Upload
howard-young
View
221
Download
0
Embed Size (px)
Citation preview
Chem 106, Prof. T. L. Heise 11
CHE 106: General Chemistry
CHAPTER ELEVEN
Copyright © Tyna L. Heise 2001
All Rights Reserved
Chem 106, Prof. T. L. Heise 22
All elements can exist in three different phases of matter
– Chemical properties remain the same
– Physical properties are different
–Many of the substances to be considered in solid and liquid phase are molecular
» Virtually all liquids are molecular
» Physical properties examined are due to the forces of attractions NOT within a molecule BUT between molecules
Molecular ComparisonMolecular Comparison of Solids and Liquids of Solids and Liquids
Chapt. 11.1
Chem 106, Prof. T. L. Heise 33
Solids, Liquids, and Gases
Chapt. 11.1
Chem 106, Prof. T. L. Heise 44
Solids, Liquids, and Gases
Chapt. 11.1
Substances can be changed from one phase to another by -heating and cooling to change the average kinetic energy
-adding and decreasing pressure changes the distance between molecules which leads to an alteration in intermolecular forces
- increasing pressure pushes molecules closer together, increasing molecular attraction
- decreasing pressure allows molecules to spread farther apart, decreasing molecular attraction
Chem 106, Prof. T. L. Heise 55
Intermolecular ForcesIntermolecular Forces
The strengths of intermolecular forces of different substances vary greatly, however they are much weaker than ionic or covalent bonds:
You are not actually breaking apart a molecule, just separating molecules from each other as you move from solid to liquid and the to a gas phase.
Physical Properties reflect the IM forces
- boiling points
- melting points
Chapt. 11.2
Chem 106, Prof. T. L. Heise 66
Intermolecular ForcesIntermolecular Forces
Three types of IM forces:
dipole - dipole forces
ion - dipole forces
London dispersion forces
hydrogen bonding
Chapt. 11.2
Chem 106, Prof. T. L. Heise 77
Ion - Dipole forcesIon - Dipole forces
Important in aqueous solutions
- important between an ion and the partial charge on the end of a polar compound
negative ion attracts to +
positive ion attracts to -
Chapt. 11.2
Chem 106, Prof. T. L. Heise 88
Dipole - Dipole forcesDipole - Dipole forces
Important in aqueous solutions
- important between two dipoles
negative end attracts to +
positive end attracts to -
**for molecules of equal size, the strengths of dipole forces increases with electronegativity
Chapt. 11.2
Chem 106, Prof. T. L. Heise 99
London Dispersion forcesLondon Dispersion forces
Important between two nonpolar molecules
- important to look at instantaneous electron arrangements
- at an instant, the position of an electron can force a temporary dipole moment.
- the temporary dipole moment can induce its’ neighbor to experience a temporary dipole moment, causing an attraction
- this force is only significant when the molecules are very close to each other
Chapt. 11.2
Chem 106, Prof. T. L. Heise 1010
London Dispersion forcesLondon Dispersion forces
Chapt. 11.2
Chem 106, Prof. T. L. Heise 1111
London Dispersion forcesLondon Dispersion forces
The amount of polarizability increases with size of molecule.
- boiling point increases with molecular weight
The more contact a molecule has with another also increases the polarizability
- n-pentane has higher boiling point than neopentane due to larger available surface area
Chapt. 11.2
Chem 106, Prof. T. L. Heise 1212
Hydrogen BondingHydrogen Bonding
A special intermolecular force which exists between the H of one molecule and the O, F, or N of a neighboring polar molecule
Chapt. 11.2
Chem 106, Prof. T. L. Heise 1313
Hydrogen BondingHydrogen Bonding
Chapt. 11.2
Chem 106, Prof. T. L. Heise 1414
Intermolecular ForcesIntermolecular Forces
Chapt. 11.2
Summary: Intermolecular Forces can be determined using 1. Composition
2. Structure
Dispersion forces are in ALL molecules
•Strengths increase with increased weight, and depend on shape
Dipole-dipole forces add to dispersion forces and are found ONLY in polar molecules
Hydrogen bonds are found between the H of one molecule and the F, O, or N of another
None of these are as strong as ionic or covalent bonds, but of these, hydrogen bonding is the strongest
Chem 106, Prof. T. L. Heise 1515
Properties of LiquidsProperties of Liquids
Chapt. 11.3
Two important properties of Liquids:
Viscosity - the resistance of a liquid to flowlow viscosity flows easilyhigh viscosity flows slowly
Surface Tension - the energy required to increase the surface area of a liquid by a unit amount.
Molecules at surface all are attracted inwards instead of in all directions, thispacks surface molecules closer together
The measurement of the inward forces that must be overcome is the surface tension
Chem 106, Prof. T. L. Heise 1616
Properties of LiquidsProperties of Liquids
Chapt. 11.3
Viscosity
Chem 106, Prof. T. L. Heise 1717
Properties of LiquidsProperties of Liquids
Chapt. 11.3
Surface Tension
Chem 106, Prof. T. L. Heise 1818
Phase ChangesPhase Changes
Chapt. 11.4
Many important properties of liquids and solids relate to the ease in which they change from one state to another:
Chem 106, Prof. T. L. Heise 1919
Phase ChangesPhase Changes
Chapt. 11.4
Each phase change is accompanied by a change in energy of the system.
Energy must be added to overcome intermolecular forces and achieve a less ordered state.
Energy must be released as intermolecular forces begin to form and a molecule achieves a more ordered state
Chem 106, Prof. T. L. Heise 2020
Phase ChangesPhase Changes
Chapt. 11.4
Heating Curves: plot of the process of changing phase of substance
Heat of Fusion: the energy needed to change a solid to a liquid
Heat of Vaporization: the amount of energy needed to change a liquid to a gas
Chem 106, Prof. T. L. Heise 2121
Phase ChangesPhase Changes
Chapt. 11.4
Special Notes:
Supercooling: reducing the heat of the liquid so fast it’s temperature falls below its
freezing point , but it remains a liquid. This occurs because the liquid never has a chance to form an ordered solid.
Critical Temperature: Highest temperature at which a substance can exist as a liquid
Critical Pressure: The pressure needed to bring about liquefaction at critical temp.
Chem 106, Prof. T. L. Heise 2222
Vapor PressureVapor Pressure
Chapt. 11.5
Molecules can escape from the surface of liquid into the gas phase by vaporization or evaporation.
Chem 106, Prof. T. L. Heise 2323
Vapor PressureVapor Pressure
Chapt. 11.5
Why?
•Molecules of liquids move at various speeds
•At any instant a molecule on the surface may possess enough energy to overcome its intermolecular forces and escape into the gas phase
•The movement of molecules between liquid and gas phases goes on continuously, eventually only so many molecules can escape in a closed container and a dynamic equilibrium will be attained
Chem 106, Prof. T. L. Heise 2424
Vapor PressureVapor Pressure
Chapt. 11.5
Open Containers
•As liquid evaporates, the gas molecules move away from surface of liquid, causing the possibility of recapture by the liquid to be very small
•Equilibrium never occurs, and the vapor continues to form until the liquid is gone
•Liquids that easily evaporate due to low intermolecular forces, creating a high vapor pressure in a closed container, are considered volatile
Chem 106, Prof. T. L. Heise 2525
Vapor PressureVapor Pressure
Chapt. 11.5
Boiling Point: the temperature at which the vapor pressure equals the external pressure acting on the surface of the liquid.
Normal Boiling Point: the boiling point of a liquid at 1 atm
Chem 106, Prof. T. L. Heise 2626
Phase DiagramsPhase Diagrams
Chapt. 11.6
Chem 106, Prof. T. L. Heise 2727
Structures of SolidsStructures of Solids
Chapt. 11.7
Crystalline Solid: particles are in a well ordered arrangement
- flat surfaces or surfaces at definite angles to each other- orderly stacking of particles causes them to have orderly shapes- examples - quartz and diamonds- melts at specific temperature
Chem 106, Prof. T. L. Heise 2828
Structures of SolidsStructures of Solids
Chapt. 11.7
Amorphous Solid: particles have no orderly structure
- mixtures of molecules that do not stack well together.- composed of large complicated molecules- examples: rubber and glass- intermolecular forces vary in strength from one part of solid to another due to irregularities in solid, melting point varies
Chem 106, Prof. T. L. Heise 2929
Structures of SolidsStructures of Solids
Chapt. 11.7
Unit Cells: the repeating unit of a crystalline solid that builds the definite patterns- three dimensional array called a lattice- generally parallelpipeds, can be described using two terms, length and angle between edges
Chem 106, Prof. T. L. Heise 3030
Structures of SolidsStructures of Solids
Chapt. 11.7
Unit Cells
Chem 106, Prof. T. L. Heise 3131
Structures of SolidsStructures of Solids
Chapt. 11.7
Sample exercise: The body-centered cubic cell off a particular crystalline form of iron is 2.8664 angstroms on each side. Calculate the density of this form of iron.
Chem 106, Prof. T. L. Heise 3232
Structures of SolidsStructures of Solids
Chapt. 11.7
Sample exercise: The body-centered cubic cell off a particular crystalline form of iron is 2.8664 angstroms on each side. Calculate the density of this form of iron.
1) D = m V
Chem 106, Prof. T. L. Heise 3333
Structures of SolidsStructures of Solids
Chapt. 11.7
Sample exercise: The body-centered cubic cell off a particular crystalline form of iron is 2.8664 angstroms on each side. Calculate the density of this form of iron.
1) D = m V
Mass = 55.845 amu
Chem 106, Prof. T. L. Heise 3434
Structures of SolidsStructures of Solids
Chapt. 11.7
Sample exercise: The body-centered cubic cell off a particular crystalline form of iron is 2.8664 angstroms on each side. Calculate the density of this form of iron.
1) D = m V
Mass = 55.845 amu 1 g 6.02 x 1023 amu
Chem 106, Prof. T. L. Heise 3535
Structures of SolidsStructures of Solids
Chapt. 11.7
Sample exercise: The body-centered cubic cell of a particular crystalline form of iron is 2.8664 angstroms on each side. Calculate the density of this form of iron.
1) D = m V
Mass = 2(55.8) amu 1 g = 1.85 x 10-22 g 6.02 x 1023 amu
Chem 106, Prof. T. L. Heise 3636
Structures of SolidsStructures of Solids
Chapt. 11.7
Sample exercise: The body-centered cubic cell of a particular crystalline form of iron is 2.8664 angstroms on each side. Calculate the density of this form of iron.
1) D = m V 3
Volume = (2.8664 angstroms) (10-10 m) 1 angstroms
Chem 106, Prof. T. L. Heise 3737
Structures of SolidsStructures of Solids
Chapt. 11.7
Sample exercise: The body-centered cubic cell of a particular crystalline form of iron is 2.8664 angstroms on each side. Calculate the density of this form of iron.
1) D = m V 3
Volume = (2.8664 angstroms) (10-10 m) = 2.36 x 10-29
1 angstroms
Chem 106, Prof. T. L. Heise 3838
Structures of SolidsStructures of Solids
Chapt. 11.7
Sample exercise: The body-centered cubic cell of a particular crystalline form of iron is 2.8664 angstroms on each side. Calculate the density of this form of iron.
1) D = m V
Volume = 2.36 x 10-29 m3 1 L
10-3 m3
Chem 106, Prof. T. L. Heise 3939
Structures of SolidsStructures of Solids
Chapt. 11.7
Sample exercise: The body-centered cubic cell of a particular crystalline form of iron is 2.8664 angstroms on each side. Calculate the density of this form of iron.
1) D = m V
Volume = 2.36 x 10-29 m3 1 L = 2.36 x 10-26 L
10-3 m3
Chem 106, Prof. T. L. Heise 4040
Structures of SolidsStructures of Solids
Chapt. 11.7
Sample exercise: The body-centered cubic cell of a particular crystalline form of iron is 2.8664 ≈ on each side. Calculate the density of this form of iron.
1) D = m = 1.85 x 10-22 g V 2.36 x 10-26 L
Chem 106, Prof. T. L. Heise 4141
Structures of SolidsStructures of Solids
Chapt. 11.7
Sample exercise: The body-centered cubic cell of a particular crystalline form of iron is 2.8664 ≈ on each side. Calculate the density of this form of iron.
1) D = m = 1.85 x 10-22 g = 7,838 g V 2.36 x 10-26 L L
Chem 106, Prof. T. L. Heise 4242
Structures of SolidsStructures of Solids
Chapt. 11.7
Close packing of spheres: structures adopted by crystalline solids are those that bring particles in closest contact to maximize intermolecular attractions
* Each sphere has 12 neighbors, known as the coordination number
Chem 106, Prof. T. L. Heise 4343
Structures of SolidsStructures of Solids
Chapt. 11.7
Sample exercise: Aluminum metal crystallizes in a cubic close-packed structure (face-centered cell)
(a) How many aluminum atoms are in the unit cell
Chem 106, Prof. T. L. Heise 4444
Structures of SolidsStructures of Solids
Chapt. 11.7
Sample exercise: Aluminum metal crystallizes in a cubic close-packed structure (face-centered cell)
(a) How many aluminum atoms are in the unit cell
1/2 atom at each of 6 faces = 3 atoms
1/8 atom at each of 8 corners = 1 atom
4 atoms total
Chem 106, Prof. T. L. Heise 4545
Structures of SolidsStructures of Solids
Chapt. 11.7
Sample exercise: Aluminum metal crystallizes in a cubic close-packed structure (face-centered cell)
(b) What is the coordination number of each aluminum atom?
Chem 106, Prof. T. L. Heise 4646
Structures of SolidsStructures of Solids
Chapt. 11.7
Sample exercise: Aluminum metal crystallizes in a cubic close-packed structure (face-centered cell)
(b) What is the coordination number of each aluminum atom?
12
Chem 106, Prof. T. L. Heise 4747
Structures of SolidsStructures of Solids
Chapt. 11.7
Sample exercise: Aluminum metal crystallizes in a cubic close-packed structure (face-centered cell)
(c) If each aluminum atom has a radius of 1.43 angstroms, what is the length of each side?
Chem 106, Prof. T. L. Heise 4848
Structures of SolidsStructures of Solids
Chapt. 11.7
Sample exercise: Aluminum metal crystallizes in a cubic close-packed structure (face-centered cell)
(c) If each aluminum atom has a radius of 1.43 angstroms, what is the length of each side?
4 radius on diagonal=4(1.43 angstroms)= 5.72 angstroms
Chem 106, Prof. T. L. Heise 4949
Structures of SolidsStructures of Solids
Chapt. 11.7
Sample exercise: Aluminum metal crystallizes in a cubic close-packed structure (face-centered cell)
(c) If each aluminum atom has a radius of 1.43 A, what is the length of each side?
4 radius on diagonal = 4(1.43 A) = 5.72 A
a2 + b2 = c2
2a2 = c22a2 = (5.72 A)2
2a2 = 32.72 A a2 = 16.36 A
a = 4.04 A
Chem 106, Prof. T. L. Heise 5050
Structures of SolidsStructures of Solids
Chapt. 11.7
Sample exercise: Aluminum metal crystallizes in a cubic close-packed structure (face-centered cell)
(d) Calculate the density of the of aluminum metal
Chem 106, Prof. T. L. Heise 5151
Structures of SolidsStructures of Solids
Chapt. 11.7
Sample exercise: Aluminum metal crystallizes in a cubic close-packed structure (face-centered cell)
(d) Calculate the density of the of aluminum metal
1) D = m V
Chem 106, Prof. T. L. Heise 5252
Structures of SolidsStructures of Solids
Chapt. 11.7
Sample exercise: Aluminum metal crystallizes in a cubic close-packed structure (face-centered cell)
(d) Calculate the density of the of aluminum metal
1) D = m V
Mass = 26.98 amu
Chem 106, Prof. T. L. Heise 5353
Structures of SolidsStructures of Solids
Chapt. 11.7
Sample exercise: Aluminum metal crystallizes in a cubic close-packed structure (face-centered cell)
(d) Calculate the density of the of aluminum metal
1) D = m V
Mass = 26.98 amu 1 g 6.02 x 1023 amu
Chem 106, Prof. T. L. Heise 5454
Structures of SolidsStructures of Solids
Chapt. 11.7
Sample exercise: Aluminum metal crystallizes in a cubic close-packed structure (face-centered cell)
(d) Calculate the density of the of aluminum metal
1) D = m V
Mass = 4(26.98) amu 1 g = 1.79 x 10-22 g 6.02 x 1023 amu
Chem 106, Prof. T. L. Heise 5555
Structures of SolidsStructures of Solids
Chapt. 11.7
Sample exercise: Aluminum metal crystallizes in a cubic close-packed structure (face-centered cell)
(d) Calculate the density of the of aluminum metal
1) D = m V 3
Volume = (4.04 A) (10-10 m) 1 L 1 A 10-3 m3
Chem 106, Prof. T. L. Heise 5656
Structures of SolidsStructures of Solids
Chapt. 11.7
Sample exercise: Aluminum metal crystallizes in a cubic close-packed structure (face-centered cell)
(d) Calculate the density of the of aluminum metal
1) D = m V
Volume = 6.59 x 10-26 L
Chem 106, Prof. T. L. Heise 5757
Structures of SolidsStructures of Solids
Chapt. 11.7
Sample exercise: Aluminum metal crystallizes in a cubic close-packed structure (face-centered cell)
(d) Calculate the density of the of aluminum metal
1) D = m = 1.79 x 10-22 g V 6.59 x 10-26 L
Chem 106, Prof. T. L. Heise 5858
Structures of SolidsStructures of Solids
Chapt. 11.7
Sample exercise: Aluminum metal crystallizes in a cubic close-packed structure (face-centered cell)
(d) Calculate the density of the of aluminum metal
1) D = m = 1.79 x 10-22 g = 2,720 g/L
V 6.59 x 10-26 L
Chem 106, Prof. T. L. Heise 5959
Bonding in SolidsBonding in Solids
Chapt. 11.7
The physical properties of crystalline solids, such as melting point and hardness, depend both on the arrangements off particles and on the attractive forces between particles in solids.
4 Types : Molecular Covalent & Network Ionic Metallic
Chem 106, Prof. T. L. Heise 6060
Bonding in SolidsBonding in Solids
Chapt. 11.7
Molecular - made up of atoms or molecules - London dispersion forces, dipole-dipole
forces, hydrogen bonding - soft, low to moderately high melting point,
poor thermal and electrical conductivity
Network Covalent - atoms connected in networks - covalent bonding - very hard, very high melting point, often poor thermal and electrical
conductivity
Chem 106, Prof. T. L. Heise 6161
Bonding in SolidsBonding in Solids
Chapt. 11.7
Ionic - made up of positively and negatively charged ions- electrostatic attractions- hard, high melting point, poor thermal and electrical conductivity
Metallic - atoms - metallic bonding - soft to very hard, low to very high melting point, excellent thermal and electrical conductivity, malleable and ductile
Chem 106, Prof. T. L. Heise 6262
Three types of Intermolecular ForcesThree types of Intermolecular Forces ViscosityViscosity Surface TensionSurface Tension Phase changesPhase changes Heats of Fusion and VaporizationHeats of Fusion and Vaporization Vapor PressureVapor Pressure Phase Diagrams, triple pointPhase Diagrams, triple point Crystalline solidsCrystalline solids Amorphous solidsAmorphous solids Unit CellsUnit Cells Bonding in SolidsBonding in Solids
Chapter Eleven; ReviewChapter Eleven; Review