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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2010

  • question_answer
    Consider the following reactions in which all the reactants and the products are in gaseous state. \[2PQ{{P}_{2}}+{{Q}_{2}};\]        \[{{K}_{1}}=2.5\times {{10}^{5}}\] \[PQ+\frac{1}{2}{{R}_{2}}PQR;\]               \[{{K}_{2}}=5\times {{10}^{-3}}\] The value of\[{{K}_{3}}\]for the equilibrium \[\frac{1}{2}{{P}_{2}}+\frac{1}{2}{{Q}_{2}}+\frac{1}{2}{{R}_{2}}PQR,\]is

    A)  \[2.5\times {{10}^{-3}}\]             

    B) \[2.5\times {{10}^{3}}\]

    C)  \[1.0\times {{10}^{-5}}\]       

    D)        \[5\times {{10}^{3}}\]

    E) \[5\times {{10}^{-3}}\]

    Correct Answer: C

    Solution :

    Given, \[2PQ{{P}_{2}}+{{Q}_{2}};\] \[{{K}_{1}}=\frac{[{{P}_{2}}][{{Q}_{2}}]}{{{[PQ]}^{2}}}=2.5\times {{10}^{5}}\] \[PQ+\frac{1}{2}{{R}_{2}}PQR;\] \[{{K}_{2}}=\frac{[PQR]}{[PQ]{{[{{R}_{2}}]}^{1/2}}}=5\times {{10}^{-3}}\] Required equilibrium is \[\frac{1}{2}{{P}_{2}}+\frac{1}{2}{{Q}_{2}}+\frac{1}{2}{{R}_{2}}PQR\] \[{{K}_{3}}=\frac{[PQR]}{{{[{{P}_{2}}]}^{1/2}}{{[{{Q}_{2}}]}^{1/2}}{{[{{R}_{2}}]}^{1/2}}}\] On multiplying and dividing by\[[PQ],\] \[{{K}_{3}}=\frac{[PQR]}{{{[{{P}_{2}}]}^{1/2}}{{[{{Q}_{2}}]}^{1/2}}{{[{{R}_{2}}]}^{1/2}}[PQ]}\] Or           \[{{K}_{3}}={{K}_{2}}\times \sqrt{\frac{1}{{{K}_{1}}}}\]                 \[=5\times {{10}^{-3}}\times \sqrt{\frac{1}{2.5\times {{10}^{5}}}}\]                 \[=\frac{5\times {{10}^{-3}}}{0.5\times {{10}^{3}}}=1\times {{10}^{-5}}\]


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