# Chemistry

posted by .

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?

• Chemistry -

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

• Chemistry -

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?

• Chemistry -

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.

• Chemistry -

OHHHH. OK! But the equllibriums for the rest is correct, right?

• Chemistry -

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

• Chemistry -

so for the Kc value it would change correct? I did this:

Kc = [PCl3][Cl2]/[PCl5]
= [0.30][0.12]/[0.72]
= 0.05

• Chemistry -

I have 0.05 also.

• Chemistry -

Thank you :)