Forces Chapter 11 Liquids, Solids, and Intermolecular

(1)

Sherril Soman

Grand Valley State University

Lecture Presentation

Chapter 11 Liquids, Solids, and Intermolecular

Forces

11.1 Climbing Geckos and Intermolecular Forces 482

11.2 Solids, Liquids, and Gases: A Molecular Comparison 484

11.3 Intermolecular Forces: The Forces That Hold Condensed States Together 487

11.4 Intermolecular Forces in Action: Surface Tension, Viscosity,

and Capil lary Action 497

11.5 Vaporization and Vapor Pressure 499 11.6 Sublimation and Fusion 509

11.7 Heating Curve for Water 511 11.8 Phase Diagrams 513

11.9 Water: An Extraordinary Substance 516 11.10 Crystalline Solids: Determining Their Structure by X-Ray Crystallogra phy 518 11.11 Crystalline Solids: Unit Cells and Basic Structures 520

11.12 Crystalline Solids: The Fundamental Types 526

11.13 Crystalline Solids: Band Theory 530

Key Learning Outcomes 534

(2)

(3)

Climbing Geckos

11.1 Climbing Geckos and Intermolecular Forces

11.1

(4)

Three Phases of Water

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 density 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. 11.2

(5)

11.2 Solids, Liquids, and Gases: A Molecular Comparison

• Fixed = keeps shape when placed in a container

• Indefinite = takes the shape of the container

Properties of the Three Phases of Matter

11.2

(6)

Solids

• Some solids have their particles arranged in an orderly geometric pattern; we call these crystalline solids.

– Salt and diamonds

• Other solids have particles that do not show a regular

geometric pattern over a long range; we call these amorphous solids.

– Plastic and glass

11.2

(7)

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.

11.2

(8)

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 particles, compared to the size of the particles.

– Therefore, the molar volume of the gas state of a material is much

larger than the molar volume of the solid or liquid states.

11.2

(9)

Phase Changes

11.2

(10)

11.3 Intermolecular Forces: The Forces That Hold Condensed States Together

• 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.

Kinds of Attractive Forces

11.3

(11)

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 forces.

– Aka 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.

11.3

(12)

Dispersion Force

11.3

(13)

Dispersion Forces Illustrated (1)

At a given instant, electron density, even in a nonpolar molecule like this one, is not perfectly uniform.

11.3

(14)

Dispersion Forces Illustrated (2)

The region of

(momentary) higher electron density attains a small (–) charge …

When another nonpolar molecule approaches …

… the other end of the molecule is slightly (+).

11.3

(15)

Dispersion Forces Illustrated (3)

… this molecule induces a tiny dipole moment …

… in this molecule.

Opposite charges ________.

11.3

(16)

Size of the Induced Dipole

• The magnitude of the induced dipole depends on several factors.

• 1. Polarizability of the electrons

– Volume of the electron cloud

– Larger molar mass = more electrons = larger electron cloud = increased

polarizability = stronger attractions

• 2. Shape of the molecule

– More surface-to-surface contact = larger induced dipole = stronger attraction

11.3

(17)

Effect of Molecular Size on Size of Dispersion Force

The Noble gases are all nonpolar atomic elements.

The stronger the attractive forces

between the molecules, the higher the boiling

point will be.

11.3

(18)

Effect of Molecular Shape on Size of Dispersion Force

11.3

small, compact, symmetrical molecules elongated molecules

(19)

Dipole–Dipole Attractions

11.3

(20)

Effect of Dipole–Dipole Attraction on Boiling and Melting Points

11.3

(21)

Dipole Movement and Boiling Point

11.3

(22)

Attractive Forces and Solubility

• 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, NH

₂

, Cl

• Nonpolar molecules dissolve in nonpolar solvents.

– Hydrophobic groups = C—H, C—C

Pentane, C

₅

H

₁₂

is a nonpolar molecule.

Water is a polar molecule.

Immiscible Liquids

11.3

(23)

O

₂

and N

₂

and both substances are nonpolar;

M.W.: O

₂

> N

₂

; O

₂

has the higher boiling point

NO has the highest boiling point of the three because of its additional permanent dipole.

Boiling point: NO > O

₂

> N

₂

Please the following substances in order of decreasing boiling point.

N

₂

NO O

₂

11.3

(24)

Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro

Example 11.1 Dipole–Dipole Forces

A molecule has dipole–dipole forces if it is polar. To determine if a molecule is polar, (1) determine if the molecule contains polar bonds and (2) determine if the polar bonds add together to form a net dipole moment (Section 9.6).

(a) CO₂

(1) Since the electronegativity of carbon is 2.5 and that of oxygen is 3.5

(Figure 9.8), CO₂ has polar bonds.

(2) The geometry of CO₂ is linear.

Consequently, the dipoles of the polar bonds cancel, so the molecule is not polar and

does not have dipole–dipole forces.

Solution

Which of these molecules have dipole–dipole forces?

a. CO

₂

b. CH

₂

CI

₂

c. CH

₄

FIGURE 9.8 Electronegativities of the Elements Electronegativity generally increases as we move across a row in the periodic table and decreases as we move down a column.

11.3

(25)

Example 11.1 Dipole–Dipole Forces

(c) CH₄

(1) Since the electronegativity of C is 2.5 and that of hydrogen is 2.1, the C—H bonds are nearly nonpolar.

(2) In addition, since the geometry of the molecule is tetrahedral, any slight polarities that the bonds might have will cancel. CH₄ is therefore nonpolar and

does not have dipole–dipole forces.

For Practice 11.1

Which molecules have dipole–dipole forces?

a. CI₄ b. CH₃Cl c. HCI Continued

(b) CH₂Cl₂

(1) The electronegativity of C is 2.5, that of H is 2.1, and that of Cl is 3.0.

Consequently, CH₂Cl₂ has two polar bonds (C—Cl) and two bonds that are nearly nonpolar (C—H).

(2) The geometry of CH₂Cl₂ is tetrahedral. Since the C—Cl bonds and the C—H bonds are different, their dipoles do not cancel but sum to a net dipole moment. The molecule is polar and

has dipole–dipole forces.

11.3

(26)

Hydrogen Bonding

• When a very electronegative atom is bonded to hydrogen, it strongly pulls the bonding

electrons toward it.

– O─H, N─H, or F─H

11.3

(27)

Boiling Points of Group 4A and 6A Compounds

11.3

(28)

Hydrogen Bonding Forces Comparison

• Trend of bonding force in single H-bond:

H—F··· H—F > H—O··· H—O > H—N··· H—N

F H

F H F

H 2 H-bond/per molecule

1 H-bond/per F atom 140o

O H H

O H H O

H H

O H H

4 H-bond/per molecule 2 H-bond/per O atom

• Trend of boiling points:

Boiling point: H ₂ O > HF > NH ₃

MW: 18 20 17 _11.3

(29)

Example 11.2 Hydrogen Bonding

One of these compounds is a liquid at room temperature. Which one and why?

Solution

The three compounds have similar molar masses:

Formaldehyde 30.03 g/mol

Fluoromethane 34.03 g/mol

Hydrogen peroxide 34.02 g/mol

So the strengths of their dispersion forces are similar. All three compounds are also polar, so they have dipole–dipole forces. Hydrogen peroxide, however, is the only one of these compounds that also contains H bonded directly to F, O, or N. Therefore, it also has hydrogen bonding and is likely to have the highest boiling point of the three. Since the example stated that only one of the compounds was a liquid, you can safely assume that hydrogen peroxide is the liquid. Note that, although fluoromethane contains both H and F, H is not directly bonded to F, so fluoromethane does not have hydrogen bonding as an intermolecular force. Similarly, formaldehyde contains both H and O, but H is not directly bonded to O, so formaldehyde does not have hydrogen bonding either.

For Practice 11.2

Which has the higher boiling point, HF or HCl? Why?

H-bond

11.3

(30)

11.4 Intermolecular Forces in Action: Surface Tension, Viscosity, and Capillary Action

Surface Tension

11.4 Spherical Water

Droplets

(31)

Viscosity

• Viscosity is the resistance of a liquid to flow.

• Larger intermolecular attractions = larger viscosity

11.4 • T cP

(32)

Capillary Action

• Capillary action is the ability of a liquid to flow up a thin tube against the influence of gravity.

– The 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 together.

– Adhesive forces attract the outer liquid molecules to the tube’s surface.

11.4

(33)

Slide 33 of 61

Meniscus formation

General Chemistry: Chapter 12

Figure 12-14 Capillary Action Figure 12-12

Wetting of a surface

co > ad

co(like) > ad(unlike) co < ad

co < ad

 FIGURE 12-13

co < ad

11.4

(34)

11.5 Vaporization and Vapor Pressure The Molecular Dance

melting (s) (l) vaporization (l) (g)

sublimation (s) (g)

freezing (l) (s) condensation (g) (l)

deposition (g) (s)

11.5

(35)

Heat of Vaporization

• The amount of heat energy required to vaporize one mole of the liquid is called the heat of

vaporization, DH _vap .

– Sometimes called the enthalpy of vaporization

• It is always endothermic; therefore, DH _vap is +.

• It is somewhat temperature dependent.

 DH condensation = −DH vaporization

liquid gas(vapor)

vaporization condensation

11.5

(36)

Example 11.3 Using the Heat of Vaporization in Calculations

Sort

Calculate the mass of water (in g) that can be vaporized at its boiling point with 155 kJ of heat.

You are given a certain amount of heat in kilojoules and asked to find the mass of water that can be vaporized.

Strategize

The heat of vaporization gives the relationship between heat absorbed and moles of water vaporized. Begin with the given amount of heat (in kJ) and convert to moles of water that can be vaporized. Then use the molar mass as a conversion factor to convert from moles of water to mass of water.

Conceptual Plan

Given: 155 kj

Find: gH₂O

11.5

(37)

Example 11.3 Using the Heat of Vaporization in Calculations

Continued

For Practice 11.3

Calculate the amount of heat (in kJ) required to vaporize 2.58 kg of water at its boiling point.

For More Practice 11.3

Suppose that 0.48 g of water at 25 °C condenses on the surface of a 55 g block of aluminum that is initially at 25 °C.

If the heat released during condensation goes only toward heating the metal, what is the final temperature (in °C) of the metal block? (The specific heat capacity of aluminum is 0.903 J/g °C.)

Solution

Relationships Used

DH_vap = 40.7 kJ/mol (at 100 °C) 18.02 g H₂O = 1 mol H₂O

Solve

Follow the conceptual plan to solve the problem.

11.5

(38)

Dynamic Equilibrium

11.5 liquid gas(vapor)

vaporization condensation

(39)

Vapor Pressure

• the weaker the attractive forces, the higher the vapor pressure.

11.5 Vapor pressure illustrated

• FIGURE 12-17

Mercury manometer

Vapor pressure of liquid

P

_vap

independent of V

_liq

P

_vap

independent of V

_gas

P

_vap

dependent on

T

(40)

Vapor Pressure Curves

11.5

(41)

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. P T

11.5

(42)

Slide 42 of 61

An Equation for Expressing Vapor Pressure Data

General Chemistry: Chapter 12

 FIGURE 12-20

 Vapor pressure data plotted as lnP versus 1/T

ln P = -A ( ) + B 1 T

A = ΔH

_vap

R

ln = - ( - ) P

₂

P

₁

1 T

₂

1 T

₁

ΔH

_vap

R

Fig11-28, nonlinear linear

11.5

(43)

Clausius–Clapeyron Equation

• The slope of the line × 8.314 J/mol ∙ K = DH _vap .

 In J/mol

ln P = -A ( ) + B 1 T

A = ΔH

_vap

R

11.5

(44)

Example 11.4 Using the Clausius–Clapeyron Equation to Determine Heat of Vaporization from Experimental

Measurements of Vapor Pressure

The vapor pressure of dichloromethane was measured as a function of temperature, and the following results were obtained:

Determine the heat of vaporization of dichloromethane.

A =

ΔH

_vap

R

11.5

(45)

Example 11.4 Using the Clausius–Clapeyron Equation to Determine Heat of Vaporization from Experimental

Measurements of Vapor Pressure

Solution

Continued

To find the heat of vaporization, use an Excel spreadsheet or a graphing calculator to make a plot of the natural log of vapor pressure (ln P) as a function of the inverse of the temperature in kelvins (1/T). Then fit the points to a line and determine the slope of the line. The slope of the best-fitting line is –3773 K. Since the slope equals –ΔH_vap/R, we find the heat of vaporization as follows:

A = ΔH

_vap

R

11.5

(46)

Example 11.5 Using the Two-Point Form of the Clausius–Clapeyron Equation to Predict the Vapor Pressure at a Given Temperature

Sort

You are given the normal boiling point of methanol (the temperature at which the vapor pressure is 760 mmHg) and the heat of vaporization. You are asked to find the vapor pressure at a specified temperature that is also given.

Methanol has a normal boiling point of 64.6 °C and a heat of vaporization (Δ Hvap) of 35.2 kJ/mol. What is the vapor pressure of methanol at 12.0 °C?

Given:

Find:

ln = - ( - ) P

₂

P

₁

1 T

₂

1 T

₁

ΔH

_vap

R

11.5

(47)

Continued

Strategize

The conceptual plan is essentially the Clausius–Clapeyron equation, which relates the given and find quantities.

(Clausius–Clapeyron equation, two-point form)

Conceptual Plan

Solve

First, convert T₁and T₂from C to K.

Then, substitute the required values into the Clausius–Clapeyron equation and solve for P₂.

Example 11.5 Using the Two-Point Form of the Clausius–Clapeyron Equation to Predict the Vapor Pressure at a Given Temperature

11.5

(48)

Solution

Continued

Example 11.5 Using the Two-Point Form of the Clausius–Clapeyron Equation to Predict the Vapor Pressure at a Given Temperature

ln = - ( - ) P

₂

P

₁

1 T

₂

1 T

₁

ΔH

_vap

R

11.5

(49)

Continued

Check

The units of the answer are correct. The magnitude of the answer makes sense because vapor pressure should be significantly lower at the lower temperature.

For Practice 11.5

Propane has a normal boiling point of –42.0 °C and a heat of vaporization (Δ H^vap) of 19.04 kJ/mol. What is the vapor pressure of propane at 25.0 °C?

Example 11.5 Using the Two-Point Form of the Clausius–Clapeyron Equation to Predict the Vapor Pressure at a Given Temperature

11.5

(50)

Supercritical Fluid

• As a liquid is heated in a sealed container, more vapor collects, causing the pressure inside the container to rise, 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.

11.5

(51)

• 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.

The Critical Point

11.5

(52)

11.6 Sublimation and Fusion

solid gas sublimation deposition

Sublimation and Deposition

melting (s) (l) vaporization (l) (g) sublimation (s) (g)

freezing (l) (s) condensation (g) (l) deposition (g) (s)

11.6

(53)

Melting and Freezing

melting (s) (l) vaporization (l) (g) sublimation (s) (g)

freezing (l) (s) condensation (g) (l) deposition (g) (s)

11.6 Melting = Fusion

(54)

Heat of Fusion

• The amount of heat energy required to melt one mole of the solid is called the heat of fusion, DH

_fus

.

– Sometimes called the enthalpy of fusion

• It is always endothermic; therefore, DH

_fus

is +.

• It is somewhat temperature dependent.

 DH

_freezing

= −DH

_fusion

 Generally much less than DH

_vap

 DH

sublimation

= DH

_fusion

+ DH

vaporization

11.6

(55)

Heats of Fusion and Vaporization

11.6

(56)

11.7 Heating Curve of Water

11.7

(57)

Segment 1 solid (Temp)

• Heating 1.00 mole of ice at −25.0 °C up to the melting point, 0.0 °C

• q = mass × C _s × DT

– Mass of 1.00 mole of ice = 18.0 g – C

_s

= 2.09 J/mol ∙ °C

11.7

(58)

Segment 2

phase change (s+l)

• Melting 1.00 mole of ice at the melting point, 0.0 °C

• q = n ∙ DH _fus

– n = 1.00 mole of ice – DH

_fus

= 6.02 kJ/mol

11.7

(59)

Segment 3

liquid(Temp)

• Heating 1.00 mole of water at 0.0 °C up to the boiling point, 100.0 °C

• q = mass × C _s × DT

– Mass of 1.00 mole of water = 18.0 g – C

_s

= 2.09 J/mol ∙ °C

11.7

(60)

Segment 4

phase change l+g

• Boiling 1.00 mole of water at the boiling point, 100.0 °C

• q = n ∙ DH _vap

– n = 1.00 mole of ice – DH

_fus

= 40.7 kJ/mol

11.7

(61)

Segment 5 gas(Temp)

• Heating 1.00 mole of steam at 100.0 °C up to 125.0 °C

• q = mass × C _s × DT

– Mass of 1.00 mole of water = 18.0 g – C

_s

= 2.01 J/mol ∙ °C

11.7

(62)

11.8 Phase Diagrams

A—B, solid-vapor equilibrium.

A—D, solid-liquid equilibrium.

A—C, liquid-vapor equilibrium.

Triple point

deposition

11.8

(63)

11.8

(64)

Phase Diagrams for Other Substances

11.8

(65)

11.9 Water – An Extraordinary Substance

11.9 • The hydrogen bonds present in water

result in a relatively high boiling point.

(66)

Hydrogen Bonding Forces Comparison

• Trend of bonding force in single H-bond:

H—F··· H—F > H—O··· H—O > H—N··· H—N

F H

F H F

H 2 H-bond/per molecule

1 H-bond/per F atom 140o

O H H

O H H O

H H

O H H

4 H-bond/per molecule 2 H-bond/per O atom

• Trend of boiling points:

Boiling point: H ₂ O > HF > NH ₃

MW: 18 20 17 _11.3

(67)

11.10 Crystalline Solids: Determining

Their Structure by X-Ray Crystallography

Diffraction from a Crystal

(68)

X-Ray Diffraction Analysis

(69)

Example 11.6 Using Bragg’s Law

Solution

When an X-ray beam of λ = 154 pm was incident on the surface of an iron crystal, it produced a maximum

reflection at an angle of θ = 32.6°. Assuming n = 1, calculate the separation between layers of iron atoms in the crystal.

To solve this problem, use Bragg’s law in the form given by Equation 11.8. The distance, d, is the separation between layers in the crystal.

For Practice 11.6

The spacing between layers of molybdenum atoms is 157 pm. Calculate the angle at which 154 pm X-rays produces a maximum reflection for n = 1.

(70)

11.11 Crystalline Solids: Unit Cells and Basic Structures

• Unit cells are three-dimensional.

– Usually containing two or three 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.

Unit Cells

(71)

Orthorhombic a ≠ b ≠ c

all 90°

Seven Unit Cells

Hexagonal a = c < b 2 faces 90°

1 face 120°

Cubic a = b = c

all 90°

Tetragonal a = c < b

all 90°

Monoclinic a ≠ b ≠ c 2 faces 90°

Rhombohedral a = b = c

no 90°

Triclinic

a ≠ b ≠ c

no 90°

(72)

The Cubic Crystalline Lattices

(73)

Simple Cubic

(74)

Body-Centered Cubic

(75)

Face-Centered Cubic

(76)

Example 11.7 Relating Density to Crystal Structure

Aluminum crystallizes with a face-centered cubic unit cell. The radius of an aluminum atom is 143 pm. Calculate the density of solid crystalline aluminum in g/cm³.

Sort

You are given the radius of an aluminum atom and its crystal structure. You are asked to find the density of solid aluminum.

Strategize

The conceptual plan is based on the definition of density.

Since the unit cell has the physical properties of the entire crystal, you can find the mass and volume of the unit cell and use these to calculate its density.

Solve

Begin by finding the mass of the unit cell. Determine the mass of an aluminum atom from its molar mass. Since the face-centered cubic unit cell contains four atoms per unit cell, multiply the mass of aluminum by 4 to get the mass of a unit cell.

Given:

Find: d

r = 143 pm, face-centered cubic

(77)

Example 11.7 Relating Density to Crystal Structure

Continued

d = m/V

m = mass of unit cell

= number of atoms in unit cell × mass of each atom V = volume of unit cell

= (edge length)³

Solution

Conceptual Plan

(78)

Example 11.7 Relating Density to Crystal Structure

Continued

Next, calculate the edge length (l) of the unit cell (in m) from the atomic radius of aluminum. For the face-centered cubic structure, l = 2 2r.

Calculate the volume of the unit cell (in cm) by converting the edge length to cm and cubing the edge length. (Use centimeters because you will want to report the density in units of g/cm³.)

Finally, calculate the density by dividing the mass of the unit cell by the volume of the unit cell.

(79)

Example 11.7 Relating Density to Crystal Structure

Continued

Check

The units of the answer are correct. The magnitude of the answer is reasonable because the density is greater than 1 g/cm³ (as you would expect for metals), but still not too high (because aluminum is a low-density metal).

For Practice 11.7

Chromium crystallizes with a body-centered cubic unit cell. The radius of a chromium atom is 125 pm. Calculate the density of solid crystalline chromium in g/cm³.

(80)

Closest-Packed Structures First 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.

(81)

Closest-Packed Structures Second 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.

(82)

Hexagonal Closest-Packed Structures

(83)

Cubic Closest-Packed Structures

(84)

11.12 Crystalline Solids: The Fundamental Types

• Crystalline solids are classified by the kinds of particles found.

• Some of the categories are subclassified by the kinds of attractive forces holding the particles together.

Classifying Crystalline

Solids

(85)

Types of Crystalline Solids

(86)

11.13 Classifying Crystalline Solids

• 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.

Band Theory

(87)