HI + Fe3O4 = H2O + I2 + FeI2

HI hydrogen iodide + FeO·Fe_2O_3 iron(II, III) oxide ⟶ H_2O water + I_2 iodine + FeI_2 ferrous iodide

Balance the chemical equation algebraically: HI + FeO·Fe_2O_3 ⟶ H_2O + I_2 + FeI_2 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 HI + c_2 FeO·Fe_2O_3 ⟶ c_3 H_2O + c_4 I_2 + c_5 FeI_2 Set the number of atoms in the reactants equal to the number of atoms in the products for H, I, Fe and O: H: | c_1 = 2 c_3 I: | c_1 = 2 c_4 + 2 c_5 Fe: | 3 c_2 = c_5 O: | 4 c_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 = 8 c_2 = 1 c_3 = 4 c_4 = 1 c_5 = 3 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 8 HI + FeO·Fe_2O_3 ⟶ 4 H_2O + I_2 + 3 FeI_2

+ ⟶ + +

hydrogen iodide + iron(II, III) oxide ⟶ water + iodine + ferrous iodide

Construct the equilibrium constant, K, expression for: HI + FeO·Fe_2O_3 ⟶ H_2O + I_2 + FeI_2 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: 8 HI + FeO·Fe_2O_3 ⟶ 4 H_2O + I_2 + 3 FeI_2 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 HI | 8 | -8 FeO·Fe_2O_3 | 1 | -1 H_2O | 4 | 4 I_2 | 1 | 1 FeI_2 | 3 | 3 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression HI | 8 | -8 | ([HI])^(-8) FeO·Fe_2O_3 | 1 | -1 | ([FeO·Fe2O3])^(-1) H_2O | 4 | 4 | ([H2O])^4 I_2 | 1 | 1 | [I2] FeI_2 | 3 | 3 | ([FeI2])^3 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 = ([HI])^(-8) ([FeO·Fe2O3])^(-1) ([H2O])^4 [I2] ([FeI2])^3 = (([H2O])^4 [I2] ([FeI2])^3)/(([HI])^8 [FeO·Fe2O3])

Construct the rate of reaction expression for: HI + FeO·Fe_2O_3 ⟶ H_2O + I_2 + FeI_2 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: 8 HI + FeO·Fe_2O_3 ⟶ 4 H_2O + I_2 + 3 FeI_2 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 HI | 8 | -8 FeO·Fe_2O_3 | 1 | -1 H_2O | 4 | 4 I_2 | 1 | 1 FeI_2 | 3 | 3 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 HI | 8 | -8 | -1/8 (Δ[HI])/(Δt) FeO·Fe_2O_3 | 1 | -1 | -(Δ[FeO·Fe2O3])/(Δt) H_2O | 4 | 4 | 1/4 (Δ[H2O])/(Δt) I_2 | 1 | 1 | (Δ[I2])/(Δt) FeI_2 | 3 | 3 | 1/3 (Δ[FeI2])/(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 = -1/8 (Δ[HI])/(Δt) = -(Δ[FeO·Fe2O3])/(Δt) = 1/4 (Δ[H2O])/(Δt) = (Δ[I2])/(Δt) = 1/3 (Δ[FeI2])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| hydrogen iodide | iron(II, III) oxide | water | iodine | ferrous iodide formula | HI | FeO·Fe_2O_3 | H_2O | I_2 | FeI_2 Hill formula | HI | Fe_3O_4 | H_2O | I_2 | FeI_2 name | hydrogen iodide | iron(II, III) oxide | water | iodine | ferrous iodide IUPAC name | hydrogen iodide | | water | molecular iodine | diiodoiron

| hydrogen iodide | iron(II, III) oxide | water | iodine | ferrous iodide molar mass | 127.912 g/mol | 231.53 g/mol | 18.015 g/mol | 253.80894 g/mol | 309.654 g/mol phase | gas (at STP) | solid (at STP) | liquid (at STP) | solid (at STP) | melting point | -50.76 °C | 1538 °C | 0 °C | 113 °C | boiling point | -35.55 °C | | 99.9839 °C | 184 °C | density | 0.005228 g/cm^3 (at 25 °C) | 5 g/cm^3 | 1 g/cm^3 | 4.94 g/cm^3 | 5.32 g/cm^3 solubility in water | very soluble | | | | slightly 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) | 0.00227 Pa s (at 116 °C) | odor | | | odorless | |