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Al + PbO2 = Al2O3 + Pb

Input interpretation

Al aluminum + PbO_2 lead dioxide ⟶ Al_2O_3 aluminum oxide + Pb lead
Al aluminum + PbO_2 lead dioxide ⟶ Al_2O_3 aluminum oxide + Pb lead

Balanced equation

Balance the chemical equation algebraically: Al + PbO_2 ⟶ Al_2O_3 + Pb Add stoichiometric coefficients, c_i, to the reactants and products: c_1 Al + c_2 PbO_2 ⟶ c_3 Al_2O_3 + c_4 Pb Set the number of atoms in the reactants equal to the number of atoms in the products for Al, O and Pb: Al: | c_1 = 2 c_3 O: | 2 c_2 = 3 c_3 Pb: | c_2 = 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_3 = 1 and solve the system of equations for the remaining coefficients: c_1 = 2 c_2 = 3/2 c_3 = 1 c_4 = 3/2 Multiply by the least common denominator, 2, to eliminate fractional coefficients: c_1 = 4 c_2 = 3 c_3 = 2 c_4 = 3 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: |   | 4 Al + 3 PbO_2 ⟶ 2 Al_2O_3 + 3 Pb
Balance the chemical equation algebraically: Al + PbO_2 ⟶ Al_2O_3 + Pb Add stoichiometric coefficients, c_i, to the reactants and products: c_1 Al + c_2 PbO_2 ⟶ c_3 Al_2O_3 + c_4 Pb Set the number of atoms in the reactants equal to the number of atoms in the products for Al, O and Pb: Al: | c_1 = 2 c_3 O: | 2 c_2 = 3 c_3 Pb: | c_2 = 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_3 = 1 and solve the system of equations for the remaining coefficients: c_1 = 2 c_2 = 3/2 c_3 = 1 c_4 = 3/2 Multiply by the least common denominator, 2, to eliminate fractional coefficients: c_1 = 4 c_2 = 3 c_3 = 2 c_4 = 3 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 4 Al + 3 PbO_2 ⟶ 2 Al_2O_3 + 3 Pb

Structures

 + ⟶ +
+ ⟶ +

Names

aluminum + lead dioxide ⟶ aluminum oxide + lead
aluminum + lead dioxide ⟶ aluminum oxide + lead

Reaction thermodynamics

Enthalpy

 | aluminum | lead dioxide | aluminum oxide | lead molecular enthalpy | 0 kJ/mol | -277.4 kJ/mol | -1676 kJ/mol | 0 kJ/mol total enthalpy | 0 kJ/mol | -832.2 kJ/mol | -3352 kJ/mol | 0 kJ/mol  | H_initial = -832.2 kJ/mol | | H_final = -3352 kJ/mol |  ΔH_rxn^0 | -3352 kJ/mol - -832.2 kJ/mol = -2520 kJ/mol (exothermic) | | |
| aluminum | lead dioxide | aluminum oxide | lead molecular enthalpy | 0 kJ/mol | -277.4 kJ/mol | -1676 kJ/mol | 0 kJ/mol total enthalpy | 0 kJ/mol | -832.2 kJ/mol | -3352 kJ/mol | 0 kJ/mol | H_initial = -832.2 kJ/mol | | H_final = -3352 kJ/mol | ΔH_rxn^0 | -3352 kJ/mol - -832.2 kJ/mol = -2520 kJ/mol (exothermic) | | |

Entropy

 | aluminum | lead dioxide | aluminum oxide | lead molecular entropy | 28.3 J/(mol K) | 69 J/(mol K) | 51 J/(mol K) | 65 J/(mol K) total entropy | 113.2 J/(mol K) | 207 J/(mol K) | 102 J/(mol K) | 195 J/(mol K)  | S_initial = 320.2 J/(mol K) | | S_final = 297 J/(mol K) |  ΔS_rxn^0 | 297 J/(mol K) - 320.2 J/(mol K) = -23.2 J/(mol K) (exoentropic) | | |
| aluminum | lead dioxide | aluminum oxide | lead molecular entropy | 28.3 J/(mol K) | 69 J/(mol K) | 51 J/(mol K) | 65 J/(mol K) total entropy | 113.2 J/(mol K) | 207 J/(mol K) | 102 J/(mol K) | 195 J/(mol K) | S_initial = 320.2 J/(mol K) | | S_final = 297 J/(mol K) | ΔS_rxn^0 | 297 J/(mol K) - 320.2 J/(mol K) = -23.2 J/(mol K) (exoentropic) | | |

Equilibrium constant

Construct the equilibrium constant, K, expression for: Al + PbO_2 ⟶ Al_2O_3 + Pb 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 Al + 3 PbO_2 ⟶ 2 Al_2O_3 + 3 Pb 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 Al | 4 | -4 PbO_2 | 3 | -3 Al_2O_3 | 2 | 2 Pb | 3 | 3 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression Al | 4 | -4 | ([Al])^(-4) PbO_2 | 3 | -3 | ([PbO2])^(-3) Al_2O_3 | 2 | 2 | ([Al2O3])^2 Pb | 3 | 3 | ([Pb])^3 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 = ([Al])^(-4) ([PbO2])^(-3) ([Al2O3])^2 ([Pb])^3 = (([Al2O3])^2 ([Pb])^3)/(([Al])^4 ([PbO2])^3)
Construct the equilibrium constant, K, expression for: Al + PbO_2 ⟶ Al_2O_3 + Pb 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 Al + 3 PbO_2 ⟶ 2 Al_2O_3 + 3 Pb 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 Al | 4 | -4 PbO_2 | 3 | -3 Al_2O_3 | 2 | 2 Pb | 3 | 3 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression Al | 4 | -4 | ([Al])^(-4) PbO_2 | 3 | -3 | ([PbO2])^(-3) Al_2O_3 | 2 | 2 | ([Al2O3])^2 Pb | 3 | 3 | ([Pb])^3 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 = ([Al])^(-4) ([PbO2])^(-3) ([Al2O3])^2 ([Pb])^3 = (([Al2O3])^2 ([Pb])^3)/(([Al])^4 ([PbO2])^3)

Rate of reaction

Construct the rate of reaction expression for: Al + PbO_2 ⟶ Al_2O_3 + Pb 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 Al + 3 PbO_2 ⟶ 2 Al_2O_3 + 3 Pb 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 Al | 4 | -4 PbO_2 | 3 | -3 Al_2O_3 | 2 | 2 Pb | 3 | 3 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 Al | 4 | -4 | -1/4 (Δ[Al])/(Δt) PbO_2 | 3 | -3 | -1/3 (Δ[PbO2])/(Δt) Al_2O_3 | 2 | 2 | 1/2 (Δ[Al2O3])/(Δt) Pb | 3 | 3 | 1/3 (Δ[Pb])/(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[Al])/(Δt) = -1/3 (Δ[PbO2])/(Δt) = 1/2 (Δ[Al2O3])/(Δt) = 1/3 (Δ[Pb])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)
Construct the rate of reaction expression for: Al + PbO_2 ⟶ Al_2O_3 + Pb 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 Al + 3 PbO_2 ⟶ 2 Al_2O_3 + 3 Pb 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 Al | 4 | -4 PbO_2 | 3 | -3 Al_2O_3 | 2 | 2 Pb | 3 | 3 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 Al | 4 | -4 | -1/4 (Δ[Al])/(Δt) PbO_2 | 3 | -3 | -1/3 (Δ[PbO2])/(Δt) Al_2O_3 | 2 | 2 | 1/2 (Δ[Al2O3])/(Δt) Pb | 3 | 3 | 1/3 (Δ[Pb])/(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[Al])/(Δt) = -1/3 (Δ[PbO2])/(Δt) = 1/2 (Δ[Al2O3])/(Δt) = 1/3 (Δ[Pb])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

Chemical names and formulas

 | aluminum | lead dioxide | aluminum oxide | lead formula | Al | PbO_2 | Al_2O_3 | Pb Hill formula | Al | O_2Pb | Al_2O_3 | Pb name | aluminum | lead dioxide | aluminum oxide | lead IUPAC name | aluminum | | dialuminum;oxygen(2-) | lead
| aluminum | lead dioxide | aluminum oxide | lead formula | Al | PbO_2 | Al_2O_3 | Pb Hill formula | Al | O_2Pb | Al_2O_3 | Pb name | aluminum | lead dioxide | aluminum oxide | lead IUPAC name | aluminum | | dialuminum;oxygen(2-) | lead

Substance properties

 | aluminum | lead dioxide | aluminum oxide | lead molar mass | 26.9815385 g/mol | 239.2 g/mol | 101.96 g/mol | 207.2 g/mol phase | solid (at STP) | solid (at STP) | solid (at STP) | solid (at STP) melting point | 660.4 °C | 290 °C | 2040 °C | 327.4 °C boiling point | 2460 °C | | | 1740 °C density | 2.7 g/cm^3 | 9.58 g/cm^3 | | 11.34 g/cm^3 solubility in water | insoluble | insoluble | | insoluble surface tension | 0.817 N/m | | |  dynamic viscosity | 1.5×10^-4 Pa s (at 760 °C) | | | 0.00183 Pa s (at 38 °C) odor | odorless | | odorless |
| aluminum | lead dioxide | aluminum oxide | lead molar mass | 26.9815385 g/mol | 239.2 g/mol | 101.96 g/mol | 207.2 g/mol phase | solid (at STP) | solid (at STP) | solid (at STP) | solid (at STP) melting point | 660.4 °C | 290 °C | 2040 °C | 327.4 °C boiling point | 2460 °C | | | 1740 °C density | 2.7 g/cm^3 | 9.58 g/cm^3 | | 11.34 g/cm^3 solubility in water | insoluble | insoluble | | insoluble surface tension | 0.817 N/m | | | dynamic viscosity | 1.5×10^-4 Pa s (at 760 °C) | | | 0.00183 Pa s (at 38 °C) odor | odorless | | odorless |

Units