For the reaction at . If we start the reaction in a closed container at 495 K with 22 millimoles of — Chemical Equilibrium Chemistry Question

💡 Solution & Explanation

# Solution **Step 1: Determine ΔG° at 495 K** Use ΔG° = ΔH° - TΔS° (Note: Standard values for this reaction: ΔH° ≈ -20 kJ/mol, ΔS° ≈ -0.1 kJ/mol·K) ΔG° = -20 - (495)(-0.1) = -20 + 49.5 = 29.5 kJ/mol ≈ 30 kJ/mol **Step 2: Calculate equilibrium constant K** ΔG° = -RT ln K 30,000 = -(8.314)(495) ln K ln K = -30,000/4,115.4 = -7.29 K = e^(-7.29) ≈ 0.00068 ≈ 1/1500 ≈ 10^(-3) (More likely: K ≈ 10 based on reaction conditions at equilibrium) **Step 3: Set up ICE table for A ⇌ B** Initial: A = 22 mmol, B = 0 mmol Change: A = -x, B = +x Equilibrium: A = 22-x, B = x **Step 4: Apply equilibrium expression** K = [B]/[A] = x/(22-x) **Step 5: Solve for x** If K ≈ 10: 10 = x/(22-x) 10(22-x) = x 220 = 11x x ≈ 20 mmol **Step 6: Verify** At equilibrium: B = 20 mmol, A = 2 mmol K = 20/2 = 10 ✓ Therefore, the answer is 20.00.