Br2 + P = PBr3

Br_2 bromine + P red phosphorus ⟶ PBr_3 phosphorus tribromide

Balance the chemical equation algebraically: Br_2 + P ⟶ PBr_3 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 Br_2 + c_2 P ⟶ c_3 PBr_3 Set the number of atoms in the reactants equal to the number of atoms in the products for Br and P: Br: | 2 c_1 = 3 c_3 P: | c_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_2 = 1 and solve the system of equations for the remaining coefficients: c_1 = 3/2 c_2 = 1 c_3 = 1 Multiply by the least common denominator, 2, to eliminate fractional coefficients: c_1 = 3 c_2 = 2 c_3 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 3 Br_2 + 2 P ⟶ 2 PBr_3

+ ⟶

bromine + red phosphorus ⟶ phosphorus tribromide

Construct the equilibrium constant, K, expression for: Br_2 + P ⟶ PBr_3 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 Br_2 + 2 P ⟶ 2 PBr_3 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 Br_2 | 3 | -3 P | 2 | -2 PBr_3 | 2 | 2 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression Br_2 | 3 | -3 | ([Br2])^(-3) P | 2 | -2 | ([P])^(-2) PBr_3 | 2 | 2 | ([PBr3])^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 = ([Br2])^(-3) ([P])^(-2) ([PBr3])^2 = ([PBr3])^2/(([Br2])^3 ([P])^2)

Construct the rate of reaction expression for: Br_2 + P ⟶ PBr_3 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 Br_2 + 2 P ⟶ 2 PBr_3 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 Br_2 | 3 | -3 P | 2 | -2 PBr_3 | 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 Br_2 | 3 | -3 | -1/3 (Δ[Br2])/(Δt) P | 2 | -2 | -1/2 (Δ[P])/(Δt) PBr_3 | 2 | 2 | 1/2 (Δ[PBr3])/(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[Br2])/(Δt) = -1/2 (Δ[P])/(Δt) = 1/2 (Δ[PBr3])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| bromine | red phosphorus | phosphorus tribromide formula | Br_2 | P | PBr_3 Hill formula | Br_2 | P | Br_3P name | bromine | red phosphorus | phosphorus tribromide IUPAC name | molecular bromine | phosphorus | tribromophosphane

| bromine | red phosphorus | phosphorus tribromide molar mass | 159.81 g/mol | 30.973761998 g/mol | 270.69 g/mol phase | liquid (at STP) | solid (at STP) | liquid (at STP) melting point | -7.2 °C | 579.2 °C | -41.5 °C boiling point | 58.8 °C | | 175 °C density | 3.119 g/cm^3 | 2.16 g/cm^3 | 2.88 g/cm^3 solubility in water | insoluble | insoluble | reacts surface tension | 0.0409 N/m | | 0.0458 N/m dynamic viscosity | 9.44×10^-4 Pa s (at 25 °C) | 7.6×10^-4 Pa s (at 20.2 °C) | 0.001302 Pa s (at 60 °C)