O2 + NH = H2O + N2

O_2 oxygen + NH ⟶ H_2O water + N_2 nitrogen

Balance the chemical equation algebraically: O_2 + NH ⟶ H_2O + N_2 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 O_2 + c_2 NH ⟶ c_3 H_2O + c_4 N_2 Set the number of atoms in the reactants equal to the number of atoms in the products for O, N and H: O: | 2 c_1 = c_3 N: | c_2 = 2 c_4 H: | c_2 = 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_1 = 1 and solve the system of equations for the remaining coefficients: c_1 = 1 c_2 = 4 c_3 = 2 c_4 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | O_2 + 4 NH ⟶ 2 H_2O + 2 N_2

+ NH ⟶ +

oxygen + NH ⟶ water + nitrogen

Construct the equilibrium constant, K, expression for: O_2 + NH ⟶ H_2O + 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: O_2 + 4 NH ⟶ 2 H_2O + 2 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 | 1 | -1 NH | 4 | -4 H_2O | 2 | 2 N_2 | 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 | 1 | -1 | ([O2])^(-1) NH | 4 | -4 | ([NH])^(-4) H_2O | 2 | 2 | ([H2O])^2 N_2 | 2 | 2 | ([N2])^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])^(-1) ([NH])^(-4) ([H2O])^2 ([N2])^2 = (([H2O])^2 ([N2])^2)/([O2] ([NH])^4)

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

| oxygen | NH | water | nitrogen formula | O_2 | NH | H_2O | N_2 Hill formula | O_2 | HN | H_2O | N_2 name | oxygen | | water | nitrogen IUPAC name | molecular oxygen | | water | molecular nitrogen

| oxygen | NH | water | nitrogen molar mass | 31.998 g/mol | 15.015 g/mol | 18.015 g/mol | 28.014 g/mol phase | gas (at STP) | | liquid (at STP) | gas (at STP) melting point | -218 °C | | 0 °C | -210 °C boiling point | -183 °C | | 99.9839 °C | -195.79 °C density | 0.001429 g/cm^3 (at 0 °C) | | 1 g/cm^3 | 0.001251 g/cm^3 (at 0 °C) solubility in water | | | | insoluble surface tension | 0.01347 N/m | | 0.0728 N/m | 0.0066 N/m dynamic viscosity | 2.055×10^-5 Pa s (at 25 °C) | | 8.9×10^-4 Pa s (at 25 °C) | 1.78×10^-5 Pa s (at 25 °C) odor | odorless | | odorless | odorless