H2O + NO + HBrO4 = HNO3 + Br2

H_2O water + NO nitric oxide + HBrO4 ⟶ HNO_3 nitric acid + Br_2 bromine

Balance the chemical equation algebraically: H_2O + NO + HBrO4 ⟶ HNO_3 + Br_2 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 H_2O + c_2 NO + c_3 HBrO4 ⟶ 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, O, N and Br: H: | 2 c_1 + c_3 = c_4 O: | c_1 + c_2 + 4 c_3 = 3 c_4 N: | c_2 = c_4 Br: | c_3 = 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_5 = 1 and solve the system of equations for the remaining coefficients: c_1 = 4/3 c_2 = 14/3 c_3 = 2 c_4 = 14/3 c_5 = 1 Multiply by the least common denominator, 3, to eliminate fractional coefficients: c_1 = 4 c_2 = 14 c_3 = 6 c_4 = 14 c_5 = 3 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 4 H_2O + 14 NO + 6 HBrO4 ⟶ 14 HNO_3 + 3 Br_2

+ + HBrO4 ⟶ +

water + nitric oxide + HBrO4 ⟶ nitric acid + bromine

Construct the equilibrium constant, K, expression for: H_2O + NO + HBrO4 ⟶ 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: 4 H_2O + 14 NO + 6 HBrO4 ⟶ 14 HNO_3 + 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 H_2O | 4 | -4 NO | 14 | -14 HBrO4 | 6 | -6 HNO_3 | 14 | 14 Br_2 | 3 | 3 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression H_2O | 4 | -4 | ([H2O])^(-4) NO | 14 | -14 | ([NO])^(-14) HBrO4 | 6 | -6 | ([HBrO4])^(-6) HNO_3 | 14 | 14 | ([HNO3])^14 Br_2 | 3 | 3 | ([Br2])^3 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])^(-4) ([NO])^(-14) ([HBrO4])^(-6) ([HNO3])^14 ([Br2])^3 = (([HNO3])^14 ([Br2])^3)/(([H2O])^4 ([NO])^14 ([HBrO4])^6)

Construct the rate of reaction expression for: H_2O + NO + HBrO4 ⟶ 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: 4 H_2O + 14 NO + 6 HBrO4 ⟶ 14 HNO_3 + 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 H_2O | 4 | -4 NO | 14 | -14 HBrO4 | 6 | -6 HNO_3 | 14 | 14 Br_2 | 3 | 3 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 | 4 | -4 | -1/4 (Δ[H2O])/(Δt) NO | 14 | -14 | -1/14 (Δ[NO])/(Δt) HBrO4 | 6 | -6 | -1/6 (Δ[HBrO4])/(Δt) HNO_3 | 14 | 14 | 1/14 (Δ[HNO3])/(Δt) Br_2 | 3 | 3 | 1/3 (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 = -1/4 (Δ[H2O])/(Δt) = -1/14 (Δ[NO])/(Δt) = -1/6 (Δ[HBrO4])/(Δt) = 1/14 (Δ[HNO3])/(Δt) = 1/3 (Δ[Br2])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| water | nitric oxide | HBrO4 | nitric acid | bromine formula | H_2O | NO | HBrO4 | HNO_3 | Br_2 name | water | nitric oxide | | nitric acid | bromine IUPAC name | water | nitric oxide | | nitric acid | molecular bromine