mullanebrianot3dpw
18.01.2020 •
Chemistry
Astudent ran the following reaction in the laboratory at 671 k: 2nh3(g) n2(g) + 3h2(g) when she introduced 7.33×10-2 moles of nh3(g) into a 1.00 liter container, she found the equilibrium concentration of h2(g) to be 0.103 m. calculate the equilibrium constant, kc, she obtained for this reaction. kc =
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Ответ:
Kc = 8.05x10⁻³
Explanation:
This is the equilibrium:
2NH₃(g) ⇄ N₂(g) + 3H₂(g)
Initially 0.0733
React 0.0733α α/2 3/2α
Eq 0.0733 - 0.0733α α/2 0.103
We introduced 0.0733 moles of ammonia, initially. So in the reaction "α" amount react, as the ratio is 2:1, and 2:3, we can know the moles that formed products.
Now we were told that in equilibrum we have a [H₂] of 0.103, so this data can help us to calculate α.
3/2α = 0.103
α = 0.103 . 2/3 ⇒ 0.0686
So, concentration in equilibrium are
NH₃ = 0.0733 - 0.0733 . 0.0686 = 0.0682
N₂ = 0.0686/2 = 0.0343
So this moles, are in a volume of 1L, so they are molar concentrations.
Let's make Kc expression:
Kc= [N₂] . [H₂]³ / [NH₃]²
Kc = 0.0343 . 0.103³ / 0.0682² = 8.05x10⁻³
Ответ:
a.Sodium hydroxide reacts with nitric acid
Naoh + Hno3 == Nano3 + h20.
acid based reaction.
b.Hydrogen + bromine
H + Br == Hbr
H2 + Br2 == 2Hbr synthesis reaction
c.Silver nitrate + aluminum chloride
3Agno3 + Alcl3 == Al (No3)3 + 3Agcl double displacement.