= Solution
At <thermal equilibrium>, <detailed balance> equates the opposing transition fluxes: $k_f p_A^{\rm eq}=k_b p_B^{\rm eq}$. Define the state <free energy> difference explicitly by $\Delta G_{AB}=G_B-G_A$. The equilibrium <Boltzmann distribution> then gives
$$
\boxed{\frac{k_f}{k_b}=\frac{p_B^{\rm eq}}{p_A^{\rm eq}}=e^{-\Delta G_{AB}/(k_BT)}.}
$$
Any 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.
\b[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 \i[forward free-energy drop] means $G_A-G_B\gg k_BT$. If instead $\Delta G_{AB}=G_B-G_A\gg k_BT$, forward transitions are unfavorable: for example $\Delta G_{AB}=10k_BT$ gives $k_f/k_b=e^{-10}$, so neglecting reverse transitions would be wrong.
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