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HNO3 + Pb3O4 = Pb(NO3)2 + H4PbO4

Input interpretation

HNO_3 nitric acid + Pb_3O_4 lead(II, IV) oxide ⟶ Pb(NO_3)_2 lead(II) nitrate + H4PbO4
HNO_3 nitric acid + Pb_3O_4 lead(II, IV) oxide ⟶ Pb(NO_3)_2 lead(II) nitrate + H4PbO4

Balanced equation

Balance the chemical equation algebraically: HNO_3 + Pb_3O_4 ⟶ Pb(NO_3)_2 + H4PbO4 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 HNO_3 + c_2 Pb_3O_4 ⟶ c_3 Pb(NO_3)_2 + c_4 H4PbO4 Set the number of atoms in the reactants equal to the number of atoms in the products for H, N, O and Pb: H: | c_1 = 4 c_4 N: | c_1 = 2 c_3 O: | 3 c_1 + 4 c_2 = 6 c_3 + 4 c_4 Pb: | 3 c_2 = c_3 + 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_1 = 4 c_2 = 1 c_3 = 2 c_4 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 4 HNO_3 + Pb_3O_4 ⟶ 2 Pb(NO_3)_2 + H4PbO4
Balance the chemical equation algebraically: HNO_3 + Pb_3O_4 ⟶ Pb(NO_3)_2 + H4PbO4 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 HNO_3 + c_2 Pb_3O_4 ⟶ c_3 Pb(NO_3)_2 + c_4 H4PbO4 Set the number of atoms in the reactants equal to the number of atoms in the products for H, N, O and Pb: H: | c_1 = 4 c_4 N: | c_1 = 2 c_3 O: | 3 c_1 + 4 c_2 = 6 c_3 + 4 c_4 Pb: | 3 c_2 = c_3 + 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_1 = 4 c_2 = 1 c_3 = 2 c_4 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 4 HNO_3 + Pb_3O_4 ⟶ 2 Pb(NO_3)_2 + H4PbO4

Structures

+ ⟶ + H4PbO4
+ ⟶ + H4PbO4

Names

nitric acid + lead(II, IV) oxide ⟶ lead(II) nitrate + H4PbO4
nitric acid + lead(II, IV) oxide ⟶ lead(II) nitrate + H4PbO4

Equilibrium constant

Construct the equilibrium constant, K, expression for: HNO_3 + Pb_3O_4 ⟶ Pb(NO_3)_2 + H4PbO4 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: 4 HNO_3 + Pb_3O_4 ⟶ 2 Pb(NO_3)_2 + H4PbO4 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 | 4 | -4 Pb_3O_4 | 1 | -1 Pb(NO_3)_2 | 2 | 2 H4PbO4 | 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 | 4 | -4 | ([HNO3])^(-4) Pb_3O_4 | 1 | -1 | ([Pb3O4])^(-1) Pb(NO_3)_2 | 2 | 2 | ([Pb(NO3)2])^2 H4PbO4 | 1 | 1 | [H4PbO4] 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])^(-4) ([Pb3O4])^(-1) ([Pb(NO3)2])^2 [H4PbO4] = (([Pb(NO3)2])^2 [H4PbO4])/(([HNO3])^4 [Pb3O4])
Construct the equilibrium constant, K, expression for: HNO_3 + Pb_3O_4 ⟶ Pb(NO_3)_2 + H4PbO4 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: 4 HNO_3 + Pb_3O_4 ⟶ 2 Pb(NO_3)_2 + H4PbO4 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 | 4 | -4 Pb_3O_4 | 1 | -1 Pb(NO_3)_2 | 2 | 2 H4PbO4 | 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 | 4 | -4 | ([HNO3])^(-4) Pb_3O_4 | 1 | -1 | ([Pb3O4])^(-1) Pb(NO_3)_2 | 2 | 2 | ([Pb(NO3)2])^2 H4PbO4 | 1 | 1 | [H4PbO4] 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])^(-4) ([Pb3O4])^(-1) ([Pb(NO3)2])^2 [H4PbO4] = (([Pb(NO3)2])^2 [H4PbO4])/(([HNO3])^4 [Pb3O4])

Rate of reaction

Construct the rate of reaction expression for: HNO_3 + Pb_3O_4 ⟶ Pb(NO_3)_2 + H4PbO4 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: 4 HNO_3 + Pb_3O_4 ⟶ 2 Pb(NO_3)_2 + H4PbO4 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 | 4 | -4 Pb_3O_4 | 1 | -1 Pb(NO_3)_2 | 2 | 2 H4PbO4 | 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 | 4 | -4 | -1/4 (Δ[HNO3])/(Δt) Pb_3O_4 | 1 | -1 | -(Δ[Pb3O4])/(Δt) Pb(NO_3)_2 | 2 | 2 | 1/2 (Δ[Pb(NO3)2])/(Δt) H4PbO4 | 1 | 1 | (Δ[H4PbO4])/(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 = -1/4 (Δ[HNO3])/(Δt) = -(Δ[Pb3O4])/(Δt) = 1/2 (Δ[Pb(NO3)2])/(Δt) = (Δ[H4PbO4])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)
Construct the rate of reaction expression for: HNO_3 + Pb_3O_4 ⟶ Pb(NO_3)_2 + H4PbO4 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: 4 HNO_3 + Pb_3O_4 ⟶ 2 Pb(NO_3)_2 + H4PbO4 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 | 4 | -4 Pb_3O_4 | 1 | -1 Pb(NO_3)_2 | 2 | 2 H4PbO4 | 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 | 4 | -4 | -1/4 (Δ[HNO3])/(Δt) Pb_3O_4 | 1 | -1 | -(Δ[Pb3O4])/(Δt) Pb(NO_3)_2 | 2 | 2 | 1/2 (Δ[Pb(NO3)2])/(Δt) H4PbO4 | 1 | 1 | (Δ[H4PbO4])/(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 = -1/4 (Δ[HNO3])/(Δt) = -(Δ[Pb3O4])/(Δt) = 1/2 (Δ[Pb(NO3)2])/(Δt) = (Δ[H4PbO4])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

Chemical names and formulas

| nitric acid | lead(II, IV) oxide | lead(II) nitrate | H4PbO4 formula | HNO_3 | Pb_3O_4 | Pb(NO_3)_2 | H4PbO4 Hill formula | HNO_3 | O_4Pb_3 | N_2O_6Pb | H4O4Pb name | nitric acid | lead(II, IV) oxide | lead(II) nitrate | IUPAC name | nitric acid | lead tetraoxide | plumbous dinitrate |
| nitric acid | lead(II, IV) oxide | lead(II) nitrate | H4PbO4 formula | HNO_3 | Pb_3O_4 | Pb(NO_3)_2 | H4PbO4 Hill formula | HNO_3 | O_4Pb_3 | N_2O_6Pb | H4O4Pb name | nitric acid | lead(II, IV) oxide | lead(II) nitrate | IUPAC name | nitric acid | lead tetraoxide | plumbous dinitrate |

Substance properties

| nitric acid | lead(II, IV) oxide | lead(II) nitrate | H4PbO4 molar mass | 63.012 g/mol | 685.6 g/mol | 331.2 g/mol | 275.2 g/mol phase | liquid (at STP) | | solid (at STP) | melting point | -41.6 °C | | 470 °C | boiling point | 83 °C | | | density | 1.5129 g/cm^3 | | | solubility in water | miscible | | | dynamic viscosity | 7.6×10^-4 Pa s (at 25 °C) | | | odor | | | odorless |
| nitric acid | lead(II, IV) oxide | lead(II) nitrate | H4PbO4 molar mass | 63.012 g/mol | 685.6 g/mol | 331.2 g/mol | 275.2 g/mol phase | liquid (at STP) | | solid (at STP) | melting point | -41.6 °C | | 470 °C | boiling point | 83 °C | | | density | 1.5129 g/cm^3 | | | solubility in water | miscible | | | dynamic viscosity | 7.6×10^-4 Pa s (at 25 °C) | | | odor | | | odorless |

Units

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