3. NH, is normally encountered as a gas with a pungent odor. It is formed from hydrogen and nitrogen (equation given below). A chemist pours a mixture of nitrogen gas and hydrogen gas in a 1.0 litre reaction vessel in 1:3 ratio. The equilibrium constant for the reaction is 9.7 × 10-3, and the equilibrium value of [H₂(g)] is 0.91 mol/L. Calculate the equilibrium value of [NH₂]. N₂(g) + 3 H₂(g) ↔ 2 NH₂(g)
3. NH, is normally encountered as a gas with a pungent odor. It is formed from hydrogen and nitrogen (equation given below). A chemist pours a mixture of nitrogen gas and hydrogen gas in a 1.0 litre reaction vessel in 1:3 ratio. The equilibrium constant for the reaction is 9.7 × 10-3, and the equilibrium value of [H₂(g)] is 0.91 mol/L. Calculate the equilibrium value of [NH₂]. N₂(g) + 3 H₂(g) ↔ 2 NH₂(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 5QRT: For the equilibrium reaction in Question 4, write the expression for the equilibrium constant. (a)...
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![3. NH, is normally encountered as a gas with a pungent odor. It is
formed from hydrogen and nitrogen (equation given below). A
chemist pours a mixture of nitrogen gas and hydrogen gas in a 1.0
litre reaction vessel in 1:3 ratio. The equilibrium constant for the
reaction is 9.7 × 10 −³, and the equilibrium value of [H₂(g)] is 0.91
mol/L. Calculate the equilibrium value of [NH₂).
N₂(g) + 3 H₂(g) → 2 NH₂(g)
(answer + units)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ffb0b435e-c9aa-43c6-8916-15cc82535c0c%2F13103a56-7dc9-4123-9f2d-1959c52d8c90%2Fhyr0kab_processed.jpeg&w=3840&q=75)
Transcribed Image Text:3. NH, is normally encountered as a gas with a pungent odor. It is
formed from hydrogen and nitrogen (equation given below). A
chemist pours a mixture of nitrogen gas and hydrogen gas in a 1.0
litre reaction vessel in 1:3 ratio. The equilibrium constant for the
reaction is 9.7 × 10 −³, and the equilibrium value of [H₂(g)] is 0.91
mol/L. Calculate the equilibrium value of [NH₂).
N₂(g) + 3 H₂(g) → 2 NH₂(g)
(answer + units)
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