Fe2O3 + CH4 = H2O + CO2 + Fe

Fe_2O_3 iron(III) oxide + CH_4 methane ⟶ H_2O water + CO_2 carbon dioxide + Fe iron

Balance the chemical equation algebraically: Fe_2O_3 + CH_4 ⟶ H_2O + CO_2 + Fe Add stoichiometric coefficients, c_i, to the reactants and products: c_1 Fe_2O_3 + c_2 CH_4 ⟶ c_3 H_2O + c_4 CO_2 + c_5 Fe Set the number of atoms in the reactants equal to the number of atoms in the products for Fe, O, C and H: Fe: | 2 c_1 = c_5 O: | 3 c_1 = c_3 + 2 c_4 C: | c_2 = c_4 H: | 4 c_2 = 2 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_2 = 1 and solve the system of equations for the remaining coefficients: c_1 = 4/3 c_2 = 1 c_3 = 2 c_4 = 1 c_5 = 8/3 Multiply by the least common denominator, 3, to eliminate fractional coefficients: c_1 = 4 c_2 = 3 c_3 = 6 c_4 = 3 c_5 = 8 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 4 Fe_2O_3 + 3 CH_4 ⟶ 6 H_2O + 3 CO_2 + 8 Fe

+ ⟶ + +

iron(III) oxide + methane ⟶ water + carbon dioxide + iron

ΔH_rxn^0 | -2895 kJ/mol - -3528 kJ/mol = 632.3 kJ/mol (endothermic)

ΔS_rxn^0 | 1277 J/(mol K) - 918 J/(mol K) = 359.5 J/(mol K) (endoentropic)

K_c = ([H2O]^6 [CO2]^3 [Fe]^8)/([Fe2O3]^4 [CH4]^3)

rate = -1/4 (Δ[Fe2O3])/(Δt) = -1/3 (Δ[CH4])/(Δt) = 1/6 (Δ[H2O])/(Δt) = 1/3 (Δ[CO2])/(Δt) = 1/8 (Δ[Fe])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| iron(III) oxide | methane | water | carbon dioxide | iron formula | Fe_2O_3 | CH_4 | H_2O | CO_2 | Fe name | iron(III) oxide | methane | water | carbon dioxide | iron