How ${O_3}$ is less thermodynamically stable than ${O_2}$? (2024)

Hint: The ozone consists of three atoms of oxygen and is in an unstable state. It decomposes into ${O_2}$ molecule with liberation of an enormous amount of heat. This heat is released in the atmosphere. Thus, the formation of oxygen molecules is thermodynamically more stable than ozone molecules.

Complete step by step answer:
The ${O_3}$ is ozone while the ${O_2}$ is called oxygen. Both are gases in nature and can interconvert into each other.
The ozone formation is endothermic and requires heat. Thus, the decomposition of ozone into oxygen releases heat making it a favourable reaction.
Further, In the diatomic oxygen molecule (${O_2}$), two lone pairs of electrons are present on each oxygen atom. So, they stabilise the oxygen instead of double bonds. But in case of ozone, there is one more oxygen atom. The third oxygen atom pushes one of its lone pairs of electrons toward one of the double bonded oxygen atoms. Thus, it interrupts the overall structural stability and thus, the molecule becomes unstable.
When an oxygen atom and one ozone molecule combine, they form an oxygen molecule. The reaction is written as-
${O_3} + O \cdot \to 2{O_2}$
So, the ${O_3}$ is thermodynamically less stable than ${O_2}$.

Note: The ozone consists of three oxygen atoms while the oxygen molecule consists of two oxygen atoms. In ozone, the four electrons are shared between three nuclei. So, the bond order is 1.5 while in case of oxygen two nuclei share this. So, the bond order is 2. For any thermodynamically unstable reaction, the value of Gibbs free energy is positive and for the ozone, the value comes out to be +163 KJ/mol.

How ${O_3}$ is less thermodynamically stable than ${O_2}$? (2024)
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