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HNO3 + PbO2 + MoS2 = H2O + H2SO4 + Pb(NO3)2 + H2Mo4

Input interpretation

HNO_3 nitric acid + PbO_2 lead dioxide + MoS_2 molybdenum disulfide ⟶ H_2O water + H_2SO_4 sulfuric acid + Pb(NO_3)_2 lead(II) nitrate + H2Mo4
HNO_3 nitric acid + PbO_2 lead dioxide + MoS_2 molybdenum disulfide ⟶ H_2O water + H_2SO_4 sulfuric acid + Pb(NO_3)_2 lead(II) nitrate + H2Mo4

Balanced equation

Balance the chemical equation algebraically: HNO_3 + PbO_2 + MoS_2 ⟶ H_2O + H_2SO_4 + Pb(NO_3)_2 + H2Mo4 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 HNO_3 + c_2 PbO_2 + c_3 MoS_2 ⟶ c_4 H_2O + c_5 H_2SO_4 + c_6 Pb(NO_3)_2 + c_7 H2Mo4 Set the number of atoms in the reactants equal to the number of atoms in the products for H, N, O, Pb, Mo and S: H: | c_1 = 2 c_4 + 2 c_5 + 2 c_7 N: | c_1 = 2 c_6 O: | 3 c_1 + 2 c_2 = c_4 + 4 c_5 + 6 c_6 Pb: | c_2 = c_6 Mo: | c_3 = 4 c_7 S: | 2 c_3 = 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_7 = 1 and solve the system of equations for the remaining coefficients: c_1 = 46 c_2 = 23 c_3 = 4 c_4 = 14 c_5 = 8 c_6 = 23 c_7 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 46 HNO_3 + 23 PbO_2 + 4 MoS_2 ⟶ 14 H_2O + 8 H_2SO_4 + 23 Pb(NO_3)_2 + H2Mo4
Balance the chemical equation algebraically: HNO_3 + PbO_2 + MoS_2 ⟶ H_2O + H_2SO_4 + Pb(NO_3)_2 + H2Mo4 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 HNO_3 + c_2 PbO_2 + c_3 MoS_2 ⟶ c_4 H_2O + c_5 H_2SO_4 + c_6 Pb(NO_3)_2 + c_7 H2Mo4 Set the number of atoms in the reactants equal to the number of atoms in the products for H, N, O, Pb, Mo and S: H: | c_1 = 2 c_4 + 2 c_5 + 2 c_7 N: | c_1 = 2 c_6 O: | 3 c_1 + 2 c_2 = c_4 + 4 c_5 + 6 c_6 Pb: | c_2 = c_6 Mo: | c_3 = 4 c_7 S: | 2 c_3 = 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_7 = 1 and solve the system of equations for the remaining coefficients: c_1 = 46 c_2 = 23 c_3 = 4 c_4 = 14 c_5 = 8 c_6 = 23 c_7 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 46 HNO_3 + 23 PbO_2 + 4 MoS_2 ⟶ 14 H_2O + 8 H_2SO_4 + 23 Pb(NO_3)_2 + H2Mo4

Structures

+ + ⟶ + + + H2Mo4
+ + ⟶ + + + H2Mo4

Names

nitric acid + lead dioxide + molybdenum disulfide ⟶ water + sulfuric acid + lead(II) nitrate + H2Mo4
nitric acid + lead dioxide + molybdenum disulfide ⟶ water + sulfuric acid + lead(II) nitrate + H2Mo4

Equilibrium constant

Construct the equilibrium constant, K, expression for: HNO_3 + PbO_2 + MoS_2 ⟶ H_2O + H_2SO_4 + Pb(NO_3)_2 + H2Mo4 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: 46 HNO_3 + 23 PbO_2 + 4 MoS_2 ⟶ 14 H_2O + 8 H_2SO_4 + 23 Pb(NO_3)_2 + H2Mo4 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 | 46 | -46 PbO_2 | 23 | -23 MoS_2 | 4 | -4 H_2O | 14 | 14 H_2SO_4 | 8 | 8 Pb(NO_3)_2 | 23 | 23 H2Mo4 | 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 | 46 | -46 | ([HNO3])^(-46) PbO_2 | 23 | -23 | ([PbO2])^(-23) MoS_2 | 4 | -4 | ([MoS2])^(-4) H_2O | 14 | 14 | ([H2O])^14 H_2SO_4 | 8 | 8 | ([H2SO4])^8 Pb(NO_3)_2 | 23 | 23 | ([Pb(NO3)2])^23 H2Mo4 | 1 | 1 | [H2Mo4] 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])^(-46) ([PbO2])^(-23) ([MoS2])^(-4) ([H2O])^14 ([H2SO4])^8 ([Pb(NO3)2])^23 [H2Mo4] = (([H2O])^14 ([H2SO4])^8 ([Pb(NO3)2])^23 [H2Mo4])/(([HNO3])^46 ([PbO2])^23 ([MoS2])^4)
Construct the equilibrium constant, K, expression for: HNO_3 + PbO_2 + MoS_2 ⟶ H_2O + H_2SO_4 + Pb(NO_3)_2 + H2Mo4 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: 46 HNO_3 + 23 PbO_2 + 4 MoS_2 ⟶ 14 H_2O + 8 H_2SO_4 + 23 Pb(NO_3)_2 + H2Mo4 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 | 46 | -46 PbO_2 | 23 | -23 MoS_2 | 4 | -4 H_2O | 14 | 14 H_2SO_4 | 8 | 8 Pb(NO_3)_2 | 23 | 23 H2Mo4 | 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 | 46 | -46 | ([HNO3])^(-46) PbO_2 | 23 | -23 | ([PbO2])^(-23) MoS_2 | 4 | -4 | ([MoS2])^(-4) H_2O | 14 | 14 | ([H2O])^14 H_2SO_4 | 8 | 8 | ([H2SO4])^8 Pb(NO_3)_2 | 23 | 23 | ([Pb(NO3)2])^23 H2Mo4 | 1 | 1 | [H2Mo4] 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])^(-46) ([PbO2])^(-23) ([MoS2])^(-4) ([H2O])^14 ([H2SO4])^8 ([Pb(NO3)2])^23 [H2Mo4] = (([H2O])^14 ([H2SO4])^8 ([Pb(NO3)2])^23 [H2Mo4])/(([HNO3])^46 ([PbO2])^23 ([MoS2])^4)

Rate of reaction

Construct the rate of reaction expression for: HNO_3 + PbO_2 + MoS_2 ⟶ H_2O + H_2SO_4 + Pb(NO_3)_2 + H2Mo4 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: 46 HNO_3 + 23 PbO_2 + 4 MoS_2 ⟶ 14 H_2O + 8 H_2SO_4 + 23 Pb(NO_3)_2 + H2Mo4 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 | 46 | -46 PbO_2 | 23 | -23 MoS_2 | 4 | -4 H_2O | 14 | 14 H_2SO_4 | 8 | 8 Pb(NO_3)_2 | 23 | 23 H2Mo4 | 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 | 46 | -46 | -1/46 (Δ[HNO3])/(Δt) PbO_2 | 23 | -23 | -1/23 (Δ[PbO2])/(Δt) MoS_2 | 4 | -4 | -1/4 (Δ[MoS2])/(Δt) H_2O | 14 | 14 | 1/14 (Δ[H2O])/(Δt) H_2SO_4 | 8 | 8 | 1/8 (Δ[H2SO4])/(Δt) Pb(NO_3)_2 | 23 | 23 | 1/23 (Δ[Pb(NO3)2])/(Δt) H2Mo4 | 1 | 1 | (Δ[H2Mo4])/(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 = -1/46 (Δ[HNO3])/(Δt) = -1/23 (Δ[PbO2])/(Δt) = -1/4 (Δ[MoS2])/(Δt) = 1/14 (Δ[H2O])/(Δt) = 1/8 (Δ[H2SO4])/(Δt) = 1/23 (Δ[Pb(NO3)2])/(Δt) = (Δ[H2Mo4])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)
Construct the rate of reaction expression for: HNO_3 + PbO_2 + MoS_2 ⟶ H_2O + H_2SO_4 + Pb(NO_3)_2 + H2Mo4 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: 46 HNO_3 + 23 PbO_2 + 4 MoS_2 ⟶ 14 H_2O + 8 H_2SO_4 + 23 Pb(NO_3)_2 + H2Mo4 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 | 46 | -46 PbO_2 | 23 | -23 MoS_2 | 4 | -4 H_2O | 14 | 14 H_2SO_4 | 8 | 8 Pb(NO_3)_2 | 23 | 23 H2Mo4 | 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 | 46 | -46 | -1/46 (Δ[HNO3])/(Δt) PbO_2 | 23 | -23 | -1/23 (Δ[PbO2])/(Δt) MoS_2 | 4 | -4 | -1/4 (Δ[MoS2])/(Δt) H_2O | 14 | 14 | 1/14 (Δ[H2O])/(Δt) H_2SO_4 | 8 | 8 | 1/8 (Δ[H2SO4])/(Δt) Pb(NO_3)_2 | 23 | 23 | 1/23 (Δ[Pb(NO3)2])/(Δt) H2Mo4 | 1 | 1 | (Δ[H2Mo4])/(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 = -1/46 (Δ[HNO3])/(Δt) = -1/23 (Δ[PbO2])/(Δt) = -1/4 (Δ[MoS2])/(Δt) = 1/14 (Δ[H2O])/(Δt) = 1/8 (Δ[H2SO4])/(Δt) = 1/23 (Δ[Pb(NO3)2])/(Δt) = (Δ[H2Mo4])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

Chemical names and formulas

| nitric acid | lead dioxide | molybdenum disulfide | water | sulfuric acid | lead(II) nitrate | H2Mo4 formula | HNO_3 | PbO_2 | MoS_2 | H_2O | H_2SO_4 | Pb(NO_3)_2 | H2Mo4 Hill formula | HNO_3 | O_2Pb | MoS_2 | H_2O | H_2O_4S | N_2O_6Pb | H2Mo4 name | nitric acid | lead dioxide | molybdenum disulfide | water | sulfuric acid | lead(II) nitrate | IUPAC name | nitric acid | | dithioxomolybdenum | water | sulfuric acid | plumbous dinitrate |
| nitric acid | lead dioxide | molybdenum disulfide | water | sulfuric acid | lead(II) nitrate | H2Mo4 formula | HNO_3 | PbO_2 | MoS_2 | H_2O | H_2SO_4 | Pb(NO_3)_2 | H2Mo4 Hill formula | HNO_3 | O_2Pb | MoS_2 | H_2O | H_2O_4S | N_2O_6Pb | H2Mo4 name | nitric acid | lead dioxide | molybdenum disulfide | water | sulfuric acid | lead(II) nitrate | IUPAC name | nitric acid | | dithioxomolybdenum | water | sulfuric acid | plumbous dinitrate |

Substance properties

| nitric acid | lead dioxide | molybdenum disulfide | water | sulfuric acid | lead(II) nitrate | H2Mo4 molar mass | 63.012 g/mol | 239.2 g/mol | 160.1 g/mol | 18.015 g/mol | 98.07 g/mol | 331.2 g/mol | 385.8 g/mol phase | liquid (at STP) | solid (at STP) | solid (at STP) | liquid (at STP) | liquid (at STP) | solid (at STP) | melting point | -41.6 °C | 290 °C | 2375 °C | 0 °C | 10.371 °C | 470 °C | boiling point | 83 °C | | | 99.9839 °C | 279.6 °C | | density | 1.5129 g/cm^3 | 9.58 g/cm^3 | 5.06 g/cm^3 | 1 g/cm^3 | 1.8305 g/cm^3 | | solubility in water | miscible | insoluble | insoluble | | 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) | | odor | | | | odorless | odorless | odorless |
| nitric acid | lead dioxide | molybdenum disulfide | water | sulfuric acid | lead(II) nitrate | H2Mo4 molar mass | 63.012 g/mol | 239.2 g/mol | 160.1 g/mol | 18.015 g/mol | 98.07 g/mol | 331.2 g/mol | 385.8 g/mol phase | liquid (at STP) | solid (at STP) | solid (at STP) | liquid (at STP) | liquid (at STP) | solid (at STP) | melting point | -41.6 °C | 290 °C | 2375 °C | 0 °C | 10.371 °C | 470 °C | boiling point | 83 °C | | | 99.9839 °C | 279.6 °C | | density | 1.5129 g/cm^3 | 9.58 g/cm^3 | 5.06 g/cm^3 | 1 g/cm^3 | 1.8305 g/cm^3 | | solubility in water | miscible | insoluble | insoluble | | 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) | | odor | | | | odorless | odorless | odorless |

Units

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