FeO = O2 + Fe

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

Balance the chemical equation algebraically: FeO ⟶ O_2 + Fe Add stoichiometric coefficients, c_i, to the reactants and products: c_1 FeO ⟶ c_2 O_2 + c_3 Fe Set the number of atoms in the reactants equal to the number of atoms in the products for Fe and O: Fe: | c_1 = c_3 O: | c_1 = 2 c_2 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 = 2 c_2 = 1 c_3 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 2 FeO ⟶ O_2 + 2 Fe

⟶ +

iron(II) oxide ⟶ oxygen + iron

| iron(II) oxide | oxygen | iron molecular enthalpy | -272 kJ/mol | 0 kJ/mol | 0 kJ/mol total enthalpy | -544 kJ/mol | 0 kJ/mol | 0 kJ/mol | H_initial = -544 kJ/mol | H_final = 0 kJ/mol | ΔH_rxn^0 | 0 kJ/mol - -544 kJ/mol = 544 kJ/mol (endothermic) | |

| iron(II) oxide | oxygen | iron molecular entropy | 61 J/(mol K) | 205 J/(mol K) | 27 J/(mol K) total entropy | 122 J/(mol K) | 205 J/(mol K) | 54 J/(mol K) | S_initial = 122 J/(mol K) | S_final = 259 J/(mol K) | ΔS_rxn^0 | 259 J/(mol K) - 122 J/(mol K) = 137 J/(mol K) (endoentropic) | |

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

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

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

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