NaOH + PbO2 + Sb = H2O + PbO + NaSbO2

Input interpretation

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NaOH sodium hydroxide + PbO_2 lead dioxide + Sb gray antimony ⟶ H_2O water + PbO lead monoxide + NaSbO2

Balanced equation

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Balance the chemical equation algebraically: NaOH + PbO_2 + Sb ⟶ H_2O + PbO + NaSbO2 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 NaOH + c_2 PbO_2 + c_3 Sb ⟶ c_4 H_2O + c_5 PbO + c_6 NaSbO2 Set the number of atoms in the reactants equal to the number of atoms in the products for H, Na, O, Pb and Sb: H: | c_1 = 2 c_4 Na: | c_1 = c_6 O: | c_1 + 2 c_2 = c_4 + c_5 + 2 c_6 Pb: | c_2 = c_5 Sb: | c_3 = c_6 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_4 = 1 and solve the system of equations for the remaining coefficients: c_1 = 2 c_2 = 3 c_3 = 2 c_4 = 1 c_5 = 3 c_6 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 2 NaOH + 3 PbO_2 + 2 Sb ⟶ H_2O + 3 PbO + 2 NaSbO2

+ + ⟶ + + NaSbO2

Names

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sodium hydroxide + lead dioxide + gray antimony ⟶ water + lead monoxide + NaSbO2

Equilibrium constant

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Construct the equilibrium constant, K, expression for: NaOH + PbO_2 + Sb ⟶ H_2O + PbO + NaSbO2 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 NaOH + 3 PbO_2 + 2 Sb ⟶ H_2O + 3 PbO + 2 NaSbO2 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 NaOH | 2 | -2 PbO_2 | 3 | -3 Sb | 2 | -2 H_2O | 1 | 1 PbO | 3 | 3 NaSbO2 | 2 | 2 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression NaOH | 2 | -2 | ([NaOH])^(-2) PbO_2 | 3 | -3 | ([PbO2])^(-3) Sb | 2 | -2 | ([Sb])^(-2) H_2O | 1 | 1 | [H2O] PbO | 3 | 3 | ([PbO])^3 NaSbO2 | 2 | 2 | ([NaSbO2])^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 = ([NaOH])^(-2) ([PbO2])^(-3) ([Sb])^(-2) [H2O] ([PbO])^3 ([NaSbO2])^2 = ([H2O] ([PbO])^3 ([NaSbO2])^2)/(([NaOH])^2 ([PbO2])^3 ([Sb])^2)

Rate of reaction

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Construct the rate of reaction expression for: NaOH + PbO_2 + Sb ⟶ H_2O + PbO + NaSbO2 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 NaOH + 3 PbO_2 + 2 Sb ⟶ H_2O + 3 PbO + 2 NaSbO2 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 NaOH | 2 | -2 PbO_2 | 3 | -3 Sb | 2 | -2 H_2O | 1 | 1 PbO | 3 | 3 NaSbO2 | 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 NaOH | 2 | -2 | -1/2 (Δ[NaOH])/(Δt) PbO_2 | 3 | -3 | -1/3 (Δ[PbO2])/(Δt) Sb | 2 | -2 | -1/2 (Δ[Sb])/(Δt) H_2O | 1 | 1 | (Δ[H2O])/(Δt) PbO | 3 | 3 | 1/3 (Δ[PbO])/(Δt) NaSbO2 | 2 | 2 | 1/2 (Δ[NaSbO2])/(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[NaOH])/(Δt) = -1/3 (Δ[PbO2])/(Δt) = -1/2 (Δ[Sb])/(Δt) = (Δ[H2O])/(Δt) = 1/3 (Δ[PbO])/(Δt) = 1/2 (Δ[NaSbO2])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)