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Chemistry ICE Table Help?
03-30-2014, 07:08 AM
Post: #1
Chemistry ICE Table Help?
I am having a hard time figuring out how to set up the I-C-E Table to this chemistry problem.

Hydrogen Iodide can decompose into hydrogen and iodide gases.

If 0.820atm HI(g) is sealed in a flask, what is the pressure of each gas when equilibrium is established?

2HI(g) <---> H2(g) + I2(g)


My teacher told plug in what we know from the I-C-E Table, but when it says 0.820 atm HI does that mean initial for 2HI? I am also having a hard time finding where to put X's in the table in preparation for the K=Formula.

All I have set up so far on the ICE table is the initial atm for the three compounds ( HI = 0.820, H2 = 0 and I2 = 0 )

Besides that I don't know how to determine what is x, 2x, 1/2x 1-1/2x and stuff like that.. I know you have to look at the ratio of moles in 2HI(g) <---> H2(g) + I2(g), but I am having a hard time doing that as well....

The answer is: HI=0.645 atm H2 = 0.083 I2= 0.083 atm

I don't know how to figure this out if someone could help me that would be greatly appreciated..

P.S. - If there is a way to do this problem without an ICE table I cannot do it because we have to show an ICE table on the test in order to get partial credit for our answer.
*****The Kp = 0.016 Sorry I did not post that the first time******

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03-30-2014, 07:12 AM
Post: #2
 
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

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