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)
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
5th Edition
ISBN:9781285199047
Author:John W. Moore, Conrad L. Stanitski
Publisher:John W. Moore, Conrad L. Stanitski
Chapter12: Chemical Equilibrium
Section: Chapter Questions
Problem 61QRT
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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)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1b66214a-9d08-49c6-bbae-58e281503e45%2Fa47fd97c-e978-49a2-885b-eac46acf4829%2F1tduz0gq_processed.png&w=3840&q=75)
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
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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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