From the following thermochemical equations, find out bond dissociation enthalpy of bond.<br><br><br — Thermodynamics and Thermochemistry Chemistry Question
Question
From the following thermochemical equations, find out bond dissociation enthalpy of $\text{CH}_3-\text{H}$ bond.<br>$\text{CH}_3\text{I(g)} \to \text{CH}_3\text{(g)} + \text{I(g)}; \Delta H = 54.0\text{ kcal}$<br>$\text{CH}_4\text{(g)} + \text{I}_2\text{(s)} \to \text{CH}_3\text{I(g)} + \text{HI(g)}; \Delta H = 29.0\text{ kcal}$<br>$\text{HI(g)} \to \text{H(g)} + \text{I(g)}; \Delta H = 79.8\text{ kcal}$<br>$\text{I}_2\text{(s)} \to 2\text{I(g)}; \Delta H = 51.0\text{ kcal}$
💡 Solution & Explanation
Required reaction: $\text{CH}_4 \to \text{CH}_3 + \text{H}$. Combine given steps: Eq2 + Eq1 + Eq3 - Eq4. $\Delta H = 29.0\text{ (Rxn2)} + 54.0\text{ (Rxn1)} + 79.8\text{ (Rxn3)} - 51.0\text{ (Rxn4)} = 162.8 - 51.0 = 111.8\text{ kcal/mol}$.