P2O5 + Ba(OH)2 = H2O + Ba3(PO4)2

P2O5 + Ba(OH)_2 barium hydroxide ⟶ H_2O water + Ba3(PO4)2

Balance the chemical equation algebraically: P2O5 + Ba(OH)_2 ⟶ H_2O + Ba3(PO4)2 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 P2O5 + c_2 Ba(OH)_2 ⟶ c_3 H_2O + c_4 Ba3(PO4)2 Set the number of atoms in the reactants equal to the number of atoms in the products for P, O, Ba and H: P: | 2 c_1 = 2 c_4 O: | 5 c_1 + 2 c_2 = c_3 + 8 c_4 Ba: | c_2 = 3 c_4 H: | 2 c_2 = 2 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 = 3 c_4 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | P2O5 + 3 Ba(OH)_2 ⟶ 3 H_2O + Ba3(PO4)2

P2O5 + ⟶ + Ba3(PO4)2

P2O5 + barium hydroxide ⟶ water + Ba3(PO4)2

Construct the equilibrium constant, K, expression for: P2O5 + Ba(OH)_2 ⟶ H_2O + Ba3(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: P2O5 + 3 Ba(OH)_2 ⟶ 3 H_2O + Ba3(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 P2O5 | 1 | -1 Ba(OH)_2 | 3 | -3 H_2O | 3 | 3 Ba3(PO4)2 | 1 | 1 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) Ba(OH)_2 | 3 | -3 | ([Ba(OH)2])^(-3) H_2O | 3 | 3 | ([H2O])^3 Ba3(PO4)2 | 1 | 1 | [Ba3(PO4)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) ([Ba(OH)2])^(-3) ([H2O])^3 [Ba3(PO4)2] = (([H2O])^3 [Ba3(PO4)2])/([P2O5] ([Ba(OH)2])^3)

Construct the rate of reaction expression for: P2O5 + Ba(OH)_2 ⟶ H_2O + Ba3(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: P2O5 + 3 Ba(OH)_2 ⟶ 3 H_2O + Ba3(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 P2O5 | 1 | -1 Ba(OH)_2 | 3 | -3 H_2O | 3 | 3 Ba3(PO4)2 | 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 P2O5 | 1 | -1 | -(Δ[P2O5])/(Δt) Ba(OH)_2 | 3 | -3 | -1/3 (Δ[Ba(OH)2])/(Δt) H_2O | 3 | 3 | 1/3 (Δ[H2O])/(Δt) Ba3(PO4)2 | 1 | 1 | (Δ[Ba3(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 = -(Δ[P2O5])/(Δt) = -1/3 (Δ[Ba(OH)2])/(Δt) = 1/3 (Δ[H2O])/(Δt) = (Δ[Ba3(PO4)2])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| P2O5 | barium hydroxide | water | Ba3(PO4)2 formula | P2O5 | Ba(OH)_2 | H_2O | Ba3(PO4)2 Hill formula | O5P2 | BaH_2O_2 | H_2O | Ba3O8P2 name | | barium hydroxide | water | IUPAC name | | barium(+2) cation dihydroxide | water |

| P2O5 | barium hydroxide | water | Ba3(PO4)2 molar mass | 141.94 g/mol | 171.34 g/mol | 18.015 g/mol | 601.92 g/mol phase | | solid (at STP) | liquid (at STP) | melting point | | 300 °C | 0 °C | boiling point | | | 99.9839 °C | density | | 2.2 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 |