Last updated: 17 August 2026
Chemical equations tell us which substances react and how many moles are involved, but they do not tell us whether the reaction warms the surroundings or cools them. To describe that energy transfer, chemists use enthalpy changes.
At A-Level, however, simply knowing whether a reaction is exothermic or endothermic is not enough. You must distinguish several precisely defined energy changes: formation, combustion, neutralisation, atomisation, hydration and solution, as well as bond energy, lattice energy and electron affinity.
These definitions look similar because many begin with the phrase “the enthalpy change when...”. The most reliable way to separate them is to ask three questions: what species do we start with, what species do we finish with, and what does the “1 mol” refer to? Once that habit becomes automatic, later Hess' Law and Born–Haber cycles become much easier to construct.
Diagnostic Quiz
Try these five questions before reading the chapter. They check the basic ideas about energy transfer, bonding and chemical equations that underpin the different enthalpy definitions.
- 5 quick prerequisite questions
- Focus on signs, bonds and standard states
- Every answer option is explained
Understanding enthalpy change, ΔH
The enthalpy of a chemical system is represented by H. In most chemical problems, we do not need its absolute value. Instead, we consider the change in enthalpy as reactants become products.
If the products have lower enthalpy than the reactants, the chemical system has lost energy to the surroundings. If the products have higher enthalpy, energy has been absorbed from the surroundings.
| Type of process | Energy transfer | Relative enthalpy | Sign of ΔH |
|---|---|---|---|
| Exothermic | Heat is released from the system to the surroundings | Products have lower enthalpy than reactants | Negative |
| Endothermic | Heat is absorbed by the system from the surroundings | Products have higher enthalpy than reactants | Positive |
Exothermic: energy leaves the chemical system, so ΔH < 0.
Endothermic: energy enters the chemical system, so ΔH > 0.
Bond breaking and bond making
Many reaction enthalpies can be understood by following what happens to chemical bonds. Breaking a bond requires energy because attractive forces between bonded atoms must be overcome. Bond breaking is therefore endothermic.
Making a bond releases energy. As a bond forms, the atoms move into a more stable, lower-energy arrangement, and energy is transferred to the surroundings.
You cannot decide the overall sign of ΔH merely by noticing that bonds are broken or formed. The reaction is exothermic only if the energy released when new bonds form is greater than the energy required to break the original bonds.
Standard conditions and standard states
Enthalpy changes depend on temperature, pressure and the physical states of the substances involved. Standard conditions give chemists a common reference so that enthalpy values can be compared.
Standard enthalpy data are normally quoted at a pressure of 1 bar and commonly at 298 K, unless another temperature is specified. For aqueous species, a concentration of 1 mol dm⁻³ is used in the A-Level treatment.
Each substance must also be in its standard state: its most stable physical form under the specified standard conditions.
| Substance | Standard state at 298 K |
|---|---|
| Hydrogen | H₂(g) |
| Oxygen | O₂(g) |
| Bromine | Br₂(l) |
| Carbon | C(s, graphite) |
| Water | H₂O(l) |
Each substance is used in whichever physical state is most stable under the standard conditions. Oxygen is O₂(g), bromine is Br₂(l), carbon is graphite and water is H₂O(l).
Standard enthalpy change of formation
The standard enthalpy change of formation, ΔH°f, is the enthalpy change when 1 mol of a compound is formed from its constituent elements in their standard states under standard conditions.
Example: formation of water
The equation forms exactly 1 mol of H₂O(l). Fractional coefficients are therefore perfectly acceptable in equations representing standard enthalpy changes.
Example: formation of carbon dioxide
The carbon must be graphite because graphite is the standard state of carbon. Oxygen must be O₂(g), not O(g), because O₂ is the standard state of elemental oxygen.
For example, ΔH°f[O₂(g)] = 0 and ΔH°f[C(s, graphite)] = 0. This does not mean that these substances contain no energy. Zero is simply the chosen reference value.
Formation of 1 mol: H₂(g) + ½O₂(g) → H₂O(l)
Not written for 1 mol: 2H₂(g) + O₂(g) → 2H₂O(l)
The second equation represents twice the standard enthalpy change of formation of water because it forms 2 mol of H₂O.
Standard enthalpy change of combustion
The standard enthalpy change of combustion, ΔH°c, is the enthalpy change when 1 mol of a substance is completely burned in oxygen under standard conditions, with the products in their standard states.
Example: combustion of methane
The definition is based on 1 mol of the substance being burned. It does not mean that 1 mol of each product must form. One mole of CH₄ produces 1 mol of CO₂ and 2 mol of H₂O.
For a hydrocarbon, complete combustion forms CO₂ and H₂O. For an organic compound that already contains oxygen, such as an alcohol, complete combustion still converts its carbon to CO₂ and its hydrogen to H₂O; the amount of external O₂ required is simply different.
Formation of CO or solid carbon because oxygen is limited does not represent the standard enthalpy change of combustion. The fuel must be completely oxidised to the specified products.
Combustion enthalpies are negative because combustion is exothermic. Overall, formation of strong bonds in the oxidised products releases more energy than is required to break bonds in the reactants.
Standard enthalpy change of neutralisation
The standard enthalpy change of neutralisation, ΔH°neut, is the enthalpy change when an acid and an alkali react to form 1 mol of water under standard conditions.
When a strong acid reacts with a strong alkali, both are essentially completely ionised in aqueous solution. The reaction responsible for the enthalpy change is:
The standard enthalpy change of neutralisation for a strong acid and strong alkali is therefore approximately −57 kJ mol⁻¹.
Why do strong acid–strong alkali reactions give similar values?
Different strong acids supply H⁺(aq), while different strong alkalis supply OH⁻(aq). The remaining ions are spectators. The net ionic reaction is therefore essentially the same regardless of which strong acid and strong alkali are used.
When a weak acid or weak alkali is involved, some energy is associated with ionising the weak electrolyte. The measured enthalpy of neutralisation can therefore be less exothermic.
For neutralisation, count the moles of water formed.
If a balanced equation forms 2 mol of H₂O, the enthalpy change for that equation corresponds to twice the standard enthalpy change of neutralisation.
Standard enthalpy change of atomisation
The standard enthalpy change of atomisation, ΔH°at, is the enthalpy change when 1 mol of gaseous atoms is formed from an element in its standard state under standard conditions.
Atomising sodium
One mole of sodium atoms is formed in the gas phase.
Atomising chlorine
Chlorine already exists as a gas under standard conditions, but it exists as Cl₂ molecules rather than individual Cl atoms. The Cl–Cl bond must therefore be broken. Half a mole of Cl₂ produces exactly 1 mol of Cl atoms.
Do not identify atomisation merely by looking for a change from solid to gas. Na(s) → Na(g) is atomisation, but ½Cl₂(g) → Cl(g) is also atomisation even though both sides are gaseous.
Bond energy
A bond energy is the average energy required to break 1 mol of a specified covalent bond in gaseous molecules.
Breaking a bond always requires energy, so bond energies are positive.
Example: breaking an H–H bond
Energy must be supplied to overcome the attraction holding the two hydrogen atoms together. The reverse process, in which two gaseous H atoms form an H–H bond, releases approximately the same amount of energy.
Using bond energies to estimate ΔH
Bonds broken are counted as positive energy requirements. Bonds formed correspond to energy release, so their bond energies are subtracted.
Worked example
Consider:
Suppose the bond energies are:
- H–H = 436 kJ mol⁻¹;
- Cl–Cl = 243 kJ mol⁻¹;
- H–Cl = 431 kJ mol⁻¹.
We first break one H–H bond and one Cl–Cl bond:
We then form two H–Cl bonds:
Therefore:
Answer: the reaction is exothermic because forming the new H–Cl bonds releases more energy than is required to break the original bonds.
Why are bond-energy calculations approximate?
The energy of a particular type of bond depends slightly on the chemical environment around it. A C–H bond in methane, for example, is not energetically identical to every C–H bond in every other molecule.
Tabulated bond energies are therefore often average bond energies obtained from several gaseous compounds. Calculated reaction enthalpies are correspondingly approximate.
Breaking bonds: energy is absorbed.
Making bonds: energy is released.
Lattice energy
In these notes, lattice energy, ΔH°latt, is defined as the enthalpy change when 1 mol of a solid ionic compound is formed from its constituent gaseous ions under standard conditions.
Example: sodium chloride
The oppositely charged gaseous ions come together to form the ionic lattice. Strong electrostatic attractions form between the ions, releasing energy.
With this lattice formation convention, lattice energy is therefore negative.
In Reviewston, ΔH°latt refers to formation of the solid from gaseous ions. The reverse process — completely separating the solid into gaseous ions — requires the same magnitude of energy but has the opposite sign.
What determines the magnitude of lattice energy?
Lattice energy depends mainly on the strength of electrostatic attraction between oppositely charged ions.
| Change | Effect on electrostatic attraction | Effect on ΔH°latt |
|---|---|---|
| Higher ionic charges | Stronger attraction | More negative |
| Smaller ionic radii | Ion centres can approach more closely | More negative |
| Lower ionic charges | Weaker attraction | Less negative |
| Larger ionic radii | Greater separation between ion centres | Less negative |
For example, Mg²⁺ and O²⁻ have larger charges than Na⁺ and Cl⁻. The electrostatic attraction between Mg²⁺ and O²⁻ is therefore much stronger, contributing to a lattice energy of much greater magnitude.
Avoid saying only “larger lattice energy” when comparing negative values.
Use phrases such as more negative lattice energy, more exothermic lattice formation or greater magnitude of lattice energy.
Electron affinity
Electron affinity describes the enthalpy change associated with adding electrons to gaseous atoms or gaseous ions.
First electron affinity
The first electron affinity, EA₁, is the enthalpy change when 1 mol of gaseous atoms each gains one electron to form 1 mol of gaseous singly charged negative ions.
For chlorine, energy is released when the electron is added because the incoming electron is attracted towards the positively charged nucleus. The first electron affinity of chlorine is therefore negative.
Second electron affinity
The second electron affinity, EA₂, corresponds to adding one electron to each gaseous 1− ion to form gaseous 2− ions.
The incoming electron is now approaching an already negatively charged ion. Energy must be supplied to overcome electrostatic repulsion. The second electron affinity is therefore endothermic and positive.
Do not write aqueous atoms or ions in an electron-affinity equation. For example, Cl(g) + e⁻ → Cl⁻(g) represents electron affinity; Cl⁻(aq) involves hydration as well.
Ionisation energy: an electron is removed from a gaseous atom or ion.
Electron affinity: an electron is added to a gaseous atom or ion.
Standard enthalpy change of hydration
The standard enthalpy change of hydration, ΔH°hyd, is the enthalpy change when 1 mol of gaseous ions becomes hydrated by water to form 1 mol of aqueous ions under standard conditions.
Example: hydration of sodium ions
The notation (aq) means that the ion is surrounded and stabilised by water molecules. Ion–dipole attractions form between the ion and the polar water molecules.
Formation of these attractive interactions releases energy, so hydration enthalpies are negative.
What determines hydration enthalpy?
A small ion with a high charge produces a stronger electric field around itself and interacts more strongly with water molecules.
| Ion property | Effect on ion–dipole attraction | Effect on ΔH°hyd |
|---|---|---|
| Higher ionic charge | Stronger attraction to H₂O | More negative |
| Smaller ionic radius | Water molecules approach the charge more closely | More negative |
| Lower ionic charge | Weaker attraction to H₂O | Less negative |
| Larger ionic radius | Weaker attraction at greater distance | Less negative |
Higher ionic charge and smaller ionic radius strengthen electrostatic interactions. In a lattice this strengthens ion–ion attractions; during hydration it strengthens ion–dipole attractions between the ion and water.
Standard enthalpy change of solution
The standard enthalpy change of solution, ΔH°sol, is the enthalpy change when 1 mol of a solute dissolves completely in sufficient solvent to form an infinitely dilute solution under standard conditions.
For an ionic solid dissolving in water, it helps to imagine the overall process as two conceptual steps.
- Separate the ionic lattice into gaseous ions. Energy must be supplied to overcome the electrostatic attractions in the lattice.
- Hydrate the gaseous ions. Energy is released when ion–dipole attractions form between the ions and water.
Because ΔH°latt in these notes is defined for formation of the lattice, separating the lattice is the reverse process and has the opposite sign.
Why is there a summation sign?
Every ion produced from one formula unit must be hydrated. The stoichiometric coefficients of the ions therefore matter.
For example:
One mole of MgCl₂ produces one mole of Mg²⁺ and two moles of Cl⁻. Therefore:
Worked example
Suppose an ionic solid has:
- ΔH°latt = −800 kJ mol⁻¹;
- total hydration enthalpy of its gaseous ions = −770 kJ mol⁻¹.
Separating the lattice reverses lattice formation:
Hydration releases 770 kJ mol⁻¹:
Answer: dissolving is endothermic because slightly more energy is required to separate the ionic lattice than is released during hydration.
Why can dissolving be exothermic or endothermic?
Hydration always provides an exothermic contribution, while separating an ionic lattice requires energy. The sign of ΔH°sol depends on the balance between these two effects.
| Energy balance | Sign of ΔH°sol | Overall process |
|---|---|---|
| Hydration releases more energy than lattice separation requires | Negative | Exothermic |
| Lattice separation requires more energy than hydration releases | Positive | Endothermic |
Do not assume that dissolving an ionic solid must be exothermic.
The hydration step is exothermic, but breaking apart the ionic lattice is endothermic. The enthalpy of solution is the result of both contributions.
Comparing the important enthalpy definitions
The definitions become easier to distinguish when you focus on what must appear at the beginning and end of each process.
| Quantity | Essential process | Key feature to check |
|---|---|---|
| ΔH°f | Elements in standard states → 1 mol of compound | Exactly 1 mol of compound is formed |
| ΔH°c | 1 mol of substance + O₂ → complete combustion products | Exactly 1 mol of the substance is burned completely |
| ΔH°neut | Acid + alkali → 1 mol H₂O | Count the amount of water formed |
| ΔH°at | Element in standard state → 1 mol gaseous atoms | Product must be individual gaseous atoms |
| Bond energy | Break 1 mol of a specified covalent bond in gaseous molecules | Bond breaking is endothermic |
| ΔH°latt | Constituent gaseous ions → 1 mol solid ionic compound | Formation convention gives a negative value |
| EA₁ | Gaseous atoms + electrons → gaseous 1− ions | An electron is added to each gaseous atom |
| ΔH°hyd | Gaseous ions → aqueous ions | Hydration is exothermic |
| ΔH°sol | 1 mol solute → infinitely dilute solution | Represents the overall dissolving process |
Ask what the “1 mol” refers to, identify the starting species and final species, and check whether gaseous, aqueous, liquid or solid states are required.
Common mistakes with enthalpy changes
| Common mistake | Why it is wrong | Correct approach |
|---|---|---|
| Giving an exothermic reaction a positive ΔH | The chemical system loses enthalpy when heat is released | Use ΔH < 0 for an exothermic process |
| Saying that breaking bonds releases energy | Energy is required to overcome the attractions holding bonded atoms together | Bond breaking is endothermic; bond making is exothermic |
| Defining formation without specifying 1 mol | ΔH°f is defined for exactly 1 mol of compound formed | Include “1 mol of a compound” in the definition |
| Using O(g) in a formation equation | O(g) is not the standard state of elemental oxygen | Use O₂(g), with a fractional coefficient if necessary |
| Using diamond as the standard state of carbon | Graphite is the standard state of carbon under the usual conditions | Use C(s, graphite) |
| Calling incomplete combustion a standard combustion process | ΔH°c refers to complete combustion | Use the fully oxidised products |
| Defining neutralisation per mole of acid | The defined amount is 1 mol of water formed | Count H₂O, not acid or alkali |
| Writing Cl₂(g) as the product of atomisation | Atomisation must produce individual gaseous atoms | Write ½Cl₂(g) → Cl(g) |
| Using bond energies without counting every bond | ΔH depends on the total numbers of bonds broken and formed | Multiply each bond energy by the number of that bond involved |
| Using aqueous ions in the definition of lattice energy | Lattice energy is defined using constituent gaseous ions | Use gaseous ions forming the solid ionic compound |
| Forgetting the lattice-energy sign convention | Formation and dissociation conventions have opposite signs | In these notes, lattice formation is negative |
| Assuming every electron affinity is negative | Adding an electron to an already negative ion requires energy | Recognise that the second electron affinity is positive |
| Defining hydration as dissolving an ionic solid | Hydration begins with gaseous ions | Use gaseous ion → aqueous ion |
| Writing ΔH°sol = ΔH°latt + ΣΔH°hyd | The lattice must be broken, reversing the lattice-formation process | Use ΔH°sol = −ΔH°latt + ΣΔH°hyd |
| Forgetting coefficients in hydration terms | Every mole of each ion produced must be hydrated | For MgCl₂, include 2ΔH°hyd(Cl⁻) |
Enthalpy-change checklist
Check the energy direction
- Exothermic means ΔH < 0.
- Endothermic means ΔH > 0.
- Bond breaking requires energy.
- Bond making releases energy.
Check the definition
- Identify what the 1 mol refers to.
- Identify the starting species.
- Identify the final species.
- Include standard conditions where required.
Check the physical states
- Formation begins with elements in standard states.
- Atomisation ends with gaseous atoms.
- Lattice energy uses gaseous ions.
- Hydration changes gaseous ions into aqueous ions.
Check the calculation
- Reverse a process by reversing the sign of ΔH.
- Multiply enthalpy terms by stoichiometric coefficients.
- For bond energies, calculate bonds broken minus bonds formed.
- For solution, include every required hydration term.
Mastery Quiz
Now apply the chapter to thermochemical equations, energy calculations and comparisons involving lattice energy, hydration and electron affinity.
- 10 application-focused questions
- Includes equations and multi-step energy reasoning
- Every answer option is explained
Frequently asked questions
Questions students often ask about enthalpy changes
These questions connect the different definitions and clear up the sign and physical-state conventions that commonly cause confusion in H2 Chemistry.
Is enthalpy the same as temperature?
No. Temperature describes the thermal state of a substance and is related to the average kinetic energy of its particles. Enthalpy is a thermodynamic quantity describing the energy of a system.
A reaction can have a fixed molar enthalpy change even though the temperature change observed experimentally depends on the quantities used, heat capacity and heat loss to the surroundings.
Why is ΔH negative when energy is released?
ΔH describes the energy change of the chemical system. When the system transfers energy to the surroundings, it loses enthalpy. The products therefore have lower enthalpy than the reactants, giving ΔH < 0.
Can fractional coefficients be used in thermochemical equations?
Yes. Fractional coefficients are often necessary because many enthalpy definitions refer to exactly 1 mol of a particular product or species.
For example, H₂(g) + ½O₂(g) → H₂O(l) is the correct formation equation for 1 mol of liquid water.
Why is the standard enthalpy of formation of an element zero?
The standard enthalpy of formation of an element in its standard state is defined as zero so that all other formation enthalpies can be measured relative to a common reference.
It does not mean that the element contains no energy.
Why is graphite used instead of diamond in standard formation equations?
Graphite is the standard state of carbon under the usual standard conditions. Diamond is another allotrope of carbon, but it is not the reference form used for standard enthalpy of formation.
Why are bond-energy calculations only approximate?
A particular bond type does not have exactly the same energy in every molecule because the surrounding atoms affect its chemical environment. Tabulated bond energies are therefore usually average values.
Is lattice energy positive or negative?
It depends on the convention being used. In Reviewston, lattice energy is defined for formation of an ionic solid from gaseous ions. This process is exothermic, so ΔH°latt is negative.
Completely separating the solid into gaseous ions is the reverse process and therefore has the same magnitude but the opposite sign.
Why do high charge and small ionic radius affect both lattice energy and hydration enthalpy?
Both quantities depend on electrostatic attraction. A highly charged, small ion produces stronger interactions at a given distance.
In the ionic lattice the interaction is between oppositely charged ions. During hydration it is between an ion and the partial charges on polar water molecules.
Does a negative hydration enthalpy mean that every ionic solid dissolves exothermically?
No. Hydrating gaseous ions releases energy, but separating the original ionic lattice requires energy. The overall enthalpy of solution depends on the balance between these two contributions.
Why is the second electron affinity positive even though the atom wants a full outer shell?
By the time the second electron is added, the species already carries a negative charge. The incoming electron is therefore repelled by the negative ion, and energy must be supplied to overcome this repulsion.
Ideas about obtaining a particular electronic configuration do not remove the electrostatic energy requirement of the actual electron-addition step.
What is the quickest way to distinguish atomisation, lattice energy and hydration?
Follow the particles and their physical states. Atomisation ends with gaseous atoms. Lattice formation changes gaseous ions into an ionic solid. Hydration changes gaseous ions into aqueous ions.
Do I need to memorise the enthalpy definitions word for word?
You should know the chemically essential parts precisely, but understanding the process is more reliable than memorising a sentence without meaning. Check what 1 mol refers to, the starting species, the final species, their physical states and the required standard conditions.
Being able to write the corresponding thermochemical equation is an excellent check that you really understand the definition.
Enthalpy terminology becomes much easier once every definition is connected to a physical process. Formation begins with elements, combustion begins with a substance and oxygen, atomisation ends with gaseous atoms, lattice formation begins with gaseous ions, and hydration ends with aqueous ions. Bond energies explain reactions in terms of bonds broken and formed, while electron affinities describe electron gain. With these definitions secure, the next step is to link them together using Hess' Law and Born–Haber cycles to calculate energy changes that cannot be measured directly.