LE CHATELIER'S PRINCIPLE

This page looks at Le Chatelier's Principle and explains how to apply it to

reactions in a state of dynamic equilibrium. It covers changes to the position of
equilibrium if you change concentration, pressure or temperature. It also explains
very briefly why catalysts have no effect on the position of equilibrium.
,

Important: If you aren't sure about the words dynamic equilibrium or position of equilibrium you
should read the introductory page before you go on


It is important in understanding everything on this page to realise that Le
Chatelier's Principle is no more than a useful guide to help you work out what
happens when you change the conditions in a reaction in dynamic equilibrium. It
doesn't explain anything. I'll keep coming back to that point!

Using Le Chatelier's Principle
A statement of Le Chatelier's Principle
If a dynamic equilibrium is disturbed by changing the conditions, the
position of equilibrium moves to counteract the change.

Using Le Chatelier's Principle with a change of concentration
Suppose you have an equilibrium established between four substances A, B, C
and D.

,

Note: In case you wonder, the reason for choosing this equation rather than having just A +B on
the left-hand side is because further down this page I need an equation which has different
numbers of molecules on each side. I am going to use that same equation throughout this page.


What would happen if you changed the conditions by increasing the
concentration of A?
According to Le Chatelier, the position of equilibrium will move in such a way as
to counteract the change. That means that the position of equilibrium will move
so that the concentration of A decreases again - by reacting it with B and turning
it into C +D. The position of equilibrium moves to the right.

This is a useful way of converting the maximum possible amount of B into C and
D. You might use it if, for example, B was a relatively expensive material
whereas A was cheap and plentiful.
What would happen if you changed the conditions by decreasing the
concentration of A?
According to Le Chatelier, the position of equilibrium will move so that the
concentration of A increases again. That means that more C and D will react to
replace the A that has been removed. The position of equilibrium moves to the
left.

This is esssentially what happens if you remove one of the products of the
reaction as soon as it is formed. If, for example, you removed C as soon as it
was formed, the position of equilibrium would move to the right to replace it. If
you kept on removing it, the equilibrium position would keep on moving
rightwards - turning this into a one-way reaction.

Important
This isn't in any way an explanation of why the position of equilibrium moves in
the ways described. All Le Chatelier's Principle gives you is a quick way of
working out what happens.
,

Note: If you know about equilibrium constants, you will find a more detailed explanation of the
effect of a change of concentration by following this link. If you don't know anything about
equilibrium constants, you should ignore this link.
If you choose to follow it, return to this page via the BACK button on your browser or via the
equilibrium menu.


Using Le Chatelier's Principle with a change of pressure
This only applies to reactions involving gases:

What would happen if you changed the conditions by increasing the
pressure?
According to Le Chatelier, the position of equilibrium will move in such a way as
to counteract the change. That means that the position of equilibrium will move
so that the pressure is reduced again.
Pressure is caused by gas molecules hitting the sides of their container. The
more molecules you have in the container, the higher the pressure will be. The
system can reduce the pressure by reacting in such a way as to produce fewer
molecules.
In this case, there are 3 molecules on the left-hand side of the equation, but only
2 on the right. By forming more C and D, the system causes the pressure to
reduce.
Increasing the pressure on a gas reaction shifts the position of equilibrium
towards the side with fewer molecules.

What would happen if you changed the conditions by decreasing the
pressure?
The equilibrium will move in such a way that the pressure increases again. It can
do that by producing more molecules. In this case, the position of equilibrium will
move towards the left-hand side of the reaction.

What happens if there are the same number of molecules on both sides of
the equilibrium reaction?
In this case, increasing the pressure has no effect whatsoever on the position of
the equilibrium. Because you have the same numbers of molecules on both
sides, the equilibrium can't move in any way that will reduce the pressure again.

Important
Again, this isn't an explanation of why the position of equilibrium moves in the
ways described. You will find a rather mathematical treatment of the explanation
by following the link below.
,

Note: You will find a detailed explanation by following this link. If you don't know anything about
equilibrium constants (particularly K
p
), you should ignore this link. The same thing applies if you
don't like things to be too mathematical! If you are a UK A' level student, you won't need this
explanation.
If you choose to follow the link, return to this page via the BACK button on your browser or via the
equilibrium menu.


Using Le Chatelier's Principle with a change of temperature
For this, you need to know whether heat is given out or absorbed during the
reaction. Assume that our forward reaction is exothermic (heat is evolved):

This shows that 250 kJ is evolved (hence the negative sign) when 1 mole of A
reacts completely with 2 moles of B. For reversible reactions, the value is always
given as if the reaction was one-way in the forward direction.
The back reaction (the conversion of C and D into A and B) would be
endothermic by exactly the same amount.


What would happen if you changed the conditions by increasing the
temperature?
According to Le Chatelier, the position of equilibrium will move in such a way as
to counteract the change. That means that the position of equilibrium will move
so that the temperature is reduced again.
Suppose the system is in equilibrium at 300C, and you increase the temperature
to 500C. How can the reaction counteract the change you have made? How can
it cool itself down again?
To cool down, it needs to absorb the extra heat that you have just put in. In the
case we are looking at, the back reaction absorbs heat. The position of
equilibrium therefore moves to the left. The new equilibrium mixture contains
more A and B, and less C and D.

If you were aiming to make as much C and D as possible, increasing the
temperature on a reversible reaction where the forward reaction is exothermic
isn't a good idea!
What would happen if you changed the conditions by decreasing the
temperature?
The equilibrium will move in such a way that the temperature increases again.
Suppose the system is in equilibrium at 500C and you reduce the temperature
to 400C. The reaction will tend to heat itself up again to return to the original
temperature. It can do that by favouring the exothermic reaction.
The position of equilibrium will move to the right. More A and B are converted
into C and D at the lower temperature.

Summary
Increasing the temperature of a system in dynamic equilibrium favours the
endothermic reaction. The system counteracts the change you have made
by absorbing the extra heat.
Decreasing the temperature of a system in dynamic equilibrium favours
the exothermic reaction. The system counteracts the change you have
made by producing more heat.

Important
Again, this isn't in any way an explanation of why the position of equilibrium
moves in the ways described. It is only a way of helping you to work out what
happens.
,

Note: I am not going to attempt an explanation of this anywhere on the site. To do it properly is
far too difficult for this level. It is possible to come up with an explanation of sorts by looking at
how the rate constants for the forward and back reactions change relative to each other by using
the Arrhenius equation, but this isn't a standard way of doing it, and is liable to confuse those of
you going on to do a Chemistry degree. If you aren't going to do a Chemistry degree, you won't
need to know about this anyway!


Le Chatelier's Principle and catalysts
Catalysts have sneaked onto this page under false pretences, because adding a
catalyst makes absolutely no difference to the position of equilibrium, and
Le Chatelier's Principle doesn't apply to them.
This is because a catalyst speeds up the forward and back reaction to the same
extent. Because adding a catalyst doesn't affect the relative rates of the two
reactions, it can't affect the position of equilibrium. So why use a catalyst?
For a dynamic equilibrium to be set up, the rates of the forward reaction and the
back reaction have to become equal. This doesn't happen instantly. For a very
slow reaction, it could take years! A catalyst speeds up the rate at which a
reaction reaches dynamic equilibrium.
,

Note: You might try imagining how long it would take to establish a dynamic equilibrium if you
took the visual model on the introductory page and reduced the chances of the colours changing
by a factor of 1000 - from 3 in 6 to 3 in 6000 and from 1 in 6 to 1 in 6000.
Starting with blue squares, by the end of the time taken for the examples on that page, you would
most probably still have entirely blue squares. Eventually, though, you would end up with the
same sort of patterns as before - containing 25% blue and 75% orange squares.

In 1884, the French Chemist Henri Le Chatelier suggested that equilibrium systems
tend to compensate for the effects of perturbing influences.
When a system at equilibrium is disturbed, the equilibrium position will shift in the
direction which tends to minimise, or counteract, the effect of the disturbance.
If the concentration of a solute reactant is increased, the equilibrium position
shifts to use up the added reactants by producing more products.
If the pressure on an equilibrium system is increased, then the equilibrium
position shifts to reduce the pressure.
If the volume of a gaseous equilibrium system is reduced (equivalent to an
increase in pressure) then the equilibrium position shifts to increase the
volume (equivalent to a decrease in pressure)
If the temperature of an endothermic equilibrium system is increased, the
equilibrium position shifts to use up the heat by producing more products.
If the temperature of an exothermic equilibrium system is increased, the
equilibrium position shifts to use up the heat by producing more reactants.
Examples
a. Changes in Concentration.
Consider the following system at equilibrium:
Fe
3+
(aq) + SCN
-
(aq)
(colourless)


FeSCN
2+
(aq)
(red)
o Increasing the concentration of either Fe
3+
(aq) or SCN
-
(aq) will result
in the equilibrium position moving to the right, using up the some of
the additional reactants and producing more FeSCN
2+
(aq).
The solution will become a darker red colour.
o Removing some of the Fe
3+
(aq) or SCN
-
(aq) will result in the
equilibrium position moving to the left to produce more Fe
3+
(aq) and
SCN
-
(aq).
The solution will become less red as FeSCN
2+
(aq) is consumed.
o Increasing the concentration of FeSCN
2+
(aq) will result in the
equilibrium position moving to the left to use up some of the additional
FeSCN
2+
(aq) and producing more Fe
3+
(aq) and SCN
-
(aq).
The solution will become less red.
o Removing some of the FeSCN
2+
(aq) will result in the equilibrium
position moving to the right to produce more FeSCN
2+
(aq)
which will lead the solution to become a darker red colour.
b. Changes in Pressure of Gaseous Equilibrium Systems
Gas pressure is related to the number of gas particles in the system, more
gas particles means more gas pressure.
Consider the following gaseous equilibrium system:
2NO
2
(g)
(red-brown)


N
2
O
4
(g)
(colourless)
o Increasing the pressure on this equilibrium system will result in the
equilibrium position shifting to reduce the pressure, that is, to the side
that has the least number of gas particles.
There are 2 gas particles on the left hand side of the reaction and 1
gas particle on the right hand side of the reaction.
Increasing the pressure on this system results in the equilibrium
position moving to the right, consuming NO
2
(g) and producing more
N
2
O
4
(g).
The system will become a lighter red-brown colour.
o Reducing the pressure on this equilibrium system will result in the
equilibrium position moving to the left, that is, to the side that has the
most gas particles, in order to increase the pressure.
The red-brown colour of the system becomes darker.
Consider the following equilibrium system:
C(s) + H
2
O(g)

CO(g) + H
2
(g)
o Increasing the pressure on this equilibrium system results in the
equilibrium position shifting to the left, CO(g) and H
2
(g) are consumed
to produce more C(s) and H
2
O(g).
There is only 1 gas particle on the left of the equation (H
2
O) and 2 gas
particles on the right hand side of the equation (CO(g) + H
2(g)
), so the
equilibrium position moves to the side that has the least number of
gas particles to reduce the pressure, that is, to the left.
o Reducing the pressure on this equilibrium system results in the
equilibrium position shifting to the side that has the most gas particles
in order to increase the pressure, that is, to the right.
C(s) and H
2
O(g) are consumed to produce more CO(g) and H
2
(g).
c. Changes in Volume of Gaseous Equilibrium Systems
Gas volume is related to gas pressure, a gas at reduced volume has a higher
pressure, a gas at increased volume has a lower pressure.
Consider the following gaseous equilibrium system:
2NO
2
(g)
(colourless)


N
2
O
4
(g)
(red-brown)
o Increasing the volume of this equilibrium system is equivalent to
reducing the pressure, so the equilibrium position moves to the side of
the reaction that has the most gas particles in order to increase the
pressure.
The equilibrium position shifts to the left producing more NO
2
(g) by
consuming N
2
O
4
(g).
The system becomes less red-brown in colour.
o Reducing the volume of this equilibrium system is equivalent to
increasing the pressure, so the equilibrium position moves to the side
of the reaction that has the least gas particles in order to reduce the
pressure.
The equilibrium position shifts to the right producing more N
2
O
4
(g) by
consuming more NO
2
(g).
The system becomes a darker red-brown.
d. Changes in Temperature
In an endothermic reaction, energy can be considered as a reactant of the
reaction.
In an exothermic reaction, energy can be considered as a product of the
reaction.
1. Endothermic Equilibrium Systems
Consider the following reaction at equilibrium:
H
2
(g)
(colourless)
+

I
2
(s)
(purple)


2HI(g)
(colourless)
H = + 52 kJ mol
-1

This reaction can also be written with the energy term incorporated
into the equation on the side with the reactants:
H
2
(g)
(colourless)
+

I
2
(s)
(purple)
+ 52 kJ



2HI(g)
(colourless)
Increasing the temperature of the equilibrium system will shift
the equilibrium position to the side that does not include the
energy term in order to reduce the temperature, that is to the
right. The increased heat will be consumed to produce more
HI(g) product, since H
2
(g) and I
2
(s) will be consumed, there will
be less purple solid in the reaction vessel.
Reducing the temperature of the equilibrium system will shift
the equilibrium to the left in order to produce more heat. HI(g)
will be consumed in order to produce more H
2
(g) and I
2
(s). The
amount of purple solid will increase.
2. Exothermic Equilibrium Systems
Consider the following reaction at equilibrium:
Ag
+
(aq) + Cl
-
(aq)
(colourless)


AgCl(s)
(white)
H = -112 kJ mol
-1


This reaction can also be written with the energy term incorporated
into the equation on the side with the the products:
Ag
+
(aq) + Cl
-
(aq)
(colourless)


AgCl(s)
(white)
+ 112 kJ

Increasing the temperature of this equilibrium system shifts the
equilibrium position to the left, consuming some of the energy
and products to produce more reactants. There will be less
white AgCl(s) in the reaction vessel.
Reducing the temperature of this equilibrium system shifts the
equilibrium position to the right, producing more heat and more
white AgCl(s).