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

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

2. Remember Enthalpy? Enthalpy is the amount of energy taken in or given out during any change (physical or chemical) under standard conditions Symbol �?� (�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.

4. ?Hi Ionisation The 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. Cl(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 = �?Hdiss E.g. in �Cl2(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?

13. A full cycle Na+(g) + Cl(g) + e-

14. Note: The Lattice enthalpy calculation assumes that the bonding is 100% ionic. It isn�t. For real lattices, the theoretical values are different from the real values. 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: ?Hat+?Hdiss+?Hi-?Hf=-?HL-?Hea But remember you may have more than one ?Hat, ?Hi, ?Hdiss or ?Hea

16. So, try this: Na(s) + �Cl2(g) ? NaCl(s) Given: 1st i.e.of Na = +496 Dissociation enthalpy of Cl = +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 enthalpy + solution enthalpy. Usually endothermic

20. This may make it clearer... Na+(g) + Cl-(g)

21. Typical problems involve.... You will be expected to find one of these, given the others, same as usual. But you�ll have to remember how. Remember: hydration enthalpy = solution enthalpy + lattice enthalpy.

22. 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

23. 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

24. INSERT BOND ENTHALPIES HERE

25. 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

26. Discuss these entropy values:

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

28. A sample question from Yahoo!answers State whether ?S 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.

29. 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?

30. Calculate: The entropy change in: CaCO3 ? CaO + CO2 Given: S? (CaCO3) = 92.9 J K-1 mol-1 S? (CaO) = 39.7 J K-1 mol-1 S? (CO2) = 213.6 J K-1 mol-1 ?S?system = 39.7+213.6-92.9 = 160.4 J K-1 mol-1

31. Calculate: The entropy change in: 2NaHCO3(s) ? Na2CO3(s) + CO2(g) + H2O(l) Given: S? (NaHCO3(s)) = 101.7 J K-1 mol-1 S? (Na2CO3(s)) = 135.0 J K-1 mol-1 S? (CO2(g)) = 213.6 J K-1 mol-1 S? (H2O(l)) = 69.9 J K-1 mol-1 . ?S?system = 135.0+213.6+69.9 � 203.4 = 215.1 J K-1 mol-1 .

32. Calculate: The entropy change in: N2(g) + 3H2(g) ? 2NH3(g) Given: S? (N2(g)) = 191.6 J K-1 mol-1 S? (H2(g)) = 130.6 J K-1 mol-1 S? (NH3(g)) = 192.3 J K-1 mol-1 ?S?system = 384.6 � 583.4 = -198.8 J K-1 mol-1 This system becomes less disorganised.

33. 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?

34. 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.

35. Examples: C(s) + O2(g) ? CO2(g) ?H = -394 kJ mol-1 ?S = +3.3 J mol-1 What is the free energy change? ?G = ?H - T?S = -394 x 103 � 298 x 3.3 J mol-1 = -395 kJ mol-1 Negligible T?S term in spite of 298

36. Examples: 2Fe(s) + 1�O2(g) ? Fe2O3(s) ?H = -825 kJ mol-1 ?S = -272 J mol-1 What is the free energy change? ?G = ?H - T?S = -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

37. Examples: 2NaHCO3(s) ? Na2CO3(s) + CO2(g) + H2O(g) ?H = +130 kJ mol-1 ?S = +335 J mol-1 What is the free energy change for 1 mole of NaHCO3? ?G = ?H - T?S = 130 x 103 � 298 x 335 J mol-1 = +30 kJ mol-1 for 2 moles, +15 kJ mol-1

38. Using the same information, prove that this reaction is feasible above around 400K Hint: using ?G = ?H � T?S you need a point where you change from not feasible to feasible, so ?G = 0

39. Melting (fusion) At equilibrium, e.g. ice at 0�C, there should be no change in G, or ?G =0 Note le Chatelier�s principle here � more heat energy would make more water from ice, so temperature does not change. This gives ?G = ?H - T?S = 0 So ?H = T?S or ?S(fus) = ?H(fus) / T(fus) Same for boiling

40. Boiling anything Same as for melting, ?G = ?H - T?S = 0 So ?H = T?S or ?S(vap) = ?H(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 ?S Therefore ?S = 6/273 = 0.22

41. 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 ?H, ?G and ?S for the production of hydrogen gas by electrolysis of water.

42. 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)

43. With water Only Na with cold water 2Na(s) + 2H2O(l) ? H2(g)+ 2Na+(aq) + 2OH-(aq) Al, Si, P, S do not react with water Mg with steam only: Mg(s) + H2O(g) ? H2(g) + MgO(s) Cl as discussed in unit 2 Ok, if I must..... Cl2(g) + H2O(l) � HCl(aq)+ HClO(aq) (sunlight) 2Cl2 + 2H2O ? 4H+ + 4Cl- + O2

44. With oxygen Na, Mg, Al, Si: 2Na(s) + �O2(g) ? Na2O(s) (yellow flame) Mg(s) + �O2(g) ? MgO(s) (white) 2Al(s) + 1�O2(g) ? Al2O3(s) (white) Si(s) + O2(g) ? SiO2(s) (white) P4(s) + 5O2(g) ? P4O10(s) (white) All very exothermic S(s) + O2(g) ? SO2(g) (pale blue)(+SO3)

45. Nature of the oxides Physical properties:

46. The oxides can be acidic, basic or amphoteric Na2O, MgO basic, make hydroxides (14,9) Al2O3 SiO2, are insoluble � no pH value (7) but we use limestone to remove SiO2 as an acid impurity in steelmaking Al2O3 is amphoteric � can neutralise acids and alkalis P4O10, SO2 (&SO3) acidic (0,3,(0)) Because: SO2 moderately soluble?weak, SO3 very soluble?strong Alkali ?????? Acid across period.

47. May need to know equations: AlCl3(s) ? Al3+ + 3Cl-(aq) followed by [Al(H2O)6]3+(aq) [Al(H2O)5(OH)]2+(aq) + H+(aq) (this is a partial hydrolysis of the aluminium ion) SiCl4(s) + 4H2O(l) ? Si(OH)4(s) + 4H+(aq) + 4Cl-(aq) PCl5(s) + 4H2O(l) ? H3PO4(aq) + 5H+(aq) + 5Cl-(aq) Total hydrolysis of the molecule in both cases.

48. Oh, no, not redox again.... OILRIG � oxidation is loss, reduction is gain Elements � oxidation state is always 0 Compounds always 0, BUT Compound ions and any other ions add up to their formal charge. Remember rules: F -1 (always) O -2 (except with F) Halogens -1 (except with F or O) Group1 +1 H +1 (but -1 in metal hydrides) Group2 +2 Remember this is a hierarchy, top rules rule and if you need to know ones that aren�t here, it�s based on electronegativity.

49. e.g. In SO42-, O is -2, so that�s -8, so the sulphur must be +6, to make the total -2 In Cr2O72-, O is -2 so that�s -14, so the two chromiums add up to +12, that�s +6 each

50. And some more... D-block elements in compounds and ions Calculate using same rules, don�t be surprised by drastic oxidation numbers, they do that E.g. in dichromate ion Cr2O72- -14 for oxygen plus the metal ions = -2 Therefore for Cr = � of 14-2 = 6 each.

51. e.g. What is the oxidation state of Cr in [CrCl2(H2O)4]+ ? Easy � we know: State of Cr + twice Cl- + 4x (H2O) = +1 Water is neutral (zero) �Cl-� is -1 So �Cr� -2 = +1 Therefore �Cr� = +3 .

52. Balancing (again) Look for the changes in oxidation state � there should be only one element changing in a half-equation, two or more in a full equation. O and H probably don�t change. Balance any oxygens by adding water Assuming acid conditions, balance any hydrogens by adding H+ Balance oxidation numbers with electrons E.g. MnO4- ? MnO2 becomes MnO4- + 4H+ + 3e- ? MnO2 + 2H2O

53. Let�s practice; balance these: VO42- ? V2+ Cr2O72- ? 2Cr3+ SO2 ? SO42- +2e-

54. Adding half-equations to get full equations The trick here is to get the electrons right Perhaps get them all on one side so they cancel E.g. displacement of copper from one of its compounds by magnesium � add the � equations ?: Mg(s) � 2e- ? Mg2+(aq) ?: Cu2+(aq) + 2e- ? + Cu(s) ? + ?: Mg(s) � 2e- + Cu2+(aq) + 2e- ? Mg2+(aq) + Cu(s) Or Mg(s) + Cu2+(aq) ? Mg2+(aq) + Cu(s)

55. Adding half-equations part 2 The alternative is to get equal numbers of electrons on each side E.g. displacement of copper from one of its compounds by magnesium � equation ?: Mg(s) � 2e- ? Mg2+(aq) � equation ?: Cu(s) � 2e- ? Cu2+(aq) ? + (in this case) the reverse of ?: Mg(s) � 2e- + Cu2+(aq) ? Mg2+(aq) + Cu(s) � 2e- Or Mg(s) + Cu2+(aq) ? Mg2+(aq) + Cu(s) Looks harder but that depends on what they give you in the question...

56. Sometimes you have to multiply: E.g. given: Cl2(g) + 2e- ? 2Cl-(g) ? And Al(s) � 3e- ? Al3+(g) ? Note that these are already balanced for charge and change of oxidation states We have to multiply ? by 3 before adding to ? doubled : 2Al(s) - 6e- + 3Cl2(g) + 6e- ? 2AlCl3(g) Or 2Al(s) + 3Cl2(g) ? 2AlCl3(g) Again, it depends on the question ......

57. Try: �What are the equations for the oxidation of sulphur dioxide to sulphate by the dichromate anion, in which chromate is reduced to a chromium(III) ion?� First know your compounds and ions and balance them: SO2 (aq) ? SO42- (aq) + 2e- And Cr2O72-(aq) +6e- ? 2Cr3+(aq)

58. Balance, multiply and add These balance to: SO2 +2H2O ? SO42- + 2e- + 4H+ And Cr2O72- + 6e- +14H+ ? 2Cr3+ + 7H2O Then you need to get the electrons equal: 3SO2 + 6H2O ? 3SO42- + 6e- + 12H+ And then add, letting the electrons cancel: 3SO2 + 6H2O + Cr2O72- +14H+ ? 3SO42- + 12H+ + 2Cr3+ + 7H2O And we can cancel the water and protons: 3SO2 + Cr2O72- + 2H+ ? 3SO42- + 2Cr3+ + H2O And I left out the state symbols on purpose to make it easier ....

59. Try this (show all working): In a titration, potassium permanganate will oxidise iron from iron(II) to iron(III) and is reduced to Mn2+(aq). Write out both half-equations Balance both, adding oxygens and H+ as needed Work out the oxidation state changes Get electrons equal in both (multiply as necessary) Add Cancel as necessary.

60. Metals and their solutions For a metal in a solution of one of its own salts, an equilibrium exists: Metal(s) � Metal+ ion + electron(s) E.g. Mg(s) � Mg2+(aq) + 2e-(aq)

61. For a reactive metal, this equilibrium lies mostly to the right, the metal is oxidised. These electrons �want� to move away from the metal, but must be replaced by electrons from the solution. The weird thing is that the electrons still end-up stuck in the solid metal electrode, which ends up being negatively charged So electrons can�t get into the solution from this metal.

62. Another metal-ion equilibrium can be connected such that it provides a way for electrons to flow. For an unreactive metal, the equilibrium lies mostly to the left: E.g. Cu(s) � Cu2+(aq) + 2e-(aq) So this equilibrium �takes� electrons from the equilibrium of the more-reactive metal. Each is a half-cell. Two half-cells make a cell.

63. Electrode potentials Conventions: Half-equations given to you are written as reductions � always adding electrons Electrodes written as reductions, ion|metal, eg: Fe2+|Fe This is writing a half-cell A full cell will be two of these, but with oxidation on the left:Zn|Zn2+||Cu2+|Cu

64. Definition: We can�t find the voltage (p.d.) of half a cell, need a reference. Standard electrode potential of an element E? is the potential difference between a standard hydrogen half-cell and a half-cell of the element under standard conditions. Standard conditions for E? : 298K 1 molar concentration solutions 1 atm. for any gases involved No current allowed (or only little ones!)

65. Standard hydrogen electrode Platinum surrounded by gaseous hydogen in a solution of H+ Gas electrode written as:Pt|H2(g)|H+(aq) if it�s left electrode, (usual) H+(aq)|H2(g)|Pt if it�s right electrode Total cell example: Pt|H2(g)|H+(aq) || Ag+(aq) | Ag

66. Left-hand cell is the more negative of the two in most diagrams, that is, it�s the one with the most negative voltage when written as a reduction. That way, you have a working cell, with the electrons going from left half-cell to right half-cell.

67. Potential of a cell is opposite polarity from electrolysis Similar: reduction occurs at the cathode and oxidation at the anode in both cells and electrolysis. Different: the anode will be negative. Think about rechargeable cells, one polarity �charges� and one �discharges�

68. For a tin/hydrogen cell, which of the following actions would change the measured cell potential? a.) Increasing the pH in the cathode compartment b.) Reducing the pH in the cathode compartment c.) Increasing [Sn2+] in the anode compartment d.) Increasing the pressure of hydrogen gas in the cathode compartment e.) All of the above