The chemical reaction is as follows:
PCl5(g)<=====>PCl3(g)+Cl2(g)
Given that; Kc = 1.80 at at 250 °C
Now write the expression for the equilibrium constant of this
reaction:
Kc = [PCl3][Cl2] / [PCl5]
Molarity of PCl5(g) = moles / volume in L= 0.268 mol/ 4.05 L
=0.066 M
PCl5(g)<=====>PCl3(g)+Cl2(g)
I
                    Â
0.066
                                  Â
0 Â Â Â Â Â Â Â 0
C
                   Â
-XÂ Â Â Â Â Â Â Â Â
                   Â
+XÂ Â Â Â Â Â +X
E
                   Â
0.066-XÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â
X Â Â Â Â Â Â Â Â X
At equilibrium
Kc= X*X/0.066-X
1.80 = X^2 / (O.066-X)
X^2-0.1188 +1.80X=0
I solved the above by the quadratic formula
X = (-b ±√ (b² - 4ac) ] / (2a)
X= 0.0637
PCl5(g)= 0.066-X= 0.066-0.0637=0.0023M
PCl3(g)= Cl2(g) =X= 0.0637
