Chapter 10
Liquids and Solids
1
2
Van Der Waals Forces
• These are
intermolecular
forces of attraction
between neutral
molecules.
• The Nobel Prize in
Physics 1910
(Johannes van der
Waals)
• "for his work on the
equation of state for
gases and liquids"
3
intER vs. intRA molecular forces
• Intramolecular forces are the forces
within a molecule or ionic compound
NaCl Ionic bond between atom of Na and atom of Cl
• Intermolecular forces are the forces
between molecules or ions and molecules
Example: Solid liquid gas
.
4
5
Intermolecular
Force
Model
Basis of
Attraction
Energy
(kJ/mol)
Example
600 – 40
Na+ &
H2O
Ion-dipole
Ion and polar
molecule
Dipole-dipole
Partial charges of 25 – 5
polar molecules
Hydrogen
bond
H bonded to N,
O, or F, and
another N, O, or
F
40 – 10
H2O &
NH3
London
dispersion
Induced dipoles
of polarizable
molecules
40 – 0.05
Xe &
Xe
HCl &
HCl
6
3 Types of van der Waals
Forces
• Dipole-Dipole forces
• London Dispersion forces
• Hydrogen bonding
8
DIPOLE-DIPOLE FORCES
• These are forces of attraction that occur between
polar molecules. (big difference in electron
negativity)
• These forces are effective only when polar
molecules are very close. As distance increase
strength of bond decreases.
• For molecules of approximately equal mass and size,
the strength of force of attraction increases as the
polarity increases.
• Radius have an effect on strength of dipole.
9
DIPOLE-DIPOLE FORCES
_
_
+
+
_
_
+
+
10
DIPOLE-DIPOLE FORCES
• Molecules with larger
dipole moments have
higher melting and boiling
points (hard to break)
than those with small
dipole moments.
• Dipole attractions are
relatively weak and tend
to be liquids or gas at
room temperature.
11
HYDROGEN BONDING
• A special type of dipole-dipole
interaction between the hydrogen atom
in a polar bond and an unshared
electron pair of an element that is very
electronegative usually a F, O, or N
atom on another molecule
• (note that all of these have very high
EN’s and small atomic radii).
12
13
FYI
14
HYDROGEN BONDING
• These types of
bonds are superhumanly strong.
• (4X stronger that diopole
dipole)
15
HYDROGEN BONDING IN
WATER
16
HYDROGEN BONDING
17
WHY HYDROGEN BONDING IS
EFFECTIVE
• F, O, & N are extremely small and very
electronegative atoms.
• Hydrogen atoms have no inner core of electrons,
therefore, the positive side of the bond dipole has
the concentrated charge of the partially exposed,
nearly bare proton of the nucleus
• …in other words, the atoms have a large difference
in electronegativity and their nuclei can get really
close.
18
IMPORTANCE OF HYDROGEN
BONDING
• Are important biologically, in stabilizing
proteins and keeping DNA together.
• Also explains why ice is less dense than
water (see text).
19
LONDON DISPERSION FORCES
• Fritz London
• These are forces that
arise as a result of
temporary dipoles
induced in the atoms or
molecules.( it’s a
temporary accident!)
• All molecules have
some degree of LD
forces
20
LONDON DISPERSION FORCES
• LD forces occur between neutral non-polar molecules.
(nobles gases and nonpolar compounds)
• LD forces are weak
• The greater the number of electrons the greater the LD
force. (ie the greater the melting and boiling pt.)
• LD force molecules have Low melting and boiling pts
22
See Graphic on next slide
• The motion of electrons in an atom or
molecule can create an instantaneous
dipole moment.
• EX: in a collection of He (g) the average
distribution of electrons about a nucleus is
spherical, the molecules are non-polar and
there is no attraction.
23
INSTANTANEOUS AND INDUCED
DIPOLES
Pg 454- 455 in text
24
LONDON DISPERSION FORCES
(CONT)
• These forces tend to increase in strength with an
increase in molecular weight (The size of the
molecule generally increases with mass and the
electrons are less tightly held…allows the
electron cloud to be more easily distorted.
• These forces are stronger in linear molecules
than comparable “bunched up” molecules.
25
LONDON DISPERSION FORCES
LD forces are generally
the WEAKEST
intermolecular forces.
Molecules with more
electrons will
experience more LD
forces
26
27
n-pentane vs neopentane
• BP = 309.4 K
BP = 282.7 K
• Same atomic masses different structure
28
Generalizations Regarding
Relative Strengths of IM Forces
• If molecules have comparable molecular
weights and shapes, dispersion forces are
approximately equal. Any difference in
attractive forces is due to dipole-dipole
attractions.
• If molecules differ widely in molecular
weight, dispersion forces are the decisive
factor. The most massive molecule has
the strongest attractions.
29
Because melting points (MPs) and boiling points (BPs) of
covalent molecules increase with the strengths of the forces
holding them together, it is common to use MPs and BPs as a
way to compare the strengths of intermolecular forces.
This is shown below, with the molecular formulas, molar masses
and normal BPs of the first five straight-chain hydrocarbons.
Molecular Formula
CH4
C 2H 6
C 3H 8
C4H10
C5H12
Molar Mass
16
30
44
58
72
Normal BP (C)
- 161.5
- 88.5
- 42.1
- 0.5
36.1
30
Which noble gas element has the
lowest boiling point?
He
Ne
Ar
Kr
Xe
31
The chemical forces between HCl
is/are
•
•
•
•
•
Dispersion
Covalent bond
Hydrogen bond
Dipol-dipole
Two of the above
All Molecules
Have
Not symmetrical
Polar
32
Consider the following list of
compounds. How many of these
have hydrogen bonding as their
principle IMF
HCl
NH3
CH3OH
H2S
CH4
PH3
Hydrogen Bonding is
between H and highley EN
atoms such as N, O, F, and H
33
Which of the following statements
are false or correct and why?
O2 is dipole dipole
FALSE London
Dispersion
symmetric/nonpolar
HCl is hydrogen bonding
CO2 is dipole dipole
NH3 is hydrogen
FALSE Dipole dipole not
symmetric/polar
FALSE London
Dispersion
symmetric/nonpolar
TRUE H + N,O, or F
34
ION-DIPOLE FORCES
• Attraction between an ion and the partial
charge on the end of a polar molecule.
35
36
ANSWER
37
A.Identify the types of bonds in
1. Glucose
2. Cyclohexane
B.Glucose is soluble in water but
cyclohexane is not. Why?
38
A.Identify the types of bonds in
1. Glucose
H, LD, VanderWal, Dip-dip
2. Cyclohexane
LD only
B.Glucose is soluble in water but
cyclohexane is not. Why?
Glucose is polar and cyclohexane is
nonpolar. Polar compounds are soluble
in polar solvent and visversa.
39
ION-DIPOLE FORCES
40
ION-DIPOLE FORCES (CONT)
• The magnitude of attraction increases as
either the charge of the ion increases or
magnitude of the dipole moment
increases.
• Ion-dipole forces are important in solutions
of ionic substances in polar liquids (e.g.
water)
41
ION-DIPOLE FORCES AND THE
SOLUTION PROCESS
42
43
Homework
• Pg 504-505
#’s : 35, 36, 37, 39 (you may need to read
10.1 for this part esp. LD portion)
44
10.2 Properties of liquids
• Liquids are vital to our lives.
•
•
•
•
•
Water is a means of food preparation
Cooling machines n industrial processes
Recreation
Cleaning
Transportation
45
CHARACTERISTICS OF LIQUIDS
• Surface tension
• Capillary action
• Viscosity
46
COHESIVE FORCES
• Intermolecular forces that bind like
molecules to one another (e.g. hydrogen
bonding).
47
ADHESIVE FORCES
• Intermolecular forces that bind a
substance to a surface.
48
SURFACE TENSION
• A measure of the
inward forces that
must be overcome in
order to expand the
surface area of a
liquid.
• The greater the forces
of attraction between
molecules of the
liquid, the greater the
surface tension.
49
Surface Tension Cont.
• Surface tension of a
liquid decreases with
increasing
temperature.
• The stronger the
intermolecular forces
the stronger the
surface tension.
Water has a high surface
tension do to hydrogen
bonding.
50
CAPILLARY ACTION
• Another way surface
tension manifests.
• The rise of liquids up
very narrow tubes.
This is limited by
adhesive and
cohesive forces.
51
Formation of meniscus
• Water : adhesive
forces are greater
than cohesive forces
• Mercury: Cohesive
are greater than
adhesive forces.
52
VISCOSITY
• The resistance of a liquid to flow.
• The less “tangled” a molecule is expected
to be, the less viscous it is.
Water = less Viscosity
syrup = high Viscosity
53
Viscosity Cont.
• Viscosity decreases with increasing
temperature (molecules gain kinetic
energy and can more easily overcome
forces of attraction).
• Viscosity Increases as pressure increases.
• Liquids with strong IMF have a higher
viscosity.
54
Homework
• Pg 505
#’s 43-45 all
55
10.3 Structure of Solids
• Two ways to categorize solids
– Crystalline
– Amorphous
56
Crystalline Solid
• Ridged and long range order of its atoms.
• Solids have flat surfaces
• Sharp melting points
• EX: Quartz, diamond, sodium Chloride.
57
Amorphous Solid
•
•
•
•
Lack a well defined arrangement
No long range order
IMF vary in strength
DO NOT have sharp melting points.
EX: rubber, glass
58
Unit Cell
• The smallest part of a crystal that will
reproduce the crystal when repeated in a
three dimensions.
• Three types
– Simple /primitive cubic cell
– Face centered cubic cell
– Body Centered
59
60
X-Ray Diffraction
• X-Ray crystallography
lead to the discovery
of DNA
Beams of light shot
at DNA and scattered
to reveal the double
helix structure.
Watson, Franklin, Crick
61
X-ray Diffraction
• Derived by the English physicists Sir W.H. Bragg
and his son Sir W.L. Bragg in 1913 to explain
why the faces of crystals appear to reflect X-ray
beams at certain angles of incidence (theta, θ).
• d is the distance (Ǻ) between atomic layers in a
crystal
• lambda λ is the wavelength (Ǻ) of the incident
X-ray beam
• n is the order of the diffraction
• Bragg’s Equation:
nλ = 2d sinθ
62
Bragg’s Equation Demo
63
Example
• The first order diffraction of x-rays from
crystal planes separated by 2.81 Ǻ occurs
at 11.8°.
a. what is the wavelength of the x-ray
64
The work
• n = 1 (first order)
• d = 2.81 Ǻ
• Θ = 11.8°
• λ=?
nλ = 2d sinθ
1(λ) = 2(2.81 Ǻ) sin 11.8°
Sin-1 = 24.14 °
λ = 24.14 °
65
Homework
• Pg 505
• #’s :47, 63, 64,
66
Changes of state
• Transformation from one state to another
Condensation
Vaporization
AKA: steam
67
Changes in state
• Liquid  Gas Vaporization Endothermic
• Gas  Liquid Condensation Exothermic
68
• Solid  Gas
Sublimation
Endothermic
• Gas  Solid
Deposition
Exothermic
69
• Solid  Liquid Melting
Endothermic
• Liquid  Solid Freezing
Exothermic
70
Changes of state
• The energy involved it phase changes is
calculated using
– Heat of fusion (solid  liquid or liquid solid)
– Heat of vaporization (liquid gas or gas liquid)
71
Energy Changes and Phase
Changes
Heat of Vaporization: Vaporization is an endothermic
process ( it requires heat). Energy is required to
overcome intermolecular forces to turn liq to gas.
Hvap is an Indicator of strength of IMF
CH4 = 9.2 kJ/mol
C3H8 = 18.1 kJ/mol
Larger molecule…greater IMF…greater
Hvap
72
Question
How much energy does it take to
vaporizer 111 g of water?
Given: Hvap water= 40.67 kJ/mol
111 g H2O 1 mol x 40.6kJ = 250kJ
18g
1mol
73
• Heat of Fusion: the enthalpy change
associated with melting. (Solid to liquid.)
• Hfusion water= 6.01 kJ/mol
• NOTE: heat of fusion is always smaller than
heat of vaporization. This makes sense think
about the level of “order” in the molecules in
these phases.
74
Heating Curve
• A plot of the temperature versus time
75
Heat of
Vaporization
Heat of Fusion
76
Example
Calculate the enthalpy change associated
with converting 1.00 mole of ice -25ºC to
water 150ºC at 1 atm. Specific heat of ice,
water, and steam are 2.09 J/g ºC and
4.184 J/g ºC, 1.84 J/g ºC . The heat of
fusion of ice is 6.01 kJ/mol and heat of
vaporization of water is 40.67 kJ/mol
77
Ice water
0°C
Water vapor
100°C
Vapor
q4
100°C
Water 0°C
q3
q2
ICE -25 °C
q1
78
1 mol ice  1 mol ice 1 mol water  1 mol water  1 mol steam
T=
-25ºC
qtotal =
0ºC
q1 +
0ºC
q2
+
100ºC
q3 +
100ºC
q4
1.)q = 2.09(18g)(-25-0)
2.) q = 6.02 KJ/mol (convert heat of fusion)
3.) q = 4.184(18g) (100-0)
4.) q = 40.7 KJ/mol (convert heat of
vaporization)
5. ) q = 1.84(18 g) (100-0)
79
Critical Stuff
• Critical Temperature: The temperature above
which it is impossible to liquefy the gas under
study no matter how high the applied pressure.
• Critical Pressure: The pressure required to
liquefy a gas as at its critical temperature
NOTE: the critical temp of a gas gives an indication
of the strength of the IMF of that gas. A substance
with weak attractive forces would have a low
critical temp.
80
Which gas can be liquefied at 25ºC
Gas
Critical Temp
ºC
Ammonia
132
Ethanol
Argon
Critical Temp above
25ºC
158
-186
Critical
Pressure
atm
112
78
Critical Temp
6 under
25ºC
81
Vapor Pressure (vp)
Vapor Pressure: Pressure
exerted by molecules that have
enough energy to escape the
surface.
As T ↑ VP ↑evaporation ↑
Liquids with high VP are volatile
(alcohol evaporates easily)
Liquids that have strong IMF have
low vapor pressures.
(take a lot of energy to overcome
IMF so it can evaporate)
82
%
o
f
• At higher temperature more
molecules have enough energy
• Higher vapor pressure.
M
o
l
e
c
u
l
e
s
T2
Kinetic energy
• Liquids with high VP
are volatile (alcohol
evaporates easily)
• Liquids that have
strong IMF have low
vapor pressures.
• (take a lot of energy to
overcome IMF so it can
evaporate)
substance
diethyl ether
C4H10O
vapor
pressure at
25oC
0.7 atm
Bromine
Br2
0.3 atm
ethyl alcohol
C2H5OH
0.08 atm
Water
H2 O
0.03 atm
84
Evaporation
• Molecules at the surface break away
and become gas.
• Only those with enough KE
escape
• Evaporation is a cooling
process.
• It requires heat.
• Endothermic.
Condensation
Change from gas to liquid
Achieves a dynamic equilibrium with
vaporization in a closed system.
What the heck is a
“dynamic equilibrium?”
Dynamic equilibrium
When first sealed the
molecules gradually escape
the surface of the liquid
As the molecules build up
above the liquid some
condense back to a liquid.
Dynamic equilibrium
As time goes by the rate of
vaporization remains constant
but the rate of condensation
increases because there are
more molecules to condense.
Equilibrium is reached when
Rate of Vaporization = Rate of Condensation
VP example
In a closed container the number of partials changing
from liquid  vapor will eventually equal the number
changing form vapor  liquid.
89
Boiling Point
The vapor pressure of the liquid = air pressure above the liquid
Note: The normal boiling point of water is 100oC. The term
normal refers to standard pressure or 1 atm, or also 101.3
kPa.
90
Boiling Pts. of H2O at Various
Elevations
Altitude compared
to Sea Level
(m)
Boiling
Point
(°C)
1609
98.3
177
100.3
91
How to make something boil
1. Increase the VP of the liquid (heat it) so
that the VP of the liquid is > that of the
atmosphere.
2. Lower the atmospheric pressure
(pressure above the liquid)
92
Boiling Point
↑ boiling pt by
↑ in IMF
Or
↓ VP
At high altitudes (low air pressure) water
boils at a lower temperature
93
Normal Boiling Point
• Temperature at which something boils
when the vp =1 atm
• Note the lower the external pressure the
lower the boiling point.
94
Freezing point/melting point
• They are the same but in opposite directions.
• When heated the particles vibrate more rapidly
until they shake themselves free of each other.
• Ionic solids have strong intermolecular forces so
a high mp.
• Covalent/molecular solids have weak
intermolecular forces so a low mp.
95
Phase Diagram
• A graphical way to summarize the
conditions under which equilibrium exists
between different states of matter.
• Allows you to predict the phase of a
substance that is stable at a given
temperature and pressure
96
Triple point = three phase are in equilibrium with
each other at the same time
1 atm
Boiling Point
Melting Point
Critical point
97
Critical Point: The temp beyond which the ,molecules
of a substance have to much kinetic energy to stick
together to form a liquid.
98
Not Water
Water
Phase diagrams of substances other than water the slope of
the solid liquid line slopes forward. (positive)
In water the slope of the solid-liquid lines slopes downward.
(negative)
99
Homework
• Pg 508
• #’s : 85, 87,89, 91,
100