Question

Comment on the thermodynamic stability of $NO(g)$, given
$\frac 12N_2(g) +\frac 12O_2(g)→NO(g); \ \ \ \ ∆_rH^Θ= 90\ kJ mol^{-1}$
$NO(g) +\frac 12 O_2(g)→NO_2(g);\ \ \ ∆_rH^Θ= –74\ kJ mol^{-1}$

Answer 1

The positive value of $∆_rH$ 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 $∆_rH$ indicates that heat is evolved during the formation of $NO_2(g)$ from $NO(g)$ and $O_2(g)$. The product, $NO_2(g)$ is stabilized with minimum energy.
Hence, unstable $NO(g)$ changes to unstable $NO_2(g)$.