If the bond dissociation energies of XY(g), `X_(2)(g) and Y_(2)(g)` are in the ratio of 1:1:0.5 and `Delta_(f)H` for the formation of XY(g) is -200kJ`*mol^(-1)`, then what will be the bond dissociation energy of `X_(2)(g)`?

Explanation

`(1)/(2)X_(2)(g)+(1)/(2)Y_(2)(g)toXY(g)`
Suppose, the bond dissociation energy of XY is x. so, the bond dissociation energies of `X_(2) and Y_(2)` will be x and `x/2`,
respectively,
for the reaction `DeltaH^(0)=DeltaH_(f)^(0)(XY)-(1)/(2)DeltaH_(f)^(0)(X_(2))-(1)/(2)DeltaH_(f)^(0)(Y_(2))=-200kJ`
In terms of bond energies:
`DeltaH^(0)=((1)/(2)x+(1)/(2)xx(x)/(2))-(x)=(3x)/(4)-x=-(x)/(4)`
Therefore, `-(x)/(4)=-200kJ`
or, x=800kJ
Thus, the bond dissociation energy of `X_(2)(g)` is 800 `kJ*mol^(-1)`.