Okay, so you may not have been waiting for this moment to happen, but we have some cases where you can simplify the algebra, and I'm going to tell you exactly how to do it. So, this falls under something called the x's small approximation, and it will let us simplify the algebra. Now, unfortunately, it won't let us do it in every case. And I'm going to give you some guidelines for figuring out if it's going to work.
But what you should take away from when can I use x is small is that what x is small means in the context of an equilibrium problem, well, x is what goes into our change row. So we're saying if the amount of change in a reaction is small, We can sometimes approximate what's happening to that reaction. And sometimes it works, sometimes it doesn't. If it doesn't, that's okay. We have another method for getting through all of this.
So what this really means is that if the reaction is already close to equilibrium, we can say that x is small and use this approximation. This happens. in a couple of standout cases, but if you can look and say x is small, then this is when you would apply this simplification. The cases that you will likely see is that if you have a small equilibrium constant and a high concentration or high pressure of reactants. Small equilibrium constant means a lot of reactant, high concentration of reactants.
is a lot of reactant, right? They match up. So you might have to make some product to get to equilibrium, but the amount that you have to make is very small.
The other case is, so the general, if things are apart by about 10 to the third, this will work. The other case, I have an approximation of about 10 to the third here, but it's harder to approximate in the other case. But you should be able to look at what kind of conditions you're under and see if you are close to equilibrium. So if you have a large Keq and a high concentration or pressure of product.
So large Keq means a lot of product. High concentration or pressure of product means a lot of product. So you may have to make some reactant to get to equilibrium, but you are basically there.
This is the... only option we have for solving higher roots. So because of the calculators that we work with, right, you cannot solve anything beyond what could be solved using quadratic formula. So if you have an x cubed, an x to the fourth, an x to the fifth, or higher, which will happen, this is the only way to solve it.
You may not have the conditions to solve it, but that's the topic for another video. So let's do an example where we can use this and show how to apply it. So I have H2S initial concentration is 0.0250 molar, and my equilibrium constant is 1.67 times 10 to the minus 7th at 800 degrees C.
I want to find the product concentration at equilibrium, and I've been given this reaction of 2H2S in equilibrium with H2 and S2. So initially, I have 0.0250H2S. I have 0H2S2.
Okay, well, my reaction has to move to the right, and I could do this. What we're going to see, this is going to give me an X cubed, which is why I'm going to need to go back and assume that X is small. Let's just say I haven't looked that far ahead yet. So I've got my initial conditions.
I know I have to move to the right. So in my change row, I'm going to lose 2x of the h2s, make 2x of the h2, and make x of the s2. So at equilibrium, I have 0.0250 minus 2x, 2x and x. My equilibrium constant expression. products over reactants.
So 2x quantity squared times x all divided by 0.0250 minus 2x whole quantity squared. So right across the top you can see that this is going to give you an x cubed. So we will not be able to solve this any other way.
The question is though can we apply x is small? My equilibrium constant is really small. And I mean, 0.0250, it doesn't seem big, but compared to the equilibrium constant, it's quite a bit bigger.
So I don't need to make much product to get to equilibrium. So the amount by which my reaction changes, right, or x is a small value. If you're doing the comparison of the powers of 10, my equilibrium constant was 10 to the minus 7th. This 0.0250, it's about 10 to the minus 2nd.
We're super generalizing here, but that's 10 to the 5th apart. So we can assume that x is small. What that means is that any place where you have added or subtracted x, just get rid of it.
Because what we are saying is that 0.0250 minus 2x, x is so small that this is basically equal to 0.0250. Like, yeah, we technically changed concentrations, but like, I took like four molecules. So your concentration is pretty much the same. Now, right looking at this, we have other x's in here.
We can't Take them all out because we still need an x to solve for. So the other thing that has to happen in order for this to work to solve is that you have to have an x somewhere by itself. So something that started with a zero and then you just added to it is really the only option.
And we will, in every single place where we have added or subtracted x, all of those are going to go away. Where this leaves us is we will have an x cubed in the top, but if we, everything else will be a number. So if we multiply all of this out, right, like, or just even rewrite it after removing the change that was so small, we're going to ignore it.
We're looking at 1.67 times 10 to the minus seventh is equal to 2x quantity squared times x divided by 0.0250 squared. So 4x cubed equals, well, 1.67 times 10 to the minus 7th times 0.025 times 0.025. They're just all numbers. So I can just solve 4x.
I can divide and then take a cubed root, which gives me x is equal to 2.97 times 10 to the minus 4th. The question is, was it a good assumption? Did this actually work?
And the way that we test it is we compare x to what we added it to or subtracted it from. And if it is small enough, which means less than or equal to 5%, then it was a good assumption. I, you know, I have kind of a janky equation here, but if it hopefully helps you, it helps you.
So this comparison means divide. So take the x value. divided by the number that it was subtracted from or added to in the original equation and multiply by 100, which will give you a percentage.
So we got 2.97 times 10 to the minus fourth, which was subtracted from 0.0250 in the original equation. If you multiply that by 100, you get 1.19%, which is great. So this was a good assumption.
We still aren't done with our problem though. So all we did was find x. Now we have to find the equilibrium concentrations.
So plugging x into concentration of S2 and concentration of H2 gives you these values. And you could also go through and find the reactant concentration, but it didn't. It didn't ask for that, so that's why it's not reported here.
Okay, so what if x is not small enough? So when you use the x is small approximation that we just talked about, and you test it, and you get a percentage that is greater than 5%. So you proved that x was not small enough to have made the assumption, but depending on what you did or what you have, You may have no other way to solve the problem.
So if it is an x cubed, x to the fourth, x to the fifth, so on, you find out that when you test it, the percent for your x was greater than 5%, so you couldn't assume that x is small. You have no other choice. What we do in this case is something called successive approximations, which is just, right, approximations are in here, and successive.
We're just going to keep re-approximating until we get closer to the correct value. So a different problem with a tiny tweak. So we have the same reaction, we have the same equilibrium constant. We're going to start with a smaller initial reactant condition.
concentration. So 2.5 times 10 to the minus fourth for the concentration of H2S, still zero for H2 and S2. I'm still going to move to the right. I have to make product, lose 2X, add 2X, add X.
I have my equilibrium row. When I do my comparison, 10 to the minus seventh versus 10 to the minus fourth, and that's 10 to the third apart. So it's kind of Right, from this I know I'm going to get an x cubed, so first of all I'm not going to have a choice, but also I'm within the range of the like, yeah you could probably assume that x is small here.
So assuming x is small, right, the only thing I'm taking out is that minus 2x. When I finish solving this and I solve for x, I get an x value of 1.38 times 10 to the minus fifth. Like, okay, that's the number.
We have to test it to find out if it was a good assumption, which is take that number, divide by the number it was added to or subtracted from, which is 2.5 times 10 to the minus fourth, multiply by 100, and you get 5.52%. It's literally so close, but it... did not meet the guidelines.
So this was not a good assumption and we have to do something else. The only thing we can do is approximate again, which we do by taking this x value that we just found, which I've labeled it as x1, and we plug it back in to where we took it from, essentially. So I've got this circled, I've got an arrow.
When I got rid of minus 2x, I found 1.38 times 10 to the minus 5th for x. I'm going to plug that back in in the place where I took x out, and I'm just going to plug it in for x. So finding x2, right, the 2x quantity squared and the x, both of those things, they... were part of the x's that I solved for.
We hadn't changed those, so those stay the same. And in the bottom, 2.5 times 10 to the minus 4th minus 2x. So minus 2 times the value that I found for x1, right?
That 1.38 times 10 to the minus 5th. And then I'm going to square all of that stuff on the bottom, and it's all equal to 1.67 times 10 to the minus 7th still. When I solve for x2...
Well, when I solve for x the second time, which I label x to, I get 1.27 times 10 to the minus 5th. So good news is that it didn't change a lot. If it changes a lot, probably something went wrong, hopefully just with the math and not with the setup.
So, I mean, now your question is probably, now what do I do with it? You plug it in again. So you take X2 and put that back in for X and you keep going. until you get values that converge. So if I take 1.27 times 10 to the minus fifth and plug it in for x, which I have it labeled here in blue so that you can see where it went in.
When I solve for x this time, I get 1.28 times 10 to the minus fifth. So these are close enough for us to say, like, oh, yeah, it's basically the same number. Like, let's just go forward with this as our x. Now this is the X that you carry forward.
If you would like to take your X and do the test again, that's one way to feel confident moving forward with it, but this should be under 5%, and if not completely under 5%, it's gotten as close as we could get. You could solve for X4, and you should get something. If I remember correctly, I think this might bounce back to 1.27.
So if you used 1.27 times some of the minus fifth, great. If you used 1.28 times some of the minus fifth, also great. Should give you less than 5%. If it doesn't, as long as it moved as close as possible to 5%, we've really done everything we can. So remember, though, this is not my final answer.
I'm going to use this x value to find my equilibrium concentrations. So I've done it for all of them here, right? 2.54 times 10 to the minus 4th minus 2x, 2x and x to get my equilibrium concentrations of everything.
And then just to note, if you have substituted for so long that you're at x6 or x7, go back in and check your work. There are occasionally problems where you will need to go to X7 or X8, but they are very rare. And especially if you are on a quiz or an exam or something like that, you shouldn't have to go beyond X6 to get the answer here.