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:9:40

Video transcript

- [Voiceover] We're gonna talk about the small x approximation. The small x approximation is an approximation method used to solve equilibrium problems. Maybe the most important thing to remember that I'll start out with is that the small x approximation doesn't always work. It works best and you should generally try to use it only when your reaction is strongly product or reactant favored. So strongly product or reactant favored. You might be asking at this point, okay, well, how do you know when your reaction is strongly favoring either products or reactants? The answer to that is it depends on the value of your equilibrium constant K. So if something is strongly favoring product, that means K is really big, and by really big, we usually mean, or I usually mean it's on the order of 10 to the fourth or larger. We're talking about Kc here where our equilibrium constant is interms of molar concentration, and similarly, when we say something is reactant favored, that means that Kc is really small. So we have mostly reactants around it and not very many products. And when Kc is very small, I mean that Kc is less than or about equal to 10 to the -4. So these are the situations where the small x approximation works really well, and then, anywhere in between where Kc is between 10 to the -4 and 10 to the 4, it actually doesn't work as well, and you'll wanna use a different method. You'll either usually wanna use the quadratic equation or you might wanna use successive approximation, which we're not covering in this video. We'll be covering those things in separate videos. So how does the small x approximation work? The first step is you assume your reaction goes fully to the side you know it's gonna go based on K. So assume reaction goes 100% to either products or reactants depending on whether K is really small or large. Our second step is to use this information from step one to make your ICE table. So initial concentration, change in concentration, and equilibrium concentration. We'll then solve for the x in our ICE table, assuming that x is really small. Hence the name of the method, the small x approximation. And we'll go into a little bit more specifics about exactly what we mean by that when we do an example. Then, the last step, which is really important, really, really, really important is to check your answer. I usually do that by plugging equilibrium concentrations back into the Kc or equilibrium constant expression to make sure the concentration that we did get as our answers make sense and we get the value of K that we thought we should get. So to make more sense of all of these things, we are going to go through an example. We're gonna go through an example of when we have a really small equilibrium constant. That means that kc is less than or equal to 10 to the -4, and that means since our equilibrium constant is really small, that we have mostly reactants and not very much product in our reaction pot when we reach equilibrium. So the example reaction that we're gonna talk about today is iodine gas, I2, and it's going to be an equilibrium with 2I-, also gas. The K value for this particular reaction at the temperature we're interested in is 5.6 times 10 to the -12. So that's a pretty small number, and it's definitely smaller than 10 to the -4. So we should be able to use the small x approximation. We have a little bit more information about our reaction. We know the initial concentration of I2 gas is 0.45 molar, and we know that the initial concentration of I- is zero. So based on this, we're gonna go through the steps for using the small x approximation. We're gonna assume reaction goes 100% to either products or reactants. Here we know that we have mostly reactants. If we start out with all reactant and no product, we assume that at equilibrium, we're still gonna have mostly reactant. It's gonna change a little bit. So some of this is gonna turn into I-. So we'll say -x for some small concentration of I2 that gets converted. And that -x will turn into 2 moles of I-, but we don't expect x to very big, and that's the key to being able to use this approximation. If we add these up to get our equilibrium concentrations, we get 0.45 molar minus x for I2, and we get 2x for our I- concentration. Now we can use our Kc expression to solve for x when we assume x is small. Kc, for this particular problem, is the concentration of our product, I- squared since we have a metric coefficient of two there divided by the concentration of I2. And if we plug in our equilibrium concentrations from our ICE table, we get this 2x squared divided by 0.45 molar minus x. So far, we haven't made any approximations. Now we can make our approximation. We're assuming x is small, and specifically, we're assuming that x is a lot smaller than 0.45 molar. What that means is we can assume that 0.45 molar minus x is approximately equal to 0.45 molar. So x doesn't really change the concentration of I2 at equilibrium very much. So that simplifies our expression for Kc quite a lot. We have 4x squared in the numerator from squaring two and squaring x, and in the denominator, we no longer have x. So this makes a much easier equation to solve, and this is all equal to 5.6 times 10 to the -12. Awesome. So now what we can do is we can multiply both sides by .45, and we can divide both sides by four, and what that gives us is that x squared is equal to 6.3 times 10 to the -13, and if we take the square root of this, we get that x is equal to 7.9 times 10 to the -7 molar. So that is what we get for x. Here is the second most important step. Besides making our approximation in the first place, we need to make sure that our approximation makes sense. We've made a change somewhere. We've decided x is small, and we've ignored it in our equation, in one part of our equation. So we have to make sure that our original assumption gave us a common sense answer. We see that x is 7.9 times 10 to the -7. We assumed that was a lot smaller than .45, which looks to be about true. It's about six orders of magnitudes smaller than .45. So, so far, so good. We can make sure that it really is right by checking our answer. We can plug in our x and calculate Kc again. So then, we get that Kc is equal to two times our x value, 7.9 times 10 to the -7 molar, and all of this squared divided by 0.45 molar minus our x value, 7.9 times 10 to the -7 molar. If you multiply this all out, what you get for Kc, or what I got for Kc is 5.6 times 10 to the -12. That matches what we had originally in our problem. So this tells us that A, our approximation was pretty good. It gave us an x value that gave us concentrations that make sense, and B, which is another reason why we wanna check our answer. B, it tells is that we didn't make any silly math errors, which is really easy to do in this kind of problem.
AP® is a registered trademark of the College Board, which has not reviewed this resource.