Entropy is maximum at equilibrium, so if we have an equilbrium chemical reaction like:

$$\ce{HI <=> H+ + I-}$$

does it mean that the container of $\ce{HI}$ has maximum entropy or the universe?

Also, constant entropy implies reversibility but I don't understand how the definition of reversibility applies to a system of $\ce{HI }$ and $\ce{H+, I-}$ in a container. I mean, there's nothing to actually reverse, is there? By reversibility I mean that the system can be brought to its original state by an infinitesimal change(s) in it's condition.

Could anyone provide an explanation for this?

Note: I have edited my full question because previously it was very unclear.


Irreversibility is more a practical than a theoretical concept, in my opinion. 'In theory' all reactions are reversible. Take this highly schematised reaction:

$$\ce{A + B <=> C + D}$$

We define the equilibrium constant as:$$K=\frac{[C][D]}{[A][B]}$$

Using Nernst we can now establish a relation between $K$ and the left-to-right change in Gibbs Free Energy, $\Delta G^0$:

$$\Delta G^0=-RT\ln K$$

(Of course, as you know $\Delta G^0=\Delta H-T\Delta S$)

Evaluate this Nernst function to conclude that the more negative $\Delta G^0$ is, the higher the value of $K$:

$$K=\mathrm e^\frac{-\Delta G^0}{RT}$$

For very negative values of $\Delta G^0$, $K\to+\infty$ and by the equilibrium constant equation, the concentration of the reagents at equilibrium is essentially $0$. Then we can write:

$$\ce{A + B -> C + D}$$

Such a reaction we would call irreversible. There is however no clear cut-off point and many reactions will have $K$ values that somewhat defy categorisation with respect to being 'reversible' or 'irreversible'.

  • 1
    $\begingroup$ There is no standard dividing line between reversible and irreversible reactions. Even from IUPAC. It is simply the case that it is more convenient to treat some reactions as irreversible (for example when the reverse reaction doesn’t contribute meaningfully to the composition of the final reaction mixture). $\endgroup$ Nov 21 '17 at 3:08

does it mean that the container of HI has maximum entropy or the universe?

At equilibrium, the entropy of the universe is maximal. There is a lot of other stuff going on in the universe, so if you can isolate your reaction and the surrounding (no matter or energy leaving the surrounding), you can also say the sum of entropy in the system and the surrounding is maximal if everything is at equilibrium.

If you would like to talk about equilibrium based on thermodynamic functions of the system, the Gibbs energy of reaction ($\Delta_r G = \frac{dG}{d\xi}$) would be appropriate for a reaction at constant temperature and pressure. Under these conditions, the reaction reaches equilibrium when the Gibbs energy is at its minimum. In the absence of non-PV work, the minimum of the Gibbs energy of reaction corresponds to maximizing the entropy of the universe.

The entropy of the reaction $\Delta_r S$, on the other hand, typically is not at its minimum when the reaction reaches equilibrium (unless $\Delta_r H$ happens to be zero, as in some transport reactions).

I don't understand how the definition of reversibility applies to a system of HI and H+,I− in a container.

Reversibility means that a process is done at near-equilibrium conditions, i.e. with minimal increase in entropy. If you are considering a process that is at equilibrium, it is certainly happening under reversible conditions. The final state is equal to the initial state - "nothing" happened, so you can certainly reverse that "nothing" without entropy increase - in fact, you don't have to do anything at all.

It is different for a process like an expansion of a gas. Depending on how you run the process, going back to the initial state will result in a large increase of entropy (irreversible) or hardy any increase (approaching a reversible cycle).

  • 1
    $\begingroup$ You probably know this, but for the OP I think it is important to note explicitly that the Gibbs free energy of the system is minimized at equilibrium only under conditions of constant T and p (since, under those conditions, minimizing G_sys corresponds to maximizing S_universe). That, of course, is why the Gibbs free energy is useful, since those are common reaction conditions. $\endgroup$
    – theorist
    Apr 23 '19 at 1:08

How do I relate the reversibility that we study in chemical equilibrium to that we study in thermodynamics?

I think I now understand what you're trying to say. The root of your problem is that the word "reversible" in chemical equilibrium is different from "reversible" in thermodynamics. They're not same as you are thinking instead.

Note that a reversible, thermodynamic, physical process is better known as a "slow" process. It is called "slow" because it takes infinite time to complete, because at each the system is in thermodynamic equilibrium. It is called a "reversible" because, if performed in the reverse direction in the same manner, you can retrieve 100% of the input energy that was input as work. This is different from "irreversible" processes, where some input energy is permanently lost.

A kinetic "reversible" process - the one you have shown - is better known as an "equilibrium" process instead, and is very different from a reversible, thermodynamic, physical process, as you may observe now.

Also, note that $\Delta S_\text{total}=0$ holds only for a thermodynamic, physical process (like a piston compressing a gas). Chemical reactions (like the one you've specified) instead always involve a change in entropy, even if they are isothermal.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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