If the bond dissociation energies of `XY,X_(2)` and `Y_(2)(` all diatomic molecules `)` are in the ratio `1:1:0.5` and `Delta_(f)H` of `XY` is `-200 kJ mol^(-1)`. The bond dissociation energy of `X_(2)` will be `:`

Options

A. 400 kJ `mol^(-1)`
B. 300 kJ `mol^(-1)`
C. 200 kJ `mol^(-1)`
D. 800 kJ `mol^(-1)`

Answer

D

Explanation

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)`