Homework Help: chemistry

Posted by Jessica on Monday, March 31, 2008 at 11:34pm.

I would also like to see Nathaniels work solved:P I have a similar question

If 1.00 mol each of carbon dioxide and hydrogen is initially injected into a 10.0-L reaction chamber at 986 degrees Celsius, what would be the concentrations of each entity at equilibrium?

CO2(g) + H2(g) „І„і CO(g) + H2O(g) k = 1.60 for 986 degrees Celsius
Heres my work:

C = n/v
= 0.1 mol/L
CO2(g) + H2(g) „І„і CO(g) + H2O(g)
I 0.1 0.1 0 0
C -x -x +x +x
E 0.1-x 0.1-x x x

Keq = [CO][H2O]/[CO2][H2]

1.60 = x(x)/(0.1 -x)(0.1 ЎV x)

This is the part I am stuck with please help thanks a lotѓє

chemistry - DrBob222, Monday, March 31, 2008 at 11:45pm
Answered.
- chemistry - Jessica, Monday, March 31, 2008 at 11:53pm
  It is incorrect, where , sadly I cannot say:(
  - chemistry - DrBob222, Monday, March 31, 2008 at 11:59pm
    It looks ok to me.
    and we can check it.
    (0.0558)^2/(0.0442)^2 = 1.594 which is so close to 1.60 (Keq in the problem) it isn't funny. If you are keying into a computer data base, then I suspect there is a problem with s.f. You must decide how many s.f. to report and key those numbers in.
    - chemistry - Jessica, Tuesday, April 1, 2008 at 12:01am
      Hmm.. I will check over if I typed in the correct numbers and I will let you know
      - chemistry - Laureta, Tuesday, April 1, 2008 at 12:33am
        I will rewrite the equaiton making it clearer
        
        CO2(g) + H2(g)<---> CO(g) + H2O(g) k = 1.60 for 986 degrees Celsius
        
        Does this change anything?
        
        chemistry - Jessica, Tuesday, April 1, 2008 at 12:37am
        For simplicity, all answers should be correct to two (2) significant figures and not expressed in scientific notation.
        
        The above was also found next to the question
chemistry - DrBob222, Tuesday, April 1, 2008 at 1:10pm
Then 0.056 for CO and H2O should be correct.
And 0.044 for CO2 and H2 should be correct.

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