Is the change of the Gibbs free energy always zero at constant pressure?

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?




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](https://chemistry.stackexchange.com/a/145989/16683). In the following discussion, one should assume constant temperature and composition.




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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.




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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.


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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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