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CHEM 1211

Lab manual

Logarithm Review

ab = c, (a > 0, a ≠ 1) logac = b

Definition

If a = 10, it is called common logarithm

log c = log10c

If a = e = 2.718281828459045 ∙ ∙ ∙, it is called natural logarithm

ln c = logecKeys on your calculator

c > 0, b can be any real number

Properties of Logarithm

ln(xy) = ln(x) + ln(y)

ln(xm) = m ln(x)

Also see Appendix I B

x > 0, y > 0

Chapter 11

Liquids, Solids andIntermolecular Forces

continued

Vapor Pressure

Surface Molecules

Temperature: T Temperature: TWhat happens to P if T is increased?

phase equilibrium

Liquid Gasvaporization

condensation

Surface Molecules

Georgia, 760 torr = 1 atm

H2O100 °C

NormalBoiling Point

Tibet, 480 torr < 1 atm

H2O85 °C

NormalBoiling Point

How are the vapor pressure P and

temperature T related exactly?

(a) The Vapor Pressure of Water, Ethanol, and Diethyl Ether as a Function of Temperature. (b) Plots of In(Pvap) versus 1/T for

Water, Ethanol, and Diethyl Ether

1/T (K−1)

T is in K!

Linear relation: y = kx + C

y

xC: intercept

slope: k = tan θ

θ

Linear relation: y = kx + C

y

xθ

k > 0

θ

k < 0

slope: k = tan θ

ln P = k(1/T) + C

Linear relation: y = kx + C

1/T (K−1)

What is the value of k?

Heat of vaporization ∆Hvap: energy needed to convert one mole

of liquid to gas. Unit: J/mol or kJ/mol.

∆Hvap > 0

slope k < 0

1ln vapHP C

R T

y x

y = kx + C

1/T (K−1)

ln P

1/T (K−1)

ln P1

1/T1

1

2ln P2

1/T2

1ln vapHP C

R T

1

2 1 2

1 1ln vapHP

P R T T

Clausius-Clapeyron Equation

1ln vapHP C

R T

The vapor pressure of water at 25 °C is 23.8 torr, and the heat

of vaporization of water is 43.9 kJ/mol. Calculate the vapor

pressure of water at 50 °C.

Five: T1, T2, P1, P2, ∆Hvap

Four known, calculate the other.

1

2 1 2

1 1ln vapHP

P R T T

Clausius-ClapeyronequationR = 8.314 J · mol−1 · K−1

Units in ideal gas law

PV = nRT

P — atm, V — L, n — mol, T — K

Option 1

R = 0.082 atm · L · mol−1 · K−1

P — Pa, V — m3, n — mol, T — KOption 2

Chem 1211

Liquid potassium has a vapor pressure of 10.00 torr at 443 °C

and a vapor pressure of 400.0 torr at 708 °C. Use these data

to calculate

(a) The heat of vaporization of liquid potassium;

(b) The normal boiling point of potassium;

(c) The vapor pressure of liquid potassium at 100. °C.

( Please try to work on this question by yourself. Will review next week)

Plan for this week’s lab

• Lab syllabus, then sign agreement

• Calculations on C-C question

• Demo on vapor pressure

• Quiz on C-C question

Thursday, IC 420Section A: 1:00 pm, Section B: 9:00 am

1

2 1 2

1 1ln vapHP

P R T T

Clausius-Clapeyron Equation

1ln vapHP C

R T

slope k < 0

1ln vapHP C

R T

y x

Linear relation: y = kx + C

y

xθ

k > 0

θ

k < 0

slope: k = tan θ

a

b

c

d

Lines tilt to the right have positive slopes (a and b), left negative(c and d). Steeper line has greater absolute value of slope. In thisgraph, the order of slopes is

ka > kb > 0 > kc > kd

y

x

What is the order of heat of vaporization for these three substances?

|kwater| > |kethanol| > |kd.e.|

kwater < kethanol < kd.e.

( ) vapH water

R

( ) vapH ethanol

R

( . .) vapH d e

R< <

( ) vapH water

R

( ) vapH ethanol

R

( . .) vapH d e

R> >

( ) vapH water ( ) vapH ethanol ( . .) vapH d e> >

Stronger intermolecular attractions

↔ Higher boiling point and ΔHvap

(Chem 1211)

H―O―H¨¨

¨H―C―

――

H

H

C―O―H

――

H

H¨

¨H―C――

―

H

H

C―O―

――

H

H¨

――

H

H

C―C―H

――

H

H

Water:

Ethanol:

Diethyl Ether:

Solids

Glass (SiO2)

Crystal

Noncrystal

Solid

Basis Crystal structure

The basis may be a single atom or molecule, or a small group of atoms, molecules, or ions.

NaCl: 1 Na+ ion and 1 Cl− ion

Cu: 1 Cu atom

Zn: 2 Zn atoms

Diamond: 2 C atoms

CO2: 4 CO2 molecules

=Use a point to represent the basis:

Lattice

Lattice point:

Unit cell: 2-D, at least a parallelogram

Unit cell is the building block of the crystal

How many kinds of 2-D unit cells

can we have?

Extend the concept of unit cell to 3-D,

the real crystals.

: 3-D, at least a parallelepiped

How many kinds of 3-D unit cells

can we have?

1. triclinic 2. monoclinic

3. orthorhombic

4. tetragonal5. rhombohedral (trigonal)

6. hexagonal7. cubic

The 14 Bravais lattices

7 crystal systems

a ≠ b ≠ cα ≠ β ≠ γ

a ≠ b ≠ c

α = β = γ = 90°

a = b ≠ cα = β = 90° ,γ = 120°

a = b = cα = β = γ = 90°

a = b ≠ cα = β = γ = 90°

a = b = c90° ≠ α = β = γ < 120°

γ

ab

ca

b

c

(Simple cubic)

Chem 1212: assume a lattice point is a single atom

• Size of the cell X-ray diffraction

Information of a cubic unit cell

The Wave Nature of LightThe Wave Nature of Light

• Number of atoms in a cell

• Size of the cell

• Size of the atoms Soon

X-ray diffraction

Now

Information of a cubic unit cell

AB

C D

AB

C D E

F

Number of atoms in a unit cell = ¼ x 4 = 1

1 2 4

Number of Atoms in a Cubic Unit Cell

The body-centered cubic unit cell of a particular crystalline

form of iron is 0.28664 nm on each side. Calculate the density

(in g/cm3) of this form of iron.

d = 7.8753 g/cm3

Closest Packing

a a

aaa

a a

a a

aa

a a

a a a a a

a

a

b b b b

b b b b

b b b b

c c c c

c c c c

c c c c

· · · abab · · ·

· · · abab · · ·

Hexagonal unit cell

1. triclinic 2. monoclinic

3. orthorhombic

4. tetragonal5. rhombohedral (trigonal)

6. hexagonal7. cubic

The 14 Bravais lattices

7 crystal systems

a ≠ b ≠ cα ≠ β ≠ γ

a ≠ b ≠ c

α = β = γ = 90°

a = b ≠ cα = β = 90° ,γ = 120°

a = b = cα = β = γ = 90°

a = b ≠ cα = β = γ = 90°

a = b = c90° ≠ α = β = γ < 120°

γ

· · · abcabc · · ·

abcabc = Cubic Closest Packing

• Number of atoms in a cell

• Size of the cell

• Size of the atoms Soon

X-ray diffraction

Now

Information of a unit cell

Now!

Example 11.7

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

r8L

L

r

2r

r

L

d = 2.71 g/cm3

What about simple cubic?

Simple Cubic

r

L

L = 2r

What about body-centeredcubic?

Body centered cubic

D

Body diagonal D = 4r

L

D

L

L F

L

Body diagonal D = 4r

r3

4L

L

Pythagorean theorem

Titanium metal has a body-centered cubic unit cell. Thedensity of titanium is 4.50 g/cm3. Calculate the edge lengthof the unit cell and a value for the atomic radius of titaniumin pm.

L = 328 pm

Ti: 47.87 g/mol

r = 142 pm

Packing Efficiency

100-mL container

50 % 70 %

50 mL 70 mL

1 2 4

Packing Efficiency: fraction of volume occupied by atoms

74 %52 % 68 %

L = 2r r3

4L r8L

prove

Quiz next week during lab session

Calculate density from a unit cell.

Relationship between the length of a cell and theradius of an atom is given.