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H2O2 + HBr = H2O + Br2

Input interpretation

H_2O_2 hydrogen peroxide + HBr hydrogen bromide ⟶ H_2O water + Br_2 bromine
H_2O_2 hydrogen peroxide + HBr hydrogen bromide ⟶ H_2O water + Br_2 bromine

Balanced equation

Balance the chemical equation algebraically: H_2O_2 + HBr ⟶ H_2O + Br_2 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 H_2O_2 + c_2 HBr ⟶ c_3 H_2O + c_4 Br_2 Set the number of atoms in the reactants equal to the number of atoms in the products for H, O and Br: H: | 2 c_1 + c_2 = 2 c_3 O: | 2 c_1 = c_3 Br: | c_2 = 2 c_4 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_1 = 1 and solve the system of equations for the remaining coefficients: c_1 = 1 c_2 = 2 c_3 = 2 c_4 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: |   | H_2O_2 + 2 HBr ⟶ 2 H_2O + Br_2
Balance the chemical equation algebraically: H_2O_2 + HBr ⟶ H_2O + Br_2 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 H_2O_2 + c_2 HBr ⟶ c_3 H_2O + c_4 Br_2 Set the number of atoms in the reactants equal to the number of atoms in the products for H, O and Br: H: | 2 c_1 + c_2 = 2 c_3 O: | 2 c_1 = c_3 Br: | c_2 = 2 c_4 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_1 = 1 and solve the system of equations for the remaining coefficients: c_1 = 1 c_2 = 2 c_3 = 2 c_4 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | H_2O_2 + 2 HBr ⟶ 2 H_2O + Br_2

Structures

 + ⟶ +
+ ⟶ +

Names

hydrogen peroxide + hydrogen bromide ⟶ water + bromine
hydrogen peroxide + hydrogen bromide ⟶ water + bromine

Reaction thermodynamics

Gibbs free energy

 | hydrogen peroxide | hydrogen bromide | water | bromine molecular free energy | -120.4 kJ/mol | -53.4 kJ/mol | -237.1 kJ/mol | 0 kJ/mol total free energy | -120.4 kJ/mol | -106.8 kJ/mol | -474.2 kJ/mol | 0 kJ/mol  | G_initial = -227.2 kJ/mol | | G_final = -474.2 kJ/mol |  ΔG_rxn^0 | -474.2 kJ/mol - -227.2 kJ/mol = -247 kJ/mol (exergonic) | | |
| hydrogen peroxide | hydrogen bromide | water | bromine molecular free energy | -120.4 kJ/mol | -53.4 kJ/mol | -237.1 kJ/mol | 0 kJ/mol total free energy | -120.4 kJ/mol | -106.8 kJ/mol | -474.2 kJ/mol | 0 kJ/mol | G_initial = -227.2 kJ/mol | | G_final = -474.2 kJ/mol | ΔG_rxn^0 | -474.2 kJ/mol - -227.2 kJ/mol = -247 kJ/mol (exergonic) | | |

Equilibrium constant

Construct the equilibrium constant, K, expression for: H_2O_2 + HBr ⟶ H_2O + Br_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: H_2O_2 + 2 HBr ⟶ 2 H_2O + Br_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 H_2O_2 | 1 | -1 HBr | 2 | -2 H_2O | 2 | 2 Br_2 | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression H_2O_2 | 1 | -1 | ([H2O2])^(-1) HBr | 2 | -2 | ([HBr])^(-2) H_2O | 2 | 2 | ([H2O])^2 Br_2 | 1 | 1 | [Br2] 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 = ([H2O2])^(-1) ([HBr])^(-2) ([H2O])^2 [Br2] = (([H2O])^2 [Br2])/([H2O2] ([HBr])^2)
Construct the equilibrium constant, K, expression for: H_2O_2 + HBr ⟶ H_2O + Br_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: H_2O_2 + 2 HBr ⟶ 2 H_2O + Br_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 H_2O_2 | 1 | -1 HBr | 2 | -2 H_2O | 2 | 2 Br_2 | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression H_2O_2 | 1 | -1 | ([H2O2])^(-1) HBr | 2 | -2 | ([HBr])^(-2) H_2O | 2 | 2 | ([H2O])^2 Br_2 | 1 | 1 | [Br2] 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 = ([H2O2])^(-1) ([HBr])^(-2) ([H2O])^2 [Br2] = (([H2O])^2 [Br2])/([H2O2] ([HBr])^2)

Rate of reaction

Construct the rate of reaction expression for: H_2O_2 + HBr ⟶ H_2O + Br_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: H_2O_2 + 2 HBr ⟶ 2 H_2O + Br_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 H_2O_2 | 1 | -1 HBr | 2 | -2 H_2O | 2 | 2 Br_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 H_2O_2 | 1 | -1 | -(Δ[H2O2])/(Δt) HBr | 2 | -2 | -1/2 (Δ[HBr])/(Δt) H_2O | 2 | 2 | 1/2 (Δ[H2O])/(Δt) Br_2 | 1 | 1 | (Δ[Br2])/(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 = -(Δ[H2O2])/(Δt) = -1/2 (Δ[HBr])/(Δt) = 1/2 (Δ[H2O])/(Δt) = (Δ[Br2])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)
Construct the rate of reaction expression for: H_2O_2 + HBr ⟶ H_2O + Br_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: H_2O_2 + 2 HBr ⟶ 2 H_2O + Br_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 H_2O_2 | 1 | -1 HBr | 2 | -2 H_2O | 2 | 2 Br_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 H_2O_2 | 1 | -1 | -(Δ[H2O2])/(Δt) HBr | 2 | -2 | -1/2 (Δ[HBr])/(Δt) H_2O | 2 | 2 | 1/2 (Δ[H2O])/(Δt) Br_2 | 1 | 1 | (Δ[Br2])/(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 = -(Δ[H2O2])/(Δt) = -1/2 (Δ[HBr])/(Δt) = 1/2 (Δ[H2O])/(Δt) = (Δ[Br2])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

Chemical names and formulas

 | hydrogen peroxide | hydrogen bromide | water | bromine formula | H_2O_2 | HBr | H_2O | Br_2 Hill formula | H_2O_2 | BrH | H_2O | Br_2 name | hydrogen peroxide | hydrogen bromide | water | bromine IUPAC name | hydrogen peroxide | hydrogen bromide | water | molecular bromine
| hydrogen peroxide | hydrogen bromide | water | bromine formula | H_2O_2 | HBr | H_2O | Br_2 Hill formula | H_2O_2 | BrH | H_2O | Br_2 name | hydrogen peroxide | hydrogen bromide | water | bromine IUPAC name | hydrogen peroxide | hydrogen bromide | water | molecular bromine

Substance properties

 | hydrogen peroxide | hydrogen bromide | water | bromine molar mass | 34.014 g/mol | 80.912 g/mol | 18.015 g/mol | 159.81 g/mol phase | liquid (at STP) | gas (at STP) | liquid (at STP) | liquid (at STP) melting point | -0.43 °C | -86.8 °C | 0 °C | -7.2 °C boiling point | 150.2 °C | -66.38 °C | 99.9839 °C | 58.8 °C density | 1.44 g/cm^3 | 0.003307 g/cm^3 (at 25 °C) | 1 g/cm^3 | 3.119 g/cm^3 solubility in water | miscible | miscible | | insoluble surface tension | 0.0804 N/m | 0.0271 N/m | 0.0728 N/m | 0.0409 N/m dynamic viscosity | 0.001249 Pa s (at 20 °C) | 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 peroxide | hydrogen bromide | water | bromine molar mass | 34.014 g/mol | 80.912 g/mol | 18.015 g/mol | 159.81 g/mol phase | liquid (at STP) | gas (at STP) | liquid (at STP) | liquid (at STP) melting point | -0.43 °C | -86.8 °C | 0 °C | -7.2 °C boiling point | 150.2 °C | -66.38 °C | 99.9839 °C | 58.8 °C density | 1.44 g/cm^3 | 0.003307 g/cm^3 (at 25 °C) | 1 g/cm^3 | 3.119 g/cm^3 solubility in water | miscible | miscible | | insoluble surface tension | 0.0804 N/m | 0.0271 N/m | 0.0728 N/m | 0.0409 N/m dynamic viscosity | 0.001249 Pa s (at 20 °C) | 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