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JIPMER Jipmer Medical Solved Paper-2001

  • question_answer
    The\[{{K}_{c}}\]for\[{{H}_{2}}(g)+{{I}_{2}}(g)\rightleftharpoons 2HI(g)\] is 64. If the volume of the container is reduced to one-half of its original volume, the value of the equilibrium constant will be:

    A)  28                                         

    B)  64                         

    C)  32                         

    D)         16

    Correct Answer: B

    Solution :

    \[\begin{matrix}    {} & {{H}_{2}}(g)+ & {{I}_{2}}(g)\rightleftharpoons  & 2HI(g)  \\    \text{at}\,\,\text{equilibrium} & (x-a) & (x-a) & (2a)  \\ \end{matrix}\] \[{{K}_{c}}=\frac{{{[HI]}^{2}}}{[{{H}_{2}}]\,[{{I}_{2}}]}=\frac{{{\left( \frac{2a}{V} \right)}^{2}}}{\frac{(x-a)}{V}\times \frac{x-a}{V}}\] \[=\frac{4{{a}^{2}}}{{{(x-a)}^{2}}}\] Thus, \[{{K}_{c}}\] is independent volume for this reaction, hence, \[{{K}_{c}}\] will remain equal to 64 (i.e., equal to initial\[{{K}_{c}}\]).


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