O2 + N2 = NO

O_2 (oxygen) + N_2 (nitrogen) ⟶ NO (nitric oxide)

Balance the chemical equation algebraically: O_2 + N_2 ⟶ NO Add stoichiometric coefficients, c_i, to the reactants and products: c_1 O_2 + c_2 N_2 ⟶ c_3 NO Set the number of atoms in the reactants equal to the number of atoms in the products for O and N: O: | 2 c_1 = c_3 N: | 2 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 = 1 c_3 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | O_2 + N_2 ⟶ 2 NO

+ ⟶

oxygen + nitrogen ⟶ nitric oxide

| oxygen | nitrogen | nitric oxide molecular enthalpy | 0 kJ/mol | 0 kJ/mol | 91.3 kJ/mol total enthalpy | 0 kJ/mol | 0 kJ/mol | 182.6 kJ/mol | H_initial = 0 kJ/mol | | H_final = 182.6 kJ/mol ΔH_rxn^0 | 182.6 kJ/mol - 0 kJ/mol = 182.6 kJ/mol (endothermic) | |

| oxygen | nitrogen | nitric oxide molecular free energy | 231.7 kJ/mol | 0 kJ/mol | 87.6 kJ/mol total free energy | 231.7 kJ/mol | 0 kJ/mol | 175.2 kJ/mol | G_initial = 231.7 kJ/mol | | G_final = 175.2 kJ/mol ΔG_rxn^0 | 175.2 kJ/mol - 231.7 kJ/mol = -56.5 kJ/mol (exergonic) | |

| oxygen | nitrogen | nitric oxide molecular entropy | 205 J/(mol K) | 192 J/(mol K) | 211 J/(mol K) total entropy | 205 J/(mol K) | 192 J/(mol K) | 422 J/(mol K) | S_initial = 397 J/(mol K) | | S_final = 422 J/(mol K) ΔS_rxn^0 | 422 J/(mol K) - 397 J/(mol K) = 25 J/(mol K) (endoentropic) | |

Construct the equilibrium constant, K, expression for: O_2 + N_2 ⟶ NO 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 + N_2 ⟶ 2 NO 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 N_2 | 1 | -1 NO | 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) N_2 | 1 | -1 | ([N2])^(-1) NO | 2 | 2 | ([NO])^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) ([N2])^(-1) ([NO])^2 = ([NO])^2/([O2] [N2])

Construct the rate of reaction expression for: O_2 + N_2 ⟶ NO 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 + N_2 ⟶ 2 NO 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 N_2 | 1 | -1 NO | 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) N_2 | 1 | -1 | -(Δ[N2])/(Δt) NO | 2 | 2 | 1/2 (Δ[NO])/(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) = -(Δ[N2])/(Δt) = 1/2 (Δ[NO])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| oxygen | nitrogen | nitric oxide formula | O_2 | N_2 | NO name | oxygen | nitrogen | nitric oxide IUPAC name | molecular oxygen | molecular nitrogen | nitric oxide

| oxygen | nitrogen | nitric oxide molar mass | 31.998 g/mol | 28.014 g/mol | 30.006 g/mol phase | gas (at STP) | gas (at STP) | gas (at STP) melting point | -218 °C | -210 °C | -163.6 °C boiling point | -183 °C | -195.79 °C | -151.7 °C density | 0.001429 g/cm^3 (at 0 °C) | 0.001251 g/cm^3 (at 0 °C) | 0.001226 g/cm^3 (at 25 °C) solubility in water | | insoluble | surface tension | 0.01347 N/m | 0.0066 N/m | dynamic viscosity | 2.055×10^-5 Pa s (at 25 °C) | 1.78×10^-5 Pa s (at 25 °C) | 1.911×10^-5 Pa s (at 25 °C) odor | odorless | odorless |