Given the bond dissociation enthalpy of bond as 103 kcal/mol and the enthalpy of formation of as –18 — Thermodynamics and Thermochemistry Chemistry Question
Question
Given the bond dissociation enthalpy of $\text{CH}_3-\text{H}$ bond as 103 kcal/mol and the enthalpy of formation of $\text{CH}_4\text{(g)}$ as –18 kcal/mol, find the enthalpy of formation of methyl radical. The dissociation energy of $\text{H}_2\text{(g)}$ into H (atoms) is 103 kcal/mol.
💡 Solution & Explanation
For the reaction $\text{CH}_4 \to \text{CH}_3^\bullet + \text{H}^\bullet$, $\Delta H = 103\text{ kcal/mol}$. $\Delta H_f(\text{H}^\bullet) = 103 / 2 = 51.5\text{ kcal/mol}$. Using Hess's Law: $103 = \Delta H_f(\text{CH}_3^\bullet) + \Delta H_f(\text{H}^\bullet) - \Delta H_f(\text{CH}_4) = \Delta H_f(\text{CH}_3^\bullet) + 51.5 - (-18)$. $103 = \Delta H_f(\text{CH}_3^\bullet) + 69.5 \implies \Delta H_f(\text{CH}_3^\bullet) = 103 - 69.5 = 33.5\text{ kcal/mol}$.