H2O + Br2 + PCl3 = HCl + H3PO4 + HBr

H_2O water + Br_2 bromine + PCl_3 phosphorus trichloride ⟶ HCl hydrogen chloride + H_3PO_4 phosphoric acid + HBr hydrogen bromide

Balance the chemical equation algebraically: H_2O + Br_2 + PCl_3 ⟶ HCl + H_3PO_4 + HBr Add stoichiometric coefficients, c_i, to the reactants and products: c_1 H_2O + c_2 Br_2 + c_3 PCl_3 ⟶ c_4 HCl + c_5 H_3PO_4 + c_6 HBr Set the number of atoms in the reactants equal to the number of atoms in the products for H, O, Br, Cl and P: H: | 2 c_1 = c_4 + 3 c_5 + c_6 O: | c_1 = 4 c_5 Br: | 2 c_2 = c_6 Cl: | 3 c_3 = c_4 P: | c_3 = 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_2 = 1 and solve the system of equations for the remaining coefficients: c_1 = 4 c_2 = 1 c_3 = 1 c_4 = 3 c_5 = 1 c_6 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 4 H_2O + Br_2 + PCl_3 ⟶ 3 HCl + H_3PO_4 + 2 HBr

+ + ⟶ + +

water + bromine + phosphorus trichloride ⟶ hydrogen chloride + phosphoric acid + hydrogen bromide

| water | bromine | phosphorus trichloride | hydrogen chloride | phosphoric acid | hydrogen bromide molecular free energy | -237.1 kJ/mol | 0 kJ/mol | -272.3 kJ/mol | -95.3 kJ/mol | -1124 kJ/mol | -53.4 kJ/mol total free energy | -948.4 kJ/mol | 0 kJ/mol | -272.3 kJ/mol | -285.9 kJ/mol | -1124 kJ/mol | -106.8 kJ/mol | G_initial = -1221 kJ/mol | | | G_final = -1516 kJ/mol | | ΔG_rxn^0 | -1516 kJ/mol - -1221 kJ/mol = -295.6 kJ/mol (exergonic) | | | | |

Construct the equilibrium constant, K, expression for: H_2O + Br_2 + PCl_3 ⟶ HCl + H_3PO_4 + HBr 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: 4 H_2O + Br_2 + PCl_3 ⟶ 3 HCl + H_3PO_4 + 2 HBr 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 H_2O | 4 | -4 Br_2 | 1 | -1 PCl_3 | 1 | -1 HCl | 3 | 3 H_3PO_4 | 1 | 1 HBr | 2 | 2 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression H_2O | 4 | -4 | ([H2O])^(-4) Br_2 | 1 | -1 | ([Br2])^(-1) PCl_3 | 1 | -1 | ([PCl3])^(-1) HCl | 3 | 3 | ([HCl])^3 H_3PO_4 | 1 | 1 | [H3PO4] HBr | 2 | 2 | ([HBr])^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 = ([H2O])^(-4) ([Br2])^(-1) ([PCl3])^(-1) ([HCl])^3 [H3PO4] ([HBr])^2 = (([HCl])^3 [H3PO4] ([HBr])^2)/(([H2O])^4 [Br2] [PCl3])

Construct the rate of reaction expression for: H_2O + Br_2 + PCl_3 ⟶ HCl + H_3PO_4 + HBr 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: 4 H_2O + Br_2 + PCl_3 ⟶ 3 HCl + H_3PO_4 + 2 HBr 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 H_2O | 4 | -4 Br_2 | 1 | -1 PCl_3 | 1 | -1 HCl | 3 | 3 H_3PO_4 | 1 | 1 HBr | 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 H_2O | 4 | -4 | -1/4 (Δ[H2O])/(Δt) Br_2 | 1 | -1 | -(Δ[Br2])/(Δt) PCl_3 | 1 | -1 | -(Δ[PCl3])/(Δt) HCl | 3 | 3 | 1/3 (Δ[HCl])/(Δt) H_3PO_4 | 1 | 1 | (Δ[H3PO4])/(Δt) HBr | 2 | 2 | 1/2 (Δ[HBr])/(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 = -1/4 (Δ[H2O])/(Δt) = -(Δ[Br2])/(Δt) = -(Δ[PCl3])/(Δt) = 1/3 (Δ[HCl])/(Δt) = (Δ[H3PO4])/(Δt) = 1/2 (Δ[HBr])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| water | bromine | phosphorus trichloride | hydrogen chloride | phosphoric acid | hydrogen bromide formula | H_2O | Br_2 | PCl_3 | HCl | H_3PO_4 | HBr Hill formula | H_2O | Br_2 | Cl_3P | ClH | H_3O_4P | BrH name | water | bromine | phosphorus trichloride | hydrogen chloride | phosphoric acid | hydrogen bromide IUPAC name | water | molecular bromine | trichlorophosphane | hydrogen chloride | phosphoric acid | hydrogen bromide