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VMMC VMMC Medical Solved Paper-2010

  • question_answer
    The first order rate constant for the decomposition of ethyl iodide by the reaction \[{{C}_{2}}{{H}_{5}}I(g)\xrightarrow{{}}{{C}_{2}}{{H}_{4}}(g)+HI(g)\] at 600 K is\[1.60\times {{10}^{-5}}{{s}^{-1}},\]. Its energy of activation is \[209\,kJ\,mo{{l}^{-1}}.\] What is the rate constant of the reaction at 700 K?

    A) \[6.36\times {{10}^{3}}{{s}^{-1}}\]                           

    B) \[6.36\times {{10}^{-3}}{{s}^{-1}}\]

    C)        \[6.36\times {{10}^{3}}s\]           

    D)        \[6.36\times {{10}^{-3}}s\]

    Correct Answer: B

    Solution :

    \[{{E}_{a}}=209\,kJ\,mo{{l}^{-1}}=209\times {{10}^{3}}J\,mo{{l}^{-1}}\] \[{{k}_{1}}=1.60\times {{10}^{-5}}{{s}^{-1}},{{T}_{1}}=600K\] \[{{k}_{2}}=?\]  \[{{T}_{2}}=700\,K\] Substituting the values \[\log \frac{{{k}_{2}}}{1.60\times {{10}^{-5}}}=\frac{209\times {{10}^{3}}}{2.303\times 8.314}\left[ \frac{700-600}{700\times 600} \right]\] \[\log \frac{{{k}_{2}}}{1.60\times {{10}^{-5}}}=2.599\] \[\frac{{{k}_{2}}}{1.60\times {{10}^{-5}}}=397.19\] \[\therefore \]  \[{{k}_{2}}=379.19\times 1.60\times {{10}^{-5}}\]                 \[=6.36\times {{10}^{-3}}{{s}^{-1}}\]


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