Br2 + FeBr2 = FeBr3

Br_2 bromine + FeBr_2 iron(II) bromide ⟶ FeBr_3 iron(III) bromide

Balance the chemical equation algebraically: Br_2 + FeBr_2 ⟶ FeBr_3 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 Br_2 + c_2 FeBr_2 ⟶ c_3 FeBr_3 Set the number of atoms in the reactants equal to the number of atoms in the products for Br and Fe: Br: | 2 c_1 + 2 c_2 = 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_1 = 1 and solve the system of equations for the remaining coefficients: c_1 = 1 c_2 = 2 c_3 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | Br_2 + 2 FeBr_2 ⟶ 2 FeBr_3

+ ⟶

bromine + iron(II) bromide ⟶ iron(III) bromide

Construct the equilibrium constant, K, expression for: Br_2 + FeBr_2 ⟶ FeBr_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: Br_2 + 2 FeBr_2 ⟶ 2 FeBr_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 Br_2 | 1 | -1 FeBr_2 | 2 | -2 FeBr_3 | 2 | 2 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression Br_2 | 1 | -1 | ([Br2])^(-1) FeBr_2 | 2 | -2 | ([FeBr2])^(-2) FeBr_3 | 2 | 2 | ([FeBr3])^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 = ([Br2])^(-1) ([FeBr2])^(-2) ([FeBr3])^2 = ([FeBr3])^2/([Br2] ([FeBr2])^2)

Construct the rate of reaction expression for: Br_2 + FeBr_2 ⟶ FeBr_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: Br_2 + 2 FeBr_2 ⟶ 2 FeBr_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 Br_2 | 1 | -1 FeBr_2 | 2 | -2 FeBr_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 Br_2 | 1 | -1 | -(Δ[Br2])/(Δt) FeBr_2 | 2 | -2 | -1/2 (Δ[FeBr2])/(Δt) FeBr_3 | 2 | 2 | 1/2 (Δ[FeBr3])/(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[Br2])/(Δt) = -1/2 (Δ[FeBr2])/(Δt) = 1/2 (Δ[FeBr3])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| bromine | iron(II) bromide | iron(III) bromide formula | Br_2 | FeBr_2 | FeBr_3 Hill formula | Br_2 | Br_2Fe | Br_3Fe name | bromine | iron(II) bromide | iron(III) bromide IUPAC name | molecular bromine | dibromoiron | tribromoiron

| bromine | iron(II) bromide | iron(III) bromide molar mass | 159.81 g/mol | 215.65 g/mol | 295.56 g/mol phase | liquid (at STP) | solid (at STP) | melting point | -7.2 °C | 684 °C | boiling point | 58.8 °C | 934 °C | density | 3.119 g/cm^3 | 4.63 g/cm^3 | solubility in water | insoluble | | surface tension | 0.0409 N/m | | dynamic viscosity | 9.44×10^-4 Pa s (at 25 °C) | | odor | | | odorless