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Br2 + Sr(OH)2 = H2O + SrBr2 + Sr(BrO3)2

Input interpretation

Br_2 bromine + Sr(OH)_2 strontium hydroxide ⟶ H_2O water + SrBr_2 strontium bromide + Sr(BrO3)2
Br_2 bromine + Sr(OH)_2 strontium hydroxide ⟶ H_2O water + SrBr_2 strontium bromide + Sr(BrO3)2

Balanced equation

Balance the chemical equation algebraically: Br_2 + Sr(OH)_2 ⟶ H_2O + SrBr_2 + Sr(BrO3)2 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 Br_2 + c_2 Sr(OH)_2 ⟶ c_3 H_2O + c_4 SrBr_2 + c_5 Sr(BrO3)2 Set the number of atoms in the reactants equal to the number of atoms in the products for Br, H, O and Sr: Br: | 2 c_1 = 2 c_4 + 2 c_5 H: | 2 c_2 = 2 c_3 O: | 2 c_2 = c_3 + 6 c_5 Sr: | 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 = 6 c_3 = 6 c_4 = 5 c_5 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 6 Br_2 + 6 Sr(OH)_2 ⟶ 6 H_2O + 5 SrBr_2 + Sr(BrO3)2
Balance the chemical equation algebraically: Br_2 + Sr(OH)_2 ⟶ H_2O + SrBr_2 + Sr(BrO3)2 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 Br_2 + c_2 Sr(OH)_2 ⟶ c_3 H_2O + c_4 SrBr_2 + c_5 Sr(BrO3)2 Set the number of atoms in the reactants equal to the number of atoms in the products for Br, H, O and Sr: Br: | 2 c_1 = 2 c_4 + 2 c_5 H: | 2 c_2 = 2 c_3 O: | 2 c_2 = c_3 + 6 c_5 Sr: | 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 = 6 c_3 = 6 c_4 = 5 c_5 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 6 Br_2 + 6 Sr(OH)_2 ⟶ 6 H_2O + 5 SrBr_2 + Sr(BrO3)2

Structures

+ ⟶ + + Sr(BrO3)2
+ ⟶ + + Sr(BrO3)2

Names

bromine + strontium hydroxide ⟶ water + strontium bromide + Sr(BrO3)2
bromine + strontium hydroxide ⟶ water + strontium bromide + Sr(BrO3)2

Equilibrium constant

Construct the equilibrium constant, K, expression for: Br_2 + Sr(OH)_2 ⟶ H_2O + SrBr_2 + Sr(BrO3)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 Br_2 + 6 Sr(OH)_2 ⟶ 6 H_2O + 5 SrBr_2 + Sr(BrO3)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 Br_2 | 6 | -6 Sr(OH)_2 | 6 | -6 H_2O | 6 | 6 SrBr_2 | 5 | 5 Sr(BrO3)2 | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression Br_2 | 6 | -6 | ([Br2])^(-6) Sr(OH)_2 | 6 | -6 | ([Sr(OH)2])^(-6) H_2O | 6 | 6 | ([H2O])^6 SrBr_2 | 5 | 5 | ([SrBr2])^5 Sr(BrO3)2 | 1 | 1 | [Sr(BrO3)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 = ([Br2])^(-6) ([Sr(OH)2])^(-6) ([H2O])^6 ([SrBr2])^5 [Sr(BrO3)2] = (([H2O])^6 ([SrBr2])^5 [Sr(BrO3)2])/(([Br2])^6 ([Sr(OH)2])^6)
Construct the equilibrium constant, K, expression for: Br_2 + Sr(OH)_2 ⟶ H_2O + SrBr_2 + Sr(BrO3)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 Br_2 + 6 Sr(OH)_2 ⟶ 6 H_2O + 5 SrBr_2 + Sr(BrO3)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 Br_2 | 6 | -6 Sr(OH)_2 | 6 | -6 H_2O | 6 | 6 SrBr_2 | 5 | 5 Sr(BrO3)2 | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression Br_2 | 6 | -6 | ([Br2])^(-6) Sr(OH)_2 | 6 | -6 | ([Sr(OH)2])^(-6) H_2O | 6 | 6 | ([H2O])^6 SrBr_2 | 5 | 5 | ([SrBr2])^5 Sr(BrO3)2 | 1 | 1 | [Sr(BrO3)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 = ([Br2])^(-6) ([Sr(OH)2])^(-6) ([H2O])^6 ([SrBr2])^5 [Sr(BrO3)2] = (([H2O])^6 ([SrBr2])^5 [Sr(BrO3)2])/(([Br2])^6 ([Sr(OH)2])^6)

Rate of reaction

Construct the rate of reaction expression for: Br_2 + Sr(OH)_2 ⟶ H_2O + SrBr_2 + Sr(BrO3)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 Br_2 + 6 Sr(OH)_2 ⟶ 6 H_2O + 5 SrBr_2 + Sr(BrO3)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 Br_2 | 6 | -6 Sr(OH)_2 | 6 | -6 H_2O | 6 | 6 SrBr_2 | 5 | 5 Sr(BrO3)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 Br_2 | 6 | -6 | -1/6 (Δ[Br2])/(Δt) Sr(OH)_2 | 6 | -6 | -1/6 (Δ[Sr(OH)2])/(Δt) H_2O | 6 | 6 | 1/6 (Δ[H2O])/(Δt) SrBr_2 | 5 | 5 | 1/5 (Δ[SrBr2])/(Δt) Sr(BrO3)2 | 1 | 1 | (Δ[Sr(BrO3)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 = -1/6 (Δ[Br2])/(Δt) = -1/6 (Δ[Sr(OH)2])/(Δt) = 1/6 (Δ[H2O])/(Δt) = 1/5 (Δ[SrBr2])/(Δt) = (Δ[Sr(BrO3)2])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)
Construct the rate of reaction expression for: Br_2 + Sr(OH)_2 ⟶ H_2O + SrBr_2 + Sr(BrO3)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 Br_2 + 6 Sr(OH)_2 ⟶ 6 H_2O + 5 SrBr_2 + Sr(BrO3)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 Br_2 | 6 | -6 Sr(OH)_2 | 6 | -6 H_2O | 6 | 6 SrBr_2 | 5 | 5 Sr(BrO3)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 Br_2 | 6 | -6 | -1/6 (Δ[Br2])/(Δt) Sr(OH)_2 | 6 | -6 | -1/6 (Δ[Sr(OH)2])/(Δt) H_2O | 6 | 6 | 1/6 (Δ[H2O])/(Δt) SrBr_2 | 5 | 5 | 1/5 (Δ[SrBr2])/(Δt) Sr(BrO3)2 | 1 | 1 | (Δ[Sr(BrO3)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 = -1/6 (Δ[Br2])/(Δt) = -1/6 (Δ[Sr(OH)2])/(Δt) = 1/6 (Δ[H2O])/(Δt) = 1/5 (Δ[SrBr2])/(Δt) = (Δ[Sr(BrO3)2])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

Chemical names and formulas

| bromine | strontium hydroxide | water | strontium bromide | Sr(BrO3)2 formula | Br_2 | Sr(OH)_2 | H_2O | SrBr_2 | Sr(BrO3)2 Hill formula | Br_2 | H_2O_2Sr | H_2O | Br_2Sr | Br2O6Sr name | bromine | strontium hydroxide | water | strontium bromide | IUPAC name | molecular bromine | strontium dihydroxide | water | strontium dibromide |
| bromine | strontium hydroxide | water | strontium bromide | Sr(BrO3)2 formula | Br_2 | Sr(OH)_2 | H_2O | SrBr_2 | Sr(BrO3)2 Hill formula | Br_2 | H_2O_2Sr | H_2O | Br_2Sr | Br2O6Sr name | bromine | strontium hydroxide | water | strontium bromide | IUPAC name | molecular bromine | strontium dihydroxide | water | strontium dibromide |

Substance properties

| bromine | strontium hydroxide | water | strontium bromide | Sr(BrO3)2 molar mass | 159.81 g/mol | 121.6 g/mol | 18.015 g/mol | 247.4 g/mol | 343.4 g/mol phase | liquid (at STP) | | liquid (at STP) | solid (at STP) | melting point | -7.2 °C | | 0 °C | 643 °C | boiling point | 58.8 °C | | 99.9839 °C | 2146 °C | density | 3.119 g/cm^3 | | 1 g/cm^3 | 4.175 g/cm^3 | solubility in water | insoluble | | | | surface tension | 0.0409 N/m | | 0.0728 N/m | | dynamic viscosity | 9.44×10^-4 Pa s (at 25 °C) | | 8.9×10^-4 Pa s (at 25 °C) | | odor | | | odorless | |
| bromine | strontium hydroxide | water | strontium bromide | Sr(BrO3)2 molar mass | 159.81 g/mol | 121.6 g/mol | 18.015 g/mol | 247.4 g/mol | 343.4 g/mol phase | liquid (at STP) | | liquid (at STP) | solid (at STP) | melting point | -7.2 °C | | 0 °C | 643 °C | boiling point | 58.8 °C | | 99.9839 °C | 2146 °C | density | 3.119 g/cm^3 | | 1 g/cm^3 | 4.175 g/cm^3 | solubility in water | insoluble | | | | surface tension | 0.0409 N/m | | 0.0728 N/m | | dynamic viscosity | 9.44×10^-4 Pa s (at 25 °C) | | 8.9×10^-4 Pa s (at 25 °C) | | odor | | | odorless | |

Units

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