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Tricky equilibrium constant question

  1. Feb 4, 2007 #1

    Roq

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    Hi, I'm completely out of ideas for this problem, and any help would be really appreciated.

    1. The problem statement, all variables and given/known data
    At 850 degrees C and 1 atm pressure, a gaseous mixture of carbon monoxide and carbon dioxide in equilibrium with solid carbon is 90.55% CO by mass. Find the equilibrium constant for this reaction at 850 C.
    C(s) + CO2(g) <--> 2CO(g)


    2. Relevant equations
    K=[products]/[reactants], PV=nRT(?)


    3. The attempt at a solution

    Since the C concentration is solid, it is constant, so it is part of the equilibrium constant, so K will be = [CO]^2/[CO]. So I just need to determine how much the CO is from being 90.55% of the mass.
    I took the atomic weights of CO (~28 amu) and CO2 (~44 amu) and made an equation that would result in CO being 90.55% of the mass.
    28x=(.9055)(28x+44y)
    X ends up being ~15 times greater than Y.
    This is where I'm stuck. I can use my X value to make a bunch of different amounts of CO and CO2 that have CO being 90.55% of the mass.
    My teacher hinted PV=nRT, but there is no given volume, and I see no way to use that equation with two unknowns.
    Also, the answer is K = .153. That doesn't help me though, since I need to show work.
     
  2. jcsd
  3. Feb 4, 2007 #2

    Borek

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    Staff: Mentor

    Do the symbolic calculations for the volume V - in the final result it should simply cancel out.
     
  4. Feb 4, 2007 #3

    Roq

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    How do I do that? I only see one step to solve for V... dividing nRT by 1 atm, which does not go anywhere.
     
  5. Feb 5, 2007 #4

    Gokul43201

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    The question is ambiguous as it does not specify whether you are asked for Kc or Kp. For the given reaction, Kp is certainly not the number given. As for Kc, it is not unitless, though the provided answer appears to be! (Clearly, something is messed up if you're written down the question and the answer correctly - you should ask you course instructor to fix it.)

    From the mole ratios you calculated (approx 1/16 and 15/16), you can write down the partial pressures, p(CO2) and p(CO), and hence calculate Kp. Then using the relationship between Kp and Kc, you can calculate Kc.
     
    Last edited: Feb 5, 2007
  6. Feb 5, 2007 #5

    Borek

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    Don't solve for V - assume volume V and do calculations as if it was given - at the end it should cancel out.

    Unless I am wrong.
     
  7. Feb 5, 2007 #6

    Roq

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    Thanks for the help guys.

    My friend gave me the answer from the solutions manual. The part I was missing was this:

    Assume there are 100 grams of CO and CO2. This will give the number of moles of CO and CO2. The number is arbitrary, but any chosen number will result in the same concentration of CO and CO2.

    I didn't think of this because I thought that picking a number of moles would change the concentration, but I guess not, since they are proportional.

    I have another question now. How can I derive Kc from Kp? I hadn't actually gone over Kp in any of my classes yet but it seemed pretty simple, but I did not see how to get Kc from it.
     
  8. Feb 5, 2007 #7

    Gokul43201

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    Consider the equilibrium:

    [tex]R_1 + R_2 + ...+R_n \leftrightharpoons P_1 + P_2 + ...+ P_m [/tex]

    where all reactants (R) and products (P) are gaseous (and the individual Rs and Ps need not be unique).

    then, [tex]K_p = \frac{p(P_1) \cdot p(P_2) \cdot \cdot\cdot p(P_m)}{p(R_1) \cdot p(R_2) \cdot \cdot \cdot p(R_n)} [/tex]

    [tex]~~~= \frac{(n(P_1)/V)RT \cdot (n(P_2)/V)RT \cdot \cdot \cdot (n(P_m)/V)RT}{(n(R_1)/V)RT \cdot (n(R_2)/V)RT \cdot \cdot \cdot (n(R_n)/V)RT} [/tex]

    [tex]~~~== \frac{(n(P_1)/V) \cdot (n(P_2)/V) \cdot \cdot \cdot (n(P_m)/V)}{(n(R_1)/V) \cdot (n(R_2)/V) \cdot \cdot \cdot (n(R_n)/V}) \cdot (RT)^{m-n}[/tex]

    [tex]~~~=K_c \cdot (RT)^{m-n} [/tex]

    Note: This uses the popular definition of Kp. The are alternative definitions used by some sources that make Kp dimensionless - these are not the popular among high school and early college classes though.
     
  9. Feb 5, 2007 #8

    Roq

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    Thank you very much.
     
  10. Feb 11, 2007 #9

    Borek

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    Note it is the same approach as using V - in both cases amounts (be it symbolic V or non-symbolic 100g) cancel out at the end.



     
  11. Feb 11, 2007 #10
    Kc = [CO]^2/[CO]
    reduces to
    Kc = [CO]

    assuming 100g total, and 90.55g CO (30.18 mol), we need to find volume and then the answer will be 30.18mol / volume

    we can use PV=nRT to find the volume

    PV = nRT
    V = nRT / P

    we know temp, pressure, R, but we gotta find n

    n = total moles. so thats 30.18 mol + carbon dioxide's moles

    now we gotta find carbon dioxide's moles

    CO2 makes up the rest of the mass, so 100g - 90.55g = 9.45g
    9.45g / 44 (molar mass) = 0.2148 mol

    so total moles = 30.18 mol + 0.2148 mol = 30.39 mol

    back to PV=nRT...

    P = 1atm
    n = 30.39 mol
    R = .082 L*atm / (K * mol)
    T = 850C = 1123K

    V = (30.39mol * 1123K * .082 L*atm*K-1*mol-1) / 1atm
    units work out (so maybe i am right...) and V = 2800 L

    Kc = [CO] = 30.18 mol / 2800 L = 0.01078

    ummmmm hope i did it right
     
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