Chem 59-250

Introductory Inorganic Chemistry

What is Inorganic Chemistry?

Chem 59-250

Chem 59-250

Chem 59-250

Chem 59-250

Chem 59-250

As: 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p3

Chem 59-250

For more information about these periodic tables visit the site where I obtained the pictures: http://chemlab.pc.maricopa.edu/periodic/default.html

                                                            

  

Chem 59-250

Chem 59-250

Classes of Inorganic Substances

Elements Ionic Compounds Covalent Compounds

Atomic/Molecular Gases

Ar, N2

Simple (binary)

NaCl

Simple (binary)

NH3, H2O, SO2

Molecular Solids

P4, S8, C60

Complex (polyatomic ions)

Na2(SO4)

Complex (polyatomic)

As(C6H5)3, organometallic compounds

Network Solids

diamond, graphite (C)

“red” phosphorus (P)

Network ions

Mg3(Si2O5)(OH)2 (talc)

DNA

Network Solids

SiO2, polymers

Solid/Liquid Metals

Hg, Ga, Na, Fe, Mg

Chem 59-250

Elements

Atomic/Molecular Gases

Ar, N2, O2 , Br2

Molecular Solids

P4, S8, C60

Network Solids

diamond, graphite (C)

“red” phosphorus (P)

Solid/Liquid Metals

Hg, Ga, Fe, Na, Mg

Chem 59-250

Ionic Compounds

Simple (binary)

NaCl

Complex (polyatomic ions)

Na2(SO4), Na2Mg(SO4)2

Network ions

Mg3(Si4O10)(OH)2 (talc)

DNA

Chem 59-250

Covalent Compounds

Simple Molecular (binary)

NH3, H2O, CO2, SO2

Complex Molecular

As(C6H5)3, organometallic compounds

Network Solids

SiO2, polymers

F

F

P F

F

F

HH

N

H

H H

O

Chem 59-250

Review of Concepts

Thermochemistry:Standard state: 298.15 K, 1 atm, unit concentration

Enthalpy Change, H°H° = H°products - H°reactants

Entropy Change, S°

Free Energy Change, GG = H - TSAt STP:G° = H° - (298.15 K)S°

Chem 59-250

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 Br2(l) Br2(g)

Standard Enthalpy of Sublimation, H°sub P4(s) P4(g)

Standard Enthalpy of Dissociation, H°d ½ Cl2(g) Cl(g)

Standard Enthalpy of Solvation, H°sol Na+(g) Na+

(aq)

Chem 59-250

Why should we care about these enthalpies?

They will provide us information about the strength ofbonding in both molecules and extended solids.

NaCl(s)

Na(s) Na(g) Na+(g)

½ Cl2(g) Cl(g) Cl-(g)H°eaH°d

H°ieH°sub

H°f

Lattice Energy, U

Chem 59-250

Free Energy Change, G = H - TS

At STP:G° = H° - (298.15 K) S°

The two factors that determine if a reaction is favourable:

If it gives off energy (exothermic)H = Hproducts - Hreactants

H < 0

If the system becomes “more disordered”S = Sproducts - Sreactants

S > 0

If G < 0, then reaction is thermodynamically favourable

Chem 59-250

G lets us predict where an equilibrium will lie throughthe relationship:G = -RT ln K

aA + bB + cC + … hH + iI + jJ + …

K [ ] [ ] [ ]

[ ] [ ] [C ]

H I J

A B

h i j

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.

Chem 59-250If both enthalpy and entropy favour the reaction:i.e. H < 0 and S > 0 then G < 0.

S(s) + O2(g) SO2(g) H° = -292.9 kJ/molTS° = 7.5 kJ/mol

G° = -300.4 kJ/mol

If enthalpy drives the reaction:i.e. H < 0 and S < 0, but |H| > |TS|, then G < 0.

N2(g) + 3 H2(g) 2 NH3(g) H° = -46.2 kJ/molTS° = -29.5 kJ/mol

G° = -16.7 kJ/mol

If entropy drives the reaction:i.e. H > 0 and S > 0, but |H| < |TS|, then G < 0.

NaCl(s) Na+(aq) + Cl-(aq) H° = 1.9 kJ/mol

TS° = 4.6 kJ/molG° = -2.7 kJ/mol