At 300 K, ozone is fifty percent dissociated. The standard free energy change at this temperature an — Chemical Equilibrium Chemistry Question

💡 Solution & Explanation

**Step 1: Set up the dissociation equilibrium** For ozone dissociation: 2O₃ ⇌ 3O₂ If initial moles of O₃ = 2, and 50% dissociates: - O₃ dissociated = 1 mole - O₃ remaining = 1 mole - O₂ formed = 1.5 moles - Total moles = 2.5 moles **Step 2: Calculate mole fractions at 1 atm** - χ(O₃) = 1/2.5 = 0.4 - χ(O₂) = 1.5/2.5 = 0.6 **Step 3: Calculate partial pressures** - P(O₃) = 0.4 × 1 = 0.4 atm - P(O₂) = 0.6 × 1 = 0.6 atm **Step 4: Calculate equilibrium constant Kp** Kp = [P(O₂)]³/[P(O₃)]² Kp = (0.6)³/(0.4)² = 0.216/0.16 = 1.35 atm **Step 5: Apply ΔG° = –RT ln Kp** ΔG° = –8.3 × 300 × ln(1.35) ΔG° = –2490 × 0.3 ΔG° = –747 J/mol Therefore, the answer is 747.