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.

  1. 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.

  2. 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.

  • 1
    $\begingroup$ In (2) you assume that ΔG is constant. In (3-4) you assume otherwise. $\endgroup$ Aug 6, 2021 at 8:42
  • 1
    $\begingroup$ You might also consider reading FAQ: How can I format math/chemistry expressions on Chemistry Stack Exchange?. Seems like you know some of it already, but note that units like kJ/mol should be typed upright, not in italics. The section on "quantities with units" has more info. $\endgroup$ Aug 6, 2021 at 9:27
  • 1
    $\begingroup$ Thanks for the suggestions @orthocresol. The edits have been made. Regarding your first comment: I had thought that the standard Gibbs free energy is strictly defined at approximately 100 kPa, 298K, which apparently isn't the case $\endgroup$
    – Heat
    Aug 6, 2021 at 9:54
  • 1
    $\begingroup$ It was pointed out before, but I went ahead and fixed the MathJax. Here is a boiler plate comment with links: On Chemistry mathematical and chemical expressions can be formatted using MathJax (and LaTeX Syntax). If you want to know more, please have a look here and here. We prefer to not use MathJax in the title field, see here for details. $\endgroup$ Aug 6, 2021 at 21:19
  • 1
    $\begingroup$ Hopefully your problem was clarified with the comments you received. If you still feel like there is an outstanding question which you'd like addressed please edit your post (it might be reopened), or post a new question. $\endgroup$
    – Buck Thorn
    Aug 7, 2021 at 8:19