IB Colourful Solutions in Chemistry

Le Chatelier's Principle states that if a dynamic equilibrium is disturbed by changing the conditions, such as concentration, temperature, or pressure, the system will respond to counteract the change and re-establish equilibrium.

Syllabus reference R2.3.4

Reactivity 2.3.4 - Le Châtelier’s principle enables the prediction of the qualitative effects of changes in concentration, temperature and pressure to a system at equilibrium.

Apply Le Chatelier’s principle to predict and explain responses to changes of systems at equilibrium.

Guidance

Include the effects on the value of K and on the equilibrium composition.
Le Chatelier’s principle can be applied to heterogeneous equilibria such as: X(g) ⇌ X(aq)

Tools and links

Reactivity 2.2 - Why do catalysts have no effect on the value of K or on the equilibrium composition?

Le Chatelier's principle
Changing conditions
Addition or removal of reactants
Change of pressure
Change of temperature
Worked examples

Le Chatelier's principle

This considers the effect of changing conditions on an equilibrium.

There is no fixed statement of the principle so it does not need to be learned as a definition.

One possible statement of the principle could be "Any system at equilibrium will respond to a change in conditions in such a way as to oppose that change."

For instance, if the concentration of a reactant in a chemical reaction is increased, the equilibrium will shift towards the formation of more products to reduce the reactant concentration. Similarly, an increase in temperature for an exothermic reaction will cause the equilibrium to shift towards the reactants, while a decrease in temperature will shift it towards the products. This principle helps predict how a system at equilibrium will react to external changes, enabling chemists to control reaction conditions to favor the desired outcome.

Example: The fundamental concept is that if a system at equilibrium is able to reduce the effect of any applied change then it will do so.

You could consider the effect of pressure change on a rubber balloon. If the balloon is subject to increasing pressure, it will respond by expanding in an attempt to decrease the pressure once again.

Although this is not a chemical system, it demonstrates the idea very clearly.

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Changing conditions

The conditions that concern us are:

Changing the amounts of reactants or products by addition or removal.
Changing the pressure
Changing the temperature

In each of these situations an equilibrium will respond, according to Le Chatelier's principle, in an attempt to reduce the effects of the change in conditions.

The process is as follows:

1 System at equilibrium
2 Change of conditions disturbs the equilibrium (i.e. system is now not at equilibrium)
3 System responds by moving forward or backwards as necessary to re-establish the equilibrium.

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Addition or removal of reactants

Addition of reactants (or products in the case of the reverse reaction) to a system at equilibrium disturbs that equiliibrium.

The system responds in such a way as to re-establish the equilibrium.

It does so by removing the added reactant, increasing the rate of the forward reaction and making more product until the equilibrium proportions are re-established.

This is an example of Le Chatelier's principle at work; the system counteracts the change in conditions, in this case the change in concentration of a reactant.

More mathematically, what is happening is that the ratio or products to reactants as given by the equilibrium law becomes not equal to the equilibrium constant. The system ceases to be at equilibrium and readjusts its reactant and product concentrations to re-establish the equilibrium.

Example: Consider the reversible reaction:

N₂O₄ ⇋ 2NO₂

The system is at equilbrium when the concentration of N₂O₄ is 0.1 mol dm^-3 and the concentration of NO₂ is 0.4 mol dm^-3. All of the reaction takes place in a 1 dm³ flask. What is the effect of adding 0.1 moles of N₂O₄ to the equilibrium?

The system products and reactants were originally in equilibrium and now more reactant has been added.

This change can only be opposed by making more product.

The system must now react to make more NO₂ and less N₂O₄. It does so by moving the reaction in the forward direction until the equilibrium proportions are reestablished.

The general principle of Le Chatelier is followed and addition to the Left Hand Side moves the reaction to the Right Hand Side.

Example Add to the left push to the right. Take from the left pull to the left.

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Change of pressure

Changing the pressure of a system at equilibrium only has an effect on the reaction if there are different numbers of moles of gases on each side of the equilibrium.

Example: Consider the following equilibrium:

N₂O₄ ⇋ 2NO₂

The equilibrium is affected by a change of pressure (1 mol of gas on the left and 2 mol of gas on the right hand side of the equilibrium)

2HI ⇋ H₂ + I₂

is NOT affected by a change of pressure (2 mol of gas on the left and 2 mol of gas on the right hand side of the equilibrium)

The question remains, why?

In order to fully understand why change of pressure affects equilibria with unequal numbers of moles on either side, it is first important to understand how the pressure of a gas can be changed.

If the temperature is to be kept constant and the number of moles of gas is constant then the only way to change the pressure of a gas is to change the volume it occupies as PV = nRT (the ideal gas law). In the ideal gas law, if n, R and T are constants then only V can change P.

The definition of concentration is moles divided by volume. So if we are changing the volume, we are also changing the concentration. Now we can apply Le Chatelier's principle. If we increase the concentration on one side the reaction moves away from that side.

But, we are increasing the concentration on both sides! This is where the number of moles on each side is important.

If there are more moles on one side then an increase in concentration will have more effect on that side in the equilibrium law ratio.

Therefore, an increase in pressure moves the reaction away from the side with more moles and towards the side with fewer moles.

Le Chateliers principle once again shows the system responding to counteract a change in conditions.

The effect of pressure on equilibria

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Change of temperature

Changing the temperature of a chemical reaction changes the rate of reaction. At equilibrium the rate of both the forward and back reactions are equal.

So how does a change in temperature affect the rate of forward and back reactions differently?

The answer is that it wouldn't if there were no energy change involved in the chemical reaction, i.e. if ΔH = 0 (not very likely).

If the reaction is exothermic then the activation energy for the forward reaction MUST be less than for the reverse reaction.

This means that an increase in temperature for the equilibrium has more of an effect on the rate of reaction involving the larger of the two activation energies, ie. the direction of endothermic change.

The mathematics are beyond the level of pre-university syllabi, but this can be demonstrated using the Arrhenius equation k = Ae^-Ea/RT.

By Le Chatelier's principle the reaction responds to a change of conditions in such a way as to oppose that change.

If the temperature is increased, there is more energy available. The reaction absorbs this extra energy by moving in the direction of endothermic change.

Example: What is the effect of an increase in temperature on the equilibrium:

N₂ + 3H₂ ⇋ 2NH₃ ΔH = -92 kJ mol^-1

The forward reaction is exothermic ( ΔH is negative). The temperature is increased, there is more energy available therefore the reaction will move in the direction of endothermic change, i.e. towards the reactants, reducing the extra energy.

This follows Le Chatelier's principle.

The extra heat applied to the system is absorbed by the endothermic direction of change.

The effect of temperature on equilibria

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Summary: The effect of changing conditions on the equilibrium constant

Change of temperature actually changes the position of equilibrium. This means that it also changes the value of the equilibrium constant.

It is the only change of conditions that has this effect.

Change in conditions	effect on reaction	change Kc
concentration	moves away from increased concentration	no
pressure	increased pressure moves away from greater moles	no
temperature	increased temperature moves towards endothermic change	yes

NOTE: Catalysts do not effect the position of equilibrium or the value of the equilibrium constant. The equilibrium is established more rapidly, but with the same quotient.

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Worked examples

Q722-01 Consider the endothermic reaction below:

5CO(g) + I₂O₅(g) ⇌ 5CO₂(g) + I₂(g)

According to Le Chatelier's principle, which change would result in an increase in the amount of CO₂?

Increasing the temperature
Decreasing the temperature
Increasing the pressure
Decreasing the pressure

Answer

As it is an endothermic reaction in the forward direction an increase in temperature pushes the equilibrium to the side of the products. There is no change in the number of moles and so the equilibrium is unaffected by pressure change.

The correct response is A.

Q722-02 The compounds N₂O₄ and NO₂ produce an equilibrium mixture according to the equation below:

N₂O₄ (g) ⇋ 2NO₂(g) ........ ∆H = +57.2 kJmol^-1

An increase in the equilibrium concentration of NO₂can be produced by increasing which of the following factors:

I. Pressure
II. Temperature

Answer

Increasing the temperature pushes the reaction in the direction of ENDOthermic change, ie to the right hand side.

Increasing the pressure pushes the reaction in the direction of the fewer number of moles of gas, ie to the left hand side in this case.

Q722-03 For the following system at equilibrium, which change will shift the position of equilibrium to the right?

CH₃COOH(l) + C₂H₅OH(l) ⇋ CH₃COOC₂H₅(l) + H₂O(l) ..... ∆H < 0

Adding a catalyst
Increasing the pressure
Removing water
Increasing the temperature

Answer

Remove from one side of an equilibrium and you pull the reaction towards that same side. If you remove one of the reacting components of the REVERSE reaction, then the rate of the reverse reaction becomes less than that of the forward reaction. The system then reacts faster in the forward direction, making more of the right hand side.

Q722-04 Which change will shift the position of equilibrium to the right in this reaction?

N₂(g) + 3H₂(g) ⇋ 2NH₃(g) ... ∆H = - 92 kJ

Increasing the temperature.
Decreasing the pressure.
Adding a catalyst.
Removing the ammonia from the equilibrium mixture.

Answer

The sign of ∆H is negative therefore it is exothermic in the forward direction. Increasing temperature favours the direction of ENDOthermic change ie to the left hand side in this case. II does not go to the right hand side.

Decreasing the pressure forces the equilibrium to move towards the side of greater moles of gas to re-establish the equilibrium concentrations.

Catalysts do NOT affect equilibria.

Removing ammonia disturbs the equilibrium concentrations and the reaction responds by making more ammonia to re-establish the equilibrium concentration proportions - this is the correct response.

Q722-05

2H₂O(l) ⇋ H₃O⁺(aq) + OH^-(aq)

The equilibrium constant for the reaction above is 1.0 x 10^-14 at 25ºC and 2.1 x 10^-14 at 35ºC. What can be concluded from this information?

[H₃O⁺] decreases as the temperature is raised
[H₃O⁺] is greater than [OH^-] at 35ºC
Water is a stronger electrolyte at 25ºC
The ionization of water is endothermic.

Answer

Kc has a larger value at the higher temperature. Thus the temperature increases and the proportion of products increases. Temperature increase favours the direction of endothermic change therefore this reaction is endothermic in the forward direction.

Q722-06 Which change increases the amount of NH₄⁺ in the below reaction?

NH₃(g) + H₂O(l) ⇋ NH₄⁺(aq) + OH^-(aq).... ∆H > 0

Decreasing the temperature
Decreasing the pressure
Removing water
Adding an acid

Answer

The sign of ∆H is positive (it is greater than 0) therefore it is endothermic in the forward direction. Increasing temperature favours the direction of ENDOthermic change ie to the right hand side in this case. Decreasing the temperature will move the equilibrium position to the left hand side.

Decreasing the pressure forces the equilibrium to move towards the side of greater moles of gas to re-establish the equilibrium concentrations, in this case the left (notice the state symbols)

Removing water disturbs the equilibrium concentrations and the reaction responds by making more water to re-establish the equilibrium concentration proportions. This pulls the reaction to the left to restore the value of Kc.

Adding acid reacts with the OH- ions from the right hand side disturbing the equilibrium proportions that are then re-established by the reaction moving to the right hand side making more NH₄⁺.

Q722-07 Consider the equilibrium reaction:

4NH₃(g) + 3O₂(g) ⇌ 2N₂(g) + 6H₂O(g) ΔH = -1268 kJ

Which change will cause the reaction to shift to the right?

Increase the temperature
Decrease the volume of the container.
Add a catalyst to speed up the reaction.
Remove the gaseous water by allowing it to react and be absorbed by KOH.

Answer

The reaction is exothermic in the forward direction, thus increasing the temperature drives the equilibrium in the reverse direction.

Decreasing the volume increases the pressure. The equilibrium has 7 moles of gas on the left and 8 moles of gas on the right, hence increasing the pressure drives the reaction to the left hand side.

Adding a catalyst has no effect on the position of equilibrium.

Removing the gasous water disturbs the equilibrium and the system must respond by making more water, i.e. by moving to the right to re-establish the equilibrium proportions.

Q722-08 Which changes will shift the position of equilibrium to the right in the following reaction?

2CO₂(g) ⇌ 2CO(g) + O₂(g)

I. adding a catalyst
II. decreasing the oxygen concentration
III. increasing the volume of the cylinder

I and II only
I and III only
II and III only
I, II and III

Answer

I - catalysts have no effect on equilibrium.

II - decreasing the oxygen concentration pulls the reaction to the right to re-establish the equilibrium proportions.

II - increasing the volume decreases the pressure. There are two moles of gas on the left and three moles of gas on the right, hence a decrease in pressure moves the reaction towards the side of the most moles (the right) to re-establish the equilibrium proportions.

Corrct response II and III only

Q722-09 The reaction below is an important step in the production of sulfuric acid. An increase in which of the following, will increase the ratio of SO₃(g) to SO₂(g) at equilibrium?

2SO₂(g) + O₂(g) ⇋ 2SO₃(g) ΔH = -197 kJ

Pressure only
Temperature
Both pressure and temperature
Neither pressure nor temperature

Answer

Increase pressure moves the reaction in the directin of fewer moles, in this case to the right. This increases the ratio of SO₃(g) to SO₂(g) at equilibrium. Increased temperature moves the position of equilibrium towards the side of endothermic change. in this case the left. This reduces the ratio of SO₃(g) to SO₂(g) at equilibrium.

Correct response A

Q722-10 The equation for a reaction used in the manufacture of nitric acid is:

4NH₃(g) + 5O₂(g) ⇋ 4NO(g) + 6H₂O(g) ΔH = -900 kJ

Which changes occur when the temperature of the reaction is increased?

	Position of equilibrium	Value of Kc
A.	shifts to the left	increases
B.	shifts to the left	decreases
C.	shifts to the right	increases
D.	shifts to the right	decreases

Answer

When the temperature is increased the position of equilibrium moves towards the direction of endothermic change, in this case it shifts to the left. The value of Kc decreases as there is less product (right hand side) and more reactants (left hand side) at equilibrium.

Correct response B

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