1. Chapter Twelve: CHEMICAL KINETICS

2. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 2 Chemical Kinetics Once thermodynamics determines that a reaction will occur… Chemical kinetics is the study of the rates of spontaneous reactions. 12.1

3. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 3 Reaction Rate Change in concentration (conc.) of a given reactant or product per unit time. Or… if no reactant or product is specified, it is the rate of the species with a coefficient of “1”. 12.1

4. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 4 Reaction Rate For 2A + B  C + 3D the rate with respect to A is: rate A = - Δ[A] Δt (reaction rates are shown as positive values) 12.1

5. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 5 Reaction Rate For 2A + B  C + 3D But if - = 0.24 mol L -1 min -1, then - = ? 0.12 mol L -1 min -1 12.1    A t    B t

6. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 6 Reaction Rate Reaction Rates can also be defined as the rate per reaction. In this case, there is only one value for the reaction rate… Which is equal to each of the individual reaction rates divided by their coefficients 12.1

7. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 7 Reaction Rate Per Rxn. As Written For 2A + B  C + 3 D Rxn rate = - 1 Δ[A] = - Δ[B] = Δ[C] = 1 Δ[D] 2 Δt Δt Δt 3 Δt 12.1

8. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 8 Reaction Rate Rxn rates generally decrease over time (So you can’t use Δ[ ]/Δtime to predict concentrations over a given time period  ) 12.1

9. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 9 Instantaneous Reaction Rate Note: by striking a tangent at any point along the curve, one can determine the rate of the reaction at that point. 12.1

10. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 10 Instantaneous Reaction Rate This is accomplished by finding the slope of that tangent line, since slope is change in y divided by change in x… in other words: Rate= slope =  [N 2 O 5 ]  time 12.1

11. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 11 Instantaneous Reaction Rate This must be done “old school”… meaning you must choose two “convenient” points anywhere on the tangent line and find the slope by  y  x 12.1

12. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 12 Reaction Rate A graph representing the following reaction: 2NO 2  2NO + O 2 Notice how the stoichiometry is evident 12.1

13. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 13 “Differential” Rate Laws For the decomposition of nitrogen dioxide: 2NO 2 (g) → 2NO(g) + O 2 (g) Rate = k[NO 2 ] n  k = rate constant  Depends on temperature and “collision factors”  n = order of the reactant  Shows how “sensitive” the reaction’s rate is to changes in concentration 12.2

14. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 14 “Differential” Rate Laws Rate = k[NO 2 ] n The concentrations of the products do not appear in the rate law 12.2

15. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 15 “Differential” Rate Laws Rate = k[NO 2 ] n The value of the exponent n must be determined by experiment; it cannot be determined from the balanced equation (unless it’s a single-step reaction… more on that, later) 12.2

16. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 16 Determining Differential Rate Law Exponents - Method of Initial Rates Rate Law exponents must be determined from actual experimental data. 12.2

17. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 17 Method of Initial Rates The basic idea is to look at how the initial rate of a reaction changes due to a change in a reactant concentration. 12.2

18. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 18 Method of Initial Rates The basic idea is to look at how the initial rate of a reaction changes due to a change in a reactant concentration. If the concentration is tripled(3x), and the rate increases by 9x, then the exponent for that reactant is 2 (since 3 2 =9) 12.2

19. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 19 Method of Initial Rates Example: Using the following data table, determine the exponents for the reactants NH 4 + and NO 2 - 12.2 NO 2 -

20. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 20 Method of Initial Rates Note that between trials 1 and 2, NH 4 + remains constant and NO 2 - doubles and the rate doubles. The exponent for NO 2 - must therefore be 1. 12.2 NO 2 -

21. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 21 Method of Initial Rates Note that between trials 2 and 3, NO 2 - remains constant and NH 4 + doubles and the rate doubles. The exponent for NH 4 + also must be 1. 12.2 NO 2 -

22. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 22 Method of Initial Rates So far, we have the differential rate law as: rate = k[NH 4 + ][NO 2 - ] All that’s left is to determine the rate constant, “k” 12.2 NO 2 -

23. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 23 Method of Initial Rates This is done by picking a trial (several, if this is actual lab data and not “book” data) and plugging in the concentration and rate values, along with the determined exponents. 12.2 NO 2 -

24. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 24 Method of Initial Rates Using trial 1: 1.35x10 -7 mol/L∙s = k(0.100mol/L) 1 (0.0050mol/L) 1 2.70x10 -4 L/mol∙s = k (Take note of the units!) 12.2 NO 2 -

25. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 25 “differential” rate laws like rate = k[A] 2 …cannot relate time elapsed to concentration remaining, since the rate constantly changes during the reaction. 12.4 The Problem with Differential Rate Laws

26. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 26 First-Order Reactant Integrated Rate Law Differential rate law: Rate = k[A] Integrated rate law: ln[A] remaining - ln[A] initial = -k ∙t elapsed or ln [A] rem = -k ∙t elapsed [A] I 1 st order is MOST common reactant order!! 12.4

27. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 27 Exercise A first order reaction is 35% complete at the end of 55 minutes. What is the value of k? 12.4

28. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 28 Exercise A first order reaction is 35% complete at the end of 55 minutes. What is the value of k? 12.4 ln [A] rem = - k ∙ t elapsed [A] initial

29. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 29 Second-Order Integrated Rate Law Differential Rate Law: rate = k[A] 2 Integrated Rate Law: 1 - 1 = kt [A] rem [A] i 12.4

30. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 30 Zero-Order Rate = k[A] 0 = k Means that changes to concentration do NOT affect the reaction rate… the conc. Vs. time graph is linear. Integrated: [A] rem - [A] i = -kt elapsed 12.4

31. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 31 Another Method (Besides “Initial Rates”) to Determine Reactant Order: Graphical Technique Given experimental concentrations at given times, we can use graphing as a diagnostic tool to determine reactant order. 12.4

32. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 32 Graphing Technique to Determine IF a reactant is 1 st order 1 st order Integrated rate law: ln[A] remaining - ln[A] initial = -k ∙t elapsed 12.4

33. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 33 Graphing Technique to Determine IF a reactant is 1 st order 1 st order Integrated rate law: ln[A] remaining - ln[A] initial = -k ∙t elapsed Rearranged: ln[A] remaining = -k ∙t elapsed + ln[A] initial y = mx + b 12.4

34. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 34 Graphing Technique to Determine IF a reactant is 1 st order ln[A] remaining = -k ∙t elapsed + ln[A] initial y = mx + b So IF a graph of the ln[A] vs. time is linear (with a negative slope), THEN the reactant is first order. 12.4

35. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 35 Graphing Technique to Determine IF a reactant is 1 st order ln[A] remaining = -k ∙t elapsed + ln[A] initial y = mx + b And if it is linear… what does the magnitude of the slope give us?? The rate constant, k k = | m | 12.4

36. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 36 Graphing Technique to Determine IF a reactant is 2 nd order 2 nd order Integrated rate law: 1 - 1 = kt [A] rem [A] i 12.4

37. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 37 Graphing Technique to Determine IF a reactant is 2 nd order 2 nd order Integrated rate law: 1 - 1 = kt [A] rem [A] i Rearranged: 1 = kt + 1 [A] rem [A] i y = mx + b 12.4

38. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 38 Graphing Technique to Determine IF a reactant is 2 nd order 1 = kt + 1 [A] rem [A] i y = mx + b So IF a graph of 1/[A] vs. time is linear with a positive slope, THEN the reactant is 2nd order, with k = | m | If it’s not linear, then it will be zero or 1 st order 12.4

39. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 39 Zero-Order Rate = k[A] 0 = k 12.4

40. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 40 Zero-Order Rate = k[A] 0 = k Integrated (rearranged): [A] rem = -kt elapsed + [A] i Graphical “diagnostic”: If zero order, a graph of [A] vs. time will be linear with a negative slope, and k = | m | 12.4

41. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 41 Half Life Amt. of time for ½ of the reactants to react away. So [A] remaining = ½ [A] initial –so ½ [A] initial is substituted for [A] rem to derive the half life integrated rate law for that particular order Most important one is 1 st order All radioactive decay is 1 st order!! 12.6

42. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 42 1 st Order Half Life Equation ln [A] rem = -k ∙t elapsed [A] I then substituting: ln ½ [A] initial = -k∙t ½ (where t ½ is the half-life of that reactant) [A] initial 12.6

43. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 43 1 st Order Half Life Equation ln ½ [A] initial = -k∙t ½ [A] initial So ln 0.5 = -k∙t ½ - 0.693 = -k∙t ½ 0.693 = k∙t ½ 12.6

44. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 44 1 st Order Half Life Equation ln ½ [A] initial = -k∙t ½ [A] initial So ln 0.5 = -k∙t ½ - 0.693 = -k∙t ½ 0.693 = k∙t ½ it’s worth your while to memorize this eqn. 12.6

45. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 45 1 st Order Half Life Equation ln ½ [A] initial = -k∙t ½ [A] initial So ln 0.5 = -k∙t ½ - 0.693 = -k∙t ½ 0.693 = k∙t ½ NOTE that for 1 st order, the half-life is independent of concentration! 12.6

46. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 46 Half Life The Half-life equations of the other orders can be derived in the same fashion, but they are rarely used… their half-lives are concentration dependent. 12.6

47. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 47 Half Life You will often be presented with the half-life of a 1 st order reactant as a means to determine the rate constant, k, to use in a related problem. 12.6

48. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 48 The “Problem” with Integrated Rate Laws They only work for “one-reactant” systems. For two-reactant systems, either one reactant must already be zero order… …Or we must “trick” the reaction into behaving like one of the reactants is zero order. 12.6

49. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 49 “Flooding” Use a comparatively high concentration of one reactant It will appear zero order! 12.6

50. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 50 “Flooding” Only one reactant affects the rate, now, so the integrated rate laws will work for that reactant. The actual order of the “flooded” reactant will be found afterward by method similar to method of initial rates once the rate constant is known. 12.6

51. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 51 Flooding Mathematics For A + F  Products Where “F” will be the flooded reactant and A is the reactant affecting the rate Rate = k real [A] x [F] y = k′[A] x Where k′ = k real [F] y “x” is determined by graphical “diagnostic”, and k′ = |slope| of the linear graph 12.6

52. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 52 Flooding Mathematics To determine the order of reactant “F”, do the experiment twice, each with different initial concentrations of “F”. You will get different k′ values corresponding to the different [F] initial values 12.6

53. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 53 Flooding Mathematics k′ rxn 1 = k real [F] y rxn 1 and k′ rxn 2 = k real [F] y rxn 2 Where the [F] values are known 12.6

54. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 54 Flooding Mathematics Substituting, rearranging k′ rxn 1 = [F] y rxn 1 k′ rxn 2 = [F] y rxn 2 12.6

55. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 55 Flooding Mathematics Substituting, rearranging k′ rxn 1 = [F] y rxn 1 k′ rxn 2 = [F] y rxn 2 Taking the log 10 of both sides: log k′ rxn 1 = y∙ log [F] rxn 1 k′ rxn 2 [F] rxn 2 12.6

56. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 56 Flooding Mathematics Solving for “y” gives the order of reactant “F” (we usually round to the nearest ½ order) 12.6

57. Chapter 12 | Slide 57 Flooding Mathematics Substitute back into k′ rxn 1 = k real [F] y rxn 1 and k′ rxn 2 = k real [F] y rxn 2 to determine k real for each trial, and then average them.

58. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 58 Reaction Mechanism Most chemical reactions occur by a series of steps. 12.6

59. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 59 Reaction Mechanism Most chemical reactions occur by a series of steps. A “Reaction Mechanism” is a listing of these steps. 12.6

60. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 60 Reaction Mechanism Most chemical reactions occur by a series of steps. A Reaction Mechanism is a listing of these steps. Each step is called an “elementary reaction” and occurs literally as the reaction equation states. 12.6

61. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 61 Reaction Mechanism Each step is a literal reaction: – the reactants of a given step collide 12.6

62. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 62 Reaction Mechanism Each step is a literal reaction: – the reactants of a given step collide – an old bond breaks while a new bond simultaneously is formed. 12.6

63. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 63 Reaction Mechanism Each step is a literal reaction: – the reactants of a given step collide – an old bond breaks while a new bond simultaneously is formed. A-B  C  A--B--C  A + B-C 12.6

64. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 64 Reaction Mechanism Each step is a literal reaction: – the reactants of a given step collide – an old bond breaks while a new bond simultaneously is formed. A-B  C  A--B--C  A + B-C This “in-between” substance is called the activated complex It is also known as the “transition state” of the reaction 12.6

65. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 65 Reaction Mechanism An intermediate is formed in one step and used up in a subsequent step and thus is never seen as a product. 12.6

66. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 66 Reaction Mechanism An intermediate is formed in one step and used up in a subsequent step and thus is never seen as a product. Catalysts are substances that are reacted away, but subsequently are reformed. 12.6

67. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 67 A Molecular Representation of the Elementary Steps in the Reaction of NO 2 and CO NO 2 + CO  NO +CO 2 12.6

68. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 68 A Molecular Representation of the Elementary Steps in the Reaction of NO 2 and CO NO 2 + CO  NO +CO 2 12.6

69. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 69 A Molecular Representation of the Elementary Steps in the Reaction of NO 2 and CO NO 2 + CO  NO +CO 2 12.6

70. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 70 Elementary Steps (Molecularity) Unimolecular – reaction involving one molecule ; first order 12.6

71. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 71 Elementary Steps (Molecularity) Unimolecular – reaction involving one molecule ; first order (no collision… just a decomposition… often initiated by some form of energy (h )) 12.6

72. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 72 Elementary Steps (Molecularity) Unimolecular – reaction involving one molecule ; first order (no collision… just a decomposition… often initiated by some form of energy (h )) Bimolecular – reaction involving the collision of two molecules… could be the same type of molecule; second order 12.6

73. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 73 Elementary Steps (Molecularity) Unimolecular – reaction involving one molecule ; first order (no collision… just a decomposition… often initiated by some form of energy (h )) Bimolecular – reaction involving the collision of two molecules… could be the same type of molecule; second order Termolecular – reaction involving the simultaneous collision of three species; third order (rare) 12.6

74. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 74 Rate-Determining Step A reaction is only as fast as its slowest step. 12.6

75. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 75 Rate-Determining Step A reaction is only as fast as its slowest step. So the “slow” step in a reaction mechanism is called the “rate- determining” step 12.6

76. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 76 Rate-Determining Step A reaction is only as fast as its slowest step. So the “slow” step in a reaction mechanism is called the “rate- determining” step The rate-determining step, therefore, is the step that determines the differential rate law! 12.6

77. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 77 Rate-Determining Step This is yet ANOTHER way to determine a rate law. 12.6

78. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 78 Using the Slow Step to Determine the Differential Rate Law For a given OVERALL reaction mA + nB  Products The general form of the Differential Rate law would be Rate = k[A] x [B] y 12.6

79. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 79 Using the Slow Step to Determine the Differential Rate Law Recall that the differential rate law exponents (x and y) are NOT determined from the balanced equation coefficients (m and n) 12.6

80. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 80 Using the Slow Step to Determine the Differential Rate Law That is because the relationship between reactants and rate is not determined by the overall equation, but by the slow step of the reaction. 12.6

81. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 81 Using the Slow Step to Determine the Differential Rate Law That is because the rate relationship between reactants and rate is not determined by the overall equation, but by the slow step of the reaction. So… the coefficients of the slow step reactants ARE the exponents in the rate law! 12.6

82. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 82 Using the Slow Step to Determine the Differential Rate Law If you know the mechanism… you know the rate law! 12.6

83. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 83 Using the Slow Step to Determine the Differential Rate Law Example: for the Reaction 2A + B  C + D Proposed Mechanism: 12.6

84. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 84 Using the Slow Step to Determine the Differential Rate Law Example: for the Reaction 2A + B  C + D Proposed Mechanism A + B  C + E slow A + E  D fast 12.6

85. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 85 Using the Slow Step to Determine the Differential Rate Law Example: for the Reaction 2A + B  C + D Proposed Mechanism A + B  C + E slow A + E  D fast The rate law would be rate = k[A][B] Since 1 A and 1 B appear as reactants in the slow step. 12.6

86. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 86 Using the Slow Step to Determine the Differential Rate Law Different Example: for the Reaction 2A + B  C + D Proposed Mechanism A + A  C + E slow B + E  D fast 12.6

87. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 87 Using the Slow Step to Determine the Differential Rate Law Different Example: for the Reaction 2A + B  C + D Proposed Mechanism A + A  C + E slow B + E  D fast The rate law would be rate = k[A] 2 since 2 “A”s appear as reactants in the slow step. 12.6

88. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 88 Using the Slow Step to Determine the Differential Rate Law Different Example: for the Reaction 2A + B  C + D Proposed Mechanism A + A  C + E slow B + E  D fast rate = k[A] 2 Note that “B” is a zero order reactant because it doesn’t “appear” in the slow step! 12.6

89. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 89 Using the Slow Step to Determine the Differential Rate Law So…. If you were asked to predict which of the previous 2 proposed mechanisms was most likely, you would have to compare the PREDICTED rate laws (from the mechanisms) to the REAL rate law (determined by experiment) 12.6

90. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 90 Reaction Mechanism Requirements 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. It must not include more than 3 species colliding… and it’s even “suspect” if 3 species collide. 12.6

91. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 91 Back to Using Reaction Mechanisms to determine Differential Rate laws Many mechanisms contain at least one step that is reversible. Many of these mechanisms end up with “Intermediates” in the slow step of the mechanism 12.6

92. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 92 Intermediates are NOT allowed to be included in rate laws, though, and must be “substituted out” in terms of actual reactants. (Catalysts ARE allowed in the rate law) 12.6

93. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 93 In AP chemistry problems, the intermediate that must be substituted out is the ONLY product of a fast equilibrium. 12.6

94. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 94 Example 1 A + B  I fast equilibrium I + A  C + D slow The “unofficial” rate law would therefore be Rate = k[ I ][A] so the “official” rate law is rate = k[A][B][A] = k[A] 2 [B] 12.6

95. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 95 A + B  I fast equilibrium I + A  C + D slow rate = k[A] 2 [B] A couple of things to note: 1. “B” is NOT in the slow step, but it is not zero order 2. Even though the rate law has 2 As and 1 B in it, it does NOT mean that 2As and 1 B are colliding simultaneously. 12.6

96. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 96 Example 2 H 2 + Br 2  2 HBr Br 2  2 Br fast equilibrium Br + H 2  HBr + H slow H + Br  HBr fast rate = k[H 2 ][Br 2 ] 1/2 12.6

97. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 97 Collision Model of Chemical Kinetics Molecules must collide to react. –Rare for more than 2 molecules to collide in a given elementary step 12.7

98. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 98 Collision Model Molecules must collide to react. -And even if the correct collision DOES occur… 12.7

99. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 99 Collision Model Molecules must collide to react. -And even if the correct collision DOES occur… Not every collision results in a reaction… 12.7

100. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 100 Requirements for an “Effective” Collision #1: The correct molecules must collide 12.7

101. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 101 Requirements for an “Effective” Collision #1: The correct molecules must collide #2: Activation Energy Consideration 12.7

102. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 102 Requirements for an “Effective” Collision Molecules must collide with enough energy 12.7

103. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 103 Molecules must collide with enough energy –Sufficient threshold collision energy is necessary to disrupt the bonds of the reactants to form the activated complex ( A---B---C)… –… called the activation energy (E A ). 12.7 Requirements for an “Effective” Collision

104. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 104 Change in Potential Energy 12.7

105. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 105 Molecules must collide with enough energy –Temperature of the reaction is the most important variable regarding collision energy… more on this to come… 12.7 Requirements for an “Effective” Collision

106. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 106 #3: The molecules must be oriented correctly in space –If the transition state must be A--B--C, then the reaction will not proceed if “C” collides with “A” instead of “B” C  A-B  No Rxn. 12.7 Requirements for an “Effective” Collision

107. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 107 12.7 Requirements for an “Effective” Collision

108. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 108 Collision Model - “Effective” Collisions So… –not only do the correct molecules need to collide…. –They need to collide with sufficient energy… –AND they need to be oriented “just right”… Or that particular collision will not work… the collision would not have been “effective” 12.7

109. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 109 Back to The Effect of Temperature on Effective Collisions 12.7

110. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 110 The Arrhenius Equation This formula relates the effect that temperature has on rate constants… …and therefore reaction rates. 12.7

111. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 111 The Arrhenius Equation The total # of collisions is symbolized by the letter “z”. 12.7

112. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 112 The Arrhenius Equation Fraction of collisions of correct orientation is symbolized by the Greek letter rho,  … this value is inherent to a given reaction. 12.7

113. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 113 The Arrhenius Equation A combination of (z∙  is known as the “frequency factor”, symbolized by “A” 12.7

114. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 114 The Arrhenius Equation The fraction of collisions exceeding the activation energy is given by e -E a /RT 12.7

115. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 115 The Arrhenius Equation So… k = z∙  ∙ e -E A /RT 12.7

116. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 116 The Arrhenius Equation So… k = z∙  ∙ e -E A /RT k = A e -E A /RT 12.7

117. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 117 The Arrhenius Equation ln k = -E A ∙ 1 + ln A R T Looks like a graphing situation to me!!! A plot of ln k vs. 1/T (K) yields a line with a slope = -E A /R 12.7

118. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 118 The Arrhenius Equation – Two-Point Version ln k 2 = - E A ( 1 - 1 ) k 1 R ( T 2 T 1 ) 12.7

119. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 119 Catalyst A substance that speeds up a reaction without being (net) consumed itself. –Homogeneous catalysts do participate in an early step in the mechanism, but are completely regenerated in one of the latter steps. –Heterogeneous catalysts (usually metals) do not participate in any elementary steps, but provide a “reactive surface” for the reaction to occur on. 12.8

120. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 120 Homogeneous Catalyst Example Atmospheric Ozone depletion: h O 3 O 2 + O∙ O 3 + Cl∙  O 2 + ∙OCl O∙ + ∙OCl  O 2 + Cl∙ 12.8

121. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 121 Heterogeneous Catalyst Example Hydrogenation of Ethene : Ni cat CH 2 =CH 2 + H 2  CH 3 -CH 3 Carbon’s electrons are attracted to the surface of the metal, thus weakening the double bond 12.8

122. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 122 Catalyst The main effect: to weaken (or even break) a bond in one or more of the reactants. This lowers the activation energy threshold needed to initiate an effective collision. So some of the collisions that didn’t used to be effective now are… thus increasing the reaction rate 12.8

123. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 123 Catalyst Catalyst addition and temperature increase are the only factors that increase the rate constant, k. 12.8

124. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 124 Potential Energy Plots for a Catalyzed and an Uncatalyzed Pathway for a Given Reaction 12.8

125. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 125 A “Maxwell-Boltzman” Look at the Effect of Catalyst Addition on Reaction Rate 12.8

126. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 126 A “Maxwell-Boltzman” Look at the Effect of Catalyst Addition on Reaction Rate 12.8

127. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 127 Factors Affecting Reaction Rates Reactant Concentration –Higher concentration, more collisions, greater statistical chance of reaction 12.8

128. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 128 Factors Affecting Reaction Rates Reaction Temperature –Higher temperature, greater average collision energy, greater number of collisions with sufficient activation energy 12.8

129. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 129 Factors Affecting Reaction Rates Catalyst Addition –Lowers activation energy threshold 12.8

130. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 130 Factors Affecting Reaction Rates Reactant Surface area (for solid reactants only) –Increased surface area leads to more of that reactant being exposed to the other reactant, thus increasing the reaction rate. 12.8

131. Copyright © Houghton Mifflin Company. All rights reserved.Chapter 12 | Slide 131 Factors Affecting Reaction Rates Physical Mixing –assures that there is not a depletion of reactant in one part of the reaction vessel 12.8