1. Topics about reaction kinetics
Rate of reaction
Order of reaction
Half-life
Reaction progress
Activation energy
Reaction mechanism

2. Chapter 13, Unnumbered Figure, Page 498

3. Figure 13.1 Reactant and Product Concentrations as a Function of Time

4. H2O2(aq) + 3 I–(aq) + 2 H+(aq) → I3(aq) + 2 H2O(l)
Example 13.1 Expressing Reaction Rates
Consider the balanced chemical equation:
H2O2(aq) + 3 I–(aq) + 2 H+(aq) → I3(aq) + 2 H2O(l)
In the first 10.0 seconds of the reaction, the concentration of I– drops from M to M.
(a) Calculate the average rate of this reaction in this time interval.
(b) Predict the rate of change in the concentration of H+ (that is, [H+]/t) during this time interval. –
SOLUTION
(a) Use Equation 13.5 to calculate the average rate of
the reaction.
(b) Use Equation 13.5 again to determine the
relationship between the rate of the reaction and
[H+]/t. After solving for [H+]/t, substitute
the calculated rate from part (a) and calculate
[H+]/t.

5. Example 13.1 Expressing Reaction Rates
Continued
For Practice 13.1
For the above reaction, predict the rate of change in the concentration of H2O2 ([H2O2]/t) and I3 ([I3]/t) during this time interval. –

6. Figure 13.1 Reactant and Product Concentrations as a Function of Time

7. Chapter 13, Unnumbered Table, Page 499

8. Thermodynamics of kinetics
Figure 13.9 The Activation Energy Barrier

9. Figure 13.2 Reactant Concentration as a Function of Time for Different Reaction Orders

10. Figure 13.3 Reaction Rate as a Function of Reactant Concentration for Different Reaction Orders

11. Figure 13.4 Sublimation

12. Chapter 13, Unnumbered Table 1, Page 504

13. Chapter 13, Unnumbered Table 2, Page 504

14. Chapter 13, Unnumbered Figure, Page 505

15. Chapter 13, Unnumbered Figure A, Page 505

16. Chapter 13, Unnumbered Figure B, Page 505

17. Chapter 13, Unnumbered Figure C, Page 505

18. Figure 13.5 First-Order Integrated Rate Law

19. Chapter 13, Unnumbered Figure 1, Page 507

20. Chapter 13, Unnumbered Figure 2, Page 507

21. Figure 13.6 Second-Order Integrated Rate Law

22. Figure 13.7 Zero-Order Integrated Rate Law

23. Chapter 13, Unnumbered Figure, Page 509

24. Chapter 13, Unnumbered Figure, Page 510

25. Example 13.4 The First-Order Integrated Rate Law: Determining the
Concentration of a Reactant at a Given Time
In Example 13.3, we learned that the decomposition of SO2Cl2 (under the given reaction conditions) is first order and has a rate constant of –4 s–1. If the reaction is carried out at the same temperature, and the initial concentration of SO2Cl2 is M, what is the SO2Cl2 concentration after 865 s?
SOLUTION
SORT You are given the rate constant of a first-order reaction and the initial concentration of the reactant. You are asked to find the concentration at 865 seconds.
STRATEGIZE Use the first-order integrated rate law to determine the information you are asked to find from the given information.
SOLVE Substitute the rate constant, the initial concentration, and the time into the integrated rate law. Solve the integrated rate law for the concentration of [SOCl2]t.
GIVEN k = –4 s–1
[SO2Cl2]0 = M
FIND [SO2Cl2] at t = 865 s
EQUATION ln [A]t = –kt + ln [A]0
SOLUTION

26. Example 13.4 The First-Order Integrated Rate Law: Determining the
Concentration of a Reactant at a Given Time
Continued
SOLUTION
CHECK
The concentration is smaller than the original concentration as expected. If the concentration were larger than the initial concentration, this would indicate a mistake in the signs of one of the quantities on the right-hand side of the equation.
For Practice 13.4
Cyclopropane rearranges to form propene in the gas phase according to the reaction:
The reaction is first order in cyclopropane and has a measured rate constant of –5 s–1 at 720 K. If the initial cyclopropane concentration is M, what is the cyclopropane concentration after minutes?

27. Example 13.5 The Second-Order Integrated Rate Law: Using Graphical
Analysis of Reaction Data
Consider the equation for the decomposition of NO2.
NO2(g) → NO(g) + O(g)
The concentration of NO2 is monitored at a fixed temperature as a function of time during the decomposition reaction and the data tabulated in the margin at the top right of the next page. Show by graphical analysis that the reaction is not first order and that it is second order. Determine the rate constant for the reaction.
SOLUTION
In order to show that the reaction is not first order, prepare a graph of ln [NO2] versus time.

28. Example 13.5 The Second-Order Integrated Rate Law: Using Graphical
Analysis of Reaction Data
Continued
SOLUTION
The plot is not linear (the straight line does not fit the data points), confirming that the
reaction is not first order. In order to show that the reaction is second order, prepare a
graph of 1/ [NO2] versus time.
This graph is linear (the data points fit well to a straight line), confirming that the reaction is indeed second order. To obtain the rate constant, determine the slope of
the best fitting line. The slope is M–1 • s–1; therefore, the rate constant is
0.255 M–1 • s–1.

29. Figure 13.8 Half-Life: Concentration versus Time for a First-Order Reaction

31. Example 13.6 Half-Life SOLUTION
Molecular iodine dissociates at 625 K with a first-order rate constant of s–1. What is the half-life of this reaction?
SOLUTION
Since the reaction is first order, the half-life is given by
Equation Substitute the value of k into the
expression and calculate t1/2.
For Practice 13.6
A first-order reaction has a half-life of 26.4 seconds. How long will it take for the concentration of the reactant in the reaction to fall to one-eighth of its initial value?

32. Figure 13.10 The Activated Complex

33. Chapter 13, Unnumbered Figure, Page 512

34. Chapter 13, Unnumbered Figure, Page 513

35. Chapter 13, Unnumbered Figure 1, Page 514

36. Chapter 13, Unnumbered Figure 2, Page 514

37. Example 13.7 Using an Arrhenius Plot to Determine Kinetic Parameters
The decomposition of ozone is important to many atmospheric reactions:
O3(g) → O2(g) + O(g)
A study of the kinetics of the reaction resulted in the following data:
Determine the value of the frequency factor and activation energy for the reaction.
SOLUTION
To find the frequency factor and activation energy, prepare a graph of the natural log of the rate constant (ln k ) versus the inverse of the temperature (1/T).

38. Example 13.7 Using an Arrhenius Plot to Determine Kinetic Parameters
Continued
SOLUTION
The plot is linear, as expected for Arrhenius behavior. The best fitting line has a slope of – K and a
y-intercept of Calculate the activation energy from the slope by setting the slope equal to –Ea/R and solving for Ea:

39. Example 13.7 Using an Arrhenius Plot to Determine Kinetic Parameters
Continued
SOLUTION
Calculate the frequency factor ( A ) by setting the intercept equal to ln A.
26.8 = ln A
A = e26.8 =
Since the rate constants were measured in units of M–1 • s–1, the frequency factor is in the same units. Consequently, we can conclude that the reaction has an activation energy of 93.1 kJ/mol and a frequency factor of M–1 • s–1.
For Practice 13.7
For the decomposition of ozone reaction in Example 13.7, use the results of the Arrhenius analysis to predict the rate constant at 298 K.

40. Figure 13.11 Thermal Energy Distribution

41. Chapter 13, Unnumbered Figure, Page 516

42. Figure 13.12 The Collision Model

43. Chapter 13, Unnumbered Figure 1, Page 517

44. Chapter 13, Unnumbered Figure 2, Page 517

45. Chapter 13, Unnumbered Figure 3, Page 517

47. Figure 13.13 Energy Diagram for a Two-Step Mechanism

48. Figure 13.14 Catalyzed and Uncatalyzed Decomposition of Ozone

49. Example 13.9 Reaction Mechanisms
Ozone naturally decomposes to oxygen by the reaction:
2 O3(g) → 3 O2(g)
The experimentally observed rate law for this reaction is:
Rate = k[O3]2[O2]–1
Show that the following proposed mechanism is consistent with the experimentally observed rate law.
SOLUTION
To determine whether the mechanism is valid, you must first determine whether the steps sum to the overall reaction. Since the steps do indeed sum to the overall reaction, the first condition is met.
The second condition is that the rate law predicted by the mechanism must be consistent with the experimentally observed rate law. Since the second step is rate limiting, write the rate law based on the second step.
Rate = k2[O3][O]

50. Example 13.9 Reaction Mechanisms
Continued
SOLUTION
Because the rate law contains an intermediate (O), you must express the concentration of the intermediate in terms of the concentrations of the reactants of the overall reaction. To do this, set the rates of the forward reaction and the reverse reaction of the first step equal to each other. Solve the expression from the previous step for [O], the concentration of the intermediate.
Finally, substitute [O] into the rate law predicted by the slow step.
CHECK Since the two steps in the proposed mechanism sum to the overall reaction, and since the rate law obtained from the proposed mechanism is consistent with the experimentally observed rate law, the proposed mechanism is valid. The –1 reaction order with respect to [O2] indicates that the rate slows down as the concentration of oxygen increases—oxygen inhibits, or slows down, the reaction.

51. Example 13.9 Reaction Mechanisms
Continued
For Practice 13.9
Predict the overall reaction and rate law that would result from the following two-step mechanism.
2 A → A Slow
A2 + B → A2B Fast

52. Figure 13.15 Homogeneous and Heterogeneous Catalysis

53. Chapter 13, Unnumbered Figure, Page 525

54. Figure 13.16 Enzyme–Substrate Binding

55. Chapter 13, Unnumbered Figure, Page 528

56. Chapter 13, Unnumbered Figure 1, Page 529

57. Chapter 13, Unnumbered Figure 2, Page 529

58. Chapter 13, Unnumbered Figure 3, Page 529

59. Chapter 13, Unnumbered Figure, Page 531

60. Chapter 13, Unnumbered Figure 1, Page 534

61. Chapter 13, Unnumbered Figure 2, Page 534

62. Chapter 13, Unnumbered Figure 1, Page 536

63. Chapter 13, Unnumbered Figure 1A, Page 536

64. Chapter 13, Unnumbered Figure 1B, Page 536

65. Chapter 13, Unnumbered Figure 1C, Page 536

66. Chapter 13, Unnumbered Figure 2, Page 536