[kimi85]

Does anyone know how to set-up an equation if the concentrations are already in equilibrium, then an amount of the reactant was added. Kc is also given?

Thank you. 

[Borek]

http://www.chembuddy.com/?left=pH-calculation&right=general-pH-calculation

Read first paragraph.

[kimi85]

thank you. I have read it.

Here's the problem:
For the process below, K = 1025 at some temperature. The system is initially at equilibrium with 0.0400 M CO, 0.100 M Cl2, and 4.10 M COCl2. If 0.060 mol CO is added, and the volume of the container is subsequently doubled, what is the molarity of CO after equilibrium is reestablished?

CO(g) + Cl2(g) <==> COCl2(g)
 
 a.0.0052 M
 b.0.050 M
c. -0.46 M
 d. 0.045 M

[firzzy87]

this should be belong to physical chemistry forum

[kimi85]

Well, it can also be in inorganic chemistry. Some topics in inorganic chemistry are also in physical chemistry. 

[vladi307]

I calculate the concentration of CO for V= 1litre initially. I find it 0.045M. Is this correct? Do you have the answer?

[kimi85]

Yes, that's correct. How did you get it? Can you show your solution? thank you very much. 

[vladi307]

Here is the solution:

Kc=[COCl2]eq/[CO]eq[Cl2]eq. If you have a initially volume of 1l, double it and put into the system  an additional amount of 0.060 mol CO then the equilibrium concentrations must be changed to balance the system.
Then Kc=(2,05-x)/(0,05-x)(0,05-x). The initial concentrations must be semidoubled because the volume doubled. If  you solve this equation you must find x=0.0052. [CO]eq=0.05-x=0.0448=0.045M

[AWK]

Then Kc=(2,05-x)/(0,05-x)(0,05-x)
??
Kc=c(COCl2) / (c(CO) x c(Cl2) ) = V x n(COCl2) / (n(CO) x n(Cl2) )
c - molar concentration, n - moles, V - volume in liters

After addition of CO and volume change
Kc = 2V x (n(COCl2)+X) / ( (n(CO)+0.06-X) x (n(Cl2)-X) )
capital X - change of reagents (in moles) to reach an equilibrium

[kimi85]

Thank you very much. But I still can't get the right answer. I used the quadratic equation. I'll just try again. Maybe something is wrong with my computation.

[kimi85]

I got it. I just forgot to divide my answer to 2 L.

[AWK]

I got 0.0495. If your result is the same, this should be a printing error

[kimi85]

I got 0.045 M.  I also followed your equation.  Thank you very much sir.

[vladi307]

Sorry, the equation is wrong. It is 2,05+x. My calculations are with the right equation but i write the wrong because I am very busy today. I am sorry.

[kimi85]

That's okay. Thank you very much.