Download - Properties of Liquids and Solids (Ch. 11)
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Properties of Liquids and Solids (Ch. 11)Inter vs. Intramolecular Forces
Intra ≈ strong forces within molecules (covalent or ionic bonds) (related to chemical reactivity of the substance) Inter ≈ weaker forces between molecules
(determine the bulk physical properties of a substance)
Coulomb’s Law—electrostatic repulsion of like charges is proportional to the amount of charge, and inversely proportional to the distance between them
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Phases of Matter (Review)
sublimation
deposition
vaporization(evaporation)
condensation
fusion (melting)
freezing
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Types of Intermolecular Forces• In pure substances:
– London forces (dispersion forces) -- very weak “instantaneous induced dipole” forces between molecules
– H-bonding -- especially strong dipole-dipole forces for compounds with H-F, H-O, or H-N bonds
– dipole-dipole forces -- between polar molecules (e.g. SO2, PF3)
• Forces within mixtures (in addition to the above):– ion-dipole -- between ionic and polar substances
– ion-induced dipole -- between ionic and non-polar substances
– dipole-induced dipole -- between polar and non-polar substances
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Dipole-Dipole Forces
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H-bonding-- especially strong dipole-dipole forces in compounds with H-F, H-O, or H-N bonds
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London Forces--very weak “instantaneous induced dipole” forces between non-polar molecules
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Sample QuestionWhat is the predominant intermolecular force in each of the
following substances?
a) Solid CO2
b) Liquid CH3CH2OH
c) Liquid SCl2
d) MgCl2 dissolved in liquid SCl2
e) CO2 dissolved in SCl2
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Sample QuestionWhat is the predominant intermolecular force in each of the
following substances?
a) Solid CO2 London forces
b) Liquid CH3CH2OH H-bonding
c) Liquid SCl2 dipole-dipole forces
d) MgCl2 dissolved in liquid SCl2 ion-dipole forces
e) CO2 dissolved in SCl2 dipole-induced dipole forces
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Bulk Properties of Liquids and Solids• (related to intermolecular forces)
– phase changes– compressibility– viscosity– resistance of a liquid to flow– capillary action—the ability of a liquid to flow up a narrow
tube against gravity (cohesive versus adhesive forces)– surface tension—the tendency of liquids to minimize their
surface area; the energy required to increase the surface area by a unit amount
e.g. arrange the following in order of increasing boiling points:NH3 PH3 CH4 SiH4
Answer:CH4 < SiH4 < PH3 < NH3
(lowest) (highest)
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Energy Changes during Changes of StateMolar Heat quantities (heat absorbed by 1 mole of substance)
– fusion (Hfusion) -- melting of solid to liquid– sublimation (Hsublimation) -- solid to gas– vaporization (Hvaporization) -- evaporation of liquid to gas– (all of these are affected by intermolecular forces)
e.g. Hvaporization values:H2O = 43.9 kJ/moleSO2 = 24.3 kJ/mole
Heating and Cooling Curves– Plots of temperature vs
amount of heat added (or removed)
– Horizontal regions are found during state changes (mp, bp, etc)
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Vaporization/Vapor PressureVapor Pressure (pressure due to gas above surface of liquid)
– Determined by strength of intermolecular forces – Related to surface tension– Volatile vs. nonvolatile
boiling point -- temp at which vapor pressure = atmospheric pressure
normal boiling point -- temp at which vapor pressure = 1 atm
•Rate of Vaporization– Increases with temp– Increases with increased surface area– Increases with weaker intermolecular forces
•Heat of Vaporization (Hvap)– Only slightly temperature-dependent– Decreases with weaker intermolecular forces– Not dependent on surface area
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Clausius-Clapeyron EquationRelationship between vapor pressure and temperature; vapor pressure increases with temperature, but not linear.
ln Pvap = –Hvap
RT + ln (do not memorize)
2-point form: ln P1
P2=
–Hvap
R
1
T1
1
T2
– memorize!
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Changes of StateDynamic Equilibrium
– Changes of state (e.g. evaporation/condensation) involve process of dynamic equilibrium:
Liquid Gas
at equilibrium: forward rate = reverse rate (no net change with
time) (e.g. rate of evaporation = rate of condensation)
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Le Chatelier’s Principle
When a dynamic equilibrium is upset by a stress, the system responds in a direction that tends to counteract the stress and, if possible, restore equilibrium.
e.g. vaporization/condensation
Liquid gas
If temp is raised (i.e. heat is added), the equilibrium shifts forward (to the right), hence, vapor pressure increases
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Phase Diagrams• Phase diagram -- pressure vs temperature plot that shows:
Three phase regions (solid, liquid, gas)
dividing lines (curves) -- equilibrium points between two phases
“triple point” -- T and P where all 3 phases coexist in equilibrium
“critical point” -- T and P upper limits on the liquid-gas curve
– supercritical fluid -- state of matter beyond the critical points (e.g. for H2O: critical T = 374 ºC and critical P = 218 atm)
H2O
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Crystal LatticeA regular, repeating 3-dimensional pattern in which the particles of a crystalline solid are arranged.Depending on the type of substance, the lattice sites can be occupied by atoms, molecules, or ions.
X-ray Diffraction–a technique for determining crystal lattice and molecular structures
Crystalline Solids
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Bragg Equationbasic mathematical tool of x-ray diffraction
n = 2dsin
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Crystal Lattice Typesunit cell -- simplest geometrical unit that defines the crystal
lattice
Common types of crystal lattices:Simple cubic
--Particle at 8 corners of a cube--Total of one particle per unit cell
Face-centered cubic (e.g. NaCl)--Simple cubic plus particle of each face-center of
the cube--Total of 4 particles per unit cell
Body-centered cubic--Simple cubic plus particle at center of cube--Total of 2 particles per unit cell
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Simple Cubic Lattice-- particle at 8 corners of a cube-- total of one particle per unit cell
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Face-Centered Cubic Lattice (e.g. NaCl)
-- simple cubic plus particle at center of each face of the cube-- total of 4 particles per unit cell
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Body-Centered Cubic Lattice-- simple cubic plus particle at center of cube-- total of 2 particles per unit cell
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Sample Problem• Gold forms face-centered cubic crystals in which the
edge dimension of the unit cell is 408 pm.a) Determine the atomic radius of gold (in pm). Use a rough
sketch of the face of a unit cell to illustrate your method.
b) Determine the specific gravity of gold.
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Sample Problem• Gold forms face-centered cubic crystals in which the
edge dimension of the unit cell is 408 ppm.a) Determine the atomic radius of gold (in pm). Use a rough
sketch of the face of a unit cell to illustrate your method.Answer: 144 pmb) Determine the specific gravity of gold.Answer: 19.3
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Closest-Packed Structures• Common for metals
Common types of closest-packing:hexagonal closest packing (hcp)
--hexagonal unit cell, not cubic--Layers have ABAB pattern
Cubic closest packing (ccp)--same as face-centered cubic! (fcc)--Layers have ABCABC pattern
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Hexagonal Closest Packing--hexagonal unit cell--layers are ABAB…
Top view:
Side view:
Unit cell:
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Cubic Closest Packing--same as fcc
--pattern is ABCABC…
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Types of Crystals
amorphous solids -- non-crystalline, glassy substances
Held together by electrostatic attractions
Held together by London forces, dipole-dipole, dipole-induced dipole, and/or H-bond
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Bonding in Solids -- Band TheoryAn energy “band” is composed of a very large number of
closely spaced energy levels that are formed by combining similar atomic orbitals of atoms throughout the substances.
• Metals and metalloids have a
“ conduction band”– Set of highly delocalized, partially filled, MO’s that extend
over the entire solid lattice structure
“band gap”– Energy difference between filled “valence band” and the
conduction band
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Energy Bands in Solid Li
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Band Gaps in Solids
Doping; adding trace impurities that bridge the band gap and allow some conductivityn-type: impurities have extra valence electronsp-type; impurities have fewer valence electrons
full full full
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Sample ProblemsThe heat of fusion of benzene (C6H6, 78.1 g/mole) is 9.92
kJ/mole. A 10.0 g cube of solid benzene at its melting point is placed in 25.0 g of water at 30.0 °C. Calculate the temperature of the water after all of the benzene has melted. (The specific heat of water is 4.184 J/g °C.)
CO2 gas can be prepared by the following reaction:CaCO3(s) + 2 HCl(aq) --> CaCl2(aq) + H2O(l) + CO2(g)
What volume of dry CO2 (in mL) at 20 °C and 745 torr can be prepared from a mixture of 12.3 g CaCO3 (100.1 g/mole) and 185 mL of 0.25 M HCl?
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Sample ProblemsThe heat of fusion of benzene (C6H6, 78.1 g/mole) is 9.92
kJ/mole. A 10.0 g cube of solid benzene at its melting point is placed in 25.0 g of water at 30.0 °C. Calculate the temperature of the water after all of the benzene has melted. (The specific heat of water is 4.184 J/g °C.)
Answer: 17.9 °C
CO2 gas can be prepared by the following reaction:CaCO3(s) + 2 HCl(aq) --> CaCl2(aq) + H2O(l) + CO2(g)
What volume of dry CO2 (in mL) at 20 °C and 745 torr can be prepared from a mixture of 12.3 g CaCO3 (100.1 g/mole) and 185 mL of 0.250 M HCl?
Answer: 567 mL
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More Sample ProblemsTitanium(IV) bromide, TiBr4, forms soft orange crystals that
melt at 39 °C to give a liquid that does not conduct electricity. The liquid boils at 230 °C. What is the most likely crystal type for TiBr4?
Of the following substancesPH3 CH4 H2O CO2 SO2
a) What is the predominant intermolecular force in each substance?
b) Which has the lowest heat of vaporization? c) Which is the best example of H-bonding? d) Which is often used as a supercritical fluid? e) Which should be the best solvent for NH4Cl?
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More Sample ProblemsTitanium(IV) bromide, TiBr4, forms soft orange crystals that
melt at 39 °C to give a liquid that does not conduct electricity. The liquid boils at 230 °C. What is the most likely crystal type for TiBr4?
Answer: molecular
Of the following substancesPH3 CH4 H2O CO2 SO2
a) What is the predominant intermolecular force in each substance?
b) Which has the lowest heat of vaporization? CH4
c) Which is the best example of H-bonding? H2Od) Which is often used as a supercritical fluid? CO2
e) Which should be the best solvent for NH4Cl? H2O
Answer: PH3 dipole-dipole, CH4 London, H2O H-bonding, CO2 London, SO2 dipole-dipole