Fe + O3 = Fe2O3

Fe iron + O_3 ozone ⟶ Fe_2O_3 iron(III) oxide

Balance the chemical equation algebraically: Fe + O_3 ⟶ Fe_2O_3 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 Fe + c_2 O_3 ⟶ c_3 Fe_2O_3 Set the number of atoms in the reactants equal to the number of atoms in the products for Fe and O: Fe: | c_1 = 2 c_3 O: | 3 c_2 = 3 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_2 = 1 and solve the system of equations for the remaining coefficients: c_1 = 2 c_2 = 1 c_3 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 2 Fe + O_3 ⟶ Fe_2O_3

+ ⟶

iron + ozone ⟶ iron(III) oxide

| iron | ozone | iron(III) oxide molecular enthalpy | 0 kJ/mol | 142.7 kJ/mol | -826 kJ/mol total enthalpy | 0 kJ/mol | 142.7 kJ/mol | -826 kJ/mol | H_initial = 142.7 kJ/mol | | H_final = -826 kJ/mol ΔH_rxn^0 | -826 kJ/mol - 142.7 kJ/mol = -968.7 kJ/mol (exothermic) | |

| iron | ozone | iron(III) oxide molecular entropy | 27 J/(mol K) | 239 J/(mol K) | 90 J/(mol K) total entropy | 54 J/(mol K) | 239 J/(mol K) | 90 J/(mol K) | S_initial = 293 J/(mol K) | | S_final = 90 J/(mol K) ΔS_rxn^0 | 90 J/(mol K) - 293 J/(mol K) = -203 J/(mol K) (exoentropic) | |

Construct the equilibrium constant, K, expression for: Fe + O_3 ⟶ 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: 2 Fe + O_3 ⟶ 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 Fe | 2 | -2 O_3 | 1 | -1 Fe_2O_3 | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression Fe | 2 | -2 | ([Fe])^(-2) O_3 | 1 | -1 | ([O3])^(-1) Fe_2O_3 | 1 | 1 | [Fe2O3] 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 = ([Fe])^(-2) ([O3])^(-1) [Fe2O3] = ([Fe2O3])/(([Fe])^2 [O3])

Construct the rate of reaction expression for: Fe + O_3 ⟶ 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: 2 Fe + O_3 ⟶ 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 Fe | 2 | -2 O_3 | 1 | -1 Fe_2O_3 | 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 Fe | 2 | -2 | -1/2 (Δ[Fe])/(Δt) O_3 | 1 | -1 | -(Δ[O3])/(Δt) Fe_2O_3 | 1 | 1 | (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 = -1/2 (Δ[Fe])/(Δt) = -(Δ[O3])/(Δt) = (Δ[Fe2O3])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| iron | ozone | iron(III) oxide formula | Fe | O_3 | Fe_2O_3 name | iron | ozone | iron(III) oxide

| iron | ozone | iron(III) oxide molar mass | 55.845 g/mol | 47.997 g/mol | 159.69 g/mol phase | solid (at STP) | gas (at STP) | solid (at STP) melting point | 1535 °C | -192.2 °C | 1565 °C boiling point | 2750 °C | -111.9 °C | density | 7.874 g/cm^3 | 0.001962 g/cm^3 (at 25 °C) | 5.26 g/cm^3 solubility in water | insoluble | | insoluble odor | | | odorless