EQUILIBRIUM CONSTANTS and LECHATELIER'S PRINCIPLE

This page looks at the relationship between equilibrium constantsand Le Chatelier's Principle. Students often get confused abouthow it is possible for the position of equilibrium to change as youchange the conditions of a reaction, although the equilibriumconstant may remain the same.

Be warned that this page assumes a good understanding of LeChatelier's Principle and how to write expressions for equilibriumconstants.

Important: If you aren't happy about the basics ofequilibrium, explore the equilibrium menu before you wasteyour time on this page.

This page should only be read when you are confident abouteverything else to do with equilibria.

Changing concentrations

The facts

Equilibrium constants aren't changed if you change theconcentrations of things present in the equilibrium. The only thingthat changes an equilibrium constant is a change of temperature.

The position of equilibrium is changed if you change theconcentration of something present in the mixture. According to LeChatelier's Principle, the position of equilibrium moves in such away as to tend to undo the change that you have made.

Suppose you have an equilibrium established between foursubstances A, B, C and D.

According to Le Chatelier's Principle, if you decrease theconcentration of C, for example, the position of equilibrium willmove to the right to increase the concentration again.

Note: The reason for choosing an equation with "2B" willbecome clearer when I deal with the effect of pressure furtherdown the page.

Explanation in terms of the constancy of the equilibriumconstant

The equilibrium constant, Kc for this reaction looks like this:

If you have moved the position of the equilibrium to the right (andso increased the amount of C and D), why hasn't the equilibriumconstant increased?

This is actually the wrong question to ask! We need to look at it theother way round.

Let's assume that the equilibrium constant mustn't change if youdecrease the concentration of C - because equilibrium constants

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are constant at constant temperature. Why does the position ofequilibrium move as it does?

If you decrease the concentration of C, the top of the Kcexpression gets smaller. That would change the value of Kc. Inorder for that not to happen, the concentrations of C and D willhave to increase again, and those of A and B must decrease. Thathappens until a new balance is reached when the value of theequilibrium constant expression reverts to what it was before.

The position of equilibrium moves - not because Le Chatelier saysit must - but because of the need to keep a constant value for theequilibrium constant.

If you decrease the concentration of C:

Changing pressure

This only applies to systems involving at least one gas.

The facts

Equilibrium constants aren't changed if you change the pressureof the system. The only thing that changes an equilibrium constantis a change of temperature.

The position of equilibrium may be changed if you change thepressure. According to Le Chatelier's Principle, the position ofequilibrium moves in such a way as to tend to undo the change thatyou have made.

That means that if you increase the pressure, the position ofequilibrium will move in such a way as to decrease the pressureagain - if that is possible. It can do this by favouring the reactionwhich produces the fewer molecules. If there are the same numberof molecules on each side of the equation, then a change ofpressure makes no difference to the position of equilibrium.

Explanation

Where there are different numbers of molecules on each sideof the equation

Let's look at the same equilibrium we've used before. This onewould be affected by pressure because there are 3 molecules onthe left but only 2 on the right. An increase in pressure would movethe position of equilibrium to the right.

Because this is an all-gas equilibriium, it is much easier to use Kp:

Once again, it is easy to suppose that, because the position of

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equilibrium will move to the right if you increase the pressure, Kpwill increase as well. Not so!

To understand why, you need to modify the Kp expression.

Remember the relationship between partial pressure, molefraction and total pressure?

Note: If you aren't happy with this, read the beginning of thepage about Kp before you go on.

Use the BACK button on your browser to return to this page.

Replacing all the partial pressure terms by mole fractions and totalpressure gives you this:

If you sort this out, most of the "P"s cancel out - but one is left at thebottom of the expression.

Now, remember that Kp has got to stay constant because thetemperature is unchanged. How can that happen if you increaseP?

To compensate, you would have to increase the terms on the top,xC and xD, and decrease the terms on the bottom, xA and xB.

Increasing the terms on the top means that you have increased themole fractions of the molecules on the right-hand side. Decreasingthe terms on the bottom means that you have decreased the molefractions of the molecules on the left.

That is another way of saying that the position of equilibrium hasmoved to the right - exactly what Le Chatelier's Principle predicts.The position of equilibrium moves so that the value of Kp is keptconstant.

Where there are the same numbers of molecules on eachside of the equation

In this case, the position of equilibrium isn't affected by a change ofpressure. Why not?

Let's go through the same process as before:

Substituting mole fractions and total pressure:

. . . and cancelling out as far as possible:

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There isn't a single "P" left in the expression. Changing thepressure can't make any difference to the Kp expression. Theposition of equilibrium doesn't need to move to keep Kp constant.

Changing temperature

The facts

Equilibrium constants are changed if you change thetemperature of the system. Kc or Kp are constant at constanttemperature, but they vary as the temperature changes.

Look at the equilibrium involving hydrogen, iodine and hydrogeniodide:

The Kp expression is:

Two values for Kp are:

temperature Kp

500 K 160

700 K 54

You can see that as the temperature increases, the value of Kpfalls.

Note: You might possibly be wondering what the units of Kpare. This particular example was chosen because in thiscase, Kp doesn't have any units. It is just a number.

The units for equilibrium constants vary from case to case. Itis much easier to understand this from a book than from a lotof maths on screen. You will find this explained in mychemistry calculations book.

This is typical of what happens with any equilibrium where theforward reaction is exothermic. Increasing the temperaturedecreases the value of the equilibrium constant.

Where the forward reaction is endothermic, increasing thetemperature increases the value of the equilibrium constant.

Note: Any explanation for this needs knowledge beyond thescope of any UK A level (or equivalent) syllabus.

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The position of equilibrium also changes if you change thetemperature. According to Le Chatelier's Principle, the position ofequilibrium moves in such a way as to tend to undo the change thatyou have made.

If you increase the temperature, the position of equilibrium willmove in such a way as to reduce the temperature again. It will dothat by favouring the reaction which absorbs heat.

In the equilibrium we've just looked at, that will be the back reactionbecause the forward reaction is exothermic.

So, according to Le Chatelier's Principle the position ofequilibrium will move to the left. Less hydrogen iodide will beformed, and the equilibrium mixture will contain more unreactedhydrogen and iodine.

That is entirely consistent with a fall in the value of the equilibriumconstant.

Adding a catalyst

The facts

Equilibrium constants aren't changed if you add (or change) acatalyst. The only thing that changes an equilibrium constant is achange of temperature.

The position of equilibrium is not changed if you add (orchange) a catalyst.

Explanation

A catalyst speeds up both the forward and back reactions byexactly the same amount. Dynamic equilibrium is establishedwhen the rates of the forward and back reactions become equal. Ifa catalyst speeds up both reactions to the same extent, then theywill remain equal without any need for a shift in position ofequilibrium.

Note: If you know about the Arrhenius equation, it isn't toodifficult to use it to show that the ratio of the rate constantsfor the forward and back reactions isn't affected by adding acatalyst. Although the activation energies for the tworeactions change when you add a catalyst, they both changeby the same amount.

I'm not going to do this bit of algebra, because it would neverbe asked at this level (UK A level or equivalent).

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Exploring some of this using a simple computerprogram

The link below will take you to a page where you can explore theeffect of changing conditions on the reaction:

The page comes from Davidson College in America. It needs youto have Java enabled in your browser.

Note: If this link stops working, please let me know usingthe address on the about this site page. If your browserdoesn't have Java enabled, then you won't see the importantpart of the page, and you will have to enable Java. I'm afraidthat is your problem - it varies from browser to browser. Youcould try reading this page about enabling Java.

You are told that the reaction is endothermic, and can changethings like the temperature, the volume of the mixture, and theamounts of all the reactants to see what would happen.

It would be best if you worked out what you expected to happenbefore you change anything. You change things by moving the greysliders.

You will notice that there is no direct way of changing the pressure.Instead, you have to change the volume. Obviously, if you decreasethe volume, keeping the amounts of everything constant, thatincreases the pressure.

If you do this to change the pressure, concentrate on the red barsshowing what happens to the number of moles of substancespresent. The blue bars are more confusing. These representconcentrations, and these will change not only because of thechange in quantities present, but also because of the change involume. That's confusing!

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 equilibrium constants and Le Chatelier's Principle

answers

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© Jim Clark 2002 (last modified May 2013)

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