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Class 11
Chemistry
Equilibrium
a sample of hi g is placed in flask at a pressure...

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 K_p 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 K_p for the given equilibrium is 4.0.

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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 K_p for the given equilibrium ?

2HI (g) ↔ H₂ (g) + I₂ (g)

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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)

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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 K_p for the given equilibrium ?

2HI (g) ↔ H₂ (g) + I₂ (g)