Le Chȃtelier’s principle: Changing concentration | Equilibrium | AP Chemistry | Khan Academy

Le Chatelier's principle says if a stress is applied to a reaction mixture at equilibrium, the net reaction goes in the direction that relieves the stress. Changing the concentration of a reactant or product is one way to place a stress on a reaction at equilibrium.

For example, let's consider the hypothetical reaction where gas A turns into gas B. Let's say the reaction is at equilibrium, and we suddenly introduce a stress, such as we increase the concentration of reactant A. According to Le Chatelier's principle, the net reaction is going to go in the direction that relieves the stress. Since we increased the concentration of A, the net reaction is going to go to the right to decrease the concentration of A.

Let's use some particulate diagrams so we can get into the details of how the reaction goes to the right. We're going to symbolize gas A by red particles and gas B by blue particles. For this hypothetical reaction, the equilibrium constant is equal to 3 at 25 degrees Celsius. Let's start by writing the reaction quotient.

Qc is equal to (the concentration of B) to the first power divided by (the concentration of A) also to the first power. Let's calculate the concentrations of B and A from our first particulate diagram. So B is represented by the blue spheres, and there are three blue spheres.

If each particle represents 0.1 moles of a substance and the volume of the container is one liter, since we have three particles, that'll be three times 0.1, which is 0.3 moles. Divided by a volume of 1 liter, this gives us 0.3 molar, so the concentration of B is 0.3 molar. For A, we have one particle, so that's 0.1 mole divided by 1 liter, which is 0.1 molar. Therefore, the concentration of A is 0.1 molar.

Now, 0.3 divided by 0.1 is equal to 3. So Qc at this moment in time is equal to 3. Notice we could have just counted our particles: three blues and one red, and said 3 over 1. That would have been a little bit faster. So Qc is equal to 3, and Kc is also equal to 3.

When Qc is equal to Kc, the reaction is at equilibrium. In this first particulate diagram, where Qc is equal to Kc, the reaction's at equilibrium. Next, we're going to introduce a stress to our reaction at equilibrium. We're going to increase the concentration of A.

Here, we're going to add four particles of A to the reaction mixture at equilibrium. The second particulate diagram shows what the reaction looks like right after we add those four red particles. We start with one red particle and we added four, so now there's a total of five red particles. We still have the same three blue particles that we had in the first particulate diagram.

Let's calculate Qc at this moment in time. Just after we introduce the stress, since there are three blue particles and five red particles, Qc is equal to 3 divided by 5, which is equal to 0.6. Since Qc is equal to 0.6 and Kc is equal to 3 at this moment in time, Qc is less than Kc. So there are too many reactants and not enough products. Therefore, the net reaction is going to go to the right, and we're going to decrease in the amount of A and increase in the amount of B.

The third particulate diagram shows what happens after the net reaction moves to the right. We said we're going to decrease the amount of A and increase the amount of B. We're going from three blues in the second particulate diagram to six blues in the third, and we're going from five reds to only two reds. Therefore, three reds must have turned into blues to get the third particulate diagram on the right.

If we calculate Qc for our third particulate diagram, it would be equal to 6 divided by 2, which is equal to 3. So at this moment in time, Qc is equal to Kc; they're both equal to 3. Thus, equilibrium has been re-established in the third particulate diagram.

It isn't always necessary to calculate Q values when doing a Le Chatelier changing concentration problem. However, for this hypothetical reaction, it's useful to calculate Q values to understand that we're starting at equilibrium and then a stress is introduced, such as changing the concentration of a reactant or product. This means the reaction is no longer at equilibrium.

Le Chatelier's principle allows us to predict which direction the net reaction will go. We could also use Q to predict the direction of the net reaction. The net reaction will continue going in that new direction until Q is equal to K again, and equilibrium has been re-established.

Let's look at another reaction. This is the synthesis of ammonia from nitrogen gas and hydrogen gas. Let's say the reaction is at equilibrium.

Let's also look at this on a graph of concentration versus time. At equilibrium, the concentrations of reactants and products are constant, which is why we see these straight lines here for the concentration of hydrogen, ammonia, and nitrogen. Let's introduce a stress to the system at equilibrium.

Right now, we are at equilibrium, and all the concentrations are constant. Let's increase the concentration of hydrogen, so we can see that on our graph. There is a sudden increase in the concentration of hydrogen. Adding hydrogen means that Q is no longer equal to K, and therefore the reaction is not at equilibrium.

Now we're not at equilibrium, and Le Chatelier's principle allows us to predict which direction the net reaction will move. Since we added a stress, the stress being the increased concentration of hydrogen, the net reaction is going to move to the right to get rid of some of that hydrogen that was added.

When the reaction goes to the right, the amount of ammonia will increase. That's what we see in the red line: the amount of ammonia is increasing. The amount of ammonia increases because nitrogen and hydrogen are reacting to form ammonia. Therefore, the amount of nitrogen and hydrogen will decrease.

Here we can see the amount of hydrogen is decreasing, and down here we can see the amount of nitrogen is decreasing. The reaction will continue to go to the right until equilibrium is re-established. That happens at the second dotted line, and we know that because we can see all of these concentrations are now constant, so the reaction has reached equilibrium.

So far, we've only talked about changing the concentration of a reactant. For example, if we increase the concentration of hydrogen, the net reaction goes to the right. We could also say it shifts to the right.

So for a reaction at equilibrium, if you increase the concentration of reactants, such as the concentration of hydrogen or the concentration of nitrogen, the reaction will shift to the right to decrease the amount of one of those reactants.

If our reaction is at equilibrium and we were to increase the amount of our product to increase the amount of ammonia, this time the stress places increased concentration of a product. Le Chatelier's principle says the net reaction is going to move in the direction that decreases the stress.

In this case, the net reaction would go to the left to decrease the amount of ammonia. If our reaction is at equilibrium and we were to decrease the concentration of our product, the net reaction would shift to the right to make more of the product.

Or if we decrease the concentration of one of our reactants, let's say of nitrogen in this case, the reaction will shift to the left to make more of our reactant.