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Pretty much the title. I've heard it said that no (see, e.g here, answer by Lukas Schaedler). To quote his argumentation:

No. Exergonic reactions may be both exothermic or endothermic. Endergonic reactions are endothermic only.

Endergonic and exergonic relate to changes in free energy (delta G), while endothermic and exothermic are related to changes in enthalpy (delta H).

Gibb’s equation written in form of free energy (G), enthalpy (H) and entropy (S) changes:

delta G = delta H - T . delta S

Therefore, it’s possible for an exergonic reaction (delta G < 0; decrease in free energy) to be both exothermic (delta H < 0; heat is released) or endothermic (delta H > 0; heat is consumed). The second scenario may happen if :

T . delta S > delta H

On the other hand, endergonic reactions are endothermic only; delta H needs to be positive and greater than T . delta S in order to delta G > 0

But it's not really clear to me why. In particular, in the example given, why is $\Delta S \ge 0$ (which in my understanding would be necessary to force $\Delta G = \Delta H - T\Delta S$ to be negative)?

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    $\begingroup$ Reactions with ΔG negative are simply non existant. $\endgroup$
    – Maurice
    Sep 1 '21 at 16:47
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    $\begingroup$ @Maurice - Huh? The definition of an endergonic reaction is one where $\Delta G^\circ > 0$. There are plenty of reactions where this is not the case. $\endgroup$
    – Todd Minehardt
    Sep 1 '21 at 17:13
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    $\begingroup$ @Damian Birchler: Posts outside chemistry.se may move (or become defunct). Thus, it is better to repost (within reason) the argumentation clearly labeled as a quotation (the > sign at the beginning of the line yields the vertical bar). $\endgroup$
    – Buttonwood
    Sep 1 '21 at 17:33
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    $\begingroup$ @Todd Minehardt. Challenge : Which reaction can you give as an example of ΔG° < 0 ? $\endgroup$
    – Maurice
    Sep 1 '21 at 19:07
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    $\begingroup$ @Maurice - Any spontaneous reaction? I mean, the answer to the OP's query is that "endergonic reactions are defined as those where $\Delta G^\circ > 0$." That's it. It's a definition. $\endgroup$
    – Todd Minehardt
    Sep 1 '21 at 20:03
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The person who posted that answer on Quora was probably confused about the significance of $\Delta S$ as it relates to spontaneity. For isolated systems in general the sign of $\Delta S$ (the change in the entropy of the system) is the relevant criterion determining whether a process is spontaneous, since the entropy of the surroundings remains constant, so changes in the entropy of the universe are limited to the system.

$\Delta G$ is of course the relevant criterion for constant temperature and pressure processes. When the change is negative (exergonic) the process is spontaneous. This places no constraints on $\Delta H$ other than that it should satisfy

$$\Delta H \le T\Delta S$$

which can be interpreted as meaning that changes in the entropy of the surroundings must more than compensate for any potential decrease in the entropy of the system: if the entropy of the system decreases (is negative) then the entropy change of the surroundings ($-\Delta H/T$) must be positive and greater than this in magnitude:

$$\Delta S_\textrm{surr} \ge -\Delta S$$

Freezing of any pure substance above its melting point is an example of a process that is not spontaneous but exothermic. By extension, the reverse is also true: melting of a substance above its freezing point is exergonic (spontaneous) yet requires an input of heat (endothermic).

Note also that $\Delta G^\circ$ determines the position of equilibrium since $\Delta G^\circ = -RT \log K_\textrm{eq}$, but does not suffice to say whether the reaction is spontaneous. For that you need to compute $\Delta G = \Delta G^\circ + RT \log Q$, that is you need information about the starting concentrations.

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