Question

Comment on the thermodynamic stability of NO (g) given:
A. \[\cfrac { 1 }{ 2 } { N }_{ 2 }(g)+\cfrac { 1 }{ 2 } { O }_{ 2 }(g)\longrightarrow NO(g);\quad { \triangle }_{ r }{ H }^{ \theta }=90KJ\quad { mol }^{ -1 }\]
B. \[NO(g)+\cfrac { 1 }{ 2 } { O }_{ 2 }(g)\longrightarrow N{ O }_{ 2 }(g);\quad { \triangle }_{ r }{ H }^{ \theta }=-74KJ\quad { mol }^{ -1 }\]

Answer 1

Hint: For determining the thermodynamic stability of a product or the spontaneity of a reaction, we should always determine the change in the Gibbs free energy associated with the product or the reaction.

Complete step by step answer:
Whether a product or a reaction will be thermodynamically stable or spontaneous respectively is determined by the change in the Gibbs free energy. If the change is negative, the product will be thermodynamically stable otherwise it will be thermodynamically unstable. Gibbs free energy change depends upon the enthalpy change and the change in entropy. The reaction is given below:

\[ \triangle G=\triangle H-T\triangle S\]

If the enthalpy change is negative and the enthalpy change is positive for a product formation, then the Gibbs free energy change will be highly negative and the product will be thermodynamically stable. Now, for the two reactions given in the question, two gaseous reactants are combining in order to give one gaseous product. Therefore we can assume that the entropy change for both of the reactions is negative even though it is not given in the question. Therefore for the products to be thermodynamically stable, the enthalpy change should be highly negative such that its magnitude is greater than the magnitude of $ T\triangle S$, making the Gibbs free energy change to be negative.
For the first reaction which involves the formation of NO gas,

\[\cfrac { 1 }{ 2 } { N }_{ 2 }(g)+\cfrac { 1 }{ 2 } { O }_{ 2 }(g)\longrightarrow NO(g);\quad { \triangle }_{ r }{ H }^{ \theta }=90KJ\quad { mol }^{ -1 }\]

The enthalpy of formation of NO gas is positive therefore the Gibbs free energy change will be positive and hence the product Nitric oxide is thermodynamically unstable.

For the second reaction which involves the formation of nitrogen dioxide gas,
\[NO(g)+\cfrac { 1 }{ 2 } { O }_{ 2 }(g)\longrightarrow N{ O }_{ 2 }(g);\quad { \triangle }_{ r }{ H }^{ \theta }=-74KJ\quad { mol }^{ -1 }\]

The enthalpy of formation of nitrogen dioxide gas is negative and therefore it is thermodynamically stable. This also tells us that nitric oxide gas (NO) has a tendency to convert into more stable nitrogen dioxide gas.

Hence nitric oxide gas (NO) is thermodynamically unstable.

Note: Always remember that the thermodynamic stability of a system is determined by both the enthalpy change of the system as well as the entropy change of the system.