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Uttarakhand PMT Uttarakhand PMT Solved Paper-2011

  • question_answer
    At a given temperature, the equilibrium constants for the reactions, \[NO(g)+\frac{1}{2}{{O}_{2}}(g)N{{O}_{2}}(g)\] and \[2N{{O}_{2}}(g)2NO(g)+{{O}_{2}}(g)\] are\[{{K}_{1}}\]and\[{{K}_{2}}\]respectively. If\[{{K}_{1}}\]is\[4\times {{10}^{-3}}\]then will be

    A)  \[8\times {{10}^{-3}}\]

    B)  \[16\times {{10}^{-3}}\]

    C)  \[6.25\times {{10}^{4}}\]

    D)  \[6.25\times {{10}^{6}}\]

    Correct Answer: C

    Solution :

     \[NO(g)+\frac{1}{2}{{O}_{2}}(g)N{{O}_{2}}(g);\]\[{{K}_{1}}=4\times {{10}^{-3}}\] \[4\times {{10}^{-3}}=\frac{[N{{O}_{2}}]}{[NO]{{[{{O}_{2}}]}^{1/2}}}\] On squaring, we get \[16\times {{10}^{-6}}=\frac{{{[N{{O}_{2}}]}^{2}}}{{{[NO]}^{2}}[{{O}_{2}}]}\] ?.(i) \[2N{{O}_{2}}(g)2NO(g)+{{O}_{2}}(g);\]       \[{{K}_{2}}=?\] \[{{K}_{2}}=\frac{{{[NO]}^{2}}[{{O}_{2}}]}{{{[N{{O}_{2}}]}^{2}}}\] ?(ii) From Eqs. (i) and (ii), \[{{K}_{2}}=\frac{1}{{{K}_{1}}}=\frac{1}{16\times {{10}^{-6}}}=6.25\times {{10}^{4}}\]


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