O2 + Pb = Pb3O4

O_2 oxygen + Pb lead ⟶ Pb_3O_4 lead(II, IV) oxide

Balance the chemical equation algebraically: O_2 + Pb ⟶ Pb_3O_4 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 O_2 + c_2 Pb ⟶ c_3 Pb_3O_4 Set the number of atoms in the reactants equal to the number of atoms in the products for O and Pb: O: | 2 c_1 = 4 c_3 Pb: | 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 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 2 O_2 + 3 Pb ⟶ Pb_3O_4

+ ⟶

oxygen + lead ⟶ lead(II, IV) oxide

Construct the equilibrium constant, K, expression for: O_2 + Pb ⟶ Pb_3O_4 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 O_2 + 3 Pb ⟶ Pb_3O_4 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 | 2 | -2 Pb | 3 | -3 Pb_3O_4 | 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 | 2 | -2 | ([O2])^(-2) Pb | 3 | -3 | ([Pb])^(-3) Pb_3O_4 | 1 | 1 | [Pb3O4] 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])^(-2) ([Pb])^(-3) [Pb3O4] = ([Pb3O4])/(([O2])^2 ([Pb])^3)

Construct the rate of reaction expression for: O_2 + Pb ⟶ Pb_3O_4 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 O_2 + 3 Pb ⟶ Pb_3O_4 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 | 2 | -2 Pb | 3 | -3 Pb_3O_4 | 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 | 2 | -2 | -1/2 (Δ[O2])/(Δt) Pb | 3 | -3 | -1/3 (Δ[Pb])/(Δt) Pb_3O_4 | 1 | 1 | (Δ[Pb3O4])/(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[O2])/(Δt) = -1/3 (Δ[Pb])/(Δt) = (Δ[Pb3O4])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| oxygen | lead | lead(II, IV) oxide formula | O_2 | Pb | Pb_3O_4 Hill formula | O_2 | Pb | O_4Pb_3 name | oxygen | lead | lead(II, IV) oxide IUPAC name | molecular oxygen | lead | lead tetraoxide

| oxygen | lead | lead(II, IV) oxide molar mass | 31.998 g/mol | 207.2 g/mol | 685.6 g/mol phase | gas (at STP) | solid (at STP) | melting point | -218 °C | 327.4 °C | boiling point | -183 °C | 1740 °C | density | 0.001429 g/cm^3 (at 0 °C) | 11.34 g/cm^3 | solubility in water | | insoluble | surface tension | 0.01347 N/m | | dynamic viscosity | 2.055×10^-5 Pa s (at 25 °C) | 0.00183 Pa s (at 38 °C) | odor | odorless | |