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H2O + ReF6 = HF + ReO2 + HReO4

Input interpretation

H_2O water + ReF_6 rhenium(VI) fluoride ⟶ HF hydrogen fluoride + ReO_2 rhenium oxide + HReO_4 perrhenic acid
H_2O water + ReF_6 rhenium(VI) fluoride ⟶ HF hydrogen fluoride + ReO_2 rhenium oxide + HReO_4 perrhenic acid

Balanced equation

Balance the chemical equation algebraically: H_2O + ReF_6 ⟶ HF + ReO_2 + HReO_4 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 H_2O + c_2 ReF_6 ⟶ c_3 HF + c_4 ReO_2 + c_5 HReO_4 Set the number of atoms in the reactants equal to the number of atoms in the products for H, O, F and Re: H: | 2 c_1 = c_3 + c_5 O: | c_1 = 2 c_4 + 4 c_5 F: | 6 c_2 = c_3 Re: | c_2 = c_4 + 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_4 = 1 and solve the system of equations for the remaining coefficients: c_1 = 10 c_2 = 3 c_3 = 18 c_4 = 1 c_5 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 10 H_2O + 3 ReF_6 ⟶ 18 HF + ReO_2 + 2 HReO_4
Balance the chemical equation algebraically: H_2O + ReF_6 ⟶ HF + ReO_2 + HReO_4 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 H_2O + c_2 ReF_6 ⟶ c_3 HF + c_4 ReO_2 + c_5 HReO_4 Set the number of atoms in the reactants equal to the number of atoms in the products for H, O, F and Re: H: | 2 c_1 = c_3 + c_5 O: | c_1 = 2 c_4 + 4 c_5 F: | 6 c_2 = c_3 Re: | c_2 = c_4 + 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_4 = 1 and solve the system of equations for the remaining coefficients: c_1 = 10 c_2 = 3 c_3 = 18 c_4 = 1 c_5 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 10 H_2O + 3 ReF_6 ⟶ 18 HF + ReO_2 + 2 HReO_4

Structures

+ ⟶ + +
+ ⟶ + +

Names

water + rhenium(VI) fluoride ⟶ hydrogen fluoride + rhenium oxide + perrhenic acid
water + rhenium(VI) fluoride ⟶ hydrogen fluoride + rhenium oxide + perrhenic acid

Equilibrium constant

Construct the equilibrium constant, K, expression for: H_2O + ReF_6 ⟶ HF + ReO_2 + HReO_4 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: 10 H_2O + 3 ReF_6 ⟶ 18 HF + ReO_2 + 2 HReO_4 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 | 10 | -10 ReF_6 | 3 | -3 HF | 18 | 18 ReO_2 | 1 | 1 HReO_4 | 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 | 10 | -10 | ([H2O])^(-10) ReF_6 | 3 | -3 | ([ReF6])^(-3) HF | 18 | 18 | ([HF])^18 ReO_2 | 1 | 1 | [ReO2] HReO_4 | 2 | 2 | ([HReO4])^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])^(-10) ([ReF6])^(-3) ([HF])^18 [ReO2] ([HReO4])^2 = (([HF])^18 [ReO2] ([HReO4])^2)/(([H2O])^10 ([ReF6])^3)
Construct the equilibrium constant, K, expression for: H_2O + ReF_6 ⟶ HF + ReO_2 + HReO_4 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: 10 H_2O + 3 ReF_6 ⟶ 18 HF + ReO_2 + 2 HReO_4 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 | 10 | -10 ReF_6 | 3 | -3 HF | 18 | 18 ReO_2 | 1 | 1 HReO_4 | 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 | 10 | -10 | ([H2O])^(-10) ReF_6 | 3 | -3 | ([ReF6])^(-3) HF | 18 | 18 | ([HF])^18 ReO_2 | 1 | 1 | [ReO2] HReO_4 | 2 | 2 | ([HReO4])^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])^(-10) ([ReF6])^(-3) ([HF])^18 [ReO2] ([HReO4])^2 = (([HF])^18 [ReO2] ([HReO4])^2)/(([H2O])^10 ([ReF6])^3)

Rate of reaction

Construct the rate of reaction expression for: H_2O + ReF_6 ⟶ HF + ReO_2 + HReO_4 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: 10 H_2O + 3 ReF_6 ⟶ 18 HF + ReO_2 + 2 HReO_4 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 | 10 | -10 ReF_6 | 3 | -3 HF | 18 | 18 ReO_2 | 1 | 1 HReO_4 | 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 | 10 | -10 | -1/10 (Δ[H2O])/(Δt) ReF_6 | 3 | -3 | -1/3 (Δ[ReF6])/(Δt) HF | 18 | 18 | 1/18 (Δ[HF])/(Δt) ReO_2 | 1 | 1 | (Δ[ReO2])/(Δt) HReO_4 | 2 | 2 | 1/2 (Δ[HReO4])/(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 = -1/10 (Δ[H2O])/(Δt) = -1/3 (Δ[ReF6])/(Δt) = 1/18 (Δ[HF])/(Δt) = (Δ[ReO2])/(Δt) = 1/2 (Δ[HReO4])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)
Construct the rate of reaction expression for: H_2O + ReF_6 ⟶ HF + ReO_2 + HReO_4 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: 10 H_2O + 3 ReF_6 ⟶ 18 HF + ReO_2 + 2 HReO_4 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 | 10 | -10 ReF_6 | 3 | -3 HF | 18 | 18 ReO_2 | 1 | 1 HReO_4 | 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 | 10 | -10 | -1/10 (Δ[H2O])/(Δt) ReF_6 | 3 | -3 | -1/3 (Δ[ReF6])/(Δt) HF | 18 | 18 | 1/18 (Δ[HF])/(Δt) ReO_2 | 1 | 1 | (Δ[ReO2])/(Δt) HReO_4 | 2 | 2 | 1/2 (Δ[HReO4])/(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 = -1/10 (Δ[H2O])/(Δt) = -1/3 (Δ[ReF6])/(Δt) = 1/18 (Δ[HF])/(Δt) = (Δ[ReO2])/(Δt) = 1/2 (Δ[HReO4])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

Chemical names and formulas

| water | rhenium(VI) fluoride | hydrogen fluoride | rhenium oxide | perrhenic acid formula | H_2O | ReF_6 | HF | ReO_2 | HReO_4 Hill formula | H_2O | F_6Re | FH | O_2Re | HO_4Re name | water | rhenium(VI) fluoride | hydrogen fluoride | rhenium oxide | perrhenic acid IUPAC name | water | hexafluororhenium | hydrogen fluoride | dioxorhenium | hydroxy-trioxorhenium
| water | rhenium(VI) fluoride | hydrogen fluoride | rhenium oxide | perrhenic acid formula | H_2O | ReF_6 | HF | ReO_2 | HReO_4 Hill formula | H_2O | F_6Re | FH | O_2Re | HO_4Re name | water | rhenium(VI) fluoride | hydrogen fluoride | rhenium oxide | perrhenic acid IUPAC name | water | hexafluororhenium | hydrogen fluoride | dioxorhenium | hydroxy-trioxorhenium

Substance properties

| water | rhenium(VI) fluoride | hydrogen fluoride | rhenium oxide | perrhenic acid molar mass | 18.015 g/mol | 300.197 g/mol | 20.006 g/mol | 218.205 g/mol | 251.21 g/mol phase | liquid (at STP) | liquid (at STP) | gas (at STP) | | solid (at STP) melting point | 0 °C | 18.8 °C | -83.36 °C | | 120 °C boiling point | 99.9839 °C | 47.6 °C | 19.5 °C | | density | 1 g/cm^3 | 6 g/cm^3 | 8.18×10^-4 g/cm^3 (at 25 °C) | 11.4 g/cm^3 | 2.16 g/cm^3 solubility in water | | | miscible | insoluble | 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 | | | |
| water | rhenium(VI) fluoride | hydrogen fluoride | rhenium oxide | perrhenic acid molar mass | 18.015 g/mol | 300.197 g/mol | 20.006 g/mol | 218.205 g/mol | 251.21 g/mol phase | liquid (at STP) | liquid (at STP) | gas (at STP) | | solid (at STP) melting point | 0 °C | 18.8 °C | -83.36 °C | | 120 °C boiling point | 99.9839 °C | 47.6 °C | 19.5 °C | | density | 1 g/cm^3 | 6 g/cm^3 | 8.18×10^-4 g/cm^3 (at 25 °C) | 11.4 g/cm^3 | 2.16 g/cm^3 solubility in water | | | miscible | insoluble | 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 | | | |

Units

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