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JCECE Medical JCECE Medical Solved Paper-2014

  • question_answer
    The bond dissociation energies of gaseous \[{{\text{H}}_{\text{2}}}\text{,C}{{\text{l}}_{\text{2}}}\]and\[\text{HCl}\] are 104, 58 and 103 kcal respectively. The enthalpy of formation of \[\text{HCl}\]gas would be

    A) \[~-\text{ }44\text{ kcal}\]

    B) \[44\text{ kcal}\] 

    C) \[~-\text{ }22\text{ kcal}\]

    D) \[\text{ }\!\!~\!\!\text{ 22 kcal}\]

    Correct Answer: C

    Solution :

     \[\frac{1}{2}{{H}_{2}}+\frac{1}{2}{{C}_{2}}\xrightarrow{{}}HCl\] \[\Delta \Eta =\sum{B{{E}_{\text{reactants}}}}-\sum{B{{E}_{\text{products}}}}\] \[=\left[ \frac{1}{2}BE({{H}_{2}})+\frac{1}{2}BE(C{{l}_{2}}) \right]-BE(HCl)\] \[=\left[ \left( \frac{1}{2}\times 104 \right)+\left( \frac{1}{2}\times 58 \right) \right]-103\] \[=(52+29)-103\] \[=-22\,kcal\]


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