general chemistry 202 chm202 general...

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General Chemistry 202 – CHM202

General Information

• Instructor

• Meeting times and places

• Text and recommended materials

• Website

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• Schedule

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• Equipment

• Instruction Method

• Study Procedures

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Policies

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• Attendance

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• General Procedures

• Data Recording

• Cleanup

• Reports


Laboratory Guidelines

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Frequently Asked Questions

Q: Where does class meet on …?

A: It’s in the syllabus.

Q: May I leave class early tonight?

A: I’ll make a note of it.

Q: I didn’t do well on the test. When can I

make it up?


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Q: I made a mistake on the test. May I have it

back, please?

A: No.

Q: Is there a test on …?


Q: What are your office hours?

A: Posted on the class website. I will only be

there by appointment.

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Q: When can I make up the test I missed

when I was absent?


Q: What if the absence wasn’t excused?


Q: How do I get my absence excused?

A: Offer me an excuse. (I may ask for

documentation.)

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Q: May I have extra time for my test?

A: Please make arrangement through the

proper channels. Such accomodations

require documentation.

Q: May I attend the earlier or later lab?

A: No.

Q: May I attend the earlier lecture?

A: By request, but space is limited

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Intermolecular Forces and

Liquids and Solids

Chapter 11

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

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A phase is a homogeneous part of the system in contact with

other parts of the system but separated from them by a well-

defined boundary.

2 Phases

Solid phase - ice

Liquid phase - water

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Intermolecular Forces

Intermolecular forces are attractive forces between molecules.

Intramolecular forces hold atoms together in a molecule.

Intermolecular vs Intramolecular

• 41 kJ to vaporize 1 mole of water (inter)

• 930 kJ to break all O-H bonds in 1 mole of water (intra)

Generally, intermolecular

forces are much weaker

than intramolecular forces.

“Measure” of intermolecular force

boiling point

melting point

DHvap

DHfus

DHsub

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Dipole-Dipole Forces

Attractive forces between polar molecules

Orientation of Polar Molecules in a Solid

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Ion-Dipole Forces

Attractive forces between an ion and a polar molecule

Ion-Dipole Interaction

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in solution

Interaction Between Water and Cations

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Dispersion Forces

Attractive forces that arise as a result of temporary

dipoles induced in atoms or molecules

ion-induced dipole interaction

dipole-induced dipole interaction

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Induced Dipoles Interacting With Each Other

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Dispersion Forces Continued

Polarizability is the ease with which the electron distribution

in the atom or molecule can be distorted.

Polarizability increases with:

• greater number of electrons

• more diffuse electron cloud

Dispersion forces usually

increase with molar mass.

Example 11.1

What type(s) of intermolecular forces exist between the

following pairs?

(a) HBr and H2S

(b) Cl2 and CBr4

(c) I2 and

(d) NH3 and C6H6

Example 11.1

Strategy Classify the species into three categories: ionic, polar

(possessing a dipole moment), and nonpolar. Keep in mind that

dispersion forces exist between all species.

Solutions

(a) Both HBr and H2S are polar molecules. Therefore, the

intermolecular forces present are dipole-dipole forces, as

well as dispersion forces.

(b) Both Cl2 and CBr4 are nonpolar, so there are only dispersion

forces between these molecules.

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Example 11.1

(c) I2 is a homonuclear diatomic molecule and therefore

nonpolar, so the forces between it and the ion, , are

ion-induced dipole forces and dispersion forces.

(d) NH3 is polar, and C6H6 is nonpolar. The forces are dipole-

induced dipole forces and dispersion forces.

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Hydrogen Bond

The hydrogen bond is a special dipole-dipole interaction

between the hydrogen atom in a polar N-H, O-H, or F-H bond

and an electronegative O, N, or F atom.

A H…B A H…Aor

A & B are N, O, or F

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Hydrogen Bond

HCOOH and water

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Why is the hydrogen bond considered a “special”

dipole-dipole interaction?

Decreasing molar mass

Decreasing boiling point

Example 11.2

Which of the following can form hydrogen bonds with water?

a) CH3OCH3

b) CH4

c) F2

d) HCOOH

e) Na+

Example 11.2

Strategy A species can form hydrogen bonds with water if it

contains one of the three electronegative elements (F, O, or N)

or it has a H atom bonded to one of these three elements.

Solution There are no electronegative elements (F, O, or N) in

either CH4 or Na+. Therefore, only CH3OCH3, F2, and HCOOH

can form hydrogen bonds with water.

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Example 11.2

Check Note that HCOOH (formic acid) can form hydrogen

bonds with water in two different ways.

HCOOH forms hydrogen bonds with two H2O molecules.

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Properties of Liquids

Surface tension is the amount of energy required to stretch

or increase the surface of a liquid by a unit area.

Strong

intermolecular

forces

High

surface

tension

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Cohesion is the intermolecular attraction between like molecules

Adhesion is an attraction between unlike molecules

Adhesion

Cohesion

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Viscosity is a measure of a fluid’s resistance to flow.

Strong

intermolecular

forces

High

viscosity

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Density of Water

Maximum Density

40C

Ice is less dense than water

Water is a Unique Substance3-D Structure of Water

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A crystalline solid possesses rigid and long-range order. In a

crystalline solid, atoms, molecules or ions occupy specific

(predictable) positions.

An amorphous solid does not possess a well-defined

arrangement and long-range molecular order.

A unit cell is the basic repeating structural unit of a crystalline

solid.

lattice

point

Unit Cell Unit cells in 3 dimensions

At lattice points:

• Atoms

• Molecules

• Ions

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Seven Basic Unit Cells

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Three Types of Cubic Unit Cells

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Arrangement of Identical Spheres in a Simple Cubic Cell

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Arrangement of Identical Spheres in a Body-Centered

Cubic Cell

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Shared by 8

unit cells

Shared by 4

unit cells

A Corner Atom, an Edge-Centered Atom and

a Face-Centered Atom

Shared by 2

unit cells

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Number of Atoms Per Unit Cell

1 atom/unit cell

(8 x 1/8 = 1)

2 atoms/unit cell

(8 x 1/8 + 1 = 2)

4 atoms/unit cell

(8 x 1/8 + 6 x 1/2 = 4)

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Relation Between Edge Length and Atomic Radius

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Closest Packing: Hexagonal and Cubic

hexagonal cubic

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Exploded Views

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Example 11.3

Gold (Au) crystallizes in a cubic close-packed structure (the

face-centered cubic unit cell) and has a density of 19.3 g/cm3.

Calculate the atomic radius of gold in picometers.

Example 11.3

Strategy We want to calculate the radius of a gold atom.

For a face-centered cubic unit cell, the relationship between

radius (r) and edge length (a), according to Figure 11.22,

is .

Therefore, to determine r of a Au atom, we need to find a. The

volume of a cube is . Thus, if we can

determine the volume of the unit cell, we can calculate a. We

are given the density in the problem.

Example 11.3

The sequence of steps is summarized as follows:

Solution

Step 1: We know the density, so in order to determine the

volume, we find the mass of the unit cell. Each unit cell

has eight corners and six faces. The total number of

atoms within such a cell, according to Figure 11.19, is

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Example 11.3

The mass of a unit cell in grams is

From the definition of density (d = m/V), we calculate the

volume of the unit cell as follows:

Example 11.3

Step 2: Because volume is length cubed, we take the cubic

root of the volume of the unit cell to obtain the edge

length (a) of the cell

Step 3: From Figure 11.22 we see that the radius of an Au

sphere (r) is related to the edge length by

Example 11.3

Therefore,

= 144 pm

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An Arrangement for Obtaining the X-ray Diffraction Pattern

of a Crystal.

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Extra distance = BC + CD = 2d sinq = nl (Bragg Equation)

Reflection of X-rays from Two Layers of Atoms

Example 11.4

X rays of wavelength 0.154 nm strike an aluminum crystal; the

rays are reflected at an angle of 19.3°.

Assuming that n = 1, calculate the spacing between the planes

of aluminum atoms (in pm) that is responsible for this angle of

reflection.

The conversion factor is obtained from 1 nm = 1000 pm.

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Example 11.4

Strategy This is an application of Equation (11.1).

Solution Converting the wavelength to picometers and using

the angle of reflection (19.3°), we write

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Types of Crystals

Ionic Crystals

• Lattice points occupied by cations and anions

• Held together by electrostatic attraction

• Hard, brittle, high melting point

• Poor conductor of heat and electricity

CsCl ZnS CaF2

Example 11.5

How many Na+ and Cl− ions are in each NaCl unit cell?

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Example 11.5

Solution NaCl has a structure based on a face-centered cubic

lattice. One whole Na+ ion is at the center of the unit cell, and

there are twelve Na+ ions at the edges. Because each edge

Na+ ion is shared by four unit cells, the total number of Na+ ions

is 1 + (12 × ¼) = 4.

Similarly, there are six Cl− ions at the face centers and eight Cl−

ions at the corners. Each face-centered ion is shared by two

unit cells, and each corner ion is shared by eight unit cells, so

the total number of Cl− ions is (6 × ½) + (8 × 1/8) = 4. Thus,

there are four Na+ ions and four Cl− ions in each NaCl unit cell.

Check This result agrees with sodium chloride’s empirical

formula.

Example 11.6

The edge length of the NaCl unit cell is 564 pm. What is the

density of NaCl in g/cm3?

Example 11.6

Strategy

To calculate the density, we need to know the mass of the unit

cell. The volume can be calculated from the given edge length

because V = a3. How many Na+ and Cl− ions are in a unit cell?

What is the total mass in amu? What are the conversion factors

between amu and g and between pm and cm?

Solution

From Example 11.5 we see that there are four Na+ ions and

four Cl− ions in each unit cell. So the total mass (in amu) of a

unit cell is

mass = 4(22.99 amu + 35.45 amu) = 233.8 amu

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Example 11.6

Converting amu to grams, we write

The volume of the unit cell is V = a3 = (564 pm)3. Converting

pm3 to cm3, the volume is given by

Finally, from the definition of density

= 2.16 g/cm3

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Types of Crystals

Covalent Crystals

• Lattice points occupied by atoms

• Held together by covalent bonds

• Hard, high melting point


diamond graphite

carbon

atoms

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Types of Crystals

Molecular Crystals

• Lattice points occupied by molecules

• Held together by intermolecular forces

• Soft, low melting point


water benzene

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Types of Crystals

Metallic Crystals

• Lattice points occupied by metal atoms

• Held together by metallic bonds

• Soft to hard, low to high melting point

• Good conductors of heat and electricity

Cross Section of a Metallic Crystal

nucleus &

inner shell e-

mobile “sea”

of e-

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Crystal Structures of Metals

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Types of Crystals

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An amorphous solid does not possess a well-defined

arrangement and long-range molecular order.

A glass is an optically transparent fusion product of inorganic

materials that has cooled to a rigid state without crystallizing

Crystalline

quartz (SiO2)

Non-crystalline

quartz glass

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Order

Least

Order

Phase Changes

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T2 > T1

Effect of Temperature on Kinetic Energy

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The equilibrium vapor pressure is the vapor pressure

measured when a dynamic equilibrium exists between

condensation and evaporation

H2O (l) H2O (g)

Rate of

condensation

Rate of

evaporation=

Dynamic Equilibrium

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Before

Evaporation

At

Equilibrium

Measurement of Vapor Pressure

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Molar heat of vaporization (DHvap) is the energy required to

vaporize 1 mole of a liquid at its boiling point.

ln P = -DHvap

RT+ C

Clausius-Clapeyron Equation

P = (equilibrium) vapor pressure

T = temperature (K)

R = gas constant (8.314 J/K•mol)

Vapor Pressure Versus Temperature

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Alternate Forms of the Clausius-Clapeyron Equation

At two temperatures

or

Example 11.7

Diethyl ether is a volatile, highly flammable organic liquid that is

used mainly as a solvent.

The vapor pressure of diethyl ether is 401 mmHg at 18°C.

Calculate its vapor pressure at 32°C.

Example 11.7

Strategy We are given the vapor pressure of diethyl ether at

one temperature and asked to find the pressure at another

temperature. Therefore, we need Equation (11.5).

Solution Table 11.6 tells us that DHvap = 26.0 kJ/mol. The data

are

From Equation (11.5) we have

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Example 11.7

Taking the antilog of both sides (see Appendix 4), we obtain

Hence

P2 = 656 mmHg

Check We expect the vapor pressure to be greater at the

higher temperature. Therefore, the answer is reasonable.

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The boiling point is the temperature at which the

(equilibrium) vapor pressure of a liquid is equal to the

external pressure.

The normal boiling point is the temperature at which a liquid

boils when the external pressure is 1 atm.

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The critical temperature (Tc) is the temperature above which

the gas cannot be made to liquefy, no matter how great the

applied pressure.

The critical pressure

(Pc) is the minimum

pressure that must be

applied to bring about

liquefaction at the

critical temperature.

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The Critical Phenomenon of SF6

T < Tc T > Tc T ~ Tc T < Tc

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H2O (s) H2O (l)

The melting point of a solid

or the freezing point of a

liquid is the temperature at

which the solid and liquid

phases coexist in

equilibrium.

Solid-Liquid Equilibrium

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Molar heat of fusion (DHfus) is the energy required to melt

1 mole of a solid substance at its freezing point.

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Heating Curve

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H2O (s) H2O (g)

Molar heat of sublimation

(DHsub) is the energy required

to sublime 1 mole of a solid.

DHsub = DHfus + DHvap

( Hess’s Law)

Solid-Gas Equilibrium

Example 11.8

Calculate the amount of energy (in kilojoules) needed to heat

346 g of liquid water from 0°C to 182°C.

Assume that the specific heat of water is 4.184 J/g · °C over the

entire liquid range and that the specific heat of steam is 1.99

J/g · °C.

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Example 11.8

Strategy The heat change (q) at each stage is given by q =

msDt (see p. 247), where m is the mass of water, s is the

specific heat, and Dt is the temperature change.

If there is a phase change, such as vaporization, then q is given

by nDHvap, where n is the number of moles of water.

Solution The calculation can be broken down in three steps.

Step 1: Heating water from 0°C to 100°C

Using Equation (6.12) we write

Example 11.8

Step 2: Evaporating 346 g of water at 100°C (a phase change)

In Table 11.6 we see DHvap = 40.79 kJ/mol for water, so

Step 3: Heating steam from 100°C to 182°C

Example 11.8

The overall energy required is given by

Check All the qs have a positive sign, which is consistent with

the fact that heat is absorbed to raise the temperature from 0°C

to 182°C. Also, as expected, much more heat is absorbed

during the phase transition.

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A phase diagram summarizes the conditions at which a

substance exists as a solid, liquid, or gas.

Phase Diagram of Water

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Phase Diagram of Carbon

Dioxide

At 1 atm

CO2 (s) CO2 (g)

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Effect of Increase in Pressure on the Melting Point

of Ice and the Boiling Point of Water

general chemistry 202 chm202 general...

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