Why isn't ΔH = TΔS in ΔG = ΔH - TΔS?

Question

Applying a Legendre transformation on $U = q - w$ we get the familiar $G = H - TS$. Making an innocent approximation delivers $\Delta G = \Delta H - T\,\Delta S$.

When one wants to predict the spontaneity of a chemical reaction, one gets the $\Delta H$ and the $\Delta S$ for the reaction in question from tables, and both refer to the system undergoing the transformation. Therefore we can write:

$\Delta G_\mathrm{sys} = \Delta H_\mathrm{sys} - T\,\Delta S_\mathrm{sys}$

In which:

$\Delta G$: Variation of the system's Gibbs energy

$\Delta H$: Variation of the system's enthalpy

$\Delta S$: Variation of the system's entropy

However: $\Delta S_\mathrm{sys} = \frac{q}{T}$, assuming the process is reversible (as usual).

So: $\Delta G_\mathrm{sys} = \Delta H_\mathrm{sys} - q$. If we are making predictions based on Gibbs energy, we are under constant pressure and so $\Delta H_\mathrm{sys}=q$.

Therefore: $\Delta G_\mathrm{sys} = q - q$, $\Delta G_\mathrm{sys} = 0$ and we are always on equilibria.

Where am I mistaken? I've already seen some say that some stuff refers to the surroundings, but that makes no sense to me. If both are at thermal equilibrium, $q$ is going to have the same effect on the entropy of the system and that of the surroundings, from $\Delta S = \frac{q}{T}$.

I'm a chemistry undergratuate and have seen no mention to anything related to this on common p-chem textbooks (Atkins, McQuarrie, Ball).

Answer 1

However: $\Delta S_{sys} = \frac{q}{T}$, assuming the process is reversible (as usual).

No. $\Delta S_{env} = \frac{q}{T}$, assuming the process is reversible (as usual). So you get $\Delta S_{universe} = -\frac{\Delta G}{T}$ for a process where the heat exchange with the environment is reversible (and there is no non-PV work).

For more background, see https://chemistry.stackexchange.com/a/114323

Answer 2

If the process is reversible, $\Delta S_{universe}=0$. If in addition the temperature is constant (the "innocent" approximation), $\Delta S_{sys}=\frac{q}{T}$ and $\Delta S_{surr}=-\frac{q}{T}$. If the pressure is also constant and there is only pV work, then $q=\Delta H$ and $$\Delta S=\frac{\Delta H}{T} \tag{system}$$ Therefore $\Delta G=0$ for a reversible process at constant T and p.

If the process is irreversible, $\Delta S_{universe}>0$. If in addition the temperature is constant, $\Delta S_{surr}=-\frac{q}{T}$, but now $\Delta S_{sys} \neq \frac{q}{T}$. If the initial and final states of the system are the same as in the reversible process, then $\Delta S_{system}$ is equal to the change for the reversible process (one reason the 2nd law of thermodynamics is useful), so that $$\begin{align}\Delta S_{univ,irrev}>&\Delta S_{univ,rev}\end{align}\\ \Delta S_{sys,irrev}+\Delta S_{surr,irrev} > \Delta S_{sys,rev}+\Delta S_{surr,rev} \\ \Delta S_{surr,irrev} > \Delta S_{surr,rev} \\ q_{surr,irrev} > q_{surr,rev} \\ q_{irrev} < q_{rev} $$

(the convention used here is that, unless noted otherwise, q is the heat exchanged by the system and is positive for an endothermic process).

So the answer is that there is no error. A reversible process is one in which the system passes through an infinite series of equilibrium states, an admittedly odd conclusion.