Consider a purely reversible elementary first-order reaction analytically defined as possessing a fo — Chemical Kinetics Chemistry Question
Question
Consider a purely reversible elementary first-order reaction analytically defined as $A \rightleftharpoons B$ possessing a forward rate constant $k_f$ and a backward rate constant $k_b$. If the initial molar concentration of A is precisely $a$ and B is $0$, the terminal equilibrium concentration of product B, $[B]_{eq}$, fundamentally evaluates to:
Answer: A
💡 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, $[B]_{eq} = \frac{k_f a}{k_f + k_b}$.
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