The bond dissociation energies of`X_(2),Y_(2)andXY` are in the ratio of`1:0.5:1.DeltaH` for the formation of `XY`is `-200kJmol^(-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 B.E. Of `x_(2), y_(2)` & xy are x kJ `mol^(-1)`, `0.5 x kJ mol^(-1)` and x kJ `mol^(-1)` respectively.
`(1)/(2)x_(2)+(1)/(2)y_(2)rarrxy, DeltaH=-200 kJ mol^(-1)`
`DeltaH=-200=Sigma(B.E)_("Reactant")-Sigma(B.E)_("Product")`
`=[(1)/(2)xx(x)+(1)/(2)xx(0.5x)]-[1xx(x)]`
`"B.E. of "X_(2)=x=800 kJ mol^(-1)`.