What is the definition of energetics in the context of Edexcel IGCSE Chemistry?

In Edexcel IGCSE Chemistry, energetics refers to the study of energy changes that occur during chemical reactions. It focuses on understanding how energy is absorbed or released in the form of heat, and how this affects the reaction process.

How do exothermic and endothermic reactions differ in energetics?

Exothermic reactions release energy to the surroundings, usually in the form of heat, resulting in an increase in temperature. In contrast, endothermic reactions absorb energy from the surroundings, leading to a decrease in temperature. These concepts are fundamental in studying energetics within the Edexcel IGCSE Chemistry curriculum.

What is enthalpy change and how is it measured in energetics?

Enthalpy change, denoted as ΔH, is the heat change at constant pressure during a chemical reaction. It is measured in kilojoules per mole (kJ/mol). In energetics, a negative ΔH indicates an exothermic reaction, while a positive ΔH signifies an endothermic reaction. This measurement is crucial for understanding energy changes in chemical processes.

Why is the concept of bond energy important in energetics?

Bond energy is the amount of energy required to break one mole of a specific type of bond in a gaseous molecule. In energetics, understanding bond energies helps predict whether a reaction is exothermic or endothermic by comparing the energy required to break bonds in reactants with the energy released when new bonds form in products.

How is a calorimeter used to study energetics in Edexcel IGCSE Chemistry?

A calorimeter is an apparatus used to measure the heat change during a chemical reaction. In Edexcel IGCSE Chemistry, students use calorimeters to determine the enthalpy change of reactions by measuring temperature changes in a controlled environment, allowing them to calculate energy changes and understand the energetics of reactions.

Why is "Using $\Delta H = \sum(\text{made}) - \sum(\text{broken})$ (the wrong way round)" a common mistake in Energetics ?

Why it happens: Confusion about the sign convention. How to avoid it: Remember: BROKEN minus MADE. Bonds-broken come FIRST (you have to break the reactants before you can make the products). If you reverse it, the sign of ΔH will be wrong. Mnemonic: 'BB-MM' (broken before made). Source: 4CH1 Examiner Reports 2022-2024

Why is "Forgetting the negative sign on ΔH for exothermic reactions" a common mistake in Energetics ?

Why it happens: Treating ΔH as just a magnitude. How to avoid it: ALWAYS state the SIGN of ΔH. Negative = exothermic (heat OUT); positive = endothermic (heat IN). Even if the calculation is perfect, dropping the sign loses a mark. Source: 4CH1 Examiner Reports 2023-2024

Why is "Forgetting to count ALL the bonds (e.g. 4 × C–H in CH₄, 2 × O=O in 2O₂)" a common mistake in Energetics ?

Why it happens: Reading the equation too quickly. How to avoid it: Draw out the FULL structural formula of every reactant and product, then COUNT each bond type. Multiply by the stoichiometric coefficient (e.g. 2H₂O has 2 × 2 = 4 O–H bonds). Methane combustion: 4(C–H) + 2(O=O) on the left, 2(C=O) + 4(O–H) on the right.

Why is "Using mass of FUEL (not mass of water) in Q = mcΔT" a common mistake in Energetics ?

Why it happens: Confusion about what is being heated. How to avoid it: In Q = mcΔT, **m is the mass of the WATER** being heated by the reaction — NOT the mass of fuel burned. The fuel mass is only used at the end to find moles burned. For neutralisation calorimetry, m is the TOTAL VOLUME of solution (acid + alkali) treated as water (1 g/cm³). Source: 4CH1 Examiner Reports 2022-2024

Why is "Converting ΔT from °C to K by adding 273 (wrong for differences)" a common mistake in Energetics ?

Why it happens: Misremembering temperature conversions. How to avoid it: A temperature DIFFERENCE in °C is NUMERICALLY EQUAL to the same difference in K (the +273 cancels). So if ΔT = 25 °C, then ΔT = 25 K (not 298 K). Use whichever unit feels natural — they're the same magnitude.

Why is "Saying exothermic reactions don't need activation energy" a common mistake in Energetics ?

Why it happens: Thinking 'exothermic = energy out, so no input needed'. How to avoid it: ALL reactions need Ea to get started (bonds in reactants must first be broken). Even strongly exothermic combustion (methane + O₂) needs a spark to ignite — at room T, almost no molecules have enough KE to reach Ea. The exothermic ΔH only describes the END point of the reaction, not the JOURNEY. Source: 4CH1 Examiner Reports 2022-2024

Physical ChemistryPearson Edexcel IGCSE Chemistry 4CH1 Higher

Energetics

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Energetics — Pearson Edexcel International GCSE Chemistry 4CH1 Study Notes (2026 onwards, Spec Issue 4)

Energy changes in chemical reactions: exothermic (releases heat, ΔH negative, T of surroundings rises) vs endothermic (absorbs heat, ΔH positive, T falls). Reaction profile diagrams with activation energy (Ea). Calculating ΔH from mean bond energies — bond breaking endo, bond making exo. Calorimetry experiments measuring Q = mcΔT and then ΔH per mole.

At a glance

Exothermic = releases heat → T of surroundings RISES → ΔH NEGATIVE. Examples: combustion, neutralisation, respiration, hand warmers.
Endothermic = absorbs heat → T of surroundings FALLS → ΔH POSITIVE. Examples: photosynthesis, thermal decomposition, dissolving NH₄NO₃ (cold packs).
Reaction profile: reactants → Ea hump → products. Exo: products LOWER than reactants. Endo: products HIGHER.
Activation energy (Ea) = minimum energy for colliding particles to react (break initial bonds).
Bond breaking is always ENDOTHERMIC (energy IN to pull atoms apart).
Bond making is always EXOTHERMIC (energy OUT as atoms reach lower-energy bonded state).
ΔH from bond energies: $\Delta H = \sum(\text{bonds broken}) - \sum(\text{bonds made})$ .
Calorimetry: $Q = mc\Delta T$ (water c = 4.18 J/g/K); $\Delta H = -Q/n$ in kJ/mol.

What you’ll learn

Mapped to the Cambridge IGCSE 4CH1 syllabus (2026 onwards).

3.1 — Know that chemical reactions in which heat energy is given out are described as exothermic, and those in which heat energy is taken in are endothermic.
3.2 — Describe simple calorimetry experiments for reactions including combustion, displacement, dissolving and neutralisation in which temperature changes can be measured.
3.3 — Draw and explain energy level diagrams to represent exothermic and endothermic reactions, including the activation energy (Ea) and the enthalpy change (ΔH).
3.4 — Know that the breaking of bonds is endothermic and the making of bonds is exothermic.
3.5 — Use bond energies to calculate the enthalpy change in a reaction using $\Delta H = \sum(\text{bonds broken}) - \sum(\text{bonds made})$ .
3.6 — Calculate the molar enthalpy change (ΔH) from the heat energy change (Q) using $Q = mc\Delta T$ where m = mass of water in g, c = 4.18 J/g/K.
3.7 — Carry out and explain simple calorimetry experiments to measure the enthalpy change of neutralisation, combustion or dissolving.
3.8 — Understand that energy changes in chemical reactions are due to the breaking and making of chemical bonds, with overall ΔH determined by the balance between these.

Exothermic vs endothermic reactions (spec 3.1, 3.2)

Heat released = exo (T rises, ΔH negative). Heat absorbed = endo (T falls, ΔH positive).

Every chemical reaction involves an energy change. Heat is either released to the surroundings or absorbed from them — these two categories define the basic energetics:

Exothermic reactions RELEASE heat energy to the surroundings.

The temperature of the surroundings (a thermometer in the mixture, your hand on the test tube) RISES.
The enthalpy change ΔH is NEGATIVE (the chemical system loses energy).
Products contain LESS chemical (potential) energy than the reactants — the surplus energy is released as heat.

Examples of exothermic reactions:

Combustion: any fuel burning in oxygen — methane, propane, ethanol, petrol, coal, candle wax. All highly exothermic — this is why we use them as fuels.
Neutralisation: HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l). Temperature rises by ~ 5-7 °C for 1 mol/dm³ acid + alkali.
Respiration: glucose + O₂ → CO₂ + H₂O + energy. Powers all living things.
Hand warmers: Fe + O₂ + H₂O → Fe₂O₃ (rusting) — slowly releases heat over hours.
Reactive metal + acid: Mg + 2HCl → MgCl₂ + H₂; test tube feels noticeably warm.

Endothermic reactions ABSORB heat energy from the surroundings.

The temperature of the surroundings FALLS.
The enthalpy change ΔH is POSITIVE (the chemical system gains energy).
Products contain MORE chemical energy than the reactants — that 'extra' energy was taken from the surroundings.

Examples of endothermic reactions:

Photosynthesis: 6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂. Plants absorb light energy from the Sun and store it as chemical energy in glucose. ΔH ≈ +2802 kJ/mol.
Thermal decomposition: CaCO₃ → CaO + CO₂ (heated to ~ 900 °C); CuCO₃ → CuO + CO₂. The reaction only proceeds while you keep adding heat.
Dissolving ammonium nitrate in water: NH₄NO₃(s) + water → NH₄⁺(aq) + NO₃⁻(aq). Temperature drops by several degrees. This is the basis of instant cold packs for sports injuries.
Electrolysis: requires electrical energy to drive the otherwise non-spontaneous decomposition (e.g. 2H₂O → 2H₂ + O₂).

Key distinction. Exothermic and endothermic refer to what happens to the SURROUNDINGS:

Surroundings	Reaction type	ΔH	Test tube feels
Get HOTTER	EXOthermic	NEGATIVE	Warm
Get COLDER	ENDOthermic	POSITIVE	Cold

(Trick to remember: surroundings 'enjoy' exothermic — they get heat — so ΔH NEGATIVE = the system LOSES it. Endothermic 'eats' heat from surroundings — ΔH POSITIVE = system GAINS it.)

Exothermic: heat OUT, T surroundings ↑, ΔH negative.
Endothermic: heat IN, T surroundings ↓, ΔH positive.
Combustion, neutralisation, respiration → exothermic.
Photosynthesis, thermal decomposition, dissolving NH₄NO₃ → endothermic.

See the full worked example for energetics →

Reaction profile (energy level) diagrams (spec 3.3)

Reactants → Ea hump → products. Exo: down. Endo: up.

A reaction profile (or 'energy level diagram') is a graph showing how the chemical (potential) energy of the molecules changes as a reaction progresses. It has two axes:

X-axis: progress of reaction (or 'reaction coordinate') — left = reactants, right = products.
Y-axis: energy of the species.

Exothermic reaction profile.

Flat horizontal line at the LEFT, labelled 'reactants', at a HIGHER energy.
Smooth peak (the 'activation-energy hump') rising above the reactants line.
Flat horizontal line at the RIGHT, labelled 'products', at a LOWER energy (below the reactants line — that's the key feature of exothermic).

Net descent from left to right → energy is released → exothermic.

Endothermic reaction profile.

Flat horizontal line at the LEFT, labelled 'reactants', at a LOWER energy.
Smooth peak rising above both reactant and product lines.
Flat horizontal line at the RIGHT, labelled 'products', at a HIGHER energy (above the reactants line).

Net ascent from left to right → energy is absorbed → endothermic.

Labelling the diagram — Ea and ΔH.

Two important quantities are marked on every reaction profile:

Activation energy (Ea) — the vertical distance from the REACTANTS line UP to the TOP of the peak. Draw a double-headed arrow from reactants line to peak; label 'Ea'.
Enthalpy change (ΔH) — the vertical distance from REACTANTS to PRODUCTS. Draw a double-headed arrow from the reactants line to the products line.
- Exothermic: arrow points DOWN; ΔH is NEGATIVE.
- Endothermic: arrow points UP; ΔH is POSITIVE.

Ea is the climb from reactants to the peak; ΔH is the reactants-to-products drop (exo) or rise (endo).

What activation energy means.

Activation energy is the minimum energy that colliding reactant particles must possess for a reaction to occur. Why is energy needed?

To break bonds in the reactants — even though new bonds will form later, the OLD bonds must break first (or weaken at the transition state).
To distort molecules into the geometry of the transition state.

Without Ea, no reaction occurs — even if the overall ΔH is highly negative. Methane and oxygen mixed at room T do not burn (Ea too high); a spark provides Ea → starts the reaction → it then sustains itself.

Effect of catalysts on the profile.

A catalyst provides an ALTERNATIVE PATHWAY with a LOWER activation energy. On the reaction profile:

The reactants and products lines stay AT THE SAME ENERGY (catalyst doesn't change ΔH).
The peak is LOWER (smaller hump) for the catalysed pathway.

This is often drawn as two overlaid peaks on the same diagram — one tall (uncatalysed Ea), one short (catalysed Ea).

Why even exothermic reactions need Ea.

This is a common point of confusion. The ΔH only tells you about the START and END energies. To get from start to end, you must climb OVER the activation-energy barrier first. Methane combustion has ΔH ≈ −890 kJ/mol (very exothermic) but Ea is high enough that methane doesn't spontaneously burn at room T — you need a spark, flame, or other ignition source to supply Ea for the first molecules. Once started, the heat released drives subsequent molecules over the barrier.

Analogy: pushing a heavy ball up a small hill. Once over the top, the ball rolls down a much steeper slope (giving back more energy than you put in). But without the initial push, the ball stays at the start.

Axes: x = progress of reaction; y = energy.
Exothermic: products LOWER than reactants. ΔH negative.
Endothermic: products HIGHER than reactants. ΔH positive.
Ea = arrow from reactants UP to peak.
ΔH = arrow from reactants to products (down for exo; up for endo).
Catalyst lowers the peak (Ea) but ΔH unchanged.

See the full worked example for energetics →

Bond breaking, bond making and ΔH from bond energies (spec 3.4, 3.5)

Breaking endo. Making exo. ΔH = bonds broken − bonds made.

Why do chemical reactions have an energy change? Because BONDS are being broken (energy IN) and made (energy OUT). Whether the overall reaction is exo or endo depends on which is bigger.

Bond breaking is always ENDOTHERMIC.

A covalent bond is a strong electrostatic attraction between bonded atoms (sharing electrons). To break it, you must PULL the atoms apart against this attraction — which requires WORK / ENERGY. The energy comes from the surroundings (heat, light, etc.).

Bond energy of a particular bond = energy needed to BREAK 1 mole of that bond.
Bond energies are always POSITIVE (e.g. C–H bond energy = 412 kJ/mol).
Breaking 1 mol of C–H bonds: $\text{CH}_4 \rightarrow \text{CH}_3\bullet + \text{H}\bullet$ , ΔH = +412 kJ/mol (endo).

Bond making is always EXOTHERMIC.

When atoms come together to form a bond, they reach a more stable, LOWER-energy state. The 'lost' energy is RELEASED to the surroundings as heat. The magnitude is the same as bond-breaking but with opposite sign.

Making 1 mol of C–H bonds: $\text{CH}_3\bullet + \text{H}\bullet \rightarrow \text{CH}_4$ , ΔH = −412 kJ/mol (exo).

Overall reaction — bonds broken minus bonds made.

A reaction involves breaking some bonds AND making others. The OVERALL ΔH is the SUM:

$\Delta H_\text{reaction} = \sum (\text{bond energies of bonds BROKEN}) - \sum (\text{bond energies of bonds MADE})$

If more energy is RELEASED by making new bonds than is ABSORBED by breaking old bonds → ΔH negative → exothermic.
If more energy is ABSORBED breaking old bonds than is RELEASED making new bonds → ΔH positive → endothermic.

ΔH = Σ(bonds broken) − Σ(bonds made): exothermic when more energy is released making bonds than is used breaking them.

Worked example: combustion of methane.

$\text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O}$

Mean bond energies (kJ/mol): C–H = 412, O=O = 496, C=O = 743, O–H = 463.

Step 1 — count bonds broken:

4 × C–H = 4 × 412 = 1648 kJ
2 × O=O = 2 × 496 = 992 kJ
TOTAL broken = 1648 + 992 = 2640 kJ

Step 2 — count bonds made:

2 × C=O = 2 × 743 = 1486 kJ
4 × O–H = 4 × 463 = 1852 kJ
TOTAL made = 1486 + 1852 = 3338 kJ

Step 3 — calculate ΔH:

$\Delta H = 2640 - 3338 = -698\ \text{kJ/mol}$

Step 4 — interpret: ΔH is NEGATIVE → EXOTHERMIC. More energy released making new bonds than absorbed breaking old bonds → 698 kJ released per mole of CH₄ burnt.

This calculated value is close to the experimental ΔH_combustion of methane (≈ −890 kJ/mol). The discrepancy comes from using MEAN bond energies (averages) rather than the exact bond energies in CH₄, CO₂ and H₂O — different molecules have slightly different bond energies depending on the chemical environment.

Mean bond energy — what does 'mean' mean.

The bond energy of C–H is quoted as 412 kJ/mol, but the EXACT energy varies depending on the molecule:

C–H in CH₄ ≈ 435 kJ/mol
C–H in CH₃Cl ≈ 423 kJ/mol
C–H in chloroform CHCl₃ ≈ 390 kJ/mol

The mean is an average over many compounds. Using it in calculations gives a reasonable estimate of ΔH, but not the exact experimental value. For exact ΔH, calorimetry (next section) is more reliable.

Sign convention shortcut. Bond energies in tables are quoted as POSITIVE values (energy to BREAK). When you USE them in ΔH = (bonds broken) − (bonds made), the SUBTRACTION takes care of the sign automatically. The 'bonds made' term gets the opposite sign because that energy is RELEASED, not absorbed.

Breaking bonds: ALWAYS endothermic (energy IN).
Making bonds: ALWAYS exothermic (energy OUT).
$\Delta H = \sum(\text{broken}) - \sum(\text{made})$ .
ΔH negative → more energy released making new bonds → exothermic.
Mean bond energies = averages over many molecules; not exact for any one compound.

See the full worked example for energetics →

Calorimetry — measuring ΔH experimentally (spec 3.6, 3.7)

Polystyrene cup or copper can; Q = mcΔT; ΔH per mol = Q ÷ n.

Calorimetry lets us MEASURE the ΔH of a reaction by measuring the temperature change it causes in a known mass of water (or solution). The principle is simple: heat released or absorbed by the reaction is transferred to/from the water, and we can calculate Q from the temperature change.

The fundamental formula.

$Q = m \times c \times \Delta T$

Q = heat energy transferred (J).
m = mass of water (or solution) being heated/cooled (g).
c = specific heat capacity of water = 4.18 J/g/K (memorise!).
ΔT = temperature change = T_final − T_initial (in °C or K — the size of 1 K = 1 °C, so the numerical value is the same).

Two key calorimetry setups in 4CH1.

Setup 1 — Neutralisation / dissolving / displacement (polystyrene cup).

Use a POLYSTYRENE CUP as the calorimeter (insulating — minimises heat loss).
Place inside a beaker or jar for extra insulation.
Add reactants directly to the cup (e.g. 50 cm³ HCl + 50 cm³ NaOH for neutralisation).
Stir with a thermometer.
Record max temperature reached → calculate ΔT.

For the mass m, ASSUME the solution has the density and specific heat capacity of water:

m (in g) = TOTAL volume of solution in cm³ (e.g. 100 g for 100 cm³).
c = 4.18 J/g/K.

Setup 2 — Combustion (spirit burner + copper can).

Use a COPPER calorimeter (a copper can or beaker) — copper CONDUCTS heat well, so the heat from the flame transfers efficiently to the water inside.
Place a known mass of water in the can.
Burn a SPIRIT BURNER containing the fuel (e.g. ethanol) directly under the can.
Measure initial and final temperature of the water → ΔT.
Measure mass of fuel burned (weigh burner before + after).

For the mass m in Q = mcΔT, use the MASS OF WATER (not mass of fuel — m refers to what's being heated).

A polystyrene cup minimises heat loss; a copper can conducts the flame's heat efficiently into the water.

Calculating ΔH per mole.

Once you have Q (in J), convert to kJ and divide by moles of reactant:

$\Delta H = -\dfrac{Q}{n}\quad \text{(in kJ/mol)}$

Q in kJ (divide J by 1000).
n = moles of the specified reactant (fuel for combustion; acid OR alkali for neutralisation, 1:1 ratio).
The NEGATIVE sign is essential when T rose (exothermic). If T fell (endothermic), use POSITIVE sign.

Worked example — neutralisation.

50.0 cm³ of 1.00 mol/dm³ HCl + 50.0 cm³ of 1.00 mol/dm³ NaOH; T rose from 21.0 °C to 27.8 °C. Find ΔH.

m = 50 + 50 = 100 g (total volume of solution treated as water).
ΔT = 27.8 − 21.0 = 6.8 °C.
Q = 100 × 4.18 × 6.8 = 2842 J = 2.84 kJ.
n(HCl) = 1.00 × 50/1000 = 0.0500 mol.
ΔH = −2.84 / 0.0500 = −56.8 kJ/mol (exothermic).

The textbook value for strong-acid–strong-alkali neutralisation is about −57 kJ/mol — close agreement.

Worked example — combustion of ethanol.

1.15 g ethanol burned; warms 200 g water by 25 °C. Find ΔH_c.

Q = 200 × 4.18 × 25 = 20,900 J = 20.9 kJ.
n(ethanol) = 1.15 / 46 = 0.0250 mol.
ΔH_c = −20.9 / 0.0250 = −836 kJ/mol (exothermic).

The DATA-BOOK value for ΔH_c (ethanol) is ≈ −1367 kJ/mol — the experimental answer is much smaller in magnitude because of HEAT LOSS to the surroundings (air, calorimeter walls, escaping draughts) and incomplete combustion. Improvements: use a draught shield, a tight-fitting copper calorimeter, and ensure good airflow.

Sources of error in calorimetry.

Heat loss to surroundings (biggest error — accounts for most of the discrepancy with data-book values).
Heat absorbed by the calorimeter itself (the cup or can warms up too).
Incomplete combustion (produces less heat than complete).
Mass / volume measurement errors (use a balance accurate to 0.01 g; pipette/burette for volume).
Thermometer resolution (use thermometer reading to 0.1 °C minimum).

Improvements:

Insulate (lid on the cup; wrap in cotton wool).
Use a polystyrene cup not glass (poor conductor).
Extrapolate the cooling curve back to the start to estimate the 'true' max temperature.
Repeat and average.

$Q = m \times c \times \Delta T$ . c (water) = 4.18 J/g/K.
m = mass of WATER being heated, NOT mass of fuel.
Polystyrene cup for neutralisation; copper can for combustion.
ΔH = -Q/n in kJ/mol. NEGATIVE if exothermic (T rose).
Experimental ΔH usually smaller magnitude than data-book due to heat loss.

See the full worked example for energetics →

Everyday applications and required practical (spec 3.7)

Hand warmers exo; cold packs endo; calorimetry measures both.

Energetics is not just textbook chemistry — exothermic and endothermic reactions are USED in practical applications all around us.

Exothermic applications.

Hand warmers / pocket warmers: contain Fe filings + NaCl + activated charcoal + sawdust + water. When opened to air, Fe oxidises: 4Fe + 3O₂ → 2Fe₂O₃ (this is RUSTING, but accelerated). Highly exothermic — releases enough heat for many hours of warmth. Reusable warmers use crystallisation of sodium acetate trihydrate (NaCH₃COO·3H₂O) — exothermic.
Self-heating cans (for drinks like soup or coffee): two-compartment can. Pull a tab to mix CaO + H₂O → Ca(OH)₂. Very exothermic. The reaction warms the food in the inner compartment.
Cooking and heating: every combustion reaction we use for cooking or heating (gas stove, oil heater, log fire) is exothermic.
Cement curing: when fresh cement reacts with water (hydration), the reaction is exothermic — a problem for very large concrete pours (heat must be dissipated).

Endothermic applications.

Instant cold packs (sports injuries, first aid): contain a sealed inner bag of water + outer area of NH₄NO₃ (solid). Squeeze the pack to break the inner bag → NH₄NO₃ dissolves: NH₄NO₃(s) + H₂O → NH₄⁺(aq) + NO₃⁻(aq), ΔH positive. Temperature drops by 10-20 °C → pack feels cold → reduces swelling and pain.
Sport ice cubes (silica-based 'ice'): use endothermic dissolving to cool drinks without dilution.
Some explosive disposal: certain endothermic reactions can absorb heat from exothermic side-reactions, controlling temperature.

4CH1 required practicals — calorimetry experiments.

You must be able to describe these:

(1) Enthalpy change of neutralisation.

Apparatus: polystyrene cup + lid + thermometer + 2 × 50 cm³ measuring cylinders.
Method: measure 50 cm³ of acid into cup; record temperature. Measure 50 cm³ of alkali separately; record temperature; pour into acid; record maximum temperature.
Calculation: Q = 100 × 4.18 × ΔT; n = cV; ΔH = −Q/n in kJ/mol.

(2) Enthalpy change of dissolving.

Apparatus: polystyrene cup + thermometer + electronic balance.
Method: 50 cm³ water in cup; record T. Add a measured mass of solid (e.g. 5.0 g NH₄NO₃ or KOH); stir; record min (or max) temperature.
Calculation: Q = 50 × 4.18 × ΔT (NH₄NO₃ dissolving: T falls → endothermic → ΔH positive).

(3) Enthalpy change of combustion.

Apparatus: copper can + thermometer + spirit burner + balance + draught shield (ideal).
Method: 100-200 g water in can; record initial T. Light spirit burner under can; let burn for measured time; record final T. Weigh burner before + after.
Calculation: Q = m_water × 4.18 × ΔT; n_fuel = m_fuel / M_r; ΔH_c = −Q/n.
Expect: experimental value is much SMALLER in magnitude than data-book due to heat loss (~ 50% loss is typical).

(4) Enthalpy change of displacement (e.g. Zn + CuSO₄).

Apparatus: polystyrene cup + thermometer.
Method: 50 cm³ CuSO₄ in cup; record T. Add excess Zn powder; stir; record max T.
Reaction: Zn + Cu²⁺ → Zn²⁺ + Cu (exothermic).

Tips for high-mark calorimetry questions.

Always state which substance is in EXCESS — needed to identify the limiting reagent (moles of which goes into n).
Insulate well; use a lid to reduce evaporation.
Stir with the thermometer to ensure even temperature.
Repeat each measurement 2-3 times and take the mean ΔT.
The cooling-curve extrapolation method (plot T vs time; extrapolate back to start of mixing) gives a more accurate 'max' temperature.

Comparing methods.

Method	Best for	Typical accuracy
Bond energies	Quick estimate	± 10-20% (uses averages)
Polystyrene-cup calorimetry	Neutralisation, dissolving, displacement	± 5-10%
Bomb calorimeter (A-level)	Combustion (eliminates heat loss)	± 1-2%
Data-book values	Exact reference	Highest

Hand warmers, self-heating cans, cooking = exothermic.
Cold packs (NH₄NO₃ dissolving) = endothermic.
4CH1 required practicals: neutralisation, dissolving, combustion calorimetry.
Always insulate to minimise heat loss; stir to even out temperature; repeat for reliability.

See the full worked example for energetics →

How it’s examined

Energetics appears on EVERY 4CH1 paper, typically worth 6-10 marks per session combined. Paper 1 (4CH1/1C) items: (a) define exothermic / endothermic with examples (3-5 marks); (b) draw and label a reaction profile, identifying Ea and ΔH (3-5 marks); (c) calorimetry calculation Q = mcΔT + ΔH per mol (5-6 marks); (d) bond energy ΔH calculation from a given equation (5-7 marks). Paper 2 (4CH1/2C) items: deeper calorimetry experiments with required-practical method descriptions (5-6 marks); explaining experimental vs data-book discrepancies (heat loss). The bond-energy calculation question is the most common 'big' question — ALWAYS count bonds carefully (4 × C–H in CH₄; 4 × O–H in 2H₂O).

Sources: Pearson Edexcel International GCSE Chemistry 4CH1 specification — Issue 4, September 2024 (valid for 2026 onwards); Pearson Edexcel 4CH1 Examiner Reports 2022-2024; 4CH1/1C May/Jun 2024 question paper and mark scheme; 4CH1/1C January 2024 question paper and mark scheme; 4CH1/1CR January 2024 question paper and mark scheme; 4CH1/2C May/Jun 2024 question paper and mark scheme; 4CH1/2C January 2024 question paper and mark scheme. Last reviewed 2026-05-12.

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Step-by-step worked examples — Energetics

Step-by-step solutions to past-paper-style questions on energetics , written exactly the way a tutor would explain them at the board.

1Calculate ΔH for the combustion of methane from bond energies (7 marks, Paper 1)
Core• Adapted from 4CH1/1C May/Jun 2024 Q7• bond energies, exothermic, spec-3.4, spec-3.5
Question
Methane burns completely in oxygen according to: $\text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O}$ . Use the mean bond energies (kJ/mol) given below to calculate the enthalpy change, ΔH, for this reaction. Bond energies: C–H = 412, O=O = 496, C=O = 743, O–H = 463. State whether the reaction is exothermic or endothermic and explain your answer. (7 marks)
Step-by-step solution
1. Step 1
  Count the bonds in the reactants (1 M). CH₄ contains 4 × C–H bonds. 2 × O₂ contains 2 × O=O bonds. Total to break:
  $\text{bonds broken} = 4(\text{C–H}) + 2(\text{O=O})$
2. Step 2
  Energy needed to break all reactant bonds (1 A). Bond breaking is ENDOTHERMIC — requires energy IN.
  $\sum E_\text{broken} = 4(412) + 2(496) = 1648 + 992 = 2640\ \text{kJ/mol}$
3. Step 3
  Count the bonds in the products (1 M). CO₂ contains 2 × C=O bonds (O=C=O). 2 × H₂O contains 2 × 2 = 4 × O–H bonds.
  $\text{bonds made} = 2(\text{C=O}) + 4(\text{O–H})$
4. Step 4
  Energy released when product bonds form (1 A). Bond making is EXOTHERMIC — releases energy OUT.
  $\sum E_\text{made} = 2(743) + 4(463) = 1486 + 1852 = 3338\ \text{kJ/mol}$
5. Step 5
  Apply the formula (1 B). $\Delta H = \sum(\text{bonds broken}) - \sum(\text{bonds made})$ .
  $\Delta H = 2640 - 3338 = -698\ \text{kJ/mol}$
6. Step 6
  Final value with correct sign and unit (1 B). ΔH = −698 kJ/mol. The negative sign is essential — it tells us heat is released.
7. Step 7
  Exothermic / endothermic conclusion (1 B). ΔH is NEGATIVE → reaction is EXOTHERMIC. Energy released when product bonds form (3338 kJ) is GREATER than energy absorbed to break reactant bonds (2640 kJ). The surplus 698 kJ is released to the surroundings as heat — this is why methane is used as a fuel.
Answer
ΔH = −698 kJ/mol. The reaction is EXOTHERMIC because more energy is released making product bonds than is absorbed breaking reactant bonds. The negative sign of ΔH always indicates an exothermic reaction.
Examiner tip
Edexcel 7-mark mark scheme: 1 each for (1) total bonds-broken expression; (2) correct total energy; (3) total bonds-made expression; (4) correct total energy; (5) ΔH formula bonds-broken minus bonds-made; (6) correct numerical answer with sign + unit; (7) exothermic conclusion with reasoning. ECF applies — even if arithmetic is wrong, the conclusion mark can still be earned for a correct sign-to-direction link. Common error: REVERSING the formula (bonds made − bonds broken) which inverts the sign.
2Drawing and labelling a reaction profile for an endothermic reaction (5 marks, Paper 1)
Core• Adapted from 4CH1/1C Jan 2024 Q5• energy profile, endothermic, spec-3.2, spec-3.3
Question
Photosynthesis is an endothermic reaction: $6\text{CO}_2 + 6\text{H}_2\text{O} \rightarrow \text{C}_6\text{H}_{12}\text{O}_6 + 6\text{O}_2$ . (a) Describe the shape of the reaction profile (energy level diagram) for this reaction, identifying the relative positions of the reactants and products. (b) On the profile, label the activation energy (Ea) and the enthalpy change (ΔH), stating the sign of ΔH. (c) Explain what is meant by activation energy. (5 marks)
Step-by-step solution
1. Step 1
  (a) Shape of the profile (1 M). The horizontal axis is the 'progress of reaction' (or 'reaction coordinate'); the vertical axis is 'energy'. The profile shows: • A flat line at the LEFT marked 'reactants' (6CO₂ + 6H₂O). • A HUMP (peak) showing the activation-energy barrier. • A flat line at the RIGHT marked 'products' (C₆H₁₂O₆ + 6O₂) — drawn HIGHER than the reactants because the reaction is endothermic.
2. Step 2
  (a) Why products higher than reactants (1 A). Endothermic = energy is ABSORBED from surroundings into the system. The chemicals gain potential (chemical) energy → products store MORE energy than the reactants did. Hence products line is drawn ABOVE reactants on the profile.
3. Step 3
  (b) Labelling Ea (1 A). The activation energy is the vertical distance from the REACTANTS line to the TOP of the hump. Draw a double-headed arrow from the reactants level UP to the peak and label it 'Ea'.
4. Step 4
  (b) Labelling ΔH and its sign (1 B). The enthalpy change is the vertical distance from the reactants line to the products line. Draw a double-headed arrow from reactants UP to products and label it 'ΔH' — for endothermic, ΔH is POSITIVE (the value would be quoted as, e.g., +2802 kJ/mol for photosynthesis).
5. Step 5
  (c) Activation energy explained (1 B). Activation energy (Ea) is the MINIMUM amount of energy that colliding particles must possess for a reaction to occur — the energy needed to BREAK the bonds in the reactants and start the reaction. Even an exothermic reaction needs Ea to get going; a spark or heat supplies this initial energy (e.g. methane needs a flame to ignite). Photosynthesis gets its Ea (and the additional ΔH energy) from sunlight absorbed by chlorophyll.
Answer
(a) Profile: flat reactants line (low) → hump (peak) → flat products line drawn HIGHER than reactants because energy is absorbed (endothermic). (b) Ea = arrow from reactants UP to the peak; ΔH = arrow from reactants UP to products, sign POSITIVE. (c) Activation energy = minimum energy needed by colliding particles to react (energy to break the initial bonds).
Examiner tip
Edexcel 5-mark mark scheme: 1 for axes / general profile shape; 1 for products ABOVE reactants (endothermic key feature); 1 for correctly labelled Ea; 1 for correctly labelled ΔH with positive sign; 1 for Ea definition (minimum energy / break bonds). Most-common error: drawing the products BELOW the reactants (that would be an exothermic profile).
3Calorimetry: measure ΔH of neutralisation (6 marks, Paper 1)
Core• Adapted from 4CH1/1C May/Jun 2024 Q8• calorimetry, neutralisation, required practical, spec-3.6, spec-3.7
Question
A student measures the enthalpy change of neutralisation between 50.0 cm³ of 1.00 mol/dm³ hydrochloric acid and 50.0 cm³ of 1.00 mol/dm³ sodium hydroxide in a polystyrene cup. The temperature of the mixture rose from 21.0 °C to 27.8 °C. Specific heat capacity of water = 4.18 J/g/K; assume the solution has the density and specific heat capacity of water. (a) Calculate the heat released, Q, in joules. (b) Calculate the moles of acid reacted. (c) Calculate ΔH per mole of acid in kJ/mol, stating whether the reaction is exothermic or endothermic. (6 marks)
Step-by-step solution
1. Step 1
  (a) Total mass of solution (1 M). Total volume = 50.0 + 50.0 = 100.0 cm³. Density of water = 1.00 g/cm³, so mass m = 100.0 g.
2. Step 2
  (a) Temperature change (1 A). $\Delta T = 27.8 - 21.0 = 6.8\ ^\circ\text{C}$ (a temperature CHANGE in °C equals the change in K).
3. Step 3
  (a) Heat released Q = mcΔT (1 A).
  $Q = mc\Delta T = 100.0 \times 4.18 \times 6.8 = 2842\ \text{J} \approx 2840\ \text{J}$
4. Step 4
  (b) Moles of acid (1 B). $n = c \times V$ where V is in dm³.
  $n(\text{HCl}) = 1.00 \times \dfrac{50.0}{1000} = 0.0500\ \text{mol}$
5. Step 5
  (c) ΔH per mole (1 B). ΔH per mol = Q ÷ n. Convert Q to kJ first.
  $\Delta H = -\dfrac{Q}{n} = -\dfrac{2.842\ \text{kJ}}{0.0500\ \text{mol}} = -56.8\ \text{kJ/mol}$
6. Step 6
  (c) Exothermic / endothermic and final answer (1 B). Temperature INCREASED → reaction released heat → EXOTHERMIC → ΔH is NEGATIVE. Final: ΔH ≈ −56.8 kJ/mol (3 sf). This is close to the textbook value for strong-acid–strong-alkali neutralisation (~ −57 kJ/mol).
Answer
(a) Q = 100.0 × 4.18 × 6.8 = 2840 J (~ 2.84 kJ). (b) n(HCl) = 1.00 × 50.0/1000 = 0.0500 mol. (c) ΔH = −2.84/0.0500 = −56.8 kJ/mol (exothermic — temperature rose, heat released).
Examiner tip
Edexcel 6-mark mark scheme: 1 for total mass (combined volumes × 1 g/cm³); 1 for ΔT; 1 for Q = mcΔT calculation; 1 for moles of acid; 1 for ΔH per mole; 1 for sign + exothermic conclusion. The NEGATIVE sign on ΔH is essential — leaving it positive loses the final mark. ECF applies throughout. Don't forget to convert Q from J to kJ before dividing.
4Calorimetry: enthalpy of combustion of ethanol (6 marks, Paper 1)
Core• Adapted from 4CH1/1CR Jan 2024 Q9• calorimetry, combustion, ΔH, spec-3.6, spec-3.7
Question
A student burns ethanol (C₂H₅OH) in a spirit burner and uses the heat to warm 200 g of water from 22.0 °C to 47.0 °C. The mass of ethanol burned = 1.15 g. (M_r of ethanol = 46; specific heat capacity of water = 4.18 J/g/K). (a) Calculate the heat absorbed by the water. (b) Calculate the moles of ethanol burned. (c) Calculate the enthalpy change of combustion of ethanol in kJ/mol. State whether the reaction is exothermic or endothermic. (6 marks)
Step-by-step solution
1. Step 1
  (a) Q = mcΔT (1 M). Heat absorbed by water:
  $Q = mc\Delta T = 200 \times 4.18 \times (47.0 - 22.0) = 200 \times 4.18 \times 25 = 20{,}900\ \text{J} = 20.9\ \text{kJ}$
2. Step 2
  (b) Moles of ethanol (1 A). n = m / M_r.
  $n(\text{C}_2\text{H}_5\text{OH}) = \dfrac{1.15}{46} = 0.0250\ \text{mol}$
3. Step 3
  (c) ΔH per mol (1 A). Divide heat by moles burned.
  $\Delta H = -\dfrac{20.9}{0.0250} = -836\ \text{kJ/mol}$
4. Step 4
  (c) Sign and conclusion (1 B). Water heated up → heat released by the reaction → EXOTHERMIC → ΔH NEGATIVE. ΔH ≈ −836 kJ/mol.
5. Step 5
  Comparison to data-book value (1 B). The accepted ΔH_c (ethanol) ≈ −1367 kJ/mol. The experimental value (−836 kJ/mol) is much LESS exothermic — heat is lost to the surroundings (not all the heat warms the water; some warms the calorimeter / air / escapes as draught) and combustion is often incomplete (CO + soot form, not just CO₂).
6. Step 6
  Improvements to the experiment (1 B). (i) Use a draught shield to reduce heat loss to air; (ii) use a copper calorimeter and lid to capture more heat; (iii) ensure complete combustion (good air supply); (iv) keep the spirit burner close to the calorimeter. These reduce systematic heat loss and bring the result closer to the data-book value.
Answer
(a) Q = 200 × 4.18 × 25 = 20,900 J = 20.9 kJ. (b) n = 1.15/46 = 0.0250 mol. (c) ΔH = −20.9/0.0250 = −836 kJ/mol (exothermic — water warmed up, heat released). The value is much less negative than the data-book −1367 kJ/mol because of heat loss to the surroundings and incomplete combustion.
Examiner tip
Edexcel 6-mark mark scheme: 1 for Q = mcΔT; 1 for moles of ethanol; 1 for ΔH per mole; 1 for negative sign + exothermic; 1 for noting heat loss / incomplete combustion; 1 for at least one improvement. The 'why experimental ≠ data-book' question is high-frequency on Paper 1. ECF applies through the calculation chain.
5Bond breaking vs bond making (5 marks, Paper 1)
Core• Adapted from 4CH1/1C Jan 2024 Q4• bond energies, spec-3.4, spec-3.5
Question
(a) Explain why bond breaking is described as endothermic and bond making as exothermic. (b) For the reaction H₂ + Cl₂ → 2HCl, the bonds broken are 1 × H–H (436 kJ/mol) and 1 × Cl–Cl (243 kJ/mol). The bonds made are 2 × H–Cl (432 kJ/mol each). Calculate ΔH and state whether the reaction is exothermic or endothermic. (5 marks)
Step-by-step solution
1. Step 1
  (a) Bond breaking is endothermic (1 M). A covalent bond is a strong electrostatic attraction between the bonded atoms. Energy must be PUT IN (absorbed from the surroundings) to overcome this attraction and pull the atoms apart. Since energy is absorbed BY the system, this is ENDOTHERMIC (ΔH for breaking is positive).
2. Step 2
  (a) Bond making is exothermic (1 A). When new bonds form, atoms 'fall' into the attractive potential energy well — they reach a more stable, lower-energy state. The 'missing' energy is RELEASED to the surroundings as heat. Since energy is released BY the system, this is EXOTHERMIC (ΔH for making is negative).
3. Step 3
  (a) Overall reaction (1 A). Whether a reaction is overall exothermic or endothermic depends on whether the energy released in making new bonds is GREATER or LESS than the energy absorbed in breaking the old bonds. • If $\sum E_\text{made} > \sum E_\text{broken}$ → net energy out → exothermic (ΔH negative). • If $\sum E_\text{broken} > \sum E_\text{made}$ → net energy in → endothermic (ΔH positive).
4. Step 4
  (b) Calculation (1 B). Bonds broken: H–H + Cl–Cl = 436 + 243 = 679 kJ/mol. Bonds made: 2 × H–Cl = 2 × 432 = 864 kJ/mol.
  $\Delta H = 679 - 864 = -185\ \text{kJ/mol}$
5. Step 5
  (b) Conclusion (1 B). ΔH = −185 kJ/mol. Negative → EXOTHERMIC. More energy is released forming the two strong H–Cl bonds than is needed to break the H–H and Cl–Cl bonds.
Answer
(a) Bond breaking absorbs energy → endothermic (you need to PULL the atoms apart against attraction). Bond making releases energy → exothermic (new bonds reach a more stable, lower-energy state). (b) ΔH = (436 + 243) − (2 × 432) = 679 − 864 = −185 kJ/mol → EXOTHERMIC.
Examiner tip
Edexcel 5-mark mark scheme: 1 each for (a) bond breaking endothermic + (b) bond making exothermic + (c) the comparison rule, plus 2 for the calculation (workings + final answer with sign + exothermic label). Common error: stating it backwards (making endo / breaking exo) — remember 'making is mending' (atoms get to a happy place) → energy out.

Key Formulae — Energetics

The formulae you need to memorise for energetics on the Pearson Edexcel IGCSE 4CH1 paper, with every variable defined in plain English and a note on when to use it.

Enthalpy change from mean bond energies (spec 3.5)
$\Delta H = \sum (\text{bonds broken}) - \sum (\text{bonds made})$
$\Delta H$
Enthalpy change of reaction (kJ/mol). Negative → exothermic. Positive → endothermic.
$\sum (\text{bonds broken})$
Sum of mean bond energies for all bonds broken in reactants (endothermic — energy IN)
$\sum (\text{bonds made})$
Sum of mean bond energies for all bonds made in products (exothermic — energy OUT)
When to use
Use when given mean bond energies and a chemical equation, and asked to calculate ΔH. Count every bond in every reactant/product molecule (don't forget × 2 for diatomic gases, × 4 for CH₄ etc.). The sign matters — negative ΔH means exothermic, positive means endothermic.
Example
Combustion of methane $\text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O}$ : bonds broken = 4(C–H) + 2(O=O); bonds made = 2(C=O) + 4(O–H); $\Delta H = (4 \times 412 + 2 \times 496) - (2 \times 743 + 4 \times 463) = 2640 - 3338 = -698\ \text{kJ/mol}$ .
Heat energy from calorimetry (spec 3.7)
$Q = mc\Delta T$
$Q$
Heat energy transferred (J)
$m$
Mass of solution (or water) in g. For a polystyrene-cup neutralisation, assume the solution has the density of water → m (in g) = total volume in cm³.
$c$
Specific heat capacity of water = 4.18 J/g/K (this value is given in the data section of 4CH1 papers)
$\Delta T$
Temperature change = T_final − T_initial, in °C or K (the size of 1 °C = 1 K, so the numerical value is the same)
When to use
Use to find the heat released or absorbed in a calorimetry experiment. m is the mass of the water (or solution if assumed to be water) being heated/cooled — NOT the mass of fuel burned. Convert Q from J to kJ at the end (÷ 1000) for the ΔH per mole step.
Example
Burning 1.15 g ethanol heats 200 g water from 22.0 °C to 47.0 °C: $Q = 200 \times 4.18 \times (47.0 - 22.0) = 200 \times 4.18 \times 25 = 20{,}900\ \text{J} = 20.9\ \text{kJ}$ .
Enthalpy change per mole (spec 3.7)
$\Delta H = -\dfrac{Q}{n}$
$\Delta H$
Enthalpy change in kJ/mol (NEGATIVE if heat released = exothermic; POSITIVE if absorbed = endothermic)
$Q$
Heat energy from calorimetry, in kJ
$n$
Moles of the SPECIFIED reactant (fuel for combustion; acid OR alkali for neutralisation — they're equal in 1:1 reactions)
When to use
Final step of any calorimetry calculation. The negative sign is conventional: if temperature went UP (exothermic), the system LOST energy, so ΔH gets a minus. If temperature went DOWN (endothermic), Q is conventionally negative and ΔH ends up positive.
Example
200 g water heated by 20.9 kJ from burning 0.0250 mol ethanol: $\Delta H = -20.9 / 0.0250 = -836\ \text{kJ/mol}$ (exothermic — temperature rose).

Model Answers — Energetics

High-scoring sample answers for energetics on the Pearson Edexcel IGCSE 4CH1 paper, with examiner-style notes mapping each response to the mark scheme and assessment objectives.

Question 1

4CH1/1C-style — exothermic / endothermic definitions5 marks

Define an exothermic reaction and an endothermic reaction. For each, give an everyday example and describe what would be observed in terms of temperature change of the surroundings. (5 marks)

Model answer

Exothermic reaction (definition). A reaction that releases heat energy to the surroundings. The enthalpy change ΔH is NEGATIVE (the products have less chemical/potential energy than the reactants, and the 'extra' energy is released as heat). The temperature of the surroundings RISES during an exothermic reaction.

Everyday examples of exothermic reactions.

Combustion of any fuel: methane burning on a stove, petrol in a car engine, candle wax burning. All combustion reactions are exothermic — they release the chemical energy stored in the fuel as heat (and light).
Neutralisation of an acid by an alkali: e.g. HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l). The temperature of the solution rises (typically ~ 6-7 °C for 1 mol/dm³ acid + alkali).
Respiration: glucose + oxygen → carbon dioxide + water + energy. Releases the energy that keeps living things warm.
Hand warmers: iron filings + air + saltwater → iron oxide; the rusting reaction releases heat slowly, warming the hands.
Reactive metal + acid: e.g. magnesium + HCl — fizzing and noticeable warming of the test tube.

Observation. Touch the reaction vessel — it feels WARM or HOT. A thermometer placed in the reaction mixture would read a HIGHER temperature than at the start.

Endothermic reaction (definition). A reaction that absorbs heat energy from the surroundings. The enthalpy change ΔH is POSITIVE (products have more chemical energy than reactants — the extra energy was taken IN from the surroundings). The temperature of the surroundings FALLS during an endothermic reaction.

Everyday examples of endothermic reactions.

Photosynthesis: 6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂. Plants absorb light (energy from the surroundings — the Sun) and convert it into chemical energy stored in glucose.
Thermal decomposition: e.g. CaCO₃ → CaO + CO₂ (heated to ~ 900 °C); copper(II) carbonate green → black on heating. The reaction only goes if you keep supplying heat.
Dissolving ammonium nitrate (or other ammonium salts) in water: NH₄NO₃(s) + H₂O → NH₄⁺(aq) + NO₃⁻(aq). The temperature DROPS — this is the basis of instant cold packs used for sports injuries (squeeze the pack to mix solid with water → cold compress).
Electrolysis: requires electrical energy input to drive the decomposition (an endothermic process).

Observation. Touch the reaction vessel — it feels COLD. A thermometer reads a LOWER temperature than at the start. (Note: this falling temperature is measured for the SURROUNDINGS — the chemicals themselves are gaining energy.)

Quick comparison table.

Feature	Exothermic	Endothermic
ΔH sign	Negative (−)	Positive (+)
Surroundings temperature	RISES	FALLS
Products vs reactants energy	Products LOWER	Products HIGHER
Examples	combustion, neutralisation, respiration, hand warmers	photosynthesis, thermal decomposition, dissolving NH₄NO₃, cold packs

Why this scores

Edexcel 5-mark mark scheme: 1 for exothermic definition (heat released, ΔH negative); 1 for endothermic definition (heat absorbed, ΔH positive); 1 for valid exothermic example (combustion / neutralisation / respiration); 1 for valid endothermic example (photosynthesis / thermal decomposition / dissolving NH₄NO₃); 1 for stating temperature change (rises vs falls). Mark scheme accepts 'gets warmer / colder' colloquially but the ΔH sign distinction often catches students out.

Question 2
4CH1/1C-style — reaction profiles5 marks
Draw and describe the energy level diagram (reaction profile) for an exothermic reaction. On your diagram, label the reactants, products, activation energy (Ea) and the enthalpy change (ΔH). Explain why the activation energy is needed even though the overall reaction releases energy. (5 marks)
Model answer
Description of the exothermic reaction profile.

Axes: horizontal axis is 'progress of reaction' (sometimes called 'reaction coordinate'); vertical axis is 'energy'.

Reactants: a flat horizontal line on the LEFT at a HIGHER energy level. Label this line 'reactants'.

Activation-energy hump: a smooth peak rising from the reactants level. The TOP of the peak represents the unstable transition state where old bonds are partly broken and new bonds are partly formed.

Products: a flat horizontal line on the RIGHT at a LOWER energy level (below the reactants line). Label this line 'products'.

Profile shape: reactants (high) → up to peak → down to products (low). Net descent → energy released → exothermic.

Labels on the diagram.

Ea (activation energy): a vertical arrow drawn from the reactants line UPWARD to the top of the peak. Label 'Ea'.

ΔH (enthalpy change): a vertical arrow drawn from the reactants line DOWNWARD to the products line. Label 'ΔH'. For an exothermic reaction, ΔH is NEGATIVE because the products are LOWER in energy than the reactants. Often the value would be quoted as, e.g., ΔH = −890 kJ/mol for the combustion of methane.

Why Ea is needed even for an exothermic reaction.

The 'overall' negative ΔH only tells us about the START and END energy levels — but to GET from reactants to products, the molecules must first travel UP and OVER the activation-energy barrier. Activation energy is the minimum energy that colliding particles must have for a reaction to occur — specifically, the energy needed to:

Break the bonds in the reactants (or partially weaken them in the transition state). Bond breaking is endothermic, so this initial step needs energy IN.

Reorient the molecules so that a successful collision (correct orientation, sufficient KE) can occur.

Once the molecules cross the barrier (form the transition state), they 'fall down the other side' and release the energy used in step 1 PLUS the extra ΔH energy. So even though the overall reaction releases energy, you need an INITIAL energy input (a spark, a flame, friction) to lift the first molecules over Ea.

Everyday illustration. Methane + oxygen is a STRONGLY exothermic reaction (ΔH ≈ −890 kJ/mol), yet methane sits in a gas pipe at room temperature without burning. Why? Because at room temperature, almost no molecules have enough KE to reach Ea. A small spark or flame supplies the initial energy → first few molecules react → they release enough heat to push the next set of molecules over Ea → the reaction is self-sustaining (a chain reaction). Without the spark, no reaction.

Analogy. Pushing a heavy box up a small hill onto a much steeper slope on the other side. You have to do work to get it to the top (Ea), but once over, it rolls down on its own and gives back MORE energy than you put in (ΔH < 0). Without the initial push, the box stays where it is.
Why this scores
Edexcel 5-mark mark scheme: 1 for axes / general shape; 1 for products LOWER than reactants (key exothermic feature); 1 for correctly labelled Ea on the up-arrow to the peak; 1 for correctly labelled ΔH with negative sign; 1 for explaining why Ea is needed (break bonds / minimum energy for collision). Common errors: drawing the peak too small (must be visible); confusing Ea and ΔH; saying 'exothermic reactions don't need activation energy' (WRONG — they all do).
Question 3
4CH1/1C-style — bond energy calculation5 marks
Hydrogen reacts with iodine in the gas phase: H₂(g) + I₂(g) → 2HI(g). Using the mean bond energies (kJ/mol): H–H = 436, I–I = 151, H–I = 299, (a) calculate the enthalpy change for this reaction and state whether it is exothermic or endothermic; (b) explain what is meant by 'mean bond energy'. (5 marks)
Model answer
(a) Bond energy calculation.

Bonds broken (in the reactants — endothermic):

1 × H–H = 436 kJ/mol

1 × I–I = 151 kJ/mol

Total broken = 436 + 151 = 587 kJ/mol

Bonds made (in the products — exothermic):

2 × H–I = 2 × 299 = 598 kJ/mol

Total made = 598 kJ/mol

Applying the formula:

$\Delta H = \sum (\text{bonds broken}) - \sum (\text{bonds made})$

$\Delta H = 587 - 598 = -11\ \text{kJ/mol}$

Sign and conclusion. ΔH is NEGATIVE → the reaction is EXOTHERMIC. More energy is released forming two H–I bonds than is absorbed breaking the H–H and I–I bonds. Net 11 kJ released per mole of reaction. (Note: this reaction is weakly exothermic and is actually reversible — H₂ + I₂ ⇌ 2HI — see Topic 3.3 on equilibria.)

(b) What 'mean bond energy' means.

A mean bond energy is the AVERAGE energy required to break one mole of a particular type of covalent bond (e.g. C–H, O–H, C=O), with all species in the gas phase. It is averaged over many different molecules in which that bond appears. For example, the C–H bond energy of 412 kJ/mol is the AVERAGE of all the C–H bonds in molecules like CH₄, C₂H₆, CH₃Cl, CH₃OH, benzene, etc.

Why it's an AVERAGE — important point. The exact energy needed to break a specific C–H bond depends on the rest of the molecule (the chemical environment) — the C–H bond in chloroform (CHCl₃) is slightly different from the C–H bond in methane. The 'mean' value is a useful compromise that works for typical molecules but is NOT perfectly accurate for any one specific molecule.

Implication for calculations. Using mean bond energies gives a CALCULATED ΔH that may differ slightly from the EXPERIMENTAL value measured by calorimetry. Bond energy calculations are quick estimates, not exact predictions. Also, bond energies only apply to gaseous reactions (state symbols matter — for liquids, you'd need to subtract the energy of liquefaction).

Sign convention. Bond energies are quoted as POSITIVE values (energy needed to BREAK the bond). When the bond is MADE, the SAME magnitude of energy is RELEASED — so in calculations, we subtract bonds-made from bonds-broken to get ΔH directly.
Why this scores
Edexcel 5-mark mark scheme: 1 for bonds-broken total; 1 for bonds-made total; 1 for ΔH = bonds broken − bonds made; 1 for correct numerical answer with sign + exothermic; 1 for explaining 'mean' bond energy (average over many molecules, gas phase only). Common error: forgetting to MULTIPLY 2 × H–I (because there are 2 product molecules each with 1 H–I bond, total 2 bonds).
Question 4
4CH1/2C-style — calorimetry experiment6 marks
Describe a calorimetry experiment to determine the enthalpy change of neutralisation between hydrochloric acid and sodium hydroxide. Include the apparatus, method, measurements taken, and how you would calculate ΔH per mole. Identify TWO sources of error and how to reduce them. (6 marks)
Model answer
Apparatus.

Polystyrene cup (expanded polystyrene 'foam' cup) — the calorimeter. Polystyrene is a poor conductor of heat, so very little heat escapes to the surroundings. The cup is placed inside a beaker or glass jar to add extra insulation and stability.

Lid with a small hole for the thermometer.

Thermometer reading to 0.1 °C (or 0.5 °C minimum).

Two 50 cm³ measuring cylinders (or two pipettes for higher accuracy).

Hydrochloric acid solution (1.00 mol/dm³, ~ 50 cm³).

Sodium hydroxide solution (1.00 mol/dm³, ~ 50 cm³).

Method.

Measure out 50.0 cm³ of HCl(aq) using a measuring cylinder and pour into the polystyrene cup.

Measure out 50.0 cm³ of NaOH(aq) into a second measuring cylinder.

Place the thermometer through the lid into the HCl in the cup. Record the temperature of the acid. (Record the temperature of the alkali in its measuring cylinder too — take the AVERAGE as the 'initial temperature'.)

QUICKLY add the alkali to the acid and replace the lid. STIR gently with the thermometer.

Record the HIGHEST temperature reached (the 'maximum temperature').

Calculate ΔT = T_max − T_initial.

Repeat for reliability and take the mean ΔT.

Calculation.

Total volume of mixture = 50.0 + 50.0 = 100.0 cm³ → mass m = 100.0 g (assuming density of solution = 1.00 g/cm³).

Heat released: $Q = m \times c \times \Delta T$ where c = 4.18 J/g/K.

Moles of acid (the limiting reagent here as 1:1 ratio, so either works): $n = c \times V = 1.00 \times 50.0/1000 = 0.0500\ \text{mol}$ .

Enthalpy change per mole: $\Delta H = -Q/n$ (in kJ/mol). The NEGATIVE sign because heat is released (exothermic, temperature rose).

Expected: ΔH ≈ −56 to −58 kJ/mol.

Two sources of error and how to reduce them.

Heat loss to the surroundings. Even with a polystyrene cup, some heat is lost to (a) the air through the open top / lid (b) the thermometer (c) the cup itself before it warms up.

Reduce by: using a tight-fitting lid, wrapping the cup in extra insulation (cotton wool, another beaker), or extrapolating the cooling curve back to the moment of mixing to get the 'true' maximum temperature.

Inaccurate volume / concentration measurements. Measuring cylinders have a ± 1 cm³ uncertainty for 50 cm³.

Reduce by: using a pipette (± 0.05 cm³) or a burette. Also use a calibrated thermometer.

Other improvements. (a) Stir to ensure uniform temperature. (b) Carry out repeats and take the mean. (c) Take temperature readings at fixed time intervals BEFORE and AFTER mixing to allow extrapolation.
Why this scores
Edexcel 6-mark mark scheme: 1 for polystyrene cup as insulating calorimeter; 1 for thermometer + measure ΔT correctly; 1 for Q = mcΔT calculation; 1 for moles of acid and ΔH per mole; 1 for ONE valid source of error; 1 for ONE valid improvement. Polystyrene cup is the expected calorimeter for neutralisation (NOT a metal one — heat would conduct away too easily). For COMBUSTION calorimetry (e.g. ethanol burner under copper can), the can IS copper to conduct heat from the flame INTO the water efficiently.
Question 5
4CH1/2C-style — bond energy / particle theory4 marks
Explain in terms of forces between atoms why bond breaking is always endothermic and bond making is always exothermic. Use the example of breaking the H–H bond in hydrogen gas. (4 marks)
Model answer
A covalent bond is a force of attraction. In a covalent bond, two atoms share a pair of electrons. The electron pair is attracted by BOTH nuclei (a 'positive nucleus pulls on the shared electrons'). This electrostatic attraction is what HOLDS the atoms together — it is the COVALENT BOND.

Bond breaking is endothermic — why.

To break the bond, you must PULL the two atoms apart so they no longer share their electron pair. To pull them apart against the electrostatic attraction, you must DO WORK on the system — supply energy (e.g. as heat or light). This energy gets transferred from the surroundings INTO the chemical system (the molecules gain potential energy as they separate).

Because energy is absorbed from the surroundings INTO the system, the process is ENDOTHERMIC. The 'bond energy' is the amount of energy needed to break one mole of that bond — for H–H, this is 436 kJ/mol.

In equation form (for hydrogen):

$\text{H}_2\text{(g)} \rightarrow 2\text{H(g)} \quad \Delta H = +436\ \text{kJ/mol}$

Positive ΔH → endothermic.

Bond making is exothermic — why.

When two atoms come together to form a bond, the electrostatic attraction PULLS them together. As they fall into the attractive 'well', they lose potential energy and reach a more stable, LOWER-energy state. The 'lost' potential energy is RELEASED to the surroundings as heat (or sometimes light or sound).

Because energy is released from the system to the surroundings, the process is EXOTHERMIC. The energy released when 1 mole of H–H bonds forms is the same magnitude as the energy needed to break them — 436 kJ/mol — but with the opposite sign.

In equation form:

$2\text{H(g)} \rightarrow \text{H}_2\text{(g)} \quad \Delta H = -436\ \text{kJ/mol}$

Negative ΔH → exothermic.

Why this is a UNIVERSAL rule. It applies to every covalent (and ionic) bond. The numbers vary (a C–H bond is 412 kJ/mol, a C=C double bond is 612 kJ/mol, an O=O bond is 496 kJ/mol etc.), but the principle is always: breaking ABSORBS energy; making RELEASES energy.

Net effect for a reaction. Whether a reaction is overall exothermic or endothermic depends on the BALANCE: if energy released making new bonds > energy absorbed breaking old bonds → exothermic; vice versa → endothermic. This is exactly what the formula $\Delta H = \sum(\text{bonds broken}) - \sum(\text{bonds made})$ captures.
Why this scores
Edexcel 4-mark mark scheme: 1 for stating a bond is an electrostatic attraction (or 'force of attraction'); 1 for breaking requires energy IN to overcome attraction → endothermic; 1 for making releases energy OUT as atoms reach lower energy state → exothermic; 1 for relating the magnitudes (same value either way for the same bond). Common error: saying 'breaking releases energy' (WRONG — that's MAKING). Mnemonic: 'Breaking = absorbing; Making = releasing'.

Key Definitions and Keywords — Energetics

Definitions to memorise and the exact keywords mark schemes credit for energetics answers — sharpened from recent examiner reports for the 2026 Pearson Edexcel IGCSE 4CH1 sitting.

Exothermic reaction
Examiner keyword
A reaction that releases heat energy to the surroundings. The temperature of the surroundings RISES. ΔH is NEGATIVE. Products have lower chemical (potential) energy than reactants. Examples: combustion, neutralisation, respiration, hand warmers.
Endothermic reaction
Examiner keyword
A reaction that absorbs heat energy from the surroundings. The temperature of the surroundings FALLS. ΔH is POSITIVE. Products have higher chemical (potential) energy than reactants. Examples: photosynthesis, thermal decomposition, dissolving ammonium nitrate (cold packs), electrolysis.
Enthalpy change (ΔH)
Examiner keyword
The energy change of a reaction at constant pressure, measured in kJ/mol. NEGATIVE for exothermic; POSITIVE for endothermic. Equal to: $\Delta H = E(\text{products}) - E(\text{reactants})$ = $\sum(\text{bonds broken}) - \sum(\text{bonds made})$ .
Activation energy (Ea)
Examiner keyword
The MINIMUM amount of energy that colliding particles must have for a reaction to occur — the energy required to break bonds in the reactants and form the transition state. Symbol Ea. Required for BOTH exothermic and endothermic reactions.
Reaction profile (energy level diagram)
Examiner keyword
A diagram showing the energy of the chemicals as a reaction progresses. X-axis = progress of reaction; Y-axis = energy. Shows reactants level, an activation-energy peak, and a products level. Products below reactants = exothermic; products above = endothermic.
Mean bond energy
Examiner keyword
The average energy required to BREAK 1 mole of a specified covalent bond in the gaseous state, averaged across many molecules containing that bond. Always POSITIVE (energy IN to break). Used to calculate ΔH from a chemical equation.
Bond breaking (energetics)
Examiner keyword
Always ENDOTHERMIC — energy must be put IN to overcome the electrostatic attraction holding the bonded atoms together. ΔH for breaking is positive (e.g. H–H bond: ΔH = +436 kJ/mol).
Bond making (energetics)
Examiner keyword
Always EXOTHERMIC — energy is RELEASED as atoms fall into the lower-energy bonded state. ΔH for making is negative (e.g. forming 1 mol H–H: ΔH = −436 kJ/mol).
Calorimetry
Examiner keyword
An experimental method for determining the heat energy transferred in a reaction by measuring the temperature change of water (or solution). Uses $Q = mc\Delta T$ where c (water) = 4.18 J/g/K. ΔH per mol = Q ÷ n.
Specific heat capacity (c)
Examiner keyword
The energy needed to raise the temperature of 1 g of a substance by 1 K (or 1 °C). For water, c = 4.18 J/g/K — used in all 4CH1 calorimetry calculations.

Common Mistakes and Misconceptions — Energetics

The traps other students keep falling into on energetics questions — taken from recent Pearson Edexcel IGCSE 4CH1 examiner reports and mark schemes — and how to avoid them.

✕Using $\Delta H = \sum(\text{made}) - \sum(\text{broken})$ (the wrong way round)
4CH1 Examiner Reports 2022-2024
Why it happens
Confusion about the sign convention.
How to avoid it
Remember: BROKEN minus MADE. Bonds-broken come FIRST (you have to break the reactants before you can make the products). If you reverse it, the sign of ΔH will be wrong. Mnemonic: 'BB-MM' (broken before made).
✕Forgetting the negative sign on ΔH for exothermic reactions
4CH1 Examiner Reports 2023-2024
Why it happens
Treating ΔH as just a magnitude.
How to avoid it
ALWAYS state the SIGN of ΔH. Negative = exothermic (heat OUT); positive = endothermic (heat IN). Even if the calculation is perfect, dropping the sign loses a mark.
✕Forgetting to count ALL the bonds (e.g. 4 × C–H in CH₄, 2 × O=O in 2O₂)
Why it happens
Reading the equation too quickly.
How to avoid it
Draw out the FULL structural formula of every reactant and product, then COUNT each bond type. Multiply by the stoichiometric coefficient (e.g. 2H₂O has 2 × 2 = 4 O–H bonds). Methane combustion: 4(C–H) + 2(O=O) on the left, 2(C=O) + 4(O–H) on the right.
✕Using mass of FUEL (not mass of water) in Q = mcΔT
4CH1 Examiner Reports 2022-2024
Why it happens
Confusion about what is being heated.
How to avoid it
In Q = mcΔT, m is the mass of the WATER being heated by the reaction — NOT the mass of fuel burned. The fuel mass is only used at the end to find moles burned. For neutralisation calorimetry, m is the TOTAL VOLUME of solution (acid + alkali) treated as water (1 g/cm³).
✕Converting ΔT from °C to K by adding 273 (wrong for differences)
Why it happens
Misremembering temperature conversions.
How to avoid it
A temperature DIFFERENCE in °C is NUMERICALLY EQUAL to the same difference in K (the +273 cancels). So if ΔT = 25 °C, then ΔT = 25 K (not 298 K). Use whichever unit feels natural — they're the same magnitude.
✕Saying exothermic reactions don't need activation energy
4CH1 Examiner Reports 2022-2024
Why it happens
Thinking 'exothermic = energy out, so no input needed'.
How to avoid it
ALL reactions need Ea to get started (bonds in reactants must first be broken). Even strongly exothermic combustion (methane + O₂) needs a spark to ignite — at room T, almost no molecules have enough KE to reach Ea. The exothermic ΔH only describes the END point of the reaction, not the JOURNEY.

Practice questions

Exam-style questions with step-by-step worked solutions. Try one before checking the method.

Past paper style quiz

Get a report showing which sub-topics you've nailed and which ones still need work.

Video lesson

Short walkthrough of the concepts students most often get stuck on.

Energetics — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Energetics

Short Study Notes in the form of Slides

Short Study Notes — Energetics

Detailed Notes

What you’ll learn

How it’s examined

1Calculate ΔH for the combustion of methane from bond energies (7 marks, Paper 1)

2Drawing and labelling a reaction profile for an endothermic reaction (5 marks, Paper 1)

3Calorimetry: measure ΔH of neutralisation (6 marks, Paper 1)

4Calorimetry: enthalpy of combustion of ethanol (6 marks, Paper 1)

5Bond breaking vs bond making (5 marks, Paper 1)

Enthalpy change from mean bond energies (spec 3.5)

Heat energy from calorimetry (spec 3.7)

Enthalpy change per mole (spec 3.7)

Exothermic reaction

Endothermic reaction

Enthalpy change (ΔH)

Activation energy (Ea)

Reaction profile (energy level diagram)

Mean bond energy

Bond breaking (energetics)

Bond making (energetics)

Calorimetry

Specific heat capacity (c)

✕Using ΔH=∑(made)−∑(broken)\Delta H = \sum(\text{made}) - \sum(\text{broken})ΔH=∑(made)−∑(broken) (the wrong way round)

✕Forgetting the negative sign on ΔH for exothermic reactions

✕Forgetting to count ALL the bonds (e.g. 4 × C–H in CH₄, 2 × O=O in 2O₂)

✕Using mass of FUEL (not mass of water) in Q = mcΔT

✕Converting ΔT from °C to K by adding 273 (wrong for differences)

✕Saying exothermic reactions don't need activation energy

Practice questions

Past paper style quiz

Assess your understanding

Video lesson

Energetics — frequently asked questions

✕Using $\Delta H = \sum(\text{made}) - \sum(\text{broken})$ (the wrong way round)