H2O + Ba3N2 = NH3 + Ba(OH)2

H_2O water + Ba3N2 ⟶ NH_3 ammonia + Ba(OH)_2 barium hydroxide

Balance the chemical equation algebraically: H_2O + Ba3N2 ⟶ NH_3 + Ba(OH)_2 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 H_2O + c_2 Ba3N2 ⟶ c_3 NH_3 + c_4 Ba(OH)_2 Set the number of atoms in the reactants equal to the number of atoms in the products for H, O, Ba and N: H: | 2 c_1 = 3 c_3 + 2 c_4 O: | c_1 = 2 c_4 Ba: | 3 c_2 = c_4 N: | 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 = 6 c_2 = 1 c_3 = 2 c_4 = 3 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 6 H_2O + Ba3N2 ⟶ 2 NH_3 + 3 Ba(OH)_2

+ Ba3N2 ⟶ +

water + Ba3N2 ⟶ ammonia + barium hydroxide

Construct the equilibrium constant, K, expression for: H_2O + Ba3N2 ⟶ NH_3 + Ba(OH)_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: 6 H_2O + Ba3N2 ⟶ 2 NH_3 + 3 Ba(OH)_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 H_2O | 6 | -6 Ba3N2 | 1 | -1 NH_3 | 2 | 2 Ba(OH)_2 | 3 | 3 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) Ba3N2 | 1 | -1 | ([Ba3N2])^(-1) NH_3 | 2 | 2 | ([NH3])^2 Ba(OH)_2 | 3 | 3 | ([Ba(OH)2])^3 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) ([Ba3N2])^(-1) ([NH3])^2 ([Ba(OH)2])^3 = (([NH3])^2 ([Ba(OH)2])^3)/(([H2O])^6 [Ba3N2])

Construct the rate of reaction expression for: H_2O + Ba3N2 ⟶ NH_3 + Ba(OH)_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: 6 H_2O + Ba3N2 ⟶ 2 NH_3 + 3 Ba(OH)_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 H_2O | 6 | -6 Ba3N2 | 1 | -1 NH_3 | 2 | 2 Ba(OH)_2 | 3 | 3 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) Ba3N2 | 1 | -1 | -(Δ[Ba3N2])/(Δt) NH_3 | 2 | 2 | 1/2 (Δ[NH3])/(Δt) Ba(OH)_2 | 3 | 3 | 1/3 (Δ[Ba(OH)2])/(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) = -(Δ[Ba3N2])/(Δt) = 1/2 (Δ[NH3])/(Δt) = 1/3 (Δ[Ba(OH)2])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| water | Ba3N2 | ammonia | barium hydroxide formula | H_2O | Ba3N2 | NH_3 | Ba(OH)_2 Hill formula | H_2O | Ba3N2 | H_3N | BaH_2O_2 name | water | | ammonia | barium hydroxide IUPAC name | water | | ammonia | barium(+2) cation dihydroxide

| water | Ba3N2 | ammonia | barium hydroxide molar mass | 18.015 g/mol | 439.995 g/mol | 17.031 g/mol | 171.34 g/mol phase | liquid (at STP) | | gas (at STP) | solid (at STP) melting point | 0 °C | | -77.73 °C | 300 °C boiling point | 99.9839 °C | | -33.33 °C | density | 1 g/cm^3 | | 6.96×10^-4 g/cm^3 (at 25 °C) | 2.2 g/cm^3 surface tension | 0.0728 N/m | | 0.0234 N/m | dynamic viscosity | 8.9×10^-4 Pa s (at 25 °C) | | 1.009×10^-5 Pa s (at 25 °C) | odor | odorless | | |