introductory inorganic chemistry
DESCRIPTION
Introductory Inorganic Chemistry. What is Inorganic Chemistry?. As: 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 3. For more information about these periodic tables visit the site where I obtained the pictures: http://chemlab.pc.maricopa.edu/periodic/default.html. - PowerPoint PPT PresentationTRANSCRIPT
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)
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)
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
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) Al2+(g) + e- H°ie = 1817 kJ/mol
Al2+(g) Al3+
(g) + e- H°ie = 2745 kJ/mol
Thus:Al(g) Al3+
(g) + e- H°ie = 5140 kJ/mol
Chem 59-250
Chem 59-250
Chem 59-250
Chem 59-250
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- O2-
(g) H°ea = 844 kJ/mol
Electron Affinity, EA = -H°ea + 5/2 RT
EA = -H°ea
Chem 59-250
Not easy to measure so many are missing
Chem 59-250
Why should we care about these enthalpies?
They will provide us information about the strength ofbonding in 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
Bond Energy, EA-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 EO-H = 459 kJ/mol
Chem 59-250H
H
NN
H
H
H2N-NH2(g) 4 H(g) + 2 N(g) H = 1724 kJ/mol
NH3(g) 3 H(g) + N(g) H = 1172 kJ/mol
Thus average N-H bond energy = 1172 / 3 EN-H = 391 kJ/mol
Since 1724 = 4 EN-H + EN-N
We can estimate N-N bond energy to be:
1724 – 4(391) = 160 kJ/mol
Chem 59-250
MeC
Me
OH H+ Me C
O
Me
H
H
EH-H = 436 kJ/mol
EC=O = 745 kJ/mol
EC-H = 414 kJ/mol
EC-O = 351 kJ/mol
EO-H = 464 kJ/mol
Hrxn = E(bonds broken) – E(bonds formed)
Hrxn = (436 + 745) – (414 + 351+ 464) kJ/mol
Hrxn = -48 kJ/mol
Chem 59-250
Remember that such calculated bond energies can change
For H2N-NH2(g): EN-N = 160 kJ/mol
For F2N-NF2(g): EN-N = 88 kJ/mol
For O2N-NO2(g): EN-N = 57 kJ/mol
They are only a rough approximation and predictions must bemade cautiously.
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
Chem 59-250
ln ln lnK
K
H
R
1
T
1
T1 2
1
2
o
1 2K K
How do people obtain these values?
Measure change in equilibrium constants with temperatureto 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 59-250Reduction-Oxidation (RedOx) reactions:
Reduction – gain of electronsOxidation – 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 involvedF = Faraday’s constant = 96.4867 kJ mol-1 V-1 (e-)-1
Note: if G° < 0, then must be E° > 0
So favourable reactions must have E° > 0
Chem 59-250
Half-Cell Reduction Potentials
Al3+(aq) + 3 e- Al(s) E° = -1.67 V
Sn4+(aq) + 2 e- Sn2+
(aq) E° = 0.15 V
thus for:
2 Al(s) + 3 Sn4+(aq) 2 Al3+
(aq) + 3 Sn2+(aq)
E° = -(-1.67 V) + (0.15 V) = 1.82 V for 6 electrons
So: G° = - nF E° = - (6 e-)F (1.82 V) = -1054 kJ/mol
Chem 59-250
Oxidation state diagrams (Frost Diagrams)
Relative Energy vs. Oxidation State (under certain conditions)
Provides:- Relative stability of oxidation states
-Energies available orrequired for RedOx reactions(the slope between reactantand product)
Chem 59-250 Oxidation state diagrams (Frost Diagrams)Some important information provided by Frost diagrams:
Chem 59-250
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.