Input interpretation
![H_2 hydrogen + Fe_2O_3 iron(III) oxide ⟶ H_2O water + FeO·Fe_2O_3 iron(II, III) oxide](../image_source/d7b8e4d333bfd126a7a5de915dfe7165.png)
H_2 hydrogen + Fe_2O_3 iron(III) oxide ⟶ H_2O water + FeO·Fe_2O_3 iron(II, III) oxide
Balanced equation
![Balance the chemical equation algebraically: H_2 + Fe_2O_3 ⟶ H_2O + FeO·Fe_2O_3 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 H_2 + c_2 Fe_2O_3 ⟶ c_3 H_2O + c_4 FeO·Fe_2O_3 Set the number of atoms in the reactants equal to the number of atoms in the products for H, Fe and O: H: | 2 c_1 = 2 c_3 Fe: | 2 c_2 = 3 c_4 O: | 3 c_2 = c_3 + 4 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_1 = 1 and solve the system of equations for the remaining coefficients: c_1 = 1 c_2 = 3 c_3 = 1 c_4 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | H_2 + 3 Fe_2O_3 ⟶ H_2O + 2 FeO·Fe_2O_3](../image_source/6abe971b0128edc97d5861fd6b340898.png)
Balance the chemical equation algebraically: H_2 + Fe_2O_3 ⟶ H_2O + FeO·Fe_2O_3 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 H_2 + c_2 Fe_2O_3 ⟶ c_3 H_2O + c_4 FeO·Fe_2O_3 Set the number of atoms in the reactants equal to the number of atoms in the products for H, Fe and O: H: | 2 c_1 = 2 c_3 Fe: | 2 c_2 = 3 c_4 O: | 3 c_2 = c_3 + 4 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_1 = 1 and solve the system of equations for the remaining coefficients: c_1 = 1 c_2 = 3 c_3 = 1 c_4 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | H_2 + 3 Fe_2O_3 ⟶ H_2O + 2 FeO·Fe_2O_3
Structures
![+ ⟶ +](../image_source/f795359934dc3f52bc65f45f9b966e0a.png)
+ ⟶ +
Names
![hydrogen + iron(III) oxide ⟶ water + iron(II, III) oxide](../image_source/44158e558e68105d702be047e3e5e823.png)
hydrogen + iron(III) oxide ⟶ water + iron(II, III) oxide
Reaction thermodynamics
Enthalpy
![| hydrogen | iron(III) oxide | water | iron(II, III) oxide molecular enthalpy | 0 kJ/mol | -826 kJ/mol | -285.8 kJ/mol | -1118 kJ/mol total enthalpy | 0 kJ/mol | -2478 kJ/mol | -285.8 kJ/mol | -2237 kJ/mol | H_initial = -2478 kJ/mol | | H_final = -2523 kJ/mol | ΔH_rxn^0 | -2523 kJ/mol - -2478 kJ/mol = -44.63 kJ/mol (exothermic) | | |](../image_source/dd7e081ec0cea00cd7ebe3cb88907bcd.png)
| hydrogen | iron(III) oxide | water | iron(II, III) oxide molecular enthalpy | 0 kJ/mol | -826 kJ/mol | -285.8 kJ/mol | -1118 kJ/mol total enthalpy | 0 kJ/mol | -2478 kJ/mol | -285.8 kJ/mol | -2237 kJ/mol | H_initial = -2478 kJ/mol | | H_final = -2523 kJ/mol | ΔH_rxn^0 | -2523 kJ/mol - -2478 kJ/mol = -44.63 kJ/mol (exothermic) | | |
Gibbs free energy
![| hydrogen | iron(III) oxide | water | iron(II, III) oxide molecular free energy | 0 kJ/mol | -742.2 kJ/mol | -237.1 kJ/mol | -1015 kJ/mol total free energy | 0 kJ/mol | -2227 kJ/mol | -237.1 kJ/mol | -2031 kJ/mol | G_initial = -2227 kJ/mol | | G_final = -2268 kJ/mol | ΔG_rxn^0 | -2268 kJ/mol - -2227 kJ/mol = -41.3 kJ/mol (exergonic) | | |](../image_source/af884b6f2ca92eacbe6104c4e2317873.png)
| hydrogen | iron(III) oxide | water | iron(II, III) oxide molecular free energy | 0 kJ/mol | -742.2 kJ/mol | -237.1 kJ/mol | -1015 kJ/mol total free energy | 0 kJ/mol | -2227 kJ/mol | -237.1 kJ/mol | -2031 kJ/mol | G_initial = -2227 kJ/mol | | G_final = -2268 kJ/mol | ΔG_rxn^0 | -2268 kJ/mol - -2227 kJ/mol = -41.3 kJ/mol (exergonic) | | |
Equilibrium constant
![Construct the equilibrium constant, K, expression for: H_2 + Fe_2O_3 ⟶ H_2O + FeO·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: H_2 + 3 Fe_2O_3 ⟶ H_2O + 2 FeO·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 H_2 | 1 | -1 Fe_2O_3 | 3 | -3 H_2O | 1 | 1 FeO·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 H_2 | 1 | -1 | ([H2])^(-1) Fe_2O_3 | 3 | -3 | ([Fe2O3])^(-3) H_2O | 1 | 1 | [H2O] FeO·Fe_2O_3 | 2 | 2 | ([FeO·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 = ([H2])^(-1) ([Fe2O3])^(-3) [H2O] ([FeO·Fe2O3])^2 = ([H2O] ([FeO·Fe2O3])^2)/([H2] ([Fe2O3])^3)](../image_source/df3cd611bc7f79b8036888bf2a5d6dec.png)
Construct the equilibrium constant, K, expression for: H_2 + Fe_2O_3 ⟶ H_2O + FeO·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: H_2 + 3 Fe_2O_3 ⟶ H_2O + 2 FeO·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 H_2 | 1 | -1 Fe_2O_3 | 3 | -3 H_2O | 1 | 1 FeO·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 H_2 | 1 | -1 | ([H2])^(-1) Fe_2O_3 | 3 | -3 | ([Fe2O3])^(-3) H_2O | 1 | 1 | [H2O] FeO·Fe_2O_3 | 2 | 2 | ([FeO·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 = ([H2])^(-1) ([Fe2O3])^(-3) [H2O] ([FeO·Fe2O3])^2 = ([H2O] ([FeO·Fe2O3])^2)/([H2] ([Fe2O3])^3)
Rate of reaction
![Construct the rate of reaction expression for: H_2 + Fe_2O_3 ⟶ H_2O + FeO·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: H_2 + 3 Fe_2O_3 ⟶ H_2O + 2 FeO·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 H_2 | 1 | -1 Fe_2O_3 | 3 | -3 H_2O | 1 | 1 FeO·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 H_2 | 1 | -1 | -(Δ[H2])/(Δt) Fe_2O_3 | 3 | -3 | -1/3 (Δ[Fe2O3])/(Δt) H_2O | 1 | 1 | (Δ[H2O])/(Δt) FeO·Fe_2O_3 | 2 | 2 | 1/2 (Δ[FeO·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 = -(Δ[H2])/(Δt) = -1/3 (Δ[Fe2O3])/(Δt) = (Δ[H2O])/(Δt) = 1/2 (Δ[FeO·Fe2O3])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)](../image_source/69d1aa55baf421e21aea80274b737844.png)
Construct the rate of reaction expression for: H_2 + Fe_2O_3 ⟶ H_2O + FeO·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: H_2 + 3 Fe_2O_3 ⟶ H_2O + 2 FeO·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 H_2 | 1 | -1 Fe_2O_3 | 3 | -3 H_2O | 1 | 1 FeO·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 H_2 | 1 | -1 | -(Δ[H2])/(Δt) Fe_2O_3 | 3 | -3 | -1/3 (Δ[Fe2O3])/(Δt) H_2O | 1 | 1 | (Δ[H2O])/(Δt) FeO·Fe_2O_3 | 2 | 2 | 1/2 (Δ[FeO·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 = -(Δ[H2])/(Δt) = -1/3 (Δ[Fe2O3])/(Δt) = (Δ[H2O])/(Δt) = 1/2 (Δ[FeO·Fe2O3])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)
Chemical names and formulas
![| hydrogen | iron(III) oxide | water | iron(II, III) oxide formula | H_2 | Fe_2O_3 | H_2O | FeO·Fe_2O_3 Hill formula | H_2 | Fe_2O_3 | H_2O | Fe_3O_4 name | hydrogen | iron(III) oxide | water | iron(II, III) oxide IUPAC name | molecular hydrogen | | water |](../image_source/3bbaf09ca58b5f68fca87a519e79e1d7.png)
| hydrogen | iron(III) oxide | water | iron(II, III) oxide formula | H_2 | Fe_2O_3 | H_2O | FeO·Fe_2O_3 Hill formula | H_2 | Fe_2O_3 | H_2O | Fe_3O_4 name | hydrogen | iron(III) oxide | water | iron(II, III) oxide IUPAC name | molecular hydrogen | | water |
Substance properties
![| hydrogen | iron(III) oxide | water | iron(II, III) oxide molar mass | 2.016 g/mol | 159.69 g/mol | 18.015 g/mol | 231.53 g/mol phase | gas (at STP) | solid (at STP) | liquid (at STP) | solid (at STP) melting point | -259.2 °C | 1565 °C | 0 °C | 1538 °C boiling point | -252.8 °C | | 99.9839 °C | density | 8.99×10^-5 g/cm^3 (at 0 °C) | 5.26 g/cm^3 | 1 g/cm^3 | 5 g/cm^3 solubility in water | | insoluble | | surface tension | | | 0.0728 N/m | dynamic viscosity | 8.9×10^-6 Pa s (at 25 °C) | | 8.9×10^-4 Pa s (at 25 °C) | odor | odorless | odorless | odorless |](../image_source/f22c4d26f50b60c9009a6e4c1beacb65.png)
| hydrogen | iron(III) oxide | water | iron(II, III) oxide molar mass | 2.016 g/mol | 159.69 g/mol | 18.015 g/mol | 231.53 g/mol phase | gas (at STP) | solid (at STP) | liquid (at STP) | solid (at STP) melting point | -259.2 °C | 1565 °C | 0 °C | 1538 °C boiling point | -252.8 °C | | 99.9839 °C | density | 8.99×10^-5 g/cm^3 (at 0 °C) | 5.26 g/cm^3 | 1 g/cm^3 | 5 g/cm^3 solubility in water | | insoluble | | surface tension | | | 0.0728 N/m | dynamic viscosity | 8.9×10^-6 Pa s (at 25 °C) | | 8.9×10^-4 Pa s (at 25 °C) | odor | odorless | odorless | odorless |
Units