Fe + HBr = H2 + FeBr3

Fe iron + HBr hydrogen bromide ⟶ H_2 hydrogen + FeBr_3 iron(III) bromide

Balance the chemical equation algebraically: Fe + HBr ⟶ H_2 + FeBr_3 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 Fe + c_2 HBr ⟶ c_3 H_2 + c_4 FeBr_3 Set the number of atoms in the reactants equal to the number of atoms in the products for Fe, Br and H: Fe: | c_1 = c_4 Br: | c_2 = 3 c_4 H: | 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 = 3 c_3 = 3/2 c_4 = 1 Multiply by the least common denominator, 2, to eliminate fractional coefficients: c_1 = 2 c_2 = 6 c_3 = 3 c_4 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 2 Fe + 6 HBr ⟶ 3 H_2 + 2 FeBr_3

+ ⟶ +

iron + hydrogen bromide ⟶ hydrogen + iron(III) bromide

Construct the equilibrium constant, K, expression for: Fe + HBr ⟶ H_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: 2 Fe + 6 HBr ⟶ 3 H_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 Fe | 2 | -2 HBr | 6 | -6 H_2 | 3 | 3 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 Fe | 2 | -2 | ([Fe])^(-2) HBr | 6 | -6 | ([HBr])^(-6) H_2 | 3 | 3 | ([H2])^3 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 = ([Fe])^(-2) ([HBr])^(-6) ([H2])^3 ([FeBr3])^2 = (([H2])^3 ([FeBr3])^2)/(([Fe])^2 ([HBr])^6)

Construct the rate of reaction expression for: Fe + HBr ⟶ H_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: 2 Fe + 6 HBr ⟶ 3 H_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 Fe | 2 | -2 HBr | 6 | -6 H_2 | 3 | 3 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 Fe | 2 | -2 | -1/2 (Δ[Fe])/(Δt) HBr | 6 | -6 | -1/6 (Δ[HBr])/(Δt) H_2 | 3 | 3 | 1/3 (Δ[H2])/(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 = -1/2 (Δ[Fe])/(Δt) = -1/6 (Δ[HBr])/(Δt) = 1/3 (Δ[H2])/(Δt) = 1/2 (Δ[FeBr3])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| iron | hydrogen bromide | hydrogen | iron(III) bromide formula | Fe | HBr | H_2 | FeBr_3 Hill formula | Fe | BrH | H_2 | Br_3Fe name | iron | hydrogen bromide | hydrogen | iron(III) bromide IUPAC name | iron | hydrogen bromide | molecular hydrogen | tribromoiron

| iron | hydrogen bromide | hydrogen | iron(III) bromide molar mass | 55.845 g/mol | 80.912 g/mol | 2.016 g/mol | 295.56 g/mol phase | solid (at STP) | gas (at STP) | gas (at STP) | melting point | 1535 °C | -86.8 °C | -259.2 °C | boiling point | 2750 °C | -66.38 °C | -252.8 °C | density | 7.874 g/cm^3 | 0.003307 g/cm^3 (at 25 °C) | 8.99×10^-5 g/cm^3 (at 0 °C) | solubility in water | insoluble | miscible | | surface tension | | 0.0271 N/m | | dynamic viscosity | | 8.4×10^-4 Pa s (at -75 °C) | 8.9×10^-6 Pa s (at 25 °C) | odor | | | odorless | odorless