OCR GCSE Chemistry A: Equilibria (J248) – Revision Notes

For full explanations and longer worked examples, use the Equilibria study guide.

Here is C5.3 Equilibria in short form, from the J248 specification for OCR GCSE (9–1) Chemistry A (Gateway Science), Version 4.0, August 2026: learning outcomes C5.3a–C5.3c and mathematical learning outcomes CM5.3i–CM5.3iv. C5.3a and C5.3b are tested at both tiers; C5.3c is Higher Tier only. Topic C5 is assessed on Paper 2 (Foundation) or Paper 4 (Higher), according to the assessment overview. Then test yourself with the Equilibria practice questions. The OCR GCSE Chemistry hub has the rest of the course, the J248 checklist lets you tick statements off, and the free 10-minute diagnostics show where to focus.

Key definitions

Term	Meaning
Reversible reaction	Products can turn back into reactants; symbol ⇌
Forward / reverse reaction	Left to right / right to left in the equation as written
Closed system	Nothing can enter or leave the container
Dynamic equilibrium	In a closed system, the forward and reverse rates are equal, so concentrations stay constant
Position of equilibrium	How much reactant and product the settled mixture holds
Le Chatelier’s principle (Higher Tier only)	A change made to a system at equilibrium makes the position shift to oppose the change

C5.3a Reversing a reaction by changing conditions

Changing the temperature or a concentration can push a reversible reaction one way or the other.

hydrated cobalt(II) chloride  ⇌  anhydrous cobalt(II) chloride + water
          pink                              blue
heat  → water driven off, pink turns blue (forward, endothermic)
add water → blue turns pink, heat released (reverse, exothermic)

ICl(l) brown + Cl₂(g)  ⇌  ICl₃(s) yellow
more chlorine → yellow solid forms
chlorine removed → brown liquid returns

If one direction is endothermic, the other direction is exothermic.

C5.3b Dynamic equilibrium

What happens in a closed system:

1. Start: only reactants. Forward rate high, reverse rate zero.
2. Reactants used up, so the forward rate falls. Products build up, so the reverse rate rises.
3. The rates become equal. Equilibrium is reached.
4. Both reactions continue. Concentrations stay constant.

Graph shapes

Graph	What you see	Equilibrium shown by
Rate against time	Forward line falls, reverse line rises	The two lines meet and run as one flat line
Concentration against time	Reactant falls, product rises	Both curves become flat (they need not meet)

Maths for this sub-topic (CM5.3i–iv)

Skill	Method
Mean rate	change in concentration (or amount) ÷ time taken
Rate at one instant	Draw a tangent; gradient = rise ÷ run
Equilibrium from a graph	Where the gradient becomes zero (curve flat)
Ratio at equilibrium	Divide one equilibrium concentration by the other; simplify
Percentage converted	(amount reacted ÷ amount at start) × 100
Proportionality	If rate ∝ concentration, the rate changes by the same factor as the concentration

Method in steps: rate from a concentration–time graph

1. Plot concentration (y-axis) against time (x-axis) and draw a smooth curve, not dot-to-dot lines.
2. For a mean rate, read two points and divide the change in concentration by the time between them.
3. For the rate at one time, draw a straight tangent that just touches the curve there.
4. Pick two points far apart on the tangent and work out rise ÷ run.
5. Give the unit: mol/dm³/s (or cm³/s if the y-axis is gas volume).

A steep curve means a fast reaction. A flat curve means no overall change: equilibrium, if the system is closed.

Worked reminder (proportion). If the forward rate is proportional to the reactant concentration and the concentration falls to one quarter, the forward rate also falls to one quarter.

Worked reminder. A product reaches 0.36 mol/dm³ at equilibrium and the reactant left is 0.12 mol/dm³. Ratio product : reactant = 0.36 : 0.12 = 3 : 1. Constant, not equal.

C5.3c Changing conditions (Higher Tier only)

Change	Position shifts	Effect on product (forward = right)
Add reactant / remove product	Right	More product
Add product / remove reactant	Left	Less product
Raise temperature	Endothermic direction	More product only if forward is endothermic
Lower temperature	Exothermic direction	More product only if forward is exothermic
Raise pressure	Side with fewer gas molecules	More product if the right has fewer gas molecules
Lower pressure	Side with more gas molecules	More product if the right has more gas molecules
Pressure, equal gas molecules each side	No shift	No change

Method in steps: predicting a shift

1. Read the equation. Note state symbols and whether the forward reaction is exothermic or endothermic.
2. Name the change (concentration, temperature or pressure).
3. Say which direction opposes it.
4. For pressure, count gas molecules on each side using the balancing numbers.
5. Finish with the effect on the substance named in the question.

Worked reminder. CO(g) + 3H₂(g) ⇌ CH₄(g) + H₂O(g), forward exothermic. Gas molecules: 4 on the left, 2 on the right. For the most methane: high pressure, low temperature, excess hydrogen.

Suggesting conditions for the most product

• Temperature: low if the forward reaction is exothermic, high if it is endothermic.
• Pressure: high if the product side has fewer gas molecules, low if it has more.
• Concentration: keep adding reactants and remove the product.

C5.3c is about the amount of product. Balancing that against rate is C6.1d.

Using data (Working Scientifically 1.2a)

If the percentage of product at equilibrium falls as the temperature rises, the forward reaction is exothermic. If it rises, the forward reaction is endothermic. Use the trend in the numbers as your evidence.

Planning a test of a prediction (Working Scientifically 1.2b, Working Scientifically 1.2c, Working Scientifically 2a, Working Scientifically 2b)

• Choose a coloured equilibrium in solution so the shift is easy to see, for example iron(III) ions with thiocyanate ions (blood red product).
• Use equal volumes in each test tube, measured with a measuring cylinder, and keep one tube as a control.
• Change one condition only, such as adding a few drops of one reactant.
• Add the same volume of water to the control so dilution is the same.
• Record colours against a white tile in a results table. Wear eye protection.

Must-know distinctions

• Constant vs equal. At equilibrium concentrations are constant; they are rarely equal.
• Rate vs amount. The rates of the two reactions are equal; the amounts of substances are not.
• Reversible vs at equilibrium. Every equilibrium involves a reversible reaction, but an open container lets a gas escape, so no equilibrium forms.
• Molecules vs atoms. Pressure rules use numbers of gas molecules, never atoms.
• Mean rate vs rate at a time. A mean rate uses two points; the rate at a time needs a tangent.

Quick self-test

1. What does the symbol ⇌ mean?
2. Blue anhydrous cobalt(II) chloride has water added. State the colour change and whether heat is given out or taken in.
3. State the two conditions for a dynamic equilibrium.
4. True or false: at equilibrium the reactions stop. Explain.
5. In 30 s, 0.24 mol/dm³ of product forms from none. Calculate the mean rate.
6. A reaction starts with 0.60 mol of reactant; 0.15 mol is left at equilibrium. What percentage reacted?
7. A forward rate is proportional to the reactant concentration. It is 0.050 mol/dm³/s at 0.80 mol/dm³. What is it at 0.40 mol/dm³?
8. A tangent passes through (0 s, 0.05 mol/dm³) and (40 s, 0.65 mol/dm³). Find the rate.
9. (Higher Tier only) PCl₅(g) ⇌ PCl₃(g) + Cl₂(g). Which way does the position shift when the pressure is raised?
10. (Higher Tier only) A forward reaction is endothermic. What happens to the amount of product when the temperature is lowered?
11. In ICl(l) + Cl₂(g) ⇌ ICl₃(s), what do you see when chlorine is removed?
12. (Higher Tier only) CO(g) + H₂O(g) ⇌ CO₂(g) + H₂(g). Does a higher pressure change the amount of hydrogen?

Answers

1. The reaction is reversible: it can go in both directions.
2. Blue to pink; heat is given out (the reverse reaction is exothermic).
3. A closed system; forward and reverse rates are equal.
4. False. Both reactions continue at the same rate, so there is no overall change.
5. 0.24 ÷ 30 = 0.008 mol/dm³/s.
6. 0.60 − 0.15 = 0.45 mol reacted; 0.45 ÷ 0.60 × 100 = 75%.
7. Concentration halves, so rate halves: 0.025 mol/dm³/s.
8. (0.65 − 0.05) ÷ 40 = 0.015 mol/dm³/s.
9. Left: 1 gas molecule on the left, 2 on the right.
10. The position shifts in the exothermic (reverse) direction, so less product.
11. Yellow solid turns back into brown liquid.
12. No change: 2 gas molecules on each side.

Where marks are usually lost

• Calling the equilibrium concentrations “equal”; the right word is “constant”.
• Saying the reactions have stopped at equilibrium.
• Leaving out “closed system” when asked for the conditions for equilibrium.
• Counting atoms, or counting a solid or liquid, when applying the pressure rule.
• Stopping at “shifts left” or “shifts right” and never saying what happens to the named product.
• Saying a higher temperature favours the exothermic direction.
• Using two data points for a “rate at 10 s” question instead of a tangent.
• Leaving a rate without a unit, or dropping the “/s” from mol/dm³/s.
• Quoting a rule from memory when the question asks you to use the data as evidence.

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.