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HNO3 + Fe = H2O + NH3 + Fe(NO3)2

Input interpretation

HNO_3 nitric acid + Fe iron ⟶ H_2O water + NH_3 ammonia + Fe(NO_3)_2 iron(II) nitrate
HNO_3 nitric acid + Fe iron ⟶ H_2O water + NH_3 ammonia + Fe(NO_3)_2 iron(II) nitrate

Balanced equation

Balance the chemical equation algebraically: HNO_3 + Fe ⟶ H_2O + NH_3 + Fe(NO_3)_2 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 HNO_3 + c_2 Fe ⟶ c_3 H_2O + c_4 NH_3 + c_5 Fe(NO_3)_2 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 + 3 c_4 N: | c_1 = c_4 + 2 c_5 O: | 3 c_1 = c_3 + 6 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_4 = 1 and solve the system of equations for the remaining coefficients: c_1 = 9 c_2 = 4 c_3 = 3 c_4 = 1 c_5 = 4 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 9 HNO_3 + 4 Fe ⟶ 3 H_2O + NH_3 + 4 Fe(NO_3)_2
Balance the chemical equation algebraically: HNO_3 + Fe ⟶ H_2O + NH_3 + Fe(NO_3)_2 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 HNO_3 + c_2 Fe ⟶ c_3 H_2O + c_4 NH_3 + c_5 Fe(NO_3)_2 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 + 3 c_4 N: | c_1 = c_4 + 2 c_5 O: | 3 c_1 = c_3 + 6 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_4 = 1 and solve the system of equations for the remaining coefficients: c_1 = 9 c_2 = 4 c_3 = 3 c_4 = 1 c_5 = 4 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 9 HNO_3 + 4 Fe ⟶ 3 H_2O + NH_3 + 4 Fe(NO_3)_2

Structures

+ ⟶ + +
+ ⟶ + +

Names

nitric acid + iron ⟶ water + ammonia + iron(II) nitrate
nitric acid + iron ⟶ water + ammonia + iron(II) nitrate

Equilibrium constant

Construct the equilibrium constant, K, expression for: HNO_3 + Fe ⟶ H_2O + NH_3 + Fe(NO_3)_2 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: 9 HNO_3 + 4 Fe ⟶ 3 H_2O + NH_3 + 4 Fe(NO_3)_2 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 | 9 | -9 Fe | 4 | -4 H_2O | 3 | 3 NH_3 | 1 | 1 Fe(NO_3)_2 | 4 | 4 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression HNO_3 | 9 | -9 | ([HNO3])^(-9) Fe | 4 | -4 | ([Fe])^(-4) H_2O | 3 | 3 | ([H2O])^3 NH_3 | 1 | 1 | [NH3] Fe(NO_3)_2 | 4 | 4 | ([Fe(NO3)2])^4 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])^(-9) ([Fe])^(-4) ([H2O])^3 [NH3] ([Fe(NO3)2])^4 = (([H2O])^3 [NH3] ([Fe(NO3)2])^4)/(([HNO3])^9 ([Fe])^4)
Construct the equilibrium constant, K, expression for: HNO_3 + Fe ⟶ H_2O + NH_3 + Fe(NO_3)_2 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: 9 HNO_3 + 4 Fe ⟶ 3 H_2O + NH_3 + 4 Fe(NO_3)_2 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 | 9 | -9 Fe | 4 | -4 H_2O | 3 | 3 NH_3 | 1 | 1 Fe(NO_3)_2 | 4 | 4 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression HNO_3 | 9 | -9 | ([HNO3])^(-9) Fe | 4 | -4 | ([Fe])^(-4) H_2O | 3 | 3 | ([H2O])^3 NH_3 | 1 | 1 | [NH3] Fe(NO_3)_2 | 4 | 4 | ([Fe(NO3)2])^4 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])^(-9) ([Fe])^(-4) ([H2O])^3 [NH3] ([Fe(NO3)2])^4 = (([H2O])^3 [NH3] ([Fe(NO3)2])^4)/(([HNO3])^9 ([Fe])^4)

Rate of reaction

Construct the rate of reaction expression for: HNO_3 + Fe ⟶ H_2O + NH_3 + Fe(NO_3)_2 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: 9 HNO_3 + 4 Fe ⟶ 3 H_2O + NH_3 + 4 Fe(NO_3)_2 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 | 9 | -9 Fe | 4 | -4 H_2O | 3 | 3 NH_3 | 1 | 1 Fe(NO_3)_2 | 4 | 4 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 | 9 | -9 | -1/9 (Δ[HNO3])/(Δt) Fe | 4 | -4 | -1/4 (Δ[Fe])/(Δt) H_2O | 3 | 3 | 1/3 (Δ[H2O])/(Δt) NH_3 | 1 | 1 | (Δ[NH3])/(Δt) Fe(NO_3)_2 | 4 | 4 | 1/4 (Δ[Fe(NO3)2])/(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 = -1/9 (Δ[HNO3])/(Δt) = -1/4 (Δ[Fe])/(Δt) = 1/3 (Δ[H2O])/(Δt) = (Δ[NH3])/(Δt) = 1/4 (Δ[Fe(NO3)2])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)
Construct the rate of reaction expression for: HNO_3 + Fe ⟶ H_2O + NH_3 + Fe(NO_3)_2 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: 9 HNO_3 + 4 Fe ⟶ 3 H_2O + NH_3 + 4 Fe(NO_3)_2 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 | 9 | -9 Fe | 4 | -4 H_2O | 3 | 3 NH_3 | 1 | 1 Fe(NO_3)_2 | 4 | 4 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 | 9 | -9 | -1/9 (Δ[HNO3])/(Δt) Fe | 4 | -4 | -1/4 (Δ[Fe])/(Δt) H_2O | 3 | 3 | 1/3 (Δ[H2O])/(Δt) NH_3 | 1 | 1 | (Δ[NH3])/(Δt) Fe(NO_3)_2 | 4 | 4 | 1/4 (Δ[Fe(NO3)2])/(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 = -1/9 (Δ[HNO3])/(Δt) = -1/4 (Δ[Fe])/(Δt) = 1/3 (Δ[H2O])/(Δt) = (Δ[NH3])/(Δt) = 1/4 (Δ[Fe(NO3)2])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

Chemical names and formulas

| nitric acid | iron | water | ammonia | iron(II) nitrate formula | HNO_3 | Fe | H_2O | NH_3 | Fe(NO_3)_2 Hill formula | HNO_3 | Fe | H_2O | H_3N | FeN_2O_6 name | nitric acid | iron | water | ammonia | iron(II) nitrate
| nitric acid | iron | water | ammonia | iron(II) nitrate formula | HNO_3 | Fe | H_2O | NH_3 | Fe(NO_3)_2 Hill formula | HNO_3 | Fe | H_2O | H_3N | FeN_2O_6 name | nitric acid | iron | water | ammonia | iron(II) nitrate

Substance properties

| nitric acid | iron | water | ammonia | iron(II) nitrate molar mass | 63.012 g/mol | 55.845 g/mol | 18.015 g/mol | 17.031 g/mol | 179.85 g/mol phase | liquid (at STP) | solid (at STP) | liquid (at STP) | gas (at STP) | melting point | -41.6 °C | 1535 °C | 0 °C | -77.73 °C | boiling point | 83 °C | 2750 °C | 99.9839 °C | -33.33 °C | density | 1.5129 g/cm^3 | 7.874 g/cm^3 | 1 g/cm^3 | 6.96×10^-4 g/cm^3 (at 25 °C) | solubility in water | miscible | insoluble | | | surface tension | | | 0.0728 N/m | 0.0234 N/m | dynamic viscosity | 7.6×10^-4 Pa s (at 25 °C) | | 8.9×10^-4 Pa s (at 25 °C) | 1.009×10^-5 Pa s (at 25 °C) | odor | | | odorless | |
| nitric acid | iron | water | ammonia | iron(II) nitrate molar mass | 63.012 g/mol | 55.845 g/mol | 18.015 g/mol | 17.031 g/mol | 179.85 g/mol phase | liquid (at STP) | solid (at STP) | liquid (at STP) | gas (at STP) | melting point | -41.6 °C | 1535 °C | 0 °C | -77.73 °C | boiling point | 83 °C | 2750 °C | 99.9839 °C | -33.33 °C | density | 1.5129 g/cm^3 | 7.874 g/cm^3 | 1 g/cm^3 | 6.96×10^-4 g/cm^3 (at 25 °C) | solubility in water | miscible | insoluble | | | surface tension | | | 0.0728 N/m | 0.0234 N/m | dynamic viscosity | 7.6×10^-4 Pa s (at 25 °C) | | 8.9×10^-4 Pa s (at 25 °C) | 1.009×10^-5 Pa s (at 25 °C) | odor | | | odorless | |

Units

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