Chemistry ICE Table Help?
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03-30-2014, 07:08 AM
Post: #1
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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****** Ads |
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03-30-2014, 07:12 AM
Post: #2
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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 Ads |
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