Pb3O4 = O2 + PbO

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

Balance the chemical equation algebraically: Pb_3O_4 ⟶ O_2 + PbO Add stoichiometric coefficients, c_i, to the reactants and products: c_1 Pb_3O_4 ⟶ c_2 O_2 + c_3 PbO Set the number of atoms in the reactants equal to the number of atoms in the products for O and Pb: O: | 4 c_1 = 2 c_2 + c_3 Pb: | 3 c_1 = 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_2 = 1 and solve the system of equations for the remaining coefficients: c_1 = 2 c_2 = 1 c_3 = 6 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 2 Pb_3O_4 ⟶ O_2 + 6 PbO

⟶ +

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

Construct the equilibrium constant, K, expression for: Pb_3O_4 ⟶ O_2 + PbO 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 Pb_3O_4 ⟶ O_2 + 6 PbO 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 Pb_3O_4 | 2 | -2 O_2 | 1 | 1 PbO | 6 | 6 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression Pb_3O_4 | 2 | -2 | ([Pb3O4])^(-2) O_2 | 1 | 1 | [O2] PbO | 6 | 6 | ([PbO])^6 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 = ([Pb3O4])^(-2) [O2] ([PbO])^6 = ([O2] ([PbO])^6)/([Pb3O4])^2

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

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

| lead(II, IV) oxide | oxygen | lead monoxide molar mass | 685.6 g/mol | 31.998 g/mol | 223.2 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 |