5.9 Pre-Equilibrium

Pre-Equilibrium Approximation

Pick a mechanism whose slow step is not first and watch the intermediate get substituted out via K₁. Each of 5 reactions derives its observed rate law in 4 lines.

Fast Pre-EquilibriumK₁ Substitution5 MechanismsObserved Rate Law
Topic 5.9

Pre-Equilibrium Approximation

Identify the rate law for a reaction from a mechanism in which the first step is not rate limiting.

If the first elementary reaction is not rate limiting, approximations such as pre-equilibrium must be made to determine a rate law expression.

The situation: a fast, reversible first step establishes an equilibrium, and a slow second step consumes the intermediate. Writing the slow step's rate law directly gives an expression containing the intermediate — which is not allowed, because an intermediate's concentration is not something you can control or measure.

The procedure:

  1. Write the rate law for the slow (rate-determining) step from its coefficients.
  2. Identify the intermediate in that expression.
  3. Write the equilibrium expression for the fast reversible step: K₁ = [products]/[reactants].
  4. Solve that for the intermediate's concentration.
  5. Substitute back, and fold all constants into a single observed k.

The result is a rate law expressed entirely in terms of species you can actually measure — and it often contains fractional or unusual orders, which is precisely the experimental signature of a pre-equilibrium mechanism.

Why the approximation is valid: the first step reaches equilibrium far faster than the slow step consumes the intermediate, so the forward and reverse rates of step 1 stay essentially equal throughout.

Key points

  • A rate law may never contain an intermediate — substitute it away.
  • A fast reversible first step supplies the substitution via its equilibrium expression.
  • Constants combine: k = k₂K₁ becomes the single observed rate constant.
  • Unusual or fractional orders often signal a pre-equilibrium mechanism.

Equations

  • not on the sheet

Common mistakes

  • Leaving the intermediate in the final rate law. The whole point of the method.
  • Using the slow step’s equilibrium instead of the fast step’s. The reversible fast step supplies the substitution.
  • Forgetting to combine constants. The observed k is k₂K₁, not k₂ alone.

Worked example

Derive the rate law for this mechanism: Step 1 (fast, reversible): 2 NO ⇌ N₂O₂ Step 2 (slow): N₂O₂ + O₂ → 2 NO₂

Step 1 — rate law of the slow step. Step 2 is elementary and bimolecular:
rate = k₂[N₂O₂][O₂]

Step 2 — identify the problem. N₂O₂ is an intermediate (produced in step 1, consumed in step 2), so it cannot remain in the rate law.

Step 3 — use the fast equilibrium. For 2 NO ⇌ N₂O₂:
K₁ = [N₂O₂] / [NO]²

Step 4 — solve for the intermediate.
[N₂O₂] = K₁[NO]²

Step 5 — substitute and combine constants.
rate = k₂K₁[NO]²[O₂]
rate = k[NO]²[O₂], where k = k₂K₁

Overall equation check: 2 NO + O₂ → 2 NO₂ ✓ (N₂O₂ cancels). The observed reaction is third order overall, second order in NO — matching what is measured for this classic reaction.

Full notes for topic 5.9 →