Li2O = O2 + Li

Li_2O lithium oxide ⟶ O_2 oxygen + Li lithium

Balance the chemical equation algebraically: Li_2O ⟶ O_2 + Li Add stoichiometric coefficients, c_i, to the reactants and products: c_1 Li_2O ⟶ c_2 O_2 + c_3 Li Set the number of atoms in the reactants equal to the number of atoms in the products for Li and O: Li: | 2 c_1 = c_3 O: | 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_2 = 1 and solve the system of equations for the remaining coefficients: c_1 = 2 c_2 = 1 c_3 = 4 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 2 Li_2O ⟶ O_2 + 4 Li

⟶ +

lithium oxide ⟶ oxygen + lithium

Construct the equilibrium constant, K, expression for: Li_2O ⟶ O_2 + 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 Li_2O ⟶ O_2 + 4 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 Li_2O | 2 | -2 O_2 | 1 | 1 Li | 4 | 4 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression Li_2O | 2 | -2 | ([Li2O])^(-2) O_2 | 1 | 1 | [O2] Li | 4 | 4 | ([Li])^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 = ([Li2O])^(-2) [O2] ([Li])^4 = ([O2] ([Li])^4)/([Li2O])^2

Construct the rate of reaction expression for: Li_2O ⟶ O_2 + 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 Li_2O ⟶ O_2 + 4 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 Li_2O | 2 | -2 O_2 | 1 | 1 Li | 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 Li_2O | 2 | -2 | -1/2 (Δ[Li2O])/(Δt) O_2 | 1 | 1 | (Δ[O2])/(Δt) Li | 4 | 4 | 1/4 (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[Li2O])/(Δt) = (Δ[O2])/(Δt) = 1/4 (Δ[Li])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| lithium oxide | oxygen | lithium formula | Li_2O | O_2 | Li name | lithium oxide | oxygen | lithium IUPAC name | dilithium oxygen(-2) anion | molecular oxygen | lithium

| lithium oxide | oxygen | lithium molar mass | 29.9 g/mol | 31.998 g/mol | 6.94 g/mol phase | | gas (at STP) | solid (at STP) melting point | | -218 °C | 180 °C boiling point | | -183 °C | 1342 °C density | 2.013 g/cm^3 | 0.001429 g/cm^3 (at 0 °C) | 0.534 g/cm^3 solubility in water | | | decomposes surface tension | | 0.01347 N/m | 0.3975 N/m dynamic viscosity | | 2.055×10^-5 Pa s (at 25 °C) | odor | | odorless |