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Lena772

  • one year ago

At 1000 K, Kc = 2.11E-2. If one starts with 1.00 mole graphite and 1.00 mole CO2 (g) in a 40.0 L vessel, what is the equilibrium concentration of CO (g)?

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  1. Lena772
    • one year ago
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    I caclulated ad got x=0.021644, but when I plugged that in to find the conc. of CO it was wrong.

  2. Lena772
    • one year ago
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    Cs, graphite + CO2 (g) ⇌ 2 CO (g)

  3. Lena772
    • one year ago
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    @Abhisar

  4. JFraser
    • one year ago
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    if you included the "concentration" of the solid graphite, you shouldn't

  5. Lena772
    • one year ago
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    i did. do i sassume 2 moles of CO as well because it doesn't explicitly say that but its 1to1to2

  6. JFraser
    • one year ago
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    the balanced reaction will create 2 moles of CO, you're correct. But you don't have 2 moles of CO at equilibrium

  7. JFraser
    • one year ago
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    remember that in equilibrium, only \(some\) of the reactants combine to form \(some\) of the products Use an ICE table approach and set the unknown concentrations equal to the Kc value

  8. Lena772
    • one year ago
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    I mean is the CO value for moles at initial 2? b/c then you would say 2mol/40.0L of soln to get the initial conc. of CO? Which is 0.05

  9. JFraser
    • one year ago
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    your initial value of CO in the flask is zero. You need to find the concentrations \(at\) \(equiliubrium\) \[K_C = \frac{[CO]^2}{[CO_2]}\]

  10. Lena772
    • one year ago
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    but kc is given

  11. JFraser
    • one year ago
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    KC is given, but the concentrations are missing. that's what you're solving for

  12. JFraser
    • one year ago
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    the starting concentration of \(CO_2\) is 0.025M, correct?

  13. Lena772
    • one year ago
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    yes

  14. Lena772
    • one year ago
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    but 0^2/0.025 is 0...

  15. JFraser
    • one year ago
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    the \(equilibrium\) concentration of \(CO_2\) will be (0.025M - x)

  16. Lena772
    • one year ago
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    yes

  17. JFraser
    • one year ago
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    the equilibrium concentration of \(CO\) is not zero

  18. JFraser
    • one year ago
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    as the \(CO_2\) is used, some CO is \(formed\)

  19. JFraser
    • one year ago
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    for every "x" amount of \(CO_2\) that is used, the amount of \(CO\) that is formed is \(twice\) that amount, because of the balanced reaction

  20. JFraser
    • one year ago
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    so the equilibrium concentration of \(CO\) is "2x"

  21. Lena772
    • one year ago
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    wouldn't it be -2x cause Q<Kc

  22. JFraser
    • one year ago
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    \(CO\) is the product \(formed\), so it's amount is "2x"

  23. JFraser
    • one year ago
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    the equilibrium equation then looks like this

  24. Lena772
    • one year ago
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    would it be 2.11e-2=(2x)^2/(0.025+x)?

  25. JFraser
    • one year ago
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    \[K_C = \frac{[CO]^2}{[CO_2]} =\frac{[2x]^2}{[0.025-x]} = 0.0211\]

  26. JFraser
    • one year ago
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    the \(CO_2\) is \(used \space up\) so its equilibrium concentration must be less than what you started with

  27. Lena772
    • one year ago
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    I just thought when Q<Kc the equilibrium shifted left :/

  28. JFraser
    • one year ago
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    it does, but that's not what this question is asking you for

  29. JFraser
    • one year ago
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    it tells you that you start with graphite \(C(s)\) and \(CO_2\). there is no \(CO\) formed yet, so the reaction \(must\) proceed forward

  30. Lena772
    • one year ago
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    0.01829034 M = CO at equilibrium?

  31. Lena772
    • one year ago
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    No i didn't. I get x=-0.0144202, x=0.00914517. I multiplied the second x value by 2, and that's what I got...

  32. Lena772
    • one year ago
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    Multiplying the first one gives a negative, and we can't have negative concentration

  33. JFraser
    • one year ago
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    did you solve it quadratically, or take the lazy shortcut?

  34. Lena772
    • one year ago
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    i graphed it.

  35. JFraser
    • one year ago
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    what did you plug in for the coefficients?

  36. Lena772
    • one year ago
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    2.11e-2=((2x)^2)/(0.025-x)

  37. JFraser
    • one year ago
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    I don't normally solve them that way, but it's probably close enough. I got 0.0229M

  38. Lena772
    • one year ago
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    are you good at lab reports?

  39. Lena772
    • one year ago
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    @JFraser

  40. JFraser
    • one year ago
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    usually, but what things you should put in it really depends on what your teacher has asked for

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