Why is it when using bond enthalpies to calculate enthalpies of reaction, they are not always accurate and enthalpies of formation are preferred?
1 Answer
Well, those are defined to be averages, so for example, a
Any bond can be weakened or strengthened due to neighboring atoms' effects.
For example:
#"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
#"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
#"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
Thus, the bond enthalpy for this particular, weaker