Consider a purely reversible elementary first-order reaction analytically defined as possessing a fo — Chemical Kinetics Chemistry Question

💡 Solution & Explanation

At equilibrium, the forward rate strictly equals the backward rate: $k_f[A]_{eq} = k_b[B]_{eq}$. Through stoichiometry, $[A]_{eq} + [B]_{eq} = a$, so $[A]_{eq} = a - [B]_{eq}$. Substituting this into the rate equality yields $k_f(a - [B]_{eq}) = k_b[B]_{eq} \Rightarrow k_f a - k_f[B]_{eq} = k_b[B]_{eq} \Rightarrow k_f a = (k_f + k_b)[B]_{eq}$. Thus, $[B]_{eq} = \frac{k_f a}{k_f + k_b}$.