Enthalpy of formation of 3 C-C double bonds
= ΔHf(cyclohexane) - ΔHf(benzene)
= -156 - (49) kJ
= -205 kJ
Now it is given that,
Cyclohexene + H2 → cyclohexane ; ΔH = -119kJ
Now, theoretical enthalpy of formation of 3 C-C double bonds in benzene ring
=3*(-119)kJ
=-357kJ
So, resonance energy of benzene is = -357 - (-205)kJ
= -152 kJmol-1
btw this was IIT 1996 question right?