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JCECE Engineering JCECE Engineering 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 equilibrium constant will be

    A) \[16\]                                   

    B) \[32\]

    C)  \[64\]                                  

    D)  \[128\]

    Correct Answer: C

    Solution :

    \[{{H}_{2}}(g)+{{I}_{2}}(g)2Hl(g)\] 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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