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I understand that entropy is defined as $\mathrm{d}S=\mathrm{d}q/T$ for a reversible change but I fail to see why $-G/T$ and $-A/T$ can be thought of as 'disguised' entropies. This is a question on my problem sheet at university and my tutor has hinted: "You need to think about the total entropy change for a process".

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  • $\begingroup$ Do you really mean $-G/A$ and not $-G/T$? I ask because I think you are talking about the Massieu Potential and the Planck Potential (see here). $\endgroup$ – Philipp Nov 23 '14 at 17:32
  • $\begingroup$ are you talking about Gibbs and Helmholz free energy? $\endgroup$ – bon Feb 17 '15 at 16:45
  • $\begingroup$ @RobChem So is it or is it not a question about free entropy? $\endgroup$ – getafix Sep 24 '16 at 2:42
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The fundamental equation of (chemical) thermodynamics $\ce{\Delta G} = \Delta H – {T\Delta S}$ may be divided by
(-T):

$$\ce{-\frac{\Delta G}{T}} = -\frac{\Delta H_{sys}}{T} + {\Delta S}_{sys}$$

In this equation $\frac{\Delta H_{sys}}{T}$ is the entropy change of the system by “heat” exchange with the surroundings, hence $\frac{\Delta H_{sys}}{T} = -\Delta S_{surroundings}$. On this basis, the above equation may be identified as:

$$\Delta S_{total} ={\Delta S_{surroundings} + \Delta S}_{sys}$$

$\Delta S_{sys}$ and $\Delta S_{surroundings}$ strive mutually to a maximum of $\Delta S_{total}$.

$\ce{\Delta G}$ is a disguised entropy change, because $\ce{\Delta H}$ is intrinsically an entropy change too, as explained above.

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