H2O + FeBr2 = HBr + Fe(OH)Br

H_2O water + FeBr_2 iron(II) bromide ⟶ HBr hydrogen bromide + FeOHBr

Balance the chemical equation algebraically: H_2O + FeBr_2 ⟶ HBr + FeOHBr Add stoichiometric coefficients, c_i, to the reactants and products: c_1 H_2O + c_2 FeBr_2 ⟶ c_3 HBr + c_4 FeOHBr Set the number of atoms in the reactants equal to the number of atoms in the products for H, O, Br and Fe: H: | 2 c_1 = c_3 + c_4 O: | c_1 = c_4 Br: | 2 c_2 = c_3 + c_4 Fe: | c_2 = c_4 Since the coefficients are relative quantities and underdetermined, choose a coefficient to set arbitrarily. To keep the coefficients small, the arbitrary value is ordinarily one. For instance, set c_1 = 1 and solve the system of equations for the remaining coefficients: c_1 = 1 c_2 = 1 c_3 = 1 c_4 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | H_2O + FeBr_2 ⟶ HBr + FeOHBr

+ ⟶ + FeOHBr

water + iron(II) bromide ⟶ hydrogen bromide + FeOHBr

Construct the equilibrium constant, K, expression for: H_2O + FeBr_2 ⟶ HBr + FeOHBr Plan: • Balance the chemical equation. • Determine the stoichiometric numbers. • Assemble the activity expression for each chemical species. • Use the activity expressions to build the equilibrium constant expression. Write the balanced chemical equation: H_2O + FeBr_2 ⟶ HBr + FeOHBr Assign stoichiometric numbers, ν_i, using the stoichiometric coefficients, c_i, from the balanced chemical equation in the following manner: ν_i = -c_i for reactants and ν_i = c_i for products: chemical species | c_i | ν_i H_2O | 1 | -1 FeBr_2 | 1 | -1 HBr | 1 | 1 FeOHBr | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression H_2O | 1 | -1 | ([H2O])^(-1) FeBr_2 | 1 | -1 | ([FeBr2])^(-1) HBr | 1 | 1 | [HBr] FeOHBr | 1 | 1 | [FeOHBr] The equilibrium constant symbol in the concentration basis is: K_c Mulitply the activity expressions to arrive at the K_c expression: Answer: | | K_c = ([H2O])^(-1) ([FeBr2])^(-1) [HBr] [FeOHBr] = ([HBr] [FeOHBr])/([H2O] [FeBr2])

Construct the rate of reaction expression for: H_2O + FeBr_2 ⟶ HBr + FeOHBr Plan: • Balance the chemical equation. • Determine the stoichiometric numbers. • Assemble the rate term for each chemical species. • Write the rate of reaction expression. Write the balanced chemical equation: H_2O + FeBr_2 ⟶ HBr + FeOHBr Assign stoichiometric numbers, ν_i, using the stoichiometric coefficients, c_i, from the balanced chemical equation in the following manner: ν_i = -c_i for reactants and ν_i = c_i for products: chemical species | c_i | ν_i H_2O | 1 | -1 FeBr_2 | 1 | -1 HBr | 1 | 1 FeOHBr | 1 | 1 The rate term for each chemical species, B_i, is 1/ν_i(Δ[B_i])/(Δt) where [B_i] is the amount concentration and t is time: chemical species | c_i | ν_i | rate term H_2O | 1 | -1 | -(Δ[H2O])/(Δt) FeBr_2 | 1 | -1 | -(Δ[FeBr2])/(Δt) HBr | 1 | 1 | (Δ[HBr])/(Δt) FeOHBr | 1 | 1 | (Δ[FeOHBr])/(Δt) (for infinitesimal rate of change, replace Δ with d) Set the rate terms equal to each other to arrive at the rate expression: Answer: | | rate = -(Δ[H2O])/(Δt) = -(Δ[FeBr2])/(Δt) = (Δ[HBr])/(Δt) = (Δ[FeOHBr])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| water | iron(II) bromide | hydrogen bromide | FeOHBr formula | H_2O | FeBr_2 | HBr | FeOHBr Hill formula | H_2O | Br_2Fe | BrH | HBrFeO name | water | iron(II) bromide | hydrogen bromide | IUPAC name | water | dibromoiron | hydrogen bromide |

| water | iron(II) bromide | hydrogen bromide | FeOHBr molar mass | 18.015 g/mol | 215.65 g/mol | 80.912 g/mol | 152.76 g/mol phase | liquid (at STP) | solid (at STP) | gas (at STP) | melting point | 0 °C | 684 °C | -86.8 °C | boiling point | 99.9839 °C | 934 °C | -66.38 °C | density | 1 g/cm^3 | 4.63 g/cm^3 | 0.003307 g/cm^3 (at 25 °C) | solubility in water | | | miscible | surface tension | 0.0728 N/m | | 0.0271 N/m | dynamic viscosity | 8.9×10^-4 Pa s (at 25 °C) | | 8.4×10^-4 Pa s (at -75 °C) | odor | odorless | | |