# Comment on the thermodynamic stability

Question:

Comment on the thermodynamic stability of $\mathrm{NO}_{(g) \text {, given }}$

$\frac{1}{2} \mathrm{~N}_{2(g)}+\frac{1}{2} \mathrm{O}_{2(g)} \rightarrow \mathrm{NO}_{(g)} ; \Delta_{r} H^{\theta}=90 \mathrm{~kJ} \mathrm{~mol}^{-1}$

$\mathrm{NO}_{(g)}+\frac{1}{2} \mathrm{O}_{2(g)} \rightarrow \mathrm{NO}_{2(g)}: \Delta_{r} H^{\theta}=-74 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Solution:

The positive value of $\Delta_{r} H$ indicates that heat is absorbed during the formation of $\mathrm{NO}_{(g)}$. This means that $\mathrm{NO}_{(g)}$ has higher energy than the reactants ( $\mathrm{N}_{2}$ and $\mathrm{O}_{2}$ ). Hence, $\mathrm{NO}_{(g)}$ is unstable.

The negative value of $\Delta, H$ indicates that heat is evolved during the formation of $\mathrm{NO}_{2(g)}$ from $\mathrm{NO}_{(g)}$ and $\mathrm{O}_{2(g)}$. The product, $\mathrm{NO}_{2(g)}$ is stabilized with minimum energy.

Hence, unstable $\mathrm{NO}_{(g)}$ changes to stable $\mathrm{NO}_{2(g)}$.