5.9 Pre-Equilibrium Approximation

Suppose the slow step is step 2, and it uses an intermediate made in step 1. Writing the rate law from step 2 would put the intermediate's concentration in it. But an intermediate exists only briefly and its concentration isn't something you control or measure easily, so an experimental rate law is always written in terms of reactants (and sometimes catalysts or products), never intermediates. You need a way to replace it.

If step 1 is fast and reversible (shown with ⇌), it goes back and forth many times while the slow step is barely getting started. So step 1 is approximately at equilibrium the whole time: its forward rate equals its reverse rate.

That gives you an equation to solve for the intermediate.

• Write the rate law for the slow step. It contains the intermediate.
• For the fast equilibrium step, set forward rate = reverse rate, using k₁ for the forward rate constant and k₋₁ for the reverse.
• Solve that equation for the intermediate's concentration.
• Substitute it into the slow step's rate law.
• Combine all the constants into one observed rate constant, k.

For A + B ⇌ I (fast), then I + C → products (slow): the slow step gives rate = k₂[I][C]. The fast equilibrium gives k₁[A][B] = k₋₁[I], so [I] = (k₁/k₋₁)[A][B]. Substituting: rate = (k₂k₁/k₋₁)[A][B][C] = k[A][B][C].

Notice the result. Here the rate law contains every reactant that enters at or before the slow step, each raised to the number of times it enters. That's a handy check on your algebra, but it only works when the fast step makes nothing except the intermediate. If the fast step also makes a product, that product ends up on the bottom of the rate law. For O₃ ⇌ O₂ + O (fast), then O + O₃ → 2O₂ (slow), [O] = (k₁/k₋₁)[O₃]/[O₂], so rate = k[O₃]²/[O₂].

This is how a third-order rate law can come from a mechanism with only bimolecular steps, avoiding an unlikely three-particle collision.

Use this approach only when the first step isn't rate limiting: a fast, reversible first step followed by a slow step. When the first step is slow, its rate law is the answer (topic 5.8), and no approximation is needed.

The first step is fast in both directions, while the second step is slow. So most of the intermediate that forms simply falls back apart into reactants long before it has a chance to go on through the slow step. Only a tiny share is drawn off by step 2, so the forward and reverse rates of step 1 stay nearly equal.

The ratio k₁/k₋₁ is the equilibrium constant for step 1. That's why the observed rate constant, k = k₂(k₁/k₋₁), combines a rate constant with an equilibrium constant, a link you'll see again in Unit 7.