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Ba(OH)2 + CIO2 = H2O + Ba(CIO2)2 + Ba(CIO3)2

Input interpretation

Ba(OH)_2 barium hydroxide + CIO2 ⟶ H_2O water + Ba(CIO2)2 + Ba(CIO3)2
Ba(OH)_2 barium hydroxide + CIO2 ⟶ H_2O water + Ba(CIO2)2 + Ba(CIO3)2

Balanced equation

Balance the chemical equation algebraically: Ba(OH)_2 + CIO2 ⟶ H_2O + Ba(CIO2)2 + Ba(CIO3)2 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 Ba(OH)_2 + c_2 CIO2 ⟶ c_3 H_2O + c_4 Ba(CIO2)2 + c_5 Ba(CIO3)2 Set the number of atoms in the reactants equal to the number of atoms in the products for Ba, H, O, C and I: Ba: | c_1 = c_4 + c_5 H: | 2 c_1 = 2 c_3 O: | 2 c_1 + 2 c_2 = c_3 + 4 c_4 + 6 c_5 C: | c_2 = 2 c_4 + 2 c_5 I: | c_2 = 2 c_4 + 2 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_4 = 1 and solve the system of equations for the remaining coefficients: c_1 = 2 c_2 = 4 c_3 = 2 c_4 = 1 c_5 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 2 Ba(OH)_2 + 4 CIO2 ⟶ 2 H_2O + Ba(CIO2)2 + Ba(CIO3)2
Balance the chemical equation algebraically: Ba(OH)_2 + CIO2 ⟶ H_2O + Ba(CIO2)2 + Ba(CIO3)2 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 Ba(OH)_2 + c_2 CIO2 ⟶ c_3 H_2O + c_4 Ba(CIO2)2 + c_5 Ba(CIO3)2 Set the number of atoms in the reactants equal to the number of atoms in the products for Ba, H, O, C and I: Ba: | c_1 = c_4 + c_5 H: | 2 c_1 = 2 c_3 O: | 2 c_1 + 2 c_2 = c_3 + 4 c_4 + 6 c_5 C: | c_2 = 2 c_4 + 2 c_5 I: | c_2 = 2 c_4 + 2 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_4 = 1 and solve the system of equations for the remaining coefficients: c_1 = 2 c_2 = 4 c_3 = 2 c_4 = 1 c_5 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 2 Ba(OH)_2 + 4 CIO2 ⟶ 2 H_2O + Ba(CIO2)2 + Ba(CIO3)2

Structures

+ CIO2 ⟶ + Ba(CIO2)2 + Ba(CIO3)2
+ CIO2 ⟶ + Ba(CIO2)2 + Ba(CIO3)2

Names

barium hydroxide + CIO2 ⟶ water + Ba(CIO2)2 + Ba(CIO3)2
barium hydroxide + CIO2 ⟶ water + Ba(CIO2)2 + Ba(CIO3)2

Equilibrium constant

Construct the equilibrium constant, K, expression for: Ba(OH)_2 + CIO2 ⟶ H_2O + Ba(CIO2)2 + Ba(CIO3)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: 2 Ba(OH)_2 + 4 CIO2 ⟶ 2 H_2O + Ba(CIO2)2 + Ba(CIO3)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 Ba(OH)_2 | 2 | -2 CIO2 | 4 | -4 H_2O | 2 | 2 Ba(CIO2)2 | 1 | 1 Ba(CIO3)2 | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression Ba(OH)_2 | 2 | -2 | ([Ba(OH)2])^(-2) CIO2 | 4 | -4 | ([CIO2])^(-4) H_2O | 2 | 2 | ([H2O])^2 Ba(CIO2)2 | 1 | 1 | [Ba(CIO2)2] Ba(CIO3)2 | 1 | 1 | [Ba(CIO3)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 = ([Ba(OH)2])^(-2) ([CIO2])^(-4) ([H2O])^2 [Ba(CIO2)2] [Ba(CIO3)2] = (([H2O])^2 [Ba(CIO2)2] [Ba(CIO3)2])/(([Ba(OH)2])^2 ([CIO2])^4)
Construct the equilibrium constant, K, expression for: Ba(OH)_2 + CIO2 ⟶ H_2O + Ba(CIO2)2 + Ba(CIO3)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: 2 Ba(OH)_2 + 4 CIO2 ⟶ 2 H_2O + Ba(CIO2)2 + Ba(CIO3)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 Ba(OH)_2 | 2 | -2 CIO2 | 4 | -4 H_2O | 2 | 2 Ba(CIO2)2 | 1 | 1 Ba(CIO3)2 | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression Ba(OH)_2 | 2 | -2 | ([Ba(OH)2])^(-2) CIO2 | 4 | -4 | ([CIO2])^(-4) H_2O | 2 | 2 | ([H2O])^2 Ba(CIO2)2 | 1 | 1 | [Ba(CIO2)2] Ba(CIO3)2 | 1 | 1 | [Ba(CIO3)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 = ([Ba(OH)2])^(-2) ([CIO2])^(-4) ([H2O])^2 [Ba(CIO2)2] [Ba(CIO3)2] = (([H2O])^2 [Ba(CIO2)2] [Ba(CIO3)2])/(([Ba(OH)2])^2 ([CIO2])^4)

Rate of reaction

Construct the rate of reaction expression for: Ba(OH)_2 + CIO2 ⟶ H_2O + Ba(CIO2)2 + Ba(CIO3)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: 2 Ba(OH)_2 + 4 CIO2 ⟶ 2 H_2O + Ba(CIO2)2 + Ba(CIO3)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 Ba(OH)_2 | 2 | -2 CIO2 | 4 | -4 H_2O | 2 | 2 Ba(CIO2)2 | 1 | 1 Ba(CIO3)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 Ba(OH)_2 | 2 | -2 | -1/2 (Δ[Ba(OH)2])/(Δt) CIO2 | 4 | -4 | -1/4 (Δ[CIO2])/(Δt) H_2O | 2 | 2 | 1/2 (Δ[H2O])/(Δt) Ba(CIO2)2 | 1 | 1 | (Δ[Ba(CIO2)2])/(Δt) Ba(CIO3)2 | 1 | 1 | (Δ[Ba(CIO3)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 = -1/2 (Δ[Ba(OH)2])/(Δt) = -1/4 (Δ[CIO2])/(Δt) = 1/2 (Δ[H2O])/(Δt) = (Δ[Ba(CIO2)2])/(Δt) = (Δ[Ba(CIO3)2])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)
Construct the rate of reaction expression for: Ba(OH)_2 + CIO2 ⟶ H_2O + Ba(CIO2)2 + Ba(CIO3)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: 2 Ba(OH)_2 + 4 CIO2 ⟶ 2 H_2O + Ba(CIO2)2 + Ba(CIO3)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 Ba(OH)_2 | 2 | -2 CIO2 | 4 | -4 H_2O | 2 | 2 Ba(CIO2)2 | 1 | 1 Ba(CIO3)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 Ba(OH)_2 | 2 | -2 | -1/2 (Δ[Ba(OH)2])/(Δt) CIO2 | 4 | -4 | -1/4 (Δ[CIO2])/(Δt) H_2O | 2 | 2 | 1/2 (Δ[H2O])/(Δt) Ba(CIO2)2 | 1 | 1 | (Δ[Ba(CIO2)2])/(Δt) Ba(CIO3)2 | 1 | 1 | (Δ[Ba(CIO3)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 = -1/2 (Δ[Ba(OH)2])/(Δt) = -1/4 (Δ[CIO2])/(Δt) = 1/2 (Δ[H2O])/(Δt) = (Δ[Ba(CIO2)2])/(Δt) = (Δ[Ba(CIO3)2])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

Chemical names and formulas

| barium hydroxide | CIO2 | water | Ba(CIO2)2 | Ba(CIO3)2 formula | Ba(OH)_2 | CIO2 | H_2O | Ba(CIO2)2 | Ba(CIO3)2 Hill formula | BaH_2O_2 | CIO2 | H_2O | C2BaI2O4 | C2BaI2O6 name | barium hydroxide | | water | | IUPAC name | barium(+2) cation dihydroxide | | water | |
| barium hydroxide | CIO2 | water | Ba(CIO2)2 | Ba(CIO3)2 formula | Ba(OH)_2 | CIO2 | H_2O | Ba(CIO2)2 | Ba(CIO3)2 Hill formula | BaH_2O_2 | CIO2 | H_2O | C2BaI2O4 | C2BaI2O6 name | barium hydroxide | | water | | IUPAC name | barium(+2) cation dihydroxide | | water | |

Substance properties

| barium hydroxide | CIO2 | water | Ba(CIO2)2 | Ba(CIO3)2 molar mass | 171.34 g/mol | 170.913 g/mol | 18.015 g/mol | 479.15 g/mol | 511.15 g/mol phase | solid (at STP) | | liquid (at STP) | | melting point | 300 °C | | 0 °C | | boiling point | | | 99.9839 °C | | density | 2.2 g/cm^3 | | 1 g/cm^3 | | surface tension | | | 0.0728 N/m | | dynamic viscosity | | | 8.9×10^-4 Pa s (at 25 °C) | | odor | | | odorless | |
| barium hydroxide | CIO2 | water | Ba(CIO2)2 | Ba(CIO3)2 molar mass | 171.34 g/mol | 170.913 g/mol | 18.015 g/mol | 479.15 g/mol | 511.15 g/mol phase | solid (at STP) | | liquid (at STP) | | melting point | 300 °C | | 0 °C | | boiling point | | | 99.9839 °C | | density | 2.2 g/cm^3 | | 1 g/cm^3 | | surface tension | | | 0.0728 N/m | | dynamic viscosity | | | 8.9×10^-4 Pa s (at 25 °C) | | odor | | | odorless | |

Units

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