Input interpretation
water + iodine + red phosphorus ⟶ hydrogen iodide + phosphorous acid
Balanced equation
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, I and P: H: | 2 c_1 = c_4 + 3 c_5 O: | c_1 = 3 c_5 I: | 2 c_2 = c_4 P: | 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 = 3/2 c_3 = 1 c_4 = 3 c_5 = 1 Multiply by the least common denominator, 2, to eliminate fractional coefficients: c_1 = 6 c_2 = 3 c_3 = 2 c_4 = 6 c_5 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 6 + 3 + 2 ⟶ 6 + 2
Structures
+ + ⟶ +
Names
water + iodine + red phosphorus ⟶ hydrogen iodide + phosphorous acid
Reaction thermodynamics
Enthalpy
| water | iodine | red phosphorus | hydrogen iodide | phosphorous acid molecular enthalpy | -285.8 kJ/mol | 0 kJ/mol | -17.6 kJ/mol | 26.5 kJ/mol | -964.4 kJ/mol total enthalpy | -1715 kJ/mol | 0 kJ/mol | -35.2 kJ/mol | 159 kJ/mol | -1929 kJ/mol | H_initial = -1750 kJ/mol | | | H_final = -1770 kJ/mol | ΔH_rxn^0 | -1770 kJ/mol - -1750 kJ/mol = -19.62 kJ/mol (exothermic) | | | |
Chemical names and formulas
| water | iodine | red phosphorus | hydrogen iodide | phosphorous acid Hill formula | H_2O | I_2 | P | HI | H_3O_3P name | water | iodine | red phosphorus | hydrogen iodide | phosphorous acid IUPAC name | water | molecular iodine | phosphorus | hydrogen iodide | phosphorous acid