H2SO4 + KClO3 + HI = H2O + HCl + I2 + KHSO4

H_2SO_4 sulfuric acid + KClO_3 potassium chlorate + HI hydrogen iodide ⟶ H_2O water + HCl hydrogen chloride + I_2 iodine + KHSO_4 potassium bisulfate

Balance the chemical equation algebraically: H_2SO_4 + KClO_3 + HI ⟶ H_2O + HCl + I_2 + KHSO_4 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 H_2SO_4 + c_2 KClO_3 + c_3 HI ⟶ c_4 H_2O + c_5 HCl + c_6 I_2 + c_7 KHSO_4 Set the number of atoms in the reactants equal to the number of atoms in the products for H, O, S, Cl, K and I: H: | 2 c_1 + c_3 = 2 c_4 + c_5 + c_7 O: | 4 c_1 + 3 c_2 = c_4 + 4 c_7 S: | c_1 = c_7 Cl: | c_2 = c_5 K: | c_2 = c_7 I: | c_3 = 2 c_6 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 = 1 c_3 = 6 c_4 = 3 c_5 = 1 c_6 = 3 c_7 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | H_2SO_4 + KClO_3 + 6 HI ⟶ 3 H_2O + HCl + 3 I_2 + KHSO_4

+ + ⟶ + + +

sulfuric acid + potassium chlorate + hydrogen iodide ⟶ water + hydrogen chloride + iodine + potassium bisulfate

| sulfuric acid | potassium chlorate | hydrogen iodide | water | hydrogen chloride | iodine | potassium bisulfate molecular enthalpy | -814 kJ/mol | -397.7 kJ/mol | 26.5 kJ/mol | -285.8 kJ/mol | -92.3 kJ/mol | 0 kJ/mol | -1161 kJ/mol total enthalpy | -814 kJ/mol | -397.7 kJ/mol | 159 kJ/mol | -857.5 kJ/mol | -92.3 kJ/mol | 0 kJ/mol | -1161 kJ/mol | H_initial = -1053 kJ/mol | | | H_final = -2110 kJ/mol | | | ΔH_rxn^0 | -2110 kJ/mol - -1053 kJ/mol = -1058 kJ/mol (exothermic) | | | | | |

| sulfuric acid | potassium chlorate | hydrogen iodide | water | hydrogen chloride | iodine | potassium bisulfate molecular free energy | -690 kJ/mol | -296.3 kJ/mol | 1.7 kJ/mol | -237.1 kJ/mol | -95.3 kJ/mol | 0 kJ/mol | -1031 kJ/mol total free energy | -690 kJ/mol | -296.3 kJ/mol | 10.2 kJ/mol | -711.3 kJ/mol | -95.3 kJ/mol | 0 kJ/mol | -1031 kJ/mol | G_initial = -976.1 kJ/mol | | | G_final = -1838 kJ/mol | | | ΔG_rxn^0 | -1838 kJ/mol - -976.1 kJ/mol = -861.8 kJ/mol (exergonic) | | | | | |

Construct the equilibrium constant, K, expression for: H_2SO_4 + KClO_3 + HI ⟶ H_2O + HCl + I_2 + KHSO_4 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: H_2SO_4 + KClO_3 + 6 HI ⟶ 3 H_2O + HCl + 3 I_2 + KHSO_4 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_2SO_4 | 1 | -1 KClO_3 | 1 | -1 HI | 6 | -6 H_2O | 3 | 3 HCl | 1 | 1 I_2 | 3 | 3 KHSO_4 | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression H_2SO_4 | 1 | -1 | ([H2SO4])^(-1) KClO_3 | 1 | -1 | ([KClO3])^(-1) HI | 6 | -6 | ([HI])^(-6) H_2O | 3 | 3 | ([H2O])^3 HCl | 1 | 1 | [HCl] I_2 | 3 | 3 | ([I2])^3 KHSO_4 | 1 | 1 | [KHSO4] 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 = ([H2SO4])^(-1) ([KClO3])^(-1) ([HI])^(-6) ([H2O])^3 [HCl] ([I2])^3 [KHSO4] = (([H2O])^3 [HCl] ([I2])^3 [KHSO4])/([H2SO4] [KClO3] ([HI])^6)

Construct the rate of reaction expression for: H_2SO_4 + KClO_3 + HI ⟶ H_2O + HCl + I_2 + KHSO_4 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: H_2SO_4 + KClO_3 + 6 HI ⟶ 3 H_2O + HCl + 3 I_2 + KHSO_4 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_2SO_4 | 1 | -1 KClO_3 | 1 | -1 HI | 6 | -6 H_2O | 3 | 3 HCl | 1 | 1 I_2 | 3 | 3 KHSO_4 | 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_2SO_4 | 1 | -1 | -(Δ[H2SO4])/(Δt) KClO_3 | 1 | -1 | -(Δ[KClO3])/(Δt) HI | 6 | -6 | -1/6 (Δ[HI])/(Δt) H_2O | 3 | 3 | 1/3 (Δ[H2O])/(Δt) HCl | 1 | 1 | (Δ[HCl])/(Δt) I_2 | 3 | 3 | 1/3 (Δ[I2])/(Δt) KHSO_4 | 1 | 1 | (Δ[KHSO4])/(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[H2SO4])/(Δt) = -(Δ[KClO3])/(Δt) = -1/6 (Δ[HI])/(Δt) = 1/3 (Δ[H2O])/(Δt) = (Δ[HCl])/(Δt) = 1/3 (Δ[I2])/(Δt) = (Δ[KHSO4])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| sulfuric acid | potassium chlorate | hydrogen iodide | water | hydrogen chloride | iodine | potassium bisulfate formula | H_2SO_4 | KClO_3 | HI | H_2O | HCl | I_2 | KHSO_4 Hill formula | H_2O_4S | ClKO_3 | HI | H_2O | ClH | I_2 | HKO_4S name | sulfuric acid | potassium chlorate | hydrogen iodide | water | hydrogen chloride | iodine | potassium bisulfate IUPAC name | sulfuric acid | potassium chlorate | hydrogen iodide | water | hydrogen chloride | molecular iodine | potassium hydrogen sulfate