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Uttarakhand PMT Uttarakhand PMT Solved Paper-2009

  • question_answer
    For a  gaseous  reaction at  300 K. \[\Delta H-\Delta E=-4.98\text{ }kJ,\]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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