Al + F2 = AlF

Al aluminum + F_2 fluorine ⟶ AlF

Balance the chemical equation algebraically: Al + F_2 ⟶ AlF Add stoichiometric coefficients, c_i, to the reactants and products: c_1 Al + c_2 F_2 ⟶ c_3 AlF Set the number of atoms in the reactants equal to the number of atoms in the products for Al and F: Al: | c_1 = c_3 F: | 2 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 = 2 c_2 = 1 c_3 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 2 Al + F_2 ⟶ 2 AlF

+ ⟶ AlF

aluminum + fluorine ⟶ AlF

Construct the equilibrium constant, K, expression for: Al + F_2 ⟶ AlF 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 Al + F_2 ⟶ 2 AlF 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 Al | 2 | -2 F_2 | 1 | -1 AlF | 2 | 2 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression Al | 2 | -2 | ([Al])^(-2) F_2 | 1 | -1 | ([F2])^(-1) AlF | 2 | 2 | ([AlF])^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 = ([Al])^(-2) ([F2])^(-1) ([AlF])^2 = ([AlF])^2/(([Al])^2 [F2])

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

| aluminum | fluorine | AlF formula | Al | F_2 | AlF name | aluminum | fluorine | IUPAC name | aluminum | molecular fluorine |

| aluminum | fluorine | AlF molar mass | 26.9815385 g/mol | 37.996806326 g/mol | 45.9799417 g/mol phase | solid (at STP) | gas (at STP) | melting point | 660.4 °C | -219.6 °C | boiling point | 2460 °C | -188.12 °C | density | 2.7 g/cm^3 | 0.001696 g/cm^3 (at 0 °C) | solubility in water | insoluble | reacts | surface tension | 0.817 N/m | | dynamic viscosity | 1.5×10^-4 Pa s (at 760 °C) | 2.344×10^-5 Pa s (at 25 °C) | odor | odorless | |