H2 + NO = H2O + N2

hydrogen + nitric oxide ⟶ water + nitrogen

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 and O: H: | 2 c_1 = 2 c_3 N: | c_2 = 2 c_4 O: | 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_4 = 1 and solve the system of equations for the remaining coefficients: c_1 = 2 c_2 = 2 c_3 = 2 c_4 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 2 + 2 ⟶ 2 +

+ ⟶ +

hydrogen + nitric oxide ⟶ water + nitrogen

| hydrogen | nitric oxide | water | nitrogen molecular enthalpy | 0 kJ/mol | 91.3 kJ/mol | -285.8 kJ/mol | 0 kJ/mol total enthalpy | 0 kJ/mol | 182.6 kJ/mol | -571.7 kJ/mol | 0 kJ/mol | H_initial = 182.6 kJ/mol | | H_final = -571.7 kJ/mol | ΔH_rxn^0 | -571.7 kJ/mol - 182.6 kJ/mol = -754.3 kJ/mol (exothermic) | | |

| hydrogen | nitric oxide | water | nitrogen molecular free energy | 0 kJ/mol | 87.6 kJ/mol | -237.1 kJ/mol | 0 kJ/mol total free energy | 0 kJ/mol | 175.2 kJ/mol | -474.2 kJ/mol | 0 kJ/mol | G_initial = 175.2 kJ/mol | | G_final = -474.2 kJ/mol | ΔG_rxn^0 | -474.2 kJ/mol - 175.2 kJ/mol = -649.4 kJ/mol (exergonic) | | |

| hydrogen | nitric oxide | water | nitrogen molecular entropy | 115 J/(mol K) | 211 J/(mol K) | 69.91 J/(mol K) | 192 J/(mol K) total entropy | 230 J/(mol K) | 422 J/(mol K) | 139.8 J/(mol K) | 192 J/(mol K) | S_initial = 652 J/(mol K) | | S_final = 331.8 J/(mol K) | ΔS_rxn^0 | 331.8 J/(mol K) - 652 J/(mol K) = -320.2 J/(mol K) (exoentropic) | | |

| hydrogen | nitric oxide | water | nitrogen Hill formula | H_2 | NO | H_2O | N_2 name | hydrogen | nitric oxide | water | nitrogen IUPAC name | molecular hydrogen | nitric oxide | water | molecular nitrogen