H2O + HNO3 + Sb = NO + H3SbO4

H_2O water + HNO_3 nitric acid + Sb gray antimony ⟶ NO nitric oxide + H3SbO4

Balance the chemical equation algebraically: H_2O + HNO_3 + Sb ⟶ NO + H3SbO4 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 H_2O + c_2 HNO_3 + c_3 Sb ⟶ c_4 NO + c_5 H3SbO4 Set the number of atoms in the reactants equal to the number of atoms in the products for H, O, N and Sb: H: | 2 c_1 + c_2 = 3 c_5 O: | c_1 + 3 c_2 = c_4 + 4 c_5 N: | c_2 = c_4 Sb: | 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_1 = 1 and solve the system of equations for the remaining coefficients: c_1 = 1 c_2 = 5/2 c_3 = 3/2 c_4 = 5/2 c_5 = 3/2 Multiply by the least common denominator, 2, to eliminate fractional coefficients: c_1 = 2 c_2 = 5 c_3 = 3 c_4 = 5 c_5 = 3 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 2 H_2O + 5 HNO_3 + 3 Sb ⟶ 5 NO + 3 H3SbO4

+ + ⟶ + H3SbO4

water + nitric acid + gray antimony ⟶ nitric oxide + H3SbO4

Construct the equilibrium constant, K, expression for: H_2O + HNO_3 + Sb ⟶ NO + H3SbO4 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 H_2O + 5 HNO_3 + 3 Sb ⟶ 5 NO + 3 H3SbO4 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 | 2 | -2 HNO_3 | 5 | -5 Sb | 3 | -3 NO | 5 | 5 H3SbO4 | 3 | 3 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression H_2O | 2 | -2 | ([H2O])^(-2) HNO_3 | 5 | -5 | ([HNO3])^(-5) Sb | 3 | -3 | ([Sb])^(-3) NO | 5 | 5 | ([NO])^5 H3SbO4 | 3 | 3 | ([H3SbO4])^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 = ([H2O])^(-2) ([HNO3])^(-5) ([Sb])^(-3) ([NO])^5 ([H3SbO4])^3 = (([NO])^5 ([H3SbO4])^3)/(([H2O])^2 ([HNO3])^5 ([Sb])^3)

Construct the rate of reaction expression for: H_2O + HNO_3 + Sb ⟶ NO + H3SbO4 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 H_2O + 5 HNO_3 + 3 Sb ⟶ 5 NO + 3 H3SbO4 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 | 2 | -2 HNO_3 | 5 | -5 Sb | 3 | -3 NO | 5 | 5 H3SbO4 | 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 H_2O | 2 | -2 | -1/2 (Δ[H2O])/(Δt) HNO_3 | 5 | -5 | -1/5 (Δ[HNO3])/(Δt) Sb | 3 | -3 | -1/3 (Δ[Sb])/(Δt) NO | 5 | 5 | 1/5 (Δ[NO])/(Δt) H3SbO4 | 3 | 3 | 1/3 (Δ[H3SbO4])/(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[H2O])/(Δt) = -1/5 (Δ[HNO3])/(Δt) = -1/3 (Δ[Sb])/(Δt) = 1/5 (Δ[NO])/(Δt) = 1/3 (Δ[H3SbO4])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| water | nitric acid | gray antimony | nitric oxide | H3SbO4 formula | H_2O | HNO_3 | Sb | NO | H3SbO4 Hill formula | H_2O | HNO_3 | Sb | NO | H3O4Sb name | water | nitric acid | gray antimony | nitric oxide | IUPAC name | water | nitric acid | antimony | nitric oxide |

| water | nitric acid | gray antimony | nitric oxide | H3SbO4 molar mass | 18.015 g/mol | 63.012 g/mol | 121.76 g/mol | 30.006 g/mol | 188.78 g/mol phase | liquid (at STP) | liquid (at STP) | solid (at STP) | gas (at STP) | melting point | 0 °C | -41.6 °C | 630 °C | -163.6 °C | boiling point | 99.9839 °C | 83 °C | 1587 °C | -151.7 °C | density | 1 g/cm^3 | 1.5129 g/cm^3 | 6.69 g/cm^3 | 0.001226 g/cm^3 (at 25 °C) | solubility in water | | miscible | | | surface tension | 0.0728 N/m | | | | dynamic viscosity | 8.9×10^-4 Pa s (at 25 °C) | 7.6×10^-4 Pa s (at 25 °C) | | 1.911×10^-5 Pa s (at 25 °C) | odor | odorless | | | |