The Kp for the production of phosgene is 18.0. If one mole each of CO and Cl2 are
mixed at the total pressure of 1 atm, what will be the partial pressure (in atm) of COCl2
after equilibrium is reached, CO(g) + Cl2(g)-->COCl2(g)
mols CO = 1
mols Cl2 = 1
total mols = 2
XCO = 1/2 = 0.5
XCl2 = 1/2 = 0.5
pCO = XCO * Ptotal = 0.5 x 1 = 0.5 atm
pCl2 = XCl2 * Ptotal = 0.5 x 1 = 0.5 atm
.......CO(g) + Cl2(g)-->COCl2(g)
I......0.5.....0.5........0
C.......-x......-x........x
E....0.5-x.....0.5-x......x
Substitute the E line into Kp expression and solve for x = pCOCl2
To find the partial pressure of COCl2 after equilibrium is reached, we will use the equation for the equilibrium constant (Kp) and apply the concept of partial pressures.
The equilibrium constant expression for the given reaction is:
Kp = (P(COCl2)) / (P(CO) * P(Cl2))
where P(COCl2), P(CO), and P(Cl2) are the partial pressures of COCl2, CO, and Cl2, respectively.
Given that Kp = 18.0, we can rearrange the equation to solve for the partial pressure of COCl2.
(P(COCl2)) / (P(CO) * P(Cl2)) = 18.0
Since the mole ratios in the balanced equation are 1:1:1, when the reaction reaches equilibrium, the moles of each gas will also be in the ratio of 1:1:1.
Let's assume that at equilibrium, the partial pressure of each gas is x atm. Therefore, we can substitute these values into the equilibrium constant expression:
(x) / (x * x) = 18.0
Simplifying the equation further:
1 / x = 18.0
Cross-multiplying:
x = 1 / 18.0
Calculating:
x ≈ 0.0556 atm
Therefore, the partial pressure of COCl2 after equilibrium is reached is approximately 0.0556 atm.