Vapour Pressure of Pure Substances
When you leave wet dishes on a draining board overnight, they will usually be dry the
next morning.
The water has evaporated – even though your kitchen was 70-80˚C below the boiling
point of the water. How is this possible?
Thermochemistry and Electrochemistry
What Makes Reactions Go?
At any given temperature, some of the molecules in the liquid will have enough energy
to escape the intermolecular forces holding the sample together.
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Vapour Pressure of Pure Substances
Vapour Pressure of Pure Substances
Heating the liquid increases the rate of vaporization by increasing the proportion
of molecules having at least this much energy.
When the vapour pressure is equal to the
atmospheric pressure, the liquid boils.
Since heating increases the extent of vaporization, this is an
__________________ process with a _________ enthalpy.
The pressure exerted by gas molecules is equal
to the pressure of the atmosphere pushing down
on the liquid’s surface.
This is consistent with rapid vaporization cooling the remaining liquid.
Hence: the boiling point is pressure-dependent.
The molecules at the surface with enough energy to evaporate do so, reducing the
average kinetic energy of the remaining liquid molecules, hence reducing the
temperature.
During slow evaporation, heat is steadily resupplied from the environment so we
don’t notice this loss.
To allow comparisons, we therefore define the temperature at which the vapour pressure of a
liquid is equal to ___________________ as a liquid’s normal boiling point.
As always there will be at least a few molecules with enough energy to evaporate,
there will be a layer of gas molecules immediately above a liquid.
In a sealed container at equilibrium, the pressure exerted by these gas molecules is
referred to as the vapour pressure.
Vapour pressure is not exclusive to liquids; many solids have a measurable vapour
pressure.
e.g. dry ice (CO2(s)), I2(s)
Reducing the pressure will decrease a liquid’s observed boiling point while increasing pressure
will increase the observed boiling point.
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Liquids with a high vapour pressure/low boiling point are referred to as __________________.
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Phase Diagrams
Phase Diagrams
Vapour pressure plotted against temperature, generates an exponential vapour pressure curve:
A complete temperature-pressure phase diagram is shown below:
Vapour pressure can be read off a
vapour pressure curve or calculated
from free energy.
e.g. The standard molar free energy
change for the evaporation of water
is +8.5 kJ/mol.
This curve is part of a phase diagram, a figure which shows the most thermodynamically stable
phases under different conditions.
Each line/curve on the diagram shows the conditions under which two phases are in
equilibrium. The triple point shows the conditions under which all three phases are in
equilibrium! At pressures and temperatures below the triple point, the liquid phase does not
form and the substance sublimes from gas to solid (or solid to gas).
For pure substances, temperature-pressure phase diagrams are common.
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Phase Diagrams
Phase Diagrams
A solid heated at a constant pressure will either:
A gas compressed at constant temperature will undergo __________to either a liquid or a solid.
Undergo ___________________ to a liquid,
Which occurs depends on the ______________________.
or
This is a normal temperature-pressure phase diagram:
Undergo _________________________ to a gas.
Which occurs depends on the ______________________.
Where the line between the solid and liquid phases has a positive slope.
What does that indicate about the substance?
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Phase Diagrams
Phase Diagrams
Notice that, on both phase diagrams, the curve between the gas and liquid phase
appears to stop.
This is the critical point.
As the temperature of a liquid is increased, its vapour pressure continues to increase,
and it is necessary to have an increasingly high pressure in order to maintain the
liquid state.
There comes a temperature at which no amount of pressure is sufficient; the kinetic
energy of the molecules is too great.
This is the critical temperature.
The point on a phase diagram at which the critical temperature and critical
pressure meet is referred to as the critical point.
The temperature-pressure phase diagram for water is not normal:
At temperatures higher than the critical temperature, high pressures give a dense gaslike state called a supercritical fluid.
where the line between the solid and liquid phases has a negative slope.
The molecules in supercritical fluids are packed together almost as tightly as in liquids
but they have enough energy to overcome intermolecular forces and move freely. As
such, they can make excellent solvents.
What does that indicate about water?
(especially supercritical CO2 – which has the added advantage of being much more
environmentally friendly than most organic solvents).
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Vapour Pressure of Solutions
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Vapour Pressure of Solutions
The vapour pressure curves for pure benzene (in red) and a
2 mol/kg solution (in blue) of a nonvolatile solute in benzene.
The surface of the pure water consists entirely of water molecules, a fraction of which
have enough energy to evaporate.
Adding the solute has lowered the vapour pressure at any
given temperature(thereby raising the boiling point)
The surface of the salt water consists primarily of water molecules but also some ions
from the salt. The same fraction of water molecules has enough energy to evaporate
but, since fewer water molecules are on the surface, fewer evaporate. Essentially, the
salt ions are “blocking the surface” of the solution.
Why does this occur?
Thus, the salt water has a lower vapour pressure than the pure water. Because of
this, the salt water will have to be heated to a higher temperature to raise its vapour
pressure to be equal to atmospheric pressure.
Consider the difference between a pure liquid and a solution with
a nonvolatile solute. e.g. pure water (right) vs. salt water (left).
Thus, the boiling point of salt water is higher than that of pure water.
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Vapour Pressure of Solutions
Equilibrium Constant and Temperature
This can be expressed mathematically. Raoult’s law defines the vapour pressure of
a solvent (“A”).
An equilibrium constant (K) is only constant as long as the temperature doesn’t
change! Why?
where PA is the vapour pressure of the solvent (in solution), PA° is the vapour
pressure of the solvent (as a pure liquid) and XA is the mole fraction of the solvent
molecules.
The enthalpy and entropy change for a reaction vary slowly with temperature;
however, the free energy change is strongly temperature-dependent:
PA  X A PA 
At any temperature:
Note that Raoult’s law only calculates the vapour pressure of the solvent. If a solute
is volatile, it will have a vapour pressure too!
G  H  TS
[ A ]  k H PA
So, as T changes, G changes and therefore K changes.
r G mo (T )  RT ln[K (T )]
where [A] is the concentration of the volatile solute “A”, PA is the vapour pressure of
the volatile solute and kH is the Henry’s law constant (solute-specific).
Unsurprisingly, this is Henry’s law. It is usually presented when discussing
solubility of gases.
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Equilibrium Constant and Temperature
Equilibrium Constant and Temperature
When comparing a reaction at temperatures T1 and T2, we can say that:
o
o
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Kw = 10-14 under standard conditions :
o
r G1 (T1 )  r H1 T1r S1  RT1 ln[K 1 ]
Hence neutral water have pH 7 at 25 °C.
and
r G 2o (T2 )  r H 2o T2r S 2o  RT2 ln[K 2 ]
What is Kw at 37.00 °C?
If we approximate that the values of H and S don’t change significantly over the
given temperature range, we can combine these equations and simplify to get:
kJ
mol
kJ
 f G ( H 2O( l ) )  285 .8
mol
 f G (OH (aq ) )  230 .0
o
or:
r H  1
1
    ln K 2  ln K 1
R T1 T 2 
K   H o
ln 2   r
R
K1 
1
1
  
T1 T 2 
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Equilibrium Constant and Temperature
Equilibrium Constant and Temperature
Recall that the boiling point of a liquid is the temperature at which its vapour
pressure equals the atmospheric pressure.
What does this tell us about the pH of water at 37.00 °C?
If the atmospheric pressure in Lethbridge is typically 90 kPa, what is the boiling
point of water on a typical day in Lethbridge?
pH   log(aH )

 vapH m (H 2O )  40.66
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kJ
mol
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Equilibrium Constant and Temp. (Solutions)
The same approach can be used to calculate the boiling point (or freezing point) of
aqueous solutions. When we salt roads, we make saturated salt solutions which
freeze below 0 C. Calculate the freezing point of NaCl(aq) prepared by dissolving 6
mol NaCl in 1 kg H2O.
 fusH m (H 2O )  6.007
kJ
mol
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