Given are the bond enthalpy data: ${\epsilon }_{\mathrm{C}-\mathrm{C}}=374\mathrm{kJ}{\mathrm{mol}}^{-1};{\epsilon }_{\mathrm{C}=\mathrm{C}}=632\mathrm{kJ}{\mathrm{mol}}^{-1};{\epsilon }_{\mathrm{C}-\mathrm{H}}=411\mathrm{kJ}{\mathrm{mol}}^{-1}$ If the enthalpy formation of benzene from gaseous atoms is $5536 kJ mo{l}^{-1}$ the resonance energy benzene is

Enthalpy of formation of benzene from the gaseous atoms using bond enthalpies data is $\mathrm{\Delta }H=-3{\epsilon }_{\mathrm{C}-\mathrm{C}}-3{\epsilon }_{\mathrm{C}=\mathrm{C}}-6{\epsilon }_{\mathrm{C}-\mathrm{H}}=[-3(374)-3(632)-6(411\left)\right]\mathrm{kJ}{\mathrm{mol}}^{-1}=-5484\mathrm{kJ}{\mathrm{mol}}^{-1}$Resonance energy$=\mathrm{\Delta }H\left(\text{ actual }\right)-\mathrm{\Delta }H$ f(from bond enthalpy)$= (-5536 + 5484) kJ mo{l}^{-1} = -52 kJ mo{l}^{-1}$