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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2007

  • question_answer
    The equlibrium constant for the reaction \[2N{{O}_{2}}(g)2NO(g)+{{O}_{2}}(g)\]is\[2\times {{10}^{-6}}\]at\[185{}^\circ C\]. Then the equilibrium constant for the reaction,\[4NO(g)+2{{O}_{2}}(g)\]\[4N{{O}_{2}}(g)\]at the same temperature would be

    A)  \[2.5\times {{10}^{-5}}\]             

    B)         \[4\times {{10}^{-12}}\]

    C)  \[2.5\times {{11}^{11}}\]            

    D)         \[2\times {{10}^{6}}\]

    E)  \[5\times {{10}^{5}}\]

    Correct Answer: C

    Solution :

    \[2N{{O}_{2}}(g)2NO(g)+{{O}_{2}}(g)\] \[K=\frac{{{[NO]}^{2}}[{{O}_{2}}]}{{{[N{{O}_{2}}]}^{2}}}=2\times {{10}^{-6}}\] \[4N{{O}_{2}}(g)+2{{O}_{2}}(g)4N{{O}_{2}}(g)\]                 \[K=\frac{{{[N{{O}_{2}}]}^{4}}}{{{[NO]}^{4}}{{[{{O}_{2}}]}^{2}}}\]                 \[=\frac{1}{{{(K)}^{2}}}=\frac{1}{{{(2\times {{10}^{-6}})}^{2}}}\] Equilibrium constant\[K=0.25\times {{10}^{12}}\] \[=2.5\times {{10}^{11}}\]


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