What is the effect of pressure on equilibrium?

Similar to concentration, pressure does not change the value of the equilibrium constant *K*. So how does pressure affect the position of chemical equilibrium of a reaction? Consider the following reaction:

The equilibrium constant for the reaction is

where *p_{i}* is the partial pressure of gas

*i*. We can also write the above equation as:

where *x_{i}* is the mole fraction of gas

*i*,

*p*is the total pressure of the system,

*n*is the number of moles of gas

_{i}*i*and

*n*is the total number of moles of the gas mixture. For ideal gases, we can substitute

*pV*=

*nRT*in eq18 to give:

If we increase the total pressure by decreasing the volume of the system at constant temperature, the denominator of the term in brackets in eq19 must increase to maintain the value of *K* (or the numerator of the term in brackets must decrease). The position of the equilibrium of the reaction therefore shifts left to increase the amount of nitrogen dioxide. This is consistent with Le Chatelier’s principle where an increase in pressure shifts the position of the equilibrium from right (3 moles of gases) to left (2 moles of gases) to minimise the change in pressure.

It is important to note that changes in pressure only affects the position of an equilibrium where species are gases, as solids and liquids are relatively incompressible.

###### Question

Consider a fixed-volume reaction vessel with the following reaction at a state of equilibrium at constant temperature:

If we increase the pressure in the vessel by adding an inert gas (assuming that all gases in the vessel are ideal gases), will the composition of the gases change?

###### Answer

No, because the partial pressures of the gases remain the same even though the total pressure of the system has increased.

Since the factor of *RT*/*V* is a constant, *p_{i }*does not change if

*n*remains the same.

_{i }