Transcript Document

Intermolecular Forces
and
Liquids and Solids
Chapter 12
Midterm II
• Any conflicts with March 20? If yes, let me
know ASAP. The original date was March
22.
Phase Diagram of Water
• Note the high critical temperature
and critical pressure:
– These are due to the strong van der
Waals forces between water
molecules.
• The slope of the solid–liquid
line is negative.
– This means that increasing the
pressure above 1 atm will raise
the boiling point and lower the
melting point.
– Lower the melting point?
Phase Diagram of Carbon Dioxide
Carbon dioxide cannot exist in the liquid state at pressures
below 5.11 atm; CO2 sublimes at normal pressures.
Phase Diagram of Carbon Dioxide
Carbon dioxide cannot exist in the liquid state at pressures
below 5.11 atm; CO2 sublimes at normal pressures.
At 1 atm, solid CO2 does not melt at any temperature.
Instead, it sublimes to form CO2 vapor. Why might it be
useful as a refrigerant?
Phase Diagram of Carbon Dioxide
Carbon dioxide cannot exist in the liquid state at pressures
below 5.11 atm; CO2 sublimes at normal pressures.
If you want to send something frozen across the country,
you can pack it in dry ice. It will be frozen when it reaches
its destination, and there will be no messy liquid left over
like you would have with normal ice.
The slope of the curve between solid and liquid is positive for
CO2 as well as almost all other substances. Why does water
differ?
Freeze-drying
Normal (right) and freeze-dried (left) spaghetti
•
Completely remove water from some material, such as food, while leaving the basic structure
and composition of the material intact
•
Two reasons
–
–
•
Keeps food from spoiling for a long period of time
Significantly reduces the total weight of the food
How?
–
–
–
Freeze the material
Lower the pressure (<0.006 atm)
Increase the temperature slightly
Freeze-drying
Normal (right) and freeze-dried (left) spaghetti
•
How?
– Freeze the material
– Lower the pressure
– Increase the temperature slightly
Physical Properties
of Solutions
Chapter 13
A solution is a homogenous mixture of 2 or
more substances
The solute is(are) the substance(s) present in the
smaller amount(s)
The solvent is the substance present in the larger
amount
13.1
A saturated solution contains the maximum amount of a
solute that will dissolve in a given solvent at a specific
temperature.
An unsaturated solution contains less solute than the
solvent has the capacity to dissolve at a specific
temperature.
A supersaturated solution contains more solute than is
present in a saturated solution at a specific temperature.
Sodium acetate crystals rapidly form when a seed crystal is
added to a supersaturated solution of sodium acetate.
13.1
Solutions
The intermolecular
forces between solute
and solvent particles
must be strong enough
to compete with those
between solute particles
and those between
solvent particles.
How Does a Solution Form?
As a solution forms, the solvent pulls solute
particles apart and surrounds, or solvates,
them.
How Does a Solution Form
If an ionic salt is
soluble in water, it is
because the iondipole interactions
are strong enough
to overcome the
lattice energy of the
salt crystal.
Energy Changes in Solution
• Simply, three processes
affect the energetics of the
process:
 Separation of solute
particles
 Separation of solvent
particles
 New interactions between
solute and solvent
Energy Changes in Solution
The enthalpy
change of the
overall process
depends on H for
each of these steps.
Three types of interactions in the solution process:
• solvent-solvent interaction
• solute-solute interaction
• solvent-solute interaction
Hsoln = H1 + H2 + H3
13.2
“like dissolves like”
Two substances with similar intermolecular forces are likely
to be soluble in each other.
•
non-polar molecules are soluble in non-polar solvents
CCl4 in C6H6
•
polar molecules are soluble in polar solvents
C2H5OH in H2O
•
ionic compounds are more soluble in polar solvents
NaCl in H2O or NH3 (l)
13.2
Concentration Units
The concentration of a solution is the amount of solute
present in a given quantity of solvent or solution.
Percent by Mass
mass of solute
x 100%
% by mass =
mass of solute + mass of solvent
mass of solute x 100%
=
mass of solution
Mole Fraction (X)
moles of A
XA =
sum of moles of all components
13.3
Concentration Units Continued
Molarity (M)
M =
moles of solute
liters of solution
Molality (m)
m =
moles of solute
mass of solvent (kg)
13.3
What is the molality of a 5.86 M ethanol (C2H5OH)
solution whose density is 0.927 g/mL?
moles of solute
m =
moles of solute
M =
mass of solvent (kg)
liters of solution
Strategy:
Find mass of solvent
Know mass of solute + mass of solvent = mass of solution
If mass of solution and mass of solute known, can calculate mass of solvent
Can calculate mass of solute from moles of solute
Can calculate mass of solution from density and volume of the solution
Solve
13.3
What is the molality of a 5.86 M ethanol (C2H5OH)
solution whose density is 0.927 g/mL?
moles of solute
moles of solute
m =
M =
mass of solvent (kg)
liters of solution
0.586 moles of solute per 1 L of solution:
5.86 moles ethanol = 270 g ethanol
927 g of solution (1000 mL x 0.927 g/mL)
mass of solvent = mass of solution – mass of solute
= 927 g – 270 g = 657 g = 0.657 kg
moles of solute
m =
mass of solvent (kg)
=
5.86 moles C2H5OH
= 8.92 m
0.657 kg solvent
13.3
Temperature and Solubility
Solid solubility and temperature
No clear correlation between ΔHsoln and the variation of solubility with temperature
solubility
solubility decreases
increases with
with
increasing temperature
13.4
Fractional crystallization is the separation of a mixture of
substances into pure components on the basis of their differing
solubilities.
Suppose you have 90 g KNO3
contaminated with 10 g NaCl.
Fractional crystallization:
1. Dissolve sample in 100 mL of
water at 600C
2. Cool solution to 00C
3. All NaCl will stay in solution
(s = 34.2g/100g)
4. 78 g of PURE KNO3 will
precipitate (s = 12 g/100g).
90 g – 12 g = 78 g
13.4
Temperature and Solubility
Gas solubility and temperature
solubility usually
decreases with
increasing temperature
13.4
Pressure and Solubility of Gases
The solubility of a gas in a liquid is proportional to the
pressure of the gas over the solution (Henry’s law).
c = kP
c is the concentration (M) of the dissolved gas
P is the pressure of the gas over the solution
k is a constant (mol/L•atm) that depends only on temperature
low P
high P
low c
high c
13.5
Colligative Properties of Nonelectrolyte Solutions
Colligative properties are properties that depend only on the
number of solute particles in solution and not on the nature of
the solute particles.
Vapor-Pressure Lowering
P1 = X1 P
0
1
Raoult’s law
P 10 = vapor pressure of pure solvent
X1 = mole fraction of the solvent
If the solution contains only one solute:
X1 = 1 – X2
P 10 - P1 = P = X2 P 10
X2 = mole fraction of the solute
13.6
Ideal Solution
PA = XA P A0
PB = XB P 0B
PT = PA + PB
PT = XA P A0 + XB P 0B
13.6
PT is greater than
predicted by Raoults’s law
PT is less than
predicted by Raoults’s law
Force
Force
Force
< A-A & B-B
A-B
Force
Force
Force
> A-A & B-B
A-B
13.6
Fractional Distillation Apparatus
13.6
Boiling-Point Elevation
Tb = Tb – T b0
T b0 is the boiling point of
the pure solvent
T b is the boiling point of
the solution
Tb > T b0
Tb > 0
Tb = Kb m
m is the molality of the solution
Kb is the molal boiling-point
elevation constant (0C/m)
13.6
Freezing-Point Depression
Tf = T 0f – Tf
T
0
Tf
f
is the freezing point of
the pure solvent
is the freezing point of
the solution
T 0f > Tf
Tf > 0
Tf = Kf m
m is the molality of the solution
Kf is the molal freezing-point
depression constant (0C/m)
13.6
13.6
What is the freezing point of a solution containing 478 g
of ethylene glycol (antifreeze) in 3202 g of water? The
molar mass of ethylene glycol is 62.01 g.
Tf = Kf m
Kf water = 1.86 0C/m
moles of solute
m =
mass of solvent (kg)
478 g x
1 mol
62.01 g
=
= 2.41 m
3.202 kg solvent
Tf = Kf m = 1.86 0C/m x 2.41 m = 4.48 0C
Tf = T 0f – Tf
Tf = T 0f – Tf = 0.00 0C – 4.48 0C = -4.48 0C
13.6
Osmotic Pressure (p)
Osmosis is the selective passage of solvent molecules through a porous
membrane from a dilute solution to a more concentrated one.
A semipermeable membrane allows the passage of solvent molecules but
blocks the passage of solute molecules.
Osmotic pressure (p) is the pressure required to stop osmosis.
dilute
more
concentrated
13.6
Osmotic Pressure (p)
High
P
Low
P
p = MRT
M is the molarity of the solution
R is the gas constant
T is the temperature (in K)
13.6
A cell in an:
isotonic
solution
hypotonic
solution
hypertonic
solution
13.6
Colligative Properties of Nonelectrolyte Solutions
Colligative properties are properties that depend only on the
number of solute particles in solution and not on the nature of
the solute particles.
Vapor-Pressure Lowering
P1 = X1 P 10
Boiling-Point Elevation
Tb = Kb m
Freezing-Point Depression
Tf = Kf m
Osmotic Pressure (p)
p = MRT
13.6
Colligative Properties of Electrolyte Solutions
0.1 m Na+ ions & 0.1 m Cl- ions
0.1 m NaCl solution
Colligative properties are properties that depend only on the
number of solute particles in solution and not on the nature of
the solute particles.
0.1 m NaCl solution
van’t Hoff factor (i) =
0.2 m ions in solution
actual number of particles in soln after dissociation
number of formula units initially dissolved in soln
i should be
nonelectrolytes
NaCl
CaCl2
1
2
3
13.7
Colligative Properties of Electrolyte Solutions
Boiling-Point Elevation
Tb = i Kb m
Freezing-Point Depression
Tf = i Kf m
Osmotic Pressure (p)
p = iMRT
13.7
A colloid is a dispersion of particles of one substance
throughout a dispersing medium of another substance.
Colloid versus solution
•
collodial particles are much larger than solute molecules
•
collodial suspension is not as homogeneous as a solution
13.8
The Cleansing Action of Soap
13.8
Chemistry In Action:
Desalination