1. Enthalpy Change of formation is the enthalpy change when one mole of a compound is formed from its constituent elements under standard conditions. Enthalpy Change of reaction is the enthalpy change that accompanies a chemical reaction in molar quantities expressed in a chemical equation. Hess’ Law: if a reaction can take place by more than one route and the initial and final conditions are the same, the total enthalpy change is independent of the route taken. In an exothermic reaction, ΔH is negative, because heat is given out to the surroundings so the reacting chemicals lose energy. In an endothermic reaction, ΔH is positive, because heat is taken in from the surroundings so the reacting chemicals gain energy. ΔH θ reaction = ΣΔH θ f (products) - ΣΔH θ f (reactants) Use the following data to calculate the enthalpy change for the reaction between hydrogen sulphide and sulphur dioxide: 2H 2 S + SO 2  3S + 2H 2 O CompoundΣΔH θ f H2SH2S-20.6 SO 2 -296.9 S0 H2OH2O-285.9 ΔH θ = products – reactants = [(3x0) + (2x-285.9)] – [(2x-20.6) + (-296.9)] = -233.7 kJmol -1

2. Lattice Enthalpy Lattice enthalpy is the enthalpy change that accompanies the formation of one mole of an ionic compound from its gaseous ions under standard conditions.  Lattice enthalpy is always an exothermic change: energy is always given out when ionic bonds are being formed from gaseous ions.  Lattice enthalpy indicates the strength of an ionic lattice and is a measure of the ionic bond strength. A large negative value shows that there are strong electrostatic forces of attraction between the oppositely charged ions in the lattice. Lattice Enthalpy cannot be determined experimentally as it is impossible to form one mole of an ionic compound from gaseous ions. If we apply Hess’ law, it is possible to determine the lattice enthalpy. Standard enthalpy change of atomisation: when one mole of gaseous atoms is formed from its element in its standard state. First ionisation energy: enthalpy change when 1 electron is removed from each atom in 1 mole of gaseous atoms to form 1 mole of gaseous 1+ ions. Second ionisation energy: enthalpy change when one electron is removed from each ion in one mole of gaseous 1+ ions to form 1 mole of gaseous 2+ ions. First electron affinity: the enthalpy change when one electron is added to each atom in one mole of gaseous atoms to form 1 mole of gaseous 1- ions. Second electron affinity: the enthalpy change when one electron is added to each ion in 1 mole of gaseous 1- ions to form 1 mole of gaseous 2- ions. The strength of the electrostatic forces depend on the ionic radius and ionic charge. The smaller the ionic radius, the higher the charge density, the stronger the electrostatic forces of attraction and the more exothermic the lattice enthalpy. The greater the ionic charge, the higher the charge density, the stronger the electrostatic forces of attraction and the more exothermic the lattice enthalpy.

3. Born Haber Cycles NaCl (s) Na (s) + ½ Cl 2 (g) Na (g) + ½ Cl 2 (g) Na (g) + Cl (g) Na + (g) + Cl(g) Na + (g) + Cl - (g) ΔH θ formation -411 ΔH θ atomisation Na +107 ΔH θ atomisation Cl +122 ΔH θ ionisation sodium +496 ΔH θ electron affinity Cl -349 ΔH θ Lattice Enthalpy Second ionisation energies are more endothermic than first ionisation energies:  Removing an electron  Less repulsion between outer electrons in Mg+  Outer electron is closer to the nucleus in Mg+  The outer electron is more strongly attracted to the nucleus in Mg+ Second electron affinities are endothermic: Adding an electron There is repulsion between the negative ion and the electron Energy is required to overcome the repulsion. Lattice Enthalpy = -(-349) – (+496) – (+122) – (+107) + (-411) = -787 kJmol -1 Use the Born Haber cycle below to calculate lattice enthalpy of sodium chloride.

4. Enthalpy Change of Hydration The standard enthalpy change of hydration is the enthalpy change when one mole of isolated gaseous ions is dissolved in water forming one mole of aqueous ions under standard conditions. When gaseous sodium ions are dissolved in water, the ions form electrostatic forces of attraction with the polar water molecules. The value of the hydration enthalpy depends on the strength of the electrostatic force of attraction. Two factors influence this: ionic radius and ionic charge. Ionic radius Smaller ionic radius (ionic radius decreases across a period) Higher charge density Stronger electrostatic forces of attraction between ions and polar water molecules More exothermic enthalpy change of hydration Ionic charge Greater ionic charge Higher charge density Stronger electrostatic forces of attraction between ions and polar water molecules More exothermic enthalpy change of hydration

5. Enthalpy Change of Solution The enthalpy change when one mole of a compound is dissolved in water under standard conditions. When an ionic compound dissolves in water, the electrostatic forces of attraction between the positive and negative ions in the giant lattice are broken and replaced by new forces of attraction between the ions and polar water molecules. When 4.24g of lithium chloride, LiCl, was dissolved in 50cm3 of water, the temperature of the water increased from 21 ° to 34.5°. Calculate the enthalpy change of solution of lithium chloride. 1.Calculate moles of compound 2.Calculate the energy change using mcΔt 3.Calculate ΔH for one mole of compound.

6. Using Born-Haber Cycles to calculate enthalpy change of solution ΔH lattice (L) = -733 ΔH hydration sodium = -406 ΔH hydration bromine = -335 Na + (g) + Br - (g) NaBr (s) Na + (aq) + Br - (g) Na + (aq) + Br - (aq) Lattice Enthalpy Hydration of sodium Hydration of bromine Enthalpy of Solution? Use the cycle below to calculate the enthalpy change of solution of sodium bromide. -(-733) + (-406) + (-335) = -8 kJmol -1

7. Enthalpy Change of Neutralisation The standard enthalpy change of neutralisation is the enthalpy change when an aqueous acid is neutralised by an aqueous base to form one mole of water under standard conditions. H + + OH -  H 2 O Enthalpy changes of neutralisation are exothermic because strong bonds are formed between H + and OH- ions. 1.Calculate energy change using mcΔt 2.Calculate moles of both acid and base 3.Write an equation to form one mole of water 4.Calculate the enthalpy change for one mole of water formed. 25cm 3 of dilute nitric acid, concentration 2moldm -3 is added to 25cm 3 of aqueous potassium hydroxide of concentration 2 moldm -3. The temperature increases from 22° to 35°. Calculate the enthalpy change of neutralisation. [take specific heat capacity to be 4.18Jg -1 K -1 ]