An empty steel container is filled with 0.0290 atm of HF. The system is allowed to reach equilibrium. If Kp = 2.76 for the reaction below, what is the equilibrium partial pressure of H2? 2 HF (g) = H2 (g) + F2 (g)

Chemistry: The Molecular Science
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Author:John W. Moore, Conrad L. Stanitski
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Chapter12: Chemical Equilibrium
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An empty steel container is filled with 0.0290 atm of HF. The system
is allowed to reach equilibrium. If Kp = 2.76 for the reaction below,
what is the equilibrium partial pressure of H2?
2 HF (g) = H2 (g) + F2 (g)
Transcribed Image Text:An empty steel container is filled with 0.0290 atm of HF. The system is allowed to reach equilibrium. If Kp = 2.76 for the reaction below, what is the equilibrium partial pressure of H2? 2 HF (g) = H2 (g) + F2 (g)
Expert Solution
Step 1

Given,

2 HF (g) ⇌ H2 (g) + F2 (g)

Equilibrium constant (Kp) = 2.76

Initial partial pressure of HF = 0.0290 atm

Equilibrium partial pressure of H2 = ?

 

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