Enthalpy of hydrogenation of one mole of benzene to cyclohexane is [Resonance energy of benzene = - 150 . 4…

[Resonance energy of benzene $=-150.4 \mathrm{kJ}/$$\mathrm{mole}.$
Enthalpy of hydrogenation of cyclohexene $=-119.5 \mathrm{kJ}/$$\mathrm{mole}$]

Reaction of cyclohexene to cyclohexane :-

${\mathrm{C}}_6{\mathrm{H}}_{12} + {\mathrm{H}}_2 \to {\mathrm{C}}_6{\mathrm{H}}_{14}$

So, the energy required to break the one double bond is $=-119.5 \mathrm{kJ}/\mathrm{mol}$

The benzene has the three double bonds. So, the energy of formation of the cyclohexane from benzene is :-

$3\times \mathrm{Enthalpy} \mathrm{of} \mathrm{hydrogenation} \mathrm{of} \mathrm{cyclohexene}=-3\times 119.5=-358.5 \mathrm{kJ}/\mathrm{mol}$

Enthalpy of hydrogenation of one mole of benzene to cyclohexane is equal to difference the enthalpy of formation of cyclohexane from benzene to the resonance energy of benzene.

Enthalpy of hydrogenation of benzene to cyclohexane = $-358.5-\left(-150.4\right)=-358.5+150.4=-208.1 \mathrm{kJ}/\mathrm{mol}$

Hence, Enthalpy of hydrogenation of one mole of benzene to cyclohexane is $-208.1 \mathrm{kJ}/\mathrm{mol}$.