NH3 + Ba = H2 + Ba(NH2)2

NH_3 ammonia + Ba barium ⟶ H_2 hydrogen + Ba(NH2)2

Balance the chemical equation algebraically: NH_3 + Ba ⟶ H_2 + Ba(NH2)2 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 NH_3 + c_2 Ba ⟶ c_3 H_2 + c_4 Ba(NH2)2 Set the number of atoms in the reactants equal to the number of atoms in the products for H, N and Ba: H: | 3 c_1 = 2 c_3 + 4 c_4 N: | c_1 = 2 c_4 Ba: | 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 = 2 c_2 = 1 c_3 = 1 c_4 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 2 NH_3 + Ba ⟶ H_2 + Ba(NH2)2

+ ⟶ + Ba(NH2)2

ammonia + barium ⟶ hydrogen + Ba(NH2)2

Construct the equilibrium constant, K, expression for: NH_3 + Ba ⟶ H_2 + Ba(NH2)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: 2 NH_3 + Ba ⟶ H_2 + Ba(NH2)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 NH_3 | 2 | -2 Ba | 1 | -1 H_2 | 1 | 1 Ba(NH2)2 | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression NH_3 | 2 | -2 | ([NH3])^(-2) Ba | 1 | -1 | ([Ba])^(-1) H_2 | 1 | 1 | [H2] Ba(NH2)2 | 1 | 1 | [Ba(NH2)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 = ([NH3])^(-2) ([Ba])^(-1) [H2] [Ba(NH2)2] = ([H2] [Ba(NH2)2])/(([NH3])^2 [Ba])

Construct the rate of reaction expression for: NH_3 + Ba ⟶ H_2 + Ba(NH2)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: 2 NH_3 + Ba ⟶ H_2 + Ba(NH2)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 NH_3 | 2 | -2 Ba | 1 | -1 H_2 | 1 | 1 Ba(NH2)2 | 1 | 1 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 NH_3 | 2 | -2 | -1/2 (Δ[NH3])/(Δt) Ba | 1 | -1 | -(Δ[Ba])/(Δt) H_2 | 1 | 1 | (Δ[H2])/(Δt) Ba(NH2)2 | 1 | 1 | (Δ[Ba(NH2)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 = -1/2 (Δ[NH3])/(Δt) = -(Δ[Ba])/(Δt) = (Δ[H2])/(Δt) = (Δ[Ba(NH2)2])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| ammonia | barium | hydrogen | Ba(NH2)2 formula | NH_3 | Ba | H_2 | Ba(NH2)2 Hill formula | H_3N | Ba | H_2 | H4BaN2 name | ammonia | barium | hydrogen | IUPAC name | ammonia | barium | molecular hydrogen |

| ammonia | barium | hydrogen | Ba(NH2)2 molar mass | 17.031 g/mol | 137.327 g/mol | 2.016 g/mol | 169.37 g/mol phase | gas (at STP) | solid (at STP) | gas (at STP) | melting point | -77.73 °C | 725 °C | -259.2 °C | boiling point | -33.33 °C | 1640 °C | -252.8 °C | density | 6.96×10^-4 g/cm^3 (at 25 °C) | 3.6 g/cm^3 | 8.99×10^-5 g/cm^3 (at 0 °C) | solubility in water | | insoluble | | surface tension | 0.0234 N/m | 0.224 N/m | | dynamic viscosity | 1.009×10^-5 Pa s (at 25 °C) | | 8.9×10^-6 Pa s (at 25 °C) | odor | | | odorless |