Disturbing a balance
The last lecture ended with a question: what happens when you disturb a system at equilibrium? Here is a real example to think about.
A sealed glass tube holds a brown gas. The brown colour comes from nitrogen dioxide, NO₂. Two NO₂ molecules can join to make dinitrogen tetroxide, N₂O₄, which is colourless. In the tube, the two are at dynamic equilibrium:
Put the tube into hot water, and the gas turns dark brown. Put it into iced water, and it turns pale. The tube is sealed, so nothing has been added or taken away. Yet the mixture inside has changed.
A rule that says "it depends" is no help in an exam. You need a way to predict which way a mixture will change, and why. In 1884, the French chemist Henry Le Chatelier gave a rule that does this for a wide range of changes. This lecture builds that rule, shows the reason behind each part of it, and shows where it stops working.
The principle
If a system at equilibrium is disturbed by a change in conditions, the position of equilibrium shifts in the direction that opposes the change.
Three phrases in this sentence need care.
"Position of equilibrium" means the mixture of reactants and products at equilibrium. A shift to the right means more products and less reactants. A shift to the left means more reactants and less products.
"Shifts" means the forward and reverse rates become unequal for a while. One direction runs faster until the rates are equal again, at a new mixture. Remember from the last lecture that both reactions keep running all the time. A shift is just a period when one of them wins.
"Opposes" does not mean "undoes". The system reduces the effect of the change. It does not cancel it. You will see this with numbers below, and it is one of the most often lost marks in the topic.
There are four kinds of change to look at: concentration, pressure, temperature and a catalyst. For each one, we will find the direction from the principle, and then check it with the two rates, so that it is a reason, not a rule.
Changing a concentration
Suppose extra N₂O₄ is pumped into the tube, at the same volume and temperature. The principle says the system will oppose the increase in N₂O₄. It can do that by using some of it up. So the equilibrium shifts to the right, making more NO₂.
Now check with the rates. Adding N₂O₄ raises its concentration. More N₂O₄ molecules means more forward reactions every second, as in the factors-affecting-rate lecture. The reverse rate has not changed yet. So for a while the forward reaction wins. NO₂ builds up, which speeds up the reverse reaction, until the two rates are equal again.
Look at the shape of this graph, because examiners often ask you to sketch one. The added substance jumps straight up at the moment it is added. Then it falls, but it never falls back to where it started. The other substance changes smoothly, with no jump. Both lines then level off at the new equilibrium.
The same reasoning covers every concentration change.
- Add a reactant, or remove a product: the equilibrium shifts right.
- Add a product, or remove a reactant: the equilibrium shifts left.
Removing a product is especially useful in industry. If a product is taken away as soon as it forms, the reverse reaction can never catch up, and the forward reaction keeps going.
Changing the pressure
Pressure matters only for gases. It works through the number of gas molecules, because in a fixed volume, more gas molecules means a higher pressure.
Count the gas molecules in the equation. On the left, 1 N₂O₄. On the right, 2 NO₂. Every forward reaction turns one gas molecule into two. Every reverse reaction turns two into one.
Now squeeze the gas into half the volume. The pressure goes up. The principle says the system will oppose the rise in pressure. It can do that by reducing the number of gas molecules. So the equilibrium shifts to the side with fewer gas molecules: here, to the left.
A real tube shows something interesting. When you push in the plunger of a sealed syringe of this gas, the colour first gets darker, because the same NO₂ molecules are now packed into half the space. Then, over the next few seconds, it gets paler, as NO₂ joins up to make N₂O₄. But it does not become as pale as it was at the start.
That is it. That is the whole idea of "opposes, not undoes". The shift makes the effect of the change smaller. It never takes the system all the way back.
For the reaction H₂(g) + I₂(g) ⇌ 2HI(g), the gas mixture is squeezed into half the volume. Which way does the equilibrium shift?
Show the answer
Neither way. There are 2 gas molecules on the left (1 H₂ and 1 I₂) and 2 on the right. A shift in either direction would leave the number of gas molecules unchanged, so it could not reduce the pressure. The concentrations of all three gases double, and the position of equilibrium stays the same. The violet colour of the iodine gets darker, simply because it is more concentrated.
One more warning. "Increase the pressure" in an exam question means "squeeze the gas into a smaller volume". If instead an unreactive gas such as argon is pumped into a sealed container of fixed volume, the total pressure rises, but the concentrations of the reacting gases do not change. Then there is no shift at all.
Changing the temperature
Temperature works differently from the other changes. To use it, you need the sign of ΔH for the forward reaction. For the N₂O₄ equilibrium:
This value comes from the enthalpies of formation of the two gases. The forward reaction breaks the bond that holds the two halves of N₂O₄ together, so it is endothermic: it takes in heat. The reverse reaction is exothermic: it gives out the same amount of heat.
Now heat the tube. The principle says the system will oppose the rise in temperature. It can do that by taking in heat, so the endothermic direction is favoured. Here that is the forward reaction. The equilibrium shifts right, more brown NO₂ forms, and the gas turns darker. That is exactly what the hot tube showed. Cooling does the opposite. The exothermic direction is favoured, the equilibrium shifts left, and the gas turns paler.
- Raise the temperature: the equilibrium shifts in the endothermic direction.
- Lower the temperature: the equilibrium shifts in the exothermic direction.
Why is temperature different in kind? A change in concentration or pressure disturbs the mixture, and the mixture then finds its way back to the same balance point. A change in temperature moves the balance point itself. This is because heating speeds up both reactions, but it speeds up the reaction with the higher activation energy more. For this equilibrium, that is the endothermic forward reaction. The next lecture puts a number on the balance point, called the equilibrium constant, and shows that temperature is the only one of these changes that alters it.
A catalyst: no shift at all
The catalysts lecture showed that a catalyst gives a new route with a lower activation energy. The reverse reaction uses the same route backwards, so its barrier is lowered too. Both rates rise by the same factor.
So a catalyst does not shift the position of equilibrium. It does not change the amount of product. It only makes the system reach equilibrium sooner. If a question asks "how does a catalyst affect the yield?", the answer is "it does not, but the yield is reached faster".
Try all six changes on the same equilibrium below.
The gas starts at equilibrium with 0.060 mol/dm³ of N₂O₄ and 0.040 mol/dm³ of NO₂. Choose a change. Predict the shift first, then watch the graph and read the new concentrations.
This table shows each change to the equilibrium mixture, which starts with 0.060 mol/dm³ of N₂O₄ and 0.040 mol/dm³ of NO₂. The numbers come from a model of the reaction, not from measurements.
| Change | Just after the change (N₂O₄ / NO₂) | New equilibrium (N₂O₄ / NO₂) | Shift |
|---|---|---|---|
| Add N₂O₄ | 0.100 / 0.040 | 0.095 / 0.050 | Right |
| Remove half the NO₂ | 0.060 / 0.020 | 0.051 / 0.037 | Right |
| Halve the volume | 0.120 / 0.080 | 0.131 / 0.059 | Left |
| Heat | 0.060 / 0.040 | 0.045 / 0.069 | Right |
| Cool | 0.060 / 0.040 | 0.069 / 0.021 | Left |
| Add a catalyst | 0.060 / 0.040 | 0.060 / 0.040 | None (reached sooner) |
In every case the amount of nitrogen is unchanged by the shift, so 2 × [N₂O₄] + [NO₂] stays the same after the change. Every shift reduces the effect of the change, but none undoes it.
Notice two things. First, after every change except the catalyst, the new equilibrium is somewhere between where the system started and where the change pushed it. Second, the heating and cooling lines have no jump at the dashed line. Temperature does not add or remove anything. It moves the balance point, and the mixture drifts towards it.
Sorting the changes
Now use the principle on several equilibria. Each card names a change. Decide which way the equilibrium shifts, and read the reason on each card.
Put each card in a box. For the Haber process, N₂(g) + 3H₂(g) ⇌ 2NH₃(g), the forward reaction is exothermic (ΔH = −92 kJ/mol). Read the reason on each card after you place it.
This list sorts nine changes by the direction of the shift, with the reason for each.
- Shifts right: Haber process: add more N₂. A reactant is added, so the system uses some of it up.
- Shifts right: Haber process: remove NH₃ as it forms. A product is removed, so the system makes more of it.
- Shifts left: Haber process: raise the temperature. The forward reaction is exothermic, so heating favours the endothermic reverse reaction.
- Shifts right: Haber process: squeeze to a higher pressure. The right side has fewer gas molecules (2, against 4 on the left).
- No shift: Haber process: add an iron catalyst. A catalyst speeds up both directions equally.
- No shift: Haber process: pump in argon at a fixed volume. The total pressure rises, but the concentrations of N₂, H₂ and NH₃ do not change.
- No shift: H₂(g) + I₂(g) ⇌ 2HI(g): squeeze to a higher pressure. Both sides have 2 gas molecules, so a shift cannot lower the pressure.
- Shifts right: N₂O₄(g) ⇌ 2NO₂(g), ΔH = +57.2 kJ/mol: heat it. The forward reaction is endothermic, so heating favours it.
- Shifts left: N₂O₄(g) ⇌ 2NO₂(g): add more NO₂. A product is added, so the system uses some of it up.
Worked examples
In the last lecture, 1.00 mol of ethanoic acid and 1.00 mol of ethanol reached equilibrium with 0.67 mol of ester. A chemist wants more of the acid to turn into ester. Suggest two ways, and explain each.
Think first. Which changes shift the equilibrium to the right?
Way 1: use more ethanol. Adding a reactant shifts the equilibrium right, so more of the acid reacts. The next lecture shows how to calculate this. With 2.00 mol of ethanol instead of 1.00 mol, the ester rises from 0.67 mol to about 0.85 mol.
Way 2: remove water as it forms. Removing a product also shifts the equilibrium right. Chemists do this with a drying agent, or by using concentrated sulfuric acid, which takes up water.
Check. Pressure cannot help here: the substances are liquids, not gases. Temperature would hardly help either, because ΔH for this reaction is small. That is why chemists use changes in concentration for this reaction.
Three sealed tubes of the N₂O₄ ⇌ 2NO₂ mixture are placed in hot water, room-temperature water and iced water. Predict and explain their colours. ΔH = +57.2 kJ/mol for the forward reaction.
Think first. The sign of ΔH tells you which direction takes in heat.
The forward reaction is endothermic. In hot water, the endothermic direction is favoured, so the equilibrium shifts right. More NO₂ forms, so that tube is the darkest brown.
In iced water, the exothermic direction (the reverse reaction) is favoured, so the equilibrium shifts left. Less NO₂ is present, so that tube is the palest.
Check. This is what demonstrations show. In a tube cooled far enough, the gas becomes almost colourless.
Predict the effect of increasing the pressure on each equilibrium. (a) N₂(g) + 3H₂(g) ⇌ 2NH₃(g). (b) H₂(g) + Cl₂(g) ⇌ 2HCl(g). (c) A sealed bottle of fizzy drink: CO₂(g) ⇌ CO₂(aq).
Think first. Count only the gas molecules on each side.
(a) 4 gas molecules on the left, 2 on the right. Higher pressure shifts it right, towards ammonia.
(b) 2 gas molecules on each side. No shift.
(c) 1 gas molecule on the left, none on the right (dissolved CO₂ is not a gas). Higher pressure shifts it right, so more CO₂ dissolves. This is why drinks are bottled under pressure.
Check. In (c), opening the bottle does the opposite. The pressure falls, the equilibrium shifts left, and CO₂ comes out of solution as bubbles. The last lecture mentioned this bottle. Now you can explain it.
Ammonia is made by the Haber process:
What Le Chatelier says. The forward reaction is exothermic, so a low temperature gives the most ammonia. The left side has 4 gas molecules and the right side 2, so a high pressure gives the most ammonia.
What plants actually use. A temperature of about 400–450 °C, a pressure of about 200 atmospheres (it varies from plant to plant), and an iron catalyst (Chemguide).
Why the temperature is not low. At a low temperature the reaction is far too slow, even with a catalyst. 400–450 °C is a compromise: less ammonia at equilibrium, but it is made quickly.
Why the pressure is not higher. Higher pressure would give more ammonia, but very high pressures need thick-walled equipment and a lot of energy for compressing the gases. 200 atmospheres is chosen on cost.
The result. Only about 15% of the nitrogen and hydrogen turns into ammonia on each pass through the reactor. So the ammonia is cooled out, which removes a product, and the unreacted gases are recycled. Over many passes, about 98% is converted in the end (Chemguide). Notice that the catalyst plays no part in the equilibrium. It is there only for the rate.
Three traps
The system opposes a change; it does not undo it. After N₂O₄ is added, there is still more N₂O₄ than at the start. After the volume is halved, the NO₂ concentration is still higher than before. Write "the equilibrium shifts to reduce the effect of the change".
Count only gas molecules, using the numbers in the balanced equation. Solids, liquids and dissolved substances do not count. And "higher pressure" means a smaller volume. Pumping in an unreactive gas at a fixed volume causes no shift.
A catalyst speeds up the forward and reverse reactions equally. It gives the same equilibrium mixture, only sooner. In an industrial process this saves time and energy, but it does not change the amount of product at equilibrium.
Work each one on paper first. Open a hint only when you are stuck.
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Sulfur trioxide is made by the equilibrium 2SO₂(g) + O₂(g) ⇌ 2SO₃(g). The forward reaction is exothermic. Predict the effect on the amount of SO₃ at equilibrium of: (a) adding more O₂, (b) raising the temperature, (c) increasing the pressure, (d) adding a catalyst.
Hint 1
For each change, ask what it adds or removes: a substance, gas molecules, or heat.
Hint 2
Count the gas molecules: 2 + 1 on the left, 2 on the right.
Full answer
(a) More SO₃: a reactant is added, so the equilibrium shifts right. (b) Less SO₃: heating favours the endothermic direction, which is the reverse reaction, so it shifts left. (c) More SO₃: the right side has fewer gas molecules (2 against 3), so it shifts right. (d) No change in the amount of SO₃: a catalyst speeds up both directions equally, so equilibrium is only reached sooner. Check: each answer names the change, the direction, and the reason.
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Methanol is made by CO(g) + 2H₂(g) ⇌ CH₃OH(g), and the forward reaction is exothermic. What temperature and pressure would give the most methanol at equilibrium? Why might a factory not use exactly those conditions?
Hint 1
Use the sign of ΔH for temperature, and count gas molecules for pressure.
Hint 2
Think about the rate at a low temperature, and the cost of a very high pressure.
Full answer
The most methanol comes from a low temperature, because the forward reaction is exothermic and cooling favours the exothermic direction. It also comes from a high pressure, because there are 3 gas molecules on the left and 1 on the right. But at a low temperature the reaction is very slow, and very high pressures are expensive and need strong equipment. So a factory uses a compromise: a moderate temperature with a catalyst, and a high but affordable pressure, as in the Haber process. Check: both conditions follow from the same two reasons as for ammonia.
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Iron(III) ions react with thiocyanate ions to form a blood-red ion: Fe³⁺(aq) + SCN⁻(aq) ⇌ FeSCN²⁺(aq). The mixture is orange-red. What colour change do you expect if (a) a few drops of concentrated Fe³⁺ solution are added, (b) a substance is added that removes Fe³⁺ from the solution?
Hint 1
The red colour comes from FeSCN²⁺. Which way does each change shift the equilibrium?
Hint 2
Adding a reactant shifts right. Removing a reactant shifts left.
Full answer
(a) Deeper red. Adding Fe³⁺, a reactant, shifts the equilibrium right, making more FeSCN²⁺. (b) Paler. Removing Fe³⁺ shifts the equilibrium left, so some FeSCN²⁺ breaks up to replace it. Check: pressure plays no part, because nothing here is a gas.
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A syringe holds N₂O₄ and NO₂ at equilibrium: 0.060 mol/dm³ of N₂O₄ and 0.040 mol/dm³ of NO₂. The volume is halved. At the new equilibrium, the NO₂ concentration is 0.059 mol/dm³. Find the new N₂O₄ concentration, and explain why the NO₂ ends up between 0.040 and 0.080 mol/dm³.
Hint 1
First work out the concentrations just after the volume is halved. Then the shift turns NO₂ into N₂O₄, 2 to 1.
Hint 2
Just after halving: 0.120 and 0.080. The NO₂ falls by 0.080 − 0.059 = 0.021, so the N₂O₄ rises by half of that.
Full answer
Halving the volume doubles both concentrations: 0.120 mol/dm³ of N₂O₄ and 0.080 mol/dm³ of NO₂. The NO₂ then falls to 0.059, a fall of 0.021 mol/dm³. Two NO₂ make one N₂O₄, so the N₂O₄ rises by 0.0105, to about 0.131 mol/dm³. The NO₂ falls below 0.080 because the equilibrium shifts left to oppose the rise in pressure. It stays above 0.040 because opposing is not undoing. Check: count nitrogen. Before the shift, . After it, . ✓
Le Chatelier stated his principle in 1884, and the German physicist Ferdinand Braun reached a similar idea independently in 1887. It is a powerful way to predict a direction. But it has limits. It says nothing about how far the equilibrium shifts. And in 1911 the physicist Paul Ehrenfest showed that loose statements of the principle can give wrong answers. You must be careful about exactly what is being changed and what is held constant. The argon case above is one example: "higher pressure" alone gives the wrong answer.
That is why chemists use Le Chatelier as a first guess and the equilibrium constant as the final word. The next lecture builds that constant. It explains every result in this lecture with one number, and shows why temperature is the only one of these changes that alters the number itself.
- Le Chatelier's principle says that when a system at equilibrium is disturbed, the position of equilibrium shifts in the direction that opposes the change.
- Adding a reactant or removing a product shifts the equilibrium right, and adding a product or removing a reactant shifts it left.
- Increasing the pressure by reducing the volume shifts the equilibrium towards the side with fewer gas molecules, and has no effect when both sides have the same number.
- Raising the temperature shifts the equilibrium in the endothermic direction and lowering it shifts it in the exothermic direction, because temperature moves the balance point itself.
- A catalyst causes no shift, and every shift reduces the effect of a change without undoing it.