K + F2 = KF2

K potassium + F_2 fluorine ⟶ KF2

Balance the chemical equation algebraically: K + F_2 ⟶ KF2 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 K + c_2 F_2 ⟶ c_3 KF2 Set the number of atoms in the reactants equal to the number of atoms in the products for K and F: K: | 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 = 1 c_3 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | K + F_2 ⟶ KF2

+ ⟶ KF2

potassium + fluorine ⟶ KF2

Construct the equilibrium constant, K, expression for: K + F_2 ⟶ KF2 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: K + F_2 ⟶ KF2 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 K | 1 | -1 F_2 | 1 | -1 KF2 | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression K | 1 | -1 | ([K])^(-1) F_2 | 1 | -1 | ([F2])^(-1) KF2 | 1 | 1 | [KF2] 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 = ([K])^(-1) ([F2])^(-1) [KF2] = ([KF2])/([K] [F2])

Construct the rate of reaction expression for: K + F_2 ⟶ KF2 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: K + F_2 ⟶ KF2 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 K | 1 | -1 F_2 | 1 | -1 KF2 | 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 K | 1 | -1 | -(Δ[K])/(Δt) F_2 | 1 | -1 | -(Δ[F2])/(Δt) KF2 | 1 | 1 | (Δ[KF2])/(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[K])/(Δt) = -(Δ[F2])/(Δt) = (Δ[KF2])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| potassium | fluorine | KF2 formula | K | F_2 | KF2 Hill formula | K | F_2 | F2K name | potassium | fluorine | IUPAC name | potassium | molecular fluorine |

| potassium | fluorine | KF2 molar mass | 39.0983 g/mol | 37.996806326 g/mol | 77.0951 g/mol phase | solid (at STP) | gas (at STP) | melting point | 64 °C | -219.6 °C | boiling point | 760 °C | -188.12 °C | density | 0.86 g/cm^3 | 0.001696 g/cm^3 (at 0 °C) | solubility in water | reacts | reacts | dynamic viscosity | | 2.344×10^-5 Pa s (at 25 °C) |