I've been investigating chemical kinetics and stumbled across the CHEMKIN program manuals and the JANAF tables. In the original CHEMKIN database they list 'standard enthalpy'. I've checked and this is not in fact enthalpy of formation as this is a separately listed value. What I'm confused about is that I cannot find any reference to molar enthalpy any where else except in the the introduction to the JANAF tables and the CHEMKIN database. What's more it lists the enthalpy of $\ce{H2O}$ and $\ce{H2O2}$ as zero and other substances have negative enthalpy. Questions:

Is specific molar enthalpy in comparison to water or some other substance?

If it is still a relative measurement of enthalpy what advantage does it provide over heat of formation in practical use?

How can the two substances both have the same enthalpy even if it is relative if one has an extra atom and an extra bond?

Am I completely wrong and the whole thing is something else entirely?

  • $\begingroup$ Just browsing... Sounds like a reference state problem. In most cases, the reference point is set to ideal gas standard conditions so that the enthalpy description (i.e. enthalpy departure calculations) are consistent with respect to chemical reactions. However, if chemical reactions are not a concern, then we could equally define the reference point to be at some temperature $T_{ref}$ in e.g. the liquid phase (at, let's say, the normal boiling point), or we could simply define the reference to be zero if we are being lazy... $\endgroup$
    – ChE Junkie
    Commented Nov 15, 2016 at 16:59
  • $\begingroup$ The key point is that we are always moving from some known state to a state of interest, and if we choose the known state well we can save some time by reducing the required steps in the calculation route (strategy). $\endgroup$
    – ChE Junkie
    Commented Nov 15, 2016 at 16:59

1 Answer 1


At a glance, the values listed in the table in the CHEMKIN database are standard enthalpies of formation at 298 K, $\Delta _fH^○(T)$, and standard entropies at 298 K. Why? If only because CHEMKIN's $H^○(298)$ for the elements in their stable states are presented as equal to zero.

You may have just mixed up different enthalpies. Along with the heats of formation, thermodynamic tables (including NIST JANAF) often present the enthalpy increments (sometimes called simply enthalpies), $H^○(T)-H^○(0)$ or $H^○(T)-H^○(298)$. These are not to be confused with the enthalpies of formation $\Delta _fH^○(T)$.

By definition, the standard enthalpy of formation (at a certain temperature) of a complex substance AB is equal to the enthalpy of a chemical process A + B = AB, where A and B are elements in their most stable modifications (at a certain temperature). By convention, $\Delta _fH^○(T)$ for the elements in their most stable modifications is zero.

The enthalpy increment $H^○(T)-H^○(0)$ is an integral of the heat capacity at constant pressure, $\int\limits_0^T C_p(T)dT$. It is no less important than the heat of formation. For example, knowing $\Delta _fH^○(298)$ for our hypothetical substance AB and all the enthalpies $H^○(T)-H^○(298)$ for A, B and AB, we can calculate $\Delta _fH^○(T)$, an enthalpy of formation at different temperature:

$\Delta _fH^○(T) = \Delta _fH^○(298) + [H^○(T)-H^○(298)]_{AB} - [H^○(T)-H^○(298)]_{A} - [H^○(T)-H^○(298)]_{B}$

Note that similar formulae can also be found in the introduction to NIST JANAF (at least to the 4th ed. of it).


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