H2SO4 + Br2 + FeO = H2O + Fe2(SO4)3 + FeBr3

H_2SO_4 sulfuric acid + Br_2 bromine + FeO iron(II) oxide ⟶ H_2O water + Fe_2(SO_4)_3·xH_2O iron(III) sulfate hydrate + FeBr_3 iron(III) bromide

Balance the chemical equation algebraically: H_2SO_4 + Br_2 + FeO ⟶ H_2O + Fe_2(SO_4)_3·xH_2O + FeBr_3 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 H_2SO_4 + c_2 Br_2 + c_3 FeO ⟶ c_4 H_2O + c_5 Fe_2(SO_4)_3·xH_2O + c_6 FeBr_3 Set the number of atoms in the reactants equal to the number of atoms in the products for H, O, S, Br and Fe: H: | 2 c_1 = 2 c_4 O: | 4 c_1 + c_3 = c_4 + 12 c_5 S: | c_1 = 3 c_5 Br: | 2 c_2 = 3 c_6 Fe: | c_3 = 2 c_5 + c_6 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_5 = 1 and solve the system of equations for the remaining coefficients: c_1 = 3 c_2 = 3/2 c_3 = 3 c_4 = 3 c_5 = 1 c_6 = 1 Multiply by the least common denominator, 2, to eliminate fractional coefficients: c_1 = 6 c_2 = 3 c_3 = 6 c_4 = 6 c_5 = 2 c_6 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 6 H_2SO_4 + 3 Br_2 + 6 FeO ⟶ 6 H_2O + 2 Fe_2(SO_4)_3·xH_2O + 2 FeBr_3

+ + ⟶ + +

sulfuric acid + bromine + iron(II) oxide ⟶ water + iron(III) sulfate hydrate + iron(III) bromide

Construct the equilibrium constant, K, expression for: H_2SO_4 + Br_2 + FeO ⟶ H_2O + Fe_2(SO_4)_3·xH_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: 6 H_2SO_4 + 3 Br_2 + 6 FeO ⟶ 6 H_2O + 2 Fe_2(SO_4)_3·xH_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 H_2SO_4 | 6 | -6 Br_2 | 3 | -3 FeO | 6 | -6 H_2O | 6 | 6 Fe_2(SO_4)_3·xH_2O | 2 | 2 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 H_2SO_4 | 6 | -6 | ([H2SO4])^(-6) Br_2 | 3 | -3 | ([Br2])^(-3) FeO | 6 | -6 | ([FeO])^(-6) H_2O | 6 | 6 | ([H2O])^6 Fe_2(SO_4)_3·xH_2O | 2 | 2 | ([Fe2(SO4)3·xH2O])^2 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 = ([H2SO4])^(-6) ([Br2])^(-3) ([FeO])^(-6) ([H2O])^6 ([Fe2(SO4)3·xH2O])^2 ([FeBr3])^2 = (([H2O])^6 ([Fe2(SO4)3·xH2O])^2 ([FeBr3])^2)/(([H2SO4])^6 ([Br2])^3 ([FeO])^6)

Construct the rate of reaction expression for: H_2SO_4 + Br_2 + FeO ⟶ H_2O + Fe_2(SO_4)_3·xH_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: 6 H_2SO_4 + 3 Br_2 + 6 FeO ⟶ 6 H_2O + 2 Fe_2(SO_4)_3·xH_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 H_2SO_4 | 6 | -6 Br_2 | 3 | -3 FeO | 6 | -6 H_2O | 6 | 6 Fe_2(SO_4)_3·xH_2O | 2 | 2 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 H_2SO_4 | 6 | -6 | -1/6 (Δ[H2SO4])/(Δt) Br_2 | 3 | -3 | -1/3 (Δ[Br2])/(Δt) FeO | 6 | -6 | -1/6 (Δ[FeO])/(Δt) H_2O | 6 | 6 | 1/6 (Δ[H2O])/(Δt) Fe_2(SO_4)_3·xH_2O | 2 | 2 | 1/2 (Δ[Fe2(SO4)3·xH2O])/(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 = -1/6 (Δ[H2SO4])/(Δt) = -1/3 (Δ[Br2])/(Δt) = -1/6 (Δ[FeO])/(Δt) = 1/6 (Δ[H2O])/(Δt) = 1/2 (Δ[Fe2(SO4)3·xH2O])/(Δt) = 1/2 (Δ[FeBr3])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| sulfuric acid | bromine | iron(II) oxide | water | iron(III) sulfate hydrate | iron(III) bromide formula | H_2SO_4 | Br_2 | FeO | H_2O | Fe_2(SO_4)_3·xH_2O | FeBr_3 Hill formula | H_2O_4S | Br_2 | FeO | H_2O | Fe_2O_12S_3 | Br_3Fe name | sulfuric acid | bromine | iron(II) oxide | water | iron(III) sulfate hydrate | iron(III) bromide IUPAC name | sulfuric acid | molecular bromine | oxoiron | water | diferric trisulfate | tribromoiron