H2O + I2 = HI + HIO4

H_2O water + I_2 iodine ⟶ HI hydrogen iodide + HIO4

Balance the chemical equation algebraically: H_2O + I_2 ⟶ HI + HIO4 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 H_2O + c_2 I_2 ⟶ c_3 HI + c_4 HIO4 Set the number of atoms in the reactants equal to the number of atoms in the products for H, O and I: H: | 2 c_1 = c_3 + c_4 O: | c_1 = 4 c_4 I: | 2 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_4 = 1 and solve the system of equations for the remaining coefficients: c_1 = 4 c_2 = 4 c_3 = 7 c_4 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 4 H_2O + 4 I_2 ⟶ 7 HI + HIO4

+ ⟶ + HIO4

water + iodine ⟶ hydrogen iodide + HIO4

Construct the equilibrium constant, K, expression for: H_2O + I_2 ⟶ HI + HIO4 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 + 4 I_2 ⟶ 7 HI + HIO4 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 I_2 | 4 | -4 HI | 7 | 7 HIO4 | 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) I_2 | 4 | -4 | ([I2])^(-4) HI | 7 | 7 | ([HI])^7 HIO4 | 1 | 1 | [HIO4] 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) ([I2])^(-4) ([HI])^7 [HIO4] = (([HI])^7 [HIO4])/(([H2O])^4 ([I2])^4)

Construct the rate of reaction expression for: H_2O + I_2 ⟶ HI + HIO4 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 + 4 I_2 ⟶ 7 HI + HIO4 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 I_2 | 4 | -4 HI | 7 | 7 HIO4 | 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) I_2 | 4 | -4 | -1/4 (Δ[I2])/(Δt) HI | 7 | 7 | 1/7 (Δ[HI])/(Δt) HIO4 | 1 | 1 | (Δ[HIO4])/(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) = -1/4 (Δ[I2])/(Δt) = 1/7 (Δ[HI])/(Δt) = (Δ[HIO4])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| water | iodine | hydrogen iodide | HIO4 formula | H_2O | I_2 | HI | HIO4 name | water | iodine | hydrogen iodide | IUPAC name | water | molecular iodine | hydrogen iodide |

| water | iodine | hydrogen iodide | HIO4 molar mass | 18.015 g/mol | 253.80894 g/mol | 127.912 g/mol | 191.91 g/mol phase | liquid (at STP) | solid (at STP) | gas (at STP) | melting point | 0 °C | 113 °C | -50.76 °C | boiling point | 99.9839 °C | 184 °C | -35.55 °C | density | 1 g/cm^3 | 4.94 g/cm^3 | 0.005228 g/cm^3 (at 25 °C) | solubility in water | | | very soluble | surface tension | 0.0728 N/m | | | dynamic viscosity | 8.9×10^-4 Pa s (at 25 °C) | 0.00227 Pa s (at 116 °C) | 0.001321 Pa s (at -39 °C) | odor | odorless | | |