It is easy to show an Arrhenius-like equation from the Eyring equation, but if you do this, you get that the activation energy is about equal to the activation enthalpy. However, the real approximation is that $E_\mathrm{a} = \Delta H^\ddagger + RT$. Why is this?


1 Answer 1


Using vibrational partition functions to define the reaction rate constant produces an equation of the form $\displaystyle k=aT^be^{-\Delta U_0^\mathrm{O}/(RT)}$ where $a,b$ are constants independent of $T$ and $\Delta U_0^\mathrm{O}$ is the difference in zero point energies at the transition state compared to reactant, this is related to what is commonly called the activation energy. This approach allows molecules to have discrete vibrational levels and so a thermal population of these. The constants $a,b$ are found from the partition functions.

The (classical) Arrhenius equation is often written as $k_A=Ae^{-E_A/(RT)}$. Taking the log of each equation and differentiating wrt. $T$ gives

$$\frac{d\ln(k)}{dT} =\frac{bRT+\Delta U_0^\mathrm{O}}{RT^2}\equiv \frac{E_A}{RT^2}$$

thus the Arrhenius activation energy is $E_A\equiv bRT+\Delta U_0^\mathrm{O}$.

(Texts on statistical thermodynamics gives details of partition functions, some book on kinetics also give this detail.)

  • $\begingroup$ how would you have got here from the simpler delta G of activation, which is more commonly seen than the difference in the zero point energies? $\endgroup$ May 18, 2020 at 12:18
  • $\begingroup$ The $\Delta U$ includes the difference in the minima of the potentials and also accounts for the difference in energy of the lowest vibrational energy level. The $\Delta G$ includes entropy changes, which can be incorporated into the pre-exponential term leaving just the $-\Delta H/RT$ you ask about in the exponential. The thermodynamic approach cannot account for any quantum effects such as zero point energy. $\endgroup$
    – porphyrin
    May 18, 2020 at 12:39

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