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)](http://pubs.acs.org/doi/abs/10.1021/ed055p679) 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).


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