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 :
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 \[X-Y=a\] then \[(BE)\] or \[(X-X)=a\] and \[(BE)\] of \[(Y-Y)=\frac{a}{2}\] \[\Delta {{H}_{f}}(X-Y)=-200kJ\] \[\therefore \] \[-400\] (for 2 moles XY) \[=a+\frac{a}{2}-2a\] \[-400=-\frac{a}{2}\] \[a=+800kJ\] The bond dissociation energy of \[{{X}_{2}}=800kJ\,mo{{l}^{-1}}\]You need to login to perform this action.
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