H2 + N2O = H2O + N2

hydrogen + nitrous 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: | 2 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_1 = 1 and solve the system of equations for the remaining coefficients: c_1 = 1 c_2 = 1 c_3 = 1 c_4 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | + ⟶ +

+ ⟶ +

hydrogen + nitrous oxide ⟶ water + nitrogen

ΔH_rxn^0 | -285.8 kJ/mol - 81.6 kJ/mol = -367.4 kJ/mol (exothermic)

ΔG_rxn^0 | -237.1 kJ/mol - 104 kJ/mol = -341.1 kJ/mol (exergonic)

ΔS_rxn^0 | 261.9 J/(mol K) - 335 J/(mol K) = -73.09 J/(mol K) (exoentropic)

| hydrogen | nitrous oxide | water | nitrogen Hill formula | H_2 | N_2O | H_2O | N_2 name | hydrogen | nitrous oxide | water | nitrogen IUPAC name | molecular hydrogen | nitrous oxide | water | molecular nitrogen