Entropy of Activation and Temperature Dependence

Question

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?

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.