1. John E. McMurry Robert C. Fay Lecture Notes Alan D. Earhart Southeast Community College Lincoln, NE General Chemistry: Atoms First Chapter 12 Chemical Kinetics Copyright © 2010 Pearson Prentice Hall, Inc.

2. Chapter 12/2 Reaction Rates

3. Copyright © 2010 Pearson Prentice Hall, Inc.Chapter 12/3 Reaction Rates Reaction Rate: Either the increase in the concentration of a product per unit time or the decrease in the concentration of a reactant per unit time. Chemical Kinetics: The area of chemistry concerned with reaction rates and the sequence of steps by which reactions occur. Rate = Concentration change Time change

4. Chapter 12/4

5. Chapter 12/5 Reaction Rates increasedecrease

6. Chapter 12/6 Reaction Rates 2N 2 O 5 (g)4NO 2 (g) + O 2 (g)

7. Copyright © 2010 Pearson Prentice Hall, Inc.Chapter 12/7 Reaction Rates 2N 2 O 5 (g)4NO 2 (g) + O 2 (g) s M = 1.9 x 10 -5 -(0.0101 M - 0.0120 M) (400 s - 300 s) = ∆t∆t ∆[N 2 O 5 ] Rate of decomposition of N 2 O 5 :

8. Copyright © 2010 Pearson Prentice Hall, Inc.Chapter 12/8 Reaction Rates a A + b Bd D + e E rate == = 1 4 - ∆[O 2 ] ∆t∆t rate = 1 2 ∆[N 2 O 5 ] ∆t∆t = ∆[NO 2 ] ∆t∆t - 1 b ∆[B] ∆t∆t =- 1 e ∆[E] ∆t∆t = 1 a ∆[A] ∆t∆t 1 d ∆[D] ∆t∆t General rate of reaction: 2N 2 O 5 (g)4NO 2 (g) + O 2 (g)

9. Chapter 12/9 Reaction Rates 2N 2 O 5 (g)4NO 2 (g) + O 2 (g)

10. Copyright © 2010 Pearson Prentice Hall, Inc.Chapter 12/10 Rate Laws and Reaction Order Rate Law: An equation that shows the dependence of the reaction rate on the concentration of each reactant. a A + b Bproducts k is the rate constant ∆[A] ∆t∆t rate = rate  k[A] m [B] n rate = k[A] m [B] n

11. Chapter 12/11 Rate Laws and Reaction Order The values of the exponents in the rate law must be determined by experiment; they cannot be deduced from the stoichiometry of the reaction.

12. Experimental Determination of a Rate Law 2NO(g) + O 2 (g)2NO 2 (g) [O 2 ] n rate = k[NO] m Compare the initial rates to the changes in initial concentrations.

13. Experimental Determination of a Rate Law [O 2 ] n rate = k[NO] 2 m = 2 The concentration of NO doubles, the concentration of O 2 remains constant, and the rate quadruples. 2 m = 4 2NO(g) + O 2 (g)2NO 2 (g)

14. [O 2 ]rate = k[NO] 2 Experimental Determination of a Rate Law n = 1 The concentration of O 2 doubles, the concentration of NO remains constant, and the rate doubles. 2 n = 2 2NO(g) + O 2 (g)2NO 2 (g)

15. Copyright © 2010 Pearson Prentice Hall, Inc.Chapter 12/15 [O 2 ]rate = k[NO] 2 Experimental Determination of a Rate Law Reaction Order with Respect to a Reactant NO:second-order O 2 :first-order Overall Reaction Order 2 + 1 = 3 (third-order) 2NO(g) + O 2 (g)2NO 2 (g)

16. Copyright © 2010 Pearson Prentice Hall, Inc.Chapter 12/16 Experimental Determination of a Rate Law =k = M s (M 2 ) (M) 1 M 2 s = rate [NO] 2 [O 2 ] [O 2 ]rate = k[NO] 2 Units for this third-order reaction: 2NO(g) + O 2 (g)2NO 2 (g)

17. Copyright © 2010 Pearson Prentice Hall, Inc.Chapter 12/17 Experimental Determination of a Rate Law Rate Law Overall Reaction OrderUnits for k Rate = kZeroth orderM/s or M s -1 Rate = k[A]First order1/s or s -1 Rate =k[A][B]Second order1/(M s) or M -1 s -1 Rate = k[A][B] 2 Third order1/(M 2 s) or M -2 s -1

18. Copyright © 2010 Pearson Prentice Hall, Inc.Chapter 12/18 Integrated Rate Law for a First- Order Reaction Aproduct(s) rate = k[A] Calculus can be used to derive an integrated rate law. ∆[A] ∆t∆t -= k[A] x y ln= ln(x) - ln(y) Using: [A] t [A] 0 ln= -kt ln[A] t = -kt + ln[A] 0 y = mx + b [A] t concentration of A at time t [A] 0 initial concentration of A

19. Copyright © 2010 Pearson Prentice Hall, Inc.Chapter 12/19 Integrated Rate Law for a First- Order Reaction y = mx + b A plot of ln[A] versus time gives a straight-line fit and the slope will be -k. ln[A] t = -kt + ln[A] 0

20. Copyright © 2010 Pearson Prentice Hall, Inc.Chapter 12/20 Integrated Rate Law for a First- Order Reaction ln[A] t = -kt + ln[A] 0

21. Integrated Rate Law for a First- Order Reaction 2N 2 O 5 (g)4NO 2 (g) + O 2 (g) Slope = -k rate = k[N 2 O 5 ]

22. Copyright © 2010 Pearson Prentice Hall, Inc.Chapter 12/22 Integrated Rate Law for a First- Order Reaction 2N 2 O 5 (g)4NO 2 (g) + O 2 (g) rate = k[N 2 O 5 ] k = 0.0017 (700 - 0) s -5.099 - (-3.912) = -0.0017 s 1 Calculate the slope: s 1

23. Copyright © 2010 Pearson Prentice Hall, Inc.Chapter 12/23 Half-Life of a First-Order Reaction Half-Life: The time required for the reactant concentration to drop to one-half of its initial value. Aproduct(s) rate = k[A] [A] t [A] 0 ln= -kt t = t 1/2 = t 1/2 [A] 2 [A] 0 = -kt 1/2 1 2 lnt 1/2 = k 0.693 or

24. Copyright © 2010 Pearson Prentice Hall, Inc.Chapter 12/24 Half-Life of a First-Order Reaction t 1/2 = k 0.693 For a first-order reaction, the half-life is independent of the initial concentration. Each successive half-life is an equal period of time.

25. Copyright © 2010 Pearson Prentice Hall, Inc.Chapter 12/25 Radioactive Decay Rates e 0 C 6 14 N 7 + ∆N∆N ∆t∆t = kNDecay rate = N is the number of radioactive nuclei k is the decay constant

26. Copyright © 2010 Pearson Prentice Hall, Inc.Chapter 12/26 Radioactive Decay Rates e 0 C 6 14 N 7 + ∆N∆N ∆t∆t = kNDecay rate = NtNt N0N0 ln= -ktt 1/2 = k 0.693

27. Copyright © 2010 Pearson Prentice Hall, Inc.Chapter 12/27 Second-Order Reactions Aproduct(s) rate = k[A] 2 Calculus can be used to derive an integrated rate law. ∆[A] ∆t∆t -= k[A] 2 [A] t concentration of A at time t [A] 0 initial concentration of A = kt + [A] 0 1 [A] t 1 y = mx + b

28. Copyright © 2010 Pearson Prentice Hall, Inc.Chapter 12/28 Second-Order Reactions 2NO 2 (g)2NO(g) + O 2 (g) Time (s)[NO 2 ]ln[NO 2 ]1/[NO 2 ] 08.00 x 10 -3 -4.828125 506.58 x 10 -3 -5.024152 1005.59 x 10 -3 -5.187179 1504.85 x 10 -3 -5.329206 2004.29 x 10 -3 -5.451233 3003.48 x 10 -3 -5.661287 4002.93 x 10 -3 -5.833341 5002.53 x 10 -3 -5.980395

29. Chapter 12/29 Second-Order Reactions 2NO 2 (g)2NO(g) + O 2 (g)

30. Copyright © 2010 Pearson Prentice Hall, Inc.Chapter 12/30 Second-Order Reactions 2NO 2 (g)2NO(g) + O 2 (g) k = 0.540 = 0.540 M s 1 Calculate the slope: (500 - 0) s (395 - 125) M 1 M s 1

31. Copyright © 2010 Pearson Prentice Hall, Inc.Chapter 12/31 = kt + [A] 0 1 [A] t 1 Second-Order Reactions Aproduct(s) rate = k[A] 2 t = t 1/2 = t 1/2 [A] 2 [A] 0 Half-life for a second-order reaction [A] 0 1 = kt 1/2 + [A] 0 2 =t 1/2 k[A] 0 1

32. Copyright © 2010 Pearson Prentice Hall, Inc.Chapter 12/32 Second-Order Reactions =t 1/2 k[A] 0 1 For a second-order reaction, the half-life is dependent on the initial concentration. Each successive half-life is twice as long as the preceding one.

34. Copyright © 2010 Pearson Prentice Hall, Inc.Chapter 12/34 Zeroth-Order Reactions Aproduct(s) rate = k[A] 0 = k ∆[A] ∆t∆t -= k For a zeroth-order reaction, the rate is independent of the concentration of the reactant. Calculus can be used to derive an integrated rate law. [A] t concentration of A at time t [A] 0 initial concentration of A y = mx + b [A] t = -kt + [A] 0

35. Copyright © 2010 Pearson Prentice Hall, Inc.Chapter 12/35 Zeroth-Order Reactions A plot of [A] versus time gives a straight-line fit and the slope will be -k.

36. Zeroth-Order Reactions rate = k[NH 3 ] 0 = k

37. Copyright © 2010 Pearson Prentice Hall, Inc.Chapter 12/37 Reaction Mechanisms Elementary Reaction (step): A single step in a reaction mechanism. Reaction Mechanism: A sequence of reaction steps that describes the pathway from reactants to products.

38. Copyright © 2010 Pearson Prentice Hall, Inc.Chapter 12/38 Reaction Mechanisms Experimental evidence suggests that the reaction between NO 2 and CO takes place by a two-step mechanism: NO 3 (g) + CO(g)NO 2 (g) + CO 2 (g) NO 2 (g) + NO 2 (g)NO(g) + NO 3 (g) NO 2 (g) + CO(g)NO(g) + CO 2 (g) elementary reaction overall reaction elementary reaction An elementary reaction describes an individual molecular event. The overall reaction describes the reaction stoichiometry and is a summation of the elementary reactions.

39. Chapter 12/39 Reaction Mechanisms NO 3 (g) + CO(g)NO 2 (g) + CO 2 (g) NO 2 (g) + NO 2 (g)NO(g) + NO 3 (g)

40. Copyright © 2010 Pearson Prentice Hall, Inc.Chapter 12/40 NO 3 (g) + CO(g)NO 2 (g) + CO 2 (g) NO 2 (g) + NO 2 (g)NO(g) + NO 3 (g) NO 2 (g) + CO(g)NO(g) + CO 2 (g) Reaction Mechanisms Experimental evidence suggests that the reaction between NO 2 and CO takes place by a two-step mechanism: elementary reaction overall reaction elementary reaction A reactive intermediate is formed in one step and consumed in a subsequent step.

41. Copyright © 2010 Pearson Prentice Hall, Inc.Chapter 12/41 Reaction Mechanisms Molecularity: A classification of an elementary reaction based on the number of molecules (or atoms) on the reactant side of the chemical equation. termolecular reaction: unimolecular reaction: bimolecular reaction: O(g) + O(g) + M(g)O 2 (g) + M(g) O3*(g)O3*(g)O 2 (g) + O(g) O 3 (g) + O(g)2 O 2 (g)

42. Copyright © 2010 Pearson Prentice Hall, Inc.Chapter 12/42 Rate Laws for Elementary Reactions The rate law for an elementary reaction follows directly from its molecularity because an elementary reaction is an individual molecular event. termolecular reaction: unimolecular reaction: bimolecular reaction: O(g) + O(g) + M(g)O 2 (g) + M(g) O3*(g)O3*(g)O 2 (g) + O(g) rate = k[O] 2 [M] rate = k[O 3 ] rate = k[O 3 ][O 2 ] O 3 (g) + O(g)2 O 2 (g)

43. Chapter 12/43 Rate Laws for Elementary Reactions

45. Copyright © 2010 Pearson Prentice Hall, Inc.Chapter 12/45 Rate Laws for Overall Reactions Rate-Determining Step: The slowest step in a reaction mechanism. It acts as a bottleneck and limits the rate at which reactants can be converted to products.

46. Copyright © 2010 Pearson Prentice Hall, Inc.Chapter 12/46 Rate Laws for Overall Reactions NO 3 (g) + CO(g)NO 2 (g) + CO 2 (g) NO 2 (g) + NO 2 (g)NO(g) + NO 3 (g) NO 2 (g) + CO(g)NO(g) + CO 2 (g) fast step overall reaction slow step Based on the slow step: rate = k 1 [NO 2 ] 2 k2k2 k1k1 Initial Slow Step

47. Copyright © 2010 Pearson Prentice Hall, Inc.Chapter 12/47 Rate Laws for Overall Reactions N 2 O(g) + H 2 (g)N 2 (g) + H 2 O(g) 2NO(g) + 2H 2 (g)N 2 (g) + 2H 2 O(g) slow step overall reaction fast step, reversible Based on the slow step: rate = k 2 [N 2 O 2 ][H 2 ] k3k3 k -1 Initial Fast Step 2NO(g)N2O2(g)N2O2(g) k1k1 N 2 O 2 (g) + H 2 (g)N 2 O(g) + H 2 O(g) k2k2 fast step

48. Copyright © 2010 Pearson Prentice Hall, Inc.Chapter 12/48 Rate Laws for Overall Reactions rate = k 2 [N 2 O 2 ][H 2 ] intermediate First step: Rate reverse = k -1 [N 2 O 2 ]Rate forward = k 1 [NO] 2 k 1 [NO] 2 = k -1 [N 2 O 2 ] [NO] 2 [N 2 O 2 ] = k -1 k1k1 Slow step: rate = k 2 [N 2 O 2 ][H 2 ]rate = k 2 [NO] 2 [H 2 ] k -1 k1k1

49. Rate Laws for Overall Reactions Procedure for Studying Reaction Mechanisms

50. Copyright © 2010 Pearson Prentice Hall, Inc.Chapter 12/50 The Arrhenius Equation The rate constant is dependent on temperature. 2N 2 O 5 (g)4NO 2 (g) + O 2 (g) rate = k[N 2 O 5 ] Typically, as the temperature increases, the rate of reaction increases.

51. Chapter 12/51

52. The Arrhenius Equation Transition State: The configuration of atoms at the maximum in the potential energy profile. This is also called the activated complex.

53. The Arrhenius Equation Collision Theory: As the average kinetic energy increases, the average molecular speed increases, and thus the collision rate increases.

54. The Arrhenius Equation Activation Energy (E a ): The minimum energy needed for reaction. As the temperature increases, the fraction of collisions with sufficient energy to react increases.

55. Chapter 12/55 The Arrhenius Equation

56. Copyright © 2010 Pearson Prentice Hall, Inc.Chapter 12/56 Using the Arrhenius Equation RT EaEa ln(k) = ln(A) - y = mx + b + ln(A) T 1 R -Ea-Ea ln(k) = ln(k) = ln(A) + ln(e -E /RT ) a rearrange the equation

57. Using the Arrhenius Equation + ln(A) T 1 R -Ea-Ea ln(k) = t (°C)T (K)k (M -1 s -1 )1/T (1/K)lnk 2835563.52 x 10 -7 0.001 80-14.860 3566293.02 x 10 -5 0.001 59-10.408 3936662.19 x 10 -4 0.001 50-8.426 4277001.16 x 10 -3 0.001 43-6.759 5087813.95 x 10 -2 0.001 28-3.231

58. Using the Arrhenius Equation R -Ea-Ea Slope = + ln(A) T 1 R -Ea-Ea ln(k) =

59. Copyright © 2010 Pearson Prentice Hall, Inc.Chapter 12/59 Catalysis Catalyst: A substance that increases the rate of a reaction without itself being consumed in the reaction. A catalyst is used in one step and regenerated in a later step.

61. Copyright © 2010 Pearson Prentice Hall, Inc.Chapter 12/61 Catalysis Catalyst: A substance that increases the rate of a reaction without itself being consumed in the reaction. A catalyst is used in one step and regenerated in a later step. H 2 O 2 (aq) + I 1- (aq)H 2 O(l) + I O 1- (aq) H 2 O 2 (aq) + I O 1- (aq)H 2 O(l) + O 2 (g) + I 1- (aq) 2H 2 O 2 (aq)2H 2 O(l) + O 2 (g)overall reaction rate-determining step fast step

62. Copyright © 2010 Pearson Prentice Hall, Inc.Chapter 12/62 H 2 O 2 (aq) + I 1- (aq)H 2 O(l) + I O 1- (aq) H 2 O 2 (aq) + I O 1- (aq)H 2 O(l) + O 2 (g) + I 1- (aq) 2H 2 O 2 (aq)2H 2 O(l) + O 2 (g) Catalysis Since the catalyst is involved in the rate-determining step, it often appears in the rate law. rate = k[H 2 O 2 ][ I 1- ] overall reaction rate-determining step fast step

63. Catalysis Note that the presence of a catalyst does not affect the energy difference between the reactants and the products.

64. Copyright © 2010 Pearson Prentice Hall, Inc.Chapter 12/64 Homogeneous and Heterogeneous Catalysts Homogeneous Catalyst: A catalyst that exists in the same phase as the reactants. Heterogeneous Catalyst: A catalyst that exists in a different phase from that of the reactants.

65. Homogeneous and Heterogeneous Catalysts