P2O5 + LiOH = H2O + Li3PO4

P2O5 + LiOH lithium hydroxide ⟶ H_2O water + Li_3PO_4 lithium phosphate

Balance the chemical equation algebraically: P2O5 + LiOH ⟶ H_2O + Li_3PO_4 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 P2O5 + c_2 LiOH ⟶ c_3 H_2O + c_4 Li_3PO_4 Set the number of atoms in the reactants equal to the number of atoms in the products for P, O, H and Li: P: | 2 c_1 = c_4 O: | 5 c_1 + c_2 = c_3 + 4 c_4 H: | c_2 = 2 c_3 Li: | c_2 = 3 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_1 = 1 and solve the system of equations for the remaining coefficients: c_1 = 1 c_2 = 6 c_3 = 3 c_4 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | P2O5 + 6 LiOH ⟶ 3 H_2O + 2 Li_3PO_4

P2O5 + ⟶ +

P2O5 + lithium hydroxide ⟶ water + lithium phosphate

Construct the equilibrium constant, K, expression for: P2O5 + LiOH ⟶ H_2O + Li_3PO_4 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: P2O5 + 6 LiOH ⟶ 3 H_2O + 2 Li_3PO_4 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 P2O5 | 1 | -1 LiOH | 6 | -6 H_2O | 3 | 3 Li_3PO_4 | 2 | 2 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression P2O5 | 1 | -1 | ([P2O5])^(-1) LiOH | 6 | -6 | ([LiOH])^(-6) H_2O | 3 | 3 | ([H2O])^3 Li_3PO_4 | 2 | 2 | ([Li3PO4])^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 = ([P2O5])^(-1) ([LiOH])^(-6) ([H2O])^3 ([Li3PO4])^2 = (([H2O])^3 ([Li3PO4])^2)/([P2O5] ([LiOH])^6)

Construct the rate of reaction expression for: P2O5 + LiOH ⟶ H_2O + Li_3PO_4 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: P2O5 + 6 LiOH ⟶ 3 H_2O + 2 Li_3PO_4 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 P2O5 | 1 | -1 LiOH | 6 | -6 H_2O | 3 | 3 Li_3PO_4 | 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 P2O5 | 1 | -1 | -(Δ[P2O5])/(Δt) LiOH | 6 | -6 | -1/6 (Δ[LiOH])/(Δt) H_2O | 3 | 3 | 1/3 (Δ[H2O])/(Δt) Li_3PO_4 | 2 | 2 | 1/2 (Δ[Li3PO4])/(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[P2O5])/(Δt) = -1/6 (Δ[LiOH])/(Δt) = 1/3 (Δ[H2O])/(Δt) = 1/2 (Δ[Li3PO4])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| P2O5 | lithium hydroxide | water | lithium phosphate formula | P2O5 | LiOH | H_2O | Li_3PO_4 Hill formula | O5P2 | HLiO | H_2O | Li_3O_4P name | | lithium hydroxide | water | lithium phosphate IUPAC name | | lithium hydroxide | water | trilithium phosphate

| P2O5 | lithium hydroxide | water | lithium phosphate molar mass | 141.94 g/mol | 23.95 g/mol | 18.015 g/mol | 115.8 g/mol phase | | solid (at STP) | liquid (at STP) | melting point | | 462 °C | 0 °C | boiling point | | | 99.9839 °C | density | | 1.46 g/cm^3 | 1 g/cm^3 | surface tension | | | 0.0728 N/m | dynamic viscosity | | | 8.9×10^-4 Pa s (at 25 °C) | odor | | odorless | odorless |