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What are reversible reactions? What is the real difference between reversible and non-reversible reactions?

To be reversible, a reaction must be quasi-static, that is for every small change that occurs in the reaction, the thermodynamic state of the system is well defined. But, such quasi-static reactions are not actually possible. So, these reactions are irreversible. But, why are they irreversible? And why do we call them irreversible?

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  • $\begingroup$ Are you talking about reversible and irreversible chemical reactions or reversible and irreversible processes? There is a difference. $\endgroup$ – Chet Miller Sep 10 '17 at 20:33
  • $\begingroup$ I have an intuition that the meaning in physics will be more intuitive $\endgroup$ – PhyEnthusiast Sep 13 '17 at 10:52
  • $\begingroup$ @ChesterMiller, Why should that matter? Chemical reactions are just one specific example of physical processes, and the general ideas are exactly the same. $\endgroup$ – Siva Mar 24 '18 at 7:33
  • $\begingroup$ @Siva It matters because the words "reversible" and "irreversible" are used in a different context for chemical reactions than for thermodynamics processes. For chemical reactions, an irreversible reaction is one that goes to completion (has an infinite equilibrium constant) and a reversible reaction is one that does not go to completion (has a finite equilibrium constant). This has nothing to do with how the reaction process is carried out (say, spontaneously compared to using a Van't Hopf equilibrium box). $\endgroup$ – Chet Miller Mar 24 '18 at 12:19
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It seems to me that the question seeks an intuitive understanding, so I'll pitch my answer from that perspective.

Very generally, the second law of thermodynamics states that entropy cannot spontaneously decrease in any process (for the system and environment combined). According to statistical mechanics, that is just exceedingly unlikely, for a system with $\sim {10}^{23}$ particles.

In the limiting case where the process generates no entropy, it is considered reversible.

If entropy increased during the process, the process could not be reversed without somehow generating order of the system. And the second law of thermodynamics states that any procedure to generate order within that system will end up generating even more entropy somewhere else.

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In a quasistatic process, the difference between the external pressure and the internal pressure is infinitesimal for each infinitesimal change in the volume of the system.

$P_{ext}=P_{int} \pm dP $,

As you know that the term for work in terms of pressure is given by, $dW=PdV$. This means that work obtained will be maximum if the pressure is maximum for each infinitesimal change in volume.

In a reversible process, the work done in each step obtained is maximum since the external pressure in only infinitesimally greater (or smaller) than the internal pressure.

This enables us to connect the internal pressure and external pressure using the ideal gas law, which in turn enables us to derive work in terms of volume change, without knowing the pressure.

For non-reversible process, such thing is not possible. This process is instantaneous. Internal pressure won't have enough time to become almost equal to external pressure. Only, external pressure is a way to find the work.

For non-reversible process,

$W=P_{ext}(V_f-V_i)$

Most processes are reversible because in real world atmospheric pressure doesn't change with respect to the internal pressure of a system. Atmospheric pressure roughly remains constant. For a process to be infinitely slow (or reversible) external pressure has to change with each infinitesimal change in internal process, which ,as mentioned before, allows us to connect external pressure to internal pressure. But since external pressure doesn't change in practice, the process can't be slow. It will be fast enough to be considered 'not reversible'.

Consider a case where you put a weight on the piston of the system. You have no idea how the internal pressure will change when the system reaches equilibrium but even under these extreme circumstances, the gas pressure at the piston face is still (by Newton's 3rd law) equal to the force per unit area exerted by the piston on the gas, so you can find out the work done by system which comes out to be negative of the work done by the weight and atmospheric pressure, $P_{ext}$ . Such a process won't be reversible.

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