Born-Haber cycle revisited — calculating lattice energy: Worked examples

Before anything else, the one equation everything below is built from. We assemble the ionic solid two ways and set them equal (that is Hess's Law):

Every symbol above is a number in kilojoules per mole ( $kJ mol^{- 1}$ ). "Positive" means the step absorbs energy (you climb the energy ladder); "negative" means it releases energy (you fall). Keep that picture in mind — the whole page is one energy ladder that must close.

Figure — Born-Haber cycle revisited — calculating lattice energy

The scenario matrix

Every Born–Haber problem you will ever meet is one of these cells. The worked examples that follow are tagged with the cell number they cover.

#	Case class	What is tricky	Example
C1	1:1 solid, solve for $U$ (Na⁺X⁻)	baseline sign bookkeeping	Ex 1
C2	$E A$ negative — subtracting a negative	double-negative flips to plus	Ex 1, Ex 2
C3	Multi-charge cation, $M X_{2}$	two $I E$ , full $D$ , $2 \times E A$	Ex 3
C4	Solve for an unknown ≠ $U$ ( $E A$ , $Δ H_{f}$ …)	rearrange, same loop	Ex 4, Ex 5
C5	Second electron affinity is POSITIVE (O²⁻)	$E A_{2}$ costs energy, sign up	Ex 6
C6	Degenerate / limiting check ( $E A = 0$ , or a step $= 0$ )	sanity behaviour	Ex 7
C7	Real-world word problem (why salt melts high)	connect $U$ to a property	Ex 8
C8	Exam twist — missing data / find it via a second cycle	combine known compounds	Ex 9

Ex 1 — C1 & C2: NaCl, solve for U

Forecast: guess the sign and rough size of $U$ before reading on. (Hint: opposite charges snapping together — release or absorb?)

Write the isolated form. $U = Δ H_{f} - Δ H_{s u b} - I E - \frac{1}{2} D - E A$ . Why this step? $U$ is the only unknown, so put it alone on the left.
Substitute, keeping each number's own sign. $U = (- 411) - (+ 108) - (+ 496) - (+ 122) - (- 349)$ Why this step? Every subtraction sign is the operator from the formula; every number keeps its measured sign. The last term is $- (- 349)$ .
Resolve the double negative (cell C2): $- (- 349) = + 349$ . Why this step? Subtracting a released-energy term means we are removing a downward step, which adds energy back.
Add up: $- 411 - 108 - 496 - 122 + 349 = - 788$ .

$U = - 788 kJ mol^{- 1}$

Verify: climb the ladder both ways — up-steps $108 + 496 + 122 = 726$ ; down-steps $E A + U = - 349 + (- 788) = - 1137$ ; net down $= 726 - 1137 = - 411 = Δ H_{f}$ . ✓ Cycle closes. Units all $kJ mol^{- 1}$ . ✓

Ex 2 — C2: KF (another double-negative, small numbers)

Forecast: F is small and very electronegative — expect $U$ bigger (more negative) than NaCl?

$U = Δ H_{f} - Δ H_{s u b} - I E - \frac{1}{2} D - E A$ . Why? Isolate the unknown.
Substitute: $U = - 567 - 89 - 419 - 79 - (- 328)$ . Why? Carry signs.
$- (- 328) = + 328$ (cell C2). Why? Removing a downward step.
$- 567 - 89 - 419 - 79 + 328 = - 826$ .

$U = - 826 kJ mol^{- 1}$

Verify: up-steps $89 + 419 + 79 = 587$ ; down-steps $- 328 - 826 = - 1154$ ; net $587 - 1154 = - 567 = Δ H_{f}$ . ✓ And $- 826 < - 788$ : the smaller F⁻ ion gives a larger lattice energy, matching $U \propto q_{+} q_{-} / r$ from Ionic bonding and Coulomb's law. ✓

Ex 3 — C3: MgCl₂ (multi-charge cation, two anions)

Forecast: Mg is +2 and small; will $U$ be near NaCl or far bigger?

Build the loop for this stoichiometry: $U = Δ H_{f} - Δ H_{s u b} - (I E_{1} + I E_{2}) - D - 2 E A$ . Why? Each term must match how many particles the formula unit contains.
Substitute: $U = - 641 - 148 - (738 + 1451) - 244 - 2 (- 349)$ . Why? $2 E A = 2 (- 349) = - 698$ ; subtracting it gives $+ 698$ .
$I E_{1} + I E_{2} = 2189$ ; then $- 641 - 148 - 2189 - 244 + 698 = - 2524$ .

$U = - 2524 kJ mol^{- 1}$

Verify: up-steps $148 + 2189 + 244 = 2581$ ; down-steps $2 E A + U = - 698 + (- 2524) = - 3222$ ; net $2581 - 3222 = - 641 = Δ H_{f}$ . ✓ Roughly $3 \times$ NaCl's magnitude — expected from higher charge and smaller radius. ✓

Ex 4 — C4: KBr, solve for electron affinity

Forecast: EA of Br — positive or negative?

Rearrange the same loop for $E A$ : $E A = Δ H_{f} - Δ H_{s u b} - I E - \frac{1}{2} D - U$ . Why? One equation, one unknown — now the unknown is $E A$ , so isolate it.
Substitute: $E A = - 394 - 89 - 419 - 112 - (- 671)$ . Why? $- (- 671) = + 671$ .
$- 394 - 89 - 419 - 112 + 671 = - 343$ .

$E A (Br) = - 343 kJ mol^{- 1}$

Verify: plug back into Ex-1-style loop: $U = Δ H_{f} - Δ H_{s u b} - I E - \frac{1}{2} D - E A = - 394 - 89 - 419 - 112 - (- 343) = - 671$ . ✓ Matches the given $U$ . Negative EA = Br releases energy accepting an electron. ✓

Ex 5 — C4: solve for enthalpy of formation

Forecast: will $Δ H_{f}$ be near NaCl's $- 411$ ?

Isolate $Δ H_{f}$ directly from the loop: $Δ H_{f} = Δ H_{s u b} + I E + \frac{1}{2} D + E A + U$ . Why? This is the un-rearranged master loop — all right-hand terms are now known.
Substitute: $Δ H_{f} = 108 + 496 + 122 + (- 349) + (- 780)$ . Why? Keep each sign.
$108 + 496 + 122 - 349 - 780 = - 403$ .

$Δ H_{f} = - 403 kJ mol^{- 1}$

Verify: back-solve $U = Δ H_{f} - Δ H_{s u b} - I E - \frac{1}{2} D - E A = - 403 - 108 - 496 - 122 + 349 = - 780$ . ✓ Close to real NaCl's $- 411$ , as expected for a near-NaCl $U$ . ✓

Ex 6 — C5: MgO, POSITIVE second electron affinity

Forecast: O²⁻ needs a second electron pushed onto an already-negative ion — does that release or cost energy?

Loop for a 1:1 solid but with a two-step EA: $U = Δ H_{f} - Δ H_{s u b} - (I E_{1} + I E_{2}) - \frac{1}{2} D - (E A_{1} + E A_{2})$ . Why? One Mg (two ionisations), one O (two electrons added).
$E A_{1} + E A_{2} = - 141 + 798 = + 657$ . Why? Add the two electron-affinity steps.
Substitute: $U = - 602 - 148 - 2189 - 249 - 657$ . Why? All terms subtracted from $Δ H_{f}$ ; note $- (+ 657)$ stays negative.
$- 602 - 148 - 2189 - 249 - 657 = - 3845$ .

$U = - 3845 kJ mol^{- 1}$

Verify: up-steps $148 + 2189 + 249 + 657 = 3243$ ; down-steps only $U = - 3845$ ; net $3243 - 3845 = - 602 = Δ H_{f}$ . ✓ Enormous magnitude from $q_{+} q_{-} = (+ 2) (- 2)$ and a tiny O²⁻ — consistent with MgO's very high melting point. ✓

Ex 7 — C6: degenerate/limiting sanity check

Forecast: removing a released-energy step — does $U$ get more or less negative?

Loop with $E A = 0$ : $U = Δ H_{f} - Δ H_{s u b} - I E - \frac{1}{2} D - 0 = - 411 - 108 - 496 - 122$ . Why? A zero term simply drops out — a clean limiting case.
$- 411 - 108 - 496 - 122 = - 1137$ .

$U_{hyp} = - 1137 kJ mol^{- 1}$

Verify: compare to real NaCl $U = - 788$ . Setting $E A$ from $- 349$ up to $0$ made $U$ more negative by exactly $349$ : $- 788 - 349 = - 1137$ . ✓ Interpretation: if the atom no longer released the $349$ during electron capture, the lattice step must supply that missing stabilisation — the sign bookkeeping self-consistently accounts for it. Limiting behaviour is sensible. ✓

Ex 8 — C7: real-world word problem

Forecast: which single step dominates the energy release?

Sum the endothermic (up) steps: $Δ H_{s u b} + I E + \frac{1}{2} D = 108 + 496 + 122 = 726 kJ mol^{- 1}$ . Why? These together are the "cost" of making the gaseous ions.
Compare with the exothermic lattice step $U = - 788$ . Why? This is the "payback" — and it's larger in magnitude than the whole cost, which is why $Δ H_{f} < 0$ at all.
Net: $726 + (- 788) + E A = 726 - 788 - 349 = - 411 = Δ H_{f}$ . The lattice + EA overwhelm the ionisation cost. Why? A large negative $U$ is what makes the solid stable and hard to melt.

$Cost = + 726, U = - 788, overall = - 411 kJ mol^{- 1}$

Verify: $726 - 788 - 349 = - 411 = Δ H_{f}$ from Ex 1. ✓ Since $U \propto q_{+} q_{-} / r$ , KI's larger ions (bigger $r$ ) give a smaller $∣ U ∣$ → weaker lattice → lower melting point, exactly as observed. See Solubility and enthalpy of hydration for the water side of the same competition. ✓

Ex 9 — C8: exam twist, missing datum via a second known cycle

Forecast: the loop has exactly one gap — which term do we isolate?

Rearrange the master loop for $Δ H_{s u b}$ : $Δ H_{s u b} = Δ H_{f} - I E - \frac{1}{2} D - E A - U$ . Why? Any single unknown can be isolated; here it's $Δ H_{s u b}$ .
Substitute: $Δ H_{s u b} = - 411 - 496 - 122 - (- 349) - (- 788)$ . Why? Two double negatives: $- (- 349) = + 349$ and $- (- 788) = + 788$ .
$- 411 - 496 - 122 + 349 + 788 = + 108$ .

$Δ H_{s u b} (Na) = + 108 kJ mol^{- 1}$

Verify: matches the standard tabulated value used in Ex 1. Plug back: $U = - 411 - 108 - 496 - 122 + 349 = - 788$ . ✓ Positive, as sublimation must absorb energy. ✓

Active Recall

Recall In Ex 1, why does the last term become

$+ 349$ ? $E A = - 349$ and the formula subtracts it: $- (- 349) = + 349$ (double negative). (Cell C2)

Recall What three things change going from NaCl to MgCl₂?

Two ionisations $I E_{1} + I E_{2}$ , full $D (Cl_{2})$ , and $2 \times E A$ . (Cell C3)

Recall Why is

$E A_{2}$ of oxygen positive? Forcing a second electron onto O⁻ works against electron–electron repulsion, so it costs energy: $E A_{2} = + 798$ . (Cell C5)

Recall Setting

$E A = 0$ in NaCl made $U$ change how? More negative by $349$ (from $- 788$ to $- 1137$ ) — the lattice must supply the stabilisation the missing EA no longer provided. (Cell C6)

Connections

Parent topic
Hess's Law and state functions
Enthalpy of formation
Ionisation energy trends
Electron affinity
Bond dissociation enthalpy
Ionic bonding and Coulomb's law
Kapustinskii equation
Solubility and enthalpy of hydration