EQUILIBRIUM CONSTANTS: K_p

This page explains equilibrium constants expressed in terms of partial pressures of gases, K_p. It covers an explanation of the terms mole fraction and partial pressure, and looks at K_p for both homogeneous and heterogeneous reactions involving gases.

The page assumes that you are already familiar with the concept of an equilibrium constant, and that you know about K_c - an equilibrium constant expressed in terms of concentrations

Defining some terms

Before we can go any further, there are two terms relating to mixtures of gases that you need to be familiar with.

Mole fraction

If you have a mixture of gases (A, B, C, etc), then the mole fraction of gas A is worked out by dividing the number of moles of A by the total number of moles of gas.

The mole fraction of gas A is often given the symbol x_A. The mole fraction of gas B would be x_B - and so on.

Pretty obvious really!

For example, in a mixture of 1 mole of nitrogen and 3 moles of hydrogen, there are a total of 4 moles of gas. The mole fraction of nitrogen is 1/4 (0.25) and of hydrogen is 3/4 (0.75).

Partial pressure

The partial pressure of one of the gases in a mixture is the pressure which it would exert if it alone occupied the whole container.

The partial pressure of gas A is often given the symbol P_A. The partial pressure of gas B would be P_B - and so on.

There are two important relationships involving partial pressures. The first is again fairly obvious.

The total pressure of a mixture of gases is equal to the sum of the partial pressures.

It is easy to see this visually:

Gas A is creating a pressure (its partial pressure) when its molecules hit the walls of its container. Gas B does the same. When you mix them up, they just go on doing what they were doing before. The total pressure is due to both molecules hitting the walls - in other words, the sum of the partial pressures.

The more important relationship is the second one:

Learn it!

That means that if you had a mixture made up of 20 moles of nitrogen, 60 moles of hydrogen and 20 moles of ammonia (a total of 100 moles of gases) at 200 atmospheres pressure, the partial pressures would be calculated like this:

Partial pressures can be quoted in any normal pressure units. The common ones are atmospheres or pascals (Pa). Pascals are exactly the same as N m^-2 (newtons per square metre).

K_p in homogeneous gaseous equilibria

A homogeneous equilibrium is one in which everything in the equilibrium mixture is present in the same phase. In this case, to use K_p, everything must be a gas.

A good example of a gaseous homogeneous equilibrium is the conversion of sulphur dioxide to sulphur trioxide at the heart of the Contact Process:

Writing an expression for K_p

We are going to start by looking at a general case with the equation:

If you allow this reaction to reach equilibrium and then measure (or work out) the equilibrium partial pressures of everything, you can combine these into the equilibrium constant, K_p.

Just like K_c, K_p always has the same value (provided you don't change the temperature), irrespective of the amounts of A, B, C and D you started with.

K_p has exactly the same format as K_c, except that partial pressures are used instead of concentrations. The gases on the right-hand side of the chemical equation are at the top of the expression, and those on the left at the bottom.

The Contact Process equilibrium

You will remember that the equation for this is:

K_p is given by:

The Haber Process equilibrium

The equation for this is:

. . . and the K_p expression is:

K_c in heterogeneous equilibria

A typical example of a heterogeneous equilibrium will involve gases in contact with solids.

Writing an expression for K_p for a heterogeneous equilibrium

Exactly as happens with K_c, you don't include any term for a solid in the equilibrium expression.

The next two examples have already appeared on the K_c page.

The equilibrium produced on heating carbon with steam

Everything is exactly the same as before in the expression for K_p, except that you leave out the solid carbon.

The equilibrium produced on heating calcium carbonate

This equilibrium is only established if the calcium carbonate is heated in a closed system, preventing the carbon dioxide from escaping.

The only thing in this equilibrium which isn't a solid is the carbon dioxide. That is all that is left in the equilibrium constant expression.

Calculations involving K_p

On the K_c page, I've already discussed the fact that the internet isn't a good medium for learning how to do calculations.

If you want lots of worked examples and problems to do yourself centred around K_p, you might be interested in my book on chemistry calculations.

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