H2O + KrF2 = O2 + HF + Kr

H_2O water + F_2Kr_1 krypton difluoride ⟶ O_2 oxygen + HF hydrogen fluoride + Kr krypton

Balance the chemical equation algebraically: H_2O + F_2Kr_1 ⟶ O_2 + HF + Kr Add stoichiometric coefficients, c_i, to the reactants and products: c_1 H_2O + c_2 F_2Kr_1 ⟶ c_3 O_2 + c_4 HF + c_5 Kr Set the number of atoms in the reactants equal to the number of atoms in the products for H, O, F and Kr: H: | 2 c_1 = c_4 O: | c_1 = 2 c_3 F: | 2 c_2 = c_4 Kr: | c_2 = c_5 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_3 = 1 and solve the system of equations for the remaining coefficients: c_1 = 2 c_2 = 2 c_3 = 1 c_4 = 4 c_5 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 2 H_2O + 2 F_2Kr_1 ⟶ O_2 + 4 HF + 2 Kr

+ ⟶ + +

water + krypton difluoride ⟶ oxygen + hydrogen fluoride + krypton

Construct the equilibrium constant, K, expression for: H_2O + F_2Kr_1 ⟶ O_2 + HF + Kr 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 + 2 F_2Kr_1 ⟶ O_2 + 4 HF + 2 Kr 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 F_2Kr_1 | 2 | -2 O_2 | 1 | 1 HF | 4 | 4 Kr | 2 | 2 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) F_2Kr_1 | 2 | -2 | ([F2Kr1])^(-2) O_2 | 1 | 1 | [O2] HF | 4 | 4 | ([HF])^4 Kr | 2 | 2 | ([Kr])^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 = ([H2O])^(-2) ([F2Kr1])^(-2) [O2] ([HF])^4 ([Kr])^2 = ([O2] ([HF])^4 ([Kr])^2)/(([H2O])^2 ([F2Kr1])^2)

Construct the rate of reaction expression for: H_2O + F_2Kr_1 ⟶ O_2 + HF + Kr 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 + 2 F_2Kr_1 ⟶ O_2 + 4 HF + 2 Kr 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 F_2Kr_1 | 2 | -2 O_2 | 1 | 1 HF | 4 | 4 Kr | 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 H_2O | 2 | -2 | -1/2 (Δ[H2O])/(Δt) F_2Kr_1 | 2 | -2 | -1/2 (Δ[F2Kr1])/(Δt) O_2 | 1 | 1 | (Δ[O2])/(Δt) HF | 4 | 4 | 1/4 (Δ[HF])/(Δt) Kr | 2 | 2 | 1/2 (Δ[Kr])/(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) = -1/2 (Δ[F2Kr1])/(Δt) = (Δ[O2])/(Δt) = 1/4 (Δ[HF])/(Δt) = 1/2 (Δ[Kr])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| water | krypton difluoride | oxygen | hydrogen fluoride | krypton formula | H_2O | F_2Kr_1 | O_2 | HF | Kr Hill formula | H_2O | F_2Kr | O_2 | FH | Kr name | water | krypton difluoride | oxygen | hydrogen fluoride | krypton IUPAC name | water | difluorokrypton | molecular oxygen | hydrogen fluoride | krypton