H2O + HNO3 + P2O3 = NO + H3PO4

H_2O water + HNO_3 nitric acid + P_2O_3 phosphorus trioxide ⟶ NO nitric oxide + H_3PO_4 phosphoric acid

Balance the chemical equation algebraically: H_2O + HNO_3 + P_2O_3 ⟶ NO + H_3PO_4 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 H_2O + c_2 HNO_3 + c_3 P_2O_3 ⟶ 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, O, N and P: H: | 2 c_1 + c_2 = 3 c_5 O: | c_1 + 3 c_2 + 3 c_3 = c_4 + 4 c_5 N: | c_2 = c_4 P: | 2 c_3 = 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 = 7/3 c_2 = 4/3 c_3 = 1 c_4 = 4/3 c_5 = 2 Multiply by the least common denominator, 3, to eliminate fractional coefficients: c_1 = 7 c_2 = 4 c_3 = 3 c_4 = 4 c_5 = 6 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 7 H_2O + 4 HNO_3 + 3 P_2O_3 ⟶ 4 NO + 6 H_3PO_4

+ + ⟶ +

water + nitric acid + phosphorus trioxide ⟶ nitric oxide + phosphoric acid

Construct the equilibrium constant, K, expression for: H_2O + HNO_3 + P_2O_3 ⟶ 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: 7 H_2O + 4 HNO_3 + 3 P_2O_3 ⟶ 4 NO + 6 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 H_2O | 7 | -7 HNO_3 | 4 | -4 P_2O_3 | 3 | -3 NO | 4 | 4 H_3PO_4 | 6 | 6 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression H_2O | 7 | -7 | ([H2O])^(-7) HNO_3 | 4 | -4 | ([HNO3])^(-4) P_2O_3 | 3 | -3 | ([P2O3])^(-3) NO | 4 | 4 | ([NO])^4 H_3PO_4 | 6 | 6 | ([H3PO4])^6 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 = ([H2O])^(-7) ([HNO3])^(-4) ([P2O3])^(-3) ([NO])^4 ([H3PO4])^6 = (([NO])^4 ([H3PO4])^6)/(([H2O])^7 ([HNO3])^4 ([P2O3])^3)

Construct the rate of reaction expression for: H_2O + HNO_3 + P_2O_3 ⟶ 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: 7 H_2O + 4 HNO_3 + 3 P_2O_3 ⟶ 4 NO + 6 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 H_2O | 7 | -7 HNO_3 | 4 | -4 P_2O_3 | 3 | -3 NO | 4 | 4 H_3PO_4 | 6 | 6 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_2O | 7 | -7 | -1/7 (Δ[H2O])/(Δt) HNO_3 | 4 | -4 | -1/4 (Δ[HNO3])/(Δt) P_2O_3 | 3 | -3 | -1/3 (Δ[P2O3])/(Δt) NO | 4 | 4 | 1/4 (Δ[NO])/(Δt) H_3PO_4 | 6 | 6 | 1/6 (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 = -1/7 (Δ[H2O])/(Δt) = -1/4 (Δ[HNO3])/(Δt) = -1/3 (Δ[P2O3])/(Δt) = 1/4 (Δ[NO])/(Δt) = 1/6 (Δ[H3PO4])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| water | nitric acid | phosphorus trioxide | nitric oxide | phosphoric acid formula | H_2O | HNO_3 | P_2O_3 | NO | H_3PO_4 Hill formula | H_2O | HNO_3 | O_3P_2 | NO | H_3O_4P name | water | nitric acid | phosphorus trioxide | nitric oxide | phosphoric acid