1. Chemical Kinetics Chapter 13

2. 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. 13.1

3. A B 13.1 rate = -  [A] tt rate = [B][B] tt time

4. Br 2 (aq) + HCOOH (aq) 2Br - (aq) + 2H + (aq) + CO 2 (g) time 393 nm light Detector  [Br 2 ]   Absorption 13.1

5. 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 13.1

6. rate  [Br 2 ] rate = k [Br 2 ] k = rate [Br 2 ] 13.1 = rate constant = 3.50 x 10 -3 s -1

7. 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 13.1

9. Reaction Rates and Stoichiometry 13.1 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

10. 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 13.1

11. The Rate Law 13.2 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

12. 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 ] 13.2

13. 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 13.2

14. Examining the effect of doubling the initial concentration gives us the order in that reactant. In this case, when the concentration doubles, the rate doubles, so m = 1. When rate is multiplied bym is ½ 10 21 42

15. 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 13.2 rate = k [S 2 O 8 2- ][I - ] 2 3 1

16. First-Order Reactions 13.3 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 exp(-kt) ln[A] = ln[A] 0 - kt

17. 13.3 2N 2 O 5 4NO 2 (g) + O 2 (g)

18. 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 = 13.3

19. First-Order Reactions 13.3 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 ln2 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 ln2 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 )

20. A product First-order reaction # of half-lives [A] = [A] 0 /n 1 2 3 4 2 4 8 16 13.3

21. Second-Order Reactions 13.3 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

22. Zero-Order Reactions 13.3 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

23. 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 ln2 k = t ½ = [A] 0 2k2k t ½ = 1 k[A] 0 13.3

24. 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?

25. 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

27. 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

28. For a zero-order reaction, a plot of [A]t versus t is linear. The y-intercept is [A]0. For a first-order reaction, a plot of ln[A] t versus t is linear. The graph crosses the origin (b = 0). For a second-order reaction, a plot of 1/[A] t versus t is linear. The y-intercept is 1/[A]0.

29. Exothermic ReactionEndothermic Reaction The activation energy (E a ) is the minimum amount of energy required to initiate a chemical reaction. 13.4 A + B AB C + D + +

30. Temperature Dependence of the Rate Constant k = A exp( -E a / RT ) 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 lnk = - EaEa R 1 T + lnA (Arrhenius equation) 13.4

31. lnk = - EaEa R 1 T + lnA

32. 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.

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

36. 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?

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

38. 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.

39. A more useful form of the Arrhenius equation is

40. 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?

41. 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.

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

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

44. 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

46. 13.4

47. 13.5 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 +

48. 2NO (g) + O 2 (g) 2NO 2 (g) 13.5

49. 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

50. Unimolecular reactionA productsrate = k [A] Bimolecular reactionA + B productsrate = k [A][B] Bimolecular reactionA + A productsrate = k [A] 2 Rate Laws and Elementary Steps 13.5 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.

51. 13.5 Sequence of Steps in Studying a Reaction Mechanism

52. 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 13.5

53. 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

54. 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…

55. 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.

56. A catalyst is a substance that increases the rate of a chemical reaction without itself being consumed. k = A exp( -E a / RT )EaEa k rate catalyzed > rate uncatalyzed E a < E a ‘ 13.6 UncatalyzedCatalyzed

57. 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 13.6

58. N 2 (g) + 3H 2 (g) 2NH 3 (g) Fe/Al 2 O 3 /K 2 O catalyst Haber Process 13.6

59. Ostwald Process Hot Pt wire over NH 3 solution 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) 13.6

60. Catalytic Converters 13.6 CO + Unburned Hydrocarbons + O 2 CO 2 + H 2 O catalytic converter 2NO + 2NO 2 2N 2 + 3O 2 catalytic converter

61. Enzyme Catalysis 13.6

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