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HBr + KBrO = H2O + Br2 + KBr

Input interpretation

HBr hydrogen bromide + KBrO ⟶ H_2O water + Br_2 bromine + KBr potassium bromide
HBr hydrogen bromide + KBrO ⟶ H_2O water + Br_2 bromine + KBr potassium bromide

Balanced equation

Balance the chemical equation algebraically: HBr + KBrO ⟶ H_2O + Br_2 + KBr Add stoichiometric coefficients, c_i, to the reactants and products: c_1 HBr + c_2 KBrO ⟶ c_3 H_2O + c_4 Br_2 + c_5 KBr Set the number of atoms in the reactants equal to the number of atoms in the products for Br, H, K and O: Br: | c_1 + c_2 = 2 c_4 + c_5 H: | c_1 = 2 c_3 K: | c_2 = c_5 O: | c_2 = c_3 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 = 2 c_2 = 1 c_3 = 1 c_4 = 1 c_5 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 2 HBr + KBrO ⟶ H_2O + Br_2 + KBr
Balance the chemical equation algebraically: HBr + KBrO ⟶ H_2O + Br_2 + KBr Add stoichiometric coefficients, c_i, to the reactants and products: c_1 HBr + c_2 KBrO ⟶ c_3 H_2O + c_4 Br_2 + c_5 KBr Set the number of atoms in the reactants equal to the number of atoms in the products for Br, H, K and O: Br: | c_1 + c_2 = 2 c_4 + c_5 H: | c_1 = 2 c_3 K: | c_2 = c_5 O: | c_2 = c_3 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 = 2 c_2 = 1 c_3 = 1 c_4 = 1 c_5 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 2 HBr + KBrO ⟶ H_2O + Br_2 + KBr

Structures

+ KBrO ⟶ + +
+ KBrO ⟶ + +

Names

hydrogen bromide + KBrO ⟶ water + bromine + potassium bromide
hydrogen bromide + KBrO ⟶ water + bromine + potassium bromide

Equilibrium constant

Construct the equilibrium constant, K, expression for: HBr + KBrO ⟶ H_2O + Br_2 + KBr 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 HBr + KBrO ⟶ H_2O + Br_2 + KBr 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 HBr | 2 | -2 KBrO | 1 | -1 H_2O | 1 | 1 Br_2 | 1 | 1 KBr | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression HBr | 2 | -2 | ([HBr])^(-2) KBrO | 1 | -1 | ([KBrO])^(-1) H_2O | 1 | 1 | [H2O] Br_2 | 1 | 1 | [Br2] KBr | 1 | 1 | [KBr] 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 = ([HBr])^(-2) ([KBrO])^(-1) [H2O] [Br2] [KBr] = ([H2O] [Br2] [KBr])/(([HBr])^2 [KBrO])
Construct the equilibrium constant, K, expression for: HBr + KBrO ⟶ H_2O + Br_2 + KBr 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 HBr + KBrO ⟶ H_2O + Br_2 + KBr 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 HBr | 2 | -2 KBrO | 1 | -1 H_2O | 1 | 1 Br_2 | 1 | 1 KBr | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression HBr | 2 | -2 | ([HBr])^(-2) KBrO | 1 | -1 | ([KBrO])^(-1) H_2O | 1 | 1 | [H2O] Br_2 | 1 | 1 | [Br2] KBr | 1 | 1 | [KBr] 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 = ([HBr])^(-2) ([KBrO])^(-1) [H2O] [Br2] [KBr] = ([H2O] [Br2] [KBr])/(([HBr])^2 [KBrO])

Rate of reaction

Construct the rate of reaction expression for: HBr + KBrO ⟶ H_2O + Br_2 + KBr 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 HBr + KBrO ⟶ H_2O + Br_2 + KBr 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 HBr | 2 | -2 KBrO | 1 | -1 H_2O | 1 | 1 Br_2 | 1 | 1 KBr | 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 HBr | 2 | -2 | -1/2 (Δ[HBr])/(Δt) KBrO | 1 | -1 | -(Δ[KBrO])/(Δt) H_2O | 1 | 1 | (Δ[H2O])/(Δt) Br_2 | 1 | 1 | (Δ[Br2])/(Δt) KBr | 1 | 1 | (Δ[KBr])/(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[HBr])/(Δt) = -(Δ[KBrO])/(Δt) = (Δ[H2O])/(Δt) = (Δ[Br2])/(Δt) = (Δ[KBr])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)
Construct the rate of reaction expression for: HBr + KBrO ⟶ H_2O + Br_2 + KBr 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 HBr + KBrO ⟶ H_2O + Br_2 + KBr 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 HBr | 2 | -2 KBrO | 1 | -1 H_2O | 1 | 1 Br_2 | 1 | 1 KBr | 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 HBr | 2 | -2 | -1/2 (Δ[HBr])/(Δt) KBrO | 1 | -1 | -(Δ[KBrO])/(Δt) H_2O | 1 | 1 | (Δ[H2O])/(Δt) Br_2 | 1 | 1 | (Δ[Br2])/(Δt) KBr | 1 | 1 | (Δ[KBr])/(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[HBr])/(Δt) = -(Δ[KBrO])/(Δt) = (Δ[H2O])/(Δt) = (Δ[Br2])/(Δt) = (Δ[KBr])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

Chemical names and formulas

| hydrogen bromide | KBrO | water | bromine | potassium bromide formula | HBr | KBrO | H_2O | Br_2 | KBr Hill formula | BrH | BrKO | H_2O | Br_2 | BrK name | hydrogen bromide | | water | bromine | potassium bromide IUPAC name | hydrogen bromide | | water | molecular bromine | potassium bromide
| hydrogen bromide | KBrO | water | bromine | potassium bromide formula | HBr | KBrO | H_2O | Br_2 | KBr Hill formula | BrH | BrKO | H_2O | Br_2 | BrK name | hydrogen bromide | | water | bromine | potassium bromide IUPAC name | hydrogen bromide | | water | molecular bromine | potassium bromide

Substance properties

| hydrogen bromide | KBrO | water | bromine | potassium bromide molar mass | 80.912 g/mol | 135 g/mol | 18.015 g/mol | 159.81 g/mol | 119 g/mol phase | gas (at STP) | | liquid (at STP) | liquid (at STP) | solid (at STP) melting point | -86.8 °C | | 0 °C | -7.2 °C | 734 °C boiling point | -66.38 °C | | 99.9839 °C | 58.8 °C | 1435 °C density | 0.003307 g/cm^3 (at 25 °C) | | 1 g/cm^3 | 3.119 g/cm^3 | 2.75 g/cm^3 solubility in water | miscible | | | insoluble | soluble surface tension | 0.0271 N/m | | 0.0728 N/m | 0.0409 N/m | dynamic viscosity | 8.4×10^-4 Pa s (at -75 °C) | | 8.9×10^-4 Pa s (at 25 °C) | 9.44×10^-4 Pa s (at 25 °C) | odor | | | odorless | |
| hydrogen bromide | KBrO | water | bromine | potassium bromide molar mass | 80.912 g/mol | 135 g/mol | 18.015 g/mol | 159.81 g/mol | 119 g/mol phase | gas (at STP) | | liquid (at STP) | liquid (at STP) | solid (at STP) melting point | -86.8 °C | | 0 °C | -7.2 °C | 734 °C boiling point | -66.38 °C | | 99.9839 °C | 58.8 °C | 1435 °C density | 0.003307 g/cm^3 (at 25 °C) | | 1 g/cm^3 | 3.119 g/cm^3 | 2.75 g/cm^3 solubility in water | miscible | | | insoluble | soluble surface tension | 0.0271 N/m | | 0.0728 N/m | 0.0409 N/m | dynamic viscosity | 8.4×10^-4 Pa s (at -75 °C) | | 8.9×10^-4 Pa s (at 25 °C) | 9.44×10^-4 Pa s (at 25 °C) | odor | | | odorless | |

Units

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