For a rate law, the rate constant $k$ is given by the equation

$$\ln k = \ln A - \frac{E_\mathrm{a}}{RT}\tag{1}$$

If $k_\mathrm{fwd}$ is the forward rate constant and $k_\mathrm{rev}$ is the reverse rate constant:

$$\ln k_\mathrm{fwd} = \ln A - \frac{E_\mathrm{a,fwd}}{RT}\tag{2}$$

$$\ln k_\mathrm{rev} = \ln A - \frac{E_\mathrm{a,rev}}{RT}\tag{3}$$

Subtracting the two equations yields

$$\ln\frac{k_\mathrm{fwd}}{k_\mathrm{rev}} = \frac{E_\mathrm{a,rev}}{RT} - \frac{E_\mathrm{a,fwd}}{RT} = \frac{1}{RT} (E_\mathrm{a,rev} - E_\mathrm{a,fwd})\tag{4}$$

If $K$ is the equilibrium constant for the reaction, then $\displaystyle K = \frac{k_\mathrm{fwd}}{k_\mathrm{rev}}$:

$$-RT\ln K = \Delta E\tag{5}$$

where $\Delta E$ is the change in potential energy on a reaction diagram, such as the one shown here.

However, we know that

$$-RT\ln K = \Delta_\mathrm{r} G^\circ\tag{6}$$

Does this mean that potential energy is equal to the standard Gibbs free energy of reaction?

  • 1
    $\begingroup$ Why did you assume that the pre-exponential factor is the same for the forward reaction as for the reverse reaction? $\endgroup$ Jul 31, 2019 at 0:38
  • $\begingroup$ Is it not? I'm not sure if it changes or not. $\endgroup$
    – Natasha A.
    Jul 31, 2019 at 0:55
  • $\begingroup$ If that were the case, the equilibrium constant for all reactions would approach 1 at high temperatures. We know that that doesn't happen. There is absolutetly no physical reason why it would be the case, however. $\endgroup$ Jul 31, 2019 at 11:06
  • $\begingroup$ @ChetMiller For a single step reaction $K=\frac{k_1}{k_{-1}}$. We also know that $K=e^{\frac{-ΔrG}{RT}}$. But $$\frac{k_1}{k_{-1}}=\frac{A_1\cdot e^{-Ea_1/RT}}{A_{-1}\cdot e^{-Ea_2/RT}}=\frac{A_1}{A_{-1}}\cdot e^{-ΔrG/RT}$$ Shouldn't pre-exponential factors be equal? $\endgroup$
    – Anton
    Apr 3, 2021 at 17:10
  • $\begingroup$ The A’s can be absorbed into the exponential $\endgroup$ Apr 3, 2021 at 18:29

1 Answer 1


Equation (1) is an empirical one that does not consider entropy. If you substitute it for the Eyring equation (considering free energy of activation $\Delta G^\ddagger$) you end up with equation (6) as expected, not equation (5). Wikipedia has a nice section on the relationship of activation energy and Gibbs energy of activation:

Although the equations look similar, it is important to note that the Gibbs energy contains an entropic term in addition to the enthalpic one. In the Arrhenius equation, this entropic term is accounted for by the pre-exponential factor A. More specifically, we can write the Gibbs free energy of activation in terms of enthalpy and entropy of activation: $ΔG‡ = ΔH‡ – T ΔS‡$. Then, for a unimolecular, one-step reaction, the approximate relationships $E_a = ΔH‡ + R T$ and $A = (k_B T/h) \exp(1 + ΔS‡/R)$ hold. Note, however, that in Arrhenius theory proper, A is temperature independent, while here, there is a linear dependence on T. For a one-step unimolecular process whose half-life at room temperature is about 2 hours, ΔG‡ is approximately 23 kcal/mol. This is also the roughly the magnitude of Ea for a reaction that proceeds over several hours at room temperature. Due to the relatively small magnitude of TΔS‡ and RT at ordinary temperatures for most reactions, in sloppy discourse, $E_a$, ΔG‡, and ΔH‡ are often conflated and all referred to as the "activation energy".

So in a more rigorous treatment, the pre-exponential factor is temperature-dependent (this is almost impossible to measure because the exponential is "very" temperature-dependent) and has an entropy component. That is why equation (5) in the question is not accurate (equation (6) is correct, and can be derived in a rigorous manner from first principles and definitions).

Related question: Y-axis of the reaction co-ordinate graph


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