In this problem, we're asked to find the Kc or the equilibrium constant for this particular reaction. We're given some information about the initial concentrations of our reactants, H2 and I2, and information about the equilibrium concentration of HI, our product. So we know this is an equilibrium process because we see we have our equilibrium arrow here. We know that because it's an equilibrium reaction, we are not going to completely convert from our reactants to our products. So what we need to find is, what are the equilibrium concentrations of the H2 and I2, so that we can solve for the value of Kc. For equilibrium problems, the way that we kind of organize and sort our data is through the use of an ICE table. And I stands for initial, change, and equilibrium. We can also set up a column here that says H2, I2, and HI. So now we can take the information given in our problem and sort it into this table. So the first thing I notice is that my initial concentrations of H2 and I2 are both 0.100, so I can put this in. And in this case, they happen to be the same, they don't have to be. Sometimes, you'll see different values. We also see that there's no mention of our initial concentration of H!, so we can assume that that is 0. The next thing we need to look at is our change row, and we do this in terms of x, so I know that if I lose some x amount of H2, I'm going to lose the same amount of I2. And I'm going to gain twice as much of my HI because as we lose reactants, we're going to products. And if I look back at the stoichiometry of my equation, what I see is that it's a 1 to 1 to 2 ratio in my balanced chemical equation. Now I can look at my equilibrium row, and all I need to do is sum the initial and change rows. So my equilibrium concentration of H2 is 0.100-x. It's the same for I2, and for HI, it's equal to 2x. So now I can go back and say, well, wait a minute, in the initial problem, it gave me what the actual equilibrium concentration of HI was. So I can use that, along with my value of 2x, to determine my value of x. So I can say HI is, at equilibrium, is equal to 0.134, and it's also equal to 2x. And so I can solve for x, and find that it equals to 0.0670. Now I can use that value of x to figure out what my equilibrium concentrations are for H2 and I2. So let's set that up to solve for those, and so, I look back at my ICE table. I look at the equilibrium row, and I see that the concentration of H2 at equilibrium was equal to 0.100-x. I substitute in the value I have for x, which is 0.0670, and what I find is that my equilibrium concentration of H2 is 0.033 M. I can do the same thing for I2, and I'm actually going to get the same value because I have the same initial concentration. And I'll also have, they're both losing x, so that value is going to be the same as well. And I get my equilibrium concentration of I2 is equal to 0.033 M. Now what I can do is I have all of my equilibrium concentrations of my substances. So now I can actually solve for the value of Kc, and I do that by setting up my law of mass action. So Kc = [HI] squared, because I have a coefficient of 2 here. So I have to put that in to my law of mass action. And I'm going to be dividing by the concentration of H2 times the concentration of I2. And because the coefficients of both of these are 1, I'm not going to have a power on either of those numbers. So now I can substitute in my concentration. So Kc equals my equilibrium concentration of HI, which is 0.134, I need to square that. And on the bottom, I’m going to have my concentration, (0.033)(0.033). And those are for the concentrations of H2 and I2. And when I solve this, I end up with a value of 16 for the equilibrium concentration. And this tells me that I'm going to have more products favor because my K value is greater than 1.
