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Chemistry equilibrium Question

  1. Feb 5, 2008 #1
    [SOLVED] Chemistry equilibrium Question

    1. The problem statement, all variables and given/known data

    K=64. at 25 degrees celsius

    N2(g) + 3H2(g) <--> 2NH3(g)

    Molarity N2= .5M H2= 1.5M. what are the equilibrium concentrations?

    i got the equilibrium table set up, and the K expression is
    (2x)(2x)
    k= ------------------------------
    (.5-x)(1.5-3x)(1.5-3x)(1.5-3x)

    how can i solve for x???
     
  2. jcsd
  3. Feb 6, 2008 #2
    I'm not sure if you raise the equilibrium concentrations to powers; if not, I have:

    K = 2x / (0.5 - x)(1.5 - 3x)
    K = 2x / (3x^2 - 3x + 0.75)
    (3x^2 - 3x + 0.75)K = 2x
    (3x^2 - 3x + 0.75)K -2x = 0
    K3x^2 - K3x + K0.75 - 2x = 0
    K = 64, so:

    192x^2 - 190x + 48 = 0

    Plug this into the quadratic equation, then substitute the answer for x for each concentration. For example, since the equilibrium concentration for NH2 is 2x, if x = 4, the equilibrium concentration would be 8M.

    Just keep in mind that I'm not sure about raising the terms to stoichiometric powers in this kind of problem...
     
  4. Feb 6, 2008 #3

    chemisttree

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    For equilibrium of a reaction of the type,

    [tex]aA + bB +cC + .... \leftrightharpoons\ dD + eE + ....[/tex]

    the equilibrium expression is:

    [tex] \frac{[D]^d[E]^e...}{[A]^a^b[C]^c...}[/tex]

    I'm not sure why you are using 'x' in your problem since it really isn't necessary.
     
    Last edited: Feb 6, 2008
  5. Feb 6, 2008 #4
    forget it i found out what to do
     
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