H2O + FeCl3 = HCl + Fe(OH)3

H_2O water + FeCl_3 iron(III) chloride ⟶ HCl hydrogen chloride + Fe(OH)_3 iron(III) hydroxide

Balance the chemical equation algebraically: H_2O + FeCl_3 ⟶ HCl + Fe(OH)_3 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 H_2O + c_2 FeCl_3 ⟶ c_3 HCl + c_4 Fe(OH)_3 Set the number of atoms in the reactants equal to the number of atoms in the products for H, O, Cl and Fe: H: | 2 c_1 = c_3 + 3 c_4 O: | c_1 = 3 c_4 Cl: | 3 c_2 = c_3 Fe: | c_2 = c_4 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 = 3 c_2 = 1 c_3 = 3 c_4 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 3 H_2O + FeCl_3 ⟶ 3 HCl + Fe(OH)_3

+ ⟶ +

water + iron(III) chloride ⟶ hydrogen chloride + iron(III) hydroxide

Construct the equilibrium constant, K, expression for: H_2O + FeCl_3 ⟶ HCl + Fe(OH)_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: 3 H_2O + FeCl_3 ⟶ 3 HCl + Fe(OH)_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 H_2O | 3 | -3 FeCl_3 | 1 | -1 HCl | 3 | 3 Fe(OH)_3 | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression H_2O | 3 | -3 | ([H2O])^(-3) FeCl_3 | 1 | -1 | ([FeCl3])^(-1) HCl | 3 | 3 | ([HCl])^3 Fe(OH)_3 | 1 | 1 | [Fe(OH)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 = ([H2O])^(-3) ([FeCl3])^(-1) ([HCl])^3 [Fe(OH)3] = (([HCl])^3 [Fe(OH)3])/(([H2O])^3 [FeCl3])

Construct the rate of reaction expression for: H_2O + FeCl_3 ⟶ HCl + Fe(OH)_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: 3 H_2O + FeCl_3 ⟶ 3 HCl + Fe(OH)_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 H_2O | 3 | -3 FeCl_3 | 1 | -1 HCl | 3 | 3 Fe(OH)_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 H_2O | 3 | -3 | -1/3 (Δ[H2O])/(Δt) FeCl_3 | 1 | -1 | -(Δ[FeCl3])/(Δt) HCl | 3 | 3 | 1/3 (Δ[HCl])/(Δt) Fe(OH)_3 | 1 | 1 | (Δ[Fe(OH)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 = -1/3 (Δ[H2O])/(Δt) = -(Δ[FeCl3])/(Δt) = 1/3 (Δ[HCl])/(Δt) = (Δ[Fe(OH)3])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| water | iron(III) chloride | hydrogen chloride | iron(III) hydroxide formula | H_2O | FeCl_3 | HCl | Fe(OH)_3 Hill formula | H_2O | Cl_3Fe | ClH | FeH_3O_3 name | water | iron(III) chloride | hydrogen chloride | iron(III) hydroxide IUPAC name | water | trichloroiron | hydrogen chloride | ferric trihydroxide

| water | iron(III) chloride | hydrogen chloride | iron(III) hydroxide molar mass | 18.015 g/mol | 162.2 g/mol | 36.46 g/mol | 106.87 g/mol phase | liquid (at STP) | solid (at STP) | gas (at STP) | melting point | 0 °C | 304 °C | -114.17 °C | boiling point | 99.9839 °C | | -85 °C | density | 1 g/cm^3 | | 0.00149 g/cm^3 (at 25 °C) | solubility in water | | | miscible | surface tension | 0.0728 N/m | | | dynamic viscosity | 8.9×10^-4 Pa s (at 25 °C) | | | odor | odorless | | |