Born-Haber Cycles

Lattice Enthalpy

In an ionic lattice, strong electrostatic forces hold oppositely charged ions together. Lattice enthalpy is a measure of the strength of these electrostatic forces. Lattice enthalpy $\Delta H^{\theta}_{LE}$ is the enthalpy change when one mole of a solid ionic substance is formed from its gaseous ions under standard conditions.

Standard Conditions

  • Pressure of $100~kPa$ (1 atmosphere)
  • $298K$ ($25°$)
  • $1~mol~dm^{−3}$ for aqueous solutions

Lattice enthalpy is an exothermic reaction as bonds are being formed. A large negative value of lattice enthalpy indicates stronger electrostatic forces between oppositely charged ions in the lattice.
Covalent structures cannot have lattice enthalpies as they are not formed of ions.

Lattice enthalpies cannot be measured directly as it is impossible to form exactly one mole of an ionic lattice from its gaseous ions. By applying Hess’s Law it is possible to calculate the lattice enthalpy through other reaction routes. Some key definitions include:

Born Haber Cycles

A Born-Haber cycle allows for the calculation of an enthalpy change which cannot be measured directly, such as lattice enthalpy. In a Born-Haber cycle, one enthalpy change can be determined from a series of other enthalpy changes. In a cycle, all $\Delta H$ values pointing upwards are endothermic while all $\Delta H$ values pointing downwards are exothermic.

A Born-Haber cycle used to determine the lattice enthalpy of caesium chloride is shown below. The following equations are used to construct the cycle:

Born-Haber cycle for calculating lattice enthalpy

$$ \Delta H^{\theta}_{f} = \text{sum of all other enthalpy changes} \\
-433 = 73 + 121 + 376 - 346 + \Delta H^{\theta}_{LE} \\
\Delta H^{\theta}_{LE} = -657~kJ~mol^{-1}$$

If the lattice enthalpy is known for a Born-Haber cycle, it is possible to calculate the values of other enthalpy changes by substituting known values into the equation and rearranging.

Enthalpy Change of Solution

The standard enthalpy change of solution, $ \Delta H^{\theta}_{s}$ is the enthalpy change when one mole of a solid ionic compound is dissolved to form its aqueous ions under standard conditions. This process can be exothermic or endothermic. For potassium chloride, the reaction is:

$$ KCl_{(s)} + aq \rightarrow K^{+}_{(aq)} + Cl^{-}_{(aq)} $$

When a substance is dissolved:

When constructing Born-Haber cycles using enthalpy change of solution and enthalpy changes of hydration, the following formula can be used to determine the lattice enthalpy:
$$ \Delta H^{\theta}_{LE} + \Delta H^{\theta}_{sol} = \Delta H^{\theta}_{hyd} + \Delta H^{\theta}_{hyd} $$

Worked Example

Using a Born-Haber cycle, calculate the lattice enthalpy of potassium chloride.
$$ \Delta H^{\theta}_{hyd} ~\text{of}~ K^{+}_{(g)} = -322~kJ mol^{-1} \\
\Delta H^{\theta}_{hyd} ~\text{of}~ Cl^{-}_{(g)} = -363~kJ mol^{-1} \\
\Delta H^{\theta}_{s} ~\text{of}~ KCl_{(s)} = +26~kJ mol^{-1} $$

Born-Haber cycle for calculating lattice enthalpy
$$ \eqalign{ \Delta H^{\theta}_{LE} + 26 &= -322 - 363 \\
\Delta H^{\theta}_{LE} &= -711~kJmol^{-1} }$$

Factors Affecting Lattice Enthalpy

An ionic solid with large electrostatic forces between oppositely charged ions has a large exothermic lattice enthalpy. The strength of the electrostatic forces depend on the ionic size and charge.

Factors Affecting $\Delta H^{\theta}_{hyd}$

Like lattice enthalpy, the ionic size and charge also affect the enthalpy change of hydration.

Trends in the Periodic Table


Entropy, $S$ is a measure of disorder in a system. The concept of entropy describes how over time, nature tends to move from a system of order to disorder. As all substances contain particles in constant motion, entropy is always a positive value. Such examples of the tendency to increase disorder include a gas spreading through a room, ice melting in a warm environment or salt dissolving in water. In these examples the energy moves from being localised to more spread out.

Some other examples of entropy changes include:

Standard Entropy Change of Reaction

The standard entropy, $S^{\ominus}$ of a substance is the entropy content of one mole of a substance under standard conditions. The units of standard entropy are $J~K^{-1}~mol^{-1}$.

The standard entropy change of reaction, $\Delta S$ can be calculated by finding the difference in entropy between the products and the reactants.

$$ \Delta S = \Sigma \Delta S_{\text{(products)}} - \Sigma \Delta S_{\text{(reactants)}} $$

As entropy is a measure of randomness, if a change makes a system more random, the value of $\Delta S$ will be positive.

Worked Example

In this equation, ammonia is formed from nitrogen and hydrogen.
$$ N_{2(g)} + 3H_{2(g)} \rightarrow 2NH_{3(g)} $$

Looking at the equation, on the left hand side there are four moles of gas while on the right there are only two moles of gas. This would indicate less disorder on the right, so the entropy would be expected to decrease ($\Delta S$ will be $-ve$).

The table below shows the values of standard entropy for the compounds involved in this reaction.

$N_{2(g)}$ $H_{2(g)}$ $NH_{3(g)}$
$S^{\ominus} / J~K^{-1}~mol^{-1}$ $+ 192$ $+ 131$ $+ 193$

It is important to account for the number of moles of each compound involved in the reaction.
$$ \eqalign{ \Delta S &= \Sigma \Delta S_{\text{(products)}} - \Sigma \Delta S_{\text{(reactants)}} \\
&= (2 \times 193) - (192 + (3 \times 131)) \\
&= -199~J~K^{-1}~mol^{-1} }$$

Spontaneous Processes

A spontaneous process is able to proceed on its own. A process is spontaneous if a chemical system becomes more stable and the overall energy decreases. Both exothermic and endothermic reactions are able to occur spontaneously.

Free Energy

Whether a process is spontaneous or not depends on three factors:

The free energy change, $\Delta G$, or change in Gibbs Free Energy is the balance between the temperature, entropy and enthalpy for a process:

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

Where $\Delta G$ is the change in Gibbs Free Energy ($kJ~mol^{-1}$), $T$ is the temperature in kelvin ($K$) and $\Delta S$ is the change in entropy ($kJ~K^{-1}~mol^{-1}$). Often, the entropy must be converted from $J~K^{-1}~mol^{-1}$ into $kJ~K^{-1}~mol^{-1}$ by dividing the entropy by $1000$.


As a process can only occur spontaneously when the overall energy decreases, $\Delta G$ must be negative for a spontaneous reaction to be feasible:

$$ \Delta G < 0 $$

Worked Example

The equation for the thermal decomposition of ammonium chloride ($NH_{4}Cl$) is shown below:
$$ NH_{4}Cl_{(s)} \rightarrow NH_{3(g)} + HCl_{(g)} \hspace{1em} \Delta H = +176~kJ mol^{-1}$$

Calculate the minimum temperature in $°C$ for thermal decomposition to occur spontaneously. The table below shows the values of standard entropy for the compounds involved in this reaction.

$NH_{4}Cl_{(s)}$ $NH_{3(g)}$ $HCl_{(g)}$
$S^{\ominus} / J~K^{-1}~mol^{-1}$ $+ 95$ $+ 192$ $+ 187$

First the entropy change must be calculated which can then be substituted in the free energy equation.
$$ \eqalign{ \Delta S &= \Sigma \Delta S_{\text{(products)}} - \Sigma \Delta S_{\text{(reactants)}} \\
&= 192 + 187 - 95 \\
&= + 284~J~K^{-1}~mol^{-1} \\
&= + 0.284~kJ~K^{-1}~mol^{-1}}$$

When the reaction first becomes feasible at the minimum temperature, $\Delta G = 0$,
$$ \eqalign{ \Delta G &= \Delta H - T \Delta S \\
0 &= \Delta H - T \Delta S \\
T &= \frac{\Delta H}{\Delta S} \\
T &= \frac{176}{0.284} \\
T &= 619.7K } $$

To convert the temperature into degrees celsius, $273$ must be subtracted from the temperature in kelvin.

$$ T = 619.7 - 273 = 346.7°C $$

© Andrew Deniszczyc, 2023