A student ran the following reaction in the laboratory at 715 K: N₂(g) + 3H₂(g) → 2NH3(g) When she introduced 0.0307 moles of N₂ (g) and 0.0522 moles of H₂(g) into a 1.00 liter container, she found the equilibrium concentration of NH3(g) to be 0.000877 M. Calculate the equilibrium constant, Ke, she obtained for this reaction. Kc = Ke
A student ran the following reaction in the laboratory at 715 K: N₂(g) + 3H₂(g) → 2NH3(g) When she introduced 0.0307 moles of N₂ (g) and 0.0522 moles of H₂(g) into a 1.00 liter container, she found the equilibrium concentration of NH3(g) to be 0.000877 M. Calculate the equilibrium constant, Ke, she obtained for this reaction. Kc = Ke
Chemistry & Chemical Reactivity
10th Edition
ISBN:9781337399074
Author:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Chapter15: Principles Of Chemical Reactivity: Equilibria
Section15.4: Using Equilibrium Constants In Calculations
Problem 15.5CYU: The decomposition of PCl5(g) to form PCl3(g) and Cl2(g) has Kc = 33.3 at a high temperature. If the...
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![A student ran the following reaction in the laboratory at 715 K:
N₂(g) + 3H2(g) → 2NH3(g)
When she introduced 0.0307 moles of N₂ (g) and 0.0522 moles of H₂(g)
into a 1.00 liter container, she found the equilibrium concentration of
NH3(g) to be 0.000877 M.
Calculate the equilibrium constant, Ke, she obtained for this reaction.
Ke
Kc =](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0a566f77-d7e8-4910-9c09-a31d7141a31f%2Fbbd5c1cb-b9a1-4052-9cdd-11164ee17d50%2Fzwyg0au_processed.jpeg&w=3840&q=75)
Transcribed Image Text:A student ran the following reaction in the laboratory at 715 K:
N₂(g) + 3H2(g) → 2NH3(g)
When she introduced 0.0307 moles of N₂ (g) and 0.0522 moles of H₂(g)
into a 1.00 liter container, she found the equilibrium concentration of
NH3(g) to be 0.000877 M.
Calculate the equilibrium constant, Ke, she obtained for this reaction.
Ke
Kc =
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