Last updated: 17 August 2026
Some enthalpy changes are difficult or impossible to measure directly. The desired reaction may be too slow, may occur together with unwanted reactions, or may involve hypothetical gaseous ions that cannot simply be mixed together in an experiment. Hess’ Law gives us a way around this problem.
Instead of insisting on one direct route, we can imagine another sequence of chemical changes connecting exactly the same initial and final states. If the enthalpy changes for those steps are known, they can be combined to obtain the enthalpy change we actually want.
This becomes particularly useful for ionic solids. We cannot conveniently measure the energy released when separate gaseous ions form one mole of an ionic lattice, but we can calculate it indirectly using a Born–Haber cycle. The key throughout this topic is to track the chemical species, stoichiometry and signs carefully rather than trying to memorise one fixed diagram.
Diagnostic Quiz
Try these five questions before reading the chapter. They check whether the basic ideas about enthalpy changes, gaseous ions and equation manipulation are already secure.
- 5 quick starting-point questions
- Focus on signs, states and energy routes
- Every answer option is explained
Hess’ Law: why the route does not matter
Hess’ Law states that the enthalpy change of a reaction is independent of the route taken, provided that the initial and final states are the same.
Suppose a chemical system can move from state A to state C either directly or by passing through an intermediate state B.
B → C ΔH₃
Since both routes begin at A and finish at C:
The intermediates do not affect the overall energy difference between the starting and finishing states. This is why a reaction that cannot conveniently be measured directly can still have its enthalpy change calculated from other reactions.
You may use any convenient sequence of reactions, but when those equations are added together, the overall equation must reduce exactly to the reaction whose enthalpy change you are trying to find.
Why does Hess’ Law work?
Enthalpy is a state function. This means that the change in enthalpy depends only on the initial and final states of the system rather than on the route by which the change occurs.
A useful analogy is travelling between two fixed elevations on a mountain. One route may be short and steep while another may wind gradually around the mountain, but the overall change in altitude is the same because the starting and finishing heights are unchanged.
Enthalpy works in the same way. Intermediate reactions may be different, but if the starting reactants and final products are identical, the overall ΔH is fixed.
Using Hess’ Law in calculations
Hess’ Law problems often provide several thermochemical equations that must be rearranged before they can be added together. The chemistry and the corresponding enthalpy values must always be changed together.
| Change to the equation | Required change to ΔH | Why? |
|---|---|---|
| Reverse the equation | Reverse the sign of ΔH | The energy transfer occurs in the opposite direction. |
| Multiply the equation by 2 | Multiply ΔH by 2 | Twice as much reaction occurs. |
| Divide the equation by 2 | Divide ΔH by 2 | Half as much reaction occurs. |
| Add equations together | Add their ΔH values | The individual reactions combine to give the required overall reaction. |
Worked example: forming carbon monoxide
Suppose the following enthalpy changes are known:
C(s) + O₂(g) → CO₂(g) ΔH = −394 kJ mol⁻¹
CO(g) + ½O₂(g) → CO₂(g) ΔH = −283 kJ mol⁻¹
We want:
CO appears on the wrong side of the second equation, so reverse that equation:
ΔH = +283 kJ mol⁻¹
Now add the two equations. CO₂ cancels from both sides, leaving the required equation.
Answer: ΔH = −111 kJ mol⁻¹.
Incorrect: Reverse the chemical equation but keep ΔH = −283 kJ mol⁻¹.
Correct: When the reaction is reversed, its enthalpy change becomes +283 kJ mol⁻¹.
A reliable way to check your equations
After manipulating the supplied reactions, add the left-hand sides and right-hand sides as though you were adding algebraic expressions. Any species appearing on both sides can be cancelled.
The remaining equation should be exactly the required reaction, including its stoichiometric coefficients and state symbols. If it is not, the enthalpy calculation is not yet ready to be performed.
What is a Born–Haber cycle?
A Born–Haber cycle is a Hess’ Law cycle used to analyse the formation of an ionic solid.
The direct route is simply the standard formation of the ionic compound from its constituent elements. The indirect route imagines converting those elements into gaseous atoms, converting the atoms into the required gaseous ions, and finally allowing those gaseous ions to form the solid lattice.
The cycle is useful because some of these energy changes, especially lattice energy, cannot conveniently be measured directly.
The enthalpy terms used in a Born–Haber cycle
| Term | Chemical meaning | Typical sign |
|---|---|---|
| Standard enthalpy change of formation, ΔH°f | Enthalpy change when 1 mol of a compound is formed from its constituent elements in their standard states under standard conditions. | Can be positive or negative |
| Standard enthalpy change of atomisation, ΔH°at | Enthalpy change when 1 mol of gaseous atoms is formed from the element in its standard state. | Positive |
| First ionisation energy, IE₁ | Energy required to remove 1 mol of electrons from 1 mol of gaseous atoms to form 1 mol of gaseous 1+ ions. | Positive |
| First electron affinity, EA₁ | Enthalpy change when 1 mol of electrons is added to 1 mol of gaseous atoms to form 1 mol of gaseous 1− ions. | Usually negative |
| Second electron affinity, EA₂ | Enthalpy change when 1 mol of electrons is added to 1 mol of gaseous 1− ions to form 1 mol of gaseous 2− ions. | Positive |
| Lattice energy, ΔH°latt | Enthalpy change when 1 mol of an ionic solid is formed from its constituent gaseous ions. | Negative |
Atomisation is defined per mole of gaseous atoms
The phrase 1 mol of gaseous atoms is important. For a diatomic element such as chlorine:
This produces exactly 1 mol of gaseous Cl atoms and therefore represents the atomisation enthalpy of chlorine.
If two moles of gaseous chlorine atoms are needed:
the corresponding enthalpy change is:
Why is the second electron affinity endothermic?
Consider oxygen. The first electron is added to a neutral gaseous O atom:
The second electron must then be added to an ion that is already negatively charged:
The incoming electron is repelled by the negatively charged O⁻ ion. Energy must therefore be supplied to overcome this electrostatic repulsion, so the second electron affinity is endothermic.
2EA₁ means two separate atoms each gain one electron. EA₁ + EA₂ means the same atom gains two electrons successively. Which expression you need depends on the ions actually present in the formula.
Building a Born–Haber cycle for NaCl
Start with the standard formation equation for sodium chloride:
ΔH°f
The Born–Haber cycle constructs an alternative route from these same elements to NaCl(s).
Step 1: form gaseous sodium atoms
ΔH°at(Na)
Ionisation energy is defined for gaseous atoms, so solid sodium must first be converted into Na(g).
Step 2: form gaseous Na⁺ ions
IE₁(Na)
Sodium forms Na⁺, so only one electron is removed and only IE₁ is required.
Step 3: form gaseous chlorine atoms
ΔH°at(Cl)
Step 4: form gaseous Cl⁻ ions
EA₁(Cl)
Step 5: form the solid ionic lattice
ΔH°latt
Hess’ Law now gives:
Worked example: calculating the lattice energy of NaCl
Suppose the following values are given:
| Enthalpy term | Value / kJ mol⁻¹ |
|---|---|
| ΔH°f[NaCl(s)] | −411 |
| ΔH°at(Na) | +108 |
| IE₁(Na) | +496 |
| ΔH°at(Cl) | +121 |
| EA₁(Cl) | −349 |
Substitute the values:
Answer: ΔH°latt = −787 kJ mol⁻¹ for the data given.
The negative sign makes chemical sense. When gaseous Na⁺ and Cl⁻ ions come together, strong electrostatic attractions form and energy is released.
Lattice formation: Na⁺(g) + Cl⁻(g) → NaCl(s) is exothermic, so the lattice energy defined in this chapter is negative.
Lattice dissociation: NaCl(s) → Na⁺(g) + Cl⁻(g) is the reverse process. It has the same magnitude but the opposite sign.
Stoichiometry in more complicated Born–Haber cycles
NaCl is relatively simple because one sodium atom loses one electron and one chlorine atom gains one electron. Compounds containing 2+ or 2− ions require more careful bookkeeping.
Example: MgCl₂
The formation equation is:
Magnesium forms Mg²⁺, so two electrons must be removed from the same Mg atom.
The ionisation-energy contribution is therefore:
MgCl₂ contains two Cl⁻ ions. Each of the two separate chlorine atoms gains only one electron:
The electron-affinity contribution is therefore:
It is not EA₁(Cl) + EA₂(Cl), because no chlorine atom is becoming Cl²⁻.
Two moles of gaseous Cl atoms are also required:
Enthalpy change = 2ΔH°at(Cl)
The complete relationship can therefore be written as:
Example: MgO
MgO contains Mg²⁺ and O²⁻. Magnesium again requires the first and second ionisation energies:
Oxygen is different from the two chlorine atoms in MgCl₂. One oxygen atom must acquire two electrons:
The electron-affinity contribution is therefore:
Ask how many atoms are present and what charge each atom must acquire. A metal forming M²⁺ requires IE₁ + IE₂. Two separate X⁻ ions require 2EA₁, whereas one X²⁻ ion requires EA₁ + EA₂.
Factors affecting lattice energy
Lattice energy reflects the strength of the electrostatic attractions formed between oppositely charged ions in an ionic lattice.
Two major factors determine how strong those attractions are:
- the charges on the ions; and
- the ionic radii.
Effect of ionic charge
Ions with larger charges attract one another more strongly. For otherwise comparable ions, an attraction between 2+ and 2− ions is stronger than an attraction between 1+ and 1− ions.
→ stronger electrostatic attraction
→ more negative lattice energy of formation
Effect of ionic radius
Smaller ions allow the centres of opposite charge to approach more closely. Electrostatic attraction therefore becomes stronger.
→ shorter distance between opposite charges
→ stronger electrostatic attraction
→ more negative lattice energy of formation
For example, when compounds containing the same anion are compared, Li⁺ generally gives stronger attraction than Na⁺ because Li⁺ has the smaller ionic radius.
| Change | Effect on attraction | Effect on lattice energy of formation |
|---|---|---|
| Higher ionic charges | Stronger | More negative |
| Lower ionic charges | Weaker | Less negative |
| Smaller ionic radii | Stronger | More negative |
| Larger ionic radii | Weaker | Less negative |
Weak explanation: “MgO has a larger lattice energy because its ions are smaller.”
Better explanation: Mg²⁺ and O²⁻ have greater ionic charges than Na⁺ and Cl⁻. The electrostatic attraction between Mg²⁺ and O²⁻ is therefore much stronger, giving MgO a much more negative lattice energy of formation.
Since lattice energy of formation is negative, it is clearer to say that one compound has a more negative lattice energy or a greater lattice-energy magnitude.
Enthalpy change of solution and hydration
Hess’ Law can also connect the energy involved in breaking an ionic lattice with the energy released when the resulting gaseous ions become surrounded by water molecules.
Standard enthalpy change of solution
The standard enthalpy change of solution, ΔH°sol, is the enthalpy change when one mole of a substance dissolves in sufficient solvent under standard conditions.
For sodium chloride:
ΔH°sol
Standard enthalpy change of hydration
The standard enthalpy change of hydration, ΔH°hyd, is the enthalpy change when one mole of gaseous ions becomes hydrated to form one mole of aqueous ions under standard conditions.
ΔH°hyd(Na⁺)
ΔH°hyd(Cl⁻)
Hydration is exothermic because ion–dipole attractions form between the ions and polar water molecules.
Connecting lattice energy, hydration and solution
Remember that lattice energy in this chapter refers to formation:
ΔH°latt
Dissolution requires the opposite process first:
Since this is the reverse of lattice formation, its enthalpy change is:
The gaseous ions are then hydrated. Therefore:
Worked example: solution enthalpy
Suppose:
ΔH°latt = −780 kJ mol⁻¹
total hydration enthalpy = −770 kJ mol⁻¹
Breaking the lattice requires the reverse of lattice formation:
Hydration then releases 770 kJ mol⁻¹:
Answer: ΔH°sol = +10 kJ mol⁻¹.
Lattice formation releases energy when gaseous ions form the ionic solid.
Lattice dissociation requires the same magnitude of energy to separate the solid into gaseous ions.
Hydration releases energy when those gaseous ions become surrounded by water molecules to form aqueous ions.
A reliable method for Born–Haber calculations
A long Born–Haber cycle can look intimidating, but the individual steps follow directly from the formula of the ionic compound.
- Write the formation equation for exactly 1 mol of the ionic solid.
- Identify the ions present in the compound and their charges.
- Atomise the elements to obtain the required numbers of gaseous atoms.
- Form the gaseous cations using the necessary successive ionisation energies.
- Form the gaseous anions using the necessary electron affinities.
- Form the ionic solid from its gaseous ions using the lattice energy.
- Apply Hess’ Law and solve for the unknown enthalpy term.
Before inserting numerical data, write every species with its correct state symbol and charge. Once the chemical changes are correct, the required energy terms and stoichiometric factors become much easier to identify.
Common mistakes with Hess’ Law and Born–Haber cycles
| Common mistake | Why it is wrong | Correct approach |
|---|---|---|
| Reversing an equation without changing the sign of ΔH | The reverse process has the opposite enthalpy change | Reverse both the equation and the sign of ΔH |
| Multiplying an equation but leaving ΔH unchanged | ΔH corresponds to the quantities in the thermochemical equation | Multiply ΔH by the same factor |
| Starting a Born–Haber cycle with an unbalanced formation equation | The entire cycle is based on forming exactly 1 mol of the compound | Balance the formation equation first |
| Using Na(s) directly in an ionisation-energy step | Ionisation energy is defined for gaseous atoms | Atomise Na(s) to Na(g) before ionising it |
| Writing Cl₂(g) → Cl(g) for chlorine atomisation | The equation is not balanced | Write ½Cl₂(g) → Cl(g) for 1 mol of gaseous Cl atoms |
| Using only IE₁ when Mg²⁺ is formed | Two electrons are removed successively from the same Mg atom | Use IE₁ + IE₂ |
| Using EA₁ + EA₂ for the two Cl⁻ ions in MgCl₂ | Each chlorine atom gains only one electron | Use 2EA₁(Cl) |
| Forgetting EA₂ when O²⁻ is formed | The same oxygen atom gains two electrons successively | Use EA₁(O) + EA₂(O) |
| Treating lattice formation and lattice dissociation as having the same sign | They are reverse processes | Check the definition of lattice energy being used |
| Saying only that one lattice energy is “larger” | Formation lattice energies are negative, so the wording can be ambiguous | Say “more negative” or “greater in magnitude” |
| Comparing lattice energies using ionic radius while ignoring ionic charge | Both charge and ion separation affect electrostatic attraction | Discuss ionic charges and ionic radii explicitly |
| Adding a negative lattice-formation value directly during dissolution | Dissolution first requires lattice dissociation | Reverse the lattice step before adding hydration enthalpies |
Hess’ Law and Born–Haber checklist
Set up Hess’ Law
- Write the target equation first.
- Reverse equations when necessary.
- Reverse the sign of ΔH with the equation.
- Scale ΔH whenever the equation is scaled.
Build the Born–Haber cycle
- Start with the elements in their standard states.
- Convert them into the required gaseous atoms.
- Form the correct gaseous cations and anions.
- Finish by forming the ionic solid from gaseous ions.
Control the stoichiometry
- Count how many gaseous atoms are required.
- Count how many electrons each individual atom loses or gains.
- Distinguish 2EA₁ from EA₁ + EA₂.
- Apply every necessary numerical coefficient.
Check the signs
- Ionisation energies are positive.
- Second electron affinity is positive.
- Lattice formation is negative using this convention.
- Reverse any enthalpy term when its process is reversed.
Mastery Quiz
Now apply the ideas to calculations and unfamiliar cycles. These questions focus on equation manipulation, Born–Haber stoichiometry, lattice-energy comparisons and solution cycles.
- 10 application-focused questions
- Includes calculations and sign reasoning
- Every answer option is explained
Frequently asked questions
Questions students often ask about Hess’ Law and Born–Haber cycles
These questions focus on the sign conventions and stoichiometric details that most often make energy cycles seem more complicated than they really are.
Is a Born–Haber cycle a different law from Hess’ Law?
No. A Born–Haber cycle is an application of Hess’ Law. It uses an indirect sequence of energy changes to connect the elements in their standard states with an ionic solid.
Should I memorise one particular Born–Haber diagram?
It is much safer to understand what each chemical step represents. Begin with the formation equation, identify the required gaseous atoms and ions, and then add the atomisation, ionisation, electron-affinity and lattice steps.
If you understand the species involved, you can reconstruct the cycle even when a question draws it in a different orientation.
Why must ionisation energy always start with gaseous atoms?
Ionisation energy is defined for isolated gaseous species. For example, IE₁(Na) refers to Na(g) → Na⁺(g) + e⁻. If the element begins as Na(s), an atomisation step is required first.
Why is ½Cl₂ used when chlorine is atomised?
Standard enthalpy of atomisation refers to the formation of 1 mol of gaseous atoms. Since 1 mol of Cl₂ molecules contains 2 mol of Cl atoms, ½ mol of Cl₂ is required to form 1 mol of Cl(g).
What is the difference between 2EA₁ and EA₁ + EA₂?
2EA₁ means that two separate neutral atoms each gain one electron. For example, the two Cl atoms in MgCl₂ each form Cl⁻, so the contribution is 2EA₁(Cl).
EA₁ + EA₂ means that the same atom gains two electrons successively. One O atom forming O²⁻ therefore requires EA₁(O) + EA₂(O).
Why is the second electron affinity endothermic?
The second electron is added to an ion that is already negatively charged. Energy must be supplied to overcome the electrostatic repulsion between the incoming electron and the existing negative ion.
Why do some sources show lattice energy as positive while this chapter uses a negative value?
Two opposite processes can be described. In this chapter, lattice energy is the enthalpy change when gaseous ions form one mole of ionic solid. This is lattice formation and is exothermic, so the value is negative.
If lattice dissociation is quoted instead, the ionic solid is separated into gaseous ions. That process is endothermic and has the same magnitude but the opposite sign. Always check the definition supplied in the question.
Does a more negative lattice energy mean stronger attraction between the ions?
Yes. A more negative lattice energy of formation means that more energy is released when the gaseous ions come together. This generally corresponds to stronger electrostatic attractions within the ionic lattice.
Which matters more for lattice energy: ionic charge or ionic radius?
Both affect electrostatic attraction. Greater ionic charges strengthen the attraction, while smaller ionic radii allow opposite charges to approach more closely.
In an exam comparison, discuss any difference in ionic charge explicitly and then consider ionic radius where relevant.
Why is lattice energy reversed in an enthalpy-of-solution cycle?
The lattice energy used here describes gaseous ions forming the solid. Dissolution begins with the solid and requires its ions to be separated. This is the reverse process, so the lattice contribution changes sign before the hydration enthalpies are added.
Hess’ Law turns many apparently different energy calculations into one simple principle: if the initial and final states are fixed, the overall enthalpy change is fixed. Born–Haber cycles apply this idea to ionic solids by connecting formation, atomisation, ionisation, electron affinity and lattice energy, while solution cycles connect lattice separation with hydration. Once the species, stoichiometric coefficients and signs are tracked carefully, these cycles become systematic chemical reasoning rather than diagrams that need to be memorised.