1. Chapter 12 Chemical Kinetics

2. Chapter 12 Table of Contents Copyright © Cengage Learning. All rights reserved 2 12.1 Reaction Rates 12.2 Rate Laws: An Introduction 12.3 Determining the Form of the Rate Law 12.4 The Integrated Rate Law 12.5 Reaction Mechanisms 12.6A Model for Chemical Kinetics 12.7Catalysis

3. Section 12.1 Reaction Rates Return to TOC Copyright © Cengage Learning. All rights reserved 3 Reaction Rate Change in concentration of a reactant or product per unit time. [A] means concentration of A in mol/L; A is the reactant or product being considered.

4. Section 12.1 Reaction Rates Return to TOC Copyright © Cengage Learning. All rights reserved 4 The Decomposition of Nitrogen Dioxide

5. Section 12.1 Reaction Rates Return to TOC Copyright © Cengage Learning. All rights reserved 5 The Decomposition of Nitrogen Dioxide

6. Section 12.1 Reaction Rates Return to TOC Copyright © Cengage Learning. All rights reserved 6 Instantaneous Rate Value of the rate at a particular time. Can be obtained by computing the slope of a line tangent to the curve at that point.

7. Section 12.2 Atomic MassesRate Laws: An Introduction Return to TOC Copyright © Cengage Learning. All rights reserved 7 Rate Law Shows how the rate depends on the concentrations of reactants. For the decomposition of nitrogen dioxide: 2NO 2 (g) → 2NO(g) + O 2 (g) Rate = k[NO 2 ] n :  k = rate constant  n = order of the reactant

8. Section 12.2 Atomic MassesRate Laws: An Introduction Return to TOC Copyright © Cengage Learning. All rights reserved 8 Rate Law Rate = k[NO 2 ] n The concentrations of the products do not appear in the rate law because the reaction rate is being studied under conditions where the reverse reaction does not contribute to the overall rate.

9. Section 12.2 Atomic MassesRate Laws: An Introduction Return to TOC Copyright © Cengage Learning. All rights reserved 9 Rate Law Rate = k[NO 2 ] n The value of the exponent n must be determined by experiment; it cannot be written from the balanced equation.

10. Section 12.2 Atomic MassesRate Laws: An Introduction Return to TOC Copyright © Cengage Learning. All rights reserved 10 Types of Rate Laws Differential Rate Law (rate law) – shows how the rate of a reaction depends on concentrations. Integrated Rate Law – shows how the concentrations of species in the reaction depend on time.

11. Section 12.2 Atomic MassesRate Laws: An Introduction Return to TOC Copyright © Cengage Learning. All rights reserved 11 Rate Laws: A Summary Because we typically consider reactions only under conditions where the reverse reaction is unimportant, our rate laws will involve only concentrations of reactants. Because the differential and integrated rate laws for a given reaction are related in a well– defined way, the experimental determination of either of the rate laws is sufficient.

12. Section 12.2 Atomic MassesRate Laws: An Introduction Return to TOC Copyright © Cengage Learning. All rights reserved 12 Rate Laws: A Summary Experimental convenience usually dictates which type of rate law is determined experimentally. Knowing the rate law for a reaction is important mainly because we can usually infer the individual steps involved in the reaction from the specific form of the rate law.

13. Section 12.3 The MoleDetermining the Form of the Rate Law Return to TOC Copyright © Cengage Learning. All rights reserved 13 Determine experimentally the power to which each reactant concentration must be raised in the rate law.

14. Section 12.3 The MoleDetermining the Form of the Rate Law Return to TOC Copyright © Cengage Learning. All rights reserved 14 Method of Initial Rates The value of the initial rate is determined for each experiment at the same value of t as close to t = 0 as possible. Several experiments are carried out using different initial concentrations of each of the reactants, and the initial rate is determined for each run. The results are then compared to see how the initial rate depends on the initial concentrations of each of the reactants.

15. Section 12.3 The MoleDetermining the Form of the Rate Law Return to TOC Copyright © Cengage Learning. All rights reserved 15 Overall Reaction Order The sum of the exponents in the reaction rate equation. Rate = k[A] n [B] m Overall reaction order = n + m k = rate constant [A] = concentration of reactant A [B] = concentration of reactant B

16. Section 12.3 The MoleDetermining the Form of the Rate Law Return to TOC Copyright © Cengage Learning. All rights reserved 16 Concept Check How do exponents (orders) in rate laws compare to coefficients in balanced equations? Why?

17. Section 12.4 The Integrated Rate Law Return to TOC Copyright © Cengage Learning. All rights reserved 17 Half-Life of Reactions

18. Section 12.4 The Integrated Rate Law Return to TOC Copyright © Cengage Learning. All rights reserved 18 Rate = k[A] Integrated: ln[A] = –kt + ln[A] o [A] = concentration of A at time t k = rate constant t = time [A] o = initial concentration of A First-Order

19. Section 12.4 The Integrated Rate Law Return to TOC Copyright © Cengage Learning. All rights reserved 19 Plot of ln[N 2 O 5 ] vs Time

20. Section 12.4 The Integrated Rate Law Return to TOC Copyright © Cengage Learning. All rights reserved 20 Half–Life: k = rate constant Half–life does not depend on the concentration of reactants. First-Order

21. Section 12.4 The Integrated Rate Law Return to TOC Copyright © Cengage Learning. All rights reserved 21 Exercise A first order reaction is 35% complete at the end of 55 minutes. What is the value of k? k = 7.8 x 10 –3 min –1

22. Section 12.4 The Integrated Rate Law Return to TOC Copyright © Cengage Learning. All rights reserved 22 Rate = k[A] 2 Integrated: [A] = concentration of A at time t k = rate constant t = time [A] o = initial concentration of A Second-Order

23. Section 12.4 The Integrated Rate Law Return to TOC Copyright © Cengage Learning. All rights reserved 23 Plot of ln[C 4 H 6 ] vs Time and Plot of 1/[C 4 H 6 ] vs Time

24. Section 12.4 The Integrated Rate Law Return to TOC Copyright © Cengage Learning. All rights reserved 24 Half–Life: k = rate constant [A] o = initial concentration of A Half–life gets longer as the reaction progresses and the concentration of reactants decrease. Each successive half–life is double the preceding one. Second-Order

25. Section 12.4 The Integrated Rate Law Return to TOC Copyright © Cengage Learning. All rights reserved 25 Exercise For a reaction aA  Products, [A] 0 = 5.0 M, and the first two half-lives are 25 and 50 minutes, respectively. a)Write the rate law for this reaction. rate = k[A] 2 b)Calculate k. k = 8.0 x 10 -3 M –1 min –1 c)Calculate [A] at t = 525 minutes. [A] = 0.23 M

26. Section 12.4 The Integrated Rate Law Return to TOC Copyright © Cengage Learning. All rights reserved 26 Rate = k[A] 0 = k Integrated: [A] = –kt + [A] o [A] = concentration of A at time t k = rate constant t = time [A] o = initial concentration of A Zero-Order

27. Section 12.4 The Integrated Rate Law Return to TOC Copyright © Cengage Learning. All rights reserved 27 Plot of [A] vs Time

28. Section 12.4 The Integrated Rate Law Return to TOC Copyright © Cengage Learning. All rights reserved 28 Half–Life: k = rate constant [A] o = initial concentration of A Half–life gets shorter as the reaction progresses and the concentration of reactants decrease. Zero-Order

29. Section 12.4 The Integrated Rate Law Return to TOC Copyright © Cengage Learning. All rights reserved 29 Concept Check How can you tell the difference among 0 th, 1 st, and 2 nd order rate laws from their graphs?

30. Section 12.4 The Integrated Rate Law Return to TOC Copyright © Cengage Learning. All rights reserved 30 Rate Laws

31. Section 12.4 The Integrated Rate Law Return to TOC Copyright © Cengage Learning. All rights reserved 31 Summary of the Rate Laws

32. Section 12.4 The Integrated Rate Law Return to TOC Copyright © Cengage Learning. All rights reserved 32 Exercise Consider the reaction aA  Products. [A] 0 = 5.0 M and k = 1.0 x 10 –2 (assume the units are appropriate for each case). Calculate [A] after 30.0 seconds have passed, assuming the reaction is: a) Zero order b) First order c) Second order 4.7 M 3.7 M 2.0 M

33. Section 12.5 Reaction Mechanisms Return to TOC Copyright © Cengage Learning. All rights reserved 33 Most chemical reactions occur by a series of elementary steps. An intermediate is formed in one step and used up in a subsequent step and thus is never seen as a product in the overall balanced reaction. Reaction Mechanism

34. Section 12.5 Reaction Mechanisms Return to TOC Copyright © Cengage Learning. All rights reserved 34 A Molecular Representation of the Elementary Steps in the Reaction of NO 2 and CO NO 2 (g) + CO(g) → NO(g) + CO 2 (g)

35. Section 12.5 Reaction Mechanisms Return to TOC Copyright © Cengage Learning. All rights reserved 35 Unimolecular – reaction involving one molecule; first order. Bimolecular – reaction involving the collision of two species; second order. Termolecular – reaction involving the collision of three species; third order. Elementary Steps (Molecularity)

36. Section 12.5 Reaction Mechanisms Return to TOC Copyright © Cengage Learning. All rights reserved 36 A reaction is only as fast as its slowest step. The rate-determining step (slowest step) determines the rate law and the molecularity of the overall reaction. Rate-Determining Step

37. Section 12.5 Reaction Mechanisms Return to TOC Copyright © Cengage Learning. All rights reserved 37 The sum of the elementary steps must give the overall balanced equation for the reaction. The mechanism must agree with the experimentally determined rate law. Reaction Mechanism Requirements

38. Section 12.5 Reaction Mechanisms Return to TOC Copyright © Cengage Learning. All rights reserved 38 Decomposition of N 2 O 5

39. Section 12.5 Reaction Mechanisms Return to TOC Copyright © Cengage Learning. All rights reserved 39 Decomposition of N 2 O 5 2N 2 O 5 (g)  4NO 2 (g) + O 2 (g) Step 1: N 2 O 5 NO 2 + NO 3 (fast) Step 2: NO 2 + NO 3 → NO + O 2 + NO 2 (slow) Step 3: NO 3 + NO → 2NO 2 (fast) 2( )

40. Section 12.5 Reaction Mechanisms Return to TOC Copyright © Cengage Learning. All rights reserved 40 Concept Check The reaction A + 2B  C has the following proposed mechanism: A + B D(fast equilibrium) D + B  C(slow) Write the rate law for this mechanism. rate = k[A][B] 2

41. Section 12.6 A Model for Chemical Kinetics Return to TOC Copyright © Cengage Learning. All rights reserved 41 Molecules must collide to react. Main Factors:  Activation energy, E a  Temperature  Molecular orientations Collision Model

42. Section 12.6 A Model for Chemical Kinetics Return to TOC Copyright © Cengage Learning. All rights reserved 42 Energy that must be overcome to produce a chemical reaction. Activation Energy, E a

43. Section 12.6 A Model for Chemical Kinetics Return to TOC Copyright © Cengage Learning. All rights reserved 43 Transition States and Activation Energy

44. Section 12.6 A Model for Chemical Kinetics Return to TOC Copyright © Cengage Learning. All rights reserved 44 Change in Potential Energy

45. Section 12.6 A Model for Chemical Kinetics Return to TOC Copyright © Cengage Learning. All rights reserved 45 Collision must involve enough energy to produce the reaction (must equal or exceed the activation energy). Relative orientation of the reactants must allow formation of any new bonds necessary to produce products. For Reactants to Form Products

46. Section 12.6 A Model for Chemical Kinetics Return to TOC Copyright © Cengage Learning. All rights reserved 46 The Gas Phase Reaction of NO and Cl 2

47. Section 12.6 A Model for Chemical Kinetics Return to TOC Copyright © Cengage Learning. All rights reserved 47 A=frequency factor E a =activation energy R=gas constant (8.3145 J/K·mol) T =temperature (in K) Arrhenius Equation

48. Section 12.6 A Model for Chemical Kinetics Return to TOC Copyright © Cengage Learning. All rights reserved 48 Linear Form of Arrhenius Equation

49. Section 12.6 A Model for Chemical Kinetics Return to TOC Copyright © Cengage Learning. All rights reserved 49 Linear Form of Arrhenius Equation

50. Section 12.6 A Model for Chemical Kinetics Return to TOC Copyright © Cengage Learning. All rights reserved 50 Exercise Chemists commonly use a rule of thumb that an increase of 10 K in temperature doubles the rate of a reaction. What must the activation energy be for this statement to be true for a temperature increase from 25°C to 35°C? E a = 53 kJ

51. Section 12.7 Catalysis Return to TOC Copyright © Cengage Learning. All rights reserved 51 A substance that speeds up a reaction without being consumed itself. Provides a new pathway for the reaction with a lower activation energy. Catalyst

52. Section 12.7 Catalysis Return to TOC Copyright © Cengage Learning. All rights reserved 52 Energy Plots for a Catalyzed and an Uncatalyzed Pathway for a Given Reaction

53. Section 12.7 Catalysis Return to TOC Copyright © Cengage Learning. All rights reserved 53 Effect of a Catalyst on the Number of Reaction-Producing Collisions

54. Section 12.7 Catalysis Return to TOC Copyright © Cengage Learning. All rights reserved 54 Most often involves gaseous reactants being adsorbed on the surface of a solid catalyst. Adsorption – collection of one substance on the surface of another substance. Heterogeneous Catalyst

55. Section 12.7 Catalysis Return to TOC Copyright © Cengage Learning. All rights reserved 55 Heterogeneous Catalysis

56. Section 12.7 Catalysis Return to TOC Copyright © Cengage Learning. All rights reserved 56 1.Adsorption and activation of the reactants. 2.Migration of the adsorbed reactants on the surface. 3.Reaction of the adsorbed substances. 4.Escape, or desorption, of the products. Heterogeneous Catalyst

57. Section 12.7 Catalysis Return to TOC Copyright © Cengage Learning. All rights reserved 57 Exists in the same phase as the reacting molecules. Enzymes are nature’s catalysts. Homogeneous Catalyst

58. Section 12.7 Catalysis Return to TOC Copyright © Cengage Learning. All rights reserved 58 Homogeneous Catalysis