§ C-24 Born–Haber Thermodynamic Cycle & Ionic Lattice Energy U_L (Grade 10–11 · Crystal Thermochemistry)
Thermodynamic balance of sublimation, dissociation, ionization, electron affinity, and ionic lattice energy Uₗat. Although transferring an electron from an isolated sodium atom to chlorine is endothermic, the enormous electrostatic energy released upon assembling the 3D ionic lattice makes salt formation strongly exothermic.
Theory Summary & Scientific Mechanism
The Born–Haber cycle applies Hess’s Law to relate the standard enthalpy of formation Δ H_f^° of an ionic solid MX(s) to its lattice energy Uₗat by constructing a closed thermodynamic path through gaseous atoms and ions.
Converting standard elements into gaseous ions requires endothermic sublimation (Δ Hₛub), bond dissociation ((1)/(2)D_X₂), and ionization (I₁), partially offset by the exothermic electron affinity (Eₑa) of the nonmetal.
The overall formation of the salt is strongly exothermic because condensing gaseous ions into a 3D crystal lattice releases a massive lattice enthalpy Uₗat = Δ Hₛub + (1)/(2)D_X₂ + I₁ + Eₑa - Δ H_f^°, which scales with ionic charges and inverse interionic distance according to the Born–Landé equation.
Everyday Analogy
Building an ionic crystal is like climbing a small hill to jump into a deep canyon: you invest energy to vaporize and ionize the elements, then get a massive exothermic payoff Uₗat when Na⁺ and Cl⁻ snap into a 3D lattice!
Interactive Question & Derivation
Why cannot the ionic lattice enthalpy Uₗat of a crystal like NaCl(s) be measured directly in a single calorimeter run, and how is it determined via the Born–Haber cycle?
- A: By Hess’s law, the closed thermodynamic cycle equates the standard enthalpy of formation Δ H_f^° to the sum of sublimation Δ Hₛub, dissociation (1)/(2)D_X₂, ionization I₁, electron affinity Eₑa, and lattice formation -Uₗat
- B: Lattice energy depends only on the melting point of the metal and is independent of ionic radii or charges
- C: Electron affinity and ionization energy are always equal in magnitude and cancel out in every salt
Study Hint: By Hess’s law, the enthalpy of formation Δ H_f^° equals the sum of sublimation, dissociation, ionization, electron affinity, and -Uₗat.
Proof & Explanation: Because gaseous ions (M⁺(g) + X⁻(g)) cannot be weighed into a calorimeter under standard conditions, Born and Haber applied Hess’s Law to construct a closed thermodynamic cycle: Δ H_f^° = Δ Hₛub + (1)/(2)D(X₂) + I₁(M) + Eₑa(X) - Uₗat. The endothermic steps of atomization and ionization are more than compensated by the huge exothermic electrostatic attraction when gaseous ions condense into the 3D crystal lattice (Uₗat ∝ (N_A A z⁺ z⁻ e²)/(4πε₀ r₀)).
Key Equations & 3D Simulation Parameters
Δ H_f^° = Δ Hₛub + (1)/(2)D_X₂ + I₁ + Eₑa - Uₗat, Uₗat = (N_A M z⁺ z⁻ e²)/(4πε₀ r₀)≤ft(1 - (1)/(n) )
- Metal sublimation enthalpy ΔHₛub(M): 75 … 180 kJ/mol (default: 108 kJ/mol)
- Halogen dissociation enthalpy ½D(X₂): 75 … 160 kJ/mol (default: 121 kJ/mol)
- Metal first ionization energy I₁(M → M⁺): 375 … 600 kJ/mol (default: 496 kJ/mol)
- Crystal lattice energy Uₗat(M⁺ + X⁻ → MX): 650 … 1050 kJ/mol (default: 787 kJ/mol)