If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content
Current time:0:00Total duration:9:22

Video transcript

Let's say we wanted to figure out the equilibrium constant for the reaction boron trifluoride in the gaseous plus 3-- so for every mole of this, we're going to have 3 moles of H2O in the liquid state-- and that's in equilibrium. It's going forward and backwards with 3 moles of hydrofluoric acid, so it's in the aqueous state. It's been dissolved in the water. If it wasn't dissolved, if it was in the solid state, you would call this hydrogen fluoride. Once it's in water, you call it hydrofluoric acid, and we'll talk more about naming in the future, hopefully. Plus 1 mole of boric acid, also in the aqueous state. It's dissolved in the water. H3BO3 in the aqueous state. So what would the expression for the equilibrium constant look like in this situation? So you might be tempted say, OK, that's easy enough, Sal. So the equilibrium constant, you just take the right-hand side. That's just the convention. There's symmetry here. I could've rewritten it either way, but let's just say you take the right-hand side and say, OK, this is dependent on the concentration of the hydrofluoric acid, the concentration of the HF, or the molarity of the HF, to the third power, times the concentration of the boric acid, H3BO3. And remember, this intuition of why you're taking this to the third power is what's the probability-- because in order for the reaction to go this way, you need to have 3 molecules of hydrofluoric acid being very close to 1 molecule of the boric acid. So if you watched the last video I just made about the intuition behind the equilibrium constant, this is indicative of the probability of this reaction happening or the probability of finding all of these molecules in the same place. Of course, you can adjust it with a constant and that's essentially what that does. But that's on the product side, or the reactant, depending on what direction you're viewing this equation, divided by the molarity of the boron trifluoride times-- and I'll do this in a different color-- the molarity of the H2O to the third power. And that's, of course, the H2O liquid. So there you go. We'll just figure this out. And my rebuttal to you is I want you to figure out the molarity of the water. What is the concentration of the water? Remember, the concentration is moles per volume, but in this case, what's happening? I'm putting some boron trifluoride gas essentially into some water, and it's creating these aqueous acids. These other molecules are dissolved completely in the water. So what's the solvent here? The solvent is H2O. This might be how the reaction happens, but pretty much, there's water everywhere. The water is in surplus. So if you were to really figure out the concentration of water, it's everywhere. I mean, you could say everything but the boron trifluoride, but it's a very high number. And if you think about it from the probability point of view, if you say, OK, in order for this reaction to happen forward, I need to figure out the probability of finding a boron trifluoride atom or molecule-- actually, molecule-- in a certain volume, and it also needs 3 moles of water in that certain volume. But you say, hey, there's water everywhere. This is the solvent. There's water everywhere, so I really just need to worry about the concentration of the boron trifluoride. So you could say the forward reaction rate, rate forward, is going to be dependent on some forward constant times just the concentration of the boron trifluoride. The water's everywhere, so you don't have to multiply it times the concentration of water, whatever that means, because the water's everywhere. So the denominator here, you do not put the solvent. So the correct answer for this one is you only put whatever is actually dissolved in the solution. Because frankly, the concentration doesn't actually makes sense for everything else, and if you think about it from the probability point of view, that also makes sense, because there's always water around. If you said, OK, what's the probability of finding water at any small volume of our fluid, it's going to be 1, so you could just multiply it by a 1 there, but that doesn't make a difference. Now, what about the following reaction? Any equilibrium where you have different states of matter is called a heterogeneous equilibrium. And so let me write another heterogeneous equilibrium. So let's say I have H2O in the gaseous state and that's essentially steam-- so it's not going to be the solvent this time-- plus carbon in the solid state. And let's say that that's an equilibrium with hydrogen in the gas state plus carbon dioxide in the gaseous state. This is a heterogeneous equilibrium because you have things in the gaseous and the solid state. And solid state, by definition, it can't be dissolved either into the gas or into the-- when we talk about solutions, we talked about colloids and suspensions and mixtures before, but we're talking about solutions. By definition, if this is in the solid state, it's not dissolved. If this was dissolved, we would write an aq here. It would be the aqueous state. So if you talk about the forward reaction, what's the forward reaction going to be dependent on? So the rate forward, well, the solid, there's a big block of carbon sitting there. There's a big cube of carbon there, and there's steam, there's water gas all around it. So if you pick any volume, especially if you pick some volume near the boundary of the carbon, you're always going to have carbon around. It's just what matters is the concentration of the water gas. That's what's going to drive the forward rate, so the forward rate is going to be dependent on some constant times the concentration of the water gas. And, of course, the backwards rate, so you need to get some H2, some molecules of-- let me draw it like that, because it has 2 hydrogen molecules plus a carbon dioxide, so maybe a carbon dioxide looks like that. So the reverse reaction, so rate, let's call that reverse, is going to be equal to some constant times the probability of finding both of these molecules in the same place. And, of course, the probability is related to or it's on a first-level approximation, depending on the concentration. So it's concentration of H2 times the concentration-- and to find both of them, you multiply the probability, because you need this and that-- times the concentration of CO. So when a reaction is in equilibrium, these two equal each other-- this is an r right here-- so this is going to be equal to the reverse rate of reaction H2 times carbon dioxide. Divide both sides by the K's, both sides by the H2O, and you get the forward coefficient or constant or whatever you want to call that, divided by the reverse constant-- I'm just dividing both sides by that-- is equal to this-- let me just copy and paste that-- is equal to that divided by this. You take that and you divide it by that. And so if we call this the equilibrium constant, because it's just two arbitrary constants, so we can just call this the equilibrium constant, you see that it actually makes a lot of sense to ignore the solid state in your equilibrium reaction. So the two takeaways here is when you're trying to calculate an equilibrium constant, you should ignore-- especially when it's in a heterogeneous equilibrium-- you should ignore the solution-- or not the solution. Ignore the solvent in that first example, where I did it with boron trifluoride with water. Water was the solvent, so I ignored it. Because water is everywhere, and you also ignore the solid state. Ignore the solid. Anyway, we'll probably use these in future things where we actually calculate the equilibrium constant. See you in the next video where we'll learn about Le Chatelier's principle.
AP® is a registered trademark of the College Board, which has not reviewed this resource.