Input interpretation
nitric acid + sulfur dioxide ⟶ sulfuric acid + nitrogen dioxide
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 Set the number of atoms in the reactants equal to the number of atoms in the products for H, N, O and S: H: | c_1 = 2 c_3 N: | c_1 = c_4 O: | 3 c_1 + 2 c_2 = 4 c_3 + 2 c_4 S: | c_2 = c_3 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 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 2 + ⟶ + 2
Structures
+ ⟶ +
Names
nitric acid + sulfur dioxide ⟶ sulfuric acid + nitrogen dioxide
Reaction thermodynamics
Gibbs free energy
| nitric acid | sulfur dioxide | sulfuric acid | nitrogen dioxide molecular free energy | -80.7 kJ/mol | -300.1 kJ/mol | -690 kJ/mol | 51.3 kJ/mol total free energy | -161.4 kJ/mol | -300.1 kJ/mol | -690 kJ/mol | 102.6 kJ/mol | G_initial = -461.5 kJ/mol | | G_final = -587.4 kJ/mol | ΔG_rxn^0 | -587.4 kJ/mol - -461.5 kJ/mol = -125.9 kJ/mol (exergonic) | | |
Entropy
| nitric acid | sulfur dioxide | sulfuric acid | nitrogen dioxide molecular entropy | 156 J/(mol K) | 248 J/(mol K) | 157 J/(mol K) | 240 J/(mol K) total entropy | 312 J/(mol K) | 248 J/(mol K) | 157 J/(mol K) | 480 J/(mol K) | S_initial = 560 J/(mol K) | | S_final = 637 J/(mol K) | ΔS_rxn^0 | 637 J/(mol K) - 560 J/(mol K) = 77 J/(mol K) (endoentropic) | | |
Chemical names and formulas
| nitric acid | sulfur dioxide | sulfuric acid | nitrogen dioxide Hill formula | HNO_3 | O_2S | H_2O_4S | NO_2 name | nitric acid | sulfur dioxide | sulfuric acid | nitrogen dioxide IUPAC name | nitric acid | sulfur dioxide | sulfuric acid | Nitrogen dioxide