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DUMET Medical DUMET Medical Solved Paper-2001

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

    A)  -44 kcal       

    B)  -88 kcal

    C)  -22kcal       

    D)  -11 kcal

    Correct Answer: C

    Solution :

    Key Idea: First write the equation for chemical change and then by using following formula enthalpy of formation of \[HCl\] gas can be calculated. \[\Delta {{H}_{f}}=\Sigma \] bond energy of reactants \[-\Sigma \] bond energy of products \[\frac{1}{2}{{H}_{2}}+\frac{1}{2}C{{l}_{2}}\to HCl\,(g)\] Given: Bond energy of \[{{H}_{2}}=104\,kcal\] Bond energy of \[C{{l}_{2}}=58\,kcal\] Bond energy of \[HCl=103\,kcal\] \[\Delta {{H}_{f}}=\frac{1}{2}(104+58)-103\] \[=81-103\] = - 22 kcal


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