yego.me
💡 Stop wasting time. Read Youtube instead of watch. Download Chrome Extension

Worked example: Using the reaction quotient to find equilibrium partial pressures | Khan Academy


4m read
·Nov 10, 2024

For the reaction of iron II 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 gases, 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 the net reaction will go to reach equilibrium.

So, let's calculate Qp. Qp is equal to... First, we think about our products, and we leave solids out of equilibrium expressions. Therefore, we also leave it out of our expression for Qp. So, we're going to leave out iron. We're going to include carbon dioxide since it's a gas. We're going to 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 raised to the first power.

Divided by... Next, we look at our reactants, and we have a solid, so we're going to leave that out. We have another gas, carbon monoxide. So this would be the partial pressure of carbon monoxide raised to the first power since there is 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. 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 ICE table for this reaction. I stands for the initial partial pressure in atmospheres. 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. 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 going to lose some of it, but we don't know how much, and therefore that's going to be represented by x.

So we're going to write minus x here for carbon dioxide, and since the coefficient is 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.

Since for Kp we're talking about the equilibrium partial pressures, we can take those directly from our ICE 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. Then we do some more algebra, and we get down to 1.26x is equal to 0.192. So, 0.192 divided by 1.26 is equal to 0.15, so x is equal to 0.15.

Now that we know that x is equal to 0.15, we can go back to our ICE table and solve for the equilibrium partial pressures. For carbon monoxide, the equilibrium partial pressure is 0.80 plus x, so that's equal to 0.80 plus 0.15, which is equal to 0.95 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 0.25 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 our 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 equilibrium partial pressures.

So this would be 0.25 atmospheres, which 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 0.26. 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.

More Articles

View All
Transitioning from counting to multiplying to find area | 3rd grade | Khan Academy
This square is one square unit. So, what is the area of rectangle A? The first thing we’re told is that each of these little squares equals one square unit, and then we’re asked to find the area of rectangle A. Here’s rectangle A, and area is the space th…
The Closer You Are to the Truth, the More Silent You Become Inside
One of the tweets that I put out a while back was: “The closer you get to the truth, the more silent you are inside.” We intuitively know this. When someone is blabbing too much, that person talks too much at the party—the court jester. You know they’re n…
The Changing World Order Has Just Begun | How To Prepare
What’s up, guys? It’s Graham here. So throughout the last week, there’s been a new topic gaining a lot of attention with over two and a half million views over these last few days. It has to do with the video posted by Ray Dalio titled Principles for Dea…
LESSONS FROM STOICISM TO STAY CALM | THE ART OF SERENITY REVEALED | STOICISM INSIGHTS
The art of temperance is the great mastery of choosing to resist rather than to respond. It is the ability to make deliberate decisions as opposed to impulsive ones. In the stoic state, along with wisdom, temperance is one of the four essential virtues. …
The Most Iconic TAG Heuer Watch of All Time | Monaco Split-Seconds Chronograph
Hey, Mr. Wonderful here, and I am in a magic zone! This is TAG. Now, this brand is legendary as a sports brand, obviously through racing, the association with racing, but it’s so much more now. And of late, for those of you that collect, we’ve expanded al…
Dipole–dipole forces | Intermolecular forces and properties | AP Chemistry | Khan Academy
So, I have these two molecules here: propane on the left and acetaldehyde here on the right. We’ve already calculated their molar masses for you, and you see that they have very close molar masses. Based on what you see in front of you, which of these do …