EQUILIBRIUM CONSTANTS: Kp

This page explains equilibrium constants expressed in terms ofpartial pressures of gases, Kp. It covers an explanation of theterms mole fraction and partial pressure, and looks at Kp for bothhomogeneous and heterogeneous reactions involving gases.

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

Important: If you have come directly to this page via asearch engine, you should first read the page on equilibriumconstants - Kc before you go on - unless you are already fullyconfident about how to write expressions for Kc.

You will find a link back to this page at the bottom of the Kcpage.

Defining some terms

Before we can go any further, there are two terms relating tomixtures 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 fractionof gas A is worked out by dividing the number of moles of A by thetotal number of moles of gas.

The mole fraction of gas A is often given the symbol xA. The molefraction of gas B would be xB - and so on.

Pretty obvious really!

For example, in a mixture of 1 mole of nitrogen and 3 moles ofhydrogen, there are a total of 4 moles of gas. The mole fraction ofnitrogen 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 thepressure which it would exert if it alone occupied the wholecontainer.

The partial pressure of gas A is often given the symbol PA. Thepartial pressure of gas B would be PB - 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 thepartial pressures.

It is easy to see this visually:

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Gas A is creating a pressure (its partial pressure) when itsmolecules hit the walls of its container. Gas B does the same.When you mix them up, they just go on doing what they were doingbefore. 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 ofnitrogen, 60 moles of hydrogen and 20 moles of ammonia (a totalof 100 moles of gases) at 200 atmospheres pressure, the partialpressures would be calculated like this:

gas mole fraction partial pressure

nitrogen 20/100 = 0.2 0.2 x 200 = 40 atm

hydrogen 60/100 = 0.6 0.6 x 200 = 120 atm

ammonia 20/100 = 0.2 0.2 x 200 = 40 atm

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

Kp in homogeneous gaseous equilibria

A homogeneous equilibrium is one in which everything in theequilibrium mixture is present in the same phase. In this case, touse Kp, everything must be a gas.

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

Writing an expression for Kp

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

If you allow this reaction to reach equilibrium and then measure (orwork out) the equilibrium partial pressures of everything, you cancombine these into the equilibrium constant, Kp.

Just like Kc, Kp always has the same value (provided you don'tchange the temperature), irrespective of the amounts of A, B, Cand D you started with.

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Kp has exactly the same format as Kc, except that partialpressures are used instead of concentrations. The gases on theright-hand side of the chemical equation are at the top of theexpression, and those on the left at the bottom.

Beware! People are sometimes tempted to write bracketsaround the individual partial pressure terms. Don't do it! Evenif you intend to write normal round brackets, it is too easy inan exam to write them as square brackets instead. Thismakes it look as if you are confusing Kp with Kc. Examinersdon't like it, and you could be penalised.

The Contact Process equilibrium

You will remember that the equation for this is:

Kp is given by:

The Haber Process equilibrium

The equation for this is:

. . . and the Kp expression is:

Kp in heterogeneous equilibria

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

Writing an expression for Kp for a heterogeneousequilibrium

Exactly as happens with Kc, you don't include any term for a solidin the equilibrium expression.

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

The equilibrium produced on heating carbon with steam

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

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The equilibrium produced on heating calcium carbonate

This equilibrium is only established if the calcium carbonate isheated in a closed system, preventing the carbon dioxide fromescaping.

The only thing in this equilibrium which isn't a solid is the carbondioxide. That is all that is left in the equilibrium constantexpression.

Calculations involving Kp

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

If you want lots of worked examples and problems to do yourselfcentred around Kp, you might be interested in my book onchemistry calculations.

Note: If you are interested in my chemistry calculationsbook you might like to follow this link.

Questions to test your understanding

If this is the first set of questions you have done, please read theintroductory page before you start. You will need to use the BACK BUTTONon your browser to come back here afterwards.

questions on Kp

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© Jim Clark 2002 (modified May 2013)

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