HNO3 + P4 = NO + NO2 + H3PO4

HNO_3 nitric acid + P_4 white phosphorus ⟶ NO nitric oxide + NO_2 nitrogen dioxide + H_3PO_4 phosphoric acid

Balance the chemical equation algebraically: HNO_3 + P_4 ⟶ NO + NO_2 + H_3PO_4 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 HNO_3 + c_2 P_4 ⟶ c_3 NO + c_4 NO_2 + 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_5 N: | c_1 = c_3 + c_4 O: | 3 c_1 = c_3 + 2 c_4 + 4 c_5 P: | 4 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_2 = 1 and solve the system of equations for the remaining coefficients: c_1 = 12 c_2 = 1 c_3 = 4 c_4 = 8 c_5 = 4 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 12 HNO_3 + P_4 ⟶ 4 NO + 8 NO_2 + 4 H_3PO_4

+ ⟶ + +

nitric acid + white phosphorus ⟶ nitric oxide + nitrogen dioxide + phosphoric acid

Construct the equilibrium constant, K, expression for: HNO_3 + P_4 ⟶ NO + NO_2 + 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: 12 HNO_3 + P_4 ⟶ 4 NO + 8 NO_2 + 4 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 | 12 | -12 P_4 | 1 | -1 NO | 4 | 4 NO_2 | 8 | 8 H_3PO_4 | 4 | 4 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression HNO_3 | 12 | -12 | ([HNO3])^(-12) P_4 | 1 | -1 | ([P4])^(-1) NO | 4 | 4 | ([NO])^4 NO_2 | 8 | 8 | ([NO2])^8 H_3PO_4 | 4 | 4 | ([H3PO4])^4 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])^(-12) ([P4])^(-1) ([NO])^4 ([NO2])^8 ([H3PO4])^4 = (([NO])^4 ([NO2])^8 ([H3PO4])^4)/(([HNO3])^12 [P4])

Construct the rate of reaction expression for: HNO_3 + P_4 ⟶ NO + NO_2 + 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: 12 HNO_3 + P_4 ⟶ 4 NO + 8 NO_2 + 4 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 | 12 | -12 P_4 | 1 | -1 NO | 4 | 4 NO_2 | 8 | 8 H_3PO_4 | 4 | 4 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 | 12 | -12 | -1/12 (Δ[HNO3])/(Δt) P_4 | 1 | -1 | -(Δ[P4])/(Δt) NO | 4 | 4 | 1/4 (Δ[NO])/(Δt) NO_2 | 8 | 8 | 1/8 (Δ[NO2])/(Δt) H_3PO_4 | 4 | 4 | 1/4 (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 = -1/12 (Δ[HNO3])/(Δt) = -(Δ[P4])/(Δt) = 1/4 (Δ[NO])/(Δt) = 1/8 (Δ[NO2])/(Δt) = 1/4 (Δ[H3PO4])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| nitric acid | white phosphorus | nitric oxide | nitrogen dioxide | phosphoric acid formula | HNO_3 | P_4 | NO | NO_2 | H_3PO_4 Hill formula | HNO_3 | P_4 | NO | NO_2 | H_3O_4P name | nitric acid | white phosphorus | nitric oxide | nitrogen dioxide | phosphoric acid IUPAC name | nitric acid | tetraphosphorus | nitric oxide | Nitrogen dioxide | phosphoric acid