HNO3 + Fe(OH)2 = H2O + NO + Fe(NO3)3

HNO_3 nitric acid + Fe(OH)_2 iron(II) hydroxide ⟶ H_2O water + NO nitric oxide + Fe(NO_3)_3 ferric nitrate

Balance the chemical equation algebraically: HNO_3 + Fe(OH)_2 ⟶ H_2O + NO + Fe(NO_3)_3 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 HNO_3 + c_2 Fe(OH)_2 ⟶ c_3 H_2O + c_4 NO + 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_2 = 2 c_3 N: | c_1 = c_4 + 3 c_5 O: | 3 c_1 + 2 c_2 = c_3 + 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_4 = 1 and solve the system of equations for the remaining coefficients: c_1 = 10 c_2 = 3 c_3 = 8 c_4 = 1 c_5 = 3 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 10 HNO_3 + 3 Fe(OH)_2 ⟶ 8 H_2O + NO + 3 Fe(NO_3)_3

+ ⟶ + +

nitric acid + iron(II) hydroxide ⟶ water + nitric oxide + ferric nitrate

Construct the equilibrium constant, K, expression for: HNO_3 + Fe(OH)_2 ⟶ H_2O + NO + 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: 10 HNO_3 + 3 Fe(OH)_2 ⟶ 8 H_2O + NO + 3 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 | 10 | -10 Fe(OH)_2 | 3 | -3 H_2O | 8 | 8 NO | 1 | 1 Fe(NO_3)_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 | 10 | -10 | ([HNO3])^(-10) Fe(OH)_2 | 3 | -3 | ([Fe(OH)2])^(-3) H_2O | 8 | 8 | ([H2O])^8 NO | 1 | 1 | [NO] Fe(NO_3)_3 | 3 | 3 | ([Fe(NO3)3])^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])^(-10) ([Fe(OH)2])^(-3) ([H2O])^8 [NO] ([Fe(NO3)3])^3 = (([H2O])^8 [NO] ([Fe(NO3)3])^3)/(([HNO3])^10 ([Fe(OH)2])^3)

Construct the rate of reaction expression for: HNO_3 + Fe(OH)_2 ⟶ H_2O + NO + 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: 10 HNO_3 + 3 Fe(OH)_2 ⟶ 8 H_2O + NO + 3 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 | 10 | -10 Fe(OH)_2 | 3 | -3 H_2O | 8 | 8 NO | 1 | 1 Fe(NO_3)_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 | 10 | -10 | -1/10 (Δ[HNO3])/(Δt) Fe(OH)_2 | 3 | -3 | -1/3 (Δ[Fe(OH)2])/(Δt) H_2O | 8 | 8 | 1/8 (Δ[H2O])/(Δt) NO | 1 | 1 | (Δ[NO])/(Δt) Fe(NO_3)_3 | 3 | 3 | 1/3 (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 = -1/10 (Δ[HNO3])/(Δt) = -1/3 (Δ[Fe(OH)2])/(Δt) = 1/8 (Δ[H2O])/(Δt) = (Δ[NO])/(Δt) = 1/3 (Δ[Fe(NO3)3])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| nitric acid | iron(II) hydroxide | water | nitric oxide | ferric nitrate formula | HNO_3 | Fe(OH)_2 | H_2O | NO | Fe(NO_3)_3 Hill formula | HNO_3 | FeH_2O_2 | H_2O | NO | FeN_3O_9 name | nitric acid | iron(II) hydroxide | water | nitric oxide | ferric nitrate IUPAC name | nitric acid | ferrous dihydroxide | water | nitric oxide | iron(+3) cation trinitrate