Born-Haber cycle revisited — calculating lattice energy: Exercises

Before we start, let us pin down the tools so nothing is used before it is named.

Definition The five energy steps (all in

$kJ mol^{- 1}$ )

$Δ H_{f}$ = enthalpy of formation — energy change making the solid directly from elements in their standard states. See Enthalpy of formation.
$Δ H_{s u b}$ = sublimation/atomisation — energy to turn 1 mol of solid metal into 1 mol of gaseous metal atoms. Always $+$ (you pull atoms apart).
$I E$ = ionisation energy — energy to strip an electron off a gaseous atom. Always $+$ . See Ionisation energy trends.
$\frac{1}{2} D$ = half the bond dissociation enthalpy — energy to break one bond of a diatomic like $Cl_{2}$ to get ONE atom. Always $+$ . See Bond dissociation enthalpy.
$E A$ = electron affinity — energy change when a gaseous atom gains an electron. Usually $-$ (energy released). See Electron affinity.
$U$ = lattice energy — energy change assembling gaseous ions into the solid. Strongly $-$ .

The energy ladder below (WHAT the whole cycle looks like) is your reference for every sign in this page: up = absorb (+), down = release (−).

Figure — Born-Haber cycle revisited — calculating lattice energy

Level 1 — Recognition

L1.1 — Name the step

For NaCl, which single symbol represents the step $Na (s) ⟶ Na (g)?$ And is it $+$ or $-$ ?

Recall Solution

Turning solid metal into gaseous atoms is sublimation/atomisation, symbol $Δ H_{s u b}$ . You are pulling atoms out of a solid lattice into free gas — that costs energy, so it is positive ( $+$ ). On the ladder (Figure s01) it is an upward step.

L1.2 — Spot the sign

A student writes $E A (Cl) = + 349 kJ mol^{- 1}$ in a Born–Haber cycle. Standard tables list "349 kJ/mol released." What sign should appear in the cycle, and why?

Recall Solution

"Released" energy is negative in an enthalpy cycle. So use $E A (Cl) = - 349 kJ mol^{- 1} .$ On the ladder this is a downward step (Figure s01, green). The word "released" tells you the direction, not the sign — you supply the minus.

Level 2 — Application

L2.1 — Lattice energy of NaCl

Given ( $kJ mol^{- 1}$ ): $Δ H_{f} = - 411$ , $Δ H_{s u b} = + 108$ , $I E = + 496$ , $\frac{1}{2} D = + 122$ , $E A = - 349$ . Find $U$ .

Recall Solution

Plug into the master equation: $U = Δ H_{f} - Δ H_{s u b} - I E - \frac{1}{2} D - E A$ $U = (- 411) - (108) - (496) - (122) - (- 349)$ The last term: subtracting a negative adds $349$ . $U = - 411 - 108 - 496 - 122 + 349 = - 788 kJ mol^{- 1}$ Large and negative — that is why NaCl is a hard, high-melting solid.

L2.2 — Lattice energy of KF

Given ( $kJ mol^{- 1}$ ): $Δ H_{f} = - 567$ , $Δ H_{s u b} (K) = + 89$ , $I E (K) = + 419$ , $\frac{1}{2} D (F_{2}) = + 79$ , $E A (F) = - 328$ . Find $U$ .

Recall Solution

$U = - 567 - 89 - 419 - 79 - (- 328)$ $U = - 567 - 89 - 419 - 79 + 328 = - 826 kJ mol^{- 1}$ More negative than NaCl because F⁻ and K⁺ pack a bit tighter — recall $U \propto q_{+} q_{-} / r$ from Ionic bonding and Coulomb's law.

L2.3 — Solve for enthalpy of formation

For AgCl: $U = - 905$ , $Δ H_{s u b} (Ag) = + 284$ , $I E (Ag) = + 731$ , $\frac{1}{2} D (Cl_{2}) = + 122$ , $E A (Cl) = - 349$ . Find $Δ H_{f}$ .

Recall Solution

This time $Δ H_{f}$ is the unknown, so use the equation in its un-rearranged form: $Δ H_{f} = Δ H_{s u b} + I E + \frac{1}{2} D + E A + U$ $Δ H_{f} = 284 + 731 + 122 + (- 349) + (- 905)$ $Δ H_{f} = 284 + 731 + 122 - 349 - 905 = - 117 kJ mol^{- 1}$

Level 3 — Analysis

L3.1 — MgCl₂: matching the stoichiometry

For MgCl₂: $Δ H_{f} = - 641$ , $Δ H_{s u b} (Mg) = + 148$ , $I E_{1} = + 738$ , $I E_{2} = + 1451$ , $D (Cl_{2}) = + 244$ (full bond), $E A (Cl) = - 349$ per Cl. Find $U$ .

Recall Solution

The formula unit needs two electrons off Mg and two Cl⁻ ions: $U = Δ H_{f} - Δ H_{s u b} - (I E_{1} + I E_{2}) - D - 2 E A$ $U = - 641 - 148 - (738 + 1451) - 244 - 2 (- 349)$ $U = - 641 - 148 - 2189 - 244 + 698 = - 2524 kJ mol^{- 1}$ Roughly three times NaCl's value: the $+ 2$ charge on Mg and its small radius make $q_{+} q_{-} / r$ far bigger.

L3.2 — Find electron affinity of Br (KBr)

For KBr: $U = - 671$ , $Δ H_{f} = - 394$ , $Δ H_{s u b} (K) = + 89$ , $I E (K) = + 419$ , $\frac{1}{2} D (Br_{2}) = + 112$ . Find $E A (Br)$ .

Recall Solution

$E A$ is the unknown, so isolate it: $E A = Δ H_{f} - Δ H_{s u b} - I E - \frac{1}{2} D - U$ $E A = - 394 - 89 - 419 - 112 - (- 671)$ $E A = - 394 - 89 - 419 - 112 + 671 = - 343 kJ mol^{- 1}$ Negative, as expected: Br gaining an electron releases energy.

L3.3 — Solve for the second ionisation energy (CaO)

For CaO: $Δ H_{f} = - 635$ , $Δ H_{s u b} (Ca) = + 178$ , $I E_{1} (Ca) = + 590$ , $\frac{1}{2} D (O_{2}) = + 249$ , combined electron affinity $E A (O \to O^{2 -}) = + 657$ (net, endothermic — see note below), $U = - 3401$ . Find $I E_{2} (Ca)$ .

Recall Solution

Ca loses two electrons, so both $I E_{1}$ and $I E_{2}$ appear. O gains two electrons; the net affinity given (+657) already covers both electron additions. Master equation: $Δ H_{f} = Δ H_{s u b} + I E_{1} + I E_{2} + \frac{1}{2} D + E A + U$ Isolate $I E_{2}$ : $I E_{2} = Δ H_{f} - Δ H_{s u b} - I E_{1} - \frac{1}{2} D - E A - U$ $I E_{2} = - 635 - 178 - 590 - 249 - 657 - (- 3401)$ $I E_{2} = - 635 - 178 - 590 - 249 - 657 + 3401 = + 1092 kJ mol^{- 1}$ Positive and larger than $I E_{1}$ — pulling a second electron off an already-positive ion is harder.

Level 4 — Synthesis

L4.1 — Chain two cycles: theoretical vs experimental $U$

For AgF the Born–Haber (experimental) lattice energy is $U_{e x p} = - 960 kJ mol^{- 1}$ , while the purely electrostatic theoretical model predicts $U_{t h eo} = - 870 kJ mol^{- 1}$ . Compute the discrepancy $U_{e x p} - U_{t h eo}$ and state what a large negative discrepancy physically means.

Recall Solution

$U_{e x p} - U_{t h eo} = (- 960) - (- 870) = - 90 kJ mol^{- 1}$ The experimental lattice is $90 kJ mol^{- 1}$ more exothermic than the ionic-only model predicts. A purely ionic (point-charge) model cannot account for this extra stability, so the "missing" energy signals partial covalent character — the ions share electron density more than pure spheres would. Silver halides are a classic case.

L4.2 — Build the whole cycle from raw data (LiF)

Only these are known for LiF ( $kJ mol^{- 1}$ ): $Δ H_{s u b} (Li) = + 159$ , $I E (Li) = + 520$ , $\frac{1}{2} D (F_{2}) = + 79$ , $E A (F) = - 328$ , and the directly measured $Δ H_{f} = - 616$ . (a) Find $U$ . (b) Then predict $Δ H_{f}$ of a hypothetical "LiF₂" if it required an extra $I E_{2} (Li) = + 7298$ and a second $\frac{1}{2} D + E A$ pair, keeping the same $U$ — comment.

Recall Solution

(a) Standard route: $U = - 616 - 159 - 520 - 79 - (- 328) = - 616 - 159 - 520 - 79 + 328 = - 1046 kJ mol^{- 1}$ (b) For imaginary LiF₂ we add $I E_{2}$ , one more $\frac{1}{2} D$ , one more $E A$ , and reuse $U = - 1046$ : $Δ H_{f} = Δ H_{s u b} + (I E_{1} + I E_{2}) + 2 \cdot \frac{1}{2} D + 2 E A + U$ $Δ H_{f} = 159 + (520 + 7298) + 2 (79) + 2 (- 328) + (- 1046)$ $Δ H_{f} = 159 + 7818 + 158 - 656 - 1046 = + 6433 kJ mol^{- 1}$ Comment: hugely positive (endothermic) — the enormous $I E_{2}$ of Li ( $7298$ , ripping into a noble-gas core) is nowhere near paid back by lattice energy. So "LiF₂" would not form; Li is stubbornly $+ 1$ . The cycle predicts non-existence quantitatively.

Level 5 — Mastery

L5.1 — Two unknowns removed by two facts (NaH)

For sodium hydride NaH you are told two things:

$Δ H_{f} (NaH) = - 56 kJ mol^{- 1}$ .
The lattice energy from a reliable calculation is $U = - 808 kJ mol^{- 1}$ .

Known: $Δ H_{s u b} (Na) = + 108$ , $I E (Na) = + 496$ , $\frac{1}{2} D (H_{2}) = + 218$ . Find the electron affinity of hydrogen, $E A (H)$ , and comment on its unusual value.

Recall Solution

One equation, one unknown ( $E A$ ): $E A = Δ H_{f} - Δ H_{s u b} - I E - \frac{1}{2} D - U$ $E A = - 56 - 108 - 496 - 218 - (- 808)$ $E A = - 56 - 108 - 496 - 218 + 808 = - 70 kJ mol^{- 1}$ Comment: $E A (H) \approx - 70$ is small in magnitude — hydrogen only weakly wants an extra electron (compare Cl at $- 349$ ). Yet NaH still forms because Na is easy to ionise and the H⁻ lattice is stabilising. The cycle lets us extract this hard-to-measure $E A (H)$ from bulk thermodynamics.

L5.2 — Full audit: find the planted error

A student's NaCl worksheet reads: $U = Δ H_{f} - Δ H_{s u b} - I E - \frac{1}{2} D - E A$ $U = - 411 - 108 - 496 - 122 - 349 = - 1486 kJ mol^{- 1}$ Data are the L2.1 values. Identify the mistake, give the corrected $U$ , and explain the size of the error.

Recall Solution

The error is the last term: $E A = - 349$ , so $- E A = - (- 349) = + 349$ , not $- 349$ . The student subtracted $349$ instead of adding it. Corrected: $U = - 411 - 108 - 496 - 122 + 349 = - 788 kJ mol^{- 1}$ The error size is $2 \times 349 = 698 kJ mol^{- 1}$ (you had the sign backwards, so you are off by twice the term). That is why the wrong answer, $- 1486$ , is nearly double the true value — a red flag that a sign flipped.

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