Pre-Equilibrium Approximation

Introduction

Some mechanisms begin with a fast reversible step that sets up an equilibrium, followed by a much slower step that siphons off a little of the equilibrium mixture to make product. Because the slow step barely disturbs the equilibrium, the intermediate's concentration can be calculated from an equilibrium constant. This pre-equilibrium approximation gives compact rate laws and explains curious observations such as reactions that slow down on heating.

Core explanation

The scheme. Consider

A + B ⇌ I (fast; k₁ forward, k₋₁ reverse) I → P (slow; k₂)

The assumption. If k₋₁ ≫ k₂, an I molecule is far more likely to revert to A + B than to go on to P. The first step therefore stays at equilibrium throughout, with

K₁ = [I]/([A][B]) = k₁/k₋₁

so [I] = K₁[A][B].

The rate law. The rate of product formation is

rate = k₂[I] = k₂K₁[A][B] = (k₁k₂/k₋₁)[A][B]

The reaction appears second order with an observed rate constant k obs = K₁k₂, a composite of an equilibrium constant and a rate constant.

Relation to the steady state. The steady-state result for the same mechanism is rate = k₁k₂[A][B]/(k₋₁ + k₂). When k₋₁ ≫ k₂, the denominator becomes k₋₁ and the two treatments agree. The pre-equilibrium approximation is therefore a limiting case of the more general steady-state treatment, but it is simpler to apply and does not require the intermediate to be scarce: I may be present at appreciable concentration if K₁ is large.

Mass-balance refinement. If K₁ is large, a significant fraction of A is tied up as I. Then the free [A] is not the total added, and the rate law becomes k₂K₁[A]total[B]/(1 + K₁[B]), which shows saturation: at high [B] the rate stops increasing and becomes first order in total A. This is the same algebraic form as Michaelis–Menten enzyme kinetics.

Temperature behaviour. Since k obs = K₁k₂, its apparent activation energy is

Eₐ,app = ΔH°₁ + Eₐ,₂

where ΔH°₁ is the enthalpy change of the pre-equilibrium. If the first step is strongly exothermic, ΔH°₁ may be more negative than Eₐ,₂ is positive, and Eₐ,app becomes negative: the reaction slows as temperature rises, because heating shifts the pre-equilibrium back towards reactants.

Classic example. The gas reaction 2NO + O₂ → 2NO₂ is third order and has a negative apparent activation energy. A widely accepted mechanism is fast pre-equilibrium dimerisation, 2NO ⇌ N₂O₂ (exothermic), followed by slow reaction N₂O₂ + O₂ → 2NO₂, giving rate = k₂K₁[NO]²[O₂].

Formulae

K₁ = k₁/k₋₁; [I] = K₁[A][B]; rate = k₂K₁[A][B]; Eₐ,app = ΔH°₁ + Eₐ,₂.

Step-by-step reasoning

1. Identify the fast reversible step and the slow product-forming step. 2. Write the equilibrium constant expression for the fast step. 3. Rearrange to express the intermediate concentration in terms of reactants. 4. Substitute into the rate equation for the slow step. 5. Combine constants into k obs and check the predicted orders against experiment.

Visual explanation

Draw a reaction profile with a shallow well for I between reactants and products. The first barrier is low, so A + B and I shuttle back and forth freely. The second barrier, from I to P, is much higher, forming a bottleneck that only occasionally lets molecules through.

Real-world analogy

A crowd mills around a large lobby, with people wandering freely in and out of the entrance. Beyond the lobby is a single slow ticket desk. The lobby population stays in balance with the street outside, and the rate of admissions is set by the ticket desk and how many people happen to be in the lobby.

Real-world example

Acid-catalysed ester hydrolysis begins with rapid, reversible protonation of the carbonyl oxygen, followed by slower attack of water on the protonated ester. The rate is proportional to [ester][H⁺] because the concentration of the protonated form is set by a fast acid–base pre-equilibrium.

Why?

Why can the slow step be ignored when calculating [I]? Only a tiny fraction of I leaks away to products in the time it takes the fast step to re-establish equilibrium many times over, so the equilibrium composition is essentially undisturbed.

Common misconception

"Only the slow step's reactants appear in the rate law." The slow step here involves I, yet the rate law contains A and B, because [I] is fixed by the preceding equilibrium. Every species before and in the rate-determining step can appear.

Worked example

Question: For a pre-equilibrium A + B ⇌ I with K₁ = 50 mol⁻¹ dm³, and I → P with k₂ = 0.020 s⁻¹, find the rate when [A] = 0.010 and [B] = 0.020 mol dm⁻³ (assuming little A is bound as I).

Reasoning: k obs = K₁k₂ = 50 × 0.020 = 1.0 mol⁻¹ dm³ s⁻¹. Rate = 1.0 × 0.010 × 0.020.

Answer: Rate = 2.0 × 10⁻⁴ mol dm⁻³ s⁻¹.

Quick check

1. What condition on the rate constants justifies the pre-equilibrium approximation for A + B ⇌ I → P? Answer: The reverse step must be much faster than product formation, k₋₁ ≫ k₂, so equilibrium is maintained.

Exam focus

Write the equilibrium constant explicitly, substitute for the intermediate and show the composite constant. When asked to compare with the steady-state approach, state the limit k₋₁ ≫ k₂ that makes them identical. Negative activation energies are a favourite discriminating question.

Advanced insight

Pre-equilibria also explain acid and base catalysis, isotope effects and saturation kinetics. In enzyme catalysis the original Michaelis–Menten treatment assumed a pre-equilibrium between enzyme and substrate, whereas the later Briggs–Haldane treatment used the steady state; the two give the same functional form with differently defined constants.

Summary

When a fast reversible step precedes a slow step, the intermediate's concentration is given by the equilibrium constant, and the rate law contains the composite constant K₁k₂. The approximation is the k₋₁ ≫ k₂ limit of the steady-state treatment. Large K₁ leads to saturation, and an exothermic pre-equilibrium can give a negative apparent activation energy.

Practice questions

1. Derive the rate law for A ⇌ I (fast, K₁) followed by I + C → P (slow, k₂). Answer: [I] = K₁[A], so rate = k₂[I][C] = k₂K₁[A][C]. 2. Why does the NO + O₂ reaction slow down on heating? Answer: The exothermic pre-equilibrium forming N₂O₂ shifts back towards NO as temperature rises, lowering [N₂O₂] more than k₂ increases. 3. Under what condition does the steady-state rate law k₁k₂[A][B]/(k₋₁ + k₂) reduce to the pre-equilibrium form? Answer: When k₋₁ ≫ k₂, the denominator becomes k₋₁ and the rate is (k₁/k₋₁)k₂[A][B] = K₁k₂[A][B]. 4. Give the apparent activation energy if ΔH°₁ = −30 kJ mol⁻¹ and Eₐ,₂ = 20 kJ mol⁻¹. Answer: Eₐ,app = −30 + 20 = −10 kJ mol⁻¹, so the rate falls as temperature rises.