KI + FeCl3 = I2 + KCl + FeCl2

KI (potassium iodide) + FeCl_3 (iron(III) chloride) ⟶ I_2 (iodine) + KCl (potassium chloride) + FeCl_2 (iron(II) chloride)

Balance the chemical equation algebraically: KI + FeCl_3 ⟶ I_2 + KCl + FeCl_2 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 KI + c_2 FeCl_3 ⟶ c_3 I_2 + c_4 KCl + c_5 FeCl_2 Set the number of atoms in the reactants equal to the number of atoms in the products for I, K, Cl and Fe: I: | c_1 = 2 c_3 K: | c_1 = c_4 Cl: | 3 c_2 = c_4 + 2 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_3 = 1 and solve the system of equations for the remaining coefficients: c_1 = 2 c_2 = 2 c_3 = 1 c_4 = 2 c_5 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 2 KI + 2 FeCl_3 ⟶ I_2 + 2 KCl + 2 FeCl_2

+ ⟶ + +

potassium iodide + iron(III) chloride ⟶ iodine + potassium chloride + iron(II) chloride

| potassium iodide | iron(III) chloride | iodine | potassium chloride | iron(II) chloride molecular enthalpy | -327.9 kJ/mol | -399.5 kJ/mol | 0 kJ/mol | -436.5 kJ/mol | -341.8 kJ/mol total enthalpy | -655.8 kJ/mol | -799 kJ/mol | 0 kJ/mol | -873 kJ/mol | -683.6 kJ/mol | H_initial = -1455 kJ/mol | | H_final = -1557 kJ/mol | | ΔH_rxn^0 | -1557 kJ/mol - -1455 kJ/mol = -101.8 kJ/mol (exothermic) | | | |

| potassium iodide | iron(III) chloride | iodine | potassium chloride | iron(II) chloride molecular free energy | -324.9 kJ/mol | -334 kJ/mol | 0 kJ/mol | -408.5 kJ/mol | -302.3 kJ/mol total free energy | -649.8 kJ/mol | -668 kJ/mol | 0 kJ/mol | -817 kJ/mol | -604.6 kJ/mol | G_initial = -1318 kJ/mol | | G_final = -1422 kJ/mol | | ΔG_rxn^0 | -1422 kJ/mol - -1318 kJ/mol = -103.8 kJ/mol (exergonic) | | | |

Construct the equilibrium constant, K, expression for: KI + FeCl_3 ⟶ I_2 + KCl + FeCl_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: 2 KI + 2 FeCl_3 ⟶ I_2 + 2 KCl + 2 FeCl_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 KI | 2 | -2 FeCl_3 | 2 | -2 I_2 | 1 | 1 KCl | 2 | 2 FeCl_2 | 2 | 2 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression KI | 2 | -2 | ([KI])^(-2) FeCl_3 | 2 | -2 | ([FeCl3])^(-2) I_2 | 1 | 1 | [I2] KCl | 2 | 2 | ([KCl])^2 FeCl_2 | 2 | 2 | ([FeCl2])^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 = ([KI])^(-2) ([FeCl3])^(-2) [I2] ([KCl])^2 ([FeCl2])^2 = ([I2] ([KCl])^2 ([FeCl2])^2)/(([KI])^2 ([FeCl3])^2)

Construct the rate of reaction expression for: KI + FeCl_3 ⟶ I_2 + KCl + FeCl_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: 2 KI + 2 FeCl_3 ⟶ I_2 + 2 KCl + 2 FeCl_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 KI | 2 | -2 FeCl_3 | 2 | -2 I_2 | 1 | 1 KCl | 2 | 2 FeCl_2 | 2 | 2 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 KI | 2 | -2 | -1/2 (Δ[KI])/(Δt) FeCl_3 | 2 | -2 | -1/2 (Δ[FeCl3])/(Δt) I_2 | 1 | 1 | (Δ[I2])/(Δt) KCl | 2 | 2 | 1/2 (Δ[KCl])/(Δt) FeCl_2 | 2 | 2 | 1/2 (Δ[FeCl2])/(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 = -1/2 (Δ[KI])/(Δt) = -1/2 (Δ[FeCl3])/(Δt) = (Δ[I2])/(Δt) = 1/2 (Δ[KCl])/(Δt) = 1/2 (Δ[FeCl2])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| potassium iodide | iron(III) chloride | iodine | potassium chloride | iron(II) chloride formula | KI | FeCl_3 | I_2 | KCl | FeCl_2 Hill formula | IK | Cl_3Fe | I_2 | ClK | Cl_2Fe name | potassium iodide | iron(III) chloride | iodine | potassium chloride | iron(II) chloride IUPAC name | potassium iodide | trichloroiron | molecular iodine | potassium chloride | dichloroiron

| potassium iodide | iron(III) chloride | iodine | potassium chloride | iron(II) chloride molar mass | 166.0028 g/mol | 162.2 g/mol | 253.80894 g/mol | 74.55 g/mol | 126.7 g/mol phase | solid (at STP) | solid (at STP) | solid (at STP) | solid (at STP) | solid (at STP) melting point | 681 °C | 304 °C | 113 °C | 770 °C | 677 °C boiling point | 1330 °C | | 184 °C | 1420 °C | density | 3.123 g/cm^3 | | 4.94 g/cm^3 | 1.98 g/cm^3 | 3.16 g/cm^3 solubility in water | | | | soluble | dynamic viscosity | 0.0010227 Pa s (at 732.9 °C) | | 0.00227 Pa s (at 116 °C) | | odor | | | | odorless |