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?


2 Answers 2


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

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.


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.

  • $\begingroup$ It's unclear what you're trying to say. According to web.mit.edu/16.unified/www/FALL/thermodynamics/notes/… the change in entropy of the system for a reversible process is in fact $\int \frac{dq}{T}$. It was just that in this case the pressure and temperature were constant because of the phase transition. So the heat flow is equal to the change in enthalpy $\Delta H_{sys}$. And for $\Delta G$ I applied the equation $\Delta G= \Delta H - T\Delta S$. $\endgroup$
    – Eyy boss
    Oct 2, 2020 at 18:00
  • $\begingroup$ Q is defined as an amount of heat transferred to out of the system. However, change of enthalpy is defined as change of heat of the system. So, $\triangle H_{system}$ must have different sign regarding to $Q$ or $\triangle H_{system} = - Q$. $\endgroup$ Oct 2, 2020 at 18:36
  • $\begingroup$ Additionally, we assume that: $\triangle S_{tot} = \triangle S_{sys} + \triangle S_{ext} = \triangle S_{sys}+ \frac{-\triangle H_{sys}}{T}$<br/>So: $\triangle H_{sys}-T\times \triangle S_{sys} = -T\times \triangle S_{tot} =-\triangle G^o_{sys}$ (let's remember $\triangle G^o$, not $\triangle G$) $\endgroup$ Oct 2, 2020 at 18:41
  • $\begingroup$ NO!! change of enthalpy IS NOT defined as change of heat of the system at all. This statement is only true for isobaric process. $\Delta H=q_p$, $\Delta H$ will not be equal to actual heat transfer in the system. Q is not defined as the amount of heat transferred out of the system. It is simply the transfer of heat in or out of the system. It doesn't matter if it is coming in or going out. $\endgroup$
    – Eyy boss
    Oct 2, 2020 at 18:43
  • $\begingroup$ $Q_p$ and $Q$ are different, my friend. These make physchem so hard. $\endgroup$ Oct 2, 2020 at 18:47

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