1. CHM 112 M. Prushan Chapter 12 Chemical Kinetics

2. CHM 112 M. Prushan Chemical Kinetics Kinetics is the study of how fast chemical reactions occur. There are 4 important factors which affect rates of reactions: –reactant concentration, –temperature, –action of catalysts, and –surface area. Goal: to understand chemical reactions at the molecular level.

3. CHM 112 M. Prushan Chemical Kinetics Reaction Rates Speed of a reaction is measured by the change in concentration with time. For a reaction A  B Suppose A reacts to form B. Let us begin with 1.00 mol A.

4. CHM 112 M. Prushan 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. CHM 112 M. Prushan A B rate = -  [A] tt rate = [B][B] tt time

6. CHM 112 M. Prushan Br 2 (aq) + HCOOH (aq) 2Br - (aq) + 2H + (aq) + CO 2 (g) time 393 nm light Detector 393 nm Br 2 (aq)

7. CHM 112 M. Prushan 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. CHM 112 M. Prushan rate  [Br 2 ] rate = k [Br 2 ] k = rate [Br 2 ] = rate constant = 3.50 x 10 -3 s -1

9. CHM 112 M. Prushan 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

10. CHM 112 M. Prushan 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

11. CHM 112 M. Prushan 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

12. CHM 112 M. Prushan 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 doublesx = 1 Quadruple [ClO 2 ] with [F 2 ] constant Rate quadruplesy = 1 rate = k [F 2 ][ClO 2 ]

13. CHM 112 M. Prushan 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

14. CHM 112 M. Prushan 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) = 0.081/M s rate = k [S 2 O 8 2- ][I - ]

15. CHM 112 M. Prushan 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 exp(-kt) ln[A] = ln[A] 0 - kt

16. CHM 112 M. Prushan 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 =

17. CHM 112 M. Prushan 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 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 = = 1216 s = 20 minutes How do you know decomposition is first order? units of k (s -1 )

18. CHM 112 M. Prushan 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

19. CHM 112 M. Prushan 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

20. CHM 112 M. Prushan 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

21. CHM 112 M. Prushan A + B C + D Exothermic Reaction Endothermic Reaction The activation energy (E a ) is the minimum amount of energy required to initiate a chemical reaction. 13.4

22. CHM 112 M. Prushan 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

23. CHM 112 M. Prushan 13.4 lnk = - EaEa R 1 T + lnA

24. CHM 112 M. Prushan 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 - ln 1.10x10 -5 L/mol*s 9.51x10 -9 L/mol*s 1 600K 1 500K -E a = - (8.314J/mol*K) E a = 1.76x10 5 J/mol = 176kJ/mol

25. CHM 112 M. Prushan 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 +

26. CHM 112 M. Prushan 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

27. CHM 112 M. Prushan 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.

28. CHM 112 M. Prushan 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

29. CHM 112 M. Prushan 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)

30. Reaction energy diagram for the two-step NO 2 -F 2 reaction Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

31. CHM 112 M. Prushan 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 uncatalyzedcatalyzed rate catalyzed > rate uncatalyzed E a < E a ‘

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

33. CHM 112 M. Prushan 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 catalyses Base catalyses

34. CHM 112 M. Prushan N 2 (g) + 3H 2 (g) 2NH 3 (g) Fe/Al 2 O 3 /K 2 O catalyst Haber Process

35. CHM 112 M. Prushan 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)

36. CHM 112 M. Prushan Catalytic Converters CO + Unburned Hydrocarbons + O 2 CO 2 + H 2 O catalytic converter NO + NO 2 N 2 + O 2 catalytic converter

37. CHM 112 Summer 2007 M. Prushan Enzyme Catalysis

38. CHM 112 M. Prushan uncatalyzed enzyme catalyzed