Pre-equilibrium then half-life

UKChO Round 2 · IChO · unreviewed

A ⇌ I fast pre-equilibrium, then slow I → P; all rate constants known. Find the half-life of A.

Worked example

A rapidly equilibrates with I (k₁ = 5.8 s⁻¹, k₋₁ = 1.7 × 10⁵ s⁻¹); I slowly gives P with k₂ = 97 s⁻¹. What is the half-life of A?

1. Divide k₁ ÷ k₋₁ → K₁ = k₁/k₋₁
= 3.412 × 10⁻⁵
2. Multiply K₁ = k₁/k₋₁ × k₂ → k_obs
= 0.003309 s⁻¹
3. Half-life (first order): solve for t½
$t_{1/2} = \dfrac{\ln 2}{k}$
= 209.4 s

Answer: t½ = 209 s

Method

Pre-equilibrium: k₁[A] = k₋₁[I], so [I] = K₁[A] with K₁ = k₁/k₋₁. Then rate = k₂[I] = K₁k₂[A], first order with k_obs = k₁k₂/k₋₁, and t½ = ln 2/k_obs.

Common mistakes

k₂ alone ignores how little I is present: the observed constant is K₁k₂ = k₁k₂/k₋₁.

Formulas used

Half-life (first order)
$t_{1/2} = \dfrac{\ln 2}{k}$

Practise this with fresh numbers

Pre-equilibrium then half-life

Worked example

Method

Common mistakes

Formulas used

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