react completely with HC3H5O2. Here, OH− is the limiting reagent. HC3H5O2 Before Change After
→
C3H5O2−
0.15 M −0.15 → 0
0.80 M +0.15 0.95
OH−
+
0.50 M −0.15 0.35
+
H2O Reacts completely
A buffer solution results since after reaction both HC3H5O2 and C3H5O2− are present. −
[C H O ] 0.95 pH = pKa + log 3 5 2 = 4.89 + log = 4.89 + 0.43 = 5.32 [HC 3 H 5 O 2 ] 0.35
c. Although solutions a and b both started out as buffers with the same pH (= 5.09), solution a is no longer a buffer after 0.15 mol NaOH was added. We added more strong base than there was weak acid present in the buffer; we overloaded the buffer system in solution a. We say that solution b has the greater buffer capacity since it has larger initial concentrations of weak acid and conjugate base. 55.
HSO3− ⇌ H+
+
Ka = 1.0 × 10−7, pKa = −log(1.0 × 10−7) = 7.00
SO32−
2−
[SO 3 ] [base] pH = pKa + log , 7.25 = 7.00 + log − [acid] [HSO 3 ]
For pH = 7.25, we need the log term to be positive. This will only happen when [SO32−] > [HSO3−]. Another way to think about this is to compare the pK a and pH values. Here, we want a pH on the “basic” side of the pKa value, i.e., we want the pH to be greater than the pKa value. To get a pH on the “basic” side of the pKa, we need more base than acid in the buffer. Hence, we need a larger concentration of SO32− than HSO3− to achieve a pH = 7.25. 7.25 = 7.00 + log
1.0 M −
, 100.25 =
−
, [HSO3−] = 0.56 M
[HSO3 ]
[HSO3 ] 56.
1.0 M
a. pKb for C6H5NH2 = −log(3.8 × 10−10) = 9.42; pKa for C6H5NH3+ = 14.00 − 9.42 = 4.58 pH = pKa + log −0.38 = log
[C 6 H 5 NH 2 ] +
[C 6 H 5 NH3 ]
0.50 M +
[C 6 H 5 NH3 ]
, 4.20 = 4.58 + log
0.50 M +
[C 6 H 5 NH3 ]
, [C6H5NH3+] = [C6H5NH3Cl] = 1.2 M
744
CHAPTER 15 b. 4.0 g NaOH ×
0.10 mol 1 mol NaOH 1 mol OH− = 0.10 mol OH−; [OH−] = = 0.10 M 1.0 L 40.00 g mol NaOH
C6H5NH3+ Before Change After
ACID-BASE EQUILIBRIA
→
C6H5NH2
0.10 M −0.10 → 0
0.50 M +0.10 0.60
OH−
+
1.2 M −0.10 1.1
+
H2 O Reacts completely
0.60 A buffer solution exists. pH = 4.58 + log = 4.32 1. 1
57.
pKb for H2NNH2 = −log(3.0 × 10−6) = 5.52; pKa for H2NNH3+ = 14.00 − 5.52 = 8.48 pH = pKa + log −0.48 = log
58.
[H 2 NNH 2 ] 1.87 M , 8.00 = 8.48 + log + [H 2 NNH 3 ] [H 2 NNH 3 + ]
1.87 M , [C6H5NH3+] = 5.6 M [H 2 NNH 3 + ]
pH = pKa + log
[CN − ] ; pKa = –log(6.2 × 10‒10) = 9.21 [HCN]
9.50 = 9.21 + log
[CN − ] [CN − ] , 0.29 = log , [CN‒] = 4.9 M 2.5 mol HCN/1.0 L 2.5 mol HCN/1.0 L
Mol Sr(CN)2 = 1.0 L ×
1 mol Sr(CN)2 4.9 mol CN − = 2.5 mol Sr(CN)2 L 2 mol CN − −
59.
pH = pKa + log
[C 2 H 3 O 2 ] ; pKa = –log(1.8 × 10 −5 ) = 4.74 [HC 2 H 3 O 2 ]
Because the buffer components, C2H3O2− and HC2H3O2, are both in the same volume of solution, the concentration ratio of [C2H3O2-] : [HC2H3O2] will equal the mole ratio of mol C2H3O2− to mol HC2H3O2. −
5.00 = 4.74 + log
0.200 mol mol C 2 H 3 O 2 ; mol HC2H3O2 = 0.5000 L × = 0.100 mol L mol HC 2 H 3 O 2 −
0.26 = log
mol C 2 H 3 O 2 mol C 2 H 3 O 2 , 0.100 mol 0.100
Mass NaC2H3O2 = 0.18 mol NaC2H3O2 × −
60.
pH = pKa + log
−
= 100.26 = 1.8, mol C2H3O2− = 0.18 mol 82.03 g = 15 g NaC2H3O2 mol −
[NO 2 ] [NO 2 ] , 3.55 = −log(4.0 × 10−4) + log [HNO 2 ] [HNO 2 ] −
− − mol NO 2 [NO 2 ] [NO 2 ] 3.55 = 3.40 + log , = 100.15 = 1.4 = mol HNO 2 [HNO 2 ] [HNO 2 ]
CHAPTER 15
ACID-BASE EQUILIBRIA
745
Let x = volume (L) of HNO2 solution needed; then 1.00 − x = volume of NaNO2 solution needed to form this buffer solution. −
Mol NO 2 = 1.4 = Mol HNO 2
0.50 mol NaNO2 0.50 − (0.50) x L = 0.50 mol HNO2 (0.50) x x L
(1.00 − x)
(0.70)x = 0.50 − (0.50)x , (1.20)x = 0.50, x = 0.42 L We need 0.42 L of 0.50 M HNO2 and 1.00 − 0.42 = 0.58 L of 0.50 M NaNO2 to form a pH = 3.55 buffer solution. 61.
C5H5NH+ ⇌ H+ + C5H5N
Kw 1.0 10 −14 = 5.9 × 10−6 = −9 Kb 1.7 10
Ka =
pKa = −log(5.9 × 10−6) = 5.23 We will use the Henderson-Hasselbalch equation to calculate the concentration ratio necessary for each buffer. pH = pKa + log
[C 5 H 5 N] [base] , pH = 5.23 + log [acid] [C 5 H 5 NH + ] [C 5 H 5 N]
a. 4.50 = 5.23 + log
+
,
[C 5 H 5 N ] = 10−0.73 = 0.19 [C 5 H 5 NH + ]
[C 5 H 5 NH ] [C 5 H 5 N] [C 5 H 5 N ] b. 5.00 = 5.23 + log , = 10−0.23 = 0.59 + + [C 5 H 5 NH ] [C 5 H 5 NH ] [C 5 H 5 N ] [C 5 H 5 N ] , = 100.0 = 1.0 + [C 5 H 5 NH ] [C 5 H 5 NH + ] [C 5 H 5 N ] [C 5 H 5 N ] d. 5.50 = 5.23 + log , = 100.27 = 1.9 + [C 5 H 5 NH ] [C 5 H 5 NH + ]
c. 5.23 = 5.23 + log
62.
NH4+ ⇌ H+ + NH3 Ka = Kw/Kb = 5.6 × 10−10; pKa = −log(5.6 × 10−10) = 9.25; we will use the Henderson-Hasselbalch equation to calculate the concentration ratio necessary for each buffer. pH = pKa + log
[NH 3 ] [base] , pH = 9.25 + log + [acid] [NH 4 ] [NH3 ]
a. 9.00 = 9.25 + log b. 8.80 = 9.25 + log
[NH 4 ] [NH 3 ] +
[NH 3 ]
,
+
[NH 4 ] [NH 3 ]
,
+
d. 9.60 = 9.25 + log
= 10−0.45 = 0.35
[NH 4 ]
[NH 4 ] c. 10.00 = 9.25 + log
= 10−0.25 = 0.56
+
[NH 3 ] +
,
[NH 3 ] +
= 100.75 = 5.6
[NH 4 ]
[NH 4 ]
[NH 3 ]
[NH 3 ]
+
[NH 4 ]
,
+
[NH 4 ]
= 100.35 = 2.2
746 63.
CHAPTER 15 pH = pKa + log
ACID-BASE EQUILIBRIA
[base] ; in a buffer, when [acid] = [base], pH = pK a. When [acid] > [base], pH [acid]
< pKa, and when [base] > [acid], pH > pK a. For the H2CO3/ HCO3− buffer, we want pH = pKa, so [H2CO3] = [HCO3−]. For the H2PO4−/ HPO42− buffer, we want pH < pKa, so [H2PO4−] > [HPO42−]. Answer b is correct. 64.
For the H2S/HS‒ buffer, pKa = ‒log(1 × 10−7) = 7.0. We want a pH > pKa, so [HS‒] > [H2S]. For the HAsO42‒/AsO43‒ buffer, pKa = ‒log(5 × 10−12) = 11.30. We want a pH < pKa, so [HAsO42-] > [AsO43-]. Answer c is correct.
65.
pH = pKa + log
−
−
[HCO 3 ] [HCO 3 ] , 7.40 = −log(4.3 × 10−7) + log 0.0012 [H 2 CO 3 ] −
−
[HCO 3 ] [HCO 3 ] log = 7.40 – 6.37 = 1.03, = 101.03, [HCO3−] = 1.3 × 10−2 M 0.0012 0.0012 −
66.
At pH = 7.40: 7.40 = −log(4.3 × 10−7) + log
[HCO 3 ] [H 2 CO 3 ] −
−
[HCO 3 ] [HCO 3 ] [H 2 CO 3 ] log = 7.40 – 6.37 = 1.03, = 101.03, = 10−1.03 = 0.093 − [H 2 CO 3 ] [H 2 CO 3 ] [HCO3 ] −
−
At pH = 7.35: log
[H 2 CO 3 ] −
[HCO 3 ] [HCO 3 ] = 7.35 – 6.37 = 0.98, = 100.98 [H 2 CO 3 ] [H 2 CO 3 ]
= 10−0.98 = 0.10
[HCO3 ] The [H2CO3] : [HCO3−] concentration ratio must increase from 0.093 to 0.10 in order for the onset of acidosis to occur. 67.
A best buffer has large and equal quantities of weak acid and conjugate base. Because [acid] [base] = [base] for a best buffer, pH = pKa + log = pKa + 0 = pKa (pH pKa for a best buffer). [acid] The best acid choice for a pH = 7.00 buffer would be the weak acid with a pK a close to 7.0 or Ka 1 × 10−7. HOCl is the best choice in Table 14.2 (Ka = 3.5 × 10−8; pKa = 7.46). To make this buffer, we need to calculate the [base] : [acid] ratio. 7.00 = 7.46 + log
[base] , [acid]
[OCl− ] = 10−0.46 = 0.35 [HOCl]
Any OCl−/HOCl buffer in a concentration ratio of 0.35 : 1 will have a pH = 7.00. One possibility is [NaOCl] = 0.35 M and [HOCl] = 1.0 M. 68.
For a pH = 5.00 buffer, we want an acid with a pK a close to 5.00. For a conjugate acid-base pair, 14.00 = pKa + pKb. So, for a pH = 5.00 buffer, we want the base to have a pK b close to
CHAPTER 15
ACID-BASE EQUILIBRIA
747
(14.0 − 5.0 =) 9.0 or a Kb close to 1 × 10−9. The best choice in Table 14.3 is pyridine (C5H5N) with Kb = 1.7 × 10−9. pH = pKa + log
K [base] 1.0 10 −14 = 5.9 × 10−6 ; Ka = w = [acid] Kb 1.7 10 −9
5.00 = −log(5.9 × 10-6) + log
[C 5 H 5 N] [base] = 10−0.23 = 0.59 , [acid] [C 5 H 5 NH + ]
There are many possibilities to make this buffer. One possibility is a solution of [C 5H5N] = 0.59 M and [C5H5NHCl] = 1.0 M. The pH of this solution will be 5.00 because the base to acid concentration ratio is 0.59 : 1. 69.
Ka for H2NNH3+ = K w /K b, H2 NNH2 = 1.0 × 10−14/3.0 × 10−6 = 3.3 × 10−9 pH = pKa + log
0.40 = −log(3.3 × 10−9) + log = 8.48 + (−0.30) = 8.18 0.80 [H 2 NNH3 ] [H 2 NNH 2 ] +
pH = pKa for a buffer when [acid] = [base]. Here, the acid (H 2NNH3+) concentration needs to decrease, while the base (H2NNH2) concentration needs to increase in order for [H2NNH3+] = [H2NNH2]. Both of these changes are accomplished by adding a strong base (like NaOH) to the original buffer. The added OH − from the strong base converts the acid component of the buffer into the conjugate base. Here, the reaction is H 2NNH3+ + OH− → H2NNH2 + H2O. Because a strong base is reacting, the reaction is assumed to go to completion. The following set-up determines the number of moles of OH−(x) that must be added so that mol H2NNH3+ = mol H2NNH2 . When mol acid = mol base in a buffer, then [acid] = [base] and pH = pK a. H2NNH3+ Before Change After
+
1.0 L × 0.80 mol/L −x 0.80 − x
OH− → x −x 0
→
H2NNH2
+
1.0 L × 0.40 mol/L +x 0.40 + x
H2 O Reacts completely
We want mol H2NNH3+ = mol H2NNH2. Therefore: 0.80 − x = 0.40 + x, 2x = 0.40, x = 0.20 mol OH− When 0.20 mol OH− is added to the initial buffer, mol H2NNH3+ is decreased to 0.60 mol, while mol H2NNH2 is increased to 0.60 mol. Therefore, 0.20 mol of NaOH must be added to the initial buffer solution to produce a solution where pH = pKa. 70.
pH = pKa + log
[OCl− ] 0.90 = −log(3.5 × 10−8) + log = 7.46 + 0.65 = 8.11 [HOCl] 0.20
pH = pKa when [HOCl] = [OCl−] (or when mol HOCl = mol OCl−). Here, the moles of the base component of the buffer must decrease, while the moles of the acid component of the buffer must increase to achieve a solution where pH = pKa. Both changes occur when a strong acid (like HCl) is added. Let x = mol H+ added from the strong acid HCl.
748
CHAPTER 15 H+ Before Change After
x −x 0
→
OCl−
+
ACID-BASE EQUILIBRIA
HOCl
1.0 L × 0.90 mol/L 1.0 L × 0.20 mol/L −x → +x Reacts completely 0.90 − x 0.20 + x
We want mol HOCl = mol OCl−. Therefore: 0.90 − x = 0.20 + x, 2x = 0.70, x = 0.35 mol H+ When 0.35 mol H+ is added, mol OCl− is decreased to 0.55 mol, while the mol HOCl is increased to 0.55 mol Therefore, 0.35 mol of HCl must be added to the original buffer solution to produce a solution where pH = pKa. 71.
The reaction OH− + CH3NH3+ → CH3NH2 + H2O goes to completion for solutions a, c, and d (no reaction occurs between the species in solution b because both species are bases). After the OH− reacts completely, there must be both CH3NH3+ and CH3NH2 in solution for it to be a buffer. The important components of each solution (after the OH− reacts completely) is(are): a. 0.05 M CH3NH2 (no CH3NH3+ remains, no buffer) b. 0.05 M OH− and 0.1 M CH3NH2 (two bases present, no buffer) c. 0.05 M OH− and 0.05 M CH3NH2 (too much OH− added, no CH3NH3+ remains, no buffer) d. 0.05 M CH3NH2 and 0.05 M CH3NH3+ (a buffer solution results) Only the combination in mixture d results in a buffer. Note that the concentrations are halved from the initial values. This is so because equal volumes of two solutions were added together, which halves the concentrations.
72.
a. No; a solution of a strong acid (HNO3) and its conjugate base (NO3−) is not generally considered a buffer solution. b. No; two acids are present (HNO3 and HF), so it is not a buffer solution. c. Yes; H+ reacts completely with F−. Since equal volumes are mixed, the initial concentrations in the mixture are 0.10 M HNO3 and 0.20 M NaF. H+ Before Change After
0.10 M –0.10 0
+
F− 0.20 M –0.10 0.10
→
HF 0
→ +0.10
Reacts completely
0.10
After H+ reacts completely, a buffer solution results; that is, a weak acid (HF) and its conjugate base (F−) are both present in solution in large quantities. d. No; a strong acid (HNO3) and a strong base (NaOH) do not form buffer solutions. They will neutralize each other to form H2O.
CHAPTER 15 73.
ACID-BASE EQUILIBRIA
749
Added OH− converts HC2H3O2 into C2H3O2−: HC2H3O2 + OH− → C2H3O2− + H2O From this reaction, the moles of C2H3O2− produced equal the moles of OH− added. Also, the total concentration of acetic acid plus acetate ion must equal 2.0 M (assuming no volume change on addition of NaOH). Summarizing for each solution: [C2H3O2−] + [HC2H3O] = 2.0 M and [C2H3O2−] produced = [OH−] added −
−
[C H O ] [C H O ] a. pH = pKa + log 2 3 2 ; for pH = pKa, log 2 3 2 =0 [HC 2 H 3 O 2 ] [HC 2 H 3 O 2 ] −
Therefore,
[C 2 H 3 O 2 ] = 1.0 and [C2H3O2−] = [HC2H3O2]. [HC 2 H 3 O 2 ]
Because [C2H3O2−] + [HC2H3O2] = 2.0 M: [C2H3O2−] = [HC2H3O2] = 1.0 M = [OH−] added To produce a 1.0 M C2H3O2− solution, we need to add 1.0 mol of NaOH to 1.0 L of the 2.0 M HC2H3O2 solution. The resulting solution will have pH = pKa = 4.74. −
b. 4.00 = 4.74 + log
−
[C 2 H 3 O 2 ] [C 2 H 3 O 2 ] , = 10−0.74 = 0.18 [HC 2 H 3 O 2 ] [HC 2 H 3 O 2 ]
[C2H3O2−] = 0.18[HC2H3O2] or [HC2H3O2] = 5.6[C2H3O2−] Because [C2H3O2−] + [HC2H3O2] = 2.0 M: [C2H3O2−] + 5.6[C2H3O2−] = 2.0 M, [C2H3O2−] =
2 .0 = 0.30 M = [OH−] added 6 .6
We need to add 0.30 mol of NaOH to 1.0 L of 2.0 M HC2H3O2 solution to produce 0.30 M C2H3O2−. The resulting solution will have pH = 4.00. −
c. 5.00 = 4.74 + log
−
[C 2 H 3 O 2 ] [C 2 H 3 O 2 ] , = 100.26 = 1.8 [HC 2 H 3 O 2 ] [HC 2 H 3 O 2 ]
1.8[HC2H3O2] = [C2H3O2−] or [HC2H3O2] = 0.56[C2H3O2−] 1.56[C2H3O2−] = 2.0 M, [C2H3O2−] = 1.3 M = [OH−] added We need to add 1.3 mol of NaOH to 1.0 L of 2.0 M HC2H3O2 to produce a solution with pH = 5.00. 74.
When H+ is added, it converts C2H3O2− into HC2H3O2: C2H3O2− + H+ → HC2H3O2. From this reaction, the moles of HC2H3O2 produced must equal the moles of H+ added and the total concentration of acetate ion + acetic acid must equal 1.0 M (assuming no volume change). Summarizing for each solution: [C2H3O2−] + [HC2H3O2] = 1.0 M and [HC2H3O2] = [H+] added
750
CHAPTER 15
ACID-BASE EQUILIBRIA
−
a. pH = pKa + log
[C 2 H 3 O 2 ] ; for pH = pKa, [C2H3O2−] = [HC2H3O2]. [HC 2 H 3 O 2 ]
For this to be true, [C2H3O2−] = [HC2H3O2] = 0.50 M = [H+] added, which means that 0.50 mol of HCl must be added to 1.0 L of the initial solution to produce a solution with pH = pKa. −
b. 4.20 = 4.74 + log
−
[C 2 H 3 O 2 ] [C 2 H 3 O 2 ] = 10 −0.54 = 0.29 , [HC 2 H 3 O 2 ] [HC 2 H 3 O 2 ]
[C2H3O2−] = 0.29[HC2H3O2]; 0.29[HC2H3O2] + [HC2H3O2] = 1.0 M [HC2H3O2] = 0.78 M = [H+] added 0.78 mol of HCl must be added to produce a solution with pH = 4.20. −
c. 5.00 = 4.74 + log
−
[C 2 H 3 O 2 ] [C 2 H 3 O 2 ] , = 100.26 = 1.8 [HC 2 H 3 O 2 ] [HC 2 H 3 O 2 ]
[C2H3O2−] = 1.8[HC2H3O2]; 1.8[HC2H3O2] + [HC2H3O2] = 1.0 M [HC2H3O2] = 0.36 M = [H+] added 0.36 mol of HCl must be added to produce a solution with pH = 5.00.
Acid-Base Titrations 75.
(f) all points after stoichiometric point
(a) and (e) pH (c)
(d)
(b) mL base
HA + OH− → A− + H2O; added OH− from the strong base converts the weak acid HA into its conjugate base A−. Initially, before any OH− is added (point d), HA is the dominant species present. After OH− is added, both HA and A− are present, and a buffer solution results (region b). At the equivalence point (points a and e), exactly enough OH - has been added to convert all the weak acid HA into its conjugate base A −. Past the equivalence point (region f), excess
CHAPTER 15
ACID-BASE EQUILIBRIA
751
OH− is present. For the answer to part b, we included almost the entire buffer region. The maximum buffer region (or the region which is the best buffer solution) is around the halfway point to equivalence (point c). At this point, enough OH − has been added to convert exactly one-half of the weak acid present initially into its conjugate base, so [HA] = [A−] and pH = pKa. A “best” buffer has about equal concentrations of weak acid and conjugate base present. 76. (d) (c) (b) pH
(a) and (e) (f) all points after stoichiometric point
mL acid
B + H+ → BH+; added H+ from the strong acid converts the weak base B into its conjugate acid BH+. Initially, before any H+ is added (point d), B is the dominant species present. After H + is added, both B and BH+ are present, and a buffered solution results (region b). At the equivalence point (points a and e), exactly enough H + has been added to convert all the weak base present initially into its conjugate acid BH+. Past the equivalence point (region f), excess H+ is present. For the answer to b, we included almost the entire buffer region. The maximum buffer region is around the halfway point to equivalence (point c), where [B] = [BH +]. Here, pH = pKa , which is a characteristic of a best buffer. 77.
This is a strong acid (HClO4) titrated by a strong base (KOH). Added OH− from the strong base will react completely with the H+ present from the strong acid to produce H2O. a. Only strong acid present. [H+] = 0.200 M; pH = 0.699 b. mmol OH− added = 10.0 mL × mmol H+ present = 40.0 mL ×
0.100 mmol OH − = 1.00 mmol OH− mL 0.200 mmol H + = 8.00 mmol H+ mL
Note: The units millimoles are usually easier numbers to work with. The units for molarity are moles per liter but are also equal to millimoles per milliliter. H+ Before Change After
8.00 mmol −1.00 mmol 7.00 mmol
+
OH−
→ H2 O
1.00 mmol −1.00 mmol 0
Reacts completely
752
CHAPTER 15 The excess H+ determines the pH. [H+]excess =
ACID-BASE EQUILIBRIA
7.00 mmol H + = 0.140 M 40.0 mL + 10.0 mL
pH = −log(0.140) = 0.854 c. mmol OH− added = 40.0 mL × 0.100 M = 4.00 mmol OH− H+ Before After [H+]excess =
OH−
+
8.00 mmol 4.00 mmol
→
H2 O
4.00 mmol 0
4.00 mmol = 0.0500 M; pH = 1.301 (40.0 + 40.0) mL
d. mmol OH− added = 80.0 mL × 0.100 M = 8.00 mmol OH−; this is the equivalence point because we have added just enough OH- to react with all the acid present. For a strong acid-strong base titration, pH = 7.00 at the equivalence point because only neutral species are present (K+, ClO4−, H2O). e. mmol OH− added = 100.0 mL × 0.100 M = 10.0 mmol OH− H+ Before After
→
OH−
+
8.00 mmol 0
H2O
10.0 mmol 2.0 mmol
Past the equivalence point, the pH is determined by the excess OH− present. 2.0 mmol [OH−]excess = = 0.014 M; pOH = 1.85; pH = 12.15 ( 40.0 + 100 .0) mL 78.
a. Only strong acid is present. pH = ‒log(0.10) = 1.00 b. mmol H+ present = 200.0 mL × mmol OH− added = 25.0 mL × H+ Before After [H+]excess =
20. mmol 10. mmol
+
0.10 mmol H+ = 20. mmol H+ mL 0.20 mmol Ca(OH)2 2 mmol OH − = 10. mmol OH− mL mmol Ca(OH)2 OH−
→
H2 O
10. mmol 0
10. mmol = 0.044 M; pH = 1.36 (200.0 + 25.0) mL
c. [H+] = 1.0 × 10−7 M at the equivalence point for a strong acid-strong base titration. The volume of Ca(OH)2 necessary to reach the equivalence point is: 20. mmol H+ ×
1 mmol Ca(OH)2 1 mmol OH − 1 mL = 50. mL mmol H + 2 mmol OH − 0.20 mmol Ca(OH)2
CHAPTER 15 79.
ACID-BASE EQUILIBRIA
753
This is a strong base, Ba(OH)2, titrated by a strong acid, HCl. The added strong acid will neutralize the OH− from the strong base. As is always the case when a strong acid and/or strong base reacts, the reaction is assumed to go to completion. a. Only a strong base is present, but it breaks up into 2 moles of OH− ions for every mole of Ba(OH)2. [OH−] = 2 × 0.100 M = 0.200 M; pOH = 0.699; pH = 13.301 b. mmol OH− present = 80.0 mL ×
0.100 mmol Ba(OH) 2 2 mmol OH− mL mmol Ba(OH) 2
= 16.0 mmol OH− mmol H+ added = 20.0 mL × OH−
0.400 mmol H mL
H+
+
+
= 8.00 mmol H+ → H2O
Before Change After
16.0 mmol −8.00 mmol 8.0 mmol
8.00 mmol −8.00 mmol 0
[OH−]excess =
8.0 mmol OH− = 0.080 M; pOH = 1.10; pH = 12.90 80.0 mL + 20.0 mL
Reacts completely
c. mmol H+ added = 30.0 mL × 0.400 M = 12.0 mmol H+ OH− Before After
H+
+
16.0 mmol 4.0 mmol
[OH−]excess =
→
H2O
12.0 mmol 0
4.0 mmol OH− = 0.036 M; pOH = 1.44; pH = 12.56 (80.0 + 30.0) mL
d. mmol H+ added = 40.0 mL × 0.400 M = 16.0 mmol H+; this is the equivalence point. Because the H+ will exactly neutralize the OH− from the strong base, all we have in solution is Ba2+, Cl−, and H2O. All are neutral species, so pH = 7.00. e. mmol H+ added = 80.0 mL × 0.400 M = 32.0 mmol H+ OH− Before After [H+]excess =
80.
H+
+
16.0 mmol 0
→ H2 O
32.0 mmol 16.0 mmol
16.0 mmol H + = 0.100 M; pH = 1.000 (80.0 + 80.0) mL
a. mmol OH− present = 50.0 mL × mmol H+ added = 50.0 mL ×
0.20 mmol OH − = 10. mmol OH− mL
0.10 mmol H+ = 5.0 mmol H+ mL
754
CHAPTER 15 OH− Before After
10. mmol 5 mmol
[OH−]excess =
→
H+
+
ACID-BASE EQUILIBRIA
H2O
5.0 mmol 0
5 mmol OH= 0.05 M; pOH = 1.3; pH = 14.0 ‒ 1.3 = 12.7 (50.0 + 50.0) mL
0.10 mmol H+ = 10. mmol H+; this is the equivalence mL point just enough H+ has been added to react with all the OH‒ present initially. For this strong base-strong acid titration, pH = 7.00 at the equivalence point.
b. mmol H+ added = 100.0 mL ×
0.10 mmol H+ = 20. mmol H+ mL + H+ → H2O
c. mmol H+ added = 200.0 mL × OH− Before After [H+]excess = 81.
10. mmol 0 mmol
20. mmol 10 mmol
10. mmol H + = 0.040 M; pH = 1.40 (50.0 + 200.0) mL
This is a weak acid (HC2H3O2) titrated by a strong base (KOH). a. Only weak acid is present. Solving the weak acid problem:
⇌
HC2H3O2 Initial Change Equil.
H+
+
C2H3O2−
0.200 M ~0 0 x mol/L HC2H3O2 dissociates to reach equilibrium −x → +x +x 0.200 − x x x
Ka = 1.8 × 10−5 =
x2 x2 , x = [H+] = 1.9 × 10−3 M 0.200 − x 0.200
pH = 2.72; assumptions good. b. The added OH− will react completely with the best acid present, HC 2H3O2. mmol HC2H3O2 present = 100.0 mL × mmol OH− added = 50.0 mL × HC2H3O2 Before Change After
20.0 mmol −5.00 mmol 15.0 mmol
+
0.200 mmol HC 2 H 3O 2 = 20.0 mmol HC2H3O2 mL
0.100 mmol OH − = 5.00 mmol OH− mL
OH−
→
5.00 mmol −5.00 mmol → 0
C2H3O2− + H2O 0 +5.00 mmol 5.00 mmol
Reacts Completely
CHAPTER 15
ACID-BASE EQUILIBRIA
755
After reaction of all the strong base, we have a buffer solution containing a weak acid (HC2H3O2) and its conjugate base (C2H3O2−). We will use the Henderson-Hasselbalch equation to solve for the pH. pH = pKa + log
− 5.00 mmol/V T [C 2 H 3 O 2 ] , where VT = = −log (1.8 × 10−5) + log [HC 2 H 3 O 2 ] 15.0 mmol/V T
total volume 5.00 pH = 4.74 + log = 4.74 + (−0.477) = 4.26 15.0
Note that the total volume cancels in the Henderson-Hasselbalch equation. For the [base]/[acid] term, the mole ratio equals the concentration ratio because the components of the buffer are always in the same volume of solution. c. mmol OH− added = 100.0 mL × (0.100 mmol OH−/mL) = 10.0 mmol OH−; the same amount (20.0 mmol) of HC2H3O2 is present as before (it doesn’t change). As before, let the OH− react to completion, then see what is remaining in solution after this reaction. HC2H3O2 Before After
+
20.0 mmol 10.0 mmol
OH−
→
10.0 mmol 0
C2H3O2− + H2O 0 10.0 mmol
A buffer solution results after reaction. Because [C 2H3O2−] = [HC2H3O2] = 10.0 mmol/total volume, pH = pKa. This is always true at the halfway point to equivalence for a weak acidstrong base titration, pH = pKa. pH = −log(1.8 × 10−5) = 4.74 d. mmol OH− added = 150.0 mL × 0.100 M = 15.0 mmol OH−. Added OH− reacts completely with the weak acid. HC2H3O2 Before After
20.0 mmol 5.0 mmol
+
OH−
→
15.0 mmol 0
C2H3O2− + H2O 0 15.0 mmol
We have a buffer solution after all the OH− reacts to completion. Using the HendersonHasselbalch equation: pH = 4.74 + log
− 15.0 mmol [C 2 H 3 O 2 ] = 4.74 + log [HC 2 H 3 O 2 ] 5.0 mmol
pH = 4.74 + 0.48 = 5.22 e. mmol OH− added = 200.00 mL × 0.100 M = 20.0 mmol OH−; as before, let the added OH− react to completion with the weak acid; then see what is in solution after this reaction.
756
CHAPTER 15 HC2H3O2 Before After
+
20.0 mmol 0
→
OH−
ACID-BASE EQUILIBRIA
C2H3O2− + H2O
20.0 mmol 0
0 20.0 mmol
This is the equivalence point. Enough OH − has been added to exactly neutralize all the weak acid present initially. All that remains that affects the pH at the equivalence point is the conjugate base of the weak acid (C2H3O2−). This is a weak base equilibrium problem. K 1.0 10 −14 C2H3O2− + H2O ⇌ HC2H3O2 + OH− Kb = w = Kb 1.8 10 −5 Initial 20.0 mmol/300.0 mL 0 0 Kb = 5.6 × 10−10 x mol/L C2H3O2 reacts with H2O to reach equilibrium Change −x → +x +x Equil. 0.0667 − x x x Kb = 5.6 × 10−10 =
x2 x2 , x = [OH−] = 6.1 × 10−6 M 0.0667 − x 0.0667
pOH = 5.21; pH = 8.79; assumptions good. f.
mmol OH− added = 250.0 mL × 0.100 M = 25.0 mmol OH− HC2H3O2 Before After
+
20.0 mmol 0
→
OH−
C2H3O2− + H2O
25.0 mmol 5.0 mmol
0 20.0 mmol
After the titration reaction, we have a solution containing excess OH − and a weak base C2H3O2−. When a strong base and a weak base are both present, assume that the amount of OH− added from the weak base will be minimal; that is, the pH past the equivalence point is determined by the amount of excess strong base. [OH−]excess = 82.
5.0 mmol 100 .0 mL + 250 .0 mL
= 0.014 M; pOH = 1.85; pH = 12.15
This is a weak acid (HCN) titrated by a strong base (KOH). a. Only weak acid is present. Solving the weak acid problem: HCN Initial Change Equil.
⇌
H+
+
CN−
0.100 M ~0 0 x mol/L HCN dissociates to reach equilibrium −x → +x +x 0.100 − x x x
Ka = 6.2 × 10−10 =
x2 x2 , x = [H+] = 7.9 × 10−6 M 0.100 − x 0.100
pH = 5.10; assumptions good.
CHAPTER 15
ACID-BASE EQUILIBRIA
757
b. The added OH− will react completely with the best acid present, HCN. mmol HCN present = 100.0 mL × mmol OH− added = 50.0 mL × HCN Before Change After
+
10.0 mmol −5.00 mmol 5.0 mmol
0.100 mmol HCN = 10.0 mmol HCN mL
0.100 mmol OH − = 5.00 mmol OH− mL
OH−
→
5.00 mmol −5.00 mmol → 0
CN− + H2O 0 +5.00 mmol 5.00 mmol
Reacts Completely
After reaction of all the strong base, we have a buffer solution containing a weak acid (HCN) and its conjugate base (CN−). We will use the Henderson-Hasselbalch equation to solve for the pH where VT = total volume of buffer solution. pH = pKa + log
5.00 mmol/V T [CN − ] = −log (6.2 × 10−10) + log 5.0 mmol/V [HCN] T
5.00 pH = 9.21 + log = 9.21 + 0.0 = 9.21 5.0
This is the halfway point to equivalence. In a weak acid-strong base titration, pH = pKa at the halfway point to equivalence because [weak acid] = [conjugate base] at this point. Also note that the total volume cancels in the Henderson-Hasselbalch equation. For the [base]/[acid] term, the mole ratio equals the concentration ratio because the components of the buffer are always in the same volume of solution. c. mmol OH− added = 75.0 mL × 0.100 M = 7.50 mmol OH−; the same amount (10.0 mmol) of HCN is present as before (it doesn’t change). Let the OH − react to completion, then see what is remaining in solution after this reaction. HCN Before After
+
10.0 mmol 2.5 mmol
OH−
→
7.50 mmol 0
CN− + H2O 0 7.50 mmol
We have a buffer solution after all the OH− reacts to completion. Using the HendersonHasselbalch equation: pH = 9.21 + log
7.50 mmol [CN − ] = 9.21 + log [HCN] 2.5 mmol
pH = 9.21 + 0.48 = 9.69 d. The equivalence point occurs at 100.0 mL of KOH. mmol OH − added = 100.00 mL × 0.100 M = 10.0 mmol OH−; as before, let the added OH− react to completion with the weak acid; then see what is in solution after this reaction.
758
CHAPTER 15 HCN Before After
→
OH−
+
10.0 mmol 0
10.0 mmol 0
ACID-BASE EQUILIBRIA
CN− + H2O 0 10.0 mmol
As expected at the equivalence point, enough OH − has been added to exactly neutralize all the weak acid present initially. All that remains that affects the pH at the equivalence point is the conjugate base of the weak acid (CN−). This is a weak base equilibrium problem. CN− Initial Change Equil.
+ H2 O
⇌
HCN
+
OH−
10.0 mmol/200.0 mL 0 0 x mol/L CN− reacts with H2O to reach equilibrium −x → +x +x 0.0500 − x x x
Kb = 1.6 × 10−5 =
Kw 1.0 10 −14 = Kb 6.2 10 −10 Kb = 1.6 × 10−5
Kb =
x2 x2 , x = [OH−] = 8.9 × 10−4 M 0.0500 − x 0.0500
pOH = 3.05; pH = 10.95; assumptions good. e. mmol OH− added = 125.0 mL × 0.100 M = 12.5 mmol OH− HCN Before After
+
10.0 mmol 0
OH−
→
CN− + H2O
12.5 mmol 2.5 mmol
0 10.0 mmol
After the titration reaction, we have a solution containing excess OH − and a weak base CN−. When a strong base and a weak base are both present, assume that the amount of OH− added from the weak base will be minimal; that is, the pH past the equivalence point is determined by the amount of excess strong base. [OH−]excess = 83.
2.5 mmol 100.0 mL + 125.0 mL
= 0.011 M; pOH = 1.96; pH = 12.04
This is a weak base (H2NNH2) titrated by a strong acid (HNO3). To calculate the pH at the various points, let the strong acid react completely with the weak base present; then see what is in solution. a. Only a weak base is present. Solve the weak base equilibrium problem. H2NNH2 + H2O ⇌ H2NNH3+ + OH− Initial Equil.
0.100 M 0.100 − x
Kb = 3.0 × 10−6 =
0 x
~0 x
x2 x2 , x = [OH−] = 5.5 × 10−4 M 0.100 − x 0.100
pOH = 3.26; pH = 10.74; assumptions good.
CHAPTER 15
ACID-BASE EQUILIBRIA
759 0.100 mmol H 2 NNH 2 = 10.0 mmol H2NNH2 mL
b. mmol H2NNH2 present = 100.0 mL × mmol H+ added = 20.0 mL × H2NNH2 Before Change After
+
10.0 mmol −4.00 mmol 6.0 mmol
0.200 mmol H + = 4.00 mmol H+ mL
H+ 4.00 mmol −4.00 mmol 0
→
H2NNH3+
→
0 +4.00 mmol Reacts completely 4.00 mmol
A buffer solution results after the titration reaction. Solving using the HendersonHasselbalch equation: pH = pKa + log
K 1.0 10 −14 [base] ; Ka = w = = 3.3 × 10−9 Kb [acid] 3.0 10 −6
6.0 mmol/V T , where VT = total volume (cancels out) pH = −log(3.3 × 10−9) + log 4.00 mmol/V T
pH = 8.48 + log(1.5) = 8.48 + 0.18 = 8.66 c. mmol H+ added = 25.0 mL × 0.200 M = 5.00 mmol H+ H2NNH2 Before After
+
10.0 mmol 5.0 mmol
→
H+ 5.00 mmol 0
H2NNH3+ 0 5.00 mmol
This is the halfway point to equivalence, where [H2NNH3+] = [H2NNH2]. At this point, pH = pKa (which is characteristic of the halfway point for any weak base-strong acid titration). pH = −log(3.3 × 10−9) = 8.48 d. mmol H+ added = 40.0 mL × 0.200 M = 8.00 mmol H+ H2NNH2 Before After
10.0 mmol 2.0 mmol
+
H+
→
8.00 mmol 0
H2NNH3+ 0 8.00 mmol
A buffer solution results. pH = pKa + log
2.0 mmol / VT [base] = 8.48 + (–0.60) = 7.88 = 8.48 + log [acid] 8.00 mmol / VT
e. mmol H+ added = 50.0 mL × 0.200 M = 10.0 mmol H+
760
CHAPTER 15 H2NNH2 Before After
→
H+
+
10.0 mmol 0
ACID-BASE EQUILIBRIA H2NNH3+
10.0 mmol 0
0 10.0 mmol
As is always the case in a weak base-strong acid titration, the pH at the equivalence point is acidic because only a weak acid (H2NNH3+) is present. Solving the weak acid equilibrium problem:
⇌
H+
10.0 mmol/150.0 mL 0.0667 − x
0 x
H2NNH3+ Initial Equil.
Ka = 3.3 × 10−9 =
+
H2NNH2 0 x
x2 x2 , x = [H+] = 1.5 × 10−5 M 0.0667 − x 0.0667
pH = 4.82; assumptions good. f.
mmol H+ added = 100.0 mL × 0.200 M = 20.0 mmol H+ H2NNH2 Before After
→
H+
+
10.0 mmol 0
H2NNH3+
20.0 mmol 10.0 mmol
0 10.0 mmol
Two acids are present past the equivalence point, but the excess H+ will determine the pH of the solution since H2NNH3+ is a weak acid. [H+]excess = 84.
10.0 mmol 100 .0 mL + 100 .0 mL
= 0.0500 M; pH = 1.301
This is a weak base (HONH2) titrated by a strong acid (HCl). To calculate the pH at the various points, let the strong acid react completely with the weak base present; then see what is in solution. a. Only a weak base is present. Solve the weak base equilibrium problem. HONH2 Initial Equil.
+ H2O ⇌ HONH3+ + OH−
0.200 M 0.200 – x
Kb = 1.1 × 10−8 =
0 x
~0 x
x2 x2 , x = [OH−] = 4.7 × 10−5 M 0.200 − x 0.200
pOH = 4.33; pH = 9.67; assumptions good. b. mmol HONH2 present = 100.0 mL × mmol H+ added = 25.0 mL ×
0.200 mmol HONH 2 = 20.0 mmol HONH2 mL
0.100 mmol H + = 2.50 mmol H+ mL
CHAPTER 15
ACID-BASE EQUILIBRIA HONH2
Before Change After
761
H+
+
20.0 mmol −2.50 mmol 17.5 mmol
2.50 mmol −2.50 mmol 0
→
HONH3+
→
0 +2.50 mmol Reacts completely 2.50 mmol
A buffer solution results after the titration reaction. Solving using the HendersonHasselbalch equation: pH = pKa + log
K 1.0 10 −14 [base] ; Ka = w = Kb [acid] 1.1 10 −8
= 9.1 × 10−7
17.5 mmol/V T , where VT = total volume (cancels out) pH = −log(9.1 × 10−7) + log 2.50 mmol/V T
pH = 6.04 + log(7.00) = 6.04 + 0.845 = 6.89 c. mmol H+ added = 70.0 mL × 0.100 M = 7.00 mmol H+ HONH2 Before After
→
H+
+
20.0 mmol 13.0 mmol
HONH3+
7.00 mmol 0
0 7.00 mmol
A buffer solution results. pH = pKa + log
13.0 mmol/V T [base] = 6.04 + (0.269) = 6.31 = 6.04 + log [acid] 7.00 mmol/V T
d. The equivalence point occurs at 200.0 mL of HCl added. mmol H + added = 200.0 mL × 0.100 M = 20.0 mmol H+ HONH2 Before After
H+
+
20.0 mmol 0
→
HONH3+
20.0 mmol 0
0 20.0 mmol
As is always the case in a weak base-strong acid titration, the pH at the equivalence point is acidic because only a weak acid (HONH 3+) is present. Solving the weak acid equilibrium problem:
⇌
H+
20.0 mmol/300.0 mL 0.0667 − x
0 x
HONH3+ Initial Equil.
Ka = 9.1 × 10−7 =
+
HONH2 0 x
x2 x2 , x = [H+] = 2.5 × 10−4 M 0.0667 − x 0.0667
pH = 3.60; assumptions good. e. mmol H+ added = 300.0 mL × 0.100 M = 30.0 mmol H+
762
CHAPTER 15 HONH2 Before After
→
H+
+
20.0 mmol 0
30.0 mmol 10.0 mmol
ACID-BASE EQUILIBRIA
HONH3+ 0 20.0 mmol
Two acids are present past the equivalence point, but the excess H+ will determine the pH of the solution since HONH3+ is a weak acid. [H+]excess = f.
85.
10.0 mmol 100.0 mL + 300.0 mL
= 0.0250 M; pH = 1.602
Notice that the pKa value for HONH3+ is equal to 6.04. This question is asking when does pH = pKa. For a weak base-strong acid titration, this always occurs at the halfway point to equivalence, where [HONH3+] = [HONH2]. It takes 200.0 mL of HCl to reach the equivalence point, so it will take 100.0 mL of HCl to reach the halfway point to equivalence where pH = pKa = 6.04.
We will do sample calculations for the various parts of the titration. All results are summarized in Table 15.1 at the end of Exercise 88. At the beginning of the titration, only the weak acid HC 3H5O3 is present. Let HLac = HC3H5O3 and Lac− = C3H5O3−. HLac
⇌
H+
+
Lac−
Ka = 10−3.86 = 1.4 × 10−4
Initial
0.100 M ~0 0 x mol/L HLac dissociates to reach equilibrium Change −x → +x +x Equil. 0.100 − x x x 1.4 × 10−4 =
x2 x2 , x = [H+] = 3.7 × 10−3 M; pH = 2.43; assumptions good. 0.100 − x 0.100
Up to the stoichiometric point, we calculate the pH using the Henderson-Hasselbalch equation. This is the buffer region. For example, at 4.0 mL of NaOH added: initial mmol HLac present = 25.0 mL × mmol OH− added = 4.0 mL ×
0.100 mmol = 2.50 mmol HLac mL
0.100 mmol = 0.40 mmol OH− mL
Note: The units millimoles are usually easier numbers to work with. The units for molarity are moles per liter but are also equal to millimoles per milliliter. The 0.40 mmol of added OH− converts 0.40 mmol HLac to 0.40 mmol Lac− according to the equation: HLac + OH− → Lac− + H2O
Reacts completely since a strong base is added.
mmol HLac remaining = 2.50 − 0.40 = 2.10 mmol; mmol Lac− produced = 0.40 mmol We have a buffer solution. Using the Henderson-Hasselbalch equation where pKa = 3.86:
CHAPTER 15
ACID-BASE EQUILIBRIA
pH = pKa + log
763
(0.40 ) [Lac− ] = 3.86 + log ( 2.10 ) [HLac]
(Total volume cancels, so we can use the ratio of moles or millimoles.) pH = 3.86 − 0.72 = 3.14 Other points in the buffer region are calculated in a similar fashion. Perform a stoichiometry problem first, followed by a buffer problem. The buffer region includes all points up to and including 24.9 mL OH− added. At the stoichiometric point (25.0 mL OH− added), we have added enough OH−to convert all the HLac (2.50 mmol) into its conjugate base (Lac−). All that is present is a weak base. To determine the pH, we perform a weak base calculation. [Lac−]0 =
2.50 mmol = 0.0500 M 25.0 mL + 25.0 mL
Lac− + H2O
⇌
HLac
+
OH−
Kb =
1.0 10 −14 = 7.1 × 10−11 1.4 10 − 4
Initial
0.0500 M 0 0 x mol/L Lac− reacts with H2O to reach equilibrium Change −x → +x +x Equil. 0.0500 − x x x Kb =
x2 x2 = 7.1 × 10−11 0.0500 − x 0.0500
x = [OH−] = 1.9 × 10−6 M; pOH = 5.72; pH = 8.28; assumptions good. Past the stoichiometric point, we have added more than 2.50 mmol of NaOH. The pH will be determined by the excess OH- ion present. An example of this calculation follows. At 25.1 mL: OH− added = 25.1 mL ×
0.100 mmol = 2.51 mmol OH− mL
2.50 mmol OH− neutralizes all the weak acid present. The remainder is excess OH −. Excess OH− = 2.51 − 2.50 = 0.01 mmol OH− [OH−]excess =
0.01 mmol (25.0 + 25.1) mL
= 2 × 10−4 M; pOH = 3.7; pH = 10.3
All results are listed in Table 15.1 at the end of the solution to Exercise 88. 86.
Results for all points are summarized in Table 15.1 at the end of the solution to Exercise 88. At the beginning of the titration, we have a weak acid problem:
764
CHAPTER 15
⇌
HOPr Initial Change Equil. Ka =
H+
+
OPr−
ACID-BASE EQUILIBRIA
HOPr = HC3H5O2, OPr− = C3H5O2−
0.100 M ~0 0 x mol/L HOPr acid dissociates to reach equilibrium −x → +x +x 0.100 − x x x
[H + ][OPr− ] x2 x2 = 1.3 × 10−5 = 0.100 − x 0.100 [HOPr]
x = [H+] = 1.1 × 10−3 M; pH = 2.96;
assumptions good.
The buffer region is from 4.0 to 24.9 mL of OH− added. We will do a sample calculation at 24.0 mL OH− added. 0.100 mmol Initial mmol HOPr present = 25.0 mL × = 2.50 mmol HOPr mL mmol OH− added = 24.0 mL ×
0.100 mmol = 2.40 mmol OH− mL
The added strong base converts HOPr into OPr−. HOPr Before Change After
+
2.50 mmol −2.40 0.10 mmol
OH−
→
2.40 mmol −2.40 → 0
OPr-
+
H2O
0 +2.40 2.40 mmol
Reacts completely
A buffer solution results. Using the Henderson-Hasselbalch equation where pKa = −log(1.3 × 10−5) = 4.89: pH = pKa + log
[OPr− ] [base] = 4.89 + log [acid] [HOPr]
2.40 pH = 4.89 + log = 4.89 + 1.38 = 6.27 0.10
(Volume cancels, so we can use the millimole ratio in the log term.) All points in the buffer region 4.0 mL to 24.9 mL are calculated this way. See Table 15.1 at the end of Exercise 88 for all the results. At the stoichiometric point (25.0 mL NaOH added), only a weak base (OPr−) is present: OPr− Initial
Change Equil.
+
H2O
2.50 mmol = 0.0500 M 50.0 mL
⇌
OH−
+ HOPr
0
0
x mol/L OPr− reacts with H2O to reach equilibrium −x → +x +x 0.0500 − x x x
CHAPTER 15 Kb =
ACID-BASE EQUILIBRIA
765
[OH− ][HOPr] K w x2 x2 = 7.7 × 10−10 = = − 0.0500 − x 0.0500 Ka [OPr ]
x = 6.2 × 10−6 M = [OH−], pOH = 5.21, pH = 8.79; assumptions good. Beyond the stoichiometric point, the pH is determined by the excess strong base added. The results are the same as those in Exercise 85 (see Table 15.1). For example, at 26.0 mL NaOH added: [OH−] = 87.
2.60 mmol − 2.50 mmol = 2.0 × 10−3 M; pOH = 2.70; pH = 11.30 (25.0 + 26.0) mL
At beginning of the titration, only the weak base NH 3 is present. As always, solve for the pH using the Kb reaction for NH3. NH3 + H2O Initial Equil. Kb =
⇌
NH4+
0.100 M 0.100 − x
0 x
+
OH−
Kb = 1.8 × 10−5
~0 x
x2 x2 = 1.8 × 10−5 0.100 − x 0.100
x = [OH−] = 1.3 × 10−3 M; pOH = 2.89; pH = 11.11; assumptions good. In the buffer region (4.0 − 24.9 mL), we can use the Henderson-Hasselbalch equation: Ka =
[NH 3 ] 1.0 10 −14 = 5.6 × 10−10; pKa = 9.25; pH = 9.25 + log −5 + 1.8 10 [NH 4 ]
We must determine the amounts of NH3 and NH4+ present after the added H+ reacts completely with the NH3. For example, after 8.0 mL HCl added: initial mmol NH3 present = 25.0 mL × mmol H+ added = 8.0 mL ×
0.100 mmol = 2.50 mmol NH3 mL
0.100 mmol = 0.80 mmol H+ mL
Added H+ reacts with NH3 to completion: NH3 + H+ → NH4+ mmol NH3 remaining = 2.50 − 0.80 = 1.70 mmol; mmol NH4+ produced = 0.80 mmol pH = 9.25 + log
1.70 = 9.58 (Mole ratios can be used since the total volume cancels.) 0.80
Other points in the buffer region are calculated in similar fashion. Results are summarized in Table 15.1 at the end of Exercise 88.
766
CHAPTER 15
ACID-BASE EQUILIBRIA
At the stoichiometric point (25.0 mL H+ added), just enough HCl has been added to convert all the weak base (NH3) into its conjugate acid (NH4+). Perform a weak acid calculation. [NH4+]0 = 2.50 mmol/50.0 mL = 0.0500 M
⇌
NH4+ Initial Equil.
0.0500 M 0.0500 - x
5.6 × 10−10 =
H+ 0 x
+
NH3
Ka = 5.6 × 10−10
0 x
x2 x2 , x = [H+] = 5.3 × 10−6 M; pH = 5.28 0.0500 − x 0.0500
Assumptions good. Beyond the stoichiometric point, the pH is determined by the excess H +. For example, at 28.0 mL of H+ added: H+ added = 28.0 mL ×
0.100 mmol = 2.80 mmol H+ mL
Excess H+ = 2.80 mmol − 2.50 mmol = 0.30 mmol excess H+ [H+]excess =
0.30 mmol = 5.7 × 10−3 M; pH = 2.24 ( 25.0 + 28.0) mL
All results are summarized in Table 15.1 at the end of Exercise 88. 88.
Initially, a weak base problem: py Initial Equil. Kb =
+
0.100 M 0.100 − x
H2O
⇌
Hpy+
+
0 x
OH−
py is pyridine.
~0 x
[Hpy+ ][OH− ] x2 x2 1.7 × 10−9 = 0.100 − x [py] 0.100
x = [OH−] = 1.3 × 10−5 M; pOH = 4.89; pH = 9.11; assumptions good. Buffer region (4.0 − 24.5 mL): Added H+ reacts completely with py: py + H+ → Hpy+. Determine the moles (or millimoles) of py and Hpy+ after reaction, then use the HendersonHasselbalch equation to solve for the pH. Ka =
[py] Kw 1.0 10 −14 = = 5.9 × 10−6; pKa = 5.23; pH = 5.23 + log −9 Kb [Hpy + ] 1.7 10
Results in the buffer region are summarized in Table 15.1, which follows this problem. See Exercise 87 for a similar sample calculation.
CHAPTER 15
ACID-BASE EQUILIBRIA
767
At the stoichiometric point (25.0 mL H+ added), this is a weak acid problem since just enough H+ has been added to convert all the weak base into its conjugate acid. The initial concentration of [Hpy+] = 0.0500 M. Hpy+ Initial Equil.
0.0500 M 0.0500 − x
5.9 × 10−6 =
⇌
py
H+
+
0 x
Ka = 5.9 × 10−6
0 x
x2 x2 , x = [H+] = 5.4 × 10−4 M; pH = 3.27; assumptions good. 0.0500 − x 0.0500
Beyond the equivalence point, the pH determination is made by calculating the concentration of excess H+. See Exercise 87 for an example. All results are summarized below in Table 15.1. Table 15.1 Summary of pH Results for Exercises 85-88 (Plot below.) Titrant
Exercise
Exercise
Exercise
Exercise
mL 85 86 87 88 ______________________________________________________ 0.0 4.0 8.0 12.5 20.0 24.0 24.5 24.9 25.0 25.1 26.0 28.0 30.0
2.43 3.14 3.53 3.86 4.46 5.24 5.6 6.3 8.28 10.3 11.30 11.75 11.96
2.96 4.17 4.56 4.89 5.49 6.27 6.6 7.3 8.79 10.3 11.30 11.75 11.96
11.11 9.97 9.58 9.25 8.65 7.87 7.6 6.9 5.28 3.7 2.71 2.24 2.04
9.11 5.95 5.56 5.23 4.63 3.85 3.5 − 3.27 − 2.71 2.24 2.04
768 89.
CHAPTER 15
ACID-BASE EQUILIBRIA
For the titration of HCl by RbOH, the pH is very acidic at the halfway point and pH = 7.00 at the equivalence point for this strong-acid strong base titration. The other titrations are weak acid-strong base titrations. For these titrations, pH = pKa at the halfway point to equivalence since a buffer solution having equal concentrations of weak acid and conjugate base are present. The acid with the smallest Ka value will have the highest pH at the halfway point. This is HCN. At the equivalence point of a weak acid-strong base titration, the major species present with acid-base properties is the conjugate base of the weak acid titrated. The titration with the highest pH (most basic pH) will have the conjugate base with the largest Kb value. For conjugate acid-base pairs, the weakest acid has the strongest conjugate base. Of the weak acids titrated, HCN is the weakest acid. So HCN will have the strongest conjugate base and the highest pH at the equivalence point.
90.
For the titration of NaOH by HNO3, the pH is very basic at the halfway point and pH = 7.00 at the equivalence point for this strong base-strong acid titration. The other titrations are weak base-strong acid titrations. At the halfway point for these titrations, a buffer solution is present that has equal concentrations of the weak base and its conjugate acid; at the halfway point, pH = pKa of the conjugate acid of the weak base titrated. The titration with the lowest pH at the halfway point has the strongest conjugate acid present. The strongest conjugate acid comes from the weakest base titrated. This is C 6H5NH2. At the equivalence point of a weak base-strong acid titration, the major species present with acid-base properties is the conjugate acid of the weak base titrated. The titration with the lowest pH (most acidic pH) will have the conjugate acid with the largest Ka value. For conjugate acidbase pairs, the weaker the base, the stronger the conjugate acid. Of the weak bases titrated, C6H5NH2 is the weakest base. So C6H5NH2 will have the strongest conjugate acid and the lowest pH at the equivalence point.
91.
75.0 mL ×
0.10 mmol 0.10 mmol = 7.5 mmol HA; 30.0 mL × = 3.0 mmol OH− added mL mL
The added strong base reacts to completion with the weak acid to form the conjugate base of the weak acid and H2O. HA Before After
+
7.5 mmol 4.5 mmol
OH− 3.0 mmol 0
→
A−
+
H2 O
0 3.0 mmol
A buffer results after the OH− reacts to completion. Using the Henderson-Hasselbalch equation: 3.0 mmol/105.0 mL [A − ] pH = pKa + log , 5.50 = pKa + log [HA] 4.5 mmol/105.0 mL pKa = 5.50 –log(3.0/4.5) = 5.50 – (–0.18) = 5.68; Ka = 10 −5.68 = 2.1 × 10 −6 92.
Mol H+ added = 0.0400 L × 0.100 mol/L = 0.00400 mol H + The added strong acid reacts to completion with the weak base to form the conjugate acid of the weak base and H2O. Let B = weak base:
CHAPTER 15
ACID-BASE EQUILIBRIA B
Before After
+
0.0100 mol 0.0060
H+
769 →
0.00400 mol 0
BH+ 0 0.0400 mol
After the H+ reacts to completion, we have a buffer solution. Using the Henderson-Hasselbalch equation: pH = pKa + log
(0.0060/V T ) [base] , 8.00 = pKa + log , (0.00400/V T ) [acid]
where VT = total volume of solution. pKa = 8.00 − log
(0.0060) = 8.00 – 0.18, pKa = 7.82 (0.00400)
For a conjugate acid-base pair, pKa + pKb = 14,00, so: pKb = 14.00 – 7.82 = 6.18; Kb = 10−6.18 = 6.6 × 10−7
Indicators 93.
HIn ⇌ In− + H+
Ka =
[In − ][H + ] = 1.0 × 10 −9 [HIn]
a. In a very acid solution, the HIn form dominates, so the solution will be yellow. b. The color change occurs when the concentration of the more dominant form is approximately ten times as great as the less dominant form of the indicator. [HIn] −
[In ]
=
10 1 ; Ka = 1.0 × 10 −9 = [H+], [H+] = 1 × 10 −8 M; pH = 8.0 at color change 1 10
c. This is way past the equivalence point (100.0 mL OH− added), so the solution is very basic, and the In− form of the indicator dominates. The solution will be blue. 94.
The color of the indicator will change over the approximate range of pH = pK a ± 1 = 5.3 ± 1. Therefore, the useful pH range of methyl red where it changes color would be about 4.3 (red) to 6.3 (yellow). Note that at pH < 4.3, the HIn form of the indicator dominates, and the color of the solution is the color of HIn (red). At pH > 6.3, the In- form of the indicator dominates, and the color of the solution is the color of In − (yellow). In titrating a weak acid with base, we start off with an acidic solution with pH < 4.3, so the color would change from red to reddish orange at pH 4.3. In titrating a weak base with acid, the color change would be from yellow to yellowish orange at pH 6.3. Only a weak base-strong acid titration would have an acidic pH at the equivalence point, so only in this type of titration would the color change of methyl red indicate the approximate endpoint.
95.
At the equivalence point, P2− is the major species. P2− is a weak base in water because it is the conjugate base of a weak acid.
770
CHAPTER 15 P2−
+
H2 O
⇌
HP−
+
ACID-BASE EQUILIBRIA OH−
Initial
0 .5 g 1 mol = 0.024 M 0 .1 L 204 .2 g
0
~0 (carry extra sig. fig.)
Equil.
0.024 − x
x
x
Kb =
Kw [HP − ][OH − ] 1.0 10 −14 x2 x2 −9 = = , 3.2 × 10 = Ka 0.024 − x 0.024 10 −5.51 P 2−
x = [OH−] = 8.8 × 10−6 M; pOH = 5.1; pH = 8.9; assumptions good. Phenolphthalein would be the best indicator for this titration because it changes color at around pH 9 (from acid color to base color). 96.
HIn ⇌ In− + H+
Ka =
[In − ][H + ] = 10 −3.00 = 1.0 × 10−3 [HIn]
At 7.00% conversion of HIn into In −, [In−]/[HIn] = 7.00/93.00. Ka = 1.0 × 10 −3 =
[In − ] 7.00 [H + ] = [H + ], [H+] = 1.3 × 10 −2 M, pH = 1.89 [HIn] 93.00
The color of the base form will start to show when the pH is increased to 1.89. 97.
98.
99.
100.
When choosing an indicator, we want the color change of the indicator to occur approximately at the pH of the equivalence point. Because the pH generally changes very rapidly at the equivalence point, we don’t have to be exact. This is especially true for strong acid-strong base titrations. The following are some indicators where the color change occurs at about the pH of the equivalence point: Exercise
pH at Eq. Pt.
77 81
7.00 8.79
Exercise
pH at Eq. Pt.
79 83
7.00 4.82
Exercise
pH at Eq. Pt.
85 87
8.28 5.28
Exercise
pH at Eq. Pt.
86 88
8.79 3.27
Indicator bromthymol blue or phenol red o-cresolphthalein or phenolphthalein Indicator bromthymol blue or phenol red bromcresol green Indicator o-cresolphthalein or phenolphthalein bromcresol green Indicator o-cresolphthalein or phenolphthalein 2,4-dinitrophenol
In the titration in Exercise 88, it will be very difficult to mark the equivalence point. The pH break at the equivalence point is too small.
CHAPTER 15
ACID-BASE EQUILIBRIA
771
101.
pH > 5 for bromcresol green to be blue. pH < 8 for thymol blue to be yellow. The pH is between 5 and 8.
102.
The pH will be less than about 0.5 because crystal violet is yellow at a pH less than about 0.5. The methyl orange result only tells us that the pH is less than about 3.5.
103.
a. yellow
104.
We want an indicator that changes color between the acid and base form at pH ≈ 4.5. From Figure 15.8, this would be bromcresol green. At pH = 4.0, the yellow acid form dominates and at pH = 5.0, the blue form dominates. At a pH between 4.0 and 5.0, both forms are present, and the color of the indicator is green, indicating a pH ≈ 4.5.
b.
green (Both yellow and blue forms are present.)
c. yellow
d. blue
Polyprotic Acids 105.
The first titration plot (from 0 − 100.0 mL) corresponds to the titration of H 2A by OH−. The reaction is H2A + OH− → HA− + H2O. After all the H2A has been reacted, the second titration (from 100.0 – 200.0 mL) corresponds to the titration of HA − by OH−. The reaction is HA− + OH− → A2− + H2O. a. At 100.0 mL of NaOH, just enough OH− has been added to react completely with all the H2A present (mol OH− added = mol H2A present initially). From the balanced equation, the mol of HA− produced will equal the mol of H2A present initially. Because mol of HA− present at 100.0 mL OH− added equals the mol of H2A present initially, exactly 100.0 mL more of NaOH must be added to react with all the HA−. The volume of NaOH added to reach the second equivalence point equals 100.0 mL + 100.0 mL = 200.0 mL. b. H2A + OH− → HA− + H2O is the reaction occurring from 0 − 100.0 mL NaOH added. HA− + OH− → A2− + H2O is the reaction occurring from 100.0 − 200.0 mL NaOH added. i.
No reaction has taken place, so H2A and H2O are the major species.
ii. Adding OH− converts H2A into HA−. The major species between 0 and 100.0 mL NaOH added are H2A, HA−, H2O, and Na+. iii. At 100.0 mL NaOH added, mol of OH− = mol H2A, so all of the H2A present initially has been converted into HA−. The major species are HA−, H2O, and Na+. iv. Between 100.0 and 200.0 mL NaOH added, the OH− converts HA− into A2−. The major species are HA−, A2− , H2O, and Na+. v. At the second equivalence point (200.0 mL), just enough OH − has been added to convert all the HA− into A2−. The major species are A2−, H2O, and Na+. vi. Past 200.0 mL NaOH added, excess OH− is present. The major species are OH−, A2−, H2O, and Na+.
772
CHAPTER 15
ACID-BASE EQUILIBRIA
c. 50.0 mL of NaOH added corresponds to the first halfway point to equivalence. Exactly one-half of the H2A present initially has been converted into its conjugate base HA−, so [H2A] = [HA−] in this buffer solution. H2A ⇌ HA− + H+
[HA − ][H + ] [H 2 A]
Ka = 1
When [HA−] = [H2A], then K a1 = [H+] or pK a1 = pH. Here, pH = 4.0, so pK a = 4.0 and K a = 10 −4.0 = 1 × 10 −4 . 1
1
150.0 mL of NaOH added correspond to the second halfway point to equivalence, where [HA−] = [A2−] in this buffer solution. HA− ⇌ A2− + H+
Ka = 2
[A 2− ][H + ] [HA − ]
When [A2−] = [HA−], then K a = [H+] or pK a = pH. 2
2
Here, pH = 8.0, so pK a = 8.0 and K a 2 = 10−8.0 = 1 × 10 −8 . 2
106.
It will take 100.0 mL of KOH to reach the first equivalence point and 200.0 mL of KOH to reach the second equivalence point in this diprotic acid titration. a. 50.0 mL of KOH added is the first halfway point to equivalence where exactly one-half of the H2A present initially has been converted into its conjugate base HA −. So [H2A] = [HA−] in this buffer solution, and pH = pK a1 : pH = ‒log(7.9 × 10‒5) = 4.10 b. 150.0 mL of KOH added is the second halfway point to equivalence where exactly onehalf of the HA‒ present at the first equivalence point has been converted into its conjugate base A2−. So [HA−] = [A2−] in this buffer solution, and pH = pK a 2 : pH = ‒log(1.6 × 10‒12) = 11.80
107.
For a triprotic titration, there are three titrations to consider. In the first titration, added OH − converts H3A into H2A−. Because the molarities of H3A and KOH are equal, the first equivalence point will be at 50.0 mL of KOH added. So from 0 – 50.0 mL of KOH added, the first proton is removed from the triprotic acid. From 50.0 – 100.0 mL of KOH added, OH− converts H2A− into HA2−, and from 100.0 – 150.0 mL of KOH added, OH − converts HA2− into A3−. a. 125.0 mL of KOH added is the third halfway point to equivalence where [HA 2−] = [A3−]. At this point we have a buffer solution where pH = pK a 3 . pH = −log(1.0 10−11) = 11.00.
CHAPTER 15 b.
ACID-BASE EQUILIBRIA
773
pK a1 = −log(5.0 10−4) = 3.30; pH = pK a1 = 3.30 at the first halfway point to equivalence. This occurs at 25.0 mL of KOH added. Here, we have a buffer solution where [H3A] = [H2A−], so pH = pK a1 .
c. 75.0 mL of KOH added is the second halfway point to equivalence where [H 2A−] = [HA2−]. At this buffer point, pH = pK a 2 . pH = −log(1.0 10−8) = 8.00; the pH will be basic at 75.0 mL of KOH added. 108.
a. At the first equivalence point, just enough NaOH has been added to react with all the triprotic acid present initially. Since the molarity of the NaOH solution is twice the molarity of the triprotic acid, it will take half the volume of the NaOH solution to deliver just enough NaOH to react with all the acid. So, it takes 50.0 mL of NaOH to reach the first equivalence point, 100.0 mL of NaOH to reach the second equivalence point, and 150.0 mL of NaOH to reach the third equivalence point. b. 125.0 mL of NaOH added is the third halfway point to equivalence point when pH = pK a3 ; pH = ‒log(5.1 × 10‒12) = 11.29 c. The K a 2 for arsenic acid is 1.7 × 10‒7. The H+ concentration will equal 1.7 × 10‒7 M when we have reached the second halfway point to equivalence where [H2AsO4‒] = [HAsO42‒]. The second hallway point to equivalence occurs at 75.0 mL of NaOH added.
ChemWork Problems 109.
NH3 + H2O ⇌ NH4+ + OH− Kb =
+
[NH 4 ][OH − ] ; taking the −log of the Kb expression: [NH 3 ] +
−log Kb = −log[OH−] − log
+
[NH 4 ] [NH 4 ] , −log[OH−] = −log Kb + log [NH 3 ] [NH 3 ]
+
pOH = pKb + log 110.
a.
[NH 4 ] [acid] or pOH = pKb + log [NH 3 ] [base]
pH = pKa = −log(6.4 × 10−5) = 4.19 since [HBz] = [Bz−], where HBz = C6H5CO2H and [Bz−] = C6H5CO2−.
b. [Bz−] will increase to 0.120 M and [HBz] will decrease to 0.080 M after OH− reacts completely with HBz. The Henderson-Hasselbalch equation is derived from the Ka dissociation reaction. pH = pKa + log c.
(0.120 ) [Bz − ] , pH = 4.19 + log = 4.37; assumptions good. (0.080 ) [HBz]
Bz− Initial 0.120 M Equil. 0.120 − x
+
H2O
⇌
HBz 0.080 M 0.080 + x
+
OH− 0 x
774
CHAPTER 15 Kb =
ACID-BASE EQUILIBRIA
Kw (0.080 + x)( x) (0.080 )( x) 1.0 10 −14 = = −5 (0.120 − x) Ka 0.120 6.4 10
x = [OH−] = 2.34 × 10−10 M (carrying extra sig. fig.); assumptions good. pOH = 9.63; pH = 4.37 d. We get the same answer. Both equilibria involve the two major species, benzoic acid and benzoate anion. Both equilibria must hold true. K b is related to Ka by Kw and [OH−] is related to [H+] by Kw, so all constants are interrelated. 111.
a. The optimum pH for a buffer is when pH = pKa. At this pH a buffer will have equal neutralization capacity for both added acid and base. As shown next, because the pK a for TRISH+ is 8.1, the optimal buffer pH is about 8.1. Kb = 1.19 × 10−6; Ka = Kw/Kb = 8.40 × 10−9; pKa = −log(8.40 × 10−9) = 8.076 b. pH = pKa + log
[TRIS] [TRIS] , 7.00 = 8.076 + log + [TRISH ] [TRISH + ]
[TRIS] = 10-1.08 = 0.083 (at pH = 7.00) [TRISH + ]
9.00 = 8.076 + log
c.
[TRIS] [TRIS] , = 100.92 = 8.3 (at pH = 9.00) + + [TRISH ] [TRISH ]
50.0 g TRIS 1 mol = 0.206 M = 0.21 M = [TRIS] 2.0 L 121.14 g 65.0 g TRISHCl 1 mol = 0.206 M = 0.21 M = [TRISHCl] = [TRISH+] 2 .0 L 157 .60 g
pH = pKa + log
(0.21) [TRIS] = 8.076 + log = 8.08 + (0.21) [TRISH ]
The amount of H+ added from HCl is: (0.50 × 10−3 L) × 12 mol/L = 6.0 × 10−3 mol H+ The H+ from HCl will convert TRIS into TRISH+. The reaction is: TRIS Before
0.21 M
Change After
−0.030 0.18
+
H+ → −3 6.0 10 = 0.030 M 0.2005 −0.030 → 0
TRISH+ 0.21 M +0.030 0.24
Reacts completely
Now use the Henderson-Hasselbalch equation to solve this buffer problem. 0.18 pH = 8.076 + log = 7.95 0.24
CHAPTER 15
ACID-BASE EQUILIBRIA −
−
112.
775
[C H O ] [C H O ] pH = pKa + log 2 3 2 , 4.00 = −log(1.8 × 10 −5 ) + log 2 3 2 [HC 2 H 3 O 2 ] [HC 2 H 3 O 2 ] −
[C 2 H 3 O 2 ] = 0.18; this is also equal to the mole ratio between C 2H3O2− and HC2H3O2. [HC 2 H 3 O 2 ]
Let x = volume of 1.00 M HC2H3O2 and y = volume of 1.00 M NaC2H3O2 x + y = 1.00 L, x = 1.00 – y x(1.00 mol/L) = mol HC2H3O2; y(1.00 mol/L) = mol NaC2H3O2 = mol C2H3O2− Thus:
y y = 0.18 or = 0.18; solving: y = 0.15 L, so x = 1.00 − 0.15 = 0.85 L. x 1.00 − y
We need 850 mL of 1.00 M HC2H3O2 and 150 mL of 1.00 M NaC2H3O2 to produce a buffer solution at pH = 4.00. 113.
A best buffer is when pH pKa; these solutions have about equal concentrations of weak acid and conjugate base. Therefore, choose combinations that yield a buffer where pH pKa; that is, look for acids whose pKa is closest to the pH. a. Potassium fluoride + HCl will yield a buffer consisting of HF (pKa = 3.14) and F−. b. Benzoic acid + NaOH will yield a buffer consisting of benzoic acid (pKa = 4.19) and benzoate anion. c. Sodium acetate + acetic acid (pKa = 4.74) is the best choice for pH = 5.0 buffer since acetic acid has a pKa value closest to 5.0. d. HOCl and NaOH: This is the best choice to produce a conjugate acid-base pair with pH = 7.0. This mixture would yield a buffer consisting of HOCl (pK a = 7.46) and OCl−. Actually, the best choice for a pH = 7.0 buffer is an equimolar mixture of ammonium chloride and sodium acetate. NH4+ is a weak acid (Ka = 5.6 × 10−10), and C2H3O2− is a weak base (Kb = 5.6 × 10−10). A mixture of the two will give a buffer at pH = 7.0 because the weak acid and weak base are the same strengths (K a for NH4+ = Kb for C2H3O2−). NH4C2H3O2 is commercially available, and its solutions are used for pH = 7.0 buffers. e. Ammonium chloride + NaOH will yield a buffer consisting of NH4+ (pKa = 9.26) and NH3.
114.
Let’s abbreviate the carboxylic acid group in alanine as RCOOH and the amino group in alanine as RNH2. The Ka reaction for the carboxylic acid group is: RCOOH ⇌ RCOO− + H+
Ka = 4.5 × 10−3
From Le Chatelier’s principle, if we have a very acidic solution, a lot of H + is present. This drives the Ka reaction to the left, and the dominant form of the carboxylic acid group will be RCOOH (an overall neutral charge). If we have a very basic solution, the excess OH − will
776
CHAPTER 15
ACID-BASE EQUILIBRIA
remove H+ from solution. As H+ is removed, the Ka reaction shifts right, and the dominant form of the carboxylic acid group will be RCOO − (an overall 1− charged ion). The Kb reaction for the amino group is: RNH2 + H2O ⇌ RNH3+ + OH− If we have a very acidic solution, the excess protons present will remove OH − from solution, and the dominant form of the amino group will be RNH3+ (an overall 1+ charged ion). If we have a very basic solution, a lot of OH − is present, and the dominant form of the amino group will be RNH2 (an overall neutral charge). In alanine, both an RCOOH group and an RNH2 group are present. The dominant form of alanine in a very acidic solution will be the form with the protons attached to the two groups that have acid-base properties. This form of alanine is:
which has an overall charge of 1+. The dominant form of alanine in a very basic solution will be in the form with the protons removed from the two groups that have acid-base properties. This form of alanine is:
which has an overall charge of 1−. 2−
115.
a. pH = pKa + log
[HPO4 ] [base] , 7.15 = −log(6.2 × 10 −8 ) + log − [acid] [H 2 PO 4 ] 2−
2−
7.15 = 7.21 + log
[HPO4 ] −
[H 2 PO 4 ]
,
[HPO4 ] −
[H 2 PO 4 ]
= 10 −0.06 = 0.9,
1 [H 2 PO4 − ] = = 1.1 1 2− 0 .9 [HPO4 ]
b. A best buffer has approximately equal concentrations of weak acid and conjugate base, so pH pKa for a best buffer. The pKa value for a H3PO4/H2PO4− buffer is −log(7.5 × 10 −3 ) = 2.12. A pH of 7.15 is too high for a H 3PO4/H2PO4− buffer to be effective. At this high of pH, there would be so little H3PO4 present that we could hardly consider it a buffer; this solution would not be effective in resisting pH changes, especially when a strong base is added.
CHAPTER 15 116.
ACID-BASE EQUILIBRIA
777
A best buffer has large and equal quantities of weak acid and conjugate base or weak base [base] and conjugate acid. Because [acid] = [base] for a best buffer, pH = pKa + log = pKa + [acid] 0 = pKa (pH pKa for a best buffer). The best acid choice for a pH = 9.0 buffer would be the weak acid with a pKa closest to 9.0 or Ka 1 × 10−9. HOBr is the best choice of the options given (K a = 2.0 × 10−9; pKa = 8.70). So, the best buffer choice for a pH = 9.00 buffer would be to use HOBr (b) and its conjugate base OBr− (h). The best acid choice for a pH = 6.00 buffer would be the weak acid with a pK a closest to 6.0 or Ka 1 × 10−6. HONH2 has a conjugate acid, HONH3+, with the pKa closest to 6.0 (Ka = Kw/Kb = 1.0 × 10‒14/1.1 × 10−8 = 9.1 × 10‒7; pKa = 6.04). So, the best buffer choice for a pH = 6.0 buffer would be to use the weak base HONH2 (g) and its conjugate acid HONH3+ (d).
117.
pH = pKa + log
(55.0 mL 0.472 M )/130.0 mL [C7 H 4 O2 F− ] = 2.90 + log [C7 H5O2 F] (75.0 mL 0.275 M )/130.0 mL
26.0 pH = 2.90 + log = 2.90 + 0.101 = 3.00 20.6
118.
The added OH− from the strong base reacts to completion with the best acid present, HF. To determine the pH, see what is in solution after the OH − reacts to completion. OH− added = 38.7 g soln
1.50 g NaOH 1 mol NaOH 1 mol OH− = 0.0145 mol OH− 100.0 g soln 40.00 g mol NaOH
For the 0.174 m HF solution, if we had exactly 1 kg of H2O, then the solution would contain 0.174 mol HF. 0.174 mol HF ×
20 .01 g = 3.48 g HF mol HF
Mass of solution = 1000.00 g H2O + 3.48 g HF = 1003.48 g Volume of solution = 1003.48 g ×
Mol HF = 250. mL × OH− Before Change After
1 mL = 912 mL 1.10 g
0.174 mol HF = 4.77 × 10 −2 mol HF 912 mL
+
0.0145 mol −0.0145 0
HF
→
0.0477 mol −0.0145 0.0332 mol
F−
+
H2O
0 +0.0145 0.0145 mol
After reaction, a buffer solution results containing HF, a weak acid, and F −, its conjugate base. Let VT = total volume of solution.
778
CHAPTER 15 pH = pKa + log
ACID-BASE EQUILIBRIA
0.0145/V T [F − ] = −log(7.2 × 10 −4 ) + log [ HF] 0.0332/V T
0.0145 pH = 3.14 + log = 3.14 + (−0.360), pH = 2.78 0.0332
119.
a. 1.00 L × 0.100 mol/L = 0.100 mol HCl added to reach stoichiometric point. The 10.00-g sample must have contained 0.100 mol of NaA.
10.00 g = 100. g/mol 0.100 mol
b. 500.0 mL of HCl added represents the halfway point to equivalence. Thus pH = pK a = 5.00 and Ka = 1.0 × 10−5. At the equivalence point, enough H+ has been added to convert all the A− present initially into HA. The concentration of HA at the equivalence point is: [HA]0 =
0.100 mol = 0.0909 M 1.10 L
HA Initial Equil.
⇌
0.0909 M 0.0909 - x
Ka = 1.0 × 10−5 =
H+
A−
+
0 x
Ka = 1.0 × 10−5
0 x
x2 x2 0.0909 − x 0.0909
x = 9.5 × 10−4 M = [H+]; pH = 3.02; assumptions good. 120.
At pH = 0.00, [H+] = 10−0.00 = 1.0 M. We begin with 1.0 L × 2.0 mol/L OH − = 2.0 mol OH−. We will need 2.0 mol HCl to neutralize the OH−, plus an additional 1.0 mol excess H+ to reduce the pH to 0.00. We need 3.0 mol HCl total assuming 1.0 L of solution.
121.
a. HC2H3O2 + OH− Keq =
⇌
C2H3O2− + H2O
K a , HC2 H 3O 2 [C2 H3O2 − ] [H + ] 1.8 10 −5 = = 1.8 × 109 = Kw [HC2 H3O2 ][OH − ] [H + ] 1.0 10 −14
Note: This reaction is the reverse of the Kb reaction for C2H3O2-. b. C2H3O2− + H+ ⇌ HC2H3O2
Keq =
[HC2 H 3 O 2 ] +
−
[H ][C 2 H 3 O 2 ]
=
1 K a , HC2 H3O 2
= 5.6 × 104
c. HCl + NaOH → NaCl + H2O Net ionic equation is H+ + OH− ⇌ H2O; Keq = 122.
1 = 1.0 × 1014 Kw
a. Because all acids are the same initial concentration, the pH curve with the highest pH at 0 mL of NaOH added will correspond to the titration of the weakest acid. This is pH curve f.
CHAPTER 15
ACID-BASE EQUILIBRIA
779
b. The pH curve with the lowest pH at 0 mL of NaOH added will correspond to the titration of the strongest acid. This is pH curve a. The best point to look at to differentiate a strong acid from a weak acid titration (if initial concentrations are not known) is the equivalence point pH. If the pH = 7.00, the acid titrated is a strong acid; if the pH is greater than 7.00, the acid titrated is a weak acid. c. For a weak acid-strong base titration, the pH at the halfway point to equivalence is equal to the pKa value. The pH curve, which represents the titration of an acid with Ka = 1.0 × 10 −6 , will have a pH = −log(1 × 10 −6 ) = 6.0 at the halfway point. The equivalence point, from the plots, occurs at 50 mL NaOH added, so the halfway point is 25 mL. Plot d has a pH 6.0 at 25 mL of NaOH added, so the acid titrated in this pH curve (plot d) has K a 1 × 10 −6 . 123.
a. This is a weak acid-strong base titration. At the halfway point to equivalence, [weak acid] = [conjugate base], so pH = pKa (always for a weak acid-strong base titration). pH = −log(6.4 × 10−5) = 4.19 mmol HC7H5O2 present = 100.0 mL × 0.10 M = 10. mmol HC7H5O2. For the equivalence point, 10. mmol of OH− must be added. The volume of OH− added to reach the equivalence point is: 10. mmol OH− ×
1 mL 0.10 mmol OH −
= 1.0 × 102 mL OH−
At the equivalence point, 10. mmol of HC7H5O2 is neutralized by 10. mmol of OH− to produce 10. mmol of C7H5O2−. This is a weak base. The total volume of the solution is 100.0 mL + 1.0 × 102 mL = 2.0 × 102 mL. Solving the weak base equilibrium problem: C7H5O2− + H2O
⇌
Initial 10. mmol/2.0 × 102 mL Equil. 0.050 − x Kb = 1.6 × 10−10 =
HC7H5O2 + OH− Kb = 0 x
1.0 10 −14 = 1.6 × 10−10 6.4 10 −5
0 x
x2 x2 , x = [OH−] = 2.8 × 10−6 M 0.050 − x 0.050
pOH = 5.55; pH = 8.45; assumptions good. b. At the halfway point to equivalence for a weak base-strong acid titration, pH = pKa because [weak base] = [conjugate acid]. Ka =
Kw 1.0 10 −14 = = 1.8 × 10−11; pH = pKa = −log(1.8 × 10−11) = 10.74 −4 Kb 5.6 10
For the equivalence point (mmol acid added = mmol base present): mmol C2H5NH2 present = 100.0 mL × 0.10 M = 10. mmol C2H5NH2
780
CHAPTER 15 mL H+ added = 10. mmol H+ ×
1 mL 0.20 mmol H
+
ACID-BASE EQUILIBRIA
= 50. mL H+
The strong acid added completely converts the weak base into its conjugate acid. Therefore, at the equivalence point, [C2H5NH3+]0 = 10. mmol/(100.0 + 50.) mL = 0.067 M. Solving the weak acid equilibrium problem: C2H5NH3+ Initial Equil.
⇌
H+
0.067 M 0.067 − x
Ka = 1.8 × 10−11 =
+
C2H5NH2
0 x
0 x
x2 x2 , x = [H+] = 1.1 × 10−6 M 0.067 − x 0.067
pH = 5.96; assumptions good. c. In a strong acid-strong base titration, the halfway point has no special significance other than that exactly one-half of the original amount of acid present has been neutralized. mmol H+ present = 100.0 mL × 0.50 M = 50. mmol H+ mL OH− added = 25 mmol OH− × H+ Before After [H+]excess =
50. mmol 25 mmol
+
1 mL = 1.0 × 102 mL OH− 0.25 mmol
OH−
→
H2 O
25 mmol 0
25 mmol = 0.13 M; pH = 0.89 (100 .0 + 1.0 10 2 ) mL
At the equivalence point of a strong acid-strong base titration, only neutral species are present (Na+, Cl−, and H2O), so the pH = 7.00. 124.
a. We will assume a base that only accepts one proton per formula unit. At the equivalence point, just enough HCl has been added to exactly react with all the base present initially. Because the strong acid is half as concentrated as the base titrated, a volume of HCl which is twice the volume of base is required to reach the equivalence point. Hence the equivalence point is at 50.0 mL HCl added. The halfway point will be 25.0 mL HCl added. Here exactly one-half of the base titrated has been converted into its conjugate acid. Only statement v is true. The equivalence point pH at 50.0 mL HCl added is 5.28, and the halfway point pH at 25.0 mL HCl added is 9.25. If the base titrated were a strong base, the pH initially would be 13.00 for a 0.10 M MOH type strong bases or 13.30 for M(OH)2 type strong bases. The initial pH before any HCl has been added is only 11.11, so the base titrated is a weak base. b. For a weak base-strong acid titration, the pH at the halfway point (9.25) is equal to the pKa of the conjugate acid for the weak base. pKa = 9.25, pKb = 14.00 – 9.25 = 4.75, Kb = 10‒4.75 = 1.8 × 10‒5
CHAPTER 15 125.
ACID-BASE EQUILIBRIA
781
In the final solution: [H+] = 10−2.15 = 7.1 × 10−3 M Beginning mmol HCl = 500.0 mL × 0.200 mmol/mL = 100. mmol HCl Amount of HCl that reacts with NaOH = 1.50 × 10 −2 mmol/mL × V 7.1 10 −3 mmol final mmol H + 100. − (0.0150)V = = mL total volume 500.0 + V
3.6 + (7.1 × 10−3)V = 100. − (1.50 × 10−2)V, (2.21 × 10−2)V = 100. − 3.6 V = 4.36 × 103 mL = 4.36 L = 4.4 L NaOH 126.
For a titration of a strong acid with a strong base, the added OH- reacts completely with the H+ present. To determine the pH, we calculate the concentration of excess H + or OH− after the neutralization reaction, and then calculate the pH. 0 mL:
[H+] = 0.100 M from HNO3; pH = 1.000
4.0 mL:
Initial mmol H+ present = 25.0 mL × mmol OH− added = 4.0 mL ×
0.100 mmol H + = 2.50 mmol H+ mL
0.100 mmol OH − = 0.40 mmol OH− mL
0.40 mmol OH− reacts completely with 0.40 mmol H+: OH− + H+ → H2O [H+]excess =
(2.50 − 0.40) mmol = 7.24 × 10 −2 M; pH = 1.140 (25.0 + 4.0) mL
We follow the same procedure for the remaining calculations. 8.0 mL:
[H+]excess =
(2.50 − 0.80) mmol = 5.15 × 10 −2 M; pH = 1.288 33.0 mL
12.5 mL:
[H+]excess =
(2.50 − 1.25) mmol = 3.33 × 10 −2 M; pH = 1.478 37.5 mL
20.0 mL:
[H+]excess =
(2.50 − 2.00) mmol = 1.1 × 10 −2 M; pH = 1.96 45.0 mL
24.0 mL:
[H+]excess =
(2.50 − 2.40) mmol = 2.0 × 10 −3 M; pH = 2.70 49.0 mL
24.5 mL:
[H+]excess =
(2.50 − 2.45) mmol = 1 × 10 −3 M; pH = 3.0 49.5 mL
24.9 mL:
[H+]excess =
(2.50 − 2.49) mmol = 2 × 10 −4 M; pH = 3.7 49.9 mL
25.0 mL:
Equivalence point; we have a neutral solution because there is no excess H+ or OH− remaining after the neutralization reaction. pH = 7.00
782
CHAPTER 15 25.1 mL:
Base in excess; [OH−]excess =
ACID-BASE EQUILIBRIA
(2.51 − 2.50) mmol = 2 × 10 −4 M; pOH = 3.7 50.1 mL
pH = 14.00 – 3.7 = 10.3
127.
26.0 mL:
[OH−]excess =
(2.60 − 2.50) mmol = 2.0 × 10 −3 M; pOH = 2.70; pH = 11.30 51.0 mL
28.0 mL:
[OH−]excess =
(2.80 − 2.50) mmol = 5.7 × 10 −3 M; pOH = 2.24; pH = 11.76 53.0 mL
30.0 mL:
[OH−]excess =
(3.00 − 2.50) mmol = 9.1 × 10 −3 M; pOH = 2.04; pH = 11.96 55.0 mL
HC2H3O2 ⇌ H+ + C2H3O2−; let C0 = initial concentration of HC2H3O2 −
From normal weak acid setup: Ka = 1.8 × 10−5 = [H+] = 10−2.68 = 2.1 × 10−3 M; 1.8 × 10−5 =
[H + ][C 2 H 3 O 2 ] [HC2 H 3 O 2 ]
=
[H + ] 2 C 0 − [H + ]
(2.1 10 −3 ) 2 , C0 = 0.25 M C 0 − (2.1 10 −3 )
25.0 mL × 0.25 mmol/mL = 6.3 mmol HC 2H3O2 Need 6.3 mmol KOH = VKOH × 0.0975 mmol/mL, VKOH = 65 mL 128.
Mol acid = 0.210 g ×
1 mol = 0.00109 mol 192 g
Mol OH− added = 0.0305 L
0.108 mol NaOH 1 mol OH− = 0.00329 mol OH− L mol NaOH
Mol OH − 0.00329 = = 3.02 Mol acid 0.00109
The acid is triprotic (H3A) because 3 mol of OH− are required to react with 1 mol of the acid; that is, the acid must have 3 mol H+ in the formula to react with 3 mol of OH−.
CHAPTER 15 129.
ACID-BASE EQUILIBRIA
783
HA + OH− → A− + H2O, where HA = acetylsalicylic acid (assuming it is a monoprotic acid). mmol HA present = 27.36 mL OH− Molar mass of HA =
0.5106 mmol OH − 1 mmol HA = 13.97 mmol HA mL OH − mmol OH −
2.51 g HA = 180. g/mol 13.97 10 −3 mol HA
To determine the Ka value, use the pH data. After complete neutralization of acetylsalicylic acid by OH−, we have 13.97 mmol of A− produced from the neutralization reaction. A − will react completely with the added H+ and re-form acetylsalicylic acid HA. mmol H+ added = 13.68 mL × A− Before Change After
0.5106 mmol H + = 6.985 mmol H+ mL
+
13.97 mmol −6.985 6.985 mmol
H+
→
6.985 mmol −6.985 → 0
HA 0 +6.985 6.985 mmol
Reacts completely
We have back titrated this solution to the halfway point to equivalence, where pH = pK a (assuming HA is a weak acid). This is true because after H + reacts completely, equal millimoles of HA and A− are present, which only occurs at the halfway point to equivalence. Assuming acetylsalicylic acid is a weak monoprotic acid, then pH = pKa = 3.48. Ka = 10−3.48 = 3.3 × 10−4. 130.
0.500 mmol = 25.0 mmol NaOH mL 0.289 mmol 1 mmol NaOH NaOH left unreacted = 31.92 mL HCl × = 9.22 mmol NaOH mL mmol HCl
NaOH added = 50.0 mL ×
NaOH reacted with aspirin = 25.0 − 9.22 = 15.8 mmol NaOH 15.8 mmol NaOH × Purity =
1 mmol aspirin 180 .2 mg = 1420 mg = 1.42 g aspirin 2 mmol NaOH mmol
1.42 g × 100 = 99.5% 1.427 g
Here, a strong base is titrated by a strong acid. The equivalence point will be at pH = 7.0. Bromthymol blue would be the best indicator since it changes color at pH 7 (from base color to acid color), although phenolphthalein is commonly used for the indicator. See Fig. 15.8 of the text. 131.
50.0 mL × 0.100 M = 5.00 mmol NaOH initially At pH = 10.50, pOH = 3.50, [OH−] = 10 −3.50 = 3.2 × 10 −4 M mmol OH− remaining = 3.2 × 10 −4 mmol/mL × 73.75 mL = 2.4 × 10 −2 mmol mmol OH− that reacted = 5.00 − 0.024 = 4.98 mmol
784
CHAPTER 15
ACID-BASE EQUILIBRIA
Because the weak acid is monoprotic, 23.75 mL of the weak acid solution contains 4.98 mmol HA. [HA]0 = 132.
4.98 mmol = 0.210 M 23.75 mL
HA + OH− → A− + H2O; it takes 25.0 mL of 0.100 M NaOH to reach the equivalence point, where mmol HA = mmol OH− = 25.0 mL(0.100 M) = 2.50 mmol. At the equivalence point, some HCl is added. The H+ from the strong acid reacts to completion with the best base present, A−. H+ Before Change After
+
13.0 mL × 0.100 M −1.3 mmol 0
A−
→
2.5 mmol −1.3 mmol 1.2 mmol
HA 0 +1.3 mmol 1.3 mmol
A buffer solution is present after the H+ has reacted completely. pH = pKa + log
1.2 mmol / VT [A − ] , where VT = total volume , 4.70 = pKa + log [HA] 1.3 mmol / VT
Because the log term will be negative [log(1.2/1.3) = −0.035)], the pKa value of the acid must be greater than 4.70 (pKa = 4.70 + 0.035 = 4.74). 133.
At equivalence point: 16.00 mL × 0.125 mmol/mL = 2.00 mmol OH − added; there must be 2.00 mmol HX present initially. HX + OH− → X− + H2O
(neutralization rection)
2.00 mL NaOH added = 2.00 mL × 0.125 mmol/mL = 0.250 mmol OH−; 0.250 mmol of OH− added will convert 0.250 mmol HX into 0.250 mmol X −. Remaining HX = 2.00 − 0.250 = 1.75 mmol HX; this is a buffer solution where [H+] = 10−6.912 = 1.22 × 10−7 M. Because total volume cancels:
Ka =
[H + ][X − ] 1.22 10 −7 (0.250/VT ) 1.22 10 −7 (0.250) = = = 1.74 10 −8 [HX] 1.75/VT 1.75
Note: We could solve for Ka using the Henderson-Hasselbalch equation. 134.
pKa = −log(1.3 × 10−7) = 6.89; the color of an indicator changes over the approximate pH range of pH = pKa 1 = 6.89 1. For cyanidin aglycone, the useful pH range where this indicator changes color is from approximately 5.9 to 7.9.
135.
Since we have added two solutions together, the concentrations of each reagent has changed. What hasn’t changed is the moles or mmoles of each reagent. Let’s determine the mmol of each reagent present by multiplying the volume in mL by the molarity in units of mmol/mL. 100.0 mL 0.100 M = 10.0 mmol NaF; 100.0 mL 0.025 M = 2.5 mmol HCl
CHAPTER 15
ACID-BASE EQUILIBRIA
785
Let the added H+ from HCl react completely with the best base present, F –. H+ + F– → HF; 2.5 mmol H+ converts 2.5 mmol F– into 2.5 mmol HF. After the reaction, a buffer solution results containing 2.5 mmol HF and (10.0 − 2.5 =) 7.5 mmol F– in 200.0 mL of solution. pH = pKa + log
7.5 mmol/200.0 mL [F − ] = 3.62; assumptions good. = 3.14 + log [HF] 2.5 mmol/200.0 mL
Challenge Problems 136.
2.43 − 3.14 ΔpH = = 0.18 ΔmL 0 − 4 .0
At 4.0 mL NaOH added:
The other points are calculated in a similar fashion. The results are summarized and plotted below. As can be seen from the plot, the advantage of this approach is that it is much easier to accurately determine the location of the equivalence point. mL
|ΔpH/ΔmL|
pH
______________________________________________
0 4.0 8.0 12.5 20.0 24.0 24.5 24.9 25.0 25.1 26.0 28.0 30.0 137.
− 0.18 0.098 0.073 0.080 0.20 0.7 2 20 20 1 0.23 0.11
2.43 3.14 3.53 3.86 4.46 5.24 5.6 6.3 8.28 10.3 11.30 11.75 11.96
0.750 mmol = 33.8 mmol HC3H5O2 mL 0.700 mmol mmol C3H5O2− present initially = 55.0 mL × = 38.5 mmol C3H5O2− mL
mmol HC3H5O2 present initially = 45.0 mL ×
The initial pH of the buffer is: 38.5 mmol 100.0 mL [C H O ] 38.5 pH = pKa + log 3 5 2 = −log(1.3 × 10−5) + log = 4.89 + log = 4.95 33.8 mmol 33.8 [HC 3 H 5 O 2 ] 100.0 mL −
Note: Because the buffer components are in the same volume of solution, we can use the mole (or millimole) ratio in the Henderson-Hasselbalch equation to solve for pH instead of using the concentration ratio of [C3H5O2−] : [HC3H5O2].
786
CHAPTER 15
ACID-BASE EQUILIBRIA
When NaOH is added, the pH will increase, and the added OH − will convert HC3H5O2 into C3H5O2−. The pH after addition of OH− increases by 2.5%, so the resulting pH is: 4.95 + 0.025(4.95) = 5.07 At this pH, a buffer solution still exists, and the millimole ratio between C 3H5O2− and HC3H5O2 is: −
pH = pKa + log
−
mmol C 3 H 5 O 2 mmol C 3 H 5 O 2 , 5.07 = 4.89 + log mmol HC 3 H 5 O 2 mmol HC 3 H 5 O 2 −
mmol C 3 H 5 O 2 = 100.18 = 1.5 mmol HC 3 H 5 O 2
Let x = mmol OH− added to increase pH to 5.07. Because OH − will essentially react to completion with HC3H5O2, the setup to the problem using millimoles is: HC3H5O2 Before Change After
→
OH−
+
33.8 mmol −x 33.8 − x
x mmol −x → 0
C3H5O2− 38.5 mmol +x 38.5 + x
Reacts completely
−
mmol C 3 H 5 O 2 38.5 + x = 1.5 = , 1.5(33.8 − x) = 38.5 + x, x = 4.9 mmol OH− added 33.8 − x mmol HC 3 H 5 O 2
The volume of NaOH necessary to raise the pH by 2.5% is: 4.9 mmol NaOH ×
1 mL = 49 mL 0.10 mmol NaOH
49 mL of 0.10 M NaOH must be added to increase the pH by 2.5%. 138.
0.400 mol/L × VNH3 = mol NH3 = mol NH4+ after reaction with HCl at the equivalence point. 0.400 VNH3 mol NH 4 = total volume 1.50 VNH3 +
At the equivalence point: [NH4+]0 = NH4+ Initial Equil. Ka =
0.267 M 0.267 − x
⇌
H+ 0 x
+
= 0.267 M
NH3 0 x
Kw 1.0 10 −14 x2 x2 −10 = 10 , 5.6 × = Kb 0.267 0.267 − x 1.8 10 −5
x = [H+] = 1.2 × 10 −5 M; pH = 4.92; assumption good. 139.
For HOCl, Ka = 3.5 × 10−8 and pKa = −log(3.5 × 10−8) = 7.46. This will be a buffer solution because the pH is close to the pKa value.
CHAPTER 15
ACID-BASE EQUILIBRIA
pH = pKa + log
787
[OCl− ] [OCl− ] [OCl− ] , 8.00 = 7.46 + log , = 100.54 = 3.5 [HOCl] [HOCl] [HOCl]
1.00 L × 0.0500 M = 0.0500 mol HOCl initially. Added OH− converts HOCl into OCl−. The total moles of OCl− and HOCl must equal 0.0500 mol. Solving where n = moles:
nOCl− + nHOCl = 0.0500 and nOCl− = (3.5)nHOCl (4.5)nHOCl = 0.0500, nHOCl = 0.011 mol; nOCl− = 0.039 mol
Need to add 0.039 mol NaOH to produce 0.039 mol OCl −. 0.039 mol = V × 0.0100 M, V = 3.9 L NaOH; Note: Normal buffer assumptions hold. 140.
50.0 mL × 0.100 M = 5.00 mmol H2SO4; 30.0 mL × 0.100 M = 3.00 mmol HOCl 25.0 mL × 0.200 M = 5.00 mmol NaOH; 10.0 mL × 0.150 M = 1.50 mmol KOH 25.0 mL × 0.100 M = 2.50 mmol Ba(OH)2 = 5.00 mmol OH−; we've added 11.50 mmol OH− total. Let OH− react completely with the best acid present (H 2SO4, then with HSO4–). 10.00 mmol OH− + 5.00 mmol H2SO4 → 0.00 mmol H2O + 5.00 mmol SO42− OH− remains after reacting completely with H2SO4. OH− will then react with the next best acid (HOCl). The remaining 1.50 mmol OH − will convert 1.50 mmol HOCl into 1.50 mmol OCl−, resulting in a solution with 1.50 mmol OCl− and (3.00 − 1.50 =) 1.50 mmol HOCl. The major species at this point are HOCl, OCl−, SO42−, and H2O plus cations that don't affect pH. SO42− is an extremely weak base (Kb = 8.3 × 10−13). We have a buffer solution composed of HOCl and OCl−. Because [HOCl] = [OCl−]: [H+] = Ka = 3.5 × 10−8 M; pH = 7.46; assumptions good. −
−
141.
pH = pKa + log
[(CH 3 ) 2 AsO 2 ] [(CH 3 ) 2 AsO 2 ] , 6.60 = 6.19 + log [(CH 3 ) 2 AsO 2 H] [(CH 3 ) 2 AsO 2 H]
−
[(CH 3 ) 2 AsO 2 ] = 100.41 = 2.6, [(CH3)2AsO2−] = 2.6 [(CH3)2AsO2H] [(CH 3 ) 2 AsO 2 H]
[(CH3)2AsO2−] + [(CH3)2AsO2H] = 0.25; 3.6 [(CH3)2AsO2H] = 0.25 [(CH3)2AsO2H] = 0.069 M and [(CH3)2AsO2−] = 0.18 M 0.500 L
0.069 mol (CH 3 ) 2 AsO 2 H 138.0 g (CH 3 ) 2 AsO 2 H = 4.8 g cacodylic acid L mol
0.500 L
0.18 mol (CH3 )2 AsO2 Na L
160.0 g (CH3 )2 AsO2 Na = 14 g sodium cacodylate mol
788 142.
CHAPTER 15
ACID-BASE EQUILIBRIA
a. At the third halfway point, pH = pK a 3 = −log(4.8 × 10−13) = 12.32. b. At third equivalence point (300.0 mL NaOH added), the reaction is: PO43− + H2O Initial
10. mmol 400 . mL
Change Equil.
−x 0.025 − x
K b = 2.1 10 − 2 =
⇌
– →
HPO42− + OH−
Kw 1.0 10 −14 = K a3 4.8 10 −13
0
0
Kb =
+x x
+x x
Kb = 2.1 × 10−2
x2 ; using the quadratic equation: 0.025 − x
x = 1.5 × 10−2 M = [OH−]; pOH = 1.82; pH = 12.18 c. The pH at the third halfway point (250.0 mL NaOH added) must be more acidic (lower pH) than the pH at the third equivalence point. Therefore, the pH at the third halfway point cannot equal 12.32. In part a we assumed that x was negligible:
⇌
HPO42− + H2O
PO43−
+
Initial 5.0 mmol/350.0 mL 5.0/350.0 Change −x → +x Equil. 0.014 − x 0.014 + x
K a 3 = 4.8 × 10−13
H3O+ 0 +x x
(0.014 + x)( x ) (0.014 )( x ) , x = 4.8 × 10−13 M (0.014 − x ) (0.014 )
This looks fine, but this is a situation where we must use the K b reaction for the weak base PO43− to solve for the pH. The [OH−] in solution is not negligible compared to 0.014 M, so the usual assumptions don’t hold here. The usual buffer assumptions don’t hold in very acidic or very basic solutions. In this very basic solution, we must use the K b reaction for PO43− and use the quadratic equation to solve: PO43− + H2O d. Initial Change Equil.
⇌
HPO42− + OH−
Kb = 2.1 × 10−2
PO43− + H2O
⇌
HPO42− + OH−
– – –
→
0.014 M +x 0.014 + x
0.014 M −x 0.014 − x
Kb = 2.1 × 10−2
0 +x x
(0.014 + x)( x) = 2.1 × 10−2; using the quadratic equation: (0.014 − x)
x = [OH−] = 7.0 × 10−3 M; pOH = 2.15; pH = 11.85 This pH answer makes more sense because it is lower than the pH at the third equivalence point calculated in part b of this problem (pH = 12.18).
CHAPTER 15 143.
ACID-BASE EQUILIBRIA
789
a. Na+ is present in all solutions. The added H+ from HCl reacts completely with CO32− to convert it into HCO3− (points A-C). After all of the CO32− is reacted (after point C, the first equivalence point), H+ then reacts completely with the next best base present, HCO 3− (points C-E). Point E represents the second equivalence point. The major species present at the various points after H+ reacts completely follow. A. CO32−, H2O, Na+
B. CO32−, HCO3−, H2O, Cl− , Na+
C. HCO3−, H2O, Cl−, Na+
D. HCO3−, CO2 (H2CO3), H2O, Cl−, Na+
E. CO2 (H2CO3), H2O, Cl−, Na+
F. H+ (excess), CO2 (H2CO3), H2O, Cl−. Na+
b. H2CO3 ⇌ HCO3− + H+
K a1 = 4.3 × 10 −7
HCO3− ⇌ CO32− + H+
K a 2 = 5.6 × 10 −11
The first titration reaction occurring between points A and C is: H+ + CO32− → HCO3− At point B, enough H+ has been added to convert one-half of the CO32− into its conjugate acid. At this halfway point to equivalence, [CO32−] = [HCO3−]. For this buffer solution, pH = pK a 2 = −log(5.6 × 10 −11 ) = 10.25 The second titration reaction occurring between points C and E is: H+ + HCO3− → H2CO3 Point D is the second halfway point to equivalence, where [HCO3−] = [H2CO3]. Here, pH = pK a1 = −log(4.3 × 10 −7 ) = 6.37. 144.
a. V1 corresponds to the titration reaction of CO32− + H+ → HCO3−; V2 corresponds to the titration reaction of HCO3− + H+ → H2CO3. Here, there are two sources of HCO3−: NaHCO3 and the titration of Na2CO3, so V2 > V1. b. V1 corresponds to two titration reactions: OH− + H+ → H2O and CO32− + H+ → HCO3−. V2 corresponds to just one titration reaction: HCO 3− + H+ → H2CO3. Here, V1 > V2 due to the presence of OH− , which is titrated in the V1 region. c. 0.100 mmol HCl/mL × 18.9 mL = 1.89 mmol H +; Because the first stoichiometric point only involves the titration of Na2CO3 by H+, 1.89 mmol of CO32− has been converted into HCO3−. The sample contains 1.89 mmol Na2CO3 × 105.99 mg/mmol = 2.00 × 102 mg = 0.200 g Na2CO3. The second stoichiometric point involves the titration of HCO 3− by H+.
790
CHAPTER 15
ACID-BASE EQUILIBRIA
0.100 mmol H + × 36.7 mL = 3.67 mmol H+ = 3.67 mmol HCO3− mL
1.89 mmol NaHCO3 came from the first stoichiometric point of the Na2CO3 titration. 3.67 − 1.89 = 1.78 mmol HCO3− came from NaHCO3 in the original mixture. 1.78 mmol NaHCO3 × 84.01 mg NaHCO3/mmol = 1.50 × 102 mg NaHCO3 = 0.150 g NaHCO3 0.200 g Mass % Na2CO3 = × 100 = 57.1% Na2CO3 (0.200 + 0.150 ) g Mass % NaHCO3 = 145.
0.150 g × 100 = 42.9% NaHCO3 0.350 g
Mmol (C2H5)3N = 100.0 mL × 0.50 M = 50. mmol Mmol H+ = 50.0 mL × 0.30 M = 15 mmol; mmol OH‒ = 50.0 mL × 0.30 M = 15 mmol Mmol (C2H5)3NH+ = 100.0 mL × 0.50 M = 50. mmol Note that K+ and ClO4‒ have no acidic and basic properties. a. Major species: (C2H5)3N, H+, and ClO4‒; let the added H+ from the strong acid react to completion with the best base present. This is (C2H5)3N. (C2H5)3N Before After
H+
+
50. mmol 35 mmol
→
15 mmol 0
(C2H5)3NH+ 0 15 mmol
A buffer solution is present after all the H+ reacts. Using the Henderson-Hasselbalch equation to solve for the pH (let VT = total volume of solution): pH = pKa + log
[base] where pKa = ‒log(1.0 × 10‒14/4.0 × 10‒4) = 10.60 [acid]
35 mmol/VT pH = 10.60 + log = 10.60 + 0.37 = 10.97 15 mmol/VT
b. Major species: (C2H5)3N, K+, OH‒, (C2H5)3NH+, and ClO4‒; let the added OH‒ from the strong base react to completion with the best acid present. This is (C 2H5)3NH+. (C2H5)3NH+ Before After
50. mmol 35 mmol
+
OH−
→
15 mmol 0
A buffer solution is present after all the OH‒ reacts.
(C2H5)3N 50. mmol 65 mmol
+ H2O
CHAPTER 15
ACID-BASE EQUILIBRIA pH = pKa + log
791
65 mmol/VT [base] = 10.60 + log = 10.60 + (0.27) = 10.87 [acid] 35 mmol/VT
c. Major species: (C2H5)3N, H+, ClO4‒, K+, OH‒, and (C2H5)3NH+; React the best acid, H+, with the best base, OH‒. Since a strong acid and a strong base is reacting, the reaction is assumed to go to completion. OH− Before After
→
H+
+
15. mmol 0
H2O
15 mmol 0
Major species after reaction: (C2H5)3N, ClO4‒, K+, and (C2H5)3NH+. We have 50. mmol of the weak base (C2H5)3N, 50. mmol of its conjugate acid (C2H5)3NH+, and some ions (ClO4‒ and K+) that have no acidic or basic properties. We have a buffer solution with equal amounts of the weak base and conjugate acid, so pH = pKa of the conjugate acid = 10.60. 146.
Mmol OCl‒ = 50.0 mL × 0.250 M = 12.5 mmol Mmol H+ = 25.0 mL × 0.500 M = 12.5 mmol; mmol OH‒ = 50.0 mL × 0.250 M = 12.5 mmol Mmol HOCl = 50.0 mL × 0.250 M = 12.5 mmol Note that solution v is composed of K+ and Br‒ ions, neither of which have acidic or basic properties. In terms of reactions, this solution can be ignored. The Na + and NO3‒ can also be ignored since they have no acidic or basic properties. a. Major species: Na+, OCl‒, H+, and NO3‒; let the H+ from the strong acid react to completion with the best base present, OCl‒. OCl‒ Before After
H+
+
12.5 mmol 0 mmol
→
12.5 mmol 0
HOCl 0 12.5 mmol
After this reaction, we have a solution with a weak acid (HOCl) and nothing else with acidic or basic properties. Solving the weak acid problem: HOCl Initial Change Equil.
⇌
H+
+
OCl−
12.5 mmol/75.0 mL ~0 0 x mol/L HOCl dissociates to reach equilibrium −x → +x +x 0.167 − x x x
Ka = 3.5 × 10−8 =
x2 x2 , x = [H+] = 7.6 × 10−5 M 0.167 − x 0.167
pH = 4.12; assumptions good.
792
CHAPTER 15
ACID-BASE EQUILIBRIA
b. Major species: H+, NO3‒, K+, OH‒, and Br‒; let the H+ from the strong acid react to completion with the OH‒ from the strong base. OH− Before After
H+
+
12.5 mmol 0
→
H2O
12.5 mmol 0
After reaction, we have a solution containing NO 3‒, K+, and Br‒. None of the ions have acidic or basic properties. pH = 7.00 of this solution. c. Major species: Na+, OCl‒, H+, NO3‒, HOCl, K+, OH‒, and Br‒; let the best acid and best base react first, then see what remains in solution. As in part b, let the H + from the strong acid react to completion with the OH‒ from the strong base. Because we have equal mmoles of H+ and OH‒, they exactly neutralize each other. After reaction, the major species are Na+, OCl‒, NO3‒, HOCl, K+, and Br‒. We have a weak acid (HOCl) and its conjugate base (OCl‒) present with other ion having no acidic or basic properties. We have a buffer solution. Because we have equal mmoles of HOCl and OCl‒ present, pH = pKa for HOCl = ‒log(3.5 × 10−8) = 7.46. 147.
An indicator changes color at pH pKa ±1. The result from each indicator tells us something about the pH. The conclusions are summarized below:
Results from
pH
bromphenol blue
≥ ~5.0
bromcresol purple
≤ ~5.0
bromcresol green *
pH pKa 4.8
alizarin
≤ ~5.5
*For bromcresol green, the resultant color is green. This is a combination of the extremes (yellow and blue). This occurs when pH pKa of the indicator. From the indicator results, the pH of the solution is about 5.0. We solve for Ka by setting up the typical weak acid problem. HX Initial Equil. Ka = Ka
1.0 M 1.0 − x
⇌
H+ ~0 x
+
X− 0 x
[H + ][X − ] x2 ; because pH 5.0, [H+] = x 1 × 10 −5 M. = 1.0 − x [HX]
(1 10 −5 ) 2 1.0 − 1 10 −5
1 × 10 −10
CHAPTER 15 148.
ACID-BASE EQUILIBRIA
793
Phenolphthalein will change color at pH 9. Phenolphthalein will mark the second end point of the titration. Therefore, we have titrated both protons on malonic acid. H2Mal + 2 OH− → 2 H2O + Mal2− where H2Mal = malonic acid 31.50 mL
[H2Mal] =
1 mmol H 2 Mal 0.0984 mmol NaOH = 1.55 mmol H2Mal mL 2 mol NaOH
1.55 mmol = 0.0620 M 25 .00 mL
Marathon Problem 149.
Major species: PO43−, OH−, H +, CN−, Na+, 5.00 mmol 5.00 mmol 15.0 mmol 7.50 mmol
K+,
Cl−,
H2O
PO43− and CN− are weak bases. PO43− + H2O
⇌
HPO42− + OH−
Kb = Kw/ K a 3 = 2.1 10−2
CN− + H2O
⇌
HCN + OH−
Kb = Kw/Ka = 1.6 10−5
One of the keys to this problem is to recognize that pK a 2 for H3PO4 = 7.21 [−log(6.2 10−8) = 7.21]. The K a 2 reaction for H3PO4 is: H2PO4−
⇌
K a 2 = 6.2 10−8; pK a 2 = 7.21
HPO42− + H+
The pH of the final solution will equal 7.21 when we have a buffer solution where [H2PO4− ] = [HPO42−]. To solve this problem, we need to determine the quantity of HNO 3 that must be added so that the final solution contains equal moles of H2PO4− and HPO42−. To start the problem, let’s see what is in solution after we let the best acid and best base react in a series of reactions. In each of the following reactions, something strong is reacting, so we assume the reactions will go to completion. The first reaction to run to completion is the strong acid reacting with the strong base: H+ + OH− → H2O. H+ Before After
+
15.0 mmol 10.0 mmol
OH−
→ H2 O
5.00 mmol 0
After all the strong base is neutralized, the next best base present is PO 43−. H+ Before After
+
10.0 mmol 5.0 mmol
PO43− 5.00 mmol 0
→
HPO42− 0 5.00 mmol
794
CHAPTER 15
ACID-BASE EQUILIBRIA
The next best base present is CN−. H+ Before After
+
5.0 mmol 0
→
CN−
7.50 mmol 2.5 mmol
HCN 0 5.0 mmol
We need to add 2.5 mmol H+ to convert all the CN− into HCN; then all that remains is 5.00 mmol HPO42− and 7.5 mmol HCN (a very weak acid with K a = 6.2 10−10). From here, we would need to add another 2.5 mmol H + in order to convert one-half of the HPO42− present into its conjugate acid so that [HPO42−] = [H2PO4−] and pH = pK a 2 = 7.21. Adding 5.0 mmol H+ to the original solution: H+ Before After
5.0 mmol 2.5 mmol H+
Before After
+
CN−
→
HCN
2.5 mmol 0 +
2.5 mmol 0
HPO42−
5.0 mmol 7.5 mmol →
5.00 mmol 2.5 mmol
H2PO4− 0 2.5 mmol
After 5.0 mmol H+ (HNO3) has been added to the original mixture, we have a final solution containing equal moles of HPO42− and H2PO4− so that pH = pK a 2 = 7.21. Note that HCN, with Ka = 6.2 10−10, is too weak of an acid to interfere with the H 2PO4−/ HPO42− buffer. Volume HNO3 = 5.0 mmol HNO3
1 mL = 50. mL HNO3 0.100 mmol HNO 3
CHAPTER 16 SOLUBILITY AND COMPLEX ION EQUILIBRIA Review Questions 1.
Ksp refers to the equilibrium reaction where a salt (ionic compound) dissolves into its ions to form a saturated solution: For example, consider Ca 3(PO4)2(s). The Ksp reaction and Ksp expression are: Ca3(PO4)2(s) ⇌ 3 Ca2+(aq) + 2 PO43−(aq)
Ksp = [Ca2+]3[PO43−]2
Ksp is just an equilibrium constant (called the solubility product constant) that refers to a specific equilibrium reaction. That reaction is always a solid salt dissolving into its ions. Because the ionic compound is a solid, it is not included in the Ksp expression. The solubility of a salt is the maximum amount of that salt that will dissolve per liter of solution. We determine solubility by solving a typical equilibrium problem using the Ksp reaction. We define our solubility (usually called s) as the maximum amount of the salt that will dissolve, then fill in our ICE table and solve. The two unknowns in the equilibrium problem are the Ksp value and the solubility. You are given one of the unknowns and asked to solve for the other. See Examples 16.1-16.3 of the text for typical calculations. In Examples 16.1 and 16.2, you are given the solubility of the salt and asked to calculate Ksp for the ionic compound. In Example 16.3, you are given the Ksp value for a salt and asked to calculate the solubility. In both types of problem, set-up the ICE table to solve. 2.
If the number of ions in the two salts is the same, the Ksp values can be compared directly to determine relative solubilities, i.e., 1 : 1 electrolytes (1 cation : 1 anion) can be compared to each other; 2 : 1 electrolytes can be compared to each other, etc. If the number of ions is the same, the salt with the largest Ksp value has the largest molar solubility. If the number of ions is different between two salts, then you must calculate the solubility of each salt to see which is more or less soluble.
3.
A common ion is when either the cation or anion of the salt in question is added from an outside source. For example, a common ion for the salt CaF 2 would be if F− (or Ca2+) was added to solution by dissolving NaF [or Ca(NO3)2]. When a common ion is present, not as much of the salt dissolves (as predicted by LeChâtelier’s Principle), so the solubility decreases. This decrease in solubility is called the common ion effect.
4.
Salts whose anions have basic properties have their solubilities increase as the solution becomes more acidic. Some examples are CaF 2, Ca3(PO4)2, CaCO3, and Fe(OH)3. In all these
795
796
CHAPTER 16
SOLUBILITY AND COMPLEX ION EQUILIBRIA
cases, the anion has basic properties and reacts with H+ in acidic solutions. This removes one of the ions from the equilibrium, so more salt dissolves to replenish the ion concentrations (as predicted by LeChâtelier’s Principle). In our example, F −, PO43−, and CO32− are all conjugate bases of weak acids so they are all weak bases. In Fe(OH)3, OH− is a strong base. For salts with no pH dependence on solubility, the ions in the salt must not have any acidic or basic properties. Examples of these salts have anions that are the conjugate bases of strong acids so they are worthless bases. Examples of these types of salts are AgCl, PbBr2, and Hg2I2. The solubility of these salts does not depend on pH because Cl−, Br−, and I− (as well as Ag+) have no basic (or acidic) properties. 5.
Q and Ksp have the same expression; the difference is the type of concentrations used. Ksp always uses equilibrium concentrations for the ions. Q, however, uses initial concentrations of the ions given in a problem. The purpose of Q is to see if a reaction is at equilibrium. If Q = Ksp, then the salt is at equilibrium with ion concentrations given in the problem. If Q Ksp, then the reaction is not at equilibrium; these concentrations must change in order for the salt to get to equilibrium with its ions. If Q > Ksp, the ion concentrations are too large. The Ksp reaction shifts left to produce the solid salt in order to reduce the ion concentrations, and to eventually get to equilibrium. If Q < Ksp, the Ksp reaction shifts right to produce more ions in order to get to equilibrium (more salt dissolves).
6.
Selective precipitation is a way to separate out ions in an aqueous mixture. One way this is done is to add a reagent that precipitates with only one of the ions in the mixture. That ion is removed by forming the precipitate. Another way to selectively precipitate out some ions in a mixture is to choose a reagent that forms a precipitate with all of the ions in the mixture. The key is that each salt has a different solubility with the ion in common. We add the common ion slowly until a precipitate starts to form. The first precipitate to form will be the least soluble precipitate. After precipitation of the first is complete, we filter off the solution and continue adding more of the common ion until precipitation of the next least soluble salt is complete. This process is continued until all of the ions have been precipitated out of solution. For the 0.10 M Mg2+, Ca2+, and Ba2+, adding F− (NaF) slowly will separate the ions due to their solubility differences with these ions. The first compound to precipitate is the least soluble CaF2 compound (it has the smallest Ksp value, 4.0 × 10 −11 ). After all the CaF2 has precipitated and been removed, we continue adding F− until the second least soluble fluoride has precipitated. This is MgF2 with the intermediate Ksp value (6.4 × 10 −9 ). Finally, the only ion left in solution is Ba2+ which is precipitated out by adding more NaF. BaF2 precipitates last because it is the most soluble fluoride salt in the mixture (Ksp = 2.4 × 10 −5 ). For the Ag+, Pb2+, Sr2+ mixture, because the phosphate salt of these ions does not all have the same number of ions, we cannot just compare Ksp values to deduce the order of precipitation. This would work for Pb3(PO4)2, Ksp = 1 × 10 −54 and Sr3(PO4)2, Ksp = 1 × 10 −31 because these contain the same number of ions (5). However, Ag 3PO4(s) only contains 4 ions per formula unit. We must calculate the [PO43−] necessary to start precipitation. We do this by setting Q = Ksp and determining the [PO43−] necessary for this to happen. Any [PO43−] greater than this calculated concentration will cause Q > Ksp and precipitation of the salt. A sample calculation is: Sr3(PO4)2(s) ⇌ 3 Sr2+(aq) + 2 PO43−(aq) Ksp = 1 × 10 −31
CHAPTER 16
SOLUBILITY AND COMPLEX ION EQUILIBRIA
797
Q = 1 × 10 −31 = [Sr2+]3[PO43−]2 = (1.0 M)3[PO43−]2, [PO43−] = 3 × 10 −16 M When [PO43−] > 3 × 10 −16 M, precipitation of Sr3(PO4)2 will occur. Doing similar calculations with the other two salts leads to the answers: Ag3PO4(s) will precipitate when [PO43−] > 1.8 × 10 −18 M. Pb3(PO4)2(s) will precipitate when [PO43−] > 1 × 10 −27 M. From the calculations, Pb3(PO4)2(s) needs the smallest [PO43−] to precipitate, so it precipitates first as we add K3PO4. The next salt to precipitate is Ag3PO4(s), and Sr3(PO4)2(s) is last to precipitate because it requires the largest [PO43−]. 7.
In step 1, the insoluble chloride salts are precipitated by adding HCl. If Ag+, Pb2+, and/or Hg22+ is/are present, these ions would be removed from solution as precipitates. In step 2, the least soluble of the insoluble sulfide salts are removed by addition of H2S. Because the solution is very acidic by addition of HCl in step 1, the amount of free S2− ions when H2S is added will be very small. This is because the presence of a lot of protons suppresses the H 2S dissociation. With only a minimal amount of S2− present, only the least insoluble sulfide salts will precipitate in this step. In the next step (step 3), NaOH is added in order to react with the excess protons. This causes more H2S to dissociate, which causes more free S 2− to be produced. As the S2− concentration increases with addition of NaOH, the insoluble sulfide salts that didn’t precipitate in step 2, will now precipitate. In the fourth and final step, ions that don’t form insoluble chloride and insoluble sulfide salts are all that remain in solution. Here, a solution of Na2CO3(aq) is added to remove Ba2+, Ca2+, and/or Mg2+ as the insoluble carbonate salts. The ions that remain after addition of Na2CO3 are those ions which do not form insoluble salts. These are the alkali metal ions and ammonium ions. Because these ions do not form insoluble salts, selective precipitation cannot be used to determine their presence.
8.
A complex ion is a charged species consisting of a metal ion surrounded by ligands. A ligand is a molecule or ion having a lone pair of electrons that can be donated to an empty orbital on the metal ion to form a covalent bond. A ligand is a Lewis base (electron pair donor). Cu2+(aq) + NH3(aq) ⇌ CuNH32+(aq)
K1 1 × 103
CuNH32+(aq) + NH3(aq) ⇌ Cu(NH3)22+(aq)
K2 1 × 104
Cu(NH3)22+(aq) + NH3(aq) ⇌ Cu(NH3)32+(aq)
K3 1 × 103
Cu(NH3)32+(aq) + NH3(aq) ⇌ Cu(NH3)42+(aq)
K4 1 × 103
Because the K values are much greater than one, all of these reactions lie far to the right (products are mostly present at equilibrium). So most of the Cu 2+ in a solution of NH3 will be converted into Cu(NH3)42+. Therefore, the [Cu(NH3)42+] will be much larger than the [Cu2+] at equilibrium.
798 9.
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Cu(OH)2(s) ⇌ Cu2+(aq) + 2 OH−(aq) As NH3 is added, it removes Cu2+ from the above Ksp equilibrium for Cu(OH)2. It removes Cu2+ by reacting with it to form Cu(NH 3)42+ mostly. As Cu2+ is removed, more of the Cu(OH)2(s) will dissolve to replenish the Cu2+. This process continues until equilibrium is reached or until all of the Cu(OH)2 dissolves. H+(aq) + NH3(aq) → NH4+(aq) The added H+ from the strong acid reacts with the weak base NH 3 to form NH4+. As NH3 is removed from solution, the complex ion equilibrium will shift left to produce more NH 3. This has the effect of also producing more Cu 2+. Eventually enough Cu2+ will be produced to react with OH− to form Cu(OH)2(s). The general effect of complex ion formation is to increase the solubility of salts that contain cations which form complex ions.
10.
The three metals that form insoluble chloride salts are Ag +, Pb2+ and Hg22+. A mixture that contains any or all of these ions is first reacted with some cold HCl(aq). All of the insoluble chloride salts form under these conditions. Next the solution is heated. PbCl 2(s) dissolves in hot HCl, while AgCl(s) and Hg2Cl2 will not dissolve upon heating. If a precipitate still is present after heating, then all we can conclude is that Ag + and/or Hg22+ could be present. The solution above the precipitate is filtered off and is then reacted with Na 2CrO4(aq). If Pb2+ is present in the original mixture, then a yellow precipitate of PbCrO 4(s) will form. If no Pb2+ ions were present, then no precipitate will form with addition of Na 2CrO4(aq). In the next step, the precipitate formed initially with cold HCl(aq) and remained after heating is examined. Some NH3(aq) is added to the precipitate, and the solution above the precipitate is filtered off. Ammonia forms a complex ion with Ag +, but does not form a complex ion with Hg22+. If Ag+ is present, then it will be in the solution above the precipitate as Ag(NH 3)2+. Also, if a precipitate remains after NH 3 is added, then the Hg22+ ion must have been present originally. If no precipitate remains, then Hg 22+ was not present. To prove Ag+ is present (or not present), the solution above the precipitate is made acidic. The added protons remove NH3 from solution. This frees up Ag+ ions to react with the Cl− ions present. So if Ag+ is present, AgCl(s) will reform when the solution is made acidic; if no precipitate forms, then no Ag+ was present in the original mixture.
Active Learning Questions 1.
The key to this question is that as long as the K sp value doesn’t change, not any more or less solute can dissolve whether you stir the solution or grind up the solute into fine particles. Once the ion concentrations are reached that satisfy the K sp constant, no more solid will dissolve. However, if the temperature is changed, this will change the value of K sp which will change the amount of solid that can dissolve to reach equilibrium.
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2.
Any procedure that allows determination of one of the ion concentrations of the salt can be used to determine the Ksp value. You don’t need to determine both ion concentrations from experiment because the stoichiometry of the salt will allow you determine one ion concentration if you know the other. Spectrophotometric experiments can be used to determine an ion concentration if one of the ions in the salt absorbs light at a certain wavelength. Titrations can also be used to determine an ion concentration. Acid-base titrations can be used to determine an ion concentration if one of the ions has acidic or basic properties. Titrations involving oxidation-reduction reaction or titrations involving complex ions can be used as well to determine one of the ion concentrations.
3.
If Ksp = 0, then none of the salt dissolves when added to water. For Ksp to equal zero, all ion concentrations must be zero when that salt is added to water.
4.
Ksp values can be used to compare the solubility of two salts only when both salts produce the same number of ions. The mathematical relationship between K sp and solubility depends on the number of ions in the solid. If two salts have the same number of ions, then they have the same mathematical relationship between Ksp and solubility. The salt with the lower Ksp value must be less soluble. However, if two salts are composed of different numbers of ions, then each salt has a different mathematical relationship between K sp and solubility. In this case, the salt with the smaller Ksp value is not necessarily least soluble.
5.
The major species present in test tube 1 are Ag +, NO3−, and H2O. When some Na2CrO4 is added, the major species initially are Ag+, NO3−, Na+, CrO42−, and H2O. Ag2CrO4 is an insoluble ionic compound, so the solid that forms in test tube 2 is silver chromate. From Table 15.1, the K sp value for Ag2CrO4 is 9.0 10−12. The relationship between the Ksp and the molar solubility is Ksp = 4s3. Solving, the molar solubility for Ag 2CrO4 is 1.3 10−4 mol/L. After reaction, the major species are Ag+, NO3−, Na+, and H2O. Because only a few drops of Na2CrO4(aq) was added, we are assuming that CrO42− is limiting in the formation of precipitate with excess Ag+ present. When some NaCl is then added, the major species in test tube 3 initially are Ag +, NO3−, Na+, Cl−, and H2O. The picture shows that a new precipitate formed. From the ions present, only silver chloride is a possibility. As Cl− reacts with Ag+ to form AgCl(s), this will cause the Ag2CrO4 precipitate to dissolve so to replenish the depleted Ag +. Assuming a sufficient amount of Cl− is present, eventually all of the silver chromate dissolves as is shown in the test tube 3 leaving only a precipitate of silver chloride. From Table 15.1, the K sp value for AgCl is 1.6 10−10. The relationship between the Ksp and the molar solubility is Ksp = s2. Solving, the molar solubility for AgCl is 1.3 10−5 mol/L. As expected, silver chloride has a smaller molar solubility than silver chromate, hence why AgCl(s) formed in the last test tube and why Ag2CrO4 dissolved. In a competition for the Ag + ion, Cl− has a great affinity for the silver ion as compared to CrO42− since silver chloride is less soluble.
6.
Many salts show an increase in solubility when the temperature is increased. For these salts, the Ksp value increases with an increase in temperature and decreases with a decrease in temperature. However, there are salts that show a decrease in solubility with an increase in temperature. There are no absolute rules regarding the effect of temperature on solubility. We must perform an experiment to determine the temperature dependence on solubility for a particular salt.
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7.
Cl− is the conjugate base of the strong acid HCl which makes Cl− a terrible base. Chloride salts are not more soluble in acidic solution because the Cl− will not react with the H+ ions in solution to any great extent. S2− is the conjugate base of the weak acid HS − which makes S2− a weak base. In acidic solution, S2− will react with H+ to form HS−. As S2− is removed from solution, more silver sulfide salt must dissolve to replenish the S2− concentration. This is necessary to reach the equilibrium ion concentrations as dictated by the K sp constant. Hence, silver sulfide is more soluble in acidic solution.
8.
The Ksp values are the same if each salt dissolves to form the same number of ions. If the salts dissolve to form different numbers of ions, like AgCl vs PbCl2, then the two salts will not have the same Ksp values. This is because the two salts would have different mathematical relationships between the solubility and the K sp value.
9.
This statement is false. If Ksp was an infinite value, then an infinite amount of NaCl would dissolve in water. This is not the case. Although NaCl is very soluble, there is a finite mass of NaCl that can dissolve per liter of solution
10.
a.
⇌
Ag2CrO4(s) Initial Change Equil.
2 Ag+(aq)
+
CrO42−(aq)
Ksp = 9.0 × 10‒12
0 0 Let s mol/L of Ag2CrO4 (s) dissolves to reach equilibrium −s → +2s +s 2s s
Ksp = 9.0 × 10‒12 = [Ag+]2[CrO42−] = (2s)2(s) = 4s3 4s3 = 9.0 × 10‒12, s = (9.0 × 10‒12/4)1/3 = 1.3 × 10‒4 mol/L [Ag+] = 2s = 2(1.3 × 10−4) = 2.6 × 10‒4 mol/L b.
Ag2CrO4(s) Initial Change Equil.
⇌
2 Ag+(aq)
+
CrO42− (aq)
0 0.50 M Let s mol/L of Ag2CrO4 (s) dissolves to reach equilibrium −s → +2s +s 2s 0.50 + s
Ksp = 9.0 × 10‒12 = [Ag+]2[CrO42−] = (2s)2(0.50 + s) ≈ 4s2(0.50) s = [9.0 × 10‒12/4(0.50)]1/2 = 2.1 × 10‒6 mol/L; assumption good [Ag+] = 2s = 4.2 × 10‒6 mol/L c. In part a, the [Ag+] = 2.6 × 10‒4 mol/L, and in part b, the [Ag+] = 4.2 × 10‒6 mol/L. In part b, a common ion was present when Ag2CrO4 disolved. The 0.50 M CrO42− common ion in part b reduced the amount of salt that can dissolve to reach equilibrium. This is called the common ion effect. 2.0 mol 1.00 mol 0.0500 L L L d. [Ag+]0 = = 1.0 M; [CrO42−]0 = = 0.50 M 0.0500 + 0.0500 L 0.1000 L 0.0500 L
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801
Note: Because we added equal volumes of each solution together, the total volume of the solution doubled, which halved the concentrations. Let Ag+ react with CrO42‒ to form Ag2CrO4(s). From the 2:1 mol ratio, 1.0 mol/L Ag+ will react exactly 0.50 mol/L CrO42− to form solid Ag2CrO4. Because both ions were limiting, there are 0 mol/L of each ion remaining after precipitation. This is impossible. Let’s solve for the solubility of Ag2CrO4(s) to determine the small amount of ions present. We did this calculation in part a, and found that 1.3 × 10‒4 mol/L of Ag2CrO4 disolves giving [Ag+] = 2(1.3 × 10‒4) = 2.6 × 10‒4 mol/L. e. In part d, we added stoichiometric quantities of each ion so they reacted completely with each other. After the ions react, we are left with water and some solid Ag2CrO4 at the bottom of the beaker. All salts dissolve to some extent, and that is what we calculated in part a. In the end both problems are the same, calculating the solubility of solid Ag2CrO4. 2.00 mol 2.0 mol 0.0500 L L L [Ag+]0 = = 1.0 M; [CrO42−]0 = = 1.0 M 0.0500 + 0.0500 L 0.1000 L 0.0500 L
f.
We have very large concentrations of the Ag + and CrO42− ions. Because the value of Ksp for Ag2CrO4 (s) is very small (Ksp << 1), the concentrations of the ions cannot both be large values. To solve this complicated problem, let the ions react to completion to form Ag2CrO4 (s), then solve a back equilibrium problem to determine the equilibrium concentrations of the ions. In the stoichiometry part of the problem, Ag + is the limiting reagent. Ag2CrO4(s)
⇌
2 Ag+
+
CrO42−
Ksp = 9.0 × 10‒12
Before
1.0 M 1.0 M Let 1.0 mol/L Ag+ react with CrO42− to completion because Ksp << 1. Change −1.0 −0.50 Reacts completely After 0 0.5 New initial s mol/L Ag2CrO4 (s) dissolves to reach equilibrium Change −s → +2s +s Equil. 2s 0.5 + s Ksp = 9.0 × 10‒12 = [Ag+]2[CrO42−] = (2s)2(0.5 + s) ≈ 4s2(0.5) s = [9.0 × 10‒12/4(0.5)]1/2 = 2 × 10‒6 mol/L; assumption good. [Ag+] = 2s = 2(2 × 10−6) = 4 × 10‒6 mol/L g. Although hidden, both parts b and f are the same problem. Once we let the ions in part f react to completion to form the precipitate, we are left with calculating the [Ag+] in 0.5 M CrO42−. This is the same calculation done in part b. 11.
a. Ag+(aq) + Cl−(aq) ⇌ AgCl(s); the first white precipitate is AgCl(s). AgCl(s) + 2 NH3(aq) ⇌ Ag(NH3)2+(aq) + Cl−(aq)
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The added ammonia forms a complex ion with Ag + having the formula Ag(NH3)2+. As Ag+ is removed from solution by forming the complex ion, some AgCl(s) dissolves to replenish the Ag+. If enough ammonia is added, the AgCl precipitate dissolves completely. Ag(NH3)2+(aq) + Br−(aq) ⇌ AgBr(s) + 2 NH3(aq), AgBr(s) is the pale yellow ppt. AgBr(s) + 2 S2O32−(aq) ⇌ Ag(S2O3)23−(aq) + Br−(aq) The added thiosulfate forms a complex ion with Ag+ having the formula Ag(S2O3)23−. As Ag+ is removed from solution by forming the complex ion, some AgBr(s) dissolves to replenish the Ag+. If enough thiosulfate is added, the AgBr precipitate dissolves completely. Ag(S2O3)23−(aq) + I−(aq) ⇌ AgI(s) + 2 S2O32−(aq), AgI(s) is the yellow precipitate. b. The least soluble salt (smallest Ksp value) must be AgI because it forms in the presence of Cl− and Br−. The most soluble salt (largest Ksp value) must be AgCl since it forms initially but never re-forms. The order of Ksp values is Ksp for AgCl > Ksp for AgBr > Ksp for AgI. Addition of S2O32− causes the AgBr(s) precipitate to dissolve, but the presence of NH 3 was unable to prevent AgBr(s) from forming. Therefore. the order of formation constants is K f for Ag(S2O3)23− > Kf for Ag(NH3)2+. This assumes concentrations are about equal.
Questions 12.
MX(s) ⇌ Mn+(aq) + Xn−(aq) Ksp = [Mn+][Xn−]; the Ksp reaction always refers to a solid breaking up into its ions. The representations all show 1 : 1 salts, i.e., the formula of the solid contains 1 cation for every 1 anion (either +1 and −1, or +2 and −2, or +3 and −3). The solution with the largest number of ions (largest [Mn+] and [Xn−]) will have the largest Ksp value. From the representations, the second beaker has the largest number of ions present, so this salt has the largest Ksp value. Conversely, the third beaker, with the fewest number of hydrated ions, will have the smallest Ksp value.
13.
Ksp values can only be directly compared to determine relative solubilities when the salts produce the same number of ions (have the same stoichiometry). Here, Ag2S and CuS do not produce the same number of ions when they dissolve, so each has a different mathematical relationship between the Ksp value and the molar solubility. To determine which salt has the larger molar solubility, you must do the actual calculations and compare the two molar solubility values.
14.
The solubility product constant (Ksp) is an equilibrium constant that has only one value for a given solid at a given temperature. Solubility, on the other hand, can have many values for a given solid at a given temperature. In pure water, the solubility is some value, yet the solubility is another value if a common ion is present. And the actual solubility when a common ion is present varies according to the concentration of the common ion. However, in all cases the product of the ion concentrations must satisfy the Ksp expression and give that one unique Ksp value at that temperature.
CHAPTER 16 15.
SOLUBILITY AND COMPLEX ION EQUILIBRIA
803
i.
This is the result when you have a salt that breaks up into two ions. Examples of these salts include AgCl, SrSO4, BaCrO4, and ZnCO3.
ii.
This is the result when you have a salt that breaks up into three ions, either two cations and one anion or one cation and two anions. Some examples are SrF 2, Hg2I2, and Ag2SO4.
iii. This is the result when you have a salt that breaks up into four ions, either three cations and one anion (Ag3PO4) or one cation and three anions (ignoring the hydroxides, there are no examples of this type of salt in Table 16.1). iv.
This is the result when you have a salt that breaks up into five ions, either three cations and two anions [Sr3(PO4)2] or two cations and three anions (no examples of this type of salt are in Table 16.1).
16.
The obvious choice is that the metal ion reacts with PO 43− and forms an insoluble phosphate salt. The other possibility is due to the weak base properties of PO43− (PO43− is the conjugate base of the weak acid HPO42−, so it is a weak base). Because PO43− is a weak base in water, OH− ions are present at a fairly large concentration. Hence the other potential precipitate is the metal ion reacting with OH− to form an insoluble hydroxide salt.
17.
For the Ksp reaction of a salt dissolving into its respective ions, a common ion would be one of the ions in the salt added from an outside source. When a common ion (a product in the K sp reaction) is present, the Ksp equilibrium shifts to the left, resulting in less of the salt dissolving into its ions (solubility decreases).
18.
S2− is a very basic anion and reacts significantly with H+ to form HS− (S2− + H+ ⇌ HS−). Thus, the actual concentration of S2− in solution depends on the amount of H+ present. In basic solutions, little H+ is present, which shifts the above reaction to the left. In basic solutions, the S2− concentration is relatively high. So, in basic solutions, a wider range of sulfide salts will precipitate. However, in acidic solutions, added H+ shifts the equilibrium to the right resulting in a lower S2− concentration. In acidic solutions, only the least soluble sulfide salts will precipitate out of solution.
19.
20.
Some people would automatically think that an increase in temperature would increase the solubility of a salt. This is not always the case as some salts show a decrease in solubility as temperature increases. The two major methods used to increase solubility of a salt both involve removing one of the ions in the salt by reaction. If the salt has an ion with basic properties, adding H+ will increase the solubility of the salt because the added H + will react with the basic ion, thus removing it from solution. More salt dissolves to make up for the lost ion. Some examples of salts with basic ions are AgF, CaCO 3, and Al(OH)3. The other way to remove an ion is to form a complex ion. For example, the Ag + ion in silver salts forms the complex ion Ag(NH3)2+ as ammonia is added. Silver salts increase their solubility as NH 3 is added because the Ag+ ion is removed through complex ion formation. PbCl2(s) ⇌ Pb2+(aq) + 2 Cl−(aq)
Ksp = [Pb2+][Cl−]2
As temperature increases, more PbCl2(s) dissolves to reach equilibrium. This is consistent with an endothermic reaction where heat is a reactant. As heat (a reactant) is added, the reaction shifts right to increase the ion concentrations present at equilibrium, resulting in a larger Ksp value.
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21.
Parts ii and ii have a common ion present before Ag2CrO4(s) dissolve. In part ii, the common ion is Ag+, and in part iii, the common is CrO42−. The presence of a common decreases the solubility of Ag2CrO4(s). Hence, the molar solubility of Ag2CrO4 in water (i) will have the largest value.
22.
Part iii has the common ion OH‒ present before Cd(OH)2(s) dissolves. A common ion decreases the solubility of a salt. Hence the solubility of Cd(OH) 2 in water will be greater than the solubility of Cd(OH)2 in 1.0 M NaOH. In part i, the presence of H+ from the strong acid will react with OH‒ to form H2O. When Cd(OH)2 dissolves, some of the OH‒ from the dissolved salt will react with H+. This causes more Cd(OH)2 to dissolve. So, the solubility in part i is the largest. The order of solubilities is iii < ii < i.
23.
For conjugate acid-base pairs, the weaker the acid, the stronger is the conjugate base. Because HX is a stronger acid (has a larger Ka value) than HY, Y− will be a stronger base than X−. In acidic solution, Y− will have a greater affinity for the H+ ions. Therefore, AgY(s) will be more soluble in acidic solution because more Y − will be removed through reaction with H+, which will cause more AgY(s) to dissolve.
24.
Salts whose anions react with H+ are more soluble in acidic solution than in water. Hydroxide salts are all more soluble in acidic solution because H + reacts with OH‒ to form water. Salts whose anions are the conjugate bases of weak acids are also more soluble in acidic solution because these anions also react with H + to form the weak acid. Salts whose anions come from strong acids do not react with H+ because the conjugate bases of strong acids are terrible bases. So, the salts that are more soluble in 1.0 M HNO3 are AgOH and AgF. AgF is more soluble because the F‒ anion is the conjugate base of the weak acid HF and will react with H +. AgCl, AgBr, and AgI are not more soluble in 1.0 M HNO3 because Cl‒ Br‒, and I‒ are conjugate bases of strong acids.
25.
The C2H3O2‒ anion is the conjugate base of the weak acid HC 2H3O2. In acidic solution, C2H3O2‒ will react with H+ to form HC2H3O2. Because of this reaction, AgC2H3O2(s) will be more soluble with addition of HNO3. Ag+ forms the complex ion Ag(NH3)2+. Because of the formation of the complex ion, AgC2H3O2(s) will be more soluble with addition of NH3.
26.
Because the formation constants are usually very large numbers, the stepwise reactions can be assumed to essentially go to completion. Thus, an equilibrium mixture of a metal ion and a specific ligand will mostly contain the final complex ion in the stepwise formation reactions.
27.
In 2.0 M NH3, the soluble complex ion Cu(NH3)42+ forms, which increases the solubility of CuCO3(s). The reaction is CuCO3 (s) + 4 NH3 ⇌ Cu(NH3)42+ + CO2−. In 2.0 M NH4NO3, NH3 is only formed by the dissociation of the weak acid NH 4+. There is not enough NH3 produced by this reaction to dissolve CuCO3(s) by the formation of the complex ion.
28.
Unlike AgCl(s), PbCl2(s) shows a significant increase in solubility with an increase in temperature. Hence add NaCl to the solution containing the metal ion to form the chloride salt precipitate, and then heat the solution. If the precipitate dissolves, then PbCl2 is present, and the metal ion is Pb2+. If the precipitate does not dissolve with an increase in temperature, then AgCl is the precipitate, and Ag+ is the metal ion present.
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SOLUBILITY AND COMPLEX ION EQUILIBRIA
805
Exercises Solubility Equilibria 29.
a. AgC2H3O2(s) ⇌ Ag+(aq) + C2H3O2−(aq) Ksp = [Ag+][C2H3O2−] b. Al(OH)3(s) ⇌ Al3+(aq) + 3 OH−(aq) Ksp = [Al3+][OH−]3 c. Ca3(PO4)2(s) ⇌ 3 Ca2+(aq) + 2 PO43−(aq) Ksp = [Ca2+]3[PO43−]2
30.
a. Ag2CO3(s) ⇌ 2 Ag+(aq) + CO32− (aq) Ksp = [Ag+]2[CO32−] b. Ce(IO3)3(s) ⇌ Ce3+(aq) + 3 IO3− (aq) Ksp = [Ce3+][IO3−]3 c. BaF2(s) ⇌ Ba2+(aq) + 2 F-(aq) Ksp = [Ba2+][F−]2
31.
In our setup, s = solubility of the ionic solid in mol/L. This is defined as the maximum amount of a salt that can dissolve. Because solids do not appear in the Ksp expression, we do not need to worry about their initial and equilibrium amounts. a.
⇌
CaC2O4(s) Initial Change Equil.
Ca2+(aq)
+
C2O42−(aq)
0 0 s mol/L of CaC2O4(s) dissolves to reach equilibrium −s → +s +s s s
From the problem, s = 4.8 × 10 −5 mol/L. Ksp = [Ca2+][C2O42−] = (s)(s) = s2, Ksp = (4.8 × 10 −5 )2 = 2.3 × 10 −9 b.
Cu(IO3)2(s) Initial Change Equil.
⇌
Cu2+(aq)
+
2 IO3−(aq)
0 0 s mol/L of Cu(IO3)2(s) dissolves to reach equilibrium −s → +s +2s s 2s
Ksp = [Cu2+][ IO3‒]2 = (s)(2s)2 = 4s3, Ksp = 4(3.3 ×10‒3)3 = 1.4 × 10‒7 32.
a.
Pb3(PO4)2(s) Initial Change Equil.
⇌
3 Pb2+(aq) + 2 PO43−(aq)
0 0 s mol/L of Pb3(PO4)2(s) dissolves to reach equilibrium = molar solubility −s → +3s +2s 3s 2s
Ksp = [Pb2+]3[PO43−]2 = (3s)3(2s)2 = 108s5, Ksp = 108(6.2 × 10−12)5 = 9.9 × 10−55
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CHAPTER 16 b.
⇌
Li2CO3(s) Initial Equil.
SOLUBILITY AND COMPLEX ION EQUILIBRIA 2 Li+(aq)
s = solubility (mol/L)
+
CO32-(aq)
0 2s
0 s
Ksp = [Li+]2[CO32−] = (2s)2(s) = 4s3, Ksp = 4(7.4 × 10 −2 )3 = 1.6 × 10 −3 33.
Solubility = s =
0.14 g Ni(OH) 2 1 mol Ni(OH) 2 = 1.5 10 −3 mol/L L 92.71 g Ni(OH) 2
Ni(OH)2(s) ⇌ Ni2+(aq) + 2 OH−(aq) Initial Change Equil.
0 1.0 × 10−7 M (from water) s mol/L of Ni(OH)2 dissolves to reach equilibrium −s → +s +2s s 1.0 × 10−7 + 2s
From the calculated molar solubility, 1.0 × 10−7 + 2s 2s. Ksp = [Ni2+][OH−]2 = s(2s)2 = 4s3, Ksp = 4(1.5 × 10−3 )3 = 1.4 × 10−8 34.
⇌
M2X3(s) Initial Change Equil.
2 M3+(aq) +
3 X2−(aq)
Ksp = [M3+]2[X2−]3
0 0 s mol/L of M2X3(s) dissolves to reach equilibrium −s → +2s +3s 2s 3s
Ksp = (2s)2(3s)3 = 108s5 ; s =
3.60 10 −7 g 1 mol M 2 X 3 = 1.25 × 10 −9 mol/L L 288 g
Ksp = 108(1.25 × 10 −9 )5 = 3.30 × 10 −43 35.
PbBr2(s) Initial Change Equil.
⇌
Pb2+(aq)
+
2 Br−(aq)
0 0 s mol/L of PbBr2(s) dissolves to reach equilibrium −s → +s +2s s 2s
From the problem, s = [Pb2+] = 2.14 × 10 −2 M. So: Ksp = [Pb2+][Br−]2 = s(2s)2 = 4s3, Ksp = 4(2.14 × 10 −2 )3 = 3.92 × 10 −5 36.
Ag2C2O4(s) Initial Equil.
s = solubility (mol/L)
⇌
2 Ag+(aq)
+
C2O42− (aq)
0 2s
0 s
From problem, [Ag+] = 2s = 2.2 × 10 −4 M, s = 1.1 × 10 −4 M Ksp = [Ag+]2[C2O42−] = (2s)2(s) = 4s3 = 4(1.1 × 10 −4 )3 = 5.3 × 10 −12
CHAPTER 16
SOLUBILITY AND COMPLEX ION EQUILIBRIA
37.
Ce(IO3)3(s) Initial Change Equil.
⇌
Ce3+(aq)
807
3 IO3−(aq)
+
0 0 s mol/L of Ce(IO3)3(s) dissolves to reach equilibrium −s → +s +3s s 3s
From the problem, 3s = [IO3−] = 5.7 × 10‒3M, s = 1.9 × 10‒3 M. Solving for Ksp: Ksp = [Ce3+][ IO3−]3 = s(3s)3 = 27s4, Ksp = 27(1.9 × 10‒3)4 = 3.5 × 10‒10 38.
BiI3(s) Initial Change Equil.
⇌
Bi3+(aq)
3 I−(aq)
+
0 0 s mol/L of BiI3 (s) dissolves to reach equilibrium −s → +s +3s s 3s
From the problem, s = [Bi3+] = 1.3 × 10‒5M. Solving for Ksp: Ksp = [Bi3+][ I−]3 = s(3s)3 = 27s4, Ksp = 27(1.3 × 10‒5)4 = 7.7 × 10‒19 39.
In our setup, s = solubility in mol/L. Because solids do not appear in the Ksp expression, we do not need to worry about their initial or equilibrium amounts. a.
Ag3PO4(s) Initial Change Equil.
⇌
3 Ag+(aq)
+
PO43− (aq)
0 0 s mol/L of Ag3PO4(s) dissolves to reach equilibrium −s → +3s +s 3s s
Ksp = 1.8 × 10 −18 = [Ag+]3[PO43−] = (3s)3(s) = 27s4 27s4 = 1.8 × 10 −18 , s = (6.7 × 10 −20 )1/4 = 1.6 × 10 −5 mol/L = molar solubility b.
CaCO3(s) Initial Equil.
⇌
s = solubility (mol/L)
Ca2+(aq) 0 s
+ CO32−(aq) 0 s
Ksp = 8.7 × 10 −9 = [Ca2+][CO32−] = s2, s = 9.3 × 10 −5 mol/L c.
Hg2Cl2(s) Initial Equil.
⇌
s = solubility (mol/L)
Hg22+(aq) + 2 Cl−(aq) 0 s
0 2s
Ksp = 1.1 × 10 −18 = [Hg22+][Cl−]2 = (s)(2s)2 = 4s3, s = 6.5 × 10 −7 mol/L
808 40.
CHAPTER 16 a.
PbI2(s) Initial Equil.
SOLUBILITY AND COMPLEX ION EQUILIBRIA
⇌
Pb2+(aq)
s = solubility (mol/L)
2 I−(aq)
+
0 s
0 2s
Ksp = 1.4 × 10 −8 = [Pb2+][I−]2 = s(2s)2 = 4s3 s = (1.4 × 10 −8 /4)1/3 = 1.5 × 10 −3 mol/L = molar solubility b.
⇌
CdCO3(s) Initial Equil.
Cd2+(aq) +
s = solubility (mol/L)
CO32−(aq)
0 s
0 s
Ksp = 5.2 × 10 −12 = [Cd2+][CO32-] = s2, s = 2.3 × 10 −6 mol/L c.
⇌
Sr3(PO4)2(s) Initial Equil.
3 Sr2+(aq) + 2 PO43-(aq)
s = solubility (mol/L)
0 3s
0 2s
Ksp = 1 × 10 −31 = [Sr2+]3[PO43-]2 = (3s)3(2s)2 = 108s5, s = 2 × 10 −7 mol/L 41.
KBT dissolves to form the potassium ion (K +) and the bitartrate ion (abbreviated as BT −). KBT(s) Initial Equil.
⇌
K+(aq) + BT−(aq)
s = solubility (mol/L)
0 s
Ksp = 3.8 × 10 −4
0 s
3.8 × 10 −4 = [K+][BT−] = s(s) = s2, s = 1.9 × 10 −2 mol/L 0.2500 L
42.
1.9 10 −2 mol KBT 188.2 g KBT = 0.89 g KBT L mol
BaSO4(s) Initial Equil.
⇌
Ba2+(aq) + SO42−(aq)
s = solubility (mol/L)
0 s
Ksp = 1.5 × 10 −9
0 s
1.5 × 10 −9 = [Ba2+][SO42−] = s2, s = 3.9 × 10 −5 mol/L 0.1000 L
43.
3.9 10 −5 mol BaSO 4 233.4 g BaSO4 = 9.1 × 10 −4 g BaSO4 L mol
Cd(OH)2(s) Initial Equil.
⇌
s = solubility (mol/L)
Cd2+(aq) 0 s
+
2 OH−(aq) 1.0 × 10 −7 M 1.0 × 10 −7 + 2s
Ksp = 5.9 × 10 −15
CHAPTER 16
SOLUBILITY AND COMPLEX ION EQUILIBRIA
Ksp = [Cd2+][OH−]2 = s(1.0 × 10 −7 + 2s)2; assume that 1.0 × 10 −7 + 2s 2s, then: Ksp = 5.9 × 10 −15 = s(2s)2 = 4s3, s = 1.1 × 10 −5 mol/L Assumption is good (1.0 × 10 −7 is 0.4% of 2s). Molar solubility = 1.1 × 10 −5 mol/L 44.
⇌
Ba(OH)2(s)
Ba2+(aq) + 2 OH−(aq) Ksp = [Ba2+][OH−]2 = 5.0 × 10 −3
Initial s = solubility (mol/L) Equil.
0 s
~0 2s
Ksp = 5.0 × 10 −3 = s(2s)2 = 4s3, s = 0.11 mol/L; assumption good. [OH−] = 2s = 2(0.11) = 0.22 mol/L; pOH = 0.66, pH = 13.34 Sr(OH)2(s)
⇌
Ksp = [Sr2+][OH−]2 = 3.2 × 10 −4
Sr2+(aq) + 2 OH−(aq)
Equil.
s
2s
Ksp = 3.2 × 10 −4 = 4s3, s = 0.043 mol/L; asssumption good. [OH−] = 2(0.043) = 0.086 M; pOH = 1.07, pH = 12.93 Ca(OH)2(s)
⇌
Equil.
Ksp = [Ca2+][OH−]2 = 1.3 × 10 −6
Ca2+(aq) + 2 OH−(aq) s
2s
Ksp = 1.3 × 10 −6 = 4s3, s = 6.9 × 10 −3 mol/L; assumption good. [OH−] = 2(6.9 × 10 −3 ) = 1.4 × 10 −2 mol/L; pOH = 1.85, pH = 12.15
45.
Let s = solubility of Al(OH)3 in mol/L. Note: Because solids do not appear in the Ksp expression, we do not need to worry about their initial or equilibrium amounts. Al(OH)3(s) Initial Change Equil.
⇌
Al3+(aq)
+
3 OH−(aq)
0 1.0 × 10-7 M (from water) s mol/L of Al(OH)3(s) dissolves to reach equilibrium = molar solubility −s → +s +3s s 1.0 × 10−7 + 3s
Ksp = 2 × 10−32 = [Al3+][OH−]3 = (s)(1.0 × 10−7 + 3s)3 s(1.0 × 10−7)3 s=
2 10 −32 = 2 × 10−11 mol/L; assumption good (1.0 × 10−7 + 3s 1.0 × 10−7). 1.0 10 − 21
809
810 46.
CHAPTER 16
SOLUBILITY AND COMPLEX ION EQUILIBRIA
Let s = solubility of Co(OH)3 in mol/L.
⇌
Co(OH)3(s) Initial Change Equil.
Co3+(aq)
+
3 OH−(aq)
0 1.0 × 10 −7 M (from water) s mol/L of Co(OH)3(s) dissolves to reach equilibrium = molar solubility −s → +s +3s s 1.0 × 10 −7 + 3s
Ksp = 2.5 × 10 −43 = [Co3+][OH−]3 = (s)(1.0 × 10 −7 + 3s)3 s(1.0 × 10 −7 )3 s=
2.5 10 −43 = 2.5 × 10 −22 mol/L; assumption good (1.0 × 10 −7 + 3s 1.0 × 10 −7 ). − 21 1.0 10
47.
M3Y4(s) Initial Change Equil.
⇌
3 M4+(aq)
+
4 Y3−(aq)
0 0 s mol/L of M3Y4(s) dissolves to reach equilibrium = molar solubility −s → +3s +4s 3s 4s
Ksp = [M4+]3[Y3−]4 = (3s)3(4s)4 = 6912s7 48.
M2Y5(s) Initial Change Equil.
⇌
2 M5+(aq)
+
5 Y2−(aq)
0 0 s mol/L of M2Y5 (s) dissolves to reach equilibrium = molar solubility −s → +2s +5s 2s 5s
Ksp = [M5+]2[Y2−]5 = (2s)2(5s)5 = 12,500s7 49.
a. Because both solids dissolve to produce three ions in solution, we can compare values of Ksp to determine relative solubility. Because the Ksp for CaF2 is the smallest, CaF2(s) has the smallest molar solubility. b. We must calculate molar solubilities because each salt yields a different number of ions when it dissolves. Ca3(PO4)2(s) Initial Equil.
⇌
s = solubility (mol/L)
3 Ca2+(aq) + 2 PO43−(aq) 0 3s
Ksp = 1.3 × 10−32
0 2s
Ksp = [Ca2+]3[PO43−]2 = (3s)3(2s)2 = 108s5, s = (1.3 × 10−32/108)1/5 = 1.6 × 10−7 mol/L FePO4(s) Initial Equil.
⇌
s = solubility (mol/L)
Fe3+(aq) 0 s
+
PO43−(aq) 0 s
Ksp = 1.0 × 10−22
CHAPTER 16
SOLUBILITY AND COMPLEX ION EQUILIBRIA
811
Ksp = [Fe3+][PO43−] = s2, s = 1.0 10 −22 = 1.0 × 10−11 mol/L FePO4 has the smallest molar solubility. 50.
a.
⇌
FeC2O4(s) Equil.
s = solubility (mol/L)
Fe2+(aq)
+
C2O42−(aq)
s
s
Ksp = 2.1 × 10 −7 = [Fe2+][C2O42−] = s2, s = 4.6 × 10 −4 mol/L Cu(IO4)2(s)
⇌
Equil.
Cu2+(aq) + 2 IO4−(aq) s 2s
Ksp = 1.4 × 10 −7 = [Cu2+][IO4−]2 = s(2s)2 = 4s3, s = (1.4 × 10−7/4)1/3 = 3.3 × 10 −3 mol/L By comparing calculated molar solubilities, FeC2O4(s) is less soluble (in mol/L). b. Each salt produces three ions in solution, so we can compare Ksp values to determine relative molar solubilities. Therefore, Mn(OH)2(s) will be less soluble (in mol/L) because it has a smaller Ksp value. 51.
Sr3(PO4)2(s) Initial Equil.
⇌
3 Sr2+(aq) + 2 PO43−(aq)
s = solubility (mol/L)
0 3s
Ksp = 1 × 10−31
0 2s
Ksp = [Sr2+]3[PO43−]2 = (3s)3(2s)2 = 108s5, s = (1 × 10−31/108)1/5 = 2 × 10−7 mol/L MnS(s) ⇌ Equil. s = solubility (mol/L)
Mn2+(aq) + s
S2−(aq) s
Ksp = 2 × 10‒13 = [Mn2+][S2−] = s2, s = 4 × 10‒7 mol/L; MnS has a greater solubility.
⇌
FeS(s) Equil.
s = solubility (mol/L)
Fe2+(aq)
+
s
S2−(aq) s
Ksp = 4 × 10‒19 = [Fe2+][S2−] = s2, s = 6 × 10‒10 mol/L FeS does not have a greater molar solubility than Sr3(PO4)2. NiS and CoS have smaller Ksp values than FeS, so they also do not have a greater molar solubility. Only MnS has a greater molar solubility than Sr3(PO4)2. 52.
MOH(s) Initial Equil.
⇌
s = solubility (mol/L)
M+(aq) 0 s
+
OH−(aq) 1 × 10 −7 M (from water) 1 × 10 −7 + s
1.0 × 10‒8 = [M+][OH−] = s(1 × 10 −7 + s) s(s), s = 1.0 × 10‒4 mol/L; assumption good
812
CHAPTER 16 M(OH)2(s)
⇌
SOLUBILITY AND COMPLEX ION EQUILIBRIA
M2+(aq) + 2 OH−(aq)
Initial s = solubility (mol/L) Equil.
0 s
Ksp = [M2+][OH−]2 = 4.0 × 10‒18
~0 2s
Ksp = 4.0 × 10‒18 = s(2s)2 = 4s3, s = 1.0 × 10‒6 mol/L The assumption that the 1.0 × 10‒7 M OH‒ from water can be ignored is bad (10% error). The actual solubility will be a little smaller than 1.0 × 10‒6 mol/L. This may be enough information to answer the question. M(OH)3(s) Initial Equil.
⇌
s = solubility (mol/L)
M3+(aq)
3 OH−(aq)
+
0 s
~0 3s
Ksp = 2.7 × × 10‒19 = [M3+][OH−]3 = (s)(3s)3 = 27s4, s = 1.0 × 10‒5 mol/L; assumption good From the previous calculations, the order from smallest molar solubility to largest is M(OH)2, then M(OH)3, with MOH having the largest molar solubility. 53.
a.
Fe(OH)3(s) Initial Change Equil.
⇌
Fe3+(aq)
+
3 OH−(aq)
0 1 × 10 −7 M (from water) s mol/L of Fe(OH)3(s) dissolves to reach equilibrium = molar solubility −s → +s +3s s 1 × 10 −7 + 3s
Ksp = 4 × 10 −38 = [Fe3+][OH−]3 = (s)(1 × 10 −7 + 3s)3 s(1 × 10−7)3 s = 4 × 10 −17 mol/L; assumption good (3s << 1 × 10−7) b.
Fe(OH)3(s) Initial Change Equil.
⇌
Fe3+(aq) + 3 OH−(aq) pH = 5.0, [OH−] = 1 × 10 −9 M
0 1 × 10−9 M (buffered) s mol/L dissolves to reach equilibrium −s → +s (assume no pH change in buffer) −9 s 1 × 10
Ksp = 4 × 10 −38 = [Fe3+][OH−]3 = (s)(1 × 10 −9 )3, s = 4 × 10 −11 mol/L = molar solubility c.
Fe(OH)3(s) Initial Change Equil.
⇌
Fe3+(aq) + 3 OH−(aq)
0 0.001 M s mol/L dissolves to reach equilibrium −s → +s s 0.001
pH = 11.0, [OH−] = 1 × 10 −3 M (buffered) (assume no pH change)
CHAPTER 16
SOLUBILITY AND COMPLEX ION EQUILIBRIA
813
Ksp = 4 × 10 −38 = [Fe3+][OH−]3 = (s)(0.001)3, s = 4 × 10 −29 mol/L = molar solubility Note: As [OH−] increases, solubility decreases. This is the common ion effect. 54.
Co(OH)2(s) Initial Equil.
⇌
Co2+(aq) + 2 OH−(aq)
s = solubility (mol/L)
pH = 11.00, [OH−] = 1.0 × 10−3 M
1.0 × 10−3 (buffered) 1.0 × 10−3 (assume no pH change)
0 s
Ksp = 2.5 × 10−16 = [Co2+][OH−]2 = s(1.0 × 10−3 )2, s = 2.5 × 10−10 mol/L 55.
a.
Ag2SO4(s) Initial Equil.
⇌
s = solubility (mol/L)
2 Ag+(aq) +
SO42-(aq)
0 2s
0 s
Ksp = 1.2 × 10 −5 = [Ag+]2[SO42−] = (2s)2s = 4s3, s = 1.4 × 10 −2 mol/L b.
Ag2SO4(s) Initial Equil.
⇌
s = solubility (mol/L)
2 Ag+(aq)
SO42−(aq)
+
0.10 M 0.10 + 2s
0 s
Ksp = 1.2 × 10 −5 = (0.10 + 2s)2(s) (0.10)2(s), s = 1.2 × 10 −3 mol/L; assumption good. c.
Ag2SO4(s) Initial Equil
⇌
s = solubility (mol/L)
2 Ag+(aq) + 0 2s
SO42−(aq) 0.20 M 0.20 + s
1.2 × 10 −5 = (2s)2(0.20 + s) 4s2(0.20), s = 3.9 × 10 −3 mol/L; assumption good. Note: Comparing the solubilities of parts b and c to that of part a illustrates that the solubility of a salt decreases when a common ion is present. 56.
a.
PbI2(s) Initial Equil.
⇌
Pb2+(aq)
s = solubility (mol/L)
2 I−(aq)
+
0 s
0 2s
Ksp = 1.4 × 10−8 = [Pb2+][I−]2 = 4s3, s = 1.5 × 10−3 mol/L b.
PbI2(s) Initial Equil.
⇌
s = solubility (mol/L)
Pb2+(aq) 0.10 M 0.10 + s
+
2 I−(aq) 0 2s
1.4 × 10−8 = (0.10 + s)(2s)2 (0.10)(2s)2 = (0.40)s2, s = 1.9 × 10−4 mol/L; assumption good.
814
CHAPTER 16 c.
PbI2(s) Initial Equil.
SOLUBILITY AND COMPLEX ION EQUILIBRIA
⇌
Pb2+(aq)
s = solubility (mol/L)
2 I−(aq)
+
0 s
0.010 M 0.010 + 2s
1.4 × 10−8 = (s)(0.010 + 2s)2 (s)(0.010)2, s = 1.4 × 10−4 mol/L; assumption good. Note that in parts b and c, the presence of a common ion decreases the solubility as compared to the solubility of PbI2(s) in water. 57. Initial Change Equil.
⇌
3 Ag+(aq) + PO43−(aq) 0 0.10 M s mol/L of Ag3PO4 (s) dissolves to reach equilibrium −s → +3s +s 3s 0.10 + s Ag3PO4(s)
Ksp = 1.8 × 10−18 = [Ag+]3[PO43−] = (3s)3(0.10 + s) Assuming 0.10 + s 0.10: 1.8 × 10−18 = (3s)3(0.10) = 27s3(0.10) Assumption good; s = molar solubility = 8.7 × 10−7 mol/L 58.
⇌
Pb3(PO4)2(s) Initial Equil.
s = solubility (mol/L)
3 Pb2+(aq) + 2 PO43−(aq) 0.10 M 0.10 + 3s
Ksp = 1 × 10−54
0 2s
1 × 10−54 = (0.10 + 3s)3(2s)2 (0.10)3(2s)2 , s = 2 × 10−26 mol/L; assumption good. 59.
Ce(IO3)3(s) Initial Equil.
⇌
s = solubility (mol/L)
Ce3+(aq) 0 s
+
3 IO3−(aq) 0.20 M 0.20 + 3s
Ksp = [Ce3+][IO3−]3 = s(0.20 + 3s)3 From the problem, s = 4.4 × 10 −8 mol/L; solving for Ksp: Ksp = (4.4 × 10 −8 )[0.20 + 3(4.4 × 10 −8 )]3 = 3.5 × 10 −10 60.
Pb(IO3)2(s) Initial Equil.
⇌
s = solubility (mol/L)
Pb2+(aq) 0 s
+ 2 IO3−(aq) 0.10 M 0.10 + 2s
Ksp = [Pb2+][IO3−]2 = (s)(0.10 + 2s)2 From the problem, s = 2.6 × 10−11 mol/L; solving for Ksp: Ksp = (2.6 × 10−11)[0.10 + 2(2.6 × 10−11)]2 = 2.6 × 10−13
CHAPTER 16 61.
SOLUBILITY AND COMPLEX ION EQUILIBRIA
If the anion in the salt can act as a base in water, then the solubility of the salt will increase as the solution becomes more acidic. Added H+ will react with the base, forming the conjugate acid. As the basic anion is removed, more of the salt will dissolve to replenish the basic anion. The salts with basic anions are Ag3PO4, CaCO3, CdCO3 and Sr3(PO4)2. Hg2Cl2 and PbI2 do not have any pH dependence because Cl− and I− are terrible bases (the conjugate bases of a strong acids). excess H
Ag3PO4(s) + H+(aq) → 3 Ag+(aq) + HPO42−(aq) CaCO3(s) + H → Ca + HCO3 +
2+
−
CdCO3(s) + H+ → Cd2+ + HCO3−
excess H
Sr3(PO4)2(s) + 2 H → 3 Sr a. AgF
2+
+
Ca2+ + H2CO3 [H2O(l) + CO2(g)] excess H
+
Cd2+ + H2CO3 [H2O(l) + CO2(g)]
+ 2 HPO42−
b. Pb(OH)2
+
3 Ag+(aq) + H3PO4(aq)
excess H +
62.
815
+
3 Sr2+ + 2 H3PO4
c. Sr(NO2)2
d. Ni(CN)2
All these salts have anions that are bases. The anions of the other choices are conjugate bases of strong acids. They have no basic properties in water and, therefore, do not have solubilities that depend on pH.
Precipitation Conditions 63.
ZnS(s) Initial Equil.
⇌
s = solubility (mol/L)
Zn2+(aq) + S2− (aq) 0.050 M 0.050 + s
Ksp = [Zn2+][S2−]
0 s
Ksp = 2.5 × 10 −22 = (0.050 + s)(s) (0.050)s, s = 5.0 × 10 −21 mol/L; assumption good. Mass ZnS that dissolves = 0.3000 L 64.
5.0 10 −21 mol ZnS 97.45 g ZnS = 1.5 × 10 −19 g L mol
For 99% of the Mg2+ to be removed, we need, at equilibrium, [Mg 2+] = 0.01(0.052 M). Using the Ksp equilibrium constant, calculate the [OH−] required to reach this reduced [Mg 2+]. Mg(OH)2(s) ⇌ Mg2+(aq) + 2 OH−(aq)
Ksp = 8.9 × 10 −12
8.9 × 10 −12 = [Mg2+][OH−]2 = [0.01(0.052 M)] [OH−]2, [OH−] = 1.3 × 10 −4 M (extra sig. fig.) pOH = −log(1.3 × 10 −4 ) = 3.89; pH = 10.11; at a pH = 10.1, 99% of the Mg 2+ in seawater will be removed as Mg(OH)2(s).
816 65.
CHAPTER 16
SOLUBILITY AND COMPLEX ION EQUILIBRIA
Sr3(PO4)2(s) ⇌ 3 Sr2+(aq) + 2 PO43−(aq)
Ksp = 1.0 × 10−31
Q = [Sr 2+ ]30 [PO34− ]02 = (1.0 × 10−6 )3(5.0 × 10−7 )2 = 2.5 × 10−31 Because Q > Ksp, Sr3(PO4)2(s) will form as a precipitate. 66.
Ag2CrO4(s) ⇌ 2 Ag+(aq) + CrO42−(aq); when Q is greater than Ksp, precipitation will occur. In the answers, the CrO42− concentration is the same in all answers. Let’s calculate the [Ag +]0 necessary for Q = Ksp. Any [Ag+]0 greater than this calculated number will cause precipitation of Ag3PO4(s). In this problem, [CrO42−]0 = [K2CrO4]0 = 1.0 × 10−4 M. Ksp = 9.0 × 10−12; Q = 9.0 × 10−12 = [Ag+ ]02 [CrO42−]0 = [Ag+ ]02 (1.0 × 10−4 M) 1/ 2
9.0 10−12 [Ag ]0 = , [Ag+]0 = 3.0 × 10−4 M −4 1.0 10 +
When [Ag+]0 = [AgNO3]0 is greater than 3.0 × 10−4 M, precipitation of Ag3PO4(s) will occur. Only answer e has [AgNO3]0 > 3.0 × 10−4 M; only in answer e will a precipitate form. 67.
PbF2(s) ⇌ Pb2+(aq) + 2 F−(aq)
[Pb2+]0 =
[F−]0 =
mmol Pb 2+ (aq) = total mL solution −
mmol F = total mL solution
Ksp = 4 × 10−8 1.0 10 −2 mmol Pb 2+ mL = 5.0 10 −3 M 100.0 mL + 100.0 mL
100.0 mL
100.0 mL
1.0 10 −3 mmol F − mL = 5.0 10 − 4 M 200.0 mL
Q = [Pb 2+ ]0 [F− ]02 = (5.0 × 10−3)(5.0 × 10−4)2 = 1.3 × 10−9 Because Q < Ksp, PbF2(s) will not form as a precipitate. 68.
Cr(OH)3(s) ⇌ Cr3+(aq) + 3 OH−(aq)
Ksp = 6.7 × 10−31
pOH = 14.0 − pH = 14.0 − 10.0 = 4.0, [OH−]0 = 10−pOH = 10−4.0 = 1.0 × 10−4 M; since equal volumes of the two reagents were added together, the initial concentrations are halved (the total volume doubled, so each molarity is halved). Q = [Sr 2+ ]0 [OH − ]30 = (1.0 × 10−3)(5.0 × 10−5)3 = 1.3 × 10−16 Because Q > Ksp, Cr(OH)3(s) will form as a precipitate. 69.
[Na2SO4] = [SO42‒] =
0.0050 mol = 0.010 M 0.5000 L
Precipitation occurs when Q > K. For each salt, let’s calculate Q, then compare this value to Ksp for each salt.
CHAPTER 16
SOLUBILITY AND COMPLEX ION EQUILIBRIA
817
For BaSO4 (Ksp = 1.5 × 10‒10): Q = [Ba2+]0[SO42‒]0 = (1.5 × 10‒3)(0.010) = 1.5 × 10‒5; Q > Ksp, so precipitate forms. For CaSO4 (Ksp = 6.1 × 10‒5): Q = [Ca2+]0[SO42‒]0 = (1.5 × 10‒3)(0.010) = 1.5 × 10‒5; Q < Ksp, so no precipitate forms. Only BaSO4(s) will form. 70.
M1V1 = M2V2; when we add equal volumes of two different solutions together, the volume doubles (V2 = 2V1). When the volume of a solution doubles, then the concentration is cut in half [M2 = (½)M1]. Therefore, the initial concentrations are [Pb 2+] = [Pb(NO3)2] = 1.2 × 10‒4 M, [I‒] = [KI] = 1.0 × 10‒4 M, [SO42‒] = [K2SO4] = 1.0 × 10‒4 M. Now determine Q for each salt to see if Q > Ksp indicating a precipitate will form. For PbI2 (Ksp = 1.4 × 10‒8): Q = [Pb2+]0[I‒]02 = (1.2 × 10‒4)( 1.0 × 10‒4)2 = 1.2 × 10‒12; Q < Ksp, no precipitate forms. For PbSO4 (Ksp = 1.3 × 10‒8): Q = [Pb2+]0[SO42‒]0 = (1.2 × 10‒4)( 1.0 × 10‒4) = 1.2 × 10‒8; Q < Ksp, no precipitate forms. Neither PbI2(s) nor PbSO4(s) will form.
71.
The concentrations of ions are large, so Q will be greater than Ksp, and BaC2O4(s) will form. To solve this problem, we will assume that the precipitation reaction goes to completion; then we will solve an equilibrium problem to get the actual ion concentrations. This makes the math reasonable. 100. mL ×
0.200 mmol K 2 C 2 O 4 = 20.0 mmol K2C2O4 mL
150. mL ×
0.250 mmol BaBr 2 = 37.5 mmol BaBr2 mL
Ba2+(aq) Before Change After
37.5 mmol −20.0 17.5
+
C2O42−(aq) → BaC2O4(s)
K = 1/Ksp >> 1
20.0 mmol −20.0 → 0
Reacts completely (K is large).
0 +20.0 20.0
New initial concentrations (after complete precipitation) are: [Ba2+] = [K+] =
17 .5 mmol = 7.00 × 10 −2 M; [C2O42−] = 0 M 250 . mL
2( 20 .0 mmol ) 2(37 .5 mmol ) = 0.160 M; [Br−] = = 0.300 M 250 . mL 250 . mL
For K+ and Br-, these are also the final concentrations. We can’t have 0 M C2O42−. For Ba2+ and C2O42−, we need to perform an equilibrium calculation.
818
CHAPTER 16 BaC2O4(s) Initial Equil.
⇌
SOLUBILITY AND COMPLEX ION EQUILIBRIA Ba2+(aq) +
C2O42−(aq)
Ksp = 2.3 × 10 −8
0.0700 M 0 s mol/L of BaC2O4(s) dissolves to reach equilibrium 0.0700 + s s
Ksp = 2.3 × 10 −8 = [Ba2+][C2O42−] = (0.0700 + s)(s) (0.0700)s s = [C2O42−] = 3.3 × 10 −7 mol/L; [Ba2+] = 0.0700 M; assumption good (s << 0.0700).
2.00 mol 3.00 mol 0.1000 L L L = 1.00 M; [IO3−]0 = = 1.50 M 0.2000 L 0.2000 L
0.1000 L 72.
[Ce3+]0 =
Note: Because we added equal volumes of each solution together, the total volume of the solution doubled, which halved the concentrations. We have very large concentrations of the Ce3+ and IO3− ions. Because the value of Ksp for Ce(IO3)3(s) is very small (Ksp << 1), the concentrations of the ions cannot both be large values. To solve this complicated problem, let the ions react to completion to form Ce(IO3)3(s), then solve a back equilibrium problem to determine the equilibrium concentrations of the ions. In the stoichiometry part of the problem, IO3− is the limiting reagent. Ce(IO3)3(s) Before
⇌
Ce3+
+
3 IO3−
Ksp = 3.2 × 10 −10
1.00 M 1.50 M 3+ Let 1.50 mol/L IO3 react with Ce to completion because Ksp << 1. Change −0.500 −1.50 Reacts completely After 0.50 0 New initial s mol/L Ce(IO3)3(s) dissolves to reach equilibrium Change −s → +s +3s Equil. 0.50 + s 3s −
Ksp = 3.2 × 10−10 = [Ce3+][ IO3−]3 = (0.50 + s)(3s)3 (0.50)27s3 Solving: s = 2.9 × 10−4 mol/L; assumption good (s is 0.058% of 0.50). [Ce3+] = 0.50 + 2.9 × 10−4 = 0.50 M; [IO3−] = 3(2.9 × 10−4 ) = 8.7 × 10−4 M 2.00 mol 3.00 mol 0.0500 L L L = 1.00 M; [CO32−]0 = = 1.50 M 0.1000 L 0.1000 L
0.0500 L
73.
[Ag+]0 =
Note: Because we added equal volumes of each solution together, the total volume of the solution doubled, which halved the concentrations. We have very large concentrations of the Ag + and CO32− ions. Because the value of Ksp for Ag2CO3(s) is very small (Ksp << 1), the concentrations of the ions cannot both be large values. To solve this complicated problem, let the ions react to completion to form Ag 2CO3(s), then
CHAPTER 16
SOLUBILITY AND COMPLEX ION EQUILIBRIA
819
solve a back equilibrium problem to determine the equilibrium concentrations of the ions. In the stoichiometry part of the problem, Ag + is the limiting reagent.
⇌
2 Ag+ + CO32− Ksp = 8.1 × 10‒12 Before 1.00 M 1.50 M Let 1.00 mol/L Ag+ react with CO32‒ to completion because Ksp << 1. Change −1.00 −0.500 Reacts completely After 0 1.00 New initial s mol/L Ag2CO3(s) dissolves to reach equilibrium Change −s → +2s +s Equil. 2s 1.00 + s Ag2CO3(s)
Ksp = 8.1 × 10−12 = [Ag+]2[ CO32−] = (2s)2( 1.00 + s) 4s2(1.00) Solving: s = 1.4 × 10−6 mol/L; assumption good (s is 1.4 × 10‒4% of 1.00). [Ag+] = 2(1.4 × 10−6) = 2.8 × 10−6 M; [CO32−] = 1.00 + 1.4 × 10−6 = 1.00 M 74.
50.0 mL × 0.10 M = 5.0 mmol Pb2+; 50.0 mL × 1.0 M = 50. mmol Cl−. For this solution, Q > Ksp, so PbCl2 precipitates. Assume precipitation of PbCl2(s) is complete. 5.0 mmol Pb2+ requires 10. mmol of Cl− for complete precipitation, which leaves 40. mmol Cl− in excess. Now, let some of the PbCl2(s) redissolve to establish equilibrium.
⇌
PbCl2(s) Initial Equil.
Pb2+(aq)
+
2 Cl−(aq)
40. mmol = 0.40 M 100 .0 mL s mol/L of PbCl2(s) dissolves to reach equilibrium s 0.40 + 2s
0
Ksp =[Pb2+][Cl−]2, 1.6 × 10 −5 = s(0.40 + 2s)2 s(0.40)2 s = 1.0 × 10 −4 mol/L; assumption good. At equilibrium: [Pb2+] = s = 1.0 × 10 −4 mol/L; [Cl−] = 0.40 + 2(1.0 × 10 −4 ) = 0.40 M 75.
Ag3PO4(s) ⇌ 3 Ag+(aq) + PO43−(aq); when Q is greater than Ksp, precipitation will occur. We will calculate the [Ag+]0 necessary for Q = Ksp. Any [Ag+]0 greater than this calculated number will cause precipitation of Ag3PO4(s). In this problem, [PO43-]0 = [Na3PO4]0 = 1.0 × 10−5 M. Ksp = 1.8 × 10 −18 ; Q = 1.8 × 10 −18 = [Ag+ ]30 [PO43−]0 = [Ag+ ]30 (1.0 × 10 −5 M) 1/ 3
1.8 10 −18 [Ag ]0 = −5 1 . 0 10 +
, [Ag+]0 = 5.6 × 10 −5 M
When [Ag+]0 = [AgNO3]0 is greater than 5.6 × 10 −5 M, precipitation of Ag3PO4(s) will occur. 76.
MgF2(s) ⇌ Mg2+(aq) + 2 F−(aq)
Ksp = 6.4 × 10−9
820
CHAPTER 16
SOLUBILITY AND COMPLEX ION EQUILIBRIA
Q = Ksp = 6.4 × 10−9 = [Mg 2+ ]0 [F− ]02 = (3.0 × 10−3) [F − ]02 , [F−]0 = 1.5 × 10 −3 M When [F−] = 1.5 × 10 −3 M, we are at equilibrium (Ksp = Q) . For a precipitate to form, we want Q > Ksp. This will occur when [F−]0 = [KF]0 > 1.5 × 10 −3 M. 77.
For each salt, determine the [OH‒]0 necessary for Q = Ksp. Precipitation occurs when Q > Ksp. This will occur when the [OH‒]0 is greater than the number calculated for Q = Ksp. The salt that requires the smallest amount of OH‒ will precipitate first as OH‒ is added gradually. For AgOH (Ksp = 2 × 10‒8): Q = Ksp = 2 × 10−8 = [Ag+ ]0 [OH− ]0 = (0.010)[OH‒]0, [OH‒]0 = 2 × 10−6 M For Sc(OH)3 (Ksp = 1 × 10‒14): Q = Ksp = 1 × 10−14= [Sc3+ ]0 [OH− ]30 = (0.010) [OH− ]30 , [OH‒]0 = 1 × 10−4 M AgOH(s) will precipitate when the OH‒ concentration is a little larger than 2 × 10−6 M, while Sc(OH)3(s) will form when the OH‒ concentration is a little larger than 1 × 10−4 M. Because precipitation of AgOH(s) requires the smaller amount of OH‒, AgOH will precipitate first and will form when the [OH‒]0 > 2 × 10−6 M.
78.
For each salt, determine the [Ag+]0 necessary for Q = Ksp. Precipitation occurs when Q > Ksp. This will occur when the [Ag+]0 is greater than the number calculated for Q = K sp. The salt that requires the smallest amount of Ag+ will precipitate first as Ag+ is added gradually. For Ag2SO4: Q = Ksp = 1.2 × 10−5 = [Ag+ ]02 [SO42− ]0 = [Ag+ ]02 [0.10]0 , [Ag+]0 = 1.1 × 10−2 M For Ag2CrO4: Q = Ksp = 9.0 × 10−12 = [Ag+ ]02 [CrO42− ]0 = [Ag+ ]02 [0.10]0 , [Ag+]0 = 9.5 × 10−6M For AgIO3: Q = Ksp = 3.1 × 10−8 = [Ag+ ]0 [IO3− ]0 = [Ag+ ]0 [0.10]0 , [Ag+]0 = 3.1 × 10−7 M For Ag3PO4: Q = Ksp = 1.8 × 10−18 = [Ag+ ]30 [PO43− ]0 = [Ag+ ]30 [0.10]0 , [Ag+]0 = 2.6 × 10−6 M AgIO3(s) will precipitate first since it requires the smallest concentration of Ag+ for Q > Ksp.
79.
For each lead salt, we will calculate the [Pb 2+]0 necessary for Q = Ksp. Any [Pb2+]0 greater than this value will cause precipitation of the salt (Q > K sp). PbF2(s) ⇌ Pb2+(aq) + 2 F−(aq) [Pb2+]0 =
4 10 −8 (1 10 − 4 ) 2
=4M
PbS(s) ⇌ Pb2+(aq) + S2−(aq) [Pb2+]0 =
7 10 −29 1 10
−4
Ksp = 4 × 10−8; Q = 4 × 10−8 = [Pb2+ ]0 [F − ]02
Ksp = 7 × 10−29; Q = 7 × 10−29 = [Pb2+]0[S2−]0
= 7 × 10−25 M
Pb3(PO4)2(s) ⇌ 3 Pb2+(aq) + 2 PO43−(aq) Ksp = 1 × 10−54
CHAPTER 16
SOLUBILITY AND COMPLEX ION EQUILIBRIA
821
Q = 1 × 10−54 = [Pb2+ ]30 [PO43− ]02 1 10 −54 [Pb ]0 = −4 2 (1 10 ) 2+
1/ 3
= 5 × 10−16 M
From the calculated [Pb2+]0, the least soluble salt is PbS(s), and it will form first. Pb3(PO4)2(s) will form second, and PbF2(s) will form last because it requires the largest [Pb2+]0 in order for precipitation to occur. 80.
From Table 16.1, Ksp for NiCO3 = 1.4 × 10 −7 and Ksp for CuCO3 = 2.5 × 10 −10 . From the Ksp values, CuCO3 will precipitate first since it has the smaller Ksp value and will be the least soluble. For CuCO3(s), precipitation begins when: [CO32−] =
K sp, CuCO 3 2+
[Cu ]
2.5 10 −10 = 1.0 × 10 −9 M CO32− 0.25 M
=
For NiCO3(s) to precipitate: [CO32−] =
K sp, NiCO3 [ Ni2+ ]
=
1.4 10 −7 = 5.6 × 10 −7 M CO32− 0.25 M
Determining the [Cu2+] when CuCO3(s) begins to precipitate: [Cu2+] =
K sp, CuCO3 2−
[CO 3 ]
=
2.5 10 −10 = 4.5 × 10 −4 M Cu2+ 5.6 10 −7 M
For successful separation, 1% Cu 2+ or less of the initial amount of Cu2+ (0.25 M) must be present before NiCO3(s) begins to precipitate. The percent of Cu2+ present when NiCO3(s) begins to precipitate is: 4.5 10 −4 M × 100 = 0.18% Cu2+ 0.25 M
Because less than 1% of the initial amount of Cu2+ remains, the metals can be separated through slow addition of Na2CO3(aq).
Complex Ion Equilibria 81.
a.
Ni2+ + CN− ⇌ NiCN+ NiCN+ + CN− ⇌ Ni(CN)2 Ni(CN)2 + CN− ⇌ Ni(CN)3− Ni(CN)3− + CN− ⇌ Ni(CN)42− Ni2+(aq) + 4 CN−(aq) ⇌ Ni(CN)42−(aq)
K1 K2 K3 K4 Kf = K1K2K3K4
Note: The various K constants are included for your information. Each CN− adds with a corresponding K value associated with that reaction. The overall formation constant Kf for the overall reaction is equal to the product of all the stepwise K values.
822
CHAPTER 16 b.
SOLUBILITY AND COMPLEX ION EQUILIBRIA
V3+ + C2O42− ⇌ VC2O4+ VC2O4+ + C2O42− ⇌ V(C2O4)2− V(C2O4)2− + C2O42− ⇌ V(C2O4)33−
K1 K2 K3
V3+(aq) + 3 C2O42−(aq) ⇌ V(C2O4)33−(aq) 82.
a.
Co3+ + F− ⇌ CoF2+ CoF2+ + F− ⇌ CoF2+ CoF2+ + F− ⇌ CoF3 CoF3 + F− ⇌ CoF4− CoF4− + F− ⇌ CoF52− CoF52− + F− ⇌ CoF63− Co3+(aq) + 6 F−(aq) ⇌ CoF63−(aq)
b.
Kf = K1K2K3
K1 K2 K3 K4 K5 K6 Kf = K1K2K3K4K5K6
Zn2+ + NH3 ⇌ ZnNH32+ ZnNH32+ + NH3 ⇌ Zn(NH3)22+ Zn(NH3)22+ + NH3 ⇌ Zn(NH3)32+ Zn(NH3)32+ + NH3 ⇌ Zn(NH3)42+
K1 K2 K3 K4
Zn2+(aq) + 4 NH3(aq) ⇌ Zn(NH3)42+(aq) 83.
Fe3+(aq) + 6 CN−(aq) K=
84.
86.
3−
K=
[Fe(CN) 6 ] [Fe 3+ ][CN − ]6
1.5 10 −3 = 1.0 × 1042 (8.5 10 − 40 )(0.11) 6
Cu2+(aq) + 4 NH3(aq) ⇌ Cu(NH3)42+(aq) K=
85.
⇌ Fe(CN)63−
Kf = K1K2K3K4
1.0 10 −3 (1.8 10
−17
)(1.5)
4
2+
K=
[Cu(NH3 ) 4 ] [Cu 2+ ][NH3 ] 4
= 1.1 1013
Hg2+(aq) + 2 I−(aq) ⇌ HgI2(s); HgI2(s) + 2 I−(aq) ⇌ HgI42−(aq) orange ppt soluble complex ion The weak base CN‒ produces OH‒ in solution. The precipitate that forms is Ni(OH)2(s). This precipitate dissolves on addition of more CN− because Ni2+ forms a soluble complex ion, Ni(CN)42−. Ni2+(aq) + 2 OH‒(aq) ⇌ Ni(OH)2(s); Ni(OH)2(s) + 4 CN‒(aq) ⇌ Ni(CN)42−(aq) + 2 OH‒(aq)
87.
The formation constant for HgI42− is an extremely large number. Because of this, we will let the Hg2+ and I− ions present initially react to completion and then solve an equilibrium problem to determine the Hg2+ concentration.
CHAPTER 16
SOLUBILITY AND COMPLEX ION EQUILIBRIA Hg2+(aq) + 4 I−(aq)
Before Change After Change Equil.
⇌
HgI42−(aq)
823
K = 1.0 × 1030
0.010 M 0.78 M 0 −0.010 −0.040 → +0.010 Reacts completely (K is large). 0 0.74 0.010 New initial x mol/L HgI42− dissociates to reach equilibrium +x +4x −x x 0.74 + 4x 0.010 − x 2−
K = 1.0 × 1030 = 1.0 × 1030 =
[HgI 4 ] (0.010 − x) ; making normal assumptions: = 2+ − 4 [Hg ][I ] ( x) (0.74 + 4 x) 4
(0.010 ) , x = [Hg2+] = 3.3 × 10−32 M ; ( x) (0.74) 4
assumptions good.
Note: 3.3 × 10 −32 mol/L corresponds to one Hg2+ ion per 5 × 107 L. It is very reasonable to approach this problem in two steps. The reaction does essentially go to completion. Ni2+(aq) + 6 NH3(aq)
88.
⇌
Ni(NH3)62+(aq)
K = 5.5 × 108
Initial
0 3.0 M 0.10 mol/0.50 L = 0.20 M 2+ x mol/L Ni(NH3)6 dissociates to reach equilibrium Change +x +6x −x Equil. x 3.0 + 6x 0.20 − x K = 5.5 × 108 = =
[Ni(NH 3 ) 62+ ] 2+
[Ni ][NH 3 ]
6
=
(0.20 − x) ( x) (3.0 + 6 x)
6
, 5.5 10 8
(0.20) ( x) (3.0) 6
x = [Ni2+] = 5.0 × 10−13 M; [Ni(NH3)62+] = 0.20 M − x = 0.20 M; assumptions good. 89.
[X−]0 = 5.00 M and [Cu+]0 = 1.0 × 10 −3 M because equal volumes of each reagent are mixed. Because the K values are much greater than 1, assume the reaction goes completely to CuX32−, and then solve an equilibrium problem. Cu+(aq) Before After Equil. K=
1.0 × 10 −3 M 0 x
+
3 X−(aq)
⇌ CuX32−(aq)
5.00 M 0 −3 5.00 − 3( 10 ) 5.00 1.0 × 10 −3 5.00 + 3x 1.0 × 10 −3 − x
(1.0 10 −3 − x)
1.0 10 −3
( x)(5.00 + 3 x)
( x)(5.00) 3
= 1.0 × 109 ≈ 3
K = K1 × K2 × K3 K = 1.0 × 109 Reacts completely
, x = [Cu+] = 8.0 × 10 −15 M
Assumptions good. [CuX32−] = 1.0 × 10 −3 − 8.0 × 10 −15 = 1.0 × 10 −3 M
824
CHAPTER 16
SOLUBILITY AND COMPLEX ION EQUILIBRIA
2−
K3 =
[CuX 3 ]
= 1.0 × 103 =
−
[CuX 2 ][X − ]
(1.0 10 −3 ) −
[CuX2−] = 2.0 × 10 −7 M
,
[CuX 2 ](5.00)
Summarizing:
90.
[CuX32−] = 1.0 × 10 −3 M
(answer a)
[CuX2−] = 2.0 × 10 −7 M
(answer b)
[Cu2+] = 8.0 × 10 −15 M
(answer c)
[Be2+]0 = 5.0 × 10−5 M and [F−]0 = 4.0 M because equal volumes of each reagent are mixed, so all concentrations given in the problem are diluted by a factor of one-half. Because the K values are large, assume all reactions go to completion, and then solve an equilibrium problem.
Before After Equil.
Be2+(aq)
+ 4 F−(aq)
⇌ BeF42−(aq)
5.0 × 10−5 M 0 x
4.0 M 4.0 M 4.0 + 4x
0 5.0 × 10−5 M 5.0 × 10−5 − x
2−
K = 7.5 × 1012 =
[BeF 4 ]
= [Be 2 + ][F − ] 4
5.0 10 −5 − x x (4.0 + 4 x) 4
K = K1K2K3K4 = 7.5 × 1012
5.0 10 −5 x (4.0) 4
x = [Be2+] = 2.6 × 10 −20 M; assumptions good. [F−] = 4.0 M; [BeF42−] = 5.0 × 10−5 M Now use the stepwise K values to determine the other concentrations. K1 = 7.9 × 10 4 = K2 = 5.8 × 10 3 =
[BeF+ ] [BeF+ ] = , [BeF+] = 8.2 × 10 −15 M − 20 2+ − (2.6 10 )(4.0) [Be ][F ] [BeF 2 ] +
−
[BeF ][F ]
[BeF 2 ]
=
(8.2 10 −15 )(4.0) −
−
K3 = 6.1 × 10 2 = 91.
a.
[BeF3 ] −
[BeF2 ][F ]
AgI(s)
=
[BeF 3 ] (1.9 10
⇌
Initial s = solubility (mol/L) Equil.
−10
Ag+(aq)
)(4.0)
+
0 s
, [BeF2] = 1.9 × 10 −10 M , [BeF3−] = 4.6 × 10 −7 M I−(aq)
Ksp = [Ag+][I−] = 1.5 × 10 −16
0 s
Ksp = 1.5 × 10 −16 = s2, s = 1.2 × 10 −8 mol/L b.
AgI(s) ⇌ Ag+ + I− Ag + 2 NH3 ⇌ Ag(NH3)2+ +
AgI(s) + 2 NH3(aq) ⇌ Ag(NH3)2+(aq) + I−(aq)
Ksp = 1.5 × 10 −16 Kf = 1.7 × 107 K = Ksp × Kf = 2.6 × 10 −9
CHAPTER 16
SOLUBILITY AND COMPLEX ION EQUILIBRIA
825
+ 2 NH3 ⇌ Ag(NH3)2+ + I− 3.0 M 0 0 s mol/L of AgI(s) dissolves to reach equilibrium = molar solubility 3.0 − 2s s s AgI(s)
Initial Equil.
+
K=
[Ag( NH3 ) 2 ][I − ] s2 s2 −9 10 = = 2.6 × , s = 1.5 × 10 −4 mol/L (3.0 − 2s) 2 (3.0) 2 [ NH3 ] 2
Assumption good. c. The presence of NH3 increases the solubility of AgI. Added NH3 removes Ag+ from solution by forming the complex ion, Ag(NH3)2+. As Ag+ is removed, more AgI(s) will dissolve to replenish the Ag+ concentration. AgBr(s) ⇌ Ag+ + Br− Ag + 2 S2O32− ⇌ Ag(S2O3)23−
92.
+
AgBr(s) + 2 S2O32− ⇌ Ag(S2O3)23− + Br-
Change Equil. K=
K = Ksp × Kf = 14.5 (Carry extra sig. figs.)
2 S2O32−(aq) ⇌ Ag(S2O3)23−(aq) + 0.500 M 0 s mol/L AgBr(s) dissolves to reach equilibrium −s −2s → +s 0.500 − 2s s AgBr(s)
Initial
Ksp = 5.0 × 10 −13 Kf = 2.9 × 1013
+
Br−(aq) 0 +s s
s2 = 14.5; taking the square root of both sides: (0.500 − 2s) 2 s = 3.81, s = 1.91 − (7.62)s, s = 0.222 mol/L 0.500 − 2 s
1.00 L ×
0.222 mol AgBr 187 .8 g AgBr = 41.7 g AgBr = 42 g AgBr L mol AgBr
AgCl(s) ⇌ Ag+ + Cl− Ag+ + 2 NH3 ⇌ Ag(NH3)2+
93.
AgCl(s) + 2 NH3(aq)
⇌ Ag(NH3)2+(aq) + Cl−(aq)
AgCl(s) + 2 NH3 Initial Equil.
Ksp = 1.6 × 10 −10 Kf = 1.7 × 107
⇌
K = Ksp × Kf = 2.7 × 10 −3
Ag(NH3)2+ + Cl−
1.0 M 0 0 s mol/L of AgCl(s) dissolves to reach equilibrium = molar solubility 1.0 – 2s s s
K = 2.7 × 10
−3
+
[Ag( NH3 ) 2 ][Cl − ] s2 = = ; taking the square root: [ NH3 ]2 (1.0 − 2s ) 2
s = (2.7 × 10 −3 )1/2 = 5.2 × 10 −2 , s = 4.7 × 10 −2 mol/L 1 .0 − 2 s
In pure water, the solubility of AgCl(s) is (1.6 × 10 −10 )1/2 = 1.3 × 10 −5 mol/L. Notice how the presence of NH3 increases the solubility of AgCl(s) by over a factor of 3500.
826 94.
CHAPTER 16 a.
⇌
CuCl(s) Initial Equil.
SOLUBILITY AND COMPLEX ION EQUILIBRIA Cu+(aq) +
Cl−(aq)
0 s
0 s
s = solubility (mol/L)
Ksp = 1.2 × 10 −6 = [Cu+][Cl−] = s2, s = 1.1 × 10 −3 mol/L b. Cu+ forms the complex ion CuCl2- in the presence of Cl-. We will consider both the Ksp reaction and the complex ion reaction at the same time. CuCl(s) ⇌ Cu+(aq) + Cl−(aq) Cu (aq) + 2 Cl− (aq) ⇌ CuCl2−(aq)
Ksp = 1.2 × 10 −6 Kf = 8.7 × 104
+
CuCl(s) + Cl−(aq) ⇌ CuCl2−(aq)
⇌
CuCl(s) + Cl− Initial Equil.
0.10 M 0.10 − s −
K = 0.10 =
[CuCl 2 ] −
[Cl ]
=
K = Ksp × Kf = 0.10
CuCl2− 0 s
where s = solubility of CuCl(s) in mol/L
s , 1.0 × 10 −2 − 0.10 s = s, s = 9.1 × 10 −3 mol/L 0.10 − s
95.
Test tube 1: Added Cl− reacts with Ag+ to form a silver chloride precipitate. The net ionic equation is Ag+(aq) + Cl−(aq) → AgCl(s). Test tube 2: Added NH3 reacts with Ag+ ions to form a soluble complex ion, Ag(NH3)2+. As this complex ion forms, Ag+ is removed from the solution, which causes the AgCl(s) to dissolve. When enough NH3 is added, all the silver chloride precipitate will dissolve. The equation is AgCl(s) + 2 NH3(aq) → Ag(NH3)2+(aq) + Cl−(aq). Test tube 3: Added H+ reacts with the weak base, NH3, to form NH4+. As NH3 is removed from the Ag(NH3)2+ complex ion, Ag+ ions are released to solution and can then react with Cl− to re-form AgCl(s). The equations are Ag(NH3)2+(aq) + 2 H+(aq) → Ag+(aq) + 2 NH4+(aq) and Ag+(aq) + Cl−(aq) → AgCl(s).
96.
In NH3, Cu2+ forms the soluble complex ion Cu(NH3)42+. This increases the solubility of Cu(OH)2(s) because added NH3 removes Cu2+ from the equilibrium causing more Cu(OH)2(s) to dissolve. In HNO3, H+ removes OH− from the Ksp equilibrium causing more Cu(OH)2(s) to dissolve. Any salt with basic anions will be more soluble in an acid solution. AgF(s) will be more soluble in either NH3 or HNO3. This is because Ag+ forms the complex ion Ag(NH3)2+, and F− is a weak base, so it will react with added H+. AgCl(s) will be more soluble only in NH3 due to Ag(NH3)2+ formation. In acid, Cl- is a horrible base, so it doesn’t react with added H +. AgCl(s) will not be more soluble in HNO3.
ChemWork Problems 97.
AgCl(s) Initial Equil.
⇌
s = solubility (mol/L)
Ag+(aq) 0 s
+
Cl−(aq) 0 s
Ksp = [Ag+][Cl−] = s(s) = s2; the expression given in the problem is incorrect.
CHAPTER 16
SOLUBILITY AND COMPLEX ION EQUILIBRIA ⇌
PbCl2(s)
Pb2+(aq)
Initial s = solubility (mol/L) Equil.
827
2 Cl−(aq)
+
0 s
0 2s
Ksp = [Pb2+][Cl−]2 = s(2s)2 = 4s3; the expression given in the problem is incorrect.
⇌
Ag3PO4(s) Initial Equil.
3 Ag+(aq)
s = solubility (mol/L)
+
0 3s
PO43− (aq) 0 s
Ksp = [Ag+]3[PO43−] = (3s)3(s) = 27s4; the expression given in the problem is incorrect.
⇌
Pb3(PO4)2(s)
3 Pb2+(aq) + 2 PO43−(aq)
Initial s = solubility (mol/L) Equil.
0 3s
0 2s
Ksp = [Pb2+]3[PO43−]2 = (3s)3(2s)2 = 108s5; the expression given in the problem is incorrect. All the expressions are incorrect. 98.
For CuS and AgI, both break up into 2 ions, so the the Ksp to molar solubility expression is Ksp = s2. (1.0 × 10‒15)2 = 1.0 × 10‒30; the compound is not CuS or AgI. Ag2S breaks up into 3 ions with an expression of Ksp = 4s3. 4(1.0 × 10‒15)3 = 4.0 × 10‒45; the compound is not Ag2S. Ag3PO4 breaks up into 4 ions with an expression of K sp = 27s4. 27(1.0 × 10‒15)4 = 2.7 × 10‒59; the compound is not Ag3PO4. The compound must Bi2S3. Let’s derive the Ksp to molar solubility expression for Bi2S3. Bi2S3 (s) Initial Change Equil.
⇌
2 Bi3+(aq)
+
3 S2−(aq)
0 0 s mol/L of Bi2S3 (s) dissolves to reach equilibrium −s → +2s +3s 2s 3s
Ksp = [Bi3+]2[ S2−]3 = (2s)2(3s)3 = 108s5, Ksp = 108(1.0 × 10‒15)5 = 1.1 × 10‒73 The compound having a molar solubility of 1.0 × 10 ‒15 mol/L is Bi2S3(s). 99.
Mol Ag+ added = 0.200 L
0.24 mol AgNO3 1 mol Ag+ = 0.048 mol Ag+ L mol AgNO3
The added Ag+ will react with the halogen ions to form a precipitate. Because the K sp values are small, we can assume these precipitation reactions go to completion. The order of precipitation will be AgI(s) first (the least soluble compound since K sp is the smallest), followed by AgBr(s), with AgCl(s) forming last [AgCl(s) is the most soluble compound listed since it has the largest Ksp]. Let the Ag+ react with I− to completion.
828
CHAPTER 16 Ag+(aq) Before Change After
+
0.048 mol −0.018 0.030 mol
SOLUBILITY AND COMPLEX ION EQUILIBRIA
I−(aq)
→
0.018 mol −0.018 0
AgI(s)
K = 1/Ksp >> 1
0 +0.018 0.018 mol
I− is limiting.
Let the Ag+ remaining react next with Br− to completion. Ag+(aq) Before Change After
+
0.030 mol −0.018 0.012 mol
→
Br−(aq) 0.018 mol −0.018 0
AgBr(s)
K = 1/Ksp >> 1
0 +0.018 0.018 mol
Br− is limiting.
Finally, let the remaining Ag+ react with Cl− to completion. Ag+(aq) Before Change After
+
0.012 mol −0.012 0
→
Cl−(aq) 0.018 mol −0.012 0.006 mol
AgCl(s)
K = 1/Ksp >> 1
0 +0.012 0.012 mol
Ag+ is limiting.
Some of the AgCl will redissolve to produce some Ag + ions; we can’t have [Ag+] = 0 M. Calculating how much AgCl(s) redissolves: AgCl(s) Initial Change Equil.
⇌
Ag+(aq)
+
Cl−(aq)
Ksp = 1.6 × 10 −10
0 0.006 mol/0.200 L = 0.03 M s mol/L of AgCl dissolves to reach equilibrium −s → +s +s s 0.03 + s
Ksp = 1.6 × 10 −10 = [Ag+][Cl−] = s(0.03 + s) (0.03)s s = 5 × 10 −9 mol/L; the assumption that 0.03 + s 0.03 is good. Mol AgCl present = 0.012 mol – 5 × 10 −9 mol = 0.012 mol Mass AgCl present = 0.012 mol AgCl
143 .4 g = 1.7 g AgCl mol AgCl
[Ag+] = s = 5 × 10 −9 mol/L 100.
Since Ksp for Mg(OH)2 << 1, not a lot of Mg(OH)2(s) will dissolve. Let’s calculate the solubility of Mg(OH)2(s) in water. Mg(OH)2(s) Initial Equil.
⇌
s = solubility (mol/L)
Mg2+(aq)
+ 2 OH−(aq)
Ksp = 8.9 × 10−12
0 s
1.0 × 10−7 M 1.0 × 10−7 + 2s
(from water)
Ksp = [Mg2+][OH−]2 = s(1.0 × 10−7 + 2s)2; assume that 1.0 × 10−7 + 2s 2s, then:
CHAPTER 16
SOLUBILITY AND COMPLEX ION EQUILIBRIA
829
Ksp = 8.9 × 10−12 = s(2s)2 = 4s3, s = 1.3 × 10−4 mol/L; assumption good. 1.3 10 −4 mol Mg(OH) 58.33 g Mg(OH)2 = 7.6 × 10−3 g Mg(OH)2 L mol Mg(OH)
Of the 10.0 g Mg(OH)2(s), only a fraction will dissolve (as expected). The moles of OH − present is: 1.3 10 −4 mol Mg(OH)2 2 mol OH − 0.5000 L = 1.3 × 10−4 mol OH− L Mg(OH)2
Even though there isn’t a lot of OH− in solution, as soon as the hydroxide ion neutralizes some stomach acid, more of the excess Mg(OH)2(s) will dissolve to replenish the OH−. This process will continue as long as Mg(OH)2(s) is present. So not a lot of magnesium hydroxide has to dissolve initially; there just has to be excess of the Mg(OH)2(s) present to replenish the hydroxide ion as it is neutralized. 101.
M3(PO4)2(s) Initial Equil.
⇌
3 M2+(aq) + 2 PO43−(aq)
s = solubility (mol/L)
0 3s
0 2s
Ksp = [M2+]3[PO43−]2 = (3s)3(2s)2 = 108s5; Ksp = 108(6.2 × 10‒12)5 = 9.9 × 10‒55. From appendix A5.4, Pb3(PO4)2 has a Ksp value closest to 9.9 × 10‒55, so M is Pb2+. Now solve the common ion part of the problem.
⇌
Pb3(PO4)2(s) Initial Equil.
3 Pb2+(aq) + 2 PO43−(aq)
s = solubility (mol/L)
0 3s
0.10 M 0.10 + 2s
Ksp = 9.9 × 10‒55 = [Pb2+]3[PO43−]2 = (3s)3(0.10 + 2s)2 27s3(0.10)2 s = 1.5 × 10‒18 mol/L; the assumption is good. 102.
MX4(s) Initial Equil.
⇌
M4+(aq)
s = solubility (mol/L)
+
4 X− (aq)
0 s
0 4s
Ksp = [M4+][X−]4 = (s)(4s)4 = 256s5; Ksp = 256(1.00 × 10‒2)5 = 2.56 × 10‒8 103.
Ca5(PO4)3OH(s) Initial Equil.
s = solubility (mol/L)
⇌
5 Ca2+ + 3 PO43− 0 5s
0 3s
OH−
+
1.0 × 10 −7 from water s + 1.0 × 10 −7 s
Ksp = 6.8 × 10 −37 = [Ca2+]5[PO43−]3[OH−] = (5s)5(3s)3(s) 6.8 × 10 −37 = (3125)(27)s9, s = 2.7 × 10 −5 mol/L;
assumption is good.
The solubility of hydroxyapatite will increase as the solution gets more acidic because both phosphate and hydroxide can react with H +.
830
CHAPTER 16
⇌
Ca5(PO4)3F(s) Initial Equil.
SOLUBILITY AND COMPLEX ION EQUILIBRIA 5 Ca2+(aq) + 3 PO43−(aq)
s = solubility (mol/L)
0 5s
+
0 3s
F−(aq) 0 s
Ksp = 1 × 10 −60 = (5s)5(3s)3(s) = (3125)(27)s9, s = 6 × 10 −8 mol/L The hydroxyapatite in tooth enamel is converted to the less soluble fluorapatite by fluoridetreated water. The less soluble fluorapatite is more difficult to remove, making teeth less susceptible to decay. 104.
1 mg F− 1g 1 mol F− = 5.3 × 10 −5 M F− = 5 × 10 −5 M F− − L 1000 mg 19.00 g F
CaF2(s) ⇌ Ca2+(aq) + 2 F−(aq) Ksp = [Ca2+][F−]2 = 4.0 × 10 −11 ; precipitation will occur when Q > Ksp. Let’s calculate [Ca2+] so that Q = Ksp. Q = 4.0 × 10 −11 = [Ca2+]0 [F − ]02 = [Ca2+]0(5 × 10 −5 )2, [Ca2+]0 = 2 × 10 −2 M CaF2(s) will precipitate when [Ca2+]0 > 2 × 10 −2 M. Therefore, hard water should have a calcium ion concentration of less than 2 × 10 −2 M to avoid CaF2(s) formation. 105.
Al(OH)3(s) ⇌ Al3+(aq) + 3 OH−(aq)
Ksp = 2 × 10−32
Q = 2 × 10 −32 = [Al3+ ]0 [OH − ]30 = (0.2) [OH − ]30 , [OH−]0 = 4.6 × 10 −11 (carrying extra sig. fig.) pOH = −log(4.6 × 10−11 ) = 10.3; when the pOH of the solution equals 10.3, K sp = Q. For precipitation, we want Q > Ksp. This will occur when [OH−]0 > 4.6 × 10−11 or when pOH < 10.3. Because pH + pOH = 14.00, precipitation of Al(OH) 3(s) will begin when pH > 3.7 because this corresponds to a solution with pOH < 10.3. 106.
CaF2(s) Initial Equil.
⇌
s = solubility (mol/L)
Ca2+(aq) + 0 s
2 F−(aq)
Ksp = [Ca2+][F−]2
0.050 M 0.050 + 2s
Ksp = 4.0 × 10−11 = s(0.050 + 2s)2 s(0.050)2, s = 1.6 × 10−8 mol/L; assumption good.
1.6 10−8 mol CaF2 = 4.8 × 10‒8 mol of CaF2 will dissolve. L 1 mol PbI 2 0.24 g PbI 2 461.0 g s = solubility = = 2.6 10 −3 M 0.2000 L 3.0 L
107.
PbI2(s) Initial Equil.
⇌
s = solubility (mol/L)
Pb2+(aq) +
2 I−(aq)
0 s
0 2s
Ksp = s(2s)2 = 4s3, Ksp = 4(2.6 × 10−3)3 = 7.0 × 10−8
Ksp = [Pb2+][I−]2
CHAPTER 16
SOLUBILITY AND COMPLEX ION EQUILIBRIA
108.
C7H5BiO4(s) + H2O(l) ⇌ C7H4O32−(aq) + Bi3+(aq) + OH−(aq) Initial Equil.
s = solubility (mol/L)
0 s
831
1.0 × 10−7 M (from water) 1.0 × 10−7 + s
0 s
K = [C7H4O32−][Bi3+][OH−] = s(s)(1.0 × 10−7 + s); from the problem, s = 3.2 × 10−19 mol/L: K = (3.2 × 10−19)2(1.0 × 10−7 + 3.2 × 10−19) = 1.0 × 10−44 109.
a.
AgOH(s)
⇌
Ag+(aq) + OH−(aq)
Initial s = solubility (mol/L) 0 Equil. s
Ksp = [Ag+][OH−] = 2.0 × 10 −8 (Ignoring OH− from water.)
~0 s
Ksp = 2.0 × 10 −8 = s(s) = s2, s = 1.4 × 10 −4 mol/L Assumption good, the amount of OH− from the autoionization of H2O can be ignored. [OH−] = s = 1.4 × 10 −4 mol/L; pOH = 3.85, pH = 10.15 b.
Cd(OH)2(s)
⇌
Cd2+(aq) + 2 OH−(aq)
Equil.
s
Ksp = [Cd2+][OH−]2 = 5.9 × 10 −15
2s
Ksp = 5.9 × 10 −15 = 4s3, s = 1.1 × 10 −5 mol/L; asssumption good. [OH−] = 2(1.1 × 10 −5 ) = 2.2 × 10 −5 M; pOH = 4.66, pH = 9.34 c.
Pb(OH)2(s)
⇌
Equil.
Pb2+(aq) + 2 OH−(aq) s
Ksp = [Pb2+][OH−]2 = 1.2 × 10 −15
2s
Ksp = 1.2 × 10 −15 = 4s3, s = 6.7 × 10 −6 mol/L; assumption good. [OH−] = 2(6.7 × 10 −6 ) = 1.3 × 10 −5 mol/L; pOH = 4.89, pH = 9.11 110.
The solution with the smallest solubility will be one of solutions with a common ion present. This is either solution b or c. Ce(IO3)3(s) Initial Equil.
⇌
s = solubility (mol/L)
Ce3+(aq) 0.10 M 0.10 + s
+
3 IO3−(aq) 0 3s
Ksp = 3.5 × 10‒10 = [Ce3+][IO3−]3 = (0.10 + s)(3s)3 (0.10)27s3 s = 5.1 × 10‒4 mol/L; assumptions good.
832
CHAPTER 16 Ce(IO3)3(s) Initial Equil.
⇌
s = solubility (mol/L)
SOLUBILITY AND COMPLEX ION EQUILIBRIA Ce3+(aq)
+
0 s
3 IO3−(aq) 0.10 M 0.10 + 3s
3.5 × 10‒10 = [Ce3+][IO3−]3 = s(0.10 + 3s)3 s(0.10)3, s = 3.5 × 10‒7 mol/L; assumptions good. The solubility in 0.10 M KIO3 is the smallest. 111.
Cu(OH)2(s) ⇌ Cu2+(aq) + 2 OH−(aq); adding either Cu2+ or OH‒ will shift this equilibrium to the left causing more precipitate to form. So, answers a and b are not possible. Cu2+(aq) + 4 NH3(aq) ⇌ Cu(NH3)42+(aq); when NH3 is added, Cu2+ is removed from solution as it reacts with NH3 to form the soluble complex ion Cu(NH3)42+. As Cu2+ is removed, this forces the Cu(OH)2(s) equilibrium to shift right to replenish the lost Cu2+. Adding NH3 will cause the precipitate to dissolve (answer c is correct). H+(aq) + OH‒(aq) → H2O(l); when H+ is added from the strong acid HNO3, it reacts with OH‒ to form H2O. As OH‒ is removed, this forces the Cu(OH)2(s) equilibrium to shift right to replenish the lost OH‒. Adding HNO3 will also cause the precipitate to dissolve (answer d is correct).
112.
a.
Ag3PO4(s) Initial Equil.
⇌
s = solubility (mol/L)
3 Ag+(aq)
+
0 3s
PO43− (aq) 0 s
Ksp = 1.8 × 10 −18 = [Ag+]3[PO43−] = (3s)3(s) = 27s4 27s4 = 1.8 × 10 −18 , s = (6.7 × 10 −20 )1/4 = 1.6 × 10‒5 mol/L; [Ag+] = 3s = 3(1.6 × 10‒5) = 4.8 × 10‒5 mol/L b. Initial Equil.
Ag3PO4(s) ⇌ 3 Ag+(aq) s = solubility (mol/L) 0 3s
+
PO43−(aq) 0.60 M 0.60 + s
Ksp = 1.8 × 10−18 = [Ag+]3[PO43−] = (3s)3(0.10 + s) (3s)3(0.60) = 27s3(0.60) s = 4.8 × 10−7 mol/L; assumption good. [Ag+] = 3s = 3(4.8 × 10‒7) = 1.4 × 10‒6 mol/L 2.40 mol 2.00 mol 1.00 L L L = 1.20 M; [PO43−]0 = = 1.00 M 2.00 L 2.00 L
1.00 L
c. [Ag+]0 =
Note: Because we added equal volumes of each solution together, the total volume of the solution doubled, which halved the concentrations. We have very large concentrations of the Ag + and PO43− ions. Because the value of Ksp for Ag3PO4(s) is very small (Ksp << 1), the concentrations of the ions cannot both be large values. To solve this complicated problem, let the ions react to completion to form Ag 2CO3 (s), then solve a back equilibrium problem to determine the equilibrium concentrations of the ions. In the stoichiometry part of the problem, Ag + is the limiting reagent.
CHAPTER 16
SOLUBILITY AND COMPLEX ION EQUILIBRIA
833
⇌
3 Ag+ + PO43− Ksp = 1.8 × 10‒18 Before 1.20 M 1.00 M Let 1.20 mol/L Ag+ react with PO43− to completion because Ksp << 1. Change −1.20 −0.400 Reacts completely After 0 0.60 New initial s mol/L Ag3PO4(s) dissolves to reach equilibrium Change −s → +3s +s Equil. 3s 0.60 + s Ag3PO4(s)
Ksp = 1.8 × 10−18 = [Ag+]3[ PO43−] = (3s)3(0.60 + s) 27s3(0.60) Solving: s = 4.8 × 10−7 mol/L; assumption good. [Ag+] = 3(4.8 × 10−7) = 1.4 × 10−6 M d. Once the ions in part c react to completion, the problem turns into calculating the solubility of Ag3PO4 in 0.6 M PO43−. This is the same problem as in part b. Mn2+ + C2O42− ⇌ MnC2O4 MnC2O4 + C2O42− ⇌ Mn(C2O4)22-
113.
K1 = 7.9 × 103 K2 = 7.9 × 101
Mn2+(aq) + 2 C2O42−(aq) ⇌ Mn(C2O4)22−(aq) Cr3+
114. Before Change After Change Equil.
1.0 × 1023 a.
CrEDTA−
+
2 H+ Buffered, constant Reacts completely New initial
Buffered
[CrEDTA − ][H + ] 2 (0.0010 − x) (1.0 10 −6 ) 2 = (x)(0.049 + x) [Cr 3+ ][H 2 EDTA 2− ]
(0.0010 ) (1.0 10 −12 ) , x = [Cr3+] = 2.0 × 10−37 M; assumptions good. x(0.049 )
Pb(OH)2(s) Initial Equil.
⇌
0.0010 M 0.050 M 0 1.0 × 10−6 M −0.0010 −0.0010 → +0.0010 No change 0 0.049 0.0010 1.0 × 10−6 − x mol/L CrEDTA dissociates to reach equilibrium +x +x −x x 0.049 + x 0.0010 − x 1.0 × 10−6
Kf = 1.0 × 1023 =
115.
+ H2EDTA2−
Kf = K1K2 = 6.2 × 105
⇌
s = solubility (mol/L)
Pb2+
+
2 OH− 1.0 × 10−7 M from water 1.0 × 10−7 + 2s
0 s
Ksp = 1.2 × 10−15 = [Pb2+][OH−]2 = s(1.0 × 10−7 + 2s)2 s(2s2) = 4s3 s = [Pb2+] = 6.7 × 10−6 M; assumption is good by the 5% rule. b.
Pb(OH)2(s) Initial Equil.
⇌
Pb2+
+
2 OH−
0 0.10 M pH = 13.00, [OH−] = 0.10 M s mol/L Pb(OH)2(s) dissolves to reach equilibrium s 0.10 (Buffered solution)
1.2 × 10−15 = (s)(0.10)2, s = [Pb2+] = 1.2 × 10−13 M
834
CHAPTER 16
SOLUBILITY AND COMPLEX ION EQUILIBRIA
c. We need to calculate the Pb2+ concentration in equilibrium with EDTA4-. Since K is large for the formation of PbEDTA2−, let the reaction go to completion and then solve an equilibrium problem to get the Pb2+ concentration. Pb2+ + EDTA4− ⇌ PbEDTA2− K = 1.1 × 1018 0.010 M 0.050 M 0 2+ 0.010 mol/L Pb reacts completely (large K) −0.010 −0.010 → +0.010 Reacts completely 0 0.040 0.010 New initial x mol/L PbEDTA2− dissociates to reach equilibrium x 0.040 + x 0.010 − x
Before Change After Equil. 1.1 × 1018 =
(0.010 − x) 0.010 , x = [Pb2+] = 2.3 × 10−19 M; assumptions good. ( x)(0.040 + x) x(0.040 )
Now calculate the solubility quotient for Pb(OH)2 to see if precipitation occurs. The concentration of OH− is 0.10 M since we have a solution buffered at pH = 13.00. Q = [Pb2+]0[OH−]02 = (2.3 × 10−19)(0.10)2 = 2.3 × 10−21 < Ksp (1.2 × 10−15) Pb(OH)2(s) will not form since Q is less than Ksp. 116.
1.0 mL ×
1.0 mmol = 1.0 mmol Cd2+ added to the ammonia solution mL
Thus [Cd2+]0 = 1.0 × 10−3 mol/L. We will first calculate the equilibrium Cd2+ concentration using the complex ion equilibrium and then determine if this Cd2+ concentration is large enough to cause precipitation of Cd(OH)2(s). Cd2+ Before Change After Change Equil.
+
4 NH3
⇌
Cd(NH3)42+
Kf = 1.0 × 107
1.0 × 10−3 M 5.0 M 0 −1.0 × 10−3 −4.0 × 10−3 → +1.0 × 10−3 0 4.996 5.0 1.0 × 10−3 x mol/L Cd(NH3)42+ dissociates to reach equilibrium +x +4x −x x 5.0 + 4x 0.0010 − x
Kf = 1.0 × 107 =
Reacts completely New initial
(0.010 − x) (0.010 ) 4 ( x)(5.0 + 4 x) ( x)(5.0) 4
x = [Cd2+] = 1.6 × 10−13 M; assumptions good. This is the maximum [Cd2+] possible. Now we will determine if Cd(OH)2(s) forms at this concentration of Cd2+. In 5.0 M NH3 we can calculate the pH:
⇌
NH3 + H2O Initial 5.0 M Equil. 5.0 − y
0 y +
Kb = 1.8 × 10−5 =
NH4+ + OH−
Kb = 1.8 × 10−5
~0 y
y2 y2 [ NH4 ][OH− ] = , y = [OH−] = 9.5 × 10−3 M [ NH3 ] 5.0 − y 5.0
CHAPTER 16
SOLUBILITY AND COMPLEX ION EQUILIBRIA
835
Assumptions good; we now calculate the value of the solubility quotient, Q: Q = [Cd2+][OH−]2 = (1.6 × 10−13)(9.5 × 10−3)2 Q = 1.4 × 10−17 < Ksp (5.9 × 10−15); 117.
therefore, no precipitate forms.
Cu(OH)2 ⇌ Cu2+ + 2 OH− Cu2+ + 4 NH3 ⇌ Cu(NH3)42+
a.
Ksp = 1.6 × 10 −19 Kf = 1.0 × 1013
Cu(OH)2(s) + 4 NH3(aq) ⇌ Cu(NH3)42+(aq) + 2 OH−(aq) K = KspKf = 1.6 × 10 −6 b.
Cu(OH)2(s) + 4 NH3
⇌ Cu(NH3)42+ +
2 OH−
K = 1.6 × 10 −6
Initial
5.0 M 0 0.0095 M s mol/L Cu(OH)2 dissolves to reach equilibrium Equil. 5.0 − 4s s 0.0095 + 2s K = 1.6 × 10 −6 =
[Cu( NH3 ) 24+ ][OH − ]2 s (0.0095 + 2s ) 2 = [ NH3 ]4 (5.0 − 4s ) 4
If s is small: 1.6 × 10 −6 =
s (0.0095 ) 2 , s = 11. mol/L (5.0) 4
Assumptions are horrible. We will solve the problem by successive approximations. scalc =
1.6 10 −6 (5.0 − 4sguess ) 4 (0.0095 + 2sguess ) 2
; the results from six trials are:
sguess:
0.10, 0.050, 0.060, 0.055, 0.056
scalc:
1.6 × 10-2, 0.071, 0.049, 0.058, 0.056
Thus the solubility of Cu(OH)2 is 0.056 mol/L in 5.0 M NH3. 118.
a.
836
CHAPTER 16
SOLUBILITY AND COMPLEX ION EQUILIBRIA
b.
c.
119.
From the problem, [Hg2+]0 = 0.088 M and [I−]0 = 5.0 M. Because the K values are large, assume all reactions go to completion, and then solve an equilibrium problem. Hg2+(aq) Before Change After Equil.
+ 4 I−(aq)
⇌ HgI42−(aq)
5.0 M −4(0.088) 4.6 M 4.6 + 4x
0 +0.088 0.088 M 0.088 − x
0.088M −0.088 0 x 2−
K = 1.0 × 10
30
=
[HgI 4 ] [Hg
2+
− 4
][I ]
=
0.088 − x x ( 4 .6 + 4 x )
4
K = K1K2K3K4 = 1.0 × 1030
Assume reacts completely.
0.088 x ( 4 .6 ) 4
x = [Hg2+] = 2.0 × 10 −34 M; assumptions good. [I−] = 4.6 M; [HgI42−] = 0.088 M a. [HgI42−] = 0.088 M;
b. [I−] = 4.6 M;
c. [Hg2+] = 2.0 × 10 −34 M
CHAPTER 16 120.
SOLUBILITY AND COMPLEX ION EQUILIBRIA
837
M: [Xe]6s24f145d10; this is mercury, Hg. Because X− has 54 electrons, X has 53 protons and is iodine, I. The identity of Q = Hg2I2. 1.98 g NaI
[I−]0 =
1 mol NaI 1 mol I − 149 .9 g mol NaI = 0.0881 mol/L 0.150 L
Hg2I2(s)
⇌
Initial s = solubility (mol/L) Equil.
Hg22+(aq) 0 s
Ksp = 4.5 × 10 −29
2 I−(aq)
+
0.0881 M 0.0881 + 2s
Ksp = 4.5 × 10 −29 = [Hg22+][ I−]2 = s(0.0881 + 2s)2 s(0.0881)2 s = 5.8 × 10 −27 mol/L; assumption good. 121.
Solubility = s =
15 g Ba(NO 3 ) 2 1 mol Ba(NO 3 ) 2 = 0.057 mol/L L 261.3 g Ba(NO 3 ) 2
Ba(NO3)2(s)
⇌
Initial s = solubility (mol/L) Equil.
Ba2+(aq) + 2 NO3−(aq) 0 s
Ksp = [Ba2+][NO3−]2
0 2s
Ksp = s(2s)2 = 4s3, Ksp = 4(0.057)3 = 7.4 × 10 −4 122.
Ksp = 6.4 × 10 −9 = [Mg2+][F−]2, 6.4 × 10 −9 = (0.00375 − y)(0.0625 − 2y)2 This is a cubic equation. No simplifying assumptions can be made since y is relatively large. Solving cubic equations is difficult unless you have a graphing calculator. However, if you don’t have a graphing calculator, one way to solve this problem is to make the simplifying assumption to run the precipitation reaction to completion. This assumption is made because of the very small value for K, indicating that the ion concentrations are very small. Once this assumption is made, the problem becomes much easier to solve.
Challenge Problems 123.
M3X2(s) Initial Equil.
⇌
s = solubility (mol/L)
3 M2+(aq)
2 X3−(aq)
+
0 3s
Ksp = [M2+]3[X3−]2
0 2s
Ksp = (3s)3(2s)2 = 108s5; total ion concentration = 3s + 2s = 5s π = iMRT, iM = total ion concentration =
RT
=
2.64 10 −2 atm 0.08206 L atm/K • mol 298 K
= 1.08 × 10 −3 mol/L 5s = 1.08 × 10 −3 mol/L, s = 2.16 × 10 −4 mol/L; Ksp = 108s5 = 108(2.16 × 10 −4 ) 5 Ksp = 5.08 × 10 −17
838
CHAPTER 16
124.
CaF2(s) ⇌ Ca2+(aq) + 2 F−(aq)
SOLUBILITY AND COMPLEX ION EQUILIBRIA Ksp = [Ca2+][F−]2
We need to determine the F− concentration present in a 1.0 M HF solution. Solving the weak acid equilibrium problem: HF(aq) ⇌ H+(aq) + F−(aq) Initial Equil.
1.0 M 1.0 − x
Ka = 7.2 × 10−4 =
~0 x
Ka =
[H + ][F− ] [HF]
0 x
x( x) x2 , x = [F−] = 2.7 × 10 −2 M; assumption good. 1.0 − x 1.0
Next, calculate the Ca2+ concentration necessary for Q = K sp, CaF2 . Q = [Ca 2+ ]0 [F − ]02 , 4.0 10 −11 = [Ca 2+ ]0 (2.7 10 −2 ) 2 , [Ca 2+ ]0 = 5.5 10 −8 mol/L Mass Ca(NO3)2 = 1.0 L
1 mol Ca(NO3 ) 2 164.10 g Ca(NO3 ) 2 5.5 10 −8 mol Ca 2+ 2+ L mol mol Ca
= 9.0 × 10 −6 g Ca(NO3 ) 2 For precipitation of CaF2(s) to occur, we need Q > Ksp. When 9.0 × 10 −6 g Ca(NO3)2 has been added to 1.0 L of solution, Q = K sp. So precipitation of CaF2(s) will begin to occur when just more than 9.0 × 10 −6 g Ca(NO3)2 has been added. 125.
a.
⇌ Cu+ + Br− Cu+ + 3 CN− ⇌ Cu(CN)32− CuBr(s) + 3 CN− ⇌ Cu(CN)32− + Br− CuBr(s)
Ksp = 1.0 × 10−5 Kf = 1.0 × 1011 K = 1.0 × 106
Because K is large, assume that enough CuBr(s) dissolves to completely use up the 1.0 M CN−; then solve the back equilibrium problem to determine the equilibrium concentrations. CuBr(s) + 3 CN− Before Change After
⇌
Cu(CN)32− + Br−
x 1.0 M 0 0 x mol/L of CuBr(s) dissolves to react completely with 1.0 M CN− −x −3x → +x +x 0 1.0 − 3x x x
For reaction to go to completion, 1.0 − 3x = 0 and x = 0.33 mol/L. Now solve the backequilibrium problem. CuBr(s) + 3 CN− Initial Change Equil.
⇌ Cu(CN)32− + Br−
0 0.33 M 0.33 M Let y mol/L of Cu(CN)32− react to reach equilibrium. +3y −y −y 3y 0.33 − y 0.33 − y
CHAPTER 16
SOLUBILITY AND COMPLEX ION EQUILIBRIA
K = 1.0 × 106 =
(0.33 − y ) 2
(3 y ) 3
839
(0.33) 2 , y = 1.6 × 10−3 M; assumptions good. 27 y 3
Of the initial 1.0 M CN−, only 3(1.6 × 10−3) = 4.8 × 10−3 M is present at equilibrium. Indeed, enough CuBr(s) did dissolve to essentially remove the initial 1.0 M CN−. This amount, 0.33 mol/L, is the solubility of CuBr(s) in 1.0 M NaCN. b. [Br−] = 0.33 − y = 0.33 − 1.6 × 10−3 = 0.33 M c. [CN−] = 3y = 3(1.6 × 10−3) = 4.8 × 10−3 M 126.
Ag+ + NH3 ⇌ AgNH3+ AgNH3+ + NH3 ⇌ Ag(NH3)2+
K1 = 2.1 × 103 K2 = 8.2 103
Ag+ + 2 NH3 ⇌ Ag(NH3)2+
K = K1K2 = 1.7 × 107
The initial concentrations are halved because equal volumes of the two solutions are mixed. Let the reaction go to completion since K is large; then solve an equilibrium problem. Ag+(aq) Before After Equil.
+
0.20 M 0 x
K = 1.7 × 107 =
2 NH3(aq)
⇌ Ag(NH3)2+(aq)
2.0 M 1.6 1.6 + 2x
0 0.20 0.20 − x
+ 0.20 − x 0.20 [Ag( NH3 ) 2 ] = , x = 4.6 × 10 −9 M 2 + 2 x (1.6 + 2 x) x (1.6) 2 [Ag ][ NH3 ]
Assumptions good; [Ag+] = x = 4.6 × 10 −9 M; [NH3] = 1.6 M; [Ag(NH3)2+] = 0.20 M Use either the K1 or K2 equilibrium expression to calculate [AgNH3+]. AgNH3+ + NH3 ⇌ Ag(NH3)2+ +
3
8.2 × 10 = 127.
a.
[Ag ( NH 3 ) 2 ] +
[AgNH 3 ][ NH 3 ]
AgBr(s) Initial Equil.
=
K2 = 8.2 × 103 0.20 +
[AgNH 3 ](1.6)
⇌
s = solubility (mol/L)
, [AgNH3+] = 1.5 × 10 −5 M
Ag+(aq) + Br−(aq) 0 s
Ksp = [Ag+][Br−] = 5.0 × 10 −13
0 s
Ksp = 5.0 × 10 −13 = s2, s = 7.1 × 10 −7 mol/L b.
AgBr(s) ⇌ Ag+ + Br− Ag+ + 2 NH3 ⇌ Ag(NH3)2+
AgBr(s) + 2 NH3(aq) ⇌ Ag(NH3)2+(aq) + Br−(aq)
Ksp = 5.0 × 10 −13 Kf = 1.7 × 107 K = Ksp × Kf = 8.5 × 10 −6
840
CHAPTER 16 AgBr(s) Initial
+ 2 NH3
⇌ Ag(NH3)2+
+ Br−
3.0 M 0 0 s mol/L of AgBr(s) dissolves to reach equilibrium = molar solubility 3.0 − 2s s s
Equil. K=
SOLUBILITY AND COMPLEX ION EQUILIBRIA
[Ag( NH3 ) +2 ][Br − ] s2 s2 −6 10 = 8.5 × , s = 8.7 × 10 −3 mol/L = [ NH3 ]2 (3.0 − 2s) 2 (3.0) 2
Assumption good. c. The presence of NH3 increases the solubility of AgBr. Added NH3 removes Ag+ from solution by forming the complex ion, Ag(NH 3)2+. As Ag+ is removed, more AgBr(s) will dissolve to replenish the Ag+ concentration. d. Mass AgBr = 0.2500 L ×
8.7 10 −3 mol AgBr 187 .8 g AgBr = 0.41 g AgBr L mol AgBr
e. Added HNO3 will have no effect on the AgBr(s) solubility in pure water. Neither H + nor NO3− react with Ag+ or Br− ions. Br− is the conjugate base of the strong acid HBr, so it is a terrible base. However, added HNO3 will reduce the solubility of AgBr(s) in the ammonia solution. NH3 is a weak base (Kb = 1.8 × 10 −5 ). Added H+ will react with NH3 to form NH4+. As NH3 is removed, a smaller amount of the Ag(NH3)2+ complex ion will form, resulting in a smaller amount of AgBr(s) that will dissolve. 128.
[NH3]0 =
3.00 M 2.00 10 −3 M = 1.50 M; [Cu2+]0 = = 1.00 × 10−3 M 2 2
Because [NH3]0 >> [Cu2+]0, and because K1, K2, K3 and K4 are all large, Cu(NH3)42+ will be the dominant copper-containing species. The net reaction will be Cu2+ + 4 NH3 → Cu(NH3)42+. Here, 1.00 × 10 −3 M Cu2+ plus 4(1.00 × 10 −3 M) NH3 will produce 1.00 × 10 −3 M Cu(NH3)42+. At equilibrium: [Cu(NH3)42+] 1.00 × 10 −3 M [NH3] = [NH3]0 – [NH3]reacted = 1.50 M – 4(1.00 × 10 −3 M) = 1.50 M Calculate [Cu(NH3)32+] from the K4 reaction: 2+
1.55 × 10 2 =
[Cu(NH3 ) 4 ] 2+
[Cu(NH3 ) 3 ][NH3 ]
=
1.00 10 −3 2+
[Cu(NH3 ) 3 ](1.50)
2+
, [Cu(NH3 ) 3 ]
= 4.30 10 −6 M Calculate [Cu(NH3)22+] from K3 reaction: 2+
3
1.00 × 10 =
[Cu(NH3 ) 3 ] 2+
[Cu(NH3 ) 2 ][NH3 ]
=
4.30 10 −6 2+
[Cu(NH3 ) 2 ](1.50)
2+
, [Cu(NH3 ) 2 ]
= 2.87 10 −9 M
CHAPTER 16
SOLUBILITY AND COMPLEX ION EQUILIBRIA
841
Calculate [Cu(NH3)32+] from the K2 reaction: 2+
3.88 × 10 3 =
[Cu(NH3 ) 2 ] 2+
[CuNH 3 ][NH3 ]
=
2.87 10 −9 2+
[CuNH 3 ](1.50)
, [CuNH3 2+ ]
= 4.93 10 −13 M Calculate [Cu2+] from the K1 reaction: 2+
4
1.86 × 10 =
[CuNH 3 ] 2+
[Cu ][NH3 ]
=
4.93 10 −13 2+
[Cu ](1.50)
, [Cu 2+ ] = 1.77 10 −17 M
The assumptions are valid. Cu(NH3)42+ is clearly the dominant copper-containing component. 129.
AgCN(s) ⇌ Ag+(aq) + CN−(aq) H (aq) + CN−(aq) ⇌ HCN(aq) +
AgCN(s) + H+(aq) ⇌ Ag+(aq) + HCN(aq) AgCN(s) + H+(aq) Initial
Ksp = 2.2 × 10 −12 K = 1/K a, HCN = 1.6 109 K = 2.2 × 10 −12 (1.6 10 9 ) = 3.5 10 −3
⇌ Ag+(aq) + HCN(aq)
1.0 M 0 0 s mol/L AgCN(s) dissolves to reach equilibrium 1.0 − s s s
Equil. 3.5 10 −3 =
[Ag + ][HCN] s( s) s2 = , s = 5.9 10 −2 1.0 − s 1.0 [H + ]
Assumption fails the 5% rule (s is 5.9% of 1.0 M). Using the method of successive approximations: 3.5 10 −3 =
s2 , s = 5.7 10 − 2 1.0 − 0.059
3.5 10 −3 =
s2 , s = 5.7 10 − 2 (consistent answer) 1.0 − 0.057
The molar solubility of AgCN(s) in 1.0 M H+ is 5.7 10 −2 mol/L. 130.
Solubility in pure water: CaC2O4(s)
⇌
Initial s = solubility (mol/L) Equil.
Ca2+(aq) + C2O42−(aq) 0 s
Ksp = 2 × 10−9
0 s
Ksp = s2 = 2 × 10−9, s = solubility = 4.47 × 10−5 = 4 × 10−5 mol/L Solubility in 0.10 M H+:
⇌ ⇌ ⇌ CaC2O4(s) + 2 H+ ⇌ CaC2O4(s) C2O42− + H+ HC2O4− + H+
Ca2+ + C2O42− HC2O4− H2C2O4
Ksp = 2 × 10−9 K = 1/ K a 2 = 1.6 104 K = 1/ K a1 = 15
Ca2+ + H2C2O4
Koverall = 5 × 10−4
842
CHAPTER 16 CaC2O4(s)
+
SOLUBILITY AND COMPLEX ION EQUILIBRIA
⇌
2 H+
Ca2+ + H2C2O4
Initial
0.10 M 0 0 s mol/L of CaC2O4(s) dissolves to reach equilibrium Equil. 0.10 – 2s s s 5 × 10−4 =
s2 s , = (5 10 − 4 )1/ 2 , s = 2 10 −3 mol / L 2 0.10 − 2s (0.10 − 2s )
Solubility in 0.10 M H + 2 10 −3 mol / L = = 50 Solubility in pure water 4 10 −5 mol / L
CaC2O4(s) is 50 times more soluble in 0.10 M H+ than in pure water. This increase in solubility is due to the weak base properties of C2O42−. 131.
H2S(aq) ⇌ H+(aq) + HS−(aq) HS−(aq) ⇌ H+(aq) + S2−(aq)
K a1 = 1.0 × 10-7 K a 2 = 1 × 10 −19
H2S(aq) ⇌ 2 H+(aq) + S2−(aq)
K = K a1 K a 2 = 1 × 10 −26 =
[H + ] 2 [S 2− ] [H 2 S]
Because K is very small, only a tiny fraction of the H 2S will react. At equilibrium, [H2S] = 0.10 M and [H+] = 1 × 10 −3 . [S2−] =
K[H 2S] (1 10 −26 )(0.10) = 1 × 10 −21 M = + 2 −3 2 [H ] (1 10 )
NiS(s) ⇌ Ni2+(aq) + S2−(aq)
Ksp = [Ni2+][S2−] = 3 × 10 −21
Precipitation of NiS will occur when Q > Ksp. We will calculate [Ni2+] for Q = Ksp. Q = Ksp = [Ni2+][S2−] = 3.0 × 10 −21 , [Ni2+] = 132.
3.0 10 −21 = 3 M = maximum concentration 1 10 − 21
We need to determine [S2−]0 that will cause precipitation of CuS(s) but not MnS(s). For CuS(s): CuS(s) ⇌ Cu2+(aq) + S2−(aq) Ksp = [Cu2+][S2−] = 8.5 × 10 −45 [Cu2+]0 = 1.0 × 10 −3 M,
K sp [Cu 2+ ] 0
=
8.5 10 −45 1.0 10 −3
= 8.5 10 − 42 M = [S 2− ]
This [S2−] represents the concentration that we must exceed to cause precipitation of CuS because if [S2−]0 > 8.5 × 10−42 M, Q > Ksp. For MnS(s): MnS(s) ⇌ Mn2+(aq) + S2−(aq)
Ksp = [Mn2+][ S2−] = 2.3 10 −13
CHAPTER 16
SOLUBILITY AND COMPLEX ION EQUILIBRIA
[Mn2+]0 = 1.0 10 −3 M,
K sp 2+
[Mn ]
=
2.3 10 −13 1.0 10
−3
843
= 2.3 10 −10 M = [S 2− ]
This value of [S2−] represents the largest concentration of sulfide that can be present without causing precipitation of MnS. That is, for this value of [S 2−], Q = Ksp, and no precipitation of MnS occurs. However, for any [S2−]0 > 2.3 × 10 −10 M, MnS(s) will form. We must have [S2−]0 > 8.5 × 10 −42 M to precipitate CuS, but [S2−]0 < 2.3 × 10 −10 M to prevent precipitation of MnS. The question asks for a pH that will precipitate CuS(s) but not MnS(s). We need to first choose an initial concentration of S2− that will do this. Let’s choose [S2−]0 = 1.0 × 10 −10 M because this will clearly cause CuS(s) to precipitate but is still less than the [S2−]0 required for MnS(s) to precipitate. The problem now is to determine the pH necessary for a 0.1 M H2S solution to have [S2−] = 1.0 × 10 −10 M. Let’s combine the K a1 and K a 2 equations for H2S to determine the required [H+]. H2S(aq) ⇌ H+(aq) + HS−(aq)
K a1 = 1.0 10 −7
HS−(aq) ⇌ H+(aq) + S2−(aq)
K a 2 = 1 10 −19
H2S(aq) ⇌ 2H+(aq) + S2−(aq)
K = K a1 K a 2 = 1 10 −26
1 10 − 26 =
[H + ] 2 (1 10 −10 ) [H + ] 2 [S2− ] = , [H + ] = 3 10 −9 M [H 2 S] 0.10
pH = − log( 3 10 −9 ) = 8.5. So, if pH = 8.5, [S2− ] = 1 10 −10 M, which will cause precipitation of CuS(s) but not MnS(s). Note: Any pH less than 8.7 would be a correct answer to this problem. 133.
Mg2+ + P3O105− ⇌ MgP3O103− K = 4.0 × 108 [Mg2+]0 =
50. 10 −3 g 1 mol = 2.1 × 10 −3 M L 24.31 g
[P3O105−]0 =
40. g Na 5 P3O10 1 mol = 0.11 M L 367 .86 g
Assume the reaction goes to completion because K is large. Then solve the back-equilibrium problem to determine the small amount of Mg2+ present. Mg2+ Before Change After Change Equil.
+
P3O105−
⇌
MgP3O103−
2.1 × 10 −3 M 0.11 M 0 −2.1 × 10 −3 −2.1 × 10 −3 → +2.1 × 10 −3 0 0.11 2.1 × 10 −3 3− x mol/L MgP3O10 dissociates to reach equilibrium +x +x −x x 0.11 + x 2.1 × 10 −3 − x
React completely New initial condition
844
CHAPTER 16 3−
8
K = 4.0 × 10 = 4.0 × 108
134.
[MgP3O10 ]
5−
[Mg 2+ ][P3O10 ]
SOLUBILITY AND COMPLEX ION EQUILIBRIA =
2.1 10 −3 − x (assume x << 2.1 × 10−3) x(0.11 + x)
2.1 10 −3 , x = [Mg2+] = 4.8 × 10 −11 M; assumptions good. x(0.11)
MX(s) ⇌ Mn+(aq) + Xn−(aq);
T =Kfm, m =
ΔT 0.028 o C = 0.015 mol/kg = Kf 1.86 o C / molal
0.015 mol 1 kg × 250 g = 0.00375 mol total solute particles (carrying extra sig. fig.) kg 1000 g
0.0375 mol = mol Mn+ + mol Xn−, mol Mn+ = mol Xn− = 0.0375/2 Because the density of the solution is 1.0 g/mL, 250 g = 250 mL of solution. [Mn+] =
(0.00375 / 2) mol M n+ (0.00375 / 2) mol X n− = 7.5 × 10 −3 M, [Xn−] = 0.25 L 0.25 L
= 7.5 × 10 −3 M Ksp = [Mn+][Xn−] = (7.5 × 10 −3 ) 2 = 5.6 × 10 −5 135.
a.
SrF2(s)
⇌
Sr2+(aq)
+ 2 F−(aq)
Initial
0 0 s mol/L SrF2 dissolves to reach equilibrium Equil. s 2s [Sr2+][F−]2 = Ksp = 7.9 × 10−10 = 4s3, s = 5.8 × 10−4 mol/L in pure water b. Greater, because some of the F− would react with water: F− + H2O ⇌ HF + OH−
Kb =
Kw = 1.4 × 10−11 K a , HF
This lowers the concentration of F−, forcing more SrF2 to dissolve. c. SrF2(s) ⇌ Sr2+ + 2 F− Ksp = 7.9 × 10−10 = [Sr2+][F−]2 Let s = solubility = [Sr2+]; then 2s = total F− concentration. Since F− is a weak base, some of the F− is converted into HF. Therefore: total F− concentration = 2s = [F−] + [HF] HF ⇌ H+ + F− Ka = 7.2 × 10−4 = 7.2 × 10−2 =
[H + ][F− ] 1.0 10 −2 [F− ] = [HF] [HF]
[F − ] , [HF] = 14[F−]; solving: [ HF]
(since pH = 2.00 buffer)
CHAPTER 16
SOLUBILITY AND COMPLEX ION EQUILIBRIA
845
[Sr2+] = s; 2s = [F−] + [HF] = [F−] + 14[F−], 2s = 15[F−], [F−] = 2s/15 2
−10
Ksp = 7.9 × 10 136.
2s = [Sr ][F ] = (s) , s = 3.5 × 10−3 mol/L in pH = 2.00 solution 15 2+
− 2
MX(s) ⇌ M+(aq) + X−(aq)
Ksp = [M+][X−]
[M+] = solubility = 3.17 10−8 mol/L in a 1.00 M H+ solution (pH = 0.000) X− + H2O ⇌ HX + OH− [OH−] =
Kw [H + ]
Kb = 10.0 =
=
Kb =
Kw 1.00 10 −14 = = 10.0 Ka 1.00 10 −15
1.00 10 −14 = 1.00 10−14 M 1.00
1.00 10 −14 [HX] , [X−] = 1.00 10−15 [HX] − [X ]
[M+] = total X− concentration = [HX] + [X−] = [HX] + 1.00 10−15 [HX] To the correct significant figures: [M+] = [HX] Thus, [X−] = 1.00 10−15 [M+] and [M+] = solubility = 3.17 10−8 M. [X−] = (1.00 10−15)(3.17 10−8) = 3.17 10−23 M Ksp = [M+][X−] = (3.17 10−8)( 3.17 10−23) = 1.00 × 10−30 137.
Major species: H+, HSO4−, Ba2+, NO3−, and H2O; Ba2+ will react with the SO42− produced from the Ka reaction for HSO4−. HSO4− ⇌ H+ + SO42− Ba2+ + SO42− ⇌ BaSO4(s)
K a 2 = 1.2 × 10−2 K = 1/Ksp = 1/(1.5 × 10−9) = 6.7 108
Ba2+ + HSO4− ⇌ H+ + BaSO4(s)
Koverall = (1.2 × 10−2) (6.7 108) = 8.0 106
Because Koverall is so large, the reaction essentially goes to completion. Because H 2SO4 is a strong acid, [HSO4−]0 = [H+]0 = 0.10 M.
Before Change After Change Equil.
Ba2+
+ HSO4−
0.30 M −0.10 0.20 +x 0.20 + x
0.10 M −0.10 0 +x x
K = 8.0 106 =
⇌ →
H+
+ BaSO4(s)
0.10 M +0.10 0.20 M −x 0.20 − x
New initial
0.20 − x 0.20 , x = 1.3 × 10−7 M; assumptions good. (0.20 + x) x 0.20( x)
846
CHAPTER 16
SOLUBILITY AND COMPLEX ION EQUILIBRIA
[H+] = 0.20 − 1.3 × 10−7 = 0.20 M; pH = −log(0.20) = 0.70 [Ba2+] = 0.20 + 1.3 × 10−7 = 0.20 M From the initial reaction essentially going to completion, 1.0 L(0.10 mol HSO4−/L) = 0.10 mol HSO4− reacted; this will produce 0.10 mol BaSO4(s). Only 1.3 × 10−7 mol BaSO4 dissolves to reach equilibrium, so 0.10 mol BaSO4(s) is produced. 0.10 mol BaSO4 × 233.4 g/mol = 23 g BaSO4 forms
Marathon Problem 138.
a. In very acidic solutions, the reaction that occurs to increase the solubility is Al(OH) 3(s) + 3H+ → Al3+(aq) + 3H2O(l). In very basic solutions, the reaction that occurs to increase solubility is Al(OH)3(s) + OH−(aq) → Al(OH)4−(aq). b. Al(OH)3(s) ⇌ Al3+ + 3 OH−; Al(OH)3(s) + OH- ⇌ Al(OH)4− S = solubility = total Al3+ concentration = [Al3+] + [Al(OH)4−] [Al3+] =
[ H + ]3 = K , because [OH−]3 = (Kw/[H+])3 sp 3 [OH − ]3 Kw
K sp
− Kw KK w [Al(OH) 4 ] = K; [OH−] = ; [Al(OH)4−] = K[OH−] = + − [H ] [H + ] [OH ]
S = [Al3+] + [Al(OH)4−] = [H+]3Ksp/Kw3 + KKw/[H+] c. Ksp = 2 × 10−32; Kw = 1.0 × 10−14; K = 40.0 S=
[H + ]3 (2 10 −32 ) 40.0 (1.0 10 −14 ) 4.0 10 −13 + 3 10 = [H ] (2 × 10 ) + + (1.0 10 −14 ) 3 [H + ] [H + ]
pH
solubility (S, mol/L)
log S
4.0 5.0 6.0 7.0 8.0 9.0 10.0 11.0 12.0
2 × 10−2 2 × 10−5 4.2 × 10−7 4.0 × 10−6 4.0 × 10−5 4.0 × 10−4 4.0 × 10−3 4.0 × 10−2 4.0 × 10−1
−1.7 −4.7 −6.38 −5.40 −4.40 −3.40 −2.40 −1.40 −0.40
As expected, the solubility of Al(OH)3(s) is increased by very acidic solutions and by very basic solutions.
CHAPTER 17 SPONTANEITY, ENTROPY, AND FREE ENERGY Review Questions 1.
a. A spontaneous process is one that occurs without any outside intervention. b. Entropy is a measure of disorder or randomness. c. Positional probability is a type of probability that depends on the number of arrangements in space that yield a particular state. The more disordered state generally has the larger positional probability. d. The system is the portion of the universe in which we are interested. e. The surroundings are everything else in the universe besides the system. f.
The universe is everything; universe = system + surroundings.
2.
Second law of thermodynamics: in any spontaneous process, there is always an increase in the entropy of the universe. Suniv = Ssys + Ssurr; When both Ssys and Ssurr are positive, Suniv must be positive (so the process is spontaneous). Suniv is always negative (so the process is nonspontaneous) when both Ssys and Ssurr are negative. When the signs of Ssys are opposite of each other [(Ssys (+), Ssurr (−) or vice versa], the process may or may not be spontaneous.
3.
Ssurr is primarily determined by heat flow. This heat flow into or out of the surroundings comes from the heat flow out of or into the system. In an exothermic process (H < 0), heat flows into the surroundings from the system. The heat flow into the surroundings increases the random motions in the surroundings and increases the entropy of the surroundings (Ssurr > 0). This is a favorable driving force for spontaneity. In an endothermic reaction (H > 0), heat is transferred from the surroundings into the system. This heat flow out of the surroundings decreases the random motions in the surroundings and decreases the entropy of the surroundings (Ssurr < 0). This is unfavorable. The magnitude of Ssurr also depends on the temperature. The relationship is inverse; at low temperatures, a specific amount of heat exchange makes a larger percent change in the surroundings than the same amount of heat flow at a higher temperature. The negative sign in the Ssurr = −H/T equation is necessary to get the signs correct. For an exothermic reaction where H is negative, this increases Ssurr so the negative sign converts the negative H value into a positive quantity. For an endothermic process where H is positive, the sign of Ssurr is negative and the negative sign converts the positive H value into a negative quantity.
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Suniv = −G/T (at constant T and P); when G is negative (Suniv > 0), the process is spontaneous. When G is positive (Suniv < 0), the process in nonspontaneous (the reverse process is spontaneous). When G = 0, the process is at equilibrium. G = H − TS; see Table 17.5 of the text for the four possible sign conventions and the temperature dependence for these sign combinations. When the signs for H and S are both the same, then temperature determines if the process is spontaneous. When H is positive (unfavorable) and S is positive (favorable), high temperatures are needed to make the favorable S term dominate, which makes the process spontaneous (G < 0). When H is negative (favorable) and S is negative (unfavorable), low temperatures are needed so the favorable H term dominates, which makes the process spontaneous (G < 0). Note that if G is positive for a process, then the reverse process has a negative G value and is spontaneous. At the phase change temperature (melting point or boiling point), two phases are in equilibrium with each other so G = 0. The G = H − TS equation reduces to H = TS at the phase change temperature. For the s → l phase change, G is negative above the freezing point because the process is spontaneous (as we know). For the l → g phase change, the sign of G is positive below the boiling point as the process is nonspontaneous (as we know).
5.
Third law of thermodynamics: The entropy of a perfect crystal at 0 K is zero. Standard entropy values (S˚) represent the increase in entropy that occurs when a substance is heated from 0 K to 298 K at 1 atm pressure. The equation to calculate S˚ for a reaction using the standard o o o entropy values is: S rxn = Σnp S products . − Σnr S reactants This equation works because entropy is a state function of the system (it is not pathwaydependent). Because of this, one can manipulate chemical reactions with known S˚ values to determine S˚ for a different reaction. The entropy change for a different reaction equals the sum of the entropy changes for the reactions added together that yield the different reaction. This is utilizing Hess’s law. The superscript ˚ indicates conditions where T = 25˚C and P = 1 atm. To predict signs for gas phase reactions, you need to realize that the gaseous state represents a hugely more disordered state (a state with much larger positional probability) as compared to the solid and liquid states. Gases dominate sign predictions for reactions. Those reactions that show an increase in the number of moles of gas as reactants are converted to products have an increase in disorder (an increase in positional probability) which translates into a positive S˚ value. S˚ values are negative when there is a decrease in the moles of gas as reactants are converted into products. When the moles of gaseous reactants and products are equal, S˚ is usually difficult to predict for chemical reactions. However, predicting signs for phase changes can be done by realizing the solid state is the most ordered phase (lowest S˚ values, smallest positional probability). The liquid state is a slightly more disordered phase (has a higher positional probability) than the solid state, with the gaseous state being the most disordered o o o phase (has the largest positional probability) by a large margin (S solid S liquid S gas ). Another process involving condensed phases whose sign is also easy to predict (usually) is the dissolution of a solute in a solvent. Here, the mixed-up solution state is usually the more disordered state (has a larger positional probability) as compared to the solute and solvent separately.
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Standard free energy change: the change in free energy that will occur for one unit of reaction if the reactants in their standard states are converted to products in their standard state. The standard free energy of formation (G of ) of a substance is the change in free energy that accompanies the formation of 1 mole of that substance from its constituent elements, with all reactants and products in their standard states. The equation that manipulates G of values to o determining Greaction is: ΔG° = Σnp ΔGfo (products) − Σnr ΔGfo (reactants) .
Because ΔG˚ is a state function (path independent), chemical reactions with known ΔG˚ values can be manipulated to determine ΔG˚ for a different reaction. ΔG˚ for the different reaction is the sum of ΔG˚ for all the steps (reactions) added together to get the different reaction. This is Hess’s law. Another way to determine ΔG° for a reaction is to utilize the G˚ = H˚ − TS˚ equation. Here, you need to know ΔH°, S˚, and the temperature, then you can use the above equation to calculate ΔG°. Of the functions ΔG˚, H˚, and S˚, ΔG˚ has the greatest dependence on temperature. The temperature is usually assumed to be 25˚C. However, if other temperatures are used in a reaction, we can estimate ΔG° at that different temperature by assuming ΔH˚ and S˚ are temperature independent (which is not always the best assumption). We calculate ΔH˚ and ΔS° values for a reaction using Appendix 4 data, then use the different temperature in the G˚ = H˚ − TS˚ equation to determine (estimate) ΔG° at that different temperature. 7.
No; when using G of values in Appendix 4, we have specified a temperature of 25°C. Further, if gases or solutions are involved, we have specified partial pressures of 1 atm and solute concentrations of 1 molar. At other temperatures and compositions, the reaction may not be spontaneous. A negative ΔG° value means the reaction is spontaneous under standard conditions. The free energy and pressure relationship is G = G° + RT ln(P). The RT ln(P) term corrects for nonstandard pressures (or concentrations if solutes are involved in the reaction). The standard pressure for a gas is 1 atm and the standard concentration for solutes is 1 M. The equation to calculate ΔG for a reaction at nonstandard conditions is: ΔG = ΔG° + RT ln Q where Q is the reaction quotient determined at the nonstandard pressures and/or concentrations of the gases and/or solutes in the reaction. The reaction quotient has the exact same form as the equilibrium constant K. The difference is that the partial pressures or concentrations used may or may not be the equilibrium concentrations. All reactions want to minimize their free energy. This is the driving force for any process. If ΔG is a negative, the process occurs. The equilibrium position represents the lowest total free energy available to any reaction system. Once equilibrium is reached, the system cannot minimize its free energy anymore. Converting from reactants to products or products to reactants will increase the total free energy of the system which reactions do not want to do.
8.
At equilibrium, ΔG = 0 and Q = K (the reaction quotient equals the equilibrium constant value). From the ΔG° = −RT ln K equation, when a reaction has K < 1, the ln K term is negative, so ΔG° is positive. When K > 1, the ln K term is positive, so ΔG° is negative. When ΔG° = 0, this tell us that K for the process is equal to one (K = 1) because ln(1) = 0.
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The sign of ΔG (positive or negative) tells us which reaction is spontaneous (the forward or reverse reaction). If ΔG < 0, then the forward reaction is spontaneous, and if ΔG > 0, then the reverse reaction is spontaneous. If ΔG = 0, then the reaction is at equilibrium (neither the forward or reverse reactions are spontaneous). ΔG° gives the equilibrium position by determining K for a reaction utilizing the equation ΔG° = −RT ln K can only be used to predict spontaneity when all reactants and products are present at standard pressures of 1 atm and/or standard concentrations of 1 M. 9.
A negative ΔG value does not guarantee that a reaction will occur. It does say that it can occur (is spontaneous), but whether it will occur also depends on how fast the reaction is (depends on the kinetics). A process with a negative ΔG may not occur because it is too slow. The example used in the text is the conversion of diamonds into graphite. Thermodynamics says the reaction can occur (ΔG > 0), but the reaction is so slow that it doesn’t occur. The rate of a reaction is directly related to temperature. As temperature increases, the rate of a reaction increases. Spontaneity, however, does not necessarily have a direct relationship to temperature. The temperature dependence of spontaneity depends on the signs of ΔH and ΔS (see Table 17.5 of the text). For example, when ΔH and ΔS are both negative, the reaction becomes more favorable thermodynamically (ΔG becomes more negative) with decreasing temperature. This is just the opposite of the kinetics dependence on temperature. Other sign combinations of ΔH and ΔS have different spontaneity temperature dependence.
10.
wmax = ΔG; when ΔG is negative, the magnitude of ΔG is equal to the maximum possible useful work obtainable from the process (at constant T and P). When ΔG is positive, the magnitude of ΔG is equal to the minimum amount of work that must be expended to make the process spontaneous. Due to waste energy (heat) in any real process, the amount of useful work obtainable from a spontaneous process is always less than wmax, and for a non-spontaneous reaction, an amount of work greater than wmax must be applied to make the process spontaneous. Reversible process: A cyclic process carried out by a hypothetical pathway, which leaves the universe the same as it was before. No real process is reversible.
Active Learning Questions 1.
Converting a liquid into a gas is an endothermic process. So the reverse process is favored energetically. The gaseous state has the higher positional probability, so the forward process is favored from a positional probability standpoint. This process will be spontaneous when the favorable positional probability term overrides the unfavorable energy term. This will happen as the temperature is increased above some value. As we know, a liquid boils above some temperature.
2.
The gaseous state is by far the state with the highest positional probability. So Sevaporation will be the larger quantity.
3.
In this reaction, we are going from weaker bonds to stronger bonds. So, this will be an exothermic reaction (H < 0). Here the products have lower potential energy, and the difference in potential energy between the reactants and products is released as thermal energy to the surroundings. This thermal energy goes to increase the kinetic energy associated with
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the random motion of the surrounding molecules which leads to Ssurr > 0. From the information in the problem, we cannot predict the sign on S, and, in turn, Suniv. 4.
ΔS at constant pressure is expected to be larger than ΔS at constant volume. At constant pressure, the volume of the container changes to keep the pressure constant. A change in volume greatly affects the available positions of the particles in the system. When the volume of a container changes, there is a larger change in positional probability than when the container volume is constant. Hence, ΔS at constant pressure should result in a larger change in positional probability; this translates to a larger magnitude for ΔS.
5.
G = H − TS; a reaction is spontaneous when G < 0. If G is negative above some temperature, then both H and S are positive. So, the reaction is endothermic. If G is positive below some temperature, then both H and S are negative. So, the reaction is exothermic.
6.
For a reaction at a certain set of concentrations and temperature, monitor the concentrations of the reactants and/or products. If the product amounts increase, then the forward reaction is spontaneous. If the reactant amounts increase, then the reverse reaction is spontaneous. If the relative amounts of reactants and products do not change, then the reaction is at equilibrium where G = 0.
7.
G = G + RTln(P); G for a substance also depends on G. Each substance has its own unique standard free energy value. So products and reactants do not have the same G value when they have the same partial pressure values because Gprod Greact. For example, in a gaseous reaction where all gases are at standard pressures of 1 atm, G = G; the reaction is not at equilibrium.
8.
For a reaction that is spontaneous at standard concentrations, G = G < 0. If a reaction is spontaneous when all gases and solutes 1 atm or 1 M, then the product gases and solutes will have equilibrium concentrations greater than 1 atm or 1 M and reactant gases and solutes will have equilibrium concentrations less than 1 atm or 1 M. This dictates that K must be a value greater than one when a reaction is spontaneous at standard concentrations. So the equation G = −RTln(K) must correct; the minus sign is necessary to ensure that when G < 0, K >1 and vice versa.
9.
a. C2H5OH(l) → C2H5OH(g); the gaseous phase has a much larger positional probability as compared to the liquid phase. Hence, S > 0 for the evaporation process. b. H2O(l) → H2O(s); the liquid phase has a larger positional probability as compared to the solid phase. Hence, S < 0 for the freezing process. c. The smaller the volume of a gas, the smaller the positional probability. So S < 0 for the compression of a gas. d. When solid NaCl dissolves, the Na+ and Cl− ions are randomly dispersed in water. The ions have access to a larger volume and a larger number of possible positions. Hence positional probability has increased and S is positive.
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10.
Ssurr > 0 when a reaction is exothermic. In an exothermic process (H < 0), heat flows into the surroundings from the system. The heat flow into the surroundings increases the random motions in the surroundings and increases the entropy of the surroundings. This is a favorable driving force for spontaneity.
11.
H2O(l) → H2O(g); at T > 100C (and 1 atm), water spontaneously boils. For this process at T > 100C, Suniv > 0 (the process is spontaneous). Since positional probability greatly increases when going form a liquid to a gas, S > 0. And to convert a liquid into a gas, energy must be added, so Ssurr < 0.
12.
High temperatures favor a reaction kinetically as the higher temperature speeds up the reaction by providing more reactants with sufficient activation energy to convert to products. However, high temperatures does not necessarily favor a reaction thermodynamically. When the signs of H and S are both negative, temperatures below some value favor spontaneity. In addition, when H is positive and S is negative (when both terms are unfavorable), a higher temperature won’t make the reaction spontaneous.
13.
G = wmax; at equilibrium, G = 0. No useful work can occur for a reaction at equilibrium. In general, a reaction needs to be far away from equilibrium to harness useful work.
14.
G = H − TS; ΔG = ΔG° + RT ln Q; G = −RTln(K); from the problem, G < 0, so this reaction is spontaneous at some temperature. If the reaction is endothermic (H > 0), then S must be positive for G to be negative at some temperature. The reaction has an increase in positional probability when reactants are converted to products. If K > 1, then G° must be negative [from the G = −RTln(K) equation]. From the problem, G > 0, so the reaction is nonspontaneous at some temperature. If H < 0 (exothermic), then S must also be negative. The reaction has a decrease in positional probability when reactants are converted to products.
15.
CH4(g) + CO2(g) → CH3CO2H(l) ΔH° = −484 − [−75 + (−393.5)] = −16 kJ; ΔS° = 160. − (186 + 214) = −240. J/K ΔG° = ΔH° − TΔS° = −16 kJ − (298 K)(−0.240 kJ/K) = 56 kJ At standard concentrations, this reaction is spontaneous only at temperatures below T = ΔH°/ΔS° = 67 K (where the favorable ΔH° term will dominate, giving a negative ΔG° value). This is not practical. Substances will be in condensed phases and rates will be very slow at this extremely low temperature. CH3OH(g) + CO(g) → CH3CO2H(l) ΔH° = −484 − [−110.5 + (−201)] = −173 kJ; ΔS° = 160. − (198 + 240.) = −278 J/K ΔG° = −173 kJ − (298 K)(−0.278 kJ/K) = −90. kJ
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This reaction also has a favorable enthalpy and an unfavorable entropy term. But this reaction, at standard concentrations, is spontaneous at temperatures below T = ΔH°/ΔS° = 622 K (a much higher temperature than the first reaction). So the reaction of CH3OH and CO will be preferred at standard concentrations. It is spontaneous at high enough temperatures that the rates of reaction should be reasonable. 16.
N2O4(g) → 2NO2(g) a. G = H − TS; at standard conditions, this reaction is not spontaneous (G° < 0), but at some nonstandard conditions, the reaction is spontaneous since G < 0. This indicates that there is a sign change on G when temperature is changed. Sign changes on G occur when either H and S are both positive or are both negative. In the reaction, moles of gas increase as reactants are converted to products, so S is positive. This dictates that H must be positive to have a sign change on G when temperature is manipulated. The reaction is endothermic. b. G = −RTln(K); because G° is positive, mathematically K must be less than 1. The ln(K) term must be a negative value for G° to be positive. This occurs when K < 1. c. At the concentrations given in the problem, the reaction is spontaneous since G is negative. To reach equilibrium, the reaction shifts right. So, the partial pressure of N2O4 will be less than 1.6 atm and the partial pressure of NO2 will be greater than 0.29 atm. d. G = wmax = ‒1000 J; the reaction can harness 1000 J of useful work in a perfect world.
Questions 17.
Living organisms need an external source of energy to carry out these processes. Green plants use the energy from sunlight to produce glucose from carbon dioxide and water by photosynthesis. In the human body, the energy released from the metabolism of glucose helps drive the synthesis of proteins. For all processes combined, ΔSuniv must be greater than zero (the second law).
18.
Dispersion increases the entropy of the universe because the more widely something is dispersed, the greater the disorder. We must do work to overcome this disorder. In terms of the second law, it would be more advantageous to prevent contamination of the environment rather than to clean it up later. As a substance disperses, we have a much larger area that must be decontaminated.
19.
As a process occurs, Suniv will increase; Suniv cannot decrease. Time, like Suniv, only goes in one direction.
20.
The introduction of mistakes is an effect of entropy. The purpose of redundant information is to provide a control to check the "correctness" of the transmitted information.
21.
This reaction is kinetically slow but thermodynamically favorable (ΔG < 0). Thermodynamics only tells us if a reaction can occur. To answer the question will it occur, one also needs to consider the kinetics (speed of reaction). The ultraviolet light provides the activation energy for this slow reaction to occur.
854 22.
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a. Thermodynamics is concerned with only the initial and final states of a reaction, and not the pathway required to get from reactants to products. It is the kinetics of a reaction that concentrates on the pathway and speed of a reaction. For these plots, the reaction with the smallest activation energy will be fastest. The activation energy is the energy reactants must have to overcome the energy barrier to convert to products. So, the fastest reaction is reaction 5 with the smallest activation. Reactions 1, 2, and 4 all should have the same speed because they have the same activation energy. (This assumes all other kinetic factors are the same.) Reaction 3 will be the slowest since it has the largest activation energy. b. If the products have lower energy than the reactants, then the reaction is exothermic. Reaction 1, 2, 3, and 5 are exothermic. If the products have higher energy than the reactants, then the reaction is endothermic. Only reaction 4 is endothermic. Note that it is the thermodynamics of a reaction that dictates whether a reaction is exothermic or endothermic. The kinetics plays no factor in these designations. c. The thermodynamics of reaction determine potential energy change. In an exothermic reaction, some of the potential energy stored in chemical bonds is converted to thermal energy (heat is released as the potential energy decreases). The exothermic reaction with the greatest loss of potential energy is the reaction with the largest energy difference (E) between reactants and products. Reactions 2 and 5 both have the same largest E values, so reactions 2 and 5 have the greatest change in potential energy. Reactions 1 and 3 both have the same smallest E values, so reactions 1 and 3 have the smallest change in potential energy.
23.
Ssurr = −H/T; heat flow (H) into or out of the system dictates Ssurr. If heat flows into the surroundings, the random motions of the surroundings increase, and the entropy of the surroundings increases. The opposite is true when heat flows from the surroundings into the system (an endothermic reaction). Although the driving force described here really results from the change in entropy of the surroundings, it is often described in terms of energy. Nature tends to seek the lowest possible energy.
24.
When solid NaCl dissolves, the Na+ and Cl− ions are randomly dispersed in water. The ions have access to a larger volume and a larger number of possible positions. Hence positional disorder has increased and S is positive. For a salt dissolving in water, H is usually a value close to zero (see Section 11.2 of the text). Sometimes H is positive and sometimes H is negative for a salt dissolving in water, but the magnitude of H is small. So the only prediction that can be made is that H for this process will be close to zero.
25
a. Positional probability increases; there is a greater volume accessible to the randomly moving gas molecules, which increases disorder. b. The positional probability doesn't change. There is no change in volume and thus no change in the numbers of positions of the molecules. c. Positional probability decreases because the volume decreases (P and V are inversely related).
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26.
At −75C, this is below the freezing temperature so solid CSe2 is the stable phase. The process is written as liquid → solid; this is a spontaneous process at temperatures below the freezing point, so Suniv must be positive (Suniv = −G/T). To convert a liquid into a solid, energy is released as more intermolecular forces are formed in the solid phase. H is negative for the freezing process, so Ssurr is positive (Ssurr = −H/T).
27.
This equation refers to breaking the bond in the H 2 molecule. To break a bond, energy must be added; this is an endothermic process. The moles of gas increase as the bond is broken to produce the individual H atoms. Because of the increase in moles of gas, S is positive. When H and S are both positive, the process is spontaneous above some temperature where the favorable S term dominates. This reaction is spontaneous at high temperatures.
28.
When H° is positive and S° is negative, it is not possible to have a spontaneous reaction at standard conditions since G° cannot be a negative value. Of the two endothermic reactions, answer b has a decrease in moles of gas while reaction c has an increase in moles of gas. Reaction b has a negative S° value, so reaction b can never be spontaneous at standard concentrations.
29.
Note that these substances are not in the solid or gaseous state but are in the aqueous state; water molecules are also present. There is an apparent increase in ordering when these ions are placed in water as compared to the separated state. The hydrating water molecules must be in a highly ordered arrangement when surrounding these anions.
30.
31.
G° = −RT ln K = H° − TS°; HX(aq) ⇌ H+(aq) + X−(aq) Ka reaction; the value of Ka for HF is less than one, while the other hydrogen halide acids have Ka > 1. In terms of G°, HF must have a positive G orxn value, while the other HX acids have Grxn < 0. The reason for the sign change in the Ka value, between HF versus HCl, HBr, and HI is entropy. S for the dissociation of HF is very large and negative. There is a high degree of ordering that occurs as the water molecules associate (hydrogen bond) with the small F − ions. The entropy of hydration strongly opposes HF dissociating in water, so much so that it overwhelms the favorable hydration energy making HF a weak acid. One can determine S° and H° for the reaction using the standard entropies and standard enthalpies of formation in Appendix 4; then use the equation G° = H° − TS°. One can also use the standard free energies of formation in Appendix 4. And finally, one can use Hess’s law to calculate G°. Here, reactions having known G° values are manipulated to determine G° for a different reaction. For temperatures other than 25°C, G° is estimated using the G° = H° − TS° equation. The assumptions made are that the H° and S° values determined from Appendix 4 data are temperature independent. We use the same H° and S° values as determined when T = 25°C; then we plug in the new temperature in Kelvin into the equation to estimate G° at the new temperature.
32.
The sign of G tells us if a reaction is spontaneous or not at whatever concentrations are present (at constant T and P). The magnitude of G equals wmax. When G < 0, the magnitude tells us how much work, in theory, could be harnessed from the reaction. When G > 0, the magnitude
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tells us the minimum amount of work that must be supplied to make the reaction occur. G° gives us the same information only when the concentrations for all reactants and products are at standard conditions (1 atm for gases, 1 M for solute). These conditions rarely occur. G° = −RT ln K; from this equation, one can calculate K for a reaction if G° is known at that temperature. Therefore, G° gives the equilibrium position for a reaction. To determine K at a temperature other than 25°C, one needs to know G° at that temperature. We assume H° and S° are temperature-independent and use the equation G° = H° − TS° to estimate G° at the different temperature. For K = 1, we want G° = 0, which occurs when H° = TS°. Again, assume H° and S° are temperature-independent; then solve for T (= H°/S°). At this temperature, K = 1 because G° = 0. This only works for reactions where the signs of H° and S° are the same (either both positive or both negative). When the signs are opposite, K will always be greater than 1 (when H° is negative and S° is positive) or K will always be less than 1 (when H° is positive and S° is negative). When the signs of H° and S° are opposite, K can never equal 1. 33.
The light source for the first reaction is necessary for kinetic reasons. The first reaction is just too slow to occur unless a light source is available. The kinetics of a reaction are independent of the thermodynamics of a reaction. Even though the first reaction is more favorable thermodynamically (assuming standard conditions), it is unfavorable for kinetic reasons. The second reaction has a negative ΔG° value and is a fast reaction, so the second reaction which occurs very quickly is favored both kinetically and thermodynamically. When considering if a reaction will occur, thermodynamics and kinetics must both be considered.
34.
Only statement a is false. At equilibrium, Suniv = 0. This will occur when Ssys = ‒Ssurr. At equilibrium, G = 0 = H − TS, so TS = H.
35.
Only statement c is false. The forward reaction is spontaneous when G < 0. This occurs when the free energy of the products is lower than the free energy of the reactants.
36.
G° = H° − TS°; G = −RTln(K); manipulating these equations gives the expression −RTln(K) = H° − TS°. Dividing by ‒RT gives the expression ln(K) = ‒ H°/RT + S°/R. This is in the form for the straight line equation y = mx + b. In the plot of ln(K) vs 1/T, the slope (m) equals ‒ H°/R and the y-intercept equals S°/R. Because the slope is positive, H° must be positive, and because the y-intercept is negative, S° must be negative.
37.
G = −RTln(K); G° = H° − TS°; because the reaction is spontaneous at standard conditions, G° (and G) must be negative. If G° < 0, then K must be greater than 1 as dictated by the G = −RTln(K) equation. Statement c is false. Because n is negative for this reaction, S° will be negative. We know G° < 0, so H° must be negative as dictated mathematically be the G° = H° − TS° equation.
38.
Using Le Chatelier's principle, a decrease in pressure (volume increases) will favor the side with the greater number of particles. Thus 2 I(g) will be favored at low pressure. Looking at ΔG: ΔG = ΔG° + RT ln (PI2 / PI2 ); ln( PI2 / PI2 ) 0 when PI = PI2 = 10 atm and ΔG is positive (not spontaneous). But at PI = PI 2 = 0.10 atm, the logarithm term is negative. If |RT ln Q| > ΔG°, then ΔG becomes negative, and the reaction is spontaneous.
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Exercises Spontaneity, Entropy, and the Second Law of Thermodynamics: Free Energy 39.
a, b, and c; from our own experiences, salt water, colored water, and rust form without any outside intervention. It takes an outside energy source to clean a bedroom, so this process is not spontaneous.
40.
c and d; it takes an outside energy source to build a house and to launch and keep a satellite in orbit, so these processes are not spontaneous.
41.
Possible arrangements for one molecule:
1 way
1 way
Both are equally probable. Possible arrangements for two molecules:
1 way
2 ways, most probable
1 way
Possible arrangement for three molecules:
1 way
3 ways
3 ways
1 way
equally most probable 42.
a. The most likely arrangement would be 3 molecules in each flask. There are 20 different ways (microstates) to achieve 3 gas molecules in each flask. Let the letters A-F represent the six molecules. The different ways to have groups of three are: ABC, ABD, ABE, ABF, ACD, ACE, ACF, ADE, ADF, AEF, BCD, BCE, BCF, BDE, BDF, BEF, CDE, CDF, CEF, and DEF. b. With six molecules and two bulbs, there are 2 6 = 64 possible arrangements. Because there are 20 ways to achieve 3 gas molecules in each bulb, the probability expressed as a percent of this arrangement is: 20 100 = 31.25% 64
858 43.
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We draw all the possible arrangements of the two particles in the three levels. 2 kJ 1 kJ 0 kJ
x xx
x x
x
Total E =
0 kJ
1 kJ
2 kJ
xx
x x
xx __
2 kJ
3 kJ
4 kJ
B
A
B A
A B
A
B
1 kJ
1 kJ
2 kJ
2 kJ
The most likely total energy is 2 kJ. 44.
2 kJ 1 kJ 0 kJ
AB AB
Etotal =
0 kJ
AB
2 kJ
4 kJ
B A
A B
3 kJ
3 kJ
The most likely total energy is 2 kJ. 45.
a. H2 at 100°C and 0.5 atm; higher temperature and lower pressure means greater volume and hence larger positional probability. b. N2 at STP; N2 at STP has the greater volume because P is smaller and T is larger. c. H2O(l) has a larger positional probability than H2O(s).
46.
Of the three phases (solid, liquid, and gas), solids are most ordered (have the smallest positional probability) and gases are most disordered (have the largest positional probability). Thus a, b, and f (melting, sublimation, and boiling) involve an increase in the entropy of the system since going from a solid to a liquid or from a solid to a gas or from a liquid to a gas increases disorder (increases positional probability). For freezing (process c), a substance goes from the more disordered liquid state to the more ordered solid state; hence, entropy decreases. Process d (mixing) involves an increase in disorder (an increase in positional probability), while separation (phase e) increases order (decreases positional probability). So, of all the processes, a, b, d, and f result in an increase in the entropy of the system.
47.
a. Boiling a liquid requires heat. Hence this is an endothermic process. All endothermic processes decrease the entropy of the surroundings (ΔS surr is negative). b. This is an exothermic process. Heat is released when gas molecules slow down enough to form the solid. In exothermic processes, the entropy of the surroundings increases (ΔS surr is positive).
48.
Ssurr = ‒ H/T a. The combustion reaction of natural gas is an exothermic process, so Ssurr is positive. b. Because the temperature of the surroundings decreases when NH4NO3 dissolves, this process is endothermic. This results in a negative value for Ssurr. c. Because the temperature of the surroundings increases when H2SO4 is added to water, this process is exothermic. This results in a positive value for Ssurr.
CHAPTER 17
SPONTANEITY, ENTROPY, AND FREE ENERGY
859
d. H2O(l) → H2O(s); when water freezes, more intermolecular forces are formed resulting in energy being released. This is an exothermic process resulting in a positive value for Ssurr. e. When a bond forms, energy is released. This is an exothermic process resulting in a positive value for Ssurr. 49.
ΔG = ΔH − TΔS; when ΔG is negative, then the process will be spontaneous. a. ΔG = ΔH − TΔS = 25 × 103 J − (300. K)(5.0 J/K) = 24,000 J; not spontaneous b. ΔG = 25,000 J − (300. K)(100. J/K) = −5000 J; spontaneous c. Without calculating ΔG, we know this reaction will be spontaneous at all temperatures. ΔH is negative and ΔS is positive (−TΔS < 0). ΔG will always be less than zero with these sign combinations for ΔH and ΔS. d. ΔG = −1.0 × 104 J − (200. K)( −40. J/K) = −2000 J; spontaneous
50.
ΔG = ΔH − TΔS; a process is spontaneous when ΔG < 0. For the following, assume ΔH and ΔS are temperature independent. a. When ΔH and ΔS are both negative, ΔG will be negative below a certain temperature where the favorable ΔH term dominates. When ΔG = 0, then ΔH = TΔS. Solving for this temperature: T=
ΔH − 18,000 J = = 3.0 × 102 K ΔS − 60. J / K
At T < 3.0 × 102 K, this process will be spontaneous (ΔG < 0). b. When ΔH and ΔS are both positive, ΔG will be negative above a certain temperature where the favorable ΔS term dominates. T=
ΔH 18,000 J = = 3.0 × 102 K ΔS 60. J / K
At T > 3.0 × 102 K, this process will be spontaneous (ΔG < 0). c. When ΔH is positive and ΔS is negative, this process can never be spontaneous at any temperature because ΔG can never be negative. d. When ΔH is negative and ΔS is positive, this process is spontaneous at all temperatures because ΔG will always be negative. 51.
At the boiling point, ΔG = 0, so ΔH = TΔS. ΔS =
52.
ΔH 27.5 kJ / mol = = 8.93 × 10 −2 kJ/K•mol = 89.3 J/K•mol T ( 273 + 35) K
At the boiling point, ΔG = 0, so ΔH = TΔS. T =
58.51 10 3 J/mol ΔH = = 629.7 K ΔS 92.92 J/K • mol
860 53.
CHAPTER 17
SPONTANEITY, ENTROPY, AND FREE ENERGY
C2H5OH(l) → C2H5OH(g); at the boiling point, G = 0 and Suniv = 0. For the vaporization process, S is a positive value, whereas H is a negative value. To calculate Ssys, we will determine Ssurr from H and the temperature; then Ssys = −Ssurr for a system at equilibrium. Ssurr =
38.7 10 3 J/mol − ΔH = −110. J/K•mol = T 351 K
Ssys = −Ssurr = − (−110.) = 110. J/K•mol 54.
a. NH3(s) → NH3(l); Ssurr = ‒H/T, ‒28.9 J/K∙mol = ‒H/196 K, H = 5.66 × 103 J/mol b. At the melting point, ΔG = 0, so H = TS. S =
H 5.66 103 J/mol = 28.9 J/K•mol = T 196 K
Note that Ssys = −Ssurr which must be the case since the system is at equilibrium where Suniv = 0.
Chemical Reactions: Entropy Changes and Free Energy 55.
a. Decrease in positional probability; ΔS° will be negative. There is only one way to arrange 12 gas molecules all in one bulb, but there are many more ways to arrange the gas molecules equally distributed in each flask. b. Decrease in positional probability; ΔS° is negative for the liquid to solid phase change. c. Decrease in positional probability; ΔS° is negative because the moles of gas decreased when going from reactants to products (3 moles → 0 moles). Changes in the moles of gas present as reactants are converted to products dictates predicting positional probability. The gaseous phase always has the larger positional probability associated with it. d. Increase in positional probability; ΔS° is positive for the liquid to gas phase change. The gas phase always has the larger positional probability.
56.
a. Decrease in positional probability (Δn < 0); ΔS°(−) b. Decrease in positional probability (Δn < 0); ΔS°(−) c. Increase in positional probability; ΔS°(+) d. Increase in positional probability; ΔS°(+)
57.
a. Cgraphite(s); diamond has a more ordered structure (has a smaller positional probability) than graphite. b. C2H5OH(g); the gaseous state is more disordered (has a larger positional probability) than the liquid state. c. CO2(g); the gaseous state is more disordered (has a larger positional probability) than the solid state.
CHAPTER 17 58.
SPONTANEITY, ENTROPY, AND FREE ENERGY
861
a. He (10 K); S = 0 at 0 K b. N2O; more complicated molecule, so has the larger positional probability. c. NH3(l); the liquid state is more disordered (has a larger positional probability) than the solid state.
59.
a. 2 H2S(g) + SO2(g) → 3 Srhombic(s) + 2 H2O(g); because there are more molecules of reactant gases than product molecules of gas (Δn = 2 − 3 < 0), ΔS° will be negative. ΔS° = n pSoproducts − n rSoreactants ΔS° = [3 mol Srhombic(s) (32 J/K•mol) + 2 mol H2O(g) (189 J/K•mol)] − [2 mol H2S(g) (206 J/K•mol) + 1 mol SO2(g) (248 J/K•mol)] ΔS° = 474 J/K − 660. J/K = −186 J/K b. 2 SO3(g) → 2 SO2(g) + O2(g); because Δn of gases is positive (Δn = 3 − 2), ΔS° will be positive. ΔS = 2 mol(248 J/K•mol) + 1 mol(205 J/K•mol) − [2 mol(257 J/K•mol)] = 187 J/K c. Fe2O3(s) + 3 H2(g) → 2 Fe(s) + 3 H2O(g); because Δn of gases = 0 (Δn = 3 − 3), we can’t easily predict if ΔS° will be positive or negative. ΔS = 2 mol(27 J/K•mol) + 3 mol(189 J/K•mol) − [1 mol(90. J/K•mol) + 3 mol (141 J/K•mol)] = 138 J/K
60.
a. H2(g) + 1/2 O2(g) → H2O(l); since Δn of gases is negative, ΔS° will be negative. ΔS° = 1 mol H2O(l) (70. J/K•mol) − [1 mol H2(g) (131 J/K•mol) + 1/2 mol O2(g) (205 J/K•mol)] ΔS° = 70. J/K − 234 J/K = −164 J/K b. 2 CH3OH(g) + 3 O2(g) → 2 CO2(g) + 4 H2O(g); because Δn of gases is positive, ΔS° will be positive. [2 mol (214 J/K•mol) + 4 mol (189 J/K•mol)] − [2 mol (240. J/K•mol) + 3 mol (205 J/K•mol)] = 89 J/K c. HCl(g) → H+(aq) + Cl−(aq); the gaseous state dominates predictions of ΔS°. Here, the gaseous state is more disordered than the ions in solution, so ΔS° will be negative. ΔS° = 1 mol H+(0) + 1 mol Cl−(57 J/K•mol) − 1 mol HCl(187 J/K•mol) = −130. J/K
862 61.
CHAPTER 17
SPONTANEITY, ENTROPY, AND FREE ENERGY
C2H2(g) + 4 F2(g) → 2 CF4(g) + H2(g); ΔS° = 2SoCF4 + SoH2 − [SoC2H2 + 4SoF2 ] −358 J/K = (2 mol) S oCF4 + 131 J/K − [201 J/K + 4(203 J/K)], S oCF4 = 262 J/K•mol
62.
o o o CS2(g) + 3 O2(g) → CO2(g) + 2 SO2(g); ΔS° = SCO + 2SSO − [3SOo 2 + SCS ] 2 2 2 o o −143 J/K = 214 J/K + 2(248 J/K) − 3(205 J/K) − (1 mol) SCS = 238 J/K•mol , SCS 2 2
63.
a. Srhombic → Smonoclinic; this phase transition is spontaneous (ΔG < 0) at temperatures above 95°C. ΔG = ΔH − TΔS; for ΔG to be negative only above a certain temperature, then ΔH is positive and ΔS is positive (see Table 17.5 of text). b. Because ΔS is positive, Srhombic is the more ordered crystalline structure (has the smaller positional probability).
64.
P4(s,α) → P4(s,β) a. At T < −76.9°C, this reaction is spontaneous, and the sign of ΔG is (−). At −76.9°C, ΔG = 0, and above −76.9 °C, the sign of ΔG is (+). This is consistent with ΔH (−) and ΔS (−). b. Because the sign of ΔS is negative, the β form has the more ordered structure (has the smaller positional probability).
65.
a.
CH4(g)
+ 2 O2(g) → CO2(g)
+ 2 H2O(g)
ΔH of
−75 kJ/mol
0
−393.5
−242
ΔG of
−51 kJ/mol
0
−394
−229
S°
186 J/K•mol
205
214
189
Data from Appendix 4
ΔH° = n p ΔHof, products − n r ΔHof, reactants ; ΔS° = n pSoproducts − n rSoreactants ΔH° = 2 mol(−242 kJ/mol) + 1 mol(−393.5 kJ/mol) − [1 mol(−75 kJ/mol)] = −803 kJ ΔS° = 2 mol(189 J/K•mol) + 1 mol(214 J/K•mol) − [1 mol(186 J/K•mol) + 2 mol(205 J/K•mol)] = −4 J/K There are two ways to get ΔG°. We can use ΔG° = ΔH° − TΔS° (be careful of units): ΔG° = ΔH° − TΔS° = −803 × 103 J − (298 K)( −4 J/K) = −8.018 × 105 J = −802 kJ or we can use ΔG of values, where ΔG° = n p ΔG of , products − n r ΔG of, reactants : ΔG° = 2 mol(−229 kJ/mol) + 1 mol(−394 kJ/mol) − [1 mol(−51 kJ/mol)] ΔG° = −801 kJ (Answers are the same within round off error.)
CHAPTER 17
SPONTANEITY, ENTROPY, AND FREE ENERGY
b.
6 CO2(g)
+
6 H2O(l)
863
→ C6H12O6(s) + 6 O2(g)
ΔH of
−393.5 kJ/mol
−286
−1275
0
S°
214 J/K•mol
70.
212
20
ΔH° = −1275 − [6−286) + 6(−393.5)] = 2802 kJ ΔS° = 6(205) + 212 − [6(214) + 6(70.)] = −262 J/K ΔG° = 2802 kJ − (298 K)( −0.262 kJ/K) = 2880. kJ P4O10(s) + 6 H2O(l) → 4 H3PO4(s)
c. ΔH of (kJ/mol)
−2984
−286
−1279
S° (J/K•mol)
229
70.
110.
ΔH° = 4 mol(−1279 kJ/mol) − [1 mol(−2984 kJ/mol) + 6 mol(−286 kJ/mol)] = −416 kJ ΔS° = 4(110.) − [229 + 6(70.)] = −209 J/K ΔG° = ΔH° − TΔS° = −416 kJ − (298 K)( −0.209 kJ/K) = −354 kJ HCl(g) + NH3(g) → NH4Cl(s)
d. ΔH of (kJ/mol)
−92
−46
−314
S° (J/K•mol)
187
193
96
ΔH° = −314 − [−92 − 46] = −176 kJ; ΔS° = 96 − [187 + 193] = −284 J/K ΔG° = ΔH° − TΔS° = −176 kJ − (298 K)( −0.284 kJ/K) = −91 kJ 66.
Appendix 4 data was used to calculate ΔH° and ΔS°, then the expression ΔG° = ΔH° − TΔS° was used to calculate ΔG°. a. C2H4(g) + O3(g) → CH3CHO(g) + O2(g) ΔH° = −166 kJ − [52 kJ + 143 kJ] = −361 kJ ΔS° = [250. J/K + 205 J/K] − [219 J/K + 239 J/K] = −3 J/K ΔG° = ΔH° − TΔS° = −361 kJ − (298 K)(−3 10−3 kJ/K) = −360. kJ b. O3(g) + NO(g) → NO2(g) + O2(g) ΔH° = 34 kJ − [143 kJ + 90. kJ] = −199 kJ ΔS° = [240. J/K + 205 J/K] − [239 J/K + 211 J/K] = −5 J/K ΔG° = ΔH° − TΔS° = −199 kJ − (298 K)(−5 10−3 kJ/K) = −198 kJ
864
CHAPTER 17
SPONTANEITY, ENTROPY, AND FREE ENERGY
c. SO3(g) + H2O(l) → H2SO4(aq) ΔH° = −909 kJ − [−396 kJ + (−286 kJ)] = −227 kJ ΔS° = [20. J/K] − [257 J/K + 70. J/K] = −307 J/K ΔG° = ΔH° − TΔS° = −227 kJ − (298 K)(−0.307 kJ/K) = −136 kJ 67.
a. ΔH° = 2(−46 kJ) = −92 kJ; ΔS° = 2(193 J/K) − [3(131 J/K) + 192 J/K] = −199 J/K ΔG° = ΔH° − TΔS° = −92 kJ − 298 K(−0.199 kJ/K) = −33 kJ b. ΔG° is negative, so this reaction is spontaneous at standard conditions. c. ΔG° = 0 when T =
− 92 kJ ΔH o = 460 K = o − 0.199 kJ / K ΔS
At T < 460 K and standard pressures (1 atm), the favorable ΔH° term dominates, and the reaction is spontaneous (ΔG° < 0). 68.
a. 2 H2S(g) + SO2(g) → 3 Srhombic(s) + 2 H2O(l) ΔH° = 2(−286 kJ) − [2(−21 kJ) + (−297 kJ)] = −233 kJ ΔS° = [3(32. J/K) + 2(70.)] − [2(206 J/K) + 248. J/K] = −424 J/K ΔG° = 2 (−237 kJ) − [2(−34 kJ) + (−300. kJ)] = −106. kJ b. The reaction is spontaneous at standard conditions since ΔG° < 0. c. ΔG° = 0 when T =
ΔHo −233 kJ = 550. K = o ΔS −0.424 kJ/K
At T < 550. K and standard pressures (1 atm), the favorable ΔH° term dominates, and the reaction is spontaneous (ΔG° < 0). 69.
C2H2(g) + H2(g) → C2H4(g); ΔG° = 0 when T =
ΔHo −174 kJ = 1550 K = 1280°C = o ΔS −0.112 kJ/K
At T < 1550 K (1280°C) and standard pressures (1 atm), the favorable ΔH° term dominates, and the reaction is spontaneous (ΔG° < 0). 70.
2 N2O5(g) → 4 NO2(g) + O2(g); ΔG° = 0 when T =
ΔHo 110.2 kJ = 242 K = o ΔS 0.455 kJ/K
At T > 242 K and standard pressures (1 atm), the favorable Δs° term dominates, and the reaction is spontaneous (ΔG° < 0). 71.
H2O(l) → H2O(g); ΔG° = 0 at the boiling point of water at 1 atm and 100. °C. ΔH° = TΔS°, ΔS° =
ΔH o 40.6 10 3 J / mol = = 109 J/K•mol T 373 K
At 90.°C: ΔG° = ΔH° − TΔS° = 40.6 kJ/mol – (363 K)(0.109 kJ/K•mol) = 1.0 kJ/mol
CHAPTER 17
SPONTANEITY, ENTROPY, AND FREE ENERGY
865
As expected, ΔG° > 0 at temperatures below the boiling point of water at 1 atm (process is nonspontaneous). At 110.°C: ΔG° = ΔH° − TΔS° = 40.6 kJ/mol – (383 K)(0.109 J/K•mol) = −1.1 kJ/mol When ΔG° < 0, the boiling of water is spontaneous at 1 atm, and T > 100. °C (as expected). 72.
I2(s) → I2(g); ΔG° = ΔH° − TΔS°, 19 kJ = 62 kJ − (298)ΔS°, ΔS° = 0.14 kJ/K At the sublimation temperature, ΔG° = 0, so ΔH° = TΔS°. T=
ΔH o 62 kJ = 440 K = 170°C = o ΔS 0.14 kJ/K
Assuming ΔH° and ΔS° are temperature independent, iodine will sublime at 170°C. 73.
CH4(g) → 2 H2(g) + C(s) 2 H2(g) + O2(g) → 2 H2O(l) C(s) + O2(g) → CO2(g) CH4(g) + 2 O2(g) → 2 H2O(l) + CO2(g)
ΔG° = − (−51 kJ) ΔG° = −474 kJ) ΔG° = −394 kJ ΔG° = −817 kJ
74.
6 C(s) + 6 O2(g) → 6 CO2(g) 3 H2(g) + 3/2 O2(g) → 3 H2O(l) 6 CO2(g) + 3 H2O(l) → C6H6(l) + 15/2 O2(g) 6 C(s) + 3 H2(g) → C6H6(l)
ΔG° = 6(−394 kJ) ΔG° = 3(−237 kJ) ΔG° = −1/2 (−6399 kJ) ΔG° = 125 kJ
75.
ΔG° = Σn p ΔGof, products − Σn r ΔGof, reactants, −374 kJ = −1105 kJ − ΔG of , SF4
ΔG of , SF4 = −731 kJ/mol 76.
−5490. kJ = 8(−394 kJ) + 10(−237 kJ) − 2 ΔGof, C4H10 , ΔGof, C4H10 = −16 kJ/mol
77.
a. ΔG° = 2 mol(0) + 3 mol(−229 kJ/mol) − [1 mol(−740. kJ/mol) + 3 mol(0)] = 53 kJ b. Because ΔG° is positive, this reaction is not spontaneous at standard conditions and 298 K. c. ΔG° = ΔH° − TΔS°, ΔS° =
H o − G o 100 . kJ − 53 kJ = = 0.16 kJ/K T 298 K
We need to solve for the temperature when ΔG° = 0: ΔG° = 0 = ΔH° − TΔS°, ΔH° = TΔS°, T =
H o S
o
=
100. kJ = 630 K 0.16 kJ/K
This reaction will be spontaneous (ΔG < 0) at T > 630 K, where the favorable entropy term will dominate.
866
CHAPTER 17
SPONTANEITY, ENTROPY, AND FREE ENERGY
78.
a. ΔG° = 2(−270. kJ) − 2(−502 kJ) = 464 kJ b. Because ΔG° is positive, this reaction is not spontaneous at standard conditions at 298 K. c. ΔG° = ΔH° − TΔS°, ΔH° = ΔG° + TΔS° = 464 kJ + 298 K(0.179 kJ/K) = 517 kJ We need to solve for the temperature when ΔG° = 0: ΔH o 517 kJ = 2890 K = o 0.179 kJ / K ΔS
ΔG° = 0 = ΔH° − TΔS°, T =
This reaction will be spontaneous at standard conditions (ΔG° < 0) when T > 2890 K. Here the favorable entropy term will dominate.
Free Energy: Pressure Dependence and Equilibrium 79.
ΔG = ΔG° + RT ln Q; for this reaction: ΔG = ΔG° + RT ln ΔG = −81.0 kJ +
2 PHCl × PBr2 2 PHBr × PCl2
8.3145 J/K • mol (0.500) 2 (0.200) (298 K) ln 1000 J/kJ (1.00) 2 (5.00)
ΔG = −81.0 kJ + (‒11.4 kJ = −92.4 kJ 80.
ΔG° = 3(0) + 2(−229) − [2(−34) + 1(−300.)] = −90. kJ ΔG = ΔG° + RT ln
PH2 O 2
PH2 S PSO 2
= −90. kJ +
2
(8.3145 ) (298) (0.030 ) 2 kJ ln −4 1000 (1.0 10 ) (0.010 )
ΔG = −90. kJ + 39.7 kJ = −50. kJ 81.
ΔG = ΔG° + RT ln Q = ΔG° + RT ln
PN 2O 4 2 PNO 2
ΔG° = 1 mol(98 kJ/mol) − 2 mol(52 kJ/mol) = −6 kJ a. These are standard conditions, so ΔG = ΔG° because Q = 1 and ln Q = 0. Because ΔG° is negative, the forward reaction is spontaneous. The reaction shifts right to reach equilibrium. b. ΔG = −6 × 103 J + 8.3145 J/K•mol (298 K) ln
0.50 (0.21) 2
ΔG = −6 × 103 J + 6.0 × 103 J = 0 Because ΔG = 0, this reaction is at equilibrium (no shift). c. ΔG = −6 × 103 J + 8.3145 J/K•mol (298 K) ln
1. 6 (0.29) 2
CHAPTER 17
SPONTANEITY, ENTROPY, AND FREE ENERGY
867
ΔG = −6 × 103 J + 7.3 × 103 J = 1.3 × 103 J = 1 × 103 J Because ΔG is positive, the reverse reaction is spontaneous, and the reaction shifts to the left to reach equilibrium. 82.
a. ΔG = ΔG° + RT ln ΔG = −34 kJ +
2 PNH 3
PN 2 PH2 2
; ΔG° = 2 ΔGof , NH3 = 2(−17) = −34 kJ
(8.3145 J/K • mol) (298 K) (50.) 2 ln 1000 J/kJ (200 .) (200 .) 3
ΔG = −34 kJ − 33 kJ = −67 kJ b. ΔG = −34 kJ +
(8.3145 J/K • mol) (298 K) (200 .) 2 ln 1000 J/kJ (200 .) (600 .) 3
ΔG = −34 kJ − 34.4 kJ = −68 kJ 83.
NO(g) + O3(g) ⇌ NO2(g) + O2(g); ΔG° = Σn p ΔGof, products − Σn r ΔGof, reactants ΔG° = 1 mol(52 kJ/mol) − [1 mol(87 kJ/mol) + 1 mol(163 kJ/mol)] = −198 kJ ΔG° = −RT ln K, K = exp
− (−1.98 10 5 J) − ΔG o 79.912 = 5.07 × 1034 = exp = e RT 8.3145 J/K • mol(298 K)
Note: When determining exponents, we will round off after the calculation is complete. This helps eliminate excessive round off error. 84.
ΔG° = 2 mol(−229 kJ/mol) − [2 mol(−34 kJ/mol) + 1 mol(−300. kJ/mol)] = −90. kJ
K = exp
85.
− ΔG o − (−9.0 10 4 J) 36.32 = 5.9 × 1015 = exp =e RT 8.3145 J/K • mol(298 K)
At 25.0°C: ΔG° = ΔH° − TΔS° = −58.03 × 10 3 J/mol − (298.2 K)(−176.6 J/K•mol) = −5.37 × 103 J/mol ΔG° = −RT ln K, ln K =
− (−5.37 10 3 J/mol) − ΔG o = exp = 2.166 RT (8.3145 J/K • mol) (298.2 K)
K = e2.166 = 8.72 At 100.0°C: ΔG° = −58.03 × 103 J/mol − (373.2 K)(−176.6 J/K•mol) = 7.88 × 103 J/mol ln K =
− (7.88 10 3 J/mol) = −2.540, K = e-2.540 = 0.0789 (8.3145 J/K • mol) (373.2 K)
Note: When determining exponents, we will round off after the calculation is complete. This helps eliminate excessive round off error.
868 86.
CHAPTER 17
SPONTANEITY, ENTROPY, AND FREE ENERGY
a. ΔG° = 3(191.2) − 78.2 = 495.4 kJ; ΔH° = 3(241.3) − 132.8 = 591.1 kJ ΔS° =
H o − G o 591 .1 kJ − 495 .4 kJ = 0.321 kJ/K = 321 J/K = T 298 K
b. ΔG° = −RT ln K, ln K =
− ΔG o − 495,400 J = = −199.942 RT 8.3145 J/K • mol(298 K)
K = e−199.942 = 1.47 × 10−87 c. Assuming ΔH° and ΔS° are temperature-independent: o = 591.1 kJ − 3000. K(0.321 kJ/K) = −372 kJ ΔG 3000
ln K = 87.
− (−372,000 J) = 14.914, K = e14.914 = 3.00 × 106 8.3145 J/K • mol(3000. K)
When reactions are added together, the equilibrium constants are multiplied together to determine the K value for the final reaction. H2(g) +O2(g) ⇌ H2O2(g) H2O(g) ⇌ H2(g) + 1/2O2(g) H2O(g) + 1/2O2(g) ⇌ H2O2(g) ΔG° = −RT ln K =
88.
a. NH3(g) O2(g) NO(g) H2O(g) NO2(g) HNO3(l) H2O(l)
K = 2.3 × 106 K = (1.8 × 1037)−1/2 K = 2.3 × 106(1.8 × 1037) −1/2 = 5.4 × 10 −13
− 8.3145 J (600. K) ln(5.4 × 10 −13 ) = 1.4 × 105 J/mol = 140 kJ/mol K • mol
ΔH of (kJ/mol)
S° (J/K•mol)
−46 0 90. −242 34 −174 −286
193 205 211 189 240. 156 70.
4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g) ΔH° = 6(−242) + 4(90.) − [4(−46)] = −908 kJ ΔS° = 4(211) + 6(189) − [4(193) + 5(205)] = 181 J/K ΔG° = −908 kJ – (298 K)(0.181 kJ/K) = −962 kJ ΔG° = − RT ln K, ln K =
− (−962 10 3 J) − ΔG o = = 388 RT 8.3145 J/K • mol 298 K
ln K = 2.303 log K, log K = 168, K = 10168 (an extremely large number)
CHAPTER 17
SPONTANEITY, ENTROPY, AND FREE ENERGY
869
2 NO(g) + O2(g) → 2 NO2(g) ΔH° = 2(34) − [2(90.)] = −112 kJ; ΔS° = 2(240.) − [2(211) + (205)] = −147 J/K ΔG° = −112 kJ − (298 K)( −0.147 kJ/K) = −68 kJ − (−68,000 J) − ΔG o 27.44 = exp K = exp = 8.3 × 1011 =e RT 8.3145 J/K • mol (298 K)
Note: When determining exponents, we will round off after the calculation is complete. 3 NO2(g) + H2O(l) → 2 HNO3(l) + NO(g) ΔH° = 2(−174) + (90.) − [3(34) + (−286)] = −74 kJ ΔS° = 2(156) + (211) − [3(240.) + (70.)] = −267 J/K ΔG° = −74 kJ − (298 K)( −0.267 kJ/K) = 6 kJ K = exp
− 6000 J − ΔG o −2.4 −2 = exp = e = 9 × 10 RT 8.3145 J/K • mol (298 K)
b. ΔG° = −RT ln K; T = 825°C = (825 + 273) K = 1098 K; we must determine ΔG° at 1098 K. o = ΔH° − TΔS° = −908 kJ − (1098 K)(0.181 kJ/K) = −1107 kJ ΔG1098
K = exp
− (−1.107 10 6 J) − ΔG o 121.258 = 4.589 × 1052 = exp =e RT 8.3145 J/K • mol (1098 K)
c. There is no thermodynamic reason for the elevated temperature because ΔH° is negative and ΔS° is positive. Thus the purpose for the high temperature must be to increase the rate of the reaction. 89.
− (−30,500 J) 5 a. ΔG° = −RT ln K, K = exp = 2.22 × 10 8.3145 J/K • mol 298 K
b. C6H12O6(s) + 6 O2(g) → 6 CO2(g) + 6 H2O(l) ΔG° = 6 mol(−394 kJ/mol) + 6 mol(−237 kJ/mol) − 1 mol(−911 kJ/mol) = −2875 kJ 2875 kJ 1 mol ATP = 94.3 mol ATP; 94.3 molecules ATP/molecule glucose mol glucose 30.5 kJ
This is an overstatement. The assumption that all the free energy goes into this reaction is false. Only 38 moles of ATP are produced by metabolism of 1 mole of glucose. 90.
a. ln K =
− 14,000 J − ΔG o = = −5.65, K = e−5.65 = 3.5 × 10−3 RT (8.3145 J/K • mol)(298 K)
870
CHAPTER 17
SPONTANEITY, ENTROPY, AND FREE ENERGY
Glutamic acid + NH3 → Glutamine + H2O ATP + H2O → ADP + H2PO4−
b.
Glutamic acid + ATP + NH3 → Glutamine + ADP + H2PO4− ln K =
91.
K=
ΔG° = 14 kJ ΔG° = −30.5 kJ ΔG° = 14 − 30.5 = −17 kJ
− (−17,000 J) − ΔG o = 6.86, K = e6.86 = 9.5 × 102 = RT 8.3145 J/K • mol(298 K)
2 PNF 3
=
PN 2 PF32
(0.48) 2 = 4.4 × 104 0.021(0.063) 3
o = −RT ln K = −8.3145 J/K•mol(800. K) ln(4.4 × 104) = −7.1 × 104 J/mol = −71 kJ/mol ΔG 800
92.
2 SO2(g) + O2(g) → 2 SO3(g); ΔG° = 2(−371 kJ) − [2(−300. kJ)] = −142 kJ ΔG° = −RT ln K, ln K =
− (−142.000 J) − ΔG o = 57.311 = RT 8.3145 J/K • mol (298 K)
K = e57.311 = 7.76 × 1024 2 PSO 3
24
K = 7.76 × 10 =
2 PSO PO 2 2
=
(2.0) 2 , PSO 2 = 1.0 × 10 −12 atm 2 PSO (0.50) 2
From the negative value of ΔG°, this reaction is spontaneous at standard conditions. There are more molecules of reactant gases than product gases, so ΔS° will be negative (unfavorable). Therefore, this reaction must be exothermic (ΔH° < 0). When ΔH° and ΔS° are both negative, the reaction will be spontaneous at relatively low temperatures where the favorable ΔH° term dominates. 93.
The equilibrium concentration of C is greater than the initial concentration, so the reaction shifts right to reach equilibrium. The general ICE table for this problem is: A(aq) Initial Change Equil
+
2 B(aq)
⇌ 3 C(aq)
1.00 M 1.00 M 1.00 M Let x mol/L of A react to reach equilibrium −x −2x → +3x 1.00 − x 1.00 − 2x 1.00 + 3x
In the problem, we are told that [C]e = 1.96 M. From the set-up, [C]e = 1.00 + 3x = 1.96 M. Solving: x = 0.32 mol/L. The other concentrations are [A] = 1.00 – 0.32 = 0.68 mol/L and [B]e = 1.00 −2(0.32) = 0.36 mol/L. Calculating K: K=
[C]3 (1.96)3 = 85 = [A][B]2 0.68(0.36) 2
ΔG° = −RT ln K = −8.3145 J/K•mol(298 K) ln(85) = −1.1 × 104 J/mol = −11 kJ/mol
CHAPTER 17
SPONTANEITY, ENTROPY, AND FREE ENERGY
94.
2 HBr(g) Initial Equil
2.00 atm 2.00 − 2x
⇌
H2(g)
+
0 x
871
Br2(g) 0 x
From the problem, PH 2 = 5.0 10−10 atm = x. K=
x( x) (5.0 10− ) 2 = 6.25 10−20 = (2.00 − 2x)2 [2.00 − (5.0 10− )]2
ΔG° = −RT ln K = −8.3145 J/K•mol(298 K) ln(6.25 10−20) = 1.10 × 105 J/mol = 110. kJ/mol ΔG° = ΔH° − TΔS, ΔS° =
95.
ΔHo − ΔG o 104 kJ − 110. kJ = ‒0.02 kJ/K = ‒20 J/K = T 298 K
− ΔH o 1 ΔSo is in the form of a straight line equation + R T R (y = mx + b). A graph of ln K versus 1/T will yield a straight line with slope = m = −ΔH°/R and a y intercept = b = ΔS°/R. The equation ln K =
From the plot: slope =
Δy 0 − 40. = = −1.3 × 104 K Δx 3.0 10 −3 K −1 − 0
−1.3 × 104 K = −ΔH°/R, ΔH° = 1.3 × 104 K × 8.3145 J/K•mol = 1.1 × 105 J/mol y intercept = 40. = ΔS°/R, ΔS° = 40. × 8.3145 J/K•mol = 330 J/K•mol As seen here, when ΔH° is positive, the slope of the ln K versus 1/T plot is negative. When ΔH° is negative as in an exothermic process, then the slope of the ln K versus 1/T plot will be positive (slope = −ΔH°/R). 96.
The ln K versus 1/T plot gives a straight line with slope = −H°/R and y intercept = S°/R. 1.352 × 104 K = −H°/R, H° = − (8.3145 J/K•mol) (1.352 × 104 K) H° = −1.124 × 105 J/mol = −112.4 kJ/mol −14.51 = S°/R, S° = (−14.51)(8.3145 J/K•mol) = −120.6 J/K•mol Note that the signs for ΔH° and ΔS° make sense. When a bond forms, ΔH° < 0 and ΔS° < 0.
97.
ΔG° = −RT ln K; ΔG° = ΔH° − TΔS°; at 25°C, K < 1 so ΔG° must be positive. At 227 °C, K > 1 so ΔG° must be negative. As temperature increased, the sign of ΔG° changed from a positive value to a negative value. This only occurs when the signs of ΔH° and ΔS° are both positive for a reaction. So, at 25°C, ΔG°, ΔH°, and ΔS° are all positive.
872
CHAPTER 17
SPONTANEITY, ENTROPY, AND FREE ENERGY
98.
ΔG° = −RT ln K; ΔG° = ΔH° − TΔS°; at the higher temperatures in the data, K < 1 which dictates that ΔG° is positive. At the lower temperatures, K > 1 so ΔG° is negative as these lower temperatures. As temperature decreased, the sign of ΔG° changed from a positive value to a negative value. This only occurs when the signs of ΔH° and ΔS° are both negative for a reaction. When both ΔH° and ΔS° are negative, as temperature decreases, the favorable enthalpy term dominates making ΔG° a negative number. The lowest temperature data is 109 °C where K > 1 and ΔG° is negative. When the temperature is lowered to 25°C, ΔG° will become more negative. So at 25°C, ΔG°, ΔH°, and ΔS° are all negative.
99.
NO(g) + 1/2 O2(g) → NO2(g) ΔG° = −RT ln K = −8.3145 J/K•mol(298 K) ln(1.50 106) = −3.52 × 104 J/mol ΔG° = ΔH° − TΔS, ΔS° =
ΔHo - ΔG o −57.0 kJ − (−35.2 kJ) = ‒0.0732 kJ/K = ‒73.2 J/K = T 298 K
2 NO(g) + O2(g) → 2 NO2(g); the coefficients of this reaction are double the original reaction. So, ΔS° = 2(‒73.2) = ‒146 J/K. 100.
ΔG° = −RT ln K = −8.3145 J/K•mol(645 K) ln(0.0150) = 2.25 × 104 J/mol = 22.5 kJ/mol ΔG° = ΔH° − TΔS, ΔS° =
ΔHo − ΔG o kJ − (22.5 kJ) = ‒0.010 kJ/K = ‒10. J/K = T 645 K
ChemWork Problems 101.
From Appendix 4, S° = 198 J/K•mol for CO(g) and S° = 27 J/K•mol for Fe(s). Let S ol = S° for Fe(CO)5(l) and S og = S° for Fe(CO)5(g). S° = −677 J/K = 1 mol(S ol ) – [1 mol (27 J/K•mol) + 5 mol(198 J/ K•mol] S ol = 340. J/K•mol S° = 107 J/K = 1 mol (S og ) – 1 mol (340. J/K•mol) S og = S° for Fe(CO)5(g) = 447 J/K•mol
102.
When an ionic solid dissolves, one would expect the disorder of the system to increase, so ΔSsys is positive. Because temperature increased as the solid dissolved, this is an exothermic process, and ΔSsurr is positive (ΔSsurr = −ΔH/T). Because the solid did dissolve, the dissolving process is spontaneous, so ΔSuniv is positive.
103.
H2O(s) → H2O(l); this is an endothermic process because energy must be added to melt an ice cube. So, ΔH is positive. There is an increase in positional probability as a solid is converted into a liquid, so ΔS is positive. At ‒10°C, melting an ice cube is not spontaneous, so ΔG is positive.
CHAPTER 17 104.
SPONTANEITY, ENTROPY, AND FREE ENERGY
873
ΔG = ΔH − TΔS = −126.4 kJ ‒ 298 K(‒0.0749 kJ/K) = ‒104.1 When ΔG° = 0, ΔH° = TΔS°, T =
ΔH −126.4 kJ = = 1690 K ΔS −0.0749 kJ/K
With negative signs for both ΔH and ΔS, the reaction will be spontaneous below some temperature. So, answer a is true. Assuming ΔH and ΔS are temperature independent, the reaction is at equilibrium at 1690 K. The reaction is not spontaneous at T > 1690 K. Answer b is false since the reaction is at equilibrium at T = 1690 K. Answer c is false since ΔG is negative at 298 K, so ΔSuniv > 0. And answer d is false since ΔS is negative. 105.
A negative value for ΔS° indicates a process that has a decrease in positional probability. Processes b, c, and d all are expected to have negative values for ΔS°. In these processes, the number of moles of gaseous molecules decreases as reactants are converted to products. Whenever this occurs, positional probability decreases. Processes a and e have positive ΔS° values. In process a, the number of gaseous molecules increase as reactants are converted to products, so positional probability increases. In process e, the liquid state has a larger positional probability as compared to the solid state.
106.
ΔS° =
n S
o p products −
n S r
o reactants ;
ΔH° =
n ΔH p
o f, products −
n ΔH r
o f, reactants ;
ΔS° = [1 mol(146 J/K•mol) + 1 mol(203 J/K•mol)] − [1 mol(300 J/K•mol)] = 49 J/K ΔH° = [1 mol(−251 kJ/mol) + 1 mol(0)] − [1 mol(−294 kJ/mol)] = 43 kJ ΔSsurr =
− (43 kJ) − ΔH o = = −0.14 kJ/K = −140 J/K T 298 K
107.
It appears that the sum of the two processes has no net change. This is not so. By the second law of thermodynamics, ΔSuniv must have increased even though it looks as if we have gone through a cyclic process.
108.
Enthalpy is not favorable, so ΔS must provide the driving force for the change. Thus, ΔS is positive. There is an increase in positional probability, so the original enzyme has the more ordered structure (has the smaller positional probability).
109.
C2H4(g) + H2O(g) → CH3CH2OH(l) ΔH° = −278 − (52 − 242) = −88 kJ; ΔS° = 161 − (219 + 189) = −247 J/K When ΔG° = 0, ΔH° = TΔS°, so T =
ΔH o − 88 10 3 J = = 360 K. − 247 J/K ΔSo
Since the signs of ΔH° and ΔS° are both negative, this reaction at standard concentrations will be spontaneous at temperatures below 360 K (where the favorable ΔH° term will dominate).
874
CHAPTER 17
SPONTANEITY, ENTROPY, AND FREE ENERGY
C2H6(g) + H2O(g) → CH3CH2OH(l) + H2(g) ΔH° = −278 − (−84.7 − 242) = 49 kJ; ΔS° = 131 + 161 − (229.5 + 189) = −127 J/K Both ΔH° and ΔS° have unfavorable signs. ΔG° can never be negative when ΔH° is positive and ΔS° is negative. So this reaction can never be spontaneous at standard conditions. Thus the reaction C2H4(g) + H2O(g) → C2H5OH(l) would be preferred at standard conditions. 110.
111.
ΔH is negative because the reaction is exothermic. Because there are more product gas molecules than reactant gas molecules (Δn > 0), ΔS will be positive. From the signs of ΔH and ΔS, this reaction is spontaneous at all temperatures. It will cost money to heat the reaction mixture. Because there is no thermodynamic reason to do this, the purpose of the elevated temperature must be to increase the rate of the reaction, that is, kinetic reasons. a.
CH2
CH2(g) + HCN(g)
CH2 CHCN(g) + H2O(l)
O ΔH° = 185.0 − 286 − (−53 + 135.1) = −183 kJ; ΔS° = 274 + 70. − (242 + 202) = −100. J/K ΔG° = ΔH° − TΔS° = −183 kJ − 298 K(−0.100 kJ/K) = −153 kJ b. HC≡CH(g) + HCN(g) → CH2=CHCN(g) ΔH° = 185.0 − (135.1 + 227) = −177 kJ; ΔS° = 274 − (202 + 201) = −129 J/K o T = 70.°C = 343 K; ΔG 343 = ΔH° − TΔS° = −177 kJ − 343 K(−0.129 kJ/K) o = −177 kJ + 44 kJ = −133 kJ ΔG 343
c. 4 CH2=CHCH3(g) + 6 NO(g) → 4 CH2=CHCN(g) + 6 H2O(g) + N2(g) ΔH° = 6(−242) + 4(185.0) − [4(20.9) + 6(90.)] = −1336 kJ ΔS° = 192 + 6(189) + 4(274) − [6(211) + 4(266.9)] = 88 J/K o T = 700.°C = 973 K; ΔG 973 = ΔH° − TΔS° = −1336 kJ − 973 K(0.088 kJ/K) o = −1336 kJ − 86 kJ = −1422 kJ ΔG 973
112.
At boiling point, ΔG = 0 so ΔS =
For hexane: ΔS =
ΔH vap T
; for methane: ΔS =
28.9 10 3 J/mol = 84.5 J/mol•K 342 K
8.20 10 3 J / mol 112 K = 73.2 J/mol•K
CHAPTER 17 Vmet =
SPONTANEITY, ENTROPY, AND FREE ENERGY
875
nRT 1.00 mol (0.08206 L atm/K • mol) (112 K) nRT = = 9.19 L; Vhex = = 28.1 L P P 1.00 atm
Hexane has the larger molar volume at the boiling point, so hexane should have the larger entropy. As the volume of a gas increases, positional disorder increases. 113.
solid I → solid II; equilibrium occurs when ΔG = 0. − 743.1 J/mol ΔG = ΔH − TΔS, ΔH = TΔS, T = ΔH/ΔS = = 43.7 K = −229.5°C − 17.0 J/K • mol
114.
a. ΔG° = −RT ln K = − (8.3145 J/K•mol)(298 K) ln 0.090 = 6.0 × 103 J/mol = 6.0 kJ/mol b. H‒O‒H + Cl‒O‒Cl → 2 H‒O‒Cl On each side of the reaction there are 2 H‒O bonds and 2 O‒Cl bonds. Both sides have the same number and type of bonds. Thus ΔH ΔH° 0. c. ΔG° = ΔH° − TΔS°, ΔS° =
ΔH o − ΔG o 0 − 6.0 10 3 J = −20. J/K = T 298 K
d. For H2O(g), ΔH of = −242 kJ/mol and S° = 189 J/K•mol. ΔH° = 0 = 2 ΔHof , HOCl − [1 mol(−242 kJ/mol) + 1 mol(80.3 J/K/mol)], ΔH of , HOCl = −81 kJ/mol −20. J/K = 2 SoHOCl − [1 mol(189 J/K•mol) + 1 mol(266.1 J/K•mol)], SoHOCl = 218 J/K•mol o e. Assuming ΔH° and ΔS° are T-independent: Δ G 500 = 0 − (500. K)( −20. J/K) = 1.0 × 104 J
− 1.0 10 4 − ΔG o −2.41 = exp ΔG° = −RT ln K, K = exp = 0.090 =e RT (8.3145)(500.) f.
ΔG = ΔG° + RT ln
2 PHOCl ; from part a, ΔG° = 6.0 kJ/mol. PH 2O PCl 2O
We should express all partial pressures in atm. However, we perform the pressure conversion the same number of times in the numerator and denominator, so the factors of 760 torr/atm will all cancel. Thus we can use the pressures in units of torr. ΔG = 6.0 kJ/mol + 115.
(8.3145 J/K • mol) (298 K) (0.10) 2 ln = 6.0 − 20. = −14 kJ/mol 1000 J/kJ (18) (2.0)
Ba(NO3)2(s) ⇌ Ba2+(aq) + 2 NO3−(aq) K = Ksp; ΔG° = −561 + 2(−109) − (−797) = 18 kJ ΔG° = −RT ln Ksp, ln Ksp = Ksp = e −7.26 = 7.0 ×10−4
− 18,000 J − ΔG o = = −7.26 8.3145 J/K • mol (298 K) RT
876 116.
CHAPTER 17
SPONTANEITY, ENTROPY, AND FREE ENERGY
2 H2O(l) → 2 H2(g) + O2(g); ΔG° = ΔH° − TΔS°; there is an increase in moles of gas as reactants are converted to products, so ΔS° is positive. Since ΔS° is positive (favorable) but ΔG° is positive, ΔH° must be positive (the reaction is endothermic). Answer d is correct. When ΔG° is positive, K < 1 so a is false. Answer b is false since ΔG° is positive. And with positive signs for both ΔH° and ΔS°, the reaction will be spontaneous at standard conditions (ΔG° < 0) when the ΔS° is positive term dominates above some temperature.
117.
118.
ΔG° = −RT ln K; ΔG = ΔH − TΔS; With K < 1, ΔG° must be positive. The reaction is spontaneous, so ΔG must be negative. This indicates that the reaction shifts right to reach equilibrium. With a negative ΔG but a positive ΔH (the reaction is endothermic), ΔS must be positive (favorable) to give a negative ΔG value. Statement c is false.
[K + ] K+(blood) ⇌ K+(muscle) ΔG° = 0; ΔG = RT ln + m ; ΔG = wmax [ K ]b ΔG =
8.3145 J 0.15 (310. K) ln , ΔG = 8.8 × 103 J/mol = 8.8 kJ/mol K mol 0.0050
At least 8.8 kJ of work must be applied to transport 1 mol K +. Other ions will have to be transported to maintain electroneutrality. Either anions must be transported into the cells, or cations (Na+) in the cell must be transported to the blood. The latter is what happens: [Na+] in blood is greater than [Na+] in cells as a result of this pumping. 119.
HgbO2 → Hgb + O2 Hgb + CO → HgbCO HgbO2 + CO → HgbCO + O2
ΔG° = − (−70 kJ) ΔG° = −80 kJ ΔG° = −10 kJ
− ΔG o − (−10 10 3 J) = exp ΔG° = −RT ln K, K = exp = 60 RT (8.3145 J/K • mol)(298 K) 120.
HF(aq) ⇌ H+(aq) + F−(aq); ΔG = ΔG° + RT ln
[H + ][F− ] [HF]
ΔG° = −RT ln K = −(8.3145 J/K•mol)(298 K) ln(7.2 × 10 −4 ) = 1.8 × 104 J/mol a. The concentrations are all at standard conditions, so ΔG = ΔG = 1.8 × 104 J/mol (Q = 1.0 and ln Q = 0). Because ΔG° is positive, the reaction shifts left to reach equilibrium. (2.7 10 −2 ) 2 b. ΔG = 1.8 × 104 J/mol + (8.3145 J/K•mol)(298 K) ln 0.98 ΔG = 1.8 × 104 J/mol − 1.8 × 104 J/mol = 0 ΔG = 0, so the reaction is at equilibrium (no shift).
CHAPTER 17
SPONTANEITY, ENTROPY, AND FREE ENERGY
c. ΔG = 1.8 × 104 J/mol + 8.3145(298 K) ln
d. ΔG = 1.8 × 104 + 8.3145(298) ln
e. ΔG = 1.8 × 104 + 8.3145(298) ln 121.
877
(1.0 10 −5 ) 2 = −1.1 × 104 J/mol; shifts right 1.0 10 −5
7.2 10 −4 (0.27) = 1.8 × 104 − 1.8 × 104 = 0; 0.27 at equilibrium 1.0 10 −3 (0.67 ) = 2 × 103 J/mol; shifts left 0.52
H2O(l) ⇌ H+(aq) + OH−(aq); ΔG = ΔG° + RT ln [H+][OH−] ΔG = 79.9 × 103 J/mol + (8.3145 J/K•mol)(298 K) ln [0.71(0.15)] ΔG = 7.99 × 104 J/mol − 5.5 × 103 J/mol = 7.44 × 104 J/mol = 74.4 kJ/mol
122.
ΔG° = 2(52 kJ) ‒ 98 kJ = 6 kJ −ΔG o −(6 103 J) = exp ΔG° = −RT ln K, K = exp = 0.09 RT (8.3145 J/K • mol)(298 K)
N2O4(g) → 2 NO2(g) K = 123.
2 PNO 2
PN2O4
=
From Section 17.9 of the text: ln K =
2 PNO 2
0.50
= 0.09, PNO2 = 0.2 atm
− ΔH o ΔSo + RT R
For two sets of K and T: ln K1 =
− ΔH o 1 ΔS − ΔH o 1 ΔSo + + ; ln K2 = R T1 R R T2 R
Subtracting the first expression from the second: ln K2 − ln K1 =
K ΔH o 1 ΔH o 1 1 1 or ln 2 = − − K1 R T1 R T1 T2 T2
Let K2/K1 = 6.67 (from the problem), so that T2 = 350.0 K and T1 = 300.0 K. ln 6.67 =
ΔH o 1 1 − 8.3145 J/K • mol 300.0 K 350.0 K
1.90 = (5.73 ×10− 5 mol/J)(ΔH°), ΔH° = 3.32 × 104 J/mol = 33.2 kJ/mol 124.
Because the partial pressure of C(g) decreased, the net change that occurs for this reaction to reach equilibrium is for products to convert to reactants. A(g) Initial Change Equil.
0.100 atm +x 0.100 + x
+
2 B(g)
⇌
C(g)
0.100 atm +2x 0.100 + 2x
0.100 atm −x 0.100 − x
878
CHAPTER 17
SPONTANEITY, ENTROPY, AND FREE ENERGY
From the problem, PC = 0.040 atm = 0.100 – x, x = 0.060 atm The equilibrium partial pressures are: P A = 0.100 + x = 0.100 + 0.060 = 0.160 atm, PB = 0.100 + 2(0.60) = 0.22 atm, and P C = 0.040 atm K=
0.040 = 5.2 0.160 (0.22 ) 2
G° = −RT ln K = −8.3145 J/K•mol(298 K) ln(5.2) = −4.1 × 103 J/mol = −4.1 kJ/mol 125.
a. False; there is a decrease in the number of moles of gaseous molecules, so ΔS° is negative (unfavorable). Because ΔG° is negative, the reaction must be exothermic (favorable). b. True c. False; heat is a product in this exothermic reaction. As heat is added (temperature increases), this reaction will shift left by Le Châtelier's principle. This results in a decrease in the value of K. At equilibrium at the higher temperature, reactant concentrations will increase and the product concentration will decrease. Therefore, the ratio of [PCl 5]/[PCl3] will decrease with an increase in temperature. d. False; when the signs of ΔH° and ΔS° are both negative, the reaction will only be spontaneous at temperatures below some value (at low temperatures) where the favorable ΔH° term dominates. e. True; from ΔG° = −RT ln K, when ΔG° < 0, K must be greater than 1.
126.
ΔG° = −RT ln Ka = −(8.3145 J/K•mol)(298 K) ln(4.5 × 10 −3 ) = 1.3 × 104 J/mol = 13 kJ/mol
127.
ΔS is more favorable (less negative) for reaction 2 than for reaction 1, resulting in K 2 > K1. In reaction 1, seven particles in solution are forming one particle in solution. In reaction 2, four particles are forming one, which results in a smaller decrease in positional probability than for reaction 1.
128.
A graph of ln K versus 1/T will yield a straight line with slope equal to −ΔH°/R and y intercept equal to ΔS°/R. Temp (°C)
T (K)
1000/T (K−1)
0 25 35 40. 50.
273 298 308 313 323
3.66 3.36 3.25 3.19 3.10
Kw
ln Kw
1.14 × 10 −15 1.00 × 10 −14 2.09 × 10 −14 2.92 × 10 −14 5.47 × 10 −14
−34.408 −32.236 −31.499 −31.165 −30.537
CHAPTER 17
SPONTANEITY, ENTROPY, AND FREE ENERGY
879
1 The straight-line equation (from a calculator) is ln K = −6.91 × 103 − 9.09. T
Slope = −6.91 × 103 K =
− ΔH o R
ΔH° = −(−6.91 × 103 K × 8.3145 J/K•mol) = 5.75 × 104 J/mol y intercept = −9.09 = 129.
ΔS o , ΔS° = −9.09 × 8.3145 J/K•mol = −75.6 J/K•mol R
ΔG° = −RT ln K; when K = 1.00, ΔG° = 0 since ln 1.00 = 0. ΔG° = 0 = ΔH° − TΔS° ΔH° = 3(−242 kJ) − [−826 kJ] = 100. kJ; ΔS° = [2(27 J/K) + 3(189 J/K)] − [90. J/K + 3(131 J/K)] = 138 J/K ΔH° = TΔS°, T =
130.
H
o
S
o
=
100. kJ = 725 K 0.138 kJ/K
Equilibrium occurs when the minimum in free energy has been reached. For this reaction, the minimum in free energy is when 1/3 of A has reacted. Because equilibrium lies closer to reactants for this reaction, the equilibrium constant will be less than 1 (K < 1) which dictates that ΔG° is greater than 0 (ΔG° > 0; ΔG° = −RTlnK). 2 A(g) Initial Change Equil.
⇌
B(g)
→
0 +x x
3.0 atm −2x 3.0 − 2x
Kp = PB PA2
From the plot, equilibrium occurs when 1/3 of the A(g) has reacted. 2x = amount A reacted to reach equilibrium = 1/3(3.0 atm) = 1.0 atm, x = 0.50 atm Kp =
x (3.0 − 2 x)
2
=
0.50 (2.0) 2
= 0.125 = 0.13 (to 2 sig. figs.)
880
CHAPTER 17
SPONTANEITY, ENTROPY, AND FREE ENERGY
Challenge Problems 131
a. Vessel 1: At 0C, this system is at equilibrium, so ΔSuniv = 0 and ΔS = ΔSsurr. Because the vessel is perfectly insulated, q = 0, so ΔSsurr = 0 = ΔSsys. b. Vessel 2: The presence of salt in water lowers the freezing point of water to a tem-perature below 0C. In vessel 2, the conversion of ice into water will be spontaneous at 0C, so ΔSuniv > 0. Because the vessel is perfectly insulated, ΔS surr = 0. Therefore, ΔSsys must be positive (ΔS > 0) in order for ΔSuniv to be positive.
132.
133.
The liquid water will evaporate at first and eventually equilibrium will be reached (physical equilibrium). •
Because evaporation is an endothermic process, ΔH is positive.
•
Because H2O(g) is more disordered (greater positional probability), ΔS is positive.
•
The water will become cooler (the higher energy water molecules leave), thus ΔT water will be negative.
•
The vessel is insulated (q = 0), so ΔSsurr = 0.
•
Because the process occurs, it is spontaneous, so ΔS univ is positive.
3 O2(g) ⇌ 2 O3(g); ΔH° = 2(143 kJ) = 286 kJ; ΔG° = 2(163 kJ) = 326 kJ ln K =
− 326 10 3 J − ΔG o = = −131.573, K = e −131.573 = 7.22 × 10 −58 RT (8.3145 J/K • mol) (298 K)
We need the value of K at 230. K. From Section 17.9 of the text: ln K =
− ΔG o ΔSo + RT R
For two sets of K and T: ln K1 =
− ΔH o 1 ΔS − ΔH o 1 ΔSo + + ; ln K2 = R T1 R T2 R R
Subtracting the first expression from the second: ln K2 − ln K1 =
K ΔH o 1 1 ΔH o 1 1 or ln 2 = − − R T1 T2 K1 R T1 T2
Let K2 = 7.22 × 10 −58 , T2 = 298 K; K1 = K230, T1 = 230. K; ΔH° = 286 × 103 J ln
7.22 10 −58 286 10 3 1 1 = − = 34.13 K 230 8.3145 230. 298
7.22 10 −58 = e34.13 = 6.6 × 1014, K230 = 1.1 × 10 −72 K 230
CHAPTER 17
SPONTANEITY, ENTROPY, AND FREE ENERGY
K230 = 1.1 × 10 −72 =
PO2 3
=
PO3 3
PO2 3 (1.0 10 −3 atm ) 3
881
, PO3 = 3.3 × 10 −41 atm
The volume occupied by one molecule of ozone is:
(1/6.022 10 23 mol)( 0.08206 L atm/K • mol)(230. K) nRT V= = = 9.5 × 1017 L P (3.3 10 − 41 atm) Equilibrium is probably not maintained under these conditions. When only two ozone molecules are in a volume of 9.5 × 1017 L, the reaction is not at equilibrium. Under these conditions, Q > K, and the reaction shifts left. But with only 2 ozone molecules in this huge volume, it is extremely unlikely that they will collide with each other. At these conditions, the concentration of ozone is not large enough to maintain equilibrium. 134.
Arrangement I and V:
S = k ln W; W = 1; S = k ln 1 = 0
Arrangement II and IV: W = 4; S = k ln 4 = 1.38 × 10 −23 J/K ln 4, S = 1.91 × 10 −23 J/K Arrangement III: 135.
W = 6; S = k ln 6 = 2.47 × 10 −23 J/K
a. From the plot, the activation energy of the reverse reaction is E a + (−ΔG°) = Ea − ΔG° (ΔG° is a negative number as drawn in the diagram). − Ea A exp − (E a − G ) k f RT − Ea = kf = A exp and kr = A exp , RT − (E a − G o ) RT k r A exp RT o
If the A factors are equal:
− E − ΔG o (E − ΔG o ) kf = exp a + a = exp kr RT RT RT
− ΔG o k ; because K and f are both equal to the From ΔG° = −RT ln K, K = exp kr RT same expression, K = kf/kr. b. A catalyst will lower the activation energy for both the forward and reverse reactions (but not change ΔG°). Therefore, a catalyst must increase the rate of both the forward and reverse reactions. 136.
At equilibrium: 1.10 1013 molecules 0.08206 L atm (298 K) 6.022 10 23 molecules/ mol K • mol nRT PH 2 = = V 1.00 L
PH 2 = 4.47 × 10 −10 atm
882
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The pressure of H2 decreased from 1.00 atm to 4.47 × 10 −10 atm. Essentially all of the H2 and Br2 has reacted. Therefore, PHBr = 2.00 atm because there is a 2 : 1 mole ratio between HBr and H2 in the balanced equation. Because we began with equal moles of H2 and Br2, we will have equal moles of H2 and Br2 at equilibrium. Therefore, PH2 = PBr2 = 4.47 × 10 −10 atm. 2 PHBr (2.00) 2 = 2.00 × 1019; assumptions good. = −10 2 PH 2 PBr2 (4.47 10 )
K=
ΔG° = −RT ln K = − (8.3145 J/K•mol)(298 K) ln(2.00 × 1019) = −1.10 × 105 J/mol ΔS° = 137.
− 103,800 J/mol − (−1.10 10 5 J/mol) ΔH o − ΔG o = 20 J/K•mol = T 298 K
a. ΔG° = G oB − G oA = 11,718 − 8996 = 2722 J
− ΔG o − 2722 J = exp K = exp = 0.333 (8.3145 J/K • mol) (298 K) RT b. When Q = 1.00 > K, the reaction shifts left. Let x = atm of B(g), which reacts to reach equilibrium. A(g) Initial Equil. K=
⇌
1.00 atm 1.00 + x
B(g) 1.00 atm 1.00 − x
PB 1.00 − x = = 0.333, 1.00−x = 0.333 + (0.333)x, x = 0.50 atm PA 1.00 + x
PB = 1.00 − 0.50 = 0.50 atm; PA = 1.00 + 0.50 = 1.50 atm c. ΔG = ΔG° + RT ln Q = ΔG° + RT ln(PB/PA) ΔG = 2722 J + (8.3145)(298) ln (0.50/1.50) = 2722 J − 2722 J = 0 (carrying extra sig. figs.) 138.
− ΔH o ΔSo + . For K at two temperatures T1 and T2, the RT R K ΔH o 1 1 − equation can be manipulated to give (see Exercise 115): ln 2 = K1 R T1 T2
From Exercise 95, ln K =
3.25 10 −2 1 ΔH o 1 = ln − 8 . 84 8.3145 J/K • mol 298 K 348 K −5.61 = (5.8 ×10-5 mol/J)(ΔH°), ΔH° = −9.7 × 104 J/mol For K = 8.84 at T = 25°C:
CHAPTER 17
SPONTANEITY, ENTROPY, AND FREE ENERGY
ln 8.84 =
883
− (−9.7 10 4 J/mol) ΔSo ΔSo = −37 + , (8.3145 J/K • mol)(298 K) 8.3145 J/K • mol 8.3145
ΔS° = −310 J/K•mol We get the same value for ΔS° using K = 3.25 × 10 −2 at T = 348 K data. ΔG° = −RT ln K; when K = 1.00, then ΔG° = 0 since ln 1.00 = 0. Assuming ΔH° and ΔS° do not depend on temperature: − 9.7 10 4 J/mol ΔH o ΔG° = 0 = ΔH° − TΔS°, ΔH° = TΔS°, T = = 310 K = − 310 J/K • mol ΔSo 139.
K = PCO 2 ; to ensure Ag2CO3 from decomposing, PCO 2 should be greater than K. From Exercise 95, ln K = ln
ΔH o ΔSo . For two conditions of K and T, the equation is: + RT R
K2 ΔH o 1 1 = + K1 R T1 T2
Let T1 = 25°C = 298 K, K1 = 6.23 × 10 −3 torr; T2 = 110.°C = 383 K, K2 = ? ln ln
K2 6.23 10 −3 torr
=
79.14 10 3 J/mol 1 1 − 8.3145 J/K • mol 298 K 383 K
K2 K2 = 7.1, = e7.1 = 1.2 × 103, K2 = 7.5 torr −3 6.23 10 6.23 10 −3
To prevent decomposition of Ag2CO3, the partial pressure of CO2 should be greater than 7.5 torr. 140.
L From the problem, χ CL 6H6 = χ CCl = 0.500. We need the pure vapor pressures (P o) in order to 4 calculate the vapor pressure of the solution. Use the thermodynamic data to determine the pure vapor pressure values.
C6H6(l) ⇌ C6H6(g) K = PC6H6 = PCo6H6 at 25°C
ΔGorxn = ΔGof , C6H6 (g) − ΔGof , C6H6 (l) = 129.66 kJ/mol − 124.50 kJ/mol = 5.16 kJ/mol ΔG° = −RT ln K, ln K =
− 5.16 10 3 J/mol − ΔG o = = −2.08 RT (8.3145 J/K • mol) (298 K)
K = PCo6H6 = e −2.08 = 0.125 atm For CCl4: ΔGorxn = ΔGof , CCl 4 (g) − ΔGof , CCl 4 (l) = −60.59 kJ/mol − (−65.21 kJ/mol) = 4.62 kJ/mol
− ΔG o − 4620 J/mol o K = PCCl = exp 4 RT = exp 8.3145 J/K • mol 298 K = 0.155 atm
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o = 0.500(0.155 atm) PC6H6 = χ CL6H6 PCo6H6 = 0.500(0.125 atm) = 0.0625 atm; PCCl 4
= 0.0775 atm
χ CV6H6 =
PC6 H 6 Ptot
=
0.0625 0.0625 atm = = 0.446 0.1400 0.0625 atm + 0.0775 atm
V = 1.000 − 0.446 = 0.554 χ CCl 4
141.
NaCl(s) ⇌ Na+(aq) + Cl−(aq)
K = Ksp = [Na+][Cl−]
ΔG° = [(−262 kJ) + (−131 kJ)] – (−384 kJ) = −9 kJ = −9000 J − (−9000 J) ΔG° = −RT ln Ksp, Ksp = exp = 38 = 40 8.3145 J/K • mol 298 K
⇌
NaCl(s) Initial Equil.
s = solubility (mol/L)
Na+(aq) + Cl−(aq) 0 s
Ksp = 40
0 s
Ksp = 40 = s(s), s = (40)1/2 = 6.3 = 6 M = [Cl−] 142.
ΔSsurr = −ΔH/T = −qP/T q = heat loss by hot water = moles × molar heat capacity × ΔT q = 1.00 × 103 g H2O × ΔSsurr =
143. Initial Equil.
1 mol H 2 O 75.4 J × (298.2 – 363.2) = −2.72 × 105 J 18.02 g K mol
− (−2.72 10 5 J ) = 912 J/K 298 .2 K
HX(aq)
⇌ H+(aq)
+ X−(aq)
0.10 M 0.10 – x
~0 x
0 x
From problem, x = [H+] = 10 −5.83 = 1.5 × 10 −6 ; Ka =
Ka =
[H + ][X − ] [HX]
(1.5 10 −6 ) 2 = 2.3 10 −11 −6 0.10 − 1.5 10
ΔG° = −RT ln K = −8.3145 J/K•mol(298 K) ln(2.3 × 10 −11 ) = 6.1 × 104 J/mol = 61 kJ/mol 144.
At the boiling point at 1 atm, ΔG = 0, so ΔH = TΔS. ΔH = TΔS = 351 K(110. J/K•mol) = 3.86 × 10 4 J/mol = 38.6 kJ/mol C2H5OH(l) → C2H5OH(g)
ΔH = 38.6 kJ; at constant pressure, ΔH = qp = 38.6 kJ
Because PV = nRT, at constant pressure and temperature: w = −PΔV = −RTΔn, where:
CHAPTER 17
SPONTANEITY, ENTROPY, AND FREE ENERGY
Δn = moles of gaseous products − moles of gaseous reactants = 1 − 0 = 1 mol w = −RTΔn = −8.3145 J/K•mol(351 K)(1 mol) = −2920 J = −2.92 kJ ΔE = q + w = 38.6 kJ + (−2.92 kJ) = 35.7 kJ 145.
ln K =
− (2.0 10 4 J) − ΔG o = −7.81, K = e−7.81 = 4.1 × 10−4 = RT 8.3145 J/K • mol(308 K)
⇌
2 NOCl(g) Initial Change Equil.
2 NO(g) +
Cl2(g)
2.0 atm 0 0 2x atm of NOCl reacts to reach equilibrium −2x → +2x +x 2.0 − 2x 2x x
K = 4.1 × 10−4 =
(2 x ) 2 ( x ) (2.0 − 2 x )
2
4x3 4 .0
, x = 7.4 × 10−2 atm
Assumption fails the 5% rule (2x is 7.4% of 2.0). Using successive approximations: 4.1 × 10−4 =
(2 x) 2 ( x) , x = 7.1 × 10−2 atm (This answers repeats itself.) [2.0 − 2(7.4 10 − 2 )]2
PNO = 2x = 2(7.1 × 10−2 atm) = 0.14 atm 146.
Use the thermodynamic data to calculate the boiling point of the solvent. At boiling point: ΔG = 0 = ΔH − TΔS, T =
33.90 10 3 J/mol ΔH = 353.3 K = ΔS 95.95 J/K • mol
ΔT = Kbm, (355.4 K −353.3 K) = 2.5 K kg/mol(m), m = Mass solvent = 150. mL ×
0.879 g 1 kg = 0.132 kg mL 1000 g
Mass solute = 0.132 kg solvent × 147.
2 .1 = 0.84 mol/kg 2 .5
0.84 mol solute 142 g = 15.7 g = 16 g solute kg solvent mol
G° = H° − TS° = −28.0 × 103 J – 298 K(−175 J/K) = 24,200 J G° = −RT ln K, ln K =
− 24,000 J − ΔG o = = −9.767 RT 8.3145 J/K • mol 298 K
K = e −9.767 = 5.73 × 10 −5 B Initial Equil.
+
0.125 M 0.125 − x
H2 O
⇌
BH+ 0 x
+
OH− ~0 x
K = Kb = 5.73 × 10 −5
885
886
CHAPTER 17 Kb = 5.73 × 10 −5 =
SPONTANEITY, ENTROPY, AND FREE ENERGY
[BH + ][OH − ] x2 x2 = , x = [OH−] = 2.68 × 10 −3 M 0.125 − x 0.125 [B]
pH = −log(2.68 × 10 −3 ) = 2.572; pOH = 14.000 – 2.572 = 11.428; assumptions good
Marathon Problem 148.
a. ΔS° will be negative because there is a decrease in the number of moles of gas. b. Because ΔS° is negative, ΔH° must be negative for the reaction to be spontaneous at some temperatures. Therefore, ΔSsurr is positive. c. Ni(s) + 4 CO(g) ⇌ Ni(CO)4(g) ΔH° = −607− [4(−110.5)] = −165 kJ; ΔS° = 417 − [4(198) + (30.)] = −405 J/K d. ΔG° = 0 = ΔH° − TΔS°, T =
ΔH o − 165 10 3 J = 407 K or 134°C = − 405 J / K ΔSo
o e. T = 50.°C + 273 = 323 K; ΔG 323 = −165 kJ − (323 K)( −0.405 kJ/K) = −34 kJ
ln K = f.
− (−34,000 J) − ΔG o = 12.66, K = e12.66 = 3.1 × 105 = RT 8.3145 J/K • mol(323 K)
T = 227°C + 273 = 500. K o ΔG 500 = −165 kJ − (500. K)( −0.405 kJ/K) = 38 kJ
ln K =
− 38,000 J = −9.14, K = e −9.14 = 1.1 × 10 −4 (8.3145 J/K • mol )(500. K)
g. The temperature change causes the value of the equilibrium constant to change from a large value favoring formation of Ni(CO)4 to a small value favoring the decomposition of Ni(CO)4 into pure Ni and CO. This is exactly what is wanted in order to purify a nickel sample. h. Ni(CO)4(l) ⇌ Ni(CO)4(g)
K = PNi( CO ) 4
At 42°C (the boiling point): ΔG° = 0 = ΔH° − TΔS° ΔS° =
29.0 10 3 J ΔH o = = 92.1 J/K T 315 K
o At 152°C: ΔG 152 = ΔH° − TΔS° = 29.0 × 103 J − 425 K(92.1 J/K) = −10,100 J
ΔG° = −RT ln K, ln K =
− (−10,100 J) = 2.858, Kp = e2.858 = 17.4 8.3145 J/K • mol(425 K)
A maximum pressure of 17.4 atm can be attained before Ni(CO)4(g) will liquefy.
CHAPTER 18 ELECTROCHEMISTRY Review Questions 1.
Redox reactions, which is shorthand for oxidation-reduction reactions, are reactions where one or more electrons are transferred. In a redox reaction, something that loses electrons is reacted with something that gains electrons. The substance reduced gains electrons and the compound/ion that contains the substance reduced is called the oxidizing agent. The substance oxidized loses electrons and the compound/ion that contains the substance oxidized is the reducing agent. Half-reactions: the two parts of an oxidation-reduction reaction, one representing oxidation, the other reduction. Overall charge must be balanced in any chemical reaction. We balance the charge in the halfreactions by adding electrons to either the reactant side (reduction half-reaction) or to the product side (oxidation half-reaction). In the overall balanced equation, the number of electrons lost by the oxidation half-reaction must exactly equal the number of electrons gained in the reduction half-reaction. Since electrons lost = electrons gained, then electrons will not appear in the overall balanced equation.
2.
Before we answer the question, here are four important terms relating to redox reactions and galvanic cells. a. Cathode: The electrode at which reduction occurs. b. Anode: The electrode at which oxidation occurs. c. Oxidation half-reaction: The half-reaction in which electrons are products. In a galvanic cell, the oxidation half-reaction always occurs at the anode. d. Reduction half-reaction: The half-reaction in which electrons are reactants. In a galvanic cell, the reduction half-reaction always occurs at the cathode. See Figures 18.2 and 18.3 for designs of galvanic cells. The electrode compartment in which reduction occurs is called the cathode and the electrode compartment in which oxidation occurs is called the anode. These compartments have electrodes (a solid surface) immersed in a solution. For a standard cell, the solution contains the reactant and product solutes and gases that are in the balanced half-reactions. The solute concentrations are all 1 M and gas partial pressures are all 1 atm for a standard cell. The electrodes are connected via a wire and a saltbridge connects the two solutions.
887
888
CHAPTER 18
ELECTROCHEMISTRY
The purpose of the electrodes is to provide a solid surface for electron transfer to occur in the two compartments. Electrons always flow from the anode (where they are produced) to the cathode (where they are reactants). The salt bridge allows counter ions to flow into the two cell compartments to maintain electrical neutrality. Without a salt bridge, no sustained electron flow can occur. In the salt bridge, anions flow into the anode to replenish the loss of negative charge as electrons are lost; cations flow into the cathode to balance the negative charge as electrons are transferred into the cathode. The “pull” or driving force on the electrons is called the cell potential (Ɛcell) or the electromotive force. The unit of electrical potential is the volt (V) which is defined as 1 joule of work per coulomb of charge transferred. It is the cell potential that can be used to do useful work. We harness the spontaneous redox reaction to produce a cell potential which can do useful work. 3.
The zero point for standard reduction potentials (Ɛ˚) is the standard hydrogen electrode. The half-reaction is: 2 H+ + 2 e− → H2. This half-reaction is assigned a standard potential of zero, and all other reduction half-reactions are measured relative to this zero point. Sub-stances less easily reduced than H+ have negative standard reduction potentials (Ɛ˚ < 0), while substances more easily reduced than H+ have positive standard reduction potentials (Ɛ˚ > 0). The species most easily reduced has the most positive Ɛ˚ value; this is F 2. The least easily reduced species is Li+ with the most negative Ɛ˚ value. When a reduction half-reaction is reversed to obtain an oxidation half-reaction, the sign of the reduction potential is reversed to give the potential for the oxidation half-reaction (Ɛ oox = − Ɛ ored ). The species oxidized are on the product side of the reduction half-reactions listed in Table 18.1. Li will have the most positive oxidation potential [Ɛ oox = − Ɛ ored = −(−3.05 V) = 3.05 V], so Li is the most easily oxidized of the species. The species most easily oxidized is the best reducing agent. The worst reducing agent is F − because it has the most negative oxidation potential (Ɛ oox = −2.87 V). For a spontaneous reaction at standard conditions, Ɛ ocell must be positive (Ɛ ocell = Ɛ ored + Ɛ oox > 0). For any two half-reactions, there is only one way to manipulate them to come up with a positive Ɛ ocell (a spontaneous reaction). Because the half-reactions do not depend on how many times the reaction occurs, half-reactions are an intensive property. This means that the value of Ɛ ored or Ɛ oox is not changed when the half-reactions are multiplied by integers to get the electrons to cross off. The line notation of the standard galvanic cell illustrated in Figure 18.5 of the text would be: Zn(s) | Zn2+(aq) || H2(g) | H+(aq) | Pt or Zn(s) | Zn2+(1.0 M) || H2(1.0 atm) | H+(1.0 M) | Pt The double line represents the salt-bridge separating the anode and cathode compartments. To the left of the double line are the pertinent anode compartment contents, and to the right are the pertinent cathode compartment contents. At each end, the electrodes are listed; to the inside, the solution contents are listed. A single line is used to separate the contents of each
CHAPTER 18
ELECTROCHEMISTRY
889
compartment whenever there is a phase change. Here in the cathode compartment, a single line is used to separate H2(g) from H+(aq) and to separate H+(aq) from Pt (the electrode). When concentrations and partial pressures are not listed, they are assumed to be standard (1.0 M for solutes and 1.0 atm for gases). For cells having nonstandard concentrations and pressures, we always include the actual concentrations and pressures in the line notation instead of the phases. 4.
ΔG˚ = −nF Ɛ˚; ΔG˚ is the standard free energy change for the overall balanced reaction, n is the number of electrons transferred in the overall balanced reaction, F is called the Faraday constant (1 F = 96,485 coulombs of charge transferred per mole of electrons), and Ɛ˚ is the standard cell potential for the reaction. For a spontaneous redox reaction, Ɛ ocell is positive while G orxn is negative. The negative sign is necessary to convert the positive Ɛ ocell value for a spontaneous reaction into a negative G orxn . The superscript ˚ indicates standard conditions. These are T = 25˚C, solute concentrations of 1.0 M, and gas partial pressures of 1.0 atm. Note that n is necessary to convert the intensive property Ɛ˚ into the extensive property ΔG˚.
5.
Ɛ = Ɛ˚ −
0.0591 RT log Q ln Q; at 25˚C, the Nernst equation is: Ɛ = Ɛ˚ − n nF
Nonstandard conditions are when solutes are not all 1.0 M and/or partial pressures of gases are not all 1.0 atm. Nonstandard conditions also occur when T 25˚C, For most problem solving, T = 25˚C is usually assumed, hence the second version of the Nernst equation is most often used. Ɛ = cell potential at the conditions of the cell; Ɛ˚ = standard cell potential; n = number of electrons transferred in the overall reaction, and Q is the reaction quotient determined at the concentrations and partial pressures of the cell contents. At equilibrium, Ɛ = 0 and Q = K. At 25˚C, Ɛ˚ = (0.0591/n)log K. The standard cell potential allows calculation of the equilibrium constant for a reaction. When K < 1, the log K term is negative, so Ɛ ocell is negative and ΔG˚ is positive. When K > 1, the log K term is positive, so Ɛ ocell is positive and ΔG˚ is negative. From the equation Ɛ˚ = (0.0591/n)log K, the value of Ɛ˚ allows calculation of the equilibrium constant K. We say that Ɛ˚ gives the equilibrium position for a reaction. Ɛ is the actual cell potential at the conditions of the cell reaction. The sign of Ɛ determines the spontaneity of the cell reaction. If Ɛ is positive, then the cell reaction is spontaneous as written (the forward reaction can be used to make a galvanic cell to produce a voltage). If Ɛ is negative, the forward reaction is not spontaneous at the conditions of cell, but the reverse reaction is spontaneous. The reverse reaction can be used to form a galvanic cell. Ɛ˚ can only be used to determine spontaneity when all reactants and products are at standard conditions (T = 25˚C, [ ] = 1.0 M, P = 1.0 atm). 6.
Concentration cell: A galvanic cell in which both compartments contain the same components, but at different concentrations. All concentration cells have Ɛ ocell = 0 because both compartments contain the same contents. The driving force for the cell is the different ion concentrations between the anode and cathode. The cell produces a voltage if the ion
890
CHAPTER 18
ELECTROCHEMISTRY
concentrations are different. Equilibrium for a concentration cell is reached (Ɛ = 0) when the ion concentrations in the two compartments are equal. The net reaction in a concentration cell is: Ma+(cathode, x M) → Ma+ (anode, y M) Ɛ ocell = 0 and the Nernst equation is: [M a + (anode)] 0.0591 0.0591 log Q = − log Ɛ = Ɛ˚ − n a [M a + (cathode)] where a is the number of electrons transferred. To register a potential (Ɛ > 0), the log Q term must be a negative value. This occurs when Ma+(cathode) > Ma+(anode). The higher ion concentration is always at the cathode and the lower ion concentration is always at the anode. The magnitude of the cell potential depends on the magnitude of the differences in ion concentrations between the anode and cathode. The larger the difference in ion concentrations, the more negative the log Q term and the more positive the cell potential. Thus, as the difference in ion concentrations between the anode and cathode compartments increase, the cell potential increases. This can be accomplished by decreasing the ion concentration at the anode and/or by increasing the ion concentration at the cathode. When NaCl is added to the anode compartment, Ag + reacts with Cl− to form AgCl(s). Adding Cl− lowers the Ag+ concentration which causes an increase in the cell potential. To determine Ksp for AgCl (Ksp = [Ag+][Cl−]), we must know the equilibrium Ag+ and Cl− concentrations. Here, [Cl−] is given and we use the Nernst equation to calculate the [Ag +] at the anode. 7.
As a battery discharges, Ɛcell decreases, eventually reaching zero. A charged battery is not at equilibrium. At equilibrium, Ɛcell = 0 and ΔG = 0. We get no work out of an equilibrium system. A battery is useful to us because it can do work as it approaches equilibrium. Both fuel cells and batteries are galvanic cells that produce cell potentials to do useful work. However, fuel cells, unlike batteries, have the reactants continuously supplied and can produce a current indefinitely.
The overall reaction in the hydrogen-oxygen fuel cell is 2 H2(g) + O2(g) → 2 H2O(l). The half-reactions are: 4 e− + O2 + 2 H2O → 4 OH−
cathode
2 H2 + 4 OH− → 4 H2O + 4 e−
anode
Utilizing the standard potentials in Table 18.1, Ɛ ocell = 0.40 V + 0.83 V = 1.23 V for the hydrogen-oxygen fuel cell. As with all fuel cells, the H2(g) and O2(g) reactants are continuously supplied. See Figure 18.16 of the text for a schematic of this fuel cell.
CHAPTER 18 8.
ELECTROCHEMISTRY
891
The corrosion of a metal can be viewed as the process of returning metals to their natural state. The natural state of metals is to have positive oxidation numbers. This corrosion is the oxidation of a pure metal (oxidation number = 0) into its ions. For corrosion of iron to take place, you must have: a. exposed iron surface – a reactant b. O2(g) – a reactant c. H2O(l) – a reactant, but also provides a medium for ion flow (it provides the salt bridge) d. ions – to complete the salt bridge Because water is a reactant and acts as a salt bridge for corrosion, cars do not rust in dry air climates, while corrosion is a big problem in humid climates. Salting roads in the winter also increases the severity of corrosion. The dissolution of the salt into ions on the surface of a metal increases the conductivity of the aqueous solution and accelerates the corrosion process. Some of the ways metals (iron) are protected from corrosion are listed below. a. Paint: Covers the metal surface so no contact occurs between the metal and air. This only works if the painted surface is not scratched. b. Durable oxide coatings: Covers the metal surface so no contact occurs between the metal and air. c. Galvanizing: Coating steel with zinc; Zn forms an effective oxide coating over steel; also, zinc is more easily oxidized than the iron in the steel. d. Sacrificial metal: Attaching a more easily oxidized metal to an iron surface; the more active metal is preferentially oxidized instead of iron. e. Alloying: Adding chromium and nickel to steel; the added Cr and Ni form oxide coatings on the steel surface. f.
9.
Cathodic protection: A more easily oxidized metal is placed in electrical contact with the metal we are trying to protect. It is oxidized in preference to the protected metal. The protected metal becomes the cathode electrode, thus, cathodic protection.
An electrolytic cell uses electrical energy to produce a chemical change. The process of electrolysis involves forcing a current through a cell to produce a chemical change for which the cell potential is negative. Electrical work is used to force a nonspontaneous reaction to occur. The units for current are amperes (A) which equal 1 coulomb of charge per second. current (A) × time (s) = coulombs of charge passed
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We use Faraday’s constant (F = 96,485 coulombs of charge per mole) to convert coulombs of charge passed into moles of electrons passed. The half-reaction gives the mole ratio between moles of electrons and moles of metal produced (or plated out). Plating means depositing the neutral metal on the electrode by reducing the metal ions in solution. In electrolysis, as with any redox reaction, the reaction that occurs first is the one most favored thermodynamically. The reduction reaction most favored thermodynamically has the largest, most positive Ɛ ored value. The oxidation reaction most likely to occur is the one with the largest, most positive Ɛ oox value. Note that for electrolytic cells, Ɛ ocell < 0, so the Ɛ ored and Ɛ oox values are commonly negative. The half-reactions that occur first as a current is applied are the ones with the least negative potentials (which are the most positive potentials). To predict the cathode half-reaction, write down the half-reaction and Ɛ ored value for all species present that can be reduced. The cathode reaction that occurs has the least negative (most positive) Ɛ ored value. The same thing is done for the anode; write down everything present that can be oxidized; the species oxidized has the least negative (most positive) Ɛ oox value. Note that we commonly assume standard conditions when predicting which half-reactions occur, so we can use the standard potentials in Table 18.1. When molten salts are electrolyzed, there is only one species present that can be oxidized (the anion in simple salts) and there is only one species that can be reduced (the cation in simple salts). When H2O is present as is the case when aqueous solutions are electrolyzed, we must consider the oxidation and reduction of water as potential reactions that can occur. When water is present, more reactions can take place, making predictions more difficult. When the voltage required to force a chemical reaction to occur is larger than expected, this is called overvoltage. The amount of overvoltage necessary to force a reaction to occur varies with the type of substance present. Because of this, Ɛ˚ values must be used cautiously when predicting the half-reactions that occur. 10.
Electrolysis is used to produce many pure metals and pure elements for commercial use. It also is used to purify metals as well as to plate out thin coatings on substances to provide protection from corrosion and to beautify objects. Another application of electrolysis is the charging of batteries. When aqueous NaCl is electrolyzed, water, with its less negative reduction potential is preferentially reduced over Na+ ions. Thus, the presence of water doesn’t allow Na+ ions to be reduced to Na. In molten NaCl, water is not present, so Na + can be reduced to Na. Purification by electrolysis is called electrorefining. See the text for a discussion of the electrorefining of copper. Electrorefining is possible because of the selectivity of electrode reactions. The anode is made up of the impure metal. A potential is applied so just the metal of interest, and all more easily oxidized metals are oxidized at the anode. The metal of interest is the only metal plated at the cathode due to the careful control of the potential applied. The metal ions that could plate out at the cathode in preference to the metal we are purifying will not be in solution, because these metals were not oxidized at the anode.
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Active Learning Questions 1.
Statement a is true. An oxidation-reduction reaction must combine a substance that loses electrons with a substance that gains electrons. You don’t have one without the other. Statement b is false as there are many reactions that do not involve a transfer of electrons; for example, acid-base reactions are proton transfer reactions, not electron transfer reactions. Not all reactions involving O2 result in the other species being reduced. When O 2(g) reacts with F2(g) to produce O2F2(s), the more electronegative fluorine in is reduced as it goes from the 0 oxidation state to the ‒1 oxidation state. This is accompanied by oxygen going from the 0 to the +1 oxidation states. Here F is reduced while O is oxidized. Statement c is false.
2.
Mass balance indicates that we have the same number and type of atoms on both sides of the equation (so that mass is conserved). Similarly, net charge must also be conserved. We cannot have a buildup of charge on one side of the reaction or the other. In redox equations, electrons are used to balance the net charge between reactants and products.
3.
See Figures 18.2 and 18.3 for designs of galvanic cells. The electrode compartment in which reduction occurs is called the cathode and the electrode compartment in which oxidation occurs is called the anode. These compartments have electrodes (a solid surface) immersed in a solution. For a standard cell, the solution contains the reactant and product solutes and gases that are in the balanced half-reactions. The solute concentrations are all 1 M and gas partial pressures are all 1 atm for a standard cell. The electrodes are connected via a wire and a saltbridge connects the two solutions. The purpose of the electrodes is to provide a solid surface for electron transfer to occur in the two compartments. Electrons always flow from the anode (where they are produced) to the cathode (where they are reactants). The salt bridge allows counter ions to flow into the two cell compartments to maintain electrical neutrality. Without a salt bridge, no sustained electron flow can occur. Therefore, the galvanic cell in Figure 18.1 doesn’t work, but the galvanic cell in Figure 18.2 does work. In the salt bridge, anions flow into the anode to replenish the loss of negative charge as electrons are lost; cations flow into the cathode to balance the negative charge as electrons are transferred into the cathode. The “pull” or driving force on the electrons is called the cell potential (Ɛcell) or the electromotive force. The unit of electrical potential is the volt (V) which is defined as 1 joule of work per coulomb of charge transferred. It is the cell potential that can be used to do useful work. We harness the spontaneous redox reaction to produce a cell potential which can do useful work.
4.
If the half- reaction for one of the compartments has a solid in it, then the electrode is generally made of that solid. All the other species in the half-reaction are put in solutions in that compartment. If none of the species in the half-reaction are solids, then an inert surface is used for the electrode. Graphite and platinum electrodes are a good choice for an inert electrode. The solution would contain all the species in the half-reaction when an inert electrode is used.
5.
The reduction half-reaction is Ni2+ + 2e− → Ni with E = −0.23 V. The two potential oxidation half-reactions are Cu → Cu2+ + 2e− with −E = −0.34 V and Zn → Zn2+ + 2e− with −E = 0.76 V. When the reduction of Ni2+ is coupled with the oxidation of Cu, the overall cell potential is negative. This tells us that Cu is not capable of reducing Ni2+ to Ni at standard concentrations. However, when the oxidation of Zn is coupled with the reduction of Ni 2+, the
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overall cell potential will be positive. This tells us that zinc can reduce Ni 2+ to Ni at standard concentrations. So, zinc should be used to plate out Ni(s). 6.
Cu → Cu2+ + 2e− −E = −0.34 V; Zn → Zn2+ + 2e− −E = 0.76 V 4 H+ + NO3− + 3 e− → NO + 2 H2O E° = 0.96 V; 2 H+ + 2 e− → H2 E° = 0.00 V The reduction potential of nitric acid is 0.96 V and the reduction potential of hydrochloric acid is 0.00 V (Cl− will not be reduced in solution, so we don’t have to worry about it). The oxidation of Cu has an oxidation potential of −0.34 V. When a copper penny is placed in a nitric acid solution, the overall standard cell potential is positive [0.76 + (−0.34) = 0.42 V]. Hence copper will react (and eventually dissolve) in nitric acid. The products of this reaction will be Cu 2+(aq), H2O(l), and NO(g). When a copper penny is placed in a hydrochloric acid solution, the overall standard cell potential is negative (−0.34 + 0.00 = −0.34 V). Thus, copper will not react in a hydrochloric acid solution. Zinc is more active metal than copper because it has a more positive oxidation potential (0.76 V vs. −0.34 V). From the potentials, zinc will react with either HNO 3 or HCl. Hence, newer pennies which contain zinc should react in either nitric acid or hydrochloric acid.
7.
(Fe2+ + 2e− → Fe) × 3 (Cr → Cr3+ + 3e−) × 2 3 Fe2+(aq) + 2 Cr(s) → 3 Fe(s) + 2 Cr3+(aq)
E = −0.44 V −E = 0.73 V E ocell = 0.29 V
The cathode compartment will contain a Fe electrode and 1.0 M Fe(NO3)2 in solution and the anode compartment with contain a Cr electrode with 1.0 M Cr(NO3)3 in solution. Electrons always flow from the anode to the cathode and the standard cell potential will be 0.29 V. The balanced cell reaction is shown above. 8.
F2 + 2 e− → 2 F− E° = 2.87 V; 2 H+ + 2 e− → H2 E° = 0.0 V; Na+ + e− → Na E° = −2.71 V The best reducing agents are the species most easily oxidized. The pertinent oxidation halfreactions are the reverse of the above reactions. Of the species oxidized (on the product side), Na has the most positive standard oxidation potential (2.71 V) so it is the best reducing agent. Note that F− is the worst reducing agent since it has the most negative standard oxidation potential (−2.87 V). The best oxidizing agents are the species most easily reduced. The species reduced are on the reactant side of the reduction half-reactions given above. The species which is the best oxidizing agent is F2 (E° = 2.87 V), followed by H+ (E° = 0.0 V), with Na+ as the worst oxidizing agent since it has the most negative standard reduction potential (−2.71 V). Only species that can be reduced can be classified as oxidizing agents. So, F−, H2, and Na are not included in this ranking because they are not reduced; they are oxidized by the reverse reaction so they are all potential reducing agents we used to answer the first question.
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895
In Table 18.1, F2 is the species most easily reduced since it has the most positive standard reduction potential; it is the best oxidizing agent. The best reducing agent in Table 18.1 is Li. It is the species with the most positive standard oxidation potential. 9.
A+ + e− → A EA° = ?; B+ + e− → B EB° = ?; If A is a better reducing agent than B, then A has a more positive oxidation potential than B (−EA° > −EB°). For this to be true, A+ must have a smaller standard reduction potential than B + (EA° < EB°). So, we can conclude that B+ must be a stronger oxidizing agent than A+ since B+ is more easily reduced.
10.
∆G = wmax = −nFE; E = E° −
0.0591 log Q n
The maximum amount of work that can be harnessed from a spontaneous reaction is equal to the magnitude of ∆G. Galvanic cells harness spontaneous redox reactions to produce a voltage which can do work. The equations ∆G = wmax = −nFE relates all these quantities to each other. 0.0591 log Q , allows one to calculate the cell potential for a The Nernst equation, E = E° − n galvanic cell at nonstandard concentrations. Q is the reaction quotient and is determined from the concentrations of the ions/partial pressures of gases in the balanced cell equation. When both the anode and cathode electrodes and solutions contain the same ions (called a concentration cell), the standard cell potential for the cell reaction must be zero (E° = 0). The cell can produce a voltage because of the Q term in the Nernst equation. If the ion concentrations between the anode and cathode are different, then Q ≠ 1 and the cell will produce a voltage. In general, the larger the difference in ion concentrations between the anode and the cathode, the larger the potential that is produced. 11.
Standard reduction potentials are intensive; they do not depend on how many times a reaction occurs. So, it doesn’t matter what are the coefficients in the balanced equation, the cell potential will be the same value. Free energy is extensive; it does depend on the coefficients in the balanced equation. The equation that relates free energy and cell potential is ∆G = −nFE. To convert the intensive quantity E to the extensive quantity ∆G, the number of electrons transferred in the balanced equation must be known. The value of n depends on the specific balanced equation.
12.
E is the cell potential for a reaction at a specific set of concentrations (assuming T = 25C). E° is the cell potential for a reaction at standard concentrations (1 M for solutes and 1 atm for gases). E = E only when all of the species present are at standard concentrations; when the concentrations are not all standard, then E ≠ E. For a reaction at equilibrium, E = 0. E = 0 when the anode and cathode compartments contain the same species. This is called a concentration cell; these types of cells only produce a voltage when the concentrations between the anode and cathode compartments are different.
13.
(Ag+ + e− → Ag) 2 Zn → Zn2+ + 2e− 2 Ag+(aq) + Zn(s) → 2 Ag(s) + Zn2+(aq)
E = 0.80 V −E = 0.76 V E ocell = 1.56 V
A galvanic cell uses a spontaneous reaction to produce a voltage. Any change that increases the tendency of the forward reaction to occur will increase the cell potential, and vice versa.
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When the product ion Zn2+ is increased, this decreases the tendency of the forward reaction to occur (increases the tendency of the reverse reaction to occur). Therefore, the cell potential will decrease. When the reactant ion Ag + increases, this increases the tendency of the forward reaction to occur which will increase the cell potential. In both cases, E ocell is a constant; it is the cell potential when all reactant and product ions and gases are at standard concentrations. 14.
Fe3+ + e− → Fe2+ E° = 0.77 V; Fe2+ + 2 e− → Fe E° = −0.44 V; Fe3+ + 3 e− → Fe E° = −0.036 V; standard reduction potentials are determined relative to a substance being oxidized. Two standard reduction potentials are only additive if one of the half-reactions is reversed. That is, a cell potential is produced when a species that is oxidized is combined with a species that is reduced, not when combining two species that are reduced. The quantity for reduction halfreactions that is additive is the free energy change for each reduction half-reaction. So, to determine the potential for third half-reaction, utilize the equation ∆G = −nFE, which can be applied to half-reactions. Fe3+ + e− → Fe2+ ∆G = −nFE = −(1 mol e−)(96,485 C/mol e−)(0.77 J/C) = −74,300 J Fe2+ + 2e− → Fe ∆G = −nFE = −(2 mol e−)(96,485 C/mol e−)(−0.44 J/C) = 84,900 J Fe3+ + 3e− → Fe
∆G = −74,300 + 84,900 = 10,600 J
(We carried extra significant figures in the calculation.) E° =
−ΔG o −(10,600 J) = −0.0366 J/C = −0.037 V = − nF (3 mol e )(96,485 C/mol e− )
Table 17.1 gives −0.036 V for the reduction potential. Round-off error explains the difference. 15.
Standard reduction potentials are relative to a zero point. The zero point for reduction potentials is the reduction of H+ to H2. Any half-reaction could be used for the zero point. By convention, 2 H+ + 2e− → H2 is assigned E = 0 V. This is not an actual potential, but an arbitrarily assigned value to help in determining relative values for the other half-reactions in Table 17.1. To use standard reduction potentials, one must couple an oxidation half-reaction with a reduction half-reaction. Only when this occurs does the zero point reference cross out and an actual standard potential can be determined.
16.
False; standard reduction potentials are all determined at standard concentrations. If the concentration is not 1 M for a solute of 1 atm for a gas, then the potential is not a standard potential. Concentration cells produce a voltage because the solute concentrations in solution are different from each other. For all concentration cells, E ocell = 0 V.
17.
Most metals are easily oxidized to higher oxidation states. This is seen in Table 18.1 as most metals have standard reduction potential less than that of oxygen gas. When any of these halfreactions are reversed (to represent the oxidation of the metal) and combined with the reduction half-reaction for oxygen, the result is a positive E° value, hence a spontaneous process at standard conditions. Aluminum forms a strong oxide coating covering the aluminum surface. If the tough oxide coating stays intact, aluminum is protected from further corrosion.
18.
In aqueous solutions, water is present which can be oxidized and reduced. In molten salts, no water is present so oxidation and reduction of H 2O can be ignored. Molten MnCl2 must be used
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if Mn(s) is to form. In aqueous MnCl2, H2O is more easily oxidized relative to Mn 2+ (Eox = −0.83 V for H2O versus Eox = −1.18 V for Mn2+). In aqueous MnCl2, Mn(s) would not form. 19.
a.
(Ni2+ + 2 e− → Ni) × 3 (Al → Al3+ + 3 e−) × 2
E° = −0.23 V −E° = 1.66 V
3 Ni2+(aq) + Al(s) → 2 Al3+(aq) + Ni(s)
E ocell = 1.43 V
A typical galvanic cell diagram is: eSalt Bridge
Anode (oxidation) ~~~~~~~
e-
cations anions ~~~~~~
~~~~~~~
Cathode (reduction) ~~~~~~
The cathode compartment has an Ni electrode and 1.0 M Ni2+ (with an appropriate counter ion like NO3‒) and the anode compartment has an Al electrode and 1.0 M Al3+ (with an appropriate counter ion). Electrons always flow from the anode to the cathode. Cations in the salt bridge always flow into the cathode compartment, and anions always flow into the anode compartment. This is required to keep each compartment electrically neutral. b. Any change in ion concentrations that increases the tendency of the forward reaction to occur will increase the cell potential, whereas any change in ion concentration that decreases the tendency of the forward reaction will decrease the cell potential. Adding Al3+ (a product in the cell reaction) shifts the reaction left which decreases the tendency of the forward reaction to occur. Hence, an increase in the Al 3+ concentration will decrease the cell potential. If the concentration of Ni2+ is lowered, this decreases the tendency of the forward reaction to occur (reaction shifts left), and the cell potential will decrease. The Ni electrode is not a reactant in the cell reaction, so solid Ni is not needed to for the cell reaction to occur. Replacing the Ni electrode with an inert platinum electrode will have no effect on the cell reaction, and thus no effect on the cell potential. The electrode is just needed in the cathode compartment for electron transfer to occur to the Ni 2+ ions. c.
Ecell = Eocell −
0.0591 [Al3+ ] 2 0.0591 [Al3+ ] 2 log log , 1.82 V = 1.43 V − 6 n (1.0) 3 [Ni 2+ ]3
log [Al3+]2 = −39.59, [Al3+]2 = 10−39.59, [Al3+] = 1.6 × 10−20 M Al(OH)3(s) ⇌ Al3+(aq) + 3 OH−(aq) Ksp = [Al3+][OH−]3; from the problem, [OH−] = 1.0 × 10−4 M. Ksp = (1.6 × 10−20)(1.0 ×10−4)3 = 1.6 × 10−32
898 20.
CHAPTER 18
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a. Fuel cells are galvanic cells for which reactants are continuously supplied. ΔG° = −RT lnK; since K > 1, ΔG° must be negative indicating a spontaneous reaction at standard concentrations. ΔG° = ΔH° − TΔS°; the moles of gas decrease in the reaction indicating a negative ΔS° value. ΔH° must be negative (an exothermic reaction) for ΔG° to be a negative value at standard conditions. b. 2 H2(g) + O2(g) → 2 H2O(l); oxygen goes from the zero oxidation state to the −2 oxidation state in H2O. Because 2 mol of O are in the balanced reaction, n = 4 moles of electrons transferred.
E ocell =
0.0591 0.0591 log K = log(1.28 × 1083), E ocell = 1.23 V n 4
c. O2 + 2 H2O + 4 e− → 4 OH− (H2 + 2 OH− → 2 H2O + 2 e−) × 2
E° = ? −E° = 0.83 V
2 H2(g) + O2(g) → 2 H2O(l)
E ocell = 1.23 V = E° + 0.83
Solving: E° = 1.23 – 0.83 = 0.40 V d. 220 min ×
1 mol O2 60 s 10.0 C 1 mol e− = 0.34 mol O2 × × × min s 96,485 C 4 mol e−
Review of Oxidation-Reduction Reactions 21.
Oxidation: increase in oxidation number; loss of electrons Reduction: decrease in oxidation number; gain of electrons
22.
See Table 4.2 in Chapter 4 of the text for rules for assigning oxidation numbers. a. H (+1), O (−2), N (+5)
b. Cl (−1), Cu (+2)
c. O (0)
d. H (+1), O (−1)
e. H(+1), O (−2), C (0)
f.
g. Pb (+2), O (−2), S (+6)
h. O (−2), Pb (+4)
i.
j.
Na (+1), O (−2), C (+3)
Ag (0)
O (−2), C (+4)
k. (NH4)2Ce(SO4)3 contains NH4+ ions and SO42− ions. Thus cerium exists as the Ce4+ ion. H (+1), N (−3), Ce (+4), S (+6), O (−2) l. 23.
O (−2), Cr (+3)
The species oxidized shows an increase in oxidation numbers and is called the reducing agent. The species reduced shows a decrease in oxidation numbers and is called the oxidizing agent. The pertinent oxidation numbers are listed by the substance oxidized and the substance reduced.
CHAPTER 18 Redox?
ELECTROCHEMISTRY
899 Substance Oxidized
Substance Reduced
CH4
CH4 (C, −4 → +2)
H2O (H, +1 → 0)
AgNO3
Cu
Cu (0 → +2)
AgNO3 (Ag, +1 → 0)
HCl
Zn
Zn (0 → +2)
HCl (H, +1 → 0)
Ox. Agent
Red. Agent
a. Yes
H2O
b. Yes c. Yes
d. No; there is no change in any of the oxidation numbers. 24.
a. 4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g) −3 +1
+2 −2
0
+1 −2
oxidation numbers
2 NO(g) + O2(g) → 2 NO2(g) +2 −2
0
+4 −2
3 NO2(g) + H2O(l) → 2 HNO3(aq) + NO(g) +4 −2
+1 −2
+1 +5 −2
+2 −2
All three reactions are oxidation-reduction reactions since there is a change in oxidation numbers of some of the elements in each reaction. b. 4 NH3 + 5 O2 → 4 NO + 6 H2O; O2 is the oxidizing agent and NH3 is the reducing agent. 2 NO + O2 → 2 NO2; O2 is the oxidizing agent and NO is the reducing agent. 3 NO2 + H2O → 2 HNO3 + NO; NO2 is both the oxidizing and reducing agent. 25.
Review Chapter 4.10 of the text for rules on balancing oxidation-reduction reactions. a. Cr → Cr3+ + 3 e−
NO3− → NO 4 H+ + NO3− → NO + 2 H2O − 3 e + 4 H+ + NO3− → NO + 2 H2O
Cr → Cr3+ + 3 e− 3 e− + 4 H+ + NO3− → NO + 2 H2O 4 H+(aq) + NO3−(aq) + Cr(s) → Cr3+(aq) + NO(g) + 2 H2O(l) b. (Ce4+ + e− → Ce3+) 6
CH3OH → CO2 H2O + CH3OH → CO2 + 6 H+ H2O + CH3OH → CO2 + 6 H+ + 6 e− 6 Ce4+ + 6 e− → 6 Ce3+ H2O + CH3OH → CO2 + 6 H+ + 6 e−
H2O(l) + CH3OH(aq) + 6 Ce4+(aq) → 6 Ce3+(aq) + CO2(g) + 6 H+(aq)
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CHAPTER 18 c.
ELECTROCHEMISTRY
SO32− → SO42− (H2O + SO32− → SO42− + 2 H+ + 2 e−) 5 MnO4− → Mn2+ (5 e− + 8 H+ + MnO4− → Mn2+ + 4 H2O) 2 5 H2O + 5 SO32− → 5 SO42− + 10 H+ + 10 e− 10 e + 16 H+ + 2 MnO4− → 2 Mn2+ + 8 H2O 6 H+(aq) + 2 MnO4−(aq) + 5 SO32−(aq) → 5 SO42−(aq) + 2 Mn2+(aq) + 3 H2O(l) −
26.
a.
PO33− → PO43− (H2O + PO33− → PO43− + 2 H+ + 2 e−) 3 MnO4− → MnO2 (3 e− + 4 H+ + MnO4− → MnO2 + 2 H2O) 2 3 H2O + 3 PO33− → 3 PO43− + 6 H+ + 6 e− 6 e− + 8 H+ + 2 MnO4− → 2 MnO2 + 4 H2O 2 H+ + 2 MnO4− + 3 PO33− → 3 PO43− + 2 MnO2 + H2O Now convert to basic solution by adding 2 OH − to both sides. 2 H+ + 2 OH− → 2 H2O on the reactant side. After converting H+ to OH−, then simplify the overall equation by crossing off 1 H2O on each side of the reaction. The overall balanced equation is: H2O(l) + 2 MnO4−(aq) + 3 PO33−(aq) → 3 PO43−(aq) + 2 MnO2(s) + 2 OH−(aq)
b.
Mg → Mg(OH)2 2 H2O + Mg → Mg(OH)2 + 2 H+ + 2 e−
OCl− → Cl− 2 e− + 2 H+ + OCl− → Cl− + H2O
2 H2O + Mg → Mg(OH)2 + 2 H+ + 2 e− 2 e + 2 H+ + OCl− → Cl− + H2O −
OCl−(aq) + H2O(l) + Mg(s) → Mg(OH)2(s) + Cl−(aq) The final overall reaction does not contain H+ so we are done. c.
H2CO → HCO3− 2 H2O + H2CO → HCO3 − + 5 H+ + 4 e−
Ag(NH3)2+ → Ag + 2 NH3 (e + Ag(NH3)2+ → Ag + 2 NH3) 4 −
2 H2O + H2CO → HCO3− + 5 H+ + 4 e− 4 e + 4 Ag(NH3)2+ → 4 Ag + 8 NH3 4 Ag(NH3)2+ + 2 H2O + H2CO → HCO3− + 5 H+ + 4 Ag + 8 NH3 −
Convert to basic solution by adding 5 OH − to both sides (5 H+ + 5 OH− → 5 H2O). Then cross off 2 H2O on both sides which gives the overall balanced equation: 5 OH− (aq) + 4 Ag(NH3)2+(aq) + H2CO(aq) → HCO3− (aq) + 3 H2O(l) + 4 Ag(s) + 8 NH3(aq)
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901
Questions 27.
Electrochemistry is the study of the interchange of chemical and electrical energy. A redox (oxidation-reduction) reaction is a reaction in which one or more electrons are transferred. In a galvanic cell, a spontaneous redox reaction occurs that produces an electric current. In an electrolytic cell, electricity is used to force a nonspontaneous redox reaction to occur.
28.
Oxidation occurs at the anode where electrons (negative charges) are lost. Reduction occurs at the cathode where electrons (negative charges) are gained. To balance the loss of negative charges at the anode, anions flow to the anode. To balance the gain of negative charges at the cathode, cations flow to the cathode.
29.
Magnesium is an alkaline earth metal; Mg will oxidize to Mg2+. The oxidation state of hydrogen in HCl is +1. To be reduced, the oxidation state of H must decrease. The obvious choice for the hydrogen product is H2(g), where hydrogen has a zero oxidation state. The balanced reaction is Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g). Mg goes from the 0 to the +2 oxidation state by losing two electrons. Each H atom goes from the +1 to the 0 oxidation state by gaining one electron. Since there are two H atoms in the balanced equation, then a total of two electrons are gained by the H atoms. Hence two electrons are transferred in the balanced reaction. When the electrons are transferred directly from Mg to H +, no work is obtained. To harness this reaction to do useful work, we must control the flow of electrons through a wire. This is accomplished by making a galvanic cell that separates the reduction reaction from the oxidation reaction to control the flow of electrons through a wire to produce a voltage.
30.
Galvanic cells use spontaneous redox reactions to produce a voltage. For a spontaneous redox reaction, the key is to have an overall positive E ocell value when manipulating the half-reactions. For any two half-reactions, the half-reaction with the most positive reduction potential will always be the cathode reaction. The remaining half-reaction (the one with the most negative E ored ) will be reversed and become the anode half-reaction. This combination will always yield an overall reaction having a positive standard cell potential that can be used to run a galvanic cell.
31.
Statements c and d are false. Reducing agents cause reduction to occur by something in the formula being oxidized. Oxidizing agents cause oxidation to occur by something in the formula being reduced. A good oxidizing agent has a large, positive reduction potential. Oxidation occurs at the anode.
32.
AgCl2 is composed of Ag2+ and Cl‒. From the potentials in Table 18.1, Ag2+ is reduced to Ag+ having a reduction potential of 1.99 V. Cl‒ is oxidized to Cl2 having an oxidation potential of ‒1.36 V. The overall potential is positive when the reduction of Ag 2+ is combined with the oxidation of Cl‒. So, Ag2+ will oxidize Cl− making AgCl2 unstable.
33.
An extensive property is one that depends directly on how many times the reaction occurs. The free energy change for a reaction depends on whether 1 mole of product is produced, or 2 moles of product is produced or 1 million moles of product is produced. This is not the case for cell potentials, which do not depend on how many times a reaction occurs. The equation that relates G to E is G = −nFE. It is the n term that converts the intensive property E into the extensive property G. n is the number of moles of electrons transferred in the balanced reaction that G is associated with.
902 34.
CHAPTER 18 E = E ocell −
ELECTROCHEMISTRY
0.0591 log Q; a concentration cell has the same anode and cathode contents; thus n
E ocell = 0 for a con-centration cell. No matter which half-reaction you choose, the opposite halfreaction is occurring in the other cell. The driving force to produce a voltage is the −log Q term in the Nernst equation. Q is determined by the concentration of ions in the anode and cathode compartments. The larger the difference in concentrations, the larger is the −log Q term, and the larger is the voltage produced. Therefore, the driving force for concentration cells is the difference in ion concentrations between the cathode and anode compartments. When the ion concentrations are equal, Q = 1 and log Q = 0, and no voltage is produced. 35.
A potential hazard when jump starting a car is the possibility for the electrolysis of H2O(l) to occur. When H2O(l) is electrolyzed, the products are the explosive gas mixture of H 2(g) and O2(g). A spark produced during jump-starting a car could ignite any H2(g) and O2(g) produced. Grounding the jumper cable far from the battery minimizes the risk of a spark nearby the battery, where H2(g) and O2(g) could be accumulating.
36.
Metals corrode because they oxidize easily. Referencing Table 18.1, most metals are associated with negative standard reduction potentials. This means that the reverse reactions, the oxidation half-reactions, have positive oxidation potentials indicating that they oxidize easily. Another key point is that the reduction of O2 (which is a reactant in corrosion processes) has a more positive E ored than most of the metals (for O2, E ored = 0.40 V). This means that when O2 is coupled with most metals, the reaction will be spontaneous since E ocell > 0, so corrosion occurs. The noble metals (Ag, Au, and Pt) all have standard reduction potentials greater than that of O2. Therefore, O2 is not capable of oxidizing these metals at standard conditions. Note: The standard reduction potential for Pt → Pt2+ + 2 e− is not in Table 18.1. As expected, its reduction potential is greater than that of O 2 ( E oPt = 1.19 V).
37.
You need to know the identity of the metal so that you know which molar mass to use. You need to know the oxidation state of the metal ion in the salt so that the moles of electrons transferred can be determined. And finally, you need to know the amount of current and the time the current was passed through the electrolytic cell. If you know these four quantities, then the mass of metal plated out can be calculated.
38.
For a galvanic cell, we want a combination of the half-reactions that give a positive overall cell potential. For these two cell compartments, the combination that gives a positive overall cell potential is: (Ag+ + e− → Ag) 2 Cu → Cu2+ + 2e−
E = 0.80 V −E = −0.34 V
2 Ag+(aq) + Cu(s) → 2 Ag(s) + Cu2+(aq)
E ocell = 0.46 V
a. From above, silver metal is a product when these two compartments make a galvanic cell ( E ocell > 0). The silver compartment is the cathode (Ag + is reduced) and the copper compartment is the anode (Cu is oxidized). As is always the case, electrons flow from the anode (copper compartment) to the cathode (silver compartment).
CHAPTER 18
ELECTROCHEMISTRY
903
b. The reverse of the spontaneous reaction produces copper metal [2 Ag(s) + Cu2+(aq) → 2 Ag+(aq) + Cu(s)]. This reverse reaction is nonspontaneous ( E ocell = −0.46 V), so this will be an electrolytic cell. For this nonspontaneous reaction, the copper compartment is the cathode and the silver compartment is the anode. Electron flow will be from the silver compartment to the copper compartment (from the anode to the cathode). c. From the initial work above, E ocell = 0.46 V when these compartments form a galvanic cell. d. From part b, 2 Ag(s) + Cu2+(aq) → 2 Ag+(aq) + Cu(s) E ocell = −0.46 V; this is the electrolytic cell reaction which tells us that a potential greater than 0.46 V must be applied to force this nonspontaneous reaction to occur. 39.
All the statements are true. Corrosion is the spontaneous oxidation of some metal (iron) coupled with the reduction of oxygen. For statement a, cars rust more easily in high-moisture areas (the humid areas) because water is a reactant in the reduction half-reaction as well as providing a medium for ion migration (a salt bridge of sorts). For statement d, paint and protective oxides are great at limiting corrosion only if the paint or protective oxide completely covers the surface. As soon as a metal surface is exposed, corrosion will occur. For statement e, salting roads adds ions to the corrosion process, which increases the conductivity of the aqueous solution and, in turn, accelerates corrosion.
40.
The oxidation potential for Mg is 2.37 V and the oxidation potential for Fe is 0.44 V. Mg is more easily oxidized than iron. Hence, Mg is preferentially oxidized, thus protecting the iron pipe from corrosion. Answer a is correct
41.
Na+ + e‒ → Na E° = ‒ 2.71 V; 2 H2O + 2 e‒ → H2 + 2 OH‒ E° = ‒0.83 V; in molten NaCl, no water is present, just Na+ and Cl‒ ions. The reaction that occurs at the cathode is the reduction of Na+ as the applied external potential is increased. In aqueous solution, H 2O is now present, and from the potentials, water is more easily reduced than Na+. In aqueous NaCl, the reduction of water occurs at the cathode producing H2(g). Answer c is correct.
42.
Mercury is hazardous when it can be oxidized to the Hg 22+ or Hg2+ ions. However, when mercury is in the solid elemental form having a zero oxidation state, it is relatively inert; it passes through the digestive system and is excreted before it can pose any risk. The reason Hg is relatively inert is because it is not easily oxidized. Only two oxidation potentials are listed in Table 18.1 that involve elemental mercury: 2 Hg → Hg22+ + 2e−
−E = −0.80 V
2 Hg + 2 Cl− → Hg2Cl2 + 2e−
−E = −0.27 V
Note that both these oxidation potentials are negative, indicating that these oxidations are relatively difficult to do. No oxidizing agents in the mouth or in the stomach have a reduction potential large enough to spontaneously oxidize mercury into its reactive and hazardous ions. So, mercury in the solid elemental form is relatively inert because it is difficult to oxidize. Note: Mercury in the vapor form is always hazardous, and specifically, inhalation of mercury vapor into the lungs is the most dangerous route of entry. Death often results from respiratory or kidney failure due to inhalation of mercury vapor.
904
CHAPTER 18
ELECTROCHEMISTRY
Exercises Galvanic Cells, Cell Potentials, Standard Reduction Potentials, and Free Energy 43.
The reducing agent causes reduction to occur; it does this by containing the species which is oxidized. Oxidation occurs at the anode, so the reducing agent will be in the anode compartment. The oxidizing agent causes oxidation to occur; it does this by containing the species which is reduced. Reduction occurs at the cathode, so the oxidizing agent will be in the cathode compartment. Electron flow is always from the anode compartment to the cathode compartment.
44.
A galvanic cell at standard conditions must have a positive overall standard cell potential ( E ocell > 0) The only combination of the half-reactions that gives a positive cell potential is: Cu2+ + 2e− → Cu Zn → Zn2+ + 2e−
E°(cathode) = 0.34 V −E°(anode) = 0.76 V
Cu2+(aq) + Zn(s) → Cu(s) + Zn2+(aq)
E ocell = 1.10 V
a. The reducing agent causes reduction to occur since it always contains the species which is oxidized. Zn is oxidized in the galvanic cell, so Zn is the reducing agent. The oxidizing agent causes oxidation to occur since it always contains the species which is reduced. Cu 2+ is reduced in the galvanic cell, so Cu2+ is the oxidizing agent. Electrons will flow from the zinc compartment (the anode) to the copper compartment (the cathode). b. From the previous work, E ocell = 1.10 V. c. The pure metal that is a product in the spontaneous reaction is copper. So, the copper electrode will increase in mass as Cu 2+(aq) is reduced to Cu(s). The zinc electrode will decrease in mass for this galvanic cell as Zn(s) is oxidized to Zn 2+(aq). 45.
A typical galvanic cell diagram is: e-
Anode (oxidation) ~~~~~~~
eSalt Bridge cations anions
~~~~~~
~~~~~~~
Cathode (reduction) ~~~~~~
The diagram for all cells will look like this. The contents of each half-cell compartment will be identified for each reaction, with all solute concentrations at 1.0 M and all gases at 1.0 atm. For Exercises 45 and 46, the flow of ions through the salt bridge was not asked for in the questions. If asked, however, cations always flow into the cathode compartment, and anions always flow into the anode compartment. This is required to keep each compartment electrically neutral.
CHAPTER 18
ELECTROCHEMISTRY
905
a. Table 18.1 of the text lists balanced reduction half-reactions for many substances. For this overall reaction, we need the Cl2 to Cl− reduction half-reaction and the Cr3+ to Cr2O72− oxidation half-reaction. Manipulating these two half-reactions gives the overall balanced equation. (Cl2 + 2 e− → 2 Cl−) × 3 7 H2O + 2 Cr3+ → Cr2O72− + 14 H+ + 6 e− 7 H2O(l) + 2 Cr3+(aq) + 3 Cl2(g) → Cr2O72− (aq) + 6 Cl−(aq) + 14 H+(aq) The contents of each compartment are: Cathode: Pt electrode; Cl2 bubbled into solution, Cl− in solution Anode: Pt electrode; Cr3+, H+, and Cr2O72− in solution We need a nonreactive metal to use as the electrode in each case, since all the reactants and products are in solution. Pt is a common choice. Another possibility is graphite. b.
Cu2+ + 2 e− → Cu Mg → Mg2+ + 2e− Cu2+(aq) + Mg(s) → Cu(s) + Mg2+(aq) Cathode: Cu electrode; Cu2+ in solution; anode: Mg electrode; Mg2+ in solution
46.
Reference the answer to Exercise 45 for a typical galvanic cell diagram. The contents of each half-cell compartment are identified below with all solute concentrations at 1.0 M and all gases at 1.0 atm. a. Reference Table 18.1 for the balanced half-reactions. 5 e− + 6 H+ + IO3− → 1/2 I2 + 3 H2O (Fe2+ → Fe3+ + e−) × 5 6 H+ + IO3− + 5 Fe2+ → 5 Fe3+ + 1/2 I2 + 3 H2O or 12 H+(aq) + 2 IO3−(aq) + 10 Fe2+(aq) → 10 Fe3+(aq) + I2(aq) + 6 H2O(l) Cathode: Pt electrode; IO3−, I2 and H2SO4 (H+ source) in solution. Note: I2(s) would make a poor electrode since it sublimes. Anode: Pt electrode; Fe2+ and Fe3+ in solution b.
(Ag+ + e− → Ag) × 2 Zn → Zn2+ + 2 e− Zn(s) + 2 Ag+(aq) → 2 Ag(s) + Zn2+(aq) Cathode: Ag electrode; Ag+ in solution; anode: Zn electrode; Zn2+ in solution
906 47.
CHAPTER 18
ELECTROCHEMISTRY
To determine E° for the overall cell reaction, we must add the standard reduction potential to the standard oxidation potential ( E ocell = E ored + E oox ) . Reference Table 18.1 for values of standard reduction potentials. Remember that E oox = − E ored and that standard potentials are not multiplied by the integer used to obtain the overall balanced equation. 45a.
E ocell = E oCl → Cl − + E oCr 3+ → Cr O 2− = 1.36 V + (−1.33 V) = 0.03 V 2
48.
2
7
45b.
E ocell = E oCu 2+ → Cu + E oMg → Mg 2+ = 0.34 V + 2.37 V = 2.71 V
46a.
E ocell = E oIO − → I + E oFe 2+ → Fe3+ = 1.20 V + (−0.77 V) = 0.43 V 3
46b.
2
E ocell = E oAg+ → Ag + E oZn → Zn 2+ = 0.80 V + 0.76 V = 1.56 V Br2 + 2 e− → 2 Br− Pb → Pb2+ + 2 e−
49.
E° = 1.09 V −E° = 0.13 V
Br2(aq) + Pb(s) → Pb2+(aq) + 2 Br−(aq)
E ocell = 1.22 V
Statements c, d, and e are true. Pb2+ is a product in the spontaneous cell reaction, so its concentration will increase as the reaction proceeds. The bromine half-cell is the cathode, and cations from the salt bridge flow to the cathode. Electrons travel from the lead half-cell to the bromine half-cell (from the anode to the cathode). (Sn2+ + 2 e− → Sn) × 3 (La → La3+ + 3 e−) × 2
E° = −0.14 V −E° = 2.37 V
3 Sn2+(aq) + 2 La(s) → 3 Sn(s) + 2 La3+(aq)
E ocell = 2.23 V
50.
Statements a and e are false. The standard cell potential is 2.23 V. The La half-cell is the anode, and the Sn half-cell is the cathode. Anions will flow into the anode, the La half-cell, and electrons will flow to the cathode, the Sn half-cell. From the overall reaction, the La3+ concentration will increase since it is a product. Sn is also a product; the Sn electrode will increase in size as Sn2+ gets reduced to Sn. 51.
Reference the answer to Exercise 45 for a typical galvanic cell design. The contents of each half-cell compartment are identified below with all solute concentrations at 1.0 M and all gases at 1.0 atm. For each pair of half-reactions, the half-reaction with the largest (most positive) standard reduction potential will be the cathode reaction, and the half-reaction with the smallest (most negative) reduction potential will be reversed to become the anode reaction. Only this combination gives a spontaneous overall reaction, i.e., a reaction with a positive overall standard cell potential. Note that in a galvanic cell as illustrated in Exercise 37 the cations in the salt bridge migrate to the cathode, and the anions migrate to the anode. a.
Cl2 + 2 e− → 2 Cl− 2 Br− → Br2 + 2 e−
E° = 1.36 V −E° = −1.09 V
Cl2(g) + 2 Br− (aq) → Br2(aq) + 2 Cl−(aq)
E ocell = 0.27 V
CHAPTER 18
ELECTROCHEMISTRY
907
The contents of each compartment are: Cathode: Pt electrode; Cl2(g) bubbled in, Cl− in solution Anode: Pt electrode; Br2 and Br− in solution (2 e− + 2 H+ + IO4− → IO3− + H2O) × 5 E° = 1.60 V − − 2+ + (4 H2O + Mn → MnO4 + 8 H + 5 e ) × 2 −E° = −1.51 V − − − + 2+ + 10 H + 5 IO4 + 8 H2O + 2 Mn → 5 IO3 + 5 H2O + 2 MnO4 + 16 H E ocell = 0.09 V
b.
This simplifies to: 3 H2O(l) + 5 IO4−(aq) + 2 Mn2+(aq) → 5 IO3− (aq) + 2 MnO4− (aq) + 6 H+(aq) E ocell = 0.09 V Cathode: Pt electrode; IO4−, IO3−, and H2SO4 (as a source of H+) in solution Anode: Pt electrode; Mn2+, MnO4−, and H2SO4 in solution 52.
Reference the answer to Exercise 45 for a typical galvanic cell design. The contents of each half-cell compartment are identified below, with all solute concentrations at 1.0 M and all gases at 1.0 atm. a. H2O2 + 2 H+ + 2 e− → 2 H2O H2O2 → O2 + 2 H+ + 2 e−
E° = 1.78 V −E° = −0.68 V
2 H2O2(aq) → 2 H2O(l) + O2(g)
E ocell = 1.10 V
Cathode: Pt electrode; H2O2 and H+ in solution Anode: Pt electrode; O2(g) bubbled in, H2O2 and H+ in solution (Fe3+ + 3 e− → Fe) × 2 (Mn → Mn2+ + 2 e−) × 3
b.
2 Fe3+(aq) + 3 Mn(s) → 2 Fe(s) + 3 Mn2+(aq)
E° = −0.036 V −E° = 1.18 V
E ocell = 1.14 V
Cathode: Fe electrode; Fe3+ in solution; anode: Mn electrode; Mn2+ in solution 53.
In standard line notation, the anode is listed first, and the cathode is listed last. A double line separates the two compartments. By convention, the electrodes are on the ends with all solutes and gases toward the middle. A single line is used to indicate a phase change. We also included all concentrations. 45a.
Pt | Cr3+ (1.0 M), Cr2O72− (1.0 M), H+ (1.0 M) || Cl2 (1.0 atm) | Cl− (1.0 M) | Pt
45b.
Mg | Mg2+ (1.0 M) || Cu2+ (1.0 M) | Cu
51a.
Pt | Br− (1.0 M), Br2 (1.0 M) || Cl2 (1.0 atm) | Cl− (1.0 M) | Pt
51b.
Pt | Mn2+ (1.0 M), MnO4− (1.0 M), H+ (1.0 M) || IO4− (1.0 M), H+ (1.0 M), IO3− (1.0 M) | Pt
908 54.
55.
CHAPTER 18
ELECTROCHEMISTRY
46a.
Pt | Fe2+ (1.0 M), Fe3+ (1.0 M) || IO3− (1.0 M), H+ (1.0 M), I2 (1.0 M) | Pt
46b.
Zn | Zn2+ (1.0 M) || Ag+ (1.0 M) | Ag
52a.
Pt | H2O2 (1.0 M), H+ (1.0 M) | O2 (1.0 atm)|| H2O2 (1.0 M), H+ (1.0 M) | Pt
52b.
Mn | Mn2+ (1.0 M) || Fe3+ (1.0 M) | Fe
Locate the pertinent half-reactions in Table 18.1, and then figure which combination will give a positive standard cell potential. In all cases, the anode compartment contains the species with the smallest standard reduction potential. For part a, the copper compartment is the anode, and in part b, the cadmium compartment is the anode. a.
b.
Au3+ + 3 e− → Au (Cu+ → Cu2+ + e−) × 3 Au3+(aq) + 3 Cu+(aq) → Au(s) + 3 Cu2+(aq)
E° = 1.50 V −E° = −0.16 V E ocell = 1.34 V
(VO2+ + 2 H+ + e− → VO2+ + H2O) × 2 Cd → Cd2+ + 2e-
E° = 1.00 V −E° = 0.40 V
2 VO2+(aq) + 4 H+(aq) + Cd(s) → 2 VO2+(aq) + 2 H2O(l) + Cd2+(aq) 56.
a.
b.
57.
a.
E ocell = 1.40 V
(H2O2 + 2 H+ + 2 e− → 2 H2O) × 3 2 Cr3+ + 7 H2O → Cr2O72− + 14 H+ + 6 e−
E° = 1.78 V −E° = −1.33 V
3 H2O2(aq) + 2 Cr3+(aq) + H2O(l) → Cr2O72−(aq) + 8 H+(aq)
E ocell = 0.45 V
(2 H+ + 2 e− → H2) × 3 (Al → Al3+ + 3 e−) × 2
E° = 0.00 V −E° = 1.66 V
6 H+(aq) + 2 Al(s) → 3 H2(g) + 2 Al3+(aq)
E ocell = 1.66 V
(5 e− + 8 H+ + MnO4− → Mn2+ + 4 H2O) × 2 E° = 1.51 V (2 I− → I2 + 2 e−) × 5 −E° = −0.54 V 16 H+(aq) + 2 MnO4−(aq) + 10 I−(aq) → 5 I2(aq) + 2 Mn2+(aq) + 8 H2O(l) E ocell = 0.97 V This reaction is spontaneous at standard conditions because E ocell > 0.
b.
(5 e− + 8 H+ + MnO4− → Mn2+ + 4 H2O) × 2 E° = 1.51 V − − (2 F → F2 + 2 e ) × 5 −E° = −2.87 V − − + 2+ 16 H (aq) + 2 MnO4 (aq) + 10 F (aq) → 5 F2(aq) + 2 Mn (aq) + 8 H2O(l) E ocell = −1.36 V This reaction is not spontaneous at standard conditions because E ocell < 0.
58.
a.
H2 → 2H+ + 2 e− H2 + 2 e− → 2H− 2H2(g) → 2H+(aq) + 2 H−(aq)
E° = 0.00 V −E° = −2.23 V
E ocell = −2.23 V
Not spontaneous ( E ocell < 0)
CHAPTER 18 b.
ELECTROCHEMISTRY
909
Au3+ + 3 e− → Au (Ag → Ag+ + e−) × 3
E° = 1.50 V −E ° = −0.80 V
Au3+(aq) + 3 Ag(s) → Au(s) + 3 Ag+(aq)
E ocell = 0.70 V Spontaneous ( E ocell > 0)
59.
Cl2 + 2 e− → 2 Cl− (ClO2− → ClO2 + e−) × 2 − 2 ClO2 (aq) + Cl2(g) → 2 ClO2(aq) + 2 Cl− (aq)
E° = 1.36 V −E° = −0.954 V E ocell = 0.41 V = 0.41 J/C
ΔG° = − nFE ocell = −(2 mol e−)(96,485 C/mol e−)(0.41 J/C) = −7.9 × 104 J = −79 kJ 60.
(4 H+ + NO3− + 3 e− → NO + 2 H2O) × 2 (Mn → Mn2+ + 2 e−) × 3
a.
E° = 0.96 V −E° = 1.18 V
3 Mn(s) + 8 H+(aq) + 2 NO3−(aq) → 2 NO(g) + 4 H2O(l) + 3 Mn2+(aq) (2 e− + 2 H+ + IO4− → IO3− + H2O) × 5 (Mn2+ + 4 H2O → MnO4− + 8 H+ + 5 e−) × 2
E ocell = 2.14 V E° = 1.60 V −E° = −1.51 V
5 IO4−(aq) + 2 Mn2+(aq) + 3 H2O(l) → 5 IO3−(aq) + 2 MnO4−(aq) + 6 H+(aq) E ocell = 0.09 V b. Nitric acid oxidation (see above for E ocell ): ΔG° = − nFE ocell = −(6 mol e−)(96,485 C/mol e−)(2.14 J/C) = −1.24 × 106 J = −1240 kJ Periodate oxidation (see above for E ocell ): ΔG° = −(10 mol e−)(96,485 C/mol e−)(0.09 J/C)(1 kJ/1000 J) = −90 kJ 61.
Because the cells are at standard conditions, wmax = ΔG = ΔG° = − nFE ocell . See Exercise 55 for the balanced overall equations and for E ocell .
62.
55a.
wmax = −(3 mol e−)(96,485 C/mol e−)(1.34 J/C) = −3.88 × 105 J = −388 kJ
55b.
wmax = −(2 mol e−)(96,485 C/mol e−)(1.40 J/C) = −2.70 × 105 J = −270. kJ
Because the cells are at standard conditions, wmax = ΔG = ΔG° = − nFE ocell . See Exercise 56 for the balanced overall equations and for E ocell .
63.
56a.
wmax = −(6 mol e−)(96,485 C/mol e−)(0.45 J/C) = −2.6 × 105 J = −260 kJ
56b.
wmax = −(6 mol e−)(96,485 C/mol e−)(1.66 J/C) = −9.61 × 105 J = −961 kJ
2 H2O + 2 e− → H2 + 2 OH− ΔG° = Σn p ΔGof, products − Σn r ΔGof, reactants = 2(−157) − [2(−237)] = 160. kJ
910
CHAPTER 18 ΔG° = −nFE°, E° =
ELECTROCHEMISTRY
− ΔG o − 1.60 10 5 J = = −0.829 J/C = −0.829 V nF (2 mol e − )(96,485 C/mol e − )
The two values agree to two significant figures (−0.83 V in Table 18.1). 64.
Fe2+ + 2 e− → Fe
E° = −0.44 V = −0.44 J/C
ΔG° = −nFE° = −(2 mol e−)(96,485 C/mol e−)(−0.44 J/C)(1 kJ/1000 J) = 85 kJ 85 kJ = 0 − [ΔG of , Fe2+ + 0] , ΔG f , Fe 2+ = −85 kJ o
We can get ΔG f , Fe 3+ two ways. Consider: Fe3+ + e− → Fe2+ o
E° = 0.77 V
ΔG° = −(1 mol e)(96,485 C/mol e−)(0.77 J/C) = −74,300 J = −74 kJ Fe2+ → Fe3+ + e− Fe → Fe2+ + 2 e−
ΔG° = 74 kJ ΔG° = −85 kJ
Fe → Fe3+ + 3 e−
ΔG° = −11 kJ,
or consider: Fe3+ + 3 e− → Fe
ΔG of , Fe 3+ = −11 kJ/mol
E° = −0.036 V
ΔG° = −(3 mol e−)(96,485 C/mol e−)(−0.036 J/C) = 10,400 J ≈ 10. kJ 10. kJ = 0 − [ΔG of , Fe3+ + 0] , ΔG f , Fe 3+ = −10. kJ/mol; round-off error explains the 1-kJ discrepancy. o
65.
G = [6 mol(−394 kJ/mol) + 6 mol(−237 kJ/mol] – [1 mol(−911 kJ/mol + 6 mol(0)] = −2875 kJ Carbon is oxidized in this combustion reaction. In C 6H12O6, H has a +1 oxidation state, and oxygen has a −2 oxidation, so 6(x) + 12(+1) + 6(−2) = 0, x = oxidation state of C in C6H12O6 = 0. In CO2, O has an oxidation state of −2, so y + 2(−2) = 0, y = oxidation state of C in CO2 = +4. Carbon goes from the 0 oxidation state in C 6H12O6 to the +4 oxidation state in CO2, so each carbon atom loses 4 electrons. Because the balanced reaction has 6 mol of carbon, 6(4) = 24 mol electrons are transferred in the balanced equation. ΔG° = −nFE°, E° =
66.
− (−2875 10 3 J) − ΔG o = 1.24 J/C = 1.24 V = nF (24 mol e − ) (96,485 C/mol e − )
CH3OH(l) + 3/2 O2(g) → CO2(g) + 2 H2O(l)
ΔG° = 2(–237) + (−394) − [−166] = −702 kJ
The balanced half-reactions are: H2O + CH3OH → CO2 + 6 H+ + 6 e− and O2 + 4 H+ + 4 e− → 2 H2O For 3/2 mol O2, 6 mol of electrons will be transferred (n = 6). ΔG° = −nFE°, E° =
− (−702,000 J) − ΔG o = = 1.21 J/C = 1.21 V nF (6 mol e − )(96,485 C/mol e − )
CHAPTER 18
ELECTROCHEMISTRY
911
67.
Good oxidizing agents are easily reduced. Oxidizing agents are on the left side of the reduction half-reactions listed in Table 18.1. We look for the largest, most positive standard reduction potentials to correspond to the best oxidizing agents. The ordering from worst to best oxidizing agents is: Mg2+ < Fe2+ < Fe3+ < Cr2O72− < Cl2 < MnO4− E(V) −2.37 −0.44 0.77 1.33 1.36 1.68
68.
Good reducing agents are easily oxidized. The reducing agents are on the right side of the reduction half-reactions listed in Table 18.1. The best reducing agents have the most negative standard reduction potentials (E°) or the most positive standard oxidation potentials (Eoox = − E°). The ordering from worst to best reducing agents is: −E(V)
69.
a. 2 H+ + 2 e− → H2
F− < Cr3+ < Fe2+ < H2 < Zn < Li −2.87 −1.33 −0.77 0.00 0.76 3.05 E° = 0.00 V; Cu → Cu2+ + 2 e−
−E° = −0.34 V
E ocell = −0.34 V; no, H+ cannot oxidize Cu to Cu2+ at standard conditions ( E ocell < 0). b. Fe3+ + e− → Fe2+
E° = 0.77 V; 2 I− → I2 + 2 e−
−E° = -0.54 V
E ocell = 0.77 − 0.54 = 0.23 V; yes, Fe3+ can oxidize I− to I2. c. H2 → 2 H+ + 2 e−
−E° = 0.00 V; Ag+ + e− → Ag E° = 0.80 V
E ocell = 0.80 V; yes, H2 can reduce Ag+ to Ag at standard conditions ( E ocell > 0). 70.
a. H2 → 2 H+ + 2 e−
−E° = 0.00 V; Ni2+ + 2 e− → Ni
E° = −0.23 V
E ocell = −0.23 V; no, H2 cannot reduce Ni2+ to Ni at standard conditions ( E ocell < 0). b. Fe2+ → Fe3+ + e−
−E° = −0.77 V; VO2+ + 2 H+ + e− → VO2+ + H2O
E° = 1.00 V
E ocell = 1.00 − 0.77 = 0.23 V; yes, Fe2+ can reduce VO2+ at standard conditions. c. Fe2+ → Fe3+ + e−
−E° = −0.77 V; Cr3+ + e− → Cr2+
E° = -0.50 V
E ocell = 0.50 − 0.77 = −1.27 V No, Fe2+ cannot reduce Cr3+ to Cr2+ at standard conditions ( E ocell < 0). 71.
The reduction of Cu2+ has E° = 0.34 V and the reduction of Al3+ has E° = ‒ 1.66 V. We need a substance that has an oxidation potential that when added to 0.34 V gives a positive value, yet when added to ‒-1.66 V, gives a negative value. From the options available, the oxidation of Fe to Fe2+ has ‒E = 0.44 V. When the oxidation of Fe2+ is coupled with the reduction of Cu2+, E ocell will be positive, so the reaction is spontaneous at standard conditions. When the oxidation of Fe2+ is coupled with the reduction of Al3+, E ocell will be negative, and this reaction will not be spontaneous at standard conditions. Answer c is correct.
912
CHAPTER 18
ELECTROCHEMISTRY
72.
Copper is oxidized, which has ‒E° = ‒0.34 V. We need a species which has E° > 0.34 V to have a spontaneous reaction at standard conditions ( E ocell > 0). From the options available, Ag+ and F2 have E° = 0.34 V, so Cu can reduce Ag+ and F2 at standard conditions (answer b).
73.
Cl2 + 2 e− → 2 Cl− Pb2+ + 2 e− → Pb Na+ + e− → Na
E° = 1.36 V E° = −0.13 V E° = −2.71 V
Ag+ + e− → Ag Zn2+ + 2 e− → Zn
E° = 0.80 V E° = −0.76 V
a. Oxidizing agents (species reduced) are on the left side of the preceding reduction halfreactions. Of the species available, Ag + would be the best oxidizing agent since it has the largest E° value. Note that Cl2 is a better oxidizing agent than Ag+, but it is not one of the choices listed. b. Reducing agents (species oxidized) are on the right side of the reduction half-reactions. Of the species available, Zn would be the best reducing agent since it has the largest −E° value. c.
SO42− + 4 H+ + 2 e− → H2SO3 + H2O E° = 0.20 V; SO42− can oxidize Pb and Zn at standard conditions. When SO42− is coupled with these reagents, E ocell is positive.
d. Al → Al3+ + 3 e− −E° = 1.66 V; Al can reduce Ag+ and Zn2+ at standard conditions because E ocell > 0. 74.
Br2 + 2 e− → 2 Br− 2 H+ + 2 e− → H2 Cd2+ + 2 e− → Cd
E° = 1.09 V E° = 0.00 V E° = −0.40 V
La3+ + 3 e− → La Ca2+ + 2 e− → Ca
E° = −2.37 V E° = −2.76 V
a. Oxidizing agents are on the left side of the preceding reduction half-reactions. Br2 is the best oxidizing agent (largest E°). b. Reducing agents are on the right side of the reduction half-reactions. Ca is the best reducing agent (largest −E°). c. MnO4− + 8 H+ + 5 e− → Mn2+ + 4 H2O
E° = 1.51 V; permanganate can oxidize Br−, H2,
Cd, and Ca at standard conditions. When MnO4− is coupled with these reagents, E cell is positive. Note: La is not one of the choices given in the question or it would have been included. d. Zn → Zn2+ + 2 e− 75.
−E° = 0.76 V; zinc can reduce Br2 and H+ because E ocell > 0.
a. 2 Br− → Br2 + 2 e− −E° = −1.09 V; 2 Cl− → Cl2 + 2 e− −E° = −1.36 V; E° > 1.09 V to oxidize Br−; E° < 1.36 V to not oxidize Cl−; Cr2O72−, O2, MnO2, and IO3− are all possible since when all of these oxidizing agents are coupled with Br-, E ocell > 0, and when coupled with Cl−, E ocell < 0 (assuming standard conditions).
CHAPTER 18
ELECTROCHEMISTRY
913
b. Mn → Mn2+ + 2 e− −E° = 1.18; Ni → Ni2+ + 2 e− −E° = 0.23 V; any oxidizing agent with −0.23 V > E° > −1.18 V will work. PbSO4, Cd2+, Fe2+, Cr3+, Zn2+, and H2O will be able to oxidize Mn but not Ni (assuming standard conditions). 76.
a. Fe3+ + e− → Fe2+ E = 0.77 V; Fe2+ + 2 e− → Fe E = −0.44 V. To reduce Fe3+ but not Fe2+, the reducing agent must have a standard oxidation potential ( E oox = −E°) between −0.77 V and 0.44 V (so E ocell is positive only for the Fe3+ reduction). The reducing agents are on the right side of the half-reactions in Table 18.1. The reagents at standard conditions which have E oox (= −E°) between −0.77 V and 0.44 V are H2O2, MnO42−, I−, Cu, OH−, Hg (in 1 M Cl−), Ag (in 1 M Cl−), H2SO3, Cu+, H2, Pb, Sn, Ni, and Cd. b. Ag+ + e− → Ag E = 0.80 V; O2 + 2 H+ + 2 e− → H2O2 E = 0.68 V; From Table 18.1, only Fe2+ will reduce Ag+ to Ag but will not reduce O2 to H2O2 (assuming standard conditions). A potential of −0.77 V is required to oxidize Fe2+ to Fe3+, which is large enough to oxidize Ag+ but small enough not to oxidize O2.
The Nernst Equation 77.
H2O2 + 2 H+ + 2 e− → 2 H2O (Ag → Ag+ + e−) × 2 H2O2(aq) + 2 H+(aq) + 2 Ag(s) → 2 H2O(l) + 2 Ag+(aq)
E° = 1.78 V −E° = −0.80 V
E ocell = 0.98 V
a. A galvanic cell is based on spontaneous redox reactions. At standard conditions, this reaction produces a voltage of 0.98 V. Any change in concentration that increases the tendency of the forward reaction to occur will increase the cell potential. Conversely, any change in concentration that decreases the tendency of the forward reaction to occur (increases the tendency of the reverse reaction to occur) will decrease the cell potential. Using Le Chatelier’s principle, increasing the reactant concentrations of H 2O2 and H+ from 1.0 to 2.0 M will drive the forward reaction further to right (will further increase the tendency of the forward reaction to occur). Therefore, E cell will be greater than E ocell . b. Here, we decreased the reactant concentration of H + and increased the product concentration of Ag+ from the standard conditions. This decreases the tendency of the forward reaction to occur, which will decrease Ecell as compared to E ocell (Ecell < E ocell ). 78.
The concentrations of Fe2+ in the two compartments are now 0.01 and 1 × 10−7 M. The driving force for this reaction is to equalize the Fe2+ concentrations in the two compartments. This occurs if the compartment with 1 × 10−7 M Fe2+ becomes the anode (Fe will be oxidized to Fe2+) and the compartment with the 0.01 M Fe2+ becomes the cathode (Fe2+ will be reduced to Fe). Electron flow, as always for galvanic cells, goes from the anode to the cathode, so electron flow will go from the right compartment ([Fe2+] = 1 × 10−7 M) to the left compartment ([Fe2+] = 0.01 M).
914 79.
CHAPTER 18
ELECTROCHEMISTRY
For concentration cells, the driving force for the reaction is the difference in ion concentrations between the anode and cathode. In order to equalize the ion concentrations, the anode always has the smaller ion concentration. The general setup for this concentration cell is: Ag+(x M) + e− → Ag Ag → Ag+ (y M) + e−
E° = 0.80 V −E° = −0.80 V
Ag+(cathode, x M) → Ag+ (anode, y M)
E ocell = 0.00 V
Cathode: Anode:
Ecell = E ocell −
[Ag + ] anode 0.0591 − 0.0591 log Q = log n 1 [Ag + ] cathode
For each concentration cell, we will calculate the cell potential using the preceding equation. Remember that the anode always has the smaller ion concentration. a. Both compartments are at standard conditions ([Ag +] = 1.0 M), so Ecell = E ocell = 0 V. No voltage is produced since no reaction occurs. Concentration cells only produce a voltage when the ion concentrations are not equal. b. Cathode = 2.0 M Ag+; anode = 1.0 M Ag+; electron flow is always from the anode to the cathode, so electrons flow to the right in the diagram. Ecell =
[Ag + ] anode − 0.0591 − 0.0591 1.0 log = log = 0.018 V + n 1 2.0 [Ag ] cathode
c. Cathode = 1.0 M Ag+; anode = 0.10 M Ag+; electrons flow to the left in the diagram. Ecell =
[Ag + ] anode − 0.0591 − 0.0591 0.10 log = log = 0.059 V + n 1 1.0 [Ag ] cathode
d. Cathode = 1.0 M Ag+; anode = 4.0 × 10−5 M Ag+; electrons flow to the left in the diagram. Ecell =
− 0.0591 4.0 10 −5 log = 0.26 V n 1.0
e. The ion concentrations are the same; thus log([Ag+]anode/[Ag+]cathode) = log(1.0) = 0 and Ecell = 0. No electron flow occurs. 80.
As is the case for all concentration cells, E ocell = 0, and the smaller ion concentration is always in the anode compartment. The general Nernst equation for the Ni | Ni2+(x M) || Ni2+(y M) | Ni concentration cell is: Ecell = E ocell −
[Ni 2+ ] anode 0.0591 − 0.0591 log Q = log n 2 [Ni 2+ ] cathode
a. Both compartments are at standard conditions ([Ni2+] = 1.0 M), and Ecell = E ocell = 0 V. No electron flow occurs. b. Cathode = 2.0 M Ni2+; anode = 1.0 M Ni2+; electron flow is always from the anode to the cathode, so electrons flow to the right in the diagram.
CHAPTER 18
ELECTROCHEMISTRY Ecell =
915
[Ni 2+ ] anode − 0.0591 − 0.0591 1.0 log = log = 8.9 × 10−3 V 2+ 2 2 2.0 [Ni ] cathode
c. Cathode = 1.0 M Ni2+; anode = 0.10 M Ni2+; electrons flow to the left in the diagram. Ecell =
− 0.0591 0.10 log = 0.030 V 2 1 .0
d. Cathode = 1.0 M Ni2+; anode = 4.0 × 10−5 M Ni2+; electrons flow to the left in the diagram. Ecell =
4.0 10 −5 − 0.0591 log = 0.13 V 2 1.0
e. Because both concentrations are equal, log(2.5/2.5) = log 1.0 = 0, and Ecell = 0. No electron flow occurs. 81.
a. As is the case for all concentration cells, E ocell = 0, and the smaller ion concentration is always in the anode compartment. Here, n = 3 as three electrons are needed to convert Al 3+ to Al and vice versa. The general Nernst equation for this Al concentration cell is: Ecell = E ocell −
Ecell =
[Al3+ ]anode 0.0591 −0.0591 log Q = log n 3 [Al3+ ]cathode
−0.0591 1.0 10−4 = 0.082 V log 3 1.5
b. The overall cell reaction is Al3+ (1.5 M) → Al3+(1.0 × 10−4 M). In general, the larger the difference in ion concentrations between the anode and cathode, the greater the cell potential. If the cathode Al3+ concentration is decreased from 1.5 M, the cell potential will potential will decrease. You can also think of this in terms of Le Châtelier’s principle. Any change that increases the tendency of the forward reaction will increase the cell potential. Decreasing the Al3+ concentration at the cathode shifts the reaction left, thus decreasing the cell potential. c. Increasing the Al3+ concentration at the anode decreases the tendency of the forward reaction to occur, thus the cell potential will decrease. d. The amount of a solid does not affect the equilibrium of a reaction. So, using a more massive Al electrode will have no effect. 82.
As is the case for all concentration cells, E ocell = 0, and the smaller ion concentration is always in the anode compartment. Here, n = 2 as two electrons are needed to convert Cu 2+ to Cu and vice versa. The general Nernst equation for this Cu concentration cell is: Ecell = E ocell −
Ecell =
[Cu 2+ ]anode 0.0591 −0.0591 log Q = log n 2 [Cu 2+ ]cathode
−0.0591 1.0 10−3 = 0.059 V log 2 0.10
916
CHAPTER 18
ELECTROCHEMISTRY
Statements d and e are true. The cell reaction is Cu 2+ (0.10 M) → Cu2+(1.0 × 10−3 M). Decreasing the anode ion concentration will increase the tendency of the forward reaction to occur (reaction shifts right more), resulting in an increase in the cell potential. For the incorrect answers, n = 2, E ocell = 0, and Q = 0.010. 83.
n = 2 for this reaction (lead goes from Pb → Pb2+ in PbSO4). E = E° −
0.0591 1 0.0591 1 log = 2.04 V − log − 2 2 + 2 2 2 ( 4.5) ( 4.5) 2 [H ] [HSO 4 ]
E = 2.04 V + 0.077 V = 2.12 V Cr2O72− + 14 H+ + 6 e− → 2 Cr3+ + 7 H2O (Al → Al3+ + 3 e−) × 2
84.
Cr2O72−(aq) + 14 H+(aq) + 2 Al(s) → 2 Al3+(aq) + 2 Cr3+(aq) + 7 H2O(l) E = E° −
3.7 10 −3
85.
E ocell = 2.99 V
0.0591 [Al 3+ ]2 [Cr 3+ ]2 0.0591 log log Q , E = 2.99 − 2− 6 n [Cr2 O 7 ][H + ]14
3.01 = 2.99 −
[H + ]14
E° = 1.33 V −E° = 1.66 V
(0.30) 2 (0.15) 2 0.0591 , log Q n (0.55)[H + ]14
3.7 10 −3 − 6(0.02) = log + 14 0.0591 [H ]
= 10−2 = 0.01, [H+]14 = 0.37, [H+] = 0.93 = 0.9 M, pH = −log(0.9) = 0.05 (Ag+ + e− → Ag) 2 Zn → Zn2+ + 2 e−
E° = 0.80 V −E° = 0.76 V
2 Ag+(aq) + Zn(s) → Zn2+(aq) + 2 Ag(s)
E ocell = 1.56 V
Because Zn2+ is a product in the reaction, the Zn2+ concentration increases from 1.00 to 1.20 M. This means that the reactant concentration of Ag+ must decrease from 1.00 to 0.60 M (from the 1 : 2 mole ratio between Zn2+ and Ag+ in the balanced reaction). Ecell = E ocell −
0.0591 0.0591 [Zn 2+ ] log Q = 1.56 V − log n 2 [Ag + ] 2
Ecell = 1.56 V − 86.
0.0591 1.20 = 1.56 V − 0.020 V = 1.54 V log 2 (0.60) 2
(Pb2+ + 2 e− → Pb) × 3 (Al → Al3+ + 3 e−) × 2 3 Pb2+(aq) + 2 Al(s) → 3 Pb(s) + 2 Al3+(aq)
E° = −0.13 V −E° = 1.66 V
E ocell = 1.53 V
From the balanced reaction, when the Al3+ has increased by 0.60 mol/L (Al3+ is a product in the spontaneous reaction), then the Pb 2+ concentration has decreased by 3/2 (0.60 mol/L) = 0.90 M.
CHAPTER 18
ELECTROCHEMISTRY
Ecell = 1.53 V −
917
(1.60) 2 0.0591 [Al3+ ] 2 0.0591 log = 1.53 − log 6 6 [Pb2+ ]3 (0.10) 3
Ecell = 1.53 V − 0.034 V = 1.50 V 87.
See Exercises 45, 47, and 51 for balanced reactions and standard cell potentials. Balanced reactions are necessary to determine n, the moles of electrons transferred. 45a.
7 H2O + 2 Cr3+ + 3 Cl2 → Cr2O72− + 6 Cl− + 14 H+ E ocell = 0.03 V = 0.03 J/C ΔG° = − nFE ocell = − (6 mol e−)(96,485 C/mol e−)(0.03 J/C) = −1.7 × 104 J = −20 kJ E cell = E ocell − E ocell =
0.0591 log Q; at equilibrium, Ecell = 0 and Q = K, so: n
0.0591 nE o 6(0.03) log K, log K = = 3.05, K = 103.05 = 1 × 103 = n 0.0591 0.0591
Note: When determining exponents, we will round off to the correct number of significant figures after the calculation is complete in order to help eliminate excessive round-off error. 45b.
ΔG° = − (2 mol e−)(96,485 C/mol e−)(2.71 J/C) = −5.23 × 105 J = −523 kJ log K =
51a.
ΔG° = − (2 mol e−)(96,485 C/mol −)(0.27 J/C) = −5.21 × 104 J = −52 kJ log K =
51b.
2(0.27 ) = 9.14, K = 1.4 × 109 0.0591
ΔG° = −(10 mol e−)(96,485 C/mol e−)(0.09 J/C) = −8.7 × 104 J = −90 kJ log K =
88.
2( 2.71) = 91.709, K = 5.12 × 1091 0.0591
10 (0.09 ) = 15.23, K = 2 × 1015 0.0591
ΔG° = − nFE ocell ; E ocell = 46a.
ΔG° = −(10 mol e−)(96,485 C/mol e−)(0.43 J/C) = −4.1 × 105 J = −410 kJ log K =
46b.
10 (0.43) = 72.76, K = 1072.76 = 5.8 × 1072 0.0591
ΔG° = −(2 mol e−)(96,485 C/mol e−)(1.56 J/C) = −3.01 × 105 J = −301 kJ log K =
52a.
0.0591 nE o log K, log K = n 0.0591
2(1.56) = 52.792, K = 6.19 × 1052 0.0591
ΔG° = − (2 mol −)(96,485 C/mol e−)(1.10 J/C) = −2.12 × 105 J = −212 kJ log K =
2(1.10) = 37.225, K = 1.68 × 1037 0.0591
918
CHAPTER 18 52b.
ΔG° = −(6 mol e−)(96,485 C/mol e−)(1.14 J/C) = −6.60 × 105 J = −660. kJ log K =
89.
a.
ELECTROCHEMISTRY
6(1.14) = 115.736, K = 5.45 × 10115 0.0591
Fe2+ + 2 e− → Fe Zn → Zn2+ + 2 e− Fe2+(aq) + Zn(s) → Zn2+(aq) + Fe(s)
E° = −0.44 V −E° = 0.76 V
E ocell = 0.32 V = 0.32 J/C
b. ΔG° = − nFE ocell = −(2 mol e−)(96,485 C/mol e−)(0.32 J/C) = −6.2 × 104 J = −62 kJ o E cell =
c.
0.0591 nE o 2(0.32) log K, log K = = = 10.83, K = 1010.83 = 6.8 × 1010 n 0.0591 0.0591
o E cell = E cell −
Ecell = 0.32 −
90.
0.0591 [Zn 2+ ] 0.0591 log Q = 0.32 V − log n n [Fe 2+ ]
0.0591 0.10 log = 0.32 − 0.12 = 0.20 V 2 1.0 10 −5
Au3+ + 3 e− → Au (Tl → Tl+ + e−) × 3
E= 1.50 V −E = 0.34 V
Au3+(aq) + 3 Tl(s) → Au(s) + 3 Tl+(aq)
E ocell = 1.84 V
a.
b. ΔG° = − nFE ocell = −(3 mol e−)(96,485 C/mol e−)(1.84 J/C) = −5.33 × 105 J = −533 kJ log K =
nE o 3(1.84) = = 93.401, K = 1093.401 = 2.52 × 1093 0.0591 0.0591
c. At 25°C, Ecell = E ocell − Ecell = 1.84 V −
0.0591 [Tl + ]3 log Q, where Q = . n [Au3+ ]
(1.0 10 −4 ) 3 [Tl + ]3 0.0591 0.0591 log = 1.84 − log 3 3 [Au3+ ] 1.0 10 − 2
Ecell = 1.84 − (−0.20) = 2.04 V 91.
Cu2+(aq) + H2(g) → 2 H+(aq) + Cu(s) E ocell = 0.34 V − 0.00 V = 0.34 V; n = 2 mol electrons
PH 2 = 1.0 atm and [H+] = 1.0 M: Ecell = E ocell − 0.0591 log n
a. Ecell = 0.34 V −
1 [Cu 2+ ]
0.0591 1 = 0.34 V − 0.11V = 0.23 V log 2 2.5 10 −4
b. 0.195 V = 0.34 V −
0.0591 1 1 = 4.91, [Cu2+] = 10−4.91 log , log 2+ 2 [Cu ] [Cu 2 + ]
= 1.2 × 10−5 M Note: When determining exponents, we will carry extra significant figures.
CHAPTER 18 92.
ELECTROCHEMISTRY
919
3 Ni2+(aq) + 2 Al(s) → 2 Al3+(aq) + 3 Ni(s) E ocell = −0.23 + 1.66 = 1.43 V; n = 6 mol electrons for this reaction. a. Ecell = 1.43 V −
(7.2 10 −3 ) 2 0.0591 0.0591 [Al3+ ] 2 = 1.43 − log log 6 6 [Ni 2+ ]3 (1.0) 3
Ecell = 1.43 V − (−0.042 V) = 1.47 V b. 1.62 V = 1.43 V −
0.0591 [Al3+ ] 2 , log [Al3+]2 = −19.29 log 6 (1.0) 3
[Al3+]2 = 10−19.29, [Al3+] = 2.3 × 10−10 M 93.
Cu2+(aq) + H2(g) → 2 H+(aq) + Cu(s)
E ocell = 0.34 V − 0.00 V = 0.34 V; n = 2
PH 2 = 1.0 atm and [H+] = 1.0 M: Ecell = E ocell − 0.0591 log 2
1 [Cu 2+ ]
Use the Ksp expression to calculate the Cu2+ concentration in the cell. Cu(OH)2(s) ⇌ Cu2+(aq) + 2 OH−(aq) Ksp = 1.6 × 10−19 = [Cu2+][OH−]2 From problem, [OH−] = 0.10 M, so: [Cu2+] = Ecell = E ocell −
1.6 10 −19 (0.10) 2
= 1.6 × 10−17 M
0.0591 1 0.0591 1 = 0.34 V − log log 2+ 2 2 1.6 10 −17 [Cu ]
Ecell = 0.34 − 0.50 = −0.16 V Because Ecell < 0, the forward reaction is not spontaneous, but the reverse reaction is spontaneous. The Cu electrode becomes the anode and Ecell = 0.16 V for the reverse reaction. The cell reaction is 2 H+(aq) + Cu(s) → Cu2+(aq) + H2(g). 94.
Pb2+(aq) + Zn(s) → Zn2+(aq) + Pb(s) E ocell = −0.13 V + 0.76 V = 0.63 V; n = 2
Ecell = Eocell −
2+
0.0591 [Zn ] 0.0591 [Zn 2+ ] log log , 1.05 V = 0.63 V − 2+ n [Pb ] 2 1.0
log [Zn2+] = −14.21, [Zn2+] = 10−14.21 = 6.2 × 10−15 M Zn(OH)2(s) ⇌ Zn2+(aq) + 2 OH−(aq) Ksp = [Zn2+][OH−]2; 6.5 × 10−17 = (6.2 × 10−15)[OH−]2, [OH‒] = 0.10 M 95.
M2+ + 2e− → M(s) M(s) → M2+ + 2e−
E° = −0.31 V − E° = 0.31 V
M2+ (cathode) → M2+ (anode)
E ocell = 0.00 V
Cathode: Anode:
920
CHAPTER 18
ELECTROCHEMISTRY
[M 2 + ] anode 0.0591 [M 2+ ] anode 0.0591 log Ecell = 0.44 V = 0.00 V − , 0.44 = − log 2 2 1.0 [M 2 + ] cathode
log [M2+]anode = −
2(0.44) = −14.89, [M2+]anode = 1.3 × 10−15 M 0.0591
Because we started with equal numbers of moles of SO 42− and M2+, [M2+] = [SO42−] at equilibrium. Ksp = [M2+][SO42−] = (1.3 × 10−15)2 = 1.7 × 10−30 96.
a. Ag+ (x M, anode) → Ag+ (0.10 M, cathode); for the silver concentration cell, E° = 0.00 (as is always the case for concentration cells) and n = 1. E = 0.76 V = 0.00 −
[Ag + ] anode 0.0591 log 1 [Ag + ] cathode
0.76 = −(0.0591)log
[Ag+ ]anode [Ag+ ]anode = 10−12.86, [Ag+]anode = 1.4 × 10−14 M , 0.10 0.10
b. Ag+(aq) + 2 S2O32−(aq) ⇌ Ag(S2O3)23−(aq) 3−
K= 97.
[ Ag (S2 O 3 ) 2 ] 2−
[ Ag + ][S2 O 3 ]2
=
1.0 10 −3 = 2.9 × 1013 −14 2 1.4 10 (0.050 )
a. Possible reaction: I2(s) + 2 Cl−(aq) → 2 I−(aq) + Cl2(g)
E ocell = 0.54 V − 1.36 V = −0.82 V
This reaction is not spontaneous at standard conditions because E ocell < 0; no reaction occurs. b. Possible reaction: Cl2(g) + 2 I−(aq) → I2(s) + 2 Cl−(aq) E ocell = 0.82 V; this reaction is o spontaneous at standard conditions because E cell > 0. The reaction will occur. Cl2(g) + 2 I−(aq) → I2(s) + 2 Cl−(aq)
E ocell = 0.82 V = 0.82 J/C
ΔG° = − nFE ocell = − (2 mol e−)(96,485 C/mol e−)(0.82 J/C) = −1.6 × 105 J = −160 kJ E° =
0.0591 nE o 2(0.82) log K, log K = = 27.75, K = 1027.75 = 5.6 × 1027 = n 0.0591 0.0591
c. Possible reaction: 2 Ag(s) + Cu2+(aq) → Cu(s) + 2 Ag+(aq) E ocell = −0.46 V; no reaction occurs. d. Fe2+ can be oxidized or reduced. The other species present are H +, SO42−, H2O, and O2 from air. Only O2 in the presence of H+ has a large enough standard reduction potential to oxidize Fe2+ to Fe3+ (resulting in E ocell > 0). All other combinations, including the possible reduction of Fe2+, give negative cell potentials. The spontaneous reaction is: 4 Fe2+(aq) + 4 H+(aq) + O2(g) → 4 Fe3+(aq) + 2 H2O(l) E ocell = 1.23 − 0.77 = 0.46 V
CHAPTER 18
ELECTROCHEMISTRY
921
ΔG° = − nFE ocell = −(4 mol e−)(96,485 C/mol e−)(0.46 J/C)(1 kJ/1000 J) = −180 kJ log K = 98.
a.
4(0.46) = 31.13, K = 1.3 × 1031 0.0591
Cu+ + e− → Cu Cu+ → Cu2+ + e− 2 Cu+(aq) → Cu2+(aq) + Cu(s)
E° = 0.52 V −E° = −0.16 V E ocell = 0.36 V; spontaneous
ΔG° = − nFE ocell = − (1 mol e−)(96,485 C/mol e−)(0.36 J/C) = −34,700 J = −35 kJ o
E ocell = b.
nE 1(0.36) 0.0591 = log K, log K = = 6.09, K = 106.09 = 1.2 × 106 0.0591 0.0591 n
Fe2+ + 2 e− → Fe (Fe2+ → Fe3+ + e−) × 2
E° = −0.44 V −E° = −0.77 V
3 Fe2+(aq) → 2 Fe3+(aq) + Fe(s)
E ocell = −1.21 V; not spontaneous
c. HClO2 + 2 H+ + 2 e− → HClO + H2O HClO2 + H2O → ClO3− + 3 H+ + 2 e− 2 HClO2(aq) → ClO3−(aq) + H+(aq) + HClO(aq)
E° = 1.65 V −E° = −1.21 V
E ocell = 0.44 V; spontaneous
ΔG° = − nFE ocell = −(2 mol e−)(96,485 C/mol e−)(0.44 J/C) = −84,900 J = −85 kJ log K =
nE o 2(0.44) = = 14.89, K = 7.8 × 1014 0.0591 0.0591
(Cr2+ → Cr3+ + e−) × 2 Co + 2 e− → Co
99.
2+
2 Cr2+(aq) + Co2+(aq) → 2 Cr3+(aq) + Co(s)
E ocell = E = E° −
0.0591 0.0591 log K = log(2.79 × 107) = 0.220 V 2 n
(2.0) 2 0.0591 0.0591 [Cr 3+ ] 2 log = 0.220 V − = 0.151 V log 2 n (0.30) 2 (0.20) [Cr 2+ ] 2 [Co 2+ ]
ΔG = −nFE = − (2 mol e−)(96,485 C/mol e−)(0.151 J/C) = −2.91 × 104 J = −29.1 kJ 100.
2 Ag+(aq) + Cu(s) → Cu2+(aq) + 2 Ag(s) E ocell = 0.80 − 0.34 = 0.46 V and n = 2 Because [Ag+] = 1.0 M, Ecell = 0.46 V − 0.99 V = 0.46 V −
0.0591 log [Cu2+]. 2
0.0591 log [Cu2+], [Cu2+] = 1.2 × 10−18 M 2
922
CHAPTER 18 Cu2+(aq) + 4 NH3(aq) ⇌ Cu(NH3)42+(aq) K =
ELECTROCHEMISTRY
[Cu(NH 3 ) 24+ ] [Cu 2 + ][NH 3 ] 4
From the problem, [NH3] = 5.0 M and [Cu(NH3)42+] = 0.010 M: 0.010 = 1.3 × 1013 (1.2 10 −18 )(5.0) 4
K= 101.
The Ksp reaction is Cd(s) ⇌ Cd2+(aq) + S2−(aq) K = Ksp. Manipulate the given equations so that when added together we get the Ksp reaction. Then we can use the value of E ocell for the reaction to determine Ksp (by using the equation log K = nE/0.0591). CdS + 2 e− → Cd + S2− Cd → Cd2+ + 2 e−
E = −1.21 V −E = 0.402 V
CdS(s) → Cd2+(aq) + S2− (aq) log Ksp =
Al3+(aq) + 6 F−(aq) → AlF63−(aq) log K = 103.
Ksp = ?
2( −0.81) nE o = = −27.41, Ksp = 10−27.41 = 3.9 × 10−28 0.0591 0.0591
Al3+ + 3 e− → Al Al + 6 F− → AlF63- + 3 e−
102.
E ocell = −0.81 V
E° = −1.66 V −E° = 2.07 V
E ocell = 0.41 V
K=?
nE o 3(0.41) = = 20.81, K = 1020.81 = 6.5 × 1020 0.0591 0.0591
e− + AgI → Ag + I− Ag → Ag+ + e− AgI(s) → Ag+(aq) + I−(aq)
E oAgI = ? −E° = −0.80 V E ocell = E oAgI − 0.80, K = Ksp = 1.5 × 10−16
For this overall reaction:
E ocell =
0.0591 0.0591 log(1.5 × 10−16) = −0.94 V log K sp = n 1
= −0.94 + 0.80 = −0.14 V E cell = − 0.94 V = E AgI − 0.80 V, E AgI
104.
CuI + e− → Cu + I− Cu → Cu+ + e− CuI(s) → Cu+(aq) + I−(aq)
E oCuI = ? −E° = −0.52 V E ocell = E oCuI − 0.52 V
For this overall reaction, K = Ksp = 1.1 × 10−12:
E ocell =
0.0591 0.0591 log Ksp = log(1.1 10 −12 ) = −0.71 V 1 n
E ocell = −0.71 V = E oCuI − 0.52, E oCuI = −0.19 V
CHAPTER 18
ELECTROCHEMISTRY
923
Electrolysis 105.
a. Al3+ + 3 e− → Al; 3 mol e− are needed to produce 1 mol Al from Al3+. 1.0 × 103 g Al ×
b. 1.0 g Ni ×
1 mol Ni 2 mol e − 96,485 C 1s = 33 s − 58.69 g Ni mol Ni 100 .0 C mol e
c. 5.0 mol Ag × 106.
1 mol e − 96,485 C 1s = 4.8 × 103 s = 1.3 hours − mol Ag 100 .0 C mol e
The oxidation state of bismuth in BiO+ is +3 because oxygen has a −2 oxidation state in this ion. Therefore, 3 moles of electrons are required to reduce the bismuth in BiO + to Bi(s). 10.0 g Bi ×
107.
1 mol Al 3 mol e − 96,485 C 1 s = 1.07 × 105 s 26.98 g Al mol Al mol e − 100 .0 C = 30. hours
15 A =
1 mol Bi 3 mol e − 96,485 C 1s = 554 s = 9.23 min − 209.0 g Bi mol Bi 25.0 C mol e
15 C 60 s 60 min = 5.4 × 104 C of charge passed in 1 hour s min h
a. 5.4 × 104 C ×
1 mol e − 1 mol Co 58.93 g Co = 16 g Co − 96,485 C mol Co 2 mol e
b. 5.4 × 104 C ×
1 mol e − 1 mol Hf 178.5 g Hf = 25 g Hf − 96,485 C mol Hf 4 mol e
c. 2 I− → I2 + 2 e−; 5.4 × 104 C ×
1 mol I 2 253 .8 g I 2 1 mol e − = 71 g I2 − 96,485 C 2 mol e mol I 2
d. CrO3(l) → Cr6+ + 3 O2−; 6 mol e− are needed to produce 1 mol Cr from molten CrO 3. 5.4 × 104 C ×
1 mol e − 1 mol Cr 52.00 g Cr = 4.9 g Cr 96,485 C 6 mol e − mol Cr
60 min 60 s 2.90 C 1 mol e− 1 mol Zr 91.22 g Zr × × = 9.18 g Zr × × h min s 96,485 C 4 mol e− mol Zr
108.
3.72 h ×
109.
Al is in the +3 oxidation in Al2O3, so 3 mol e− are needed to convert Al3+ into Al(s). 2.00 h ×
110.
60 min 60 s 1.00 10 6 C 1 mol e − 1 mol Al 26.98 g Al = 6.71 × 105 g h min s 96,485 C 3 mol e − mol Al
Au is in the +3 oxidation state in HAuCl4. To form Au, 3 mol of electrons are transferred. 0.50 g Au ×
1 mol Au 3 mol e− 96, 485 C 1s 1h = 2.0 hours − 197.0 g 1 mol Au mol e 0.10 C 3600 s
924 111.
CHAPTER 18
From the MCl2 formula, M forms +2 ions. Two mol of e− are transferred to form the metal M. Moles of M = 748 s Molar mass of M =
112.
5.00 C 1 mol e − 1 mol M 1 mol e − = 1.94 × 10−2 mol M s 96,485 C 2 mol e − 96,485 C
0.471 g M = 24.3 g/mol; M is magnesium, Mg. 1.94 10 − 2 mol M
M3+ + 3 e− → M; 1397 s Molar mass =
113.
ELECTROCHEMISTRY
6.50 C 1 mol e − 1 mol M = 3.14 × 10−2 mol M s 96,485 C 3 mol e −
1.41 g M 3.14 10
−2
mol M
=
44.9 g ; the element is scandium, Sc. mol
60 s 2.50 C 1 mol e − = 7.77 × 10−2 mol e− min s 96,485 C 1 mol Ru Moles of Ru = 2.618 g Ru × = 2.590 × 10−2 mol Ru 101 .1 g Ru
Moles of e− = 50.0 min
Moles of e − 7.77 10 −2 mol e − = 3.00; the charge on the ruthenium ions is +3. = Moles of Ru 2.590 10 − 2 mol Ru
(Ru3+ + 3 e− → Ru) 114.
3600 s 2.0 C 1 mol e= 7.5 × 10−2 mol e− × × h s 96,485 C 1 mol Mn Moles of Mn = 1.02 g Mn × = 0.0186 mol Mn 54.94 g Mn Moles of e− = 1.0 h ×
Moles of e− 7.5 10−2 mol e− = 4.0 = Moles of Mn 0.0186 mol Mn The charge on the Mn ions is +4, so MnCl4 is the formula. 115.
F2 is produced at the anode: 2 F− → F2 + 2 e− 2.00 h ×
60 min 60 s 10.0 C 1 mol e − = 0.746 mol e− h min s 96,485 C
0.746 mol e− × V=
1 mol F2 2 mol e
−
= 0.373 mol F2; PV = nRT, V =
nRT P
(0.373 mol)( 0.08206 L atm/K • mol) (298 K) = 9.12 L F2 1.00 atm
K is produced at the cathode: K+ + e− → K 0.746 mol e− ×
1 mol K 39.10 g K = 29.2 g K − mol K mol e
CHAPTER 18 116.
ELECTROCHEMISTRY
925
The half-reactions for the electrolysis of water are: (2 e− + 2 H2O → H2 + 2 OH−) × 2 2 H2O → 4 H+ + O2 + 4 e− 2 H2O(l) → 2 H2(g) + O2(g)
15.0 min
Note: 4 H+ + 4 OH− react to give 4 H2O, and n = 4 for this reaction.
2 mol H 2 60 s 2.50 C 1 mol e − = 1.17 × 10-2 mol H2 min s 96,485 C 4 mol e −
At STP, 1 mol of an ideal gas occupies a volume of 22.42 L (see Chapter 5 of the text).
117.
1.17 × 10 −2 mol H2 ×
22.42 L = 0.262 L = 262 mL H2 mol H 2
1.17 × 10 −2 mol H2 ×
1 mol O 2 22.42 L = 0.131 L = 131 mL O2 2 mol H 2 mol O 2
Al3+ + 3 e− → Al; 3 mol e− are needed to produce Al from Al3+ 2000 lb Al
453 .6 g 1 mol Al 3 mol e − 96,485 C = 1 × 1010 C of electricity needed lb 26.98 g mol Al mol e −
1 1010 C 1h 1 min = 1 × 105 C/s = 1 × 105 A 24 h 60 min 60 s 118.
Barium is in the +2 oxidation state in BaCl2. Ba2+ + 2e− → Ba 1.00 10 6 g Ba
1 mol Ba 2 mol e − 96,485 C = 1.41 10 9 C of electricit y needed − 137.3 g mol Ba mol e
1.41 10 9 C 1h = 9.79 10 4 A 4.00 h 3600 s
119.
2.30 min ×
60 s 2.00 C 1 mol e − 1 mol Ag = 138 s; 138 s = 2.86 × 10−3 mol Ag − min s 96,485 C mol e
[Ag+] = 2.86 × 10 −3 mol Ag+/0.250 L = 1.14 × 10−2 M 120.
Cu2+ + 2 e‒ → Cu; 1380 s ×
[Cu2+] = 121.
2.00 C 1 mol e1 mol Cu 2+ = 1.43 × 10−2 mol Cu2+ × × s 96,485 C 2 mol e-
0.0143 mol Cu 2+ = 0.572 M 0.0250 L
Au3+ + 3 e− → Au
E° = 1.50 V
Ni2+ + 2 e− → Ni
E° = −0.23 V
Ag+ + e− → Ag
E° = 0.80 V
Cd2+ + 2 e− → Cd
E° = −0.40 V
2 H2O + 2e− → H2 + 2 OH− E° = −0.83 V (Water can also be reduced.) Au(s) will plate out first since it has the most positive reduction potential, followed by Ag(s), which is followed by Ni(s), and finally Cd(s) will plate out last since it has the most negative reduction potential of the metals listed. Water will not interfere with the plating process.
926 122.
CHAPTER 18
ELECTROCHEMISTRY
The reduction reactions and the cell potentials are: Cu2+ + 2 e− → Cu E = E = 0.34 V (standard conditions) Ag+ + e− → Ag
E = E −
0.0591 1 1 = 0.56 V log = 0.80 − 0.0591 log + n [Ag ] (1.0 10 − 4 )
It is easier to plate out Ag from 1.0 10−4 M Ag+ since this process has a more positive reduction potential than Cu2+ to Cu. 123.
The species present in solution are Ca2+, I‒, and H2O. Ca2+ and H2O can be reduced, while I‒ and H2O can be oxidized. For the cathode reaction, H2O has a more positive reduction potential than Ca2+ (‒0.83 V vs. ‒2.37 V), so the cathode reaction will be the reduction of water. For the anode reaction, I‒ has a more positive oxidation potential than H2O (‒0.54 V vs. ‒1.23 V), so the anode reaction will be the oxidation of I‒. The reactions that will occur are: Cathode: 2 H2O + 2 e− → H2 + 2 OH‒; anode: 2 I− → I2 + 2 e− Answer c is correct as I2 will be produced at the anode and H2 and OH‒ will be produced at the cathode.
124.
The species present in solution are Ni2+, F‒, and H2O. Ni2+ and H2O can be reduced, while F‒ and H2O can be oxidized. For the cathode reaction, Ni2+ has a more positive reduction potential than H2O (‒0.23 V vs. ‒0.83 V), so the cathode reaction will be the reduction of Ni2+. For the anode reaction, H2O has a more positive oxidation potential than F‒ (‒1.23 V vs. ‒2.87 V), so the anode reaction will be the oxidation of H2O. The reactions that will occur are: Cathode: Ni2+ + 2 e− → Ni ; anode: anode: 2 H2O → O2(g) + 4 H+ + 4 e− Answer d is correct as O2 will be produced at the anode. Ni will be produced at the cathode.
125.
Species present: Na+, SO42−, and H2O. H2O and Na+ can be reduced and H2O and SO42− can be oxidized. From the potentials, H2O is the most easily reduced and the most easily oxidized species present. This is the case because water, of the species present, has the most positive reduction potential as well as the most positive oxidation potential. The reactions are: Cathode:
126.
2 H2O + 2 e− → H2(g) + 2 OH−; anode: 2 H2O → O2(g) + 4 H+ + 4 e−
a. The spoon is where Cu2+ is reduced to Cu, so the spoon will be the cathode. The anode will be the copper strip where Cu is oxidized to Cu 2+. b. Cathode reaction: Cu2+ + 2 e− → Cu; anode reaction: Cu → Cu2+ + 2 e−
127.
Reduction occurs at the cathode, and oxidation occurs at the anode. First, determine all the species present; then look up pertinent reduction and/or oxidation potentials in Table 18.1 for all these species. The cathode reaction will be the reaction with the most positive reduction potential, and the anode reaction will be the reaction with the most positive oxidation potential. a.
Species present: Ni2+ and Br−; Ni2+ can be reduced to Ni, and Br− can be oxidized to Br2 (from Table 18.1). The reactions are:
CHAPTER 18
ELECTROCHEMISTRY Cathode: Ni2+ + 2e− → Ni Anode:
b.
−E° = −1.09 V
Cathode: Al3+ + 3 e− → Al
E° = −1.66 V
2 F− → F2 + 2 e−
−E° = −2.87 V
Species present: Mn2+ and I−; Mn2+ can be reduced, and I− can be oxidized. The reactions are: Cathode: Mn2+ + 2 e− → Mn Anode:
128.
E° = −0.23 V
Species present: Al3+ and F−; Al3+ can be reduced, and F− can be oxidized. The reactions are:
Anode: c.
2 Br− → Br2 + 2 e−
927
2 I− → I2 + 2 e−
E° = −1.18 V −E° = −0.54 V
Reduction occurs at the cathode, and oxidation occurs at the anode. First, determine all the species present; then look up pertinent reduction and/or oxidation potentials in Table 18.1 for all these species. The cathode reaction will be the reaction with the most positive reduction potential, and the anode reaction will be the reaction with the most positive oxidation potential. a. Species present: K+ and F−; K+ can be reduced to K, and F− can be oxidized to F2 (from Table 18.1). The reactions are: Cathode: K+ + e− → K
E° = −2.92 V
Anode: 2 F− → F2 + 2 e−
−E° = −2.87 V
b. Species present: Cu2+ and Cl-; Cu2+ can be reduced, and Cl− can be oxidized. The reactions are: Cathode: Cu2+ + 2 e− → Cu
E° = 0.34 V
Anode: 2 Cl− → Cl2 + 2 e−
−E° = −1.36 V
c. Species present: Mg2+ and I−; Mg2+ can be reduced, and I− can be oxidized. The reactions are:
129.
Cathode: Mg2+ + 2 e− → Mg
E° = −2.37 V
Anode: 2 I− → I2 + 2 e−
−E° = −0.54 V
These are all in aqueous solutions, so we must also consider the reduction and oxidation of H2O in addition to the potential redox reactions of the ions present. For the cathode reaction, the species with the most positive reduction potential will be reduced, and for the anode reaction, the species with the most positive oxidation potential will be oxidized. a. Species present: Ni2+, Br−, and H2O. Possible cathode reactions are: Ni2+ + 2e− → Ni 2 H2O + 2 e− → H2 + 2 OH−
E° = −0.23 V E° = −0.83 V
928
CHAPTER 18
ELECTROCHEMISTRY
Because it is easier to reduce Ni2+ than H2O (assuming standard conditions), Ni2+ will be reduced by the preceding cathode reaction. Possible anode reactions are: 2 Br− → Br2 + 2 e−
−E° = −1.09 V
2 H2O → O2 + 4 H+ + 4 e−
−E° = −1.23 V
Because Br− is easier to oxidize than H2O (assuming standard conditions), Br− will be oxidized by the preceding anode reaction. b. Species present: Al3+, F−, and H2O; Al3+ and H2O can be reduced. The reduction potentials are E° = −1.66 V for Al3+ and E° = −0.83 V for H2O (assuming standard conditions). H2O will be reduced at the cathode (2 H2O + 2 e− → H2 + 2 OH−). F− and H2O can be oxidized. The oxidation potentials are −E° = −2.87 V for F− and −E° = −1.23 V for H2O (assuming standard conditions). From the potentials, we would predict H2O to be oxidized at the anode (2 H2O → O2 + 4 H+ + 4 e−). c. Species present: Mn2+, I−, and H2O; Mn2+ and H2O can be reduced. The possible cathode reactions are: Mn2+ + 2 e−→ Mn
E° = −1.18 V
2 H2O + 2 e− → H2 + 2 OH−
E° = −0.83 V
Reduction of H2O will occur at the cathode since E oH 2O is most positive. I− and H2O can be oxidized. The possible anode reactions are: 2 I− → I2 + 2 e− 2 H2O → O2 + 4 H+ + 4 e−
−E° = −0.54 V −E° = −1.23 V
Oxidation of I− will occur at the anode since − E oI− is most positive. 130.
These are all in aqueous solutions, so we must also consider the reduction and oxidation of H2O in addition to the potential redox reactions of the ions present. For the cathode reaction, the species with the most positive reduction potential will be reduced, and for the anode reaction, the species with the most positive oxidation potential will be oxidized. a. Species present: K+, F−, and H2O. Possible cathode reactions are: K+ + e− → K 2 H2O + 2 e− → H2 + 2 OH−
E° = −2.92 V E° = −0.83 V
Because it is easier to reduce H2O than K+ (assuming standard conditions), H2O will be reduced by the preceding cathode reaction. Possible anode reactions are:
CHAPTER 18
ELECTROCHEMISTRY 2 F− → F2 + 2 e−
929 −E° = −2.87 V
2 H2O → 4 H+ + O2 + 4 e−
−E° = −1.23 V
Because H2O is easier to oxidize than F− (assuming standard conditions), H2O will be oxidized by the preceding anode reaction. b. Species present: Cu2+, Cl−, and H2O; Cu2+ and H2O can be reduced. The reduction potentials are E° = 0.34 V for Cu 2+ and E° = −0.83 V for H2O (assuming standard conditions). Cu2+ will be reduced to Cu at the cathode (Cu 2+ + 2 e− → Cu). Cl− and H2O can be oxidized. The oxidation potentials are −E° = −1.36 V for Cl− and −E° = −1.23 V for H2O (assuming standard conditions). From the potentials, we would predict H2O to be oxidized at the anode (2 H 2O → 4 H+ + O2 + 4 e−). Note: In real life, Cl− is oxidized to Cl2 when water is present due to a phenomenon called overvoltage (see Section 18.8 of the text). Because overvoltage is difficult to predict, we will generally ignore it when solving problems like these. c. Species present: Mg2+, I−, and H2O: The only possible cathode reactions are: 2 H2O + 2 e− → H2 + 2 OH−
E° = −0.83 V
Mg + 2 e → Mg
E° = −2.37 V
2+
−
Reduction of H2O will occur at the cathode since E oH 2O is more positive. The only possible anode reactions are: 2 I− → I2 + 2 e−
−E° = −0.54 V
2 H2O → O2 + 4 H+ + 4 e−
−E° = −1.23 V
Oxidation of I− will occur at the anode because − E oI− is more positive.
ChemWork Problems 131.
The half-reaction for the SCE is: Hg2Cl2 + 2 e− → 2 Hg + 2 Cl−
ESCE = 0.242 V
For a spontaneous reaction to occur, Ecell must be positive. Using the standard reduction potentials in Table 18.1 and the given the SCE potential, deduce which combination will produce a positive overall cell potential. a. Cu2+ + 2 e− → Cu
E° = 0.34 V
Ecell = 0.34 − 0.242 = 0.10 V; SCE is the anode. b. Fe3+ + e− → Fe2+
E° = 0.77 V
Ecell = 0.77 − 0.242 = 0.53 V; SCE is the anode.
930
CHAPTER 18
ELECTROCHEMISTRY
c. AgCl + e− → Ag + Cl− E° = 0.22 V Ecell = 0.242 − 0.22 = 0.02 V; SCE is the cathode. d. Al3+ + 3 e− → Al
E° = −1.66 V
Ecell = 0.242 + 1.66 = 1.90 V; SCE is the cathode. e. Ni2+ + 2 e− → Ni
E° = −0.23 V
Ecell = 0.242 + 0.23 = 0.47 V; SCE is the cathode. 132.
The potential oxidizing agents are NO3− and H+. Hydrogen ion cannot oxidize Pt under either condition. Nitrate cannot oxidize Pt unless there is Cl- in the solution. Aqua regia has both Cl− and NO3−. The overall reaction is: (NO3− + 4 H+ + 3 e− → NO + 2 H2O) × 2 (4 Cl− + Pt → PtCl42− + 2 e−) × 3
E° = 0.96 V −E° = −0.73 V
12 Cl−(aq) + 3 Pt(s) + 2 NO3−(aq) + 8 H+(aq) → 3 PtCl42−(aq) + 2 NO(g) + 4 H2O(l) E ocell = 0.23 V 133.
Ag+(aq) + Cu(s) → Cu2+(aq) + 2 Ag(s) E ocell = 0.80 − 0.34 V = 0.46 V; a galvanic cell produces a voltage as the forward reaction occurs. Any stress that increases the tendency of the forward reaction to occur will increase the cell potential, whereas a stress that decreases the tendency of the forward reaction to occur will decrease the cell potential. a. Added Cu2+ (a product ion) will decrease the tendency of the forward reaction to occur, which will decrease the cell potential. b. Added NH3 removes Cu2+ in the form of Cu(NH3)42+. Because a product ion is removed, this will increase the tendency of the forward reaction to occur, which will increase the cell potential. c. Added Cl− removes Ag+ in the form of AgCl(s). Because a reactant ion is removed, this will decrease the tendency of the forward reaction to occur, which will decrease the cell potential. d. Q1 =
Q2 =
[Cu 2+ ]0 [Ag + ]02
; as the volume of solution is doubled, each concentration is halved.
1/2 [Cu 2 + ] 0 (1/2 [Ag + ] 0 ) 2
=
2[Cu 2+ ] 0 [Ag + ] 02
= 2Ql
The reaction quotient is doubled because the concentrations are halved. Because reactions are spontaneous when Q < K, and because Q increases when the solution volume doubles, the reaction is closer to equilibrium, which will decrease the cell potential.
CHAPTER 18
ELECTROCHEMISTRY
931
e. Because Ag(s) is not a reactant in this spontaneous reaction, and because solids do not appear in the reaction quotient expressions, replacing the silver electrode with a platinum electrode will have no effect on the cell potential. (Al3+ + 3 e− → Al) × 2 (M → M2+ + 2 e−) × 3
134.
E° = −1.66 V −E° = ?
3 M(s) + 2 Al3+(aq) → 2 Al(s) + 3 M2+(aq)
E ocell = −E° − 1.66 V
ΔG° = −nFE°, −411 × 103 J = −(6 mol e−)(96,485 C/mol e−) E ocell , E ocell = 0.71 V
E ocell = −E° − 1.66 V = 0.71 V, −E° = 2.37 or E° = −2.37 From Table 18.1, the reduction potential for Mg2+ + 2 e− → Mg is −2.37 V, which fits the data. Hence, the metal is magnesium. 135.
(Fe2+ + 2 e− → Fe) × 3 (La → La3+ + 3 e−) × 2
a.
3 Fe2+(aq) + 2 La(s) → 3 Fe(s) + 2 La3+(aq)
E° = −0.44 V −E° = 2.37 V
E ocell = 1.93 V
The standard cell potential would be 1.93 V. b. The oxidizing agent is Fe2+. c. The anode would be composed of a La electrode and 1.0 M La3+. d. The electrons flow from the La/La3+ compartment to the Fe2+/Fe compartment (from the anode to the cathode.) e. Six electrons are transferred per unit of cell reaction. f.
E = E° −
0.0591 0.0591 [La 3+ ] 2 log Q , Ecell = 1.93 V − log n 6 [Fe 2+ ]3
Ecell = 1.93 −
(3.00 10 −3 ) 2 0.0591 , Ecell = 1.93 – 0.0596 = 1.87 V log 6 (2.00 10 − 4 ) 3
(M4+ + 4 e− → M) × 3 (N → N3+ + 3 e−) × 4
E° = 0.66 V −E° = −0.39V
3 M4+(aq) + 4 N(s) → 3 M(s) + 4 N3+(aq)
E ocell = 0.27 V
136.
ΔG° = − nFE ocell = − (12 mol e−)(96,485 C/mol e−)(0.27 J/C) = −3.1 × 105 J = −310 kJ E ocell =
137.
a.
0.0591 12(0.27 ) nE o = log K, log K = = 54.82, K = 1054.82 = 6.6 × 1054 n 0.0591 0.0591
(Au3+ + 3 e− → Au) × 2 (Mg → Mg2+ + 2 e−) × 3 2 Au3+(aq) + 3 Mg(s) → 2 Au(s) + 3 Mg2+(aq)
E° = 1.50 V −E° = 2.37 V
E ocell = 3.87 V
932
CHAPTER 18 b. Ecell = 3.87 V − (1.0 10 −5 ) 3 [Au 3+ ] 2
138.
a.
−5 3
(1.0 10 ) 0.0591 [Mg 2+ ]3 0.0591 log log , 4.01 = 3.87 − 3+ 2 6 6 [Au 3+ ] 2 [Au ]
= 10−14.21, [Au3+] = 0.40 M
(In+ + e− → In) × 2 In+ → In3+ + 2 e−
E° = −0.126 V −E° = 0.444 V
3 In+(aq) → In3+(aq) + 2 In(s)
E ocell = 0.318
log K =
ELECTROCHEMISTRY
nE o 2(0.318 ) = = 10.761, K = 1010.761 = 5.77 × 1010 0.0591 0.0591
b. G° = −nFE° = −(2 mol e−)(96,485 C/mol e−)(0.318 J/C) = −6.14 × 105 J = −61.4 kJ
ΔG orxn = −61.4 kJ = [2(0) + 1(−97.9 kJ] – 3 ΔG of , In + , ΔG of , In + = −12.2 kJ/mol 139.
a. ΔG° = n p ΔGof, products − n r ΔGof, reactants = 2(−480.) + 3(86) − [3(−40.)] = −582 kJ From oxidation numbers, n = 6. ΔG° = −nFE°, E° = log K =
− (−582,000 J) − ΔG o = 1.01 V = nF 6(96,485) C
nE o 6(1.01) = 102.538, K = 10102.538 = 3.45 × 10102 = 0.0591 0.0591
2 e− + Ag2S → 2 Ag + S2−) × 3 (Al → Al3+ + 3 e−) × 2
b.
E oAg2S = ?
−E° = 1.66 V
3 Ag2S(s) + 2 Al(s) → 6 Ag(s) + 3 S2−(aq) + 2 Al3+(aq)
E ocell = 1.01 V = E oAg2S + 1.66V
E oAg2S = 1.01 V − 1.66 V = −0.65 V 140.
E ocell =
0.0591 0.0591 log(8.23 10 4 ) = 0.291 V log K = 1 n
Au+ + e− → Au Au + 2 Cl− → AuCl2‒ + e−
E = 1.69 V −Eº = ?
Au+(aq) + 2 Cl−(aq) ⇌ AuCl2−(aq)
E ocell = 0.291 V
E ocell = 0.291 = 1.69 ‒ Eº, Eº = 1.40 V AuCl2−(aq) + e− → Au(s) + 2 Cl−(aq) 141.
E = 1.40 V
0.0591 log Q n 0.0591 0.120 0.180 = 0.160 − log(9.32 × 10−3) 0.020 = , n=6 n n
E ocell = 0.400 V – 0.240 V = 0.160 V; E = E° −
Six moles of electrons are transferred in the overall balanced reaction. We now must figure out how to get 6 mol e− into the overall balanced equation. The two possibilities are to have ion charges of +1 and +6 or +2 and +3; only these two combinations yield a 6 when common
CHAPTER 18
ELECTROCHEMISTRY
933
multiples are determined when adding the reduction half-reaction to the oxidation half-reaction. Because N forms +2 charged ions, M must form for +3 charged ions. The overall cell reaction can now be determined. (M3+ + 3 e− → M) × 2 (N → N2+ + 2 e−) × 3
E°= 0.400 V − E°= −0.240 V
2 M3+(aq) + 3 N(s) → 3 N2+(aq) + 2 M(s) −3
Q = 9.32 × 10 =
[ N 2 + ]0 [ M 3+ ]0
3
= 2
E ocell = 0.160 V
(0.10) 3 , [M3+] = 0.33 M [ M 3+ ] 2
wmax = G = −nFE = −6(96,485)(0.180) = −1.04 × 105 J = −104 kJ The maximum amount of work this cell could produce is 104 kJ. 142.
2 Ag+(aq) + Cu(s) → Cu2+(aq) + 2 Ag(s) E ocell = 0.80 − 0.34 = 0.46 V and n = 2 Because [Ag+] = 1.0 M, Ecell = 0.46 V −
0.0591 log [Cu2+]. 2
Use the equilibrium reaction to calculate the Cu 2+ concentration in the cell. Cu (aq) + 4 NH3(aq) ⇌ 2+
Cu(NH3)42+(aq)
K=
[Cu(NH 3 ) 24+ ] [Cu 2 + ][NH 3 ] 4
= 1.0 × 1013
From the problem, [NH3] = 10.00 M and [Cu(NH3)42+] = 0.0024 M: 1.0 × 1013 = Ecell = 0.46 − 143.
0.0024 , [Cu2+] = 2.4 × 10−20 M [Cu ](10.00) 4 2+
0.0591 log (2.4 × 10−20) = 0.46 − (−0.58) = 1.04 V 2
a. In a concentration cell, the anode always has the smaller ion concentration. So the cell with AgCl(s) at the bottom has the smaller [Ag +] and is the anode, while the compartment with [Ag+] = 1.0 M is the cathode. Electron flow is from the anode to the cathode. b. AgCl(s) ⇌ Ag+(aq) + Cl−(aq) Ksp = [Ag+][Cl−]; use the concentration cell potential of 0.58 V to calculate the silver ion concentration in the anode compartment. For a concentration cell, E = 0, and for Ag+/Ag half-reactions, n = 1. E = 0.58 V = 0 −
[Ag + ] anode 0.0591 log 1 [Ag + ] cathode
− 0.58 [Ag + ]anode = log , [Ag+]anode = 1.5 10−10 M 0.0591 1. 0
At the anode, we have [Ag+] = 1.5 10−10 M and [Cl−] = 1.0 M: Ksp = [Ag+][Cl−] = 1.5 10−10(1.0) = 1.5 10−10
934 144.
CHAPTER 18
ELECTROCHEMISTRY
Zn → Zn2+ + 2 e− −E° = 0.76 V; Fe → Fe2+ + 2 e− −E° = 0.44 V It is easier to oxidize Zn than Fe, so the Zn would be preferentially oxidized, protecting the iron of the Monitor's hull.
145.
146.
Aluminum forms a durable oxide coating over its surface. Once the HCl dissolves this oxide coating, Al is exposed to H+ and is easily oxidized to Al3+, i.e., the Al foil disappears after the oxide coating is dissolved. o = Pb(s) + 2 Ag+(aq) → 2 Ag(s) + Pb2+(aq); E cell
0.0591 log K, where n = 2 n
E ocell = E oAg+ → Ag + E oPb → Pb2+ = 0.80 V + 0.13 V = 0.93 V log K = 147.
2(0.93) nE o = = 31.47, K = 1031.47 = 3.0 1031 0.0591 0.0591
H2O2 as an oxidizing agent: H2O2 + 2 H+ + 2 e− → 2 H2O H2O2 as a reducing agent: H2O2 → O2 + 2 H+ + 2 e−
E° = 1.78 V
−E° = −0.68 V
From the more positive reduction potential, H2O2 is a better oxidizing agent than it is a reducing agent at standard conditions. 148.
Cu2+(aq) + H2(g) → 2 H+(aq) + Cu(s)
E ocell = 0.34 V − 0.00V = 0.34 V and n = 2
Since PH 2 = 1.0 atm and [H+] = 1.0 M: Ecell= E ocell −
0.0591 1 log 2 [Cu 2+ ]
Ecell = E ocell − (0.0591/2) log(1/[Cu2+]) = E ocell + 0.0296 log[Cu2+]; this equation is in the form of a straight-line equation, y = mx + b. A graph of Ecell versus log[Cu2+] will yield a straight line with slope equal to 0.0296 V or 29.6 mV. 149.
For C2H5OH, H has a +1 oxidation state, and O has a −2 oxidation state. This dictates a −2 oxidation state for C. For CO2, O has a −2 oxidation state, so carbon has a +4 oxidation state. Six moles of electrons are transferred per mole of carbon oxidized (C goes from −2 → +4). Two moles of carbon are in the balanced reaction, so n = 12. wmax = −1320 kJ = G = −nFE, −1320 × 103 J = −nFE = −(12 mol e−)(96,485 C/mol e−)E E = 1.14 J/C = 1.14 V
150.
2 H2(g) + O2(g) → 2 H2O(l); oxygen goes from the zero oxidation state to the −2 oxidation state in H2O. Because 2 mol of O are in the balanced reaction, n = 4 moles of electrons transferred. a.
E ocell =
0.0591 0.0591 log K = log(1.28 × 1083), E ocell = 1.23 V n 4
ΔG° = − nFE ocell = −(4 mol e−)(96,485 C/mol e−)(1.23 J/C) = −4.75 × 105 J = −475 kJ
CHAPTER 18
ELECTROCHEMISTRY
935
b. Because the moles of gas decrease as reactants are converted into products, ΔS° will be negative (unfavorable). Because the value of ΔG° is negative, ΔH° must be negative to override the unfavorable ΔS° (ΔG° = ΔH° − TΔS°). c. ΔG = wmax = ΔH − TΔS. Because ΔS is negative, as T increases, ΔG becomes more positive (closer to zero). Therefore, wmax will decrease as T increases. 151.
O2 + 2 H2O + 4 e− → 4 OH− (H2 + 2 OH− → 2 H2O + 2 e−) × 2 2 H2(g) + O2(g) → 2 H2O(l)
E° = 0.40 V −E° = 0.83 V
E ocell = 1.23 V = 1.23 J/C
Because standard conditions are assumed, wmax = ΔG° for 2 mol H2O produced. ΔG° = − nFE ocell = −(4 mol e−)(96,485 C/mol e−)(1.23 J/C) = −475,000 J = −475 kJ For 1.00 × 103 g H2O produced, wmax is: 1.00 × 103 g H2O ×
1 mol H 2 O − 475 kJ = −13,200 kJ = wmax 18.02 g H 2 O 2 mol H 2 O
The work done can be no larger than the free energy change. The best that could happen is that all of the free energy released would go into doing work, but this does not occur in any real process because there is always waste energy in a real process. Fuel cells are more efficient in converting chemical energy into electrical energy; they are also less massive. The major disadvantage is that they are expensive. In addition, H2(g) and O2(g) are an explosive mixture if ignited; much more so than fossil fuels. 152.
Cadmium goes from the zero oxidation state to the +2 oxidation state in Cd(OH) 2. Because 1 mol of Cd appears in the balanced reaction, n = 2 mol electrons transferred. At standard conditions: wmax = ΔG° = −nFE° wmax = − (2 mol e−)(96,485 C/mol e−)(1.10 J/C) = −2.12 × 105 J = −212 kJ
153.
(CO + O2− → CO2 + 2 e−) × 2 O2 + 4 e− → 2 O2− CO + O2 → 2 CO2 ΔG = −nFE, E =
154.
− (−380 10 3 J) − ΔG o = = 0.98 V nF (4 mol e − )(96,485 C/mol e − )
For a spontaneous process, Ecell > 0. In each electron transfer step, we need to couple a reduction half-reaction with an oxidation half-reaction. To determine the correct order, each step must have a positive cell potential to be spontaneous. The only possible order for spontaneous electron transfer is: Step 1:
cyt a(Fe3+) + e− → cyt a(Fe2+) E = 0.385 V cyt c(Fe2+) → cyt c(Fe3+) + e− −E = −0.254 V cyt a(Fe3+) + cyt c(Fe2+) → cyt a(Fe2+) + cyt c((Fe3+) Ecell = 0.131 V
936
CHAPTER 18 Step 2:
ELECTROCHEMISTRY
cyt c(Fe3+) + e− → cyt c(Fe2+) cyt b(Fe2+) → cyt b(Fe3+) + e−
E = 0.254 V −E = −0.030 V
cyt c(Fe3+) + cyt b(Fe2+) → cyt c(Fe2+) + cyt b((Fe3+)
Ecell = 0.224 V
Step 3 would involve the reduction half-reaction of cyt b(Fe3+) + e− → cyt b(Fe2+) coupled with some oxidation half-reaction. This is the only order that utilizes all three cytochromes and has each step with a positive cell potential. Therefore, electron transport through these cytochromes occurs from cytochrome a to cytochrome c to cytochrome b to some other substance and eventually to oxygen in O 2. 155.
150 . 10 3 g C 6 H 8 N 2 1 mol C 6 H 8 N 2 1h 1 min 2 mol e − 96,485 C h 60 min 60 s 108 .14 g C 6 H 8 N 2 mol C 6 H 8 N 2 mol e −
= 7.44 × 104 C/s, or a current of 7.44 × 104 A 156.
If the metal M forms 1+ ions, then the atomic mass of M would be: 1.25 C 1 mol e − 1 mol M = 1.94 × 10−3 mol M s 96,485 C 1 mol e −
mol M = 150. s Atomic mass of M =
0.109 g M 1.94 10 −3 mol M
= 56.2 g/mol
From the periodic table, the only metal with an atomic mass close to 56.2 g/mol is iron, but iron does not form stable 1+ ions. If M forms 2+ ions, then the atomic mass would be: mol M = 150. s Atomic mass of M =
1.25 C 1 mol e − 1 mol M = 9.72 × 10−4 mol M s 96,485 C 2 mol e − 0.109 g M
9.72 10 − 4 mol M
= 112 g/mol
Cadmium has an atomic mass of 112.4 g/mol and does form stable 2+ ions. Cd 2+ is a much more logical choice than Fe+. 157.
15000 J h 60 s 60 min = 5.4 × 107 J or 5.4 × 104 kJ (Hall-Heroult process) s min h 1 mol Al 10.7 kJ To melt 1.0 kg Al requires: 1.0 × 103 g Al × = 4.0 × 102 kJ 26.98 g mol Al
15 kWh =
It is feasible to recycle Al by melting the metal because, in theory, it takes less than 1% of the energy required to produce the same amount of Al by the Hall-Heroult process. 158.
In the electrolysis of aqueous sodium chloride, H2O is reduced in preference to Na+, and Cl− is oxidized in preference to H2O. The anode reaction is 2 Cl− → Cl2 + 2 e−, and the cathode reaction is 2 H2O + 2 e− → H2 + 2 OH−. The overall reaction is: 2 H2O(l) + 2 Cl−(aq) → Cl2(g) + H2(g) + 2 OH−(aq)
CHAPTER 18
ELECTROCHEMISTRY
937
From the 1 : 1 mol ratio between Cl2 and H2 in the overall balanced reaction, if 257 L of Cl2(g) is produced, then 257 L of H2(g) will also be produced because moles and volume of gas are directly proportional at constant T and P (see Chapter 5 of text). 159.
Pd is in the +2 oxidation state in PdCl2. Pd2+ + 2e− → Pd
0.1064 g Pd Current = 160.
1 mol Pd 2 mol e − 96,485 C = 193 .0 C of electricit y needed 106.4 g mol Pd mol e −
193 .0 C = 3.97 A 48.6 s
0.50 L × 0.010 mol Pt4+/L = 5.0 × 10−3 mol Pt4+ To plate out 99% of the Pt4+, we will produce 0.99 × 5.0 × 10−3 mol Pt. 0.99 × 5.0 × 10−3 mol Pt
161.
4 mol e − 96,485 C 1s = 480 s − mol Pt 4.00 C mol e
Chromium(III) nitrate [Cr(NO3)3] has chromium in the +3 oxidation state. 1.15 g Cr ×
1 mol Cr 3 mol e − 96 , 485 C = 6.40 × 103 C of charge − 52 .00 g mol Cr mol e
For the Os cell, 6.40 × 103 C of charge also was passed. 3.15 g Os
1 mol Os 1 mol e − = 0.0166 mol Os; 6.40 × 103 C × = 0.0663 mol e− 190 .2 g 96,485 C
0.0663 Mol e − = = 3.99 4 0.0166 Mol Os
This salt is composed of Os4+ and NO3− ions. The compound is Os(NO3)4, osmium(IV) nitrate. For the third cell, identify X by determining its molar mass. Two moles of electrons are transferred when X2+ is reduced to X. Molar mass =
2.11 g X = 63.6 g/mol 1 mol e − 1 mol X 3 6.40 10 C 96,485 C 2 mol e −
This is copper (Cu), which has an electron configuration of [Ar]4s 13d10.
Challenge Problems 162.
To begin plating out Pd: 4
(1.0) 0.0591 0.0591 [Cl − ] 4 log E = 0.62 − = 0.62 − log 2− 2 0.020 2 [PdCl 4 ]
938
CHAPTER 18
ELECTROCHEMISTRY
E = 0.62 V − 0.050 V = 0.57 V When 99% of Pd has plated out, [PdCl4−] = E = 0.62 −
0.020 = 0.00020 M. 100
(1.0) 4 0.0591 = 0.62 V − 0.11V = 0.51 V log 2 2.0 10 − 4
To begin Pt plating: E = 0.73 V −
(1.0) 4 0.0591 = 0.73 − 0.050 = 0.68 V log 2 0.020
When 99% of Pt plated: E = 0.73 − To begin Ir plating: E = 0.77 V −
(1.0) 4 0.0591 = 0.73 − 0.11 = 0.62 V log 2 2.0 10 − 4
(1.0) 4 0.0591 = 0.77 − 0.033 = 0.74 V log 3 0.020
When 99% of Ir plated: E = 0.77 − 0.0591 log 3
(1.0) 4 = 0.77 − 0.073 = 0.70 V 2.0 10 − 4
Yes, because the range of potentials for plating out each metal does not overlap, we should be able to separate the three metals. The exact potential to apply depends on the oxidation reaction. The order of plating will be Ir(s) first, followed by Pt(s), and finally, Pd(s) as the potential is gradually increased. 163.
a. 3 e− + 4 H+ + NO3− → NO + 2 H2O
E° = 0.96 V
Nitric acid can oxidize Co to Co2+ ( E ocell > 0), but is not strong enough to oxidize Co to Co3+ ( E ocell < 0). Co2+ is the primary product assuming standard conditions. b. Concentrated nitric acid is about 16 mol/L. [H+] = [NO3−] = 16 M; assume PNO = 1 atm. E = 0.96 V −
0.0591 1 P 0.0591 log log + 4 NO − = 0.96 − 3 (16) 5 3 [H ] [ NO3 ]
E = 0.96 − (−0.12) = 1.08 V ; no, concentrated nitric acid still will only be able to oxidize Co to Co2+. 164.
O2 + 2 H2O + 4 e- → 4 OH− (H2 + 2 OH- → 2 H2O + 2 e−) 2 2 H2(g) + O2(g) → 2 H2O(l)
E = 0.40 V −E = 0.83 V
E ocell = 1.23 V = 1.23 J/C
ΔG° = −nFE° = −(4 mol e−)(96,485 C/mol e−)(1.23 J/C) = −4.75 105 J = −475 kJ ΔH° = n p ΔHof, products − n r ΔHof, reactants ; ΔS° = n pSoproducts − n rSoreactants ΔH = [2 mol(−286 kJ/mol)] − [2(0) + 1(0)] = −572 kJ ΔS = [2 mol (70. J/K•mol)] − [2 mol (131 J/K•mol) + 1 mol (205 J/K•mol)] = −327 J/K At 0°C: ΔG° = ΔH° − TΔS°, ΔG° = −572 kJ − (273 K)(−0.327 kJ/K) = −483 kJ
CHAPTER 18
ELECTROCHEMISTRY
ΔG° = −nFE°, E =
939
− (−483 10 3 J) − ΔG o = = 1.25 J/C = 1.25 V nF (4 mol e − )(96,485 C/mol e − )
At 90.°C: ΔG° = ΔH° − TΔS°, ΔG° = −572 kJ − (363 K)(−0.327 kJ/K) = −453 kJ ΔG° = −nFE°, E =
165.
− (−453 10 3 J) − ΔG o = 1.17 J/C = 1.17 V = nF (4 mol e − )(96,485 C/mol e − )
ΔG° = −nFE° = ΔH° − TΔS°, E° =
TΔSo ΔH o − nF nF
If we graph E° versus T we should get a straight line (y = mx + b). The slope of the line is equal to ΔS°/nF, and the y intercept is equal to −ΔH°/nF. From the preceding equation, E° will have a small temperature dependence when ΔS° is close to zero. 166.
a. We can calculate ΔG° from ΔG° = ΔH° − TΔS° and then E° from ΔG° = −nFE°, or we can use the equation derived in the answer to Exercise 165. For this reaction, n = 2 (from oxidation states).
E o− 20 = b.
TS o − ΔH o (253 K)(263.5 J/K) − (−315.9 10 3 J) = = 1.98 J/C = 1.98 V nF (2 mol e − )(96,485 C/mol e − )
E − 20 = E o− 20 −
RT 1 RT ln Q = 1.98 V − ln + 2 − nF nF [H ] [HSO4 ]2
E−20 = 1.98 V −
(8.3145 J/ K • mol )(253 K) −
−
(2 mol e )(96,485 C/mol e )
ln
1 = 1.98 V − (−0.066 V) (4.5) (4.5) 2 2
E−20 = 2.05 V c. From Exercise 83, E = 2.12 V at 25C. As the temperature decreases, the cell potential decreases. Also, oil becomes more viscous at lower temperatures, which adds to the difficulty of starting an engine on a cold day. The combination of these two factors results in batteries failing more often on cold days than on warm days. (Ag+ + e− → Ag) × 2 Pb → Pb2+ + 2 e−
E° = 0.80 V −E° = −(−0.13)
2 Ag+(aq) + Pb(s) → 2 Ag(s) + Pb2+(aq)
E ocell = 0.93 V
167.
E = E° − log
0.0591 [Pb2+ ] (1.8) 0.0591 log , 0.83 V = 0.93 V − log + 2 n n [Ag + ] 2 [Ag ]
(1.8) 0.10(2) (1.8) = = 3.4, = 103.4, [Ag+] = 0.027 M + 2 0.0591 [ Ag + ]2 [Ag ]
Ag2SO4(s)
⇌
Initial s = solubility (mol/L) Equil.
2 Ag+(aq) + SO42−(aq) 0 2s
Ksp = [Ag+]2[SO42−]
0 s
From problem: 2s = 0.027 M, s = 0.027/2; Ksp = (2s)2(s) = (0.027)2(0.027/2) = 9.8 × 10−6
940 168.
CHAPTER 18
ELECTROCHEMISTRY
a. Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s) E ocell = 1.10 V Ecell = 1.10 V −
0.0591 [ Zn 2+ ] log 2 [Cu 2+ ]
Ecell = 1.10 V −
0.0591 0.10 log = 1.10 V − (−0.041 V) = 1.14 V 2 2.50
b. 10.0 h ×
60 min 60 s 10.0 C 1 mol e − 1 mol Cu = 1.87 mol Cu produced h min s 96,485 C 2 mol e −
The Cu2+ concentration decreases by 1.87 mol/L, and the Zn2+ concentration will increase by 1.87 mol/L. [Cu2+] = 2.50 − 1.87 = 0.63 M; [Zn2+] = 0.10 + 1.87 = 1.97 M Ecell = 1.10 V −
0.0591 1.97 log = 1.10 V − 0.015 V = 1.09 V 2 0.63
c. 1.87 mol Zn consumed ×
65.38 g Zn = 122 g Zn mol Zn
Mass of electrode = 200. − 122 = 78 g Zn 1.87 mol Cu formed ×
63 .55 g Cu = 119 g Cu mol Cu
Mass of electrode = 200. + 119 = 319 g Cu d. Three things could possibly cause this battery to go dead: (1) All the Zn is consumed. (2) All the Cu2+ is consumed. (3) Equilibrium is reached (Ecell = 0). We began with 2.50 mol Cu2+ and 200. g Zn × 1 mol Zn/65.38 g Zn = 3.06 mol Zn. Because there is a 1 : 1 mole relationship between Cu 2+ and Zn in the balanced cell equation, Cu2+ is the limiting reagent and will run out first. To react all the Cu 2+ requires: 2.50 mol Cu2+
2 mol e − 96,485 C 1s 1h = 13.4 h 2+ − 10.0 C 3600 s mol Cu mol e
For equilibrium to be reached: E = 0 = 1.10 V −
0.0591 [ Zn 2+ ] log 2 [Cu 2+ ]
[ Zn 2+ ] = K = 102(1.10)/0.0591 = 1.68 × 1037 [Cu 2+ ]
This is such a large equilibrium constant that virtually all the Cu 2+ must react to reach equilibrium. So, the battery will go dead in 13.4 hours.
CHAPTER 18 169.
ELECTROCHEMISTRY
941
2 H+ + 2 e− → H2 Fe → Fe2+ + 2e−
E = 0.000 V −E = −(−0.440V)
2 H+(aq) + Fe(s) → H2(g) + Fe3+(aq)
E ocell = 0.440 V
Ecell = E ocell −
P [Fe 3+ ] 0.0591 log Q, where n = 2 and Q = H 2 + 2 n [H ]
To determine Ka for the weak acid, first use the electrochemical data to determine the H + concentration in the half-cell containing the weak acid. 0.333 V = 0.440 V −
0.0591 1.00 atm(1.00 10 −3 M ) log 2 [ H + ]2
0.107 (2) 1.0 10 −3 1.0 10 −3 = 103.621 = 4.18 × 103, [H+] = 4.89 × 10−4 M = log , 0.0591 [ H + ]2 [ H + ]2
Now we can solve for the Ka value of the weak acid HA through the normal setup for a weak acid problem. HA(aq) Initial Equil. Ka = 170.
⇌
H+(aq)
1.00 M 1.00 − x
A−(aq)
+
~0 x
Ka =
[H + ][A − ] [HA]
0 x
(4.89 10 −4 ) 2 x2 , where x = [H+] = 4.89 × 10−4 M, Ka = = 2.39 × 10−7 −4 1.00 − x 1.00 − 4.89 10
a. Nonreactive anions are present in each half-cell to balance the cation charges. e-
eSalt Bridge
Anode
~~~~~~~ Pt ~~~~~~
~~~~~~~ Au ~~~~~~
1.0 M Fe2+ 1.0 M Fe3+
1.0 M Au 3+
b. Au3+(aq) + 3 Fe2+(aq) → 3 Fe3+(aq) + Au(s) Ecell = E ocell −
Cathode
E ocell = 1.50 − 0.77 = 0.73 V
0.0591 0.0591 [Fe 3+ ]3 log Q = 0.73 V − log n 3 [Au 3+ ][Fe 2+ ]3
Because [Fe3+] = [Fe2+] = 1.0 M: 0.31 V = 0.73 V −
0.0591 1 log 3 [Au 3+ ]
3( −0.42) 1 = − log , log [Au3+] = −21.32, [Au3+] = 10−21.32 = 4.8 × 10−22M 0.0591 [Au 3+ ]
942
CHAPTER 18
ELECTROCHEMISTRY
Au3+(aq) + 4 Cl−(aq) ⇌ AuCl4−(aq); because the equilibrium Au3+ concentration is so small, assume [AuCl4−] [Au3+]0 1.0 M, i.e., assume K is large, so the reaction essentially goes to completion. −
K= 171.
1.0 [AuCl4 ] = = 2.1 × 1025; assumption good (K is large). 3+ − 4 (4.8 10 − 22 ) (0.10) 4 [Au ][Cl ]
a. Ecell = Eref + 0.05916 pH, 0.480 V = 0.250 V + 0.05916 pH 0.480 − 0.250 = 3.888; uncertainty = ±1 mV = ± 0.001 V 0.05916 0.481 − 0.250 0.479 − 0.250 pHmax = = 3.905; pHmin = = 3.871 0.05916 0.05916
pH =
Thus, if the uncertainty in potential is ±0.001 V, then the uncertainty in pH is ±0.017, or about ±0.02 pH units. For this measurement, [H +] = 10−3.888 = 1.29 × 10−4 M. For an error of +1 mV, [H+] = 10−3.905 = 1.24 × 10−4 M. For an error of −1 mV, [H+] = 10−3.871 = 1.35 × 10−4 M. So the uncertainty in [H+] is ±0.06 × 10−4 M = ±6 × 10−6 M. b. From part a, we will be within ±0.02 pH units if we measure the potential to the nearest ±0.001 V (±1 mV). 172.
a. Ecell = E ocell −
1.0 10 −4 [Cu 2+ ] anode 0.0591 0.0591 log = 0 − log 1.00 , Ecell = 0.12 V n 2 [Cu 2+ ] cathode
b. Cu2+(aq) + 4 NH3(aq) ⇌ Cu(NH3)42+(aq) K = 1.0 1013; because K is so large, this reaction lies far to the right. Assume the reaction goes to completion, then solve an equilibrium problem for the new Cu2+ concentration at the anode.
Before Change After
Cu2+(aq)
+ 4 NH3(aq)
⇌
Cu(NH3)42+(aq)
1.0 × 10−4 M −1.0 × 10−4 0
2.0 M −4.0 × 10−4 2.0
→
0 +1.0 × 10−4 1.0 × 10−4
Now allow the reaction to get to equilibrium. Cu2+(aq) + 4 NH3(aq) Initial 0 Change +x Equil. x (1.0 10 −4 − x) x(2.0 + 4 x) 4
2.0 M +4x 2.0 + 4x
1.0 10 −4 x(2.0) 4
⇌
Cu(NH3)42+(aq)
1.0 × 10−4 M −x 1.0 × 10−4 − x
= 1.0 × 1013, x = [Cu2+] = 6.3 × 10−19 M; assumptions good.
In the concentration cell, [Cu2+]anode = 6.3 × 10−19 M. Solving for the cell potential: E=0−
6.3 10 −19 0.0591 , E = 0.54 V log 2 1.00
CHAPTER 18
ELECTROCHEMISTRY
173.
(Ag+ + e− → Ag) × 2 Cu → Cu2+ + 2 e−
E = 0.80 V −E = −0.34 V
2 Ag+(aq) + Cu(s) → 2 Ag(s) + Cu2+(aq)
E ocell = 0.46 V
a.
Ecell = E ocell −
943
[Cu 2+ ] 0.0591 log Q, where n = 2 and Q = . n [Ag+ ]2
To calculate Ecell, we need to use the Ksp data to determine [Ag+]. AgCl(s) Initial Equil.
⇌
s = solubility (mol/L)
Ag+(aq) 0 s
+
Cl−(aq)
Ksp = 1.6 × 10−10 = [Ag+][Cl−]
0 s
Ksp = 1.6 × 10−10 = s2, s = [Ag+] = 1.3 × 10−5 mol/L Ecell = 0.46 V −
0.0591 2. 0 log = 0.46 V − 0.30 = 0.16 V 2 (1.3 10 −5 ) 2
b. Cu2+(aq) + 4 NH3(aq) ⇌ Cu(NH4)42+(aq)
K = 1.0 × 1013 =
[Cu ( NH3 ) 24 + ] [Cu 2 + ][ NH3 ]4
Because K is very large for the formation of Cu(NH 3)42+, the forward reaction is dominant. At equilibrium, essentially all the 2.0 M Cu2+ will react to form 2.0 M Cu(NH3)42+. This reaction requires 8.0 M NH3 to react with all the Cu2+ in the balanced equation. Therefore, the moles of NH3 added to 1.0-L solution will be larger than 8.0 mol since some NH 3 must be present at equilibrium. To calculate how much NH3 is present at equilibrium, we need to use the electrochemical data to determine the Cu2+ concentration. Ecell = E ocell − log
0.0591 0.0591 [Cu 2+ ] log Q, 0.52 V = 0.46 V − log n 2 (1.3 10 −5 ) 2
[Cu 2+ ] [Cu 2+ ] − 0.06(2) = −2.03, = 10−2.03 = 9.3 × 10−3 = 0.0591 (1.3 10 −5 ) 2 (1.3 10 −5 ) 2
[Cu2+] = 1.6 × 10−12 = 2 × 10−12 M (We carried extra significant figures in the calculation.) Note: Our assumption that the 2.0 M Cu2+ essentially reacts to completion is excellent because only 2 × 10−12 M Cu2+ remains after this reaction. Now we can solve for the equilibrium [NH3]. K = 1.0 × 1013 =
[Cu ( NH3 ) 24 + ] ( 2 .0 ) = , [NH3] = 0.6 M ( 2 10 −12 ) [ NH 3 ]4 [Cu 2 + ][ NH3 ]4
Because 1.0 L of solution is present, 0.6 mol NH 3 remains at equilibrium. The total moles of NH3 added is 0.6 mol plus the 8.0 mol NH 3 necessary to form 2.0 M Cu(NH3)42+. Therefore, 8.0 + 0.6 = 8.6 mol NH3 was added.
944 174.
CHAPTER 18
ELECTROCHEMISTRY
Standard reduction potentials can only be manipulated and added together when electrons in the reduction half-reaction exactly cancel with the electrons in the oxidation half-reaction. We will solve this problem by applying the equation G° = −nFE° to the half-reactions. M3+ + 3 e− → M
G° = −nFE° = −3(96,485)(0.10) = −2.9 × 104 J
Because M and e− have ΔG of = 0: −2.9 × 104 J = − ΔG of , M3+ , ΔG of , M3+ = 2.9 × 104 J M2+ + 2 e− → M
G° = −nFE° = −2(96,485)(0.50) = −9.6 × 104 J
−9.6 × 104 J = − ΔG of , M 2+ , ΔGof, M2+ = 9.6 × 104 J M3+ + e− → M2+ E° =
175.
G° = 9.6 × 104 J – (2.9 × 104 J) = 6.7 × 104 J
− ΔG o − (6.7 10 4 ) = = −0.69 V; M3+ + e− → M2+ E = −0.69 V nF (1)(96,485)
2 Ag+(aq) + Ni(s) → Ni2+(aq) + Ag(s); the cell is dead at equilibrium (E = 0).
E ocell = 0.80 V + 0.23 V = 1.03 V 0 = 1.03 V −
0.0591 log K, K = 7.18 × 1034 2
K is very large. Let the forward reaction go to completion. 2 Ag+(aq) + Ni(s) → Before Change After
1.0 M −1.0 0
Ni2+(aq) + 2 Ag(s)
K = [Ni2+]/[Ag+]2 = 7.18 × 1034
1.0 M → +0.50 1.5 M
Now solve the back-equilibrium problem. 2 Ag+(aq) + Ni(s) ⇌ Ni2+(aq) + 2 Ag(s) Initial 0 1.5 M Change +2x −x Equil. 2x 1.5 − x K = 7.18 × 1034 =
1.5 − x 1 .5 ; solving, x = 2.3 10−18 M. Assumptions good. 2 (2 x) 2 (2 x)
[Ag+] = 2x = 4.6 × 10−18 M; [Ni2+] = 1.5 − 2.3 × 10−18 = 1.5 M 176.
a. Ag2CrO4(s) + 2 e− → 2 Ag(s) + CrO42−(aq) Hg2Cl2 + 2 e− → 2 Hg + 2 Cl−
E° = 0.446 V ESCE = 0.242 V
SCE will be the oxidation half-reaction with Ecell = 0.446 − 0.242 = 0.204 V. ΔG = −nFEcell = −2(96,485)(0.204)J = −3.94 × 104 J = −39.4 kJ
CHAPTER 18
ELECTROCHEMISTRY
945
b. In SCE, we assume all concentrations are constant. Therefore, only CrO 42− appears in the Q expression, and it will appear in the numerator since CrO 42− is produced in the reduction half-reaction. To calculate Ecell at nonstandard CrO42- concentrations, we use the following equation. 0.0591 0.0591 log[CrO42−] = 0.204 V − log[CrO42−] 2 2 0.0591 c. Ecell = 0.204 − log(1.00 × 10−5) = 0.204 V − (−0.148 V) = 0.352 V 2
Ecell = E ocell −
d. 0.504 V = 0.204 V − (0.0591/2)log[CrO42−] log[CrO42−] = −10.152, [CrO42−] = 10−10.152 = 7.05 × 10−11 M e.
Ag2CrO4 + 2 e− → 2 Ag + CrO42− (Ag → Ag+ + e−) × 2
E° = 0.446 V −E° = −0.80 V
Ag2CrO4(s) → 2 Ag+(aq) + CrO42−(aq)
E ocell =
E ocell = −0.35 V; K = Ksp = ?
(−0.35 V) (2) 0.0591 log Ksp, log Ksp = = −11.84, Ksp = 10−11.84 = 1.4 × 10−12 0.0591 n
(Ag+ + e− → Ag) × 2 Cd → Cd2+ + 2 e−
E° = 0.80 V −E° = 0.40 V
2 Ag+(aq) + Cd(s) → Cd2+(aq) + 2 Ag(s)
E ocell = 1.20 V
177.
Overall complex ion reaction: Ag+(aq) + 2 NH3(aq) → Ag(NH3)2+(aq)
K = K1K2 = 2.1 103(8.2 103) = 1.7 107
Because K is large, we will let the reaction go to completion and then solve the backequilibrium problem. Ag+(aq) Before After Change Equil.
+
1.00 M 0 x x
2 NH3(aq) 15.0 M 13.0 +2x 13.0 + 2x
⇌
Ag(NH3)2+(aq)
0 1.00 −x 1.00 − x
+
K=
[Ag( NH3 ) 2 ] 1.00 − x 1.00 ; 1.7 10 7 = + 2 2 [Ag ][ NH3 ] x(13.0 + 2 x) x(13.0) 2
Solving: x = 3.5 10−10 M = [Ag+]; assumptions good. E = E −
0.0591 1.0 0.0591 [Cd 2+ ] log = 1.20 V − log + 2 −10 2 2 2 [Ag ] (3.5 10 )
E = 1.20 − 0.56 = 0.64 V
K = 1.7 107 New initial
946 178.
CHAPTER 18 3 × (e− + 2 H+ + NO3− → NO2 + H2O) 2 H2O + NO → NO3− + 4 H+ + 3 e−
a.
ELECTROCHEMISTRY E°= 0.775 V −E°= −0.957 V
2 H+(aq) + 2 NO3−(aq) + NO(g) → 3 NO2(g) + H2O(l) log K =
E ocell = −0.182 V, K = ?
Eo 3( −0.182 ) = = −9.239, K = 10−9.239 = 5.77 × 10−10 0.0591 0.0591
b. Let C = concentration of HNO3 = [H+] = [NO3−]. 3 PNO 2
5.77 × 10−10 =
= − 2
PNO [H + ] 2 [NO3 ]
3 PNO 2
PNO C 4
If 0.20% NO2 by moles and Ptotal = 1.00 atm: PNO2 =
0.20 mol NO 2 × 1.00 atm = 2.0 × 10−3 atm; PNO = 1.00 − 0.0020 = 1.00 atm 100 . mol total
5.77 × 10−10 =
(2.0 10 −3 ) 3 , C = 1.9 M HNO3 (1.00)C 4
Marathon Problems 179.
a.
Cu2+ + 2 e− → Cu V → V2+ + 2 e−
E = 0.34 V −E = 1.20 V
Cu2+(aq) + V(s) → Cu(s) + V2+(aq)
E ocell = 1.54 V
Ecell = E ocell −
0.0591 [V 2+ ] [V 2+ ] log Q, where n = 2 and Q = . = n [Cu 2 + ] 1.00 M
To determine Ecell, we must know the initial [V2+], which can be determined from the stoichiometric point data. At the stoichiometric point, moles H2EDTA2− added = moles V2+ present initially. Mol V2+ present initially = 0.5000 L ×
0.0800 mol H 2 EDTA 2− 1 mol V 2+ L mol H 2 EDTA 2−
= 0.0400 mol V2+ 2+
0.0400 mol V = 0.0400 M 1.00 L 0.0591 0.0400 log Ecell = 1.54 V − = 1.54 V − (−0.0413) = 1.58 V 2 1.00
[V2+]0 =
b. Use the electrochemical data to solve for the equilibrium [V 2+]. Ecell = E ocell −
0.0591 [V 2 + ] log , n [Cu 2 + ]
1.98 V = 1.54 V −
[V2+] = 10−(0.44)(2)/0.0591 = 1.3 × 10−15 M
0.0591 [V 2+ ] log 2 1.00 M
CHAPTER 18
ELECTROCHEMISTRY
H2EDTA2−(aq) + V2+(aq) ⇌ VEDTA2−(aq) + 2 H+(aq)
947 K=
[VEDTA2− ][H + ]2 [H 2 EDTA2− ][V 2+ ]
In this titration reaction, equal moles of V2+ and H2EDTA2− are reacted at the stoichiometric point. Therefore, equal moles of both reactants must be present at equilibrium, so [H2EDTA2−] = [V2+] = 1.3 × 10−15 M. In addition, because [V2+] at equilibrium is very small compared to the initial 0.0400 M concentration, the reaction essentially goes to completion. The moles of VEDTA2− produced will equal the moles of V2+ reacted (= 0.0400 mol). At equilibrium, [VEDTA2−] = 0.0400 mol/(1.00 L + 0.5000 L) = 0.0267 M. Finally, because we have a buffer solution, the pH is assumed not to change, so [H +] = 10−10.00 = 1.0 × 10−10 M. Calculating K for the reaction: K=
(0.0267 )(1.0 10 −10 ) 2 [VEDTA2− ][H + ]2 = = 1.6 × 108 (1.3 10 −15 )(1.3 10 −15 ) [H 2 EDTA2− ][V 2+ ]
c. At the halfway point, 250.0 mL of H2EDTA2− has been added to 1.00 L of 0.0400 M V2+. Exactly one-half the 0.0400 mol of V2+ present initially has been converted into VEDTA 2−. Therefore, 0.0200 mol of V2+ remains in 1.00 + 0.2500 = 1.25 L solution. Ecell = 1.54 V −
0.0591 (0.0200 / 1.25) 0.0591 [V 2+ ] log = 1.54 − log 2 1.00 2 [Cu 2+ ]
Ecell = 1.54 − (−0.0531) = 1.59 V 180.
Begin by choosing any reduction potential as 0.00 V. For example, let’s assume B2+ + 2 e− → B
E° = 0.00 V
From the data, when B/B2+ and E/E2+ are together as a cell, E° = 0.81 V. E2+ + 2 e− → E must have a potential of –0.81 V or 0.81 V since E may be involved in either the reduction or the oxidation half-reaction. We will arbitrarily choose E to have a potential of −0.81 V. Setting the reduction potential at −0.81 for E and 0.00 for B, we get the following table of potentials. B2+ + 2 e− → B
0.00 V
E +2e →E
−0.81 V
D2+ + 2 e → D
0.19 V
C2+ + 2 e− → C
−0.94 V
A +2e →A
0.53 V
2+
−
−
2+
−
From largest to smallest: D2+ + 2 e− → D
0.19 V
B +2e →B
0.00 V
2+
−
A +2e →A
−0.53 V
E2+ + 2 e− → E
−0.81 V
C +2e →C
−0.94 V
2+
2+
−
−
948
CHAPTER 18
ELECTROCHEMISTRY
A2+ + 2 e− → A is in the middle. Let’s call this 0.00 V. The other potentials would be: D2+ + 2 e− → D 0.72 V B2+ + 2 e− → B 0.53 V A2+ + 2 e− → A 0.00 V − 2+ E + 2 e → E −0.28 V C2+ + 2 e− → C −0.41 V Of course, since the reduction potential of E could have been assumed to 0.81 V instead of −0.81 V, we can also get: C2+ + 2 e− → C
0.41 V
E +2e →E
0.28 V
A +2e →A
0.00 V
B +2e →B
−0.53 V
D2+ + 2 e → D
−0.72 V
2+
2+
2+
−
-−
−
−
One way to determine which table is correct is to add metal C to a solution with D 2+ present, as well as to add metal D to a different solution with C2+ present. If metal D forms in the first solution, then the first table is correct. If metal C forms in the second solution, then the second table is correct.
CHAPTER 19 THE NUCLEUS: A CHEMIST'S VIEW Review Questions 1.
a. Thermodynamic stability: The potential energy of a particular nucleus as compared to the sum of the potential energies of its component protons and neutrons. b. Kinetic stability: The probability that a nucleus will undergo decomposition to form a different nucleus. c. Radioactive decay: A spontaneous decomposition of a nucleus to form a different nucleus. d. Beta-particle production: A decay process for radioactive nuclides where an electron is produced; the mass number remains constant and the atomic number changes. e. Alpha-particle production: A common mode of decay for heavy radioactive nuclides, where a helium nucleus is produced causing the mass number to change. f.
Positron production: A mode of nuclear decay in which a particle is formed having the same mass as an electron but opposite in charge.
g. Electron capture: A process in which one of the inner-orbital electrons in an atom is captured by the nucleus. h. Gamma-ray emissions: The production of high-energy photons called gamma rays that frequently accompany nuclear decays and particle reactions. A is the mass number and is equal to the number of protons plus neutrons in a nuclei; the sum of all the mass number values must be the same on both sides of the equation (A is conserved). Z is the atomic number and is equal to the number of protons in a nucleus; the sum of the atomic number values must also be the same on both sides of the equation (Z is conserved). 2.
The zone of stability is the area encompassing the stable nuclides on a plot of their positions as a function of the number of protons and the number of neutrons in the nucleus. The neutron/proton ratio increases to a number greater than one as elements become heavier. Nuclides with too many neutrons undergo beta-particle production to decrease the neutron/proton ratio to a more stable value. Position production, electron capture and alphaparticle production all increase the neutron/proton ratio, so these occur for nuclides having too many protons.
3.
First-order kinetics is where there is a direct relationship between the rate of decay and the number of nuclides in a given sample. The rate law for all radioactive decay is rate = kN.
949
950
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THE NUCLEUS: A CHEMIST’S VIEW
Because of the direct relationship between rate and N, as the number of nuclides is halved, the rate is also halved. The first-order rate law and the integrated first-order rate law are: N = −kt rate = kN and ln No
k is the first-order rate constant, N is the number of nuclides present at some time t, No is the initial number of nuclides present at t = 0, and t is the time the decay process has been occurring. The half-life equation is: t1/2 = ln 2/k. The half-life for all radioactive decay is independent of the number of nuclides present; it is a constant. From the half-life equation, t1/2 is inversely related to the rate constant, k. As k increases, t1/2 decreases and vice versa. 4.
Nuclear transformation: The change of one element into another. Like all nuclear processes, the reaction must be mass number balanced and atomic number balanced. Particle accelerators are devices used to accelerate nuclear particles to very high speeds. Because of the electrostatic repulsion between the target nucleus and a positive ion, accelerators are needed for bombardment of like charged ions in order to overcome the electrostatic repulsion.
5.
Geiger counter: An instrument that measures the rate of radioactive decay based on the ions and electrons produced as a radioactive particle passes through a gas-filled chamber. The instrument takes advantage of the fact that the high-energy particles from radioactive decay processes produce ions when they travel through matter. The formation of ions and electrons by the passage of high-energy particles allows a momentary current to flow. Electronic devices detect the current flow and the number of these events can be counted. This gives the decay rate of the radioactive sample. Scintillation counter: An instrument that measures radioactive decay by sensing flashes of light produced in a substance by the radiation. Certain substances, such as zinc sulfide, give off light when they are struck by high energy radiation. A photocell senses the flashes of light which is a measure of the number of decay events per unit of time. Radiotracer: A radioactive nuclide, introduced into an organism for diagnostic purposes, whose pathway can be traced by monitoring its radioactivity. 14C and 31P work well as radiotracers because the molecules in the body contain carbon and/or phosphorus; they will be incorporated into the worker molecules of the body easily, which allows monitoring of the pathways of these worker molecules. 131I works well for thyroid problems because iodine concentrates in the thyroid. To study chemical equilibrium, a nonradioactive substance can be put in equilibrium with a radioactive substance. The two materials can then be checked to see if all the radioactivity remains in the original material or if it has been scrambled by the equilibrium. The scrambling of the radioactive substance is proof that equilibrium is dynamic.
6.
Plants take in CO2 in the photosynthesis process, which incorporates carbon, including 14C, into its molecules. If the plant is alive, the 14C/12C ratio in the plant will equal the ratio in the atmosphere. When the plant dies, 14C is not replenished as 14C decays by beta-particle production. By measuring the 14C activity today in the artifact and comparing this to the assumed 14C activity when the plant died to make the artifact, an age can be determined for the artifact. The assumptions are that the 14C level in the atmosphere is constant or that the 14C
CHAPTER 19
THE NUCLEUS: A CHEMIST’S VIEW
951
level at the time the plant died can be calculated. A constant 14C level is a pure assumption, and accounting for variation is complicated. Another problem is that some of the material must be destroyed to determine the 14C level. 238
U has a half-life of 4.5 × 109 years. In order to be useful, we need a significant number of decay events by 238U to have occurred. With the extremely long half-life of 238U, the period required for a significant number of decay events is on the order of 10 8 years. This is the time frame of when the earth was formed. 238U is worthless for aging 10,000 year-old objects or less because a measurable quantity of decay events has not occurred in 10,000 years or less. 14C is good at dating these objects because 14C has a half-life on the order of 103 years. 14C is worthless for dating ancient objects because of the relatively short half-life; no discernible amount of 14C will remain after 108 years. 7.
Mass defect: The change in mass occurring when a nucleus is formed from its component nucleons. Binding energy: The energy required to decompose a nucleus into its component nucleons. The mass defect is determined by summing the masses of the individual neutrons and protons that make up a nuclide and comparing this to the actual mass of the nuclide. The difference in mass is the mass defect. This quantity of mass determines the energy released when a nuclide is formed from its protons and neutrons. This energy is called the binding energy. The equation E = mc2 allows conversion of the mass defect into the binding energy. 56Fe, with the largest binding energy per nucleon of any nuclide, is the most stable nuclide. This is because when 56 Fe is formed from its protons and neutrons, it has the largest relative mass defect. Therefore, 56 Fe is the most stable nuclei because it would require the largest amount of energy per nucleon to decompose the nucleus.
8.
Fission: Splitting of a heavy nucleus into two (or more) lighter nuclei. Fusion: Combining two light nuclei to form a heavier nucleus. The energy changes for these nuclear processes are typically millions of times larger than those associated with chemical reactions. The fusion of 235U produces about 26 million times more energy than the combustion of methane. The maximum binding energy per nucleon occurs at Fe. Nuclei smaller than Fe become more stable by fusing to form heavier nuclei closer in mass to Fe. Nuclei larger than Fe form more stable nuclei by splitting to form lighter nuclei closer in mass to Fe. In each process, more stable nuclei are formed. The difference in stability is released as energy. For fusion reactions, a collision of sufficient energy must occur between two positively charged particles to initiate the reaction. This requires high temperatures. In fission, an electrically neutral neutron collides with the positively charged nucleus. This has a much lower activation energy.
9.
235
U splits into lighter elements when it absorbs a neutron; neutrons are also produced in the fission reaction. These neutrons produced can go on to react with other 235U nuclides, thus continuing the reaction. This self-sustaining fission process is called a chain reaction.
CHAPTER 19
952
THE NUCLEUS: A CHEMIST’S VIEW
For a fission process to be self-sustaining, at least one neutron from each fission event must go on to split another nucleus. If, on average, less than one neutron causes another fission event, the process dies out and the reaction is said to be subcritical. A reaction is critical when exactly one neutron from each fission event causes another event to occur. For supercritical reactions, more than one neutron produced causes another fission event to occur. Here, the process escalates rapidly, and the heat build-up causes a violent explosion. The critical mass is the mass of fissionable material required to produce a self-sustaining chain reaction, not too small and not too large. Reference Figure 19.14 of the text for a schematic of a nuclear power plant. The energy produced from controlled fission reactions is used to heat water to produce steam to run turbine engines. This is how coal-burning power plants generate energy. Moderator: Slows the neutrons to increase the efficiency of the fission reaction. Control rods: Absorbs neutrons to slow or halt the fission reaction. Some problems associated with nuclear reactors are radiation exposure to workers, disposal of wastes, nuclear accidents including a meltdown, and potential terrorist targets. Another problem is the supply of 235U, which is not endless. There may not be enough 235U on earth to make fission economically feasible in the long run. Breeder reactors produce fissionable fuel as the reactor runs. The current breeder reactors convert the abundant 238U isotope (which is non-fissionable) into fissionable plutonium. The reaction involves absorption of a neutron. Breeder reactors, however, have the additional hazard of handling plutonium, which flames on contact with air and is very toxic. 10.
Some factors for the biological effects of radiation exposure are: a. The energy of the radiation. The higher the energy, the more damage it can cause. b. The penetrating ability of radiation. The ability of specific radiation to penetrate human tissue where it can do damage must be considered. c. The ionizing ability of the radiation. When biomolecules are ionized, there function is usually disturbed. d. The chemical properties of the radiation source. Specifically, can the radioactive substance be readily incorporated into the body, or is the radiation source inert chemically so it passes through the body relatively quickly. 90
Sr will be incorporated into the body by replacing calcium in the bones. Once incorporated, 90Sr can cause leukemia and bone cancer. Krypton is chemically inert so it will not be incorporated into the body. Even though gamma rays penetrate human tissue very deeply, they are very small and cause only occasional ionization of biomolecules. Alpha particles, because they are much more massive, are very effective at causing ionization of biomolecules; alpha particles produce a dense trail of damage once they get inside an organism.
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953
Active Learning Questions 1.
a. For light nuclei, the number of neutrons and protons are about equal. As the mass number increases, the number of neutrons in stable nuclei become greater than the number of protons. See Figure 19.1 for a plot of the number of neutrons vs. the number of protons. b. Radioactive nuclei having too many neutrons typically undergo -particle decay. An example is: 3 3 0 1 H → 2 He + −1 e
Radioactive nuclei having too many protons typically undergo positron decay. An example is: 8 8 0 5 B → 4 Be + +1 e
Alpha particle production also increases the neutron to proton ratio, but the change is not as dramatic as with positron decay. Alpha particle decay generally occurs for heavy nuclei. An example is: 238 4 234 92 U → 2 He + 90Th
c. Positron production has the net effect of turning a proton into a neutron. Beta-particle production has the net effect of turning a neutron into a proton. d. Electron capture is a process in which an inner-orbital electron is captured by the nucleus. Electron capture increases the neutron to proton ratio. An example is: 7 0 7 4 Be + −1 e → 3 Li
Gamma rays are high energy electromagnetic radiation. Gamma rays have no mass or charge; they have no effect on the neutron to proton ratio. 2.
a. Binding energy: the energy required to decompose a nucleus into its component nucleons. Fission: splitting of a heavy nucleus into two (or more) lighter nuclei. Fusion: combining two light nuclei to form a heavier nucleus. b. Large nuclei have a relatively small binding energy. When a heavier nucleus is split into two smaller nuclei, the smaller nuclei have a binding energy greater than that of the heavier nuclei. So heavier nuclei become more stable when they split up. The difference in binding energies between the smaller nuclei and the heavier nucleus corresponds to the energy released in a fission process. Looking at Figure 19.9, nuclei larger than 56Fe have binding energies that decrease steadily from 56Fe. If a nucleus smaller than 56Fe is split into two lighter nuclei, these lighter nuclei have smaller binding energies. Because the lighter nuclei are less stable than the nuclei split, energy must be absorbed to make this happen; the process is endothermic.
CHAPTER 19
954
THE NUCLEUS: A CHEMIST’S VIEW
c. Light nuclei have a relatively small binding energy as compared to nuclei up to 56Fe (see Figure 19.9). When two small nuclei are combined to form a heavier nucleus, the binding energy of the product nuclei is greater than the binding energy of the smaller nuclei combined. The difference in binding energies corresponds to the energy released in a fusion process. Combining two nuclei larger than 56Fe form a heavier nucleus with a smaller binding energy than the nuclei combined. Because the heavier nuclei are less stable than the ones combined, energy must be absorbed to make this happen; the process is endothermic. d. For fusion reactions, there is a huge activation energy required to bring two similarly charged nuclei together. To overcome the activation energy, extremely high temperatures are required. These extreme temperatures are reached in the core of the sun but are not yet attainable in a laboratory setting. Some problems associated with fission reactions and nuclear reactors are radiation exposure to workers, disposal of wastes, nuclear accidents including a meltdown, and potential terrorist targets. Another problem is the supply of 235U is not endless. There may not be enough 235U on earth to make fission economically feasible in the long run.
Questions 3.
Characteristic frequencies of energies emitted in a nuclear reaction suggest that discrete energy levels exist in the nucleus. The extra stability of certain numbers of nucleons and the predominance of nuclei with even numbers of nucleons suggest that the nuclear structure might be described by using quantum numbers.
4.
210 a. Alpha particle ( 42 He) production is required to convert 214 84 Po into 82 Pb.
210 b. Beta particle ( −01 e) production is required to convert 210 82 Pb into 83 Bi. 202 c. Beta particle ( −01 e) production is required to convert 202 79 Au into 80 Hg.
5.
Radiotracers generally have short half-lives. The radioactivity from the radiotracers is monitored to study a specific area of the body. However, we don’t want long-lived radioactivity in order to minimize potential damage to healthy tissue by the radioactivity.
6.
No, coal-fired power plants also pose risks. A partial list of risks is: Coal Air pollution Coal mine accidents Health risks to miners (black lung disease)
7.
Nuclear Radiation exposure to workers Disposal of wastes Meltdown Terrorists Public fear
Beta-particle production has the net effect of turning a neutron into a proton. Radioactive nuclei having too many neutrons typically undergo -particle decay. Positron production has the net effect of turning a proton into a neutron. Nuclei having too many protons typically undergo positron decay.
CHAPTER 19
8.
THE NUCLEUS: A CHEMIST’S VIEW
955
First order kinetics refers to the rate law for the reaction. The rate law for radioactive decay is Rate = kN, where N is the number of radioactive nuclides. Because the rate law depends on N to the first power, we call this first order kinetics. The integrated rate law and half-life expression for first order kinetics are: N ln 2 = −kt , t1/2 = ln k N0
The half-life is the time it takes for ½ of the radioactive nuclides to decay. At the half-life, N = ½ N0, where N0 is the initial number of radioactive nuclides. Substituting ½ N0 for N in the integrated rate law, the N0 terms cross off leaving ln(1/2). Rearranging gives the t1/2 expression above which has no dependence on the number of nuclei one starts with. 9.
The deuterium nuclei are positively charges particles. For the two nuclei to bind together and release energy, they must get very close together. Because the deuterium nuclei are both positively charged, they repel each other electrostatically. The increase in energy as two deuterium nuclei are brought together is due to the electrostatic repulsion of the two positively charge nuclei for each other. To overcome the electrostatic barrier, the two nuclei must be travelling at extremely fast velocities when they collide. Temperatures of 4 × 107 K are required to give the nuclei enough energy upon collision to bind together.
10.
a. Nothing; binding energy is related to thermodynamic stability; it is not related to kinetics. Binding energy indicates nothing about how fast or slow a specific nucleon decays. b.
56
Fe has the largest binding energy per nucleon, so it is the most stable nuclide. 56Fe has the greatest mass loss per nucleon when the protons and neutrons are brought together to form the 56Fe nucleus. The least stable nuclide shown, having the smallest binding energy per nucleon, is 2H.
c. Fusion refers to combining two light nuclei having relatively small binding energies per nucleon to form a heavier nucleus which has a larger binding energy per nucleon. The difference in binding energies per nucleon is related to the energy released in a fusion reaction. Nuclides to the left of 56Fe can undergo fusion. Nuclides to the right of 56Fe can undergo fission. In fission, a heavier nucleus having a relatively small binding energy per nucleon is split into two smaller nuclei having larger binding energy per nucleons. The difference in binding energies per nucleon is related to the energy released in a fission reaction. 11.
Carbon-14 has 8 neutrons. Some other nuclides with 8 protons are 13B, 15N, 16O, and 17F.
12.
Neutrons are uncharged. Because of this, they are not repelled by target nuclei. In addition, neutrons are readily absorbed by many nuclei, which go onto produce other nuclides.
13.
The transuranium elements are the elements having more protons than uranium. They are synthesized by bombarding heavier nuclei with neutrons and positive ions in a particle accelerator.
CHAPTER 19
956
THE NUCLEUS: A CHEMIST’S VIEW
14.
All radioactive decay follows first-order kinetics. A sample is analyzed for the 176Lu and 176Hf content, from which the first-order rate law can be applied to determine the age of the sample. The reason 176Lu decay is valuable for dating very old objects is the extremely long half-life. Substances formed a long time ago that have short half-lives have virtually no nuclei remaining. On the other hand, 176Lu decay hasn’t even approached one half-life when dating 5-billionyear-old objects.
15.
E = mc2; the key difference is the mass change when going from reactants to products. In chemical reactions, the mass change is indiscernible. In nuclear processes, the mass change is discernible. It is the conversion of this discernible mass change into energy that results in the huge energies associated with nuclear processes.
16.
Effusion is the passage of a gas through a tiny orifice into an evacuated container. Graham’s law of effusion says that the effusion of a gas in inversely proportional to the square root of the mass of its particle. The key to effusion, and to the gaseous diffusion process, is that they are both directly related to the velocity of the gas molecules, which is inversely related to the molar mass. The lighter 235UF6 gas molecules have a faster average velocity than the heavier 238UF6 gas molecules. The difference in average velocity is used in the gaseous diffusion process to enrich the 235U content in natural uranium.
17.
The temperatures of fusion reactions are so high that all physical containers would be destroyed. At these high temperatures, most of the electrons are stripped from the atoms. A plasma of gaseous ions is formed that can be controlled by magnetic fields.
18.
Penetration power refers to the ability of radioactive decay particles or rays to penetrate human tissue. Gamma radiation easily penetrates human tissue, beta particles can penetrate about 1 cm, and alpha particles are stopped by the skin, so they are the least penetrating.
19.
Somatic damage is immediate damage to the organism itself, resulting in sickness or death. Genetic damage is damage to the genetic material which can be passed on to future generations. The organism will not feel immediate consequences from genetic damage, but its offspring may be damaged.
20.
The linear model postulates that damage from radiation is proportional to the dose, even at low levels of exposure. Thus any exposure is dangerous. The threshold model, on the other hand, assumes that no significant damage occurs below a certain exposure, called the threshold exposure. A recent study supported the linear model.
Exercises Radioactive Decay and Nuclear Transformations 21.
All nuclear reactions must be charge balanced and mass balanced. To charge balance, balance the sum of the atomic numbers on each side of the reaction, and to mass balance, balance the sum of the mass numbers on each side of the reaction. a.
3 3 0 1 H → 2 He + −1 e
b.
8 0 3 Li → 4 Be + −1 e 8 4 4 Be → 2 2 He
8
8 4 0 3 Li → 2 2 He + −1 e
CHAPTER 19 c. 22.
23.
THE NUCLEUS: A CHEMIST’S VIEW
7 0 7 4 Be + −1 e → 3 Li
d.
957
8 8 0 5 B → 4 Be + +1 e
All nuclear reactions must be charge balanced and mass balanced. To charge balance, balance the sum of the atomic numbers on each side of the reaction, and to mass balance, balance the sum of the mass numbers on each side of the reaction. a.
60 60 0 27 Co → 28 Ni + −1 e
b.
97 43Tc
c.
99 99 0 43Tc → 44 Ru + −1 e
d.
239 235 4 94 Pu → 92 U + 2 He
+ −01 e → 97 42 Mo
All nuclear reactions must be charge balanced and mass balanced. To charge balance, balance the sum of the atomic numbers on each side of the reaction, and to mass balance, balance the sum of the mass numbers on each side of the reaction. a.
238 4 234 92 U → 2 He + 90Th ; this is alpha-particle production.
b.
234 234 0 90Th → 91Pa + −1 e ; this is -particle production.
24.
a.
51 0 51 24 Cr + −1 e → 23V
25.
a.
68 Ga 31
+ −01 e
c.
212 Fr 87
→
a.
73 Ga 31
→
c.
205 Bi 83
26.
→
b.
131 0 131 53 I → −1e + 54 Xe
c.
68 30 Zn
b.
62 Cu 29
→ +01 e +
4 He 2
+ 208 85 At
d.
129 Sb 51
→
0 −1 e
73 32 Ge
+ −01 e
b.
192 Pt 78
→
188 Os 76
0 → 205 82 Pb + +1 e
d.
241 Cm 96
+ −01 e → 241 95 Am
32 0 32 15 P → −1e + 16 S
62 Ni 28
+ 129 52 Te + 42 He
76 76 0 As + 01 n → 76 33 As → 34 Se + −1 e; 34 Se is the final product.
27.
75 33
28.
214 0 214 4 210 0 210 0 210 83 Bi →−1 e + 84 Po → 2 He + 82 Pb → −1e + 83Bi → −1e + 84 Po
210 210 210 The products of the various steps are 214 84 Po, 82 Pb, 83Bi, and 84 Po.
29.
235 92 U
→
207 82 Pb +
? 42 He + ? −01 e
From the two possible decay processes, only alpha-particle decay changes the mass number. So the mass number change of 28 from 235 to 207 must be done in the decay series by seven alpha particles. The atomic number change of 10 from 92 to 82 is due to both alpha-particle production and beta-particle production. However, because we know that seven alpha-particles are in the complete decay process, we must have four beta-particle decays in order to balance the atomic number. The complete decay series is summarized as: 235 92 U
4 0 → 207 82 Pb + 7 2 He + 4 −1 e
CHAPTER 19
958 30.
31.
THE NUCLEUS: A CHEMIST’S VIEW
4 0 → 206 82 Pb + ? 2 He + ? −1 e; the change in mass number (242 − 206 = 36) is due exclusively to the alpha-particles. A change in mass number of 36 requires 9 42 He particles to be produced. The atomic number only changes by 96 − 82 = 14. The 9 alpha-particles change the atomic number by 18, so 4 −01 e (4 beta-particles) are also produced in the decay series of 242 Cm to 206Pb. 242 96 Cm
a.
241 Am 95
→
b.
241 95 Am
209 → 8 42 He + 4 −01 e + 209 83 Bi; the final product is 83 Bi.
c.
241 95 Am
229 233 233 225 → 237 90 Th + → 93 Np + → 92 U + → 88 Ra + 91 Pa + → 213 213 217 221 225 85 At + 87 Fr + 84 Po + 83 Bi + 89 Ac + 209 209 82 Pb + → 83 Bi +
4 2 He
+ 237 93 Np
The intermediate radionuclides are: 237 93 Np,
32.
229 217 209 233 225 225 221 213 213 233 91 Pa, 92 U, 90 Th, 88 Ra, 89 Ac, 87 Fr, 85 At, 83 Bi, 84 Po, and 82 Pb
The complete decay series is: 232 90 Th
4 4 228 228 224 0 0 → 228 88 Ra + 2 He → 89 Ac + −1 e → 90 Th + −1 e → 88 Ra + 2 He
212 0 −1 e + 84 Po
212 0 −1 e + 83 Bi
42 He
4 4 216 220 + 212 82 Pb 2 He + 84 Po 86 Rn + 2 He
4 208 82 Pb + 2 He
33.
Reference Table 19.2 of the text for potential radioactive decay processes. 8B and 9B contain too many protons or too few neutrons. Electron capture or positron production are both possible decay mechanisms that increase the neutron to proton ratio. Alpha particle production also increases the neutron to proton ratio, but it is not likely for these light nuclei. 12 B and 13B contain too many neutrons or too few protons. Beta-particle production lowers the neutron to proton ratio, so we expect 12B and 13B to be β-emitters.
34.
Reference Table 19.2 of the text for potential radioactive decay processes. 17F and 18F contain too many protons or too few neutrons. Electron capture and positron production are both possible decay mechanisms that increase the neutron to proton ratio. Alpha-particle production also increases the neutron-to-proton ratio, but it is not likely for these light nuclei. 21F contains too many neutrons or too few protons. Beta-particle production lowers the neutronto-proton ratio, so we expect 21F to be a beta-emitter.
35.
a.
249 98 Cf
263 + 188 O → 106 Sg + 4 01 n
b.
259 104 Rf ;
263 106 Sg
→ 42 He + 259 104 Rf
CHAPTER 19 36.
THE NUCLEUS: A CHEMIST’S VIEW
a.
240 95 Am
c.
249 98 Cf
959
1 + 42 He → 243 97 Bk + 0 n
b.
238 92 U
1 + 126 C → 244 98 Cf + 6 0 n
1 + 157 N → 260 105 Db + 4 0 n
d.
249 98 Cf
1 + 105 B → 257 103 Lr + 2 0 n
Kinetics of Radioactive Decay 37.
For t1/2 = 12,000 yr: k=
ln 2 1 yr 1d 1h 0.693 = = 1.8 10−12 s−1 t 1/2 12,000 yr 365 d 24 h 3600 s
Rate = kN = 1.8 10−12 s−1 6.02 1023 nuclei = 1.1 1012 decays/s For t1/2 = 12 h: k=
ln 2 1h 0.693 = = 1.6 10−5 s−1 t 1/2 12 h 3600 s
Rate = kN = 1.6 10−5 s−1 6.02 1023 nuclei = 9.6 1018 decays/s For t1/2 = 12 s: Rate = kN =
0.693 6.02 1023 nuclei = 3.5 1022 decays/s 12 s 32
38.
a. 120 g K3PO4
1 mol K 3 PO4 32
213.3 mg K 3 PO4
1 mol 32 P 32
mol K 3 PO4
6.02 10 23 nuclei mol = 3.4 1023 P-32 nuclei
k=
ln 2 1h 0.69315 = = 1.35 10−5 s−1 t 1/2 14.3 h 3600 s
Rate = kN = 1.35 10−5 s−1 3.4 1023 nuclei b. k =
Ci = 1.2 108 Ci 3.7 10 decays/s 10
ln 2 1 yr 1d 1h 0.693 = = 9.2 10−13 s−1 t 1/2 24,000 yr 365 d 24 h 3600 s
Rate = kN = 9.2 10−13 s−1 6.02 1023 nuclei
Ci 1000 mCi Ci 3.7 10 decays/s 10
Rate = 1.5 104 mCi 39.
All radioactive decay follows first-order kinetics where t1/2 = (ln 2)/k. t1/2 =
ln 2 0.6931 = = 1.41 1010 yr k 4.91 10−11 yr −1
CHAPTER 19
960 40.
k=
THE NUCLEUS: A CHEMIST’S VIEW
ln 2 0.69315 1 yr 1d 1h = = 5.08 × 10 −11 s −1 t1 / 2 433 yr 365 d 24 h 3600 s
Rate = kN = 5.08 × 10 −11 s −1 × 5.00 g
1 mol 6.022 10 23 nuclei 241 g mol = 6.35 × 1011 decays/s
6.35 × 1011 alpha particles are emitted each second from a 5.00-g 241Am sample. 41.
12.5% of 40K will remain after 3 half-lives: 100%
t1/2
50%
t1/2
25%
t1/2
12.5%
So, each half-life is 3.9 109 /3 = 1.3 109 yr. 42.
Kr-81 is most stable because it has the longest half-life. Kr-73 is hottest (least stable); it decays most rapidly because it has the shortest half-life. 12.5% of each isotope will remain after 3 half-lives: 100%
t1/2
50%
t1/2
25%
t1/2
12.5%
For Kr-73: t = 3(27 s) = 81 s; for Kr-74: t = 3(11.5 min) = 34.5 min For Kr-76: t = 3(14.8 h) = 44.4 h; for Kr-81: t = 3(2.1 × 105 yr) = 6.3 × 105 yr 43.
1.0 day ×
24 h 1 half-life = 4 half-lives; day 6h
After 4 half-lives, 6.25% of the sample remains: 0.0625(150.0 mg) = 9.4 mg 99Tc remains 44.
After 4 half-lives, 6.25% of the sample remains:
After 4 half-lives, 1/16 of the original sample remains. The mass of the original sample is 16(2.5 mg) = 40. mg 13N. 45.
Units for N and N0 are usually number of nuclei but can also be grams if the units are identical for both N and N0. In this problem, m0 = the initial mass of 47Ca2+ to be ordered. k=
5.0 μg Ca 2+ − 0.693(2.0 d) N ln 2 − (0.693) t = = − kt = ; ln , ln = − 0.31 t1 / 2 t1 / 2 m 4 . 5 d 0 N0
5 .0 = e−0.31 = 0.73, m0 = 6.8 µg of 47Ca2+ needed initially m0
6.8 µg 47Ca2+ ×
107.0 μg 47 CaCO3 47.0 μg
47
Ca
2+
= 15 µg 47CaCO3 should be ordered at the minimum.
CHAPTER 19 46.
a. k =
THE NUCLEUS: A CHEMIST’S VIEW
961
ln 2 0.6931 1d 1h = = 6.27 × 10 −7 s −1 t1 / 2 12.8 d 24 h 3600 s
1 mol 6.022 10 23 nuclei b. Rate = kN = 6.27 × 10 −7 s −1 28.0 10 −3 g 64.0 g mol Rate = 1.65 × 1014 decays/s c. 25% of the 64Cu will remain after 2 half-lives (100% decays to 50% after one half-life, which decays to 25% after a second half-life). Hence 2(12.8 days) = 25.6 days is the time frame for the experiment. 47.
t = 78.0 yr; k =
N N ln 2 −(0.6931) 78.0 yr = e−1.87 = 0.154 = −kt = ; ln = −1.87, t1/ 2 28.9 yr N N 0 0
15.4% of the 90Sr remains as of July 16, 2023. 48.
Assuming 2 significant figures in 1/100: ln(N/N0) = −kt; N = (0.010)N0; t1/2 = (ln 2)/k ln(0.010) =
49.
− (ln 2) t − (0.693) t = , t = 53 days t1 / 2 8 .0 d
175 mg Na332PO4
32.0 mg 32P 32
165 .0 mg Na 3 PO 4
= 33.9 mg 32P; k =
ln 2 t1/ 2
N m − (0.6931) t − 0.6931(35.0 d ) = − kt = = ln , ln ; carrying extra sig. figs.: t1/ 2 14.3 d 33.9 mg N0
ln(m) = −1.696 + 3.523 = 1.827, m = e1.827 = 6.22 mg 32P remains 50.
a. 0.0100 Ci × Rate = kN,
ln 2 3.7 1010 decays/ s = 3.7 × 108 decays/s; k = t1/ 2 Ci
3.7 10 8 decays 0.6931 1h × N, N = 5.5 × 1012 atoms of 38S = s 2 . 87 h 3600 s 38
5.5 × 1012 atoms 38S ×
1 mol Na 2 SO 4 1 mol 38S = 9.1 × 10 −12 mol Na238SO4 23 6.02 10 atoms mol 38S
9.1 × 10 −12 mol Na238SO4 ×
38
148 .0 g Na 2 SO 4 38
mol Na 2 SO 4
= 1.3 × 10 −9 g = 1.3 ng Na238SO4
− (0.6931) t 0.01 b. 99.99% decays, 0.01% left; ln , t = 38.1 hours 40 hours = −kt = 2.87 h 100
CHAPTER 19
962
51.
1 .0 g N − (ln 2) t = = −kt = ln , ln t1 / 2 N0 m0
THE NUCLEUS: A CHEMIST’S VIEW 24 h 60 min − 0.693 3.0 d d h 1.0 10 3 min
1 .0 g 1 .0 = −3.0, = e −3.0 , m0 = 20. g 82Br needed ln m0 m0
20. g 82Br 52.
1 mol 82 Br 1 mol Na 82Br 105 .0 g Na 82Br = 26 g Na82Br 82.0 g mol 82Br mol Na 82Br
Assuming the current year is 2023, t = 77 yr. N − (0.693)t −0.693(77 yr) 0.072 decay events N = −kt = , N= ln , ln = min • 100. g water 12.3 yr t 1/2 5.5 N0
53.
N = 180 lb
453.6 g 18 g C 1.6 10 −10 g 14 C 1 mol 14 C lb 100 g body 100 g C 14 g 14 C
Rate = kN; k =
6.022 10 23 nuclei 14 C mol 14 C
= 1.0 1015 carbon-14 nuclei
ln 2 0.693 1 yr 1d 1h = = 3.8 10 −12 s −1 t1/2 5730 yr 365 d 24 h 3600 s
Rate = kN; k = 3.8 10 −12 s −1 (1.0 1015 14C nuclei ) = 3800 decays/s A typical 180 lb person produces 3800 beta particles each second. 54.
N = 180 lb ×
453.6 g 0.34 g K 0.012 g 40 K 1 mol 40 K lb 100 g body 100 g K 40. g 40 K ×
Rate = kN; k =
6.022 1023 nuclei 40 K = 5.0 × 1020 potassium-40 nuclei mol 40 K
ln 2 0.693 1 yr 1d 1h = × × × = 1.7 × 10−17 s −1 9 t1/2 1.3 10 yr 365 d 24 h 3600 s
Rate = kN; k = 1.7 ×10−17 s−1 (5.0× 1020 40 K nuclei) = 8500 decays/s 55.
k=
N ln 2 − 0.693(15,0 00 yr) − (0.693)t N = −kt = ; ln = −1.8 , ln = t1/ 2 t 1/2 5730 yr 13.6 N0
N = e −1.8 = 0.17, N = 13.6 × 0.17 = 2.3 counts per minute per g of C 13 .6
If we had 10. mg C, we would see:
CHAPTER 19
THE NUCLEUS: A CHEMIST’S VIEW
10. mg ×
963
1g 2.3 counts 0.023 counts = 1000 mg min g min
It would take roughly 40 min to see a single disintegration. This is too long to wait, and the background radiation would probably be much greater than the 14C activity. Thus 14C dating is not practical for very small samples. 56.
57.
N − (0.6931)t 1.2 − (0.6931)t = −kt = ln , ln , t = 2.0 × 104 yr = t 5730 yr N 13 . 6 1/2 0
Assuming 1.000 g 238U present in a sample, then 0.688 g 206Pb is present. Because 1 mol 206Pb is produced per mol 238U decayed: 238
U decayed = 0.688 g Pb ×
1 mol Pb 1 mol U 238 g U = 0.795 g 238U 206 g Pb mol Pb mol U
Original mass 238U present = 1.000 g + 0.795 g = 1.795 g 238U N 1.000 g − (ln 2)t − 0.693(t ) = −kt = = ln , ln , t = 3.8 × 109 yr 9 t 1 . 795 N g 4 . 5 10 yr 1/ 2 0
58.
a. The decay of 40K is not the sole source of 40Ca. b. Decay of 40K is the sole source of 40Ar and no 40Ar is lost over the years. c.
0.95 g 40Ar = current mass ratio 1.00 g 40K
0.95 g of 40K decayed to 40Ar. 0.95 g of 40K is only 10.7% of the total 40K that decayed, or: 0.107(m) = 0.95 g, m = 8.9 g = total mass of 40K that decayed Mass of 40K when the rock was formed was 1.00 g + 8.9 g = 9.9 g.
1.00 g 40K − (ln 2) t − (0.6931) t = −kt = , t = 4.2 × 109 years old = ln 40 9 t 1.27 10 yr 1/ 2 9.9 g K d. If some 40Ar escaped, then the measured ratio of 40Ar/40K is less than it should be. We would calculate the age of the rock to be less than it actually is.
Energy Changes in Nuclear Reactions 59.
ΔE = Δmc2, Δm =
3.9 10 23 kg m 2 /s 2 ΔE = = 4.3 × 106 kg 2 8 2 c (3.00 10 m/s)
The sun loses 4.3 × 106 kg of mass each second. Note: 1 J = 1 kg m2/s2
CHAPTER 19
964 60.
THE NUCLEUS: A CHEMIST’S VIEW
1.8 1014 kJ 1000 J 3600 s 24 h = 1.6 × 1022 J/day s kJ h day
ΔE = Δmc2, Δm =
ΔE 1.6 10 22 J = = 1.8 × 105 kg c2 (3.00 10 8 m/s) 2
1.8 × 105 kg of solar material provides 1 day of solar energy to the earth. 1.6 × 1022 J
1 kJ 1g 1 kg = 5.0 × 1014 kg 1000 J 32 kJ 1000 g
5.0 × 1014 kg of coal is needed to provide the same amount of energy as 1 day of solar energy. 61.
For 127I (53e, 53p, and 74n), mass defect = Δm = [mass of 127I nucleus] − [mass of 53 protons + mass of 73 neutrons]. The atomic masses given include the mass of the electrons. Let me = mass of electron. Δm = (126.9004 u − 53me) − [53(1.00782 − me) + 74(1.00866)]; mass of electrons (me) cancel. Δm = 126.9004 − [53(1.00782) + 74(1.00866)] = −1.1549 u = mass defect for 127I For 1 mol of 127I, the mass defect is ‒1.1549 g.
62.
For 75As (33e, 33p, and 42n), mass defect = Δm = [mass of 75As nucleus] − [mass of 33 protons + mass of 42 neutrons]. The atomic masses given include the mass of the electrons. Let x = atomic mass of 75As and me = mass of electron. For a single 75As nucleus, the mass defect is −0.70018 u. Δm = [x − 33me] − [33(1.00782 − me) + 42(1.00866)]; mass of electrons cancel. Δm = ‒0.70018 = x − [33(1.00782) + 42(1.00866)], x = atomic mass of 75As = 74.9216 u
63.
We need to determine the mass defect Δm between the mass of the nucleus and the mass of the individual parts that make up the nucleus. Once Δm is known, we can then calculate ΔE (the binding energy) using E = mc2. Note: 1 J = 1 kg m2/s2. For 232 94 Pu (94 e, 94 p, 138 n): mass of 232Pu nucleus = 3.85285 × 10 −22 g − mass of 94 electrons mass of 232Pu nucleus = 3.85285 × 10 −22 g − 94(9.10939 × 10 −28 ) g = 3.85199 × 10 −22 g Δm = 3.85199 × 10 −22 g − (mass of 94 protons + mass of 138 neutrons) Δm = 3.85199 × 10 −22 g − [94(1.67262 × 10 −24 ) + 138(1.67493 × 10 −24 )] g = −3.168 × 10 −24 g For 1 mol of nuclei: Δm = −3.168 × 10 −24 g/nuclei × 6.0221 × 1023 nuclei/mol = −1.908 g/mol ΔE = Δmc2 = (−1.908 × 10 −3 kg/mol)(2.9979 × 108 m/s)2 = −1.715 × 1014 J/mol
CHAPTER 19
THE NUCLEUS: A CHEMIST’S VIEW
965
For 231 91 Pa (91 e, 91 p, 140 n): mass of 231Pa nucleus = 3.83616 × 10 −22 g − 91(9.10939 × 10-28) g = 3.83533 × 10 −22 g Δm = 3.83533 × 10 −22 g − [91(1.67262 × 10 −24 ) + 140(1.67493 × 10 −24 )] g = −3.166 × 10 −24 g − 3.166 10 − 27 kg 6.0221 10 23 nuclei 2.9979 10 8 m ΔE = Δmc = nuclei mol s
2
2
= −1.714 × 1014 J/mol 64.
The atomic masses given include the mass of electrons. Let me = mass of electron. Iron-56 contains 26 protons, 26 electrons, and 30 neutrons. Mass of 56 26 Fe nucleus = mass of atom − mass of electrons = 55.9349 − 26(me). Δm = [55.9349 − 26(me)] − [26(1.00782 ‒ me) + 30(1.00866)] u = −0.5282 u ΔE = Δmc2 = −0.5282 u
1.6605 10 −27 kg (2.9979 × 108 m/s)2 = −7.883 × 10−11 J u
Binding energy 7.883 × 10-11 J = = 1.408 × 10−12 J/nucleon Nucleon 56 nucleons 65.
For 12C (6e, 6p, and 6n), mass defect = Δm = [mass of 12C nucleus] − [mass of 6 protons + mass of 6 neutrons]. The atomic masses given include the mass of the electrons. Let me = mass of electron. Δm = (12.00000 u − 6me) − [6(1.00782 − me) + 6(1.00866)]; mass of electrons cancel. Δm = 12.00000 − [6(1.00782) + 6(1.00866)] = −0.09888 u 1.6605 10 −27 kg (2.9979 × 108 m/s)2 = −1.476 × 10−11 J u Binding energy 1.476 10 −11 J = = 1.230 × 10−12 J/nucleon Nucleon 12 nucleons
ΔE = Δmc2 = −0.09888 u
For 235U (92e, 92p, and 143n): Δm = (235.0439 − 92me) − [92(1.00782 − me) + 143(1.00866)] = −1.9139 u ΔE = Δmc2 = −1.9139 ×
1.66054 10 −27 kg (2.99792 × 108 m/s)2 = −2.8563 × 10−10 J u
Binding energy 2.8563 10 −10 J = = 1.2154 × 10−12 J/nucleon Nucleon 235 nucleons
Because 56Fe is the most stable known nucleus, the binding energy per nucleon for 56Fe (1.408 × 10−12 J/nucleon) will be larger than that of 12C or 235U (see Figure 19.9 of the text).
CHAPTER 19
966 66.
THE NUCLEUS: A CHEMIST’S VIEW
For 21 H , mass defect = Δm = mass of 21 H nucleus − mass of proton − mass of neutron. The mass of the 2H nucleus will equal the atomic mass of 2H minus the mass of the electron in an 2 H atom. From the text, the pertinent masses are me = 5.49 × 10−4 u, mp = 1.00728 u, and mn = 1.00866 u. Δm = (2.01410 u − 0.000549 u) − (1.00728 u + 1.00866 u) = −2.39 × 10−3 u ΔE = Δmc2 = −2.39 × 10−3 u ×
1.6605 10 −27 kg × (2.998 × 108 m/s)2 = −3.57 × 10−13 J u
Binding energy 3.57 10 −13 J = = 1.79 × 10−13 J/nucleon Nucleon 2 nucleons
For 13 H : Δm = (3.01605 − 0.000549) − [1.00728 + 2(1.00866)] = −9.10 × 10−3 u ΔE = −9.10 × 10−3 u ×
1.6605 10 −27 kg × (2.998 × 108 m/s)2 = −1.36 × 10−12 J u
Binding energy 1.36 10 −12 J = = 4.53 × 10−13 J/nucleon Nucleon 3 nucleons
67.
ΔE = Δmc2, Δm =
− 3.086 1012 J ΔE = = −3.434 × 10 −5 kg c2 (2.9979 10 8 m/s) 2
The mass defect for 1 mol of 6Li is −3.434 × 10 −5 kg = −0.03434 g. The mass defect for one 6 Li nuclei is −0.03434 u. Let mLi = mass of 6Li nucleus; an 6Li nucleus has 3p and 3n. Mass defect = −0.03434 u = mLi − (3mp + 3mn) = mLi − [3(1.00728 u) + 3(1.00866 u)] mLi = 6.01348 u Mass of 6Li atom = 6.01348 u + 3me = 6.01348 + 3(5.49 × 10 −4 u) = 6.01513 u (includes mass of 3 e−) 68.
Binding energy =
1.326 10 −12 J × 27 nucleons = 3.580 × 10 −11 J for each 27Mg nucleus nucleon
ΔE = Δmc2, Δm =
ΔE − 3.580 10 −11 J = = −3.983 10 −28 kg (2.9979 108 m/s ) 2 c2
Δm = −3.983 10 −28 kg ×
1u 1.6605 10 − 27 kg
= −0.2399 u = mass defect
Let mMg = mass of 27Mg nucleus; an 27Mg nucleus has 12 p and 15 n. Mass defect = −0.2399 u = mMg − (12mp + 15mn) = mMg − [12(1.00728 u) + 15(1.00866 u)] mMg = 26.9764 u Mass of 27Mg atom = 26.9764 u + 12me, 26.9764 + 12(5.49 × 10 −4 u) = 26.9830 u (includes mass of 12 e−)
CHAPTER 19 69.
THE NUCLEUS: A CHEMIST’S VIEW
2 1 1 1H + 1 H → 1 H
967
0
+ +1 e; Δm = (2.01410 u − me + me) − 2(1.00782 u − me)
Δm = 2.01410 − 2(1.00782) + 2(0.000549) = −4.4 × 10 −4 u for two protons reacting When 2 mol of protons undergoes fusion, Δm = −4.4 × 10 −4 g. ΔE = Δmc2 = −4.4 × 10 −7 kg × (3.00 × 108 m/s)2 = −4.0 × 1010 J − 4.0 1010 J 1 mol = −2.0 × 1010 J/g of hydrogen nuclei 2 mol protons 1.01 g
70.
→ 42 He + 0 n; using atomic masses, the masses of the electrons cancel when determining Δm for this nuclear reaction. 3 2 1 H + 1H
1
Δm = [4.00260 + 1.00866 − (2.01410 + 3.01605)] u = −1.889 × 10 −2 u To produce 1 mol of 42 He : Δm = −1.889 × 10 −2 g = −1.889 × 10 −5 kg ΔE = Δmc2 = −1.889 × 10-5 kg × (2.9979 × 108 m/s)2 = −1.698 × 1012 J/mol For 1 nucleus of 42 He :
− 1.698 10 12 J 1 mol = −2.820 × 10 −12 J/nucleus mol 6.0221 10 23 nuclei
Detection, Uses, and Health Effects of Radiation 71.
The Geiger-Müller tube has a certain response time. After the gas in the tube ionizes to produce a "count," sometime must elapse for the gas to return to an electrically neutral state. The response of the tube levels off because at high activities, radioactive particles are entering the tube faster than the tube can respond to them.
72.
Not all of the emitted radiation enters the Geiger-Müller tube. The fraction of radiation entering the tube must be constant.
73.
Water is produced in this reaction by removing an OH group from one substance and an H from the other substance. There are two ways to do this:
Because the water produced is not radioactive, methyl acetate forms by the first reaction where all the oxygen-18 ends up in methyl acetate.
CHAPTER 19
968 74.
THE NUCLEUS: A CHEMIST’S VIEW
The only product in the fast-equilibrium step is assumed to be N16O18O2, where N is the central atom. However, this is a reversible reaction where N16O18O2 will decompose to NO and O2. Because any two oxygen atoms can leave N 16O18O2 to form O2, we would expect (at equilibrium) one-third of the NO present in this fast equilibrium step to be N 16O and two-thirds to be N18O. In the second step (the slow step), the intermediate N 16O18O2 reacts with the scrambled NO to form the NO2 product, where N is the central atom in NO2. Any one of the three oxygen atoms can be transferred from N 16O18O2 to NO when the NO2 product is formed. The distribution of 18O in the product can best be determined by forming a probability table. N16O (1/3)
N18O (2/3)
16
O (1/3) from N16O18O2
N16O2 (1/9)
N18O16O (2/9)
18
O (2/3) from N16O18O2
N16O18O (2/9)
N18O2 (4/9)
From the probability table, 1/9 of the NO2 is N16O2, 4/9 of the NO2 is N18O2, and 4/9 of the NO2 is N16O18O (2/9 + 2/9 = 4/9). Note: N16O18O is the same as N18O16O. In addition, N16O18O2 is not the only NO3 intermediate formed; N16O218O and N18O3 can also form in the fast-equilibrium first step. However, the distribution of 18O in the NO2 product is the same as calculated above, even when these other NO3 intermediates are considered. 75.
76.
77.
90 1 0 + 01 n → 144 58 Ce + 38 Sr + ? 0 n + ? −1 e; to balance the atomic number, we need 4 beta-particles, and to balance the mass number, we need 2 neutrons. 235 92 U
239 0 + 01 n → 239 92 U → −1 e + 93 Np → fissionable material in breeder reactors.
238 92 U
0 −1 e
+ 239 94 Pu; plutonium-239 is the
Release of Sr is probably more harmful. Xe is chemically unreactive. Strontium is in the same family as calcium and could be absorbed and concentrated in the body in a fashion similar to Ca. This puts the radioactive Sr in the bones; red blood cells are produced in bone marrow. Xe would not be readily incorporated into the body. The chemical properties determine where a radioactive material may be concentrated in the body or how easily it may be excreted. The length of time of exposure and what is exposed to radiation significantly affects the health hazard. (See Exercise 64 for a specific example.)
78.
(i) and (ii) mean that Pu is not a significant threat outside the body. Our skin is sufficient to keep out the alpha-particles. If Pu gets inside the body, it is easily oxidized to Pu 4+ (iv), which is chemically similar to Fe3+ (iii). Thus Pu4+ will concentrate in tissues where Fe3+ is found. One of these is the bone marrow, where red blood cells are produced. Once inside the body, alpha-particles cause considerable damage.
ChemWork Problems 79.
The most abundant isotope is generally the most stable isotope. The periodic table predicts that the most stable isotopes for exercises a-d are 39K, 56Fe, 23Na, and 204Tl. (Reference Table 19.2 of the text for potential decay processes.)
CHAPTER 19
THE NUCLEUS: A CHEMIST’S VIEW
969
a. Unstable; 45K has too many neutrons and will undergo beta-particle production. b. Stable c. Unstable; 20Na has too few neutrons and will most likely undergo electron capture or positron production. Alpha-particle production makes too severe of a change to be a likely decay process for the relatively light 20Na nuclei. Alpha-particle production usually occurs for heavy nuclei. d. Unstable; 194Tl has too few neutrons and will undergo electron capture, positron production, and/or alpha-particle production. 80.
a. Cobalt is a component of vitamin B12. By monitoring the cobalt-57 decay, one can study the pathway of vitamin B12 in the body. b. Calcium is present in the bones in part as Ca3(PO4)2. Bone metabolism can be studied by monitoring the calcium-47 decay as it is taken up in bones. c. Iron is a component of hemoglobin found in red blood cells. By monitoring the iron-59 decay, one can study red blood cell processes.
81.
82.
The equations for the nuclear reactions are:
232 90
239 235 4 94 Pu → 92U + 2 He ;
214 214 0 82 Pb → 83 Bi + −1 e ;
99 99 0 43Tc → 44 Ru + −1 e ;
239 239 0 93 Np → 94 Pu + −1 e
60 60 0 27 Co → 28Ni + −1e
4 0 Th → 208 82 Pb + ? 2 He + ? −1 e; the change in mass number (232 − 208 = 24) is due exclusively
to the alpha-particles. A change in mass number of 24 requires 6 42 He particles to be produced. The atomic number only changes by 90 − 82 = 8. The 6 alpha-particles change the atomic number by 12, so 4 −01 e (4 beta-particles) are also produced in the decay series of 232Cm to 208Pb. 83.
→ 267 107 Bh + ?; this equation is charge balanced, but it is not mass balanced. The products are off by 4 mass units. The only possibility to account for the 4 mass units is to have 4 neutrons produced. The balanced equation is: 249 22 97 Bk + 10 Ne
249 22 97 Bk + 10 Ne
1 → 267 107 Bh + 4 0 n
N − (0.6931) t − (0.6931) t 11 = −kt = ln , ln , t = 62.7 s = t 15 .0 s N 199 1/ 2 0
Bh: [Rn]7s25f146d5 is the expected electron configuration. 84.
a. t1/2 = (ln 2)/k, k =
ln 2 1h 0.69315 = = 6.42 × 10 −5 s −1 t 1/2 3.00 h 3600 s
b. Rate = kN = 6.42 × 10 −5 s −1 × 1.000 mol
6.022 10 23 nuclei = 3.87 × 1019 decays/s mol
CHAPTER 19
970 85.
THE NUCLEUS: A CHEMIST’S VIEW
t1/2 = 5730 yr; k = (ln 2)/t1/2; ln(N/N0) = −kt; ln
11.2 − (ln 2) t = , t = 2580 yr 15.3 5730 yr
The age of the icon is 2580 yr. 86.
t1/2 = 5730 yr; k = (ln 2)/t1/2; ln(N/N0) = −kt; ln
15.1 − (ln 2) t = , t = 109 yr 15.3 5730 yr
No; from 14C dating, the painting was produced during the early 1900s. 87.
The third-life will be the time required for the number of nuclides to reach one-third of the original value (N0/3). N − (0.6931)t 1 − (0.6931)t = −kt = , t = 49.8 yr ln , ln = t 1/2 31.4 yr 3 N0
The third-life of this nuclide is 49.8 years. 88.
ln(N/N0) = −kt; k = (ln 2)/t1/2 ; N = 0.001 × N0 0.001 N 0 − (ln 2) t = , ln(0.001) = −(2.88 × 10−5)t, t = 2 × 105 yr = 200,000 yr ln 24 , 100 yr N 0
89.
0.17 N 0 N − (ln 2) t = −kt = = −(5.64 × 10 −2 )t, t = 31.4 yr , ln ln 12 .3 yr N0 N0
It takes 31.4 years for the tritium to decay to 17% of the original amount. Hence the watch stopped fluorescing enough to be read in 1975 (1944 + 31.4). 90.
Because 1 mol 87Sr is produced per mol 87Rb decayed: 87
Rb decayed = 3.1 μg Sr ×
1 mol Sr 1 mol Rb 86.90919 g Rb = 3.1 μg 87Rb 86.90888 g Sr mol Sr mol Rb
Original mass 87Rb present = 109.7 μg + 3.1 μg = 112.8 μg 87Rb N 109.7 μg − (ln 2)t − 0.693(t) = −kt = = ln , ln , t = 1.9 × 109 yr t 1/2 4.7 10 10 yr 112.8 μg N0
The rock is 1.9 × 109 yr old. 91.
92.
Iron-56 has the largest mass defect per nucleon (see Fig 19.9). The mass defect per nucleon decreases from 56Fe as nuclei get lighter and get heavier. Of the nuclei listed, 34S is the nuclei with a mass closest to 56Fe, so it will have the largest mass defect per nucleon. Answer e is correct. 24 For 24 12 Mg : mass defect = Δm = mass of 12 Mg nucleus − mass of 12 protons − mass of 12 neutrons. The mass of the 24Mg nucleus will equal the atomic mass of 24Mg minus the mass of the 12 electrons in an 24Mg atom.
Δm = [23.9850 u − 12(0.0005486 u)] − [12(1.00728 u) + 12(1.00866 u)] = −0.2129 u
CHAPTER 19
THE NUCLEUS: A CHEMIST’S VIEW
ΔE = Δmc2 = −0.2129 u ×
971
1.6605 10 −27 kg × (2.9979 × 108 m/s)2 = −3.177 × 10−11 J u
The binding energy of 24Mg is 3.177 × 10−11 J. 93.
60 → 27 Co + ?; in order to balance the equation, the missing particle has no mass and a charge of 1−; this is an electron. 58 1 26 Fe + 2 0 n
60 An atom of 27 Co has 27 e, 27 p, and 33 n. The mass defect of the 60Co nucleus is:
m = (59.9338 – 27me) – [27(1.00782 – me) + 33(1.00866)] = − 0.5631 u E = mc2 = − 0.5631 u ×
1.6605 10 −27 kg × (2.9979 × 108 m/s)2 = − 8.403 × 10 −11 J u
Binding energy 8.403 10 −11 J = = 1.401 × 10 −12 J/nucleon Nucleon 60 nucleons
The emitted particle was an electron, which has a mass of 9.109 × 10 −31 kg. The deBroglie wavelength is: λ =
94.
h 6.626 10 −34 J s = = 2.7 × 10 −12 m −31 8 mv 9.109 10 kg (0.90 2.998 10 m/s)
Δm = −2(5.486 × 10-4 u) = −1.097 × 10 −3 u ΔE = Δmc2 = −1.097 × 10 −3 u ×
1.6605 10 −27 kg × (2.9979 × 108 m/s)2 u
= −1.637 × 10 −13 J
Ephoton = 1/2(1.637 × 10 −13 J) = 8.185 × 10 −14 J = hc/λ λ=
95.
hc 6.6261 10 −34 J s 2.9979 10 8 m/s = = 2.427 × 10 −12 m = 2.427 × 10 −3 nm −14 E 8.185 10 J
20,000 ton TNT ×
4 10 9 J 1 mol 235U 235 g 235U = 940 g 235U 900 g 235U ton TNT 2 1013 J mol 235U
This assumes that all of the 235U undergoes fission. 96.
To sustain a nuclear chain reaction, the neutrons produced by the fission must be contained within the fissionable material so that they can go on to cause other fissions. The fissionable material must be closely packed together to ensure that neutrons are not lost to the outside. The critical mass is the mass of material in which exactly one neutron from each fission event causes another fission event so that the process sustains itself. A supercritical situation occurs when more than one neutron from each fission event causes another fission event. In this case, the process rapidly escalates, and the heat build-up causes a violent explosion.
CHAPTER 19
972 97.
THE NUCLEUS: A CHEMIST’S VIEW
Mass of nucleus = atomic mass – mass of electron = 2.01410 u – 0.000549 u = 2.01355 u
3 RT urms = M KEavg =
1/2
3(8.3145 J/K • mol)(4 10 7 K) = 2.01355 g(1 kg/1000 g)
1/2
= 7 × 105 m/s
1.66 10 −27 kg 1 1 (7 × 105 m/s)2 = 8 × 10 −16 J/nuclei mu 2 = 2.01355 u 2 2 u
We could have used KEave = (3/2)RT to determine the same average kinetic energy. 98.
1 1 1 1 1 H + 0 n → 2 1H + 0 n
+ −11H; mass −11H = mass 11H = 1.00728 u = mass of proton = mp
Δm = 3mp + mn − (mp + mn) = 2mp = 2(1.00728) = 2.01456 u ΔE = Δmc2 = 2.01456 amu ×
1.66056 10 −27 kg × (2.997925 × 108 m/s)2 amu
ΔE = 3.00660 × 10−10 J of energy is absorbed per nuclei, or 1.81062 × 1014 J/mol nuclei. The source of energy is the kinetic energy of the proton and the neutron in the particle accelerator. 99.
All evolved oxygen in O2 comes from water and not from carbon dioxide.
100.
Sr-90 is an alkaline earth metal having chemical properties like calcium. Sr-90 can collect in bones, replacing some of the calcium. Once embedded inside the human body, beta- particles can do significant damage. Rn-222 is a noble gas, so one would expect Rn to be unreactive and pass through the body quickly; it does. The problem with Rn-222 is the rate at which it produces alpha-particles. With a short half-life, the few moments that Rn-222 is in the lungs, a significant number of decay events can occur; each decay event produces an alpha-particle that is very effective at causing ionization and can produce a dense trail of damage.
Challenge Problems 101.
k=
ln 2 ; t1 / 2
N − (0.693)t = − kt = ln t1/2 N0
N − (0.693)(4. 5 10 9 yr) N = = − 0.693, = e −0.693 = 0.50 For 238U: ln 9 N N 4.5 10 yr 0 0 N − (0.693)(4. 5 10 9 yr) N = = − 4.39, = e − 4.39 = 0.012 For 235U: ln 8 N0 7.1 10 yr N0
If we have a current sample of 10,000 uranium nuclei, 9928 nuclei of 238U and 72 nuclei of 235U are present. Now let’s calculate the initial number of nuclei that must have been present 4.5 × 109 years ago to produce these 10,000 uranium nuclei.
CHAPTER 19 For 238U:
THE NUCLEUS: A CHEMIST’S VIEW
973
N N 9928 nuclei = 0.50, N 0 = = = 2.0 10 4 238U nuclei N0 0.50 0.50
For 235U: N 0 =
N 72 nuclei = = 6.0 × 103 235 U nuclei 0.012 0.012
So 4.5 billion years ago, the 10,000-nuclei sample of uranium was composed of 2.0 × 104 238U nuclei and 6.0 × 103 235U nuclei. The percent composition 4.5 billion years ago would have been:
2.0 10 4 238 U nuclei (6.0 10 + 2.0 10 ) total nuclei 3
102.
4
100 = 77% 238U and 23% 235U
Total activity injected = 86.5 × 10−3 Ci Activity withdrawn =
3.6 10 −6 Ci 1.8 10 −6 Ci = 2.0 mL H 2 O mL H 2 O
Assuming no significant decay occurs, then the total volume of water in the body multiplied by 1.8 × 10−6 Ci/mL must equal the total activity injected. V×
1.8 10 −6 Ci = 8.65 × 10−2 Ci, V = 4.8 × 104 mL H2O mL H 2 O
Assuming a density of 1.0 g/mL for water, the mass percent of water in this 150-lb person is:
1.0 g H 2 O 1 lb mL 453.6 g 100 = 71% 150 lb
4.8 10 4 mL H 2 O
103.
Assuming that the radionuclide is long-lived enough that no significant decay occurs during the time of the experiment, the total counts of radioactivity injected are: 0.10 mL ×
5.0 10 3 cpm = 5.0 × 102 cpm mL
Assuming that the total activity is uniformly distributed only in the rat’s blood, the blood volume is: V× 104.
48 cpm = 5.0 × 102 cpm, V = 10.4 mL = 10. mL mL
a. From Table 18.1: 2 H2O + 2 e− → H2 + 2 OH− E° = − 0.83 V
Eocell = E oH2O − EoZr = − 0.83 V + 2.36 V = 1.53 V Yes, the reduction of H2O to H2 by Zr is spontaneous at standard conditions because E ocell > 0. b.
(2 H2O + 2 e− → H2 + 2 OH−) × 2 Zr + 4 OH− → ZrO2•H2O + H2O + 4 e− 3 H2O(l) + Zr(s) → 2 H2(g) + ZrO2•H2O(s)
CHAPTER 19
974
THE NUCLEUS: A CHEMIST’S VIEW
c. ΔG° = −nFE° = −(4 mol e−)(96,485 C/mol e−)(1.53 J/C) = −5.90 × 105 J = −590. kJ E = E° − E° =
0.0591 log Q; at equilibrium, E = 0 and Q = K. n
0.0591 4(1.53) log K, log K = = 104, K 10104 0.0591 n
d. 1.00 × 103 kg Zr ×
2 mol H 2 1000 g 1 mol Zr = 2.19 × 104 mol H2 kg 91.22 g Zr mol Zr
2.19 × 104 mol H2 ×
2.016 g H 2 = 4.42 × 104 g H2 mol H 2
(2.19 10 4 mol)( 0.08206 L atm/mol • K)(1273 K) nRT = 2.3 × 106 L H2 = P 1.0 atm
V=
e. Probably yes; less radioactivity overall was released by venting the H2 than what would have been released if the H2 had exploded inside the reactor (as happened at Chernobyl). Neither alternative is pleasant, but venting the radioactive hydrogen is the less unpleasant of the two alternatives. 105.
a.
12
b.
13
C; it takes part in the first step of the reaction but is regenerated in the last step. not consumed, so it is not a reactant.
12
C is
N, 13C, 14N, 15O, and 15N are the intermediates.
c. 4 11 H → 42 He + 2 +01 e ; Δm = 4.00260 u − 2 me + 2 me − [4(1.00782 u − me)] Δm = 4.00260 − 4(1.00782) + 4(0.000549) = −0.02648 u for four protons reacting For 4 mol of protons, Δm = −0.02648 g, and ΔE for the reaction is: ΔE = Δmc2 = −2.648 × 10 −5 kg × (2.9979 × 108 m/s)2 = −2.380 × 1012 J For 1 mol of protons reacting: 106.
a.
− 2.380 1012 J 4 mol 1 H
= −5.950 × 1011 J/mol 1H
4 0 → 222 86 Rn + ? 2 He + ? −1 e ; to account for the mass number change, four alphaparticles are needed. To balance the number of protons, two beta-particles are needed. 238 92 U
222 86 Rn
222 → 42 He + 218 Rn decays. 84 Po; polonium-218 is produced when
b. Alpha-particles cause significant ionization damage when inside a living organism. Because the half-life of 222Rn is relatively short, a significant number of alpha-particles will be produced when 222Rn is present (even for a short period of time) in the lungs. c.
218 4 214 → 42 He + 218 84 Po; 84 Po → 2 He + 82 Pb; polonium-218 is produced when radon-222 decays. 218Po is a more potent alpha-particle producer since it has a much shorter half-life than 222Rn. In addition, 218Po is a solid, so it can get trapped in the lung 222 86 Rn
CHAPTER 19
THE NUCLEUS: A CHEMIST’S VIEW
975
tissue once it is produced. Once trapped, the alpha-particles produced from polonium218 (with its very short half-life) can cause significant ionization damage. d. Rate = kN; rate =
4.0 pCi 1 10 −12 Ci 3.7 1010 decays/sec = 0.15 decays/s•L L pCi Ci
k=
ln 2 0.6391 1d 1h = = 2.10 × 10 −6 s −1 t1 / 2 3.82 d 24 h 3600 s
N=
0.15 decays/s • L rate = = 7.1 × 104 222Rn atoms/L − 6 −1 k 2.10 10 s
7.1 10 4 222 Rn atoms 1 mol 222 Rn = 1.2 × 10 −19 mol 222Rn/L 23 L 6.02 10 atoms 107.
Moles of I− = [I−] =
33 counts 1 mol I • min = 6.6 × 10 −11 mol I− 11 min 5.0 10 counts
6.6 10 −11 mol I − = 4.4 × 10 −10 mol/L 0.150 L
Hg2I2(s)
⇌
Initial s = solubility (mol/L) Equil.
Hg22+(aq)
+
0 s
2 I−(aq)
Ksp = [Hg22+][I−]2
0 2s
From the problem, 2s = 4.4 × 10 −10 mol/L, s = 2.2 × 10 −10 mol/L. Ksp = (s)(2s)2 = (2.2 × 10 −10 )(4.4 × 10 −10 )2 = 4.3 × 10 −29 108.
2 1H
E=
+ 21 H →
4 2 He;
Q for 21 H = 1.6 × 10 −19 C; mass of deuterium = 2 u.
9.0 10 9 J • m/C 2 (Q1Q 2 ) 9.0 10 9 J • m/C 2 (1.6 10 −19 C) 2 = r 2 10 −15 m = 1 × 10 −13 J per alpha particle
KE = 1/2 mv2; 1 × 10 −13 J = 1/2 (2 u × 1.66 × 10 −27 kg/u)v2, v = 8 × 106 m/s From the kinetic molecular theory discussed in Chapter 5: 1/ 2
3RT urms = M
, where M = molar mass in kilograms = 2 × 10 −3 kg/mol for deuterium
3(8.3145 J/K • mol)(T) 8 × 10 m/s = 2 10 −3 kg 6
1/2
, T = 5 × 109 K
CHAPTER 20 THE REPRESENTATIVE ELEMENTS Review Questions 1.
Oxygen and silicon are the two most abundant elements in the earth’s crust, oceans, and atmosphere. Oxygen is found in the atmosphere as O 2, in the oceans in H2O, and in the earth’s crust primarily in silicate and carbonate minerals. Because oxygen is everywhere, it is not too surprising that it is the most abundant element. The second most abundant element, silicon, is found throughout the earth’s crust in silica and silicate minerals that form the basis of most sand, rocks, and soils. Again, it is not too surprising that silicon is the second most abundant element, as it is involved in the composition of much of the earth. The four most abundant elements in the human body are oxygen, carbon, hydrogen, and nitrogen. Not surprisingly, these elements form the basis for all biologically important molecules in the human body. They should be abundant.
2.
Hydrogen forms many compounds in which the oxidation state is +1, as do the Group 1A elements. For example, H2SO4 and HCl compared to Na2SO4 and NaCl. On the other hand, hydrogen forms diatomic H2 molecules and is a nonmetal, while the Group 1A elements are metals. Hydrogen also forms compounds with a 1 oxidation state, which is not characteristic of Group 1A metals, e.g., NaH. Alkali metals have a ns1 valence shell electron configuration. Alkali metals lose this valence electron with relative ease to form M+ cations when in ionic compounds. They all are easily oxidized. Therefore, to prepare the pure metals, alkali metals must be produced in the absence of materials (H2O, O2) that are capable of oxidizing them. The method of preparation is electrochemical processes, specifically, electrolysis of molten chloride salts and reduction of alkali salts with Mg and H2. In all production methods, H2O and O2 must be absent. Table 20.5 of the text lists some reactions of alkali metals. From Table 20.5, MF, M 2S, M3P, MH, and MOH are the predicted formulas when alkali metals react with F 2, S, P4, H2, and H2O, respectively. These formulas exhibit the typical oxidation states as predicted by the periodic table.
3.
Alkaline earth metals have ns2 for valence electron configurations. They are all very reactive, losing their two valence electrons to nonmetals to form ionic compounds containing M2+ cations. Alkaline earth metals, like alkali metals, are easily oxidized. Their preparation as pure metals must be done in the absence of O 2 and H2O. The method of preparation is electrolysis of molten alkaline earth halides. Table 20.7 of the text summarizes the formulas of alkaline earth metals with typical nonmetals. With a rare exception or two, there aren’t many surprises. The typical formulas are MF 2, MO, MS, M3N2, MH2, and M(OH)2, when alkaline earth metals are reacted with F 2, O2, S, N2, H2, and H2O, respectively.
976
CHAPTER 20 4.
THE REPRESENTATIVE ELEMENTS
977
The valence electron configuration of Group 3A elements is ns2np1. The lightest Group 3A element, boron, is a nonmetal as most of its compounds are covalent. Aluminum, although commonly thought of as a metal, does have some nonmetallic properties as its bonds to other nonmetals have significant covalent character. The other Group 3A elements have typical metal characteristics; its compounds formed with nonmetals are ionic. From this discussion, metallic character increases as the Group 3A elements get larger. As discussed above, boron is a nonmetal in both properties and compounds formed. However aluminum has physical properties of metals like high thermal and electrical conductivities and a lustrous appearance. The compounds of aluminum with other nonmetals, however, do have some nonmetallic properties as the bonds have significant covalent character. The formulas that aluminum form with nonmetals (as well as the formulas of other heavier Group 3A elements) are summarized in Table 20.9. The compounds formed between aluminum and F2, O2, S, and N2 follow what would be predicted for ionic compounds containing Al3+. However, there is some covalent character in these compounds. The formulas are AlF 3, Al2O3, Al2S3, and AlN.
5.
The valence electron configuration of Group 4A elements is ns2np2. The two most important elements on earth are Group 4A elements. They are carbon, found in all biologically important molecules, and silicon, found in most of the compounds that make up the earth’s crust. They are important because they are so prevalent in compounds necessary for life and the geologic world. As with Group 3A, Group 4A shows an increase in metallic character as the elements get heavier. Carbon is a typical nonmetal, silicon and germanium have properties of both metals and nonmetals so they are classified as semimetals, while tin and lead have typical metallic characteristics. The two major allotropic forms of carbon are graphite and diamond. See Section 10.5 of the text for their structures and a description of their properties. Germanium is a relatively rare element and is classified as a semimetal. Its main uses are in the manufacture of semiconductors. Tin is a metal and is used to form alloys with other metals. Some alloys containing tin are bronze, solder, and pewter. Tin’s major current use is as a protective coating for steel which helps prevent the corrosion of iron in steel. Lead is a metal but has a relatively low melting point. Lead is very toxic and the use of lead paints and lead additives to gasoline are not allowed in the United States. The major use of lead is for electrodes in the lead storage battery used in automobiles. Ge forms GeF4 when reacted with F2 and forms GeO2 when reacted with O2. Both these compounds have Ge in the predicted +4 oxidation state as determined from its position in the periodic table. However, the fluorine compound is strictly covalent in nature, while the oxygen compound has more ionic character.
6.
Group 5A: ns2np3; As with groups 3A and 4A, metallic character increases going down a group. Nitrogen is strictly a nonmetal in properties, while bismuth, the heaviest Group 5A element, has mostly metallic physical properties. The trend of increasing metallic character going down the group is due in part to the decrease in electronegativity. Nitrogen, with its high electronegativity, forms covalent compounds as nonmetals do. Bismuth and antimony, with
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much lower electronegativities, exhibit ionic character in most of their compounds. Bismuth and antimony exist as +3 metal cations in these ionic compounds. This is due to nitrogen’s ability to form strong bonds whereas heavier group 5A elements do not form strong bonds. Therefore, P2, As2, and Sb2 do not form since two bonds are required to form these diatomic substances. White phosphorus consists of discrete tetrahedral P 4 molecules. The bond angles in the P 4 tetrahedrons are only 60°, which makes P 4 very reactive, especially towards oxygen. Red and black phosphorus are covalent network solids. In red phosphorus, the P 4 tetrahedra are bonded to each other in chains, making them less reactive than white phosphorus. They need a source of energy to react with oxygen, such as when one strikes a match. Black phosphorus is crystalline, with the P atoms tightly bonded to each other in the crystal and is fairly unreactive towards oxygen. 7.
N: 1s22s22p5; The extremes of the oxidation states for N can be rationalized by examining the electron configuration of N. Nitrogen is three electrons short of the stable Ne electron configuration of 1s22s22p6. Having an oxidation state of −3 makes sense. The +5 oxidation state corresponds to N “losing” its 5 valence electrons. In compounds with oxygen, the N−O bonds are polar covalent, with N having the partial positive end of the bond dipole. In the world of oxidation states, electrons in polar covalent bonds are assigned to the more electronegative atom; this is oxygen in N−O bonds. N can form enough bonds to oxygen to give it a +5 oxidation state. This loosely corresponds to losing all of the valence electrons. NH3: fertilizers, weak base properties, can form hydrogen bonds; N 2H4: rocket propellant, blowing agent in manufacture of plastics, can form hydrogen bonds; NH2OH: weak base properties, can form hydrogen bonds; N 2: makes up 78% of air, very stable compound with a very strong triple bond, is inert chemically; N2O: laughing gas, propellant in aerosol cans, effect on earth’s temperature being studied; NO: toxic when inhaled, may play a role in regulating blood pressure and blood clotting, one of the few odd electron species that forms; N 2O3, least common of nitrogen oxides, a blue liquid that readily dissociates into NO(g) and NO2(g); NO2: another odd electron species, dimerizes to form N2O4, plays a role in smog production; HNO3: important industrial chemical, used to form nitrogen-based explosives, strong acid and a very strong oxidizing agent. Hydrazine also can hydrogen bond because it has covalent N−H bonds as well as having a lone pair of electrons on each N. The high boiling point for hydrazine’s relatively small size supports this. Even though phosphine and ammonia have identical Lewis structures, the bond angles of PH 3 are only 94˚, well below the predicted tetrahedral bond angles of 109.5˚. PH3 is an unusual exception to the VSEPR model.
8.
Group 6A: ns2np4; as expected from the trend in other groups, oxygen has properties which are purely nonmetal. Polonium, on the other hand, has some metallic properties. The most significant property differences are radioactivity and toxicity. Polonium is only composed of radioactive isotopes, unlike oxygen, and polonium is highly toxic, unlike oxygen.
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THE REPRESENTATIVE ELEMENTS
979
The two allotropic forms of oxygen are O2 and O3. O2, 2(6) = 12 e−
O3, 3(6) = 18 e−
Ozone has a V-shape molecular structure with bond angles of 117˚, slightly less than the predicted 120˚ trigonal planar bond angle. Both rhombic and monoclinic sulfur exist in S8 rings. The difference between the two is that the S8 rings are stacked together differently giving different solid structures. Oxygen forms strong bonds and, because of this, exists in nature as O 2 molecules. Sulfur forms much stronger sigma bonds than bonds. Therefore, elemental sulfur is found in nature singly bonded to other sulfur atoms. We assume SO doesn’t form because of the difference in ability of oxygen and sulfur to form bonds. Sulfur forms relatively weak bonds as compared to oxygen. SO2(aq) + H2O(l) → H2SO3(aq); SO3(aq) + H2O(l) → H2SO4(aq); SO2 and SO3 dissolve in water to form the acids H2SO3 and H2SO4, respectively. The molecular structure of SO2 is bent with a ~120˚ bond angle. The molecular structure of SO 3 is trigonal planar with 120 bond angles. A dehydrating agent is one that has a high affinity for water. Sulfuric acid grabs water whenever it can. When it reacts with sugar (C12H22O11), it removes the hydrogen and oxygen in a 2 : 1 mole ratio even though there are no H 2O molecules in sugar. H2SO4 is indeed a powerful dehydrating agent. 9.
Group 7A: ns2np5; the diatomic halogens (X2) are nonpolar, so they only exhibit London dispersion (LD) intermolecular forces. The strength of LD forces increases with size. The boiling points and melting points steadily increase from F 2 to I2 because the strength of the intermolecular forces are increasing. Fluorine is the most reactive of the halogens because it is the most electronegative atom and the bond in the F2 molecule is very weak. HF exhibits the relatively strong hydrogen bonding intermolecular forces, unlike the other hydrogen halides. HF has a high boiling point due to its ability to form these hydrogen bonding interactions. The halide ion is the −1 charged ion that halogens form when in ionic compounds. As can be seen from the positive standard reduction potentials in Table 20.17 of the text, the halogens energetically favor the X− form over the X2 form. Because the reduction potentials are so large, this give an indication of the relative ease to which halogens will grab electrons to form the halide ion. In general, the halogens are highly reactive; that is why halogens exist as cations in various minerals and in seawater as opposed to free elements in nature. Some compounds of chlorine exhibiting the −1 to +7 oxidation state are: HCl (−1), HOCl (+1), HClO2 (+3), HClO3 (+5), and HClO4 (+7).
980 10.
CHAPTER 20
THE REPRESENTATIVE ELEMENTS
The noble gases have filled s and p valence orbitals (ns2np6 = valence electron configuration). They don’t need to react like other representative elements to achieve the stable ns2np6 configuration. Noble gases are unreactive because they do not want to lose their stable valence electron configuration. Noble gases exist as free atoms in nature. They only exhibit London dispersion forces in the condensed phases. Because LD forces increase with size, as the noble gas gets bigger, the strength of the intermolecular forces get stronger leading to higher melting and boiling points. In Mendeleev's time, none of the noble gases were known. Since an entire family was missing, no gaps seemed to appear in the periodic arrangement. Mendeleev had no evidence to predict the existence of such a family. The heavier members of the noble gases are not inert. Xe and Kr have been shown to react and form compounds with other elements. XeF2: 180˚, dsp3; XeO2F2: ~90˚ and ~120˚, dsp3; XeO3: < 109.5˚, sp3; XeO4: 109.5˚, sp3; XeF4: 90˚, d2sp3; XeO3F2: 90˚ and 120˚, dsp3; XeO2F4, 90˚, d2sp3
Active Learning Questions 1.
In ionic compounds, fluorine forms −1 charged ions. Covalent compounds with F are harder to predict. Use your experience with drawing Lewis structures to come up with example covalent compounds. a. HF
2.
b. NaF
f.
PF3 and PF5;
i.
XeF2, XeF4, and XeF6
c. CaF2
g. SF2, SF4, and SF6
d. GaF3
e. CF4 , CF2CF2, + others
h. BrF, BrF3, and BrF5
a. Carbon makes up 18% of the human body with a huge variety of compounds. The study of carbon and its compounds is called organic chemistry. b. Silicon is found throughout the earth’s crust in silica and silicate minerals that form the basis for sand, rocks, and soil. c. Oxygen is found in O2 and H2O; two compounds essential for life to exist. Oxygen is also important to the chemistry of silicon. Silicon does not form strong bonds to itself but does form strong bonds to oxygen. The compounds of silicon are dominated by Si−O bonds. d.
Carbon has the unusual ability to form strong bonds to itself leading to chains and rings of carbon atoms. The chains of carbon atoms can be short or extremely long, and anywhere in between.
e. Silicon forms stronger bonds to oxygen than to itself. The silica and silicate minerals that make up sand, rocks, and soil consist of SiO4 tetrahedra linked together with Si−O−Si bridges. f.
Because oxygen forms strong sigma bonds to hydrogen, water (H‒O‒H) is a very stable compound. O2 is another stable compound containing oxygen. The bonding in O2 requires the oxygen atoms to form a π bond. When parallel unhybridized p atomic orbitals from two O atoms combine to form the π bond, there is good overlap of the orbitals forming a strong π bond.
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THE REPRESENTATIVE ELEMENTS
981
Questions 3.
The gravity of the earth is not strong enough to keep the light H2 molecules in the atmosphere.
4.
(1) Ammonia production and (2) hydrogenation of vegetable oils.
5.
Calcium is found in the structural material that make up bones and teeth. Magnesium plays a vital role in metabolism and in muscle function.
6.
Size decreases from left to right and increases going down the periodic table. So, going one element right and one element down would result in a similar size for the two elements diagonal to each other. The ionization energies will be similar for the diagonal elements since the periodic trends also oppose each other. Electron affinities are harder to predict, but atoms with similar size and ionization energy should also have similar electron affinities.
7.
For Groups 1A-3A, the small sizes of H (as compared to Li), Be (as compared to Mg), and B (as compared to Al) seem to be the reason why these elements have nonmetallic properties, while others in the Groups 1A-3A are strictly metallic. The small sizes of H, Be, and B also cause these species to polarize the electron cloud in nonmetals, thus forcing a sharing of electrons when bonding occurs. For Groups 4A-6A, a major difference between the first and second members of a group is the ability to form bonds. The smaller elements form stable bonds, while the larger elements are not capable of good overlap between parallel p orbitals and, in turn, do not form strong bonds. For Group 7A, the small size of F as compared to Cl is used to explain the low electron affinity of F and the weakness of the F−F bond.
8.
SiC would have a covalent network structure similar to diamond.
9.
Solids have stronger intermolecular forces than liquids. In order to maximize the hydrogen bonding in the solid phase, ice is forced into an open structure. This open structure is why H2O(s) is less dense than H2O(l).
10.
Nitrogen fixation is the process of transforming atmospheric N2 to other nitrogen-containing compounds. Some examples are: N2(g) + 3 H2(g) → 2 NH3(g) N2(g) + O2(g) → 2 NO(g) N2(g) + 2 O2(g) → 2 NO2(g)
11.
Group 1A and 2A metals are all easily oxidized. They must be produced in the absence of materials (H2O, O2) that are capable of oxidizing them.
12.
All the choices factor in when explaining the property differences between P and N. Nitrogen forms strong pi bonds. The large size of P does not allow for effective overlap of parallel p atomic orbitals required to form strong pi bonds. The large electronegativity of N allows it to form a very polar bond to H, which gives rise to hydrogen bonding intermolecular forces. P-H bonds are nonpolar. The existence of empty valence d orbitals allows phosphorus to expand its octet when forming compounds.
982 13.
CHAPTER 20
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The reaction N2(g) + 3H2(g) ⇌ 2NH3(g) is exothermic. Thus, the value of the equilibrium constant K decreases as the temperature increases. Lower temperatures are favored for maximum yield of ammonia. However, at lower temperatures the rate is slow; without a catalyst the rate is too slow for the process to be feasible. The discovery of a catalyst increased the rate of reaction at a lower temperature favored by thermodynamics.
14.
In all these compounds, the phosphorus atoms are in a tetrahedral arrangement. P 4O6 and P4O10 are formed by adding oxygen atoms to the base tetrahedral structure of P 4.
15.
Boranes are covalent compounds made up of boron and hydrogen. The B−H bonds in boranes are relatively weak and are very reactive. When boranes are reacted with oxygen, the product bonds are much stronger than the reactant bonds. When this is the case, very exothermic reactions result which is a common trait for good rocket fuels.
16.
The xenon fluorides react with oxygen and water to form oxycompounds with xenon like XeO3, XeO4, and XeOF4; some of which are explosive.
17.
Atomic size decreases from left to right across the periodic table. This means the alkali metals have the largest radius. The electronegativity trend is to increase from left to right across the periodic table. The noble gases are the group furthest to the right in the periodic table. However, electronegativity is a bonding term. Since noble gases are generally unreactive, they don’t have electronegativity values. The group having the highest electronegativity values is the halogens.
18.
Pi bonds are formed from overlap of parallel p atomic orbitals. The parallel p orbitals on C, N, and O overlap effectively to form strong π bonds. The larger size for Si, P, and S limits how effectively parallel p atomic orbitals can overlap with each other. Because of the poor side-toside overlap, Si, P, and S do not form pi bonds.
Exercises Group 1A Elements 19.
a. ΔH° = −110.5 − [−75 + (−242)] = 207 kJ; ΔS° = 198 + 3(131) − [186 + 189] = 216 J/K b. ΔG° = ΔH° − TΔS°; ΔG° = 0 when T =
ΔH o 207 10 3 J = = 958 K o 216 J / K ΔS
At T > 958 K and standard pressures, the favorable ΔS° term dominates, and the reaction is spontaneous (ΔG° < 0). 20.
a. ΔH° = 2(−46 kJ) = −92 kJ; ΔS° = 2(193 J/K) − [3(131 J/K) + 192 J/K] = −199 J/K ΔG° = ΔH° − TΔS° = −92 kJ − 298 K(−0.199 kJ/K) = −33 kJ b. Because ΔG° is negative, this reaction is spontaneous at standard conditions. c. ΔG° = 0 when T =
ΔH o − 92 kJ = = 460 K o − 0.199 kJ / K ΔS
At T < 460 K and standard pressures, the favorable ΔH° term dominates, and the reaction is spontaneous (ΔG° < 0).
CHAPTER 20
THE REPRESENTATIVE ELEMENTS
983
21.
The atomic radius of H is much smaller than the atomic radius of Li. The small H atom has a much greater attraction for electrons than Li. This attraction for electrons results in H acting as a nonmetal when it forms compounds. Li acts as a typical metal when forming compounds; Li loses an electron to form 1+ charged cations in its ionic compounds.
22.
Lithium has a higher melting point than sodium and potassium. When Na and K are added to water, the enthalpy change for the reaction is large energy to melt the Na and K, making it easier for water to react with them. With Li having the higher melting point, the enthalpy change for the reaction with water is not large enough to melt Li; it remains a solid as it reacts with water. This limits the surface area contact with water resulting in a slower reaction.
23.
4 Li(s) + O2(g) → 2 Li2O(s) 2 Li(s) + S(s) → Li2S(s); 2 Li(s) + Cl2(g) → 2 LiCl(s) 12 Li(s) + P4(s) → 4 Li3P(s); 2 Li(s) + H2(g) → 2 LiH(s) 2 Li(s) + 2 H2O(l) → 2 LiOH(aq) + H2(g); 2 Li(s) + 2 HCl(aq) → 2 LiCl(aq) + H2(g)
24.
We need another reactant beside NaCl(aq) because oxygen and hydrogen are in some of the products. The obvious choice is H2O. 2 NaCl(aq) + 2 H2O(l) → Cl2(g) + H2(g) + 2 NaOH(aq) Note that hydrogen is reduced, and chlorine is oxidized in this electrolysis process.
25.
When lithium reacts with excess oxygen, Li2O forms, which is composed of Li+ and O2− ions. This is called an oxide salt. When sodium reacts with oxygen, Na2O2 forms, which is composed of Na+ and O22−ions. This is called a peroxide salt. When potassium (or rubidium or cesium) reacts with oxygen, KO2 forms, which is composed of K+ and O2− ions. For your information, this is called a superoxide salt. So the three types of alkali metal oxides that can form differ in the oxygen anion part of the formula (O 2− versus O22− versus O2−). Each of these anions has unique bonding arrangements and oxidation states.
26.
The first illustration is an example of a covalent hydride like H 2O. Covalent hydrides are just binary covalent compounds formed between hydrogen and some other nonmetal and exist as individual molecules. The middle illustration represents interstitial (or metallic) hydrides. In interstitial hydrides, hydrogen atoms occupy the holes of a transition metal crystal. These hydrides are more like solid solutions than true compounds. The third illustration represents ionic (or salt-like) hydrides like LiH. Ionic hydrides form when hydrogen reacts with a metal from Group 1A or 2A. The metals lose electrons to form cations and the hydrogen atoms gain electrons to form the hydride anions (H−). These are just ionic compounds formed between a metal and hydrogen.
27.
The small size of the Li+ cation results in a much greater attraction to water. The attraction to water is not so great for the other alkali metal ions. Thus, lithium salts tend to absorb water.
28.
Counting over in the periodic table, the next alkali metal will be element 119. It will be located under Fr. One would expect the physical properties of element 119 to follow the trends shown in Table 20.4. Element 119 should have the smallest ionization energy, the most negative
984
CHAPTER 20
THE REPRESENTATIVE ELEMENTS
standard reduction potential, the largest radius, and the smallest melting point of all the alkali metals listed in Table 20.4. It should also be radioactive like Fr.
Group 2A Elements 29.
CaCO3(s) + H2SO4(aq) → CaSO4(aq) + H2O(l) + CO2(g)
30.
2 Sr(s) + O2(g) → 2 SrO(s); Sr(s) + S(s) → SrS(s) Sr(s) + Cl2(g) → SrCl2(s); 6 Sr(s) + P4(s) → 2 Sr3P2(s) Sr(s) + H2(g) → SrH2(s); Sr(s) + 2 H2O(l) → Sr(OH)2(aq) + H2(g) Sr(s) + 2 HCl(aq) → SrCl2(aq) + H2(g)
31.
Mg2+ + 2 e− → Mg 1.00 × 106 g Mg ×
32.
1 mol Mg 2 mol e− 96, 465 C 1s 1h = 44.1 hours − 4 24.31 g mol Mg mol e 5.00 10 C 3600 s
Alkaline earth metal chlorides (MCl2) are composed of M2+ and Cl− ions. In the electrolysis, the metal ions are reduced to the pure metal (M 2+ + 2 e− → M) and the Cl− ions are oxidized to chlorine gas (2 Cl− → Cl2 + 2 e−). So the metal will be produced at the cathode and Cl2(g) will be produced at the anode. Mol M = 748 s
5.00 C 1 mol e − 1 mol M = 1.94 × 10−2 mol M s 96,485 C 2 mol e −
Molar mass of M =
0.471 g M 1.94 10 − 2 mol M
= 24.3 g/mol; M is Mg, so MgCl2 was electrolyzed.
33.
Beryllium has a small size and a large electronegativity as compared to the other alkaline earth metals. The electronegativity of Be is so high that it does not readily give up electrons to nonmetals, as is the case for the other alkaline earth metals. Instead, Be has significant covalent character in its bonds; it prefers to share valence electrons rather than give them up to form ionic bonds.
34.
The alkaline earth ions that give water the hard designation are Ca 2+ and Mg2+. These ions interfere with the action of detergents and form unwanted precipitates with soaps. Large-scale water softeners remove Ca2+ by precipitating out the calcium ions as CaCO 3. In homes, Ca2+ and Mg2+ (plus other cations) are removed by ion exchange. See Figure 20.6 for a schematic of a typical cation exchange resin.
35.
CaCO3(s) Initial Equil.
⇌
s = solubility (mol/L)
Ca2+(aq) 0 s
+ CO32−(aq) 0 s
Ksp = 8.7 × 10 −9 = [Ca2+][CO32−] = s2, s = 9.3 × 10 −5 mol/L
CHAPTER 20 36.
THE REPRESENTATIVE ELEMENTS
985
1 mg F − 1g 1 mol F− = 5.3 × 10−5 M F− = 5 × 10−5 M F− L 1000 mg 19.0 g F−
CaF2(s) ⇌ Ca2+(aq) + 2 F−(aq) Ksp = [Ca2+][F−]2 = 4.0 × 10−11; precipitation will occur when Q > Ksp. Let’s calculate [Ca2+] so that Q = Ksp. Q = 4.0 × 10−11 = [Ca2+]0[F−]02 = [Ca2+]0(5 × 10−5)2, [Ca2+]0 = 2 × 10−2 M CaF2(s) will precipitate when [Ca2+]0 > 2 × 10−2 M. Therefore, hard water should have a calcium ion concentration of less than 2 × 10−2 M in order to avoid precipitate formation.
Group 3A Elements 37.
Nh: [Rn]7s25f146d107p1; Nh falls below Tl in the periodic table. Like Tl, we would expect Nh to form +1 and +3 oxidation states in its compounds.
38.
Tl2O3, thallium(III) oxide; Tl2O, thallium(I) oxide; InCl3, indium(III) chloride; InCl, indium(I) chloride
39.
B2H6(g) + 3 O2(g) → 2 B(OH)3(s)
40.
B2O3(s) + 3 Mg(s) → 3 MgO(s) + 2 B(s)
41.
2 Ga(s) + 3 F2(g) → 2 GaF3(s); 4 Ga(s) + 3 O2(g) → 2 Ga2O3(s) 2Ga(s) + 3 S(s) → Ga2S3(s); 2 Ga(s) + 6 HCl(aq) → 2 GaCl3(aq) + 3 H2(g)
42.
2 Al(s) + 2 NaOH(aq) + 6 H2O(l) → 2 Al(OH)4−(aq) + 2 Na+(aq) + 3 H2(g)
43.
An amphoteric substance is one that can behave as either an acid or as a base. Al 2O3 dissolves in both acidic and basic solutions. The reactions are: Al2O3(s) + 6 H+(aq) → 2Al3+(aq) + 3 H2O(l) Al2O3(s) + 2 OH−(aq) + 3 H2O(l) → 2 Al(OH)4−(aq)
44.
In solution, Al3+ forms a hydrated ion having the formula Al(H2O)63+. The Ka for Al(H2O)63+ is 1.4 × 10−5 and Ka reaction is: Al(H2O)63+ ⇌ Al(H2O)5(OH)2+ + H+ The high charge of the Al3+ ion polarizes the O‒H bonds in the hydrated waters, giving the hydrated ion the ability to donate a proton as a weak acid.
45.
Compounds called boranes have three-centered bonds. Three-centered bonds occur when a single H atom forms bridging bonds between two boron atoms. The bonds have two electrons bonding all three atoms together. The bond is electron-deficient and makes boranes very reactive.
986 46.
CHAPTER 20
THE REPRESENTATIVE ELEMENTS
Gallium has a relatively low melting point of 30°C, but a relatively high boiling point of 2400°C. Gallium is a liquid over the largest temperature range of any element. This makes gallium a good choice for high temperature thermometers.
Group 4A Elements 47.
Compounds containing Si‒Si single and multiple bonds are rare, unlike compounds of carbon. The bond strengths of the Si‒Si and C‒C single bonds are similar. The difference in bonding properties must be for other reasons. One reason is that silicon does not form strong π bonds, unlike carbon. Another reason is that silicon forms particularly strong sigma bonds to oxygen, resulting in compounds with Si‒O bonds instead of Si‒Si bonds.
48.
CO2 is a molecular substance composed of individual CO 2 molecules. SiO2 does not exist as discreet molecules. Instead, SiO2 is the empirical formula for quartz, which is composed of a network of SiO4 tetrahedra with shared oxygen atoms between the various tetrahedra. The major reason for the difference in structures is that carbon can form bonds, whereas silica does not form stable bonds. To form discrete CO2 molecules, bonds must form. Because silicon does not form stable bonds, silicon atoms achieve a noble gas configuration by forming several Si‒O single bonds. These Si‒O single bonds extend in all directions, giving the network structure of quartz.
49. The darker green orbitals about carbon are sp hybrid orbitals. The lighter green orbitals about each oxygen are sp2 hybrid orbitals, and the gold orbitals about all of the atoms are unhybridized p atomic orbitals. In each double bond in CO2, one sigma and one bond exists. The two carbon-oxygen sigma bonds are formed from overlap of sp hybrid orbitals from carbon with a sp2 hybrid orbital from each oxygen. The two carbon-oxygen bonds are formed from side-to-side overlap of the unhybridized p atomic orbitals from carbon with an unhybridized p atomic orbital from each oxygen. These two bonds are oriented perpendicular to each other as illustrated in the figure. 50.
The formulas for the stable oxides of carbon are CO, CO2, and C3O2. CO, 4 + 6 = 10 e−;
CO2, 4 + 2(6) = 16 e−;
C3O2, 3(4) + 2(6) = 24 e−
For the bonding in CO, both carbon and oxygen are sp hybridized. In the CO triple bond, 1 sigma and 2 bonds exist. Overlap of sp hybrid orbitals from carbon and oxygen form the sigma bond. The two bonds are formed from side-to-side overlap of unhybridized p atomic orbitals from carbon and oxygen. These bonds are oriented perpendicular to each other. For the bonding in C3O2, the carbon atoms are all sp hybridized while the terminal oxygen atoms are sp2 hybridized. In each double bond in C3O2, one sigma and one bond exists. The sigma bonds between the carbon and oxygen atoms are formed from overlap of an sp hybrid orbital from carbon with an sp2 hybrid orbital from oxygen. The sigma bond between the
CHAPTER 20
THE REPRESENTATIVE ELEMENTS
987
carbon atoms are both formed from overlap of sp hybrid orbitals from each carbon. All four bonds in the C3O2 molecule are formed from side-to-side overlap of unhybridized p atomic orbitals from the various atoms. 51.
a. SiO2(s) + 2 C(s) → Si(s) + 2 CO(g) b. SiCl4(l) + 2 Mg(s) → Si(s) + 2 MgCl2(s) c. Na2SiF6(s) + 4 Na(s) → Si(s) + 6 NaF(s)
52.
Sn(s) + 2 Cl2(g) → SnCl4(s); Sn(s) + O2(g) →SnO2(s) Sn(s) + 2 HCl(aq) → SnCl2(aq) + H2(g)
53.
Pb3O4: we assign −2 for the oxidation state of O. The sum of the oxidation states of Pb must be +8. We get this if two of the lead atoms are Pb(II) and one is Pb(IV). Therefore, the mole ratio of lead(II) to lead(IV) is 2 : 1.
54.
Sn(s) + 2F2(g) → SnF4(s), tin(IV) fluoride; Sn(s) + F2(g) → SnF2(s), tin(II) fluoride
55.
The electronegativity of Pb4+ is larger than the electronegativity of Pb2+. With the higher electronegativity, Pb4+ forms covalent compounds with nonmetals, unlike the ionic compounds formed by Pb2+. PbCl4: 4 + 4(7) = 32 valence electrons:
The central Pb atom exhibits a tetrahedral arrangement of the bonded atoms giving expected bond angles of 109.5º. 56.
Sn(s) + 2 H+(aq) → Sn2+(aq) + H2(g)
Group 5A Elements 57.
NO43− 3-
O N O
O O
Both NO43− and PO43− have 32 valence electrons, so both have similar Lewis structures. From the Lewis structure for NO43−, the central N atom has a tetrahedral arrangement of electron pairs. N is small. There is probably not enough room for all 4 oxygen atoms around N. P is larger; thus PO43− is stable.
988
CHAPTER 20 PO3− O P O
58.
a. PF5;
O
THE REPRESENTATIVE ELEMENTS
PO3− and NO3− each have 24 valence electrons, so both have similar Lewis structures. From the Lewis structure for PO3−, PO3− has a trigonal planar arrangement of electron pairs about the central P atom (two single bonds and one double bond). P=O bonds are not particularly stable, while N=O bonds are stable. Thus NO3− is stable.
N is too small and doesn't have low-energy d-orbitals to expand its octet to form NF5.
b. AsF5; I is too large to fit 5 atoms of I around As. c. NF3; N is too small for three large bromine atoms to fit around it. 59.
Production of bismuth: 2 Bi2S3(s) + 9 O2(g) → 2 Bi2O3(s) + 6 SO2(g); 2 Bi2O3(s) + 3 C(s) → 4 Bi(s) + 3 CO2(g) Production of antimony: 2 Sb2S3(s) + 9 O2(g) → 2 Sb2O3(s) + 6 SO2(g); 2 Sb2O3(s) + 3 C(s) → 4 Sb(s) + 3 CO2(g)
60.
4 As(s) + 3 O2(g) → As4O6(s); 4 As(s) + 5 O2(g) → As4O10(s) As4O6(s) + 6 H2O(l) → 4 H3AsO3(aq); As4O10(s) + 6 H2O(l) → 4 H3AsO4(aq)
61.
NH3, 5 + 3(1) = 8 e−
Trigonal pyramid; sp3
AsCl5, 5 + 5(7) = 40 e−
Trigonal bipyramid; dsp3
PF6−, 5 + 6(7) + 1 = 48 e−
Octahedral; d2sp3 Nitrogen does not have low-energy d orbitals it can use to expand its octet. Both NF 5 and NCl6− would require nitrogen to have more than 8 valence electrons around it; this never happens.
CHAPTER 20 62.
THE REPRESENTATIVE ELEMENTS
989
AsF3Cl2, 5 + 5(7) = 40 e−
AsCl4+, 5 + 4(7) − 1 = 32 e−
AsF6−, 5 + 6(7) + 1 = 48 e−
+
Cl As Cl
Cl Cl
63.
Nitrogen can form strong pi bonds with itself. This is not the case for the other larger atoms in Group 5A which do not form strong pi bonds because of poor side-to-side overlap of the larger valence p atomic orbitals. The P4, As4, and Sb4 molecules are composed entirely of single bonds.
64.
MO model (where B.O. = bond order): NO+: (σ2s)2(σ2s*)2(π2p)4(σ2p)2; B.O = (8 − 2)/2 = 3, 0 unpaired e− (diamagnetic) NO: (σ2s)2(σ2s*)2(π2p)4(σ2p)2(π2p*)1; B.O. = 2.5, 1 unpaired e− (paramagnetic) NO-: (σ2s)2(σ2s*)2(π2p)4(σ2p)2(π2p*)2; B.O. = 2, 2 unpaired e− (paramagnetic) N
Lewis structures: NO+:
NO:
NO−:
N
O
N
O
N
+
O
N
O
O
The two models give the same results only for NO + (a triple bond with no unpaired electrons). Lewis structures are not adequate for NO and NO −. The MO model gives a better representation for all three species. For NO, Lewis structures are poor for odd electron species. For NO −, both models predict a double bond, but only the MO model correctly predicts that NO − is paramagnetic.
990
CHAPTER 20
THE REPRESENTATIVE ELEMENTS
65.
1/2 N2(g) + 1/2 O2(g) → NO(g) ΔG° = ΔG of , NO = 87 kJ/mol; by definition, ΔG of for a compound equals the free energy change that would accompany the formation of 1 mol of that compound from its elements in their standard states. NO (and some other oxides of nitrogen) have weaker bonds as compared to the triple bond of N 2 and the double bond of O2. Because of this, NO (and some other oxides of nitrogen) have higher (positive) standard free energies of formation as compared to the relatively stable N 2 and O2 molecules.
66.
ΔH° = 2(90. kJ) − (0 + 0) = 180. kJ; ΔS° = 2(211 J/K) − (192 + 205) = 25 J/K ΔG° = 2(87 kJ) − (0) = 174 kJ At the high temperatures in automobile engines, the reaction N 2 + O2 → 2 NO becomes spontaneous since the favorable ΔS° term will become dominate. In the atmosphere, even though 2 NO → N2 + O2 is spontaneous at the cooler temperatures of the atmosphere, it doesn't occur because the rate is slow. Therefore, higher concentrations of NO are present in the atmosphere as compared to what is predicted by thermodynamics.
67.
The pollution provides sources of nitrogen and phosphorus nutrients so that the algae can grow. The algae consume oxygen, which decrease the dissolved oxygen levels below that required for other aquatic life to survive, and fish die.
68.
For a buffer solution: pH = pKa + log
2−
−
From the problem,
[H 2 PO4 ] 2−
[HPO4 ]
[ HPO 4 ] [base] ; pH = −log(6.2 × 10 −8 ) + log − [acid] [ H 2 PO 4 ]
= 1.1, so: pH = 7.21 + log
1 , pH = 7.21 – 0.041 = 7.17 1 .1
69.
The acidic hydrogens in the oxyacids of phosphorus all are bonded to oxygen. The hydrogens bonded directly to phosphorus are not acidic. H 3PO4 has three oxygen-bonded hydrogens, and it is a triprotic acid. H3PO3 has only two of the hydrogens bonded to oxygen, and it is a diprotic acid. The third oxyacid of phosphorus, H3PO2, has only one of the hydrogens bonded to an oxygen; it is a monoprotic acid.
70.
TSP = Na3PO4; PO43− is the conjugate base of the weak acid HPO 42− (Ka = 4.8 × 10 −13 ). All conjugate bases of weak acids are effective bases (K b = Kw/Ka = 1.0 × 10 −14 /4.8 × 10 −13 = 2.1 × 10 −2 ). The weak base reaction of PO43− with H2O is PO43− + H2O ⇌ HPO42− + OH− Kb = 2.1 × 10 −2 .
Group 6A Elements 71.
O=O‒O → O=O + O Break O‒O bond: ΔH =
146 kJ 1 mol = 2.42 × 10 −22 kJ = 2.42 × 10 −19 J mol 6.022 10 23
A photon of light must contain at least 2.42 × 10 −19 J to break one O‒O bond. Ephoton =
(6.626 10 −34 J s) (2.998 10 8 m/s) hc , λ = = 8.21 × 10 −7 m = 821 nm λ 2.42 10 −19 J
CHAPTER 20
THE REPRESENTATIVE ELEMENTS
72.
From Figure 7.2 in the text, light from violet to green will work.
73.
H2SeO4(aq) + 3 SO2(g) → Se(s) + 3 SO3(g) + H2O(l)
74.
a. 2 SO2(g) + O2(g) → 2 SO3(g);
991
b. SO3(g) + H2O(l) → H2SO4(aq)
c. C12H22O11(s) + 11 H2SO4(conc) → 12 C(s) + 11 H2SO4•H2O(l) 75.
In the upper atmosphere, O3 acts as a filter for ultraviolet (UV) radiation:
O3 is also a powerful oxidizing agent. It irritates the lungs and eyes, and at high concentration, it is toxic. The smell of a "spring thunderstorm" is O3 formed during lightning discharges. Toxic materials don't necessarily smell bad. For example, HCN smells like almonds. 76.
Chlorine is a good oxidizing agent. Similarly, ozone is a good oxidizing agent. After chlorine reacts, residues of chloro compounds are left behind. Long-term exposure to some chloro compounds may cause cancer. Ozone would not break down and form harmful substances. The major problem with ozone is that because virtually no ozone is left behind after initial treatment, the water supply is not protected against recontamination. In contrast, for chlorination, significant residual chlorine remains after treatment, thus reducing (eliminating) the risk of recontamination.
77.
ΔG = ΔH − TΔS; Srhombic(s) → Smonoclinic(s); from the problem, rhombic sulfur converts to monoclinic sulfur only at high temperatures. For this to be true, the signs on ΔH and ΔS must both be positive. At high enough temperatures, the favorable ΔS term starts to dominate over the unfavorable ΔH term, and the process becomes spontaneous. Because the sign of ΔS is positive for the process Srhombic(s) → Smonoclinic(s), Smonoclinic has the more disordered structure (has the larger positional probability).
78.
S(s) + O2(g) → SO2(g), combustion of coal 2 SO2(g) + O2(g) → 2 SO3(g), reaction with atmospheric O2 SO3(g) + H2O(l) → H2SO4(aq), reaction with atmospheric H2O Dust particles and water droplets catalyze the conversion of SO2 into SO3 in the atmosphere.
79.
O2: (σ2s)2(σ2s*)2(σ2p)2(π2p)4(π2p*)2; the MO electron configuration of O2 has two unpaired electrons in the degenerate antibonding (π*2 p ) orbitals. A substance with unpaired electrons is paramagnetic (see Figure 9.39).
80.
SO2, 6 + 2(6) = 18 e−
SO3, 6 + 3(6) = 24 e−
992
CHAPTER 20
THE REPRESENTATIVE ELEMENTS
The molecular structure of SO2 is bent with a 119 bond angle (close to the predicted 120 trigonal planar geometry). The molecular structure of SO3 is trigonal planar with 120 bond angles. Both SO2 and SO3 exhibit resonance. Both sulfurs in SO2 and SO3 are sp2 hybridized. To explain the equal bond lengths that occur in SO 2 and SO3, the molecular orbital model assumes that the electrons are delocalized over the entire surface of the molecule. The orbitals that form the delocalized bonding system are unhybridized p atomic orbitals from the sulfurs and oxygens in each molecule. When all the p atomic orbitals overlap together, there is a cloud of electron density above and below the entire surface of the molecule. Because the electrons are delocalized over the entire surface of the molecule in SO 2 and SO3, all of the S−O bonds in each molecule are equivalent.
Group 7A Elements 81.
O2F2 has 2(6) + 2(7) = 26 valence e−; from the following Lewis structure, each oxygen atom has a tetrahedral arrangement of electron pairs. Therefore, bond angles are 109.5° and each O is sp3 hybridized. F
O
O
F
Formal Charge Formal charge
00
00
00
00
Oxid. Number
-1
+1
+1
-1
−1
Oxidation state
+1
+1
−1
Oxidation states are more useful. We are forced to assign +1 as the oxidation state for oxygen. Oxygen is very electronegative, and +1 is not a stable oxidation state for O. 82.
OF2, 6 + 2(7) = 20 e−
V-shaped; <109.5; sp3 Because fluorine is more electronegative than oxygen, each fluorine atom in OF2 has a −1 oxidation number, which results in oxygen having a +2 oxidation number. Oxygen is very electronegative, so +2 is an extremely unstable oxidation state for this element. One would expect OF2, with oxygen in the +2 oxidation state, to be an even stronger oxidizing agent than O2F2, which has oxygen in the +1 oxidation state. 83.
SF2, 6 + 2(7) = 20 e−
SF4, 6 + 4(7) = 34 e−
SF6, 6 + 6(7) = 48 e−
V-shaped; <109.5
See-saw; 90, 120
Octahedral; 90
CHAPTER 20
THE REPRESENTATIVE ELEMENTS
993
OF4 would have the same Lewis structure as SF 4. To form OF4, the central oxygen atom must expand its octet. O is too small and doesn’t have low-energy d orbitals available to expand its octet. Therefore, OF4 would not be a stable compound. 84.
Selenium should form compounds like those that sulfur forms because both are group 6A nonmetals. Because sulfur forms covalent compounds with halogens having SX 2, SX4, and SX6 formulas, one would predict selenium and chlorine to form covalent compounds having the formulas SeCl2, SeCl4, and SeCl6.
85.
Sn4+ has a relatively large electronegativity compared to Sn 2+. The compounds of Sn4+ have significant covalent character unlike the ionic compounds of Sn 2+. The ionic forces that hold ionic compounds together in the condensed phases are always much stronger than the covalent intermolecular forces that hold covalent compounds together. Hence the covalent SnCl 4 compound has a much smaller boiling point as compared to the ionic SnCl2 compound.
86.
A disproportion reaction is an oxidation-reduction reaction in which one species will act as both the oxidizing agent and the reducing agent. The species reacts with itself, forming products with higher and lower oxidation states. Here, chlorine is in the zero-oxidation state in Cl2 and goes to the +1 oxidation state in HOCl and the ‒1 oxidation state in Cl ‒.
87.
The oxyacid strength increases as the number of oxygens in the formula increase. Therefore, the order of the oxyacids from weakest to strongest acid is HOCl < HClO2 < HClO3 < HClO4.
88.
One reason is that the H‒F bond is stronger than the other hydrohalides, making it more difficult to form H+ and F−. The main reason HF is a weak acid is entropy. When F −(aq) forms from the dissociation of HF, there is a high degree of ordering that takes place as water molecules hydrate this small ion. Entropy is considerably more unfavorable for the formation of hydrated F- than for the formation of the other hydrated halides. The result of the more unfavorable ΔS° term is a positive ΔG° value that leads to a K a value less than one.
Group 8A Elements 89.
Helium is formed from the α-decay of radioactive elements. The α particles pick up electrons from the environment to form helium atoms.
90.
Inert means unreactive. No compounds containing noble gases were know before 1962. In 1962, the first stable compound containing a noble gas was discovered.
91.
Xe has one more valence electron than I. Thus, the isoelectric species will have I plus one extra electron substituted for Xe, giving a species with a net minus one charge. a. IO4−
92.
b. IO3−
a. KrF2, 8 + 2(7) = 22 e−
c. IF2−
d. IF4−
e. IF6−
b. KrF4, 8 + 4(7) = 36 e− F
F
Kr
F
Linear; 180°; dsp3
F Kr
F
F
Square planar; 90°; d2sp3
994
CHAPTER 20
THE REPRESENTATIVE ELEMENTS
c. XeO2F2, 8 + 2(6) + 2(7) = 34 e− F
O
O
O
F or
Xe
F or
Xe
O
F F
Xe O
O
F
All are see-saw; 90° and 120°; dsp3 d.
XeO2F4, 8 + 2(6) + 4(7) = 48 e− O F
F F or
Xe F
F
F
O Xe
F
O
O F
All are octahedral; 90°; d2sp3 93.
Helium is unreactive and doesn't combine with any other elements. It is a very light gas and would easily escape the earth's gravitational pull as the planet was formed.
94.
Because Ar and Rn are both noble gases, both species will be relatively unreactive. However, all nuclei of Rn are radioactive, unlike most nuclei of Ar. The radioactive decay products of Rn can cause biological damage when inhaled.
95.
One would expect RnF2, RnF4, and maybe RnF6 to form in a fashion similar to XeF2, XeF4, and XeF6.
96.
RnF2, 8 + 2(7) = 22 e−
RnF4, 8 + 4(7) = 36 e−
Linear; 180
Square planar; 90
RnF6, 8 + 6(7) = 50 e−
The structure for RnF6 is difficult to predict. For six electron pairs about a central atom, the geometry is octahedral with 90 bond angles. RnF6 has seven electron pairs about the central Rn atom, so the structure is not octahedral. We will call the molecular structure of RnF 6 a distorted octahedral structure with exact bond angles that are hard to predict.
ChemWork Problems 97.
a. ΔH° = ‒110.5 ‒ (‒242) = 132 kJ; ΔS° = 198 + 131 ‒ [6 + 189] = 134 J/K
CHAPTER 20
THE REPRESENTATIVE ELEMENTS
b. ΔG° = ΔH° − TΔS°; ΔG° = 0 when T =
995
ΔH o 132 10 3 J = = 985 K 134 J/K ΔSo
At T > 985 K and standard pressures, the favorable ΔS° term dominates, and the reaction is spontaneous (ΔG° < 0). 98.
99.
Hard water contains Ca2+ and Mg2+ ions. To soften water, the Ca2+ and Mg2+ ions are removed. This occurs when hard water is passed through an ion exchange resin. The resin has many ionic (charged) sites, which have Na+ ions bound to the negative charged sites on the resin. When hard water is passed over the resin, Ca2+ and Mg2+ bind to the resin in place of Na+. So, Na+ ions replace the Ca2+ and Mg2+ ions in the hard water, resulting in “softened” water. H
H N
H
(l)
N
+
O
O (g)
N
N (g) + 2 H
O
H (g)
H
Bonds broken:
Bonds formed:
1 N‒N (160. kJ/mol)
1 N≡N (941 kJ/mol)
4 N‒H (391 kJ/mol)
2 × 2 O‒H (467 kJ/mol)
1 O=O (495 kJ/mol) ΔH = 160. + 4(391) + 495 − [941 + 4(467)] = 2219 kJ − 2809 kJ = −590. kJ 100.
The inert pair effect refers to the difficulty of removing the pair of s electrons from some of the elements in the fifth and sixth periods of the periodic table. As a result, multiple oxidation states are exhibited for the heavier elements of Groups 3A and 4A. In +, In3+, Tl+, and Tl3+ oxidation states are all important to the chemistry of In and Tl.
101.
Ga(I): [Ar]4s23d10, no unpaired e−; Ga(III): [Ar]3d10, no unpaired e− Ga(II): [Ar]4s13d10, 1 unpaired e−; note that the s electrons are lost before the d electrons. If the compound contained Ga(II), it would be paramagnetic, and if the compound contained Ga(I) and Ga(III), it would be diamagnetic. This can be determined easily by measuring the mass of a sample in the presence and in the absence of a magnetic field. Paramagnetic compounds will have an apparent increase in mass in a magnetic field.
102.
The π electrons are free to move in graphite, thus giving it greater conductivity (lower resistance). The electrons in graphite have the greatest mobility within sheets of carbon atoms, resulting in a lower resistance in the plane of the sheets (basal plane). Electrons in diamond are not mobile (high resistance). The structure of diamond is uniform in all directions; thus resistivity has no directional dependence in diamond.
103.
NaH(s) + H2O(l) → Na+(aq) + OH−(aq) + H2(g); NaH is an ionic compound composed of Na+ and H− ions. Oxidation-reduction: The oxidation state of hydrogen is −1 in NaH, +1 in
996
CHAPTER 20
THE REPRESENTATIVE ELEMENTS
H2O, and zero in H2. Hydrogen is oxidized when it goes from NaH to H2 (from −1 → 0) and hydrogen is reduced when it goes from H 2O to H2 (from +1 → 0). In this reaction, an electron is transferred from the hydride ion to a hydrogen in water when forming H 2. Acid-base: A proton is transferred from an acid, H2O, to a base, H−, forming the conjugate base of water, OH−, and the conjugate acid of H−, H2. 104.
a. Ca2+ + 2 e− → Ca 5.52 103 g Ca
1 mol Ca 2 mol e − 96,485 C = 2.66 107 C − 40.08 g Ca 1 mol Ca 1 mol e
1h 2.66 10 7 C = 924 C/s = 924 A 8.00 h 3600 s b. Ca2+ + 2 Cl− → Ca + Cl2 5.52 103 g Ca ×
105.
1 mol Ca 1 mol Cl 2 70.90 g Cl 2 = 9.76 103 g = 9.76 kg Cl2 40.08 g mol Ca mol Cl 2
pH = 9.42, pOH = 14.00 − 9.42 = 4.58, [OH−] = 10−4.58 = 2.6 10−5 M
⇌
Mg(OH)2(s) Initial Equil.
s = solubility (mol/L)
Mg2+(aq) 0 s
+ 2 OH−(aq) 2.6 × 10−5 M 2.6 × 10−5 M
(from buffer) (pH constant)
Ksp = [Mg2+][OH−]2, 8.9 10−12 = s(2.6 × 10−5)2, s = solubility = 0.013 mol/L Pb2+
106.
Before 0.0050 M Change −0.0050
0.075 M −0.0050
0 → +0.0050
1.0 × 10−7 M (Buffer, [H+] constant) No change Reacts completely
Change
0 0.070 0.0050 1.0 × 10−7 2− x mol/L PbEDTA dissociates to reach equilibrium +x +x −x
Equil.
x
After
K = 6.7 × 1021 = 6.7 × 1021
107.
+ H2EDTA2− ⇌ PbEDTA2− + 2 H+
0.070 + x
0.0050 − x
1.0 × 10−7
New initial conditions
(Buffer)
(0.0050 − x)(1.0 10 −7 ) 2 [PbEDTA2− ][H + ]2 = ( x)(0.070 + x) [Pb2+ ][H 2 EDTA2− ]
(0.0050 )(1.0 10 −14 ) , x = [Pb2+] = 1.1 × 10 −37 M; assumptions good. ( x)(0.070 )
a. AgCl(s) Ag(s) + Cl; the reactive chlorine atom is trapped in the crystal. When light is removed, Cl reacts with silver atoms to re-form AgCl; i.e., the reverse reaction occurs. In pure AgCl, the Cl atoms escape, making the reverse reaction impossible. b. Over time, chlorine is lost, and the dark silver metal is permanent.
CHAPTER 20
THE REPRESENTATIVE ELEMENTS
ClO− + H2O + 2 e− → 2 OH− + Cl− 2 NH3 + 2 OH− → N2H4 + 2 H2O + 2 e−
108.
997 E = 0.90 V −E = 0.10 V
ClO−(aq) + 2 NH3(aq) → Cl−(aq) + N2H4(aq) + H2O(l)
E ocell = 1.00 V
Because E ocell is positive for this reaction, ClO−, at standard conditions, can spontaneously oxidize NH3 to the somewhat toxic N2H4. 109.
Tl3+ + 2 e− → Tl+ 3 I− → I3− + 2 e−
E = 1.25 V −E = −0.55 V
Tl3+ + 3 I− → Tl+ + I3−
E ocell = 0.70 V
(Spontaneous because E ocell > 0.)
In solution, Tl3+can oxidize I− to I3−. Thus we expect TlI3 to be thallium(I) triiodide. 110.
See Table 20.14 of the text for the Lewis structures of N 2O3 and N2. a. NO+: 10 valence electrons The nitrogen atom in NO+ has a linear arrangement of electron pairs which dictates sp hybridization. b. Both nitrogen atoms in N2O3 have a trigonal planar arrangement of electron pairs which dictates sp2 hybridization. c.
NO2−: 18 valence electrons
The nitrogen atom in NO2− has a trigonal planar arrangement of electron pairs which dictates sp2 hybridization. d. Each nitrogen atom in N2 has a linear arrangement of electron pairs which dictates sp hybridization. 111.
a. SF6, 48 valence electrons
b. ClF3, 28 valence electrons
S is d2sp3 hybridized.
Cl is dsp3 hybridized.
998
CHAPTER 20 c. GeCl4, 32 valence electrons
THE REPRESENTATIVE ELEMENTS d. XeF4, 36 valence electrons
Ge is sp3 hybridized. 112.
Ptotal = 94.0 kPa ×
Xe is d2sp3 hybridized.
1 atm 760 torr = 705 torr 101.3 kPa atm
Ptotal = PN 2O + PH 2O , 705 torr = PN 2O + 21 torr, PN 2O = 684 torr × 8.68 g NH4NO3 ×
V=
n N 2O RT PN 2O
1 atm = 0.900 atm 760 torr
1 mol NH 4 NO 3 1 mol N 2 O = 0.108 mol N2O 80.05 g mol NH 4 NO 3
0.108 mol =
0.08206 L atm 295 K K mol = 2.90 L N2O 0.900 atm
113.
Strontium and calcium are both alkaline earth metals, so both have similar chemical properties. Because milk is a good source of calcium, strontium could replace some calcium in milk without much difficulty.
114.
a. 7.26 m × 8.80 m × 5.67 m = 362 m3; assume P = 1.0 atm and T = 25C for both parts.
1L 9.0 10 −6 L Xe 10 dm 362 m3 = 3.3 × 10−2 L of Xe in the room 3 m 100 L air dm 3
PV = nRT, n =
(1.0 atm)(3.3 10 −2 L) PV = 1.3 × 10−3 mol Xe = RT (0.08206 L atm/K • mol) (298 K)
1.3 × 10−3 mol Xe ×
131.3 g Xe = 0.17 g Xe in the room mol Xe
b. A 2-L breath contains: 2 L air × n =
9.0 10 −6 L Xe = 2 × 10-7 L Xe 100 L air
(1.0 atm)(2 10 −7 L) PV = = 8 × 10-9 mol Xe RT (0.08206 L atm/K • mol) (298 K)
8 × 10−9 mol Xe ×
6.022 10 23 atoms = 5 × 1015 atoms of Xe in a 2-L breath mol
CHAPTER 20 115.
THE REPRESENTATIVE ELEMENTS
999
+6 oxidation state: SO42−, SO3, SF6 +4 oxidation state: SO32−, SO2, SF4 +2 oxidation state: SCl2 0 oxidation state: S8 and all other elemental forms of sulfur −2 oxidation state: H2S, Na2S
116.
ClF, 7 + 7 = 14 e−
ClF3, 7 + 3(7) = 28 e−
Linear; no bond angle present
T-shaped; 90
ClF5, 7 + 5(7) = 42 e−
Square pyramid; 90 To form FCl3, F would have to expand its octet of electrons. Fluorine is too small and doesn’t have low-energy d orbitals available to expand its octet. Therefore, FCl 3 would not be a stable compound. 117.
a. −307 kJ = (−1136 + x) – [(−254 kJ) + 3(−96 kJ)], x = ΔHof , NI3 = 287 kJ/mol b. IF2+, 7 + 2(7) – 1 = 20 e−
BF4−, 3 + 4(7) + 1 = 32 e−
V-shaped; sp3 118.
1.75 × 108 g pitchblende
Tetrahedral; sp3 1 metric ton 1.0 g Ra 1 mol Ra 6 7.0 metric tons 226 g Ra 1.0 10 g
6.022 10 23 atoms Ra = 6.7 × 1022 atoms Ra mol Ra
1000
CHAPTER 20
THE REPRESENTATIVE ELEMENTS
Radioactive decay follows first-order kinetics. N N − 0.6931(100 . yr) − (ln 2) t = −kt = = ln ; ln , N = 14.4 mg Ra t1/ 2 1.60 10 3 yr 15.0 mg N0
14.4 × 10 −3 g Ra
1 mol Ra 6.022 10 23 atoms Ra = 3.84 × 1019 atoms Ra 226 g Ra mol Ra
Challenge Problems 119.
The reaction is X(s) + 2H2O(l) → H2(g) + X(OH)2(aq). Mol X = mol H2 =
Molar mass X =
PV 1.00 atm 6.10 L = 0.249 mol = 0.08206 L atm RT 298 K K mol
10 .00 g X = 40.2 g/mol; X is Ca. 0.249 mol X
Ca(s) + 2 H2O(l) → H2(g) + Ca(OH)2(aq); Ca(OH)2 is a strong base. 10.00 g Ca
[OH−] =
1 mol Ca 1 mol Ca(OH)2 2 mol OH − 40.08 g mol Ca mol Ca(OH)2 = 0.0499 M 10.0 L
pOH = −log(0.0499) = 1.302, pH = 14.000 – 1.302 = 12.698 120.
White tin is stable at normal temperatures. Gray tin is stable at temperatures below 13.2°C. Thus, for the phase change Sn(gray) → Sn(white), ΔG is (−) at T > 13.2°C, and ΔG is (+) at T < 13.2°C. This is only possible if ΔH is (+) and ΔS is (+). Thus, gray tin has the more ordered structure (has the smaller positional probability).
121.
PbX4 → PbX2 + X2; from the equation, mol PbX4 = mol PbX2. Let x = molar mass of the halogen. Setting up an equation where mol PbX 4 = mol PbX2: 25.00 g 16.12 g = ; solving, x = 127.1; the halogen is iodine, I. 207 .2 + 4 x 207 .2 + 2 x
122.
In order to form a π bond, the d and p orbitals must overlap “side to side” instead of “head to head” as in sigma bonds. A representation of the side-to-side overlap follows. For a bonding orbital to form, the phases of the lobes must match (positive to positive and negative to negative).
d
+
p
CHAPTER 20 123.
THE REPRESENTATIVE ELEMENTS
1001
For the reaction: O N
N
O
NO2 + NO
O
the activation energy must in some way involve breaking a nitrogen-nitrogen single bond. For the reaction: O N
N
O
O2 + N 2O
O
at some point nitrogen-oxygen bonds must be broken. N‒N single bonds (160. kJ/mol) are weaker than N‒O single bonds (201 kJ/mol). In addition, resonance structures indicate that there is more double-bond character in the N‒O bonds than in the N‒N bond. Thus, NO2 and NO are preferred by kinetics because of the lower activation energy. 124.
a. The sum of the two steps gives the overall balanced equation. O3(g) + NO(g) → NO2(g) + O2(g) NO2(g) + O(g) → NO(g) + O2(g) O3(g) + O(g) → 2 O2(g)
overall equation
b. NO is the catalyst. NO is present in the first step of the mechanism on the reactant side, but it is not a reactant because it is regenerated in the second step and does not appear in the overall balanced equation. c. NO2 is an intermediate. It is produced in the first step, but is consumed in the second step. Intermediates also never appear in the overall balanced equation. In a mechanism, intermediates always appear first on the product side, while catalysts always appear first on the reactant side. d. The rate of the slow step in a mechanism gives the rate law for the reaction. From the problem, the rate determining step (the slow step) is step 1. The derived rate law is: Rate = k[O3][NO] Because NO is a catalyst and not a proposed intermediate, it can appear in the rate law. e. The mechanism for the chlorine-catalyzed destruction of ozone is: O3(g) + Cl(g) → O2(g) + ClO(g) ClO(g)+ O(g) → O2(g) + Cl(g) O3(g) + O(g) → 2 O2(g)
slow step 1 fast step 2 overall equation
1002 125.
CHAPTER 20 NH3 + NH3 ⇌ NH4+ + NH2−
THE REPRESENTATIVE ELEMENTS
K = [NH4+][NH2−] = 1.8 × 10 −12
NH3 is the solvent, so it is not included in the K expression. In a neutral solution of ammonia: [NH4+] = [NH2−]; 1.8 × 10 −12 = [NH4+]2, [NH4+] = 1.3 × 10 −6 M = [NH2−] We could abbreviate this autoionization as: NH3 ⇌ H+ + NH2−, where [H+] = [NH4+]. This abbreviation is synonymous with the abbreviation used for the autoionization of water (H2O ⇌ H+ + OH−). So pH = pNH4+ = −log(1.3 × 10 −6 ) = 5.89. 126.
The representation indicates that we have an equimolar mixture of N 2(g) and H2(g) (6 molecules of each are shown). To solve the problem, let’s assume a reaction between 3.00x moles of N2 and 3.00x moles of H2 (equimolar). The reaction going to completion is summarized in the following table. Note that H2 is limiting. N2(g) Before Change After
+
3.00x mol −1.00x mol 2.00x mol
3 H2(g)
→
3.00x mol −3.00x mol → 0
2 NH3(g) 0 +2.00x mol 2.00x mol
When an equimolar mixture is reacted, the total moles of gas present decreases from 6.00x moles initially to 4.00x moles after completion. a. The total pressure in the piston apparatus is a constant 1.00 atm. After the reaction, we have 2.00x moles of N2 and 2.00x moles of NH3. One-half of the moles of gas present are NH 3 molecules, so one-half of the total pressure is due to the NH3 molecules. PNH3 = 0.500 atm. b.
NH3 =
moles NH 3 2.00 x mol = = 0.500 total moles ( 2.00 x + 2.00 x) mol
c. At constant P and T, volume is directly proportional to n, the moles of gas present. Because n decreased from 6.00x to 4.00x moles, the volume will decrease by the same factor. Vfinal = 15.0 L (4/6) = 10.0 L 127.
Let n SO 2 = initial moles SO2 present. The reaction is summarized in the following table (O 2 is in excess). 2 SO2
+
O2(g)
→
2 SO3(g)
Initial
n SO 2
2.00 mol
0
Change
− n SO 2
− n SO 2 /2
+ n SO 2
Final
0
2.00 − n SO 2 /2
n SO 2
CHAPTER 20
THE REPRESENTATIVE ELEMENTS
1003
Density = d = mass/volume; let di = initial density of gas mixture and df = final density of gas mixture after reaction. Because mass is conserved in a chemical reaction, mass i = massf. df mass f /Vf V = = i di mass i /Vi Vf
At constant P and T, V n, so
df V n = i = i ; setting up an equation: di Vf nf
n SO2 + 2.00 n + 2.00 df n 0.8471 g/L = = 1.059, 1.059 = i = = SO2 di 0.8000 g/L nf (2.00 − n SO2 /2) + n SO2 2.00 + n SO2 /2 Solving: n SO 2 = 0.25 mol; so, 0.25 moles of SO3 is formed. 0.25 mol SO3 × 128.
80.07 g SO 3 = 20. g SO3 mol
Mg2+ + P3O105− ⇌ MgP3O103− pK = −8.60; [Mg2+]0 = [P3O105−]0 =
50. 10 −3 g 1 mol = 2.1 × 10−3 M L 24.3 g
40. g Na 5 P3O10 1 mol = 0.11 M L 367 .9 g
Assume the reaction goes to completion because K is very large (K = 108.60 = 4.0 × 108). Then solve the back-equilibrium problem to determine the small amount of Mg 2+ present. Mg2+ Before Change After Change Equil.
129.
P3O105−
⇌
MgP3O103−
2.1 × 10−3 M 0.11 M 0 −2.1 × 10−3 −2.1 × 10−3 → +2.1 × 10−3 Reacts completely −3 0 0.11 2.1 × 10 New initial x mol/L MgP3O103− dissociates to reach equilibrium +x +x −x x 0.11 + x 2.1 × 10−3 − x
K = 4.0 × 108 = 4.0 × 108
+
3− [MgP3 O10 ] 5− [Mg 2+ ][P3 O10 ]
=
2.1 10 −3 − x (assume x << 2.1 × 10−3) x(0.11 + x)
2.1 10 −3 , x = [Mg2+] = 4.8 × 10−11 M; assumptions good. x(0.11)
Table 20.2 lists the mass percents of various elements in the human body. If we consider the mass percents through sulfur, that will cover 99.5% of the body mass, which is fine for a reasonable estimate. 150 lb 454 g/lb = 68,000 g. We will carry an extra significant figure in some of the following calculations. Moles of O = 0.650 × 68,000 g × 1 mol O/16.00 g O = 2760 mol Moles of C = 0.180 × 68,000 g × 1 mol C/12.01 g C = 1020 mol Moles of H = 0.100 × 68,000 g × 1 mol H/1.008 g H = 6750 mol
1004
CHAPTER 20
THE REPRESENTATIVE ELEMENTS
Moles of N = 0.030 × 68,000 g × 1 mol N/14.01 g N = 150 mol Moles of Ca = 0.014 × 68,000 g × 1 mol Ca/40.08 g Ca = 24 mol Moles of P = 0.010 × 68,000 g × 1 mol P/30.97 g P = 22 mol Moles of Mg = 0.0050 × 68,000 g × 1 mol Mg/24.31 g Mg = 14 mol Moles of K = 0.0034 × 68,000 g × 1 mol K/39.10 g K = 5.9 mol Moles of S = 0.0026 × 68,000 g × 1 mol S/32.07 g S = 5.5 mol Total moles of elements in 150-lb body = 10,750 mol atoms 6.022 10 23 atoms = 6.474 × 1027 atoms 6.5 × 1027 atoms mol atoms
10,750 mol atoms ×
130.
a. Moles of In(CH3)3 = Moles of PH3 =
2.00 atm 2.56 L PV = = 0.0693 mol RT 0.08206 L atm/K • mol 900. K
3.00 atm 1.38 L PV = = 0.0561 mol RT 0.08206 L atm/K • mol 900. K
Because the reaction requires a 1 : 1 mole ratio between these reactants, the reactant with the small number of moles (PH3) is limiting. 0.0561 mol PH3
1 mol InP 145 .8 g InP = 8.18 g InP mol PH 3 mol InP
The actual yield of InP is: 0.87 × 8.18 g = 7.1 g InP b.
λ =
6.626 10 −34 J s 2.998 10 8 m/s hc = = 9.79 × 10 −7 m = 979 nm E 2.03 10 −19 J
From the Figure 7.2 of the text, visible light has wavelengths between 4 × 10 −7 and 7 × 10 −7 m. Therefore, this wavelength is not visible to humans; it is in the infrared region of the electromagnetic radiation spectrum. c. [Kr]5s24d105p4 is the electron configuration for tellurium, Te. Because Te has more valence electrons than P, this would form an n-type semiconductor (n-type doping). 131.
a. Because the hydroxide ion has a 1− charge, Te has a +6 oxidation state. b. Assuming Te is limiting: (0.545 cm)3
6.240 g cm
3
1 mol Te 1 mol TeF6 = 7.92 × 10 −3 mol TeF6 127.6 mol Te
Assuming F2 is limiting: Mol F2 = n =
1.06 atm 2.34 L PV = = 0.101 mol F2 RT 0.08206 L atm/K • mol 298 K
CHAPTER 20
THE REPRESENTATIVE ELEMENTS 0.101 mol F2
1005
1 mol TeF6 = 3.37 × 10 −2 mol TeF6 3 mol F2
Because Te produces the smaller amount of product, Te is limiting and 7.92 × 10 −3 mol TeF6 can be produced. From the first equation given in the question, the moles of TeF6 reacted equals the moles of Te(OH)6 produced. So 7.92 10−3 mol Te(OH)6 can be produced. [Te(OH)6] =
7.92 10 −3 mol Te(OH)6 = 6.89 × 10 −2 M 0.115 L
Because K a1 K a 2 , the amount of protons produced by the K a 2 reaction will be insignificant. Te(OH)6 ⇌ Te(OH)5O− + Initial 0.0689 M Equil. 0.0689 − x
K a1 = 2.1 × 10 −8 =
0 x
H+
K a1 = 10 −7.68 = 2.1 × 10 −8
~0 x
x2 x2 , x = [H+] = 3.8 × 10 −5 M 0.0689 − x 0.0689
pH = −log(3.8 × 10 −5 ) = 4.42; assumptions good.
Marathon Problems 132.
The answers to the clues are: (1) HI has the second highest boiling point; (2) HF is the weak hydrogen halide acid; (3) He was first discovered from the sun’s emission spectrum; (4) of the elements in Table 20.13, Bi will have the most metallic character (metallic character increases down a group); (5) Te is a semiconductor; (6) S has both rhombic and monoclinic solid forms; (7) Cl2 is a yellowgreen gas; (8) O is the most abundant element in and near the earth’s crust; (9) Se has a 4s24p4 valence shell configuration; (10) Kr is the smallest noble gas that forms compounds such as KrF2 and KrF4 (the symbol in reverse order is rk); (11) As forms As4 molecules; (12) N2 is a major inert component of air, and N is often found in fertilizers and explosives. Filling in the blank spaces with the answers to the clues, the message is “If he bites, close ranks.”
133.
The answer to the clues are: (1) BeO is amphoteric; (2) From Table 20.2, N makes up about 3.0% of the human body; (3) Fr has the 7s1 valence electron configuration; (4) Na has the least negative E° value (the symbol in reverse is an); (5) in intracellular fluids, K+ is the more concentrated alkali metal ion; (6) only Li forms Li3N; (7) In is the first Group 3A element to form stable +1 and +3 ions in its compounds (the second letter of the symbol is n). Inserting the symbols into the blanks gives Ben Franklin for the name of the American scientist.
CHAPTER 21 TRANSITION METALS AND COORDINATION CHEMISTRY Review Questions 1.
Chromium ([Ar]:4s03d5) and copper [Ar]:4s13d10) have electron configurations which are different from that predicted from the periodic table. Other exceptions to the predicted filling order are transition metal ions. These all lose the s electrons before they lose the d electrons. In neutral atoms, the ns and (n − 1)d orbitals are very close in energy, with the ns orbitals slightly lower in energy. However, for transition metal ions, there is an apparent shifting of energies between the ns and (n − 1)d orbitals. For transition metal ions, the energy of the (n − 1)d orbitals are significantly less than that of the ns electrons. So when transition metal ions form, the highest energy electrons are removed, which are the ns electrons. For example, Mn2+ has the electron configuration [Ar]:4s03d5 and not [Ar]:4s23d3. Most transition metals have unfilled d orbitals, which creates a large number of other electrons that can be removed. Stable ions of the representative metals are determined by how many s and p valence electrons can be removed. In general, representative metals lose all of the s and p valence electrons to form their stable ions. Transition metals generally lose the s electron(s) to form +1 and +2 ions, but they can also lose some (or all) of the d electrons to form other oxidation states as well. 2. a.
Coordination compound: A compound composed of a complex ion (see b) and counter ions (see c) sufficient to give no net charge.
b. Complex ion: A charged species consisting of a metal ion surrounded by ligands (see e). c. Counter ions: Anions or cations that balance the charge on a complex ion in a coordination compound. d. Coordination number: The number of bonds formed between the metal ion and the ligands (see e) in a complex ion.
e. Ligand: Species that donates a pair of electrons to form a covalent bond to a metal ion. Ligands act as Lewis bases (electron pair donors). f.
Chelate: Ligand that can form more than one bond to a metal ion.
g. Bidentate: Ligand that forms two bonds to a metal ion.
Because transition metals form bonds to species that donate lone pairs of electrons, transition metals are Lewis acids (electron pair acceptors). The Lewis bases in coordination compounds are the ligands, all of which have an unshared pair of electrons to donate. The coordinate covalent bond between the ligand and the transition metal just indicates that both electrons in the bond originally came from one of the atoms in the bond. Here, the electrons in the bond come from the ligand.
1006
CHAPTER 21 3.
TRANSITION METALS AND COORDINATION CHEMISTRY
1007
Linear geometry (180˚ bond angles) is observed when the coordination number is 2. Tetrahedral geometry (109.5˚ bond angles) or square planar geometry (90˚ bond angles) is observed when the coordination number is 4. Octahedral geometry (90˚ bond angles) is observed when the coordination number is 6. For the following complex ions, see Table 21.13 if you don’t know the formula, the charge, or the number of bonds the ligands form. a. Ag(CN)2−; Ag+: [Kr]4d10
b. Cu(H2O)4+; Cu+: [Ar]3d10
c. Mn(C2O4)2−; Mn2+: [Ar]3d5
d. Pt(NH3)42+; Pt2+: [Xe]4f145d8
e. Fe(EDTA)−; Fe3+: [Ar]3d5; Note: EDTA has an overall 4− charge and is a six coordinate ligand. f.
Co(Cl)64−; Co2+: [Ar]3d7
g. Cr(en)33+ where en = ethylenediamine (NH2CH2CH2NH2); Cr3+: [Ar]3d3 4.
See Section 21.3 of the text for a summary of the nomenclature rules. a. The correct name is tetraamminecopper(II) chloride. The complex ion is named incorrectly in several ways. b. The correct name is bis(ethylenediamine)nickel(II) sulfate. The ethylenediamine ligands are neutral and sulfate has a 2− charge. Therefore, Ni 2+ is present, not Ni4+. c. The correct name is potassium diaquatetrachlorochromate(III). Because the complex ion is an anion, the –ate suffix ending is added to the name of the metal. Also, the ligands were not in alphabetical order (a in aqua comes before c in chloro). d. The correct name is sodium tetracyanooxalatocobaltate(II). The only error is that tetra should be omitted in front of sodium. That four sodium ions are needed to balance charge is deduced from the name of the complex ion.
5.
a. Isomers: Species with the same formulas but different properties; they are different compounds. See the text for examples of the following types of isomers. b. Structural isomers: Isomers that have one or more bonds that are different. c. Stereoisomers: Isomers that contain the same bonds but differ in how the atoms are arranged in space. d. Coordination isomers: Structural isomers that differ in the atoms that make up the complex ion. e. Linkage isomers: Structural isomers that differ in how one or more ligands are attached to the transition metal. f.
Geometric isomers (cis-trans isomerism): Stereoisomers that differ in the positions of atoms with respect to a rigid ring, bond, or each other.
1008
CHAPTER 21 TRANSITION METALS AND COORDINATION CHEMISTRY g. Optical isomers: Stereoisomers that are nonsuperimposable mirror images of each other; that is, they are different in the same way that our left and right hands are different. The trans form of Cr(en)Cl2 is not optically active, but the the cis form is optically active. See Figure 21.17 of the text for illustrations showing the cis and trans forms for a similar compound; shown also is the optical activity of the cis form. The only difference between the complex in this question, and the complex in Figure 21.17, is that Cr 2+ has replaced Co3+. Note that not all cis isomers are optically active. For example, the cis isomer of Cr(NH3)4Cl2 is not optically active because the mirror image is superimposable (prove it to yourself). In Figure 21.17, a plane of symmetry exists through the square planar orientation of the two en ligands in the trans form. Other planes of symmetry also exist in the trans isomer. In the cis isomer in Figure 21.17, no plane of symmetry exists, so this cis form is optically active (as we know).
6.
The crystal field model focuses on the energies of the d orbitals and what happens to the energies of these d orbitals as negative point charges (the ligands) approach (and repel) the electrons in the d orbitals. For octahedral geometry, six ligands are bonded to the metal ion. Because of the different orientations of the d orbitals, not all d orbitals are affected the same when six negative point charges (ligands) approach the metal ion along the x, y, and z axis. It turns out that the dxy, dxz, and dzy orbitals are all destabilized by the same amount from the octahedrally arranged point charges, as are the d x 2 − y 2 and d z 2 orbitals. These are the two sets that the d orbitals split into. The dxy, dxz, and dyz set is called the t2g set, while the d x 2 − y 2 and
d z 2 set is called the eg set. Another major point for the octahedral crystal field diagram is that the eg set of orbitals is destabilized more than the t2g set. This is because the t2g orbital set (dxy, dxz, and dzy) points between the point charges while the eg orbital set ( d x 2 − y 2 and d z 2 ) points directly at the point charges. Hence, there is more destabilization in the eg orbital set, and they are at a higher energy. a. Weak field ligand: Ligand that will give complex ions with the maximum number of unpaired electrons. b. Strong-field ligand: Ligand that will give complex ions with the minimum number of unpaired electrons. c. Low-spin complex: Complex ion with a minimum number of unpaired electrons (low-spin = strong-field). d. High-spin complex: Complex ion with a maximum number of unpaired electrons (highspin = weak-field). In both cobalt complex ions, Co3+ exists which is a d6 ion (6 d electrons are present). The difference in magnetic properties is that Co(NH3)63+ is a strong-field (low-spin) complex having a relatively large , while CoF63− is a weak-field (high-spin) complex having a relatively small
CHAPTER 21
TRANSITION METALS AND COORDINATION CHEMISTRY
1009
. The electron configurations for Co 3+ in a strong field vs. a weak field is shown in Figure 21.22. The strong-field d6 ion is diamagnetic because all electrons are paired. This is the diagram for Co(NH3)63+. The weak field d6 ion is paramagnetic because it has unpaired electrons (4 total). This is the diagram for CoF 63−. Looking at Figure 21.22, d1, d2, and d3 metal ions would all have the same number of unpaired electrons. This won’t happen again until we get all the way up to d8, d9, and d10 metal ions (prove it to yourself that d4, d5, d6, and d7 metal ions have different d orbital electron configurations depending on a strong-field or a weak-field). V3+ is a d2 ion (2 unpaired electrons in the t2g set). It has the same diagram no matter how strong the field strength. The same is true for the d8 Ni2+ ion (filled t2g set and half-filled eg set). However, Ru2+, a d6 ion, will have different diagrams depending on a strong-field or a weak-field. If a weak field is present, then there are four unpaired electrons. In the strong-field case, all six d electrons are in the t2g set and all are paired (no unpaired electrons are present). 7.
The valence d electrons for the metal ion in the complex ion are placed into the octahedral crystal field diagram. If electrons are all paired, then the complex is predicted to be diamagnetic. If there are unpaired electrons, then the complex is predicted to be paramagnetic. Color results by the absorption of specific wavelengths of light. The d-orbital splitting, , is on the order of the energies of visible light. The complex ion absorbs the wavelength of light that has energy equal to the d-orbital splitting, . The color we detect for the substance is not the color of light absorbed. We detect (see) the complementary color to that color of light absorbed. See Table 21.15 for observed colors of substances given the color of light absorbed. From Table 21.15, if a complex appears yellow then it absorbs blue light on the order of ~450 nm. Therefore, Cr(NH3)63+ absorbs blue light. The spectrochemical series places ligands in order of their ability to split the d-orbitals. The strongest field ligands (large ) are on one side of the series with the weakest field ligands (small ) on the other side. The series was developed from studies of the light absorbed by many octahedral complexes. From the color of light absorbed, one can determine the d-orbital splitting. Strong-field ligands absorb higher energy light (violet light, for example, with ~400 nm), while weak-field ligands absorb lower energy light (red light, for example, with ~650 nm). The higher the charge on the metal ion, the larger the d-orbital splitting. Thus, the Co3+ complex ion [Co(NH3)63+], would absorb higher energy (shorter wavelength) light than a Co2+ complex ion (assuming the ligands are the same).
8.
The crystal field diagrams are different because the geometries of where the ligands point is different. The tetrahedrally oriented ligands point differently in relationship to the d-orbitals than do the octahedrally oriented ligands. Plus, we have more ligands in an octahedral complex. See Figure 21.27 for the tetrahedral crystal field diagram. Notice that the orbitals are reverse of that in the octahedral crystal field diagram. The degenerate d z 2 and d x 2 − y 2 are at a lower energy than the degenerate dxy, dxz, and dyz orbitals. Again, the reason for this is that tetrahedral
1010
CHAPTER 21 TRANSITION METALS AND COORDINATION CHEMISTRY ligands are oriented differently than octahedral field ligands, so the interactions with specifically oriented d-orbitals are different. Also notice that the difference in magnitude of the d-orbital splitting for the two geometries. The d-orbital splitting in tetrahedral complexes is less than one-half the d-orbital splitting in octahedral complexes. There are no known ligands powerful enough to produce the strong-field case, hence all tetrahedral complexes are weakfield or high spin. See Figure 21.28 for the descriptions of the square planar and linear crystal field diagrams. Each is unique which is not surprising. The ligands for any specific geometry will point differently relative to the orientations of the five d-orbitals. Different interactions result giving different crystal field diagrams.
9.
Each hemoglobin molecule can bind four O 2 molecules. It is an Fe2+ ion in hemoglobin that binds an individual O2 molecule, and each hemoglobin molecule has four of these Fe2+ binding sites. The Fe2+ ion at the binding site is six-coordinate. Five of the coordination sites come from nitrogen atoms in the hemoglobin molecule. The sixth site is available to attach an O 2 molecule. When the O2 molecule is released, H2O takes up the sixth position around the Fe2+ ion. O2 is a strong field ligand, unlike H2O, so in the lungs, O2 readily replaces the H2O ligand. With four sites, each hemoglobin molecule has a total of four O2 molecules attached when saturated with O2 from the lungs. In the cells, O2 is released by the hemoglobin and the O2 site is replaced by H2O. The oxygen binding is pH dependent, so changes in pH in the cells as compared to blood, causes the release of O2. Once the O2 is released and replaced by H2O, the hemoglobin molecules return to the lungs to replenish with the O2.
10.
The definitions follow. See section 21.8 for examples. a. Roasting: Converting sulfide minerals to oxides by heating in air below their melting points. b. Smelting: Reducing metal ions to the free metal. c. Flotation: Separation of mineral particles in an ore from the unwanted impurities. This process depends on the greater wetability of the mineral particles as compared to the unwanted impurities. d. Leaching: The extraction of metals from ores using aqueous chemical solutions. e. Gangue: The impurities (such as clay, sand, or rock) in an ore. Advantages of hydrometallurgy: cheap energy cost; less air pollution; Disadvantages of hydrometallurgy: chemicals used in hydrometallurgy are expensive and sometimes toxic. In zone refining, a bar of impure metal travels through a heater. The impurities present are more soluble in the molten metal than in the solid metal. As the molten zone moves down a metal, the impurities are swept along with the liquid, leaving behind relatively pure metal.
CHAPTER 21
TRANSITION METALS AND COORDINATION CHEMISTRY
1011
Active Learning Questions 1.
The Cl− ions must all be ligands; if any were counter ions, a precipitate would have formed with AgNO3. The charge on platinum will be 4+ and the complex ion is Pt(Cl) 62−. The compound formula is K2[Pt(Cl)6], which is composed of K+ and Pt(Cl)62− ions.
2.
In both complexes, nickel is in the +2 oxidation state: Ni 2+: [Ar]3d8. The differences in unpaired electrons must be due to differences in molecular structure. Ni(NH3)42− is a tetrahedral complex, and Ni(SCN)42− is a square planar complex. The corresponding d-orbital splitting diagrams are:
2-
N iCl 4
Ni(NH3)42−
3.
2-
N i(CN )4
Ni(SCN)42−
Fe2+: [Ar]3d6; a d6 octahedral crystal field diagram can either be low spin (0 unpaired electrons) or high spin (4 unpaired electrons).
To be paramagnetic, we need a weak field. From the spectrochemical series, H 2O is more likely to produce a weak field as compared to CN − ligands. So the compound most likely to be paramagnetic is Fe(H2O)62+. 4.
A d4 metal ion has 4 unpaired electrons in the high spin case vs. 2 unpaired electrons in the low spin case. Thus, the compound could contain a d4 transition metal ion like Cr2+.
High Spin, small
Low Spin, large
1012
CHAPTER 21 TRANSITION METALS AND COORDINATION CHEMISTRY The metal ion could also be a d7 ion like Co2+ which have 3 unpaired electrons in the high spin case vs. 1 unpaired electron in the low spin case.
High spin, small
5.
Low spin, large
a. The coordination compound has the formula [Co(H 2O)6]Cl2. The complex ion is Co(H2O)62+, and the counterions are the Cl− ions. The geometry would be octahedral, and the electron configuration of Co2+ is [Ar]3d7. The crystal field diagram depends on whether a weak or strong field is present. The diagrams for a weak filed and strong filed are:
High spin, small
Low spin, large
b. The coordination compound is Na3[Ag(S2O3)2].The compound consists of Na+ counterions and the Ag(S2O3)23− complex ion. The complex ion is linear, and the electron configuration of Ag+ is [Kr]4d10. The crystal filed diagram for a d10 metal ion would have all d-orbitals filled with electrons. c. The reactant coordination compound is [Cu(NH 3)4]Cl2. The complex ion is Cu(NH3)42+, and the counterions are Cl− ions. The complex ion is tetrahedral (given in the question), and the electron configuration of Cu 2+ is [Ar]3d9. The product coordination compound is [Cu(NH3)4]Cl. The complex ion is Cu(NH3)4+ with Cl− counter ions. The complex ion is tetrahedral, and the electron configuration of Cu + is [Ar]3d10. The crystal filed diagram for a d10 metal ion would have all d-orbitals filled with electrons. 6.
The transition metal ion must form octahedral complex ions; only with the octahedral geometry are two different arrangements of d electrons possible in the split d orbitals. These two arrangements depend on whether a weak field or strong field is present. For four unpaired electrons, the two possible weak field cases are for transition metal ions with 3d 4 or 3d6 electron configurations: small
small
d4
d6
Of these two, only d6 ions have no unpaired electron in the strong field case. large
CHAPTER 21
TRANSITION METALS AND COORDINATION CHEMISTRY
1013
Therefore, the transition metal ion has a 3d6 arrangement of electrons. Two possible metal ions that are 3d6 are Fe2+ and Co3+. Thus, one of these ions is present in the four coordination compounds, and each of these complex ions has a coordination number of 6. The colors of the compounds are related to the magnitude of (the d-orbital splitting value). The weak-field compounds will have the smallest , so the of light absorbed will be longest. Using Table 21.16, the green solution (absorbs 650-nm light) and the blue solution (absorbs 600-nm light) absorb the longest-wavelength light; these solutions contain the complex ions that are the weak-field cases with four unpaired electrons. The red solution (absorbs 490-nm light) and yellow solution (absorbs 450-nm light) contain the two strong- field case complex ions because they absorb the shortest-wavelength (highest-energy) light. These complex ions are diamagnetic. For the green and yellow solutions that contain weak field complex ions, we will use the lower charged Fe2+ transition metal ion along with two weak field ligands like I‒ and Cl‒. A possible compound producing a green solution is Na4[FeI6] and a possible compound producing the blue solution is Na4[FeCl6]. Note that we used the weakest field ligand I‒ for the complex ion producing the green solution since this splitting is the smallest of all the solutions. For the red and yellow solutions that contain strong field complex ions, we will use the higher charged Co3+ ion along with two strong field ligands like CN ‒ and ethylenediamine (en). A possible compound producing a red solution would be Na3[Co(en)3] and a possible compound producing the yellow solution would be Na3[Co(CN)6]. Note that we used strongest field ligand (CN‒) for the complex ion producing the yellow solution since this splitting is the largest of all the solutions.
Questions 7.
The lanthanide elements are located just before the 5d transition metals. The lanthanide contraction is the steady decrease in the atomic radii of the lanthanide elements when going from left to right across the periodic table. As a result of the lanthanide contraction, the sizes of the 4d and 5d elements are very similar. This leads to a greater similarity in the chemistry of the 4d and 5d elements in each vertical group.
8.
Only the Cr3+ ion can form four different compounds with H2O ligands and Cl− ions. The Cr2+ ion could form only three different compounds, while the Cr4+ ion could form five different compounds. The Cl− ions that form precipitates with Ag+ are the counterions, not the ligands in the complex ion. The four compounds and mol AgCl precipitate that would form with 1 mol of compound are: Compound [Cr(H2O)6]Cl3 [Cr(H2O)5Cl]Cl2 [Cr(H2O)4Cl2]Cl [Cr(H2O)3Cl3]
Mol AgCl(s) 3 mol 2 mol 1 mol 0 mol
1014
CHAPTER 21 TRANSITION METALS AND COORDINATION CHEMISTRY
9. +
Cl H3 N
NH3
Cl Cl
NH3 Cl
trans (mirror image is superimposable)
Cl H3 N
NH3 Co
Co H3 N
+
Cl Co
NH3
H3 N
+
NH3
cis
NH3
H3 N H3 N
mirror
The mirror image of the cis isomer is also superimposable.
No; both the trans and the cis forms of Co(NH 3)4Cl2+ have mirror images that are superimposable. For the cis form, the mirror image only needs a 90 rotation to produce the original structure. Hence neither the trans nor cis form is optically active. 10.
Octahedral Cr2+ complexes should be used. Cr2+: [Ar]3d4; high-spin (weak-field) Cr2+ complexes have 4 unpaired electrons, and low-spin (strong-field) Cr2+ complexes have 2 unpaired electrons. Ni2+: [Ar]3d8; octahedral Ni2+ complexes will always have 2 unpaired electrons, whether high or low spin. Therefore, Ni2+ complexes cannot be used to distinguish weak- from strong-field ligands by examining magnetic properties. Alternatively, the ligand field strengths can be measured using visible spectra. Either Cr2+ or Ni2+ complexes can be used for this method.
11.
Fe2O3(s) + 6 H2C2O4(aq) → 2 Fe(C2O4)33−(aq) + 3 H2O(l) + 6 H+(aq); the oxalate anion forms a soluble complex ion with iron in rust (Fe2O3), which allows rust stains to be removed.
12.
a.
Low Spin, large
High Spin, small
A d6 octahedral crystal field diagram can either be low spin (0 unpaired electrons) or high spin (4 unpaired electrons). The diagram in the question is for the low spin d6 crystal field. b.
Low Spin, large
High Spin, small
CHAPTER 21
TRANSITION METALS AND COORDINATION CHEMISTRY
1015
There is only one possible crystal field diagram for a d8 species. Hence, one cannot tell from the diagram whether it is low spin or high spin; both have 2 unpaired electrons in the eg orbitals. c.
Low Spin, large
High Spin, small
There are two possible octahedral crystal field diagrams for a d 4 species. Low spin has 2 unpaired electrons and high spin has 4 unpaired electrons. The diagram in the question is for a high spin d4 crystal field. 13.
a. CoCl42−; Co2+: 4s03d7; all tetrahedral complexes are a weak field (high spin).
CoCl42− is an example of a weak-field case having three unpaired electrons. small b. Co(CN)63−: Co3+ : 4s03d6; because CN− is a strong-field ligand, Co(CN)63− will be a strong-field case (low-spin case). CN− is a strong-field ligand, so Co(CN)63− will be a low-spin case having zero unpaired electrons. large 14.
Transition compounds exhibit the color complementary to that absorbed. From Table 21.16, Ni(H2O)6Cl2 absorbs red light and Ni(NH3)6Cl2 absorbs yellow-green light. Ni(NH3)6Cl2 absorbs the shorter wavelength light which is the higher energy light (E = hc/λ). Therefore, Δ is larger for Ni(NH3)6Cl2, which means that NH3 is a stronger field ligand than H2O. This is consistent with the spectrochemical series.
15.
From Table 21.15, the red octahedral Co(H2O)62+ complex ion absorbs blue-green light ( 490 nm), whereas the blue tetrahedral CoCl42− complex ion absorbs orange light ( 600 nm). Because tetrahedral complexes have a d-orbital splitting much less than octahedral complexes, one would expect the tetrahedral complex to have a smaller energy difference between split d orbitals. This translates into longer-wavelength light absorbed (E = hc/) for tetrahedral complex ions compared to octahedral complex ions. Information from Table 21.15 confirms this.
1016
CHAPTER 21 TRANSITION METALS AND COORDINATION CHEMISTRY
16.
Co2+: [Ar]3d7; the corresponding d-orbital splitting diagram for tetrahedral Co 2+ complexes is:
All tetrahedral complexes are high spin since the d-orbital splitting is small. Ions with two or seven d electrons should give the most stable tetrahedral complexes because they have the greatest number of electrons in the lower-energy orbitals as compared with the number of electrons in the higher-energy orbitals. 17.
Linkage isomers differ in the way that the ligand bonds to the metal. SCN − can bond through the sulfur or through the nitrogen atom. NO2− can bond through the nitrogen or through the oxygen atom. OCN− can bond through the oxygen or through the nitrogen atom. N 3−, NH2CH2CH2NH2, and I− are not capable of linkage isomerism.
18.
Cu2+: [Ar]3d9; Cu+: [Ar]3d10; Cu(II) is d9 and Cu(I) is d10. Color is a result of the electron transfer between split d orbitals. This cannot occur for the filled d orbitals in Cu(I). Cd2+, like Cu+, is also d10. We would not expect Cd(NH3)4Cl2 to be colored since the d orbitals are filled in this Cd2+ complex.
19.
Sc3+ has no electrons in d orbitals. Ti3+ and V3+ have d electrons present. The color of transition metal complexes results from electron transfer between split d orbitals. If no d electrons are present, no electron transfer can occur, and the compounds are not colored.
20.
The maximum number of unpaired electrons is 5. This would be the case for a weak-field (high spin) d5 transition metal ion. The crystal field diagram would be:
5 2+ 5 3+ 5 2+ Some potential ionssmall would High d spin, be Mn : [Ar]3d and Fe : [Ar]3d . For a weak field, the Mn ion is the better choice because it has the smaller charge. A good choice for the ligands would be I− because it is a weak-field ligand. A potential formula for an octahedral complex ion with 5 unpaired electrons would be MnI64−.
21.
The compound is composed of CO and Ni. From the name, the formula would be Ni(CO) 4. Since the CO ligands are neutral in charge, nickel has an oxidation state of zero in Ni(CO) 4.
22.
Metals are easily oxidized by oxygen and other substances to form the metal cations. Because of this, metals are found in nature combined with nonmetals such as oxygen, sulfur, and the halogens. These compounds are called ores. To recover and use the metals, we must separate them from their ores and reduce the metal ions. Then, because most metals are unsuitable for use in the pure state, we must form alloys with the metals in order to form materials having desirable properties.
CHAPTER 21
TRANSITION METALS AND COORDINATION CHEMISTRY
1017
23.
At high altitudes, the oxygen content of air is lower, so less oxyhemoglobin is formed, which diminishes the transport of oxygen in the blood. A serious illness called high-altitude sickness can result from the decrease of O2 in the blood. High-altitude acclimatization is the phenomenon that occurs with time in the human body in response to the lower amounts of oxyhemoglobin in the blood. This response is to produce more hemoglobin and hence, increase the oxyhemoglobin in the blood. High-altitude acclimatization takes several weeks to take hold for people moving from lower altitudes to higher altitudes.
24.
CN− and CO form much stronger complexes with Fe2+ than O2. Thus, O2 cannot be transported by hemoglobin in the presence of CN − or CO because the binding sites prefer the toxic CN − and CO ligands.
25.
Roasting is the process of converting sulfide minerals to oxides by heating in air at temperatures below their melting points. Smelting is the process of reducing metal cations to their free metals.
26.
Cyanidation is the process in which crushed gold ore is treated with an aqueous cyanide solution in the presence of air to dissolve the gold by forming Au(CN)2‒. Pure gold is recovered by the reduction of the Au+ ion to Au.
Exercises Transition Metals and Coordination Compounds 27.
28.
29.
30.
a. Ni: [Ar]4s23d8
b. Cd: [Kr]5s24d10
c. Zr: [Kr]5s24d2
d. Os: [Xe]6s24f145d6
Transition metal ions lose the s electrons before the d electrons. a. Ni2+: [Ar]3d8
b. Cd2+: [Kr]4d10
c. Zr3+: [Kr]4d1; Zr4+: [Kr]
d. Os2+: [Xe]4f145d6; Os3+: [Xe]4f145d5
Transition metal ions lose the s electrons before the d electrons. a. Ti: [Ar]4s23d2
b. Re: [Xe]6s24f145d5
c. Ir: [Xe]6s24f145d7
Ti2+: [Ar]3d2
Re2+: [Xe]4f145d5
Ir2+: [Xe]4f145d7
Ti4+: [Ar] or [Ne]3s23p6
Re3+: [Xe]4f145d4
Ir3+: [Xe]4f145d6
Cr and Cu are exceptions to the normal filling order of electrons. a. Cr: [Ar]4s13d5
b. Cu: [Ar]4s13d10
c. V: [Ar]4s23d3
Cr2+: [Ar]3d4
Cu+: [Ar]3d10
V2+: [Ar]3d3
Cr3+: [Ar]3d3
Cu2+: [Ar]3d9
V3+: [Ar]3d2
1018 31.
CHAPTER 21 TRANSITION METALS AND COORDINATION CHEMISTRY a. With K+ and CN− ions present, iron has a 3+ charge. Fe3+: [Ar]3d5 b. With a Cl− ion and neutral NH3 molecules present, silver has a 1+ charge. Ag+: [Kr]4d10 c. With Br− ions and neutral H2O molecules present, nickel has a 2+ charge. Ni2+: [Ar]3d8 d. With NO2− ions, an I− ion, and neutral H2O molecules present, chromium has a 3+ charge. Cr3+: [Ar]3d3
32.
a. With NH4+ ions, Cl− ions, and neutral H2O molecules present, iron has a 2+ charge. Fe2+: [Ar]3d6 b. With I− ions and neutral NH3 and NH2CH2CH2NH2 molecules present, cobalt has a 2+ charge. Co2+: [Ar]3d7 c. With Na+ and F− ions present, tantalum has a 5+ charge. Ta5+: [Xe]4f14 (expected) d. Each platinum complex ion must have an overall charge if the two complex ions are counterions to each. Knowing that platinum forms 2+ and 4+ charged ions, we can deduce that the six-coordinate complex ion has a 4+ charged platinum ion and the four coordinate complex ion has a 2+ charged ion. With I− ions and neutral NH3 molecules present, the two complex ions are [Pt(NH3)4I2]2+ and [PtI4]2− . Pt2+: [Xe]4f145d8; Pt4+: [Xe]4f145d6
33.
a. molybdenum(IV) sulfide; molybdenum(VI) oxide b. MoS2, +4; MoO3, +6; (NH4)2Mo2O7, +6; (NH4)6Mo7O24•4H2O, +6
34.
a. 4 O ions on faces 1/2 O/face = 2 O ions, 2 O ions inside body; total: 4 O ions 8 Ti ions on corners 1/8 Ti/corner + 1 Ti ion/body center = 2 Ti ions Formula of the unit cell is Ti2O4. The empirical formula is TiO2. +4 −2
0
0
+4 −1
+4 −2
0
+4 −2
+2 −2
b. 2 TiO2 + 3 C + 4 Cl2 → 2 TiCl4 + CO2 + 2 CO; Cl is reduced, and C is oxidized. Cl2 is the oxidizing agent, and C is the reducing agent. +4 −1
0
TiCl4 + O2 → TiO2 + 2 Cl2; O is reduced, and Cl is oxidized. O 2 is the oxidizing agent, and TiCl4 is the reducing agent. 35.
NH3 is a weak base which produces OH− ions in solution. The white precipitate is Cu(OH)2(s). Cu2+(aq) + 2 OH−(aq) → Cu(OH)2(s) With excess NH3 present, Cu2+ forms a soluble complex ion, Cu(NH3)42+. Cu(OH)2(s) + 4 NH3(aq) → Cu(NH3)42+(aq) + 2 OH−(aq)
CHAPTER 21 36.
TRANSITION METALS AND COORDINATION CHEMISTRY
1019
CN− is a weak base, so OH− ions are present that lead to precipitation of Ni(OH)2(s). As excess CN− is added, the Ni(CN)42− complex ion forms. The two reactions are: Ni2+(aq) + 2 OH−(aq) → Ni(OH)2(s); the precipitate is Ni(OH)2(s). Ni(OH)2(s) + 4 CN−(aq) → Ni(CN)42−(aq) + 2 OH−(aq); Ni(CN)42− is a soluble species.
37.
Only [Cr(NH3)6]Cl3 will form a precipitate since only this compound will have Cl− ions in solution. The Cl− ions in the other compounds are ligands and are bound to the central Cr 3+ ion. The Cl− ions in [Cr(NH3)6]Cl3 are counter ions needed to produce a neutral compound, while the NH3 molecules are the ligands bound to Cr3+.
38.
BaCl2 gives no precipitate, so SO42− must be in the coordination sphere (BaSO4 is insoluble). A precipitate with AgNO3 means the Cl− is not in the coordination sphere. Because there are only four ammonia molecules in the coordination sphere, SO 42− must be acting as a bidentate ligand. The structure is: + H 3N
NH3
O
Co H 3N
39.
NH3
O Cl -
S O
O
To determine the oxidation state of the metal, you must know the charges of the various common ligands (see Table 21.12 of the text). a. hexacyanomanganate(II) ion b. cis-tetraamminedichlorocobalt(III) ion c. pentaamminechlorocobalt(II) ion
40.
41.
To determine the oxidation state of the metal, you must know the charges of the various common ligands (see Table 21.12 of the text). a. pentaamminechlororuthenium(III) ion
b. hexacyanoferrate(II) ion
c. tris(ethylenediamine)manganese(II) ion
d. pentaamminenitrocobalt(III) ion
e. tetracyanonicklate(II) ion
f.
g. tris(oxalato)ferrate(III) ion
h. tetraaquadithiocyanatocobalt(III) ion
a. hexaamminecobalt(II) chloride
b. hexaaquacobalt(III) iodide
c. potassium tetrachloroplatinate(II)
d. potassium hexachloroplatinate(II)
e. pentaamminechlorocobalt(III) chloride
f.
tetraamminedichlorochromium(III) ion
triamminetrinitrocobalt(III)
1020 42.
43.
44.
45.
CHAPTER 21 TRANSITION METALS AND COORDINATION CHEMISTRY a. pentaaquabromochromium(III) bromide
b. sodium hexacyanocobaltate(III)
c. bis(ethylenediamine)dinitroiron(III) chloride
d. tetraamminediiodoplatinum(IV) tetraiodoplatinate(II)
a. K2[CoCl4]
b. [Pt(H2O)(CO)3]Br2
c. Na3[Fe(CN)2(C2O4)2]
d. [Cr(NH3)3Cl(H2NCH2CH2NH2)]I2
a. FeCl4−
b. [Ru(NH3)5H2O]3+
c. [Cr(CO)4(OH)2]+
d. [Pt(NH3)Cl3] −
a. O C
C
O
O
O
C
O
H 2O
C
O O
OH 2
O
O
C
Co
Co H 2O
O O
O
OH 2
C
O
O
C C
O
O
cis
trans
Note: C2O42− is a bidentate ligand. Bidentate ligands bond to the metal at two positions that are 90° apart from each other in octahedral complexes. Bidentate ligands do not bond to the metal at positions 180° apart. b. 2+
2+
I
I
H 3N
I
H 3N
Pt H 3N
NH3
H 3N
NH3
trans
cis
trans
Cl H 3N
Cl Cl
H 3N
Ir H 3N
NH3 I
cis
c.
NH3 Pt
Cl Ir
Cl
H 3N
NH3
NH3
Cl
cis
trans
CHAPTER 21
TRANSITION METALS AND COORDINATION CHEMISTRY
1021
d. +
+
N
N
H 3N
N
I
Cr
I
N
N Cr
Cr
I
NH3
I
H 3N
Note: en = N N (H2NCH2CH2NH2). 46.
+ N H 3N
I
I
NH3
are abbreviations for the bidentate ligand ethylenediamine
a.
b. H 2C
H 2N
H 2C
Cl
Cl
H 2N
Pt H 2C
NH2
CH 2
NH2
CH 2
Co Cl
H 2N
H 2C
c.
H 2N
Cl
d. NH3 H3N
Cl
Cl
H3N
NH3
Co
Co
H3N
NO2
H3N
NH3
NH3
ONO
e. 2+
H 2C
H 2N
OH 2
NH2
CH 2
NH2
CH 2
Cu H 2C
47.
monodentate
H 2N
OH 2
bidentate O
M
NH 3
O
C
M
O M
O
bridging
C O
O C
O M
O
O
1022
CHAPTER 21 TRANSITION METALS AND COORDINATION CHEMISTRY
48.
O O
C
M = transition metal ion
M
CH2 H2N
O
O C
O
NH2
C
CH2
H2N
O
and
Cu H2C
O
O
C
NH2
CH2
Cu H2C
C
O
H2N
O
49.
a. 2; forms bonds through the lone pairs on the two oxygen atoms. b. 3; forms bonds through the lone pairs on the three nitrogen atoms. c. 4; forms bonds through the two nitrogen atoms and the two oxygen atoms. d. 4; forms bonds through the four nitrogen atoms.
50.
M = transition metal ion CH2SH
CH2OH HS
HS
CH
HS
CH
M
M HS
M HO
CH2
CH2
CH2
HO
CH CH2
51. H3N H3N
NO2 Co NO2
NH3
O2N
NH3
H3N
NH3
H3N
NH3
H3N
ONO ONO Co H3N
NH3
NO2 Co NH3
ONO NH3
H3N
NH3
H3N
NH3
O2N
NH3
H3N
ONO Co NO2
Co ONO
NH3 NH3
ONO Co NH3
NH3 NH3
SH
CHAPTER 21
TRANSITION METALS AND COORDINATION CHEMISTRY
1023
52. H 3N
SCN
H 3N
Pt
SCN
H 3N
Pt
N CS
H 3N
NH3
SCN
SCN
H 3N
Pt
NH3
SCN
N CS
H 3N
Pt
H 3N
SCN
H 3N
N CS Pt NH3
N CS Pt
SCN
H 3N
N CS
53.
The formula for the complex ion is Co(H2O)4F2+ (one of the fluorides is a counter ion to the +1 complex ion). The compound can have cis and trans isomers, depending on the relative positions of the fluoride ligands in the complex ion. If the fluorides are 180° apart, this is the trans isomer. If the fluorides are 90° apart, this is the cis isomer. Both the cis and trans complex ions have superimposable mirror images, so they do not have optical isomers. Lastly, linkage isomerism is not possible with H2O and F‒ ligands.
54.
The formula for the complex ion is Fe(H 2NCH2CH2NH2)2I2+ (one of the iodides is a counter ion to the +1 complex ion). The compound can have cis and trans isomers, depending on the relative positions of the iodide ligands in the complex ion. If the iodides are 180° apart, this is the trans isomer. If the iodides are 90° apart, this is the cis isomer. The cis has a nonsuperimposable mirror image, so it has an optical isomer (like co(en)2Cl2+ in Figure 21.17b). The trans isomer has a superimposable mirror image, so it does not have an optical isomer. Lastly, linkage isomerism is not possible with H2NCH2CH2NH2 and I‒ ligands.
55.
Like the molecules discussed in Figures 21.16 and 21.17 of the text, Cr(acac) 3 and cisCr(acac)2(H2O)2 are optically active. The mirror images of these two complexes are nonsuperimposable. There is a plane of symmetry in trans-Cr(acac)2(H2O)2, so it is not optically active. A molecule with a plane of symmetry is never optically active because the mirror images are always superimposable. A plane of symmetry is a plane through a molecule where one side reflects the other side of the molecule.
56.
There are five geometrical isomers (labeled i-v). Only isomer v, where the CN−, Br−, and H2O ligands are cis to each other, is optically active. The nonsuperimposable mirror image is shown for isomer v. i
ii
CN Br Pt Br CN
iv Br
OH2
OH2
Br
OH2
Br
Br H2O H2O
Br
CN Pt CN Br NC
CN Br
CN
NC
Pt Br
OH2
CN OH2
Pt NC
iii CN
Pt
v CN
OH2
OH2 OH2 optically mirror active
Br Pt
H2O
Br H2O mirror image of v (nonsuperimposable)
1024
CHAPTER 21 TRANSITION METALS AND COORDINATION CHEMISTRY
Bonding, Color, and Magnetism in Coordination Compounds 57.
NH3 and H2O are neutral charged ligands, while chloride and bromide are 1− charged ligands. The metal ions in the three compounds are Cr3+: [Ar]d3, Co3+: [Ar]d6, and Fe3+: [Ar]d5. a. With five electrons each in a different orbital, this diagram is for the weak-field [Fe(H2O)6]3+ complex ion. b. With three electrons, this diagram is for the [Cr(NH3)5Cl]2+ complex ion. c. With six electrons all paired up, this diagram is for the strong-field [Co(NH3)4Br2]+ complex ion.
58.
The metal ions are both d5 (Fe3+: [Ar]d5 and Mn2+: [Ar]d5). One of the diagrams (a) is for a weak-field (high-spin) d5 complex ion while the other diagram (b) is for a strong-field (lowspin) d5 complex ion. From the spectrochemical series, CN − is a strong-field ligand while H2O is in the middle of the series. Because the iron complex ion has CN − for the ligands as well as having a higher metal ion charge (+3 vs. +2), one would expect [Fe(CN)6]3− to have the strongfield diagram in b, while [Mn(H2O)6]2+ would have the weak-field diagram in a.
59.
a. Fe2+: [Ar]3d6
High spin, small
b. Fe3+: [Ar]3d5
Low spin, large
a. Zn2+: [Ar]3d10
c. Ni2+: [Ar]3d8
High spin, small
60.
CHAPTER 21
TRANSITION METALS AND COORDINATION CHEMISTRY
1025
b. Co2+: [Ar]3d7
High spin, small
Low spin, large
c. Ti3+: [Ar]3d1
61.
CN‒ is a strong field ligand. The metal ions in the complexes are Ni2+: [Ar]3d8, Ti3+: [Ar]3d1, Fe3+: [Ar]3d5, Co3+: [Ar]3d6, and Cr3+: [Ar]3d3. A metal ion must have an even number of d electrons to have a possibility of being diamagnetic. So, we only need to consider Co(CN)63‒ which has a d6 metal ion and Ni(CN)64‒ which has a d8 metal ion. Co3+: [Ar]3d6, no unpaired e-
Ni2+: [Ar]3d8, 2 unpaired e
Low spin, large
Only Co(CN)63‒ will be diamagnetic (answer d). 62.
The metal ions are Cr2+: [Ar]3d4, Co3+: [Ar]3d6, Ti3+: [Ar]3d1, Mn2+ [Ar]3d5, and V2+: [Ar]3d3. The Ti3+ complex ion has 1 unpaired electron since it only has 1 d electron.
The order of increasing unpaired electrons is Ti(H 2O)63+ (1 unpaired electron) < Cr(en)32+ (2) < VCl64‒ (3) < CoF63‒ (4) < MnBr64‒ (5 unpaired electrons).
1026
CHAPTER 21 TRANSITION METALS AND COORDINATION CHEMISTRY
63.
Because fluorine has a −1 charge as a ligand, chromium has a +2 oxidation state in CrF 64−. The electron configuration of Cr2+ is [Ar]3d4. For four unpaired electrons, this must be a weakfield (high-spin) case where the splitting of the d-orbitals is small and the number of unpaired electrons is maximized. The crystal field diagram for this ion is:
64.
small
NH3 and H2O are neutral ligands, so the oxidation states of the metals are Co 3+ and Fe2+. Both have six d electrons ([Ar]3d6). To explain the magnetic properties, we must have a strong-field for Co(NH3)63+ and a weak-field for Fe(H2O)62+. Co3+: [Ar]3d6
Fe2+: [Ar]3d6
large
small
Only this splitting of d-orbitals gives a diamagnetic Co(NH3)63+ (no unpaired electrons) and a paramagnetic Fe(H2O)62+ (unpaired electrons present). 65.
To determine the crystal field diagrams, you need to determine the oxidation state of the transition metal, which can only be determined if you know the charges of the ligands (see Table 21.12). The electron configurations and the crystal field diagrams follow. a. Ru2+: [Kr]4d6, no unpaired e-
b. Ni2+: [Ar]3d8, 2 unpaired e-
Low spin, large
c. V3+: [Ar]3d2, 2 unpaired e-
Note: Ni2+ must have 2 unpaired electrons, whether high-spin or low-spin, and V3+ must have 2 unpaired electrons, whether high-spin or low-spin.
CHAPTER 21
TRANSITION METALS AND COORDINATION CHEMISTRY
1027
66.
In both compounds, iron is in the +3 oxidation state with an electron configuration of [Ar]3d 5. Fe3+ complexes have one unpaired electron when a strong-field case and five unpaired electrons when a weak-field case. Fe(CN)62− is a strong-field case, and Fe(SCN)63− is a weak-field case. Therefore, cyanide (CN−) is a stronger-field ligand than thiocyanate (SCN−).
67.
Co: [Ar]4s23d7; cobalt is a d6 is Co(NH3)63+ and a d7 in Co(NH3)62+.
From the diagrams, Co(NH3)63+ has 0 unpaired electrons while Co(NH3)62+ has 3 unpaired electrons. The complex ion Co(NH3)62+ has 3 more unpaired electrons than Co(NH3)63+ (asnswer c). 68.
Complex ions having d4 and d7 metal ions have two more unpaired electrons in a weak field vs a strong field.
Answers a (Co2+: [Ar]3d7) and e (Cr2+: [Ar]3d4) are possible ions. 69.
All have octahedral Co3+ ions, so the difference in d orbital splitting and the wavelength of light absorbed only depends on the ligands. From the spectrochemical series, the order of the ligands from strongest to weakest field is CN− > en > H2O > I−. The strongest-field ligand produces the greatest d-orbital splitting (largest ) and will absorb light having the smallest
1028
CHAPTER 21 TRANSITION METALS AND COORDINATION CHEMISTRY wavelength. The weakest-field ligand produces the smallest and absorbs light having the longest wavelength. The order is: Co(CN)63− < Co(en)33+ < Co(H2O)63+ < CoI63− shortest absorbed longest absorbed
70.
Replacement of water ligands by ammonia ligands resulted in shorter wavelengths of light being absorbed. Energy and wavelength are inversely related, so the presence of the NH3 ligands resulted in a larger d-orbital splitting (larger ). Therefore, NH3 is a stronger-field ligand than H2O.
71.
From Table 21.15 of the text, the violet complex ion absorbs yellow-green light (λ 570 nm), the yellow complex ion absorbs blue light (λ 450 nm), and the green complex ion absorbs red light (λ 650 nm). The spectrochemical series shows that NH3 is a stronger-field ligand than H2O, which is a stronger-field ligand than Cl−. Therefore, Cr(NH3)63+ will have the largest d-orbital splitting and will absorb the lowest-wavelength electromagnetic radiation (λ 450 nm) since energy and wavelength are inversely related (λ = hc/E). Thus the yellow solution contains the Cr(NH3)63+ complex ion. Similarly, we would expect the Cr(H2O)4Cl2+ complex ion to have the smallest d-orbital splitting since it contains the weakest-field ligands. The green solution with the longest wavelength of absorbed light contains the Cr(H 2O)4Cl2+ complex ion. This leaves the violet solution, which contains the Cr(H2O)63+ complex ion. This makes sense because we would expect Cr(H2O)63+ to absorb light of a wavelength between that of Cr(NH3)63+ and Cr(H2O)4Cl2+.
72.
All these complex ions contain Co3+ bound to different ligands, so the difference in d-orbital splitting for each complex ion is due to the difference in ligands. The spectrochemical series indicates that CN− is a stronger-field ligand than NH3 which is a stronger-field ligand than F−. Therefore, Co(CN)63− will have the largest d-orbital splitting and will absorb the lowestwavelength electromagnetic radiation (λ = 290 nm) since energy and wavelength are inversely related (λ = hc/E). Co(NH3)63+ will absorb 440-nm electromagnetic radiation, while CoF63− will absorb the longest-wavelength electromagnetic radiation (λ = 770 nm) since F − is the weakestfield ligand present.
73.
CoBr64− has an octahedral structure, and CoBr42− has a tetrahedral structure (as do most Co2+ complexes with four ligands). Coordination complexes absorb electromagnetic radiation (EMR) of energy equal to the energy difference between the split d-orbitals. Because the tetrahedral d-orbital splitting is less than one-half the octahedral d-orbital splitting, tetrahedral complexes will absorb lower energy EMR, which corresponds to longer-wavelength EMR (E = hc/λ). Therefore, CoBr62− will absorb EMR having a wavelength shorter than 3.4 × 10 −6 m.
74.
In both complexes, nickel is in the +2 oxidation state: Ni2+: [Ar]3d8. The differences in unpaired electrons must be due to differences in molecular structure. NiCl 42− is a tetrahedral complex, and Ni(CN)42− is a square planar complex. The corresponding d-orbital splitting diagrams are:
CHAPTER 21
TRANSITION METALS AND COORDINATION CHEMISTRY
2NiCl42− N iCl 4
75.
1029
22−
NNi(CN) i(CN )44
Assuming the four ligands lie along the x and y-axis, the square planar crystal field diagram is:
Answer b is true. The four ligands point directly at the four lobes of d x 2 − y 2 orbital, making it the highest energy orbital. Since the d x 2 − y 2 orbital is by far the largest energy orbital, d 8 complex ions are diamagnetic.
76.
The tetrahedral crystal field diagram is:
Answers c-e are false. The splitting is much smaller in the tetrahedral arrangement because none of the d orbitals point directly at the ligands. All tetrahedral complexes are weak field or high spin. The arrangement of the orbitals in the diagram are the exact opposite of the octahedral field diagram, so the d z 2 and d x 2 − y 2 are the lowest energy orbitals in the tetrahedral field. The dxy, dxz, and dyx orbitals are all degenerate, but they are the highest energy orbitals in the tetrahedral field.
1030 77.
CHAPTER 21 TRANSITION METALS AND COORDINATION CHEMISTRY Because the ligands are Cl−, iron is in the +3 oxidation state. Fe3+: [Ar]3d5
78.
Because all tetrahedral complexes are high spin, there are 5 unpaired electrons in FeCl4−.
Pd is in the +2 oxidation state in PdCl42−; Pd2+: [Kr]4d8. If PdCl42− were a tetrahedral complex, then it would have 2 unpaired electrons and would be paramagnetic (see diagram below). Instead, PdCl42− has a square planar molecular structure with the d-orbital splitting diagram also shown below. Note that all electrons are paired in the square planar diagram; this explains the diamagnetic properties of PdCl42−.
tetrahedral d8
square planar d8
Metallurgy 79.
a. To avoid fractions, let’s first calculate ΔH for the reaction: 6 FeO(s) + 6 CO(g) → 6 Fe(s) + 6 CO2(g) 6 FeO + 2 CO2 → 2 Fe3O4 + 2 CO 2 Fe3O4 + CO2 → 3 Fe2O3 + CO 3 Fe2O3 + 9 CO → 6 Fe + 9 CO2
ΔH° = −2(18 kJ) ΔH° = −(−39 kJ) ΔH° = 3(−23 kJ)
6 FeO(s) + 6 CO(g) → 6 Fe(s) + 6 CO2(g)
ΔH° = −66 kJ
So for: FeO(s) + CO(g) → Fe(s) + CO2(g)
ΔH° =
− 66 kJ = −11 kJ 6
b. ΔH° = 2(−110.5 kJ) − (−393.5 kJ + 0) = 172.5 kJ ΔS° = 2(198 J/K) − (214 J/K + 6 J/K) = 176 J/K ΔG° = ΔH° − TΔS°, ΔG° = 0 when T =
172.5 kJ ΔH o = = 980. K o 0.176 kJ/K ΔS
Due to the favorable ΔS° term, this reaction is spontaneous at T > 980. K. From Figure 21.36 of the text, this reaction takes place in the blast furnace at temperatures greater than 980. K, as required by thermodynamics.
CHAPTER 21 80.
TRANSITION METALS AND COORDINATION CHEMISTRY
1031
a. H = 2(−1117) + (−393.5) − [3(−826) + (−110.5)] = −39 kJ S = 2(146) + 214 − [3(90.) + 198] =38 J/K b. G = H − TS; T = 800. + 273 = 1073 K G = −39 kJ − 1073 K(0.038 kJ/K) = −39 kJ − 41 kJ = −80. kJ
81.
Fe2O3: iron has a +3 oxidation state; Fe3O4: iron has a +8/3 oxidation state. The three iron ions in Fe3O4 must have a total charge of +8. The only combination that works is to have two Fe 3+ ions and one Fe2+ ion per formula unit. This makes sense from the other formula for magnetite, FeO•Fe2O3. FeO has an Fe2+ ion, and Fe2O3 has two Fe3+ ions.
82.
3 Fe + C → Fe3C; ΔH° = 21 − [3(0) + 0] = 21 kJ; ΔS° = 108 − [3(27) + 6] = 21 J/K ΔG° = ΔH° − TΔS°; when ΔH° and ΔS° are both positive, the reaction is spontaneous at high temperatures, where the favorable ΔS° term becomes dominant. Thus, to incorporate carbon into steel, high temperatures are needed for thermodynamic reasons but will also be beneficial for kinetic reasons (as the temperature increases, the rate of the reaction will increase). The relative amount of Fe3C (cementite) that remains in the steel depends on the cooling process. If the steel is cooled slowly, there is time for the equilibrium to shift back to the left; small crystals of carbon form, giving a relatively ductile steel. If cooling is rapid, there is not enough time for the equilibrium to shift back to the left; Fe3C is still present in the steel, and the steel is more brittle. Which cooling process occurs depends on the desired properties of the steel. The process of tempering fine-tunes the steel to the desired properties by repeated heating and cooling.
83.
Review Section 4.11 for balancing reactions in basic solution by the half-reaction method. (2 CN− + Ag → Ag(CN)2− + e−) × 4 4 e− + O2 + 4 H+ → 2 H2O 8 CN− + 4 Ag + O2 + 4 H+ → 4 Ag(CN)2− + 2 H2O Adding 4 OH− to both sides and canceling out 2 H2O on both sides of the equation gives the balanced equation: 8 CN−(aq)+ 4 Ag(s) + O2(g) + 2 H2O(l) → 4 Ag(CN)2−(aq) + 4 OH−(aq)
84.
Mn + HNO3 → Mn2+ + NO2 Mn → Mn2+ + 2 e−
HNO3 → NO2 HNO3 → NO2 + H2O (e− + H+ + HNO3 → NO2 + H2O) × 2
Mn → Mn2+ + 2 e− 2 e + 2 H + 2 HNO3 → 2 NO2 + 2 H2O −
+
2 H+(aq) + Mn(s) + 2 HNO3(aq) → Mn2+(aq) + 2 NO2(g) + 2 H2O(l)
1032
CHAPTER 21 TRANSITION METALS AND COORDINATION CHEMISTRY Mn2+ + IO4− → MnO4− + IO3− (4 H2O + Mn2+ → MnO4− + 8 H+ + 5 e−) × 2
(2 e− + 2 H+ + IO4− → IO3− + H2O) × 5
8 H2O + 2 Mn2+ → 2 MnO4− + 16 H+ + 10 e− 10 e−+ 10 H+ + 5 IO4− → 5 IO3− + 5 H2O 3 H2O(l) + 2 Mn2+(aq) + 5 IO4−(aq) → 2 MnO4−(aq) + 5 IO3−(aq) + 6 H+(aq)
ChemWork Problems 85.
Hg2+(aq) + 2 I−(aq) → HgI2(s), orange precipitate HgI2(s) + 2 I−(aq) → HgI42−(aq), soluble complex ion Hg2+ is a d10 ion. Color is the result of electron transfer between split d orbitals. This cannot occur for the filled d orbitals in Hg 2+. Therefore, we would not expect Hg 2+ complex ions to form colored solutions.
86.
Color of transition metal complexes is the result of the electron transfer between split d orbitals. This cannot occur when all the d orbitals are filled with electrons which occurs for d10 complex ions or when there are no d orbitals electrons. Zn2+ and Cu+ ions are both d10 ions, while Ti4+ and Sc3+ both are d0 ions. The only ion in the answers that isn’t a d0 or d10 is Mn2+ which is 4s03d6. So only Mn2+ octahedral complex ions are expected to be colored (answer b).
87.
0.112 g Eu2O3 ×
0.0967 g 304.0 g Eu = 0.0967 g Eu; mass % Eu = × 100 = 33.8% Eu 352.0 g Eu 2 O 3 0.286 g
Mass % O = 100.00 − (33.8 + 40.1 + 4.71) = 21.4% O Assuming 100.00 g of compound: 33.8 g Eu × 4.71 g H ×
1 mol 1 mol = 0.222 mol Eu; 40.1 g C × = 3.34 mol C 152 .0 g 12.01 g
1 mol 1 mol = 4.67 mol H; 21.4 g O × = 1.34 mol O 1.008 g 16.00 g
3.34 4.67 1.34 = 15.0, = 21.0, = 6.04 0.222 0.222 0.222
The molecular formula is EuC15H21O6. Because each acac− ligand has a formula of C5H7O2-, an abbreviated molecular formula is Eu(acac)3. 88.
Assuming 100.0 g of compound, 100.0 ‒ 29.9 = 70.1 g is CuSO4. Mol CuSO4 = 70.1 g CuSO 4 Mol NH3 = 29.9 g NH3 ×
1 mol CuSO 4 = 0.439 mol 159.62
1 mol NH 3 1.76 = 1.76 mol; = 4.01 0.439 17.03 g
The formula is Cu(NH3)4SO4, so x = 4.
CHAPTER 21 89.
TRANSITION METALS AND COORDINATION CHEMISTRY
1033
The five possible geometrical isomers for MA2B2C2 are:
The first three isomers have 90° bond angles between two of the three different ligands and a 180° bond angle between the third ligand. The other two isomers have 90° bond angles between the various ligands or 180° bond angles between the various ligands. 90.
The complex ion is PtCl42−, which is composed of Pt2+ and four Cl− ligands. Pt2+: [Xe]4f145d8. With square planar geometry, geometric (cis-trans) isomerism is possible. Cisplatin is the cis isomer of the compound and has the following structural formula.
(Au(CN)2− + e− → Au + 2 CN−) × 2 Zn + 4 CN− → Zn(CN)42− + 2 e−
E° = −0.60 V −E° = 1.26 V
2 Au(CN)2−(aq) + Zn(s) → 2 Au(s) + Zn(CN)42−(aq)
o = 0.66 V E cell
91.
o ΔG° = − nFE cell = −(2 mol e−)(96,485 C/mol e−)(0.66 J/C) = −1.3 × 105 J = −130 kJ
E° =
0.0591 nE o 2(0.66) = = 22.34, K = 1022.34 = 2.2 × 1022 log K, log K = n 0.0591 0.0591
Note: We carried extra significant figures to determine K. 92.
a. In the following structures, we omitted the 4 NH3 ligands coordinated to the outside cobalt atoms. 6+
6+
Co
H
H
O
O
OH
HO
HO
Co
Co
Co
O
HO
OH
Co O
OH Co
Co
OH
HO H
H
mirror
Co
1034
CHAPTER 21 TRANSITION METALS AND COORDINATION CHEMISTRY b. All are Co(III). The three “ligands” each contain 2 OH − and 4 NH3 groups. If each cobalt is in the +3 oxidation state, then each ligand has a +1 overall charge. The 3+ charge from the three ligands, along with the 3+ charge of the central cobalt atom, gives the overall complex a +6 charge. This is balanced by the 6− charge of the six Cl− ions. c. Co3+: [Ar]3d6; there are zero unpaired electrons if a low-spin (strong-field) case.
large
93.
There are four geometrical isomers (labeled i-iv). Isomers iii and iv are optically active, and the nonsuperimposable mirror images are shown.
i.
+ Br
H 3N
NH2
ii.
CH 2
+ Br
NH3
Cr H 3N
Cl
NH2
CH 2
NH2
CH 2
Cr NH2
CH 2
Cl
NH3
+
iii. Br
Cl
NH2
+
CH 2
H 2C
H 2N
Cr H 3N
Br
Cl
Cr NH2
CH 2
H 2C
H 2N
NH3
optically active
mirror
iv.
H 3N
NH3
mirror image of iii (nonsuperimposable)
+ Br
Cl
NH2
+
CH 2
H 2C
H 2N
Cr H 3N
NH3
Cl
Br
Cr NH2
optically active
CH 2
H 2C
mirror
H 2N
H 3N
NH3
mirror image of iv (nonsuperimposable)
CHAPTER 21 94.
TRANSITION METALS AND COORDINATION CHEMISTRY
1035
a. Be(tfa)2 exhibits optical isomerism. A representation for the tetrahedral optical isomers is: CH3 C
O
HC
O
CF3
CF3
C
C
Be C
O
CH O
CF3
CH3 O
HC
O
C
Be
C
C
CH3
CH3
O
CH O
C CF3
mirror
Note: The dotted line indicates a bond pointing into the plane of the paper, and the wedge indicates a bond pointing out of the plane of the paper. b. Square planar Cu(tfa)2 molecules exhibit geometric isomerism. In one geometric isomer, the CF3 groups are cis to each other, and in the other isomer, the CF3 groups are trans. CF3 C
O
HC
O
CF3
C
C
Cu C
O
CH3 cis
95.
CF3
CH O
CH3 O
HC
O
C
Cu
C
C
CH3
CH3
O
CH O
C CF3
trans
Ti2+: [Ar]3d2; Ti4+: [Ar]3d0; Cu+: [Ar]3d10; Cu2+: [Ar]3d9; Pd2+: [Ar]4d8; V2+: [Ar]3d3; Cr2+: [Ar]3d4; Cr3+: [Ar]3d3; consider the crystal field diagram for V2+ and Pd2+:
Any octahedral complex ion having 3 or fewer d electrons will have the same numbered of unpaired electrons in a weak or a strong field. The same is true for complex ions have 8 or more d electrons. Except for Cr2+, which is a d4, all the ions have the same number of unpaired electrons whether in a strong or a weak field.
1036 96.
CHAPTER 21 TRANSITION METALS AND COORDINATION CHEMISTRY a. The structure of cis-Pt(NH3)Cl2:
Its mirror image will be superimposable, so cis-Pt(NH3)Cl2 is not optically active. b. Like the trans isomer examined in Figure 21.17a of the text, trans-Ni(en)2(Br)2 will not be optically active because its mirror image is superimposable. c. Like the cis isomer examined in Figure 21.17b, cis-Ni(en)2(Br)2 will be optically active because its mirror image is not superimposable. d. When a tetrahedral central atom has four different groups bonded to it, the molecule or ion will always be optically active. So yes, this tetrahedral ion with Co2+ bonded to F−, Cl−, Br−, and I− will be optically active (the mirror image will not be superimposable). 97.
a. Zn2+ has a 3d10 electron configuration. This diagram is for a d 10 octahedral complex ion, not a d10 tetrahedral complex as is given. So, this diagram is incorrect. b. This diagram is correct. In this complex ion, Mn has a +3 oxidation state and an [Ar]3d 4 electron configuration. This strong field diagram for a d 4 ion is shown. c. This diagram is correct. In this complex ion, Ni2+ is present which has a 3d8 electron configuration. The diagram shown is correct for a diamagnetic square planar complex.
98.
a. sodium tris(oxalato)nickelate(II)
b. potassium tetrachlorocobaltate(II)
c. tetraamminecopper(II) sulfate d. chlorobis(ethylenediamine)thiocyanatocobalt(III) chloride 99.
100.
a. [Co(C5H5N)6]Cl3
b. [Cr(NH3)5I]I2
d. K2[Ni(CN)4]
e. [Pt(NH3)4Cl2][PtCl4]
c. [Ni(NH2CH2CH2NH2)3]Br2
a. Ru(phen)32+ exhibits optical isomerism [like Co(en)33+ in Figure 21.16 of the text]. b. Ru2+: [Kr]4d6; because there are no unpaired electrons, Ru 2+ is a strong-field (low-spin) case.
large
CHAPTER 21
TRANSITION METALS AND COORDINATION CHEMISTRY
1037
101.
Because each compound contains an octahedral complex ion, the formulas for the compounds are [Co(NH3)6]I3, [Pt(NH3)4I2]I2, Na2[PtI6], and [Cr(NH3)4I2]I. Note that in some cases the I− ions are ligands bound to the transition metal ion as required for a coordination number of 6, while in other cases the I− ions are counter ions required to balance the charge of the complex ion. The AgNO3 solution will only precipitate the I− counterions and will not precipitate the I− ligands. Therefore, 3 moles of AgI will precipitate per mole of [Co(NH 3)6]I3, 2 moles of AgI will precipitate per mole of [Pt(NH3)4I2]I2, 0 moles of AgI will precipitate per mole of Na2[PtI6], and l mole of AgI will precipitate per mole of [Cr(NH3)4I2]I.
102.
The oxidation state of iron in Fe(CNS)64− and Fe(CN)64− is +2. The electron configuration of Fe2+ is [Ar]3d6. The two possible crystal field diagrams for octahedral d 6 ions are:
High spin, small
Low spin, large
Because both Fe(CNS)64− and Fe(CN)64− have 0 unpaired electrons, they both are strong field (low spin) cases. CoCl42− is a tetrahedral complex ion. All tetrahedral complex ions are weak field. In this complex ion, cobalt has a +2 oxidation state with a [Ar]3d 7 electron configuration. The tetrahedral d7 crystal field diagram showing 3 unpaired electrons is:
Fe(H2O)63+ has iron in the +3 oxidation state with an [Ar]3d5 electron configuration. For this complex ion to have 5 unpaired electrons, this must be a weak field (high spin) case. The crystal field diagram is:
High spin, small
103.
Octahedral complex ions with d5 or d6 metal ions have 4 more unpaired electrons in the highspin (weak filed) as compared to the low-spin (weak field) case:
1038
CHAPTER 21 TRANSITION METALS AND COORDINATION CHEMISTRY
Mn2+ and Fe3+ both have [Ar]3d5 configurations, and Fe2+ and Co3+ both have [Ar]3d6 configurations. These four ions have 4 more unpaired electrons in a high spin vs a low spin case. The other ions are not d5 or d6 metal ions. 104.
a. Major species: Co(H2O)63+ (Ka = 1.0 × 10 −5 ), Cl− (neutral), and H2O (Kw = 1.0 × 10 −14 ); Co(H2O)63+ will determine the pH because it is a stronger acid than water. Solving the weak acid problem in the usual manner: Co(H2O)63+ Initial Equil.
⇌
0.10 M 0.10 – x
Ka = 1.0 × 10 −5 =
Co(H2O)5(OH)2+ 0 x
+
H+
Ka = 1.0 × 10 −5
~0 x
x2 x2 , x = [H+] = 1.0 × 10 −3 M; assumptions good. 0.10 0.10 − x
pH = –log(1.0 × 10 −3 ) = 3.00 b. As the charge on the metal ion increases, acid strength increases. Because of the lower charge, Co2+(aq) will not be as strong an acid as Co 3+(aq), so a solution of cobalt(II) nitrate will be less acidic (have a higher pH) than a solution with the same concentration of cobalt(III) nitrate. c. As the charge on the metal ion increases, the splitting of the d orbitals increases. Co 3+ ions ([Ar]d6) produce a strong-field case (low spin) and have 0 unpaired electrons as shown below. The lower charged Co2+ ions ([Ar]3d7) produce a weak-field case (high spin) and have 3 unpaired electrons.
d6 low spin, large
d7 high spin, small
CHAPTER 21 105.
TRANSITION METALS AND COORDINATION CHEMISTRY
HbO2 → Hb + O2 Hb + CO → HbCO HbO2(aq) + CO(g) → HbCO(aq) + O2(g)
1039
ΔG° = −(−70 kJ) ΔG° = −80 kJ ΔG° = −10 kJ
− ΔG o − (−10 10 3 J) = exp ΔG° = −RT ln K, K = exp = 60 RT (8.3145 J/K • mol)(298 kJ) 106.
2+
Cl H3N
NH3 Co
H3N
NH3 NH3
To form the trans isomer, Cl− would replace the NH3 ligand that is bold in the structure above. If any of the other four NH3 molecules are replaced by Cl−, the cis isomer results. Therefore, the expected ratio of the cis:trans isomer in the product is 4 : 1. 107.
a. Because O is in the −2 oxidation state, iron must be in the +6 oxidation state. Fe6+: [Ar]3d2. b. Using the half-reaction method of balancing redox reactions, the balanced equation is: 10 H+(aq) + 2 FeO42−(aq) + 2 NH3(aq) → 2 Fe3+(aq) + N2(g) + 8 H2O(l) 0.0250 L ×
0.243 mol FeO4 2− 1 mol N 2 = 3.04 × 10 −3 mol N2 L 2 mol FeO4 2− (if FeO42− limiting)
0.0550 L ×
1.45 mol NH 3 1 mol N 2 = 3.99 × 10 −2 mol N2 (if NH3 limiting) L 2 mol NH 3
Because the FeO42− reagent produces the smaller quantity of N2, FeO42− is limiting and 3.04 × 10 −3 mol N2 can be produced.
VN 2 =
nRT = P
0.08206 L atm 298 K K mol 1.50 atm
3.04 10 −3 mol N 2
VN 2 = 0.0496 L = 49.6 mL N2 108.
a. =
hc 6.626 10 −34 J s 2.998 10 8 m / s = = 5.72 × 10 −7 m = 572 nm − 23 E 1 . 986 10 J 1.75 10 4 cm−1 cm−1
b. There are three resonance structures for NCS −. From a formal charge standpoint, the following resonance structure is best.
1040
CHAPTER 21 TRANSITION METALS AND COORDINATION CHEMISTRY The N in this resonance structure is sp hybridized. Because the sp hybrid orbitals are 180 apart, one would expect that when the lone pair in an sp hybrid orbital on N is donated to the Cr3+ ion, the 180 bond angle would stay intact between Cr, N, C, and S. Similar to Co(en)2Cl2+ discussed in Figure 21.17 of the text, [Co(en)2(NCS)2]+ would exhibit cis-trans isomerism (geometric isomerism), and only the cis form would exhibit optical isomerism. For [Co(en)2(NCS)2]+, NCS− just replaces the Cl− ions in the isomers drawn in Figure 21.17. The trans isomer would not exhibit optical isomerism.
Challenge Problems 109.
Ni2+ = d8; if ligands A and B produced very similar crystal fields, the cis-[NiA2B4]2+ complex ion would give the following octahedral crystal field diagram for a d 8 ion:
This is paramagnetic.
Because it is given that the complex ion is diamagnetic, the A and B ligands must produce different crystal fields, giving a unique d-orbital splitting diagram that would result in a diamagnetic species. 110.
Cr3+: [Ar]d3; looking at Figure 21.20 of the text, the d z 2 orbital will be destabilized the most when stronger field ligands are on the z-axis. The dxy orbital will be destabilized the least by this arrangement of ligands. Assuming the A ligands produce a significantly stronger field than the B ligands, we will assume the dxz and dyz orbitals will be destabilized the second most of all the orbitals due to the lobes of these orbitals lying in the z-axis plane. This leaves the d x 2 − y 2 orbitals, whose relative position is very difficult to predict. We will assume the d x 2 − y 2 orbital is destabilized more than the dxy orbital because the lobes point directly at the B ligands, but is not destabilized as much as the other orbitals which all have lobes in the z-plane. Making these assumptions, the d-orbital splitting diagram is (assuming a low-spin case):
dz2
d xz , d yz
dx2 − y2
d xy Note: It could be that the relative positions of the d x 2 − y 2 with the dxz and dyz orbitals are switched.
CHAPTER 21 111.
TRANSITION METALS AND COORDINATION CHEMISTRY
1041
a. Consider the following electrochemical cell: Co3+ + e− → Co2+ Co(en)32+ → Co(en)33+ + e− Co3+ + Co(en)32+ → Co2+ + Co(en)33+
E° = 1.82 V −E° = ?
E ocell = 1.82 − E°
The equilibrium constant for this overall reaction is: Co3+ + 3 en → Co(en)33+ Co(en)32+ → Co2+ + 3 en
K1 = 2.0 × 1047 K2 = 1/1.5 × 1012
Co3+ + Co(en)32+ → Co(en)33+ + Co2+
K = K1K2 =
2.0 10 47 = 1.3 × 1035 1.5 1012
From the Nernst equation for the overall reaction:
E ocell =
0.0591 0.0591 log K = log(1.3 10 35 ), E ocell = 2.08 V n 1
E ocell = 1.82 − E° = 2.08 V, E° = 1.82 V − 2.08 V = −0.26 V b. The stronger oxidizing agent will be the more easily reduced species and will have the more positive standard reduction potential. From the reduction potentials, Co3+ (E° = 1.82 V) is a much stronger oxidizing agent than Co(en)33+ (E° = −0.26 V). c. In aqueous solution, Co3+ forms the hydrated transition metal complex Co(H2O)63+. In both complexes, Co(H2O)63+ and Co(en)33+, cobalt exists as Co3+, which has 6 d electrons. Assuming a strong-field case for each complex ion, the d-orbital splitting diagram for each is: eg
t2g
When each complex gains an electron, the electron enters a higher energy eg orbital. Since en is a stronger-field ligand than H2O, the d-orbital splitting is larger for Co(en)33+, and it takes more energy to add an electron to Co(en)33+ than to Co(H2O)63+. Therefore, it is more favorable for Co(H2O)63+ to gain an electron than for Co(en)33+ to gain an electron. II
112.
III
III
II
(H2O)5Cr‒Cl‒Co(NH3)5 → (H2O)5Cr‒Cl‒Co(NH3)5 → Cr(H2O)5Cl2+ + Co(II) complex Yes; after the oxidation, the ligands on Cr(III) won't exchange. Since Cl− is in the coordination sphere, it must have formed a bond to Cr(II) before the electron transfer occurred (as proposed through the formation of the intermediate).
113.
No; in all three cases, six bonds are formed between Ni2+ and nitrogen, so ΔH values should be similar. ΔS° for formation of the complex ion is most negative for 6 NH3 molecules reacting with a metal ion (7 independent species become 1). For penten reacting with a metal ion, 2 independent species become 1, so ΔS° is least negative of all three of the reactions. Thus the
1042
CHAPTER 21 TRANSITION METALS AND COORDINATION CHEMISTRY chelate effect occurs because the more bonds a chelating agent can form to the metal, the less unfavorable ΔS° becomes for the formation of the complex ion, and the larger the formation constant.
114.
____
L M
L
z axis (pointing out of the plane of the paper)
____ ____
dz2
____
L
d x 2 − y 2 , d xy
____
d xz , d yz
The d x 2 − y 2 and dxy orbitals are in the plane of the three ligands and should be destabilized the most. The amount of destabilization should be about equal when all the possible interactions are considered. The d z 2 orbital has some electron density in the xy plane (the doughnut) and should be destabilized a lesser amount than the d x 2 − y 2 and dxy orbitals. The dxz and dyz orbitals have no electron density in the plane and should be lowest in energy. 115.
____
L L M L L
L
z axis (pointing out of the plane of the paper)
dz2
____
____
d x 2 − y 2 , d xy
____
____
d xz , d yz
The d z 2 orbital will be destabilized much more than in the trigonal planar case (see Exercise 114). The d z 2 orbital has electron density on the z axis directed at the two axial ligands. The d x 2 − y 2 and dxy orbitals are in the plane of the three trigonal planar ligands and should be
destabilized a lesser amount than the d z 2 orbital; only a portion of the electron density in the d x 2 − y 2 and dxy orbitals is directed at the ligands. The dxz and dyz orbitals will be destabilized
the least since the electron density is directed between the ligands. 116.
For a linear complex ion with ligands on the x axis, the d x 2 − y 2 orbital will be destabilized the most, with the lobes pointing directly at the ligands. The dyz orbital has the fewest interactions with x-axis ligands, so it is destabilized the least. The dxy and dxz orbitals will have similar destabilization but will have more interactions with x-axis ligands than the dyz orbital. Finally, the d z 2 orbital with the doughnut of electron density in the xy plane will probably be destabilized more than the dxy and dxz orbitals but will have nowhere near the amount of destabilization that occurs with the d x 2 − y 2 orbital. The only difference that would occur in the diagram if the ligands were on the y axis is the relative positions of the dxy, dxz, and dyz orbitals. The dxz will have the smallest destabilization of all these orbitals, whereas the d xy and dyz orbitals will be degenerate since we expect both to be destabilized equivalently from y-axis ligands. The d-orbital splitting diagrams are:
CHAPTER 21
TRANSITION METALS AND COORDINATION CHEMISTRY
a.
b.
dx 2 − y2
dx 2 − y2
dz2
dz2
d xy , d xz
d xy , d yz
d yz
d xz
linear y-axis ligands
linear x-axis ligands 117.
1043
Ni2+: [Ar]3d8; the coordinate system for trans-[Ni(NH3)2(CN)4]2− is shown below. Because CN− produces a much stronger crystal field, it will dominate the d-orbital splitting. From the coordinate system, the CN− ligands are in a square planar arrangement. Therefore, the diagram will most likely resemble the square planar diagram given in Figure 21.28. Note that the relative position of d z 2 orbital is hard to predict. With the NH3 ligands on the z axis, we will assume the dz2 orbital is destabilized more than the dxy orbital. However, this is only an assumption. It could be that the dxy orbital is destabilized more. NH3 NC
y
dx 2 − y2
CN
dz2
Ni NC
CN
dxy
x
NH3
dxz
dyz
z
118.
We need to calculate the Pb2+ concentration in equilibrium with EDTA4-. Because K is large for the formation of PbEDTA2-, let the reaction go to completion; then solve an equilibrium problem to get the Pb2+ concentration. Pb2+ Before Change After Equil. 1.1 × 1018 =
+
EDTA4−
⇌ PbEDTA2−
0.010 M 0.050 M 0 0.010 mol/L Pb2+ reacts completely (large K) −0.010 −0.010 → +0.010 0 0.040 0.010 x mol/L PbEDTA2− dissociates to reach equilibrium x 0.040 + x 0.010 − x
K = 1.1 × 1018
Reacts completely New initial condition
(0.010 − x ) (0.010 ) , x = [Pb2+] = 2.3 × 10 −19 M; assumptions good. ( x)(0.040 + x) x (0.040 )
Now calculate the solubility quotient for Pb(OH)2 to see if precipitation occurs. The concentration of OH− is 0.10 M because we have a solution buffered at pH = 13.00 (pOH = 14.00 − 13.00 = 1.00, [OH−] = 10−1.00).
1044
CHAPTER 21 TRANSITION METALS AND COORDINATION CHEMISTRY Q = [Pb2+ ]0 [OH− ]0 2 = (2.3 × 10 −19 )(0.10)2 = 2.3 × 10 −21 < Ksp (1.2 × 10 −15 ) Pb(OH)2(s) will not form because Q is less than Ksp. AgBr(s) ⇌ Ag+ + Br‒ Ag + 2 NH3 ⇌ Ag(NH3)2+
119.
Ksp = 5.0 × 10−13 Kf = 1.7 × 107
+
AgBr(s) + 2 NH3(aq) ⇌ Ag(NH3)2+(aq) + Br‒(aq) AgBr(s)
+
2 NH3
⇌
K = Ksp × Kf = 8.5 × 10−6
Ag(NH3)2+ + Br−
Initial
3.0 M 0 0 If 0.10 mol of AgBr dissolves in 1.0 L, then the concentration changes would be: Change −0.10 −0.20 → +0.10 +0.10 After 2.8 M 0.10 M 0.10 M To check to see if all the solid dissolves, let’s see what the reaction quotient indicates using these concentrations. Q=
[Ag(NH3 ) +2 ] o [Br − ] o 2
[NH3 ] O
(0.10)(0.10) = 1.3 × 10−3 2 (2.8)
=
Since Q > K (8.5 × 10−6), the reaction will shift left to establish equilibrium, causing AgBr(s) to form. No, 0.10 mol of AgBr will not dissolve in 1.0 L of 3.0 M NH3. 120.
All of the ions in the answers break up into a 2+ charged complex ion and two 1‒ counter ions for 3 ions total. So the van’t Hoff factor, i, is 3. π = iMRT, 3.14 atm = 3 × M × Mol compound = 0.3000 L × Molar mass of compound =
0.08206 L atm × 298 K, M = 0.0428 mol/L K mol
0.0428 mol = 0.0128 mol L
2.58 g = 202 g/mol 0.0128 mol
The cobalt compound is [Co(NH3)5F]F2 (answer b) because it has a molar mass of 201 g/mol. The others have a molar mass significantly greater than 202 g/mol.
121.
0.0203 g CrO3 ×
0.0106 g 52.00 g Cr = 0.0106 g Cr; % Cr = × 100 = 10.1% Cr 100.0 g CrO 3 0.105 g
ii. 32.93 mL HCl ×
0.100 mmol HCl 1 mmol NH 3 17.03 mg NH 3 = 56.1 mg NH3 mL mmol HCl mmol
i.
% NH3 =
56.1 mg × 100 = 16.5% NH3 341 mg
iii. 73.53% + 16.5% + 10.1% = 100.1%; the compound must be composed of only Cr, NH3, and I.
CHAPTER 21
TRANSITION METALS AND COORDINATION CHEMISTRY
1045
Out of 100.00 g of compound: 10.1 g Cr ×
1 mol = 0.194 mol; 52.00 g
0.194 = 1.00 0.194
1 mol = 0.969 mol; 17.03 g
0.969 = 4.99 0.194
1 mol = 0.5794 mol; 126 .9 g
0.5794 = 2.99 0.194
16.5 g NH3 × 73.53 g I ×
Cr(NH3)5I3 is the empirical formula. If we assume an octahedral complex ion, then compound A is made of the octahedral [Cr(NH 3)5I]2+ complex ion and two I− ions as counterions; the formula is [Cr(NH3)5I]I2. Let’s check this proposed formula using the freezing-point data. iv. ΔTf = iKfm; for [Cr(NH3)5I]I2, i = 3.0 (assuming complete dissociation). Molality = m =
0.601 g complex 1 mol complex −2 517 .9 g complex 1.000 10 kg H 2 O
= 0.116 mol/kg
ΔTf = 3.0 × 1.86 °C kg/mol × 0.116 mol/kg = 0.65°C Because ΔTf is close to the measured value, this is consistent with the formula [Cr(NH3)5I]I2. So our assumption of an octahedral complex ion is probably a good assumption.
Marathon Problem 122.
CrCl3•6H2O contains nine possible ligands, only six of which are used to form the octahedral complex ion. The three species not present in the complex ion will either be counterions to balance the charge of the complex ion and/or waters of hydration. The number of counter- ions for each compound can be determined from the silver chloride precipitate data, and the number of waters of hydration can be determined from the dehydration data. In all experiments, the ligands in the complex ion do not react. Compound I: mol CrCl3•6H2O = 0.27 g ×
1 mol = 1.0 × 10−3 mol CrCl3•6H2O 266 .5 g
mol waters of hydration = 0.036 g H2O ×
1 mol = 2.0 × 10−3 mol H2O 18.02 g
mol waters of hydration 2.0 10 −3 mol = = 2.0 mol compound 1.0 10 −3 mol
In compound I, two of the H2O molecules are waters of hydration, so the other four water molecules are present in the complex ion. Therefore, the formula for compound I must be [Cr(H2O)4Cl2]Cl•2H2O. Two of the Cl− ions are present as ligands in the octahedral
1046
CHAPTER 21 TRANSITION METALS AND COORDINATION CHEMISTRY complex ion, and one Cl− ion is present as a counterion. The AgCl precipitate data that refer to this compound are the one that produces 1430 mg AgCl: 0.100 mol [Cr(H 2 O) 4 Cl 2 ]Cl • 2H 2 O L − 1 mol Cl × = 0.0100 mol Cl− mol [Cr(H 2 O) 4 Cl 2 ]Cl • 2H 2 O
mol Cl− from compound I = 0.1000 L ×
mass AgCl produced = 0.0100 mol Cl− ×
1 mol AgCl 143 .4 g AgCl = 1.43 g − mol AgCl mol Cl = 1430 mg AgCl
Compound II: 1 mol mol waters of hydration 18.02 g = = 1.0 −3 mol compound 1.0 10 mol compound 0.018 g H 2 O
The formula for compound II must be [Cr(H2O)5Cl]Cl2•H2O. The 2870-mg AgCl precipitate data refer to this compound. For 0.0100 mol of compound II, 0.0200 mol Cl− is present as counterions: mass AgCl produced = 0.0200 mol Cl− ×
1 mol AgCl 143 .4 g = 2.87 g − mol mol Cl = 2870 mg AgCl
Compound III: This compound has no mass loss on dehydration, so there are no waters of hydration present. The formula for compound III must be [Cr(H 2O)6]Cl3. 0.0100 mol of this compound produces 4300 mg of AgCl(s) when treated with AgNO3. 0.0300 mol Cl−
1 mol AgCl 143 .4 g AgCl = 4.30 g = 4.30 × 103 mg AgCl − mol AgCl mol Cl
The structural formulas for the compounds are: Compound I +
Cl H2O
OH2 Cr
H2O
OH2 Cl
+
Cl H2O Cl • 2 H2O
or
Cl Cr
H2O
OH2 OH2
Cl • 2 H2O
CHAPTER 21
TRANSITION METALS AND COORDINATION CHEMISTRY
Compound II
1047
Compound III
H2O
H2O
OH2 Cr
H2O
OH2 OH2
Cl2 • H2O
3+
OH2
2+
Cl
OH2 Cr
H2O
OH2
Cl3
OH2
From Table 21.15 of the text, the violet compound will be the one that absorbs light with the shortest wavelength (highest energy). This should be compound III. H 2O is a stronger-field ligand than Cl−; compound III with the most coordinated H2O molecules will have the largest d-orbital splitting and will absorb the higher-energy light. The magnetic properties would be the same for all three compounds. Cr 3+ is a d3 ion. With only three electrons present, all Cr3+complexes will have three unpaired electrons, whether strong field or weak field. If Cr2+was present with the d4 configuration, then the magnetic properties might be different for the complexes and could be worth examining.
CHAPTER 22 ORGANIC AND BIOLOGICAL MOLECULES Review Questions 1. A hydrocarbon is a compound composed of only carbon and hydrogen. A saturated hydrocarbon has only carbon-carbon single bonds in the molecule. An unsaturated hydrocarbon has one or more carbon-carbon multiple bonds but may also contain carbon-carbon single bonds. A normal hydrocarbon has one chain of consecutively bonded carbon atoms, with each carbon atom in the chain bonded to one or two other carbon atoms. A branched hydrocarbon has at least one carbon atom in the structure that forms bonds to three or four other carbon atoms; the structure is not one continuous chain of carbon atoms. An alkane is a saturated hydrocarbon composed of only C−C and C−H single bonds. Each carbon in an alkane is bonded to four other atoms (either C or H atoms). If the compound contains a ring in the structure and is composed of only C−C and C−H single bonds, then it is called a cyclic alkane. Alkanes: general formula = CnH2n + 2; all carbons are sp3 hybridized; bond angles = 109.5˚ Cyclic alkanes: general formula CnH2n (if only one ring is present in the compound); all carbons are sp3 hybridized; prefers 109.5˚ bond angles, but rings with three carbons or four carbons or five carbons are forced into bond angles less than 109.5˚. In cyclopropane, a ring compound made up of three carbon atoms, the bond angles are forced into 60˚ in order to form the three-carbon ring. With four bonds to each carbon, the carbons prefer 109.5˚ bond angles. This just can’t happen for cyclopropane. Because cyclopropane is forced to form bond angles smaller than it prefers, it is very reactive. The same is true for cyclobutane. Cyclobutane is composed of a four-carbon ring. To form a ring compound with four carbons, the carbons in the ring are forced to form 90˚ bond angles; this is much smaller than the preferred 109.5˚ bond angles. Cyclopentane (five-carbon rings) also has bond angles slightly smaller than 109.5˚, but they are very close (108˚), so cyclopentane is much more stable than cyclopropane or cyclobutane. For rings having six or more carbons, the observed bonds are all 109.5˚. Straight chain hydrocarbons just indicates that there is one chain of consecutively bonded Catoms in the molecule. They are not in a straight line which infers 180˚ bond angles. The bond angles are the predicted 109.5˚. To determine the number of hydrogens bonded to the carbons in cyclic alkanes (or any alkane where they may have been omitted), just remember that each carbon has four bonds. In cycloalkanes, only the C−C bonds are shown. It is assumed you know that the remaining bonds on each carbon are C−H bonds. The number of C−H bonds is that number required to give the carbon four total bonds.
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CHAPTER 22 2.
ORGANIC AND BIOLOGICAL MOLECULES
1049
Alkenes are unsaturated hydrocarbons that contain a carbon-carbon double bond. Carboncarbon single bonds may also be present. Alkynes are unsaturated hydrocarbons that contain a carbon-carbon triple bond. Alkenes: CnH2n is the general formula. The carbon atoms in the C=C bond exhibit 120˚ bond angles. The double-bonded carbon atoms are sp2 hybridized. The three sp2 hybrid orbitals form three sigma bonds to the attached atoms. The unhybridized p atomic orbital on each sp2 hybridized carbon overlap side to side to form the bond in the double bond. Because the p orbitals must overlap parallel to each other, there is no rotation in the double bond (this is true whenever bonds are present). See Figure 22.7 for the bonding in the simplest alkene, C 2H4. Alkynes: CnH2n – 2 is the general formula. The carbon atoms in the C−C bond exhibit 180˚ bond angles. The triple bonded carbons are sp hybridized. The two sp hybrid orbitals form two sigma bonds to the bonded atoms. The two unhybridized p atomic orbitals overlap with two unhybridized p atomic orbitals on the other carbon in the triple bond, forming two bonds. If the z-axis is the internuclear axis, then one bond would form by parallel overlap of py orbitals on each carbon and the other bond would form by parallel overlap of px orbitals. As is the case with alkenes, alkynes have restricted rotation due to the bonds. See Figure 22.10 for the bonding in the simplest alkyne, C2H2. Any time a multiple bond or a ring structure is added to a hydrocarbon, two hydrogens are lost from the general formula. The general formula for a hydrocarbon having one double bond and one ring structure would lose four hydrogens from the alkane general formula. The general formula would be CnH2n – 2.
3.
Aromatic hydrocarbons are a special class of unsaturated hydrocarbons based on the benzene ring. Benzene has the formula C6H6. It is a planar molecule (all atoms are in the same plane). The bonding in benzene is discussed in detail in Section 9.5 of the text. Figures 9.46, 9.47, and 9.48 detail the bonding in benzene. C6H6, has 6(4) + 6(1) = 30 valence electrons. The two resonance Lewis structures for benzene are:
These are abbreviated as:
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CHAPTER 22
ORGANIC AND BIOLOGICAL MOLECULES
Each carbon in benzene is attached to three other atoms; it exhibits trigonal planar geometry with 120° bond angles. Each carbon is sp2 hybridized. The sp2 hybrid orbitals form three sigma bonds to each carbon. The unhybridized p atomic orbital on each carbon overlap side to side with unhybridized p orbitals on adjacent carbons to form the bonds. All six of the carbons in the six-membered ring have one unhybridized p atomic orbital. All six of the unhybridized p orbitals overlap side to side to give a ring of electron density above and below the sixmembered ring of benzene. The six electrons in the bonds in benzene can roam about above and below the entire ring surface; these electrons are delocalized. This is important because all six carbon-carbon bonds in benzene are equivalent in length and strength. The Lewis structures say something different (three of the bonds are single and three of the bonds are double). This is not correct. To explain the equivalent bonds, the bonds can’t be situated between two carbon atoms as is the case in alkenes and alkynes; that is, the bonds can’t be localized. Instead, the six electrons can roam about over a much larger area; they are delocalized over the entire surface of the molecule. All this is implied in the following shorthand notation for benzene.
4.
A short summary of the nomenclature rules for alkanes, alkenes, and alkynes follows. See the text for details. a. Memorize the base names of C1–C10 carbon chains (see Table 22.1). When the C1–C10 carbon chains are named as a substituent, change the –ane suffix to–yl. b. Memorize the additional substituent groups in Table 22.2. c. Names are based on the longest continuous carbon chain in the molecule. Alkanes use the suffix –ane, alkenes end in –ene, and alkynes end in –yne. d. To indicate the position of a branch or substituent, number the longest chain of carbons consecutively to give the lowest numbers to the substituents or branches. Identify the number of the carbon that the substituent is bonded to by writing the number in front of the name of the substituent. e. Name substituents in alphabetical order. f.
Use a prefix (di–, tri–, tetra–, etc.) to indicate the number of a substituent if more than one is present. Note that if, for example, three methyl substituent groups are bonded to carbons on the longest chain, use the tri-prefix but also include three numbers indicating the positions of the methyl groups on the longest chain. Also note that prefixes like di–, tri–, tetra–, etc. are not considered when determining the alphabetical order of the substituent groups.
g. A cyclic hydrocarbon is designated by the prefix cyclo–.
CHAPTER 22
ORGANIC AND BIOLOGICAL MOLECULES
1051
h. For alkenes and alkynes, the position of the double or triple bond is indicated with a number placed directly in front of the base name of the longest chain. If more than one multiple bond is present, the number of multiple bonds is indicted in the base name using the prefix di–, tri–, tetra–, etc., but also a number for the position of each multiple bond is indicated in front of the base name. When numbering the longest chain, if double or triple bonds are present, give the multiple bonds the lowest number possible (not the substituent groups). This is a start. As you will find out, there are many interesting situations that can come up which aren’t covered by these rules. We will discuss them as they come up. For aromatic nomenclature rules, reference Section 22.3 of the text. The errors in the names are discussed below. a. The longest chain gives the base name. b. The suffix –ane indicates only alkanes. Alkenes and alkynes have different suffixes as do other “types” of organic compounds. c. Smallest numbers are used to indicate the position of substituents. d. Numbers are required to indicate the positions of double or triple bonds. e. Multiple bonds (double or triple) get the lowest number. f.
5.
The term ortho– in benzene nomenclature indicates substituents in the benzene ring bonded to C–1 and C–2. The term meta– describes C–1 and C–3 substituent groups, while para– is used for C–1 and C–4 substituent groups.
See Table 22.4 for the types of bonds and atoms in the functional groups listed in a-h of this question. Examples are also listed in Table 22.4. a. Halohydrocarbons: name the halogens as substituents, adding o to the end of the name of the halogen. Assuming no multiple bonds, all carbons and the halogens are sp3 hybridized because the bond angles are all 109.5˚. b. Alcohols: –ol suffix; the oxygen is sp3 hybridized because the predicted bond angle about O is 109.5˚. c. Ethers: name the two R-groups bonded to O as substituent groups (in alphabetical order), and then end the name in ether. These are the common nomenclature rules for ethers. The oxygen in an ether is sp3 hybridized due to the predicted 109.5˚ bond angle. d. Aldehydes: –al suffix; the carbon doubly bonded to oxygen is sp2 hybridized because this carbon exhibits 120˚ bond angles. The oxygen in the double bond is also sp2 hybridized because it has three effective pairs of electrons around it (two lone pairs and the bonded carbon atom). e. Ketones: –one suffix; the carbon doubly bonded to oxygen is sp2 hybridized because the bond angles about this carbon are 120˚. The O in the double bond is sp2 hybridized.
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CHAPTER 22 f.
ORGANIC AND BIOLOGICAL MOLECULES
Carboxylic acids: –oic acid is added to the end of the base name; the carbon doubly bonded to oxygen is sp2 hybridized (120˚ bond angles), and the oxygen with two single bonds is sp3 hybridized (predicted 109.5˚ bond angles). The oxygen in the double bond is sp2 hybridized.
g. Esters: the alcohol part of ester is named first as a substituent using the –yl suffix; the carboxylic acid part of the ester gives the base name using the suffix –oate. In the common nomenclature rules, the carboxylic acid part is named using common names for the carboxylic acid ending in the suffix –ate. The bonding and bond angles are the same as discussed previously with carboxylic acids. h. Amines: like ethers, the R-groups are named as substituent groups (in alphabetical order), and then end the name in amine (common rules). The nitrogen in amines is sp3 hybridized due to predicted 109.5˚ bond angles. The difference between a primary, secondary, and tertiary alcohol is the number of R-groups (other carbons) that are bonded to the carbon with the OH group. Primary: 1 R-group; secondary: 2 R-groups; tertiary: 3 R-groups. A number is required to indicate the location of a functional group only when that functional group can be present in more than one position in the longest chain. Halohydrocarbons, alcohols, and ketones require a number. The aldehyde group must be on C–1 in the longest chain, and the carboxylic acid group must also be on C–1; no numbers are used for aldehyde and carboxylic acid nomenclature. In addition, no numbers are used for nomenclature of simple ethers, simple esters, and simple amines. carboxylic acid
aldehyde
The R designation may be a hydrogen but is usually a hydrocarbon fragment. The major point in the R-group designation is that if the R-group is a hydrocarbon fragment, then the first atom in the R-group is a carbon atom. What the R-group has after the first carbon is not important to the functional group designation. 6.
Resonance: All atoms are in the same position. Only the positions of electrons are different. Isomerism: Atoms are in different locations in space. Isomers are distinctly different substances. Resonance is the use of more than one Lewis structure to describe the bonding in a single compound. Resonance structures are not isomers. Structural isomers: Same formula but different bonding, either in the kinds of bonds present or the way in which the bonds connect atoms to each other.
CHAPTER 22
ORGANIC AND BIOLOGICAL MOLECULES
1053
Geometrical isomers: Same formula and same bonds but differ in the arrangement of atoms in space about a rigid bond or ring. To distinguish isomers from molecules that differ by rotations about some bonds, name them. If two structures have different names, they are different isomers (different compounds). If the two structures have the same name, then they are the same compound. The two compounds may look different, but if they have the same names, they are the same compounds that only differ by some rotations about single bonds in the molecule. An alkene and a cyclic alkane having the C4H8 formula are:
For cis-trans isomerism (geometric isomerism), you must have at least two carbons with restricted rotation (double bond or ring) that each have two different groups bonded to it. The cis isomer will generally have the largest groups bonded to the two carbons with restricted rotation on the same side of the double bond or ring. The trans isomer generally has the largest groups bonded to the two carbons with restricted rotation on opposite sides of the double bond or ring. For alcohols and ethers, consider the formula C3H8O. An alcohol and an ether that have this formula are:
For aldehydes and ketones, consider the formula C 4H8O. An aldehyde and a ketone that have this formula are:
Esters are structural isomers of carboxylic acids. An ester and a carboxylic acid having the formula C2H4O2 are:
carboxylic acid
ester
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CHAPTER 22
ORGANIC AND BIOLOGICAL MOLECULES
Optical isomers: The same formula and the same bonds, but the compounds are nonsuperimposable mirror images of each other. The key to identifying optical isomerism in organic compounds is to look for a tetrahedral carbon atom with four different substituents attached. When four different groups are bonded to a carbon atom, then a nonsuperimposable mirror image does exist.
7.
1−bromo−1−chloroethane
1−bromo−2−chloroethane
The carbon with the asterisk has 4 different groups bonded to it (1−Br; 2−Cl; 3−CH3; 4−H). This compound has a nonsuperimposable mirror image.
Neither of the two carbons have four different groups bonded to it. The mirror image of this molecule will be superimposable (it does not exhibit optical isomerism).
Hydrocarbons only have nonpolar C−C and C−H bonds; they are always nonpolar compounds having only London dispersion forces. The strength of the London dispersion (LD) forces is related to size (molar mass). The larger the compound, the stronger the LD forces. Because nheptane (C7H16) is a larger molecule than n-butane (C4H10), n-heptane has the stronger LD forces holding the molecule together in the liquid phase and will have a higher boiling point. Another factor affecting the strength of intermolecular forces is the shape of the molecule. The strength of LD forces also depends on the surface area contact among neighboring molecules. As branching increases, there is less surface area contact among neighboring molecules, leading to weaker LD forces and lower boiling points. All the function groups in Table 22.4 have a very electronegative atom bonded to the carbon chain in the compound. This creates bond dipoles in the molecule leading to a polar molecule which leads to additional dipole-dipole forces. Most of the functional groups have carbonoxygen polar bonds leading to a polar molecule. In halohydrocarbons, the polar bond is C−X where X is a halogen. In amines, the polar bond is C−N. Note that CF4, even though it has 4 polar C−F bonds, is nonpolar. The bond dipoles are situated about carbon, so they all cancel each other out. This type of situation is atypical in hydrocarbons. Alcohols, carboxylic acids and amines exhibit a special type of dipole force. That force is the relatively strong hydrogen-bonding interaction. These compounds have an O−H or N−H bond, which is a requirement for H−bonding. Reference the isomers of C3H8O in Review question 6. The alcohol can H−bond, the ether cannot (no O−H bonds exist in the ether). Because the alcohol can H−bond, it will have a significantly higher boiling point than the ether.
CHAPTER 22
ORGANIC AND BIOLOGICAL MOLECULES
1055
The same holds true for carboxylic acids and esters, which are structural isomers. Even though the isomers have the same formula, the bonds are arranged differently. In a carboxylic acid, there is an O−H bond, so it can H−bond. The ester does not have an O−H bond. Therefore, carboxylic acids boil at a higher temperature than same size esters.
Because these compounds are about the same size (molar mass: 44-46 g/mol), they all have about the same strength LD forces. However, three of the compounds have additional intermolecular forces, hence they boil at a higher (and different) temperature than the nonpolar CH3CH2CH3. Among the polar compounds, the two compounds which H−bond will have higher boiling points than the aldehyde. Between the two compounds which can H−bond, the carboxylic acid wins out because it has additional dipole forces from the polar C=O bond. The alcohol does not have this. The order of boiling points is: CH3CH2CH3 < CH3CHO < CH3CH2OH < HCOOH lowest boiling point highest boiling point 8.
Substitution: An atom or group is replaced by another atom or group. catalyst
e.g., H in benzene is replaced by Cl. C6H6 + Cl2 → C6H5Cl + HCl Addition: Atoms or groups are added to a molecule. e.g., Cl2 adds to ethene. CH2=CH2 + Cl2 → CH2Cl ‒ CH2Cl To react Cl2 with an alkane, ultraviolet light must be present to catalyze the reaction. To react Cl2 with benzene, a special iron catalyst is needed. Its formula is FeCl3. For both hydrocarbons, if no catalyst is present, there is no reaction. This is not the case for reacting Cl2 with alkenes or alkynes. In these two functional groups, the electrons situated above and below the carboncarbon bond are easily attacked by substances that are attracted to the negative charge of the electrons. . Hence, the bonds in alkenes and alkynes are why these are more reactive. Note that even though benzene has electrons, it does not want to disrupt the delocalized bonding. When Cl2 reacts with benzene, it is the C−H bond that changes, not the bonding. A combustion reaction just means reacting something with oxygen (O 2) gas. For organic compounds made up of C, H, and perhaps O, the assumed products are CO 2(g) and H2O(g).
1056
9.
CHAPTER 22
ORGANIC AND BIOLOGICAL MOLECULES
a. Addition polymer: A polymer that forms by adding monomer units together (usually by reacting double bonds). Teflon, polyvinyl chloride and polyethylene are examples of addition polymers. b. Condensation polymer: A polymer that forms when two monomers combine by eliminating a small molecule (usually H2O or HCl). Nylon and Dacron are examples of condensation polymers. c. Copolymer: A polymer formed from more than one type of monomer. Nylon and Dacron are copolymers. d. Homopolymer: A polymer formed from the polymerization of only one type of monomer. Polyethylene, Teflon, and polystyrene are examples of homopolymers. e. Polyester: A condensation polymer whose monomers link together by formation of the ester functional group. f.
Polyamide: A condensation polymer whose monomers link together by formation of the amide functional group. Nylon is a polyamide as are proteins in the human body.
A thermoplastic polymer can be remelted; a thermoset polymer cannot be softened once it is formed. The physical properties depend on the strengths of the intermolecular forces among adjacent polymer chains. These forces are affected by chain length and extent of branching: longer chains = stronger intermolecular forces; branched chains = weaker intermolecular forces. Crosslinking makes a polymer more rigid by bonding adjacent polymer chains together.
CHAPTER 22
ORGANIC AND BIOLOGICAL MOLECULES
1057
The regular arrangement of the methyl groups in the isotactic chains allows adjacent polymer chains to pack together very closely. This leads to stronger intermolecular forces among chains as compared to atactic polypropylene where the packing of polymer chains is not as efficient. Polyvinyl chloride contains some polar C−Cl bonds compared to only nonpolar C−H bonds in polyethylene. The stronger intermolecular forces would be found in polyvinyl chloride since there are dipole-dipole forces present in PVC that are not present in polyethylene. 10-12. These questions are meant to guide you as you read Section 22.6 of the text on Natural Polymers. The three specific natural polymers discussed are proteins, carbohydrates, and nucleic acids, all of which are essential polymers found and utilized by our bodies. Read the questions to familiarize yourself with the important terms and concepts covered in Section 22.6, and then review the Natural Polymer section to help you answer these questions
Active Learning Questions 1.
The three structural isomers of C5H12 are:
a. The isomer is 2,2-dimethylpropane and the monobromination product is 1-bromo-2,2dimethylpropane. b. The isomer is pentane (or n-pentane). The three monobromination products are 1bromopentane, 2-bromopentane, and 3-bromopentane. c. The isomer is 2-methylbutane. The four isomers are 1-bromo-2-methylbutane, 2-bromo2-methylbutane, 2-bromo-3-methylbutane, and 1-bromo-3-methylbutane. d. These isomers are all nonpolar substances having only London dispersion forces. All the isomers have the same molar mass, so we can’t use molar mass to differentiate the strength of the LD forces. However, the strength of LD forces also depends on the surface area contact among neighboring molecules. As branching increases, there is less surface area contact among neighboring molecules, leading to weaker LD forces and lower boiling points. The more elongated pentane will have the strongest London dispersion forces; it has the 36°C boiling point. 2,2-dimethylpropane has the most branching; it will have the weakest LD forces and the lowest boiling point of 9.5°C. This leaves 2-methylbutane having the 28°C boiling point.
1058
CHAPTER 22
2.
ORGANIC AND BIOLOGICAL MOLECULES
a. CH2
.
Pt
CH2 + H2
H
H
CH2
CH2
ethene
b.
CH3
CH
ethane
CH
CH3
+
HCl
CH3
Cl
H
CH
CH
2-butene (cis or trans)
2-chlorobutane
2-methyl-1-propene (or 2-methylpropene)
1,2-dichloro-2-methylpropane
c.
d.
ethyne
1,1,2,2-tetrabromoethane
Another possibility would be:
H
C
C
Br
Br
1,2-dibromoethene
H
+
Br2
Br
Br
HC
CH
Br
Br
1,1,2,2-tetrabromoethane
e.
benzene
chlorobenzene
CH3
CHAPTER 22 f.
ORGANIC AND BIOLOGICAL MOLECULES Cr2O3 CH3CH3 ethane
500oC
CH2
CH2 + H2
ethene
1059
1060
CHAPTER 22
ORGANIC AND BIOLOGICAL MOLECULES
Questions 3.
Carbon has the unusual ability to form bonds to itself, whether they be single, double, or triple bonds, as well as the ability to form long chains or rings of carbon atoms. Carbon also forms strong bonds to other nonmetals, such as hydrogen, oxygen, phosphorus, sulfur, and the halogens. Because of this bonding ability, carbon can form millions of compounds, including biomolecules which are the basis for the existence of life.
4.
Cyclopropane (C3H6) has, in theory, 60o bond angles, cyclobutane (C4H8) has, in theory, 90o bond angles, methane (CH4, and any alkane) exhibits 109.5o bond angles, ethene (C2H4) exhibits 120o bond angles, and ethyne (C2H2) exhibits 180o bond angles. The structures are below.
5.
a. 1-sec-butylpropane
b. 4-methylhexane CH3 CH3CH2CH2CHCH2CH3
3-methylhexane is correct. c. 2-ethylpentane
3-methylhexane is correct. d. 1-ethyl-1-methylbutane
CH2CH3 CH3CHCH2CH2CH3
CHCH2CH2CH3 CH2CH3
3-methylhexane is correct.
CH3 3-methylhexane is correct.
CHAPTER 22
ORGANIC AND BIOLOGICAL MOLECULES
e. 3-methylhexane
f.
CH3CH2CHCH2CH2CH3
1061
4-ethylpentane CH3CH2CH2CHCH3
CH3
CH2CH3
3-methylhexane is correct. All six of these compounds are the same. They only differ from each other by rotations about one or more carbon-carbon single bonds. Only one isomer of C7H16 is present in all these names, 3-methylhexane. 6.
a. 2-Chloro-2-butyne would have 5 bonds to the second carbon. Carbon never expands its octet. Cl CH3
C
CCH3
b. 2-Methyl-2-propanone would have 5 bonds to the second carbon. O CH3
C
CH3
CH3
c. Carbon-1 in 1,1-dimethylbenzene would have 5 bonds. CH3
CH3
d. You cannot have an aldehyde functional group off a middle carbon in a chain. Aldehyde groups: O C
H
can only be at the beginning and/or the end of a chain of carbon atoms. e. You cannot have a carboxylic acid group off a middle carbon in a chain. Carboxylic groups: O C
OH
must be at the beginning and/or the end of a chain of carbon atoms.
1062
CHAPTER 22 f.
ORGANIC AND BIOLOGICAL MOLECULES
In cyclobutanol, the 1 and 5 positions refer to the same carbon atom. 5,5-Dibromo-1cyclobutanol would have five bonds to carbon-1. This is impossible; carbon never expands its octet. Br OH Br
7.
Hydrocarbons are nonpolar substances exhibiting only London dispersion forces. Size and shape are the two most important structural features relating to the strength of London dispersion forces. For size, the bigger the molecule (the larger the molar mass), the stronger are the London dispersion forces, and the higher is the boiling point. For shape, the more branching present in a compound, the weaker are the London dispersion forces, and the lower is the boiling point.
8.
To hydrogen-bond, the compound must have at least one N−H, O−H or H−F covalent bond in the compound. In Table 22.4, alcohols and carboxylic acids have an O−H covalent bond, so they can hydrogen-bond. In addition, primary and secondary amines have at least one N−H covalent bond, so they can hydrogen-bond. H
F C
H
C F
CH2CF2 cannot form hydrogen bonds because it has no hydrogens covalently bonded to the fluorine atoms. 9.
Compounds with ring structures or compounds with double and/or triple bonds have restricted rotation about some of the bonds in the compounds. Only alkanes, compounds composed of C-C and C-H single bonds, have free rotation about all bonds in the compound. Ethane (answer b) is the only compound listed that has free rotation about every bond.
10.
Alkenes (answer b) and alkynes (answer c) undergo addition reactions. Alkanes and aromatic compounds undergo substitution reactions; they do not undergo addition reactions.
11.
The general formula for cycloalkanes is C nH2n and the general formula for alkanes is C nH2n+2. Since the general formulas for alkanes and cycloalkanes are different, they are not structural isomers of each other. Answer b is the false statement. All the other statements are true. Cyclobutane is forced into 90° bond angles and is reactive because of this. Alkanes are composed of only single bonded atoms; atoms in these single bonds can rotate. All carbon atoms in alkanes exhibit tetrahedral bond angles, so all carbon atoms in alkanes are sp3 hybridized. With the restricted rotation of the ring atoms is cycloalkanes, they can exhibit cis/trans isomerism.
12.
Cyclohexane exhibits 109.5° bond angles. Because of the tetrahedral bond angles, cyclohexane is not a planar molecule (answer c is the false statement). All the other statements are true. The structural formula for 2-hexene is CH3CH=CHCH2CH2CH3. Both 2-hexene and cyclohexane have C6H12 formulas along with different bonds, so they are structural isomers.
CHAPTER 22
13.
ORGANIC AND BIOLOGICAL MOLECULES
The amide functional group is:
O
H
C
N
1063
When the amine end of one amino acid reacts with the carboxylic acid end of another amino acid, the two amino acids link together by forming an amide functional group. A polypeptide has many amino acids linked together, with each linkage made by the formation of an amide functional group. Because all linkages result in the presence of the amide functional group, the resulting polymer is called a polyamide. For nylon, the monomers also link together by forming the amide functional group (the amine end of one monomer reacts with the carboxylic acid end of another monomer to give the amide functional group linkage). Hence nylon is also a polyamide. The correct order of strength is: H
H
H
H
C
C
C
C
O
<
C
O R
C
O
R'
O
n H
H
H
n
polyester
H
O
polyhydrocarbon weakest fibers
<
C
R
H
O
N
C
H
R
N
n
polyamide strongest fibers
The difference in strength is related to the types of intermolecular forces present. All these types of polymers have London dispersion forces. However, the polar ester group in polyesters and the polar amide group in polyamides give rise to additional dipole forces. The polyamide has the ability to form relatively strong hydrogen-bonding interactions, hence why it would form the strongest fibers. 14.
Polystyrene is an addition polymer formed from the monomer styrene.
n CH2
CH
CH2CHCH2CH n
a. Syndiotactic polystyrene has all of the benzene ring side groups aligned on alternate sides of the chain. This ordered alignment of the side groups allows individual polymer chains of polystyrene to pack together efficiently, maximizing the London dispersion forces. Stronger London dispersion forces translate into stronger polymers. b. By copolymerizing with butadiene, double bonds exist in the carbon backbone of the polymer. These double bonds can react with sulfur to form crosslinks (bonds) between individual polymer chains. The crosslinked polymer is stronger.
1064
CHAPTER 22
ORGANIC AND BIOLOGICAL MOLECULES
c. The longer the chain of polystyrene, the stronger are the London dispersion forces between polymer chains. d. In linear (versus branched) polystyrene, chains pack together more efficiently, resulting in stronger London dispersion forces. 15.
a. A polyester forms when an alcohol functional group reacts with a carboxylic acid functional group. The monomer for a homopolymer polyester must have an alcohol functional group and a carboxylic acid functional group present within the structure of the monomer. b. A polyamide forms when an amine functional group reacts with a carboxylic acid functional group. For a copolymer polyamide, one monomer would have at least two amine functional groups present, and the other monomer would have at least two carboxylic acid functional groups present. For polymerization to occur, each monomer must have two reactive functional groups present. c. To form an addition polymer, a carbon-carbon double bond must be present. Polyesters and polyamides are condensation polymers. To form a polyester, the monomer would need the alcohol and carboxylic acid functional groups present. To form a polyamide, the monomer would need the amine and carboxylic acid functional groups present. The two possibilities are for the monomer to have a carbon-carbon double bond, an alcohol functional group, and a carboxylic acid functional group present or to have a carbon-carbon double bond, an amine functional group, and a carboxylic acid functional group present.
16.
Proteins are polymers made up of monomer units called amino acids. One of the functions of proteins is to provide structural integrity and strength for many types of tissues. In addition, proteins transport and store oxygen and nutrients, catalyze many reactions in the body, fight invasion by foreign objects, participate in the body’s many regulatory systems, and transport electrons in the process of metabolizing nutrients. Carbohydrate polymers, such as starch and cellulose, are composed of the monomer units called monosaccharides or simple sugars. Carbohydrates serve as a food source for most organisms. Nucleic acids are polymers made up of monomer units called nucleotides. Nucleic acids store and transmit genetic information and are also responsible for the synthesis of various proteins needed by a cell to carry out its life functions.
17.
Denaturation is the breakdown of the three-dimensional structure of the protein. Both the secondary and the tertiary structures of the protein are disrupted in denaturation. The primary structure (the order in which the amino acids link to each other) is not affected by denaturation.
18.
Starch is a polymer of α-D-glucose, which contain many –OH groups off the polymer chain. Hydrogen bonding will be the primary intermolecular force between water and starch.
CHAPTER 22
ORGANIC AND BIOLOGICAL MOLECULES
1065
Exercises Hydrocarbons 19.
i. CH3
CH2
CH2
CH2
CH2
CH3
ii. CH3 CH3
CH
CH2
CH2
CH3
iii.
iv.
v.
CH 3
CH 3
CH 3
CH
CH
CH 3
All other possibilities are identical to one of these five compounds. 20.
See Exercise 19 for the structures. The names of structures i-v respectively, are hexane (or nhexane), 2-methylpentane, 3-methylpentane, 2,2-dimethylbutane, and 2,3-dimethylbutane.
21.
A difficult task in this problem is recognizing different compounds from compounds that differ by rotations about one or more C‒C bonds (called conformations). The best way to distinguish different compounds from conformations is to name them. Different name = different compound; same name = same compound, so it is not an isomer but instead is a conformation. a. CH3 CH3CHCH2CH2CH2CH2CH3 2-methylheptane
CH3 CH3CH2CHCH2CH2CH2CH3 3-methylheptane
CH3 CH3CH2CH2CHCH2CH2CH3 4-methylheptane
1066
CHAPTER 22
ORGANIC AND BIOLOGICAL MOLECULES
b. CH3 CH3 CH3
C
C
CH3
CH3 CH3 2,2,3,3-tetramethylbutane
22.
a. CH3
CH3
CH3CCH2CH2CH2CH3 CH3
CH3CHCHCH2CH2CH3 CH3
2,2-dimethylhexane
CH3
2,3-dimethylhexane
CH3
CH3CHCH2CHCH2CH3
CH3CHCH2CH2CHCH3
CH3
CH3
2,4-dimethylhexane
2,5-dimethylhexane
CH3
CH3
CH3CH2CCH2CH2CH3
CH3CH2CHCHCH2CH3
CH3 3,3-dimethylhexane CH2CH3 CH3CH2CHCH2CH2CH3 3-ethylhexane
CH3 3,4-dimethylhexane
CHAPTER 22
ORGANIC AND BIOLOGICAL MOLECULES
1067
b. H3C
CH3
C
CH
CH3
CH3 CH2
CH3
CH3
CH3
C
CH3 CH2
CH
CH3
CH3
2,2,3-trimethylpentane
2,2,4-trimethylpentane
CH3 CH3
CH3 CH3 CH3
CH3
CH
C
CH2
CH3
CH3
CH
CH
CH
CH3
CH3 2,3,3-trimethylpentane
2,3,4-trimethylpentane
CH3 CH2CH3 CH3 CH
CH
CH2
CH2CH3 CH3
CH3
CH2
C
CH2
CH3
CH3 3-ethyl-2-methylpentane
3-ethyl-3-methylpentane CH3
CH3
23.
a.
CH3CHCH3
b.
CH3
CH3
c.
d. CH3CHCH2CH2CH3
CH3CHCH2CH2CH2CH3
CH3
CH3
24.
CH3CHCH2CH3
a.
b. CH3CCH2CH2CH2CH2CH3
CH3CHCHCH2CH2CH2CH3 CH3
CH3
CH3
CH3
c.
d. CH3CH2CCH2CH2CH2CH3 CH3
CH3CHCH2CHCH2CH2CH3 CH3
1068 25.
CHAPTER 22
ORGANIC AND BIOLOGICAL MOLECULES
The structures of the compounds are:
The correct name for the first structure is 3,5-dimethyloctane, and the correct name for the second structure is 4-ethyl-2-methylheptane. 26.
The structures of the compounds are:
The correct name for the first structure is 2,2,3,5-tetramethylheptane and the correct name for the second structure is 3,4-dibromo-2,3,4,5-tetramethylhexane. 27.
a. 2,2,4-trimethylhexane
b. 5-methylnonane
c. 2,2,4,4-tetramethylpentane
d. 3-ethyl-3-methyloctane Note: For alkanes, always identify the longest carbon chain for the base name first, then number the carbons to give the lowest overall numbers for the substituent groups. 28.
a. pentane or n-pentane
b. 3-ethyl-2,5-dimethylhexane
c. 4-ethyl-5-isopropyloctane 29.
The hydrogen atoms in ring compounds are commonly omitted. In organic compounds, carbon atoms satisfy the octet rule of electrons by forming four bonds to other atoms. Therefore, add C-H bonds to the carbon atoms in the ring in order to give each C atom four bonds. You can also determine the formula of these cycloalkanes by using the general formula C nH2n. a. isopropylcyclobutane; C7H14 b. 1-tert-butyl-3-methylcyclopentane; C10H20 c. 1,3-dimethyl-2-propylcyclohexane; C11H22
CHAPTER 22 30.
ORGANIC AND BIOLOGICAL MOLECULES
a. iodocyclopropane
1069
b. 2-chloro-1,3,5-trimethylcyclohexane
c. 1-chloro-2-ethylcyclopentane 31. CH3
CH2
CH2
CH3
H
H
H
C
C
H
H
C
C
H
H
H
Each carbon is bonded to four other carbon and/or hydrogen atoms in a saturated hydrocarbon (only single bonds are present). 32.
An unsaturated hydrocarbon has at least one carbon-carbon double and/or triple bond in the structure. 33.
a. 1-butene
b. 4-methyl-2-hexene
c. 2,5-dimethyl-3-heptene
Note: The multiple bond is assigned the lowest number possible. 34.
35.
a.
2,3-dimethyl-2-butene
b. 4-methyl-2-hexyne
c.
2,3-dimethyl-1-pentene
a.
CH3‒CH2‒CH=CH‒CH2‒CH3
b. CH3‒CH=CH‒CH=CH‒CH2CH3
c. CH3 CH3
CH
CH
CH
CH2CH2CH2CH3
36. CH 3 a.
HC
C
CH 2
CH
CH 3 CH 3 CH 3
b.
H 2C
C
C CH 3
CH 2CH 3 c.
CH 3CH 2
CH
CH
CH
CH 2CH 2CH 2CH 2CH 3
CH 2CH 2CH 3
1070 37.
CHAPTER 22 a.
ORGANIC AND BIOLOGICAL MOLECULES b.
CH 3 CH 2CH 3 H 3C
c.
CH 3
CH 3
C
C
CH 3
CH 3
CH 3
d. CH 2CH 3
CH 2
CH 2CH 3
CH
CH
CH 3
38.
isopropylbenzene or 2-phenylpropane
39.
a.
1,3-dichlorobutane
b.
1,1,1-trichlorobutane
c.
2,3-dichloro-2,4-dimethylhexane
d.
1,2-difluoroethane
40.
a. 2,2-dibromo-3-methylpentane
b. 4-ethyl-3-iodo-2-methylhexane
c. 4-fluoro-2-methylnonane 41.
a. 3-chloro-l-butene
b. 1-ethyl-3-methycyclopentene
c. 3-chloro-4-propylcyclopentene
d. 1,2,4-trimethylcyclohexane
e. 2-bromotoluene (or o-bromotoluene or 1-bromo-2-methylbenzene) f.
1-bromo-2-methylcyclohexane
g. 4-bromo-3-methylcyclohexene
Note: If the location of the double bond is not given in the name, it is assumed to be located between C1 and C2. Also, when the base name can be numbered in equivalent ways, give the first substituent group the lowest number; e.g., for part f, 1-bromo-2-methylcyclohexane is preferred to 2-bromo-1-methycyclohexane. 42.
a. 2-methyl-1-butene
b. 2,4-dimethyl-1,4-pentadiene
c. 6-ethyl-2-methyl-4-octene
d. 3-bromo-1-heptene
e. 7-chloro-2,5,5-trimethyl-3-heptyne
f.
4-ethyl-3-methyl-1-octyne
CHAPTER 22
ORGANIC AND BIOLOGICAL MOLECULES
1071
Isomerism 43.
CH2Cl‒CH2Cl, 1,2-dichloroethane: In this compound, there is free rotation about the C‒C single bond that doesn't lead to different compounds. CHCl=CHCl, 1,2-dichloroethene: This compound, however, has no free rotation about the C=C double bond. This creates the cis and trans isomers, which are different compounds.
44.
a. All of these structures have the formula C5H8. The compounds with the same physical properties will be the compounds that are identical to each other, i.e., compounds that only differ by rotations of C−C single bonds. To recognize identical compounds, name them. The names of the compounds are: i.
trans-1,3-pentadiene
iii. cis-1,3-pentadiene
ii. cis-1,3-pentadiene iv. 2-methyl-1,3-butadiene
Compounds ii and iii are identical compounds, so they would have the same physical properties. b. Compound i is a trans isomer because the bulkiest groups off the C3=C4 double bond are on opposite sides of the double bond. c. Compound iv does not have carbon atoms in a double bond that each have two different groups attached. Compound iv does not exhibit cis-trans isomerism. 45.
All the compounds have a C6H12 formula. The cis isomer has the bulkiest groups bonded to the carbons of the double on the same side of the double bond. The names of the various substances are: I. trans-3-methyl-2-pentene; II. 2-methyl-1-pentene; III. trans-2-hexene; iv. cis-2-hexene Only statement d is false. Structures I and IV are structural isomers of each other, by they are not geometric isomers of each other. Geometric isomers have the same bonds, but different spatial arrangement of some groups in the compound. Geometric isomers will always have the same base name. For example, structures III and IV have the same base name, but they have a different spatial arrangement of the groups bonded to the carbons with the double bond. Structures III and IV are geometric isomers of each other
46.
All the compounds have a C5H9F formula. The trans isomer has the bulkiest groups bonded to the carbons of the double on opposite sides of the double bond. The names of the various substances are: I. trans-1-fluoro-2-methyl-2-butene; II. cis-1-fluoro-2-methyl-2-butene III. cis-4-fluoro- 2-pentene; IV. 2-fluoro-3-methyl-2-butene; V. trans-1-fluoro-2-methyl-2-butene Statement e is false. Compounds I and V are the same compound. They only differ by a rotation.
1072 47.
CHAPTER 22
ORGANIC AND BIOLOGICAL MOLECULES
To exhibit cis-trans isomerism, each carbon in the double bond must have two structurally different groups bonded to it. In Exercise 33, this occurs for compounds b and c. The cis isomer has the bulkiest groups on the same side of the double bond while the trans isomer has the bulkiest groups on opposite sides of the double bond. The cis and trans isomers for 33b and 33c are: 33b. CH2CH3 H
H C
H
C
H3C
CHCH3 C
CHCH3
C
H3C
H
CH2CH3 cis
trans
33c. CH 3 H
H C
H
C
CH 3CH 2CH
C CH CH 3
CH 3
CH CH 3 C
CH 3CH 2CH
CH 3
H
CH 3
cis
trans
Similarly, all the compounds in Exercise 35 exhibit cis-trans isomerism. In compound a of Exercise 33, the first carbon in the double bond does not contain two different groups. The first carbon in the double bond contains two H atoms. To illustrate that this compound does not exhibit cis-trans isomerism, let’s look at the potential cis-trans isomers. H
H C
H
H
C
CH 2CH 3 C
CH 2CH 3
H
C H
These are the same compounds; they only differ by a simple rotation of the molecule. Therefore, they are not isomers of each other but instead are the same compound. 48.
In Exercise 34, none of the compounds can exhibit cis-trans isomerism since none of the carbons with the multiple bond have two different groups bonded to each. In Exercise 36, only 3-ethyl-4-decene can exhibit cis-trans isomerism since C4 and C5 in this compound each have two different groups bonded to them.
CHAPTER 22 49.
ORGANIC AND BIOLOGICAL MOLECULES
1073
To help distinguish the different isomers, we will name them. Cl
CH3 C
H
CH3
C
H
C H
C
Cl
cis-1-chloro-1-propene
H
trans-1-chloro-1-propene Cl
CH2
C
CH3
CH2
CH
CH2
Cl
Cl
2-chloro-1-propene
50.
3-chloro-1-propene
chlorocyclopropane
HCBrCl‒CH=CH2 H 3C
Cl C
C
Br
H
Cl
C H
Cl
Cl C
H3C
C
H
C
Br
Br C
H C
H3C
Cl
H 3C
C
H 3C
Br
H
H
Br
H3C
C
Br C
C Cl
1074
CHAPTER 22
ORGANIC AND BIOLOGICAL MOLECULES
The cyclic isomers of bromochloropropene (C3H4BrCl) are: Br
Cl
Cl
Br
Br
Cl
trans
cis
51. F
CH2CH3 C
H
C
H
C H
CH2
C
F
F CH2
CHCHCH3
F
CH3
H3C
H3C
H
F
F
CH3
CH
CCH3
C
C
CCH2CH3
H
F CH2
F
CH2CH3
H2CF C H
CHCH2CH2
CH3 C
CH3 C H
H
C
CH3 C
H
H2CF
C H
CH3 CH2
CCH2 F
52.
The cis isomer has the CH3 groups on the same side of the ring. The trans isomer has the CH3 groups on opposite sides of the ring. CH 3
H
H
H H
CH 3
CH 3
CH 3 trans
cis
The cyclic structural and geometric isomers of C 4H7F are: F
F
CH2F
CH3 CH3
F
CH3 cis
F trans
CHAPTER 22 53.
a.
ORGANIC AND BIOLOGICAL MOLECULES
H 3C C
H 3C
H C
C
H
54.
b.
CH 2CH 2CH 3
a. cis-1-bromo-1-propene
H 3C
c.
C
H
H
1075 CH 2CH 3 C
CH 3
Cl
C Cl
b. cis-4-ethyl-3-methyl-3-heptene
c. trans-1,4-diiodo-2-propyl-1-pentene Note: In general, cis-trans designations refer to the relative positions of the largest groups. In compound b, the largest group off the first carbon in the double bond is CH2CH3, and the largest group off the second carbon in the double bond is CH2CH2CH3. Because their relative placement is on the same side of the double bond, this is the cis isomer. 55.
There are many possible structural isomers for the formula C 6H12. Two structural isomers are: H
H
H
H
H H
H H
H
CH2=CHCH2CH2CH2CH3 1-hexene
H H
H
cyclohexane The structural isomer 2-hexene (plus others) exhibits geometric isomerism. H3C
H
CH2CH2CH3 C
C
C
H
CH2CH2CH3
H3C
H
cis
C H
trans
The structural isomer 3-methyl-1-pentene exhibits optical isomerism (the asterisk marks the chiral carbon). CH3
CH2
CH
C
*
CH2CH3
H
For this compound, the chiral carbon has four different groups bonded to it; the mirror image of this compound will be non-superimposable. Optical isomerism is also possible with some of the cyclobutane and cyclopropane structural isomers.
1076
CHAPTER 22
ORGANIC AND BIOLOGICAL MOLECULES
56.
Two of the many possible structural isomers of C5H11F are:
1-fluoropentane
2-fluoropentane
Two of the optically active isomers having a C 5H11F formula are:
2-fluoro-3-methylbutane
2-fluoropentane
The asterisks in each compound mark the chiral carbon (a carbon with four different groups bonded to it). For each of these compounds, non-superimposable mirror images can be drawn. No isomers of C5H12O exhibit geometric isomerism because no double bonds or ring structures are possible with 12 total hydrogen and fluorine atoms bonded to the 5 carbons. 57.
C5H10 has the general formula for alkenes and cycloalkanes, C nH2n. The alkene isomers of C5H10 are: CH 2
CCH 2CH 3 CH 3
CH CH 3
CH 3
2-methyl-1-butene CH 3CH CH
CH 3C
2-methyl-2-butene
CH 2
CH 3 3-methyl-1-butene CH 2
CH CH 2CH 2CH 3
1-pentene
The cycloalkane isomers of C5H10 are:
CH 3CH
CH CH 2CH 3 2-pentene
CHAPTER 22
ORGANIC AND BIOLOGICAL MOLECULES
1077
a. 2-pentene exhibits cis/trans isomerism. b. 1-pentene, 2-methyl-1-butene, 2-methyl-2-butene, and 3-methyl-1-butene do not exhibit cis/trans isomerism. c. 1,2-dimethylcyclopropane exhibits cis/trans isomerism. d. 1,1-dimethylcyclopropane, methylcyclobutane, and cyclopentane do not exhibit cis/trans isomerism. 58.
a. 3-iodo-1-butene is optically active at the third carbon.
b. 1-iodo-1-butene, 1-iodo-2-butene, and 2-iodo-2-butene all exhibit is/trans isomerism.
c. 1-iodo-2-methylcyclopropane exhibits cis/trans isomerism.
59.
a.
*
CH *
CH2
3
CH *
CH2
2
CH3
There are three different types of hydrogens in npentane (see asterisks). Thus there are three monochloro isomers of n-pentane (1-chloropentane, 2-chloropentane, and 3-chloropentane).
b. CH3 * CH CH2*
CH3*
CH3*
There are four different types of hydrogens in 2methylbutane, so four monochloro isomers of 2methylbutane are possible.
c. CH3 * CH CH2*
CH * 3
d.
CH3 CH
CH3
There are three different types of hydrogens, so three monochloro isomers are possible.
CH3* *
H2C
C
*HC
CH2
2
H*
There are four different types of hydrogens, so four monochloro isomers are possible.
1078 60.
CHAPTER 22 a.
ORGANIC AND BIOLOGICAL MOLECULES
Cl
Cl
Cl
Cl
Cl Cl ortho
meta
para
b. There are three trichlorobenzenes (1,2,3-trichlorobenzene, 1,2,4-trichlorobenzene, and 1,3,5-trichlorobenzene). c. The meta isomer will be very difficult to synthesize. d. 1,3,5-Trichlorobenzene will be the most difficult to synthesize since all Cl groups are meta to each other in this compound.
Functional Groups 61.
62.
Reference Table 22.4 for the common functional groups. a. ketone
b. aldehyde
c. carboxylic acid
d. amine
a.
b. alcohol OH CH3 CH3
O ketone
CHAPTER 22
ORGANIC AND BIOLOGICAL MOLECULES
1079
c. amide O H2N amine
ester
O C
CH
C
NH
CH2
C
OH
OCH3
CHCH2
carboxylic acid
O
R
Note: The amide functional group
O
R'
C
N
R"
is not covered in Section 22.4
of the text. We point it out for your information. 63.
a. H
H C
O ketone
C
C
N C
H
H
C
O alcohol
H
H
N amine
C
C
C
H
H
O
O H carboxylic acid
amine
H
b. 5 carbons in the ring and the carbon in ‒CO2H: sp2; the other two carbons: sp3 c. 24 sigma bonds; 4 pi bonds 64.
Hydrogen atoms are usually omitted from ring structures. In organic compounds, the carbon atoms form four bonds. With this in mind, the following structure has the missing hydrogen atoms included in order to give each carbon atom the four bond requirement. H
H
H H
H
a b
H
H N
N
N
H H
H
H
d
H
c N
H
e O
H
N H
1080
CHAPTER 22
ORGANIC AND BIOLOGICAL MOLECULES
a. Minoxidil would be more soluble in acidic solution. The nitrogen atoms with lone pairs can be protonated, forming a water soluble cation. b. The two nitrogen atoms in the ring with double bonds are sp2 hybridized. The other three N's are sp3 hybridized. c. The five carbon atoms in the ring with one nitrogen are all sp 3 hybridized. The four carbon atoms in the other ring with double bonds are all sp 2 hybridized. d. Angles a and b 109.5°; angles c, d, and e 120° e. 31 sigma bonds f. 65.
3 pi bonds
a. 3-chloro-1-butanol; because the carbon containing the OH group is bonded to just 1 other carbon (1 R group), this is a primary alcohol. b. 3-methyl-3-hexanol; because the carbon containing the OH group is bonded to three other carbons (3 R groups), this is a tertiary alcohol. c. 2-methylcyclopentanol; secondary alcohol (2 R groups bonded to carbon containing the OH group). Note: In ring compounds, the alcohol group is assumed to be bonded to C 1, so the number designation is commonly omitted for the alcohol group.
66.
OH a.
CH2
CH2
CH2
CH3
primary alcohol
OH b.
c.
CH3
CH
OH
CH3
CH2
CH
CH2 CH3
secondary alcohol
CH2
primary alcohol
CH3
CH3 d.
CH3
C OH
CH2
CH3
tertiary alcohol
CHAPTER 22
ORGANIC AND BIOLOGICAL MOLECULES
1081
67. OH HO
CH 2
CH 2
CH 2
CH 3
CH 3
1-butanol
CH
CH 2
CH 3
2-butanol CH 3
HO
CH 2
CH
CH 3 CH 3
CH 3
C
CH 3
OH 2-methyl-1-propanol
2-methyl-2-propanol
There are three possible ethers with the formula C 4H10O. They are: CH 3 CH 3CH 2
O
CH 2CH 3
CH 3
O
CH 2CH 2CH 3
CH 3
O
CH CH 3
diethyl ether
68.
methylpropyl ether OH
OH
isopropylmethyl ether OH
CH3CH2CH2CH2CH2
CH3CH2CH2CHCH3
CH3CH2CHCH2CH3
1-pentanol
2-pentanol
3-pentanol
OH
OH
CH3CH2CHCH2
CH3CHCH2CH2
CH3 2-methyl-1-butanol
CH3
CH3CHCHCH3 CH3 3-methyl-2-butanol
CH3
CH3CH2CCH3 CH3
3-methyl-1-butanol
OH
OH
CH3
OH
C
CH2
2-methyl-2-butanol
CH3 2,2-dimethyl-1-propanol
There are six isomeric ethers with formula C5H12O. The structures follow:
1082
CHAPTER 22
ORGANIC AND BIOLOGICAL MOLECULES CH3
CH3
O
CH2CH2CH2CH3
CH3
O
CH3
CHCH2CH3
CH3
O
CH2CHCH3
CH3 CH3
O
C
CH3 CH3
CH3CH2
O
CH2CH2CH3
CH3CH2
O
CH
CH3
69.
CH3
Two possible aldehydes: O H
O
C
CH 2
CH 2
CH 3
H
C
CH
CH 3
CH 3 butanal
2-methylpropanal
One possible ketone:
O CH 3
C
CH 2
CH 3
2-butanone 70.
There are four aldehydes and three ketones with formula C 5H10O. The structures follow: O
O
O
CH3CH2CH2CH2CH
CH3CH2CHCH
CH3CHCH2CH
CH3 pentanal
2-methylbutanal
CH3 O CH3
C
C
O H
CH3CH2CH2CCH3
CH3 2,2-dimethylpropanal
2-pentanone
O
O
CH3CH2CCH2CH3
CH3CHCCH3 CH3
3-pentanone
3-methyl-2-butanone
CH3 3-methylbutanal
CHAPTER 22 71.
ORGANIC AND BIOLOGICAL MOLECULES
a. 4,5-dichloro-3-hexanone
1083
b. 2,3-dimethylpentanal
c. 3-methylbenzaldehyde or m-methylbenzaldehyde 72.
a.
b. O H
C
O H
CH3CH2CH2CCH2CH2CH3
c.
d. O
O H
CH3
CH3 CCH2CH2CCH3
CCH2CHCH3
CH3
Cl
73.
a. 4-chlorobenzoic acid or p-chlorobenzoic acid b. 3-ethyl-2-methylhexanoic acid c. methanoic acid (common name = formic acid)
74.
a.
b. CH3
O
O
CH3CH2CHCH2C
CH3CH2
OH
c.
CH
d. O C
75.
O
CH3
OCH3
CH3CH2CH
CH3 CH
CHC
Cl
O
OH
a. 1-pentanol; 2-methyl-2,4-pentadiol b. trans-1,2-cyclohexadiol; 4-methyl-1-penten-3-ol (does not exhibit cis-trans isomerism)
76.
a. heptanal
b. 3-ethyl-5-methylhexanal
d. 4,5-dimethyl-3-hexanone 77.
c. 3-hexanone
e. 3,4-dimethylhexanoic acid
Only statement d is false. The other statements refer to compounds having the same formula but different attachment of atoms; they are structural isomers.
1084
CHAPTER 22 a.
ORGANIC AND BIOLOGICAL MOLECULES
O
Both have a formula of C5H10O2.
CH3CH2CH2CH2COH
b.
O CH3CHCCH2CH3
Both have a formula of C6H12O.
CH3
c. OH
Both have a formula of C5H12O.
CH3CH2CH2CHCH3 O
d. HCCH
2-Butenal has a formula of C4H6O while the alcohol has a formula of C4H8O.
CHCH3
e. CH3NCH3
Both have a formula of C3H9N. CH3
78.
a.
CH3
trans-2-butene:
H C
H H
H
H or
H
H H
CH3
H
H H
, formula = C4H8
C
H
H
CH3
H
H
O b.
propanoic acid:
CH3CH2C
O
OH, formula = C3H6O2
O
CH3C
O
CH3
or
HC
O
CH2CH3
O c.
butanal:
CH3CH2CH2CH,
O CH3CH2CCH3
formula = C4H8O
CHAPTER 22
d.
ORGANIC AND BIOLOGICAL MOLECULES
butylamine: CH3CH2CH2CH2NH2,
1085
formula = C4H11N:
A secondary amine has two R groups bonded to N. CH3
N
CH3
H
H
CH3CH2
CH3CHCH3
CH2CH2CH3 e.
N
N
H
CH2CH3
A tertiary amine has three R groups bonded to N. (See answer d for structure of butylamine.) CH3
N
CH3
CH2CH3 CH3 f.
2-methyl-2-propanol: CH3CCH3 , formula = C4H10O OH CH3
O
CH2CH2CH3
CH3
CH3 O
CH
CH3CH2
O
CH2CH3
CH3
g.
A secondary alcohol has two R groups attached to the carbon bonded to the OH group. (See answer f for the structure of 2-methyl-2-propanol.) OH CH3CHCH2CH3
Reactions of Organic Compounds 79.
The two monochlorination products are 1-chloro-2-methylpentane and 2-chloro-2methylpentane. All the other possible products differ from one of these by a rotation.
80.
The three monochlorination products are 1-chloro-2,4-dimethylpentane, 2-chloro-2,4dimethylpentane, and 3-chloro-2,4-dimethylpentane.
81.
H a.
c.
CH3CH
H CHCH3
Cl + HCl
b.
Cl
Cl
Cl
Cl
CH2
CHCHCH
CH
CH3
CH3
1086 82.
CHAPTER 22
ORGANIC AND BIOLOGICAL MOLECULES
a. The two possible products for the addition of HOH to this alkene are: OH
H
OH
CH3CH2CH
CH2
H
CH3CH2CH
CH2
major product
minor product
We would get both products in this reaction. Using the rule given in the problem, the first compound listed is the major product. In the reactant, the terminal carbon has more hydrogens bonded to it (2 versus 1), so H forms a bond to this carbon, and OH forms a bond to the other carbon in the double bond for the major product. We will list only the major product for the remaining parts to this problem. b.
c. Br
Br
H
CH3CH2CH
CH3CH2C
CH2
Br
c.
CH H
e. CH3 OH H
83.
H
CH3CH2
Cl
H
C
C
CH3
CH3 H
The reaction is:
84.
The major product has the Cl add to the carbon with the fewest H atoms attached, and the H goes to the carbon with the most H atoms attached.
CHAPTER 22
ORGANIC AND BIOLOGICAL MOLECULES
1087
85. CH3
CH3 H
H
H 3+ catalyst
Fe
+ 2 Cl 2 H
H
H
H
H
H
Cl
ortho
para
Cl
CH 2
H
H
H
H + 2 HCl
H
CH 3
+ Cl 2
H +
H
H
CH3 Cl
H
light
+ HCl H
H
H H
H
To substitute for the benzene ring hydrogens, an iron(III) catalyst must be present. Without this special iron catalyst, the benzene ring hydrogens are unreactive. To substitute for an alkane hydrogen, light must be present. For toluene, the light-catalyzed reaction substitutes a chlorine for a hydrogen in the methyl group attached to the benzene ring. 86.
When CH2=CH2 reacts with HCl, there is only one possible product, chloroethane. When Cl2 is reacted with CH3CH3 (in the presence of light), there are six possible products because any number of the six hydrogens in ethane can be substituted for by Cl. The light-catalyzed substitution reaction is very difficult to control; hence it is not a very efficient method of producing monochlorinated alkanes.
87.
Primary alcohols (a, d, and f) are oxidized to aldehydes, which can be oxidized further to carboxylic acids. Secondary alcohols (b, e, and f) are oxidized to ketones, and tertiary alcohols (c and f) do not undergo this type of oxidation reaction. Note that compound f contains a primary, secondary, and tertiary alcohol. For the primary alcohols (a, d, and f), we listed both the aldehyde and the carboxylic acid as possible products. O a.
H
O CH2CHCH3 +
C
HO
C
CH3
CH2CHCH3 CH3
O b.
CH3
C
CHCH3 CH3
c.
No reaction
1088
CHAPTER 22
ORGANIC AND BIOLOGICAL MOLECULES O
O d.
C
+
H
C
OH
O e.
CH3
O
OH CH3
f.
C
O H
OH CH3
+
C
O
88.
OH
O
a.
b. CH3
O CH3CH2C
CH3CH2CHCH
OH
O C
OH
CH3
c.
CH3CH2 O C
OH
89.
Primary alcohols and aldehydes are oxidized to carboxylic acids. Only 1-butanol is a primary alcohol and there are no aldehydes in the choices. So answer a (1-butanol) is correct.
90.
The reaction is:
A secondary alcohol (2-pentanol) is the major product when water is added to 1-pentene. The secondary alcohol is then oxidized to a ketone (2-pentanone).
CHAPTER 22 91.
ORGANIC AND BIOLOGICAL MOLECULES
1089
CH3CH=CH2 + Br2 → CH3CHBrCH2Br (addition reaction of Br2 with propene)
a. b.
The oxidation of 2 propanol will yield acetone. c.
Oxidation of 1-propanol will eventually yield propanoic acid. The aldehyde propanal is an intermediate product in the reaction. 92.
a. CH2=CHCH2CH3 will react with Br2 without any catalyst present. CH3CH2CH2CH3 reacts with Br2 only when ultraviolet light is present. O CH3CH2CH2COH is an acid, so this compound should react positively with a base like NaHCO3. The other compound is a ketone, which will not react with a base.
b.
c. CH3CH2CH2OH can be oxidized with KMnO4 to propanoic acid. 2-Propanone (a ketone) will not react with KMnO4. d. CH3CH2NH2 is an amine, so it behaves as a base in water. Dissolution of some of this base in water will produce a solution with a basic pH. The ether, CH3OCH3, will not produce a basic pH when dissolved in water. 93.
Reaction of a carboxylic acid with an alcohol can produce these esters. O
O CH3C
OH + HOCH2(CH2)6CH3
ethanoic acid (acetic acid)
octanol
O CH3CH2C
O
CH2(CH2)6CH3 + H2O
n-octylacetate
O OH + HOCH2(CH2)4CH3
propanoic acid
94.
CH3C
CH3CH2C
O
CH2(CH2)4CH3 + H2O
hexanol
When an alcohol is reacted with a carboxylic acid, an ester is produced. a.
1090
CHAPTER 22
ORGANIC AND BIOLOGICAL MOLECULES
b.
Polymers 95.
The backbone of the polymer contains only carbon atoms, which indicates that Kel-F is an addition polymer. The smallest repeating unit of the polymer and the monomer used to produce this polymer are: F
F
C
C
Cl
F
n
F
F
C
C
Cl
F
Note: Condensation polymers generally have O or N atoms in the backbone of the polymer. 96.
a.
repeating unit: CHF
CH 2
monomer: CHF=CH2
n
b. O
repeating unit:
monomer: HO‒CH2CH2‒CO2H
OCH 2CH 2C n
c. repeating unit: H N
CH2CH2
H
O
N
C
O CH2CH2
copolymer of: H2NCH2CH2NH2 and HO2CCH2CH2CO2H
C n
d. monomer: CH 3
e. monomer: C
CH 2
CH
CH CH 3
CHAPTER 22 f.
ORGANIC AND BIOLOGICAL MOLECULES
1091
copolymer of:
Addition polymers: a, d, and e; condensation polymers: b, c, and f; copolymer: c and f 97.
CN
CN
C
CH 2
C
CH 2
C
OCH 3
C
OCH 3
n
O
O
Super glue is an addition polymer formed by reaction of the C=C bond in methyl cyanoacrylate. 98.
a. 2-methyl-1,3-butadiene b. H 2C
CH 2 C
CH 2
C
CH 2 C
H 3C
H
CH 2
C
H 3C
CH 2 C
H
C
H 3C
H
H 3C
CH 2
n
cis-polyisoprene (natural rubber)
H 3C
CH 2 C
CH 2
C
H C
CH 2
H
C
H 3C
C CH 2
C
CH 2
H n
trans-polyisoprene (gutta percha) 99.
H2O is eliminated when Kevlar forms. Two repeating units of Kevlar are: H
H
O
O
H
H
O
O
N
N
C
C
N
N
C
C
n
1092 100.
CHAPTER 22
ORGANIC AND BIOLOGICAL MOLECULES
This condensation polymer forms by elimination of water. The ester functional group repeats, hence the term polyester. CH3 O O
CH
C
CH3 O O
CH
CH3 O
C
O
CH
C n
101.
This is a condensation polymer, where two molecules of H2O form when the monomers link together.
H 2N
102.
CO2H
H O2C
CO2H
and
NH2
H O2C
H O2C
CO2H and
H 2N
NH2
CO2H
103.
Divinylbenzene has two reactive double bonds that are used during formation of the polymer. The key is for the double bonds to insert themselves into two different polymer chains during the polymerization process. When this occurs, the two chains are bonded together (are crosslinked). The chains cannot move past each other because of the crosslinks, making the polymer more rigid.
104.
a.
O O
CH 2CH 2
O
O
C
CH
CH
C n
b.
O OCH 2CH 2OC
O CH
CH C n
CH 2 CH O OCH 2CH 2OC O
CH
CH C CH 2 CH
n
CHAPTER 22 105.
ORGANIC AND BIOLOGICAL MOLECULES
1093
a. The polymer formed using 1,2-diaminoethane will exhibit relatively strong hydrogenbonding interactions between adjacent polymer chains. Hydrogen bonding is not present in the ethylene glycol polymer (a polyester polymer forms), so the 1,2-diaminoethane polymer will be stronger. b. The presence of rigid groups (benzene rings or multiple bonds) makes the polymer stiffer. Hence the monomer with the benzene ring will produce the more rigid polymer. c. Polyacetylene will have a double bond in the carbon backbone of the polymer.
The presence of the double bond in polyacetylene will make polyacetylene a more rigid polymer than polyethylene. Polyethylene doesn’t have C =C bonds in the backbone of the polymer (the double bonds in the monomers react to form the polymer). 106.
At low temperatures, the polymer is coiled into balls. The forces between poly(lauryl methacrylate) and oil molecules will be minimal, and the effect on viscosity will be minimal. At higher temperatures, the chains of the polymer will unwind and become tangled with the oil molecules, increasing the viscosity of the oil. Thus the presence of the polymer counteracts the temperature effect, and the viscosity of the oil remains relatively constant.
Natural Polymers 107.
a. Serine, tyrosine, and threonine contain the -OH functional group in the R group. b. Aspartic acid and glutamic acid contain the -COOH functional group in the R group. c. An amine group has a nitrogen bonded to other carbon and/or hydrogen atoms. Histidine, lysine, arginine, and tryptophan contain the amine functional group in the R group. d. The amide functional group is:
R
O
R'
C
N
R ''
This functional group is formed when individual amino acids bond together to form the peptide linkage. Glutamine and asparagine have the amide functional group in the R group. 108.
Crystalline amino acids exist as zwitterions, +H3NCRHCOO−, held together by ionic forces. The ionic interparticle forces are strong. Before the temperature gets high enough to melt the solid, the amino acid decomposes.
1094
CHAPTER 22
109.
a. Aspartic acid and phenylalanine make up aspartame.
H 2N
H
O
C
C
ORGANIC AND BIOLOGICAL MOLECULES
OH
H
N
H
O
N
C
COH
amide bond forms here
CH 2
CH 2
CO2H
b. Aspartame contains the methyl ester of phenylalanine. This ester can hydrolyze to form methanol: RCO2CH3 + H2O ⇌ RCO2H + HOCH3 110.
O -
O
O
O
O
CCHCH2CH2C
NHCHC
NHCHC
-
O
NH3 +
CH2SH
H
glutamic acid
cysteine
glycine
Glutamic acid, cysteine, and glycine are the three amino acids in glutathione. Glutamic acid uses the −COOH functional group in the R group to bond to cysteine instead of the carboxylic acid group bonded to the α-carbon. The cysteine-glycine bond is the typical peptide linkage. 111.
O H 2N CH C
O N H CH CO2H
CH 2
H 2N CH C
CH 3
N H CH CO2H
CH 3
CH 2
OH
OH
ser - ala
112.
ala - ser
O
O
O
H2NCHC
NHCHC
NHCHCOH
H gly
CH3 ala
CH2OH ser
O
O
H2NCHC
NHCHC
NHCHCOH
CH3 ala
H gly
CH2OH ser
O
There are six possible tripeptides with gly, ala, and ser. The other four tripeptides are gly-serala, ser-gly-ala, ala-gly-ser, and ala-ser-gly.
CHAPTER 22 113.
ORGANIC AND BIOLOGICAL MOLECULES
1095
a. Six tetrapeptides are possible. From NH2 to CO2H end: phe-phe-gly-gly, gly-gly-phe-phe, gly-phe-phe-gly, phe-gly-gly-phe, phe-gly-phe-gly, gly-phe-gly-phe b. Twelve tetrapeptides are possible. From NH 2 to CO2H end: phe-phe-gly-ala, phe-phe-ala-gly, phe-gly-phe-ala, phe-gly-ala-phe, phe-ala-phe-gly, phe-ala-gly-phe, gly-phe-phe-ala, gly-phe-ala-phe, gly-ala-phe-phe ala-phe-phe-gly, ala-phe-gly-phe, ala-gly-phe-phe
114.
There are 5 possibilities for the first amino acid, 4 possibilities for the second amino acid, 3 possibilities for the third amino acid, 2 possibilities for the fourth amino acid, and 1 possibility for the last amino acid. The number of possible sequences is: 5 × 4 × 3 × 2 × 1 = 5! = 120 different pentapeptides
115.
The secondary structure of a protein describes the arrangement in space of the protein’s polypeptide chain. The most common secondary structures are the -helix and the pleated sheet. The -helix secondary structure gives a protein elasticity; examples are wool, hair, and tendons. The pleated sheet secondary structure has hydrogen bonding interactions between protein chains. As several protein chains interact with each other through hydrogen bonding, fibers result that are very strong and are resistant to stretching. Examples of proteins having pleated sheet secondary structures are silk and muscle fibers.
116.
The tertiary structure describes the overall shape of the protein. Long and narrow or globular are typical terms to describe the tertiary structure. The secondary structure describes the arrangement in space of the protein’s polypeptide chain, such as -helical or pleated sheet. The tertiary structure describes how the -helix and/or pleated sheets are arranged amongst themselves, such as long and narrow or globular.
117.
a. Ionic: Need NH2 on side chain of one amino acid with CO2H on side chain of the other amino acid. The possibilities are: NH2 on side chain = His, Lys, or Arg; CO2H on side chain = Asp or Glu
1096
CHAPTER 22
ORGANIC AND BIOLOGICAL MOLECULES
b. Hydrogen bonding: Need N‒H or O‒H bond present in side chain. The hydrogen bonding interaction occurs between the X‒ H bond and a carbonyl group from any amino acid. X‒H · · · · · · · O = C (carbonyl group) Ser Asn Glu Thr Tyr Asp His Gln Arg Lys
Any amino acid
c. Covalent: Cys‒Cys (forms a disulfide linkage) d. London dispersion: All amino acids with nonpolar R groups. They are: Gly, Ala, Pro, Phe, Ile, Trp, Met, Leu, and Val e. Dipole-dipole: Need side chain with OH group. Tyr, Thr and Ser all could form this specific dipole-dipole force with each other since all contain an OH group in the side chain. 118.
Reference Exercise 117 for a more detailed discussion of these various interactions. a. Covalent
b. Hydrogen bonding
c. Ionic
d. London dispersion
119.
Glutamic acid: R = ‒CH2CH2CO2H; valine: R = ‒CH(CH3)2; a polar side chain is replaced by a nonpolar side chain. This could affect the tertiary structure of hemoglobin and the ability of hemoglobin to bind oxygen.
120.
Glutamic acid: R = ‒CH2CH2COOH; glutamine: R = ‒CH2CH2CONH2; the R groups only differ by OH versus NH2. Both of these groups are capable of forming hydrogen-bonding interactions, so the change in intermolecular forces is minimal. Thus this change is not critical because the secondary and tertiary structures of hemoglobin should not be greatly affected.
121.
See Figures 22.29 and 22.30 of the text for examples of the cyclization process. CH 2OH CH 2OH O H H
H H OH
H OH
OH
O H
H
OH
D-Ribose
OH
HO
OH H
H
D-Mannose
CHAPTER 22 122.
ORGANIC AND BIOLOGICAL MOLECULES
1097
The chiral carbon atoms are marked with asterisks. A chiral carbon atom has four different substituent groups attached. O
O
C
H
H
*C
H H
C
H
OH
HO * C
H
*C
OH
HO
*C
H
*C
OH
H
*C
OH
CH 2OH
H
*C
OH
CH 2OH D-Ribose D-Mannose
123.
The aldohexoses contain 6 carbons and the aldehyde functional group. Glucose, mannose, and galactose are aldohexoses. Ribose and arabinose are aldopentoses since they contain 5 carbons with the aldehyde functional group. The ketohexose (6 carbons + ketone functional group) is fructose, and the ketopentose (5 carbons + ketone functional group) is ribulose.
124.
This is an example of Le Châtelier’s principle at work. For the equilibrium reactions among the various forms of glucose, reference Figure 22.30 of the text. The chemical tests involve reaction of the aldehyde group found only in the open-chain structure. As the aldehyde group is reacted, the equilibrium between the cyclic forms of glucose, and the open-chain structure will shift to produce more of the open-chain structure. This process continues until either the glucose or the chemicals used in the tests run out.
125.
The α and β forms of glucose differ in the orientation of a hydroxy group on one specific carbon in the cyclic forms (see Figure 22.30 of the text). Starch is a polymer composed of only α-D-glucose, and cellulose is a polymer composed of only β-D-glucose.
126.
Humans do not possess the necessary enzymes to break the β-glycosidic linkages found in cellulose. Cows, however, do possess the necessary enzymes to break down cellulose into the β-D-glucose monomers and therefore can derive nutrition from cellulose.
127.
A chiral carbon has four different groups attached to it. A compound with a chiral carbon is optically active. Isoleucine and threonine contain more than the one chiral carbon atom (see asterisks). H
H
H 3C
C* CH 2CH 3
H 3C
C*
OH
H 2N
C*
H 2N
C*
CO2H
CO2H
H
H
isoleucine
threonine
1098
CHAPTER 22
ORGANIC AND BIOLOGICAL MOLECULES
128.
There is no chiral carbon atom in glycine since it contains no carbon atoms with four different groups bonded to it.
129.
Only one of the isomers is optically active. The chiral carbon in this optically active isomer is marked with an asterisk. Cl C*
H
Br
CH
CH 2
130.
See Exercise 22.51 for the noncyclic structural isomers of C4H7F. Of the 11 noncyclic structural isomers, only 3 fluoro-1-butene is optically active; only this isomer has a carbon atom with four different groups bonded to it. The chiral carbon is marked with an asterisk.
131.
Aspartame has two chiral carbons (marked with an *). Only these two carbons have four different groups bonded to each of them.
132.
OH H3C
*
C *
H
OH * *
O
OH
The compound has four chiral carbon atoms. The fourth group bonded to the three chiral carbon atoms in the ring is a hydrogen atom.
CHAPTER 22 133.
ORGANIC AND BIOLOGICAL MOLECULES
1099
The chiral carbons are marked with an asterisk. A chiral carbon has four different groups bonded to it. The compound in answer a has only two chiral carbon atoms.
134.
The formula for estradiol is C18H24O2. Note that the C-H bonds are missing. Each carbon in the structure has 4 bonds, so any missing bonds to a carbon are C-H bonds. The asterisks mark the 5 chiral carbons. 135.
The complementary base pairs in DNA are cytosine (C) and guanine (G), and thymine (T) and adenine (A). The complementary sequence is C−C−A−G−A−T−A−T−G
136.
For each letter, there are 4 choices: A, T, G, or C. Hence the total number of codons is 4 × 4 × 4 = 64.
1100 137.
CHAPTER 22
ORGANIC AND BIOLOGICAL MOLECULES
Uracil will hydrogen bond to adenine. The dashed lines represent the H-bonding interactions. H uracil
O N
H
N
N
H
N
N N
N O
sugar
138.
sugar
adenine
The tautomer could hydrogen bond to guanine, forming a G‒T base pair instead of A‒T.
H 3C
O N
N
O
H H
N
N N
N O
sugar
H
sugar
N H
139.
Base pair: RNA
DNA
A........ T G........ C C........ G U........ A a. Glu: CTT, CTC
Val: CAA, CAG, CAT, CAC
Met: TAC
Trp: ACC
Phe: AAA, AAG
Asp: CTA, CTG
b. DNA sequence for trp-glu-phe-met: ACC −CTT −AAA −TAC or or CTC AAG
CHAPTER 22 c.
ORGANIC AND BIOLOGICAL MOLECULES
1101
Due to glu and phe, there is a possibility of four different DNA sequences. They are: ACC−CTT−AAA−TAC or ACC−CTC−AAA−TAC or ACC−CTT−AAG−TAC or ACC−CTC−AAG −TAC
d.
T
A
C
C
met e. 140.
T
G
A
asp
A
G
phe
TAC−CTA−AAG; TAC−CTA−AAA; TAC−CTG−AAA
In sickle cell anemia, glutamic acid is replaced by valine. DNA codons: Glu: CTT, CTC; Val: CAA, CAG, CAT, CAC; replacing the middle T with an A in the code for Glu will code for Val. CTT → CAT or CTC → CAC Glu Val Glu Val
ChemWork Problems 141.
We omitted the hydrogens for clarity. The number of hydrogens bonded to each carbon is the number necessary to form four bonds. a.
b.
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C C
2,3,5,6-tetramethyloctane c.
2,2,3,5-tetramethylheptane d.
C C
C
C
C
C
C
C
C
C
2,3,4-trimethylhexane
C C
C
C
C
3-methyl-1-pentyne
1102 142.
CHAPTER 22
ORGANIC AND BIOLOGICAL MOLECULES
a. Only one monochlorination product can form (1-chloro-2,2-dimethylpropane). The other possibilities differ from this compound by a simple rotation, so they are not different compounds. CH3 CH3
C
CH2
CH3 Cl
b. Three different monochlorination products are possible (ignoring cis-trans isomers). Cl Cl
CH2
H3C
CH3
CH3
H3C
H3C
Cl
c. Two different monochlorination products are possible (the other possibilities differ by a simple rotation of one of these two compounds). CH3 CH3
CH
Cl CH
CH3
CH2
CH3
Cl
CH
C CH3
CH3
143.
CH3
Cl
O
Cl
Cl
O
Cl
There are many possibilities for isomers. Any structure with four chlorines replacing four hydrogens in any four of the numbered positions would be an isomer; i.e., 1,2,3,4-tetrachlorodibenzo-p-dioxin is a possible isomer. 144.
a. Two monochloro products are formed: CH 2ClCH2CH3 and CH3CHClCH3 The names are 1-chloropropane and 2-chloropropane. b. Four dichloro products are formed: CHCl2CH2CH3, CH3CCl2CH3, CH2ClCHClCH3 and CH2ClCH2CH2Cl The names are 1,1-dichloropropane, 2,2-dichloropropane, 1,2-dichloropropane, and 1,3dichloropropane.
CHAPTER 22
ORGANIC AND BIOLOGICAL MOLECULES
145.
The line notation for the five structural isomers of C 6H14 are:
146.
The line notation for the 9 structural isomers of C7H14 are:
1103
1104 147.
CHAPTER 22
ORGANIC AND BIOLOGICAL MOLECULES
All the structures have a C8H18 formula. To tell if a structure is a conformation or a different isomer, name it. If a structure has the same name, it is a conformation. If the compound has a different name, it is a different isomer. The names of the compounds are: I. 2.4-dimethylhexane; II. 2,5-dimethylhexane; III. 2,3-dimethylhexane IV. 2,4-dimethylhexane; V. 2,4-dimethylhexane Structures I, IV, and V are all conformations of 2,4-dimethylhexane.
148.
Answer b is correct. Alkenes are nonpolar substances, so they only exhibit London dispersion forces. Ketones have the polar C=O group in the structure, so they have additional dipole forces. If a ketone and an alkene have similar molar mass, both have about the same strength LD forces, but the ketone has additional dipole forces, so the ketone has the higher boiling point. Alcohols have an O-H bond, so they can form hydrogen bonding intermolecular forces. Alcohols of similar molar mass will have a higher boiling point than a ketone which will have a higher boiling point than an alkene. Answer a is in reverse order. Answer c has carboxylic acids misplaced. Carboxylic acids can hydrogen bond, so they would have the highest boiling point in this grouping. For answer d, the alkanes would have the lowest boiling point in this grouping. Note in answer d, it would be difficult to differentiate aldehydes and ketone boiling points. They both exhibit same type of forces, LD forces as well as dipole forces. For answer e, alkanes would have the lowest boiling point.
149.
The isomers are: CH3
O
CH3
dimethyl ether, −23C
CH3CH2OH
ethanol, 78.5C
Ethanol, with its ability to form the relatively strong hydrogen-bonding interactions, boils at the higher temperature. 150.
The carboxylic acid group in compound A is more polar than any of the functional groups found in the other compounds. In addition compound A can form the relatively strong hydrogen bonding forces, so compound A is expected to be most soluble in water. Compound C will be the next most soluble because it has a polar carbonyl group in the structure. Compound B will not be soluble in water because it is composed entirely of nonpolar bonds so it is a nonpolar molecule. Therefore, the expected water solubility order from least to most soluble is compound B < compound C < compound A.
151.
Alcohols consist of two parts, the polar OH group and the nonpolar hydrocarbon chain attached to the OH group. As the length of the nonpolar hydrocarbon chain increases, the solubility of the alcohol decreases in water, a very polar solvent. In methyl alcohol (methanol), the polar OH group overrides the effect of the nonpolar CH 3 group, and methyl alcohol is soluble in water. In stearyl alcohol, the molecule consists mostly of the long nonpolar hydrocarbon chain, so it is insoluble in water.
CHAPTER 22
ORGANIC AND BIOLOGICAL MOLECULES
1105
152.
CH3CH2CH2CH2CH2CH2CH2COOH + OH− → CH3‒(CH2)6‒COO− + H2O; octanoic acid is more soluble in 1 M NaOH. Added OH− will remove the acidic proton from octanoic acid, creating a charged species. As is the case with any substance with an overall charge, solubility in water increases. When morphine is reacted with H+ +, the amine group is protonated, creating a positive charge on morphine (R 3 N + H + → R 3 N H). By treating morphine with HCl, an ionic compound results that is more soluble in water and in the bloodstream than the neutral covalent form of morphine.
153.
The structures, the types of intermolecular forces exerted, and the boiling points for the compounds are: O CH3CH2CH2COH
CH3CH2CH2CH2CH2OH
butanoic acid, 164C LD + dipole + H-bonding
1-pentanol, 137C LD + H-bonding
O CH3CH2CH2CH2CH
CH3CH2CH2CH2CH2CH3
pentanal, 103C LD + dipole
n-hexane, 69C LD only
All these compounds have about the same molar mass. Therefore, the London dispersion (LD) forces in each are about the same. The other types of forces determine the boiling-point order. Since butanoic acid and 1-pentanol both exhibit hydrogen bonding interactions, these two compounds will have the two highest boiling points. Butanoic acid has the highest boiling point since it exhibits H bonding along with dipole-dipole forces due to the polar C=O bond. 154.
For a ring structure to exhibit cis/trans isomerism, there must be two carbons in the ring that have two different groups attached. This occurs for 1-chloro-2-methylcyclobutene and for 1chloro-3-methylcyclobutane. The other two monochlorination products do not have two carbons in the ring with two different groups bonded to them.
155.
The necessary carboxylic acid is 2-chloropropanoic acid.
156.
85.63 g C ×
1 mol C 1 mol H = 7.130 mol C; 14.37 g H × = 14.26 mol H 12.01 g C 1.008 g H
Because the mol H to mol C ratio is 2 : 1 (14.26/7.130 = 2.000), the empirical formula is CH 2. The empirical formula mass 12 + 2(1) = 14. Since 4 × 14 = 56 puts the molar mass between 50 and 60, the molecular formula is C4H8. The isomers of C4H8 are:
1106
CHAPTER 22
ORGANIC AND BIOLOGICAL MOLECULES CH3
CH2
CH3CH
CHCH2CH3
CHCH3
CH
2-butene
1-butene
CHCH3
2-methyl-1-propene
CH3
cyclobutane
methylcyclopropane
Only the alkenes will react with H2O to produce alcohols, and only 1-butene will produce a secondary alcohol for the major product and a primary alcohol for the minor product. H CH2 CHCH2CH3 + H2O
OH
CH2 CHCH2CH3 2 alcohol, major product o
CH2 CHCH2CH3 + H2O
OH
H
CH2
CHCH2CH3
o
1 alcohol, minor product
2-Butene will produce only a secondary alcohol when reacted with H2O, and 2-methyl-1propene will produce a tertiary alcohol as the major product and a primary alcohol as the minor product. 157.
KMnO4 will oxidize primary alcohols to aldehydes and then to carboxylic acids. Secondary alcohols are oxidized to ketones by KMnO4. Tertiary alcohols and ethers are not oxidized by KMnO4. The three isomers and their reactions with KMnO 4 are: CH3
O
KMnO4
CH2CH3
no reaction
ether OH CH3
CH
O CH3
KMnO4
o
2 alcohol
o
1 alcohol
C
CH3
2-propanone (acetone)
OH CH3CH2CH2
CH3
O
O KMnO4
CH3CH2CH propanal
KMnO4
CH3CH2C OH propanoic acid
The products of the reactions with excess KMnO 4 are 2-propanone and propanoic acid.
CHAPTER 22 158.
ORGANIC AND BIOLOGICAL MOLECULES
1107
When addition polymerization of monomers with C=C bonds occurs, the backbone of the polymer chain consists of only carbon atoms. Because the backbone contains oxygen atoms, this is not an addition polymer; it is a condensation polymer. Because the ester functional group is present, we have a polyester condensation polymer. To form an ester functional group, we need the carboxylic acid and alcohol functional groups present in the monomers. From the structure of the polymer, we have a copolymer formed by the following monomers.
HO
O
O
C
C
OH
HO
CH2
CH2
OH
159.
In nylon, hydrogen-bonding interactions occur due to the presence of N‒H bonds in the polymer. For a given polymer chain length, there are more N‒H groups in Nylon-46 as compared to Nylon-6. Hence Nylon-46 forms a stronger polymer compared to Nylon-6 due to the increased hydrogen-bonding interactions.
160.
The monomers for nitrile are CH2=CHCN (acrylonitrile) and CH2=CHCH=CH2 (butadiene). The structure of the polymer nitrile is:
CH2
CH C
161.
CH2
CH
CH
CH2 n
N
a. H 2N
NH2
and
H O2C
CO2H
b. Repeating unit:
The two polymers differ in the substitution pattern on the benzene rings. The Kevlar chain is straighter, and there is more efficient hydrogen-bonding between Kevlar chains than between Nomex chains.
1108 162.
CHAPTER 22
ORGANIC AND BIOLOGICAL MOLECULES
Polyacrylonitrile: CH2
CH
N
C
n
The CN triple bond is very strong and will not easily break in the combustion process. A likely combustion product is the toxic gas hydrogen cyanide, HCN(g). 163.
a. The bond angles in the ring are about 60°. VSEPR predicts bond angles close to 109°. The bonding electrons are closer together than they prefer, resulting in strong electronelectron repulsions. Thus ethylene oxide is unstable (reactive). b. The ring opens up during polymerization; the monomers link together through the formation of O‒C bonds. O
CH2CH2 O
CH2CH2 O
CH2CH2
n
164.
Two linkages are possible with glycerol. A possible repeating unit with both types of linkages is shown above. With either linkage, there are unreacted OH groups on the polymer chains. These unreacted OH groups on adjacent polymer chains can react with the acid groups of phthalic acid to form crosslinks (bonds) between various polymer chains. 165.
Glutamic acid: H2N
Monosodium glutamate:
CH
CO2H
H2N
One of the two acidic protons in the carboxylic acid groups is lost to form MSG. Which proton is lost is impossible for you to predict. a.
H2N
CO2H
CH2CH2CO2-Na+
CH2CH2CO2H
166.
CH
In MSG, the acidic proton from the carboxylic acid in the R group is lost, allowing formation of the ionic compound.
CH2
CO2H + H2N O
CH2
CO2H
H2N
CH2
CH2
CO2H + H
C
N H
O
H
CHAPTER 22
ORGANIC AND BIOLOGICAL MOLECULES
Bonds broken:
1109
Bonds formed:
1 C‒O (358 kJ/mol)
1 C‒N (305 kJ/mol)
1 H‒N (391 kJ/mol)
1 H‒O (467 kJ/mol)
ΔH = 358 + 391 − (305 + 467) = −23 kJ b. ΔS for this process is negative (unfavorable) because the dipeptide has a more ordered structure (positional probability decreases). c. ΔG = ΔH − TΔS; although ΔH is favorable (negative), the magnitude of ΔH is very small. In this process, the unfavorable entropy change will dominate, giving a positive ΔG value. The reaction is not spontaneous at normal temperatures. 167.
ΔG = ΔH − TΔS; for the reaction, we break a P‒O and O‒H bond and form a P‒O and O‒H bond, so ΔH 0 based on bond dissociation energies. ΔS for this process is negative (unfavorable) because positional probability decreases. Thus, ΔG > 0 due to the unfavorable ΔS term, and the reaction is not expected to be spontaneous.
168.
Both proteins and nucleic acids must form for life to exist. From the simple analysis, it looks as if life can't exist, an obviously incorrect assumption. A cell is not an isolated system. There is an external source of energy to drive the reactions. A photosynthetic plant uses sunlight, and animals use the carbohydrates produced by plants as sources of energy. When all processes are combined, ΔSuniv must be greater than zero, as is dictated by the second law of thermodynamics.
169.
Alanine can be thought of as a diprotic acid. The first proton to leave comes from the carboxylic acid end with Ka = 4.5 × 10 −3 . The second proton to leave comes from the protonated amine end (Ka for R‒NH3+ = Kw/Kb = 1.0 × 10 −14 /7.4 × 10 −5 = 1.4 × 10 −10 ). In 1.0 M H+, both the carboxylic acid and the amine end will be protonated since H + is in excess. The protonated form of alanine is below. In 1.0 M OH−, the dibasic form of alanine will be present because the excess OH− will remove all acidic protons from alanine. The dibasic form of alanine follows.
1.0 M H+ :
+ H3N
CH3 O CH
C
CH3 O OH
1.0 M OH
-:
H2N
protonated form
170.
CH
dibasic form
The number of approximate base pairs in a DNA molecule is: 4.5 10 9 g / mol = 8 × 106 base pairs 600 g / mol
The approximate number of complete turns in a DNA molecule is: 8 × 106 base pairs ×
0.34 nm 1 turn = 8 × 105 turns base pair 3.4 nm
C
-
O
1110
CHAPTER 22
ORGANIC AND BIOLOGICAL MOLECULES
171.
For denaturation, heat is added so it is an endothermic process. Because the highly ordered secondary structure is disrupted, positional probability increases, so entropy will increase. Thus H and S are both positive for protein denaturation.
172.
a.
+
H3NCH2COO− + H2O ⇌ H2NCH2CO2− + H3O+
Keq = Ka (‒NH3+) =
Kw 1.0 10 −14 = 1.7 × 10 −10 = K b (− NH2 ) 6.0 10 −5
b. H2NCH2CO2− + H2O ⇌ H2NCH2CO2H + OH− Keq = Kb (‒CO2−) = c.
+
Kw 1.0 10 −14 = 2.3 × 10 −12 = −3 K a (−CO 2 H) 4.3 10
H3NCH2CO2H ⇌ 2 H+ + H2NCH2CO2−
Keq = Ka(‒CO2H) × Ka(‒NH3+) = (4.3 × 10 −3 )(1.7 × 10 −10 ) = 7.3 × 10 −13 173.
a. Zn2+ has the [Ar]3d10 electron configuration, and zinc does form 2+ charged ions. Mass % Zn =
mass of 1 mol Zn 65.38 g × 100 = × 100 = 37.50% Zn mass of 1 mol CH 3CH 2 ZnBr 174 .34 g
b. The reaction is: OH
O C*
CH3CH2CH
CH3
CH3
CH3CH2CH CH3
C*
CH3
CH2CH3
The hybridization changes from sp2 to sp3. c. 3,4-dimethyl-3-hexanol 174.
a. 0.5063 g CO2 Mass % C =
1 mol CO 2 1 mol C 12.01 g C = 0.1382 g C 44.01 g mol CO 2 mol C
0.1382 g C × 100 = 95.31% C 0.1450 g compound
Mass % H = 100.00 – 95.31 = 4.69% H Assuming 100.00 g compound: 95.31 g C ×
1 mol C = 7.936 mol C/4.653 = 1.706 mol C 12.01 g C
4.69 g H ×
1 mol H = 4.653 mol H/4.653 = 1 mol H 1.008 g H
CHAPTER 22
ORGANIC AND BIOLOGICAL MOLECULES
1111
Multiplying by 10 gives the empirical formula C 17H10. b. Moles of helicene = 0.0125 kg × Molar mass =
0.0175 mol helicene = 2.19 × 10 −4 mol helicene kg solvent
0.0938 g = 428 g/mol 2.19 10 − 4 mol
Empirical formula mass 17(12) + 10(1) = 214 g/mol Because
428 = 2.00, the molecular formula is (C17H10) × 2 = C34H20 214
c. C34H20(s) + 39 O2(g) → 34 CO2(g) + 10 H2O(l)
Challenge Problems 175.
Out of 100.00 g: 71.89 g C
1 mol C = 5.986 mol 6 mol C 12.01 g C
12.13 g H
1 mol H = 12.03 mol 12 mol H 1.008 g H
15.98 g O
1 mol O = 0.9988 mol 1 mol O 16.00 g O
The empirical formula is C 6H12O.
The general reaction for this hydrolysis reaction is: O R1
C
O O
R2 + H 2O
R1
C
OH
+ HOCH 2CH 3
R2 must be CH3CH2 because CH3CH2OH is one of the products. The molar mass of CO2H is 45 g/mol, so the molar mass of R1 is 172 − 45 = 127 g/mol. Because R1 is a straight chain alkane, its general formula is CH3(CH2)n. So 15 + n(14) = 127 and n = 112/14 = 8. Ethyl caprate is:
This compound has a molecular formula of C12H24O2 with an empirical formula of C6H12O, which agrees with the mass percent data calculation above. 176.
For the reaction: 3 CH2=CH2(g) + 3 H−H(g) → 3 CH3−CH3(g) Bonds broken: 3 C=C (614 kJ/mol) 3 H−H (432 kJ/mol)
Bonds formed: 3 C−C (347 kJ/mol) 6 C− H (413 kJ/mol)
1112
CHAPTER 22
ORGANIC AND BIOLOGICAL MOLECULES
ΔH = 3(614) + 3(432) − [3(347) + 6(413)] = −381 kJ From enthalpies of formation: ΔH = 3 ΔHf , C2H6 − 3 ΔHf , C2H4 = 3(−84.7) − 3(52) = −410. kJ The two values agree fairly well. For C6H6(g) + 3 H2(g) → C6H12(g) and starting with one of the Lewis structures for benzene, we would get the same ΔH from bond energies as the first reaction since the same number and type of bonds are broken and formed as in the previous reaction. So ΔH = −381 kJ. From enthalpies of formation: ΔH = −90.3 kJ − (82.9 kJ) = −173.2 kJ There is about a 208 kJ discrepancy between the two calculated ΔH values. Benzene is more stable by about 208 kJ/mol (lower in energy) than we expect from bond energies. This extra stability is evidence for resonance stabilization. Whenever resonance structures can be drawn, there is extra stability associated with this. 177.
For the reaction: +
H3NCH2CO2H ⇌ 2 H+ + H2NCH2CO2− Keq = 7.3 × 10 −13 = Ka (−CO2H) × Ka (−NH3+) 7.3 × 10 −13 =
−
[H + ]2 [H 2 NCH2 CO 2 ] = [H+]2, [H+] = (7.3 × 10 −13 )1/2 [ + H 3 NCH2 CO 2 H]
[H+] = 8.5 × 10 −7 M; pH = −log[H+] = 6.07 = isoelectric point 178.
a. The new amino acid is most similar to methionine due to its ‒CH2CH2SCH3 R group. b. The new amino acid replaces methionine. The structure of the tetrapeptide is:
CH2SCH3
H2N
CH2
O
CH
C
CH O H2C NH C C
O NH
CH
C
CH2
NH
CH CH2
CH2 CH2 NH C HN
O
NH2
C
NH2
CHAPTER 22
ORGANIC AND BIOLOGICAL MOLECULES
1113
c. The chiral carbons are indicated with an asterisk. H
H
H 2N
H *
H O2C
179.
* CH 2SCH 3
a. Even though this form of tartaric acid contains 2 chiral carbon atoms (see asterisks in the following structure), the mirror image of this form of tartaric acid is superimposable. Therefore, it is not optically active. An easier way to identify optical activity in molecules with two or more chiral carbon atoms is to look for a plane of symmetry in the molecule. If a molecule has a plane of symmetry, then it is never optically active. A plane of symmetry is a plane that bisects the molecule where one side exactly reflects on the other side.
b. The optically active forms of tartaric acid have no plane of symmetry. The structures of the optically active forms of tartaric acid are:
These two forms of tartaric acid are nonsuperimposable. 180.
Assuming 1.000 L of the hydrocarbon (CxHy), then the volume of products will be 4.000 L, and the mass of products (H2O + CO2) will be: 1.391 g/L × 4.000 L = 5.564 g products Moles of CxHy = n C x H y =
PV 0.959 atm 1.000 L = = 0.0392 mol 0.08206 L atm RT 298 K K mol
1114
CHAPTER 22 Moles of products = np =
ORGANIC AND BIOLOGICAL MOLECULES
PV 1.51 atm 4.000 L = = 0.196 mol 0.08206 L atm RT 375 K K mol
CxHy + oxygen → x CO2 + y/2 H2O; setting up two equations: 0.0392x + 0.0392(y/2) = 0.196 (moles of products) 0.0392x(44.01 g/mol) + 0.0392(y/2)(18.02 g/mol) = 5.564 g (mass of products) Solving: x = 2 and y = 6, so the formula of the hydrocarbon is C2H6, which is ethane. 181. O HC
C
C
C
CH
C
CH
CH
CH
CH
CH
CH 2
C
13
12
11
10
9
8
7
6
5
4
3
2
1
182. H
CH 3
H O2C
C C
C
C
H
H O2C C
H
H C
H O2C
C
C
CH 3
trans-2-trans-4-hexadienoic acid H C
CH 3
C
H
H
CH 3 C
H
cis-2-cis-4-hexadienoic acid
183.
C
H
H C
H C
H
H O2C
C
H
cis-2-trans-4-hexadienoic acid
H
H
C H
trans-2-cis-4-hexadienoic acid
a. The three structural isomers of C5H12 are: CH3 CH3CH2CH2CH2CH3
n-pentane
CH3CHCH2CH3
CH3
C
CH3
CH3
CH3
2-methylbutane
2,2-dimethylpropane
OH
CHAPTER 22
ORGANIC AND BIOLOGICAL MOLECULES
1115
n-Pentane will form three different monochlorination products: 1-chloropentane, 2chloropentane, and 3-chloropentane (the other possible monochlorination products differ by a simple rotation of the molecule; they are not different products from the ones listed). 2,2-Dimethylpropane will only form one monochlorination product: 1-chloro-2,2dimethylpropane. 2-Methylbutane is the isomer of C5H12 that forms four different monochlorination products: 1-chloro-2-methylbutane, 2-chloro-2-methyl-butane, 3chloro-2-methylbutane (or we could name this compound 2-chloro-3-methylbutane), and 1-chloro-3-methylbutane. b. The structure of 1-chloro-1-methylcyclohexane is: Cl
CH3
The addition reaction of HCl with an alkene is a likely choice for this reaction (see Exercise 82). The two isomers of C7H12 that produce 1-chloro-1-methylcyclohexane as the major product are: CH3
CH2
H
H
H
H
H H
H
H H
H H
H
H
H H
H
H H
H
c. Working backwards, 2° alcohols produce ketones when they are oxidized (1° alcohols produce aldehydes, then carboxylic acids). The easiest way to produce the 2° alcohol from a hydrocarbon is to add H2O to an alkene (with H+ present). The alkene reacted is 1propene (or propene). OH CH2
CHCH3 + H2O
propene
d.
CH3CCH3
O oxidation
CH3CCH3 acetone
The C5H12O formula has too many hydrogens to be anything other than an alcohol (or an unreactive ether). 1° Alcohols are first oxidized to aldehydes, then to carboxylic acids. Therefore, we want a 1° alcohol. The 1° alcohols with formula C5H12O are:
1116
CHAPTER 22 CH3
CH3
CH2CHCH2CH3
CH3CHCH2CH2
CH2
C
OH
OH
CH3
OH CH2CH2CH2CH2CH3
ORGANIC AND BIOLOGICAL MOLECULES
OH 1-pentanol
2-methyl-1-butanol
CH3
3-methyl-1-butanol
CH3
2,2-dimethyl-1-propanol
There are other alcohols with formula C5H12O, but they are all 2° or 3° alcohols, which do not produce carboxylic acids when oxidized. 184.
a. CH 3 O
O
C
O
C
CH 3 O
O
C
CH 3
O
C
n
CH 3
b. Condensation; HCl is eliminated when the polymer bonds form. 185. O
O
O
N COCH 2CH 2OCN
NC
H
H
O OCH 2CH 2OCN H
186.
H
n
a.
CN H2C
CH
CH2
acrylonitrile
CH
CH
CH2
CH2
butadiene
CH
styrene
The structure of ABS plastic assuming a 1 : 1 : 1 mole ratio is:
CN CH2
CH
CN CH2
CH
CH2
CH
CH2
CH
CH2
CH
CH2
CH
CH2
CH
CH2
CH
Note: Butadiene does not polymerize in a linear fashion in ABS plastic (unlike other butadiene polymers). There is no way for you to be able to predict this.
n
CHAPTER 22
ORGANIC AND BIOLOGICAL MOLECULES
1117
b. Only acrylonitrile contains nitrogen. If we have 100.00 g of polymer: 8.80 g N ×
1 mol C3 H 3 N 53.06 g C3 H 3 N = = 33.3 g C3H3N 14.01 g N 1 mol C3 H 3 N
Mass % C3H3N =
33.3 g C3 H 3 N = 33.3% C3H3N 100.00 g polymer
Br2 adds to double bonds of alkenes (benzene’s delocalized π bonds in the styrene monomer will not react with Br2 unless a special catalyst is present). Only butadiene in the polymer has a reactive double bond. From the polymer structure in part a, butadiene will react in a 1 : 1 mol ratio with Br2. 0.605 g Br2 ×
1 mol C 4 H 6 54.09 g C 4 H 6 1 mol Br 2 = 0.205 g C4H6 159 .8 g Br 2 mol Br 2 mol C 4 H 6
Mass % C4H6 =
0.205 g × 100 = 17.1% C4H6 1.20 g
Mass % styrene (C8H8) = 100.0 − 33.3 − 17.1 = 49.6% C8H8. c. If we have 100.0 g of polymer: 33.3 g C3H3N ×
1 mol C 3 H 3 N = 0.628 mol C3H3N 53 .06 g
17.1 g C4H6 ×
1 mol C 4 H 6 = 0.316 mol C4H6 54 .09 g C 4 H 6
49.6 g C8H8 ×
1 mol C8 H 8 = 0.476 mol C8H8 104 .14 g C8 H 8
Dividing by 0.316:
0.316 0.476 0.628 = 1.99; = 1.00; = 1.51 0.316 0.316 0.316
This is close to a mole ratio of 4 : 2 : 3. Thus there are 4 acrylonitrile to 2 butadiene to 3 styrene molecules in this polymer sample, or (A4B2S3)n. 187.
a.
The temperature of the rubber band increases when it is stretched.
b. Exothermic because heat is released. c. As the polymer chains that make up the rubber band are stretched, they line up more closely together, resulting in stronger London dispersion forces between the chains. Heat is released as the strength of the intermolecular forces increases.
1118
CHAPTER 22
ORGANIC AND BIOLOGICAL MOLECULES
d. Stretching is not spontaneous, so ΔG is positive. ΔG = ΔH − TΔS; since ΔH is negative, ΔS must be negative in order to give a positive ΔG. e. The structure of the stretched polymer chains is more ordered (has a smaller positional probability). Therefore, entropy decreases as the rubber band is stretched.
188.
a
Step 1:
OH
H
CH2
CHCH 2CH 3
H+
CH2
1-butanol CH2
Step 2:
CHCH 2CH 3 + H 2O
1-butene Pt
CHCH 2CH 3 + H 2
CH 3CH2 CH2 CH 3
1-butene
Step 1:
OH
H
CH 2
CHCH 2CH3
butane
H+
CHCH 2CH3 + H 2 O
CH 2
1-butanol
1-butene
b. Step 2:
CH 2
CHCH 2CH3 + H 2 O
H
1-butene
CH 3CHCH2 CH3
2-butanol
189.
4.2 × 10 −3 g K2CrO7
OH
CH 2
CHCH 2CH3
2-butanol (major product)
OH
Step 3:
H +
O
oxidation
CH 3CCH 2CH 3
2-butanone 2−
1 mol K 2 Cr2 O 7 1 mol Cr2 O 7 3 mol C 2 H 5 OH 2− 294.20 g mol K 2 Cr2 O 7 2 mol Cr2 O 7 = 2.1 × 10 −5 mol C2H5OH
CHAPTER 22
ORGANIC AND BIOLOGICAL MOLECULES
1 atm 750. mm Hg 0.500 L 760 mm Hg PV = 0.0198 mol breath n breath = = 0.08206 L atm RT 303 K K mol Mole % C2H5OH =
2.1 10 −5 mol C 2 H 5OH × 100 = 0.11% alcohol 0.0198 mol total
Marathon Problems 190.
a. urea, ammonium cyanate d. straight-chain or normal g. longest j. substitution m. aromatic p. carbon monoxide s. oxidation
b. e. h. k. n. q. t.
saturated bonds number addition functional fermentation carboxyl
c. f. i. l. o. r. u.
tetrahedral −ane combustion hydrogenation primary carbonyl esters, alcohol
191.
a. statement (17) d. statement (12) g. statement (16) j. statement (10) m. statement (14) p. statement (1)
b. e. h. k. n. q.
statement (13) statement (8) statement (2) statement (11) statement (3) statement (5)
c. f. i. l. o.
statement (15) statement (9) statement (4) statement (7) statement (6)
192.
a. deoxyribonucleic acid d. ester g. gene
b. nucleotides e. complementary h. transfer, messenger
c. ribose f. thymine, guanine i. DNA
1119