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- 1. Learning check <ul><ul><li>What are the three standard conditions? </li></ul></ul><ul><ul><li>What is the standard state of bromine? </li></ul></ul><ul><ul><li>What is the standard state of silicon? </li></ul></ul><ul><ul><li>What is the standard state of oxygen? </li></ul></ul><ul><li>The standard enthalpy change of formation ∆H f Θ of a compound is defined as the amount of energy exchanged with the surroundings when 1 mole of the compound is formed from its elements in their standard states under standard condition . </li></ul>
- 2. Write chemical equations for the following reactions – which have ∆H f Θ as their enthalpy change – including all state symbols <ul><ul><li>The formation of sodium chloride </li></ul></ul><ul><ul><li>The formation of magnesium oxide </li></ul></ul><ul><ul><li>The formation of methane </li></ul></ul><ul><ul><li>The formation of sulphur dioxide. </li></ul></ul><ul><ul><li>The formation of carbon dioxide </li></ul></ul>
- 3. Δ H c θ <ul><li>The standard enthalpy change of combustion ∆H c Θ is defined as the amount of energy given to the surroundings when 1 mole of a compound undergoes complete combustion </li></ul>
- 4. <ul><li>Write chemical equations for the following reactions – which have ∆H c Θ as their enthalpy change – including all state symbols. </li></ul><ul><ul><li>The combustion of methane </li></ul></ul><ul><ul><li>The combustion of graphite </li></ul></ul><ul><ul><li>The combustion of ethanol </li></ul></ul><ul><li>Extension: Which chemical equation shows both the enthalpy change of formation of a compound and the enthalpy change of combustion of an element? </li></ul>
- 5. HESS’S LAW <ul><li>According to the German scientist Hess, the total enthalpy change for a chemical reaction is independent of the route by which the reaction takes place. </li></ul><ul><li>Explain why Hess’s Law is just a specific case of the Law of Conservation of Energy. </li></ul><ul><li>Use Hess’s Law to answer the following question </li></ul><ul><li>The standard enthalpy change of formation of Al2O3(s) is −1669 kJ mol−1 and the standard enthalpy change of formation of Fe2O3(s) is −822 kJ mol−1. Use these values to calculate ∆ H ο for the following reaction. </li></ul><ul><li>Fe 2 O 3 (s) + 2Al(s) -> 2Fe(s) + Al 2 O 3 (s) </li></ul><ul><li>State whether the reaction is exothermic or endothermic. </li></ul>
- 6. ENTHALPY CYCLES & HESS'S LAW – using enthalpy changes of combustion to work out enthalpy changes of formation <ul><li>Instead of manipulating the chemical equations and their enthalpy changes, it is possible to put the chemical reactions into a “triangular diagram” or enthalpy cycle. These enthalpy cycles are useful because they enable a value for an enthalpy change to be determined for a reaction which cannot be determined directly from experiment. </li></ul><ul><li>For example: The enthalpy change of formation of methane cannot be determined directly by experiment. It is possible, however, to determine the enthalpy changes of combustion of carbon, hydrogen and methane. The key idea is that the total enthalpy change for one (short) route is the same as the total enthalpy change for an alternative (longer) route. </li></ul>
- 7. Use your Chemistry data booklet to work out the enthalpy change of formation of methane from the enthalpy changes of combustion of hydrogen, carbon and methane .
- 8. ENTHALPY CYCLES & HESS'S LAW -using enthalpy changes of formation to determine enthalpy change of reaction <ul><li>It is not possible to determine the enthalpy change for the reaction between silicon tetrachloride and water that makes silica and hydrogen chloride gas directly by experiment but you can work it out by using the enthalpy changes of formation of the reactants and products. </li></ul>
- 10. 15.2 Born-Haber Cycle (2.5h) <ul><li>15.2.1 Define and apply the terms lattice enthalpy and electron affinity . </li></ul><ul><li>The sign of ∆Hlattice indicates whether the lattice is being formed or broken . </li></ul><ul><li>15.2.2 Explain how the relative sizes and the charges of ions affect the lattice enthalpies of different ionic compounds. </li></ul><ul><li>The relative value of the theoretical lattice enthalpy increases with higher ionic charge and smaller ionic radius due to increased attractive forces. </li></ul><ul><li>15.2.3 Construct a Born–Haber cycle for group 1 and 2 oxides and chlorides and use it to calculate an enthalpy change. </li></ul><ul><li>15.2.4 Discuss the difference between theoretical and experimental lattice enthalpy values of ionic compounds in terms of their covalent character. </li></ul><ul><li>A significant difference between the two values indicates covalent character </li></ul>
- 11. THE BORN-HABER CYCLE This is a special form of a Hess cycle that deals with ionic compounds <ul><li>THE LATTICE ENERGY </li></ul><ul><li>The lattice energy is the standard enthalpy change that occurs when one mole of the solid ionic lattice forms from its gaseous ions. </li></ul><ul><li>After a class discussion, draw a Born-Haber cycle for the formation of sodium chloride and add all of the enthalpy changes associated the cycle. </li></ul>
- 12. <ul><li>Use your data book to look up the values for all of the enthalpy changes and use them to calculate the value of the lattice enthalpy for sodium chloride. </li></ul><ul><li>Choose one of the following compounds and construct a fully labelled Born-Haber cycle for it: </li></ul><ul><ul><li>Potassium oxide </li></ul></ul><ul><ul><li>Magnesium chloride </li></ul></ul><ul><ul><li>Magnesium oxide </li></ul></ul><ul><li>When you come to revise, complete the other Born-Haber cycles and use your Chemistry data booklet or any other source to calculate the lattice energy of each of the three ionic compounds. </li></ul>
- 13. Born-Haber cycle calculations
- 14. Via an Energy level cycle. This is the method always required in examinations The sum of energies for route 1 = sum of energies for route 2
- 15. <ul><li>Although you never see a test tube in this section of the course, the lattice energy is obtained from experimental data via the Born-Haber cycle. </li></ul><ul><li>The value of the lattice energy gives an indication of the size of the forces of attraction holding the ions together in the crystal. </li></ul><ul><li>L.E. for LiF is -1031 kJ/mol +1 cation/-1 anion </li></ul><ul><li>L.E. for BeO is -4443 kJ/mol +2 cation/-2 anion </li></ul><ul><li>From common sense: doubling the charge on an ion is likely to double the force holding it next to an oppositely charged ion. </li></ul><ul><li>doubling charge on both cation and anion will lead to the force being x4. </li></ul><ul><li>Large ions can't get as close to one and other as small ions, hence you would expect ionic solids formed by large ions, e.g. Rb and I ions, to have a smaller L.E. than crystals formed form small ions e.g. Li and F ions. </li></ul>
- 16. The lattice energy can also be calculated using a computer programme and laws of electrostatics (Ref; Born-Meyer equation and the Madelung constant). <ul><li>This theoretical value assumes that the ions are spherical and not polarized. </li></ul><ul><li>The results from experiment and theory can be compared . </li></ul><ul><li>Good agreement: If the Born-Haber experimental value agrees well with the theoretical value, the assumptions made in the computer calculation are valid and the bonding in the crystal can be thought of as being almost 100% ionic . </li></ul><ul><li>Poor agreement: If there is a big difference between the two values, the bonding in the crystal has a significant covalent character and the ions are not spherical . </li></ul>

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