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Main content
Current time:0:00Total duration:6:36
AP.Chem:
TRA‑7 (EU)
,
TRA‑7.D (LO)
,
TRA‑7.D.1 (EK)
,
TRA‑7.D.2 (EK)
,
TRA‑7.D.3 (EK)

Video transcript

- [Instructor] An equilibrium constant has one value for a particular reaction at a certain temperature. For example, for this reaction, we have oxalic acid turning into two H plus ions and the oxalate anion. The equilibrium constant, K C, for this reaction is equal to 3.8 times 10 to the negative six at 25 degrees Celsius. However, the value of the equilibrium constant changes if we change how we write the balanced equation. For example, let's reverse everything. So, instead of having oxalic acid on the reactant side, let's have it on the product side. And instead of having H plus ions in the product side and oxalate on the product side, let's put them on the reactant side. Since we've reversed the reaction, there's a new equilibrium constant. So we're going to write K C for the reverse reaction, is equal to, and to figure out the value for the new equilibrium constant, since we reversed everything, we take the inverse of the original equilibrium constant. So to find the new K C value, we take one over the original equilibrium constant of 3.8 times 10 to the negative six. And this is equal to 2.6 times 10 to the fifth, also at 25 degrees Celsius. Let's look at another example of how changing how we write the equation changes the value for the equilibrium constant. And we start by looking at the ionization of hydrofluoric acid, turning into an H plus ion and a fluoride anion. The K C value for this reaction is equal to 6.8 times 10 to the negative fourth at 25 degrees Celsius. This time we're going to multiply everything through by a factor of two. So it would be two H F turning into two H plus, plus two F minus. Our goal is to find the value for the new equilibrium constant K C. And since we multiplied everything through by a factor of two, we're going to take the old equilibrium constant and we're going to raise it to the second power. So this is K C is equal to 6.8 times 10 to the negative fourth. And we're going to square that. So this is equal to 4.6 times 10 to the negative seventh at 25 degrees Celsius. So since we multiplied through by a factor of two, we raised the equilibrium constant to the second power. If we had multiplied through by a factor of three, we would have raised the old equilibrium constant to the third power. Let's use the two reactions we've just looked at here. So we have oxalic acid turning into H plus and oxalate. And the other one we looked at was hydrofluoric acid ionizing to turn into H plus and F minus. Let's use those two reactions to calculate the equilibrium constant for a third reaction. And this third reaction shows oxalate reacting with hydrofluoric acid to form oxalic acid and the fluoride anion. Our first step is to get these first two reactions to look more like the third reaction. For example, we have oxalic acid on the reactant side, in the first reaction here, and oxalic acid is on the product side for the third reaction. Also notice, oxalate is on the product side for the first and over here for the third, it's on the reactant side. So we need to reverse the first reaction to make it look more like the third one. Here we have that reaction reversed. And since we reversed the reaction, we would have to take the inverse of the equilibrium constant, which we did in the earlier example. So let's go ahead and cross out this first one here, because we're not going to need it anymore. And let's move on to our next reaction and try to make it look more like the third one. So here we have hydrofluoric acid turning into H plus and F minus. And for our third reaction, we have hydrofluoric acid on the reactant side, but notice there is a coefficient of two in the balanced equation, and we have only a one in how we have it written. So we need to multiply everything through by a factor of two to give us two H F and two H plus and two F minus. And since we multiplied everything through by a factor of two, we would need to square the equilibrium constant for the original reaction. And once again, this is from our previous example. So we already know that equilibrium constant value. Next, let's add these two reactions together and make sure they give us the third reaction. So we have all of our reactants on one side and we have all of our products on the other side here. And we look for things that are the same on both the reactants and the products side. Well, there's two H plus on the left and there's two H plus on the right, so that would cancel out. And now, notice that we would have oxalate plus two H F going to oxalic acid plus two F minus. So adding our two reactions together does give us our third reaction. Once we've confirmed that adding our two reactions together gives us our third reaction, we can use the equilibrium constants for the first two reactions to figure out the equilibrium constant for the third. So the equilibrium constant for this first reaction we're gonna call K one, for the second reaction we're gonna call K two, and the equilibrium constant that we're trying to find for the third reaction we'll call K C. To find the equilibrium constant for the third reaction K C, we need to multiply the equilibrium constants for the first two reactions together. So K C is equal to K one times K two. We've already calculated the value for our K one from an earlier example, it was 2.6 times 10 to the fifth, and we also calculated K two, and it was 4.6 times 10 to the negative seven. So to find K C, the equilibrium constant for the third reaction, we simply multiply those two together and we get that K C is equal to point one two at 25 degrees Celsius. So whenever you add reactions together to get a new reaction, to find the equilibrium constant for the new reaction, you simply need to multiply the equilibrium constants of the reactions that you added together.
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