H2 + NO2 = H2O + NH3

H_2 hydrogen + NO_2 nitrogen dioxide ⟶ H_2O water + NH_3 ammonia

Balance the chemical equation algebraically: H_2 + NO_2 ⟶ H_2O + NH_3 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 H_2 + c_2 NO_2 ⟶ c_3 H_2O + c_4 NH_3 Set the number of atoms in the reactants equal to the number of atoms in the products for H, N and O: H: | 2 c_1 = 2 c_3 + 3 c_4 N: | c_2 = c_4 O: | 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_2 = 1 and solve the system of equations for the remaining coefficients: c_1 = 7/2 c_2 = 1 c_3 = 2 c_4 = 1 Multiply by the least common denominator, 2, to eliminate fractional coefficients: c_1 = 7 c_2 = 2 c_3 = 4 c_4 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 7 H_2 + 2 NO_2 ⟶ 4 H_2O + 2 NH_3

+ ⟶ +

hydrogen + nitrogen dioxide ⟶ water + ammonia

| hydrogen | nitrogen dioxide | water | ammonia molecular enthalpy | 0 kJ/mol | 33.2 kJ/mol | -285.8 kJ/mol | -45.9 kJ/mol total enthalpy | 0 kJ/mol | 66.4 kJ/mol | -1143 kJ/mol | -91.8 kJ/mol | H_initial = 66.4 kJ/mol | | H_final = -1235 kJ/mol | ΔH_rxn^0 | -1235 kJ/mol - 66.4 kJ/mol = -1302 kJ/mol (exothermic) | | |

| hydrogen | nitrogen dioxide | water | ammonia molecular free energy | 0 kJ/mol | 51.3 kJ/mol | -237.1 kJ/mol | -16.4 kJ/mol total free energy | 0 kJ/mol | 102.6 kJ/mol | -948.4 kJ/mol | -32.8 kJ/mol | G_initial = 102.6 kJ/mol | | G_final = -981.2 kJ/mol | ΔG_rxn^0 | -981.2 kJ/mol - 102.6 kJ/mol = -1084 kJ/mol (exergonic) | | |

| hydrogen | nitrogen dioxide | water | ammonia molecular entropy | 115 J/(mol K) | 240 J/(mol K) | 69.91 J/(mol K) | 193 J/(mol K) total entropy | 805 J/(mol K) | 480 J/(mol K) | 279.6 J/(mol K) | 386 J/(mol K) | S_initial = 1285 J/(mol K) | | S_final = 665.6 J/(mol K) | ΔS_rxn^0 | 665.6 J/(mol K) - 1285 J/(mol K) = -619.4 J/(mol K) (exoentropic) | | |

Construct the equilibrium constant, K, expression for: H_2 + NO_2 ⟶ H_2O + NH_3 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 H_2 + 2 NO_2 ⟶ 4 H_2O + 2 NH_3 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 H_2 | 7 | -7 NO_2 | 2 | -2 H_2O | 4 | 4 NH_3 | 2 | 2 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression H_2 | 7 | -7 | ([H2])^(-7) NO_2 | 2 | -2 | ([NO2])^(-2) H_2O | 4 | 4 | ([H2O])^4 NH_3 | 2 | 2 | ([NH3])^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 = ([H2])^(-7) ([NO2])^(-2) ([H2O])^4 ([NH3])^2 = (([H2O])^4 ([NH3])^2)/(([H2])^7 ([NO2])^2)

Construct the rate of reaction expression for: H_2 + NO_2 ⟶ H_2O + NH_3 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 H_2 + 2 NO_2 ⟶ 4 H_2O + 2 NH_3 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 H_2 | 7 | -7 NO_2 | 2 | -2 H_2O | 4 | 4 NH_3 | 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 H_2 | 7 | -7 | -1/7 (Δ[H2])/(Δt) NO_2 | 2 | -2 | -1/2 (Δ[NO2])/(Δt) H_2O | 4 | 4 | 1/4 (Δ[H2O])/(Δt) NH_3 | 2 | 2 | 1/2 (Δ[NH3])/(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[H2])/(Δt) = -1/2 (Δ[NO2])/(Δt) = 1/4 (Δ[H2O])/(Δt) = 1/2 (Δ[NH3])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| hydrogen | nitrogen dioxide | water | ammonia formula | H_2 | NO_2 | H_2O | NH_3 Hill formula | H_2 | NO_2 | H_2O | H_3N name | hydrogen | nitrogen dioxide | water | ammonia IUPAC name | molecular hydrogen | Nitrogen dioxide | water | ammonia

| hydrogen | nitrogen dioxide | water | ammonia molar mass | 2.016 g/mol | 46.005 g/mol | 18.015 g/mol | 17.031 g/mol phase | gas (at STP) | gas (at STP) | liquid (at STP) | gas (at STP) melting point | -259.2 °C | -11 °C | 0 °C | -77.73 °C boiling point | -252.8 °C | 21 °C | 99.9839 °C | -33.33 °C density | 8.99×10^-5 g/cm^3 (at 0 °C) | 0.00188 g/cm^3 (at 25 °C) | 1 g/cm^3 | 6.96×10^-4 g/cm^3 (at 25 °C) solubility in water | | reacts | | surface tension | | | 0.0728 N/m | 0.0234 N/m dynamic viscosity | 8.9×10^-6 Pa s (at 25 °C) | 4.02×10^-4 Pa s (at 25 °C) | 8.9×10^-4 Pa s (at 25 °C) | 1.009×10^-5 Pa s (at 25 °C) odor | odorless | | odorless |