The bond enthalpy of C–C bond (kJ/mol) is — Thermodynamics and Thermochemistry Chemistry Question

💡 Solution & Explanation

For propane ($\text{C}_3\text{H}_8$): $\Delta H_{\text{atomization}} = 3\Delta H_{\text{sub}}(\text{C}) + 4\Delta H_{\text{diss}}(\text{H}_2) - \Delta H_f(\text{C}_3\text{H}_8) = 3(719) + 4(435) - (-85) = 2157 + 1740 + 85 = 3982\text{ kJ/mol}$. Propane strictly has 2 C-C bonds and 8 C-H bonds. $2 E_{\text{C-C}} + 8(414) = 3982 \implies 2 E_{\text{C-C}} = 3982 - 3312 = 670 \implies E_{\text{C-C}} = 335.0\text{ kJ/mol}$.