Let's look at another equilibrium example. The equilibrium constant for the given reaction is 0.561 at 800 K. If the initial concentration of HI is 0.500 M, what are the equilibrium concentrations of each species? Now this is an equilibrium problem. The first thing we need to do is set up an ice table. So we have I for initial, C for change, E for equilibrium. We're told we have an initial concentration of HI = 0.500 M. And since it doesn't mention anything about initial concentrations of I2 and H2, we're going to assume they're 0. Now this problem is a little bit different because here, we're starting with the product, or something on the right side of the equation, rather than the left side. It really doesn't matter what we start with because it's an equilibrium reaction, the process is actually going in both directions. For our change row, we still have to do it in terms of x. We'll still be gaining H2, so +x, and gaining I2 so +x there as well. On the HI, we'll be losing HI, but we have to put -2x. Because if we look at the coefficients in our balanced chemical equation, we see that we have a coefficient of 2 in front of the HI. Our equilibrium row is simply the sum of the initial and change row. So we have x, x, and 0.500-2x. Now, we can set up our log mass action, so K = [HI]. Remember we have to square that because we have a coefficient of 2, over [H2] and [I2]. Now because all of these are in pressures, we could have done this as a KP or KC. However, the values were given in terms of molarity, and our K is given in terms of concentration. So we want to make sure we do this in terms of molarity. Now I can substitute in what I know, I have 0.561 = my HI concentration at equilibrium, which is 0.500-2x, and then I need to square that. And on the bottom, I have x and x, so I'm going to write that as x squared. Now this might the point at which you're saying, where can I make that simplifying assumption? And I can't make this simplifying assumption because of the value of K. However, there is a solution that makes this problem a little bit easier. Notice that on the fraction, we have both of our terms squared. That means I can take the square root of both sides. This will greatly simplify the calculation I need to do. When I take the square root of 0.561, I get 0.749. And what that leaves me with on the right side is (0.500-2x) / x. Because I took the square root of everything. Now I can rearrange, and simplify my expression a little bit, combine some terms. And then I can solve for x and found that x = 0.182. Now, I can use that information to figure out what my equilibrium concentrations are. For H2 and I2, my equilibrium concentrations will be just that, 0.182 M. For HI, I'll have to do a little bit of a calculation, I have to do 0.500-2x. And I end up with 0.464 M for my HI concentration. So on those problems where you can't make the simplifying assumption, always look to see if there's a mathematical solution, which will make the problem a little bit easier.