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JEE Main & Advanced JEE Main Paper (Held On 12 May 2012)

  • question_answer
    The difference between the reaction enthalpy change \[({{\Delta }_{r}}H)\]arid reaction internal energy change \[({{\Delta }_{r}}U)\] for the reaction: \[2{{C}_{6}}{{H}_{6}}(l)+15{{O}_{2}}(g)\xrightarrow[{}]{{}}12C{{O}_{2}}(g)+6{{H}_{2}}O(l)\]at 300 K is \[(R=8.314\,J\,mo{{l}^{-1}}\,{{K}^{-1}})\]   JEE Main Online Paper (Held On 12 May 2012)

    A) \[0\,J\,mo{{l}^{-1}}\,\]

    B)                        \[2490\,J\,mo{{l}^{-1}}\,\]

    C)                        \[-2490\,J\,mo{{l}^{-1}}\,\]         

    D)                        \[-7482\,J\,mo{{l}^{-1}}\,\]

    Correct Answer: D

    Solution :

                     \[\Delta H=\Delta U+\Delta {{n}_{g}}RT\] For the reaction \[\Delta {{n}_{g}}=12-15=-3\] \[\Delta H-\Delta U=-3\times 8.314\times 300\] \[=-7482\,J\,mo{{l}^{-1}}\]


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