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.