If for reversible process, ΔS= 0 for universe (system+surroundings), why would there be non-zero change in gibbs energy?

Question

As Second Law says, the entropy must increase for universe

$$\Delta S_\textrm{universe}\ge 0$$

Now, we know, $$\Delta G= -T\Delta S_\textrm{universe}\le 0^\dagger$$

For reversible process, $$\Delta S_\textrm{reversible, universe} = 0$$

That would mean $$\Delta G_\textrm{reversible}= 0$$

But does it happen so?

Peter Atkins, in his book writes:

At constant temperature and pressure, for a reversible process: $\mathrm dG = \mathrm dw'_\rm{rev}\;.$

Now, isn't it contradicting that $\mathrm dG\ne 0$ for reversible process? If $\Delta G= -T\Delta S_\textrm{universe}$, then wouldn't $\mathrm dG= 0$ for $\mathrm dS_\textrm{universe}= 0$ during reversible process?

---

\begin{align}^\dagger \Delta S_\textrm{universe}&=\Delta S_\textrm{system}+ \Delta S_\textrm{surroundings} \\ &= \Delta S_\text{system}+ \frac{-\Delta H_\text{system}}{T} \\ \implies -T\Delta S_\text{universe} &= \Delta H_\text{system} - T\Delta S_\text{system} \\&=\Delta G \end{align}

Answer 1

The equation in Atkins' book refers to the system only, and not to the universe, and, moreover, in this equation, w' specifically refers to non-PV work, like electrical.

Answer 2

The proper equation is $$\mathrm dG=\mathrm dH-\mathrm d(TS)$$ Since $H=U+pV$, we rewrite the above equation as $$\mathrm dG=\mathrm dU+(p\mathrm dV+V\mathrm dp)-T\mathrm dS-S\mathrm dT$$ Furthermore, we know from the first law that $$\mathrm dU=\delta Q + \delta W$$ and $$\Delta S=\int{\frac{\delta Q_\text{rev}}{T}}$$ which is equivalent to writing $$T\mathrm dS=\delta Q_\text{rev}$$ so, putting it all together and remembering that temperature is constant, we have $$\mathrm dG=T\mathrm dS+\delta W_\text{rev}-T\mathrm dS$$ Then, $$\mathrm dG=\delta W_\text{rev}$$ where the pressure-volume work went to zero because, as you point out in one of your comments, all of this is only valid for zero expansion work.

I hope this kind of answers your question, but you seem to have a pretty good grasp on this already.

---

Also, it probably shouldn't be written that $$\mathrm dG=\mathrm dW_\text{rev}$$ because this seems to imply that integrating this expression will have you take an integral with respect to work, but this cannot be done because work is not a state function. You would figure this out, however, when trying to figure out what values of work to use as the boundaries in that integral.