Visual walkthrough — Born-Haber cycle revisited — calculating lattice energy

We are chasing one number: the lattice energy $U$, the energy released when free floating gas ions snap together into a solid crystal. We cannot measure it directly, so we will trap it inside a loop of steps we can measure. Let us draw that loop.

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Step 1 — Two dots, two roads

WHAT. Put a starting point and an ending point on the page. Start = the plain elements (a lump of sodium metal, a jar of chlorine gas), each in its ordinary everyday state. End = the finished ionic solid, a crystal of table salt NaCl.

WHY. A "cycle" needs a start and a finish that both roads agree on. If two different journeys begin at the same place and end at the same place, and if the thing we are tracking (energy) only cares about where you are, not how you got there — then the two journeys must cost the same total. That single idea is Hess's Law, and it is the whole engine of this page.

PICTURE. The bottom dot is the elements; the top-right dot is the solid. The short thick arrow straight across is the easy road; the long arrow curving up and around (drawn faint for now) is the hard road we will build step by step.

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Step 2 — The direct road: enthalpy of formation

WHAT. Take the elements and let them react directly to form one mole of the solid. The single energy for this whole trip is the enthalpy of formation, written $\Delta H_f$.

$\underset{\text{metal lump}}{\underbrace{\text{Na}(s)}}+\underset{{\text{half a Cl}}_2\text{molecule}}{\underbrace{\frac{1}{2}{\text{Cl}}_2(g)}}⟶\underset{\text{the crystal}}{\underbrace{\text{NaCl}(s)}}\Delta H=\Delta H_f$

Reading the symbols: $\Delta$ means "change in", $H$ is heat content, and the little $f$ tags it as formation from elements. The $\frac{1}{2}$ in front of ${\text{Cl}}_2$ is not a fudge — one NaCl unit needs only one chlorine atom, and chlorine gas comes in pairs (${\text{Cl}}_2$), so we take half a molecule.

WHY. This is the ONE step on the easy road, and — crucially — it is measurable in a calorimeter. It is our known anchor. See Enthalpy of formation.

PICTURE. One bold amber arrow straight from the elements-dot to the solid-dot. It points downward because forming salt releases heat ($\Delta H_f=-411$ for NaCl, a negative number = downhill).

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Step 3 — The long road, part 1: make gaseous atoms

WHAT. Now we start the scenic route. First we must turn our raw elements into lonely gas-phase atoms, because the ions we eventually want live in the gas phase.

• The metal: heat solid sodium until single atoms boil off. This is the enthalpy of sublimation, $\Delta H_{sub}$.
• The non-metal: snap the ${\text{Cl}}_2$ bond to free one chlorine atom. Breaking a whole ${\text{Cl}}_2$ bond costs $D$, the Bond dissociation enthalpy; since we need only one Cl, we pay half: $\frac{1}{2}D$.

$\text{Na}(s)\stackrel{\Delta H_{sub}}{\to }\text{Na}(g),\frac{1}{2}{\text{Cl}}_2(g)\stackrel{\frac{1}{2}D}{\to }\text{Cl}(g)$

Term by term: $\Delta H_{sub}$ is the cost to vaporise one mole of the metal into free atoms; $\frac{1}{2}D$ is the cost to break enough ${\text{Cl}}_2$ bonds to hand us one mole of Cl atoms.

WHY. Both are endothermic — you must spend energy to tear atoms out of a solid and to snap a chemical bond. Nature never gives these for free.

PICTURE. Two separate upward arrows (energy going in), one for the metal ($\Delta H_{sub}=+108$) and one for the half-bond ($\frac{1}{2}D=+122$), rising off the elements baseline.

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Step 4 — The long road, part 2: charge the atoms up

WHAT. Neutral gas atoms are not ions yet. We must strip an electron off the metal and stuff it onto the non-metal.

• Rip an electron off sodium: this costs the ionisation energy $IE$.
• Give that electron to chlorine: this is the Electron affinity $EA$.

$\text{Na}(g)\stackrel{IE}{\to }{\text{Na}}^{+}(g)+e^{-},\text{Cl}(g)+e^{-}\stackrel{EA}{\to }{\text{Cl}}^{-}(g)$

Symbol check: $IE$ is always positive (endothermic) — it takes a hard tug to pull an electron off an atom, so this arrow goes up. $EA$ for chlorine is negative (exothermic) — an incoming electron slots into a hungry chlorine and releases energy, so this arrow goes down.

WHY. The whole point of the crystal is that it is built from charged particles. No charges, no electrostatic snap-together. So we pay to create the $+$ and $-$ ions.

PICTURE. From the atom level, one tall up arrow ($IE=+496$) creates ${\text{Na}}^{+}$; a short down arrow ($EA=-349$) creates ${\text{Cl}}^{-}$. The electron literally hops across from the sodium arrow's top to the chlorine.

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Step 5 — The final drop: the lattice energy itself

WHAT. We now have a cloud of free gaseous ions, ${\text{Na}}^{+}(g)$ and ${\text{Cl}}^{-}(g)$. Let them rush together into the crystal. The energy released is exactly our target, the lattice energy $U$.

${\text{Na}}^{+}(g)+{\text{Cl}}^{-}(g)\stackrel{U}{\to }\text{NaCl}(s)$

WHY. Opposite charges attract; when they collapse from far apart into a tightly packed lattice, they release a huge amount of energy — $U$ is strongly negative. This is Coulomb's law in action: $U\propto \frac{q_{+}{q}_{-}}{r}$, where $q_{+},q_{-}$ are the ion charges and $r$ is the distance between ion centres. Bigger charges or smaller ions ⇒ a steeper, deeper drop.

PICTURE. From the high ledge of gaseous ions, one enormous down arrow plunges all the way to the solid-dot we placed in Step 2. This is the step we could never measure directly — but now it is the only gap left in a closed loop.

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Step 6 — Close the loop and read off $U$

WHAT. Both roads now meet at the solid. Hess's Law says their totals are equal:

$\underset{\text{easy road}}{\underbrace{\Delta H_f}}=\underset{\text{long road, all five arrows added}}{\underbrace{\Delta H_{sub}+IE+\frac{1}{2}D+EA+U}}$

Every arrow that pointed up enters with a $+$ (energy spent); every arrow that pointed down carries its own negative sign. Now solve for the one thing we cannot measure — isolate $U$:

$\boxed{U=\Delta H_f-\Delta H_{sub}-IE-\frac{1}{2}D-EA}$

WHY. This is the payoff of the whole cycle: one equation, one unknown. Every term on the right is in a data book; $U$ is the leftover.

PICTURE. The full energy ladder: the elements baseline, the up-steps stacking to the ceiling of gaseous ions, then the giant $U$ drop, with the single $\Delta H_f$ arrow bridging bottom to solid. Total up = total down; the ladder closes.

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Step 7 — The degenerate & edge cases

WHAT. The cycle bends but never breaks when the compound is more complicated. Three cases to watch:

1. A $+2$ metal (e.g. Mg). Losing two electrons means paying two ionisation energies: $I{E}_1+I{E}_2$. On the ladder, the ionisation step becomes two stacked up-arrows.
2. Two anions (e.g. MgCl₂). Two Cl atoms are needed, so use the full $D({\text{Cl}}_2)$ (not $\frac{1}{2}D$) and two electron affinities, $2EA$.
3. Solving for a different unknown. If $U$ is already known (from theory / the Kapustinskii equation) but $EA$ is not, rearrange the same loop for $EA$ instead.

$U_{{\text{MgCl}}_2}=\Delta H_f-\Delta H_{sub}-(I{E}_1+I{E}_2)-D-2EA$ $EA=\Delta H_f-\Delta H_{sub}-IE-\frac{1}{2}D-U(\text{when }U\text{ is the known})$

WHY. The loop's logic never changes — Hess's Law is path-independent no matter how many electrons hop or how many bonds break. You only ever match the stoichiometry of the formula unit.

PICTURE. The NaCl ladder redrawn for MgCl₂: the ionisation section is now a taller double step ($I{E}_1$ then a much larger $I{E}_2$), and the dissociation and EA steps are drawn doubled. The final $U$ drop is far deeper (about $-2524$) because the $+2$ charge makes Coulomb's $q_{+}{q}_{-}/r$ much stronger.

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The one-picture summary

Everything above, compressed into a single labelled ladder: elements at the floor, five measurable up/down steps building to the gaseous-ion ceiling, the unmeasurable lattice drop closing the loop, and the direct $\Delta H_f$ arrow proving both roads agree.

Recall Feynman retelling — the whole walkthrough in plain words

I want to know how hard it is to rip a salt crystal back into a fog of separate charged specks — call that the lattice energy. I can't do that rip in the lab. So instead I take a completely different journey to the same salt crystal: I boil the sodium into single atoms (costs energy), snap a chlorine bond to free one chlorine (costs energy), yank an electron off the sodium (costs a lot), let that electron drop onto the chlorine (gives a little back), and finally let all those charged specks crash together into the crystal (gives back a huge amount — that's the number I want). Because the total energy of a journey depends only on start and finish, this long trip must cost exactly the same as the one short trip where the raw elements just react into salt directly. I know the cost of that short trip and every step of the long trip except the crash-together, so I make the two totals match and read the missing crash energy straight off. Money and energy both add up the same no matter which road you take.

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Connections

• 2.5.10 Born-Haber cycle revisited — calculating lattice energy (Hinglish)
• 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