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Determining final concentrations of a Reaction given an equilibrium K

  1. Feb 16, 2014 #1
    1. The problem statement, all variables and given/known data

    Rxn: 2SO2 (g) + O2 (g) [itex]\Leftrightarrow[/itex] 2SO3 (g)

    If the initial concentration of SO3 is .500 M (moles/liter) @1500K, what are the final concentrations of [SO2], [O2] and [SO3]. Kc = .15

    Stop when you get to the cubic equation

    2. Relevant equations

    RICE table

    Kc= [SO3]2/ [O2][SO2]2 This is equilibrium quotient for forward reaction. I will flip it to get reverse reaction Kc equation



    3. The attempt at a solution

    This is the Kc given for the forward reaction. If I flip the reaction around I invert the Kc and it becomes = 6.6667 (for the reverse reaction)

    I then used a "rice" table with the initial concentration of SO3 and "x" variable for the SO2 and O2[

    Plugging these values in I get the kc equation:

    6.6667= [2x]2[x]/[.500-2x]

    I'm not getting the correct cubic equation when expanding this quotient. I'm sure of my math, so I was wondering if i'm setting up the quotient wrong? - Thanks!
     
  2. jcsd
  3. Feb 16, 2014 #2

    Borek

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    Staff: Mentor

    You don't need to reverse the reaction (you flipped and calculated the inverse of the Kc, which effectively cancels out).

    What is the "correct" answer you are expecting to get? At first sight I don't see anything wrong with your work.
     
  4. Feb 21, 2014 #3

    epenguin

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    Homework Helper
    Gold Member

    Looks OK to me too. Maybe the problem is this is a fairly complete conversion so don't round down too much your x as this could introduce considerable error in your (0.5 - 2x) when you back calculate to check?

    (I get x = 0.24566)
     
  5. Feb 21, 2014 #4
    Thanks for the input Borek and epenguin.
    If turns out there was an error on the answer sheet I was given.
    My cubic equation was correct after all.
     
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