Which of these things is not like the other? Enthalpy of formation, enthalpy of atom combination and bond energy

I have had many questions about the difference between enthalpy of formation (ΔHf) values and enthalpy of atom combination (ΔHac) values.  I’ve also been told there is not much about ΔHac­ values on the internet (which is fine because it would probably be wrong).  Two deal with both issues simultaneously I’m writing this blog post. There are many ways to calculate the change in enthalpy (ΔH) of a chemical reaction.  This is because enthalpy (H) is a state function – the path taken from start to finish does not matter, only the starting materials (reactants) and ending materials (products). The most common ways of calculating ΔH are:

  • Experimentally (from doing the reaction and measuring the heat)
  • Estimating with bond energies (broken bond absorb energy, formed bonds release energy)
  • Using an equation based on the enthalpy of formation (ΔHf): ΔH=ΣΔHf(products) − ΣΔHf(reactants)

ΔHf values are based on the reaction of taking a compound’s elements as they exist at room temperature and atmospheric pressure and measuring the heat released or absorbed when the compound is formed.  In other words, the ΔHf for liquid water is the ΔH of the reaction

H2 (g) + ½ O2(g) →  H2O(l)             ΔH=ΔHf = − 286 kJ/mole

Notice that we have H2 and O2 in the reactants here because oxygen exists as O2 gas  and hydrogen exists at H2 gas molecules at room temperature and atmospheric pressure.  To make liquid water from its element, those elements have to be broken up into atoms then recombined into the compound. Most people use ΔHf values because it gets you the right answer and because… that’s what everyone else does.  Enthalpy of formation values have become one of those things you do not because you really understand it or because it’s the best way but because that’s how it’s always been done.

It turns out there is a better way to do this which is with enthalpy of atom combination values, ΔHac.  The ΔHac values are a measure of the change in enthalpy going from gas phase atoms to a compound.  For liquid water, the corresponding reaction is:

2H (g) + O(g) →  H2O(l)                 ΔH=ΔHac = − 970 kJ/mole

Why?  What difference does it make?  In terms of plugging into the equation, none.  It’s still products minus reactant values.  The units are kJ/mole, so you still have to multiply by the stoichiometric coefficient.  Why bother then? It’s worth bothering because the ΔHac values actually reveal the total bond energy present in 1 mole of a substance.  For gas phase water, the ΔHac value is – 926 kJ/mole, which is two times the O to H bond energy value of 463 kJ/mole.  The extra 44 kJ/mole for liquid water is the hydrogen bonding energy. The enthalpy of formation values do not give you the bond energy.  Enthalpy of formation values only tell you the relative enthalpy change going from the substances elements, which are almost always either molecules or solids.  As such, the enthalpy of breaking down the bonds in those elements is folded into ΔHf, making the value itself useless.  ΔHac values can be plugged in the equation and have intrinsic value. Bond energy is the real concept worth emphasizing here.  Most people don’t understand bond energy and, as a practicing chemist, bond energy is the far more useful concept for understanding new situations, if for no other reason than it doesn’t require tracking down obscure ΔHf or ΔHac­ values.  I’m pretty sure I used ΔHf values once in my entire physical chemistry graduate career.

One more example: carbon dioxide. ΔHac for CO2 = −1609 kJ/mole ΔHf for diamond: −394 kJ/mole What does the ΔHf tell you about bonding in CO2?  Basically nothing because mixed into that number is the breaking of bonds from graphite and O2.  Meanwhile, the ΔHac value tells you that the bond energy of the C to O double bond in CO2 is 805 kJ/mole.

Comments?  Typos?  Still confused?  Tell me in the comment section.

Here is the reference for the paper that first proposed using ΔHac values: Gillespie, R. J., Spencer, J. N., Moog, R. S.  “An Approach to Reaction Thermodynamics through Enthalpies, Entropies, and Free Energies of Atomization.”  J. Chem. Educ., 1996, 73 (7), p 631. http://pubs.acs.org/doi/abs/10.1021/ed073p631


5 thoughts on “Which of these things is not like the other? Enthalpy of formation, enthalpy of atom combination and bond energy

  1. 2H (g) + O2(g) → H2O(l) ΔH=ΔHac = − 970 kJ/mole

    Is there a difference between the delta Hac equation having a gaseous O2 versus gaseous (mono) O. I thought, since it’s just combining the atoms needed to make the compounds, the equation would look more like 2 Hydrogen atoms plus 1 Oxygen atom yields water. Because gaseous O2 implies a bonded molecule rather than just a single oxygen atom???

  2. Yes, there is a difference.

    for O2(g) the equation is:
    2O (g) -> O2 (g) ΔH=ΔHac = −498 kJ/mole

    For O(g) the equation is
    O(g) -> O (g) ΔH=ΔHac = 0

    No bond is formed in making O (monatomic gas), so it’s ΔHac is 0. For O2 (g) a bond is formed (O=O) so energy is released. Does that help?

  3. Okay, I understand that is true for oxygen atoms and oxygen gas. I guess I’m just confused why O2 is used for the atomic combination equation for water? I though oxygen would be in its natural diatomic state for the enthalpy formation equation, but would just be an individual oxygen atom for the atomic combination? Because if water only needs one oxygen atom, then the atomic combination formula would have a single oxygen atom as opposed to breaking apart an O2 molecule.
    Basically I don’t understand if
    2H (g) + O2(g) → H2O(l) ΔH=ΔHac = − 970 kJ/mole
    means the same as
    2H (g) + O(g) → H2O(l) ΔH=ΔHac = − 970 kJ/mole

  4. Okay, good, I was really confused for a sec there.
    Thanks for the oxygen atoms/oxygen gas explanation though, that was helpful.

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