DUMET Medical DUMET Medical Solved Paper-2002

  • question_answer
    Calculate the standard heat of formation of carbon disulphide in liquid state. Given that the standard heat of combustion of carbon (s) sulphur (s) and carbon disulphide \[(l)\] are  393.3, - 29a72 and -1108.76 \[kJ\,mo{{l}^{-1}}\]respectively is:

    A)  + 128.02 \[kJ\,mo{{l}^{-1}}\]

    B)  -12802 \[kJ\,mo{{l}^{-1}}\]

    C)  + 12.802 \[kJ\,mo{{l}^{-1}}\]

    D)  -128.02 \[kJ\,mo{{l}^{-1}}\]

    Correct Answer: A

    Solution :

    Key Idea: Use Hesss law to solve the problem. Given : \[C(s)+{{O}_{2}}(g)\to C{{O}_{2}}(g)\Delta {{H}_{1}}=-393.3\,kJ\] ... (i) \[S(s)+{{O}_{2}}(g)\to S{{O}_{2}}(g)\Delta {{H}_{2}}=-293.72\,kJ\]...(ii) \[C{{S}_{2}}(l)+3{{O}_{2}}(g)\to C{{O}_{2}}(g)+2S{{O}_{2}}(g)\] \[\Delta {{H}_{3}}=-1108.76\,kJ\] ... (iii) Required equation \[C(s)+2S(s)-C{{S}_{2}}(g),\,\Delta H=?\] Multiply Eq. (ii) by 2 and add to Eq. (i) \[C(s)+2S(s)-3{{O}_{2}}(g)\to C{{O}_{2}}(g)+2S{{O}_{2}}(g)\] \[\Delta {{H}_{4}}=-392.3+2\times -293.72\] \[=-979.14kJ\] ... (iv) Subtract Eq. (iii) from Eq. (iv) \[C(s)+2S(s)\to C{{S}_{2}}\] \[\Delta H=-979.14-(-1108.76)=128.02\,kJ\,mo{{l}^{-1}}\]


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