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. `100 kJ "mol"^(-1)`
B. `200 kJ "mol"^(-1)`
C. `800 kJ "mol"^(-1)`
D. `400 kJ "mol"^(-1)`

Answer

C

Explanation

Let the bond dissociation energy of `XY, X_(2)` and `Y_(2)` be x, x and 0.5 x respectively
`1/2 X_(2) + 1/52Y_(2) to XY Delta_(f)H = -200 kJ "mol"^(-1)`
`DeltaH = 1/2 B.E(X_(2)) + 1/2 B.E (Y_(2)) - B.E(XY)`
` -200 = 1/(2x) + (0.5x)/(2)-x`
or `-0.25x = -200`
`x = (200)/(0.25) = 800 kJ "mol"^(-1)`