# Relationship between Gibbs free energy and potential energy (related to reaction coordinate diagrams)

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?

• Why did you assume that the pre-exponential factor is the same for the forward reaction as for the reverse reaction? – Chet Miller Jul 31 at 0:38
• Is it not? I'm not sure if it changes or not. – Natasha A. Jul 31 at 0:55
• 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. – Chet Miller Jul 31 at 11:06

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".