The equilibrium constant (Kp) for the interconversion of PCl5 and PCl3 is 0.0121:

Question

PCl5 (g) ↔ PCl3 (g) + Cl2 (g)

A vessel is charged with PCl5, giving an initial pressure of 0.123 atm. At equilibrium, the partial pressure of PCl3 is __________ atm.

I have a hit a wall with this question, this is where I got to:

PCl5 (g) ↔ PCl3 (g) + Cl2 (g)
I: 0.123 0 0
C: -2x. x x
E: 0.123-2x x x

Kp= [PCl3][Cl2]/[PCl5]
Kp=[x][x]/[0.123-2x]
0.0121=x^2/0.123

and thats where I have gotten stuck can't seem to get the correct answer, I know is 0.0330 but I can not figure out how to exactly get to it. Thanks for any help.

Answer 1

First, that C line is -x.........x.......x and not -2x
ane the E line is 0.123-x.......x.......x

Then you have Kp = 0.0121 = (x)(x)/(0.123-x) and solve that quadratic. I get 0.032996 which rounds to 0.0330