CHAPTER 11:
INTERMOLECULAR
FORCES AND LIQUIDS
AND SOLIDS
Chemistry 1411
Joanna Sabey
Forces
• Phase: homogeneous part of the system in contact with
other parts of the system but separated from them by a
well-defined boundary.
State
volume/shape
Density
Compressibility
Motion of molecules
Gas
Assumes the
volume and
shape of its
container
Low
Very
compressible
Very free motion
Liquid Has a definite
volume but
assumes the
shape of its
container
High
Only slightly
compressible
Slide past one
another freely
Solid
High
Virtually
Incompressible
Vibrate about fixed
positions
Has a definite
shape and
volume
Forces
• Intermolecular Forces: attractive forces between molecules
• Measure of Intermolecular forces:
• melting point
• Boiling point
• ΔH vaporization
• ΔH fusion
• ΔH sublimation
• Intramolecular Forces: holds atoms together in a molecule,
chemical bondings.
• In general intermolecular forces are weaker than
intramolecular forces.
Intermolecular Forces
• Dipole-Dipole Forces: attractive forces between polar
molecules.
Intermolecular Forces
• Ion-Dipole Forces: attract and ion and a polar molecule to
each other.
Intermolecular Forces
• Dispersion Forces: attractive forces that arise as a result of
temporary dipoles induced in atoms or molecules.
• Induced dipole: the separation of positive and negative
charges in the atom, non-polar molecule, is due to the
proximity of an ion or polar molecule.
Intermolecular Forces
• Polarizability: the ease with which the electron distribution
in the atom or molecule can be distorted. Increases with
greater number of electrons and a more diffuse electron
cloud.
• Dispersion forces usually increase with molar mass.
Intermolecular Forces
• What type(s) of intermolecular forces exist between the
following pairs?
• HBr and H2S
• Dipole-dipole and dispersion forces
• Cl2 and CBr4
• Dispersion forces
• I2 and NO3• Ion-induced dipole and dispersion forces
• NH3 and C6H6
• Dipole induced dipole forces and dispersion forces
Intermolecular Forces
• Hydrogen Bond: a special dipole-dipole interaction between
the hydrogen atom in a polar N-H, O-H, or F-H bond and an
electronegative O, N, or F atom.
Hydrogen Bonding
Hydrogen Bonding
• Which of the following can form hydrogen bonds with
water?
• CH3OCH3
• Yes
• CH4
• No
• F2
• Yes
• HCOOH
• Yes
• Na+
• No
Properties of Liquids
• Surface tension: the amount of energy required to stretch
or increase the surface of a liquid by a unit area. Strong
intermolecular forces make strong surface tension
• Cohesion: the intermolecular attraction between like
molecules.
• Adhesion: an attraction between unlike molecules.
• Viscosity: a measure of a fluid’s resistance to flow. Strong
Intermolecular forces make high viscosity solution.
Crystal Structure
• Crystalline solid: posses rigid and long-range order; its atoms,
molecules, or ions occupy specific positions.
• Amorphous solids: do not possess a well-defined arrangement
and long-range molecular order.
• Unit Cell: the basic repeating structural unit of a crystalline solid
lattice
point
At lattice points:
•
Atoms
•
Molecules
•
Ions
Seven Basic Unit Cells
Cubic Unit Cells
Atoms
Shared by 8 unit
cells
Shared by 4 unit
cells
Shared by 2 unit
cells
Closet Packing
• Closet Packing: the most efficient arrangement of spheres.
hexagonal
cubic
Structure
• Gold (Au) crystallizes in a cubic close-packed structure (the face-centered
cubic unit cell) and has a density of 19.3 g/cm3. Calculate the atomic radius of
gold in picometers.
• Calculate the Mass of Gold Au using the structure:
• Mass=(4 atoms/1unit cell)(1 mol/ 6.02X1023 atoms)(197.0 g/mol) =1.31 X 10-21 g/unit cell
• Calculate the volume of unit cell:
• D=m/V V=m/D
• V = (1.31 X 10-21 g/unit cell)/ (19.3 g/cm3) = 6.97X10-23 cm3
• Determine the edge length(a) of the cell:
• Determine the radius:
• R = 144 pm
X-Ray Diffraction
• X-ray diffraction: the scattering of X rays by the units of a
crystalline solid.
X-Ray Diffraction
• Bragg Equation:
2d sinθ = nλ
X-Ray Diffraction
• X rays of wavelength 0.154 nm strike an aluminum crystal;
the rays are reflected at an angle of 19.3°. Assuming that n =
1, calculate the spacing between the planes of aluminum
atoms (in pm) that is responsible for this angle of reflection.
The conversion factor is obtained from 1 nm = 1000 pm.
• Convert the wavelength to pm:
• 0.154nm X (1000pm/1nm) = 154 pm
• Rearrange Equation:
• 2d sinθ = nλ
• d= n λ / 2 sin θ
• Plug in values to solve for the spacing:
• d= (1)( 154 pm) / 2 sin (19.3°)
• d= 233 pm
Crystals
• Ionic Crystals: composed of charged species and the anions
and cations are generally quite different in size.
• Held together by electrostatic attraction
•
Hard, brittle, high melting point
•
Poor conductor of heat and electricity
CsCl
ZnS
CaF2
Crystals
• Covalent Crystals: atoms are held together in an extensive
three-dimensional network entirely by covalent bonds.
• Hard, high melting point
•
Poor conductor of heat and electricity
diamond
graphite
Crystals
• Molecular Crystals: the lattice points are occupied by
molecules and the attractive forces between them are van
der Waals forces and/or hydrogen bonding.
• Held together by intermolecular forces
•
Soft, low melting point
•
Poor conductor of heat and electricity
Crystals
• Metallic Crystals: every lattice point in a crystal is occupied
by an atom of the same metal.
• Soft to hard, low to high melting point
•
Good conductors of heat and electricity
nucleus &
inner shell e-
mobile “sea”
of e-
Metallic Crystals
Amorphous Solids
• Amorphous Solids: lack a regular three-dimensional
arrangement of atoms.
• Glass: An optically transparent fusion product of inorganic
materials that has cooled to a rigid state without
crystallizing.
Crystalline
quartz (SiO2)
Non-crystalline
quartz glass
Crystals
Phase Changes
Least
Order
Greatest
Order
Liquid-Vapor Equilibrium
• Equilibrium vapor pressure: the vapor pressure measured
when a dynamic equilibrium exists between condensation
and evaporation.
• Evaporation, vaporization: process in which a liquid is
transformed into a gas
• Condensation: the change from the gas phase to the liquid
phase.
• Dynamic Equilibrium: the rate of a forward process is
exactly balance by the rate of the reverse process.
Vaporization
• Molar heat of vaporization (∆Hvap): the energy required to
vaporize 1 mole of a liquid at its boiling point.
Clausius-Clapeyron Equation
ln P = -
∆Hvap
+C
RT
• P = (equilibrium) vapor pressure
• T = temperature (K)
• R = gas constant (8.314 J/K•mol)
Vapor Pressure Versus
Temperature
Vaporization
Vaporization
• Diethyl ether is a volatile, highly flammable organic liquid that is
used mainly as a solvent. The vapor pressure of diethyl ether is
401 mmHg at 18°C. Calculate its vapor pressure at 32°C.
• Plug in given values into the second equation, convert Celsius to
Kelvin.
• P= 656 mmHg
Critical Temperature and Pressure
• Critical Temperature (Tc): the temperature above which the
gas cannot be made to liquefy, no matter how great the
applied pressure
• Critical pressure (Pc): the minimum pressure that must be
applied to bring about liquefaction at the critical temperature.
T < Tc
T > Tc
T ~ Tc
T < Tc
Liquid-Solid Equilibrium
• Melting point of solid /freezing
point of a liquid: A liquid is the
temperature at which solid and
liquid phases coexist at
equilibrium.
• Molar heat of fusion (∆Hfus): the
energy required to melt 1 mole of
a solid substance at its freezing
point.
Heating Curve
Solid-Gas Equilibrium
• Molar heat of sublimation (∆Hsub):
the energy required to sublime 1
mole of a solid.
•
∆Hsub = ∆Hfus + ∆hvap
• Sublimation: the process in which
molecules go directly from the
solid into the vapor phase
• Deposition: molecules make the
transition from vapor to solid
directly.
Solid-Gas Equilibrium
• Calculate the amount of energy (in kilojoules) needed to
heat 346 g of liquid water from 0°C to 182°C. Assume that
the specific heat of water is 4.184 J/g · °C over the entire
liquid range and that the specific heat of steam is 1.99 J/g ·
°C.
• Find the heat change(q) at each stage:
• Heating of water:
• q1=msΔT
• =(346g)(4.184 J/ g X°C)(182°C-0°C) = 145 kJ
• Evaporating of water:
• q2= (346g) (1 mol /18.02 g) (40.97 kJ / mol) = 783 kJ
• Heating steam from 100°C to 182°C:
• q3= (346g)(1.99J/ J/g · °C) (182 °C- 100 °C) = 56.5 kJ
• Add the heat changes together:
• q = 145 kJ + 783 kJ + 56.5 kJ = 985 kJ
Phase Diagram
• Phase Diagram: summarizes
the conditions at which a
substance exists as a solid,
liquid, or gas.
• Triple Point: the only
condition under which all
three phases can be in
equilibrium with each other.
Phase Diagram of Water