O2 + NH3 = H2O + N2O4

O_2 oxygen + NH_3 ammonia ⟶ H_2O water + N_2O_4 dinitrogen tetroxide

Balance the chemical equation algebraically: O_2 + NH_3 ⟶ H_2O + N_2O_4 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 O_2 + c_2 NH_3 ⟶ c_3 H_2O + c_4 N_2O_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 + 4 c_4 H: | 3 c_2 = 2 c_3 N: | c_2 = 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_4 = 1 and solve the system of equations for the remaining coefficients: c_1 = 7/2 c_2 = 2 c_3 = 3 c_4 = 1 Multiply by the least common denominator, 2, to eliminate fractional coefficients: c_1 = 7 c_2 = 4 c_3 = 6 c_4 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 7 O_2 + 4 NH_3 ⟶ 6 H_2O + 2 N_2O_4

+ ⟶ +

oxygen + ammonia ⟶ water + dinitrogen tetroxide

| oxygen | ammonia | water | dinitrogen tetroxide molecular enthalpy | 0 kJ/mol | -45.9 kJ/mol | -285.8 kJ/mol | 11.1 kJ/mol total enthalpy | 0 kJ/mol | -183.6 kJ/mol | -1715 kJ/mol | 22.2 kJ/mol | H_initial = -183.6 kJ/mol | | H_final = -1693 kJ/mol | ΔH_rxn^0 | -1693 kJ/mol - -183.6 kJ/mol = -1509 kJ/mol (exothermic) | | |

| oxygen | ammonia | water | dinitrogen tetroxide molecular free energy | 231.7 kJ/mol | -16.4 kJ/mol | -237.1 kJ/mol | 99.8 kJ/mol total free energy | 1622 kJ/mol | -65.6 kJ/mol | -1423 kJ/mol | 199.6 kJ/mol | G_initial = 1556 kJ/mol | | G_final = -1223 kJ/mol | ΔG_rxn^0 | -1223 kJ/mol - 1556 kJ/mol = -2779 kJ/mol (exergonic) | | |

| oxygen | ammonia | water | dinitrogen tetroxide molecular entropy | 205 J/(mol K) | 193 J/(mol K) | 69.91 J/(mol K) | 304.3 J/(mol K) total entropy | 1435 J/(mol K) | 772 J/(mol K) | 419.5 J/(mol K) | 608.6 J/(mol K) | S_initial = 2207 J/(mol K) | | S_final = 1028 J/(mol K) | ΔS_rxn^0 | 1028 J/(mol K) - 2207 J/(mol K) = -1179 J/(mol K) (exoentropic) | | |

Construct the equilibrium constant, K, expression for: O_2 + NH_3 ⟶ H_2O + N_2O_4 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: 7 O_2 + 4 NH_3 ⟶ 6 H_2O + 2 N_2O_4 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 | 7 | -7 NH_3 | 4 | -4 H_2O | 6 | 6 N_2O_4 | 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 | 7 | -7 | ([O2])^(-7) NH_3 | 4 | -4 | ([NH3])^(-4) H_2O | 6 | 6 | ([H2O])^6 N_2O_4 | 2 | 2 | ([N2O4])^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])^(-7) ([NH3])^(-4) ([H2O])^6 ([N2O4])^2 = (([H2O])^6 ([N2O4])^2)/(([O2])^7 ([NH3])^4)

Construct the rate of reaction expression for: O_2 + NH_3 ⟶ H_2O + N_2O_4 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: 7 O_2 + 4 NH_3 ⟶ 6 H_2O + 2 N_2O_4 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 | 7 | -7 NH_3 | 4 | -4 H_2O | 6 | 6 N_2O_4 | 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 | 7 | -7 | -1/7 (Δ[O2])/(Δt) NH_3 | 4 | -4 | -1/4 (Δ[NH3])/(Δt) H_2O | 6 | 6 | 1/6 (Δ[H2O])/(Δt) N_2O_4 | 2 | 2 | 1/2 (Δ[N2O4])/(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 = -1/7 (Δ[O2])/(Δt) = -1/4 (Δ[NH3])/(Δt) = 1/6 (Δ[H2O])/(Δt) = 1/2 (Δ[N2O4])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| oxygen | ammonia | water | dinitrogen tetroxide formula | O_2 | NH_3 | H_2O | N_2O_4 Hill formula | O_2 | H_3N | H_2O | N_2O_4 name | oxygen | ammonia | water | dinitrogen tetroxide IUPAC name | molecular oxygen | ammonia | water |

| oxygen | ammonia | water | dinitrogen tetroxide molar mass | 31.998 g/mol | 17.031 g/mol | 18.015 g/mol | 92.01 g/mol phase | gas (at STP) | gas (at STP) | liquid (at STP) | gas (at STP) melting point | -218 °C | -77.73 °C | 0 °C | -15 °C boiling point | -183 °C | -33.33 °C | 99.9839 °C | 21.2 °C density | 0.001429 g/cm^3 (at 0 °C) | 6.96×10^-4 g/cm^3 (at 25 °C) | 1 g/cm^3 | 1.45 g/cm^3 (at 20 °C) solubility in water | | | | reacts surface tension | 0.01347 N/m | 0.0234 N/m | 0.0728 N/m | 0.0275 N/m dynamic viscosity | 2.055×10^-5 Pa s (at 25 °C) | 1.009×10^-5 Pa s (at 25 °C) | 8.9×10^-4 Pa s (at 25 °C) | 4.18×10^-4 Pa s (at 20 °C) odor | odorless | | odorless |