Pre-Equilibrium Approximation

Introduction

A mechanism may begin with a fast reversible association that establishes an approximately stable ratio among reactants and an intermediate. If a later step forms product more slowly, the fast stage can be treated as a pre-equilibrium. This relation lets us replace an unmeasured intermediate concentration in the product-forming rate law.

Core explanation

Take A+B ⇌ I as a fast first step, followed by I → P as a slower step. If the first step remains close to equilibrium, define K₁=[I]/([A][B]) using a simple concentration model. Then [I]=K₁[A][B]. The slow elementary product step has rate≈k₂[I], so substitution gives rate≈k₂K₁[A][B]. The measured overall rate is first order in each reactant under these assumptions, even though the product-forming elementary event involves only I.

For the fast stage, the relation can also be written K₁=k₁/k₋₁ if forward and reverse elementary steps obey simple mass-action laws: forward flux k₁[A][B], reverse flux k₋₁[I]. Approximate equality gives [I]≈(k₁/k₋₁)[A][B]. A thermodynamic equilibrium constant is dimensionless when activities are used, whereas concentration-based K₁ may carry concentration units depending on how it is defined. Introductory formulas often suppress this standard-state distinction; dimensional checks remain useful.

The approximation requires the reversible first step to equilibrate faster than the slow step drains I. If product formation consumes I as quickly as it forms, the first step cannot remain near equilibrium. An initial transient may also occur before the pre-equilibrium establishes. The derived law therefore may apply after a short settling time and within a defined concentration range.

If B is held in large excess, [B] is nearly constant and the measured law can look pseudo-first-order in A: rate≈kobs[A] with kobs=k₂K₁[B]. A sequence of experiments varying B can reveal the hidden dependence. This connects mechanism approximation with the previous pseudo-order topic.

Not all pre-equilibria are simple associations. A fast protonation, ligand binding or conformational change can make an intermediate. If several species share the reactant pool, the fraction present in the reactive form may have a denominator, giving saturation or inhibition rather than a simple product of concentrations. The method is to write the relevant equilibrium and material balance, not memorize one universal final law.

Checking the net equation is still required. The first and second steps add to A+B → P after I cancels. Agreement with net stoichiometry and rate data supports the proposal, but alternative mechanisms can give the same apparent law. Independent evidence may be needed to identify I or verify a fast equilibrium.

Step-by-step reasoning

1. Write the fast reversible and slower product-forming steps. 2. Sum them to confirm the overall equation. 3. Write a concentration or activity equilibrium relation for the fast step. 4. Solve that relation for the intermediate concentration. 5. Substitute into the slower step rate and state validity conditions.

Visual explanation

Draw A+B ⇌ I with thick double arrows labeled fast, then I → P with a thinner arrow labeled slow. Beneath the first write [I]=K₁[A][B], then show substitution into rate=k₂[I]. Shade the first-arrow balance to emphasize its approximate nature.

Real-world analogy

A busy waiting room may rapidly balance arrivals and departures with a nearby lobby, while a slow service desk removes people from the room. If lobby exchange is much faster, the room occupancy follows an approximate ratio set by the lobby even while service proceeds.

Real-world example

Fast binding of a reactant to a catalyst can precede a slower chemical conversion. Under low-coverage conditions, binding equilibrium may make product rate proportional to reactant concentration; at high coverage, a full site balance gives saturation instead.

Why?

Why does the intermediate not need to be measured directly in this approximation? The fast reversible step relates its concentration to measurable reactants through an equilibrium ratio, allowing algebraic elimination.

Common misconception

“Any reversible first step is automatically at equilibrium.” It must be fast relative to intermediate removal; a reversible arrow alone is not enough to justify the approximation.

Worked example

Suppose A+B ⇌ I has concentration relation K₁=5.0 M⁻¹ and I → P has k₂=0.20 s⁻¹. At [A]=0.10 M and [B]=0.20 M, [I]≈5.0(0.10)(0.20)=0.10 M in this simplified model. Product rate≈0.20×0.10=0.020 M s⁻¹. Equivalently k₂K₁[A][B]=0.20×5.0×0.10×0.20=0.020 M s⁻¹. The calculation requires the fast-stage approximation to be justified.

Quick check

1. What relation eliminates I for fast A+B ⇌ I? Answer: [I]≈K₁[A][B] in the simple concentration model.

Exam focus

Show the equilibrium relation, intermediate substitution, net equation and timescale assumption. Keep concentration-based K units consistent and avoid calling any reversible step automatically equilibrated.

Advanced insight

When catalyst or reactant mass balance matters, free [A] may differ from total analytical concentration. Solving both equilibrium and conservation equations can yield a rational, saturating rate law rather than a pure [A][B] product.

Summary

A fast reversible step can maintain a pre-equilibrium while slower product formation proceeds. Its equilibrium relation expresses an intermediate through measurable species, producing a testable approximate rate law under clear timescale conditions.

Practice questions

1. For fast A+B⇌I and slow I→P, what simple law results? Answer: rate≈k₂K₁[A][B] if [I]≈K₁[A][B]. 2. What invalidates the pre-equilibrium assumption? Answer: Product formation draining I as rapidly as the reversible stage can equilibrate it. 3. If B is held in large excess, what observed order in A results from the simple law? Answer: Pseudo-first-order in A, with kobs=k₂K₁[B] approximately constant.