Question

When 4.00 moles of PCl5 (g) is put in a 1.00 L vessel the equilibrium: PCl3 (g) + Cl2 (g) <--> PCl5 (g) is established. At equilibrium [Cl2]e = 5.20 x 10^-3 M. What is Kc for this equilibrium as written?

a.) 9.61 x 10^2

b.) 6.77 x 10^-6

c.) 1.48 x 10^5

d.) 9.0 x 10^-6

e.) impossible to determine

Answer 1

             PCl3 (g) + Cl2 (g) <--> PCl5

I            0                  0                   4

C          0.005           0.005             -0.005

E        0.005             0.005            3.995

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

        = 3.995/0.005*0.005

        = 1.59*10⁵

c.) 1.48 x 10^5 >>>>> answer