Apparent activation energy of a pre-equilibrium mechanism

UKChO Round 2 · IChO · unreviewed

Pre-equilibrium followed by a slow step; combine the enthalpy and activation energy, then use the two-point Arrhenius equation.

Practise this with fresh numbers

Worked example

For the mechanism A + B ⇌ I (fast, ΔH° = −16 kJ mol⁻¹), I → P (slow, Eₐ = 70 kJ mol⁻¹), k_obs = K₁k₂ = 0.14 dm³ mol⁻¹ s⁻¹ at 298 K. Find k_obs at 320 K, assuming ΔH° and Eₐ do not vary with temperature.

1. Add ΔH°(step 1) + Eₐ(step 2) → Eₐ(apparent) = ΔH°₁ + Eₐ₂
= 54.00 kJ mol⁻¹
2. Arrhenius equation (two-point): solve for k_obs at T₂
$k_2 = k_1\exp\!\left[-\dfrac{E_\mathrm{a}}{R}\left(\dfrac{1}{T_2}-\dfrac{1}{T_1}\right)\right]$
= 0.6264 dm³ mol⁻¹ s⁻¹

Answer: k_obs at T₂ = 0.626 dm³ mol⁻¹ s⁻¹

Method

k_obs = K₁k₂. Using van ’t Hoff for K₁ and Arrhenius for k₂, ln k_obs falls with 1/T with slope −(ΔH°₁ + Eₐ₂)/R, so the apparent activation energy is ΔH°₁ + Eₐ₂. If the pre-equilibrium is exothermic enough, this is negative and the reaction slows on heating, as for 2NO + O₂.

Common mistakes

k_obs = K₁k₂, so its temperature dependence includes the pre-equilibrium: Eₐ(apparent) = ΔH°₁ + Eₐ₂.

Formulas used

Arrhenius equation (two-point)
$\ln\dfrac{k_2}{k_1} = -\dfrac{E_\mathrm{a}}{R}\left(\dfrac{1}{T_2}-\dfrac{1}{T_1}\right)$