# New relationship between exponential factors and entropy of a reaction?

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

• Your expression seems to be correct. I have never seen it before. – Maurice Feb 5 at 16:59

$$k=\bigg(\frac{k_B T}{h}e^{\frac{\Delta^\ddagger S^\circ}{R}}\bigg)\cdot e^{-\frac{\Delta^\ddagger H^\circ}{RT}}$$
• 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. – Andrew Feb 6 at 12:35