${\text{C}}_{\text{6}}{\text{H}}_{\text{6}}\underset{ }{\overset{ }{\to }}{\text{C}}_{\text{6}}{\text{H}}_{12},\mathrm{\Delta }\text{\hspace{0.17em}H}=-204\text{\hspace{0.17em}KJ}$. Heat of hydrogenation of each C = C (assumed to be) in benzene is [in KJ]

 Heat of hydrogenation of three double bonds = – 204 
Therefore heat of hydrogenation of each double bond = $\frac{-204}{3}=-68$