Q.Comment on the thermodynamic stability of NO(g), given: $\tfrac{1}{2} N_2(g) + \tfrac{1}{2} O_2(g) \rightarrow NO(g)$; $\Delta_r H = 90$ kJ mol$^{-1}$; $NO(g) + \tfrac{1}{2} O_2(g) \rightarrow NO_2(g)$; $\Delta_r H = -74$ kJ mol$^{-1}$.

Concept: Standard Enthalpy of Formation and Thermodynamic Stability

A compound is thermodynamically stable relative to its elements if its formation from elements is exothermic ($\Delta_f H^\circ < 0$). The given data directly provides $\Delta_f H^\circ[\text{NO}(g)] = +90 \text{ kJ mol}^{-1}$.

Reasoning:

1. The first reaction shows formation of NO from its elements in their standard states ($\text{N}_2$ and $\text{O}_2$). Since $\Delta_r H = +90 \text{ kJ mol}^{-1} > 0$, the process is endothermic.

2. An endothermic formation means NO lies higher in energy than its constituent elements. The positive enthalpy of formation indicates NO is thermodynamically unstable with respect to decomposition into $\text{N}_2$ and $\text{O}_2$. …

The question hands us two reactions and asks us to comment on stability. Thermodynamic stability hinges on whether a substance sits at a lower energy than alternative arrangements of its atoms. A positive standard enthalpy of formation tells us the compound is endothermic — it contains more energy than the elements from which it formed, so it has a natural tendency to decompose back. But tendency is not fate: kinetics often freezes thermodynamically unstable species in place.

Let's decode what the data reveal.

Understanding the Given Reactions

The first reaction is precisely the formation of NO from its elements in their standard states:

This $\Delta_r H^\circ$ is the standard enthalpy of formation of NO, $\Delta_f H^\circ[\text{NO}(g)]$. A positive value means energy must be pumped in to make NO from nitrogen and oxygen — the molecule is thermodynamically unstable relative to the elements.

The second reaction shows NO reacting further with oxygen:

This is exothermic: NO readily oxidizes to $NO_2$ when oxygen is available, releasing energy.

Stability with Respect to the Elements

1. Positive $\Delta_f H^\circ$ signals instability. Since forming NO from $N_2$ and $O_2$ requires $+90$ kJ mol$^{-1}$, the reverse decomposition

is exothermic and thermodynamically favored. In principle, NO should fall apart into its elements.

2. Why does NO exist at all? The decomposition has a high activation energy. At room temperature, NO molecules lack the kinetic energy to surmount the barrier, so the reaction is kinetically hindered. NO is a classic example of a kinetically stable but thermodynamically unstable species.

Stability with Respect to Further Oxidation

3. Disproportionation tendency. We can combine the two given reactions to explore whether NO might disproportionate. Reverse the formation reaction and add it to the oxidation:

Sum these (canceling $\tfrac{1}{2} O_2$ on both sides):

$2\,NO(g) \rightarrow \tfrac{1}{2} N_2(g) + NO_2(g), \quad \Delta H = -164 \text{ kJ per 2 mol NO}$ …