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JIPMER Jipmer Medical Solved Paper-2001

  • question_answer
    For the reaction\[2N{{O}_{2}}(g)\rightleftharpoons 2NO(g)+{{O}_{2}}(g)\] \[{{K}_{c}}=1.8\times {{10}^{-6}}\]at\[185{}^\circ C.\] At \[185{}^\circ C,\]the \[{{K}_{c}}\]for reaction\[NO(g)+\frac{1}{2}({{O}_{2}})\,(g)\rightleftharpoons N{{O}_{2}}(g)is:\]

    A)  \[1.95\times {{10}^{-3}}\]                           

    B)  \[1.95\times {{10}^{3}}\]            

    C)         \[7.5\times {{10}^{2}}\]               

    D)         \[0.9\times {{10}^{6}}\]

    Correct Answer: C

    Solution :

    \[2N{{O}_{2}}(g)\rightleftharpoons 2NO(g)+{{O}_{2}}(g)\] \[{{K}_{c}}=\frac{{{[NO]}^{2}}[{{O}_{2}}]}{{{[N{{O}_{2}}]}^{2}}}=1.8\times {{10}^{-6}}\] For reaction \[\frac{1}{2}{{O}_{2}}(g)+NO(g)\rightleftharpoons N{{O}_{2}}(g)\] \[K_{c}^{}=\frac{[N{{O}_{2}}]}{{{[{{O}_{2}}]}^{1/2}}[NO]}=\sqrt{\frac{1}{{{K}_{c}}}}\] \[=\sqrt{\frac{1}{1.8\times {{10}^{-6}}}}=\sqrt{0.56\times {{10}^{-6}}}\] \[=0.75\times {{10}^{3}}=7.5\times {{10}^{2}}\]


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