HCl + KBr + K2CrO4 = H2O + KCl + Br2 + CrCl3

HCl hydrogen chloride + KBr potassium bromide + K_2CrO_4 potassium chromate ⟶ H_2O water + KCl potassium chloride + Br_2 bromine + CrCl_3 chromic chloride

Balance the chemical equation algebraically: HCl + KBr + K_2CrO_4 ⟶ H_2O + KCl + Br_2 + CrCl_3 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 HCl + c_2 KBr + c_3 K_2CrO_4 ⟶ c_4 H_2O + c_5 KCl + c_6 Br_2 + c_7 CrCl_3 Set the number of atoms in the reactants equal to the number of atoms in the products for Cl, H, Br, K, Cr and O: Cl: | c_1 = c_5 + 3 c_7 H: | c_1 = 2 c_4 Br: | c_2 = 2 c_6 K: | c_2 + 2 c_3 = c_5 Cr: | c_3 = c_7 O: | 4 c_3 = 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_3 = 1 and solve the system of equations for the remaining coefficients: c_1 = 8 c_2 = 3 c_3 = 1 c_4 = 4 c_5 = 5 c_6 = 3/2 c_7 = 1 Multiply by the least common denominator, 2, to eliminate fractional coefficients: c_1 = 16 c_2 = 6 c_3 = 2 c_4 = 8 c_5 = 10 c_6 = 3 c_7 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 16 HCl + 6 KBr + 2 K_2CrO_4 ⟶ 8 H_2O + 10 KCl + 3 Br_2 + 2 CrCl_3

+ + ⟶ + + +

hydrogen chloride + potassium bromide + potassium chromate ⟶ water + potassium chloride + bromine + chromic chloride

Construct the equilibrium constant, K, expression for: HCl + KBr + K_2CrO_4 ⟶ H_2O + KCl + Br_2 + CrCl_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: 16 HCl + 6 KBr + 2 K_2CrO_4 ⟶ 8 H_2O + 10 KCl + 3 Br_2 + 2 CrCl_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 HCl | 16 | -16 KBr | 6 | -6 K_2CrO_4 | 2 | -2 H_2O | 8 | 8 KCl | 10 | 10 Br_2 | 3 | 3 CrCl_3 | 2 | 2 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression HCl | 16 | -16 | ([HCl])^(-16) KBr | 6 | -6 | ([KBr])^(-6) K_2CrO_4 | 2 | -2 | ([K2CrO4])^(-2) H_2O | 8 | 8 | ([H2O])^8 KCl | 10 | 10 | ([KCl])^10 Br_2 | 3 | 3 | ([Br2])^3 CrCl_3 | 2 | 2 | ([CrCl3])^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 = ([HCl])^(-16) ([KBr])^(-6) ([K2CrO4])^(-2) ([H2O])^8 ([KCl])^10 ([Br2])^3 ([CrCl3])^2 = (([H2O])^8 ([KCl])^10 ([Br2])^3 ([CrCl3])^2)/(([HCl])^16 ([KBr])^6 ([K2CrO4])^2)

Construct the rate of reaction expression for: HCl + KBr + K_2CrO_4 ⟶ H_2O + KCl + Br_2 + CrCl_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: 16 HCl + 6 KBr + 2 K_2CrO_4 ⟶ 8 H_2O + 10 KCl + 3 Br_2 + 2 CrCl_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 HCl | 16 | -16 KBr | 6 | -6 K_2CrO_4 | 2 | -2 H_2O | 8 | 8 KCl | 10 | 10 Br_2 | 3 | 3 CrCl_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 HCl | 16 | -16 | -1/16 (Δ[HCl])/(Δt) KBr | 6 | -6 | -1/6 (Δ[KBr])/(Δt) K_2CrO_4 | 2 | -2 | -1/2 (Δ[K2CrO4])/(Δt) H_2O | 8 | 8 | 1/8 (Δ[H2O])/(Δt) KCl | 10 | 10 | 1/10 (Δ[KCl])/(Δt) Br_2 | 3 | 3 | 1/3 (Δ[Br2])/(Δt) CrCl_3 | 2 | 2 | 1/2 (Δ[CrCl3])/(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 = -1/16 (Δ[HCl])/(Δt) = -1/6 (Δ[KBr])/(Δt) = -1/2 (Δ[K2CrO4])/(Δt) = 1/8 (Δ[H2O])/(Δt) = 1/10 (Δ[KCl])/(Δt) = 1/3 (Δ[Br2])/(Δt) = 1/2 (Δ[CrCl3])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| hydrogen chloride | potassium bromide | potassium chromate | water | potassium chloride | bromine | chromic chloride formula | HCl | KBr | K_2CrO_4 | H_2O | KCl | Br_2 | CrCl_3 Hill formula | ClH | BrK | CrK_2O_4 | H_2O | ClK | Br_2 | Cl_3Cr name | hydrogen chloride | potassium bromide | potassium chromate | water | potassium chloride | bromine | chromic chloride IUPAC name | hydrogen chloride | potassium bromide | dipotassium dioxido-dioxochromium | water | potassium chloride | molecular bromine | trichlorochromium