For the equilibrium N₂O₄(g) $\rightleftharpoons$ 2NO₂(g), at 298 K, K_p = 0.15. For this reaction system, it is found that the partial pressure of N₂O₄ is 3.7 × 10^-2 atm at equilibrium. What is the partial pressure of NO₂ at equilibrium?

Question

Quizplus · Accepted Answer

The equilibrium constant expression for the reaction is Kp = (P_NO2)^2 / (P_N2O4). Given Kp = 0.15 and P_N2O4 = 0.037 atm, solving for P_NO2 gives P_NO2 = sqrt(0.15 * 0.037) = 0.074 atm.

For the Equilibrium N2O4(g) ⇌\rightleftharpoons⇌ 2NO2(g), at 298 K, Kp = 0

For the Equilibrium N₂O₄(g) $\rightleftharpoons$ 2NO₂(g), at 298 K, K_p = 0