For a strictly reversible first-order elementary reaction , the forward rate constant is and the bac — Chemical Kinetics Chemistry Question
Question
For a strictly reversible first-order elementary reaction $A \rightleftharpoons B$, the forward rate constant $k_f$ is $1.0 \times 10^{-3} \text{ s}^{-1}$ and the backward rate constant $k_b$ is $2.0 \times 10^{-3} \text{ s}^{-1}$. If the initial concentration of $A$ is $a$ and $B$ is $0$, what is the exact time required for the concentration of $B$ to reach exactly half of its ultimate equilibrium value?
💡 Solution & Explanation
The integrated rate law for a reversible first-order reaction approaching equilibrium is $(k_f + k_b) = \frac{1}{t} \ln \left(\frac{x_{eq}}{x_{eq} - x}\right)$. We are given $x = \frac{x_{eq}}{2}$. Substituting this gives $(k_f + k_b) = \frac{1}{t} \ln \left(\frac{x_{eq}}{x_{eq} - x_{eq}/2}\right) = \frac{1}{t} \ln 2$. Therefore, $t = \frac{\ln 2}{k_f + k_b} = \frac{0.693}{(1.0 + 2.0) \times 10^{-3}} = \frac{0.693}{0.003} = 231 \text{ s}$.