1
$\begingroup$

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

$\endgroup$
1
  • $\begingroup$ Your expression seems to be correct. I have never seen it before. $\endgroup$ – Maurice Feb 5 at 16:59
1
$\begingroup$

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:

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

$\endgroup$
1
  • 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 Feb 6 at 12:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.