Let me preface this question by saying that I probably made a mistake somewhere. However, I can't figure out what it is.
I was messing around with the Gibbs free energy equation to determine how the value of $K$ (equilibrium constant) varies with temperature.
Given \begin{align} K &= \exp\left({-\Delta G^° \over RT} \right). \label{5-7}\tag 1\\ \end{align}
From this it can be observed that when $\mathrm{d}G^\circ < 0$, $K > 1$. Conversely if $\mathrm{d}G^\circ > 0$, $K < 1$. This is also logical from a chemistry perspective.
Now if I differentiate $K$ over $T$, \begin{align} \frac{\mathrm{d}K}{\mathrm{d}T} &= \frac{\Delta G^\circ\mathrm{e}^{-\frac{\Delta G^\circ}{RT}}}{RT^2} \label{two}\tag 2\\ \end{align}
It can then be observed that the sign of the derivative is dependent only on the sign of $\Delta G^\circ$. That is, when $\Delta G^\circ < 0$, $\frac{\mathrm{d}K}{\mathrm{d}T} < 0$. This also makes sense from a chemistry perspective, because that means that as $T$ increases, $K$ decreases and therefore $[\text{reactants}]$ also decreases.
Now that last part reminded me of Le Chatelier's principle, so now by considering $$\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$$ and substituting this into equation \eqref{5-7}, \begin{align} K &= \exp\left(\frac{\Delta S^\circ}{R} -\frac{\Delta H^\circ}{RT}\right) \tag 3\\ \end{align} and once again differentiating with respect to $T$, \begin{align} \frac{\mathrm{d}K}{\mathrm{d}T} &= \frac{\Delta H^\circ\exp\left( \frac{\Delta S^°}{R}-\frac{\Delta H^°}{RT} \right)}{RT^2} \label{four}\tag 4\\ \end{align}
And now it seems that the derivative is only dependent on the sign of $\Delta H^\circ$. Therefore, when $\Delta H^\circ < 0$, $\frac{\mathrm{d}K}{\mathrm{d}T} < 0$.
Here's the problem: By considering all the equations, when $\frac{\mathrm{d}K}{\mathrm{d}T} < 0$, both $\Delta G^\circ < 0$ and $\Delta H^\circ < 0$. From a chemistry perspective, this means that if equilibrium $[\mathrm{reactants}]$ increases when temperature increases, I can conclude that $\Delta G^\circ < 0$ and also go further to say $\Delta H^\circ < 0$.
The last sentence above doesn't sit right with me for several reasons.
Given that $\Delta G^° = \Delta H^\circ - T\Delta S^\circ$, simply having a negative $\Delta G^\circ$ value should not necessitate a negative $\Delta H^\circ$ value, because a positive $\Delta G^\circ$ can be offset by a large increase in entropy.
Consulting a table of standard thermodynamic values (https://owl.oit.umass.edu/departments/Chemistry/appendix/thermodynamic.html) shows that this isn't true. For example, the process $$\ce{Cl2 (g) <=> Cl2 (aq)}$$ Has the calculated values $\Delta H^\circ = \pu{-23.0 kJ//mol}$ and $\Delta S^\circ = \pu{-102.1 J//K*mol}$. The given $\Delta G^\circ$ value is $\Delta G^\circ = \pu{+7.0 kJ//mol}$. (Calculating it using the equation $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$ gives $\Delta G^\circ = \pu{+7.44 kJ/mol}$, which is similar to the value in the table.) From this example one can observe a process where $\Delta G^\circ > 0$, $\Delta H^\circ < 0$. From equation \eqref{two}, this means that as temperature $T$ increases, $K$ increases and $\ce{[Cl2(aq)]}$ increases.
However, this would contradict equation \eqref{four}. Not only that, it seems to violate Le Chatelier's principle, because given that the forward reaction is exothermic, an increase in temperature should favor the backwards reaction and equilibrium $\ce{[Cl2 (g)]}$ should increase instead.
Overall, I am rather confused. I have a feeling there could be a simple arithmetic error somewhere, which would make this long question rather embarrassing. Either that or there are some implicit thermodynamic conditions that are not being considered. In any case I would appreciate any mathematical corrections but I would also like the answer/comments to use chemistry principles to address any chemistry-based misconceptions.
