These numbers are estimates so you need to go through and recalculate all of them.
(Cl2) = mols/L = 2.60/71/10 = about 0.004M
(CO) = 0.004 (actually closer to 0.00366).
..........CO + Cl2 ==> COCl1
I.....0.004..0.004.......0
C........-x.....-x.......x
E.....0.004-x.0.004-x....x
Kc = (COCl2)/(CO)(Cl2)
1.23E3 = (x)/(0.004-x)^2
solve for x and I obtained about 0.002
At the 1st equilibrium (CO) = 0.004-0.002 = bout 0.002 and (Cl2) = about 0.002 with (COCl2) = about 0.002
So now set up a second ICE chart like this
..........CO + Cl2 ==> COCl2
I.......0.002..0.002...0.002
add....+0.004..0.004.....0..
I.......0.006..0.006....0.002
C.......-x.......-x.......x
E.......0.006-x..0.006-x...x
Substitute into Kc and solve for x = M COCl2.
Then convert to mols COCl2 and use PV = nRT to solve for pressure.
Post your work if you get stuck. Remember those numbers I've used are estimates and I've rounded extensively.
A chemist wants to prepare phosgene, COCl2, by the following reaction: CO(g) + Cl2(g) COCl2(g) He places 2.60 g of chlorine, Cl2, and an equal molar amount of carbon monoxide, CO, into a 10.00 L reaction vessel at 395 °C. After the reaction comes to equilibrium, he adds another 2.60 g of chlorine to the vessel in order to push the reaction to the right to get more product. What is the partial pressure of phosgene when the reaction again comes to equilibrium? Kc = 1.23E+3.
5 answers
Thank you but i'm not understanding how you got .002 for x
You solve the equation.
mols Cl2 = 2.60/71 = 0.0366
M Cl2 = mols/L = 0.0366/10 = 0.00366
M CO = 0.00366
Kc = (COCl2)/(CO)(Cl2)
1.23E3 = [(x)/(0.00366-x)^2]
1.23E3*(0.00366-x)^2 = x
1.23E3*(0.00366-x)(0.00366-x) = x
and go from there.
x = 0.0023 I believe although I've thrown my work away.
Then (CO) = (Cl2) = 0.00366-x = 0.00366-0.0023 = 0.00136M
Check these numbers correctly. I worked it through the first equilibrium and those numbers are what I remember. Perhaps I just don't remember correctly. Then you go through the second equilibrium as I've shown and finally use PV = nRT
mols Cl2 = 2.60/71 = 0.0366
M Cl2 = mols/L = 0.0366/10 = 0.00366
M CO = 0.00366
Kc = (COCl2)/(CO)(Cl2)
1.23E3 = [(x)/(0.00366-x)^2]
1.23E3*(0.00366-x)^2 = x
1.23E3*(0.00366-x)(0.00366-x) = x
and go from there.
x = 0.0023 I believe although I've thrown my work away.
Then (CO) = (Cl2) = 0.00366-x = 0.00366-0.0023 = 0.00136M
Check these numbers correctly. I worked it through the first equilibrium and those numbers are what I remember. Perhaps I just don't remember correctly. Then you go through the second equilibrium as I've shown and finally use PV = nRT
How do you get moles of COCl2 from the [COCl2].
Instead of using PV=nRT to find the partial pressure, shouldn't I use Kp=Kc(RT)^delta(n)
Instead of using PV=nRT to find the partial pressure, shouldn't I use Kp=Kc(RT)^delta(n)
After you go through the equilibrium the second time you end up with M COCl2.
You know M = mols/L. You know M from your calculation and you know L (10L) so
mols = M x L = ? = n and use PV = nRT to solve for pressure. No, you don't go through the Kp thing because there is no pressure in that equation. If you want pressure, the only way to get it is to use PV = nRT. You could have converted Kc to Kp if you wish but since you are given concentrations it is easier to use Kc from the beginning, especially since that is what is given.
You know M = mols/L. You know M from your calculation and you know L (10L) so
mols = M x L = ? = n and use PV = nRT to solve for pressure. No, you don't go through the Kp thing because there is no pressure in that equation. If you want pressure, the only way to get it is to use PV = nRT. You could have converted Kc to Kp if you wish but since you are given concentrations it is easier to use Kc from the beginning, especially since that is what is given.
2 answers