TOPIC 17 EQUILIBRIUM HL
17. 1 Liquid-vapour equilibrium
17.1.1 Describe the equilibrium established between a liquid and its own vapour and how it is affected by temperature
changes.
17.1.2.Sketch graphs showing the relationship between vapour pressure and temperature and explain them in terms
of the kinetic theory.
17.1.3 State and explain the relationship between enthalpy of vaporization, boiling point and intermolecular forces.
A phase equilibrium is established between two different phases or states of matter of the same substance.
Such a phase equilibrium provides information about the structure and the bonding of the substance.
A phase is any part of a system that is uniform in chemical composition and physical properties e.g. solid
water, liquid water and water vapour are 3 different phases. We will limit our study of phase equilibria by
just considering a one component system (one substance only) in a liquid/vapour equilibrium only; it is an
equilibrium which is achieved when the rate of condensation equals the rate of evaporation.
Application of kinetic theory to explain the equilibrium between a liquid and its own vapour
Important points:

System needs to be closed otherwise vapour particles simply diffuse away and no equilibrium is
established.

In any liquid at any particular temperature there are a number of particles on the surface of the liquid
that posses sufficient energy to escape the forces of attraction of its neighbouring particles; these
particles escape from the liquid phase and form the vapour above the liquid, which in a closed system,
cannot escape;

Vapour particles accumulate above the liquid and collide with the walls of the contained producing
pressure inside the vessel.

Some vapour particles also collide with the surface of the liquid and become trapped again by the liquid
returning to it; this is condensation.

The rate at which particles evaporate is determined by the liquid’s temperature which in a closed
system is being kept constant, therefore rate of evaporation is constant. The higher the temperature,
the more particles there are that have enough energy to escape, the higher the rate of evaporation.

The rate of condensation depends on the concentration of the particles in the gaseous state: the more
molecules per unit of volume, the more collisions they will make with the surface of the liquid and the
higher the rate of condensation.

In the beginning when a liquid is placed in a container, there are no molecules in the gaseous state.
The rate of evaporation is zero.

Evaporation (which takes place at a constant rate) produces a net movement of particles into the gas or
vapour phase. As the number of particles in the gas phase increases the rate of condensation also
increases. This continues until the rate of condensation equals or catches up with the rate of
condensation. When this happens, the number of particles in the gas phase stays the same and the
pressure also remains the same. The level of the liquid also remains the same.
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rate of
evaporation
dynamic equilibrium
achieved
amount of vapour
+ level of liquid remain the same
rate
rate of condensation
vapour pressure is reached
Vapour pressure (in closed system)
See simulation: http://www.chem.purdue.edu/gchelp/liquids/vpress.html
 saturated vapour pressure (equilibrium vapour pressure) = the pressure exerted by the vapour particles
above a liquid when a dynamic equilibrium has been established between the liquid and its vapour at a
given temperature i.e. the rate of evaporation = rate of condensation; when the saturated vapour
pressure is reached the amount of gas and liquid remain constant (see diagram above)
 vapour pressure is also an indication of the position of equilibrium (but in this case of a reversible
physical process); the higher the vapour pressure the more the equilibrium lies towards the vapour
phase; the more the forward process (i.e. evaporation) is favoured by the conditions.

measurement of saturated vapour pressure: (from http://www.chemguide.co.uk/physical/phaseeqia/vapourpress.html)
If you have a mercury barometer tube
in a trough of mercury, at 1 atmosphere
pressure the column will be 760 mm
tall. 1 atmosphere is sometimes quoted
as 760 mmHg ("millimetres of
mercury").
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If you squirt a few drops of liquid into the tube, it will rise to form
a thin layer floating on top of the mercury. Some of the liquid will
evaporate and you will get the equilibrium we've just been
talking about - provided there is still some liquid on top of the
mercury. It is only an equilibrium if both liquid and vapour are
present. The saturated vapour pressure of the liquid will force
the mercury level down a bit. You can measure the drop - and
this gives a value for the saturated vapour pressure of the liquid
at this temperature. In this case, the mercury has been forced
down by a distance of 760 - 630 mm. The saturated vapour
pressure of this liquid at the temperature of the experiment is
130 mmHg.
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Factors which affect vapour pressure
Vapour pressure depends on 2 principle factors which each affect the rate of evaporation. The rate of
evaporation determines the vapour pressure at which the rate of condensation catches up; the higher the
rate of evaporation, the larger the rate of condensation needs to be for it to equal the rate of evaporation; a
large rate of condensation can only be achieved if there is a large amount of vapour molecules.

temperature: Using the apparatus above the vapour pressure at various temperatures can be
measured; the data can be used to draw a vapour pressure curve
(from http://www.chem.purdue.edu/gchelp/liquids/vpress.html):
The higher the temperature; the higher the constant rate of evaporation; the molecular kinetic energy is
greater at a higher temperature so a greater amount of molecules will be able to escape the surface
and vapourize.
As long as the rate of evaporation outstrips the rate of condensation, the vapour pressure increases.
The rate of condensation catches up with the rate of evaporation at a higher vapour pressure (a higher
rate of evaporation means a higher concentration of vapour particles is needed); the equilibrium will be
further towards the vapour side.
See also:


http://www.chm.davidson.edu/vce/phasechanges/vaporpressure.html
http://hyperphysics.phy-astr.gsu.edu/hbase/kinetic/vappre.html
 nature of attractive forces: the stronger the intermolecular forces, the lower the vapour pressure at any
given temperature as fewer molecules evaporate at that temperature lowering the rate of evaporation
and needing less particles in the vapour phase to let the rate of condensation catch up. Liquids with low
vapour pressures have stronger intermolecular forces.
Factors which do not affect vapour pressure
This is because pressure is force per unit of area:
 the volume of container in which the homogenous system is kept; this is because pressure is force per
unit of area:
 amount of liquid or vapour (equilibrium can be reached from both directions!) in the system;
 surface area of the liquid phase in the system; This is because pressure is force per unit of area:
Vapour pressure in an open system: prediction of boiling point
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We can also use the vapour pressure curve to predict the boiling point of a substance.
During the heating of a liquid in an open system, bubbles form within the liquid. These bubbles contain
vapour which has a certain vapour pressure (there is an equilibrium between the vapour within the bubble
and the liquid surrounding the bubble). The vapour pressure in the bubble depends on the temperature of
the liquid.
As we continue heating the liquid, the vapour pressure in the bubbles increases as more liquid evaporates
into it; the bubble gets bigger.
At any given moment:
 if the vapour pressure inside the bubble equals the atmospheric pressure on top of the liquid than the
bubbles rise and will burst at the surface of the liquid allowing the vapour to escape into the air above
the liquid;
 if the pressure is lower than the atmospheric pressure than the bubble is squashed before it gets to the
surface; to get to the surface the bubble needs to push away the liquid above it which is pushed down
by the atmosphere at a higher pressure; a higher atmospheric pressure makes the bubble collapse
before it gets to the surface. Further heating causes the equilibrium in the bubble to shift more to the
vapour side until the vapour pressure inside the bubble equals the atmospheric pressure and the bubble
stays and rises.
So a liquid boils when its vapour pressure equals the pressure on the liquid surface or atmospheric
pressure (usually 1 atm) if the system is open.
The temperature at which its vapour pressure equals the atmospheric pressure when it is 1 atm is its
normal boiling temperature.
The boiling point can be derived from a vapour pressure-temperature curve/phase diagram by finding the
temperature at which the vapour pressure is 1 atmosphere.
Boiling point is the temperature at which the vapour pressure of the liquid is equal to the atmospheric
pressure (usually 1 atm).
The diagram below shows the vapour curves of different liquids and how they can be used to determine
their boiling points. (from http://www.algebralab.org/practice/practice.aspx?file=reading_vaporpressurecurves.xml)
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Vapour pressure curves can also be used to predict which phase is the most stable for a specific set of
conditions.
1 atm
atmospheric
pressure (in
atm)
Tx = boiling point at pressure Px
vapour pressure of the liquid
liquid
Pvpz
Pz
Px
Z
gas
X
Pvp
Ty
Tx
Tz
boiling point at 1 atm pressure
temperature (in K)
For instance:
 at point X, conditions = pressure is Px and the temperature is Ty:.
at that atmospheric pressure, Px , and temperature Ty, the vapour pressure of the liquid is only Pvp which
is a lot lower than the atmospheric pressure, Px, and therefore the liquid does not boil and is more
stable as a liquid;
 at point Z, conditions = pressure is Px and the temperature is Tz:.
at point Z: at that atmospheric pressure, Pz , and temperature Tz, the vapour pressure, Pvpz , of the liquid
is greater than the atmospheric pressure, Pz, and therefore the substance is more stable in its vapour
state;
 at any combination of pressures or temperatures along the line the substance is in equilibrium between
the liquid and the gaseous state.
Relationship between boiling point, enthalpy of vaporisation, intermolecular forces and vapour
pressure
 standard enthalpy of vaporisation = amount of heat needed to convert 1 mole of liquid to a vapour at a
pressure of 1 atm. It is the difference between the enthalpy of the vapour and the enthalpy of the liquid
as shown by the enthalpy level diagram below.
It measures the amount of heat needed to overcome the intermolecular forces; it indicates how strongly
the molecules are held together in a liquid;
A (g)
 Hvapour
A (l)
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 most liquids at room temperature are covalently bonded molecules; there is a linear relationship directly proportional - between the boiling point and the enthalpy of vaporisation both of which are
determined by the size of the intermolecular forces (=electrostatic force as a result of polarisation within
a molecule; only acts over a short distance)
compound
CH4
C2H4
C3H8
C4H10
C6H14
C8H18
C10H22
F2
Cl2
Br2

Hvapour (kJ/mol)
9.2
14
18.1
22.3
28.6
33.9
35.8
6.52
20.4
30.7
Boiling point (in C)
-161
-89
-30
0
68
125
160
-188
-34.6
59
compound
HF
HCl
HBr
HI
H2O
H2S
NH3
PH3
SiH3
Hvapour (kJ/mol)
30.2
15.1
16.3
18.2
40.6
18.8
23.6
14.6
12.3
Boiling point (in C)
17
-84
-70
-37
100
-61
-33
-88
-112
the greater the enthalpy of vaporisation, the higher the boiling point.
 the effect of the intermolecular forces increases as the number of electrons within a molecule increases
as it are the electrons that cause either permanent or instantaneous polarity; the larger the number of
electrons, the more the molecules are polar or can be polarised instantaneously generating larger
intermolecular forces;
 at any given temperature, substances with large molecular forces (molecules attracted strongly) have
low vapour pressures (few molecules have enough energy to escape) as they have a low rate of
vaporisation; the equilibrium is more to the left;
strong intermolecular attraction = high enthalpy of vaporisation = high boiling point = low vapour pressure
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17. 2 The equilibrium law
17.2.1 Solve homogeneous equilibrium problems using the expression for Kc.
Calculating Kc or equilibrium concentrations when given equilibrium concentrations and Kc.
Exercises.
1. A mixture of nitrogen and hydrogen was sealed into a steel vessel and held at 1000K until equilibrium
was reached. The contents were then analysed and the results are in the table below;
equilibrium concentration in mol/dm-3
0.142
1.84
1.36
substance
N2 (g)
H2 (g)
NH3 (g)
N2 (g) + 3H2 (g)  2NH3 (g)
(a) Write an expression for Kc for this reaction:
(b) Calculate a value for Kc. Remember to give units.
2. For the reaction
S2 (g) + 2H2 (g)  2H2S (g)
Kc was found to be 9.4 x 105 mol-1 dm3 at 1020 K. Equilibrium concentrations were measured as
[H2(g)] = 0.234 mol dm-3 and [H2S (g) ] = 0.442 mol dm-3.
(a) Write an expression for Kc for this reaction.
(b) What is the equilibrium concentration of S2 (g) in the above mixture?
3. At 25 C the equilibrium constant Kc = 1.7 x 104 for the reaction CO (g) + 2H2 (g)  CH3OH (g).
What is the value of Kc at this temperature for the following reactions:
(a) CH3OH (g)  CO (g) + 2H2 (g)
(b) 2CO (g) + 4H2 (g)  2CH3OH (g)
4. At 460 C, Kc = 85.0 for the reaction
SO2 (g) + NO2 (g)  NO (g) + SO3 (g)
A mixture of these gases has the following concentrations of the reactants and products:
[SO2] = 0.040 mol dm-3, [NO2] = 0.50 mol dm-3, [NO] = 0.30 mol dm-3, [SO3] = 0.02 mol dm-3.
Is this system at equilibrium? If not, in which direction must the reaction proceed to reach equilibrium?
5. An equilibrium mixture of nitrogen, hydrogen and ammonia had the following concentrations: 0.1207
mol dm-3 H2, 0.0402 mol dm-3 N2, and 0.00272 mol dm-3 NH3 ,
Calculate Kc for the following reaction:
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N2 (g) + 3H2 (g)  2NH3 (g)
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6. Methanol can be produced from carbon monoxide and hydrogen according to the equation:
CO (g) + 2H2 (g)  CH3OH (g).
In a certain equilibrium mixture, [CO(g)] = 0.2 mol dm-3, [H2] = 0.1 mol dm-3, [CH3OH] = 2 mol dm-3.
What is the value of for this reaction?
7. At a temperature of 500 C the equilibrium constant, Kc, for the production of ammonia
N2 (g) + 3H2 (g)  2NH3 (g)
has a value of 6.0 x 10-2. If, in a particular reaction vessel at this temperature, there are 0.250 mol dm-3
of H2 and 0.050 mol dm-3 of NH3 present at equilibrium, what is the concentration of N2?
Calculating K when [initial], one equilibrium concentration and the equation are given.
Method:
1. Tabulate the known initial and equilibrium concentrations of all species involved in the equilibrium.
2. For those species for which both the initial and equilibrium concentrations are known, calculate the
change in concentration that occurs as the system reaches equilibrium.
3. Use the stoichiometry of the equation to calculate the changes in concentration for all the other species.
4. From the initial concentrations and the changes in concentration, calculate the equilibrium
concentrations. Calculate Kc.
Example:
A mixture of 5.00 x 10-3 mol of H2 and 1.00 x 10-2 mol of I2 were placed in a 5.00 L container at 448 C and
allowed to come to equilibrium. Analysis of the equilibrium shows that the concentration of HI is 1.87 x 10-3
mol dm-3. Calculate Kc at 448C.
Answer:
1. Tabulate the known initial and equilibrium concentrations of all species involved in the equilibrium.
H2 (g)
initial
change
equilibrium
+
1.00 x 10-3 mol dm-3

I2 (g)
2 HI (g)
2.00 x 10-3 mol dm-3
0 mol dm-3
1.87 x 10-3 mol dm-3
2. For those species for which both the initial and equilibrium concentrations are known, calculate the
change in concentration that occurs as the system reaches equilibrium.
That change is 1.87 x 10-3 mol dm-3 which is the change of HI (g).
3. Use the stoichiometry of the equation to calculate the changes in concentration for all the other
species. The ratio of HI to the other two species, H2 (g) and I2 (g), is 2:1 so their change will be
1.87 x 10-3 mol dm-3 / 2 = 0.935 x 10-3 mol dm-3
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4. From the initial concentrations and the changes in concentration, calculate the equilibrium
concentrations.
H2 (g)
+
I2 (g)

2 HI (g)
initial
change
equilibrium
1.00 x 10-3 mol dm-3
- 0.935 x 10-3 mol dm-3
0.065 x 10-3 mol dm-3
2.00 x 10-3 mol dm-3
- 0.935 x 10-3 mol dm-3
1.065 x 10-3 mol dm-3
0 mol dm-3
+ 1.87 x 10-3 mol dm-3
1.87 x 10-3 mol dm-3
5. Calculate Kc.
Kc =
[HI]2
[H2 (g)] [I2 (g)]
=
1.87 x 10-3 mol dm-3
0.065 x 10 mol dm-3 x 1.065 x 10-3 mol dm-3
-3
= 51.
Exercises:
1. Sulfur trioxide decomposes at high temperature in a sealed container:
2SO2 (g) + O2 (g)  2SO3 (g).
Initially the vessel is charged at 1000 K with SO3 (g) at concentration of 6.99 x 10-3 mol dm-3. At
equilibrium, the SO3 (g) concentration is 2.44 x 10-3 mol dm-3.
Calculate the Kc value at 1000K.
2. Brown NO2 gas and the colourless N2O4 gas exist in equilibrium:
2NO2 
N2O4
In an experiment, 0.625 mol of N2O4 was introduced into a 5.00 dm-3 vessel at a certain temperature.
The N2O4 gradually decomposes until it reached equilibrium with NO2. At equilibrium, the concentration
of N2O4 was 0.075 mol dm-3. What is Kc for this reaction at this temperature?
3. Calculate the equilibrium constant at 25 C for the reaction
2NOCl (g)

2NO (g) + Cl2(g)
In one experiment, 2.00 moles of NOCl were placed in a 1.00 dm3 flask, and the concentration of NO
after equilibrium was achieved was 0.66 mol dm-3.
4. 1.00 mole of Br2 is placed in a 1.00 dm-3 vessel and heated to 1756 K, a temperature at which the
halogen molecules dissociate to Br atoms.
Br2 (g)  2Br (g)
If 1% of Br2 (g) is dissociated at this temperature, calculate Kc.
5. Equal numbers of moles of H2 (g) and I2 (g) are mixed in a closed flask and are heated to 700 C. The
initial partial concentration of each gas is 0.70 mol dm-3 and 78.6 % of the I2 (g) has been consumed
when the following reaction comes to equilibrium.
H2 (g) + I2 (g) 
2HI (g)
Calculate Kc.
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6. When 2.0 moles of HI (g) are placed in a 1.0 L flask at 25 C and allowed to dissociate according to the
equation 2HI (g)  H2 (g) + I2 (g), it is found that 20 % of the HI has dissociated at equilibrium.
Calculate Kc.
7. A mixture of 1.374 g of H2 and 70.31 g of Br2 is heated in a 2.00 dm-3 vessel at 700 K. These
substances react as follows:
H2 (g) + Br2 (g)  2HBr (g)
At equilibrium, the vessel is found to contain 0.566g of H2.
(a) Calculate the equilibrium concentrations of H2, Br2 and HBr.
(b) Calculate Kc.
8. A mixture of 0.100 mol of NO, 0.050 mol of H2, and 0.100 mol of H2O is placed in a 1.00 dm-3 vessel.
The following equilibrium is established:
2NO (g) + 2H2 (g)  N2 (g) + 2H2O (g)
At equilibrium [NO] = 0.062 mol dm-3.
(a) Calculate the equilibrium concentrations of H2, N2 and H2O.
(b) Calculate Kc.
9. 23.0 g (0.50 mol) of ethanol was reacted with 60.0 g (1.0 mol) of ethanoic acid and the reaction allowed
to reach equilibrium at 373 K. 37.0 g (0.42 mol) of ethyl ethanoate was found to be present in the
equilibrium mixture.
Calculate Kc.
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