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JEE Main & Advanced AIEEE Solved Paper-2005

  • 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     AIEEE  Solved  Paper-2005

    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)        None of these

    Correct Answer: D

    Solution :

    Formation of\[XY\]is shown as \[{{X}_{2}}+{{Y}_{2}}\xrightarrow{{}}2XY\] \[\Delta H={{(BE)}_{X-X}}+{{(BE)}_{Y-Y}}-2{{(BE)}_{X-Y}}\] If (BE) of\[XY=a\] Then (BE) of\[(XX)=a\] and (BE)of \[(Y-Y)=\frac{a}{2}\] \[\therefore \]  \[\Delta {{H}_{f}}(X-Y)\,\,=-200\,kJ\] \[\therefore \]\[-400\](for 2 mol\[XY\])\[=a+\frac{a}{2}-2a\] \[-400=\frac{a}{2}\] \[a=+800kJ\] The bond dissociation energy \[{{X}_{2}}=800\text{ }kJ\,mo{{l}^{-1}}\]


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