Born-Haber Cycles and Lattice Enthalpy

Sign Conventions

• Atomisation: always endothermic (positive)
• Ionisation energy: always endothermic (positive)
• First electron affinity: usually exothermic (negative) for most elements
• Second electron affinity: always endothermic (positive) — adding to already negative ion
• Lattice formation enthalpy: always exothermic (negative)

A Born-Haber cycle for NaCl:

$\text{Na}(s) + \frac{1}{2}\text{Cl}_2(g) \xrightarrow{\Delta_f H} \text{NaCl}(s)$

Alternative route (going up then down):

1. $\text{Na}(s) \rightarrow \text{Na}(g)$ — atomisation of Na ($\Delta_{at} H$)
2. $\frac{1}{2}\text{Cl}_2(g) \rightarrow \text{Cl}(g)$ — atomisation of Cl ($\frac{1}{2}$ bond dissociation energy or $\Delta_{at} H$)
3. $\text{Na}(g) \rightarrow \text{Na}^+(g) + e^-$ — first ionisation energy
4. $\text{Cl}(g) + e^- \rightarrow \text{Cl}^-(g)$ — first electron affinity
5. $\text{Na}^+(g) + \text{Cl}^-(g) \rightarrow \text{NaCl}(s)$ — lattice formation enthalpy

By Hess's law:

$\Delta_f H = \Delta_{at} H(\text{Na}) + \Delta_{at} H(\text{Cl}) + IE_1(\text{Na}) + EA_1(\text{Cl}) + \Delta_{LE} H$

Example: NaCl

Given data:

• $\Delta_f H(\text{NaCl}) = -411$ kJ/mol
• $\Delta_{at} H(\text{Na}) = +107$ kJ/mol
• $\Delta_{at} H(\text{Cl}) = +122$ kJ/mol
• $IE_1(\text{Na}) = +496$ kJ/mol
• $EA_1(\text{Cl}) = -349$ kJ/mol

$\Delta_f H = \Delta_{at}(\text{Na}) + \Delta_{at}(\text{Cl}) + IE_1 + EA_1 + \Delta_{LE} H$

$-411 = 107 + 122 + 496 + (-349) + \Delta_{LE} H$

$-411 = 376 + \Delta_{LE} H$

$\Delta_{LE} H = -411 - 376 = -787 \text{ kJ mol}^{-1}$

The lattice enthalpy of NaCl is −787 kJ/mol (formation).

Lattice enthalpy becomes more exothermic (larger magnitude) with:

1. Smaller ionic radii → ions closer together → stronger attraction
2. Higher ionic charges → stronger electrostatic attraction

$\text{Lattice enthalpy} \propto \frac{q^+ \times q^-}{r^+ + r^-}$

MgO has a much larger lattice enthalpy than NaCl because of the 2+ and 2− charges and the smaller ionic radii.

Enthalpy of Solution ($\Delta_{sol} H$)

The enthalpy change when 1 mol of solute dissolves completely in water:

$\Delta_{sol} H = -\Delta_{LE} H + \Delta_{hyd} H(\text{cation}) + \Delta_{hyd} H(\text{anion})$

Or equivalently: $\Delta_{sol} H = \sum \Delta_{hyd} H - \Delta_{LE} H_{\text{dissociation}}$

Enthalpy of Hydration ($\Delta_{hyd} H$)

The enthalpy change when 1 mol of gaseous ions is surrounded by water molecules:

$\text{X}^{n+}(g) + aq \rightarrow \text{X}^{n+}(aq)$

Always exothermic (ion-dipole attractions release energy).

Factors Affecting Hydration Enthalpy

• Smaller ions: more exothermic (charge density higher)
• Higher charge: more exothermic (stronger attraction to water)

Example

Question: Calculate $\Delta_{sol} H$ for NaCl given:

• Lattice dissociation enthalpy = +787 kJ/mol
• $\Delta_{hyd} H$(Na⁺) = −406 kJ/mol
• $\Delta_{hyd} H$(Cl⁻) = −363 kJ/mol

Solution: $\Delta_{sol} H = (+787) + (-406) + (-363) = +18 \text{ kJ mol}^{-1}$

The enthalpy of solution is slightly endothermic (+18 kJ/mol), which is why NaCl dissolving in water causes a very slight temperature decrease.

A reaction is thermodynamically feasible (spontaneous) if the Gibbs free energy change is negative:

$\Delta G = \Delta H - T\Delta S$

where:

• $\Delta G$ = Gibbs free energy change (kJ/mol)
• $\Delta H$ = enthalpy change (kJ/mol)
• $T$ = temperature (K)
• $\Delta S$ = entropy change (J/K/mol) — convert to kJ!

Conditions for Feasibility

Example

Question: A reaction has $\Delta H = -50$ kJ/mol and $\Delta S = -100$ J/K/mol. Below what temperature is it feasible?

$\Delta G = \Delta H - T\Delta S = 0 \text{ at the boundary}$ $0 = -50 - T(-0.100)$ $0 = -50 + 0.100T$ $T = 500 \text{ K}$

Feasible below 500 K (below this, $\Delta G < 0$).

• Draw Born-Haber cycles with clear labels and correct signs
• Remember: atomisation and IE are always positive; first EA is usually negative
• For lattice enthalpy calculations, be careful about the sign convention (formation vs dissociation)
• Convert $\Delta S$ from J to kJ when using the Gibbs equation
• $\Delta G < 0$ means feasible, but doesn't say anything about rate

• Born-Haber cycles use Hess's law to calculate lattice enthalpies
• Lattice enthalpy depends on ionic charge and radius
• Enthalpy of solution = lattice dissociation + hydration enthalpies
• Hydration enthalpy depends on charge density (charge/radius)
• Gibbs free energy: $\Delta G = \Delta H - T\Delta S$; feasible when $\Delta G < 0$