PbO2 + BrF3 = O2 + Br2 + PbF4

PbO_2 lead dioxide + BrF_3 bromine trifluoride ⟶ O_2 oxygen + Br_2 bromine + PbF_4 lead(IV) fluoride

Balance the chemical equation algebraically: PbO_2 + BrF_3 ⟶ O_2 + Br_2 + PbF_4 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 PbO_2 + c_2 BrF_3 ⟶ c_3 O_2 + c_4 Br_2 + c_5 PbF_4 Set the number of atoms in the reactants equal to the number of atoms in the products for O, Pb, Br and F: O: | 2 c_1 = 2 c_3 Pb: | c_1 = c_5 Br: | c_2 = 2 c_4 F: | 3 c_2 = 4 c_5 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_4 = 1 and solve the system of equations for the remaining coefficients: c_1 = 3/2 c_2 = 2 c_3 = 3/2 c_4 = 1 c_5 = 3/2 Multiply by the least common denominator, 2, to eliminate fractional coefficients: c_1 = 3 c_2 = 4 c_3 = 3 c_4 = 2 c_5 = 3 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 3 PbO_2 + 4 BrF_3 ⟶ 3 O_2 + 2 Br_2 + 3 PbF_4

+ ⟶ + +

lead dioxide + bromine trifluoride ⟶ oxygen + bromine + lead(IV) fluoride

Construct the equilibrium constant, K, expression for: PbO_2 + BrF_3 ⟶ O_2 + Br_2 + PbF_4 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: 3 PbO_2 + 4 BrF_3 ⟶ 3 O_2 + 2 Br_2 + 3 PbF_4 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 PbO_2 | 3 | -3 BrF_3 | 4 | -4 O_2 | 3 | 3 Br_2 | 2 | 2 PbF_4 | 3 | 3 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression PbO_2 | 3 | -3 | ([PbO2])^(-3) BrF_3 | 4 | -4 | ([BrF3])^(-4) O_2 | 3 | 3 | ([O2])^3 Br_2 | 2 | 2 | ([Br2])^2 PbF_4 | 3 | 3 | ([PbF4])^3 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 = ([PbO2])^(-3) ([BrF3])^(-4) ([O2])^3 ([Br2])^2 ([PbF4])^3 = (([O2])^3 ([Br2])^2 ([PbF4])^3)/(([PbO2])^3 ([BrF3])^4)

Construct the rate of reaction expression for: PbO_2 + BrF_3 ⟶ O_2 + Br_2 + PbF_4 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: 3 PbO_2 + 4 BrF_3 ⟶ 3 O_2 + 2 Br_2 + 3 PbF_4 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 PbO_2 | 3 | -3 BrF_3 | 4 | -4 O_2 | 3 | 3 Br_2 | 2 | 2 PbF_4 | 3 | 3 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 PbO_2 | 3 | -3 | -1/3 (Δ[PbO2])/(Δt) BrF_3 | 4 | -4 | -1/4 (Δ[BrF3])/(Δt) O_2 | 3 | 3 | 1/3 (Δ[O2])/(Δt) Br_2 | 2 | 2 | 1/2 (Δ[Br2])/(Δt) PbF_4 | 3 | 3 | 1/3 (Δ[PbF4])/(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 = -1/3 (Δ[PbO2])/(Δt) = -1/4 (Δ[BrF3])/(Δt) = 1/3 (Δ[O2])/(Δt) = 1/2 (Δ[Br2])/(Δt) = 1/3 (Δ[PbF4])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| lead dioxide | bromine trifluoride | oxygen | bromine | lead(IV) fluoride formula | PbO_2 | BrF_3 | O_2 | Br_2 | PbF_4 Hill formula | O_2Pb | BrF_3 | O_2 | Br_2 | F_4Pb name | lead dioxide | bromine trifluoride | oxygen | bromine | lead(IV) fluoride IUPAC name | | | molecular oxygen | molecular bromine | tetrafluoroplumbane

| lead dioxide | bromine trifluoride | oxygen | bromine | lead(IV) fluoride molar mass | 239.2 g/mol | 136.9 g/mol | 31.998 g/mol | 159.81 g/mol | 283.2 g/mol phase | solid (at STP) | liquid (at STP) | gas (at STP) | liquid (at STP) | solid (at STP) melting point | 290 °C | 8.77 °C | -218 °C | -7.2 °C | 600 °C boiling point | | 125.8 °C | -183 °C | 58.8 °C | density | 9.58 g/cm^3 | 2.803 g/cm^3 | 0.001429 g/cm^3 (at 0 °C) | 3.119 g/cm^3 | 6.7 g/cm^3 solubility in water | insoluble | decomposes | | insoluble | surface tension | | 0.0363 N/m | 0.01347 N/m | 0.0409 N/m | dynamic viscosity | | | 2.055×10^-5 Pa s (at 25 °C) | 9.44×10^-4 Pa s (at 25 °C) | odor | | | odorless | |