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VMMC VMMC Medical Solved Paper-2009

  • question_answer
    For  a  gaseous  reaction  at  300  K. \[\Delta H=\Delta E=-4.98\,kJ,\] assuming that\[R=8.3\,J{{K}^{-1}}\,mo{{l}^{-1}},\,\Delta {{n}_{(g)}}\]is

    A) 1                                             

    B) 2                             

    C) \[-2\]                    

    D) 0

    Correct Answer: C

    Solution :

    We know that, \[\Delta H=\Delta E+\Delta {{n}_{g}}RT\] \[\Delta H-\Delta E=\Delta {{n}_{g}}RT\]                 \[\Rightarrow \]               \[-4.98=\Delta {{n}_{g}}\times 8.3\times {{10}^{-3}}\times 300\]                                 \[\Delta {{n}_{g}}=\frac{-4.98}{8.3\times {{10}^{-3}}\times 300}\]                 \[\therefore \]  \[\Delta {{n}_{g}}=-2\]


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