1 chapter 10 liquids, solids, and phase changes. 2 gases, liquids and solids gases have little or no...
TRANSCRIPT
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Chapter 10
Liquids, Solids, and Phase Changes
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Gases, Liquids and Solids
•Gases have little or no interactions.
•Liquids and solids have significant interactions.
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Gases, Liquids and Solids
•Liquids and solids have well-defined volume.
•Liquid molecules “flow,” while solids are held “rigid.”
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Intermolecular Forces
Intermolecular forces are attractive forces between molecules.
Intramolecular forces hold atoms together in a molecule.
Intermolecular vs Intramolecular
• 41 kJ to vaporize 1 mole of water (inter)
• 930 kJ to break all O-H bonds in 1 mole of water (intra)
Generally, intermolecular forces are much weaker than intramolecular forces.
“Measure” of intermolecular force
boiling point
melting point
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Dipole Moments 01
•Polar covalent bonds form between atoms of different electronegativity. This is described as a bond dipole.
Intermolecular Forces
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Dipole Moments 02•Dipole Moment (µ): The measure of net molecular polarity or charge separation.
µ = Q r
r = distance between charges
+ = Q, – = –Q
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Dipole Moments: µ = Q r
• Q = Charge of electron: Q = 1.60 x 10 -19 C,• r = bond length, m• μ , dipole moments are expressed in debyes (D) where
1 D = 3.336 x 10–30 C·m What is the dipole moment if one proton separated from one electron by a distance of 100 PM
μ = Q x r = (1.60 x 10 -19 C) (100 x 10 -12 m) (1D/3.336 x 10-30 C . m) = 4.80 D
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Dipole Moments 03
•Polarity can be illustrated with an electrostatic potential map. These show electron-rich groups as red and electron-poor groups as blue-green.
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Dipole Moments 04
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Intermolecular Forces 01
•Attractive forces between molecules and ions.
•Several types of forces:
–Dipole–dipole
–Instantaneous induced dipole (dispersion forces)
–Ion–dipole
–Hydrogen “bonds.”
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Intermolecular Forces
Dipole-Dipole Forces
Attractive forces between polar molecules
Orientation of Polar Molecules in a Solid
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Intermolecular Forces
Ion-Dipole Forces
Attractive forces between an ion and a polar molecule
Ion-Dipole Interaction
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Intermolecular Forces 04
•London Dispersion Forces: Attraction is due to instantaneous, temporary dipoles formed due to electron motions.
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Intermolecular ForcesDispersion Forces Continued
Polarizability is the ease with which the electron distribution in the atom or molecule can be distorted.
Polarizability increases with:
• greater number of electrons
• more diffuse electron cloud
Dispersion forces usually increase with molar mass.
2XXXXX XXXX
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SO
O
What type(s) of intermolecular forces exist between each of the following molecules?
HBrHBr is a polar molecule: dipole-dipole forces. There are also dispersion forces between HBr molecules.
CH4
CH4 is nonpolar: dispersion forces.
SO2
SO2 is a polar molecule: dipole-dipole forces. There are also dispersion forces between SO2 molecules.
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Hydrogen Bond
•Hydrogen Bond: Molecules containing N–H, O–H, or F–H groups, and an electronegative O, N, or F.
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Why is the hydrogen bond considered a “special” dipole-dipole interaction?
Decreasing molar massDecreasing boiling point
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Which of the following molecules can hydrogen bond with itself?
• 1, 2• 2, 3• 3, 4• 1, 2, 3• 1, 2, 3, 4
CH2F2 NH3 CH3-O-H CH3C CH3
O
1 2 3 4
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Which of the following molecules can hydrogen bond with itself?
• 1, 2• 2, 3• 3, 4• 1, 2, 3• 1, 2, 3, 4
CH2F2 NH3 CH3-O-H CH3C CH3
O
1 2 3 4
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Of the following substances, predict which has the lowest boiling point based on London dispersion forces.
1. He2. Ne3. Ar4. Kr5. Xe
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Correct Answer:
More massive species have more polarizability and stronger London dispersion forces; consequently, amongst the noble gases He has the lowest boiling point.
1. He2. Ne3. Ar4. Kr5. Xe
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Of the following substances, predict which has the highest boiling point based upon intermolecular forces?
1. CH4
2. H2O3. H2S4. SiH4
5. H2Se
NH ……. O=C
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Correct Answer:
Of these, only H2O has any hydrogen bonding. Hydrogen bonding substantially increases the intermolecular forces, and hence the boiling point.
1. CH4
2. H2O3. H2S4. SiH4
5. H2Se
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With nerves as steady as a chemical bond.
© 2003 John Wiley and Sons Publishers
Courtesy Ken Karp
26Figure 13.1: “Floating” a tack on water.
© 2003 John Wiley and Sons Publishers
27Figure 13.2: Place a tack on the surface of a glass of water.
© 2003 John Wiley and Sons Publishers
Courtesy Ken Karp
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Surface Tension
Water strider walks on a pond without penetrating the surface
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Surface Tension
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Properties of Liquids
Surface tension is the amount of energy required to stretch or increase the surface of a liquid by a unit area.
Strong intermolecular
forces
High surface tension
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Viscosity
Viscosity is the measure of a liquid’s resistance to flow and is related to the ease with which molecules move around, and thus to the intermolecular forces.
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Surface Tension
Tensiometer
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Intermolecular Forces 09•Viscosity is the measure of a liquid’s resistance to flow and is related to the ease with which molecules move around, and thus to the intermolecular forces.
Another unit of viscosity is kg/m.s
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Phase Changes 01
Phase ChangesPhase Change (State Change): A change in physical form but not the chemical identity of a substance.
liquid to solid
gas to liquid
gas to solid
Freezing:
Condensation:
Deposition:
solid to liquid
liquid to gas
solid to gas
Fusion (melting):
Vaporization :
Sublimation:
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Phase Changes 02
kJ/mol (∆Hvap )(∆Hfus)
(∆Hfus) = 6.01 KJ/mol (∆Hvap ) = 40.67 KJ/mol
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Phase Changes
• Vapor Pressure: The pressure exerted by gaseous molecules above a liquid.
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Molar heat of vaporization (Hvap) is the energy required to vaporize 1 mole of a liquid.
ln P = -Hvap
RT+ C
Clausius-Clapeyron EquationP = (equilibrium) vapor pressure
T = temperature (K)
R = gas constant (8.314 J/K•mol)
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Vapor Pressure
• The boiling point of a liquid is the temperature at which its vapor pressure equals atmospheric pressure.
• The normal boiling point is the temperature at which its vapor pressure is 760 torr.
Evaporation, Vapor Pressure, and Boiling Point
+ C1
T-
Hvap
Rln Pvap =
m x b+y =
Evaporation, Vapor Pressure, and Boiling Point
Vapor Pressure
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•By taking measurements at two temps, we get:
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vap
2
1 11ln
TTR
H
PP
ln P = -Hvap
RT+ C
Clausius-Clapeyron Equation
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The normal boiling point of benzene is 80.1 °C, and ΔHvap = 30.8 kJ/mol, what is boiling point of benzene on top of Mount Everest, where P = 260 mm Hg
• P1 = 760 mm Hg; P2 = 260 mm Hg; t1 = 80.1oC, T2 = ?
• ΔHvap = 30.8 kJ/mol
• , R = 8.3145 J / K . mol
• Solve for T2 (the boiling point for benzene at 260 mm Hg).
T2 = 320 K ; t = 47oC
(boiling point is lower at lower pressure)
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vap
2
1 11ln
TTR
H
PP
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Which statement is
true? • Boiling point ~120°C• Boiling point ~95°C • Boiling point ~75°C• Melting point ~95°C• Melting point ~75°C
0
200
400
600
800
Vap
or
Pre
ssu
re (
mm
Hg
)
25 50 75 100Temperature (° C)
0
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Which statement is
true? • Boiling point ~120°C• Boiling point ~95°C • Boiling point ~75°C• Melting point ~95°C• Melting point ~75°C
0
200
400
600
800
Vap
or
Pre
ssu
re (
mm
Hg
)
25 50 75 100Temperature (° C)
0
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An amorphous solid does not possess a well-defined arrangement and long-range molecular order they do not have a fixed sharp melting point. .
A glass is an optically transparent fusion product of inorganic materials that has cooled to a rigid state without crystallizing
Crystallinequartz (SiO2)
Non-crystallinequartz glass
Solids
•Structure of a crystalline solid is based on the unit cell, a basic
• repeating structural unit.
Kinds of SolidsAmorphous Solids: Particles are randomly arranged and have no ordered long-range structure. Example - rubber.
Crystalline Solids: Particles have an ordered arrangement extending over a long range.
• ionic solids• molecular solids• covalent network solids• metallic solids
Kinds of Solids
Ionic Solids: Particles are ions ordered in a regular three-dimensional arrangement and held together by ionic bonds. Example - sodium chloride.
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Unit Cell
latticepoint
Unit cells in 3 dimensions
At lattice points:
• Atoms
• Molecules
• Ions
Simple CubicPacking
Body-Centered CubicPacking
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Crystal StructureSimple CubeSimple Cube Body-Centered Cube:Body-Centered Cube:
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Crystal Structure 04
Face-Centered Cube:Face-Centered Cube:
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Crystalline SolidsWe can determine the empirical formula of an ionic solid by determining how many ions of each element fall within the unit cell.
Hexagonal Closest Pack
A-B-A-B- Space used 74%
Cubic Closes Pack
• Space used 74%
• A-B-C-A-B-C
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Unit Cells and the Packing of Spheres in Crystalline Solids
Unit Cell: A small repeating unit that makes up a crystal.
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Types of Crystals
What are the empirical formulas for these compounds?(a) Orange:chlorine; Gray:cesium(b) Blue:sulfur; Gray: zinc(c) Green:fluorine, Gray: calcium
CsCl ZnS CaF2
(a) (b) (c)
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Types of Crystal 04
•Carbon:
Carbon Allotropes
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Types of Crystals
Covalent Crystals• Lattice points occupied by atoms• Held together by covalent bonds• Hard, high melting point
diamond graphite
carbonatoms
lattice points:
• Atoms
• Molecules
• Ions
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Types of Crystals
Molecular Crystals• Lattice points occupied by molecules• Held together by intermolecular forces• Soft, low melting point
11.6
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Types of Crystals
Metallic Crystals• Lattice points occupied by metal atoms• Held together by metallic bonds• Soft to hard, low to high melting point• Good conductors of heat and electricity
Cross Section of a Metallic Crystal
nucleus &inner shell e-
mobile “sea”of e-
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• Ionic Crystals• Covalent Crystals• Molecular Crystals• Metallic Crystals
66Body-centered cubic Face-centered cubicSimple cubic cell
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When silver crystallizes, it forms face-centered cubic cells. The unit cell edge length is 409 pm. Calculate the density of silver.
d = mV
V = a3 = (409 pm)3 = 6.83 x 10-23 cm3
4 atoms/unit cell in a face-centered cubic cell
m = 4 Ag atoms107.9 gmole Ag
x1 mole Ag
6.022 x 1023 atomsx = 7.17 x 10-22 g
d = mV
7.17 x 10-22 g6.83 x 10-23 cm3
= = 10.5 g/cm3
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X-Ray Crystallography
• Diffraction is the scattering of radiation by an object containing regularly spaced lines, with a spacing that is equivalent to the wavelength of radiation.
• Diffraction is due to interference between two waves passing through the same region of space at the same time.
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70Extra distance = BC + CD = 2d sin = n (Bragg Equation)
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X rays of wavelength 0.154 nm are diffracted from a crystal at an angle of 14.170. Assuming that n = 1, what is the distance (in pm) between layers in the crystal?
n = 2d sin n = 1 = 14.170 = 0.154 nm = 154 pm
d =n
2sin=
1 x 154 pm
2 x sin14.17= 315 pm
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Titanium metal has a density of 4.54 g/cm3, and an atomic radius of 144.8 pm. What is the structure of cubic unit cell?
• mass of one Ti atom = 7.951 x 10 -23 g/atom (problem 10.78)
• r = 144.8 pm = 144.8 x 10 -12 m• r = 144.8 x 10 -12 m = 1.448 x 10 -8 cm• Calculate the volume and then the density for Ti assuming
it is Simple (primitive) cubic, body-centered cubic, and face-centered cubic. Compare the calculated density with the actual density to identify the unit cell.
• For primitive cubic:• a = 2r; volume = a3 = [2(1.448 x 10 -8 cm)]3 = 2.429 x 10 -23
cm3
• density = m/v = 3.273 g/cm3
atoms 10 x 6.022
mol 123
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• For face-centered cubic:• a = √8. r; volume = a3 = [2√2 . (1.448 x 10- 8
cm)]3 = 6.870 x 10 -23 cm3
• density = = 4.630 g/cm3
• For body-centered cubic:• a = 4r/√3; volume = a3 = 3.739 x 10 23 cm3
• density = = 4.253 g/cm3• The calculated density for a face-centered cube
(4.630 g/cm3) is closest to the actual density of 4.54 g/cm3. Ti crystallizes in the face-centered cubic unit cell.
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Phase Diagrams 01
Water
Phase Diagrams
Normal BP: Occurs at 1 atm.
Critical Point: A combination of temperature and pressure beyond which a gas cannot be liquefied.
• Critical Temperature: The temperature beyond which a gas cannot be liquefied regardless of the pressure.
• Critical Pressure: The pressure beyond which a liquid cannot be vaporized regardless of the temperature.
Supercritical Fluid: A state of matter beyond the critical point that is neither liquid nor gas.
Triple Point: A point at which three phases coexist in equilibrium.
Supercritical CO2
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Caffeine extraction from green coffee with supercritical CO2
Application of Supercritical CO2 in dry cleaning
Phase DiagramsCarbon Dioxide
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Phase Diagrams 05
• Approximately, what is the normal boiling point and what is the normal melting point of the substance?
• What is the physical state when:i. T = 150 K, P = 0.5 atm
ii. T = 325 K, P = 0.9 atm
iii. T = 450 K, P = 265 atm
The End
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Coordination Numbers
81
82