Input interpretation
![HNO_3 (nitric acid) + Fe (iron) ⟶ H_2O (water) + NO_2 (nitrogen dioxide) + Fe(NO_3)_3 (ferric nitrate)](../image_source/df08af60db8ce396233c07e1385630eb.png)
HNO_3 (nitric acid) + Fe (iron) ⟶ H_2O (water) + NO_2 (nitrogen dioxide) + Fe(NO_3)_3 (ferric nitrate)
Balanced equation
![Balance the chemical equation algebraically: HNO_3 + Fe ⟶ H_2O + NO_2 + Fe(NO_3)_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(NO_3)_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 + 3 c_5 O: | 3 c_1 = c_3 + 2 c_4 + 9 c_5 Fe: | c_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_2 = 1 and solve the system of equations for the remaining coefficients: c_1 = 6 c_2 = 1 c_3 = 3 c_4 = 3 c_5 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 6 HNO_3 + Fe ⟶ 3 H_2O + 3 NO_2 + Fe(NO_3)_3](../image_source/861c93fe0a941c9af0ffc4867edbc57f.png)
Balance the chemical equation algebraically: HNO_3 + Fe ⟶ H_2O + NO_2 + Fe(NO_3)_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(NO_3)_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 + 3 c_5 O: | 3 c_1 = c_3 + 2 c_4 + 9 c_5 Fe: | c_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_2 = 1 and solve the system of equations for the remaining coefficients: c_1 = 6 c_2 = 1 c_3 = 3 c_4 = 3 c_5 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 6 HNO_3 + Fe ⟶ 3 H_2O + 3 NO_2 + Fe(NO_3)_3
Structures
![+ ⟶ + +](../image_source/6c40353c54ec909d96e86640d66ea440.png)
+ ⟶ + +
Names
![nitric acid + iron ⟶ water + nitrogen dioxide + ferric nitrate](../image_source/512cfb39cce7c3bd27fdfb358932b432.png)
nitric acid + iron ⟶ water + nitrogen dioxide + ferric nitrate
Equilibrium constant
![Construct the equilibrium constant, K, expression for: HNO_3 + Fe ⟶ H_2O + NO_2 + Fe(NO_3)_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 + Fe ⟶ 3 H_2O + 3 NO_2 + Fe(NO_3)_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 | 1 | -1 H_2O | 3 | 3 NO_2 | 3 | 3 Fe(NO_3)_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 | 1 | -1 | ([Fe])^(-1) H_2O | 3 | 3 | ([H2O])^3 NO_2 | 3 | 3 | ([NO2])^3 Fe(NO_3)_3 | 1 | 1 | [Fe(NO3)3] 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])^(-1) ([H2O])^3 ([NO2])^3 [Fe(NO3)3] = (([H2O])^3 ([NO2])^3 [Fe(NO3)3])/(([HNO3])^6 [Fe])](../image_source/2faed003a0694c8ade520885668de7de.png)
Construct the equilibrium constant, K, expression for: HNO_3 + Fe ⟶ H_2O + NO_2 + Fe(NO_3)_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 + Fe ⟶ 3 H_2O + 3 NO_2 + Fe(NO_3)_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 | 1 | -1 H_2O | 3 | 3 NO_2 | 3 | 3 Fe(NO_3)_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 | 1 | -1 | ([Fe])^(-1) H_2O | 3 | 3 | ([H2O])^3 NO_2 | 3 | 3 | ([NO2])^3 Fe(NO_3)_3 | 1 | 1 | [Fe(NO3)3] 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])^(-1) ([H2O])^3 ([NO2])^3 [Fe(NO3)3] = (([H2O])^3 ([NO2])^3 [Fe(NO3)3])/(([HNO3])^6 [Fe])
Rate of reaction
![Construct the rate of reaction expression for: HNO_3 + Fe ⟶ H_2O + NO_2 + Fe(NO_3)_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 + Fe ⟶ 3 H_2O + 3 NO_2 + Fe(NO_3)_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 | 1 | -1 H_2O | 3 | 3 NO_2 | 3 | 3 Fe(NO_3)_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 | 1 | -1 | -(Δ[Fe])/(Δt) H_2O | 3 | 3 | 1/3 (Δ[H2O])/(Δt) NO_2 | 3 | 3 | 1/3 (Δ[NO2])/(Δt) Fe(NO_3)_3 | 1 | 1 | (Δ[Fe(NO3)3])/(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) = -(Δ[Fe])/(Δt) = 1/3 (Δ[H2O])/(Δt) = 1/3 (Δ[NO2])/(Δt) = (Δ[Fe(NO3)3])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)](../image_source/0e503ebaecaa619163b002a1ba29cd1b.png)
Construct the rate of reaction expression for: HNO_3 + Fe ⟶ H_2O + NO_2 + Fe(NO_3)_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 + Fe ⟶ 3 H_2O + 3 NO_2 + Fe(NO_3)_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 | 1 | -1 H_2O | 3 | 3 NO_2 | 3 | 3 Fe(NO_3)_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 | 1 | -1 | -(Δ[Fe])/(Δt) H_2O | 3 | 3 | 1/3 (Δ[H2O])/(Δt) NO_2 | 3 | 3 | 1/3 (Δ[NO2])/(Δt) Fe(NO_3)_3 | 1 | 1 | (Δ[Fe(NO3)3])/(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) = -(Δ[Fe])/(Δt) = 1/3 (Δ[H2O])/(Δt) = 1/3 (Δ[NO2])/(Δt) = (Δ[Fe(NO3)3])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)
Chemical names and formulas
![| nitric acid | iron | water | nitrogen dioxide | ferric nitrate formula | HNO_3 | Fe | H_2O | NO_2 | Fe(NO_3)_3 Hill formula | HNO_3 | Fe | H_2O | NO_2 | FeN_3O_9 name | nitric acid | iron | water | nitrogen dioxide | ferric nitrate IUPAC name | nitric acid | iron | water | Nitrogen dioxide | iron(+3) cation trinitrate](../image_source/8517c8bd5079c563760a2b2cf541ec1b.png)
| nitric acid | iron | water | nitrogen dioxide | ferric nitrate formula | HNO_3 | Fe | H_2O | NO_2 | Fe(NO_3)_3 Hill formula | HNO_3 | Fe | H_2O | NO_2 | FeN_3O_9 name | nitric acid | iron | water | nitrogen dioxide | ferric nitrate IUPAC name | nitric acid | iron | water | Nitrogen dioxide | iron(+3) cation trinitrate
Substance properties
![| nitric acid | iron | water | nitrogen dioxide | ferric nitrate molar mass | 63.012 g/mol | 55.845 g/mol | 18.015 g/mol | 46.005 g/mol | 241.86 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 | 35 °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) | 1.7 g/cm^3 solubility in water | miscible | insoluble | | reacts | very soluble 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 | |](../image_source/64f09e0537f47729a5605db700d2506a.png)
| nitric acid | iron | water | nitrogen dioxide | ferric nitrate molar mass | 63.012 g/mol | 55.845 g/mol | 18.015 g/mol | 46.005 g/mol | 241.86 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 | 35 °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) | 1.7 g/cm^3 solubility in water | miscible | insoluble | | reacts | very soluble 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 | |
Units