Reaction Mechanism The reaction mechanism is the series

Reaction Mechanism The reaction mechanism is the series of elementary steps by which a chemical reaction occurs. ØThe sum of the elementary steps must give the overall balanced equation for the reaction Ø The mechanism must agree with the experimentally

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

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

Slow Initial Step NO 2 (g) + CO (g) NO (g) + CO 2 (g) 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. � © 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. © 2009, Prentice-Hall, Inc.

Fast Initial Step 2 NO (g) + Br 2 (g) 2 NOBr (g) � 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. © 2009, Prentice. Hall, Inc.

Fast Initial Step �A proposed mechanism is Step 1: NO + Br 2 NOBr 2 Step 2: NOBr 2 + NO 2 NOBr (fast) (slow) Step 1 includes the forward and reverse reactions. © 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]? © 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, Ratef = Rater © 2009, Prentice. Hall, Inc.

Fast Initial Step � Because Ratef = Rater , k 1 [NO] [Br 2] = k− 1 [NOBr 2] � Solving for [NOBr 2] gives us k 1 [NO] [Br ] = [NOBr ] 2 2 k− 1 © 2009, Prentice. Hall, Inc.

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

Identifying the Rate-Determining Step For the reaction: 2 H 2(g) + 2 NO(g) N 2(g) + 2 H 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) + 2 NO(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

Identifying Intermediates For the reaction: 2 H 2(g) + 2 NO(g) N 2(g) + 2 H 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) + 2 NO(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) 2 H 2(g) + 2 NO(g) N 2(g) + 2 H 2 O(g) N 2 O(g) is an intermediate