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The heat of atomisation of `PH_(3(g))` is `228kcal mol^(-1)` and that of `P_(2)H_(2)` is `355kcal mol^(-1)`. Calculate the average bond energy of `P-P` bond. |
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Answer» `PH_(3(g))rarr P_((g))+3H_((g)), DeltaH=228kcal//mol` `H_(2)P=PH_(2)rarr 2P_((g))+4H_((g)),DeltaH=355kcal//mol` `:. DeltaH=e_(P-P)+4e_(P-H)` or `355-e_(P-P)+4xx(228)/(3)( :. e_(P-H)=(228)/(3))` `:. e_(P-P)=51kcal//mol` |
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