Input interpretation
![HNO_3 nitric acid + Fe iron ⟶ H_2O water + NO_2 nitrogen dioxide + Fe_2O_3 iron(III) oxide](../image_source/a173ef776fbdeab06cc6d3d7c050264e.png)
HNO_3 nitric acid + Fe iron ⟶ H_2O water + NO_2 nitrogen dioxide + Fe_2O_3 iron(III) oxide
Balanced equation
![Balance the chemical equation algebraically: HNO_3 + Fe ⟶ H_2O + NO_2 + Fe_2O_3 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 HNO_3 + c_2 Fe ⟶ c_3 H_2O + c_4 NO_2 + c_5 Fe_2O_3 Set the number of atoms in the reactants equal to the number of atoms in the products for H, N, O and Fe: H: | c_1 = 2 c_3 N: | c_1 = c_4 O: | 3 c_1 = c_3 + 2 c_4 + 3 c_5 Fe: | c_2 = 2 c_5 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_5 = 1 and solve the system of equations for the remaining coefficients: c_1 = 6 c_2 = 2 c_3 = 3 c_4 = 6 c_5 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 6 HNO_3 + 2 Fe ⟶ 3 H_2O + 6 NO_2 + Fe_2O_3](../image_source/8e97a484e8a619d96d44e0c7b36871e6.png)
Balance the chemical equation algebraically: HNO_3 + Fe ⟶ H_2O + NO_2 + Fe_2O_3 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 HNO_3 + c_2 Fe ⟶ c_3 H_2O + c_4 NO_2 + c_5 Fe_2O_3 Set the number of atoms in the reactants equal to the number of atoms in the products for H, N, O and Fe: H: | c_1 = 2 c_3 N: | c_1 = c_4 O: | 3 c_1 = c_3 + 2 c_4 + 3 c_5 Fe: | c_2 = 2 c_5 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_5 = 1 and solve the system of equations for the remaining coefficients: c_1 = 6 c_2 = 2 c_3 = 3 c_4 = 6 c_5 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 6 HNO_3 + 2 Fe ⟶ 3 H_2O + 6 NO_2 + Fe_2O_3
Structures
![+ ⟶ + +](../image_source/4a8649b41d824731629dead033cdacad.png)
+ ⟶ + +
Names
![nitric acid + iron ⟶ water + nitrogen dioxide + iron(III) oxide](../image_source/84b6d8ea5d393a6d4addba8aa1e6ed18.png)
nitric acid + iron ⟶ water + nitrogen dioxide + iron(III) oxide
Reaction thermodynamics
Entropy
![| nitric acid | iron | water | nitrogen dioxide | iron(III) oxide molecular entropy | 156 J/(mol K) | 27 J/(mol K) | 69.91 J/(mol K) | 240 J/(mol K) | 90 J/(mol K) total entropy | 936 J/(mol K) | 54 J/(mol K) | 209.7 J/(mol K) | 1440 J/(mol K) | 90 J/(mol K) | S_initial = 990 J/(mol K) | | S_final = 1740 J/(mol K) | | ΔS_rxn^0 | 1740 J/(mol K) - 990 J/(mol K) = 749.7 J/(mol K) (endoentropic) | | | |](../image_source/93dda11e50026280064817918be47675.png)
| nitric acid | iron | water | nitrogen dioxide | iron(III) oxide molecular entropy | 156 J/(mol K) | 27 J/(mol K) | 69.91 J/(mol K) | 240 J/(mol K) | 90 J/(mol K) total entropy | 936 J/(mol K) | 54 J/(mol K) | 209.7 J/(mol K) | 1440 J/(mol K) | 90 J/(mol K) | S_initial = 990 J/(mol K) | | S_final = 1740 J/(mol K) | | ΔS_rxn^0 | 1740 J/(mol K) - 990 J/(mol K) = 749.7 J/(mol K) (endoentropic) | | | |
Equilibrium constant
![Construct the equilibrium constant, K, expression for: HNO_3 + Fe ⟶ H_2O + NO_2 + 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: 6 HNO_3 + 2 Fe ⟶ 3 H_2O + 6 NO_2 + 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 HNO_3 | 6 | -6 Fe | 2 | -2 H_2O | 3 | 3 NO_2 | 6 | 6 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 HNO_3 | 6 | -6 | ([HNO3])^(-6) Fe | 2 | -2 | ([Fe])^(-2) H_2O | 3 | 3 | ([H2O])^3 NO_2 | 6 | 6 | ([NO2])^6 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 = ([HNO3])^(-6) ([Fe])^(-2) ([H2O])^3 ([NO2])^6 [Fe2O3] = (([H2O])^3 ([NO2])^6 [Fe2O3])/(([HNO3])^6 ([Fe])^2)](../image_source/ca35a7d361f152c85f2cd1f0a32869ef.png)
Construct the equilibrium constant, K, expression for: HNO_3 + Fe ⟶ H_2O + NO_2 + 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: 6 HNO_3 + 2 Fe ⟶ 3 H_2O + 6 NO_2 + 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 HNO_3 | 6 | -6 Fe | 2 | -2 H_2O | 3 | 3 NO_2 | 6 | 6 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 HNO_3 | 6 | -6 | ([HNO3])^(-6) Fe | 2 | -2 | ([Fe])^(-2) H_2O | 3 | 3 | ([H2O])^3 NO_2 | 6 | 6 | ([NO2])^6 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 = ([HNO3])^(-6) ([Fe])^(-2) ([H2O])^3 ([NO2])^6 [Fe2O3] = (([H2O])^3 ([NO2])^6 [Fe2O3])/(([HNO3])^6 ([Fe])^2)
Rate of reaction
![Construct the rate of reaction expression for: HNO_3 + Fe ⟶ H_2O + NO_2 + 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: 6 HNO_3 + 2 Fe ⟶ 3 H_2O + 6 NO_2 + 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 HNO_3 | 6 | -6 Fe | 2 | -2 H_2O | 3 | 3 NO_2 | 6 | 6 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 HNO_3 | 6 | -6 | -1/6 (Δ[HNO3])/(Δt) Fe | 2 | -2 | -1/2 (Δ[Fe])/(Δt) H_2O | 3 | 3 | 1/3 (Δ[H2O])/(Δt) NO_2 | 6 | 6 | 1/6 (Δ[NO2])/(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 = -1/6 (Δ[HNO3])/(Δt) = -1/2 (Δ[Fe])/(Δt) = 1/3 (Δ[H2O])/(Δt) = 1/6 (Δ[NO2])/(Δt) = (Δ[Fe2O3])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)](../image_source/40baaf7cea1c417788bbee10750129b9.png)
Construct the rate of reaction expression for: HNO_3 + Fe ⟶ H_2O + NO_2 + 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: 6 HNO_3 + 2 Fe ⟶ 3 H_2O + 6 NO_2 + 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 HNO_3 | 6 | -6 Fe | 2 | -2 H_2O | 3 | 3 NO_2 | 6 | 6 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 HNO_3 | 6 | -6 | -1/6 (Δ[HNO3])/(Δt) Fe | 2 | -2 | -1/2 (Δ[Fe])/(Δt) H_2O | 3 | 3 | 1/3 (Δ[H2O])/(Δt) NO_2 | 6 | 6 | 1/6 (Δ[NO2])/(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 = -1/6 (Δ[HNO3])/(Δt) = -1/2 (Δ[Fe])/(Δt) = 1/3 (Δ[H2O])/(Δt) = 1/6 (Δ[NO2])/(Δt) = (Δ[Fe2O3])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)
Chemical names and formulas
![| nitric acid | iron | water | nitrogen dioxide | iron(III) oxide formula | HNO_3 | Fe | H_2O | NO_2 | Fe_2O_3 name | nitric acid | iron | water | nitrogen dioxide | iron(III) oxide IUPAC name | nitric acid | iron | water | Nitrogen dioxide |](../image_source/b1550b8e3f9d2cf140dc8e63f3d7f33e.png)
| nitric acid | iron | water | nitrogen dioxide | iron(III) oxide formula | HNO_3 | Fe | H_2O | NO_2 | Fe_2O_3 name | nitric acid | iron | water | nitrogen dioxide | iron(III) oxide IUPAC name | nitric acid | iron | water | Nitrogen dioxide |
Substance properties
![| nitric acid | iron | water | nitrogen dioxide | iron(III) oxide molar mass | 63.012 g/mol | 55.845 g/mol | 18.015 g/mol | 46.005 g/mol | 159.69 g/mol phase | liquid (at STP) | solid (at STP) | liquid (at STP) | gas (at STP) | solid (at STP) melting point | -41.6 °C | 1535 °C | 0 °C | -11 °C | 1565 °C boiling point | 83 °C | 2750 °C | 99.9839 °C | 21 °C | density | 1.5129 g/cm^3 | 7.874 g/cm^3 | 1 g/cm^3 | 0.00188 g/cm^3 (at 25 °C) | 5.26 g/cm^3 solubility in water | miscible | insoluble | | reacts | insoluble surface tension | | | 0.0728 N/m | | dynamic viscosity | 7.6×10^-4 Pa s (at 25 °C) | | 8.9×10^-4 Pa s (at 25 °C) | 4.02×10^-4 Pa s (at 25 °C) | odor | | | odorless | | odorless](../image_source/a5e7a019b6783bdf4c2d05265133d1cd.png)
| nitric acid | iron | water | nitrogen dioxide | iron(III) oxide molar mass | 63.012 g/mol | 55.845 g/mol | 18.015 g/mol | 46.005 g/mol | 159.69 g/mol phase | liquid (at STP) | solid (at STP) | liquid (at STP) | gas (at STP) | solid (at STP) melting point | -41.6 °C | 1535 °C | 0 °C | -11 °C | 1565 °C boiling point | 83 °C | 2750 °C | 99.9839 °C | 21 °C | density | 1.5129 g/cm^3 | 7.874 g/cm^3 | 1 g/cm^3 | 0.00188 g/cm^3 (at 25 °C) | 5.26 g/cm^3 solubility in water | miscible | insoluble | | reacts | insoluble surface tension | | | 0.0728 N/m | | dynamic viscosity | 7.6×10^-4 Pa s (at 25 °C) | | 8.9×10^-4 Pa s (at 25 °C) | 4.02×10^-4 Pa s (at 25 °C) | odor | | | odorless | | odorless
Units