KOH + NO = H2O + N2 + KNO2

potassium hydroxide + nitric oxide ⟶ water + nitrogen + potassium nitrite

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, K, O and N: H: | c_1 = 2 c_3 K: | c_1 = c_5 O: | c_1 + c_2 = c_3 + 2 c_5 N: | c_2 = 2 c_4 + 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_4 = 1 and solve the system of equations for the remaining coefficients: c_1 = 4 c_2 = 6 c_3 = 2 c_4 = 1 c_5 = 4 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 4 + 6 ⟶ 2 + + 4

+ ⟶ + +

potassium hydroxide + nitric oxide ⟶ water + nitrogen + potassium nitrite

| potassium hydroxide | nitric oxide | water | nitrogen | potassium nitrite molecular enthalpy | -424.6 kJ/mol | 91.3 kJ/mol | -285.8 kJ/mol | 0 kJ/mol | -369.8 kJ/mol total enthalpy | -1698 kJ/mol | 547.8 kJ/mol | -571.7 kJ/mol | 0 kJ/mol | -1479 kJ/mol | H_initial = -1151 kJ/mol | | H_final = -2051 kJ/mol | | ΔH_rxn^0 | -2051 kJ/mol - -1151 kJ/mol = -900.3 kJ/mol (exothermic) | | | |

| potassium hydroxide | nitric oxide | water | nitrogen | potassium nitrite molecular free energy | -379.4 kJ/mol | 87.6 kJ/mol | -237.1 kJ/mol | 0 kJ/mol | -306.6 kJ/mol total free energy | -1518 kJ/mol | 525.6 kJ/mol | -474.2 kJ/mol | 0 kJ/mol | -1226 kJ/mol | G_initial = -992 kJ/mol | | G_final = -1701 kJ/mol | | ΔG_rxn^0 | -1701 kJ/mol - -992 kJ/mol = -708.6 kJ/mol (exergonic) | | | |

| potassium hydroxide | nitric oxide | water | nitrogen | potassium nitrite Hill formula | HKO | NO | H_2O | N_2 | KNO_2 name | potassium hydroxide | nitric oxide | water | nitrogen | potassium nitrite IUPAC name | potassium hydroxide | nitric oxide | water | molecular nitrogen | potassium nitrite