Le Châtelier's Principle

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 $$Q_c$$ is increased, making it greater than the Equilibrium Constant $$K_c$$. 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

$N_2 + 3H_2 \rightleftharpoons 2NH_3$

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.

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)

1. Adding products makes $$Q_c$$ larger than $$K_c$$. This creates a net change in the reverse direction, toward reactants. The opposite occurs when adding more reactants. (2)
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)
3. 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" ($$\Delta{H} > 0$$), then the reaction favors the formation of products at elevated temperature. Similarly, if temperature is a "product" ($$\Delta{H} > 0$$), 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

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.

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.

Example 3

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 colors 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.

Problems

1. Varying Concentration
What will happen to the equilibrium when more 2SO2 (g) is added to the system?

2SO2(g) + O2 (g) ↔ 2SO3 (g)

Solution:
Adding more reactants will shift the equilibrium in the direction of the products; therefore, the equilibrium will shift to the right.
Overall, the concentration of 2SO2 from initial equilibrium to final equilibrium will increase because only a portion of the added amount of 2SO2 will be consumed.
The concentration of O2 will decrease because as the equilibrium is reestablished, O2 is consumed with the 2SO2 to create more 2SO3. The concentration of 2SO3 will be greater because none of it is lost and more is being generated.

2. Varying Pressure
What will happen to the equilibrium when the volume of the system is decreased?

2SO2(g) + O2 (g) ↔ 2SO3 (g)

Solution:
Decreasing the volume leads to an increase in pressure which will cause the equilibrium to shift towards the side with fewer moles. In this example there are 3 moles on the reactant side and 2 moles on the product side, so the new equilibrium will shift towards the products (to the right).

3. Varying Temperature
What will happen to the equilibrium when the temperature of the system is decreased?

N2(g) + O2 (g) ↔ 2NO (g)     ΔH= 180.5 kJ

Solution
Since ΔH is positive, the reaction is endothermic in the forward direction. Removing heat from the system forces the equilibrium to shift towards the exothermic reaction, so the reverse reaction will occur and more reactants will be produced.

References

1. Pauling, L., College Chemistry, 3rd ed., Freeman, San Francisco, CA, 1964.
2. Petrucci, R., Harwood, W., Herring, F., Madura, J., General Chemistry, 9th ed., Pearson, New Jersey, 1993.
3. http://www.jce.divched.org/Journal/I...2N08/p1190.pdf

Contributors

00:37, 12 Feb 2014