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J & K CET Engineering J and K - CET Engineering Solved Paper-2009

  • question_answer
    In a reversible reaction, the enthalpy change and the, activation energy in the forward direction  are  respectively \[-x\,\text{kJ}\,\text{mo}{{\text{l}}^{-1}}\] and \[y\,kJ\,mo{{l}^{-1}},\]Therefore, the energy of activation in the backward direction in \[kJmo{{l}^{-1}},\]is

    A) \[(y-x)\]

    B) \[(x+y)\]

    C)  \[(x-y)\]

    D) \[-(x+y)\]

    Correct Answer: A

    Solution :

     In case of exothermic reaction, \[{{E}_{a(back)}}={{E}_{a(forward)}}+\Delta H\] \[=y-x\]


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