40% of HI undergoes decomposition to H and I at 300 K. ∆G for this decomposition reaction at one atm — Chemical Equilibrium Chemistry Question

💡 Solution & Explanation

**Step 1: Set up the equilibrium problem** For the decomposition: 2HI ⇌ H₂ + I₂ Initial: 2 mol HI 40% decomposes: 0.8 mol HI remains, 0.4 mol products (0.2 mol H₂ + 0.2 mol I₂) Total moles = 0.8 + 0.2 + 0.2 = 1.2 mol **Step 2: Calculate partial pressures** At 1 atm total pressure: - P(HI) = 0.8/1.2 = 2/3 atm - P(H₂) = 0.2/1.2 = 1/6 atm - P(I₂) = 0.2/1.2 = 1/6 atm **Step 3: Calculate reaction quotient Q** Q = [P(H₂) × P(I₂)] / [P(HI)]² Q = (1/6 × 1/6) / (2/3)² = (1/36) / (4/9) = 1/16 **Step 4: Apply the ΔG equation** ΔG° = −RT ln K At equilibrium: ΔG = −RT ln Q ΔG = −8.31 × 300 × ln(1/16) ΔG = −2493 × ln(1/16) = −2493 × (−ln 16) ΔG = 2493 × ln(16) = 2493 × 4 ln(2) ΔG = 2493 × 4 × 0.693 = 2493 × 2.772 ≈ 2739 J/mol Therefore, the answer is 2739.