Chemistry · 6 Chemical reactions

6.3 Reversible reactions and equilibrium

Before you start

Rate of reaction and the five factors that change it (6.2).
Exothermic reactions give out heat; endothermic reactions take it in.
Reading state symbols in an equation: (g), (l), (s), (aq).

Syllabus objectives covered (11)

State that some chemical reactions are reversible as shown by the symbol ⇌
Describe how changing the conditions can change the direction of a reversible reaction for: (a) the effect of heat on hydrated compounds (b) the addition of water to anhydrous compounds limited to copper(II) sulfate and cobalt(II) chloride
State that a reversible reaction in a closed system is at equilibrium when: (a) the rate of the forward reaction is equal to the rate of the reverse reaction (b) the concentrations of reactants and products are no longer changing
Predict and explain, for a reversible reaction, how the position of equilibrium is affected by: (a) changing temperature (b) changing pressure (c) changing concentration (d) using a catalyst using information provided
State the symbol equation for the production of ammonia in the Haber process, N2(g) + 3H2(g) ⇌ 2NH3(g)
State the sources of the hydrogen (methane) and nitrogen (air) in the Haber process
State the typical conditions in the Haber process as 450 °C, 20 000 kPa / 200 atm and an iron catalyst
State the symbol equation for the conversion of sulfur dioxide to sulfur trioxide in the Contact process, 2SO2(g) + O2(g) ⇌ 2SO3(g)
State the sources of the sulfur dioxide (burning sulfur or roasting sulfide ores) and oxygen (air) in the Contact process
State the typical conditions for the conversion of sulfur dioxide to sulfur trioxide in the Contact process as 450 °C, 200 kPa / 2 atm and a vanadium(V) oxide catalyst
Explain, in terms of rate of reaction and position of equilibrium, why the typical conditions stated are used in the Haber process and in the Contact process, including safety considerations and economics

Part 1 of 7

Reactions that go both ways

Most reactions you have met go to completion — the reactants turn into products and that is that. Some do not: the products react to re-form the reactants at the same time.
These are reversible reactions, and they are written with the symbol $\overset{}{⇌}$ instead of an arrow. Two half-arrows pointing opposite ways, meaning both directions are happening.
Read the equation left to right for the forward reaction, right to left for the reverse. The same equation describes both.
One consequence matters throughout this topic: a reversible reaction can never give a 100% yield, because some product is always turning back.

Part 2 of 7

Two reversible reactions you must know

Copper(II) sulfate. Heating the blue hydrated crystals drives off the water and leaves a white anhydrous powder. Add water back and the blue returns, with heat given out.
$CuSO_{4} \cdot 5 H_{2} O \overset{}{⇌} CuSO_{4} + 5 H_{2} O$
In words: blue hydrated copper(II) sulfate $\overset{}{⇌}$ white anhydrous copper(II) sulfate + water. Heat drives it forwards; adding water drives it back.
Cobalt(II) chloride. The same pattern with different colours: pink hydrated crystals become blue anhydrous on heating, and turn pink again with water.
$CoCl_{2} \cdot 6 H_{2} O \overset{}{⇌} CoCl_{2} + 6 H_{2} O$
Get the colours the right way round, because it is a one-mark question that people reverse. Copper: blue is wet. Cobalt: blue is dry. They are opposites, which is exactly why they get muddled.
Both are used as tests for the presence of water: anhydrous copper(II) sulfate turning from white to blue, and anhydrous cobalt(II) chloride turning from blue to pink.
Neither test tells you the water is pure. For that you also need a boiling point of exactly 100 °C or a freezing point of exactly 0 °C. Questions often ask for both parts.

Part 3 of 7

What equilibrium actually is

Put a reversible reaction in a closed system — nothing enters, nothing escapes — and leave it. It reaches equilibrium.
At equilibrium two things are true, and the syllabus wants both: the rate of the forward reaction equals the rate of the reverse reaction, and the concentrations of reactants and products are no longer changing.
The word for this is dynamic. Both reactions are still going on, at full speed — they simply cancel out. Nothing has stopped.
This is the point most often lost in exams. 'The reaction has stopped' is wrong. 'The forward and reverse reactions are still occurring but at equal rates' is right.
Equal rates does not mean equal amounts. Equilibrium can sit far to the right, with mostly products, or far to the left, with mostly reactants. That is what 'position of equilibrium' means.
The closed system matters. In an open beaker a gas escapes, the reverse reaction has nothing to work with, and no equilibrium is ever established.

Part 4 of 7

Shifting the position of equilibrium

The governing idea: change a condition and the equilibrium shifts in whichever direction opposes the change. Work out what would oppose it, and you have your answer.
Concentration. Increase the concentration of a reactant and the equilibrium shifts to the right, using it up and making more product. Remove a product and it also shifts right, replacing what you took.
Pressure (gases only). Increasing the pressure shifts the equilibrium towards the side with fewer molecules of gas, because that reduces the pressure.
So count the gas molecules on each side from the balanced equation. In $N_{2} (g) + 3 H_{2} (g) \overset{}{⇌} 2 NH_{3} (g)$ there are 4 on the left and 2 on the right, so high pressure favours ammonia.
If both sides have the same number of gas molecules, pressure has no effect on the position at all — a favourite exam trap.
Temperature. Increasing the temperature shifts the equilibrium in the endothermic direction, because that absorbs the extra heat. Decreasing it shifts the equilibrium in the exothermic direction.
You will be told which direction is exothermic, or given the sign of the energy change. Both industrial reactions here are exothermic forwards, so lower temperature would give a better yield.
A catalyst does not shift the position of equilibrium at all. It speeds up the forward and reverse reactions equally, so equilibrium is reached sooner, with exactly the same composition.

Part 5 of 7

The Haber process

The Haber process makes ammonia, mostly for fertilisers.
$N_{2} (g) + 3 H_{2} (g) \overset{}{⇌} 2 NH_{3} (g)$
The nitrogen comes from the air, which is about 78% nitrogen. The hydrogen comes from methane — natural gas.
The conditions to state: 450 °C, 20 000 kPa (200 atm), and an iron catalyst.
Learn the three numbers as a set. A question asking you to State the conditions is asking for all three, and two out of three is not full marks.

Part 6 of 7

The Contact process

The Contact process makes sulfuric acid; the step examined here is the conversion of sulfur dioxide to sulfur trioxide.
$2 SO_{2} (g) + O_{2} (g) \overset{}{⇌} 2 SO_{3} (g)$
The sulfur dioxide comes from burning sulfur, or from roasting sulfide ores. The oxygen comes from the air.
The conditions to state: 450 °C, 200 kPa (2 atm), and a vanadium(V) oxide catalyst.
Same temperature as the Haber process, but a hundred times lower pressure and a different catalyst. The temperature coincidence makes the pressure difference easy to lose — check which process you are answering about.

Part 7 of 7

Why those conditions, and not the ideal ones

Both forward reactions are exothermic, and both have fewer gas molecules on the right. So for the best yield you would want low temperature and high pressure. Neither process uses the ideal.
Temperature is a compromise. A low temperature gives a better yield but a hopelessly slow rate. 450 °C sacrifices some yield to get the ammonia made in a usable time — and a catalyst recovers some of the lost rate.
That word, compromise, is what the mark scheme is looking for. Say what is gained and what is given up.
Pressure for the Haber process is a compromise too, but with cost and safety rather than rate. Higher pressure gives more ammonia, but demands stronger vessels and more energy for compression, and raises the risk of leaks or rupture. 200 atm is where the extra yield stops being worth the expense.
Pressure for the Contact process is genuinely different, and worth understanding rather than memorising. The equilibrium already lies far to the right at only 2 atm, giving about 96% conversion — so paying for high pressure would buy almost nothing.
The catalyst in both cases increases the rate without shifting the equilibrium. It allows a lower temperature than would otherwise be workable, which indirectly improves the yield.
So the shape of a full answer is always: the ideal condition for yield, the practical reason it cannot be used, and the compromise actually chosen.

Worked examples

Example 1. For the equilibrium $2 SO_{2} (g) + O_{2} (g) \overset{}{⇌} 2 SO_{3} (g)$ , the forward reaction is exothermic. Predict and explain the effect on the yield of sulfur trioxide of (i) increasing the temperature and (ii) increasing the pressure.

Show the working

Take temperature first. Raising the temperature shifts the equilibrium in the endothermic direction, which here is the reverse reaction.
Shifting to the left means less sulfur trioxide, so the yield falls.
Now pressure. Count the gas molecules: $2 + 1 = 3$ on the left, 2 on the right.
Raising the pressure shifts the equilibrium towards the side with fewer gas molecules, which is the right.
Shifting to the right means more sulfur trioxide, so the yield rises.

Answer: (i) The yield decreases — higher temperature favours the endothermic reverse reaction. (ii) The yield increases — higher pressure favours the right-hand side, which has fewer gas molecules (2 against 3).

Example 2. The forward reaction in the Haber process is exothermic. Explain why a temperature of $450^{\circ} C$ is used rather than a much lower temperature.

Show the working

State the ideal first, so the compromise has something to be measured against. Since the forward reaction is exothermic, a lower temperature would shift the equilibrium to the right and give a higher yield of ammonia.
Now the problem. At a low temperature the particles have little kinetic energy, few collisions exceed the activation energy, and the rate is far too slow to be economic.
Name the trade-off explicitly: 450 °C is a compromise between an acceptable yield and an acceptable rate.
Finish with the catalyst, which is part of the same decision — the iron catalyst raises the rate further, so a lower temperature can be tolerated than would otherwise be possible.

Answer: A lower temperature would give a higher yield, because the forward reaction is exothermic, but the rate would be uneconomically slow. 450 °C is a compromise between yield and rate, and the iron catalyst raises the rate further so that this moderate temperature is workable.

Example 3. A student heats blue hydrated copper(II) sulfate crystals, then allows the solid to cool and adds a few drops of water. Describe what is seen at each stage, and write the equation for the reaction.

Show the working

On heating, the blue crystals turn white as water is driven off, and droplets of water condense at the cool end of the tube.
The white solid is anhydrous copper(II) sulfate — the same compound without its water of crystallisation.
Adding water reverses the change: the solid turns blue again, and the tube becomes warm because the reverse reaction is exothermic.
Write the equation with the reversible symbol, since both directions have been demonstrated.
$CuSO_{4} \cdot 5 H_{2} O \overset{}{⇌} CuSO_{4} + 5 H_{2} O$
Mention the heat given out on adding water — it is a marking point that is easy to observe and easy to forget.

Answer: On heating, blue turns to white and water condenses at the top of the tube. On adding water, the white solid turns blue and heat is given out. $CuSO_{4} \cdot 5 H_{2} O \overset{}{⇌} CuSO_{4} + 5 H_{2} O$

Where people go wrong

Mistake: Saying that at equilibrium 'the reaction has stopped'.

Instead: Both reactions continue at equal rates — equilibrium is dynamic.

It is the most frequently examined point in the subtopic, and 'stopped' is treated as a contradiction that cancels the mark.

Mistake: Thinking equal rates means equal amounts of reactants and products.

Instead: Equal rates only means concentrations stop changing. The position may lie far to either side.

The rates depend on both concentration and the reaction itself, so they can balance at any composition.

Mistake: Saying a catalyst increases the yield at equilibrium.

Instead: A catalyst speeds up both directions equally, so equilibrium is reached sooner with the same composition.

It changes when you arrive, not where you arrive.

Mistake: Swapping the copper and cobalt colours.

Instead: Copper(II) sulfate: blue hydrated, white anhydrous. Cobalt(II) chloride: pink hydrated, blue anhydrous.

Both use blue for opposite states, which is exactly what makes them confusable. Remember 'copper: blue is wet'.

Mistake: Quoting the Haber pressure for the Contact process.

Instead: Haber: 200 atm. Contact: 2 atm. Both 450 °C.

The shared temperature makes people assume the pressures match too. The Contact process needs almost no pressure because the yield is already about 96%.

Mistake: Answering 'why 450 °C' with only 'to make it faster'.

Instead: Say that a lower temperature would give a higher yield but too slow a rate, so 450 °C is a compromise.

The question is about the trade-off. An answer naming only one side of it cannot score the compromise mark.

Check yourself

What does the symbol $\overset{}{⇌}$ tell you about a reaction?
Answer
That it is reversible — the forward and reverse reactions both take place.
State the colour changes when hydrated cobalt(II) chloride is heated and then has water added.
Answer
Pink to blue on heating, and blue back to pink on adding water.
State the two conditions that are true at equilibrium.
Answer
The rate of the forward reaction equals the rate of the reverse reaction, and the concentrations of reactants and products are no longer changing.
In an equilibrium with 3 gas molecules on the left and 2 on the right, which way does increasing the pressure shift it, and why?
Answer
To the right, towards the side with fewer gas molecules, because that opposes the increase in pressure.
Give the equation for the Haber process and the source of each reactant.
Answer
$N_{2} (g) + 3 H_{2} (g) \overset{}{⇌} 2 NH_{3} (g)$ . Nitrogen from the air, hydrogen from methane (natural gas).
State the conditions used in the Haber process and in the Contact process.
Answer
Haber: 450 °C, 20 000 kPa (200 atm), iron catalyst. Contact: 450 °C, 200 kPa (2 atm), vanadium(V) oxide catalyst.
Give the equation for the conversion of sulfur dioxide to sulfur trioxide, and the source of the sulfur dioxide.
Answer
$2 SO_{2} (g) + O_{2} (g) \overset{}{⇌} 2 SO_{3} (g)$ . The sulfur dioxide comes from burning sulfur or from roasting sulfide ores.
Why does the Contact process use only 2 atm when higher pressure would increase the yield?
Answer
The yield is already about 96% at 2 atm, so the extra yield from high pressure would not justify the cost of compression and stronger equipment.

Exam questions

Try each part first. Open the mark scheme after you have written your answer.

Question 1. Ammonia is manufactured by the Haber process. $N_{2} (g) + 3 H_{2} (g) \overset{}{⇌} 2 NH_{3} (g)$ The forward reaction is exothermic.

(a) State the source of the nitrogen and the source of the hydrogen used in this process. 2 marks No calculator

Mark scheme

B1 nitrogen from the air
B1 hydrogen from methane / natural gas

(b) State the typical conditions used in this process. 3 marks No calculator

Mark scheme

B1 450 °C
B1 20 000 kPa / 200 atm
B1 iron catalyst

(c) Explain why a high pressure increases the yield of ammonia. 2 marks No calculator

Mark scheme

B1 there are fewer molecules of gas on the right (2) than on the left (4)
B1 increasing the pressure shifts the position of equilibrium to the right / towards the side with fewer gas molecules

(d) Suggest two reasons why a pressure much higher than 200 atm is not used. 2 marks No calculator

Mark scheme

B1 the cost of compression / stronger vessels and pipework is high
B1 safety: greater risk of leaks or of the vessel rupturing

(e) Explain the effect of the iron catalyst on the yield of ammonia and on the time taken to reach equilibrium. 2 marks No calculator

Mark scheme

B1 no effect on the yield / position of equilibrium
B1 equilibrium is reached more quickly, because both forward and reverse rates increase equally

Question 2. A sealed flask contains the equilibrium $2 SO_{2} (g) + O_{2} (g) \overset{}{⇌} 2 SO_{3} (g)$ , in which the forward reaction is exothermic.

(a) Describe what is meant by a reaction being at equilibrium. 2 marks No calculator

Mark scheme

B1 the rate of the forward reaction equals the rate of the reverse reaction
B1 the concentrations of reactants and products remain constant. CON if the reaction is said to have stopped

(b) Predict the effect on the amount of sulfur trioxide of decreasing the temperature, and justify your answer. 2 marks No calculator

Mark scheme

B1 the amount of SO3 increases
B1 lowering the temperature shifts the equilibrium in the exothermic direction, which is the forward reaction

(c) Explain why this reaction must be carried out in a closed system for an equilibrium to be established. 1 mark No calculator

Mark scheme

B1 so that no gas escapes and the reverse reaction can occur, owtte

Question 3. A student is given a colourless liquid and asked whether it is water.

(a) Describe a chemical test to show that the liquid contains water, including the result. 2 marks No calculator

Mark scheme

B1 add the liquid to anhydrous copper(II) sulfate (or anhydrous cobalt(II) chloride)
B1 white turns blue (or blue turns pink)

(b) Explain why a positive result in this test does not prove the liquid is pure water. 2 marks No calculator

Mark scheme

B1 the test only shows that water is present, not that nothing else is
B1 purity is shown by a boiling point of exactly 100 °C (or a freezing point of exactly 0 °C) at atmospheric pressure

Words to know

Reversible reaction: A reaction in which the products can react to re-form the reactants, shown by $\overset{}{⇌}$ .
Closed system: A system from which nothing can enter or escape, so an equilibrium can be established.
Dynamic equilibrium: The state in which forward and reverse reactions continue at equal rates and concentrations stay constant.
Position of equilibrium: How far the equilibrium lies towards products or reactants.
Anhydrous: Containing no water of crystallisation.
Hydrated: Containing water of crystallisation.

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