More Enthalpy, more periodic table and the joy of redox potentials

1. More Enthalpy, more periodic table and the joy of redox potentials Hydration Bond enthalpies Entropy Free energy Period 3 Redox potentials Note: this PowerPoint may need the Royal Society of Chemistry font installed. It can be found at the RSC website: http://www.rsc.org/education/teachers/learnnet/RSCfont.htm

2. Remember Enthalpy? • Enthalpy is the amount of energy taken in or given out during any change (physical or chemical) under standard conditions • Symbol “ʅ” (“plimsoll”) • Assume standard conditions • 100kPa • 298K • concentration 1 mole dm-3 • Standard amount – mole

3. ∆Hf Formation • The enthalpy change involved in the production of one mole of a compound from its elements under standard conditions, reactants and products being in their standard states. • E.g. ½ H2 + ½ Cl2 HCl • Note, this is NOT the normal balanced equation.

4. ∆Hi Ionisation • The molar enthalpy change involved in the removal of an electron from a species in the gas phase to form a positive ion and an electron, both also in the gas phase. • Keywords cation and anion • E.g. Na(g) Na+(g) + e-(g) • Can omit (g) for electron

5. ∆Hea Electron affinity • The standard molar enthalpy change when an electron is added to an isolated atom in the gas phase. • E.g. ½ Cl2(g) + e- Cl-(g) • ∆Hea = -364 kJ mol-1 • Strongly electronegative elements attract electrons, they “like” being negative, and so ∆Hea is negative.

6. ∆Hdiss Bond dissociation enthalpy • The standard molar enthalpy change which accompanies the breaking of a covalent bond in a gaseous molecule to form two free radicals, also in the gas phase. • E.g. Cl2(g)  2Cl•(g) ∆H = +242 kJ mol-1 • “homolytic fission” • The dot for the radicals is often omitted in thermodynamic equations, but you have to remember it should be there

7. ∆Hat Enthalpy of atomisation • The standard enthalpy change which accompanies the formation of one mole of gaseous atoms. • For an atomic solid, this is the same as the standard enthalpy of sublimation • E.g. Na(s) Na(g) • This is always endothermic • If the element is already a gas, this should be zero and ignored. (See over)

8. Note that for a diatomic molecule, you get 2 moles of atoms, so ∆Hat = ½∆Hdiss • E.g. in ½Cl2(g) Cl(g) • ∆Hat = +121 kJ mol-1

9. ∆HL (or ∆Hlatt) Lattice enthalpy • The enthalpy of lattice dissociation is the standard enthalpy change which accompanies the separation of one mole of a solid ionic lattice into its gaseous ions. • NaCl(s) Na+(g) + Cl-(g)∆HL = +771 kJ mol-1 • If you write this as above, then it is dissociation and enthalpy is always positive.

10. If, however, you write it the other way round, then this value is the enthalpy of formation of the lattice: • Na+(g) + Cl-(g) NaCl(s)∆HL = -771 kJ mol-1

11. Born-Haber Diagrams….. • Show all of the enthalpy values involved in the creation of a compound, used to find the lattice enthalpy, or anything else you don’t know • It’s basically Hess’s law (remember it?) • (Not copy, unless you’re desperate for reassurance!) • It’s just a bunch of numbers again, but there’s lots of them. It’s a six-variable problem, one unknown and five known. • (occasionally eight-variable, but you would know seven)

12. What’s in it? Positive (metal) and negative (non-metal)gaseous ions • Jobs to do: • Atomise the metalΔHat or ΔHsub • Ionise the metalΔHie1 • atomise the gasΔHat • Add electron to non-metalΔHea • Join ions to make solid compoundΔHL Elements join to make solid compound ΔHf Solid ionic compound

13. A full cycle “Highest enthalpy level” Na+(g) + Cl(g) + e- Electron affinity of non-metal ∆Hea Ionisation of metal ∆Hi Na(g) + Cl(g) Na+(g) + Cl-(g) Atomisation of non-metal ½∆Hdiss(same as ∆Hat) Na(g) + ½Cl2(g) Lattice formation of crystal Atomisation or sublimation of the metal ∆Hsub ∆HL Na(s) + ½Cl2(g) ∆Hf Formation NaCl(s) “Lowest enthalpy level”

14. Note: • Lattice enthalpy determined experimentally is not the same as the theoretical value. • The Lattice enthalpy calculation assumes that the bonding is 100% ionic. • It isn’t. • Don’t forget to ionise G2 metals (Mg etc.) twice, Al three times and oxygen has two electron affinities. • Think about the ion sizes and charges – they affect lattice enthalpy

15. Look at the diagram – notice: • Endothermic (as drawn) – ionisation, atomisation, sublimation • Exothermic (as drawn) – Formation, electron affinity, lattice formation • And all of the relationships follow Hess’s law, any two paths will have the same enthalpy. • As one equation, LHS = RHS: • ΔHsub + ΔHi + ΔHat – ΔHf = – ΔHL – ΔHea • But remember you may have more than one ΔHat, ΔHi, ΔHdiss or ΔHea

16. Trends • The LE is larger for: • Smaller anions like F- • Closer ions • More attraction (e.g. 2+ cation)

17. So, try this: Na(s) + ½Cl2(g) NaCl(s) • Given: • 1st i.e.of Na = +496 • Dissociation enthalpy of Cl2 = +242 • Atomisation enthalpy of Na = +107 • Lattice enthalpy of NaCl = -786 • Electron affinity of Cl = -349 • Find the value for enthalpy of formation for NaCl

18. ∆Hhyd Hydration enthalpy • The enthalpy of hydration is the standard molar enthalpy change for the process: • X±(g) X±(aq)∆H = ∆Hhyd • In which 1 mole of gaseous ions is completely hydrated in water to infinite dilution, under standard conditions • X± represents any cation (X+) or anion (X-) • Usually exothermic

19. ∆Hsol Solution enthalpy • The enthalpy of solution is the standard molar enthalpy change for the process in which 1 mole of an ionic solid dissolves in enough water to ensure that the ions are separated and do not interact • NaCl(s) Na+(aq) + Cl-(aq) ∆Hsol=+2 kJ mol-1 • So, hydration enthalpy = lattice formation enthalpy + solution enthalpy. • Usually endothermic

20. Na+(g) and Cl-(g) ∆Hhyd for Na+ Na+(aq) + Cl-(g) ∆HL for NaCl ∆Hhyd for Cl- Na+(aq) + Cl-(aq) ∆Hsol for NaCl NaCl(s) Typical problems involve.... • You will be expected to find one of these, given the others, as usual. But you’ll have to remember how. • Remember: hydration enthalpy = solution enthalpy + lattice formation enthalpy.

21. Na+(g) and Cl-(g) ∆Hhyd Na+ Na+(aq) + Cl-(g) ∆HL for NaCl ∆Hhyd Cl- Na+(aq) + Cl-(aq) ∆Hsol NaCl NaCl(s) E.g. • Given: • ∆HL for NaCl = +771 kJmol-1 • And ∆Hhyd for Na+ = -405 kJmol-1 • And ∆Hhyd for Cl- = -364 kJmol-1 • Calculate the enthalpy change of solution for NaCl • Draw a cycle: • Calculate ∆Hsol= +771 +(-405) + (-364) • = +2 kJmol-1

22. You do: • Given: • ∆HL for KCl = +701 kJmol-1 • And ∆Hhyd for K+ = -322 kJmol-1 • And ∆Hhyd for Cl- = -364 kJmol-1 • Calculate the enthalpy change of solution for KCl

23. Bonds and reaction energies

24. Bonds • Energy is needed to break a bond between two atoms (endothermic) • Making new bonds releases energy (exothermic) • Total reaction energy = energy of bonds broken – energy of bonds made

25. Or • Etotal = Ereactants – Eproducts • Unit : kilojoules per mole • kJ per mole • kJ mol-1

26. H-H = 436, Br-Br = 193, H-Br = 368 e.g. • H2 + Br2 2HBr • broken: made: • H-H H-Br x 2 • Br-Br Total = (436 + 193) – (368 + 368) = -107  exothermic

27. Steps: • Obtain balanced equation • Draw molecules for reactant and product • Count each bond • Total each bond • Final calculation

28. H-H = 436 O=O = 500 H-O = 464 e.g. • 2H2 + O2 2H2O • broken: made: • H-H (twice) H-O x 4 • O=O Total = (436 *2 + 500) – (464 x 4) = 872 + 500 – 1856 = - 484  exothermic ( For one water it is -242 kJ mol-1)

29. H-H = 436 I-I = 151 H-I = 297 Try this: • H2 + I2 2HI • Think – what bonds are in the reactant molecules, what bonds are in the product molecules • Broken: Made: • H-H, I-I H-I x 2 • Total = (436 + 151) – (297 *2) • = -7 kJ mol-1 for the reaction as shown • (-3.5 kJ mol-1 for each mole of HI)

30. C-H = 414 O=O = 500 H-O = 464 C=O = 805 Try this: • CH4 + 2O2 CO2 +2H2O • Think – what bonds are broken,what bonds are made. • Broken: Made: • C-H x4, O=O x2 C=O x2, H-O x 4 • Total = (1656 +1000) – (1610 + 1856) • = 2656 – 3466 • = - 810 kJ mol-1

31. C-H = 414 O=O = 500 H-O = 464 C=O = 805 C=C = 613 Try this: • C2H4 + 3O2 2CO2 +2H2O • Think – what bonds are broken,what bonds are made. • Broken: Made: • C-H x4, O=O x3, C=C C=O x4, H-O x 4 • Total = (1656 +1500 + 613) – (3220 + 1856) • = 3769 – 5076 • = - 1307 kJ mol-1

32. Exothermic Endothermic atoms atoms Products Reactants ΔHreact ΔHreact Reactants Products

33. A new quantity - Entropy • Degree of “disorder” in a system. • Zero entropy at 0K • Absolute scale – not just increments like ∆H • Depends on temperature and solution. • Gases most, then liquids, then solids • Simple substances – low entropy values • Not zero for elements – depends on temperature • Different entropy values for same chemical • e.g. CO2 as solid or gas, • Water as solid, liquid or gas

34. Discuss these entropy values: Consider: which is the most/least organised; how big each molecule is; what type of intermolecular forces there are; how strong they are, etc.

35. Continued.... • Symbol S – units J K-1 mol-1 • Yes, Joules, not kJ • So units will be a problem (and catch you out!) • Sɵ - usualconditions – can be looked-up • E.g. CO2(g), S=214 • For NaCl(s), S=72 • For C(diamond), S=2.4 • ∆S = Sɵ (products) - Sɵ (reactants) • Note direction of subtraction

36. A sample question from Yahoo!answers • State whether ΔS is positive, negative, or zero for each of the following processes:(a) 46 g of liquid water goes from 60°C to 50°C at 1 atm pressure.(b) 7 g of N2(g) goes from 50°C and 2.0 atm pressure to 50°C and 1.0 atm pressure.Please give an explanation and the general relationship. Thank you.

37. A simple question • Each answer may be used once, more than once, or not at all. • a) CO(g) + H2O(g) = CO2(g) + H2(g) • b) KCl(s) = K+(aq) + Cl-(aq) • c) Na2CO3(s) = Na2O(s) + CO2(g) • d) N2(g) + 3H2(g) = 2NH3(g) • e) H2O(s) = H2O(l) • In which reaction would the entropy change be closest to zero? • In which reaction(s) would the entropy change have a positive value? • Which would be most positive? • In which reaction(s) would the entropy change have a negative value?

38. Calculate: • The entropy change in: • CaCO3 CaO + CO2 • Given: Sʅ (CaCO3) = 92.9 J K-1 mol-1 • Sʅ (CaO) = 39.7 JK-1 mol-1 • Sʅ (CO2) = 213.6 JK-1 mol-1 • ∆Sʅsystem = 39.7+213.6-92.9 • = 160.4 JK-1 mol-1

39. Calculate: • The entropy change in: • 2NaHCO3(s) Na2CO3(s) + CO2(g) + H2O(l) • Given: Sʅ (NaHCO3(s)) = 101.7 J K-1 mol-1 • Sʅ (Na2CO3(s)) = 135.0 JK-1 mol-1 • Sʅ (CO2(g)) = 213.6 JK-1 mol-1 • Sʅ (H2O(l)) = 69.9 JK-1 mol-1 . • ∆Sʅsystem = 135.0+213.6+69.9 – 203.4 • = 215.1 JK-1 mol-1 .

40. Calculate: • The entropy change in: • N2(g) + 3H2(g) 2NH3(g) • Given: Sʅ (N2(g)) = 191.6 J K-1 mol-1 • Sʅ (H2(g)) = 130.6 J K-1 mol-1 • Sʅ (NH3(g)) = 192.3 J K-1 mol-1 • ∆Sʅsystem = 384.6 – 583.4 • = -198.8 JK-1 mol-1 • This system becomes less disorganised.

41. Why do reactions happen? • E.g. • HCl + NaOH Exothermic neutralisation • 4Fe + 3O2 Exothermic rusting • 2Mg + O2 Exothermic combustion • CaCO3 CaO + CO2 Endothermic • but • NaCl + H2O  Na+(aq) + Cl-(aq) Endothermic • 2HCl + NaHCO3 Endothermic So why do these two happen spontaneously?

42. So how can we know which reactions happen? • A “balance” – more entropy could “beat” a positive enthalpy value • New quantity – Gibbs free energy change: • ∆G = ∆H - T∆S • A negative Gibbs value means the reaction should be spontaneous (feasible) • It has nothing to do with kinetics, a feasible reaction still might not noticeably happen, high Ea being a likely obstacle.

43. Examples: • C(s) + O2(g) CO2(g) • ΔH = -394 kJ mol-1 • ΔS = +3.3 J mol-1 • What is the free energy change? • ΔG = ΔH - TΔS • = -394 x 103 – 298 x 3.3 J mol-1 • = -395 kJ mol-1 • Negligible TΔS term in spite of 298 Could do in kJ instead

44. Examples: • 2Fe(s) + 1½O2(g) Fe2O3(s) • ΔH = -825 kJ mol-1 • ΔS = -272 J mol-1 • What is the free energy change? • ΔG = ΔH - TΔS • = -825 x 103 – 298 x -272 J mol-1 • = -744 kJ mol-1 • Very feasible. Negative entropy term as gas goes to solid. Large enthalpy term

45. Examples: • 2NaHCO3(s) Na2CO3(s) + CO2(g) + H2O(g) • ΔH = +130 kJ mol-1 • ΔS = +335 J mol-1 • What is the free energy change for 1 mole of NaHCO3? • ΔG = ΔH - TΔS • = 130 x 103 – 298 x 335 J mol-1 • = +30 kJ mol-1 for 2 moles, +15 kJ mol-1 Could do in kJ instead

46. Using the same information, prove that this reaction is feasible above around 400K • Hint: • using ΔG = ΔH – TΔS you need a point where you change from not feasible to feasible, so ΔG = 0

47. Melting (fusion) • At equilibrium, e.g. ice at 0°C, there should be no change in G, or ΔG =0 • Note le Chatelier’s principle here – more heat energy would make more water from ice, so temperature does not change. • This gives ΔG = ΔH - TΔS = 0 • So ΔH = TΔS or ΔS(fus) = ΔH(fus) / T(fus) • Same for boiling

48. Boiling anything • Same as for melting, ΔG = ΔH - TΔS = 0 • So ΔH = TΔS or ΔS(vap) = ΔH(vap) / T(vap) • Given that the enthalpy change of fusion of ice is 6.0 kJ mol-1, calculate the entropy change involved and explain why it is small. • 6 = 273 ΔS • Therefore ΔS = 6/273 = 0.22

49. Try this: • Hydrogen gas is a non-polluting fuel. Hydrogen gas may be prepared by electrolysis of water. 2H2O(l) 2H2(g) + O2(g) • Predict the signs of ΔH, ΔG and ΔS for the production of hydrogen gas by electrolysis of water.

50. Periodicity • Chemical reactions of period 3 elements • (Because they ring all of the bells) • Na, Mg, Al, Si, P, S, Cl, Ar • Some obvious facts: • Ar – no reactions • Na, Mg, Al metals (Al is a metalloid really)