# Le Chatelier's Principle

Henry Louis Le Châtelier, a French chemist in the late 19th to early 20th century, noticed the behavior of reversible reactions and their tenancy to reach equilibrium. We have discussed some of the mathematical relationships involving equilibrium and explored their behaviors quantitatively. Le Châtelier’s Principle (the Equilibrium Law) is a qualitative description of the behaviors arising from equilibrium.

When a chemical system at equilibrium is disturbed, the system will respond in such a way to return to equilibrium.

The above statement is the general concept at the foundation of Le Châtelier’s Principle.

## Changing Concentrations

Assume the following reaction is at equilibrium

$\mathrm{Zn}(s) + 2\mathrm{HCl}(aq) \rightleftharpoons \mathrm{ZnCl_2}(aq) + \mathrm{H_2}(aq) \qquad K = \dfrac{[\mathrm{ZnCl_2]}[\mathrm{H_2}]}{[\mathrm{HCl}]^2}$

I have included the equilibrium expression next to the reaction.

Let us disturb this equilibrium by adding HCl to this system. The concentration of HCl will increase and we will no longer be at equilibrium. You should be able to recognize that by adding HCl (a reactant), the denominator in the equilibrium expression will now be larger giving rise to a Q that is less than K! As previously discussed, if Q < K, the reaction will proceed to the right.
Therefore,

Adding reactant to a reaction at equilibrium will cause the reaction to proceed (or shift) to the right.

Conversely,

Adding product to a reaction at equilibrium will cause the reaction to proceed (or shift) to the left.

If we add reactant to a reaction at equilibrium, the reaction will respond in such a way to undo what we did. Here, if we add reactant, the reaction will try to undo that by removing reactant. The only way for a reaction to do this is by consuming reactant to produce product. If we add product to a reaction at equilibrium, the reaction will try to remove the added product by consuming it to make more reactant!

Removing reactant to a reaction at equilibrium will cause the reaction to proceed (or shift) to the left.

Removing product to a reaction at equilibrium will cause the reaction to proceed (or shift) to the right

## Changing Pressure

Consider the following reaction that involves gases

$4\mathrm{NH}_{3}(g) + 7\mathrm{O}_{2}(g) \rightleftharpoons 4\mathrm{NO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)$

We already can predict which way the reaction will shift if we add reactant or product. Now consider a case where this reaction is in a sealed, rigid vessel at equilibrium. If we reduce the volume of the vessel, the reaction mixture will be compressed. Which way will the reaction shift?

Compressing the mixture at a constant temperature reduces the total volume that the mixture occupies. If the volume decreases, the total pressure of the system will increase according to the Ideal Gas Law. How can the reaction respond in such a way to undo what we did? There is no way for the reaction to increase its volume given that it is stuck inside a sealed, rigid vessel. Therefore, the only stress that the reaction can try and counteract is the pressure.

Since the pressure increased due to compression, how can the reaction try to reduce the pressure? The reaction will shift towards the side of the equation that has the fewest number of moles of gas. Why does it do this? Consider the ideal gas law!

\begin{align*} PV = nRT \\[1ex] P = \dfrac{nRT}{V} \end{align*}

We know that the reaction cannot change its total volume, V, and the temperature, T, is constant. Therefore, the number of moles of gas, n, is the only thing that it can change. If we increase P, the reaction will try to reduce P by reducing the number of moles of gas, n. Therefore,

Compressing a gaseous reaction (at constant T) will drive the reaction to the side of the equation with the fewest number of moles of gas

Decompressing a gaseous reaction (at constant T) will drive the reaction to the side of the equation with the largest number of moles of gas

Another way to think about it is using Dalton’s law of partial pressures. If you compress the reaction vessel, the partial pressures of each gas will change differently and we will no longer be at K.

Finally, what if we were to add an inert gas to the gaseous reaction at equilibrium? (An inert gas is simply a gas that is not involved in the reaction.) Adding an inert gas will not cause the reaction to shift in either direction (i.e. nothing will happen). The reason for this is due to the fact that even though the total pressure of the reaction will increase, the partial pressures of each gas will remain the same. Therefore,

Adding an inert gas to a gaseous reaction at equilibrium will not affect the equilibrium

## Changing Temperature

Consider the following reaction

$\mathrm{SO_2}(g) + \dfrac{1}{2}\mathrm{O_2}(g) \rightleftharpoons \mathrm{SO_3}(g) \qquad \Delta H = -98.9~\mathrm{kJ~mol^{-1}}$

Here I have included the enthalpy of reaction. Since $$\Delta H$$ is negative, this reaction is exothermic; that is, heat is given off when the reaction is performed. Think about that. Heat ($$\Delta$$) is a product of the reaction. I can rewrite the equation like this

$\mathrm{SO_2}(g) + \dfrac{1}{2}\mathrm{O_2}(g) \rightleftharpoons \mathrm{SO_3}(g) + \Delta$

If we increase the temperature of this reaction at equilibrium, the reaction will respond in such a way to remove the added heat. Since heat is being treated as a product, we rationalize that the reaction will proceed to the left to consume the added heat.

Increasing the temperature of an exothermic reaction at equilibrium will drive the reaction to the left.

Decreasing the temperature of an exothermic reaction at equilibrium will drive the reaction to the right.

Finally, consider an endothermic reaction.

$\mathrm{H_2O}(g) \rightleftharpoons \mathrm{H_2}(g) + \dfrac{1}{2}\mathrm{O_2}(g) \qquad \Delta H = 248.1~\mathrm{kJ~mol^{-1}}$

Heat is a reactant (for an endothermic process to occur, heat must be absorbed). We can rewrite the above reaction as

$\Delta + \mathrm{H_2O}(g) \rightleftharpoons \mathrm{H_2}(g) + \dfrac{1}{2}\mathrm{O_2}(g)$

You should be able to rationalize the shift if temperature is increased and decreased.

Increasing the temperature of an endothermic reaction at equilibrium will drive the reaction to the right.

Decreasing the temperature of an exothermic reaction at equilibrium will drive the reaction to the left.

## Catalysts

Catalysts do not affect the equilibrium of a reaction.

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