H2O + XeF6 = HF + XeO3

H_2O water + F_6Xe_1 xenon hexafluoride ⟶ HF hydrogen fluoride + XeO_3 xenon trioxide

Balance the chemical equation algebraically: H_2O + F_6Xe_1 ⟶ HF + XeO_3 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 H_2O + c_2 F_6Xe_1 ⟶ c_3 HF + c_4 XeO_3 Set the number of atoms in the reactants equal to the number of atoms in the products for H, O, F and Xe: H: | 2 c_1 = c_3 O: | c_1 = 3 c_4 F: | 6 c_2 = c_3 Xe: | 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 = 6 c_4 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 3 H_2O + F_6Xe_1 ⟶ 6 HF + XeO_3

+ ⟶ + XeO_3

water + xenon hexafluoride ⟶ hydrogen fluoride + xenon trioxide

Construct the equilibrium constant, K, expression for: H_2O + F_6Xe_1 ⟶ HF + XeO_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 H_2O + F_6Xe_1 ⟶ 6 HF + XeO_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 H_2O | 3 | -3 F_6Xe_1 | 1 | -1 HF | 6 | 6 XeO_3 | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression H_2O | 3 | -3 | ([H2O])^(-3) F_6Xe_1 | 1 | -1 | ([F6Xe1])^(-1) HF | 6 | 6 | ([HF])^6 XeO_3 | 1 | 1 | [XeO3] 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 = ([H2O])^(-3) ([F6Xe1])^(-1) ([HF])^6 [XeO3] = (([HF])^6 [XeO3])/(([H2O])^3 [F6Xe1])

Construct the rate of reaction expression for: H_2O + F_6Xe_1 ⟶ HF + XeO_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 H_2O + F_6Xe_1 ⟶ 6 HF + XeO_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 H_2O | 3 | -3 F_6Xe_1 | 1 | -1 HF | 6 | 6 XeO_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 H_2O | 3 | -3 | -1/3 (Δ[H2O])/(Δt) F_6Xe_1 | 1 | -1 | -(Δ[F6Xe1])/(Δt) HF | 6 | 6 | 1/6 (Δ[HF])/(Δt) XeO_3 | 1 | 1 | (Δ[XeO3])/(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[H2O])/(Δt) = -(Δ[F6Xe1])/(Δt) = 1/6 (Δ[HF])/(Δt) = (Δ[XeO3])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| water | xenon hexafluoride | hydrogen fluoride | xenon trioxide formula | H_2O | F_6Xe_1 | HF | XeO_3 Hill formula | H_2O | F_6Xe | FH | O_3Xe_1 name | water | xenon hexafluoride | hydrogen fluoride | xenon trioxide IUPAC name | water | hexafluoroxenon | hydrogen fluoride |