Why is dinitrogen tetraoxide less stable at high temperature? I have calculated the enthalpy of dimerisation for the reaction at low temperature:

$$\ce{2NO2 &lt;=&gt; N2O4}$$
$$\Delta_\mathrm{dimerisation}\ce{H}= \pu{-13.87 kcal mol^{-1}}= \pu{-58.03  kJ mol^{-1}}$$

Also for this reaction, $\ce{\Delta S}$ is negative due to presence of more gaseous moles on LHS. 

My book mentions that $\ce{N2O4}$ is stable only at low temperature. Why is it so? I couldn't find any reliable sources on the internet.
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.