Zn + F2 = ZnF2

Zn zinc + F_2 fluorine ⟶ ZnF_2 zinc fluoride

Balance the chemical equation algebraically: Zn + F_2 ⟶ ZnF_2 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 Zn + c_2 F_2 ⟶ c_3 ZnF_2 Set the number of atoms in the reactants equal to the number of atoms in the products for Zn and F: Zn: | 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: | | Zn + F_2 ⟶ ZnF_2

+ ⟶

zinc + fluorine ⟶ zinc fluoride

| zinc | fluorine | zinc fluoride molecular enthalpy | 0 kJ/mol | 0 kJ/mol | -764.4 kJ/mol total enthalpy | 0 kJ/mol | 0 kJ/mol | -764.4 kJ/mol | H_initial = 0 kJ/mol | | H_final = -764.4 kJ/mol ΔH_rxn^0 | -764.4 kJ/mol - 0 kJ/mol = -764.4 kJ/mol (exothermic) | |

| zinc | fluorine | zinc fluoride molecular entropy | 42 J/(mol K) | 202.8 J/(mol K) | 73.7 J/(mol K) total entropy | 42 J/(mol K) | 202.8 J/(mol K) | 73.7 J/(mol K) | S_initial = 244.8 J/(mol K) | | S_final = 73.7 J/(mol K) ΔS_rxn^0 | 73.7 J/(mol K) - 244.8 J/(mol K) = -171.1 J/(mol K) (exoentropic) | |

Construct the equilibrium constant, K, expression for: Zn + F_2 ⟶ ZnF_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: Zn + F_2 ⟶ ZnF_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 Zn | 1 | -1 F_2 | 1 | -1 ZnF_2 | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression Zn | 1 | -1 | ([Zn])^(-1) F_2 | 1 | -1 | ([F2])^(-1) ZnF_2 | 1 | 1 | [ZnF2] 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 = ([Zn])^(-1) ([F2])^(-1) [ZnF2] = ([ZnF2])/([Zn] [F2])

Construct the rate of reaction expression for: Zn + F_2 ⟶ ZnF_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: Zn + F_2 ⟶ ZnF_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 Zn | 1 | -1 F_2 | 1 | -1 ZnF_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 Zn | 1 | -1 | -(Δ[Zn])/(Δt) F_2 | 1 | -1 | -(Δ[F2])/(Δt) ZnF_2 | 1 | 1 | (Δ[ZnF2])/(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[Zn])/(Δt) = -(Δ[F2])/(Δt) = (Δ[ZnF2])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| zinc | fluorine | zinc fluoride formula | Zn | F_2 | ZnF_2 Hill formula | Zn | F_2 | F_2Zn name | zinc | fluorine | zinc fluoride IUPAC name | zinc | molecular fluorine | difluorozinc

| zinc | fluorine | zinc fluoride molar mass | 65.38 g/mol | 37.996806326 g/mol | 103.4 g/mol phase | solid (at STP) | gas (at STP) | solid (at STP) melting point | 420 °C | -219.6 °C | 872 °C boiling point | 907 °C | -188.12 °C | 1500 °C density | 7.14 g/cm^3 | 0.001696 g/cm^3 (at 0 °C) | 4.95 g/cm^3 solubility in water | insoluble | reacts | dynamic viscosity | | 2.344×10^-5 Pa s (at 25 °C) | odor | odorless | |