Comment on the thermodynamic stability of NO(g), given
\[\frac{\text{1}}{\text{2}}\text{NO(g)+}\frac{\text{1}}{\text{2}}{{\text{O}}_{\text{2}}}\text{(g)}\to \text{N}{{\text{O}}_{\text{2}}}\text{(g):}{{\text{ }\!\!\Delta\!\!\text{ }}_{\text{r}}}{{\text{H}}^{\text{ }\!\!\theta\!\!\text{ }}}\text{=90kJmo}{{\text{l}}^{\text{-1}}}\]
\[\text{N}{{\text{O}}_{\text{(g)}}}\text{+}\frac{\text{1}}{\text{2}}{{\text{O}}_{\text{2(g)}}}\to {{\text{O}}_{\text{2(g)}}}\text{:}{{\text{ }\!\!\Delta\!\!\text{ }}_{\text{r}}}{{\text{H}}^{\text{ }\!\!\theta\!\!\text{ }}}\text{=-74 }\!\!~\!\!\text{ kJ }\!\!~\!\!\text{ mo}{{\text{l}}^{\text{-1}}}\]
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Ans: The positive value of ${{\Delta }_{r}}H$ indicates that heat is absorbed during the formation of NO(g). This means that NO(g) has higher energy than the reactants (${{N}_{2}}$ and${{O}_{2}}$). Hence, NO(g) is unstable. The negative value of ${{\Delta }_{r}}H$indicates that heat is evolved during the formation of $N{{O}_{2}}(g)$ from NO(g) and ${{O}_{2}}$(g). The product, $N{{O}_{2}}(g)$is stabilized with minimum energy.
Hence, unstable NO(g) changes to unstable $N{{O}_{2}}(g)$.