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					   LIQUIDS, SOLIDS, and
  Intermolecular Forces and Some Proper-
     ties of Liquids
  Vaporization of Liquids: Vapor Pressure
  Some Properties of Solids
  Phase Diagrams
  Van der Waals Forces
  Hydrogen Bonding
 Chemical Bonds as Intermolecular Forces
 Crystal Structures
 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
liquid rises.
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.
 • 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 .

                                1 kJ
                      767 J ---------
                               1000 J
ΔHvap = ------------------------------------ = 25.5 kJ/mol
                                  1 mol
        1.75 g (CH3)2 CO -----------
                                 58.08 g
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

                    heat reqd
          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 equals
atmospheric pressure.
     pressure exerted by molecules of the atmosphere;
    - happens when a liquid is heated in a container open
      to the atmosphere. Vaporization occurs throughout
      the liquid.
- In boiling , temp is constant until all the liquid has
  boiled away.
    • 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
                     0.525 L
                   760 mm Hg
 = 0.370 atm x                       = 281 mm Hg
                     1 atm
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 –
 Clapeyron eqtn.
              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
equation ,
     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
  sublimation pressure.
Enthalpy of Sublimation – quantity of heat needed
to convert a solid to vapor. At melting point ,
    sublimation is equivalent to melting followed by
     vaporization.Thus ,
                ΔHsub = ΔHfus + ΔHvap

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