Pearson Edexcel International GCSE Chemistry 4CH1: Physical chemistry – Revision Notes

Condensed for the final weeks. For full explanations and worked examples, use the Physical chemistry study guide, then test yourself with the Physical chemistry practice questions.

These notes cover Topic 3 of the Pearson Edexcel International GCSE Chemistry (4CH1) specification, Issue 3 (September 2024): points 3.1–3.22 across (a) Energetics, (b) Rates of reaction and (c) Reversible reactions and equilibria. The qualification is untiered. Points with a C reference are Paper 2 only; the rest can be tested on either paper. Course links: 4CH1 Chemistry hub, printable checklist and the free all-topics diagnostic.

(a) Energetics

Definitions (3.1, 3.6C)

Term	Meaning
Exothermic	Heat energy given out; surroundings get hotter; ΔH negative
Endothermic	Heat energy taken in; surroundings get colder; ΔH positive
Bond breaking	Endothermic (Paper 2 only)
Bond making	Exothermic (Paper 2 only)

Formulas

Quantity	Formula	Units
Heat energy change	Q = m × c × ΔT	J
Molar enthalpy change	ΔH = Q ÷ n (sign added)	kJ/mol
From bond energies (Paper 2 only)	ΔH = bonds broken − bonds made	kJ/mol

m is the mass of the water or solution heated, not of the fuel or solid. c is given in the question.

Method in steps: ΔH from an experiment (3.3, 3.4)

1. Q = m × c × ΔT, with m = mass of water or total volume of solutions (1 cm³ = 1 g).
2. Divide by 1000 to get kJ.
3. Find the moles of the reactant that was not in excess (mass ÷ Mr, or concentration × volume in dm³).
4. ΔH = Q ÷ moles.
5. Add the sign: temperature rose → negative; temperature fell → positive.

Calorimetry set-ups (3.2, 3.8)

• Solutions (neutralisation, displacement, dissolving): polystyrene cup, thermometer, stir, record the highest or lowest temperature.
• Combustion: spirit burner weighed before and after; copper can of water above it; record the temperature rise.
• Main error: heat loss to the surroundings, so measured ΔH is less negative than the data-book value. Also incomplete combustion of the fuel.

Energy level diagrams (3.5C) – Paper 2 only

• Exothermic: products below reactants; ΔH arrow points down.
• Endothermic: products above reactants; ΔH arrow points up.
• Label the vertical axis “energy”, and label reactants, products and ΔH.

Method in steps: bond energies (3.7C) – Paper 2 only

1. Draw displayed formulae for every molecule.
2. Count each bond type broken on the left, multiplied by the number of molecules.
3. Count each bond type made on the right, the same way.
4. ΔH = total broken − total made.
5. Negative answer: exothermic. Positive answer: endothermic.

Small reminder: in 2H₂ + O₂ → 2H₂O, two H–H bonds and one O=O bond are broken, and four O–H bonds are made.

(b) Rates of reaction

Collision theory in one line

Particles must collide with energy at least equal to the activation energy. More frequent successful collisions = faster rate.

Effects table (3.10, 3.11)

Increase in…	Rate	Reason
Surface area of a solid	Up	More particles exposed → more frequent collisions
Concentration of a solution	Up	More particles per unit volume → more frequent collisions
Pressure of a gas	Up	Particles closer together → more frequent collisions
Temperature	Up	Faster particles → more frequent collisions and more collisions with energy ≥ Ea
Catalyst added	Up	Alternative pathway with lower activation energy

Catalysts (3.12, 3.13)

• Increases the rate; chemically unchanged at the end (same mass can be recovered).
• Works by an alternative pathway with lower activation energy.

Practicals (3.9, 3.15, 3.16)

• Marble chips + dilute HCl: measure mass loss (cotton wool plug lets CO₂ out but stops acid spray) or gas volume at regular times. Change one variable: chip size or acid concentration. Keep mass of chips, volume of acid and temperature the same.
• H₂O₂ decomposition: same volume and concentration of H₂O₂, same mass of each solid; collect O₂ in a gas syringe; compare volumes in the same time.
• Graph: steeper curve = faster. Levels off when a reactant runs out. Mean rate = change in amount ÷ time.

Reaction profiles (3.14C) – Paper 2 only

• Activation energy: from the reactants line up to the peak.
• ΔH: from the reactants line to the products line.
• A catalysed profile has a lower peak, same start and end levels.

(c) Reversible reactions and equilibria

Reversible reactions (3.17, 3.18)

Reaction	Forward (heat)	Reverse
CuSO₄·5H₂O ⇌ CuSO₄ + 5H₂O	Blue → white; endothermic	Add water: white → blue; exothermic, gets hot
NH₄Cl(s) ⇌ NH₃(g) + HCl(g)	White solid disappears on heating	White solid re-forms at the cool top of the tube

Dynamic equilibrium (3.19C, 3.20C) – Paper 2 only

Reached in a sealed container. Two characteristics, learn them word for word:

• forward and reverse reactions occur at the same rate
• concentrations of reactants and products remain constant.

Shifting the position (3.21C, 3.22C) – Paper 2 only

Change	Position of equilibrium moves…
Temperature up	In the endothermic direction
Temperature down	In the exothermic direction
Pressure up	To the side with fewer moles of gas
Pressure down	To the side with more moles of gas
Catalyst added	No change: forward and reverse rates increase equally; equilibrium is reached sooner

Le Chatelier’s principle does not need to be named.

Method in steps: predicting a shift

1. Is the forward reaction exothermic or endothermic? (ΔH negative = exothermic forward.)
2. Count moles of gas on each side (ignore solids and liquids).
3. Apply the table. Say which way the position moves and what happens to the amount of product.

Must-know distinctions

• Activation energy vs ΔH: Ea is reactants to peak; ΔH is reactants to products.
• Rate vs position: a catalyst changes the rate at which equilibrium is reached, not the position.
• Equal rates vs equal concentrations: at equilibrium the rates are equal; the concentrations are constant, not equal.
• Q vs ΔH: Q is in J for this experiment; ΔH is in kJ per mole with a sign.
• More collisions vs more successful collisions: only temperature and catalysts change the proportion of collisions that succeed.

Quick self-test

1. 200 g of water is heated and its temperature rises by 15.0 °C. Calculate Q. (c = 4.2 J/g/°C)
2. Excess zinc powder is added to 50.0 cm³ of 0.200 mol/dm³ copper(II) sulfate solution. The temperature rises by 10.2 °C. Calculate ΔH in kJ per mole of CuSO₄. (c = 4.2 J/g/°C)
3. Burning 0.43 g of hexane (Mr = 86) heats 150 g of water by 20.0 °C. Calculate ΔH of combustion. (c = 4.2 J/g/°C)
4. (Paper 2 only) Calculate ΔH for N₂ + 3H₂ → 2NH₃. Bond energies (kJ/mol): N≡N 945, H–H 436, N–H 391.
5. (Paper 2 only) Calculate ΔH for 2HBr → H₂ + Br₂. Bond energies (kJ/mol): H–Br 366, H–H 436, Br–Br 193. Is it exothermic or endothermic?
6. State two reasons why raising the temperature increases the rate of a reaction.
7. A catalyst is added to a reaction. What happens to the activation energy and to ΔH?
8. (Paper 2 only) H₂(g) + I₂(g) ⇌ 2HI(g). State the effect of increasing the pressure on the position of equilibrium.
9. (Paper 2 only) 2NO₂(g) ⇌ N₂O₄(g). The forward reaction is exothermic. State the effect of increasing the temperature on the amount of NO₂.
10. Water is added to white anhydrous copper(II) sulfate. State two observations.
11. 45 cm³ of gas is collected in the first 30 s of a reaction. Calculate the mean rate over this time.
12. (Paper 2 only) State the two characteristics of a reaction at dynamic equilibrium.

Answers

1. Q = 200 × 4.2 × 15.0 = 12 600 J (12.6 kJ).
2. Q = 50.0 × 4.2 × 10.2 = 2142 J = 2.142 kJ. Moles CuSO₄ = 0.0500 × 0.200 = 0.0100 mol. ΔH = −2.142 ÷ 0.0100 = −214 kJ/mol.
3. Q = 150 × 4.2 × 20.0 = 12 600 J = 12.6 kJ. Moles = 0.43 ÷ 86 = 0.0050 mol. ΔH = −12.6 ÷ 0.0050 = −2520 kJ/mol. (Less negative than the data-book value because of heat loss.)
4. Broken: 945 + 3(436) = 2253. Made: 6(391) = 2346. ΔH = 2253 − 2346 = −93 kJ/mol.
5. Broken: 2(366) = 732. Made: 436 + 193 = 629. ΔH = 732 − 629 = +103 kJ/mol: endothermic.
6. Particles move faster so collide more often; a greater proportion of collisions have energy equal to or greater than the activation energy.
7. Activation energy is lowered; ΔH is unchanged.
8. No effect: 2 moles of gas on each side.
9. Position shifts to the left (endothermic direction), so the amount of NO₂ increases.
10. Turns from white to blue; the mixture gets hot.
11. 45 ÷ 30 = 1.5 cm³/s.
12. Forward and reverse reactions occur at the same rate; concentrations of reactants and products remain constant.

Where marks are usually lost

• Putting the mass of fuel, or the mass of solid added, into Q = mcΔT.
• For mixed solutions, using only one volume (25 g) instead of the total (50 g).
• Dividing Q by the moles of the reactant in excess instead of the limiting one.
• Missing the sign on ΔH, or giving ΔH in J/mol.
• Bond energy sums done as made − broken, or with a molecule’s multiplier missed.
• Reaction profiles with the activation energy arrow drawn from the axis instead of the reactants line.
• “The particles have more energy so they collide more”: this gets only one of the two temperature marks.
• Describing equilibrium as “the reaction has stopped” or “the concentrations are equal”.
• Counting moles of all substances instead of moles of gas when deciding the effect of pressure.
• Saying a catalyst increases the yield.

Official syllabus

Pearson Edexcel International GCSE in Chemistry (4CH1), Specification, Issue 3, September 2024, published by Pearson Education Limited. Topic 3: Physical chemistry, points 3.1–3.22.