OFB Chapter 6 lecture notes - School of Chemistry and...

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1/26/2004 OFB Chapter 6 1 OFB Chapter 6 Condensed Phases and Phase Transitions 6-1 Intermolecular Forces: Why Condensed Phases Exist 6-2 The Kinetic Theory of Liquids and Solids 6-3 Phase Equilibrium 6-4 Phase Transitions 6-5 Phase Diagrams 6-6 Colligative Properties of Solutions 6-7 Mixtures and Distillation 6-8 Colloidal Dispersions

Transcript of OFB Chapter 6 lecture notes - School of Chemistry and...

1/26/2004 OFB Chapter 6 1

OFB Chapter 6

Condensed Phases and

Phase Transitions

6-1 Intermolecular Forces: Why Condensed Phases Exist

6-2 The Kinetic Theory of Liquids and Solids

6-3 Phase Equilibrium

6-4 Phase Transitions

6-5 Phase Diagrams

6-6 Colligative Properties of Solutions

6-7 Mixtures and Distillation

6-8 Colloidal Dispersions

1/26/2004 OFB Chapter 6 2

Intermolecular Forces: Why Condensed Phases Exist

• Intramolecular Forces– Chemical bonds

• Strong• Directional• Short Range (relative)

• Intermolecular Forces• Weaker than chemical bonds, usually much

weaker• Less directional than covalent bonds, more

directional than ionic bonds• Longer range than covalent bonds but at

shorter range than ionic bonds• Condensed Phases

– Solids and Liquids– Intermolecular forces: mutual

attractions hold the molecules closer together than gases

• Potential Energy Curves– Distinction between intramolecular and

intermolecular Forces

1/26/2004 OFB Chapter 6 3

Potential Energy

• At Rest

• Stretched, then relaxed

• Compressed, then relaxed

1/26/2004 OFB Chapter 6 4

Potential Energy Curves

1/26/2004 OFB Chapter 6 5

Potential EnergyCurves

• Atoms or molecules approach at large distance (zero PE (energy of position)

• PE goes negative as intermolecular forces come into play

• PE minimum energy well is a balance of Intermolecular forces and repulsive forces

• PE goes positive as repulsive forces dominate

1/26/2004 OFB Chapter 6 6

Types of Non-Bonded (Intermolecular) Attractions

1. Dipole-Dipole Interactions – e.g., Hydrogen Bonding

2. Ion-Dipole Interactions3. Induced Dipole Attractions4. Dispersive Forces

• temporary fluctuations in electron distribution

1/26/2004 OFB Chapter 6 7

Types of Non-Bonded (Intermolecular) Attractions

1. Dipole-Dipole Interactions – e.g., Hydrogen Bonding

1/26/2004 OFB Chapter 6 8

Types of Non-Bonded (Intermolecular) Attractions

1. Dipole-Dipole Interactions – e.g., Hydrogen Bonding

2. Ion-Dipole Interactions

1/26/2004 OFB Chapter 6 9

Types of Non-Bonded (Intermolecular) Attractions

1. Dipole-Dipole Interactions – e.g., Hydrogen Bonding

2. Ion-Dipole Interactions3. Induced Dipole Attractions

Argon Atom

Argon Atom

1/26/2004 OFB Chapter 6 10

Types of Non-Bonded (Intermolecular) Attractions

1. Dipole-Dipole Interactions – e.g., Hydrogen Bonding

2. Ion-Dipole Interactions3. Induced Dipole Attractions4. Dispersive Forces

• temporary fluctuations in electron distribution

1/26/2004 OFB Chapter 6 11

Types of Non-Bonded (Intermolecular) Attractions

1. Dipole-Dipole Interactions – e.g., Hydrogen Bonding– Between H2O to H2O or – R-OH to H2O or – R-OH to R-OH – (R is an organic unit)

1/26/2004 OFB Chapter 6 12

Kinetic Theory of Liquids and Solids

1/26/2004 OFB Chapter 6 13

6-3 Phase Equilibrium

• Coexisting phase equilibrium– Vapor pressure

Phase: A sample of matter that is uniform throughout, both in its chemical constitution and in its physical state.

1/26/2004 OFB Chapter 6 14

Correcting for the Vapor Pressure of Water in “Wet Gases”

• Many chemical reactions conducted in aqueous solution often generate gaseous products

• The gas collected will be “humid” or “wet” due to the coexistence of the gas and water vapor.

• A correction is made to account for the water vapor present

1/26/2004 OFB Chapter 6 15

Exercise page 6-2:

Passage of an electric current through a dilute aqueous solution of sodium sulfate produces a mixture of gaseous hydrogen and oxygen according to the following equation:

2 H2O(l) → 2 H2(g) + O2(g)

One liter of the mixed gases is collected over water at 25oC and under a total pressure of 755.3 torr. Calculate the mass of oxygen that is present. The vapor pressure of water is 23.8 torr at 25oC.

1/26/2004 OFB Chapter 6 16

For ideal gas mixtures

P of a gas is proportional to the number of moles of gas

2 H2O(l) → 2 H2(g) + O2(g)

Exercise page 6-2:

atm9625.0P

atm 0.0313atm 0.9938P

PPP

gases

gases

O2

Htotalgases

=

−=

−=

atm0.32083 /atm 0.962

P 1/3H moles 2O 1mole

O1moleP gases22

2

2O

==

=+

=

1/26/2004 OFB Chapter 6 17

atm9625.0P

atm 0.0313atm 0.9938P

PPP

gases

gases

O2

Htotalgases

=

−=

−=

atm0.32083 /atm 0.962

P 1/3H moles 2O 1mole

O1moleP gases22

2

2O

==

=+

=

gramsm

m

Mmn

molesKLatm

RT

VPn

RTnVP

O

O

O

O

OO

421.032

01316.0

01316.0)298)(082.0()0.1)(32.0(

2

2

2

2

22

2

=

===

=

==

=

1/26/2004 OFB Chapter 6 18

6-6 Colligative Properties of Solutions

• For some properties, the amount of difference between a pure solvent and dilute solution depend only on the number of solute particles present and not on their chemical identify.

• Called Colligative Properties• Examples

1. Vapor Pressure2. Melting Point3. Boiling Point4. Osmotic Pressure

1/26/2004 OFB Chapter 6 19

Colligative Properties of Solutions

Mass percentage (weight percentage):

mass percentage of the component =

× 100%mass of component total mass of mixture

Mole fraction: The chemical amount (in moles) divided by the total amount (in moles)

X1 = n1/(n1 + n2)

X2 = n2/(n1 + n2) = 1 - X1

1/26/2004 OFB Chapter 6 20

Molality

msolute =

moles solute per kilograms solvent

= mol kg-1

Molarity

csolute =

moles solute per volume solution

= mol L-1

1/26/2004 OFB Chapter 6 21

Exercise 6-7:

Suppose that 32.6 g of acetic acid, CH3COOH, is dissolved in 83.8 g of water, giving a total solution volume of 112.1 mL. Calculate the molality and molarity of acetic acid (molar mass 60.05 g mol-1) in this solution.

1solute

1solute

kg mol 6.48kg 0.0838

mol 0.543

kgg1000

g 83.8

molg60.05

g 32.6

m

kg molsolvent kg

solutemolesmMolality

==

=

===

1/26/2004 OFB Chapter 6 22

Exercise 6-7:

Suppose that 32.6 g of acetic acid, CH3COOH, is dissolved in 83.8 g of water, giving a total solution volume of 112.1 mL. Calculate the molality and molarity of acetic acid (molar mass 60.05 g mol-1) in this solution.

1solute

1solute

L mol 84.40.1121

mol 0.543

Lml1000

ml 112.1

molg60.05

g 32.6

c

L molsolvent volumesolutemolesc Molarity

==

=

===

1/26/2004 OFB Chapter 6 23

Ideal Solutions and Raoult’s Law

• Consider a non-volatile solute in a solvent

• X1 = mole fraction of solvent

P1=X1 P°1

Negative deviation

= solute-solvent attractions > solvent-solvent attractions

Positive deviation

= solute-solvent attractions < solvent-solvent attractions

1/26/2004 OFB Chapter 6 24

1. Lowering of Vapor Pressure– Vapor Pressure of a solvent above a

dilute solution is always less than the vapor pressure above the pure solvent.

2. Elevation of Boiling Point– The boiling point of a solution of a

non-volatile solute in a volatile solvent always exceeds the boiling point of a pure solvent

Ideal Solutions and Raoult’s Law

1/26/2004 OFB Chapter 6 25

Elevation of Boiling Point

Where m = molality(moles of solute per kilogram of solvent)

The Effect of Dissociation

i = the number of particles released into the solution per formula unit of solute

e.g., NaCl dissociates into i = e.g., Na2SO4 dissociates into i =

(2 Na+ + 1 SO4-2)

1/26/2004 OFB Chapter 6 26

3. Depression of Freezing Point∆Tf = − m Kf

∆Tf = − i m Kf

1/26/2004 OFB Chapter 6 27

Osmotic Pressure• Fourth Colligative Property• Important for transport of molecules

across cell membranesOsmotic Pressure = Π = g d h

Π = c RTΠV = n RTΠV = i c RT

1/26/2004 OFB Chapter 6 28

Exercise 6-17

A dilute aqueous solution of a non-dissociating compound contains 1.19 g of the compound per liter of solution and has an osmotic pressure of 0.0288 atm at a temperature of 37°C. Compute the molar mass of the compound

cGiven

Strategy

===

=

=

lmole

lg

moleg M Rearrange .)3

massMolar g molethat Recall 2.)

mol/L in c the find to cRT Π use 1.)

1/26/2004 OFB Chapter 6 29

Exercise 6-17

A dilute aqueous solution of a non-dissociating compound contains 1.19 g of the compound per liter of solution and has an osmotic pressure of 0.0288 atm at a temperature of 37°C. Compute the molar mass of the compound

g/mol1.05x10mol/l 1.132x10

lg1.19

lmole

lg

moleg M Rearrange 3.)

mol/l 1.132x10c

273.15K))(37Kmol atm L (0.0820atm 0.0288

RTΠc

RTΠcor cRT Π use 1.)

olution

33

3

11

==

==

=

+==

==

−−

M

S

1/26/2004 OFB Chapter 6 30

6-8 Colloidal Dispersions• Colloids are large particles dispersed

in solution– 1nm to 1000 nm in size– E.g., Globular proteins 500nm

• Examples– Opal (water in solid SiO2)– Aerosols (liquids in Gas)– Smoke (solids in Air)– Milk (fat droplets & solids in water)– Mayonnaise (water droplets in oil)– Paint (solid pigments in liquid)– Biological fluids (proteins & fats in

water)• Characteristics

– Large particle size colloids: translucent, cloudy, milky)

– Small particle size colloids: can be clear

1/26/2004 OFB Chapter 6 31

6-8 Colloidal Dispersions– Tyndall Effect

• Light Scattering

1/26/2004 OFB Chapter 6 32

Chapter 6Condensed Phases and Phase

Transitions

Examples / Exercises– All (6-1 thru 6-17, skip 6-18)

HW Problems– 4, 20, 26, 38, 44, 52, (62 skip)

Skip sections 6-7 and 6-8