Chapter 12 Intermolecular Forces of...
Transcript of Chapter 12 Intermolecular Forces of...
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