O2 + NH3 = H2O + NO + N2

O_2 oxygen + NH_3 ammonia ⟶ H_2O water + NO nitric oxide + N_2 nitrogen

Balance the chemical equation algebraically: O_2 + NH_3 ⟶ H_2O + NO + N_2 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 O_2 + c_2 NH_3 ⟶ c_3 H_2O + c_4 NO + c_5 N_2 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 = c_3 + c_4 H: | 3 c_2 = 2 c_3 N: | c_2 = c_4 + 2 c_5 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_5 = 1 and solve the system of equations for the remaining coefficients: c_2 = (4 c_1)/5 + 4/5 c_3 = (6 c_1)/5 + 6/5 c_4 = (4 c_1)/5 - 6/5 c_5 = 1 The resulting system of equations is still underdetermined, so an additional coefficient must be set arbitrarily. Set c_1 = 4 and solve for the remaining coefficients: c_1 = 4 c_2 = 4 c_3 = 6 c_4 = 2 c_5 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 4 O_2 + 4 NH_3 ⟶ 6 H_2O + 2 NO + N_2

+ ⟶ + +

oxygen + ammonia ⟶ water + nitric oxide + nitrogen

| oxygen | ammonia | water | nitric oxide | nitrogen molecular enthalpy | 0 kJ/mol | -45.9 kJ/mol | -285.8 kJ/mol | 91.3 kJ/mol | 0 kJ/mol total enthalpy | 0 kJ/mol | -183.6 kJ/mol | -1715 kJ/mol | 182.6 kJ/mol | 0 kJ/mol | H_initial = -183.6 kJ/mol | | H_final = -1532 kJ/mol | | ΔH_rxn^0 | -1532 kJ/mol - -183.6 kJ/mol = -1349 kJ/mol (exothermic) | | | |

| oxygen | ammonia | water | nitric oxide | nitrogen molecular free energy | 231.7 kJ/mol | -16.4 kJ/mol | -237.1 kJ/mol | 87.6 kJ/mol | 0 kJ/mol total free energy | 926.8 kJ/mol | -65.6 kJ/mol | -1423 kJ/mol | 175.2 kJ/mol | 0 kJ/mol | G_initial = 861.2 kJ/mol | | G_final = -1247 kJ/mol | | ΔG_rxn^0 | -1247 kJ/mol - 861.2 kJ/mol = -2109 kJ/mol (exergonic) | | | |

| oxygen | ammonia | water | nitric oxide | nitrogen molecular entropy | 205 J/(mol K) | 193 J/(mol K) | 69.91 J/(mol K) | 211 J/(mol K) | 192 J/(mol K) total entropy | 820 J/(mol K) | 772 J/(mol K) | 419.5 J/(mol K) | 422 J/(mol K) | 192 J/(mol K) | S_initial = 1592 J/(mol K) | | S_final = 1033 J/(mol K) | | ΔS_rxn^0 | 1033 J/(mol K) - 1592 J/(mol K) = -558.5 J/(mol K) (exoentropic) | | | |

Construct the equilibrium constant, K, expression for: O_2 + NH_3 ⟶ H_2O + NO + N_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: 4 O_2 + 4 NH_3 ⟶ 6 H_2O + 2 NO + N_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 O_2 | 4 | -4 NH_3 | 4 | -4 H_2O | 6 | 6 NO | 2 | 2 N_2 | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression O_2 | 4 | -4 | ([O2])^(-4) NH_3 | 4 | -4 | ([NH3])^(-4) H_2O | 6 | 6 | ([H2O])^6 NO | 2 | 2 | ([NO])^2 N_2 | 1 | 1 | [N2] 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])^(-4) ([NH3])^(-4) ([H2O])^6 ([NO])^2 [N2] = (([H2O])^6 ([NO])^2 [N2])/(([O2])^4 ([NH3])^4)

Construct the rate of reaction expression for: O_2 + NH_3 ⟶ H_2O + NO + N_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: 4 O_2 + 4 NH_3 ⟶ 6 H_2O + 2 NO + N_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 O_2 | 4 | -4 NH_3 | 4 | -4 H_2O | 6 | 6 NO | 2 | 2 N_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 O_2 | 4 | -4 | -1/4 (Δ[O2])/(Δt) NH_3 | 4 | -4 | -1/4 (Δ[NH3])/(Δt) H_2O | 6 | 6 | 1/6 (Δ[H2O])/(Δt) NO | 2 | 2 | 1/2 (Δ[NO])/(Δt) N_2 | 1 | 1 | (Δ[N2])/(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 = -1/4 (Δ[O2])/(Δt) = -1/4 (Δ[NH3])/(Δt) = 1/6 (Δ[H2O])/(Δt) = 1/2 (Δ[NO])/(Δt) = (Δ[N2])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| oxygen | ammonia | water | nitric oxide | nitrogen formula | O_2 | NH_3 | H_2O | NO | N_2 Hill formula | O_2 | H_3N | H_2O | NO | N_2 name | oxygen | ammonia | water | nitric oxide | nitrogen IUPAC name | molecular oxygen | ammonia | water | nitric oxide | molecular nitrogen