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O2 + Fe = Fe2O3 + FeO

Input interpretation

O_2 oxygen + Fe iron ⟶ Fe_2O_3 iron(III) oxide + FeO iron(II) oxide
O_2 oxygen + Fe iron ⟶ Fe_2O_3 iron(III) oxide + FeO iron(II) oxide

Balanced equation

Balance the chemical equation algebraically: O_2 + Fe ⟶ Fe_2O_3 + FeO Add stoichiometric coefficients, c_i, to the reactants and products: c_1 O_2 + c_2 Fe ⟶ c_3 Fe_2O_3 + c_4 FeO Set the number of atoms in the reactants equal to the number of atoms in the products for O and Fe: O: | 2 c_1 = 3 c_3 + c_4 Fe: | c_2 = 2 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_3 = 1 and solve the system of equations for the remaining coefficients: c_2 = 2 c_1 - 1 c_3 = 1 c_4 = 2 c_1 - 3 The resulting system of equations is still underdetermined, so an additional coefficient must be set arbitrarily. Set c_1 = 2 and solve for the remaining coefficients: c_1 = 2 c_2 = 3 c_3 = 1 c_4 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: |   | 2 O_2 + 3 Fe ⟶ Fe_2O_3 + FeO
Balance the chemical equation algebraically: O_2 + Fe ⟶ Fe_2O_3 + FeO Add stoichiometric coefficients, c_i, to the reactants and products: c_1 O_2 + c_2 Fe ⟶ c_3 Fe_2O_3 + c_4 FeO Set the number of atoms in the reactants equal to the number of atoms in the products for O and Fe: O: | 2 c_1 = 3 c_3 + c_4 Fe: | c_2 = 2 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_3 = 1 and solve the system of equations for the remaining coefficients: c_2 = 2 c_1 - 1 c_3 = 1 c_4 = 2 c_1 - 3 The resulting system of equations is still underdetermined, so an additional coefficient must be set arbitrarily. Set c_1 = 2 and solve for the remaining coefficients: c_1 = 2 c_2 = 3 c_3 = 1 c_4 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 2 O_2 + 3 Fe ⟶ Fe_2O_3 + FeO

Structures

 + ⟶ +
+ ⟶ +

Names

oxygen + iron ⟶ iron(III) oxide + iron(II) oxide
oxygen + iron ⟶ iron(III) oxide + iron(II) oxide

Reaction thermodynamics

Enthalpy

 | oxygen | iron | iron(III) oxide | iron(II) oxide molecular enthalpy | 0 kJ/mol | 0 kJ/mol | -826 kJ/mol | -272 kJ/mol total enthalpy | 0 kJ/mol | 0 kJ/mol | -826 kJ/mol | -272 kJ/mol  | H_initial = 0 kJ/mol | | H_final = -1098 kJ/mol |  ΔH_rxn^0 | -1098 kJ/mol - 0 kJ/mol = -1098 kJ/mol (exothermic) | | |
| oxygen | iron | iron(III) oxide | iron(II) oxide molecular enthalpy | 0 kJ/mol | 0 kJ/mol | -826 kJ/mol | -272 kJ/mol total enthalpy | 0 kJ/mol | 0 kJ/mol | -826 kJ/mol | -272 kJ/mol | H_initial = 0 kJ/mol | | H_final = -1098 kJ/mol | ΔH_rxn^0 | -1098 kJ/mol - 0 kJ/mol = -1098 kJ/mol (exothermic) | | |

Entropy

 | oxygen | iron | iron(III) oxide | iron(II) oxide molecular entropy | 205 J/(mol K) | 27 J/(mol K) | 90 J/(mol K) | 61 J/(mol K) total entropy | 410 J/(mol K) | 81 J/(mol K) | 90 J/(mol K) | 61 J/(mol K)  | S_initial = 491 J/(mol K) | | S_final = 151 J/(mol K) |  ΔS_rxn^0 | 151 J/(mol K) - 491 J/(mol K) = -340 J/(mol K) (exoentropic) | | |
| oxygen | iron | iron(III) oxide | iron(II) oxide molecular entropy | 205 J/(mol K) | 27 J/(mol K) | 90 J/(mol K) | 61 J/(mol K) total entropy | 410 J/(mol K) | 81 J/(mol K) | 90 J/(mol K) | 61 J/(mol K) | S_initial = 491 J/(mol K) | | S_final = 151 J/(mol K) | ΔS_rxn^0 | 151 J/(mol K) - 491 J/(mol K) = -340 J/(mol K) (exoentropic) | | |

Equilibrium constant

Construct the equilibrium constant, K, expression for: O_2 + Fe ⟶ Fe_2O_3 + FeO 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 O_2 + 3 Fe ⟶ Fe_2O_3 + FeO 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 | 2 | -2 Fe | 3 | -3 Fe_2O_3 | 1 | 1 FeO | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression O_2 | 2 | -2 | ([O2])^(-2) Fe | 3 | -3 | ([Fe])^(-3) Fe_2O_3 | 1 | 1 | [Fe2O3] FeO | 1 | 1 | [FeO] 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])^(-2) ([Fe])^(-3) [Fe2O3] [FeO] = ([Fe2O3] [FeO])/(([O2])^2 ([Fe])^3)
Construct the equilibrium constant, K, expression for: O_2 + Fe ⟶ Fe_2O_3 + FeO 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 O_2 + 3 Fe ⟶ Fe_2O_3 + FeO 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 | 2 | -2 Fe | 3 | -3 Fe_2O_3 | 1 | 1 FeO | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression O_2 | 2 | -2 | ([O2])^(-2) Fe | 3 | -3 | ([Fe])^(-3) Fe_2O_3 | 1 | 1 | [Fe2O3] FeO | 1 | 1 | [FeO] 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])^(-2) ([Fe])^(-3) [Fe2O3] [FeO] = ([Fe2O3] [FeO])/(([O2])^2 ([Fe])^3)

Rate of reaction

Construct the rate of reaction expression for: O_2 + Fe ⟶ Fe_2O_3 + FeO 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 O_2 + 3 Fe ⟶ Fe_2O_3 + FeO 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 | 2 | -2 Fe | 3 | -3 Fe_2O_3 | 1 | 1 FeO | 1 | 1 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 | 2 | -2 | -1/2 (Δ[O2])/(Δt) Fe | 3 | -3 | -1/3 (Δ[Fe])/(Δt) Fe_2O_3 | 1 | 1 | (Δ[Fe2O3])/(Δt) FeO | 1 | 1 | (Δ[FeO])/(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[O2])/(Δt) = -1/3 (Δ[Fe])/(Δt) = (Δ[Fe2O3])/(Δt) = (Δ[FeO])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)
Construct the rate of reaction expression for: O_2 + Fe ⟶ Fe_2O_3 + FeO 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 O_2 + 3 Fe ⟶ Fe_2O_3 + FeO 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 | 2 | -2 Fe | 3 | -3 Fe_2O_3 | 1 | 1 FeO | 1 | 1 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 | 2 | -2 | -1/2 (Δ[O2])/(Δt) Fe | 3 | -3 | -1/3 (Δ[Fe])/(Δt) Fe_2O_3 | 1 | 1 | (Δ[Fe2O3])/(Δt) FeO | 1 | 1 | (Δ[FeO])/(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[O2])/(Δt) = -1/3 (Δ[Fe])/(Δt) = (Δ[Fe2O3])/(Δt) = (Δ[FeO])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

Chemical names and formulas

 | oxygen | iron | iron(III) oxide | iron(II) oxide formula | O_2 | Fe | Fe_2O_3 | FeO name | oxygen | iron | iron(III) oxide | iron(II) oxide IUPAC name | molecular oxygen | iron | | oxoiron
| oxygen | iron | iron(III) oxide | iron(II) oxide formula | O_2 | Fe | Fe_2O_3 | FeO name | oxygen | iron | iron(III) oxide | iron(II) oxide IUPAC name | molecular oxygen | iron | | oxoiron

Substance properties

 | oxygen | iron | iron(III) oxide | iron(II) oxide molar mass | 31.998 g/mol | 55.845 g/mol | 159.69 g/mol | 71.844 g/mol phase | gas (at STP) | solid (at STP) | solid (at STP) | solid (at STP) melting point | -218 °C | 1535 °C | 1565 °C | 1360 °C boiling point | -183 °C | 2750 °C | |  density | 0.001429 g/cm^3 (at 0 °C) | 7.874 g/cm^3 | 5.26 g/cm^3 | 5.7 g/cm^3 solubility in water | | insoluble | insoluble | insoluble surface tension | 0.01347 N/m | | |  dynamic viscosity | 2.055×10^-5 Pa s (at 25 °C) | | |  odor | odorless | | odorless |
| oxygen | iron | iron(III) oxide | iron(II) oxide molar mass | 31.998 g/mol | 55.845 g/mol | 159.69 g/mol | 71.844 g/mol phase | gas (at STP) | solid (at STP) | solid (at STP) | solid (at STP) melting point | -218 °C | 1535 °C | 1565 °C | 1360 °C boiling point | -183 °C | 2750 °C | | density | 0.001429 g/cm^3 (at 0 °C) | 7.874 g/cm^3 | 5.26 g/cm^3 | 5.7 g/cm^3 solubility in water | | insoluble | insoluble | insoluble surface tension | 0.01347 N/m | | | dynamic viscosity | 2.055×10^-5 Pa s (at 25 °C) | | | odor | odorless | | odorless |

Units