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Pb(NO3)2 = O2 + NO2 + PbO

Input interpretation

Pb(NO_3)_2 (lead(II) nitrate) ⟶ O_2 (oxygen) + NO_2 (nitrogen dioxide) + PbO (lead monoxide)
Pb(NO_3)_2 (lead(II) nitrate) ⟶ O_2 (oxygen) + NO_2 (nitrogen dioxide) + PbO (lead monoxide)

Balanced equation

Balance the chemical equation algebraically: Pb(NO_3)_2 ⟶ O_2 + NO_2 + PbO Add stoichiometric coefficients, c_i, to the reactants and products: c_1 Pb(NO_3)_2 ⟶ c_2 O_2 + c_3 NO_2 + c_4 PbO Set the number of atoms in the reactants equal to the number of atoms in the products for N, O and Pb: N: | 2 c_1 = c_3 O: | 6 c_1 = 2 c_2 + 2 c_3 + c_4 Pb: | c_1 = 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 = 2 c_2 = 1 c_3 = 4 c_4 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 2 Pb(NO_3)_2 ⟶ O_2 + 4 NO_2 + 2 PbO
Balance the chemical equation algebraically: Pb(NO_3)_2 ⟶ O_2 + NO_2 + PbO Add stoichiometric coefficients, c_i, to the reactants and products: c_1 Pb(NO_3)_2 ⟶ c_2 O_2 + c_3 NO_2 + c_4 PbO Set the number of atoms in the reactants equal to the number of atoms in the products for N, O and Pb: N: | 2 c_1 = c_3 O: | 6 c_1 = 2 c_2 + 2 c_3 + c_4 Pb: | c_1 = 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 = 2 c_2 = 1 c_3 = 4 c_4 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 2 Pb(NO_3)_2 ⟶ O_2 + 4 NO_2 + 2 PbO

Structures

⟶ + +
⟶ + +

Names

lead(II) nitrate ⟶ oxygen + nitrogen dioxide + lead monoxide
lead(II) nitrate ⟶ oxygen + nitrogen dioxide + lead monoxide

Equilibrium constant

Construct the equilibrium constant, K, expression for: Pb(NO_3)_2 ⟶ O_2 + NO_2 + PbO 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: 2 Pb(NO_3)_2 ⟶ O_2 + 4 NO_2 + 2 PbO 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 Pb(NO_3)_2 | 2 | -2 O_2 | 1 | 1 NO_2 | 4 | 4 PbO | 2 | 2 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression Pb(NO_3)_2 | 2 | -2 | ([Pb(NO3)2])^(-2) O_2 | 1 | 1 | [O2] NO_2 | 4 | 4 | ([NO2])^4 PbO | 2 | 2 | ([PbO])^2 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 = ([Pb(NO3)2])^(-2) [O2] ([NO2])^4 ([PbO])^2 = ([O2] ([NO2])^4 ([PbO])^2)/([Pb(NO3)2])^2
Construct the equilibrium constant, K, expression for: Pb(NO_3)_2 ⟶ O_2 + NO_2 + PbO 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: 2 Pb(NO_3)_2 ⟶ O_2 + 4 NO_2 + 2 PbO 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 Pb(NO_3)_2 | 2 | -2 O_2 | 1 | 1 NO_2 | 4 | 4 PbO | 2 | 2 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression Pb(NO_3)_2 | 2 | -2 | ([Pb(NO3)2])^(-2) O_2 | 1 | 1 | [O2] NO_2 | 4 | 4 | ([NO2])^4 PbO | 2 | 2 | ([PbO])^2 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 = ([Pb(NO3)2])^(-2) [O2] ([NO2])^4 ([PbO])^2 = ([O2] ([NO2])^4 ([PbO])^2)/([Pb(NO3)2])^2

Rate of reaction

Construct the rate of reaction expression for: Pb(NO_3)_2 ⟶ O_2 + NO_2 + PbO 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: 2 Pb(NO_3)_2 ⟶ O_2 + 4 NO_2 + 2 PbO 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 Pb(NO_3)_2 | 2 | -2 O_2 | 1 | 1 NO_2 | 4 | 4 PbO | 2 | 2 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 Pb(NO_3)_2 | 2 | -2 | -1/2 (Δ[Pb(NO3)2])/(Δt) O_2 | 1 | 1 | (Δ[O2])/(Δt) NO_2 | 4 | 4 | 1/4 (Δ[NO2])/(Δt) PbO | 2 | 2 | 1/2 (Δ[PbO])/(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 = -1/2 (Δ[Pb(NO3)2])/(Δt) = (Δ[O2])/(Δt) = 1/4 (Δ[NO2])/(Δt) = 1/2 (Δ[PbO])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)
Construct the rate of reaction expression for: Pb(NO_3)_2 ⟶ O_2 + NO_2 + PbO 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: 2 Pb(NO_3)_2 ⟶ O_2 + 4 NO_2 + 2 PbO 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 Pb(NO_3)_2 | 2 | -2 O_2 | 1 | 1 NO_2 | 4 | 4 PbO | 2 | 2 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 Pb(NO_3)_2 | 2 | -2 | -1/2 (Δ[Pb(NO3)2])/(Δt) O_2 | 1 | 1 | (Δ[O2])/(Δt) NO_2 | 4 | 4 | 1/4 (Δ[NO2])/(Δt) PbO | 2 | 2 | 1/2 (Δ[PbO])/(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 = -1/2 (Δ[Pb(NO3)2])/(Δt) = (Δ[O2])/(Δt) = 1/4 (Δ[NO2])/(Δt) = 1/2 (Δ[PbO])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

Chemical names and formulas

| lead(II) nitrate | oxygen | nitrogen dioxide | lead monoxide formula | Pb(NO_3)_2 | O_2 | NO_2 | PbO Hill formula | N_2O_6Pb | O_2 | NO_2 | OPb name | lead(II) nitrate | oxygen | nitrogen dioxide | lead monoxide IUPAC name | plumbous dinitrate | molecular oxygen | Nitrogen dioxide |
| lead(II) nitrate | oxygen | nitrogen dioxide | lead monoxide formula | Pb(NO_3)_2 | O_2 | NO_2 | PbO Hill formula | N_2O_6Pb | O_2 | NO_2 | OPb name | lead(II) nitrate | oxygen | nitrogen dioxide | lead monoxide IUPAC name | plumbous dinitrate | molecular oxygen | Nitrogen dioxide |

Substance properties

| lead(II) nitrate | oxygen | nitrogen dioxide | lead monoxide molar mass | 331.2 g/mol | 31.998 g/mol | 46.005 g/mol | 223.2 g/mol phase | solid (at STP) | gas (at STP) | gas (at STP) | solid (at STP) melting point | 470 °C | -218 °C | -11 °C | 886 °C boiling point | | -183 °C | 21 °C | 1470 °C density | | 0.001429 g/cm^3 (at 0 °C) | 0.00188 g/cm^3 (at 25 °C) | 9.5 g/cm^3 solubility in water | | | reacts | insoluble surface tension | | 0.01347 N/m | | dynamic viscosity | | 2.055×10^-5 Pa s (at 25 °C) | 4.02×10^-4 Pa s (at 25 °C) | 1.45×10^-4 Pa s (at 1000 °C) odor | odorless | odorless | |
| lead(II) nitrate | oxygen | nitrogen dioxide | lead monoxide molar mass | 331.2 g/mol | 31.998 g/mol | 46.005 g/mol | 223.2 g/mol phase | solid (at STP) | gas (at STP) | gas (at STP) | solid (at STP) melting point | 470 °C | -218 °C | -11 °C | 886 °C boiling point | | -183 °C | 21 °C | 1470 °C density | | 0.001429 g/cm^3 (at 0 °C) | 0.00188 g/cm^3 (at 25 °C) | 9.5 g/cm^3 solubility in water | | | reacts | insoluble surface tension | | 0.01347 N/m | | dynamic viscosity | | 2.055×10^-5 Pa s (at 25 °C) | 4.02×10^-4 Pa s (at 25 °C) | 1.45×10^-4 Pa s (at 1000 °C) odor | odorless | odorless | |

Units

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