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If the bond dissociation energies of `XY,X_(2)` and `Y_(2)(` all diatomic molecules `)` are in the ratio `1:1:0.5` and `Delta_(f)H` of `XY` is `-200 kJ mol^(-1)`. The bond dissociation energy of `X_(2)` will be `:`A. `800kJ mol^(-1)`B. `200kJ mol^(-1)`C. `300 kJ mol^(-1)`D. `400kJ mol^(-1)` |
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Answer» Correct Answer - A Given `(1)/(2)X_(2)+(1)/(2)Y_(2) rarr XY,DeltaH=-200kJ mol^(-1)` Let bond dissociation energy of `X_(2),Y_(2)` and `XY` be `a:(a)/(2)a(` the ratio given `)` `X_(2) rarr 2X, DeltaH=a ….(1)` `Y_(2) rarr 2Y, DeltaH=(a)/(2) …(2)` `XY rarr X+Y, DeltaH=a .....(3)` By `eq. (1)+(2)-(3)xx2` `X_(2)+Y_(2) rarr 2XY` `Delta H=a+(a)/(2)-2a` or `(1)/(2)X_(2)+(1)/(2)Y_(2) rarr XY,` `DeltaH=(a)/(2)+(a)/(2)-a=-(a)/(4)` `:. -(a)/(4)=-200` `:. a=800kJ mol^(-1)` |
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