Dissociation constant of molecular dinitrogen as a function of temperature?

I'm looking for the dissociation constant of molecular dinitrogen ($\ce{N2}$) as a function of temperature,

$$\ce{N2 <=>[K] 2N} \quad K=\frac{[\ce{N}]^2}{[\ce{N2}]}$$

Casas & Bermelo (1978) reported three data points for nitrogen (I couldn't access their references):

$$K_{298}=2\cdot 10^{-160},\ K_{2000}=10^{-3},\ K_{20000}=1.29\cdot 10^5$$

1. Are these the accepted values today? Is there a more complete database online?
2. Would it be correct to fit an exponential function using these data points to extract $K$ for other temperatures?

Update: If you plot the $\log(K_p)$ vs temperature it does not give a straight line over such a large temperature range so you are unlikely to be successful.
However, you can estimate $K_p$ as the molecular properties of dinitrogen are known so you can calculate the partition functions Z and so the equilibrium constant at any temperature. You need the vibrational frequency (and possibly anharmonicity), rotational constant and dissociation energy. Work out the partition functions for translation, rotation & vibration for dinitrogen and translation for N atoms and that is all you need.
The equations needed are given in most phys. chem. textbooks and specialist books on statistical mechanics. At a standard state of $1$ bar pressure $$k_P= \frac{Z_{trans}(N)^2}{NZ_{trans}(N_2)Z_{rot}(N_2)Z_{vib}(N_2)}\exp(-E_0/(kT))$$
The experimental values at low temperature are likely to have the biggest error since dinitrogen is such a stable molecule, and it probably does not matter much how small the $K_p$ is its always tiny at low temperatures. Using the method above I calculated values as $2.4 10^{-160}, ~ 8.4 10^{-19}$ and $62400$ at the three temperatures you quote (using the dissociation energy as 9.756 eV).
• The NIST data is for the rate constant not the equilibrium constant. When calculating values the atom Z has to be multiplied by its multiplicity, 4 in the case of N atoms ($^4S_{3/2}$) and the rotational Z divided by 2 due to symmetry. – porphyrin Apr 6 '17 at 19:59