HNO3 + HF + B = H2O + NO2 + HBF4

HNO_3 nitric acid + HF hydrogen fluoride + B boron ⟶ H_2O water + NO_2 nitrogen dioxide + HBF_4 fluoboric acid

Balance the chemical equation algebraically: HNO_3 + HF + B ⟶ H_2O + NO_2 + HBF_4 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 HNO_3 + c_2 HF + c_3 B ⟶ c_4 H_2O + c_5 NO_2 + c_6 HBF_4 Set the number of atoms in the reactants equal to the number of atoms in the products for H, N, O, F and B: H: | c_1 + c_2 = 2 c_4 + c_6 N: | c_1 = c_5 O: | 3 c_1 = c_4 + 2 c_5 F: | c_2 = 4 c_6 B: | c_3 = c_6 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 = 3 c_2 = 4 c_3 = 1 c_4 = 3 c_5 = 3 c_6 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 3 HNO_3 + 4 HF + B ⟶ 3 H_2O + 3 NO_2 + HBF_4

+ + ⟶ + +

nitric acid + hydrogen fluoride + boron ⟶ water + nitrogen dioxide + fluoboric acid

Construct the equilibrium constant, K, expression for: HNO_3 + HF + B ⟶ H_2O + NO_2 + HBF_4 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 HNO_3 + 4 HF + B ⟶ 3 H_2O + 3 NO_2 + HBF_4 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 | 3 | -3 HF | 4 | -4 B | 1 | -1 H_2O | 3 | 3 NO_2 | 3 | 3 HBF_4 | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression HNO_3 | 3 | -3 | ([HNO3])^(-3) HF | 4 | -4 | ([HF])^(-4) B | 1 | -1 | ([B])^(-1) H_2O | 3 | 3 | ([H2O])^3 NO_2 | 3 | 3 | ([NO2])^3 HBF_4 | 1 | 1 | [HBF4] 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])^(-3) ([HF])^(-4) ([B])^(-1) ([H2O])^3 ([NO2])^3 [HBF4] = (([H2O])^3 ([NO2])^3 [HBF4])/(([HNO3])^3 ([HF])^4 [B])

Construct the rate of reaction expression for: HNO_3 + HF + B ⟶ H_2O + NO_2 + HBF_4 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 HNO_3 + 4 HF + B ⟶ 3 H_2O + 3 NO_2 + HBF_4 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 | 3 | -3 HF | 4 | -4 B | 1 | -1 H_2O | 3 | 3 NO_2 | 3 | 3 HBF_4 | 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_3 | 3 | -3 | -1/3 (Δ[HNO3])/(Δt) HF | 4 | -4 | -1/4 (Δ[HF])/(Δt) B | 1 | -1 | -(Δ[B])/(Δt) H_2O | 3 | 3 | 1/3 (Δ[H2O])/(Δt) NO_2 | 3 | 3 | 1/3 (Δ[NO2])/(Δt) HBF_4 | 1 | 1 | (Δ[HBF4])/(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[HNO3])/(Δt) = -1/4 (Δ[HF])/(Δt) = -(Δ[B])/(Δt) = 1/3 (Δ[H2O])/(Δt) = 1/3 (Δ[NO2])/(Δt) = (Δ[HBF4])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| nitric acid | hydrogen fluoride | boron | water | nitrogen dioxide | fluoboric acid formula | HNO_3 | HF | B | H_2O | NO_2 | HBF_4 Hill formula | HNO_3 | FH | B | H_2O | NO_2 | BF_4H name | nitric acid | hydrogen fluoride | boron | water | nitrogen dioxide | fluoboric acid IUPAC name | nitric acid | hydrogen fluoride | boron | water | Nitrogen dioxide | hydrogen(+1) cation; tetrafluoroboron