# Why is it when using bond enthalpies to calculate enthalpies of reaction, they are not always accurate and enthalpies of formation are preferred?

Jul 13, 2016

Well, those are defined to be averages, so for example, a $\text{C"-"C}$ bond enthalpy is considered to be an estimate, and you know that estimates are inherently less accurate.

Any bond can be weakened or strengthened due to neighboring atoms' effects.

For example:

$\text{H"_3stackrel(color(red)(2))"C"-stackrel(color(red)(1))"C""H"_2-"Br}$

Bromine is an electron-withdrawing group, so electron density is drawn towards it, making carbon-1 more electropositive (more partially-positively-charged) than without $\text{Br}$:

$\text{H"_3"C"-stackrel(color(red)(delta^(+)))"C""H"_2-stackrel(color(red)(delta^(-)))"Br}$

To stabilize itself, a more electropositive atom draws electron density from a less electropositive atom, like carbon-2 --- which, though it has become electropositive, is less so because it is farther away from $\text{Br}$.

$\text{H"_3stackrel(color(red)(delta^(+)))"C"-stackrel(color(red)(delta^(+)))stackrel(color(red)(delta^(+)))"C""H"_2-stackrel(color(red)(delta^(-)))stackrel(color(red)(delta^(-)))"Br}$

That makes the electron sharing in the $\text{C"-"C}$ bond uneven, weakening the bond.

Thus, the bond enthalpy for this particular, weaker $\text{C"-"C}$ bond is lower than average because less energy is required to break it.