At 45 ∘C, Kc = 0.619 for the reaction N2O4(g)⇌2NO2(g). If 48.2 g of N2O4 is introduced into an empty 2.08 L container, what are the partial pressures of NO2 and N2O4 after equilibrium has been achieved at 45∘C?

P N₂O₄ = 3,05 atm

P NO₂ = 7,04 atm

Explanation:

48,2g of N₂O₄ are:

48,2g ₓ (1mol / 92,011g) = 0,524mol of N₂O₄ / 2,08L = 0,252M

Based on the reaction

N₂O₄(g) ⇌ 2NO₂(g) Kc = 0,619 = [NO₂]² / [N₂O₄] (1)

Concentrations in equilibrium are:

[N₂O₄] = 0,252M - X

[NO₂] = 2X

Replacing in (1):

0,619 = [2X]² / [0,252-X]

0,156 - 0,619X - 4X² = 0

Solving for X:

X = -0,289 → False answer, there is no negative concentrations

X = 0,135

Replacing:

[N₂O₄] = 0,252M - 0,135

[N₂O₄] = 0,117M

[NO₂] = 2X

[NO₂] = 2×0,135 = 0,270M

using:

P = M×R×T

Where P is pressure, M is molarity, R is gas constant (0,082atmL/molK) and T is temperature (45 + 273,15 = 318,15K). Pressure of N₂O₄ and NO₂ are:

P N₂O₄ = 0,117M×0,082atmL/molK×318,15K = 3,05atm

P NO₂ = 0,270M×0,082atmL/molK×318,15K = 7,04atm

I hope it helps!

the answer is true.

the answer for sodium is 0.186

the answer for rubidium is 0.251