H2O + O2 + FeSO4 = (FeOH)SO4

H_2O water + O_2 oxygen + FeSO_4 duretter ⟶ FeOHSO4

Balance the chemical equation algebraically: H_2O + O_2 + FeSO_4 ⟶ FeOHSO4 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 H_2O + c_2 O_2 + c_3 FeSO_4 ⟶ c_4 FeOHSO4 Set the number of atoms in the reactants equal to the number of atoms in the products for H, O, Fe and S: H: | 2 c_1 = c_4 O: | c_1 + 2 c_2 + 4 c_3 = 5 c_4 Fe: | c_3 = c_4 S: | 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 = 4 c_4 = 4 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 2 H_2O + O_2 + 4 FeSO_4 ⟶ 4 FeOHSO4

+ + ⟶ FeOHSO4

water + oxygen + duretter ⟶ FeOHSO4

Construct the equilibrium constant, K, expression for: H_2O + O_2 + FeSO_4 ⟶ FeOHSO4 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 + 4 FeSO_4 ⟶ 4 FeOHSO4 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 FeSO_4 | 4 | -4 FeOHSO4 | 4 | 4 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) FeSO_4 | 4 | -4 | ([FeSO4])^(-4) FeOHSO4 | 4 | 4 | ([FeOHSO4])^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])^(-2) ([O2])^(-1) ([FeSO4])^(-4) ([FeOHSO4])^4 = ([FeOHSO4])^4/(([H2O])^2 [O2] ([FeSO4])^4)

Construct the rate of reaction expression for: H_2O + O_2 + FeSO_4 ⟶ FeOHSO4 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 + 4 FeSO_4 ⟶ 4 FeOHSO4 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 FeSO_4 | 4 | -4 FeOHSO4 | 4 | 4 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) FeSO_4 | 4 | -4 | -1/4 (Δ[FeSO4])/(Δt) FeOHSO4 | 4 | 4 | 1/4 (Δ[FeOHSO4])/(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) = -1/4 (Δ[FeSO4])/(Δt) = 1/4 (Δ[FeOHSO4])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| water | oxygen | duretter | FeOHSO4 formula | H_2O | O_2 | FeSO_4 | FeOHSO4 Hill formula | H_2O | O_2 | FeO_4S | HFeO5S name | water | oxygen | duretter | IUPAC name | water | molecular oxygen | iron(+2) cation sulfate |

| water | oxygen | duretter | FeOHSO4 molar mass | 18.015 g/mol | 31.998 g/mol | 151.9 g/mol | 168.91 g/mol phase | liquid (at STP) | gas (at STP) | | melting point | 0 °C | -218 °C | | boiling point | 99.9839 °C | -183 °C | | density | 1 g/cm^3 | 0.001429 g/cm^3 (at 0 °C) | 2.841 g/cm^3 | 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) | | odor | odorless | odorless | |