Why does increasing temperature for an endothermic reaction always increase equilibrium constant?

Question

According to Le Chatelier's Principle, increasing temperature for an endothermic reaction shifts the reaction towards the products.

However, unless there are things that I am missing, I believe there are three ways to imagine this phenomenon, albeit they all arise from the same equation/concept.

1. Classic $\Delta G = \Delta H - T \Delta S$

2. $\Delta S_{\text{sys}} + \Delta S_{\text{sur}} = \Delta S_{tot\text{}}$

3. Van't Hoff Isochore: $\ln\left(\frac{K_2}{K_1}\right) = \frac{-\Delta H}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)$

In the second case, it makes sense that an endothermic reaction would favor an increase in temperature because $\Delta S_{\text{sur}} = \frac{-\Delta H_{\text{sys}}}{T}$ assuming constant pressure. Thus an increase in temperature would mean an increase in total entropy change, indicating a more favorable change.

Van't Hoff's isochore also predicts that an increase in temperature favors the forward reaction of an endothermic reaction as well if you do out the math.

However, in the first case of the classic equation, the equation suggests that there is actually an entropy dependence rather than an enthalpy dependence of the gibbs free energy, which is related to the equilibrium constant by $-RT \ln(K) = \Delta G$.

It further confused me when considering that case 1 derives from case 2, and case 3 derives from case 1. I need help identifying my mistake or rectifying the confusion on why equilibrium is not entropy dependent.

Answer 1

The first equation, the corresponding change in G, and the equilibrium constant is supposed to be related to the following two initial and final states:

State 1: Pure reactants in separate containers at temperature T and pressure 1 bar

State 2: Pure products in separate containers at temperature T and pressure 1 bar

The $\Delta G$ you are talking about refers to the change between these two thermodynamic equilibrium states. But please note that these thermodynamic equilibrium states that are not in chemical reaction equilibrium with each other. So, to get the equilibrium constant for the reaction, you are using the $\Delta G$ for these two states which are not in chemical equilibrium. To understand how this plays out, you need to review the derivation of the relationship between this $\Delta G$ (actually denoted $\Delta G^0$ to distinguish the specific states 1 and 2) and the equilibrium constant K.

Now, when you change the temperature, all three quantities in the equation for $\Delta G^0$ change: T, $\Delta H^0$, and $\Delta S^0$, not just T. This is why the Van't Hopf equation gives the correct result for the effect of temperature on the equilibrium constant, and the other approach you described does not.

Answer 2

Saying that $- RT \ln K = \Delta H - T\Delta S$ shows that the dependence of $\ln K$ to $T$ is related to $\Delta S$ is flawed since both terms are multiplied times $-T$. Hence:

$$ \ln K = -\frac{\Delta H}{RT} + \frac{\Delta S}{R}$$

(which is part of the derivation of van't Hoff's equation, which results after derivation against $T$).

From here, you can see that the dependence of $\ln K$ with $T$ is associated with $\Delta H$, not $\Delta S$.