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HNO3 + Sb2S3 = H2O + H2SO4 + NO + HSbO3

Input interpretation

HNO_3 nitric acid + Sb_2S_3 antimony(III) sulfide ⟶ H_2O water + H_2SO_4 sulfuric acid + NO nitric oxide + HSbO3
HNO_3 nitric acid + Sb_2S_3 antimony(III) sulfide ⟶ H_2O water + H_2SO_4 sulfuric acid + NO nitric oxide + HSbO3

Balanced equation

Balance the chemical equation algebraically: HNO_3 + Sb_2S_3 ⟶ H_2O + H_2SO_4 + NO + HSbO3 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 HNO_3 + c_2 Sb_2S_3 ⟶ c_3 H_2O + c_4 H_2SO_4 + c_5 NO + c_6 HSbO3 Set the number of atoms in the reactants equal to the number of atoms in the products for H, N, O, S and Sb: H: | c_1 = 2 c_3 + 2 c_4 + c_6 N: | c_1 = c_5 O: | 3 c_1 = c_3 + 4 c_4 + c_5 + 3 c_6 S: | 3 c_2 = c_4 Sb: | 2 c_2 = c_6 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 = 14 c_2 = 3/2 c_3 = 1 c_4 = 9/2 c_5 = 14 c_6 = 3 Multiply by the least common denominator, 2, to eliminate fractional coefficients: c_1 = 28 c_2 = 3 c_3 = 2 c_4 = 9 c_5 = 28 c_6 = 6 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 28 HNO_3 + 3 Sb_2S_3 ⟶ 2 H_2O + 9 H_2SO_4 + 28 NO + 6 HSbO3
Balance the chemical equation algebraically: HNO_3 + Sb_2S_3 ⟶ H_2O + H_2SO_4 + NO + HSbO3 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 HNO_3 + c_2 Sb_2S_3 ⟶ c_3 H_2O + c_4 H_2SO_4 + c_5 NO + c_6 HSbO3 Set the number of atoms in the reactants equal to the number of atoms in the products for H, N, O, S and Sb: H: | c_1 = 2 c_3 + 2 c_4 + c_6 N: | c_1 = c_5 O: | 3 c_1 = c_3 + 4 c_4 + c_5 + 3 c_6 S: | 3 c_2 = c_4 Sb: | 2 c_2 = c_6 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 = 14 c_2 = 3/2 c_3 = 1 c_4 = 9/2 c_5 = 14 c_6 = 3 Multiply by the least common denominator, 2, to eliminate fractional coefficients: c_1 = 28 c_2 = 3 c_3 = 2 c_4 = 9 c_5 = 28 c_6 = 6 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 28 HNO_3 + 3 Sb_2S_3 ⟶ 2 H_2O + 9 H_2SO_4 + 28 NO + 6 HSbO3

Structures

+ ⟶ + + + HSbO3
+ ⟶ + + + HSbO3

Names

nitric acid + antimony(III) sulfide ⟶ water + sulfuric acid + nitric oxide + HSbO3
nitric acid + antimony(III) sulfide ⟶ water + sulfuric acid + nitric oxide + HSbO3

Equilibrium constant

Construct the equilibrium constant, K, expression for: HNO_3 + Sb_2S_3 ⟶ H_2O + H_2SO_4 + NO + HSbO3 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: 28 HNO_3 + 3 Sb_2S_3 ⟶ 2 H_2O + 9 H_2SO_4 + 28 NO + 6 HSbO3 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 | 28 | -28 Sb_2S_3 | 3 | -3 H_2O | 2 | 2 H_2SO_4 | 9 | 9 NO | 28 | 28 HSbO3 | 6 | 6 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression HNO_3 | 28 | -28 | ([HNO3])^(-28) Sb_2S_3 | 3 | -3 | ([Sb2S3])^(-3) H_2O | 2 | 2 | ([H2O])^2 H_2SO_4 | 9 | 9 | ([H2SO4])^9 NO | 28 | 28 | ([NO])^28 HSbO3 | 6 | 6 | ([HSbO3])^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 = ([HNO3])^(-28) ([Sb2S3])^(-3) ([H2O])^2 ([H2SO4])^9 ([NO])^28 ([HSbO3])^6 = (([H2O])^2 ([H2SO4])^9 ([NO])^28 ([HSbO3])^6)/(([HNO3])^28 ([Sb2S3])^3)
Construct the equilibrium constant, K, expression for: HNO_3 + Sb_2S_3 ⟶ H_2O + H_2SO_4 + NO + HSbO3 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: 28 HNO_3 + 3 Sb_2S_3 ⟶ 2 H_2O + 9 H_2SO_4 + 28 NO + 6 HSbO3 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 | 28 | -28 Sb_2S_3 | 3 | -3 H_2O | 2 | 2 H_2SO_4 | 9 | 9 NO | 28 | 28 HSbO3 | 6 | 6 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression HNO_3 | 28 | -28 | ([HNO3])^(-28) Sb_2S_3 | 3 | -3 | ([Sb2S3])^(-3) H_2O | 2 | 2 | ([H2O])^2 H_2SO_4 | 9 | 9 | ([H2SO4])^9 NO | 28 | 28 | ([NO])^28 HSbO3 | 6 | 6 | ([HSbO3])^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 = ([HNO3])^(-28) ([Sb2S3])^(-3) ([H2O])^2 ([H2SO4])^9 ([NO])^28 ([HSbO3])^6 = (([H2O])^2 ([H2SO4])^9 ([NO])^28 ([HSbO3])^6)/(([HNO3])^28 ([Sb2S3])^3)

Rate of reaction

Construct the rate of reaction expression for: HNO_3 + Sb_2S_3 ⟶ H_2O + H_2SO_4 + NO + HSbO3 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: 28 HNO_3 + 3 Sb_2S_3 ⟶ 2 H_2O + 9 H_2SO_4 + 28 NO + 6 HSbO3 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 | 28 | -28 Sb_2S_3 | 3 | -3 H_2O | 2 | 2 H_2SO_4 | 9 | 9 NO | 28 | 28 HSbO3 | 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 HNO_3 | 28 | -28 | -1/28 (Δ[HNO3])/(Δt) Sb_2S_3 | 3 | -3 | -1/3 (Δ[Sb2S3])/(Δt) H_2O | 2 | 2 | 1/2 (Δ[H2O])/(Δt) H_2SO_4 | 9 | 9 | 1/9 (Δ[H2SO4])/(Δt) NO | 28 | 28 | 1/28 (Δ[NO])/(Δt) HSbO3 | 6 | 6 | 1/6 (Δ[HSbO3])/(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 = -1/28 (Δ[HNO3])/(Δt) = -1/3 (Δ[Sb2S3])/(Δt) = 1/2 (Δ[H2O])/(Δt) = 1/9 (Δ[H2SO4])/(Δt) = 1/28 (Δ[NO])/(Δt) = 1/6 (Δ[HSbO3])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)
Construct the rate of reaction expression for: HNO_3 + Sb_2S_3 ⟶ H_2O + H_2SO_4 + NO + HSbO3 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: 28 HNO_3 + 3 Sb_2S_3 ⟶ 2 H_2O + 9 H_2SO_4 + 28 NO + 6 HSbO3 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 | 28 | -28 Sb_2S_3 | 3 | -3 H_2O | 2 | 2 H_2SO_4 | 9 | 9 NO | 28 | 28 HSbO3 | 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 HNO_3 | 28 | -28 | -1/28 (Δ[HNO3])/(Δt) Sb_2S_3 | 3 | -3 | -1/3 (Δ[Sb2S3])/(Δt) H_2O | 2 | 2 | 1/2 (Δ[H2O])/(Δt) H_2SO_4 | 9 | 9 | 1/9 (Δ[H2SO4])/(Δt) NO | 28 | 28 | 1/28 (Δ[NO])/(Δt) HSbO3 | 6 | 6 | 1/6 (Δ[HSbO3])/(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 = -1/28 (Δ[HNO3])/(Δt) = -1/3 (Δ[Sb2S3])/(Δt) = 1/2 (Δ[H2O])/(Δt) = 1/9 (Δ[H2SO4])/(Δt) = 1/28 (Δ[NO])/(Δt) = 1/6 (Δ[HSbO3])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

Chemical names and formulas

| nitric acid | antimony(III) sulfide | water | sulfuric acid | nitric oxide | HSbO3 formula | HNO_3 | Sb_2S_3 | H_2O | H_2SO_4 | NO | HSbO3 Hill formula | HNO_3 | S_3Sb_2 | H_2O | H_2O_4S | NO | HO3Sb name | nitric acid | antimony(III) sulfide | water | sulfuric acid | nitric oxide | IUPAC name | nitric acid | thioxo-(thioxostibanylthio)stibane | water | sulfuric acid | nitric oxide |
| nitric acid | antimony(III) sulfide | water | sulfuric acid | nitric oxide | HSbO3 formula | HNO_3 | Sb_2S_3 | H_2O | H_2SO_4 | NO | HSbO3 Hill formula | HNO_3 | S_3Sb_2 | H_2O | H_2O_4S | NO | HO3Sb name | nitric acid | antimony(III) sulfide | water | sulfuric acid | nitric oxide | IUPAC name | nitric acid | thioxo-(thioxostibanylthio)stibane | water | sulfuric acid | nitric oxide |

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