KOH + I2 = H2O + KI + KIO3

KOH (potassium hydroxide) + I_2 (iodine) ⟶ H_2O (water) + KI (potassium iodide) + KIO_3 (potassium iodate)

Balance the chemical equation algebraically: KOH + I_2 ⟶ H_2O + KI + KIO_3 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 KOH + c_2 I_2 ⟶ c_3 H_2O + c_4 KI + c_5 KIO_3 Set the number of atoms in the reactants equal to the number of atoms in the products for H, K, O and I: H: | c_1 = 2 c_3 K: | c_1 = c_4 + c_5 O: | c_1 = c_3 + 3 c_5 I: | 2 c_2 = c_4 + 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_5 = 1 and solve the system of equations for the remaining coefficients: c_1 = 6 c_2 = 3 c_3 = 3 c_4 = 5 c_5 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 6 KOH + 3 I_2 ⟶ 3 H_2O + 5 KI + KIO_3

+ ⟶ + +

potassium hydroxide + iodine ⟶ water + potassium iodide + potassium iodate

| potassium hydroxide | iodine | water | potassium iodide | potassium iodate molecular enthalpy | -424.6 kJ/mol | 0 kJ/mol | -285.8 kJ/mol | -327.9 kJ/mol | -501.4 kJ/mol total enthalpy | -2548 kJ/mol | 0 kJ/mol | -857.5 kJ/mol | -1640 kJ/mol | -501.4 kJ/mol | H_initial = -2548 kJ/mol | | H_final = -2998 kJ/mol | | ΔH_rxn^0 | -2998 kJ/mol - -2548 kJ/mol = -450.8 kJ/mol (exothermic) | | | |

| potassium hydroxide | iodine | water | potassium iodide | potassium iodate molecular free energy | -379.4 kJ/mol | 0 kJ/mol | -237.1 kJ/mol | -324.9 kJ/mol | -418.4 kJ/mol total free energy | -2276 kJ/mol | 0 kJ/mol | -711.3 kJ/mol | -1625 kJ/mol | -418.4 kJ/mol | G_initial = -2276 kJ/mol | | G_final = -2754 kJ/mol | | ΔG_rxn^0 | -2754 kJ/mol - -2276 kJ/mol = -477.8 kJ/mol (exergonic) | | | |

Construct the equilibrium constant, K, expression for: KOH + I_2 ⟶ H_2O + KI + KIO_3 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: 6 KOH + 3 I_2 ⟶ 3 H_2O + 5 KI + KIO_3 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 KOH | 6 | -6 I_2 | 3 | -3 H_2O | 3 | 3 KI | 5 | 5 KIO_3 | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression KOH | 6 | -6 | ([KOH])^(-6) I_2 | 3 | -3 | ([I2])^(-3) H_2O | 3 | 3 | ([H2O])^3 KI | 5 | 5 | ([KI])^5 KIO_3 | 1 | 1 | [KIO3] 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 = ([KOH])^(-6) ([I2])^(-3) ([H2O])^3 ([KI])^5 [KIO3] = (([H2O])^3 ([KI])^5 [KIO3])/(([KOH])^6 ([I2])^3)

Construct the rate of reaction expression for: KOH + I_2 ⟶ H_2O + KI + KIO_3 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: 6 KOH + 3 I_2 ⟶ 3 H_2O + 5 KI + KIO_3 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 KOH | 6 | -6 I_2 | 3 | -3 H_2O | 3 | 3 KI | 5 | 5 KIO_3 | 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 KOH | 6 | -6 | -1/6 (Δ[KOH])/(Δt) I_2 | 3 | -3 | -1/3 (Δ[I2])/(Δt) H_2O | 3 | 3 | 1/3 (Δ[H2O])/(Δt) KI | 5 | 5 | 1/5 (Δ[KI])/(Δt) KIO_3 | 1 | 1 | (Δ[KIO3])/(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 = -1/6 (Δ[KOH])/(Δt) = -1/3 (Δ[I2])/(Δt) = 1/3 (Δ[H2O])/(Δt) = 1/5 (Δ[KI])/(Δt) = (Δ[KIO3])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| potassium hydroxide | iodine | water | potassium iodide | potassium iodate formula | KOH | I_2 | H_2O | KI | KIO_3 Hill formula | HKO | I_2 | H_2O | IK | IKO_3 name | potassium hydroxide | iodine | water | potassium iodide | potassium iodate IUPAC name | potassium hydroxide | molecular iodine | water | potassium iodide | potassium iodate

| potassium hydroxide | iodine | water | potassium iodide | potassium iodate molar mass | 56.105 g/mol | 253.80894 g/mol | 18.015 g/mol | 166.0028 g/mol | 214 g/mol phase | solid (at STP) | solid (at STP) | liquid (at STP) | solid (at STP) | solid (at STP) melting point | 406 °C | 113 °C | 0 °C | 681 °C | 560 °C boiling point | 1327 °C | 184 °C | 99.9839 °C | 1330 °C | density | 2.044 g/cm^3 | 4.94 g/cm^3 | 1 g/cm^3 | 3.123 g/cm^3 | 1.005 g/cm^3 solubility in water | soluble | | | | surface tension | | | 0.0728 N/m | | dynamic viscosity | 0.001 Pa s (at 550 °C) | 0.00227 Pa s (at 116 °C) | 8.9×10^-4 Pa s (at 25 °C) | 0.0010227 Pa s (at 732.9 °C) | odor | | | odorless | |