O2 + PbO = Pb2PbO4

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

Balance the chemical equation algebraically: O_2 + PbO ⟶ Pb_3O_4 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 O_2 + c_2 PbO ⟶ 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 + c_2 = 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_1 = 1 and solve the system of equations for the remaining coefficients: c_1 = 1 c_2 = 6 c_3 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | O_2 + 6 PbO ⟶ 2 Pb_3O_4

+ ⟶

oxygen + lead monoxide ⟶ lead(II, IV) oxide

Construct the equilibrium constant, K, expression for: O_2 + PbO ⟶ 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: O_2 + 6 PbO ⟶ 2 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 | 1 | -1 PbO | 6 | -6 Pb_3O_4 | 2 | 2 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) PbO | 6 | -6 | ([PbO])^(-6) Pb_3O_4 | 2 | 2 | ([Pb3O4])^2 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) ([PbO])^(-6) ([Pb3O4])^2 = ([Pb3O4])^2/([O2] ([PbO])^6)

Construct the rate of reaction expression for: O_2 + PbO ⟶ 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: O_2 + 6 PbO ⟶ 2 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 | 1 | -1 PbO | 6 | -6 Pb_3O_4 | 2 | 2 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) PbO | 6 | -6 | -1/6 (Δ[PbO])/(Δt) Pb_3O_4 | 2 | 2 | 1/2 (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 = -(Δ[O2])/(Δt) = -1/6 (Δ[PbO])/(Δt) = 1/2 (Δ[Pb3O4])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

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

| oxygen | lead monoxide | lead(II, IV) oxide molar mass | 31.998 g/mol | 223.2 g/mol | 685.6 g/mol phase | gas (at STP) | solid (at STP) | melting point | -218 °C | 886 °C | boiling point | -183 °C | 1470 °C | density | 0.001429 g/cm^3 (at 0 °C) | 9.5 g/cm^3 | solubility in water | | insoluble | surface tension | 0.01347 N/m | | dynamic viscosity | 2.055×10^-5 Pa s (at 25 °C) | 1.45×10^-4 Pa s (at 1000 °C) | odor | odorless | |