Input interpretation
nitric acid + sodium ⟶ water + nitrogen dioxide + sodium nitrate
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, N, O and Na: H: | c_1 = 2 c_3 N: | c_1 = c_4 + c_5 O: | 3 c_1 = c_3 + 2 c_4 + 3 c_5 Na: | c_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_2 = 1 and solve the system of equations for the remaining coefficients: c_1 = 2 c_2 = 1 c_3 = 1 c_4 = 1 c_5 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 2 + ⟶ + +
Structures
+ ⟶ + +
Names
nitric acid + sodium ⟶ water + nitrogen dioxide + sodium nitrate
Reaction thermodynamics
Entropy
| nitric acid | sodium | water | nitrogen dioxide | sodium nitrate molecular entropy | 156 J/(mol K) | 51 J/(mol K) | 69.91 J/(mol K) | 240 J/(mol K) | 116 J/(mol K) total entropy | 312 J/(mol K) | 51 J/(mol K) | 69.91 J/(mol K) | 240 J/(mol K) | 116 J/(mol K) | S_initial = 363 J/(mol K) | | S_final = 425.9 J/(mol K) | | ΔS_rxn^0 | 425.9 J/(mol K) - 363 J/(mol K) = 62.91 J/(mol K) (endoentropic) | | | |
Chemical names and formulas
| nitric acid | sodium | water | nitrogen dioxide | sodium nitrate Hill formula | HNO_3 | Na | H_2O | NO_2 | NNaO_3 name | nitric acid | sodium | water | nitrogen dioxide | sodium nitrate IUPAC name | nitric acid | sodium | water | Nitrogen dioxide | sodium nitrate