This page looks at Le Châtelier's Principle and explains how to apply it to reactions in a state of dynamic equilibrium. Le Châtelier's Principle states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium moves to counteract the change. If a
chemical reaction is at equilibrium and experiences a change such as
pressure, temperature, or concentration of products or reactants the
equilibrium will shift in the direction to accommodate the change.
It covers changes to the position of equilibrium if you change
concentration, pressure or temperature and explains very briefly why
catalysts have no effect on the position of equilibrium.
Introduction
An
action that tends to change the temperature, pressure, or
concentrations of reactants in a system at equilibrium stimulates a
response that partially offsets the change while a new equilibrium
condition is established (2). Hence, Le Châtelier's
principle states that any change to a system at equilibrium will adjust
to compensate for that change. In 1884 the French chemist and engineer
Henry-Louis Le Châtelier proposed one of the
central concepts of chemical equilibria, which describes what happens to
a system when something briefly removes it from a state of equilibrium.
It is important in understanding that Le Châtelier's
Principle is only a useful guide to help you work out what happens when
you change the conditions in a reaction in dynamic equilibrium, but it
does not explain the microscopic reason for the changes.
Concentration Changes
Le Châtelier's
principle states that if the system is changed in a way that increases
the concentration of one of the reacting species, it must favor the
reaction in which that species is consumed. In other words, if there is
an increase in products, the Reaction Quotient Qc is increased, making it greater than the Equilibrium Constant Kc . Suppose you have an equilibrium established between four substances A, B, C and D.
What would happen if you changed the conditions by increasing the concentration of A ?
According to Le Châtelier,
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 equilibrium moves to the right (green arrow below).
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.
Decreasing the concentration of A?
According to Le Châtelier, 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 essentially 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.
Pressure Changes
This only applies to reactions involving gases.
Increasing the pressure
According to Le Châtelier,
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 three 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.
| Example 1: Haber Process Equilibrium |
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If
this mixture is transferred from a 1.5 L flask to a 5 L flask, in which
direction does a net change occur to return to equilibrium? Because we
are increasing volume (and therefore reducing the pressure), the shift
occurs in the direction that produces more moles of gas. To restore
equilibrium the shift needs to occur to the left, in the direction of
the reverse reaction.
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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 cannot move in any way that
will reduce the pressure again. 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 detailed explanation .
Summary
Three
ways to change the pressure of an equilibrium mixture are: 1. Add or
remove a gaseous reactant or product, 2. Add an inert gas to the
constant-volume reaction mixture or 3. Change the volume of the system.
(2)
- Adding products makes
Qc larger thanKc . This creates a net change in the reverse direction, toward reactants. The opposite occurs when adding more reactants. (2) - Adding an inert gas into a gas-phase equilibrium at constant volume does not result in a shift. This is because the addition of a non-reactive gas does not change the partial pressures of the other gases in the container. While the total pressure of the system increases, the total pressure does not have any effect on the equilibrium constant. (1)
- When the volume of a mixture is reduced, a net change occurs in the direction that produces fewer moles of gas. When volume is increased the change occurs in the direction that produces more moles of gas.
Temperature Changes
For this, you need to know whether heat is released 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.
The main effect of temperature on equilibrium is in changing the value of the equilibrium constant.
Warning:
It is not uncommon that textbooks and instructors to consider
temperature as a independent "species" in a reaction. While this is
rigorously incorrect since once cannot "add or remove
temperature" to a reaction as with species, it serves as a convenient
mechanism to predict the shift of reactions with changing temperature.
For example, if temperature is a "reactant" (ΔH>0 ), then the reaction favors the formation of products at elevated temperature. Similarly, if temperature is a "product" (ΔH>0 ), then the reaction favors the formation of reactants. A more accurate, and hence preferred, description is discussed below.
Increasing the temperature
According to Le Châtelier,
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 300°C, and you increase the temperature to 500°C. 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!
Decreasing the temperature?
The
equilibrium will move in such a way that the temperature increases
again. Suppose the system is in equilibrium at 500°C and you reduce the
temperature to 400°C. 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
with more A and B are converted into C and D at the lower temperature.
| Example 2: I see you |
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O2 + 2H2 → 2H2O ΔH= -125.7 kJ
1. What side of the reaction is favored? Because the heat is a product of the reaction, the reactants are favored.
2. Would the conversion of O2 and H2 to H2O be favored with heat as a product or as a reactant? Heat as a product would shift the reaction forward, creating H2O. The more heat added to the reaction, the more H2O created.
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Summary
- Increasing the temperature of a system in dynamic equilibrium favors 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 favors the exothermic reaction. The system counteracts the change you have made by producing more heat.
Again,
this is not 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.
Catalysts
Adding a catalyst makes absolutely no difference to the position of equilibrium and Le Châtelier's
Principle does not apply to them. This is because a catalyst speeds up
the forward and back reaction to the same extent and adding a catalyst
does not affect the relative rates of the two reactions, it cannot
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 does not happen
instantly and for very slow reactions, it may take years! A catalyst only speeds up the rate at which a reaction reaches dynamic equilibrium.

