HCl + KI + BaCrO4 = H2O + I2 + KCl + CrCl3 + BaCl2

HCl hydrogen chloride + KI potassium iodide + BaCrO_4 barium chromate ⟶ H_2O water + I_2 iodine + KCl potassium chloride + CrCl_3 chromic chloride + BaCl_2 barium chloride

Balance the chemical equation algebraically: HCl + KI + BaCrO_4 ⟶ H_2O + I_2 + KCl + CrCl_3 + BaCl_2 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 HCl + c_2 KI + c_3 BaCrO_4 ⟶ c_4 H_2O + c_5 I_2 + c_6 KCl + c_7 CrCl_3 + c_8 BaCl_2 Set the number of atoms in the reactants equal to the number of atoms in the products for Cl, H, I, K, Ba, Cr and O: Cl: | c_1 = c_6 + 3 c_7 + 2 c_8 H: | c_1 = 2 c_4 I: | c_2 = 2 c_5 K: | c_2 = c_6 Ba: | c_3 = c_8 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 = 3/2 c_6 = 3 c_7 = 1 c_8 = 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 = 3 c_6 = 6 c_7 = 2 c_8 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 16 HCl + 6 KI + 2 BaCrO_4 ⟶ 8 H_2O + 3 I_2 + 6 KCl + 2 CrCl_3 + 2 BaCl_2

+ + ⟶ + + + +

hydrogen chloride + potassium iodide + barium chromate ⟶ water + iodine + potassium chloride + chromic chloride + barium chloride

Construct the equilibrium constant, K, expression for: HCl + KI + BaCrO_4 ⟶ H_2O + I_2 + KCl + CrCl_3 + BaCl_2 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 KI + 2 BaCrO_4 ⟶ 8 H_2O + 3 I_2 + 6 KCl + 2 CrCl_3 + 2 BaCl_2 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 KI | 6 | -6 BaCrO_4 | 2 | -2 H_2O | 8 | 8 I_2 | 3 | 3 KCl | 6 | 6 CrCl_3 | 2 | 2 BaCl_2 | 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) KI | 6 | -6 | ([KI])^(-6) BaCrO_4 | 2 | -2 | ([BaCrO4])^(-2) H_2O | 8 | 8 | ([H2O])^8 I_2 | 3 | 3 | ([I2])^3 KCl | 6 | 6 | ([KCl])^6 CrCl_3 | 2 | 2 | ([CrCl3])^2 BaCl_2 | 2 | 2 | ([BaCl2])^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) ([KI])^(-6) ([BaCrO4])^(-2) ([H2O])^8 ([I2])^3 ([KCl])^6 ([CrCl3])^2 ([BaCl2])^2 = (([H2O])^8 ([I2])^3 ([KCl])^6 ([CrCl3])^2 ([BaCl2])^2)/(([HCl])^16 ([KI])^6 ([BaCrO4])^2)

Construct the rate of reaction expression for: HCl + KI + BaCrO_4 ⟶ H_2O + I_2 + KCl + CrCl_3 + BaCl_2 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 KI + 2 BaCrO_4 ⟶ 8 H_2O + 3 I_2 + 6 KCl + 2 CrCl_3 + 2 BaCl_2 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 KI | 6 | -6 BaCrO_4 | 2 | -2 H_2O | 8 | 8 I_2 | 3 | 3 KCl | 6 | 6 CrCl_3 | 2 | 2 BaCl_2 | 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) KI | 6 | -6 | -1/6 (Δ[KI])/(Δt) BaCrO_4 | 2 | -2 | -1/2 (Δ[BaCrO4])/(Δt) H_2O | 8 | 8 | 1/8 (Δ[H2O])/(Δt) I_2 | 3 | 3 | 1/3 (Δ[I2])/(Δt) KCl | 6 | 6 | 1/6 (Δ[KCl])/(Δt) CrCl_3 | 2 | 2 | 1/2 (Δ[CrCl3])/(Δt) BaCl_2 | 2 | 2 | 1/2 (Δ[BaCl2])/(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[KI])/(Δt) = -1/2 (Δ[BaCrO4])/(Δt) = 1/8 (Δ[H2O])/(Δt) = 1/3 (Δ[I2])/(Δt) = 1/6 (Δ[KCl])/(Δt) = 1/2 (Δ[CrCl3])/(Δt) = 1/2 (Δ[BaCl2])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| hydrogen chloride | potassium iodide | barium chromate | water | iodine | potassium chloride | chromic chloride | barium chloride formula | HCl | KI | BaCrO_4 | H_2O | I_2 | KCl | CrCl_3 | BaCl_2 Hill formula | ClH | IK | BaCrO_4 | H_2O | I_2 | ClK | Cl_3Cr | BaCl_2 name | hydrogen chloride | potassium iodide | barium chromate | water | iodine | potassium chloride | chromic chloride | barium chloride IUPAC name | hydrogen chloride | potassium iodide | barium(+2) cation; dioxido-dioxochromium | water | molecular iodine | potassium chloride | trichlorochromium | barium(+2) cation dichloride