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DeltaH for the formation of XY is -200 k...

`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.

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Let the bond enthalpy of `X _(2)` is a, `Y_(2)` is (a/2) and XY is `a. (1)/(2) X _(2) + (1)/(T) Y_(2) to XY: Delta H =- 200 kJ`
Heat of reaction `, Delta H = ` (Enthlpy of bond dissociation ) - (Enthalpy of bond forming)
`= ((a)/(2) +(a)/(4)) - (a) =- 200 kJ =- (a)/(4) (or) a =800 kJ`
The bond enthalpy of `X _(2) = 800 kJ mol ^(-1), Y_(2) = 400 kJ mol ^(-1) and XY = 800 kJ mol ^(-1)`
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