1. 1 Chemical Kinetics Chapter 13 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

2. Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 14– 2 Reaction Rates Chemical kinetics is the study of reaction rates, how reaction rates change under varying conditions, and what molecular events occur during the overall reaction. Concentration of reactants. Concentration of a catalyst Temperature at which the reaction occurs. Surface area of a solid reactant or catalyst. –What variables affect reaction rate?

3. Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 14– 3 Definition of Reaction Rate The reaction rate is the increase in molar concentration of a product of a reaction per unit time. –It can also be expressed as the decrease in molar concentration of a reactant per unit time.

4. 4 Chemical Kinetics Thermodynamics – does a reaction take place? Kinetics – how fast does a reaction proceed? Reaction rate is the change in the concentration of a reactant or a product with time (M/s). A B rate = -  [A] tt rate =  [B] tt  [A] = change in concentration of A over time period  t  [B] = change in concentration of B over time period  t Because [A] decreases with time,  [A] is negative.

5. 5 A B rate = -  [A] tt rate = [B][B] tt

6. 6 Br 2 (aq) + HCOOH (aq) 2Br - (aq) + 2H + (aq) + CO 2 (g) time 393 nm light Detector  [Br 2 ]   Absorption red-brown t 1 < t 2 < t 3

7. 7 Br 2 (aq) + HCOOH (aq) 2Br - (aq) + 2H + (aq) + CO 2 (g) average rate = -  [Br 2 ] tt = - [Br 2 ] final – [Br 2 ] initial t final - t initial slope of tangent slope of tangent slope of tangent instantaneous rate = rate for specific instance in time

8. 8 rate  [Br 2 ] rate = k [Br 2 ] k = rate [Br 2 ] = rate constant = 3.50 x 10 -3 s -1

9. 9 2H 2 O 2 (aq) 2H 2 O (l) + O 2 (g) PV = nRT P = RT = [O 2 ]RT n V [O 2 ] = P RT 1 rate =  [O 2 ] tt RT 1 PP tt = measure  P over time

10. 10

11. 11 Reaction Rates and Stoichiometry 2A B Two moles of A disappear for each mole of B that is formed. rate =  [B] tt rate = -  [A] tt 1 2 aA + bB cC + dD rate = -  [A] tt 1 a = -  [B] tt 1 b =  [C] tt 1 c =  [D] tt 1 d

12. 12 Write the rate expression for the following reaction: CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O (g) rate = -  [CH 4 ] tt = -  [O 2 ] tt 1 2 =  [H 2 O] tt 1 2 =  [CO 2 ] tt

13. Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 14– 13 Dependence of Rate on Concentration A rate law is an equation that relates the rate of a reaction to the concentration of reactants (and catalyst) raised to various powers. –The rate constant, k, is a proportionality constant in the relationship between rate and concentrations.

14. 14 The Rate Law The rate law expresses the relationship of the rate of a reaction to the rate constant and the concentrations of the reactants raised to some powers. aA + bB cC + dD Rate = k [A] x [B] y Reaction is xth order in A Reaction is yth order in B Reaction is (x +y)th order overall

15. Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 14– 15 Dependence of Rate on Concentration As a more general example, consider the reaction of substances A and B to give D and E. –You could write the rate law in the form –The exponents m, n, and p are frequently, but not always, integers. They must be determined experimentally and cannot be obtained by simply looking at the balanced equation.

16. Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 14– 16 Dependence of Rate on Concentration Reaction Order –The reaction order with respect to a given reactant species equals the exponent of the concentration of that species in the rate law, as determined experimentally. –The overall order of the reaction equals the sum of the orders of the reacting species in the rate law.

17. Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 14– 17 Dependence of Rate on Concentration Reaction Order –Consider the reaction of nitric oxide with hydrogen according to the following equation. –Thus, the reaction is second order in NO, first order in H 2, and third order overall. –The experimentally determined rate law is

18. Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 14– 18 Dependence of Rate on Concentration Reaction Order –Zero and negative orders are also possible.. –The concentration of a reactant with a zero- order dependence has no effect on the rate of the reaction. –Although reaction orders frequently have whole number values (particularly 1 and 2), they can be fractional.

19. Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 14– 19 Dependence of Rate on Concentration Determining the Rate Law. –If doubling the concentration of a reactant has a doubling effect on the rate, then one would deduce it was a first-order dependence. –If doubling the concentration had a quadrupling effect on the rate, one would deduce it was a second-order dependence. –A doubling of concentration that results in an eight-fold increase in the rate would be a third- order dependence.

20. 20 F 2 (g) + 2ClO 2 (g) 2FClO 2 (g) rate = k [F 2 ] x [ClO 2 ] y Double [F 2 ] with [ClO 2 ] constant Rate doubles x = 1 Quadruple [ClO 2 ] with [F 2 ] constant Rate quadruples y = 1 rate = k [F 2 ][ClO 2 ]

21. 21 F 2 (g) + 2ClO 2 (g) 2FClO 2 (g) rate = k [F 2 ][ClO 2 ] Rate Laws Rate laws are always determined experimentally. Reaction order is always defined in terms of reactant (not product) concentrations. The order of a reactant is not related to the stoichiometric coefficient of the reactant in the balanced chemical equation. 1

22. 22 Determine the rate law and calculate the rate constant for the following reaction from the following data: S 2 O 8 2- (aq) + 3I - (aq) 2SO 4 2- (aq) + I 3 - (aq) Experiment [S 2 O 8 2- ][I - ] Initial Rate (M/s) 10.080.0342.2 x 10 -4 20.080.0171.1 x 10 -4 30.160.0172.2 x 10 -4 rate = k [S 2 O 8 2- ] x [I - ] y Double [I - ], rate doubles (experiment 1 & 2) y = 1 Double [S 2 O 8 2- ], rate doubles (experiment 2 & 3) x = 1 k = rate [S 2 O 8 2- ][I - ] = 2.2 x 10 -4 M/s (0.08 M)(0.034 M) = 0.08/M s rate = k [S 2 O 8 2- ][I - ]

23. 23 First-Order Reactions A product rate = -  [A] tt rate = k [A] k = rate [A] = 1/s or s -1 M/sM/s M =  [A] tt = k [A] - [A] is the concentration of A at any time t [A] 0 is the concentration of A at time t=0 [A] = [A] 0 e −kt ln[A] = ln[A] 0 - kt

24. 24 2N 2 O 5 4NO 2 (g) + O 2 (g) Graphical Determination of k

25. 25 The reaction 2A B is first order in A with a rate constant of 2.8 x 10 -2 s -1 at 80 0 C. How long will it take for A to decrease from 0.88 M to 0.14 M ? ln[A] = ln[A] 0 - kt kt = ln[A] 0 – ln[A] t = ln[A] 0 – ln[A] k = 66 s [A] 0 = 0.88 M [A] = 0.14 M ln [A] 0 [A] k = ln 0.88 M 0.14 M 2.8 x 10 -2 s -1 =

26. 26 First-Order Reactions The half-life, t ½, is the time required for the concentration of a reactant to decrease to half of its initial concentration. t ½ = t when [A] = [A] 0 /2 ln [A] 0 [A] 0 /2 k = t½t½ ln 2 k = 0.693 k = What is the half-life of N 2 O 5 if it decomposes with a rate constant of 5.7 x 10 -4 s -1 ? t½t½ ln 2 k = 0.693 5.7 x 10 -4 s -1 = = 1200 s = 20 minutes How do you know decomposition is first order? units of k (s -1 )

27. 27 A product First-order reaction # of half-lives [A] = [A] 0 /n 1 2 3 4 2 4 8 16

28. 28 Second-Order Reactions A product rate = -  [A] tt rate = k [A] 2 k = rate [A] 2 = 1/M s M/sM/s M2M2 =  [A] tt = k [A] 2 - [A] is the concentration of A at any time t [A] 0 is the concentration of A at time t=0 1 [A] = 1 [A] 0 + kt t ½ = t when [A] = [A] 0 /2 t ½ = 1 k[A] 0

29. 29 Zero-Order Reactions A product rate = -  [A] tt rate = k [A] 0 = k k = rate [A] 0 = M/s  [A] tt = k - [A] is the concentration of A at any time t [A] 0 is the concentration of A at time t = 0 t ½ = t when [A] = [A] 0 /2 t ½ = [A] 0 2k2k [A] = [A] 0 - kt

30. 30 Summary of the Kinetics of Zero-Order, First-Order and Second-Order Reactions OrderRate Law Concentration-Time Equation Half-Life 0 1 2 rate = k rate = k [A] rate = k [A] 2 ln[A] = ln[A] 0 - kt 1 [A] = 1 [A] 0 + kt [A] = [A] 0 - kt t½t½ ln 2 k = t ½ = [A] 0 2k2k t ½ = 1 k[A] 0

31. Ammonium nitrite is unstable because ammonium ion reacts with nitrite ion to produce nitrogen: NH 4 + (aq) + NO 2 - (aq)  N 2 (g) + 2H 2 O(l) In a solution that is 10.0 M in NH 4 +, the reaction is first order in nitrite ion (at low concentrations), and the rate constant at 25°C is 3.0 × 10 3 /s. What is the half-life of the reaction?

32. The rate is first order in nitrite, NO 2 - : k = 3.0 × 10 -3 /s For first-order reactions: kt ½ = 0.693 t½ = 2.3 × 10 2 s = 3.9 min

34. Let’s look again at the integrated rate laws: In each case, the rate law is in the form of y = mx + b, allowing us to use the slope and intercept to find the values. y = mx + b

35. 35 Exothermic ReactionEndothermic Reaction The activation energy (E a ) is the minimum amount of energy required to initiate a chemical reaction. A + B AB C + D + +

36. 36 Temperature Dependence of the Rate Constant E a is the activation energy (J/mol) R is the gas constant (8.314 J/Kmol) T is the absolute temperature A is the frequency factor ln k = - EaEa R 1 T + lnA (Arrhenius equation) Alternate format:

37. 13.4 lnk = - EaEa R 1 T + lnA

38. The potential energy diagram for a reaction visually illustrates the changes in energy that occur. The next slide shows the diagram for the reaction NO + Cl 2  NOCl 2 ‡  NOCl + Cl where NOCl 2 ‡ indicates the activated complex.

40. The reaction of NO with Cl 2 is an endothermic reaction. The next slide shows the curve for a generic exothermic reaction.

42. Sketch a potential energy diagram for the decomposition of nitrous oxide. N 2 O(g)  N 2 (g) + O(g) The activation energy for the forward reaction is 251 kJ;  H o = +167 kJ. Label your diagram appropriately. What is the reverse activation energy?

43. Reactants Products Ea= 251 kJ  H = 167 kJ Ea(reverse) = 84 kJ EnergypermolEnergypermol Progress of reaction N2O N2 + O

44. Rate constants for most chemical reactions closely follow an equation in the form This is called the Arrhenius equation. A is the frequency factor, a constant.

45. A more useful form of the Arrhenius equation is

46. In a series of experiments on the decomposition of dinitrogen pentoxide, N 2 O 5, rate constants were determined at two different temperatures: At 35°C, the rate constant was 1.4 × 10 -4 /s. At 45°C, the rate constant was 5.0 × 10 - 4 /s. What is the activation energy? What is the value of the rate constant at 55°C?

47. This is actually two problems. First, we will use the Arrhenius equation to find E a. Then, we will use the Arrhenius equation with E a to find the rate constant at a new temperature.

48. T 1 = 35°C = 308 KT 2 = 45°C = 318 K k 1 = 1.4 × 10 -4 /sk 2 = 5.0 × 10 -4 /s

49. We will solve the left side and rearrange the right side.

50. T 1 = 35°C = 308 KT 2 = 55°C = 328 K k 1 = 1.4 × 10 -4 /sk 2 = ? E a = 1.04 × 10 5 J/mol

52. 13.4

53. 53 Reaction Mechanisms The overall progress of a chemical reaction can be represented at the molecular level by a series of simple elementary steps or elementary reactions. The sequence of elementary steps that leads to product formation is the reaction mechanism. 2NO (g) + O 2 (g) 2NO 2 (g) N 2 O 2 is detected during the reaction! Elementary step:NO + NO N 2 O 2 Elementary step:N 2 O 2 + O 2 2NO 2 Overall reaction:2NO + O 2 2NO 2 +

54. 54 2NO (g) + O 2 (g) 2NO 2 (g) Mechanism:

55. 55 Elementary step:NO + NO N 2 O 2 Elementary step:N 2 O 2 + O 2 2NO 2 Overall reaction:2NO + O 2 2NO 2 + Intermediates are species that appear in a reaction mechanism but not in the overall balanced equation. An intermediate is always formed in an early elementary step and consumed in a later elementary step. The molecularity of a reaction is the number of molecules reacting in an elementary step. Unimolecular reaction – elementary step with 1 molecule Bimolecular reaction – elementary step with 2 molecules Termolecular reaction – elementary step with 3 molecules

56. 56 Unimolecular reactionA productsrate = k [A] Bimolecular reactionA + B productsrate = k [A][B] Bimolecular reactionA + A productsrate = k [A] 2 Rate Laws and Elementary Steps Writing plausible reaction mechanisms: The sum of the elementary steps must give the overall balanced equation for the reaction. The rate-determining step should predict the same rate law that is determined experimentally. The rate-determining step is the slowest step in the sequence of steps leading to product formation.

57. 57 Sequence of Steps in Studying a Reaction Mechanism

58. 58 The experimental rate law for the reaction between NO 2 and CO to produce NO and CO 2 is rate = k[NO 2 ] 2. The reaction is believed to occur via two steps: Step 1:NO 2 + NO 2 NO + NO 3 Step 2:NO 3 + CO NO 2 + CO 2 What is the equation for the overall reaction? NO 2 + CO NO + CO 2 What is the intermediate? NO 3 What can you say about the relative rates of steps 1 and 2? rate = k[NO 2 ] 2 is the rate law for step 1 so step 1 must be slower than step 2

59. Mechanism with an initial slow step: NO 2 (g) + CO(g)  NO(g) + CO 2 (g) It is found experimentally that Rate Law = k[NO 2 ] 2 We can propose a mechanism consistent with this rate law: Step 1: NO 2 (g) + NO 2 (g)  NO 3 (g) + NO(g); k 1 (slow) Step 2: NO 3 (g) + CO(g)  NO 2 (g) + CO 2 (g); k 2 (fast) Overall: NO2(g) + CO(g)  NO(g) + CO2(g) Step 1, the slowest bimolecular step, is the rate-determining step. Thus the rate law is : Rate = k[NO 2 ] 2

60. Mechanism with an initial fast step: 2NO(g) + Br 2 (g)  2NOBr(g) The experimental rate law is Rate = k[NO] 2 [Br 2 ] Termolecular processes are so rare, we need to consider another pathway that does not involve termolecular steps: The rate of the overall reaction is governed by step 2: Rate = k 2 [NOBr 2 ][NO] By definition of equilibrium: k 1 [NO][Br 2 ] = k -1 [NOBr 2 ] Therefore, the overall rate law becomes…

61. Note the final rate law is consistent with the experimentally observed rate law: Rate = k[NO] 2 [Br 2 ] In general, whenever a fast step precedes a slow one, we can solve for the concentration of an intermediate by assuming that an equilibrium is established in the fast step.

62. 62 A catalyst is a substance that increases the rate of a chemical reaction without itself being consumed. EaEa k rate catalyzed > rate uncatalyzed E a < E a ′ UncatalyzedCatalyzed

63. 63 In heterogeneous catalysis, the reactants and the catalysts are in different phases. In homogeneous catalysis, the reactants and the catalysts are dispersed in a single phase, usually liquid. Haber synthesis of ammonia Ostwald process for the production of nitric acid Catalytic converters Acid catalysis Base catalysis

64. 64 N 2 (g) + 3H 2 (g) 2NH 3 (g) Fe/Al 2 O 3 /K 2 O catalyst Haber Process

65. 65 Ostwald Process Pt-Rh catalysts used in Ostwald process 4NH 3 (g) + 5O 2 (g) 4NO (g) + 6H 2 O (g) Pt catalyst 2NO (g) + O 2 (g) 2NO 2 (g) 2NO 2 (g) + H 2 O (l) HNO 2 (aq) + HNO 3 (aq)

66. 66 Catalytic Converters CO + Unburned Hydrocarbons + O 2 CO 2 + H 2 O catalytic converter 2NO + 2NO 2 2N 2 + 3O 2 catalytic converter

67. 67 Enzyme Catalysis

68. uncatalyzed enzyme catalyzed 13.6 rate =  [P] tt rate = k [ES]