HNO3 + Fe = H2O + NO2 + Fe(NO3)

HNO_3 nitric acid + Fe iron ⟶ H_2O water + NO_2 nitrogen dioxide + Fe(NO_3)_3 ferric nitrate

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

+ ⟶ + +

nitric acid + iron ⟶ water + nitrogen dioxide + ferric nitrate

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])

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)

| 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