O2 + Fe = Fe3O4

O_2 (oxygen) + Fe (iron) ⟶ FeO·Fe_2O_3 (iron(II, III) oxide)

Balance the chemical equation algebraically: O_2 + Fe ⟶ FeO·Fe_2O_3 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 O_2 + c_2 Fe ⟶ c_3 FeO·Fe_2O_3 Set the number of atoms in the reactants equal to the number of atoms in the products for O and Fe: O: | 2 c_1 = 4 c_3 Fe: | c_2 = 3 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_3 = 1 and solve the system of equations for the remaining coefficients: c_1 = 2 c_2 = 3 c_3 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 2 O_2 + 3 Fe ⟶ FeO·Fe_2O_3

+ ⟶

oxygen + iron ⟶ iron(II, III) oxide

| oxygen | iron | iron(II, III) oxide molecular enthalpy | 0 kJ/mol | 0 kJ/mol | -1118 kJ/mol total enthalpy | 0 kJ/mol | 0 kJ/mol | -1118 kJ/mol | H_initial = 0 kJ/mol | | H_final = -1118 kJ/mol ΔH_rxn^0 | -1118 kJ/mol - 0 kJ/mol = -1118 kJ/mol (exothermic) | |

Construct the equilibrium constant, K, expression for: O_2 + Fe ⟶ 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: 2 O_2 + 3 Fe ⟶ 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 O_2 | 2 | -2 Fe | 3 | -3 FeO·Fe_2O_3 | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression O_2 | 2 | -2 | ([O2])^(-2) Fe | 3 | -3 | ([Fe])^(-3) FeO·Fe_2O_3 | 1 | 1 | [FeO·Fe2O3] 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 = ([O2])^(-2) ([Fe])^(-3) [FeO·Fe2O3] = ([FeO·Fe2O3])/(([O2])^2 ([Fe])^3)

Construct the rate of reaction expression for: O_2 + Fe ⟶ 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: 2 O_2 + 3 Fe ⟶ 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 O_2 | 2 | -2 Fe | 3 | -3 FeO·Fe_2O_3 | 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 O_2 | 2 | -2 | -1/2 (Δ[O2])/(Δt) Fe | 3 | -3 | -1/3 (Δ[Fe])/(Δt) FeO·Fe_2O_3 | 1 | 1 | (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 = -1/2 (Δ[O2])/(Δt) = -1/3 (Δ[Fe])/(Δt) = (Δ[FeO·Fe2O3])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| oxygen | iron | iron(II, III) oxide formula | O_2 | Fe | FeO·Fe_2O_3 Hill formula | O_2 | Fe | Fe_3O_4 name | oxygen | iron | iron(II, III) oxide IUPAC name | molecular oxygen | iron |

| oxygen | iron | iron(II, III) oxide molar mass | 31.998 g/mol | 55.845 g/mol | 231.53 g/mol phase | gas (at STP) | solid (at STP) | solid (at STP) melting point | -218 °C | 1535 °C | 1538 °C boiling point | -183 °C | 2750 °C | density | 0.001429 g/cm^3 (at 0 °C) | 7.874 g/cm^3 | 5 g/cm^3 solubility in water | | insoluble | surface tension | 0.01347 N/m | | dynamic viscosity | 2.055×10^-5 Pa s (at 25 °C) | | odor | odorless | |