# Relation Between Degree Of Dissociation And Vapour Density

## Degree of Dissociation:

Degree of dissociation may be defined as the fraction of a mole of the reactant undergoing dissociation. It is usually denoted by 'α'.

a = number of moles dissociate/number of moles present initially = Percentage Dissociation/100

For example, let us consider the reaction

2SO3(g) ⇄ 2SO2(g) + 1/2O2(g)

Let the initial moles of SO3 be 'a' and the moles of SO3 dissociated at equilibrium be 'x'.

2SO3(g) ⇄ 2SO2(g) + 1/2O2(g)

Initially              a       0        0

At equilibrium  a–x     x      x/2

Now if x moles dissociate from a moles of SO3, then degree of dissociation of SO3
would be x/a.

If the moles dissociated would have been 2x', then

2SO3(g) ⇄ 2SO2(g) + 1/2O2(g)

Initially                          a            0         0

At equilibrium            a–2x '       2x '      x'

α = 2X/a'

Let us now derive the relation between vapour density and the degree of dissociation.

Assuming gaseous components at equilibrium behaves ideally, we can apply ideal gas equation.

PV = nRT = W/m *RT

M = w/v x RT/P = ρRT/P

V. D. = ρRT/2nRT x v = ρV/2n [since molar mass = 2 × V.D.]

Since P = nRT/V

V.D. = ρRT/2nRT x V = ρV/2n

For a reaction at equilibrium, volume of the vessel (V) and density of gaseous mixture (ρ) is a constant.

V. D. ∝ 1/n

Total moles at equilibrium/Initial number of moles = Vapour density initial/Vapour Density at equilibrium

Let us consider a general equilibrium reaction,

A(g) ⇄ nB(g)

Initially                    c           0

At equilibrium    c(1−α)      cnα

Total moles at equilibrium = c − cα + cnα = c[1 + α(n−1)]

Let the initial vapour density and the vapour density at equilibrium be D and d respectively.

c[1 + α (n - 1)]/c = D/d

1 + α(n−1) = D/d

α = (D-d)/(n-1)d

Knowing the value of D and d, α can be calculated.

For example,

SO3(g) ⇄ SO2(g) + ½O2(g)

Initially                        c              0             0

At equilibrium         c − cα       cα             cα/2

Total moles at equilibrium =

c(1 + α/2)

c(1 + α/2)/c = D/d

α = (2(D-d)/d)

[Note: This method is valid only for those equilibrium reactions whose Kp exists and Δn ≠ 0 i.e. equilibrium having at least one gas and no solution component]