Question

sample of pure NO2 is heated to 339 ∘C at which temperature it partially dissociates according to the equation 2NO2(g)⇌2NO(g)+O2(g) At equilibrium the density of the gas mixture is 0.515 g/L at 0.755 atm . Calculate Kc for the reaction.

Answer 1

Given info : A sample of pure NO2 is heated to 339°C at which temperature it partially dissociates according to the equation

2NO2(g)⇌2NO(g) + O2(g)

To find : At equilibrium the density of the gas mixture is 0.515 g/L at 0.755 atm, The Kc for the reaction is...

solution : here density of the gas mixture, d = 0.52 g/L

let mass of has mixture is m and M is molecular weight of gas.

so, d = m/V

⇒V = m/d

so, PV = nRT

⇒Pm/d = m/M × RT

⇒M = dRT/P

= 0.52 × 0.082 × 612/0.75 = 36.24 g/mol

[ here T = 339° = 612 K, ]

from reaction,

2NO2(g)⇌2NO(g) + O2(g)

at t = 0. 1 0. 0

at eql. 1 - 2x 2x x

total moles = 1 - 2x + 2x + x = 1 + x

molecular weight of mixture is M = (1 - 2x)/(1 + x) × $M_{NO_2}$ + 2x/(1 + x) $M_{NO}$ + x/(1 + x) × $O_2$

⇒36.24 = (1 - 2x)/(1 + x) × 46 + 2x/(1 + x) × 30 + x/(1 + x) × 32

⇒x = 0.22 mol

now [NO2] = 1 - 2x = 1 - 0.44 = 0.66 mol

[NO] = 2x = 0.44 mol

[O2] = x = 0.22 mol

Therefore equilibrium constant, Kc = [NO]²[O2]/[NO2]²

= (0.44)²(0.22)/(0.66)²

= (2/3)² × 0.22

= 0.0977 ≈ 0.1

Therefore the equilibrium constant would be 0.1 [approximately]