IB Chemistry HLTopic 4 — Energy CyclesPaper 1 & 2Core skill~13 min read
Born-Haber Cycle Calculations
Once the staircase is drawn, the calculation is one line of arithmetic. The marks are won and lost somewhere else: remembering to double the halogen, remembering that a 2+ ion needs two ionisation energies, and getting the sign of the formation step the right way round.
📘 What you need to know
The full relationship: ΔHf = ΔHat + ΔHat + ΔHie + ΔHea − ΔHlatt
Group the middle terms into one number, ΔH1, the total cost of turning the elements into gaseous ions.
That simplifies to ΔHf = ΔH1 − ΔHlatt, so ΔHlatt = −ΔHf + ΔH1
Multiply every step by its stoichiometry before adding: MgCl2 doubles both chlorine steps.
A 2+ cation needs ΔHie1 and ΔHie2; a 2− anion needs ΔHea1 and ΔHea2.
The same equation, rearranged, finds any missing step — not just the lattice enthalpy.
Where the equation comes from
Forget the formula for a moment and look at the shape of the cycle. There are two ways to get from the elements to the gaseous ions.
Start at the ionic solid. Travel backwards along the formation arrow, which flips its sign, then climb the whole staircase. That is the entire method.
The full version
ΔHf = ΔHat(metal) + ΔHat(non-metal) + ΔHie + ΔHea − ΔHlatt
The version you should actually use
ΔHlatt = −ΔHf + ΔH1 where ΔH1 = the sum of every atomisation, ionisation and electron affinity step
Do not memorise the long version. Memorise the triangle. If you can see that the formation arrow must be travelled backwards, you can rebuild the equation from scratch in five seconds, and you will never be caught out by an unusual compound.
The method
🧩 Every Born-Haber calculation, in five steps
Write out the ions in the formula. NaF gives Na+ and F−. CaCl2 gives Ca2+ and 2Cl−. This decides everything that follows.
List every step with its multiplier. Two chlorides means atomisation × 2 and electron affinity × 2. A 2+ cation means two ionisation energies.
Add them all up to get ΔH1, keeping the signs exactly as given.
Apply ΔHlatt = −ΔHf + ΔH1, with brackets around every value.
Sanity check. Lattice enthalpies for 1+/1− compounds are roughly 600 to 1000 kJ mol−1. Bring in a 2+ or 2− ion and it jumps to a few thousand.
WORKED EXAMPLE
Lattice enthalpy of sodium fluoride
Use the data below to calculate ΔHlatt for NaF.
Enthalpy change
Value / kJ mol−1
Enthalpy of atomisation of Na
+107
Enthalpy of atomisation of F
+79
First ionisation energy of Na
+496
First electron affinity of F
−328
Enthalpy of formation of NaF
−574
Step 1: the ions are Na⁺ and F⁻, one eachNothing needs doubling, and sodium only loses one electron.Step 2: add up the staircase, ΔH(1)(+107) + (+79) + (+496) + (−328)ΔH(1) = +354 kJ mol−¹Step 3: apply the formulaΔHlatt = −(−574) + (+354)ΔHlatt = 574 + 354ΔHlatt = +928 kJ mol−¹positive, in the hundreds — exactly what a 1+/1− lattice should be
When the stoichiometry bites: calcium chloride
CaCl2 contains one Ca2+ and two Cl−. That means two changes to the simple cycle, and both of them are places students lose marks.
Calcium loses two electrons, so you need ΔHie1and ΔHie2 as separate steps.
You need two chloride ions, so the atomisation of chlorine and the electron affinity of chlorine both happen twice.
Count the electrons on each level. Two leave the calcium, so 2e− travel with it until the two chlorine atoms take one each.
WORKED EXAMPLE
Lattice enthalpy of calcium chloride
Use the data below to calculate ΔHlatt for CaCl2.
Enthalpy change
Value / kJ mol−1
Enthalpy of atomisation of Ca
+178
Enthalpy of atomisation of Cl
+121
First ionisation energy of Ca
+590
Second ionisation energy of Ca
+1145
First electron affinity of Cl
−349
Enthalpy of formation of CaCl2
−796
Step 1: identify the ions and the multipliersCa²⁺ and 2Cl⁻. So: both ionisation energies, and chlorine steps twice.Step 2: build ΔH(1) step by stepatomise Ca: +178ionise twice: (+590) + (+1145) = +1735atomise Cl twice: 2 × (+121) = +242electron affinity twice: 2 × (−349) = −698ΔH(1) = 178 + 1735 + 242 − 698 = +1457Step 3: apply the formulaΔHlatt = −(−796) + (+1457)ΔHlatt = +2253 kJ mol−¹over twice the NaF value — that is the 2+ charge on calcium doing the work
Why the number jumps so much. Lattice enthalpy depends on the product of the ionic charges and on how close the ions get. Swapping Na+ for Ca2+ doubles one of the charges, so the attraction and therefore the lattice enthalpy roughly doubles too. If your CaCl2 answer came out under 1000, you almost certainly missed a doubling.
Compounds with a 2− ion
Oxides and sulfides need two electron affinity steps, because the anion picks up two electrons one at a time. The first is exothermic and points down. The second is endothermic and points back up, because you are forcing an electron onto something already negative.
Try it yourself. For Na2O, using ΔHat(Na) = +107, ΔHie1(Na) = +496, ΔHat(O) = +249, ΔHea1(O) = −141, ΔHea2(O) = +798 and ΔHf = −414 kJ mol−1, you should get ΔHlatt = +2526 kJ mol−1. Remember that the sodium steps are both doubled, because there are two Na+ ions.
Finding a step other than the lattice enthalpy
Nothing about this equation is special to lattice enthalpy. If a question gives you the lattice value and hides something else, rearrange for that instead. The safest approach is to write the full relationship out with x in the gap and solve it like any other equation.
WORKED EXAMPLE
Finding the electron affinity of iodine
For potassium iodide: ΔHlatt = +649, ΔHf = −328, ΔHat(K) = +89, ΔHat(I) = +107 and ΔHie1(K) = +419 kJ mol−1. Calculate the first electron affinity of iodine.
Step 1: write the full relationship with x for the unknownΔHf = ΔHat(K) + ΔHat(I) + ΔHie1 + x − ΔHlattStep 2: substitute everything you have−328 = (+89) + (+107) + (+419) + x − (+649)Step 3: collect the known numbers89 + 107 + 419 − 649 = −34−328 = −34 + xStep 4: solvex = −328 + 34ΔHea(I) = −294 kJ mol−¹negative, as a first electron affinity should be — a positive answer means a sign slipped
Always finish by asking whether the sign is plausible. Lattice enthalpies and ionisation energies must be positive. First electron affinities are almost always negative. Second electron affinities are always positive. If your answer breaks one of those rules, go back and hunt for the missing bracket.
Sense-checking your answers
Compound type
Example
Typical ΔHlatt / kJ mol−1
1+ with 1−, large ions
KI
about 650
1+ with 1−, small ions
NaF
about 930
2+ with two 1−
CaCl2
about 2250
Two 1+ with one 2−
Na2O
about 2500
2+ with 2−
MgO
about 3800
The pattern is worth understanding rather than memorising: bigger charges and smaller ions both push lattice enthalpy up, because both make the electrostatic attraction stronger. A single-charge lattice in the hundreds, a double-charge lattice in the thousands.
💡 Exam tip
Write the ions down first. Na+ and F−, or Ca2+ and 2Cl−. Every multiplier in the question follows from this one line.
Do the multiplying in a separate column before you add anything. Mixing multiplication and sign handling in one line is where errors creep in.
Brackets around every value with its own sign. −(−796) is clear; −−796 is an accident waiting to happen.
Show ΔH1 as an intermediate result. If you slip at the last step, the examiner can still award the earlier marks.
Check the data booklet carefully — ionisation energies, electron affinities and formation enthalpies are in different sections, and it is easy to grab the wrong column under pressure.
Finish with a plausibility check against the table above.
⚠ Common mix-up
Forgetting to double the halogen steps in MX2 compounds. This is the most frequently lost mark in the whole topic.
Using only the first ionisation energy for a 2+ ion. You need the second as well, and it is much larger.
Getting the sign of ΔHf wrong. The formula needs minus ΔHf, so a negative formation enthalpy makes a positive contribution.
Doubling ΔHf or ΔHlatt. Both are already quoted per mole of the whole compound, so they are never multiplied.
Making the second electron affinity negative. It is endothermic because of electron-electron repulsion.
Reporting a negative lattice enthalpy. On the IB definition it is dissociation, so the answer must be positive.
Up next: Entropy and Spontaneity — enthalpy is only half the story of why reactions happen, and the other half is about disorder.
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