We consider a unimolecular equilibrium reaction $$\text{A}\rightleftharpoons \text{B}$$ between chemical species A and B. This actually involves two simultaneous reactions, namely

forward $\displaystyle \text{A}\stackrel{k_1}{\longrightarrow} \text{B}$ with reaction rate constant $k_1$

backward $\displaystyle \text{B}\stackrel{k_{2}}{\longrightarrow} \text{A}$ with rate constant $k_{2}$

Species A is not only converted into B, but it is also formed from species B by the reverse reaction. Similarly substance B is not only formed, but disappears again by the reverse reaction. When we assume elementary chemical reactions in this system, then the following two coupled differential equations describe the reaction kinetics: $$\left\{\begin{aligned}\frac{\dd[\text{A}]}{\dd t}&=-k_1[\text{A}]+k_{2}[\text{B}]\\ \\ \frac{\dd[\text{B}]}{\dd t}&=k_1[\text{A}]-k_{2} [\text{B}]\end{aligned}\right.$$ Under the assumption that the total concentration of A and B is constant, we can uncouple the differential equations: if $[\text{A}]+[\text{B}]=K$ for some constant $K$, then you can rewrite the equations as $$\begin{aligned}\frac{\dd[\text{A}]}{\dd t}&=-k_1 [\text{A}]+k_{2} (K-[\text{A}])\\ \\ \frac{\dd[\text{B}]}{\dd t}&=k_1 (K-[\text{B}])-k_{2}[\text{B}]\end{aligned}$$ So: $$\begin{aligned}\frac{\dd[\text{A}]}{\dd t}&=k_{2} K-(k_1+k_{2}) [\text{A}]\\ \\ \frac{\dd[\text{B}]}{\dd t}&=k_1 K-(k_1+k_{2}) [\text{B}]\end{aligned}$$ Both equations correspond with restricted exponential growth, and an equilibrium settles in after sufficient time.

There is chemical equilibrium when $$\frac{\dd[\text{A}]}{\dd t}=0, \frac{\dd[\text{B}]}{\dd t}=0$$ The equilibrium concentrations $[\text{A}]_{\infty}$ and $[\text{B}]_{\infty}$ can be determined herewith; verify that the follwoing is true when the system is in equilibrium: $$[\text{A}]_{\infty}=\frac{k_{-1}\cdot c}{(k_1+k_{-1})},\quad [\text{B}]_{\infty}=\frac{k_{1}\cdot c}{(k_1+k_{-1})}$$ In equilibrium, the concentrations are related as quotient of the forward and reverse reaction rate constants: $$\frac{[\text{A}]_{\infty}}{[\text{B}]_{\infty}} = \frac{k_{-1}}{k_{1}}$$