Question

Carbon monoxide 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 345 torr of CO and 452 torr of Cl2, what is the mole fraction of COCl2 when equilibrium is reached? please be sure of your answer its my last attempt

Answer 1

Kp = P_COCl2 P_CO P_Cl2
P_COCl2 gains the stoichiometric partial pressure p from an initial pressure of 0, whereas both P_CO and P_Cl2 lose partial pressure. Therefore, the equation for Kp becomes:
3.10 = p (345-p) (452-p)
which rearranges to the quadratic equation,
3.1p² - 2470.7p + 483414 = 0

Using the quadratic formula, we get two answers, p = 345 torr and p = 452 torr. Since Cl₂ can't lose any more than 345 torr, the former value is taken. Hence at equilibrium,

P_COCl2 = p = 345 torr
P_CO = 345 - 345 = 0 torr
P_Cl2 = 452 - 345 = 107 torr

The equilibrium composition is checked by substituting these partial pressures back into the equilibrium equation to recalculate Kp.
The mole fraction of COCl₂ in the equilibrium composition can be calculated using the ideal gas law:
PV=nRT.

If V and T are constant, the number of moles of a gas is directly proportional to its partial pressure in the mixture.

Therefore, the mole fraction of COCl₂ is:

X_COCl2 = 345 (345 + 0 + 107) = 0.7633, or about 76% by mole.