InterviewSolution
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InterviewSolution
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The `C-H` bond of the side chain in toluene, `C_(6)H_(5)-CH_(3)`, has a dissociation energy of `77.5 kcal mol^(-1)`. Calculate `Delta_(f)H^(Theta)` of benzy`1` radical and the strength of the central bond in dibenzy`1 C_(6)H_(5)-CH_(2)-CH_(2)-C_(6)H_(5)` given that `Delta_(f)H^(Theta)` to toluene vapour in `12 kcal mol^(-1)` and that of dibenzy`1` vapour is `27.8 kcal mol^(-1). BE` of `H_(2) = 104 kcal mol^(-1)`. |
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Answer» `BE of H -H = 104 kcal mol^(-1)` `H -H =2H` `C_(6)H_(5)CH_(3)(g) rarr C_(6)H_(5)CH_(2)+H DeltaH = 77.5 ….(i)` `DeltaH_(1) = [Delta_(f)C_(6)H_(5) +Delta_(f)H H] -[ Delta_(f)H(Toluene)]` `2(C_(6)H_(5)CH_(2)) rarr underset(("Dibenzyl"))(C_(6)H_(5)-CH_(2) -CH_(2)-PH)` ....(ii) `DeltaH_(2) = 2_(f)H ("Dibenzyl") -2Delta_(f)H^(Theta) (C_(6)H_(5)CH_(2))` `Delta_(f)H^(Theta)H =(104)/(2) = 52` `[(1)/(2)H_(2) rarr H]` From reaction (i), `77.5 = (Delta_(f)H(C_(6)H_(5)CH_(2)) +52) -12` `rArr Delta_(f)H^(Theta)C_(6)H_(5)CH_(2) = 37.5 kcal mol^(-1)` From reaction (ii), `DeltaH_(2) = (27.8 -2 xx 37.5) = - 47.2 kcal mol^(-1)` The bond strength of `PhCH_(2) -CH_(2)Ph = 47.2 kcal mol^(-1)` |
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