H2O + O2 + FeCO3 = CO2 + Fe(OH)3

H_2O water + O_2 oxygen + FeCO_3 iron(II) carbonate ⟶ CO_2 carbon dioxide + Fe(OH)_3 iron(III) hydroxide

Balance the chemical equation algebraically: H_2O + O_2 + FeCO_3 ⟶ CO_2 + Fe(OH)_3 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 H_2O + c_2 O_2 + c_3 FeCO_3 ⟶ c_4 CO_2 + c_5 Fe(OH)_3 Set the number of atoms in the reactants equal to the number of atoms in the products for H, O, C and Fe: H: | 2 c_1 = 3 c_5 O: | c_1 + 2 c_2 + 3 c_3 = 2 c_4 + 3 c_5 C: | c_3 = c_4 Fe: | c_3 = c_5 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 = 6 c_2 = 1 c_3 = 4 c_4 = 4 c_5 = 4 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 6 H_2O + O_2 + 4 FeCO_3 ⟶ 4 CO_2 + 4 Fe(OH)_3

+ + ⟶ +

water + oxygen + iron(II) carbonate ⟶ carbon dioxide + iron(III) hydroxide

Construct the equilibrium constant, K, expression for: H_2O + O_2 + FeCO_3 ⟶ CO_2 + 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: 6 H_2O + O_2 + 4 FeCO_3 ⟶ 4 CO_2 + 4 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 H_2O | 6 | -6 O_2 | 1 | -1 FeCO_3 | 4 | -4 CO_2 | 4 | 4 Fe(OH)_3 | 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 | 6 | -6 | ([H2O])^(-6) O_2 | 1 | -1 | ([O2])^(-1) FeCO_3 | 4 | -4 | ([FeCO3])^(-4) CO_2 | 4 | 4 | ([CO2])^4 Fe(OH)_3 | 4 | 4 | ([Fe(OH)3])^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])^(-6) ([O2])^(-1) ([FeCO3])^(-4) ([CO2])^4 ([Fe(OH)3])^4 = (([CO2])^4 ([Fe(OH)3])^4)/(([H2O])^6 [O2] ([FeCO3])^4)

Construct the rate of reaction expression for: H_2O + O_2 + FeCO_3 ⟶ CO_2 + 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: 6 H_2O + O_2 + 4 FeCO_3 ⟶ 4 CO_2 + 4 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 H_2O | 6 | -6 O_2 | 1 | -1 FeCO_3 | 4 | -4 CO_2 | 4 | 4 Fe(OH)_3 | 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 | 6 | -6 | -1/6 (Δ[H2O])/(Δt) O_2 | 1 | -1 | -(Δ[O2])/(Δt) FeCO_3 | 4 | -4 | -1/4 (Δ[FeCO3])/(Δt) CO_2 | 4 | 4 | 1/4 (Δ[CO2])/(Δt) Fe(OH)_3 | 4 | 4 | 1/4 (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 = -1/6 (Δ[H2O])/(Δt) = -(Δ[O2])/(Δt) = -1/4 (Δ[FeCO3])/(Δt) = 1/4 (Δ[CO2])/(Δt) = 1/4 (Δ[Fe(OH)3])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| water | oxygen | iron(II) carbonate | carbon dioxide | iron(III) hydroxide formula | H_2O | O_2 | FeCO_3 | CO_2 | Fe(OH)_3 Hill formula | H_2O | O_2 | CFeO_3 | CO_2 | FeH_3O_3 name | water | oxygen | iron(II) carbonate | carbon dioxide | iron(III) hydroxide IUPAC name | water | molecular oxygen | ferrous carbonate | carbon dioxide | ferric trihydroxide