HNO3 + B = NO2 + H3BO3

HNO_3 nitric acid + B boron ⟶ NO_2 nitrogen dioxide + B(OH)_3 boric acid

Balance the chemical equation algebraically: HNO_3 + B ⟶ NO_2 + B(OH)_3 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 HNO_3 + c_2 B ⟶ c_3 NO_2 + c_4 B(OH)_3 Set the number of atoms in the reactants equal to the number of atoms in the products for H, N, O and B: H: | c_1 = 3 c_4 N: | c_1 = c_3 O: | 3 c_1 = 2 c_3 + 3 c_4 B: | c_2 = c_4 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 = 3 c_2 = 1 c_3 = 3 c_4 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 3 HNO_3 + B ⟶ 3 NO_2 + B(OH)_3

+ ⟶ +

nitric acid + boron ⟶ nitrogen dioxide + boric acid

Construct the equilibrium constant, K, expression for: HNO_3 + B ⟶ NO_2 + B(OH)_3 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 + B ⟶ 3 NO_2 + B(OH)_3 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 B | 1 | -1 NO_2 | 3 | 3 B(OH)_3 | 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) B | 1 | -1 | ([B])^(-1) NO_2 | 3 | 3 | ([NO2])^3 B(OH)_3 | 1 | 1 | [B(OH)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 = ([HNO3])^(-3) ([B])^(-1) ([NO2])^3 [B(OH)3] = (([NO2])^3 [B(OH)3])/(([HNO3])^3 [B])

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

| nitric acid | boron | nitrogen dioxide | boric acid formula | HNO_3 | B | NO_2 | B(OH)_3 Hill formula | HNO_3 | B | NO_2 | BH_3O_3 name | nitric acid | boron | nitrogen dioxide | boric acid IUPAC name | nitric acid | boron | Nitrogen dioxide | boric acid

| nitric acid | boron | nitrogen dioxide | boric acid molar mass | 63.012 g/mol | 10.81 g/mol | 46.005 g/mol | 61.83 g/mol phase | liquid (at STP) | solid (at STP) | gas (at STP) | solid (at STP) melting point | -41.6 °C | 2075 °C | -11 °C | 160 °C boiling point | 83 °C | 4000 °C | 21 °C | density | 1.5129 g/cm^3 | 2.34 g/cm^3 | 0.00188 g/cm^3 (at 25 °C) | solubility in water | miscible | insoluble | reacts | dynamic viscosity | 7.6×10^-4 Pa s (at 25 °C) | | 4.02×10^-4 Pa s (at 25 °C) | odor | | | | odorless