The bond dissociation energies of `X_(2), Y_(2)` and XY are in the ratio of ` 1 : 0.5 : 1. DeltaH` for the formation of XY is `-200 kJ mol^(-1)`.The bond dissociation energy of `X_(2)` will be

Options

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

Answer

C

Explanation

Let bond dissociation energy of `X_(2)` be x kJ `mol^(-1)`
Then
B.E. Of `Y_(2) = 0.5 x kJ mol^(-1)`
B.E of XY = x kJ `mol^(-1)`
`1/2 X_(2) + 1/2Y_(2) to XY DeltaH = -200 kJ mol^(-1)`
`DeltaH = Sigma`B.E (reactants) - `Sigma`B.E(products)
`DeltaH = 1/2 B.E.(X_(2)) + 1/2 B.E. (Y_(2)) - B.E.(XY)`
` -200 = 1/2(x) + 1/2 (0.5 x) - 1(x)`
`-200 = -0.25x`
`therefore x = (200)/(0.25) = 800 kJ "mol"^(-1)`.