IB Chemistry HLTopic 4 — Energy CyclesPaper 1 & 2Core idea~13 min read
Born-Haber Cycles
Lattice enthalpy is the one number in ionic bonding you can never measure. You cannot separate a crystal into a cloud of free gaseous ions and put a thermometer in it. A Born-Haber cycle is Hess’s Law dressed up as a staircase, and it lets you get at that value by going the long way round.
📘 What you need to know
A Born-Haber cycle is just Hess’s Law applied to an ionic compound, drawn as energy levels rather than a triangle.
Its usual purpose is to find the lattice enthalpy, which cannot be measured directly.
Endothermic steps point up. Exothermic steps point down. Only the direction matters, not the length of the arrow.
You must know the definitions of: lattice enthalpy, enthalpy of atomisation, first and second ionisation energy, first and second electron affinity, and enthalpy of formation.
Each step needs its own arrow, its own equation with state symbols and electrons, and its own value.
If the formula needs two of an ion, that step’s value is doubled.
The idea behind the staircase
You have two ways of making solid potassium bromide from potassium metal and liquid bromine.
The direct way. Just let them react. One step, and the energy released is the enthalpy of formation, ΔHf.
The absurdly long way. Vaporise the potassium, split the bromine into atoms, strip an electron off the potassium, hand it to the bromine, then let the resulting cloud of gaseous ions collapse into a crystal.
Nobody would ever do the second one in a lab. But Hess’s Law says the two routes must total the same, and every step of the long route except the last one has been measured. So the last one — the lattice enthalpy — falls out as the only unknown.
You will lose no marks for drawing a 500 kJ step and a 2500 kJ step the same length. You will lose marks for pointing one the wrong way.
The definitions you have to know
Lattice enthalpy, ΔHlatt
The energy change when one mole of an ionic compound is separated into its gaseous ions. Pulling oppositely charged ions apart against their attraction costs energy, so lattice enthalpy is endothermic and positive, and its arrow points up.
KBr(s) → K+(g) + Br−(g) ΔHlatt = +689 kJ mol−1
Watch the wording. Some textbooks define lattice enthalpy the other way round, as gaseous ions coming together, which makes it negative. IB uses the dissociation version, so on your paper lattice enthalpy is positive. If a question hands you a negative lattice value, read the equation it is attached to before you use it.
Enthalpy of atomisation, ΔHat
The enthalpy change when one mole of gaseous atoms is formed from an element in its standard state. It covers both melting/vaporising the element and breaking any bonds in it, all in one number. Making free atoms always costs energy, so it is endothermic and points up.
Note the ½ in front of Br2. Atomisation is defined per mole of atoms produced, so you only take half a mole of the diatomic molecule. Forgetting this halving is a favourite exam trap.
Ionisation energy, ΔHie
The first ionisation energy is the enthalpy change when one mole of gaseous atoms each lose one electron to become 1+ ions. The second takes 1+ ions to 2+ ions. Removing an electron from a positive ion is harder than from a neutral atom, so the second value is always bigger than the first. Both are endothermic and point up.
The first electron affinity is the enthalpy change when one mole of gaseous atoms each gain one electron to become 1− ions. A neutral atom attracting an electron usually releases energy, so first electron affinities are normally exothermic and point down.
The second electron affinity is different, and this catches people out. You are now pushing an electron onto a particle that is already negative. The two repel, so energy must be supplied. Second electron affinities are always endothermic and point up.
An electron is pulled away from an attracting nucleus
Second ionisation energy
+ (larger)
Up
Now removing from an already positive ion
First electron affinity
Usually −
Down
The nucleus attracts the incoming electron
Second electron affinity
+
Up
Electron and 1− ion repel each other
Lattice enthalpy (IB definition)
+
Up
Ions are pulled apart against strong attraction
Enthalpy of formation
Usually −
Usually down
Most ionic compounds are stable relative to their elements
Building the cycle: potassium bromide
🧩 How to draw it, in order
Put the elements in their standard states on a horizontal line about a third of the way up. K(s) + ½Br2(l).
Drop down to the ionic solid with an arrow labelled ΔHf. This is the bottom of the diagram.
Climb to the gaseous atoms with one atomisation arrow per element. Order does not matter.
Climb again to strip the electrons off the metal. Show the electrons in the equation.
Move across and down for electron affinity. Shifting this step to the right of the diagram keeps it readable.
Join the ionic solid to the gaseous ions with one long arrow: the lattice enthalpy.
The green arrows are exothermic and drop; the red ones are endothermic and climb. The purple lattice arrow is the only quantity here that could not be measured in a lab.
Look at the sizes. Ionisation energy alone costs +419 and atomisation another +201, yet forming KBr still releases 394 kJ overall. The lattice enthalpy of +689 is what pays for all of it — ionic bonding is powerful precisely because the lattice term is so large.
When the formula is not one-to-one
Everything above assumed one cation and one anion. Real compounds are often not so tidy, and this is where most Born-Haber marks are dropped. Two rules cover it:
A 2+ ion needs both ionisation energies. Magnesium going to Mg2+ uses ΔHie1and ΔHie2, as two separate steps with two separate arrows. You cannot merge them.
Two of an ion means the step is doubled. MgCl2 contains 2Cl−, so you need to atomise chlorine twice and add an electron twice. Both values get multiplied by 2.
WORKED EXAMPLE
Writing the equation for every step
Write the equation for each step in the Born-Haber cycle of KBr, and state whether its arrow points up or down.
Enthalpy of formation, downK(s) + ½Br₂(l) → KBr(s)Atomisation of potassium, upK(s) → K(g)Atomisation of bromine, up½Br₂(l) → Br(g)First ionisation energy, upK(g) → K⁺(g) + e⁻First electron affinity, downBr(g) + e⁻ → Br⁻(g)Lattice enthalpy, upKBr(s) → K⁺(g) + Br⁻(g)show the electrons! leaving e⁻ out of the ionisation and affinity steps loses marks
WORKED EXAMPLE
Planning a cycle for magnesium chloride
List the steps needed for the Born-Haber cycle of MgCl2, stating how many times each value is used.
Step 1: work out the ionsMgCl₂ contains one Mg²⁺ and two Cl⁻.Step 2: the magnesium sideΔHat(Mg) × 1ΔHie1(Mg) × 1, then ΔHie2(Mg) × 1two separate arrows, because Mg loses two electrons in two stagesStep 3: the chlorine sideΔHat(Cl) × 2ΔHea(Cl) × 2two chlorides means everything chlorine does happens twiceStep 4: closing the cycleΔHf(MgCl₂) × 1 and ΔHlatt(MgCl₂) × 17 arrows in totalthe top level is Mg²⁺(g) + 2e⁻ + 2Cl(g)
A quick check for any Born-Haber diagram. Count the electrons. If your metal has released two, there must be two loose e− written on the top level, and two anions must eventually take them. If the electrons do not balance, an arrow is missing.
💡 Exam tip
You will not be asked to draw a whole cycle from a blank page. You will be asked to complete a partly drawn one, so practise slotting steps into gaps.
Every arrow needs three things: a label such as ΔHie1, a balanced equation with state symbols, and a value.
Always show the electrons. Writing K(g) → K+(g) without the e− is an unbalanced equation and it does cost marks.
Do not worry about scale. Direction is marked, relative arrow length is not.
No energy axis is needed, though a rough vertical arrow labelled “enthalpy” never hurts.
Halve diatomic elements for atomisation: ½Br2(l), ½Cl2(g), ½O2(g).
⚠ Common mix-up
Making lattice enthalpy negative. On the IB definition it is dissociation, so it is positive and its arrow points up.
Using one combined ionisation energy for a 2+ ion. First and second are separate values and need separate arrows.
Forgetting to double the halogen steps in compounds like MgCl2 or CaBr2.
Assuming all electron affinities are exothermic. The second one is endothermic because of the repulsion, and its arrow points up.
Forgetting to halve the diatomic element during atomisation, which doubles that step by accident.
Leaving the electrons out of the top energy level. If Mg2+ has formed, 2e− must be written alongside it.
Up next: Born-Haber Cycle Calculations — putting real numbers into the staircase and getting a lattice enthalpy out.
Want this explained one-to-one?
Book a free session with an experienced IB Chemistry tutor and get your trickiest topics made simple.
