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.
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.
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
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)
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.
To find the partial pressure of phosgene, COCl2, when the reaction comes to equilibrium, we need to use the equilibrium constant, Kc, and the given information about the reactants and products.
First, let's determine the number of moles of chlorine, Cl2, and carbon monoxide, CO, in the 2.60 g of chlorine.
1. Calculate the molar mass of chlorine, Cl2:
Molar mass of Cl2 = 35.45 g/mol (atomic mass of Cl) * 2 = 70.90 g/mol
2. Calculate the number of moles of Cl2:
Moles of Cl2 = mass of Cl2 / molar mass of Cl2
= 2.60 g / 70.90 g/mol
≈ 0.0366 moles
Since the reaction is 1:1 for Cl2 and CO, the number of moles of CO is also 0.0366 moles.
Next, let's calculate the initial partial pressures of CO and Cl2 in the reaction vessel.
3. Calculate the initial partial pressures of CO and Cl2:
Partial pressure = (moles / total volume) * (ideal gas constant * temperature)
For CO:
Partial pressure of CO = (moles of CO / total volume) * (ideal gas constant * temperature)
= (0.0366 moles / 10.00 L) * (0.0821 L*atm/(mol*K) * (395 + 273) K
≈ 0.156 atm
For Cl2:
Partial pressure of Cl2 = (moles of Cl2 / total volume) * (ideal gas constant * temperature)
= (0.0366 moles / 10.00 L) * (0.0821 L*atm/(mol*K) * (395 + 273) K
≈ 0.156 atm
Since the initial reaction is at equilibrium, the partial pressure of phosgene, COCl2, can be calculated using the equilibrium constant, Kc.
4. Calculate the partial pressure of COCl2 at equilibrium using Kc:
Kc = (Partial pressure of COCl2) / (Partial pressure of CO * Partial pressure of Cl2)
Rearranging the equation:
Partial pressure of COCl2 = Kc * (Partial pressure of CO * Partial pressure of Cl2)
Plugging in the given value for Kc:
Partial pressure of COCl2 = 1.23E+3 * (0.156 atm * 0.156 atm)
≈ 30.722 atm
Therefore, the partial pressure of phosgene, COCl2, when the reaction again comes to equilibrium is approximately 30.722 atm.