| 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
Correct Answer: B
Solution :
| 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. |
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