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I2 + Ba(OH)2 = H2O + BaI2 + Ba(IO)3

Input interpretation

I_2 iodine + Ba(OH)_2 barium hydroxide ⟶ H_2O water + BaI_2 barium iodide + Ba(IO)3
I_2 iodine + Ba(OH)_2 barium hydroxide ⟶ H_2O water + BaI_2 barium iodide + Ba(IO)3

Balanced equation

Balance the chemical equation algebraically: I_2 + Ba(OH)_2 ⟶ H_2O + BaI_2 + Ba(IO)3 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_2 + c_5 Ba(IO)3 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 + 3 c_5 Ba: | c_2 = c_4 + c_5 H: | 2 c_2 = 2 c_3 O: | 2 c_2 = c_3 + 3 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 = 7/2 c_2 = 3 c_3 = 3 c_4 = 2 c_5 = 1 Multiply by the least common denominator, 2, to eliminate fractional coefficients: c_1 = 7 c_2 = 6 c_3 = 6 c_4 = 4 c_5 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 7 I_2 + 6 Ba(OH)_2 ⟶ 6 H_2O + 4 BaI_2 + 2 Ba(IO)3
Balance the chemical equation algebraically: I_2 + Ba(OH)_2 ⟶ H_2O + BaI_2 + Ba(IO)3 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_2 + c_5 Ba(IO)3 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 + 3 c_5 Ba: | c_2 = c_4 + c_5 H: | 2 c_2 = 2 c_3 O: | 2 c_2 = c_3 + 3 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 = 7/2 c_2 = 3 c_3 = 3 c_4 = 2 c_5 = 1 Multiply by the least common denominator, 2, to eliminate fractional coefficients: c_1 = 7 c_2 = 6 c_3 = 6 c_4 = 4 c_5 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 7 I_2 + 6 Ba(OH)_2 ⟶ 6 H_2O + 4 BaI_2 + 2 Ba(IO)3

Structures

+ ⟶ + + Ba(IO)3
+ ⟶ + + Ba(IO)3

Names

iodine + barium hydroxide ⟶ water + barium iodide + Ba(IO)3
iodine + barium hydroxide ⟶ water + barium iodide + Ba(IO)3

Equilibrium constant

Construct the equilibrium constant, K, expression for: I_2 + Ba(OH)_2 ⟶ H_2O + BaI_2 + Ba(IO)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: 7 I_2 + 6 Ba(OH)_2 ⟶ 6 H_2O + 4 BaI_2 + 2 Ba(IO)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 I_2 | 7 | -7 Ba(OH)_2 | 6 | -6 H_2O | 6 | 6 BaI_2 | 4 | 4 Ba(IO)3 | 2 | 2 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression I_2 | 7 | -7 | ([I2])^(-7) Ba(OH)_2 | 6 | -6 | ([Ba(OH)2])^(-6) H_2O | 6 | 6 | ([H2O])^6 BaI_2 | 4 | 4 | ([BaI2])^4 Ba(IO)3 | 2 | 2 | ([Ba(IO)3])^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 = ([I2])^(-7) ([Ba(OH)2])^(-6) ([H2O])^6 ([BaI2])^4 ([Ba(IO)3])^2 = (([H2O])^6 ([BaI2])^4 ([Ba(IO)3])^2)/(([I2])^7 ([Ba(OH)2])^6)
Construct the equilibrium constant, K, expression for: I_2 + Ba(OH)_2 ⟶ H_2O + BaI_2 + Ba(IO)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: 7 I_2 + 6 Ba(OH)_2 ⟶ 6 H_2O + 4 BaI_2 + 2 Ba(IO)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 I_2 | 7 | -7 Ba(OH)_2 | 6 | -6 H_2O | 6 | 6 BaI_2 | 4 | 4 Ba(IO)3 | 2 | 2 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression I_2 | 7 | -7 | ([I2])^(-7) Ba(OH)_2 | 6 | -6 | ([Ba(OH)2])^(-6) H_2O | 6 | 6 | ([H2O])^6 BaI_2 | 4 | 4 | ([BaI2])^4 Ba(IO)3 | 2 | 2 | ([Ba(IO)3])^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 = ([I2])^(-7) ([Ba(OH)2])^(-6) ([H2O])^6 ([BaI2])^4 ([Ba(IO)3])^2 = (([H2O])^6 ([BaI2])^4 ([Ba(IO)3])^2)/(([I2])^7 ([Ba(OH)2])^6)

Rate of reaction

Construct the rate of reaction expression for: I_2 + Ba(OH)_2 ⟶ H_2O + BaI_2 + Ba(IO)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: 7 I_2 + 6 Ba(OH)_2 ⟶ 6 H_2O + 4 BaI_2 + 2 Ba(IO)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 I_2 | 7 | -7 Ba(OH)_2 | 6 | -6 H_2O | 6 | 6 BaI_2 | 4 | 4 Ba(IO)3 | 2 | 2 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 | 7 | -7 | -1/7 (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_2 | 4 | 4 | 1/4 (Δ[BaI2])/(Δt) Ba(IO)3 | 2 | 2 | 1/2 (Δ[Ba(IO)3])/(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 = -1/7 (Δ[I2])/(Δt) = -1/6 (Δ[Ba(OH)2])/(Δt) = 1/6 (Δ[H2O])/(Δt) = 1/4 (Δ[BaI2])/(Δt) = 1/2 (Δ[Ba(IO)3])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)
Construct the rate of reaction expression for: I_2 + Ba(OH)_2 ⟶ H_2O + BaI_2 + Ba(IO)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: 7 I_2 + 6 Ba(OH)_2 ⟶ 6 H_2O + 4 BaI_2 + 2 Ba(IO)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 I_2 | 7 | -7 Ba(OH)_2 | 6 | -6 H_2O | 6 | 6 BaI_2 | 4 | 4 Ba(IO)3 | 2 | 2 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 | 7 | -7 | -1/7 (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_2 | 4 | 4 | 1/4 (Δ[BaI2])/(Δt) Ba(IO)3 | 2 | 2 | 1/2 (Δ[Ba(IO)3])/(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 = -1/7 (Δ[I2])/(Δt) = -1/6 (Δ[Ba(OH)2])/(Δt) = 1/6 (Δ[H2O])/(Δt) = 1/4 (Δ[BaI2])/(Δt) = 1/2 (Δ[Ba(IO)3])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

Chemical names and formulas

| iodine | barium hydroxide | water | barium iodide | Ba(IO)3 formula | I_2 | Ba(OH)_2 | H_2O | BaI_2 | Ba(IO)3 Hill formula | I_2 | BaH_2O_2 | H_2O | BaI_2 | BaI3O3 name | iodine | barium hydroxide | water | barium iodide | IUPAC name | molecular iodine | barium(+2) cation dihydroxide | water | barium(+2) cation diiodide |
| iodine | barium hydroxide | water | barium iodide | Ba(IO)3 formula | I_2 | Ba(OH)_2 | H_2O | BaI_2 | Ba(IO)3 Hill formula | I_2 | BaH_2O_2 | H_2O | BaI_2 | BaI3O3 name | iodine | barium hydroxide | water | barium iodide | IUPAC name | molecular iodine | barium(+2) cation dihydroxide | water | barium(+2) cation diiodide |

Substance properties

| iodine | barium hydroxide | water | barium iodide | Ba(IO)3 molar mass | 253.80894 g/mol | 171.34 g/mol | 18.015 g/mol | 391.136 g/mol | 566.037 g/mol phase | solid (at STP) | solid (at STP) | liquid (at STP) | solid (at STP) | melting point | 113 °C | 300 °C | 0 °C | 740 °C | boiling point | 184 °C | | 99.9839 °C | | density | 4.94 g/cm^3 | 2.2 g/cm^3 | 1 g/cm^3 | 5.15 g/cm^3 | surface tension | | | 0.0728 N/m | | dynamic viscosity | 0.00227 Pa s (at 116 °C) | | 8.9×10^-4 Pa s (at 25 °C) | | odor | | | odorless | |
| iodine | barium hydroxide | water | barium iodide | Ba(IO)3 molar mass | 253.80894 g/mol | 171.34 g/mol | 18.015 g/mol | 391.136 g/mol | 566.037 g/mol phase | solid (at STP) | solid (at STP) | liquid (at STP) | solid (at STP) | melting point | 113 °C | 300 °C | 0 °C | 740 °C | boiling point | 184 °C | | 99.9839 °C | | density | 4.94 g/cm^3 | 2.2 g/cm^3 | 1 g/cm^3 | 5.15 g/cm^3 | surface tension | | | 0.0728 N/m | | dynamic viscosity | 0.00227 Pa s (at 116 °C) | | 8.9×10^-4 Pa s (at 25 °C) | | odor | | | odorless | |

Units

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