H2 + Pb3O4 = H2O + Pb

H_2 hydrogen + Pb_3O_4 lead(II, IV) oxide ⟶ H_2O water + Pb lead

Balance the chemical equation algebraically: H_2 + Pb_3O_4 ⟶ H_2O + Pb Add stoichiometric coefficients, c_i, to the reactants and products: c_1 H_2 + c_2 Pb_3O_4 ⟶ c_3 H_2O + c_4 Pb Set the number of atoms in the reactants equal to the number of atoms in the products for H, O and Pb: H: | 2 c_1 = 2 c_3 O: | 4 c_2 = c_3 Pb: | 3 c_2 = c_4 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 = 4 c_2 = 1 c_3 = 4 c_4 = 3 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 4 H_2 + Pb_3O_4 ⟶ 4 H_2O + 3 Pb

+ ⟶ +

hydrogen + lead(II, IV) oxide ⟶ water + lead

Construct the equilibrium constant, K, expression for: H_2 + Pb_3O_4 ⟶ H_2O + Pb 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: 4 H_2 + Pb_3O_4 ⟶ 4 H_2O + 3 Pb 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 H_2 | 4 | -4 Pb_3O_4 | 1 | -1 H_2O | 4 | 4 Pb | 3 | 3 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression H_2 | 4 | -4 | ([H2])^(-4) Pb_3O_4 | 1 | -1 | ([Pb3O4])^(-1) H_2O | 4 | 4 | ([H2O])^4 Pb | 3 | 3 | ([Pb])^3 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 = ([H2])^(-4) ([Pb3O4])^(-1) ([H2O])^4 ([Pb])^3 = (([H2O])^4 ([Pb])^3)/(([H2])^4 [Pb3O4])

Construct the rate of reaction expression for: H_2 + Pb_3O_4 ⟶ H_2O + Pb 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: 4 H_2 + Pb_3O_4 ⟶ 4 H_2O + 3 Pb 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 H_2 | 4 | -4 Pb_3O_4 | 1 | -1 H_2O | 4 | 4 Pb | 3 | 3 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 H_2 | 4 | -4 | -1/4 (Δ[H2])/(Δt) Pb_3O_4 | 1 | -1 | -(Δ[Pb3O4])/(Δt) H_2O | 4 | 4 | 1/4 (Δ[H2O])/(Δt) Pb | 3 | 3 | 1/3 (Δ[Pb])/(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 = -1/4 (Δ[H2])/(Δt) = -(Δ[Pb3O4])/(Δt) = 1/4 (Δ[H2O])/(Δt) = 1/3 (Δ[Pb])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| hydrogen | lead(II, IV) oxide | water | lead formula | H_2 | Pb_3O_4 | H_2O | Pb Hill formula | H_2 | O_4Pb_3 | H_2O | Pb name | hydrogen | lead(II, IV) oxide | water | lead IUPAC name | molecular hydrogen | lead tetraoxide | water | lead

| hydrogen | lead(II, IV) oxide | water | lead molar mass | 2.016 g/mol | 685.6 g/mol | 18.015 g/mol | 207.2 g/mol phase | gas (at STP) | | liquid (at STP) | solid (at STP) melting point | -259.2 °C | | 0 °C | 327.4 °C boiling point | -252.8 °C | | 99.9839 °C | 1740 °C density | 8.99×10^-5 g/cm^3 (at 0 °C) | | 1 g/cm^3 | 11.34 g/cm^3 solubility in water | | | | insoluble surface tension | | | 0.0728 N/m | dynamic viscosity | 8.9×10^-6 Pa s (at 25 °C) | | 8.9×10^-4 Pa s (at 25 °C) | 0.00183 Pa s (at 38 °C) odor | odorless | | odorless |