Lattice Enthalpy

Giant Ionic Compounds: Giant ionic compounds have strong ionic bonds- electrostatic
attractions between oppositely- charged ions. Must overcome to melt.
 Lattice Enthalpy: Lattice enthalpy is the enthalpy change that accompanies formation of one
mole of an ionic compound from its gaseous ions under standard conditions. I.e. the strength of
ionic bond. Lattice enthalpy is exothermic as form ionic bonds. Negative value.
Mg2+ (g) + 2Cl- (g) -> MgCl2 (s) Gaseous ions -> solid ionic compound.

Born- Haber Cycle: Lattice enthalpy cannot be measured directly. Indirectly determine from
Born- Haber cycle. According to Hess’ Law, both routes equal each other. Routes must follow
arrows.
 Route 2: Route 2 is the enthalpy change of formation ∆fH, exothermic.
 Remember exothermic arrows points downwards.
 Remember enthalpy values are relative to number in front, so will have to halve or double when
numbers in front of element.
 Standard Enthalpy Change of Formation ∆fH is the enthalpy change when one mole of
compound is formed from its elements under standard conditions, with all reactants and
products in standard states. Na (g) + ½ Cl2 (g) -> NaCl (s). Na2O (s) -> 2Na (s) + ½ O2 (g).
 Standard Enthalpy Change of Atomisation ∆atH
θ
is the enthalpy change for formation of one
mole of gaseous atoms from elements in standard state. Endothermic as bonds broken.
Cl2 (g) -> 2Cl (g) this one here is two moles, not one. Na (s) -> Na (g).
 First Ionisation Energy ∆IEH
θ
is the enthalpy change required to remove one electron from each
atom in one mole of gaseous atoms to form one mole of gaseous 1+ ions. Na (g) -> Na+(g) + e-.
– Endothermic because energy is required to remove electrons which are attracted to nucleus.
– Second ionisation energy is more endothermic than first ionisation energy because electron
lost closer to nucleus, so Ca+ is smaller than Ca so outer electron has greater nuclear
attraction.
 First Electron Affinity ∆EAH
θ
is the enthalpy change when one electron added in one mole of
gaseous atoms to form one mole of gaseous 1- ions. EA opposite to IE. Exothermic as electron
being added is attracted to nucleus. Cl (g) + e-  Cl- (g)
 An Example: Will be asked to construct Born- Haber cycle.
 If asked to label Born Haber cycle and 2Na, write 2x enthalpy change of atomisation of Na. Also
label which element/ compound the enthalpy change is of.
 Done separately for ∆atH
θ
(Na) and ∆atH
θ
(Cl). However still rewrite the other species as the
same.
 Route 1 = route 2 according to Hess’ Law. Substiute numbers and find unknown.
 If given Mg, to get to Mg2+ need to do first and second ionisation energy ∆IE 1H
θ
and ∆IE 2H
θ
, so
another energy level needed.
 Below are 2Cl- or Cl2, so must multiply enthalpy change by 2 when calculating. Ignore if ½ O2.
Notice how CaO is converted into ½ O2 and then into O.

Successive Electron Affinities: Second electron affinities are endothermic unlike first EA. Oxide
ion O- and electron are both negative. Therefore energy is required to add electron by
overcoming repulsion.
Other species ‘takes’ electron.
 Difficult to determine lattice enthalpy of sodium carbonate because cycle would need formation
of CO3
2- from C and O. So better to use enthalpy change of solution.
Enthalpy Changes in Solution
 Dissolving Ionic Compounds: The δ- oxygen in water attracted to positive metal ion. The δ+
hydrogen atoms in water attracted to negative non- metal ion. Name the ions. So the ionic
lattice breaks down.
 Enthalpy Change of Solution ∆solH
θ
is the enthalpy change when one mole of solute dissolves in
a water. Can either be endothermic or exothermic. Forms aqueous ions- NaCl (s) + aq -> Na+ (aq)
+ Cl- (aq).
 Experimental Determination of Enthalpy Change of Solution: Weigh ionic compound. Water in
plastic cup in beaker. Measure temperature of water. Quickly tip ionic compound in and stir with
thermometer- do this until all solid dissolves and temperature constant.
 Calculation for Enthalpy Change of Solution:
1) Use q = mc∆T. Where t is temperature change. C is specific heat capacity, usually 4.18. M is
mass of solution = mass of water + mass of ionic compound. OR sometimes just given mass
of final solution so use that instead. (As the solution is changing temperature). Q is in kJ
(heat energy). Divide by 1000 to get into kJ.

Find moles of ionic compound dissolved using mass/ Mr.
3) Calculate ∆solH
θ
. Divide step 1 answer/ step 2 answer (as want to find out energy gained
from 1 mole).
4) Remember to add on negative sign as exothermic. Or don’t if endothermic.
 If mass doubled, then temperature change is doubled but enthalpy change of solution is the
same since value is per mole.
 Can do opposite of above. If given an energy value in kJ and enthalpy change of reaction, but
need to find mass of a reactant. Divide energy by enthalpy change to get one mol of reactant.
Multiply this value by balancing number in equation. Then use Mr Moles = mass.
 Two Types of Energy when ionic compounds dissolve… 1) Ionic lattice broken up forming
gaseous ions, the opposite energy change from lattice energy. 2) Gaseous ions interact with
polar water molecules to form aqueous ions- called enthalpy change of hydration.
 Enthalpy Change of Hydration ∆hydH
θ
is the enthalpy change when one mole of gaseous ions in
water form one mole of aqueous ions. Always exothermic.
 An Example: Same rules as Born- Haber cycle above.
– If 2Br-, multiply enthalpy by 2.
– So hydration = LE + solution. Route 1 = route 2. Rearrange to find unknown.
– Hydration is change of (g) to (aq) one at a time.
– Solution is dissolving (s) to (aq) straight away.
– If enthalpy change exothermic, point arrow downwards.
– Lattice enthalpy is (s) to (g).
– Learn these two types of energy cycles below- usually the only two that can come up.
Although orientation of whole graph might be different.
– Gases always on top.