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.

DeltaH for the formation of XY is -200 kJ mol^(-1).The bond enthalpies

1.	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.
Answer» Solution :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 ofbond 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)`