Cl2 + Fe = FeCl3

Cl_2 (chlorine) + Fe (iron) ⟶ FeCl_3 (iron(III) chloride)

Balance the chemical equation algebraically: Cl_2 + Fe ⟶ FeCl_3 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 Cl_2 + c_2 Fe ⟶ c_3 FeCl_3 Set the number of atoms in the reactants equal to the number of atoms in the products for Cl and Fe: Cl: | 2 c_1 = 3 c_3 Fe: | c_2 = c_3 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 = 3/2 c_2 = 1 c_3 = 1 Multiply by the least common denominator, 2, to eliminate fractional coefficients: c_1 = 3 c_2 = 2 c_3 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 3 Cl_2 + 2 Fe ⟶ 2 FeCl_3

+ ⟶

chlorine + iron ⟶ iron(III) chloride

| chlorine | iron | iron(III) chloride molecular enthalpy | 0 kJ/mol | 0 kJ/mol | -399.5 kJ/mol total enthalpy | 0 kJ/mol | 0 kJ/mol | -799 kJ/mol | H_initial = 0 kJ/mol | | H_final = -799 kJ/mol ΔH_rxn^0 | -799 kJ/mol - 0 kJ/mol = -799 kJ/mol (exothermic) | |

Construct the equilibrium constant, K, expression for: Cl_2 + Fe ⟶ FeCl_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 Cl_2 + 2 Fe ⟶ 2 FeCl_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 Cl_2 | 3 | -3 Fe | 2 | -2 FeCl_3 | 2 | 2 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression Cl_2 | 3 | -3 | ([Cl2])^(-3) Fe | 2 | -2 | ([Fe])^(-2) FeCl_3 | 2 | 2 | ([FeCl3])^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 = ([Cl2])^(-3) ([Fe])^(-2) ([FeCl3])^2 = ([FeCl3])^2/(([Cl2])^3 ([Fe])^2)

Construct the rate of reaction expression for: Cl_2 + Fe ⟶ FeCl_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 Cl_2 + 2 Fe ⟶ 2 FeCl_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 Cl_2 | 3 | -3 Fe | 2 | -2 FeCl_3 | 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 Cl_2 | 3 | -3 | -1/3 (Δ[Cl2])/(Δt) Fe | 2 | -2 | -1/2 (Δ[Fe])/(Δt) FeCl_3 | 2 | 2 | 1/2 (Δ[FeCl3])/(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[Cl2])/(Δt) = -1/2 (Δ[Fe])/(Δt) = 1/2 (Δ[FeCl3])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| chlorine | iron | iron(III) chloride formula | Cl_2 | Fe | FeCl_3 Hill formula | Cl_2 | Fe | Cl_3Fe name | chlorine | iron | iron(III) chloride IUPAC name | molecular chlorine | iron | trichloroiron

| chlorine | iron | iron(III) chloride molar mass | 70.9 g/mol | 55.845 g/mol | 162.2 g/mol phase | gas (at STP) | solid (at STP) | solid (at STP) melting point | -101 °C | 1535 °C | 304 °C boiling point | -34 °C | 2750 °C | density | 0.003214 g/cm^3 (at 0 °C) | 7.874 g/cm^3 | solubility in water | | insoluble |