Al + Fe3O4 = Fe + Al2O4

Al aluminum + FeO·Fe_2O_3 iron(II, III) oxide ⟶ Fe iron + Al2O4

Balance the chemical equation algebraically: Al + FeO·Fe_2O_3 ⟶ Fe + Al2O4 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 Al + c_2 FeO·Fe_2O_3 ⟶ c_3 Fe + c_4 Al2O4 Set the number of atoms in the reactants equal to the number of atoms in the products for Al, Fe and O: Al: | c_1 = 2 c_4 Fe: | 3 c_2 = c_3 O: | 4 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_2 = 1 and solve the system of equations for the remaining coefficients: c_1 = 2 c_2 = 1 c_3 = 3 c_4 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 2 Al + FeO·Fe_2O_3 ⟶ 3 Fe + Al2O4

+ ⟶ + Al2O4

aluminum + iron(II, III) oxide ⟶ iron + Al2O4

Construct the equilibrium constant, K, expression for: Al + FeO·Fe_2O_3 ⟶ Fe + Al2O4 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: 2 Al + FeO·Fe_2O_3 ⟶ 3 Fe + Al2O4 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 Al | 2 | -2 FeO·Fe_2O_3 | 1 | -1 Fe | 3 | 3 Al2O4 | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression Al | 2 | -2 | ([Al])^(-2) FeO·Fe_2O_3 | 1 | -1 | ([FeO·Fe2O3])^(-1) Fe | 3 | 3 | ([Fe])^3 Al2O4 | 1 | 1 | [Al2O4] 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 = ([Al])^(-2) ([FeO·Fe2O3])^(-1) ([Fe])^3 [Al2O4] = (([Fe])^3 [Al2O4])/(([Al])^2 [FeO·Fe2O3])

Construct the rate of reaction expression for: Al + FeO·Fe_2O_3 ⟶ Fe + Al2O4 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: 2 Al + FeO·Fe_2O_3 ⟶ 3 Fe + Al2O4 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 Al | 2 | -2 FeO·Fe_2O_3 | 1 | -1 Fe | 3 | 3 Al2O4 | 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 Al | 2 | -2 | -1/2 (Δ[Al])/(Δt) FeO·Fe_2O_3 | 1 | -1 | -(Δ[FeO·Fe2O3])/(Δt) Fe | 3 | 3 | 1/3 (Δ[Fe])/(Δt) Al2O4 | 1 | 1 | (Δ[Al2O4])/(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 = -1/2 (Δ[Al])/(Δt) = -(Δ[FeO·Fe2O3])/(Δt) = 1/3 (Δ[Fe])/(Δt) = (Δ[Al2O4])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| aluminum | iron(II, III) oxide | iron | Al2O4 formula | Al | FeO·Fe_2O_3 | Fe | Al2O4 Hill formula | Al | Fe_3O_4 | Fe | Al2O4 name | aluminum | iron(II, III) oxide | iron |

| aluminum | iron(II, III) oxide | iron | Al2O4 molar mass | 26.9815385 g/mol | 231.53 g/mol | 55.845 g/mol | 117.96 g/mol phase | solid (at STP) | solid (at STP) | solid (at STP) | melting point | 660.4 °C | 1538 °C | 1535 °C | boiling point | 2460 °C | | 2750 °C | density | 2.7 g/cm^3 | 5 g/cm^3 | 7.874 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 | | |