CO + HIO3 = H2O + CO2 + I2

CO (carbon monoxide) + HIO_3 (iodic acid) ⟶ H_2O (water) + CO_2 (carbon dioxide) + I_2 (iodine)

Balance the chemical equation algebraically: CO + HIO_3 ⟶ H_2O + CO_2 + I_2 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 CO + c_2 HIO_3 ⟶ c_3 H_2O + c_4 CO_2 + c_5 I_2 Set the number of atoms in the reactants equal to the number of atoms in the products for C, O, H and I: C: | c_1 = c_4 O: | c_1 + 3 c_2 = c_3 + 2 c_4 H: | c_2 = 2 c_3 I: | c_2 = 2 c_5 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_3 = 1 and solve the system of equations for the remaining coefficients: c_1 = 5 c_2 = 2 c_3 = 1 c_4 = 5 c_5 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 5 CO + 2 HIO_3 ⟶ H_2O + 5 CO_2 + I_2

+ ⟶ + +

carbon monoxide + iodic acid ⟶ water + carbon dioxide + iodine

| carbon monoxide | iodic acid | water | carbon dioxide | iodine molecular enthalpy | -110.5 kJ/mol | -230.1 kJ/mol | -285.8 kJ/mol | -393.5 kJ/mol | 0 kJ/mol total enthalpy | -552.5 kJ/mol | -460.2 kJ/mol | -285.8 kJ/mol | -1968 kJ/mol | 0 kJ/mol | H_initial = -1013 kJ/mol | | H_final = -2253 kJ/mol | | ΔH_rxn^0 | -2253 kJ/mol - -1013 kJ/mol = -1241 kJ/mol (exothermic) | | | |

Construct the equilibrium constant, K, expression for: CO + HIO_3 ⟶ H_2O + CO_2 + 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 CO + 2 HIO_3 ⟶ H_2O + 5 CO_2 + 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 CO | 5 | -5 HIO_3 | 2 | -2 H_2O | 1 | 1 CO_2 | 5 | 5 I_2 | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression CO | 5 | -5 | ([CO])^(-5) HIO_3 | 2 | -2 | ([HIO3])^(-2) H_2O | 1 | 1 | [H2O] CO_2 | 5 | 5 | ([CO2])^5 I_2 | 1 | 1 | [I2] 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 = ([CO])^(-5) ([HIO3])^(-2) [H2O] ([CO2])^5 [I2] = ([H2O] ([CO2])^5 [I2])/(([CO])^5 ([HIO3])^2)

Construct the rate of reaction expression for: CO + HIO_3 ⟶ H_2O + CO_2 + 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 CO + 2 HIO_3 ⟶ H_2O + 5 CO_2 + 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 CO | 5 | -5 HIO_3 | 2 | -2 H_2O | 1 | 1 CO_2 | 5 | 5 I_2 | 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 CO | 5 | -5 | -1/5 (Δ[CO])/(Δt) HIO_3 | 2 | -2 | -1/2 (Δ[HIO3])/(Δt) H_2O | 1 | 1 | (Δ[H2O])/(Δt) CO_2 | 5 | 5 | 1/5 (Δ[CO2])/(Δt) I_2 | 1 | 1 | (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[CO])/(Δt) = -1/2 (Δ[HIO3])/(Δt) = (Δ[H2O])/(Δt) = 1/5 (Δ[CO2])/(Δt) = (Δ[I2])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| carbon monoxide | iodic acid | water | carbon dioxide | iodine formula | CO | HIO_3 | H_2O | CO_2 | I_2 name | carbon monoxide | iodic acid | water | carbon dioxide | iodine IUPAC name | carbon monoxide | iodic acid | water | carbon dioxide | molecular iodine