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AMU Medical AMU Solved Paper-2004

  • question_answer
    For \[CaC{{O}_{3}}(s)CaO(s)+C{{O}_{2}}(g)\] at\[{{927}^{o}}C\], \[\Delta H=176\,kJ;\] then \[\Delta E\] is

    A)  180 kJ                                  

    B)  186.4 kJ

    C)  166.0 kJ                              

    D)  160 kJ

    Correct Answer: C

    Solution :

                     Key Idea The relation between \[\Delta H\] and \[\Delta E\] is given by                 \[\Delta H=\Delta E+\Delta {{n}_{g}}\,RT\]where \[\Delta {{n}_{g}}\] change in the number of gaseous moles. For the reaction                 \[CaC{{O}_{3}}(s)CaCO(s)+C{{O}_{2}}(g)\]                 \[\Delta H=176\,\,kJ\,mol\]                 \[\Delta \,{{n}_{g}}=1-0=1\]                 \[=273+927=1200\,K\]                 \[R=8.3\times {{10}^{-3}}\,kJ\,mo{{l}^{-1}}{{K}^{-1}}\] Put these value in the above equation                 \[176=\Delta E+1\times 8.31\times {{10}^{-3}}\times 1200\]                 \[=\Delta E+9.972\]                 \[\Delta E=176-9.972=166.03\,\,kJ\]


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