Pearson Edexcel International GCSE Chemistry 4CH1: Physical chemistry – Study Guide

This guide teaches Topic 3, Physical chemistry, of the Pearson Edexcel International GCSE Chemistry (4CH1) specification, Issue 3 (September 2024). It covers specification points 3.1–3.22: (a) Energetics, (b) Rates of reaction and (c) Reversible reactions and equilibria. The qualification is untiered. Statements with a “C” reference (bold in the specification) are assessed on Paper 2 only; everything else can appear on Paper 1 or Paper 2. This guide labels those points Paper 2 only.

Useful links: the 4CH1 Chemistry course hub, the printable 4CH1 checklist, the condensed Physical chemistry revision notes and the Physical chemistry practice questions. To find your weak spots first, take the free 4CH1 all-topics diagnostic.

Paper 1 (2 hours, 110 marks) assesses points without a C reference. Paper 2 (1 hour 15 minutes, 70 marks) assesses all content. A calculator may be used on both.

What this topic covers

Spec point	What you must be able to do	Paper
3.1	Define exothermic and endothermic	1 and 2
3.2, 3.8	Describe calorimetry for combustion, displacement, dissolving and neutralisation	1 and 2
3.3–3.4	Calculate Q = mcΔT, then ΔH in kJ/mol	1 and 2
3.5C–3.7C	Energy level diagrams; bond breaking and making; bond energy calculations	Paper 2 only
3.9–3.13, 3.15–3.16	Describe and explain rate experiments; catalysts	1 and 2
3.14C	Reaction profile diagrams showing ΔH and activation energy	Paper 2 only
3.17–3.18	Reversible reactions and the ⇌ symbol	1 and 2
3.19C–3.22C	Dynamic equilibrium; effect of catalyst, temperature and pressure	Paper 2 only

(a) Energetics

Exothermic and endothermic (3.1)

An exothermic reaction gives out heat energy to the surroundings, so the temperature of the surroundings rises. ΔH is negative. Combustion, neutralisation and most displacement reactions are exothermic.

An endothermic reaction takes in heat energy from the surroundings, so the temperature falls. ΔH is positive. Dissolving ammonium chloride or ammonium nitrate in water is endothermic.

Calorimetry experiments (3.2, 3.8)

Reactions in solution (neutralisation, displacement, dissolving):

1. Measure a known volume of solution into an insulated polystyrene cup and record its temperature.
2. Add the second reactant (the other solution, a measured mass of metal powder, or a solid to dissolve). Stir.
3. Record the highest (or lowest) temperature reached.

Combustion:

1. Weigh a spirit burner containing the fuel. Put a measured mass of water in a copper can above it and record the temperature.
2. Light the burner and heat, stirring, until the temperature has risen by about 20–30 °C.
3. Record the final temperature and reweigh the burner. The loss in mass is the mass of fuel burned.

Heat lost to the surroundings makes measured ΔH values less negative than data-book values. A lid, insulation and a draught shield reduce this.

Calculating Q and ΔH (3.3, 3.4)

Q = m × c × ΔT
Q  = heat energy change (J)
m  = mass of the substance being heated (g), usually the water or solution
c  = specific heat capacity (J/g/°C), given in the question
ΔT = temperature change (°C)

Then:

ΔH = −Q ÷ moles of the reactant   (temperature rose: exothermic, ΔH negative)
ΔH = +Q ÷ moles of the reactant   (temperature fell: endothermic, ΔH positive)
Convert Q from J to kJ (divide by 1000) so ΔH is in kJ/mol.

For a dilute solution, assume 1 cm³ has a mass of 1 g and c is the same as for water.

Worked example (neutralisation). 25.0 cm³ of 1.00 mol/dm³ hydrochloric acid is mixed with 25.0 cm³ of 1.00 mol/dm³ sodium hydroxide. The temperature rises by 6.8 °C. Take c = 4.2 J/g/°C. Calculate ΔH per mole of water formed.

Mass of solution = 25.0 + 25.0 = 50 g
Q = 50 × 4.2 × 6.8 = 1428 J = 1.428 kJ
Moles of HCl = 25.0/1000 × 1.00 = 0.0250 mol  → 0.0250 mol of water formed
ΔH = −1.428 ÷ 0.0250 = −57.12 → −57 kJ/mol

The mass is the total volume of both solutions, not just the acid.

Worked example (dissolving). 5.35 g of ammonium chloride (Mr = 53.5) dissolves in 50.0 g of water. The temperature falls by 7.0 °C. c = 4.2 J/g/°C.

Q = 50.0 × 4.2 × 7.0 = 1470 J = 1.47 kJ
Moles of NH₄Cl = 5.35 ÷ 53.5 = 0.100 mol
ΔH = +1.47 ÷ 0.100 = +14.7 kJ/mol   (positive: the temperature fell)

Energy level diagrams (3.5C) – Paper 2 only

An energy level diagram shows the energy of the reactants and products on a vertical axis.

Exothermic                          Endothermic
Energy                              Energy
 |  reactants ____                   |              ____ products
 |               |  ΔH negative      |             ↑
 |               ↓                   |  ΔH positive|
 |           ____ products           |  reactants ____
 +--------------------               +--------------------

For an exothermic reaction the products are lower than the reactants: energy has been released to the surroundings. For an endothermic reaction the products are higher. Label ΔH with an arrow from reactants to products.

Bond breaking and bond making (3.6C, 3.7C) – Paper 2 only

Breaking a bond needs energy: it is endothermic. Making a bond releases energy: it is exothermic. The overall ΔH depends on which is bigger.

ΔH = total energy of bonds broken − total energy of bonds made

Worked example. Calculate ΔH for CH₄ + 2O₂ → CO₂ + 2H₂O using these bond energies (kJ/mol): C–H 412, O=O 496, C=O 743, O–H 463.

Bonds broken: 4 × C–H + 2 × O=O = 4(412) + 2(496) = 1648 + 992 = 2640 kJ/mol
Bonds made:   2 × C=O + 4 × O–H = 2(743) + 4(463) = 1486 + 1852 = 3338 kJ/mol
ΔH = 2640 − 3338 = −698 kJ/mol

More energy is released making bonds than is taken in breaking them, so the reaction is exothermic. Draw displayed formulae first so you count every bond.

(b) Rates of reaction

Experiments (3.9, 3.15, 3.16)

Follow the volume of gas given off (gas syringe) or the loss in mass as gas escapes (flask on a balance).

Marble chips and hydrochloric acid (3.15). CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂. To test surface area, use the same mass of large chips, then small chips, with the same volume and concentration of acid at the same temperature. To test concentration, keep the chip size and mass the same and change only the acid concentration. Record the mass loss or gas volume every 30 s. To test temperature (3.9), warm the acid to different temperatures in a water bath, changing nothing else.

Catalysts for hydrogen peroxide (3.16). 2H₂O₂ → 2H₂O + O₂. Add the same mass of different solids (for example manganese(IV) oxide, copper(II) oxide, zinc oxide) to the same volume of hydrogen peroxide solution and collect the oxygen in a gas syringe. The faster the gas is produced, the better the catalyst. Filter, dry and reweigh the solid afterwards to show a catalyst is unchanged in mass.

Reading a graph. On a graph of gas volume against time, the steeper the curve, the faster the reaction. The curve levels off when one reactant is used up. The mean rate over a time interval is the change in volume divided by the time taken. If 36 cm³ is collected in the first 20 s, the mean rate is 36 ÷ 20 = 1.8 cm³/s. If the total is 54 cm³ at 40 s, the mean rate from 20 s to 40 s is 18 ÷ 20 = 0.9 cm³/s: the reaction slows as the reactants are used up.

Effects on rate and collision theory (3.10, 3.11)

Particles must collide with at least the activation energy to react. Anything that makes successful collisions more frequent increases the rate.

Change	Effect on rate	Explanation
Larger surface area (smaller pieces, powder)	Increases	More particles exposed, so more frequent collisions
Higher concentration of a solution	Increases	More particles in the same volume, so more frequent collisions
Higher pressure of a gas	Increases	Particles closer together, so more frequent collisions
Higher temperature	Increases	Particles move faster, so collide more often; more collisions have energy ≥ activation energy
Adding a catalyst	Increases	Alternative pathway with lower activation energy

Temperature has two effects and a full answer gives both.

Catalysts (3.12, 3.13)

A catalyst increases the rate of a reaction but is chemically unchanged at the end. It works by providing an alternative pathway with a lower activation energy, so a larger proportion of collisions are successful.

Reaction profile diagrams (3.14C) – Paper 2 only

A reaction profile shows energy against progress of reaction. The activation energy is the height from the reactants up to the peak. ΔH is the difference between reactants and products.

Energy        ___                 Exothermic profile
 |           /   \   ← peak
 |   Ea ↑   /     \
 | reactants       \
 |  ‾‾‾‾‾‾          \   ΔH (reactants → products, down)
 |                   \____ products
 +------------------------------ progress of reaction

With a catalyst the peak is lower but ΔH does not change. For an endothermic profile the products end higher than the reactants. Start the Ea arrow at the reactants line, not the axis.

(c) Reversible reactions and equilibria

Reversible reactions (3.17, 3.18)

Some reactions can go in both directions. The symbol ⇌ shows this.

Hydrated copper(II) sulfate. Heating blue hydrated copper(II) sulfate drives off water and leaves white anhydrous copper(II) sulfate. Adding water turns it blue again and the mixture gets hot.

CuSO₄·5H₂O(s) ⇌ CuSO₄(s) + 5H₂O(l)
    blue           white

The forward reaction is endothermic; the reverse is exothermic.

Ammonium chloride. Heating the white solid breaks it into two gases, which recombine as white solid near the cooler top of the tube.

NH₄Cl(s) ⇌ NH₃(g) + HCl(g)

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

In a sealed container nothing can escape, so a reversible reaction can reach dynamic equilibrium. At equilibrium:

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

“Dynamic” means both reactions are still happening. The concentrations are constant, not equal.

Changing the conditions (3.21C, 3.22C) – Paper 2 only

Catalyst. A catalyst speeds up the forward and reverse reactions by the same amount, so the position of equilibrium does not change. Equilibrium is simply reached sooner.

Temperature. Increasing the temperature shifts the position of equilibrium in the direction of the endothermic reaction. Decreasing it shifts it in the exothermic direction.

Pressure (gases). Increasing the pressure shifts the position of equilibrium in the direction that produces fewer moles of gas. Decreasing it favours more moles of gas. If both sides have the same number of moles of gas, pressure has no effect on the position.

You do not need to name Le Chatelier’s principle; the specification says references to it are not required. Use the two rules above.

Worked example. 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), forward reaction exothermic.

Moles of gas: left 2 + 1 = 3, right 2
Higher pressure → shifts right (fewer moles of gas) → more SO₃
Higher temperature → shifts left (endothermic direction) → less SO₃
Adding a catalyst → no change in the amount of SO₃ at equilibrium

Common errors in this topic

• Using the mass of the fuel or the solid in Q = mcΔT instead of the mass of water or solution.
• Leaving ΔH in J/mol, or forgetting to divide by the moles.
• Giving a positive ΔH for a reaction where the temperature rose.
• In bond energy sums, subtracting the wrong way round (made − broken) or missing the “2 ×” for a molecule that appears twice.
• Saying a catalyst “gives particles more energy”. It lowers the activation energy by providing another route.
• Saying higher temperature only makes particles “collide more”. Add that more collisions have energy equal to or greater than the activation energy.
• Writing that at equilibrium the concentrations are equal, or that the reaction has stopped.

Next steps

Test yourself with the Physical chemistry practice questions, then use the revision notes for the final weeks. Moles are in the Principles of chemistry study guide. The next topic is Organic chemistry.

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.