H3PO4 + Li = H2 + Li3PO4

H_3PO_4 phosphoric acid + Li lithium ⟶ H_2 hydrogen + Li_3PO_4 lithium phosphate

Balance the chemical equation algebraically: H_3PO_4 + Li ⟶ H_2 + Li_3PO_4 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 H_3PO_4 + c_2 Li ⟶ c_3 H_2 + c_4 Li_3PO_4 Set the number of atoms in the reactants equal to the number of atoms in the products for H, O, P and Li: H: | 3 c_1 = 2 c_3 O: | 4 c_1 = 4 c_4 P: | c_1 = c_4 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 = 3 c_3 = 3/2 c_4 = 1 Multiply by the least common denominator, 2, to eliminate fractional coefficients: c_1 = 2 c_2 = 6 c_3 = 3 c_4 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 2 H_3PO_4 + 6 Li ⟶ 3 H_2 + 2 Li_3PO_4

+ ⟶ +

phosphoric acid + lithium ⟶ hydrogen + lithium phosphate

Construct the equilibrium constant, K, expression for: H_3PO_4 + Li ⟶ H_2 + 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: 2 H_3PO_4 + 6 Li ⟶ 3 H_2 + 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 H_3PO_4 | 2 | -2 Li | 6 | -6 H_2 | 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 H_3PO_4 | 2 | -2 | ([H3PO4])^(-2) Li | 6 | -6 | ([Li])^(-6) H_2 | 3 | 3 | ([H2])^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 = ([H3PO4])^(-2) ([Li])^(-6) ([H2])^3 ([Li3PO4])^2 = (([H2])^3 ([Li3PO4])^2)/(([H3PO4])^2 ([Li])^6)

Construct the rate of reaction expression for: H_3PO_4 + Li ⟶ H_2 + 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: 2 H_3PO_4 + 6 Li ⟶ 3 H_2 + 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 H_3PO_4 | 2 | -2 Li | 6 | -6 H_2 | 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 H_3PO_4 | 2 | -2 | -1/2 (Δ[H3PO4])/(Δt) Li | 6 | -6 | -1/6 (Δ[Li])/(Δt) H_2 | 3 | 3 | 1/3 (Δ[H2])/(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 = -1/2 (Δ[H3PO4])/(Δt) = -1/6 (Δ[Li])/(Δt) = 1/3 (Δ[H2])/(Δt) = 1/2 (Δ[Li3PO4])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| phosphoric acid | lithium | hydrogen | lithium phosphate formula | H_3PO_4 | Li | H_2 | Li_3PO_4 Hill formula | H_3O_4P | Li | H_2 | Li_3O_4P name | phosphoric acid | lithium | hydrogen | lithium phosphate IUPAC name | phosphoric acid | lithium | molecular hydrogen | trilithium phosphate

| phosphoric acid | lithium | hydrogen | lithium phosphate molar mass | 97.994 g/mol | 6.94 g/mol | 2.016 g/mol | 115.8 g/mol phase | liquid (at STP) | solid (at STP) | gas (at STP) | melting point | 42.4 °C | 180 °C | -259.2 °C | boiling point | 158 °C | 1342 °C | -252.8 °C | density | 1.685 g/cm^3 | 0.534 g/cm^3 | 8.99×10^-5 g/cm^3 (at 0 °C) | solubility in water | very soluble | decomposes | | surface tension | | 0.3975 N/m | | dynamic viscosity | | | 8.9×10^-6 Pa s (at 25 °C) | odor | odorless | | odorless |