The bond dissoication energies of X_(2), Y_92) and XY are in the ratio of 1: 0.5 : 1 . DeltaH for the formation of XY is -200 kJ mol^(-1). The bond dissociation energy ofX_(2) will be

The bond dissoication energies of X_(2), Y_92) and XY are in the ratio

1.	The bond dissoication energies of X_(2), Y_92) and XY are in the ratio of 1: 0.5 : 1 . DeltaH for the formation of XY is -200 kJ mol^(-1). The bond dissociation energy ofX_(2) will be
Answer» `200 kJ mol^(-1)` `100 kJ mol^(-1)` ` 800 kJ mol^(-1)` ` 400 kJ mol^(-1)` Solution :Suppose the BOND dissociation energy of`X_(2) = 'a' kJ mol^(-1), i.e., BE(X_(2))= akJmol^(-1)`then `BE(Y_(2)) =0.5 a` andBE of XY `=akJmol^(-1)` Given `=(1)/(2) X_(2)+(1)/(Y_(2)) rarr XY ,DELTAH = - 200 kJ mol^(-1)` `Delta_(r) H=` BE (Reactants) - BE( Products) `= [ (1)/(2) BE(X_(2))+(1)/(2) BE(Y_(2))]-BE(XY)` `:. - 200 = (a)/(2)+ ( 0.5)/( 2) -a =0.5a+ 0.25a-a =-0.25 a` `:. a=( 200)/( 0.25) = 80 kJ mol^(-1)`