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Manipal Medical Manipal Medical Solved Paper-2012

  • question_answer
    If the bond dissociation energies of\[XY,{{X}_{2}}\]and \[{{Y}_{2}}\] (all diatomic molecules) are in the ratio of \[1:1:0.5\] and\[\Delta {{H}_{f}}\]for the formation of XY is\[-200\text{ }kJ\text{ }mo{{l}^{-1}},\]the bond dissociation energy of\[{{X}_{2}}\]will be .

    A)  \[400\text{ }kJ\text{ }mo{{l}^{-1}}\]

    B)  \[300\text{ }kJ\text{ }mo{{l}^{-1}}\]

    C)  \[200\text{ }kJ\text{ }mo{{l}^{-1}}\]

    D)  \[800\text{ }kJ\text{ }mo{{l}^{-1}}\]

    Correct Answer: D

    Solution :

     \[{{X}_{2}}+{{Y}_{2}}\xrightarrow[{}]{{}}2XY\] \[\Delta H={{(BE)}_{X-X}}+{{(BE)}_{Y-Y}}-2{{(BE)}_{X-Y}}\] If BE of \[(XY)=a\] Then BE of \[(XX)=a\] and BE of \[(Y-Y)=\frac{a}{2}\] \[\Delta {{H}_{f}}=(X-Y)=-200\,kJ\] \[\therefore \]\[-400\] (for 2 moles\[XY\])\[=a+\frac{a}{2}-2a\] \[-400=-\frac{a}{2}\] or              \[a=+800\text{ }kJ\]


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