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

The Chatelier's principle says that if a stress is applied to a reaction mixture at equilibrium, the net reaction goes in the direction that relieves the stress. One possible stress that we could do is to change the volume on a reaction at equilibrium.

Let's say we have a hypothetical reaction where solid A, which is symbolized by red in our particular diagram, turns into solid B, which is symbolized by blue, and also C, which is a gas, and C will be symbolized by white spheres. The equilibrium constant Kc for this reaction is equal to 0.4 at 25 degrees Celsius. The first particular diagram shows the reaction at equilibrium, and we can see there's some solid A and some solid B at the bottom of the container, and there's also some gaseous particles of C.

Let's introduce a stress to our reaction mixture at equilibrium. Let's decrease the volume of the container. So looking from the first particular diagram to the second, we can see there's been a decrease in the volume of the container. That's going to cause an increase in the pressure because pressure comes from these gas particles slamming into the sides of the container. If we decrease the volume, now there's less distance for these particles to travel before they slam into the sides of the container, which means we increase the collision frequency and therefore, the pressure increases.

According to Le Chatelier's principle, the net reaction is going to go in the direction that relieves the stress. So if the stress is increased pressure, the net reaction is going to try to move in the direction that decreases the pressure. Looking at the equation for this hypothetical reaction, the two solids aren't really contributing to the pressure, so it's only gas C that we need to worry about. There is one mole of gas C on the product side, and there are zero moles of gas on the reactant side.

So if the net reaction went to the right, we'd be going from zero moles of gas to one mole of gas. Going to the right would increase the moles of gas, which would increase the pressure. However, that's not what the reaction wants to do. The goal of the reaction is to relieve the stress, and therefore, the reaction wants to decrease the pressure. So the reaction is going to go to the left to get rid of some of that gas.

Decreasing the amount of gas will decrease the pressure; therefore, we're relieving the stress. If the net reaction moves to the left, we're going to lose some of our products. So we're going to decrease the amount of C, and we're going to decrease the amount of B, and we're going to gain some of our reactants. So we're going to increase the amount of A.

We can see all that in the third particulate diagram. So going from the second particular diagram to the third particular diagram, we've decreased the amount of C. We've gone from four particles of C to only two particles of C. We've also decreased the amount of B; you can see that this blue solid has gotten smaller, and we've increased the amount of A. So A is this red sphere here, and you can see how it's gotten bigger from the second particulate diagram to the third.

By going from four particles of C to only two particles of C, we've decreased the amount of the gas, and therefore, we've decreased the pressure. To better understand what happens to a reaction mixture at equilibrium when a stress is placed on it, let's calculate the reaction quotient Q for these three particular diagrams for the same hypothetical reaction.

The expression for the reaction quotient Qc has the same form as the equilibrium constant expression for Kc. Since solids are left out of the equilibrium constant expression, we only need to include the concentration of the gas. Since there's a coefficient of 1 in front of C, Qc is equal to the concentration of C raised to the first power. Since there are four particles of C in the first particulate diagram, and if each particle represents 0.1 moles of C, 4 times 0.1 is equal to 0.4 moles of C.

To find the concentration of C, we take the moles and divide that by the volume of the container, which is 1.0 liter. So, 0.4 divided by 1.0 is equal to 0.4 molar. So that's the concentration of C in the first particular diagram. We plug that into our expression for Q, so Qc is equal to 0.4. Notice that Kc is also equal to 0.4, so at this moment in time Qc is equal to Kc, which tells us the reaction is at equilibrium.

Next, we think about the stress that was applied to the reaction at equilibrium. We decreased the volume, and if we look at the volumes here, we're going from 1.0 liters to 0.5 liters, so we're decreasing the volume by a factor of 2, which would cause an increase in the pressure by a factor of 2. Changing the volume would change the concentration, so instead of 0.4 divided by 1.0, it would be 0.4 divided by 0.5, which is equal to 0.8 molar.

So the concentration has doubled. If we calculate Q for our second particulate diagram, we plug in the concentration of C, which is 0.8 molar, so Qc is equal to 0.8. K is equal to 0.4, so Q is not equal to K. Therefore, the reaction is not at equilibrium for our second particulate diagram.

Let me write in here: not at equilibrium. In this case, Qc is greater than Kc, which tells us we have too many products and not enough reactants; therefore, the net reaction goes to the left to get rid of some of the products and to increase the amount of reactants. The net reaction keeps going to the left until we reach equilibrium again.

If we calculate the concentration of C in the third particular diagram, here there are only two particles, so that would be 0.2 moles divided by a volume of 0.5 liters. So, 0.2 divided by 0.5 is equal to 0.4 molar. Thus, Qc is equal to 0.4. Since Qc is equal to Kc, which is still 0.4, Qc is equal to Kc, and we're at equilibrium.

So equilibrium has been re-established. Since the reaction is at equilibrium in the third particular diagram, the net reaction stops going to the left, and the concentration of C remains constant.

Let's apply what we've learned to another reaction: the synthesis of ammonia from nitrogen gas and hydrogen gas. If we have a mixture of these gases at equilibrium and we introduce the stress to the system, like a decrease in volume, the decrease in the volume of the container would cause an increase in the pressure. According to Le Chatelier's principle, the net reaction is going to go in the direction that relieves the stress that was placed on the system.

So if the stress is increased pressure, the net reaction says, “I want to move in the direction that decreases that pressure.” So the net reaction moves to the right because there are four moles of gas on the left and only two moles of gas on the right. By moving to the right, that goes from 4 moles of gas to 2 moles of gas, which decreases the amount of gas and causes a decrease in the pressure.

If we had a mixture of these gases at equilibrium and we increased the volume, an increase in the volume would cause a decrease in the pressure. So the stress, this time, is decreased pressure. To relieve the stress, the net reaction wants to move in the direction that increases the pressure. Therefore, the net reaction is going to move to the left because if it moves to the left, we're going from two moles of gas on the right to four moles of gas on the left. So that's an increase in the moles of gas, and an increase in the amount of gas causes an increase in the pressure.

Now, let's see what happens when we have equal amounts of moles of gas on both sides of the equation. For example, for the hypothetical reaction where gas A turns into gas B, there's one mole of gas on the reactant side, and there's one mole of gas on the product side. Let's use particulate diagrams and reaction quotients to understand what's going on here.

So for our first particular diagram, here's the Qc expression: it's equal to the concentration of B to the first power divided by the concentration of A to the first power. The concentration of B, since B is blue, there's two blue spheres in this first particulate diagram. So, 2 times 0.1 moles is 0.2 moles of blue divided by a volume of 1 liter. Therefore, the concentration of B is equal to 0.2.

The concentration of A would also be equal to 0.2 molar because there's two particles of A in here. So, 0.2 divided by 0.2 is equal to 1, and since K is equal to 1 for this reaction, Kc is equal to 1 at 25 degrees Celsius. Qc is equal to Kc at this moment in time; therefore, the reaction is at equilibrium in this first particular diagram.

If we introduce a stress to our system at equilibrium, let's decrease the volume by a factor of two. So we're going from one liter to 0.5 liters. That's going to increase the pressure by a factor of 2, and it's also going to double the concentrations. So the concentration of B is now 0.4 molar, and the concentration of A is also 0.4 molar. So, 0.4 divided by 0.4 is equal to 1.

Qc is still equal to Kc; the reaction is still at equilibrium. Since changing the volume didn't change the value for Q, Qc is still equal to Kc. Therefore, changing the volume on a reaction at equilibrium when there are equal amounts of moles of gas on both sides has no effect on the composition of the equilibrium mixture.

Let's go back to the reaction for the synthesis of ammonia gas from nitrogen gas and hydrogen gas, and let's say this reaction is at equilibrium, and we add in some helium gas to the reaction mixture at equilibrium. Helium is an inert gas, which means it doesn't react with any of the gases that we have present.

It's tempting to say adding this inert gas would increase the total pressure and therefore the net reaction would shift to the right to get rid of that increased pressure. However, notice that helium is not a part of the expression for Qc, and therefore after you add it, you're not actually changing any of these concentrations. So, Qc is still equal to Kc after the addition of helium, and since Qc is equal to Kc, the reaction is still at equilibrium, and there's no shift either direction.

Therefore, adding an inert gas to a reaction mixture at equilibrium has no effect on the composition of the reaction mixture.