Le Châtelier analysis: For the gas-phase equilibrium N2 + O2 ⇌ 2 NO with ΔH = +80 kJ per kmol of NO formed (endothermic forward), which change favors the decomposition of NO?

Introduction / Context:
Nitric oxide formation and decomposition are central to high-temperature combustion and NOx control. Le Châtelier’s principle provides qualitative guidance on how temperature and pressure shifts affect equilibria, especially for endothermic versus exothermic directions.

Given Data / Assumptions:

• Reaction: N2 + O2 ⇌ 2 NO.
• Enthalpy change: ΔH > 0 in the forward direction (endothermic formation of NO).
• Total moles: 2 → 2 (no net mole change).

Concept / Approach:
For endothermic forward reactions, lowering temperature favors the reverse (exothermic) direction—here, the decomposition of NO to N2 and O2. Pressure has little or no effect when there is no net change in total moles of gas (2 → 2), assuming ideal behavior. Changing the concentration of a reactant such as N2 does not directly force decomposition of NO; it tends instead to drive formation when O2 is present.

Step-by-Step Solution:

Verification / Alternative check:
Equilibrium constant K(T) for an endothermic reaction increases with temperature (van ’t Hoff). Thus, at lower T, K decreases, favoring reactants (N2, O2) and NO decomposes.

Why Other Options Are Wrong:

• Increase/decrease in pressure: negligible effect because Δn = 0.
• Increasing [N2]: tends to promote NO formation if O2 is available.
• Adding inert at constant pressure: mainly dilutes; with Δn = 0 the effect on equilibrium position is minor.

Common Pitfalls:
Assuming pressure always influences gas equilibria; you must check Δn_gas. Misapplying van ’t Hoff by ignoring the sign of ΔH.

Final Answer:
Decrease in temperature