Cl2 + KOH + Br2 = H2O + KCl + KBrO3

Cl_2 (chlorine) + KOH (potassium hydroxide) + Br_2 (bromine) ⟶ H_2O (water) + KCl (potassium chloride) + KBrO_3 (potassium bromate)

Balance the chemical equation algebraically: Cl_2 + KOH + Br_2 ⟶ H_2O + KCl + KBrO_3 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 Cl_2 + c_2 KOH + c_3 Br_2 ⟶ c_4 H_2O + c_5 KCl + c_6 KBrO_3 Set the number of atoms in the reactants equal to the number of atoms in the products for Cl, H, K, O and Br: Cl: | 2 c_1 = c_5 H: | c_2 = 2 c_4 K: | c_2 = c_5 + c_6 O: | c_2 = c_4 + 3 c_6 Br: | 2 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_3 = 1 and solve the system of equations for the remaining coefficients: c_1 = 5 c_2 = 12 c_3 = 1 c_4 = 6 c_5 = 10 c_6 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 5 Cl_2 + 12 KOH + Br_2 ⟶ 6 H_2O + 10 KCl + 2 KBrO_3

+ + ⟶ + +

chlorine + potassium hydroxide + bromine ⟶ water + potassium chloride + potassium bromate

| chlorine | potassium hydroxide | bromine | water | potassium chloride | potassium bromate molecular enthalpy | 0 kJ/mol | -424.6 kJ/mol | 0 kJ/mol | -285.8 kJ/mol | -436.5 kJ/mol | -360.2 kJ/mol total enthalpy | 0 kJ/mol | -5095 kJ/mol | 0 kJ/mol | -1715 kJ/mol | -4365 kJ/mol | -720.4 kJ/mol | H_initial = -5095 kJ/mol | | | H_final = -6800 kJ/mol | | ΔH_rxn^0 | -6800 kJ/mol - -5095 kJ/mol = -1705 kJ/mol (exothermic) | | | | |

| chlorine | potassium hydroxide | bromine | water | potassium chloride | potassium bromate molecular free energy | 0 kJ/mol | -379.4 kJ/mol | 0 kJ/mol | -237.1 kJ/mol | -408.5 kJ/mol | -2712 kJ/mol total free energy | 0 kJ/mol | -4553 kJ/mol | 0 kJ/mol | -1423 kJ/mol | -4085 kJ/mol | -5424 kJ/mol | G_initial = -4553 kJ/mol | | | G_final = -10932 kJ/mol | | ΔG_rxn^0 | -10932 kJ/mol - -4553 kJ/mol = -6379 kJ/mol (exergonic) | | | | |

Construct the equilibrium constant, K, expression for: Cl_2 + KOH + Br_2 ⟶ H_2O + KCl + KBrO_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: 5 Cl_2 + 12 KOH + Br_2 ⟶ 6 H_2O + 10 KCl + 2 KBrO_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 Cl_2 | 5 | -5 KOH | 12 | -12 Br_2 | 1 | -1 H_2O | 6 | 6 KCl | 10 | 10 KBrO_3 | 2 | 2 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression Cl_2 | 5 | -5 | ([Cl2])^(-5) KOH | 12 | -12 | ([KOH])^(-12) Br_2 | 1 | -1 | ([Br2])^(-1) H_2O | 6 | 6 | ([H2O])^6 KCl | 10 | 10 | ([KCl])^10 KBrO_3 | 2 | 2 | ([KBrO3])^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 = ([Cl2])^(-5) ([KOH])^(-12) ([Br2])^(-1) ([H2O])^6 ([KCl])^10 ([KBrO3])^2 = (([H2O])^6 ([KCl])^10 ([KBrO3])^2)/(([Cl2])^5 ([KOH])^12 [Br2])

Construct the rate of reaction expression for: Cl_2 + KOH + Br_2 ⟶ H_2O + KCl + KBrO_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: 5 Cl_2 + 12 KOH + Br_2 ⟶ 6 H_2O + 10 KCl + 2 KBrO_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 Cl_2 | 5 | -5 KOH | 12 | -12 Br_2 | 1 | -1 H_2O | 6 | 6 KCl | 10 | 10 KBrO_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 Cl_2 | 5 | -5 | -1/5 (Δ[Cl2])/(Δt) KOH | 12 | -12 | -1/12 (Δ[KOH])/(Δt) Br_2 | 1 | -1 | -(Δ[Br2])/(Δt) H_2O | 6 | 6 | 1/6 (Δ[H2O])/(Δt) KCl | 10 | 10 | 1/10 (Δ[KCl])/(Δt) KBrO_3 | 2 | 2 | 1/2 (Δ[KBrO3])/(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 = -1/5 (Δ[Cl2])/(Δt) = -1/12 (Δ[KOH])/(Δt) = -(Δ[Br2])/(Δt) = 1/6 (Δ[H2O])/(Δt) = 1/10 (Δ[KCl])/(Δt) = 1/2 (Δ[KBrO3])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| chlorine | potassium hydroxide | bromine | water | potassium chloride | potassium bromate formula | Cl_2 | KOH | Br_2 | H_2O | KCl | KBrO_3 Hill formula | Cl_2 | HKO | Br_2 | H_2O | ClK | BrKO_3 name | chlorine | potassium hydroxide | bromine | water | potassium chloride | potassium bromate IUPAC name | molecular chlorine | potassium hydroxide | molecular bromine | water | potassium chloride | potassium bromate

| chlorine | potassium hydroxide | bromine | water | potassium chloride | potassium bromate molar mass | 70.9 g/mol | 56.105 g/mol | 159.81 g/mol | 18.015 g/mol | 74.55 g/mol | 167 g/mol phase | gas (at STP) | solid (at STP) | liquid (at STP) | liquid (at STP) | solid (at STP) | solid (at STP) melting point | -101 °C | 406 °C | -7.2 °C | 0 °C | 770 °C | 350 °C boiling point | -34 °C | 1327 °C | 58.8 °C | 99.9839 °C | 1420 °C | density | 0.003214 g/cm^3 (at 0 °C) | 2.044 g/cm^3 | 3.119 g/cm^3 | 1 g/cm^3 | 1.98 g/cm^3 | 3.218 g/cm^3 solubility in water | | soluble | insoluble | | soluble | surface tension | | | 0.0409 N/m | 0.0728 N/m | | dynamic viscosity | | 0.001 Pa s (at 550 °C) | 9.44×10^-4 Pa s (at 25 °C) | 8.9×10^-4 Pa s (at 25 °C) | | odor | | | | odorless | odorless |