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## AP®︎/College Chemistry

### Unit 7: Lesson 6

Using the reaction quotient

# Worked example: Using the reaction quotient to find equilibrium partial pressures

In some equilibrium problems, we first need to use the reaction quotient to predict the direction a reaction will proceed to reach equilibrium. Once we know this, we can build an ICE table, which we can then use to calculate the concentrations or partial pressures of the reaction species at equilibrium. Created by Jay.

## Video transcript

- [Tutor] For the reaction of iron two oxide plus carbon monoxide goes to solid iron and carbon dioxide. The equilibrium constant Kp is equal to 0.26 at 1000 Kelvin. Our goal is to find the equilibrium partial pressures of our two gasses, carbon monoxide and carbon dioxide. If the initial partial pressures are 0.80 atmospheres for carbon monoxide and 0.40 atmospheres for carbon dioxide, we can use the reaction quotient Q, to predict which direction that reaction will go to reach equilibrium. So let's calculate Qp and Qp is equal to, first we think about our products and we leave solids out of equilibrium expressions, and therefore we also leave it out of our expression for Qp. So we're gonna leave out iron. We're going to include carbon dioxide since it's a gas. So we're gonna write the partial pressure of carbon dioxide, and since we have a coefficient of one in front of carbon dioxide, it's the partial pressure raise to the first power divided by, next we look at our reactants, and we have a solid, so we're gonna leave that out. So we have another gas, carbon monoxide. So this will be the partial pressure of carbon monoxide raised to the first power, since there's also a coefficient of one. Next, we plug in our partial pressures at this moment in time. The partial pressure of carbon dioxide is 0.40, and the partial pressure of carbon monoxide is 0.80 atmospheres. So we plug those into our expression for Qp and 0.40 divided by 0.80 is equal to 0.50. So Qp at this moment in time is equal to 0.50. Since QP is not equal to Kp at this moment in time, the reaction is not at equilibrium. So Qp is equal to 0.50 and Kp is equal to 0.26. So Qp is greater than Kp. And when Qp is greater than Kp, there are too many products and not enough reactants. Therefore the net reaction is going to move to the left. Next let's fill out our I.C.E table for this reaction. I stands for the initial partial pressure in atmospheres. And so the initial partial pressure of carbon monoxide was 0.80 atmospheres. And the initial partial pressure of carbon dioxide is 0.40 atmospheres. C stands for change. And E stands for the equilibrium partial pressure. Calculating Qp allowed us to realize that the net reaction moves to the left. And if the net reaction moves to the left, we're going to lose some carbon dioxide and we're going to gain some carbon monoxide. So first let's think about carbon dioxide. We're gonna lose some of it, but we don't know how much and therefore that's gonna be represented by X. So we're gonna write minus X here for carbon dioxide. And since the coefficient is a one in front of carbon dioxide, and it's also one in front of carbon monoxide, if we lose X for carbon dioxide, we're going to gain X for carbon monoxide. Therefore the equilibrium partial pressure for carbon monoxide would be 0.80 plus X. And the equilibrium partial pressure for carbon dioxide would be 0.40 minus X. Our next step is to write an equilibrium constant expression for this reaction. So Kp is equal to the partial pressure of carbon dioxide divided by the partial pressure of carbon monoxide. So the expressions for Kp and Qp look the same, but the difference is for Kp, it would be the equilibrium partial pressures only. Whereas for Qp, it's the partial pressures at any moment in time. And since for Kp, we're talking about the equilibrium partial pressures, we can take those directly from our I.C.E table and plug them in. So we can plug in the equilibrium partial pressure of CO2 and the equilibrium partial pressure of carbon monoxide. Here we can see our two equilibrium partial pressures plugged into our Kp expression and also the equilibrium constant Kp is equal to 0.26 for this reaction, so that's plugged in as well. Our next step is to solve for X. So we multiply both sides by 0.80 plus X, and we get this and then we do some more algebra and we get down to 1.26 X is equal to 0.192. So 0.192 divided by 1.26 is equal to 0.15. So X is equal to 0. 1 5. Now that we know that X is equal to 0.15, we can go back to our I.C.E table and solve for the equilibrium partial pressures. So for carbon monoxide, the equilibrium partial pressure is 0.80 plus X. So that's equal to 0.80 plus 0. 1 5, which is equal to 0. 9 5 atmospheres. So that's the equilibrium partial pressure of carbon monoxide. For carbon dioxide, the equilibrium partial pressure is 0.40 minus X. So 0.40 minus 0.15 is equal to 0. 2 5 atmospheres. So that's the equilibrium partial pressure for carbon dioxide. Finally, we can use the reaction quotient Qp to make sure that these two answers, for equilibrium partial pressures are correct. So we can write that Qp is equal to the partial pressure of CO2 divided by the partial pressure of CO. And we can plug in those, those equilibrium partial pressures. So this would be 0.25 atmospheres, was the equilibrium partial pressure of carbon dioxide and 0.95 was the equilibrium partial pressure of carbon monoxide. And 0.25 divided by 0.95 is equal to 0. 2 6. And since Kp is also equal to 0.26 at this moment in time, Qp is equal to Kp and the reaction is at equilibrium. Therefore we know we have the correct equilibrium partial pressures.