Steady state: A ⇌ I → P

UKChO Round 2 · IChO · unreviewed

A ⇌ I → P with all three rate constants known; steady state on I.

Worked example

A reacts through an intermediate: A ⇌ I (k₁ = 0.58 s⁻¹, k₋₁ = 1700 s⁻¹), then I → P (k₂ = 1100 s⁻¹). Apply the steady-state approximation to I and find k_eff.

1. Multiply k₁ × k₂ → k₁k₂
= 638.0 s⁻²
2. Add k₋₁ + k₂
= 2800 s⁻¹
3. Divide k₁k₂ ÷ (k₋₁ + k₂) → k_eff
= 0.2279 s⁻¹

Answer: k_eff = 0.228 s⁻¹

Method

d[I]/dt = k₁[A] − k₋₁[I] − k₂[I] = 0 gives [I] = k₁[A]/(k₋₁ + k₂), so rate = k₂[I] = k₁k₂[A]/(k₋₁ + k₂): k_eff = k₁k₂/(k₋₁ + k₂).

Common mistakes

That is the pre-equilibrium result k₁k₂/k₋₁, which needs k₋₁ ≫ k₂. Here they are comparable: use the steady-state expression.
That takes the first step as rate-determining, which only holds when k₂ ≫ k₋₁.

Practise this with fresh numbers

Steady state: A ⇌ I → P

Worked example

Method

Common mistakes

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