Question:medium

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

Updated On: Jan 19, 2026
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Solution and Explanation

Given Data

Formation of NO(g) from elements:

N2(g)} + O2(g) NO(g) = +90kJ mol}^{-1}

Oxidation of NO(g) to NO₂(g):

NO(g) + O2(g) – > NO2(g),= -74 kJ mol

Interpretation of Enthalpy Changes

  • Formation of NO from N2 and O2 is endothermic (+90 kJ mol⁻¹). This means NO has a higher enthalpy (is less stable) than its constituent elements in their standard states.
  • Conversion of NO to NO₂ is exothermic (–74 kJ mol⁻¹). NO₂ lies at a lower enthalpy than NO, so NO tends to get oxidised further to NO₂ in the presence of oxygen.

Comment on Stability

  • Because its formation from N2 and O2 absorbs heat, NO(g) is thermodynamically unstable relative to the mixture of N2(g) and O2(g).
  • Also, NO(g) can spontaneously (exothermically) convert to NO₂(g); therefore NO is less stable than NO₂ as well.
  • Conclusion: NO(g) is a thermodynamically unstable species, existing mainly because it can be formed under high‑temperature conditions and then persists kinetically, but it is not the most stable form of N–O combination.
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