Liquids and Solids
AP Chemistry – Chapter 10
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Liquids and Solids
„ Gases are much easier to study because
molecules move independent of each other.
„ In liquids and solids forces between molecules
become very important and they differ greatly
from one substance to another substance.
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10.1 – Intermolecular Forces
„ Intramolecular forces –
„ Intermolecular forces –
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Types of Intermolecular Forces:
1) London Dispersion Forces
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Types of Intermolecular Forces:
2) Dipole-Dipole Forces
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Types of Intermolecular Forces:
2) Dipole-Dipole Forces
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Types of Intermolecular Forces:
Hydrogen Bonding
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Applications of Intermolecular
Forces:
Alcohols have ________ boiling points
compared to their alkane derivatives
CH3OH
C2H5OH
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CH4
C2H6
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Applications of Intermolecular
Forces:
• If comparing two polar substances, the one
with ____________________ typically has
the higher b.p.
Ex. HCl vs. HBr vs. HI
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Applications of Intermolecular
Forces – Covalent Hydrides
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IMF SUMMARY QUESTIONS:
1. Why is the boiling point of O2 (-183°C)
higher than N2 (-196°C)?
2. Why is the boiling point of NO (-151°C)
higher than both O2 and N2 even though it
has approximately the same molar mass?
3. What has a higher boiling point – Cl2 or ICl?
______ Why?
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IMF SUMMARY QUESTIONS:
4. Which of the following would have hydrogen
bonding?
Ethyl alcohol
Acetic acid
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dimethyl ether
acetone
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hydrazine, N2H4
IMF SUMMARY QUESTIONS:
5. What types of intermolecular forces are present
in a sample each of the following?
a. H2
b. CCl4
c. OCS
d. NH3
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10.3 – An Introduction to Structure
and Types of Solids
„ amorphous solids –
„ crystalline solids –
lattice
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10.3 – An Introduction to Structure
and Types of Solids
„ x-ray diffraction used to determine crystalline
structure
diffraction – when beams of light are scattered from a
regular array of points in which spacing between the
components are comparable with the wavelengths of
light
(due to constructive and destructive interference)
See fig. 10.10,11 pg. 433
xy + yz = nλ
xy + yz = 2d sin Ө
nλ = 2d sin Ө
Bragg Equation
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10.3 – An Introduction to Structure
and Types of Solids
xy + yz = nλ
xy + yz = 2d sin Ө
nλ = 2d sin Ө
Bragg Equation
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Types of Crystalline Solids:
Ionic Compounds
Examples - NaCl, Fe(NO3)3 - Salts - most metal compounds
Properties 1) ___________ Melting points <<not as high as network>>
- Strong forces between oppositely charges ions.
2) Non-conductor as solid
Conductor as liquid or dissolved
3) Often soluble in water
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Types of Crystalline Solids:
Ionic Compounds
Ionic Bond Strength and Lattice Energy varies by Coulomb’s
Law.
(Lattice Energy is ΔH to form solid from gaseous ions)
Coulomb’s Law:
E = k x Q1 x Q2
d
Q1 and Q2 = ion charge
d = inter nuclear distance
„ Bond strength and lattice energies increase as ion charge
_________ and ion size _____________.
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Molecular Substance Properties
(
bonded molecules)
1. Nonconductors of electricity when pure.
2. Usually insoluble in water but soluble in
nonpolar solvents such as CCl4 or benzene.
3. Volatile, with appreciable vapor pressures at
room temperature.
4. Low melting and boiling points.
* The stronger the intermolecular forces, the
___________ the boiling point.
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Atomic Solids
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Network Covalent Substances:
Examples - Diamond, Graphite, Quartz, Mica,
Asbestos, SiO2
Properties 1) _______ melting points often above
1000 °C
- Covalent bonds must be broken
2) Non-conductors
3) Insoluble in water and all common solvents
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Summary
Atomic
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X
X
X
X
X
X
X
X
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10.2 – The Liquid State
Properties:
„
low compressibility
„
lack rigidity
„
high density
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10.2 – The Liquid State
Properties:
„
low compressibility
„
lack rigidity
„
high density
„
surface tension –
„
capillary action –
meniscus
„
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viscosity –
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Summary – Types of Solids
a. CO2
g. KBr
b. SiO2
h. H2O
c. Si
i. NaOH
d. CH4
j. U
e. Ru
k. CaCO3
f. I2
l. PH3
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Summary – Types of Solids
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Summary – Types of Solids
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10.4 – Structure and Bonding in Metals
Properties –
1. durable
2. _______ b.p. ( _________ of melting points)
3. thermal/ electrical conductor
4. malleable
5. ductile
6. Insoluble in water
Bonding –
(see models below)
(accounts for physical properties!)
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10.4 – Structure and Bonding in Metals
Structure See fig. 10.14,15 pg. 437 (next slide)
1) hexagonal Closest packed (hcp) structure
Ex. Mg, Zn (Ca)
2) cubic closest packed (ccp) structure
Ex. Al, Fe, Cu, Co Ni
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10.4 – Structure and Bonding in Metals
Structure See fig. 10.13,14,15 pg. 437
1) hexagonal Closest packed (hcp) structure
Ex. Mg, Zn (Ca)
74.04% packing efficiency
aba arrangement; hexagonal prism
2) cubic closest packed (ccp) structure
Ex. Al, Fe, Cu, Co Ni
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74.04% packing efficiency
abc arrangement;
face-centered
cubic
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10.4 – Structure and Bonding in Metals
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10.4 – Structure and Bonding in Metals
Unit Cells
See fig 10.9 pg. 432
„ simple cubic (Po)
ƒ Seldom found in nature
ƒ Only 52.3% packing efficiency
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10.4 – Structure and Bonding in Metals
Unit Cells
See fig 10.9 pg. 432
„ body-centered (alkali metals, U)
ƒ 68.02% packing efficiency
ƒ 20% metals have this packing
ƒ Found for group IA metals and Barium
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10.4 – Structure and Bonding in Metals
Unit Cells
See fig 10.9 pg. 432
„ face-centered (Au)
ƒ 73.04% packing efficiency (ccp)
ƒ 40% metals have this packing
ƒ Found in Calcium and Strontium
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Unit Cell Types
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Unit Cell Types
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Counting Atoms in Unit Cells….
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10.4 – Structure and Bonding in Metals
Unit Cells
See fig 10.9 pg. 432
„ face-centered (Au)
ƒ 73.04% packing efficiency (ccp)
ƒ 40% metals have this packing
ƒ Found in Calcium and Strontium
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10.4 – Structure and Bonding in Metals
Unit Cells
See fig 10.9 pg. 432
„ body-centered (alkali metals, U)
ƒ 68.02% packing efficiency
ƒ 20% metals have this packing
ƒ Found for group IA metals and Barium
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10.4 – Structure and Bonding in Metals
Unit Cells
See fig 10.9 pg. 432
„ simple cubic (Po)
ƒ Seldom found in nature
ƒ Only 52.3% packing efficiency
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Unit Cell Types – Metal Radii
S = 2r
(4r)2 = s2 + s2
4r = s √2
4r = √s2 +s2 + s2
4r = s √3
„ Important to know the # atoms per cell &
arrangement (ex. to calc. density)
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Calculations - density
1. Titanium has a body-centered cubic unit cell.
The density of titanium is 4.50 g/cm3.
Calculate the edge length of the unit cell and
a value for the atomic radius of titanium.
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Calculations - density
2. The radius of gold is 144 pm, and the
density if 19.32 g/cm3. Does elemental gold
have a face-centered cubic structure or a
body-centered cubic structure.
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Bonding Models for Metals
„ electron sea model:
„ band model/ molecular orbital (MO) model:
electrons are assumed to travel around the
metal crystal in molecular orbitals formed
from the valence atomic orbitals of the metal
atoms
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Bonding Models for Metals
„ Li2 – two widely spaced MO Energy levels
result
„ In a metal crystal the large # of resulting
MO’s become more closely spaced and form
a virtual continuum of levels called bands
See fig. 10.19
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Bonding Models for Metals
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Alloys
Alloy – a substance that contains a mixture of
elements and has metallic properties
Substitutional alloy –
Ex. Brass (Cu & 1/3 Zn), Sterling Silver (93% Ag & 7% Cu), pewter (85%
Sn, 7% Cu, 6% Bi, 2% Sb)
Interstitial alloy –
Ex. Steel (Fe w/C)
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Conduction of Carbon and Silicon
„ Fig 10.23 Diamond vs. typical metals MO E’s
„ Graphite – has closely spaced π molecular
orbitals
„ Silicon
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Semiconductors
n-type –
p-type –
A silicon semiconductor can be doped with Arsenic (each
having one more valence e- than Si), the extra electrons
become available for conduction.
A silicon semiconductor can be doped with Boron (each
having one less valence e- than Si), leaving
vacancies/holes where e- would’ve been.
This leaves unpaired e- which can serve as conduction e-.
p-n junction –
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involved contact of a p-type & an n-type semiconductor
b) reverse bias – no current flows; the junction resists the imposed
current flow
c) foward bias – current flows; the movement of electrons and holes
is in the favored
direction.
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Semiconductors
See fig 10.31
electrons
(-)
(+)
holes
Negative
terminal
Positive
terminal
(-)
(+)
p
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n
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(+)
Positive
terminal
(-)
Negative
terminal
10.7 Ionic Solids
NOTE: Both __________ and __________ exist in a
lattice structure.
„
„
In _______________________ all atoms/ions have the
same radii.
In _______________________ there are different size
ions.Key: understand fundamental principles
governing their structures (the rest is just details).
3 hole types:
1) trigonal
2) tetrahedral
3) octahedral
trigonal < tetrahedral < octahedral
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10.7 Ionic
Solids
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10.7 Ionic Solids
* the type of hole used depends on the relative
cation and anion sizes.
Ex. ZnS
1. What type of packing do the gray atoms in (a) have? _______________
2. How many tetrahedral holes, as shown in (b), are there in this unit cell? ____
3. How many Zinc ions in the unit cell in (c)? _____ Sulfide ions? ______
4. If
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Cl- ions
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were used instead, how many of the tetrahedral holes would be used?
10.7 Ionic Solids
Key: understand
fundamental
principles
governing their
structures (the
rest is just
details).
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1. Ionic compounds have
electrical neutrality
(positive and negative
must equal).
2. Typically the larger ions
(usually the neg. ions)
are packed in hcp or
ccp, and the smaller
cations fit into holes
among the negative
ions
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10.7 Ionic Solids
This unit cell uses octahedral holes in a
________________________ cubic unit cell.
Is this more likely sodium chloride or sodium oxide?
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10.7 Ionic Solids
1. The structures of some common crystalline
substances are shown below. Show that the net
composition of each unit cell corresponds to the correct
formula of each substance.
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10.7 Ionic Solids
1. The structures of some common crystalline
substances are shown below. Show that the net
composition of each unit cell corresponds to the correct
formula of each substance.
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10.7 Ionic Solids
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10.7 Ionic Solids
* the type of hole used depends
on the relative cation and anion
sizes.
Ex. ZnS
NaCl
<<see table 10.7 pg. 458>>
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Key: understand
fundamental
principles
governing their
structures (the
rest is just
details).
10.8 – Vapor Pressure and Changes
of State
equilibrium vapor pressure:
„ the stronger the IMF, the __________
equilibrium vapor pressure
„ the higher the temperature, the __________
equilibrium vapor pressure
<<see fig. 10.41 pg. 461 & table 10.8>>
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10.8 – Vapor Pressure and Changes
of State
Patm = Pvap + PHg column
„ As long as both liquid and vapor are present,
the pressure exerted by the vapor is
independent of the volume of the container.
„ This is different than a normal gas – why???
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10.8 – Vapor Pressure and Changes
of State
PROBLEM: Given 1.00 g of H2O, at 35 °C Pvap
= 42 mm Hg
a) How much liquid water remains in a 1.00 L
flask when equilibrium is established?
b) How many grams of H2O(l) remain in a 5.0 L
flask at equilibrium?
c) How large a flask is needed to evaporate all
the liquid water?
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10.8 – Vapor Pressure and Changes
of State
PROBLEM: Given 1.00 g of H2O, at 35 °C Pvap
= 42 mm Hg
a) How much liquid water remains in a 1.00 L
flask when equilibrium is established?
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10.8 – Vapor Pressure and Changes
of State
PROBLEM: Given 1.00 g of H2O, at 35 °C Pvap
= 42 mm Hg
b) How many grams of H2O(l) remain in a 5.0 L
flask at equilibrium?
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10.8 – Vapor Pressure and Changes
of State
PROBLEM: Given 1.00 g of H2O, at 35 °C Pvap
= 42 mm Hg
c) How large a flask is needed to evaporate all
the liquid water?
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10.8 – Vapor Pressure and Changes
of State
„ The increase in equilibrium vapor pressure
with temp is a logarithmic relationship
between absolute temperature and the
molar heat of vaporization of the liquid
(ΔHvap).
ln Pvap = -ΔHvap 1 + C
R
T
Clausius-Clapeyron Equation:
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R is the ideal gas
law constant
8.31 J/mol K
10.8 – Vapor Pressure and Changes
of State
PROBLEMS:
1. Water has a Hvap = 40.7 kJ/mol and at 35.0
°C the vapor pressure = 42.2 mm Hg. Find
the vapor pressure at 55.0 °C.
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Heating Curve
Boiling point – a liquid will boil
at a temp. when the vapor
pressure of the liquid becomes
equal to the pressure above its
surface
Normal boiling point – b.p.
when Patm = 1 atm
Super cooling
Super heating
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Melting point – a substance will
melt when the solid and liquid
have the same equilibrium vapor
pressure
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(see fig. 10.45 & cases 1-3)
Phase Diagrams - Water
„ Critical temperature –
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the highest temperature
a substance can exist as
a liquid (above this
temperature it will only
exist as a gas no matter
how much pressure is
applied)
„ Critical pressure – the
pressure needed to
condense a substance at
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Phase Diagrams
„ Carbon Dioxide
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Phase diagrams
represent closed
systems and cannot
be used to explain
thing occurring in
nature.
Phase
Diagram
for Carbon
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