Introduction

Le Châtelier’s principle, despite its intimidating name, is actually a very simple idea and quite important to AP® Chemistry. It simply means that if a reaction at equilibrium is disturbed in a way that would shift the balance to one side, the reaction will move to restore the equilibrium.

It’s sort of like if you were on a balancing beam with your arms stretched out at your sides, and your friend suddenly tied a weight to your right arm. Two things would happen: first, you would bitterly curse your ex-friend, and second, you would shift your weight over to your left side to counteract the weight on your right arm, thus restoring equilibrium.

Now, in the context of chemical reactions, what kind of disturbances can happen? Essentially, we can change concentration/amount of reactants, pressure/volume, and temperature. We’ll go over each of them in turn in this section of the AP® Chemistry Crash Course.

Disturbance 1: Concentration/Amount of Reactants

Fortunately, this is the simplest of the three to understand. In fact, the balancing beam analogy was an example of this disturbance: if we view your right side as reactants and your left side as products, then your friend simply added more reactants, causing you to shift some of your reactants into products.

On a more concrete level, let’s look at the Haber Process. Here is the equation:

N_2 + 3H_2 \rightleftharpoons 2NH_3

If we increase the amount of hydrogen, we will cause more ammonia (NH_3) to be produced, according to Le Châtelier’s principle. Of course, this will also mean a decrease in nitrogen.

If we instead add ammonia, the reverse will be true. More nitrogen and hydrogen will be produced.

On the flip side, you can also cause a shift by taking product or reactant away from the reaction vessel. If you remove ammonia from the system, equilibrium will shift right (toward the products) and result in more ammonia being formed.

Disturbance 2: Pressure/Volume

Those two terms are combined because, of course, the only way to increase pressure while keeping the amount of gas and temperature constant is to reduce the volume, as specified by the ideal gas law.

Going back to our Haber Process:

N_2 + 3H_2 \rightleftharpoons 2NH_3

Let’s say we have this nitrogen-hydrogen-ammonia mixture in a box, and we squeeze the box to increase the pressure. Now, pressure is in itself a component of the equilibrium, so it wants to get back to equilibrium, according to Le Châtelier’s principle. How can we reduce the pressure?

Well, if we look at the equation, we see that there are 4 \text{ moles} of gas on the left side and only 2 \text{ moles} of gas on the right side. So, in the extreme cases, if the mixture were all reactants, it would be under twice as much pressure as it would be if it were all products.

So, if we increase the pressure, the equilibrium will shift to relieve the pressure; the equilibrium will shift to the right, and we’ll have more products.

An increase in pressure causes the reaction to shift toward the side with the fewest moles of gas. Be careful not to include moles of solid or liquid in your determination.

Does that make sense? Pressure is probably the hardest to understand, so if you’re struggling, take some extra time. And if you’re not struggling, sit back and relax because temperature is actually pretty simple.

Disturbance 3: Temperature

Guess what? We’ll be using the Haber Process again. However, there’s a new term to add to the reaction: heat. The reaction is exothermic, which means that heat is produced by the reaction. The exact amount of heat is irrelevant for this part of the discussion, so we’ll just write heat.

N_2 + 3H_2 \rightleftharpoons 2NH_3 + \text{heat}

How can we use this additional piece to disturb the equilibrium? Well, heat has essentially become one of our products. So, taking our lesson from the first disturbance method, adding more of a product will drive the equilibrium to the left; taking away some of the product will drive the equilibrium to the right.

So, if we want to get more ammonia, we can simply cool down the container we’re using to store the mixture. This cooling will remove the heat, one of the products, and thus will cause more ammonia and heat to be produced, in accordance with Le Châtelier’s principle.

On the other hand, if we instead want to break down ammonia to make nitrogen and hydrogen, we need to increase the ambient temperature. Then we’ll have more products and thus we can drive the equilibrium to the left.

Got all the disturbances down? Are you sure? Let’s test your knowledge.

Application of Le Châtelier’s Principle: Increasing Yield

We’re going to switch to a different reaction just to make this more interesting.

H_2 + I_2 \rightleftharpoons 2HI + \text{heat}

Let’s say that this reaction is currently at equilibrium, and you have complete control over the container. Now we tell you that you need more hydrogen iodide. What can you do to obtain more HI?

Let’s go through each disturbance in order and see if any of them will help you.

Disturbance 1: This is the obvious choice. Just add more hydrogen and iodine. The added reactants will push the equilibrium to the right, and we’ll get the hydrogen iodide. Yay! So we’re done…not so fast!

It turns out we don’t have any more hydrogen and iodine. You’re going to have to think of something else.

Disturbance 1 (again): Well, if you can’t add hydrogen and iodine, you can at least take out the hydrogen iodide that has formed and give it to us immediately. It’ll start going back to equilibrium immediately, but that’s our problem, not yours. And once the hydrogen iodide is out of the container, more will be produced.

But no. It turns out we lied about the “complete control” thing. You’re not allowed to open the container or tamper with its contents directly. You have to increase the amount of hydrogen iodide some other way.

Disturbance 2: We could still crush the box to increase the pressure, as long as we’re very careful not to open it. But wait, will that actually help? Remember that pressure only shifts the equilibrium if the reaction has some way of relieving the pressure. In this particular reaction, there are two moles of gas on each side. Crushing the box won’t help us at all.

Disturbance 3: Our last hope. Fortunately, there’s heat on one side of the equation, so we know this can work. Heat is a product, so to shift the equilibrium to the left, you need to remove heat. So, you can place the whole container in the freezer.

Hooray! That works! Now you know the basic idea of Le Châtelier’s principle, the three major kinds of disturbances to equilibrium, and how to use those disturbances to increase the yield of a reaction. Good luck on the AP® Chem exam!

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