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Homework Help: Chem: Equilibrium Constant Question

  1. Jan 31, 2009 #1
    1. The problem statement, all variables and given/known data
    For the equilibrium PCl5 <---> PCl3 + Cl2, all gases, Keq = 4.16e-2. The equilibrium concentration of PCl5 is 1.0 M, and both PCl3 and Cl2 are 0.204 M. If the volume of this closed system is doubled, what are the concentrations of each gas when a new equilibrium is reached?

    2. Relevant equations
    Keq = ([product 1]^n * [product 2]^n) / ([reactant 1]^n * [reactant 2]^n)

    3. The attempt at a solution
    old E: 1 M, 0.204 M, 0.204 M
    new E: PCl5: 1-x, PCl3: 0.204+x, Cl2: 0.204+x
    since the equilibrium would shift to the right where there's more moles.
    Now, when I try to solve it, I run into an obstacle:
    4.16e-2 = ((0.204+x)(0.204+x))/(1-x)
    4.16e-2 - 4.16e-2(x) = 0.0416+0.408x+x^2
    0 = x^2 + 0.4496x
    0 = x(x+0.4496)
    But x cannot be negative!
  2. jcsd
  3. Jan 31, 2009 #2


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    The volume doubles for for the new E you'd have to use 0.5-x and 0.102+x as your concentrations.
  4. Feb 1, 2009 #3
    Ooh that makes sense now... so the initial concentration is halved AND the equilibrium shifts.
    Would have never figured that out. Thanks.
    Last edited: Feb 1, 2009
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