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)
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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Welcome to the NCERT Solutions for Class 11 Chemistry - Chapter . This page offers a step-by-step solution to the specific question from Excercise 1 , Question 11: A sample of HI(g) is placed in flask at a pressure of 0.2 atm. At equilibrium the partial pressure o....
Comments
Correction needed in the equilibrium concentration of I2.