O2 + MoS2 = SO2 + MoO3

oxygen + molybdenum disulfide ⟶ sulfur dioxide + molybdenum trioxide

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 O, Mo and S: O: | 2 c_1 = 2 c_3 + 3 c_4 Mo: | c_2 = c_4 S: | 2 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 = 7/2 c_2 = 1 c_3 = 2 c_4 = 1 Multiply by the least common denominator, 2, to eliminate fractional coefficients: c_1 = 7 c_2 = 2 c_3 = 4 c_4 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 7 + 2 ⟶ 4 + 2

+ ⟶ +

oxygen + molybdenum disulfide ⟶ sulfur dioxide + molybdenum trioxide

| oxygen | molybdenum disulfide | sulfur dioxide | molybdenum trioxide molecular enthalpy | 0 kJ/mol | -235.1 kJ/mol | -296.8 kJ/mol | -745.1 kJ/mol total enthalpy | 0 kJ/mol | -470.2 kJ/mol | -1187 kJ/mol | -1490 kJ/mol | H_initial = -470.2 kJ/mol | | H_final = -2677 kJ/mol | ΔH_rxn^0 | -2677 kJ/mol - -470.2 kJ/mol = -2207 kJ/mol (exothermic) | | |

| oxygen | molybdenum disulfide | sulfur dioxide | molybdenum trioxide molecular free energy | 231.7 kJ/mol | -225.9 kJ/mol | -300.1 kJ/mol | -668 kJ/mol total free energy | 1622 kJ/mol | -451.8 kJ/mol | -1200 kJ/mol | -1336 kJ/mol | G_initial = 1170 kJ/mol | | G_final = -2536 kJ/mol | ΔG_rxn^0 | -2536 kJ/mol - 1170 kJ/mol = -3707 kJ/mol (exergonic) | | |

| oxygen | molybdenum disulfide | sulfur dioxide | molybdenum trioxide Hill formula | O_2 | MoS_2 | O_2S | MoO_3 name | oxygen | molybdenum disulfide | sulfur dioxide | molybdenum trioxide IUPAC name | molecular oxygen | dithioxomolybdenum | sulfur dioxide | trioxomolybdenum