Question

What is the partial pressure of SO2Cl2 in this mixture?

Express your answer in torr using three significant figures.

Consider the reaction:

SO2Cl2(g)?SO2(g)+Cl2(g),    Kp=2.91×10³ at 298 K

In a reaction at equilibrium, the partial pressure of SO2 is 0.134 torr and that of Cl2 is 0.366 torr .

Answer 1

Ans :- Partial pressure of SO₂ at equilibrium = P_SO2Cl2 = 1.69 x 10^-5

Explanation :-

Given,

Partial pressure of SO₂ at equilibrium = P_SO2 = 0.134 torr

Partial pressure of SO₂ at equilibrium = P_Cl2 = 0.366 torr and

Partial pressure of SO₂ at equilibrium = P_SO2Cl2 = ?

Given reaction is :

SO₂Cl₂(g) <-------------> SO₂(g)+Cl₂(g),    K_p = 2.91×10³ at 298 K

Now, Expression of Pressure equilibrium constant i.e. K_p is :

K_p= P_SO2.P_Cl2 / P_SO2Cl2

K_p= (0.134 ) (0.366) / P_SO2Cl2

2.91 x 10³ = 0.049044 / P_SO2Cl2

P_SO2Cl2 = 0.049044 / 2.91 x 10³

P_SO2Cl2 = 1.69 x 10^-5

Hence, Partial pressure of SO₂ at equilibrium = P_SO2Cl2 = 1.69 x 10^-5