For $A+B\rightleftharpoons C+D$, the forward and reverse rates are
$$
r_f=k_fn_An_B,
\qquad
r_r=k_rn_Cn_D.
$$
At <thermochemical equilibrium>, $r_f=r_r$ by <detailed balance>, while
$$
K_c=\frac{n_Cn_D}{n_An_B}
=\exp\left(-\frac{\Delta_rG^\circ}{RT}\right)
$$
after the appropriate standard-concentration factors are included. The standard reaction <Gibbs free energy> is
$$
\Delta_rG^\circ=G_C^\circ+G_D^\circ-G_A^\circ-G_B^\circ.
$$
Consequently
$$
\boxed{k_r=\frac{k_f}{K_c}
=k_f\exp\left(\frac{\Delta_rG^\circ}{RT}\right)}.
$$
For other stoichiometries, the same argument uses the corresponding activity product and its standard-state factors.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 315 / 4 / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-315/4/b/solution