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

2. Chemical Kinetics Thermodynamics – Does a reaction take place? (Chapter 18) Kinetics – How fast does a reaction proceed? Concentration Temperature Catalyst GOAL - Understand behavior of reaction and how we can control it Determine Rate Law Determine Reaction Mechanism 13.1

3. 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 Reaction rate units = concentration time concentration usually in units of Molarity or moles time usually in units of seconds or minutes

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

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

6. 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

7. Since the slope is constantly changing, it may be hard to get an accurate measure of rate from a plot (there is error in measuring the slope of experimental data) => Scientists look for ways to plot data so that a straight line is produced Remember the Clausius-Clapeyron Equation - plot ln P vs 1/T NOT P vs T => Different reactions may need to be plotted in different ways

8. rate  [Br 2 ] rate = k [Br 2 ] k = rate [Br 2 ] 13.1 = rate constant = 3.50 x 10 -3 s -1 Look for trends in data as [Br 2 ] is cut in half, so is rate => linear relationship

9. The units of the rate constant (k) depend on the order of the reaction Rate = k [A] x [B] y [C] z or k = Rate. [A] x [B] y [C] z Rate is units of concentration time concentration = M or moles time = seconds or minutes Reaction Order Rate k units 0 k M/s 1 k [A] (M/s)/M = 1/s 2 k [A] 2 (M/s)/M 2 = 1/Ms 3 k [A] 3 (M/s)/M 3 = 1/M 2 s 4 k [A] 4 (M/s)/M 4 = 1/M 3 s Always have time units in denominator

10. 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

11. 2H 2 O 2 (aq) 2H 2 O (l) + O 2 (g) 13.1

12. 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

13. 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

14. 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

15. F 2 (g) + 2ClO 2 (g) 2FClO 2 (g) rate = k [F 2 ] x [ClO 2 ] y design experiments to reveal trends in data 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 ] units of k = 1/Ms 13.2 Initial rate = rate at time = 0

16. 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

17. 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 13.2 rate = k [S 2 O 8 2- ][I - ]

18. 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 Rearrange:  [A] = -k  t [A] Integrate: t = 0 => t and [A] 0 = [A] t [A] t t d[A] = -k dt [A] [A] 0 0 ln [A] t - ln [A] 0 = - k t or ln [A] t = - k t + ln [A] 0 y = mx + b

19. First-Order Reactions 13.3 A product [A] = [A] 0 exp(-kt)ln[A] = ln[A] 0 - kt ln [A] t - ln [A] 0 = - k t or ln [A] t = - k t + ln [A] 0 y = mx + b

20. Decomposition of N 2 O 5 13.3

21. First-Order Reactions A product ln [A] t - ln [A] 0 = - k t or ln [A] t = - k t [A] 0. at t 1/2 [A] t 1/2 = 1/2 [A] 0 half-life or t 1/2 = ln 2 k half-life is independent of initial concentration

22. 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

23. 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 = = 1216 s = 20 minutes How do you know decomposition is first order? units of k (s -1 )

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

26. 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 Rearrange:  [A] = -k  t [A] 2 Integrate: t = 0 => t and [A] 0 => [A] t [A] t t d[A] = - k dt [A] 2 [A] 0 0 1 [A] t = 1 [A] 0 kt + -- y = mx + b 1 = 1 [A] 0 kt ½ + [A] 0 /2 Half-life is inversely proportional to initial concentration Note negative signs

27. 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] t = [A] 0 /2 t ½ = [A] 0 2k2k [A] t = - kt + [A] 0 Rearrange:  [A] = -k  t Integrate: t = 0 => t and [A] 0 => [A] t [A] t t d[A] = - k dt [A] 0 0 [A] t - [A] 0 = - kt y = mx + b [A] 0 /2 = - kt ½ + [A] 0 Half-life is proportional to initial concentration

28. 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