Fe + NaBr = Na + FeBr3

Fe iron + NaBr sodium bromide ⟶ Na sodium + FeBr_3 iron(III) bromide

Balance the chemical equation algebraically: Fe + NaBr ⟶ Na + FeBr_3 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 Fe + c_2 NaBr ⟶ c_3 Na + 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 Na: Fe: | c_1 = c_4 Br: | c_2 = 3 c_4 Na: | 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 = 3 c_3 = 3 c_4 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | Fe + 3 NaBr ⟶ 3 Na + FeBr_3

+ ⟶ +

iron + sodium bromide ⟶ sodium + iron(III) bromide

Construct the equilibrium constant, K, expression for: Fe + NaBr ⟶ Na + 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: Fe + 3 NaBr ⟶ 3 Na + 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 | 1 | -1 NaBr | 3 | -3 Na | 3 | 3 FeBr_3 | 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) NaBr | 3 | -3 | ([NaBr])^(-3) Na | 3 | 3 | ([Na])^3 FeBr_3 | 1 | 1 | [FeBr3] 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) ([NaBr])^(-3) ([Na])^3 [FeBr3] = (([Na])^3 [FeBr3])/([Fe] ([NaBr])^3)

Construct the rate of reaction expression for: Fe + NaBr ⟶ Na + 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: Fe + 3 NaBr ⟶ 3 Na + 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 | 1 | -1 NaBr | 3 | -3 Na | 3 | 3 FeBr_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 Fe | 1 | -1 | -(Δ[Fe])/(Δt) NaBr | 3 | -3 | -1/3 (Δ[NaBr])/(Δt) Na | 3 | 3 | 1/3 (Δ[Na])/(Δt) FeBr_3 | 1 | 1 | (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[Fe])/(Δt) = -1/3 (Δ[NaBr])/(Δt) = 1/3 (Δ[Na])/(Δt) = (Δ[FeBr3])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| iron | sodium bromide | sodium | iron(III) bromide formula | Fe | NaBr | Na | FeBr_3 Hill formula | Fe | BrNa | Na | Br_3Fe name | iron | sodium bromide | sodium | iron(III) bromide IUPAC name | iron | sodium bromide | sodium | tribromoiron

| iron | sodium bromide | sodium | iron(III) bromide molar mass | 55.845 g/mol | 102.89 g/mol | 22.98976928 g/mol | 295.56 g/mol phase | solid (at STP) | solid (at STP) | solid (at STP) | melting point | 1535 °C | 755 °C | 97.8 °C | boiling point | 2750 °C | 1396 °C | 883 °C | density | 7.874 g/cm^3 | 3.2 g/cm^3 | 0.968 g/cm^3 | solubility in water | insoluble | soluble | decomposes | dynamic viscosity | | | 1.413×10^-5 Pa s (at 527 °C) | odor | | | | odorless