2. Reaction Mechanism The reaction mechanism is the series of elementary steps by which a chemical reaction occurs.  The sum of the elementary steps must give the overall balanced equation for the reaction  The mechanism must agree with the experimentally determined rate law

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

4. Rate-Determining Step In a multi-step reaction, the slowest step is the rate-determining step. It therefore determines the rate of the reaction. The experimental rate law must agree with the 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. © 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)

8. © 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)

9. © 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 ]?

10. © 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

11. © 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 k1k−1k1k−1 [NO] [Br 2 ] = [NOBr 2 ]

12. © 2009, Prentice-Hall, Inc. 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 ]

13. Identifying the Rate-Determining Step For the reaction: 2H 2 (g) + 2NO(g)  N 2 (g) + 2H 2 O(g) The experimental rate law is: R = k[NO] 2 [H 2 ] Which step in the reaction mechanism is the rate-determining (slowest) step? Step #1 H 2 (g) + 2NO(g)  N 2 O(g) + H 2 O(g) Step #2 N 2 O(g) + H 2 (g)  N 2 (g) + H 2 O(g) Step #1 agrees with the experimental rate law

14. Identifying Intermediates For the reaction: 2H 2 (g) + 2NO(g)  N 2 (g) + 2H 2 O(g) Which species in the reaction mechanism are intermediates (do not show up in the final, balanced equation?) Step #1 H 2 (g) + 2NO(g)  N 2 O(g) + H 2 O(g) Step #2 N 2 O(g) + H 2 (g)  N 2 (g) + H 2 O(g) 2H 2 (g) + 2NO(g)  N 2 (g) + 2H 2 O(g)  N 2 O(g) is an intermediate