`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 = -200 kJ`
Heat of reaction, `DeltaH=`(Enthalpy of bond dissociation) - (Enthalpy of bond formation)
`=(a/2 + a/4) - (aq) = -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).`