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HNO3 + FeS2 = H2O + H2SO4 + NO + Fe2(NO3)3

Input interpretation

HNO_3 nitric acid + FeS_2 pyrite ⟶ H_2O water + H_2SO_4 sulfuric acid + NO nitric oxide + Fe2(NO3)3
HNO_3 nitric acid + FeS_2 pyrite ⟶ H_2O water + H_2SO_4 sulfuric acid + NO nitric oxide + Fe2(NO3)3

Balanced equation

Balance the chemical equation algebraically: HNO_3 + FeS_2 ⟶ H_2O + H_2SO_4 + NO + Fe2(NO3)3 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 HNO_3 + c_2 FeS_2 ⟶ c_3 H_2O + c_4 H_2SO_4 + c_5 NO + c_6 Fe2(NO3)3 Set the number of atoms in the reactants equal to the number of atoms in the products for H, N, O, Fe and S: H: | c_1 = 2 c_3 + 2 c_4 N: | c_1 = c_5 + 3 c_6 O: | 3 c_1 = c_3 + 4 c_4 + c_5 + 9 c_6 Fe: | c_2 = 2 c_6 S: | 2 c_2 = c_4 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_6 = 1 and solve the system of equations for the remaining coefficients: c_1 = 12 c_2 = 2 c_3 = 2 c_4 = 4 c_5 = 9 c_6 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 12 HNO_3 + 2 FeS_2 ⟶ 2 H_2O + 4 H_2SO_4 + 9 NO + Fe2(NO3)3
Balance the chemical equation algebraically: HNO_3 + FeS_2 ⟶ H_2O + H_2SO_4 + NO + Fe2(NO3)3 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 HNO_3 + c_2 FeS_2 ⟶ c_3 H_2O + c_4 H_2SO_4 + c_5 NO + c_6 Fe2(NO3)3 Set the number of atoms in the reactants equal to the number of atoms in the products for H, N, O, Fe and S: H: | c_1 = 2 c_3 + 2 c_4 N: | c_1 = c_5 + 3 c_6 O: | 3 c_1 = c_3 + 4 c_4 + c_5 + 9 c_6 Fe: | c_2 = 2 c_6 S: | 2 c_2 = c_4 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_6 = 1 and solve the system of equations for the remaining coefficients: c_1 = 12 c_2 = 2 c_3 = 2 c_4 = 4 c_5 = 9 c_6 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 12 HNO_3 + 2 FeS_2 ⟶ 2 H_2O + 4 H_2SO_4 + 9 NO + Fe2(NO3)3

Structures

+ ⟶ + + + Fe2(NO3)3
+ ⟶ + + + Fe2(NO3)3

Names

nitric acid + pyrite ⟶ water + sulfuric acid + nitric oxide + Fe2(NO3)3
nitric acid + pyrite ⟶ water + sulfuric acid + nitric oxide + Fe2(NO3)3

Equilibrium constant

Construct the equilibrium constant, K, expression for: HNO_3 + FeS_2 ⟶ H_2O + H_2SO_4 + NO + Fe2(NO3)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: 12 HNO_3 + 2 FeS_2 ⟶ 2 H_2O + 4 H_2SO_4 + 9 NO + Fe2(NO3)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 | 12 | -12 FeS_2 | 2 | -2 H_2O | 2 | 2 H_2SO_4 | 4 | 4 NO | 9 | 9 Fe2(NO3)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 | 12 | -12 | ([HNO3])^(-12) FeS_2 | 2 | -2 | ([FeS2])^(-2) H_2O | 2 | 2 | ([H2O])^2 H_2SO_4 | 4 | 4 | ([H2SO4])^4 NO | 9 | 9 | ([NO])^9 Fe2(NO3)3 | 1 | 1 | [Fe2(NO3)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 = ([HNO3])^(-12) ([FeS2])^(-2) ([H2O])^2 ([H2SO4])^4 ([NO])^9 [Fe2(NO3)3] = (([H2O])^2 ([H2SO4])^4 ([NO])^9 [Fe2(NO3)3])/(([HNO3])^12 ([FeS2])^2)
Construct the equilibrium constant, K, expression for: HNO_3 + FeS_2 ⟶ H_2O + H_2SO_4 + NO + Fe2(NO3)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: 12 HNO_3 + 2 FeS_2 ⟶ 2 H_2O + 4 H_2SO_4 + 9 NO + Fe2(NO3)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 | 12 | -12 FeS_2 | 2 | -2 H_2O | 2 | 2 H_2SO_4 | 4 | 4 NO | 9 | 9 Fe2(NO3)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 | 12 | -12 | ([HNO3])^(-12) FeS_2 | 2 | -2 | ([FeS2])^(-2) H_2O | 2 | 2 | ([H2O])^2 H_2SO_4 | 4 | 4 | ([H2SO4])^4 NO | 9 | 9 | ([NO])^9 Fe2(NO3)3 | 1 | 1 | [Fe2(NO3)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 = ([HNO3])^(-12) ([FeS2])^(-2) ([H2O])^2 ([H2SO4])^4 ([NO])^9 [Fe2(NO3)3] = (([H2O])^2 ([H2SO4])^4 ([NO])^9 [Fe2(NO3)3])/(([HNO3])^12 ([FeS2])^2)

Rate of reaction

Construct the rate of reaction expression for: HNO_3 + FeS_2 ⟶ H_2O + H_2SO_4 + NO + Fe2(NO3)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: 12 HNO_3 + 2 FeS_2 ⟶ 2 H_2O + 4 H_2SO_4 + 9 NO + Fe2(NO3)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 | 12 | -12 FeS_2 | 2 | -2 H_2O | 2 | 2 H_2SO_4 | 4 | 4 NO | 9 | 9 Fe2(NO3)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 | 12 | -12 | -1/12 (Δ[HNO3])/(Δt) FeS_2 | 2 | -2 | -1/2 (Δ[FeS2])/(Δt) H_2O | 2 | 2 | 1/2 (Δ[H2O])/(Δt) H_2SO_4 | 4 | 4 | 1/4 (Δ[H2SO4])/(Δt) NO | 9 | 9 | 1/9 (Δ[NO])/(Δt) Fe2(NO3)3 | 1 | 1 | (Δ[Fe2(NO3)3])/(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) = -1/2 (Δ[FeS2])/(Δt) = 1/2 (Δ[H2O])/(Δt) = 1/4 (Δ[H2SO4])/(Δt) = 1/9 (Δ[NO])/(Δt) = (Δ[Fe2(NO3)3])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)
Construct the rate of reaction expression for: HNO_3 + FeS_2 ⟶ H_2O + H_2SO_4 + NO + Fe2(NO3)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: 12 HNO_3 + 2 FeS_2 ⟶ 2 H_2O + 4 H_2SO_4 + 9 NO + Fe2(NO3)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 | 12 | -12 FeS_2 | 2 | -2 H_2O | 2 | 2 H_2SO_4 | 4 | 4 NO | 9 | 9 Fe2(NO3)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 | 12 | -12 | -1/12 (Δ[HNO3])/(Δt) FeS_2 | 2 | -2 | -1/2 (Δ[FeS2])/(Δt) H_2O | 2 | 2 | 1/2 (Δ[H2O])/(Δt) H_2SO_4 | 4 | 4 | 1/4 (Δ[H2SO4])/(Δt) NO | 9 | 9 | 1/9 (Δ[NO])/(Δt) Fe2(NO3)3 | 1 | 1 | (Δ[Fe2(NO3)3])/(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) = -1/2 (Δ[FeS2])/(Δt) = 1/2 (Δ[H2O])/(Δt) = 1/4 (Δ[H2SO4])/(Δt) = 1/9 (Δ[NO])/(Δt) = (Δ[Fe2(NO3)3])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

Chemical names and formulas

| nitric acid | pyrite | water | sulfuric acid | nitric oxide | Fe2(NO3)3 formula | HNO_3 | FeS_2 | H_2O | H_2SO_4 | NO | Fe2(NO3)3 Hill formula | HNO_3 | FeS_2 | H_2O | H_2O_4S | NO | Fe2N3O9 name | nitric acid | pyrite | water | sulfuric acid | nitric oxide | IUPAC name | nitric acid | bis(sulfanylidene)iron | water | sulfuric acid | nitric oxide |
| nitric acid | pyrite | water | sulfuric acid | nitric oxide | Fe2(NO3)3 formula | HNO_3 | FeS_2 | H_2O | H_2SO_4 | NO | Fe2(NO3)3 Hill formula | HNO_3 | FeS_2 | H_2O | H_2O_4S | NO | Fe2N3O9 name | nitric acid | pyrite | water | sulfuric acid | nitric oxide | IUPAC name | nitric acid | bis(sulfanylidene)iron | water | sulfuric acid | nitric oxide |

Substance properties

| nitric acid | pyrite | water | sulfuric acid | nitric oxide | Fe2(NO3)3 molar mass | 63.012 g/mol | 120 g/mol | 18.015 g/mol | 98.07 g/mol | 30.006 g/mol | 297.7 g/mol phase | liquid (at STP) | | liquid (at STP) | liquid (at STP) | gas (at STP) | melting point | -41.6 °C | | 0 °C | 10.371 °C | -163.6 °C | boiling point | 83 °C | | 99.9839 °C | 279.6 °C | -151.7 °C | density | 1.5129 g/cm^3 | 4.89 g/cm^3 | 1 g/cm^3 | 1.8305 g/cm^3 | 0.001226 g/cm^3 (at 25 °C) | solubility in water | miscible | | | very soluble | | surface tension | | | 0.0728 N/m | 0.0735 N/m | | dynamic viscosity | 7.6×10^-4 Pa s (at 25 °C) | | 8.9×10^-4 Pa s (at 25 °C) | 0.021 Pa s (at 25 °C) | 1.911×10^-5 Pa s (at 25 °C) | odor | | odorless | odorless | odorless | |
| nitric acid | pyrite | water | sulfuric acid | nitric oxide | Fe2(NO3)3 molar mass | 63.012 g/mol | 120 g/mol | 18.015 g/mol | 98.07 g/mol | 30.006 g/mol | 297.7 g/mol phase | liquid (at STP) | | liquid (at STP) | liquid (at STP) | gas (at STP) | melting point | -41.6 °C | | 0 °C | 10.371 °C | -163.6 °C | boiling point | 83 °C | | 99.9839 °C | 279.6 °C | -151.7 °C | density | 1.5129 g/cm^3 | 4.89 g/cm^3 | 1 g/cm^3 | 1.8305 g/cm^3 | 0.001226 g/cm^3 (at 25 °C) | solubility in water | miscible | | | very soluble | | surface tension | | | 0.0728 N/m | 0.0735 N/m | | dynamic viscosity | 7.6×10^-4 Pa s (at 25 °C) | | 8.9×10^-4 Pa s (at 25 °C) | 0.021 Pa s (at 25 °C) | 1.911×10^-5 Pa s (at 25 °C) | odor | | odorless | odorless | odorless | |

Units

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