BHU PMT BHU PMT Solved Paper-2003

  • question_answer
    \[{{C}_{(diamond)}}{{O}_{2}}\xrightarrow[{}]{{}}C{{O}_{2}};\]\[\Delta H=-395.4\,kJ/mol\] \[{{C}_{(graphite)}}+{{O}_{2}}\xrightarrow[{}]{{}}C{{O}_{2}}\Delta H;=-393.5\,kJ/mol\] \[{{C}_{(graphite)}}\xrightarrow[{}]{{}}{{C}_{(dia\operatorname{mo}nd)}};\Delta H=?:\]

    A)  \[-3.8\]                               

    B) \[-1.9\]

    C)  \[+3.8\]                              

    D) \[+1.9\]

    Correct Answer: D

    Solution :

                     Key Idea: Use Hess law which states that total heat changes during a chemical reaction are independent of path. \[{{C}_{(diamond)}}+{{O}_{2}}\xrightarrow{{}}C{{O}_{2}}\] \[\Delta H=-395.4\text{ }kJ/mol\]   ...(1) \[{{C}_{(graphite)}}+{{O}_{2}}\xrightarrow{{}}C{{O}_{2}}\] \[\Delta H=-393.5\text{ }kJ/mol\]  ...(2) \[{{C}_{(graphite)}}\xrightarrow{{}}{{C}_{(diamond)}}\Delta H=?\] Subtracting Eq. (1) from Eq. (2) we get \[{{C}_{(graphite)}}\xrightarrow{{}}{{C}_{(diamond)}},\Delta H=1.9\,kJ/mol\]


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