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Al2O3 = O2 + Al

Input interpretation

Al_2O_3 (aluminum oxide) ⟶ O_2 (oxygen) + Al (aluminum)
Al_2O_3 (aluminum oxide) ⟶ O_2 (oxygen) + Al (aluminum)

Balanced equation

Balance the chemical equation algebraically: Al_2O_3 ⟶ O_2 + Al Add stoichiometric coefficients, c_i, to the reactants and products: c_1 Al_2O_3 ⟶ c_2 O_2 + c_3 Al Set the number of atoms in the reactants equal to the number of atoms in the products for Al and O: Al: | 2 c_1 = c_3 O: | 3 c_1 = 2 c_2 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_1 = 1 and solve the system of equations for the remaining coefficients: c_1 = 1 c_2 = 3/2 c_3 = 2 Multiply by the least common denominator, 2, to eliminate fractional coefficients: c_1 = 2 c_2 = 3 c_3 = 4 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: |   | 2 Al_2O_3 ⟶ 3 O_2 + 4 Al
Balance the chemical equation algebraically: Al_2O_3 ⟶ O_2 + Al Add stoichiometric coefficients, c_i, to the reactants and products: c_1 Al_2O_3 ⟶ c_2 O_2 + c_3 Al Set the number of atoms in the reactants equal to the number of atoms in the products for Al and O: Al: | 2 c_1 = c_3 O: | 3 c_1 = 2 c_2 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_1 = 1 and solve the system of equations for the remaining coefficients: c_1 = 1 c_2 = 3/2 c_3 = 2 Multiply by the least common denominator, 2, to eliminate fractional coefficients: c_1 = 2 c_2 = 3 c_3 = 4 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 2 Al_2O_3 ⟶ 3 O_2 + 4 Al

Structures

 ⟶ +
⟶ +

Names

aluminum oxide ⟶ oxygen + aluminum
aluminum oxide ⟶ oxygen + aluminum

Reaction thermodynamics

Enthalpy

 | aluminum oxide | oxygen | aluminum molecular enthalpy | -1676 kJ/mol | 0 kJ/mol | 0 kJ/mol total enthalpy | -3352 kJ/mol | 0 kJ/mol | 0 kJ/mol  | H_initial = -3352 kJ/mol | H_final = 0 kJ/mol |  ΔH_rxn^0 | 0 kJ/mol - -3352 kJ/mol = 3352 kJ/mol (endothermic) | |
| aluminum oxide | oxygen | aluminum molecular enthalpy | -1676 kJ/mol | 0 kJ/mol | 0 kJ/mol total enthalpy | -3352 kJ/mol | 0 kJ/mol | 0 kJ/mol | H_initial = -3352 kJ/mol | H_final = 0 kJ/mol | ΔH_rxn^0 | 0 kJ/mol - -3352 kJ/mol = 3352 kJ/mol (endothermic) | |

Entropy

 | aluminum oxide | oxygen | aluminum molecular entropy | 51 J/(mol K) | 205 J/(mol K) | 28.3 J/(mol K) total entropy | 102 J/(mol K) | 615 J/(mol K) | 113.2 J/(mol K)  | S_initial = 102 J/(mol K) | S_final = 728.2 J/(mol K) |  ΔS_rxn^0 | 728.2 J/(mol K) - 102 J/(mol K) = 626.2 J/(mol K) (endoentropic) | |
| aluminum oxide | oxygen | aluminum molecular entropy | 51 J/(mol K) | 205 J/(mol K) | 28.3 J/(mol K) total entropy | 102 J/(mol K) | 615 J/(mol K) | 113.2 J/(mol K) | S_initial = 102 J/(mol K) | S_final = 728.2 J/(mol K) | ΔS_rxn^0 | 728.2 J/(mol K) - 102 J/(mol K) = 626.2 J/(mol K) (endoentropic) | |

Equilibrium constant

Construct the equilibrium constant, K, expression for: Al_2O_3 ⟶ O_2 + Al 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 Al_2O_3 ⟶ 3 O_2 + 4 Al 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_2O_3 | 2 | -2 O_2 | 3 | 3 Al | 4 | 4 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression Al_2O_3 | 2 | -2 | ([Al2O3])^(-2) O_2 | 3 | 3 | ([O2])^3 Al | 4 | 4 | ([Al])^4 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 = ([Al2O3])^(-2) ([O2])^3 ([Al])^4 = (([O2])^3 ([Al])^4)/([Al2O3])^2
Construct the equilibrium constant, K, expression for: Al_2O_3 ⟶ O_2 + Al 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 Al_2O_3 ⟶ 3 O_2 + 4 Al 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_2O_3 | 2 | -2 O_2 | 3 | 3 Al | 4 | 4 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression Al_2O_3 | 2 | -2 | ([Al2O3])^(-2) O_2 | 3 | 3 | ([O2])^3 Al | 4 | 4 | ([Al])^4 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 = ([Al2O3])^(-2) ([O2])^3 ([Al])^4 = (([O2])^3 ([Al])^4)/([Al2O3])^2

Rate of reaction

Construct the rate of reaction expression for: Al_2O_3 ⟶ O_2 + Al 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 Al_2O_3 ⟶ 3 O_2 + 4 Al 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_2O_3 | 2 | -2 O_2 | 3 | 3 Al | 4 | 4 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_2O_3 | 2 | -2 | -1/2 (Δ[Al2O3])/(Δt) O_2 | 3 | 3 | 1/3 (Δ[O2])/(Δt) Al | 4 | 4 | 1/4 (Δ[Al])/(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[Al2O3])/(Δt) = 1/3 (Δ[O2])/(Δt) = 1/4 (Δ[Al])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)
Construct the rate of reaction expression for: Al_2O_3 ⟶ O_2 + Al 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 Al_2O_3 ⟶ 3 O_2 + 4 Al 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_2O_3 | 2 | -2 O_2 | 3 | 3 Al | 4 | 4 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_2O_3 | 2 | -2 | -1/2 (Δ[Al2O3])/(Δt) O_2 | 3 | 3 | 1/3 (Δ[O2])/(Δt) Al | 4 | 4 | 1/4 (Δ[Al])/(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[Al2O3])/(Δt) = 1/3 (Δ[O2])/(Δt) = 1/4 (Δ[Al])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

Chemical names and formulas

 | aluminum oxide | oxygen | aluminum formula | Al_2O_3 | O_2 | Al name | aluminum oxide | oxygen | aluminum IUPAC name | dialuminum;oxygen(2-) | molecular oxygen | aluminum
| aluminum oxide | oxygen | aluminum formula | Al_2O_3 | O_2 | Al name | aluminum oxide | oxygen | aluminum IUPAC name | dialuminum;oxygen(2-) | molecular oxygen | aluminum

Substance properties

 | aluminum oxide | oxygen | aluminum molar mass | 101.96 g/mol | 31.998 g/mol | 26.9815385 g/mol phase | solid (at STP) | gas (at STP) | solid (at STP) melting point | 2040 °C | -218 °C | 660.4 °C boiling point | | -183 °C | 2460 °C density | | 0.001429 g/cm^3 (at 0 °C) | 2.7 g/cm^3 solubility in water | | | insoluble surface tension | | 0.01347 N/m | 0.817 N/m dynamic viscosity | | 2.055×10^-5 Pa s (at 25 °C) | 1.5×10^-4 Pa s (at 760 °C) odor | odorless | odorless | odorless
| aluminum oxide | oxygen | aluminum molar mass | 101.96 g/mol | 31.998 g/mol | 26.9815385 g/mol phase | solid (at STP) | gas (at STP) | solid (at STP) melting point | 2040 °C | -218 °C | 660.4 °C boiling point | | -183 °C | 2460 °C density | | 0.001429 g/cm^3 (at 0 °C) | 2.7 g/cm^3 solubility in water | | | insoluble surface tension | | 0.01347 N/m | 0.817 N/m dynamic viscosity | | 2.055×10^-5 Pa s (at 25 °C) | 1.5×10^-4 Pa s (at 760 °C) odor | odorless | odorless | odorless

Units