Question

Gaseous n2o4 dissociates into gaseous no2 according to the reaction n2o4 (g) 2no2 (g) at 300 k and 1 atm pressure, the degree of dissociation of n2o4 is 0.2. If one mole of n2o4 gas is contained in a vessel, then the density of the equilibrium mixture is

Answer 1

reaction is ... $N_2O_4\rightarrow 2NO_2$
at t = 0, mole of N2O4 = 1 
mole of NO2 = 0

at t = equilibrium, 
mole of N2O4 = 1 - a 
mole of NO2 = 2a 

where a is degree of dissociation

now, molecular weight of mixture = {(1 - a) × molar wt. of N2O4 + 2a × molar wt. of NO2}/(1 - a + 2a) 
= {(1 - 0.2) × (28 + 64) + 2 × 0.2 × (14 + 32)}/(1 + 0.2) 
= {0.8 × 92 + 0.4 × 46}/1.2 
= 76.66 

now, Molar wt.of gaseous mixture , M = 76.66
pressure, P = 1atm
Temperature , T = 300K
use formula, d = PM/RT 
where, d is density of gaseous mixture.

so, d = 1 × 76.66/0.082 × 300 
d =3.11 g/L