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Unit 3:

Chapter 11/12: Liquids, Solids and Phase Changes Homework: Read Chapter 11 and 12

Keep up with assignments

Liquids and solids are quite different

from gases due to their attractive

forces between the particles.

Intermolecular dispersion forces

increase attractions as the surface

area increases. The foot pads of a

gecko demonstrate this and allow a

gecko to climb on vertical glass or

even ceilings. Millions of microhairs

(setae) line the toes. Each seta

branches to hundred flattened tips called spatula. The unusually close contact with a

surface allows for the short distance intermolecular forces to attract. Geckos can adhere to

almost any surface

Review Lewis Structures/how to determine polarity of substance: ion, polar, nonpolar.

Polar verses Nonpolar particles:

Polar molecules have an overall dipole moment ( that we learned about earlier.

This dipole moment occurs when polar covalent bonds within a molecule do not

cancel each other out.

Nonpolar molecules or particles like the noble gases have zero dipole moment.

In general, like will dissolve in like. Two polar liquids are generally miscible with

each other. Two nonpolar liquids are miscible as well. But, when you take a polar

liquid (vinegar) and a nonpolar liquid (vegetable oil) the two will be immiscible, and

due to different densities one floats on the other.

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Intramolecular Forces:

Intramolecular forces occur within a substance (molecule, ionic compound or

network solid). These very strong forces include ionic bonding and polar and

nonpolar covalent bonding. It takes high energy and/or temperatures to break apart

intramolecular bonds.

Bond energies vary around 150 to 1000kJ/mol.

Ionic lattice energies vary from around 600 to16000 kJ/mol.

When water is vaporized from a liquid to a solid the intramolecular bonding

between H-O-H does not break. H2O (l) H2O (g)

When water undergoes electrolysis: 2 H2O (l) 2 H2 (g) + O2 (g), the

intramolecular bonds between H-O-H break apart while new intramolecular bonds

H-H and O=O form.

Intermolecular Forces:

Intermolecular forces are weak attractions that hold molecules and noble gas

particles close together in a liquid or solid form. An alternative name for

intermolecular forces is the van der Waals forces. They include London Dispersion

Forces, dipole-dipole forces, and hydrogen bonds. An additional attraction is the

ion-dipole forces that occur between polar molecules and ions, as in saltwater.

Ion-Dipole Forces:

Fairly strong intermolecular attractions. Energies range 10-50 kJ/mol

Electrical interactions attract between the charge of an ion and the

opposite partial charge on the polar molecule solvent.

Solvent cage (polar water) surrounds an ion (Na+1 or Cl-1) in a way that

opposite charges attract. (NaCl and water)

Interaction energy, E = z/r2, where z is the ion charge, is the dipole

moment on the ion and r is the distance between.

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London Dispersion Forces:

Weaker of the intermolecular attractions. Energies range 1-10 kJ/mol

LD Forces exist in all molecules (polar and nonpolar) and in noble

gases.

Atoms develop temporary dipole arrangement of charges as an electron

moves around the nucleus.

This instantaneous dipole induces a similar dipole on neighbors

Interatomic interaction is weak and short-lived

The magnitude of the induced dipole depends on several factors

larger the molecule, greater volume of electron cloud

greater the molecular weight, more electrons, larger e-1 cloud

more spread out the shape, maximizing the surface area, surface to

surface contact

… the GREATER the attraction.

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Dipole - Dipole Forces:

Intermolecular attractions found in polar molecules. Typical energy

range 3-4 kJ/mol

DD forces are stronger as the distance between the opposite partial

charges are closer.

DD forces are stronger as the charge differentials between the opposite

partial charges (higher dipole moment, ) are greater.

Hydrogen Bonds:

Stronger of the weak intermolecular attractions. Energies 10-40 kJ/mol

H-bonding is a special case of DD forces in which the distance is small

and the dipole moment is large

H-bonding occurs between molecules which have H directly bonded

to F, O, or N within the molecule.

F, O and N all have nonbonding electrons that play a role in creating

the H-bond.

Water has especially strong attractions since it can create two H-bonds

from each O and an H bond from each H in a water molecule.

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Example 2:

Identify the expected type of intermolecular attractions in the following.

a) NH4Br (aq)

b) C2H4

c) HF

d) Ne

e) C6H5NH2

f) H2S

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Example 3:

Choose the substance that has the stronger intermolecular attraction in each,

explain your choice.

a) C4H10 vs C3H7OH

b) C2H6 vs Ne

c) H2S vs H2O

d) HCl vs HI

e) Normal pentane, C5H12 vs 2,2-dimethylpropane, C5H12

Example 4:

Identify the type of attractions (intermolecular for molecules or noble gas, or

intramolecular for ionic and network solids)

Order to the best of your ability the lowest to highest expected melting point.

NaCl, C(diamond), He, C2H6, H2O, CH2O, KI, MgS

Properties of Liquids:

Surface Tension: Resistance to spreading out and increasing surface area.

Molecules at the surface feel attractive forces on only one side and are pulled

toward the liquid causing the liquid to bead up (spherical droplets)

Small, nonpolar molecules with weak intermolecular attractions have low

surface tension

Viscosity: Resistance to flow.

Small, nonpolar molecules with weak intermolecular attractions have low

viscosity (CCl4)

Large, spread out, molecules with stronger intermolecular forces have higher

viscosity (maple syrup)

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Capillary Action: ability of liquid to rise up inside a narrow tube.

For liquid to rise in a capillary, both adhesive (between the liquid and the tube

surface) attraction and cohesive (liquid to itself) attraction must be present.

Narrower tubes will have a greater rise. Water will rise inside a glass tube

well since it has both strong adhesive and cohesive forces. Mercury lacks

adhesive attraction and will not rise in a capillary tube.

Evaporation Rate: Rate in which liquid will vaporize in an open container.

Vapor Pressure: The vapor pressure above a liquid inside a closed container

Normal Melting Point: the temperature at 1 atmosphere pressure that the solid

changes to liquid.

Normal Boiling Point: the temperature at 1 atmosphere pressure that the liquid

converts to gas.

Enthalpy of Fusion: Energy required converting 1 mole of solid to liquid at its

melting point.

Enthalpy of Vaporization: Energy required converting 1 mole of liquid to gas at its

melting point.

The Clausius Clapeyron Equation:

Vapor pressure exponentially rises as the temperature increases. The Clausius-

Clapeyron equation below shows the relationship between vapor pressure,

temperature and enthalpy of vaporization.

ln(P2/P1) = (Hvap/R)(1/T1 -1/T2 )

Example 6:

Chloroform (CHCl3) has a normal boiling point of 62.0˚C. Its enthalpy of

vaporization is 29.2 kJ/mol. Determine the vapor pressure of chloroform in a closed

container at 24˚C.

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Solution miscibility: Solubility depends, in part, on the attractive forces of the

solute and solvent molecules, like dissolves like, miscible liquids will always

dissolve in each other.

Polar substances dissolve in polar solvents

hydrophilic groups = OH, CHO, C=O, COOH, NH2, Cl

Nonpolar molecules dissolve in nonpolar solvents

hydrophobic 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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Phase Changes:

Melting (fusion), freezing, vaporization, condensation, sublimation, deposition.

Heating or Cooling Curves and calculating the energy involved.

Q = msT for heating or cooling a constant phase. (s in J/mol°C)

Q = nH for phase changes (n for moles, H is generally kJ/mol)

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Example 5:

How much energy is required to change 9.00 grams of ice (H2O solid) at -25.0˚C to

steam (H2O gas) at 125.0˚C?

Given: specific heat capacities:

2.03 J/g˚C for H2O solid, 4.184 J/g˚C for H2O liquid, 2.0 J/g˚C for H2O gas

Hfus = 6.01 kJ/mol; Hvap = 40.67 kJ/mol

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Phase Diagrams:

Pressure (y axis) verses Temperature (x axis)

Solid, liquid, gas, supercritical fluid, triple point normal melting and boiling points.

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Phase diagram for CO2

Try This:

Draw a phase diagram with Pressure on the y axis and Temperature on the x axis that

corresponds to the following information…

The triple point is 0.50 atm, 40 ˚C

The normal melting point is 42˚C and the normal boiling point is 170˚C

The supercritical point is 12 atm and 220˚C

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Solids:

Amorphous: no long range order (obsidian, glass)

Crystalline: has a repeating pattern of angles, distances between atoms.

Unit Cells, the Structure of Crystalline Solids:

Simple cells, Coordination number

Simple cubic, 6

Body centered cubic, 8

Closest packing: With spheres it is more efficient to offset rows of atoms…

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Hexagonal closest packed, 12, abababab

Cubic closest packed, 12, abcabcabc, face centered cubic

Counting atoms in a unit cell:

body = 1

face = ½

edge = ¼

corner = 1 8⁄

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Example 7:

Draw a NaCl cell. Each corner and face has a sodium ion. Chloride ions are

placed interstitially on 12 edges and in the center. How many Na and Cl

atoms make up one unit cell?

Types of Solids:

Ionic

Lattice sites occupied by ions

Held together by attractions between oppositely charged ions, 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

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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

Molecular (polar and nonpolar)

The lattice sites are occupied by molecules: CO2, H2O, C12H22O11

The molecules are held together by intermolecular attractive forces:

dispersion forces, dipole–dipole attractions, and H-bonds

Because the attractive forces are weak, they tend

to have low melting points, generally < 300 °C

Covalent Network

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 points, generally >

1000 °C

Dimensionality of the network affects other

physical properties

Quartz (SiO2) is shown. Melts at ~1600 °C, Very hard

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 structure, maximizes attractive

forces and minimizes energy

Metallic

Solid held together by metallic bonds, strength varies with sizes and charges

of cations; coulombic attractions

Melting point varies

Mostly closest-packed arrangements of the lattice points, cations and sea of

electrons

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Sea of electrons

band theory

Metal Alloys

substitutional alloys

brass: 2/3 Cu, 1/3 Zn

sterling silver: 93% Ag, 7% Cu

pewter: 96% Sn, 4% Cu

interstitial alloys

steel: Fe with 0-1.5% C

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

• 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

• HOMO, highest occupied molecular orbital

• LUMO, lowest unoccupied molecular orbital

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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 gap

• The larger the band gap, the fewer electrons there are with enough energy to make the jump

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

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• 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.

• 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

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Solids and Liquids: Example Problems:

1. Identify for the following if there is a dipole moment or not.

CO2 SF4 BrF5 BeH2 H2O NH3 C3H8

2. The dipole moment of ClF is observed to be 0.88 D. Its bond length is 163 pm. Solve

for the calculated dipole moment if it was “ ionic” [Cl]+1[F]-1 using = Q x r. Solve

for the percent ionic character in the actual Cl-F bond. (1 D =

3.336 x 10-30 C·m. The charge of a single electron is 1.60 x 10-19 C.

3. Identify all of the following which exhibit ion-dipole forces.

NaCl (s) NaCl (aq) Na (s) Cl2 (g) K2CO3 (aq)

4. Which is expected to have the largest London Dispersion Forces?

a) C2H6 b) C8H18 c) N2 d) CO2 e) C3H8

5. Describe the essential nature of the following and include example molecules.

a) London Dispersion Forces

b) Dipole-dipole attractions

c) H- Bonding

d) Ion-dipole attractions

6. Identify the types of intermolecular attractions (London Dispersion, Dipole-dipole,

Hydrogen bonding) and/or intramolecular attractions (Covalent or Ionic bonding) that

are present in each of the following.

NaBr CH2Br2 BeH2 H2O Al2S3 C3H8 SiC

7. Arrange the following in order of expected increasing boiling point.

CH3CH2OH CH3CH2CH3 H3C–O–CH3 CH3CH2NH2

8. Briefly describe each of the following terms and predict what is expected for the

following as liquids as the attraction forces get stronger.

a) Will viscosity increase or decrease?

b) Will surface tension increase or decrease?

c) Will normal melting point increase or decrease?

d) Will normal boiling point increase or decrease?

e) Will enthalpy of vaporization (Hvap) increase or decrease?

f) Will vapor pressure increase or decrease?

g) Will the evaporation rate increase or decrease?

9. Ether has a normal boiling point of 34.6°C and the enthalpy of vaporization for ether

is 28.6kJ/mol. Calculate the vapor pressure of ether at 20.0C?

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10. A kitchen pressure cooker operates at 1.50 atm. The Hvap of water is 40.7 kJ/mole.

What is the boiling point of water in the pressure cooker in ˚C? The normal boiling

point of water is 100˚C.

11. When a substance melts from solid to liquid at the normal melting point, will the sign

for Hfus be positive or negative? And will the Sfus be positive or negative?

12. A 250-gram sample of liquid mercury at 30C is added to a large quantity of liquid

nitrogen kept at its normal boiling point of -196C in a thermally insulated container.

Information:

Normal freezing point for Hg = -39C

Specific heat for liquid Hg = 0.138 J/gC

Specific heat of solid Hg = 0.126 J/gC

Enthalpy of fusion for Hg = 11.5 J/g

Normal boiling point liquid N2= -196C

Enthalpy of Vaporization N2 = 199 J/g

a) Solve for the total joules of energy lost by the mercury as it goes from a liquid at 30C to a solid at

-196C

b) Remembering the law of conservation of energy states that the energy lost is equal but opposite in

sign as the energy gained. What mass of liquid nitrogen must be vaporized as Hg is brought to the

temperature of liquid N2 (-196C)?

13. How many atoms are inside a body centered cubic cell made up of atoms in each

corner and one in the center.

14. How many atoms are inside a face centered cubic cell made up of atoms in each corner

and one on each face.

15. A binary ionic compound MxAy, crystallizes in a cubic structure that contains eight

anions entirely within the unit cell and the metal cations are on each corner and on

each face. What is the empirical formula?

16. Describe the essential nature/characteristics of the following types of solids and

include example molecules.

a) Polar molecule

b) Nonpolar molecule

c) Ionic

d) Covalent network

e) Metallic

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17. Use the Phase Diagram for questions below.

atm.

1.0 0.8 0.6 0.4 0.2 0.0

2.0 1.8 1.6 1.4 1.2P

T °C

0 10 20 30 40 50 60 70 80 90 100 110

AT

CD

EF

G

H

I

J

a) What phase is represented by E F G

b) What is T and why is it significant?

c) What is C and why is it significant?

d) What two phases are in equilibrium at H I J

e) Name the phase changes occurring on the lines TD

TC TA

f) Estimate the normal melting point

g) Estimate the normal boiling point

h) What phase exists at 0.80 atm and 40°C?

i) Starting at point J if the temperature increases 50°C the phase exists.

By increasing the pressure at this new point by 1 atm the new phase is