JEE Main & Advanced Chemistry Chemical Kinetics / रासायनिक बलगतिकी Sample Paper Topic Test - Chemical Kinetics

  • 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

    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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