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Chemistry: A Molecular Approach , 2nd Ed. Nivaldo Tro. Chapter 11 Liquids, Solids, and Intermolecular Forces. Roy Kennedy Massachusetts Bay Community College Wellesley Hills, MA. Climbing Geckos. Geckos can adhere to almost any surface - PowerPoint PPT Presentation
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Copyright 2011 Pearson Education, Inc.
Chapter 11Liquids,
Solids, and Intermolecular
Forces
Roy KennedyMassachusetts Bay Community College
Wellesley Hills, MA
Chemistry: A Molecular Approach, 2nd Ed.Nivaldo Tro
Copyright 2011 Pearson Education, Inc.
Climbing Geckos• Geckos can adhere to almost any surface• Recent studies indicate that this amazing ability
is related to intermolecular attractive forces• Geckos have millions of tiny hairs on their feet
that branch out and flatten out on the endsetae = hairs, spatulae = flat ends
• This structure allows the gecko to have unusually close contact to the surface – allowing the intermolecular attractive forces to act strongly
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Properties of the three Phases of Matter
• Fixed = keeps shape when placed in a container • Indefinite = takes the shape of the container
State Shape Volume Density
Compressible?
Will it Flow?
Strength of
Intermolecular
Attractions
Solid fixed fixed high No No very strongLiquid indefinite fixed high No Yes intermediate
Gas indefinite indefinite low Yes Yes weak
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Notice that the densities of ice and liquid water are much larger than the density of steam
Notice that the densities and molar volumes of ice and liquid water are much closer to each other than to steam
Notice that the densities of ice is larger than the density of liquid water. This is not the norm, but is vital to the development of life as we know it.
Three Phases of Water
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Degrees of Freedom• Particles may have one or several types of
freedom of motionand various degrees of each type
• Translational freedom is the ability to move from one position in space to another
• Rotational freedom is the ability to reorient the particles direction in space
• Vibrational freedom is the ability to oscillate about a particular point in space
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Solids• The particles in a solid are packed
close together and are fixed in position though they may vibrate
• The close packing of the particles results in solids being incompressible
• The inability of the particles to move around results in solids retaining their shape and volume when placed in a new container, and prevents the solid from flowing
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Solids
• Some solids have their particles arranged in an orderly geometric pattern – we call these crystalline solidssalt and diamonds
• Other solids have particles that do not show a regular geometric pattern over a long range – we call these amorphous solidsplastic and glass
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Liquids• The particles in a liquid are
closely packed, but they have some ability to move around
• The close packing results in liquids being incompressible
• But the ability of the particles to move allows liquids to take the shape of their container and to flow – however, they don’t have enough freedom to escape or expand to fill the container
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Gases• In the gas state, the particles have
complete freedom of motion and are not held together
• The particles are constantly flying around, bumping into each other and the container
• There is a large amount of space between the particlescompared to the size of the particlestherefore the molar volume of the
gas state of a material is much larger than the molar volume of the solid or liquid states
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Gases• Because there is a lot of empty
space, the particles can be squeezed closer together – therefore gases are compressible
• Because the particles are not held in close contact and are moving freely, gases expand to fill and take the shape of their container, and will flow
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Compressibility
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Kinetic – Molecular Theory
• What state a material is in depends largely on two major factors
1. the amount of kinetic energy the particles possess
2. the strength of attraction between the particles• These two factors are in competition with
each other
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States and Degrees of Freedom• The molecules in a gas have complete freedom of
motion their kinetic energy overcomes the attractive forces
between the molecules• The molecules in a solid are locked in place, they
cannot move around though they do vibrate, they don’t have enough kinetic
energy to overcome the attractive forces• The molecules in a liquid have limited freedom –
they can move around a little within the structure of the liquid they have enough kinetic energy to overcome some of
the attractive forces, but not enough to escape each other
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Kinetic Energy
• Increasing kinetic energy increases the motion energy of the particles
• The more motion energy the molecules have, the more freedom they can have
• The average kinetic energy is directly proportional to the temperatureKEavg = 1.5 kT
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Attractive Forces• The particles are attracted to each other by
electrostatic forces• The strength of the attractive forces varies,
some are small and some are large• The strength of the attractive forces depends
on the kind(s) of particles• The stronger the attractive forces between the
particles, the more they resist movingthough no material completely lacks particle motion
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Kinetic–Molecular Theoryof Gases
• When the kinetic energy is so large it overcomes the attractions between particles, the material will be a gas
• In an ideal gas, the particles have complete freedom of motion – especially translational
• This allows gas particles to expand to fill their containergases flow
• It also leads to there being large spaces between the particles therefore low density and compressibility
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Gas Structure
Gas molecules are rapidly moving in random straight lines, and are free from sticking to each other.
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Kinetic–Molecular Theory of Solids
• When the attractive forces are strong enough so the kinetic energy cannot overcome it at all, the material will be a solid
• In a solid, the particles are packed together without any translational or rotational motionthe only freedom
they have is vibrational motion
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Kinetic–Molecular Theoryof Liquids
• When the attractive forces are strong enough so the kinetic energy can only partially overcome them, the material will be a liquid
• In a liquid, the particles are packed together with only very limited translational or rotational freedom
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Explaining the Properties of Liquids
• Liquids have higher densities than gases and are incompressible because the particles are in contact
• They have an indefinite shape because the limited translational freedom of the particles allows them to move around enough to get to the container walls
• It also allows them to flow• But they have a definite volume because the limit
on their freedom keeps the particles from escaping each other
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Phase Changes• Because the attractive forces between the molecules are
fixed, changing the material’s state requires changing the amount of kinetic energy the particles have, or limiting their freedom
• Solids melt when heated because the particles gain enough kinetic energy to partially overcome the attactive forces
• Liquids boil when heated because the particles gain enough kinetic energy to completely overcome the attractive forces the stronger the attractive forces, the higher you will need to raise
the temperature• Gases can be condensed by decreasing the temperature
and/or increasing the pressure pressure can be increased by decreasing the gas volume reducing the volume reduces the amount of translational freedom
the particles have
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Phase Changes
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Intermolecular Attractions
• The strength of the attractions between the particles of a substance determines its state
• At room temperature, moderate to strong attractive forces result in materials being solids or liquids
• The stronger the attractive forces are, the higher will be the boiling point of the liquid and the melting point of the solidother factors also influence the melting point
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Why Are Molecules Attracted to Each Other? • Intermolecular attractions are due to
attractive forces between opposite charges + ion to − ion + end of polar molecule to − end of polar
molecule H-bonding especially strong
even nonpolar molecules will have temporary charges
• Larger charge = stronger attraction• Longer distance = weaker attraction• However, these attractive forces are
small relative to the bonding forces between atoms generally smaller charges generally over much larger distances
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Trends in the Strength of Intermolecular Attraction
• The stronger the attractions between the atoms or molecules, the more energy it will take to separate them
• Boiling a liquid requires we add enough energy to overcome all the attractions between the particlesHowever, not breaking the covalent bonds
• The higher the normal boiling point of the liquid, the stronger the intermolecular attractive forces
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Kinds of Attractive Forces
• Temporary polarity in the molecules due to unequal electron distribution leads to attractions called dispersion forces
• Permanent polarity in the molecules due to their structure leads to attractive forces called dipole–dipole attractions
• An especially strong dipole–dipole attraction results when H is attached to an extremely electronegative atom. These are called hydrogen bonds.
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Dispersion Forces• Fluctuations in the electron distribution in atoms
and molecules result in a temporary dipole region with excess electron density has partial (─)
charge region with depleted electron density has partial (+)
charge• The attractive forces caused by these temporary
dipoles are called dispersion forcesaka London Forces
• All molecules and atoms will have them• As a temporary dipole is established in one
molecule, it induces a dipole in all the surrounding molecules
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Dispersion Force
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Size of the Induced Dipole• The magnitude of the induced dipole
depends on several factors• Polarizability of the electrons
volume of the electron cloud larger molar mass = more electrons =
larger electron cloud = increased polarizability = stronger attractions
• Shape of the moleculemore surface-to-surface contact =
larger induced dipole = stronger attraction
+ + + + + + +
- - - - - - -
++ + +
+
-- - -
-
molecules that are flat have more surface
interaction than spherical ones
+ + + + + + +++
+ + + ++
− −− −− −− −− −−− −
larger molecules have more electrons, leading
to increased polarizability
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The Noble gases are all nonpolar atomic elements
As the molar mass increases, the number of electrons increases. Therefore the strength of the dispersion forces increases.
Effect of Molecular Sizeon Size of Dispersion Force
The stronger the attractive forces between the molecules, the higher the boiling point will be.
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Properties of Straight Chain AlkanesNonPolar Molecules
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Boiling Points of n-Alkanes
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Effect of Molecular Shapeon Size of Dispersion Force
35
the larger surface-to- surface contact between molecules in n-pentane
results in stronger dispersion force
attractions
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Alkane Boiling Points
• Branched chains have lower BPs than straight chains
• The straight chain isomers have more surface-to-surface contact
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a) CH4 C4H10
b) C6H12 C6H12
Practice – Choose the Substance in Each Pair with the Higher Boiling Point
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Practice – Choose the Substance in Each Pair with the Higher Boiling Point
a) CH4 CH3CH2CH2CH3
b) CH3CH2CH=CHCH2CH3 cyclohexane
Both molecules are nonpolarlarger molar mass
Both molecules are nonpolar, but the flatter ring molecule has larger surface-to-surface contact
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Dipole–Dipole Attractions• Polar molecules have a permanent dipole
because of bond polarity and shapedipole momentas well as the always present induced dipole
• The permanent dipole adds to the attractive forces between the molecules raising the boiling and melting points relative to
nonpolar molecules of similar size and shape
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Effect of Dipole–Dipole Attraction on Boiling and Melting Points
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replace with the figure 11.8
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Example 11.1b: Determine if dipole–dipole attractions occur between CH2Cl2 molecules
molecules that have dipole–dipole attractions must be polar
CH2Cl2, EN C = 2.5, H = 2.1, Cl = 3.0If there are dipole–dipole attractions
Solution:
Conceptual Plan:
Relationships:
Given:Find:
EN Difference Shape
Lewis Structure
BondPolarity
MoleculePolarityFormula
Cl—C3.0−2.5 = 0.5polarC—H2.5−2.1 = 0.4nonpolar
polar molecule; therefore dipole–dipole attractions
42
polar bonds andtetrahedral shape = polar molecule
4 bonding areasno lone pairs = tetrahedral shape
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or
Practice – Choose the substance in each pair with the higher boiling point
a) CH2FCH2F CH3CHF2
b)
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Practice – Choose the substance in each pair with the higher boiling point
more polara) CH2FCH2F CH3CHF2
b)
44
or
polar nonpolar
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Hydrogen Bonding• When a very electronegative atom is bonded to
hydrogen, it strongly pulls the bonding electrons toward itO─H, N─H, or F─H
• Because hydrogen has no other electrons, when its electron is pulled away, the nucleus becomes deshieldedexposing the H proton
• The exposed proton acts as a very strong center of positive charge, attracting all the electron clouds from neighboring molecules
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H-Bonding
HF
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H-Bonding in Water
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H-Bonds
• Hydrogen bonds are very strong intermolecular attractive forcesstronger than dipole–dipole or dispersion forces
• Substances that can hydrogen bond will have higher boiling points and melting points than similar substances that cannot
• But hydrogen bonds are not nearly as strong as chemical bonds2 to 5% the strength of covalent bonds
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Effect of H-Bonding on Boiling Point
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For nonpolar molecules, such as the hydrides of Group 4, the intermolecular attractions are due to dispersion forces. Therefore they increase down the column, causing the boiling point to increase.
HF, H2O, and NH3 have unusually strong dipole-dipole attractions, called hydrogen bonds. Therefore they have higher boiling points than you would expect from the general trends.
Polar molecules, such as the hydrides of Groups 5–7, have both dispersion forces and dipole–dipole attractions. Therefore they have higher boiling points than the corresponding Group 4 molecules.
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Step 3. Evaluate the strengths of the total intermolecular attractive forces. The substance with the strongest will be the liquid.Formaldehyde:
dispersion forces: MM 30.03, trigonal planardipole–dipole: very polar C=O bond uncancelledH-bonding: no O–H, N–H, or F–H therefore no H-bond
Fluoromethane: dispersion forces: MM 34.03, tetrahedraldipole–dipole: very polar C–F bond uncancelledH-bonding: no O–H, N–H, or F–H therefore no H-bond
Hydrogen peroxide: dispersion forces: MM 34.02, tetrahedral bentdipole–dipole: polar O–H bonds uncancelledH-bonding: O–H, therefore H-bond
Because the molar masses are similar, the size of the dispersion force attractions should be similar
Because only hydrogen peroxide has the additional very strong H-bond additional attractions, its intermolecular attractions will be the strongest. We therefore expect hydrogen peroxide to be the liquid.
Example 11.2: One of these compounds is a liquid at room temperature (the others are
gases). Which one and why?
51
Step 1. Determine the kinds of intermolecular attractive forces
MM = 30.03PolarNo H-Bonds
MM = 34.03PolarNo H-Bonds
MM = 34.02PolarH-Bonds
Step 2. Compare intermolecular attractive forces
Because all the molecules are polar, the size of the dipole–dipole attractions should be similar
Only the hydrogen peroxide also has additional hydrogen bond attractions
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a) CH3OH CH3CHF2
b) CH3-O-CH2CH3 CH3CH2CH2NH2
Practice – Choose the substance in each pair that is a liquid at room temperature (the other is a gas)
can H-bond
can H-bond
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Attractive Forces and Solubility• Solubility depends, in part, on the attractive forces
of the solute and solvent moleculeslike dissolves likemiscible liquids will always dissolve in each other
• Polar substances dissolve in polar solventshydrophilic groups = OH, CHO, C=O, COOH, NH2, Cl
• Nonpolar molecules dissolve in nonpolar solventshydrophobic groups = C-H, C-C
• Many molecules have both hydrophilic and hydrophobic parts – solubility in water becomes a competition between the attraction of the polar groups for the water and the attraction of the nonpolar groups for their own kind
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Immiscible Liquids
54
Pentane, C5H12 is a nonpolar molecule.
Water is a polar molecule.
The attractive forces between the water molecules is much stronger than their attractions for the pentane molecules. The result is the liquids are immiscible.
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Dichloromethane(methylene chloride)
Polar Solvents
55
WaterEthanol
(ethyl alcohol)
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Nonpolar Solvents
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Ion–Dipole Attraction• In a mixture, ions from an ionic compound are
attracted to the dipole of polar molecules• The strength of the ion–dipole attraction is one
of the main factors that determines the solubility of ionic compounds in water
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a) CH3OH CH3CHF2
b) CH3CH2CH2CH2CH3 CH3Cl
Practice – Choose the substance in each pair that is more soluble in water
can H-bond with H2O
more polar
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Summary
• Dispersion forces are the weakest of the intermolecular attractions
• Dispersion forces are present in all molecules and atoms
• The magnitude of the dispersion forces increases with molar mass
• Polar molecules also have dipole–dipole attractive forces
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Summary (cont’d)• Hydrogen bonds are the strongest of the
intermolecular attractive forcesa pure substance can have
• Hydrogen bonds will be present when a molecule has H directly bonded to either O , N, or F atomsonly example of H bonded to F is HF
• Ion–dipole attractions are present in mixtures of ionic compounds with polar molecules.
• Ion–dipole attractions are the strongest intermolecular attraction
• Ion–dipole attractions are especially important in aqueous solutions of ionic compounds
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Liquids
Properties &Structure
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Surface Tension• Surface tension is a property of liquids that
results from the tendency of liquids to minimize their surface area
• To minimize their surface area, liquids form drops that are sphericalas long as there
is no gravity
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Surface Tension• The layer of molecules on the surface behave
differently than the interior because the cohesive forces on the surface
molecules have a net pull into the liquid interior• The surface layer acts like an elastic skin
allowing you to “float” a paper clip even though steel is denser than water
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Surface Tension
• Because they have fewer neighbors to attract them, the surface molecules are less stable than those in the interiorhave a higher potential energy
• The surface tension of a liquid is the energy required to increase the surface area a given amountsurface tension of H2O = 72.8 mJ/m2
at room temperaturesurface tension of C6H6 = 28 mJ/m2
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Factors Affecting Surface Tension
• The stronger the intermolecular attractive forces, the higher the surface tension will be
• Raising the temperature of a liquid reduces its surface tensionraising the temperature of the liquid increases the
average kinetic energy of the moleculesthe increased molecular motion makes it easier to
stretch the surface
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Viscosity• Viscosity is the resistance of a liquid to flow
1 poise = 1 P = 1 g/cm∙soften given in centipoise, cP
H2O = 1 cP at room temperature
• Larger intermolecular attractions = larger viscosity
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Factors Affecting Viscosity
• The stronger the intermolecular attractive forces, the higher the liquid’s viscosity will be
• The more spherical the molecular shape, the lower the viscosity will bemolecules roll more easily less surface-to-surface contact lowers attractions
• Raising the temperature of a liquid reduces its viscosity raising the temperature of the liquid increases the average
kinetic energy of the molecules the increased molecular motion makes it easier to overcome
the intermolecular attractions and flow69Tro: Chemistry: A Molecular Approach, 2/e
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Insert Table 11.6
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Capillary Action
• Capillary action is the ability of a liquid to flow up a thin tube against the influence of gravitythe narrower the tube, the higher the liquid rises
• Capillary action is the result of two forces working in conjunction, the cohesive and adhesive forces cohesive forces hold the liquid molecules togetheradhesive forces attract the outer liquid molecules to
the tube’s surface
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Capillary Action• The adhesive forces pull the surface liquid up the
side of the tube, and the cohesive forces pull the interior liquid with it
• The liquid rises up the tube until the force of gravity counteracts the capillary action forces
• The narrower the tube diameter, the higher the liquid will rise up the tube
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Meniscus• The curving of the liquid surface in a
thin tube is due to the competition between adhesive and cohesive forces
• The meniscus of water is concave in a glass tube because its adhesion to the glass is stronger than its cohesion for itself
• The meniscus of mercury is convex in a glass tube because its cohesion for itself is stronger than its adhesion for the glassmetallic bonds are stronger than
intermolecular attractions
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The Molecular Dance• Molecules in the liquid are constantly in
motionvibrational, and limited rotational and
translational• The average kinetic energy is proportional
to the temperature• However, some molecules have more
kinetic energy than the average, and others have less
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Vaporization• If these high energy molecules
are at the surface, they may have enough energy to overcome the attractive forcestherefore – the larger the surface
area, the faster the rate of evaporation
• This will allow them to escape the liquid and become a vapor
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Distribution of Thermal Energy• Only a small fraction of the molecules in a liquid
have enough energy to escape• But, as the temperature increases, the fraction of
the molecules with “escape energy” increases• The higher the temperature, the faster the rate
of evaporation
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Condensation
• Some molecules of the vapor will lose energy through molecular collisions
• The result will be that some of the molecules will get captured back into the liquid when they collide with it
• Also some may stick and gather together to form droplets of liquidparticularly on surrounding surfaces
• We call this process condensation
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Evaporation vs. Condensation• Vaporization and condensation are opposite processes• In an open container, the vapor molecules generally
spread out faster than they can condense• The net result is that the rate of vaporization is greater
than the rate of condensation, and there is a net loss of liquid
• However, in a closed container, the vapor is not allowed to spread out indefinitely
• The net result in a closed container is that at some time the rates of vaporization and condensation will be equal
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Effect of Intermolecular Attraction on Evaporation and Condensation
• The weaker the attractive forces between molecules, the less energy they will need to vaporize
• Also, weaker attractive forces means that more energy will need to be removed from the vapor molecules before they can condense
• The net result will be more molecules in the vapor phase, and a liquid that evaporates faster – the weaker the attractive forces, the faster the rate of evaporation
• Liquids that evaporate easily are said to be volatilee.g., gasoline, fingernail polish remover liquids that do not evaporate easily are called nonvolatile
e.g., motor oil
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Energetics of Vaporization
• When the high energy molecules are lost from the liquid, it lowers the average kinetic energy
• If energy is not drawn back into the liquid, its temperature will decrease – therefore, vaporization is an endothermic processand condensation is an exothermic process
• Vaporization requires input of energy to overcome the attractions between molecules
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Heat of Vaporization• The amount of heat energy required to vaporize
one mole of the liquid is called the heat of vaporization, DHvapsometimes called the enthalpy of vaporization
• Slways endothermic, therefore DHvap is +• Somewhat temperature dependent· DHcondensation = −DHvaporization
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Example 11.3: Calculate the mass of water that can be vaporized with 155 kJ of heat at 100 °C
82
because the given amount of heat is almost 4x the DHvap, the amount of water makes sense
1 mol H2O = 40.7 kJ, 1 mol = 18.02 g
155 kJg H2O
Check:
Solution:
Conceptual Plan:
Relationships:
Given:Find:
kJ mol H2O g H2O
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Practice – Calculate the amount of heat needed to vaporize 90.0 g of C3H7OH at its boiling point
(DHvap = 39.9 kJ/mol)
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Practice – Calculate the amount of heat needed to vaporize 90.0 g of C3H7OH at its boiling point
84
because the given amount of C3H7OH is more than 1 mole the amount of heat makes sense
1 mol C3H7OH = 39.9 kJ, 1 mol = 60.09 g
90.0 gkJ
Check:
Solution:
Conceptual Plan:
Relationships:
Given:Find:
g mol kJ
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Dynamic Equilibrium• In a closed container, once the rates of
vaporization and condensation are equal, the total amount of vapor and liquid will not change
• Evaporation and condensation are still occurring, but because they are opposite processes, there is no net gain or loss of either vapor or liquid
• When two opposite processes reach the same rate so that there is no gain or loss of material, we call it a dynamic equilibrium this does not mean there are equal amounts of vapor
and liquid – it means that they are changing by equal amounts
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Dynamic Equilibrium
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Vapor Pressure• The pressure exerted by the vapor when it is in
dynamic equilibrium with its liquid is called the vapor pressure remember using Dalton’s Law of Partial Pressures to
account for the pressure of the water vapor when collecting gases by water displacement?
• The weaker the attractive forces between the molecules, the more molecules will be in the vapor
• Therefore, the weaker the attractive forces, the higher the vapor pressure the higher the vapor pressure, the more volatile the
liquid
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Vapor–Liquid Dynamic Equilibrium• If the volume of the chamber is increased, it will
decrease the pressure of the vapor inside the chamber at that point, there are fewer vapor molecules in a given
volume, causing the rate of condensation to slow• Therefore, for a period of time, the rate of vaporization
will be faster than the rate of condensation, and the amount of vapor will increase
• Eventually enough vapor accumulates so that the rate of the condensation increases to the point where it is once again as fast as evaporation equilibrium is reestablished
• At this point, the vapor pressure will be the same as it was before
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Changing the Container’s Volume Disturbs the Equilibrium
Initially, the rate of vaporization and
condensation are equal and the system is in dynamic equilibrium
When the volume is increased, the rate of vaporization becomes faster than the rate of
condensation
When the volume is decreased, the rate of vaporization becomes slower than the rate of
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Dynamic Equilibrium
• A system in dynamic equilibrium can respond to changes in the conditions
• When conditions change, the system shifts its position to relieve or reduce the effects of the change
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Vapor Pressure vs. Temperature
• Increasing the temperature increases the number of molecules able to escape the liquid
• The net result is that as the temperature increases, the vapor pressure increases
• Small changes in temperature can make big changes in vapor pressurethe rate of growth depends on strength of the
intermolecular forces
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Vapor Pressure Curves
760 mmHg
normal BP100 °C
BP Ethanol at 500 mmHg68.1°C
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a) waterb) TiCl4c) etherd) ethanole) acetone
Practice – Which of the following is the most volatile?
a) waterb) TiCl4c) etherd) ethanole) acetone
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a) waterb) TiCl4c) etherd) ethanole) acetone
Practice – Which of the following has the strongest Intermolecular attractions?
a) waterb) TiCl4c) etherd) ethanole) acetone
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Boiling Point
• When the temperature of a liquid reaches a point where its vapor pressure is the same as the external pressure, vapor bubbles can form anywhere in the liquidnot just on the surface
• This phenomenon is what is called boiling and the temperature at which the vapor pressure = external pressure is the boiling point
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Boiling Point
• The normal boiling point is the temperature at which the vapor pressure of the liquid = 1 atm
• The lower the external pressure, the lower the boiling point of the liquid
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a) waterb) TiCl4c) etherd) ethanole) acetone
Practice – Which of the following has the highest normal boiling point?
a) waterb) TiCl4c) etherd) ethanole) acetone
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Heating Curve of a Liquid• As you heat a liquid, its
temperature increases linearly until it reaches the boiling pointq = mass x Cs x DT
• Once the temperature reaches the boiling point, all the added heat goes into boiling the liquid – the temperature stays constant
• Once all the liquid has been turned into gas, the temperature can again start to rise
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• The logarithm of the vapor pressure vs. inverse absolute temperature is a linear function
• A graph of ln(Pvap) vs. 1/T is a straight line • The slope of the line x 8.314 J/mol∙K = DHvap
in J/mol
Clausius–Clapeyron Equation
• The graph of vapor pressure vs. temperature is an exponential growth curve
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Example 11.4: Determine the DHvap of dichloromethane given the vapor pressure vs.
temperature data• Enter the data into a spreadsheet and calculate
the inverse of the absolute temperature and natural log of the vapor pressure
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Example 11.4: Determine the DHvap of dichloromethane given the vapor pressure vs.
temperature data• Graph the inverse of the absolute temperature vs.
the natural log of the vapor pressure
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Example 11.4: Determine the DHvap of dichloromethane given the vapor pressure vs.
temperature data• Add a trendline, making sure the display equation
on chart option is checked off
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Example 11.4: Determine the DHvap of dichloromethane given the vapor pressure vs.
temperature data• Determine the slope of the line
−3776.7 ≈ −3800 K
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Example 11.4: Determine the DHvap of dichloromethane given the vapor pressure vs.
temperature data• Use the slope of the line to determine the heat of
vaporizationslope ≈ −3800 K, R = 8.314 J/mol∙K
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Clausius–Clapeyron Equation2-Point Form
• The equation below can be used with just two measurements of vapor pressure and temperature however, it generally gives less precise results
fewer data points will not give as precise an average because there is less averaging out of the errorso as with any other sets of measurements
• It can also be used to predict the vapor pressure if you know the heat of vaporization and the normal boiling point remember: the vapor pressure at the normal boiling point is 760
torr
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T1 = BP = 64.6 °C, P1 = 760 torr, DHvap = 35.2 kJ/mol, T2 = 12.0 °C
P2, torr
T1 = BP = 337.8 K, P1 = 760 torr, DHvap = 35.2 kJ/mol, T2 = 285.2 K
P2, torr
Solution:
Example 11.5: Calculate the vapor pressure of methanol at 12.0 °C
106
the units are correct, the size makes sense because the vapor pressure is lower at lower temperatures
Check:
Conceptual Plan:
Relationships:
Given:
Find:
P1, T1, DHvap P2
T(K) = T(°C) + 273.15
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Practice – Determine the vapor pressure of water at 25 C (normal BP = 100.0 C, DHvap= 40.7 kJ/mol)
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T1 = BP = 100.0 °C, P1 = 760 torr, DHvap = 40.7 kJ/mol, T2 = 25.0 °C
P2, torr
T1 = BP = 373.2 K, P1 = 760 torr, DHvap = 40.7 kJ/mol, T2 = 298.2 K
P2, torr
Practice – Determine the vapor pressure of water at 25 C
108
the units are correct, the size makes sense because the vapor pressure is lower at lower temperatures
Check:
Solution:
Conceptual Plan:
Relationships:
Given:
Find:
P1, T1, DHvap P2
T(K) = T(°C) + 273.15
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Supercritical Fluid• As a liquid is heated in a sealed container, more vapor
collects, causing the pressure inside the container to rise and the density of the vapor to increase and the density of the liquid to decrease
• At some temperature, the meniscus between the liquid and vapor disappears and the states commingle to form a supercritical fluid
• Supercritical fluids have properties of both gas and liquid states
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The Critical Point• The temperature required to produce a
supercritical fluid is called the critical temperature
• The pressure at the critical temperature is called the critical pressure
• At the critical temperature or higher temperatures, the gas cannot be condensed to a liquid, no matter how high the pressure gets
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Sublimation and Deposition• Molecules in the solid have thermal energy that
allows them to vibrate• Surface molecules with sufficient energy may
break free from the surface and become a gas – this process is called sublimation
• The capturing of vapor molecules into a solid is called deposition
• The solid and vapor phases exist in dynamic equilibrium in a closed containerat temperatures below the melting point therefore, molecular solids have a vapor pressure
solid gassublimation
deposition
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Sublimation
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Melting = Fusion
• As a solid is heated, its temperature rises and the molecules vibrate more vigorously
• Once the temperature reaches the melting point, the molecules have sufficient energy to overcome some of the attractions that hold them in position and the solid melts (or fuses)
• The opposite of melting is freezing
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Heating Curve of a Solid• As you heat a solid, its
temperature increases linearly until it reaches the melting point q = mass x Cs x DT
• Once the temperature reaches the melting point, all the added heat goes into melting the solid – the temperature stays constant
• Once all the solid has been turned into liquid, the temperature can again start to rise ice/water will always have a
temperature of 0 °C at 1 atm
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Energetics of Melting• When the high energy molecules are lost from
the solid, it lowers the average kinetic energy• If energy is not drawn back into the solid its
temperature will decrease – therefore, melting is an endothermic processand freezing is an exothermic process
• Melting requires input of energy to overcome the attractions between molecules
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Heat of Fusion• The amount of heat energy required to melt one mole of the
solid is called the Heat of Fusion, DHfus sometimes called the enthalpy of fusion
• Always endothermic, therefore DHfus is +• Somewhat temperature dependent· DHcrystallization = −DHfusion
· Generally much less than DHvap
· DHsublimation = DHfusion + DHvaporization
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Heats of Fusion and Vaporization
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Heating Curve of Water
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Segment 1
• Heating 1.00 mole of ice at −25.0 °C up to the melting point, 0.0 °C
• q = mass x Cs x DTmass of 1.00 mole of ice = 18.0 gCs = 2.09 J/mol∙°C
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Segment 2
• Melting 1.00 mole of ice at the melting point, 0.0 °C
• q = n∙DHfus
n = 1.00 mole of iceDHfus = 6.02 kJ/mol
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Segment 3
• Heating 1.00 mole of water at 0.0 °C up to the boiling point, 100.0 °C
• q = mass x Cs x DTmass of 1.00 mole of water = 18.0 gCs = 2.09 J/mol∙°C
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Segment 4
• Boiling 1.00 mole of water at the boiling point, 100.0 °C
• q = n∙DHvap
n = 1.00 mole of iceDHfus = 40.7 kJ/mol
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Segment 5
• Heating 1.00 mole of steam at 100.0 °C up to 125.0 °C
• q = mass x Cs x DTmass of 1.00 mole of water = 18.0 gCs = 2.01 J/mol∙°C
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Practice – How much heat, in kJ, is needed to raise the temperature of a 12.0 g benzene
sample from −10.0 °C to 25.0 °C?
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12.0 g benzene, seg 1 (T1 = −10.0 °C, T2 = 5.5 °C), seg 2 = melting, seg 3 (T1 = 5.5 °C, T2 = 25.0 °C)kJ
12.0 g benzene, seg 1 = 0.2325 kJ,seg 2 = 1.51 kJ, seg 3 (T1 = 5.5 °C, T2 = 25.0 °C)kJ
Practice – How much heat is needed to raise the temperature of a 12.0 g benzene sample from −10.0 °C to 25.0 °C?
125
DHfus 9.8 kJ/mol, 1 mol = 78.11 g, 1 kJ = 1000 J, q = m∙Cs∙DTCs,sol = 1.25 J/g°C, Cs,liq = 1.70 J/g°C
Solution:
Conceptual Plan:
Relationships:
Given:Find:
g J kJSeg 1
12.0 g benzene, seg 1 = 0.2325 kJ,seg 2 = melting, seg 3 (T1 = 5.5 °C, T2 = 25.0 °C)kJ
g J kJSeg 3 g mol kJSeg 2
12.0 g benzene, seg 1 = 0.2325 kJ,seg 2 = 1.51 kJ, seg 3 = 0.3978 kJkJ
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Phase Diagrams• Phase diagrams describe the different states and
state changes that occur at various temperature/pressure conditions
• Regions represent states• Lines represent state changes
liquid/gas line is vapor pressure curveboth states exist simultaneouslycritical point is the furthest point on the vapor
pressure curve• Triple point is the temperature/pressure condition
where all three states exist simultaneously• For most substances, freezing point increases as
pressure increases126Tro: Chemistry: A Molecular Approach, 2/e
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Phase DiagramsP
ress
ure
Temperature
vaporization
condensation
criticalpoint
triplepoint
Solid Liquid
Gas
1 atm
normalmelting pt.
normalboiling pt.
Fusion Curve
Vapor PressureCurve
SublimationCurve
melting
freezing
sublimation
deposition
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Phase Diagram of Water
Temperature
Pre
ssur
e
criticalpoint
374.1 °C217.7 atm
triplepoint
Ice Water
Steam
1 atm
normalboiling pt.
100 °C
normalmelting pt.
0 °C
0.01 °C0.006 atm
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Morphic Forms of Ice
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Phase Diagram of CO2P
ress
ure
Temperature
criticalpoint
31.0 °C72.9 atm
triplepoint
Solid Liquid
Gas1 atm
-56.7 °C5.1 atm
normalsublimation pt.
-78.5 °C
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• 20.0 °C, 72.9 atm
• −56.7 °C, 5.1 atm
• 10.0 °C, 1.0 atm
• −78.5 °C, 1.0 atm
• 50.0 °C, 80.0 atm
• 20.0 °C, 72.9 atm liquid
• −56.7 °C, 5.1 atm solid, liquid, gas
• 10.0 °C, 1.0 atm gas
• −78.5 °C, 1.0 atm solid, gas
• 50.0 °C, 80.0 atm scf
Practice – Consider the phase diagram of CO2
shown. What phase(s) is/are present at each of the following conditions?
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Water – An Extraordinary Substance• Water is a liquid at room temperature
most molecular substances with similar molar masses are gases at room temperature
e.g. NH3, CH4 due to H-bonding between molecules
• Water is an excellent solvent – dissolving many ionic and polar molecular substances because of its large dipole moment even many small nonpolar molecules have some solubility in water
e.g. O2, CO2
• Water has a very high specific heat for a molecular substance moderating effect on coastal climates
• Water expands when it freezes at a pressure of 1 atm
about 9% making ice less dense than liquid water
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SolidsProperties &
Structure
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Crystal Lattice• When allowed to cool slowly, the particles in a
liquid will arrange themselves to give the maximum attractive forcestherefore minimize the energy
• The result will generally be a crystalline solid• The arrangement of the particles in a
crystalline solid is called the crystal lattice• The smallest unit that shows the pattern of
arrangement for all the particles is called the unit cell
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Unit Cells• Unit cells are 3-dimensional
usually containing 2 or 3 layers of particles• Unit cells are repeated over and over to give the
macroscopic crystal structure of the solid• Starting anywhere within the crystal results in the same
unit cell• Each particle in the unit cell is called a lattice point• Lattice planes are planes connecting equivalent points in
unit cells throughout the lattice
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7 Unit Cells
Cubica = b = c
all 90°
a
bc
Tetragonala = c < b
all 90°
a
b
c
Orthorhombica ¹ b ¹ c
all 90°
a
b
c
Monoclinica ¹ b ¹ c
2 faces 90°
a b
c
Hexagonala = c < b
2 faces 90°1 face 120°
c
a
b
Rhombohedrala = b = cno 90°
ab
c
Triclinica ¹ b ¹ c
no 90°
a
bc
140
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Unit Cells• The number of other particles each particle is in
contact with is called its coordination number for ions, it is the number of oppositely charged ions an
ion is in contact with• Higher coordination number means more
interaction, therefore stronger attractive forces holding the crystal together
• The packing efficiency is the percentage of volume in the unit cell occupied by particles the higher the coordination number, the more efficiently
the particles are packing together
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Cubic Unit Cells• All 90° angles between corners of the unit cell• The length of all the edges are equal• If the unit cell is made of spherical particles
⅛ of each corner particle is within the cube½ of each particle on a face is within the cube¼ of each particle on an edge is within the cube
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Cubic Unit Cells - Simple Cubic
• Eight particles, one at each corner of a cube
• 1/8th of each particle lies in the unit celleach particle part of eight cellstotal = one particle in each unit
cell8 corners x 1/8
• Edge of unit cell = twice the radius
• Coordination number of 6
2r
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Simple Cubic
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Cubic Unit Cells - Body-Centered Cubic
• Nine particles, one at each corner of a cube + one in center
• 1/8th of each corner particle lies in the unit celltwo particles in each unit cell
8 corners x 1/8 + 1 center
• Edge of unit cell = (4/Ö 3) times the radius of the particle
• Coordination number of 8146
4r3
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Body-Centered Cubic
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Cubic Unit Cells - Face-Centered Cubic
• 14 particles, one at each corner of a cube + one in center of each face
• 1/8th of each corner particle + 1/2 of face particle lies in the unit cell4 particles in each unit cell
8 corners x 1/8 + 6 faces x 1/2
• Edge of unit cell = 2 2 times the radius of the particle
• Coordination number of 12148
2r 2
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Face-Centered Cubic
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Solution:fcc = 4 atoms/uc, Al = 26.982 g/mol, 1 mol = 6.022 x 1023 atoms
Example 11.6: Calculate the density of Al if it crystallizes in a fcc and has a radius of 143 pm
150
the accepted density of Al at 20°C is 2.71 g/cm3, so the answer makes sense
face-centered cubic, r = 143 pm
density, g/cm3
Check:
Conceptual Plan:
Relationships:
Given:Find:
1 cm = 102 m, 1 pm = 10−12 m
lr Vmassfcc
dm, V# atoms x mass of 1 atom
V = l 3, l = 2r√2, d = m/Vd = m/V
l = 2r√2 V = l 3
face-centered cubic, r = 1.43 x 10−8 cm, m = 1.792 x 10−22 g
density, g/cm3
face-centered cubic, V = 6.618 x 10−23 cm3, m =1.792 x 10−22g
density, g/cm3
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Practice – Estimate the density of Rb if it crystallizes in a body-centered cubic unit cell and has an atomic
radius of 247.5 pm
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Solution:bcc = 2 atoms/uc, Rb = 85.47 g/mol, 1 mol = 6.022 x 1023 atoms
Practice – Estimate the density of Rb if it crystallizes in a bcc and has a radius of 247.5 pm
the accepted density of Rb at 20°C is 1.53 g/cm3, so the answer makes sense
body-centered cubic, r = 247.5 pm
density, g/cm3
Check:
Conceptual Plan:
Relationships:
Given:Find:
1 cm = 102 m, 1 pm = 10−12 m
lr Vmassbcc
dm, V# atoms x mass of 1 atom
V = l 3, l = 4r/√3, d = m/Vd = m/V
l = 4r/√3 V = l 3
body-centered cubic, r = 2.475 x 10−8 cm, m = 2.839 x 10−22 g
density, g/cm3
body-centered cubic, V= 1.868 x 10−22 cm3, m= 2.839 x 10−22 g
density, g/cm3
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Closest-Packed StructuresFirst Layer
• With spheres, it is more efficient to offset each row in the gaps of the previous row than to line up rows and columns
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Closest-Packed StructuresSecond Layer
• The second layer atoms can sit directly over the atoms in the first layer– called an AA pattern
• Or the second layer can sit over the holes in the first layer – called an AB pattern
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Closest-Packed StructuresThird Layer – with Offset 2nd Layer
• The third layer atoms can align directly over the atoms in the first layer– called an ABA pattern
• Or the third layer can sit over the uncovered holes in the first layer– called an ABC pattern
Hexagonal Closest-PackedCubic Closest-PackedFace-Centered Cubic
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Hexagonal Closest-Packed Structures
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Cubic Closest-Packed Structures
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Classifying Crystalline Solids
• Crystalline solids are classified by the kinds of particles found
• Some of the categories are sub-classified by the kinds of attractive forces holding the particles together
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Classifying Crystalline Solids• Molecular solids are solids whose composite
particles are molecules• Ionic solids are solids whose composite
particles are ions• Atomic solids are solids whose composite
particles are atomsnonbonding atomic solids are held together by
dispersion forcesmetallic atomic solids are held together by metallic
bondsnetwork covalent atomic solids are held together by
covalent bonds159Tro: Chemistry: A Molecular Approach, 2/e
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Molecular Solids
• The lattice sites are occupied by moleculesCO2, H2O, C12H22O11
• The molecules are held together by intermolecular attractive forcesdispersion forces, dipole–dipole attractions, and
H-bonds• Because the attractive forces are weak, they
tend to have low melting pointsgenerally < 300 °C
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Ionic Solids• Lattice sites occupied by ions• Held together by attractions between oppositely charged ions
nondirectional therefore every cation attracts all anions around it, and vice-versa
• The coordination number represents the number of close cation–anion interactions in the crystal
• The higher the coordination number, the more stable the solid lowers the potential energy of the solid
• The coordination number depends on the relative sizes of the cations and anions that maintains charge balance generally, anions are larger than cations the number of anions that can surround the cation is limited by the
size of the cation the closer in size the ions are, the higher the coordination number is
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Ionic Crystals
CsClcoordination number = 8
Cs+ = 167 pmCl─ = 181 pm
NaClcoordination number = 6
Na+ = 97 pmCl─ = 181 pm
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Lattice Holes
Simple CubicHole
OctahedralHole
TetrahedralHole
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Lattice Holes• In hexagonal closest-packed or cubic closest-
packed lattices there are eight tetrahedral holes and four octahedral holes per unit cell
• In a simple cubic lattice there is one cubic hole per unit cell
• Number and type of holes occupied determines formula (empirical) of the salt
= Octahedral
= Tetrahedral
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Cesium Chloride Structures
• Coordination number = 8• ⅛ of each Cl─ (184 pm) inside
the unit cell• Whole Cs+ (167 pm) inside the
unit cellcubic hole = hole in simple
cubic arrangement of Cl─ ions• Cs:Cl = 1: (8 x ⅛), therefore
the formula is CsCl
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Rock Salt Structures• Coordination number = 6• Cl─ ions (181 pm) in a face-centered
cubic arrangement ⅛ of each corner Cl─ inside the unit cell ½ of each face Cl─ inside the unit cell
• Na+ (97 pm) in holes between Cl─
octahedral holes 1 in center of unit cell 1 whole particle in every octahedral hole ¼ of each edge Na+ inside the unit cell
• Na:Cl = (¼ x 12) + 1: (⅛ x 8) + (½ x 6) = 4:4 = 1:1,
• Therefore the formula is NaCl
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Zinc Blende Structures• Coordination number = 4• S2─ ions (184 pm) in a face-centered
cubic arrangement ⅛ of each corner S2─ inside the unit cell ½ of each face S2─ inside the unit cell
• Each Zn2+ (74 pm) in holes between S2─ tetrahedral holes 1 whole particle in ½ the holes
• Zn:S = (4 x 1) : (⅛ x 8) + (½ x 6) = 4:4 = 1:1,
• Therefore the formula is ZnS
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Fluorite Structures• Coordination number = 4• Ca2+ ions (99 pm) in a face-centered
cubic arrangement ⅛ of each corner Ca2+ inside the unit cell ½ of each face Ca2+ inside the unit cell
• Each F─ (133 pm) in holes between Ca2+ tetrahedral holes 1 whole particle in all the holes
• Ca:F = (⅛ x 8) + (½ x 6): (8 x 1) = 4:8 = 1:2,
• Therefore the formula is CaF2 fluorite structure common for 1:2 ratio
• Usually get the antifluorite structure when the cation:anion ratio is 2:1 the anions occupy the lattice sites and
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Practice – Gallium arsenide crystallizes in a cubic closest-packed array of arsenide ions with gallium ions in ½ the tetrahedral holes. What is the ratio of gallium ions to arsenide ions in the structure and the empirical
formula of the compound?
As = cpp = 4 atoms per unit cell
Ga = ½ (8 tetrahedral holes per unit cell)Ga = 4 atoms per unit cell
Ga:As = 4 atoms :4 atoms per unit cell = 1:1The formula is GaAs
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Nonbonding Atomic Solids• Noble gases in solid form• Solid held together by weak dispersion forces
very low melting• Tend to arrange atoms in closest-packed
structureeither hexagonal cp or cubic cpmaximizes attractive forces and minimizes energy
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Metallic Atomic Solids
• Solid held together by metallic bondsstrength varies with sizes and charges of cations
coulombic attractions• Melting point varies• Mostly closest-packed arrangements of the
lattice pointscations
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Metallic Structure
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Metallic Bonding• Metal atoms release their valence
electrons• Metal cation “islands” fixed in a “sea” of
mobile electrons
e-
e- e-
e-
e-
e-
e-
e-
e-
e-
e-
e-
e-
e-
e-
e-
+ + + + + + + + +
+ + + + + + + + +
+ + + + + + + + +
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Network Covalent Solids• Atoms attached to their nearest neighbors by
covalent bonds• Because of the directionality of the covalent
bonds, these do not tend to form closest-packed arrangements in the crystal
• Because of the strength of the covalent bonds, these have very high melting pointsgenerally > 1000 °C
• Dimensionality of the network affects other physical properties
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The Diamond Structure:a 3-Dimensional Network
• The carbon atoms in a diamond each have four covalent bonds to surrounding atomssp3
tetrahedral geometry• This effectively makes each crystal one giant
molecule held together by covalent bondsyou can follow a path of covalent bonds from any
atom to every other atom
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Properties of Diamond • Very high melting point, ~3800 °C
need to overcome some covalent bonds• Very rigid
due to the directionality of the covalent bonds
• Very hard due to the strong covalent bonds holding
the atoms in position used as abrasives
• Electrical insulator• Thermal conductor
best known • Chemically very nonreactive
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The Graphite Structure:a 2-Dimensional Network
• In graphite, the carbon atoms in a sheet are covalently bonded together forming six-member flat rings fused together
similar to benzenebond length = 142 pm
sp2 each C has three sigma and one pi bond
trigonal-planar geometryeach sheet a giant molecule
• The sheets are then stacked and held together by dispersion forcessheets are 341 pm apart
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Properties of Graphite• Hexagonal crystals• High melting point, ~3800 °C
need to overcome some covalent bonding• Slippery feel
because there are only dispersion forces holding the sheets together, they can slide past each other
glide planes lubricants
• Electrical conductor parallel to sheets
• Thermal insulator• Chemically very nonreactive
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Silicates
• ~90% of Earth’s crust• Extended arrays of SiO
sometimes with Al substituted for Si – aluminosilicates• Glass is the amorphous form
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Quartz• SiO2 in pure form
impurities add color• 3-dimensional array of Si covalently bonded
to 4 O tetrahedral
• Melts at ~1600 °C• Very hard
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Micas• There are various kinds of mica that have slightly
different compositions – but are all of the general form X2Y4-6Z8O20(OH,F)4 X is K, Na, or Ca or less commonly Ba, Rb, or Cs Y is Al, Mg, or Fe or less commonly Mn, Cr, Ti, Li, etc.Z is chiefly Si or Al but also may include Fe3+ or Ti
• Minerals that are mainly 2-dimensional arrays of Si bonded to Ohexagonal arrangement of atoms
• Sheets• Chemically stable• Thermal and electrical insulator
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a) KCl ionic SCl2 molecular
b) C(s, graphite) cov. network S8 molecular
c) Kr atomic K metallic
d) SrCl2 ionic SiO2 (s, quartz) cov. network
Practice – Pick the solid in each pair with the highest melting point
a) KCl SCl2
b) C(s, graphite)
c) Kr K
d) SrCl2 SiO2 (s, quartz)
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Band Theory
• The structures of metals and covalent network solids result in every atom’s orbitals being shared by the entire structure
• For large numbers of atoms, this results in a large number of molecular orbitals that have approximately the same energy; we call this an energy band
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Band Theory• When two atomic orbitals combine they
produce both a bonding and an antibonding molecular orbital
• When many atomic orbitals combine they produce a band of bonding molecular orbitals and a band of antibonding molecular orbitals
• The band of bonding molecular orbitals is called the valence band
• The band of antibonding molecular orbitals is called the conduction band
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Molecular Orbitals of Polylithium
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Band Gap• At absolute zero, all the electrons will occupy
the valence band• As the temperature rises, some of the electrons
may acquire enough energy to jump to the conduction band
• The difference in energy between the valence band and conduction band is called the band gapthe larger the band gap, the fewer electrons there
are with enough energy to make the jump
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Types of Band Gaps andConductivity
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Band Gap and Conductivity• The more electrons at any one time that a substance has in
the conduction band, the better conductor of electricity it is• If the band gap is ~0, then the electrons will be almost as
likely to be in the conduction band as the valence band and the material will be a conductor metals the conductivity of a metal decreases with temperature
• If the band gap is small, then a significant number of the electrons will be in the conduction band at normal temperatures and the material will be a semiconductor graphite the conductivity of a semiconductor increases with temperature
• If the band gap is large, then effectively no electrons will be in the conduction band at normal temperatures and the material will be an insulator
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Doping Semiconductors• Doping is adding impurities to the semiconductor’s
crystal to increase its conductivity• Goal is to increase the number of electrons in the
conduction band • n-type semiconductors do not have enough
electrons themselves to add to the conduction band, so they are doped by adding electron-rich impurities
• p-type semiconductors are doped with an electron-deficient impurity, resulting in electron “holes” in the valence band. Electrons can jump between these holes in the valence band, allowing conduction of electricity.
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Diodes
• When a p-type semiconductor adjoins an n-type semiconductor, the result is an p-n junction
• Electricity can flow across the p-n junction in only one direction – this is called a diode
• This also allows the accumulation of electrical energy – called an amplifier
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