O2 + Li = LiO2

O_2 oxygen + Li lithium ⟶ LiO2

Balance the chemical equation algebraically: O_2 + Li ⟶ LiO2 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 O_2 + c_2 Li ⟶ c_3 LiO2 Set the number of atoms in the reactants equal to the number of atoms in the products for O and Li: O: | 2 c_1 = 2 c_3 Li: | c_2 = 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_1 = 1 and solve the system of equations for the remaining coefficients: c_1 = 1 c_2 = 1 c_3 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | O_2 + Li ⟶ LiO2

+ ⟶ LiO2

oxygen + lithium ⟶ LiO2

Construct the equilibrium constant, K, expression for: O_2 + Li ⟶ LiO2 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: O_2 + Li ⟶ LiO2 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 O_2 | 1 | -1 Li | 1 | -1 LiO2 | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression O_2 | 1 | -1 | ([O2])^(-1) Li | 1 | -1 | ([Li])^(-1) LiO2 | 1 | 1 | [LiO2] 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 = ([O2])^(-1) ([Li])^(-1) [LiO2] = ([LiO2])/([O2] [Li])

Construct the rate of reaction expression for: O_2 + Li ⟶ LiO2 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: O_2 + Li ⟶ LiO2 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 O_2 | 1 | -1 Li | 1 | -1 LiO2 | 1 | 1 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 O_2 | 1 | -1 | -(Δ[O2])/(Δt) Li | 1 | -1 | -(Δ[Li])/(Δt) LiO2 | 1 | 1 | (Δ[LiO2])/(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 = -(Δ[O2])/(Δt) = -(Δ[Li])/(Δt) = (Δ[LiO2])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| oxygen | lithium | LiO2 formula | O_2 | Li | LiO2 name | oxygen | lithium | IUPAC name | molecular oxygen | lithium |

| oxygen | lithium | LiO2 molar mass | 31.998 g/mol | 6.94 g/mol | 38.94 g/mol phase | gas (at STP) | solid (at STP) | melting point | -218 °C | 180 °C | boiling point | -183 °C | 1342 °C | density | 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 | |