Chemistry
posted by Lena .
Initially, 1.68 mol of PCl5(g) and 0.36 mol of PCl3(g) are in mixed in a 2.00 l container. It is later found that 1.44 mol of PCl5 are present when the system has reached equillibrium. Calculate the value of the equllibrium.
This is what I did:
Equation:
PCl5(g) <> PCl3(g) + Cl2(g)
PCl5(g): Intial Concentration = 1.68 mol/2.00l = 0.84 mol/L
Change in concentration =0.84  (1.44 mol/2.00 L) = 0.84 mol/L 0.72 mol/L= 0.12 mol/L
Equallibrium = 0.12 mol/L
PCl3(g): Initial concentration = 0.36mol/2.00 L = 0.18 mol/L
Change in concentration = 0.12 mol/L + 0.18 mol/L = 0.30 mol/L
Equillibrium = 0.30 mol/L
Cl2(g):
Initial Concentration = 0 mol/L
Change in conc. = 0.12 mol/L
Equillibrium = 0.12 mol/L
K= [0.30][0.12]/[0.12]
= 0.3
Did I do this correctly?

I'm sure you intended to divide by (PCl5) which is 0.72, not 0.12. Otherwise it looks ok to me.

Divide what :S
I actually guessed how to do this.
Im not too sure what I did.
How would I get the 0.72 for the equllibrium of PCl5? 
The problem tells you that the equilibrium concentration of PCl5 is 1.44 moles and that divided by 2 L = 0.72 moles/L. You used the following:
PCl5(g): Intial Concentration = 1.68 mol/2.00l = 0.84 mol/L
Change in concentration =0.84  (1.44 mol/2.00 L) = 0.84 mol/L 0.72 mol/L= 0.12 mol/L
Equallibrium = 0.12 mol/L
0.72 to subtract from 0.84 to find the change of 0.12 but 0.72 is the final concn of PCl5. 
OHHHH. OK! But the equllibriums for the rest is correct, right?

The rest of it looks ok. I would have done the PCl5 part this way: You had
PCl5(g): Intial Concentration = 1.68 mol/2.00l = 0.84 mol/L is ok.
Change in concentration =0.84  (1.44 mol/2.00 L) = 0.84 mol/L 0.72 mol/L= 0.12 mol/L is ok.
Equallibrium = 0.12 mol/L should read equilibrium 0.72 mol/L.

so for the Kc value it would change correct? I did this:
Kc = [PCl3][Cl2]/[PCl5]
= [0.30][0.12]/[0.72]
= 0.05 
I have 0.05 also.

Thank you :)