Data given for the following reaction is as follows: Substance ∆H°(kJ mol ) ∆S°(J mol K ) FeO –266.3 — Thermodynamics and Thermochemistry Chemistry Question
Question
Data given for the following reaction is as follows: Substance ∆H°(kJ mol ) ∆S°(J mol K ) FeO –266.3 57.49 C 0 5.74 Fe 0 27.28 CO –110.5 197.6 The minimum temperature in K at which the reaction becomes spontaneous is ________ . (Integer answer) –1 –1 –1 (s) (graphite) (S) (g)
💡 Solution & Explanation
**Step 1: Identify the reaction** FeO(s) + C(graphite) → Fe(s) + CO(g) **Step 2: Calculate ∆H°rxn using Hess's Law** ∆H°rxn = [∆H°(Fe) + ∆H°(CO)] − [∆H°(FeO) + ∆H°(C)] ∆H°rxn = [0 + (−110.5)] − [(−266.3) + 0] ∆H°rxn = −110.5 + 266.3 = +155.8 kJ/mol **Step 3: Calculate ∆S°rxn** ∆S°rxn = [S°(Fe) + S°(CO)] − [S°(FeO) + S°(C)] ∆S°rxn = [27.28 + 197.6] − [57.49 + 5.74] ∆S°rxn = 224.88 − 63.23 = +161.65 J mol⁻¹K⁻¹ **Step 4: Apply the spontaneity condition** For spontaneity: ∆G° ≤ 0 Using ∆G° = ∆H° − T∆S° At the minimum temperature (equilibrium point): ∆G° = 0 0 = ∆H° − T∆S° T = ∆H°/∆S° **Step 5: Calculate minimum temperature** Convert ∆H° to J/mol: 155.8 kJ/mol = 155,800 J/mol T = 155,800 J/mol ÷ 161.65 J mol⁻¹K⁻¹ T = 963.5 K ≈ 964 K Therefore, the answer is 964.00.