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RAJASTHAN PMT Rajasthan - PMT Solved Paper-2008

  • question_answer
    For the reaction equilibrium,                 \[{{N}_{2}}{{O}_{4}}(g)2NO(g)\] the concentrations of \[{{N}_{2}}{{O}_{4}}\] and \[N{{O}_{2}}\] at equilibrium are \[4.8\times {{10}^{-2}}\] and \[1.2\times {{10}^{-2}}mol\,\,{{L}^{-1}}\]respectively. The value of \[{{K}_{c}}\] for the reaction is

    A) \[3.3\times {{10}^{2}}mol\,\,{{L}^{-1}}\]               

    B)        \[3\times {{10}^{-1}}mol\,\,{{L}^{-1}}\]

    C)        \[3\times {{10}^{-3}}mol\,\,{{L}^{-1}}\] 

    D)        \[3\times {{10}^{3}}mol\,\,{{L}^{-1}}\]

    Correct Answer: C

    Solution :

    \[{{K}_{c}}=\frac{{{[N{{O}_{2}}]}^{2}}}{[{{N}_{2}}{{O}_{4}}]}=\frac{{{(1.2\times {{10}^{-2}})}^{2}}}{4.8\times {{10}^{-2}}}\]                          \[=3\times {{10}^{-3}}mol\,\,{{L}^{-1}}\]


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