CHAPTER 14:

KINETICS

The following information is often presented on the first page of exams covering this chapter under the heading: “potentially useful information.”

[X] = concentration of X at equilibrium, in moles per liter

(X) = concentration of X at any moment in time, in moles per liter

k = ZeEa/RT             R = 8.314 J/mol-K

First-order reaction: ln [(X)/(X)0] = -kt

Second-order reaction: 1/(X) - 1/(X)0 = kt

Rate Laws

14-1. The rate of a zero-order reaction:

(a) increases as reactant is consumed.

(b) depends on the concentration of products.

(c) decreases as reactant is consumed.

(d) is independent of temperature.

(e) is independent of the concentration of reactants and products.

Answer: (e)

14-2. Which of the following choices is a correct expression for the rate of the forward reaction for the following reaction?

2 NO(g) + O2(g) ➝ 2 NO2(g)

(a) d[NO]2[O2]/dt                (b) -d[NO]2[O2]/dt

(c) -d[NO]2/dt                      (d) -d[O2]/dt

(e) d[O2]/dt

Answer: (d)

14-3. For the reaction:

2 NO2(g) + Cl2(g) ➝ 2 NO2Cl,

the rate of the forward reaction is defined by which of the following equations?

(a) -d(Cl2)/dt                       (b) -d(NO2)/dt × -d(Cl2)/dt

(c) -d(NO2)2/dt                   (d) -1/2 d(Cl2)/dt

(e) -2 d(Cl2)/dt

Answer: (a)

14-4. What is the correct rate law for the reaction:

2 NO(g) + O2(g) ➝ 2 NO2(g)?

(a) rate = k[NO]2               (b) rate = k[NO]2[O2]

(c) rate = k[NO][O2]           (d) rate = k[NO]2

(e) The rate law is impossible to determine from the information given.

Answer: (e)

14-5. What is the correct rate law for the following reaction?

2 NO2(g) + F2(g) ➝ 2 NO2F(g)

(a) rate = kf(NO)2                          (b) rate = kf(NO2)2(F2)

(c) rate = -kf(NO2)2(F2)                 (d) rate = -kf(F2)

(e) Not enough information is available to determine the correct rate law for this reaction.

Answer: (e)

Rate Constants and Equilibrium Constants

14-6. For a one-step second-order reaction, the forward reaction is found to have a rate constant, kf = 5 x 1013 M-1 s-1. For this process, the equilibrium constant has a value of 1 x 1025. What is the rate constant in M-1 s-1 for the reverse process?

(a) 5 x 1013 (b) 1 x 1025 (c) 5 x 1012 (d) 5 x 1025 (e) 5 x 10-12

Answer: (e)

14-7. The following is a one-step reaction for which the forward rate constant is 0.52 x 10-6 M-1 s-1 .

CH3Cl + I- ➝ CH3I + Cl-

The equilibrium constant for this reaction is 3.5 x 104. What is the rate constant for the reverse reaction, in units of M-1 s-1?

(a) 1.5 x 10-11 (b) 0.52 x 10-6 (c) 2.9 x 10-5 (d) 6.7 x 1010 (e) none of these

Answer: (a)

14-8. For the reaction: CO(g) + Cl2(g) ➝ COCl2(g), kf is 1 x 107 M-1 s-1and kr = 2 x 102 s-1. What is the equilibrium constant for this reaction?

(a) 2 x 10-5 (b) 4.70 (c) 5 x 104 (d) 2 x 109

(e) impossible to determine from this information

Answer: (c)

Initial Instantaneous Rate of Reaction Data

Questions 9 through 11 refer to the reaction:

CH3I(aq) + OH-(aq) ➝ CH3OH(aq) + I-(aq)

for which the following data were obtained.

initial             initial             initial instantaneous

(CH3I)           (OH-)            rate of reaction

Trial 1           1.35 M          0.10 M          8.8 x 10-6 M/s

Trial 2           1.00 M          0.10 M          6.5 x 10-6 M/s

Trial 3           0.85 M          0.10 M          5.5 x 10-6 M/s

Trial 4           0.85 M          0.15 M          8.3 x 10-6 M/s

Trial 5           0.85 M          0.25 M          1.4 x 10-5 M/s

14-9. The rate law for this reaction is:

(a) zero-order in CH3I        (b) half-order in CH3I

(c) first-order in CH3I                    (d) second-order in CH3I

(e) none of the above.

Answer: (c)

14-10. The rate law for this reaction is:

(a) zero-order in OH-                    (b) half-order in OH-

(c) first-order in OH-                      (d) second-order in OH-

(e) none of the above.

Answer: (c)

14-11. The rate constant (k) for this reaction is:

(a) less than 1 x 10-6 (b) between 10-6 and 10-4 (c) between 10-4 and 10-2

(d) between 10-2 and 1 (e) more than 1

Answer: (b)

14-12. Use the following data on the initial rate of reaction to determine the overall order of the reaction.

BrO3-(aq) + 5 Br-(aq) + 6 H+(aq) ➝ 3 Br2(aq) + 3 H2O(l)

Initial rate

of reaction    Initial [BrO3-]  Initial [Br-]    Initial [H+]

9.0 x 10-4      0.10 M           0.10 M         0.10 M

1.4 x 10-3      0.10 M           0.15 M         0.10 M

3.5 x 10-3      0.25 M           0.15 M         0.10 M

8.8 x 10-3      0.25 M           0.15 M         0.25 M

(a) first-order overall          (b) second-order overall

(c) third-order overall         (d) fourth-order overall

(e) none of these

Answer: (c)

14-13. What are the units of the specific rate constant (k) for the reaction in the previous question?

(a) M-2 s (b) M-3 s-1 (c) M-3 s (d) M-4 s-1 (e) none of these

Answer: (a)

Questions 14-17 refer to the reaction:

NO(g) + O3(g) ➝ NO2(g) + O2(g)

for which the following rate data were obtained.

Initial (NO)               Initial (O3)                Initial Rate of Reaction

Trial 1           2.1 x 10-6 M             2.1 x 10-6 M              1.6 x 10-5 M s1

Trial 2           4.2 x 10-6 M             2.1 x 10-6 M              3.2 x 10-5 M s1

Trial 3           6.3 x 10-6 M             2.1 x 10-6 M              4.8 x 10-5 M s1

Trial 4           6.3 x 10-6 M             4.2 x 10-6 M              9.6 x 10-5 M s1

Trial 5           6.3 x 10-6 M             6.3 x 10-6 M             14.4 x 10-5 M s1

14-14. The rate law for the reaction would be:

(a) zero-order in NO                      (b) first-order in NO

(c) second-order in NO      (d) third-order in NO

(e) none of the above

Answer: (b)

14-15. The rate law for the reaction would be:

(a) zero-order in O3             (b) first-order in O3

(c) second-order in O3      (d) third-order in O3

(e) none of the above

Answer: (b)

14-16. If these data were obtained by watching the rate at which NO disappears, what would be the initial instantaneous rate of disappearance of O3 in Trial 3?

(a) 1.6 x 10-5 M/s (b) 3.2 x 10-5 M/s (c) 4.8 x 10-5 M/s

(d) 9.6 x 10-5 M s (e) 14.4 x 10-5 M/s

Answer: (c)

14-17. If a trial was run in which the initial (NO) was 3.15 x 10-6 M and the initial (O3) was 3.15 x 10-6 M, the initial rate of reaction would be:

(a) 1.6 x 10-5 M/s (b) 3.2 x 10-5 M/s (c) 3.6 x 10-5 M/s

(d) 4.8 x 10-5 M/s (e) 7.2 x 10-5 M/s

Answer: (c)

Questions 18 through 22 refer to the reaction

2 NO(g) + Br2(g) ➝ 2 NOBr(g)

for which the following data were obtained.

initial (NO) initial (Br2) initial rate of reaction

Trial 1           0.025 M        0.040 M         1.2 M/s

Trial 2           0.025 M        0.080 M         2.4 M/s

Trial 3           0.025 M        0.120 M         3.6 M/s

Trial 4           0.050 M        0.040 M         4.8 M/s

Trial 5           0.075 M        0.040 M        10.8 M/s

14-18. The rate law for this reaction would be:

(a) zero-order in NO (b) half-order in NO (c) first-order in NO

(d) second-order in NO (e) none of the above.

Answer: (d)

14-19. The rate law for this reaction would be:

(a) zero-order in Br2 (b) half-order in Br2 (c) first-order in Br2

(d) second-order in Br2 (e) none of the above.

Answer: (c)

14-20. If these data were obtained by watching the rate at which NO disappears, what would be the initial instantaneous rate of disappearance of Br2 in Trial 1?

(a) 0.30 M/s (b) 0.60 M/s (c) 1.2 M/s

(d) 2.4 M/s (e) 4.8 M/s

Answer: (b)

14-21. The magnitude of the rate constant, k, for this reaction is:

(a) less than 1 x 10-4 (b) between 10-4 and 10-2 (c) between 0.01 and 100

(d) between 100 and 10,000 (e) more than 10,000

Answer: (e)

14-22. A plot of (NO) versus time would most closely resemble.

(a) a straight line with a positive slope.

(b) a straight line with a negative slope.

(c) a straight line with a slope of zero.

(d) a curve in which the NO concentration increases rapidly at first and then slows down until eventually a maximum concentration is achieved.

(e) a curve in which the NO concentration decreases rapidly at first and then slows down until eventually a minimum concentration isachieved.

Answer: (e)

Question 23 through 27 refer to the reaction

Cr2O72-(aq) + 6 I-(aq) + 14 H+(aq) ➝ 2 Cr3+(aq) + 3 I2(aq) + 7 H2O(l)

for which the following initial instantaneous rate of reaction were obtained.

initial (Cr2O72-)        initial (I-)        initial (H+)      initial rate

Trial 1           0.0040 M                 0.010 M        0.020 M         5 x 10-4 M/s

Trial 2           0.0040 M                 0.020 M        0.020 M         20 x 10-4 M/s

Trial 3           0.0040 M                 0.030 M        0.020 M         46 x 10-4 M/s

Trial 4           0.0040 M                 0.030 M        0.040 M        179 x 10-4 M/s

Trial 5           0.0080 M                 0.030 M        0.040 M        355 x 10-4 M/s

14-23. If the initial rate data are for the rate at which I- disappears — - d(I-)/dt — what would be the rate of disappearance of Cr2O72- in Trial 1?

(a) 0.83 x 10-4 (b) 2.5 x 10-4 (c) 5 x 10-4 (d) 10 x 10-4 (e) 30 x 10-4 M/s

Answer: (a)

14-24. The rate law for this reaction would be:

(a) zero-order in Cr2O72- (b) half-order in Cr2O72-

(c) first-order in Cr2O72- (d) second-order in Cr2O72-

(e) none of the above

Answer: (c)

14-25. The rate law for this reaction would be:

(a) first-order in both I- and H+

(b) first-order in I- and second-order in H+

(c) first-order in H+ and second-order in I-

(d) second-order in both H+ and I-

(e) none of the above

Answer: (d)

14-26. The rate constant, k, for this reaction is:

(a) less than 10-6 (b) between 10-6 and 10-3 (c) between 10-3 and 103

(d) between 103 and 106 (e) more than 106

Answer: (e)

14-27. A graph of (Cr2O72-) versus time would most closely resemble:

(a) a straight line with a negative slope.

(b) a straight line with a slope of zero.

(c) a straight line with a positive slope.

(d) a curve in which the Cr2O72- concentration decreases rapidly at first, and then the rate at which Cr2O72- disappears slows down with time.

(e) a curve in which the Cr2O72- concentration increases rapidly at first, and then the rate at which Cr2O72- appears slows down with time.

Answer: (d)

Relative Rates of Reaction

14-28. Which of the following equations correctly describes the relationship between the rate at which NO2 and Cl2 are consumed in the following reaction?

2 NO2(g) + Cl2(g) ➝ 2 NO2Cl(g)

(a) - d(NO2)/dt = 1/2 [-d(Cl2)/dt]    (b) - d(NO2)/dt = 1/2 [d(Cl2)/dt]

(c) - d(NO2)/dt = -d(Cl2)/dt             (d) - d(NO2)/dt = 2 [-d(Cl2)/dt]

(e) - d(NO2)/dt = 2 [d(Cl2)/dt]

Answer: (d)

14-29. Which statement is true?

(a) the rate of appearance of products of a chemical reaction is always equal to the rate of disappearance of reactants.

(b) if a reaction follows a second-order rate law, it must have two steps in its reaction mechanism.

(c) the half-life for a first-order reaction is always larger than the half-life for a second-order reaction.

(d) the half-life for a first-order reaction is independent of the initial concentration of the reactant.

(e) the half-life for a second-order reaction is independent of the initial concentration of the reactant.

Answer: (d)

14-30. The instantaneous rate of appearance of water from the reaction,

4 NH3(g) + 5 O2(g) ➝ 4 NO(g) + 6 H2O(g)

at some moment in time is 21.3 mmHg per minute. What is the instantaneous rate of disappearance of NH3 at the same moment in time?

(a) 10.7 mmHg/min (b) 14.2 mmHg/min (c) 21.3 mmHg/min

(d) 32.0 mmHg/min (e) none of these

Answer: (b)

14-31. When NH3 is treated with O2 at elevated temperatures, the rate of disappearance of NH3 is found to be 3.5 x 10-2 M-1 s-1 The equation for the reaction is:

4 NH3(g) + 5 O2(g) ➝ 4 NO(g) + 6 H2O(g).

Calculate the rate of appearance of NO and H2O.

Answer: NO = 0.035 M-1 s-1; H2O = 0.053 M-1 s-1

14-32. For which reactant or product in the following reaction will the rate of change of concentration with time be the largest?

6 Fe2+ + Cr2O72- + 14 H+ ➝ 2 Cr3+ + 6 Fe3+ + 7 H2O

(a) Cr3+ (b) Cr2O72- (c) Fe2+ (d) H+ (e) H2O

Answer: (d)

14-33. The rate of disappearance of MnO4- in the reaction:

2 MnO4- + 10 I- + 16 H+ ➝ 2 Mn2+ + 5 I2 + 8 H2O

is 8.9 x 10-3 M/s. What is the rate of appearance of I2 in M/s?

(a) -8.9 x 10-3 (b) 3.56 x 10-3 (c) 8.9 x 10-3

(d) 22.2 x 10-3 (e) 44.5 x 10-3

Answer: (d)

14-34. In the following reaction

2 VO43- + 3 Zn + 16 H+ ➝ 2 V2+ + 3 Zn2+ + 8 H2O

the initial rate of disappearance of VO43- was found to be 0.56 M/s. What is the initial rate of appearance of Zn2+?

(a) -0.56 M/s (b) 0.37 M/s (c) 0.56 M/s (d) 0.84 M/s (e) 1.12 M/s

Answer: (d)

Integrated Rate Laws

14-35. Which of the following statements correctly describes a reaction for which a plot of log (X) versus time produces a straight line?

(a) the rate constant for the reaction can be obtained from the value of the intercept on the vertical axis.

(b) the rate constant is proportional to slope of the line.

(c) the rate of the reaction does not depend on the concentration of X.

(d) the initial concentration of X can be calculated from the intercept on the horizontal axis.

(e) all of the above statements are true.

Answer: (b)

14-36. If a plot of 1/[A] versus time produces a straight line, which of the following is true?

(a) this reaction is first-order in A.

(b) the reaction is second-order in A.

(c) the reaction is first-order in two reactants.

(d) the rate of reaction does not depend on the concentration of A.

(e) none of the above

Answer: (b)

14-37. The reaction: N2O5 ➝ NO2 + 1/2 O2 is first-order in N2O5 with a half-life of 19.25 minutes. How long would it take for the N2O5 concentration to decrease from 0.050 M to 0.030 M?

Answer: 14.2 min

14-38. The reaction: 2 N2O5 ➝ 4 NO2 + O2 obeys the rate law: rate = k(N2O5). What is the relationship between the rate at which N2O5 disappears and O2 appears? If half of the initial N2O5 reacts in 1403 seconds, what is the value of the rate constant, k? If the initial concentration of N2O5 is 1.00 M, what will be the concentration of O2 when t = 701 seconds?

Answer: 0.71 M

Use the following data to answer questions 39-40. These data were obtained by monitoring the rate at which the OCl- ion was consumed in the presence of a large excess of the I- ion.

OCl-(aq) + I-(aq) ➝ OI-(aq) + Cl-(aq)

[OCl-], M 0.00250 0.00137 0.00075 0.00041

time, s 0 3 6 9

14-39. What is the rate constant (k) for this reaction?

(a) between 10-2 and 10-1  (b) larger than 10-1 and smaller than 1

(c) between 1 and 10         (d) larger than 10 and smaller than 100

(e) between 100 and 1000

Answer: (a)

14-40. What is the half-life in seconds?

(a) between 0.01 and 0.1 seconds (b) between 0.1 and 1 second

(c) between 1 and 10 seconds      (d) between 10 and 100 seconds

(e) none of the above

Answer: (c)

14-41. What is the concentration of OCl- after 12 seconds?

(a) between 1 x 10-4 and 2 x 10-4 (b) between 2 x 10-4 and 3 x 10-4

(c) between 3 x 10-4 and 4 x 10-4 (d) between 4 x 10-4 and 5 x 10-4

(e) between 5 x 10-4 and 6 x 10-4

Answer: (b)

14-42. The decomposition of H2O2 is first-order in H2O2 with a rate constant in the presence of a platinum catalyst of 2 x 10-2 s-1. If 10 gallons of H2O2 are used to fuel a rocket, how long is it before only 5 gallons remain?

(a) 10-2 s (b) 0.05 s (c) 34.7 s (d) 138.6 s (e) 200 s

Answer: (c)

The following data table should be used to answer questions 43-45.

Reaction:      A ➝ B

(A), M           0.380 0.319 0.268 0.225 0.189 0.158 0.133

time, s            0     10 20 30 40 50 60

14-43. What is the rate constant, k, for this reaction?

(a) 0.0175 (b) 0.0560 (c) 0.0815 (d) 0.112 (e) none of these

Answer: (a)

14-44. What is the half-life for this reaction, in seconds?

(a) 29 (b) 39.6 (c) 47 (d) 150 (e) none of these

Answer: (b)

14-45. What will be the concentration of A when t = 120 seconds?

(a) 0.047 (b) 0.10 (c) 0.13 (d) 0.21 (e) none of these

Answer: (a)

The following data were obtained for the decomposition of H2O2.

2 H2O2(aq) ➝ 2 H2O(l) + O2(g)

(H2O2), M:   10.00 8.25 6.80 5.61 4.63

time, s:           0 500 1000 1500 2000

14-46. The rate law for this reaction is:

(a) zero-order in H2O2       (b) first-order in H2O2

(c) second-order in H2O2 (d) third-order in H2O2

(e) none of these

Answer: (b)

14-47. The half-life for this reaction is:

(a) between 0 and 50 s                (b) between 50 and 500 s

(c) between 500 and 1500 s        (d) between 1500 and 2000 s

(e) the half-life can't be obtained from these data

Answer: (d)

14-48. What will be the H2O2 concentration after 4000 seconds?

(a) less than 1.5 M             (b) between 1.5 and 2.5 M

(c) between 2.5 and 3.0 M (d) between 3.0 and 3.5 M

(e) between 3.5 and 4.5 M

Answer: (b)

Questions 49 through 51 refer to the reaction:

2 N2O5(g) ➝ 4 NO2(g) + O2(g)

for which the following data were obtained.

(N2O5), M:   5.00 3.52 2.48 1.75 1.23

time, s:         0 500 1000 1500 2000

14-49. The rate law for this reaction would be:

(a) zero-order in N2O5       (b) half-order in N2O5

(c) first-order in N2O5         (d) second-order in N2O5

(e) none of the above.

Answer: (c)

14-50. The half-life for this reaction is:

(a) between 0 and 12 s                (b) between 12 and 120 s

(c) between 120 and 1200 s        (d) between 1200 and 12,000 s

(e) between 12,000 and 120,000 s

Answer: (c)

14-51. What will be the concentration of N2O5 after 5000 seconds?

(a) between 0.001 and 0.010 M   (b) between 0.010 and 0.10 M

(c) between 0.10 and 0.4 M                    (d) between 0.4 and 0.8 M

(e) between 0.8 and 1.2 M

Answer: (c)

Questions 52 through 53 refer to the reaction

Pt

2 H2O2(aq) ➝ 2 H2O(l) + O2(g)

for which the following data were obtained at 25°C.

(H2O2), M:    1.000 0.549 0.301 0.165 0.091 0.050 0.027

time, s:         0    30 60 90 120 150 180

14-52. The rate law for this reaction is:

(a) zero-order in hydrogen peroxide        (b) half-order in hydrogen peroxide

(c) first-order in hydrogen peroxide        (d) second-order in hydrogen peroxide

(e) none of the above

Answer: (c)

14-53. When will the concentration of H2O2 reach 0.010 M?

(a) before 200 s                  (b) between 200 and 240 s

(c) between 240 and 280 s (d) between 280 and 320 s

(e) after 320 s

Answer: (b)

Half-life of a Chemical Reaction

14-54. The half-life for the first-order decomposition of nitramide, NH2NO2, into nitrous oxide, N2O, and water is 123 min at 15°C.

NH2NO2 ➝ N2O + H2O

What is the value of the rate constant (k) for this reaction?

Answer: 5.64 x 10-3 min-1

14-55. How long will it take for 2.0 g of nitramide to decompose until only 0.20 g remains?

Answer: 410 min

14-56. For a second-order reaction, it takes 15 seconds for the initial concentration of a reactant to decrease from 0.60 M to 0.53 M. What is the initial half-life of this reaction?

(a) 47 s (b) 84 s (c) 114 s (d) 128 s

Answer: (c)

14-57. A certain second-order reaction is found to have a rate constant of 0.135 M-1 s-1 What is the half-life of the reaction?

(a) 5.1 s (b) 6.5 s (c) 7.4 s (d) none of these

(e) impossible to determine from this information

Answer: (e)

14-58. Which of the following statements about the half-life of a reaction is true?

(a) the half-life doesn’t depend on the order of the reaction.

(b) the half-life of a first-order reaction increases with time.

(c) the half-life of a second-order reaction is independent of concentration

(d) a zero-order reaction doesn’t have a half-life.

(e) none of the above are true.

Answer: (e)

Kinetics of Radioactive Decay

14-59. The rate constant, k, for the first-order decay of 14C is 1.21 x 10-4 yr. If analysis of a piece of paper, parchment, or papyri suggests that 97.4% of the 14C that was present initially still remains in the sample, it would most likely be associated with which of the following battles?

(a) Battle of Actium, 31 BC, Octarian defeating Mark Anthony.

(b) Battle of Hastings, 1066 AD, William of Normandy defeating King Harold II of                      England.

(c) Battle of Yorktown, 1781 AD, Marquis de Lafayette defeating Lord Cornwallis.

(d) Battle of Waterloo, 1815 AD, Wellington defeating Napoleon.

Answer: (c)

14-60. The decay of radioactive nuclei is a first-order kinetic process. Determine the half-life for the decay of 238U to 206Pb if the first-order rate constant is 4.87 x 10-18 s-1

(a) 3.37 x 10-18 s (b) 39.5 s (c) 1.42 x 1017 s (d) 4.11 x 1017 s (e) none of these

Answer: (c)

14-61. The half-life of 228Ra is 1590 years. How long will it take for a sample of this nuclide to decay to 1% of its original activity?

(a) 15.90 years (b) 1590 years (c) 4064 years

(d) 10,568 years (e) none of these

Answer: (d)

Mechanism of a Reaction

14-62. The overall rate of a chemical reaction is determined by:

(a) the fastest step in the reaction mechanism.

(b) the first step in the reaction mechanism.

(c) the slowest step in the reaction mechanism.

(d) the last step in the reaction mechanism.

(e) ΔG for the overall reaction.

Answer: (c)

14-63. Which of the following rate laws suggests that the reaction probably occurs in a single step?

(a) NO(g) + O2(g) ➝ NO2(g) + O(g)                 rate = k[NO][O2]

(b) H2(g) + Br2(g) ➝ 2 HBr(g)                          rate = k[H2][Br2]1/2

(c) (CH3)3CBr + OH- ➝ (CH3)3COH + Br-         rate = k[(CH3)3CBr]

(d) H2O2 + 3 I- + 2 H+ ➝ I3- + 2 H2O               rate = kl[H2O2][I-] + K2[H2O2][I-][H+]

Answer: (a)

14-64. Use the rate laws given below to determine which of the following reactions most likely occurs in a single step.

(a) 2 NO2(g) + F2(g) ➝ 2 NO2F(g)                     rate = k(NO2)(F2)

(b) 2 NOCl(g) ➝ 2 NO(g) + 2Cl2g)                     rate = k(NOCl)2

(c) NO2(g) + CO(g) ➝ NO(g) + CO(g)                rate = k(NO2)2

(d) 2 O3(g) ➝ 3 O2(g)                                         rate = k(O3)

(e) H2(g) + Br2(g) ➝ 2 HBr(g)                            rate = k(Br2)1/2(H2)

Answer: (b)

14-65. Nitrogen oxide reacts with hydrogen to form nitrogen and water vapor.

2 NO(g) + 2 H2(g) ➝ N2(g) + 2 H2O(g)

The following mechanism has been proposed for this reaction.

Step 1: 2 NO + H2 ➝ NO + H2O2         (slow step)

Step 2: H2O2 + H2 ➝ 2 H2O                (fast step)

Use this mechanism to predict the rate law for this reaction.

Answer: rate = k(NO)2(H2)

14-66. The oxidation of iodide ion by hypochlorite ion in aqueous solution

ClO- + I- ➝ Cl- + IO-

has been postulated to occur by the following three-step mechanism.

1. ClO- + H2O ➝ HClO + OH-       (fast, equilibrium)

2. I- + HClO ➝ HIO + Cl-              (slow, equilibrium)

3. OH- + HIO ➝ H2O + IO-           (fast, equilibrium)

Determine the rate law required by this mechanism.

Answer: rate = k(I-)(ClO-)(OH-)-1

14-67. The reaction:

2 NO2(g) + F2(g) ➝ 2 NO2F(g)

has the following experimental rate law.

rate = k(NO2)(F2)

Which of the following mechanisms provides the best explanation of the observed rate law?

(a) 2 NO2 + F2 ➝ 2 NO2F             (one step)

(b) NO2 + F2 ➝ NO2F + F (fast)

NO2 + F ➝ NO2F                      (slow)

(c) NO2 + F2 ➝ NO2F + F            (slow)

NO2 + F ➝ NO2F                      (fast)

(d) F2 ➝ 2 F                                  (slow)

2 NO2 + 2 F ➝ 2 NO2F             (fast)

(e) none of these mechanism explains the observed rate law.

Answer: (c)

14-68. The reaction:

3 NO(g) ➝ N2O(g) + NO2(g)

has the following experimental rate law.

rate = k(NO)3

Which of the following mechanisms is consistent with this rate law?

(a) 2 NO ➝ N2O2                          (slow, equilibrium step)

N2O2 + NO ➝ N2O + NO2       (fast step)

(b) 2 NO ⇌ N2O2                      (fast step)

N2O2 + NO ➝ N2O + NO2       (slow, equilibrium step)

(c) Both of these mechanisms are consistent with this rate law.

(d) Neither of these mechanisms are consistent with this rate law.

Answer: (b)

Activation Energies

14-69. If the rate constant increases from 0.40 M-1 s-1 at 25°C to 0.80 M-1 s-1 at 35°C, what is the activation energy for this reaction?

(a) between 0 and 40 kJ mol       (b) between 41 and 80 kJ/mol

(c) between 81 and 120 kJ/mol    (d) between 121 and 160 kJ/mol

(e) between 161 and 200 kJ/mol

Answer: (b)

14-70. Which of the following correctly describes the variation of rate constant, kf, with temperature?

(a) kf = ln A - Ea/RT           (b) ln kf = ln A + e-(Ea/RT)

(c) ln kf = ln A - Ea/RT        (d) ln kf = ln A + Ea/RT

(e) kf = ln A - Ea/RT

Answer: (c)

14-71. The decomposition of N2O5 occurs with a rate constant of 4.3 x 10-3 s-1 at 65°C and 3.0 x 10-5 s-1 at 25°C. What is the activation energy for the process?

(a) 1.0 x 10-5 J/mol (b) 1.0 x 10-3 J/mol (c) 1.0 x 103 J/mol

(d) 1.0 x 105 J/mol (e) 1.0 x 106 J/mol

Answer: (d)

14-72. Looking at the units of the rate constants in the previous question, can you say what the order of the reaction is?

(a) 1/2 (b) 1st (c) 2nd (d) 3rd  (e) not enough information to say

Answer: (b)

14-73. The activation energy for the formation of NO2F(g) from NO2(g) and F2(g) is 32 kcal/mol. The reaction is endothermic with a ΔH of 2.5 kcal/mol. What is the activation energy of the reverse reaction, in kcal/mole?

(a) 34.5 (b) 32 (c) -34.5 (d) -29.5 (e) 29.5

Answer: (e)

14-74. The rate constant for the acid hydrolysis of sucrose (cane sugar) is 2.12 x 10-4 M-1 s-1 at 27°C and 8.46 x 10-4 M-1 s-1 at 37°C. What is the activation energy for this reaction?

(a) between 0 and 50 kJ/mol       (b) between 50 and 100 kJ/mol

(c) between 100 and 150 kJ/mol (d) between 150 and 200 kJ/mol

(e) between 200 and 250 kJ/mol

Answer: (c)

14-75. The rate constant for the reaction:

2 ICl + H2 ➝ I2 + 2 HCl

is 0.150 M-1 s-1 at 127°C and 1.25 x 103 M-1 s-1 at 227°C What is the activation energy for this reaction?

(a) between 0 and 40 kJ/mol        (b) between 40 and 80 kJ/mol

(c) between 80 and 120 kJ/mol    (d) between 120 and 160 kJ/mol

(e) more than 160 kJ/mol

Answer: (d)