CHEMICAL EQUILIBRIUM
Equilibrium and chemical reactions
Consider the following system (water in a conical flask)
When the container is open, the water level drops as the water evaporates
If the container is closed, the water level also begins to drop, but then it reaches
A point when the level remains constant
When the level is constant, there is no noticeable change. This indicates that the rate of
evaporation is equal to the rate of condensation. The liquid is in equilibrium with the
vapour
- Equilibrium: The rate of the forward reaction = rate of the reverse reaction
- Dynamic: The reaction doesn’t stop but continues in both directions
- Dynamic equilibrium…
“At equilibrium, the reaction processes do not stop. The forward and reverse reactions
Still continue in both directions at the same rate”
Note: At equilibrium the concentration of the products and reactants don’t change!
Open and closed systems
- A closed system is one in which there is no interaction between the system and its
surroundings (one in which no substances can be removed or escape)
- In an open system, the substances can escape to the surroundings
Reversible reactions
- Some reactions can only take place in one direction
- Other reactions (such as the one below) can go in two directions
CuSO4.5H2O(s)  CuSO4(aq) + 5H2O(l)
- Blue copper sulfate crystals become white when heated
- If some water is added to the white crystals, they become blue again…
CuSO4(aq) + 5H2O(l)  CuSO4.5H2O(s)
- This reaction is said to be reversible (indicated with double arrows
thus be written…
CuSO4.5H2O(s)
CuSO4(aq) + 5H2O(l)
) and can
Note: When a reaction is reversible, both reactions don’t have to occur at the same rate,
but… for a DYNAMIC equilibrium, the forward and reverse reactions need to take
place simultaneously and at the same rate!
Reaching an equilibrium
- Consider the heating of a gaseous mixture of H2 and I2 gas
- The molecules will collide with each other to produce HI molecules
H2(g) + I2(g)
2HI(g)
- At the beginning of the reaction, there is maximum reactant concentration
- There is no product formed yet (product concentration = 0 mol.dm-3)
- As the reaction begins to proceed, the reactant concentration deceases, while the product
concentration increases RAPIDLY
- The reaction will start slowing down as the reactant concentration decreases
- As soon as some HI is formed, the reverse reaction can begin (but slowly)
- The more HI that is produced, the faster the reverse reaction reaction
- A stage is reached where THE RATE OF THE FORWARD REACTION IS EQUAL TO THE
RATE OF THE REVERSE REACTION! (equilibrium)
- It will appear as if the reactions have stopped, but they actually continue
Calculating the amount of substance and concentration at equilibrium
Example: A mixture of 10mol H2 and 12mol I2 is placed in a sealed container of 2dm3 at a
temperature of 500OC. Equilibrium is reached after some time. At equilibrium
there is 4mol gaseous HI in the container. Calculate the concentrations of H2
and I2 at equilibrium.
H2 + I2
2HI
It is a good idea to use a table…
H2
I2
2HI
Initial mol
Mol
reacted/formed
Mol at
equilibrium
Equilibrium
concentration
1) Write all the changes (in terms of x), place balancing coefficient in front of the x
(“-” must be included for substances used up, “+” for substances formed)
2) Calculate the value of x (0 + 2x = 4, therefore x = 2)
3) Insert these into the table and calculate the number of mol at equilibrium
4) Calculate the equilibrium concentrations (C = n/V)
Note: The concentration of a solid cannot change (once it’s formed)
EQUILIBRIUM CONSTANT
- At equilibrium, there is no change in the concentration of the reaction substances
- Thus the product of the concentrations of the reactants is a constant (k1)
- and the product of the concentrations of the products is also a constant (k2)
- Consider the following…
A2 + B2
2AB
- Thus we have…
[A2] x [B2] = k1
[AB] x [AB] = k2
- And therefore…
[AB] x [AB] = k2
[A2] x [B2] = k1
= Kc
- Kc is the equilibrium constant for the reaction (at a particular temperature)
- The subscript “c” is used to indicate that concentration was used to calculate the constant
- The law of MASS ACTION was used to calculate the equilibrium constant…
[C]c x [D]d
Kc=
[A]a x [B]b
For the general reaction:
aA + bB
cC + dD
Note:
1) The value of Kc is only a number (NO UNITS)
2) Kc is a constant at a PARTICULAR TEMPERATURE
3) PRODUCT concentration appears in the numerator, REACTANT concentration is inserted
in the denominator
4) The concentration of each substance is raised to the power of the balancing number
5) If Kc > 1, then the product concentration is greater than the reactant concentration
If Kc < 1, the reactant concentration is in the majority
If Kc = 0, there is equal concentrations of reactants and products
Note: In industry, a high Kc value is desired (why?)
HOMOGENEOUS and HETEROGENEOUS systems
-
When substances, in the system, are in the same phase  Homogeneous system
-
When substances, in the system, are in different phases  Heterogeneous system
-
Note: When a reaction involves a heterogeneous system (solids and liquids appear)
the concentration of these substances remain constant and therefore must not
be included in the law of mass action equation!!! (solvents, such as water, are
given a value of 1 (since their concentrations don’t change appreciably)
CHANGE OF EQUILIBRIUM CONDITIONS (Le Chatelier’s principle)
- Equilibrium can be influenced by the following factors…
1) Concentration 2) Temperature 3) Pressure
- The influence of these factors can be predicted using Le Chatelier’s principle…
“If the equilibrium in a closed system is disturbed, the system will react in such a way
that it will try to restore the equilibrium by working against the disturbing factor”
- The changes can be summarised as follows (see spreadsheet)
The effect of pressure and temperature on the gaseous N2O4 – NO2 equilibrium
- NO2(g) is a red-brown gas
- N2O4 is a colourless gas
- At room temperature, there is a mixture of the two gases in equilibrium…
2NO2(g)
N2O4(g)
DH<0
- Consider changing the pressure of the system…
Disturbance
Result
= Pressure increase (press down on closed syringe)
= Favour reaction that produces less mols
Favour reaction = Forward
Result
= Becomes colourless
- Consider changing the temperature of the system…
Disturbance
Result
= Temperature increase (place in warm water)
= Favour endothermic reaction
Favour reaction = Reverse
Result
= Becomes red-brown
Equilibrium mixtures in solution
- Since liquids cannot be compressed, a change in pressure has no effect on the
equilibrium of solutions
- Therefore, only temperature and concentration will be influencing factors
Common ion effect
- Consider a cobalt chloride crystal (CoCl2.6H2O) dissolved in ethanol
- This results in a blue solution of CoCl42- ions
- By adding water, the solution will turn pink (formation of Co(H2O)62+ ions)
CoCl42-(aq) + 6H2O(l)
Co(H2O)62+(aq) + 4Cl-(aq)
DH<0
- Now that the mixture is in equilibrium, consider concentration and temperature effects
- Adding HCl or NaCl to the mixture will increase the [Cl-] (common ion effect)
- This will cause the equilibrium to favour the reverse reaction!
- Adding AgNO3 will result in Ag+ ions being present
- This will result in the removal of Cl- ions (Ag+ + Cl-  AgCl)
- Causing the forward reaction to be favoured! (solution turns pink)
- Adding H2O will increase [H2O], thus the forward reaction will be favoured
- Adding H2SO4 (a dehydrating agent) will remove water (ie. [H2O] will decrease)
- Therefore reverse reaction will be favoured (solution turns blue)
- Increasing the temperature (ENDOthermic reaction favoured)
- Thus the reverse reaction is favoured (solution turns blue)
- Decrease in temperature (EXOthermic reaction favoured)
- Thus the forward reaction is favoured (solution turns pink)
Equilibrium principals in industry
1) Haber process (industrial preparation of ammonia)
- H2 gas and N2 gas is purified, dried and mixed
- The gas mixture is placed under pressure
- A catalyst is added (Fe or FeO)
N2(g) + H2(g)
2NH3(g)
(DH = -92kJ)
- The forward reaction is exothermic (DH<0)
- Thus a decrease in temperature will favour the forward reaction
- There are less number of mols on the right of the reaction
- So an increase in pressure (decrease in volume) will favour the forward reaction
Ideal reaction conditions are thus…
1) Low temperature
2) High pressure
Problems…
1) - The reaction is exothermic (so a high temperature will favour the reverse reaction)
- This will imply a poor yield of NH3 (product)
- So the temperature of the reaction must be decreased (to favour forward reaction)
- BUT… a low temperature will slow the reaction down (reaction kinetics)
- Answer: A moderate temperature!
2) - A high pressure increases the yield, but strong reaction vessels are expensive
- So it is more economical to operate at a relatively low pressure
Conditions for most economic and favourable yield…
- Increasing pressure (200x atmospheric pressure)
- Using a moderate temperature (5000C)
- Using a catalyst (Fe or FeO)
- Increasing the area of the catalyst
2) The contact process (industrial preparation of sulfuric acid)
- The important equilibrium reaction that is important in the determination of the
production rate and yield is the one that occurs in the converter between SO2 and O2
in the presence of the catalyst (vanadium pentoxide, V2O5)
2SO2(g) + O2(g)
2SO3(g)
DH = -98kJ
- The forward reaction is exothermic (DH<0)
- It will be favoured by a decrease in temperature
- There are more mols of gas of reactants
- So the forward reaction will be favoured by an increase in pressure
Conditions for a large yield are thus…
1) A Low temperature
2) A high pressure
Since the same conditions for a good yield apply for this process as they did for the Haber
process, the same problems will arise! (write down the ideal reaction conditions)
http://www.tutorvista.com/content/chemistry/chemistry-ii/chemical-equilibrium/chemical-equilibrium-animation.php
Initial mol
Mol
reacted/formed
Mol at equilibrium
Equilibrium
concentration