NH3 + NiF2 = N2 + NH4F + Ni3N

NH_3 ammonia + NiF_2 nickel(II) fluoride ⟶ N_2 nitrogen + NH_4F ammonium fluoride + Ni3N

Balance the chemical equation algebraically: NH_3 + NiF_2 ⟶ N_2 + NH_4F + Ni3N Add stoichiometric coefficients, c_i, to the reactants and products: c_1 NH_3 + c_2 NiF_2 ⟶ c_3 N_2 + c_4 NH_4F + c_5 Ni3N Set the number of atoms in the reactants equal to the number of atoms in the products for H, N, F and Ni: H: | 3 c_1 = 4 c_4 N: | c_1 = 2 c_3 + c_4 + c_5 F: | 2 c_2 = c_4 Ni: | c_2 = 3 c_5 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_3 = 1 and solve the system of equations for the remaining coefficients: c_1 = 16 c_2 = 6 c_3 = 1 c_4 = 12 c_5 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 16 NH_3 + 6 NiF_2 ⟶ N_2 + 12 NH_4F + 2 Ni3N

+ ⟶ + + Ni3N

ammonia + nickel(II) fluoride ⟶ nitrogen + ammonium fluoride + Ni3N

Construct the equilibrium constant, K, expression for: NH_3 + NiF_2 ⟶ N_2 + NH_4F + Ni3N 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: 16 NH_3 + 6 NiF_2 ⟶ N_2 + 12 NH_4F + 2 Ni3N 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 | 16 | -16 NiF_2 | 6 | -6 N_2 | 1 | 1 NH_4F | 12 | 12 Ni3N | 2 | 2 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression NH_3 | 16 | -16 | ([NH3])^(-16) NiF_2 | 6 | -6 | ([NiF2])^(-6) N_2 | 1 | 1 | [N2] NH_4F | 12 | 12 | ([NH4F])^12 Ni3N | 2 | 2 | ([Ni3N])^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])^(-16) ([NiF2])^(-6) [N2] ([NH4F])^12 ([Ni3N])^2 = ([N2] ([NH4F])^12 ([Ni3N])^2)/(([NH3])^16 ([NiF2])^6)

Construct the rate of reaction expression for: NH_3 + NiF_2 ⟶ N_2 + NH_4F + Ni3N 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: 16 NH_3 + 6 NiF_2 ⟶ N_2 + 12 NH_4F + 2 Ni3N 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 | 16 | -16 NiF_2 | 6 | -6 N_2 | 1 | 1 NH_4F | 12 | 12 Ni3N | 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 NH_3 | 16 | -16 | -1/16 (Δ[NH3])/(Δt) NiF_2 | 6 | -6 | -1/6 (Δ[NiF2])/(Δt) N_2 | 1 | 1 | (Δ[N2])/(Δt) NH_4F | 12 | 12 | 1/12 (Δ[NH4F])/(Δt) Ni3N | 2 | 2 | 1/2 (Δ[Ni3N])/(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 = -1/16 (Δ[NH3])/(Δt) = -1/6 (Δ[NiF2])/(Δt) = (Δ[N2])/(Δt) = 1/12 (Δ[NH4F])/(Δt) = 1/2 (Δ[Ni3N])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| ammonia | nickel(II) fluoride | nitrogen | ammonium fluoride | Ni3N formula | NH_3 | NiF_2 | N_2 | NH_4F | Ni3N Hill formula | H_3N | F_2Ni | N_2 | FH_4N | NNi3 name | ammonia | nickel(II) fluoride | nitrogen | ammonium fluoride | IUPAC name | ammonia | difluoronickel | molecular nitrogen | |

| ammonia | nickel(II) fluoride | nitrogen | ammonium fluoride | Ni3N molar mass | 17.031 g/mol | 96.6902 g/mol | 28.014 g/mol | 37.037 g/mol | 190.087 g/mol phase | gas (at STP) | solid (at STP) | gas (at STP) | solid (at STP) | melting point | -77.73 °C | 1370 °C | -210 °C | 160 °C | boiling point | -33.33 °C | | -195.79 °C | | density | 6.96×10^-4 g/cm^3 (at 25 °C) | 4.72 g/cm^3 | 0.001251 g/cm^3 (at 0 °C) | 1.0092 g/cm^3 | solubility in water | | | insoluble | | surface tension | 0.0234 N/m | | 0.0066 N/m | | dynamic viscosity | 1.009×10^-5 Pa s (at 25 °C) | | 1.78×10^-5 Pa s (at 25 °C) | | odor | | | odorless | |