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) DNA 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) DNA

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 Why should we care about these enthalpies? They will provide us information about the strength of bonding in both molecules and extended 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 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