**Degree of Dissociation (α): **

Let us consider the reaction; 2NH_{3} (g) <—> N_{2} (g) + 3H_{2} (g)

Let the initial moles of NH_{3} (g) be ‘ a ‘

Let x moles of NH_{3} dissociate at equilibrium.

2NH_{3} (g) <—> N_{2} (g) + 3H_{2} (g)

Initial moles a 0 0

At equilibrium a – x x/2 3x/2

Degree of dissociation (α) of NH_{3} is defined as the number of moles of NH_{3} dissociated per mole of NH_{3} .

∴ if x moles dissociate from ‘ a ‘ moles of NH_{3} , then, the degree of dissociation of NH_{3} would be . We can also look at the reaction in the following manner.

2NH_{3} (g) <—> N_{2} (g) + 3H_{2} (g)

Initial moles: a 0 0

At equilibrium : a(1 – α) aα/2 3aα/2

or a – 2x’ x’ 3 x’

where α = 2x’/a

Here total no. of moles at equilibrium is

a – 2x’ + x’ + 3x’ = a + 2x’

Mole fraction of NH3 = (a – 2x’)/(a + 2x’)

Mole fraction of N_{2} = x’/(a + 2x’)

Mole fraction of H_{2} = 3x’/(a + 2x’)

The expression of Kp is

$\large K_P = \frac{(\frac{x’}{a+2x’})P_T \times (\frac{3x’}{a+2x’})P_T^3}{(\frac{a-2x’}{a+2x’})^2 \times P_T^2}$

In this way you should calculate the basic equation. So my advice to you is that, while solving problem follow the method given below:

1. Write the balanced chemical reaction (mostly it will be given)

2. Under each component write the initial no. of moles.

3. Do the same for equilibrium condition.

4. Then derive the expression.

### Dependence of Degree of Dissociation

**Dependence of Degree of Dissociation on Density Measurements :**

The following is the method of calculating the degree of dissociation of a gas using vapour densities. This method is valid only for reactions whose K_{P} exist ,

i.e., reactions having at least one gas and having no solution

Since PV = nRT

$\large P V = (\frac{w}{M}) R T$

$\large M = (\frac{w R T}{P V}) = \frac{\rho R T}{P}$

∴ VD = ρRT/2P

Since P = nRT/V

∴ VD = ρV/2n

For a reaction at eqb., V is a constant and ρ is a constant.

∴ vapour Density ∝ (1/n)

$\large \frac{Total \; Moles \;at\; Equilibrium}{Initial \;total\; moles} = \frac{vapour\; density \;initial}{vapour\; density \;at \;equilibrium} = \frac{D}{d} = \frac{M}{m}$

(∴ molecular weight = 2 × V.D)

Here M = molecular weight initial

m = molecular weight at equilibrium