Al + FeBr3 = Fe + AlBr3

Al aluminum + FeBr_3 iron(III) bromide ⟶ Fe iron + AlBr_3 aluminum tribromide

Balance the chemical equation algebraically: Al + FeBr_3 ⟶ Fe + AlBr_3 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 Al + c_2 FeBr_3 ⟶ c_3 Fe + c_4 AlBr_3 Set the number of atoms in the reactants equal to the number of atoms in the products for Al, Br and Fe: Al: | c_1 = c_4 Br: | 3 c_2 = 3 c_4 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_1 = 1 and solve the system of equations for the remaining coefficients: c_1 = 1 c_2 = 1 c_3 = 1 c_4 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | Al + FeBr_3 ⟶ Fe + AlBr_3

+ ⟶ +

aluminum + iron(III) bromide ⟶ iron + aluminum tribromide

Construct the equilibrium constant, K, expression for: Al + FeBr_3 ⟶ Fe + AlBr_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: Al + FeBr_3 ⟶ Fe + AlBr_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 Al | 1 | -1 FeBr_3 | 1 | -1 Fe | 1 | 1 AlBr_3 | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression Al | 1 | -1 | ([Al])^(-1) FeBr_3 | 1 | -1 | ([FeBr3])^(-1) Fe | 1 | 1 | [Fe] AlBr_3 | 1 | 1 | [AlBr3] 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])^(-1) ([FeBr3])^(-1) [Fe] [AlBr3] = ([Fe] [AlBr3])/([Al] [FeBr3])

Construct the rate of reaction expression for: Al + FeBr_3 ⟶ Fe + AlBr_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: Al + FeBr_3 ⟶ Fe + AlBr_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 Al | 1 | -1 FeBr_3 | 1 | -1 Fe | 1 | 1 AlBr_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 Al | 1 | -1 | -(Δ[Al])/(Δt) FeBr_3 | 1 | -1 | -(Δ[FeBr3])/(Δt) Fe | 1 | 1 | (Δ[Fe])/(Δt) AlBr_3 | 1 | 1 | (Δ[AlBr3])/(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 = -(Δ[Al])/(Δt) = -(Δ[FeBr3])/(Δt) = (Δ[Fe])/(Δt) = (Δ[AlBr3])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| aluminum | iron(III) bromide | iron | aluminum tribromide formula | Al | FeBr_3 | Fe | AlBr_3 Hill formula | Al | Br_3Fe | Fe | AlBr_3 name | aluminum | iron(III) bromide | iron | aluminum tribromide IUPAC name | aluminum | tribromoiron | iron | tribromoalumane

| aluminum | iron(III) bromide | iron | aluminum tribromide molar mass | 26.9815385 g/mol | 295.56 g/mol | 55.845 g/mol | 266.69 g/mol phase | solid (at STP) | | solid (at STP) | solid (at STP) melting point | 660.4 °C | | 1535 °C | 96 °C boiling point | 2460 °C | | 2750 °C | 265 °C density | 2.7 g/cm^3 | | 7.874 g/cm^3 | 3.205 g/cm^3 solubility in water | insoluble | | insoluble | reacts surface tension | 0.817 N/m | | | dynamic viscosity | 1.5×10^-4 Pa s (at 760 °C) | | | odor | odorless | odorless | |