H2O + KOH + PbO = K2Pb(OH)4

H_2O water + KOH potassium hydroxide + PbO lead monoxide ⟶ K2Pb(OH)4

Balance the chemical equation algebraically: H_2O + KOH + PbO ⟶ K2Pb(OH)4 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 H_2O + c_2 KOH + c_3 PbO ⟶ c_4 K2Pb(OH)4 Set the number of atoms in the reactants equal to the number of atoms in the products for H, O, K and Pb: H: | 2 c_1 + c_2 = 4 c_4 O: | c_1 + c_2 + c_3 = 4 c_4 K: | 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_1 = 1 and solve the system of equations for the remaining coefficients: c_1 = 1 c_2 = 2 c_3 = 1 c_4 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | H_2O + 2 KOH + PbO ⟶ K2Pb(OH)4

+ + ⟶ K2Pb(OH)4

water + potassium hydroxide + lead monoxide ⟶ K2Pb(OH)4

Construct the equilibrium constant, K, expression for: H_2O + KOH + PbO ⟶ K2Pb(OH)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: H_2O + 2 KOH + PbO ⟶ K2Pb(OH)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 H_2O | 1 | -1 KOH | 2 | -2 PbO | 1 | -1 K2Pb(OH)4 | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression H_2O | 1 | -1 | ([H2O])^(-1) KOH | 2 | -2 | ([KOH])^(-2) PbO | 1 | -1 | ([PbO])^(-1) K2Pb(OH)4 | 1 | 1 | [K2Pb(OH)4] 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])^(-1) ([KOH])^(-2) ([PbO])^(-1) [K2Pb(OH)4] = ([K2Pb(OH)4])/([H2O] ([KOH])^2 [PbO])

Construct the rate of reaction expression for: H_2O + KOH + PbO ⟶ K2Pb(OH)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: H_2O + 2 KOH + PbO ⟶ K2Pb(OH)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 H_2O | 1 | -1 KOH | 2 | -2 PbO | 1 | -1 K2Pb(OH)4 | 1 | 1 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 | 1 | -1 | -(Δ[H2O])/(Δt) KOH | 2 | -2 | -1/2 (Δ[KOH])/(Δt) PbO | 1 | -1 | -(Δ[PbO])/(Δt) K2Pb(OH)4 | 1 | 1 | (Δ[K2Pb(OH)4])/(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[H2O])/(Δt) = -1/2 (Δ[KOH])/(Δt) = -(Δ[PbO])/(Δt) = (Δ[K2Pb(OH)4])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| water | potassium hydroxide | lead monoxide | K2Pb(OH)4 formula | H_2O | KOH | PbO | K2Pb(OH)4 Hill formula | H_2O | HKO | OPb | H4K2O4Pb name | water | potassium hydroxide | lead monoxide |

| water | potassium hydroxide | lead monoxide | K2Pb(OH)4 molar mass | 18.015 g/mol | 56.105 g/mol | 223.2 g/mol | 353.4 g/mol phase | liquid (at STP) | solid (at STP) | solid (at STP) | melting point | 0 °C | 406 °C | 886 °C | boiling point | 99.9839 °C | 1327 °C | 1470 °C | density | 1 g/cm^3 | 2.044 g/cm^3 | 9.5 g/cm^3 | solubility in water | | soluble | insoluble | surface tension | 0.0728 N/m | | | dynamic viscosity | 8.9×10^-4 Pa s (at 25 °C) | 0.001 Pa s (at 550 °C) | 1.45×10^-4 Pa s (at 1000 °C) | odor | odorless | | |