# 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).

• You suppose that the number of moles is constant. It is not the case in chemistry. In a chemical reaction ΔG° is measured by ΔG° = -zEF or by ΔG° = -nRTlnK. And ΔH is measured by ΔH = mCΔT. ΔG changes a lot with temperature. ΔH does not change so much. Both curves (ΔG° vs. T and ΔH vs T) intersects at O K. The difference is TΔS. – Maurice Feb 14 '20 at 22:32
• I believe my answer here directly explains why it's not the case that "$\Delta G_{sys} = q - q$, $\Delta G_{sys} = 0$ and we are always on equilibria.", while at the same time addressing your question about the entropy of the system vs. the entropy of the surroundings: chemistry.stackexchange.com/questions/124412/… – theorist Feb 15 '20 at 0:53
• The following thread in another forum specifically addresses your question is great detail: physicsforums.com/threads/… DrDu in post #2 does a masterful job of answering. – Chet Miller Feb 15 '20 at 2:34

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.

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

• Why tho? Why is it that the entropy change of the system is not equal to its enthalpy change over temperature but the surrounding's is? If I were to insert some heat into the system wouldn't its entropy change be given by heat/temperature? It seems that there is some not clear distinction here. What I got is that the heat of one can be used to calculate the change in entropy of the other and vice versa but not the change in itself entropy. – BananaAsker Feb 15 '20 at 0:24
• @BananaAsker If your argument would hold then there could be no change of entropy in an insulated system. q would always be zero, so no change in entropy. However, all you have to do is to mix two substances, and the entropy increases. So there is a flaw in the argument. – Karsten Theis Feb 15 '20 at 3:48