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9. At equilibrium a 2 L vessel contains 0.360
mol of H2, 0.110 mol of Br, and 37.0 mol of
HBr. What is the equilibrium constant for the
reaction at this temperature?
H2(g) + Brz(8) 22HBr(8)

Respuesta :

Answer:

Ke = 34570.707

Explanation:

  • H2(g) + Br2(g) → 2 HBr(g)

equilibrium constant (Ke):

⇒ Ke = [HBr]² / [Br2] [H2]

∴ [HBr] = (37.0 mol) / (2 L) = 18.5 mol/L

∴ [Br2] = (0.110 mol) / (2 L) = 0.055 mol/L

∴ [H2] = (0.360 mol) / (2 L) = 0.18 mol/L

⇒ Ke = (18.5 mol/L)² / (0.055 mol/L)(0.18 mol/L)

⇒ Ke = 34570.707

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