Punjab Medical Punjab - MET Solved Paper-2008

  • question_answer
    Calculate the difference between \[\Delta E\] and \[\Delta H\]for the following reaction at \[{{27}^{o}}C\](in kcal).                 \[{{C}_{(graphite)}}+{{H}_{2}}(g)\xrightarrow{{}}C{{H}_{4}}(g)\]

    A) \[-0.6\]                                

    B) \[-1.2\]

    C) \[+0.6\]                                               

    D) \[+1.2\]

    Correct Answer: C

    Solution :

    \[{{C}_{(graphite)}}+2{{H}_{2}}(g)\xrightarrow{{}}C{{H}_{4}}(g)\]                 \[\Delta H=\Delta E+\Delta {{n}_{g}}RT\]      \[\Delta E-\Delta H=-\Delta {{n}_{g}}RT\]               \[\Delta {{n}_{g}}=1-2=-1\]      \[\Delta E-\Delta H=-(-1)\times 2\times 300\]                        \[=600\,\,cal\]                       \[=0.6\,\,kcal\]


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