Rob Lederer: Now you can be given questions where you know what the K value is for a reaction, so let's take iodine gas, vaporized iodine and turn into vaporous monoatomic iodine, which as bounce for I2 breaks down to 2I gases. We know the K value, we know the initial concentration of the iodine, the question can we find out the equilibrium concentrations? Okay. You have to setup the box, we did before with initial change like an equilibrium. I-C-E, some people call up the ice box. Now when you get the spots all done here, you will be able to plug into the expression, let's go through it. Well, and the implication is if you have 0.2, this we have none of this to begin with. So if the chemical is not even mentioned as having a concentration, it's zero. Now what we have to do? Well, we have to make something here, something here in order to get to equilibrium. So we have to lose from here and we lose X here, because it is the 1 in front, so it's a 1X, but what we do gain here? We take the coefficient in front, put in front of X. We gained 2X here. So an equilibrium, we add the lines together. 0.20-X is the concentration of the I2 at equilibrium. That makes sense, doesn't it? This is going to be well, two times whatever X is. Plug it into the expression. Here is the expression. I-squared divided by the I2. The K is going to be 2X and it's going to be square divided by 0.20 minus X. And we know what that K value is, so we only rank K, we just rank 3.8 times 10-to the negative 5. So what we're looking for? X, and once we find X, we'll be able to substitute it into this line here, the equilibrium line. We can put two times X there, that is the concentration of that equilibrium, and 0.2 minus X is the concentration of that at equilibrium. So we're just going to work out just lazy. But you know what? 2x-square is actually 4X-square, isn't it? 2X-squared is 4X2. Oh, boy, X-square and X term in that odd manner, we're going to have to rearrange this in dual quadratic formula, not until -- but here is the thing. There are ways of being able to do this question little sneaky, but you know what, it's just - really it's understanding what math is all about. Okay. Now here is the thing. Know the K value suggesting here. Remember we talked about this before. With a very, very small value of K, you don't make a heck of a lot of product. So two times not a heck of a lot is actually going to be not a very big number, and that very, very small number, which is what X is going to be because K value is also small, tending it away from 0.2 is essentially going to still give you two significant digits, probably 0.20. Sure, something is going to be lost here but this concentration will end up in 0.19995. Well, to two significant digits, it's still 0.2. You know you can do any time X. X is added to or substrated from a number. You have the ability to perhaps disregard that X if the K value is low enough. How do you know that? If you take the initial concentration of 0.2 and divided by the K value, so 3.8 times 10-to the negative 5, if you get a number, that is going to be here; greater than or equal to 1,000. That's what we call -- it's essentially called the five percent rule, but here is the thing. If you get initial concentration divided by K, greater than or equal to a 1,000, you can just rely on X in the calculation minutes. Add it to or subtract it from a number -- big. What does that mean for this? It means that this X here is relevant, it's going to too small to make a difference. By the way, that number is greater than or equal to 1,000, it's really, really large. So when we do the math here, you actually get 4X-squared equals, now it's going to be this number 4X-squared equals this times this. We multiply each time by 0.2 now, and this is gone. Then you get here 7.6 times 10-to the negative 6. So when you do that at here, X equals 1.4 times 10-to the negative, and it better be three, and of course it is, 10-to the negative 3. That's the value