F2 + B = BF3

F_2 fluorine + B boron ⟶ BF_3 boron trifluoride

Balance the chemical equation algebraically: F_2 + B ⟶ BF_3 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 F_2 + c_2 B ⟶ c_3 BF_3 Set the number of atoms in the reactants equal to the number of atoms in the products for F and B: F: | 2 c_1 = 3 c_3 B: | 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_2 = 1 and solve the system of equations for the remaining coefficients: c_1 = 3/2 c_2 = 1 c_3 = 1 Multiply by the least common denominator, 2, to eliminate fractional coefficients: c_1 = 3 c_2 = 2 c_3 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 3 F_2 + 2 B ⟶ 2 BF_3

+ ⟶

fluorine + boron ⟶ boron trifluoride

| fluorine | boron | boron trifluoride molecular enthalpy | 0 kJ/mol | 0 kJ/mol | -1136 kJ/mol total enthalpy | 0 kJ/mol | 0 kJ/mol | -2272 kJ/mol | H_initial = 0 kJ/mol | | H_final = -2272 kJ/mol ΔH_rxn^0 | -2272 kJ/mol - 0 kJ/mol = -2272 kJ/mol (exothermic) | |

Construct the equilibrium constant, K, expression for: F_2 + B ⟶ BF_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: 3 F_2 + 2 B ⟶ 2 BF_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 F_2 | 3 | -3 B | 2 | -2 BF_3 | 2 | 2 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression F_2 | 3 | -3 | ([F2])^(-3) B | 2 | -2 | ([B])^(-2) BF_3 | 2 | 2 | ([BF3])^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 = ([F2])^(-3) ([B])^(-2) ([BF3])^2 = ([BF3])^2/(([F2])^3 ([B])^2)

Construct the rate of reaction expression for: F_2 + B ⟶ BF_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: 3 F_2 + 2 B ⟶ 2 BF_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 F_2 | 3 | -3 B | 2 | -2 BF_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 F_2 | 3 | -3 | -1/3 (Δ[F2])/(Δt) B | 2 | -2 | -1/2 (Δ[B])/(Δt) BF_3 | 2 | 2 | 1/2 (Δ[BF3])/(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 = -1/3 (Δ[F2])/(Δt) = -1/2 (Δ[B])/(Δt) = 1/2 (Δ[BF3])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| fluorine | boron | boron trifluoride formula | F_2 | B | BF_3 name | fluorine | boron | boron trifluoride IUPAC name | molecular fluorine | boron | trifluoroborane

| fluorine | boron | boron trifluoride molar mass | 37.996806326 g/mol | 10.81 g/mol | 67.81 g/mol phase | gas (at STP) | solid (at STP) | gas (at STP) melting point | -219.6 °C | 2075 °C | -127 °C boiling point | -188.12 °C | 4000 °C | -100 °C density | 0.001696 g/cm^3 (at 0 °C) | 2.34 g/cm^3 | 0.002772 g/cm^3 (at 25 °C) solubility in water | reacts | insoluble | surface tension | | | 0.0172 N/m dynamic viscosity | 2.344×10^-5 Pa s (at 25 °C) | | 1.701×10^-5 Pa s (at 25 °C)