For the gas phase reaction SO3(g) SO2 (g) + ½ O2 (g) if the gaseous mixture in a closed container is allowed to come at equilibrium and the degree of dissociation (α) at equilibrium is found to be 2/3 at 400 K and 1 atm pressure then calculate the vapour density of the equilibrium mixture.40013060

To calculate the vapor density of the equilibrium mixture for the reaction:

Reaction Overview

The reaction is:

SO₃(g) ⇌ SO₂(g) + ½ O₂(g)

Understanding Degree of Dissociation

The degree of dissociation (α) is given as 2/3. This means that 2/3 of the SO₃ has dissociated at equilibrium.

Initial Moles

• Let the initial moles of SO₃ be 1.

Moles at Equilibrium

At equilibrium, the moles of each component can be calculated as follows:

• Moles of SO₃ remaining = 1 - α = 1 - 2/3 = 1/3
• Moles of SO₂ formed = α = 2/3
• Moles of O₂ formed = ½ * α = ½ * (2/3) = 1/3

Total Moles at Equilibrium

The total moles at equilibrium (n_total) is:

n_total = moles of SO₃ + moles of SO₂ + moles of O₂ = (1/3) + (2/3) + (1/3) = 4/3

Molar Mass Calculation

Next, we calculate the molar mass of the mixture:

• Molar mass of SO₃ = 80 g/mol
• Molar mass of SO₂ = 64 g/mol
• Molar mass of O₂ = 32 g/mol

Average Molar Mass of the Mixture

The average molar mass (M_avg) can be calculated as:

M_avg = (moles of SO₃ * molar mass of SO₃ + moles of SO₂ * molar mass of SO₂ + moles of O₂ * molar mass of O₂) / n_total

M_avg = [(1/3 * 80) + (2/3 * 64) + (1/3 * 32)] / (4/3)

M_avg = [(26.67 + 42.67 + 10.67)] / (4/3) = 80 / (4/3) = 60 g/mol

Vapor Density Calculation

The vapor density (D) is given by:

D = M_avg / 22.4

D = 60 g/mol / 22.4 L/mol = 2.68 g/L

Final Result

The vapor density of the equilibrium mixture is approximately 2.68 g/L.