HNO3 + HNO2 = H2O + NO2

HNO_3 nitric acid + HNO_2 nitrous acid ⟶ H_2O water + NO_2 nitrogen dioxide

Balance the chemical equation algebraically: HNO_3 + HNO_2 ⟶ H_2O + NO_2 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 HNO_3 + c_2 HNO_2 ⟶ c_3 H_2O + c_4 NO_2 Set the number of atoms in the reactants equal to the number of atoms in the products for H, N and O: H: | c_1 + c_2 = 2 c_3 N: | c_1 + c_2 = c_4 O: | 3 c_1 + 2 c_2 = c_3 + 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_1 = 1 and solve the system of equations for the remaining coefficients: c_1 = 1 c_2 = 1 c_3 = 1 c_4 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | HNO_3 + HNO_2 ⟶ H_2O + 2 NO_2

+ ⟶ +

nitric acid + nitrous acid ⟶ water + nitrogen dioxide

Construct the equilibrium constant, K, expression for: HNO_3 + HNO_2 ⟶ H_2O + NO_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: HNO_3 + HNO_2 ⟶ H_2O + 2 NO_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_3 | 1 | -1 HNO_2 | 1 | -1 H_2O | 1 | 1 NO_2 | 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 | 1 | -1 | ([HNO3])^(-1) HNO_2 | 1 | -1 | ([HNO2])^(-1) H_2O | 1 | 1 | [H2O] NO_2 | 2 | 2 | ([NO2])^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])^(-1) ([HNO2])^(-1) [H2O] ([NO2])^2 = ([H2O] ([NO2])^2)/([HNO3] [HNO2])

Construct the rate of reaction expression for: HNO_3 + HNO_2 ⟶ H_2O + NO_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: HNO_3 + HNO_2 ⟶ H_2O + 2 NO_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_3 | 1 | -1 HNO_2 | 1 | -1 H_2O | 1 | 1 NO_2 | 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 | 1 | -1 | -(Δ[HNO3])/(Δt) HNO_2 | 1 | -1 | -(Δ[HNO2])/(Δt) H_2O | 1 | 1 | (Δ[H2O])/(Δt) NO_2 | 2 | 2 | 1/2 (Δ[NO2])/(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 = -(Δ[HNO3])/(Δt) = -(Δ[HNO2])/(Δt) = (Δ[H2O])/(Δt) = 1/2 (Δ[NO2])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| nitric acid | nitrous acid | water | nitrogen dioxide formula | HNO_3 | HNO_2 | H_2O | NO_2 name | nitric acid | nitrous acid | water | nitrogen dioxide IUPAC name | nitric acid | nitrous acid | water | Nitrogen dioxide

| nitric acid | nitrous acid | water | nitrogen dioxide molar mass | 63.012 g/mol | 47.013 g/mol | 18.015 g/mol | 46.005 g/mol phase | liquid (at STP) | | liquid (at STP) | gas (at STP) melting point | -41.6 °C | | 0 °C | -11 °C boiling point | 83 °C | | 99.9839 °C | 21 °C density | 1.5129 g/cm^3 | | 1 g/cm^3 | 0.00188 g/cm^3 (at 25 °C) solubility in water | miscible | | | reacts surface tension | | | 0.0728 N/m | dynamic viscosity | 7.6×10^-4 Pa s (at 25 °C) | | 8.9×10^-4 Pa s (at 25 °C) | 4.02×10^-4 Pa s (at 25 °C) odor | | | odorless |