Question

Carbon monoxide and chlorine gas react to form phosgene:
CO(g)+Cl2(g)?COCl2(g) Kp = 3.10 at700 K

If a reaction mixture initially contains 172torr of CO and 357torr of Cl2, what is the mole fraction of COCl2when equilibrium is reached?

Answer 1

CO (g) + Cl₂(g) ------> COCl₂(g)

$K_{P}=\frac{P_{COCl_{2}}}{P_{CO}\times P_{Cl_{2}}}$

K_p = 3.10

P_CO = 172 torr = 170/760 atm = 0.224 atm

P_Cl2 = 357 torr = 375/760 atm =0.493 atm

$3.10=\frac{P_{COCl_{2}}}{0.224atm\times 0.493atm}$

P_COCl2 = 0.342 atm

Total pressure at equilibrium is the sum of partial pressures of the gases CO, Cl₂ and COCl₂.

P_tot = 0.224 + 0.493 + 0.342 = 1.059 atm

For COCl_2, the partialpressure is the product of its mole fraction and total pressure.

P_COCl2= X_COCl2.P_tot

(0.342 atm) = X_COCl2(1.059 atm)

X_COCl2 = 0.322

Therefore, molefraction of COCl₂ at equilibrium is 0.322.