P + F2 = PF5

P red phosphorus + F_2 fluorine ⟶ PF_5 phosphorus pentafluoride

Balance the chemical equation algebraically: P + F_2 ⟶ PF_5 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 P + c_2 F_2 ⟶ c_3 PF_5 Set the number of atoms in the reactants equal to the number of atoms in the products for P and F: P: | c_1 = c_3 F: | 2 c_2 = 5 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 = 5/2 c_3 = 1 Multiply by the least common denominator, 2, to eliminate fractional coefficients: c_1 = 2 c_2 = 5 c_3 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 2 P + 5 F_2 ⟶ 2 PF_5

+ ⟶

red phosphorus + fluorine ⟶ phosphorus pentafluoride

Construct the equilibrium constant, K, expression for: P + F_2 ⟶ PF_5 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 P + 5 F_2 ⟶ 2 PF_5 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 P | 2 | -2 F_2 | 5 | -5 PF_5 | 2 | 2 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression P | 2 | -2 | ([P])^(-2) F_2 | 5 | -5 | ([F2])^(-5) PF_5 | 2 | 2 | ([PF5])^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 = ([P])^(-2) ([F2])^(-5) ([PF5])^2 = ([PF5])^2/(([P])^2 ([F2])^5)

Construct the rate of reaction expression for: P + F_2 ⟶ PF_5 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 P + 5 F_2 ⟶ 2 PF_5 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 P | 2 | -2 F_2 | 5 | -5 PF_5 | 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 P | 2 | -2 | -1/2 (Δ[P])/(Δt) F_2 | 5 | -5 | -1/5 (Δ[F2])/(Δt) PF_5 | 2 | 2 | 1/2 (Δ[PF5])/(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[P])/(Δt) = -1/5 (Δ[F2])/(Δt) = 1/2 (Δ[PF5])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| red phosphorus | fluorine | phosphorus pentafluoride formula | P | F_2 | PF_5 Hill formula | P | F_2 | F_5P name | red phosphorus | fluorine | phosphorus pentafluoride IUPAC name | phosphorus | molecular fluorine | pentafluorophosphorane

| red phosphorus | fluorine | phosphorus pentafluoride molar mass | 30.973761998 g/mol | 37.996806326 g/mol | 125.96577781 g/mol phase | solid (at STP) | gas (at STP) | liquid (at STP) melting point | 579.2 °C | -219.6 °C | -83 °C boiling point | | -188.12 °C | density | 2.16 g/cm^3 | 0.001696 g/cm^3 (at 0 °C) | solubility in water | insoluble | reacts | decomposes dynamic viscosity | 7.6×10^-4 Pa s (at 20.2 °C) | 2.344×10^-5 Pa s (at 25 °C) |