The enthalpy of combustion of propane, graphite and dihydrogen at 298 K are: –2220.0 kJ mol , – 393. — Thermodynamics and Thermochemistry Chemistry Question

💡 Solution & Explanation

**Step 1: Write the combustion equation for propane** C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(l) ΔH_combustion = –2220.0 kJ mol⁻¹ **Step 2: Apply Hess's Law** The enthalpy of formation of propane can be found using: ΔH_combustion = Σ ΔH_f(products) – Σ ΔH_f(reactants) **Step 3: Set up the equation** –2220.0 = [3 × ΔH_f(CO₂) + 4 × ΔH_f(H₂O)] – [ΔH_f(C₃H₈) + 5 × ΔH_f(O₂)] **Step 4: Substitute known values** From the given data: - ΔH_f(CO₂) = –393.5 kJ mol⁻¹ - ΔH_f(H₂O) = –285.8 kJ mol⁻¹ - ΔH_f(O₂) = 0 kJ mol⁻¹ (element in standard state) –2220.0 = [3(–393.5) + 4(–285.8)] – [ΔH_f(C₃H₈) + 0] **Step 5: Calculate** –2220.0 = [–1180.5 + (–1143.2)] – ΔH_f(C₃H₈) –2220.0 = –2323.7 – ΔH_f(C₃H₈) ΔH_f(C₃H₈) = –2323.7 + 2220.0 = –103.7 kJ mol⁻¹ **Step 6: Find magnitude** |ΔH_f(C₃H₈