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JCECE Medical JCECE Medical Solved Paper-2014

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

    A)  16               

    B)  32

    C)  64               

    D)  128

    Correct Answer: C

    Solution :

     \[{{H}_{2}}(g)+{{l}_{2}}(g)2Hl(g)\] For this reaction, \[\Delta {{n}_{g}}=0\] \[\therefore \]The reaction and its equilibrium constant is not affected by change in volume. Moreover, equilibrium   constant  depends  only  on temperature.


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