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The Born-Haber cycle is Hess's law with five or six steps, and it is one of the most reliable sources of lost marks in Class 11 thermodynamics. The chemistry is simple; the bookkeeping is not. This post builds the NCERT cycle for sodium chloride step by step, uses it twice more, and lists the sign errors to watch for.
What is lattice enthalpy?
The NCERT Class 11 chapter Thermodynamics defines the lattice enthalpy of an ionic compound as the enthalpy change when one mole of it dissociates into its ions in the gaseous state:
NaCl(s) → Na⁺(g) + Cl⁻(g), ΔlatticeH = +788 kJ mol⁻¹
It is positive because separating ions takes energy. The Chemical Bonding chapter uses the same definition and explains why it matters: the stability of an ionic compound is measured by its enthalpy of lattice formation, not simply by its ions reaching an octet. Both the JEE (Main) 2026 syllabus and the NEET (UG)-2026 syllabus list "calculation of lattice enthalpy" under ionic bonding.
Why do we need a cycle?
NCERT points out that lattice enthalpy cannot be measured directly by experiment. Instead you go round a loop. Start from the elements, reach the gaseous ions by steps you can measure, then close the loop through the solid. Because enthalpy is a state function, the enthalpy changes round a closed cycle add to zero. Our post on Hess's law covers the general method.
Worked example 1: lattice enthalpy of NaCl
NCERT's data, all in kJ mol⁻¹:
Sublimation of sodium, Na(s) → Na(g): ΔsubH = +108.4
Ionisation of sodium, Na(g) → Na⁺(g) + e⁻: ΔiH = +496
Half the dissociation of chlorine, ½Cl₂(g) → Cl(g): ½ΔbondH = +121
Electron gain by chlorine, Cl(g) + e⁻ → Cl⁻(g): ΔegH = −348.6
Formation of NaCl from its elements, Na(s) + ½Cl₂(g) → NaCl(s): ΔfH = −411.2
Route A goes from NaCl(s) to the ions directly: that is ΔlatticeH. Route B goes from NaCl(s) back to the elements (reverse of step 5, so +411.2), then through steps 1 to 4. Both routes start and end in the same place, so:
ΔlatticeH = 411.2 + 108.4 + 121 + 496 − 348.6
Add in stages: 411.2 + 108.4 = 519.6; plus 121 is 640.6; plus 496 is 1136.6; minus 348.6 is 788.0 kJ mol⁻¹. This matches NCERT.
NCERT adds that the internal energy change is smaller by 2RT, because two moles of gas form from a solid: 2 × 8.314 × 298 ≈ 4955 J, about 5 kJ, giving about +783 kJ mol⁻¹.
Worked example 2: run the cycle backwards
Exam questions often give you the lattice enthalpy and ask for another step. Suppose you know ΔlatticeH = +788 and all the steps except the electron gain enthalpy of chlorine.
ΔegH = ΔlatticeH − (411.2 + 108.4 + 121 + 496)
= 788 − 1136.6 = −348.6 kJ mol⁻¹
The negative sign is right: a chlorine atom releases energy when it gains an electron. If your answer is positive, recheck your signs.
Worked example 3: will the salt dissolve with heat change?
NCERT links lattice enthalpy to the enthalpy of solution: ΔsolH = ΔlatticeH + ΔhydH, where ΔhydH is the enthalpy of hydration of the gaseous ions. With ΔhydH = −784 kJ mol⁻¹ from NCERT:
ΔsolH = +788 − 784 = +4 kJ mol⁻¹
So dissolving NaCl involves very little heat, which matches everyday experience. NCERT notes that if the lattice enthalpy is very high, the compound may not dissolve at all.
Why is the lattice enthalpy of MgO much larger?
The attraction between ions grows with the product of their charges and falls as they get further apart. Mg²⁺ and O²⁻ carry twice the charge of Na⁺ and Cl⁻, so the charge product is four times larger, and both ions are smaller, so they sit closer. Both effects raise the lattice enthalpy. This reasoning explains why small, highly charged ions form very stable lattices, and it is the reverse side of the polarisation rules in our post on Fajans' rules.
Which mistakes cost marks?
Sign convention. NCERT defines lattice enthalpy for breaking the lattice (positive). Some books define it for forming the lattice (negative). Read the question's definition and stick to it.
Full bond enthalpy for chlorine. You need one Cl atom, so use half the Cl–Cl bond enthalpy: 121, not the full 243 listed in NCERT's Table 5.3.
Sign of electron gain enthalpy. For chlorine it is negative.
Missing sublimation. Sodium starts as a solid; it must become a gas before it can ionise.
Mixing eV and kJ. Convert everything to kJ mol⁻¹ first.
A practice routine
Draw the NaCl cycle from memory, with arrows and signs, until it takes under three minutes.
Rebuild it for KCl or MgO using data your teacher or reference book gives, and identify the extra step MgO needs (a second ionisation, and electron gain to O²⁻).
Do two "find the missing step" problems, one for ΔfH and one for ΔegH.
Write one line explaining each answer's sign.
If the cycle keeps falling apart halfway, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning rather than printing an answer.
Today, close this page and reproduce the 788 kJ mol⁻¹ calculation from the five steps alone.
