HNO2 + HBrO3 = H2O + HNO3 + Br2

HNO_2 nitrous acid + HO_3Br bromic acid ⟶ H_2O water + HNO_3 nitric acid + Br_2 bromine

Balance the chemical equation algebraically: HNO_2 + HO_3Br ⟶ H_2O + HNO_3 + Br_2 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 HNO_2 + c_2 HO_3Br ⟶ c_3 H_2O + c_4 HNO_3 + c_5 Br_2 Set the number of atoms in the reactants equal to the number of atoms in the products for H, N, O and Br: H: | c_1 + c_2 = 2 c_3 + c_4 N: | c_1 = c_4 O: | 2 c_1 + 3 c_2 = c_3 + 3 c_4 Br: | c_2 = 2 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_3 = 1 and solve the system of equations for the remaining coefficients: c_1 = 5 c_2 = 2 c_3 = 1 c_4 = 5 c_5 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 5 HNO_2 + 2 HO_3Br ⟶ H_2O + 5 HNO_3 + Br_2

+ ⟶ + +

nitrous acid + bromic acid ⟶ water + nitric acid + bromine

Construct the equilibrium constant, K, expression for: HNO_2 + HO_3Br ⟶ H_2O + HNO_3 + Br_2 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 HNO_2 + 2 HO_3Br ⟶ H_2O + 5 HNO_3 + Br_2 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 HNO_2 | 5 | -5 HO_3Br | 2 | -2 H_2O | 1 | 1 HNO_3 | 5 | 5 Br_2 | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression HNO_2 | 5 | -5 | ([HNO2])^(-5) HO_3Br | 2 | -2 | ([H1O3Br1])^(-2) H_2O | 1 | 1 | [H2O] HNO_3 | 5 | 5 | ([HNO3])^5 Br_2 | 1 | 1 | [Br2] 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 = ([HNO2])^(-5) ([H1O3Br1])^(-2) [H2O] ([HNO3])^5 [Br2] = ([H2O] ([HNO3])^5 [Br2])/(([HNO2])^5 ([H1O3Br1])^2)

Construct the rate of reaction expression for: HNO_2 + HO_3Br ⟶ H_2O + HNO_3 + Br_2 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 HNO_2 + 2 HO_3Br ⟶ H_2O + 5 HNO_3 + Br_2 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 HNO_2 | 5 | -5 HO_3Br | 2 | -2 H_2O | 1 | 1 HNO_3 | 5 | 5 Br_2 | 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 HNO_2 | 5 | -5 | -1/5 (Δ[HNO2])/(Δt) HO_3Br | 2 | -2 | -1/2 (Δ[H1O3Br1])/(Δt) H_2O | 1 | 1 | (Δ[H2O])/(Δt) HNO_3 | 5 | 5 | 1/5 (Δ[HNO3])/(Δt) Br_2 | 1 | 1 | (Δ[Br2])/(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[HNO2])/(Δt) = -1/2 (Δ[H1O3Br1])/(Δt) = (Δ[H2O])/(Δt) = 1/5 (Δ[HNO3])/(Δt) = (Δ[Br2])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| nitrous acid | bromic acid | water | nitric acid | bromine formula | HNO_2 | HO_3Br | H_2O | HNO_3 | Br_2 Hill formula | HNO_2 | BrHO_3 | H_2O | HNO_3 | Br_2 name | nitrous acid | bromic acid | water | nitric acid | bromine IUPAC name | nitrous acid | bromic acid | water | nitric acid | molecular bromine