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Uttarakhand PMT Uttarakhand PMT Solved Paper-2006

  • question_answer
    \[C(dia)+{{O}_{2}}\xrightarrow[{}]{{}}C{{O}_{2}};\]\[\Delta H=-395.4\text{ }kJ/mol\] \[C(gr)+{{O}_{2}}\xrightarrow[{}]{{}}C{{O}_{2}}\] \[\Delta H=-393.5\text{ }kJ/mol\] \[C(gr)\xrightarrow[{}]{{}}C(dia);\Delta H=?:\]

    A)  \[-3.8\]             

    B)  \[-1.9\]

    C)  \[+3.8\]             

    D)  \[+1.9\]

    Correct Answer: D

    Solution :

     \[C(dia)+{{O}_{2}}\to C{{O}_{2}}\] \[(\Delta H=-395.4kJ/mol)\]              ...(i) \[C(gr)+{{O}_{2}}\to C{{O}_{2}}\] \[(\Delta H=-393.5kJ/mol)\]             ...(2) Subtracting Eq. (i) from (ii), we have \[C(gr)\xrightarrow[{}]{{}}C(dia);\] \[(\Delta H=+1.9kJ/mol)\] [\[\because \]\[\Delta H=-393.5-(395.4)=+1.9kJ\,mo{{l}^{-1}}\]]


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