H2O + Fe = H2 + Fe3O4

H_2O (water) + Fe (iron) ⟶ H_2 (hydrogen) + FeO·Fe_2O_3 (iron(II, III) oxide)

Balance the chemical equation algebraically: H_2O + Fe ⟶ H_2 + FeO·Fe_2O_3 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 H_2O + c_2 Fe ⟶ c_3 H_2 + c_4 FeO·Fe_2O_3 Set the number of atoms in the reactants equal to the number of atoms in the products for H, O and Fe: H: | 2 c_1 = 2 c_3 O: | c_1 = 4 c_4 Fe: | c_2 = 3 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 = 4 c_2 = 3 c_3 = 4 c_4 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 4 H_2O + 3 Fe ⟶ 4 H_2 + FeO·Fe_2O_3

+ ⟶ +

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

| water | iron | hydrogen | iron(II, III) oxide molecular enthalpy | -285.8 kJ/mol | 0 kJ/mol | 0 kJ/mol | -1118 kJ/mol total enthalpy | -1143 kJ/mol | 0 kJ/mol | 0 kJ/mol | -1118 kJ/mol | H_initial = -1143 kJ/mol | | H_final = -1118 kJ/mol | ΔH_rxn^0 | -1118 kJ/mol - -1143 kJ/mol = 24.92 kJ/mol (endothermic) | | |

Construct the equilibrium constant, K, expression for: H_2O + Fe ⟶ H_2 + 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: 4 H_2O + 3 Fe ⟶ 4 H_2 + 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 H_2O | 4 | -4 Fe | 3 | -3 H_2 | 4 | 4 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 H_2O | 4 | -4 | ([H2O])^(-4) Fe | 3 | -3 | ([Fe])^(-3) H_2 | 4 | 4 | ([H2])^4 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 = ([H2O])^(-4) ([Fe])^(-3) ([H2])^4 [FeO·Fe2O3] = (([H2])^4 [FeO·Fe2O3])/(([H2O])^4 ([Fe])^3)

Construct the rate of reaction expression for: H_2O + Fe ⟶ H_2 + 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: 4 H_2O + 3 Fe ⟶ 4 H_2 + 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 H_2O | 4 | -4 Fe | 3 | -3 H_2 | 4 | 4 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 H_2O | 4 | -4 | -1/4 (Δ[H2O])/(Δt) Fe | 3 | -3 | -1/3 (Δ[Fe])/(Δt) H_2 | 4 | 4 | 1/4 (Δ[H2])/(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 = -1/4 (Δ[H2O])/(Δt) = -1/3 (Δ[Fe])/(Δt) = 1/4 (Δ[H2])/(Δt) = (Δ[FeO·Fe2O3])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| water | iron | hydrogen | iron(II, III) oxide formula | H_2O | Fe | H_2 | FeO·Fe_2O_3 Hill formula | H_2O | Fe | H_2 | Fe_3O_4 name | water | iron | hydrogen | iron(II, III) oxide IUPAC name | water | iron | molecular hydrogen |

| water | iron | hydrogen | iron(II, III) oxide molar mass | 18.015 g/mol | 55.845 g/mol | 2.016 g/mol | 231.53 g/mol phase | liquid (at STP) | solid (at STP) | gas (at STP) | solid (at STP) melting point | 0 °C | 1535 °C | -259.2 °C | 1538 °C boiling point | 99.9839 °C | 2750 °C | -252.8 °C | density | 1 g/cm^3 | 7.874 g/cm^3 | 8.99×10^-5 g/cm^3 (at 0 °C) | 5 g/cm^3 solubility in water | | insoluble | | surface tension | 0.0728 N/m | | | dynamic viscosity | 8.9×10^-4 Pa s (at 25 °C) | | 8.9×10^-6 Pa s (at 25 °C) | odor | odorless | | odorless |