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Suppose a reaction has $\Delta S > 0$ and $\Delta H < 0$.

The equation $\Delta G = \Delta H - T\Delta S$ predicts that as the temperature increases, the reaction becomes more favorable because the $T\Delta S$ term becomes larger and $\Delta G$ becomes more negative.

However, the equation $$\ln(\frac{K_2}{K_1}) = -\frac{\Delta H}{R}(\frac{1}{T_2}-\frac{1}{T_1})$$ as well as Le Châtelier's principle, predict that as the temperature increases, K will decrease (and the reaction shift to the left, thereby becoming less favorable and $\Delta G$ becomes more positive).

These two claims seem to be inconsistent, but I can't identify the error in my reasoning.

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You should use the van't Hoff equation together with the standard enthalpy of reaction $\Delta H^\circ$ for the process to predict the response to a temperature change. This should provide you with a prediction consistent with Le Châtelier's principle that an endothermic reaction is pushed in the direction of products as $T$ increases.

The general equation for the Gibbs' free energy $$\Delta G = \Delta H - T \Delta S$$ should not be relied upon to determine the response of the reaction equilibrium constant to $T$. The expression is valid at constant $T$, but this is not the problem with using this equation, rather it is its generality: it is not limited to describing reactions at equilibrium. Note that $\Delta G = 0$ for a reaction at equilibrium.

On the other hand, you could consider using $$\Delta G^\circ = \Delta H^\circ - T \Delta S^\circ$$to guide you, and the result should be consistent with analysis based on the van't Hoff equation. It is simpler however to work with the van't Hoff equation directly. In fact, assuming $\Delta H^\circ$ and $\Delta S^\circ$ are independent of $T$ you can use fairly straightforward mathematical manipulations to arrive at the usual integrated form of the van't Hoff equation.

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  • $\begingroup$ Why shouldn't the equation for the Gibb's free energy be used to determine the response to T? $\endgroup$ – DrPepper Mar 10 at 18:54
  • $\begingroup$ @DrPepper Well, I give one reason in the answer: you can use $\Delta G^\circ$, but for a reaction at equilibrium $\Delta G = 0$ which is not very useful. $\endgroup$ – Buck Thorn Mar 10 at 18:57
  • $\begingroup$ If you were given delta H and delta S, and wanted to calculate delta G at a specific temperature, then you would use delta G = delta H - T * delta S. Doesn't this determine the equation's response to temperature? Why is it incorrect? $\endgroup$ – DrPepper Mar 10 at 19:00
  • $\begingroup$ @DrPepper Well, $\frac{\partial \Delta G}{\partial T} = - \Delta S$, so no.... $\endgroup$ – Buck Thorn Mar 10 at 19:04
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    $\begingroup$ $dG = dH - TdS -SdT$. The equation $dG = dH - TdS$ holds when T is constant. $\endgroup$ – Buck Thorn Mar 10 at 19:13
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The equation $\Delta G = \Delta H - T\Delta S$ predicts that as the temperature increases, the reaction becomes more favorable because the $T\Delta S$ term becomes larger and $\Delta G$ becomes more negative.

It does predict that $\Delta G$ becomes more negative (for example for the standard state). This also means that the maximal work available from the reaction becomes larger. If you equate favorability with $\Delta G$ or maximum work, you could also say the reaction becomes more favorable. It does not mean, however, that the equilibrium constant increases. This is because the equilibrium constant depends on both $\Delta G$ and the temperature. Both terms change because we are changing the temperature and $\Delta G$ is temperature-dependent:

$$ \mathrm{ln}(K) = - \frac{\Delta G^\circ}{RT}$$

It is fine to equate more negative $\Delta G^\circ$ with larger equilibrium constant when comparing two processes at the same temperature, though.

However, the equation $$\ln(\frac{K_2}{K_1}) = -\frac{\Delta H}{R}(\frac{1}{T_2}-\frac{1}{T_1})$$ as well as Le Châtelier's principle, predict that as the temperature increases, K will decrease (and the reaction shift to the left, thereby becoming less favorable and $\Delta G$ becomes more positive).

This is all true except for the last bit about $\Delta G$ becoming more positive. $\Delta G$ at equilibrium concentrations will be zero at both temperatures, and $\Delta G^\circ$ will become more negative, as you correctly stated in your question (assuming we can neglect the temperature dependence of enthalpy and entropy of reaction). Again, the misconception is linking changes in $\Delta G$ to changes of K at two different temperatures.

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