Calculating a Salt's Lattice Enthalpy

Introduction

The Born–Haber cycle becomes a quantitative tool when all but one enthalpy are known. For NaCl, the direct standard formation enthalpy equals sodium atomization, sodium ionization, half chlorine bond dissociation, chlorine electron gain and lattice formation. Rearranging the signed sum gives the lattice term.

Core explanation

Start with the overall reaction Na(s) + ½Cl₂(g) → NaCl(s). Label its standard formation enthalpy Δ fH°. The alternative route is Na(s) → Na(g), then Na(g) → Na⁺(g) + e⁻, then ½Cl₂(g) → Cl(g), then Cl(g) + e⁻ → Cl⁻(g), then Na⁺(g) + Cl⁻(g) → NaCl(s). Hess's law gives Δ fH° = ΔH atom(Na) + IE₁(Na) + ½D(Cl₂) + ΔH eg(Cl) + ΔH lattice,formation.

Solve algebraically: ΔH lattice,formation = Δ fH° − ΔH atom − IE₁ − ½D − ΔH eg. The electron gain term is often negative, so subtracting it adds its magnitude in the rearranged arithmetic. A reliable method is to sum all four known preparatory steps first with their signs, then subtract that sum from the formation enthalpy.

Use illustrative rounded data at one common reference condition: Δ fH°[NaCl(s)] = −411 kJ mol⁻¹, Na atomization +108, Na first ionization +496, Cl₂ full bond dissociation +242, and Cl first electron gain −349 kJ mol⁻¹. Half the Cl₂ dissociation is +121. The known-step sum is 108 + 496 + 121 − 349 = +376 kJ mol⁻¹. Therefore lattice formation is −411 − (+376) = −787 kJ mol⁻¹. The reverse lattice dissociation is +787 kJ mol⁻¹ under this illustrative enthalpy convention.

The magnitude says the crystal is substantially stabilized relative to separated gaseous ions. It does not mean NaCl formation from elements releases 787 kJ mol⁻¹; the actual overall formation enthalpy is −411 kJ mol⁻¹ in the example because several steps cost energy. Nor is the +787 kJ dissociation value the heat of dissolving NaCl in water. Hydration changes the endpoint and energy balance.

Different tables may give slightly different values because of temperature, 1 bar versus older 1 atm convention, measured versus rounded enthalpies, or whether they report lattice energy ΔU instead of lattice enthalpy ΔH. Do not mix a positive dissociation number from one source with a negative formation convention in the same equation without changing its sign.

To check the result, substitute it back: +108 + 496 + 121 − 349 − 787 = −411 kJ mol⁻¹. The arithmetic and net chemical equation both close. This back-substitution is often faster than searching for an arithmetic sign error after the fact.

The same method generalizes to salts with higher charges, but each extra electron and halogen atom introduces an additional ionization or electron-gain contribution. The NaCl example is valuable because all species and coefficients are simple enough to reveal the logic.

Step-by-step reasoning

1. Write overall NaCl formation and each gas-phase cycle step. 2. Confirm the lattice term is for gaseous ions forming solid. 3. Halve the full Cl₂ dissociation value. 4. Sum known signed steps and subtract from Δ fH°. 5. Substitute back and reverse sign if dissociation is requested.

Visual explanation

Draw a vertical energy ladder with +108, +496, +121 and −349 steps leading from elements to gaseous ions. A large downward lattice arrow reaches NaCl(s). Place the direct −411 arrow beside the ladder. The unknown lattice arrow must make both routes end at the same enthalpy level.

Real-world analogy

If a final account balance is −411 after known transactions totaling +376, the missing transaction must be −787. A Born–Haber calculation is the same signed-balance logic, with the added requirement that each transaction correspond to a correctly balanced chemical state change.

Real-world example

A textbook may quote NaCl lattice enthalpy as a positive separation cost. A student solving the formation cycle must insert the negative crystal-formation value instead. Writing the ion-to-solid arrow above the number prevents a convention error.

Why?

Why is the lattice-formation term strongly negative? Opposite gaseous ions become stabilized throughout a three-dimensional ionic lattice, releasing substantial energy as they assemble.

Common misconception

“The lattice term alone equals the salt's formation enthalpy.” Atomization, ionization, chlorine bond splitting and electron gain also contribute; the sum gives formation enthalpy.

Worked example

Using the stated rounded values, known steps total +376 kJ mol⁻¹. Let L be gas-ion-to-crystal enthalpy. Hess's law is −411 = +376 + L, so L = −787 kJ mol⁻¹. If the exam defines lattice enthalpy as solid-to-gas-ion dissociation, report +787 kJ mol⁻¹ instead. Both values describe opposite directions of the same state change under this data set.

Quick check

1. What chlorine bond term belongs in a one-mole NaCl cycle if D(Cl₂) is for Cl₂ → 2Cl? Answer: Half the full dissociation enthalpy, because ½Cl₂ produces one Cl atom.

Exam focus

Show all signed steps, then solve one algebraic equation. Check units and the definition of lattice direction. Back-substitute to confirm both the numerical sum and the overall formation equation.

Advanced insight

An energy-based lattice quantity can differ from enthalpy because the gaseous-ion side has gas pressure-volume contribution. For one mole of NaCl formed from two moles of separated ideal gaseous ions, the ΔH–ΔU relation involves a negative two-mole gas-count term at the stated temperature. Precise sources must specify which quantity they report.

Summary

The NaCl Born–Haber balance subtracts the signed atomization, ionization, half bond-dissociation and electron-gain terms from Δ fH° to obtain lattice formation. In the illustrative data set it is −787 kJ mol⁻¹; dissociation is +787. The cycle and sign convention must be explicit.

Practice questions

1. If Δ fH° = −400 and all known steps sum to +350 kJ mol⁻¹, find lattice formation enthalpy. Answer: −750 kJ mol⁻¹, because −400 = +350 + L. 2. What is lattice dissociation enthalpy if lattice formation is −750 kJ mol⁻¹? Answer: +750 kJ mol⁻¹ for the reverse matched process. 3. Why cannot +750 kJ mol⁻¹ automatically be called the heat of dissolving the salt? Answer: Dissolution ends with hydrated aqueous ions, whereas lattice dissociation ends with separated gaseous ions.