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Bond disociation enegry of `XY,X_(2)` and `Y_(2)` (all diatomic molecules) are in the ratio of `1: 1:0.5` and `DeltaH_(f) of XY` is `-100 kJ mol^(-1)`. The bond dissociation enegry of `X_(2)` is `100x`. Find the value of `x`. |
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Answer» `(1)/(2)X_(2) +(1)/(2)Y_(2) rarr XY, DeltaH = - 100 kJ` Let the bond dissociation energy of `X_(2),Y_(2)`, and `XY` be `a: (a)/(2):a` (the given ratio) `kJ mol^(-1)` respectively. `:. (a)/(2)+(a)/(4) -a = - 100` `:. A= 400` `100 x = 400` `x = 4` |
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