Monday

April 21, 2014

April 21, 2014

Posted by **Ken** on Wednesday, June 25, 2008 at 1:55pm.

This is what I did. I changed all the moles ot M. Then I subbed all the M's into an ICE chart, and then I subbed all of that information into the Kc formula, and then tried to work the quadradic equation, but nothing looks right with the answer, not to mention that none of them match up with the multiple choice answers.

- Chem II -
**DrBob222**, Wednesday, June 25, 2008 at 2:20pmI don't see anything wrong with what you said you did. Did you equation look like this?

(HI)^2/(H2)(I2) =

[(0.369+2x)^2/(0.2975-x)(0.410-x)] = 54.3.

Solve for x.

- Chem II -
**Ken**, Wednesday, June 25, 2008 at 3:34pmOk, that is what I did with the quad. equation, twice. The answer that I got was .518 and .246. Then I took .246 and subbed it back into the ICE chart, and got the H2 concentration of .052. According to my teacher, that answer is wrong. The correct answer for H2 should be .070 M H2, and 0.182 M I2. What am I doing wrong? I have an exam on this tomorrow!!

- Chem II -
**DrBob222**, Wednesday, June 25, 2008 at 4:12pmYou must be making a simple arithmetic mistake, such as slipping a sign change or something like that. I worked it through and obtained (H2) = 0.0699 which rounds to 0.070 M and (I2) = 0.182 M. If you solve the equation I wrote in my last response, x is 0.228 (0.2276) and 0.567 (0.56658). Throw away the 0.567 since that is larger than H2 and I2 initially so subtracting that would lead to negative concentrations and that is meaningless. You may post your work if you wish and I can find the error. Good luck on your test.

- Chem II -
**Ken**, Wednesday, June 25, 2008 at 4:46pmOk, here is my attempt at my work

54.3= 0.136 = 4x^2/(0.2975-x)(0.410-x)

after foiling everything I get

54.3(x^2-0.71x + 6.5 = 0.136+4x^2

which I get

50.3x^2 - 38.4x + 6.4 = 0

plug everything into the quad. and I keep getting .245. I subtract that from .2975 for H2 and my answer still doesn't add up. HELP!!

- Chem II -
**DrBob222**, Wednesday, June 25, 2008 at 5:11pmI think your first error is in expanding the term (0.369+2x) = 0.136 + 1.48x + 4x^2 (in round numbers). The 1.48x term missing in your expansion makes a difference at the end of

50.3x^2 -39.9 + 6.5.

I think a second error is in not carrying out to enough decimals. When dealing with small numbers like this it is easy to make rounding errors and never know it. For example, if I solve the quadratic above using 50.3x^2 - 39.9x + 6.5, I end up with 0.229 which gives an answer of 0.2975-0.229 = 0.0685 which is very close to 0.07 but not quite. I ALWAYS carry at least one more place, and more than likely two or three during the stages of a problem, then round at the end. For those who want to argue that the accuracy isn't that good, I will agree, BUT I don't have rounding errors that way. The way I look at it is that I can ALWAYS throw the numbers away at the end but I can NEVER add them back in during the middle if I throw them away there.

- Chem II -
**Ken**, Wednesday, June 25, 2008 at 9:42pmOk I finally got the right answer. Thanks for all your help, as always

**Related Questions**

Chem II - Equilibrium - For the reaction at 430 degree C, H2 + I2 <=> 2HI ...

chemistry - H2+I2->2HI If 3 moles of H2, I2, and HI are in a 3L flask, what ...

Chemistry - The following system is at equilibrium,at 699K in a 5L container. H2...

Chemistry - The decomposition of HI(g) is represented by the equation: 2HI(g) &...

Chem - I stuck on this problem. Can you help me to solve it? A reaction mixture ...

Chemistry - A student ran the following reaction in the laboratory at 650 K: H2(...

Chemistry II - At a certain temperature, Kc = 33 for the reaction: H2(g) + I2(g...

Chemistry 12 - given the reacting system: H2 (G) + I2(G) ---> 2HI(g) Keq= 64 ...

Chemistry 12 - given the reacting system: H2 (G) + I2(G) ---> 2HI(g) Keq= 64 ...

Chemistry - The equilibrium constant for HI decomposition at 500 °C is 5.8×10-3...