1. What an enthalpy change is
The system is the reacting chemicals. The surroundings are everything else: the solvent, the container, the air and the thermometer. Energy moves between the two as heat.
An enthalpy change, ΔH, is the heat energy change of a reaction measured at constant pressure. For a specified sample, the enthalpy change is in J or kJ. A molar enthalpy change, as quoted in these notes, is in kJ mol−1, and the mole basis (per mole of a named substance or of the equation as written) must be stated. We cannot measure the absolute enthalpy H of a substance, only changes in it.
Temperature is not enthalpy. Temperature tells you how hot something is. Enthalpy is about energy stored in the chemicals. A thermometer in a reaction mixture measures the surroundings warming or cooling, and we use that to work out ΔH.
2. Exothermic and endothermic
| Exothermic | Endothermic | |
|---|---|---|
| Heat flow | System → surroundings | Surroundings → system |
| Sign of ΔH | Negative | Positive |
| Enthalpy of products | Lower than reactants | Higher than reactants |
| Surroundings | Get warmer (temperature rises) | Get cooler (temperature falls) |
| Examples | Combustion, neutralisation, respiration | Thermal decomposition of CaCO3, photosynthesis, dissolving NH4NO3 |
3. Reaction profiles and activation energy
The activation energy, Ea, is the minimum energy colliding particles need for a reaction to happen. On a profile it is measured from the reactants up to the peak. ΔH is measured from the reactants to the products.
Diagram placeholder
Enthalpy profiles: exothermic (left) and endothermic (right)
Labels to include:
- Vertical axis: Enthalpy, H (no numbers)
- Horizontal axis: Progress of reaction
- Reactants level and products level as flat lines
- Exothermic: products below reactants; endothermic: products above reactants
- Ea: vertical arrow from the reactants level up to the peak
- ΔH: vertical arrow from the reactants level to the products level (pointing down = negative, up = positive)
- Dashed lower curve labelled 'with catalyst' and a shorter Ea arrow; reactant and product levels unchanged
Both profiles rise from the reactants to a single peak and then fall to the products. In the exothermic profile the products finish lower than the reactants, so the ΔH arrow points down. In the endothermic profile they finish higher, so the ΔH arrow points up and Ea must be larger than ΔH.
A catalyst provides an alternative route with a lower activation energy. It does not change the enthalpies of the reactants or products, so ΔH stays the same.
4. Standard conditions and states
The symbol ⦵ (often typed as °) means standard conditions:
- a pressure of 100 kPa;
- a stated temperature, which is usually 298 K but does not have to be (so write ΔH⦵298 when you need to be specific);
- each substance in its standard state: its normal physical state at 100 kPa and that temperature (for example H2O(l) and CO2(g) at 298 K);
- solutions at a concentration of 1 mol dm−3 (the A-level convention).
Always include state symbols in thermochemical equations: H2O(l) and H2O(g) have different enthalpies, so the ΔH changes if the state changes.
5. Standard enthalpy of formation, ΔfH⦵
The enthalpy change when 1 mole of a compound is formed from its elements in their standard states under standard conditions.
Fractions are allowed (and often needed) because the product side must be exactly 1 mol.
ΔfH⦵ of an element in its reference (standard) state is zero, because forming it from itself is no change. This is only true for that state: ΔfH⦵[C(s, graphite)] = 0, but diamond is not zero; ΔfH⦵[O2(g)] = 0, but ozone, O3(g), is not zero.
6. Standard enthalpy of combustion, ΔcH⦵
The enthalpy change when 1 mole of a substance is completely burned in oxygen, with all reactants and products in their standard states under standard conditions.
At 298 K the water must be written as H2O(l) for a standard enthalpy of combustion. Combustion values are always negative. Note that ΔcH⦵ of hydrogen is the same reaction as ΔfH⦵ of liquid water.
7. Enthalpy of neutralisation
The enthalpy change when an acid and a base react to form 1 mole of water under stated conditions. It is defined per mole of water, not per mole of acid.
8. Scaling and reversing equations
A ΔH value belongs to the equation it is written with: “per mole” means per mole of the reaction as written.
- Multiply the equation by a number → multiply ΔH by the same number.
- Reverse the equation → change the sign of ΔH, keep the size the same.
Common mistakes
- Defining formation with “from its elements” but forgetting “in their standard states” or “1 mole”.
- Writing combustion without “completely” or “in oxygen”.
- Writing H2O(g) in a standard combustion equation at 298 K.
- Saying a catalyst changes ΔH.
Quick checks
These are Finesse practice questions. The step-by-step answers are indicative worked solutions, not official AQA mark allocations.
Q1. Write the equation that represents the standard enthalpy of formation of propan-1-ol, C3H7OH(l).Show answer
Step 1: product is exactly 1 mol of C3H7OH(l).
Step 2: elements in standard states: C(s, graphite), H2(g), O2(g). Count atoms: 3 C, 8 H (4 H2), 1 O (½O2).
3C(s) + 4H2(g) + ½O2(g) → C3H7OH(l)
Q2. Write the equation for the standard enthalpy of combustion of pentane, C5H12(l).Show answer
Step 1: 5 C → 5CO2; 12 H → 6H2O(l).
Step 2: O atoms needed = 10 + 6 = 16, so 8O2.
C5H12(l) + 8O2(g) → 5CO2(g) + 6H2O(l)
Q3. ΔfH⦵[NH3(g)] = −46 kJ mol−1. Give ΔH for (a) N2(g) + 3H2(g) → 2NH3(g) and (b) 2NH3(g) → N2(g) + 3H2(g).Show answer
(a) The equation forms 2 mol, so 2 × (−46) = −92 kJ mol−1.
(b) Reverse of (a): +92 kJ mol−1.
Q4. Explain why ΔfH⦵ of O2(g) is zero but ΔfH⦵ of O3(g) is not.Show answer
O2(g) is the standard state of oxygen, so forming it from oxygen in its standard state involves no change.
O3(g) is a different form of the element; making it from O2(g) is a real change with an enthalpy change (it is endothermic).
Q5. 0.0500 mol of H2SO4 is neutralised by excess NaOH(aq). ΔHneut = −57.0 kJ per mole of water. How much heat energy is released?Show answer
Step 1: each H2SO4 forms 2 H2O, so n(H2O) = 0.100 mol.
Step 2: heat released = 0.100 × 57.0 = 5.70 kJ (ΔH is negative; the quantity released is 5.70 kJ).
Sources
Sources and examiner guidance (reviewed 1 October 2026)
- Chemrevise — AQA 1.4 Energetics revision guide (N. Goalby) — Primary checklist for definitions and standard states (Oct 2025 edition).
- AQA 7405 specification — 3.1.4 Energetics (incl. Required Practical 2) — 3.1.4.1 Enthalpy change.
- AQA 7404/2 mark scheme, November 2020 series (footer reads June 2020) — Q01.1: formation definition points.
- AQA 7404/2 mark scheme, November 2021 series (footer reads June 2021) — Q08.1: enthalpy change as heat energy change at constant pressure.
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