Chemistry - Chemical Equilibrium Question with Solution | TestHub

The reaction is

${\text{N}}_2{\text{O}}_4\rightleftharpoons 2{\text{NO}}_2$

mol. wt. of N₂O₄ = 14 x 2 + 16 x 4 = 92

$\therefore$Vapour density (D) of N₂O₄= 92/2 = 46

We know that,

$\alpha =\frac{\text{D}-\text{d}}{\left(\text{n}-1\right)\text{d}}=\frac{46-30}{\left(2-1\right)30}=\frac{16}{30}$

where,

D = Initial vapour density of reactant

d = vapour density of equilibrium mixture

$\alpha$= degree of dissociation

$\therefore \alpha =\text{0.533}$

$= \text{53.3 \%}$