Consider the reaction $\ce{A <=> B}$. Let $k_f$, $E_A$, and $A$ denote the forwards rate constant, activation energy, and pre exponential factor, respectively. Let $k_r$, $E_A'$, and $A'$ denote the backwards rate constant, activation energy, and pre exponential factor, respectively. I seem to have determined the following relationship:

$\ln{\frac{A}{A'}} = \frac{\Delta S}{R}$

First, is this a correct equation? Second, is this a known equation? I have not seen it in any of the textbooks that I have read. Third, is this a useful equation? I imagine it could be used to determine the ratio between the pre exponential factors from the entropy.

My derivation is below:

The equilibrium constant $K$ is equal the ratio of the forward and backwards rate constants $\frac{k_f}{k_r}$, so $\ln{(K)} = \ln{\left(\frac{k_f}{k_r}\right)}$. Furthermore, the Arrhenius equation expresses forward and backwards rate constants as $k_f = Ae^{-E_a/RT}$ and $k_r = A'e^{-E_a'/RT}$, where $A$ is the pre-exponential factor of the forward reaction and $A'$ is the pre-exponential factor of the reverse reaction. Substituting these expressions yields $\ln{(K)} = \ln{\left(\frac{Ae^{-E_a/RT}}{A'e^{-E_a'/RT}}\right)} = \ln{\frac{A}{A'}}-\frac{(E_a-E_a')}{RT}$.

In Chapter 5 of Atkins' Chemical Principles 7th edition, the difference in activation energies of the forwards and backwards reactions $E_a-E_a'$ is equated to the reaction enthalpy $\Delta H$. Applying this relationship, we obtain: $\ln{(K)} = \ln{\frac{A}{A'}}-\frac{\Delta H}{RT}$.

It is also known, that since $\Delta G = -RT\ln{K} = \Delta H - T\Delta S$, $\ln{(K)} = -\frac{\Delta H}{RT} + \frac{\Delta S}{R}$.

Combining these two equations yields: $\frac{\Delta S}{R} = \ln{\frac{A}{A'}}$.

  • $\begingroup$ Your expression seems to be correct. I have never seen it before. $\endgroup$
    – Maurice
    Commented Feb 5, 2021 at 16:59

1 Answer 1


Yes, your expression is correct. It's nothing new though. Entropy of activation can be interpreted as part of the pre-exponential factor in the Eyring equation:

$$k=\bigg(\frac{k_B T}{h}e^{\frac{\Delta^\ddagger S^\circ}{R}}\bigg)\cdot e^{-\frac{\Delta^\ddagger H^\circ}{RT}}$$

  • 2
    $\begingroup$ In case it is not obvious to OP - note that $\Delta S_{rxn}$ is the difference between the two entropies of activation, so if you use this equation to rewrite the $\frac{A}{A'}$ term in the original question, the right side reduces to the left side. The Eyring form is much more informative and useful than the derived equation. $\endgroup$
    – Andrew
    Commented Feb 6, 2021 at 12:35

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