Chemistry

A chemist wants to prepare phosgene, COCl2, by the following reaction: CO(g) + Cl2(g) COCl2(g) He places 2.60 g of chlorine, Cl2, and an equal molar amount of carbon monoxide, CO, into a 10.00 L reaction vessel at 395 °C. After the reaction comes to equilibrium, he adds another 2.60 g of chlorine to the vessel in order to push the reaction to the right to get more product. What is the partial pressure of phosgene when the reaction again comes to equilibrium? Kc = 1.23E+3.

  1. 7
asked by Nina
  1. These numbers are estimates so you need to go through and recalculate all of them.
    (Cl2) = mols/L = 2.60/71/10 = about 0.004M
    (CO) = 0.004 (actually closer to 0.00366).
    ..........CO + Cl2 ==> COCl1
    I.....0.004..0.004.......0
    C........-x.....-x.......x
    E.....0.004-x.0.004-x....x

    Kc = (COCl2)/(CO)(Cl2)
    1.23E3 = (x)/(0.004-x)^2
    solve for x and I obtained about 0.002
    At the 1st equilibrium (CO) = 0.004-0.002 = bout 0.002 and (Cl2) = about 0.002 with (COCl2) = about 0.002

    So now set up a second ICE chart like this
    ..........CO + Cl2 ==> COCl2
    I.......0.002..0.002...0.002
    add....+0.004..0.004.....0..
    I.......0.006..0.006....0.002
    C.......-x.......-x.......x
    E.......0.006-x..0.006-x...x

    Substitute into Kc and solve for x = M COCl2.
    Then convert to mols COCl2 and use PV = nRT to solve for pressure.
    Post your work if you get stuck. Remember those numbers I've used are estimates and I've rounded extensively.

    posted by DrBob222
  2. Thank you but i'm not understanding how you got .002 for x

    posted by Nina
  3. You solve the equation.
    mols Cl2 = 2.60/71 = 0.0366
    M Cl2 = mols/L = 0.0366/10 = 0.00366
    M CO = 0.00366

    Kc = (COCl2)/(CO)(Cl2)
    1.23E3 = [(x)/(0.00366-x)^2]
    1.23E3*(0.00366-x)^2 = x
    1.23E3*(0.00366-x)(0.00366-x) = x
    and go from there.
    x = 0.0023 I believe although I've thrown my work away.
    Then (CO) = (Cl2) = 0.00366-x = 0.00366-0.0023 = 0.00136M
    Check these numbers correctly. I worked it through the first equilibrium and those numbers are what I remember. Perhaps I just don't remember correctly. Then you go through the second equilibrium as I've shown and finally use PV = nRT

    posted by DrBob222
  4. How do you get moles of COCl2 from the [COCl2].

    Instead of using PV=nRT to find the partial pressure, shouldn't I use Kp=Kc(RT)^delta(n)

    posted by Andrew P.
  5. After you go through the equilibrium the second time you end up with M COCl2.
    You know M = mols/L. You know M from your calculation and you know L (10L) so
    mols = M x L = ? = n and use PV = nRT to solve for pressure. No, you don't go through the Kp thing because there is no pressure in that equation. If you want pressure, the only way to get it is to use PV = nRT. You could have converted Kc to Kp if you wish but since you are given concentrations it is easier to use Kc from the beginning, especially since that is what is given.

    posted by DrBob222

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