H3PO4 + P2O5 + SrO = H2O + Sr3(PO4)2

H_3PO_4 phosphoric acid + P2O5 + SrO strontium oxide ⟶ H_2O water + Sr3(PO4)2

Balance the chemical equation algebraically: H_3PO_4 + P2O5 + SrO ⟶ H_2O + Sr3(PO4)2 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 H_3PO_4 + c_2 P2O5 + c_3 SrO ⟶ c_4 H_2O + c_5 Sr3(PO4)2 Set the number of atoms in the reactants equal to the number of atoms in the products for H, O, P and Sr: H: | 3 c_1 = 2 c_4 O: | 4 c_1 + 5 c_2 + c_3 = c_4 + 8 c_5 P: | c_1 + 2 c_2 = 2 c_5 Sr: | c_3 = 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_2 = 1 and solve the system of equations for the remaining coefficients: c_2 = 1 c_3 = (3 c_1)/2 + 3 c_4 = (3 c_1)/2 c_5 = c_1/2 + 1 The resulting system of equations is still underdetermined, so an additional coefficient must be set arbitrarily. Set c_1 = 2 and solve for the remaining coefficients: c_1 = 2 c_2 = 1 c_3 = 6 c_4 = 3 c_5 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 2 H_3PO_4 + P2O5 + 6 SrO ⟶ 3 H_2O + 2 Sr3(PO4)2

+ P2O5 + ⟶ + Sr3(PO4)2

phosphoric acid + P2O5 + strontium oxide ⟶ water + Sr3(PO4)2

Construct the equilibrium constant, K, expression for: H_3PO_4 + P2O5 + SrO ⟶ H_2O + Sr3(PO4)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 H_3PO_4 + P2O5 + 6 SrO ⟶ 3 H_2O + 2 Sr3(PO4)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 H_3PO_4 | 2 | -2 P2O5 | 1 | -1 SrO | 6 | -6 H_2O | 3 | 3 Sr3(PO4)2 | 2 | 2 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression H_3PO_4 | 2 | -2 | ([H3PO4])^(-2) P2O5 | 1 | -1 | ([P2O5])^(-1) SrO | 6 | -6 | ([SrO])^(-6) H_2O | 3 | 3 | ([H2O])^3 Sr3(PO4)2 | 2 | 2 | ([Sr3(PO4)2])^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 = ([H3PO4])^(-2) ([P2O5])^(-1) ([SrO])^(-6) ([H2O])^3 ([Sr3(PO4)2])^2 = (([H2O])^3 ([Sr3(PO4)2])^2)/(([H3PO4])^2 [P2O5] ([SrO])^6)

Construct the rate of reaction expression for: H_3PO_4 + P2O5 + SrO ⟶ H_2O + Sr3(PO4)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 H_3PO_4 + P2O5 + 6 SrO ⟶ 3 H_2O + 2 Sr3(PO4)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 H_3PO_4 | 2 | -2 P2O5 | 1 | -1 SrO | 6 | -6 H_2O | 3 | 3 Sr3(PO4)2 | 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 H_3PO_4 | 2 | -2 | -1/2 (Δ[H3PO4])/(Δt) P2O5 | 1 | -1 | -(Δ[P2O5])/(Δt) SrO | 6 | -6 | -1/6 (Δ[SrO])/(Δt) H_2O | 3 | 3 | 1/3 (Δ[H2O])/(Δt) Sr3(PO4)2 | 2 | 2 | 1/2 (Δ[Sr3(PO4)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[H3PO4])/(Δt) = -(Δ[P2O5])/(Δt) = -1/6 (Δ[SrO])/(Δt) = 1/3 (Δ[H2O])/(Δt) = 1/2 (Δ[Sr3(PO4)2])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| phosphoric acid | P2O5 | strontium oxide | water | Sr3(PO4)2 formula | H_3PO_4 | P2O5 | SrO | H_2O | Sr3(PO4)2 Hill formula | H_3O_4P | O5P2 | OSr | H_2O | O8P2Sr3 name | phosphoric acid | | strontium oxide | water | IUPAC name | phosphoric acid | | oxostrontium | water |

| phosphoric acid | P2O5 | strontium oxide | water | Sr3(PO4)2 molar mass | 97.994 g/mol | 141.94 g/mol | 103.6 g/mol | 18.015 g/mol | 452.8 g/mol phase | liquid (at STP) | | solid (at STP) | liquid (at STP) | melting point | 42.4 °C | | 2430 °C | 0 °C | boiling point | 158 °C | | | 99.9839 °C | density | 1.685 g/cm^3 | | 4.7 g/cm^3 | 1 g/cm^3 | solubility in water | very soluble | | slightly soluble | | surface tension | | | | 0.0728 N/m | dynamic viscosity | | | | 8.9×10^-4 Pa s (at 25 °C) | odor | odorless | | | odorless |