Negative and Zero Enthalpy?
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).