01Substituting intermediates
When a fast reversible step precedes the slow step, assume it reaches equilibrium: k_f[A][B] = k_r[I]. Solve for [I] and substitute.
This produces rate laws with orders that look strange but match experiment.
02Notebook box
[I] = (k_f/k_r)[A][B] → plug into slow step.
Worked example: Fast 2NO ⇌ N₂O₂; slow N₂O₂ + O₂ → 2NO₂ → rate = k[NO]²[O₂].