What is wrong in this argument that dG must always be zero?

Question

Under constant temperature and pressure, the change of the Gibbs free energy can be written as

$$\mathrm dG_\textrm{sys} = \mathrm dH_\mathrm{sys} - T\,\mathrm dS_\mathrm{sys}$$

And in the textbook that I have (Theodore L. Brown, Pearson), when deriving this equation, they use the equation

$$\mathrm dS_\mathrm{univ} = \mathrm dS_\mathrm{sys} - \mathrm dH_\mathrm{sys}/T$$

I can understand to this step but the question occurs. Can't we write $\mathrm dS_\mathrm{sys}$ as $\mathrm dH_\mathrm{sys}/T$ also? If this T and the T that is used in the equation

$$\mathrm dS_\mathrm{univ} = \mathrm dS_\mathrm{sys} - \mathrm dH_\mathrm{sys}/T$$

are the same, than shouldn't $\mathrm dG$ always be zero? At which point am I wrong?

Answer 1

Your textbook's derivation is done under the assumption of constant $T$, which means $T_{sys} = T_{surr} =T$. However, this does not mean $dG_{sys}$ is always zero. Let's start with the following:

$$dS_{univ}=dS_{sys}+dS_{surr}= \frac{\text{đ}q_{rev, sys}}{T_{sys}}+\frac{\text{đ}q_{rev, surr}}{T_{surr}}$$

Since heat flow always affects the surroundings reversibly (the surroundings are in the limit of being infinitely large, and thus heat flow affects them only infinitesimally), the reversible heat flow into the surrondings equals the actual heat flow, which is the negative of the actual heat flow into the system. And since $T_{sys} = T_{surr} =T$, we have:

$$\frac{\text{đ}q_{rev, surr}}{T}=-\frac{\text{đ}q_{sys}}{T}$$

Which gives:

$$dS_{univ}=dS_{sys}+dS_{surr}= \frac{\text{đ}q_{rev, sys}}{T}+\frac{\text{đ}q_{rev, surr}}{T}=\frac{\text{đ}q_{rev, sys}}{T} -\frac{\text{đ}q_{sys}}{T}$$

Under the restrictions of constant $p$, and no non-$pV$ work,

$$\text{đ}q_{sys}=dH_{sys}$$

Hence:

$$dS_{univ}=\frac{\text{đ}q_{rev, sys}}{T} -\frac{dH_{sys}}{T} = dS_{sys} - \frac{dH_{sys}}{T},$$

which is what your textbook provides.

Note, however, that

$$\frac{dH_{sys}}{T} = \frac{\text{đ}q_{sys}}{T} \ne \frac{\text{đ}q_{rev,sys}}{T}=dS_{sys},$$ unless the process is reversible, i.e., non-spontaneous, in which case we could replace the "$\ne$" with "=", and $dG_{sys}$ would equal $0$.

This connects directly to the great beauty of using $\Delta G_{sys}$ at constant $T \text{ and } p$, when there is no non-$pV$ work: Under those conditions, $\Delta G_{sys} = -T \Delta S_{univ}$. I.e., under those conditions, we can use $\Delta G_{sys}$ (something we can calculate) as a surrogate $\Delta S_{univ}$, whose value is the criterion for spontaneity (and is something we typically can't calculate directly).

Here's how we get there:

If the system is at constant $T$: $$\Delta G_{sys} = \Delta H_{sys} - T \Delta S_{sys}$$

That's because: $G=H-TS => dG = dH - d(TS) = dH -TdS -SdT$; at const $T, dG= dH -TdS $; note that a constant-$p$ restriction is not required to write this.

As we mentioned above, if we add the restrictions of constant $p$, and no non-$pV$ work, we obtain:

$$\Delta H_{sys} = -T \Delta S_{surr}$$

This in turn gives:

$$\Delta G_{sys} = -T \Delta S_{surr} - T \Delta S_{sys} = -T \Delta S_{univ}$$

Answer 2

For a microscopic step at constant $T$ and $p$ $$\mathrm dG=0\tag{constant $T$ and $p$}$$ implies:

1. reversibility (equilibrium)
2. $\mathrm dS_\mathrm{univ} = 0$
3. $\mathrm dH_\mathrm{sys} = T\,\mathrm dS_\mathrm{sys}$ since $\mathrm dG = \mathrm dH_\mathrm{sys} - T\,\mathrm dS_\mathrm{sys} \tag{constant $T$ and $p$}$

The derivation you suggest seems strange.

A derivation that shows why the Gibbs free energy serves as a condition of spontaneity under constant $T$ and $p$ could start from the expression

$$\begin{align} \mathrm dS_\mathrm{univ} &= \mathrm dS_\mathrm{sys} + \mathrm dS_\mathrm{surr} \\ &= \mathrm dS_\mathrm{sys} - \frac{\mathrm dq}{T} \\ &\ge 0 \end{align}$$

Combining this result with the 1st Law leads to the condition

$$\mathrm dG\le 0 $$