1. 16-1 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Chapter 14 Kinetics: Rates and Mechanisms of Chemical Reactions

2. 16-2 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 16.1 Reaction rate: the central focus of chemical kinetics

3. 16-3 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 16.2 The wide range of reaction rates.

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. Five factors can be controlled during the reaction: 1.Concentration - molecules must collide to react; 2.Physical state - molecules must mix to collide; 3.Surface area – if reactants are in different phases, reaction occurs where these phases meet 4. Temperature - molecules must collide with enough energy to react; 5. The use of a catalyst.

5. 16-5 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 16.3 The effect of surface area on reaction rate.

6. 16-6 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 16.4 Collision energy and reaction rate.

7. 16-7 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

8. 16-8 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

9. 16-9 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 16.5 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 =

10. 16-10 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 16.6 Plots of [C 2 H 4 ] and [O 2 ] vs. time. Tools of the Laboratory

11. 16-11 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 )

12. 16-12 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).

13. 16-13 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.

14. 16-14 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

15. 16-15 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 )

16. 16-16 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

17. 16-17 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

18. 16-18 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 overall rxn order for 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)

19. 16-19 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.