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

Answer

A

Explanation

Given `(1)/(2)X_(2)+(1)/(2)Y_(2) rarr XY,DeltaH=-200kJ mol^(-1)` Let bond dissociation energy of `X_(2),Y_(2)` and `XY` be `a:(a)/(2)a(` the ratio given `)`
`X_(2) rarr 2X, DeltaH=a ….(1)`
`Y_(2) rarr 2Y, DeltaH=(a)/(2) …(2)`
`XY rarr X+Y, DeltaH=a .....(3)`
By `eq. (1)+(2)-(3)xx2`
`X_(2)+Y_(2) rarr 2XY`
`Delta H=a+(a)/(2)-2a`
or `(1)/(2)X_(2)+(1)/(2)Y_(2) rarr XY,`
`DeltaH=(a)/(2)+(a)/(2)-a=-(a)/(4)`
`:. -(a)/(4)=-200`
`:. a=800kJ mol^(-1)`