HI + HIO3 = H2O + I2

HI hydrogen iodide + HIO_3 iodic acid ⟶ H_2O water + I_2 iodine

Balance the chemical equation algebraically: HI + HIO_3 ⟶ H_2O + I_2 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 HI + c_2 HIO_3 ⟶ c_3 H_2O + c_4 I_2 Set the number of atoms in the reactants equal to the number of atoms in the products for H, I and O: H: | c_1 + c_2 = 2 c_3 I: | c_1 + c_2 = 2 c_4 O: | 3 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 = 5 c_2 = 1 c_3 = 3 c_4 = 3 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 5 HI + HIO_3 ⟶ 3 H_2O + 3 I_2

+ ⟶ +

hydrogen iodide + iodic acid ⟶ water + iodine

| hydrogen iodide | iodic acid | water | iodine molecular enthalpy | 26.5 kJ/mol | -230.1 kJ/mol | -285.8 kJ/mol | 0 kJ/mol total enthalpy | 132.5 kJ/mol | -230.1 kJ/mol | -857.5 kJ/mol | 0 kJ/mol | H_initial = -97.6 kJ/mol | | H_final = -857.5 kJ/mol | ΔH_rxn^0 | -857.5 kJ/mol - -97.6 kJ/mol = -759.9 kJ/mol (exothermic) | | |

Construct the equilibrium constant, K, expression for: HI + HIO_3 ⟶ H_2O + I_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: 5 HI + HIO_3 ⟶ 3 H_2O + 3 I_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 | 5 | -5 HIO_3 | 1 | -1 H_2O | 3 | 3 I_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 | 5 | -5 | ([HI])^(-5) HIO_3 | 1 | -1 | ([HIO3])^(-1) H_2O | 3 | 3 | ([H2O])^3 I_2 | 3 | 3 | ([I2])^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])^(-5) ([HIO3])^(-1) ([H2O])^3 ([I2])^3 = (([H2O])^3 ([I2])^3)/(([HI])^5 [HIO3])

Construct the rate of reaction expression for: HI + HIO_3 ⟶ H_2O + I_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: 5 HI + HIO_3 ⟶ 3 H_2O + 3 I_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 | 5 | -5 HIO_3 | 1 | -1 H_2O | 3 | 3 I_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 | 5 | -5 | -1/5 (Δ[HI])/(Δt) HIO_3 | 1 | -1 | -(Δ[HIO3])/(Δt) H_2O | 3 | 3 | 1/3 (Δ[H2O])/(Δt) I_2 | 3 | 3 | 1/3 (Δ[I2])/(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 = -1/5 (Δ[HI])/(Δt) = -(Δ[HIO3])/(Δt) = 1/3 (Δ[H2O])/(Δt) = 1/3 (Δ[I2])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| hydrogen iodide | iodic acid | water | iodine formula | HI | HIO_3 | H_2O | I_2 name | hydrogen iodide | iodic acid | water | iodine IUPAC name | hydrogen iodide | iodic acid | water | molecular iodine

| hydrogen iodide | iodic acid | water | iodine molar mass | 127.912 g/mol | 175.91 g/mol | 18.015 g/mol | 253.80894 g/mol phase | gas (at STP) | solid (at STP) | liquid (at STP) | solid (at STP) melting point | -50.76 °C | 110 °C | 0 °C | 113 °C boiling point | -35.55 °C | | 99.9839 °C | 184 °C density | 0.005228 g/cm^3 (at 25 °C) | 4.629 g/cm^3 | 1 g/cm^3 | 4.94 g/cm^3 solubility in water | very soluble | 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) | 0.00227 Pa s (at 116 °C) odor | | | odorless |