# 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. Feb 14, 2020 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/… Feb 15, 2020 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. Feb 15, 2020 at 2:34

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. Feb 15, 2020 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
Feb 15, 2020 at 3:48

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.