`DeltaH` for the formation of XY is `-200 kJ mol^(-1)`.The bond enthalpies of `X_2, Y_2,` and XY are in the ratio 1:0.5:1. Then determine the bond enthalpies.

Explanation

Let, the bond enthalpy of `X_(2)` is a, `Y_(2)` is `(a/2)and XY` is a
`1/2X_(2) +1/2Y_(2) to XY,DeltaH=-200KJ`
Heat of reaction , `DeltaH` = (Enthalpy of bond dissociation)-(Enthalpy of bond fomation ) `(a/2+a/4)-(a) =200KJ=-a/4`(or)`a=800KJ`
The bond entalpy of `X_(2)=800KJmol^(-1),Y_(2)=400KJmol^(-1)`and`XY=800KJmol^(-1)`.