Why isn't ΔH = TΔS in ΔG = ΔH - TΔS?

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).





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


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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. 


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