Fe + Al2O3 = Al + Fe3O4

Fe iron + Al_2O_3 aluminum oxide ⟶ Al aluminum + FeO·Fe_2O_3 iron(II, III) oxide

Balance the chemical equation algebraically: Fe + Al_2O_3 ⟶ Al + FeO·Fe_2O_3 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 Fe + c_2 Al_2O_3 ⟶ c_3 Al + c_4 FeO·Fe_2O_3 Set the number of atoms in the reactants equal to the number of atoms in the products for Fe, Al and O: Fe: | c_1 = 3 c_4 Al: | 2 c_2 = c_3 O: | 3 c_2 = 4 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_4 = 1 and solve the system of equations for the remaining coefficients: c_1 = 3 c_2 = 4/3 c_3 = 8/3 c_4 = 1 Multiply by the least common denominator, 3, to eliminate fractional coefficients: c_1 = 9 c_2 = 4 c_3 = 8 c_4 = 3 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 9 Fe + 4 Al_2O_3 ⟶ 8 Al + 3 FeO·Fe_2O_3

+ ⟶ +

iron + aluminum oxide ⟶ aluminum + iron(II, III) oxide

| iron | aluminum oxide | aluminum | iron(II, III) oxide molecular enthalpy | 0 kJ/mol | -1676 kJ/mol | 0 kJ/mol | -1118 kJ/mol total enthalpy | 0 kJ/mol | -6704 kJ/mol | 0 kJ/mol | -3355 kJ/mol | H_initial = -6704 kJ/mol | | H_final = -3355 kJ/mol | ΔH_rxn^0 | -3355 kJ/mol - -6704 kJ/mol = 3349 kJ/mol (endothermic) | | |

Construct the equilibrium constant, K, expression for: Fe + Al_2O_3 ⟶ Al + FeO·Fe_2O_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: 9 Fe + 4 Al_2O_3 ⟶ 8 Al + 3 FeO·Fe_2O_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 | 9 | -9 Al_2O_3 | 4 | -4 Al | 8 | 8 FeO·Fe_2O_3 | 3 | 3 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression Fe | 9 | -9 | ([Fe])^(-9) Al_2O_3 | 4 | -4 | ([Al2O3])^(-4) Al | 8 | 8 | ([Al])^8 FeO·Fe_2O_3 | 3 | 3 | ([FeO·Fe2O3])^3 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 = ([Fe])^(-9) ([Al2O3])^(-4) ([Al])^8 ([FeO·Fe2O3])^3 = (([Al])^8 ([FeO·Fe2O3])^3)/(([Fe])^9 ([Al2O3])^4)

Construct the rate of reaction expression for: Fe + Al_2O_3 ⟶ Al + FeO·Fe_2O_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: 9 Fe + 4 Al_2O_3 ⟶ 8 Al + 3 FeO·Fe_2O_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 | 9 | -9 Al_2O_3 | 4 | -4 Al | 8 | 8 FeO·Fe_2O_3 | 3 | 3 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 | 9 | -9 | -1/9 (Δ[Fe])/(Δt) Al_2O_3 | 4 | -4 | -1/4 (Δ[Al2O3])/(Δt) Al | 8 | 8 | 1/8 (Δ[Al])/(Δt) FeO·Fe_2O_3 | 3 | 3 | 1/3 (Δ[FeO·Fe2O3])/(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 = -1/9 (Δ[Fe])/(Δt) = -1/4 (Δ[Al2O3])/(Δt) = 1/8 (Δ[Al])/(Δt) = 1/3 (Δ[FeO·Fe2O3])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| iron | aluminum oxide | aluminum | iron(II, III) oxide formula | Fe | Al_2O_3 | Al | FeO·Fe_2O_3 Hill formula | Fe | Al_2O_3 | Al | Fe_3O_4 name | iron | aluminum oxide | aluminum | iron(II, III) oxide IUPAC name | iron | dialuminum;oxygen(2-) | aluminum |

| iron | aluminum oxide | aluminum | iron(II, III) oxide molar mass | 55.845 g/mol | 101.96 g/mol | 26.9815385 g/mol | 231.53 g/mol phase | solid (at STP) | solid (at STP) | solid (at STP) | solid (at STP) melting point | 1535 °C | 2040 °C | 660.4 °C | 1538 °C boiling point | 2750 °C | | 2460 °C | density | 7.874 g/cm^3 | | 2.7 g/cm^3 | 5 g/cm^3 solubility in water | insoluble | | insoluble | surface tension | | | 0.817 N/m | dynamic viscosity | | | 1.5×10^-4 Pa s (at 760 °C) | odor | | odorless | odorless |