Why is the O²⁻ step so endothermic?

Adding one electron to O(g) is exothermic, but adding a second electron to O⁻ is strongly opposed by electron–electron repulsion; textbooks often give one combined positive enthalpy for O(g)+2e⁻→O²⁻(g) in the cycle.

Is U the same as lattice energy from the Born–Landé equation?

Same conceptual quantity at 0 K vs enthalpy at 298 K differ slightly (Δ(pV) and zero-point motion). This page uses the cycle enthalpy closure, not a Madelung-sum model.

Can I edit the numbers?

Not on this page — it is a reference closure for two classic examples. Adjusting one step would require re-solving the cycle consistently.

Born–Haber Cycle (NaCl & CaO)

The Born–Haber cycle is a Hess-law bookkeeping loop that relates the standard enthalpy of formation of an ionic solid to the enthalpies of converting elements in their reference states into gas-phase ions plus the lattice enthalpy U (conventionally quoted as a positive magnitude for separating one mole of solid into gaseous ions). For NaCl the path uses sodium sublimation, half the chlorine bond energy, sodium ionization, chlorine electron affinity, and finally Na⁺(g)+Cl⁻(g)→NaCl(s) with ΔH = −U. For CaO the same idea uses calcium sublimation, first-plus-second ionization, half the oxygen bond energy, and a single tabulated endothermic bundle for O(g)+2e⁻→O²⁻(g) reflecting the very unfavorable second electron attachment. With formation enthalpy and all other steps taken from tables, U follows from closure: ΔH_f° equals the sum of the listed steps including −U. This page fixes rounded textbook-style numbers for NaCl and CaO, tabulates each contribution, computes U, and draws a cumulative enthalpy ladder to the formation level.

Who it's for: General chemistry thermochemistry after Hess’s law; complements lattice-energy discussions in inorganic and solid-state introductions.

Key terms

Born–Haber cycle
lattice enthalpy
electron affinity
ionization energy
bond dissociation
standard formation enthalpy

Step	ΔH° (kJ/mol)
Na(s) → Na(g) (sublimation)	+107.3
½ Cl₂(g) → Cl(g)	+121.7
Na(g) → Na⁺(g) + e⁻	+495.8
Cl(g) + e⁻ → Cl⁻(g)	-348.8
Mⁿ⁺(g) + Xᵐ⁻(g) → MX(s) (−U)	−787.2
Net → standard formation	-411.2

Numbers are rounded literature values for teaching; O²⁻ step bundles first + second electron attachment.

Live graphs

How it works

The Born–Haber cycle closes standard enthalpies of formation for ionic solids from sublimation, bond dissociation, ionization, electron gain (often a combined O + 2e⁻ → O²⁻ term), and lattice enthalpy U (endothermic to break the crystal into gas ions). Here U = Σ(ΔH steps before lattice) − ΔH_f° with literature-rounded numbers for NaCl and CaO; the diagram shows cumulative H along the cycle down to ΔH_f° (green dashed).

Key equations

ΔH_f° = Σ ΔH(cycle to gas ions) − U

U = Σ ΔH(gas-ion steps) − ΔH_f°

Frequently asked questions

Why is the O²⁻ step so endothermic?: Adding one electron to O(g) is exothermic, but adding a second electron to O⁻ is strongly opposed by electron–electron repulsion; textbooks often give one combined positive enthalpy for O(g)+2e⁻→O²⁻(g) in the cycle.
Is U the same as lattice energy from the Born–Landé equation?: Same conceptual quantity at 0 K vs enthalpy at 298 K differ slightly (Δ(pV) and zero-point motion). This page uses the cycle enthalpy closure, not a Madelung-sum model.
Can I edit the numbers?: Not on this page — it is a reference closure for two classic examples. Adjusting one step would require re-solving the cycle consistently.

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