Lattice enthalpy

In a nutshell

An ionic solid is held together by strong electrostatic attraction between oppositely charged ions. Lattice enthalpy puts a number on how strong that attraction is.

You cannot measure lattice enthalpy directly, so you find it indirectly with a Born-Haber cycle. A second kind of cycle links lattice enthalpy to what happens when an ionic solid dissolves: the enthalpy changes of solution and hydration.

Assumed knowledge: Bonding and structure: ionic bonding and giant ionic lattices; Periodicity: first ionisation energy; Enthalpy changes: Δ_fH^⦵ and Hess' law cycles.

Core content

Lattice enthalpy: definition and ionic bond strength

Lattice enthalpy, Δ_LEH: the enthalpy change that accompanies the formation of one mole of an ionic compound from its gaseous ions (under standard conditions).

Na⁺(g) + Cl^-(g) → NaCl(s)    Δ_LEH = -787 kJ mol^-1

Mg²⁺(g) + 2Cl^-(g) → MgCl₂(s)

Three things the definition and equation must show:

• one mole of the ionic compound is formed (not one mole of ions)
• the starting ions are gaseous, so every ion on the left has (g)
• the product is the solid lattice, (s)

Lattice enthalpy is always exothermic (negative). Gaseous ions are far apart; when they come together into a lattice, strong attractions form and energy is released.

Lattice enthalpy is a measure of the strength of ionic bonding in a giant ionic lattice:

• the more exothermic (more negative) the lattice enthalpy, the stronger the electrostatic attraction between the ions
• stronger attraction means more energy is needed to overcome it, so the compound has a higher melting point

Born-Haber and related enthalpy cycles: the energy terms

A Born-Haber cycle uses five kinds of enthalpy change. Learn the equation for each, with state symbols, and its sign.

Only two of these need word-for-word definitions. Both must include every condition:

Standard enthalpy change of formation, Δ_fH^⦵: the enthalpy change that takes place when one mole of a compound is formed from its elements, under standard conditions, with all reactants and products in their standard states.

First ionisation energy: the energy required to remove one electron from each atom in one mole of gaseous atoms to form one mole of gaseous 1+ ions.

For the others you must understand the term and write its equation, but a formal definition is not required:

• Enthalpy change of atomisation: forms one mole of gaseous atoms from the element in its standard state. For chlorine that is ½Cl₂(g) → Cl(g), so it is half the Cl-Cl bond enthalpy.
• First electron affinity: each atom in one mole of gaseous atoms gains one electron to form one mole of gaseous 1- ions. It is exothermic because the incoming electron is attracted to the nucleus.
• Second electron affinity: each gaseous 1- ion gains another electron. It is endothermic, because the electron is being added to an ion that is already negative, so energy is needed to overcome the repulsion.
• Second ionisation energy: needed for any metal that forms a 2+ ion (Mg²⁺, Ca²⁺), on top of the first.

Born-Haber and related enthalpy cycles: constructing a Born-Haber cycle

Lattice enthalpy cannot be measured directly: you cannot make a mole of separate gaseous ions and let them combine. A Born-Haber cycle finds it indirectly by conservation of energy (Hess' law): the route from the elements straight to the compound (Δ_fH^⦵) has the same enthalpy change as the long route through gaseous atoms and gaseous ions.

How to build any Born-Haber cycle:

1. Start at the elements in their standard states (energy zero). Write Δ_fH^⦵ as the arrow from here down to the ionic solid.
2. Atomise the metal, then ionise it: endothermic steps, arrows up.
3. Atomise the non-metal: arrow up.
4. Add electrons to the non-metal atoms (electron affinity): an exothermic first EA is an arrow down; a second EA is an arrow up.
5. The lattice enthalpy is the big arrow down from the gaseous ions to the solid.
6. Every level must show all the species present, with state symbols (including the electrons while they are "in transit").

Going round the cycle, both routes from the elements to NaCl(s) must give the same total:

$\Delta_f H^{\ominus} = \Delta_{at}H(\text{Na}) + \text{1st IE(Na)} + \Delta_{at}H(\text{Cl}) + \text{1st EA(Cl)} + \Delta_{LE}H$

Rearranging gives the lattice enthalpy:

$\Delta_{LE}H = \Delta_f H^{\ominus} - \left[\Delta_{at}H(\text{Na}) + \text{1st IE(Na)} + \Delta_{at}H(\text{Cl}) + \text{1st EA(Cl)}\right]$

$\Delta_{LE}H = -411 - \left[(+107) + (+496) + (+122) + (-349)\right] = -411 - (+376) = -787\ \text{kJ mol}^{-1}$

Still don't get it? · why a Born-Haber cycle works

Imagine a building where the lift to one floor is broken. You want to know how far below the ground floor the basement is, but you cannot take the direct stairs. You do know the height of every other flight: up to floor 1, up to floor 3, down to floor 2, and so on.

If you know where the basement sits compared with the ground floor by another route, you can work out the one flight you never walked.

1. The ground floor is the elements, Na(s) and ½Cl₂(g).
2. The basement is NaCl(s). Its height below the ground floor is Δ_fH^⦵, which can be measured.
3. The flights you can measure are atomisation, ionisation energy and electron affinity. They take you from the ground floor to the gaseous ions.
4. The one flight you cannot walk is gaseous ions down to the solid: the lattice enthalpy.
5. Total height change by the long route must equal the total by the short route, because energy is conserved.

Exam version: Δ_fH^⦵ = (sum of atomisation, ionisation and electron affinity terms) + Δ_LEH, so Δ_LEH = Δ_fH^⦵ - (sum of the other terms). Write the expression out with every sign before you put numbers in.

Born-Haber and related enthalpy cycles: scaling for MgCl₂ and oxides

The formula of the compound decides how many times each step happens.

Both kinds of cycle appear in Worked examples 1 and 3.

Born-Haber and related enthalpy cycles: enthalpy change of solution and hydration

Enthalpy change of solution, Δ_solH: the enthalpy change when one mole of a solute (here, an ionic compound) dissolves in water.

NaCl(s) + aq → Na⁺(aq) + Cl^-(aq)

MgCl₂(s) + aq → Mg²⁺(aq) + 2Cl^-(aq)

Enthalpy change of hydration, Δ_hydH: the enthalpy change when one mole of gaseous ions dissolves in water to form aqueous ions.

Na⁺(g) + aq → Na⁺(aq)

Cl^-(g) + aq → Cl^-(aq)

Signs:

• Δ_hydH is always exothermic: attractions form between the ion and polar water molecules, releasing energy.
• Δ_solH can be exothermic or endothermic, depending on the balance between breaking up the lattice and hydrating the ions.

Δ_solH can be measured in the lab by calorimetry (dissolve a weighed solid in water, then use q = mcΔT). Lattice enthalpy and Δ_hydH cannot, so they come from cycles.

Born-Haber and related enthalpy cycles: enthalpy cycles for dissolving

Dissolving can be split into two imaginary steps:

1. Break up the lattice into gaseous ions. This is the reverse of lattice enthalpy, so its value is -Δ_LEH (endothermic).
2. Hydrate each gaseous ion. Exothermic.

The final level sits 124 kJ mol^-1 below the solid, so Δ_solH(MgCl₂) = -124 kJ mol^-1 with these data (Worked example 4).

$\Delta_{sol}H = -\Delta_{LE}H + \sum \Delta_{hyd}H$

Remember to multiply each Δ_hydH by the number of those ions in the formula.

The same expression rearranges to find any one unknown: a lattice enthalpy, or the hydration enthalpy of one ion (Worked example 5).

Still don't get it? · why the lattice enthalpy changes sign in a solution cycle

Think of a Lego model. Building it from loose bricks gives you a finished model; taking it apart is exactly the same job run backwards. If building it "releases" 787 units, taking it apart "costs" 787 units.

1. Lattice enthalpy, as OCR defines it, is the building direction: gaseous ions → solid. It is negative.
2. Dissolving starts from the solid. The first imaginary step is taking the lattice apart: solid → gaseous ions.
3. That is the building step in reverse, so it has the same size but the opposite sign: -Δ_LEH, a positive number.
4. Then each gaseous ion is surrounded by water molecules: the hydration steps, each negative.
5. Add the steps up: (lattice taken apart) + (all the hydrations) = Δ_solH.

Exam version: Δ_solH = -Δ_LEH + ΣΔ_hydH. If you put the lattice enthalpy in with its own negative sign, you are going round the cycle the wrong way.

Born-Haber and related enthalpy cycles: effect of ionic charge and ionic radius

Both lattice enthalpy and hydration enthalpy come from electrostatic attraction, so both follow the same two rules.

Lattice enthalpy

• Smaller ionic radius: the ions are closer together, so there is stronger attraction between the oppositely charged ions, more energy is released, and Δ_LEH is more exothermic (more negative).
• Greater ionic charge: stronger attraction between the oppositely charged ions, so Δ_LEH is more exothermic.

Down Group 1 the cation gets larger, so the lattice enthalpies of the chlorides become less exothermic. Doubling both charges (NaF to MgO) makes the lattice enthalpy roughly four times as exothermic.

Enthalpy change of hydration

Water is a polar molecule. Each ion attracts the oppositely charged end of the water molecules around it:

• a positive ion attracts the δ- oxygen atoms of water molecules
• a negative ion attracts the δ+ hydrogen atoms of water molecules

• Smaller ionic radius: stronger attraction between the ion and water molecules, so Δ_hydH is more exothermic.
• Greater ionic charge: stronger attraction between the ion and water molecules, so Δ_hydH is more exothermic.

Na⁺ and Mg²⁺ show both effects at once: Mg²⁺ has a greater charge and a smaller radius, so its hydration enthalpy is far more exothermic.

You never need to memorise these numbers: questions give the data. You do need to explain the trends.

Worked examples

All data are given, as they would be in a question. Values are in kJ mol^-1, to the nearest whole number, so answers are given to the nearest whole number too.

Worked example 1: lattice enthalpy of MgCl₂

Calculate Δ_LEH of magnesium chloride.

Step 1: write the cycle expression, scaling the chlorine terms by 2 and including both ionisation energies.

$\Delta_{LE}H = \Delta_f H^{\ominus} - \left[\Delta_{at}H(\text{Mg}) + \text{1st IE} + \text{2nd IE} + 2\Delta_{at}H(\text{Cl}) + 2\,\text{1st EA(Cl)}\right]$

Step 2: add up the terms in the bracket, keeping each sign.

$(+148) + (+738) + (+1451) + 2(+122) + 2(-349)\ \text{kJ mol}^{-1}$ $= 148 + 738 + 1451 + 244 - 698 = +1883\ \text{kJ mol}^{-1}$

Step 3: subtract from Δ_fH^⦵.

$\Delta_{LE}H = -641 - (+1883)\ \text{kJ mol}^{-1}$ $\Delta_{LE}H = -2524\ \text{kJ mol}^{-1}$

Where marks go: forgetting the 2nd IE gives -1073; forgetting to double the chlorine terms gives -2751. Both are common.

Worked example 2: finding an electron affinity

Use the data to calculate the first electron affinity of chlorine.

Step 1: convert the bond enthalpy to an atomisation enthalpy. Breaking one mole of Cl-Cl bonds makes two moles of Cl atoms, but atomisation is per one mole of atoms.

$\Delta_{at}H(\text{Cl}) = \frac{+242}{2} = +121\ \text{kJ mol}^{-1}$

Step 2: write the cycle expression.

$\Delta_f H^{\ominus} = \Delta_{at}H(\text{K}) + \text{1st IE(K)} + \Delta_{at}H(\text{Cl}) + \text{1st EA(Cl)} + \Delta_{LE}H$

Step 3: rearrange for the unknown.

$\text{1st EA(Cl)} = \Delta_f H^{\ominus} - \Delta_{at}H(\text{K}) - \text{1st IE(K)} - \Delta_{at}H(\text{Cl}) - \Delta_{LE}H$

Step 4: substitute.

$\text{1st EA(Cl)} = (-437) - (+89) - (+419) - (+121) - (-718)\ \text{kJ mol}^{-1}$ $\text{1st EA(Cl)} = -437 - 89 - 419 - 121 + 718 = -348\ \text{kJ mol}^{-1}$

Where marks go: using the full bond enthalpy (+242) gives -469. Subtracting -718 means adding 718; a sign slip here gives -1784.

Worked example 3: lattice enthalpy of an oxide (MgO)

Calculate Δ_LEH of magnesium oxide.

Step 1: add up every term between the elements and the gaseous ions. Δ_atH(O) is used once: MgO has one oxygen atom, made from ½O₂.

$(+148) + (+738) + (+1451) + (+249) + (-141) + (+798)\ \text{kJ mol}^{-1} = +3243\ \text{kJ mol}^{-1}$

Step 2: subtract from Δ_fH^⦵.

$\Delta_{LE}H = -602 - (+3243)\ \text{kJ mol}^{-1}$ $\Delta_{LE}H = -3845\ \text{kJ mol}^{-1}$

Where marks go: treating the 2nd EA as negative (-798) gives -2249. The 2nd EA is the one upward step after the ions start forming; it is endothermic.

Worked example 4: enthalpy change of solution of MgCl₂

Use Δ_LEH(MgCl₂) = -2524 kJ mol^-1, Δ_hydH(Mg²⁺) = -1920 kJ mol^-1 and Δ_hydH(Cl^-) = -364 kJ mol^-1 to calculate Δ_solH(MgCl₂). This is the cycle drawn in Core content.

Step 1: write the expression. Two chloride ions are hydrated.

$\Delta_{sol}H = -\Delta_{LE}H + \Delta_{hyd}H(\text{Mg}^{2+}) + 2\Delta_{hyd}H(\text{Cl}^-)$

Step 2: substitute.

$\Delta_{sol}H = -(-2524) + (-1920) + 2(-364)\ \text{kJ mol}^{-1}$ $\Delta_{sol}H = +2524 - 1920 - 728\ \text{kJ mol}^{-1}$ $\Delta_{sol}H = -124\ \text{kJ mol}^{-1}$

Where marks go: using +2524 the wrong way round (adding -2524) gives -5172; hydrating only one Cl^- gives +240.

Worked example 5: lattice enthalpy from a solution cycle

Δ_solH(LiCl) = -37 kJ mol^-1, Δ_hydH(Li⁺) = -519 kJ mol^-1 and Δ_hydH(Cl^-) = -364 kJ mol^-1. Calculate Δ_LEH(LiCl).

Step 1: rearrange the solution expression for the lattice enthalpy.

$\Delta_{sol}H = -\Delta_{LE}H + \sum \Delta_{hyd}H$ $\Delta_{LE}H = \sum \Delta_{hyd}H - \Delta_{sol}H$

Step 2: sum the hydration enthalpies.

$\sum \Delta_{hyd}H = (-519) + (-364) = -883\ \text{kJ mol}^{-1}$

Step 3: substitute.

$\Delta_{LE}H = -883 - (-37)\ \text{kJ mol}^{-1}$ $\Delta_{LE}H = -846\ \text{kJ mol}^{-1}$

Check the sign: a lattice enthalpy must be negative. A positive answer means the rearrangement went wrong (the common wrong answer is +846 or -920).

Worked example 6: explaining a difference in lattice enthalpy

Explain why the lattice enthalpy of KCl is less exothermic than that of NaCl.

A full answer makes three linked points, about ions:

1. K⁺ has a larger ionic radius than Na⁺ (the chloride ion is the same in both).
2. So there is weaker attraction between the K⁺ and Cl^- ions than between Na⁺ and Cl^- ions.
3. So less energy is released when the KCl lattice forms: Δ_LEH is less exothermic (less negative).

The same three-step pattern (ion size or charge, then attraction, then energy released) works for comparing hydration enthalpies, with "attraction between the ion and water molecules" in step 2.

Common exam mistakes

• Defining lattice enthalpy as formation of a compound "from its elements": that is Δ_fH^⦵. Lattice enthalpy starts from gaseous ions.
• Leaving out "gaseous" from the lattice enthalpy or hydration definition, or writing (aq) or no state symbols on the ions in the equation.
• Writing "one mole of ions" in the lattice enthalpy definition; it is one mole of the ionic compound (one mole of MgCl₂ contains three moles of ions).
• Describing lattice enthalpy as the energy needed to break up the lattice. In OCR's definition the lattice is formed, so the value is negative.
• Using the full Cl-Cl bond enthalpy as the atomisation enthalpy of chlorine: atomisation makes one mole of Cl atoms, so it is half the bond enthalpy.
• Writing atomisation as Cl₂(g) → 2Cl(g), or the electron affinity as starting from Cl₂. The equations are ½Cl₂(g) → Cl(g) and Cl(g) + e^- → Cl^-(g).
• For MgCl₂, forgetting the 2nd ionisation energy of magnesium, or forgetting to double the atomisation and electron affinity of chlorine.
• Giving the second electron affinity of oxygen a negative sign, or drawing its arrow downwards. It is endothermic because the electron is repelled by the negative O^- ion.
• Bracket and sign slips when rearranging: Δ_LEH = Δ_fH^⦵ minus the sum of the other terms, and subtracting a negative electron affinity means adding.
• In a solution cycle, using +Δ_LEH instead of -Δ_LEH for breaking up the lattice.
• Hydrating only one chloride ion for MgCl₂ (it needs 2 × Δ_hydH(Cl^-)).
• Defining Δ_hydH without "one mole of gaseous ions", or defining Δ_solH without "one mole of solute" dissolving in water.
• Explaining lattice enthalpy trends with atoms, nuclear charge or shielding. The argument is about the ions: their size and charge, and the attraction between oppositely charged ions.
• Calling the forces in an ionic lattice "intermolecular forces" or talking about "molecules" of NaCl. It is ionic bonding in a giant ionic lattice.
• Writing "larger lattice enthalpy" or "higher lattice enthalpy" when comparing. Say more exothermic or more negative so the direction is unambiguous.
• Saying a positive ion attracts the hydrogen atoms of water. It attracts the δ- oxygen; a negative ion attracts the δ+ hydrogen.
• Leaving the sign or the units (kJ mol^-1) off a final answer.

Key definitions

• Lattice enthalpy, Δ_LEH: the enthalpy change that accompanies the formation of one mole of an ionic compound from its gaseous ions (under standard conditions).
• Standard enthalpy change of formation, Δ_fH^⦵: the enthalpy change that takes place when one mole of a compound is formed from its elements, under standard conditions, with all reactants and products in their standard states.
• First ionisation energy: the energy required to remove one electron from each atom in one mole of gaseous atoms to form one mole of gaseous 1+ ions.
• Enthalpy change of solution, Δ_solH: the enthalpy change when one mole of a solute dissolves in water.
• Enthalpy change of hydration, Δ_hydH: the enthalpy change when one mole of gaseous ions dissolves in water.
• Enthalpy change of atomisation, Δ_atH: the enthalpy change when one mole of gaseous atoms is formed from the element in its standard state (a formal definition is not required).
• First electron affinity: the enthalpy change when each atom in one mole of gaseous atoms gains one electron to form one mole of gaseous 1- ions (a formal definition is not required).

Specification

• I can define lattice enthalpy (formation of 1 mol of ionic lattice from gaseous ions) and write its equation with state symbols. (5.2.1(a))
• I can explain why lattice enthalpy is a measure of the strength of ionic bonding in a giant ionic lattice. (5.2.1(a))
• I can define first ionisation energy and enthalpy change of formation, and write equations for atomisation, ionisation energy and electron affinity. (5.2.1(b))
• I can construct a Born-Haber cycle for a simple ionic solid such as NaCl or MgCl₂, with every species and state symbol. (5.2.1(b)(i))
• I can calculate a lattice enthalpy, or any other unknown term, from a Born-Haber cycle. (5.2.1(b)(ii))
• I can define enthalpy change of solution and enthalpy change of hydration. (5.2.1(c))
• I can construct an enthalpy cycle linking Δ_solH, Δ_hydH and lattice enthalpy. (5.2.1(d)(i))
• I can calculate Δ_solH, a lattice enthalpy or a hydration enthalpy from that cycle. (5.2.1(d)(ii))
• I can explain qualitatively how ionic charge and ionic radius affect the exothermic value of a lattice enthalpy and of an enthalpy change of hydration. (5.2.1(e))

Related notes

• Bonding and structure: ionic bonding and the properties of giant ionic lattices
• Periodicity: first and successive ionisation energies
• Enthalpy changes: Δ_fH^⦵, Hess' law cycles and measuring ΔH with q = mcΔT
• Enthalpy and entropy: why a reaction or a dissolving process with a positive ΔH can still happen