Consider an equilibrium reaction
The rate of the forward reaction, \(r _{ f }\) is equal to the rate of the backward reaction, \(r_{r}\)
\(r_{f}=k_{1}[A], r_{r}=k_{2}[B]\)
where \(k_{1}\) and \(k_{2}\) are the rate constants of the forward and reverse reactions respectively.
\(\Longrightarrow k_{1}[A]=k_{2}[B]\)
\(\frac{k_{1}}{k_{2}}=\frac{[B]}{[A]}=k f, k f=\frac{[B]}{[A]}\)y the same approach, \(k_{r}=\frac{[ A ]}{[ B ]}\)
where \(k_{f}\) and \(k_{r}\) are the equilibrium constants of the forward and reverse reactions respectively.
\begin{aligned}
k_{f} &=\frac{[B]}{[A] \times[B 1 /[B]}=\frac{[B]}{[A] /[B] \times[B]} \\
&=\frac{1}{[A / / B]}=\frac{1}{k_{r}}
\end{aligned}
i.e. \(k_{f}\) is the multiplicative inver-
se of \(k_{r}\)