Chemical Equilibrium and Le Chatelier's Principle
Le Chatelier's principle states that when the equilibrium in a closed system is disturbed, the system will re-instate a new equilibrium by favouring the reaction that opposes the disturbance. The word doing the work is opposes. Add a reactant and the system uses it up. Heat it and the system absorbs the heat. This page explains the principle and the four things that shift an equilibrium, then gives two Grade 12 experiments that show it happening as a colour change.
Quick answers
| Question | Answer |
|---|---|
| Does a catalyst shift the equilibrium? | No. It speeds the forward and reverse reactions equally, so equilibrium arrives sooner but in exactly the same position. This is the most-failed question on the topic |
| Does the system return to where it started? | No. It moves to a new position that only partly offsets the change |
| What does raising the temperature do? | Favours the endothermic direction, because that absorbs the added heat |
| What does raising the pressure do? | Favours the side with fewer gas molecules |
| Is anything still happening at equilibrium? | Yes. Both reactions continue at equal rates. That is why it is called dynamic |
Where this fits in the curriculum
| Subject | Physical Sciences |
|---|---|
| Grade | 12 |
| Term | 2 |
| Topic | Chemical change, chemical equilibrium |
| Status | Not a prescribed formal assessment, but heavily examinable |
What chemical equilibrium is
Most reactions you meet before Grade 12 go to completion: they run until a reactant is used up and then stop. Reversible reactions do not. They settle at a point where both reactants and products are present together.
Equilibrium is reached when the forward and reverse reactions are going at the same rate. Nothing appears to change, because the concentrations stay constant, but both reactions are still running. That is why it is called a dynamic equilibrium.
Three conditions have to hold: the system must be closed, the temperature must be constant, and the two rates must be equal.
Le Chatelier's principle
When the equilibrium in a closed system is disturbed, the system will re-instate a new equilibrium by favouring the reaction that opposes the disturbance.
Two words carry the marks. Closed, because the principle does not apply to an open beaker losing gas to the room. And opposes, because the system does not fix the change, it partly offsets it.
That second point is worth dwelling on. The equilibrium never goes back to where it was. It moves to a new position that cancels part of what you did. Add extra reactant and some of it is used up, but not all of it, which is why the new mixture stays different from the old one.
The four factors
Concentration
Add a reactant and the forward reaction is favoured, using it up. Add a product and the reverse reaction is favoured. Remove something and the system makes more of it.
Temperature
Raising the temperature favours the endothermic direction, because absorbing heat opposes the increase. Lowering it favours the exothermic direction.
Temperature is the only factor that changes the value of Kc. Concentration and pressure shift the position of the equilibrium but leave the constant alone, which is why a temperature must always be quoted with a Kc value.
Pressure
For gases only. Increasing the pressure favours the side with fewer gas molecules, because that reduces the pressure again. If both sides have the same number of gas molecules, pressure changes nothing.
Catalyst
A catalyst has no effect on the position of an equilibrium. It lowers the activation energy for the forward and reverse reactions by the same amount, so both speed up equally. Equilibrium is reached sooner, at exactly the same place.
Learners lose this mark more than any other on the topic. A catalyst improves the rate, never the yield.
The Haber process, and why it is a compromise
3H2(g) + N2(g) ⇌ 2NH3(g) ; ΔH = −92 kJ
This is examined almost every year, and the interesting part is that the two conditions pull against each other.
| Change | Effect on yield | Why |
|---|---|---|
| Increase pressure | Increases | 4 mol of gas on the left, 2 mol on the right. The forward reaction reduces the pressure |
| Increase temperature | Decreases | The forward reaction is exothermic, so heat favours the reverse reaction |
| Add a catalyst | No change | Equilibrium arrives sooner, in the same place |
So a high yield wants high pressure and low temperature. But a low temperature makes the reaction unusably slow, and very high pressure is expensive and dangerous. The industrial process runs at a compromise, around 450 degrees and high pressure, accepting a lower yield in exchange for getting there fast enough to matter.
That trade-off between yield and rate is the whole point of the question.
Experiment 1: changing a concentration
Fe3+(aq) + SCN−(aq) ⇌ FeSCN2+(aq)
pale yellow · colourless · blood red
A warning about the equation. Some printed versions give the product as FeSCN−. That is wrong. Iron(III) is 3+ and thiocyanate is 1−, so the complex carries a 2+ charge. A learner who copies the minus sign into an exam loses the mark.
Method
- Make up 0,1 mol·dm−3 iron(III) chloride, about 2,7 g in 100 cm³, and 0,1 mol·dm−3 potassium thiocyanate, about 1,0 g in 100 cm³.
- Put 5 drops of each into 100 cm³ of water. This diluted mixture is what you work with.
- Put about 10 cm³ of it into each of three test tubes.
- Leave tube 1 alone as the control.
- Add 5 drops of the iron(III) solution to tube 2.
- Add 5 drops of the thiocyanate solution to tube 3.
The step most methods leave out
The dilution in step 2 is what decides whether this practical works. Most printed versions tell you to mix the two stock solutions directly. Do that and the result is almost opaque dark red, and nothing appears to change when you add more of either ion.
You want a clear, transparent blood-red, like weak cordial. If it looks like ink, dilute it further.
What you should see
Both tubes go a deeper red than the control. That surprises learners, and it is the point: Fe3+ and SCN− are both reactants, so adding either one drives the equilibrium the same way.
Experiment 2: changing the temperature
[Cu(H2O)6]2+ + 4Cl− ⇌ [CuCl4]2− + 6H2O
blue ⇌ yellow-green, forward reaction endothermic
Dissolve about 1 g of copper(II) chloride in 20 cm³ of water. Add concentrated hydrochloric acid drop by drop until the colour sits midway between blue and green. Divide between three tubes, keep one as a control, stand one in hot water and one in iced water.
Hot goes green. Cold goes blue. Heating favours the endothermic forward reaction, which absorbs the added heat. Put the hot tube into the cold water and it changes back, which is the most convincing demonstration of reversibility a class will see.
Why not the usual ones
Most textbooks demonstrate temperature using nitrogen dioxide in sealed syringes. It works well, but the gas is toxic, it needs a fume cupboard, and the teacher has to prepare the syringes. Most schools cannot run it.
The other common version uses cobalt(II) chloride, which is now being withdrawn from school laboratories as a suspected carcinogen.
Copper(II) chloride shows the same principle with neither problem.
Safety precautions
Potassium thiocyanate must never come into contact with acid. Thiocyanates release hydrogen cyanide gas in acidic conditions. Keep the two experiments on separate benches, with separate tubes, and never pour the two wastes into the same container.
- Goggles throughout, both experiments
- Iron(III) chloride stains skin and clothing and is mildly corrosive. Wear gloves
- Copper(II) chloride is harmful to aquatic life. Dilute waste heavily
- The second experiment needs near-boiling water. Use the beaker, not your hands
- Concentrated hydrochloric acid is corrosive. It is not something to hand to a class
If it does not work
| What you see | What caused it |
|---|---|
| The red is so dark nothing changes | The commonest failure by far. The stock solutions were mixed directly instead of diluted into 100 cm³ of water |
| The red fades during the lesson | Iron(III) chloride solution hydrolyses over time. Make it fresh |
| The iron solution looks cloudy or orange-brown | Same cause. It has hydrolysed |
| Copper solution stays blue whatever you do | Not enough chloride. Add more acid before you start heating |
| Copper solution is already green cold | Too much acid. Dilute with a little water |
| Nothing reverses on cooling | Not left long enough. Give it several minutes in the iced water |
The mistake worth teaching to
Learners think the equilibrium moves back to where it was. It does not. It moves to a new position that partly offsets the change, which is why tube 2 stays darker than the control rather than fading back to match it.
Two tubes side by side make that argument better than any explanation.
What you need
| Item | Qty |
|---|---|
| Test tubes, borosilicate | 8 |
| Test tube rack | 2 |
| Beakers, 250 ml and 100 ml | 3 |
| Measuring cylinder, 25 ml | 1 |
| Thermometer | 1 |
| Disposable pipettes | 20 |
| Spatula | 1 |
Chemicals
- Iron(III) chloride
- Potassium thiocyanate
- Copper(II) chloride
You supply
Distilled water, a kettle, ice, and concentrated hydrochloric acid for the second experiment. We do not ship concentrated acid to schools. If your lab does not hold it, run the first experiment on its own, or do the second as a demonstration with your own acid.
Free worksheet and marking memo
Both free, no sign up, straight to the PDF.
- Learner worksheet, 40 marks, covering both experiments plus the theory, the Haber process and the catalyst question
- Marking memorandum, with the mark allocation, model answers and a note on the six places learners most often drop marks
Buy this experiment
We are putting together a complete Chemical Equilibrium Kit for Grade 12 with the glassware, thermometer and the three chemicals in one box, plus an annual refill. Both coming shortly.
In the meantime every item in the apparatus table above is in stock and sold separately.