13.8 g of N2O4 was placed in a 1 L reaction vessel at 400 K and allow

13.8 g of N2O4 was placed in a 1 L reaction vessel at 400 K and allowed to attain equilibrium- N2O4(g) ⇌ 2NO2(g)The total pressure at equilibrium was found to be 9.15 bar. Calculate Kc . Kp and partial pressure at equilibrium.

Answer»

Given,

Total volume (V) = 1L

Mass of N₂O₄ = 13.8g

Molar mass of N₂O₄ = 14(2) + 16(4) = 92g

Number of moles N₂O₄(n) = \(\frac{13.8g}{92g}\) = 0.15

Gas constant (R) = 0.083 bar L mol⁻¹ K⁻¹

Temperature (T) = 400 K

According to ideal gas equation,

pV = nRT

p × 1 L = 0.15 mol × 0.083 bar L mol⁻¹ K⁻¹x 400 K

\(\therefore\) p = 4.98 bar

For this reaction,

N₂O₄ (g) ⇌ 2NO₂ (g)

Initial pressure	4.98 bar	0
At equilibrium	(4.98 – x) bar	2x bar

Hence,

P_totalat equilibrium = \(P_{N_2O_4}\) + \(P_{NO_2}\)

9.15 = (4.98 - x) + 2x

9.15 = 4.98 + x

\(\therefore\) x = 9.15 - 4.98 = 4.17 bar

Partial pressures at equilibrium are,

\(P_{N_2O_4}\) = 4.98 - 4.17 = 0.81 bar

\(P_{NO_2}\) = 2x = 2 x 4.17 = 8.34 bar

K_p = \(\frac{(P_{NO_2})^2}{P_{N_2O_4}}\)

= \(\frac{(8.34)^2}{0.81}\) = 85.87

K_p = K_c (0.083 x 4.00)1

\(\therefore\) K_c = \(\frac{85.87}{0.083\times400}\) = 2.586