H2 + Fe2O3 = H2O + FeO

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

Balance the chemical equation algebraically: H_2 + Fe_2O_3 ⟶ H_2O + FeO Add stoichiometric coefficients, c_i, to the reactants and products: c_1 H_2 + c_2 Fe_2O_3 ⟶ c_3 H_2O + c_4 FeO Set the number of atoms in the reactants equal to the number of atoms in the products for H, Fe and O: H: | 2 c_1 = 2 c_3 Fe: | 2 c_2 = c_4 O: | 3 c_2 = c_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_1 = 1 and solve the system of equations for the remaining coefficients: c_1 = 1 c_2 = 1 c_3 = 1 c_4 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | H_2 + Fe_2O_3 ⟶ H_2O + 2 FeO

+ ⟶ +

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

| hydrogen | iron(III) oxide | water | iron(II) oxide molecular enthalpy | 0 kJ/mol | -826 kJ/mol | -285.8 kJ/mol | -272 kJ/mol total enthalpy | 0 kJ/mol | -826 kJ/mol | -285.8 kJ/mol | -544 kJ/mol | H_initial = -826 kJ/mol | | H_final = -829.8 kJ/mol | ΔH_rxn^0 | -829.8 kJ/mol - -826 kJ/mol = -3.83 kJ/mol (exothermic) | | |

| hydrogen | iron(III) oxide | water | iron(II) oxide molecular free energy | 0 kJ/mol | -742.2 kJ/mol | -237.1 kJ/mol | -255 kJ/mol total free energy | 0 kJ/mol | -742.2 kJ/mol | -237.1 kJ/mol | -510 kJ/mol | G_initial = -742.2 kJ/mol | | G_final = -747.1 kJ/mol | ΔG_rxn^0 | -747.1 kJ/mol - -742.2 kJ/mol = -4.9 kJ/mol (exergonic) | | |

| hydrogen | iron(III) oxide | water | iron(II) oxide molecular entropy | 115 J/(mol K) | 90 J/(mol K) | 69.91 J/(mol K) | 61 J/(mol K) total entropy | 115 J/(mol K) | 90 J/(mol K) | 69.91 J/(mol K) | 122 J/(mol K) | S_initial = 205 J/(mol K) | | S_final = 191.9 J/(mol K) | ΔS_rxn^0 | 191.9 J/(mol K) - 205 J/(mol K) = -13.09 J/(mol K) (exoentropic) | | |

Construct the equilibrium constant, K, expression for: H_2 + Fe_2O_3 ⟶ H_2O + FeO 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: H_2 + Fe_2O_3 ⟶ H_2O + 2 FeO 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_2 | 1 | -1 Fe_2O_3 | 1 | -1 H_2O | 1 | 1 FeO | 2 | 2 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression H_2 | 1 | -1 | ([H2])^(-1) Fe_2O_3 | 1 | -1 | ([Fe2O3])^(-1) H_2O | 1 | 1 | [H2O] FeO | 2 | 2 | ([FeO])^2 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 = ([H2])^(-1) ([Fe2O3])^(-1) [H2O] ([FeO])^2 = ([H2O] ([FeO])^2)/([H2] [Fe2O3])

Construct the rate of reaction expression for: H_2 + Fe_2O_3 ⟶ H_2O + FeO 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: H_2 + Fe_2O_3 ⟶ H_2O + 2 FeO 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_2 | 1 | -1 Fe_2O_3 | 1 | -1 H_2O | 1 | 1 FeO | 2 | 2 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_2 | 1 | -1 | -(Δ[H2])/(Δt) Fe_2O_3 | 1 | -1 | -(Δ[Fe2O3])/(Δt) H_2O | 1 | 1 | (Δ[H2O])/(Δt) FeO | 2 | 2 | 1/2 (Δ[FeO])/(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 = -(Δ[H2])/(Δt) = -(Δ[Fe2O3])/(Δt) = (Δ[H2O])/(Δt) = 1/2 (Δ[FeO])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| hydrogen | iron(III) oxide | water | iron(II) oxide formula | H_2 | Fe_2O_3 | H_2O | FeO name | hydrogen | iron(III) oxide | water | iron(II) oxide IUPAC name | molecular hydrogen | | water | oxoiron