Any factor that can affect the rate of either the forward or reverse reaction relative to the other
can potentially affect the equilibrium position. The following factors can change the chemical
equilibrium position of a reaction:
concentration
temperature
pressure (for gaseous reactants)
It is important to understand what effect a change in one of these factors will have on a system that
is in chemical equilibrium. However, performing an experiment every time to find out would waste
a lot of time. Towards the end of the 1800s the French chemist Henry Louis Le Chatelier came up
with principle to predict those effects.
Henry Louis Le Chatelier.
Le Chatelier's Principle helps to predict what effect a change in temperature,
concentration or pressure will have on the position of the equilibrium in a chemical reaction.
This is very important, particularly in industrial applications, where yields must be accurately
predicted and maximised.
When an external stress (change in pressure, temperature or concentration) is applied
to a system in chemical equilibrium, the equilibrium will change in such a way
as to reduce the effect of the stress.
So if the concentration of one (or more) of the reactants or products is increased the equilibrium
will shift to decrease the concentration. Or if the temperature is decreased the equilibrium
will shift to increase the temperature by favouring the exothermic reaction. Le Chatelier's
principle is that:
If you change the \(\color{blue}{\textbf{concentration}}\) of a reactant then the
position of the equilibrium will shift to counteract that change.
If you change the \(\color{red}{\text{temperature}}\) of the
reaction the equilibrium will shift to counteract that change.
If you change the \(\color{orange}{\textbf{pressure}}\) of the system the position of
the equilibrium will shift to counteract that change.
Each of these concepts is discussed in detail in the following pages.
The following video gives an example of Le Chatelier's principle in action.
The effect of concentration on equilibrium (ESCNP)
If the concentration of a substance is changed, the equilibrium will shift to minimise the
effect of that change.
If the concentration of a \(\color{blue}{\textbf{reactant}}\) is
increased the equilibrium will shift in the direction
of the reaction that uses the reactants, so
that the reactant concentration decreases. The forward reaction
is favoured.
The forward reaction is also favoured if the concentration of the
\(\color{red}{\textbf{product}}\) is decreased, so that
more product is formed.
If the concentration of a \(\color{blue}{\textbf{reactant}}\) is
decreased the equilibrium will shift in the direction
of the reaction that produces the reactants,
so that the reactant concentration increases. The reverse
reaction is favoured.
The reverse reaction is also favoured if the concentration of the
\(\color{red}{\textbf{product}}\) is increased, so that
product is used.
For example, in the reaction between sulfur dioxide and oxygen to produce sulfur trioxide:
If the \(\color{blue}{\text{SO}_{2}}\) or \(\color{blue}{\text{O}_{2}}\)
concentration was increased:
Le Chatelier's principle predicts that equilibrium will shift
to decrease the concentration of reactants.
Increasing the rate of the forward reaction
will mean a decrease in reactants.
So some of the sulfur dioxide or oxygen is used to produce
sulfur trioxide.
Equilibrium shifts to the right. That is,
when a new equilibrium is reached (when the rate of
forward and reverse reactions are equal again), there
will be more product than before.
When the concentration of reactants is increased, the
equilibrium shifts to the right and there will
be more product than before. There will also be
more reactants than before (more reactants were
added). Once equilibrium has been reestablished
(the rate of the forward and reverse reactions
are equal again), \(\text{K}_{\text{c}}\) will
be the same as it was before the change to the
system. This concept is explained in more detail
later in this chapter.
If the \(\color{blue}{\text{SO}_{2}}\) or \(\color{blue}{\text{O}_{2}}\)
concentration was decreased:
Le Chatelier's principle predicts that the equilibrium will
shift to increase the concentration of reactants.
Increasing the rate of the reverse reaction
will mean an increase in reactants.
So some sulfur trioxide would change back to sulfur dioxide
and oxygen to restore equilibrium.
Equilibrium shifts to the left. That is,
when a new equilibrium is reached there will be less
product than before.
If [\(\color{red}{\text{SO}_{3}}\)] decreases:
Le Chatelier's principle predicts that the equilibrium will
shift to increase the concentration of products.
Increasing the rate of the forward reaction
will mean an increase in products.
So some sulfur dioxide or oxygen is used to produce sulfur
trioxide.
Equilibrium shifts to the right. That is,
when a new equilibrium is reached there will be more
product than before.
If [\(\color{red}{\text{SO}_{3}}\)] increases:
Le Chatelier's principle predicts that the equilibrium will
shift to decrease the concentration of products.
Increasing the rate of the reverse reaction
will mean a decrease in products.
So some of the sulfur trioxide would change back to sulfur
dioxide and oxygen to restore equilibrium.
Equilibrium shifts to the left. That is,
when a new equilibrium is reached there will be less
product than before.
Although not required by CAPS the common-ion effect is a useful concept for the
students to know if there is time.
Common-ion effect
In solutions the change in equilibrium position can come about due to the
common-ion effect. The common-ion effect is where one substance
releases ions (upon dissociating or dissolving) which are already present in the
equilibrium reaction.
If solid sodium chloride is added to an aqueous solution and dissolves, the following
dissociation occurs:
The added \(\text{Cl}^{-}\) ion (common-ion) interferes with the equilibrium by
raising the concentration of the \(\text{Cl}^{-}\) ion. According to Le
Chatelier's principle the reverse reaction speeds up as it tries to reduce the
effect of the added \(\text{Cl}^{-}\). As a result the equilibrium position
shifts to the left.
temp text
The effect of temperature on equilibrium (ESCNQ)
If the temperature of a reaction mixture is changed, the equilibrium will shift to minimise
that change.
If the temperature is increased the equilibrium will shift
to favour the reaction which will reduce the
temperature. The endothermic
reaction is favoured.
If the temperature is decreased the equilibrium will shift
to favour the reaction which will increase the
temperature. The exothermic
reaction is favoured.
For example the forward reaction shown below is exothermic (shown by the negative value for
\(\Delta H\)). This means that the forward reaction, where nitrogen and hydrogen react
to form ammonia, gives off heat, increasing the temperature (the forward
reaction is exothermic). In the reverse reaction, where ammonia
decomposes into hydrogen and nitrogen gas, heat is taken in by the reaction, cooling the
vessel (the reverse reaction is endothermic).
Remember that heat is released during an exothermic reaction. It is a
product of the reaction.
Favours the endothermic reaction because it
takes in energy (cools the container).
The reverse reaction is endothermic, so the reverse
reaction is favoured.
The yield of ammonia \((\text{NH}_{3})\)
will decrease.
A decrease in temperature:
Favours the exothermic reaction because it
releases energy (warms the container).
The forward reaction is exothermic, so the forward
reaction is favoured.
The yield of \(\text{NH}_{3}\) will
increase.
In the informal experiment on Le Chatelier's principle, the solution should be purple
to start. To achieve this \(\text{CoCl}_{2}\) must be dissolved in ethanol and a
few drops of water must be added. This solution is toxic, and
all the usual laboratory precautions should be taken.
The hot water will make the solution a deep blue, the cold water will make the
solution a pink/red colour. If necessary the test tube can be gently shaken to
ensure mixing.
Le Chatelier's principle
Aim
To determine the effect of a change in concentration and temperature on
chemical equilibrium
Apparatus
\(\text{0,2}\) \(\text{mol·dm$^{-3}$}\) purple
\(\text{CoCl}_{2}\) in ethanol solution, concentrated
\(\text{HCl}\), water
test tube, tongs
ice-bath, water-bath, hot-plate or bunsen burner
Method
Place the water bath on the hot-plate and heat.
During each step observe and record the colour change that
takes place.
Put \(\text{4}\) – \(\text{5}\) drops of \(\text{0,2}\)
\(\text{mol·dm$^{-3}$}\) \(\text{CoCl}_{2}\)
solution
into the test tube.
Add \(\text{10}\) – \(\text{12}\) drops of water.
Add \(\text{20}\) – \(\text{25}\) drops of concentrated
\(\text{HCl}\).
Place the test tube in the water-bath on the hot-plate (use
tongs). Leave for \(\text{1}\) – \(\text{2}\)
minutes. Record your observations.
Place the test tube in the ice-bath. Leave for \(\text{1}\)
– \(\text{2}\) minutes. Record your observations.
The equation for the reaction that takes place is:
Complete your observations in the table below, noting the colour changes that
take place, and also indicating whether the concentration of each of the
ions in solution increases or decreases.
If the pressure of a gaseous reaction mixture is changed the equilibrium will shift to
minimise that change.
If the pressure is increased the equilibrium will shift to
favour a decrease in pressure.
If the pressure is decreased the equilibrium will shift to
favour an increase in pressure.
When the volume of a system is decreased (and the temperature is constant), the pressure will
increase. There are more collisions with the walls of the container. If there are fewer
gas molecules there will be fewer collisions, and therefore lower pressure. The
equilibrium will shift in a direction that reduces the number of gas molecules so that
the pressure is also reduced. So, to predict in which direction the equilibrium will
shift to change pressure you need to \(\color{darkgreen}{\text{look at the number of gas
molecules in the balanced}}\) \(\color{darkgreen}{\text{reactions}}\).
The ratio in the balanced equation is \(1:3:2\). That is, for every \(\text{1}\)
\(\text{molecule}\) of \(\color{blue}{\text{N}_{2}\text{ gas}}\) there are \(\text{3}\)
\(\text{molecules}\) of \(\color{blue}{\text{H}_{2}\text{ gas}}\) and \(\text{2}\)
\(\text{molecules}\) of \(\color{red}{\text{NH}_{3}\text{ gas}}\) (from the balanced
equation). Therefore the ratio is \(\color{blue}{\textbf{4 molecules of reactant gas}}\)
to \(\color{red}{\textbf{2 molecules of product gas}}\).
An increase in pressure will:
Favour the reaction that decreases the
number of gas molecules.
There are \(\color{red}{\textbf{fewer molecules of product
gas}}\) than reactant gas, so the forward
reaction is favoured.
The equilibrium will shift to the right and
the yield of
\(\color{red}{\textbf{NH}_{3}}\) will
increase.
A decrease in pressure will:
Favour the reaction that increases the
number of gas molecules.
There are \(\color{blue}{\textbf{more molecules of reactant
gas}}\), so the reverse reaction is
favoured.
The equilibrium will shift to the left and
the yield of
\(\color{red}{\textbf{NH}_{3}}\) will
decrease.
Figure 8.2: (a) A decrease in the pressure of this reaction
favours the
reverse reaction (more gas molecules), the equilibrium shifts to the
left. (b) An increase in the pressure of this reaction
favours the forward reaction (fewer gas molecules), the equilibrium
shifts to the right.
Figure 8.2 shows how
changing the pressure of a system results in a shift in the equilibrium to counter that
change. In the original system there are 12 molecules in total:
\({\color{skyblue}{6{\text {H}}_{2}}} + {\color{blue}{2{\text{N}}_{2}}}
\rightleftharpoons {\color{red}{4{\text{NH}}_{3}}}\)
If you decrease the pressure (shown by an increase in volume), the
equilibrium will shift to increase the number of gas molecules. That
shift is to the left and the number of \(\text{H}_{2}\) and \(\text{N}_{2}\) molecules
will increase while the number of \(\text{NH}_{3}\) molecules will decrease:
If you increase the pressure (shown by a decrease in volume), the
equilibrium will shift to decrease the number of gas molecules. That
shift is to the right and the number of \(\text{H}_{2}\) and \(\text{N}_{2}\) molecules
will decrease while the number of \(\text{NH}_{3}\) molecules will increase:
Note that the total number of nitrogen and hydrogen atoms remains the same in all three
situations. Equations (a) and (b) are not balanced equations.
Another example is the reaction between sulfur dioxide and oxygen:
In this reaction, \(\color{red}{\textbf{two molecules of product gas}}\) are formed for every
\(\color{blue}{\textbf{three molecules of}}\) \(\color{blue}{\textbf{reactant gas}}\).
An increase in pressure will:
Favour the reaction that decreases the
number of gas molecules.
There are \(\color{red}{\textbf{fewer molecules of product
gas}}\) than reactant gas, so the
forward reaction is favoured.
The equilibrium will shift to the right and
the yield of
\(\color{red}{\textbf{SO}_{3}}\) will
increase.
A decrease in pressure will:
Favour the reaction that increases the
number of gas molecules.
There are \(\color{blue}{\textbf{more molecules of reactant
gas}}\) than product gas, so the
reverse reaction is favoured.
The equilibrium will shift to the left and
the yield of
\(\color{red}{\textbf{SO}_{3}}\) will
decrease.
temp text
The effect of a catalyst on equilibrium (ESCNS)
If a catalyst is added to a reaction, both the forward and
reverse reaction rates will be increased. If both rates are increased
then the concentrations of the reactants and products will remain the
same. This means that a catalyst has no effect on the equilibrium
position.
However, a catalyst will affect how quickly equilibrium is reached. This is very
important in industry where the longer a process takes, the more money it costs. So if a
catalyst reduces the amount of time it takes to form specific products, it also reduces
the cost of production.
Factors that influence the value of \(\text{K}_{\text{c}}\) (ESCNT)
Concentration, pressure, and temperature all affect the equilibrium position of a reaction,
and a catalyst affects reaction rates. However, only
\(\color{red}{\textbf{temperature}}\) affects the value of \(\text{K}_{\text{c}}\).
Changing \(\color{blue}{\text{concentration}}\):
Changing the concentration of a reactant or product results
in one of the reactions (forward or reverse) being
favoured.
This change in reaction rate minimises the effect of the
change and restores the concentration ratio between
reactants and products. There will just be more
reactants and products.
\(\color{blue}{\text{K}_{\text{c}}}\)
\(\color{blue}{\text{will remain the same}}\).
Changing \(\color{orange}{\text{pressure}}\):
Changing the pressure of the system will change the ratio
between the reactant and product concentrations.
The equilibrium then shifts to minimise the effect of the
change and restores the ratio between reactant and
product concentrations.
\(\color{orange}{\text{K}_{\text{c}}} \color{orange}{\text{
will remain the same}}\).
Adding a \(\color{purple}{\text{catalyst}}\) to the system:
Both the forward and reverse reactions rates are increased.
Therefore the ratio between reactant and product
concentrations will remain the same.
\(\color{purple}{\text{K}_{\text{c}}}\color{purple}{\text{
will remain the same}}\).
Changing \(\color{red}{\text{temperature}}\):
Changing the temperature will favour either the endothermic
or exothermic reaction.
The ratio between the concentration of the reactants and
products will change.
So make sure that when comparing \(\text{K}_{\text{c}}\) values for different
reactions, the different reactions took place at the same temperature.
Using Le Chatelier's Principle (ESCNV)
When a system is in chemical equilibrium, and there has been a change in conditions (e.g.
concentration, pressure, temperature) the following steps are suggested:
Identify the disturbance or stress on the system.
\(\color{blue}{\text{For example, there is an }\textbf{increase}\text{ in the
concentration of reactant}}\).
Use Le Chatelier's principle to decide how the system will respond.
\(\color{blue}{\text{Le Chatelier predicts a shift to
}\textbf{decrease}\text{ the concentration of reactant}}\).
Look at the given equation and decide whether the rate of the forward
reaction or the rate of the reverse reaction is increased. State the
shift in equilibrium.
\(\color{blue}{\text{The }\textbf{forward reaction will be favoured}\text{.
The equilibrium will shift }\textbf{to the right}}\).
Where appropriate, link equilibrium shift to any observed change in the
system.
\(\color{blue}{\text{This might result in a }\textbf{colour change}}\).
Worked example 6: Using Le Chatelier's principle
Table salt is added to the (purple) solution in equilibrium:
Use Le Chatelier's principle to predict the change in
equilibrium position.
What would be observed?
Identify the disturbance or stress on the system
Adding \(\text{NaCl}\) produces \(\text{Na}^{+}\) ions and \(\text{Cl}^{-}\)
ions as the salt dissolves. Looking at the given equilibrium
\(\text{Cl}^{-}\) is in the equation and the disturbance is the increase
in concentration of the \(\text{Cl}^{-}\) ion.
Use Le Chatelier's principle to decide how the system will
respond
By Le Chatelier's principle, the equilibrium position will shift to reduce
the concentration of \(\text{Cl}^{-}\) ions.
Decide whether the rate of the forward reaction or the rate of
the reverse reaction is increased and state the resulting shift in
equilibrium
The reverse reaction uses \(\text{Cl}^{-}\) ions and hence the rate of the
reverse reaction will increase. The reverse reaction is favoured and the
equilibrium will shift to the left.
What would the colour change be due to this equilibrium shift?
The solution will appear more blue as more blue \(\text{CoCl}_{4}^{2-}\) ions
are formed.
Graphs and Le Chatelier's principle
Graphs can be used to represent data about equilibrium reactions. The following are
some points to keep in mind when presented with a graph.
Identify the type of graph by looking at the label on the y-axis. You
will find either:
a rate-time graph
a mole-time or concentration-time graph
For rate-time graphs, when the rate for the forward reaction and the
rate for the reverse reaction are equal, the system is in
equilibrium.
For concentration-time graphs or mole-time graphs equilibrium occurs
where the concentration or number of moles of the reactants and
products are constant. These values need not be
equal to one another.
When calculating \(\text{K}_{\text{c}}\) make sure you only take
values from the sections of the graph where the y-value is
constant. \(\text{K}_{\text{c}}\) can only be calculated when
the system is in equilibrium.
Rate-time graphs
A change in concentration of a substance will
favour the reaction that counteracts that change.
An increase in concentration of
either the reactants
or the products will result in both the forward
and reverse reaction rates being
higher once
equilibrium is re-established.
An increase in reactant
concentration
will favour the
forward reaction.
The forward
reaction rate will
increase
sharply, and then gradually decrease until
equilibrium
is re-established. At the same time the reverse
reaction
rate will gradually increase.
An increase in product
concentration
will favour the
reverse reaction.
The reverse reaction rate will increase sharply,
and
then gradually decrease until equilibrium is
re-established. At the same time the forward
reaction
rate will gradually increase.
A decrease in concentration of
either the reactants
or the products will result in both the forward
and reverse reaction rates being
lower once
equilibrium is re-established.
A decrease in reactant
concentration
will favour the
reverse reaction.
The forward
reaction rate will
decrease
sharply, and then gradually increase until
equilibrium
is re-established. At the same time the reverse
reaction
rate will gradually decrease.
A decrease in product
concentration
will favour the
forward reaction.
The reverse reaction rate will decrease sharply,
and
then gradually increase until equilibrium is
re-established. At the same time the forward
reaction
rate will gradually decrease.
A change in pressure of a reaction will
favour the reaction that counteracts that change.
An increase in pressure will
sharply increase both the forward and
reverse reaction rates.
The reaction that will reduce the pressure (fewer gas
particles) will be
favoured and will initially increase more. That
reaction rate will then decrease gradually until
equilibrium is re-established.
The other reaction will initially increase less, and
then gradually continue to increase until
equilibrium is re-established.
A decrease in pressure will
sharply decrease both the forward and
reverse reaction rates.
The reaction that will increase the pressure (more
gas particles) will be
favoured and will initially decrease less. That
reaction rate will then continue to decrease
gradually until
equilibrium is re-established.
The other reaction will initially decrease more, and
then gradually increase until
equilibrium is re-established.
A change in temperature of a reaction will
favour the reaction that counteracts that change.
An increase in temperature will
sharply increase both the forward and
reverse reaction rates.
The reaction that will decrease the temperature (the
endothermic reaction) will be
favoured and will initially increase more. That
reaction rate will then decrease gradually until
equilibrium is re-established.
The other reaction will initially increase less, and
then gradually continue to increase until
equilibrium is re-established.
A decrease in temperature will
sharply decrease both the forward and
reverse reaction rates.
The reaction that will increase the temperature (the
exothermic reaction) will be
favoured and will initially decrease less. That
reaction rate will then continue to decrease
gradually until
equilibrium is re-established.
The other reaction will initially decrease more, and
then gradually increase until
equilibrium is re-established.
The addition of a catalyst will favour both the
forward and reverse reactions by the same amount.
Worked example 7: Rate-time graphs
For the reaction \(2\text{AB}(\text{g})\) \(\rightleftharpoons\)
\(2\text{A}(\text{g}) + \text{B}_{2}(\text{g})\), \(\Delta\)H =
\(\text{26}\) \(\text{kJ}\)
the following graph can be plotted:
What stress has occurred in this system? Label the graph with what is
happening at each stage.
Check the axes so that you know what the variables are
on this graph
The axes are labelled rate and time. Therefore this is a rate-time
graph.
Are the rates both affected in the same way or is one
rate increased and the other decreased when the stress is
applied?
Both rates are affected in the same way (increased) therefore the
stress must be a catalyst or a change in temperature.
(A change in pressure or concentration would favour one reaction
direction only)
Are both rates affected equally?
No, the forward rate is increased more than the reverse rate.
Therefore the stress must be a change in temperature.
(A catalyst would increase both rates equally)
Was the temperature increased or decreased? (Use Le
Chatelier's principle)
The forward reaction is endothermic (\(\Delta\)H is positive). The
forward reaction was favoured more than the reverse reaction. An
increase in temperature will favour the reaction that cools the
reaction vessel (the endothermic reaction). Therefore the stress
must have been an increase in temperature.
Label what is occuring at each stage on the graph
Worked example 8: Rate-time graphs
What is responsible for the change at t = \(\text{10}\)
\(\text{minutes}\) in the graph below?
Check the axes so that you know what the variables are
on this graph
The axes are labelled rate and time. Therefore this is a rate-time
graph.
Are the rates both affected in the same way or is one
rate increased and the other decreased when the stress is
applied?
Both rates are affected in the same way (increased) therefore the
stress must be a catalyst or a change in temperature.
(A change in pressure or concentration would favour one reaction
direction only)
Are both rates affected equally?
Yes, both rates are increased by the same amount.
What was responsible for the change at t =
\(\text{10}\) \(\text{minutes}\)?
The addition of a catalyst (a change in temperature would affect both
rates, but unequally).
Concentration-time and mole-time graphs
A change in concentration of a substance would
appear as a sharp increase or decrease in the concentration or
number of moles of that substance and a gradual change in the
other substances.
A change in temperature would affect both the
forward and reverse reactions. However, one reaction would be
affected more than the other. So the reactants and products
would be affected gradually, in the opposite direction (one
increased, the other decreased).
A change in pressure of the reaction would cause a
sharp increase or decrease in all the reactants and products.
For an increase in pressure, if the forward reaction is then
favoured the reactant concentrations will decrease, and if the
reverse reaction is then favoured the product concentrations
will decrease.
The addition of a catalyst would increase both the
forward and reverse reaction rates, meaning the equilibrium is
reached faster. So if the reaction is already at equilibrium
there will be no effect on a concentration-time or moles-time
graph.
Worked example 9: Concentration-time graphs
Consider the following chemical equation and graph
and answer the
questions that follow.
How much time was necessary for the system to reach
equilibrium for the first time?
How do the rates of the forward and reverse reactions
compare at the following times:
\(t = \text{5}\) \(\text{s}\)
\(t = \text{17}\) \(\text{s}\)
\(t = \text{23}\) \(\text{s}\)
Determine the equilibrium constants for the system at
\(t = \text{17 s}\), \(\text{27 s}\), and
\(\text{42 s}\).
What happens at \(t = \text{20 s}\)?
Explain your answer by referring to Le
Chatelier's principle.
What effect does the stress at \(t = \text{20 s}\)
have on \(\text{K}_{\text{c}}\)?
Check the axes so that you know what the variables are
on this graph
The axes are labelled concentration and time. Therefore this is a
concentration-time graph.
How much time was necessary for the system to reach
equilibrium for the first time?
The concentration of all three compounds becomes constant at t =
\(\text{15}\) \(\text{s}\). This means that the reaction has
reached equilibrium.
How can you determine which rate is faster from
concentration-time graphs?
If the concentrations of the reactants are steadily
decreasing and
the concentrations of the products are steadily
increasing then the forward reaction is using
reactants faster
than the reverse reaction is using products. The rate of
the forward reaction must be faster
than the rate of the reverse
reaction.
If the concentrations of the reactants are steadily
increasing and
the concentrations of the products are steadily
decreasing then the forward reaction must be
using reactants
slower than the reverse reaction is using products. The
rate of the forward reaction must be
slower than the rate of the
reverse reaction.
If the concentrations of the reactants and products are
constant, the
forward reaction must be using reactants at
the same rate as the reverse reaction is using products. The
rate of the forward reaction must be equal
to the rate of
the reverse reaction (the system is in equilibrium).
Compare the forward and reverse reaction rates at \(t
=
\text{5} \text{ s}\), \(t = \text{17} \text{ s}\) and \(t
= \text{23} \text{ s}\)
At \(t = \text{5} \text{ s}\) and at \(t = \text{23}
\text{ s}\) the concentrations of the reactants are decreasing
and the concentration of the product is increasing. Therefore
the rate of the forward reaction is faster than the rate of the
reverse reaction.
At \(t = \text{17} \text{s}\) the concentrations of the reactants
and products are constant (unchanging). Therefore the reaction
is in equilibrium and the rate of the forward reaction equals
the rate of the reverse reaction.
What are the concentrations of \(\text{CO}\),
\(\text{Cl}_{2}\) and \(\text{COCl}_{2}\) at \(t = \text{17}
\text{ s}\), \(t = \text{27 s}\) and \(t = \text{42 s}\)?
At \(t = \text{17 s}\): \(\;\text{K}_{\text{c}} =
\dfrac{\text{0,9}}{(\text{1,8})(\text{1,1})} =\)
\(\text{0,45}\)
At \(t = \text{27 s}\): \(\;\text{K}_{\text{c}} =
\dfrac{\text{1,1}}{(\text{2,7})(\text{0,9})} =\)
\(\text{0,45}\)
At \(t = \text{42 s}\): \(\;\text{K}_{\text{c}} =
\dfrac{\text{1,5}}{(\text{2,3})(\text{0,5})} =\)
\(\text{1,30}\)
Note that \(\text{K}_{\text{c}}\) is different at \(\text{42 s}\).
This
means that the stress at \(\text{30 s}\) must have been a change
in temperature.
What happens at \(t = \text{20 s}\)?
The concentration of \(\text{CO}\) increases sharply. The
concentrations of the other compounds do not change
dramatically. Therefore \(\text{CO}\) must have been added to
the system.
After this addition of \(\text{CO}\) there is a shift to reduce the
amount of \(\text{CO}\), that is in the forward direction.
Therefore the concentration of the product increases while the
concentrations of the reactants decrease.
What effect will this have on \(\text{K}_{\text{c}}\)?
Only a change in temperature affects \(\text{K}_{\text{c}}\),
therefore this will have no effect on \(\text{K}_{\text{c}}\).
Worked example 10: Concentration-time graphs
Consider the following chemical equilibrium and graph and answer the
questions that follow.
After how many seconds does the system reach
equilibrium?
Calculate the value of the equilibrium constant.
Explain what happened at t = \(\text{20}\)
\(\text{s}\).
If the change at t = \(\text{35}\) \(\text{s}\) is
due to an increase in temperature, is the
reaction exothermic or endothermic? Explain your
answer.
Check the axes so that you know what the variables are
on this graph
The axes are labelled concentration and time. Therefore this is a
concentration-time graph.
After how many seconds does the system reach
equilibrium?
The concentration of all three compounds becomes constant at t =
\(\text{10}\) \(\text{s}\). This means that the system takes
\(\text{10}\) \(\text{s}\) to reach equilibrium.
What are the concentrations of \(\text{H}_{2}\),
\(\text{I}_{2}\), and \(\text{HI}\) at t = \(\text{10}\)
\(\text{s}\), \(\text{25}\) \(\text{s}\), and \(\text{45}\)
\(\text{s}\)?
Reading off the graph you can see that at \(\text{10}\) \(\text{s}\):
\([\text{H}_{2}]\) and \([\text{I}_{2}]\) = \(\text{0,75}\)
\(\text{mol·dm$^{-3}$}\), \([\text{HI}]\) = \(\text{6,0}\)
\(\text{mol·dm$^{-3}$}\)
At \(\text{25}\) \(\text{s}\):
\([\text{H}_{2}]\) and \([\text{I}_{2}]\) = \(\text{0,6}\)
\(\text{mol·dm$^{-3}$}\), \([\text{HI}]\) = \(\text{4,8}\)
\(\text{mol·dm$^{-3}$}\)
At \(\text{45}\) \(\text{s}\):
\([\text{H}_{2}]\) and \([\text{I}_{2}]\) = \(\text{1,5}\)
\(\text{mol·dm$^{-3}$}\), \([\text{HI}]\) = \(\text{2,75}\)
\(\text{mol·dm$^{-3}$}\)
Write an expression for the equilibrium constant of
this reaction and calculate \(\text{K}_{\text{c}}\).
The different \(\text{K}_{\text{c}}\) at \(\text{45}\) \(\text{s}\)
means that the event at t = \(\text{35}\) \(\text{s}\) must be a
change in temperature.
Explain what happened at t = \(\text{20}\)
\(\text{s}\)?
The concentration of \(\text{HI}\) decreases sharply, as a result
there is a slight decrease in the concentration of
\(\text{H}_{2}\) and \(\text{I}_{2}\). Therefore \(\text{HI}\)
must have been removed from the system.
After this there is a shift to increase the amount of \(\text{HI}\),
that is in the forward direction.
If the change at t = \(\text{35}\) \(\text{s}\) is due
to an increase in temperature, is the reaction exothermic or
endothermic? Explain your answer.
An increase in temperature caused the concentration of the product to
decrease and the concentrations of the reactants to increase.
This means that the reverse reaction has been favoured.
An increase in temperature will favour the reaction that takes heat
in and cools the reaction vessel (endothermic). Therefore the
reverse reaction must be endothermic and the forward reaction
must be exothermic. The reaction is exothermic.
Summary (ESCNW)
The following rules will help in predicting the changes that take place in equilibrium
reactions:
If the concentration of a reactant (on the left) is
increased, then some of it must be used to form the
products (on the right) for equilibrium to be maintained. The
equilibrium position will shift to the right.
\(\text{K}_{\text{c}}\) is unchanged.
If the concentration of a reactant (on the left) is
decreased, then some of the products (on the right)
must be used to form reactants for equilibrium to be maintained. The
equilibrium position will shift to the left.
\(\text{K}_{\text{c}}\) is unchanged.
If the forward reaction is endothermic,
then an increase in temperature will favour this
reaction and the product yield and
\(\text{K}_{\text{c}}\) will increase. A
decrease in temperature will decrease product
yield and likewise decrease
\(\text{K}_{\text{c}}\).
If the forward reaction is exothermic, then
an decrease in temperature will favour this reaction
and the product yield and
\(\text{K}_{\text{c}}\) will increase. An
increase in temperature will decrease product
yield and likewise decrease
\(\text{K}_{\text{c}}\).
Increasing the pressure favours the reaction that will
decrease the number of gas molecules. This is shown
in the balanced chemical equation. This rule applies in reactions with
one or more gaseous reactants or products. \(\text{K}_{\text{c}}\) is
unchanged.
Decreasing the pressure favours the the reaction that will
increase the number of gas molecules. This rule applies
in reactions
with one or more gaseous reactants or products. \(\text{K}_{\text{c}}\)
is unchanged.
A catalyst does not affect the equilibrium position of a
reaction. It only influences the rate of the reaction, in other
words, how quickly equilibrium is reached.
The following simulation will help you to understand these concepts.
The forward reaction is exothermic (\(\Delta{H} < 0\)) so
the reverse reaction must be endothermic. A decrease in
temperature will cause the equilibrium to shift to
favour the exothermic reaction. Therefore the reverse
reaction rate will decrease sharply, and then gradually
increase until equilibrium is re-established.
The addition of a catalyst will speed up both the forward and
reverse reactions.
\(\text{NO}_{2}\) is a product. Therefore the addition of
more \(\text{NO}_{2}\) will increase the rate of the
formation of reactants. The rate of the reverse reaction
will therefore increase sharply, and then gradually
decrease until equilibrium is re-established.
The equilibrium position:
A decrease in temperature will favour the exothermic reaction
and the forward reaction is exothermic. Therefore the
equilibrium position will shift to the right.
The addition of a catalyst will have no effect on the
equilibrium position as both the forward and reverse
reactions rates would be increased equally.
The addition of more \(\text{NO}_{2}\) will favour the
formation of the reactants and so the equilibrium will
shift to the left.
Worked example 12: Graphs of equilibrium
Study the graph and answer the questions that follow:
Does the equilibrium favour the reactants or the products?
Determine the value of \(\text{K}_{\text{c}}\) if the coefficients of
the balanced equation all equal 1.
The forward reaction has \(\Delta{H} > 0\). What effect will an
increase in temperature have on [A], [B] and [C]?
Determine which compounds are the reactants and which are the products
The concentration of reactants decreases from the start of the reaction to
equilibrium. Therefore A and B are reactants.
The concentration of products increases from the start of the reaction to
equilibrium. Therefore C is a product.
We are told that all coefficients in the balanced equation equal 1. Therefore the
general equation is: \(\text{A} + \text{B}\) \(\rightleftharpoons\) \(\text{C}\)
Do reactants or products have higher concentrations at equilibrium?
A and B (the reactants) have higher concentrations at equilibrium.
Does the equilibrium favour the reactants or products?
The reactants have higher concentrations than the products, therefore the equilibrium
must favour the reactants.
Determine the equilibrium concentration values of A, B and C
Determine which reaction is exothermic and which is endothermic
The forward reaction has \(\Delta{H} > 0\). This means that the forward reaction
is endothermic. The reverse reaction must therefore be exothermic.
Which reaction is favoured by an increase in temperature?
The endothermic reaction would be favoured by an increase in temperature (to lower
the temperature). This is the forward reaction.
How would [A], [B] and [C] change?
The forward reaction is favoured, therefore the equilibrium would shift to the right.
This means that the reactant concentrations ([A] and [B]) would decrease and the
product concentration ([C]) would increase.
Equilibrium
Textbook Exercise
8.3
The following reaction has reached equilibrium in a closed container:
The pressure of the system is then decreased. How will the
concentration of the \(\text{H}_{2}\)(g) and the value of
\(\text{K}_{\text{c}}\) be affected when the new equilibrium is
established? Assume that the temperature of the system remains
unchanged.
Hydrogen concentration
\(\text{K}_{\text{c}}\)
(a)
increases
increases
(b)
increases
stays the same
(c)
stays the same
stays the same
(d)
decreases
stays the same
(IEB Paper 2, 2004)
According to Le Chatelier's principle, the reaction that increases
the number of gas molecules (and therefore the pressure) will be
favoured. That is the forward reaction, and so the concentration
of hydrogen will
increase. Only temperature affects the value of
\(\text{K}_{\text{c}}\), therefore \(\text{K}_{\text{c}}\)
remains the same.
b) Hydrogen concentration increases and equilibrium
constant stays the same
During a classroom experiment copper metal reacts with concentrated
nitric acid to produce \(\text{NO}_{2}\) gas.
The \(\text{NO}_{2}\) is collected in a gas syringe.
When enough gas has collected in the syringe, the delivery tube is
clamped so that no gas can escape. The brown \(\text{NO}_{2}\)
gas collected reaches an equilibrium with colourless
\(\text{N}_{2}\text{O}_{4}\) gas as represented by the following
equation:
\[2\text{NO}_{2}(\text{g}) \rightleftharpoons
\text{N}_{2}\text{O}_{4}(\text{g}), \quad \Delta H < 0\]
Once this equilibrium has been established, there are \(\text{0,01}\)
\(\text{moles}\) of \(\text{NO}_{2}\) gas and \(\text{0,03}\)
\(\text{moles}\) of \(\text{N}_{2}\text{O}_{4}\) gas present in
the syringe.
A learner, noticing that the colour of the gas mixture in the
syringe is no longer changing, comments that all
chemical reactions in the syringe must have stopped. Is
this assumption correct? Explain.
No. The learner is not correct. A dynamic chemical
equilibrium has been reached and the products are
changing into reactants at the same rate as reactants
are changing into products. This results in no colour
change being observed in the syringe.
The gas in the syringe is cooled. The volume of the gas is
kept constant during the cooling process. Will the gas
be lighter or darker at the lower temperature? Explain
your answer.
The forward reaction is exothermic (\(\Delta H < 0\)). A
decrease in temperature will favour the forward
reaction. \(\text{N}_{2}\text{O}_{4}(\text{g})\) is
colourless, therefore the gas will be lighter at the
lower temperature.
The volume of the syringe is now reduced (at constant
temperature) to \(\text{75}\) \(\text{cm$^{3}$}\) by
pushing the plunger in and holding it in the new
position. There are \(\text{0,032}\) \(\text{moles}\) of
\(\text{N}_{2}\text{O}_{4}\) gas present once the
equilibrium has been re-established at the reduced
volume (\(\text{75}\) \(\text{cm$^{3}$}\)). Calculate
the value of the equilibrium constant for this
equilibrium.
(IEB Paper 2, 2004)
The volume is reduced, so the pressure is increased. This
means that the foward reaction will be favoured to
reduce the number of gas molecules.
There are \(\text{0,01}\) moles of \(\text{NO}_{2}\) and
\(\text{0,03}\) moles of \(\text{N}_{2}\text{O}_{4}\)
initially.
For every 2x moles of \(\text{NO}_{2}\) used, 1x moles of
\(\text{N}_{2}\text{O}_{4}\) are produced.
Reaction
\(2\text{NO}_{2}(\text{g})\)
\(1\text{N}_{2}\text{O}_{4}(\text{g})\)
Initial quantity
(mol)
\(\text{0,01}\)
\(\text{0,03}\)
Change (mol)
-2x
+x
Equilibrium quantity
(mol)
\(\text{0,01}\)-2x
\(\text{0,03}\)+x
Equilibrium
concentration
(\(\text{mol·dm$^{-3}$}\))
There are \(\text{0,032}\) \(\text{mol}\)
\(\text{N}_{2}\text{O}_{4}\) at the new equilibrium.
Therefore \(\text{0,03}\) \(\text{mol}\) + x =
\(\text{0,032}\) \(\text{mol}\)
x = \(\text{0,032}\) - \(\text{0,03}\) \(\text{mol}\) =
\(\text{0,002}\) \(\text{mol}\)
Gases X and Y are pumped into a \(\text{2}\) \(\text{dm$^{3}$}\)
container. When the container is sealed, \(\text{4}\)
\(\text{moles}\) of gas X and \(\text{4}\) \(\text{moles}\) of
gas Y are present. The following equilibrium is established:
At \(\text{70}\) \(\text{s}\) the temperature is increased.
Is the forward reaction endothermic or exothermic?
Explain in terms of Le Chatelier's Principle.
Exothermic. The amount of product decreases (and amount of
reactants increases) when the temperature is increased
indicating that the reverse reaction is favoured. Le
Chatelier's principle states that an increase in
temperature will favour the endothermic reaction
(cooling the reaction vessel). Therefore the reverse
reaction must be endothermic and the forward reaction is
exothermic.
How will this increase in temperature affect the value of the
equilibrium constant?
A change in temperature is the only thing that affects the
equilibrium constant. The amount of product decreased
and the amount of reactants increased, therefore the
equilibrium constant will decrease.
Consider the following hypothetical reaction that takes place in a
closed \(\text{2}\) \(\text{dm$^{3}$}\) flask
at \(\text{298}\) \(\text{K}\).
Calculate the concentration of each of the reactants and the
product using figures from the graph between
\(\text{5}\) \(\text{minutes}\) and \(\text{10}\)
\(\text{minutes}\) and hence calculate the equilibrium
constant \(\text{K}_{\text{c}}\) for this reaction at
\(\text{298}\) \(\text{K}\)
State what a low value of \(\text{K}_{\text{c}}\) indicates
about the yield of product for a reaction.
A value of \(\text{K}_{\text{c}}\) between \(\text{0}\) and
\(\text{1}\) indicates that the equilibrium lies to the
left. Therefore the yield of product is low (more
reactants than products).
Why is it not possible to calculate \(\text{K}_{\text{c}}\)
using figures from the graph during the first
\(\text{5}\) \(\text{minutes}\)
\(\text{K}_{\text{c}}\) is the equilibrium constant,
therefore it cannot be calculated except when the
reaction is in equilibrium. During the first
\(\text{5}\) \(\text{minutes}\) the reaction is not in
equilibrium.
State Le Chatelier's principle.
When an external stress (change in pressure, temperature or
concentration) is applied to a system in chemical
equilibrium, the equilibrium will change in such a way
as to reduce the effect of the stress.
At \(\text{10}\) \(\text{minutes}\) the temperature of the
flask was increased. Using Le Chatelier's principle,
determine if the production of \(\text{AB}_{2}\) is
exothermic or endothermic?
\(\text{AB}_{2}\) is produced through the forward reaction.
Le Chatelier's principle states that if the temperature
is increased the equilibrium will change to decrease the
temperature of the vessel. That is the reaction that
takes in heat (endothermic) will be favoured.
The number of moles of the reactants decreases after the
temperature increase, while the number of moles of the
product increases. Therefore the temperature increase
favours the forward reaction. This means that the
forward reaction is endothermic.
How would the equilibrium constant be
affected by each of the following changes:
(state either increase, decrease or no effect)
Increasing the pressure of the flask with no change
to temperature.
no effect
Adding a catalyst to the flask.
no effect
Increasing the temperature of the flask.
increase
Industrial applications (ESCNX)
In industrial processes, it is important to get the product as quickly and as efficiently as
possible. The less expensive the process the better.
The Haber process is a good example of an industrial process which uses the
equilibrium principles that have been discussed. The equation for the process is as
follows:
Since the forward reaction is exothermic, to produce a lot of product and favour the forward
reaction the system needs to be colder. However, cooling a system slows down all
chemical reactions and so the system can't be too cold. This process is carried out at a
much higher temperature to ensure the speed of production.
Because high temperature favours the reverse reaction, the \(\text{NH}_{3}\) product is
actually removed as it is made (product concentration decreased) to prevent ammonia
being used in the reverse reaction. The decrease of product concentration favours the
forward reaction.
High pressure is also used to ensure faster reaction time and to favour the production of
\(\text{NH}_{3}\). The forward reaction is favoured by higher pressures because there
are \(\text{2}\) gas molecules of product for every \(\text{4}\) gas molecules of
reactant.
Refer to Chapter 14 for more information on the Haber process and other industrial
applications.
Applying equilibrium principles
Textbook Exercise 8.4
Look at the values of \(\text{K}_{\text{c}}\) calculated for
the Haber process reaction at different temperatures,
and then answer the questions that follow:
\(T\left(℃\right)\)
\(\text{K}_{\text{c}}\)
\(\text{25}\)
\(\text{6,4} \times \text{10}^{\text{2}}\)
\(\text{200}\)
\(\text{4,4} \times \text{10}^{-\text{1}}\)
\(\text{300}\)
\(\text{4,3} \times \text{10}^{-\text{3}}\)
\(\text{400}\)
\(\text{1,6} \times \text{10}^{-\text{4}}\)
\(\text{500}\)
\(\text{1,5} \times \text{10}^{-\text{5}}\)
What happens to the value of \(\text{K}_{\text{c}}\)
as the temperature increases?
The value of \(\text{K}_{\text{c}}\) decreases
Which reaction is being favoured when the temperature
is \(\text{300}\) \(\text{℃}\)?
A value of \(\text{K}_{\text{c}}\) between 0 and 1
means that there are more reactants than
products. Therefore the reverse reaction is
favoured.
According to this table, which temperature would be
best if you wanted to produce as much ammonia as
possible? Explain.
\(\text{25}\) \(\text{℃}\). At this temperature
the forward reaction is favoured and so the the
maximum yield of ammonia is achieved.
