HCl + Fe3O4 = H2O + FeCl3 + FeCl2

HCl hydrogen chloride + FeO·Fe_2O_3 iron(II, III) oxide ⟶ H_2O water + FeCl_3 iron(III) chloride + FeCl_2 iron(II) chloride

Balance the chemical equation algebraically: HCl + FeO·Fe_2O_3 ⟶ H_2O + FeCl_3 + FeCl_2 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 HCl + c_2 FeO·Fe_2O_3 ⟶ c_3 H_2O + c_4 FeCl_3 + c_5 FeCl_2 Set the number of atoms in the reactants equal to the number of atoms in the products for Cl, H, Fe and O: Cl: | c_1 = 3 c_4 + 2 c_5 H: | c_1 = 2 c_3 Fe: | 3 c_2 = c_4 + c_5 O: | 4 c_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_2 = 1 and solve the system of equations for the remaining coefficients: c_1 = 8 c_2 = 1 c_3 = 4 c_4 = 2 c_5 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 8 HCl + FeO·Fe_2O_3 ⟶ 4 H_2O + 2 FeCl_3 + FeCl_2

+ ⟶ + +

hydrogen chloride + iron(II, III) oxide ⟶ water + iron(III) chloride + iron(II) chloride

| hydrogen chloride | iron(II, III) oxide | water | iron(III) chloride | iron(II) chloride molecular enthalpy | -92.3 kJ/mol | -1118 kJ/mol | -285.8 kJ/mol | -399.5 kJ/mol | -341.8 kJ/mol total enthalpy | -738.4 kJ/mol | -1118 kJ/mol | -1143 kJ/mol | -799 kJ/mol | -341.8 kJ/mol | H_initial = -1857 kJ/mol | | H_final = -2284 kJ/mol | | ΔH_rxn^0 | -2284 kJ/mol - -1857 kJ/mol = -427.3 kJ/mol (exothermic) | | | |

| hydrogen chloride | iron(II, III) oxide | water | iron(III) chloride | iron(II) chloride molecular free energy | -95.3 kJ/mol | -1015 kJ/mol | -237.1 kJ/mol | -334 kJ/mol | -302.3 kJ/mol total free energy | -762.4 kJ/mol | -1015 kJ/mol | -948.4 kJ/mol | -668 kJ/mol | -302.3 kJ/mol | G_initial = -1778 kJ/mol | | G_final = -1919 kJ/mol | | ΔG_rxn^0 | -1919 kJ/mol - -1778 kJ/mol = -140.9 kJ/mol (exergonic) | | | |

Construct the equilibrium constant, K, expression for: HCl + FeO·Fe_2O_3 ⟶ H_2O + FeCl_3 + FeCl_2 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: 8 HCl + FeO·Fe_2O_3 ⟶ 4 H_2O + 2 FeCl_3 + FeCl_2 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 HCl | 8 | -8 FeO·Fe_2O_3 | 1 | -1 H_2O | 4 | 4 FeCl_3 | 2 | 2 FeCl_2 | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression HCl | 8 | -8 | ([HCl])^(-8) FeO·Fe_2O_3 | 1 | -1 | ([FeO·Fe2O3])^(-1) H_2O | 4 | 4 | ([H2O])^4 FeCl_3 | 2 | 2 | ([FeCl3])^2 FeCl_2 | 1 | 1 | [FeCl2] 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 = ([HCl])^(-8) ([FeO·Fe2O3])^(-1) ([H2O])^4 ([FeCl3])^2 [FeCl2] = (([H2O])^4 ([FeCl3])^2 [FeCl2])/(([HCl])^8 [FeO·Fe2O3])

Construct the rate of reaction expression for: HCl + FeO·Fe_2O_3 ⟶ H_2O + FeCl_3 + FeCl_2 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: 8 HCl + FeO·Fe_2O_3 ⟶ 4 H_2O + 2 FeCl_3 + FeCl_2 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 HCl | 8 | -8 FeO·Fe_2O_3 | 1 | -1 H_2O | 4 | 4 FeCl_3 | 2 | 2 FeCl_2 | 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 HCl | 8 | -8 | -1/8 (Δ[HCl])/(Δt) FeO·Fe_2O_3 | 1 | -1 | -(Δ[FeO·Fe2O3])/(Δt) H_2O | 4 | 4 | 1/4 (Δ[H2O])/(Δt) FeCl_3 | 2 | 2 | 1/2 (Δ[FeCl3])/(Δt) FeCl_2 | 1 | 1 | (Δ[FeCl2])/(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 = -1/8 (Δ[HCl])/(Δt) = -(Δ[FeO·Fe2O3])/(Δt) = 1/4 (Δ[H2O])/(Δt) = 1/2 (Δ[FeCl3])/(Δt) = (Δ[FeCl2])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| hydrogen chloride | iron(II, III) oxide | water | iron(III) chloride | iron(II) chloride formula | HCl | FeO·Fe_2O_3 | H_2O | FeCl_3 | FeCl_2 Hill formula | ClH | Fe_3O_4 | H_2O | Cl_3Fe | Cl_2Fe name | hydrogen chloride | iron(II, III) oxide | water | iron(III) chloride | iron(II) chloride IUPAC name | hydrogen chloride | | water | trichloroiron | dichloroiron