PbSO4 = O2 + PbSO3

PbSO_4 lead(II) sulfate ⟶ O_2 oxygen + 2PbO·PbSO_3 dibasic lead(II) sulfite

Balance the chemical equation algebraically: PbSO_4 ⟶ O_2 + 2PbO·PbSO_3 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 PbSO_4 ⟶ c_2 O_2 + c_3 2PbO·PbSO_3 Set the number of atoms in the reactants equal to the number of atoms in the products for O, Pb and S: O: | 4 c_1 = 2 c_2 + 3 c_3 Pb: | c_1 = c_3 S: | 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 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 2 PbSO_4 ⟶ O_2 + 2 2PbO·PbSO_3

⟶ +

lead(II) sulfate ⟶ oxygen + dibasic lead(II) sulfite

Construct the equilibrium constant, K, expression for: PbSO_4 ⟶ O_2 + 2PbO·PbSO_3 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 PbSO_4 ⟶ O_2 + 2 2PbO·PbSO_3 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 PbSO_4 | 2 | -2 O_2 | 1 | 1 2PbO·PbSO_3 | 2 | 2 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression PbSO_4 | 2 | -2 | ([PbSO4])^(-2) O_2 | 1 | 1 | [O2] 2PbO·PbSO_3 | 2 | 2 | ([2PbO·PbSO3])^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 = ([PbSO4])^(-2) [O2] ([2PbO·PbSO3])^2 = ([O2] ([2PbO·PbSO3])^2)/([PbSO4])^2

Construct the rate of reaction expression for: PbSO_4 ⟶ O_2 + 2PbO·PbSO_3 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 PbSO_4 ⟶ O_2 + 2 2PbO·PbSO_3 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 PbSO_4 | 2 | -2 O_2 | 1 | 1 2PbO·PbSO_3 | 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 PbSO_4 | 2 | -2 | -1/2 (Δ[PbSO4])/(Δt) O_2 | 1 | 1 | (Δ[O2])/(Δt) 2PbO·PbSO_3 | 2 | 2 | 1/2 (Δ[2PbO·PbSO3])/(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[PbSO4])/(Δt) = (Δ[O2])/(Δt) = 1/2 (Δ[2PbO·PbSO3])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| lead(II) sulfate | oxygen | dibasic lead(II) sulfite formula | PbSO_4 | O_2 | 2PbO·PbSO_3 Hill formula | O_4PbS | O_2 | O_3PbS name | lead(II) sulfate | oxygen | dibasic lead(II) sulfite IUPAC name | | molecular oxygen | lead(+2) cation sulfite

| lead(II) sulfate | oxygen | dibasic lead(II) sulfite molar mass | 303.3 g/mol | 31.998 g/mol | 287.3 g/mol phase | solid (at STP) | gas (at STP) | melting point | 1087 °C | -218 °C | boiling point | | -183 °C | density | 6.29 g/cm^3 | 0.001429 g/cm^3 (at 0 °C) | solubility in water | slightly soluble | | surface tension | | 0.01347 N/m | dynamic viscosity | | 2.055×10^-5 Pa s (at 25 °C) | odor | | odorless |