For complete combustion of methanol the amount of heat produced as measured by bomb calorimeter is 7 — Thermodynamics and Thermochemistry Chemistry Question

💡 Solution & Explanation

**Step 1: Write the combustion reaction of methanol** CH₃OH(l) + 3/2 O₂(g) → CO₂(g) + 2H₂O(l) **Step 2: Identify the relationship between heat of combustion and enthalpy** The heat measured in a bomb calorimeter at constant volume is ΔU (internal energy change), not ΔH (enthalpy change). **Step 3: Apply the relationship between ΔH and ΔU** ΔH = ΔU + Δ(nRT) = ΔU + ΔnRT where Δn = (moles of gaseous products) – (moles of gaseous reactants) **Step 4: Calculate Δn** Δn = 1 (CO₂) – 3/2 (O₂) = 1 – 1.5 = –0.5 mol **Step 5: Calculate the correction term** ΔnRT = (–0.5) × 8.3 × 300 = –0.5 × 2490 = –1245 J/mol = –1.245 kJ/mol **Step 6: Calculate ΔH** ΔH = ΔU + ΔnRT ΔH = –726 + (–1.245) ΔH = –727.245 kJ/mol ΔH ≈ –727 kJ/mol Therefore, the answer is 727.