Input interpretation
sodium hydroxide + fluorine ⟶ water + sodium fluoride + oxygen difluoride
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, Na, O and F: H: | c_1 = 2 c_3 Na: | c_1 = c_4 O: | c_1 = c_3 + c_5 F: | 2 c_2 = c_4 + 2 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 = 2 c_2 = 2 c_3 = 1 c_4 = 2 c_5 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 2 + 2 ⟶ + 2 +
Structures
+ ⟶ + +
Names
sodium hydroxide + fluorine ⟶ water + sodium fluoride + oxygen difluoride
Reaction thermodynamics
Enthalpy
| sodium hydroxide | fluorine | water | sodium fluoride | oxygen difluoride molecular enthalpy | -425.8 kJ/mol | 0 kJ/mol | -285.8 kJ/mol | -576.6 kJ/mol | 109 kJ/mol total enthalpy | -851.6 kJ/mol | 0 kJ/mol | -285.8 kJ/mol | -1153 kJ/mol | 109 kJ/mol | H_initial = -851.6 kJ/mol | | H_final = -1330 kJ/mol | | ΔH_rxn^0 | -1330 kJ/mol - -851.6 kJ/mol = -478.4 kJ/mol (exothermic) | | | |
Gibbs free energy
| sodium hydroxide | fluorine | water | sodium fluoride | oxygen difluoride molecular free energy | -379.7 kJ/mol | 0 kJ/mol | -237.1 kJ/mol | -546.3 kJ/mol | 105.3 kJ/mol total free energy | -759.4 kJ/mol | 0 kJ/mol | -237.1 kJ/mol | -1093 kJ/mol | 105.3 kJ/mol | G_initial = -759.4 kJ/mol | | G_final = -1224 kJ/mol | | ΔG_rxn^0 | -1224 kJ/mol - -759.4 kJ/mol = -465 kJ/mol (exergonic) | | | |
Entropy
| sodium hydroxide | fluorine | water | sodium fluoride | oxygen difluoride molecular entropy | 64 J/(mol K) | 202.8 J/(mol K) | 69.91 J/(mol K) | 51.1 J/(mol K) | 216.4 J/(mol K) total entropy | 128 J/(mol K) | 405.6 J/(mol K) | 69.91 J/(mol K) | 102.2 J/(mol K) | 216.4 J/(mol K) | S_initial = 533.6 J/(mol K) | | S_final = 388.5 J/(mol K) | | ΔS_rxn^0 | 388.5 J/(mol K) - 533.6 J/(mol K) = -145.1 J/(mol K) (exoentropic) | | | |
Chemical names and formulas
| sodium hydroxide | fluorine | water | sodium fluoride | oxygen difluoride Hill formula | HNaO | F_2 | H_2O | FNa | F_2O name | sodium hydroxide | fluorine | water | sodium fluoride | oxygen difluoride IUPAC name | sodium hydroxide | molecular fluorine | water | sodium fluoride | fluoro hypofluorite