At 35°C, Kp = 0.315 for the reaction N₂O₄(g) ⇌ 2 NO₂(g), if the initial pressure of NO₂(g) in a container is 3.00 atm, what is the equilibrium pressure of N₂O₄(g)?

Question

Quizplus · Accepted Answer

The equilibrium expression for the reaction is Kp = (P_NO2)^2 / (P_N2O4). Let x be the change in pressure of N2O4 that forms at equilibrium. Then, P_NO2 = 3.00 - 2x and P_N2O4 = x. Substituting into the Kp expression and solving for x gives the equilibrium pressure of N2O4 as 1.19 atm.