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)

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