Ba(IO3)2 = O2 + I2 + Ba5(IO6)2

BaI_2O_6 barium iodate ⟶ O_2 oxygen + I_2 iodine + Ba5(IO6)2

Balance the chemical equation algebraically: BaI_2O_6 ⟶ O_2 + I_2 + Ba5(IO6)2 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 BaI_2O_6 ⟶ c_2 O_2 + c_3 I_2 + c_4 Ba5(IO6)2 Set the number of atoms in the reactants equal to the number of atoms in the products for Ba, I and O: Ba: | c_1 = 5 c_4 I: | 2 c_1 = 2 c_3 + 2 c_4 O: | 6 c_1 = 2 c_2 + 12 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 = 5 c_2 = 9 c_3 = 4 c_4 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 5 BaI_2O_6 ⟶ 9 O_2 + 4 I_2 + Ba5(IO6)2

⟶ + + Ba5(IO6)2

barium iodate ⟶ oxygen + iodine + Ba5(IO6)2

Construct the equilibrium constant, K, expression for: BaI_2O_6 ⟶ O_2 + I_2 + Ba5(IO6)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 BaI_2O_6 ⟶ 9 O_2 + 4 I_2 + Ba5(IO6)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 BaI_2O_6 | 5 | -5 O_2 | 9 | 9 I_2 | 4 | 4 Ba5(IO6)2 | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression BaI_2O_6 | 5 | -5 | ([BaI2O6])^(-5) O_2 | 9 | 9 | ([O2])^9 I_2 | 4 | 4 | ([I2])^4 Ba5(IO6)2 | 1 | 1 | [Ba5(IO6)2] 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 = ([BaI2O6])^(-5) ([O2])^9 ([I2])^4 [Ba5(IO6)2] = (([O2])^9 ([I2])^4 [Ba5(IO6)2])/([BaI2O6])^5

Construct the rate of reaction expression for: BaI_2O_6 ⟶ O_2 + I_2 + Ba5(IO6)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 BaI_2O_6 ⟶ 9 O_2 + 4 I_2 + Ba5(IO6)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 BaI_2O_6 | 5 | -5 O_2 | 9 | 9 I_2 | 4 | 4 Ba5(IO6)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 BaI_2O_6 | 5 | -5 | -1/5 (Δ[BaI2O6])/(Δt) O_2 | 9 | 9 | 1/9 (Δ[O2])/(Δt) I_2 | 4 | 4 | 1/4 (Δ[I2])/(Δt) Ba5(IO6)2 | 1 | 1 | (Δ[Ba5(IO6)2])/(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[BaI2O6])/(Δt) = 1/9 (Δ[O2])/(Δt) = 1/4 (Δ[I2])/(Δt) = (Δ[Ba5(IO6)2])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| barium iodate | oxygen | iodine | Ba5(IO6)2 formula | BaI_2O_6 | O_2 | I_2 | Ba5(IO6)2 Hill formula | BaI_2O_6 | O_2 | I_2 | Ba5I2O12 name | barium iodate | oxygen | iodine | IUPAC name | barium(+2) cation diiodate | molecular oxygen | molecular iodine |

| barium iodate | oxygen | iodine | Ba5(IO6)2 molar mass | 487.13 g/mol | 31.998 g/mol | 253.80894 g/mol | 1132.43 g/mol phase | | gas (at STP) | solid (at STP) | melting point | | -218 °C | 113 °C | boiling point | | -183 °C | 184 °C | density | 4.998 g/cm^3 | 0.001429 g/cm^3 (at 0 °C) | 4.94 g/cm^3 | surface tension | | 0.01347 N/m | | dynamic viscosity | | 2.055×10^-5 Pa s (at 25 °C) | 0.00227 Pa s (at 116 °C) | odor | | odorless | |