Activation energy from two rate constants

Worked example

A rate constant is 4.53 s⁻¹ at 310 K and 130 s⁻¹ at 62 °C. Calculate the activation energy.

1. 1. Arrhenius equation (two-point): solve for Eₐ

   $E_\mathrm{a} = -\dfrac{R\ln(k_2/k_1)}{1/T_2 - 1/T_1}$

   = 115.3 kJ mol⁻¹

Answer: Eₐ = 115 kJ mol⁻¹

Method

ln(k₂/k₁) = −(Eₐ/R)(1/T₂ − 1/T₁). Use the natural log and temperatures in kelvin.

Formulas used

• Arrhenius equation (two-point)

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

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