JEE Main & Advanced JEE Main Online Paper (Held On 08 April 2018)

  • question_answer
    The combustion of benzene (l) gives \[\text{C}{{\text{O}}_{2}}(g)\] and\[{{\text{H}}_{\text{2}}}\text{O(l)}\]. Given that heat of combustion of benzene at constant volume is \[\text{-3263}\text{.9 kJ mo}{{\text{l}}^{\text{-1}}}\] at \[\text{25 }\!\!{}^\circ\!\!\text{ C}\]; heat of combustion (in\[\text{kJ mo}{{\text{l}}^{\text{-1}}}\]) of benzene at constant pressure will be:                       [JEE Main Online 08-04-2018] \[\text{(R=8}\text{.314 J}{{\text{K}}^{\text{-1}}}\text{ mo}{{\text{l}}^{\text{-1}}}\text{)}\]

    A)  3260              

    B)  -3267.6

    C)  4152.6                       

    D)  -452.46

    Correct Answer: B

    Solution :

    \[{{C}_{6}}{{H}_{6}}(\ell )+\frac{15}{2}{{O}_{2}}(g)\xrightarrow{{}}6C{{O}_{2}}(g)+3{{H}_{2}}O(\ell )\] \[\Delta {{n}_{g}}=6-7.5=-1.5\] \[\Delta H=\Delta U+\Delta {{n}_{g}}RT\] \[\Delta H=-3263.9-\frac{1.5\times 8.314\times 298}{1000}=-3267.6\,\,kJ\,\,mo{{l}^{-1}}\]


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