Question

If the bond dissociation energies of `XY`,`X_(2)` and `Y_(2)` are in the ratio of `1:1:0.5` and `DeltaH_(f)` for the formation of `Xy` is `-200 KJ//mol`. The bond dissociation energy of `X_(2)` will be `:-`
A. `800 " kJ mol"^(-1)`
B. `100 " kJ mol"^(-1)`
C. `200 " kJ mol"^(-1)`
D. `400 " kJ mol"^(-1)`

Answer 1

Correct Answer - A
Key Concept Relation between heat of reaction `(Delta,H)` and bond energies (BE) of reactants and products is given by
`Delta_(r )H=sum BE_("Reactants")-sum BE_("Products")`
The reaction of formation of XY is
`(1)/(2)X_(2)(g)+(1)/(2)Y_(2)(g)to XY(g), DeltaH=-200 "kJ mol"^(-1)`
Given, the bond dissociation energies of `X_(2),Y_(2)` and XY are in the ratio 1 : 0.5 : 1. Let the bond dissociation energies of `X_(2),Y_(2)` and XY are a kJ `"mol"^(-1),0.5 "a kJ mol"^(-1) " and a kJ mol"^(-1)`, respectively.
`therefore Delta_(r )H=sumBE_("Reactants")-Delta BE_("Products")`
`=[(1)/(2)xxa+(1)/(2)xx0.5a]-[1xxa]`
`-200=(a)/(2)+(a)/(4)-a`
`-200=(2a+a-4a)/(4)=(-a)/(4)`
`a=800 " kJ mol"^(-1)`
`therefore` The bond dissociation energy of
`X_(2)= " a kJ mol"^(-1)`
`=800 " kJ mol"^(-1)`