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5.9 Pre-Equilibrium Approximation

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Sometimes a mechanism begins with a fast reversible step that reaches an equilibrium before a slower step occurs. In this case, the first step is treated as a pre-equilibrium, and the slow step happens after that equilibrium is established. This is called the pre-equilibrium approximation. It is useful because it allows you to write an equilibrium expression for an intermediate formed in the fast step, then substitute that expression into the rate law for the slow step. This lets you rewrite the rate law entirely in terms of reactants instead of intermediates.

The main reason this method matters is that intermediates usually should not appear in a final experimentally measurable rate law. If the slow step depends on an intermediate, you can use the equilibrium relationship from the first fast step to eliminate it. So the pre-equilibrium approximation is really a bridge between equilibrium ideas and kinetics ideas. It helps explain more complicated rate laws that cannot be understood from a single slow elementary step alone.

Example 1: Suppose the mechanism is

Step 1 (fast equilibrium):
A+B⇌CA + B \rightleftharpoons C
Step 2 (slow):
C+D→EC + D \rightarrow E

The slow step gives Rate = k2[C][D]{k_2[C][D]}

From the fast equilibrium,
K=[C][A][B]K = \frac{[C]}{[A][B]}

So [C]=K[A][B]{[C] = K[A][B]}

Substitute into the rate law: Rate = k2(K[A][B])[D]{k_2(K[A][B])[D]}
Rate = k[A][B][D]{k[A][B][D]}

where k=k2K{k = k_2K}

Example 2: Suppose the mechanism is

Step 1 (fast equilibrium):
NO+NO⇌N2O2NO + NO \rightleftharpoons N_2O_2
Step 2 (slow):
N2O2+H2→N2O+H2ON_2O_2 + H_2 \rightarrow N_2O + H_2O
The slow step gives Rate =
k2[N2O2][H2]k_2[N_2O_2][H_2]
From the fast equilibrium, K =
[N2O2][NO]2\frac{[N_2O_2]}{[NO]^2}
So
[N2O2]=K[NO]2[N_2O_2] = K[NO]^2
Substitute into the rate law: Rate =
k2(K[NO]2)[H2]k_2(K[NO]^2)[H_2]
Rate =
k[NO]2[H2]k[NO]^2[H_2]

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