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HNO3 + (NH4)2S = H2O + S + NO + NH4NO3

Input interpretation

HNO_3 nitric acid + (NH_4)_2S diammonium sulfide ⟶ H_2O water + S mixed sulfur + NO nitric oxide + NH_4NO_3 ammonium nitrate
HNO_3 nitric acid + (NH_4)_2S diammonium sulfide ⟶ H_2O water + S mixed sulfur + NO nitric oxide + NH_4NO_3 ammonium nitrate

Balanced equation

Balance the chemical equation algebraically: HNO_3 + (NH_4)_2S ⟶ H_2O + S + NO + NH_4NO_3 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 HNO_3 + c_2 (NH_4)_2S ⟶ c_3 H_2O + c_4 S + c_5 NO + c_6 NH_4NO_3 Set the number of atoms in the reactants equal to the number of atoms in the products for H, N, O and S: H: | c_1 + 8 c_2 = 2 c_3 + 4 c_6 N: | c_1 + 2 c_2 = c_5 + 2 c_6 O: | 3 c_1 = c_3 + c_5 + 3 c_6 S: | 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_2 = 1 and solve the system of equations for the remaining coefficients: c_2 = 1 c_3 = (7 c_1)/2 - 8 c_4 = 1 c_5 = 4 c_1 - 10 c_6 = 6 - (3 c_1)/2 Multiply by the least common denominator, 5, to eliminate fractional coefficients: c_2 = 5 c_3 = (7 c_1)/2 - 40 c_4 = 5 c_5 = 4 c_1 - 50 c_6 = 30 - (3 c_1)/2 The resulting system of equations is still underdetermined, so an additional coefficient must be set arbitrarily. Set c_1 = 14 and solve for the remaining coefficients: c_1 = 14 c_2 = 5 c_3 = 9 c_4 = 5 c_5 = 6 c_6 = 9 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 14 HNO_3 + 5 (NH_4)_2S ⟶ 9 H_2O + 5 S + 6 NO + 9 NH_4NO_3
Balance the chemical equation algebraically: HNO_3 + (NH_4)_2S ⟶ H_2O + S + NO + NH_4NO_3 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 HNO_3 + c_2 (NH_4)_2S ⟶ c_3 H_2O + c_4 S + c_5 NO + c_6 NH_4NO_3 Set the number of atoms in the reactants equal to the number of atoms in the products for H, N, O and S: H: | c_1 + 8 c_2 = 2 c_3 + 4 c_6 N: | c_1 + 2 c_2 = c_5 + 2 c_6 O: | 3 c_1 = c_3 + c_5 + 3 c_6 S: | 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_2 = 1 and solve the system of equations for the remaining coefficients: c_2 = 1 c_3 = (7 c_1)/2 - 8 c_4 = 1 c_5 = 4 c_1 - 10 c_6 = 6 - (3 c_1)/2 Multiply by the least common denominator, 5, to eliminate fractional coefficients: c_2 = 5 c_3 = (7 c_1)/2 - 40 c_4 = 5 c_5 = 4 c_1 - 50 c_6 = 30 - (3 c_1)/2 The resulting system of equations is still underdetermined, so an additional coefficient must be set arbitrarily. Set c_1 = 14 and solve for the remaining coefficients: c_1 = 14 c_2 = 5 c_3 = 9 c_4 = 5 c_5 = 6 c_6 = 9 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 14 HNO_3 + 5 (NH_4)_2S ⟶ 9 H_2O + 5 S + 6 NO + 9 NH_4NO_3

Structures

+ ⟶ + + +
+ ⟶ + + +

Names

nitric acid + diammonium sulfide ⟶ water + mixed sulfur + nitric oxide + ammonium nitrate
nitric acid + diammonium sulfide ⟶ water + mixed sulfur + nitric oxide + ammonium nitrate

Equilibrium constant

Construct the equilibrium constant, K, expression for: HNO_3 + (NH_4)_2S ⟶ H_2O + S + NO + NH_4NO_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: 14 HNO_3 + 5 (NH_4)_2S ⟶ 9 H_2O + 5 S + 6 NO + 9 NH_4NO_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 | 14 | -14 (NH_4)_2S | 5 | -5 H_2O | 9 | 9 S | 5 | 5 NO | 6 | 6 NH_4NO_3 | 9 | 9 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression HNO_3 | 14 | -14 | ([HNO3])^(-14) (NH_4)_2S | 5 | -5 | ([(NH4)2S])^(-5) H_2O | 9 | 9 | ([H2O])^9 S | 5 | 5 | ([S])^5 NO | 6 | 6 | ([NO])^6 NH_4NO_3 | 9 | 9 | ([NH4NO3])^9 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])^(-14) ([(NH4)2S])^(-5) ([H2O])^9 ([S])^5 ([NO])^6 ([NH4NO3])^9 = (([H2O])^9 ([S])^5 ([NO])^6 ([NH4NO3])^9)/(([HNO3])^14 ([(NH4)2S])^5)
Construct the equilibrium constant, K, expression for: HNO_3 + (NH_4)_2S ⟶ H_2O + S + NO + NH_4NO_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: 14 HNO_3 + 5 (NH_4)_2S ⟶ 9 H_2O + 5 S + 6 NO + 9 NH_4NO_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 | 14 | -14 (NH_4)_2S | 5 | -5 H_2O | 9 | 9 S | 5 | 5 NO | 6 | 6 NH_4NO_3 | 9 | 9 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression HNO_3 | 14 | -14 | ([HNO3])^(-14) (NH_4)_2S | 5 | -5 | ([(NH4)2S])^(-5) H_2O | 9 | 9 | ([H2O])^9 S | 5 | 5 | ([S])^5 NO | 6 | 6 | ([NO])^6 NH_4NO_3 | 9 | 9 | ([NH4NO3])^9 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])^(-14) ([(NH4)2S])^(-5) ([H2O])^9 ([S])^5 ([NO])^6 ([NH4NO3])^9 = (([H2O])^9 ([S])^5 ([NO])^6 ([NH4NO3])^9)/(([HNO3])^14 ([(NH4)2S])^5)

Rate of reaction

Construct the rate of reaction expression for: HNO_3 + (NH_4)_2S ⟶ H_2O + S + NO + NH_4NO_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: 14 HNO_3 + 5 (NH_4)_2S ⟶ 9 H_2O + 5 S + 6 NO + 9 NH_4NO_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 | 14 | -14 (NH_4)_2S | 5 | -5 H_2O | 9 | 9 S | 5 | 5 NO | 6 | 6 NH_4NO_3 | 9 | 9 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 | 14 | -14 | -1/14 (Δ[HNO3])/(Δt) (NH_4)_2S | 5 | -5 | -1/5 (Δ[(NH4)2S])/(Δt) H_2O | 9 | 9 | 1/9 (Δ[H2O])/(Δt) S | 5 | 5 | 1/5 (Δ[S])/(Δt) NO | 6 | 6 | 1/6 (Δ[NO])/(Δt) NH_4NO_3 | 9 | 9 | 1/9 (Δ[NH4NO3])/(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 = -1/14 (Δ[HNO3])/(Δt) = -1/5 (Δ[(NH4)2S])/(Δt) = 1/9 (Δ[H2O])/(Δt) = 1/5 (Δ[S])/(Δt) = 1/6 (Δ[NO])/(Δt) = 1/9 (Δ[NH4NO3])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)
Construct the rate of reaction expression for: HNO_3 + (NH_4)_2S ⟶ H_2O + S + NO + NH_4NO_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: 14 HNO_3 + 5 (NH_4)_2S ⟶ 9 H_2O + 5 S + 6 NO + 9 NH_4NO_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 | 14 | -14 (NH_4)_2S | 5 | -5 H_2O | 9 | 9 S | 5 | 5 NO | 6 | 6 NH_4NO_3 | 9 | 9 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 | 14 | -14 | -1/14 (Δ[HNO3])/(Δt) (NH_4)_2S | 5 | -5 | -1/5 (Δ[(NH4)2S])/(Δt) H_2O | 9 | 9 | 1/9 (Δ[H2O])/(Δt) S | 5 | 5 | 1/5 (Δ[S])/(Δt) NO | 6 | 6 | 1/6 (Δ[NO])/(Δt) NH_4NO_3 | 9 | 9 | 1/9 (Δ[NH4NO3])/(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 = -1/14 (Δ[HNO3])/(Δt) = -1/5 (Δ[(NH4)2S])/(Δt) = 1/9 (Δ[H2O])/(Δt) = 1/5 (Δ[S])/(Δt) = 1/6 (Δ[NO])/(Δt) = 1/9 (Δ[NH4NO3])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

Chemical names and formulas

| nitric acid | diammonium sulfide | water | mixed sulfur | nitric oxide | ammonium nitrate formula | HNO_3 | (NH_4)_2S | H_2O | S | NO | NH_4NO_3 Hill formula | HNO_3 | H_8N_2S | H_2O | S | NO | H_4N_2O_3 name | nitric acid | diammonium sulfide | water | mixed sulfur | nitric oxide | ammonium nitrate IUPAC name | nitric acid | diammonium sulfide | water | sulfur | nitric oxide |
| nitric acid | diammonium sulfide | water | mixed sulfur | nitric oxide | ammonium nitrate formula | HNO_3 | (NH_4)_2S | H_2O | S | NO | NH_4NO_3 Hill formula | HNO_3 | H_8N_2S | H_2O | S | NO | H_4N_2O_3 name | nitric acid | diammonium sulfide | water | mixed sulfur | nitric oxide | ammonium nitrate IUPAC name | nitric acid | diammonium sulfide | water | sulfur | nitric oxide |

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