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

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
up vote 10 down vote accepted

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 $\Delta G$.


[EDIT following OP comment]

This means that, in the oxygen-ozone equilibrium, both forward and backward reactions are happening, at the same rates, but the rate constants are such that the equilibrium concentration (or partial pressure) of O$_3$ is much lower than that of O$_2$.

This is not how things happen in the stratosphere, nor in most methods for preparing ozone. Generally, these involve a multi-step process, in which the first step is the dissociation of O$_2$ into two oxygen atoms, by absorbing a photon of UV light, or through an electrical discharge (for example). Wikipedia provides a reasonable summary of the Ozone-oxygen cycle, or Chapman cycle, with links to further resources, while various production methods are described on the Ozone page. The above equilibrium thermodynamics relations cannot be applied to processes like this, because of the injection of energy.

Instead, in the case of the Chapman cycle, a steady-state kinetic scheme can be set up for creation and destruction of O$_3$, and O atoms, where some of the rate constants depend on the intensity of UV light. The net result is a steady (small) population of O$_3$, and O atoms, accompanied by the conversion of energy from UV light into heat which is dissipated in the atmosphere. In real life, the situation is more complicated still, because of the roles of other species and reactions: books such as RP Wayne's Chemistry of Atmospheres give a lot more detail.


[2nd EDIT following further OP comments]

For the equilibrium between oxygen and ozone, the standard Gibbs free energy change for the reaction $\Delta_r G^\circ$ may be related to the equilibrium constant $K$ (see e.g. here) $$ \Delta_r G^\circ = -RT \ln K $$ where $K$ is expressed (for ideal gases) in terms of the partial pressures relative to the standard pressure $p^\circ$ (1 bar) $$ K = \frac{(p_{\text{O}_3}/p^\circ)^2}{(p_{\text{O}_2}/p^\circ)^3} $$ For large positive $\Delta_r G^\circ$, $K$ is extremely small, so the equilibrium partial pressure of ozone is extremely small, if we are given that the partial pressure of oxygen has a reasonable value. Indeed, if we take (for the reaction as written, producing two moles of ozone) $$ \Delta_r G^\circ=2\times\Delta_{\text{f}} G^\circ (\text{O}_3) = 326.4 \, \text{kJ}\,\text{mol}^{-1} $$ then $K=\exp(-\Delta_r G^\circ/RT)\approx 6\times10^{-58}$; if the partial pressure of oxygen is $\approx 0.21$ bar (as in the atmosphere around us), then the equilibrium partial pressure of ozone is $p_{\text{O}_3}\approx 2.4\times 10^{-30}$ bar.

It is wrong to conclude that the reaction producing ozone "cannot happen" because the free energy change is positive. $\Delta_r G^\circ$ is simply the free energy change associated with the complete conversion of reactants into products. In reality, an equilibrium is always established somewhere in between. In this case, it is very far over to one side, but not 100%. If $\Delta_r G^\circ$ were not so large (but still positive) the equilibrium would not be so far in favour of reactants. Thermodynamically, you can think of the position of equilibrium as being determined by both $\Delta_r G^\circ$ and the entropy of mixing of the reactants with the products: this will always result in an equilibrium position that is neither 100% reactants nor 100% products.

It is also possible to give a dynamical interpretation of chemical equilibria in terms of forward and reverse reaction rates, but I emphasize that this argument only applies, at least in this simplified form, to one-step chemical reactions. The oxygen-ozone equilibrium certainly involves multiple steps (nobody is pretending that three molecules of oxygen miraculously collide together at once). To see why this more complicated mechanism does not affect the argument above about the equilibrium constant, see How is it that the equilibrium constant does not depend on the mechanism? and you might also find Transition state and free energy useful. So consider the dynamical equilibrium $$ \text{A} + \text{B} \leftrightharpoons \text{C} + \text{D} $$ and assume that the forward and reverse reactions are simple one-step processes obeying the rate equations $$ \text{Rate}(\rightarrow) = k_\rightarrow [\text{A}][\text{B}], \qquad\text{and}\qquad \text{Rate}(\leftarrow) = k_\leftarrow [\text{C}][\text{D}]. $$ At equilibrium both forward and reverse reactions are happening at the same rate because otherwise the concentrations would be changing. We can write $$ \text{Rate}(\rightarrow) = \text{Rate}(\leftarrow) \qquad\Rightarrow\qquad K = \frac{[\text{C}][\text{D}]}{[\text{A}][\text{B}]} = \frac{k_\rightarrow}{k_\leftarrow} $$ So, if $\Delta_r G^\circ$ is large and positive, $K\ll 1$, and $k_\leftarrow \gg k_\rightarrow$. The rate constant for the reverse reaction is much larger than that for the forward reaction. However, both reactions are still happening, and the rates of these reactions are equal.

As I said, you can't apply this argument directly to a multi-step reaction scheme, but the two links I gave above show how the same ideas apply in such a case. Also, most of the above material is covered in standard physical chemistry texts.

  • Yes, I realised that I had badly misread my book. Could you please tell me how the reaction happens despite having positive gibbs free energy? – Abcd Oct 23 at 5:44
  • I've added a few lines to my answer, to try and address this. Please let me know if it is not clear. – LonelyProf Oct 23 at 10:09
  • I m not sure your edit answered my query. My question is how can a reaction with positive value of Gibbs free energy happen. – Abcd Oct 24 at 4:59
  • I think negative value of Gibbs free energy is essential for anything to happen – Abcd Oct 24 at 5:00
  • I've added some more material which I hope covers this. There's quite a lot of detail. I suggest that you take some time to digest this, have a look at the links I've given, possibly consult a physical chemistry textbook to confirm the details, and then let me know if there are still any points that are not clear. – LonelyProf Oct 24 at 10:57

Your Answer

 
discard

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.