I am trying to understand the extrapolation of enthalpy $\Delta H^{\ddagger}$ and entropy of activation $\Delta S^{\ddagger}$ from the Eyring equation. It's typically cast as:

$$\ln\left(\frac{k}{T}\right) = \frac{-\Delta H^{\ddagger}}{RT} + \frac{\Delta S^{\ddagger}}{R} + \ln\left(\frac{k_\mathrm{B}}{h}\right)$$

where $k$ is the rate, $T$ is the temperature, $R$ is the ideal gas constant, $k_\mathrm{B}$ is the Bolztmann constant, and $h$ is the Planck constant. However, the entropy itself also has a temperature dependence in principle (i.e. harmonic oscillator entropy is a function of temperature). Is there a way to determine the temperature dependence of the entropy of activation from the Eyring equation for a real experiment?

  • $\begingroup$ That would depend on the temperature range; the traditional experiment for this sort of thing fits the left side to a linear function of $1/T$ which works out on a small enough temperature range that the enthalpy and entropy changes can be assumed constant (or close enough to constant that other errors dominate). Without that assumption it seems that you will need much finer scale information to get anywhere. $\endgroup$
    – Ian
    Jul 12, 2017 at 18:51

1 Answer 1


The equation

$$\ln\left(\frac{k}{T}\right) = \frac{-\Delta H^{\ddagger}}{RT} + \frac{\Delta S^{\ddagger}}{R} + \ln\left(\frac{k_\mathrm{B}}{h}\right)$$

does not assume that $\Delta S^{\ddagger}$ is temperature independent. To evaluate its T dependence you might proceed as follows:

$$\left(\frac{\partial \Delta G^{\ddagger} }{\partial T}\right)_p = -\Delta S^{\ddagger}$$

$$\rightarrow\left(\frac{\partial [RT\ln K^{\ddagger}] }{\partial T}\right)_p = \Delta S^{\ddagger}$$

$$\rightarrow \Delta S^{\ddagger}=\left(\frac{ \partial [RT \ln (hk/k_\mathrm{B}T)] }{\partial T}\right)_p $$

It follows that you can evaluate $\Delta S^{\ddagger}(T)$, $\Delta S^{\ddagger}$ at different temperatures, from the change of $[RT \ln (hk/k_\mathrm{B}T)]$ with respect to temperature.


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