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Homework Help: Chemical Equilibrium (answer check)

  1. Apr 25, 2006 #1
    Consider the following reaction

    A(g) <---> 2B(g) + C(g)

    When 1.00 mol of A is placed in a 4.00L 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?

    Here is my answer:

    The balanced equation shows that for each mol of C formed, 2 mol of B is formed and 1 mol of A is consumed. We are told that, at equilibrium, 0.050 mol of C has been produced. This means that 0.050 mol of B must also have been produced and that 0.050 mol of A was consumed. Therefore, at equilibrium,

    0.050 mol C x 2mol B / 1mol C = 0.1 mol B

    1.00 mol A - [0.050 mol C x 1mol A / 2mol C]

    = 0.975 / 0.98 mol A

    and the equilibrium concentrations are:

    [C] = 0.050 mol / 4.00 L = 0.0125 mol/L

    2 = 0.1 mol / 4.00 L = 0.025 mol/L

    [A] = 0.98 mol / 4.00 L = 0.245 mol/L

    Substitute the equilibrium concentrations in the equilibrium law expression and solve for Ke:

    Ke = 2[C] / [A]

    = 0.025 mol/L x 0.0125 mol/L
    0.245 mol/L

    = 0.0013 mol/L
     
  2. jcsd
  3. Oct 21, 2016 #2

    Bystander

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    It goes "south" from here; mixed stoichiometric and concentrative expressions. Salvageable? Maybe.
     
  4. Oct 21, 2016 #3

    epenguin

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    You're given final [C]. → [ B]

    [C] → moles C. Moles C, initial moles A, → equilibrium moles A. Is in 4 L → [A]

    Then you have all you need to calculate K.

    I made it 0.4.
     
  5. Oct 22, 2016 #4

    Bystander

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    [A]Init = 0.25: [A]Final = 0.2; (B)Final = 0.1, and [C]Final = 0.05, and 5 x 10-4 over 2 x 10-1 = 2.5 x 10-3?

    OK, how do I get conc. of B to render?
     
  6. Oct 22, 2016 #5

    epenguin

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    BTW OP's expression for the equilibrium law and constant is wrong.
     
  7. Oct 22, 2016 #6

    Bystander

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    The problem statement itself?
     
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