OCR GCSE Chemistry A: Equilibria (J248)

This guide teaches C5.3 Equilibria, part of Topic C5 (Monitoring and controlling chemical reactions) in the J248 specification for OCR GCSE (9–1) Chemistry A (Gateway Science), Version 4.0, August 2026. It covers learning outcomes C5.3a to C5.3c and the mathematical learning outcomes CM5.3i to CM5.3iv. Outcomes C5.3a and C5.3b are assessed at both tiers. Outcome C5.3c, predicting how conditions move the position of equilibrium, is Higher Tier only. The specification’s assessment overview puts Topic C5 on Paper 2 (Foundation) and Paper 4 (Higher).

After this guide, shorten your notes using the Equilibria revision notes and test yourself with the Equilibria practice questions. The OCR GCSE Chemistry hub lists the whole course, and the printable J248 checklist helps you track every statement. A free 10-minute diagnostic points to your weakest areas.

What this sub-topic covers

Statement	Your task	Tier
C5.3a	Recall that some reactions can be reversed by changing the reaction conditions	Both tiers
C5.3b	Recall that dynamic equilibrium happens in a closed system when the forward and reverse rates are equal	Both tiers
C5.3c	Use Le Chatelier’s principle (concentration, temperature, pressure) to predict shifts and suggest conditions that give as much of a product as possible	Higher Tier only
CM5.3i	Arithmetic and ratio when measuring rates	Both tiers
CM5.3ii	Draw and interpret graphs of data to find rates	Both tiers
CM5.3iii	Use gradients, including tangents to curves, as a measure of rate	Both tiers
CM5.3iv	Use proportionality when comparing factors that affect rate	Both tiers

The specification lists no practical suggestion for C5.3. Its working scientifically links (hypotheses, planning, choosing apparatus, recording observations) are built into the examples below.

Two ideas from earlier topics help here: writing balanced equations with state symbols (see Introducing chemical reactions) and the meaning of exothermic and endothermic (see Energetics). Rate graphs are taught in Controlling reactions.

C5.3a Reversible reactions

In a reversible reaction, the products can turn back into the reactants. The symbol ⇌ marks this. Left to right is the forward reaction; right to left is the reverse reaction. Changing the conditions can change which direction is favoured.

Hydrated and anhydrous cobalt(II) chloride

hydrated cobalt(II) chloride  ⇌  anhydrous cobalt(II) chloride + water
           pink                              blue

Heat the pink solid and water is driven off, leaving blue anhydrous cobalt(II) chloride. Add water to the blue solid and it turns pink again. The forward reaction takes in heat (endothermic), so the reverse reaction gives out heat (exothermic).

Iodine monochloride and chlorine

ICl(l)       +  Cl₂(g)  ⇌  ICl₃(s)
brown liquid                 yellow solid

With plenty of chlorine, brown ICl turns to yellow crystals of ICl₃. Let the chlorine out and the yellow solid turns back into brown liquid. Here the condition changed is the concentration of chlorine.

Worked example 1

A student heats some pink crystals in a test tube. They turn blue. Droplets of colourless liquid form near the cool mouth of the tube. When the tube is cool, the student pours the liquid back on. The crystals turn pink and the tube feels warm. Explain these observations.

1. Heating breaks down hydrated cobalt(II) chloride into anhydrous cobalt(II) chloride (blue) and water vapour.
2. The water vapour condenses on the cool part of the tube.
3. Adding the water back reverses the reaction, re-forming pink hydrated cobalt(II) chloride.
4. The forward reaction is endothermic, so the reverse is exothermic: energy is released and the tube warms up.

C5.3b Dynamic equilibrium

Picture a reversible reaction in a closed system, a container from which nothing can escape and to which nothing is added.

• At the start there are only reactants, so the forward reaction is fast and the reverse reaction has not started.
• As reactants are used up, the forward rate falls.
• As products build up, the reverse rate rises.
• Eventually the two rates become equal. From then on, every time some product forms, the same amount turns back into reactant.

This state is called dynamic equilibrium:

• dynamic, because both reactions are still happening
• equilibrium, because the concentrations of reactants and products stay constant.

Three misconceptions the specification highlights:

• Constant does not mean equal. There may be much more product than reactant, or the reverse.
• The reactions have not stopped. The net change is zero because the rates match.
• It is one reaction, not two. A dynamic equilibrium is a single reversible reaction going both ways at once.

A closed system matters: if a gaseous product escapes from an open flask, equilibrium is never reached.

Graphs at equilibrium (CM5.3ii)

Plot rate against time and the forward-rate line starts high and curves down; the reverse-rate line starts at zero and curves up. They meet and then run together as one horizontal line. That point is where equilibrium is reached.

On a graph of concentration against time, the reactant curve falls and the product curve rises. Both then level off. They do not have to meet; they just become flat.

Worked example 2 (CM5.3i, CM5.3ii, CM5.3iii)

Substance A changes into substance B in a sealed flask: A(g) ⇌ B(g). The table shows the concentration of B. At the start the concentration of A is 1.00 mol/dm³ and there is no B.

Time (s)	0	10	20	30	40	50	60
[B] (mol/dm³)	0.00	0.44	0.61	0.67	0.69	0.70	0.70

(a) Calculate the mean rate of formation of B in the first 10 s, and between 10 s and 20 s.

mean rate = change in concentration ÷ time taken
0–10 s:   (0.44 − 0.00) ÷ 10 = 0.044 mol/dm³/s
10–20 s:  (0.61 − 0.44) ÷ 10 = 0.017 mol/dm³/s

The rate falls: less A is left, and more B turns back into A.

(b) Use the data to say when equilibrium is reached. [B] stops changing at 0.70 mol/dm³ from about 50 s, so equilibrium is reached at about 50 s. The curve is flat from there.

(c) A tangent drawn to the curve at 10 s passes through (0 s, 0.19 mol/dm³) and (30 s, 0.96 mol/dm³). Find the rate at 10 s.

gradient = rise ÷ run = (0.96 − 0.19) ÷ (30 − 0) = 0.77 ÷ 30
         = 0.026 mol/dm³/s (2 s.f.)

A tangent gives the rate at one instant. After 50 s the tangent is horizontal, so the net rate is zero, though both reactions continue.

(d) Find the ratio of B to A at equilibrium, and the percentage of A converted.

[A] at equilibrium = 1.00 − 0.70 = 0.30 mol/dm³
B : A = 0.70 : 0.30 = 7 : 3
percentage of A converted = (0.70 ÷ 1.00) × 100 = 70%

Proportionality and rate (CM5.3iv)

Questions may tell you that a rate is directly proportional to a concentration. Then if the concentration falls to a fraction of its value, the rate falls to the same fraction.

In worked example 2, suppose you are told that the forward rate is proportional to [A] and starts at 0.070 mol/dm³/s when [A] = 1.00 mol/dm³. At equilibrium [A] = 0.30 mol/dm³, so

forward rate = 0.070 × (0.30 ÷ 1.00) = 0.021 mol/dm³/s

At equilibrium the reverse rate must also be 0.021 mol/dm³/s. Work out the scale factor, then multiply.

C5.3c Changing the position of equilibrium (Higher Tier only)

The position of equilibrium tells you how much reactant and how much product the settled mixture holds. “Shifts to the right” means more products; “shifts to the left” means more reactants.

Le Chatelier’s principle: when you alter a condition for a system at equilibrium, the position moves in whichever direction opposes the alteration. The specification asks you to apply it to three conditions.

Concentration

• Add more of a reactant, or remove a product: the position shifts right, making more product.
• Add more of a product, or remove a reactant: the position shifts left.

For ICl + Cl₂ ⇌ ICl₃, pumping in more chlorine shifts the position right, and more yellow ICl₃ forms. Removing chlorine shifts it left, so the solid turns back to brown liquid.

Temperature

• Raise the temperature: the position shifts in the endothermic direction, which takes in heat and opposes the rise.
• Lower the temperature: the position shifts in the exothermic direction.

So if the forward reaction is exothermic, a higher temperature gives less product at equilibrium.

Pressure (reactions with gases)

• Raise the pressure: the position shifts to the side with fewer molecules of gas.
• Lower the pressure: it shifts to the side with more molecules of gas.
• Equal numbers of gas molecules on the left and right: pressure leaves the position unchanged.

Count gas molecules from the balancing numbers; ignore solids and liquids.

Worked example 3

Methanol is manufactured by this reaction:

CO(g) + 2H₂(g)  ⇌  CH₃OH(g)      forward reaction exothermic

Predict the effect of each change on the amount of methanol at equilibrium. Then suggest conditions that give as much methanol as possible.

Change	Shift	Reason	Amount of methanol
Raise the pressure	Right	Left side has 1 + 2 = 3 gas molecules; right side has 1	Increases
Raise the temperature	Left	Reverse reaction is endothermic and takes in the extra heat	Decreases
Add more hydrogen	Right	Opposes the increase in a reactant	Increases
Remove methanol as it forms	Right	Replaces the product that was taken away	Increases

Conditions for the most methanol: high pressure, low temperature, plenty of CO and H₂, and remove methanol as it forms.

C5.3c asks only about getting as much product as possible. The trade-off between yield and rate in industry is a separate statement, C6.1d (also Higher Tier only), covered in Improving processes and products.

Worked example 4: using data to form a hypothesis

In a sealed vessel, gases R and S react: R(g) + S(g) ⇌ 2T(g). The percentage of T in the equilibrium mixture is measured at several temperatures.

Temperature (°C)	150	250	350
T at equilibrium (%)	31	44	58

The amount of T rises with temperature, and a higher temperature favours the endothermic direction. So the hypothesis is that the forward reaction is endothermic. There are 2 gas molecules on each side, so pressure should not change the percentage of T; you could test that at a different pressure and the same temperature.

Planning an experiment

To show a concentration effect in the lab, you can use a coloured equilibrium in solution:

Fe³⁺(aq) + SCN⁻(aq)  ⇌  FeSCN²⁺(aq)
yellow       colourless       blood red

1. Mix dilute iron(III) solution with dilute thiocyanate solution to make a red mixture.
2. Pour equal volumes into four test tubes with a measuring cylinder.
3. Keep one as a control. Add a few drops of iron(III) solution to the second, a few drops of thiocyanate solution to the third, and the same volume of water to the control.
4. Compare the colours against a white tile and record them in a table.
5. Wear eye protection.

Prediction from Le Chatelier’s principle: adding either reactant shifts the position right, so the colour becomes a deeper red.

Common errors

• Writing that the concentrations are equal at equilibrium. They are constant.
• Writing that the reactions stop. Both continue at equal rates.
• Forgetting the closed system, so a gas escapes and equilibrium is never set up.
• Counting atoms, or counting solids and liquids, in a pressure question. Count gas molecules only.
• Stating “shifts right” without saying what happens to the substance the question names.
• Linking a higher temperature to the exothermic direction. A higher temperature always favours the endothermic direction.

Where to go next

Fix the key words with the Equilibria revision notes, then try the Equilibria practice questions.

Official syllabus

OCR GCSE (9–1) Chemistry A (Gateway Science) J248 specification, Version 4.0 (August 2026), for first assessment in 2018, published by OCR. Sub-topic C5.3 Equilibria, learning outcomes C5.3a–C5.3c and mathematical learning outcomes CM5.3i–CM5.3iv.