Fe2O3 + HI = H2O + FeI3

Fe_2O_3 iron(III) oxide + HI hydrogen iodide ⟶ H_2O water + FeI3

Balance the chemical equation algebraically: Fe_2O_3 + HI ⟶ H_2O + FeI3 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 Fe_2O_3 + c_2 HI ⟶ c_3 H_2O + c_4 FeI3 Set the number of atoms in the reactants equal to the number of atoms in the products for Fe, O, H and I: Fe: | 2 c_1 = c_4 O: | 3 c_1 = c_3 H: | c_2 = 2 c_3 I: | 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_1 = 1 and solve the system of equations for the remaining coefficients: c_1 = 1 c_2 = 6 c_3 = 3 c_4 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | Fe_2O_3 + 6 HI ⟶ 3 H_2O + 2 FeI3

+ ⟶ + FeI3

iron(III) oxide + hydrogen iodide ⟶ water + FeI3

Construct the equilibrium constant, K, expression for: Fe_2O_3 + HI ⟶ H_2O + FeI3 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: Fe_2O_3 + 6 HI ⟶ 3 H_2O + 2 FeI3 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 Fe_2O_3 | 1 | -1 HI | 6 | -6 H_2O | 3 | 3 FeI3 | 2 | 2 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression Fe_2O_3 | 1 | -1 | ([Fe2O3])^(-1) HI | 6 | -6 | ([HI])^(-6) H_2O | 3 | 3 | ([H2O])^3 FeI3 | 2 | 2 | ([FeI3])^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 = ([Fe2O3])^(-1) ([HI])^(-6) ([H2O])^3 ([FeI3])^2 = (([H2O])^3 ([FeI3])^2)/([Fe2O3] ([HI])^6)

Construct the rate of reaction expression for: Fe_2O_3 + HI ⟶ H_2O + FeI3 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: Fe_2O_3 + 6 HI ⟶ 3 H_2O + 2 FeI3 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 Fe_2O_3 | 1 | -1 HI | 6 | -6 H_2O | 3 | 3 FeI3 | 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 Fe_2O_3 | 1 | -1 | -(Δ[Fe2O3])/(Δt) HI | 6 | -6 | -1/6 (Δ[HI])/(Δt) H_2O | 3 | 3 | 1/3 (Δ[H2O])/(Δt) FeI3 | 2 | 2 | 1/2 (Δ[FeI3])/(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[Fe2O3])/(Δt) = -1/6 (Δ[HI])/(Δt) = 1/3 (Δ[H2O])/(Δt) = 1/2 (Δ[FeI3])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| iron(III) oxide | hydrogen iodide | water | FeI3 formula | Fe_2O_3 | HI | H_2O | FeI3 name | iron(III) oxide | hydrogen iodide | water |

| iron(III) oxide | hydrogen iodide | water | FeI3 molar mass | 159.69 g/mol | 127.912 g/mol | 18.015 g/mol | 436.558 g/mol phase | solid (at STP) | gas (at STP) | liquid (at STP) | melting point | 1565 °C | -50.76 °C | 0 °C | boiling point | | -35.55 °C | 99.9839 °C | density | 5.26 g/cm^3 | 0.005228 g/cm^3 (at 25 °C) | 1 g/cm^3 | solubility in water | insoluble | very soluble | | surface tension | | | 0.0728 N/m | dynamic viscosity | | 0.001321 Pa s (at -39 °C) | 8.9×10^-4 Pa s (at 25 °C) | odor | odorless | | odorless |