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O2 + FeS = Fe2O3 + SO3

Input interpretation

O_2 oxygen + FeS ferrous sulfide ⟶ Fe_2O_3 iron(III) oxide + SO_3 sulfur trioxide
O_2 oxygen + FeS ferrous sulfide ⟶ Fe_2O_3 iron(III) oxide + SO_3 sulfur trioxide

Balanced equation

Balance the chemical equation algebraically: O_2 + FeS ⟶ Fe_2O_3 + SO_3 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 O_2 + c_2 FeS ⟶ c_3 Fe_2O_3 + c_4 SO_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 = 3 c_3 + 3 c_4 Fe: | c_2 = 2 c_3 S: | c_2 = 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 = 9/2 c_2 = 2 c_3 = 1 c_4 = 2 Multiply by the least common denominator, 2, to eliminate fractional coefficients: c_1 = 9 c_2 = 4 c_3 = 2 c_4 = 4 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: |   | 9 O_2 + 4 FeS ⟶ 2 Fe_2O_3 + 4 SO_3
Balance the chemical equation algebraically: O_2 + FeS ⟶ Fe_2O_3 + SO_3 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 O_2 + c_2 FeS ⟶ c_3 Fe_2O_3 + c_4 SO_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 = 3 c_3 + 3 c_4 Fe: | c_2 = 2 c_3 S: | c_2 = 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 = 9/2 c_2 = 2 c_3 = 1 c_4 = 2 Multiply by the least common denominator, 2, to eliminate fractional coefficients: c_1 = 9 c_2 = 4 c_3 = 2 c_4 = 4 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 9 O_2 + 4 FeS ⟶ 2 Fe_2O_3 + 4 SO_3

Structures

 + ⟶ +
+ ⟶ +

Names

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

Reaction thermodynamics

Gibbs free energy

 | oxygen | ferrous sulfide | iron(III) oxide | sulfur trioxide molecular free energy | 231.7 kJ/mol | -100.4 kJ/mol | -742.2 kJ/mol | -373.8 kJ/mol total free energy | 2085 kJ/mol | -401.6 kJ/mol | -1484 kJ/mol | -1495 kJ/mol  | G_initial = 1684 kJ/mol | | G_final = -2980 kJ/mol |  ΔG_rxn^0 | -2980 kJ/mol - 1684 kJ/mol = -4663 kJ/mol (exergonic) | | |
| oxygen | ferrous sulfide | iron(III) oxide | sulfur trioxide molecular free energy | 231.7 kJ/mol | -100.4 kJ/mol | -742.2 kJ/mol | -373.8 kJ/mol total free energy | 2085 kJ/mol | -401.6 kJ/mol | -1484 kJ/mol | -1495 kJ/mol | G_initial = 1684 kJ/mol | | G_final = -2980 kJ/mol | ΔG_rxn^0 | -2980 kJ/mol - 1684 kJ/mol = -4663 kJ/mol (exergonic) | | |

Equilibrium constant

Construct the equilibrium constant, K, expression for: O_2 + FeS ⟶ Fe_2O_3 + SO_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: 9 O_2 + 4 FeS ⟶ 2 Fe_2O_3 + 4 SO_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 | 9 | -9 FeS | 4 | -4 Fe_2O_3 | 2 | 2 SO_3 | 4 | 4 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression O_2 | 9 | -9 | ([O2])^(-9) FeS | 4 | -4 | ([FeS])^(-4) Fe_2O_3 | 2 | 2 | ([Fe2O3])^2 SO_3 | 4 | 4 | ([SO3])^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 = ([O2])^(-9) ([FeS])^(-4) ([Fe2O3])^2 ([SO3])^4 = (([Fe2O3])^2 ([SO3])^4)/(([O2])^9 ([FeS])^4)
Construct the equilibrium constant, K, expression for: O_2 + FeS ⟶ Fe_2O_3 + SO_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: 9 O_2 + 4 FeS ⟶ 2 Fe_2O_3 + 4 SO_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 | 9 | -9 FeS | 4 | -4 Fe_2O_3 | 2 | 2 SO_3 | 4 | 4 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression O_2 | 9 | -9 | ([O2])^(-9) FeS | 4 | -4 | ([FeS])^(-4) Fe_2O_3 | 2 | 2 | ([Fe2O3])^2 SO_3 | 4 | 4 | ([SO3])^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 = ([O2])^(-9) ([FeS])^(-4) ([Fe2O3])^2 ([SO3])^4 = (([Fe2O3])^2 ([SO3])^4)/(([O2])^9 ([FeS])^4)

Rate of reaction

Construct the rate of reaction expression for: O_2 + FeS ⟶ Fe_2O_3 + SO_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: 9 O_2 + 4 FeS ⟶ 2 Fe_2O_3 + 4 SO_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 | 9 | -9 FeS | 4 | -4 Fe_2O_3 | 2 | 2 SO_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 O_2 | 9 | -9 | -1/9 (Δ[O2])/(Δt) FeS | 4 | -4 | -1/4 (Δ[FeS])/(Δt) Fe_2O_3 | 2 | 2 | 1/2 (Δ[Fe2O3])/(Δt) SO_3 | 4 | 4 | 1/4 (Δ[SO3])/(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 = -1/9 (Δ[O2])/(Δt) = -1/4 (Δ[FeS])/(Δt) = 1/2 (Δ[Fe2O3])/(Δt) = 1/4 (Δ[SO3])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)
Construct the rate of reaction expression for: O_2 + FeS ⟶ Fe_2O_3 + SO_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: 9 O_2 + 4 FeS ⟶ 2 Fe_2O_3 + 4 SO_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 | 9 | -9 FeS | 4 | -4 Fe_2O_3 | 2 | 2 SO_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 O_2 | 9 | -9 | -1/9 (Δ[O2])/(Δt) FeS | 4 | -4 | -1/4 (Δ[FeS])/(Δt) Fe_2O_3 | 2 | 2 | 1/2 (Δ[Fe2O3])/(Δt) SO_3 | 4 | 4 | 1/4 (Δ[SO3])/(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 = -1/9 (Δ[O2])/(Δt) = -1/4 (Δ[FeS])/(Δt) = 1/2 (Δ[Fe2O3])/(Δt) = 1/4 (Δ[SO3])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

Chemical names and formulas

 | oxygen | ferrous sulfide | iron(III) oxide | sulfur trioxide formula | O_2 | FeS | Fe_2O_3 | SO_3 Hill formula | O_2 | FeS | Fe_2O_3 | O_3S name | oxygen | ferrous sulfide | iron(III) oxide | sulfur trioxide IUPAC name | molecular oxygen | | | sulfur trioxide
| oxygen | ferrous sulfide | iron(III) oxide | sulfur trioxide formula | O_2 | FeS | Fe_2O_3 | SO_3 Hill formula | O_2 | FeS | Fe_2O_3 | O_3S name | oxygen | ferrous sulfide | iron(III) oxide | sulfur trioxide IUPAC name | molecular oxygen | | | sulfur trioxide

Substance properties

 | oxygen | ferrous sulfide | iron(III) oxide | sulfur trioxide molar mass | 31.998 g/mol | 87.9 g/mol | 159.69 g/mol | 80.06 g/mol phase | gas (at STP) | solid (at STP) | solid (at STP) | liquid (at STP) melting point | -218 °C | 1195 °C | 1565 °C | 16.8 °C boiling point | -183 °C | | | 44.7 °C density | 0.001429 g/cm^3 (at 0 °C) | 4.84 g/cm^3 | 5.26 g/cm^3 | 1.97 g/cm^3 solubility in water | | insoluble | insoluble | reacts surface tension | 0.01347 N/m | | |  dynamic viscosity | 2.055×10^-5 Pa s (at 25 °C) | 0.00343 Pa s (at 1250 °C) | | 0.00159 Pa s (at 30 °C) odor | odorless | | odorless |
| oxygen | ferrous sulfide | iron(III) oxide | sulfur trioxide molar mass | 31.998 g/mol | 87.9 g/mol | 159.69 g/mol | 80.06 g/mol phase | gas (at STP) | solid (at STP) | solid (at STP) | liquid (at STP) melting point | -218 °C | 1195 °C | 1565 °C | 16.8 °C boiling point | -183 °C | | | 44.7 °C density | 0.001429 g/cm^3 (at 0 °C) | 4.84 g/cm^3 | 5.26 g/cm^3 | 1.97 g/cm^3 solubility in water | | insoluble | insoluble | reacts surface tension | 0.01347 N/m | | | dynamic viscosity | 2.055×10^-5 Pa s (at 25 °C) | 0.00343 Pa s (at 1250 °C) | | 0.00159 Pa s (at 30 °C) odor | odorless | | odorless |

Units