F2 + Xe = XeF6

F_2 fluorine + Xe xenon ⟶ F_6Xe_1 xenon hexafluoride

Balance the chemical equation algebraically: F_2 + Xe ⟶ F_6Xe_1 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 F_2 + c_2 Xe ⟶ c_3 F_6Xe_1 Set the number of atoms in the reactants equal to the number of atoms in the products for F and Xe: F: | 2 c_1 = 6 c_3 Xe: | 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 c_2 = 1 c_3 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 3 F_2 + Xe ⟶ F_6Xe_1

+ ⟶

fluorine + xenon ⟶ xenon hexafluoride

Construct the equilibrium constant, K, expression for: F_2 + Xe ⟶ F_6Xe_1 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 + Xe ⟶ F_6Xe_1 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 Xe | 1 | -1 F_6Xe_1 | 1 | 1 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) Xe | 1 | -1 | ([Xe])^(-1) F_6Xe_1 | 1 | 1 | [F6Xe1] 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) ([Xe])^(-1) [F6Xe1] = ([F6Xe1])/(([F2])^3 [Xe])

Construct the rate of reaction expression for: F_2 + Xe ⟶ F_6Xe_1 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 + Xe ⟶ F_6Xe_1 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 Xe | 1 | -1 F_6Xe_1 | 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 F_2 | 3 | -3 | -1/3 (Δ[F2])/(Δt) Xe | 1 | -1 | -(Δ[Xe])/(Δt) F_6Xe_1 | 1 | 1 | (Δ[F6Xe1])/(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) = -(Δ[Xe])/(Δt) = (Δ[F6Xe1])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| fluorine | xenon | xenon hexafluoride formula | F_2 | Xe | F_6Xe_1 Hill formula | F_2 | Xe | F_6Xe name | fluorine | xenon | xenon hexafluoride IUPAC name | molecular fluorine | xenon | hexafluoroxenon

| fluorine | xenon | xenon hexafluoride molar mass | 37.996806326 g/mol | 131.293 g/mol | 245.283 g/mol phase | gas (at STP) | gas (at STP) | melting point | -219.6 °C | -111.8 °C | boiling point | -188.12 °C | -108 °C | density | 0.001696 g/cm^3 (at 0 °C) | 0.0059 g/cm^3 (at 0 °C) | solubility in water | reacts | slightly soluble | dynamic viscosity | 2.344×10^-5 Pa s (at 25 °C) | 2.306×10^-5 Pa s (at 25 °C) | odor | | odorless |