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

  • question_answer
    The reaction cis\[-X\text{trans}-X\] is first order in both directions. At \[\text{25}{}^\circ \text{C,}\] the equilibrium constant is 0.10 and the rate constant \[{{k}_{f}}=3\times {{10}^{-4}}{{s}^{-1}}\]. In an experiment starting with the pure cis-form, how long would it take for half of the equilibrium amount of the trans-isomer to be formed?

    A) 150 s                            

    B) 200 s

    C) 240 s                

    D) 210 s

    Correct Answer: D

    Solution :

    \[{{K}_{(aq)}}=\frac{{{k}_{f}}}{{{k}_{b}}};{{k}_{b}}=\frac{3\times {{10}^{-4}}}{0.1}=3\times {{10}^{-3}}\]
    Initial a 0
    At time t \[a-x\] x
    At equilibrium \[a-{{x}_{e}}\] \[{{x}_{e}}\]
    As we know,
    \[\left( {{k}_{f}}+{{k}_{b}} \right)=\frac{1}{t}\ln \,\left( \frac{{{x}_{e}}}{{{x}_{e}}-x} \right)\]
    Given \[x=\frac{{{x}_{e}}}{2}\]
    \[\therefore \]      \[\,\left( {{k}_{f}}+{{k}_{b}} \right)=\frac{1}{2}\ln 2\]
    or         \[\left( 3\times {{10}^{-3}}+3\times {{10}^{-4}} \right)=\frac{0.693}{t}\]

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