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# Le Chȃtelier’s principle: Changing concentration

AP.Chem:
TRA‑7 (EU)
,
TRA‑7.F (LO)
,
TRA‑7.F.1 (EK)
,
TRA‑8 (EU)
,
TRA‑8.A (LO)
,
TRA‑8.A.1 (EK)
,
TRA‑8.B (LO)
,
TRA‑8.B.1 (EK)

## Video transcript

- [Instructor] 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. Change in 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. And 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 gonna go in the direction that relieves the stress. And since we increase the concentration of A, the net reaction is gonna go to the right to decrease the concentration of A. Let's use some particular diagrams so we can get into the details of how the reaction goes to the right. So we're gonna symbolize gas A by red particles and gas B by blue particles. And for this hypothetical reaction, the equilibrium constant is equal to three at 25 degrees Celsius. Let's start by writing the reaction quotient. Qc is equal to, and we get that from our balanced equation. That would be 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 particular 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 litter, since we have three particles, that'd be three times 0.1, which is 0.3 moles divided by a volume of one litter is 0.3 molar.. So the concentration of B is 0.3 molar. For A, we have one particles, that's 0.1 mole divided by one litter, which is 0.1 molar. So the concentration of A is 0.1 molar. And 0.3 divided by 0.1 is equal to three. So Qc at this moment in time is equal to three. Notice we could have just counted our particles, three blues and one red and said three over one. That would have been a little bit faster. So Qc is equal to three and Kc is also equal to three. So I should have written a C in here. So when Qc is equal to Kc, the reaction is at equilibrium. So in this first particular diagram here where Qc is equal to Kc, the reactions are at equilibrium. Next, we're gonna introduce a stress to our reaction at equilibrium. We're going to increase the concentration of A. So here, we're gonna 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. So we started with one red particle and we added four. So now there's a total of five red particles. And we still have the same three blue particles that we had in the first particular diagram. Let's calculate Qc at this moment in time. So just after we introduced the stress. Since there are three blue particles and five red particles, Qc is equal to three divided by five, which is equal to 0.6. Since Qc is equal to 0.6, and Kc is equal to three, 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 we're gonna increase in the amount of B. The third particular diagram shows what happens after the net reaction moves to the right. So we said, we're gonna decrease the amount of A and increase in the amount of B. We're going from three blues in the second particular 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 particular diagram on the right. And if we calculate Qc for our third particular diagram, it'd be equal to six divided by two, which is equal to three. So at this moment in time, Qc is equal to Kc. They're both equal to three. So equilibrium has been reestablished in the third particular diagram. It isn't always necessary to calculate Q values when doing a Le Chatelier's 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 reaction or product. And that means the reaction is no longer at equilibrium. Le Chatelier's principle allows us to predict which direction the net reaction will go or 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 reestablished, let's look at another reaction. This is the synthesis of ammonia from nitrogen gas and hydrogen gas. And let's see the reaction is at equilibrium. So 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. And let's introduce a stress to the system at equilibrium. So right now we are at equilibrium and all the concentrations are constant. And let's increase the concentration of hydrogen. So we can see that on our graph. So right here, there's 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. So let's go ahead and write over here, Now we're not at equilibrium. And Le Chatelier's principle allows us to predict which direction the net reaction will move. So since we added a stress, the stress being increased concentration of hydrogen, the net reaction is gonna move to the right to get rid of some of that hydrogen that was added. And when the reaction goes to the right, the amount of ammonia will increase. And that's what we can see right here on this red line here, the amount of ammonia is increasing. And 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 reestablished. And that happens at the second dotted line here. 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 change in the concentration of a reactant so for example, if we increase the concentration of hydrogen, the net reaction goes to the right. We could also say 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. And if our reaction is at equilibrium and we were to increase the amount of our product, increase the amount of ammonia, so this time the stress place is increased concentration of a product, Le Chatelier's principle says the net reaction is going to move in the direction that decreases the stress. So in this case, the net reaction would go to the left to decrease the amount of ammonia. And 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.
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