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H2O + OsF6 = HF + OsO4 + OsO2

Input interpretation

H_2O water + F_6Os_1 osmium(VI) fluoride ⟶ HF hydrogen fluoride + OsO_4 osmium tetroxide + OsO_2 osmium(IV) oxide
H_2O water + F_6Os_1 osmium(VI) fluoride ⟶ HF hydrogen fluoride + OsO_4 osmium tetroxide + OsO_2 osmium(IV) oxide

Balanced equation

Balance the chemical equation algebraically: H_2O + F_6Os_1 ⟶ HF + OsO_4 + OsO_2 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 H_2O + c_2 F_6Os_1 ⟶ c_3 HF + c_4 OsO_4 + c_5 OsO_2 Set the number of atoms in the reactants equal to the number of atoms in the products for H, O, F and Os: H: | 2 c_1 = c_3 O: | c_1 = 4 c_4 + 2 c_5 F: | 6 c_2 = c_3 Os: | 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 = 6 c_2 = 2 c_3 = 12 c_4 = 1 c_5 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 6 H_2O + 2 F_6Os_1 ⟶ 12 HF + OsO_4 + OsO_2
Balance the chemical equation algebraically: H_2O + F_6Os_1 ⟶ HF + OsO_4 + OsO_2 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 H_2O + c_2 F_6Os_1 ⟶ c_3 HF + c_4 OsO_4 + c_5 OsO_2 Set the number of atoms in the reactants equal to the number of atoms in the products for H, O, F and Os: H: | 2 c_1 = c_3 O: | c_1 = 4 c_4 + 2 c_5 F: | 6 c_2 = c_3 Os: | 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 = 6 c_2 = 2 c_3 = 12 c_4 = 1 c_5 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 6 H_2O + 2 F_6Os_1 ⟶ 12 HF + OsO_4 + OsO_2

Structures

+ ⟶ + +
+ ⟶ + +

Names

water + osmium(VI) fluoride ⟶ hydrogen fluoride + osmium tetroxide + osmium(IV) oxide
water + osmium(VI) fluoride ⟶ hydrogen fluoride + osmium tetroxide + osmium(IV) oxide

Equilibrium constant

Construct the equilibrium constant, K, expression for: H_2O + F_6Os_1 ⟶ HF + OsO_4 + OsO_2 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: 6 H_2O + 2 F_6Os_1 ⟶ 12 HF + OsO_4 + OsO_2 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 | 6 | -6 F_6Os_1 | 2 | -2 HF | 12 | 12 OsO_4 | 1 | 1 OsO_2 | 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 | 6 | -6 | ([H2O])^(-6) F_6Os_1 | 2 | -2 | ([F6Os1])^(-2) HF | 12 | 12 | ([HF])^12 OsO_4 | 1 | 1 | [OsO4] OsO_2 | 1 | 1 | [OsO2] 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])^(-6) ([F6Os1])^(-2) ([HF])^12 [OsO4] [OsO2] = (([HF])^12 [OsO4] [OsO2])/(([H2O])^6 ([F6Os1])^2)
Construct the equilibrium constant, K, expression for: H_2O + F_6Os_1 ⟶ HF + OsO_4 + OsO_2 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: 6 H_2O + 2 F_6Os_1 ⟶ 12 HF + OsO_4 + OsO_2 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 | 6 | -6 F_6Os_1 | 2 | -2 HF | 12 | 12 OsO_4 | 1 | 1 OsO_2 | 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 | 6 | -6 | ([H2O])^(-6) F_6Os_1 | 2 | -2 | ([F6Os1])^(-2) HF | 12 | 12 | ([HF])^12 OsO_4 | 1 | 1 | [OsO4] OsO_2 | 1 | 1 | [OsO2] 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])^(-6) ([F6Os1])^(-2) ([HF])^12 [OsO4] [OsO2] = (([HF])^12 [OsO4] [OsO2])/(([H2O])^6 ([F6Os1])^2)

Rate of reaction

Construct the rate of reaction expression for: H_2O + F_6Os_1 ⟶ HF + OsO_4 + OsO_2 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: 6 H_2O + 2 F_6Os_1 ⟶ 12 HF + OsO_4 + OsO_2 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 | 6 | -6 F_6Os_1 | 2 | -2 HF | 12 | 12 OsO_4 | 1 | 1 OsO_2 | 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 | 6 | -6 | -1/6 (Δ[H2O])/(Δt) F_6Os_1 | 2 | -2 | -1/2 (Δ[F6Os1])/(Δt) HF | 12 | 12 | 1/12 (Δ[HF])/(Δt) OsO_4 | 1 | 1 | (Δ[OsO4])/(Δt) OsO_2 | 1 | 1 | (Δ[OsO2])/(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 = -1/6 (Δ[H2O])/(Δt) = -1/2 (Δ[F6Os1])/(Δt) = 1/12 (Δ[HF])/(Δt) = (Δ[OsO4])/(Δt) = (Δ[OsO2])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)
Construct the rate of reaction expression for: H_2O + F_6Os_1 ⟶ HF + OsO_4 + OsO_2 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: 6 H_2O + 2 F_6Os_1 ⟶ 12 HF + OsO_4 + OsO_2 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 | 6 | -6 F_6Os_1 | 2 | -2 HF | 12 | 12 OsO_4 | 1 | 1 OsO_2 | 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 | 6 | -6 | -1/6 (Δ[H2O])/(Δt) F_6Os_1 | 2 | -2 | -1/2 (Δ[F6Os1])/(Δt) HF | 12 | 12 | 1/12 (Δ[HF])/(Δt) OsO_4 | 1 | 1 | (Δ[OsO4])/(Δt) OsO_2 | 1 | 1 | (Δ[OsO2])/(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 = -1/6 (Δ[H2O])/(Δt) = -1/2 (Δ[F6Os1])/(Δt) = 1/12 (Δ[HF])/(Δt) = (Δ[OsO4])/(Δt) = (Δ[OsO2])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

Chemical names and formulas

| water | osmium(VI) fluoride | hydrogen fluoride | osmium tetroxide | osmium(IV) oxide formula | H_2O | F_6Os_1 | HF | OsO_4 | OsO_2 Hill formula | H_2O | F_6Os | FH | O_4Os | O_2Os_1 name | water | osmium(VI) fluoride | hydrogen fluoride | osmium tetroxide | osmium(IV) oxide IUPAC name | water | hexafluoroosmium | hydrogen fluoride | tetraoxoosmium |
| water | osmium(VI) fluoride | hydrogen fluoride | osmium tetroxide | osmium(IV) oxide formula | H_2O | F_6Os_1 | HF | OsO_4 | OsO_2 Hill formula | H_2O | F_6Os | FH | O_4Os | O_2Os_1 name | water | osmium(VI) fluoride | hydrogen fluoride | osmium tetroxide | osmium(IV) oxide IUPAC name | water | hexafluoroosmium | hydrogen fluoride | tetraoxoosmium |

Substance properties

| water | osmium(VI) fluoride | hydrogen fluoride | osmium tetroxide | osmium(IV) oxide molar mass | 18.015 g/mol | 304.22 g/mol | 20.006 g/mol | 254.23 g/mol | 222.23 g/mol phase | liquid (at STP) | | gas (at STP) | solid (at STP) | melting point | 0 °C | | -83.36 °C | 40 °C | boiling point | 99.9839 °C | | 19.5 °C | 130 °C | density | 1 g/cm^3 | | 8.18×10^-4 g/cm^3 (at 25 °C) | 1.04 g/cm^3 | 11.4 g/cm^3 solubility in water | | | miscible | | 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 | osmium(VI) fluoride | hydrogen fluoride | osmium tetroxide | osmium(IV) oxide molar mass | 18.015 g/mol | 304.22 g/mol | 20.006 g/mol | 254.23 g/mol | 222.23 g/mol phase | liquid (at STP) | | gas (at STP) | solid (at STP) | melting point | 0 °C | | -83.36 °C | 40 °C | boiling point | 99.9839 °C | | 19.5 °C | 130 °C | density | 1 g/cm^3 | | 8.18×10^-4 g/cm^3 (at 25 °C) | 1.04 g/cm^3 | 11.4 g/cm^3 solubility in water | | | miscible | | 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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