OCR Advanced GCE in Chemistry/Lattice enthalpy

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Lattice enthalpy is the enthalpy change when one molecule of an ionic crystal is formed from its constituent ions in their gaseous state at standard conditions (101kPa pressure and 298K temperature).

Lattice enthalpy[edit]

Lattice enthalpy is simply the change in Enthalpy associated with the formation of one mole of an ionic compound from it's oppositely charged ions in their standard states under standard conditions.

e.g. Mg2+(g) + 2Cl-(g) ---> MgCl2(s)ΔHlatt = -2526 kj mol-1

Lattice-enthalpy-NaCl-3D-ionic.png

Therefore:

  • The enthalpy change will always be exothermic (negative)
  • A more negative value shows greater electostatic attraction and therefore a stronger bond in the solid

Definition: The enthalpy change when 1 mole of an ionic compound is formed from its gaseous ions under standard conditions

Lattice enthalpy cannot be measured because gaseous ions do not combine directly to form a compound.

Born-Haber cycles[edit]

As lattice enthalpy cannot be measured, another way of working out lattice enthalpy values is needed. To do this we must show the complete pathway for the reaction, including lattice enthalpy (Route 2 below) and compare this to the enthalpy change of formation (Route 1). We can do this because Hess's law states:

enthalpy change of formation = atomisation enthalpies + electron affinity + lattice enthalpy

BornHaberLiF.PNG

The lattice enthalpy is step 5 in route 2 of the example. First the elements must be atomised and ionised in steps 1-3:

Step 1 - Atomisation of solid Lithium to the gaseous state

Definition: ΔHatThe standard enthalpy change of atomisation of an element is the enthalpy change when one mole of gaseous atoms are formed from the element in its standard state.

Step 2 - Ionisation of gaseous Lithium atoms to 1+ ions

Definition: ΔHi.e.The standard enthalpy change accompanying the removal of one electron from an atom in the gas phase.

Step 3 - Atomisation of fluorine gas (Hint: fluorine still needs to be atomised here even though it is gaseous because it is an Fl2 molecule)

Step 4 - Electron affinity of fluorine. This is basically making the gaseous fluorine atoms into ions by adding an electron.

Definition: ΔHea1The first electron affinity is the enthalpy change when one electron is added to each gaseous atom in one mole, to form one mole of 1- ions.

Step 5 - Lattice enthalpy! can now be worked out. The easiest way to understand how the cycle works is to remember that route 1 and route 2 are equal. So using Hess's law (as above) lattice enthalpy can be calculated when the other values are known.

Things to watch out for:

  • Ions requiring two additions or subtractions of electrons e.g. Mg2+

For these use the values for each, DO NOT simply double the first value.

  • Ionic solids that have more than one of a single ion e.g. MgCl2

Also try and learn the definitions because these are easy marks.

Trends in Lattice enthalpy[edit]

There are two factors that affect the lattice enthalpy of an ionic compound:

  • The size of the ions
  • The charge of the ions

This is because these are also the factors that affect charge density. Charge density of an ion will determine it's "attracting power" and therefore the greater the charge density of the ions, the greater the electrostatic forces between them. So a larger ion will decrease the charge density and a more higly charged ion will increase the charge density.

Thermal decomposition of group 2 carbonates[edit]

The Group 2 carbonates MgCO3, CaCO3, SrCO3 and BaCO3 all react in the same way when heated:

A Carbonate ion

MgCO3(s) ---> MgO(s) + CO2(g)

CaCO3(s) ---> CaO(s) + CO2(g)

SrCO3(s) ---> SrO(s) + CO2(g)

BaCO3(s) ---> BaO(s) + CO2(g)


The decomposition temperatures for these reactions are as follows:

MgCO3 = 350 °C

CaCO3 = 832 °C

SrCO3 = 1340 °C

BaCO3 = 1450 °C

Polarisation of carbonate ion

A trend of increasing temperature down the group is evident. This may seem confusing because the magnesium cation is the smallest and they all have the same charge. Magnesium therefore has the greatest lattice enthalpy and the greatest attraction to the carbonate ion. However, this does not result in a higher decomposition temperature - the stronger pull of a smaller cation combined with the diffuse electron cloud of a carbonate ion causes much more polarisation of the carbonate ion to occur.

The change of shape is where the electrons are pulled onto one of the Oxygen atoms causing distortion in the carbonate ion. This change in shape reduces strength and causes the carbonate to break up when heated.

Magnesium Oxide is a particularly favourable product due to its high lattice enthalpy, this contributes to the ease of decomposition of magnesium carbonate and is why magnesium oxide is used as a refractory lining.

Questions[edit]

1.Explain why the lattice enthalpy of NaBr is much less exothermic than MgCl2

2.How is Hess's law used to work out lattice enthalpy from born-haber cycle?

3.What are the factors affecting lattice enthalpy?

4.What is the definition of 1st electron affinity?

5.What is the trend in decomposition temperatures of group two carbonates? why is this?

Answers