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Chhattisgarh PMT Chhattisgarh PMT Solved Paper-2004

  • question_answer
    The equilibrium constant for the reversible reaction, \[{{N}_{2}}+3{{H}_{2}}\rightleftharpoons 2N{{H}_{3}}\] is K and for the reaction \[\frac{1}{2}{{N}_{2}}+\frac{3}{2}{{H}_{2}}\rightleftharpoons N{{H}_{3}},\] the equilibrium constant is \[K'\]. K and \[K'\] will be related as

    A)  \[K=K'\]

    B)  \[K=\sqrt{K}\]

    C)  \[K=\sqrt{K'}\]

    D)  \[K\times K'=1\]

    Correct Answer: B

    Solution :

    \[{{N}_{2}}(g)+3{{H}_{2}}(g)\rightleftharpoons 2N{{H}_{3}}(g)\] From law of mass action,                 \[K=\frac{{{[N{{H}_{3}}]}^{2}}}{[{{N}_{2}}]{{[{{H}_{2}}]}^{3}}}\]                 ?..(i) Similarly, for reaction                 \[\frac{1}{2}{{N}_{2}}+\frac{3}{2}{{H}_{2}}\rightleftharpoons N{{H}_{3}}\]                 \[K'=\frac{[N{{H}_{3}}]}{{{[{{N}_{2}}]}^{1/2}}{{[{{H}_{2}}]}^{3/2}}}\] So, from Eqs.(i) and (ii)                 \[K'=\sqrt{K}\]


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