HF + V = H2 + VF3

HF hydrogen fluoride + V vanadium ⟶ H_2 hydrogen + VF_3 vanadium trifluoride

Balance the chemical equation algebraically: HF + V ⟶ H_2 + VF_3 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 HF + c_2 V ⟶ c_3 H_2 + c_4 VF_3 Set the number of atoms in the reactants equal to the number of atoms in the products for F, H and V: F: | c_1 = 3 c_4 H: | c_1 = 2 c_3 V: | c_2 = 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_2 = 1 and solve the system of equations for the remaining coefficients: c_1 = 3 c_2 = 1 c_3 = 3/2 c_4 = 1 Multiply by the least common denominator, 2, to eliminate fractional coefficients: c_1 = 6 c_2 = 2 c_3 = 3 c_4 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 6 HF + 2 V ⟶ 3 H_2 + 2 VF_3

+ ⟶ +

hydrogen fluoride + vanadium ⟶ hydrogen + vanadium trifluoride

Construct the equilibrium constant, K, expression for: HF + V ⟶ H_2 + VF_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: 6 HF + 2 V ⟶ 3 H_2 + 2 VF_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 HF | 6 | -6 V | 2 | -2 H_2 | 3 | 3 VF_3 | 2 | 2 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression HF | 6 | -6 | ([HF])^(-6) V | 2 | -2 | ([V])^(-2) H_2 | 3 | 3 | ([H2])^3 VF_3 | 2 | 2 | ([VF3])^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 = ([HF])^(-6) ([V])^(-2) ([H2])^3 ([VF3])^2 = (([H2])^3 ([VF3])^2)/(([HF])^6 ([V])^2)

Construct the rate of reaction expression for: HF + V ⟶ H_2 + VF_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: 6 HF + 2 V ⟶ 3 H_2 + 2 VF_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 HF | 6 | -6 V | 2 | -2 H_2 | 3 | 3 VF_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 HF | 6 | -6 | -1/6 (Δ[HF])/(Δt) V | 2 | -2 | -1/2 (Δ[V])/(Δt) H_2 | 3 | 3 | 1/3 (Δ[H2])/(Δt) VF_3 | 2 | 2 | 1/2 (Δ[VF3])/(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 = -1/6 (Δ[HF])/(Δt) = -1/2 (Δ[V])/(Δt) = 1/3 (Δ[H2])/(Δt) = 1/2 (Δ[VF3])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| hydrogen fluoride | vanadium | hydrogen | vanadium trifluoride formula | HF | V | H_2 | VF_3 Hill formula | FH | V | H_2 | F_3V name | hydrogen fluoride | vanadium | hydrogen | vanadium trifluoride IUPAC name | hydrogen fluoride | vanadium | molecular hydrogen | trifluorovanadium

| hydrogen fluoride | vanadium | hydrogen | vanadium trifluoride molar mass | 20.006 g/mol | 50.9415 g/mol | 2.016 g/mol | 107.937 g/mol phase | gas (at STP) | solid (at STP) | gas (at STP) | solid (at STP) melting point | -83.36 °C | 1890 °C | -259.2 °C | 1406 °C boiling point | 19.5 °C | 3380 °C | -252.8 °C | density | 8.18×10^-4 g/cm^3 (at 25 °C) | 6.11 g/cm^3 | 8.99×10^-5 g/cm^3 (at 0 °C) | 3.363 g/cm^3 solubility in water | miscible | insoluble | | dynamic viscosity | 1.2571×10^-5 Pa s (at 20 °C) | | 8.9×10^-6 Pa s (at 25 °C) | odor | | | odorless |