 Solution

The products lie below the reactants, while the transition state lies above both. On an energy-versus-reaction-coordinate plot, the forward activation energy is the vertical separation $E_{\mathrm{TS}}-E_{\mathrm{R}}$, not the difference between product and reactant energies. For an exothermic elementary path, $\Delta E=E_{\mathrm{P}}-E_{\mathrm{R}}<0$, and the reverse barrier is $E_{a,-}=E_{a,+}-\Delta E>E_{a,+}$.