From Vapour Density Measurements: Equilibrium Constant

A state of equilibrium is defined as a state of balance between two chemically reacting compounds. The equilibrium constant of a reversible chemical reaction is described as the ratio of the concentration of the products to the concentration of the reactants. 

The degree of dissociation in an ionisation reaction is defined as the fraction of moles of electrolytes that dissociate after the state of equilibrium. The relationship between the vapour density and degree of dissociation can be established by calculating electrolytes’ molecular mass and molar density.

 Gas density or vapor related to hydrogen, oxygen, or air. If we take hydrogen as a reference, the vapor pressure is a measure of the mass of a certain volume of gas and the equivalent volume of hydrogen under the same pressure and temperature conditions.

 Continue reading to know the importance of vapour density measurements.

Equilibrium Constant

The law of chemical equilibrium gives the expression for the equilibrium constant. 

 For a general chemical reaction, the chemical equilibrium is given by, 

a.A + b.B ⇌ c.C + d.D,

Kc = [C]c [D]d / [A]a [B]b (Kc = Equilibrium Constant)

 Degree of Dissociation 

The degree of dissociation of an electrolyte reactant is defined as the fraction of the total number of moles of the electrolyte that dissociate and give its ions after achieving the state of equilibrium. The degree of dissociation of an electrolyte is represented by α. 

In a general chemical reaction, the degree of dissociation is calculated by,

α = (total moles of dissociated substance) / (total amount of substance)

Solved Example

1. The degree of dissociation of electrolyte HF in the 0.04 M HF solution is unknown.

Calculate the concentration of all the electrolytes present (H3O+, F–, and HF) in the solution and its pH value. (Here, ionisation constant = 3.2 × 10–4)

 Solution:

From the given question the following ionisation reactions are possible,

• HF + H2O ⇌ H3O+ + F– (Ka = 3.2 x 10-4)

• H2O + H2O ⇌ H3O+ + OH– (Kw = 1 x 10-4)

From the above two reactions, it is clear that the former is the primary reaction (Ka >> Kw),

HF + H2O ⇌ H3O+ + F–

Initial concentration

0.04 0 0

Change in concentration,

-0.04α +0.04α +0.04α 

Concentration at equilibrium,

0.04-0.04α 0.04α 0.04α

Substituting the concentrations in the equilibrium reactions,

Ka = (0.04)2 / (0.04-0.04α)

= 0.04.α2/(1 –α) = 3.2 × 10–4

The above expression gives a quadratic equation,

= α2 + 0.8α x 10-2– 0.8 x 10-2

Upon solving the above equation the value of α will be,

α = 0.08,

α = -0.09

Because, -ve value can not be considered, α = 0.08.

Therefore, the concentration of other components in the equation,

 [H3O+] = [F–] = c.α = 0.04 × 0.08

= 3.2 x 10-3 Moles.

And, [HF] = c.(1 – α) = 0.04.(1 – 0.08)

36.8 x 10-3 Moles.

The pH, value is given by,

pH = – log[H+] = –log(3.2 × 10–3) = 2.49.

 Relationship between Vapour Density and Degree of Dissociation

Consider a reversible chemical reaction, 

A  →  y.B

 Initial moles,

1 0

In Equilibrium,

(1-x) y.x

Here, let’s say x is the degree of dissociation,

Total number of moles at equilibrium,

1-x + y.x,

1 + x.(y – 1),

If the initial volume of the reactant is V,

Then the volume at equilibrium,

1 + x.(y – 1).V

The molar density of before dissociation is given by,

D = molecular weight/ Volume,

= m/V

Molar density after dissociation is given by,

d = m/[1 + x.(y – 1).V].

D/d = 1 + x (y – 1).

x = D – d/[d.(y – 1)]

In terms of molecular mass,

x = (M – m)/ (y- 1).m

Solved Examples

1. Consider the following chemical reaction,

N2O4 ⇌ 2.NO2

Calculate the degree of dissociation of the above equation if the vapour density is 36.5 at 32oC.

Solution: 

The given reaction is, 

N2O4 ⇌ 2.NO2

 The degree of dissociation,

 α = (M – m)/ (y- 1).m

 m = 2 x vapour density

 m = 2 x 36.5 = 73 

M = M.N2O4

 = 92.

 Here, y = 2

 From the formula,

 = (92 – 73)/(2 – 1). 73.

 = 0.26.

 2. Consider the following chemical reaction, 

P.Cl5 ⇌ PCl3 + Cl2

 Calculate the degree of dissociation of the above equation if the vapour density is 30 at 32oC, and y = 2.

 Solution: 

The given reaction is,

 P.Cl5 ⇌ PCl3 + Cl2

 The degree of dissociation, 

α = (M – m)/ (y- 1).m

 m = 2 x vapour density

m = 2 x 30 = 60

M = M.PCl5

= 1 x 30.9 + 5 x 35.45

= 208.15

Here, y = 2

From the formula,

= (208.15 – 60)/(2 – 1). 60.

= 2.47.

Conclusion

The vapour density measurement (equilibrium constant) can be concluded by establishing a relationship between molar density and the degree of dissociation of electrolytes in a state of equilibrium. In the above article, we derive the formula of the degree of dissociation in a chemical reaction that has attained the state of equilibrium. 

The degree of dissociation is defined as the fraction of the moles in the reactant of chemical equilibrium. The vapour density of a molecule in a state of chemical equilibrium is given with respect to the total hydrogen [H+] ions in the reaction. Below are some important questions related to the vapour density measurement (equilibrium constant).