The following reaction is at equilibrium at 298 K: 2(g, 0.00001 bar) + (g, 0.01 bar) ⇌ 2NOCl(g, 0.01 — Thermodynamics and Thermochemistry Chemistry Question

💡 Solution & Explanation

At equilibrium, Δ G = 0 => Δ G° = -R * T * ln $K_p$. $K_p$ = P_NOCl^2 / (P_NO^2 * P_Cl2) = (0.01)^2 / ((10^-5)^2 * 0.01) = 10^-4 / 10^-12 = 10^8. Δ G° = -8.314 * 298 * ln(10^8) = -8.314 * 298 * 8 * 2.303 ≈ -45.65 kJ.