Question: 138 gm of N₂O₄(g) is placed in 8.2L container at 300 K. The equilibrium vapour density of mixture wa...

Question

138 gm of N₂O₄(g) is placed in 8.2L container at 300 K. The equilibrium vapour density of mixture was found to be 30.67. Then (R = 0.082 L atm mol⁻¹ K⁻¹)

Answer 1

Solution

The problem involves the dissociation of N₂O₄ gas into NO₂ gas and requires calculating several parameters at equilibrium.

The reaction is: N₂O₄(g) ⇌ 2NO₂(g)

Given data: Mass of N₂O₄ = 138 gm Volume of container (V) = 8.2 L Temperature (T) = 300 K Equilibrium vapour density of mixture = 30.67 R = 0.082 L atm mol⁻¹ K⁻¹

1. Calculate initial moles of N₂O₄: Molar mass of N₂O₄ (M_theoretical) = (2 × 14) + (4 × 16) = 28 + 64 = 92 gm/mol Initial moles of N₂O₄ (n_initial) = Mass / Molar mass = 138 gm / 92 gm/mol = 1.5 mol

2. Calculate the observed molar mass of the equilibrium mixture (M_obs): M_obs = 2 × Vapour density = 2 × 30.67 = 61.34 gm/mol

3. Calculate the degree of dissociation (α): For the dissociation reaction A(g) ⇌ nB(g), the relationship between theoretical molar mass (M_theoretical) and observed molar mass (M_obs) is: M_theoretical / M_obs = 1 + (n - 1)α Here, n = 2 (since 1 mole of N₂O₄ produces 2 moles of NO₂). 92 / 61.34 = 1 + (2 - 1)α 1.50 = 1 + α α = 1.50 - 1 = 0.50

Therefore, option (A) α = 0.25 is incorrect.

4. Calculate the total pressure at equilibrium (P_total): Initial moles: N₂O₄ = 1.5 mol, NO₂ = 0 mol At equilibrium: Moles of N₂O₄ = 1.5(1 - α) = 1.5(1 - 0.5) = 1.5 × 0.5 = 0.75 mol Moles of NO₂ = 2 × 1.5α = 3α = 3 × 0.5 = 1.5 mol Total moles at equilibrium (n_total) = 0.75 + 1.5 = 2.25 mol

Using the ideal gas law, PV = nRT: P_total = (n_total × R × T) / V P_total = (2.25 mol × 0.082 L atm mol⁻¹ K⁻¹ × 300 K) / 8.2 L P_total = (2.25 × 0.082 × 300) / 8.2 P_total = (2.25 × 300) / 100 P_total = 2.25 × 3 = 6.75 atm

Therefore, option (C) Total pressure at equilibrium = 6.75 atm is correct.

5. Calculate Kₚ of the reaction: Partial pressure of N₂O₄ (P_N₂O₄) = (Moles of N₂O₄ / Total moles) × P_total P_N₂O₄ = (0.75 / 2.25) × 6.75 atm = (1/3) × 6.75 atm = 2.25 atm

Partial pressure of NO₂ (P_NO₂) = (Moles of NO₂ / Total moles) × P_total P_NO₂ = (1.5 / 2.25) × 6.75 atm = (2/3) × 6.75 atm = 4.5 atm

Kₚ = (P_NO₂)² / P_N₂O₄ Kₚ = (4.5)² / 2.25 Kₚ = 20.25 / 2.25 = 9 atm

Therefore, option (B) Kₚ of N₂O₄ ⇌ 2NO₂(g) will be 9 atm is correct.

6. Calculate the density of the equilibrium mixture: Density (ρ) = Total mass / Total volume Total mass of the mixture is conserved and is equal to the initial mass of N₂O₄ = 138 gm. Total volume = 8.2 L ρ = 138 gm / 8.2 L = 16.829... gm/L ≈ 16.83 gm/L

Therefore, option (D) The density of equilibrium mixture will be 16.83 gm/litre is correct.

Conclusion: Options (B), (C), and (D) are correct.