Input interpretation
oxygen + diborane ⟶ water + boron oxide
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 O, B and H: O: | 2 c_1 = c_3 + 3 c_4 B: | 2 c_2 = 2 c_4 H: | 6 c_2 = 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 = 3 c_2 = 1 c_3 = 3 c_4 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 3 + ⟶ 3 +
Structures
+ ⟶ +
Names
oxygen + diborane ⟶ water + boron oxide
Reaction thermodynamics
Enthalpy
| oxygen | diborane | water | boron oxide molecular enthalpy | 0 kJ/mol | 36.4 kJ/mol | -285.8 kJ/mol | -1274 kJ/mol total enthalpy | 0 kJ/mol | 36.4 kJ/mol | -857.5 kJ/mol | -1274 kJ/mol | H_initial = 36.4 kJ/mol | | H_final = -2131 kJ/mol | ΔH_rxn^0 | -2131 kJ/mol - 36.4 kJ/mol = -2167 kJ/mol (exothermic) | | |
Gibbs free energy
| oxygen | diborane | water | boron oxide molecular free energy | 231.7 kJ/mol | 87.6 kJ/mol | -237.1 kJ/mol | -1194 kJ/mol total free energy | 695.1 kJ/mol | 87.6 kJ/mol | -711.3 kJ/mol | -1194 kJ/mol | G_initial = 782.7 kJ/mol | | G_final = -1906 kJ/mol | ΔG_rxn^0 | -1906 kJ/mol - 782.7 kJ/mol = -2688 kJ/mol (exergonic) | | |
Chemical names and formulas
| oxygen | diborane | water | boron oxide Hill formula | O_2 | B_2H_6 | H_2O | B_2O_3 name | oxygen | diborane | water | boron oxide IUPAC name | molecular oxygen | diborane | water |