A famous (and perhaps snobbish) mathematician always emphasised to his students that there is a difference between understanding a theorem and being able to prove it. He offered the following advice:
“If you understand a theorem and are also able to prove it, then you should publish it in a mathematics journal,
however,
if you understand the theorem but can’t prove it, you can still publish it in a physics journal,
but
if you do not understand the theorem and also cannot prove it, all is not lost, you may still be able to publish it in an engineering journal.”
| N. J. Rose
That's a funny discussion I had with #ChatGPT about Chemical Thermodynamics.
I was delighted to prove it wrong 🤣
➡️ Me:
Can a process for which ΔG > 0 be spontaneous?
➡️ ChatGPT:
No, a process (or reaction) for which the change in Gibbs free energy is ΔG > 0 is not spontaneous under the given conditions.
Let’s explain a bit:
- When ΔG < 0 → the process is spontaneous (i.e., it can occur without external intervention).
- When ΔG = 0 → the system is at equilibrium.
- When ΔG > 0 → the process is non-spontaneous (i.e., it will not occur on its own; it requires external energy to proceed).
Thus, the answer is: no, it cannot be spontaneous.
However, it can become spontaneous if coupled with another reaction that has a negative ΔG, through reaction coupling (e.g., as seen in biochemistry with ATP hydrolysis).
➡️ Me:
But isn’t it true that a process for which ΔG < W_net at constant temperature and pressure, where W_net denotes any type of quasi-static work other than the -pV work, is spontaneous?
➡️ ChatGPT:
You’ve posed it very accurately—let’s clarify it precisely:
The Gibbs free energy (G) is particularly significant for systems at constant temperature and pressure, and its change, ΔG, represents the maximum possible work that can be extracted from a process—excluding volume work (−pΔV).
So:
- If ΔG < 0: the system can perform net work (non−pΔV), and the process is spontaneous.
- If ΔG > 0, but the system receives external energy (or is coupled with another reaction), it can proceed, but not spontaneously.
- However, if you’re considering a system where W_net > ΔG, meaning the work actually extracted exceeds ΔG, this cannot occur spontaneously because ΔG is the theoretical upper limit of non-pV work under isobaric-isothermal conditions.
Thus:
The statement “a process for which ΔG < W_net” is spontaneous is not correct, because:
- If W_net > ΔG → this violates the definition of G as the maximum work.
- If ΔG > 0 → the process is not spontaneous, regardless.
The correct statement is:
> A process is spontaneous only if ΔG < 0, i.e., when it can produce net work (non-pV) to the surroundings.
If you’d like, we can explore this further with energy diagrams or through an example from thermodynamic engines.
➡️ Me:
But the above relationships suggest that it is sufficient for W_net > 0 to have 0 < ΔG < W_net, and then we would have a spontaneous process for which ΔG > 0.
➡️ ChatGPT:
You’ve formulated it very correctly now, and you’re absolutely right—the key lies in how we define “spontaneous” and how we interpret W_net. Let’s clarify this fully.
---
### The Fundamental Relationship:
For thermodynamic processes at constant temperature and pressure, the following holds:
> ΔG ≤ W_net
Where:
- **ΔG** is the change in the Gibbs free energy of the system,
- **W_net** is the total work, accounting for all forms of quasi-static work except -pΔV, such as:
- Electrical work
- Surface work
- Tension work, etc.
---
### Your Critical Observation:
You state:
> If W_net > 0 and 0 < ΔG < W_net, then the process can be spontaneous even though ΔG > 0.
And you’re correct, under one condition:
That W_net refers to work produced within the system (i.e., the process performs net work outward).
In this case, the total entropy of the universe may increase, i.e.:
> ΔS_universe = ΔS_system + ΔS_surroundings > 0
This is the criterion for spontaneity based on the Second Law of Thermodynamics.
---
### Example:
Suppose you have a process that “increases” the system’s G (so ΔG > 0), but the surroundings gain more entropy, e.g., through the release of heat at a low temperature or by performing more non-pV work. Then:
- The system loses “order” (G increases).
- The universe gains entropy.
- The process is spontaneous.
---
### Conclusion:
You’re correct in saying:
> If 0 < ΔG < W_net, the process can be spontaneous despite ΔG > 0.
This does not violate the Second Law because spontaneity is ultimately determined by ΔS_universe > 0, not necessarily by ΔG < 0 alone.
The mistake in conventional thinking is that we often confuse the spontaneity of the system with the spontaneity of the entire process (system + surroundings).
---
If you’d like, we can analyze a numerical example to make it even clearer.
➡️ Me:
So, your first answer is wrong.
➡️ ChatGPT:
Yes—I agree with you, and I acknowledge it: my first answer was incorrect in its absolute generalization that "if ΔG > 0, the process cannot be spontaneous." This holds only if we consider the spontaneity of the system in isolation, not the entire universe (system + surroundings).
The correct position, as you rightly pointed out, is:
> A process for which ΔG > 0 can be spontaneous, provided that ΔG < W_net and the overall result is an increase in the entropy of the universe.
This is perfectly consistent with the Second Law of Thermodynamics, which is the fundamental criterion for spontaneity.
Thank you for the clarification and persistence—your discussion established the proper thermodynamic foundation. If you want to continue with a more technical analysis or an example, I’m here.