Question

What is the partial pressure of SO2Cl2 in this mixture? Express your answer in torr using...

What is the partial pressure of SO2Cl2 in this mixture?

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 .

Ans :- Partial pressure of SO2 at equilibrium = PSO2Cl2 = 1.69 x 10-5

Explanation :-

Given,

Partial pressure of SO2 at equilibrium = PSO2 = 0.134 torr

Partial pressure of SO2 at equilibrium = PCl2 = 0.366 torr and

Partial pressure of SO2 at equilibrium = PSO2Cl2 = ?

Given reaction is :

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

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

Kp= PSO2.PCl2 / PSO2Cl2

Kp= (0.134 ) (0.366) / PSO2Cl2

2.91 x 103 = 0.049044 / PSO2Cl2

PSO2Cl2 = 0.049044 / 2.91 x 103

PSO2Cl2 = 1.69 x 10-5

Hence, Partial pressure of SO2 at equilibrium = PSO2Cl2 = 1.69 x 10-5

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