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NEET Sample Paper NEET Sample Test Paper-81

  • question_answer
    Given that \[{{\operatorname{C}}_{diamond}}\,+{{O}_{2}}\,\,\xrightarrow{{}}\,\,C{{O}_{2}},\,\,\Delta H=395.3\,\,kJ/mol\] \[{{\operatorname{C}}_{graphite}}\,+{{O}_{2}}\,\,\xrightarrow{{}}\,\,C{{O}_{2}},\,\,\Delta H=-\,393.4\,\,kJ/mol\] Then for reaction\[{{C}_{graphite}}~\xrightarrow{{}}\,\,{{C}_{diamond}}\] What is the value of \[\Delta H\]?

    A)  - 3.8                

    B)  + 3.8

    C)  - 1.9                

    D)  + 1.9

    Correct Answer: D

    Solution :

    Given that \[{{C}_{diamond}}+{{O}_{2}}\xrightarrow[{}]{}\,\,C{{O}_{2}},\,\,\Delta H=395.3\,kJ/mol\]\[{{C}_{graphite}}+{{O}_{2}}\xrightarrow[{}]{}\,\,C{{O}_{2}},\,\,\Delta H=-393.4\,kJ/mol\] eqs. (i)-(ii) \[{{C}_{diamond}}\xrightarrow{{}}{{C}_{graphite}}\] \[\Delta H = \Delta {{H}_{1}} - \Delta {{H}_{2}}\] \[= \,395.3 - 393.4 = 1.9 kJ mo{{l}^{-}}^{1}\]


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