N2 + K = K3N

N_2 nitrogen + K potassium ⟶ K3N

Balance the chemical equation algebraically: N_2 + K ⟶ K3N Add stoichiometric coefficients, c_i, to the reactants and products: c_1 N_2 + c_2 K ⟶ c_3 K3N Set the number of atoms in the reactants equal to the number of atoms in the products for N and K: N: | 2 c_1 = c_3 K: | c_2 = 3 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 = 6 c_3 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | N_2 + 6 K ⟶ 2 K3N

+ ⟶ K3N

nitrogen + potassium ⟶ K3N

Construct the equilibrium constant, K, expression for: N_2 + K ⟶ K3N 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: N_2 + 6 K ⟶ 2 K3N 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 N_2 | 1 | -1 K | 6 | -6 K3N | 2 | 2 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression N_2 | 1 | -1 | ([N2])^(-1) K | 6 | -6 | ([K])^(-6) K3N | 2 | 2 | ([K3N])^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 = ([N2])^(-1) ([K])^(-6) ([K3N])^2 = ([K3N])^2/([N2] ([K])^6)

Construct the rate of reaction expression for: N_2 + K ⟶ K3N 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: N_2 + 6 K ⟶ 2 K3N 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 N_2 | 1 | -1 K | 6 | -6 K3N | 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 N_2 | 1 | -1 | -(Δ[N2])/(Δt) K | 6 | -6 | -1/6 (Δ[K])/(Δt) K3N | 2 | 2 | 1/2 (Δ[K3N])/(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[N2])/(Δt) = -1/6 (Δ[K])/(Δt) = 1/2 (Δ[K3N])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| nitrogen | potassium | K3N formula | N_2 | K | K3N name | nitrogen | potassium | IUPAC name | molecular nitrogen | potassium |

| nitrogen | potassium | K3N molar mass | 28.014 g/mol | 39.0983 g/mol | 131.302 g/mol phase | gas (at STP) | solid (at STP) | melting point | -210 °C | 64 °C | boiling point | -195.79 °C | 760 °C | density | 0.001251 g/cm^3 (at 0 °C) | 0.86 g/cm^3 | solubility in water | insoluble | reacts | surface tension | 0.0066 N/m | | dynamic viscosity | 1.78×10^-5 Pa s (at 25 °C) | | odor | odorless | |