6.21   Comment on the thermodynamic stability of NO(g), given$1/2 N_2(g) + 1/2 O_2(g)\rightarrow NO(g) ; \Delta _ rH ^ \ominus = 90 kJ mol ^{-1}\\\\ .\: \: \: NO(g) + 1/2 O_2(g) \rightarrow NO_ 2(g) : \Delta _ r H^ \ominus = -74 kJ mol ^{-1}$

The formation of $NO$ is unstable because $\Delta _rH$ is positive it means heat is absorbed during the reaction. So,  $NO$(g) has higher energy than its reactants $N_{2}$ and $O_{2}$.
On the other hand, $NO_{2}$ is stable because $\Delta _rH$ is negative means heat is released during the formation of $NO_{2}$. It is stablised with minimum energy. Hence unstable $NO$ changes to stable $NO_{2}$.