O2 + FeS = SO2 + Fe2O3

O_2 (oxygen) + FeS (ferrous sulfide) ⟶ SO_2 (sulfur dioxide) + Fe_2O_3 (iron(III) oxide)

Balance the chemical equation algebraically: O_2 + FeS ⟶ SO_2 + Fe_2O_3 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 O_2 + c_2 FeS ⟶ c_3 SO_2 + c_4 Fe_2O_3 Set the number of atoms in the reactants equal to the number of atoms in the products for O, Fe and S: O: | 2 c_1 = 2 c_3 + 3 c_4 Fe: | c_2 = 2 c_4 S: | c_2 = c_3 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_4 = 1 and solve the system of equations for the remaining coefficients: c_1 = 7/2 c_2 = 2 c_3 = 2 c_4 = 1 Multiply by the least common denominator, 2, to eliminate fractional coefficients: c_1 = 7 c_2 = 4 c_3 = 4 c_4 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 7 O_2 + 4 FeS ⟶ 4 SO_2 + 2 Fe_2O_3

+ ⟶ +

oxygen + ferrous sulfide ⟶ sulfur dioxide + iron(III) oxide

| oxygen | ferrous sulfide | sulfur dioxide | iron(III) oxide molecular enthalpy | 0 kJ/mol | -100 kJ/mol | -296.8 kJ/mol | -826 kJ/mol total enthalpy | 0 kJ/mol | -400 kJ/mol | -1187 kJ/mol | -1652 kJ/mol | H_initial = -400 kJ/mol | | H_final = -2839 kJ/mol | ΔH_rxn^0 | -2839 kJ/mol - -400 kJ/mol = -2439 kJ/mol (exothermic) | | |

| oxygen | ferrous sulfide | sulfur dioxide | iron(III) oxide molecular free energy | 231.7 kJ/mol | -100.4 kJ/mol | -300.1 kJ/mol | -742.2 kJ/mol total free energy | 1622 kJ/mol | -401.6 kJ/mol | -1200 kJ/mol | -1484 kJ/mol | G_initial = 1220 kJ/mol | | G_final = -2685 kJ/mol | ΔG_rxn^0 | -2685 kJ/mol - 1220 kJ/mol = -3905 kJ/mol (exergonic) | | |

| oxygen | ferrous sulfide | sulfur dioxide | iron(III) oxide molecular entropy | 205 J/(mol K) | 67 J/(mol K) | 248 J/(mol K) | 90 J/(mol K) total entropy | 1435 J/(mol K) | 268 J/(mol K) | 992 J/(mol K) | 180 J/(mol K) | S_initial = 1703 J/(mol K) | | S_final = 1172 J/(mol K) | ΔS_rxn^0 | 1172 J/(mol K) - 1703 J/(mol K) = -531 J/(mol K) (exoentropic) | | |

Construct the equilibrium constant, K, expression for: O_2 + FeS ⟶ SO_2 + Fe_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: 7 O_2 + 4 FeS ⟶ 4 SO_2 + 2 Fe_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 O_2 | 7 | -7 FeS | 4 | -4 SO_2 | 4 | 4 Fe_2O_3 | 2 | 2 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression O_2 | 7 | -7 | ([O2])^(-7) FeS | 4 | -4 | ([FeS])^(-4) SO_2 | 4 | 4 | ([SO2])^4 Fe_2O_3 | 2 | 2 | ([Fe2O3])^2 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])^(-7) ([FeS])^(-4) ([SO2])^4 ([Fe2O3])^2 = (([SO2])^4 ([Fe2O3])^2)/(([O2])^7 ([FeS])^4)

Construct the rate of reaction expression for: O_2 + FeS ⟶ SO_2 + Fe_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: 7 O_2 + 4 FeS ⟶ 4 SO_2 + 2 Fe_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 O_2 | 7 | -7 FeS | 4 | -4 SO_2 | 4 | 4 Fe_2O_3 | 2 | 2 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 | 7 | -7 | -1/7 (Δ[O2])/(Δt) FeS | 4 | -4 | -1/4 (Δ[FeS])/(Δt) SO_2 | 4 | 4 | 1/4 (Δ[SO2])/(Δt) Fe_2O_3 | 2 | 2 | 1/2 (Δ[Fe2O3])/(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 = -1/7 (Δ[O2])/(Δt) = -1/4 (Δ[FeS])/(Δt) = 1/4 (Δ[SO2])/(Δt) = 1/2 (Δ[Fe2O3])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| oxygen | ferrous sulfide | sulfur dioxide | iron(III) oxide formula | O_2 | FeS | SO_2 | Fe_2O_3 Hill formula | O_2 | FeS | O_2S | Fe_2O_3 name | oxygen | ferrous sulfide | sulfur dioxide | iron(III) oxide IUPAC name | molecular oxygen | | sulfur dioxide |

| oxygen | ferrous sulfide | sulfur dioxide | iron(III) oxide molar mass | 31.998 g/mol | 87.9 g/mol | 64.06 g/mol | 159.69 g/mol phase | gas (at STP) | solid (at STP) | gas (at STP) | solid (at STP) melting point | -218 °C | 1195 °C | -73 °C | 1565 °C boiling point | -183 °C | | -10 °C | density | 0.001429 g/cm^3 (at 0 °C) | 4.84 g/cm^3 | 0.002619 g/cm^3 (at 25 °C) | 5.26 g/cm^3 solubility in water | | insoluble | | insoluble surface tension | 0.01347 N/m | | 0.02859 N/m | dynamic viscosity | 2.055×10^-5 Pa s (at 25 °C) | 0.00343 Pa s (at 1250 °C) | 1.282×10^-5 Pa s (at 25 °C) | odor | odorless | | | odorless