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

Options

A. `200 kJ mol^(-1)`
B. `100 kJ mol^(-1)`
C. ` 800 kJ mol^(-1)`
D. ` 400 kJ mol^(-1)`

Answer

C

Explanation

Suppose the bond dissociation energy of`X_(2) = 'a' kJ mol^(-1), i.e., BE(X_(2))= akJ mol^(-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)`