Solution :-
Balanced reaction equation
PCl5 <-----> PCl3 + Cl2
Kc= 1.80
Lets calculate the initial concentration of the PCl5 at the
initial
Molarity = moles / volume in liter
[PCl5]= 0.135 mol / 2.15 L
         = 0.0628
M
Now lets set up the ICE table
PCl5Â Â Â Â <----->
    PCl3    + Â
Cl2
0.0628
MÂ Â Â Â Â Â Â Â Â Â Â Â Â Â
0Â Â Â Â Â Â Â Â Â Â Â Â
0
-x                          Â
+x          Â
+x
0.0628-x             Â
x            Â
x
Kc= [PCl3][Cl2]/[PCl5]
1.80= [x][x]/[0.0628-x]
1.80*[0.0628-x] =x^2
Solving for the x we get
0.0607 M=x
So the equilibrium concentrations of the PCl3 and Cl2 are as
follows
[PCl3]eq = x = 0.0607 M
[Cl2]eq = x = 0.0607 M
