Why is dinitrogen tetraoxide less stable at high temperature?
I figured out the answer using Van't Hoff Equation:
$$\ln\left(\frac{\ce{K_2}}{\ce{K_1}}\right)=-\frac{\ce{\Delta H}}{\ce{R}}\left({\frac{1}{\ce{T_2}}-\frac{1}{\ce{T_1}}}\right)$$

Let's assume that $\ce{\Delta H}$ is constant for a limited range of temperature. 

Since $\ce{\Delta H}$ is negative for the reaction $(=\pu{-13.87 kcal mol^{-1}})$ if we increase the temperature i.e. $\ce{T_2&gt;T_1} \to\ce{K_1&gt; K_2}$. Thus, equilibrium shifts towards the left as the temperature increases implying the decreasing stability of $\ce{N2O4}$ with increasing temperature.