Chem: Equilibrium Constant Question

[BioBabe91]

Homework Statement

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?

Homework Equations

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

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!

 

[Ygggdrasil]

The volume doubles for for the new E you'd have to use 0.5-x and 0.102+x as your concentrations.

 

[BioBabe91]

Ooh that makes sense now... so the initial concentration is halved AND the equilibrium shifts.
Would have never figured that out. Thanks.