When N₂O₅(g) is heated to 600 K, it dissociates as: N₂O₅(g) ⇌ N₂O₃(g) + O₂(g); K_c = 2.5 M. Simultan — Chemical Equilibrium Chemistry Question
Question
When N₂O₅(g) is heated to 600 K, it dissociates as: N₂O₅(g) ⇌ N₂O₃(g) + O₂(g); K_c = 2.5 M. Simultaneously, N₂O₃(g) decomposes as: N₂O₃(g) ⇌ N₂O(g) + O₂(g). When initially 4.0 moles of N₂O₅(g) is taken in a 2.0 l flask and allowed to attain equilibrium, the equilibrium concentration of O₂(g) is found to be 2.5 M. The equilibrium concentration of N₂O(g) (in M) is
💡 Solution & Explanation
Initial [N₂O₅] = 4.0 / 2.0 = 2.0 M. Let x be the concentration of N₂O₅ that dissociates, and y be the concentration of N₂O₃ that decomposes. Equilibrium concentrations: [N₂O₅] = 2.0 - x, [N₂O₃] = x - y, [N₂O] = y, [O₂] = x + y. Given [O₂] = x + y = 2.5 M. For the first reaction, K_c = ([N₂O₃][O₂]) / [N₂O₅] = ((x - y)(2.5)) / (2.0 - x) = 2.5. This simplifies to (x - y) = 2.0 - x, giving 2x - y = 2.0. We have a system: x + y = 2.5 and 2x - y = 2.0. Adding them gives 3x = 4.5 => x = 1.5 M. Substituting x back: 1.5 + y = 2.5 => y = 1.0 M. The equilibrium concentration of N₂O is y = 1 M. Therefore, correct answer is 1.