# The Wright Stuff

 Gliding Weather with Attitude Computing Web Help About Me

# Le Chatelier's Principle

This is one of the most useful ways of predicting how the position of equilibrium in a chemical reaction will change when the conditions are changed. Le Chatelier's Principle is stated as follows: -

When the conditions of a system at equilibrium are changed, the position of equilibrium moves in such a way as to oppose the change.

Great, but what does that mean? Le Chatelier's Principle says: -

• If we add more of any reactant or product chemical, the system reacts in such a way to try to use it up.
• If we remove some reactant or product chemical, the system tries to replace it by moving the equilibrium.
• If we heat up the system, it tries to cool down again.
• If we cool it down, it tries to warm up again.
• For reactions involving gases, if we increase the pressure, the system then tries to reduce the pressure.
• For reactions involving gases, if we reduce the pressure, it then tries to raise the pressure.

In every case the position of equilibrium moves to try to bring about the appropriate change. It doesn't return to the exact same concentration or pressure or temperature before the change was made, it just tries to do this and partly succeeds.

Note that the equilibrium constant doesn't change, (except in the cases where the temperature is changed of course). Instead, the amounts of the reactants and products present change because the equilibrium constant is a constant at any given temperature. The equilibrium constant controls the concentrations of the reactants and products that are present when the system comes to equilibrium.

Le Chatelier's Principle allows you to predict which way the equilibrium will move when you change the reaction conditions, and helps provide ways to increase the yield in a chemical reaction. In fact every chemist has at some time in some reaction applied Le Chatelier's Principle to improve the yield they got. That's how important it is. Le Chatelier's Principle is a popular topic in exams, so have a good look at these examples below, which cover the type of things you are expected to know.

If you want to learn more about the principle's discoverer, the Henri Le Chatelier page describes some of the background work, and about half way down the page you get his life history.

### Simple Examples

Here's some simple examples, where A, B, C, D, etc. are some reactants.

For any reaction

 A + B = C + D (reaction 1)

If we add more A or B, the position of equilibrium moves to the right, more C and D are formed and some of the extra A or B is used up. (This is why some reactions use an excess of one reactant, to drive the equilibrium to the right. Usually it's the cheapest reactant that is in excess.)

If we add more C or D, the position of equilibrium moves to the left, more A and B are formed and there is a bit less C and D present. This can be a problem if water is a product and is used as the solvent. Solution? Don't use water as the solvent.

This is summarized as add more reactants, equilibrium moves to the right, add more products, it moves to the left.

If we remove some of a product, the equilibrium moves to the right. So if a product is a liquid and we try to distill it out (assuming for the moment that it has the lowest boiling point of anything present), the reaction makes more of it - i.e. moves to the right. Along the same lines, concentrated sulphuric acid is used in the reaction between an organic acid and an alcohol to produce an ester and water. The conc. sulphuric acid acts as a catalyst, but it also absorbs some of the water formed by reacting with it to form solvated ions, thus reducing the amount of free water in the system. Thus it pulls the equilibrium over to the right.

### Gas Examples

If one of the products is a gas (e.g. acid + carbonate = salt, water and CO2), it bubbles out of the reaction vessel. So it is continually removing itself, and it never reaches the required equilibrium concentration. Therefore the equilibrium keeps moving to the right, and it appears to go to completion. (It is never actually allowed to reach the equilibrium concentration of CO2!)

With gases we have to be a bit more careful. Strictly speaking, the equilibrium constant refers to a system where all the reactants are in the same phase e.g. all in solution together or all as gases together. Le Chatelier's Principle let's us understand what happens when the gases bubble off. With the acid - carbonate example, if we let the gas escape it can't build up enough of a concentration dissolved in solution to reach equilibrium - in may be an extremely high value, impossible to achieve at normal pressures.

When dealing with a system where all the reactants are gases, things are easier in some cases, and trickier in others. First assume for a reaction where everything is a gas that it occurs in a sealed container (otherwise it can't come to equilibrium). Then the earlier "add more - remove some" rule applies perfectly.

Now the sneakier bit. If we increase the pressure, the system tries to reduce it. It does this by moving the position of equilibrium to the side of the equation with less gas molecules shown in the balanced equation. So there isn't a simple left or right rule as with the first example. (If there are the same number of gas molecules given on each side of the balanced equation, then changing the pressure has no effect on the position of equilibrium.)

Here are a couple of examples.

 X(g) + Y(g) = Z(g) (reaction 2)

If we raise the pressure, le Chatelier's Principle says the equilibrium moves to the side of the balanced equation with the least number of gas molecules. So it moves to the right in this case. We are not interested in the actual moles of each gas present at equilibrium, just the numbers given in the balanced equation.

 Q(g) = R(g) + T(g) (reaction 3)

In reaction 3, above, an increase in pressure would move the equilibrium to the left hand side.

 3R(g) = D(g) + Z(g) (reaction 4)

Here, in reaction 4, the left hand side has more gas molecules than the right hand side, even though two substances appear on the right hand side.

If we lower the pressure, the equilibrium moves to the side with the most gas molecules in the balanced equation. In reaction 2, it would move to the left. In reaction 3, it would move to the right. In reaction 4, it would move to the left. Check for yourself that these are the sides with the least number of gas molecules present in the balanced equation.

A little trick they play on you in the exam - "what if the number of moles is the same but the volume of the container is reduced / increased, perhaps via a piston..." Increasing the container volume reduces the pressure, decreasing the container volume increases the pressure. Then just use the pressure rules above.

### Solids and Gases

A special case involves a reaction where solids and gases are involved. Because of the "same phase" rule, we can't use the concentration of the solid, we have to use the solid's vapour pressure, often called its partial pressure, at that temperature. But the partial pressure of a solid at any given temperature is a constant, so it is usually removed from the problem as shown below.

 A(s) + Q(g) = R(s) + T(g) (reaction 5)

The equilibrium constant would be

 K = [R] x [T] [A] x [Q]

Remove the concentrations of the solids by multiplying the opposite side by their inverse.

 K x [A] = [T] [R] [Q]

Then as these concentrations are a constant, we just incorporate their value into the quoted value of K and call it Kp.

 Kp = [T] [Q]

And the concentration of T and Q are then quoted in pressure units of some sort. We can even simply use the number of moles present, as the pressure is directly proportional to the moles present. (It's a sealed container, remember, so moles over volume = concentration, and the volume is fixed.)

If this sounds a bit complicated just say that a far as you are concerned the answer is
"in heterogeneous reactions, the concentrations of the solids do not appear in the equilibrium equation".
They often try to trick you with this one in exam questions, by having a set of questions on equilibrium in solutions then throwing in a solid-gas equilibrium. So watch out for it. No solids in the final equation, it becomes a gas type example..

### Exothermic and Endothermic Reactions

If a reaction gives out heat (i.e. the reaction mixture gets hotter), it is called exothermic. Delta H, the heat of reaction, is negative.

If a reaction absorbs heat (i.e. the reaction mixture gets colder) it is called endothermic. Delta H, the heat of reaction, is positive.

We are not talking about whether or not you have to heat the reaction or cool it, but about what it would do if it occurred by itself. Sometimes we have to heat it to get it started, even if it is exothermic (usually we heat endothermic reactions to keep them going). Sometimes we carry out exothermic reactions in a cooled container to stop it overheating, but sometimes we do a reaction in a cooled container to stop the product decomposing. You will be told in any question that requires you to know if it is exothermic or endothermic which of the two it is, although sometimes this is done by stating the value or sign of delta H, the heat of reaction.

 A + B = C + D (reaction 1)

Let's suppose that reaction 1 is an exothermic reaction.

If we raise the temperature, the system tries to lower it again, according to le Chatelier's Principle. To do this the equilibrium moves to the left. This is because if a reaction as written is exothermic, its reverse is endothermic, and moving the equilibrium to the endothermic side will absorb some of the heat we added.

If we lower the temperature, this equilibrium will move to the right in an attempt to raise it again. More energy will be given out as this occurs, which is how it tries to raise the temperature again. Remember it will not succeed in getting back to the original temperature, it just changes the final temperature achieved.

 X(g) + Y(g) = Z(g) (reaction 2)

Let's suppose that reaction 2 is endothermic.

If we raise the temperature for this reaction, the equilibrium will move to the right. This will result in more of the energy absorbing reaction occurring, thus trying to reduce the temperature.

If we lower the temperature, the equilibrium will move to the left. Thus more of the reverse exothermic reaction occurs, trying to raise the temperature

These are the opposite of the changes for the exothermic reaction.

So you have to learn this summary for exothermic and endothermic reactions:-

Exothermic - raise the temperature and the equilibrium moves to the left, lower it and the equilibrium moves to the right.
Endothermic - raise the temperature and the equilibrium moves to the left, lower it and the equilibrium moves to the right.

There is a final catch with exothermic and endothermic reactions. Industry doesn't always use the obvious temperature to get the results you'd expect. Sometimes a reaction is just too slow to do at the temperature that gives the highest yield. So rather than take six months to get a 90% yield of products, industry accepts a 1% yield achieved in five minutes by doing the reaction at a higher temperature than le Chatelier's Principle would say was best. One of the best know industrial examples of this is the Haber Process for the production of NH3, where they use a temperature that is much higher than you'd expect, to speed things up. In industry, the cost of using very high temperatures also affects how they will carry out a reaction. This can confuse students, as the Haber Process is often used when discussing le Chatelier and his work.

Note that in exams which cover equilibria and rates of reactions (e.g. How Far, How Fast as one course calls it), they tend to deliberately alternate between equilibria and rate questions to try to catch you out. Don't answer an equilibria question as if it was a rate question or vice versa. Reading the question before you answer it is always a good start.