Equilibrium
Learning Objectives
• A state of equilibrium is reached in a closed system when the rates of the
forward and reverse reactions are equal.
• The equilibrium law describes how the equilibrium constant (Kc) can be
determined for a particular chemical reaction.
• The magnitude of the equilibrium constant indicates the extent of a reaction
at equilibrium and is temperature dependent.
• The reaction quotient (Q) measures the relative amount of products and
reactants present during a reaction at a particular point in time. Q is the
equilibrium expression with non-equilibrium concentrations. The position of
the equilibrium changes with changes in concentration, pressure, and
temperature.
• A catalyst has no effect on the position of equilibrium or the equilibrium
constant
Paint the picture
• Reversible reactions:
are reactions that can occur in both directions. Normally we describe the
reaction from left to right as: the forward reaction. The reaction from right to
left as: the backward or reverse reaction.
this symbol is used to show the reaction is an equilibrium
reaction (is reversible)
• We know, due to kinetics, that all particles are in constant motion, vibration
and rotation cause by energy
Paint the picture
• Energy is not distributed evenly, distributed in a completely random fashion.
• It can be seen in the diagram that there are very few particles with low energy and the
• We know,
due
to kinetics,
all particles
constant
motion,
number
of particles
with that
high energy
tails offare
in ain
curve.
The average
or mean energy
vibration
rotation
cause
by energyto the absolute temperature.
of theand
particles
is directly
proportional
Paint the picture
• http://ibchem.com/IB16/07.11.
htm#a1
• What happens to the particles
in an open system over time?
• E.g. leaving water to stand 
will slowly evaporate over time
• On the Maxwell-Boltzmann
distribution of energies over
particles, some particles will
always have enough energy to
‘escape’ the attraction of other
particles and become vapor
• E.g. snow will eventually
evaporate over time, even
when it is still cold.
Closed systems?
• A closed system is one in which neither matter nor
energy can enter or leave
• Example: water in a closed container = closed system,
• Initially, particles of water in the liquid leave to form
vapour. Occasionally, due to collisions, some of these
vapour particles will lose energy and re-join the body of
liquid. So 2 processes are happening simultaneously.
• Liquid particles  vapour and vapour particles
liquid.
• It has establish a dynamic equilibrium:
Macroscopic properties are constant (concentration of all reactants and
product remain constant) and the rate of the forward reaction is equal
to the rate of the revers reaction
Example #2
• Decomposition of CaCO3 is reversible, at steady temp. in sealed
container:
• The calcium carbonate converts to calcium oxide and carbon dioxide is
formed. The carbon dioxide can then react with the calcium oxide to
reform calcium carbonate
• After some time the amounts of calcium carbonate, calcium oxide and
carbon dioxide stay the same  the reaction appears to have stopped, it
has reached a state of equilibrium.
• The reaction is actually proceeding in both directions at equal rates, and
all concentrations remain constant
Reaction is in dynamic equilibrium
This is an example of a chemical system equilibria – equilibrium
established as a result of chemical reactions.
The rate of reaction
• H2(g) + I2(g) <꓿> 2HI(g)
• In this reaction we start with hydrogen and iodine vapour in a closed
system at a certain temperature and follow how the concentration of
hydrogen and hydrogen iodine change with time, we obtain a graph:
• The concentration of H2
decreases at first until it
levels off when equilibrium
is established
• Concentration of HI is
initially zero but it increases
until it flattens off and does
not change after equilibrium
is reached
• If we plot a graph of rate of
reaction against time for the
forward and reverse
reaction, we get the
following graph
• As the reaction progresses, the
concentration of H2 and I2 decrease and
hence rate of forward reaction slows down.
• concentration of hydrogen iodide increases
and therefore the rate of backward reaction
increases.
• A stage is reached when the rate of
backward reaction becomes equal to the
rate of forward reaction and the system
attains equilibrium.
• After this no change in concentration
occurs provided the temperature of the
reaction mixture is kept constant.
• Thus at equilibrium the reaction does not
stop but the system acquires constant
observable properties because of the equal
rates of forward and backward reactions.
• Thus the equilibrium is dynamic in nature
The characteristics of the equilibrium state
1. Macroscopic properties are constant at equilibrium:
At Equilib. The concentrations of all reactants and products remain constant.
2. At equilib. The rate of the forward reaction is equal to the rate of the reverse
reaction:
3. equilib. Can be attained only in a closed system:
A closed system has no exchange of mater with the surroundings, so equilib. Is
achieved where both reactants and products can react and recombine with each
other
4. All species in the chemical equation are present in the equlib. Reaction mixture:
The concentrations of reactants and products are constant, as they are being
produced and destroyed at an equal rate
5. equlib. Can be attained from either direction:
The same equlib. Mixture will result under the same conditions no matter whether
the reaction is started with all reactants, all products, or a mixture of both.
Physical equilibrium
• This is equilibria involving a change in state, for example, the
equilibrium between a liquid and its vapour
• Evaporation: when a liquid is in an open container it evaporates. At
a molecular level, particles must overcome the forces holding them
in the liquid (intermolecular forces) in order to escape into the gas
phase – evaporation is endothermic
• Consider a volatile liquid e.g. bromine in a closed container (has a
boiling point close to room temp.
• A lot of particles will have enough energy to evaporate into vapor
• Some vapor molecules will collide with the surface of the liquid and
lose energy  turn back to liquid = condensation
Position of equilibrium
• This refers to the relative amounts of reactants and products present
at equilibrium
• Reactions where the mixture contains predominantly products are
said to : ‘lie to the right’
• Reactions with predominantly reactants are said to : ‘lie to the left’
• You can capture this information mathematically to compare the
equilibrium mixtures of different reactions and the effect of different
conditions
Examples of different positions of equilib.
•
•
•
•
•
•
•
2NO(g)  N2(g) + O2(g)
At 700K the position of equilibrium lies a long way to the right
There is a large amount of N2 and O2 and not much NO at equilibrium
This reaction is almost to completion
H2(g) + CO2(g)  H2O(g) + CO(g)
At 1100K the total number of H2 and CO2( molecules at equilibrium is roughly equal
to the total number of H2O and CO molecules
The equilibrium is evenly balanced
H2O(l)  H+ (aq) + OH-(aq)
AT 298K the number of water molecules present at equilibrium is over 250million
times greater than the total number of H+ and OH- ions.
The position of equilibrium lies a long way to the left- there are not many ions
present
Check for understanding
1.
I.
II.
III.
a)
b)
c)
d)
Which statements are correct for a reaction at equilibrium?
The forward and reverse reactions both continue
The rates of the forward and reverse reactions are equal
The concentration of reactants and products are equal
I & II
I & III
II & III
I, II & III
Check for understanding
1. Which statement is always true for a chemical reaction that has
reached equilibrium at constant temperature?
a) The yield of product(s) is greater than 50%
b) The rate of the reverse reaction is lower than that of the forward
reaction
c) The amounts of reactants and products do not change
d) Both the forward and the reverse reactions have stopped
C
Check for understanding
1. Which statement is not true for a mixture of ice and water at
equilibrium at constant temperature?
a) The rate of melting and freezing are equal
b) The amounts of ice and water are equal
c) The same position of equilibrium can be reached by cooling water
and by heating ice
d) There is no observable change in the system
B
Le Chatelier’s Principle
• Henry Louis Le Chatelier , born Oct. 8, 1850, Paris, France—died Sept. 17,
1936, French chemist who proposed one of the central concepts of chemical
equilibria
• States :
If a system at equilibrium is subjected to a change, the position of
equilibrium will shift in order to minimise the effect of the change
• Means that if we take a particular system at equilib. Under a certain set of
conditions and change one of those conditions, i.e temperature or pressure,
the system will move to a new position of equilib.
• This principle allows us to predict the position the equilib. will shift
Change in temperature
2HI(g)  H2(g) + I2(g) ΔH= +10kJ/mol
3.4%
96.6%
At room temp. (298K) the number of molecules of HI is roughly 28 times the total number of
molecules of H2 and I2
• When the system reaches equilib. At 700K the total number of HI molecules is only
approximately 7 times the total number of H2 and I2 molecules
2HI(g)  H2(g) + I2(g)
88%
12%
Here it means that increasing the temperature, shifted the equilibrium to the right (there is
more product present)
In terms of Le Chatelier’s:
• given the positive ΔH, the forward reaction is endothermic, the reverse is exothermic
• To minimise the effect of the increase in temperature, the position of the equilibrium shifts
in the endothermic direction to take the heat that is added, (convert it to chemical energy)
• The endothermic direction is to the right, so equilib. position shifts to the right to produce
relatively more products
N2(g) +3H2(g)  2NH3(g)
ΔH= -92kJ/mol
85%
15%
At 300K and 10atm pressure we have : 85% : 15%
And at 700K and same pressure we have: 99.8% : 0.2%
N2(g) +3H2(g)  2NH3(g)
99.8%
0.2%
Increasing the temperature causes the position of equilib. To be shifted to the left, i.e.
there is less ammonia present at equilibrium at the higfher temp.
In terms of Le Chatelier’s:
• given the positive ΔH, the forward reaction is exothermic, the reverse is endothermic
• To minimise the effect of the increase in temperature, the position will shift in the
endothermic direction, to take the heat that is added, (convert it to chemical energy)
• The endothermic direction is to the left, so equilib. position shifts to the left to
produce relatively more reactants
Rule of Thumb
• HEAT reaction mixture: position of equilibrium is shifted
in the endothermic direction
• COOL reaction mixture: position of equilibrium is shifted
in the exothermic direction
Effect of Pressure
• Equilibria involving gases will be affected by a change in pressure if the
reaction involves a change in the number of molecules.
• If the equilibrium has an increase in pressure the system responds to decrease
this pressure by favouring the side with the smaller number of molecules.
• Conversely, a decrease in pressure will cause a shift in the equilibrium position
to the side with the larger number of molecules of gas.
• Consider this equilibrium
CO(g) +2H2(g) CH3OH(g)
There are 3 molecules of gas on the left and 1 molecule of gas on the right
High pressure will shift the equilibrium to the right; in favour of the smaller
number of gas molecules, which will increase the yield of product
H2(g) + I2(g)  2HI(g)
Increasing pressure will increase the rate of reaction but will not shift the
equilibrium
Effect of concentration
Le Chatelier’s principle:
• addition of reactant causes the system to respond by removing
reactant- this favours the forward reaction and so shifts the
equilibrium to the right
• Removal of product causes the system to respond by making more
product- this favours the forward reaction and so shifts the
equilibrium to the right
• When will the concentration change cause an equilibrium shift the
left?
• An increase in concentration of product or a decrease in
concentration of reactant
Equilibrium Constant
• The concentrations of reactants and products are constant at equilibrium.
This means that their ratio can be expressed as a simple mathematical
formula:
(for any reaction)
aA +bB  cC +dD
Kc =[C]c[D]d Products
[A]a[B]b Reactants
• [A] = the concentration of component A,
• [B] = the concentration of component B, etc.
• Kc is called the equilibrium constant with respect to the concentration of
the reactants and products (that's why there is a subscript 'c' following the
letter traditionally used for a constant value, 'K'.
the coefficients of the equation appear in the equilibrium expression as
powers to which the concentrations of the reactants and products are
raised.
Example
H2(g) + I2(g) 2HI(g)
Kc is given by the product concentrations raised to the power of their coefficients, divided by
the reactants concentrations raised to the power of their coefficients:
N2O4(g)  2NO2(g)
KC= [NO2(g)]2
[N2O4(g)]
N2O5(g) + NO(g) 3NO2(g)
KC= [NO2(g)]3
[N2O5(g)][NO(g)]
CO(g) +3H2(g) CH4(g) + H2O(g)
KC= [CH4(g)][H2O(g)]
[CO(g)][H2(g)]3
Solids and liquids in Kc
• Only those substances that appear in the equilibrium in a form that can be expressed in
terms of concentration can appear in the equilibrium expression.
• This means that solids NEVER appear and liquids only appear if they are part of
a homogeneous system (all reactants and products in the same physical state)
• Example: Show the expression for the equilibrium that exists between iron, water, hydrogen
and iron(III) oxide at elevated temperatures:
2Fe(s) + 3H2O(g)
Fe2O3(s) + 3H2(g)
• Inspection of the equation reveals that both iron and iron oxide are in the solid states, i.e.
they do not have a concentration. This means that they cannot appear in the equilibrium
expression, therefore:
Equilibrium constant can be predicted from reactions
stoichiometry
H2(g) + I2(g) 2HI(g)
• You can carry out a series of experiments on this reaction with different starting
concentrations of each compound, wait until each reaction reached equilibrium and then
measure the composition of each equilibrium mixture. Results obtained at 440°C:
#1
Initial
Equilibrium
concentration/mol/dm3 concentration/ mol/dm3
H2
0.100
0.0222
#3 Initial
concentration/mol/dm3
I2
0.100
0.0222
H2
0.0150
0.0150
HI
0.000
0.156
I2
0.000
0.0135
HI
0.127
0.100
#2
Initial
concentration/mol/dm3
Equilibrium
concentration/ mol/dm3
H2
0.000
0.0350
I2
0.0100
0.0450
HI
0.350
0.280
Equilibrium
concentration/ mol/dm3
At first glance it appears as if there is not pattern. However a
predictable relationship is amount the different compositions
of these equilibrium mixtures and you can discover it using
the stoichiometry of the reaction and the concentrations at
equilibrium
H2(g) + I2(g) 2HI(g)
• Experiment #1:
(0.156)2
=49.9
0.0222x0.0222
• Experiment #2:
(0.280)2
=49.8
0.035x0.0450
• Experiment #3:
(0.100)2
=49.4
0.0150x0.0135
• As you can see, when you process the
equilibrium data, it produces a constant value
within the limits of experimental accuracy (all 3
Kc are extremely close together).
• As always, the equilibrium constant, Kc, has a
fixed value for the specific reaction at a
specified temperature, as shown in the
calculations.
• Take not that Kc does NOT have a unit
What factors can affect the equilibrium constant?
• Concentration?
Change in concentration of reactants and products alter the equilibrium position and so
change the composition of the equilibrium mixture, but the Kc stays the same
• Pressure?
Changes in pressure or volume will affect the position of equilibrium of a reaction if it
involves a change in the number of gas molecules, but the values of Kc remains the same
• Temperature?
Kc is temperature dependant so changing the temperature will change Kc. However in
order to predict how it will change you must look at the enthalpy changes of the forward
and backwards reactions.
An increase in temperature increases the value of Kc for an endothermic reaction and
decreases the value of Kc for an exothermic reaction
Example:if the forward reaction is endothermic (+ve ΔH) decreasing the temp. with favor
the exothermic backwards reaction; the shift will be to the left to favor the reactants and
Kc will decrease
What use is an equilibrium constant?
• Different reactions have different values of Kc.
The equilibrium constant provides information about how far a reaction proceeds
at a particular temperature
• if K is a large number, it means that the equilibrium concentration of the
products is large. In this case, the reaction as written will proceed to the right
(resulting in an increase in the concentration of products)
• If K is a small number, it means that the equilibrium concentration of the
reactants is large. In this case, the reaction as written will proceed to the left
(resulting in an increase in the concentration of reactants)
Examples
• H2(g) + I2(g)  HI(g)
Kc= 2
• H2(g) + Br2(g) 2HBr(g) Kc=1010
• H2(g) + Cl2(g) 2HCl(g)
Kc= 1018
• the large range in their Kc values tells us about the different extents of these
reactions.
• We can deduce that the reaction between H2 + Cl2 has taken place the most fully at
this temp., and the position of the equilibrium lies to the right
• H2(g) + I2 have reacted the least
• N2(g) + O2(g)  2NO(g) Kc=10-31
• At 298K, this value is very much less than 1, indicating that the reaction hardly
proceeds at all towards the products – the position of equilibrium lies a long way to
the left
Reaction Quotient
Kc =[C]c[D]d
[A]a[B]b
• The reaction quotient, Q, enables us the predict the direction of the reaction at any
given time.
Remember that the value for Kc is calculated from substituting the equilibrium
concentrations of all the reactants and products into the equilibrium constant
expression. So the only values used are at equilibrium
• If we take the concentrations of reactants and products at any moment when the
reaction is not at equlib. and substitute those concentrations into the equlib.
expression we get a value called the reaction quotient Q.
• It helps us predict the direction in which the reaction will proceed because as time
passes/reaction continues, all of the components will eventually reach equilibrium,
regardless of which direction the reaction is shifting.
• Summery of the use of reaction quotient Q in determining the direction of reaction:
1. If Q=Kc, reaction is at equilibrium, no net reaction occurs
2. If Q<Kc, reaction proceeds to the right on favour of products
3. If Q> Kc, reaction proceeds to the left in favour of reactants
Example
• H2(g) + I2(g)  2HI(g)
Kc= 49.5 at 440°C
#1
Experiment 1:
concentration at time t
/mol/dm3
Experiment 2:
concentration at time t/
mol/dm3
H2
00500
0.0250
I2
0.0500
0.0350
HI
0.100
0.300
Experiment #1, time t:
Q= (0.100)2
(0.0500x0.0500)
=4.00
Experiment #2, time t:
Q= (0.300)2
(0.0250 x0.0350)
=103
• In experiment 1, Q< Kc and so Q must increase as the reaction moves towards equilibrium.
This means that the net reaction must be to the right, in favour of products
• In experiment 2, Q>Kc, and so Q must decrease as the reaction moves towards equilibrium.
This means that the net reaction must be to the left, favour of the reactants
Relationship between Kc for different equations of
a reaction
Because Kc is defined as an expression with stoichiometry, we CAN
manipulate its value according to the changes made to the reaction
For examples we will use the generic reaction: (aA +bB  cC +dD)
1. Kc for the inverse (backwards) reaction:
cC +dDaA +bB we write the equilibrium constant as Kc’: Kc’ =[A]a[B]b
[C]c[D]d
Here the equilibrium constant for a reaction is the reciprocal of the
equilibrium constant for its inverse reaction.
Kc’ = 1/Kc or
Kc’ =Kc-1
Kc for a multiple of a reaction
Consider the reaction: 2aA +2bB  2cC +2dD
We will write the equilibrium constant as Kcx
Kcx =[C]2c[D]2d
[A]2a[B]2b
You can see that each term has been squared
So Kcx = Kc2
Similarly you can conclude that tripling of the stochiometric coefficients
would lead to cubing the value of Kc
Halving the stochiometric coefficients would lead to square rooting the
value of Kc and so on.
Summary
Inversing the reaction
doubling the reaction
coefficients
Tripling the reacting
coefficients
Halving the reaction
coefficients
Adding together two
reactions
Effect on equilibrium
expression
Inverts the expression
Squares the
expression
Cubes the expression
Effect on Kc
Square roots the
expression
Multiplies the two
expressions
√Kc
1/Kc or Kc-1
Kc2
Kc3
Kci x Kcii
Example
KC= [H2(g)][I2(g)]
• The equilibrium constant for the reaction:
2
[HI(g)]
2HI(g)  H2(g) + I2(g)
Is 0.04 at a certain temperature. what would be the value of the equilibrium
constant Kc’ for the following reaction at the same temperature?
½ H2(g) + ½ I2(g)  HI(g)
Write the equilibrium expression:
Note the reaction is reversed and the coefficients in the equation are halved.
Overall the value of KC’ is the square root of the value of KC
So KC’ = 1/√KC or √(KC )-1
KC ’ = [HI(g)]
KC’ =√1/0.04 = 5.0
[H (g)]1/2 [I (g)]1/2
2
2
Addition of catalysts
• We know that catalyst speed up a reaction by providing an alternative
pathway that has a transition state with a lower activation energy. This in
turn increases the number of particles that have the sufficient energy to
react without raising the temperature
• The catalyst will have no effect on the position of equilibrium or on the Kc
value. It will increase the speed at which the equilibrium is achieved.
• Because the forward and backward reaction pass through the same
transition state, a catalyst lowers the activation energy of both the
forward and reverse reaction by the same amount.
• The catalysts are not involved in the reaction, so it is not shown in the
reaction equation and has no effect on equilibrium concentration
Haber Process
•
https://www.youtube.com/watch?v=o1_D4FscMnU https://www.youtube.com/watch?v=ztzKHU2oaF8
• All reactants and products are gases
• There are 4 molecules on the left and 2 molecules on the right
• The forward reaction is exothermic and the backwards reaction in endothermic
• We can use Le Chatelier’s principle to give us the optimum conditions for a high yield of products:
1. Concentration: the reactants N and H are supplied in the molar ratio of 1:3. you can remove the ammonia as
its formed to help shift the equilibrium to the right: increasing the yield.
2. Pressure: the forward reaction involved a decrease in gas molecules, so an increase in pressure will favour the
yield of product. = 2x107 Pa
3. Temperature: the forward reaction is exothermic so will be favoured by a decrease in temp. although if you go
too low, the reaction will occur too slowly, so optimum is about 450°C
4. Catalyst: the use will help speed up the reaction to compensate for the moderate temp used. = finely divided
iron with small amounts of aluminium and magnesium oxides to improve activity. Recently ruthenium has
been used as it has reduced the energy requirements.
This process achieves a conversion of H2 and N2 into NH3 of only about 10-20% per pass through the reactor. So
once the product is separated the unconverted reactant is recycled back into the reactor to obtain an overall yield
of 95%.
This recycling allows processes with low equilibrium yield to be made commercially viable.