NO + O3 = O2 + NO2

NO nitric oxide + O_3 ozone ⟶ O_2 oxygen + NO_2 nitrogen dioxide

Balance the chemical equation algebraically: NO + O_3 ⟶ O_2 + NO_2 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 NO + c_2 O_3 ⟶ c_3 O_2 + c_4 NO_2 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_4 O: | c_1 + 3 c_2 = 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_3 = (3 c_2)/2 - 1/2 c_4 = 1 The resulting system of equations is still underdetermined, so an additional coefficient must be set arbitrarily. Set c_2 = 1 and solve 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: | | NO + O_3 ⟶ O_2 + NO_2

+ ⟶ +

nitric oxide + ozone ⟶ oxygen + nitrogen dioxide

| nitric oxide | ozone | oxygen | nitrogen dioxide molecular enthalpy | 91.3 kJ/mol | 142.7 kJ/mol | 0 kJ/mol | 33.2 kJ/mol total enthalpy | 91.3 kJ/mol | 142.7 kJ/mol | 0 kJ/mol | 33.2 kJ/mol | H_initial = 234 kJ/mol | | H_final = 33.2 kJ/mol | ΔH_rxn^0 | 33.2 kJ/mol - 234 kJ/mol = -200.8 kJ/mol (exothermic) | | |

| nitric oxide | ozone | oxygen | nitrogen dioxide molecular free energy | 87.6 kJ/mol | 163.2 kJ/mol | 231.7 kJ/mol | 51.3 kJ/mol total free energy | 87.6 kJ/mol | 163.2 kJ/mol | 231.7 kJ/mol | 51.3 kJ/mol | G_initial = 250.8 kJ/mol | | G_final = 283 kJ/mol | ΔG_rxn^0 | 283 kJ/mol - 250.8 kJ/mol = 32.2 kJ/mol (endergonic) | | |

| nitric oxide | ozone | oxygen | nitrogen dioxide molecular entropy | 211 J/(mol K) | 239 J/(mol K) | 205 J/(mol K) | 240 J/(mol K) total entropy | 211 J/(mol K) | 239 J/(mol K) | 205 J/(mol K) | 240 J/(mol K) | S_initial = 450 J/(mol K) | | S_final = 445 J/(mol K) | ΔS_rxn^0 | 445 J/(mol K) - 450 J/(mol K) = -5 J/(mol K) (exoentropic) | | |

Construct the equilibrium constant, K, expression for: NO + O_3 ⟶ O_2 + 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: NO + O_3 ⟶ O_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 NO | 1 | -1 O_3 | 1 | -1 O_2 | 1 | 1 NO_2 | 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) O_3 | 1 | -1 | ([O3])^(-1) O_2 | 1 | 1 | [O2] NO_2 | 1 | 1 | [NO2] 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) ([O3])^(-1) [O2] [NO2] = ([O2] [NO2])/([NO] [O3])

Construct the rate of reaction expression for: NO + O_3 ⟶ O_2 + 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: NO + O_3 ⟶ O_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 NO | 1 | -1 O_3 | 1 | -1 O_2 | 1 | 1 NO_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 NO | 1 | -1 | -(Δ[NO])/(Δt) O_3 | 1 | -1 | -(Δ[O3])/(Δt) O_2 | 1 | 1 | (Δ[O2])/(Δt) NO_2 | 1 | 1 | (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[NO])/(Δt) = -(Δ[O3])/(Δt) = (Δ[O2])/(Δt) = (Δ[NO2])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| nitric oxide | ozone | oxygen | nitrogen dioxide formula | NO | O_3 | O_2 | NO_2 name | nitric oxide | ozone | oxygen | nitrogen dioxide IUPAC name | nitric oxide | ozone | molecular oxygen | Nitrogen dioxide

| nitric oxide | ozone | oxygen | nitrogen dioxide molar mass | 30.006 g/mol | 47.997 g/mol | 31.998 g/mol | 46.005 g/mol phase | gas (at STP) | gas (at STP) | gas (at STP) | gas (at STP) melting point | -163.6 °C | -192.2 °C | -218 °C | -11 °C boiling point | -151.7 °C | -111.9 °C | -183 °C | 21 °C density | 0.001226 g/cm^3 (at 25 °C) | 0.001962 g/cm^3 (at 25 °C) | 0.001429 g/cm^3 (at 0 °C) | 0.00188 g/cm^3 (at 25 °C) solubility in water | | | | reacts surface tension | | | 0.01347 N/m | dynamic viscosity | 1.911×10^-5 Pa s (at 25 °C) | | 2.055×10^-5 Pa s (at 25 °C) | 4.02×10^-4 Pa s (at 25 °C) odor | | | odorless |