H2O + O2 + Pb = Pb(OH)2

H_2O water + O_2 oxygen + Pb lead ⟶ Pb(OH)_2 lead(II) hydroxide

Balance the chemical equation algebraically: H_2O + O_2 + Pb ⟶ Pb(OH)_2 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 H_2O + c_2 O_2 + c_3 Pb ⟶ c_4 Pb(OH)_2 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_4 O: | c_1 + 2 c_2 = 2 c_4 Pb: | c_3 = 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 = 2 c_2 = 1 c_3 = 2 c_4 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 2 H_2O + O_2 + 2 Pb ⟶ 2 Pb(OH)_2

+ + ⟶

water + oxygen + lead ⟶ lead(II) hydroxide

Construct the equilibrium constant, K, expression for: H_2O + O_2 + Pb ⟶ Pb(OH)_2 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 H_2O + O_2 + 2 Pb ⟶ 2 Pb(OH)_2 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_2O | 2 | -2 O_2 | 1 | -1 Pb | 2 | -2 Pb(OH)_2 | 2 | 2 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression H_2O | 2 | -2 | ([H2O])^(-2) O_2 | 1 | -1 | ([O2])^(-1) Pb | 2 | -2 | ([Pb])^(-2) Pb(OH)_2 | 2 | 2 | ([Pb(OH)2])^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 = ([H2O])^(-2) ([O2])^(-1) ([Pb])^(-2) ([Pb(OH)2])^2 = ([Pb(OH)2])^2/(([H2O])^2 [O2] ([Pb])^2)

Construct the rate of reaction expression for: H_2O + O_2 + Pb ⟶ Pb(OH)_2 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 H_2O + O_2 + 2 Pb ⟶ 2 Pb(OH)_2 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_2O | 2 | -2 O_2 | 1 | -1 Pb | 2 | -2 Pb(OH)_2 | 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 H_2O | 2 | -2 | -1/2 (Δ[H2O])/(Δt) O_2 | 1 | -1 | -(Δ[O2])/(Δt) Pb | 2 | -2 | -1/2 (Δ[Pb])/(Δt) Pb(OH)_2 | 2 | 2 | 1/2 (Δ[Pb(OH)2])/(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[H2O])/(Δt) = -(Δ[O2])/(Δt) = -1/2 (Δ[Pb])/(Δt) = 1/2 (Δ[Pb(OH)2])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| water | oxygen | lead | lead(II) hydroxide formula | H_2O | O_2 | Pb | Pb(OH)_2 Hill formula | H_2O | O_2 | Pb | H_2O_2Pb name | water | oxygen | lead | lead(II) hydroxide IUPAC name | water | molecular oxygen | lead | plumbous dihydroxide

| water | oxygen | lead | lead(II) hydroxide molar mass | 18.015 g/mol | 31.998 g/mol | 207.2 g/mol | 241.2 g/mol phase | liquid (at STP) | gas (at STP) | solid (at STP) | melting point | 0 °C | -218 °C | 327.4 °C | boiling point | 99.9839 °C | -183 °C | 1740 °C | density | 1 g/cm^3 | 0.001429 g/cm^3 (at 0 °C) | 11.34 g/cm^3 | solubility in water | | | insoluble | surface tension | 0.0728 N/m | 0.01347 N/m | | dynamic viscosity | 8.9×10^-4 Pa s (at 25 °C) | 2.055×10^-5 Pa s (at 25 °C) | 0.00183 Pa s (at 38 °C) | odor | odorless | odorless | |