H2O + Br2 + KI = HBr + KIO3

water + bromine + potassium iodide ⟶ hydrogen bromide + potassium iodate

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

+ + ⟶ +

water + bromine + potassium iodide ⟶ hydrogen bromide + potassium iodate

| water | bromine | potassium iodide | hydrogen bromide | potassium iodate molecular enthalpy | -285.8 kJ/mol | 0 kJ/mol | -327.9 kJ/mol | -36.3 kJ/mol | -501.4 kJ/mol total enthalpy | -857.5 kJ/mol | 0 kJ/mol | -327.9 kJ/mol | -217.8 kJ/mol | -501.4 kJ/mol | H_initial = -1185 kJ/mol | | | H_final = -719.2 kJ/mol | ΔH_rxn^0 | -719.2 kJ/mol - -1185 kJ/mol = 466.2 kJ/mol (endothermic) | | | |

| water | bromine | potassium iodide | hydrogen bromide | potassium iodate molecular free energy | -237.1 kJ/mol | 0 kJ/mol | -324.9 kJ/mol | -53.4 kJ/mol | -418.4 kJ/mol total free energy | -711.3 kJ/mol | 0 kJ/mol | -324.9 kJ/mol | -320.4 kJ/mol | -418.4 kJ/mol | G_initial = -1036 kJ/mol | | | G_final = -738.8 kJ/mol | ΔG_rxn^0 | -738.8 kJ/mol - -1036 kJ/mol = 297.4 kJ/mol (endergonic) | | | |

| water | bromine | potassium iodide | hydrogen bromide | potassium iodate Hill formula | H_2O | Br_2 | IK | BrH | IKO_3 name | water | bromine | potassium iodide | hydrogen bromide | potassium iodate IUPAC name | water | molecular bromine | potassium iodide | hydrogen bromide | potassium iodate