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InterviewSolution
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Calculate the resonance energy of `C_(6)H_(6)` using kekule formula of `C_(6)H_(6)` from the following data. a. `Delta_(f)H^(Theta)` for `C_(6)H_(6) =- 358.5 kJ mol^(-1)` b. Heat of atomisation of `C =716.8 kJ mol^(-1)` c. Bond enegry of `C-H, C-C, C=C`and `H-H` are `490, 620, 436.9 kJ "mole"^(-1)` respectively. |
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Answer» `6C(s) +3H_(2)(g) rarr C_(6)H_(6) Delta_(f)H_((exp)) =- 358.5 kJ` `Delta_(f)H_(cal) = BE(Reactants) -BE(Products)` `BE(Reactants) = 6C_((srarrg)) +3(H-H)` `= (6 xx 716.8 +3 xx 436.9)` `= 4300.8 +1310.7 = 5611.5` `BE(Products) = 3(C-C) +3(C =C) +6(C-H)` `=3 xx 340 +3 xx 620 +6 xx 490 = 5820 kJ` `DeltaH_(cal) = 5611.5 - 5820 =- 208.5 kJ mol^(-1)` `RE = Delta_(f)H_((Exp)) - Delta_(f)H_((Cal))` `=- 358.5 -(-208.5) =- 150.0 kJ mol^(-1)` |
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