 Solution

ID: past-exam-of-the-biological-physics-course-of-the-university-of-oxford/2022/4/a/solution

Detailed balance means that at thermodynamic equilibrium each elementary transition and its reverse carry equal probability flux. For normalized state probabilities

A_{eq}

and

B_{eq}

k_{1} A_{eq} = k_{2} B_{eq}, \frac{B _{eq}}{A _{eq}} = e^{- (G_{B} - G_{A}) / (k_{B} T)} .

With the printed convention

Δ G_{A B} = G_{A} - G_{B}

, the sign is therefore

\frac{k _{1}}{k _{2}} = e^{Δ G_{A B} / (k_{B} T)} .

The

G

values are free energies per system, including internal degeneracies; for molar Gibbs free energies, replace

k_{B} T

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

\dot{B} = k_{1} (1 - B) - k_{2} B

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}, 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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