1. 12 Thermodynamics 12.1 Types of Enthalpy Change 12.2 Born-Haber Cycles 12.3 Enthalpy Changes – Enthalpy of Solution 12.4 Mean Bond Enthalpy 12.5 Entropy

2. 12.1 Enthalpy Change – Ionic Compounds Learning Objectives: 1.Describe what is meant by the term enthalpy change. 2.Describe the different types of enthalpy changes (formation, atomisation, ionisation energy, electron affinity, lattice formation, hydration, solution, bond enthalpy). 3.Calculate the enthalpy changes on forming ionic compounds.

3. Enthalpy Review Enthalpy change is the heat change at constant pressure. Standard conditions: 100kPa, 298 K (starting temperature) Remember that heat and temperature are not the same. Heat is a type of energy and is measured in joules and heat changes lead to temperature changes, which is measure in Kelvins.

4. Types of Enthalpy Changes Enthalpy of Formation Enthalpy of Atomisation First Ionisation Energy/Second Ionisation Energy First Electron Affinity/Second Electron Affinity Lattice Enthalpy of Formation Enthalpy of Lattice Dissociation Enthalpy of Hydration Enthalpy of Solution Mean Bond Enthalpy Write down the symbol and the definition

5. Standard Enthalpy of Formation ∆H f Enthalpy change when one mole of a compound is formed from its constituent elements under standard conditions all reactants and products in their standard states. change standard conditions formation

6. Standard Enthalpy of Atomisation ∆H at Enthalpy change when one mole of gaseous atoms is formed from the element In it’s standard state under standard conditions

7. First Ionisation Energy First IE Enthalpy change when one mole of gaseous atoms is converted into one mole of gaseous +1 ions under standard conditions

8. Second Ionisation Energy Second IE Enthalpy change when one mole of gaseous +1 ions is converted into one mole of gaseous +2 ions under standard conditions

9. First Electron Affinity First ∆H ea Enthalpy change when one mole of gaseous atoms is converted into one mole of gaseous -1 ions under standard conditions

10. Second Electron Affinity Second ∆H ea Enthalpy change when one mole of gaseous -1 ions is converted into one mole of gaseous -2 ions under standard conditions

11. Lattice Formation Enthalpy ∆H L Enthalpy change when one mole of solid ionic compound is formed from it’s gaseous ions under standard conditions (always negative, energy released)

12. Enthalpy of Lattice Dissociation -∆H L Enthalpy change when one mole of solid ionic compound dissociates into it’s gaseous ions under standard conditions (always positive, energy is absorbed)

13. Standard Enthalpy of Hydration ∆H hyd Enthalpy change when one mole of gaseous atoms is surrounded by water molecules under standard conditions

14. Standard Enthalpy of Solution ∆H sol Enthalpy change when one mole of solute completely dissolves in sufficient solvent to form a solution in which the molecules are ions do not interact under standard conditions

15. Mean Bond Enthalpy ∆H diss Enthalpy change when one mole of gaseous molecules breaks a covalent bond forming two free radicals averaged over a range of compounds at standard conditions

16. For each type… a)Write an equation to represent the chemical reaction being described b)Tell me if the process is likely to be positive or negative c)Explain why.

17. Standard Enthalpy of Formation

18. Standard Enthalpy of Atomisation

19. First Ionisation Energy First IE Na (g)  Na + (g) + e - Positive Removing an electron takes energy

20. Second Ionisation Energy Second IE Na + (g)  Na 2+ (g) + e - Very positive Removing electron from positive ion require a lot of energy.

21. First Electron Affinity First ∆H ea O (g) + e -  O - (g) Usually Negative Energy is gained when electrons are added.

22. Second Electron Affinity Second ∆H ea O - (g) + e -  O 2- (g) Usually Positive Because of repulsion, adding the second electron requires more energy than is gained.

23. Lattice Formation Enthalpy ∆H L Na + (g) + Cl - (g)  NaCl (s) Always negative Bond making releases energy, more stable in lattice form.

24. Enthalpy of Lattice Dissociation -∆H L NaCl (s)  Na + (g) + Cl - (g) Always positive This is opposite of lattice formation, breaking bonds requires energy.

25. Standard Enthalpy of Hydration ∆H hyd Na + (g) + aq  Na + (aq) Cl - (g) + aq  Cl - (aq) Usually negative Water molecules stabilise the charges of the ions.

26. Standard Enthalpy of Solution ∆H sol NaCl (s) + aq  Na + (aq) + Cl - (aq) Usually slightly positive Breaking the bonds of the lattice requires energy, however, the water molecules stabilise the ions so overall only small amount of energy absorbed.

27. Mean Bond Enthalpy ∆H diss CH 4 (g)  C (g) + 4H (g) Always positive Bond breaking requires energy.

28. 12.2 Born-Haber Cycles Learning Objectives: 1.Describe Hess’ Law. 2.Use Born-Haber Cycles to calculate enthalpy changes

29. Hess’s Law of Thermodynamics The enthalpy change for a reaction is the same, no matter what route is taken. For example: CH 4 (g) + O 2 (g)  CO 2 (g) + H 2 O (g) C (s) + H 2 (g) + O 2 (g)

30. Born-Haber Cycles Born-Haber Cycles are just another method to solve for the unknown enthalpy change of a chemical reaction by using enthalpy changes that we DO know. It uses a diagram to represent the enthalpy changes on a vertical scale. Increases in energy are UP ( ) arrows, decreases in energy are DOWN ( )arrows.

31. Molly started out with £0. Then she received £100 for her birthday. She went out to dinner, this costed £30. Then she bought some new shoes. At the end of the day to had spent all of her birthday money. How much did her new shoes cost? With Birthday Money After Dinner Broke ∆M bd = +£100 ∆M din = -£30 ∆M shu = ? = -£70

32. Formation of an Ionic Compound Electrons are transferred to atoms to form ions. Ions then attract and are arranged into an ionic lattice. This is how ionic lattices are formed.

33. Enthalpy Change in Formation of Ionic Compounds

36. Born-Haber Cycle Formation of NaCl Na (g) + Cl (g) Na + (g) + Cl (g) Na + (g) + Cl - (g) NaCl (s) ∆H at = +108 kJ/mol ∆H at = +122 kJ/mol First IE = +496 kJ/mol First EA = -349 kJ/mol ∆H L = -788 kJ/mol ∆H f = ?

37. Born-Haber Cycle Formation of NaCl Na (g) + Cl (g) Na + (g) + Cl (g) Na + (g) + Cl - (g) NaCl (s) ∆H at = +108 kJ/mol ∆H at = +122 kJ/mol First IE = +496 kJ/mol First EA = -349 kJ/mol ∆H L = -788 kJ/mol ∆H f = -411 kJ/mol

38. Example: Lattice Formation Enthalpy of MgCl 2 Write out the overall equation for the formation of magnesium chloride. Write equations for all of the steps in the formation of magnesium chloride. HINT: there are six steps

39. Draw a Born-Haber Diagram for MgCl 2 HINT: there is a “trick” step, can you catch it? Remember your definitions ∆H at Mg = +148 kJ/mol ∆H at Cl = +122 kJ/mol First IE Mg= +738 kJ/mol Second IE Mg = +1451 kJ/mol First EA Cl = -349 kJ/mol ∆H f MgCl 2 = -641 kJ/mol Use your Born-Haber Diagram to Calculate the Lattice Formation Enthalpy

40. Mg (s) + Cl 2 (g) Mg (g) + Cl 2 (g) Mg (g) + 2Cl (g) Mg 2+ (g) + 2Cl (g) Mg 2+ (g) + 2Cl - (g) MgCl 2 (s) ∆H at = +148 kJ/mol 2 x ∆H at = +122 kJ/mol x 2 = +244 kJ/mol First IE = +738 kJ/mol 2 x First EA = -349 kJ/mol x 2 = -698 kJ/mol ∆H L = -2524 kJ/mol ∆H f = ? Second IE = +1451 kJ/mol Mg + (g) + 2Cl (g)

41. Mg (s) + Cl 2 (g) Mg (g) + Cl 2 (g) Mg (g) + 2Cl (g) Mg 2+ (g) + 2Cl (g) Mg 2+ (g) + 2Cl - (g) MgCl 2 (s) ∆H at = +148 kJ/mol 2 x ∆H at = +122 kJ/mol x 2 = +244 kJ/mol First IE = +738 kJ/mol 2 x First EA = -349 kJ/mol x 2 = -698 kJ/mol ∆H L = -2524 kJ/mol ∆H f = -641 kJ/mol Second IE = +1451 kJ/mol Mg + (g) + 2Cl (g)

42. 12.3 More Enthalpy Changes Learning Objectives: 1.Calculate enthalpy change of solution. 2.Describe how lattice enthalpy calculations support models for ionic bonding. 3.Explain how ions can become polarised.

43. Enthalpy of Solution Ionic solids can dissolve in polar solvents. This is called hydration if the solvent is water. Hydration is when the water molecules surround ions. What are the steps for process of forming a solution? 1.Breaking the ionic lattice (enthalpy of lattice dissociation). 2.Hydrating the positive ions (enthalpy of hydration). 3.Hydrating the negative ions (enthalpy of hydration).

44. Example: NaCl

45. Ionic Bonding Models For most ionic compounds the theoretical values calculated from Born-Haber cycles agrees with experimental values. This proves that the model for ionic bonding (lattice) is correct. However, some ionic compounds have theoretical and experimental values that DO NOT agree. Another model needed to be found to explain these discrepancies.

46. Polarisation ZnSe experimental lattice formation enthalpy = -3611 kJ/mol theoretical lattice formation enthalpy = -3305 kJ/mol WHY? Zn 2+ is very small and has a high + charge Se 2- is very large and has a high – charge Zn 2+ moves closely to electron density of Se 2- and attracts the e - Since Se 2- is large, the e - are far from the nucleus and easily pulled away This distorts the electron cloud surrounding Se 2-

47. Polarisation The distortion causes their to be some electron density shared between the two ions (slightly covalent nature). The Se 2- ion is said to be polarised. This causes the enthalpy change to be greater than expected.

48. When does polarisation happen? Cation = small size, high charge Anion = large size, high charge

49. 12.4 Mean Bond Enthalpy Learning Objectives: 1.Explain the term mean bond enthalpy. 2.Calculate enthalpy changes using mean bond enthalpy. 3.Explain why this method is not as accurate.

50. Mean Bond Enthalpy The average bond enthalpy term is the average amount of energy needed to break a specific covalent bond, measured over a wide variety of different molecules. A measure of strength of a covalent bond. In comparison, lattice enthalpy is a measure of the strength of an ionic bond.

51. Predicting reactions Mean bond enthalpies can be used to predict how molecules may react. We can predict which bonds may be more likely to break. Which bond is most likely to break? C-H 413 kJ/mol C-C 437 kJ/mol C-Br 290 kJ/mol

52. Predicting reactions Mean bond enthalpies can also be used to compare reactivities of different molecules. Which haloalkane is more reactive? C-F467 kJ/mol C-Cl346 kJ/mol C-Br290 kJ/mol C-I228 kJ/mol

53. Calculating Approximate Enthalpy Changes Hess’s Law can be applied. One possible route to products would be to break all bonds in the reactants and then form all of the bonds for the products. The enthalpies for these two processes can then be summed up to find the total enthalpy change. Remember: bond breaking requires energy (+ value) bond formation releases energy (- value)

56. 12.5 Why do chemical reactions take place? Learning Objectives: 1.Explain the concept of entropy. 2.Calculate using enthalpy and entropy whether a reaction will spontaneously occur. 3.Analyse the effects of temperature on feasibility of a reaction.

57. Is a reaction feasible or spontaneous? Reactions that will take place on their own are called spontaneous. If it is possible for a reaction to take place on their own, the reaction is feasible. What determines if a reaction is feasible? If ΔH (enthalpy) is negative, the reaction is exothermic If ΔS (entropy) is positive, the reaction increases in randomness

58. Entropy Entropy is a mathematical measure of the randomness of a system. Change in entropy is represented as ΔS. The universe prefers randomness (higher entropy) and is always moving towards disorder. Values for entropy of different substances are determined mathematically, you will not be expected to calculate these, only how to use them. (see pg. 179)

59. Calculating Entropy Changes Calculate the difference in entropy from reactants to products to find the ΔS of a reaction. ΔS = S products – S reactants If ΔS is positive, entropy is increasing, disorder is increasing. The products are more disordered than the reactants. If ΔS is negative, entropy is decreasing, disorder is decreasing. The products are less disordered than the reactants.

60. Gibbs Free Energy ΔG represents the Gibbs free energy and combines both enthalpy and entropy. It is used to determine whether or not a reaction is feasible. ΔG = ΔH – TΔS If ΔG is negative (-) the reaction is feasible. If ΔG is positive (+) the reaction is NOT feasible.

61. What happens if ΔG = 0? There will be a temperature where ΔG = 0. This is the temperature at which the reaction is just feasible. In a closed system an equilibrium between products and reactants occur. ΔG = 0 can also be used to calculate ΔS. Cases where both forms are equally likely (ie melting point), ΔG = 0.

62. Thermodynamics does not predict the rate of a reaction Thermodynamics = Kinetics Thermodynamics only predicts whether a reaction is feasible. It DOES NOT predict how quickly the reaction may take place. Kinetics is the branch of chemistry dealing with rate of reaction.