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JEE Main & Advanced JEE Main Paper (Held On 11 April 2015)

  • question_answer
    For the equilibrium, \[A(g)\rightleftharpoons B(g),\Delta H\]is -40 kJ/mol. If the ratio of the activation energies of the forward \[({{E}_{f}})\] and reverse \[({{E}_{b}})\]reactions is \[\frac{2}{3}\]then : [JEE Main Online Paper (Held On 11 April 2015)]  

    A) \[{{E}_{f}}=60\,kJ/mol;{{E}_{b}}=100kJ/mol\]

    B) \[{{E}_{f}}=30\,kJ/mol;{{E}_{b}}=70kJ/mol\]

    C) \[{{E}_{f}}=80\,kJ/mol;{{E}_{b}}=120kJ/mol\]

    D) \[{{E}_{f}}=70\,kJ/mol;{{E}_{b}}=30kJ/mol\]

    Correct Answer: C

    Solution :

    \[Ef=80kj/mole;\,Eb=120kj/mole\] given, \[A\left( g \right)B\left( g \right)\] \[\Delta H=-40\,kg/mole\] \[\frac{Ef}{Eb}=\frac{2}{3}\]we know that \[Ef-Eb=\Delta H\] \[Ef-Eb=-40\] \[Eb\frac{2}{3}-Eb=-40\] Eb = 120 kg/mole Ef = 80 kg/mole


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