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H2O + HBrO3 + B = HBr + H3BO3

Input interpretation

H_2O water + HO_3Br bromic acid + B boron ⟶ HBr hydrogen bromide + B(OH)_3 boric acid
H_2O water + HO_3Br bromic acid + B boron ⟶ HBr hydrogen bromide + B(OH)_3 boric acid

Balanced equation

Balance the chemical equation algebraically: H_2O + HO_3Br + B ⟶ HBr + B(OH)_3 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 H_2O + c_2 HO_3Br + c_3 B ⟶ c_4 HBr + c_5 B(OH)_3 Set the number of atoms in the reactants equal to the number of atoms in the products for H, O, Br and B: H: | 2 c_1 + c_2 = c_4 + 3 c_5 O: | c_1 + 3 c_2 = 3 c_5 Br: | c_2 = c_4 B: | c_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_2 = 1 and solve the system of equations for the remaining coefficients: c_1 = 3 c_2 = 1 c_3 = 2 c_4 = 1 c_5 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 3 H_2O + HO_3Br + 2 B ⟶ HBr + 2 B(OH)_3
Balance the chemical equation algebraically: H_2O + HO_3Br + B ⟶ HBr + B(OH)_3 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 H_2O + c_2 HO_3Br + c_3 B ⟶ c_4 HBr + c_5 B(OH)_3 Set the number of atoms in the reactants equal to the number of atoms in the products for H, O, Br and B: H: | 2 c_1 + c_2 = c_4 + 3 c_5 O: | c_1 + 3 c_2 = 3 c_5 Br: | c_2 = c_4 B: | c_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_2 = 1 and solve the system of equations for the remaining coefficients: c_1 = 3 c_2 = 1 c_3 = 2 c_4 = 1 c_5 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 3 H_2O + HO_3Br + 2 B ⟶ HBr + 2 B(OH)_3

Structures

+ + ⟶ +
+ + ⟶ +

Names

water + bromic acid + boron ⟶ hydrogen bromide + boric acid
water + bromic acid + boron ⟶ hydrogen bromide + boric acid

Equilibrium constant

Construct the equilibrium constant, K, expression for: H_2O + HO_3Br + B ⟶ HBr + B(OH)_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: 3 H_2O + HO_3Br + 2 B ⟶ HBr + 2 B(OH)_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 H_2O | 3 | -3 HO_3Br | 1 | -1 B | 2 | -2 HBr | 1 | 1 B(OH)_3 | 2 | 2 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression H_2O | 3 | -3 | ([H2O])^(-3) HO_3Br | 1 | -1 | ([H1O3Br1])^(-1) B | 2 | -2 | ([B])^(-2) HBr | 1 | 1 | [HBr] B(OH)_3 | 2 | 2 | ([B(OH)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 = ([H2O])^(-3) ([H1O3Br1])^(-1) ([B])^(-2) [HBr] ([B(OH)3])^2 = ([HBr] ([B(OH)3])^2)/(([H2O])^3 [H1O3Br1] ([B])^2)
Construct the equilibrium constant, K, expression for: H_2O + HO_3Br + B ⟶ HBr + B(OH)_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: 3 H_2O + HO_3Br + 2 B ⟶ HBr + 2 B(OH)_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 H_2O | 3 | -3 HO_3Br | 1 | -1 B | 2 | -2 HBr | 1 | 1 B(OH)_3 | 2 | 2 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression H_2O | 3 | -3 | ([H2O])^(-3) HO_3Br | 1 | -1 | ([H1O3Br1])^(-1) B | 2 | -2 | ([B])^(-2) HBr | 1 | 1 | [HBr] B(OH)_3 | 2 | 2 | ([B(OH)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 = ([H2O])^(-3) ([H1O3Br1])^(-1) ([B])^(-2) [HBr] ([B(OH)3])^2 = ([HBr] ([B(OH)3])^2)/(([H2O])^3 [H1O3Br1] ([B])^2)

Rate of reaction

Construct the rate of reaction expression for: H_2O + HO_3Br + B ⟶ HBr + B(OH)_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: 3 H_2O + HO_3Br + 2 B ⟶ HBr + 2 B(OH)_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 H_2O | 3 | -3 HO_3Br | 1 | -1 B | 2 | -2 HBr | 1 | 1 B(OH)_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 H_2O | 3 | -3 | -1/3 (Δ[H2O])/(Δt) HO_3Br | 1 | -1 | -(Δ[H1O3Br1])/(Δt) B | 2 | -2 | -1/2 (Δ[B])/(Δt) HBr | 1 | 1 | (Δ[HBr])/(Δt) B(OH)_3 | 2 | 2 | 1/2 (Δ[B(OH)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 = -1/3 (Δ[H2O])/(Δt) = -(Δ[H1O3Br1])/(Δt) = -1/2 (Δ[B])/(Δt) = (Δ[HBr])/(Δt) = 1/2 (Δ[B(OH)3])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)
Construct the rate of reaction expression for: H_2O + HO_3Br + B ⟶ HBr + B(OH)_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: 3 H_2O + HO_3Br + 2 B ⟶ HBr + 2 B(OH)_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 H_2O | 3 | -3 HO_3Br | 1 | -1 B | 2 | -2 HBr | 1 | 1 B(OH)_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 H_2O | 3 | -3 | -1/3 (Δ[H2O])/(Δt) HO_3Br | 1 | -1 | -(Δ[H1O3Br1])/(Δt) B | 2 | -2 | -1/2 (Δ[B])/(Δt) HBr | 1 | 1 | (Δ[HBr])/(Δt) B(OH)_3 | 2 | 2 | 1/2 (Δ[B(OH)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 = -1/3 (Δ[H2O])/(Δt) = -(Δ[H1O3Br1])/(Δt) = -1/2 (Δ[B])/(Δt) = (Δ[HBr])/(Δt) = 1/2 (Δ[B(OH)3])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

Chemical names and formulas

| water | bromic acid | boron | hydrogen bromide | boric acid formula | H_2O | HO_3Br | B | HBr | B(OH)_3 Hill formula | H_2O | BrHO_3 | B | BrH | BH_3O_3 name | water | bromic acid | boron | hydrogen bromide | boric acid
| water | bromic acid | boron | hydrogen bromide | boric acid formula | H_2O | HO_3Br | B | HBr | B(OH)_3 Hill formula | H_2O | BrHO_3 | B | BrH | BH_3O_3 name | water | bromic acid | boron | hydrogen bromide | boric acid

Substance properties

| water | bromic acid | boron | hydrogen bromide | boric acid molar mass | 18.015 g/mol | 128.91 g/mol | 10.81 g/mol | 80.912 g/mol | 61.83 g/mol phase | liquid (at STP) | | solid (at STP) | gas (at STP) | solid (at STP) melting point | 0 °C | | 2075 °C | -86.8 °C | 160 °C boiling point | 99.9839 °C | | 4000 °C | -66.38 °C | density | 1 g/cm^3 | | 2.34 g/cm^3 | 0.003307 g/cm^3 (at 25 °C) | solubility in water | | | insoluble | miscible | surface tension | 0.0728 N/m | | | 0.0271 N/m | dynamic viscosity | 8.9×10^-4 Pa s (at 25 °C) | | | 8.4×10^-4 Pa s (at -75 °C) | odor | odorless | | | | odorless
| water | bromic acid | boron | hydrogen bromide | boric acid molar mass | 18.015 g/mol | 128.91 g/mol | 10.81 g/mol | 80.912 g/mol | 61.83 g/mol phase | liquid (at STP) | | solid (at STP) | gas (at STP) | solid (at STP) melting point | 0 °C | | 2075 °C | -86.8 °C | 160 °C boiling point | 99.9839 °C | | 4000 °C | -66.38 °C | density | 1 g/cm^3 | | 2.34 g/cm^3 | 0.003307 g/cm^3 (at 25 °C) | solubility in water | | | insoluble | miscible | surface tension | 0.0728 N/m | | | 0.0271 N/m | dynamic viscosity | 8.9×10^-4 Pa s (at 25 °C) | | | 8.4×10^-4 Pa s (at -75 °C) | odor | odorless | | | | odorless

Units

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