Br2 + H2O2 = HBr + HBrO2

Br_2 bromine + H_2O_2 hydrogen peroxide ⟶ HBr hydrogen bromide + HBrO2

Balance the chemical equation algebraically: Br_2 + H_2O_2 ⟶ HBr + HBrO2 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 Br_2 + c_2 H_2O_2 ⟶ c_3 HBr + c_4 HBrO2 Set the number of atoms in the reactants equal to the number of atoms in the products for Br, H and O: Br: | 2 c_1 = c_3 + c_4 H: | 2 c_2 = c_3 + c_4 O: | 2 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 = 1 c_3 = 1 c_4 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | Br_2 + H_2O_2 ⟶ HBr + HBrO2

+ ⟶ + HBrO2

bromine + hydrogen peroxide ⟶ hydrogen bromide + HBrO2

Construct the equilibrium constant, K, expression for: Br_2 + H_2O_2 ⟶ HBr + HBrO2 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: Br_2 + H_2O_2 ⟶ HBr + HBrO2 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 | 1 | -1 H_2O_2 | 1 | -1 HBr | 1 | 1 HBrO2 | 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 | 1 | -1 | ([Br2])^(-1) H_2O_2 | 1 | -1 | ([H2O2])^(-1) HBr | 1 | 1 | [HBr] HBrO2 | 1 | 1 | [HBrO2] 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])^(-1) ([H2O2])^(-1) [HBr] [HBrO2] = ([HBr] [HBrO2])/([Br2] [H2O2])

Construct the rate of reaction expression for: Br_2 + H_2O_2 ⟶ HBr + HBrO2 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: Br_2 + H_2O_2 ⟶ HBr + HBrO2 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 | 1 | -1 H_2O_2 | 1 | -1 HBr | 1 | 1 HBrO2 | 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 | 1 | -1 | -(Δ[Br2])/(Δt) H_2O_2 | 1 | -1 | -(Δ[H2O2])/(Δt) HBr | 1 | 1 | (Δ[HBr])/(Δt) HBrO2 | 1 | 1 | (Δ[HBrO2])/(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[Br2])/(Δt) = -(Δ[H2O2])/(Δt) = (Δ[HBr])/(Δt) = (Δ[HBrO2])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| bromine | hydrogen peroxide | hydrogen bromide | HBrO2 formula | Br_2 | H_2O_2 | HBr | HBrO2 Hill formula | Br_2 | H_2O_2 | BrH | HBrO2 name | bromine | hydrogen peroxide | hydrogen bromide | IUPAC name | molecular bromine | hydrogen peroxide | hydrogen bromide |

| bromine | hydrogen peroxide | hydrogen bromide | HBrO2 molar mass | 159.81 g/mol | 34.014 g/mol | 80.912 g/mol | 112.91 g/mol phase | liquid (at STP) | liquid (at STP) | gas (at STP) | melting point | -7.2 °C | -0.43 °C | -86.8 °C | boiling point | 58.8 °C | 150.2 °C | -66.38 °C | density | 3.119 g/cm^3 | 1.44 g/cm^3 | 0.003307 g/cm^3 (at 25 °C) | solubility in water | insoluble | miscible | miscible | surface tension | 0.0409 N/m | 0.0804 N/m | 0.0271 N/m | dynamic viscosity | 9.44×10^-4 Pa s (at 25 °C) | 0.001249 Pa s (at 20 °C) | 8.4×10^-4 Pa s (at -75 °C) |