HNO3 + Br2 = H2O + NO2 + HBrO3

HNO_3 nitric acid + Br_2 bromine ⟶ H_2O water + NO_2 nitrogen dioxide + HO_3Br bromic acid

Balance the chemical equation algebraically: HNO_3 + Br_2 ⟶ H_2O + NO_2 + HO_3Br Add stoichiometric coefficients, c_i, to the reactants and products: c_1 HNO_3 + c_2 Br_2 ⟶ c_3 H_2O + c_4 NO_2 + c_5 HO_3Br 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 = 2 c_3 + c_5 N: | c_1 = c_4 O: | 3 c_1 = c_3 + 2 c_4 + 3 c_5 Br: | 2 c_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_2 = 1 and solve the system of equations for the remaining coefficients: c_1 = 10 c_2 = 1 c_3 = 4 c_4 = 10 c_5 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 10 HNO_3 + Br_2 ⟶ 4 H_2O + 10 NO_2 + 2 HO_3Br

+ ⟶ + +

nitric acid + bromine ⟶ water + nitrogen dioxide + bromic acid

Construct the equilibrium constant, K, expression for: HNO_3 + Br_2 ⟶ H_2O + NO_2 + 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: 10 HNO_3 + Br_2 ⟶ 4 H_2O + 10 NO_2 + 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 HNO_3 | 10 | -10 Br_2 | 1 | -1 H_2O | 4 | 4 NO_2 | 10 | 10 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 HNO_3 | 10 | -10 | ([HNO3])^(-10) Br_2 | 1 | -1 | ([Br2])^(-1) H_2O | 4 | 4 | ([H2O])^4 NO_2 | 10 | 10 | ([NO2])^10 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 = ([HNO3])^(-10) ([Br2])^(-1) ([H2O])^4 ([NO2])^10 ([H1O3Br1])^2 = (([H2O])^4 ([NO2])^10 ([H1O3Br1])^2)/(([HNO3])^10 [Br2])

Construct the rate of reaction expression for: HNO_3 + Br_2 ⟶ H_2O + NO_2 + 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: 10 HNO_3 + Br_2 ⟶ 4 H_2O + 10 NO_2 + 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 HNO_3 | 10 | -10 Br_2 | 1 | -1 H_2O | 4 | 4 NO_2 | 10 | 10 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 HNO_3 | 10 | -10 | -1/10 (Δ[HNO3])/(Δt) Br_2 | 1 | -1 | -(Δ[Br2])/(Δt) H_2O | 4 | 4 | 1/4 (Δ[H2O])/(Δt) NO_2 | 10 | 10 | 1/10 (Δ[NO2])/(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 = -1/10 (Δ[HNO3])/(Δt) = -(Δ[Br2])/(Δt) = 1/4 (Δ[H2O])/(Δt) = 1/10 (Δ[NO2])/(Δt) = 1/2 (Δ[H1O3Br1])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| nitric acid | bromine | water | nitrogen dioxide | bromic acid formula | HNO_3 | Br_2 | H_2O | NO_2 | HO_3Br Hill formula | HNO_3 | Br_2 | H_2O | NO_2 | BrHO_3 name | nitric acid | bromine | water | nitrogen dioxide | bromic acid IUPAC name | nitric acid | molecular bromine | water | Nitrogen dioxide | bromic acid