Melting points of solids
Chapter 12
Solids are held together by IMFs that are
strong enough that the atoms and
molecules can only vibrate. However, as
temperature increases the vibrations
become more pronounced.
Eventually, the temperature reaches a
point where the vibrations are large
enough to overcome the IMFs – the solid
melts at this melting point temperature!
Intermolecular Forces: Liquids and Solids
Dr. Peter Warburton
[email protected]
http://www.chem.mun.ca/zcourses/1050.php
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Freezing points of solids
Enthalpy of fusion
Obviously the opposite process can occur.
As a liquid is cooled, the molecules loose
energy and the IMFs “take hold”, keeping
the atoms and molecules so close they
can only vibrate.
The freezing point temperature for a
substance is the same as it’s melting point
temperature.
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However, the scientific term for a solid
becoming a liquid is not “melting”, but
rather, it’s called fusion.
The amount of energy it takes to convert
one mole of a solid into a liquid is the
enthalpy of fusion ∆Hfus.
Generally, the stronger the IMFs, the
larger the enthalpy of fusion7
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Enthalpy of fusion
Phase changes occur at constant T!
As we change the phase of a substance,
all of the energy that comes into or out of
the system is directly a result of the
change of the average IMFs between the
molecules.
Therefore, none of the energy change
reflects a change in the average kinetic
energy of the molecules. During phase
changes, the temperature is CONSTANT!
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Sublimation
Phase changes occur at constant T!
It is possible to go from the solid phase
directly to the gas phase for certain
substances at certain conditions. This
process is called sublimation. The
opposite process is called deposition.
We saw examples of these processes in
the picture of iodine in a beaker on slide
46.
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Enthalpy of sublimation
Enthalpy of sublimation
The enthalpy of sublimation ∆Hsub is the
energy required to convert one mole of a
substance from the solid to the gas phase.
Also, since enthalpy is a state function, the
enthalpy of sublimation MUST be the sum
of the enthalpies of fusion and
vaporization, since going from a solid to a
liquid, then to a gas does not make a
difference to a state function!
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∆Hsub = ∆Hfus + ∆Hvap
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Phase transitions
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Clausius-Clapeyron equation
Because the sublimation pressure is a
function of intermolecular forces, just like
vapor pressure, we can use the ClausiusClapeyron equation for sublimation
pressure as a function of temperature too!
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−∆‫ܪ‬௦௨௕ 1
1
ܲଶ
=
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ܶଶ ܶଵ
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Phase diagrams
Phase diagrams
On the phase diagram
seen here, we see
there are very specific
regions of T and P
where a substance
exists in ONLY ONE
specific phase. There
also appears to be
noticeable dividing
lines between the
regions.
We can plot a graph
where the points show
the phase (state of
matter) of a given
substance as a
function of the
pressure and the
temperature.
These are called
phase diagrams.
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Phase diagrams
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Phase diagrams
The dividing lines
between the regions
represent sets of
conditions (T and P)
where two phases of
the substance can
exist in equilibrium with
each other. There’s
also a special point
where these lines
intersect!
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The point where these
lines intersect is called
a triple point. It is a
ONE very specific T
and P (the triple point
temperature and
pressure) where ALL
THREE PHASES of
the substance exist in
equilibrium with each
other!
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Phase diagram of iodine
Phase diagram of iodine
The slopes of the phase
coexistence curves are
related to the enthalpy for
the phase change being
considered. As the IMFs
get stronger, these curves
tend to become more
steep, because the
enthalpy of the phase
change increases with
IMF strength!
Here we see the phase
diagram for iodine. The
line BO is the solid-gas
coexistence line, the line
OD is the solid-liquid
coexistence line, and the
line OC is the liquid-gas
coexistence line (or
vapor pressure curve like slide 50!). Point O is
the triple point.
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Phase diagram of iodine
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Phase diagram of iodine
We see a dotted line
coming across the
graph at P = 1 atm.
If we travel across this
line starting from T = 0,
we see we hit the solidliquid coexistence line
at 113.6 °C – the
normal melting point of
iodine!
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If we keep travelling
across this P = 1 atm
line, we see we hit the
liquid-gas coexistence
line at 184.4 °C – the
normal boiling point of
iodine!
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Phase diagram of CO2
Phase diagram of CO2
Here is the phase diagram
for carbon dioxide. Here,
we see if we travel along the
P = 1 atm line from T = 0,
we see the first curve we
cross is the solid-gas
coexistence curve. At
normal pressure dry ice
sublimes to gas if we are
above the normal
sublimation temperature of
-78.5 °C.
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It turns out that if we
want to ever see liquid
carbon dioxide then we
need to be above the
triple point pressure of
5.1 atm, and above the
triple point temperature
of -56.7 °C. If we were
at 298 K, we’ll only see
liquid CO2 at pressures
greater than ~ 70 atm!
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Phase diagram of CO2
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Supercritical fluids
Supercritical fluids
often have properties
much different from the
liquid or gas phase of
the same substance.
For example, caffeine
dissolves readily in
supercritical CO2, so
we use this substance
to make decaffeinated
coffee!
There is a point C on this
diagram and on the phase
diagram for iodine we
haven’t discussed yet.
This is the critical point
where the liquid and vapor
become indistinguishable
from each other. If we are
above the critical pressure
and temperature, we get a
supercritical fluid!
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Phase diagram of water
Phase diagram of water
The next interesting
thing is that there is NOT
ONE solid form of water.
Rather, the diagram
shows us SEVEN
different forms of ice and
ELEVEN forms are
actually known!
This means there can be
solid-solid phase
transitions and
coexistence curves!
Here we see the phase
diagram for water. The
first interesting thing to
note is that the solidliquid coexistence line
OD has a negative
slope. This doesn’t
happen for too many
substances, as it means
increasing the
pressure decreases the
melting point!
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2.
As we increase temperature at a
constant pressure (an isobar), the
enthalpy increases since we are putting
heat into the system.
As we increase the pressure at constant
temperature (an isotherm) the volume
of the substance decreases, meaning
the density increases. Melting and
boiling points increase with P as well –
except for water – see slide 89!
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Problem
General features of phase diagrams
1.
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Using the phase diagram of water from the
previous slide, describe the phase
changes a sample of water will undergo if
it starts at the conditions of point R, is then
brought to point P by lowering the
temperature at constant pressure, and
then the pressure is increased at constant
temperature to get to point Q.
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Problem answer
Born-Fajans-Haber cycles
At point R the water is a gas. As the
temperature is lowered, we condense the
gas to a liquid, and then the liquid freezes
before we get to point P. At point P the
water is solid (Ice I, which is what we think
of as “ice” in our every day life). If we then
increase the pressure to get to point Q, we
cross the ice-liquid coexistence curve, so
the ice melts and we have liquid at point Q!
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Born-Fajans-Haber cycles
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Born-Fajans-Haber cycles
To get the lattice energy
of NaCl we combine a
series of steps for which
the enthalpy changes
are well known! First
we start from the
elements in their
reference states with
correct stoichiometry to
give NaCl – Na (s) and
½ Cl2
However, we can calculate the lattice
energy quite accurately anyway, because
enthalpy is a state function. It doesn’t
matter how we “make” the solid ionic
crystal. As long as our steps take us from
gas phase ions to solid phase ions, all the
energy changes add together to give us
the lattice energy.
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When ionic solids are formed from ions in
the gas phase, a large amount of energy is
released as the ions are attracted to each
other by very strong ionic bonds.
This is often called the lattice energy of the
ionic solid crystal.
Lattice energies are hard to evaluate
experimentally, though!
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Born-Fajans-Haber cycles
Born-Fajans-Haber cycles
It takes an energy input
of 107 kJ mol-1 to turn the
solid Na into Na gas (the
enthalpy of sublimation).
It then takes 122 kJ mol-1
(half of the Cl2 bond
dissociation energy - see
Table 10.3) to give us a
single Cl atom in the gas
phase from the ½ Cl2
molecule.
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Born-Fajans-Haber cycles
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Born-Fajans-Haber cycles
For Cl and an e- to
become Cl- gives off 349
kJ mol-1 of energy – the
electron affinity of Cl (see
Figure 9.11!)
Therefore, to turn the Na
and Cl atoms in the gas
phase into Na+ and Clions in the gas phase
requires the net input of
496 – 349 = 147 kJ mol-1
of energy.
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So to get to Na and Cl
atoms in the gas phase
requires adding 107 + 122
= 229 kJ mol-1 to the
elements in their standard
states. Now we need to
turn these atoms into ions.
For Na to become Na+
and an e- requires an input
of 496 kJ mol-1 of energy –
the ionization energy of
Na (see Table 9.3!)
So now the total
input of energy to
go from the
elements in their
reference states to
ions in the gas
phase is 229 +
147 = 376 kJ mol-1
of energy.
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Born-Fajans-Haber cycles
Problem
Finally, since the enthalpy
of formation of solid NaCl
is known to be -411 kJ
mol-1, then the lattice
energy must be the
difference in energy
between the gas phase
ions and this enthalpy of
formation for the solid, that
is -411 – (+376) =
The enthalpy of sublimation of cesium is
78.2 kJ mol-1 and the enthalpy of formation
of solid cesium chloride is -442.8 kJ mol-1.
With this info and other info from the text,
what is the lattice energy for solid cesium
chloride?
- 787 kJ mol-1
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lattice energy is – 669.2 kJ mol-1
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