HCl + HNO3 + FeCl2 = H2O + FeCl3 + N2O

HCl hydrogen chloride + HNO_3 nitric acid + FeCl_2 iron(II) chloride ⟶ H_2O water + FeCl_3 iron(III) chloride + N_2O nitrous oxide

Balance the chemical equation algebraically: HCl + HNO_3 + FeCl_2 ⟶ H_2O + FeCl_3 + N_2O Add stoichiometric coefficients, c_i, to the reactants and products: c_1 HCl + c_2 HNO_3 + c_3 FeCl_2 ⟶ c_4 H_2O + c_5 FeCl_3 + c_6 N_2O Set the number of atoms in the reactants equal to the number of atoms in the products for Cl, H, N, O and Fe: Cl: | c_1 + 2 c_3 = 3 c_5 H: | c_1 + c_2 = 2 c_4 N: | c_2 = 2 c_6 O: | 3 c_2 = c_4 + c_6 Fe: | c_3 = 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_6 = 1 and solve the system of equations for the remaining coefficients: c_1 = 8 c_2 = 2 c_3 = 8 c_4 = 5 c_5 = 8 c_6 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 8 HCl + 2 HNO_3 + 8 FeCl_2 ⟶ 5 H_2O + 8 FeCl_3 + N_2O

+ + ⟶ + +

hydrogen chloride + nitric acid + iron(II) chloride ⟶ water + iron(III) chloride + nitrous oxide

| hydrogen chloride | nitric acid | iron(II) chloride | water | iron(III) chloride | nitrous oxide molecular free energy | -95.3 kJ/mol | -80.7 kJ/mol | -302.3 kJ/mol | -237.1 kJ/mol | -334 kJ/mol | 104 kJ/mol total free energy | -762.4 kJ/mol | -161.4 kJ/mol | -2418 kJ/mol | -1186 kJ/mol | -2672 kJ/mol | 104 kJ/mol | G_initial = -3342 kJ/mol | | | G_final = -3754 kJ/mol | | ΔG_rxn^0 | -3754 kJ/mol - -3342 kJ/mol = -411.3 kJ/mol (exergonic) | | | | |

Construct the equilibrium constant, K, expression for: HCl + HNO_3 + FeCl_2 ⟶ H_2O + FeCl_3 + N_2O 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: 8 HCl + 2 HNO_3 + 8 FeCl_2 ⟶ 5 H_2O + 8 FeCl_3 + N_2O 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 HCl | 8 | -8 HNO_3 | 2 | -2 FeCl_2 | 8 | -8 H_2O | 5 | 5 FeCl_3 | 8 | 8 N_2O | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression HCl | 8 | -8 | ([HCl])^(-8) HNO_3 | 2 | -2 | ([HNO3])^(-2) FeCl_2 | 8 | -8 | ([FeCl2])^(-8) H_2O | 5 | 5 | ([H2O])^5 FeCl_3 | 8 | 8 | ([FeCl3])^8 N_2O | 1 | 1 | [N2O] 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 = ([HCl])^(-8) ([HNO3])^(-2) ([FeCl2])^(-8) ([H2O])^5 ([FeCl3])^8 [N2O] = (([H2O])^5 ([FeCl3])^8 [N2O])/(([HCl])^8 ([HNO3])^2 ([FeCl2])^8)

Construct the rate of reaction expression for: HCl + HNO_3 + FeCl_2 ⟶ H_2O + FeCl_3 + N_2O 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: 8 HCl + 2 HNO_3 + 8 FeCl_2 ⟶ 5 H_2O + 8 FeCl_3 + N_2O 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 HCl | 8 | -8 HNO_3 | 2 | -2 FeCl_2 | 8 | -8 H_2O | 5 | 5 FeCl_3 | 8 | 8 N_2O | 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 HCl | 8 | -8 | -1/8 (Δ[HCl])/(Δt) HNO_3 | 2 | -2 | -1/2 (Δ[HNO3])/(Δt) FeCl_2 | 8 | -8 | -1/8 (Δ[FeCl2])/(Δt) H_2O | 5 | 5 | 1/5 (Δ[H2O])/(Δt) FeCl_3 | 8 | 8 | 1/8 (Δ[FeCl3])/(Δt) N_2O | 1 | 1 | (Δ[N2O])/(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 = -1/8 (Δ[HCl])/(Δt) = -1/2 (Δ[HNO3])/(Δt) = -1/8 (Δ[FeCl2])/(Δt) = 1/5 (Δ[H2O])/(Δt) = 1/8 (Δ[FeCl3])/(Δt) = (Δ[N2O])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| hydrogen chloride | nitric acid | iron(II) chloride | water | iron(III) chloride | nitrous oxide formula | HCl | HNO_3 | FeCl_2 | H_2O | FeCl_3 | N_2O Hill formula | ClH | HNO_3 | Cl_2Fe | H_2O | Cl_3Fe | N_2O name | hydrogen chloride | nitric acid | iron(II) chloride | water | iron(III) chloride | nitrous oxide IUPAC name | hydrogen chloride | nitric acid | dichloroiron | water | trichloroiron | nitrous oxide