Chem 59-250 Introductory Inorganic Chemistry What is Inorganic Chemistry?

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Presentation transcript:

Chem Introductory Inorganic Chemistry What is Inorganic Chemistry?

Chem

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

Chem For more information about these periodic tables visit the site where I obtained the pictures:

Chem

Classes of Inorganic Substances ElementsIonic CompoundsCovalent 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 diamond, graphite (C  ) “red” phosphorus (P  ) Network ions Mg 3 (Si 2 O 5 )(OH) 2 (talc) Network Solids SiO 2, polymers Solid/Liquid Metals Hg, Ga, Na, Fe, Mg

Chem 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

Chem 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)

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

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

Chem 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 ½ Cl 2(g)  Cl (g) Standard Enthalpy of Solvation,  H° sol Na + (g)  Na + (aq)

Chem 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 -  H° ie = 578 kJ/mol Al + (g)  Al 2+ (g) + e -  H° ie = 1817 kJ/mol Al 2+ (g)  Al 3+ (g) + e -  H° ie = 2745 kJ/mol Thus: Al (g)  Al 3+ (g) + e -  H° ie = 5140 kJ/mol

Chem

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)  H° ea = -142 kJ/mol O - (g) + e -  O 2- (g)  H° ea = 844 kJ/mol Electron Affinity, EA = -  H° ea + 5/2 RT EA = -  H° ea

Chem Not easy to measure so many are missing

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

Chem 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)  H = 497 kJ/mol O-H (g)  H (g) + O (g)  H = 421 kJ/mol Thus: H-O-H (g)  2 H (g) + O (g)  H = 918 kJ/mol Average O-H bond energy = 918 / 2 E O-H = 459 kJ/mol

Chem H 2 N-NH 2(g)  4 H (g) + 2 N (g)  H = 1724 kJ/mol NH 3(g)  3 H (g) + N (g)  H = 1172 kJ/mol 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: 1724 – 4(391) = 160 kJ/mol

Chem 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 =  E(bonds broken) –  E(bonds formed)  H rxn = ( ) – ( ) kJ/mol  H rxn = -48 kJ/mol

Chem 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.

Chem 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 =  H products -  H reactants  H < 0 If the system becomes “more disordered”  S =  S products -  S reactants  S > 0 If  G < 0, then reaction is thermodynamically favourable

Chem  G lets us predict where an equilibrium will lie through the relationship:  G = -RT ln K aA + bB + cC + … hH + iI + jJ + … So if  G 1 and equilibrium lies to the right. There are three possible ways that this can happen with respect to  H and  S.

Chem If both enthalpy and entropy favour the reaction: i.e.  H 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 |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

Chem How do people obtain these values? Measure change in equilibrium constants with temperature to get  H° using the relationship: Measure the equilibrium constant for the equilibrium, then determine  G° using the relationship ? :  G° = -RT ln K Often not that easy…

Chem 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° = - n F  E° where, n = number of electrons involved F = Faraday’s constant = kJ mol -1 V -1 (e - ) -1 Note: if  G° 0 So favourable reactions must have  E° > 0

Chem 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° = - n F  E° = - (6 e - ) F (1.82 V) = kJ/mol

Chem 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)

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

Chem The diagram for Mn displays many of these features. Oxidation state diagrams (Frost Diagrams) 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.