, what is the equilibrium concentration of Cl2 in the same system?

Initial concentrations:
[PCl₃]ₒ = [Cl₂]ₒ = (0.40 mole)/(2 L) = 0.20 M

          PCl₅(g)  ⇌  PCl₃(g)  +  Cl₂(g)    Kc = 0.040
Initial:      0 M     0.20 M   0.20 M
Change:    +y M     - y M    -y M
Eqm:       y M     (0.20 - y)M  (0.20 - y) M

At equilibrium:
Kc = [PCl₃] [Cl₂] / [PCl₅]
0.040 = (0.20 - y)² / y
0.040y = 0.040 - 0.40y + y²
y² - 0.44y + 0.040 = 0
y = 0.13  or  y = 0.31 (rejected for 0.20 - y < 0)

[Cl₂] at equilibrium = (0.20 - 0.13) M = 0.07 M

The answer: (a) 0.07 M

kc = 0.40 for the reaction
PCL5 (g) ==== PCL3 (g) + CL2
if reaction is initiated with 0.40 mol of CL2 and 0.40 mol PCL3 in a 2.0 liter container, what is the equilibrium concentration of CL2

    Construct ICE table
    
     PCL5 (g) ==== PCL3 (g) + CL2

I   0.4 mol            0.4 mol
   _______           ________
   
     2.0 L                 2.0L

C   -x                      -x                x


E  0.2 -x               0.2 - x           x


Calculate the Kc for the reaction as follow:


Kc = [PCL5 (g) ]/[PCl3 (g)][Cl2 (g)]

- 0.4 = [x]/[0.2 - x][0.2-x]

   x = -0.019

therefore, equilibrium concentration is 0.2 + 0.019

0.219M