AQA A-Level Chemistry: Energetics (7405)

This study guide covers section 3.1.4 Energetics of the AQA AS and A-level Chemistry specification (7404/7405), version 1.2, July 2026, for AS and A-level exams June 2016 onwards. It teaches all four subsections: 3.1.4.1 Enthalpy change, 3.1.4.2 Calorimetry (including Required practical 2), 3.1.4.3 Applications of Hess’s law and 3.1.4.4 Bond enthalpies. None of this content is marked “A-level only”, so it is examined in both the AS and the A-level.

Use it with the Energetics revision notes and the Energetics practice questions. The AQA A-level Chemistry hub lists every topic, and the printable checklist lets you tick off each outcome. Every calculation here uses moles, so make sure you are secure on relative mass and the mole first. The bond-breaking ideas link back to Bonding.

What this topic covers

Spec section	What you must be able to do	Status
3.1.4.1 Enthalpy change	Recognise exothermic and endothermic reactions; know ΔH is the heat energy change at constant pressure; know standard conditions (100 kPa and a stated temperature); define ΔcH⦵ and ΔfH⦵	AS and A-level
3.1.4.2 Calorimetry	Use q = mcΔT to calculate a molar enthalpy change and in related calculations; Required practical 2: measurement of an enthalpy change	AS and A-level
3.1.4.3 Applications of Hess’s law	Use Hess’s law to calculate ΔH, including from enthalpies of combustion or of formation	AS and A-level
3.1.4.4 Bond enthalpies	Define mean bond enthalpy; use it to find an approximate ΔH for gas-phase reactions; explain why these values differ from Hess’s law values	AS and A-level

The specification places 3.1.4 in AS Paper 1 and AS Paper 2, and in A-level Paper 1 and Paper 2. A-level Paper 3 can test any content, including the practical work. At A-level, 3.1.8 Thermodynamics (A-level only) builds directly on this topic.

3.1.4.1 Enthalpy change

Exothermic and endothermic

• An exothermic reaction transfers heat energy to the surroundings. The surroundings get warmer and ΔH is negative. Combustion and neutralisation are exothermic.
• An endothermic reaction takes in heat energy from the surroundings. The surroundings cool and ΔH is positive. Thermal decomposition is endothermic.

Enthalpy change (ΔH) is the heat energy change measured under conditions of constant pressure. Its units are kJ mol⁻¹ when quoted per mole of the reaction as written.

Standard conditions

A standard enthalpy change, written with the symbol ⦵, refers to standard conditions: 100 kPa and a stated temperature. The temperature is shown as a subscript, for example ΔH⦵₂₉₈ for 298 K. Each substance must also be in its standard state: its normal physical state under those conditions (for example H₂O(l), Br₂(l) and C(s) at 298 K).

The two definitions you must learn

Standard enthalpy of combustion, ΔcH⦵: the enthalpy change when 1 mole of a substance is completely burned in oxygen under standard conditions, with all reactants and products in their standard states.

Standard enthalpy of formation, ΔfH⦵: the enthalpy change when 1 mole of a compound is formed from its elements under standard conditions, with all reactants and products in their standard states.

Two consequences follow from the definitions:

• ΔfH⦵ of an element in its standard state is zero, because forming it from itself is no change. So ΔfH⦵ of O₂(g) is 0, but ΔfH⦵ of Br₂(g) is not, because bromine’s standard state is liquid.
• The equation must show exactly 1 mole of the substance named, so fractions are allowed elsewhere.

ΔcH⦵ of methanol:   CH₃OH(l) + 1½O₂(g) → CO₂(g) + 2H₂O(l)
ΔfH⦵ of ethanol:    2C(s) + 3H₂(g) + ½O₂(g) → C₂H₅OH(l)

3.1.4.2 Calorimetry

The heat change, q, is given by

q = mcΔT

• m is the mass of the substance that changes temperature (usually the water or solution, not the mass of the reactant), in g.
• c is its specific heat capacity, in J g⁻¹ K⁻¹. You are not expected to recall c; the question gives it (4.18 J g⁻¹ K⁻¹ is the value usually given for water and dilute solutions).
• ΔT is the temperature change, in K. A change of 1 °C equals a change of 1 K.

With these units, q comes out in joules. Divide by 1000 for kJ.

To turn q into a molar enthalpy change:

ΔH = −q ÷ n        (q in kJ, n = moles of the reactant that limits the reaction)

The minus sign is there because a temperature rise means heat has left the reaction (exothermic, ΔH negative). If the temperature falls, ΔH is positive.

Worked example 1: neutralisation

50.0 cm³ of 1.00 mol dm⁻³ HCl is mixed with 50.0 cm³ of 1.10 mol dm⁻³ NaOH in a polystyrene cup. The temperature rises by 6.7 K. Take the solution’s mass as 100.0 g and c = 4.18 J g⁻¹ K⁻¹. Calculate the enthalpy change per mole of water formed.

q = mcΔT = 100.0 × 4.18 × 6.7 = 2800.6 J = 2.8006 kJ
n(HCl) = 1.00 × 50.0/1000 = 0.0500 mol   (NaOH is in excess, so HCl limits)
n(H₂O) = 0.0500 mol
ΔH = −2.8006 ÷ 0.0500 = −56.0 kJ mol⁻¹

The temperature change was given to only 2 significant figures, so report ΔH = −56 kJ mol⁻¹. The specification expects you to report results only to the limits of the least accurate measurement.

Worked example 2: combustion of an alcohol

A spirit burner of butan-1-ol (Mr = 74.0) loses 0.592 g as it heats 150 g of water. The water temperature rises by 25.0 K. c = 4.18 J g⁻¹ K⁻¹. Calculate ΔcH of butan-1-ol.

q = 150 × 4.18 × 25.0 = 15 675 J = 15.675 kJ
n(butan-1-ol) = 0.592 ÷ 74.0 = 0.00800 mol
ΔcH = −15.675 ÷ 0.00800 = −1959 kJ mol⁻¹ ≈ −1960 kJ mol⁻¹ (3 s.f.)

Here m is the mass of water, not of fuel. The value is less exothermic than a data-book value because of heat loss to the surroundings, incomplete combustion and evaporation of alcohol.

Required practical 2: measurement of an enthalpy change

The specification suggests reactions such as dissolving potassium chloride or sodium carbonate, neutralising NaOH with HCl, the displacement reaction between CuSO₄ and Zn, and combustion of alcohols. For a reaction in solution, a typical method is:

1. Measure a known volume of one solution into a polystyrene cup with a lid, inside a beaker for stability.
2. Record its temperature every 30 seconds for a few minutes to get a steady starting value.
3. At a recorded time, add the second reagent (or a weighed solid) and stir.
4. Keep recording the temperature every 30 seconds until it has clearly peaked and started to fall.
5. Plot temperature against time. Draw a best-fit line through the cooling points and extrapolate it back to the time of mixing. The difference between this extrapolated value and the starting temperature is ΔT.

Extrapolating corrects for heat lost while the mixture was heating up. The polystyrene and lid reduce heat transfer.

3.1.4.3 Applications of Hess’s law

Hess’s law: the enthalpy change for a reaction is independent of the route taken, provided the initial and final conditions are the same.

This lets you find ΔH for reactions you cannot measure directly. You build a cycle in which the direct route and an indirect route start and finish at the same substances.

Using enthalpies of formation

Both reactants and products can be formed from the same elements, so the cycle gives:

ΔH = ΣΔfH⦵(products) − ΣΔfH⦵(reactants)

Worked example 3. Calculate ΔH for Fe₂O₃(s) + 3CO(g) → 2Fe(s) + 3CO₂(g). Data (kJ mol⁻¹): ΔfH⦵ Fe₂O₃(s) = −824, CO(g) = −111, CO₂(g) = −394.

Products:  2(0) + 3(−394) = −1182
Reactants: (−824) + 3(−111) = −1157
ΔH = −1182 − (−1157) = −25 kJ mol⁻¹

Fe(s) is an element in its standard state, so its ΔfH⦵ is zero.

Using enthalpies of combustion

Both reactants and products burn to the same combustion products, so the arrows point down to the combustion products and the cycle gives:

ΔH = ΣΔcH⦵(reactants) − ΣΔcH⦵(products)

Worked example 4. Calculate ΔfH⦵ of propane: 3C(s) + 4H₂(g) → C₃H₈(g). Data (kJ mol⁻¹): ΔcH⦵ C(s) = −394, H₂(g) = −286, C₃H₈(g) = −2220.

Reactants: 3(−394) + 4(−286) = −1182 − 1144 = −2326
Product:   −2220
ΔfH⦵ = −2326 − (−2220) = −106 kJ mol⁻¹

The subtraction is the opposite way round to the formation rule, so draw the cycle every time.

Hess’s law with calorimetry

The specification’s suggested practicals include the hydration of CuSO₄ or MgSO₄ and the thermal decomposition of NaHCO₃. You cannot measure these directly in a cup: it is hard to add exactly the right amount of water to a solid and measure its temperature change, and a decomposition needs continuous heating. Instead you measure two reactions that you can do in a cup and combine them.

Worked example 5. For CuSO₄(s) + 5H₂O(l) → CuSO₄·5H₂O(s):

ΔH₁: CuSO₄(s) + aq → CuSO₄(aq)                  measured: −67 kJ mol⁻¹
ΔH₂: CuSO₄·5H₂O(s) + aq → CuSO₄(aq) + 5H₂O(l)    measured: +12 kJ mol⁻¹
Route: anhydrous → solution directly = anhydrous → hydrated → solution
ΔH₁ = ΔH(hydration) + ΔH₂
ΔH(hydration) = ΔH₁ − ΔH₂ = −67 − (+12) = −79 kJ mol⁻¹

3.1.4.4 Bond enthalpies

Mean bond enthalpy: the enthalpy change when 1 mole of a stated covalent bond is broken in gaseous molecules, averaged over a range of different compounds that contain that bond.

Breaking bonds is always endothermic and making bonds is always exothermic, so:

ΔH = Σ(bond enthalpies of bonds broken) − Σ(bond enthalpies of bonds made)

This gives an approximate ΔH and only applies to reactions in the gaseous phase.

Worked example 6. Use mean bond enthalpies to estimate ΔH for C₂H₄(g) + H₂(g) → C₂H₆(g). Data (kJ mol⁻¹): C=C 612, C–C 348, C–H 412, H–H 436.

Broken: 1 C=C + 4 C–H + 1 H–H = 612 + 1648 + 436 = 2696
Made:   1 C–C + 6 C–H         = 348 + 2472       = 2820
ΔH = 2696 − 2820 = −124 kJ mol⁻¹

Cancelling the four C–H bonds present on both sides gives the same answer: 1048 − 1172 = −124.

Why bond enthalpy answers differ from Hess’s law answers

• Mean values. Each bond enthalpy is averaged over many compounds. The actual strength of a bond in one particular molecule differs from that average, because it depends on the rest of the molecule.
• States. Bond enthalpies apply to gases. If a substance in the real reaction is a liquid or solid (such as H₂O(l)), the bond enthalpy method ignores the enthalpy change of the change of state.

ΔfH⦵ and ΔcH⦵ data refer to the actual substances, so the Hess’s law value is more accurate.

Common errors

• Using the mass of the fuel or solid as m in q = mcΔT. Use the mass of the water or solution.
• Forgetting to convert J to kJ before dividing by moles, giving an answer 1000 times too big.
• Leaving out the sign of ΔH, or giving a positive value when the temperature rose.
• Dividing by moles of the reagent in excess rather than the limiting one.
• Writing a formation equation from compounds (such as CO₂ or H₂O) instead of elements, or forming more than 1 mole of the product.
• Giving ΔfH⦵ of an element a non-zero value, or giving O₃(g) or Br₂(g) a value of zero.
• Reversing the subtraction in combustion-based Hess’s law calculations.
• Miscounting bonds: draw displayed formulas and multiply by each coefficient.
• Quoting more significant figures than the least precise measurement allows.

Next steps

Test yourself on the practice questions, read exam preparation, or try the free diagnostics.

Official syllabus

AQA, AS and A-level Chemistry specification (7404/7405), version 1.2, July 2026, for AS and A-level exams June 2016 onwards: section 3.1.4 Energetics.