1. © 2009, Prentice-Hall, Inc. Reaction Mechanisms The sequence of events that describes the actual process by which reactants become products is called the reaction mechanism.

2. © 2009, Prentice-Hall, Inc. Reaction Mechanisms Reactions may occur all at once or through several discrete steps. Each of these processes is known as an elementary reaction or elementary process.

3. © 2009, Prentice-Hall, Inc. Reaction Mechanisms The molecularity of a process tells how many molecules are involved in the process.

4. © 2009, Prentice-Hall, Inc. Multistep Mechanisms In a multistep process, one of the steps will be slower than all others. The overall reaction cannot occur faster than this slowest, rate-determining step.

5. © 2009, Prentice-Hall, Inc. Slow Initial Step The rate law for this reaction is found experimentally to be Rate = k [NO 2 ] 2 CO is necessary for this reaction to occur, but the rate of the reaction does not depend on its concentration. This suggests the reaction occurs in two steps. NO 2 (g) + CO (g)  NO (g) + CO 2 (g)

6. © 2009, Prentice-Hall, Inc. Slow Initial Step A proposed mechanism for this reaction is Step 1: NO 2 + NO 2  NO 3 + NO (slow) Step 2: NO 3 + CO  NO 2 + CO 2 (fast) The NO 3 intermediate is consumed in the second step. As CO is not involved in the slow, rate-determining step, it does not appear in the rate law.

7. Sample Exercise 14.12 Determining Molecularity and Identifying Intermediates It has been proposed that the conversion of ozone into O 2 proceeds by a two-step mechanism: (a) Describe the molecularity of each elementary reaction in this mechanism. (b) Write the equation for the overall reaction. (c) Identify the intermediate(s). Solution

8. Sample Exercise 14.12 Determining Molecularity and Identifying Intermediates For the reaction the proposed mechanism is (a) Is the proposed mechanism consistent with the equation for the overall reaction? (b) What is the molecularity of each step of the mechanism? (c) Identify the intermediate(s). Answer: (a) Yes, the two equations add to yield the equation for the reaction. (b) The first elementary reaction is unimolecular, and the second one is bimolecular. (c) Mo(CO) 5 Practice Exercise

9. Sample Exercise 14.13 Predicting Rate Law for an Elementary Reaction If the following reaction occurs in a single elementary reaction, predict its rate law: Solution Comment: Experimental studies of this reaction show that the reaction actually has a very different rate law: Because the experimental rate law differs from the one obtained by assuming a single elementary reaction, we can conclude that the mechanism cannot occur by a single elementary step. It must, therefore, involve two or more elementary steps.

10. Consider the following reaction: 2 NO(g) + Br 2 (g) → 2 NOBr(g). (a) Write the rate law for the reaction, assuming it involves a single elementary reaction. (b) Is a single step mechanism likely for this reaction? Answer: (a) Rate = k[NO] 2 [Br 2 ] (b) No, because termolecular reactions are very rare. Practice Exercise

11. Sample Exercise 14.14 Determining the Rate Law for a Multistep Mechanism The decomposition of nitrous oxide, N 2 O, is believed to occur by a two-step mechanism: (a) Write the equation for the overall reaction. (b) Write the rate law for the overall reaction. Solution Ozone reacts with nitrogen dioxide to produce dinitrogen pentoxide and oxygen: Practice Exercise

12. Sample Exercise 14.14 Determining the Rate Law for a Multistep Mechanism The reaction is believed to occur in two steps: The experimental rate law is rate = k[O 3 ][NO 2 ]. What can you say about the relative rates of the two steps of the mechanism? Answer: Because the rate law conforms to the molecularity of the first step, that must be the rate-determining step. The second step must be much faster than the first one. Practice Exercise (continued)

13. © 2009, Prentice-Hall, Inc. Fast Initial Step The rate law for this reaction is found to be Rate = k [NO] 2 [Br 2 ] Because termolecular processes are rare, this rate law suggests a two-step mechanism. 2 NO (g) + Br 2 (g)  2 NOBr (g)

14. © 2009, Prentice-Hall, Inc. Fast Initial Step A proposed mechanism is Step 2: NOBr 2 + NO  2 NOBr (slow) Step 1 includes the forward and reverse reactions. Step 1: NO + Br 2 NOBr 2 (fast)

15. © 2009, Prentice-Hall, Inc. Fast Initial Step The rate of the overall reaction depends upon the rate of the slow step. The rate law for that step would be Rate = k 2 [NOBr 2 ] [NO] But how can we find [NOBr 2 ]?

16. © 2009, Prentice-Hall, Inc. Fast Initial Step NOBr 2 can react two ways: – With NO to form NOBr – By decomposition to reform NO and Br 2 The reactants and products of the first step are in equilibrium with each other. Therefore, Rate f = Rate r

17. © 2009, Prentice-Hall, Inc. Fast Initial Step Because Rate f = Rate r, k 1 [NO] [Br 2 ] = k −1 [NOBr 2 ] Solving for [NOBr 2 ] gives us k 1 k −1 [NO] [Br 2 ] = [NOBr 2 ]

18. © 2009, Prentice-Hall, Inc. Fast Initial Step Substituting this expression for [NOBr 2 ] in the rate law for the rate-determining step gives k 2 k 1 k −1 Rate =[NO] [Br 2 ] [NO] = k [NO] 2 [Br 2 ]

19. Sample Exercise 14.15 Deriving the Rate Law for a Mechanism with a Fast Initial Step Solution The second step is rate determining, so the overall rate is We solve for the concentration of the intermediate N 2 O 2 by assuming that an equilibrium is established in step 1; thus, he rates of the forward and reverse reactions in step 1 are equal:

20. Sample Exercise 14.15 Deriving the Rate Law for a Mechanism with a Fast Initial Step The first step of a mechanism involving the reaction of bromine is What is the expression relating the concentration of Br(g) to that of Br 2 (g)? Practice Exercise