Born-Haber Cycles (Edexcel International A Level (IAL) Chemistry): Revision Note

Exam code: YCH11

Philippa Platt

Written by: Philippa Platt

Reviewed by: Caroline Carroll

Updated on

Born-Haber Cycles - Construction

  • A Born-Haber cycle is a specific application of Hess’ Law for ionic compounds and enable us to calculate lattice enthalpy which cannot be found by experiment

  • The basic principle of drawing the cycle is to construct a diagram in which energy increases going up the diagram

Energy axis for Born-Haber cycles: endothermic reactions go up because the energy of the substances increases; exothermic reactions go down because it decreases

The basic principle of a Born-Haber cycle

  • The cycle shows all the steps needed to turn atoms into gaseous ions and from gaseous ions into the ionic lattice

  • The alternative route to the ionic lattice begins from the enthalpy of formation of the elements in their standard states

Drawing the cycle for sodium chloride

  • A good starting point is to draw the elements with their state symbols about a third of the way up the diagram

  • This is shown as the left hand side of the equation for the process indicated

  • The location is marked by drawing a horizontal bar or line which represents the starting energy level

Start of the sodium chloride Born-Haber cycle: Na(s) + 1/2 Cl2(g) drawn a third of the way up, with the gaseous ions above and the ionic solid below

Drawing a Born-Haber cycle step 1

  • Next, we need to create the gaseous ions

  • This is a two step process of first creating the gaseous atoms and then turning them into ions

  • Creating gaseous atoms is a bond breaking process, so arrows must be drawn upwards

  • It doesn't matter whether you start with sodium or chlorine

  • The enthalpy of atomisation of sodium is

Na (s) → Na (g)           ΔatHꝋ = +108 kJ mol -1

  • The enthalpy of atomisation of chlorine is

½Cl2 (g) → Cl (g)       ΔatHꝋ = +121 kJ mol -1

  • We can show the products of the process on the horizontal lines and the energy value against a vertical arrow connecting the energy levels

Born-Haber cycle step 2: atomisation of sodium (+108) and of chlorine (+121) take Na(s) + 1/2 Cl2(g) up to Na(g) + Cl(g)

Drawing a Born-Haber cycle step 2 - creating the gaseous atoms

  • Now the ions are created

  • The sodium atom loses an electron, so this energy change is the first ionisation energy for sodium

Na (g) → Na+ (g) + e–          ΔieHꝋ = +500 kJ mol-1

  • The change is endothermic so the direction continues upwards

  • The chlorine atom gains an electron, so this is electron affinity

Cl (g) + e– → Cl- (g)           ΔeaHꝋ = -364 kJ mol-1

  • The exothermic change means this is downwards

  • The change is displaced to the right to make the diagram easier to read

Born-Haber cycle step 3: first ionisation energy of sodium (+500) up to Na+(g) + Cl(g) + e-, then electron affinity of chlorine (-364) down to Na+(g) + Cl-(g)

Drawing a Born-Haber cycle step 3 - creating the gaseous ions

  • The two remaining parts of the cycle can now be completed

  • The enthalpy of formation of sodium chloride is added at the bottom of the diagram

Na (s) + ½Cl2 (g) → NaCl (s)            ΔfHꝋ = -411 kJ mol -1

  • This is an exothermic change for sodium chloride so the arrow points downwards

  • Enthalpy of formation can be exothermic or endothermic, so you may need to show it above the elements ( and displaced to the right) for a endothermic change

  • The final change is lattice enthalpy, which is usually shown as a formation. For sodium chloride the equation is

Na+(g) + Cl-(g) → NaCl (s)  ΔlattHꝋ 

 

Completed Born-Haber cycle for sodium chloride: enthalpy of formation (-411) down to NaCl(s) and lattice enthalpy down from Na+(g) + Cl-(g) to NaCl(s)

Drawing a Born-Haber cycle step 4 - completing the cycle

  • The cycle is now complete

  • The cycle is usually used to calculate the lattice enthalpy of an ionic solid, but can be used to find other enthalpy changes if you are given the lattice enthalpy

Worked Example

Constructing a Born-Haber cycle for KCl

Construct a Born-Haber Cycle which can be used to calculate the lattice energy of potassium chloride

Question table: steps to build the Born-Haber cycle for KCl with the equation and enthalpy change for each step, ending with adding values to get delta H1 and applying Hess law for the lattice enthalpy

Answer

Born-Haber cycle for KCl showing atomisation of K and Cl, first ionisation energy of K, first electron affinity of Cl, lattice enthalpy and enthalpy of formation, with the direct and indirect routes marked

Born-Haber cycle for KCl, from the bottom upwards: K(s) + ½Cl2(g) is the starting level. Up (endothermic): K(s) → K(g), ΔatH(K); ½Cl2(g) → Cl(g), ΔatH(Cl); K(g) → K+(g) + e-, IE1, giving K+(g) + Cl(g) + e-. Down (exothermic): Cl(g) + e- → Cl-(g), EA1, giving K+(g) + Cl-(g); then K+(g) + Cl-(g) → KCl(s), ΔlattH, which is the direct route in the diagram. ΔfH is K(s) + ½Cl2(g) → KCl(s), downwards.

Worked Example

Constructing a Born-Haber cycle for MgO

Construct a Born-Haber Cycle which can be used to calculate the lattice energy of magnesium oxide

Question table: steps to build the Born-Haber cycle for MgO, including IE1 and IE2 of magnesium and EA1 and EA2 of oxygen

Answer

Born-Haber cycle for MgO showing atomisation of Mg and O, IE1 and IE2 of Mg, EA1 and EA2 of O, lattice enthalpy and enthalpy of formation

Born-Haber cycle for MgO, from the bottom upwards: Mg(s) + ½O2(g) is the starting level. Up: Mg(s) → Mg(g), ΔatH(Mg); ½O2(g) → O(g), ΔatH(O); Mg(g) → Mg+(g) + e-, IE1; Mg+(g) → Mg2+(g) + e-, IE2. Then down: O(g) + e- → O-(g), EA1; up: O-(g) + e- → O2-(g), EA2 (endothermic); then Mg2+(g) + O2-(g) → MgO(s), ΔlattH, downwards. ΔfH is Mg(s) + ½O2(g) → MgO(s), downwards.

Born-Haber Cycles - Calculations

  • Once a Born-Haber cycle has been constructed, it is possible to calculate the lattice energy (ΔHlattꝋ) by applying Hess’s law and rearranging:

ΔfHꝋ = ΔatHꝋ + ΔatHꝋ + IE + EA + ΔlattHꝋ

  • If we simplify this into three terms, this makes the equation easier to see:

    • ΔHlattꝋ

    • ΔHfꝋ

    • ΔH1ꝋ (the sum of all of the various enthalpy changes necessary to convert the elements in their standard states to gaseous ions)

  • The simplified equation becomes

ΔfHꝋ = Δ1Hꝋ + ΔlattHꝋ

      So, if we rearrange to calculate the lattice energy, the equation becomes

ΔlattHꝋ = ΔfHꝋ - Δ1Hꝋ

  • When calculating the ΔHlattꝋ, all other necessary values will be given in the question

  • A Born-Haber cycle could be used to calculate any stage in the cycle

    • For example, you could be given the lattice energy and asked to calculate the enthalpy change of formation of the ionic compound

    • The principle would be exactly the same

    • Work out the direct and indirect route of the cycle (the stage that you are being asked to calculate will always be the direct route)

    • Write out the equation in terms of enthalpy changes and rearrange if necessary to calculate the required value

  • Remember: sometimes a value may need to be doubled or halved, depending on the ionic solid involved

    • For example, with MgCl2 the value for the first electron affinity of chlorine would need to be doubled in the calculation, because there are two moles of chlorine atoms

    • Therefore, you are adding 2 moles of electrons to 2 moles of chlorine atoms, to form 2 moles of Cl- ions

Worked Example

Calculating the lattice energy of KCl

Given the data below, calculate the ΔlattHꝋ of potassium chloride (KCl)  

Data table for KCl: enthalpy of atomisation K +90, Cl +122; IE1 K +418; EA1 Cl -349; enthalpy of formation of KCl -437 kJ mol-1

Data (kJ mol-1): ΔatH K +90; ΔatH Cl +122; IE1 K +418; EA1 Cl -349; ΔfH KCl -437.

Answer

Step 1: The corresponding Born-Haber cycle is:

 

Born-Haber cycle for KCl showing atomisation of K and Cl, first ionisation energy of K, first electron affinity of Cl, lattice enthalpy and enthalpy of formation, with the direct and indirect routes marked

Step 2: Applying Hess’ law, the lattice energy of KCl is:

ΔlattHꝋ = ΔfHꝋ - Δ1Hꝋ

ΔlattHꝋ = ΔfHꝋ - [(ΔatHꝋ K) + (ΔatHꝋ Cl) + (IE1 K) + (EA1 Cl)]

Step 3: Substitute in the numbers:

ΔlattH = (-437) - [(+90) + (+122) + (+418) + (-349)] = -718 kJ mol-1

Worked Example

Calculating the lattice energy of MgO

Given the data below, calculate the ΔlattHꝋ of magnesium oxide (MgO)

Data table for MgO: enthalpy of atomisation Mg +148, O +248; IE1 +736, IE2 +1450 for Mg; EA1 -142, EA2 +770 for O; enthalpy of formation of MgO -602 kJ mol-1

Data (kJ mol-1): ΔatH Mg +148; ΔatH O +248; IE1 Mg +736; IE2 Mg +1450; EA1 O -142; EA2 O +770; ΔfH MgO -602.

Answer

Step 1: The corresponding Born-Haber cycle is:

 

Born-Haber cycle for MgO showing atomisation of Mg and O, IE1 and IE2 of Mg, EA1 and EA2 of O, lattice enthalpy and enthalpy of formation

Step 2: Applying Hess’ law, the lattice energy of MgO is:

ΔlattHꝋ = ΔfHꝋ - Δ1Hꝋ

ΔlattHꝋ = ΔfHꝋ - [(ΔatHꝋ Mg) + (ΔatHꝋ O) + (IE1 Mg) + (IE2 Mg) + (EA1 O) + (EA2 O)]

Step 3: Substitute in the numbers:

ΔlattHꝋ = (-602) - [(+148) + (+248) + (+736) + (+1450) + (-142) + (+770)]

= -3812 kJ mol-1

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Philippa Platt

Author: Philippa Platt

Expertise: Chemistry Content Creator

Philippa has worked as a GCSE and A level chemistry teacher and tutor for over thirteen years. She studied chemistry and sport science at Loughborough University graduating in 2007 having also completed her PGCE in science. Throughout her time as a teacher she was incharge of a boarding house for five years and coached many teams in a variety of sports. When not producing resources with the chemistry team, Philippa enjoys being active outside with her young family and is a very keen gardener

Caroline Carroll

Reviewer: Caroline Carroll

Expertise: Head of Content Delivery

Caroline graduated from the University of Nottingham with a degree in Chemistry and Molecular Physics. She spent several years working as an Industrial Chemist in the automotive industry before retraining to teach. Caroline has over 12 years of experience teaching GCSE and A-level chemistry and physics. She is passionate about delivering high-quality resources to help students achieve their full potential.