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O2 + F2 = OF2

Input interpretation

O_2 oxygen + F_2 fluorine ⟶ F_2O oxygen difluoride
O_2 oxygen + F_2 fluorine ⟶ F_2O oxygen difluoride

Balanced equation

Balance the chemical equation algebraically: O_2 + F_2 ⟶ F_2O Add stoichiometric coefficients, c_i, to the reactants and products: c_1 O_2 + c_2 F_2 ⟶ c_3 F_2O Set the number of atoms in the reactants equal to the number of atoms in the products for O and F: O: | 2 c_1 = c_3 F: | 2 c_2 = 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_1 = 1 and solve the system of equations for the remaining coefficients: c_1 = 1 c_2 = 2 c_3 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: |   | O_2 + 2 F_2 ⟶ 2 F_2O
Balance the chemical equation algebraically: O_2 + F_2 ⟶ F_2O Add stoichiometric coefficients, c_i, to the reactants and products: c_1 O_2 + c_2 F_2 ⟶ c_3 F_2O Set the number of atoms in the reactants equal to the number of atoms in the products for O and F: O: | 2 c_1 = c_3 F: | 2 c_2 = 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_1 = 1 and solve the system of equations for the remaining coefficients: c_1 = 1 c_2 = 2 c_3 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | O_2 + 2 F_2 ⟶ 2 F_2O

Structures

 + ⟶
+ ⟶

Names

oxygen + fluorine ⟶ oxygen difluoride
oxygen + fluorine ⟶ oxygen difluoride

Reaction thermodynamics

Enthalpy

 | oxygen | fluorine | oxygen difluoride molecular enthalpy | 0 kJ/mol | 0 kJ/mol | 109 kJ/mol total enthalpy | 0 kJ/mol | 0 kJ/mol | 218 kJ/mol  | H_initial = 0 kJ/mol | | H_final = 218 kJ/mol ΔH_rxn^0 | 218 kJ/mol - 0 kJ/mol = 218 kJ/mol (endothermic) | |
| oxygen | fluorine | oxygen difluoride molecular enthalpy | 0 kJ/mol | 0 kJ/mol | 109 kJ/mol total enthalpy | 0 kJ/mol | 0 kJ/mol | 218 kJ/mol | H_initial = 0 kJ/mol | | H_final = 218 kJ/mol ΔH_rxn^0 | 218 kJ/mol - 0 kJ/mol = 218 kJ/mol (endothermic) | |

Gibbs free energy

 | oxygen | fluorine | oxygen difluoride molecular free energy | 231.7 kJ/mol | 0 kJ/mol | 105.3 kJ/mol total free energy | 231.7 kJ/mol | 0 kJ/mol | 210.6 kJ/mol  | G_initial = 231.7 kJ/mol | | G_final = 210.6 kJ/mol ΔG_rxn^0 | 210.6 kJ/mol - 231.7 kJ/mol = -21.1 kJ/mol (exergonic) | |
| oxygen | fluorine | oxygen difluoride molecular free energy | 231.7 kJ/mol | 0 kJ/mol | 105.3 kJ/mol total free energy | 231.7 kJ/mol | 0 kJ/mol | 210.6 kJ/mol | G_initial = 231.7 kJ/mol | | G_final = 210.6 kJ/mol ΔG_rxn^0 | 210.6 kJ/mol - 231.7 kJ/mol = -21.1 kJ/mol (exergonic) | |

Entropy

 | oxygen | fluorine | oxygen difluoride molecular entropy | 205 J/(mol K) | 202.8 J/(mol K) | 216.4 J/(mol K) total entropy | 205 J/(mol K) | 405.6 J/(mol K) | 432.8 J/(mol K)  | S_initial = 610.6 J/(mol K) | | S_final = 432.8 J/(mol K) ΔS_rxn^0 | 432.8 J/(mol K) - 610.6 J/(mol K) = -177.8 J/(mol K) (exoentropic) | |
| oxygen | fluorine | oxygen difluoride molecular entropy | 205 J/(mol K) | 202.8 J/(mol K) | 216.4 J/(mol K) total entropy | 205 J/(mol K) | 405.6 J/(mol K) | 432.8 J/(mol K) | S_initial = 610.6 J/(mol K) | | S_final = 432.8 J/(mol K) ΔS_rxn^0 | 432.8 J/(mol K) - 610.6 J/(mol K) = -177.8 J/(mol K) (exoentropic) | |

Equilibrium constant

Construct the equilibrium constant, K, expression for: O_2 + F_2 ⟶ F_2O 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: O_2 + 2 F_2 ⟶ 2 F_2O 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 O_2 | 1 | -1 F_2 | 2 | -2 F_2O | 2 | 2 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression O_2 | 1 | -1 | ([O2])^(-1) F_2 | 2 | -2 | ([F2])^(-2) F_2O | 2 | 2 | ([F2O])^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 = ([O2])^(-1) ([F2])^(-2) ([F2O])^2 = ([F2O])^2/([O2] ([F2])^2)
Construct the equilibrium constant, K, expression for: O_2 + F_2 ⟶ F_2O 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: O_2 + 2 F_2 ⟶ 2 F_2O 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 O_2 | 1 | -1 F_2 | 2 | -2 F_2O | 2 | 2 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression O_2 | 1 | -1 | ([O2])^(-1) F_2 | 2 | -2 | ([F2])^(-2) F_2O | 2 | 2 | ([F2O])^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 = ([O2])^(-1) ([F2])^(-2) ([F2O])^2 = ([F2O])^2/([O2] ([F2])^2)

Rate of reaction

Construct the rate of reaction expression for: O_2 + F_2 ⟶ F_2O 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: O_2 + 2 F_2 ⟶ 2 F_2O 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 O_2 | 1 | -1 F_2 | 2 | -2 F_2O | 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 O_2 | 1 | -1 | -(Δ[O2])/(Δt) F_2 | 2 | -2 | -1/2 (Δ[F2])/(Δt) F_2O | 2 | 2 | 1/2 (Δ[F2O])/(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 = -(Δ[O2])/(Δt) = -1/2 (Δ[F2])/(Δt) = 1/2 (Δ[F2O])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)
Construct the rate of reaction expression for: O_2 + F_2 ⟶ F_2O 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: O_2 + 2 F_2 ⟶ 2 F_2O 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 O_2 | 1 | -1 F_2 | 2 | -2 F_2O | 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 O_2 | 1 | -1 | -(Δ[O2])/(Δt) F_2 | 2 | -2 | -1/2 (Δ[F2])/(Δt) F_2O | 2 | 2 | 1/2 (Δ[F2O])/(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 = -(Δ[O2])/(Δt) = -1/2 (Δ[F2])/(Δt) = 1/2 (Δ[F2O])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

Chemical names and formulas

 | oxygen | fluorine | oxygen difluoride formula | O_2 | F_2 | F_2O name | oxygen | fluorine | oxygen difluoride IUPAC name | molecular oxygen | molecular fluorine | fluoro hypofluorite
| oxygen | fluorine | oxygen difluoride formula | O_2 | F_2 | F_2O name | oxygen | fluorine | oxygen difluoride IUPAC name | molecular oxygen | molecular fluorine | fluoro hypofluorite

Substance properties

 | oxygen | fluorine | oxygen difluoride molar mass | 31.998 g/mol | 37.996806326 g/mol | 53.996 g/mol phase | gas (at STP) | gas (at STP) | gas (at STP) melting point | -218 °C | -219.6 °C | -223.9 °C boiling point | -183 °C | -188.12 °C | -145 °C density | 0.001429 g/cm^3 (at 0 °C) | 0.001696 g/cm^3 (at 0 °C) | 0.002207 g/cm^3 (at 25 °C) solubility in water | | reacts |  surface tension | 0.01347 N/m | |  dynamic viscosity | 2.055×10^-5 Pa s (at 25 °C) | 2.344×10^-5 Pa s (at 25 °C) |  odor | odorless | |
| oxygen | fluorine | oxygen difluoride molar mass | 31.998 g/mol | 37.996806326 g/mol | 53.996 g/mol phase | gas (at STP) | gas (at STP) | gas (at STP) melting point | -218 °C | -219.6 °C | -223.9 °C boiling point | -183 °C | -188.12 °C | -145 °C density | 0.001429 g/cm^3 (at 0 °C) | 0.001696 g/cm^3 (at 0 °C) | 0.002207 g/cm^3 (at 25 °C) solubility in water | | reacts | surface tension | 0.01347 N/m | | dynamic viscosity | 2.055×10^-5 Pa s (at 25 °C) | 2.344×10^-5 Pa s (at 25 °C) | odor | odorless | |

Units