KBr + F2 = Br2 + KF

KBr potassium bromide + F_2 fluorine ⟶ Br_2 bromine + KF potassium fluoride

Balance the chemical equation algebraically: KBr + F_2 ⟶ Br_2 + KF Add stoichiometric coefficients, c_i, to the reactants and products: c_1 KBr + c_2 F_2 ⟶ c_3 Br_2 + c_4 KF Set the number of atoms in the reactants equal to the number of atoms in the products for Br, K and F: Br: | c_1 = 2 c_3 K: | c_1 = c_4 F: | 2 c_2 = c_4 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_2 = 1 and solve the system of equations for the remaining coefficients: c_1 = 2 c_2 = 1 c_3 = 1 c_4 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 2 KBr + F_2 ⟶ Br_2 + 2 KF

+ ⟶ +

potassium bromide + fluorine ⟶ bromine + potassium fluoride

| potassium bromide | fluorine | bromine | potassium fluoride molecular enthalpy | -393.8 kJ/mol | 0 kJ/mol | 0 kJ/mol | -567.3 kJ/mol total enthalpy | -787.6 kJ/mol | 0 kJ/mol | 0 kJ/mol | -1135 kJ/mol | H_initial = -787.6 kJ/mol | | H_final = -1135 kJ/mol | ΔH_rxn^0 | -1135 kJ/mol - -787.6 kJ/mol = -347 kJ/mol (exothermic) | | |

| potassium bromide | fluorine | bromine | potassium fluoride molecular free energy | -380.7 kJ/mol | 0 kJ/mol | 0 kJ/mol | -537.8 kJ/mol total free energy | -761.4 kJ/mol | 0 kJ/mol | 0 kJ/mol | -1076 kJ/mol | G_initial = -761.4 kJ/mol | | G_final = -1076 kJ/mol | ΔG_rxn^0 | -1076 kJ/mol - -761.4 kJ/mol = -314.2 kJ/mol (exergonic) | | |

Construct the equilibrium constant, K, expression for: KBr + F_2 ⟶ Br_2 + KF 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 KBr + F_2 ⟶ Br_2 + 2 KF 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 KBr | 2 | -2 F_2 | 1 | -1 Br_2 | 1 | 1 KF | 2 | 2 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression KBr | 2 | -2 | ([KBr])^(-2) F_2 | 1 | -1 | ([F2])^(-1) Br_2 | 1 | 1 | [Br2] KF | 2 | 2 | ([KF])^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 = ([KBr])^(-2) ([F2])^(-1) [Br2] ([KF])^2 = ([Br2] ([KF])^2)/(([KBr])^2 [F2])

Construct the rate of reaction expression for: KBr + F_2 ⟶ Br_2 + KF 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 KBr + F_2 ⟶ Br_2 + 2 KF 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 KBr | 2 | -2 F_2 | 1 | -1 Br_2 | 1 | 1 KF | 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 KBr | 2 | -2 | -1/2 (Δ[KBr])/(Δt) F_2 | 1 | -1 | -(Δ[F2])/(Δt) Br_2 | 1 | 1 | (Δ[Br2])/(Δt) KF | 2 | 2 | 1/2 (Δ[KF])/(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[KBr])/(Δt) = -(Δ[F2])/(Δt) = (Δ[Br2])/(Δt) = 1/2 (Δ[KF])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| potassium bromide | fluorine | bromine | potassium fluoride formula | KBr | F_2 | Br_2 | KF Hill formula | BrK | F_2 | Br_2 | FK name | potassium bromide | fluorine | bromine | potassium fluoride IUPAC name | potassium bromide | molecular fluorine | molecular bromine | potassium fluoride

| potassium bromide | fluorine | bromine | potassium fluoride molar mass | 119 g/mol | 37.996806326 g/mol | 159.81 g/mol | 58.0967 g/mol phase | solid (at STP) | gas (at STP) | liquid (at STP) | solid (at STP) melting point | 734 °C | -219.6 °C | -7.2 °C | 858 °C boiling point | 1435 °C | -188.12 °C | 58.8 °C | 1505 °C density | 2.75 g/cm^3 | 0.001696 g/cm^3 (at 0 °C) | 3.119 g/cm^3 | 1.89 g/cm^3 solubility in water | soluble | reacts | insoluble | surface tension | | | 0.0409 N/m | dynamic viscosity | | 2.344×10^-5 Pa s (at 25 °C) | 9.44×10^-4 Pa s (at 25 °C) |