An equilibrium mixture of I2 (g), Cl2 (g), and ICl (g) at 298 K has partial pressures of 0.0100 atm, 0.0100 atm, and 0.0900 atm, respectively. What is ΔG° at 298 K for the reaction I2 (g) + Cl2 (g) ⇌ 2 ICl (g)?

• A. -10.9 kJ/mol
• B. -16.9 kJ/mol
• C. 10.9 kJ/mol
• D. 16.9 kJ/mol

Answer: (A)

Gibbs free energy change under standard conditions for a gas reaction is linked to the equilibrium constant by ΔG° = -RT ln Kp. For the reaction I2 + Cl2 ⇌ 2 ICl, the equilibrium constant in terms of partial pressures is Kp = (P_ICl)^2 / (P_I2 P_Cl2) because two moles of ICl are formed.

Plugging in the given equilibrium partial pressures: P_ICl = 0.0900 atm, P_I2 = 0.0100 atm, P_Cl2 = 0.0100 atm, gives Kp = (0.0900)^2 / (0.0100 × 0.0100) = 0.0081 / 0.0001 = 81.

Then ΔG° = -RT ln Kp = -(8.314 J/mol·K)(298 K) ln(81). With ln(81) ≈ 4.394 and RT ≈ 2477.7 J/mol, ΔG° ≈ -2477.7 × 4.394 ≈ -1.09 × 10^4 J/mol ≈ -10.9 kJ/mol.

So the standard Gibbs free energy change is about -10.9 kJ/mol. The negative value indicates the reaction is favored under standard conditions, which aligns with the relatively large Kp (>1) and the observed product-heavy equilibrium.