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Question in regards to the equilibrium law and expressions

  1. Mar 13, 2005 #1
    Considering the following equilibrium reaction

    [tex]A(g) \longleftrightarrow 2B(g) + C(g)[/tex]

    And the question:

    When 1.00 mol of A is placed in a 4.00 L container at temperature t, the concentration of C at equilibrium is 0.050 mol/L. What is the equilibrium constant for the reaction at temperature t?

    Does this mean that for every x mol of A consumed, 2(C) mol of B is produced and 0.050 mol of C is produced.

    For example

    We get the number of mols of C within the 4.00 L container:

    0.050 mol/L C x 4.00 L = 0.20 mol C

    Then we use that number to calculate the number of mols produced and consumed at equilibrium for the other gases:

    0.20 mol C x 2 mol B/ 1 mol C = 0.20 x 2 mol B = 0.40 mol B

    1.00 mol A - [0.20 mol C x 1 mol A / 1 mol C] = 0.80 mol A

    Then for the concentrations at equilibrium:

    A = 0.80 mol / 4.00 L = 0.20 mol/L
    B = 0.40 mol / 4.00 L = 0.10 mol/L
    C = 0.20 mol / 4.00 L = 0.050 mol/L

    then the equilibrium constant:

    Ke = ^2 [C] / [A]
    =[0.10]^2 [0.050] / [0.20]
    =0.0025 mol/L

    0.0025 mol/L seems like a small number for the equilibrium constant which is why I've come here to see if anyone can check the work I've done and point out any mistakes if there are any. I think I might be making a mistake with the 2B, I'm not sure if I'm handling that properly.
    Last edited: Mar 13, 2005
  2. jcsd
  3. Mar 13, 2005 #2
  4. Mar 14, 2005 #3


    User Avatar

    Staff: Mentor

    No. It means that for every x mol of A 2x mol of B and x mol of C is produced, just like you have in you reaction equation.

    But the rest seems OK. 0.0025 is nothing unusuall.

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    Last edited: Mar 14, 2005
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