Equilibrium Chem Problem?

Question

I need some help on this problem I've been trying to work out. Any help appreciated. thanks!


Suppose that 0.155 mol PCl5 is placed in a reaction vessel of volume 500.0 mL and allowed to reach equilibrium according to the following reaction:

PCl5 <-> PCl3 + Cl2 (Kc = 1.80 at 250 °C)

What are the equilibrium concentrations of each reactant and product?
(All three substances are gases at 250 °C)

Answer 1

PCl5 <-> PCl3 + Cl2
0.31M -x......x ........x

Using the quadratic equation x^2 +1.80x - 0.558 =0
and solving for x
x = 0.269 or 0.27 M = PCl3 = Cl2
0.31 - 0.27 = 0.04M = PCl5

....... dots added to separate the x values.

Answer 2

At equilibrium x [PCl3] forms = [Cl2]

Depletion of original [PCl5] = 0.155/.5 - 2x

so Kc = [Cl2][PCl3]/[PCl5] = x2(0.155/.5-2x) = 1.8

gives

x2 + 3.6x - 0.558 = 0

x = 0.149 M

Equilibrium concentrations:

[Cl2]=[PCl3] = 0.149 M

[PCl5] = 0.012 M

Answer 3

Kc = [PCl3][Cl2]/[PCl5] = 1.80

Let [PCl3] = [Cl2] = x

Then [PCl5] = (0.310 - x) (Because 0.155mol/500mL x 1000mL/1L)

x^2/(0.310 - x) = 1.80

x^2 = 0.558 - 1.80x

x^2 + 1.80x - 0.558 = 0 and quadratic equation from there.