At 35 °C,K_p = 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

To find the equilibrium pressure of N₂O₄, we set up the expression for Kp: Kp = (P_NO₂)² / (P_N₂O₄). Let x be the change in pressure for N₂O₄, then at equilibrium, P_NO₂ = 3.00 - 2x and P_N₂O₄ = x. Solving the equation 0.315 = ((3.00 - 2x)²) / x gives x = 1.19 atm.

At 35 °C,Kp = 0

At 35 °C,K_p = 0