Question

At 27°C, N2O4(g) dissociates 40% into NO2(g) the equilibrium vapour density of the mixture is​

Answer 1

Answer:

Explanation dissociation constant=D-d/(n-1)d

D-VD befor dissociation

d-VD of mixture @ equilibruim

n-no of moles before dissociation

D=Molecular mass of n2o4/2

=92/2=46

n=2

On substituting values on above equation

40/100=46-d/d

d=32.85

Answer 2

The equilibrium vapor density of the mixture is​ 32.85

Explanation:

The dissociation constant is given by the formula:

α = (D - d)/((n - 1)d))

Where,

D = Vapor density before dissociation

d = Vapor density after equilibrium

n = Number of moles before dissociation = 2

Now,

2 × Vapor density before dissociation  = Molecular mass of N₂O₄

2 × Vapor density before dissociation  = 92

∴ Vapor density before dissociation  = 46 = D

Now,

40/100 = (46 - d)/((2 - 1)d))

4/10 = (46 - d)/d

4d = 460 - 10d

14d = 460

∴ d = 32.85