VMMC VMMC Medical Solved Paper-2011

  • question_answer
    For the third order reaction, \[\text{3A}\to \]Products, with 0.1 M as the initial concentration of \[\text{A,}{{\text{t}}_{\text{1/2}}}\]is 8 h 20 min. The rate constant of the reaction is

    A) \[50\,{{L}^{2}}mo{{l}^{-2}}{{s}^{-1}}\]    

    B) \[5\times {{10}^{-3}}L\,mo{{l}^{-1}}{{s}^{-1}}\]

    C)         \[5\times {{10}^{-2}}{{L}^{2}}\,mo{{l}^{-2}}{{s}^{-1}}\]

    D)         \[5\times {{10}^{-3}}{{L}^{2}}mo{{l}^{-2}}{{s}^{-1}}\]

    Correct Answer: D

    Solution :

    For a third order reaction, \[k=\frac{3}{2\times {{t}_{1/2}}\times {{a}^{2}}}\] where, k = rate constant a = initial concentration of reactant Given, \[{{t}_{1/2}}=8\,h\,20\,\min =30000\,s\] \[a=0.1\,M\]                 \[\therefore \]  \[k=\frac{3}{2\times 30000\times 0.1\times 0.1}\]                                 \[=5.0\times {{10}^{-3}}{{L}^{2}}\,mo{{l}^{-2}}{{s}^{-1}}\]


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