1. 16-1 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Chapter 16 Kinetics: Rates and Mechanisms of Chemical Reactions

2. 16-2 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Kinetics: Rates and Mechanisms of Chemical Reactions 16.1 Factors That Influence Reaction Rates 16.2 Expressing the Reaction Rate 16.3 The Rate Law and Its Components 16.4 Integrated Rate Laws: Concentration Changes over Time 16.7 Reaction Mechanisms: Steps in the Overall Reaction 16.8 Catalysis: Speeding Up a Chemical Reaction 16.5 The Effect of Temperature on Reaction Rate 16.6 Explaining the Effects of Concentration and Temperature

3. 16-3 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 16.1 Reaction rate: the central focus of chemical kinetics.

4. 16-4 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Factors That Influence Reaction Rate Under a specific set of conditions, every reaction has its own characteristic rate, which depends upon the chemical nature of the reactants. Four factors can be controlled during the reaction: 1.Concentration - molecules must collide to react; 2.Physical state - molecules must mix to collide; 3.Temperature - molecules must collide with enough energy to react; 4.The use of a catalyst.

5. 16-5 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 16.2 Collision energy and reaction rate.

6. 16-6 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Expressing the Reaction Rate reaction rate - changes in the concentrations of reactants or products per unit time reactant concentrations decrease while product concentrations increase rate of reaction = - for A B change in concentration of A change in time = - conc A 2 -conc A 1 t 2 -t 1  (conc A) - tt

7. 16-7 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Table 16.1 Concentration of O 3 at Various Time in its Reaction with C 2 H 4 at 303K Time (s) Concentration of O 3 (mol/L) 0.0 20.0 30.0 40.0 50.0 60.0 10.0 3.20x10 -5 2.42x10 -5 1.95x10 -5 1.63x10 -5 1.40x10 -5 1.23x10 -5 1.10x10 -5  (conc A) - tt

8. 16-8 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 16.3 The concentrations of O 3 vs. time during its reaction with C 2 H 4 -  [C 2 H 4 ] tt rate = -  [O 3 ] tt =

9. 16-9 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 16.4 Plots of [C 2 H 4 ] and [O 2 ] vs. time.

10. 16-10 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. In general, for the reaction aA + bB cC + dD rate = 1 a -= -  [A] tt 1 b  [B] tt 1 c  [C] tt = + 1 d  [D] tt = + The numerical value of the rate depends upon the substance that serves as the reference. The rest is relative to the balanced chemical equation.

11. 16-11 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Sample Problem 16.1 PLAN: SOLUTION: Expressing Rate in Terms of Changes in Concentration with Time PROBLEM:Because it has a nonpolluting product (water vapor), hydrogen gas is used for fuel aboard the space shuttle and may be used by Earth-bound engines in the near future. 2H 2 ( g ) + O 2 ( g ) 2H 2 O( g ) (a) Express the rate in terms of changes in [H 2 ], [O 2 ], and [H 2 O] with time. (b) When [O 2 ] is decreasing at 0.23 mol/L*s, at what rate is [H 2 O] increasing? Choose [O 2 ] as a point of reference since its coefficient is 1. For every molecule of O 2 which disappears, 2 molecules of H 2 disappear and 2 molecules of H 2 O appear, so [O 2 ] is disappearing at half the rate of change of H 2 and H 2 O. - 1 2  [H 2 ] tt = -  [O 2 ] tt = +  [H 2 O] tt 1 2 0.23mol/L*s = +  [H 2 O] tt 1 2 ; = 0.46mol/L*s  [H 2 O] tt rate = (a)  [O 2 ] tt -= -(b)

12. 16-12 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Sample Problem 16.2 SOLUTION: Determining Reaction Order from Rate Laws PROBLEM:For each of the following reactions, determine the reaction order with respect to each reactant and the overall order from the given rate law. (a) 2NO( g ) + O 2 ( g ) 2NO 2 ( g ); rate = k[NO] 2 [O 2 ] (b) CH 3 CHO( g ) CH 4 ( g ) + CO( g ); rate = k[CH 3 CHO] 3/2 (c) H 2 O 2 ( aq ) + 3 I - ( aq ) + 2H + ( aq ) I 3 - ( aq ) + 2H 2 O( l ); rate = k[H 2 O 2 ][ I - ] PLAN:Look at the rate law and not the coefficients of the chemical reaction. (a) The reaction is 2nd order in NO, 1st order in O 2, and 3rd order overall. (b) The reaction is 3/2 order in CH 3 CHO and 3/2 order overall. (c) The reaction is 1st order in H 2 O 2, 1st order in I - and zero order in H +, while being 2nd order overall.

13. 16-13 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Table 16.2 Initial Rates for a Series of Experiments in the Reaction Between O 2 and NO Experiment Initial Reactant Concentrations (mol/L) Initial Rate (mol/L*s) 1 2 3 4 5 O2O2 NO 1.10x10 -2 1.30x10 -2 3.21x10 -3 1.10x10 -2 3.90x10 -2 28.8x10 -3 2.20x10 -2 1.10x10 -2 3.30x10 -2 1.30x10 -2 2.60x10 -2 1.30x10 -2 6.40x10 -3 12.8x10 -3 9.60x10 -3 2NO( g ) + O 2 ( g ) 2NO 2 ( g )

14. 16-14 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Determining Reaction Orders Using initial rates - Run a series of experiments, each of which starts with a different set of reactant concentrations, and from each obtain an initial rate. See Table 16.2 for data on the reaction O 2 ( g ) + 2NO( g ) 2NO 2 ( g )rate = k [O 2 ] m [NO] n Compare 2 experiments in which the concentration of one reactant varies and the concentration of the other reactant(s) remains constant. k [O 2 ] 2 m [NO] 2 n k [O 2 ] 1 m [NO] 1 n = rate 2 rate 1 = [O 2 ] 2 m [O 2 ] 1 m = 6.40x10 -3 mol/L*s 3.21x10 -3 mol/L*s [O 2 ] 2 [O 2 ] 1 m = 1.10x10 -2 mol/L 2.20x10 -2 mol/L m ;2 = 2 m m = 1 Do a similar calculation for the other reactant(s).

15. 16-15 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Sample Problem 16.3 PLAN: SOLUTION: Determining Reaction Order from Initial Rate Data PROBLEM:Many gaseous reactions occur in a car engine and exhaust system. One of these is rate = k[NO 2 ] m [CO] n Use the following data to determine the individual and overall reaction orders. ExperimentInitial Rate(mol/L*s)Initial [NO 2 ] (mol/L)Initial [CO] (mol/L) 1 2 3 0.0050 0.080 0.0050 0.10 0.40 0.10 0.20 Solve for each reactant using the general rate law using the method described previously. rate = k [NO 2 ] m [CO] n First, choose two experiments in which [CO] remains constant and the [NO 2 ] varies.

16. 16-16 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Sample Problem 16.3Determining Reaction Order from Initial Rate Data continued 0.080 0.0050 rate 2 rate 1 [NO 2 ] 2 [NO 2 ] 1 m = k [NO 2 ] m 2 [CO] n 2 k [NO 2 ] m 1 [CO] n 1 = 0.40 0.10 = m ;16 = 4 m and m = 2 k [NO 2 ] m 3 [CO] n 3 k [NO 2 ] m 1 [CO] n 1 [CO] 3 [CO] 1 n = rate 3 rate 1 = 0.0050 = 0.20 0.10 n ;1 = 2 n and n = 0 The reaction is 2nd order in NO 2. The reaction is zero order in CO. rate = k [NO 2 ] 2 [CO] 0 = k [NO 2 ] 2

17. 16-17 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Integrated Rate Laws rate = -  [A] tt = k [A]rate = -  [A] tt = k [A] 0 rate = -  [A] tt = k [A] 2 first order rate equation second order rate equation zero order rate equation ln [A] 0 [A] t = - ktln [A] 0 = -kt + ln [A] t 1 [A] t 1 [A] 0 - = kt 1 [A] t 1 [A] 0 += kt [A] t - [A] 0 = - kt

18. 16-18 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Table 16.3 Units of the Rate Constant k for Several Overall Reaction Orders Overall Reaction OrderUnits of k (t in seconds) 0mol/L*s (or mol L -1 s -1 ) 11/s (or s -1 ) 2L/mol*s (or L mol -1 s -1 ) 3L 2 / mol 2 *s (or L 2 mol -2 s -1 )

19. 16-19 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Sample Problem 16.4 PLAN: SOLUTION: Determining Reaction Concentration at a Given Time PROBLEM:At 1000 0 C, cyclobutane (C 4 H 8 ) decomposes in a first-order reaction, with the very high rate constant of 87s -1, to two molecules of ethylene (C 2 H 4 ). (a) If the initial C 4 H 8 concentration is 2.00M, what is the concentration after 0.010 s? (b) What fraction of C 4 H 8 has decomposed in this time? Find the [C 4 H 8 ] at time, t, using the integrated rate law for a 1st order reaction. Once that value is found, divide the amount decomposed by the initial concentration. ; ln 2.00 [C 4 H 8 ] = (87s -1 )(0.010s) [C 4 H 8 ] = 0.83mol/L ln [C 4 H 8 ] 0 [C 4 H 8 ] t = kt (a) (b)[C 4 H 8 ] 0 - [C 4 H 8 ] t [C 4 H 8 ] 0 = 2.00M - 0.87M 2.00M = 0.58

20. 16-20 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 16.5 Integrated rate laws and reaction order. ln[A] t = -kt + ln[A] 0 1/[A] t = kt + 1/[A] 0 [A] t = -kt + [A] 0

21. 16-21 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 16.8 Graphical determination of the reaction order for the decomposition of N 2 O 5.

22. 16-22 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 16.9 A plot of [N 2 O 5 ] vs. time for three half-lives. t 1/2 = for a first-order process ln 2 k 0.693 k =

23. 16-23 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Sample Problem 16.5 PLAN: SOLUTION: Determining the Half-Life of a First-Order Reaction PROBLEM:Cyclopropane is the smallest cyclic hydrocarbon. Because its 60 0 bond angles allow poor orbital overlap, its bonds are weak. As a result, it is thermally unstable and rearranges to propene at 1000 0 C via the following first-order reaction: The rate constant is 9.2s -1, (a) What is the half-life of the reaction? (b) How long does it take for the concentration of cyclopropane to reach one-quarter of the initial value? Use the half-life equation, t 1/2 = 0.693 k, to find the half-life. One-quarter of the initial value means two half-lives have passed. t 1/2 = 0.693/9.2s -1 = 0.075s(a)2 t 1/2 = 2(0.075s) = 0.150s(b)

24. 16-24 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Table 16.4 An Overview of Zero-Order, First-Order, and Simple Second-Order Reactions Zero OrderFirst OrderSecond Order Plot for straight line Slope, y-intercept Half-life Rate law rate = k rate = k [A]rate = k [A] 2 Units for k mol/L*s 1/sL/mol*s Integrated rate law in straight-line form [A] t = k t + [A] 0 ln[A] t = -k t + ln[A] 0 1/[A] t = k t + 1/[A] 0 [A] t vs. t ln[A] t vs. t1/[A] t = t k, [A] 0 -k, ln[A] 0 k, 1/[A] 0 [A] 0 /2kln 2/k1/k [A] 0

25. 16-25 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 16.10 Dependence of the rate constant on temperature

26. 16-26 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. The Arrhenius Equation The Effect of Temperature on Reaction Rate ln k = ln A - E a /RT ln k2k2 k1k1 = EaEa RT - 1 T2T2 1 T1T1 - where k is the kinetic rate constant at T E a is the activation energy R is the energy gas constant T is the Kelvin temperature A is the collision frequency factor

27. 16-27 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 16.11 Graphical determination of the activation energy ln k = -E a /R (1/T) + ln A

28. 16-28 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Sample Problem 16.6 PLAN: SOLUTION: Determining the Energy of Activation PROBLEM:The decomposition of hydrogen iodide, 2H I ( g ) H 2 ( g ) + I 2 ( g ) has rate constants of 9.51x10 -9 L/mol*s at 500. K and 1.10x10 -5 L/mol*s at 600. K. Find E a. Use the modification of the Arrhenius equation to find E a. ln k2k2 k1k1 = EaEa - R 1 T2T2 1 T1T1 - E a = - Rln k2k2 k1k1 1 T2T2 1 T1T1 - 1 600K 1 500K - ln 1.10x10 -5 L/mol*s 9..51x10 -9 L/mol*s E a = - (8.314J/mol*K) E a = 1.76x10 5 J/mol = 176 kJ/mol

29. 16-29 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 16.12 Information sequence to determine the kinetic parameters of a reaction. Series of plots of concentra- tion vs. time Initial rates Reaction orders Rate constant (k) and actual rate law Integrated rate law (half-life, t 1/2 ) Rate constant and reaction order Activation energy, E a Plots of concentration vs. time Find k at varied T Determine slope of tangent at t 0 for each plot Compare initial rates when [A] changes and [B] is held constant and vice versa Substitute initial rates, orders, and concentrations into general rate law: rate = k [A] m [B] n Use direct, ln or inverse plot to find order Rearrange to linear form and graph Find k at varied T

30. 16-30 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 16.13 The dependence of possible collisions on the product of reactant concentrations. AA AA BB BB AA AA BB BB AA 4 collisions Add another molecule of A 6 collisions Add another molecule of B AA AA BB BB AABB

31. 16-31 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Table 16.5 The Effect of E a and T on the Fraction (f) of Collisions with Sufficient Energy to Allow Reaction E a (kJ/mol)f (at T = 298 K) 501.70x10 -9 757.03x10 -14 1002.90x10 -18 Tf (at E a = 50 kJ/mol) 25 0 C(298K)1.70x10 -9 35 0 C(308K)3.29x10 -9 45 0 C(318K)6.12x10 -9

32. 16-32 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 16.14 The effect of temperature on the distribution of collision energies

33. 16-33 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 16.15 Energy-level diagram for a reaction REACTANTS PRODUCTS ACTIVATED STATE Collision Energy E a (forward) E a (reverse) The forward reaction is exothermic because the reactants have more energy than the products.

34. 16-34 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 16.16 An energy-level diagram of the fraction of collisions exceeding E a.

35. 16-35 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 16.17 The importance of molecular orientation to an effective collision. NO + NO 3 2 NO 2 A is the frequency factor A = pZ where Z is the collision frequency p is the orientation probability factor

36. 16-36 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 16.18 Nature of the transition state in the reaction between CH 3 Br and OH -. CH 3 Br + OH - CH 3 OH + Br - transition state or activated complex

37. 16-37 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 16.19 Reaction energy diagram for the reaction of CH 3 Br and OH -.

38. 16-38 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 16.20 Reaction energy diagrams and possible transition states.

39. 16-39 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Sample Problem 16.7 SOLUTION: Drawing Reaction Energy Diagrams and Transition States PROBLEM:A key reaction in the upper atmosphere is O 3 ( g ) + O( g ) 2O 2 ( g ) The E a(fwd) is 19 kJ, and the  H rxn for the reaction is -392 kJ. Draw a reaction energy diagram for this reaction, postulate a transition state, and calculate E a(rev). PLAN:Consider the relationships among the reactants, products and transition state. The reactants are at a higher energy level than the products and the transition state is slightly higher than the reactants. E a = 19kJ  H rxn = -392kJ E a(rev) = (392 + 19)kJ = 411kJ transition state

40. 16-40 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Table 16.6 Rate Laws for General Elementary Steps Elementary StepMolecularityRate Law A product 2A product A + B product 2A + B product Unimolecular Bimolecular Termolecular Rate = k [A] Rate = k [A] 2 Rate = k [A][B] Rate = k [A] 2 [B] REACTION MECHANISMS

41. 16-41 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Sample Problem 16.8 PLAN: SOLUTION: Determining Molecularity and Rate Laws for Elementary Steps PROBLEM:The following two reactions are proposed as elementary steps in the mechanism of an overall reaction: (1)NO 2 Cl( g )NO 2 ( g ) + Cl ( g ) (2)NO 2 Cl( g ) + Cl ( g )NO 2 ( g ) + Cl 2 ( g ) (a) Write the overall balanced equation. (b) Determine the molecularity of each step. (a) The overall equation is the sum of the steps. (b) The molecularity is the sum of the reactant particles in the step. 2NO 2 Cl( g )2NO 2 ( g ) + Cl 2 ( g ) (c) Write the rate law for each step. rate 2 = k 2 [NO 2 Cl][Cl] (1)NO 2 Cl( g )NO 2 ( g ) + Cl ( g ) (2)NO 2 Cl( g ) + Cl ( g )NO 2 ( g ) + Cl 2 ( g ) (a) Step(1) is unimolecular. Step(2) is bimolecular. (b) rate 1 = k 1 [NO 2 Cl](c)

42. 16-42 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. The Rate-Determining Step of a Reaction Mechanism The overall rate of a reaction is related to the rate of the slowest, or rate-determining step. Correlating the Mechanism with the Rate Law The elementary steps must add up to the overall equation. The elementary steps must be physically reasonable. The mechanism must correlated with the rate law.

43. 16-43 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 16.21 Reaction energy diagram for the two-step reaction of NO 2 and F 2.

44. 16-44 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. CATALYSTS Each catalyst has its own specific way of functioning. In general a catalyst lowers the energy of activation. Lowering the E a increases the rate constant, k, and thereby increases the rate of the reaction A catalyst increases the rate of the forward AND the reverse reactions. A catalyzed reaction yields the products more quickly, but does not yield more product than the uncatalyzed reaction. A catalyst lowers E a by providing a different mechanism, for the reaction through a new, lower energy pathway.

45. 16-45 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 16.22 Reaction energy diagram of a catalyzed and an uncatalyzed process.

46. 16-46 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 16.23 Mechanism for the catalyzed hydrolysis of an organic ester.

47. 16-47 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

48. 16-48 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 16.24 The metal-catalyzed hydrogenation of ethylene

49. 16-49 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 16.24

50. 16-50 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Tools of the Laboratory Figure B16.1 Spectrophotometric monitoring of a reaction.

51. 16-51 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure B16.2 Figure B16.3 Conductometric monitoring of a reaction Manometric monitoring of a reaction Tools of the Laboratory

52. 16-52 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure B16.4 The widely separated amino acid groups that form an active site.

53. 16-53 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure B16.5 Two models of enzyme action

54. 16-54 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure B16.6 The increasing size of the Antarctic ozone hole O 2 2O O + O 3 2O 2 (ozone breakdown) O + O 2 O 3 (ozone formation)