Al + Br2 = AlBr3

Al (aluminum) + Br_2 (bromine) ⟶ AlBr_3 (aluminum tribromide)

Balance the chemical equation algebraically: Al + Br_2 ⟶ AlBr_3 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 Al + c_2 Br_2 ⟶ c_3 AlBr_3 Set the number of atoms in the reactants equal to the number of atoms in the products for Al and Br: Al: | c_1 = c_3 Br: | 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 Al + 3 Br_2 ⟶ 2 AlBr_3

+ ⟶

aluminum + bromine ⟶ aluminum tribromide

| aluminum | bromine | aluminum tribromide molecular enthalpy | 0 kJ/mol | 0 kJ/mol | -527.2 kJ/mol total enthalpy | 0 kJ/mol | 0 kJ/mol | -1054 kJ/mol | H_initial = 0 kJ/mol | | H_final = -1054 kJ/mol ΔH_rxn^0 | -1054 kJ/mol - 0 kJ/mol = -1054 kJ/mol (exothermic) | |

Construct the equilibrium constant, K, expression for: Al + Br_2 ⟶ 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: 2 Al + 3 Br_2 ⟶ 2 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 | 2 | -2 Br_2 | 3 | -3 AlBr_3 | 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) Br_2 | 3 | -3 | ([Br2])^(-3) AlBr_3 | 2 | 2 | ([AlBr3])^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) ([Br2])^(-3) ([AlBr3])^2 = ([AlBr3])^2/(([Al])^2 ([Br2])^3)

Construct the rate of reaction expression for: Al + Br_2 ⟶ 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: 2 Al + 3 Br_2 ⟶ 2 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 | 2 | -2 Br_2 | 3 | -3 AlBr_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 Al | 2 | -2 | -1/2 (Δ[Al])/(Δt) Br_2 | 3 | -3 | -1/3 (Δ[Br2])/(Δt) AlBr_3 | 2 | 2 | 1/2 (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 = -1/2 (Δ[Al])/(Δt) = -1/3 (Δ[Br2])/(Δt) = 1/2 (Δ[AlBr3])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| aluminum | bromine | aluminum tribromide formula | Al | Br_2 | AlBr_3 name | aluminum | bromine | aluminum tribromide IUPAC name | aluminum | molecular bromine | tribromoalumane

| aluminum | bromine | aluminum tribromide molar mass | 26.9815385 g/mol | 159.81 g/mol | 266.69 g/mol phase | solid (at STP) | liquid (at STP) | solid (at STP) melting point | 660.4 °C | -7.2 °C | 96 °C boiling point | 2460 °C | 58.8 °C | 265 °C density | 2.7 g/cm^3 | 3.119 g/cm^3 | 3.205 g/cm^3 solubility in water | insoluble | insoluble | reacts surface tension | 0.817 N/m | 0.0409 N/m | dynamic viscosity | 1.5×10^-4 Pa s (at 760 °C) | 9.44×10^-4 Pa s (at 25 °C) | odor | odorless | |