Lindemann–Hinshelwood fall-off

UKChO Round 2 · IChO · unreviewed

Lindemann mechanism with k₁, k₋₁, k₂ and [M] known; find the effective unimolecular rate constant.

Worked example

For A + M ⇌ A* + M (k₁ = 3.2 × 10⁵ dm³ mol⁻¹ s⁻¹, k₋₁ = 6.7 × 10⁹ dm³ mol⁻¹ s⁻¹), A* → P (k₂ = 6.8 × 10⁵ s⁻¹), calculate the effective first-order rate constant at [M] = 4.7 × 10⁻⁵ mol dm⁻³.

1. Multiply k₋₁ × [M] → k₋₁[M]
= 3.149 × 10⁵ s⁻¹
2. Add k₋₁[M] + k₂
= 9.949 × 10⁵ s⁻¹
3. Multiply k₁ × [M] → k₁[M]
= 15.04 s⁻¹
4. Multiply k₁[M] × k₂ → k₁k₂[M]
= 1.023 × 10⁷ s⁻²
5. Divide k₁k₂[M] ÷ (k₋₁[M] + k₂) → k_uni
= 10.28 s⁻¹

Answer: k_uni = 10.3 s⁻¹

Method

d[A*]/dt = k₁[A][M] − k₋₁[A*][M] − k₂[A*] = 0, so [A*] = k₁[A][M]/(k₋₁[M] + k₂) and rate = k₂[A*]: k_uni = k₁k₂[M]/(k₋₁[M] + k₂). It tends to k₁k₂/k₋₁ at high [M] and k₁[M] at low [M].

Common mistakes

That is the high-pressure limit k₁k₂/k₋₁. At this [M], deactivation and reaction of A* compete.
That is the low-pressure limit k₁[M]. At this [M], deactivation of A* is not negligible.

Practise this with fresh numbers

Lindemann–Hinshelwood fall-off

Worked example

Method

Common mistakes

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