What is the Kc for the following equilibrium at 500∘C, N2(g)+3H2(g)⇌2NH3(g) [H2] = 0.35 M , [N2] = 0.40 M , [NH3] = 1.9 M ?
回答 (1)
2016-06-30 5:54 pm
N₂(g) + 3H₂(g) ⇌ 2NH₃(g)
Kc = [NH₃]²/([N₂][H₂]³) = 1.9²/(0.40 × 0.35³) = 210 (L²/mol²)
Kc = [NH₃]²/([N₂][H₂]³) = 1.9²/(0.40 × 0.35³) = 210 (L²/mol²)
收錄日期: 2021-05-01 13:05:53
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