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
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