Al + LiO = Al2O3 + Li

Al aluminum + LiO ⟶ Al_2O_3 aluminum oxide + Li lithium

Balance the chemical equation algebraically: Al + LiO ⟶ Al_2O_3 + Li Add stoichiometric coefficients, c_i, to the reactants and products: c_1 Al + c_2 LiO ⟶ c_3 Al_2O_3 + c_4 Li Set the number of atoms in the reactants equal to the number of atoms in the products for Al, Li and O: Al: | c_1 = 2 c_3 Li: | c_2 = c_4 O: | c_2 = 3 c_3 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 c_3 = 1 c_4 = 3 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 2 Al + 3 LiO ⟶ Al_2O_3 + 3 Li

+ LiO ⟶ +

aluminum + LiO ⟶ aluminum oxide + lithium

Construct the equilibrium constant, K, expression for: Al + LiO ⟶ Al_2O_3 + Li 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 + 3 LiO ⟶ Al_2O_3 + 3 Li 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 | 2 | -2 LiO | 3 | -3 Al_2O_3 | 1 | 1 Li | 3 | 3 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression Al | 2 | -2 | ([Al])^(-2) LiO | 3 | -3 | ([LiO])^(-3) Al_2O_3 | 1 | 1 | [Al2O3] Li | 3 | 3 | ([Li])^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])^(-2) ([LiO])^(-3) [Al2O3] ([Li])^3 = ([Al2O3] ([Li])^3)/(([Al])^2 ([LiO])^3)

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

| aluminum | LiO | aluminum oxide | lithium formula | Al | LiO | Al_2O_3 | Li name | aluminum | | aluminum oxide | lithium IUPAC name | aluminum | | dialuminum;oxygen(2-) | lithium

| aluminum | LiO | aluminum oxide | lithium molar mass | 26.9815385 g/mol | 22.94 g/mol | 101.96 g/mol | 6.94 g/mol phase | solid (at STP) | | solid (at STP) | solid (at STP) melting point | 660.4 °C | | 2040 °C | 180 °C boiling point | 2460 °C | | | 1342 °C density | 2.7 g/cm^3 | | | 0.534 g/cm^3 solubility in water | insoluble | | | decomposes surface tension | 0.817 N/m | | | 0.3975 N/m dynamic viscosity | 1.5×10^-4 Pa s (at 760 °C) | | | odor | odorless | | odorless |