VAPOR PRESSURE
The term "vapor" is applied to the gas of any compound that
would normally be found as a liquid at room temperature and
pressure
For example, water, gasoline, rubbing alcohol, and finger nail
polish remover (ethyl acetate) are all normally liquids, but they
all evaporate to give a gas.
The molecules that are found on the upper side have only about
half as many neighbors as do molecules inside the liquid so to
begin with the forces holding them in the liquid are slightly
lower than molecules in the bulk liquid. The rate at which a
given volume of liquid will evaporate is determined by the
surface area.
In order for a molecule to break free of the liquid and go into
the gas phase (evaporate) its kinetic energy must be greater
than the forces holding it into the liquid. We recall that the
distribution of kinetic energies in a collection of molecules is
described by the Maxwell-Boltzmann relationship, so that at a
given temperature only the fraction of molecules with kinetic
energies greater than the intermolecular forces (as shown
below) will be able to break free.
NOTE:
*the number of molecules with energies sufficient to
break free will be proportional to the temperature. As the
temperature increases a greater fraction of the molecules
may break free.
*At a given temperature, a liquid with low intermolecular
forces will have a larger fraction of its molecules with
sufficient kinetic energy to evaporate than a liquid with higher
intermolecular forces. Another way of saying this is that the
liquid with low intermolecular forces has a higher vapor
pressure than a liquid with higher intermolecular forces.
Equilibrium Vapor Pressure
If we put the beaker in a closed chamber a slightly different phenomenon
occurs. Now molecules from the liquid evaporate as before, but some of
these evaporated molecules may also return to the liquid.
Boltzman curve
Boltzman curve in which we assume that only those molecules with
energies in excess of some value (which we loosely relate to the
Intermolecular Force) will be able to break free and go into the vapor
phase. Obviously, surface area is important, but it doesn't affect the
equilibrium vapor pressure, only the rate at which equilibrium is
achieved.
A simple Boltzman curve is illustrated below:
Recall that we think of the area under the curve as representing
all of the molecules in a container. If we had 1,000,000
molecules, then the area would represent these 1,000,000
molecules.
Imagine we are going to make a solution in which we're going
to dissolve some nonvolatile compound into our liquid. In real
life we'd just put the compound into our liquid, but since we're
trying to focus on the effect on the Boltzman curve, imagine
we remove 1,000 solvent molecules and put in 1,000 solute
molecules. This way we still have 1,000,000 molecules
represented by the curve. Since the properties of the solvent
and solute molecules should be additive, we can think of the
Boltzman curve as being the sums of a Boltzman curve for the
solvent and one for the solute.
When a nonvolatile solute is added
to a liquid to form a solution, the
vapor pressure above the solution
decreases
The reasons
. Liquid molecules at the surface of a liquid can escape to the gas phase
when they have a sufficient amount of energy to break free of the liquid's
intermolecular forces. That vaporization process is reversible. Gaseous
molecules coming into contact with the surface of a liquid can be trapped
by intermolecular forces in the liquid. Eventually the rate of escape will
equal the rate of capture to establish a constant, equilibrium vapor pressure
above the pure liquid.
If we add a nonvolatile solute to that liquid, the amount of surface area
available for the escaping solvent molecules is reduced because some of
that area is occupied by solute particles. Therefore, the solvent molecules
will have a lower probability to escape the solution than the pure solvent.
That fact is reflected in the lower vapor pressure for a solution relative to
the pure solvent. That statement is only true if the solvent is nonvolatile. If
the solute has its own vapor pressure, then the vapor pressure of the
solution may be greater than the vapor pressure of the solvent.
On the surface of the pure solvent (shown on the left) there
are more solvent molecules at the surface than in the righthand solution flask. Therefore, it is more likely that
solvent molecules escape into the gas phase on the left
than on the right. Therefore, the solution should have a
lower vapor pressure than the pure solvent.
The French chemist Francois Raoult discovered the law
that mathematically describes the vapor pressure lowering
phenomenon.
Raoult's law states that the vapor pressure of a solution, P,
equals the mole fraction of the solvent, csolvent, multiplied by
the vapor pressure of the pure solvent, Po.
P = csolvent * Po
Raoult's law, is easy to understand. When the solvent is pure, and the
mole fraction of the solvent is equal to 1, P is equal to Po. As the mole
fraction of the solvent becomes smaller, the vapor pressure of the solvent
escaping from the solution also becomes smaller.
Let's assume, for the moment, that the solvent is the only component of
the solution that is volatile enough to have a measurable vapor pressure.
If this is true, the vapor pressure of the solution will be equal to the
vapor pressure of the solvent escaping from the solution. Raoult's law
suggests that the difference between the vapor pressure of the pure
solvent and the solution increases as the mole fraction of the solvent
decreases.
Solutions that obey Raoult's law are called ideal solutions because they
behave exactly as we would predict. Solutions that show a deviation from
Raoult's law are called non-ideal solutions because they deviate from the
expected behavior.
When a solute is added to the solvent, some of the solute molecules occupy the
space near the surface of the liquid. When a solute is dissolved in a solvent, the
number of solvent molecules near the surface decreases, and the vapor pressure
of the solvent decreases.
This has no effect on the rate at which solvent molecules in the gas phase
condense to form a liquid. But it decreases the rate at which the solvent
molecules in the liquid can escape into the gas phase. As a result, the vapor
pressure of the solvent escaping from a solution should be smaller than the
vapor pressure of the pure solvent.
The change in the vapor pressure that
occurs when a solute is added to a solvent
is therefore a colligative property. If it
depends on the mole fraction of the solute,
then it must depend on the ratio of the
number of particles of solute to solvent in
the solution but not the identity of the
solute.
Colligative Properties are those properties of a liquid
that may be altered by the presence of a solute.
Examples of Colligative Properties
•vapor pressure
•melting point
•boiling point
•osmotic pressure
Colligative Properties
•
•
•
•
Freezing point depression
Boiling point elevation
Osmotic pressure
Vapor pressure lowering
Colligative Properties
Colligative properties of solutions
depend only on the amount of
solute!
The type of solute plays no role
other than being an electrolyte or
non-electrolyte!
Freezing Point Depression
A solute will lower the freezing point of a solvent by lowering the
vapor pressure of the solvent. The amount by which the vapor
pressure is lowered can be calculated using Raoult’s Law.
Freezing Point Depression
The vapor pressure of a solution depends on the mole fraction
of the solute, Xa, and the vapor pressure of the pure solvent.
We assume that the vapor pressure of the solute is negligible.
Freezing Point Depression
To calculate the change in the freezing point of a solvent with
a solute added, you need to know:
–
–
–
–
The freezing point depression constant, Kf
The molality of the solution
The nature of the solute: electrolyte or nonelectrolyte
The freezing point of the pure solvent
Freezing Point Depression
The change in freezing point, t, for any solute/solvent
system can be calculated using this relationship:
Δt = i · Kf · m
Where I = dissociation factor
Kf = freezing point depression constant
m = molality of the solution
Δt = change in freezing temperature
Freezing Point Depression
The molality of a solution is defined to be:
m=
moles solute
kg solvent
Freezing Point Depression
The dissociation factor is one (1) for any nonelectrolyte. A non-electrolyte is a compound made of
nonmetals and bonded by covalent bonds. Some
examples of non-electrolytes are:
1. carbon dioxide
2. ethanol
3. sugars
Freezing Point Depression
The dissociation factor is for an electrolyte, which is a
compound formed by a metal and nonmetal and bonded by
ionic bonds, is determined by the number of ions that will be
formed when dissolved in the solvent.
For example, NaCl will form 2 ions in water and has an i
value of 2. AlCl3 would form 4 ions in solution and has an i
value of 4.
Freezing Point Depression
The change in freezing point, t, is calculated using this
equation:
Δt = TFinal - TInitial
TFinal = the new freezing temperature of the solvent
TInitial = the freezing point of the pure solvent
Freezing Point Depression
The freezing point depression constant
is different for every solvent and must
be provided (unless that is what is
being asked in the question) For
water, Kf = - 1.86C/m
Freezing Point Depression
Example: Calculate the freezing point of a solution made by adding 1.00
mole of sucrose to 1.00 kg water.
Δt = i x Kf x m
i = 1 for sucrose (covalent, nonelectrolyte)
Kf = - 1.86C/m
m = 1.00 m = 1.00 mole
1.00 kg
Freezing Point Depression
Δt = i x Kf x m
Δt = (1) x (-1.86C/m) x (1.00 m)
Δt = - 1.86C
Freezing Point Depression
Since Δt = - 1.86C, the actual freezing point of the solution can be calculated
using:
Δt = TFinal – TInitial
and
- 1.86C = TFinal – 0C (the freezing
point of pure water)
Freezing Point Depression
The final answer is:
- 1.86C = TFinal