Question #44221

Consider the homolytic cleavage of the $C_3-H$ bond in 1,4-pentadiene, \ce{C5H8(g)}, to form the pentadienyl radical, \ce{C5H7^\bullet(g)}, and a hydrogen atom, \ce{H^\bullet(g)}, at $298 \text{ K}$. You are given the following thermodynamic data at $298 \text{ K}$: Standard enthalpy of formation of 1,4-pentadiene, $\Delta H_f^\circ(\ce{C5H8}, g) = 106 \text{ kJ mol}^{-1}$ Standard enthalpy of formation of a hydrogen atom, $\Delta H_f^\circ(\ce{H^\bullet}, g) = 218 \text{ kJ mol}^{-1}$ Standard localized $C(sp^3)-H$ bond dissociation enthalpy $= 415 \text{ kJ mol}^{-1}$ Resonance energy of the pentadienyl radical $= 82 \text{ kJ mol}^{-1}$ (Assume the resonance energy of 1,4-pentadiene is negligible). Calculate the standard enthalpy of formation of the pentadienyl radical, $\Delta H_f^\circ(\ce{C5H7^\bullet}, g)$.