Consider 1.40 mol of carbon monoxide and 3.20 mol of chlorine sealed in a 8.00 L container at 476 oC. The equilibrium constant, Kc, is 2.50 (in M-1) for
CO(g) + Cl2(g) ↔ COCl2(g)
Calculate the equilibrium molar concentration of CO.
So I tried this question by finding the equilibrium concentrations then using the quadratic equation to get .187 which wasn't right. Help?
Chemists studied the formation of phosgene by sealing 0.74 atm of carbon monoxide and 0.92 atm of chlorine in a reactor at a certain temperature. The pressure dropped smoothly to 1.36 atm as the system reached equilbrium. Calculate Kp (in atm-1) for
CO(g) + Cl2(g) ↔ COCl2(g)
I found the final pressure of the gases to get 94.44 which was really off because I think the answer is supposed to be either single digits or a decimal. Please help!
Thanks in advance!
1.40 mol/8.0 L = 0.175M = (CO)
3.20 mol/8.0L = 0.400 M = (Cl2)
............CO + Cl2 ==> COCl2
I........0.175...0.400.....0
C..........-x.....-x.......x
E.......0.175-x..0.400-x....x
Kc = 2.5 = (COCl2)/(CO)(Cl2).
Solve for x.
If you will post your work I will find the error. You know 0.187 can't be right; that's more CO than you started with.
.........CO + Cl2 ==> COCl2
I.......0.74..0.92.....0
C........-p....-p.......p
E.....0.74-p.0.92-p.....p
Ptotal = 1.36 = 0.74-p + 0.92-p + p
Solve for p = 0.30 and substitute into Kp expression. Solve for Kp.
thank you!
To solve both of these questions, we need to use the equilibrium expression and the given values to set up an equation. Let's start with the first question:
1. For the reaction CO(g) + Cl2(g) ↔ COCl2(g), the equilibrium constant expression Kc is given as 2.50 (in M^-1).
To calculate the equilibrium molar concentration of CO, let's assume the equilibrium molar concentration of CO is x M. Since the initial molar concentration of CO is 1.40 mol and the change at equilibrium will be -x (as CO is being consumed in the forward reaction), the equilibrium molar concentration of CO will be (1.40 - x) M.
The equilibrium molar concentration of Cl2 is 3.20 mol, and the equilibrium molar concentration of COCl2 will also be (1.40 - x) M (as one mole of COCl2 is produced per mole of CO consumed in the forward reaction).
Using the equilibrium constant expression, we can set up the equation:
Kc = [COCl2] / ([CO][Cl2])
Substituting the equilibrium molar concentrations, we have:
2.50 = (1.40 - x) / ((1.40 - x)(3.20))
Simplifying the equation:
2.50 = 1 / 3.20
Now, solve for x to find the equilibrium molar concentration of CO. Multiply both sides by (1.40 - x)(3.20):
2.50(1.40 - x)(3.20) = 1
7.00 - 2.50x = 1
2.50x = 7.00 - 1
2.50x = 6.00
x = 6.00 / 2.50
x = 2.40 M
So, the equilibrium molar concentration of CO is 2.40 M.
Now let's move on to the second question:
2. For the reaction CO(g) + Cl2(g) ↔ COCl2(g), we need to calculate Kp (in atm^-1) using the given pressure values.
Let's assume the equilibrium molar pressure of CO is x atm. Since the initial molar pressure of CO is 0.74 atm and the change at equilibrium will be -x, the equilibrium molar pressure of CO will be (0.74 - x) atm.
The equilibrium molar pressure of Cl2 is 0.92 atm, and the equilibrium molar pressure of COCl2 will also be (0.74 - x) atm.
Using the equilibrium constant expression in terms of partial pressures, we have:
Kp = (PCOCl2) / (PCO)(PCl2)
Substituting the equilibrium molar pressures, we have:
Kp = (0.74 - x) / ((0.74 - x)(0.92))
Simplifying the equation:
Kp = 1 / 0.92
Now, solve for x to find the equilibrium molar pressure of CO. Multiply both sides by (0.74 - x)(0.92):
Kp(0.74 - x)(0.92) = 1
(0.74 - x) / (0.74 - x) = 1 / Kp(0.92)
0.74 - x = 1 / Kp(0.92)
x = 0.74 - 1 / Kp(0.92)
x = 0.74 - 1.36
x = -0.62 atm (which is not possible)
It seems there may be an error in the given values or calculations for the second question. Please double-check the values and calculations.