 Solution

At thermal equilibrium, detailed balance equates the opposing transition fluxes: $k_f{p}_A^{\mathrm{eq}}=k_b{p}_B^{\mathrm{eq}}$. Define the state free energy difference explicitly by $\Delta G_{AB}=G_B-G_A$. The equilibrium Boltzmann distribution then givesAny state degeneracy is included in the state free energies. Detailed balance is stronger than simply requiring a time-independent total occupancy: it equates each pair of equilibrium transition fluxes.

Source illustration TODO: the source's reaction coordinate diagram is missing. The definition above fixes the sign independently; no barrier heights or ordering of drawn energy levels are inferred. Below, a large forward free-energy drop means $G_A-G_B\gg k_{B}T$. If instead $\Delta G_{AB}=G_B-G_A\gg k_{B}T$, forward transitions are unfavorable: for example $\Delta G_{AB}=10k_{B}T$ gives $k_f/k_b=e^{-10}$, so neglecting reverse transitions would be wrong.