AP Chemistry / Unit 5: Kinetics
5.9 Pre-Equilibrium Approximation
Notes
When a fast reversible step comes before the slow step, assume the fast step is at equilibrium and substitute to remove intermediates from the rate law.
Practice
Fast: $2NO\rightleftharpoons N_2O_2$. Slow: $N_2O_2+O_2\to2NO_2$. The rate law:
$[N_2O_2]\propto[NO]^2$.
Why can't intermediates appear in the final rate law?
Express everything in reactants.
The pre-equilibrium approach assumes the fast step:
Forward and reverse rates are equal.
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