**Hess' Law**

At

At A2 Unit 5 we use

**we used thermodynamic cycles whilst using Hess' law on enthalpy change.**__AS Unit 2__At A2 Unit 5 we use

**Born-Haber cycles**.The above image shows the Born-Haber cycle for the reaction

**Na(s) + 1/2Cl2(g) → NaCl(s)**This shows that the enthalpy of formation (blue arrow) for this reaction is equivalent to:

- The
**atomisation enthalpy**for**Na**and(Beware about diatomic elements). This is the enthalpy change to get the elements into gaseous state__Cl__ - The
**first ionisation enthalpy**for**Na**(In group 2 elements you will need the second ionisation enthalpy) - The
**electron affinity enthalpy**for**Cl** - The
**Lattice association enthalpy**for NaCl from its gaseous ions

When working with the born-haber cycle you can work out questions about working out an enthalpy in the reaction there are different ways you can solve this sort of question.

Rearrange for the enthalpy you wish to find

The Modulus of a number is its

So the modulus of -5 is 5.

This can be applied when using Born-Haber cycles.

Both sides of the cycle should have the same modulus value.

**Mathematical equation:****form(NaCl) =***ΔH***at(Na) +***ΔH***at(1/2Cl2) +***ΔH***1st ion(Na) +***ΔH***ea(Cl) +***ΔH***La(NaCl)***ΔH**(or***-***ΔHLattice dissociation(NaCl)**)*Rearrange for the enthalpy you wish to find

**Modulus Method:**The Modulus of a number is its

**magnitude or value**. This**ignores any sign**put in front of itSo the modulus of -5 is 5.

This can be applied when using Born-Haber cycles.

Both sides of the cycle should have the same modulus value.

Using this method we can find out the lattice association (or lattice formation) enthalpy

Below is what happens with a Group 2 element

**Enthalpy Definitions**

*Lattice association/formation enthalpy -*The enthalpy change when

**one mole**of

**crystalline/solid ionic substance**is formed from its

**constituent ions in their gaseous state**under standard conditions

*Lattice dissociation enthalpy -*The enthalpy change when

**one mole**of

**crystalline/solid ionic substance**breaks up into its

**constituent ions in their gaseous state**under standard conditions

*First Ionisation Energy -*The enthalpy change when

**one mole**of

**gaseous atoms loses one electron**per atom under standard conditions

*Second Ionisation Energy -*The enthalpy change when

**one mole**of

**gaseous +1 ions lose one electron**per atom under standard conditions

*First Electron Affinity Enthalpy -*The Enthalpy change when

**one mole**of

**gaseous atoms gains one electron**per atom under standard conditions

*Second Electron Affinity Enthalpy -*The Enthalpy change when

**one mole**of

**gaseous -1 ions gain an electron**per atom under standard conditions

*Hydration Enthalpy -*The Enthalpy change when

**one mole**of

**gaseous ions become hydrated or aqueous**

*Enthalpy of Solution -*The Enthalpy change when

**one mole**of an

**ionic solid dissolves**into enough water to

**completely separate all the ions**

*Enthalpy of Formation -*The Enthalpy change when

**one mole**of a substance is

**formed from its constituent elements**in their standard states under standard conditions

*Enthalpy of Combustion -*The Enthalpy change when

**one mole**of a substance is

**completely combusted in oxygen**in their standard states under standard conditions

*Enthalpy of Atomisation -*The Enthalpy change when

**one mole**of

**gaseous atoms**are produced from

**one mole**of

**standard state elements**

*Enthalpy of Bond Dissociation -*Enthalpy change when

**one mole**of a

**covalent bond**is

**broken in the gaseous state**

**Entropy**

**Entropy is the measure of disorder.**

Gas is more disordered than a liquid which is more disordered than a solid

If there is an increases in particles then there is an increase in entropy

It is measured in

**JK^-1mol^-1**and the symbol

**S**

For a reaction to occur the total entropy must be positive

To work this out the entropy change of the surroundings must also be deduced as well as the entropy of the reaction.

The Entropy change of a reaction is worked out in the same way as the enthalpy change of a reaction

**Gibbs Free-energy**

For a spontaneous reaction to be feasible then Gibbs free-energy must be zero or negative

Gibbs free-energy is worked out using the below equation but

Gibbs free-energy is worked out using the below equation but

**beware of the units**If you are asked to work at what temperature a reaction becomes feasible remember to make Gibbs free-energy equal zero. From there you can rearrange the equation to work out whatever you want.