The enthalpies of formation of and are –65.0 and –197.0 kcal/mol, respectively. A mixture of the two — Thermodynamics and Thermochemistry Chemistry Question
Question
The enthalpies of formation of $\text{FeO(s)}$ and $\text{Fe}_2\text{O}_3\text{(s)}$ are –65.0 and –197.0 kcal/mol, respectively. A mixture of the two oxides contains FeO and $\text{Fe}_2\text{O}_3$ in the mole ratio 2:1. If by oxidation it is changed in to a 1:2 mole ratio mixture, how much of thermal energy will be released per mole of the initial mixture?
💡 Solution & Explanation
Initial 1 mole mixture: 2/3 mol FeO, 1/3 mol $\text{Fe}_2\text{O}_3$ (Total Fe = 4/3 mol). Final 1:2 mixture conserving Fe contains 4/15 mol FeO and 8/15 mol $\text{Fe}_2\text{O}_3$. Thus, $2/3 - 4/15 = 6/15 = 2/5$ moles of FeO is oxidized to $\text{Fe}_2\text{O}_3$. The reaction $2\text{FeO} + 0.5\text{O}_2 \to \text{Fe}_2\text{O}_3$ has $\Delta H = -197 - 2(-65) = -67\text{ kcal}$. For 2/5 moles of FeO, heat released $= (2/5 / 2) \times 67 = 13.4\text{ kcal}$.