I am having trouble conceptualizing endergonic reactions and their difference from endothermic reactions. An endergonic reaction is one that has a positive Gibbs free energy change, such that energy is taken in for the reaction to take place; in biological systems the very exergonic breakdown of ATP is a major source. The expression for Gibbs free energy (for a reaction at constant temperature and pressure) is given by: $$\Delta G = \Delta H - T\Delta S$$ It is known to me that some endothermic reactions can in fact be exergonic (at certain temperature thresholds) as the entropic increase will be able to counteract the enthalpy change. However, if the reaction is endothermic and associated with a decrease in entropy, the reaction is endergonic at all temperatures. I have a hard time understanding this; isn't heat energy a source of energy as well? If we raise the temperature high enough, surely there would be enough energy to drive the reaction. Intuitively, I imagine that heat energy can't flow into the system due to a thermodynamic law, but I can't seem to figure it out. Does it have to do with entropy or anything related to transfer of heat?
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$\begingroup$ if the reaction is endothermic ($ \Delta H> 0 $) and there is a decrease in entropy (I suppose you mean $ \Delta S <0 $) then we necessarily have $ \Delta G> 0 $ so the reaction is endergonic $\endgroup$– NicolasFeb 21, 2021 at 9:52
1 Answer
I think you can look at this entirely from the perspective of entropy (forget about the Gibbs free energy for a moment). You can divide the universe into two parts, the system and its surroundings.
When a change happens in your system, and some heat is exchanged with the surroundings then the change in entropy of the surroundings is: $$\Delta S_\mathrm{surr}=-\frac{\Delta H}{T}\tag{1}$$
(minus sign because if $\Delta H$ is positive, that means heat is taken in from the surroundings so the entropy of the surroundings must go down).
The change in entropy of the system itself is: ${\Delta S_\mathrm{sys}}$ (which you are referring to as $\Delta S$ in your question).
Now the total entropy change of the universe is: $$\Delta S_\mathrm{univ}=\Delta S_\mathrm{surr}+ \Delta S_\mathrm{sys}=-\frac{\Delta H}{T}+\Delta S.\tag{2}$$
If you have a reaction where $\Delta H$ is positive, then the first term (i.e. $\Delta S_\mathrm{surr}$) is negative. If the entropy change in the system is also negative (i.e. $\Delta S_\mathrm{sys} < 0)$ then the total entropy change of the universe becomes negative. Of course, the universe is an isolated system, so the second law forbids any change that reduces the entropy of the universe.
Now what happens when you increase $T$? Well… nothing. Assuming that the signs of enthalpy and entropy change of the system stay negative regardless of rising temperature, the total entropy change is going to stay negative. All you do by raising $T$ is decrease the magnitude of the first term.
If you look at the third equation, and multiply both sides by $T$ and change signs, you will basically get the expression for Gibbs free energy. In the end, the sign of Gibbs free energy basically tells you if the total entropy of the universe is going down or not when you do that change in the system.
So even if you raise the temperature, an endergonic reaction will not go, and heat won't flow into the system because that would mean reducing the total entropy of universe.
As an aside, I should mention here that all of this applies to one system. In real world, there are many things you can do to drive an endergonic reaction. For example you can couple it with another reaction/change that makes more entropy than it consumes, and that can drive the endergonic reaction. You can supply electricity (think of it as the entropy that was produced during the generation of electricity feeding the endergonic reaction).
Lastly, even if a reaction has $\Delta_\mathrm{r} G < 0$ that does not mean that you won't get any product. It just means that the equilibrium lies to the side of the reactants. The values of $\Delta_\mathrm{r} G,$ $\Delta_\mathrm{r} H,$ $\Delta_\mathrm{r} S$ that we write consider only the reaction happening, it does not include the entropy of mixing that comes from mixing of the products and reactants. So, even though the reaction is disfavoured, you might get a tiny amount of the product.
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1$\begingroup$ You may find IUPAC's brief guide on using italic and roman letters (PDF) useful. $\endgroup$– andselisk ♦Apr 18, 2021 at 17:25