Fe + K3PO4 = K + Fe3(PO4)2

Fe iron + K3PO4 ⟶ K potassium + Fe3(PO4)2

Balance the chemical equation algebraically: Fe + K3PO4 ⟶ K + Fe3(PO4)2 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 Fe + c_2 K3PO4 ⟶ c_3 K + c_4 Fe3(PO4)2 Set the number of atoms in the reactants equal to the number of atoms in the products for Fe, K, P and O: Fe: | c_1 = 3 c_4 K: | 3 c_2 = c_3 P: | c_2 = 2 c_4 O: | 4 c_2 = 8 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_4 = 1 and solve the system of equations for the remaining coefficients: c_1 = 3 c_2 = 2 c_3 = 6 c_4 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 3 Fe + 2 K3PO4 ⟶ 6 K + Fe3(PO4)2

+ K3PO4 ⟶ + Fe3(PO4)2

iron + K3PO4 ⟶ potassium + Fe3(PO4)2

Construct the equilibrium constant, K, expression for: Fe + K3PO4 ⟶ K + Fe3(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: 3 Fe + 2 K3PO4 ⟶ 6 K + Fe3(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 Fe | 3 | -3 K3PO4 | 2 | -2 K | 6 | 6 Fe3(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 Fe | 3 | -3 | ([Fe])^(-3) K3PO4 | 2 | -2 | ([K3PO4])^(-2) K | 6 | 6 | ([K])^6 Fe3(PO4)2 | 1 | 1 | [Fe3(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 = ([Fe])^(-3) ([K3PO4])^(-2) ([K])^6 [Fe3(PO4)2] = (([K])^6 [Fe3(PO4)2])/(([Fe])^3 ([K3PO4])^2)

Construct the rate of reaction expression for: Fe + K3PO4 ⟶ K + Fe3(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: 3 Fe + 2 K3PO4 ⟶ 6 K + Fe3(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 Fe | 3 | -3 K3PO4 | 2 | -2 K | 6 | 6 Fe3(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 Fe | 3 | -3 | -1/3 (Δ[Fe])/(Δt) K3PO4 | 2 | -2 | -1/2 (Δ[K3PO4])/(Δt) K | 6 | 6 | 1/6 (Δ[K])/(Δt) Fe3(PO4)2 | 1 | 1 | (Δ[Fe3(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 = -1/3 (Δ[Fe])/(Δt) = -1/2 (Δ[K3PO4])/(Δt) = 1/6 (Δ[K])/(Δt) = (Δ[Fe3(PO4)2])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| iron | K3PO4 | potassium | Fe3(PO4)2 formula | Fe | K3PO4 | K | Fe3(PO4)2 Hill formula | Fe | K3O4P | K | Fe3O8P2 name | iron | | potassium |

| iron | K3PO4 | potassium | Fe3(PO4)2 molar mass | 55.845 g/mol | 212.26 g/mol | 39.0983 g/mol | 357.47 g/mol phase | solid (at STP) | | solid (at STP) | melting point | 1535 °C | | 64 °C | boiling point | 2750 °C | | 760 °C | density | 7.874 g/cm^3 | | 0.86 g/cm^3 | solubility in water | insoluble | | reacts |