My teacher said that

The decrease in Gibbs free energy is the maximum non-mechanical work that can be obtained when the process occurs reversibly at constant temperature and pressure

But for a reversible process always $ΔG = 0$, so doesn't it contradict it? I mean he's just telling that maximum useful work in any process is zero.


3 Answers 3


You said:

But for a reversible process delta G is always zero.

This is not true; $\Delta G$ is always zero for a reversible process if it is carried out at constant temperature and pressure and the only kind of work involved is the $p-V$ work.

If the only kind of work involved is $p-V$, in the above conditions, it is true that the maximum work obtained is zero. In a general case, try to replace in your notes the work $w$ with $w = w_{pV} + w_{\text{no}\,pV}$. Where $w_{pV}$ is the $p-V$ work and $w_{\text{no}\,pV}$ is the work of any other types, to get the complete history.

In response to your comment:

Teoretically, every thermodynamic problem can be solved by using internal energy function. Many times, in practice is convenient to use others variables. It can be done by obtaining new functions that depends of the variables that we want to. We do that by apling Legendre transformations (http://en.wikipedia.org/wiki/Legendre_transformation) of the internal energy function. One posibility is to get the Gibbs energy.

The energy energy function naturally takes different forms depending on the system. I mean, if there are works others than $p-V$ they must to be included to account for energy changes.

The general form of a kind of work is: $X\cdot dx$, where $X$ is a generlized force (an intensive variable) and $x$ a generalized variable of displacement (an extensive variable), so we have for the change in internal energy ($dU$):

$ dU = \sum _{i=1} ^k X_i\cdot dx_i $

As the energy function change when we add non $p-V$ work, its Legendre transforms corresponding to $G$ also change, so in this case: $ G \neq H - TS$.

Finally the way for calculating $\Delta G$ depends on the works involved and the specific problem.

  • $\begingroup$ Thanks! But then, for reversible process invloving non mechanical work how ΔG is calculated? $\endgroup$ Feb 20, 2015 at 19:37
  • 2
    $\begingroup$ You are welcome! I reply you in the answer above. $\endgroup$ Feb 21, 2015 at 0:19

Reversible process vs reversible reaction

A reversible reaction just means that forward and reverse rate are significant. The reaction will show net changes in concentrations until it reaches a state of equilibrium, where the forward and reverse rate are the same and ΔG = 0.

A reversible process is one where there is no change in entropy, i.e. you carefully choose conditions so that no objects fall off a cliff, no cold objects come in contact with hot objects, and you don't mix solutions of different concentrations.

This is an ideal situation where nothing would happen. You approximate this by imposing very small changes at a time so that the things you are interested in do happen but with very little entropy production.

But for a reversible process always ΔG=0, so doesn't it contradict it?

A reversible process has ΔS(universe) = 0, not ΔG = 0. If ΔG were zero, you would not be able to do any work. The way you can achieve this is, for example, in an electrochemical reaction where you keep the current very low. Under these circumstances, you can measure the electromotive force (emf), and can calculate the work an electron is capable of doing when going from anode to cathode. The chemical reaction is not at equilibrium, but you are running the process in a reversible way (which means you could reverse the direction of the electron by putting in just about the same amount of work as you got out when it went the other direction).

Why does it have to be a reversible process?

A closed system (such as a chemical reaction in a closed reaction vessel) can exchange energy with the surroundings in the form of work or heat. If you run the process in an irreversible way, more energy will dissipate as heat, and the system will do less work. That is why you would not measure the maximal work in this case.

The process is called irreversible because you can't reverse the process. Some of the work you need has not been transferred in the forward process as work, but as heat. So that energy is now unavailable. (In a simpler argument, the entropy of the universe increased in one direction. To get back to the initial state, you would have to decrease the entropy of the universe. This would be a contradiction with the 2nd law of thermodynamics).

  • $\begingroup$ Hi Karsten! I am actually confused when original PO said:' The decrease in Gibbs energy is...when the process occurs reversibly at constant p And T.' Is this true? Must the process be reversible for its delta G to be maximum non-expansion work? If we work from G=H-TS until dG=dw(add,max), I see the two constraints imposed are constant T and constant p only, but no constraint on the process must be reversible. $\endgroup$
    – TheLearner
    May 30, 2019 at 1:36
  • $\begingroup$ @The99sLearner If the entropy of the universe increases, it means you have "wasted" some of the available Gibbs energy (in the form of heat instead of work). See chemistry.stackexchange.com/questions/112900/… $\endgroup$ May 30, 2019 at 1:45

The term "reversible process" does not mean that $\Delta G = 0$.

Reversible process is an imaginary or idealized process which occurs in infinitesimally small steps, such that the system very close to equilibrium throughout the process.

However, the infinite number of infinitesimally small steps would add up to an finite change.

  • $\begingroup$ In a reversible process ΔS total = 0, and ΔG = - TΔS total. So why ΔG is not zero in reversible process? I mean what's wrong in this? $\endgroup$ Feb 19, 2015 at 17:08
  • $\begingroup$ "total" means system + surroundings. The statement your teacher told you is about how much work the system can do (on the surroundings). $\endgroup$
    – DavePhD
    Feb 19, 2015 at 17:09
  • $\begingroup$ In normal language, a reversible process is one that will undo (and possibly redo) itself, that is, at roughly equal parts equilibrium. Such a 'reversible process' 'tends' to have a very small DG, largely because of DG= -RT ln K and the closer to DG=0 is the process is, the closer to K=1 is. However, strictly speaking, 'reversible process' means, chemically, a process that may be reversed. In this context, there is no explicity requirement that DG = 0. $\endgroup$
    – Lighthart
    Feb 19, 2015 at 17:10
  • $\begingroup$ I disagree with @Lighthart $\endgroup$
    – DavePhD
    Feb 19, 2015 at 17:19
  • $\begingroup$ @DavePhD: About my interpretation of the common language meaning of reversible or something else? $\endgroup$
    – Lighthart
    Feb 19, 2015 at 17:20

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