Question

The reaction: PCl₅(g) <-> PCl₃(g) + Cl₂(g) has K_c=0.0900. A 0.1000 mol sammple of PCl₅ is placed in an empty 1.00 flask and the above reaction is allowed to come to equilibrium at a certain temp. How manny moles of PCl_5, PCl₃ and Cl₂, respectively, are present at equilibrium?

Answer 1

        PCl₅(g) <-> PCl₃(g) + Cl₂(g)

I       0.1                 0             0

C     -x                   +x            +x

E     0.1-x               +x           +x

     Kc = [PCl3][Cl2]/[PCl5]

    0.09   = x*x/(0.1-x)

    0.09*(0.1-x) = x^2

         x      = 0.06

   [PCl3]   = x = 0.06moles

   [Cl2]     =   x   = 0.06 moles

[PCl5]     = 0.1-x = 0.1-0.06   = 0.04 moles