From ΔG‡ to a reaction time at another temperature

Worked example

A unimolecular rearrangement has ΔG‡ = 90.7 kJ mol⁻¹ at 298 K and Eₐ = 82.6 kJ mol⁻¹. Taking [A]₀ = 1, how long does it take at 358 K for [A] to fall to 0.50?

1. 1. Eyring equation (ΔG‡): solve for k

   $k = \dfrac{k_\mathrm{B}T}{h}\,e^{-\Delta G^\ddagger/RT}$

   = 7.854 × 10⁻⁴ s⁻¹

2. 2. Arrhenius equation (two-point): solve for k₂

   $k_2 = k_1\exp\!\left[-\dfrac{E_\mathrm{a}}{R}\left(\dfrac{1}{T_2}-\dfrac{1}{T_1}\right)\right]$

   = 0.2097 s⁻¹

3. 3. First-order integrated rate law: solve for t

   $t = \dfrac{1}{k}\ln\dfrac{[\mathrm{A}]_0}{[\mathrm{A}]}$

   = 3.305 s

Answer: t = 3.31 s

Method

Eyring gives k at T₁; two-point Arrhenius moves it to T₂; then t = ln([A]₀/[A])/k.

Common mistakes

• That is the time at T₁. Use the activation energy to get k at T₂ first.

Formulas used

• Eyring equation (ΔG‡)

  $k = \dfrac{k_\mathrm{B}T}{h}\,e^{-\Delta G^\ddagger/RT}$

• 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)

• First-order integrated rate law

  $\ln\dfrac{[\mathrm{A}]_0}{[\mathrm{A}]} = kt$

More kinetics questions

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