O2 + NH3 = H2O + NO2

oxygen + ammonia ⟶ water + nitrogen dioxide

Balance the chemical equation algebraically: + ⟶ + Add stoichiometric coefficients, c_i, to the reactants and products: c_1 + c_2 ⟶ c_3 + c_4 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 + 2 c_4 H: | 3 c_2 = 2 c_3 N: | c_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_2 = 1 and solve the system of equations for the remaining coefficients: c_1 = 7/4 c_2 = 1 c_3 = 3/2 c_4 = 1 Multiply by the least common denominator, 4, to eliminate fractional coefficients: c_1 = 7 c_2 = 4 c_3 = 6 c_4 = 4 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 7 + 4 ⟶ 6 + 4

+ ⟶ +

oxygen + ammonia ⟶ water + nitrogen dioxide

| oxygen | ammonia | water | nitrogen dioxide molecular enthalpy | 0 kJ/mol | -45.9 kJ/mol | -285.8 kJ/mol | 33.2 kJ/mol total enthalpy | 0 kJ/mol | -183.6 kJ/mol | -1715 kJ/mol | 132.8 kJ/mol | H_initial = -183.6 kJ/mol | | H_final = -1582 kJ/mol | ΔH_rxn^0 | -1582 kJ/mol - -183.6 kJ/mol = -1399 kJ/mol (exothermic) | | |

| oxygen | ammonia | water | nitrogen dioxide molecular free energy | 231.7 kJ/mol | -16.4 kJ/mol | -237.1 kJ/mol | 51.3 kJ/mol total free energy | 1622 kJ/mol | -65.6 kJ/mol | -1423 kJ/mol | 205.2 kJ/mol | G_initial = 1556 kJ/mol | | G_final = -1217 kJ/mol | ΔG_rxn^0 | -1217 kJ/mol - 1556 kJ/mol = -2774 kJ/mol (exergonic) | | |

| oxygen | ammonia | water | nitrogen dioxide molecular entropy | 205 J/(mol K) | 193 J/(mol K) | 69.91 J/(mol K) | 240 J/(mol K) total entropy | 1435 J/(mol K) | 772 J/(mol K) | 419.5 J/(mol K) | 960 J/(mol K) | S_initial = 2207 J/(mol K) | | S_final = 1379 J/(mol K) | ΔS_rxn^0 | 1379 J/(mol K) - 2207 J/(mol K) = -827.5 J/(mol K) (exoentropic) | | |

K_c = ([H2O]^6 [NO2]^4)/([O2]^7 [NH3]^4)

rate = -1/7 (Δ[O2])/(Δt) = -1/4 (Δ[NH3])/(Δt) = 1/6 (Δ[H2O])/(Δt) = 1/4 (Δ[NO2])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| oxygen | ammonia | water | nitrogen dioxide Hill formula | O_2 | H_3N | H_2O | NO_2 name | oxygen | ammonia | water | nitrogen dioxide IUPAC name | molecular oxygen | ammonia | water | Nitrogen dioxide