Worked example: Calculating the equilibrium total pressure after a change in volume | Khan Academy

Phosphorus pentachloride will decompose into phosphorus trichloride and chlorine gas. Kp for this reaction is equal to 0.500 at 500 Kelvin. Let's say that this reaction is at equilibrium in a reaction vessel that has a volume of 2.0 liters.

The equilibrium partial pressure of PCl5 is equal to 0.980 atmospheres. The equilibrium partial pressure of PCl3 is equal to 0.700 atmospheres, and the equilibrium partial pressure of chlorine gas is equal to 0.700 atmospheres. If we add up those three partial pressures, we get the total pressure of the gas mixture at equilibrium, which is equal to 2.38 atmospheres, and we're going to call this total pressure P1.

If we decrease the volume from 2 liters down to 1 liter and we keep the temperature constant at 500 Kelvin, we decrease the volume by a factor of 2, which means we increase the pressure by a factor of two. So, all the partial pressures of our gases double, and there's a new total pressure which is twice the original total pressure, and this new total pressure is equal to 4.76 atmospheres. From now on, we'll call this total pressure P2.

So, when the volume was 2 liters, the reaction started out at equilibrium, and by decreasing the volume to 1 liter and by doubling the pressure, we've introduced a stress to the system. At this moment in time, when these are the partial pressures, the reaction is not at equilibrium. Le Chatelier's principle says the net reaction will move in the direction that decreases the stress.

If the stress is an increase in the pressure, the net reaction will move in the direction that decreases the pressure. Looking at the balanced equation, there's one mole of gas on the reactant side and there are two moles of gas on the product side. So if the net reaction goes from the products to the reactants, if the net reaction goes to the left, the net reaction is going to the side with the smaller number of moles of gas, which would decrease the pressure and relieve the stress.

The net reaction keeps moving to the left until equilibrium is re-established. When equilibrium is re-established, there'll be a new total pressure which we'll call P3. Our goal is to calculate P3 so we can compare it to P1 and P2, and we're going to do this in a quantitative way and in a more qualitative way.

Let's use an ICE table to help us figure out the final total pressure P3. In an ICE table, I stands for the initial partial pressure, C is the change, and E is the equilibrium partial pressure. The initial partial pressure of PCl5 after the volume was reduced to 1 liter was 1.96 atmospheres, and the partial pressures of PCl3 and Cl2 were both 1.40 atmospheres.

We already used Le Chatelier's principle to realize the net reaction is going to go to the left, which means we're going to decrease in the amount of our products and we're going to increase the amount of the reactants. So if we're going to increase the amount of PCl5, we don't know how much we're going to increase; we're going to call that x. But we know it's going to increase, so we write plus x under the change part on the ICE table.

Since our mole ratio of PCl5 to PCl3 is one to one, if we're gaining x for PCl5, we must be losing x for PCl3, and the same goes for Cl2 since there's a coefficient of one. So we write minus x in our ICE table. Therefore, the equilibrium partial pressure of PCl5 would be 1.96 plus x, the equilibrium partial pressure of PCl3 would be 1.40 minus x, and the equilibrium partial pressure of Cl2 would also be 1.40 minus x.

Next, we can plug in the equilibrium partial pressures into our Kp expression, so we can plug in 1.40 minus x for the equilibrium partial pressure of PCl3, 1.40 minus x for the equilibrium partial pressure of Cl2, and 1.96 plus x for the equilibrium partial pressure of PCl5. We can also plug in the value for the equilibrium constant Kp.

Here's what our equilibrium constant expression looks like with everything plugged in. Next, we would need to solve for x, which would involve the use of a quadratic equation. When you do all that math, you find that x is equal to 0.330.

Now that we know x is equal to 0.33, we can solve for the equilibrium partial pressures. 1.96 plus 0.33 is equal to 2.29; therefore, the equilibrium partial pressure of PCl5 is 2.29 atmospheres. For PCl3, it's 1.40 minus x; 1.40 minus 0.33 is equal to 1.07. Therefore, the equilibrium partial pressure of PCl3 is equal to 1.07 atmospheres.

It's the same math for Cl2, so the equilibrium partial pressure of Cl2 is also 1.07 atmospheres. To find the total pressure P3, we simply need to add up the individual partial pressures: 2.29 plus 1.07 plus 1.07 is equal to 4.43 atmospheres; therefore, P3 is equal to 4.43 atmospheres.

Doing the math helps us realize that x is not a very large number, and the reason why x is not a very large number is because the Kp value is equal to 0.500 for this reaction. When K is close to 1, there's an appreciable amount of both reactants and products at equilibrium. We can see that with our equilibrium partial pressures; there's an appreciable amount of both of them.

Since there has to be a decent amount of both reactants and products at equilibrium, we're not going to see a huge change from these initial partial pressures. So, there will definitely be a shift to the left to decrease the pressure. When these were the partial pressures, remember P2 was equal to 4.76, so there's going to be a decrease of pressure.

There's going to be a decrease, so the pressure is going to go down from 4.76, but since there's not a large change, it's not going to be a huge change. That's why we saw P3 only drop to 4.43 atmospheres. So, if we go back to the original problem and our goal is to figure out P3 in relation to P1 and to P2 without doing all that math, we could think to ourselves, okay, so we decreased the volume by a factor of two which doubled the total pressure, but then the net reaction moves to the left to decrease the pressure.

Since it's not going to move much to the left, it's not going to decrease the pressure by a lot. Therefore, the final pressure P3 is going to be a little less than 4.76 but greater than 2.38 atmospheres.