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InterviewSolution
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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. 200 kJ `mol^(-1)`B. 100 kJ `mol^(-1)`C. 800 kJ `mol^(-1)`D. 300 kJ `mol^(-1)` |
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Answer» Correct Answer - C Given `(1)/(2)X_(2)+(1)/(2)(Y)/(2)rarr XY , Delta H=-200 kJ mol^(_1)` Let the bond dissociation energy of XY = a The ratio of bond dissociation energies of `X_(2), Y_(2)` and XY be `a:(a)/(2):a` (the ratio given) For given equation : `Delta H=(B.E.)_(R )-(B.E.)_(P)` `-200=[(1)/(2)xx a+(1)/(2)xx(a)/(2)]-a` `therefore a=800 kJ mol^(-1)` |
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