Fe + Br2 = FeBr2

Fe (iron) + Br_2 (bromine) ⟶ FeBr_2 (iron(II) bromide)

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

+ ⟶

iron + bromine ⟶ iron(II) bromide

| iron | bromine | iron(II) bromide molecular enthalpy | 0 kJ/mol | 0 kJ/mol | -249.8 kJ/mol total enthalpy | 0 kJ/mol | 0 kJ/mol | -249.8 kJ/mol | H_initial = 0 kJ/mol | | H_final = -249.8 kJ/mol ΔH_rxn^0 | -249.8 kJ/mol - 0 kJ/mol = -249.8 kJ/mol (exothermic) | |

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

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

| iron | bromine | iron(II) bromide formula | Fe | Br_2 | FeBr_2 Hill formula | Fe | Br_2 | Br_2Fe name | iron | bromine | iron(II) bromide IUPAC name | iron | molecular bromine | dibromoiron

| iron | bromine | iron(II) bromide molar mass | 55.845 g/mol | 159.81 g/mol | 215.65 g/mol phase | solid (at STP) | liquid (at STP) | solid (at STP) melting point | 1535 °C | -7.2 °C | 684 °C boiling point | 2750 °C | 58.8 °C | 934 °C density | 7.874 g/cm^3 | 3.119 g/cm^3 | 4.63 g/cm^3 solubility in water | insoluble | insoluble | surface tension | | 0.0409 N/m | dynamic viscosity | | 9.44×10^-4 Pa s (at 25 °C) |