If for reversible process, ΔS= 0 for universe (system+surroundings), why would there be non-zero change in gibbs energy? 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}&amp;=\Delta S_\textrm{system}+ \Delta S_\textrm{surroundings} \\ &amp;= \Delta S_\text{system}+ \frac{-\Delta H_\text{system}}{T} \\ \implies   -T\Delta S_\text{universe} &amp;= \Delta H_\text{system} - T\Delta S_\text{system} \\&amp;=\Delta G  \end{align}
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.