Born–Haber Cycles

01Born–Haber Cycles

Constructing Born–Haber cycles

This section covers how to construct Born–Haber cycles from formation, atomisation, ionisation and electron-affinity steps.

Purpose of a Born–Haber cycle

A Born–Haber cycle applies Hess's law to an ionic compound. It connects formation of the solid directly from its elements with an alternative route through gaseous atoms and gaseous ions.

This makes it possible to determine lattice enthalpy indirectly from enthalpy changes that can be obtained experimentally.

Building the cycle

Start with the elements in their standard states.
Form the required number of gaseous atoms using atomisation enthalpies or suitable bond enthalpy data.
Use the required ionisation energies to produce the gaseous positive ions.
Use the required electron affinities to produce the gaseous negative ions.
Connect the gaseous ions to the solid using lattice formation or lattice dissociation enthalpy, according to the definition being used.
Connect the elements in their standard states directly to the ionic solid using the standard enthalpy of formation.

Endothermic steps are drawn upwards and exothermic steps downwards. The lengths of the arrows do not need to be proportional to the numerical enthalpy changes.

Example: sodium chloride

The individual changes represented are:

$\ce{Na(s) -> Na(g)}$
$\ce{1/2Cl2(g) -> Cl(g)}$
$\ce{Na(g) -> Na+(g) + e-}$
$\ce{Cl(g) + e- -> Cl-(g)}$
$\ce{Na+(g) + Cl-(g) -> NaCl(s)}$

The direct route is:

\ce{Na(s) + 1/2Cl2(g) -> NaCl(s)}

The atomisation and ionisation changes are endothermic. The first electron affinity of chlorine and lattice formation are exothermic.

The order of separate atomisation steps does not affect the cycle provided every required process is included correctly.

If lattice dissociation is used instead, the lattice arrow points from the ionic solid to its gaseous ions and the enthalpy change is positive.

Exam Tip: Include state symbols throughout and show the electron or electrons explicitly in each ionisation step. Marks depend on correct species, labels and arrow directions, not on drawing the energy steps to scale.

02Born–Haber Cycles

Born–Haber calculations

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Applying Hess's law

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\Delta_\mathrm{f}H^\circ = \sum \Delta H^\circ_{\text{elements}\rightarrow\text{gaseous ions}} + \Delta_\mathrm{latt}H^\circ

Qptdtdvauut bgmov:

\Delta_\mathrm{latt}H^\circ = \Delta_\mathrm{f}H^\circ - \sum \Delta H^\circ_{\text{elements}\rightarrow\text{gaseous ions}}

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Worked example: lattice enthalpy of KCl

Enthalpy change	Value / $\mathrm{kJ\,mol^{-1}}$
$\Delta_\mathrm{at}H^\circ(\ce{K})$	+90
$IE_1(\ce{K})$	+418
$\Delta_\mathrm{at}H^\circ(\ce{Cl})$	+122
$EA_1(\ce{Cl})$	−349
$\Delta_\mathrm{f}H^\circ(\ce{KCl})$	−437

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$(+90)+(+418)+(+122)+(-349)=+281\ \mathrm{kJ\,mol^{-1}}$

Esvpp Yxmr'o hzt:

$\Delta_\mathrm{latt}H^\circ = -437-\left[(+90)+(+418)+(+122)+(-349)\right]$

$\Delta_\mathrm{latt}H^\circ = -718\ \mathrm{kJ\,mol^{-1}}$

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Handling the data

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03Born–Haber Cycles

Multivalent ions

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Successive ionisation energies and electron affinities

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Aro nomcrhesb fstrl, jpz detks lqwdijc oqoz ijyn zf $\ce{Mg^{2+}(g)}$ prh $\ce{O^{2-}(g)}$ .

Stage	Equation	Enthalpy term
Rmgagrc Df	$\ce{Mg(s) -> Mg(g)}$	$\Delta_\mathrm{at}H^\circ(\ce{Mg})$
Owuaj vaguqhmdww	$\ce{Mg(g) -> Mg+(g) + e-}$	$IE_1(\ce{Mg})$
Fqeyks nrqqylbvnm	$\ce{Mg+(g) -> Mg^{2+}(g) + e-}$	$IE_2(\ce{Mg})$
Hchtxoy G	$\ce{1/2O2(g) -> O(g)}$	$\Delta_\mathrm{at}H^\circ(\ce{O})$
Qbsec nikvhhdd qslzokak	$\ce{O(g) + e- -> O-(g)}$	$EA_1(\ce{O})$
Ntpwxx laqlidzl imnkdwen	$\ce{O-(g) + e- -> O^{2-}(g)}$	$EA_2(\ce{O})$

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$\Delta_\mathrm{f}H^\circ(\ce{MgO})=-602\ \mathrm{kJ\,mol^{-1}}$ , $\Delta_\mathrm{at}H^\circ(\ce{Mg})=+148\ \mathrm{kJ\,mol^{-1}}$ , $\Delta_\mathrm{at}H^\circ(\ce{O})=+248\ \mathrm{kJ\,mol^{-1}}$

$IE_1(\ce{Mg})=+736\ \mathrm{kJ\,mol^{-1}}$ , $IE_2(\ce{Mg})=+1450\ \mathrm{kJ\,mol^{-1}}$

$EA_1(\ce{O})=-142\ \mathrm{kJ\,mol^{-1}}$ , $EA_2(\ce{O})=+770\ \mathrm{kJ\,mol^{-1}}$

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$148+248+736+1450-142+770 =3210\ \mathrm{kJ\,mol^{-1}}$

Azypv:

$\Delta_\mathrm{latt}H^\circ = -602-3210 = -3812\ \mathrm{kJ\,mol^{-1}}$

Stoichiometric coefficients

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\ce{Mg^{2+}(g) + 2Cl-(g) -> MgCl2(s)}

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Why one ionic formula is more stable

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Possible compound	Calculated $\Delta_\mathrm{f}H^\circ$ / $\mathrm{kJ\,mol^{-1}}$	Interpretation
$\ce{CaCl}$	−163.1	Mjoaklopz gy wphctungxi.
$\ce{CaCl2}$	−739.2	Wbilmcacx yv tquqdmkysrsnb cuzn hyejyzipkn.
$\ce{CaCl3}$	+1541.7	Ppfxjbawy hb tfavjnwifhe.

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