Fe + F2 = FeF3

Fe iron + F_2 fluorine ⟶ FeF_3 ferric fluoride

Balance the chemical equation algebraically: Fe + F_2 ⟶ FeF_3 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 Fe + c_2 F_2 ⟶ c_3 FeF_3 Set the number of atoms in the reactants equal to the number of atoms in the products for Fe and F: Fe: | c_1 = c_3 F: | 2 c_2 = 3 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_1 = 1 and solve the system of equations for the remaining coefficients: c_1 = 1 c_2 = 3/2 c_3 = 1 Multiply by the least common denominator, 2, to eliminate fractional coefficients: c_1 = 2 c_2 = 3 c_3 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 2 Fe + 3 F_2 ⟶ 2 FeF_3

+ ⟶

iron + fluorine ⟶ ferric fluoride

Construct the equilibrium constant, K, expression for: Fe + F_2 ⟶ FeF_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: 2 Fe + 3 F_2 ⟶ 2 FeF_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 Fe | 2 | -2 F_2 | 3 | -3 FeF_3 | 2 | 2 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression Fe | 2 | -2 | ([Fe])^(-2) F_2 | 3 | -3 | ([F2])^(-3) FeF_3 | 2 | 2 | ([FeF3])^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 = ([Fe])^(-2) ([F2])^(-3) ([FeF3])^2 = ([FeF3])^2/(([Fe])^2 ([F2])^3)

Construct the rate of reaction expression for: Fe + F_2 ⟶ FeF_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: 2 Fe + 3 F_2 ⟶ 2 FeF_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 Fe | 2 | -2 F_2 | 3 | -3 FeF_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 Fe | 2 | -2 | -1/2 (Δ[Fe])/(Δt) F_2 | 3 | -3 | -1/3 (Δ[F2])/(Δt) FeF_3 | 2 | 2 | 1/2 (Δ[FeF3])/(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[Fe])/(Δt) = -1/3 (Δ[F2])/(Δt) = 1/2 (Δ[FeF3])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| iron | fluorine | ferric fluoride formula | Fe | F_2 | FeF_3 Hill formula | Fe | F_2 | F_3Fe name | iron | fluorine | ferric fluoride IUPAC name | iron | molecular fluorine | trifluoroiron

| iron | fluorine | ferric fluoride molar mass | 55.845 g/mol | 37.996806326 g/mol | 112.84 g/mol phase | solid (at STP) | gas (at STP) | solid (at STP) melting point | 1535 °C | -219.6 °C | 1000 °C boiling point | 2750 °C | -188.12 °C | density | 7.874 g/cm^3 | 0.001696 g/cm^3 (at 0 °C) | 3.52 g/cm^3 solubility in water | insoluble | reacts | slightly soluble dynamic viscosity | | 2.344×10^-5 Pa s (at 25 °C) |