If enthalpy of hydrogenation of C 6 H 6 l into C 6 H 12 l is – 205 k J and resonance energy of C 6 H 6 l is – 152 k J m o l – 1 then enthalpy of hydrogenation of l to C 6 H 12 l is? [Assume enthalpiies of vapourisation of all liquids involved are equal].

Step-by-step solution

C 6 H 6 l ⁡ + 3 H 2 g ⟶ C 6 H 1 2 l ⁡ Δ H hyd C 6 H 6 l ⁡ ∘ = - 2 0 5 kJ Benzene on reactant side so add resonance energy of benzene with sign. Heat evolved for hydrogenation of three double bonds = − 205 − 152 = − 357 kJ mol − 1 Δ H hyd ∘ = - 3 5 7 3 = - 1 1 9 kJ / mole