The compound oxygen difluoride is quite reactive, giving oxygen and HF when treated with water
OF2(g) + H2O(g) → O2(g) + 2HF(g)
E rxn = –318 kJ
Using bond energies, calculate the bond dissociation energy of the O–F bond in OF2
To calculate the bond dissociation energy of the O–F bond in OF2, we need to use the given reaction and the bond energies of the other bonds involved in the reaction.
First, let's find the bond dissociation energies of the other involved bonds:
- O–H bond in H2O has a bond energy of 463 kJ/mol.
- H–F bond in HF has a bond energy of 565 kJ/mol.
Now, let's write an equation for the energy change in the reaction, which involves bond breaking and bond formation:
E_reaction = Bond breaking - Bond forming
The bond breaking involves breaking one O–H bond in H2O and two O–F bonds in OF2:
E_breaking = 1 * (O–H) + 2 * (O–F)
The bond forming involves forming one O=O bond in O2 and two H–F bonds in HF:
E_forming = 1 * (O=O) + 2 * (H–F)
The energy of the O=O bond can be found in reference books, and it's approximately 498 kJ/mol. Plugging the values into the equation:
-318 kJ = (1 * 463 + 2 * O–F) - (1 * 498 + 2 * 565)
Now, solve for the O–F bond energy:
-318 = 463 + 2O–F - 498 - 1130
-318 = -35 + 2O–F
O–F = (318 - 35) / 2
O–F ≈ 141.5 kJ/mol
Thus, the bond dissociation energy of the O–F bond in OF2 is approximately 141.5 kJ/mol.