H2O + Br2 + As2O3 = H3AsO4 + BrH

H_2O water + Br_2 bromine + As_2O_3 arsenic trioxide ⟶ H_3AsO_4 arsenic acid, solid + HBr hydrogen bromide

Balance the chemical equation algebraically: H_2O + Br_2 + As_2O_3 ⟶ H_3AsO_4 + HBr Add stoichiometric coefficients, c_i, to the reactants and products: c_1 H_2O + c_2 Br_2 + c_3 As_2O_3 ⟶ c_4 H_3AsO_4 + c_5 HBr Set the number of atoms in the reactants equal to the number of atoms in the products for H, O, Br and As: H: | 2 c_1 = 3 c_4 + c_5 O: | c_1 + 3 c_3 = 4 c_4 Br: | 2 c_2 = c_5 As: | 2 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_3 = 1 and solve the system of equations for the remaining coefficients: c_1 = 5 c_2 = 2 c_3 = 1 c_4 = 2 c_5 = 4 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 5 H_2O + 2 Br_2 + As_2O_3 ⟶ 2 H_3AsO_4 + 4 HBr

+ + ⟶ +

water + bromine + arsenic trioxide ⟶ arsenic acid, solid + hydrogen bromide

Construct the equilibrium constant, K, expression for: H_2O + Br_2 + As_2O_3 ⟶ H_3AsO_4 + HBr 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: 5 H_2O + 2 Br_2 + As_2O_3 ⟶ 2 H_3AsO_4 + 4 HBr 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 | 5 | -5 Br_2 | 2 | -2 As_2O_3 | 1 | -1 H_3AsO_4 | 2 | 2 HBr | 4 | 4 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression H_2O | 5 | -5 | ([H2O])^(-5) Br_2 | 2 | -2 | ([Br2])^(-2) As_2O_3 | 1 | -1 | ([As2O3])^(-1) H_3AsO_4 | 2 | 2 | ([H3AsO4])^2 HBr | 4 | 4 | ([HBr])^4 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])^(-5) ([Br2])^(-2) ([As2O3])^(-1) ([H3AsO4])^2 ([HBr])^4 = (([H3AsO4])^2 ([HBr])^4)/(([H2O])^5 ([Br2])^2 [As2O3])

Construct the rate of reaction expression for: H_2O + Br_2 + As_2O_3 ⟶ H_3AsO_4 + HBr 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: 5 H_2O + 2 Br_2 + As_2O_3 ⟶ 2 H_3AsO_4 + 4 HBr 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 | 5 | -5 Br_2 | 2 | -2 As_2O_3 | 1 | -1 H_3AsO_4 | 2 | 2 HBr | 4 | 4 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 | 5 | -5 | -1/5 (Δ[H2O])/(Δt) Br_2 | 2 | -2 | -1/2 (Δ[Br2])/(Δt) As_2O_3 | 1 | -1 | -(Δ[As2O3])/(Δt) H_3AsO_4 | 2 | 2 | 1/2 (Δ[H3AsO4])/(Δt) HBr | 4 | 4 | 1/4 (Δ[HBr])/(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 = -1/5 (Δ[H2O])/(Δt) = -1/2 (Δ[Br2])/(Δt) = -(Δ[As2O3])/(Δt) = 1/2 (Δ[H3AsO4])/(Δt) = 1/4 (Δ[HBr])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| water | bromine | arsenic trioxide | arsenic acid, solid | hydrogen bromide formula | H_2O | Br_2 | As_2O_3 | H_3AsO_4 | HBr Hill formula | H_2O | Br_2 | As_2O_3 | AsH_3O_4 | BrH name | water | bromine | arsenic trioxide | arsenic acid, solid | hydrogen bromide IUPAC name | water | molecular bromine | 2, 4, 5-trioxa-1, 3-diarsabicyclo[1.1.1]pentane | arsoric acid | hydrogen bromide