Dissociation of PCl5 - Equilibrium constants in terms of degree of dissociation

Degree of dissociation (x)

In the study of dissociation equilibrium, it is easier to derive the equilibrium constant expression in terms of degree of dissociation (x). It is considered as the fraction of total molecules that actually, dissociate into the simpler molecules x has no units. If x is the degree of dissociation then for completely dissociating molecules x = 1.0. For all dissociations involving equilibrium state, x is a fractional value. If x is known, K_c or K_p can be calculated and vice-versa.

Equilibrium constants in terms of degree of dissociation

Dissociation of PCl₅

Phosphorus pentachloride dissociates in gas phase to give PCl₃ and Cl₂. This is an example of gaseous homogeneous equilibrium. It can be represented as

PCl5(g)  -- > < --  PCl3(g)  + Cl2(g)

For this reaction ∆n=1    K_p=K_c(RT)

Let us consider that one mole of PCl5 is present initially in a vessel of volume V dm³. At equilibrium let x mole dissociates to give x mole of PCl₃ and x mole of Cl₂. The equilibrium concentrations of the components can be given as follows :

                                             PCl5(g)   PCl3(g)  Cl2(g)

Initial number of moles         1          0          0

Number of moles reacted         x       -          -

Number of moles at equilibrium 1-x           x        x

Equilibrium concentration       1-x/V x/V    x/V

According to law of mass action,

 Kc = [PCl3][Cl2] / [PCl5]

Substituting  the  values  of  equilibrium  concentrations  in  the  above equation we get,                Kc = (x/V)(x/V)  / (1-x)/V

 = x²V/V²(1-x)

= x²/ (1-x)V

Derivation of Kp in terms of x

Let us consider that one mole of PCl5 is present initially. At equilibrium, let us assume that x mole of PCl5 dissociates to give x mole of PCl₃ and x mole of Cl₂. Let the total pressure at equilibrium be P atmosphere. The number of moles of PCl₅, PCl₃ and Cl₂ present at equilibrium can be given as follows :

                   PCl_5(g)          PCl_3(g)          Cl_2(g)

Initial number of moles  1        0        0

Number of moles reacted         x        -        -

Number of moles at equilibrium       1-x    X       x

 Total number of moles at equilibrium  =   1 - x + x + x

          =       1 + x

We know that partial pressure is the product of mole fraction and the total pressure. Mole fraction is the number of moles of that component divided by the total number of moles in the mixture. Therefore

P_PCl4 =(1-x)P / 1+x

P_PCl3 =(x)P / 1+x

P_PCl2 =(x)P / 1+x

We know that K_p = P_PCL3.P_CL2/P_PCL5

substituting the values of partial pressures in this expression

K_p = x².P²/(1+x)²  .  (1+x)/(1+x)  .  1/P

K_p = x²P / 1-x²

When x < < 1, x² value can be neglected when compared to one.

\ K_p  ~ x²P

This equation can be used to predict the influence of pressure on this equilibrium.

 Influence of pressure : The expression for Kc contains the volume term and the expression for Kp contains the pressure term. Therefore this equilibrium is affected by the total pressure. According to the above equation, increase in the value of P will tend to increase the value of Kp. But Kp is a constant at constant temperature. Therefore, in order to maintain the constancy of K_p the value of x should decrease. Thus, increase in total pressure favours the reverse reaction and decreases the value of x.

 Influence of concentration : Increase in the concentration of PCl₅ favours the forward reaction, while increase in the concentration of either PCl₃ or Cl₂ favours the reverse reaction. Increase in the concentration of a substance in a reversible reaction will favour the reaction in that direction in which the substance is used up.

 Influence of catalyst : A catalyst will affect the rates of the forward and reverse reactions to the same extent. It does not change either the amount of reactants and products present at equilibrium or the numerical value of Kp or Kp. However, the equilibrium is obtained quickly in the presence of a catalyst.