LIQUIDS, SOLIDS, and
Intermolecular Forces and Some Proper-
ties of Liquids
Vaporization of Liquids: Vapor Pressure
Some Properties of Solids
Van der Waals Forces
Chemical Bonds as Intermolecular Forces
Energy Changes in the Formation of Ionic
I. Intermolecular Forces and some
Properties of Liquids
At high pressures and low temp, inter-
molecular forces cause gas behavior to
deviate from ideality. When these forces
are sufficiently strong, a gas condenses to a
Intermolecular Forces :
a.) Cohesive forces – intermolecular forces
between like molecules; Ex. H2O, oil.
b.) Adhesive forces – between unlike
molecules.Ex. H2O- wood and Hg-glass
Properties of Liquids:
a.) Surface Tension – the energy required
to increase the surface area of a liquid by a
unit of area.
- It results from an imbalance of intermole -
cular forces, the cohesive forces between
- represented by gamma (γ ) ; has units
energy/area , joules/m2.
- it decreases with increased temperature.
The molecules at the surface
of this sample of liquid water
are not surrounded by other
water molecules. The
molecules inside the sample
are surrounded by other
The unbalanced attraction of molecules at the surface of a
liquid tends to pull the molecules back into the bulk liquid
leaving the minimum number of molecules on the surface. It
required energy to increase the surface area of a liquid because
a larger surface area contains more molecules in the unbalanced
b.) The difference in strength between cohesive
forces and adhesive forces determine the
behavior of a liquid in contact with a solid surface.
• Water does not wet waxed surfaces because the
cohesive forces within the drops are stronger than
the adhesive forces between the drops and the wax.
• Water wets glass and spreads out on it because the
adhesive forces between the liquid and the glass are
stronger than the cohesive forces within the water.
c.) Formation of meniscus
Mercury confined in a tube
Liquid water confined in a has its surface (meniscus)
tube has its surface (menis- convex shaped because the
cus) concave shaped because cohesive forces in liquid
it wets the surface and creeps mercury tend to draw it into a
up the side. drop.
d.) Capillary Action - rise of a liquid that wets
a tube up the inside of a small diameter tube (i.e.,
a capillary) immersed in the liquid.
• The liquid creeps up the inside of the tube ( result of
adhesive forces between the liquid and the inner walls
of the tube) until the adhesive and cohesive forces of
the liquid are balanced by the weight of the liquid.
• The smaller the diameter of the tube, the higher the
e.)Viscosity - the liquid’s resistance to flow;
• Stronger intermolecular attraction, greater
viscosity. Cohesive forces within liquid creates
“internal friction” which reduces liquid flow.
Ex: honey & heavy motor oil.
• Weaker cohesive forces, low viscosity.
Ex: water & ethyl alcohol.
II. VAPORIZATION OF LIQUIDS :
• Vaporization ( Evaporation ) - passage of
molecules from the surface of the liquid into the
gaseous or vapor state; molecules have sufficiently
high energies to overcome intermolecular forces
and escape from a liquid; increases with tempe –
rature and weaker intermolecular forces.
a.) Enthalpy of Vaporization – quantity of
heat to be absorbed if certain quantity of
liquid is vaporized at constant temp.
Stated mathematically as,
ΔHvaporization = Hvapor – Hliquid
- ΔHvap is measurable (table 13.1 p.423);
positive since vaporization is endothermic.
Ex. Determining an Enthalpy of Vapor. To
vaporize 1.75 g of acetone , (CH3)2CO , at 298 K,
767 J of heat is required. What is the enthalpy of
vaporization of acetone , in kJ/ mole ?
Sol: Determine the no. of moles of acetone in 1.75 g.
Convert 767 J to kJ and divide this quantity of
heat by the moles of acetone .
767 J ---------
ΔHvap = ------------------------------------ = 25.5 kJ/mol
1.75 g (CH3)2 CO -----------
Ex 2. How much heat is reqd to vaporize a 2.35 g
sample of diethyl ether at 298 K ?
Using the data in table of enthalpies p 423 , and if
ΔHvap = ---------- , then
heat reqd = ΔHvap ( mol ) . So
(29.1 kJ / mol) ( 2.35 g )
= ------------------------------ = 0.923 kJ
74.12 g / mol
• condensation - conversion of gas to liquid .It
is the reverse of vaporization. It is opposite in
sign but equal in magnitude to ΔHvap thus ,
ΔHcon is always negative. Condensation is
b.) Vapor Pressure – the pressure exerted by a
vapor in dynamic equilibrium with its liquid.
Dynamic equilibrium – a condition at which
two opposing processes occur simultaneously
and at equal rates. ( Refer to fig 13.6 , p. 424 of
the book )
- dependent only on the particular liquid and its
temperature , not on the amount of the liquid.
- Liquids with high vapor pressures at room temp are
said to be volatile , whose intermolecular forces are
weaker. Liquids with low vapor pressure are
nonvolatile , intermolecular forces great.
c.) Boiling and the Boiling Point
Boiling – process at which the pressure exerted by
the escaping molecules of a liquid equals that of the
The normal boiling
point is the temperature at
which its vapor pressure
is 760 torr.
The boiling point of a
liquid is the temperature
at which its vapor
pressure exerted by molecules of the atmosphere;
- happens when a liquid is heated in a container open
to the atmosphere. Vaporization occurs throughout
- In boiling , temp is constant until all the liquid has
• Normal boiling point – temp at which vapor pressure
of a liquid equal to the atmospheric pressure. The
normal b.p. of several liquids can be determined
from fig 13.8 ( vapor pressure curve of liquid.)
d.) Critical Point – the point at which the liquid and
vapor pressure of a pure stable substance have
the same density, surface tension zero and
meniscus between vapor and liquid disappears
Happens if a liquid is heated in a sealed
container where boiling does not occur.
Instead , temp and vapor pressure rise
continuously. The temp at critical point ,
critical temp Tc and the pressuure is critical
pressure , Pc . ( Table 13.3)
e.) Using Vapor Pressure Data
1.) Section 6-6 concerns collecting gases
over liquids, water.
2.) Predicting whether a substance exists solely
as as a vapor or as a liquid and vapor in
equilibrium.( Refer fig 13-11, p.427 )
Ex 13.2. As a result of a chem rxn , 0.132 g H2O is
produced and maintained at a temp of 50.0oC in
closed flask of 525 - ml volume. Will the water
be present as vapor only or liquid & vapor in
Sol : First use the ideal gas eqtn to calculate the
pressure that would exist if the water were
present as vapor only.
P = nRT ÷ V
1 mol 0.08206 L atm
= 0.132 g x -------- x ---------------- x 323.3 K
18.02 g mol K
760 mm Hg
= 0.370 atm x = 281 mm Hg
Now obtain the vapor pressure of water at 50.0oC from
table 13.2. Because the calculated pressure exceeds the
data from table, some of the vapor must condense to
liquid. The water is present as liquid and vapor in equi-
librium at 92.5 mm Hg.
f.) An Equation for Expressing Vapor Pressure Data
A particularly common form of vapor pressure
eqtn is shown below, which expresses the natural loga
rithm(ln) of vapor pressure as a function of the reci –
procal of the Kelvin temp (1/T). The relationship is
that of a straight line and the plots are drawn in fig
13.13, p 429 .
ln P = - A ( 1/ T ) + B
eqnt of straight line y = m x + b
The constant A is related to the enthalpy of vapo -
rization of liquid : A = ΔHvap/ R , where ΔHvap is in
J/mol and R is 8.3145 J mol-1 K-1. It is customary
to eliminate B ( Refer to Appendix A – 4)
by rewriting the eqtn in the form called Clausius –
P2 ΔHvap 1 1
ln = --- - ---
P1 R T1 T2
Ex. Applying the Clausius – Clapeyron Eqtn.
Calculate the vapor pressure of water at 35.00C
with data from Table 13.1 and 13.2
Sol: Let P1 = unknown vapor pressure
T1 = 35.00C + 273.2 = 308.2 K
From table 13.2, choose for P2 and T2 known data at
temp close to 35.00C. At T2 = 40.00C = 313.2 K,
P2 = 55.3 mm Hg. Assuming ΔHvap is independent of
temp, its value from table 13.2 is 44.4 kJ/mol, which
we convert to J/mol. Using the Clausius – Clapeyron
55.3 mm Hg 44.0 x 103 J mol-1 1 1
Ln = - K-1
P1 8.3145 J mol-1 K-1 308.2 313.2
= 5.29 x 103 ( 0.003245 – 0.003193 ) = 0.28
Determine that e0.28 = 1.32 ( Appendix A) . Thus,
55.3 mm Hg
= e0.28 = 1.32
P1 = 55.3 mm Hg / 1.32 = 41.9 mm Hg
III. Some Properties of Solids
a.) Melting, Melting Point and Heat of Fusion
Melting(Fusion) – solid to liquid
Freezing(solidification) – liquid to solid
Melting pt of solid and freezing pt of liquid are iden-
tical. Solid and liquid coexist in equilibrium .
Enthalpy of fusion (ΔHfus )) – heat required to melt
a solid. Expressed in kJ/mol , table 13.4.
Supercooling – the condition at which the tempe -
rature may drop below the freezing point without
any solid appearing.
Supercooling – the condition at which the tempe -
rature may drop below the freezing point
without any solid appearing. Liquid must
contain some small particles ( suspended dust
particles ) on which crystals can form.; the
lesser the number of particles the crystals can
grow, it may super cool for a time before
(Refer to fig 13-14 for diagram.)
a.) Sublimation – direct passage of molecules from
solid to the vapor state. Reverse process is called
Deposition. When both processes occur, dynamic
equilibrium exists between solid and its vapor.
The vapor exerts a characteristic pressure called
Enthalpy of Sublimation – quantity of heat needed
to convert a solid to vapor. At melting point ,
sublimation is equivalent to melting followed by
ΔHsub = ΔHfus + ΔHvap