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H2S + Bi(NO3)3 = HNO3 + Bi2S3

Input interpretation

H_2S hydrogen sulfide + Bi(NO3)3 ⟶ HNO_3 nitric acid + Bi_2S_3 bismuth sulfide
H_2S hydrogen sulfide + Bi(NO3)3 ⟶ HNO_3 nitric acid + Bi_2S_3 bismuth sulfide

Balanced equation

Balance the chemical equation algebraically: H_2S + Bi(NO3)3 ⟶ HNO_3 + Bi_2S_3 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 H_2S + c_2 Bi(NO3)3 ⟶ c_3 HNO_3 + c_4 Bi_2S_3 Set the number of atoms in the reactants equal to the number of atoms in the products for H, S, Bi, N and O: H: | 2 c_1 = c_3 S: | c_1 = 3 c_4 Bi: | c_2 = 2 c_4 N: | 3 c_2 = c_3 O: | 9 c_2 = 3 c_3 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_4 = 1 and solve the system of equations for the remaining coefficients: c_1 = 3 c_2 = 2 c_3 = 6 c_4 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 3 H_2S + 2 Bi(NO3)3 ⟶ 6 HNO_3 + Bi_2S_3
Balance the chemical equation algebraically: H_2S + Bi(NO3)3 ⟶ HNO_3 + Bi_2S_3 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 H_2S + c_2 Bi(NO3)3 ⟶ c_3 HNO_3 + c_4 Bi_2S_3 Set the number of atoms in the reactants equal to the number of atoms in the products for H, S, Bi, N and O: H: | 2 c_1 = c_3 S: | c_1 = 3 c_4 Bi: | c_2 = 2 c_4 N: | 3 c_2 = c_3 O: | 9 c_2 = 3 c_3 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_4 = 1 and solve the system of equations for the remaining coefficients: c_1 = 3 c_2 = 2 c_3 = 6 c_4 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 3 H_2S + 2 Bi(NO3)3 ⟶ 6 HNO_3 + Bi_2S_3

Structures

+ Bi(NO3)3 ⟶ +
+ Bi(NO3)3 ⟶ +

Names

hydrogen sulfide + Bi(NO3)3 ⟶ nitric acid + bismuth sulfide
hydrogen sulfide + Bi(NO3)3 ⟶ nitric acid + bismuth sulfide

Equilibrium constant

Construct the equilibrium constant, K, expression for: H_2S + Bi(NO3)3 ⟶ HNO_3 + Bi_2S_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: 3 H_2S + 2 Bi(NO3)3 ⟶ 6 HNO_3 + Bi_2S_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 H_2S | 3 | -3 Bi(NO3)3 | 2 | -2 HNO_3 | 6 | 6 Bi_2S_3 | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression H_2S | 3 | -3 | ([H2S])^(-3) Bi(NO3)3 | 2 | -2 | ([Bi(NO3)3])^(-2) HNO_3 | 6 | 6 | ([HNO3])^6 Bi_2S_3 | 1 | 1 | [Bi2S3] 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 = ([H2S])^(-3) ([Bi(NO3)3])^(-2) ([HNO3])^6 [Bi2S3] = (([HNO3])^6 [Bi2S3])/(([H2S])^3 ([Bi(NO3)3])^2)
Construct the equilibrium constant, K, expression for: H_2S + Bi(NO3)3 ⟶ HNO_3 + Bi_2S_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: 3 H_2S + 2 Bi(NO3)3 ⟶ 6 HNO_3 + Bi_2S_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 H_2S | 3 | -3 Bi(NO3)3 | 2 | -2 HNO_3 | 6 | 6 Bi_2S_3 | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression H_2S | 3 | -3 | ([H2S])^(-3) Bi(NO3)3 | 2 | -2 | ([Bi(NO3)3])^(-2) HNO_3 | 6 | 6 | ([HNO3])^6 Bi_2S_3 | 1 | 1 | [Bi2S3] 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 = ([H2S])^(-3) ([Bi(NO3)3])^(-2) ([HNO3])^6 [Bi2S3] = (([HNO3])^6 [Bi2S3])/(([H2S])^3 ([Bi(NO3)3])^2)

Rate of reaction

Construct the rate of reaction expression for: H_2S + Bi(NO3)3 ⟶ HNO_3 + Bi_2S_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: 3 H_2S + 2 Bi(NO3)3 ⟶ 6 HNO_3 + Bi_2S_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 H_2S | 3 | -3 Bi(NO3)3 | 2 | -2 HNO_3 | 6 | 6 Bi_2S_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 H_2S | 3 | -3 | -1/3 (Δ[H2S])/(Δt) Bi(NO3)3 | 2 | -2 | -1/2 (Δ[Bi(NO3)3])/(Δt) HNO_3 | 6 | 6 | 1/6 (Δ[HNO3])/(Δt) Bi_2S_3 | 1 | 1 | (Δ[Bi2S3])/(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 = -1/3 (Δ[H2S])/(Δt) = -1/2 (Δ[Bi(NO3)3])/(Δt) = 1/6 (Δ[HNO3])/(Δt) = (Δ[Bi2S3])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)
Construct the rate of reaction expression for: H_2S + Bi(NO3)3 ⟶ HNO_3 + Bi_2S_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: 3 H_2S + 2 Bi(NO3)3 ⟶ 6 HNO_3 + Bi_2S_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 H_2S | 3 | -3 Bi(NO3)3 | 2 | -2 HNO_3 | 6 | 6 Bi_2S_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 H_2S | 3 | -3 | -1/3 (Δ[H2S])/(Δt) Bi(NO3)3 | 2 | -2 | -1/2 (Δ[Bi(NO3)3])/(Δt) HNO_3 | 6 | 6 | 1/6 (Δ[HNO3])/(Δt) Bi_2S_3 | 1 | 1 | (Δ[Bi2S3])/(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 = -1/3 (Δ[H2S])/(Δt) = -1/2 (Δ[Bi(NO3)3])/(Δt) = 1/6 (Δ[HNO3])/(Δt) = (Δ[Bi2S3])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

Chemical names and formulas

| hydrogen sulfide | Bi(NO3)3 | nitric acid | bismuth sulfide formula | H_2S | Bi(NO3)3 | HNO_3 | Bi_2S_3 Hill formula | H_2S | BiN3O9 | HNO_3 | Bi_2S_3 name | hydrogen sulfide | | nitric acid | bismuth sulfide IUPAC name | hydrogen sulfide | | nitric acid | thioxo-(thioxobismuthanylthio)bismuthane
| hydrogen sulfide | Bi(NO3)3 | nitric acid | bismuth sulfide formula | H_2S | Bi(NO3)3 | HNO_3 | Bi_2S_3 Hill formula | H_2S | BiN3O9 | HNO_3 | Bi_2S_3 name | hydrogen sulfide | | nitric acid | bismuth sulfide IUPAC name | hydrogen sulfide | | nitric acid | thioxo-(thioxobismuthanylthio)bismuthane

Substance properties

| hydrogen sulfide | Bi(NO3)3 | nitric acid | bismuth sulfide molar mass | 34.08 g/mol | 394.99 g/mol | 63.012 g/mol | 514.14 g/mol phase | gas (at STP) | | liquid (at STP) | solid (at STP) melting point | -85 °C | | -41.6 °C | 763 °C boiling point | -60 °C | | 83 °C | density | 0.001393 g/cm^3 (at 25 °C) | | 1.5129 g/cm^3 | 7.7 g/cm^3 solubility in water | | | miscible | insoluble dynamic viscosity | 1.239×10^-5 Pa s (at 25 °C) | | 7.6×10^-4 Pa s (at 25 °C) |
| hydrogen sulfide | Bi(NO3)3 | nitric acid | bismuth sulfide molar mass | 34.08 g/mol | 394.99 g/mol | 63.012 g/mol | 514.14 g/mol phase | gas (at STP) | | liquid (at STP) | solid (at STP) melting point | -85 °C | | -41.6 °C | 763 °C boiling point | -60 °C | | 83 °C | density | 0.001393 g/cm^3 (at 25 °C) | | 1.5129 g/cm^3 | 7.7 g/cm^3 solubility in water | | | miscible | insoluble dynamic viscosity | 1.239×10^-5 Pa s (at 25 °C) | | 7.6×10^-4 Pa s (at 25 °C) |

Units

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