Input interpretation
bromine trifluoride + boron oxide ⟶ oxygen + bromine + boron trifluoride
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 + c_5 Set the number of atoms in the reactants equal to the number of atoms in the products for Br, F, B and O: Br: | c_1 = 2 c_4 F: | 3 c_1 = 3 c_5 B: | 2 c_2 = c_5 O: | 3 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 = 2 c_2 = 1 c_3 = 3/2 c_4 = 1 c_5 = 2 Multiply by the least common denominator, 2, to eliminate fractional coefficients: c_1 = 4 c_2 = 2 c_3 = 3 c_4 = 2 c_5 = 4 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 4 + 2 ⟶ 3 + 2 + 4
Structures
+ ⟶ + +
Names
bromine trifluoride + boron oxide ⟶ oxygen + bromine + boron trifluoride
Reaction thermodynamics
Gibbs free energy
| bromine trifluoride | boron oxide | oxygen | bromine | boron trifluoride molecular free energy | -2405 kJ/mol | -1194 kJ/mol | 231.7 kJ/mol | 0 kJ/mol | -1119 kJ/mol total free energy | -9620 kJ/mol | -2389 kJ/mol | 695.1 kJ/mol | 0 kJ/mol | -4478 kJ/mol | G_initial = -12009 kJ/mol | | G_final = -3783 kJ/mol | | ΔG_rxn^0 | -3783 kJ/mol - -12009 kJ/mol = 8226 kJ/mol (endergonic) | | | |
Chemical names and formulas
| bromine trifluoride | boron oxide | oxygen | bromine | boron trifluoride Hill formula | BrF_3 | B_2O_3 | O_2 | Br_2 | BF_3 name | bromine trifluoride | boron oxide | oxygen | bromine | boron trifluoride IUPAC name | | | molecular oxygen | molecular bromine | trifluoroborane