Standard enthalpy of formation of gaseous ethane, ethene and benzene from gaseous atoms are –2839, – — Thermodynamics and Thermochemistry Chemistry Question

💡 Solution & Explanation

Calculate $\text{BE(C-C)}$ from ethane: $2839 - 6(412) = 2839 - 2472 = 367\text{ kJ/mol}$. Calculate $\text{BE(C=C)}$ from ethene: $2275 - 4(412) = 2275 - 1648 = 627\text{ kJ/mol}$. Theoretical atomization of Kekule benzene (3 C-C, 3 C=C, 6 C-H): $3(367) + 3(627) + 6(412) = 1101 + 1881 + 2472 = 5454\text{ kJ/mol}$. Given experimental atomization $= 5506\text{ kJ/mol}$. Resonance energy $= 5506 - 5454 = 52\text{ kJ/mol}$.