Lattice enthalpies and Born-Haber cycles

In a nutshell

The lattice enthalpy change of formation is the enthalpy change when gaseous ions form one mole of a solid ionic compound under standard conditions. Born-Haber cycles can be drawn to calculate the lattice enthalpy change of formation. The theoretical values of lattice enthalpy may not match the experimental values.

Equations

There are general chemical equations for different enthalpy changes you should be able to use with any given reaction.

$atomelement\>in\>its\>standard\>state→one\>mole\>of\>gaseous \>atom$

$atomsone\>mole\>of\>compound \>in\>its\>standard\>state→\>gaseous\>atoms$

$atomsgaseous\>compound\>→\>gaseous\>atoms$

$one\>mole\>of\>gaseous\>atoms\>→one\>mole\>of\>gaseous\>ions^{1+}+electron$

$one\>mole\>of\>gaseous\>ions^{1+}\>→one\>mole\>of\>gaseous\>ions^{2+}+electron$

$one\>mole\>of\>gaseous\>atoms\>+\>electron\>→one\>mole\>of\>gaseous\>ions^{1-}$

$one\>mole\>of\>gaseous\>ions^{1-}\>+\>electron\>→one\>mole\>of\>gaseous\>ions^{2-}$

$ionsone\>mole\>of\>gaseous\>ions\>→\>one\>mole\>of\>aqueous\>ions$

$soluteone\>mole\>of\>solute\>→\>aqueous\>solute$

$compoundelements\>in\>standard\>states\>→\>one\>mole\>of\>compound$

Lattice enthalpy

Formation and dissociation

Positive and negative ions are held together via electrostatic attractions to form giant ionic lattices. The energy released when the solid lattice is formed from gaseous ions is called lattice energy. The lattice enthalpy of formation, $△fH\triangle_{f}H$ , is the enthalpy change when gaseous ions form one mole of a solid ionic compound under standard conditions.

The lattice enthalpy measures how strong an ionic bond is. The more negative it is (more exothermic), the more stronger the ionic bond is.

Example

The following lattice formation enthalpies are given below. Which species has the strongest ionic bonds?

$K^+(g)+Cl^-(g)→KCl(s)$

$mol−1\triangle_{lattice}H^{⦵}=-698\>kJ\>mol^{-1}$

$Ca^{2+}(g)+2Cl^-(g)→CaCl_2(s)$

$mol−1\triangle_{lattice}H^{⦵}=-1145\>kJ\>mol^{-1}$

$CaCl_2$ has stronger ionic bonds as the standard lattice energy is more exothermic.

The lattice enthalpy of dissociation is the enthalpy change under standard conditions when one mole of the solid ionic compound dissociates into its gaseous ions completely. It is endothermic, opposite to the enthalpy of formation.

Example

The lattice dissociation enthalpy of potassium chloride is given below.

$KCl(s)→K^+(g)+Cl^-(g)$

$mol−1\triangle_{lattice}H^{⦵}=698\>kJ\>mol^{-1}$

Factors affecting lattice energy

The lattice energy is affected by ionic charge and ion size. The larger the ionic charge, the larger the lattice energy. This is because more energy is released upon formation of the ionic lattice. The smaller the radius of the ion, the lattice energy will be more exothermic. This is because the ions are held closer together in the lattice.

Example

Predict which lattice energy is larger, $K Cl$ or $CaCl_2$ .

A potassium ion is singly charged and has a larger ionic radius. A calcium ion is doubly charged and has a smaller ionic radius. The charge density for calcium chloride is larger. Therefore, calcium chloride has a larger lattice energy.

Enthalpy change

The standard enthalpy change, $△H⦵\triangle H^{⦵}$ , is the transfer of heat energy in a reaction under standard conditions. A negative enthalpy change indicates an exothermic reaction, heat energy is released. A positive enthalpy change indicates an endothermic reaction, heat energy is absorbed.

There are many types of enthalpy changes. You must know all of these definitions and be able to identify them.

Enthalpy change of atomisation of an element

Enthalpy change of atomisation of an element, $△atH\triangle_{at}H$ , is the enthalpy change when an element in its standard state forms one mole of its gaseous atoms.

$atomelement\>in\>its\>standard\>state→one\>mole\>of\>gaseous \>atom$

Example

$atomone\>gaseous\>bromine\>element\>→one\>mole\>of\>gaseous\>bromine\>atom$

$12Br2(g)→Br(g)\dfrac{1}{2}Br_2(g)→Br(g)$

Enthalpy change of atomisation of a compound

Enthalpy change of atomisation of a compound, $△atH\triangle_{at}H$ , is the enthalpy change when gaseous atoms are formed from one mole of a compound in its standard state.

$atomsone\>mole\>of\>compound \>in\>its\>standard\>state→\>gaseous\>atoms$

Example

$\>potassium\>chloride→\>gaseous\>potassium\>atom\>+\>gaseous\>chlorine\>atom$

$K Cl (s) \to K (g) + Cl (g)$

Bond dissociation enthalpy

Bond dissociation enthalpy, $△dissH\triangle_{diss}H$ , is the enthalpy change when one mole of gaseous bonds of all the same type are broken.

$atomsgaseous\>compound\>→\>gaseous\>atoms$

Example

$atomsgaseous\>bromine\>→\>gaseous\>bromine\>atoms$

$Br_2(g)→2Br(g)$

First ionisation energy

First ionisation energy, $△ie1H\triangle_{ie1}H$ , is the enthalpy change when one mole of gaseous atoms forms one mole of gaseous $1 +$ ions.

$one\>mole\>of\>gaseous\>atoms\>→one\>mole\>of\>gaseous\>ions^{1+}+electron$

Example

$one\>mole\>of\>gaseous\>calcium\>→one\>mole\>of\>gaseous\>calcium\>ion^{1+}+electron$

$Ca(g)→Ca^{1+}(g)+e^-$

Second ionisation energy

Second ionisation energy, $△ie2H\triangle_{ie2}H$ , is the enthalpy change when one mole of gaseous $1 +$ ions forms one mole of gaseous $2 +$ ions.

$one\>mole\>of\>gaseous\>ions^{1+}\>→one\>mole\>of\>gaseous\>ions^{2+}+electron$

Example

$one\>mole\>of\>gaseous\>calcium\>ion^{1+}→one\>mole\>of\>gaseous\>calcium\>ion^{2+}+electron$

$Ca^{1+}(g)→Ca^{2+}(g)+e^-$

First electron affinity

First electron affinity, $△ea1H\triangle_{ea1}H$ , is the enthalpy change when one mole of gaseous atoms form one mole of gaseous $1 -$ ions.

$one\>mole\>of\>gaseous\>atoms\>+\>electron\>→one\>mole\>of\>gaseous\>ions^{1-}$

Example

$one\>mole\>of\>gaseous\>sulfur\>+\>electron→one\>mole\>of\>gaseous\>sulfur\>ion^{1-}$

$S(g)+e^-→S^{1-}(g)$

Second electron affinity

Second electron affinity, $△ea2H\triangle_{ea2}H$ , is the enthalpy change when one mole of gaseous $1 -$ ions form one mole of gaseous $2 -$ ions.

$one\>mole\>of\>gaseous\>ions^{1-}\>+\>electron\>→one\>mole\>of\>gaseous\>ions^{2-}$

Example

$one\>mole\>of\>gaseous\>sulfur^{1-}\>+\>electron→one\>mole\>of\>gaseous\>sulfur\>ion^{2-}$

$S^{1-}(g)+e^-→S^{2-}(g)$

Enthalpy change of hydration

Enthalpy change of hydration, $△hydH\triangle_{hyd}H$ , is the enthalpy change when one mole of gaseous ions form one mole of aqueous ions.

$ionsone\>mole\>of\>gaseous\>ions\>→\>one\>mole\>of\>aqueous\>ions$

Example

$ionone\>mole\>of\>gaseous\>\>potassium\>ion\>→\>one\>mole\>of\>aqueous\>potassium\>ion$

$K^+(g)→K^+(aq)$

Enthalpy change of solution

Enthalpy change of solution, $△solutionH\triangle_{solution}H$ , is the enthalpy change when sufficient solvent dissolves one mole of solute where there is not any further enthalpy change upon further dilution.

$soluteone\>mole\>of\>solute\>→\>aqueous\>solute$

Example

$chloridesolid\>potassium\>chloride\>→\>aqueous\>potassium\>chloride$

$K Cl (s) \to K Cl (a q)$

Enthalpy change of formation

Enthalpy change of formation, $△fH\triangle_{f}H$ , is the enthalpy change when elements in their standard state forms one mole of a compound under standard conditions.

$compoundelements\>in\>standard\>states\>→\>one\>mole\>of\>compound$

Example

$waterhydrogen\>+\>oxygen\>→\>one\>mole\>of\>water$

$H2(g)+12O2(g)→H2O(l)H_2(g)+\dfrac{1}{2}O_2(g)→H_2O(l)$

Born-Haber cycles

Born-Haber cycles are used to calculate lattice enthalpies.

procedure

1.	Draw the Born-Haber cycle.
2.	Find a route to calculate the lattice enthalpy.
3.	Calculate lattice enthalpy by adding all arrows facing in the correct direction and subtracting all arrows not facing in the correct direction.

Example

The Born-Haber cycle for calcium chloride is given below. Calculate the lattice enthalpy of formation for calcium chloride.

Chemistry; Energetics II; KS5 Year 12; Lattice enthalpies and Born-Haber cycles

$7=△LEHLabel\>1=\triangle _{ie2}H \newline Label\>2=\triangle _{ie1}H \newline Label\>3=\triangle _{at}H \>(Ca) \newline Label\>4=2 \times \triangle _{at}H\>(Cl) \newline Label\>5=\triangle _{f}H \newline Label\>6=2 \times \triangle _{ea}H \newline Label\>7=\triangle _{LE}H \newline$

The enthalpies values are given below.

Type of enthalpy	Symbol	Value
Enthalpy of formation	$△fH\triangle_{f}H$	$795.8\>kJ\>mol^{-1}$
Enthalpy of atomisation for chlorine	$△atH(Cl)\triangle_{at}H(Cl)$	$121\>kJ\>mol^{-1}$
Enthalpy of atomisation for calcium	$△atH(Ca)\triangle_{at}H(Ca)$	$178\>kJ\>mol^{-1}$
Enthalpy of first ionisation energy for calcium	$△ie1H\triangle_{ie1}H$	$589.8\>kJ\>mol^{-1}$
Enthalpy of second ionisation energy for calcium ion	$△ie2H\triangle_{ie2}H$	$1145.5\>kJ\>mol^{-1}$
Enthalpy of electron affinity for chlorine	$△eaH\triangle_{ea}H$	$346\>kJ\>mol^{-1}$

Find a route with known values to calculate the lattice enthalpy. Subtract all arrows facing in the opposite direction and add all arrows facing in the right direction:

$△LEH=−2(△eaH)−△ie2H−△ie1H−△atH(Ca)−2(△atH(Cl))+△fH\triangle_{LE}H=-2(\triangle_{ea}H)-\triangle_{ie2}H-\triangle_{ie1}H-\triangle_{at}H(Ca)-2(\triangle_{at}H(Cl))+\triangle_{f}H$

The atomisation and electron affinity of chlorine are multiplied by two as there are two chlorine atoms.

Calculate the lattice enthalpy by inserting known the values:

$mol−1\triangle_{LE}H=-2(-346)-1145.5-589.8-178-2(121)+(-795.8)=-2259.1\>kJ\>mol^{-1}$

The lattice enthalpy of calcium chloride is $mol−1‾\underline{-2259.1\>kJ\>mol^{-1}}$ .

Theoretical and experimental

The purely ionic model of a lattice can be used to calculate the the theoretical lattice enthalpy. This model uses the assumption that all ions are spherical because they have an evenly distributed charge. However, this isn't true in real life.

Some ionic compounds have covalent character. Some ions have high polarising character that arises covalent character. This distorts the ideal spherical ionic model. This results in the experimental lattice enthalpy value being more exothermic.