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)` .

Explanation

To calculate the resonance energy of benzene using the given enthalpies of hydrogenation, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Enthalpy of Hydrogenation for Ethene and Benzene:**
- The enthalpy of hydrogenation for ethene (C₂H₄) is given as ΔH₁ = -136.68 kJ/mol.
- The enthalpy of hydrogenation for benzene (C₆H₆) is given as ΔH₂ = -205.65 kJ/mol.

...