Bi(OH)3 = H2O + Bi2O3

Bi(OH)3 ⟶ H_2O water + Bi_2O_3 bismuth trioxide

Balance the chemical equation algebraically: Bi(OH)3 ⟶ H_2O + Bi_2O_3 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 Bi(OH)3 ⟶ c_2 H_2O + c_3 Bi_2O_3 Set the number of atoms in the reactants equal to the number of atoms in the products for Bi, O and H: Bi: | c_1 = 2 c_3 O: | 3 c_1 = c_2 + 3 c_3 H: | 3 c_1 = 2 c_2 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_3 = 1 and solve the system of equations for the remaining coefficients: c_1 = 2 c_2 = 3 c_3 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 2 Bi(OH)3 ⟶ 3 H_2O + Bi_2O_3

Bi(OH)3 ⟶ +

Bi(OH)3 ⟶ water + bismuth trioxide

Construct the equilibrium constant, K, expression for: Bi(OH)3 ⟶ H_2O + Bi_2O_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 Bi(OH)3 ⟶ 3 H_2O + Bi_2O_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 Bi(OH)3 | 2 | -2 H_2O | 3 | 3 Bi_2O_3 | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression Bi(OH)3 | 2 | -2 | ([Bi(OH)3])^(-2) H_2O | 3 | 3 | ([H2O])^3 Bi_2O_3 | 1 | 1 | [Bi2O3] 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 = ([Bi(OH)3])^(-2) ([H2O])^3 [Bi2O3] = (([H2O])^3 [Bi2O3])/([Bi(OH)3])^2

Construct the rate of reaction expression for: Bi(OH)3 ⟶ H_2O + Bi_2O_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 Bi(OH)3 ⟶ 3 H_2O + Bi_2O_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 Bi(OH)3 | 2 | -2 H_2O | 3 | 3 Bi_2O_3 | 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 Bi(OH)3 | 2 | -2 | -1/2 (Δ[Bi(OH)3])/(Δt) H_2O | 3 | 3 | 1/3 (Δ[H2O])/(Δt) Bi_2O_3 | 1 | 1 | (Δ[Bi2O3])/(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[Bi(OH)3])/(Δt) = 1/3 (Δ[H2O])/(Δt) = (Δ[Bi2O3])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| Bi(OH)3 | water | bismuth trioxide formula | Bi(OH)3 | H_2O | Bi_2O_3 Hill formula | H3BiO3 | H_2O | Bi_2O_3 name | | water | bismuth trioxide IUPAC name | | water | oxo-oxobismuthanyloxybismuthane

| Bi(OH)3 | water | bismuth trioxide molar mass | 260.001 g/mol | 18.015 g/mol | 465.958 g/mol phase | | liquid (at STP) | solid (at STP) melting point | | 0 °C | 825 °C boiling point | | 99.9839 °C | 1890 °C density | | 1 g/cm^3 | 8.9 g/cm^3 solubility in water | | | decomposes surface tension | | 0.0728 N/m | dynamic viscosity | | 8.9×10^-4 Pa s (at 25 °C) | odor | | odorless | odorless