H3PO3 = H3PO4 + PH3

HP(O)(OH)_2 phosphorous acid ⟶ H_3PO_4 phosphoric acid + PH_3 phosphine

Balance the chemical equation algebraically: HP(O)(OH)_2 ⟶ H_3PO_4 + PH_3 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 HP(O)(OH)_2 ⟶ c_2 H_3PO_4 + c_3 PH_3 Set the number of atoms in the reactants equal to the number of atoms in the products for H, O and P: H: | 3 c_1 = 3 c_2 + 3 c_3 O: | 3 c_1 = 4 c_2 P: | c_1 = c_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_3 = 1 and solve the system of equations for the remaining coefficients: c_1 = 4 c_2 = 3 c_3 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 4 HP(O)(OH)_2 ⟶ 3 H_3PO_4 + PH_3

⟶ +

phosphorous acid ⟶ phosphoric acid + phosphine

Construct the equilibrium constant, K, expression for: HP(O)(OH)_2 ⟶ H_3PO_4 + PH_3 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: 4 HP(O)(OH)_2 ⟶ 3 H_3PO_4 + PH_3 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 HP(O)(OH)_2 | 4 | -4 H_3PO_4 | 3 | 3 PH_3 | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression HP(O)(OH)_2 | 4 | -4 | ([HP(O)(OH)2])^(-4) H_3PO_4 | 3 | 3 | ([H3PO4])^3 PH_3 | 1 | 1 | [PH3] 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 = ([HP(O)(OH)2])^(-4) ([H3PO4])^3 [PH3] = (([H3PO4])^3 [PH3])/([HP(O)(OH)2])^4

Construct the rate of reaction expression for: HP(O)(OH)_2 ⟶ H_3PO_4 + PH_3 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: 4 HP(O)(OH)_2 ⟶ 3 H_3PO_4 + PH_3 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 HP(O)(OH)_2 | 4 | -4 H_3PO_4 | 3 | 3 PH_3 | 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 HP(O)(OH)_2 | 4 | -4 | -1/4 (Δ[HP(O)(OH)2])/(Δt) H_3PO_4 | 3 | 3 | 1/3 (Δ[H3PO4])/(Δt) PH_3 | 1 | 1 | (Δ[PH3])/(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 = -1/4 (Δ[HP(O)(OH)2])/(Δt) = 1/3 (Δ[H3PO4])/(Δt) = (Δ[PH3])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| phosphorous acid | phosphoric acid | phosphine formula | HP(O)(OH)_2 | H_3PO_4 | PH_3 Hill formula | H_3O_3P | H_3O_4P | H_3P name | phosphorous acid | phosphoric acid | phosphine

| phosphorous acid | phosphoric acid | phosphine molar mass | 81.995 g/mol | 97.994 g/mol | 33.998 g/mol phase | solid (at STP) | liquid (at STP) | gas (at STP) melting point | 74 °C | 42.4 °C | -132.8 °C boiling point | | 158 °C | -87.5 °C density | 1.597 g/cm^3 | 1.685 g/cm^3 | 0.00139 g/cm^3 (at 25 °C) solubility in water | | very soluble | slightly soluble dynamic viscosity | | | 1.1×10^-5 Pa s (at 0 °C) odor | | odorless |