Born-Haber cycle revisited — calculating lattice energy: Question bank

Before we start, one reminder of the master equation you are being tested against: $U = Δ H_{f} - Δ H_{s u b} - I E - \frac{1}{2} D - E A$ Here $U$ is lattice energy (energy to build the solid from gaseous ions), $Δ H_{f}$ is enthalpy of formation, $Δ H_{s u b}$ is metal atomisation, $I E$ is ionisation energy, $\frac{1}{2} D$ is half a bond dissociation enthalpy, and $E A$ is electron affinity.

True or false — justify

True or false: Lattice energy is always exothermic (negative).

True — pulling oppositely charged gaseous ions together always releases energy, so by the "gaseous ions → solid" convention

U

is negative; only under the reverse (lattice-breaking) convention would you quote it as positive.

True or false: The Born–Haber cycle would fail if enthalpy were not a state function.

True — the whole method relies on both paths giving the same total energy; that path-independence IS the definition of a state function, so without it you couldn't equate the two routes.

True or false: A larger (more negative) lattice energy always means a more stable compound overall.

False — stability of the compound is governed by

Δ H_{f}

, which also pays for sublimation, ionisation and dissociation; a huge

U

can still be outweighed by huge ionisation costs (why high-charge compounds sometimes don't form).

True or false: You can measure lattice energy directly in a calorimeter.

False — free gaseous

M^{+}

and

X^{-}

ions cannot be bottled and condensed, so

U

is only ever obtained indirectly through the cycle.

True or false: For MgCl₂ you use two electron affinities because Mg loses two electrons.

False — you double

E A

because there are two Cl atoms each gaining one electron; the two electrons lost by Mg are counted by

I E_{1} + I E_{2}

, a separate matter.

True or false: Electron affinity of an atom is always exothermic.

False — adding the first electron is usually exothermic, but adding a second (e.g. forming

O^{2 -}

) is endothermic because you push an electron onto an already-negative ion against repulsion.

True or false: Swapping

Δ H_{s u b}

for a bond dissociation term for a metal is harmless if the numbers are close.

False — metals form monatomic solids and are sublimed, not dissociated; using the wrong physical step means the cycle no longer represents real processes even if a coincidence of numbers looks right.

True or false: The order in which you list the Born–Haber steps affects the final value of

U

False — because enthalpy is a state function only the set of steps and their signs matter, not their sequence; the "Some Idiots Don't Enjoy Lattices" order is just a memory aid.

Spot the error

Spot the error: "

U = Δ H_{f} - Δ H_{s u b} - I E - \frac{1}{2} D + E A

The last sign is wrong — every Path-2 step is subtracted when isolating

U

, so it must be

- E A

; the "+" would double-count the electron-affinity contribution.

Spot the error: For NaCl a student writes

U = (- 411) - 108 - 496 - 122 - 349

They forgot that

E A (Cl) = - 349

is already negative, so

- E A = - (- 349) = + 349

; subtracting a positive

349

instead flips the answer badly.

Spot the error: For MgCl₂ a student uses

\frac{1}{2} D (Cl_{2})

MgCl₂ needs two Cl atoms, so the full bond

D (Cl_{2})

is required; halving it under-counts the atomisation of chlorine by one bond's worth.

Spot the error: "Since EA releases energy, adding it makes the cycle more endothermic."

Contradiction — energy released lowers the total, so a negative

E A

makes the descent more exothermic, not more endothermic; the sign and the words disagree.

Spot the error: "

Δ H_{f}

of NaCl is positive because forming bonds releases energy."

The reasoning is backwards: releasing energy makes

Δ H_{f}

negative (

- 411

kJ/mol for NaCl); a positive value would mean formation absorbs energy.

Spot the error: A cycle diagram shows the lattice-energy arrow pointing up the energy ladder.

Building the lattice releases energy, so its arrow must point down; an upward arrow implies the exothermic step absorbs energy and the ladder will fail to close.

Spot the error: "For MgO we use

I E_{1}

of Mg and

E A_{1}

of O only."

MgO involves

Mg^{2 +}

and

O^{2 -}

, so you need

I E_{1} + I E_{2}

for magnesium and

E A_{1} + E A_{2}

for oxygen (the second EA being endothermic).

Why questions

Why do we subtract every Path-2 term when solving for

U

, even the endothermic ones?

Because

Δ H_{f} = Δ H_{s u b} + I E + \frac{1}{2} D + E A + U

; isolating

U

moves all the others to the other side, and moving any term across an equals sign flips its sign regardless of whether it was endo- or exothermic.

Why is the lattice energy of MgCl₂ far more negative than that of NaCl?

The Mg²⁺ charge is double Na⁺ and the ion is smaller, so via Coulomb's law

U \propto \frac{q _{+} q _{-}}{r}

the attraction is much stronger, giving a far larger release.

Why must the atomisation term for a metal differ from that for a diatomic gas?

A metal's atoms are held in a solid lattice and are freed by sublimation (

Δ H_{s u b}

), whereas diatomic gas atoms are freed by breaking a covalent bond (

D

); these are physically distinct processes with different energies.

Why can the same master equation solve for

E A

instead of

U

The cycle is one linear equation in several knowns and one unknown; algebra doesn't care which symbol is unknown, so you isolate whichever term (e.g.

E A

) is missing.

Why does the energy ladder have to "close"?

Because both paths connect the same start and end states and enthalpy is path-independent, the total upward (endothermic) steps must exactly equal the total downward (exothermic) steps; a mismatch signals a sign or stoichiometry mistake.

Why is a positive quoted electron affinity a common source of error?

Many textbooks report EA as a positive "energy released" magnitude, but a Born–Haber enthalpy cycle needs the signed value: energy released is negative, so you must insert

- 349

for Cl, not

+ 349

Why doesn't a large lattice energy alone guarantee a compound will form?

Formation must "pay" for sublimation, ionisation and dissociation first; if those endothermic costs exceed the combined release from

E A

and

U

, then

Δ H_{f}

is positive and the compound is thermodynamically unfavourable.

Edge cases

Edge case: What happens in the cycle for an ion needing a second, endothermic electron affinity like O²⁻?

The second EA term is positive and must be added into Path 2; forgetting its endothermic sign wrongly inflates the predicted lattice energy.

Edge case: How does the cycle change for a metal that is a liquid or gas at standard state?

You still need an atomisation term to reach gaseous atoms, but it may be smaller (a gas needs none, a liquid needs only vaporisation) — always use the actual enthalpy to reach isolated gaseous atoms.

Edge case: For Br₂ (a liquid) what does

\frac{1}{2} D

actually include?

It must include both vaporising the liquid to gaseous Br₂ and breaking the bond, per mole of Br atoms needed; using only the gas-phase bond energy under-counts the atomisation.

Edge case: If a Born–Haber calculation gives a positive lattice energy, what does that signal?

Almost certainly a sign error — real lattice energies (gaseous-ions → solid convention) are negative, so a positive result means a term's sign was mishandled, not that physics changed.

Edge case: What if the experimental and Born–Haber lattice energies disagree markedly?

The bonding is not purely ionic — significant covalent character (polarisation) makes the real lattice more stable than a point-charge model predicts, revealing a limitation of the ionic assumption behind the theoretical estimate.

Edge case: How does the cycle connect to whether a salt dissolves?

A very large lattice energy makes the crystal hard to break apart, so unless the enthalpy of hydration of the ions is comparably large, dissolution is enthalpically unfavourable.

Recall One-line self-test: what single fact makes the whole cycle possible?

Enthalpy is a state function — its change depends only on start and end states, so two different routes from elements to the ionic solid must give equal totals.

Connections

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