The effect of pressure change on the equilibrium constant
It is difficult to consider the effect of a pressure change without a full treatment of the partial pressures involved in an equilibrium. It becomes slightly easier if we consider the case of Kc the concentration equilibrium constant
A change in pressure MUST involve a change in volume at constant temperature if the system is closed. This change in volume will of course affect the concentrations of the gas:
for example,
If the gas pressure is doubled then the volume will be halved with a consequent
doubling of the concentration of the gas involved.
Why is this important?
For an equilibrium with a change in the number of moles of gas the numerator and denominator in the equilibrium law equation will then change by different quantities, and hence the value of Kc would be momentarily different. The equilibrium then shifts so as to reestablish the original value of Kc only now with the new (and different) concentrations.
Consider the simple equilibrium:
2NO_{2}  N_{2}O_{4} 
Kc =  [ N_{2}O_{4} ]  
[ NO_{2} ]2 
Let equilibrium concentrations be:
[N_{2}O_{4} ] = a
[ NO_{2} ] = b
The value of the equilibrium constant Kc now equals a/b2
If the pressure is double then the volume is halved then the concentration of each gas (at equilibrium) is doubled. Hence the new concentrations are:
[N_{2}O_{4} ] = 2a [ NO_{2} ] = 2b
And the value of the equilibrium constant is (temporarily) = 2a/4b2
or = 1/2 x a/b2
In other words it now has half the original value.
To compensate for this the equilibrium now shifts to the right hand side to produce more N_{2}O_{4} which increases the concentration of N_{2}O_{4} and decreases the concentration of NO_{2} ,
restoring the original value of Kc.

