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?