Applying a Legendre transformation on $U = q - w$ we get the familiar $G = H - TS$. Making an innocent approximation delivers $\Delta G = \Delta H - T\,\Delta S$.

When one wants to predict the spontaneity of a chemical reaction, one gets the $\Delta H$ and the $\Delta S$ for the reaction in question from tables, and both refer to the system undergoing the transformation. Therefore we can write:

$\Delta G_\mathrm{sys} = \Delta H_\mathrm{sys} - T\,\Delta S_\mathrm{sys}$

In which:

$\Delta G$: Variation of the system's Gibbs energy

$\Delta H$: Variation of the system's enthalpy

$\Delta S$: Variation of the system's entropy

However: $\Delta S_\mathrm{sys} = \frac{q}{T}$, assuming the process is reversible (as usual).

So: $\Delta G_\mathrm{sys} = \Delta H_\mathrm{sys} - q$. If we are making predictions based on Gibbs energy, we are under constant pressure and so $\Delta H_\mathrm{sys}=q$.

Therefore: $\Delta G_\mathrm{sys} = q - q$, $\Delta G_\mathrm{sys} = 0$ and we are always on equilibria.

Where am I mistaken? I've already seen some say that some stuff refers to the surroundings, but that makes no sense to me. If both are at thermal equilibrium, $q$ is going to have the same effect on the entropy of the system and that of the surroundings, from $\Delta S = \frac{q}{T}$.

I'm a chemistry undergratuate and have seen no mention to anything related to this on common p-chem textbooks (Atkins, McQuarrie, Ball).

  • $\begingroup$ You suppose that the number of moles is constant. It is not the case in chemistry. In a chemical reaction ΔG° is measured by ΔG° = -zEF or by ΔG° = -nRTlnK. And ΔH is measured by ΔH = mCΔT. ΔG changes a lot with temperature. ΔH does not change so much. Both curves (ΔG° vs. T and ΔH vs T) intersects at O K. The difference is TΔS. $\endgroup$
    – Maurice
    Commented Feb 14, 2020 at 22:32
  • $\begingroup$ I believe my answer here directly explains why it's not the case that "$\Delta G_{sys} = q - q$, $\Delta G_{sys} = 0$ and we are always on equilibria.", while at the same time addressing your question about the entropy of the system vs. the entropy of the surroundings: chemistry.stackexchange.com/questions/124412/… $\endgroup$
    – theorist
    Commented Feb 15, 2020 at 0:53
  • $\begingroup$ The following thread in another forum specifically addresses your question is great detail: physicsforums.com/threads/… DrDu in post #2 does a masterful job of answering. $\endgroup$ Commented Feb 15, 2020 at 2:34

2 Answers 2


However: $\Delta S_{sys} = \frac{q}{T}$, assuming the process is reversible (as usual).

No. $\Delta S_{env} = \frac{q}{T}$, assuming the process is reversible (as usual). So you get $\Delta S_{universe} = -\frac{\Delta G}{T}$ for a process where the heat exchange with the environment is reversible (and there is no non-PV work).

For more background, see https://chemistry.stackexchange.com/a/114323

  • $\begingroup$ Why tho? Why is it that the entropy change of the system is not equal to its enthalpy change over temperature but the surrounding's is? If I were to insert some heat into the system wouldn't its entropy change be given by heat/temperature? It seems that there is some not clear distinction here. What I got is that the heat of one can be used to calculate the change in entropy of the other and vice versa but not the change in itself entropy. $\endgroup$ Commented Feb 15, 2020 at 0:24
  • $\begingroup$ @BananaAsker If your argument would hold then there could be no change of entropy in an insulated system. q would always be zero, so no change in entropy. However, all you have to do is to mix two substances, and the entropy increases. So there is a flaw in the argument. $\endgroup$
    – Karsten
    Commented Feb 15, 2020 at 3:48

If the process is reversible, $\Delta S_{universe}=0$. If in addition the temperature is constant (the "innocent" approximation), $\Delta S_{sys}=\frac{q}{T}$ and $\Delta S_{surr}=-\frac{q}{T}$. If the pressure is also constant and there is only pV work, then $q=\Delta H$ and $$\Delta S=\frac{\Delta H}{T} \tag{system}$$ Therefore $\Delta G=0$ for a reversible process at constant T and p.

If the process is irreversible, $\Delta S_{universe}>0$. If in addition the temperature is constant, $\Delta S_{surr}=-\frac{q}{T}$, but now $\Delta S_{sys} \neq \frac{q}{T}$. If the initial and final states of the system are the same as in the reversible process, then $\Delta S_{system}$ is equal to the change for the reversible process (one reason the 2nd law of thermodynamics is useful), so that $$\begin{align}\Delta S_{univ,irrev}>&\Delta S_{univ,rev}\end{align}\\ \Delta S_{sys,irrev}+\Delta S_{surr,irrev} > \Delta S_{sys,rev}+\Delta S_{surr,rev} \\ \Delta S_{surr,irrev} > \Delta S_{surr,rev} \\ q_{surr,irrev} > q_{surr,rev} \\ q_{irrev} < q_{rev} $$

(the convention used here is that, unless noted otherwise, q is the heat exchanged by the system and is positive for an endothermic process).

So the answer is that there is no error. A reversible process is one in which the system passes through an infinite series of equilibrium states, an admittedly odd conclusion.


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