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HNO3 + PbO2 + Mn(OH)3 = H2O + Pb(NO3)2 + HMnO4

Input interpretation

HNO_3 nitric acid + PbO_2 lead dioxide + Mn(OH)3 ⟶ H_2O water + Pb(NO_3)_2 lead(II) nitrate + HMnO4
HNO_3 nitric acid + PbO_2 lead dioxide + Mn(OH)3 ⟶ H_2O water + Pb(NO_3)_2 lead(II) nitrate + HMnO4

Balanced equation

Balance the chemical equation algebraically: HNO_3 + PbO_2 + Mn(OH)3 ⟶ H_2O + Pb(NO_3)_2 + HMnO4 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 HNO_3 + c_2 PbO_2 + c_3 Mn(OH)3 ⟶ c_4 H_2O + c_5 Pb(NO_3)_2 + c_6 HMnO4 Set the number of atoms in the reactants equal to the number of atoms in the products for H, N, O, Pb and Mn: H: | c_1 + 3 c_3 = 2 c_4 + c_6 N: | c_1 = 2 c_5 O: | 3 c_1 + 2 c_2 + 3 c_3 = c_4 + 6 c_5 + 4 c_6 Pb: | c_2 = c_5 Mn: | c_3 = c_6 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_3 = 1 and solve the system of equations for the remaining coefficients: c_1 = 4 c_2 = 2 c_3 = 1 c_4 = 3 c_5 = 2 c_6 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 4 HNO_3 + 2 PbO_2 + Mn(OH)3 ⟶ 3 H_2O + 2 Pb(NO_3)_2 + HMnO4
Balance the chemical equation algebraically: HNO_3 + PbO_2 + Mn(OH)3 ⟶ H_2O + Pb(NO_3)_2 + HMnO4 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 HNO_3 + c_2 PbO_2 + c_3 Mn(OH)3 ⟶ c_4 H_2O + c_5 Pb(NO_3)_2 + c_6 HMnO4 Set the number of atoms in the reactants equal to the number of atoms in the products for H, N, O, Pb and Mn: H: | c_1 + 3 c_3 = 2 c_4 + c_6 N: | c_1 = 2 c_5 O: | 3 c_1 + 2 c_2 + 3 c_3 = c_4 + 6 c_5 + 4 c_6 Pb: | c_2 = c_5 Mn: | c_3 = c_6 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_3 = 1 and solve the system of equations for the remaining coefficients: c_1 = 4 c_2 = 2 c_3 = 1 c_4 = 3 c_5 = 2 c_6 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 4 HNO_3 + 2 PbO_2 + Mn(OH)3 ⟶ 3 H_2O + 2 Pb(NO_3)_2 + HMnO4

Structures

+ + Mn(OH)3 ⟶ + + HMnO4
+ + Mn(OH)3 ⟶ + + HMnO4

Names

nitric acid + lead dioxide + Mn(OH)3 ⟶ water + lead(II) nitrate + HMnO4
nitric acid + lead dioxide + Mn(OH)3 ⟶ water + lead(II) nitrate + HMnO4

Equilibrium constant

Construct the equilibrium constant, K, expression for: HNO_3 + PbO_2 + Mn(OH)3 ⟶ H_2O + Pb(NO_3)_2 + HMnO4 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 + 2 PbO_2 + Mn(OH)3 ⟶ 3 H_2O + 2 Pb(NO_3)_2 + HMnO4 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 PbO_2 | 2 | -2 Mn(OH)3 | 1 | -1 H_2O | 3 | 3 Pb(NO_3)_2 | 2 | 2 HMnO4 | 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) PbO_2 | 2 | -2 | ([PbO2])^(-2) Mn(OH)3 | 1 | -1 | ([Mn(OH)3])^(-1) H_2O | 3 | 3 | ([H2O])^3 Pb(NO_3)_2 | 2 | 2 | ([Pb(NO3)2])^2 HMnO4 | 1 | 1 | [HMnO4] 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) ([PbO2])^(-2) ([Mn(OH)3])^(-1) ([H2O])^3 ([Pb(NO3)2])^2 [HMnO4] = (([H2O])^3 ([Pb(NO3)2])^2 [HMnO4])/(([HNO3])^4 ([PbO2])^2 [Mn(OH)3])
Construct the equilibrium constant, K, expression for: HNO_3 + PbO_2 + Mn(OH)3 ⟶ H_2O + Pb(NO_3)_2 + HMnO4 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 + 2 PbO_2 + Mn(OH)3 ⟶ 3 H_2O + 2 Pb(NO_3)_2 + HMnO4 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 PbO_2 | 2 | -2 Mn(OH)3 | 1 | -1 H_2O | 3 | 3 Pb(NO_3)_2 | 2 | 2 HMnO4 | 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) PbO_2 | 2 | -2 | ([PbO2])^(-2) Mn(OH)3 | 1 | -1 | ([Mn(OH)3])^(-1) H_2O | 3 | 3 | ([H2O])^3 Pb(NO_3)_2 | 2 | 2 | ([Pb(NO3)2])^2 HMnO4 | 1 | 1 | [HMnO4] 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) ([PbO2])^(-2) ([Mn(OH)3])^(-1) ([H2O])^3 ([Pb(NO3)2])^2 [HMnO4] = (([H2O])^3 ([Pb(NO3)2])^2 [HMnO4])/(([HNO3])^4 ([PbO2])^2 [Mn(OH)3])

Rate of reaction

Construct the rate of reaction expression for: HNO_3 + PbO_2 + Mn(OH)3 ⟶ H_2O + Pb(NO_3)_2 + HMnO4 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 + 2 PbO_2 + Mn(OH)3 ⟶ 3 H_2O + 2 Pb(NO_3)_2 + HMnO4 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 PbO_2 | 2 | -2 Mn(OH)3 | 1 | -1 H_2O | 3 | 3 Pb(NO_3)_2 | 2 | 2 HMnO4 | 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) PbO_2 | 2 | -2 | -1/2 (Δ[PbO2])/(Δt) Mn(OH)3 | 1 | -1 | -(Δ[Mn(OH)3])/(Δt) H_2O | 3 | 3 | 1/3 (Δ[H2O])/(Δt) Pb(NO_3)_2 | 2 | 2 | 1/2 (Δ[Pb(NO3)2])/(Δt) HMnO4 | 1 | 1 | (Δ[HMnO4])/(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) = -1/2 (Δ[PbO2])/(Δt) = -(Δ[Mn(OH)3])/(Δt) = 1/3 (Δ[H2O])/(Δt) = 1/2 (Δ[Pb(NO3)2])/(Δt) = (Δ[HMnO4])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)
Construct the rate of reaction expression for: HNO_3 + PbO_2 + Mn(OH)3 ⟶ H_2O + Pb(NO_3)_2 + HMnO4 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 + 2 PbO_2 + Mn(OH)3 ⟶ 3 H_2O + 2 Pb(NO_3)_2 + HMnO4 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 PbO_2 | 2 | -2 Mn(OH)3 | 1 | -1 H_2O | 3 | 3 Pb(NO_3)_2 | 2 | 2 HMnO4 | 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) PbO_2 | 2 | -2 | -1/2 (Δ[PbO2])/(Δt) Mn(OH)3 | 1 | -1 | -(Δ[Mn(OH)3])/(Δt) H_2O | 3 | 3 | 1/3 (Δ[H2O])/(Δt) Pb(NO_3)_2 | 2 | 2 | 1/2 (Δ[Pb(NO3)2])/(Δt) HMnO4 | 1 | 1 | (Δ[HMnO4])/(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) = -1/2 (Δ[PbO2])/(Δt) = -(Δ[Mn(OH)3])/(Δt) = 1/3 (Δ[H2O])/(Δt) = 1/2 (Δ[Pb(NO3)2])/(Δt) = (Δ[HMnO4])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

Chemical names and formulas

| nitric acid | lead dioxide | Mn(OH)3 | water | lead(II) nitrate | HMnO4 formula | HNO_3 | PbO_2 | Mn(OH)3 | H_2O | Pb(NO_3)_2 | HMnO4 Hill formula | HNO_3 | O_2Pb | H3MnO3 | H_2O | N_2O_6Pb | HMnO4 name | nitric acid | lead dioxide | | water | lead(II) nitrate | IUPAC name | nitric acid | | | water | plumbous dinitrate |
| nitric acid | lead dioxide | Mn(OH)3 | water | lead(II) nitrate | HMnO4 formula | HNO_3 | PbO_2 | Mn(OH)3 | H_2O | Pb(NO_3)_2 | HMnO4 Hill formula | HNO_3 | O_2Pb | H3MnO3 | H_2O | N_2O_6Pb | HMnO4 name | nitric acid | lead dioxide | | water | lead(II) nitrate | IUPAC name | nitric acid | | | water | plumbous dinitrate |

Substance properties

| nitric acid | lead dioxide | Mn(OH)3 | water | lead(II) nitrate | HMnO4 molar mass | 63.012 g/mol | 239.2 g/mol | 105.96 g/mol | 18.015 g/mol | 331.2 g/mol | 119.94 g/mol phase | liquid (at STP) | solid (at STP) | | liquid (at STP) | solid (at STP) | melting point | -41.6 °C | 290 °C | | 0 °C | 470 °C | boiling point | 83 °C | | | 99.9839 °C | | density | 1.5129 g/cm^3 | 9.58 g/cm^3 | | 1 g/cm^3 | | solubility in water | miscible | insoluble | | | | surface tension | | | | 0.0728 N/m | | dynamic viscosity | 7.6×10^-4 Pa s (at 25 °C) | | | 8.9×10^-4 Pa s (at 25 °C) | | odor | | | | odorless | odorless |
| nitric acid | lead dioxide | Mn(OH)3 | water | lead(II) nitrate | HMnO4 molar mass | 63.012 g/mol | 239.2 g/mol | 105.96 g/mol | 18.015 g/mol | 331.2 g/mol | 119.94 g/mol phase | liquid (at STP) | solid (at STP) | | liquid (at STP) | solid (at STP) | melting point | -41.6 °C | 290 °C | | 0 °C | 470 °C | boiling point | 83 °C | | | 99.9839 °C | | density | 1.5129 g/cm^3 | 9.58 g/cm^3 | | 1 g/cm^3 | | solubility in water | miscible | insoluble | | | | surface tension | | | | 0.0728 N/m | | dynamic viscosity | 7.6×10^-4 Pa s (at 25 °C) | | | 8.9×10^-4 Pa s (at 25 °C) | | odor | | | | odorless | odorless |

Units

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