H2O + Br2 + HClO = HCl + HBrO3

H_2O (water) + Br_2 (bromine) + HOCl (hypochlorous acid) ⟶ HCl (hydrogen chloride) + HO_3Br (bromic acid)

Balance the chemical equation algebraically: H_2O + Br_2 + HOCl ⟶ HCl + HO_3Br Add stoichiometric coefficients, c_i, to the reactants and products: c_1 H_2O + c_2 Br_2 + c_3 HOCl ⟶ c_4 HCl + c_5 HO_3Br Set the number of atoms in the reactants equal to the number of atoms in the products for H, O, Br and Cl: H: | 2 c_1 + c_3 = c_4 + c_5 O: | c_1 + c_3 = 3 c_5 Br: | 2 c_2 = c_5 Cl: | c_3 = 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 = 5 c_4 = 5 c_5 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | H_2O + Br_2 + 5 HOCl ⟶ 5 HCl + 2 HO_3Br

+ + ⟶ +

water + bromine + hypochlorous acid ⟶ hydrogen chloride + bromic acid

Construct the equilibrium constant, K, expression for: H_2O + Br_2 + HOCl ⟶ HCl + 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: H_2O + Br_2 + 5 HOCl ⟶ 5 HCl + 2 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 H_2O | 1 | -1 Br_2 | 1 | -1 HOCl | 5 | -5 HCl | 5 | 5 HO_3Br | 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 | 1 | -1 | ([H2O])^(-1) Br_2 | 1 | -1 | ([Br2])^(-1) HOCl | 5 | -5 | ([HOCl])^(-5) HCl | 5 | 5 | ([HCl])^5 HO_3Br | 2 | 2 | ([H1O3Br1])^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])^(-1) ([Br2])^(-1) ([HOCl])^(-5) ([HCl])^5 ([H1O3Br1])^2 = (([HCl])^5 ([H1O3Br1])^2)/([H2O] [Br2] ([HOCl])^5)

Construct the rate of reaction expression for: H_2O + Br_2 + HOCl ⟶ HCl + 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: H_2O + Br_2 + 5 HOCl ⟶ 5 HCl + 2 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 H_2O | 1 | -1 Br_2 | 1 | -1 HOCl | 5 | -5 HCl | 5 | 5 HO_3Br | 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 | 1 | -1 | -(Δ[H2O])/(Δt) Br_2 | 1 | -1 | -(Δ[Br2])/(Δt) HOCl | 5 | -5 | -1/5 (Δ[HOCl])/(Δt) HCl | 5 | 5 | 1/5 (Δ[HCl])/(Δt) HO_3Br | 2 | 2 | 1/2 (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 = -(Δ[H2O])/(Δt) = -(Δ[Br2])/(Δt) = -1/5 (Δ[HOCl])/(Δt) = 1/5 (Δ[HCl])/(Δt) = 1/2 (Δ[H1O3Br1])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| water | bromine | hypochlorous acid | hydrogen chloride | bromic acid formula | H_2O | Br_2 | HOCl | HCl | HO_3Br Hill formula | H_2O | Br_2 | ClHO | ClH | BrHO_3 name | water | bromine | hypochlorous acid | hydrogen chloride | bromic acid IUPAC name | water | molecular bromine | hypochlorous acid | hydrogen chloride | bromic acid