AQA A-Level Chemistry: Energetics (7405) – Revision Notes

For full explanations and worked examples, read the Energetics study guide first. These notes condense 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. They cover 3.1.4.1 to 3.1.4.4 and Required practical 2. Nothing in this section is “A-level only”: it is examined in AS Papers 1 and 2 and A-level Papers 1 and 2, and A-level Paper 3 can test any content.

Course links: AQA A-level Chemistry hub | printable checklist | Energetics practice questions | the mole revision notes | free diagnostics

3.1.4.1 Definitions to learn word for word

Term	Definition
Exothermic	Heat energy is transferred to the surroundings; ΔH is negative
Endothermic	Heat energy is taken in from the surroundings; ΔH is positive
Enthalpy change, ΔH	The heat energy change measured under conditions of constant pressure
Standard conditions	100 kPa and a stated temperature, e.g. ΔH⦵₂₉₈ for 298 K
Standard enthalpy of combustion, ΔcH⦵	The enthalpy change when 1 mole of a substance is completely burned in oxygen under standard conditions, 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, all reactants and products in their standard states
Mean bond enthalpy	The enthalpy change to break 1 mole of a stated covalent bond in gaseous molecules, averaged over a range of compounds
Hess’s law	The enthalpy change for a reaction is independent of the route taken, provided the initial and final conditions are the same

Writing the equations

• Formation: elements in their standard states on the left, exactly 1 mol of the compound on the right. Example: C(s) + 2H₂(g) → CH₄(g).
• Combustion: exactly 1 mol of the fuel on the left, with enough O₂ (fractions allowed) for complete combustion. Example: C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(l).
• Elements in their standard states have ΔfH⦵ = 0 (O₂(g), C(s), Br₂(l), Na(s)). Br₂(g) and O₃(g) are not in their standard states, so their ΔfH⦵ is not zero.

Formula table

Use	Formula	Units and notes
Heat change	q = mcΔT	m in g (water or solution), c in J g⁻¹ K⁻¹, ΔT in K; q in J
Molar enthalpy change	ΔH = −q ÷ n	q in kJ; n = moles of limiting reactant
Hess’s law, formation data	ΔH = ΣΔfH⦵(products) − ΣΔfH⦵(reactants)	multiply each value by its coefficient
Hess’s law, combustion data	ΔH = ΣΔcH⦵(reactants) − ΣΔcH⦵(products)	note the reversed order
Bond enthalpies	ΔH = Σ(bonds broken) − Σ(bonds made)	gases only; gives an approximate value

Learn the two Hess’s law shortcuts as “formation: products first” and “combustion: reactants first”. Better still, sketch the cycle so you can see the arrow directions.

3.1.4.2 Calorimetry: method in steps

1. q: q = mcΔT, with m = mass of water or solution (assume 1 cm³ of dilute solution has a mass of 1 g if told to).
2. Convert q from J to kJ (÷ 1000).
3. Moles: n = mass ÷ Mr, or n = concentration × volume (dm³). Use the limiting reactant.
4. Divide: ΔH = q ÷ n.
5. Sign: temperature went up → ΔH negative; temperature went down → ΔH positive.
6. Significant figures: match the least precise measurement in the data.

Required practical 2: measurement of an enthalpy change

• Suggested reactions in the specification: dissolving KCl or Na₂CO₃, NaOH + HCl, CuSO₄ + Zn, combustion of alcohols.
• Solution reactions: polystyrene cup with a lid; record temperature every 30 s before mixing to get a steady start value, then after mixing until the temperature has peaked and started to fall.
• Plot temperature against time and extrapolate the cooling line back to the time of mixing. This corrects for heat lost while the temperature was rising.
• Combustion of alcohols: weigh the spirit burner before and after, use a copper can of known mass of water, stir, and shield from draughts. Main errors: heat loss, incomplete combustion, evaporation of alcohol from the wick.

3.1.4.3 Hess’s law: method in steps

1. Write the target equation with state symbols.
2. Decide which data you have: ΔfH⦵ or ΔcH⦵.
3. Draw the cycle. For formation, arrows go up from the elements. For combustion, arrows go down to the combustion products.
4. Multiply each value by the number of moles in the equation. Elements in their standard states contribute zero to formation sums.
5. Follow the arrows: going against an arrow reverses its sign.

Calorimetry plus Hess’s law. When a reaction cannot be done directly in a cup (for example hydration of MgSO₄ or CuSO₄, or thermal decomposition of NaHCO₃), measure two reactions that can, such as dissolving the anhydrous and the hydrated salt, then combine them in a cycle: ΔH(hydration) = ΔH(dissolve anhydrous) − ΔH(dissolve hydrated).

3.1.4.4 Bond enthalpies: method in steps

1. Draw displayed formulas so you can count every bond.
2. Add up all bonds broken in the reactants (endothermic, positive).
3. Add up all bonds made in the products (exothermic).
4. ΔH = broken − made.
5. You may cancel bonds that are identical on both sides first.

Why bond enthalpy values differ from Hess’s law values (two points):

• Mean bond enthalpies are averages over many compounds, not the actual values for the bonds in these molecules.
• The method assumes every substance is gaseous; if any are liquids or solids, enthalpy changes of state are ignored.

Must-know distinctions

• ΔH vs q. q is heat energy in J for the amount used in the experiment. ΔH is per mole, in kJ mol⁻¹, with a sign.
• m in q = mcΔT vs mass of reactant. m is what changes temperature (the water or solution). The reactant’s mass is only used to find moles.
• Standard state vs “gas”. H₂O’s standard state at 298 K is liquid. Using H₂O(g) changes the value.
• Formation vs combustion cycles. Different arrow directions, so the subtraction is reversed.
• Bond enthalpy vs Hess value. Bond enthalpy gives an approximate value for gases; Hess’s law with formation or combustion data gives a more accurate value for the actual states.
• Exothermic temperature change vs ΔH sign. The surroundings get hotter, but ΔH is negative.

Quick self-test

1. Define standard enthalpy of formation.
2. Which of these has ΔfH⦵ = 0: O₂(g), O₃(g), H₂O(l)?
3. 25.0 g of water is heated by 12.0 K. c = 4.18 J g⁻¹ K⁻¹. Calculate q in kJ.
4. That heat comes from a reaction in which 0.0250 mol of the limiting reactant reacted, and the temperature rose. Calculate ΔH.
5. Calculate ΔcH⦵ of ethanol, C₂H₅OH(l) + 3O₂(g) → 2CO₂(g) + 3H₂O(l), given ΔfH⦵ (kJ mol⁻¹): C₂H₅OH(l) −277, CO₂(g) −394, H₂O(l) −286.
6. Calculate ΔH for C₂H₄(g) + H₂(g) → C₂H₆(g) from ΔcH⦵ (kJ mol⁻¹): C₂H₄ −1411, H₂ −286, C₂H₆ −1560.
7. Estimate ΔH for H₂(g) + Cl₂(g) → 2HCl(g). Mean bond enthalpies (kJ mol⁻¹): H–H 436, Cl–Cl 243, H–Cl 432.
8. Give two reasons why a bond enthalpy estimate differs from a value found using Hess’s law.
9. State the units of c in q = mcΔT, and what ΔT must be measured in.
10. State the pressure used for standard enthalpy changes.
11. 2.24 g of KCl (0.0300 mol) dissolves in 50.0 g of water and the temperature falls by 3.4 K. c = 4.18 J g⁻¹ K⁻¹. Calculate the enthalpy change of solution.
12. Why do you extrapolate a temperature-time graph back to the time of mixing?

Answers

1. 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.
2. O₂(g) only. O₃ is not the standard state of oxygen, and H₂O is a compound.
3. q = 25.0 × 4.18 × 12.0 = 1254 J = 1.25 kJ.
4. ΔH = −1.254 ÷ 0.0250 = −50.2 kJ mol⁻¹ (negative because the temperature rose).
5. 2(−394) + 3(−286) − (−277) = −788 − 858 + 277 = −1369 kJ mol⁻¹.
6. (−1411 + −286) − (−1560) = −137 kJ mol⁻¹ (reactants minus products).
7. Broken: 436 + 243 = 679. Made: 2 × 432 = 864. ΔH = 679 − 864 = −185 kJ mol⁻¹.
8. Bond enthalpies are mean values over many compounds; and the method assumes all species are gases.
9. J g⁻¹ K⁻¹; ΔT in K (a change of 1 °C is the same as 1 K).
10. 100 kPa.
11. q = 50.0 × 4.18 × 3.4 = 710.6 J = 0.7106 kJ. ΔH = +0.7106 ÷ 0.0300 = +23.7, so +24 kJ mol⁻¹ (2 s.f., positive because the temperature fell).
12. To correct for heat lost to the surroundings while the temperature was changing, giving a ΔT closer to the value with no heat loss.

Where marks are usually lost

• Using the mass of solid or fuel as m in q = mcΔT instead of the mass of water or solution.
• Not converting J to kJ before dividing by moles.
• Missing or wrong sign on ΔH, especially for endothermic dissolving.
• Using moles of the reagent in excess instead of the limiting reagent.
• Definitions missing “1 mole”, “standard conditions” or “standard states”, or saying “burned in air” rather than “completely burned in oxygen”.
• Formation equations written with 2 moles of product, or from compounds rather than elements.
• Subtracting the wrong way round in combustion-data Hess’s law questions.
• Forgetting to multiply bond enthalpies or ΔfH⦵ values by the coefficients in the equation.
• Defining mean bond enthalpy without “gaseous” or without “averaged over a range of compounds”.
• Quoting answers to more significant figures than the data justify.

Practise all of this on the Energetics practice questions.

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.