Lecture_Ch12_Review+Supplement_chem162pikul

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Intermolecular Attractions & the
Properties of Liquids & Solids
CHAPTER 12
Chemistry: The Molecular Nature of Matter, 6th edition
By Jesperson, Brady, & Hyslop
REVIEW
CHAPTER 12 Concept Review
Strength of Intermolecular Forces
London Dispersion Forces
Weakest
Dipole-Dipole Forces
Hydrogen Bonds
(a type of Dipole-Dipole Force)
Ion-Dipole or Ion-Induced Dipole Forces
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
Strongest
2
CHAPTER 12 Concept Review
Strength of Intermolecular Forces
London Dispersion Forces: minimized surface area
London Dispersion Forces: maximized surface area
Weakest
Dipole-Dipole Forces: small overall dipole moment
Dipole-Dipole Forces: large overall dipole moment
Hydrogen Bonds: with 1 H-bond per molecule
Hydrogen Bonds: with multiple H-bonds per molecule
Strongest
Ion-Dipole or Ion-Induced Dipole Forces
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
3
CHAPTER 12 Concept Review
Strength of Intermolecular Forces
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
4
Property of s, l, g
Boiling Point
Melting Point
Compressibility
Diffusion
Retention of V &
Shape
Surface Tension
Wetting
Viscosity
Increases
Decreases
Example
Property of s, l, g
Increases
Decreases
Boiling Point
increasing total
intermolecular forces
decreasing total
intermolecular forces
Melting Point
increasing total
intermolecular forces
decreasing total
intermolecular forces
Compressibility
Diffusion
Retention of V &
Shape
Surface Tension
Wetting
Viscosity
increasing distance
between collisions with
other particles
with increasing kinetic
energy & increased
distance between
collisions
decreasing distance
between collisions with
other particles
with deceasing kinetic
energy & decreased
distance between
collisions
Decreasing
Increasing intermolecular intermolecular forces,
forces and decreasing T & and increasing kinetic
P
energy of particles or T &
P
with increasing
intermolecular forces
with decreasing
intermolecular forces
Example
Water has a high boiling point because it has H-bonding,
dipole, and dispersion forces. It is close to heptane
(C7H16), a heavier molecule that only experiences
dispersion forces .
The melting point of ionic solids is extremely high
compared to water which experiences all other
intermolecular forces, but not ion-dipole forces. (NaCl is
1074 K and water is 273 K)
Gases are very compressible because the particles have
higher kinetic energies, and great distances between
particles.
Diffusion is much slower in a solid or in a liquid and much
faster in a gas.
Gases will fill the volume and shape of the container that
holds them, while solids will retain their own shape and
volume regardless of the container.
The molecules on the surface have less neighbors (and
therefore less stabilizing intermolecular forces) and so have a
higher potential energy, which the material will try to reduce
with its shape (sphere): water beading.
When the intermolecular Water beads on a greasy surface rather then wetting because
when there are fewer
forces in the liquid are
the dipole forces and hydrogen bonds are so much stronger
intermolecular attractions
then the dispersion forces that water experiences with the
stronger then the
to overcome in order to
surface. If the surface is clean it can experience dipole forces
intermolecular forces
interact with the surface
and hydrogen bonds with the oxygen in SiO2.
with the surface
increasing intermolecular decreasing intermolecular Not just a property of liquids, also gases and solids.
Amorphous solids change shape over time because of their
forces and decreasing
forces and decreasing
viscosity.
temperature
temperature
Phase Changes = changes of physical state with
temperature ( α to KE)
fusion
SOLID
evaporation
LIQUID
freezing
GAS
condensation
deposition
sublimation
endothermic
exothermic
7
System absorbs energy from surrounds in the form of heat
o Requires the addition of heat
System releases energy into surrounds in the form of heat or light
o Requires heat to be decreased
HEATING CURVE
gas
TEMPERATURE
l <--> g
liquid
evaporation
or vaporization
ΔHvap
endothermic
s <--> l
solid
fusion
ΔHfus
endothermic
HEAT ADDED
Equilibrium & Phase Diagrams
T1 = 78°C
P1 = 330 atm
To increase
T2 = 100°C
The system must
respond by increasing
P2 = 760 to restore
equilibrium:
o T is higher
o Volume of liquid is
lower
o P of vapor higher
9
Le Chatelier’s Principle
Liquid + Heat  Vapor
If you increase either the liquid
or the heat the reaction is
driven to the right to reestablish equilibrium.
If you increase vapor the
reaction will be driven to the
left to re-establish equilibrium.
Liquid
+ Heat  Vapor
Liquid + Heat  Vapor
Liquid + Heat 
Vapor
Liquid + Heat  Vapor
3-D Simple Cubic Lattice
Unit
Cell
Portion of lattice—
open view
11
Space filling
model
Other Cubic Lattices
Face Centered
Cubic
12
Body Centered
Cubic
Counting Atoms in Unit Cells
Site
Body
Face
Edge
Corner
Counts as Shared by X unit cells
1
1/2
1/4
1/8
1
2
4
8
Interpreting Diffraction Data
Bragg Equation
• nλ=2d sinθ
– n = integer (1, 2, …)
–  = wavelength of
X rays
– d = interplane
spacing in crystal
–  = angle of incidence
and angle of
reflectance of
X rays to various
crystal planes
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Example: Using Diffraction Data
X-ray diffraction measurements reveal that
copper crystallizes with a face-centered cubic
lattice in which the unit cell length is 362 pm.
What is the radius of a copper atom expressed
in picometers?
This is basically a geometry problem.
15
Ex. Using Diffraction Data (cont.)
Pythagorean theorem: a2 + b2 = c2
Where a = b = 362 pm sides and c = diagonal
2a2 = c2
and
c  2a 2  2a
diagonal = 2 ´ (362 pm) = 512 pm
diagonal = 4  rCu = 512 pm
rCu = 128 pm
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