HNO3 + Sb = H2O + NO + Sb2O3

HNO_3 nitric acid + Sb gray antimony ⟶ H_2O water + NO nitric oxide + Sb_2O_3 antimony trioxide

Balance the chemical equation algebraically: HNO_3 + Sb ⟶ H_2O + NO + Sb_2O_3 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 HNO_3 + c_2 Sb ⟶ c_3 H_2O + c_4 NO + c_5 Sb_2O_3 Set the number of atoms in the reactants equal to the number of atoms in the products for H, N, O and Sb: H: | c_1 = 2 c_3 N: | c_1 = c_4 O: | 3 c_1 = c_3 + c_4 + 3 c_5 Sb: | c_2 = 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 = 2 c_3 = 1 c_4 = 2 c_5 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 2 HNO_3 + 2 Sb ⟶ H_2O + 2 NO + Sb_2O_3

+ ⟶ + +

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

Construct the equilibrium constant, K, expression for: HNO_3 + Sb ⟶ H_2O + NO + Sb_2O_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: 2 HNO_3 + 2 Sb ⟶ H_2O + 2 NO + Sb_2O_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 HNO_3 | 2 | -2 Sb | 2 | -2 H_2O | 1 | 1 NO | 2 | 2 Sb_2O_3 | 1 | 1 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) Sb | 2 | -2 | ([Sb])^(-2) H_2O | 1 | 1 | [H2O] NO | 2 | 2 | ([NO])^2 Sb_2O_3 | 1 | 1 | [Sb2O3] 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) ([Sb])^(-2) [H2O] ([NO])^2 [Sb2O3] = ([H2O] ([NO])^2 [Sb2O3])/(([HNO3])^2 ([Sb])^2)

Construct the rate of reaction expression for: HNO_3 + Sb ⟶ H_2O + NO + Sb_2O_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: 2 HNO_3 + 2 Sb ⟶ H_2O + 2 NO + Sb_2O_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 HNO_3 | 2 | -2 Sb | 2 | -2 H_2O | 1 | 1 NO | 2 | 2 Sb_2O_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 HNO_3 | 2 | -2 | -1/2 (Δ[HNO3])/(Δt) Sb | 2 | -2 | -1/2 (Δ[Sb])/(Δt) H_2O | 1 | 1 | (Δ[H2O])/(Δt) NO | 2 | 2 | 1/2 (Δ[NO])/(Δt) Sb_2O_3 | 1 | 1 | (Δ[Sb2O3])/(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) = -1/2 (Δ[Sb])/(Δt) = (Δ[H2O])/(Δt) = 1/2 (Δ[NO])/(Δt) = (Δ[Sb2O3])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| nitric acid | gray antimony | water | nitric oxide | antimony trioxide formula | HNO_3 | Sb | H_2O | NO | Sb_2O_3 Hill formula | HNO_3 | Sb | H_2O | NO | O_3Sb_2 name | nitric acid | gray antimony | water | nitric oxide | antimony trioxide IUPAC name | nitric acid | antimony | water | nitric oxide | oxo-oxostibanyloxystibane