For the equilibrium N₂O₄(g) 2NO₂(g),at 298 K,K_p = 0.15.For this reaction system,it is found that the partial pressure of N₂O₄ is 3.5 × 10^-2 atm at equilibrium.What is the partial pressure of NO₂ at equilibrium? (R = 0.0821 L ∙ atm/(K ∙ mol))

Question

Quizplus · Accepted Answer

Use the equilibrium expression: K_p = 0.15 = (pN₂O₄)/(pNO₂)^2 Rearranging for pNO₂: pNO₂ = sqrt((pN₂O₄)/(0.15)) = sqrt((3.5 × 10^-2)/(0.15)) = 0.072 atm. Therefore, the answer is E.

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

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