InterviewSolution
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InterviewSolution
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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 beA. `200KJmol^(-1)`B. `100KJmol^(-1)`C. `800KJmol^(-1)`D. `400 KJ mol^(-1)` |
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Answer» Correct Answer - A Let `B.E.` of `x_(2),y_(2)` & `xy` are `xKJmol^(-1) , 0.5xKJmol^(-1)` and `xKJmol^(-1)` respectivaly `(1)/(2)x_(2)+(1)/(2)y_(2)rarrxy,Delta=-200KJmol^(-1)` `DeltaH=-200=Sigma(B.E)_("Reactant")=Sigma(B.E.)_("Product")` `=(1)/(2)B.E_(x-x)+(1)/(2)B.E_(y-y)-B.E_(x-y)` `=[(1)/(2)xx(x)+(1)/(2)xx(0.5x)]-[1xx(x)]` `B.E.` of `x_(2)=x=800KJmol^(-1)` . |
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