chemistry

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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) „²„³ 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) „²„³ 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 ¡V x)

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

  • chemistry -

    Answered.

  • chemistry -

    It is incorrect, where , sadly I cannot say:(

  • chemistry -

    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 -

    Hmm.. I will check over if I typed in the correct numbers and I will let you know

  • chemistry -

    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 -

    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 -

    Then 0.056 for CO and H2O should be correct.
    And 0.044 for CO2 and H2 should be correct.

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