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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2001

  • question_answer
    The heats of combustion of graphite and carbon monoxide, respectively are, \[-\text{ }393.5\text{ }kJ\text{ }mo{{l}^{-1}}\]and\[-283\text{ }kJ\text{ }mo{{l}^{-1}}\]. Therefore, the heat of formation of carbon monoxide in, \[kJ\text{ }mo{{l}^{-1}}\]is:

    A) +172.5                                  

    B) \[-110.5\]

    C) \[-1070\]             

    D)        \[-676.5\]

    E) + 110.5

    Correct Answer: B

    Solution :

    Combustion of graphite is as follows: \[C(s)+{{O}_{2}}(g)\xrightarrow[{}]{{}}C{{O}_{2}}(g)+393.5\,kJ\]                ...(i) Combustion of carbon monoxide is, \[CO(g)+\frac{1}{2}(g)\xrightarrow[{}]{{}}C{{O}_{2}}(g)+283\,kJ\]...(ii) For obtaining\[CO\]as a product we reverse the Eq. (ii) as \[C{{O}_{2}}(g)\xrightarrow[{}]{{}}CO(g)+\frac{1}{2}{{O}_{2}}(g)-283\,kJ\]...(iii) \[C(g)+{{O}_{2}}(g)\xrightarrow[{}]{{}}C{{O}_{2}}(g)+393.5\,kJ\]...(iv) On adding Eq. (i) and Eq. (iii) \[C+\frac{1}{2}{{O}_{2}}\xrightarrow[{}]{{}}CO+(393.5-283)\] \[=+110.5\text{ }kJ\] \[\therefore \] Heat of formation of CO is \[=+\text{ }110.5\text{ }kJ\] \[\therefore \] \[\Delta H=-110.5\text{ }kJ\]


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