Question

Enthalpic of hydrogenation of ethene `(C_(2)H_(4))` and benzene `(C_(6)H_(6))` are -136.68 and `205.65 kJ mo1^(-1)` respectively
Calculate the resonance energy of benzene
(a) `C_(2) H_(4(g)) + H_(2(g)) rarrC_(2) H_(6), DeltaH_(1) = - 136.68 "kJ mo1"^(-1)`
(b) `C_(6)H_(6_((1))) + 3H_(2(g)) rarr C_(6) H_(12),DeltaH_(2) = - 205.65 "kJ mo1"^(-1)` .

Answer 1

if benzene `(C_(6)H_(6))` had three isolated `(C =C)` double bonds `DeltaH^(Θ)` hydrogenation would be close to three times `DeltaH` hydrogenation of ethene `(H_(2) C =CH_(2))` with one double bond I,e ` - 136.68 xx 3 = -410.4 KJ mo1^(-1)`
`DeltaH` hydrogenation of benzene ` = - 205. 65 kJ mo1^(-1)`
hydrogenation of benzene is less exothermic by Resonance energy = `DeltaH^(Θ)` hydrogentaion of `C_(6)H_(6) - DeltaH^(Θ)`
Hydrogenation of `C_(6)H_(4)`
` = - 205.65 - (-410.04) kJ mo1^(-1)`
` =204.39 kJ mo1^(-1)`
It means the benzene has been stabilised by `204 kJ mo1^(-1)` .