O2 + H2 + (FeOH)3 = Fe(OH)3

O_2 oxygen + H_2 hydrogen + (FeOH)3 ⟶ Fe(OH)_3 iron(III) hydroxide

Balance the chemical equation algebraically: O_2 + H_2 + (FeOH)3 ⟶ Fe(OH)_3 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 O_2 + c_2 H_2 + c_3 (FeOH)3 ⟶ c_4 Fe(OH)_3 Set the number of atoms in the reactants equal to the number of atoms in the products for O, H and Fe: O: | 2 c_1 + 3 c_3 = 3 c_4 H: | 2 c_2 + 3 c_3 = 3 c_4 Fe: | 3 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_3 = 1 and solve the system of equations for the remaining coefficients: c_1 = 3 c_2 = 3 c_3 = 1 c_4 = 3 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 3 O_2 + 3 H_2 + (FeOH)3 ⟶ 3 Fe(OH)_3

+ + (FeOH)3 ⟶

oxygen + hydrogen + (FeOH)3 ⟶ iron(III) hydroxide

Construct the equilibrium constant, K, expression for: O_2 + H_2 + (FeOH)3 ⟶ Fe(OH)_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: 3 O_2 + 3 H_2 + (FeOH)3 ⟶ 3 Fe(OH)_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 O_2 | 3 | -3 H_2 | 3 | -3 (FeOH)3 | 1 | -1 Fe(OH)_3 | 3 | 3 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression O_2 | 3 | -3 | ([O2])^(-3) H_2 | 3 | -3 | ([H2])^(-3) (FeOH)3 | 1 | -1 | ([(FeOH)3])^(-1) Fe(OH)_3 | 3 | 3 | ([Fe(OH)3])^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 = ([O2])^(-3) ([H2])^(-3) ([(FeOH)3])^(-1) ([Fe(OH)3])^3 = ([Fe(OH)3])^3/(([O2])^3 ([H2])^3 [(FeOH)3])

Construct the rate of reaction expression for: O_2 + H_2 + (FeOH)3 ⟶ Fe(OH)_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: 3 O_2 + 3 H_2 + (FeOH)3 ⟶ 3 Fe(OH)_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 O_2 | 3 | -3 H_2 | 3 | -3 (FeOH)3 | 1 | -1 Fe(OH)_3 | 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 O_2 | 3 | -3 | -1/3 (Δ[O2])/(Δt) H_2 | 3 | -3 | -1/3 (Δ[H2])/(Δt) (FeOH)3 | 1 | -1 | -(Δ[(FeOH)3])/(Δt) Fe(OH)_3 | 3 | 3 | 1/3 (Δ[Fe(OH)3])/(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 = -1/3 (Δ[O2])/(Δt) = -1/3 (Δ[H2])/(Δt) = -(Δ[(FeOH)3])/(Δt) = 1/3 (Δ[Fe(OH)3])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| oxygen | hydrogen | (FeOH)3 | iron(III) hydroxide formula | O_2 | H_2 | (FeOH)3 | Fe(OH)_3 Hill formula | O_2 | H_2 | H3Fe3O3 | FeH_3O_3 name | oxygen | hydrogen | | iron(III) hydroxide IUPAC name | molecular oxygen | molecular hydrogen | | ferric trihydroxide

| oxygen | hydrogen | (FeOH)3 | iron(III) hydroxide molar mass | 31.998 g/mol | 2.016 g/mol | 218.56 g/mol | 106.87 g/mol phase | gas (at STP) | gas (at STP) | | melting point | -218 °C | -259.2 °C | | boiling point | -183 °C | -252.8 °C | | density | 0.001429 g/cm^3 (at 0 °C) | 8.99×10^-5 g/cm^3 (at 0 °C) | | surface tension | 0.01347 N/m | | | dynamic viscosity | 2.055×10^-5 Pa s (at 25 °C) | 8.9×10^-6 Pa s (at 25 °C) | | odor | odorless | odorless | |