I2 + F2 = IF5

I_2 iodine + F_2 fluorine ⟶ F_5I_1 iodine pentafluoride

Balance the chemical equation algebraically: I_2 + F_2 ⟶ F_5I_1 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 I_2 + c_2 F_2 ⟶ c_3 F_5I_1 Set the number of atoms in the reactants equal to the number of atoms in the products for I and F: I: | 2 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 c_3 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | I_2 + 5 F_2 ⟶ 2 F_5I_1

+ ⟶

iodine + fluorine ⟶ iodine pentafluoride

Construct the equilibrium constant, K, expression for: I_2 + F_2 ⟶ F_5I_1 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: I_2 + 5 F_2 ⟶ 2 F_5I_1 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 I_2 | 1 | -1 F_2 | 5 | -5 F_5I_1 | 2 | 2 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression I_2 | 1 | -1 | ([I2])^(-1) F_2 | 5 | -5 | ([F2])^(-5) F_5I_1 | 2 | 2 | ([F5I1])^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 = ([I2])^(-1) ([F2])^(-5) ([F5I1])^2 = ([F5I1])^2/([I2] ([F2])^5)

Construct the rate of reaction expression for: I_2 + F_2 ⟶ F_5I_1 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: I_2 + 5 F_2 ⟶ 2 F_5I_1 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 I_2 | 1 | -1 F_2 | 5 | -5 F_5I_1 | 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 I_2 | 1 | -1 | -(Δ[I2])/(Δt) F_2 | 5 | -5 | -1/5 (Δ[F2])/(Δt) F_5I_1 | 2 | 2 | 1/2 (Δ[F5I1])/(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[I2])/(Δt) = -1/5 (Δ[F2])/(Δt) = 1/2 (Δ[F5I1])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| iodine | fluorine | iodine pentafluoride formula | I_2 | F_2 | F_5I_1 Hill formula | I_2 | F_2 | F_5I name | iodine | fluorine | iodine pentafluoride IUPAC name | molecular iodine | molecular fluorine | pentafluoro-$l^{5}-iodane

| iodine | fluorine | iodine pentafluoride molar mass | 253.80894 g/mol | 37.996806326 g/mol | 221.89649 g/mol phase | solid (at STP) | gas (at STP) | melting point | 113 °C | -219.6 °C | boiling point | 184 °C | -188.12 °C | density | 4.94 g/cm^3 | 0.001696 g/cm^3 (at 0 °C) | solubility in water | | reacts | dynamic viscosity | 0.00227 Pa s (at 116 °C) | 2.344×10^-5 Pa s (at 25 °C) |