K p for COCl₂ (g) CO (g) + Cl₂ (g) is 1.34*10^{- 5} .
What is the partial pressure, in atm, of Cl₂ (g) when 4.2 atm of COCl₂ (g) is placed in a reactor and allowed to reach equilibrium?

Question

K p for COCl₂ (g) CO (g) + Cl₂ (g) is 1.34*10^{- 5} . What is the partial pressure, in atm, of Cl₂ (g) when 4.2 atm of COCl₂ (g) is placed in a reactor and allowed to reach equilibrium?

Quizplus · Accepted Answer

We can use the reaction quotient, Qp, to determine the direction in which the reaction will shift in order to reach equilibrium. At equilibrium, Qp = Kp.

Qp = (P_CO)(P_Cl2)/(P_COCl2)

We are given that Kp = 1.34*10^-5 and P_COCl2 = 4.2 atm.

Let x be the partial pressure of Cl2 (g) at equilibrium. Then (4.2 - x) is the partial pressure of CO (g) at equilibrium. Substituting these into the expression for Qp:

1.34*10^-5 = (4.2 - x)*x/4.2

Solving for x:

x = 7.5*10^-4 atm

Therefore, the partial pressure of Cl2 (g) at equilibrium is 7.5×10^-4 atm, which corresponds to choice (C).

K P for COCl₂ (G) CO (G) + Cl₂