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AP Chemistry
Solids, Liquids and Solutions
Phase Changes
Melting Freezing Vaporization Condensation Sublimation Deposition –
Vaporization and Vapor Pressure
- Evaporation = Vaporization
- Molecules vaporize because they have enough energy to overcome the intermolecular
forces between themselves and neighbouring molecules.
- As the more energetic molecules vaporize, the average kinetic energy of the remaining
molecules decreases.
- Consequently, the temperature of the liquid decreases.
- When a liquid vapourizes on your hand, your hand feels cold.
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Enthalpy (Heat) of Vaporization
To maintain a constant temperature when it vaporizes, a liquid must absorb heat to
replace the kinetic energy carried away by the vaporizing molecules.
The Enthalpy of Vaporization is the quantity of heat that must be absorbed to vaporize a
given amount of liquid at a constant temperature.
For water:
H 2O(l )  H 2O( g )
H vapn  44.0kJ / mol
This means it takes 44.0kJ to vaporize 1 mol of water
Condensation is the opposite of vaporization.
For water:
H condn  44.0kJ / mol
Overall:
H vapn  H condn  0
Example:
How much heat in kilojoules is required to vaporize 125g of methanol at 25ºC if
H vapn  38.0kJ / mol for methanol?
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Vapor Pressure
When we place a liquid in an open container, eventually all the liquid will escape into the
gaseous state and become dispersed in the atmosphere. When we place a liquid in a
closed container, the system reaches a dynamic equilibrium. The rate of vapourization
and condensation are equal.
Liquid  Gas
The Vapour Pressure of a liquid is the partial pressure exerted by its vapor in dynamic
equilibrium with liquid at a constant temperature.
The Vapour Pressure of a liquid increases with temperature. At a higher temperature
more molecules have enough kinetic energy to escape from the liquid state. At higher
temperature both the rate of vapourization and condensation are increased (but still
equal).
Example:
What is the vapor pressure of hexane in atmospheres at 298.15 K if the density of the
vapour is 0.701g/L?
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Example:
How much heat in kilojoules is required to convert 85.2g of water from liquid at 250C to
vapour at 470C?
Note:
For water
DH vapn = 44.0kJ / mol
Specific Heat = 4.180J.g-1.0 C -1
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Example:
An ice cube with a mass of 25.5g at a temperature of 0.0 0C is added to 125mL of water
at 26.50C in an insulated container. What is the final temperature of the water after the
ice has melted?
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Generally:
If the intermolecular forces within a liquid are weak, many molecules escape from the
surface. The vapour pressure is high and the liquid is volatile.
If the intermolecular forces within a liquid are strong, few molecules escape from the
surface. The vapour pressure is low and the liquid is not volatile.
Boiling Point
The Boiling Point of a liquid is the temperature at which its vapour pressure becomes
equal to the atmospheric pressure.
Once boiling begins, the temperature of the liquid cannot rise above the boiling point.
The energy provided by the heating goes into converting liquid to vapor, not into raising
the temperature.
The Normal Boiling Point of a liquid is its boiling point at 1 atm..
Critical Temperature and Pressure
The critical temperature Tc  is the highest temperature at which a liquid and vapour can
co-exist in equilibrium.
The vapour pressure at this temperature is called the critical pressure Pc  .
The condition corresponding to a temperature of Tc and pressure of Pc is called the critical
point.
Alternatively, the critical temperature is the highest temperature at which a gas can be
condensed into a liquid solely by increasing the pressure of the gas.
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Melting, Melting Point and Heat of Fusion
The conversion of a solid to a liquid is called melting or fusion.
The quantity of heat required to melt a given amount of solid is called the Enthalpy
(Heat) of Fusion ( H fusion ).
Melting is an endothermic process:
H 2O(s)  H 2O(l )
H  6.01kJ / mol
Freezing is an exothermic process:
H 2O(l )  H 2O(s)
H  6.01kJ / mol
Sublimation
The passage of molecules directly from the solid state to the vapour state is called
sublimation. The reverse process is called deposition. A dynamic equilibrium is reached
when the rates of sublimation and deposition become equal.
Like vapour in equilibrium with a liquid, vapour in equilibrium with a solid exerts a
pressure, often called a sublimation pressure.
A plot of vapour pressure of a solid versus temperature is called a sublimation curve.
The Triple Point
The triple point is the only temperature at which the three states of matter can co-exist at
equilibrium.
It is the point where the vapour pressure curve of a substance and its sublimation curve
meet.
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Phase Diagrams
A phase diagram is a graphical representation of the conditions of temperature and
pressure under which a substance exists as a solid liquid or gas.
Consider the following Phase Diagram:
Point A –
Curve AD –
Curve AB –
Curve AC –
Point C –
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Intermolecular Forces
Dispersion Forces (London Forces)
-
Found in all substances
Caused by random movement of electrons
Increase in strength with increasing molecular size
Increase in strength with increasing molecular elongation
example:
Octane (Elongated)
Iso-octane (Compact)
Dipole- Dipole Forces
-
Found in Polar Substances
The more polar the substance, the greater the dipole moment, the greater the
intermolecular force
Intermolecular forces are USUALLY stronger in polar molecules
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Hydrogen Bonds
-
Occur in polar molecules where H is bonded to N, O or F
Water a good example:
-
each water molecule is bonded to four other water molecules
Hydrogen bonding is the reason ice is less dense than water
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Surface Tension
-
Molecules within the bulk of a liquid are attracted more strongly to neighbouring
molecules than are surface molecules
-
It takes energy to increase the surface area of a liquid
-
Surface Tension is the amount of energy required to extend the surface of a
liquid
-
Surface Tension decreases with increase in temperature
-
Why does water have a greater Surface Tension than hexane?
-
Surface tension allows insects to “walk on water”. (Their weight is not large
enough to break the surface of the water)
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Adhesive Forces: Intermolecular forces between unlike molecules
Cohesive Forces: Intermolecular forces between like molecules
If the adhesive forces between a liquid and a surface are greater than the cohesive
forces within a liquid, the liquid will wet the surface
Examples:
Glass is wet by water:
Glass coated with an oil film is not wet by water:
The meniscus of water and mercury in a glass tube:
Viscosity
The Viscosity of a liquid is a measure of its resistance to flow
Honey has a high viscosity
Water has a low viscosity
The viscosity of a liquid is related to the strength of intermolecular forces
Viscosity decreases with increase in temperature
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Network Covalent Solids
Diamond
Structure:
Properties:
Graphite
Structure:
Properties:
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Solutions
Solute:
Solvent:
Problems
(1) Find the percent by mass of NaCl in a solution made by dissolving 124g NaCl in 768g
of water.
(2) Find the percent by volume of ethanol in a solution made by mixing 45.0mL of
ethanol with 85.0mL of water.
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Molarity Versus Molality
The Molarity (mol/L) of a solution varies with temperature, because the volume of a
solution increases with increasing temperature.
The Molality is another unit of concentration that is independent of temperature:
Amount of solute (mol)
Molality (m) = ______________________
Mass of solvent (kg)
Example:
What is the molality of a solution prepared by dissolving 5.05g of C10 H 8 ( s) in 75.0mL of
Benzene ( C6 H 6 ), which has a density = 0.879 g/mL?
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Mole Fraction and Mole Percent
Consider the following example:
A solution is made by mixing 4 mol methanol and 6 mol of ethanol.
The mole fractions are as follows:
xCH 3OH =4/(4+6) = 0.40
xCH 3CH 2OH =6/(4+6) = 0.60
The mole percents are as follows:
40 mol% methanol
60 mol% ethanol
Solution Formation
Why do some substances dissolve in a given solvent and others do not?
To answer this question, we need to consider two things, the enthalpy of solution and
intermolecular forces.
Enthalpy of Solution
Consider the steps required to obtain a solution:
1) Pure Solvent  Separated solvent molecules
2) Pure Solute  Separated solute molecules
3) Separated solvent and solute molecules  Solution
Net: Pure Solvent + Pure Solute  Solution
H1
H 2
H 3
(>0)
(>0)
(<0)
H solution  H 1  H 2  H 3
If H solution < 0, it is likely a solute will dissolve in a solvent.
Generally if the forces between solute and solvent particles are stronger than the forces
between the solvent particles and the solute particles, the solute will dissolve in the
solvent.
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Aqueous Solutions of Ionic Compounds
The forces that cause an ionic substance to dissolve in water are ion-dipole forces, the
attraction between water dipoles, and cations and anions.
The extent to which an ionic solid dissolves in water is determined largely by the strength
of these forces compared to the strength of the ion-ion attractions in the ionic solid.
It can be difficult to predict the solubility of a particular ionic compound. We can refer to
a solubility table.
Saturated Solutions
Saturated Solutions reach a dynamic equilibrium.
Example:
NaCl(s)  Na  (aq)  Cl  (aq)
The rate of dissolving equals the rate of crystallizing.
Solubility as a Function of Temperature
Solubility varies with temperature. For 95% of all ionic compounds, solubility increases
significantly with temperature. (Sulfates and Acetates can be an exception).
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The Solubility of Gases
Gases are somewhat soluble in water
Examples:
Carbon Dioxide in Coke
Air in lakes so fish can breathe
The solubility of a gas in a liquid depends on both temperature and pressure.
Temperature
Most gases become less soluble in liquids as temperature increases.
Pressure
At constant temperature, the solubility of a gas is directly proportional to the pressure of
the gas. The greater the pressure of the gas the greater the solubility.
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Vapour Pressure of Solutions
Raoult found that the presence of a solute lowers the vapor pressure of the solvent in a
solution.
Raoult’s Law:
The vapour pressure of the solvent above a solution ( Psolv ) is the product of the vapour
pressure of the pure solvent ( P 0 solv ) and the mole fraction of the solvent in solution
( x solv ).
Psolv  x solv  P 0 solv
(Textbook)
OR
PA  X A  PT
(Exam Formula Sheet)
Raoult’s Law is only strictly followed by ideal solutions. It also works for non-ideal
solutions that are very dilute.
If molecules A and B are in a solution and are almost identical chemically, then the
solution will be ideal. Since the interaction energies between A and B are the same, it
follows that there is no overall energy change when the substances are mixed.
Examples:
A solution of 1-butanol and 2-butanol will be ideal and will obey Raoult’s Law.
A solution of water and octane will not be ideal and will not obey Raoult’s Law.
Problem:
The vapour pressure of pure water at 20ºC is 17.5mmHg. What is the vapour pressure if
a solution has 1.50mol of sucrose (C12 H 22O11 ) dissolved per kilogram of water?
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Freezing Point Depression and Boiling Point Elevation
In this section, we will focus on solutions where the solvent is volatile and the solute is
non-volatile and a non-electrolyte.
We know that the presence of such a solute lowers the vapour pressure of a solvent in a
solution because the mole fraction of the solvent is reduced.
The presence of such a solute also lowers the freezing point of the solvent and raises the
boiling point of the solvent.
Generally:
T f  T f ( solution )  T f ( solvent )   K f  m
Tb  Tb (solution )  Tb (solvent )  K b  m
Where:
T f - freezing point depression (change in freezing point)
Tb - boiling point elevation (change in boiling point)
K f , K b - constants
m- molality (mol solute/kg solvent)
Generally, only the solvent freezes, and only the solvent escapes as vapour during
boiling.
As a result, the concentration of the solute increases as freezing or boiling occurs, causing
the freezing point to drop still more, and the boiling point to rise higher.
Note for water:
K f =1.86 0 Cm 1
K b =0.512 0 Cm 1
(you do not need to remember these constants)
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Example:
What is the freezing point of an aqueous sucrose solution that has 25.0g of sucrose per
100.0g of water?
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Example:
An aqueous solution of sorbitol contains 1.00 g of sorbitol in 100.0 g of water. It has a
freezing point of -0.102 degrees celcius. The sorbitol contains 39.56% C, 7.75% H and
52.7% O by mass. What are the molar mass and molecular formula of sorbitol?
Freezing point depression and boiling point elevation of non-volatile solutions have
practical applications such as radiator coolant (ethyl glycol in water) in cars.
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Osmotic Pressure
Filter paper is permeable to water, but impermeable to coffee grounds.
There are some materials that are semi-permeable, they allow solvent molecules to
pass but not solute molecules. These materials are called semi-permeable
membranes.
Osmosis is the net flow of solvent molecules through semi-permeable membrane
from a solution of lower concentration to a solution of higher concentration.
The pressure required to stop osmosis is called osmotic pressure.
For dilute solutions:
Osmotic Pressure ( ) 
n
RT
V
 = Osmotic Pressure (atm)
V = Volume (L)
R = Gas Constant
T = Temperature (K)
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Example:
1.50g of hemocyanin is dissolved in 0.250L of water. The solution has an Osmotic
Pressure of 0.00342 atm at 277 K. What is the molar mass of hemocyanin? What is the
freezing point of the solution?
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Solutions of Electrolytes
So far, we have looked at solutions that contain non-electrolytes. We know, however,
that electrolytes can also impact the boiling and freezing point of solutions.
Consider NaCl and CaCl2 . Both of these substances are strong electrolytes, they
dissociate relatively easily in water.
As long as the concentration of a solution is low, 1 mole of NaCl forms 2 moles of ions
while 1 mole of CaCl2 forms 3 moles of ions. We can use the van’t Hoff factor (i) to
find the freezing point depression and boiling point elevation.
For a dilute solution of NaCl , i=2.
(in theory)
For a dilute solution of CaCl2 , i=3.
(in theory)
We can modify our equations:
T f  iK f m
Tb  iK b m
Compared to NaCl , a solution of CaCl2 should have more ions in solution for the same
concentration. This should lead to increased freezing point depression and boiling point
elevation.
Hence CaCl2 works better for de-icing roads than NaCl .
Note that if we are considering a weaker electrolyte (example: CH 3COOH ) or a
somewhat concentrated solution (more than 0.01M), i values tend to be considerably
lower than might be expected.
Examples:
0.01M CH 3COOH , i=1.04
(might expect i=2)
0.01M NaCl , i=1.94 (might expect i=2)
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