HCl + K2Cr2O7 + SnCl2 = H2O + CrCl3 + SnCl4 + K

HCl hydrogen chloride + K_2Cr_2O_7 potassium dichromate + SnCl_2 stannous chloride ⟶ H_2O water + CrCl_3 chromic chloride + SnCl_4 stannic chloride + K potassium

Balance the chemical equation algebraically: HCl + K_2Cr_2O_7 + SnCl_2 ⟶ H_2O + CrCl_3 + SnCl_4 + K Add stoichiometric coefficients, c_i, to the reactants and products: c_1 HCl + c_2 K_2Cr_2O_7 + c_3 SnCl_2 ⟶ c_4 H_2O + c_5 CrCl_3 + c_6 SnCl_4 + c_7 K Set the number of atoms in the reactants equal to the number of atoms in the products for Cl, H, Cr, K, O and Sn: Cl: | c_1 + 2 c_3 = 3 c_5 + 4 c_6 H: | c_1 = 2 c_4 Cr: | 2 c_2 = c_5 K: | 2 c_2 = c_7 O: | 7 c_2 = c_4 Sn: | c_3 = c_6 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 = 14 c_2 = 1 c_3 = 4 c_4 = 7 c_5 = 2 c_6 = 4 c_7 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 14 HCl + K_2Cr_2O_7 + 4 SnCl_2 ⟶ 7 H_2O + 2 CrCl_3 + 4 SnCl_4 + 2 K

+ + ⟶ + + +

hydrogen chloride + potassium dichromate + stannous chloride ⟶ water + chromic chloride + stannic chloride + potassium

Construct the equilibrium constant, K, expression for: HCl + K_2Cr_2O_7 + SnCl_2 ⟶ H_2O + CrCl_3 + SnCl_4 + K 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: 14 HCl + K_2Cr_2O_7 + 4 SnCl_2 ⟶ 7 H_2O + 2 CrCl_3 + 4 SnCl_4 + 2 K 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 | 14 | -14 K_2Cr_2O_7 | 1 | -1 SnCl_2 | 4 | -4 H_2O | 7 | 7 CrCl_3 | 2 | 2 SnCl_4 | 4 | 4 K | 2 | 2 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression HCl | 14 | -14 | ([HCl])^(-14) K_2Cr_2O_7 | 1 | -1 | ([K2Cr2O7])^(-1) SnCl_2 | 4 | -4 | ([SnCl2])^(-4) H_2O | 7 | 7 | ([H2O])^7 CrCl_3 | 2 | 2 | ([CrCl3])^2 SnCl_4 | 4 | 4 | ([SnCl4])^4 K | 2 | 2 | ([K])^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])^(-14) ([K2Cr2O7])^(-1) ([SnCl2])^(-4) ([H2O])^7 ([CrCl3])^2 ([SnCl4])^4 ([K])^2 = (([H2O])^7 ([CrCl3])^2 ([SnCl4])^4 ([K])^2)/(([HCl])^14 [K2Cr2O7] ([SnCl2])^4)

Construct the rate of reaction expression for: HCl + K_2Cr_2O_7 + SnCl_2 ⟶ H_2O + CrCl_3 + SnCl_4 + K 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: 14 HCl + K_2Cr_2O_7 + 4 SnCl_2 ⟶ 7 H_2O + 2 CrCl_3 + 4 SnCl_4 + 2 K 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 | 14 | -14 K_2Cr_2O_7 | 1 | -1 SnCl_2 | 4 | -4 H_2O | 7 | 7 CrCl_3 | 2 | 2 SnCl_4 | 4 | 4 K | 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 | 14 | -14 | -1/14 (Δ[HCl])/(Δt) K_2Cr_2O_7 | 1 | -1 | -(Δ[K2Cr2O7])/(Δt) SnCl_2 | 4 | -4 | -1/4 (Δ[SnCl2])/(Δt) H_2O | 7 | 7 | 1/7 (Δ[H2O])/(Δt) CrCl_3 | 2 | 2 | 1/2 (Δ[CrCl3])/(Δt) SnCl_4 | 4 | 4 | 1/4 (Δ[SnCl4])/(Δt) K | 2 | 2 | 1/2 (Δ[K])/(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 = -1/14 (Δ[HCl])/(Δt) = -(Δ[K2Cr2O7])/(Δt) = -1/4 (Δ[SnCl2])/(Δt) = 1/7 (Δ[H2O])/(Δt) = 1/2 (Δ[CrCl3])/(Δt) = 1/4 (Δ[SnCl4])/(Δt) = 1/2 (Δ[K])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| hydrogen chloride | potassium dichromate | stannous chloride | water | chromic chloride | stannic chloride | potassium formula | HCl | K_2Cr_2O_7 | SnCl_2 | H_2O | CrCl_3 | SnCl_4 | K Hill formula | ClH | Cr_2K_2O_7 | Cl_2Sn | H_2O | Cl_3Cr | Cl_4Sn | K name | hydrogen chloride | potassium dichromate | stannous chloride | water | chromic chloride | stannic chloride | potassium IUPAC name | hydrogen chloride | dipotassium oxido-(oxido-dioxochromio)oxy-dioxochromium | dichlorotin | water | trichlorochromium | tetrachlorostannane | potassium