HNO3 + H3PO3 = H2O + NO + H3PO4

HNO_3 nitric acid + HP(O)(OH)_2 phosphorous acid ⟶ H_2O water + NO nitric oxide + H_3PO_4 phosphoric acid

Balance the chemical equation algebraically: HNO_3 + HP(O)(OH)_2 ⟶ H_2O + NO + H_3PO_4 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 HNO_3 + c_2 HP(O)(OH)_2 ⟶ c_3 H_2O + c_4 NO + c_5 H_3PO_4 Set the number of atoms in the reactants equal to the number of atoms in the products for H, N, O and P: H: | c_1 + 3 c_2 = 2 c_3 + 3 c_5 N: | c_1 = c_4 O: | 3 c_1 + 3 c_2 = c_3 + c_4 + 4 c_5 P: | c_2 = 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_3 = 1 and solve the system of equations for the remaining coefficients: c_1 = 2 c_2 = 3 c_3 = 1 c_4 = 2 c_5 = 3 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 2 HNO_3 + 3 HP(O)(OH)_2 ⟶ H_2O + 2 NO + 3 H_3PO_4

+ ⟶ + +

nitric acid + phosphorous acid ⟶ water + nitric oxide + phosphoric acid

Construct the equilibrium constant, K, expression for: HNO_3 + HP(O)(OH)_2 ⟶ H_2O + NO + H_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 HNO_3 + 3 HP(O)(OH)_2 ⟶ H_2O + 2 NO + 3 H_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 HNO_3 | 2 | -2 HP(O)(OH)_2 | 3 | -3 H_2O | 1 | 1 NO | 2 | 2 H_3PO_4 | 3 | 3 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression HNO_3 | 2 | -2 | ([HNO3])^(-2) HP(O)(OH)_2 | 3 | -3 | ([HP(O)(OH)2])^(-3) H_2O | 1 | 1 | [H2O] NO | 2 | 2 | ([NO])^2 H_3PO_4 | 3 | 3 | ([H3PO4])^3 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 = ([HNO3])^(-2) ([HP(O)(OH)2])^(-3) [H2O] ([NO])^2 ([H3PO4])^3 = ([H2O] ([NO])^2 ([H3PO4])^3)/(([HNO3])^2 ([HP(O)(OH)2])^3)

Construct the rate of reaction expression for: HNO_3 + HP(O)(OH)_2 ⟶ H_2O + NO + H_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 HNO_3 + 3 HP(O)(OH)_2 ⟶ H_2O + 2 NO + 3 H_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 HNO_3 | 2 | -2 HP(O)(OH)_2 | 3 | -3 H_2O | 1 | 1 NO | 2 | 2 H_3PO_4 | 3 | 3 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 HNO_3 | 2 | -2 | -1/2 (Δ[HNO3])/(Δt) HP(O)(OH)_2 | 3 | -3 | -1/3 (Δ[HP(O)(OH)2])/(Δt) H_2O | 1 | 1 | (Δ[H2O])/(Δt) NO | 2 | 2 | 1/2 (Δ[NO])/(Δt) H_3PO_4 | 3 | 3 | 1/3 (Δ[H3PO4])/(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[HNO3])/(Δt) = -1/3 (Δ[HP(O)(OH)2])/(Δt) = (Δ[H2O])/(Δt) = 1/2 (Δ[NO])/(Δt) = 1/3 (Δ[H3PO4])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| nitric acid | phosphorous acid | water | nitric oxide | phosphoric acid formula | HNO_3 | HP(O)(OH)_2 | H_2O | NO | H_3PO_4 Hill formula | HNO_3 | H_3O_3P | H_2O | NO | H_3O_4P name | nitric acid | phosphorous acid | water | nitric oxide | phosphoric acid