JEE Main & Advanced Sample Paper JEE Main Sample Paper-4

  • question_answer
    For the complete combustion of ethanol amount of heat produced as measured in bomb calorimeter, is \[1364.47\text{ kJ }mo{{l}^{-1}}\] at \[25{}^\circ C\]. Assuming ideality the enthalpy of combustion, \[\Delta {{H}_{C}},\] for the reaction will be\[(R=8.314J{{K}^{-1}}mo{{l}^{-1}})\]

    A) \[=1366.95\,kJ\,mo{{l}^{-1}}\]   

    B) \[=1361.95\,kJ\,mo{{l}^{-1}}\]

    C) \[-1460.50kJ\,mo{{l}^{-1}}\]       

    D) \[-1350.50\,kJ\,mo{{l}^{-1}}\]

    Correct Answer: A

    Solution :

    \[{{\Delta }_{r}}{{G}^{0}}={{\Delta }_{f}}{{G}^{0}}\](products)\[-{{\Delta }_{f}}{{G}^{0}}\](reactants) \[=-394.4-(-137.2)=-257.2kJ<0\] The above negative value of\[\Delta G\]indicates that the process is spontaneous. Also \[\Delta {{G}^{0}}=\Delta {{H}^{0}}-T\Delta {{G}^{0}}\Rightarrow \Delta {{H}^{0}}=\Delta {{G}^{0}}+T\Delta {{S}^{0}}\] \[=-257.2+300(-0.094)=-285.4kJ<0\]

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