At -80°C,K for the reaction N₂O₄(g) 2NO₂(g) Is 4.66 $\times$ 10^-⁸.We introduce 0.047 mole of N₂O₄ into a 1.0-L vessel at -80°C and let equilibrium be established.The total pressure in the system at equilibrium will be:

Question

Quizplus · Accepted Answer

First, we need to write the balanced chemical equation and the expression for equilibrium constant (K): N₂O₄(g) ⇌ 2NO₂(g) K = [NO₂]^2/[N₂O₄] Now, we are given the initial concentration of N₂O₄ as 0.047 mol in a 1.0 L vessel. Therefore, the initial concentration of NO₂ is 0 mol. Let the concentration of N₂O₄ that reacts be x mol, then the concentration of NO₂ that is formed is 2x mol. At equilibrium, the concentration of N₂O₄ is (0.047 - x) mol and the concentration of NO₂ is 2x mol. Using the equilibrium constant expression, we can write: K = (2x)^2/(0.047 - x) At equilibrium, K = 4.66x10^10. Substituting this and solving for x, we get x ≈ 3.3x10^-8 mol. The total pressure in the system at equilibrium is the sum of the partial pressures of the gases: P(total) = P(NO₂) + P(N₂O₄) Since these gases are in a 2:1 ratio, the partial pressure of NO₂ is twice that of N₂O₄: P(NO₂) = 2xRT/V = 2(3.3x10^-8)(0.0821)(193) / 1 = 1.25x10^-5 atm P(N₂O₄) = xRT/V = (3.3x10^-8)(0.0821)(193) / 1 = 6.45x10^-6 atm Therefore, the total pressure at equilibrium: P(total) = P(NO₂) + P(N₂O₄) ≈ 1.9x10^-5 atm ≈ 0.74 atm The closest choice is B.

At -80°C,K for the Reaction N2O4(g) 2NO2(g) Is 4 ×\times×

At -80°C,K for the Reaction N₂O₄(g) 2NO₂(g)
Is 4 $\times$