Did you notice that in the formula that doesn't mention bonds, ΔH is a standard heat of formation, not a bond energy?
Understanding the heat of reaction based on bonds is more direct, clearer, OK, fine for a first attempt. But it is inconvenient for practical uses, because
- Forming a molecule from individual atoms releases a huge energy, both at the reactants and products, but the difference of both, which is the interesting heat of reaction, is much smaller. Computing from the bond energies would then be inaccurate.
- It would need to know the energy necessary to atomize all elements, but this is poorly known. Even for carbon: vaporizing it is difficult, and what proportion of C, C2, C4 and others the gas contains is still debated. How to make a measure then?
- The bond energy depends not only on the two atoms, but on their neighbours, the strain, and so on. Pretty unmanageable.
This has lend to measure and tabulate a different energy, more convenient and accurate: the standard enthalpy of formation. It tells the heat of reaction to form a compound from the elements in their standard state, that is, the state they take commonly at 1atm and 298K.
The energy to form a molecule from its atoms would be the sum of all bond energies (with signs possibly), but here "standard state" implies to form it from "molecules" of elements most often. Nitrogen is the gaseous molecule N2
, carbon is solid graphite written only C but where the atoms belong to huge molecules, iron written just Fe involves the atoms in metallic bonds, and so on. This makes a big numerical change, as these molecular elements have released much energy when formed from atoms.
The standard heat of formation answers the drawbacks. Heats of reaction are measured, tabulated and computed for reactions from molecules to molecules, which avoids the differences of big and poorly known values.
A resulting subtlety is that heats of formation have a sign, as opposed to bond energies. Depending of whether the molecule has stronger or weaker bonds than its molecular elements, the heat of formation is negative or positive. In the example you chose:
ΔH=0 because it's the element in its normal state
ΔH=0 same reason
HF ΔH=-568kJ/mol (minus because the formation releases heat)
so the enthalpy for the reaction is computed as
-0 -0 +2*(-568k) = 2*(-568k)
and it is negative.
My suggestion about the signs: it's necessary to understand them once, but only computers can manage them without errors. Humans think more reliably by wondering for each compound of a reaction whether to add or subtract, rather than applying formulas.