= Solution
<Detailed balance> means that at <thermodynamic equilibrium> each elementary transition and its reverse carry equal <probability> flux. For normalized state <probabilities> $A_{\rm eq}$ and $B_{\rm eq}$,
$$
k_1A_{\rm eq}=k_2B_{\rm eq},\qquad
\frac{B_{\rm eq}}{A_{\rm eq}}=e^{-(G_B-G_A)/(k_BT)}.
$$
With the printed convention $\Delta G_{AB}=G_A-G_B$, the sign is therefore
$$
\boxed{\frac{k_1}{k_2}=e^{\Delta G_{AB}/(k_BT)}}.
$$
The $G$ values are free energies per system, including internal degeneracies; for molar <Gibbs free energies>, replace $k_BT$ by $RT$. If B has lower <free energy>, its forward-to-reverse rate ratio exceeds one, as it should.
For constant rates, conservation gives $A=1-B$, so the <master equation> is $\dot B=k_1(1-B)-k_2B$. Its <steady state> is $B_0=k_1/(k_1+k_2)$, and separation of the decaying deviation yields
$$
B(t)=B_0+[B(0)-B_0]e^{-(k_1+k_2)t},\qquad
\boxed{B(t)=\frac{k_1}{k_1+k_2}(1-e^{-(k_1+k_2)t})}
$$
for the specified initial state A. The relaxation <time constant> is $(k_1+k_2)^{-1}$.
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