H2O + NO = O2 + NH3

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

Balance the chemical equation algebraically: H_2O + NO ⟶ O_2 + NH_3 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 H_2O + c_2 NO ⟶ c_3 O_2 + c_4 NH_3 Set the number of atoms in the reactants equal to the number of atoms in the products for H, O and N: H: | 2 c_1 = 3 c_4 O: | c_1 + 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 = 3/2 c_2 = 1 c_3 = 5/4 c_4 = 1 Multiply by the least common denominator, 4, to eliminate fractional coefficients: c_1 = 6 c_2 = 4 c_3 = 5 c_4 = 4 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 6 H_2O + 4 NO ⟶ 5 O_2 + 4 NH_3

+ ⟶ +

water + nitric oxide ⟶ oxygen + ammonia

| water | nitric oxide | oxygen | ammonia molecular enthalpy | -285.8 kJ/mol | 91.3 kJ/mol | 0 kJ/mol | -45.9 kJ/mol total enthalpy | -1715 kJ/mol | 365.2 kJ/mol | 0 kJ/mol | -183.6 kJ/mol | H_initial = -1350 kJ/mol | | H_final = -183.6 kJ/mol | ΔH_rxn^0 | -183.6 kJ/mol - -1350 kJ/mol = 1166 kJ/mol (endothermic) | | |

| water | nitric oxide | oxygen | ammonia molecular free energy | -237.1 kJ/mol | 87.6 kJ/mol | 231.7 kJ/mol | -16.4 kJ/mol total free energy | -1423 kJ/mol | 350.4 kJ/mol | 1159 kJ/mol | -65.6 kJ/mol | G_initial = -1072 kJ/mol | | G_final = 1093 kJ/mol | ΔG_rxn^0 | 1093 kJ/mol - -1072 kJ/mol = 2165 kJ/mol (endergonic) | | |

| water | nitric oxide | oxygen | ammonia molecular entropy | 69.91 J/(mol K) | 211 J/(mol K) | 205 J/(mol K) | 193 J/(mol K) total entropy | 419.5 J/(mol K) | 844 J/(mol K) | 1025 J/(mol K) | 772 J/(mol K) | S_initial = 1263 J/(mol K) | | S_final = 1797 J/(mol K) | ΔS_rxn^0 | 1797 J/(mol K) - 1263 J/(mol K) = 533.5 J/(mol K) (endoentropic) | | |

Construct the equilibrium constant, K, expression for: H_2O + NO ⟶ O_2 + 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: 6 H_2O + 4 NO ⟶ 5 O_2 + 4 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_2O | 6 | -6 NO | 4 | -4 O_2 | 5 | 5 NH_3 | 4 | 4 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression H_2O | 6 | -6 | ([H2O])^(-6) NO | 4 | -4 | ([NO])^(-4) O_2 | 5 | 5 | ([O2])^5 NH_3 | 4 | 4 | ([NH3])^4 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 = ([H2O])^(-6) ([NO])^(-4) ([O2])^5 ([NH3])^4 = (([O2])^5 ([NH3])^4)/(([H2O])^6 ([NO])^4)

Construct the rate of reaction expression for: H_2O + NO ⟶ O_2 + 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: 6 H_2O + 4 NO ⟶ 5 O_2 + 4 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_2O | 6 | -6 NO | 4 | -4 O_2 | 5 | 5 NH_3 | 4 | 4 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_2O | 6 | -6 | -1/6 (Δ[H2O])/(Δt) NO | 4 | -4 | -1/4 (Δ[NO])/(Δt) O_2 | 5 | 5 | 1/5 (Δ[O2])/(Δt) NH_3 | 4 | 4 | 1/4 (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 = -1/6 (Δ[H2O])/(Δt) = -1/4 (Δ[NO])/(Δt) = 1/5 (Δ[O2])/(Δt) = 1/4 (Δ[NH3])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

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

| water | nitric oxide | oxygen | ammonia molar mass | 18.015 g/mol | 30.006 g/mol | 31.998 g/mol | 17.031 g/mol phase | liquid (at STP) | gas (at STP) | gas (at STP) | gas (at STP) melting point | 0 °C | -163.6 °C | -218 °C | -77.73 °C boiling point | 99.9839 °C | -151.7 °C | -183 °C | -33.33 °C density | 1 g/cm^3 | 0.001226 g/cm^3 (at 25 °C) | 0.001429 g/cm^3 (at 0 °C) | 6.96×10^-4 g/cm^3 (at 25 °C) surface tension | 0.0728 N/m | | 0.01347 N/m | 0.0234 N/m dynamic viscosity | 8.9×10^-4 Pa s (at 25 °C) | 1.911×10^-5 Pa s (at 25 °C) | 2.055×10^-5 Pa s (at 25 °C) | 1.009×10^-5 Pa s (at 25 °C) odor | odorless | | odorless |