HBrO = HBr + HBrO3

HOBr hypobromous acid ⟶ HBr hydrogen bromide + HO_3Br bromic acid

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

⟶ +

hypobromous acid ⟶ hydrogen bromide + bromic acid

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

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

| hypobromous acid | hydrogen bromide | bromic acid formula | HOBr | HBr | HO_3Br Hill formula | BrHO | BrH | BrHO_3 name | hypobromous acid | hydrogen bromide | bromic acid

| hypobromous acid | hydrogen bromide | bromic acid molar mass | 96.91 g/mol | 80.912 g/mol | 128.91 g/mol phase | | gas (at STP) | melting point | | -86.8 °C | boiling point | | -66.38 °C | density | | 0.003307 g/cm^3 (at 25 °C) | solubility in water | | miscible | surface tension | | 0.0271 N/m | dynamic viscosity | | 8.4×10^-4 Pa s (at -75 °C) |