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JCECE Engineering JCECE Engineering Solved Paper-2014

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

    A) \[-44\,\,kcal\]                   

    B) \[44\,\,kcal\]

    C) \[-22\,\,kcal\]                   

    D) \[22\,\,kcal\]

    Correct Answer: C

    Solution :

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


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