P + Li = Li3P

P red phosphorus + Li lithium ⟶ Li3P

Balance the chemical equation algebraically: P + Li ⟶ Li3P Add stoichiometric coefficients, c_i, to the reactants and products: c_1 P + c_2 Li ⟶ c_3 Li3P Set the number of atoms in the reactants equal to the number of atoms in the products for P and Li: P: | c_1 = c_3 Li: | 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 = 3 c_3 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | P + 3 Li ⟶ Li3P

+ ⟶ Li3P

red phosphorus + lithium ⟶ Li3P

Construct the equilibrium constant, K, expression for: P + Li ⟶ Li3P 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: P + 3 Li ⟶ Li3P 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 P | 1 | -1 Li | 3 | -3 Li3P | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression P | 1 | -1 | ([P])^(-1) Li | 3 | -3 | ([Li])^(-3) Li3P | 1 | 1 | [Li3P] 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 = ([P])^(-1) ([Li])^(-3) [Li3P] = ([Li3P])/([P] ([Li])^3)

Construct the rate of reaction expression for: P + Li ⟶ Li3P 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: P + 3 Li ⟶ Li3P 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 P | 1 | -1 Li | 3 | -3 Li3P | 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 P | 1 | -1 | -(Δ[P])/(Δt) Li | 3 | -3 | -1/3 (Δ[Li])/(Δt) Li3P | 1 | 1 | (Δ[Li3P])/(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[P])/(Δt) = -1/3 (Δ[Li])/(Δt) = (Δ[Li3P])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| red phosphorus | lithium | Li3P formula | P | Li | Li3P name | red phosphorus | lithium | IUPAC name | phosphorus | lithium |

| red phosphorus | lithium | Li3P molar mass | 30.973761998 g/mol | 6.94 g/mol | 51.8 g/mol phase | solid (at STP) | solid (at STP) | melting point | 579.2 °C | 180 °C | boiling point | | 1342 °C | density | 2.16 g/cm^3 | 0.534 g/cm^3 | solubility in water | insoluble | decomposes | surface tension | | 0.3975 N/m | dynamic viscosity | 7.6×10^-4 Pa s (at 20.2 °C) | |