$$\ce{3O2(g) -> 2O3(g)} \qquad \Delta H^\circ (\pu{298 K}) = \pu{+142 kJ mol^-1}$$

Since the RHS of the equation contains less number of gaseous molecules, naturally I expected the $\Delta S$ to be negative. In fact, while solving many thermodynamics questions relating to sign of $\Delta S$, I used this concept to get the correct answers (we were even taught this, that's why).

However, this appears to be an exception as this reaction does happen (thermodynamically possible due to negative $\Delta G$). It's given in my textbook (page 186, NCERT Chemistry) and the textbook accounts for it stating that the reaction has $\Delta S > 0$ without mentioning the reason for it.

Why is the entropy of ozone more that that of oxygen? Is there any simple explanation to it?

  • 3
    Not clear what you mean by "why?" If you're asking why, intuitively, why the relationship in nature exists, I suspect it's because you are placing too much emphasis on the change in number of molecules: you're getting new rotational states in ozone that you don't have in oxygen. – Zhe Sep 13 at 19:22
  • 2
    And vibrational states as well... – Jon Custer Sep 13 at 20:30
  • 3
    To put it more plainly - it's molar entropy and mole of O3 is more then mole of O2. – Mithoron Sep 13 at 21:57

You are mixing up two different quantities (entropy of reaction, and molar entropies of the two gases involved in the reaction) and either your text book is in error, or you have misread it somehow. For that reaction, as written, the formation of ozone from oxygen, $\Delta G$ is positive and $\Delta S$ is negative (the opposite of what you wrote in your question). It is unfavourable on both enthalpic and entropic grounds. Ozone is thermodynamically unstable with respect to oxygen under standard conditions.

Your experience, regarding the sign of $\Delta S$ and its relation to the change in the number of molecules in a gas-phase reaction, is by-and-large a good guide, because the translational contribution to the entropy typically dominates over the rotational and vibrational contributions in a gas. This reaction is consistent with this rule of thumb.

The relevant thermodynamic quantities are the standard Gibbs free energy, enthalpy, and entropy, of formation of ozone, since the oxygen is in its standard state. Taking the figures from this copy of CRC we have $$ \Delta_{\text{f}} G^\circ (\text{O}_3) = 163.2 \, \text{kJ}\,\text{mol}^{-1} \quad \Delta_{\text{f}} H^\circ (\text{O}_3) = 142.7 \, \text{kJ}\,\text{mol}^{-1} $$ and for the standard entropies $$ S^\circ (\text{O}_3) = 238.9 \, \text{J}\,\text{K}^{-1}\text{mol}^{-1} \quad S^\circ (\text{O}_2) = 205.2 \, \text{J}\,\text{K}^{-1}\text{mol}^{-1} $$ So indeed the standard molar entropy of ozone is a little higher than that of oxygen (as the comments say, if you want to explain this precisely you need to take into account internal degrees of freedom, as well as properly calculating the translational contributions), but this does not lead to a positive entropy of reaction. To calculate this, i.e. the entropy of formation of ozone, we need $$ \Delta_{\text{f}} S^\circ (\text{O}_3) = S^\circ (\text{O}_3) - \tfrac{3}{2} S^\circ (\text{O}_2) = -68.9 \, \text{J}\,\text{K}^{-1}\text{mol}^{-1} $$ remembering that all these standard quantities of formation are per mole of ozone. Equivalently you can calculate this number from $[\Delta_{\text{f}} H^\circ (\text{O}_3)-\Delta_{\text{f}} G^\circ (\text{O}_3)]/T$ giving the same answer.

So the reaction has a negative $\Delta S$ and a positive $\Delta G$.

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