Let's look at an example dealing with pressures, instead of concentrations. If I2 has an initial pressure of 1.00 atm, what are the equilibrium pressures of I2 and I? We're given our balanced chemical equation and the value of Kp. Just like we have for other equilibrium problems, we're going to set up an ice table, I for initial, C for change, and E for equilibrium. I know my initial pressure of iodine is 1.00. There's no mention of atomic iodine. So I'm going to assume that the pressure of that is 0. Now for the change row, I have to use my coefficients from my balanced chemical equation, and in doing so, I'll see I'm going to lose some of the I2 and I'm going to gain I. But I'm going to gain 2 times the amount of I because of the 2 in my balanced chemical equation. For my equilibrium, I simply add up the initial and change rows, so I have 1.00- x and 2x. So now I have my equilibrium values with respect to x, my equilibrium pressures with respect to x. Now I can set up my KP expression, so I know that KP = PI, elemental iodine, squared / PI2. Now I can substitute in the values that I know. I have 2.91 x 10 to the -4th for my KP value. That's going to be equal to the pressure of iodine, which note that is 2x. I still have to square that because the pressure is 2x. The law of mass action tells me that whatever that term is, it must also be squared. Note that the squared does go outside the parenthesis, so that we square both the 2 and the x. That's a common mistake that students make. And on the bottom, we have the pressure by 2, which is 1.00- x. And we're going to, again, make our simplifying assumption that x is much, much less than 1.00. And when we do, we can write out 2.91 x 10 to the -4th = 4x squared. We could show the 1 in the denominator, but it doesn't change the value in our problem, so we're going to leave it out. Now I can divide both sides by 4 and take the square root. And when I do that, I find that x = 8.53 x 10 to the -3rd. Now I can use this value of x to find my equilibrium pressures by substituting this value m back into what I have in my equilibrium row. And what I find is the equilibrium pressure of I2 is equal to 0.991, and the equilibrium pressure of elemental iodine is 0.0171. And these are both in units of atmospheres.