Gaseous nitrosyl chloride (NOCl) and N₂ are taken in a flask, sealed and heated to some temperature — Chemical Equilibrium Chemistry Question
Question
Gaseous nitrosyl chloride (NOCl) and N₂ are taken in a flask, sealed and heated to some temperature where the total pressure would have been 1.0 bar had not the following equilibrium been established 2NOCl(g) ⇌ 2NO(g) + Cl₂(g). But the actual pressure was found to be 1.2 bar. Now, into the equilibrium mixture, some Cl₂ gas was introduced at constant volume and temperature so that the total pressure would have been 8.3 bar had no further reaction occurred but the actual pressure was found to be 8.2 bar. The equilibrium constant, K_p (in bar), for the decomposition reaction under the given experimental condition is (Answer by multiplying the K_p value with 10)
💡 Solution & Explanation
Initial equivalent pressure = 1.0. Let initial P_NOCl = p₀. Eq 1: 2x reacts. Total P = 1.0 + x = 1.2 => x = 0.2 bar. P_NO = 0.4, P_Cl₂ = 0.2. P_NOCl = p₀ - 0.4. When Cl₂ is added, theoretical P jumps to 8.3. Cl₂ added = 8.3 - 1.2 = 7.1 bar. Actual P drops to 8.2, meaning P drops by 0.1 bar as reaction shifts backward. 2NO + Cl₂ → 2NOCl. For a 0.1 bar drop, y = 0.1 bar of Cl₂ reacts. New eq: P_NO = 0.4 - 2(0.1) = 0.2 bar, P_Cl₂ = 0.2 + 7.1 - 0.1 = 7.2 bar, P_NOCl = p₀ - 0.4 + 2(0.1) = p₀ - 0.2. K_p = P_NO² P_Cl₂ / P_NOCl² = (0.4)²(0.2) / (p₀ - 0.4)² = (0.2)²(7.2) / (p₀ - 0.2)². 0.032 / (p₀ - 0.4)² = 0.288 / (p₀ - 0.2)². 9(p₀ - 0.4)² = (p₀ - 0.2)². 3(p₀ - 0.4) = -(p₀ - 0.2) (valid root). 3p₀ - 1.2 = -p₀ + 0.2 => 4p₀ = 1.4 => p₀ = 0.35. Wait, 3(p₀ - 0.4) = p₀ - 0.2 => 2p₀ = 1.0 => p₀ = 0.5. K_p = 0.032 / (0.5 - 0.4)² = 0.032 / 0.01 = 3.2. Multiply by 10 gives 32. Therefore, correct answer is 0032.