2. 1 © 2009 Brooks/Cole - Cengage Chemical Kinetics Chapter 12 H 2 O 2 decomposition in an insect H 2 O 2 decomposition catalyzed by MnO 2

3. 2 © 2009 Brooks/Cole - Cengage Objectives Understand rates of reaction and the conditions affecting rates. Derive a rate of reaction, rate constant, and reaction order from experimental data Use integrated rate laws Discuss collision theory and the role of activation energy in a reaction Discuss reaction mechanisms and their effect on rate law

4. 3 © 2009 Brooks/Cole - Cengage We can use thermodynamics to tell if a reaction is product- or reactant-favored.We can use thermodynamics to tell if a reaction is product- or reactant-favored. But this gives us no info on HOW FAST reaction goes from reactants to products.But this gives us no info on HOW FAST reaction goes from reactants to products. KINETICS — the study of REACTION RATES and their relation to the way the reaction proceeds, i.e., its MECHANISM.KINETICS — the study of REACTION RATES and their relation to the way the reaction proceeds, i.e., its MECHANISM. Chemical Kinetics

5. 4 © 2009 Brooks/Cole - Cengage Reaction Mechanisms A reaction mechanism is a sequence of events at the molecular level that control the speed and outcome of a reaction.A reaction mechanism is a sequence of events at the molecular level that control the speed and outcome of a reaction. The reaction mechanism is our goal!The reaction mechanism is our goal!

6. 5 © 2009 Brooks/Cole - Cengage Reaction rate = change in concentration of a reactant or product with time.Reaction rate = change in concentration of a reactant or product with time. Three “types” of ratesThree “types” of rates –initial rate –average rate –instantaneous rate Reaction Rates Section 12.1

7. 6 © 2009 Brooks/Cole - Cengage Factors Affecting Reaction Rates Physical State of the Reactants –G–Gas, liquid or solid – how molecules are able to interact with each other –S–Solids react faster when surface area is greater so fine powders react faster than big chunks Concentration of Reactants –A–As the concentration of reactants increases, so does the likelihood that reactant molecules will collide. Temperature –A–At higher temperatures, reactant molecules have more kinetic energy, move faster, and collide more often and with greater energy. Catalysts –S–Speed up reaction by changing mechanism. –C–Catalysts don’t get used up themselves

8. 7 © 2009 Brooks/Cole - Cengage Concentrations & Rates 0.3 M HCl6 M HCl Mg(s) + 2 HCl(aq)  MgCl 2 (aq) + H 2 (g)

9. 8 © 2009 Brooks/Cole - Cengage Physical state of reactantsPhysical state of reactants Factors Affecting Rates

10. 9 © 2009 Brooks/Cole - Cengage Catalysts: catalyzed decomp of H 2 O 2 2 H 2 O 2  2 H 2 O + O 2 2 H 2 O 2  2 H 2 O + O 2 Factors Affecting Rates

11. 10 © 2009 Brooks/Cole - Cengage TemperatureTemperature Factors Affecting Rates Bleach at 54 ˚CBleach at 22 ˚C

12. 11 © 2009 Brooks/Cole - Cengage Determining a Reaction Rate Blue dye is oxidized with bleach. Its concentration decreases with time. The rate — the change in dye conc with time — can be determined from the plot. Blue dye is oxidized with bleach. Its concentration decreases with time. The rate — the change in dye conc with time — can be determined from the plot. Dye Conc Time

13. 12 © 2009 Brooks/Cole - Cengage Determining a Reaction Rate

14. 13 © 2009 Brooks/Cole - Cengage Reaction Rates The average rate of the reaction over each interval is the change in concentration divided by the change in time: C 4 H 9 Cl (aq) + H 2 O (l)  C 4 H 9 OH (aq) + HCl (aq) Average Rate, M/s

15. 14 © 2009 Brooks/Cole - Cengage Reaction Rates Note that the average rate decreases as the reaction proceeds. This is because as the reaction goes forward, there are fewer collisions between reactant molecules. C 4 H 9 Cl (aq) + H 2 O (l)  C 4 H 9 OH (aq) + HCl (aq)

16. 15 © 2009 Brooks/Cole - Cengage Reaction Rates C 4 H 9 Cl (aq) + H 2 O (l)  C 4 H 9 OH (aq) + HCl (aq)

17. 16 © 2009 Brooks/Cole - Cengage Reaction Rates The reaction slows down with time because the concentration of the reactants decreases. C 4 H 9 Cl (aq) + H 2 O (l)  C 4 H 9 OH (aq) + HCl (aq)

18. 17 © 2009 Brooks/Cole - Cengage Reaction Rates and Stoichiometry In this reaction, the ratio of C 4 H 9 Cl to C 4 H 9 OH is 1:1. Thus, the rate of disappearance of C 4 H 9 Cl is the same as the rate of appearance of C 4 H 9 OH. C 4 H 9 Cl (aq) + H 2 O (l)  C 4 H 9 OH (aq) + HCl (aq) Rxn Rate = -  [C 4 H 9 Cl]  t =  [C 4 H 9 OH]  t

19. 18 © 2009 Brooks/Cole - Cengage Reaction Rates In this reaction, the concentration of nitrogen dioxide, NO 2, was measured at various times, t. 2NO 2 (g)  2NO (g) + O 2 (g) [C 4 H 9 Cl] M

20. 19 © 2009 Brooks/Cole - Cengage Reaction Rates and Stoichiometry What if the ratio is not 1:1? 2NO 2 (g)  2NO(g) + O 2 (g)

21. 20 © 2009 Brooks/Cole - Cengage Reaction Rates But since the coefficient for oxygen is ½ of the other two, it’s rate is of production rate is half as fast. Or 2 x rate O 2 = rate of NO

22. 21 © 2009 Brooks/Cole - Cengage Reaction Rates

23. 22 © 2009 Brooks/Cole - Cengage Reaction Rates Practice 1.What is the relative rate of disappearance of the reactants and relative rate of appearance of the products for each reaction? A.2O 3 (g)  3O 2 (g) B.2HOF(g)  2HF(g) + O 2 (g) 2.In the synthesis of ammonia, if Δ[H 2 ]/Δt =-4.5 x 10 -4 mol/Lmin, what is rate with respect to NH 3 ? wrt N 2 ? N 2 (g) + 3H 2 (g)  2NH 3 (g)

24. 23 © 2009 Brooks/Cole - Cengage Concentrations & Rates Rate of reaction is proportional to [NO 2 ] We express this as a RATE LAW Rate of reaction = k [NO 2 ] n where k = rate constant k is independent of conc. but increases with T n is the order of the reactant 2NO 2 (g)  2NO(g) + O 2 (g)

25. 24 © 2009 Brooks/Cole - Cengage Concentrations, Rates, & Rate Laws In general, for a A + b B  x X with a catalyst C Rate = k [A] m [B] n [C] p The exponents m, n, and p: sum of m, n and p are the reaction ordersum of m, n and p are the reaction order can be 0, 1, 2 or fractions like 3/2can be 0, 1, 2 or fractions like 3/2 must be determined by experiment! They are not simply related to stoichiometry!must be determined by experiment! They are not simply related to stoichiometry!

26. 25 © 2009 Brooks/Cole - Cengage Ch. 12.3 - Interpreting Rate Laws Determining Rate by inspection (quick method): Rate = k [A] m [B] n [C] p If m = 1, rxn. is 1st order in AIf m = 1, rxn. is 1st order in A Rate = k [A] 1 If [A] doubles, then rate goes up by factor of _2_ If [A] doubles, then rate goes up by factor of _2_ If m = 2, rxn. is 2nd order in A.If m = 2, rxn. is 2nd order in A. Rate = k [A] 2 Rate = k [A] 2 Doubling [A] increases rate by ___4_____ If m = 0, rxn. is zero order.If m = 0, rxn. is zero order. Rate = k [A] 0 Rate = k [A] 0 If [A] doubles, rate _Stays the same_

27. 26 © 2009 Brooks/Cole - Cengage The method of Initial Rates This method requires that a reaction be run several times. The initial concentrations of the reactants are varied. The reaction rate is measured just after the reactants are mixed. Eliminates the effect of the reverse reaction.

28. 27 © 2009 Brooks/Cole - Cengage Deriving Rate Laws Expt. [CH 3 CHO]Disappear of CH 3 CHO (mol/L)(mol/Lsec) 10.100.020 20.200.081 30.300.182 40.400.318 Derive rate law and k for CH 3 CHO(g)  CH 4 (g) + CO(g) from experimental data for rate of disappearance of CH 3 CHO

29. 28 © 2009 Brooks/Cole - Cengage Deriving Rate Laws Determination of Rate by inspection: Here the rate goes up by __4___ when initial conc. doubles. Therefore, we say this reaction is ________2 nd _________ order. So, the Rate of rxn = k [CH 3 CHO] 2 Now determine the value of k. Use expt. #3 data— 0.182 mol/Ls = k (0.30 mol/L) 2 k = 2.0 (L / mols) k = 2.0 (L / mols) Using k you can calc. rate at other values of [CH 3 CHO] at same T.

30. 29 © 2009 Brooks/Cole - Cengage Deriving Rate Laws

31. 30 © 2009 Brooks/Cole - Cengage Determination of Rate Law Practice For the reaction BrO 3 - + 5Br - + 6H +  3Br 2 + 3H 2 O The general form of the Rate Law is Rate = k[BrO 3 - ] n [Br - ] m [H + ] p We use experimental data to determine the values of n, m, and p We will choose trials that vary only one of the concentrations to determine the order based upon that reactant.

32. 31 © 2009 Brooks/Cole - Cengage Initial concentrations (M) Rate (M/s) BrO 3 - Br - H+H+H+H+ 0.100.100.100.8 x 10 -3 0.100.100.100.8 x 10 -3 0.200.100.101.6 x 10 -3 0.200.100.101.6 x 10 -3 0.200.200.103.2 x 10 -3 0.200.200.103.2 x 10 -3 0.100.100.203.2 x 10 -3 0.100.100.203.2 x 10 -3 1)Determine the rate law wrt each reactant 2)Determine the rate constant, k (with units) 3)Determine the overall order of reaction Determination of Rate Law Practice

33. 32 © 2009 Brooks/Cole - Cengage Ch. 12.4 – Integrated Rate Law (Conc vs time) What is concentration of reactant as function of time? Consider FIRST ORDER REACTIONS The rate law is And the units of k are sec -1.

34. 33 © 2009 Brooks/Cole - Cengage Integrated Rate Law Expresses the reaction concentration as a function of time. Form of the equation depends on the order of the rate law (differential). Changes Rate =  [A] n  t We will only work with n=0, 1, and 2

35. 34 © 2009 Brooks/Cole - Cengage Concentration/Time Relations Integrating - (∆ [A] / ∆ time) = k [A], we get ln is natural logarithm [A] 0 at time = 0 [A] / [A] 0 =fraction remaining after time t has elapsed. Called the integrated first-order rate law.

36. 35 © 2009 Brooks/Cole - Cengage Concentration/Time Relations Sucrose decomposes to simpler sugars Rate of disappearance of sucrose = k [sucrose] Glucose If k = 0.21 hr -1 and [sucrose] = 0.010 M How long to drop 90% (to 0.0010 M)?

37. 36 © 2009 Brooks/Cole - Cengage Concentration/Time Relations Rate of disappear of sucrose = k [sucrose], k = 0.21 hr -1. If initial [sucrose] = 0.010 M, how long to drop 90% or to 0.0010 M? Use the first order integrated rate law ln (0.100) = - 2.3 = - (0.21 hr -1 )(time) time = 11 hours

38. 37 © 2009 Brooks/Cole - Cengage Using the Integrated Rate Law The integrated rate law suggests a way to tell the order based on experiment. 2 N 2 O 5 (g)  4 NO 2 (g) + O 2 (g) Time (min)[N 2 O 5 ] (M) ln [N 2 O 5 ] 01.000 1.00.705-0.35 2.00.497-0.70 5.00.173-1.75 Rate = k [N 2 O 5 ]

39. 38 © 2009 Brooks/Cole - Cengage Using the Integrated Rate Law 2 N 2 O 5 (g)  4 NO 2 (g) + O 2 (g) Rate = k [N 2 O 5 ] Data of conc. vs. time plot do not fit straight line. Plot of ln [N 2 O 5 ] vs. time is a straight line!

40. 39 © 2009 Brooks/Cole - Cengage Using the Integrated Rate Law All 1st order reactions have straight line plot for ln [A] vs. time. Plot of ln [N 2 O 5 ] vs. time is a straight line! Eqn. for straight line: y = mx + b y = mx + b Plot of ln [N 2 O 5 ] vs. time is a straight line! Eqn. for straight line: y = mx + b y = mx + b ln[N 2 O 5 ] = -kt + ln[N 2 O 5 ] 0    conc at rate const conc at time t =-slope time=0

41. 40 © 2009 Brooks/Cole - Cengage Half-Life HALF-LIFE is the time it takes for 1/2 a sample is disappear. For 1st order reactions, the concept of HALF-LIFE is especially useful. See Active Figure 15.9

42. 41 © 2009 Brooks/Cole - Cengage Reaction is 1st order decomposition of H 2 O 2.Reaction is 1st order decomposition of H 2 O 2. Half-Life

43. 42 © 2009 Brooks/Cole - Cengage Half-Life Reaction after 1 half-life.Reaction after 1 half-life. 1/2 of the reactant has been consumed and 1/2 remains.1/2 of the reactant has been consumed and 1/2 remains.

44. 43 © 2009 Brooks/Cole - Cengage Half-Life After 2 half-lives 1/4 of the reactant remains.After 2 half-lives 1/4 of the reactant remains.

45. 44 © 2009 Brooks/Cole - Cengage Half-Life A 3 half-lives 1/8 of the reactant remains.A 3 half-lives 1/8 of the reactant remains.

46. 45 © 2009 Brooks/Cole - Cengage Half-Life After 4 half-lives 1/16 of the reactant remains.After 4 half-lives 1/16 of the reactant remains.

47. 46 © 2009 Brooks/Cole - Cengage Half-Life Sugar is fermented in a 1st order process (using an enzyme as a catalyst). Sugar is fermented in a 1st order process (using an enzyme as a catalyst). sugar + enzyme  products sugar + enzyme  products Rate of disappear of sugar = k[sugar] k = 3.3 x 10 -4 sec -1 What is the half-life of this reaction?

48. 47 © 2009 Brooks/Cole - Cengage Half-Life Solution [A] / [A] 0 = fraction remaining when t = t 1/2 then fraction remaining = _______ Therefore, ln (1/2) = - k · t 1/2 - 0.693 = - k · t 1/2 t 1/2 = 0.693 / k t 1/2 = 0.693 / k So, for sugar, t 1/2 = 0.693 / k = 2100 sec = 35 min Rate = k[sugar] and k = 3.3 x 10 -4 sec -1. What is the half- life of this reaction? NOTE: For a first-order process, the half-life does not depend on [A] 0.

49. 48 © 2009 Brooks/Cole - Cengage Half-Life Solution 2 hr and 20 min = 4 half-lives Half-life Time ElapsedMass Left 1st35 min2.50 g 2nd701.25 g 3rd1050.625 g 4th1400.313 g Rate = k[sugar] and k = 3.3 x 10 -4 sec -1. Half-life is 35 min. Start with 5.00 g sugar. How much is left after 2 hr and 20 min (140 min)?

50. 49 © 2009 Brooks/Cole - Cengage Second Order Rate = -Δ[A]/Δt = k[A] 2 integrated rate law 1/[A] = kt + 1/[A] 0 y= 1/[A] m = k x= tb = 1/[A] 0 A straight line if 1/[A] vs t is plotted Knowing k and [A] 0 you can calculate [A] at any time t

51. 50 © 2009 Brooks/Cole - Cengage Determining rxn order The decomposition of NO 2 at 300°C is described by the equation NO 2 (g) NO (g) + 1/2 O 2 (g) and yields these data: Time (s)[NO 2 ], M 0.00.01000 50.00.00787 100.00.00649 200.00.00481 300.00.00380

52. 51 © 2009 Brooks/Cole - Cengage Graphing ln [NO 2 ] vs. t yields: Time (s)[NO 2 ], Mln [NO 2 ] 0.00.01000-4.610 50.00.00787-4.845 100.00.00649-5.038 200.00.00481-5.337 300.00.00380-5.573 The plot is not a straight line, so the process is not first-order in [A]. Determining rxn order Does not fit:

53. 52 © 2009 Brooks/Cole - Cengage Second-Order Processes A graph of 1/[NO 2 ] vs. t gives this plot. Time (s)[NO 2 ], M1/[NO 2 ] 0.00.01000100 50.00.00787127 100.00.00649154 200.00.00481208 300.00.00380263 This is a straight line. Therefore, the process is second-order in [NO 2 ].

54. 53 © 2009 Brooks/Cole - Cengage Zero Order Rate Law Rate = k[A] 0 = k Rate does not change with concentration. Integrated [A] = -kt + [A] 0 When [A] = [A] 0 /2 then t = t 1/2 t 1/2 = [A] 0 /2k

55. 54 © 2009 Brooks/Cole - Cengage Most often when reaction happens on a surface because the surface area stays constant.Most often when reaction happens on a surface because the surface area stays constant. Also applies to enzyme chemistry.Also applies to enzyme chemistry. Zero Order Rate Law

56. 55 © 2009 Brooks/Cole - Cengage Time ConcentrationConcentration  A]/  t tt k =  A]

57. 56 © 2009 Brooks/Cole - Cengage Rate Laws Summary Zero Order First Order Second Order Rate Law Rate = kRate = k[A]Rate = k[A] 2 Integrated Rate Law [A] = -kt + [A] 0 ln[A] = -kt + ln[A] 0 Plot that produces a straight line [A] versus tln[A] versus t Relationship of rate constant to slope of straight line Slope = -k Slope = k Half-Life

58. 57 © 2009 Brooks/Cole - Cengage Ch. 12.5 - Reaction Mechanisms Ch. 12.5 - Reaction Mechanisms Mechanism: how reactants are converted to products at the molecular level. RATE LAW  MECHANISM experiment  theory

59. 58 © 2009 Brooks/Cole - Cengage Reaction Mechanisms The molecularity of a process tells how many molecules are involved in the process. The rate law for an elementary step is written directly from that step.

60. 59 © 2009 Brooks/Cole - Cengage Reaction Mechanisms If a reaction is an elementary step, then we know its rate law. AllAll of the molecules involved in an elementary step are defined in the rate law. The full molecularity has to be included. The sum of the elementary steps in a mechanism must give the overall balanced equation for the reaction.

61. 60 © 2009 Brooks/Cole - Cengage Molecularity Example Given the reactions and their rate laws below, which is the only one that can be an elementary step? A) 2A  PRate = k[A] B) A + B  PRate = k[A][B] C) A + 2B  PRate = k[A] 2 D) A + B + C  PRate = k[A][C]

62. 61 © 2009 Brooks/Cole - Cengage Mechanisms Most rxns. involve a sequence of elementary steps. 2 I - + H 2 O 2 + 2 H +  I 2 + 2 H 2 O 2 I - + H 2 O 2 + 2 H +  I 2 + 2 H 2 O Rate = k [I - ] [H 2 O 2 ] Rate = k [I - ] [H 2 O 2 ]NOTE 1.Rate law comes from experiment 2.Order and stoichiometric coefficients not necessarily the same! 3.Rate law reflects all chemistry down to and including the slowest step in multistep reaction.

63. 62 © 2009 Brooks/Cole - Cengage Mechanisms Proposed Mechanism Step 1 — slowHOOH + I -  HOI + OH - Step 2 — fastHOI + I -  I 2 + OH - Step 3 — fast2 OH - + 2 H +  2 H 2 O Rate of the reaction controlled by slow step — RATE DETERMINING STEP, rds. RATE DETERMINING STEP, rds. Rate can be no faster than rds! Most rxns. involve a sequence of elementary steps. 2 I - + H 2 O 2 + 2 H +  I 2 + 2 H 2 O 2 I - + H 2 O 2 + 2 H +  I 2 + 2 H 2 O Rate = k [I - ] [H 2 O 2 ] Rate = k [I - ] [H 2 O 2 ]

64. 63 © 2009 Brooks/Cole - Cengage Mechanisms Elementary Step 1 is bimolecular and involves I - and HOOH. Therefore, this predicts the rate law should be Rate  [I - ] [H 2 O 2 ] — as observed!! The species HOI and OH - are reaction intermediates. More on intermediates coming up. 2 I - + H 2 O 2 + 2 H +  I 2 + 2 H 2 O Rate = k [I - ] [H 2 O 2 ] Rate = k [I - ] [H 2 O 2 ] Step 1 — slowHOOH + I -  HOI + OH - Step 2 — fastHOI + I -  I 2 + OH - Step 3 — fast2 OH - + 2 H +  2 H 2 O

65. 64 © 2009 Brooks/Cole - Cengage Rate Laws and Mechanisms NO 2 + CO reaction: Rate = k[NO 2 ] 2 Single step Two possible mechanisms Two steps: step 1 Two steps: step 2

66. 65 © 2009 Brooks/Cole - Cengage Slow Initial Step The rate law for this reaction is found experimentally to be Rate = k [NO 2 ] 2 CO is necessary for this reaction to occur, but the rate of the reaction does not depend on its concentration. This suggests the reaction occurs in two steps. NO 2 (g) + CO (g)  NO (g) + CO 2 (g)

67. 66 © 2009 Brooks/Cole - Cengage Slow Initial Step A proposed mechanism for this reaction is Step 1: NO 2 + NO 2  NO 3 + NO (slow) Step 2: NO 3 + CO  NO 2 + CO 2 (fast) The NO 3 intermediate is consumed in the second step. As CO is not involved in the slow, rate-determining step, it does not appear in the rate law.

68. 67 © 2009 Brooks/Cole - Cengage Fast Initial Step The rate law for this reaction is found (experimentally) to be Because termolecular (aka trimolecular) processes are rare, this rate law suggests a two- step mechanism. 67

69. 68 © 2009 Brooks/Cole - Cengage Fast Initial Step A proposed mechanism is Step 1 is an equilibrium- it includes the forward and reverse reactions.

70. 69 © 2009 Brooks/Cole - Cengage Fast Initial Step The rate of the overall reaction depends upon the rate of the slow step. The rate law for that step would be But how can we find [NOBr 2 ]?

71. 70 © 2009 Brooks/Cole - Cengage Intermediates intermediate[NOBr 2 ] is an intermediate, an unstable molecule of low, unknown concentration Intermediates can’t appear in the rate law. Remember: only reactants (and products, rarely) can appear in a rate law. Slow step determines the rate and the rate law.

72. 71 © 2009 Brooks/Cole - Cengage Intermediates Use the reactions that form them If the reactions are fast and irreversible the concentration of the intermediate is based on stoichiometry. If it is formed by a reversible reaction set the rates equal to each other.

73. 72 © 2009 Brooks/Cole - Cengage Fast Initial Step NOBr 2 can react two ways: –With NO to form NOBr –By decomposition to reform NO and Br 2 The reactants and products of the first step are in equilibrium with each other. Therefore, Rate f = Rate r

74. 73 © 2009 Brooks/Cole - Cengage Fast Initial Step Because Rate f = Rate r, k 1 [NO] [Br 2 ] = k −1 [NOBr 2 ] Solving for [NOBr 2 ] gives us k 1 /k -1 [NO] [Br 2 ] = [NOBr 2 ]

75. 74 © 2009 Brooks/Cole - Cengage Fast Initial Step Substituting this expression for [NOBr 2 ] in the rate law for the rate-determining step gives 74

76. 75 © 2009 Brooks/Cole - Cengage Ch. 12.6 – A Model for Kinetics Molecules need a minimum amount of energy to react. Visualized as an energy barrier - activation energy, E a. Reaction coordinate diagram

77. 76 © 2009 Brooks/Cole - Cengage MECHANISMS & Activation Energy Conversion of cis to trans-2-butene requires twisting around the C=C bond. Rate = k [trans-2-butene]

78. 77 © 2009 Brooks/Cole - Cengage MECHANISMS Activation energy barrier CisTrans Transition state

79. 78 © 2009 Brooks/Cole - Cengage Energy involved in conversion of trans to cis butene trans cis energy Activated Complex +262 kJ -266 kJ MECHANISMS 4 kJ/mol

80. 79 © 2009 Brooks/Cole - Cengage Collision Theory Reactions require (a) activation energy and (b) correct orientation. O 3 (g) + NO(g)  O 2 (g) + NO 2 (g) O 3 (g) + NO(g)  O 2 (g) + NO 2 (g) 2. Activation energy and geometry 1. Activation energy

81. 80 © 2009 Brooks/Cole - Cengage Mechanisms O 3 + NO reaction occurs in a single ELEMENTARY step. Most others involve a sequence of elementary steps. Adding elementary steps gives NET reaction.

82. 81 © 2009 Brooks/Cole - Cengage Mechanisms Reaction passes thru a TRANSITION STATE where there is an activated complex that has sufficient energy to become a product.Reaction passes thru a TRANSITION STATE where there is an activated complex that has sufficient energy to become a product. ACTIVATION ENERGY, E a = energy req’d to form activated complex. Here E a = 262 kJ/mol

83. 82 © 2009 Brooks/Cole - Cengage Also note that trans-butene is MORE STABLE than cis-butene by about 4 kJ/mol. Therefore, cis  trans is EXOTHERMIC This is the connection between thermo- dynamics and kinetics. MECHANISMS

84. 83 © 2009 Brooks/Cole - Cengage Effect of Temperature Reactions generally occur slower at lower T.Reactions generally occur slower at lower T. Iodine clock reaction.  H 2 O 2 + 2 I - + 2 H +  2 H 2 O + I 2 Room temperature In ice at 0 o C

85. 84 © 2009 Brooks/Cole - Cengage Activation Energy and Temperature Reactions are faster at higher T because a larger fraction of reactant molecules have enough energy to convert to product molecules. In general, differences in activation energy cause reactions to vary from fast to slow.

86. 85 © 2009 Brooks/Cole - Cengage Mechanisms 1.Why is trans-butene cis-butene reaction observed to be 1st order? 1.Why is trans-butene  cis-butene reaction observed to be 1st order? As [trans] doubles, number of molecules with enough E also doubles. As [trans] doubles, number of molecules with enough E also doubles. 2.Why is the trans cis reaction faster at higher temperature? 2.Why is the trans  cis reaction faster at higher temperature? Fraction of molecules with sufficient activation energy increases with T. Fraction of molecules with sufficient activation energy increases with T.

87. 86 © 2009 Brooks/Cole - Cengage More About Activation Energy Arrhenius equation — Rate constant Temp (K) 8.31 x 10 -3 kJ/Kmol Activation energy Frequency factor Frequency factor related to frequency of collisions with correct geometry. Plot ln k vs. 1/T straight line. slope = -E a /R Plot ln k vs. 1/T  straight line. slope = -E a /R

88. 87 © 2009 Brooks/Cole - Cengage Arrhenius Equation Example Calculate the Ea from the data below.Calculate the Ea from the data below. T( ⁰C) k(s -1 ) 189.72.52 X 10 -5 198.95.25 X 10 -5 230.36.30 X 10 -4 251.23.16 X 10 -3

89. 88 © 2009 Brooks/Cole - Cengage Arrhenius Equation Example 1 st step  change T(⁰C) to T(K), find 1/T and change k to ln k. T(K)1/Tln(k) 462.90.002160-10.589 472.10.002118-9.855 503.50.001986-7.370 524.40.001907-5.757

90. 89 © 2009 Brooks/Cole - Cengage

91. 90 © 2009 Brooks/Cole - Cengage Arrhenius Equation Example 2 nd Step  find the slope (If not using graphing calculator or excel) Slope = (-5.757 – (-9.855))/(0.001907 – 0.002118) = 4.098/(-0.000211) = -19422 K 3 rd Step  Multiply the slope by -8.314 J/mol K to get E a = -19422 s -1 * -8.314 J/mol K = 161472 J = 161 kJ

92. 91 © 2009 Brooks/Cole - Cengage Determining the Activation Energy If we do not have a lot of data, then we recognize More on the Arrhenius Equation

93. 92 © 2009 Brooks/Cole - Cengage Example Problem The Ea of a certain reaction is 65.7 kJ/mol. How many times faster is the rxn at 50 ⁰ C than at 0 ⁰ C?

94. 93 © 2009 Brooks/Cole - Cengage Ch. 12.7 - CATALYSIS Catalysts speed up reactions by altering the mechanism to lower the activation energy barrier. Dr. James Cusumano, Catalytica Inc. What is a catalyst? Catalysts and society Catalysts and the environment PLAY MOVIE

95. 94 © 2009 Brooks/Cole - Cengage CATALYSIS In auto exhaust systems — Pt, NiO 2 CO + O 2  2 CO 2 2 NO  N 2 + O 2

96. 95 © 2009 Brooks/Cole - Cengage CATALYSIS 2.Polymers: H 2 C=CH 2  polyethylene 3.Acetic acid: CH 3 OH + CO  CH 3 CO 2 H 4. Enzymes — biological catalysts

97. 96 © 2009 Brooks/Cole - Cengage CATALYSIS Catalysis and activation energy Uncatalyzed reaction Catalyzed reaction MnO 2 catalyzes decomposition of H 2 O 2 2 H 2 O 2  2 H 2 O + O 2

98. 97 © 2009 Brooks/Cole - Cengage Iodine-Catalyzed Isomerization of cis-2-Butene See Figure 15.15

99. 98 © 2009 Brooks/Cole - Cengage Iodine-Catalyzed Isomerization of cis-2-Butene

100. 99 © 2009 Brooks/Cole - Cengage AP Exam Practice 2010 AP Exam #3b-d 2009B AP Exam #2 2009 AP exam #3d-e 2008 AP Exam #3d-f 2008B AP Exam #2a-g 2005B AP Exam #32005B AP Exam #3 2005 AP Exam #3 2003 AP Exam #3a-c