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 3.6 $\times$ 10^-2 atm at equilibrium.What is the partial pressure of NO₂ at equilibrium?

Question

Quizplus · Accepted Answer

The equilibrium constant expression(Kp) for the given reaction is:

Kp = (P(N2O)/(P(NS2O))^2 * (P(O2)/P(SO2))^14 * (P(NS2O4)/P(NS2O))^12

At equilibrium,

Kp = 0.15 = (P(N2O)/(3.6x10^-10))^2 * (1/1)^14 * (1/ (P(NS2O)/(3.6x10^-10)))^12

Rearranging the equation,

P(N2O) = sqrt(0.15 x (3.6x10^-10)^2 / (1.6x10^-12)) = 0.073 atm (approx)

Therefore, the partial pressure of N2O at equilibrium is 0.073 atm.

Hence, the correct answer is E.

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

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