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The pre-equilibrium approximation

When a fast, reversible step comes before the slow one, the intermediate is trapped in the slow step's rate law. The pre-equilibrium trick uses the fast step's balance to swap that intermediate out for real reactants.

§1

Using a fast equilibrium.

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The pre-equilibrium approximation applies to a mechanism where a fast, reversible step comes before the slow rate-determining step. The fast step reaches equilibrium almost immediately.

The slow step's rate law often contains an intermediate produced by the fast step. Because that intermediate is not directly measurable, it cannot stay in the final rate law.

The fix: use the fast step's equilibrium relationship to express the intermediate in terms of the actual reactants, then substitute it into the slow step's rate law. The result is a rate law written entirely in measurable reactant concentrations.

UNIT 5 TOPIC 5.9 • PRE-EQUILIBRIUM APPROXIMATION PRE-EQUILIBRIUM APPROXIMATION TWO-STEP MECHANISM A B I P + FAST EQUILIBRIUM A + B ⇌ I SLOW (rate-determining) I → P STEP 1 — FAST EQUILIBRIUM A + B ⇌ I K = [I] / ([A][B]) [I] = K [A][B] STEP 2 — SLOW (RDS) I → P The slow step sets the rate: rate = k₂ [I] SUBSTITUTE [I] rate = k₂ (K [A][B]) Combine constants k = k₂K : rate = k [A][B] THE INTERMEDIATE I • Forms in a fast, reversible step • Consumed in the slower, later step • Present in small concentration • Not measured directly — solved via K USE IT WHEN A fast, reversible first step reaches equilibrium BEFORE the slow, rate-determining step — so the pre-equilibrium fixes [I] and the rate law follows from K. KEY IDEA The intermediate is made quickly in tiny amounts, then consumed slowly. Its concentration is locked by the fast equilibrium ([I] = K[A][B]), letting us write the overall rate law in measurable [A] and [B]. AP Chemistry · Unit 5 · Kinetics
Fig. 5.9.1 The pre-equilibrium approximation handles a fast reversible step before a slow one. Because the fast step reaches equilibrium quickly, its equilibrium expression lets you replace an intermediate in the slow step's rate law with measurable reactant concentrations.
§2

Applying the approximation.

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Write the slow-step rate law, then eliminate the intermediate.

  1. Identify the fast equilibrium and the slow step. A reversible fast step precedes the slow, rate-determining step.
  2. Write the slow step's rate law. From its molecularity; it will contain the intermediate.
  3. Use the fast-step equilibrium. Set the forward and reverse rates of the fast step equal to solve for the intermediate.
  4. Substitute. Replace the intermediate in the slow-step rate law with its expression in reactant concentrations.
§3

The pieces you'll meet.

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A fast equilibrium eliminates the intermediate.

pre-eq
Pre-equilibrium
Fast reversible step reaches equilibrium before the slow step.
fast step
Fast reversible step
Comes to equilibrium quickly; supplies the intermediate.
slow step
Slow step
Rate-determining; its rate law contains the intermediate.
intermediate
Intermediate
Expressed via the fast equilibrium, then substituted out.
equilibrium
Equilibrium relation
Forward = reverse rate of the fast step; used to solve for the intermediate.
measurable
Measurable form
Final rate law is in reactant concentrations only.
§4

Worked example: substitute out the intermediate.

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Mechanism. Fast: A + B ⇌ I (intermediate). Slow: I + C → products.

Slow-step rate law. rate = k[I][C] — but I is an intermediate that cannot remain.

Fast equilibrium. Setting the fast step's forward and reverse rates equal gives [I] proportional to [A][B].

Substitute. Replacing [I] gives rate = k'[A][B][C], now entirely in measurable reactants. The pre-equilibrium of the fast step is what let us eliminate the intermediate.

§5

Mistakes that cost real points.

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Pitfall · 01

"You can leave the intermediate in the final rate law."

An intermediate is not directly measurable, so it cannot appear in the final rate law. The whole point of the pre-equilibrium approximation is to replace it with reactant concentrations using the fast step's equilibrium.

Fix. Always substitute the intermediate out via the fast equilibrium; the final rate law must contain only reactants.

Pitfall · 02

"The pre-equilibrium step is the rate-determining step."

The fast, reversible pre-equilibrium step is not the rate-determining step — the slow step is. The fast step is used only to express the intermediate; the slow step still sets the rate.

Fix. Keep the roles straight: the slow step determines the rate; the fast equilibrium supplies the intermediate substitution.

Pitfall · 03

"Copy the overall coefficients for the rate law, ignoring the mechanism."

The rate law comes from the slow step (with the intermediate substituted), not the overall equation's coefficients. Ignoring the mechanism and copying overall coefficients gives the wrong orders.

Fix. Derive the rate law from the slow step and the fast-equilibrium substitution, not from the overall equation.

§6

Skill Check.

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