• # question_answer Two consecutive irreversible first order reaction can be represented by $A\xrightarrow[{}]{{{k}_{1}}}B\xrightarrow[{}]{{{k}_{2}}}C$. The rate equation for A is integrated to obtain ${{[A]}_{t}}={{[A]}_{0}}{{e}^{-{{k}_{1}}t}}$ and ${{[B]}_{t}}=\frac{{{k}_{1}}[{{A}_{0}}]}{{{k}_{2}}-{{k}_{1}}}[{{e}^{-{{k}_{1}}t}}-{{e}^{-{{k}_{2}}t}}]$. At what time will B be present in the greatest concentration? A) ${{t}_{\max }}=\frac{1}{{{k}_{1}}+{{k}_{2}}}\ln \frac{{{k}_{2}}}{{{k}_{1}}}$      B) ${{t}_{\max }}=\frac{1}{{{k}_{1}}+{{k}_{2}}}\ln \frac{{{k}_{2}}}{{{k}_{1}}}$ C) ${{t}_{\max }}=\frac{1}{{{k}_{2}}+{{k}_{1}}}\ln \frac{{{k}_{1}}}{{{k}_{2}}}$      D) None of these

 Idea This problem includes concept of rate law expression for consecutive reaction and determination of ${{t}_{\max }}$. While solving this problem students are advised to write the rate expression and solve the problem using differentiation. For maximum concentration of $B,\frac{d[B]}{dt}=0$ $\frac{d}{dt}\left[ \frac{{{k}_{1}}{{[A]}_{0}}}{{{k}_{2}}-{{k}_{1}}}({{e}^{-{{k}_{1}}t}}-{{e}^{-{{k}_{2}}t}}) \right]=0$ $\frac{{{k}_{1}}{{[A]}_{0}}}{{{k}_{2}}-{{k}_{1}}}\left[ \frac{d}{dt}({{e}^{-{{k}_{1}}t}}-{{e}^{-{{k}_{2}}t}}) \right]=0$ Solving (differentiating), we get ${{t}_{\max }}=\frac{1}{{{k}_{2}}-{{k}_{1}}}\ln \frac{{{k}_{1}}}{{{k}_{2}}}$ TEST Edge Similar problem including concept of rate law expression for parallel reaction using arrhenius equation may be asked, so students are advised to go through a great understanding of these topics.