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HNO3 + Fe = H2O + NO + Fe(NO3)3 + NH4NO3

Input interpretation

HNO_3 nitric acid + Fe iron ⟶ H_2O water + NO nitric oxide + Fe(NO_3)_3 ferric nitrate + NH_4NO_3 ammonium nitrate
HNO_3 nitric acid + Fe iron ⟶ H_2O water + NO nitric oxide + Fe(NO_3)_3 ferric nitrate + NH_4NO_3 ammonium nitrate

Balanced equation

Balance the chemical equation algebraically: HNO_3 + Fe ⟶ H_2O + NO + Fe(NO_3)_3 + NH_4NO_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 + c_5 Fe(NO_3)_3 + c_6 NH_4NO_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 + 4 c_6 N: | c_1 = c_4 + 3 c_5 + 2 c_6 O: | 3 c_1 = c_3 + c_4 + 9 c_5 + 3 c_6 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_2 = (4 c_1)/15 - 1/15 c_3 = (3 c_1)/10 + 4/5 c_4 = 1 c_5 = (4 c_1)/15 - 1/15 c_6 = c_1/10 - 2/5 The resulting system of equations is still underdetermined, so an additional coefficient must be set arbitrarily. Set c_1 = 34 and solve for the remaining coefficients: c_1 = 34 c_2 = 9 c_3 = 11 c_4 = 1 c_5 = 9 c_6 = 3 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 34 HNO_3 + 9 Fe ⟶ 11 H_2O + NO + 9 Fe(NO_3)_3 + 3 NH_4NO_3
Balance the chemical equation algebraically: HNO_3 + Fe ⟶ H_2O + NO + Fe(NO_3)_3 + NH_4NO_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 + c_5 Fe(NO_3)_3 + c_6 NH_4NO_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 + 4 c_6 N: | c_1 = c_4 + 3 c_5 + 2 c_6 O: | 3 c_1 = c_3 + c_4 + 9 c_5 + 3 c_6 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_2 = (4 c_1)/15 - 1/15 c_3 = (3 c_1)/10 + 4/5 c_4 = 1 c_5 = (4 c_1)/15 - 1/15 c_6 = c_1/10 - 2/5 The resulting system of equations is still underdetermined, so an additional coefficient must be set arbitrarily. Set c_1 = 34 and solve for the remaining coefficients: c_1 = 34 c_2 = 9 c_3 = 11 c_4 = 1 c_5 = 9 c_6 = 3 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 34 HNO_3 + 9 Fe ⟶ 11 H_2O + NO + 9 Fe(NO_3)_3 + 3 NH_4NO_3

Structures

+ ⟶ + + +
+ ⟶ + + +

Names

nitric acid + iron ⟶ water + nitric oxide + ferric nitrate + ammonium nitrate
nitric acid + iron ⟶ water + nitric oxide + ferric nitrate + ammonium nitrate

Equilibrium constant

Construct the equilibrium constant, K, expression for: HNO_3 + Fe ⟶ H_2O + NO + Fe(NO_3)_3 + NH_4NO_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: 34 HNO_3 + 9 Fe ⟶ 11 H_2O + NO + 9 Fe(NO_3)_3 + 3 NH_4NO_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 | 34 | -34 Fe | 9 | -9 H_2O | 11 | 11 NO | 1 | 1 Fe(NO_3)_3 | 9 | 9 NH_4NO_3 | 3 | 3 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression HNO_3 | 34 | -34 | ([HNO3])^(-34) Fe | 9 | -9 | ([Fe])^(-9) H_2O | 11 | 11 | ([H2O])^11 NO | 1 | 1 | [NO] Fe(NO_3)_3 | 9 | 9 | ([Fe(NO3)3])^9 NH_4NO_3 | 3 | 3 | ([NH4NO3])^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])^(-34) ([Fe])^(-9) ([H2O])^11 [NO] ([Fe(NO3)3])^9 ([NH4NO3])^3 = (([H2O])^11 [NO] ([Fe(NO3)3])^9 ([NH4NO3])^3)/(([HNO3])^34 ([Fe])^9)
Construct the equilibrium constant, K, expression for: HNO_3 + Fe ⟶ H_2O + NO + Fe(NO_3)_3 + NH_4NO_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: 34 HNO_3 + 9 Fe ⟶ 11 H_2O + NO + 9 Fe(NO_3)_3 + 3 NH_4NO_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 | 34 | -34 Fe | 9 | -9 H_2O | 11 | 11 NO | 1 | 1 Fe(NO_3)_3 | 9 | 9 NH_4NO_3 | 3 | 3 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression HNO_3 | 34 | -34 | ([HNO3])^(-34) Fe | 9 | -9 | ([Fe])^(-9) H_2O | 11 | 11 | ([H2O])^11 NO | 1 | 1 | [NO] Fe(NO_3)_3 | 9 | 9 | ([Fe(NO3)3])^9 NH_4NO_3 | 3 | 3 | ([NH4NO3])^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])^(-34) ([Fe])^(-9) ([H2O])^11 [NO] ([Fe(NO3)3])^9 ([NH4NO3])^3 = (([H2O])^11 [NO] ([Fe(NO3)3])^9 ([NH4NO3])^3)/(([HNO3])^34 ([Fe])^9)

Rate of reaction

Construct the rate of reaction expression for: HNO_3 + Fe ⟶ H_2O + NO + Fe(NO_3)_3 + NH_4NO_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: 34 HNO_3 + 9 Fe ⟶ 11 H_2O + NO + 9 Fe(NO_3)_3 + 3 NH_4NO_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 | 34 | -34 Fe | 9 | -9 H_2O | 11 | 11 NO | 1 | 1 Fe(NO_3)_3 | 9 | 9 NH_4NO_3 | 3 | 3 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 | 34 | -34 | -1/34 (Δ[HNO3])/(Δt) Fe | 9 | -9 | -1/9 (Δ[Fe])/(Δt) H_2O | 11 | 11 | 1/11 (Δ[H2O])/(Δt) NO | 1 | 1 | (Δ[NO])/(Δt) Fe(NO_3)_3 | 9 | 9 | 1/9 (Δ[Fe(NO3)3])/(Δt) NH_4NO_3 | 3 | 3 | 1/3 (Δ[NH4NO3])/(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 = -1/34 (Δ[HNO3])/(Δt) = -1/9 (Δ[Fe])/(Δt) = 1/11 (Δ[H2O])/(Δt) = (Δ[NO])/(Δt) = 1/9 (Δ[Fe(NO3)3])/(Δt) = 1/3 (Δ[NH4NO3])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)
Construct the rate of reaction expression for: HNO_3 + Fe ⟶ H_2O + NO + Fe(NO_3)_3 + NH_4NO_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: 34 HNO_3 + 9 Fe ⟶ 11 H_2O + NO + 9 Fe(NO_3)_3 + 3 NH_4NO_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 | 34 | -34 Fe | 9 | -9 H_2O | 11 | 11 NO | 1 | 1 Fe(NO_3)_3 | 9 | 9 NH_4NO_3 | 3 | 3 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 | 34 | -34 | -1/34 (Δ[HNO3])/(Δt) Fe | 9 | -9 | -1/9 (Δ[Fe])/(Δt) H_2O | 11 | 11 | 1/11 (Δ[H2O])/(Δt) NO | 1 | 1 | (Δ[NO])/(Δt) Fe(NO_3)_3 | 9 | 9 | 1/9 (Δ[Fe(NO3)3])/(Δt) NH_4NO_3 | 3 | 3 | 1/3 (Δ[NH4NO3])/(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 = -1/34 (Δ[HNO3])/(Δt) = -1/9 (Δ[Fe])/(Δt) = 1/11 (Δ[H2O])/(Δt) = (Δ[NO])/(Δt) = 1/9 (Δ[Fe(NO3)3])/(Δt) = 1/3 (Δ[NH4NO3])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

Chemical names and formulas

| nitric acid | iron | water | nitric oxide | ferric nitrate | ammonium nitrate formula | HNO_3 | Fe | H_2O | NO | Fe(NO_3)_3 | NH_4NO_3 Hill formula | HNO_3 | Fe | H_2O | NO | FeN_3O_9 | H_4N_2O_3 name | nitric acid | iron | water | nitric oxide | ferric nitrate | ammonium nitrate IUPAC name | nitric acid | iron | water | nitric oxide | iron(+3) cation trinitrate |
| nitric acid | iron | water | nitric oxide | ferric nitrate | ammonium nitrate formula | HNO_3 | Fe | H_2O | NO | Fe(NO_3)_3 | NH_4NO_3 Hill formula | HNO_3 | Fe | H_2O | NO | FeN_3O_9 | H_4N_2O_3 name | nitric acid | iron | water | nitric oxide | ferric nitrate | ammonium nitrate IUPAC name | nitric acid | iron | water | nitric oxide | iron(+3) cation trinitrate |

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