Question

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?

Answer 1

N₂O₄ = 3.1 atm and NO₂ = 7 atm.

Step-by-step explanation:

First, the balanced equation of Reaction is given below;

N₂O₄(g) ⇌ 2NO₂(g).

One mole of N₂O₄(g) produces two moles of 2NO₂(g).

STEP ONE: find the concentration of N₂O₄.

The molarity of N₂O₄ = 92 g/mol, volume =2.08 L and we are given the mass of N₂O₄ to be 48.2 g.

Therefore, number of moles,n = mass/molar mass.

==> 48.2g / 92 g= 0.524mol.

The concentration of reaction = number of moles,n ÷ volume.

==> 0.524mol/ 2.08 L.

= 0.252M

The initial concentration of N₂O₄(g) and

NO₂(g) at time,t = 0 is 0.252M and OM.

The concentration at time,t = t of N₂O₄(g) and NO₂(g) is 0.252M - X and 2X respectively.

STEP TWO: Find the value of X at equilibrium

At equilibrium we have the concentration of N₂O₄ = 0.252M - X, and the concentration of NO₂ = 2X

Therefore, kc= [NO2]² / N₂O₄ .

==> 0.619 = [2X]² / [0.252-X]

0,M.156 - 0.619X - 4X² = 0

After solving the quadratic equation to find the value of X, we have X = 0.135

Therefore, the concentration of N₂O₄ = 0.252M - 0.135 = 0.117M. And that of NO₂ =0.270M (as 2 × 0.135).

STEP THREE: Calculate the pressure. This can be done by using the formula below;

Partial Pressure,P = Concentration,C × gas constant(0.082atmL/mol.k) × temperature, T.

N₂O₄ partial pressure = 0.117M × 0.082 ×318 = 3.1 atm

NO₂ partial pressure = 0.270M × 0.082 × 318= 7atm.

Answer 2

P N₂O₄ = 3,05 atm

P NO₂ = 7,04 atm

Step-by-step 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!