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The bond dissociation energies of X2, Y2...

The bond dissociation energies of `X_2, Y_2 and XY` are in the ratio of `1 : 0.5 : 1. DeltaH` for the formation of XY is -200 kJ mol^(-1)`. The bond dissociation energy of `X_2` will be

A

400 kJ `mol^(-1)`

B

300 kJ `mol^(-1)`

C

200 kJ `mol^(-1)`

D

800 kJ `mol^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
D
Formation of XY is shown as
`X_(2)+Y_(2)to2XY`
`DeltaH=(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)=`(a)/(2)`
`DeltaH_(f)(X-Y)=-200kJ`
`therefore-400` (for 2 moles XY)=`a+(a)/(2)-2a`
`-400=-(a)/(2)`
a=+800kJ
The bond dissociation energy of `X_(2)=800kJ" "mol^(-1)`
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