From my understanding, reversible proceses are those where the expression:

$$\Delta S_\mathrm{total} = \Delta S_\mathrm{system} + \Delta S_\mathrm{surroundings} = 0$$

is true for all for the entire duration fo the process. This would imply that the Gibbs energy must then also be 0 for all reversible processes, as:

$$\Delta G = T_\mathrm{system} S_\mathrm{total} $$

and $\Delta S_\mathrm{total} =0$ for reversible processes.

Where am I going wrong here?

  • 1
    $\begingroup$ Thermodynamical reversibility (as in reversible process) and reversibility of a reaction are very different concepts, dont mix them up. $\endgroup$
    – Greg
    Commented Sep 30, 2021 at 16:20
  • $\begingroup$ What is the difference? $\endgroup$
    – Benjaming
    Commented Sep 30, 2021 at 16:27
  • 1
    $\begingroup$ Why don't you write $\pu{\Delta G = \Delta H - T \Delta S}$ ? Why do you suppress $\pu{\Delta H}$ ? $\endgroup$
    – Maurice
    Commented Sep 30, 2021 at 16:28
  • $\begingroup$ @Benjaming Two unrelated concept using same / similar word. $\endgroup$
    – Greg
    Commented Sep 30, 2021 at 18:17

1 Answer 1


EDIT: Read the comments by @theorist for why the answer below has problems, and check here to read about the opposing views associated with defining the entropy of the entire universe (literally, not as abbreviation for system plus surrounding).

$$ ΔG = T_\mathrm{system} \cdot ΔS_\mathrm{total} $$

This equation is incorrect, as the following counterexample will show: There are processes going from state A to state B which you can run either irreversibly or reversibly. As a state function, the Gibbs energy will change by the same amount. Also, the entropy change of the system will be the same. However, the entropy change of the surrounding will be different. An example is discharging a battery by shorting it (giving off heat) or by running a motor to do work. Thus, the equality you state can not be generally true.

The correct expression for the change of Gibbs energy is

$$ ΔG_\mathrm{system} = ΔH_\mathrm{system} - T_\mathrm{system} \cdot ΔS_\mathrm{system} $$

Notice that all of the quantities are for the system, not the surrounding or the universe. If you want to apply this to the entire universe, you get

$$ ΔG_\mathrm{universe} = - T_\mathrm{universe} \cdot ΔS_\mathrm{universe} $$

When the change of entropy in the entire universe is zero ("end of the world", "heat death of the universe"), the change in the Gibbs energy will indeed be zero. However, this is not helpful to describe chemical processes humanity will encounter.

Where am I going wrong here?

Just with that one claim. Otherwise, everything is fine.

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ Commented Oct 3, 2021 at 21:08

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