A 250 ml flask and 100 ml flask are separated by a stopcock. At 350 K, the nitric oxide in the large — Chemical Equilibrium Chemistry Question
Question
A 250 ml flask and 100 ml flask are separated by a stopcock. At 350 K, the nitric oxide in the larger flask exerts a pressure of 0.4 atm, and the smaller one contains oxygen at 0.8 atm. The gases are mixed by opening the stopcock. The reactions occurring are:<br>2NO + O₂ → 2NO₂ ⇌ N₂O₄<br>The first reaction is complete while the second one is at equilibrium. Assuming all the gases to behave ideally, calculate the K_p for the second reaction if the total pressure is 0.3 atm.
💡 Solution & Explanation
Total V = 350 mL. Using PV units: NO = 0.4 × 250 = 100 units, O₂ = 0.8 × 100 = 80 units. 2NO + O₂ → 2NO₂. NO is limiting. 100 NO reacts with 50 O₂ to form 100 NO₂. Remaining O₂ = 30 units. Now, 2NO₂ ⇌ N₂O₄. Let 2x units of NO₂ react to form x units of N₂O₄. Total gas units = O₂ + NO₂ + N₂O₄ = 30 + (100 - 2x) + x = 130 - x. Total pressure = 0.3 atm, so total units = 0.3 × 350 = 105. 130 - x = 105 => x = 25. Thus, N₂O₄ = 25 units, NO₂ = 50 units. Partial pressures: P_N₂O₄ = 25/350 = 1/14 atm, P_NO₂ = 50/350 = 1/7 atm. K_p = P_N₂O₄ / (P_NO₂)² = (1/14) / (1/7)² = 49 / 14 = 3.5 atm⁻¹. Therefore, correct answer is A.