O2 + HNO2 = HNO3

O_2 oxygen + HNO_2 nitrous acid ⟶ HNO_3 nitric acid

Balance the chemical equation algebraically: O_2 + HNO_2 ⟶ HNO_3 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 O_2 + c_2 HNO_2 ⟶ c_3 HNO_3 Set the number of atoms in the reactants equal to the number of atoms in the products for O, H and N: O: | 2 c_1 + 2 c_2 = 3 c_3 H: | c_2 = c_3 N: | c_2 = c_3 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 = 2 c_3 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | O_2 + 2 HNO_2 ⟶ 2 HNO_3

+ ⟶

oxygen + nitrous acid ⟶ nitric acid

Construct the equilibrium constant, K, expression for: O_2 + HNO_2 ⟶ HNO_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: O_2 + 2 HNO_2 ⟶ 2 HNO_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 O_2 | 1 | -1 HNO_2 | 2 | -2 HNO_3 | 2 | 2 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression O_2 | 1 | -1 | ([O2])^(-1) HNO_2 | 2 | -2 | ([HNO2])^(-2) HNO_3 | 2 | 2 | ([HNO3])^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 = ([O2])^(-1) ([HNO2])^(-2) ([HNO3])^2 = ([HNO3])^2/([O2] ([HNO2])^2)

Construct the rate of reaction expression for: O_2 + HNO_2 ⟶ HNO_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: O_2 + 2 HNO_2 ⟶ 2 HNO_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 O_2 | 1 | -1 HNO_2 | 2 | -2 HNO_3 | 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 O_2 | 1 | -1 | -(Δ[O2])/(Δt) HNO_2 | 2 | -2 | -1/2 (Δ[HNO2])/(Δt) HNO_3 | 2 | 2 | 1/2 (Δ[HNO3])/(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[O2])/(Δt) = -1/2 (Δ[HNO2])/(Δt) = 1/2 (Δ[HNO3])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| oxygen | nitrous acid | nitric acid formula | O_2 | HNO_2 | HNO_3 name | oxygen | nitrous acid | nitric acid IUPAC name | molecular oxygen | nitrous acid | nitric acid

| oxygen | nitrous acid | nitric acid molar mass | 31.998 g/mol | 47.013 g/mol | 63.012 g/mol phase | gas (at STP) | | liquid (at STP) melting point | -218 °C | | -41.6 °C boiling point | -183 °C | | 83 °C density | 0.001429 g/cm^3 (at 0 °C) | | 1.5129 g/cm^3 solubility in water | | | miscible surface tension | 0.01347 N/m | | dynamic viscosity | 2.055×10^-5 Pa s (at 25 °C) | | 7.6×10^-4 Pa s (at 25 °C) odor | odorless | |