Dissociation constant of molecular dinitrogen as a function of temperature?
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 &amp; 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).