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Manipal Engineering Manipal Engineering Solved Paper-2010

  • question_answer
    The activation energy of exothermic reaction \[A\to B\]is\[80\,\,kJ\,\,mo{{l}^{-1}}\]. The heat of reaction is\[200\,\,kJ\,\,mo{{l}^{-1}}\]. The activation energy for the reaction\[B\to A\]\[(in\,\,kJ\,\,mo{{l}^{-1}})\]will be

    A) 80                                          

    B) 120

    C) 40                          

    D)        280

    Correct Answer: D

    Solution :

    \[{{E}_{a}}(B\to A)=80\,\,kJ\,\,mo{{l}^{-1}}\] Heat of reaction\[(A\to B)=200kJ\,\,mo{{l}^{-1}}\] For\[(B\to A)\]backward reaction, \[{{E}_{a}}(B\to A)={{E}_{a}}(A\to B)+\]heat of reaction \[=80+200=280\,\,kJ\,\,mo{{l}^{-1}}\]


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