Small x approximation for large Kc

- [Voiceover] In this video, we're going to be talking about using the small x approximation to solve equilibrium problems when Kc is large. And by large, I mean that Kc is greater than or approximately equal to 10 to the fourth. So we have another video before this that talks about using the small x approximation for when Kc is small, and I would say that's probably a simpler place to start if this is the first time you're seeing it. And in this video, we'll be talking about the opposite situation. As we talked about in the previous video, the steps for solving this kind of problem are fourfold. So there are four steps. The first step is to assume that the reaction goes 100% in the favored direction. So the favored direction when K is really large is that means we're going all the way to products. We're assuming we have almost no starting materials left. The second step is to set up an ICE table. And then to solve for x in our ICE table, assuming that x is small, hence small x approximation. And then the last and possibly the most important step is to check our answer. So make sure that the size of x is actually small compared to what we said it was smaller than, and also to make sure that it gives us the right answer when we plug it back in to calculate Kc. In this video, we're going to go through an example problem. The example is the reaction of NO gas reacting with chlorine gas to make 2NOCl gas. And this particular reaction has a K value that is equal to 6.25 times 10 to the fourth. So that is indeed on the order of 10 to the fourth, so we should be able to use the small x approximation. We're going to do it two ways. So we're gonna set up our ICE table, and we know the initial concentrations are 2.0 molar for NO, and 2.0 molar for Cl2. And we have no NOCl at the beginning. So if we are setting up an ICE table as we normally did, we would say, "Okay, we'll, we're going to make "some amount of NOCl." So we'll just say minus 2x for our NO since for every one Cl2 we use up, we'll use up 2NO, which means we'll have minus x for Cl2, and we'll make 2x molar concentration of our product. And in the end, once we reach equilibrium, we'll have to 2.0 molar minus 2x, and we just get that by adding the initial concentration with the change. In the same way, we get 2.0 molar minus x for chlorine gas. And for a product, we just get 2x at equilibrium. Let's look back at our original steps that we talked about at the very beginning of this video. We just made an ICE table, so that's step two. Did we assume that the reaction goes to 100% in our favored direction? It turns out we didn't. We actually skipped step one to set up our ICE table in what seems like a pretty natural way. And it turns out that's not good. So we'll see what happens when we skip step one and keep going. So if we keep going, we can solve for x, assuming x is small, and we'll get for our equilibrium expression, we get Kc is equal to 2x squared divided by 2.0 molar minus x, which our Cl2 concentration, and multiply that by 2.0 molar minus 2x, and that's all squared since we have that stoichiometric coefficient of two in front. Just to point out, we already started with a balanced reaction, that's pretty important, or else our Kc expression would be wrong. So make sure everything is balanced before you get started. So we've written our expression for Kc, and now we are going to erroneously assume, we're just going to be like, "Oh, I don't care." (chuckles) "I'm just going to assume "x is small. "I don't care if that's a good assumption at all." So if x is small, what we're really seeing is x is a lot smaller than two molar, here and here. So we're really seeing 2x is a lot smaller than two molar. So we're seeing this is approximately equal to 2x squared divided by 2.0 molar. Since x is a lot smaller than two molar, we'll just ignore it entirely, and x is a lot smaller than 2x, or sorry, x is a lot smaller than 2.0 molar even if you multiply x by two, so this is going to be 2.0 molar squared. And so if we multiply this out, we get that this is equal to x squared over this two squared is canceled out by this two, so we actually get x squared over two is equal to our Kc, which is 6.25 times 10 to the fourth. So that's what we get for x squared. So if we multiply both sides by two, we get that x squared is equal to 1.25 times 10 to the fifth, which means x is equal to the square root of that, which is 354. So this is where... Well, okay. So we've gone through steps two and three, and now we're going to check our answer. We're going to see if our answer makes sense. So we're saying that the change in concentration here is 354. That clearly doesn't make sense because it actually gives us a negative concentration here. So we're getting negative concentrations for NO and Cl2, so that's bad. So that doesn't make sense. The other thing is that this already tells us that our assumption was really bad. We assumed to solve our equation for Kc. We assumed that K is a lot smaller than two, or sorry, we assumed that x is a lot smaller than two, but then we got that x is 350, which is clearly not smaller than two. So it looks like skipping step one was bad. So let's give this another try. Try number two. What was step one? Step one says assume the reaction goes 100% in the favored direction. And since Kc is really large for this particular problem, we're saying it's going all the way to products. What this basically means is that we're gonna set up our ICE table again, except that this time, our initial concentrations are going to be assuming that our reaction already went all the way to products. And we can figure that out, we can figure out those initial concentrations using stoichiometry. So in the beginning, we started out with two molar NO and two molar Cl2, and we know that the NO and Cl2 react in a 2:1 ratio. So since we have the same amount of both of these, our limiting reactant will be the NO. And so that will get used up completely when we get to equilibrium. So that means if we assume that it goes all the way to products, we'll have zero molar NO left, and will make 2.0 molar of our product. And since we had our Cl2 in excess, we'll use one molar of it to react with two molar of our NO, and we'll have one molar left over. So this is the special step where we actually followed step one and assumed we got 100% of product. And the reason why we made this assumption was because we know that K is really large. So we know that at equilibrium, we should have all product or mostly product. So now let's go through the other steps. We assume that it's going to all product, but we're not quite at equilibrium. So if we're going to equilibrium, that means we assume that we'll see a little bit of NO at equilibrium, so that'll actually be plus 2x. And we'll expect to see a little bit more Cl2 at equilibrium. So that'll be plus x because of the stoichiometry, and that would give us minus 2x for the NOCl. We expect a little of this to get used up to go in the reverse reaction. So then if we add everything together from the initial and change, then we get 2x for our NO concentration, we get x for our Cl2 concentration, and we get 2.0 minus 2x for our product concentration. So far, so good. So now we're going to set up our Kc expression just like we did before. So Kc is still equal to 6.25 times 10 to the fourth, and this is equal to the concentration of NOCl squared. So that is 2.0 minus 2x squared divided by 2x squared, which is our concentration of NO squared times x. Oh, oops, I made a mistake. This is actually 1.0 plus x, sorry. So this should be x plus 1.0 molar. So that is our full expression for Kc using our equilibrium concentrations, and we have made zero approximations of R. But now we're gonna assume that we can assume, (chuckles) now we're gonna assume that x is small. We're gonna assume that assume. So if x is small, what we really mean is that x is a lot smaller than one molar. And we're also saying it's a lot smaller than two molar. So then we're saying that our numerator is approximately equal to 2.0 molar squared, because we're saying that this is small. I'm gonna make this an approximately equal sign. The 2x squared stays the same. Since we're assuming x is a lot more than 1.0 molar, this just becomes 1.0 molar. So now if we multiply this all out, we get that four divided by 4x squared is equal to Kc, and the fours cancel out. So x squared is equal to one over Kc, or 6.25 times 10 to the fourth. And then if we take the square root of both sides here, we get that x is equal to, let's see, we get x is equal to 4.0 times 10 to the minus three molar. So this is where common sense is needed to make sure that, well, everything worked this time around. So first of all, we can ask ourselves, "Okay, is this x actually a lot smaller than the numbers "we said it was smaller than?" So now we're comparing x to one, which, it's about three orders of magnitude smaller. So that's good. And we can also compare 2x to two. And again, it's about three orders of magnitude smaller. So, so far, so good. But our final test will be to plug it back in our Kc expression. So if we plug in our value of x, we get that Kc is equal to 2.0 molar minus two times 4.0 times 10 to the minus three molar, all squared, divided by two times 4.0 times 10 to the minus three molar, and that's also squared, four, and that's the NO concentration at equilibrium. And then the last part is our Cl2 concentration, which is 4.0 times 10 to the minus three molar plus 1.0 molar. And so if we multiply that all out, what you get is that Kc is equal to 6.23 times 10 to the fourth. And so we can compare that to the value of Kc we started out our problem with, which is, let's see. It is 6.25 times 10 to the fourth. And this is actually pretty good. We did make an approximation, so our answer isn't exactly right. But it's pretty close. And so if we wanted to get it even closer, there are other methods we could use. But for most purposes, this is actually, this tells us that our approximation was good. So we can see that when Kc is really large, what we need to do is assume that we have 100% product when we're setting up our ICE table, and that'll help us safely assume that x is small.