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HNO3 + FeI2 = H2O + NO2 + HIO3 + Fe(NO3)2

Input interpretation

HNO_3 nitric acid + FeI_2 ferrous iodide ⟶ H_2O water + NO_2 nitrogen dioxide + HIO_3 iodic acid + Fe(NO_3)_2 iron(II) nitrate
HNO_3 nitric acid + FeI_2 ferrous iodide ⟶ H_2O water + NO_2 nitrogen dioxide + HIO_3 iodic acid + Fe(NO_3)_2 iron(II) nitrate

Balanced equation

Balance the chemical equation algebraically: HNO_3 + FeI_2 ⟶ H_2O + NO_2 + HIO_3 + Fe(NO_3)_2 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 HNO_3 + c_2 FeI_2 ⟶ c_3 H_2O + c_4 NO_2 + c_5 HIO_3 + c_6 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, Fe and I: H: | c_1 = 2 c_3 + c_5 N: | c_1 = c_4 + 2 c_6 O: | 3 c_1 = c_3 + 2 c_4 + 3 c_5 + 6 c_6 Fe: | c_2 = c_6 I: | 2 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 = 14 c_2 = 1 c_3 = 6 c_4 = 12 c_5 = 2 c_6 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 14 HNO_3 + FeI_2 ⟶ 6 H_2O + 12 NO_2 + 2 HIO_3 + Fe(NO_3)_2
Balance the chemical equation algebraically: HNO_3 + FeI_2 ⟶ H_2O + NO_2 + HIO_3 + Fe(NO_3)_2 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 HNO_3 + c_2 FeI_2 ⟶ c_3 H_2O + c_4 NO_2 + c_5 HIO_3 + c_6 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, Fe and I: H: | c_1 = 2 c_3 + c_5 N: | c_1 = c_4 + 2 c_6 O: | 3 c_1 = c_3 + 2 c_4 + 3 c_5 + 6 c_6 Fe: | c_2 = c_6 I: | 2 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 = 14 c_2 = 1 c_3 = 6 c_4 = 12 c_5 = 2 c_6 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 14 HNO_3 + FeI_2 ⟶ 6 H_2O + 12 NO_2 + 2 HIO_3 + Fe(NO_3)_2

Structures

+ ⟶ + + +
+ ⟶ + + +

Names

nitric acid + ferrous iodide ⟶ water + nitrogen dioxide + iodic acid + iron(II) nitrate
nitric acid + ferrous iodide ⟶ water + nitrogen dioxide + iodic acid + iron(II) nitrate

Equilibrium constant

Construct the equilibrium constant, K, expression for: HNO_3 + FeI_2 ⟶ H_2O + NO_2 + HIO_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: 14 HNO_3 + FeI_2 ⟶ 6 H_2O + 12 NO_2 + 2 HIO_3 + 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 | 14 | -14 FeI_2 | 1 | -1 H_2O | 6 | 6 NO_2 | 12 | 12 HIO_3 | 2 | 2 Fe(NO_3)_2 | 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 | 14 | -14 | ([HNO3])^(-14) FeI_2 | 1 | -1 | ([FeI2])^(-1) H_2O | 6 | 6 | ([H2O])^6 NO_2 | 12 | 12 | ([NO2])^12 HIO_3 | 2 | 2 | ([HIO3])^2 Fe(NO_3)_2 | 1 | 1 | [Fe(NO3)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 = ([HNO3])^(-14) ([FeI2])^(-1) ([H2O])^6 ([NO2])^12 ([HIO3])^2 [Fe(NO3)2] = (([H2O])^6 ([NO2])^12 ([HIO3])^2 [Fe(NO3)2])/(([HNO3])^14 [FeI2])
Construct the equilibrium constant, K, expression for: HNO_3 + FeI_2 ⟶ H_2O + NO_2 + HIO_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: 14 HNO_3 + FeI_2 ⟶ 6 H_2O + 12 NO_2 + 2 HIO_3 + 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 | 14 | -14 FeI_2 | 1 | -1 H_2O | 6 | 6 NO_2 | 12 | 12 HIO_3 | 2 | 2 Fe(NO_3)_2 | 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 | 14 | -14 | ([HNO3])^(-14) FeI_2 | 1 | -1 | ([FeI2])^(-1) H_2O | 6 | 6 | ([H2O])^6 NO_2 | 12 | 12 | ([NO2])^12 HIO_3 | 2 | 2 | ([HIO3])^2 Fe(NO_3)_2 | 1 | 1 | [Fe(NO3)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 = ([HNO3])^(-14) ([FeI2])^(-1) ([H2O])^6 ([NO2])^12 ([HIO3])^2 [Fe(NO3)2] = (([H2O])^6 ([NO2])^12 ([HIO3])^2 [Fe(NO3)2])/(([HNO3])^14 [FeI2])

Rate of reaction

Construct the rate of reaction expression for: HNO_3 + FeI_2 ⟶ H_2O + NO_2 + HIO_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: 14 HNO_3 + FeI_2 ⟶ 6 H_2O + 12 NO_2 + 2 HIO_3 + 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 | 14 | -14 FeI_2 | 1 | -1 H_2O | 6 | 6 NO_2 | 12 | 12 HIO_3 | 2 | 2 Fe(NO_3)_2 | 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 | 14 | -14 | -1/14 (Δ[HNO3])/(Δt) FeI_2 | 1 | -1 | -(Δ[FeI2])/(Δt) H_2O | 6 | 6 | 1/6 (Δ[H2O])/(Δt) NO_2 | 12 | 12 | 1/12 (Δ[NO2])/(Δt) HIO_3 | 2 | 2 | 1/2 (Δ[HIO3])/(Δt) Fe(NO_3)_2 | 1 | 1 | (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 = -1/14 (Δ[HNO3])/(Δt) = -(Δ[FeI2])/(Δt) = 1/6 (Δ[H2O])/(Δt) = 1/12 (Δ[NO2])/(Δt) = 1/2 (Δ[HIO3])/(Δt) = (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 + FeI_2 ⟶ H_2O + NO_2 + HIO_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: 14 HNO_3 + FeI_2 ⟶ 6 H_2O + 12 NO_2 + 2 HIO_3 + 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 | 14 | -14 FeI_2 | 1 | -1 H_2O | 6 | 6 NO_2 | 12 | 12 HIO_3 | 2 | 2 Fe(NO_3)_2 | 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 | 14 | -14 | -1/14 (Δ[HNO3])/(Δt) FeI_2 | 1 | -1 | -(Δ[FeI2])/(Δt) H_2O | 6 | 6 | 1/6 (Δ[H2O])/(Δt) NO_2 | 12 | 12 | 1/12 (Δ[NO2])/(Δt) HIO_3 | 2 | 2 | 1/2 (Δ[HIO3])/(Δt) Fe(NO_3)_2 | 1 | 1 | (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 = -1/14 (Δ[HNO3])/(Δt) = -(Δ[FeI2])/(Δt) = 1/6 (Δ[H2O])/(Δt) = 1/12 (Δ[NO2])/(Δt) = 1/2 (Δ[HIO3])/(Δt) = (Δ[Fe(NO3)2])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

Chemical names and formulas

| nitric acid | ferrous iodide | water | nitrogen dioxide | iodic acid | iron(II) nitrate formula | HNO_3 | FeI_2 | H_2O | NO_2 | HIO_3 | Fe(NO_3)_2 Hill formula | HNO_3 | FeI_2 | H_2O | NO_2 | HIO_3 | FeN_2O_6 name | nitric acid | ferrous iodide | water | nitrogen dioxide | iodic acid | iron(II) nitrate IUPAC name | nitric acid | diiodoiron | water | Nitrogen dioxide | iodic acid |
| nitric acid | ferrous iodide | water | nitrogen dioxide | iodic acid | iron(II) nitrate formula | HNO_3 | FeI_2 | H_2O | NO_2 | HIO_3 | Fe(NO_3)_2 Hill formula | HNO_3 | FeI_2 | H_2O | NO_2 | HIO_3 | FeN_2O_6 name | nitric acid | ferrous iodide | water | nitrogen dioxide | iodic acid | iron(II) nitrate IUPAC name | nitric acid | diiodoiron | water | Nitrogen dioxide | iodic acid |

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