Question

Carbon Minoxide and Chlorine gas react to form phosgene: CO(g) + Cl2(g) <==> COCl2(g) Kp=3.10 at 700 K

If a reaction mixture initially contains 404 torr of CO and 257 torr of Cl2, what is the mole fraction of COCl2 when equilibrium is reached?

Mole Fraction COCl2= ?

Answer 1

K_p = P(COCl₂) / P(CO) P(Cl₂)
In going to equilibrium, P(COCl₂) gains the stoichiometric partial pressure p from an initial pressure of 0, while both P(CO) and P(Cl₂) lose p pressure. The K_p equation becomes
3.10 = p / (404-p) (257-p)
which rearranges to the quadratic
3.1p² - 2048.1p + 321866.8 = 0.
Using the quadratic formula, we get two answers, p = 256.57 torr and p = 403.11 torr. Since Cl₂ can't lose any more than 257 torr, the former value is the one we want. At equilibrium then:
P(COCl₂) = p = 256.57 torr
P(CO) = 404-p = 147.43 torr
P(Cl₂) = 257-p = 0.43 torr.
The equilibrium composition is checked by substituting these partial pressures back into the equilibrium equation to recalculate K_p.
The question asks for mole fraction of COCl₂ in the equilibrium composition; we can do this easily by remembering the ideal gas law
PV=nRT.
This tells us that, if V and T are constant, the number of moles of a gas is directly proportional to its partial pressure in the mixture. The mole fraction of COCl₂ is therefore:
X(COCl₂) = 256.57 / (256.57 + 147.43 + 0.43) = 0.634, or about 63.4% by mole.