H2O + O2 + Fe2(SO4)3 + CuFeS2 = H2SO4 + CuSO4 + FeSO4

H_2O water + O_2 oxygen + Fe_2(SO_4)_3·xH_2O iron(III) sulfate hydrate + CuFeS_2 copper(II) ferrous sulfide ⟶ H_2SO_4 sulfuric acid + CuSO_4 copper(II) sulfate + FeSO_4 duretter

Balance the chemical equation algebraically: H_2O + O_2 + Fe_2(SO_4)_3·xH_2O + CuFeS_2 ⟶ H_2SO_4 + CuSO_4 + FeSO_4 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 H_2O + c_2 O_2 + c_3 Fe_2(SO_4)_3·xH_2O + c_4 CuFeS_2 ⟶ c_5 H_2SO_4 + c_6 CuSO_4 + c_7 FeSO_4 Set the number of atoms in the reactants equal to the number of atoms in the products for H, O, Fe, S and Cu: H: | 2 c_1 = 2 c_5 O: | c_1 + 2 c_2 + 12 c_3 = 4 c_5 + 4 c_6 + 4 c_7 Fe: | 2 c_3 + c_4 = c_7 S: | 3 c_3 + 2 c_4 = c_5 + c_6 + c_7 Cu: | c_4 = 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_1 = 1 and solve the system of equations for the remaining coefficients: c_1 = 1 c_3 = 1 c_4 = c_2/4 + 1/8 c_5 = 1 c_6 = c_2/4 + 1/8 c_7 = c_2/4 + 17/8 Multiply by the least common denominator, 2, to eliminate fractional coefficients: c_1 = 2 c_3 = 2 c_4 = c_2/4 + 1/4 c_5 = 2 c_6 = c_2/4 + 1/4 c_7 = c_2/4 + 17/4 The resulting system of equations is still underdetermined, so an additional coefficient must be set arbitrarily. Set c_2 = 15 and solve for the remaining coefficients: c_1 = 2 c_2 = 15 c_3 = 2 c_4 = 4 c_5 = 2 c_6 = 4 c_7 = 8 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 2 H_2O + 15 O_2 + 2 Fe_2(SO_4)_3·xH_2O + 4 CuFeS_2 ⟶ 2 H_2SO_4 + 4 CuSO_4 + 8 FeSO_4

+ + + CuFeS_2 ⟶ + +

water + oxygen + iron(III) sulfate hydrate + copper(II) ferrous sulfide ⟶ sulfuric acid + copper(II) sulfate + duretter

Construct the equilibrium constant, K, expression for: H_2O + O_2 + Fe_2(SO_4)_3·xH_2O + CuFeS_2 ⟶ H_2SO_4 + CuSO_4 + FeSO_4 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 + 15 O_2 + 2 Fe_2(SO_4)_3·xH_2O + 4 CuFeS_2 ⟶ 2 H_2SO_4 + 4 CuSO_4 + 8 FeSO_4 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 | 15 | -15 Fe_2(SO_4)_3·xH_2O | 2 | -2 CuFeS_2 | 4 | -4 H_2SO_4 | 2 | 2 CuSO_4 | 4 | 4 FeSO_4 | 8 | 8 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 | 15 | -15 | ([O2])^(-15) Fe_2(SO_4)_3·xH_2O | 2 | -2 | ([Fe2(SO4)3·xH2O])^(-2) CuFeS_2 | 4 | -4 | ([CuFeS2])^(-4) H_2SO_4 | 2 | 2 | ([H2SO4])^2 CuSO_4 | 4 | 4 | ([CuSO4])^4 FeSO_4 | 8 | 8 | ([FeSO4])^8 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])^(-15) ([Fe2(SO4)3·xH2O])^(-2) ([CuFeS2])^(-4) ([H2SO4])^2 ([CuSO4])^4 ([FeSO4])^8 = (([H2SO4])^2 ([CuSO4])^4 ([FeSO4])^8)/(([H2O])^2 ([O2])^15 ([Fe2(SO4)3·xH2O])^2 ([CuFeS2])^4)

Construct the rate of reaction expression for: H_2O + O_2 + Fe_2(SO_4)_3·xH_2O + CuFeS_2 ⟶ H_2SO_4 + CuSO_4 + FeSO_4 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 + 15 O_2 + 2 Fe_2(SO_4)_3·xH_2O + 4 CuFeS_2 ⟶ 2 H_2SO_4 + 4 CuSO_4 + 8 FeSO_4 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 | 15 | -15 Fe_2(SO_4)_3·xH_2O | 2 | -2 CuFeS_2 | 4 | -4 H_2SO_4 | 2 | 2 CuSO_4 | 4 | 4 FeSO_4 | 8 | 8 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 | 15 | -15 | -1/15 (Δ[O2])/(Δt) Fe_2(SO_4)_3·xH_2O | 2 | -2 | -1/2 (Δ[Fe2(SO4)3·xH2O])/(Δt) CuFeS_2 | 4 | -4 | -1/4 (Δ[CuFeS2])/(Δt) H_2SO_4 | 2 | 2 | 1/2 (Δ[H2SO4])/(Δt) CuSO_4 | 4 | 4 | 1/4 (Δ[CuSO4])/(Δt) FeSO_4 | 8 | 8 | 1/8 (Δ[FeSO4])/(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) = -1/15 (Δ[O2])/(Δt) = -1/2 (Δ[Fe2(SO4)3·xH2O])/(Δt) = -1/4 (Δ[CuFeS2])/(Δt) = 1/2 (Δ[H2SO4])/(Δt) = 1/4 (Δ[CuSO4])/(Δt) = 1/8 (Δ[FeSO4])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| water | oxygen | iron(III) sulfate hydrate | copper(II) ferrous sulfide | sulfuric acid | copper(II) sulfate | duretter formula | H_2O | O_2 | Fe_2(SO_4)_3·xH_2O | CuFeS_2 | H_2SO_4 | CuSO_4 | FeSO_4 Hill formula | H_2O | O_2 | Fe_2O_12S_3 | CuFeS_2 | H_2O_4S | CuO_4S | FeO_4S name | water | oxygen | iron(III) sulfate hydrate | copper(II) ferrous sulfide | sulfuric acid | copper(II) sulfate | duretter IUPAC name | water | molecular oxygen | diferric trisulfate | | sulfuric acid | copper sulfate | iron(+2) cation sulfate