Chemistry

HCONH2(g) <-> NH3(g) + CO2(g) Kc= 4.84 at 400 K2(g)

If 0.186 mol of HCONH2(g) dissciates in 2.16L at 400K, what will be the total pressure at equilibrium?

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asked by K
  1. ............HCONH2 ==> NH3 + CO2
    initial.....0.186 mols..0.....0
    change.......-x.........x......x
    equil.......0.186-x.....x......x

    Substitute the above ICE chart into the Kc expression for the reaction and solve for x. That will give you moles NH3 and moles CO2. Add the moles together and substitute into PV = nRT to solve for pressure.

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  2. When I am solving for X my formula comes out to be

    4.84 = X^2/ (.186-X)

    Can I assume that x << .05 and eliminate it from the denominator and just solve for the equation 4.84 = X^2/.186?

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    posted by K
  3. No, you may not make that assumption; however, I may have made an error. Let me think about this awhile. I'll post something different and erase what is there now or post a note that the original response stands.

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  4. Okay thanks so much.

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    posted by K
  5. OK. I made an error and the problem can't be solved that way. First, we convert moles to molarity.
    (HCONH2) = 0.186/2.16L = 0.08611M

    ............HCONH2 ==> NH3 + CO2
    initial.....0.08611M....0.....0
    change.......-xM.........xM......xM
    equil......0.08611-x M.....xM....xM

    Substitute the above ICE chart into the Kc expression for the reaction and solve for x. That will give you M NH3, M CO2 and M HCONH2. Convert M to moles for each (M x L = moles), add the moles together and substitute into PV = nRT to solve for pressure. I will leave the original post there for the time being so you can make comparisons if you wish; however, I'll erase it before I go to bed tonight. I don't like to leave incorrect posts on the board.

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  6. Ok thanks so much.

    So when I solve for X I got

    4.48(0.086 - x) = x^2
    0.385 - 4.48x - x^2 = 0

    Now i'd substitute that into the quadratic formula? Or is there an easier way to solve for X?

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    posted by K
  7. That K is 4.84 and not 4.48 so you need to redo that part. When you solve that quadratic that gives you x (and I don't know of an easier way to do it.) Most calculators now have that built in to solve those things. I programmed mine to do that. That gives you x, use that to determine (HCONH2), (NH3), and (CO2), convert concns to moles, add the moles to find the total, then use PV = nRT to solve for P.

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  8. Ok this helps so much. Thanks a lot I appreciate it.

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    posted by K

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