I reviewed several questions already, but none address the same point except the following:

Is Gibbs energy minimized for processes at constant temperature are pressure only?

However, this question remains unanswered until today. The derivations and equations below are from Robin Smith's Chemical Process Design (Chapter 5) and Kolt's Chemical Thermodynamics (Chapter 6 and 7)

For the calculation of the equilibrium constant, it is the case the $\Delta G$ is set to zero at the equilibrium condition. However, doing so seem improper from what I know, and I am wondering what I am getting wrong.

I am considering a reaction of ideal gases, in which the total pressure is changing while the reaction is progressing. As the reaction progress, Gibbs energy can be described by the following correct equation:

$$\Delta G_{\mathrm{r}} = \Delta G - \Delta G^0=RT\ln{K}$$

However, it is claimed that at equilibrium $\Delta G = 0$. Therefore,

$$\Delta G^0=-RT\ln{K_{\mathrm{eq}}}$$

This latter equation is the one I am concerned with due to the imposition that $\Delta G = 0$, which does not seem proper as the pressure is changing while the reaction is progressing. My concern stems from the logic detailed below.

The standard thermodynamic condition for equilibrium is $\mathrm{d}S_{\mathrm{tot}} = 0$. This then can lead to an equilibrium condition involving Gibbs energy as follows:

$$\mathrm{d}S_{\mathrm{total}} = \mathrm{d}S_{\mathrm{env}} + \mathrm{d}S_{\mathrm{sys}}$$

Rewriting the equilibrium condition in terms of system variables only gives:

$$\mathrm{d}S_{\mathrm{sys}} = \frac{DQ_{\mathrm{sys}}}{T_{sys}}$$

This assumes $T_{\mathrm{env}} = T_{\mathrm{sys}}$.

Embedding the first law of thermodynamics and assuming only $PV$ work leaves us with:

$$\mathrm{d}U + P\mathrm{d}v - T\mathrm{d}S = 0$$

Imposing the assumption of constant temperature and pressure allows adding P and T differentials since they are zeros.

$$\mathrm{d}U + P\mathrm{d}V + V\mathrm{d}P - T\mathrm{d}S - S\mathrm{d}T = 0$$

Once integrated, this gives:

$$\Delta U + \Delta (PV) - T\Delta S = 0 $$

or equivalently

$$\Delta H - T \Delta S = 0 \text{with} \Delta G = \Delta H - T \Delta S$$

However, this final result only came to be with the assumption that $\mathrm{d}P$ and $\mathrm{d}T$ are zero. If this assumption is removed, the equilibrium condition differential would be as follows:

$$\mathrm{d}U + P\mathrm{d}v + V\mathrm{d}P - T\mathrm{d}S - S\mathrm{d}T = -V\mathrm{d}P - S\mathrm{d}T$$


$$\mathrm{d}G = V\mathrm{d}P - S\mathrm{d}T$$

This latter equation does not produce $\Delta G = 0$. In fact, the $\Delta G$ and $K$ relation is derived from it. The end result is that we are applying an equation for constant pressure and another with non-constant pressure for the same system at the same time; this does not make sense to me, as pressure is either constant or not.

How is it then that $\Delta G^0=-RT\ln{K_{\mathrm{eq}}}$?

  • $\begingroup$ See also constant T,V and minimum Helmholtz free energy as equilibrium condition. $\endgroup$
    – Poutnik
    Mar 25, 2023 at 13:29

5 Answers 5


You are dealing with a problem of chemical equilibrium for a single reaction in a homogeneous phase. The Gibbs free energy, in this case, is a function of pressure, temperature, and moles of all the chemical species.

The only thing the following equation will tell you $$ \Delta_r G = 0 \longrightarrow K_\pu{eq} = \prod_j a_j^{\nu_j } \tag{1} $$ is that there exist one and only one state, described by the macroscopic variables $(P,T,y_k)$ once you reach chemical equilibrum.

So I will clarify:

As the reaction progress, Gibbs energy can be described by the following correct equation.

When the reaction progresses, you are not in chemical equilibrium, this is why your system "moves". In this state, there are dissipative forces/driving forces that will modify the macroscopic variables in space and time of your system. If you want to describe the system in this state, you resort to transport phenomena in order to obtain the relationship between those macroscopic variables with space and time.

This latter equation is the one I am concerned with due to the imposition that $\Delta_r G=0$, which does not seem proper as the pressure is changing while the reaction is progressing.

Indeed you are right, since there is one and only one state described by $(P,T,y_k)$ achieved for a certain set of initial conditions. I understand your question, because in books they ask you things like:

  1. If the pressure is increased, what happens to equilibrium?
  2. If the temperature is increased, what happens to equilibrium?

So you are crushing your head to the wall, wondering why they are talking about changes and equilibrium at the same time. The fundamental idea is the following: it does not matter how you reach that final state, what interests you is that $ \Delta_r G = 0 $ will tell that new state described by a new set of $(P,T,y_k)$. So the question should be reformulated as:

  1. If at equilibrium we have a system at $(P_1,T_1,y_{k1})$, what will happen to the new equilibrium established if now the temperature is $T_2$?
  2. If at equilibrium we have a system at $(P_1,T_1,y_{k1})$, what will happen to the new equilibrium established if now the pressure is $P_2$?

For example, you may already know that if there is no change of moles in the chemical reaction, nothing will happen if the pressure is changed to $P_2$. You can prove it with Eq. (1), considering ideal gases, and noting that Eq. (1) will be satisfied regardless the value of the pressure.

The derivation you made is related to a closed system for a pure substance. In that case, the internal energy is only a function of $2$ variables $ U = U(P,T) $ if you like. This is not the case we are analyzing, dealing with a chemical reaction.


Your equation for $\mathrm{d}G$ is incorrect. It should read $$\mathrm{d}G = -S\mathrm{d}T + V\mathrm{d}P + \sum{\mu_jdn_j},$$ where the $\mu$s are the chemical potentials of the various species and $dn$'s are the changes in the amounts of substances.


Metal Storm did provide a correct answer. My initial confusion indeed stemmed from talking about equilibrium and changes (to the equilibrium) simultaneously. I will re-write their answer to address my concerns specifically in the way I thought about them previously.

At Equilibrium, the reaction is no longer changing position. Therefore, both $T$ and $P$ are constant at the equilibrium (there is nothing changing them). Therefore, $\mathrm{d}G = 0$ and $\Delta G = 0$ per the derivation in the original posted question.

Now, if the pressure was getting changed somehow under constant temperature, what will become of the equilibrium condition?

The total differential for $\Delta G$ is:

\begin{align} \mathrm{d}\Delta G &= \Delta V\mathrm{d}P - \Delta S\mathrm{d}T & \text{with } \left.\frac{\delta \Delta G}{\delta P}\right|_{T} &= \Delta V \end{align}

Thus, assuming an ideal gas:

$$G_{\Delta P_{\mathrm{final}}} - G_{\Delta P_{\mathrm{initial}}} = \int \Delta V\mathrm{d}P = \int \frac{nRT}{\Delta P}\mathrm{d}P = nRT \ln\left(\frac{\Delta P_{\mathrm{final}}}{\Delta P_{\mathrm{initial}}}\right)$$

Or, once we incorporate multiple components in a reaction (derivation skipped),

$$\Delta G_{\mathrm{components}, \Delta P_{\mathrm{final}}} - \Delta G_{\mathrm{components}, \Delta P_{\mathrm{initial}}} = RT \ln(K)$$

This latter equation is correct regardless of whether equilibrium is achieved on the final state or not, and is correct regardless of the chosen initial and final conditions are. It has no relation to spontaneity and it makes no claims regarding equilibrium. It did assume a reversible pathway connecting the two pressures though.

However, if we are concerned with a final state that is at equilibrium, the equation can be re-written as:

$$\Delta G_{\mathrm{components}, \mathrm{equil}} - \Delta G_{\mathrm{components}, \mathrm{standard}} = RT \ln(K_{\mathrm{eq}})$$

Per the initial discussion, at equilibrium, $\mathrm{d}G = 0$ and $\Delta G = 0$. Therefore,

$$- \Delta G_{\mathrm{components}, \mathrm{standard}} = RT \ln(K_{eq})$$


This answer was deleted with no explanation. I am resubmitting trusting the moderator will explain the problem.

As you finally seem to conclude when equilibrium is attained T, P, and activities of all participating species are constant and these conditions satisfy an equilibrium constant at the given T and P. When T and P are standard Conditions the Gibbs [free] Energy between the condition of all activities equal to 1. and the equilibrium condition is the standard Delta G of the reaction, Delta G[0]. Delta G[0] can be measured or calculated from thermal or electrochemical data or from Hess's Law calculations or calculated from a measured equilibrium constant. Standard free energies are simply a way to express equilibrium constants in a compact, recognizable, form [energy] similar to pKa and pH in acidity. At equilibrium Delta G and Delta A = zero and the chemical activities satisfy the equilibrium constant. It took Gibbs, Helmholtz et al. to realize that both energy and entropy are involved in attaining equilibrium and determining the availability of work from a reaction.

Equilibrium at other than standard conditions results in a different Delta G[0] and Keq. To reduce to standard conditions energy is removed or added as appropriate. Kinetics may or not allow equilibrium under all conditions.


At constant temperature and pressure, the graph of dependence of the system Gibbs energy as function of the reaction quotient has minimum at the reaction equilibrium.

With constant volume instead of pressure, the same is valid for the Helmholtz energy.

If the above $G=\mathrm{f}(Q)_{T,P}$ or $A=\mathrm{g}(Q)_{T,V}$ are plotted for variable temperature, pressure, or volume, the respective graphs would be changed and the function minimum does not generally match the equilibrium any more.


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