Fe2O3 + HBr = H2O + FeBr3

Fe_2O_3 iron(III) oxide + HBr hydrogen bromide ⟶ H_2O water + FeBr_3 iron(III) bromide

Balance the chemical equation algebraically: Fe_2O_3 + HBr ⟶ H_2O + FeBr_3 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 Fe_2O_3 + c_2 HBr ⟶ c_3 H_2O + c_4 FeBr_3 Set the number of atoms in the reactants equal to the number of atoms in the products for Fe, O, Br and H: Fe: | 2 c_1 = c_4 O: | 3 c_1 = c_3 Br: | c_2 = 3 c_4 H: | c_2 = 2 c_3 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 = 6 c_3 = 3 c_4 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | Fe_2O_3 + 6 HBr ⟶ 3 H_2O + 2 FeBr_3

+ ⟶ +

iron(III) oxide + hydrogen bromide ⟶ water + iron(III) bromide

Construct the equilibrium constant, K, expression for: Fe_2O_3 + HBr ⟶ H_2O + FeBr_3 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: Fe_2O_3 + 6 HBr ⟶ 3 H_2O + 2 FeBr_3 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 Fe_2O_3 | 1 | -1 HBr | 6 | -6 H_2O | 3 | 3 FeBr_3 | 2 | 2 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression Fe_2O_3 | 1 | -1 | ([Fe2O3])^(-1) HBr | 6 | -6 | ([HBr])^(-6) H_2O | 3 | 3 | ([H2O])^3 FeBr_3 | 2 | 2 | ([FeBr3])^2 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 = ([Fe2O3])^(-1) ([HBr])^(-6) ([H2O])^3 ([FeBr3])^2 = (([H2O])^3 ([FeBr3])^2)/([Fe2O3] ([HBr])^6)

Construct the rate of reaction expression for: Fe_2O_3 + HBr ⟶ H_2O + FeBr_3 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: Fe_2O_3 + 6 HBr ⟶ 3 H_2O + 2 FeBr_3 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 Fe_2O_3 | 1 | -1 HBr | 6 | -6 H_2O | 3 | 3 FeBr_3 | 2 | 2 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 Fe_2O_3 | 1 | -1 | -(Δ[Fe2O3])/(Δt) HBr | 6 | -6 | -1/6 (Δ[HBr])/(Δt) H_2O | 3 | 3 | 1/3 (Δ[H2O])/(Δt) FeBr_3 | 2 | 2 | 1/2 (Δ[FeBr3])/(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[Fe2O3])/(Δt) = -1/6 (Δ[HBr])/(Δt) = 1/3 (Δ[H2O])/(Δt) = 1/2 (Δ[FeBr3])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| iron(III) oxide | hydrogen bromide | water | iron(III) bromide formula | Fe_2O_3 | HBr | H_2O | FeBr_3 Hill formula | Fe_2O_3 | BrH | H_2O | Br_3Fe name | iron(III) oxide | hydrogen bromide | water | iron(III) bromide IUPAC name | | hydrogen bromide | water | tribromoiron

| iron(III) oxide | hydrogen bromide | water | iron(III) bromide molar mass | 159.69 g/mol | 80.912 g/mol | 18.015 g/mol | 295.56 g/mol phase | solid (at STP) | gas (at STP) | liquid (at STP) | melting point | 1565 °C | -86.8 °C | 0 °C | boiling point | | -66.38 °C | 99.9839 °C | density | 5.26 g/cm^3 | 0.003307 g/cm^3 (at 25 °C) | 1 g/cm^3 | solubility in water | insoluble | miscible | | surface tension | | 0.0271 N/m | 0.0728 N/m | dynamic viscosity | | 8.4×10^-4 Pa s (at -75 °C) | 8.9×10^-4 Pa s (at 25 °C) | odor | odorless | | odorless | odorless