Chemistry · Ch 7 — Thermodynamics

Lattice Energy and the Born-Haber Cycle

7.9

Lattice Energy and the Born-Haber Cycle

Lattice energy, $\Delta H_{lattice}$ (also called lattice enthalpy), is defined as the amount of energy required to completely remove the constituent ions of one mole of a crystal from its crystal lattice, out to infinite separation. For sodium chloride:

NaCl(s)\rightarrow Na^+(g)+Cl^-(g),\qquad \Delta H_{lattice}=+788\ \text{kJ mol}^{-1}

This tells us directly that 788 kJ of energy has to be supplied to pull the Na $^+$ and Cl $^-$ ions apart from just one mole of solid NaCl.

The Born-Haber cycle. Lattice energy cannot be measured directly by experiment -- there is no calorimeter that can watch a crystal fly apart into gaseous ions. Instead, it is calculated indirectly using the Born-Haber cycle, named after the two German scientists Max Born and Fritz Haber who developed it. The cycle applies Hess's law to the formation of a simple ionic solid, such as an alkali-metal halide $MX$ , formed from the reaction of a metal with a halogen (or another non-metallic element such as oxygen), by breaking the overall formation into five distinct, individually-measurable steps:

$\Delta H_1$ -- enthalpy of SUBLIMATION, $M(s)\rightarrow M(g)$
$\Delta H_2$ -- enthalpy of DISSOCIATION of $\tfrac{1}{2}X_2(g)\rightarrow X(g)$
$\Delta H_3$ -- IONISATION energy, $M(g)\rightarrow M^+(g)$
$\Delta H_4$ -- ELECTRON AFFINITY, the enthalpy for $X(g)\rightarrow X^-(g)$
$U$ -- the LATTICE ENTHALPY for the formation of solid $MX$ from its gaseous ions

Since reactants start as pure elements and products end in their standard states at 1 bar, the overall enthalpy change of this five-step cycle is simply the compound's enthalpy of formation, $\Delta H_f$ , related to the five individual steps by Hess's law:

\Delta H_f = \Delta H_1+\Delta H_2+\Delta H_3+\Delta H_4+U

Worked example: lattice energy of NaCl. Using $\Delta H_f=-411.3$ kJ mol $^{-1}$ (heat of formation of NaCl), $\Delta H_1=+108.7$ kJ mol $^{-1}$ (heat of sublimation of Na(g)), $\Delta H_2=+495.0$ kJ mol $^{-1}$ (ionisation energy of Na(g)), $\Delta H_3=+122$ kJ mol $^{-1}$ (half the dissociation energy of Cl $_2$ (g), which is $244$ kJ mol $^{-1}$ for a full mole of Cl $_2$ ), and $\Delta H_4=-349.0$ kJ mol $^{-1}$ (electron affinity of Cl): …

Figure born-haber-generalBorn-Haber cycle (general, for MX)

What this figure shows. A rectangular cycle. Top: $M(s)+\tfrac{1}{2}X_2(g)\xrightarrow{\Delta H_f}MX(s)$ (the direct formation, top-left to top-right). Left side, going down from $M(s)+\tfrac{1}{2}X_2(g)$ : $\Delta H_1$ (sublimation) reaches $M(g)$ at bottom-left. Middle, going down from the same starting point: $\Delta H_2$ (dissociation of $\tfrac{1}{2}X_2$ ) reaches $X(g)$ , drawn just above the bottom row. Bottom row: $M(g)\xrightarrow{\Delta H_3}M^+(g)$ (ionisation, left to right along the bottom). Middle row: $X(g)\xrightarrow{\Delta H_4}X^-(g)$ (electron affinity), with a '+' sign showing $X^-(g)$ joins $M^+(g)$ . Right side, going up: $U$ (lattice enthalpy) car …

Figure born-haber-naclBorn-Haber cycle for NaCl (worked, with numbers)

What this figure shows. The same cycle shape specialised to sodium chloride: $Na(s)+\tfrac{1}{2}Cl_2\xrightarrow{\Delta H_f=-411.3}NaCl(s)$ across the top; down the left, $\Delta H_1=108.7$ (sublimation of Na) to $Na(g)$ ; down the middle, $\Delta H_2=495.0$ (this cycle's own labelling calls this step the ionisation of Na, differing from the general cycle's $\Delta H_2$ -for-dissociation labelling) reaching a middle stage; along the bottom and middle rows, $\Delta H_3=244$ (dissociation of Cl $_2$ ) and $\Delta H_4=-349.0$ (electron affinity of Cl) lead to $Na^+(g)+Cl^-(g)$ ; and $U$ (the lattice energy being solved for) closes the cycle back up to $ …

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