In our chemical bonding class we were given the following problem:


I am confused by how the question says that you need to consider the heat of vaporization/sublimation for HBr and HI.

Bond dissociation enthalpy is defined as "the energy needed to break one mole of the bond to give separated atoms - everything being in the gas state."

So if you can find the heat of formation of that molecule in the gas phase, why would you need anything else?

My attempt:


The above answer obtained is not the one provided by the professor so I assume I'm missing something here. The answer provided is-

364 kJ/mol


1 Answer 1


Well, I think that you have to look up the actual way bond dissociation enthalpy is defined in your textbook/lecture notes, since it is a not quite standard term. IUPAC defines only bond-dissociation energy but not enthalpy.

bond-dissociation energy, $D$

The enthalpy (per mole) required to break a given bond of some specific molecular entity by homolysis, e.g. for $\ce{CH4 -> .CH3 + H.}$, symbolized as $D(\ce{CH3−H})$ (cf. heterolytic bond dissociation energy).

Many resources on the Internet claim that the IUPAC definition of the bond-dissociation energy refers to $\pu{0 K}$ (see e.g. Wikipedia, UC Davis ChemWiki) and that it differs from bond dissociation enthalpy, which refers to the enthalpy change at room temperature ($\ce{298K}$).


The bond-dissociation energy is sometime also called the bond-dissociation enthalpy (or bond enthalpy), but these terms are not strictly equivalent, as they refer to the above reaction enthalpy at standard conditions, and differ from $D_0$ by about $\pu{1.5 kcal mol-1} \pu{(6 kJ mol-1)}$ in the case of a bond to hydrogen in a large organic molecule.

UC Davis ChemWiki

Officially, the IUPAC definition of bond dissociation energy refers to the energy change that occurs at $\pu{0 K}$, and the symbol is $D_0$. However, it is commonly referred to as BDE, the bond dissociation energy, and it is generally used, albeit imprecisely, interchangeably with the bond dissociation enthalpy, which generally refers to the enthalpy change at room temperature ($\pu{298K}$). Although there are technically differences between BDEs at $\pu{0 K}$ and $\pu{298 K}$, those difference are not large and generally do not affect interpretations of chemical processes.

If we follow this definitions, then yeah, at standard conditions $\ce{HBr}$ is liquid so we have to add up $\pu{12 kJ mol-1}$ for its enthalpy of vaporization, and (almost) get the desired $\pu{364 kJ mol-1}$. The small discrepancy is probably due to slightly wrong value of enthalpy of vaporization which I just quickly found somewhere.

  • $\begingroup$ Great thank you for your explanation. I believe you're correct as this yields the professor's answer $\endgroup$
    – Plex ASM
    Commented Sep 18, 2015 at 16:09
  • $\begingroup$ I think he wants it at 0 K which would make it a solid. If you take the heat of fusion and vaporization into account, the correct answer is given. Thanks again $\endgroup$
    – Plex ASM
    Commented Sep 18, 2015 at 16:36

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