Introductory Inorganic Chemistry

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1 Introductory Inorganic Chemistry What is Inorganic Chemistry?

6 As: 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 3

8 Classes of Inorganic Substances Elements Ionic Compounds Covalent Compounds Atomic/Molecular Gases Ar, N 2 Simple (binary) NaCl Simple (binary) NH 3, H 2 O, SO 2 Molecular Solids P 4, S 8, C 60 Complex (polyatomic ions) Na 2 (SO 4 ) Complex (polyatomic) As(C 6 H 5 ) 3, organometallic compounds Network Solids Network ions Network Solids diamond, graphite (C ) Mg 3 (Si 2 O 5 )(OH) 2 (talc) SiO 2, polymers red phosphorus (P ) Solid/Liquid Metals Hg, Ga, Na, Fe, Mg

9 Elements Atomic/Molecular Gases Ar, N 2, O 2, Br 2 Molecular Solids P 4, S 8, C 60 Network Solids diamond, graphite (C ) red phosphorus (P ) Solid/Liquid Metals Hg, Ga, Fe, Na, Mg

10 Ionic Compounds Simple (binary) NaCl Complex (polyatomic ions) Na 2 (SO 4 ), Na 2 Mg(SO 4 ) 2 Network ions Mg 3 (Si 4 O 10 )(OH) 2 (talc)

11 H O H H N H F F H P F Covalent Compounds F Simple Molecular (binary) F NH 3, H 2 O, CO 2, SO 2 Complex Molecular As(C 6 H 5 ) 3, organometallic compounds Network Solids SiO 2, polymers

12 Review of Concepts Thermochemistry: Standard state: K, 1 atm, unit concentration Enthalpy Change, H H = ΣH products - ΣH reactants Entropy Change, S Free Energy Change, G G = H - T S At STP: G = H - ( K) S

13 Standard Enthalpy of Formation, H f H for the formation of a substance from its constituent elements Standard Enthalpy of Fusion, H fus Na (s) Na (l) Standard Enthalpy of Vapourization, H vap Br 2(l) Br 2(g) Standard Enthalpy of Sublimation, H sub P 4(s) P 4(g) Standard Enthalpy of Dissociation, H d ½Cl 2(g) Cl (g) Standard Enthalpy of Solvation, H sol Na + (g) Na + (aq)

14 Ionization Enthalpy, H ie The enthalpy change for ionization by loss of electron(s) Na (g) Na + (g) + e - H ie = 502 kj/mol Al (g) Al + (g) + e - Al + (g) Al 2+ (g) + e - Al 2+ (g) Al 3+ (g) + e - Thus: Al (g) Al 3+ (g) + e - H ie = 578 kj/mol H ie = 1817 kj/mol H ie = 2745 kj/mol H ie = 5140 kj/mol

18 Electron Attachment Enthalpy, H ea The enthalpy change for the gain of an electron Cl (g) + e - Cl - (g) H ea = -349 kj/mol O (g) + e - O - (g) O - (g) + e - O 2- (g) H ea = -142 kj/mol H ea = 844 kj/mol Electron Affinity, EA = - H ea + 5/2 RT EA = - H ea

19 Not easy to measure so many are missing

20 Why should we care about these enthalpies? They will provide us information about the strength of bonding in solids. H sub H ie Na (s) Na (g) Na + (g) ½Cl 2(g) Cl (g) Cl - (g) H d H ea Lattice Energy, U H f NaCl (s)

21 Bond Energy, E A-B Diatomic: H-Cl (g) H (g) + Cl (g) H = 431 kj/mol Polyatomic: H-O-H (g) H (g) + O-H (g) O-H (g) H (g) + O (g) Thus: H-O-H (g) 2 H (g) + O (g) H = 497 kj/mol H = 421 kj/mol H = 918 kj/mol Average O-H bond energy = 918 / 2 E O-H = 459 kj/mol

22 H H H 2 N-NH 2(g) 4 H (g) + 2 N (g) H = 1724 kj/mol N N NH 3(g) 3 H (g) + N (g) H = 1172 kj/mol H H Thus average N-H bond energy = 1172 / 3 E N-H = 391 kj/mol Since 1724 = 4 E N-H + E N-N We can estimate N-N bond energy to be: (391) = 160 kj/mol

23 O O Me C + H H Me C Me H H Me E H-H = 436 kj/mol E C=O = 745 kj/mol E C-H = 414 kj/mol E C-O = 351 kj/mol E O-H = 464 kj/mol H rxn = ΣE(bonds broken) ΣE(bonds formed) H rxn = ( ) ( ) kj/mol H rxn = -48 kj/mol

24 Remember that such calculated bond energies can change For H 2 N-NH 2(g) : E N-N = 160 kj/mol For F 2 N-NF 2(g) : E N-N = 88 kj/mol For O 2 N-NO 2(g) : E N-N = 57 kj/mol They are only a rough approximation and predictions must be made cautiously.

25 Free Energy Change, G = H - T S At STP: G = H - ( K) S The two factors that determine if a reaction is favourable: If it gives off energy (exothermic) H = ΣH products - ΣH reactants H < 0 If the system becomes more disordered S = ΣS products - ΣS reactants S > 0 If G < 0, then reaction is thermodynamically favourable

26 G lets us predict where an equilibrium will lie through the relationship: G = -RT ln K aa + bb + cc + hh + ii + jj + K = [ ] h [][ i ] j H I J [ A] [ B] [C] a b c So if G < 0, then K > 1 and equilibrium lies to the right. There are three possible ways that this can happen with respect to H and S.

27 If both enthalpy and entropy favour the reaction: i.e. H < 0 and S > 0 then G < 0. S (s) + O 2(g) SO 2(g) H = kj/mol T S = 7.5 kj/mol G = kj/mol If enthalpy drives the reaction: i.e. H < 0 and S < 0, but H > T S, then G < 0. N 2(g) + 3 H 2(g) 2 NH 3(g) H = kj/mol T S = kj/mol G = kj/mol If entropy drives the reaction: i.e. H > 0 and S > 0, but H < T S, then G < 0. NaCl (s) Na + (aq) + Cl- (aq) H = 1.9 kj/mol T S = 4.6 kj/mol G = -2.7 kj/mol

28 How do people obtain these values? Measure change in equilibrium constants with temperature to get H using the relationship: ln ln ln K o 1 H 1 K1 K2 = = K2 R T 1 T 1 2 Measure the equilibrium constant for the equilibrium, then determine G using the relationship? : Often not that easy G = -RT ln K

29 Reduction-Oxidation (RedOx) reactions: Reduction gain of electrons Oxidation loss of electrons E, the standard potential for an equilibrium, gives access to G through the following relationship: G = - nf E where, n = number of electrons involved F = Faraday s constant = kj mol -1 V -1 e --1 Note: if G < 0, then must be E > 0 So favourable reactions must have E > 0

30 Half-Cell Reduction Potentials Al 3+ (aq) + 3 e - Al (s) E = V Sn 4+ (aq) + 2 e - Sn 2+ (aq) E = 0.15 V thus for: 2 Al (s) + 3 Sn 4+ (aq) 2 Al 3+ (aq) + 3 Sn 2+ (aq) E = -(-1.67 V) + (0.15 V) = 1.82 V for 6 electrons So: G = - nf E = - (6 e - )F (1.82 V) = kj/mol

31 Oxidation state diagrams (Frost Diagrams) Relative Energy vs. Oxidation State (under certain conditions) Provides: - Relative stability of oxidation states -Energies available or required for RedOx reactions (the slope between reactant and product)

32 Oxidation state diagrams (Frost Diagrams) Some important information provided by Frost diagrams:

33 Oxidation state diagrams (Frost Diagrams) The diagram for Mn displays many of these features. The most useful aspect of Frost diagrams is that they allow us to predict whether a RedOx reaction will occur for a given pair of reagents and what the outcome of the reaction will be. This is described in the handout.

Lecture 2. Review of Basic Concepts

Lecture 2. Review of Basic Concepts Lecture 2 Review of Basic Concepts Thermochemistry Enthalpy H heat content H Changes with all physical and chemical changes H Standard enthalpy (25 C, 1 atm) (H=O for all elements in their standard forms