The Basics
• Elements, Molecules, Compounds, Ions
• Parts of the Periodic Table
• How to Name
1
Classification of Elements
•Metalloids
semi-metals)
– along
••Nonmetals
Metals – found
–(or
found
on the
on the
left-side
right-hand
of thethe
sidestair-step
of
line
the
Periodic
Periodic
Table
Table
Properties are intermediate between metals and
nonmetals
2
Elements , Molecules, Compounds, and Ions
• Element – one single type of atoms
– Al Cu He
– Naturally occurring elements that are Diatomic are
still elements
– N2 O2 F2 Cl2 Br2 I2 H2
– How many elements are in Mn(SO4)2 ? 3
– How many Atoms? 9
• Molecule - smallest electrically neutral unit of a
substance that still has the properties of that
substance, 2 or more different elements
3
Elements , Molecules, Compounds, and Ions
• Compounds – groups of atoms
– Ionic and Molecular
• Molecular Compounds – share electrons
typically 2 or more non-metals (hydrocarbons)
• Example H2S CO2 C5H10
• Ionic Compound (salts) – transfer electrons
typically metal and non metal (watch for poly
atomic ions)
• FeS Mg(OH)2 (NH4)3PO4
4
Ions
• Ions have either lost or gained electrons
• Typically Metals lose electrons to become
positive
• Example cations
• Mono-valent Mg -> Mg2+ + 2e– Group 1A = Metal1+, 2A =Metal2+ and 3A = metal3+
– Multi-valent Fe --> Fe2+ + 2e- and Fe --> Fe3+ + 3e-
• Non-metals gain electrons - anion
• S + 2e- -> S25
Poly Atomic Ions
• A molecular compound
with a charge
• NH4+ • Ammonium
• CO32- • Carbonate
• SO42- • Sulfate
• NO3- • Nitrate
• OH- • Hydroxide
• H3O+ • Hydronium
6
Acids and Bases From Ions H+ or OH• Acids look for hydrogen up front (HA) or as COOH
•
•
•
•
•
Example HF H3PO4 C4H6COOH
Strong Acids
HCl, H2SO4, HBr, HI, HClO3,HNO3
Base look for hydroxide or NH group
Example KOH C4H6NH2
7
Naming Compounds
1. Ionic or Covalent
2. Ionic – two ions or
Poly atomic ions
Covalent
2 non-metals
Or a hydrocarbon
Type of Metal
Mono-valent metals groups
1A 2A 3A
Multi-valent Metal
Transitions metals and under the stairs
Name the Metal
Name the Nonmetal + ide
(if PAI use its name)
Find the Charge on the Metal
To make the compound neutral
Write the Charge with roman numeral
Name nonmetal + ide (if PAI use its name)
CaF2
RbNO3
Al(OH)3
calcium fluoride
rubidium nitrate
aluminum hydroxide
Ni+ClPb2+SO42Pb4+(SO4)22-
nickel(I) chloride
lead(II) sulfate
lead(IV) sulfate
8
Ionic naming
• Name to Formula– Final compound must be
neutral based on subscripts and charges
• Magnesium Fluoride Mg2+ + F-  MgF2
• Ammonium Sulfide NH4+ + S2- (NH4)2S
• Tin(II) Carbonate Sn2+ + CO32-  SnCO3
• Iron(III) Oxide Fe3+ + O2-  Fe2O3
• Iron(II) Oxide Fe2+ + O2-  FeO
9
Ionic or Covalent
Ionic – two ions or
Poly atomic ions
Covalent
2 non-metals
Or a hydrocarbon
Use the prefix to tell how many of
each atom there are
Mono is never used with the first element
Example
PBr3
CCL4
P2O5
CO
Phosphorous tribromide
Carbon tetrachloride
diphosphorous pentoxide
carbon monoxide
Hydrogen up front
Most likely an Acid you
should have memorized
Hydrocarbons
Look for how many
carbons
HCl - Hydrochloric acid
HI - Hydroiodic acid
HBr - Hydrobromic acid
HNO3 Nitric Acid
H2SO4 - Sulrufic acid
HClO - Hypochlorous acid
One – methane CH4
Two – ethane C2H6
Three – propane C3H8
Four – butane C4H10
10
Molecular Compounds
• Name to formula – charge does not matter
this time, just use the prefixes or memorize
• Tetraarsenic hexoxide
As4O6
• Sulfur hexafluroide
SF6
• Butane
C4H10
• Nitric acid HNO3
11
Lewis Dot Structures
• Rules
1. Fewest number of atoms goes in the middle
or C if present
2. Connect remaining elements with single
bonds
3. Make sure all elements have 8 electrons (H only 2)
4. Count the number of electrons in structure
5. Add up valence electrons from PT
– Too many e- in structure: remove 2 adjacent
pairs fill in with one bond
– Too few e- in structure: add to central atom
12
EXAMPLES:
•
•
•
•
•
•
CH4
1.
(1) C + (4) H
(1)(4e-) + (4)(1e-) = 8e2. Spatial order
H
H C H
3. Draw bonds
H
4. Octet rule satisfied?
5. # of e- match?
13
EXAMPLES:
•
•
•
CO2
1.
(1) C
(1)(4e-)
2. Spatial order
+
+
•
3. Draw Bonds
•
•
4. Octet rule satisfied?
5. # of e- match?
(2) O
(2)(6e-) = 16e-
O C O
14
EXAMPLES:
•
•
NH3
1.
•
(1) N +
(1)(5e-) +
2. Spatial order
(3)H
(3)(1e-) =
•
3. Draw bonds
•
•
4. Octet rule satisfied?
5. # of e- match?
8e-
H N H
H
15
EXAMPLES:
•
•
CCl4
1.
•
(1) C + (4) Cl
(1)(4e-) + (4)(7e-) = 32e2. Spatial Order
Cl
Cl C Cl
3. Draw bonds
Cl
•
•
4. Octet rule satisfied?
5. # of e- match?
•
16
EXAMPLES:
•
•
NH4+
1. (1) N + (4) H
•
(1)(5e-)+ (4)(1e-) 2. Spatial order
-
(1)(+)
(1)(1e-) = 8eH
H N H
H
[
•
3. Draw bonds
•
•
4. Octet rule satisfied?
5. # of e- match?
+
]
17
EXAMPLES:
•
•
•
•
•
•
SO421. (1) S + (4) O + (2)(-)
(1)(6e-)+ (4)(6e-) + (2)(1e-) = 32e2. Spatial Order
O
2O S O
3. Draw bonds
O
[
4. Octet rule satisfied?
5. # of e- match?
]
18
EXAMPLES:
•
•
CN1. (1) C + (1) N +
•
(1)(4e-) + (1)(5e-)+
2. Spatial order
•
3. Draw Bonds
•
•
4. Octet rule satisfied?
5. # of e- match?
(1)(-)
(1)(1e-) = 10e-
[C N ]
19
EXAMPLES:
•
•
•
CO321. (1) C + (3) O + (2)(-)
(1)(4e-)+ (3)(6e-) + (2)(1e-) = 24e2. Spatial Order
2O C O
3. Draw bonds
O
•
•
4. Octet rule satisfied?
5. # of e- match?
•
[
]
20
VSEPR:
• Regions of electron density (where pairs of electrons are
found) can be used to determine the shape of the
molecule.
• CO2
O C O
• Here there are two regions of electron density.
• The regions want to be as far apart as possible, so it is
linear.
21
EXAMPLES:
4
1
H
• CH4
H C H
• There are four electron pairs.
2
H
• You would expect that the bond angles
3 would be 90°
but…
• Because the molecule is three-dimensional, the
angles are 109.5°.
• The molecule is of tetrahedral arrangement.
22
EXAMPLES:
• NH3
4
H N H
• Four regions of electron density
H
• But one of the electron pairs is a3 lone pair
• The shape is called trigonal pyramidal
1
2
23
EXAMPLES:
•
•
•
•
4
1
H2O
H O H
Four regions of electron density
But two are lone pairs
3
This structure is referred to as bent
2
24
EXAMPLES:
• CO32-
1
3
[
O C O
O
2-
]
• Three regions of electron density
• This structure is referred to as trigonal planar
2
25
Practice determining molecular
shape:
• H2S
– 4 regions of e- density
– 2 lone pairs
– bent
H S H
H
S
H
26
Practice determining molecular
shape:
• SO2
– 3 areas of e- density
– 1 lone pair
– bent
O S O
O
S
O
27
Practice determining molecular
shape:
• CCl4
– 4 areas of e- density
– tetrahedral
Cl
Cl C Cl
Cl
3d
28
Practice determining molecular
shape:
• BF3
– 3 areas of e- density
– trigonal planar
F B F
F
F
F
B
F
29
Practice determining molecular
shape:
• NF3
– 4 areas of e- density
– 1 lone pair
– pyramidal
F N F
F
3d
30
16.3 Polar Bonds and Molecules
• In covalent bonds, the sharing of electrons can
be equal
• or it can be unequal.
31
Nonpolar Covalent Bonds
• Nonpolar covalent bond - This is a covalent bond in
which the electrons are shared equally.
• EXAMPLES:
• H2
• Br2
• O2
• N2
• Cl2
• I2
Br
Cl
H
N
FIO
OBr
Cl
H
N
FI
• F2
32
Polar Bonds and Molecules
• If the sharing is unequal, the bond is referred
to as a dipole.
• A dipole has 2 separated, equal but opposite
charges.
• “∂” means partial
+
_
33
Polar Bonds and Molecules
• Polar covalent bond - a covalent bond that
has a dipole
• It usually occurs when 2 different elements
form a covalent bond.
• EXAMPLE:
• H +
Cl

H Cl
34
Polar Bonds and Molecules
• Electronegativity - This is the measure of the
attraction an atom has for a shared pair of electrons
in a bond.
• Electronegativity values increase across a period
and up a group.
35
Examples:
• Identify the type of bond for each of the
following compounds:
• HBr
Br =
2.8
H
=
2.1
.1 < 0.7< 1.9
Polar Covalent
H
Br
36
Examples:
• NaF
F
Na
• N2
N
N
=
=
=
=
4.0
0.9
3.1 > 2
3.0
3.0
0.0
Na
F
Ionic
Non-Polar
Covalent
N
N
37
Molecular Polarity
• If there is only one bond in the molecule, the bond type
and polarity will be the same.
• If the molecule consists of more than 2 atoms, you
must consider the shape. To determine its polarity,
consider the following:
– Lone pairs on central atom
• If so… it is polar
– Spatial arrangement of atoms
• Do bonds cancel each other out (symmetrical)?
– If so… nonpolar
• Do all bonds around the central element have the same
difference of electronegativity?
– If so… nonpolar
38
Polar Molecules
• If the molecule is symmetrical it will be
nonpolar.
– Exception hydrocarbons are nonpolar
• If the molecule is not symmetrical it
will be polar, with a different atom or with lone
pair(s)
Br
Cl C Cl
Cl
39
Attractions Between Molecules
• Van der Waals forces – the weakest of the
intermolecular forces. These include London
dispersion and dipole-dipole forces.
– London dispersion forces – between nonpolar
molecules and is caused by movement of electrons
40
Attractions Between Molecules
• van der Waals forces(cont.)– Dipole interactions – between polar molecules
and is caused by a difference in electronegativity.
41
Attractions Between Molecules
• Hydrogen bonds – attractive forces in which hydrogen,
covalently bonded to a very electronegative atom (N, O, or
F) is also weakly bonded to an unshared (lone) pair of
electrons on another electronegative atom.
H
O
H
H
H
O
H
O
H
42
Attractions Between Molecules
• Ionic Bonding-occurs between metals and
nonmetals when electron are transferred from
one atom to another.
• These bonds are very strong.
43
Summary of the Strengths of Attractive
Forces
Ionic bonds
hydrogen bonds
dipole-dipole attractions
LDF
44
Writing and Balancing Chemical Equations
Example:
Write the equation for the formation of sodium
hydroxide and hydrogen, from the reaction of
sodium with water.
45
Write the equation for the formation of sodium
hydroxide and hydrogen, from the reaction of
sodium with water.
1. Write the formulas of all reactants to the left of the
arrow and all products to the right of the arrow.
Sodium + water
sodium hydroxide + hydrogen
Translate the equation and be sure the formulas are
correct.
Na + H2O  NaOH + H2
46
Write the equation for the formation of sodium
hydroxide and hydrogen, from the reaction of
sodium with water.
2. Once the formulas are correctly written, DO NOT change
them. Use coefficients (numbers in front of the
formulas), to balance the equation. DO NOT CHANGE
THE SUBSCRIPTS!
_____Na + _____H2O  ____NaOH + _____H2
47
3. Begin balancing with an element that occurs only once
on each side of the arrow.
Ex: Na
2
2 2O  ____NaOH
2
_____Na
+ _____H
+ _____H2
When you are finished, you should have equal numbers
of each element on either side of the equation
Na
2
Na
2
H
4
H
4
O
2
O
2
48
4. To determine the number of atoms of a given element
in one term of the equation, multiply the coefficient by
the subscript of the element.
Ex: In the previous equation (below), how many
hydrogen atoms are there?
4
2
2
2
____Na
+ _____H
2O  ____NaOH + _____H2
49
• Balance elements one at a time.
• Balance polyatomic ions that appear on both
sides of the equation as single units. (Ex:
Count sulfate ions, not sulfur and oxygen
separately)
(NH•4)2Balance
SO4 (aq) +HBaCl
(aq)
 BaSO
(s) +one
2NHthat
and2 O
last.
Save4 the
is in
4Cl (aq)
the most places for last…
• Use Pencil!
50
Practice:
• Balance the equation for the formation of
magnesium nitride from its elements.
Mg2+
N3Mg3N2
3
____Mg
+ ____N2 
____Mg3N2
51
Ex: NH3 + O2  NO2 + H2O
• H can be balanced by placing a 2 in front of
NH3 and a 3 in front of H2O. Then put a 2 in
front of NO2 for nitrogen to balance.
2
_____NH
3 + _____O2 
3
2
____NO
2 + ____H
2O
52
2
7/2 2 
_____NH
3 + _____O
2
3 2O
____NO
2 + ____H
• Now all that is left to balance is the oxygen.
There are 2 O on the reactant side and 7 on
the product side. Our only source of oxygen is
the O2. Any whole number we place in front
of the O2 will result in an even number of
atoms. The only way to balance the equation
is to use a coefficient of 7/2.
53
Stoichiometry – study of
calculations of quantities in
chemical reactions using balanced
chemical equations.
2Mg +
2 moles Mg
O2
+

2MgO
1mole O2 
2 moles MgO
54
2Mg + O2 
2MgO
• The mole ratios can be obtained from the
coefficients in the balanced chemical equation.
• What are the mole ratios in this problem?
• Mole ratios can be used as conversion factors to
predict the amount of any reactant or product
involved in a reaction if the amount of another
reactant and/or product is known.
55
What’s that mean?
Well, a stoichiometry problem gives you an
amount of one chemical and asks you to solve for
a different chemical.
To get from one type of chemical to
another, a MOLE RATIO must be
found between the two chemicals.
You get the MOLE RATIO from the
BALANCED CHEMICAL EQUATION!
56
A balanced chemical equation tells the quantity
of reactants and products as well as what they
are.
2 mol
1 mol
2 mol
*the coefficients are*
57
The MOLE RATIO is your mechanism of transition
between the chemical that is your starting given
and the chemical you are solving for.
The MOLE RATIO is the bridge
between the two different
chemicals
moles
moles
(given)
given
(want)
Mole
Ratio
? want
58
EXAMPLE
• How many grams of KClO3 must decompose to
produce KCl and 1.45 moles O2?
2KClO3
1.45 moles O2
→
2KCl + 3O2
2 mol KClO3
122.6 g KClO3
3 mol O2
1 mol KClO3
mol-mol ratio
=
119 g KClO3
GFM
59
• CaCO3, limestone, is heated to produce calcium oxide,
CaO, and CO2. What mass of limestone is required to
produce 156.0 g of CaO?
CaCO3 (s)  CaO (s) + CO2 (g)
156.0 g CaO
GFM
1 mol CaO
1 mol CaCO3
100.1 g CaCO3
56.1 g CaO
1 mol CaO
1 mol CaCO3
mol-mol ratio
GFM
=
278.4 g CaCo3
60
EXAMPLE
• Calculate the number of joules released by the
oxidation of 5.00 moles of Na completely react
with oxygen gas. ΔH = -416 kJ/mol
4Na + O2  2Na2O
1000 J
5.00 mol Na 2 mol Na2O -416 kJ
4 mol Na 1 mol Na2O 1 kJ
mol-mol ratio
enthalpy
1.04x106 J
61
Atomic orbital – the region in space
where the electron is likely to be
found
A quantum mechanical model of a hydrogen atom, which has
one electron, in its state of lowest energy. The varying density
of the spots indicates the relative likelihood of finding the
62
electron in any particular region.
Electrons can be described by
a series of 4 quantum
numbers.
You must be familiar with
all of these!
63
1. Principle quantum number (n)
-describes the principal energy level an
electron occupies
-values of 1,2,3,4,etc
1
2
3
4
5
6
7
64
2. Azimuthal quantum number (l)
-describes the shape of atomic orbitals
-s orbitals are spherical, p orbitals are
peanut shaped, d orbitals are daisies and f
orbitals are fancy
-designates a sublevel
-values of 0 up to and including n-1
0=s, 1=p, 2=d, 3=f
65
66
67
68
69
Where do I find the orbital shapes?
70
3. magnetic quantum number (ml)
-Designates the spatial orientation of an
atomic orbital in space
-values of -l to +l
so s has 1 orbital
p has 3 orbitals (x, y, and z)
d had 5 orbitals (xy, xz, yz, x2-y2 and z2)
f has 7 orbitals
71
4. spin quantum number
-describes the orientation of the
individual electrons; values of +1/2
and -1/2
-each orbital can hold 2 electrons with
opposite spins
72
Orbitals Electrons
Symbol
Shape
s
sphere
1
2
p
peanut
3
6
d
Daisy/
donut
5
10
f
fancy
7
14
73
Now, try
letsthe
try three
to do in
anyour
electron
notes.
configuration for carbon.
Begin
He
1s2with the first quantum number and use the periodic table to write
the configuration.
Na
1s2 2s2 2p6 3s1
Ti 1s2 2s2 2p6 3s2 3p6 4s2 3d2
Quick and Easy Electron Configuration
1
2
3
4
5
6
7
3
4
5
6
4
5
74
For a shorter way to write electron configuration, write the nearest
noble gas and then continue. AKA “Shorthand Notation”.
Ex: Ti can be written as
OR
1s2 2s2 2p6 3s2 3p6 4s2 3d2
[Ar] 4s2 3d2
75
Electron Configuration of Ions
• For the loss of an electron remove electrons from
the last orbital
Ge 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p2
Ge2+ 1s2 2s2 2p6 3s2 3p6 4s2 3d10
For the gain of electrons add them in the last orbital
filled
At [Xe] 6s2 4f14 5d10 6p5
At- [Xe] 6s2 4f14 5d10 6p6
76
Increasing Atomic Radius
Atomic Radius
Increasing Atomic Radius
77
• Ion Size
• Anions are larger than the atoms from which they
were formed.
• The negative charge means more electrons are
present causing the size of the ion to be larger.
• Cations are smaller than the atoms from which they
were formed.
• The positive charge means fewer electrons are
surrounding the nucleus, thus pulling the existing
electrons closer and causing the ion to be smaller.
• Trend: ionic radius increases from right to left and top
to bottom
78
Increasing Ionic Radius
Ionic Radius
Increasing Ionic Radius
79
• Ionization energy - energy required to
overcome the attraction of the nuclear charge
and remove an electron from a gaseous atom
• 1st ionization energy: the energy required to
remove the first electron
• 2nd ionization energy: the energy required to
remove the second electron
• 3rd ionization energy: the energy required
removing the third electron
• Trend: ionization energy increases from left to
right and bottom to top
80
Increasing Ionization Energy
Increasing Ionization Energy
Ionization Energy
81
• Electronegativity – the tendency for the atoms
of the element to attract electrons when they
are chemically combined with atoms of
another element
• Note: Noble gases don’t have values for
electronegativity because their outer orbitals
are full and they do not need to gain or lose
electrons to be stable.
• Trend: electronegativity increases from left to
right and bottom to top
82
Increasing Electronegativity
Increasing Electronegativity
Electronegativity
83
We can
measure not
only the
length of
each wave of
light, but
also its
frequency of
occurrence.
Parts of a wave:
Amplitude
amplitude- height of the wave from
the origin to the crest
wavelength -  - distance between the
crests
frequency -  - the number of wave
cycles to pass a given point per unit of
time.
The units of frequency are 1/s, s-1, or
Hertz (Hz)
 = c/
where c = speed of light
c= 3.00 x 108 m/s
As  increases,  decreases.
Ex. A certain wavelength of yellow
16
-1
light has a frequency of 2.73 x 10 s .
Calculate its wavelength.
 = c/
= c/
= 3.00 x 108m/s
2.73 x 1016s-1
= 1.10 x 10-8 m
Spectrum- series of colors produced
when sunlight is separated by a
diffraction gradient.
ROY G. BIV
Red: has the longest wavelength,
lowest frequency  Lowest energy
Violet: has the shortest wavelength,
highest frequency  highest energy
Kinetic theory- The tiny particles in all forms of
matter are in constant motion.
1. A gas is composed of particles, usually
molecules or atoms. We treat them as, Hard
spheres, Insignificant volume, and Far from
each other
2. The particles in a gas move rapidly in
constant random motion.
3. All collisions are perfectly elastic.
92
Gas Laws
•
•
•
•
One single set of conditions
PV = nRT
Two Sets of conditions
P1 V1 = P2 V2
T1
T2
93
Ex.
If a helium-filled balloon has a volume of
3.40 L at 25.0oC and 120.0 kPa, what is its
volume at STP = 0°C, 1 ATM?
V1 = 3.40L
V2 = ?
P1 = 120.0 kPa P2 = 1 atm = 101.3 kPa
T1 = 25.0oC + 273 = 298K
T2 = 0oC + 273 = 273K
P1V1 = P2V2 (120.0 kPa)(3.40L) = (101.3 kPa)V2
T1
T2
298K
273K
V2 = 3.69L
94
Ex.
A 5.0 L flask contains 0.60 g O2 at a
temperature of 22oC. What is the pressure
(in atm) inside the flask?
P=? V= 5.0L R = 0.0821 (L.atm/mol.K)
T = 22°C +273= 295K
n = 0.60g O2 1 mol O2 = 0.01875 mol
32.0g O2
PV = nRT
P(5.0L) = (0.01875mol)(0.0821Latm/molK)(295K)
P = 0.091 atm
95
Factors affecting solubility:
• The nature of the solute and the solvent
– “like dissolves like”
• Miscible - liquids that are soluble in each
other
– Ex. ethanol and water
• Immiscible- liquids that are not soluble in
each other
– Ex. oil and water
Molarity(M) = moles of solute
liters of solution
• A 1 M solution contains 1 mole of solute per 1
L of solution. A 0.5 M NaCl solution has 0.5
mol NaCl in 1 L total solution.
EXAMPLES:
• What is the concentration in molarity of a
solution made with 1.25 mol NaOH in 4.0 L of
# mol
solution?
M=
M=?
#mol = 1.25 mol
V = 4.0 L
M = 1.25 mol
4.0 L
#L
= 0.3125 M
= 0.31 M
Exothermic reactions have -  H
Endothermic reactions have +  H
100
Enthalpy (H)the amount of heat in
a system at a given
temperature
Enthalpy change:
H =q=mCT
101
Ex. A 25 g sample of a metal at 75.0oC is placed in a
calorimeter containing 25 g of H2O at 20.0oC. The
temperature stopped changing at 29.4oC. What is the
specific heat of the metal?
103
Standard Heat of Formation of a compound
( Hfo)
* Hfo of a free element in its standard state is
zero.
 H =  Hfo products -  Hforeactants
104
Ex.
1. Calculate  H for the following reaction:
(endo or exo thermic?)
CaCO3(s)  CaO(s) + CO2(g)
∆Hfo values:
CaCO3 = -1207.0 kJ/mol
CaO = -635.1 kJ/mol
CO2 = -393.5 kJ/mol
∆H = [-635.1 + (-393.5)] – [-1207.0]
∆H = 178.4 kJ
endothermic, heat going in
105
2.
Calculate the heat of reaction for the
following reaction: (endo or exo thermic?)
2H2(g) + O2(g)  2H2O(g)
∆Hfo values:
H2O(g) = -241.8 kJ/mol
∆H = [2(-241.8)] – [2(0) + 0]
∆H = -483.6 kJ
Remember to multiply heat
values by coefficients!!!
Exothermic
Heat leaving
106
Entropy Calculation Example:
Mn(s) + 2O2(g)  MnO4(s)
S° J mol-1 K-1 Mn (s) = 32.8 O2 = 205.0 MnO4 = 120.5
∆S = [S° Products] – [S° Reactants]
∆S = [120.5] – [2(205.0) + 32.8]
∆S = -322.3 J/mol K
107
Entropy, S - a measure of randomness
or disorder
• associated with probability (There are more ways for
something to be disorganized than organized.)
• Entropy increases going from a solid to a liquid to a gas.
• Entropy increases when solutions are formed.
• Entropy increases in a reaction when more atoms or molecules
are formed.
• The entropy of a substance increases with temperature.
108
Gibbs free energy, G
• Energy available to do work
• Go = standard free energy change
– change in free energy that occurs if the reactants in their standard
states are converted to products in their standard states
Go =Ho -T So
When Go is negative the reaction is spontaneous
is the forward direction
When Go is positive the reaction is
nonspontaneous is the forward direction
109
• What temperature would a reaction be
spontaneous if ΔH = 9500 J/mol and ΔS =
6.5J/mol K?
Go =Ho -T So
9500 = T(6.5)
T = 1461 K
Above 1461 K this reaction will be spontaneous
and below it will not
110
• Will this reaction be spontaneous at 100°C?
ΔH = -18 KJ/mol and ΔS = 94.3 J/mol K
Go =Ho -T So
Yes at all temperatures
Go =(Ho) -T (So)
Go =(-) - (+)(+)
Go = can only be negative
Don’t believe me try putting in the values and
craze high/low number for temperature
111
Go =Ho -T So
• A spontaneous reaction has a negative G. For
example, when ice melts H is positive (endothermic),
S is positive and G = 0 at 0oC.
• If...
Entropy, ΔS
Enthalpy, ΔH
Spontaneity
Positive
Positive
Yes at high temp
Negative
Positive
Never spontaneous
Positive
Negative
Always spontaneous
Negative
Negative
Yes at low temp
112
Collision Theory:
• In order to react, two or more
particles must collide with
sufficient energy (called the
activation energy) and with
the proper molecular
orientation. If the colliding
particles do not have either of
these two prerequisites, no
product is formed.
113
Factors that affect reaction rate:
• Temperature- Reactions go faster at higher
temperatures. Particles have more kinetic energy.
More colliding particles have enough energy to
overcome the activation energy barrier.
114
Factors that affect reaction rate:
• Concentration- Increasing the concentration of
reactants usually increases the reaction rate. If
there are more particles to collide, there should be
a greater number of collisions that produce
products.
115
Factors that affect reaction rate:
• Catalysts- A catalyst is a substance that speeds up a
reaction by lowering the activation energy barrier.
It is not a product or reactant and it is not used up
or changed itself.
116
Rate Law
• equation that is written that expresses how
the reaction rate of a particular reaction is
dependent upon the concentrations of its
reactants.
• For the reaction aA + bB  cC + dD, the
general form of the rate law would be:
Rate = k [A]a[B]b
117
Rate = k
x
y
[A] [B]
• Rate is usually expressed as mol/L∙time.
• k is the specific rate constant. It is constant for a given
reaction at a given temperature. The faster a reaction, the
larger the k value.
• [A] and [B] represent the concentrations of reactants A and
B in moles per liter (M).
• x and y are the order of the reactant. They can only be
determined by analyzing experimental data. These
exponents are usually positive integers.
118
EXAMPLES
2A + B 2C
Ex.
[A]
[B]
Rate
1
0.1 0.2
0.10
2
0.1 0.4
0.20
3
0.2 0.4
0.80
Determine the rate law:
3
[A]
x 0.1
0.1
( 0.2) = ( 0.8)
[B]
x 0.1
0.2
( 0.4) = 0.2
23 =8
1
rate = k [A] [B]
119
Half-life
Based on how much time does it take ½ of the
substance to change into products
x = time / half-life (number of half-lives)
mf = (mi)(1/2)x
120
If you start with 2.00 g of nitrogen-13 how many
grams will remain after 4 half lives?
mf = 2.00 (1/2)4
mf =0.125 g
Phosphorous-32 has a half-life of 14.3 yr. How
many grams remain after 57.2 yr from a 4.0 g
sample
mf = 4.00 (1/2)(57.2/14.3)
mf = 0.25 g
121
BrØnstedLowry
• Acid: hydrogen-ion donor (proton donor)
• Base: hydrogen-ion acceptor (proton acceptor)
NH3 + H2O  NH4+ + OHbase acid
BrØnstedLowry
• Acid: hydrogen-ion donor (proton donor)
• Base: hydrogen-ion acceptor (proton acceptor)
NH3 + H2O  NH4+ + OHbase acid
Working Ka and Kb problems
• 1st: Write the equation
• 2nd: Set up a reaction
diagram (RICE diagram)
• 3rd: Set up Ka or Kb expression
• 4th: Substitute values into Ka
expression
• 5th: Solve Ka expression for X.
• 6th: Calculate pH from H+ or OHconcentration.
R = reaction
I = initial
C = change
E = equilibrium
Example:
Calculate the pH of a 0.10 M solution of
acetic acid. The Ka for acetic acid is 1.8 x
10-5.
R HC2H3O2  H+ + C2H3O2I
.1
0
0
-x
+x
+x
C
E
.1- x
x
x
Ka =
[H+] [C2H3O2-]
[HC2H3O2]
[x] [x]
=
[0.1-x]
= 1.8 x 10-5
[x] [x]
-5
1.8
x
10
=
[.1-x]
x2
= 1.8 x 10-5
.1-x
x2
-5
1.8
x
10
=
.1
x2 = 1.8 x 10-6
x = 1.3 x 10-3 = [H+]
pH = -log [H+]
pH = -log(1.3 x 10-3)
pH= 2.87
Example:
Calculate the pH of a 0.25 M solution of HCN.
The Ka for HCN is 6.2 x 10-10.
R HCN  H+ + CNI
.25
0
0
C
-x
+x
+x
E .25 - x
x
x
+][CN-]
[H
[x][x]
-10
=
Ka = 6.2 x 10 =
[HCN]
.25 – x
Ka = 6.2 x 10-10 = [H+][CN-] = [x][x]
[HCN]
.25 – x
X2
= 6.2 x 10-10
.25
X2 = 1.55 x 10-10
x = 1.24 x 10-5
= [H+]
pH = -log [H+]
pH = -log(1.24 x 10-5)
pH= 4.9
• Calculate the pH of a 0.15 M solution of
ammonia. The Ka of ammonia is 1.8 x 10-9.
R
I
C
E
NH3 + H2O (l)  NH4+ + OH.15
0
0
-x
+x
+x
.15 –x
Ka = [NH4+] [OH-]
[NH3]
x
[x][x]
= .15 – x
x
Ka = [NH4+] [OH-] = [x][x]
[NH3]
.15 – x
pH = -log (H+)
X2 = 1.8 x 10-9
.15
X2 = 2.7 x 10-10
X = 0.000016= [H+]
pH = -log (0.000016)
pH = 4.78
LeChatelier's Principle
• When a stress is
applied to a system,
the equilibrium will
shift in the direction
that will relieve the
stress.
Henry Louis Le Chatelier
Changes in concentration
• An increase in concentration of:
– a reactant will cause equilibrium to shift to the right
to form more products.
– a product will cause equilibrium to shift to the left to
form more reactants.
• A decrease in concentration of:
– a product will cause equilibrium to shift to the right
to form more products.
– a reactant will cause equilibrium to shift to the left to
make more reactants.
A+BC+D
•
•
•
•
Remove A or B
Add C or D
Remove C or D
Add A or B
←
←
→
→
Changes in temperature
• Treat energy as a product or reactant and temperature
changes work just like changes in concentration!
• An increase in temperature of an exothermic reaction (H
is negative) will cause equilibrium to shift to the left. A
decrease in temperature of an exothermic reaction will
cause equilibrium to shift to the right.
• An increase in temperature of an endothermic reaction
(H is positive) will cause equilibrium to shift to the right.
A decrease in temperature of an endothermic reaction will
cause equilibrium to shift to the left.
Example:
181 kJ + N2 + O2  2NO
H = 181 kJ (endothermic)
• addition of heat
• lower temperature
Example:
2SO2 + O2  2SO3
H= -198 kJ (exothermic)
• increase temperature
• remove heat
+ 181 kJ
Example:
CaCO3 + 556 kJ  CaO + CO2
• decrease temperature
Changes in pressure
• Changes in pressure only affect equilibrium systems
having gaseous products and/or reactants.
• Increasing the pressure of a gaseous system will cause
equilibrium to shift to the side with fewer gas particles.
• Decreasing the pressure of a gaseous system will cause
equilibrium to shift to the side with more gas particles.
Addition of a catalyst
• Adding a catalyst does not affect equilibrium.
Catalysts speed up the forward and reverse reactions
equally.
Example:
P4(s) + 6Cl2(g)  4PCl3(l)
• increase container volume
– Shifts to side with more gas
• decrease container volume
– Shifts to side with less gass
• add a catalyst
N/C
– Inert gases have no effect on equilibrium
Example:
Consider the reaction:
2NO2(g)  N2(g) + 2O2(g) + HEAT
which is exothermic
• NO2 is added
• N2 is removed
• The volume is halved
N/C
• He (g) is added
• The temperature is increased
N/C
• A catalyst is added
Mechanisms
• Uni molecular
A→B
AB → A + B
A2 → 2A
• Bi molecular
A + B → AB
AB + C → AC + B
Only one thing changing at a time
142
Overall reactions
NO2 + F2  NO2F + F (slow step)
NO2 + F  NO2F (fast step)
Simplify F on both sides
2NO2 + F2  NO2F
Intermediate F
Rate = k[NO2]1[F2]1
Rate depends on both NO2 and F2
143
Overall reactions
PO2 + Cl  PClO2 (fast step)
PClO2 + PO2  P2O4 + Cl (slow step)
Simplify Cl and PClO2 on both sides
2PO2 P2O4
Intermediate PClO2
Catalyst Cl
Rate = k[PO2]2
Rate depends on only PO2
144
Oxidation Reduction Reactions
• Oxidation-reduction reactions- chemical
changes that occur when electrons are
transferred between reactants.
• Also called REDOX reactions
Oxidation Reduction Reactions
•
•
•
•
•
•
Oxidation
Modern definition - loss of electrons
Examples
4Fe + 3O2  2Fe2O3 (rusting of iron)
C + O2  CO2 (burning of carbon)
C2H5OH + 3O2  2CO2 + 3H2O (burning of
ethanol)
To help remember these definitions, use
one of these mnemonic devices:
Oxidation
Is
Loss
Reduction
Is
Gain
LEO (Lose Electrons-Oxidation)
the lion goes
GER (Gain Electrons-Reduction)
Formation of Ions
• Ex. 2Na + S  Na2S
• Sodium goes from the neutral atom to the 1+
ion. Therefore, it has lost an electron (It was
oxidized). Sulfur goes from the neutral atom
to the 2- ion. Therefore, it has gained two
electrons. (It was reduced)
Question
• Lead loses four electrons. It take on a
charge of ___. Does this mean that it is
oxidized or reduced?
REDOX Reaction Examples
• Identify the element oxidized, the element
reduced, the oxidizing agent and the
reducing agent for each of the following:
+4
-2
+1 -1
+2
-1
0
+1
MnO2 + 4HCl  MnCl2 + Cl2 + 2H2O
Mn +4
+2
Gained 2 e-
-2
Reduced
If it was Reduced, then the reactant that contains Mn acts as the “Oxidizing
Agent”. (MnO2)
Cl
-1
0
Lost 1 e-
Oxidized
If it was Oxidized, then the reactant that contains Cl acts as the “Reducing
Agent”. (HCl)
Batteries
• Electrochemical cells that
convert chemical energy into
electrical energy are called
voltaic cells.
• The energy is produced by
spontaneous redox reactions.
• Voltaic cells can be separated
into two half cells.
• A half cell consists of a metal
rod or strip immersed in a
solution of its ions.
Batteries
• We write half reactions to show what happens
in each part of the cell.
– Example Write the half reactions that occur in the
Fe2+/Ni2+ cell.
– Oxidation
Fe
 Fe2+ + 2e– Reduction
Ni2+ + 2e-  Ni
Diagram of voltaic cell for the reaction of
zinc and copper.
• Diagram of voltaic cell for the reaction of
zinc and copper.
• Oxidized: Zn
 Zn2+ + 2e• Reduced: Cu2+ + 2e-  Cu Direction of electron flow
From anode to cathode
anode
cathode
Solution of
Zinc ions
Solution of
Copper ions
Half-Cells
• The half cells are connected by a salt bridge. A salt
bridge is a tube containing a solution of ions.
• Ions pass through the salt bridge to keep the
charges balanced.
• Electrons pass through an external wire.
• The metal rods in voltaic cells are called electrodes.
• Oxidation occurs at the anode and reduction
occurs at the cathode. (An Ox and Red Cat)
• The direction of electron flow is from the anode to
the cathode. (FAT CAT )
Calculating the Charge of a Battery
• The potential charge of a battery can be
calculated with a set of values from a table of
reduction potentials.
– To do this, write the oxidation and reduction half
reactions.
– Look up the cell potentials from the data table.
– Flip the sign of the cell potential for oxidation.
– Add the potentials together.
Example A common battery is made with
nickel and cadmium. What is the cell potential
of this battery? (E0Cd = -0.40V, E0Ni = -0.25V)
Oxidation
Reduction
Cd  Cd2+ + 2eE0 = 0.40V
Ni2+ + 2e-  Ni + E0 = -0.25V
Total =
E0 = 0.15V
Spontaneous Electrochemical
Reactions
ΔG = -nFE
Recall the NiCd battery
Total =
E0 = 0.15V
Positive Voltage, -ΔG, spontaneous reaction
Negative Voltage, +ΔG, nonspontaneous reaction
157
The word, polymer, implies that polymers are constructed
from pieces (monomers) that can be easily connected into
long chains (polymer). When you look at the above shapes,
your mind should see that they could easily fit together.
There are two types of
polyethylene polymers
(plastics). One is when
the polyethylene exists as
long straight chains. The
picture here shows the
chains of one carbon with
two hydrogen atoms
repeating. The chain can
be as long as 20,000
carbons to 35,000
carbons. This is called
high density polyethylene
(HDPE).
Low density polyethylene (LDPE) is made by causing the long chains
of ethylene to branch. That way they cannot lie next each other,
which reduces the density and strength of the polyethylene. This
makes the plastic lighter and more flexible.