The equilibrium constant for this reaction is 60.2 at 250 kelvin. If the initial concentrations of hydrogen and iodine are both 0.250, what is the equilibrium concentration of HI? And we're also given the balanced equation H2 plus I2 yields 2HI. Since this is an equilibrium problem, the first thing I want to do is set up my ice table that gives me a way to summarize the information that's given and figure out how I'm going to get to the equilibrium concentration to our product. So I put in I for initial, C for change and E for equilibrium. When I look back at the problem I see that my initial concentration of H2 0.0250 my initial concentration of I2 is also 0.250 and there's no mention of HI being present initially. And so I assume that concentration initially is 0. Now I need to look at the change row and to do that I have to look at two things. One, the stoichiometry of the balance chemical equation and I have to determine which side is going to be losing substance and which side is going to be gaining substance. Because I start with no HI, the only thing I can do on that side is gain. So that's going to have to be the gain side. And what I'm going to see is on the reactive side, I'm going to lose H2 and I'm going to lose I2. Then I put this in terms of the balance chemical equation, and I say well if I lose one M of H2 and I represent that as X, I will also lose one M of I2 and I will gain 2M of HI. Now I can take my initial row and my change row, and add them together to get the equilibrium row, so I end up with 0.250- x for H2, 0.250- x for I2 and 2x for HI. Now, I want to set up the Kc expression and so I say that Kc equals the concentration of HI, and I square that because I have a 2 in my balance chemical equation over the concentration of H2 times the concentration of I2. Now I can plug in the values that I know here. I know my value of KC is 60.2. I know my equilibrium concentration of HI is represented with 2x and I have to square that. Then I'm going to put that over 0.250- x and this term is also squared because what I see is that the equilibrium concentration of H2 and I2 are exactly the same. Because the value of K is so large I can't make any simplifying assumptions. However, I can make a mathematical simplification by taking the square root of both sides. And this works because I have a square on both the numerator and the denominator. And when I do that, I find that I get 7.76 equals 2x over 0.250 minus x. And I can use my algebra skills to solve for x, and so I end up with 1.94 minus 7.76x equals 2x. Then 1.94 equals 9.76x. And finally, the value of x if equal to 1.99. Now I'm not quite finished with the problem because I see that the value of x is 1.99 and that gives me the x that in my ice table, but I'm asked for the equilibrium concentration of the HI and I noticed that when I look at my ice table, the equilibrium concentration of HI is equal 2 times x. So the concentration of HI = to 2x and therefore the concentration of HI will be = to 2 times 0.199 and that will make it equal to 0.398 M. And these types of moles you can also check them work we can go back if we find a concentration of HI, and we can substitute the value on xn for what the HI, and we can solve and we should get a value fairly close to 60.2. There may be a slight variation due to rounding along the way.