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

  • question_answer
    The acid catalyzed hydrolysis of an organic, compound 'A' at \[30{}^\circ C\] has a time for half change of 100 minute when carried out in a buffer solution at \[(pH=5)\] and 10 minute when carried out at \[(pH=4)\]. Both times of half change are independent of the initial concentration of A. If rate constant K is defined by \[\frac{-d[A]}{dt}=K{{[A]}^{a}}{{[{{H}^{+}}]}^{b}},\] the values of a and b respectively are

    A) 1, 1  

    B) 1, 2

    C) 0, 1                  

    D) 1, 0

    Correct Answer: A

    Solution :

    During any experiment, pH is constant, hence
    \[\frac{-d[A]}{dt}=K'{{[A]}^{a}}\] where \[K'=K{{[{{H}^{+}}]}^{b}}\]
    As \[{{t}_{1/2}}\] is independent of initial conc. So \[a=1\]
    Also K' can be given by \[=\frac{0.693}{{{t}_{1/2}}}\]
    \[\frac{{{({{t}_{1/2}})}_{1}}}{{{({{t}_{1/2}})}_{2}}}=\frac{K_{2}^{'}}{K_{1}^{'}}=\frac{K[{{H}^{+}}]_{2}^{b}}{K[{{H}^{+}}]_{1}^{b}}=\frac{[{{H}^{+}}]_{2}^{b}}{[{{H}^{+}}]_{1}^{b}}\]
    Now \[\frac{100}{10}=\left( \frac{{{10}^{-4}}}{{{10}^{-5}}} \right)\text{or}\,b=1\]


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