Born–Haber Cycles

A2 · 13 min

A Born–Haber cycle is a Hess's law energy cycle that links the enthalpy change of formation of an ionic solid to the energy changes involved in making its gaseous ions. Its main job is to find the lattice energy, which cannot be measured directly, but the same cycle can be rearranged to find any one unknown term: an electron affinity, an ionisation energy or an enthalpy change of formation. Born–Haber questions appear on almost every Paper 4, typically worth 4 to 6 marks: draw or complete the cycle, then calculate.

The idea: two routes from elements to solid

Imagine forming sodium chloride from sodium metal and chlorine gas. There are two routes.

Direct route: the elements react to give the solid. Its enthalpy change is the enthalpy change of formation, $\Delta H^{\ominus}_{\text{f}}$ .
Indirect route: turn the elements into gaseous atoms, then into gaseous ions, then let the ions come together to form the lattice.

Hess's law says the total enthalpy change is the same by either route, because enthalpy depends only on the start and end states. If you know every term except one, you can find the missing one.

Definition

The standard enthalpy change of formation, $\Delta H^{\ominus}_{\text{f}}$ , is the enthalpy change when one mole of a compound is formed from its elements in their standard states under standard conditions.

The steps in the indirect route

For a compound $\ce{MX}$ of a Group 1 metal and a Group 17 element, the indirect route has four steps before the lattice forms.

Step	Equation for NaCl	Name	Sign
1	$\ce{Na(s) -> Na(g)}$	enthalpy change of atomisation of Na	+
2	$\ce{1/2Cl2(g) -> Cl(g)}$	enthalpy change of atomisation of Cl	+
3	$\ce{Na(g) -> Na^+(g) + e-}$	first ionisation energy of Na	+
4	$\ce{Cl(g) + e- -> Cl^-(g)}$	first electron affinity of Cl	–
5	$\ce{Na^+(g) + Cl^-(g) -> NaCl(s)}$	lattice energy of NaCl	–

Steps 1 to 5 together convert $\ce{Na(s) + 1/2Cl2(g)}$ into $\ce{NaCl(s)}$ , exactly like the direct route.

The Born–Haber equation

\Delta H^{\ominus}_{\text{f}} = \Delta H^{\ominus}_{\text{at}}(\text{metal}) + \sum IE + \Delta H^{\ominus}_{\text{at}}(\text{non-metal}) + \sum EA + \Delta H^{\ominus}_{\text{latt}}

Rearranged for the lattice energy:

\Delta H^{\ominus}_{\text{latt}} = \Delta H^{\ominus}_{\text{f}} - \left[\Delta H^{\ominus}_{\text{at}}(\text{metal}) + \sum IE + \Delta H^{\ominus}_{\text{at}}(\text{non-metal}) + \sum EA\right]

Each term must be multiplied by the number of moles of that particle in the formula.

Drawing the cycle as an energy-level diagram

Exam questions usually present the cycle as an energy-level diagram. Upward arrows are endothermic steps, downward arrows are exothermic steps, and each horizontal line is labelled with the species present at that stage. The elements in their standard states sit at zero.

Born–Haber cycle for sodium chloride (energies in kJ mol⁻¹, drawn roughly to scale). Upward arrows are endothermic, downward arrows exothermic. The dashed arrow is the unknown lattice energy.

Two features of the diagram are worth noticing. First, the electron removed from sodium in step 3 is still written on the line ( $\ce{e-}$ ) until it is used up in step 4: electrons are conserved on every level. Second, the lattice energy arrow is very long. Most of the stability of an ionic solid comes from the lattice.

Constructing and using a Born–Haber cycle

Write the formula of the solid and identify the ions, including their charges (for example $\ce{Mg^{2+}}$ and $\ce{2Cl-}$ ).
Start at zero with the elements in their standard states, using the right amounts (for $\ce{MgCl2}$ : $\ce{Mg(s) + Cl2(g)}$ ).
Atomise the metal, then atomise the non-metal, multiplying by the number of atoms needed.
Ionise the metal: add first, second (and so on) ionisation energies until you reach the cation's charge.
Add electrons to the non-metal atoms: first electron affinity, then second if the anion is 2–, multiplied by the number of anions.
Draw the lattice energy arrow down from the gaseous ions to the solid, and the formation arrow from the elements to the solid.
Apply Hess's law: the sum of the indirect steps equals $\Delta H^{\ominus}_{\text{f}}$ . Rearrange for the unknown and keep every sign.

Cations with a 2+ charge and anions with a 2– charge

The syllabus limits Born–Haber cycles to 1+ and 2+ cations and 1– and 2– anions. Each extra charge adds one more term.

A 2+ cation needs both the first and second ionisation energies: $\ce{Mg(g) -> Mg^+(g) + e-}$ then $\ce{Mg^+(g) -> Mg^{2+}(g) + e-}$ .
A 2– anion needs both electron affinities: $EA_1$ (usually exothermic) and $EA_2$ (always endothermic).
A formula with two anions per cation, such as $\ce{MgCl2}$ , needs twice the atomisation enthalpy and twice the electron affinity of chlorine.
A formula with two cations per anion, such as $\ce{Na2O}$ , needs twice the atomisation and twice the ionisation energy of sodium.

Watch out

For $\ce{MgCl2}$ the atomisation term for chlorine is $2 \times \Delta H^{\ominus}_{\text{at}}(\ce{Cl})$ , which equals the $\ce{Cl-Cl}$ bond energy once ( $242\ \text{kJ mol}^{-1}$ ). Do not double it again. Equally, students often forget to double the electron affinity. Write out every term with its multiplier before adding.

Worked examples

Lattice energy of sodium chloride

Use the data to calculate the lattice energy of sodium chloride.

Term	Value / kJ mol⁻¹
$\Delta H^{\ominus}_{\text{f}}(\ce{NaCl(s)})$	–411
$\Delta H^{\ominus}_{\text{at}}(\ce{Na})$	+107
$\Delta H^{\ominus}_{\text{at}}(\ce{Cl})$	+121
first ionisation energy of Na	+494
first electron affinity of Cl	–349

Solution

Sum of the indirect steps before the lattice forms:

107 + 121 + 494 + (-349) = +373\ \text{kJ mol}^{-1}

Hess's law:

\Delta H^{\ominus}_{\text{f}} = 373 + \Delta H^{\ominus}_{\text{latt}}

\Delta H^{\ominus}_{\text{latt}} = -411 - 373 = -784\ \text{kJ mol}^{-1}

Data book values are close to $-787\ \text{kJ mol}^{-1}$ ; the small difference comes from rounding in the input data.

Magnesium oxide: two ionisation energies and two electron affinities

Calculate the lattice energy of magnesium oxide.

Term	Value / kJ mol⁻¹
$\Delta H^{\ominus}_{\text{f}}(\ce{MgO(s)})$	–602
$\Delta H^{\ominus}_{\text{at}}(\ce{Mg})$	+148
first and second ionisation energies of Mg	+736, +1450
$\Delta H^{\ominus}_{\text{at}}(\ce{O})$	+249
first and second electron affinities of O	–141, +798

Solution

Indirect steps:

148 + 736 + 1450 + 249 + (-141) + 798 = +3240\ \text{kJ mol}^{-1}

\Delta H^{\ominus}_{\text{latt}} = -602 - 3240 = -3842\ \text{kJ mol}^{-1}

The cycle shows why $\ce{MgO}$ forms despite an energy cost of $+3240\ \text{kJ mol}^{-1}$ to make the gaseous ions: the lattice energy releases even more.

Magnesium chloride: counting the chlorines

$\Delta H^{\ominus}_{\text{f}}(\ce{MgCl2(s)}) = -641\ \text{kJ mol}^{-1}$ . Using the data above and $\Delta H^{\ominus}_{\text{at}}(\ce{Cl}) = +121$ , $EA_1(\ce{Cl}) = -349\ \text{kJ mol}^{-1}$ , calculate the lattice energy of $\ce{MgCl2}$ .

Solution

Write each term with its multiplier:

Step	Contribution / kJ mol⁻¹
atomise Mg	+148
$IE_1 + IE_2$ of Mg	+736 + 1450 = +2186
atomise 2 Cl	2 × 121 = +242
2 × $EA_1$ of Cl	2 × (–349) = –698
total	+1878

\Delta H^{\ominus}_{\text{latt}} = -641 - 1878 = -2519\ \text{kJ mol}^{-1}

Finding an electron affinity

Calculate the first electron affinity of bromine from these data (kJ mol⁻¹): $\Delta H^{\ominus}_{\text{f}}(\ce{KBr}) = -394$ ; $\Delta H^{\ominus}_{\text{at}}(\ce{K}) = +89$ ; first ionisation energy of K $= +418$ ; $\Delta H^{\ominus}_{\text{at}}(\ce{Br}) = +112$ ; $\Delta H^{\ominus}_{\text{latt}}(\ce{KBr}) = -689$ .

Solution

\Delta H^{\ominus}_{\text{f}} = \Delta H^{\ominus}_{\text{at}}(\ce{K}) + IE_1 + \Delta H^{\ominus}_{\text{at}}(\ce{Br}) + EA_1 + \Delta H^{\ominus}_{\text{latt}}

-394 = 89 + 418 + 112 + EA_1 + (-689)

EA_1 = -394 - 89 - 418 - 112 + 689 = -324\ \text{kJ mol}^{-1}

The answer is negative, as expected for a first electron affinity of a halogen, and slightly less exothermic than that of chlorine.

Why magnesium chloride is MgCl₂ and not MgCl (exam-hard)

A student estimates the lattice energy of the hypothetical compound $\ce{MgCl(s)}$ , containing $\ce{Mg+}$ ions, as $-753\ \text{kJ mol}^{-1}$ .

(a) Using the data from the previous examples, calculate $\Delta H^{\ominus}_{\text{f}}(\ce{MgCl(s)})$ . (b) Calculate the enthalpy change for $\ce{2MgCl(s) -> Mg(s) + MgCl2(s)}$ and comment on your answer. (c) Suggest why the lattice energy of $\ce{MgCl2}$ is so much more exothermic than that of $\ce{MgCl}$ .

Solution

(a) Only the first ionisation energy is needed for $\ce{Mg+}$ , and only one chlorine:

\Delta H^{\ominus}_{\text{f}}(\ce{MgCl}) = 148 + 736 + 121 + (-349) + (-753) = -97\ \text{kJ mol}^{-1}

(b) Using enthalpies of formation (the element $\ce{Mg(s)}$ is zero):

\Delta H = \Delta H^{\ominus}_{\text{f}}(\ce{MgCl2}) - 2\Delta H^{\ominus}_{\text{f}}(\ce{MgCl}) = -641 - 2(-97) = -447\ \text{kJ mol}^{-1}

$\ce{MgCl}$ would be energetically unstable with respect to disproportionation into $\ce{Mg}$ and $\ce{MgCl2}$ , so even if it formed it would convert to $\ce{MgCl2}$ . This is why magnesium always forms $\ce{Mg^{2+}}$ in its compounds.

(c) $\ce{Mg^{2+}}$ has double the charge of $\ce{Mg+}$ and is smaller, and there are two chloride ions per cation. The attraction between the ions is much stronger, and the extra lattice energy released more than repays the second ionisation energy ( $+1450\ \text{kJ mol}^{-1}$ ).

Reading a Born–Haber cycle for information

Once a cycle is drawn, you can read off more than the lattice energy.

Which steps are endothermic? Atomisation, ionisation and second electron affinity. Which are exothermic? First electron affinity (for non-metals), lattice energy and, usually, formation.
The biggest single cost in forming a 2+ ionic compound is normally the ionisation of the metal. The biggest single release is the lattice energy.
Comparing compounds: a more exothermic lattice energy is the main reason a compound with highly charged, small ions has a very negative enthalpy change of formation.

Common mistakes

Using the formation definition of lattice energy but then reversing it in the cycle. Keep one direction: gaseous ions to solid, exothermic.
Forgetting the second ionisation energy for a 2+ cation, or the second electron affinity for a 2– anion.
Not multiplying atomisation and electron affinity by 2 for $\ce{MgCl2}$ or $\ce{CaF2}$ , or ionisation and atomisation by 2 for $\ce{Na2O}$ .
Using the bond energy instead of the atomisation enthalpy for chlorine in $\ce{NaCl}$ : $\Delta H^{\ominus}_{\text{at}}(\ce{Cl})$ is half the bond energy.
Sign errors when rearranging. Write the full equation first, substitute with brackets around negative values, then rearrange.
Leaving electrons off the energy-level diagram. On every level, the species must balance the elements you started with.

Exam technique

Exam tip

When asked to complete a Born–Haber cycle, label every level with the species and state symbols, and include $\ce{e-}$ where needed. Marks are often given per correctly labelled level.
Show the Hess's law equation in words or symbols before substituting. If you make an arithmetic slip, the method mark is still available.
Give the answer with sign and units: " $-2519\ \text{kJ mol}^{-1}$ ". A lattice energy without a minus sign is marked wrong.
Questions often follow up with "explain why the lattice energy of X is more exothermic than that of Y". Use ionic charge and ionic radius, as in lattice energy and electron affinity.
Check the magnitude: lattice energies of 1+/1– compounds are typically $-600$ to $-1050\ \text{kJ mol}^{-1}$ ; 2+/1– compounds around $-2000$ to $-2900$ ; 2+/2– compounds around $-3000$ to $-4000$ .

Summary

A Born–Haber cycle applies Hess's law: $\Delta H^{\ominus}_{\text{f}}$ = atomisation of metal + ionisation energies + atomisation of non-metal + electron affinities + lattice energy.
It is used to find lattice energies, which cannot be measured directly, or any single unknown term.
Energy-level diagrams put endothermic steps as upward arrows and exothermic steps as downward arrows, starting from the elements at zero.
2+ cations need $IE_1 + IE_2$ ; 2– anions need $EA_1 + EA_2$ ; multiply terms by the number of each particle in the formula.
Lattice energy is the dominant exothermic term that makes ionic compounds stable relative to their elements.

Practice questions

Question

Write the equation, with state symbols, for each step of the Born–Haber cycle for potassium chloride.
Calculate the lattice energy of $\ce{KCl}$ : $\Delta H^{\ominus}_{\text{f}}(\ce{KCl}) = -437$ ; $\Delta H^{\ominus}_{\text{at}}(\ce{K}) = +89$ ; $IE_1(\ce{K}) = +418$ ; $\Delta H^{\ominus}_{\text{at}}(\ce{Cl}) = +121$ ; $EA_1(\ce{Cl}) = -349$ (all kJ mol⁻¹).
Calculate the lattice energy of $\ce{CaCl2}$ : $\Delta H^{\ominus}_{\text{f}}(\ce{CaCl2}) = -796$ ; $\Delta H^{\ominus}_{\text{at}}(\ce{Ca}) = +178$ ; $IE_1 = +590$ , $IE_2 = +1150$ ; $\Delta H^{\ominus}_{\text{at}}(\ce{Cl}) = +121$ ; $EA_1(\ce{Cl}) = -349$ .
Calculate the enthalpy change of formation of $\ce{MgF2}$ : $\Delta H^{\ominus}_{\text{at}}(\ce{Mg}) = +148$ ; $IE_1 + IE_2 = +2186$ ; $E(\ce{F-F}) = 158$ ; $EA_1(\ce{F}) = -328$ ; $\Delta H^{\ominus}_{\text{latt}}(\ce{MgF2}) = -2957$ .
Calculate the lattice energy of $\ce{Na2O}$ : $\Delta H^{\ominus}_{\text{f}}(\ce{Na2O}) = -414$ ; $\Delta H^{\ominus}_{\text{at}}(\ce{Na}) = +107$ ; $IE_1(\ce{Na}) = +494$ ; $\Delta H^{\ominus}_{\text{at}}(\ce{O}) = +249$ ; $EA_1(\ce{O}) = -141$ ; $EA_2(\ce{O}) = +798$ .
Calculate the first electron affinity of iodine: $\Delta H^{\ominus}_{\text{f}}(\ce{NaI}) = -288$ ; $\Delta H^{\ominus}_{\text{at}}(\ce{Na}) = +107$ ; $IE_1(\ce{Na}) = +494$ ; $\Delta H^{\ominus}_{\text{at}}(\ce{I}) = +107$ ; $\Delta H^{\ominus}_{\text{latt}}(\ce{NaI}) = -704$ .
Calculate the enthalpy change of atomisation of calcium from: $\Delta H^{\ominus}_{\text{f}}(\ce{CaO}) = -635$ ; $IE_1 + IE_2(\ce{Ca}) = +1740$ ; $\Delta H^{\ominus}_{\text{at}}(\ce{O}) = +249$ ; $EA_1 = -141$ ; $EA_2 = +798$ ; $\Delta H^{\ominus}_{\text{latt}}(\ce{CaO}) = -3459$ .
In a Born–Haber cycle for $\ce{MgO}$ , identify the step with the largest endothermic contribution and the step with the largest exothermic contribution, and explain why $\ce{MgO}$ has a negative enthalpy change of formation.
A student suggests the compound $\ce{NaCl2}$ , containing $\ce{Na^{2+}}$ ions, and estimates its lattice energy as $-2500\ \text{kJ mol}^{-1}$ . The second ionisation energy of sodium is $+4560\ \text{kJ mol}^{-1}$ . Using other data from question 2 and the sodium values above, calculate $\Delta H^{\ominus}_{\text{f}}(\ce{NaCl2})$ and explain why the compound does not exist.
Sketch, as an energy-level diagram, the Born–Haber cycle for calcium fluoride, labelling each level with species and state symbols. State which arrows point upwards.

Answers

$\ce{K(s) -> K(g)}$ ; $\ce{1/2Cl2(g) -> Cl(g)}$ ; $\ce{K(g) -> K^+(g) + e-}$ ; $\ce{Cl(g) + e- -> Cl^-(g)}$ ; $\ce{K^+(g) + Cl^-(g) -> KCl(s)}$ ; overall $\ce{K(s) + 1/2Cl2(g) -> KCl(s)}$ .
Indirect steps: $89 + 418 + 121 - 349 = +279$ . $\Delta H^{\ominus}_{\text{latt}} = -437 - 279 = -716\ \text{kJ mol}^{-1}$ .
Indirect steps: $178 + 590 + 1150 + 2(121) + 2(-349) = 178 + 1740 + 242 - 698 = +1462$ . $\Delta H^{\ominus}_{\text{latt}} = -796 - 1462 = -2258\ \text{kJ mol}^{-1}$ .
Atomising two fluorine atoms needs $2 \times \tfrac{1}{2}(158) = +158$ . $\Delta H^{\ominus}_{\text{f}} = 148 + 2186 + 158 + 2(-328) + (-2957) = -1121\ \text{kJ mol}^{-1}$ .
Two sodium atoms: $2(107) + 2(494) = 214 + 988 = 1202$ . Oxygen: $249 - 141 + 798 = 906$ . Total $= +2108$ . $\Delta H^{\ominus}_{\text{latt}} = -414 - 2108 = -2522\ \text{kJ mol}^{-1}$ .
$-288 = 107 + 494 + 107 + EA_1 - 704$ , so $EA_1 = -288 - 708 + 704 = -292\ \text{kJ mol}^{-1}$ .
$-635 = \Delta H^{\ominus}_{\text{at}}(\ce{Ca}) + 1740 + 249 - 141 + 798 - 3459$ . The known terms sum to $-813$ , so $\Delta H^{\ominus}_{\text{at}}(\ce{Ca}) = -635 + 813 = +178\ \text{kJ mol}^{-1}$ .
Largest endothermic: the second ionisation energy of magnesium ( $+1450$ ), with the second electron affinity of oxygen ( $+798$ ) also large. Largest exothermic: the lattice energy (about $-3800$ ). Because the ions are small and doubly charged, the lattice energy released is greater than the total energy needed to form the gaseous ions, so $\Delta H^{\ominus}_{\text{f}}$ is negative.
$\Delta H^{\ominus}_{\text{f}} = 107 + 494 + 4560 + 2(121) + 2(-349) + (-2500) = +2205\ \text{kJ mol}^{-1}$ . The second electron would be removed from the inner 2p subshell of sodium, close to the nucleus and poorly shielded, so $IE_2$ is enormous. Even the extra lattice energy from a 2+ ion cannot compensate, so $\ce{NaCl2}$ would be hugely endothermic and does not form.
Levels from bottom: $\ce{CaF2(s)}$ ; $\ce{Ca(s) + F2(g)}$ at zero; $\ce{Ca(g) + F2(g)}$ ; $\ce{Ca(g) + 2F(g)}$ ; $\ce{Ca^+(g) + 2F(g) + e-}$ ; $\ce{Ca^{2+}(g) + 2F(g) + 2e-}$ ; then down to $\ce{Ca^{2+}(g) + 2F^-(g)}$ ( $2 \times EA_1$ ); then down to $\ce{CaF2(s)}$ (lattice energy). Upward arrows: atomisation of Ca, atomisation of 2F, first and second ionisation energies of Ca. Downward: $2EA_1$ of F, lattice energy, and $\Delta H^{\ominus}_{\text{f}}$ (from the elements down to the solid).

The idea: two routes from elements to solid

The steps in the indirect route

Drawing the cycle as an energy-level diagram

Cations with a 2+ charge and anions with a 2– charge

Worked examples

Reading a Born–Haber cycle for information

Exam technique

Summary

Practice questions

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