At constant pressure, which is the usual condition for chemical reactions, heat absorbed/released by the system is the same thing as enthalpy change. According to the Gibbs equation,

$$\mathrm dG = \mathrm dH - T\,\mathrm dS = \mathrm dH - đq$$But, $\mathrm dH=\mathrm đq$ at constant a pressure. So is the Gibbs energy change $\mathrm{d}G$ simply zero for any reaction?


2 Answers 2


You wrote:

$$\mathrm{d}G = \mathrm{d}H - T\,\mathrm{d}S = \mathrm{d}H - đq$$

There are several issues with this. I don't wish to co-opt other answers and comments, but allow me to mention these here, since my answer stands as the accepted (which I have no power to change):

  • You haven't assumed constant temperature, which means that there is an additional term $-S\,\mathrm{d}T$ which you did not take into account;
  • You haven't assumed constant composition, which likewise leads to another term $\sum_i \mu_i\,\mathrm{d}n_i$.

In fact, these considerations (particularly the second) are more important than my original point, which is retained below. See Kexanone's answer for discussion of these points. In the following discussion, one should assume constant temperature and composition.

In the second equality, where you equate $đq$ with $T\,\mathrm{d}S$, this assumes that the process is reversible. Without going into too much detail (it can be found easily elsewhere), entropy is defined by the following equation:

$$\mathrm{d}S = \frac{đq_\mathrm{rev}}{T}$$

where the subscript $\mathrm{rev}$ indicates a reversible process. It can be shown that for an irreversible process,

$$\frac{đq_\mathrm{irrev}}{T} < \frac{đq_\mathrm{rev}}{T} = \mathrm{d}S$$

This result is the second law of thermodynamics. Therefore, for an irreversible process, $T\mathrm{d}S > đq$.

You are right in saying that at constant pressure, $\mathrm{d}H = đq$. Putting it all together, we have:

$$\mathrm{d}G = \mathrm{d}H - T\mathrm{d}S = đq - T\mathrm{d}S < 0 \qquad \text{(irrev. process; closed system; }p, T\text{ const.)}$$

Therefore, the direct answer to the question is: $\mathrm{d}G$ is, in general, not equal to $0$ because $T\mathrm{d}S \neq đq$.

In fact, you may notice that the condition $\mathrm{d}G < 0$ is the condition for spontaneity under constant $p$ and $T$. This is no coincidence. The concepts of spontaneity (in the context of $G$) and irreversibility (in the context of $S$) are extremely similar, in that they both refer to a tendency for a process to proceed.

For a reversible process, $T\mathrm{d}S = đq$. Following the logic that you used in your question, this means that $\mathrm{d}G = 0$, which is correct. If $\mathrm{d}G = 0$, the forward process cannot be spontaneous (that would require $\mathrm{d}G < 0$) and the reverse process cannot be spontaneous (that would require $\mathrm{d}G > 0$).

This can only mean one thing: the system is in equilibrium. Indeed, the condition for equilibrium at constant $p$ and $T$ is $\mathrm{d}G = 0$. And at this point, it should not come as a surprise that a reversible process is, by definition, one that is at equilibrium throughout.

  • 1
    $\begingroup$ @orthocresol What if we want to include the number of particles? I mean $$dH=TdS + Vdp + \sum_i \mu_i dn_i $$ so now $dq\neq dH$. $\endgroup$
    – ado sar
    Mar 5, 2020 at 16:04
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    $\begingroup$ For isobaric heat transfer: $$dS = \frac{dH}{T} \geq \frac{\delta q}{T}$$ $dS = \frac{dH}{T}$ is correct even for an irreversible processes. $\endgroup$
    – Kexanone
    Feb 7, 2021 at 9:33

The accepted answer is incorrect. It has nothing to do with reversibility. The underlining issue is that your $dG$ is incomplete: $$dG = dH - TdS - SdT$$

You are missing $-SdT$, so at constant pressure you don't end up with zero.

As "ado sar" has commented in the accepted answer:

$$dH = TdS + Vdp + \sum_i \mu_i \cdot dn_i$$

Hence, you end up with: $$dG = Vdp - SdT + \sum_i \mu_i \cdot dn_i$$

The chemical potentials $\mu_i$ describe the change of Gibbs energy when particle $i$ get removed/added, e.g. by chemical reactions.

The nice thing about Gibbs energy is that at constant pressure and temperature, you only have to worry about the chemical potentials, i.e. the chemical reaction: $$dG = \sum_i \mu_i \cdot dn_i$$

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    $\begingroup$ The OP was explicitly talking about chemical reactions and did not realize that they are described by the chemical potentials and he also forgot the $-SdT$ term. I agree, the nice thing about orthocresol's answer is that he explains why OP can't just set $TdS$ equal to $\delta q$ in general, but he fails to realize that $dH=TdS$ even for an irreversible isobaric heat transfer. $\endgroup$
    – Kexanone
    Feb 7, 2021 at 10:07
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    $\begingroup$ Good point: $dH=dU+pdV =TdS$ when: constant p; only pV work; constant composition $\endgroup$
    – Buck Thorn
    Feb 7, 2021 at 18:02
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    $\begingroup$ By the way it is important to note that "dH=TdS" for a microscopic (infinitely small) isobaric heat transfer. As soon as the transfer is finite, there are no guarantees about reversibility since isobaric might refer to the end points being at the same pressure, but the path itself may not be isobaric. $\endgroup$
    – Buck Thorn
    Feb 8, 2021 at 0:49
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    $\begingroup$ OK, I added a note to recommend yours at the top of my answer. I don't mean to "steal" the points you raise, but my answer is accepted which means that (for better or for worse) people will see it first. $\endgroup$
    – orthocresol
    Feb 8, 2021 at 1:24
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    $\begingroup$ I don't care about points. I want people that see this question to have a more complete picture, which referring to my part in your answer exactly does. Thx for the edit. $\endgroup$
    – Kexanone
    Feb 8, 2021 at 2:19

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