1. Reaction Mechanisms Chapter 12, Section 6

2. Reaction Mechanisms The sequence of events that describes the actual process by which reactants become products is called the reaction mechanism. Proposed Mechanism: Step 1: NO 2 + NO 2  NO 3 + NO Step 2: NO 3 + CO  NO 2 + CO 2 NO 2 + CO  NO + CO 2 Rate = k[NO 2 ] 2

3. 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.

4. Reaction Mechanisms The molecularity of a process tells how many molecules are involved in the process. Termolecular elementary reactions seldom occur.

5. 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.

6. Reaction Mechanisms must satisfy 2 requirements: 1.The sum of the elementary steps must give the overall balanced equation. 2.The mechanism must agree with the experimentally determined rate law.

7. 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)

8. 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) Does it make sense? What is the intermediate in this reaction?

9. 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)

10. 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)

11. 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 ]?

12. 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

13. Fast Initial Step Because Rate f = Rate r, k 1 [NO] [Br 2 ] = k −1 [NOBr 2 ] Solving for [NOBr 2 ] gives us k1k−1k1k−1 [NO] [Br 2 ] = [NOBr 2 ]

14. Fast Initial Step Substituting this expression for [NOBr 2 ] in the rate law for the rate-determining step gives k2k1k−1k2k1k−1 Rate =[NO] [Br 2 ] [NO] = k [NO] 2 [Br 2 ]

15. Catalysts A catalyst speeds up a reaction without being consumed. Identify the catalyst in the reaction mechanism below: Cl (g) + O 3(g)  ClO (g) + O 2(g) O (g) + ClO (g)  Cl (g) + O 2(g) How is it different than an intermediate?