NO + NO2 = N2O3

NO nitric oxide + NO_2 nitrogen dioxide ⟶ N_2O_3 nitrogen trioxide

Balance the chemical equation algebraically: NO + NO_2 ⟶ N_2O_3 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 NO + c_2 NO_2 ⟶ c_3 N_2O_3 Set the number of atoms in the reactants equal to the number of atoms in the products for N and O: N: | c_1 + c_2 = 2 c_3 O: | c_1 + 2 c_2 = 3 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 = 1 c_3 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | NO + NO_2 ⟶ N_2O_3

+ ⟶

nitric oxide + nitrogen dioxide ⟶ nitrogen trioxide

| nitric oxide | nitrogen dioxide | nitrogen trioxide molecular enthalpy | 91.3 kJ/mol | 33.2 kJ/mol | 86.6 kJ/mol total enthalpy | 91.3 kJ/mol | 33.2 kJ/mol | 86.6 kJ/mol | H_initial = 124.5 kJ/mol | | H_final = 86.6 kJ/mol ΔH_rxn^0 | 86.6 kJ/mol - 124.5 kJ/mol = -37.9 kJ/mol (exothermic) | |

| nitric oxide | nitrogen dioxide | nitrogen trioxide molecular free energy | 87.6 kJ/mol | 51.3 kJ/mol | 142.4 kJ/mol total free energy | 87.6 kJ/mol | 51.3 kJ/mol | 142.4 kJ/mol | G_initial = 138.9 kJ/mol | | G_final = 142.4 kJ/mol ΔG_rxn^0 | 142.4 kJ/mol - 138.9 kJ/mol = 3.5 kJ/mol (endergonic) | |

Construct the equilibrium constant, K, expression for: NO + NO_2 ⟶ N_2O_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: NO + NO_2 ⟶ N_2O_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 NO | 1 | -1 NO_2 | 1 | -1 N_2O_3 | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression NO | 1 | -1 | ([NO])^(-1) NO_2 | 1 | -1 | ([NO2])^(-1) N_2O_3 | 1 | 1 | [N2O3] 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 = ([NO])^(-1) ([NO2])^(-1) [N2O3] = ([N2O3])/([NO] [NO2])

Construct the rate of reaction expression for: NO + NO_2 ⟶ N_2O_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: NO + NO_2 ⟶ N_2O_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 NO | 1 | -1 NO_2 | 1 | -1 N_2O_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 NO | 1 | -1 | -(Δ[NO])/(Δt) NO_2 | 1 | -1 | -(Δ[NO2])/(Δt) N_2O_3 | 1 | 1 | (Δ[N2O3])/(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[NO])/(Δt) = -(Δ[NO2])/(Δt) = (Δ[N2O3])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| nitric oxide | nitrogen dioxide | nitrogen trioxide formula | NO | NO_2 | N_2O_3 name | nitric oxide | nitrogen dioxide | nitrogen trioxide IUPAC name | nitric oxide | Nitrogen dioxide | nitramide

| nitric oxide | nitrogen dioxide | nitrogen trioxide molar mass | 30.006 g/mol | 46.005 g/mol | 76.011 g/mol phase | gas (at STP) | gas (at STP) | gas (at STP) melting point | -163.6 °C | -11 °C | -111 °C boiling point | -151.7 °C | 21 °C | -27 °C density | 0.001226 g/cm^3 (at 25 °C) | 0.00188 g/cm^3 (at 25 °C) | 1.4 g/cm^3 (at 2 °C) solubility in water | | reacts | very soluble dynamic viscosity | 1.911×10^-5 Pa s (at 25 °C) | 4.02×10^-4 Pa s (at 25 °C) |