Chapter 12

Intermolecular Forces of

Attraction

Intermolecular Forces – Attractive or Repulsive Forces between molecules.

Molecule - - - - - - Molecule

Intramolecular Forces – bonding forces within the molecule.

H – O – H

Review: Characteristics of Physical States

Review: Phase Changes - Changes in Physical States

Solid Liquid Melting (Fusion)

Liquid Solid Freezing

Liquid Gas Vaporization

Gas Liquid Condensation

Solid Gas Sublimation

Gas Solid Deposition

Which of the physical changes are endothermic or exothermic?

Endothermic Exothermic

Melting FreezingSublimation DepositionVaporization Condensation

H of a Phase Change: The amount of energy required to produce a phase change for one mole of a substance (kJ/mol).

Note: The VALUE of Hvap. = Hcond.

However, the sign is different depending on exothermic or endothermic processes.

Phase Changes of pure substances require a

specific amount of energy per mole ( H)

Examples

H2O(l) H2O(g) Hvap = 40.7 kJ/mol

H2O(g) H2O(l) Hcond. = - 40.7

kJ/mol

H2O(s) H2O(l) Hfus = 6.02 kJ/mol

H2O(l) H2O(s) Hfreez = - 6.02 kJ/mol

Heats of Vaporization & Fusion for

Various Substances

Quantitative Aspects of Phase Changes

Calculating the amount of energy for a substance to

undergo phase changes.

If you had 25.00 grams of water at 130.0oC, how much

energy would be released when the water cooled to

-40.0oC.

Some important facts to know:

Specific Heat of H2O(g): 33.1 J/mol-oC

Specific Heat of H2O(l) : 75.4 J/mol-oC (4.184 J/go C)

Specific Heat of H2O(s): 37.6 J/mol-oC

Stage 1 Stage 5

GAS – LIQUID

LIQUID

H 0vap

LIQUID–

SOLID

Stage 4Stage 3Stage 2

Tem

pera

ture

(ºC

)

SOLID

GAS

100

0

130

– 40

H 0fus

Heat removed

Tem

pera

ture

(ºC

)

GAS

100

0

130

– 40Heat removed

Stage 1

Calculating the amount of heat absorbed or released when a substance undergoes a temperature change:

q = mc T or q = nc T

n = (25.00gH2O)(1mol H2O)

(18.02 g H2O)

n = 1.387 mol H2O

q = (1.387 mol)(33.1J/moloC)(100.0oC -130.0oC)

q = -1377.291 J = -1.377 kJ

GAS – LIQUID

H 0vap

Tem

pera

ture

(ºC

)

GAS

100

0

130

– 40Heat removed

Stage 1 Stage 2

Calculating the amount of heat gained or

lost during a phase change.

q = n H?

q = 1.387 mol (- 40.7 kJ/mol)

q = -56.5 kJ

GAS – LIQUID

LIQUID

H 0vap

Tem

pera

ture

(ºC

)

GAS

100

0

130

– 40Heat removed

Stage 1 Stage 2 Stage 3

q = nc T

q = 1.387mol (75.4 J/moloC)(0.0oC – 100.0oC)

q = -10457.98 J = -10.5 kJ

GAS – LIQUID

LIQUID

H 0vap

LIQUID–

SOLID

Tem

pera

ture

(ºC

)

GAS

H 0fus

100

0

130

– 40Heat removed

Stage 1 Stage 4Stage 3Stage 2

q = n Hfus

q = 1.387mol(-6.02 kJ/mol)

q = -8.35 kJ

GAS – LIQUID

LIQUID

H 0vap

LIQUID–

SOLID

Heat removed

Tem

pera

ture

(ºC

)

SOLID

GAS

H 0fus

100

0

130

– 40

Stage 1 Stage 5Stage 4Stage 3Stage 2

q = nc T

q = 1.387mol (37.6 J/moloC)(-40.0oC – 0.0oC)

q = -2086.048 J = -2.09 kJ

Total Heat Change

q = (heat lost by steam + heat of

condensation + heat lost by liquid + heat

of freezing + heat lost by solid)

Remember, you can’t add kJ with J!

1,000 J = 1 kJ

-1.377 kJ

-56.5 kJ

-10.5 kJ

-8.35 kJ

-2.09 kJ

-78.8 kJ

Final Answer: - 78.8 kJ of heat are released

Exam Question(s)

You should be able to calculate the amount

of heat released or absorbed for any

substance given H’s, C’s, and temps.

Phase Changes and Equilibrium

Equilibrium – The rate of the forward reaction (or

process) is equal to the rate of the reverse

reaction (or process).

Example: At 0oC and 1 atm

H2O(s) H2O(l)

Rate(forward) = Rate(reverse) @ Equilibrium

Example

Open Container(Evaporation)

Closed Container(Equilibrium)

Disturbing Equilibrium (Placing a

Stress on the System)1. Add more vapor to the container

2. Remove some of the vapor

3. Add heat

4. Remove heat

Note: If left alone the system will eventually return to equilibrium.

The pressure, at a particular temperature,

exerted by a vapor (gas) in equilibrium

with its liquid in a closed container is

called the vapor pressure.

The temperature at which the vapor

pressure is equal to the external

(atmospheric pressure) is the boiling

point of a substance.

The Effects of Temperature on

Equilibrium

As the temperature is increased so will the

vapor pressure

At a particular temperature, the weaker the

intermolecular forces of attraction of a

substance, the higher the vapor pressure.

Clausius-Clapeyron Equation

lnP2

P1

= - Hvap

R1 1

T2 T1

-

R = 8.314 J/mol-K

T must be in Kelvin

Problem Solving

Calculate the vapor pressure of water at

85oC.

Hint: Remember that water’s vapor

pressure is equal to 1.00 atm (760. torr) at

its b.p.

Answer

P1 = 760. torr

P2 = ?

T1 = 373.15 K

T2 = 358.15 K

lnP2

760.=

-

8.3141 1

358.15 K 373.15 K-

J/mol-K

J/mol

torr

P2 = 439 torr or 0.577 atm

Phase Diagrams

Regions - Various phases

Lines between regions – Equilibrium of phases

Critical Point – Point at which the substance has the density of a gas and solvent capability of a liquid.

Triple Point – Point at which all three physical states exist in equilibrium

Types of Intermolecular Forces