H2O + XeF4O = HF + XeO3

H_2O water + XeF4O ⟶ HF hydrogen fluoride + XeO_3 xenon trioxide

Balance the chemical equation algebraically: H_2O + XeF4O ⟶ HF + XeO_3 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 H_2O + c_2 XeF4O ⟶ 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, Xe and F: H: | 2 c_1 = c_3 O: | c_1 + c_2 = 3 c_4 Xe: | c_2 = c_4 F: | 4 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 = 2 c_2 = 1 c_3 = 4 c_4 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 2 H_2O + XeF4O ⟶ 4 HF + XeO_3

+ XeF4O ⟶ + XeO_3

water + XeF4O ⟶ hydrogen fluoride + xenon trioxide

Construct the equilibrium constant, K, expression for: H_2O + XeF4O ⟶ 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: 2 H_2O + XeF4O ⟶ 4 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 | 2 | -2 XeF4O | 1 | -1 HF | 4 | 4 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 | 2 | -2 | ([H2O])^(-2) XeF4O | 1 | -1 | ([XeF4O])^(-1) HF | 4 | 4 | ([HF])^4 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])^(-2) ([XeF4O])^(-1) ([HF])^4 [XeO3] = (([HF])^4 [XeO3])/(([H2O])^2 [XeF4O])

Construct the rate of reaction expression for: H_2O + XeF4O ⟶ 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: 2 H_2O + XeF4O ⟶ 4 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 | 2 | -2 XeF4O | 1 | -1 HF | 4 | 4 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 | 2 | -2 | -1/2 (Δ[H2O])/(Δt) XeF4O | 1 | -1 | -(Δ[XeF4O])/(Δt) HF | 4 | 4 | 1/4 (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 = -1/2 (Δ[H2O])/(Δt) = -(Δ[XeF4O])/(Δt) = 1/4 (Δ[HF])/(Δt) = (Δ[XeO3])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| water | XeF4O | hydrogen fluoride | xenon trioxide formula | H_2O | XeF4O | HF | XeO_3 Hill formula | H_2O | F4OXe | FH | O_3Xe_1 name | water | | hydrogen fluoride | xenon trioxide

| water | XeF4O | hydrogen fluoride | xenon trioxide molar mass | 18.015 g/mol | 223.286 g/mol | 20.006 g/mol | 179.3 g/mol phase | liquid (at STP) | | gas (at STP) | melting point | 0 °C | | -83.36 °C | boiling point | 99.9839 °C | | 19.5 °C | density | 1 g/cm^3 | | 8.18×10^-4 g/cm^3 (at 25 °C) | 4.55 g/cm^3 solubility in water | | | miscible | soluble surface tension | 0.0728 N/m | | | dynamic viscosity | 8.9×10^-4 Pa s (at 25 °C) | | 1.2571×10^-5 Pa s (at 20 °C) | odor | odorless | | |