Transcript Chapter 10

Liquids and Solids



In Chapters 8 and 9 we studied how atoms
form stable units called molecules by sharing
electrons. This is call intramolecular forces.
Now we will look to see how intermolecular
forces form attractions between condensed
states of matter (liquids and solids).
Key components to IMFs in liquids are:
Dipole-dipole attractions
 Hydrogen bonding (which isn’t really a chemical
bond)
 London dispersion forces

Forces between nonpolar and noble gases
London dispersion forces are the weakest.
Dipole-dipole interactions
occur between polar molecules
Hydrogen bonding occurs between hydrogens
bound to highly electronegative atoms like
nitrogen, oxygen and fluorine- (FON). These bonds
are very strong.





Surface tension: The resistance of a liquid to an
increase in its surface area.
Adhesive forces are forces between liquid
molecules and their containers.
Cohesive forces are the intermolecular forces
among the molecules of the liquid.
Capillary action is the spontaneous rising of a
liquid in a narrow tube.
Viscosity is the resistance to flow. (molasses is
more viscous than water)


Crystalline solids have highly regular
arrangements of their components represented
by a lattice. The smallest repeating unit of a
lattice is called a unit cell.
Amorphous solids have considerable disorder
to their structures.

An X-ray which reflects from the
surface of a substance has
travelled less distance than an Xray which reflects from a plane
of atoms inside the crystal. The
penetrating X-ray travels down
to the internal layer, reflects, and
travels back over the same
distance before being back at the
surface. The distance travelled
depends on the separation of the
layers and the angle at which the
X-ray entered the material. For
this wave to be in phase with the
wave which reflected from the
surface it needs to have travelled
a whole number of wavelengths
while inside the material. Bragg
expressed this in an equation
now known as Bragg's Law.




n = an integer (1,2,3 etc.)
θ is the angle of incident rays
and surface of the crystal
d is the spacing between the
layers of atoms.
λ is the wavelength




X-rays from a copper X-ray
tube (λ = 154 pm) were
diffracted at an angle of
14.22 degrees by a crystal
of silicon. Assuming firstorder diffraction (n=1)
what is the interplanar
spacing in the silicon?
nλ = 2d sin θ
d= nλ ÷ 2 sin θ
d = 313 pm = 3.13 x 10-10m


X-rays of wavelength
1.54 Å were used to
analyze an aluminum
crystal. A reflection
was produced at θ =
19.3 degrees.
Assuming n = 1,
calculate the distance
between the planes of
atoms producing this
reflection.
233 pm






Atomic solids: atoms at the lattice points-form from
London dispersion forces between noble gases but only
at very low temperatures—ie: solid Argon.
Molecular solid: Have lower melting and boiling
points than other solids. Ex: ice
Ionic solid: attraction of ions to one another to form
the crystal lattice. Ex: sodium chloride
Metallic solid: Held together by valence electrons. Are
malleable, ductile, shiny and good conductors of heat
and electricity. Ex: silver
Network solids: huge crystals held together by valence
electrons. Ex: ruby, diamond, amethyst
Amorphous solids: do not have a crystalline structure.
Ex: glass, plastic

A metallic crystal can be seen as containing spherical
atoms packed together and bonded to each other
equally in all directions. This is called closest packing.
There are two types, both having spheres with 12
neighbors. These arrangements are why metals have
high thermal/electrical conductivity, are malleable and
ductile.

Knowing the net number of spheres (atoms) in a unit
cell is important to applications involving solids. The
number is defined by the centers of the spheres on each
cube’s corners. 8 cubes share a given sphere so 1/8 of
this sphere lies inside each unit cell. 8 x 1/8 = 1. The
cubes at the center of each face are shared by 2 unit
cells so ½ of each lies inside a particular unit cell.
There are 6 sides so 6 x ½ or enough to make 3 whole
spheres. The net number of spheres in a face-centered
cube is (8 x 1/8) + (6 x ½ ) = 4







Density = mass ÷ volume
For a face-centered cube, calculate the density of a unit cell of silver.
Since the atoms touch along the diagonal we’ll use that length. r will be
the radius of the spheres. r + 2r + r = 4r Use the Pythagorean theorem:
d2 + d2 = (4r)2 where d is the depth of the sides. 2d2 = 16r2 , d2 = 8r2
d = √8r2 = r √8 since r = 144 pm for a Ag atom,
d = (144 pm)(√8 ) = 407 pm
The volume of the unit cell is d3 = (407pm)3 = 6.74 x 107 pm3
6.74 x 107 pm3 x (1.00 x 10-10 cm/pm)3 = 6.74 x 10-23 cm3
D = (4 atoms)(107.9 g/mol)(1 mol/6.022 x 1023 atoms)
6.74 x 10-23 cm3
= 10.6 g/cm3

Calcium has a cubic closest packed structure as a
solid. Assuming that calcium has an atomic radius
of 197 pm, calculate the density of solid calcium in
cm3.
r = 1.97 x 10-8cm
d = (1.97 x 10-8 cm)(√8) = 5.57 x10-8cm
V=d3=(5.57 x 10-8)3 = 1.73 x 10-22 cm3

D = (4 atoms)(40.08 g/mol)(1mol/6.022 x 1023atoms



1.73 x 10-22 cm3

1.54 g/cm3






Vaporization is an endothermic
process.
The heat of vaporization or
enthalpy of vaporization is the
amount of heat it takes to
vaporize 1 mole of a liquid at 1
atm. ΔHvap
Consider the graph to the right.
Which of the following would
be true?
This liquid has weaker IMF’s
than water.
The liquid’s normal boiling
point is around 85oC.
The liquid boils at room
temperature when the pressure
is dropped to about 200 mm
Hg.


Describe what is
happening to the liquid
in the sealed
Erhlenmeyer flask over
time.
Did you know that a
significant portion of
the sun’s energy that
reaches earth is spent
evaporating water from
the oceans, lakes, and
rivers rather than
warming the earth?






Patm = Pvapor + PHg
Liquids with a high vapor
pressure are volatile,
meaning they vaporize
readily.
The vapor pressure of a
liquid depends on its IMFs.
Those with high IMFs have
low vapor pressures.
Of course with increased
temperatures comes
increased vapor pressures.
Notice the nonlinear increase
in v.p. for all the liquids as
temperature increases. How
can we get a straight line?
You didn’t really think we
were done with math stuff…





T = Kelvin temperature
ΔHvap = enthalpy of
vaporization
R= universal gas
constant
R = 8.3145 J/K· mol
ln(Pvap) = natural log of
vapor pressure




y = ln(Pvap)
x = 1/T
m = slope = - ΔHvap ÷ R
b = intercept = C





Using the plots in the figure
to the right, determine
whether water, diethyl ether
or ethanol has the larger
enthalpy of vaporization.
Plot ln(Pvap) on y and 1/T on
the x.
If you plot this, then ether
ends up with a slope of -3230,
ethanol -4450 and water -4504.
All three have a negative
slope with ether having the
smaller slope. Thus ether has
a smaller ΔHvap.
Does that make sense?
Consider the hydrogen
bonding in water as you
justify your answer.


ln(Pvap) = - ΔHvap ÷ R (1/T) + C
If you know the values of ΔHvap and Pvap at one
temperature you can use the above equation to calculate
Pvap at another temperature. Since C is a constant and is
not dependent on temperature, you can write the equality:

ln(Pvap, T1) + ΔHvap1 ÷ RT1 = C = ln(Pvap, T2) + ΔHvap2 ÷ RT2

Rearrange:

ln(Pvap, T1) - ln(Pvap, T2) = ΔHvap ÷ R [ 1/T2 – 1/T1]

OR

ln [Pvap, T1 ÷ Pvap, T2] = ΔHvap ÷ R [ 1/T2 – 1/T1]
The vapor pressure of water at 25oC is 23.8 torr and the
heat of vaporization of water at 25o C is 43.9 kJ/mol.
Calculate the vapor pressure of water at 50oC.
ln[23.8torr/PvapT2] = 43,900J/8.3145 J/mol K ( 1/298K – 1/323K)
ln (23.8 torr/ PvapT2) = -1.37
Take antilog of both sides:
ln (23.8 torr/ PvapT2) = 0.254
PvapT2 = 23.8/0.254 = 93.7 torr

In Breckenridge, Colorado the typical
atmospheric pressure is 520. torr. What is the
boiling point of water in Breckenridge? ΔHvap
40.7 kJ/mol.

ln [520 torr/760 torr] = 40700J/8.3145 J/mol K(1/373K – 1/T2)

T2 = 362 K or 89.3oC
=





Carbon tetrachloride has a vapor pressure of
213 torr at 40.oC and 836 torr at 80oC. What is
the normal boiling point of CCl4
Solve for ΔHvap first.
ΔHvap = 3.1 x 104 J/mol
Using ΔHvap , 1.00 atm and 760. torr, solve for
T.
Answer: 350.K or 77oC






A-B specific heat of ice: 2.09
J/g oC
B-C heat of fusion: 334 J/g
C-D specific heat of water: 4.184
J/g oC
D-E heat of vaporization:
2260 J/g
E-F specific heat of water
vapor: 1.01 J/g oC
Calculate the amount of
energy it would take to
change the temperature of
50.0 grams of water from
-20oC to 120oC



The melting and boiling points
of a substance are determined
by the vapor pressures of the
solid and liquid phases.
Notice how when temperature
is increased, the vapor
pressure of the ice increases
more rapidly than the liquid.
When both the vapor pressure
of the solid and liquid match,
that is the melting point.

When a liquid is raised to a
temperature above its
boiling point and doesn’t
actually boil, it is super
heated. A superheated
liquid has a vapor pressure
in the liquid greater than the
atmospheric pressure.
When a bubble does form,
since the internal pressure is
greater than the external
pressure, it can burst before
rising to the surface,
blowing the surrounding
REALLY HOT liquid out of
the container.



At this temperature, the solid requires a higher
pressure than the liquid does to be in equilibrium
with the vapor. As the vapor is released from the
solid the liquid will absorb it in an attempt to
reduce the vapor pressure to its equilibrium value.
The solid will decrease, the volume of liquid will
increase and finally there will only be the liquid
which will come to equilibrium with the vapor.
The temperature must be above the melting point
of ice, since only the liquid state can exist.

This is the opposite of the last scenario. To
reach equilibrium, the liquid is going to have to
obtain a higher pressure than the solid in order
to reach equilibrium with the vapor. The liquid
will gradually disappear and the amount of ice
will increase. Finally, only the solid will
remain, which will achieve equilibrium with
the vapor. The temperature must be below the
melting point of ice since only the solid state
can exist.

In this case, the solid and liquid states have the
same vapor pressure, so they can coexist in the
apparatus at equilibrium simultaneously with
the vapor. The temperature would be at the
freezing point where both the solid and liquid
states can exist.



The melting point is the temperature at which the
solid and liquid states have the same vapor
pressure under conditions where the total pressure
is 1 atm.
The normal boiling point of a liquid is the
temperature at which the vapor pressure of the
liquid is exactly 1 atm.
Note: At temperatures where the vapor pressure
of the liquid is less than 1 atm, no bubbles of vapor
can form because the pressure on the surface of the
liquid is greater than the pressure in any spaces in
the liquid where the bubbles are trying to form.






In a closed system a phase diagram is a convenient
way of representing the phases of a substance with
respect to pressure and temperature.
Line AB is the liquid-vapor line, showing the vapor
pressure of the liquid. It represents the equilibrium
between the liquid and gas phase.
Line AC is the solid-vapor line, representing the
variation in the vapor pressure of the solid as it
sublimes at different temperatures.
Line AD is the solid-liquid line, representing the
change in melting point of the solid with increasing
pressure. This line usually slopes slightly to the right
as pressure increases, because the solid phase of a
substance is usually more dense than the liquid
phase.
The critical temperature is the point at which any
temperature above that is where the vapor cannot
be liquefied no matter how much pressure is
applied.
The critical pressure is the pressure required to
liquefy the vapor at the critical temperature.






a) Approximately what is
the normal boiling point
and what is the normal
melting point of the
substance?
300K
What is the physical state
of the substance under the
conditions in the table:
Solid
Vapor
Gas as a supercritical fluid
because you’re above the
critical pt. which is 260 atm.
and 450 K.
Temperature
Pressure
150 K
0.5 atm
325 K
0.9 atm
450 K
165 atm