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Question:

A sample of HI(g) is placed in flask at a pressure of 0.2 atm. At equilibrium the partial pressure of HI(g) is 0.04 atm. What is Kp for the given equilibrium ?

2HI (g) ↔ H2 (g) + I2 (g)

Answer:

The initial concentration of HI is 0.2 atm. At equilibrium, it has a partial pressure of 0.04 atm. Therefore, a decrease in the pressure of HI is 0.2 - 0.04 = 0.16. The given reaction is:

                                  2HI (g)           ↔             H2 (g)       +        I2 (g)

Intial concentration       0.2 atm                           0                         0

At equilibrium              0.04 atm                       0.16/2                   2.15/2

Therefore,

 

Therefore, Hence, the value of Kp for the given equilibrium is 4.0.


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