Degree of dissociation, mole fraction and associated equilibrium

Degree of dissociation, mole fraction and associated equilibrium:

The number of molecules dissociated out of total number of molecules originally taken is called the degree of dissociation.

Mathematically,

Degree of dissociation = α = (Number of molecules dissociated / total number of molecules) X 100

It is expressed in percentage or as fraction or decimals. If 40 molecules are dissociated out of 100 molecules than the degree of dissociation is expressed as 40% or 0.4.

            The ratio of number of moles of a component to the total number of moles of the entire component in a given system is called the mole fraction. If nA and nB are the numbers of moles of two components A and B of a system then the total number of moles = nA + nB. Thus the mole fraction of component A = xA or XA (read as chi A) = nA / (nA + nB). Similiarly, the molefraction of B = xB or XB = nB / (nA + nB).

            The mole fraction of a component multiplied by the total pressure inside a container due to all components gives the partial pressure of that component. Using the partial pressures of various components the equilibrium constant can be calculated.

            pA = partial pressure of the component A = xA (total pressure) and

            pB = partial pressure of the component B = xB (total pressure)

Example: At 300k and 1 atm pressure N₂O₄ is 20% dissociated. What will be the degree of dissociation or percentage of dissociation at 300k and 0.1 atm pressure.

Solution:             

                                                       

Total moles:   1 — α + 2 α = 1 + α

Mole fraction of N₂O₄ at eq. = xN₂O₄ = number of moles N₂O₄ / total moles = 1 — α / 1 + α

Mole fraction of NO₂ at equilibrium = xNO₂ = number of moles NO₂ / total moles = 2 α / 1 + α

If the total pressure is = P

The partial pressure of N₂O₄ = pN₂O₄ = P. xN₂O₄ = P X (1 — α / 1 + α)

The partial pressure of NO₂ = pNO₂ = P. xNO₂ = P X (2 α / 1 + α)

Now Kp = (pNO₂)² / pN₂O₄ = { P X (2 α / 1 + α)}² / { P X (1 — α / 1 + α)} = {4α² / 1 — α²} P

At the first equilibrium, α = 20% 0r 0.2 and the total pressure, P = 1 atm

Thus, Kp = {4 (0.2)² / 1 — (0.2)²} P = 0.166, this is constant and remains constant as long as the temperature remains constant. Thus when a new equilibrium is established due to change in pressure, the value of equilibrium constant remains the same though the degree of dissociation changes.

Let the new degree of dissociation is = β, when pressure (P) changes to 0.1 atm.

Thus, Kp = 0.166 = {4β ² / 1 — β ²} 0.01, solving which we get β = 0.54 or 54%