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. `300 KJ mol^(-1)`
D. `800 KJ mol^(-1)`

Answer

D

Explanation

let the bond dissociation energy of XY, `X_(2)` and `Y_(2)` be x,x and x,`(x)/(2)` KJ/mol respectively,
`(1)/(2)X_(2) + (1)/(2)Y_(2) rarr XY , DeltaHf =-200 KJ mol^(-1)`
`DeltaH` reaction = [(sum of bond dissociation energy of all reactants) - (sum of bond dissociation energy of all product)]
`=[(1)/(2)DeltaH_(X2) + (1)/(2) DeltaH_(Y2) - DeltaH_(XY)] " "=(x)/(2) + (0.5x)/(2) -x =-200`
`therefore " "x=(200)/(0.25) = 800 KJ mol^(-1).`