Regarding reversible phase transition of a real gas

Question

According to change in entropy for a reversible phase transition, say from liquid to gas at constant pressure $p$ and temperature $T$

: $\Delta S =\frac{\int dq_p}{T} $ which comes out to be $\Delta S =\frac{\Delta H_{vap}}{T_b}$. This means that this process is spontaneous since Entropy change is positive. But the Gibb's Energy change for this process is $0$. Why is that?

Change in entropy for this process suggests that this is a spontaneous process but Gibb's Energy change suggests it is not. Why does irregularity occur?

Answer 1

$$\Delta S_\mathrm{sys} =\frac{\Delta H_\mathrm{vap}}{T_b}$$

is the entropy change of the system.

$$\frac {\Delta G}{T_\mathrm{b}} = \Delta S_\mathrm{tot} = \Delta S_\mathrm{ext} + \Delta S_\mathrm{sys} = \Delta S_\mathrm{ext} + \frac{\Delta H_\mathrm{vap}}{T_b} = 0$$

is proportional to the total entropy change.

Answer 2

There are 2 unreasonable points your question raised.
Firstly, you made a mistake when you transformed $\triangle S = \frac{\int dq_p}{T}$, because $dq$ should equals $-dH_{sys}$. The above answer replicated that mistake too. Because $\triangle S_{sys}$ is state function, it can not be transformed like that, actually $\triangle S_{ext} = \frac{\int -dH_{sys}}{T}$, so $\triangle S_{tot} = \triangle S_{sys} - \frac{\triangle H_{sys}}{T}$. (those equations can be found in any physical chemistry books).
Secondly, you can not distinguish between standard-state Gibbs energy and Gibbs energy. We have that: $\triangle G = \triangle G^o + RTlnQ$ with Q is the proportion of component. In order to make one reaction spontaneous, Gibbs energy $(\triangle G)$ should equal 0, even though whatever $\triangle G^o$ is.