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 `DeltaH_(f)` for the formation 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. 20 kJ `mol^(-1)`
D. none of these

Answer

D

Explanation

Formation of XY is known as
`X_(2)+Y_(2) to 2XY`
`DeltaH=(BE)_(X-X)+(BE)_(Y-Y)-2(BE)_(X-Y)` ,brgt If `(BE)` of `X-Y=a`
then (B E) of (X-X)=a
and (B E) of (Y-Y)=`(a)/(2)`
`therefore DeltaH_(f)(X-Y)=-200kJ`
`therefore-400` (for 2 mol XY)`=a+(a)/(2)-2a`
`-400=-(a)/(2)impliesa=+800kJ`
The bond dissociation energy of `X_(2)=800" kJ "mol^(-1)`