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A problem asks, if you have a positive $\Delta S$ (positive change in entropy) for an exothermic (meaning a negative $\Delta H$, or negative change in enthalpy) reaction and temperature increased, what would happen to $K_{\mathrm{eq}}$ and $K_{\mathrm{eq}}/Q$ (where $K_\mathrm{eq}$ is the equilibrium constant and $Q$ is the reaction quotient)?

The answer states that $K_{\mathrm{eq}}/Q$ and $K_{\mathrm{eq}}$ would increase. This is because if you use the thermodynamic equation, $\Delta G = \Delta H - T \Delta S$ (where $\Delta G$ is the change in Gibbs free energy and $T$ is for temperature), as you increase temperature you get a more negative number (since $\Delta S$ is positive and $T\Delta S$ is being subtracted). This would lead to a more negative $\Delta G$. Then, using another thermodynamic equation, $\Delta G = RT\ln(Q/K_{\mathrm{eq}})$ (where $R$ is the ideal gas constant), if $\Delta G$ is decreasing (becoming more negative), then the term $(Q/K_{\mathrm{eq}})$ must also be going down. This means that $K_{\mathrm{eq}}/Q$ must be going up and therefore $K_{\mathrm{eq}}$ increases, as temperature increases for this exothermic reaction.

So my issue is, I thought for an exothermic reaction, where heat can be represented as a product, I thought increasing the temperature would shift the reaction to the left and decrease $K_{\mathrm{eq}}$. Could someone clarify this for me? I tried looking this up with Chemistry Libretexts, but it didn't seem to clarify anything. The table seemed to suggest that we do use heat as a component of the reaction in endothermic or exothermic reactions to predict the $K_\mathrm{eq}$ change.

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    $\begingroup$ Heat is not a product. $\endgroup$
    – Zhe
    Commented Apr 14, 2021 at 15:54
  • $\begingroup$ @Zhe Some intro chemistry texts suggest that students think of thermal energy as if it were a product or reactant to help understand Le Chatelier's Principle. It's not an unreasonable anzats. $\endgroup$
    – theorist
    Commented Apr 15, 2021 at 6:21
  • $\begingroup$ To get better understanding think of it as, "The reaction wants to release heat but if the ambient temperature is too high than it would be difficult for the reaction to do so, whereas at lower ambient temperature it will be easy for the reaction to liberate heat. " This is not accurate but just a way in which you can think intuitively about such conditions. $\endgroup$
    – Nisarg Bhavsar
    Commented Apr 15, 2021 at 6:24
  • $\begingroup$ @Jay The problem is wrong. The temperature-dependence of $K_{eq}$ is determined solely by the sign of $\Delta H$. To understand this, note that $\Delta G = -RT \ln K_{eq}= \Delta H - T \Delta S => \ln K_{eq} = \frac{\Delta S}{R}-\frac{\Delta H}{RT} $. I.e., you were looking at the T-dependence of $T \ln K_{eq}$ instead of $K_{eq}$. When you look at the latter, you see the T-dependence switches from the entropy term to the enthalpy term. $\endgroup$
    – theorist
    Commented Apr 15, 2021 at 6:31
  • $\begingroup$ @theorist I was taught that myself. It may give the correct answer, but the reasoning is wrong (obviously, since your comment nicely highlights the correct relationship). So I oppose it on those grounds. $\endgroup$
    – Zhe
    Commented Apr 15, 2021 at 16:26

2 Answers 2

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An exothermic reaction has a reduced equilibrium constant at higher $T$ because while the contribution of the change in the entropy of the system is a fixed quantity (for a small $T$ change), the effect of transferring heat to the surroundings is reduced at higher $T$ (because it causes a smaller change in the entropy of the surroundings). Mathematically this is described by the van't Hoff equation: $$\left(\frac{\partial \ln K_\textrm{eq}}{\partial T}\right)_p = \frac{\Delta H^\circ}{RT^2}$$

On the other hand the change in $\Delta G$ with temperature is given by $\Delta S$: $$\left(\frac{\partial\Delta G}{\partial T}\right)_p = -\Delta S$$

Since $\Delta G =\Delta G^\circ +RT\ln Q = -RT\ln (K/Q)$ one can write

$$\left(\frac{\partial\Delta G}{\partial T}\right)_p = \left(\frac{\partial\Delta G^\circ}{\partial T}\right)_p +R\ln Q\\ = -\Delta S^\circ+R\ln Q$$

Therefore if you want to see how $ K_\textrm{eq}$ depends on $T$ you should focus on $\Delta H^\circ$. If you want to see how $\Delta G$ depends on $T$ you should inspect $-\Delta S^\circ+R\ln Q$.

It is somehow unfortunate that the temperature dependence of $\Delta G$ and the equilibrium constant are different, because it leads to endless mix-ups, but it reminds you that they describe different things.

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  • $\begingroup$ Thank you for that response! So, are you saying for an exothermic reaction at higher temperature, that Keq decreases, while Keq/Q increases, because of the reduced effect of transferring heat at higher temperatures (meaning we disregard T*delta S). It's hard to deduce that from delta G = delta H - T * delta S . Would that mean that for Keq/Q to increase, while Keq decreases that Q decreases by a greater amount? Do you have a good educational website/webpage that would describe this scenario in great detail? This is very confusing. $\endgroup$
    – Jay
    Commented Apr 14, 2021 at 17:27
  • $\begingroup$ If you like the answer, please upvote and/or accept the answer. $\endgroup$
    – Mathew Mahindaratne
    Commented Apr 14, 2021 at 17:38
  • $\begingroup$ I still had two additional questions. $\endgroup$
    – Jay
    Commented Apr 14, 2021 at 17:46
  • $\begingroup$ @Buck Thorn: I'm just curious that van't Hoff equation is about standard enthalpy change, isn't it? $\endgroup$
    – Mathew Mahindaratne
    Commented Apr 14, 2021 at 17:46
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    $\begingroup$ @MathewMahindaratne You're right, thank you. $\endgroup$
    – Buck Thorn ♦
    Commented Apr 14, 2021 at 18:51
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You have a right to be confused. The first enthalpy is a product of a reaction and it is unjustifiably ignored because it is usually removed and discarded.

At equilibrium,

$$\Delta G = \Delta H -T\Delta S =0$$

For an exothermic reaction, $\Delta H$ and $\Delta S$ are both negative. This means that raising the temperature will decrease the equilibrium constant. This is exactly what le Chatelier's principle predicts.

The confusion generated by the different signs is just from the definition that negative energy is more favorable and positive $K_\mathrm{eq}$ is more favorable. Also adding to the confusion is that standard free energy an enthalpy changes are calculated with the reactants and products in their standard states. Both sets of concentrations are changed to the equilibrium conditions here.

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  • $\begingroup$ You can use MathJax to format here.. A full list of what you can do $\endgroup$
    – Safdar Faisal
    Commented Apr 15, 2021 at 6:12
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    $\begingroup$ In an exothermic reaction, $\Delta H$ is negative. By contrast, whether a reaction is exothermic places no constraint on $\Delta S$. It can be positive, negative, or zero. $\endgroup$
    – theorist
    Commented Apr 15, 2021 at 6:23
  • $\begingroup$ @NisargBhavsar What do you mean "for equilibrium, Δ𝐻 and Δ𝑆 both must be negative"? Remember that when we say that Δ𝐻 is negative for an exothermic rxn, we are referring to the standard enthalpy change: "exothermic reaction: A reaction for which the overall standard enthalpy change ΔH⚬ is negative." goldbook.iupac.org/terms/view/E02269 $\endgroup$
    – theorist
    Commented Apr 15, 2021 at 6:58
  • $\begingroup$ @NisargBhavsar The temperature-dependence is determined by the standard enthalpy change, $\Delta H^\circ$, which occurs in the expression $-RT \ln K_{eq} = \Delta G^\circ = \Delta H^\circ - T \Delta S^\circ$. When you write $\Delta G = \Delta H - T \Delta S=0$ at equilibrium, that is the actual reaction $\Delta G$, not the standard $\Delta G^\circ$. If a reaction is exothermic, then $\Delta H^\circ < 0$. However, $\Delta H^\circ < 0$ doesn't necessarily mean $\Delta H< 0$ at equilibrium. $\endgroup$
    – theorist
    Commented Apr 15, 2021 at 8:07

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