Main content

## AP®︎/College Chemistry

### Unit 7: Lesson 6

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

AP.Chem:

TRA‑8 (EU)

, TRA‑8.B (LO)

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.

## Want to join the conversation?

- I dont see the point of comparing the reaction quotient with the equilibrium pressure, cant you just use an ICE table assuming +x on reactant side and -x on product side, and when you solve for x the signs will balance out to get the equilibrium partial pressures?(1 vote)
- We need to know the reaction quotient because we need to know whether the production of reactants or products is favored. Essentially we need to know if the reactants/products are increasing or decreasing. Knowing the reaction quotient allows us to know reactant concentrations will increase as opposed to them decreasing. This ultimately gives us the correct information for the ICE table.

Hope that helps.(1 vote)

- Sorry to ask something completely unrelated to chemistry, but at4:34shouldn't it be -0.192? Because 0.40 - 0.208 = -0.192.(0 votes)
- 0.40 - 0.208 is the same as 0.40 + (-0.208). And when adding positive numbers to negative numbers then the larger of the two determines the sign of the answer. 0.40 > 0.208, so the answer should be positive 0.192

Hope that helps.(2 votes)

## 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.