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BCECE Engineering BCECE Engineering Solved Paper-2010

  • question_answer
    The enthalpies of combustion of carbon and carbon monoxide are \[~-393.5\] and \[-283\text{ kJ mo}{{\text{l}}^{-1}}\]respectively. The enthalpy of formation of carbon monoxide per mole is

    A) \[\text{110}\text{.5kJ}\]                              

    B)  \[\text{676}\text{.5 kJ}\]                            

    C) \[-\text{676}\text{.5 kJ}\]       

    D)        \[-\text{110}\text{.5kJ}\]

    Correct Answer: A

    Solution :

    (i) \[C(s)+{{O}_{2}}(g)\xrightarrow{{}}C{{O}_{2}}(g);\] \[\Delta H=-393.5\,kJ\] (ii) \[CO(g)+\frac{1}{2}{{O}_{2}}(g)\xrightarrow{{}}C{{O}_{2}}(g);\]                                 \[\Delta H=-283.0\,kJ\] On subtracting Eq. (II) from Eq. (I), we get (iii) \[C(s)+\frac{1}{2}{{O}_{2}}(g)\xrightarrow{{}}CO(g);\]                                 \[\Delta \Eta =-110.5\,kJ\] The equation III also represents formation of 1 mole of CO and thus, enthalpy change, ie, \[-110.5\,kJ,\]is the heat of formation of \[CO(g).\]


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