CHAPTER 11: STATES OF
MATTER; LIQUIDS & SOLIDS
Vanessa N. Prasad-Permaul
Valencia Community College
CHM 1046
1
STATES OF MATTER
 In a gas, the particles of matter are far apart and
are uniformly distributed throughout the
container.
 Gases have an indefinite shape and assume the
shape of their container.
 Gases can be compressed and have an indefinite
volume.
 Gases have the most energy of the three states
of matter.
 Gases are in constant random motion (KineticMolecular Theory).
2
STATES OF MATTER
 In a liquid, the particles of matter are loosely
packed and are free to move past one another.
 Liquids have an indefinite shape and assume the
shape of their container.
 Liquids cannot be compressed and have a definite
volume.
 Liquids have less energy than gases but more
energy than solids.
 Liquids are also in constant random motion, but are
more tightly packed (less free space).
3
STATES OF MATTER
 In a solid, the particles of matter are tightly packed
together.
 Solids have a definite, fixed shape.
 Solids cannot be compressed and have a definite
volume.
 Solids have the least energy of the three states of
matter.
 Solids do NOT move about freely, but oscillate or
vibrate about fixed sites, explaining the rigidity.
4
STATES OF MATTER
5
STATES OF MATTER
RECALL that gases normally follow the
Ideal Gas Law: PV = nRT
Van der Waals Equation: (P + n2a)*(V-nb) = nRT
V2
Where: a is intermolecular forces
b is molecular size
Liquids and solids are different from gases in that
they have strong attractive forces between
particles
6
PHASE TRANSITIONS
7
PHASE TRANSITIONS
 When a solid changes to a liquid, the phase change is
called melting.
 A substance melts as the temperature increases.
H2O(s)
H2O(l)
(melting, fusion)
 When a liquid changes to a solid, the phase change is
called freezing.
 A substance freezes as the temperature decreases.
H2O(l)
H2O(s)
(freezing)
8
PHASE TRANSITIONS
 When a liquid changes to a gas, the phase change is
called vaporization.
H2O(l)
H2O(g)
(vaporization)
 A substance vaporizes as the
temperature increases.
 When a gas changes to a liquid,
the phase change is called
condensation.
H2O(l)
H2O(g)
(condensation)
 A substance condenses as the
temperature decreases.
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PHASE TRANSITIONS
When a solid changes directly to a gas, the phase
change is called sublimation.
H2O(s)
H2O(g)
(sublimation)
A substance sublimes as the
temperature increases.
When a gas changes
directly to a solid, the phase
change is called deposition.
H2O(g)
H2O(s)
(deposition)
A substance undergoes
deposition as the
temperature decreases.
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PHASE TRANSITIONS
 Physical form changes but chemical identity does not
change :
Fusion (melting)
Freezing
Vaporization
Condensation
Sublimation
solid  liquid
liquid  solid
liquid  gas
gas  liquid
solid  gas
Deposition
gas  solid
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PHASE TRANSITIONS
Vapor pressure of a liquid is the partial pressure of the
vapor over the liquid, measured at equilibrium at a given
temperature.
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PHASE TRANSITIONS
Numerical value of Vapor Pressure depends on:
a)
Magnitude of intermolecular forces
The smaller the forces the higher the vapor
pressure, loosely held molecules escape
easily
b)
Temperature
The higher the temperature, the higher the
vapor pressure, larger fraction of molecules
have sufficient kinetic energy to escape
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PHASE TRANSITIONS
The higher the temperature and the lower the
boiling point of the substance the greater the
fraction of molecules in the sample that have
sufficient kinetic energy to break free from the
surface of the liquid and escape into the vapor.
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PHASE TRANSITIONS
15
PHASE TRANSITIONS
Boiling Point: The temperature at which the vapor
pressure of liquid equals the pressure exerted on the
liquid .
H2O(l)
H2O(g)
Freezing Point: The temperature at which a pure liquid
changes to a crystalline solid, or freezes.
Solid
Liquid
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PHASE TRANSITIONS
Heat of Fusion: the heat needed for melting of a solid,
(Hfus).
H2O(s)
H2O(l) ; Hfus = 6.01kj/mol
Heat of Vaporization: the heat needed for the
evaporation of liquid, (Hvap).
H2O(s)
H2O(l) ; Hvap = 40.7kj/mol
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PHASE TRANSITIONS
HEATING CURVE FOR WATER
18
EXAMPLE 11.1: A particular refrigerator cools by
evaporating liquefied dichlorodifluoromethane. How
many kg of this liquid must be evaporated to freeze a
tray of water @ 0oC. The mass of the water is 525g, the
heat of fusion of ice is 6.01kj/mol, and the heat of
vaporization of CCl2F2 is 17.4kj/mol.
525g H2O x 1mol H2O x -6.01kj = -175kj
18.0g H2O 1 mol H2O
175kj x 1 mol CCl2F2 x 121g CCl2F2 =
1.22 x 103g CCl2F2
1.22kg of CCl2F2 must be evaporated in order to
freeze 525g of water
19
EXERCISE 11.1: The heat of vaporization of ammonia is
23.4kj/mol. How much heat is required to vaporize
1.00kg of ammonia? How many grams of water @ 0oC
could be frozen to ice at 0oC by the evaporation of this
amount of ammonia?
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PHASE TRANSITIONS
The Clausius-Clapeyron Equation: Relating vapor
pressure and liquid temperature
ln Pvap
Y
= - Hvap + C
RT
= m x + b
Where:
m is the slope
b is the y-intercept
R is the gas constant 8.314 J/K mol
C is a constant characteristic of each specific substance
T temperature in Kelvin
21
PHASE TRANSITIONS
The Clausius-Clapeyron Equation
ln P2 = Hvap
P1
R
1 - 1
T 1 T2
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PHASE TRANSITIONS
 This equation makes it possible to calculate the heat
of vaporization of a liquid by measuring its vapor
pressure at several temperatures and then plotting
the results to obtain the slope
 Once the heat of vaporization and the vapor pressure
at one temperature are known, the vapor pressure of
the liquid at any other temperature can be calculated.
23
EXAMPLE 11.2: Estimate the vapor pressure of water @
85oC. Note that the normal boiling point of water is
100oC and it’s heat of vaporization is 40.7kj/mol.
Using the Clausius-Clapeyron Equation:
ln P2 = Hvap
P1
R
ln
P2
760mmHg
1 - 1
T1
T2
=
40.7kj/mol
8.314J/(K.mol)
1 1
373K
358K
= (4898K)(-1.123x10-41/K)
= -0.550
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EXAMPLE 11.2 cont…
P2
= antiln (-0.550)
760mmHg
REMEMBER e-0.550 = 0.577
P2
= 0.577
760mmHg
P2 = 0.577 x 760mmHg
= 439mmHg
25
EXERCISE 11.2: Carbon disulfide has a normal boiling
point of 46oC and a heat of vaporization of 28.6kJ/mol.
What is the vapor pressure of CS2 at 35oC?
26
EXAMPLE 11.3: Calculate the heat of vaporization of
diethyl ether C4H10O, from the following vapor
pressures: 400mmHg @ 18oC and 760mmHg @ 35oC.
Using the Clausius-Clapeyron Equation:
ln P2 = Hvap
P1
R
1 T1
ln 760mmHg
400mmHg
=
1
T2
Hvap
1 8.314J/(K.mol) 291K
1
308K
0.642 = (-2.28 x 10-5) x Hvap/(J/mol)
Hvap = 2.82 x 104 J/mol
= 28.2kJ/mol
27
EXERCISE 11.3: Selenium tetrafluoride is a colorless
liquid. It has a vapor pressure of 757mmHg @ 105oC and
522mmHg @ 95oC. What is the heat of vaporization of
SeF4?
28
Phase Diagrams
 PHASE DIAGRAM: a graphical way to summarize the
conditions under which the different states of a
substance are stable.
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Phase Diagrams
 Shows which phase is stable at different
combinations of pressure and temperature.
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Phase Diagrams
Triple Point: The only condition under which all three
phases can be in equilibrium with one another.
Critical Temperature (Tc): The temperature at and
above which vapor of the substance cannot be
liquefied, no matter how much pressure is applied.
Critical Pressure (Pc) : The minimum pressure required
to liquefy a gas at its critical temp.
Supercritical Fluid: Neither true liquid nor true gas
Normal boiling and melting point always at 1 atm
31
EXAMPLE 11.4: The critical temperatures of ammonia
and nitrogen are 132oC and -147oC, respectively. Explain
why ammonia can be liquefied at room temperature by
merely compressing the gas to a high enough pressure,
whereas the compression of nitrogen requires a low
temperature as well.
Above the critical temperature, a gas cannot be
liquefied, so the critical temperature of NH3 is way
above room temperature (~25oC)  if NH3 is
compressed @ room temperature, it will liquefy.
Nitrogen, however, has a critical temperature well below
room temperature, so it cannot liquefy at room
temperature simply by applying pressure.
32
EXERCISE 11.4: Describe how you could liquefy the
following gases:
A. Methyl chloride (CH3Cl - critical point 144oC, 66atm)
B. Oxygen (O2 - critical point -119oC, 50atm)
33
Properties of Liquids
Surface Tension
1. The resistance of a liquid to spread out and increase
its surface area
2. Caused by differences in intermolecular forces
experienced by molecules at the surface and the
interior
3. Surface molecules feel attractive forces on only one
side and are drawn in toward the liquid
4. Interior molecules are drawn equally in all directions
5. Higher in liquids that have stronger intermolecular
forces
34
Properties of Liquids
Viscosity
1. The measure of a liquids resistance to flow
2. Related to the ease with which individual molecules
move around in the liquid and thus to the
intermolecular forces present
3. Substances with small non-polar molecules have
weak intermolecular forces and low viscosities (free
flowing)
4. More polar substances have stronger intermolecular
forces and have higher viscosities
35
Intermolecular Forces
 Van der Waals forces – intermolecular forces as a
whole, all are electrical in origin and result from
the mutual attraction of unlike charge or mutual
repulsion of like charges.
4 main types
 Dipole-dipole
 Ion-dipole
 London Dispersion forces
 Hydrogen bonding
36
Intermolecular Forces
37
Polar Molecules
 Polar Molecules
 Just as bonds can be polar, molecules as a whole
can be polar
 Net sum of individual bond polarities and lone-
pair contributions
38
Polar Covalent Bonds
 Polar Covalent Bonds
 Form between a non-metal/non-metal of different
electronegativities
EN = ENA – ENB
When EN  1.7 ionic bond
When EN < 1.7 polar covalent bond
When EN < .5 non-polar covalent bond
39
Dipole-dipole
a) Neutral but polar molecules experience dipole-dipole
forces as a result of electrical interactions among
dipoles on neighboring molecules.
b) Forces can be attractive or repulsive, depending on the
orientation of the molecules.
c) These forces are weak 3-4 kJ/mol and only significant if
molecules are close
40
HH
H
H
O
+
O Na O
O
H
H
H
H
H
O
Ion-dipole
H
Cl
H
O
H
a) Result of electrical interactions between an ion and the partial
charges on a polar molecule
b) Particularly important in aqueous solutions of ionic substances
such as NaCl, in which polar water molecules surround the ions
41
London Dispersion Forces
a) Result from the motion of electrons
b) At any given time more electrons may be in a particular area of
the molecule
c) This gives the molecule an instantaneous dipole
d) This short lived dipole can affect the electron distribution in
neighboring molecules and induce temporary dipoles in them
e) More electrons a molecule has the stronger the dispersion
forces
42
Hydrogen Bonding
a) Attractive interaction between a hydrogen atom bonded to a
very electronegative atom (O, N, F) and an unshared electron
pair on another electronegative atom
b)
Hydrogen bonds arise because O-H, N-H, and F-H bonds are
highly polar with partial positive charge on the hydrogen and
partial negative on the electronegative atom.
c)
Hydrogen has no core electrons to shield its nucleus and it is
small so it can be approached closely by other molecules
d) The dipole-dipole attraction between the hydrogen and an
unshared electron pair on a nearby atom is usually strong
43
Hydrogen Bonding
e) Water is able to form a vast 3D network of hydrogen bonds
because each H2O molecule has two hydrogens and two
electron pairs
44
INTERMOLECULAR FORCES
•
Vapor pressure depends on intermolecular forces
•
l
•
When intermolecular forces are strong in a liquid,
g depends on the attraction to other molecules
vapor pressure will be low
•
Van der Waals forces
•
•
London forces exist in every molecule
Dipole dipole forces are usually appreciable
only in small polar molecules or large
molecules having very large dipole moments
45
INTERMOLECULAR FORCES
•
London Forces increase with increasing molecular
mass
•
•
More electrons, more polarizable
With molecules of the same molecular mass, but
different arrangements (isomers)
•
•
More compact, less polarizable
London forces are smaller
Lower vapor pressure
•
Lower Hvap
•
46
EXAMPLE 11.5: What kind of intermolecular forces
(London, dipole-dipole, hydrogen bonding) are expected
in the following substances?
A. CH4, methane
B. CHCl3, tricholoromethane
C. CH3CH2CH2CH2OH, butanol
A. Methane is a nonpolar molecule so the only
intermolecular forces are the London Dispersion Forces.
B. Tricholormethane is an unsymmetrical molecule with
polar bonds, so there will be dipole-dipole and London
forces.
C. Butanol has a hydrogen atom attached to an oxygen
atom,  there will be hydrogen bonding. There will also
be dipole-dipole and London forces.
47
EXERCISE 11.5: List the different intermolecular forces
you would expect for each of the following compounds:
A. Propanol, CH3CH2CH2OH
B. Carbon Dioxide, CO2
C. Sulfur Dioxide, SO2
48
EXAMPLE 11.6: For each of the following pairs, choose
the substance you expect to have a lower vapor pressure
at a given temperature:
A. CO2 or SO2
B. Dimethyl ether CH3OCH3 or ethanol CH3CH2OH
The molecular mass of CO2 is 44amu and the molecule is
non-polar, so there are no dipole-dipole forces, only
London forces exist.
49
The molecular mass of SO2 is 64amu and the molecule
is polar, so dipole-dipole forces as well as London
forces exist.
 Sulfur dioxide has a lower vapor pressure
Experimental values confirm that the vapor pressure
of CO2 @ 20oC is 56.3atm and SO2 is 3.3atm.
50
Ethanol
Dimethyl Ether
The molecular masses are equal (46amu) so the London
forces are approximately the same. There is a strong
hydrogen bonding in ethanol  ethanol should have a
lower vapor pressure.
Experimental values confirm that the vapor pressure @
20oC is 4.88atm for dimethyl ether and 0.056atm for
Ethanol.
51
EXERCISE 11.6: Arrange the following hydrocarbons in
order of increasing vapor pressure:
Ethane, C2H6
Propane, C3H8
Butane, C4H10
52
EXERCISE 11.7: At the same temperature, methyl
chloride CH3Cl, has a vapor pressure of 1490mmHg.
Ethanol has a vapor pressure of 42mmHg. Explain why
this is to be expected.
53
SOLID STATE
Types of solids
1. Molecular solid
-held together by intermolecular forces
-H2O(s), CO2(s)
2. Metallic solid
-positively charged atomic cores
surrounded by delocalized electrons
-Fe, Cu, Ag
54
SOLID STATE
Types of solids
3. Ionic solid
-cations and anions held together by
electrical attraction of opposite charges
-NaCl
4. Covalent network solid
-atoms held together in large networks by
covalent bonds
-diamonds
55
EXAMPLE 11.7: Which of the four basic types of solids
would you expect the following substances to be?
A.
B.
C.
D.
Solid ammonia
Cesium
Cesium iodide
Silicon
A. Ammonia NH3 has a molecular structure,  it freezes as
a molecular solid.
B. Cesium (Cs) is a metal, it is a metallic solid.
C. Cesium Iodide, CsI, is n ionic substance, so it exists as an
ionic solid.
D. Silicon (Si) atoms might be expected to form covalent
bonds to other silicon atoms, as carbon does in diamond.
A covalent network solid would result.
56
EXERCISE 11.8: Classify each of the following solids
according to the forces of attraction that exist between
the structural units:
A.
B.
C.
D.
Zinc
Sodium iodide
Silicon carbide
Methane
57
SOLID STATE
PHYSICAL PROPERTIES:
 MELTING POINT AND STRUCTURE
 The stronger the intermolecular forces, the higher the
melting points
 HARDNESS AND STRUCTURE
 Molecular crystals with weak intermolecular forces are
soft compared to ionic crystals in which the attractive
forces are stronger
 Generally brittle whereas metallic crystals are malleable
 ELECTRICAL CONDUCTIVITY AND STRUCTURE
 Metals are good electrical conductors
 Covalent and ionic solids are non-conductors
58
EXAMPLE 11.9: Arrange the following elements in order
of increasing melting point: silicon, hydrogen and
lithium. Explain your reasoning.
59
EXERCISE 11.9: Decide what type of solid is formed for
each of the following substances:
C2H5OH, CH4, CH3Cl, MgSO4. On the basis of the type
of solid and the expected magnitude of intermolecular
forces (for molecular crystals), arrange these substances
in order of increasing melting point. Explain your
reasoning.
60
Example 1
Chloromethane
Cl
H
H
H
a) Calculate the dipole moment
b) Calculate the % ionic character of the bond
Experimentally measured dipole moment = 1.87 D
C-Cl bond distance = 178 pm = 178 x 10-12 m
If we assume that the contributions of the nonpolar C-H bonds
are small, then most of the chloromethane dipole
moment is due to the C-Cl bond
61
Example 2
Hydrochloric acid
Cl
H
Calculate the % ionic charcater
1. Distance between atoms is 127 pm
2. Experimentally measured dipole
moment = 1.03 D
62
Example 3
Tell which of the following compounds are likely to have
a dipole moment and show the direction of each.
a) SF6
b) CHCl3
c) CH2Cl2
d) CH2CH2
63
Example 4
Identify the likely kinds of intermolecular forces in the
following
A) HCl
H Cl
H H
B) CH3CH3
H
H
H H
H
C) CH3NH2
H
H
N
H
H
D) Kr
64
Example 5
Of the substances Ar, Cl2, CCl4 and HNO3 which has:
a) The largest dipole-dipole forces?
b) The largest hydrogen-bond forces?
c) The smallest dispersion forces?
65
Example 6
Chloroform has Hvap = 29.2 kJ/mol and Svap =
87.5 J/K mol. What is the boiling point of
chloroform?
66
Example 7
The vapor pressure of ethanol at 34.7C is 100.0 mm
Hg, and the heat of vaporization of ethanol is 38.6
kJ/mol. What is the vapor pressure of ethanol in mm
Hg at 65.0C?
67
Example 8
The normal boiling point of benzene is 80.1 C and the
heat of vaporization is 30.8 kJ/mol. What is the boiling
point of benzene (in C) on top of Mt. Everest where P
= 260 mm Hg?
68
Example 9
Can you label the following?
a) solid region
b) Liquid region
c) Gas region
d) Normal boiling point
e) Normal melting point
f) Triple point
g) Supercritical fluid region
h) Critical point, what is the
critical pressure and
temperature
69