BF3 + Li2SO3 = LiF + B2(SO3)3

BF_3 boron trifluoride + Li2SO3 ⟶ LiF lithium fluoride + B2(SO3)3

Balance the chemical equation algebraically: BF_3 + Li2SO3 ⟶ LiF + B2(SO3)3 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 BF_3 + c_2 Li2SO3 ⟶ c_3 LiF + c_4 B2(SO3)3 Set the number of atoms in the reactants equal to the number of atoms in the products for B, F, Li, S and O: B: | c_1 = 2 c_4 F: | 3 c_1 = c_3 Li: | 2 c_2 = c_3 S: | c_2 = 3 c_4 O: | 3 c_2 = 9 c_4 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_4 = 1 and solve the system of equations for the remaining coefficients: c_1 = 2 c_2 = 3 c_3 = 6 c_4 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 2 BF_3 + 3 Li2SO3 ⟶ 6 LiF + B2(SO3)3

+ Li2SO3 ⟶ + B2(SO3)3

boron trifluoride + Li2SO3 ⟶ lithium fluoride + B2(SO3)3

Construct the equilibrium constant, K, expression for: BF_3 + Li2SO3 ⟶ LiF + B2(SO3)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 BF_3 + 3 Li2SO3 ⟶ 6 LiF + B2(SO3)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 BF_3 | 2 | -2 Li2SO3 | 3 | -3 LiF | 6 | 6 B2(SO3)3 | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression BF_3 | 2 | -2 | ([BF3])^(-2) Li2SO3 | 3 | -3 | ([Li2SO3])^(-3) LiF | 6 | 6 | ([LiF])^6 B2(SO3)3 | 1 | 1 | [B2(SO3)3] 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 = ([BF3])^(-2) ([Li2SO3])^(-3) ([LiF])^6 [B2(SO3)3] = (([LiF])^6 [B2(SO3)3])/(([BF3])^2 ([Li2SO3])^3)

Construct the rate of reaction expression for: BF_3 + Li2SO3 ⟶ LiF + B2(SO3)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 BF_3 + 3 Li2SO3 ⟶ 6 LiF + B2(SO3)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 BF_3 | 2 | -2 Li2SO3 | 3 | -3 LiF | 6 | 6 B2(SO3)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 BF_3 | 2 | -2 | -1/2 (Δ[BF3])/(Δt) Li2SO3 | 3 | -3 | -1/3 (Δ[Li2SO3])/(Δt) LiF | 6 | 6 | 1/6 (Δ[LiF])/(Δt) B2(SO3)3 | 1 | 1 | (Δ[B2(SO3)3])/(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[BF3])/(Δt) = -1/3 (Δ[Li2SO3])/(Δt) = 1/6 (Δ[LiF])/(Δt) = (Δ[B2(SO3)3])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| boron trifluoride | Li2SO3 | lithium fluoride | B2(SO3)3 formula | BF_3 | Li2SO3 | LiF | B2(SO3)3 Hill formula | BF_3 | Li2O3S | FLi | B2O9S3 name | boron trifluoride | | lithium fluoride | IUPAC name | trifluoroborane | | lithium fluoride |

| boron trifluoride | Li2SO3 | lithium fluoride | B2(SO3)3 molar mass | 67.81 g/mol | 93.9 g/mol | 25.94 g/mol | 261.8 g/mol phase | gas (at STP) | | solid (at STP) | melting point | -127 °C | | 845 °C | boiling point | -100 °C | | 1676 °C | density | 0.002772 g/cm^3 (at 25 °C) | | 2.64 g/cm^3 | surface tension | 0.0172 N/m | | | dynamic viscosity | 1.701×10^-5 Pa s (at 25 °C) | | |