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  • Comment on the thermodynamic stability of NO(g), given 1/2 N2(g) + 1/2  O2(g) gives NO(g) ; delta r H = 90 kJ mol–1 NO(g) + 1/2 O2(g) gives NO2(g) : delta rH= –74 kJ mol–1

6.21. Comment on the thermodynamic stability of NO (g), given

1/2 N_2(g) + 1/2 O_2(g)\rightarrow NO(g) ; \Delta _ rH ^ \ominus = 90 kJ mol ^{-1}\\\\ .\: \: \: NO(g) + 1/2 O_2(g) \rightarrow NO_ 2(g) : \Delta _ r H^ \ominus = -74 kJ mol ^{-1}


 

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The formation of NO is unstable because \Delta _rH is positive, which means heat is absorbed during the reaction. So, NO(g) has higher energy than its reactants N_{2} and O_{2}.
On the other hand, NO_{2} is stable because \Delta _rH is negative, which means heat is released during the formation of NO_{2}. It is stabilised with minimum energy. Hence, unstable NO changes to stable NO_{2}.

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