Chemistry ICE Table Help?

[george]

Your ICE table will look like this:

2HI........<-->...H2......+... I2
0.820..............0..............0
-x...................+1/2x.......+1/2x
0.820-x...........+1/2x.......+1/2x

You start with 0.820 and you have no hydrogen or iodine at the beginning. An x amount of HI will decompose. Since 2 moles of HI produce 1 mole of H2 and I2, x will produce 1/2x of H2 and I2 or if you prefer, 2x of HI will produce x of H2 and x of I2.

An easy way of determining what should be x or 2x or 1/2x is this. In the above case, we let the HI decomposed equal to x. Now we know that 2HI --> 1H so,

2HI / x = 1H / # (the hashtag represents what the representation of H should be. Please note, this is not a mathematical equation, I'm just trying to illustrate a way for you to work this stuff out).

So we "solve" for #:

# = (1H * x) / 2HI = 1/2x (we drop the atoms, just work with the numbers).

Your equation will now be:

Kc = (1/2x) x (1/2x) / (0.820-x)^2 (HI final concentration to power of 2 because 2 moles of HI)

You have the value for Kc and you can now solve for x.

(1/2x)^2 / (0.820-x)^2 = 0.016