LE CHATELIER'S PRINCIPLE

This page looks at Le Chatelier's Principle and explains how toapply it to reactions in a state of dynamic equilibrium. It coverschanges to the position of equilibrium if you change concentration,pressure or temperature. It also explains very briefly why catalystshave no effect on the position of equilibrium.

Important: If you aren't sure about the words dynamicequilibrium or position of equilibrium you should read theintroductory page before you go on

It is important in understanding everything on this page to realisethat Le Chatelier's Principle is no more than a useful guide to helpyou work out what happens when you change the conditions in areaction in dynamic equilibrium. It doesn't explain anything. I'llkeep coming back to that point!

Using Le Chatelier's Principle

A statement of Le Chatelier's Principle

If a dynamic equilibrium is disturbed by changing theconditions, the position of equilibrium moves to counteractthe change.

Using Le Chatelier's Principle with a change ofconcentration

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

Note: In case you wonder, the reason for choosing thisequation rather than having just A + B on the left-hand side isbecause further down this page I need an equation which hasdifferent numbers of molecules on each side. I am going touse that same equation throughout this page.

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What would happen if you changed the conditions byincreasing the concentration of A?

According to Le Chatelier, the position of equilibrium will move insuch a way as to counteract the change. That means that theposition of equilibrium will move so that the concentration of Adecreases again - by reacting it with B and turning it into C + D.The position of equilibrium moves to the right.

This is a useful way of converting the maximum possible amount ofB into C and D. You might use it if, for example, B was a relativelyexpensive material whereas A was cheap and plentiful.

What would happen if you changed the conditions bydecreasing the concentration of A?

According to Le Chatelier, the position of equilibrium will move sothat the concentration of A increases again. That means that moreC and D will react to replace the A that has been removed. Theposition of equilibrium moves to the left.

This is esssentially what happens if you remove one of theproducts of the reaction as soon as it is formed. If, for example,you removed C as soon as it was formed, the position ofequilibrium would move to the right to replace it. If you kept onremoving it, the equilibrium position would keep on movingrightwards - turning this into a one-way reaction.

Important

This isn't in any way an explanation of why the position ofequilibrium moves in the ways described. All Le Chatelier'sPrinciple gives you is a quick way of working out what happens.

Note: If you know about equilibrium constants, you will find amore detailed explanation of the effect of a change ofconcentration by following this link. If you don't know anythingabout equilibrium constants, you should ignore this link.

If you choose to follow it, return to this page via the BACKbutton on your browser or via the equilibrium menu.

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Using Le Chatelier's Principle with a change of pressure

This only applies to reactions involving gases:

What would happen if you changed the conditions byincreasing the pressure?

According to Le Chatelier, the position of equilibrium will move insuch a way as to counteract the change. That means that theposition of equilibrium will move so that the pressure is reducedagain.

Pressure is caused by gas molecules hitting the sides of theircontainer. The more molecules you have in the container, thehigher the pressure will be. The system can reduce the pressure byreacting in such a way as to produce fewer molecules.

In this case, there are 3 molecules on the left-hand side of theequation, but only 2 on the right. By forming more C and D, thesystem causes the pressure to reduce.

Increasing the pressure on a gas reaction shifts the position ofequilibrium towards the side with fewer molecules.

What would happen if you changed the conditions bydecreasing the pressure?

The equilibrium will move in such a way that the pressureincreases again. It can do that by producing more molecules. Inthis case, the position of equilibrium will move towards the left-hand side of the reaction.

What happens if there are the same number of molecules onboth sides of the equilibrium reaction?

In this case, increasing the pressure has no effect whatsoever onthe position of the equilibrium. Because you have the samenumbers of molecules on both sides, the equilibrium can't move inany way that will reduce the pressure again.

Important

Again, this isn't an explanation of why the position of equilibriummoves in the ways described. You will find a rather mathematicaltreatment of the explanation by following the link below.

Note: You will find a detailed explanation by following thislink. If you don't know anything about equilibrium constants(particularly Kp), you should ignore this link. The same thingapplies if you don't like things to be too mathematical! If youare a UK A' level student, you won't need this explanation.

If you choose to follow the link, return to this page via theBACK button on your browser or via the equilibrium menu.

Using Le Chatelier's Principle with a change of temperature

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For this, you need to know whether heat is given out or absorbedduring the reaction. Assume that our forward reaction isexothermic (heat is evolved):

This shows that 250 kJ is evolved (hence the negative sign) when1 mole of A reacts completely with 2 moles of B. For reversiblereactions, the value is always given as if the reaction was one-wayin the forward direction.

The back reaction (the conversion of C and D into A and B) wouldbe endothermic by exactly the same amount.

What would happen if you changed the conditions byincreasing the temperature?

According to Le Chatelier, the position of equilibrium will move insuch a way as to counteract the change. That means that theposition of equilibrium will move so that the temperature isreduced again.

Suppose the system is in equilibrium at 300°C, and you increasethe temperature to 500°C. How can the reaction counteract thechange you have made? How can it cool itself down again?

To cool down, it needs to absorb the extra heat that you have justput in. In the case we are looking at, the back reaction absorbsheat. The position of equilibrium therefore moves to the left. Thenew equilibrium mixture contains more A and B, and less C and D.

If you were aiming to make as much C and D as possible,increasing the temperature on a reversible reaction where theforward reaction is exothermic isn't a good idea!

What would happen if you changed the conditions bydecreasing the temperature?

The equilibrium will move in such a way that the temperatureincreases again.

Suppose the system is in equilibrium at 500°C and you reduce thetemperature to 400°C. The reaction will tend to heat itself up againto return to the original temperature. It can do that by favouring theexothermic reaction.

The position of equilibrium will move to the right. More A and B areconverted into C and D at the lower temperature.

Summary

Increasing the temperature of a system in dynamic

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equilibrium favours the endothermic reaction. The systemcounteracts the change you have made by absorbing theextra heat.

Decreasing the temperature of a system in dynamicequilibrium favours the exothermic reaction. The systemcounteracts the change you have made by producing moreheat.

Important

Again, this isn't in any way an explanation of why the position ofequilibrium moves in the ways described. It is only a way of helpingyou to work out what happens.

Note: I am not going to attempt an explanation of thisanywhere on the site. To do it properly is far too difficult forthis level. It is possible to come up with an explanation ofsorts by looking at how the rate constants for the forward andback reactions change relative to each other by using theArrhenius equation, but this isn't a standard way of doing it,and is liable to confuse those of you going on to do aChemistry degree. If you aren't going to do a Chemistrydegree, you won't need to know about this anyway!

Le Chatelier's Principle and catalysts

Catalysts have sneaked onto this page under false pretences,because adding a catalyst makes absolutely no difference tothe position of equilibrium, and Le Chatelier's Principle doesn'tapply to them.

This is because a catalyst speeds up the forward and backreaction to the same extent. Because adding a catalyst doesn'taffect the relative rates of the two reactions, it can't affect theposition of equilibrium. So why use a catalyst?

For a dynamic equilibrium to be set up, the rates of the forwardreaction and the back reaction have to become equal. This doesn'thappen instantly. For a very slow reaction, it could take years! Acatalyst speeds up the rate at which a reaction reaches dynamicequilibrium.

Note: You might try imagining how long it would take toestablish a dynamic equilibrium if you took the visual modelon the introductory page and reduced the chances of thecolours changing by a factor of 1000 - from 3 in 6 to 3 in 6000and from 1 in 6 to 1 in 6000.

Starting with blue squares, by the end of the time taken forthe examples on that page, you would most probably stillhave entirely blue squares. Eventually, though, you would endup with the same sort of patterns as before - containing 25%blue and 75% orange squares.

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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 Le Chatelier's Principle

answers

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© Jim Clark 2002 (modified April 2013)

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