The standard molar enthalpies of formation at $25^\circ ext{C}$ for liquid cyclohexane and liquid b

Question

The standard molar enthalpies of formation at $25^\circ	ext{C}$ for liquid cyclohexane and liquid benzene are $-156	ext{ kJ mol}^{-1}$ and $+49	ext{ kJ mol}^{-1}$, respectively. The standard molar enthalpy of catalytic hydrogenation of liquid cyclohexene to liquid cyclohexane at $25^\circ	ext{C}$ is $-119	ext{ kJ mol}^{-1}$. Using these thermochemical data, calculate the magnitude of the resonance stabilization energy (in $	ext{kJ mol}^{-1}$) of benzene.

Professor PC Thomas Classes · Answer

The hydrogenation reaction of cyclohexene is: $	ext{C}_6	ext{H}_{10}(l) + 	ext{H}_2(g) ightarrow 	ext{C}_6	ext{H}_{12}(l); \Delta H_1^\circ = -119	ext{ kJ mol}^{-1}$. This value represents the energy released upon hydrogenation of a single isolated carbon-carbon double bond in a six-membered ring. If benzene were a hypothetical non-resonant structure containing three localized double bonds (1,3,5-cyclohexatriene), the theoretical enthalpy of hydrogenation of liquid benzene to liquid cyclohexane would be three times that of cyclohexene: $\Delta H^\circ_{	ext{theoretical}} = 3 	imes (-119	ext{ kJ mol}^{-1}) = -357	ext{ kJ mol}^{-1}$. Now, we determine the actual experimentally observed enthalpy of hydrogenation of benzene: $	ext{C}_6	ext{H}_6(l) + 3	ext{H}_2(g) ightarrow 	ext{C}_6	ext{H}_{12}(l)$. Using enthalpies of formation: $\Delta H^\circ_{	ext{actual}} = \Delta H^\circ_f[	ext{cyclohexane}(l)] - \Delta H^\circ_f[	ext{benzene}(l)] = -156 - (+49) = -205	ext{ kJ mol}^{-1}$. The resonance stabilization energy is the difference between the theoretical and actual hydrogenation enthalpies: $	ext{Resonance Energy} = \Delta H^\circ_{	ext{theoretical}} - \Delta H^\circ_{	ext{actual}} = -357 - (-205) = -152	ext{ kJ mol}^{-1}$. Therefore, the magnitude of the resonance energy is $152	ext{ kJ mol}^{-1}$.