H2O + N2O4 = HNO3 + HNO2

H_2O water + N_2O_4 dinitrogen tetroxide ⟶ HNO_3 nitric acid + HNO_2 nitrous acid

Balance the chemical equation algebraically: H_2O + N_2O_4 ⟶ HNO_3 + HNO_2 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 H_2O + c_2 N_2O_4 ⟶ c_3 HNO_3 + c_4 HNO_2 Set the number of atoms in the reactants equal to the number of atoms in the products for H, O and N: H: | 2 c_1 = c_3 + c_4 O: | c_1 + 4 c_2 = 3 c_3 + 2 c_4 N: | 2 c_2 = c_3 + 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 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | H_2O + N_2O_4 ⟶ HNO_3 + HNO_2

+ ⟶ +

water + dinitrogen tetroxide ⟶ nitric acid + nitrous acid

Construct the equilibrium constant, K, expression for: H_2O + N_2O_4 ⟶ HNO_3 + HNO_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: H_2O + N_2O_4 ⟶ HNO_3 + HNO_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 | 1 | -1 N_2O_4 | 1 | -1 HNO_3 | 1 | 1 HNO_2 | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression H_2O | 1 | -1 | ([H2O])^(-1) N_2O_4 | 1 | -1 | ([N2O4])^(-1) HNO_3 | 1 | 1 | [HNO3] HNO_2 | 1 | 1 | [HNO2] 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])^(-1) ([N2O4])^(-1) [HNO3] [HNO2] = ([HNO3] [HNO2])/([H2O] [N2O4])

Construct the rate of reaction expression for: H_2O + N_2O_4 ⟶ HNO_3 + HNO_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: H_2O + N_2O_4 ⟶ HNO_3 + HNO_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 | 1 | -1 N_2O_4 | 1 | -1 HNO_3 | 1 | 1 HNO_2 | 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 H_2O | 1 | -1 | -(Δ[H2O])/(Δt) N_2O_4 | 1 | -1 | -(Δ[N2O4])/(Δt) HNO_3 | 1 | 1 | (Δ[HNO3])/(Δt) HNO_2 | 1 | 1 | (Δ[HNO2])/(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[H2O])/(Δt) = -(Δ[N2O4])/(Δt) = (Δ[HNO3])/(Δt) = (Δ[HNO2])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| water | dinitrogen tetroxide | nitric acid | nitrous acid formula | H_2O | N_2O_4 | HNO_3 | HNO_2 name | water | dinitrogen tetroxide | nitric acid | nitrous acid

| water | dinitrogen tetroxide | nitric acid | nitrous acid molar mass | 18.015 g/mol | 92.01 g/mol | 63.012 g/mol | 47.013 g/mol phase | liquid (at STP) | gas (at STP) | liquid (at STP) | melting point | 0 °C | -15 °C | -41.6 °C | boiling point | 99.9839 °C | 21.2 °C | 83 °C | density | 1 g/cm^3 | 1.45 g/cm^3 (at 20 °C) | 1.5129 g/cm^3 | solubility in water | | reacts | miscible | surface tension | 0.0728 N/m | 0.0275 N/m | | dynamic viscosity | 8.9×10^-4 Pa s (at 25 °C) | 4.18×10^-4 Pa s (at 20 °C) | 7.6×10^-4 Pa s (at 25 °C) | odor | odorless | | |