UNIT IX
SOLIDS, LIQUIDS HEAT
PROBLEMS
CHAPTER 16 PART1 AND
CHAPTER 14
INTERMOLECULAR FORCES


Forces between molecules
Not as strong as within molecules
(covalent and ionic)
van der Waals Forces (Intramolecular Force)

Dispersion Forces (London Forces)
 Exists between non-polar
molecules
 weakest I.M.F.
 Due to temporary shifts in
electron cloud density
 Examples
 CH4
 O2

Dipole-Dipole Forces
• slightly polar
• Example:
 CHCl3

HYDROGEN BONDING
• VERY polar
• Strongest
• Examples
 NH3 (N -- H)
 H2O (O -- H)
 HF
(F-- H)
 HCl (Cl -- H)
SOLIDS AND LIQUIDS
SOLIDS

Orderly rigid and cohesive
Particles that vibrate around fixed
points

SOLIDS

CRYSTAL
• true solids
• particles are arranged in an
orderly repeating 3-D pattern
SOLIDS
CRYSTALS (cont)
– consists of a MEMBER
o one particle (ion, atom,
molecule
SOLIDS
several
members together make
up UNIT CELL
 simplest repeating unit
 retains its shape
SOLIDS
several unit cells together make up a
CRYSTAL LATTICE
 3-D arrangement of unit cells
repeated over and over
SOLIDS
VocabANHYDROUS (without water) - compound
containing no water of hydration
HYDRATE-compound with water molecules
attached
(CuSO4 * 6H2O)
SOLIDS
AMORPHOUS
– solid
– no definite repeating pattern
– no true melting point
– no plateau
EXAMPLES: glass, butter, tar,
plastic
LIQUIDS
DEFINITION
– particles vibrate around a moving
point
– non-orderly, non-rigid, cohesive
– more space between particles than
a solid
– exert a vapor pressure
– Fluid – ability to flow
LIQUIDS
UNITS
Temperature
 average kinetic energy (KE)
°C
°F
K (Kelvin)
LIQUIDS
VAPOR PRESSURE
Definition
 pressure exerted by vapor
molecules above a liquid when
dynamic equilibrium is reached
LIQUIDS
Pressure
 measure of force with which gas
molecules hit the side of container
 normal atmospheric pressure at sea
level
 Standard Pressure Units =760 torrs
= 760 mmHg = 101.3 kilopascals
(kPa)
LIQUIDS
VAPOR PRESSURE
Dynamic equilibrium - 2 opposite
processes occurring at same time
and same rate
VAPOR
LIQUID
LIQUIDS
VAPOR PRESSURE
Dynamic Equilibrium
depends upon:
 Temperature increase temperature,
increase vapor pressure
VAPOR
T VP
LIQUID
LIQUIDS
•
•
Strength of inter-molecular
forces; hydrogen
bonding(such as water) is
strongest.
increase forces; decrease
vapor pressure
VAPOR
VAPOR
LIQUID
LIQUID
WATER
ALCOHOL
IMF VP
LIQUIDS
Viscosity measure of resistance to flow (how
thick)
Example – Molasses (syrup) has
a high viscosity
 Volatility - how easily a liquid
evaporates

LIQUIDS

Very volatile:
 high vapor pressure
 low IMF
 low boiling point
 EXAMPLES: alcohol,
perfume
VAPORVAPOR
LIQUID
ALCOHOL
LIQUIDS
Not
volatile
 low vapor pressure
 high IMF
 high boiling point
 Examples: molasses,
water
VAPOR
VAPOR
LIQUID
WATER
CHANGES IN STATE OR PHASES

Sublimation– solid changes directly into gas
without going through the liquid
state
Examples: solid iodine, solid
air fresheners, "dry" ice
CHANGES IN STATE OR
PHASES
Melting / Freezing
– goes from solid to liquid or liquid
to solid
CHANGES IN STATE OR PHASES

Vaporization • evaporation
 occurs only on the surface
 at room temperature
 cooling process
 Sweat
• boiling
 occurs throughout the liquid
 requires energy
CHANGES IN STATE OR
PHASES

Boiling Point:
• vapor pressure = atmospheric (outside)
pressure (for any boiling point)
• normal boiling point
 vapor pressure = standard pressure
 standard pressure = 1 atm, 760 torrs,
760 mm Hg,101.3 kPa

Boiling Point:
• different altitudes
 higher altitudes have lower air
pressures
 Denver has a lower boiling point
95 °C than Houston has (100 °C)
 Foods take longer to cook in
Denver than Houston.
VAPOR PRESSURE DIAGRAMS
1000
900
800
760 700
600
500
400
300
200
100
20
40
60
Temperature ( °C)
80
100
PHASE DIAGRAMS
Graphs that show conditions
(temperature and pressure) under
which a substance will exist as a
solid, liquid, or gas.
PHASE DIAGRAMS
Z
800
760
700
600
500
400
X - Triple point
X-Y
line
- These
Z
Critical
temp.
and
All three states are in
are
sublimation
pressure.
A
gas
can't
equilibrium at this
points.
be
liquified
above
this
temperature and pressure.
point.
X
300
200
100
40
60
100
120
80
Temperature (°C)
140
160
180
PHASE DIAGRAMS
Z
800
760
700
600
500
Lines represent 2
phases in equilibrium.
LIQUID
X
400
300
200
GAS
100
40
60
100
120
80
Temperature (°C)
140
160
180
PHASE DIAGRAMS
Z
800
760
700
600
500
400
300
Normal boiling point
(condensation) occurs
here.
when standard pressure
crosses liquid / gas
line
X
200
100
40
60
100
120
80
Temperature (°C)
140
160
180
PHASE DIAGRAMS
Z
800
760
700
600
500
400
300
Normal melting point
(freezing) occurs where
standard pressure crosses
Normal
melting
liquid
/ solid
line.point
(freezing) occurs here
X
200
100
40
60
100
120
80
Temperature (°C)
140
160
180
PHASE DIAGRAMS
Z
800
760
700
600
500
400
X
300
200
100
40
60
100
120
80
Temperature ( °C)
140
160
180
UNIQUE PROPERTIES OF
WATER
STRONG HYDROGEN BONDING
CAUSES:
– high boiling point and melting
point
– high specific heat capacity
– high surface tension
needle floats
– Water droplets are spherical
HEAT

VS.
TEMPERATURE
Energy transferred  Average Kinetic
from one body to
Energy
another because of a  Written as KE
difference in
temperature
HEAT

VS.
UNITS
– calories (c)
– kCal - C
• (1000 calories)
– Joules - J
• energy for one
heartbeat
– 1 cal = 4.18 J
– 1 kCal = 4180 J
TEMPERATURE

UNITS
– °C - celsius
–
°F -Fahrenheit
–
K - kelvin
(no degree sign!)
HEAT

VS.
Measured by:
– indirectly by a
calorimeter
TEMPERATURE

Measured by:
– thermometer
HEAT

VS.
DEPENDS UPON
– mass
 more mass means
more heat
– Cp (S) - specific heat
 type of matter
 some hold heat
better than others
– DT - change in
temperature
TEMPERATURE

DEPENDS UPON
– amount of
movement of the
particles in the
substance
HEAT



VS.
FORMULA
 q=energy (J)
 m=mass (g)
q = (m) (DT) (Cp)
q = (m) (T2-T1) (Cp)
TEMPERATURE
Specific Heat or Heat Capacity
Amount of heat needed to raise 1 gram of a
substance 1 degree Celsius
 Units

–
–

(J/goC)
(cal/goC)
Examples
–
–
–
water --- 4.18 J/goC or 1 cal/goC
Au --- 0.129 cal/goC
alcohol --- 2.45 J/goC
Calorie


Amount of heat needed to raise one gram of
water one degree of celsius
It takes one calorie to raise one gram of
water one degree of Celsius
Heat of Fusion - Hf
Amount of heat needed to melt one gram of
a substance at its melting point
 Units

(cal/g)

Examples
–
–
water (Hf) = 334 J/g or 76.4 cal/g
Ag = 88 J/g
HEAT OF VAPORIZATION - Hv
 Amount of heat needed to vaporize one
gram of a substance at its boiling point
Examples
– water (Hv) = 2260 J/g or 539 cal/g
– Pb = 858 J/g
PHASE CHANGE DIAGRAMS
TEMPERATURE ( C)
WATER
Heat of vaporization
Boiling point- substance is
becoming a liquid
100
Heat of fusion Melting point substance is becoming
a liquid
0
GAS
LIQUID
SOLID
HEAT (cal/g) OR TIME
Heat Calculations - Formulas
The state remains the same and there is
no change in temperature.
q= joules
m=grams
Cp=J/g or J/c
q= (m) (Cp)
q = Heat
Example of Non-Changing State
Melting/freezing at melting point
Vaporizing/condensing at boiling point
How much energy does it take to melt
55g of gold at its melting point?
Cp = 64.5 J/g
q= (m) (Cp)
= (55g)(64.5 J/g)
= 3547.5 J
HEAT EQUATION
One substance with a temperature
change
q = (m) (Cp) (T2-T1)
q=joules (J)
m= mass (g)
Cp = specific heat
capacity (J/g °C) (J/c °C)
T2 = final temperature
T1 = initial temperature
HEAT EQUATION EXAMPLE
***Heating or cooling with no change
in state***
How much energy is released as 33 g
of solid silver cools from 95 °C to
60°C?
Cp of silver = 0.236 J/g °C
HEAT TRANSFER EQUATION
How a substance changes the
temperature of another substance
used in calorimeter calculations
Energy LOST = Energy GAINED
(m1) (Cp1) (T2-T1) = (m2) (Cp2) (T2-T1)
Warm substance
losing energy
Cool substance
gaining energy
HEAT TRANSFER EQUATION
EXAMPLE
A piece of metal is dropped into a beaker of
o
boiling water whose temperature is 95 C.
The 5g piece of metal is put into 100g of cold
water at 20 oC. The temperature of the water
rises to 30 oC. What is the specific heat of
the metal?
o
Cp(water) = 4.18 J/g C
EQUATION FOR CHANGING
TEMPERATURE AND STATES
Draw the phase change diagram
TEMPERATURE ( C)
CHANGING STATES AND
TEMPERATURE
100
0
Use the following equations:
q = (m) (Cp)
q = (m) (Cp) (T2-T1)
HEAT (cal/g) OR TIME
TEMPERATURE ( C)
CHANGING STATES AND
TEMPERATURE
100
1. Heat solid to
melting point
0
q = (m) (Cp) (T2-T1)
HEAT (cal/g) OR TIME
TEMPERATURE ( C)
CHANGING STATES AND
TEMPERATURE
100
2. Melting solid
to liquid
0
q = (m) (Cp)
HEAT (cal/g) OR TIME
TEMPERATURE ( C)
CHANGING STATES AND
TEMPERATURE
100
3. Heat liquid
to boiling point
q = (m) (Cp) (T2-T1)
0
HEAT (cal/g) OR TIME
TEMPERATURE ( C)
CHANGING STATES AND
TEMPERATURE
4. Change liquid
to gas
100
q = (m) (Cp)
0
HEAT (cal/g) OR TIME
TEMPERATURE ( C)
CHANGING STATES AND
TEMPERATURE
5. Heating gas
100
q = (m) (Cp) (T2-T1)
0
HEAT (cal/g) OR TIME
CHANGING STATES AND TEMPERATURES
When to use which equations:
1. Heat solid to melting point : q = (m) (Cp) (T2-T1)
KE
2. Melt solid to liquid:
q = (m) (Cp)
PE
3. Heat liquid to boiling point:
q = (m) (Cp) (T2-T1)
KE
4. Change liquid to gas:
q = (m) (Cp)
PE
5. Heat gas:
KE
q = (m) (Cp) (T2-T1)
CHANGING TEMPERATURE AND
CHANGING STATES EXAMPLE
How much energy is needed to
change 30g of ice at -5 °C to steam at
120 °C?