H2O + Br2 + AgBrO3 = HBrO3 + AgBr

H_2O water + Br_2 bromine + AgBrO_3 silver bromate ⟶ HO_3Br bromic acid + AgBr silver bromide

Balance the chemical equation algebraically: H_2O + Br_2 + AgBrO_3 ⟶ HO_3Br + AgBr Add stoichiometric coefficients, c_i, to the reactants and products: c_1 H_2O + c_2 Br_2 + c_3 AgBrO_3 ⟶ c_4 HO_3Br + c_5 AgBr Set the number of atoms in the reactants equal to the number of atoms in the products for H, O, Br and Ag: H: | 2 c_1 = c_4 O: | c_1 + 3 c_3 = 3 c_4 Br: | 2 c_2 + c_3 = c_4 + c_5 Ag: | c_3 = c_5 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 = 5/3 c_4 = 2 c_5 = 5/3 Multiply by the least common denominator, 3, to eliminate fractional coefficients: c_1 = 3 c_2 = 3 c_3 = 5 c_4 = 6 c_5 = 5 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 3 H_2O + 3 Br_2 + 5 AgBrO_3 ⟶ 6 HO_3Br + 5 AgBr

+ + ⟶ +

water + bromine + silver bromate ⟶ bromic acid + silver bromide

Construct the equilibrium constant, K, expression for: H_2O + Br_2 + AgBrO_3 ⟶ HO_3Br + AgBr 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 H_2O + 3 Br_2 + 5 AgBrO_3 ⟶ 6 HO_3Br + 5 AgBr 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 | 3 | -3 Br_2 | 3 | -3 AgBrO_3 | 5 | -5 HO_3Br | 6 | 6 AgBr | 5 | 5 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression H_2O | 3 | -3 | ([H2O])^(-3) Br_2 | 3 | -3 | ([Br2])^(-3) AgBrO_3 | 5 | -5 | ([AgBrO3])^(-5) HO_3Br | 6 | 6 | ([H1O3Br1])^6 AgBr | 5 | 5 | ([AgBr])^5 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])^(-3) ([Br2])^(-3) ([AgBrO3])^(-5) ([H1O3Br1])^6 ([AgBr])^5 = (([H1O3Br1])^6 ([AgBr])^5)/(([H2O])^3 ([Br2])^3 ([AgBrO3])^5)

Construct the rate of reaction expression for: H_2O + Br_2 + AgBrO_3 ⟶ HO_3Br + AgBr 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 H_2O + 3 Br_2 + 5 AgBrO_3 ⟶ 6 HO_3Br + 5 AgBr 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 | 3 | -3 Br_2 | 3 | -3 AgBrO_3 | 5 | -5 HO_3Br | 6 | 6 AgBr | 5 | 5 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 | 3 | -3 | -1/3 (Δ[H2O])/(Δt) Br_2 | 3 | -3 | -1/3 (Δ[Br2])/(Δt) AgBrO_3 | 5 | -5 | -1/5 (Δ[AgBrO3])/(Δt) HO_3Br | 6 | 6 | 1/6 (Δ[H1O3Br1])/(Δt) AgBr | 5 | 5 | 1/5 (Δ[AgBr])/(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[H2O])/(Δt) = -1/3 (Δ[Br2])/(Δt) = -1/5 (Δ[AgBrO3])/(Δt) = 1/6 (Δ[H1O3Br1])/(Δt) = 1/5 (Δ[AgBr])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| water | bromine | silver bromate | bromic acid | silver bromide formula | H_2O | Br_2 | AgBrO_3 | HO_3Br | AgBr Hill formula | H_2O | Br_2 | AgBrO_3 | BrHO_3 | AgBr name | water | bromine | silver bromate | bromic acid | silver bromide IUPAC name | water | molecular bromine | | bromic acid | bromosilver