A 1.000 L vessel is filled with 2.000 moles of
N2, 1.000 mole of H2, and 2.000 moles of NH3.
When the reaction
N2(g) + 3 H2(g) ⇀↽ 2 NH3(g)
comes to equilibrium, it is observed that the
concentration of H2 is 2.47 moles/L. What is
the numerical value of the equilibrium constant
Kc?
AP Chemistry Lost?
2017-01-11 10:51 am
回答 (1)
2017-01-11 11:03 am
N₂(g) + 3 H₂(g) ⇌ 2 NH₃(g) …. Kc
Initial concentrations :
[N₂]ₒ = 2.000 mol/L
[H₂]ₒ = 1.000 mol/L
[NH₃]ₒ = 2.000 mol/L
When equilibrium is established, increase in [H₂] = (2.47 - 1.000) = 1.47 mol/L
Equilibrium concentrations :
[N₂] = {2.000 + 1.47 × (1/3)} mol/L = 2.49 mol/L
[H₂] = 2.47 mol/L
[NH₃] = {2.000 - 1.47 × (2/3)} mol/L = 1.02 mol/L
Kc = [NH₃]² / ([N₂] [H₂]³) = 1.02² / (2.49 × 2.47³) = 0.0277
Initial concentrations :
[N₂]ₒ = 2.000 mol/L
[H₂]ₒ = 1.000 mol/L
[NH₃]ₒ = 2.000 mol/L
When equilibrium is established, increase in [H₂] = (2.47 - 1.000) = 1.47 mol/L
Equilibrium concentrations :
[N₂] = {2.000 + 1.47 × (1/3)} mol/L = 2.49 mol/L
[H₂] = 2.47 mol/L
[NH₃] = {2.000 - 1.47 × (2/3)} mol/L = 1.02 mol/L
Kc = [NH₃]² / ([N₂] [H₂]³) = 1.02² / (2.49 × 2.47³) = 0.0277
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https://hk.answers.yahoo.com/question/index?qid=20170111025147AAgZ3jD