The bond dissociation energies of X_(2), Y_(2) and XY are in the ratio of1 : 0.5 : 1. DeltaH for the formation of XY is -200 kJ mol^(-1).The bond dissociation energy of X_(2) will be

The bond dissociation energies of X_(2), Y_(2) and XY are in the ratio

1.	The bond dissociation energies of X_(2), Y_(2) and XY are in the ratio of1 : 0.5 : 1. DeltaH for the formation of XY is -200 kJ mol^(-1).The bond dissociation energy of X_(2) will be
Answer» `200 kJ"MOL"^(-1)` `100 kJ"mol"^(-1)` `800 kJ"mol"^(-1)` `400 kJ"mol"^(-1)` Solution :Let bond DISSOCIATION energy of `X_(2)` be x kJ `mol^(-1)` Then B.E. Of `Y_(2) = 0.5 x kJ mol^(-1)` B.E of XY = x kJ `mol^(-1)` `1/2 X_(2) + 1/2Y_(2) to XY DeltaH = -200 kJ mol^(-1)` `DeltaH = Sigma`B.E (reactants) - `Sigma`B.E(PRODUCTS) `DeltaH = 1/2 B.E.(X_(2)) + 1/2 B.E. (Y_(2)) - B.E.(XY)` ` -200 = 1/2(x) + 1/2 (0.5 x) - 1(x)` `-200 = -0.25x` `therefore x = (200)/(0.25) = 800 kJ"mol"^(-1)`.