I2 + Ba(OH)2 = H2O + Ba(IO3)2 + BaI2

I_2 (iodine) + Ba(OH)_2 (barium hydroxide) ⟶ H_2O (water) + BaI_2O_6 (barium iodate) + BaI_2 (barium iodide)

Balance the chemical equation algebraically: I_2 + Ba(OH)_2 ⟶ H_2O + BaI_2O_6 + BaI_2 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 I_2 + c_2 Ba(OH)_2 ⟶ c_3 H_2O + c_4 BaI_2O_6 + c_5 BaI_2 Set the number of atoms in the reactants equal to the number of atoms in the products for I, Ba, H and O: I: | 2 c_1 = 2 c_4 + 2 c_5 Ba: | c_2 = c_4 + c_5 H: | 2 c_2 = 2 c_3 O: | 2 c_2 = c_3 + 6 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 = 6 c_2 = 6 c_3 = 6 c_4 = 1 c_5 = 5 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 6 I_2 + 6 Ba(OH)_2 ⟶ 6 H_2O + BaI_2O_6 + 5 BaI_2

+ ⟶ + +

iodine + barium hydroxide ⟶ water + barium iodate + barium iodide

Construct the equilibrium constant, K, expression for: I_2 + Ba(OH)_2 ⟶ H_2O + BaI_2O_6 + BaI_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: 6 I_2 + 6 Ba(OH)_2 ⟶ 6 H_2O + BaI_2O_6 + 5 BaI_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 I_2 | 6 | -6 Ba(OH)_2 | 6 | -6 H_2O | 6 | 6 BaI_2O_6 | 1 | 1 BaI_2 | 5 | 5 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression I_2 | 6 | -6 | ([I2])^(-6) Ba(OH)_2 | 6 | -6 | ([Ba(OH)2])^(-6) H_2O | 6 | 6 | ([H2O])^6 BaI_2O_6 | 1 | 1 | [BaI2O6] BaI_2 | 5 | 5 | ([BaI2])^5 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 = ([I2])^(-6) ([Ba(OH)2])^(-6) ([H2O])^6 [BaI2O6] ([BaI2])^5 = (([H2O])^6 [BaI2O6] ([BaI2])^5)/(([I2])^6 ([Ba(OH)2])^6)

Construct the rate of reaction expression for: I_2 + Ba(OH)_2 ⟶ H_2O + BaI_2O_6 + BaI_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: 6 I_2 + 6 Ba(OH)_2 ⟶ 6 H_2O + BaI_2O_6 + 5 BaI_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 I_2 | 6 | -6 Ba(OH)_2 | 6 | -6 H_2O | 6 | 6 BaI_2O_6 | 1 | 1 BaI_2 | 5 | 5 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 I_2 | 6 | -6 | -1/6 (Δ[I2])/(Δt) Ba(OH)_2 | 6 | -6 | -1/6 (Δ[Ba(OH)2])/(Δt) H_2O | 6 | 6 | 1/6 (Δ[H2O])/(Δt) BaI_2O_6 | 1 | 1 | (Δ[BaI2O6])/(Δt) BaI_2 | 5 | 5 | 1/5 (Δ[BaI2])/(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[I2])/(Δt) = -1/6 (Δ[Ba(OH)2])/(Δt) = 1/6 (Δ[H2O])/(Δt) = (Δ[BaI2O6])/(Δt) = 1/5 (Δ[BaI2])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| iodine | barium hydroxide | water | barium iodate | barium iodide formula | I_2 | Ba(OH)_2 | H_2O | BaI_2O_6 | BaI_2 Hill formula | I_2 | BaH_2O_2 | H_2O | BaI_2O_6 | BaI_2 name | iodine | barium hydroxide | water | barium iodate | barium iodide IUPAC name | molecular iodine | barium(+2) cation dihydroxide | water | barium(+2) cation diiodate | barium(+2) cation diiodide